Structural Design of Steelwork to EN 1993 and EN 1994 Third edition
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Structural Design of Steelwork to EN 1993 and EN 1994 Third edition

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Structural Design of Steelwork to EN 1993 and EN 1994 Third edition

L.H. Martin Bsc, PhD, CEng FICE J.A. Purkiss Bsc (Eng), PhD

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Butterworth-Heinemann is an imprint of Elsevier

Butterworth-Heinemann is an imprint of Elsevier Linacre House, Jordan Hill, Oxford OX2 8DP, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA First edition published by Edward Arnold 1984 Second edition 1992 Third edition 2008 Copyright © 2008, L.H. Martin and J.A. Purkiss. All rights reserved The right of L.H. Martin and J.A. Purkiss to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [email protected] Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN13: 978-0-7506-5060-1 For information on all Butterworth-Heinemann publications visit our web site at books.elsevier.com Typeset by Charon Tec Ltd (A Macmillan Company), Chennai, India www.charontec.com Printed and bound in Great Britan 07 08 09 10

10 9 8 7 6 5 4 3 2 1

Contents Preface Acknowledgements Principal Symbols

ix xi xiii

CHAPTER

1

General 1.1 1.2 1.3 1.4

Description of steel structures Development, manufacture and types of steel Structural design Fabrication of steelwork

1 1 5 7 12

CHAPTER

2

Mechanical Properties of Structural Steel 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Variation of material properties Characteristic strength Design strength Other design values for steel Corrosion and durability of steelwork Brittle fracture Residual stresses Fatigue Stress concentrations Failure criteria for steel

17 17 18 19 19 20 22 22 23 24 24

CHAPTER

3

Actions 3.1 3.2 3.3 3.4 3.5

Description Classification of actions Actions varying in time Design values of actions Actions with spatial variation

27 27 27 28 29 31

vi

•

Contents

CHAPTER

4

Laterally Restrained Beams 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Structural classification of sections Elastic section properties and analysis in bending Elastic shear stresses Elastic torsional shear stresses Plastic section properties and analysis Effect of shear force on the plastic moment of resistance Lateral restraint Resistance of beams to transverse forces

35 35 38 54 62 67 73 77 78

CHAPTER

5

Laterally Unrestrained Beams

91

5.1 5.2 5.3

91 127 132

Lateral torsional buckling of rolled sections symmetric about both axes Pure torsional buckling Plate girders

CHAPTER

6

Axially Loaded Members 6.1 6.2 6.3 6.4

Axially loaded tension members Combined bending and axial force – excluding buckling Buckling of axially loaded compression members Combined bending and axial force – with buckling

175 175 177 180 190

CHAPTER

7

Structural Joints 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

Introduction The ideal structural joint Welded joints Bolted joints Plate thicknesses for joint components Joints subject to shear forces Joints subject to eccentric shear forces Joints with end bearing ‘Pinned’ joints ‘Rigid’ joints

198 198 199 199 207 218 224 225 226 228 230

Contents 7.11 7.12

Joint rotational stiffness Frame-to-joint stiffness

•

vii 275 277

CHAPTER

8

Frames and Framing 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14

Single storey structures Multi-storey construction Influence of connection design and detailing Structural actions Single storey structures under horizontal loading Multi-storey construction Behaviour under accidental effects Transmission of loading Design of bracing Fire performance Additional design constraints Design philosophies Design issues for multi-storey structures Portal frame design

282 282 283 286 286 288 293 294 298 300 301 303 306 310 314

CHAPTER

9

Trusses 9.1 9.2

Triangulated trusses Non-triangulated trusses

341 341 347

CHAPTER

10

Composite Construction 10.1 10.2 10.3 10.4

Composite slabs Design of decking Composite beams Composite columns

358 358 359 370 395

CHAPTER

11

Cold-formed Steel Sections 11.1 11.2

Analytical model Local buckling

413 414 419

viii

•

Contents 11.3 11.4 11.5 11.6 11.7 11.8 11.9

Distortional buckling Lateral–torsional buckling Calculation of deflections Finite strip methods Design methods for beams partially restrained by sheeting Working examples Chapter conclusions

424 430 434 435 442 445 453

ANNEX

458

INDEX

469

Preface This book conforms to the latest recommendations for the design of steel and composite steel–concrete structures as described in Eurocode 3: Design of steel structures and Eurocode 4: Design of composite steel–concrete structures. References to relevant clauses of the Codes are given where appropriate. Note that for normal steelwork design, including joints, three sections of EN 1993 are required: • Part 1–1 General rules and rules for buildings • Part 1–5 Plated structural elements • Part 1–8 Design of joints Additionally if design for cold formed sections is carried out from first principles then Part 1–3 Cold formed thin gauge members and sheeting is also required. Whilst it has not been assumed that the reader has a knowledge of structural design, a knowledge of structural mechanics and stress analysis is a prerequisite. However, as noted below certain specialist areas of analysis have been covered in detail since the Codes do not provide the requisite information. Thus the book contains detailed explanations of the principles underlying steelwork design and provides appropriate references and suggestions for further reading. The text should prove useful to students reading for engineering degrees at University, especially for design projects. It will also aid designers who require an introduction to the new Eurocodes. For those familiar with current practice, the major changes are: (1) There is need to refer to more than one part of the various codes with calculations generally becoming more extensive and complex. (2) Steelwork design stresses are increased as the gamma values on steel are taken as 1,0, and the strength of high yield reinforcement is 500 MPa albeit with a gamma factor of 1,15. (3) A deeper understanding of buckling phenomena is required as the Codes do not supply the relevant formulae. (4) Flexure and axial force interaction equations are more complex, thus increasing the calculations for column design. (5) The checking of webs for in-plane forces is more complex. (6) Although tension field theory (or its equivalent) may be used for plate girders, the calculations are simplified compared to earlier versions of the Code. (7) Joints are required to be designed for both strength and stiffness. (8) More comprehensive information is given on thin-walled sections.

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Acknowledgements Lawrence Martin and John Purkiss would like to thank Long-yuan Li and Xiao-ting Chu for writing Chapter 11 on Cold-formed steel sections. The second author would like to thank Andrew Orton (Corus) for help with problems over limiting critical lengths for lateral–torsional buckling of rolled sections. The authors further wish to thank the following for permission to reproduce material: • • • •

Albion Sections Ltd for Fig. 11.1 www.access-steel.com for Figs 11.24 and 11.25 Karoly Zalka and the Institution of Civil Engineers for Fig. 8.13 BSI BSI Ref.

Book Ref.

BS 5950: Part 1: 1990

Tables 15–17

Annex A7

BS 5950: Part 1: 2000

Table 14

Table 5.3

EN 1993-1-3

Fig. 5.6 Fig. 5.7a Fig. 10.2

Fig. 11.8 Fig. 11.9 Fig. 11.20

EN 1994-1-1

Fig. 9.8

Fig. 10.3

British Standards may be obtained from BSI Customer Services, 389 Chiswick High Road, London W4 4AL. Tel: +44 (0)20 8996 9001. e-mail: [email protected]

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Principal Symbols Listed below are the symbols and suffixes common to European Codes

LATIN UPPER AND LOWER CASE A a B b c d E e F f G H h i I k L l M N n p Q q R r S s T t uu vv

accidental action; area distance; throat thickness of a weld bolt force breadth outstand depth of web; diameter modulus of elasticity edge distance; end distance action; force strength of a material permanent action; shear modulus of steel total horizontal load or reaction; warping constant of section height radius of gyration second moment of area stiffness length; span; buckling length effective buckling length; torsion constant; warping constant bending moment axial force number pitch; spacing variable action; prying force uniformly distributed action resistance; reaction radius; root radius; number of redundancies stiffness staggered pitch; distance; bearing length torsional moment thickness principal major axis principal minor axis

xiv

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Principal Symbols V v W w

shear force; total vertical load or reaction shear stress section modulus deflection

GREEK LOWER CASE α β γ δ ε η θ λ μ ν ρ σ τ φ χ ψ

coefficient of linear thermal expansion; angle; ratio; factor angle; ratio; factor partial safety factor deflection; deformation strain; coefficient (235/fy )1/2 where fy is in MPa distribution factor; shear area factor; critical buckling mode; buckling imperfection coefficient angle; slope slenderness ratio; ratio slip factor Poisson’s ratio unit mass; factor normal stress; standard deviation shear stress rotation; slope; ratio reduction factor for buckling stress ratio

SUFFIXES Ed design strength el elastic f flange j joint o initial; hole p plate pl plastic Rd resistance strength t torsion u ultimate strength v shear w web; warping x x-x axis y y-y axis; yield strength z z-z axis

Chapter

1 / General

1.1 DESCRIPTION OF STEEL STRUCTURES 1.1.1 Shapes of Steel Structures The introduction of structural steel, circa 1856, provided an additional building material to stone, brick, timber, wrought iron and cast iron. The advantages of steel are high strength, high stiffness and good ductility combined with relative ease of fabrication and competitive cost. Steel is most often used for structures where loads and spans are large and therefore is not often used for domestic architecture. Steel structures include low-rise and high-rise buildings, bridges, towers, pylons, floors, oil rigs, etc. and are essentially composed of frames which support the self-weight, dead loads and external imposed loads (wind, snow, traffic, etc.). For convenience load bearing frames may be classified as: (a) Miscellaneous isolated simple structural elements (e.g. beams and columns) or simple groups of elements (e.g. floors). (b) Bridgeworks. (c) Single storey factory units (e.g. portal frames). (d) Multi-storey units (e.g. tower blocks). (e) Oil rigs. A real structure consists of a load bearing frame, cladding and services as shown in Fig. 1.1(a). A load bearing frame is an assemblage of members (structural elements) arranged in a regular geometrical pattern in such a way that they interact through structural connections to support loads and maintain them in equilibrium without excessive deformation. Large deflections and distortions in structures are controlled by the use of bracing which stiffens the structure and can be in the form of diagonal structural elements, masonry walls, reinforced concrete lift shafts, etc. A load bearing steel frame is idealized, for the purposes of structural design, as center lines representing structural elements which intersect at joints, as shown in Fig. 1.1(b). Other shapes of load bearing frames are shown in Figs 1.1 (c) to (e). Structural elements are required to resist forces and displacements in a variety of ways, and may act in tension, compression, flexure, shear, torsion or in any combination of

2

•

Chapter 1 / General Dead and snow loading

Connection Wind loading

Services Cladding

Load bearing frame Connection (a) Real structure

(b) Idealized load bearing frame

Pinned connection

Bracing (c)

(d) Rigid connection

(e)

FIGURE 1.1 Typical load bearing frames

these forces. The structural behaviour of a steel element depends on the nature of the forces, the length and shape of the cross section of the member, the elastic properties and the magnitude of the yield stress. For example a tie behaves in a linear elastic manner until yield is reached. A slender strut behaves in a non-linear elastic manner until first yield is attained, provided that local buckling does not occur first. A laterally supported beam behaves elastically until a plastic hinge forms, while an unbraced beam fails by elastic torsional buckling. These modes of behaviour are considered in detail in the following chapters. The structural elements are made to act as a frame by connections. These are composed of plates, welds and bolts which are arranged to resist the forces involved. The connections are described for structural design purposes as pinned, semi-rigid and rigid, depending on the amount of rotation, and are described, analysed and designed in detail in Chapter 7.

Structural Design of Steelwork to EN 1993 and EN 1994

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1.1.2 Standard Steel Sections The optimization of costs in steel construction favours the use of structural steel elements with standard cross-sections and common bar lengths of 12 or 15 m. The billets of steel are hot rolled to form bars, flats, plates, angles, tees, channels, I sections and hollow sections as shown in Fig. 1.2. The detailed dimensions of these sections are given in BS 4, Pt 1 (2005), BSEN 10056-1 (1990), and BSEN 10210-2 (1997). Where thickness varies, for example, Universal beams, columns and channels, sections are identified by the nominal size, that is, ‘depth × breadth × mass per unit length × shape’. Where thickness is constant, for example, tees and angle sections, the identification is ‘breadth × depth × thickness × shape’. In addition a section is identified by the grade of steel. To optimize on costs steel plates should be selected from available stock sizes. Thicknesses are in the range of 6, 8, 10, 12,5, 15 mm and then in 5 mm increments. Thicknesses of less than 6 mm are available but because of lower strength and poorer corrosion resistance their use is limited to cold formed sections. Stock plate widths are in the range 1, 1,25, 1,5, 2, 2,5 and 3 m, but narrow plate widths are also available. Stock plate lengths are in the range 2, 2,5, 3, 4, 5, 6, 10 and 12 m. The adoption of stock widths and lengths avoids work in cutting to size and also reduces waste. The application of some types of section is obvious, for example, when a member is in tension a round or flat bar is the obvious choice. However, a member in tension may

Universal beam (UB)

Universal column (UC)

Circular hollow section

Retangular hollow section

FIGURE 1.2 Standard steel sections

Channel

Angle

Bars

Structural tee from UB

Plate

4

•

Chapter 1 / General be in compression under alternative loading and an angle, tee, or tube is often more appropriate. The connection at the end of a bar or tube, however, is more difficult to make. If a structural element is in bending about one axis then the ‘I’ section is the most efficient because a large proportion of the material is in the flanges, that is, at the extreme fibres. Alternatively, if a member is in bending about two axes at right angles and also supports an axial load then a tube, or rectangular hollow section, is more appropriate. Other steel sections available are cold formed from steel plate into a variety of cross sections for use as lightweight lattice beams, glazing bars, shelf racks, etc. Not all these sections are standardized because of the large variety of possible shapes and uses, however, there is a wide range of sections listed in BSEN 10162 (2003). Local buckling can be a problem and edges are stiffened using lips. Also when used as beams the relative thinness of the material may lead to web crushing, shear buckling and lateral torsional buckling. Although the thickness of the material (1–3 mm) is less than that of the standard sections the resistance to corrosion is good because of the surface finish obtained by pickling and oiling. After degreasing this surface can be protected by galvanizing, or painting, or plastic coating. The use in building of cold formed sections in light gauge plate, sheet and strip steel 6 mm thick and under is dealt with in BSEN 5950 (2001) and EN 1993-1-1 (2005).

1.1.3 Structural Classification of Steel Sections (cl 5.5. EN 1993-1-1 (2005)) A section, or element of a member, in compression due to an axial load may fail by local buckling. Local buckling can be avoided by limiting the width to thickness ratios (b/tf or d/tw ) of each element of a cross-section. The use of the limiting values given in Table 5.2, EN 1993-1-1 (2005) avoids tedious and complicated calculations. Depending on the b/tf or d/tw ratios standard or built-up sections are classified for structural purposes as: • Class 1: Low values of b/tf or d/tw where a plastic hinge can be developed with sufficient rotation capacity to allow redistribution of moments within the structure. • Class 2: Full plastic moment capacity can be developed but local buckling may prevent development of a plastic hinge with sufficient rotation capacity to permit plastic design. • Class 3: High values of b/tf and d/tw , where stress at the extreme fibres can reach design strength but local buckling may prevent the development of the full plastic moment. • Class 4: Local buckling may prevent the stress from reaching the design strength. Effective widths are used to allow for local buckling (cl 5.5.2(2), EN 1993-1-1 (2005)).

Structural Design of Steelwork to EN 1993 and EN 1994

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1.1.4 Structural Joints (EN 1993-1-8 (2005)) Structural elements are connected together at joints which are not necessarily at the ends of members. A structural connection is an assembly of components (plates, bolts, welds, etc.) arranged to transmit forces from one member to another. A connection may be subject to any combination of axial force, shear force and bending moment in relation to three perpendicular axes, but for simplicity, where appropriate, the situation is reduced to forces in one plane. There are other types of joints in structures which are not structural connections. For example a movement joint is introduced into a structure to take up the free expansion and contraction that may occur on either side of the joint due to temperature, shrinkage, expansion, creep, settlement, etc. These joints may be detailed to be watertight but do not generally transmit forces. Detailed recommendations are given by Alexander and Lawson (1981). Another example is a construction joint which is introduced because components are manufactured to a convenient size for transportation and need to be connected together on site. In some cases these joints transmit forces but in other situations may only need to be waterproof.

1.2 DEVELOPMENT, MANUFACTURE AND TYPES OF STEEL 1.2.1 Outline of Developments in Design Using Ferrous Metals Prior to 1779, when the Iron Bridge at Coalbrookdale on the Severn was completed, the most important materials used for load bearing structures were masonry and timber. Ferrous materials were only used for fastenings, armaments and chains. The earliest use of cast iron columns in factory buildings (circa 1780) enabled relatively large span floors to be constructed. Due to a large number of disastrous fires around 1795, timber beams were replaced by cast iron with the floors carried on brick jack arches between the beams. This mode of construction was pioneered by Strutt in an effort to attain a fire proof construction technique. Cast iron, however, is weak in tension and necessitates a tension flange larger than the compression flange and consequently cast iron was used mainly for compression members. Large span cast iron beams were impractical, and on occasions disastrous as in the collapse of the Dee bridge designed by Robert Stephenson in 1874. The last probable use of cast iron in bridge works was in the piers of the Tay bridge in 1879 when the bridge collapsed in high winds due to poor design and unsatisfactory supervision during construction. In an effort to overcome the tensile weakness of cast iron, wrought iron was introduced in 1784 by Henry Cort. Wrought iron enabled the Victorian engineers to produce the following classic structures. Robert Stephenson’s Brittania Bridge was the first box girder bridge and represented the first major collaboration between engineer, fabricator (Fairburn) and scientist (Hodgkinson). I.K. Brunel’s Royal Albert Bridge

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Chapter 1 / General at Saltash combined an arch and suspension bridge. Telford’s Menai suspension bridge used wrought iron chains which have sine been replaced by steel chains. Telford’s Pont Cysyllte is a canal aqueduct near Llangollen. The first of the four structures was replaced after a fire in 1970. The introduction of wrought iron revolutionized ship building and enabled Brunel to produce the S.S. Great Britain. Steel was first produced in 1740, but was not available in large quantities until Bessemer invented the converter in 1856. The first major structure to use the new steel exclusively was Fowler and Baker’s railway bridge at the Firth of Forth. The first steel rail was rolled in 1857 and installed at Derby where it was still in use 10 years later. Cast iron rails in the same position lasted about 3 months. Steel rails were in regular production at Crewe under Ramsbottom from 1866. By 1840 standard shapes in wrought iron, mainly rolled flats, tees and angles, were in regular production and were appearing in structures about 10 years later. Compound girders were fabricated by riveting together the standard sections. Wrought iron remained in use until around the end of the nineteenth century. By 1880 the rolling of steel ‘I’ sections had become widespread under the influence of companies such as Dorman Long. Riveting continued in use as a fastening method until around 1950 when it was superseded by welding. Bessemer steel production in Britain ended in 1974 and the last open hearth furnace closed in 1980. Further information on the history of steel making can be found in Buchanan (1972), Cossons (1975), Derry and Williams (1960), Pannel (1964) and Rolt (1970).

1.2.2 Manufacture of Steel Sections The manufacture of standard steel sections, although now a continuous process, can be conveniently divided into three stages: (1) Iron production (2) Steel production (3) Rolling. Iron production is a continuous process and consists of chemically reducing iron ore in a blast furnace using coke and crushed limestone. The resulting material, called cast iron, is high in carbon, sulphur and phosphorus. Steel production is a batch process and consists in reducing the carbon, sulphur and phosphorus levels and adding, where necessary, manganese, chromium, nickel, vanadium, etc. This process is now carried out using a Basic Oxygen Converter, which consists of a vessel charged with molten cast iron, scrap steel and limestone through which oxygen is passed under pressure to reduce the carbon content by oxidation. This is a batch process which typically produces about 250–300 tons every 40 min. The alternative electric arc furnace is in limited use (approximately 5% of the UK steel production), and is generally used for special steels such as stainless steel.

Structural Design of Steelwork to EN 1993 and EN 1994

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From the converter the steel is ‘teemed’ into ingots which are then passed to the rolling mills for successive reduction in size until the finished standard section is produced. The greater the reduction in size the greater the work hardening, which produces varying properties in a section. The variation in cooling rates of different thicknesses introduces residual stresses which may be relieved by the subsequent straightening process. Steel plate is now produced using a continuous casting procedure which eliminates, ingot casting, mould stripping, heating in soaking pits and primary rolling. Continuous casting permits, tighter control, improved quality, reduced wastage and lower costs.

1.2.3 Types of Steel The steel used in structural engineering is a compound of approximately 98% iron and small percentages of carbon, silicon, manganese, phosphorus, sulphur, niobium and vanadium as specified in BS 4360 (1990). Increasing the carbon content increases strength and hardness but reduces ductility and toughness. Carbon content therefore is restricted to between 0,25% and 0,2% to produce a steel that is weldable and not brittle. The niobium and vanadium are introduced to raise the yield strength of the steel; the manganese improves corrosion resistance; and the phosphorus and sulphur are impurities. BS 4360 (1990) also specifies tolerances, testing procedure and specific requirements for weldable structural steel. Steels used in practice are identified by letters and number, for example, S235 is steel with a tensile yield strength of 235 MPa (Table 3.1, EN 1993-1-1 (2005)).

1.3 STRUCTURAL DESIGN 1.3.1 Initiation of a Design The demand for a structure originates with the client. The client may be a private person, private or public firm, local or national government, or a nationalized industry. In the first stage preliminary drawings and estimates of costs are produced, followed by consideration of which structural materials to use, that is, reinforced concrete, steel, timber, brickwork, etc. If the structure is a building, an architect only may be involved at this stage, but if the structure is a bridge or industrial building then a civil or structural engineer prepares the documents. If the client is satisfied with the layout and estimated costs then detailed design calculations, drawings and costs are prepared and incorporated in a legal contract document. The design documents should be adequate to detail, fabricate and erect the structure. The contract document is usually prepared by the consultant engineer and work is carried out by a contractor who is supervised by the consultant engineer. However, larger firms, local and national government, and nationalized industries, generally employ their own consultant engineer.

8

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Chapter 1 / General The work is generally carried out by a contractor, but alternatively direct labour may be used. A further alternative is for the contractor to produce a design and construct package, where the contractor is responsible for all parts and stages of the work.

1.3.2 The Object of Structural Design The object of structural design is to produce a structure that will not become unserviceable or collapse in its lifetime, and which fulfils the requirements of the client and user at reasonable cost. The requirements of the client and user may include any or all of the following: (a) The structure should not collapse locally or overall. (b) It should not be so flexible that deformations under load are unsightly or alarming, or cause damage to the internal partitions and fixtures; neither should any movement due to live loads, such as wind, cause discomfort or alarm to the occupants/users. (c) It should not require excessive repair or maintenance due to accidental overload, or because of the action of weather. (d) In the case of a building, the structure should be sufficiently fire resistant to, give the occupants time to escape, enable the fire brigade to fight the fire in safety and to restrict the spread of fire to adjacent structures. The designer should be conscious of the costs involved which include: (a) The initial cost which includes fees, site preparation, cost of materials and construction. (b) Maintenace costs (e.g. decoration and structural repair). (c) Insurance chiefly against fire damage. (d) Eventual demolition. It is the responsibility of the structural engineer to design a structure that is safe and which conforms to the requirements of the local bye-laws and building regulations. Information and methods of design are obtained from Standards and Codes of Practice and these are ‘deemed to satisfy’ the local bye-laws and building regulations. In exceptional circumstances, for example, the use of methods validated by research or testing, an alternative design may be accepted. A structural engineer is expected to keep up to date with the latest research information. In the event of a collapse or malfunction where it can be shown that the engineer has failed to reasonably anticipate the cause or action leading to collapse, or has failed to apply properly the information at his disposal, that is, Codes of Practice, British Standards, Building Regulations, research or information supplied by the manufacturers, then he may be sued for professional negligence. Consultants and contractors carry liability insurance to mitigate the effects of such legal action.

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1.3.3 Limit State Design (cl 2.2, EN 1993-1-1 (2005)) It is self-evident that a structure should be ‘safe’ during its lifetime, that is, free from the risk of collapse. There are, however, other risks associated with a structure and the term safe is now replaced by the term ‘serviceable’. A structure should not during its lifetime become ‘unserviceable’, that is, it should be free from risk of collapse, rapid deterioration, fire, cracking, excessive deflection, etc. Ideally it should be possible to calculate mathematically the risk involved in structural safety based on the variation in strengths of the material and variation in the loads. Reports, such as the CIRIA Report 63 (1977), have introduced the designer to elegant and powerful concept of ‘structural reliability’. Methods have been devised whereby engineering judgement and experience can be combined with statistical analysis for the rational computation of partial safety factors in codes of practice. However, in the absence of complete understanding and data concerning aspects of structural behaviour, absolute values of reliability cannot be determined. It is not practical, nor is it economically possible, to design a structure that will never fail. It is always possible that the structure will contain material that is less than the required strength or that it will be subject to loads greater than the design loads. If actions (forces) and resistance (strength of materials) are determined statistically then the relationship can be represented as shown in Fig. 1.3. The design value of resistance (Rd ) must be greater than the design value of the actions (Ad ). It is therefore accepted that 5% of the material in a structure is below the design strength, and that 5% of the applied loads are greater than the design loads. This does not mean therefore that collapse is inevitable, because it is extremely unlikely that the weak material and overloading will combine simultaneously to produce collapse. The philosophy and objectives must be translated into a tangible form using calculations. A structure should be designed to be safe under all conditions of its useful life Frequency

Resistance R

Actions A

Resistance or actions

Ad Rd Design value of actions

Design value of resistance

Rd − Ad > 0

FIGURE 1.3 Statistical relationship between actions and resistance

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Chapter 1 / General and to ensure that this is accomplished certain distinct performance requirements, called ‘limit states’, have been identified. The method of limit state design recognizes the variability of loads, materials, construction methods and approximations in the theory and calculations. Limit states may be at any stage of the life of a structure, or at any stage of loading and are important for the design of steelwork. To reduce the number of load cases to be considered only serviceability and ultimate limit states are specified. Each of these sections is subdivided although some may not be critical in every design. Calculations for limit states involve loads and load factors (Chapter 3), and material factors and strengths (Chapter 2). Stability, an ultimate limit state, is the ability of a structure, or part of a structure, to resist overturning, overall failure and sway. Calculations should consider the worst realistic combination of loads at all stages of construction. All structures, and parts of structures, should be capable of resisting sway forces, for example, by the use of bracing, ‘rigid’ joints, or shear walls. Sway forces arise from horizontal loads, for example, winds, and also from practical imperfections, for example, lack of verticality. The sway forces from practical imperfections are difficult to quantify and advice is given in cl 5.3.3, EN 1993-1-1 (2005). Also involved in limit state design is the concept of structural integrity. Essentially this means that the structure should be tied together as a whole, but if damage occurs, it should be localized. Deflection is a serviceability limit state. Deflections should not impair the efficiency of a structure, or its components, nor cause damage to the finishes. Generally the worst realistic combination of unfactored imposed loads is used to calculate elastic deflections. These values are compared with empirical values related to the length of a member or height. Dynamic effects to be considered at the serviceability limit state are vibrations caused by machines, and oscillations caused by harmonic resonance, for example, wind gusts on buildings. The natural frequency of the building should be different from the exciting source to avoid resonance. Fortunately there are few structural failures and when they do occur they are often associated with human error involved in design calculations, or construction, or in the use of the structure.

1.3.4 Structural Systems Structural frame systems may be described as: (a) simple frames, (b) continuous frames, (c) semi-continuous frames.

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These titles refer to the types of joints and whether bracing is included. Simple design assumes that ‘pin joints’ connect the members and joint rotations are prevented by bracing. Historically this method was popular because parts of the structure could be designed in isolation and calculations could be done by hand. With the advent of the computer calculations are less onerous but the method is still in use. Continuous frames assume that the connections between members are rigid and therefore the angles between members can be maintained without the use of bracing. Calculations for the design of members and connections are more complicated and a computer is generally used. Global analysis of the frame is based on elastic, plastic, or elastic–plastic analysis assuming full continuity. Semi-continuous frames acknowledges that in reality, end moments and rotations exist at the connections. Global analysis using the computer is based on the moment– rotation and force displacement characteristics of the connections. Bracing is often necessary for this type of frame to reduce sway.

1.3.5 Errors The consequences of an error in structural design can lead to loss of life and damage to property, and it is necessary to appreciate where errors can occur. Small errors in design calculations can occur in the rounding off of figures but these generally do not lead to failures. The common sense advice is that the accuracy of the calculation should match the accuracy of the values given in the European Code. Errors that occur in structural design calculations and which affect structural safety are: (1) Ignorance of the physical behaviour of the structure under load and which consequently introduces errors in the basic assumptions used in the theoretical analysis. (2) Errors in estimating the loads, especially the erection forces. (3) Numerical errors in the calculations. These should be eliminated by checking, but when speed is paramount checks are often ignored. (4) Ignorance of the significance of certain effects (e.g. residual stresses, fatigue, etc). (5) Introduction of new materials, or methods, which have not been proved by tests. (6) Insufficient allowance for tolerances or temperature strains. (7) Insufficient information (e.g. in erection procedures). Errors that can occur in workshops or on construction sites are: (1) Using the wrong grade of steel, and when welding using the wrong type of electrode. (2) Using the wrong weight of section. A number of sections are the same nominal size but differ in web or flange thickness. (3) Errors in manufacture (e.g. holes in the wrong position).

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Chapter 1 / General Errors that occur in the life of a structure and also affect safety are: (1) Overloading (2) Removal of structural material (e.g. to insert service ducts) (3) Poor maintenance.

1.4 FABRICATION OF STEELWORK 1.4.1 Drawings Detailed design calculations are essential for any steel work design but the sizes of the members, dimensions and geometrical arrangement are usually presented as drawings. Initially the drawings are used by the fabricator and eventually by the contractor on site. General arrangement drawings are often drawn to scale of 1:100, while details are drawn to a scale of 1:20 or 1:10. Special details are drawn to larger scales where necessary. Drawings should be easy to read and should not include superfluous detail. Some important notes are: (a) Members and components should be identified by logically related mark numbers, for example, related to the grid system used in the drawings. (b) The main members should be presented by a bold outline (0,4 mm wide) and dimension lines should be unobtrusive (0,1 mm wide). (c) Dimensions should be related to centre lines, or from one end; strings of dimensions should be avoided. Dimensions should appear once only so that ambiguity cannot arise when revisions occur. Fabricators should not be put in the position of having to do arithmetic in order to obtain an essential dimension. (d) Tolerances for erection purposes should be clearly shown. (e) The grade of steel to be used should be clearly indicated. (f) The size, weight and type of section to be used should be clearly stated. (g) Detailing should take account of possible variations due to rolling margins and fabrication variations. (h) Keep the design and construction as simple as possible. Where possible use simple connections, avoid stiffeners, use the minimum number of sections and avoid changes in section along the length of a member. (i) Site access, transport and use of cranes should be considered.

1.4.2 Tolerances (cl 3.2.5, EN 1993-1-1 (2005)) Tolerances are limits places on unintentional inaccuracies that occur in dimensions which must be allowed for in design if structural elements and components are to fit together. In steelwork variations occur in the rolling process, marking out, cutting and drilling during fabrication, and in setting out during erection.

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In the rolling process the allowable tolerances for length, width, thickness and flatness for plates are given in BS 4360 (1990). Length and width tolerances are positive while those for thickness and flatness are negative and positive. The dimensional and weight tolerances for sections are given in BS 4 Pt 1 (2005), or BS 10056-1 (1990), as appropriate. During fabrication there is a tendency for members and components to increase rather than reduce, and the tolerance is therefore often specified as negative; it is often cheaper and simpler to insert packing rather than shorten a member, provided that the packing is not excessive. Where concrete work is associated with steelwork variations in dimensions are likely to be greater. When casting concrete, for example, errors in dimensions may arise from shrinkage or from warping of the shuttering, especially when it is re-used. Therefore, by virtue of the construction method, larger tolerances are specified for work involving concrete. To facilitate erection all members and connections should be provided with the maximum tolerance that is acceptable from structural and architectural considerations. A typical example is a connection between a steel column and a reinforced concrete base. It would produce great difficulties if the base were set too high and a tolerance of approximately 50 mm is often included in the design, with provision for grouting under the base. Tolerances are also provided to allow lateral adjustment of the foundation bolts. Tolerances between concrete and steelwork are also important because two different contractors are involved.

1.4.3 Fabrication, Assembly and Erection of Steelwork The drawings produced by the structural designer are used first by the steel fabricator and later by the contractor on site. The steel fabricator obtains the steel either direct from the rolling mills or from the steel stockiest, and then cuts, drills and welds the steel components to form the structural elements as shown on the drawings. In general, for British practice, the welding is confined to the workshop and the connections on site are made using bolts. In American, however, site welding is common practice. When marking out, the measurements of length for overall size, position of holes, etc. can be done by hand, but if there are several identical components then wooden or cardboard templates are made and repeated measurements avoided. Now automatic machines, controlled by a computer, or punched paper tape, are used to cut and drill standard sections. When completed, the steel work should be marked clearly and manufactured to the accepted tolerances. When fabricated, parts of the structure are delivered to the site in the largest pieces that can be transported and erected. For example a lattice girder may be sent fully assembled to a site in this country, but sent in pieces to fit a standard transport container

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Chapter 1 / General for erection abroad. All components should be assembled within the tolerances and cambers specified, and should not be bent or twisted or otherwise damaged. On site the general contractor may be responsible for the assembly, erection, connections, alignment and leveling of the complete structure. Alternatively the erection work may be done by the steel fabricator, or sublet to a specialist steel erector. The objective of the erection process is to assemble the steelwork in the most cost-effective method whilst maintaining the stability of individual members, and/or part or complete structure. To do this it may be necessary to introduce cranes and temporary bracing which must also be designed to resist the loads involved. During assembly on site it is inevitable that some components will not fit, despite the tolerances that have been allowed. A typical example is that the faying surfaces for a friction grip joint are not in contact when the bolts are stressed. Other examples are given by Mann and Morris (1981). The correction of some faults and the consequent litigation can be expensive.

1.4.4 Testing of Steelwork (cl 2.5, EN 1993-1-1 (2005)) Steel is routinely sampled and tested during production to maintain quality. However, occasionally new methods of construction are suggested and there may be some doubt as to the validity of the assumptions of behaviour of the structure. Alternatively if the structure collapses there may be some dispute as to the strength of a component, or member, of the structure. In such cases testing of components, or part of the structure, may be necessary. However it is generally expensive because of, the accuracy required, cost of material, cost of fabrication, necessity to repeat tests to allow for variations and to report accurately. Tests may be classified as: (a) acceptance tests – non-destructive for confirming structural performance, (b) strength tests – used to confirm the calculated capacity of a component or structure, (c) tests to failure – to determine the real mode of failure and the true capacity of a specimen, (d) check tests – where the component assembly is designed on the basis of tests. The size, shape, position of the gauges, and method of testing of small sample pieces of steel is given in BS 4360 (1990) and BSEN 10002-1 (2001). The tensile test is most frequently employed, and gives values of, Young’s modulus, limit of proportionality, yield stress or proof stress, percentage elongation and ultimate stress. Methods of destructive testing fusion welded joints and weld metal in steel are given in numerous Standards. The Charpy V-notch test for impact resistance is used to measure toughness, that is, the total energy, elastic and plastic, which can be absorbed by a specimen before

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fracture. The test specimen is a small beam of rectangular cross section with a ‘V’ notch at mid-length. The beam is fractured by a blow from a swinging pendulum, and the amount of energy absorbed is calculated from the loss of height of the pendulum swing after fracture. Details of the test specimen and procedure are given in BS 4360 (1990) and BSEN 10045-1 (1990). The Charpy V-notch test is often used to determine the temperature at which transition from brittle to ductile behaviour occurs. Structures which are unconventional, and/or method of design which are unusual or not fully validated by research, should be subject to acceptance tests. Essentially these consist of loading the structure to ensure that it has adequate strength to support, for example, 1 (test dead load) + 1,15 (remainder of dead load) + 1,25 (imposed load). Where welds are of vital importance, for example, in pressure vessels, they should be subject to non-destructive tests. The defects that can occur in welds are: slag inclusions, porosity, lack of penetration and sidewall fusion, liquation, solidification, hydrogen cracking, lamellar tearing and brittle fracture. A surface crack in a weld may be detected visibly but alternatively a dye can be sprayed onto the joint which seeps into the cracks. After removing any surplus dye the weld is resprayed with a fine chalk suspension and the crack then shows as a coloured line on the white chalk background. A variant of this technique is to use fluorescent dye and a crack then shows as a bright green line in ultra violet light. A surface crack may also be detected if the weld joint area is magnetized and sprayed with iron powder. The powder congregates along a crack, which shows as a black line. Other weld defects cannot be detected on the surface and alternative methods must be used. Radiographic methods use an X-ray, or gamma-ray, source on one side of the weld and a photographic film on the other. Rays are absorbed by the weld metal, but if there is a hole or crack there is less absorbtion which shows as a dark area on the film. Not all defects are detected by radiography since the method is sensitive to the orientation of the flaw, for example, cracks at right angles to the X-ray beam are not detected. Radiography also requires access to both sides of the joint. The method is therefore most suitable for in-line butt weld for plates. An alternative method to detect hidden defects in welds uses ultrasonics. If a weld contains a flaw then high frequency vibrations are reflected. The presence of a flaw can therefore be indicated by monitoring the reduction of transmission of ultrasonic vibrations, or by monitoring the reflections. The reflection method is extremely useful for welds where access is only possible from one side. Further details can be obtained from Gourd (1980).

REFERENCES Alexander S.J. and Lawson R.M. (1981). Movement design in buildings. Technical Note 107. Construction Industry Research and Information Association Publication.

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Chapter 1 / General BS 4 (2005). Specification for hot rolled sections Pt1. BSI. BSEN 10002-1 (2001). Methods for tensile testing of metals. BSI. BSEN 10045-1 (1990). The Charpy V-notch impact test on metals. BSI. BSEN 10162 (2003). Cold rolled sections. BSI. BS 4360 (1990) Specification for weldable structural steels. BSI. BSEN 10056-1 (1990). Hot rolled structural steel sections. BSI. BSEN 10210-2 (1997). Hot rolled structural steel hollow sections. BSI. BSEN 5950 (2001 and 1998). Structural use of steel work in buildings. BSI. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Buchanan R.A. (1972). Industrial archeology in Britain. Penguin. CIRIA (1977). Rationalisation of safety and serviceability factors in structural codes. Report 63. Construction Industry Research and Information Association Publication. Cossons N. (1975). The BP book of industrial archeology. David and Charles. Derry T.K. and Williams T.I. (1960). A short history of technology. Oxford University Press. Gourd L.M. (1980). Principles of welding technology. Edward Arnold. Mann A.P. and Morris L.J. (1981). Lack of fit in steel structures, Technical Report 87. Construction Industry Research and Information Association Publication. Pannel J.P.M. (1964). An illustrated history of civil engineering. Thames and Hudson. Rolt, L.T.C. (1970). Victorian engineering. Penguin.

Chapter

2 / Mechanical Properties of Structural Steel

2.1 VARIATION OF MATERIAL PROPERTIES All manufactured material properties vary because the molecular structure of the material is not uniform and because of inconsistencies in the manufacturing process. The variations that occur in the manufacturing process are dependent on the degree of control. Variations in material properties must be recognized and incorporated into the design process. The material properties that are of most importance for structural design using steel are strength and Young’s modulus. Other properties which are of lesser importance are hardness, impact resistance and melting point. If a number of samples are tested for a particular property, for example, strength, and the number of specimens with the same strength (frequency) plotted against the strength, then the results approximately fit a normal distribution curve as shown in Fig. 2.1. This curve can be expressed mathematically by the equation shown in Fig. 2.1 which can be used to define ‘safe’ values for design purposes.

y

Frequency

⎛

y=

⎛2

x−x 1 e−½ ⎜⎝ s ⎜⎝ s √2 p

5% of results

x Characteristic strength

1,64s

Strength Mean strength

x

FIGURE 2.1 Variation in material properties

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Chapter 2 / Mechanical Properties of Structural Steel

2.2 CHARACTERISTIC STRENGTH (cl 3.1, EN 1993-1-1 (2005)) A strength to be used as a basis for design must be selected from the variation in values shown in Fig. 2.1. This strength, when defined, is called the characteristic strength. If the characteristic strength is defined as the mean strength, then clearly from Fig. 2.1, 50% of the material is below this value. This is not acceptable. Ideally the characteristic strength should include 100% of the samples, but this is also impractical because it is a low value and results in heavy and costly structures. A risk is therefore accepted and it is therefore recognized that 5% of the samples fall below the characteristic strength. The characteristic strength is calculated from the equation fy = fmean − 1,64σ where for n samples the standard deviation σ=

( fmean − f )2 (n − 1)

1/2

Nominal values of the characteristic yield strengths and ultimate tensile strengths are given in Table 3.1, EN 1993-1-1 (2005) and some examples are given in Table 2.1.

TABLE 2.1 Some nominal values of yield strength for hot rolled steel (Table 3.1, EN 1993-1-1 (2005)). Standard EN 10025-2: Grade S235 S275 S355

Nominal thickness of material t ≤ 40 mm

40 < t ≤ 80 mm

fy (MPa)

fu (MPa)

fy (MPa)

fu (MPa)

235 275 355

360 430 510

215 255 335

360 410 470

TABLE 2.2 Some partial safety factors (cl 6.1, EN 1993-1-1 (2005) and Table 2.1, EN 1993-1-8 (2005)). Situation

Symbol

Value

Buildings

γM0 γM1 γM2 γM3 γM4 γM5 γM7

1,00 1,00 1,25 1,25 1,00 1,00 1,10

Joints

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2.3 DESIGN STRENGTH (cl 6.1, EN 1993-1-1 (2005)) The characteristic strength of steel is the value obtained from tests at the rolling mills, but by the time the steel becomes part of a finished structure this strength may be reduced (e.g. by corrosion or accidental damage). The strength to be used in design calculations is therefore the characteristic strength divided by a partial safety factor (γM ) (Table 2.2).

2.4 OTHER DESIGN VALUES FOR STEEL (cl 3.2.6, EN 1993-1-1 (2005)) The elastic modulus for steel (E) is obtained from the relationship between stress and strain as shown in Fig. 2.2. This is a material property and therefore values from a set of samples vary. However, the variation for steel is very small and the European Code assumes E = 210 GPa. The elastic shear modulus (G) is related to Young’s modulus by the expression E = 2G(1 + ν) where Poissons ratio ν = 0,3 in the elastic range and is used in calculations involving plates. The thermal coefficient of expansion for steel is given as α = 12E-6/K for T ≤ 100◦ C and is used in calculations involving temperature changes.

Fracture Fracture

Yield

Stress

Stress

Yield

Strain Plastic

Strain hardening

0,002

Strain

Elastic (a) Low-strength grade S 235

FIGURE 2.2 Tensile stress–strain relationships for steel

(b) High-strength grade S 450

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Chapter 2 / Mechanical Properties of Structural Steel Hardness is material property that is occasionally of importance in structural steel design. It is measured by the resistance the surface of the steel offers to, the indentation of a hardened steel ball (Brinell test), a square-based diamond pyramid (Vickers test) or a diamond cone (Rockwell test). Higher strength often correlates with greater hardness but this relationship is not infallible. Ductility may be described as the ability of a material to change its shape without fracture. This is measured by the percentage elongation, that is, 100 × (change in length)/(original length). Values of 20% can be obtained for mild steel but it is less for high-strength steel. A high value is advantageous because it allows the redistribution of stresses at ultimate load and the formation of plastic hinges.

2.5 CORROSION AND DURABILITY OF STEELWORK (cl 4, EN 1993-1-1 (2005)) Durability is a service limit state and the following factors should be considered at the design stage: (a) (b) (c) (d) (e) (f)

Environment, Degree of exposure, Shape of the members and details, Quality of workmanship and control, Protective measures, Maintenance.

Methods of protecting steel work are given in BSEN ISO 12944 (1998) and the specification for weather resistant steel is given is BS 7668 (1994). Corrosion of steel work reduces the cross-section of members and thus affects safety. Corrosion, which occurs on the surface of steel, is a chemical reaction between iron, water and oxygen, which produces a hydrated iron oxide called rust. Electrons are liberated in the reaction and a small electrical current flows from the corroded area to the uncorroded area. The elimination of water, oxygen or the electrical current, reduces the rate of corrosion. In contrast pollutants in the air, for example, sulphur dioxides from industrial atmospheres and salt from marine atmospheres, increase the electrical conductivity of water and accelerate the corrosion reaction. Steel is particularly susceptible to atmospheric corrosion which is often severe in coastal or industrial environments and the corrosion may reduce the section size due to pitting or flaking of the surface. Modern rolling techniques and higher-strength steels result in less material being used, for example, the web of an ‘I’ section may be only 6 mm thick. Generally in structural engineering 8 mm is the minimum thickness

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used for exposed steel, and 6 mm for unexposed steel. For sealed hollow sections these limits are reduced to 4 and 3 mm, respectively. Corrosion of steel usually takes the form of rust which is a complex oxide of iron. The rust builds up a deposit on the surface and may eventually flake off. The coating of rust does not inhibit corrosion, except in special steels, and corrosion progresses beneath the rust forming conical pits and the thickness of the metal is reduced. The conical pits can act as stress raisers, that is, centres of high local stress, and in cases where there are cyclic reversals of stress, may become the initiating points of fatigue cracks or brittle fracture. The corrosion resistance of unprotected steel is dependent on its chemical composition, the degree of pollution in the atmosphere, and the frequency of wetting and drying. Low-strength carbon steels are inexpensive but are particularly susceptible to atmospheric corrosion which is often greatest in industrial or coastal environments. High-strength low-alloy steels (Cr–Si–Cu–P) do not pit as severely as carbon steels and the rust that forms becomes a protective coating against further deterioration. These steels therefore have several times the corrosion resistance of carbon steels. The longer steel remains wet the greater the corrosion and therefore the detailing of steelwork should include drainage holes, avoid pockets and allow the free flow of air for rapid drying. The most common, and cheapest form of protection process is to clean the surface by sand or shot blasting, and then to paint with a lead primer, generally in the workshop prior to delivery on site. Joint contact surfaces need not be protected unless specified. On site the steel is erected and protection is completed with an undercoat and finishing coat, or coats, of paint. In the case of surfaces to be welded steel should not be painted, nor metal coated, within a suitable distance of any edges to be welded, if the paint specified or the metal coating is likely to be harmful to welders or impair the quality of the welds. Welds and adjacent parent metal should not be painted prior to de-slagging, inspection and approval. Encasing steel in concrete provides an alkaline environment and no corrosion will take place unless water diffuses through the concrete carrying with it SO2 and CO2 gases from the air in the form of weak acids. The resulting corrosion of the steel and the increase in pressure spalls the concrete. Parts to be encased in concrete should not be painted nor oiled, and where friction grip fasteners are used protective treatment should not be applied to the faying surfaces. A more expensive protection is zinc, or aluminium spray coating which is sometimes specified in corrosive atmospheres. Further improvements are hot dip zinc galvanizing, or the use of stainless steels. These and other forms of protection are described in BSEN ISO 12944 (1998). Recently zinc coated highly stressed steel has been shown to be susceptible to hydrogen cracking.

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2.6 BRITTLE FRACTURE (cl 3.2.3, EN 1993-1-1 (2005)) Brittle fracture is critical at the ultimate limit state. Evidence of brittle fracture is a small crack, which may or may not be visible, and which extends rapidly to produce a sudden failure with few signs of plastic deformation. This type of fracture is more likely to occur in welded structures (Stout et al., 2000). The essential conditions leading to brittle fracture are: (a) There must be a tensile stress in the material but it need not be very high, and may be a residual stress from welding. (b) There must be a notch, or defect, or hole in the material which produces a stress concentration. (c) The temperature of the material must be below the transition temperature (generally below room temperature). At low temperatures crack initiation and propagation is more likely because of lower ductility. The mechanism of failure is that the notch, defect or hole raises the local tensile stresses to values as high as three times the average tensile stress. The material which generally fails by a shearing mechanism now tends to fail by a brittle fracture cleavage mechanism which exhibits considerably less plastic deformation. A drop in the temperature encourages the cleavage failure. A ductile material which has an extensive plastic range is more likely to resist brittle fracture and a test used as a guide to resistance to brittle fracture is the Charpy V-notch impact test (BS 7668 (1994)). The importance of brittle fracture was shown by the failure of the welded ‘liberty’ cargo ships mass produced by the USA during the Second World War. The ships broke apart in harbour and at sea during the cold weather. Brittle fracture is considered only where tensile stresses exist. The mode of failure is mainly dependent on the following: (a) (b) (c) (d) (e) (f)

Steel strength grade Thickness of material Loading speed Lowest service temperature Material toughness T ype of structural element.

No further check for brittle fracture need to be made if the conditions given in EN 1993-1-10 (2005) are satisfied for lowest temperature. For further information see NDAC (1970).

2.7 RESIDUAL STRESSES Residual stresses are present in steel due to uneven heating and cooling. The stresses are induced in steel during, rolling, welding which constrains the structure to a

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particular geometry, force fitting of individual components, lifting and transportation. These stresses may be relieved by subsequent reheating and slow cooling but the process is expensive. The presence of residual stresses adversely affects the buckling of columns, introduces premature yielding, fatigue resistance and brittle fracture. Welding raises the local temperature of the steel which expands relative to the surrounding metal. When it cools it contracts inducing tensile stresses in the weld and the immediately adjacent metal. These tensile stresses are balanced by compressive stresses in the metal on either side. During rolling the whole of the steel section is initially at a uniform temperature, but as the rolling progresses some parts of the cross-section become thinner than others and consequently cool more quickly. Thus, as in the welded joint, the parts which cool last have a residual tensile stress and the parts which cool first may be in compression. Since the cooling rate also affects the yield strength of the steel, the thinner sections tend to have a higher yield stress than the thicker sections. A tensile test piece cut from the thin web of a Universal Beam will probably have a higher yield stress than one cut from the thicker flange. The residual stress and yield stress in rolled sections are also affected by the cold straightening which is necessary for many rolled sections before leaving the mills. Residual stresses are not considered directly in the European Code but are allowed for in material factors. For further information see Ogle (1982).

2.8 FATIGUE This is an ultimate limit state. The term fatigue is generally associated with metals and is the reduction in strength that occurs due to progressive development of existing small pits, grooves or cracks when subject to fluctuating loads. The rate of development of these cracks depends on the size of the crack and on the magnitude of the stress variation in the material and also the metallurgical properties. The number of stress variations, or cycles of stress, that a material will sustain before failure is called fatigue life and there is a linear experimental relationship between the log of the stress range and the log of the number of cycles. Welds are susceptible to a reduction in strength due to fatigue because of the presence of small cracks, local stress concentrations and abrupt changes of geometry. Research into the fatigue strength of welded structures is described by Munse (1984). Other references are BS 5400 (1980), Grundy (1985) and ECSS. All structures are subject to varying loads but the variation may not be significant. Stress changes due to fluctuations in wind loading need not be considered, but windinduced oscillations must not be ignored. The variation in stress depends on the ratio of dead load to imposed load, or whether the load is cyclic in nature, for example, where machinery is involved. For bridges and cranes fatigue effects are

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Chapter 2 / Mechanical Properties of Structural Steel more likely to occur because of the cyclic nature of the loading which causes reversals of stress. Generally calculations are only required for: (a) (b) (c) (d)

Lifting appliances or rolling loads, Vibrating machinery, Wind-induced oscillations, Crowd-induced oscillations.

The design stress range spectrum must be determined, but simplified design calculations for loading may be based on equivalent fatigue loading if more accurate data is not available. The design strength of the steel is then related to the number and range of stress cycles. For further information see EN 1993-1-9 (2005).

2.9 STRESS CONCENTRATIONS Structural elements and connections often have abrupt changes in geometry and also contain holes for bolts. These features produce stress concentrations, which are localized stresses greater than the average stress in the element, for example, tensile stresses adjacent to a hole are approximately three times the average tensile stress. If the average stress in a component is low then the stress concentration may be ignored, but if high then appropriate methods of structural analysis must be used to cater for this effect. The effect of stress concentrations has been shown to be critical in plate web girders in recent history. Stress concentrations are also associated with fatigue and can affect brittle fracture. Formulae for stress concentrations are given in Roark and Young (1975).

2.10 FAILURE CRITERIA FOR STEEL The structural behaviour of a metal at or close to failure may be described as ductile or brittle. A typical brittle metal is cast iron which exhibits a linear load–displacement relationship until fracture occurs suddenly with little or no plastic deformation. In contrast mild steel is a ductile material which also exhibits a linear load–displacement relationship, but at yield large plastic deformations occur before fracture. The nominal yield strength is a characteristic strength in the European Code and is therefore an important failure criterion for steel. The tensile yield condition can be related to various stress situations, for example, tension, compression, shear or various combinations of stresses. There are four generally acceptable theoretical yield criteria as follows: (1) The maximum stress theory, which states that yield occurs when the maximum principal stress reaches the uniaxial tensile stress.

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(2) The maximum strain theory, which states that yield occurs when the maximum principal tensile strain reaches the uniaxial tensile strain at yield. (3) The maximum shear stress theory, which states that yield occurs when the maximum shear stress reaches half of the yield stress in uniaxial tension. (4) The distortion strain energy theory, or shear strain energy theory, which states that yielding occurs when the shear strain energy reaches the shear strain energy in simple tension. For a material subject to principal stresses σ1 , σ2 and σ3 it is shown (Timoshenko, 1946) that this occurs when (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2fy2

(2.1)

This theory was originally developed by Huber, Von-Mises and Hencky. Alternatively Eq. (2.1) can be expressed in terms of direct stresses σb , σbc and σbt , and shear stress τ on two mutually perpendicular planes. It can be shown from Mohr’s circle of stress that the principal stresses 1/2 (σb + σbc ) (σb − σbc )2 2 σ1 = (2.2) − +τ 2 4 and

1/2 (σb − σbc ) (σb − σbc )2 σ2 = + + τ2 2 4

(2.3)

If Eqs (2.2) and (2.3) are inserted in Eq. (2.1) with σ3 = 0 and fy is equal to the design stress fy /γM then 2 ( fy /γM )2 = σbc + σb2 − σbc σb + 3τ 2

(2.4)

If σbc is replaced by σbt with a change in sign then 2 ( fy /γM )2 = σbc + 3τ 2

(2.5)

This equation is a yield criteria which is applicable in some design situations, for example the design of welds (EN 1993-1-8 (2005)).

REFERENCES BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-2 (1993), BSEN 10113-3 (1993) and BS 10210-1 (1994). Specification for weldable structural steels. BSI. BS 5400 (1980). Code of practice for fatigue, Pt 10. BSI. BSEN ISO 12944-1 to 8 (1998) and BSEN ISO 14713 (1999). Code of practice for protection of iron and steel structures. BSI. ECCS (1981). Recommendations for fatigue design of steel structures. European Convention for Structural Steelwork. Construction Press. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. EN 1993-1-9 (2005). Design of steel structures: Fatigue strength of steel structures. BSI.

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Chapter 2 / Mechanical Properties of Structural Steel EN 1993-1-10 (2005). Design of steel structures: Selection of steel for fracture toughness and through thickness properties. BSI. Grundy, P. (1985). Fatigue limit design for steel structures. Civil Engineering Transactions, Institution of Civil Engineers, Australia, CE27, No. 1. Munse, W.H. (1984). Fatigue of welded structures, Welding Research Council. NDAC (1970). Brittle fracture in steel structures, Navy Department Advisory Committee on Structural Steels (ed G.M. Boyd). Butterworth. Ogle, M.H. (1982). Residual stresses in a steel box-girder bridge. Tech Note 110. Construction Industry Research and Information Association Publication. Roark, J.R. and Young, W.C. (1975). Formula for Stress and Strain (5th edition). McGraw-Hill. Stout, R.D., Tor, S.S. and Ruzek, J.M. (1951). The effect of fabrication procedures on steels used in pressure vessels, Welding Journal 30. Timoshenko, S. (1946). Strength of Materials, Pt II. D. Van Nostrand.

Chapter

3 / Actions

3.1 DESCRIPTION Actions are a set of forces (loads) applied to a structure, or/and deformations produced by temperature, settlement or earthquakes (EN 1990 (2002)). Values of actions are obtained by determining characteristic or representative values of loads or forces. Ideally, loads applied to a structure during its working life, should be analysed statistically and a characteristic load is determined. The characteristic load might then be defined as the load above which no more than 5% of the loads exceed, as shown in Fig. 3.1. However, data is not available and the characteristic value of an action is given as a mean value, an upper value or a nominal value.

3.2 CLASSIFICATION OF ACTIONS Actions are classified as: Permanent Variable Accidental Seismic

Frequency

(1) (2) (3) (4)

5% of results

Loads Mean load

1,64s

Characteristic load

FIGURE 3.1 Variation in loads

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Chapter 3 / Actions In addition actions can be classified by: (1) Variation in time: (a) Permanent actions (G), for example, self-weight and fixed equipment. (b) Variable actions (Q), for example, imposed loads, wind actions or snow loads. (2) Spacial variation: (a) Fixed actions, for example, structures sensitive to self-weight. (b) Free actions which result in different arrangements of actions, for example, movable imposed loads, wind actions and snow loads.

3.3 ACTIONS VARYING IN TIME 3.3.1 Permanent Actions (G) Permanent actions are due to the weight of the structure, that is, walls, permanent partitions, floors, roofs, finishes and services. The actual weights of the materials (Gk ) should be used in the design calculations, but if these are unknown values of density in kN/m3 may be obtained from EN 1991-1-1 (2002). Also included in this group are water and soil pressures, prestressing force and the indirect actions such as settlement of supports.

3.3.2 Variable Actions (Q) (a) Imposed floor loads are variable actions and for various dwellings are given in EN 1991-1-1 (2002). These loads include a small allowance for impact and other dynamic effects that may occur in normal occupancy. They do not include forces resulting from the acceleration and braking of vehicles or movement of crowds. The loads are usually given in the form of a distributed load or an alternative concentrated load. The one that gives the most severe effect is used in design calculations. When designing a floor it is not necessary to consider the concentrated load if the floor is capable of distributing the load and for the design of the supporting beams the distributed load is always used. When it is known that mechanical stacking of materials is intended, or other abnormal loads are to be applied to the floor, then actual values of the loads should be used, not those obtained from EN 1991-1-1 (2002). In multi-storey buildings the probability that all the floors will simultaneously be required to support the maximum loads is remote and reductions to column loads are therefore allowed. (b) Snow roof loads are variable actions and are related to access for maintenance. They are specified in EN 1991-1-3 (2002) and, as with floor loads, they are expressed as a uniformly distributed load on plan, or as an alternative concentrated load. The magnitude of the loads decrease as the roof slope increases and in special situations, where roof shapes are likely to result in drifting snow, then loads are increased.

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29

(c) Wind actions are variable but for convenience they are expressed as static pressures in EN 1991-1-4 (2002). The pressure at any point on a structure is related to the shape of the building, the basic wind speed, topography and ground roughness. The effects of vibration, such as resonance in tall buildings must be considered separately. (d) Thermal effects need to be considered for chimneys, cooling towers, tanks, hot and cold storage, and services. They are classed as indirect variable actions. Elements of structures which are restrained or highly redundant introduce stresses which need to be determined (EN 1991-1-5 (2003)). (e) EN 1991-1-2 (2002) covers the actions to be taken into account in the structural design of buildings which are required to give adequate performance in fire.

3.3.3 Accidental Actions (A) (a) Accidental actions during execution include scaffolding, props and bracing (EN 1991-1-6 (2002)). These may involve consideration of construction loads, instability and collapse prior to the completion of the project. Erection forces are of great importance in steelwork construction because prefabrication is normal practice. Compression members which will be restrained in a completed structure may buckle during erection when subject to relatively minor forces. Joints which are rigid when fully bolted may, during erection, act as a pin and induce collapse of the structure. Suspension points for members or parts of structures may have to be specified to avoid damage to components. It is extremely difficult to anticipate all possible erection forces and the contractor is responsible for erection which should be carried out with due care and attention. Nevertheless a designer should have knowledge of the most likely method of erection and design accordingly. If necessary temporary stiffening or supports should be specified, and/or instructions given. (b) Accidental actions include impact and explosions which are covered in EN 1991-17 (2004). No structure can be expected to resist all actions but it must be designed so that it does not suffer extreme damage from probable actions, for example, vehicle collisions in a multi-storey car park. Local damage from accidental actions is acceptable. (c) When designing for earthquakes the inertial forces must be calculated as described in EN 1998-8 (2004). This is not of major importance in the UK. Actions induced by cranes and machinery are dealt with in EN 1991-3 (2004).

3.4 DESIGN VALUES OF ACTIONS Partial safety factors allow for the probability that there will be a variation in the effect of the action, for example, a variable action is more likely to vary than a permanent action. The values also allow for inaccurate modelling of the actions, uncertainties

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Chapter 3 / Actions in the assessment of the effects of actions, and uncertainties in the assessment of the limit state considered. The design value of an action is obtained by multiplying the characteristic value by a partial safety factor, for example, for a permanent action the design value Gd = γG Gk . For a variable action the design value Qd = ψ0 Qk or ψ1 Qk or ψ2 Qk . These represent combination, frequency and quasi-permanent values. The combination value (ψ0 Qk ) allows for the reduced probability that unfavourable independent actions occur simultaneously at the ultimate limit state. The frequency value (ψ1 Qk ) involves accidental actions and reversible ultimate limit states. The quasi-permanent value (ψ2 Qk ) also involves accidental actions and reversible serviceability limit states. Recommended values of ψ0 , ψ1 and ψ2 are given in EN 1990 (2002).

3.4.1 Combination of Design Actions For the ultimate limit state three alternative combinations of actions, modified by appropriate partial safety factors (γ), must be investigated: (a) Fundamental: a combination of all permanent actions including self-weight (Gk ), the prestressing action (P), the dominant variable action (Qk ) and combination values of all other variable actions (ψ0 Qk ). (b) Accidental: a combination value of the dominant variable actions (ψ0 Qk ). This combination assumes that accidents (explosions, fire or vehicular impact) of short duration have a low probability of occurrence. (c) Seismic: reduces the permanent action partial safety factors (γG ) with a reduction factor (ξ) between 0,85 and 1. For the serviceability limit state three alternative types of combination of actions must be investigated: (a) The characteristic rare combination occurring in cases when exceeding a limit state causes permanent local damage or deformation. (b) The combination which produces large deformations or vibrations which are temporary. (c) Quasi-permanent combinations used mainly when long-term effects are important. A combination of actions can be symbolically represented for design purposes, for example, for one of three conditions at the ultimate limit state:

γG Gk + γP P +

γ Q ψ 0 Qk

Similar equations can be formed for the other two conditions at ultimate limit state and for the three conditions at the serviceability limit state (EN 1990 (2002)).

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31

3.5 ACTIONS WITH SPACIAL VARIATION 3.5.1 Pattern Loading All possible actions relevant to a structure should be considered in design calculations. The actions should be considered separately and in realistic combinations to determine which is most critical for strength and stability of the structure. For continuous structures, connected by rigid joints or continuous over the supports, vertical actions should be arranged in the most unfavourable but realistic pattern for each element. Permanent actions need not be varied when considering such pattern loading, but should be varied when considering stability against overturning. Where horizontal actions are being considered pattern loading of vertical actions need not be considered. For the design of a simply supported beam it is obvious that the critical condition for strength is when the beam supports the maximum permanent action and maximum variable action at the ultimate limit state. The size of the beam is then determined from this condition and checked for deflection at the serviceability limit state. A more complicated structure is a simply supported beam with a cantilever as shown in Fig. 3.2(a). Assuming that the beam is of uniform section and that the permanent actions are uniformly applied over the full length of the beam, it is necessary to consider various combinations of the variable actions as shown in Figs. 3.2(b)–(d). Although partial

A

B

C

(a) Imposed design load Dead design load (b) Imposed design load Dead design load (c) Imposed design load Dead design load (d)

FIGURE 3.2 Pattern loading

3m

Chapter 3 / Actions

3m

Imposed loads

b

3m

a

3m

•

3m

32

5m

5m

5m

FIGURE 3.3 Multi-storey frame

loading of spans is possible this is not generally considered except in special cases of rolling actions (e.g. a train on a bridge span). For a particular section it is not immediately apparent which combination of actions is most critical because it depends on the relative span dimensions and magnitude of the actions. Therefore calculations are necessary to determine the condition and section for maximum bending moment and shear force at the ultimate limit state. Analysis of a multi-storey building is more complicated as shown by Holicky (1996). Where loads on other storeys affect a particular span they may be considered as uniformly distributed (EN 1991-1-1 (2002)). However, the critical positioning of the loads may be different, for example, Fig. 3.3 shows the load positions for verification of the bending resistance at points a and b. In other situations, for example, when checking the overturning of a structure, the critical combination of actions may be the minimum permanent action, minimum imposed action and maximum wind action.

3.5.2 Design Envelopes The effect of pattern loading can be seen by constructing a design envelope. This is a graph showing, at any point on a structural member, the most critical effect that results from various realistic combinations of actions. Generally the most useful envelopes

Structural Design of Steelwork to EN 1993 and EN 1994 A

B

(2) 91,0 81,7

(1) (3)66,4

11,7

20,0

•

33

C

Case 1 Case 2 Case 3 Envelope 38,3 99,0 108,3(1)

Shear force (kN) 66,4 (1)(3)

2,34 m

20 F

E 6,9

0,21 m

87,9 109(2) 2,40 m Bending moment (kNm)

FIGURE 3.4 Example: Design envelope

are for shear force and bending moment at the ultimate limit state. The formation and use of a design envelope is demonstrated by the following example.

EXAMPLE 3.1 Example of a design envelope. The beam ABC in Fig. 3.4-carries the following characteristic loads: Dead load Gk = 10 kN/m on both spans; Imposed loads Qk = 15 kN/m on span AB, 12 kN/m on span BC. Sketch the design envelope for the bending moment and shear force at the ultimate limit state. Indicate all the maximum values and positions of zero bending moment (points of contraflexure). The maximum and minimum design loads on the spans are: Maximum on AB = γG Gk + γQ Qk = 1,4 × 10 + 1,6 × 15 = 38 kN/m. Maximum on BC = γG Gk + γQ Qk = 1,4 × 10 + 1,6 × 12 = 33, 2 kN/m. Minimum on AB or BC = γG Gk = 1,0 × 10 = 10 kN/m.

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Chapter 3 / Actions Consider the following design load cases: (1) Maximum on AB and BC. (2) Maximum on AB, minimum on BC. (3) Minimum on AB, maximum on BC. The bending moment and shear force diagrams are shown in Fig. 3.4.

Comments (a) Only the numerical value of the shear force is required in design, the sign however, may be important in the analysis of the structure. (b) Positive (sagging) bending moments indicate that the bottom of the beam will be in tension; negative (hogging) moments indicate that the top of the beam will be in tension. (c) The envelope, shown as a heavy line, indicates the maximum values produced by any of the load cases. Note that on AB the envelope for shear force changes from case (2) to case (1) at the point where the numerical values of the shear force are equal. This process is tedious and an experienced designer knows the critical action combinations and the positions of the critical values and avoids some of the work involved. Alternatively the diagrams can be generated from input data using computer graphics. In more complicated structures and loading situations, envelopes are useful in determining where a change of member size could occur and where splices could be inserted. In other situations wind action is a further alternative to combinations of permanent and variable actions.

REFERENCES EN 1990 (2002). Basis of structural design. BSI. EN 1991-1-1 (2002). Densities, self weight and imposed loads. BSI. EN 1991-1-2 (2002). Actions on structures exposed to fire. BSI. EN 1991-1-3 (2002). Snow loads. BSI. EN 1991-1-4 (2002). Wind loads. BSI. EN 1991-1-5 (2003). Thermal actions. BSI. EN 1991-1-6 (2002). Actions during execution. BSI. EN 1991-1-7 (2004). Accidental actions due to impact and explosions. BSI. EN 1998-8 (2004). Design of structures for earthquake resistance. BSI. EN 1991-3(2004). Actions induced by cranes and machinery. BSI. Holicky, M. (1996). Densities, self weight and imposed loads on buildings. Czech Technical University, Prague.

Chapter

4 / Laterally Restrained Beams

4.1 STRUCTURAL CLASSIFICATION OF SECTIONS (CL 5.5, EN 1993-1-1 (2005)) Chapters 4 and 5 are concerned with the design of members which are predominantly in bending, that is, where axial loads, if any, are small and transverse shear forces are not excessive. Chapter 4 contains basic theoretical work on section properties and the design of laterally restrained beams using Class 1 standard sections and plastic methods of analysis. Sections of steel beams in common use are shown in Fig. 4.1. The rolled sections shown at (a) are used most often and of these the ‘I’ section is used widely. Some sections are of uniform thickness while others are of different thickness for the web and flange. The rolled sections are generally in stock, are lowest in cost, require less design and connections are straightforward. Hollow sections are not as efficient in bending but corrosion resistance is better and aesthetically they may be more acceptable. Cold formed sections are thinner and are therefore more susceptible to corrosion unless protected, however they are very economical for use as puriins. Fabricated sections are used when a suitable rolled section is not available, but costs are higher and delivery times are longer. Castellated sections are used for large spans with relatively low loads and where transverse shear forces are not excessive. Tapered beams are efficient in resisting bending moments but must be checked for shear forces. Composite steel–concrete sections are used for floors. The four classes of cross-section of steel ‘I’ beams are described in cl 5.5.2, EN 1993-1-1 (2005). To allow for flange buckling sections are reduced to effective sections. All members subject to bending should be checked for the following at critical sections: (a) (b) (c) (d) (e)

A combination of bending and shear force Deflection Lateral restraint Local buckling Web bearing and buckling.

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Chapter 4 / Laterally Restrained Beams

(a) Hot rolled

(b) Hollow

(c) Cold formed, light gauge

Crane rail

Channel welded to an ‘I’ beam (d) Fabricated

Weld

Cut

Standard ‘I’ beam (e) Tapered Weld

Cut

Standard ‘I’ beam

60°

(f) Castellated Shear connectors

Reinforced concrete

Reinforced concrete (g) Composite

Steel beam encased in concrete

FIGURE 4.1 Types of steel beams

This chapter is concerned with members which are predominantly subject to bending and where lateral torsional buckling and local buckling of the compression flange are prevented. It is important to recognize the characteristics of these two forms of buckling shown in Fig. 4.2. Lateral torsional buckling exhibits vertical movement (bending about about the y–y axis), lateral displacement (bending about the z–z axis),

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37

Compression flange

Local buckling

Moment

z x y

y x

Load

Moment

z (a) Lateral torsional buckling of a cantilever

(b) Local buckling of a flange

FIGURE 4.2 Buckling of beams

and torsional rotation (rotation about the x–x axis). Local buckling exhibits local deformation of an outstand, for example, a flange of an ‘I’ beam. Lateral torsional buckling occurs when the buckling resistance about the z–z axis and the torsional resistance about the x–x axis are low. The buckling resistance about the z–z axis can be improved by lateral restraints, for example, transverse members which prevent lateral movement of the compression flange. Local buckling occurs when the flange outstand to thickness ratio (b/tf ) is high and is avoided by choosing Class 1 sections (cl 5.5.2, EN 1993-1-1 (2005)). Where both types of buckling are prevented, as for Class 1 and Class 2 sections, then the section can be stressed to the maximum design stresses in bending, i.e. plastic methods of analysis and design can be used. If Class 3 or Class 4 sections are used the plastic moment capacity is reduced to an elastic moment capacity. If a steel ‘I’ section is used as a simply supported beam and loaded with a uniformly distributed load then the bending moment distribution varies parabolically. If the section is bent about a major axis then the stress distribution at centre span at various stages of loading is shown in Fig. 4.3(c). In the early stages of loading the stress distribution is elastic, then elastic–plastic and finally fully plastic. The corresponding moment curvature relationship is shown in Fig. 4.3(b). The fully plastic stage corresponds to the condition for the tensile stress–strain relationship for the steel shown in Fig. 4.3(a). Theoretically the load cannot be increased beyond this plastic condition but strain hardening occurs and this increases the resistance. Note that for full plasticity large strains occur, of the order of 20%, which makes mild steel ideal, while other steels with less plastic strain behave in a more brittle fashion. Although bending is the predominant design criteria checks must be made for the magnitude of the shear stresses. Shear stresses are introduced from vertical shear forces, or torsion moments (cls 6.2.6 and 6.2.7, EN 1993-1-1 (2005)).

Chapter 4 / Laterally Restrained Beams

Moment

•

Stress

38

Plastic Elastic

Elastic–plastic

Plastic Strain hardening

Elastic Elastic–plastic

Strain (a) Tensile stress–strain relationship

Strain hardening

Curvature (1/R ) (b) Moment–curvature relationship

sy

Neutral axis

(c) Change of bending stress distribution as a plastic hinge forms

(d) Development of a plastic hinge

FIGURE 4.3 Development of a plastic hinge

For the design of beams calculations are required in the elastic stage of behaviour, for example, stresses and deflections, and also at the fully plastic stage (e.g. collapse load). The calculations involve certain basic section properties.

4.2 ELASTIC SECTION PROPERTIES AND ANALYSIS IN BENDING 4.2.1 Sectional Axes and Sign Conventions For all standard sections rectangular centroidal axes y–y and z–z are defined parallel to the main faces of the section, as shown in Fig. 4.4. The position of these axes is given in Section Tables. For angles, and other sections where the rectangular and principal axes do not coincide, the principal axes are denoted by u–u and v–v. The major axis u–u is conventionally inclined to the y–y axis by an angle α, as shown in Figs 4.4(e) and (f). For equal angles, α = 45◦ . For problems involving simple uniaxial or biaxial bending of symmetrical sections a strict sign convention is not necessary, but for the solution of complex problems it is desirable. In this chapter the positive conventions of sagging curvature and downward deflections are adopted; and the direction of the angle α is anti-clockwise, consistent with the Section Tables. Fig. 4.5(a) shows the coordinates of a point P in the positive

•

Structural Design of Steelwork to EN 1993 and EN 1994 z

z

z

y

y

y

y

y

y

z

z (a) ‘I’ section

z

(b) Tee

v

z

(c) Hollow section

z

y

a y

y

z

(d) Channel

u

v

(e) Unequal angle

u a 45° y

y

u z

z

v

u y

39

z

v

(f) Equal angle

FIGURE 4.4 Sectional axes

Fv

Fz Mz

Mv

α

z

y v

Mu My

O

Pu

u

O

Fu

y

Mz Vy

Vy

O Fy

My y

y

z

Vz

z

x

Vz z

v

(a) Coordinates and external loads

x (b) Stress resultants

FIGURE 4.5 Sign conventions

quadrant of a section and the positive directions for the externally applied forces and couples. The positive directions of the corresponding stress resultants (shear forces and bending moments) are shown for the horizontal and vertical planes in Fig. 4.5(b). Positive directions relative to the u–u and v–v axes can be inferred. The convention for the moments has been chosen so that positive moments give tensile stresses in the positive quadrant of the section.

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Chapter 4 / Laterally Restrained Beams Normally the coordinates of points in a section relative to the rectangular axes are known, or can easily be obtained. The coordinates relative to the principal axes are given by u = y cos α + z sin α v = z cos α − y sin α

(4.1)

External forces and shear forces transform in exactly the same way, thus Fu = Fy cos α + Fz sin α Fv = Fz cos α − Fy sin α

(4.2)

However the directions chosen for the moments are consistent with the rules for a right hand set of axes, which gives rise to changes in sign, thus Mu = My cos α − Mz sin α Mv = Mz cos α + My sin α

(4.3)

4.2.2 Elastic Second Moments of Area This property is derived from the simple theory of elastic bending (Croxton and Martin Vol 1 (1987 and 1989)). In design it is used to calculate stresses and deflections in the elastic stage of behaviour, that is, at service loads. Second moments of area for all standard sections are given in Section Tables but for fabricated sections they must be calculated. The procedure involves application of the theorems of parallel axes which, for the single element of area A in Fig. 4.6, can be stated as follows: Iy = Ia + Az2 Iz = Ib + Ay2

(4.4)

Iyz = Iab + Ayz where Iy = elemental second moment of area about y–y Iz = elemental second moment of area about z–z Iyz = elemental product moment of area about y–y and z–z a–a and b–b are centroidal axes through the element, parallel to y–y and z–z, respectively. For the determination of Iyz , which can be either positive or negative, the correct signs must be allocated to the coordinates y and z. The positive directions are indicated by arrows in Fig. 4.6.

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41

O

y b z c

a

a y

Area A b

z

FIGURE 4.6 Parallel axes for an element

When second moments of area about the rectangular axes have been computed, the direction of the principal axes can be obtained from tan 2α = 2Iyz /(Iz − Iy )

(4.5)

The principal second moments of area are then give by Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α

(4.6)

If Iy is arranged to be greater than Iz , then α will be less than 45◦ and Iu will be the major principal second moment of area. A negative result for Eq. (4.5) indicates that α is to be measured clockwise from y–y.

4.2.3 Elastic Section Moduli These values are derived from the second moments of area by dividing by the distance to the extreme fibres (i.e. Wel = I/z). Values of section moduli are given in Section Tables. For structural tees two values of Wel are given, referring to the extreme fibres in the table and the stalk.

4.2.4 Elastic Bending of Symmetrical Sections When either of the rectangular axes is an axis of symmetry the normal bending stress at any point in the section is given by σ=

Mx z Mz y + Iy Iz

(4.7)

If the directions of the bending moments and the coordinates are in accordance with the sign convention of Fig. 4.5, a positive result indicates that the stress is tensile.

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Chapter 4 / Laterally Restrained Beams For simple bending about the y–y axis, which is the most common event, the stress in the extreme fibres σmax =

My zmax My = Iy Wely

(4.8)

Similarly for bending about the z–z axis σmax =

My zmax My = Iy Wely

(4.9)

4.2.5 Elastic Bending of Unsymmetrical Sections When a section is subject to bending about an axis which is not a principal axis the effect is the same as if the section were subject to the components of the bending moment acting about the principal axis. In other words the bending is biaxial. For standard rolled angles the principal second moments of area and the directions of the principal axes are given in Section Tables. Transforming bending moments and coordinates to the principal axes by means of Eqs (4.3) and (4.1). Bending stress σ=

Mu v Mv u + Iu Iv

(4.10)

This is the same as Eq. (4.7) but with all the terms treated to the principal axes. If the sign convention of Fig. 4.5 is observed, a positive result indicates tension. In other cases the additional calculations required for the solution of problems by principal axes can be avoided by the use of ‘effective bending moments’. These are modified bending moments which can be considered to act about the rectangular axes of the section. The bending stress is then given by an expression having exactly the same from as Eq. (4.7) σ=

Mey z Mez y + Iy Iz

(4.11)

where Mey and Mez are effective bending moments about y–y and z–z axes, respectively and are given by Mey =

My − Mz Iyz /Iz 1 − Iyz2 /(Iy Iz )

Mez =

Mz − My Iyz /Iy 1 − Iyz2 /(Iy Iz )

(4.12)

These expressions are derived from the application of conventional elastic bending theory to curvature in both the yx and zx planes. One such derivation is given by Megson (1980).

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By successive differentiation with respect to x, the longitudinal dimension, similar expressions for the effective shear force and effective load intensity can be obtained, thus Vey =

Vy − Vz Iyz /Iy 1 − Iyz2 /(Iy Iz )

Vez =

Vz − Vy Iyz /Iz 1 − Iyz2 /(Iy Iz )

(4.13)

and fey =

( fy − fz Iyz /Iy ) 1 − Iyz2 /(Iy Iz )

fez =

( fz − fy Iyz /Iz ) 1 − Iyz2 /(Iy Iz )

(4.14)

It should be noted that the quantities Iy and Iz are interchanged in Eqs (4.13) and (4.14). This is because the expressions for the shear force and the load intensity in the y direction are obtained by successive differentiation of bending moments along the z–z axis and vice versa. All bending moment problems with unsymmetrical sections can be solved simply by replacing ordinary loads, shears, and bending moments by their effective counterparts. Note however that these effective counterparts have values related to both the y–y and z–z axes, even if the section is only loaded in the direction of one of the rectangular axes.

EXAMPLE 4.1 Principal axes for an unequal angle section. Find the directions of the principal axes and the values of the principal second moments of area for the angle section in Fig. 4.7(a). For the calculation of section properties the work is simplified considerably, with insignificant loss of accuracy, by using the dimensions of the section profile, that is, the shape formed by the centre line of the elements, as shown in Fig. 4.7(b). The position of the centroid O is found by taking moments of the area about the centre lines of each leg in turn

Areas

mm2

A B A C

140 × 20 = 2800 290 × 20 = 5800 Total = 8600

Chapter 4 / Laterally Restrained Beams 150

140 c z

cz

A

c y

B

cy

O a 290

y

O 300

•

20

44

u

20

z (a) Actual section

c

v

(b) Section profile

FIGURE 4.7 Example: principal axes for an unequal angle Taking moments about A B 8600cy =

5800 × 290 2

hence cy = 97,8 mm

Taking moments about A C 8600cz =

2800 × 140 2

hence cz = 22,8 mm

Hence for the full section (Fig. 4.7(a)) cy = 107,8 mm

and

cz = 32,8 mm

Coordinates of the centroids of the legs AB and AC are therefore given by Leg A B

140 − cz = 47,2 mm 2 z = −cy = −97,8 mm

Leg A C

y = −cz = −22,8 mm

y=

z=

290 − cy = 47,2 mm 2

The second moments of area about the rectangular axes are obtained in the usual way be applying the parallel axes formula to each leg. bh3 + A(leg)z2 12 where b and h are dimensions of the leg parallel to the y–y and z–z axes, respectively. Iy (leg) =

Structural Design of Steelwork to EN 1993 and EN 1994

Leg A B

•

45

203 + 2800 × (−97,8)2 = 26,87E6 mm4 12 2903 + 5800 × 47,22 = 53,57E6 mm4 20 × 12

140 ×

Leg A C

Iy = 80,44E6 mm4 Iz (leg) =

hb3 + A(leg)y2 12

Leg A B Leg A C

20 ×

290 ×

1403 + 2800 × 47,22 = 10,81E6 mm4 12

203 + 5800 × (−22,8)2 = 3,21E6 mm4 12 Iz = 14,02E6 mm4

The product moment of area Iyz is obtained by applying the parallel axis formula to each leg Iyz (leg) = Iab + A(leg)yz For each leg the term Iab is equal to zero, because the parallel axes through the centroid of the leg are principal axes. Leg A B

2800 × 42,7 × (−97,8) = −12,93E6 mm4

Leg A C

5800 × (−22,8) × 47,2 = −6,24E6 mm4 Iyz = −19,17E6 mm4

Direction of the principal axes from Eq. (4.5) tan 2α =

2Iyz Iz − I y

2 × (−19, 17) 2α = arctan 14,02 − 80,44

hence α = 15◦

Principal second moments of area from Eq. (4.6) Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α Substituting values Iu = 85,58E6 mm4

and

Iv = 8,88E6 mm4 .

Chapter 4 / Laterally Restrained Beams bf tf

z

P cy

•

R y

y

h

46

100 kN m tw Q 50 kN m z

FIGURE 4.8 Example: structural tee in biaxial bending

As a check on the transformation Iy + Iz = Iu + Iv which is correct.

EXAMPLE 4.2 Structural tee in biaxial bending. Calculate the maximum extreme fibre stresses in a standard 292 × 419 × 113 kg structural tee cut from a Universal Beam. The tee is loaded by two moments as shown in Fig. 4.8. From Section Tables bf = 293,8 mm, h = 425,5 mm, tw = 16,1 mm, tf = 26,8 mm, Cy = 108 mm, Iy = 246,6E6 mm4 , Iz = 56,76E6 mm4 , Wely (flange) = 2,277E6 mm3 , Wely (toe) = 0,7776E6 mm3 , Welz = 0,3865E6 mm3 . By inspection the maximum compressive stress occurs at a point P because the stresses from both moments are compressive. σ= =

My Mz + Wely (flange) Welz 50E6 100E6 + = 173 MPa 2,277E6 0,3865E6

The maximum tensile stress can occur at point Q or point R, depending on the relative magnitude of the bending moments. It is necessary to check both points. Using the sign convention of Fig. 4.5 both bending moments are positive and the coordinates of the points are given by Point Q, Point R,

tw 16,1 = = 8,05 mm, z = h − cy = 425,5 − 108 = 317,5 mm. 2 2 bf 293,8 y= = = 146,9 mm, z = tf − cy = 26,8 − 108 = −81,2 mm. 2 2

y=

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From Eq. (4.7) the stresses My z Mz y + Iy Iz

σ=

100 × 317,5 50 × 8,05 + = 135,8 MPa 246,6 56,76 100 × (−81,2) 50 × 146,9 σR = + = 96,5 MPa 246,6 56,76

σQ =

Summarizing, the maximum stresses are: At Q, σQ = +135,8 MPa (tension) At P, σP = −173 MPa (compression)

EXAMPLE 4.3 Bending stresses in an unequal angle section. Calculate the bending stresses in the angle section shown in Fig. 4.9 where Iy = 80,44E6 Iz = 14,02E6

Iyz = −19,17E6 mm4

Effective moments from Eq. (4.12) Mey =

(My − Mz Iyz /Iz ) = 74,92 kNm (1 − Iyz2 /Iy Iz )

Mez =

(Mz − My Iyz /Iy ) = 32,86 kNm (1 − Iyz2 /Iy Iz )

Maximum compressive stress at A σA =

Mey zA Mez yA + Iy Iz

74,92E6 × (−107,8) 32,86E6 × (−32,8) + 80,44E6 14,02E6 = −177,3 MPa (compression)

=

Check for maximum tensile stress at B, and at C σB = =

Mey zB Mez yB + Iy Iz 74,92E6 × (−87,8) 32,86E6 × 117,2 + 80,44E6 14,02E6

= +192,9 MPa (tension) σC = =

Mey zC Mez yC + Iz Iy 74,92E6 × 192,2 32,86E6 × (−12,8) + 80,44E6 14,02E6

= +149,0 MPa (tension)

Chapter 4 / Laterally Restrained Beams 150 117,2

107,8

32,8 A

B 30 kN m

300

y

15 kN m

192,2 C 12,8

(a)

z

Mz

183,6 B 78,4

Mz y

My z

144,5 A

155

•

87,8

48

My Neutral axis x

(b) Positive bending moments

(c)

C125,6

FIGURE 4.9 Example: bending stresses in an unequal angle

Summarizing, the maximum stresses are: At B, σB = +192,9 MPa (tension) At A, σA = −177,3 MPa (compression) Note that although the position of the centroid and the values of the second moment of area can be calculated without significant error from the profile dimensions of the section, the same is not true of the stresses.

EXAMPLE 4.4 Bending about principal axes of an angle section. Recalculate the stresses at points A, B and C in the previous example considering bending about the principal axes.

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The coordinates of the points are transformed in accordance with Eq. (4.1), u = y cos α + z sin α sin α = 0, 2588 and Point

y/z axes y −32,8 117,2 −12,8

A B C

and v = z cos α − y sin α. cos α = 0, 9659, u/v axes

z u −107,8 −59,6 −87,8 90,5 192,2 37,5

v −95,6 −115,1 189,0

Bending moments transform in accordance with Eq. (4.3) Mu = My cos α − Mz sin α = 25,10 kNm Mv = Mz cos α + My sin α = 22,25 kNm Bending stresses from Eq. (4.10) σ=

Mu v Mv u + , Iu Iv

σA = −177,3

σB = 192,9

σC = 149,0 MPa

which are the same as obtained previously.

4.2.6 Elastic Analysis of Beams The elastic analysis of simply supported beams with examples of shear force and bending moment diagrams and deflection calculations are given in many introductory books on structural analysis (e.g. Croxton and Martin Vol. 1 (1987 and 1989)). The analysis of continuous beams is more complicated but there is a choice of methods such as area– moment, moment distribution, slope deflection, and matrix methods. These methods are also covered by Croxton and Martin Vol. 2 (1987 and 1989). with a computer program for analysis using matrix methods.

4.2.7 Elastic Deflections of Beams (cl 7.2.1, EN 1993-1-1 (2005)) The deflections under serviceability loads of a building or part should not impair the strength or efficiency of the structure or its components or cause damage to the finishes. When checking for deflections the most adverse realistic combination and arrangement of serviceability loads should be assumed, and the structure may assumed to be elastic. The theory and the methods of calculating deflections for static and hyperstatic structures are given in Croxton and Martin Vols. 1 and 2 (1987 and 1989). For simple beams standard cases can be superimposed and some useful cases are shown in Figs 4.10 and 4.11.

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Chapter 4 / Laterally Restrained Beams Maximum deflection (at free end) Q a 2(3 a)QL3/(6EI ) aL L (a) q

a 3(4 a)qL4/(24EI )

aL (b) q a 3(5 a)qL4/(120EI ) aL

FIGURE 4.10 Deflections of cantilevers

(c)

For simply supported beams the central deflection rather than the maximum is given, so that deflections from individual cases can be added. For most loading cases the central deflection only differs by a small percentage from the maximum. In case (a) of Fig. 4.11, for example, the difference is always within 2,5%. A notable exception is the case of equal end moments acting in the same direction, when the central deflection is zero. However in such a case the deflection at other points along the beam are likely to be small. A more accurate analysis can be formed if it is suspected that the deflection is likely to exceed the limit. Recommendations for limiting values for deflections are given in cl 7.2.1, EN 1993-1-1 (2005).

EXAMPLE 4.5 Deflections for a hyperstatic structure. The size of the members for the symmetrical structure shown in Fig. 4.12 have been determined and the structure requires to be checked for deflections in the elastic range of behaviour. The imposed variable characteristic loads are shown in Fig. 4.12 and the second moment of area is Iy = 127,56E6 mm4 . The bending moments (positive clockwise) at the joints are given in the following table. Joint

Span

Moment (kNm)

B B C C C G

AB BC BC CG CD CG

+60 −60 −6 +72 −66 +36

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Central deflection, and rotation at supports Q End 1

End 2 For a 0.5 d (3a 4a3)QL3/(48EI) u1 (2a 3a2 a3)QL2/(6EI ) u2 (a3 a)QL2/(6EI )

aL

(a)

L

q

d 5qL4/(384EI ) u1 qL3/(24EI ) u2 u1

(b)

q

For a 0.5 d (3a2 2a4)qL4/(96EI) u1 (a4 4a3 4a2)qL3/(24EI ) u2 (a4 2a2)qL3/(24EI)

(c) aL

d qL4/(120EI) u1 5qL3/(192EI ) u2 u1

q (d)

M1

M2 d (M1M2)L2/(16EI ) u1 (2M1M2)L /(6EI ) u2 (2M2M1)L /(6EI )

(e)

24 A

30 B

24 36

C

FIGURE 4.11 Displacements of simply supported beams

E 30

D 8

G 4

18

H 3

6

3

18

4

Dimensions in m; loads in kN.

FIGURE 4.12 Example: deflections of a symmetrical continuous structure

F

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Chapter 4 / Laterally Restrained Beams For all beams EIy = 210E3 × 127,56E6 = 26,79E12 N mm2

Span CD 1. Uniform load (Fig. 4.11(b)) ∂ul = =

5qL4 5QL3 = 384EIy 384EIy 5 × 36E3 × 12E33 = 30,2 mm 384 × 26,79E12

2. Concentrated loads (Fig. 4.11(a)) ∂cl = =

2[3a − 4a3 ]QL3 48EIy 2[3 × (3/12) − 4 × (3/12)3 ] × 24E3 × 12E33 = 44,3 mm. 48 × 26,79E12

3. End moments (Fig. 4.11(e)) ∂M = L2

(M1 − M2 ) 16EIy

= 12E32 ×

(−66 − 66)E6 = −44,3 mm (upwards) 16 × 26,79E12

Total deflection = ∂ul + ∂cl + ∂M = 30,2 + 44,3 + (−44,3) = 30,2 mm A limit for beams with plaster finish = L/350 = 12E3/350 = 34,3 mm

Span BC End moments (Fig. 4.11(e)) ∂ 3 = L2

(M1 − M2 ) 16EIy

= 18E32 ×

(−60 + 6)E6 = −40,8 mm (upwards) 16 × 26,79E12

An accurate analysis gives −42,4 mm at 7,35 m from B, A limit for beams with plaster finish = L/350 = 18E3/350 = 51,43 mm

Cantilever span AB For this span the deflection is due to the flexure of the cantilever, assuming the beam is horizontal at B, plus the effect of the anti-clockwise rotation of the beam at B, that is, −θ1 for span BC.

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End moments for span BC (Fig. 4.11(e)) θ1 = L ×

(2M1 − M2 ) 6EIy

= 18E3 ×

(−2 × 60 + 6)E6 = −0,01277 rad 6 × 26,79E12

1. Deflection at A due to rotation ∂R = −Lθ1 = −4E3 × (−0,01277) = 51,07 mm 2. Deflection due to load (Fig. 4.10(b)) ∂ul = =

a3 (4 − a)qL4 QL3 = 24EI 8EIy 30E3 × 4E33 = 8,96 mm 8 × 26,79E12

Total deflection = ∂R + ∂ul = 51,1 + 9,0 = 60,1 mm A limit for a cantilever beam with plaster finish 2L 2 × 4E3 = = 22,9 mm 350 350 If the deflection of the cantilever exceeds the limit and stiffening is required then increase the size of the section, or add flange plates. It is also necessary to stiffen spans AB and BC because the deflection is dependent on both.

4.2.8 Span/Depth Ratios for Simply Supported Beams An initial estimate for the depth of a simply supported ‘I’ beam carrying a uniformly distributed load can be obtained by using the deflection limit. If σ max is the maximum elastic bending stress at service load then from elastic bending theory σmax =

Mz QL h/2 = × Iy 8 Iy

rearranging Q=

16σmax Iy Lh

(i)

Assuming a deflection limit for beams with plaster finish of L 5QL3 = 350 384EIy

(ii)

Eliminating Q by combining Eqs (i) and (ii) and putting E = 210 GPa, the span depth ratio L 2880 = h σmax

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Chapter 4 / Laterally Restrained Beams If a beam is laterally restrained so that lateral torsional buckling does not occur, then if the maximum bending stress is approximately fy /1,5, the span/depth ratios for different grades of steel and different limits are: TABLE 4.1 Span/depth ratios for beams. Span/depth ratio Grade

Characteristic stress (fy )

L/350

L/250

S235 S275 S355

235 275 355

18,4 15,7 12,2

25,7 22,0 17,0

Note that since Youngs modulus (E) is a constant for all grades of steel, the stiffness of a beam does not increase with a higher grade of steel. If a design is governed by deflection then there is no advantage in using a higher grade of steel.

4.3 ELASTIC SHEAR STRESSES 4.3.1 Elastic Shear Stress Distribution for a Symmetrical Section When a beam is bent elastically by a system of transverse loads, plane sections no longer remain plane after bending, but are warped by shear strains. In most cases the effect is small and the errors introduced in the use of conventional bending theory are negligible. Important exceptions are discussed briefly in Section 4.3.2. Formulae for the calculation of shear stresses in an elastic beam are derived by considering the variation in bending stresses along a short length of beam. Consider the very short length of beam in Fig. 4.13(a). At a point S in the web the shear stresses on the vertical and the horizontal section are complementary and are given by the established formula (Eq. (6.20), EN 1993-1-1 (2005)) vs =

VAzc It

(4.15)

where V = the vertical shear force on the section A = the hatched area, that is, the part of the section between point S and the extreme fibres zc = the distance from the centroid of the area A to the neutral axis I = the second moment of area of the whole section about the neutral axis t = the thickness of the section at the point S.

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b

P

tf

tw

zc

NA

vs S

vs bf

Area A

vp tf

P

(a) Shear flow h

tw S

(b) Stress distribution

vs

FIGURE 4.13 Shear stresses in an ‘I’ beam

The formula cannot be used to obtain the vertical shear stress in the outstanding parts of the flange. However as this must be equal to zero at the top and bottom faces, it must be very small. In fact the resistance of the section to vertical shear is provided almost entirely by the web. The resultant of the longitudinal shear stress in the web is in equilibrium with the change in the normal tensile force on the area A due to the variation in bending moment along the beam. Similar longitudinal stresses exist in the flanges and give rise to horizontal complementary stresses in the directions shown. For example, at point P in the top flange. A = bf tf ,

tf = t,

zc =

(h − tf ) 2

and Eq. (4.15) becomes vp =

Vb(h − tf ) 2I

(4.16)

This expression is linear with respect to the variable b, and vp has a maximum value at the centre of the flange where b = bf /2, that is vp(max) =

Vbf (h − tf ) 4I

(4.17)

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Chapter 4 / Laterally Restrained Beams The complete distribution of shear stress on the cross-section is shown in Fig. 4.13(b). Equation (4.15) can be expressed in terms of the shear flow, which is the product of the shear stress and the thickness of the section, thus v = tvs =

VAzc I

(4.18)

In the longitudinal sense the shear flow is equal to the shear force per unit length of beam, and is a convenient quantity for the calculation of the shear force to be resisted by bolts or welds in a fabricated section.

4.3.2 Elastic Shear Stresses in Thin Walled Open Sections The shear stress distribution for a rectangular cross-section subject to a transverse shear force is shown in Fig. 4.14(a) and can be calculated using Eq. (4.15). The shear stress distribution for the open cross-section is different as shown in Fig. 4.14(b). the distribution for an ‘I’ section is shown in Fig. 4.14(c). Steel sections are usually composed of relatively thin elements, for which the analysis can be simplified by: (a) referring all dimensions to the profile of the section; (b) assuming that the shear stress does not vary across the thickness; (c) ignoring any shear stresses acting at right angles to the section profile. As these are equal to zero at each outside surface they must always be very small in a thin walled section. If it is further assumed that the load is applied in such a way that no twisting of the beam occurs, the shear flow at a point S on the profile of the section is given by s s Vey Vez vs = vo − tz ds − ty ds (4.19) Iy Iz o o here Vey and Vez are the effective shear forces obtained from Eq. (4.13) or by applying the effective loads obtained from Eq. (4.14). The variable s is the distance around the profile to the point of interest, starting from any point at which the shear flow

(a)

(b)

(c)

FIGURE 4.14 Distribution of shear stresses

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vo is known. At any open end, such as the end of a flange, the value of vo is zero. The direction of s can be chosen arbitrarily and, provided that the sign convention of Fig. 4.5 is adopted, a positive sign for vs indicates that the shear flow is in the direction chosen for s.

4.3.3 Elastic Shear Stresses in Thin Walled Closed Sections The shear stress and shear flow in a symmetrical closed section can be obtained directly from Eqs (4.15) and (4.18), respectively. For an unsymmetrical section Eq. (4.19) can be used, but analysis is complicated by the fact that that vo is not known at any point. The problem can be solved by first cutting the section at some point and finding the position of the shear centre in the resulting open section. The shear flow in the closed cross-section then results from the combined action of the applied shear loads transferred to the shear centre of the cut (open) section, and the torque on the closed section due to the transference of loads. Examples of shear stress distribution are given in Fig. 4.14. A similar approach is used to find the position of the shear centre of the closed section.

4.3.4 Elastic Shear Lag (cl 6.2.2.3, EN 1993-1-1 (2005)) The simple theory of bending is based on the assumption that plane sections remain plane after bending. In reality shear strains cause the section to warp. The effect in the flanges is to modify the bending stresses obtained by the simple theory, producing higher stresses near the junction of a web and lower stresses at points remote from it as shown in Fig. 4.15. This effect is described as ‘shear lag’. The discrepancies produced by shear lag are minimal in rolled sections, which have relatively narrow and thick flanges. However in plate girders, or box sections, having wide thin flanges the effects can be significant when they are subjected to high shear forces, especially in the vicinity of concentrated loads where the sudden change in shear force produces highly incompatible warping distortions. Shear lag effects can be allowed for by using effective widths (cl 3, EN 1993-1-5 (2003)).

Shear force

Bending stress non-uniform Warping displacements

FIGURE 4.15 Shear lag effects for an ‘I’ section

58

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Chapter 4 / Laterally Restrained Beams

(a)

(b) Shear centre

(c)

(d)

Centroid

FIGURE 4.16 Position of shear centre

Vz y′ Vy N B

C D

A

Shear centre Centroid

FIGURE 4.17 Shear centre–unsymmetrical section

4.3.5 Elastic Shear Centre Equation (4.19) is only valid if no twisting of the beam occurs at the section considered. Torsion in a section can be generated by a transverse load if the resultants of the shear stresses in the elements of the section produce a torque. To counteract this the line of action of the applied load must pass through the shear centre. In a symmetrical section the shear centre lies on an axis of symmetry, and loads applied along such an axis do not cause twisting. In some sections the position of the shear centre can be inferred from the direction of the shear flow (Fig. 4.16). In (a) the shear center lies at the intersection of the two axes of symmetry and is coincident with the centroid; in (b) and (c) it lies at the intersection of lines of shear flow; in (d), if the flanges are of the same size, the shear stresses in them set up opposing torques about the centroid which is therefore the shear centre. For the general case of an unsymmetrical thin walled open section subject to biaxial bending, with shear forces Vy and Vz , the position of the shear centre can be found by determination of the shear flow from Eqs (4.18) or (4.19), applying Vy and Vz in turn, assuming that they pass through the shear centre. Consider, for example, the section profile in Fig. 4.17. If point B is chosen as the fulcrum it is only necessary to find the resultant shear forces in the leg CD due to Vy and Vz in turn. These forces produce torques equal to Vy z and Vz y , respectively. By taking moments about B the values of

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59

z and y can be obtained. There is no need to calculate the shear stresses in AB or BC because their lines of action pass through the point B, and generate no moment. The resultant shear forces in CD are obtained by integrating the shear stresses obtained by Eq. (4.19) along the leg. The above process is tedious since, for each value of Vy and Vz , the corresponding effective shear forces Vey and Vez must be calculated and applied. If there is an axis of symmetry Eq. (4.18) can be used and the analysis is simplified.

EXAMPLE 4.6

Distribution of shear stresses for an angle section. Calculate the shear stresses in the simply supported angle section shown in Fig. 4.18(a). Iy = 80,44E6, Iz = 14,02E6, Iyz = −19,17E6 mm4 . To the left of mid-span the shear forces are Vy = 5

and

Vz = 10 kN

5

140 20 kN 117,2

s1

97,8

10

10 kN

22,8

y

s2

sp

192,2

6m

290

5

10

an

(a)

(b)

z

78,4 Vy y

2,4 (max)

20,3 0,9

z

155,2

Vy Vz

2,8 (max)

Vz (c)

(d)

FIGURE 4.18 Example: distribution of shear stresses for an unsymmetrical section

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Chapter 4 / Laterally Restrained Beams Effective shear forces from Eq. (4.13) Vey =

(Vy − Vz Iyz /Iy ) = 10,95 kN 1 − Iyz2 /Iy Iz

Vez =

(Vz − Vy Iyz /Iz ) = 24,97 kN 1 − Iyz2 /Iy Iz

Shear flow from Eq. (4.19) s s Vey Vez vs = vo − tz ds − ty ds Iy Iz o o

(i)

For the horizontal leg starting from the left hand end vo = 0,

s = s1 ,

y = (117,2 − s1 )

and

z = −97,8 mm

Substituting these values in (i) and integrating vs1 = 0,00781s21 − 1,224s1

(ii)

This equation shows that vs1 = 0 only when s1 = 0. Differentiating with respect to s1 and equating to zero gives a turning point at s1 = 78,4 mm. Hence from (ii) vs1 (max) = −47,96 N/mm, and s1 = 140 mm, vs1 = −18,28 N/mm. The negative signs indicate that the shear flow is in the opposite direction to s1 . For the vertical leg s = s2 ,

y = −22,8 mm,

z = (s2 − 97,8) mm,

vo = vA = −18,28 N/mm.

Substitution of these values into (i) gives vs2 = −18,28 + 0,9633s2 − 0,003104s22

(iii)

Solving (iii) for s2 shows that when vs2 = 0, s2 = 20,3 mm. There is also a turning point at s2 = 155,2 mm. Hence from (iii) vs2 (max) = 56,46 N/mm. The positive sign indicates that the shear flow is in the direction of s2 . As a check, putting s2 = 290 mm gives vs2 = 0, which is correct. The shear stresses in MPa are obtained by dividing the shear flows by the thickness, that is, 20 mm, and are plotted for the whole section in Fig. 4.18(d). This example demonstrates the method of analysis but the shear stresses are very low and do not justify such a detailed treatment. As a rough guide as to whether detailed analysis is required the shear forces are divided by the area of the appropriate leg. If the shear stress is only required at particular points in the section Eq. (4.15) can be used with the effective shear forces, taking each axis in turn and superimposing the results. Integration is avoided but the directions of the stresses have to be found by inspection. It can be seen from the position of the neutral axis in Example 4.3 that the maximum shear stresses occur where the neutral axis intersects the profile of the

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61

vz

b

A

y

d

y

t Shear centre

FIGURE 4.19 Example: shear centre of a channel

section, as in a symmetrical section. If these points have previously been found, then the maximum shear stress can be calculated directly as above, using Eq. (4.15).

EXAMPLE 4.7 Shear centre for a channel section. Find the position for the shear centre for the channel shown in Fig. 4.19 which has a uniform thickness. As the shear centre lies on the axis of symmetry y–y, there is no need to consider Vy . If point A at the intersection of axis y–y with the centre line of the web is taken as the fulcrum, then only the shear force in the flanges need to be considered since the resultant shear force in the web produces no moment about A. Equation (4.18) is vs =

VAzc I

The distribution of shear flow in the flanges in linear with zero at the ends and maximum at the web centre line. Hence, for maximum shear flow A = bt,

V = Vz ,

vs(max) =

Vz btd 2Iy

zc =

d 2

and

I = Iy , which gives

The resultant shear force is equal to half the maximum shear flow multiplied by the flange width, that is Force =

Vz b2 td 4Iy

The torque about point A from both flanges is equivalent to the torque produced by the applied shear force when it passes through the shear centre, thus V z y =

Vz b2 td2 4Iy

from which y =

b2 d2 t 4Iy

(4.20)

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4.4 ELASTIC TORSIONAL SHEAR STRESSES (CL 6.2.7, EN 1993-1-1 (2005)) 4.4.1 Elastic Torsion Generally torsion is not a major problem in the design of beams except for special cases. Torsional shear stresses arise from a variety of causes including, beams cranked or curved in plan, and loads whose line of action does not pass through the shear centre of the section. It is important to realize that in torsional problems the lever arm for the torque is measured from the shear centre, not the centroid. In particular, the distributed loads on unsymmetrical sections, such as angles or channels, usually act through the centroid of the section, as in the case of self-weight, although generally self-weight does not produce large torsional stresses. However, if other loads are offset from the shear centre the effect can be considerable, for example, for a 381 × 102 channel section the lever arm for loads acting through the centroid is approximately 50 mm. The total torsional moment at any section is the sum of the elastic torsion (St. Venant) plus the internal warping torsion. The warping torsion consists of a bi-moment plus warping.

4.4.2 Uniform and Non-uniform Elastic Torsion In general the cross-sections of members subject to torsion do not remain plane, but tend to warp. Warping is the change in geometric shape of the member so that crosssections do not remain plane as shown in Fig. 4.20. The degree of warping that takes place depends on the shape of the section, and is most pronounced in thin walled channels. In some sections, such as angles and tees, solid and hollow circular sections and square box sections of uniform thickness, warping is virtually non existent, while in others, such as closed box sections of general shape, its effect is small. In ‘I’ sections most of the warping takes place in the flanges and its effect on the web is very small and can be ignored. If torque is applied only at the ends of the member and warping is not restrained, the flanges remain virtually straight and maintain their original shape as shown in Fig. 4.20(a). The result is that the sectional planes of the flanges rotate in opposite directions, producing warping displacements which are constant along the whole length of the member. Under these circumstances the member is said to be in the state of uniform, or St. Venant, torsion. If warping is prevented, for example, by rigid supports at the ends of the member, the torsional stiffness is increased and longitudinal tensile and compressive stresses are induced. In practice warping is not so obvious but may arise from the action of structural connections, or from the incompatibility of warping displacements that occur when the torque is not uniform along the length of the member. Warping restraint increases the torsional stiffness of a member, and at any point along its length the applied torque is resisted by the two components, one due to the St. Venant torsion,

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Mb

(a) tf

H

h Mx H

Mb

(c)

T

(b)

T Stress (d)

Stress (e)

FIGURE 4.20 Torsion of thin walled sections

and the other to warping torsion from the effects of the restraints. The proportions of the two components depend on the type of loading and the distance from a restraint. Both components of torsion produce shear stresses parallel to the walls of the section, and their combined effects can locally be greater than the effect of St. Venant alone. However in members other than channels with very thin walls the increase in shear stresses can usually be ignored in design. In beams the maximum shear stress can be obtained approximately by combining the effects of transverse shears, using Eq. (4.15), with the shear stresses from torsion, assuming that the whole of the applied torque results in St. Venant torsion. A more significant effect of warping restraint in the design of beams is the introduction of longitudinal stresses. The effect is illustrated for an ‘I’ beam in Fig. 4.20(b). In this case the warping displacements are confirmed to the flanges, whose positions, if warping were allowed to occur freely, are shown by dotted outlines. Bi-moments Mb are induced in the planes of the flanges when warping is restrained, and these give rise to tensile and compressive stresses as shown. A full treatment of the analysis of members subject to warping torsion is beyond the scope of this book and the reader is referred to Zbirohowski-Koscia (1967). The analysis is tedious, but for ‘I’ beams a conservative estimate of the longitudinal stresses due to warping torsion can be obtained by assuming that each flange acts independently and is bent in its own plane by an analogous system of lateral loads which replace the applied torques, as in Fig. 4.20(c). The value of the equal and opposite lateral loads H, analogous to the applied torque Mx is H=

Mx h − tf

(4.21)

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Chapter 4 / Laterally Restrained Beams The ends of the flanges can be assumed to be either fixed or simply supported, depending on whether or not warping is restrained by the structural connections. The results obtained by this method are conservative because in reality the warping stresses are produced only by the warping component of the applied torque, not the whole torque as assumed. However, Eq. (4.21) can be useful in preliminary designs where it is necessary to assess whether the effects of torsion are likely to be significant.

4.4.3 Elastic Torsion of Circular Sections The elastic (St. Venant) theory of torsion of prismatic members with solid or hollow sections can be expressed by the well established formula T v Gθ = = It r L

(4.22)

where T is the torsional moment It is the torsion constant which, for a circular section, is equal to the polar second moment of area v is the shear stress at radius r G is the shear modulus θ is the angle of twist L is the length of the member The polar second moment of area for a solid section of radius R is It =

πR4 2

(4.23)

For a thin walled tube of mean radius Rm and wall thickness t, an approximate formula is It = 2πR3m t

(4.24)

The error is below 3% for t/Rm ratios of 1/3 or less, and is on the safe side. The polar second moment of area (It ) is twice the second moment of area (I) about a diameter, values for which are given in Section Tables. The elastic distribution of shear stress along the radius of a solid circular section is linear, with zero at the centre and a maximum at the outside surface. For a thin walled tube, the stress varies linearly across the wall thickness and in the range of standard structural tubes unsafe errors in the shear stress of up to about 18% are introduced by the use of Rm in Eq. (4.22) instead of the outside radius.

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4.4.4 Elastic Torsion of Thin Walled Open Sections The torsional constant It for a thin rectangle of width b and thickness t is It =

bt3 3

(4.25)

This formula is accurate when b/t is infinite and gives unsafe errors of 6% when b/t = 10, and 10% when b/t = 6. The b/t ratios are typical of the flanges of Universal Column and Beam sections. Most sections in steelwork design are composed of thin rectangles, and for a complete section the torsional constant can be obtained by summing the torsional constants for each rectangular element, that is

3 bt It = (4.26) 3 Values of It are given in Section Tables. In standard rolled sections the root fillets at the junctions of the web and the flanges give additional torsional stiffness. The shear stresses in an open section under St. Venant torsion vary from zero on the centre line of the wall to a maximum on the outside surface (Fig. 4.20(d)), and their direction is reversed on each side of the centre line. The shear flow constitutes a closed loop. The maximum stress in any element of thickness t is v=

Tt It

(4.27)

The maximum shear stress in a section therefore occurs in the thickest element. At re-entrant corners the flow lines are crowded together, giving rise to very high stress concentrations. The effect is reduced by fillet radii. The shear stress is zero at the outside corners. The angle of twist θ=

TL GIt

(4.28)

where It is calculated from Eq. (4.26).

4.4.5 Elastic Torsion of Thin Walled Closed Sections The shear stress distribution for closed sections is shown in Fig. 4.20(e). The flow is unidirectional with respect to the profile, contrasting with open sections. Variations in stress across the thickness of the section are ignored. The shear flow is constant at all points on the profile and is given by v=

T 2A

where A is the cross-sectional area of the profile.

(4.29)

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Chapter 4 / Laterally Restrained Beams The shear stress is a maximum in the thinnest part of the section and is obtained by dividing the shear flow by the thickness. This is in direct contrast to the open section where maximum shear stress occurs in the thickest part. Angle of twist TL θ= 4A2 G

ds t

(4.30)

As in open sections additional stresses are introduced when warping is restrained. Equations (4.29) and (4.30) are derived from the Bredt–Batho hypothesis in which it is assumed that the shape of the section remains unchanged. To ensure that this assumption remains valid it may be necessary to stiffen the section with internal diaphragms at intervals along its length, and at points where concentrated loads are applied.

EXAMPLE 4.7 Torsion and transverse shear in a box section. Find the maximum shear stress in the box section shown in Fig. 4.21 which is subject to a torque T = 200 kNm and a shear force V = 500 kN. Assume that the section is adequately stiffened to prevent distortion of the profile and ignore the effects of warping restraints. Calculate the angle of twist per metre length. The torsion and shear may be considered separately and the resulting shear flows are shown in Figs 4.21(a) and (b). For both calculations the profile dimensions shown in Fig. 4.21(c) can be used. The shear modulus is 80 GPa. First consider torsion. The area enclosed by the profile A = 790 × 380 = 0,3002E6 mm2 . From Eq. (4.29) the shear flow v=

T 200E6 = = 333 N/mm 2A 2 × 0,3002E6

The shear stress is maximum in the web where the section is thinnest vt =

333 v = = 33,3 MPa t 10

500 kN

800

790

20 10

T

380

400

T 200 kNm (a) Torsion

(b) Shear

(c) Profile

FIGURE 4.21 Example: torsion and transverse shear in a box section

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From Eq. (4.30) the angle of twist TL ds θ= 2 t 4A G 200E6 × 1E3 790 380 = × 2 + 20 10 (4 × 0,30022 × 1E12 × 80 × 1E3) = 0,00107 rad/m This is small which shows that the box is very stiff torsionally. Now consider the direct shear force. The maximum shear stress is in the webs at the neutral axis, and the first moment of areas 380 = 3,002E6 mm3 2 380 380 = 2 × 10 × × = 0,361E6 mm3 2 4

Azc(flange) = 20 × 790 × Azc(webs)

Total = 3,363E6 mm3 Second moment of area 3803 2 I = 2 10 × + 20 × 790 × 190 = 1232E6 mm4 12 Direct shear stress from Eq. (4.15) vs =

VAzc 500E3 × 3,363E6 = = 68,2 MPa It 1232E6 × 2 × 10

Combing torsional and direct shear stresses v(combined) = vt + vs = 33,3 + 68,2 = 101,5 MPa Information on combining shear and torsion in design calculations is given in cl 6.2.7, EN 1993-1-1 (2005).

4.5 PLASTIC SECTION PROPERTIES AND ANALYSIS 4.5.1 Plastic Section Modulus Plastic global analysis may be used in the design of steel structures for class 1 crosssections (cl 5.6, EN 1993-1-1 (2005)). For the general case of a steel section symmetrical about the plane of bending, the stress distributions in the elastic and fully plastic state are shown in Fig. 4.3(c). For equilibrium of normal forces, the tensile and compressive forces must be equal. In the elastic state, when the bending stress varies from zero at the neutral axis to a maximum at the extreme fibres, this condition is achieved when the neutral axis passes through the centroid of the section. In the fully plastic state,

Chapter 4 / Laterally Restrained Beams b f 420,5

A4/2

z A2 tw 21,5 z2

y

A3 A1

z1

y

tf 36,6

•

h 920,5

68

y3 tw/4

A4/2 y4

z (a)

(b)

(c)

FIGURE 4.22 Example: plastic section modulus for an ‘I’ section

because the stress is uniformly equal to the yield stress, equilibrium is obtained when the neutral axis divides the section into two equal areas. Mpl = (first moment of area about the plastic NA)fy

(4.31)

EXAMPLE 4.8 Plastic section moduli for an ‘I’ section. Determine the plastic section moduli about the y–y and z–z axes for the ‘I’ section shown in Fig. 4.22(a). The section is for a 914 × 419 × 388 kg Universal Beam with the root radius omitted. To determine the plastic section modulus about the y–y axis divide the section into A1 and A2 as shown in Fig. 4.22(b) where h 920,5 A1 = tw = 21,5 = 9895,375 mm2 2 2 A2 = (bf − tw )tf = (420,5 − 21,5)36,6 = 14603,5 mm2 and h 920,5 = = 230,125 mm 4 4 h tf 920,5 − 36,6 z2 = − = = 441,95 mm 2 2 2 z1 =

Plastic section modulus Wply = 2(A1 z1 + A2 z2 ) = 2(9895,375 × 230,125 + 14603,4 × 441,95) = 17,462E6 mm3 The value obtained from Section Tables is 17,657E6 mm3 which is slightly greater because of the additional material at the root radius.

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Similarly for the plastic section modulus about the z–z axis divide the section into areas A3 and A4 as shown is Fig. 4.22(c) where (h − 2tf )tw (920,5 − 2 × 36,6)21,5 = = 9108,475 mm2 2 2 bf 420,5 A4 = 2 tf = 2 36,6 = 15390,3 mm2 2 2 A3 =

and tw 21,5 = = 5,375 mm 4 4 bf 420,5 y4 = = = 105,125 mm 4 4

y3 =

Plastic section modulus Wplz = 2(A3 y3 + A4 y4 ) = 2(9108,475 × 5,375 + 15390,3 × 105,125) = 3,334E6 mm3 . The value obtained from Section Tables is 3,339E6 mm3 which is slightly greater because of the additional material at the root radius. From Section Tables the ratio of plastic/elastic section modulus (shape factor) for this section about the x–x axis is Wply /Wely = 17,657E6/15,616E6 = 1,1307 The value of the shape factor for bending about the y–y axis generally quoted for ‘I’ sections in current use in design is 1,15. The corresponding value of the shape factor for bending about the z–z axis is Wplz /Welz = 3,339E6/2,160E6 = 1,55 which is typical for an ‘I’ section.

EXAMPLE 4.9 Plastic section moduli for a channel section. Determine the plastic section moduli about the y–y and z–z axes for the channel section shown in Fig. 4.23(a). The section is for a 432 × 102 × 65,54 kg channel with the root radius omitted. To determine the plastic section modulus about the y–y axis divide the section into A1 and A2 and shown in Fig. 4.23(b) where h 431,8 A1 = tw = 12,2 = 2633,98 mm2 2 2 A2 = (bf − tw )tf = (101,6 − 12,2)16,8 = 1501,92 mm2 and h 431,8 = = 107,95 mm 4 4 431,8 − 16,8 h tf z2 = − = = 207,5 mm 2 2 2 z1 =

Chapter 4 / Laterally Restrained Beams bf 101,6

A2

A5

yn

y

y

A1

A3

z4

z1

z2

tw 12,2

z5

A4

tf 16,8

•

h 431,8

70

(a)

(b)

(c)

FIGURE 4.23 Example: plastic section modulus for a channel section

Plastic section modulus Wply = 2(A1 z1 + A2 z2 ) = 2(2633,98 × 107,95 + 1501,92 × 207,5) = 1,192E6 mm3 . The value obtained from Section Tables is 1,207E6 mm3 which is slightly different because of the additional material at the root radius and the fact that the flanges taper. Similarly for the plastic section modulus about the z–z axis divide the section into areas A3 , A4 and A5 and shown in Fig. 4.23(c). It is first necessary to determine the position of the neutral axis z–z. Since the axis z–z divides the total area into two equal parts yn =

A1 + A 2 2633,98 + 1501,92 = = 9,578 mm h 431,8

Areas

h 431,8 A3 = yn = 9,578 = 2067,89 mm2 2 2 h 431,8 A4 = − tf (tw − yn ) = − 16,8 (12,2 − 9,578) = 522,04 mm2 2 2 A5 = (bf − yn )tf = (101,6 − 9,578)16,8 = 1545,97 mm2

and yn 9,578 = = 4,789 mm 2 2 tw − yn 12,2 − 9,578 y4 = = = 1,311 mm 2 2 bf − y n 101,6 − 9,578 y5 = = = 46,011 mm 2 2

y3 =

•

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Plastic section modulus Wpz = 2(A3 y3 + A4 y4 + A5 y5 ) = 2(2067,89 × 4,789 + 522,04 × 1,311 + 1545,97 × 46,011) = 0,1634E6 mm3 The value obtained from Section Tables is 0,1531E6 mm3 which is slightly different because of the additional material at the root radius and the taper on the flanges.

4.5.2 Plastic Methods of Analysis A plastic collapse mechanism depends on the formation of a plastic hinge(s) and this will now be considered in detail. The tensile stress–strain curve for mild steel is shown in Fig. 4.3(a). The curve is idealized into three stages namely elastic, plastic, and strain hardening stages of deformation. The moment–curvature for a beam made of the same material is shown in Fig. 4.3(b) with the corresponding distribution of stress at various loading stages shown in Fig. 4.3(c). The spread of the plastic hinge along the length of the beam is shown in Fig. 4.3(d). The amount of rotation that can take place at a plastic hinge is determined by the length of the yield plateau shown in Fig. 4.3(b). For mild steel the length is considerable and the work hardening stage is ignored. For higher grade steels work hardening occurs immediately after yielding and there is no plateau. A plastic hinge does form but with an increasing moment of resistance. In design this increase in resistance due to work hardening is ignored which errs on the side of safety. Plastic methods of analysis and design consider a structure at collapse when sufficient plastic hinges have formed to produce a mechanism. Examples of collapse mechanisms are shown in Fig. 4.24. In simple situations, are shown in Fig. 4.24(a), the position of the plastic hinge is obvious and it is simple to calculate the collapse load using the method of virtual work. q/unit length

L

Pin

θ

q/unit length

L

Pin

Plastic hinge

Plastic hinge (a)

(b)

FIGURE 4.24 Examples of plastic collapse mechanisms

θ1 x

θ2 (c)

Pin

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Chapter 4 / Laterally Restrained Beams The method must only be applied to structures where the material becomes plastic at the yield stress and is capable of accommodating large plastic deformations. It must therefore not be applied to brittle materials such as cast iron. However it can be applied to reinforced concrete because the steel reinforcement behaves plastically at collapse but care must taken to check the rotation at the hinge for heavily reinforced sections. The plastic method can be seen as a more rational method for design because all parts of the structure can be given the same safety factor against collapse. In contrast for elastic methods the safety factor varies. Intrinsically the plastic method of analysis is simpler than the elastic method because there is no need to satisfy elastic strain compatibilty conditions. However calculations for instability and elastic deflections require careful consideration when using the plastic method, but nevertheless it is very popular for the design of some structures (e.g. beams and portal frames). The method of analysis demonstrated in this chapter is based on the principle of virtual work. This states that if a structure, which is equilibrium, is given a set of small displacements then the work done by the external loads on the external displacements is equal to the work done by the internal forces on the internal displacements. More concisely, external work equals internal work. The displacements need not be real, they can be arbitrary, which explains the use of the word ‘virtual’. However the external and internal geometry must be compatible. It is tacitly assumed that collapse is due to the formation of plastic hinges at certain locations and that other possible causes of failure, for example, local or general instability, axial or shear forces, are prevented from occurring. It is also important to understand that at collapse: (a) the structure is in equilibrium, that is, the forces and moments, externally and internally, balance, (b) no bending moment exceeds the plastic moment of resistance of a member, (c) there are sufficient hinges to form a collapse mechanism. These three conditions lead to three theorems for plastic analysis. (1) Lower bound theorem: if only conditions (a) and (b) are satisfied then the solution is less than or equal to the collapse load. (2) Upper bound theorem: if only conditions (a) and (c) are satisfied then solution is greater than or equal to the collapse load. (3) Uniqueness theorem: if conditions (a), (b) and (c) are satisfied then the solution is equal to the collapse load. Settlement of the supports has no effect on the solution at collapse because the only effect is to change the amount of rotation required. This is in contrast to elastic methods of analysis where settlement calculations must be included. Plastic hinges form in a member at the maximum bending moment. However at the intersection of two members, where the bending moment is the same, the hinge forms

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in the weaker member. Generally the locations of hinges are at restrained ends, intersection of members and at point loads. The hinges may not form simultaneously as loading increases but this is not important for calculating the final collapse load. Generally the number of plastic hinges n=r+1 where r is the number of redundancies. However there are exceptions, for example, partial collapse of a beam in a structure. Equating external to internal work for the member shown in Fig. 4.24(a) L wL θ = Mpl (2θ) hence Mpl = wL2/8 4 In other conditions, as shown in Fig. 4.24(c), the position of the plastic hinge for minimum collapse load is not obvious and the calculations are more complicated. Equating external to internal work for the member shown in Fig. 4.24(c). xθ1 θ2 qx (i) + q(L − x)2 = Mpl (2θ1 + θ2 ) 2 2 From geometry xθ1 = (L − x)θ2 Combining (i) and (ii) 2 1 qL = 2Mpl − x L−x

(ii)

(iii)

Differentiating (iii) to determine the value of x for which q is a minimum √ x2 − 4Lx + 2L2 = 0 hence x = L(2 − 2)

(iv)

Combining (iii) and (iv) √ Mpl = (1,5 − 2)qL2

(v)

The method can be applied to a variety of structures. Further explanation and examples are given in Moy (1981), and Croxton and Martin Vol. 2 (1987 and 1989).

4.6 EFFECT OF A SHEAR FORCE ON THE PLASTIC MOMENT OF RESISTANCE (CLS 5.6 AND 6.2.6, EN 1993-1-1 (2005)) In general the effect of a transverse shear force is to reduce the plastic moment of resistance but the reduction for an ‘I’ section is small and may be ignored (cl 6.2.8, EN 1993-1-1 (2005)) if VEd ≤ 0,5 Vpl,Rd

(4.32)

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Chapter 4 / Laterally Restrained Beams where VEd is the design shear force, and the plastic shear resistance and (Eq. (6.18), EN 1993-1-1 (2005))

Av fy /31/2 Vpl,Rd = (4.33) γM0 The areas resisting shear (Av ) for various sections are given in cl 6.2.6(3), EN 1993-1-1 (2005) and in Section Tables. The area resisting shear for an ‘I’ section is Av = A − 2btf + (tw + 2r)tf

(4.34)

This is a slight increase on the web area (htw ) which has been used in the past. The maximum shear stress of fy /31/2 is based on the failure criterion expressed in Eq. (2.5). There is no reduction in the plastic moment of resistance for plastic or compact sections provided that the design shear force does not exceed 50% of the plastic shear resistance. This recommendation is related to the work of Morris and Randall (1979) who stated that shear can be ignored unless the average shear stress in the web exceeds fy /3, or fy /4 when the ratio of the overall depth to flange width (h/bf ratio) exceeds 2.5. Where the design shear force exceeds 50% of the plastic shear resistance the European Code recommends that the yield strength is reduced to (1 − ρ)fy where for shear (Eq. (6.29), EN 1993-1-1 (2005)) 2 VEd ρ= 2 −1 (4.35) Vpl,Rd and for torsion 2 VEd ρ= 2 −1 Vpl,T,Rd

(4.35a)

These recommendations may be compared with theory by Horne (1971). If a shear force is applied to an ‘I’ section most of the shear force is resisted by the web. If it is assumed that all of the shear force is resisted by the web the effect on the plastic moment of resistance is shown in Fig. 4.25(a). The outer fibres in bending are at yield while the inner fibres are linear elastic. It is not possible to resist shear forces on the outer fibres and thus the inner fibres resist all of the shear force. Plastic moment of resistance of the web dp2 dp2 dp2 d2 d2 Mplw = tw − tw + tw fy = tw − tw fy 4 4 6 4 12

(i)

From the parabolic distribution of shear stress V =

2 tw dp τy 3

(ii)

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fy

tw

d

h

Parabolic distribution of shear stress τy

dp

(a) 0 V/Vmax 2/3 fy τ σ

τy

(b) 2/3 V/Vmax 1 1,0

Factor

0,8

Eq. (4.36)

0,6 Eq. (4.37) 0,4

0,2

0

0,2

0,4 0,6 V /Vmax

0,8

1,0

(c)

FIGURE 4.25 Effect of a shear force on the plastic section modulus of the web of an ‘I’ section

Defining Vmax = tw hτy Combining (i) to (iii) 2 tw d2 fy 3 V Mplw = 1− 4 4 Vmax This expression is valid for 0 ≤ V/Vmax ≤ 2/3.

(iii)

(4.36)

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Chapter 4 / Laterally Restrained Beams When 2/3 ≤ V/Vmax ≤ 1 then assuming the stress distributions shown in Fig. 4.25(b) the plastic moment of resistance of the web tw d2 σ 6 Applied shear force Mplw =

(i)

2 V = dtw τ + dtw (τy − τ) 3 If Vmax = dtw τy , ratio 1 τ V/Vmax = 2+ 3 τy

(ii)

(iii)

Adopting a failure criterion of the form shown Eq. (2.5) 2 2 σ τ + =1 fy τy

(iv)

Combining (i) to (iv) Mplw

⎤ ⎡ 2 tw d2 fy 2 3V ×⎣ = 1− −2 ⎦ 4 3 Vmax

(4.37)

The expressions in the square brackets in Eqs (4.36) and (4.37) are plotted in Fig. 4.25(c) expressed as Mplw = factor(tw d2 fy /4). This theory is approximate and conservative but it does give a general appreciation of the effect of a shear force on the plastic moment of resistance. A fuller description and less conservative theories are given in Horne (1958). In addition when the ratio hw /tw > 72ε/η (cl 6.2.6, EN 1993-1-1 (2005)) the web should be checked for shear buckling. The limiting value of hw /tw is related to experimental work by Longbottom and Heyman (1956), and later work by Horne (1958).

EXAMPLE 4.10

Plastic section modulus reduced by shear. Determine the plastic modulus for a 762 × 267 × 197 kg UB grade S275, using (a) Horne’s method, (b) European Code method, when subject to a design value of shear force VEd = 1145 kN. Ignore the material factor for this example. (a) Horne’s method: VEd VEd 1145E3 = = = 0,6 1/2 Vmax htw fy /3 769,6 × 15,6 × 275/31/2 According to Horne, dividing Eq. (4.36) by fy Wpl(web)

td2 3 VEd 2 = 1− 4 4 Vmax = 0,73(td2/4),

that is 27% loss in the web.

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Reduced plastic section modulus of the complete section Wpl(reduced) = Wpl(whole) − Wpl(web) (1 − factor) 15,6 × 685,82 = 7,164E6 − (1 − 0,73) = 6,672E6 mm3 4 Percentage reduction in plastic modulus for the whole section =

100(7,167E6 − 6,672E6) = 6,91% 7,167E6

(b) EN 1993-1-1 (2005) method: VEd VEd = Vpl,Rd (Av fy /31/2 ) =

1145E3 = 0,568 (12,7E3 × 275/31/2 )

where Av = 12,7E3 is obtained from Section Tables. According to cl 6.29, EN 1993-1-1 (2005) there is a reduction in the plastic modulus if VEd > 0,5 Vpl,Rd From Eq. (4.35a) 2 VEd ρ= 2 − 1 = (2 × 0,568 − 1)2 = 0,0185 Vpl,Rd and Wpl(reduced) =

Mv,Rd ρA2v = Wpl − fy 4tw

= 7,164E6 − 0,0185 ×

12,7E32 4 × 15,6

= 7,116E6 mm3 (7,164E6 − 7,116E6) × 100 Percentage reduction = = 0,67%. 7,164E6

4.7 LATERAL RESTRAINT (CL 6.3.5, EN 1993-1-1 (2005)) The full rotation required at a plastic hinge in a beam may not be realized unless lateral support is provided at the hinge position. It may also be necessary to provide lateral support at other points along the span to ensure that lateral torsional buckling does not occur. Lateral torsional buckling is considered to be prevented if the compression flange is prevented from moving laterally, either by an intersecting member, or by frictional restraint from intersecting floor units.

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4.8 RESISTANCE OF BEAMS TO TRANSVERSE FORCES The first check for transverse forces is the shear stress in the web at the neutral axis in the elastic stage of behaviour (cl 6.2.6(4), EN 1993-1-1 (2005)). In the plastic stage of behaviour the strength of the web is determined using values given in cl 6.2.6(2), EN 1993-1-1 (2005). Also it is necessary to consider web plate shear buckling at the ultimate limit state by checking if 72ε/η > hw /tw (cl 5.1(2), EN 1993-1-5(2003)). Design calculations are also required for concentrated transverse forces applied to girders from supports, cross beams, columns, etc. (Fig. 4.26). The concentrated loads are dispersed through plates, angles and flanges to the web of the supporting girder. The deformations that occur to the supporting beam are shown in Fig. 4.27 and include yielding of the flange and local buckling of the web as shown in experiments (Astill et al. (1980)). The design resistance is expressed simply as (cl 6.2, Eq. (6.1), EN 1993-1-5(2003)) FRd =

fyw Leff tw γM1

The effective bearing length (Leff ) is an extension of the stiff bearing length (ss ) which assumes a 45◦ dispersion through plates, flanges and angles as shown in Fig. 4.26. The root radius of a section increases the length of the stiff bearing by (2 − 21/2 )r.

Clearance

ss

ss = h + 2t p

h tp

Ss

‘I’ beam

45° 45°

ss = 2ta + (2 − √2 )ra − Clearance

ra

‘I’ beam

Angle support ta (a)

(b)

‘I’ beam

tf

ss 45°

ss = 2t f + t w + 2(2 − √2 )rb

rb ‘I’ beam tw (c)

FIGURE 4.26 Stiff bearing lengths

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tf

Yielding of flange

ss

Buckling of web

Leff

tw

FIGURE 4.27 Transverse concentrated load The extension of the stiff bearing length is based on theory of flange yielding and buckling of the web as shown in Examples 4.11 and 4.12. If the transverse resistance of an unstiffened web is insufficient stiffeners are required (cl 9.4(1), EN 1993-1-5 (2003)) designed according to cl 6.3.3 or 6.3.4, EN 1993-1-1 (2005) with a buckling length of not less than 0,75hw and using buckling curve c (Fig. 6.4, EN 1993-1-1 (2005)).

EXAMPLE 4.11 Simply supported beam carrying a uniformly distributed load and laterally restrained. The floor of an office building consists of 125 mm precast concrete units, with a mass of 205 kg/m2 , topped with a 40 mm concrete screed and 20 mm wood blocks. Lightweight partitions supported by the floor are equivalent to a superficial load of 1,0 kN/m2 and the suspended ceiling has a mass of 40 kg/m2 . The floor rests on the top flanges of simply supported steel beams spanning 8 m and at a pitch of 3,75 m. Characteristic loads Dead load Self-weight of steel beam (assumed) 125 mm precast units 40 mm concrete screed 2400 × 0,04 20 mm wood blocks 900 × 0,02 Suspended ceiling

kg/m2

20 205 96 18 40 379 × 9,81/1E3 =

kN/m2

Lightweight partitions

3,72 1,00

Total dead load Imposed load for an office (EN 1991-1-1 (2002))

4,72 2,50

Maximum design bending moment at mid-span

kNm

Permanent load BM = γG QL/8 = 1,35 × 4,72 × 82 × 3,75/8 = Variable load BM = γQ QL/8 = 1,5 × 2,5 × 82 × 3,75/8 =

191,2 112,5

Total BM

303,7

80

•

Chapter 4 / Laterally Restrained Beams Using grade S275 steel, fy = 275 MPa, the plastic section modulus required Wply =

M 303,7E6 = 1,215E6 mm3 . = fy /γM 275/1,1

From Table 4.1 for L/250, L/h = 22, hence h = 8E3/22 = 364 mm. From Section Tables, try 457 × 152 × 60 kg UB, Wply = 1,284E6 mm3 , tf = 13,3 mm, tw = 8 mm, d = 407,7 mm, Av = 3890 mm2 , bf = 152,9 mm, h = 454,7 mm, hw = (h − 2tf ) = (454,7 − 2 × 13,3) = 428,1 mm, r = 10,2 mm. Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) d 407,8 = = 50,96 tw 8 235 1/2 235 1/2 72ε = 72 = 72 = 66,6 > 50,96 satisfactory. fy 275 Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(152,9 − 8)/2 − 10,2] = = = 4,67 tf tf 13,3 235 1/2 235 1/2 9ε = 9 × =9 = 8,32 > 4,67 satisfactory. fy 275 This is a Class 1 section and calculations can be reduced by using Section Tables. Check deflection at the service limit state Permanent and variable load on span = 2,5 × 8 × 3,75 = 75 kN 5QL3 384EI 5 × 75 × (8E3)3 = 9,35 mm = 384 × 210 × 254,64E6

Maximum deflection = δ1 + δ2 =

Deflection limit from Table 4.1 δmax =

L 8E3 = = 32 > 9,35 mm satisfactory. 250 250

Check shear resistance of web at the ultimate limit state. Design shear force at the support VEd =

QL (1,35 × 4,72 + 1,5 × 2,5)8 × 3,75 = = 151,8 kN. 2 2

Area of web (cl 6.2.6(3), EN 1993-1-1 (2005)) Av = A − 2btf + (tw + 2r)tf = 7580 − 2 × 152,9 × 13,3 + (8 + 2 × 10,2)13,3 = 3890 mm2 , or obtain value from Section Tables.

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Design plastic shear resistance (cl 6.2.6(1), EN 1993-1-1 (2005)) Vpl,Rd =

Av τy Av (fy /31/2 ) = γM0 γM0

= 3890 ×

(275/31/2 ) = 561,5 kN 1,1 × 1E3

151,8 VEd = = 0,27 < 1,0 satisfactory. Vpl,Rd 561,5 Check web plate buckling from shear at the ultimate limit state (cl 5.1 (2), EN 1993-1-5 (2003)) 72ε 72(235/275)1/2 = = 55,5 η 1, 2 hw 428,1 = 53,5 < 55,5 satisfactory. = tw 8 Assuming 150 × 75 × 10 mm angle supports at the ends of the beam (Fig. 4.26) the transverse shear buckling load (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)) FRd =

fyw Leff tw 275 × 77,53 × 8 = γM1 1,0 × 1E3

= 170,6 > VEd = 151,8 kN satisfactory no stiffener required. The effective length (Leff ) in the previous equation is obtained as follows: The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), for a 150 × 75 × 10 mm angle support (Fig. 4.26) ss = 2ta + (2 − 21/2 )ra − clearance = 2 × 10 + (2 − 21/2 ) × 11 − 3 = 23,44 < hw = 428,1 mm Buckling co-efficient for the load application (cl 6.1(4), Type(c), EN 1993-1-5 (2003)) kF = 2 +

6(ss + c) 6(23,44 + 0) =2+ = 2,33 < 6 satisfactory. hw 428,1

Effective load length (cl 6.5, Eq. (6.13), EN 1993-1-5 (2003)) le =

kF Etw2 2fyw hw

2,33 × 210E3 × 82 2 × 275 × 428,1 = 133,0 > (ss + c = 23,44 + 0) use 23,44 mm. =

Force (cl 6.4 Eq. (6.5) EN 1993-1-5 (2003)) Fcr =

0,9 kF Etw3 0,9 × 2,33 × 210E3 × 83 = 526,7 kN. = hw (428,1 × 1E3)

82

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Chapter 4 / Laterally Restrained Beams The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003)) m1 =

fyf bf 275 × 152,9 = 19,11 = fyw tw 275 × 8

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003)) m2 = 0,02

hw tf

2

= 0,02

428,1 13,3

2 = 20,72

Yield length (cl 6.5, Eq. (6.12), EN 1993-1-5 (2003)) ly = le + tf (m1 + m2 )1/2 = 23,44 + 13,3(19,11 + 20,72)1/2 = 107,4 mm An alternative yield length (cl 6.5, Eq. (6.11), EN 1993-1-5 (2003)) 1/2 2 m1 le ly = le + tf + + m2 2 tf 1/2 19,11 23,44 2 = 23,44 + 13,3 + + 20,72 2 13,3 = 100,3 mm use minimum. An alternative yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003)) ly = ss + 2tf [1 + (m1 + m2 )1/2 ] = 23,44 + 2 × 13,3[1 + (19,11 + 20,72)1/2 ] = 217,9 mm. Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003)) ly tw fyw 1/2 100,3 × 8 × 275 1/2 λF = = = 0,647 Fcr 526,7E3 Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003)) χF =

0,5 0,5 = = 0,773 0,647 λF

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003)) Leff = χF ly = 0,773 × 100,3 = 77,53 < hw = 428,1 mm. Check self-weight of steel beam = 60/3,75 = 16 < 20 kg/m2 assumed in loading calculations, acceptable.

EXAMPLE 4.12 Support for a conveyor. Part of the support for a conveyor consists of a pair of identical beams as shown in Fig. 4.28. Each beam is connected to a stanchion at end A by a cleat and is supported on a cross beam at D by bolting through the connecting flanges. Lateral restraint is provided by transverse beams at A, B and E connected to rigid supports. The loads shown are at the ultimate limit state.

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•

83

450 kN

15

203 × 203 × 46 kg UC 225 kN

A

B

C

D

150 × 75 × 10 mm angle

2m

150 kN

533 × 210 × 101 kg UB

E

610 × 229 × 140 kg UB

3m

3m

2m 300

Bending moment (kNm) 600 825 300 75

150

375

Shear force (kN)

FIGURE 4.28 Example: support for a conveyor

Assuming pin joints at A and D. Reactions are determined by taking moments about A, then about D. 2 × 225 + 5 × 450 + 10 × 150 = 525 kN 8 6 × 225 + 3 × 450 − 2 × 150 RA = = 300 kN 8 RD =

Important bending moments are MB = RA × 2 = 300 × 2 = 600 kNm MC = RA × 5 − 225 × 3 = 300 × 5 − 225 × 3 = 825 kNm MD = −150 × 2 = −300 kNm The shear force and bending moment diagrams at the ultimate limit state are shown in Fig. 4.28.

84

•

Chapter 4 / Laterally Restrained Beams Using Grade S355 steel with a characteristic strength fy = 355 MPa. The plastic section modulus required Wply =

Mmax 825E6 = 2,56E6 mm3 . = fy /γM0 355/1,1

Deflection limits (Table 4.1), L/250, L/h = 17, hence h = 8E3/17 = 470 mm. From Section Tables try 533 × 210 × 101 kg UB, a Class 1 section, Wply = 2,619E6 mm3 , tf = 17,4 mm, tw = 10,9 mm, Av = 6250 mm2 , d = 476,5 mm. Check deflections at service load Using area and area moment methods the deflections due to the imposed loads at service loads are: mid-span in ABCD 11,3 mm (downwards) end of cantilever DE −6,5 mm (upwards) Deflection limits (Table 4.1) L 8E3 = = 32 > 11,3 mm satisfactory. 250 250 2L 2 × 2E3 span DE = = = 16 > 6,5 mm satisfactory. 250 250

span ABCD =

Check web shear resistance between C and D at the ultimate limit state. Design shear force VEd = 375 kN. Plastic shear resistance Vpl,Rd =

Av ( fy /31/2 ) γM

6240 × (355/31/2 ) 1,1 × 1E3 = 1163 > 375 kN satisfactory. =

Shear area (cl 6.2.6, EN 1993-1-1 (2005)) Av = A − 2btf + (tw + 2r)tf = 129,3 × 100 − 2 × 210,1 × 17,4 + (10,9 + 2 × 12,7) × 17,4 = 6250 mm2 or obtain a value from Section Tables. VEd /Vpl,Rd = 375/1163 = 0,322 < 0,5 therefore no reduction in the plastic section modulus (cl 6.2.8, EN 1993-1-1 (2005)). Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) c 476,5 = = 43,7 tw 10,9 235 1/2 235 1/2 72ε = 72 = 72 = 58,6 > 43,7 satisfactory. fy 355

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Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(210,1 − 10,9)/2 − 12,7] = = = 4,99 tf tf 17,4 235 1/2 235 1/2 9ε = 9 × =9× = 7,32 > 4,99 satisfactory. fy 355 This is a Class 1 section and calculations can be reduced by using Section Tables. Assuming a 150 × 75 × 10 mm angle support at A at the end of the beam (Fig. 4.26) the transverse shear buckling strength (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)) is determined as shown previously in Example 4.11. At D (Fig. 4.28) two UBs intersect which have the following dimensions. From Section Tables the 533 × 210 × 101 kg upper load carrying beam ABCDE. h = 536,7 mm, bf = 210,1 mm, hw = (h − 2tf ) = (536,7 − 2 × 17,4) = 501,9 mm, tf =17,4 mm, tw = 10,9 mm, r = 12,7 mm. From Section Tables the 610 × 229 × 140 kg lower support beam at D. h = 617 mm, bf = 230,1 mm, hw = (h − 2tf ) = (617 − 2 × 22,1) = 572,8 mm, 22,1 mm, tw =13,1 mm, r = 12,7 mm.

tf =

The transverse shear buckling strength of the upper 533 × 210 × 101 kg load carrying beam (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)). FRd =

fyw Leff tw 355 × 501,9 × 10,9 = γM1 1,0 × 1E3

= 1942 > RD = 525 kN satisfactory no stiffener required. The previous calculation for FRd includes the effective bearing length (Leff ) which is calculated as follows: The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), provided by the flange and web of the lower 610 × 229 × 140 UB ss = 2tf + tw + (2 − 21/2 )rb = 2 × 22,1 + 13,1 + (2 − 21/2 ) × 12,7 = 64,74 < hw = 501,9 mm Buckling co-efficient for the load application (cl 6.1(4), Type (a), EN 1993-1-5 (2003)) assuming a is large kF = 6 + 2

hw a

2 =6

Force (cl 6.4, Eq. (6.5), EN 1993-1-5 (2003)) Fcr =

0,9 kF Etw3 0,9 × 6 × 210E3 × 10,93 = = 2926 kN hw 501,9 × 1E3

86

•

Chapter 4 / Laterally Restrained Beams The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003)) m1 =

fyf bf 355 × 210,1 = 19,28 = fyw tw 355 × 10,9

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003)) m2 = 0,02

hw tf

2

= 0,02

501,9 17,4

2 = 16,64

Yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003)) ly = ss + 2tf [1 + (m1 + m2 )1/2 ] = 64,74 + 2 × 17,4[1 + (19,28 + 16,64)1/2 ] = 308,1 mm Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003)) λF =

ly tw fyw Fcr

1/2

=

308,1 × 10,9 × 355 2926E3

1/2 = 0,638

Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003)) χF =

0,5 0,5 = = 0,784 < 1 0,638 λF

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003)) Leff = χF ly = 0,784 × 308,1 = 241,6 < hw = 501,9 mm Similar calculations are required for the web buckling strength of the upper 533 × 210 × 101 kg UB at B, C and E. Stiffeners are not required for the 533 × 210 × 101 kg UB but in the past many designers have inserted them where large point loads are applied. Assuming that one is required at D(RD = 525 kN) the following calculations are required. Assume a symmetrical stiffener of 10 mm thickness welded to the web and the flanges. The non-dimensional slenderness ratio of the stiffener (Eq. (6.50), EN 1993-1-1 (2005)) λz = =

Afy Ncr

1/2 =

(Lcr /iy ) × 1 λ1

(0,75 × 572,8/15E3) × 1 = 0,375E-3 [93,9 × (235/355)1/2 ]

assuming a stiffener on both sides of the web, breadth of stiffener bs = bf − tw − 2 (weld leg) = 210,1 − 10,9 − 2 × 6 = 187,2 (use 180) mm iz = bs ts3 /12 = 180 × 103/12 = 15E3 mm4

Structural Design of Steelwork to EN 1993 and EN 1994 Q 85 kN

q = 42,5 kN/m A

B

C

E 3m

3m

478,1

478,1

Bending moment (kNm)

441

441

3m

87

Q 85 kN

D

11 m

•

Collapse of span AB

Collapse of span BC 329,4 201,9

190,3 116,9

Shear force (kN) 10,6 95,6 223,1

277,2

FIGURE 4.29 Example: two span beam

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) with λz = 0,375E-3 the reduction factor χz = 1,0 and the design buckling resistance χz Afy 1,0 × 180 × 10 × 355 = λM 1,1 × 1E3 = 581 > RD = 525 kN satisfactory.

Nb,Rd =

The design resistance of four vertical 6 mm fillet welds connecting the stiffener to both sides of the web (cl 4.5.3.3(3), Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = 4hw fu /(31/2 βγM2 ) × 0,7a = 4 × 572,8 × 510/(31/2 × 0,9 × 1,25) × 0,7 × 6/1E3 = 2519 > RD = 525 kN satisfactory.

EXAMPLE 4.13 Two span beam. Determine the size of Universal Beam required to support the design loads at the ultimate limit state as shown in Fig. 4.29. Assume that the compression flange is fully restrained and that lateral torsional buckling does not occur.

88

•

Chapter 4 / Laterally Restrained Beams Plastic analysis of the beam produces the following. Collapse of span AB considered as a propped cantilever √ √ Mpl = 1,5 − 2 qL2 = 1,5 − 2 42,5 × 112 = 441 kNm. Collapse of span BC with plastic hinges assumed to be at E and B (Fig. 4.29) External work = internal work 2L 2L θ1 2L L L L Q θ1 + Q +q θ1 + q θ2 3 3 2 3 3 3 6 = Mpl (θ1 + θ1 + θ2 ) and from geometry 2L L θ1 = θ2 3 3

hence θ2 = 2θ1

(i)

(ii)

Combining eqs (i) and (ii) and rearranging QL qL2 85 × 9 42,5 × 92 + = + = 478,1 kNm 4 12 4 12

Mpl =

This value of Mpl is the greater than the value for span AB and therefore is used to determine the size of a section which is continuous for two spans. The assumption that a plastic hinge is at E is not correct, it actually occurs at 3,25 m from C and Mpl = 479,1 kNm. However the error is small and is ignored. Used Grade S355 steel and assuming a characteristic strength fy = 355 MPa the plastic section modulus required Wply =

Mmax 478,1E6 = 1,347E6 mm3 = fy /γM 355/1,0

From Section Tables try 457 × 191 × 67 kg UB which is a Class 1 section bending about the y–y axis, Wply = 1,471E6 mm3 , tf = 12,7 mm, tw = 8,5 mm, Av = 4100 mm2 , d = 407,8 mm, bf = 189,9 mm. Check deflections at service load. Using area and area moment methods the deflections due to the imposed load are 20,7 mm for AB and 25 mm for BC. Deflection limit (Table 4.1) AB is

L 11E3 = = 44 > 20,7 mm, satisfactory 250 250

BC is

L 9E3 = = 36 > 25 mm, satisfactory. 250 250

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Sketch the shear force diagram (Fig. 4.29) at the ultimate limit state for the collapse of span BC. For EC

E + 478,1 + 42,5 × 3 × 1,5 − 3RC = 0

hence RC = 223,1 kN

For AB

B + 478,1 −

42,5 × 11 × 11 + 11RA = 0 2

hence RA = 190,3 kN

For ABC ↑ + RA + RB + RC −

qL −

Q=0

+190,3 + RB + 223,1 − 42,5 × 20 − 2 × 85 = 0

hence RB = 606,6 kN

Check web shear resistance at B at the ultimate limit state. Design shear force VEd = 329,4 kN (Fig. 4.29) Design plastic shear resistance Vpl,Rd =

Av τy Av (fy /31/2 ) 4100 × (355/31/2 ) = = γM γM 1,0 × 1E3

= 840 > 329,4 kN satisfactory. VEd /Vpl,Rd = 329,4/840 = 0,392 < 0,5 therefore no reduction in the plastic section modulus (cl 6.2.8, EN 1993-1-1 (2005)). Check for buckling of the web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) c 407,9 = 48,0 = tw 8,5 235 1/2 235 1/2 72ε = 72 = 72 = 58,6 > 48,0 satisfactory. fy 355 Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(189,9 − 8,5)/2 − 10,2] = = = 6,33 tf tf 12,7 235 1/2 235 1/2 9ε = 9 × =9× = 7,32 > 6,33 satisfactory. fy 355 This is a Class 1 section and calculations can be reduced by using Section Tables.

90

•

Chapter 4 / Laterally Restrained Beams A continuous 457 × 191 × 67 kg UB can be used for both spans. Alternatively if a minimum weight design is required then: (a) A smaller section could be used for span AB but there would have to a splice which would increase fabrication costs. (b) A smaller section could be used for span AB, with flange plates added to increase the moment of resistance for span BC, but again this would increase fabrication costs. Calculations, similar to Example 4.12, are required for the web buckling strength of the 457 × 191 × 67 kg UB at A, B, and C.

REFERENCES Astill, A.W., Holmes, M. and Martin, L.H. (1980). Web buckling of steel ‘I’ beams. CIRIA Tech. note 102. Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving problems in structures Vols. 1 and 2. Longman Scientific and Technical. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-5 (2003). Plated structural elements. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Horne, M.R. (1971). Plastic theory of structures. Nelson. Horne, M.R. (1958). The full plastic moment of sections subjected to shear force and axial loads, British Welding Journal, 5, 170. Longbottom, E. and Heyman, J. (1956). Experimental verification of the strength of plate girders designed in accordance with the revised BS153: tests on full size and on model plate girders, Proceedings of the ICE, 5(III), 462. Megson, T.H.G. (1980). Strength of materials for civil engineers. Nelson. Morris L.J. and Randall A.L. (1979) Plastic Design, Constrado. Moy, S.S. (1981). Plastic methods for steel and concrete structures. Macmillan Press. Zbirohowski-Koscia, K. (1967). Thin walled beams. crosby Lockwood.

Chapter

5 / Laterally Unrestrained Beams

In the previous chapter it was assumed that the compression flange of a beam was fully restrained, that is, the flange is unable to move laterally under the effect of loads or actions. In practice, this is rarely the case and although compression flanges may be restrained at discrete points the flange is still capable of buckling between restraints. Such buckling reduces the moment capacity of the member.

5.1 LATERAL TORSIONAL BUCKLING OF ROLLED SECTIONS SYMMETRIC ABOUT BOTH AXES 5.1.1 Basic Theory A beam under the action of flexure alone due to the application of point moments producing single curvature is considered as the basic case. The supports allow rotation about a vertical axis but do not allow relative displacement of the top and bottom flanges, that is, twisting is not allowed (Fig. 5.1). The governing equation for flexure about the minor axis is EIz

d2 v = −Mφ dx2

(5.1)

where EI z is the flexural rigidity about the minor axis. The governing equation for torsion is given by GIt

dφ d3 φ dv − EIw 3 = M dx dx dx

(5.2)

where GI t is the torsional rigidity, EI w the warping rigidity and the final term is the disturbing torque. Eliminate the term dv/dx between Eqs (5.1) and (5.2) to give GIt

d2 φ d4 φ M2 − EI = − φ w EIz dx2 dx4

(5.3)

•

Chapter 5 / Laterally Unrestrained Beams M Pins

Pins M

L (a) Elevation of beam loaded in single curvature

Unbuckled position

Buckled position (b) Support arrangement with f =

d2f dx 2

=0

L dv dx

x

V

92

cL of unbuckled

Beam cL after buckling (c) Plan view of buckled beam

V

Original position Buckled position

f

(d) Section

FIGURE 5.1 Lateral torsional buckling of beams

beam

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93

The solution to Eq. (5.3) is given by the first term of a sine series φ = φ0 sin

πx L

(5.4)

where φ0 is the angle of twist at mid-span. Substituting Eq. (5.4) into Eq. (5.3) with the boundary condition that φ = 0 at x = L gives the elastic critical moment Mcr as π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 (5.5)

or 1/2 √ π EIz GIt π2 EIw Mcr = 1+ 2 L L GIt

(5.6)

The full derivation of the elastic critical moment is given in Kirby and Nethercot (1979) or Trahair and Bradford (1988). In practice, real beams do not achieve full elastic buckling except at very high slenderness ratios and are also subject to the limit imposed by the section plastic capacity. Reductions in elastic buckling capacity are caused by the existence of residual stresses and any lack of initial straightness. The residual stresses within the section arise in the case of rolled sections due to differential rates of cooling in the web and flanges after hot rolling, and of plate girders due to both the preparation of the web and flange plates and the welding procedure (Nethercot, 1974a). These residual stresses do not affect the plastic capacity (Nethercot, 1974b; Kirby and Nethercot, 1979). The value of the critical moment given by Eq. (5.5) or (5.6) takes no account of major axis flexure. Trahair and Bradford (1988) indicate the critical moment obtained from Eq. (5.5) or (5.6) should be divided by a factor K where K is given by K=

1−

EIz EIy

1−

GIt π2 EIw 1+ EIy GIt L2

(5.7)

In practice for I or H sections GIt /EIy 1 + π2 EIw /GIt L2 is negligible compared to unity and thus K may be taken as K=

EIz 1− = EIy

1−

Iz Iy

(5.8)

For British Universal Beams the value of K is between 0,94 and 0,97 thus increasing Mcr by around 3% to 6%. This enhancement may therefore be neglected for Universal Beams. For Universal Columns, however, K is between 0,80 and 0,83, thus increasing Mcr by around 17% to 20%, and should therefore possibly be taken into account.

Chapter 5 / Laterally Unrestrained Beams Full plastic action

Elastic buckling curve (Eq.(5.5) or Eq.(5.6))

1.0

Perfect beam behaviour MpI

•

Mb,Rd

94

Experimental results 0.5

Design curve (Lower bound to experimental results)

0

0.4

0.8

1.2

1.6

⎛ M ⎛½ pl Effective slenderness ⎜ ⎜ ⎝ M cr ⎝

FIGURE 5.2 Beam behaviour with lateral torsional buckling

5.1.2 Interaction between Plastic Moment Capacity, Elastic Critical Moment and Allowable Bending Moment If the results for beam strength against effective slenderness are plotted in nondimensional format the results appear as in Fig. 5.2. The performance of beams is reduced below the perfect conditions due to the residual stresses and initial imperfections. A convenient lower bound to test data is given by a Perry–Robertson type approach to modelling the interaction. The imperfection co-efficient makes allowance both for geometric imperfections and residual stress levels. It should be noted that unlike columns there is no theoretical justification for such an approach.

5.1.2.1 Design Interaction between Buckling and Plasticity For any type of beam, the interaction equation is given by (Mb,Rd − Mcr )(Mb,Rd − Mpl,Rd ) = ηLT Mcr Mb,Rd

(5.9)

where ηLT is the imperfection factor for lateral torsional buckling. Equation (5.9) may be written in a normalized form to give Mb,Rd Mb,Rd Mcr Mb,Rd Mcr − − 1 = ηLT Mpl,Rd Mpl,Rd Mpl,Rd Mpl,Rd Mpl,Rd

(5.10)

It is convenient to define two parameters χLT and λLT given as χLT =

Mb,Rd Mpl,Rd

(5.11)

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TABLE 5.1 Values of αLT for the generalized lateral torsional buckling case. Cross section

Size limits (h/b)

αLT

Rolled I sections

≤2 >2 ≤2 >2

0,21 0,34 0,49 0,76 0,76

Welded I sections Other cross-sections

λLT

Mpl,Rd = Mcr

1/2 (5.12)

Equation (5.10) then becomes (χLT − 1)(χLT − (λLT )2 ) = ηLT

χLT (λLT )2

(5.13)

The solution to which is given by χLT =

1 − (λLT )2 ]1/2

LT + [2LT

(5.14)

where LT is given by LT = 0,5[1 + ηLT + (λLT )2 ]

(5.15)

with ηLT given as ηLT = αLT (λLT − 0,2)

(5.16)

The values of αLT are dependant upon the type of section and whether rolled or welded. The appropriate values are given in Table 5.1 (cl 6.3. 2.1. EN 1993-1-1) In design, full plastic moment capacity may be mobilized when λLT is less than λLT,0 which may be taken as 0,4 or when MEd /Mcr ≤ (λLT,0 )2 .

5.1.2.2 Rolled Sections and Equivalent Welded Sections (cl 6.3.2.3 EN 1993-1-1) For rolled sections and welded sections which are by implication symmetric, an alternative formulation of the interaction formula may be used, χLT =

LT + [2LT

1 − β(λLT )2 ]1/2

(5.17)

where LT is given by LT = 0,5[1 + ηLT + β(λLT )2 ]

(5.18)

with ηLT as ηLT = αLT (λLT − λLT,0 )

(5.19)

96

•

Chapter 5 / Laterally Unrestrained Beams TABLE 5.2 Values of αLT for lateral torsional buckling of rolled and welded sections. Cross section

Size limits (h/b)

αLT

Rolled I sections

≤2 >2 ≤2 >2

0,34 0,49 0,49 0,76

Welded I sections

The parameter χLT is subject to the limits χLT ≤ 1,0 and ≤ (λLT )2 . The code recommends that the maximum value of λLT,0 should be 0,4 and the minimum value of β as 0,75. The UK National Annex will adopt these values but will effectively specify the approach should be limited to rolled sections, and that, therefore, the generalized method must be used for welded sections. Again the values of αLT are dependant upon the type of section and whether rolled or welded. The appropriate values are given in Table 5.2. Additionally, if this method is used, then the calculation of Mb,Rd needs modifying by using a factor χLT,mod instead of χLT where χLT,mod =

χLT f

(5.20)

The factor f is given by f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] ≤ 1,0

(5.21)

The factor kc is determined from the type of loading. For loading only at points of restraint, kc =

1 1,33 − 0,33ψ

(5.22)

where ψ is the ratio between the moments at restraint points subject to the condition that −1 ≤ ψ ≤ 1. For loading between restraints values of kc are given in Table 6.6 of EN 1993-1-1. The additional factor is due to the critical moment being determined elastically but failure will be by generation of plasticity not necessarily at the point of maximum deflection for buckling.

5.1.2.3 Simplified Assessment Methods for Building Structures (cl 6.3.2.4 EN 1993-1-1) If there are discrete restraints to the compression flange, then lateral torsional buckling will not occur if the length Lc between restraints or the resultant slenderness λf satisfies the following equation λf =

Mc,Rd kc Lc = λc,0 ifz λ1 My,Ed

(5.23)

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where Mc,Rd is the flexural capacity of the section, kc is the factor to correct for moment gradient, if,z is the radius of gyration of the effective compression flange which comprises the actual compression flange together with one-third of the part of the web which is in compression and λ1 is given by 93,9(235/fy )1/2 and λC,0 is the normalized slenderness ratio of the effective compression flange.

5.1.2.4 Modifications Dependant upon Section Classification (cl 6.3.2.1 (3) EN 1993-1-1) For sections other than Classes 1 and 2, a further modification is required as for Class 3 sections the capacity is based on the elastic section modulus, and Class 4 an effective section modulus is used. The modification is made by defining the flexural capacity as Wel,y fy for Class 3 and Weff,y fy for Class 4. Also the buckling capacity Mb,Rd is defined as χLT Wpl,y fy/γM1 for Class 1 or Class 2, χLT Wel,y fy/γM1 for Class 3, χLT Weff,y fy/γM1 for Class 4. Having established the basic determination of the allowable moment capacity, Mb,Rd , two additional modifications due to varying support conditions and non-uniform flexural loading need to be examined.

5.1.3 Effect of Support Conditions If the beam has a rotational end restraint of R (where R takes a value of 0 for no restraint and 1 for full restraint), then the effective length kL is given by Eq. (5.24) (Trahair and Bradford, 1988). R π π = − cot 1−R 2k 2k

(5.24)

With a little loss in accuracy Eq. (5.24) can be written as k = 1 − 0,5R

(5.25)

Both Eqs (5.24) and (5.25) are plotted in Fig. 5.3. The results from Eq. (5.25) represent ideal conditions therefore in practice the values of kL used are higher at more complete fixity as absolute rigid joints do not exist. Pillinger (1988) gave some indications of the relationship between practical end conditions and effective lengths. Figure 5.4 gives typical end conditions in terms of connection type and degree of restraint. For beams the effective length L between restraints can be taken as lying between 0,7 and 1,0 times the actual length with the lower factor implying full rotational restraint to both flanges and the higher factor with both flanges free to rotate in plan. If the rotational restraint parameter R can be assessed, then Eq. (5.25) can be modified to give k = 1 − 0,3R

(5.26)

•

Chapter 5 / Laterally Unrestrained Beams 1.2 1 Effective length factor (k)

98

0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

1.2

Degree or restraint (R) Exact solution

Linear appproximation

Practical values

FIGURE 5.3 Values of effective length factor (k)

The values of kL for beams should be increased by 20% for destabilizing loads (BS 5950-1: Part 1 namely 1990 and 2000). For cantilevers the situation can be more complex as the effective length will depend on both the restraint at the encastré end and at the tip. Table 5.3 (from BS 5950-1 (2000)) gives suitable values. It should be noted that when the loading is destabilizing the effective length factors can be extremely high. Guidance is also given in Nethercot and Lawson (1992). Although it was suggested that for beams it is generally conservative to take the actual length, this may not be appropriate where the beam is supported solely on its bottom flange with no web or top flange restraint (Fig. 5.5). This type of situation will produce low lateral torsion buckling resistance, and Bradford (1989) suggests that in this case the system length L should be taken as, ⎛ ⎞ b hs tw 3 ⎝ 1 + hs ⎠ L=1+α (5.27) 6 tf 2 where the beam is under a moment gradient, or ⎛ ⎞ 3/2 1 + b h s tw hs ⎠ ⎝ L = 1 + 10 6 tf 2 under a central point load.

(5.28)

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Stability cleat Beam compression flange free to rotate on plan

Full torsional end restraint to beam

(a) Thin end plate to web only

(b) Bottom flange cleat

Beam compression flange free to rotate on plan Compression flange free to rotate. No torsional restraint

(c) Web cleats bolted to beam

Full torsional end restraint to beam

(d) Flange bolted to padstone

Compression flange rotation reduced. Torsional end restraint

(e) Full depth end plate welded to web and flange

(f) Flange bolted to padstone, web stiffener Full torsional restraint. No rotation of compression flange (g) Full end stiffener

FIGURE 5.4 Typical connection and support detail In both cases l is the span, hs the distance between the centroids of the flanges, b the flange width and tf and tw the thickness of the flange and web respectively, and α is given by α = 4 + 7ψ + 4ψ2

(5.29)

where ψ is the ratio of the end moments.

5.1.4 Intermediate Restraints Consider the beam in Fig. 5.6 but with a spring restraints giving a horizontal restraint stiffness of αt (force/per unit displacement) and rotational restraint stiffness of αr

TABLE 5.3 Effective length LE for cantilevers without intermediate restraint. Restraint conditions

Loading conditions

At support

At tip

Normal

Destabilizing

(a) Continuous, with lateral restraint to top flange

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

3.0L 2.7L

7.5L 7.5L

2.4L 2.1L

4.5L 3.6L

(b) Continuous, with partial (1) Free torsional restraint (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

2.0L 1.8L

5.0L 5.0L

1.6L 1.4L

3.0L 2.4L

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

1.0L 0.9L

2.5L 2.5L

0.8L 0.7L

1.5L 1.2L

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

0.8L 0.7L

1.4L 1.4L

0.6L 0.5L

0.6L 0.5L

L

L

(c) Continuous, with lateral and torsional restraint

L

(d) Restrained laterally, torsionally and against rotation on plan

L

(1) Free

(not braced on plan) Source: BS 5950 Part 1:2000

Tip restraint conditions (2) Lateral restraint to (3) Torsional restraint top flange

(4) Lateral and torsional restraint

(braced on plan in at least one bay)

(braced on plan in at least one bay)

(not braced on plan)

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A

A

cL of bolts

Section AA

FIGURE 5.5 Beam supported on bottom flange with no restraint to top flange

at mid-span (Fig 5.6(a)) at a height of bs above the centroidal axis. It can be shown (Mutton and Trahair, 1973) that the elastic critical moment is given by Eq. (5.6) or (5.7) with kL (the effective, or system, length) substituted for L, with k related to αt and αr by ⎛ L3

αt ⎝1 + 16EIz αr L3 16EIw 2b 2b 1 − h s h0

2bs h 2b0 h

⎞ ⎠=

π 3 π cot 2k 2k π π cot 2k −1 2k

(5.30)

=

π 3 π cot 2k 2k π π cot 2k −1 2k

(5.31)

where the distance of the centre of rotation below the shear centre b0 is given by b0 =

Mcr EI

π2 (kL)z 2

2 h k = 1+ 2 K

(5.32)

and K is defined K=

π2 EIw L2 GIt

1/2 (5.33)

•

Chapter 5 / Laterally Unrestrained Beams x

φL

Axis of buckled beam

2 ⎛

vL

⎛

α1 ⎜VL bs φL ⎜

2

⎝ 2

2⎝

bs

αeφL 2

y

z

FIGURE 5.6(a) Beam with elastic intermediate restraints

1.2

1 Effective length factor (k)

102

Braced mode 1 buckling 0.8

0.6 Mode 2 buckling

0,5 0.4

0.2

0 0

5

π2 10

15 20 Normalized restraint factor

25

30

FIGURE 5.6(b) Effect of spring restraint on effective length factor

The relationships from Eqs (5.30) and (5.31) are plotted in Fig 5.6(b). The value of k changes from 1 (when the beam buckles as an entity) to k = 0,5 when the beam buckles as two half waves when the right hand sides of Eq. (5.30) or (5.31) reach a value of π2 . From Eq. (5.32) the minimum value of b0 is h/2. From Eq. (5.30) the maximum value of αt applied at the top flange (bs = h/2) to force a change into second mode is given by αt =

4Mcr 4Pf = Lh L

where Pf is the force in the flange.

(5.34)

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5.1.5 Loading The effect of the applied loading on the beam system needs considering under two headings. The first is concerned with how the load is applied whether at restraint points or between restraints. The second is concerned with any possible destabilizing effects of the load.

5.1.5.1 Load Pattern The background to any amendments to the value of Mcr to allow for load pattern is given in Nethercot and Rockey (1971), or Trahair and Bradford (1988). For loading at points of lateral torsional restraint, the critical moment Mcr is modified by a factor C1 which is dependant solely on the moment gradient within the section of the beam being considered. The moment gradient is defined by the ratio of the applied moments at either end of the beam segment. The small moment within the segment caused by self-weight is not taken into account in this calculation. From Kirby and Nethercot (1979) C1 is given by 1 = 0,57 + 0,33β1 + 0,1β12 ≥ 0,43 C1

(5.35)

where β1 = 1,0 for single curvature (or −1 for double curvature). And from Trahair and Bradford (1988) as either, C1 = 1,75 + 1,05β2 + 0,3β22 ≤ 2,56

(5.36)

1 = 0,6 − 0,4β2 ≥ 0,4 C1

(5.37)

or,

where β2 = −1 for single curvature (and +1 for double curvature). Rewrite Eqs (5.35) to (5.37) using ψ rather than β1 or β2 , defining ψ = 1 for single curvature to give 1 = 0,57 + 0,33ψ + 0,1ψ2 ≤ 0,43 C1

(5.38)

C1 = 1,75 − 1,05ψ + 0,3ψ2 ≤ 2,56

(5.39)

1 = 0,6 + 0,4ψ ≥ 0,4 C1

(5.40)

A comparison between the values of C1 from Eqs (5.38) to (5.39) is given in Table 5.4 and Fig. 5.7 where it is observed that there is little difference between the values, and that this difference is reduced as the value of the normalized slenderness ratio requires 1/2 the value of C1 . For the examples herein Eq. (5.40) will be used.

Chapter 5 / Laterally Unrestrained Beams TABLE 5.4 Comparison of values of C 1 . ψ

Eq. (5.36)

Eq. (5.37)

Eq. (5.38)

1,0 0,75 0,5 0,25 0 −0,25 −0,5 −0,75 −1,0

1,00 1,14 1,32 1,52 1,75 2,03 2,33 2,33 2,33

1,00 1,13 1,30 1,51 1,75 2,03 2,35 2,56 2,56

1,00 1,11 1,25 1,43 1,67 2,00 2,50 2,50 2,50

3

2.5

2

1.5

1

0.5

2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1

1

0.

0

0.

1

2

0.

3

0.

4

0.

5

0.

6

0.

7

0.

8

0.

9

0.

1

0

0.

•

C1

104

Moment ratio Eq. (5.36)

Eq. (5.37)

Eq. (5.38)

FIGURE 5.7 Values of C 1 For loads between restraints use the n factor method from Tables 15 to 17 of BS 5950-1 (1990) should be used. In this case C1 is defined by 1 C1 = √ n

(5.41)

Tables 15 to 17 of BS 5950: Part 1: 1990 are reproduced in Annexe A7.

5.1.5.2 Destabilizing and Stabilizing Loads A destabilizing load is one which is applied to the compression flange and is free to move as the flange buckles laterally. Such a load has the effect of reducing the elastic critical moment as an additional disturbing torque is introduced (Anderson and Trahair, 1972; Trahair and Bradford, 1988). A stabilizing load, therefore, is one

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105

Additional torque

No additional torque

Decreased torque

FIGURE 5.8 Destabilizing loads applied in such a way that the effect counterbalances the buckling effect and thus enhances the elastic critical moment. Thus for simply supported beams if a load free to move is applied above the shear centre it is destablizing, and below the shear centre it is stabilizing (Fig. 5.8). The reverse is true for a cantilever! For loading applied at other than the centroid of the section it is only possible to determine the critical moment under a given load pattern. Timoshenko and Gere (1961) give the value of the critical load for a simply supported I beam under either a uniformly distributed load (UDL) or central point load as √ EIz GIt (5.42) Pcr = γ2 L2 where γ2 is dependant upon the position of the load (top flange, centroid or bottom flange) and the factor L2 GI t /EI w . Values of γ2 are plotted in Fig. 5.9 for values of beam stiffness 1/K 2 (where K is given by Eq. (5.33)).

5.1.5.3 Cantilevers and Beams Cantilevering Over Supports • Cantilevers Trahair (1983) demonstrated that for cantilevers built in at the support that the elastic critical moment Mcr could be given with little loss of accuracy as √ EIz GIt Mcr = (1,6 + 0,8K) (5.43) L √ where K is given by Eq. (5.33) as (π2 EI w /(L2 GI t )). It will be noted that this differs from the more usual solution given in Eq. (5.6) where the elastic critical moment is proportional to (1 + K 2 )1/2 .

Chapter 5 / Laterally Unrestrained Beams 160 140 120 100 Gamma 2

•

80 60 40 20 0 0

50

100

150

200

250

300

350

400

450

400

450

Beam stiffness (1/K 2) Load at top

Load at centre

Load at bottom

FIGURE 5.9(a) Factor for point load not applied at shear centre 250

200

Gamma 2

106

150

100

50

0 0

50

100

150

200

250

300

350

Beam stiffness (1/K 2) Load at top

Load at centre

Load at bottom

FIGURE 5.9(b) Factor for UDL not applied at shear centre For a point load Pcr at the free end of the cantilever Trahair gives the following expression, √ 1,2ε 1,2(ε − 0,1) EIz GIt Pcr = 11 1 + + 4(K − 2) 1 + L2 1 + 1,22 (ε − 0,1)2 1 + 1,22 ε2 (5.44)

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where the parameter ε defines the position relative to the centroidal axis of the point of application of the load, and is given by a EIz 2a K ε= = (5.45) L GIt h π where a is the distance of the point of application of the load For a UDL of qcr applied to whole length of the cantilever, √ 2 EIz GIt 1,4 (ε − 0,1) qcr = 27 1 + L3 1 + 1,42 (ε − 0,1)2 1,3(ε − 0,1) + 10(K − 2) 1 + 1 + 1,32 (ε − 0,1)2 For loading applied at the centroid, ε = 0, and Eq. (5.44) reduces to √ EIz GIt [3,96 + 3,52K] Pcr = L2 The moment Mcr,P due to this load is given by √ EIz GIt [3,96 + 3,52K] Mcr,P = Pcr L = L Equation (5.46) becomes when loading is applied at the centroid √ 2 EIz GIt [5,83 + 8,71K] qcr = L3 The moment Mcr,q due to this load is given by √ qcr L2 EIz GIt [5,83 + 8,71K] Mcr,q = = 2 L

(5.46)

(5.47)

(5.48)

(5.49)

(5.50)

Equations (5.43), (5.48) and (5.50) are plotted in Fig. 5.10, where it is observed that as the loading progresses from uniform moment through a point load at the end to a UDL, the equivalent elastic moment increases. This is due to the fact that for the UDL the moment quickly drops off from the support and thus has less of an effect. The moment due to the point load also drops off, but more slowly. It is therefore to be noted that the general UK practice of ignoring the beneficial effects of moment gradients on a cantilever (i.e. C1 = 1,0) is extremely conservative. • Overhanging beams For the situation shown in Fig. 5.11(a), where a beam overhangs the supports symmetrically and is loaded by end point loads, then the lateral torsional buckling of such a system is controlled by the restraint at the internal supports which may often be of the type shown in Fig. 5.11(b), where such restraints can be considered elastic. Such restraints increase the torsional flexibility of the beam and thereby decrease the buckling moment. The decrease P/P0 is dependant upon the stiffness

•

Chapter 5 / Laterally Unrestrained Beams 35 Normalized elastic critical moment

108

30 25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

Stiffness parameter K Pure moment

Point load

UDL

FIGURE 5.10 Normalized elastic critical moment for a cantilever

P

P

L

γL

L

(a) Basic geometry Over hanging beam

Bolts

Supporting beam

(b) Type support detail

FIGURE 5.11 Overhanging beams

at the support αR , and Trahair (1983) gives the following equation for beams loaded centroidally, αR L P 2a GIt = 1 − Kβ2 (1 − β) (5.51) αR L 4K 2 P0 h 5 + 1+K 2 + GI t

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109

20 kN Permanent 30 kN Variable

A

D B 3.0 m

C 3.0 m

3.0 m

Notes: (1) All loads are characteristic loads (2) Lateral torsional restraints exists at A, B, C and D

FIGURE 5.12 Design data for Example 5.1

where β is defined by β=

αR L GIt αR L 1 + GI t

(5.52)

It should be noted for situations where the internal span is greater than the sum of the overhangs, then the internal span dominates the behaviour.

EXAMPLE 5.1 Beam with loading applied at restraints. Prepare a design in Grade S355 steel for the beam for which the data are given in Fig. 5.12. Factored actions at ULS: at B: 1,35 × 40 + 1,5 × 70 = 159 kN at C: 1,35 × 20 + 1,5 × 30 = 72 kN Total load = 231 kN The BM and SF diagrams are drawn in Fig. 5.13. The critical section for design is the central section BC as the moment gradient is the least. Try a 406 × 178 × 74 UKB Mpl,Rd = Wpl,y

fy 355 = 1501000 × 10−6 = 533 kNm γM0 1,0

MSd = 390 kNm, beam satisfies the plastic capacity criterion. Section classification: Compression flange: c = 0,5[b − 2r − tw ] = 0,5[179,5 − 2 × 10,2 − 9,5] = 74,8 mm c/tf = 74,8/16 = 4,68

110

•

Chapter 5 / Laterally Unrestrained Beams A

B

C

D

303 390 (a) BMD (kNm) 130

B A

C

D

−29 −101 (b) SFD (kN)

FIGURE 5.13 BMD and SFD for Example 5.1

Maximum value for a Class 1 flange is 9ε = 9(235/355)1/2 = 7,32. Web: c/t = d/tw = 37,9. Maximum value for a Class 1 web is 9ε = 72(235/355)1/2 = 58,6. Thus a 406 × 178 × 74 UKB Grade S355 is Class 1. Shear check: fy 1 1 355 Vpl,Rd = √ Av = √ 4260 × 10−3 = 873 kN 1,0 3 γM0 3 By inspection, the moment capacity is not reduced due to shear. Calculation of Mcr : E = 210 GPa; G = 81 GPa. The most foolproof method to determine Mcr given the use of the cm and dm in section property tables is to work in kN and m. Use Eq. (5.5) to determine Mcr . System length L = 3 m. Iw 0,608 × 10−6 = = 0,03924 m2 Iz 1545 × 10−8 π2 EIz π2 × 210 × 106 × 1545 × 10−8 = = 3558 kN L2 32

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L2 GIt GIt 81 × 106 × 62,8 × 10−8 = 0,0143 m2 = = 2 π EIz 3558 π2 EIz 2 L

π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 = 3558 [0,03924 + 0,0143]1/2 = 823 kNm

Determine the value of C1 using Eq. (5.40): The moment ratio ψ is given by ψ=

MEd,C 303 = 0,777 = 390 MEd,B

1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0,777 = 0,911 C1 or C1 =

1 = 1,098 0,911

The value of Mcr to be used in Eq. (5.12) in the calculation of the normalized slenderness ratio is C1 Mcr , thus λLT is given as Wy fy 1501000 × 355 λLT = = = 0,768 Mcr 1,098 × 823 × 106 Both methods available to determine the strength reduction factors due to lateral torsional buckling will be used in this example to demonstrate any differences. (a) General case The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.1, αLT = 0,34 (curve b). Use Eqs (5.15) and (5.16) to determine LT : LT = 0,5[1 + ηLT + (λLT )2 ] = 0,5[1 + 0,34(0,768 − 0,2) + 0,7682 ] = 0,891 Use Eq. (5.14) to determine χLT : χLT =

1 1 = = 0,745 2 − 0,7682 ]1/2 2 1/2 0,891 + [0,891 − (λLT ) ]

LT + [2LT

Mb,Rd = χLT Wpl,y

fy 355 = 0,745 × 1501000 × 10−6 = 397 kNm γM1 1,0

This is greater than the moment at B (390 kNm). Clearly the end section AB does not need checking as the system length is the same as BC (therefore the basic value of Mcr does not change and since ψ = 0, C1 now becomes 1,75 with the effect of reducing the value of λLT and of increasing the value of χLT and hence Mb,Rd . (b) Method for rolled sections The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.2, αLT = 0,49 (curve c).

112

•

Chapter 5 / Laterally Unrestrained Beams Determine ηLT from Eq. (5.19) ηLT = αLT (λLT − λLT,0 ) = 0,49(0,768 − 0,4) = 0,180 Determine LT from Eq. (5.18) LT = 0,5[1 + ηLT + β(λLT )2 ] = 0,5[1 + 0,180 + 0,75 × 0,7682 ] = 0,811 Determine χLT from Eq. (5.17) χLT =

1 1 = = 0,784 0,811 + [0,8112 − 0,75 × 0,7682 ] LT + [2LT − β(λLT )2 ]1/2

Initially ignore the correction factor f , to give Mb,Rd as Mb,Rd = χLT Wpl,y

fy 355 = 0,784 × 1501000 × 10−6 = 418 kNm γM1 1,0

Determine f : The moment ratio ψ = 303/390 = 0,777, so from Eq. (5.33), kc =

1 1 = = 0,931 1,33 − 0,33ψ 1,33 − 0,33 × 0,777

From Eq. (5.32) f is given by f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] ≤ 1,0 = 1 − 0,5(1 − 0,931)[1 − 2,0(0,768 − 0,8)2 ] = 0,966 Using Eq. (5.31) Mb,Rd now becomes 418/0,966 = 433 kNm. The effect of the factor f is not terribly significant. Even without the factor, f lateral torsional buckling clauses for rolled sections produces a marginally higher value of Mb,Rd by around 20 kNm (or around 5%). Deflection (under service loading): EI = 210 × (27430 × 104 ) × 10−6 = 57603 kNm2 Mid-span deflection due to an asymmetric point load is given by 3 WL3 b b δ= 3 −4 48EI L L Variable action deflection: Load at B: W = 70 kN, b = 3 m, l = 9 m, δ = 0,016 m Load at C: W = 30 kN, b = 3 m, L = 9 m, δ = 0,007 m Deflection under service variable actions = 0,023 m. This deflection is equivalent to span/390, and is therefore acceptable.

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113

Deflection under total actions: Load at B: W = 110 kN, b = 3 m, l = 9 m, δ = 0,025 m Load at C: W = 50 kN, b = 3 m, L = 9 m, δ = 0,011 m Total deflection = 0,036 m. This deflection is equivalent to span/250, and is therefore acceptable. Web check (refer to Section 4.8 and cl 6 EN 1993-1-5): The web capacity under transverse forces needs checking at A and B. Note as a rolled section is being used, fyw = fyf = 355 MPa. Since at either point the length of stiff bearing ss is not known, set ss = 0 and determine the value required should the check fail At A: FSd = RA = 130 kN. For an end support with c = ss = 0, kF = 2 (type c) Determine m1 : m1 =

fyf bf bf 179,5 = = = 18,9 fyw tw tw 9,5

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 18,9 = 16,0 = 49,2 mm 2 2

The depth of the web hw has been taken as d, the depth between fillets. tw3 9,53 = 0,9 × 2 × 210 = 889 kN hw 360,4 ly tw fyw 49,2 × 9,5 × 355 λF = = = 0,432 FCR 889 × 103

FCR = 0,9kF E

As λF < 0.5, m2 = 0. χF =

0,5 0,5 = = 1,16 0,432 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 49,2 = 49,2 mm FRd = Leff tw

fyw 355 × 10−3 = 166 kN = 49,2 × 9,5 γM1 1,0

As FRd is greater than FEd (=RA = 130 kN), η2 < 1,0, therefore the web resistance at A is satisfactory without a stiff bearing.

114

•

Chapter 5 / Laterally Unrestrained Beams Check at B: The applied load is 159 kN, and the applied moment is 390 kNm. For the situation where the load is applied through the top flange, kF = 6 (type a, with the stiffener spacing a effectively taken as infinity) Determine m1 : m1 =

fyf bf bf 179,5 = = = 18,9 fyw tw tw 9,5

As m2 is dependant upon λF initially assume m2 = 0. As ss has been assumed to be zero, then the value of ly is given by √ ly = 2tf (1 + m1 ) = 2 × 16,0(1 + 18,9) = 171 mm The depth of the web hw has been taken as d, the depth between fillets. tw3 9,53 = 0,9 × 6 × 210 = 2698 kN hw 360,4 ly tw fyw 171 × 9,5 × 355 λF = = = 0,462 FCR 2698 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. 0,5 0,5 χF = = = 1,08 0,462 λF The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 171 = 171 mm FRd = Leff tw η2 =

fyw 355 = 171 × 9,5 × 10−3 = 577 kN γM1 1,0

FEd fyw Leff tw γM1

=

FEd 159 = = 0,276 FRd 577

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing. However an interaction equation needs checking owing to the co-existence of shear and bending moment: η2 + 0,8η1 ≤ 1,4 As there is no axial force and no shift in the neutral axis as the section is Class 1, the equation for η1 reduces to η1 =

MEd f

y Wpl γM1

=

390 × 106 1501 × 103 355 1,0

= 0,732

η2 + 0,8η1 = 0,276 + 0,8 × 0,732 = 0,862 ≤ 1,4 The web at B is therefore satisfactory.

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EXAMPLE 5.2 Beam design with loads applied between lateral torsional restraints. Prepare a design in Grade S355 steel for the beam whose data are given in Fig. 5.14 Factored actions: Factored UDL: 1,35 × 10 + 1,5 × 20 = 43,5 kN/m Factored point load: 1,5 × 15 = 22,5 kN Figures 5.15 (b) and (c) show the resultant BM and SF diagrams. Try a 457 × 191 × 98 UKB Grade S355 Section classification: Flanges: c = 0,5(b − 2r − tw ) = 0,5(192,8 − 2 × 10,2 − 11,4) = 80,5 mm c 80,5 = = 4,11 tf 19,6 Class 1 limit: 9ε = 9

235 = 7,32 355

Flanges are Class 1. Web: c d 407,6 = = = 35,8 tw tw 11,4 Class 1 limit: 72ε = 72

235 = 58,56 355

Web is Class 1, therefore the section classification is Class 1.

A

20 kN/m Variable 10 kN/m Permanent

15 kN Variable B C

7.0 m

Notes: (1) All loading is characteristic loading (2) Lateral torsional restraints exist at A and B (3) The TOP flange is restrained at C

FIGURE 5.14 Design data for Example 5.2

3.5 m

116

•

Chapter 5 / Laterally Unrestrained Beams 43,5 kN/m

22,5 kN

A

B C

(a) Beam loading at ULS 2,37 174,7 102,9

D

22,5

B

C

−201,6 (b) SFD (kN) −192,3

A

D

B C

345,2 (c) BMD (kNm)

FIGURE 5.15 BMD and SFD for Example 5.2 Plastic moment capacity: Mpl,Rd = Wpl,y

fy 355 = 2232 × 103 × 10−6 = 793 kNm γM0 1,0

This exceeds the maximum applied moment of 345,2 kNm. Shear capacity: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 1146 kN = √ 5590 γ 1,0 3 3 M0

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By inspection, there is no reduction in moment capacity for the effect of shear at the cantilever support. System length AB: The system length is taken as the span AB, not the distance between the points of contraflexure calculated from the applied loading as these will not form nodal points in the post-buckled shape of the beam. Such nodal points may only form at points of lateral restraint since at a nodal point the lateral deflection must be zero (Kirby and Nethercot, 1979). Determination of C1 : As the loading is between restraints at A and C, the ‘n’ factor method from BS 5950: Part 1: 1990 must be used (Annexe A7). From Table 17, the BMD is that of the fifth diagram with β = 0 (as the BM at A is zero), and γ is negative. The bending moment M0 assuming the beam to be simply supported between A and C is given by qult L2 43,5 × 72 = = 266,4 kNm 8 8 −345,2 M = γ= = −1,3 M0 266,4

M0 =

From Table 16 with β = 0 and γ = −1,3, n = 0,53 1 1 C1 = √ = √ = 1,374 n 0,53 Determine Mcr from Eq. (5.5): Iw 1,18 × 10−6 = = 0,0503 m2 Iz 2347 × 10−8 L = 7,0 m. π2 EIz π2 × 210 × 106 × 2347 × 10−8 = = 993 kN L2 72 L2 GIt GIt 81 × 106 × 121 × 10−8 = 2 = = 0,0987 m2 2 π EIz 993 π EIz 2 L

Mcr =

π2 EI L2

z

Iw L2 GIt + 2 Iz π EIz

1/2 = 993[0,0503 + 0,0987]1/2 = 383 kNm

Determine λLT using the moment gradient modified value of Mcr in Eq. (5.12), 1/2 Wpl,y fy 1/2 2232 × 103 × 355 × 10−6 λLT = = = 1,227 1,374 × 383 C1 Mcr

118

•

Chapter 5 / Laterally Unrestrained Beams As in the previous example calculate the strength reduction factor using both methods: • General method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.1, αLT = 0,34 Determine LT from Eqs (5.15) and (5.16) LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,34(1,227 − 0,2) + 1,2272 ] = 1,427 Determine χLT from Eq. (5.14) χLT =

LT +

1

2LT

Mb,Rd = χLT Wpl,y

− (λLT

1/2 )2

=

1 = 0,464 1,427 + [1,4272 − 1,2272 ]1/2

fy 355 = 0,464 × 2232 × 103 × 10−6 = 368 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. • Rolled section method: h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.2, αLT = 0,49 Determine LT from Eqs (5.18) and (5.19) LT = 0,5[1 + αLT (λLT − λLT,0 ) + β(λLT )2 ] = 0,5[1 + 0,49(1,227 − 0,4) + 0,75 × 1,2272 ] = 1,267 Determine χLT from Eq. (5.17) χLT =

LT + [2LT

1 1 = = 0,511 2 2 1/2 1,267 + [1,267 − 0,75 × 1,2272]1/2 − β(λLT) ]

Without the correction factor f : Mb,Rd = χLT Wpl,y

fy 355 = 0,511 × 2232 × 103 × 10−6 = 405 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. From Table 6.6 of EN 1993-1-1, kc = 0,91 Determine f from Eq. (5.32): f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] = 1 − 0,5(1 − 0,91)[1 − 2,0(1,227 − 0,8)2 ] = 0,971 Thus the corrected value of Mb,Rd given by Eq. (5.31) is Mb,Rd =

405 = 417 kNm 0,971

It is again noted that the rolled section approach even without the factor f gives slightly higher values of Mb,Rd than the general case.

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System length BC. Take the conventional approach and adopt C1 = 1,0 Determine Mcr from Eq. (5.5): Iw 1,18 × 10−6 = = 0,0503 m2 Iz 2347 × 10−8 L = 3,5 m π2 EIz π2 × 210 × 106 × 2347 × 10−8 = = 3971 kN L2 3,52 L2 GIt GIt 81 × 106 × 121 × 10−8 = 2 = = 0,0247 m2 2 π EIz 3971 π EIz L2

π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 = 3971[0,0503 + 0,0247]1/2 = 1088 kNm

Determine λLT from Eq. (5.12), 1/2 Wpl,y fy 1/2 2232 × 103 × 355 × 10−6 λLT = = = 0,853 C1 Mcr 1088 As in the previous example calculate the strength reduction factor using both methods: • General method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.1, αLT = 0,34 Determine LT from Eqs (5.15) and (5.16) LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,34(0,853 − 0,2) + 0,8532 ] = 0,975 Determine χLT from Eq. (5.14) χLT =

1

LT + 2LT − (λLT )2

Mb,Rd = χLT Wpl,y

1/2 =

1 = 0,691 0,975 + [0,9752 − 0,8532 ]1/2

fy 355 = 0,691 × 2232 × 103 × 10−6 = 548 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. • Rolled section method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.2, αLT = 0,49 Determine LT from Eqs (5.18) and (5.19) LT = 0,5[1 + αLT (λLT − λLT,0 ) + β(λLT )2 ] = 0,5[1 + 0,49(0,853 − 0,4) + 0,75 × 0,8532 ] = 0,884

120

•

Chapter 5 / Laterally Unrestrained Beams A

B

C

El constant a Deflection: At mid-span (span AB) At C

b (qa 2/384 EI)(5a 2 − 12b 2) (q b/24 EI)(3b 3 + 4 ab2 − a 3) (a) UDL P

A

B C

El constant a

b

Deflection: At mid-span (span AB) −Pba 2/16EI At C Pb 2(a + b)/3EI (b) Point load

FIGURE 5.16 Deflection formulae for Example 5.2

Determine χLT from Eq. (5.17) χLT = =

LT + [2LT

1 − β(λLT )2 ]1/2

1 = 0,730 0,884 + [0,8842 − 0,75 × 0,8532 ]1/2

As there appears to be no consideration given to cantilevers in Table 6.6, f will be taken as 1,0. Mb,Rd = χLT Wpl,y

fy 355 = 0,730 × 2232 × 103 × 10−6 = 578 kNm 1,0 γM1

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. It is again noted that the rolled section approach even without the factor f gives slightly higher values of Mb,Rd than the general case. For this particular design case, the span AC is critical. Deflection check: EI = 210 × 10−6 × 45730 × 104 = 96033 kNm2 . The relevant formulae are given in Fig. 5.16. (a) Variable action check: UDL: 20 kN/m

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121

Span AB, central deflection δ=

qa2 20 × 72 (5a2 − 12b2 ) = (5 × 72 − 12 × 3,52 ) = 0,0026 m 384EI 384 × 96033

Span BC, at C qb 20 × 3,5 (3b3 + 4ab2 − a3 ) = (3 × 3,53 + 4 × 7 × 3,52 − 73 ) 24EI 24 × 96033 = 0,0039 m

δ=

Point load at C Span AB, mid-span δ=−

Pba2 15 × 3,5 × 72 =− = −0,0017 m 16EI 16 × 96033

Span BC, at C δ=

P(a + b)b2 15(7 + 3,5)3,52 = = 0,0067 m 3EI 3 × 96033

Net deflections: at mid-span = 0,0026 − 0,0017 = 0,0009 m Span deflection ratio is 7/0,0009 = 7780. This is more than acceptable. at C = 0,0039 + 0,0067 = 0,0106 m Span deflection ratio (based on twice the span) is 2 × 3,5/0,0106 = 660. This is acceptable. (b) Check under total actions. Deflection due to point load as above. Total UDL of 30 kN/m: Mid-span, δ = 0,0039 m; at C, δ = 0,0059 m. Total deflection at mid-span = 0,0039 − 0,0017 = 0,0022 m Span deflection ratio: 7/0,0022 = 3180. This is satisfactory. Total deflection at C = 0,0059 + 0,0067 = 0,0126 m Span deflection ratio (based on twice the span) is 2 × 3,5/0,0126 = 556. This is satisfactory. Web check at B (refer to Section 4.8 and cl 6 EN 1993-1-5): This is the only point that needs checking, as the other reaction point has a much lower force (even allowing for reduced dispersion length) and no coincident moment. RSd = 124,7 kN (Reaction at B); M = 114,5 kNm Ignore any stiff bearing (ss = 0). For the situation where the load is applied through the top flange, kF = 6 (type a, with the stiffener spacing a effectively taken as infinity)

122

•

Chapter 5 / Laterally Unrestrained Beams Determine m1 : m1 =

fyf bf bf 192,8 = = = 16,9 fyw tw tw 11,4

As m2 is dependant upon λF initially assume m2 = 0. As ss has been assumed to be zero, then the value of ly is given by √ ly = 2tf (1 + m1 ) = 2 × 19,6(1 + 16,9) = 200 mm The depth of the web hw has been taken as d, the depth between fillets. tw3 11,43 = 0,9 × 6 × 210 = 4122 kN hw 407,6 ly tw fyw 200 × 11,4 × 355 λF = = = 0,443 FCR 4122 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. χF =

0.5 0,5 = 1,13 = 0,443 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 200 = 200 mm FRd = Leff tw η2 =

fyw 355 = 200 × 11,4 × 10−3 = 809 kN γM1 1,0

FEd fyw Leff tw γ M1

=

FEd 124,7 = = 0,154 FRd 809

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing. However an interaction equation needs checking owing to the co-existence of shear and bending moment: η2 + 0,8η1 ≤ 1,4 As there is no axial force and no shift in the neutral axis as the section is Class 1, the equation for η1 reduces to η1 =

MEd fy

Wpl γ

M1

=

114,5 × 106 2232 × 103 355 1,0

= 0,145

η2 + 0,8η1 = 0,154 + 0,8 × 0,145 = 0,27 ≤ 1,4 The web at B is therefore satisfactory.

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123

5.1.6 Other Section Profiles A number of special cases need considering, hollow sections as the earlier equations for critical moment are not applicable, rectangular sections as the warping stiffness is zero, and T sections.

5.1.6.1 Rolled Hollow Sections Rees (1990) indicates that thin wall tubes of circular and triangular cross with uniform thickness cannot warp. For rectangular thin walled tubes only those whose wall thicknesses are in a constant ratio the length of the sides do not warp. These tubes are known as Neuber tubes. If there is no warping then lateral torsional buckling can only be resisted by torsion. For conventional hollow sections where the wall thickness is constant, then lateral torsional buckling is in part resisted by warping. It will be conservative to neglect the warping stiffness, and, therefore as a result, Eq. (5.3) can be reduced to GIt

d2 φ M2 = − φ EIz dx2

(5.53)

The resultant value of Mcr (with no allowance for major axis bending) is given from Eq. (5.6) as √ π EIz GIt (5.54) Mcr = L The factor K from Eq. (5.7) with the warping constant Iw set equal to zero becomes, EIz GIt K= 1− 1− (5.55) EIy EIy Combining Eqs (5.54) and (5.55) gives the elastic critical moment as √ π EIz GIt Mcr = EI GI L 1 − EIz 1 − EI t y

(5.56)

y

or the normalized slenderness ratio, λLT , is given from Eq. (5.12) with the introduction of the moment gradient factor C1 as t Wpl,y fy L 1 − IIz 1 − GI EI y y Wpl,y fy λLT = = (5.57) √ C1 Mcr C1 π EIz GIt The normalized lateral torsional buckling slenderness ratio λLT is also given by λLT =

λLT λ1

(5.58)

124

•

Chapter 5 / Laterally Unrestrained Beams or λLT = λLT λ1 = π

E λLT fy

t πEWpl,y L 1 − IIz 1 − GI EIy y = √ C1 EIz GIt

(5.59)

Rewrite Eq. (5.59) as

λLT =

1

t 1/2 Wpl,y L 1 − IIz 1 − GI EI y y E π √ G Iz It

1/2

C1

(5.60)

Define the slenderness ratio λ as λ=

L L =! iz Iz

(5.61)

A

when Eq. (5.62) becomes

λLT

t 1/2 Wpl,y λ 1 − IIz 1 − GI EIy y 1 E = 1/2 π √ G AIt C1

(5.62)

or λLT is given as λLT =

1 1/2

C1

E π G

1/2 (φb λ)1/2

(5.63)

where φb is defined as by " #" # ⎞1/2 I GI 2 1 − Iz 1 − EI t Wpl,y y y ⎠ φb = ⎝ AIt ⎛

(5.64)

This is the equation given in Section B2.6.1 of BS 5950-1. The term in rectangular parentheses in Eq. (5.63) has a value of 2,25. An alternative approach avoiding the calculation of lateral torsion buckling is to determine the critical length lcrit (in mm) corresponding to a value of λLT equalling 0,4 (below which buckling will not occur) (Rondal, et al., 1992). This value is given by

lcrit =

113400(h − t) fy

b−t 2 h−t b−t 1 + 3 h−t

3 + b−t h−t 1+

b−t h−t

(5.64a)

It should be noted that there is an apparent anomaly in Eq. (5.64) in that for a square section (b = h), lcrit remains finite, whereas Eq. (5.57) indicates φb (and hence λLT ) = 0.

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125

The reason is that the values from Eq. (5.64) in Rondal, et al. are also given in tabular form for discrete values of (h − t)/(b − t), and that interpolation would not be possible if the value for a square section were given as infinity. It should be noted that there is an error in the formula quoted in Rondal, et al., but that the values in Table 15 in the same publication are correct. Eq. (5.168) has been corrected (Kaim, 2006). It should be noted that the values of lcrit are extremely safe. Kaim (2006) suggests the normalized slenderness limit λz,lim is given by 25

λz,lim = h b

235 fy

(5.64b)

5.1.6.2 Rectangular Sections For rectangular sections of width b and depth h the equation for critical moment Mcr given in Eq. (5.6) reduces to Mcr =

π EIz GIt L

(5.65)

as the warping constant Iw is zero. (a) Thin sections For thin sections, It = hb3 /3 and Iz = hb3 /12, so Eq. (5.65) reduces to Mcr =

hb3 πE L 72(1 + ν)

(5.66)

(b) Thick sections In this case It is no longer given by ht3 /3. The following approximate formula can be used hb3 b 1 b 4 It = (5.67) 1 − 0,63 1− 3 h 12 h If h/b > 2, Eq. (5.67) can with little loss in accuracy be reduced to hb3 b It = 1 − 0,63 3 h

(5.68)

However the major axis bending is now important and to ignore it would be too conservative, thus the parameter K from Eq. (5.8) must be introduced to give Mcr as Mcr =

πhb3 √ 6L

1 − 0,63 b h EG 2 b 1− h

(5.69)

Chapter 5 / Laterally Unrestrained Beams

t1

b1

z0

zc

Axis through shear centre

hs

Elastic neutral axis

tw

t2

•

d

126

FIGURE 5.17 Calculation of monosymmetry index

b2

5.1.6.3 Monosymmetric Beams For beams with only one axis of symmetry (usually the minor axis), the centroid and the shear centre do not coincide, thus an additional disturbing torque occurs due to the longitudinal flexural stresses. From Trahair and Bradford (1988) the elastic critical moment Mcr for the monosymmetric beam in Fig. 5.17 is given by ⎧ ⎫ ⎨ 2 2 π π EIw πγM ⎬ πγM Mcr = EIz GIt 1+ + + (5.70) ⎩ L 2 ⎭ 2 GIt L2 where γM is given by βy EIz γM = L GIt

(5.71)

The monosymmetry parameter for the section βy is given by βy = where βy1

βy2

+ 1* βy1 − βy2 + βy3 − 2z0 Iy

(5.72)

b3 t2 = (hs − zc ) 2 + b2 t2 (hs − zc ) 12 b31 t1 = zc + b1 t1 zc2 12

(5.73)

(5.74)

Structural Design of Steelwork to EN 1993 and EN 1994

βy3

tw = 4

t2 hs − zc − 2

4

t1 − zc − 2

•

127

4 (5.75)

where hs = d −

t 1 + t2 2

(5.76)

t

b2 t2 hs + 2w (d − t1 − t2 )(d − t2 ) zc = b1 t1 + b2 t2 + (d − t1 − t2 )tw

(5.77)

z0 = αhs − zc

(5.78)

α=

1 b

1 + b1 2

3

t1 t2

(5.79)

The warping constant Iw is now given by Iw = α(1 − α)Iz h2s

(5.80)

5.1.6.4 ‘T’ Beams For ‘T’ beams, it should be first checked that the value of K from Eq. (5.8) is real as a large number of commercial ‘T’ beams have Iy > Iz in which case lateral torsional buckling cannot occur. Where Mcr needs calculating, it should be noted Iw is zero as α = 0 (Eq. (5.80)) and z0 = −zc . The position of the centroid zc and Iy are tabulated in section property tables.

5.1.6.5 Parallel Flange Channels The procedure for calculating Mcr follows that for ‘I’ beams except that Iw is calculated as Iw =

tf b3f h2 3bf tf + 2htw 12 6bf tf + htw

(5.81)

where bf and tf are the width and thickness of the flange and h and tw are the height and thickness of the web. For channels with tapered flanges tf may be taken as the mean thickness of the flange (Kirby and Nethercot, 1979).

5.2 PURE TORSIONAL BUCKLING This form of failure can only occur in open sections and is most likely to only where the sections are thin walled. There can then be an interaction between strut buckling or buckling in pure torsion.

128

•

Chapter 5 / Laterally Unrestrained Beams σ x Thickness (t ) b

b

b

b

x (a) Basic layout

cL

(b) Cross-section

s

dv

φ φ

dH

dθ dd

(c) Deflected geometry

FIGURE 5.18 Cruciform strut

5.2.1 Interaction between Torsional Buckling and Strut Buckling This is best illustrated by considering the case of a cruciform strut loaded by a uniform stress σ over its cross-section. To demonstrate the principles a thin section strut is considered (Fig. 5.18). The load induced by the stress σ remains parallel to the x axis, and therefore induces a lateral force in the yz plane. If the vertical component in Fig 5.18(c) is dV then the horizontal component dH is given by +dH = dV tan φ = φdV = s

dθ dV dx

(5.82)

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129

as φ is small, and where dθ/dx is the angle of twist per unit length and s is a distance measured from the x axis. The force dV is given by dV = σdA = σtds

(5.83)

where t is the thickness. The incremental torque dT is given by dT = sdH

(5.84)

Substitute Eqs (5.81) and (5.83) into Eq. (5.84) to give dT = σ

dθ 2 ts ds dx

(5.85)

Integrate Eq. (5.85) over the four arms of the strut to give h dθ dθ 4 3 T = 4σ t s2 ds = σ b t dx 0 dx 3

(5.86)

where 2b is the width of the strut. Equation (5.86) may be rewritten as T =σ

dθ Ix dx

(5.87)

where Ix is the polar second moment of area about the x axis, and is given by Ix =

4 3 b t 3

(5.88)

The second moment of area of the strut about either the z or y axis is given by Iz = Iy =

1 2 (2b)3 t = b3 t 12 3

(5.89)

as the other arm of the strut has negligible second moment of area (bt3 /6) compared with the other direction. Note that

2 3 4 Ix = Iy + Iz = 2 b t = b3 t 3 3

(5.90)

From torsion theory the torque that may be carried by the section is given by T = GIt

dθ dx

(5.91)

where G is the shear modulus and Tt is the torsional second moment of area. For the cruciform thin walled section It is given by 4 1 3 bt = bt3 It = 4 (5.92) 3 3

•

Chapter 5 / Laterally Unrestrained Beams

L b Strut buckling

π2E 6s0

130

Yield

Torsional buckling b t

E 2(1v)σ0

FIGURE 5.19 Interaction diagram

Substitute Eq. (5.92) into Eq. (5.91) to give T=

4 3 dθ bt G 3 dx

(5.93)

For the section to be able to sustain its twisted shape the two values of the torques from Eqs (5.87) and (5.93) must be equal, so σ=G

2 t b

(5.94)

There also exists the possibility that the strut can undergo normal Euler buckling under a stress σcr given by σcr =

π2 E 23 b3 t π2 EIz π2 E = = 6 AL2 4btL2

2 b L

(5.95)

Assuming no interaction, there are three possibilities of behaviour; the stress produces yield; torsional buckling occurs, or strut buckling occurs (Fig. 5.19). The stress which causes torsional buckling and strut buckling to occur simultaneously is given when the stresses from Eqs (5.94) and (5.95) are equal, or 2 t π2 E b 2 G = b 6 L

(5.96)

Structural Design of Steelwork to EN 1993 and EN 1994 or

L b = b t

π2 E b = 6G t

π2 (1 + v) 3

•

131

(5.97)

as G = E/2(1 + v). The transition from yield to torsional buckling occurs when the stress in Eq. (5.94) equals the yield stress σ0 , or b G E = (5.98) = t σ0 2(1 + ν)σ0 and from yield to Euler buckling when the stress in Eq. (5.95) equals σ0 , or L π2 E = b 6σ0

(5.99)

Note, other sections such as thin walled angles may also suffer similar behaviour, but the analysis is more complex as buckling is about the principal axes.

5.2.2 Torsional Buckling Interaction The critical axial load for torsional buckling Ncr is given in Chapman and Buhagiar (1993) as 1 π2 EIw Ncr,T = 2 GIt + (5.100) is lT2 where lT is the buckling length, and is the polar radius of gyration given by i2s = i2z + i2y + y20 + z02

(5.101)

where iz and iy are the flexural radii of gyration and z0 and y0 are distances from the shear centre to the geometric centroid. For a section whose centroid and shear centre co-incide y0 and z0 are zero. Alternatively the critical stress σcr,T is given by 1 π2 EIw σcr,T = GIt + (5.102) I0 lT2 where I0 is the polar moment of area. Timoshenko and Gere (1961) give the following interaction equation between strut buckling and torsion buckling i2s (N − Ncr,z )(N − Ncr,y )(N − Ncr,T ) − N 2 z02 (N − Ncr,y ) − N 2 y20 (N − Ncr,z ) = 0 (5.103) where N is the critical value of the axial load and Ncr,z and Ncr,y are the Euler buckling loads about the zz and xx axes.

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Chapter 5 / Laterally Unrestrained Beams For a section whose centroid and shear centre co-incide, Eq. (5.103) reduces to (N − Ncr,z )(N − Ncr,y )(N − Ncr,T ) = 0

(5.104)

That is, N is therefore the least of Ncr,x , Ncr,y and Ncr,T . Chapman and Buhagiar (1993) also indicate that where buckling can occur about both axes (i.e. where the buckling lengths and second moments of area are approximately equal, or where the critical loads are similar), then the imperfection factor η in the standard strut buckling interaction equation should be taken as twice its normal value to allow for torsional buckling.

5.3 PLATE GIRDERS Plate girders are used either on long spans where a rolled section would need to be spliced and as a result may be inefficient, or to support heavy loads such as on a bridge structure. It is important to note that although plate girders may be lighter than other forms of compound beams, fabrication costs are likely to be much higher. Also as Corus now roll UKB’s with a depth of 1016 mm, the use of plate girders in building structures unless spans are extremely high is less likely. Plate girders are built up from two flange plates and a web plate, generally from the same grade of steel. Continuous automatic electric arc or submerged gas welding is used to form the fillet welds between the flange and web (Fig. 5.20). Such welding is generally performed as a double pass one on either side of the girder. It should be noted that this process may cause very high residual stresses to exist in the flanges and web. The exact magnitudes will depend also on whether the flange plates were sheared or flame cut to size (Nethercot, 1974a). The stiffeners are then welded in place often manually. Such stiffeners are needed either to help combat the effects of web buckling or to provide support to any concentrated load or reaction. Only straight girders with equal flanges and vertical stiffeners are considered in this text. Flange plate

Web plate

Welds on 2nd pass

Fillet welds

1st pass

FIGURE 5.20 Plate girder fabrication

Welds from 1st pass

2nd pass

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5.3.1 Minimum Web Thickness With no web stiffeners the web should be sized to avoid the flange undergoing local buckling due to the web being unable to support the flange. This is known as flange induced buckling (cl. 8 EN 1993-1-5). In pure bending the flanges are subjected to equal and opposite forces. The force per unit length is given by σv tw =

fyf Afc R

(5.105)

where σv is the vertical stress in the web, Afc is the area of the compression flange and R is the radius of curvature. The curvature is dependant upon the variation of strain εf occurring at the mid-depth of the flanges. The residual strain due to fabrication εf is assumed to have a value of 0,5εy , and εf is the sum of the yield strain εy and the residual strain εf , that is, a total of 1,5εy . The radius of curvature R is then given by R=

0,5hw εf

(5.106)

where for simplicity hw has been taken as the depth between the centroids of the flanges rather than the clear depth of the web. It should be noted any error will be small. From Eqs (5.105) and (5.106), σv = 3

2 Afc fyf Aw E

(5.107)

The stress σv cannot exceed the elastic critical buckling stress for a simply supported thin plate which is given by (Bulson, 1970) as π2 E σv = 12(1 − ν2 )

tw hw

2 (5.108)

Equating Eqs (5.107) and (5.108) gives with a slight change in notation with Ac replacing Afc hw E Aw ≤ 0,55 (5.109) tw fyf Ac EN 1993-1-5 modifies the co-efficient of 0,55 to allow for situations where higher strains are required. Thus the critical hw /tw ratio is given by hw E Aw ≤k (5.110) tw fyf Ac where fyf is the yield strength of the compression flange, Afc is the effective area of the compression flange and Aw is the area of the web. The parameter k takes values of 0,3 where plastic hinge rotation is utilized, 0,4 if the plastic resistance is utilized

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Chapter 5 / Laterally Unrestrained Beams and 0,55 if the elastic resistance is utilized. Thus for rigid (continuous) design k = 0,3 unless the analysis is elastic with no redistribution. For simply supported beams k may be taken as 0,4.

5.3.2 Bending Resistance The section classification is determined in the same manner as rolled sections.

5.3.2.1 Compression Flange Restrained (i.e. Lateral Torsional Buckling cannot Occur) There are two methods that can be used for girder design: (1) The flanges carrying the bending moment and the web the shear force This is probably best used where the maximum bending moment and maximum shear force are not coincident, as the ability of the flange to contribute towards shear capacity may be utilized. Thus for a restrained beam under a UDL, this method may be advantageous. If the maximum bending moment and maximum shear are coincident in either a simply supported beam under point loading or in a continuous beam at the internal support, then the flange capacity will not be able to be utilized to resist shear, thus probably necessitating a thicker web. With this method, the moment capacity is only dependant on the section classification of the flanges as the web does not carry compression. (2) The girder carrying the forces as an entity This method of design is more complex as the beam is likely to be Class 4 and may not show any resultant economies over the first method, but should be utilized where maximum moment and maximum shear are co-incident.

5.3.2.2 Lateral Buckling May Occur In this case the second method must be used. The design then follows that for rolled beams, except that the general method for calculating χLT should be used with the value of αLT appropriate for welded sections. Section properties will need to be calculated from first principles.

5.3.3 Basic Dimensioning One method of dimensioning is to consider a minimum weight solution. It must be noted that a minimum weight solution is not synonymous with a minimum cost solution (Gibbons, 1995). Method 1: Assuming the moment to be resisted by the flanges alone, then MRd = fyd bf tf hw

(5.111)

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where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. Equation (5.111) is slightly conservative for beams of Classes 1 to 3. The cross-sectional area A is given by A = 2bf tf + hw t

(5.112)

Eliminate bf tf between Eqs (5.111) and (5.112) to give A=

2MRd + hw t hw fyd

(5.113)

Define the web slenderness ratio hw /t as λ, then Eq. (5.113) becomes A=

2MRd + λt2 λtfyd

(5.114)

For an optimum solution, dA/dt = 0, so Eq. (5.114) becomes, dA 2MRd + 2λt = 0 =− dt λfyd t2 or,

t=

3

and

MRd λ2 fyd

hw =

3

(5.115)

(5.116)

λMRd fyd

(5.117)

The area of the web, Aw is then given by 2 MRd 3 Aw = 2 λfyd Using Eq. (5.112), the flange area, Af is given by 2 MRd MRd 3 Af = bf tf = = 2 λfyd λM fyd 3 f Rd

(5.118)

(5.119)

yd

Thus the area of a single flange is equal to that of the web. Method 2 (Classes 1 and 2): It is recognized that this is not a likely case but is included for completeness. The moment is resisted by the complete section, when the moment capacity is given by that due to the flanges (Eq. (5.111)) and the additional plastic capacity of the web MRd = fyd bf tf hw + fyd

th2w 4

(5.120)

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•

Chapter 5 / Laterally Unrestrained Beams where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. The cross-sectional area A is given by Eqs (5.112), thus from Eqs (5.112) and (5.120), bf tf is given by bf tf =

MRd hw t − fyd hw 4

(5.121)

Eliminate bf tf between Eqs (5.112) and (5.121) to give A=

2MRd hw t hw t 2MRd + hw t − 2 + = hw fyd 4 hw fyd 2

(5.122)

Define the web slenderness ratio hw /t as λ, then Eq. (5.122) becomes A=

2MRd λt2 + λtfyd 2

(5.123)

For an optimum solution, dA/dt = 0, so Eq. (5.123) becomes, dA 2MRd + λt = 0 =− dt λfyd t2

(5.124)

or, t=

3

and

2MRd λ2 fyd

hw =

3

(5.125)

2λMRd fyd

(5.126)

The area of the web, Aw is then given by 2 4MRd 3 Aw = 2 λfyd Using Eq. (5.121), the flange area, Af is given by 2 MRd 3 1 1 3 3 Af = bf tf = − 2 2 16 λfyd or, 3

Af = Aw

2 MRd 2 λfyd

! 3

3

1 2

−

2 4MRd 2 λfyd

! 3

1 16

(5.127)

(5.128)

=

1 4

Thus the area of the web is equal four times that of a single flange.

(5.129)

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Method 2 (Class 3): The moment is resisted by the complete section, when the moment capacity is given by that due to the flanges (Eq. (5.111)) and the additional elastic capacity of the web MRd = fyd bf tf hw + fyd

th2w 6

(5.130)

where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. The cross-sectional area A is given by Eq. (5.112), thus from Eqs (5.112) and (5.130), bf tf is given by bf tf =

MRd hw t − fyd hw 6

(5.131)

Eliminate bf tf between Eqs (5.112) and (5.131) to give A=

2MRd hw t 2hw t 2MRd + hw t − 2 + = hw fyd 6 hw fyd 3

(5.132)

Define the web slenderness ratio hw /t as λ, then Eq. (5.132) becomes A=

2MRd 2λt2 + λtfyd 3

(5.133)

For an optimum solution, dA/dt = 0, so Eq. (5.133) becomes, dA 2MRd 4 =− + λt = 0 2 dt 3 λfyd t

(5.134)

or, t=

3

3MRd 2λ2 fyd

(5.135)

and hw =

3

3λMRd 2fyd

(5.136)

The area of the web, Aw is then given by 2 9MRd 3 Aw = 2 4λfyd

(5.137)

Using Eq. (5.131), the flange area, Af is given by 2 MRd 3 2 1 3 Af = bf tf = − 2 3 6 λfyd

3

9 4

(5.138)

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•

Chapter 5 / Laterally Unrestrained Beams Thus 3

Af = Aw

2 MRd 2 λfyd

! 3

2 3

− 16

! 3

2 3 9MRd

9 4

=

1 2

(5.139)

4λfyd

Thus the area of the web is equal twice that of a single flange. It should be noted that a plate girder may well be Class 4 in which case an effective section needs calculating, and thus minimum weight optimization is not directly possible, although it can be simulated by increasing the applied moment (see Example 5.4).

5.3.4 Web Design Experimental work (Basler, 1961; Porter et al., 1975; Rockey et al., 1978; Davies and Griffith, 1999) showed that after web buckling occurred there was still a reserve of strength in the web. This additional reserve of strength in the web is due to a tension field forming in the central diagonal portion of the web (Fig. 5.21). The shear capacity in the web is determined using a shear buckling slenderness λw which is dependant upon the critical shear strength τcr (cl. A.1 EN 1993-1-5) The critical shear strength τcr is given by τcr = kτ σE

(5.140)

where kτ is a shear buckling co-efficient dependant upon the aspect ratio of a web panel and σE is the elastic critical stress, From classical plate buckling theory (Bulson, 1970) σE =

π2 E 12(1 − ν2 )

t hw

2

= 19000

t hw

2 (5.141)

where t is the thickness and hw the depth. The parameter kτ is given as (Bulson, 1970), for a/hw ≥ 1,0 kτ = 5,34 + 4,00

hw a

2 (5.142)

for a/hw < 1,0 kτ = 4,00 + 5,34

hw a

2 (5.143)

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τcr

hw

a

τcr

(a) Buckled shape

τ Buckled panel

Stiffener

Tension panel

Buckled panel

(b) Panel showing tension field

FIGURE 5.21 Shear failure of a plate girder

The non-dimensionalized web slenderness ratio λw is defined as (cl. 5.3 EN 1993-1-5) ⎤1/2 ⎡ f √yw

⎥ ⎢ λw = ⎣ 3 ⎦ τcr

= 0,76

fyw τcr

(5.144)

For webs with transverse stiffeners at the supports and either intermediate transverse or longitudinal stiffeners, the normalized web slenderness ratio λw is obtained by

140

•

Chapter 5 / Laterally Unrestrained Beams substituting Eqs (5.140) and (5.141) into (5.144) λw =

hw √ 37,4εt kτ

(5.145)

For webs with transverse stiffeners only at the supports, a/hw is large, hence with little loss in accuracy kτ = 5,34 from Eq. (5.142). Thus λw is given by λw =

hw hw = √ 86,4εt 37,4εt 5,34

(5.146)

Clearly the upper limit to the value the shear force that may be carried by the web is the plastic shear capacity. The limiting value of web slenderness beyond which buckling need to be considered may be derived as follows. From Eq. (5.142) for an infinitely long web, kI = 5,34, thus from Eq. (5.140), the limiting value of hw /t is given when √ τcr = τyw = fyk / 3, or hw = t

5,34π2 E 235 × 12(1 − ν2 )

235 = fyk

5,34π2 × 210 × 103 ε = 65,7ε 12(1 − 0,32 ) × 235

(5.147)

√ The code uses a lower limit of 72ε/η for unstiffened webs and 31ε kτ /η (cl 5.1.(2), EN 1993-1-5). The recommended value of η for steel grades of S460 or lower is 1,2. For steel grades higher than S460, η = 1,0. With η = 1,2, the lower limit for steel grades up to and including S460 is 72ε/1,2 = 60ε which is slightly more conservative than the figure derived in Eq. (5.147). Basler (1961) and Rockey and Škaloud (1971) recognized that actual web behaviour could be categorized by three regimes: pure shear, elastic buckling at the extremes and a transition phase between the two limits. (1) Non-rigid end post In the case of a non-rigid end post (which cannot generate post-buckling strength as it is incapable of resisting the additional horizontal forces) EN 1993-1-5 only defines two zones for the contribution of the web to shear buckling resistance χw where a tension field cannot be generated as the anchorage force is unable to be sustained, λw ≤ 0,83/η χw = η λw > 0,83/η 0,83 χw = λw

(5.148)

(5.149)

(2) Rigid end post Where there is a rigid end stiffener, the anchorage force from tension field theory can be sustained. The original theory behind tension field theory was outlined by Porter et al. (1975) and Rockey et al. (1978). EN 1993-1-5 has adopted Höglund’s rotating stress–field theory (Davies and Griffith, 1999) which is easier to apply

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than the original tension field theory. Höglund’s rotating stress–field theory also mobilizes post-buckling behaviour, but only if there is a rigid end post. In this case the relationships between normalized web slenderness λw and χw are given by λw ≤ 0, 83/η χw = η

(5.150)

0,83/η ≤ λw < 1,08 χw =

0,83 λw

(5.151)

λw ≥ 1,08 χw =

1,37 0,7 + λw

(5.152)

It will be observed that the difference between the two methods is that the web shear parameter χw is enhanced for λw ≥ 1,08. The design resistance of a web Vb,Rd whether stiffened or unstiffened (cl 5.2 EN 1993-1-5) is given by Vb,Rd = χv hw t

f √yw 3

γM1

(5.153)

where shear co-efficient χv is given by χv = χw + χf ≤ η

(5.154)

The parameter χf represents the contribution to shear resistance from the flanges for MEd < Mf,Rd and is given by √ bf tf2 fyf 3 MEd 2 χf = 1− (5.155) cthw fyw Mf,Rd where bf is the width of the flange taken as not greater than 15εtf on each side of the web, tf is the web thickness, Mf,Rd is design moment of resistance of the cross-section determined using the effective flanges only and c is the width of the portion of the web between the plastic hinges (see Fig. 5.22) and is given by Mpl,f 1,6bf tf2 fyf c = a 0,25 + 1,6 (5.156) = a 0,25 + Mpl,w th2w fyw Equation (5.155) is derived by determining the shear that may be carried by the portion of both flanges of length c with the yield strength fyf reduced by considering the effect of induced axial forces in the flanges. The background to the simplified method adopted in the Code is given in Davies and Griffith (1999).

•

Chapter 5 / Laterally Unrestrained Beams Position of hinges

Flange

Web

d

142

c a (a) Basic geometry of web panel

(b) Failure mechanism used to derive post buckling strength of web

FIGURE 5.22 Overall behaviour of web at failure The calculations for web shear capacity are iterative as both the shear factors and the applied moment are dependant upon the stiffener spacing. The calculations for shear capacity in the examples following were performed on a spreadsheet.

5.3.5 Stiffeners 5.3.5.1 Rigid End Post (cl. 9.3.1 EN 1993-1-5) This may either be a set of flats welded at the end and above the support with the centroids a distance e apart, or the end post may comprise a rolled section in which case e is the distance between the flange centroids. The horizontal stress σh for large slenderness ratios can be given as 0.43 σh = fy (5.157) λw Substitute the value of λw from Eq. (5.145) into Eq. (5.157) to give an upper bound for the UDL qh as √ fy t 2 ε k τ qh = σh t = 16.1 (5.158) hw

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Taking the maximum value of ε as 1,0 and the maximum value of kτ as 9,34, the value of qh becomes qh = 49

t2 fy hw

(5.159)

The distributed load is not uniform over the depth of the girder and the theoretical value of σh is high, the co-efficient of 49 in Eq. (5.159) is replaced by 32. Assuming the end post is simply supported, the maximum moment Mmax is given by Mmax =

qh h2w 8

(5.160)

The section modulus, ignoring any contribution from the web, W is given by Amin e, thus assuming the maximum stress is given by fy , then 32t2 fy h2w

σmax

Mmax h 8 = fy = = w W Amin e

(5.161)

Thus the minimum cross-sectional area of each pair Amin is given by Amin =

4h2w t e

(5.162)

with e > 0,1hw . The restriction on e would appear to be a detailing requirement. The end panel may be designed as non-rigid shear panel carrying the whole of the applied shear. This was originally proposed by Basler (1961). From Eq. (5.153) with χf = 0, the required value of χw is given by √ VEd 3γM1 χw = fyw hw t

(5.163)

From Eq. (5.149), the normalized web slenderness λw is given by λw =

0,83 χw

(5.164)

The buckling parameter kτ is given from Eq. (5.145) as kτ =

hw 37,4tελw

2 (5.165)

As a/hw < 1, kτ is given by Eq. (5.143), or the required panel width a is given as 5,35 a = hw (5.166) kτ − 4,00 Note, this will give an upper bound to the value of a.

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5.3.5.2 Transverse Stiffeners (cl 9.2.1 EN 1993-1-5) These should be checked as a simply supported beam with an initial sinusoidal imperfection w0 given by Eq. (5.169) together with any eccentricities. The transverse stiffener should carry the deviation forces from the adjacent panels assuming the adjacent transverse stiffeners are rigid and straight. A second order analysis should be used to determine that the maximum stress does not exceed fyd nor any additional deflection b/300. This will be more critical for single sided stiffeners. As double sided plate stiffeners have been used in the ensuing examples, only the stress criterion has been checked. In the absence of transverse loads or axial forces in the stiffeners, then the strength and deflection criteria are satisfied if they have a second moment of area given by σM b 4 300 Ist = 1 + w0 u (5.167) E π b where σM is given by σM

σcr,c NEd = σcr,p b

1 1 + a1 a2

(5.168)

where a1 and a2 are the panel lengths either side of the stiffener under consideration, NEd is the larger compressive force in the adjacent panels, b is the height of the stiffener. The initial imperfection w0 is given as w0 =

1 LEAST(a1 , a2 , b) 300

(5.169)

The parameter u is given by u=

π2 Eemax 300bfy γM1

≥ 1,0

(5.170)

The distance emax is taken from the extreme fibre of the stiffener to the centroid of the stiffener. The critical stress for plate between vertical stiffeners σcr,c is given by σcr,c =

π2 Et2 12(1 − ν2 )a2

(5.171)

The critical stress σcr,p is given by kσ,p σE . The value of σE is given by Eq. (5.141). The value for plates with longitudinal stiffeners in Annex A of EN 1993-1-5. For unstiffened plates σcr,p = σcr,c To avoid lateral torsional buckling of the stiffener, fy IT ≥ 5,3 Ip E

(5.172)

where IT and Ip are St. Venant torsional constant for the stiffener alone and Ip is the polar second moment of area about the edge fixed to the plate. Eq. (5.172) can be

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derived as follows. The critical buckling stress σcr for open section stiffeners (with negligible warping stiffness) is given by Eq. (5.102) with Iw = 0, σcr = G

IT Ip

(5.173)

From cl 9.2.1(8) of EN 1993-1-5, the critical stress σcr is limited by σcr ≥ θfy

(5.174)

For plate stiffeners, θ is taken as 2,0, thus Eq. (5.173) becomes fy IT ≥ 5,2 Ip E

(5.175)

The code replaces the co-efficient 5,2 by 5,3.

5.3.5.3 Intermediate Transverse Stiffeners (cl. 9.3.3 EN 1993-1-5) The force Ns,Rd to be resisted by a stiffener is given by Ns,Rd = VEd − χw hw t

f √yw 3

γM1

(5.176)

Note, χw is calculated for the web panel between adjacent stiffeners assuming the stiffener under consideration is removed. In the case of variable shear, then the check is performed at a distance 0,5hw from the edge of the pane with the larger shear force. To determine the buckling resistance of the stiffener a portion of the web may taken into account (Rockey et al., 1981). A section of the web in length equal to 15εt either side of the stiffener may be considered (cl 9.1, EN 1933-1-3) (Fig. 5.23) For a symmetric stiffener, the effective area Ae is given by Aequiv = Ast + 30εt2

(5.177)

and the effective second moment of area Iequiv by Iequiv = Ist +

1 30εt4 12

(5.178)

where Ast is the area of the stiffener and Ist is the second moment of area of the stiffener. For end stiffeners the co-efficient of 30 in Eqs (5.177) and (5.178) should be replaced by 15. The effective length of the stiffener may be taken as 0,75hw and buckling curve ‘c’ used to determine the strength reduction factor (cl 9.4 EN 1993-1-5). In order to provide adequate restraint against buckling it was found that the stiffeners need to possess a minimum second moment of area (Rockey et al., 1981).

146

•

Chapter 5 / Laterally Unrestrained Beams

A

A

Elevation

t

15 εt

15 εt

Section AA

FIGURE 5.23 Stiffener geometry The minimum second moment of area Is is given by √ for a/hw < 2 Ist ≥ 1,5 for a/hw ≥

√

h3w t3 a2

(5.179)

2

Ist ≥ 0,75hw t3

(5.180)

It can be demonstrated that for compression buckling a change from 1 to 2 half sine √ waves occurs at a/hw = 2, and that thereafter the buckling co-efficient is sensibly independent of the aspect ratio of the panel. Thus Eq. (5.180) is determined from √ Eq. (5.179) by substituting a = hw 2 (cl 9.3.3 (3) EN 1993-1-5).

5.3.5.4 Plate Splices (cl 9.2.3 EN 1993-1-5) The splice whether in the web or flanges, should ideally occur at a transverse stiffener. If not then the stiffener should be at a distance no greater than b0 /2 along the thinner plate where b0 is the depth of the web (or the least spacing of longitudinal stiffeners).

5.3.5.5 Longitudinal Welds (Web to Flange) (cl 9.3.5 EN 1993-1-5) The weld between the web and flange(s) should be designed for a shear flow of VEd /hw , provided VEd < χw hw t

f √yw 3

γM1

(5.181)

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If the condition in Eq. (5.181) is not satisfied, the welds should be designed under a √ shear flow of ηt(fyw / 3)/γM1 .

EXAMPLE 5.3 Design of a laterally restrained plate girder Design a plate girder in Grade S355 steel to carry a characteristic variable load of 150 kN/m over a span of 20 m. The compression flange is fully restrained against lateral torsional buckling. For a plate girder span/depth ratios are generally around 8 or 10 to 1. The higher this ratio is the lower the flange size but at the probable expense of a thicker web to overcome buckling. To calculate the imposed bending moment, assume the total weight of the beam is 100 kN. Total loading on the beam = 1,5 × 150 + 1,35 × (100/20) = 232 kN/m MSd = 232 × 202 /8 = 11,6 MNm The critical slenderness ratio for the web can be controlled by flange-induced buckling with k = 0,4 as plastic rotation is not utilized. If the flanges resist the bending moment, Aw = Af (from Eq. (5.119)), thus the critical hw /t ratio is given by Eq. (5.110) hw E ≤k t fyf

Aw 210 × 103 √ = 0,4 1 = 237 Ac 355

From Eq. (5.117) calculate hw hw =

3

λMRd 237 × 11,6 × 109 = = 1978 mm 355 fyd 1,0

The thickness t is given as t=

hw 1978 = = 8,34 λ 237

Use t = 9 mm Maximum ratio flange outstand to flange thickness beyond the weld for a Class 1 section is 9ε where ε = (235/355)1/2 (=0,814). So maximum flange outstand is 9 × tf × (235/355)1/2 = 7,33tf So flange area is 14,66tf2 (ignoring the effect of weld width). Mpl,Rd ≈ Af

fy hw γM0

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•

Chapter 5 / Laterally Unrestrained Beams or Af =

MEd fy hw γ M0

=

11,6 × 109 1978 355 1,0

= 16, 520 mm2

or 14,66tf2 = 16,520 or Inset tf = 33,6 mm Use 35 mm thick plate with a width of 500 mm. Overall depth, h: h = 2 × 35 + 1978 = 2048 mm. Use h = 2000 mm. Check actual hw /t ratio: Actual web slenderness: hw 1930 = = 214 t 9 Web is Class 4. However as the webs do not carry any compression, then the section may be treated dependant upon the classification of the flanges (cl 5.5.2 (12) EN 1993-1-1). Maximum hw /t ratio: hw E Aw 210 × 103 =k = 0,4 t fyf Af 355

1930 × 9 = 236 500 × 35

The actual value is below the allowable, and is therefore satisfactory. Plastic moment of resistance of the flanges, Mpl,Rd : Mpl,Rd = Af

fy 355 (2000 − 35) = 12,2 MNm (h − tf ) = 35 × 500 γM0 1,0

Vpl,Rd = 232 × 10 = 2320 kN (at the support) hw = h − 2tf = 2000 − 2 × 35 = 1930 mm The web will be designed both ways, non-rigid and rigid end post: The determined flange and web contributions take the maximum shear and maximum moment in a web panel, even though they are at opposite ends of the panel. This will be conservative. Determination of χf : To determine χf , the flange width is limited to 15εtf on either side of the web:

235 15εtf = 15 × 35 355

1/2 = 427 mm

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Actual flange width = 0,5(500 − 9) = 245 mm. This is less, therefore use actual width, so Mf,Rd = 12,2 MNm (from above). Both the web and flanges have a yield strength of 355 MPa. Non-rigid end post: First panel from the support: Intermediate stiffener is 1,2 m from the support. Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

MEd = 2320 × 1,2 − 232

1,6 × 500 × 352 × 355 = 1200 0,25 + 9 × 19302 × 355

= 335 mm

1,22 = 2617 kNm 2

Determine the flange contribution factor χf from Eq. (5.155) √ √ bf tf2 fyf 3 MEd 2 500 × 352 × 355 × 3 2617 2 χf = 1− = 1− cthw fyw Mf,Rd 335 × 9 × 1930 × 355 12200 = 0,174 hw 1930 = = 1,608 a 1200 a = 0,622 hw For a/hw < 1,0, kτ is given by Eq. (5.143), or

hw kτ = 4 + 5, 34 a

2 = 4 + 5,34 × 1,6082 = 17,81

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 1,670 ! √ = √ 235 37,4εt kτ 37,4 355 × 9 17,81

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,497 1,670 λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,174 + 0,497 = 0,671

150

•

Chapter 5 / Laterally Unrestrained Beams TABLE 5.5 Shear capacity calculations with non-rigid end post (Example 5.3). Distance from Support (m) 1,2 2,55 4,15 6,25 10 Panel width (m) 1,2 1,35 1,6 2,1 3,75 2617 5162 7630 9969 11600 Moment M Ed (kNm) 2320 2042 1728 1357 870 Shear V Ed (kN) c (mm) 335 377 447 586 1047 0,174 0,133 0,083 0,034 0,006 χf 0,622 0,699 0,829 1,088 1,943 a/hw 17,81 14,91 11,77 8,719 6,400 kτ λw 1,670 1,825 2,054 2,387 2,786 0,497 0,455 0,404 0,378 0,298 χw 0,671 0,588 0,487 0,382 0,304 χv 2389 2093 1735 1361 1081 V Rd (kN)

The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t

f √yw 3

γM1

355 0,671 × 1930 × 9 √3 = = 2389 kN 1000 1,0

The calculations for subsequent panels are summarized in Table 5.5 (together with the first panel) Intermediate stiffeners: Check strength: First stiffener: The axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5 hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(1,2 − 0,5 × 1,930) = 2265 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1200 + 1350 = 2550 mm hw 1930 = = 0,757 a 2550 a = 1,321 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or 2 hw kτ = 5,34 + 4 = 5,34 + 4 × 0,7572 = 7,63 a

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Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,55 ! √ √ = 37,4εt kτ 37,4 × 9 235 7,63 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = 0,325 = 2,55 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 2265 − 0,325 × 1930 × 9 × 10−3

355 √ 3

1,0

= 1108 kN

Second stiffener: Axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(2,55 − 0,5 × 1,930) = 1952 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1350 + 1600 = 2950 mm hw 1930 = = 0,654 a 2950 a = 1,528 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or kτ = 5,34 + 4

hw a

2 = 5,34 + 4 × 0,6542 = 7,05

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,65 ! √ √ = 37,4εt kτ 37,4 × 9 235 7,05 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,313 2,65 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1952 − 0,313 × 1930 × 9 × 10

−3

355 √ 3

1,0

= 838 kN

152

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Chapter 5 / Laterally Unrestrained Beams Third stiffener: Axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(4,15 − 0,5 × 1,930) = 1581 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1600 + 2100 = 3700 mm hw 1930 = = 0,522 a 3700 a = 1,917 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or

hw kτ = 5,34 + 4 a

2 = 5,34 + 4 × 0,5222 = 6,43

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,78 ! √ √ = 37,4εt kτ 37,4 × 9 235 6,43 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,299 2,78 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1581 − 0,299 × 1930 × 9 × 10−3

355 √ 3

1,0

= 517 kN

Fourth stiffener: Axial force NEd is given by Eq. (5.176) NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(6,25 − 0,5 × 1,930) = 1094 kN

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The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 2100 + 3750 = 5850 mm hw 1930 = = 0,330 a 5850 a = 3,03 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or kτ = 5,34 + 4

hw a

2 = 5,34 + 4 × 0,332 = 5,78

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,93 ! √ √ = 37,4εt kτ 37,4 × 9 235 5,78 355

As λw > 1,08, χw is given by Eq. (5.145) as χw =

0,83 0,83 = = 0,283 2,93 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1094 − 0,283 × 1930 × 9 × 10

−3

355 √ 3

1,0

= 86 kN

This value is small, thus there is no need to check the centre stiffener. The minimum stiffness requirement for all but the two panels either side of the central √ stiffener is given by the case a < 2hw , so design on the least value of a: Is = 1,5

h3w t3 19303 × 93 = 1,5 = 5,46 × 106 mm4 2 a 12002

Use 9 mm thick plate, then the total breadth of the stiffener b is given by b=

3

12 × 5,46 × 106 = 194 mm 9

Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as NRd = 2 × 200 × 9

355 × 10−3 = 1278 kN. 1,0

Buckling check: Effective length = 0,75 × 1930 = 1448 mm From Eq. (5.177), the effective area Aequiv is given as Aequiv = Ast + 30εt2 = 200 × 9 + 30 × 9

235 = 2020 mm2 355

154

•

Chapter 5 / Laterally Unrestrained Beams From Eq. (5.178), the effective area Iequiv is given as Iequiv = Ist + 30εt4 =

235 1 1 2003 × 9 + 30 × 94 = 6,013 × 106 mm4 12 12 355

π2 EIequiv π2 × 210 × 6,013 × 106 = = 5944 kN L2 14482 Aequiv fy 2020 × 355 × 10−3 λ= = 0,121 = Ncr 5944

Ncr =

As λ ≤ 0,2 strut buckling need not be checked (cl 6.3.1.2 (4) (EN 1993-1-1)). Thus the minimum stiffener size will be adequate for all the intermediate stiffeners. End stiffener: Try a 500 wide by 15 thick plate: The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 15εt = 15 × 9 (235/355) = 110 mm As,eff = As + Aweb = 500 × 15 + 110 × 9 = 8490 mm2 15 × 5003 110 × 93 Is,eff = Is + Iweb = + = 0,156 × 109 mm4 12 12 Is,eff 0,156 × 109 i= = = 135,6 mm As,eff 8490 Use an effective length of 0,75hw = 0,75 × 1930 = 1448 mm π2 EI 210 × 0,156 × 109 π2 = = 154210 kN l2 14482 Afy 8490 × 355 λ= = = 0,14 Ncr 154210 × 103

Ncr =

From cl 6.3.1.2 (4) (EN 1993-1-1), there is no reduction for strut buckling as λ < 0,2. So, fy 355 × 10−3 = 3014 kN = 8490 γM0 1.0 This exceeds the reaction of 2320 kN. NRd = A

Check cl 9.2.1 (7) Determine Ip : The second moment of area about the web centre line, Iy : Iy =

15 × 5003 bh3 = = 0,156 × 109 mm4 12 12

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The second moment of area normal to the web centre line about one edge, Iy : Iy =

500 × 153 bh3 = = 0,563 × 106 mm4 3 3

Ip = Ix + Iy = 0,156 × 109 + 0,563 × 106 = 0,157 × 109 mm4 bh3 = 0,563 × 106 mm4 3 IT 0,563 × 106 = = 3,59 × 10−3 Ip 0,157 × 109 IT =

Limiting value: 5,3

fy 355 = 5,3 = 8,96 × 10−3 E 210 × 103

The actual value is less than the limiting value thus the stiffener size must be increased. A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT /IP = 0,01) (and will clearly satisfy the strength and buckling criteria). Flange to web welds: √ From Table 5.5 VEd > hw t( fyw / 3)/γM1 as the flange contribution χf has been mobilized. Thus welds should be designed for a shear flow of

ηt

f √yw 3

γM1

= 1,2 × 9

355 √ 3

1,0

= 2214 N/mm

The final layout of the girder whose self-weight is 91,1 kN is given in Fig 5.24(a). Rigid end post: First panel from the support: Intermediate stiffener is 1,45 m from the support. Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

1,6 × 500 × 352 × 355 = 1450 0,25 + 9 × 19302 × 355

= 405 mm MEd = 2320 × 1,45 − 232

1,452 = 3120 kNm 2

Determine the flange contribution factor χf from Eq. (5.155) √ √ bf tf2 fyf 3 MEd 2 500 × 352 × 355 × 3 3120 2 χf = 1− = 1− cthw fyw Mf,Rd 405 × 9 × 1930 × 355 12200 = 0,141

Chapter 5 / Laterally Unrestrained Beams cL

35 500 flanges 9 mm web

2,0

•

End stiffeners 500 25

1,2 2,55 4,15

Intermediate stiffeners 2 200 9

6,25 (a) Non-rigid end post

cL

Intermediate 2 200 9

2,0

156

End post 2 500 15

1,45 0,3

3,25 6,0 (b) Rigid end post

FIGURE 5.24 Final layout for the beam in Example 5.3

1930 hw = = 1,331 a 1450 a = 0,751 hw For a/hw < 1,0, kτ is given by Eq. (5.142), or 2 hw kτ = 4 + 5,34 = 4 + 5,34 × 1,3312 = 13,46 a Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 1,920 ! √ = √ 37,4εt kτ 37,4 235 × 9 13,46 355

As λw > 1,08, χw is given by Eq. (5.152) as χw =

1,37 1,37 = = 0,523 0,7 + 1,92 0,7 + λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,141 + 0,523 = 0,664

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The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t

f √yw 3

γM1

355 0,664 × 1930 × 9 √3 = 2363 kN = 1000 1,0

The resultant spacing of stiffeners and a summary of the remaining calculations is given in Table 5.6 The final result for the calculation of stiffener forces are also given in Table 5.6. The minimum stiffness requirement for all but the two panels either side of the central √ stiffener is given by the case a < 2hw , so design on the least value of a: Is = 1,5

h3w t3 19303 × 93 = 1,5 = 3,74 × 106 mm4 a2 14502

Use 9 mm thick plate, then the total breadth of the stiffener b is given by 6 3 12 × 3,74 × 10 b= = 171 mm 9 Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as NRd = 2 × 200 × 9

355 × 10−3 = 1278 kN. 1,0

This exceeds the values of NEd in the last line of Table 5.6. Rigid end post: Use Eqs (5.163)–(5.166) to determine the maximum panel size in accordance with cl 9.3.1(4) (EN 1993-1-5) Determine χw from Eq. (5.163) √ √ VEd 3γM1 2320 × 103 3 × 1,0 χw = = = 0,652 fyw hw t 355 × 1930 × 9

TABLE 5.6 Shear capacity calculations with rigid end post (Example 5.3). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) Stiffener force N Rd (kN)

1,45 1,45 3120 2320 405 0,141 0,751 13,46 1,920 0,523 0,664 2363 1118

3,25 1,80 6315 1984 503 0,089 0,933 10,14 2,213 0,470 0,559 1991 758

6,0 2,75 9744 1566 768 0,029 1,425 7,310 2,607 0,414 0,443 1578 154

10 4,00 11 600 928 1117 0,005 2,073 6,271 2,814 0,390 0,395 1407 −826

158

•

Chapter 5 / Laterally Unrestrained Beams The required normalized web slenderness λw from Eq. (5.164) is given by λw =

0,83 0,83 = = 1,273 χw 0,652

The buckling parameter kτ is given by Eq. (5.165) as kτ =

hw 37,4tελw

2

⎛ ⎜ =⎝

⎞2 1930 ⎟ ! ⎠ = 30,65 235 37,4 × 9 × 355 × 1,273

As a/hw < 1, kτ is given by Eq. (5.166), or the required panel width a is given as a = hw

5,35 5,35 = 1930 = 865 mm kτ − 4,00 30,65 − 4

Centre to centre distance of the pair of double stiffeners should be greater than 0,1 hw (=193 mm). Use a centre to centre distance of 300 mm (The shear resistance will be greater than 2320 kN). Minimum cross-sectional area of each plate is 4hw t2 4 × 1930 × 92 = = 2084 mm2 e 300 As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is 2084/500 (=4 mm). As for the non-rigid end post, a stiffener 500 mm by 15 mm suffices for resisting the reaction, use the same here. The layout of the beam whose self-weight is 93 kN is given in Fig 5.24 (b). The selfweight per unit length of the beam is 4,56 kN/m for the non-rigid end post and 4,51 for the rigid end post (i.e. almost identical). It would be possible to further optimize the weight of the girder by decreasing the web thickness towards the centre and the flange width and/or thickness towards the ends of the beam. Note, however, that so doing may also increase the stiffener requirements.

EXAMPLE 5.4 Design of a plate girder with lateral torsional buckling. Design a plate girder in Grade S355 steel to carry a characteristic variable load of 1500 kN at the centre of a span of 20 m. Lateral torsional restraints exist at the supports and the load. To calculate the imposed bending moment, assume the total weight of the beam is 150 kN. MEd =

1,35 × 150 × 20 1,5 × 1500 × 20 + = 11760 kNm 8 4

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To size the beam allowing for the fact that its classification will be Class 4 (needing the determination of an effective section) and lateral torsion buckling will occur, increase the moment by 50% and use the optimization equations for Class 3. MEd =

1,5 × 1500 × 20 1,35 × 200 × 20 + = 11925 kNm 4 8

Approximate design: As the moment resistance is calculated on an elastic resistance utilization, the factor k in Eq. (5.110) may be taken as 0,55, with the flange area equal to half the web area for an optimal design under elastic stress distribution (Eq. (5.139)), so hw E Aw 210 × 103 √ λ= =k = 0,55 2 = 460 t fyf Af 355 To determine the optimal solution increase MEd by 50% to 1,5 × 11925 (=17888 kNm). From Eq. (5.135), t is given by 6 3M 3 3 × 17888 × 10 Ed t= 3 2 = = 7,1 mm 2 2λ fyd 2 × 460 × 355 Use 8 mm plate. Determine hw from Eq. (5.136) 3λMEd 3 × 460 × 17888 × 106 hw = = = 3264 mm 2fyd 2 × 355 As the plate thickness has been rounded up from 7,1 to 8 mm, then the web height may be rounded down to 3240 mm. The area of the flange Af is given by Eq. (5.139) Af =

8 × 3240 Aw = = 12960 mm2 2 2

Use the limit of the flange outstand as a Class 1 section (i.e. 7,33tf ) so the area of the flange is 14,66tf2 . Thus 14,66tf2 = 12960, or tf = 29,7 mm Use a flange plate 30 mm thick. bf ≈ 14,66tf = 14,66 × 30 = 440 mm Use a flange plate 500 mm wide (Af = 15000 mm2 ). The reason for keeping the flange classification as low as possible is in order to mobilize as much of the flange capacity as possible to resist shear.

160

•

Chapter 5 / Laterally Unrestrained Beams Actual web slenderness hw 3240 = = 405 t 8 Allowable web slenderness: hw 210 × 103 E Aw = 0,55 =k t fyf Af 355

3240 × 8 = 427 500 × 30

Thus flange-induced buckling will not occur. From the actual web slenderness the web (and therefore the complete section) is Class 4. Determination of effective section properties (cl 4.4 EN 1993-1-5): The process is iterative as the amount of web not considered is a function of the stresses at the top and bottom of the web. Note, the calculation takes compressive stresses positive rather than the normal stress analysis convention of compressive stresses negative. First Iteration: Determine the stresses at the top and bottom of the web with no loss of section: Gross Area, A A = 2bf tf + hw t = 2 × 500 × 30 + 3240 × 8 = 55920 mm2 Gross Iy : Iy =

1 1 [b(hw + 2tf )3 − (b − t)h3w ] = [500 × 33003 − 492 × 32403 ] 12 12

= 102, 88 × 109 mm4 Stress at the top of the web, σ1 : 11 925 × 106 × 1620 = 188 MPa 102,88 × 109 Stress at the top of the web, σ2 : σ1 =

11 925 × 106 × 1620 = −188 MPa 102,88 × 109 Determine the stress ratio, ψ: −188 σ2 = −1,0 = ψ= σ1 188 From Table 4.1 of EN 1993-1-5, the buckling factor kσ = 23,9 for ψ = −1,0. σ2 = −

The normalized slenderness ratio λp is given by λp =

b 3240 = 3,585 ! √ √ = 28,4tε kσ 28,4 × 8 235 23,9 355

where b is the depth of the web.

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The reduction factor ρ for an internal compression member is given by ρ=

λp − 0,055(3 + ψ) (λp

)2

=

3,585 − 0,055(3 + (−1)) = 0,271 3,5852

The effective depth beff is given by beff =

ρhw 0,271 × 3240 = = 439 mm 1−ψ 1 − (−1)

The depth of web left at the top be1 : be1 = 0,4beff = 0,4 × 439 = 176 mm The depth of web left at the bottom (above the centroidal axis), be2 : be1 = 0,6beff = 0,6 × 439 = 263 mm The ineffective portion of web has a length lw = 1620 − 176 − 263 = 1181 mm The net loss of area of the web, Aw is given by Aw = 1181 × 8 = 9448 mm2 Effective area of section, Aeff : Aeff = A − Aw = 59920 − 9448 = 50472 mm2 Position of effective centroid, zeff : l A h2 − Aw zeff,prev + be2 + 2w zeff = Aeff 3300 1181 59920 3300 − 9448 + 263 + 2 2 2 = = 1490 mm 50472 Note, zeff,prev is the neutral axis position at the previous iteration. For the first iteration, zeff,prev = h/2. Effective second moment of area, Iy,eff : 2 tlw3 lw − Aw zeff,prev + be2 + − zeff 12 2 2 3300 8 × 11813 = 102,88 × 109 + 59920 − 1490 − 2 12 2 1181 3300 + 263 + − 1490 − 9448 2 2

Iy,eff = Iy + A

h − zeff 2

2

−

= 102,88 × 109 + 1,53 × 109 − 1,10 × 109 − 9,70 × 109 = 93,61 × 109 mm4

162

•

Chapter 5 / Laterally Unrestrained Beams Stress at the top of the web, σ1 : 11925 × 106 × (3270 − 1490) = 227 MPa 93,61 × 109

σ1 =

Stress at the bottom of the web, σ2 : σ2 = −

11925 × 106 × (1490 − 30) = −186 MPa 93,61 × 109

Determine the stress ratio, ψ: σ2 −186 = = −0,819 σ1 227

ψ=

From Table 4.1 of EN 1993-1-5, the buckling factor kσ for ψ = −0,794 is given by kσ = 7,81 − 6,29ψ + 9,78ψ2 = 7,81 − 6,29(−0,819) + 9,78(−0,819)2 = 19,5 The normalized slenderness ratio λp is given by λp =

b

√

28,4tε kσ

=

3240 = 3,97 ! √ 28,4 × 8 235 19,5 355

where b is the depth of the web. The reduction factor ρ for an internal compression member is given by ρ=

λp − 0,055(3 + ψ) (λp

)2

=

3,97 − 0,055(3 + (−0,819)) = 0,244 3,972

The effective depth beff is given by beff =

ρhw 0,244 × 3240 = = 435 mm 1−ψ 1 − (−0,819)

The depth of web left at the top be1 : be1 = 0,4beff = 0,4 × 435 = 174 mm The depth of web left at the bottom (above the centroidal axis), be2 : be1 = 0,6beff = 0,6 × 435 = 261 mm The ineffective portion of web has a length lw = (3240 − 1490) − 174 − 261 = 1315 mm The net loss of area of the web, Aw is given by Aw = 1315 × 8 = 10520 mm2 Effective area of section, Aeff : Aeff = A − Aw = 59920 − 10520 = 49400 mm2

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Position of effective centroid, zeff : l A h2 − Aw zeff,prev + be2 + 2w zeff = Aeff 1315 59920 3300 − 10520 1490 + 261 + 2 2 = = 1488 mm 49400 Effective second moment of area, Iy,eff :

2 tlw3 lw − Aw zeff,prev + be2 + − zeff 12 2 2 3300 8 × 13153 = 102,88 × 109 + 59920 − 1488 − 2 12 2 1315 − 1488 − 10520 1490 + 261 + 2

Iy,eff = Iy + A

h − zeff 2

2

−

= 102,88 × 109 + 1,57 × 109 − 1,52 × 109 − 8,91 × 109 = 94,02 × 109 mm4 Stress at the top of the web, σ1 : σ1 =

11925 × 106 × (3270 − 1488) = 226 MPa 94,02 × 109

Stress at the bottom of the web, σ2 : σ2 = −

11925 × 106 × (1488 − 30) = −185 MPa 94,02 × 109

As these stresses are virtually identical to those on the first iteration, there is no need to continue. Note, that as the changes in be1 and be2 were small the iteration could have stopped without further calculations of σ1 and σ2 (ψ = −0,819) The lesser elastic section modulus Weff,y is given as Ieff 94,02 × 109 = 64,49 × 106 mm3 = zeff − tf 1488 − 30 fy 355 × 10−6 = 22900 kNm = Weff,y = 64,49 × 106 1,0 γM0

Weff,y = MRd

Lateral torsional buckling check Moment gradient factor, C1 from Eq. (5.40), with ψ = 0: 1 = 0,4ψ + 0,6 = 0,6 C1

164

•

Chapter 5 / Laterally Unrestrained Beams The gross section is used to calculate the section properties required for Mcr . 1 1 [2tf b3f + hw t3 ] = [2 × 0,030 × 0,53 + 3,24 × 0,0083 ] 12 12

Iz =

= 0,625 × 10−3 m4 Iw = Iz

h2s 3,2702 = 0,625 × 10−3 = 1.67 × 10−3 m6 4 4

Note: hs is the distance between the centroids of the flanges. The torsional second moment of area may be calculated on the thin plate assumption. It =

1 1 [2bf tf3 + hw t3 ] = [2 × 0,5 × 0,0303 + 3,24 × 0,0083 ] = 9,55 × 10−6 m4 3 3

Use Eq. (5.5) to determine Mcr 1/2 π2 EIz Iw L2 GIt Mcr = + 2 Iz L2 π EIz L = 10 m. π2 EIz π2 × 210 × 106 × 0,625 × 10−3 = = 12954 kN L2 102 GIt = 81 × 106 × 9,55 × 10−6 = 774 kNm2 GIt π2 EI

z

L2

=

774 = 0,060 m2 12954

Iw 1,67 × 10−3 = = 2,672 m2 Iz 0,625 × 10−3 Mcr = 12954[2,672 + 0,060]1/2 = 21410 kNm From Eq. (5.12), λLT is given by fy Weff,y 355 × 64,49 × 106 × 10−6 λLT = = 0,6 = 0,801 C1 Mcr 21410 h/b > 2, so αLT = 0,76 LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,76(0,801 − 0,2) + 0,8012 ] = 1,049 χLT =

1 1 = = 0,579 2 − 0,8012 )1/2 2 1/2 1,049 + (1,049 − (λLT ) )

LT + (2LT

Mb,Rd = χLT Weff,y

fy 355 = 0,579 × 64,49 × 106 × 10−6 = 13260 kNm γM1 1,0

This exceeds the applied moment of 11925 kN. The beam is therefore satisfactory for flexure.

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Deflection check: Igross = 102,88 × 109 mm4 For a worst case scenario assume that all the variable load contributes to the deflection, δ=

1 WL3 1 1500 × 203 = 0,012 m = 48 EI 48 210 × 106 × 102,88 × 10−3

This is equivalent to span/1667, which is satisfactory. Web design: As the web carries compression due to flexure, then the following interaction equation must be satisfied (cl 7.1(1), EN 1993-1-5), Mf,Rd η1 + 1 − (2η3 − 1)2 ≤ 1,0 (5.182) Mpl,Rd where Mf,Rd is the plastic moment resistance of the flanges and Mpl,Rd is the plastic moment of resistance of the section (both calculations are irrespective of classification of the section). The shear contribution factor η3 is defined as (cl. 5.5 (1), EN 1993-1-5) η3 =

VEd ≤ 1, 0 √ χv hw t[(fyw / 3)/γM1 ]

(5.183)

as there is no axial force the equation for η1 from cl 4.6 (1) (EN 1993-1-5) reduces to Med η1 = f W ≤ 1,0 y eff

(5.184)

γM0

If η3 < 0,5 there is no reduction in moment capacity (cl 7.1 (1), EN 1993-1-5) Mpl,Rd = Wpl

fy 500 × 33002 − 492 × 32402 355 × 10−6 = 24870 kNm = γM0 4 1,0

Mf,Rd = Af hs 1−

Mf,Rd Mpl,Rd

= 1−

fy 355 = 500 × 30 × (3300 − 30) × 10−6 = 17410 kNm γM0 1,0

17410 = 0,3 24870

Flexible end post: Three intermediate stiffeners are required at 2,8, 5,6 and 7,8 m from the support, together with a load bearing stiffener at the centre. Stiffener at 2,8 m: MEd = 1260 × 2,8 − 6,75 × 2,82 = 3475 kNm VEd = 1260 kN (at support)

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Chapter 5 / Laterally Unrestrained Beams Determine c from Eq. (5.156) 1,6bf tf2 fyf 1,6 × 500 × 302 × 355 c = a 0,25 + = 2800 0,25 + = 724 mm th2w fyw 8 × 32402 × 355 Determine the flange contribution factor χf from Eq. (5.155) √ bf tf2 fyf 3 MEd 2 χf = 1− cthw fyw Mf,Rd √ 500 × 302 × 355 × 3 3475 2 = = 0,040 1− 724 × 8 × 3240 × 355 17410 hw 3240 = = 1,157 a 2800 a = 0,864 hw For a/hw < 1,0, kτ is given by Eq. (5.143), or 2 hw kτ = 4 + 5,34 = 4 + 5,34 × 1,1572 = 11,15 a Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 3240 = 3.985 ! √ = √ 235 37.4εt kτ 37.4 355 × 8 11.15

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,208 3,985 λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,040 + 0,208 = 0,248 The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t η3 =

f √yw 3

γM1

355 0,248 × 3240 × 8 √3 = = 1318 kN 1000 1,0

1260 VEd = = 0,956 √ 1318 χv hw t[(fyw / 3)/γM1 ]

As η3 > 0,5 the interaction equation between moment and shear must be considered (Eq. (5.152)) Med 3475 η1 = f W = = 0,152 y eff 22900 γM0

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TABLE 5.7 Shear capacity calculations with non-rigid end post (Example 5.4). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) η1 η3 Interaction equation Stiffener force (kN)

2,8 2,8 3475 1260 724 0,040 0,864 11,15 3,990 0,208 0,248 1318 0,152 0,956 0,401 1079

5,6 2,8 6844 1241 724 0,035 0,864 11,15 3,990 0,208 0,243 1293 0,209 0,960 0,553 210

7,8 2,2 9417 1222 568 0,038 0,679 15,58 3,371 0,246 0,284 1506 0,411 0,811 0,527 −134

10 2,2 11 925 1207 568 0,028 0,679 15,58 3,371 0,246 0,274 1457 0,521 0,829 0,650 −932

Mf,Rd (2η3 − 1)2 = 0,152 + 0,3(2 × 0,956 − 1)2 = 0,402 ≤ 1,0 η1 + 1 − Mpl,Rd The calculations for the remaining stiffeners are carried out in Table 5.7, together with the calculation of the forces on the stiffeners. Intermediate stiffeners: √ √ For all the panels, a ≤ 2hw (=3,24 2 = 4,58 m), thus based on the lesser value of a (=2,2 m) the stiffener requirement, Ist is given by Ist = 1,5

h3w t3 32403 × 83 = 1,5 = 5,4 × 106 mm4 2 a 22002

Use 8 mm plate, so 6 3 12 × 5,4 × 10 b= = 200 mm 8 Use double intermediate stiffeners of 200 × 8 mm plates either side of the web. Load capacity of stiffeners NRd (with no buckling): NRd = Ast

fy 355 × 10−3 = 1136 kN = 2 × 200 × 8 1,0 γM0

This exceeds the maximum stiffener force of 1079 kN. As in earlier examples strut buckling is not critical, it will not be checked in this example. End post: VEd = 1260 kN Use a 500 wide by 10 mm thick plate.

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Chapter 5 / Laterally Unrestrained Beams The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 15εt = 15 × 8 (235/355) = 98 mm From Eq. (5.177) Aequiv is given by Aequiv = Ast + 15εt2 = 500 × 10 + 98 × 8 = 5784 mm2 and Iequiv by 1 10 × 5003 98 × 83 15εt4 = + = 0,104 × 109 mm4 12 12 12 Iequiv 0,104 × 109 i= = = 134,1 mm Aequiv 5784

Iequiv = Ist +

Use an effective length of 0,75hw = 0,75 × 3240 = 2430 mm π2 EI 210 × 0,104 × 109 π2 = = 36500 kN l2 24302 Afy 5784 × 355 λ= = = 0,237 Ncr 36500 × 103

Ncr =

Use buckling curve ‘c’ (α = 0,49) = 0,5[1 + α(λ − 0,2) + (λ)2 ] = 0,5[1 + 0,49(0,237 − 0,2) + 0,2372 ] = 0,537 χ=

+

NRd = χA

!

1 2 − (λ)2

=

0,537 +

1 0,5372 − 0,2372

= 0,981

fy 355 × 10−3 = 2014 kN = 0,981 × 5784 1,0 γM0

This exceeds the reaction of 1260 kN. Check cl 9.2.1 (7) Determine Ip : The second moment of area about the web centre line, Iy : bh3 10 × 5003 = = 0,104 × 109 mm4 12 12 The second moment of area normal to the web centre line about one edge, Iy : Iy =

Iy =

bh3 500 × 103 = = 0,042 × 106 mm4 3 3

Ip = Ix + Iy = 0,104 × 109 + 0,042 × 106 = 0,104 × 109 mm4 bh3 = 0,042 × 106 mm4 3 IT 0,042 × 106 = = 0,4 × 10−3 Ip 0,104 × 109 IT =

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Limiting value: 5,3

fy 355 = 8,96 × 10−3 = 5,3 E 210 × 103

The actual value is less than the limiting value thus the stiffener size must be increased. A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT /IP = 0,01) (and will clearly satisfy the strength and buckling criteria). Central stiffener: NEd = 1,5 × 1500 = 2250 kN Use two 240 wide by 15 mm thick plates. The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 30εt = 30 × 8 (235/355) = 195 mm Aequiv = Ast + 30εt2 = 2 × 240 × 15 + 195 × 8 = 8760 mm2 15 × (240 + 8 + 240)3 195 × 83 Isequiv = Ist + 30εt4 = + = 0,145 × 109 mm4 12 12 Iequiv 0,145 × 109 i= = = 128,7 mm Aequiv 8760 Use an effective length of 0,75 hw = 0,75 × 3240 = 2430 mm π2 EI 210 × 0,145 × 109 π2 = = 50895 kN l2 24302 Afy 8760 × 355 λ= = = 0,247 Ncr 50895 × 103

Ncr =

Use buckling curve ‘c’ (α = 0,49): = 0,5[1 + α(λ − 0,2) + (λ)2 ] = 0,5[1 + 0,49(0,247 − 0,2) + 0,2472 ] = 0,542 χ=

+

NRd = χA

!

1 2 − (λ)2

=

0,542 +

1 0,5422 − 0,2472

fy 355 × 10−3 = 3035 kN = 0,976 × 8760 γM0 1,0

This exceeds the applied load of 2250 kN.

= 0,976

•

Chapter 5 / Laterally Unrestrained Beams

cL

500 30 flanges 8 mm web 3,3

170

Intermediate 2 200 8 End 500 25 Central 2 240 15

2,8 5,6 7,8 (a) Non-rigid end post

cL

End post 500 25 Central 2 240 15

0,4

Intermediate 2 150 8

5,0

(b) Rigid end post

FIGURE 5.25 Final layout for the girder of Example 5.4

Flange to web welds: √ From Table 5.7, VEd > hw t[(fyw / 3)/γM1 ] as the flange contribution χf has been mobilized. Thus welds should be designed for a shear flow of ηt

f √yw 3

γM1

= 1.2 × 8

355 √ 3

1.0

= 1968 N/mm

The final layout is given in Fig 5.25 (a). The self-weight of the beam is 101 kN (or 5,05 kN/m). (b) Rigid end post First panel from the support: Intermediate stiffener is 5 m from the support. MEd = 1260 × 5 − 6,75 × 52 = 6131 kNm VEd = 1260 kN (at the support)

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Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

1,6 × 500 × 302 × 355 = 5000 0,25 + 8 × 32402 × 355

Determine the flange contribution factor χf from Eq. (5.155) √ bf tf2 fyf 3 MEd 2 χf = 1− cthw fyw Mf,Rd √ 500 × 302 × 355 × 3 6131 2 = = 0,020 1− 1293 × 8 × 3240 × 355 17410 hw 3240 = = 0,648 a 5000 a = 1,543 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or

hw kτ = 5,34 + 4 a

2 = 5,34 + 4 × 0,6482 = 7,02

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 3240 = 5,02 ! √ = √ 37,4εt kτ 37,4 235 × 8 7,02 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

1,37 1,37 = = 0,240 0,7 + 5,02 0,7 + λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,020 + 0,240 = 0,260 The shear capacity Vb,Rd is determined from Eq. (5.153) f √yw 3

355

0,260 × 3240 × 8 √3 = 1380 kN γM1 1000 1,0 VEd 1260 η3 = = 0,913 = √ 1380 χv hw t[(fyw / 3)/γM1 ]

Vb,Rd = χv hw t

=

= 1293 mm

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•

Chapter 5 / Laterally Unrestrained Beams As η3 > 0,5 an interaction equation between moment and shear must be considered, Mf,Rd η1 + 1 − (2η3 − 1)2 ≤ 1,0 Mpl,Rd Med 6131 η1 = f W = = 0,268 y eff 22900 γM0

Mf,Rd η1 + 1 − (2η3 − 1)2 = 0,268 + 0,3(2 × 0,913 − 1)2 = 0,473 ≤ 1,0 Mpl,Rd The resultant spacing of stiffeners and a summary of the remaining calculations is given in Table 5.8. The final result for the calculation of stiffener forces are also given in Table 5.8. Note, it is possible to eliminate the intermediate stiffener by increasing the web thickness to 9 mm. This may be more economic when considering the overall materials and fabrication costs. The minimum stiffness requirement for both panels either side of the central stiffener √ is given by the case a > 2hw , so design on the least value of a: Ist = 0,75hw t2 = 0,75 × 3240 × 92 = 0,197 × 106 mm4 Use 8 mm thick plate, then the total breadth of the stiffener b is given by 6 3 12 × 0,197 × 10 b= = 67 mm 8 Minimum area to carry a stiffener force of 757 kN is given as A=

757 × 103 = 2132 mm2 355

TABLE 5.8 Shear capacity calculations with rigid end post (Example 5,4). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) η1 η3 Interaction equation Stiffener force N Rd (kN)

5 5 6131 1260 1293 0,020 1,543 7,02 5,02 0,239 0,259 1380 0,268 0,913 0,473 757

10 5 11 925 1226 1293 0,012 1,543 7,02 5,02 0,239 0,261 1337 0,521 0,917 0,729 −419

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This would require a total width of 2132/8 = 267 mm. Use two plates 150 mm by 8 mm as intermediate stiffeners. Design of rigid end post. Use Eqs (5.163) to (5.166) to determine the maximum panel size in accordance with cl 9.3.1(4) (EN 1993-1-5) Determine χw from Eq. (5.163) χw =

√ √ VEd 3γM1 1260 × 103 3 × 1.0 = 0,237 = fyw hw t 355 × 3240 × 8

The required normalized web slenderness λw from Eq. (5.164) is given by λw =

0,83 0,83 = = 3,502 χw 0,237

The buckling parameter kτ is given by Eq. (5.165) as kτ =

hw 37,4tελw

2

⎛ ⎜ =⎝

⎞2 3240 ⎟ ! ⎠ = 11,44 37,4 × 8 × 235 × 3,502 355

As a/hw < 1, kτ is given by Eq. (5.143), or the required panel width a is given as a = hw

5,35 5,35 = 3240 = 2750 mm kτ − 4,00 11,44 − 4

Centre to centre distance of the pair of double stiffeners should be greater than 0.1 hw (=324 mm). Use a centre to centre distance of 400 mm (the shear resistance will be greater than 1260 kN). Minimum cross-sectional area of each plate is 4hw t2 4 × 3240 × 82 = = 2074 mm2 e 400 As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is 2074/500 (=4 mm). As for the non-rigid end post, a stiffener 500 mm by 25 mm is needed. Stiffener under the point load: The calculations are the same as those for the non-rigid end post. The final layout is given in Fig 5.25 (b). The self-weight is 105 kN (5,06 kN/m)

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REFERENCES Anderson, J.M. and Trahair, N.S. (1972). Stability of mono-symmetric beams and cantilevers, Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, 98, 269–286. Basler, K. (1961). Strength of plate girders in shear, Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, 87, 151–180. Bradford, M.A. (1989). Buckling of beams supported on seats, Structural Engineer, 69(23), 411–414. Bulson, P.S. (1970). The stability of flat plates. Chatto and Windus. Chapman, J.C. and Buhagiar, D. (1993). Application of Young’s buckling equation to design against torsional buckling. Proceedings of the Institution of Civil Engineers, Structures and Buildings, 99, 359–369. Davies, A.W. and Griffiths, D.S.C. (1999). Shear strength of steel plate girders, Proceedings of the Institution of Civil Engineers, 134, 147–157. Gibbons, C. (1995). Economic steelwork design, Structural Engineer, 73(15), 250–253. Kaim, P. (2006). Buckling of members with rectangular hollow sections, In Tubular structures XI (eds Packer and Willibald). Taylor and Francis Group, 443–449. Kirby, P.A. and Nethercot, D.A. (1979). Design for structural stability. Granada. Mutton, B.R. and Trahair, N.S. (1973). Stiffness requirements for lateral bracing, Journal of the Structural Division, ASCE, 99, 2167. Nethercot, D.A. (1974a). Residual stresses and their influence upon the lateral buckling of rolled steel beams, Structural Engineer, 52(3), 89–96. Nethercot, D.A. (1974b). Buckling of welded beams and girders, IABSE, 34, 163–182. Nethercot, D.A.and Lawson, R.M. (1992). Lateral stability of steel beams and columns – common cases of restraint, Publication 093. Steel Construction Institute. Nethercot, D.A. and Rockey, K.C. (1971). A unified approach to the elastic critical buckling of beams, Structural Engineer, 49(7), 321–330. Pillinger, A.H. (1988). Structural steelwork: a flexible approach to the design of joints in simple construction, Structural Engineer, 66(19), 316–321. Porter, D.M., Rockey, K.C. and Evans, H.R. (1975). The collapse of plate girders loaded in shear, Structural Engineer, 53(8), 313–325. Rees, D.W.A. (1990). Mechanics of solids and structures. McGraw Hill. Rockey, K.C. and Škaloud, M. (1971). The ultimate load behaviour of plate girders loaded in shear, IABSE Colloquium – design of plate girders for ultimate strength. London, 1–19. Rockey, K.C., Evans, H.R. and Porter, D.M. (1978). A design method for predicting the collapse behaviour of plate girders, Proceedings of The Institution of Civil Engineers, 65, 85–112. Rockey, K.C., Valtinat, G. and Tang, K.H. (1981). The design of transverse stiffeners on webs loaded in shear – an ultimate load approach, Proceedings of The Institution of Civil Engineers, 71(2), 1069–1099. Rondal, J., Würker, K-G., Dutta, D., Wardenier, J. and Yeomans, N. (1992). Structural stability of hollow sections. Verlag TÜV Rheinland. Timoshenko, S.T. and Gere, J.M. (1961). Theory of elastic stability. McGraw Hill. Trahair, N.S. (1983). Lateral buckling of overhanging beams, In Instability and plastic collapse of steel structures (ed L.J. Morris). Granada, 503–518. Trahair, N.S. and Bradford, M.A. (1988). The behaviour and design of steel structures (2nd Edition). Chapman and Hall. BS 5950-1: Structural use of steelwork in building – Part 1: Code of practice for design – Rolled and welded sections. BSI. EN 1993-1-1. Eurocode 3: Design of steel structures, Part 1-1: General rules and rules for buildings. CEN/BSI. EN 1993-1-5. Eurocode 3: Design of steel structures, Part 1-5: Plated structural elements. CEN/BSI.

Chapter

6 / Axially Loaded Members

6.1 AXIALLY LOADED TENSION MEMBERS A member subject to axial tension extends and tends to remain straight or, if there is a small initial curvature, to straighten out as the axial load is increased. Tension members (ties) occur in trusses, bracing and hangers for floor beams. A flat can be used as a tie, but this is generally impractical because it buckles if it goes into compression. Tie sections are therefore angles and tees for small loads and ‘I’ sections for larger loads. In situations where the load is not applied axially then the member is designed to resist an axial force plus a bending moment. A tension member extends when subject to an axial load and is deemed to have failed when the yield or ultimate stress is reached. The failure load is independent of the length of the member which is in contrast to an axially loaded compression member which fails by buckling. A member which is purely in tension does not buckle locally or overall and is therefore not affected by the classification of sections. The characteristic stress is not reduced, for design purposes, except by the material factor.

6.1.1 Angles as Tension Members (cl 4.13, EN 1993-1-8 (2005)) Generally angles are connected by one leg at the end of the member and this introduces an eccentric load. Where an angle is connected to one side of a gusset plate (as in a truss) bending moments are introduced in addition to the direct axial force. For a tension member these moments produce lateral deflections which reduce the eccentricity of the load near the middle of the member. Thus under increasing load the bending stresses become concentrated more towards the ends of the member. For angles connected by one leg the principal sectional axes are inclined to the plane containing the bending moment. Secondary deflections therefore occur normal to the plane of bending and, because of the restraints provided by the gusset plates, twisting also takes place. Generally eccentrically loaded members are designed to resist an axial load and bending moment. However angle and tee experiments (Nelson (1953); Regan and Salter

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•

Chapter 6 / Axially Loaded Members (1984)) demonstrated that the above effects could be compensated for in design by reducing the cross-sectional area of the member. If there are holes then these also reduce the area of the cross-section. Angles may be treated as axially loaded members provided that the net area is reduced to the effective area (cl 4.13, EN 1993-1-8 (2005)). For an equal angle, or an unequal angle connected by the larger leg the effective area is the gross area. For an unequal angle connected by the smaller leg the effective area is twice that of the smaller leg.

6.1.2 Design Value of a Tension Member (cl 6.2.3, EN 1993-1-1 (2005)) The design value of the tensile force NEd at each cross-section should satisfy (Eq. (6.5), EN 1993-1-1 (2005)) NEd ≤1 Nt,Rd

(6.1)

where Nt,Rd is the design tension resistance taken as the smaller of: (a) The design plastic resistance of the gross cross-section (Eq. (6.6), EN 1993-1-1 (2005)) Npl,Rd =

Afy γM0

(6.2a)

(b) The design ultimate resistance of the net cross-section at holes for fasteners (Eq. (6.7), EN 1993-1-1 (2005)) Nu,Rd = 0,9

Anet fu γM2

(6.2b)

(c) The design ultimate resistance of the net cross-section at holes for fasteners which are preloaded or non-preloaded (Eq. (6.8), EN 1993-1-1 (2005) and cl 3.4.2.1(1), EN 1993-1-8 (2005)) Nnet,Rd =

Anet fy γM0

(6.2c)

EXAMPLE 6.1 An angle in tension connected by one leg. A 100 × 65 × 6 mm single angle tie is connected through the smaller leg by two 20 mm diameter bolts in line with a pitch of 2,5 d0 . Determine the design ultimate resistance of the angle assuming S275 steel and material factors of γM2 = 1,25 and γM0 = 1. Net area of angle connected by the smaller leg and allowing for holes Anet = (b − do )t + bt = (65 − 22) × 6 + 65 × 6 = 648 mm2 Compare this value with the gross area = 1120 mm2 (Section Tables).

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Design ultimate tensile resistance of the net cross-section for a two bolt angle connection (Eq. (6.7), EN 1993-1-1 (2005)) Nu,Rd = 0,9

Anet fu 0,9 × 648 × 430 = = 200,6 kN. γM2 1,25 × 1E3

or Npl,Rd =

Afy (2 × 65 × 6) × 275 = = 214,5 > 200,6 kN. γM0 1,0 × 1E3

A check must be made for the strength of the bolts as shown in Chapter 7.

6.2 COMBINED BENDING AND AXIAL FORCE – EXCLUDING BUCKLING (cl 6.2.9, EN 1993-1-1 (2005)) Combining an axial force (tension or compression) and bending moment induces stresses which vary across a ductile steel section. At a certain point the stresses may combine to produce a yield stress but this does not produce collapse of the member because collapse only occurs if the entire section is at yield stress (i.e. the section is plastic). If the axial force is compressive it is assumed that buckling does not occur.

6.2.1 Rectangular Sections (cl 6.2.9.1(3), EN 1993-1-1 (2005)) A rectangular section subject to an axial force and bending moment produces a stress diagram as shown in Fig. 6.1. The neutral axis is displaced from the equal area position and the stress diagram can be represented in two parts, one for the axial load and one for the reduced plastic modulus. Plastic moment of resistance about y–y axis My = fy bh1 (h − h1 )

(6.3)

Axial load N = fy bh(h − 2h1 )

(6.4) fy

fy

fy

h1

b

y

h

y

fy

fy

FIGURE 6.1 Effect of axial force on the plastic moment of resistance of a rectangular section

Chapter 6 / Axially Loaded Members bf

fy

fy

y

y

N

2c

tf

•

h

178

A

tw

(a) Neutral axis (NA) in web

y

y N

A

(b) Neutral axis in flange

FIGURE 6.2 Effect of axial force on the plastic moment of resistance of an ‘I’ section Combining Eqs (6.3) and (6.4) and eliminating h1 My /(bh2 fy /4) + [N/(bhfy )]2 = 1

(6.5a)

This relationship is plotted on Fig. 6.3. The term (bh2 fy /4) is the plastic moment of resistance for a rectangular section. This form of the equation is given in Eq. (6.32), EN 1993-1-1 (2005) and is applicable to rectangular solid sections without holes. NEd 2 MN,Rd = Mpl,Rd 1 − (6.5b) Npl,Rd

6.2.2 ‘I’ Sections The relationship between bending moment and axial force for an ‘I’ section is more complicated. An ‘I’ section subject to an axial load and bending moment produces a stress diagram as shown in Fig. 6.2. The neutral axis is displaced from the equal area position. This stress diagram can be represented in two parts, one for the axial load and one for the reduced plastic modulus. A convenient ratio to determine the reduced plastic section modulus is σ n= fy where σ is the mean axial stress fy is the specified minimum yield strength of steel. If the axial force is small the neutral axis is in the web. Alternatively if the axial force is large the neutral axis is in the flange as shown in Fig. 6.2. For bending about the y–y axis, the neutral axis moves from the web into the flange when n > nc where h − 2tf nc = tw A A is the total area of the section.

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Subtracting the plastic modulus of the hatched area from the whole section. Wyr = Wy − W(shaded area) W(shaded area) =

tw (2c)2 4

From equilibrium σA = (2c)tw fy and since by definition n = σ/fy When n ≤ nc Wyr = Wy − an2 When n > nc Wyr = b(1 − n)(c + n) where a=

A2 4tw

A2 4bf 2bf h c= −1 A

b=

For bending about the y–y axis the change point for n is nc =

tw h A

and the values of A2 4h A2 b= 8tf 4tf bf c= −1 A

a=

Values of a, b, c and nc for ‘I’ sections are given in Section Tables. The plastic section modulus determined in this way is applicable in tension and for short columns. For longer columns instability effects due to deflection of the column reduce this value. The relationship between N/Np and M/Mp for a typical ‘I’ section is plotted in Fig. 6.3.

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•

Chapter 6 / Axially Loaded Members

N/N p

1.0

0

M/Mp

1.0

FIGURE 6.3 Relationship between N/Np and M/Mp at collapse

The above theory for ‘I’ and ‘H’ sections is simplified in the European Code by the following recommendations (cl 6.2.9.1(4), EN 1993-1-1 (2005)). (a) For Class 1 and Class 2 cross-sections the plastic moment of resistance is not reduced if axial loads are limited. Bending about the y–y axis (Eqs (6.33) and (6.44), EN 1993-1-1 (2005)) NEd ≤ 0,25Npl,Rd NEd ≤ 0,5

h w t w fy γM0

Bending about the z–z axis NEd ≤

hw tw fy γM0

(b) For Class 3 and Class 4 cross-sections the maximum longitudinal stress is limited to fy /γM0 . In the case of Class 4 sections the effective area of the section is used.

6.3 BUCKLING OF AXIALLY LOADED COMPRESSION MEMBERS Compression members are present in many structures, for example, trusses, bracing and columns. They are generally greater in cross-sectional area than tension members because they may fail in buckling (Fig. 6.4). If buckling is likely to occur then sections must be capable of resisting bending moments.

6.3.1 Compression Members (cl 6.3.1, EN 1993-1-1 (2005)) An efficient cross-section for a strut is a hot finished tube because residual stresses are a minimum and the buckling resistance is the same for all axes of bending. However the use of a tube is not always practical because of the difficulties of making connections.

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N

L /2

L /2

NE

x y

NE (a) Euler strut

L /2

a L /2

•

O

Δ0

y

y0

x

N (b) Practical strut

FIGURE 6.4 Buckling behaviour based on Euler strut and practical strut

Connections to Universal Column ‘I’ sections are simpler but the section is not as efficient because of the weaker z–z axis of bending. Where loads are relatively small angles and ‘T’ sections are used as struts (e.g. in roof trusses). The strength of a compression member is not reduced significantly by welding or holes with fasteners but they must be arranged sensibly. Short steel compression members fail by squashing at the yield stress, while long, or more accurately slender members, fail by buckling. Buckling may occur at an axial stress which is less than the yield stress and is related to slenderness ratio, lack of straightness and non-axial loads. As buckling progresses the load becomes progressively more eccentric to the longitudinal axis of the member and a bending moment is introduced as shown in Fig. 6.4(a).

6.3.2 Buckling Theory The Euler theory was the first attempt to produce a rational explanation of buckling behaviour of a strut. It was based on the differential equation, related to Fig. 6.4(a), which shows the final deflected form of a pin-ended strut EI

d2 y = M = NE (a − y) dx2

(6.6)

The solution of Eq. (6.6) shows that an axially loaded pin-ended strut becomes elastically unstable and buckles at the Euler critical stress fE =

π2 E λ2

where E Young’s elastic modulus

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•

Chapter 6 / Axially Loaded Members L length of a pin-ended strut i radius of gyration λ slenderness ratio = L/i The Euler buckling theory makes no allowance for: (a) (b) (c) (d) (e) (f) (g) (h) (i)

homogeneity of column material, isotropy of column material, variation of E value and elastic–plastic behaviour of column material, loading not axial, residual stresses, lack of straightness of a column, cross-section of a column not rectangular, local buckling, alternative end conditions to a column.

These factors are present in practice and reduce the Euler buckling load. The Euler differential equation can be modified in a number of ways (Bleich (1952)) to take account of these factors. One method produces the Perry–Robertson formula which is used in the European Code and is related to the Euler buckling stress. Small unavoidable eccentricities of loading and lack of initial straightness can be simulated mathematically by assuming an initial curvature which produces a small central deflection 0 (Fig. 6.4(b)). When a load N is applied the deflection at x is increased by y and the differential equation of bending similar to Eq. (6.6) is EI

d2 y = M = −N[ y + y0 ] dx2

Adopting a sinusoidal function for the initial curvature y0 = 0 cos (πx/L) and putting μ2 = N/EI then πx # " d2 y 2 + μ cos y + =0 0 L dx2 The solution to this equation is y = A sin μx + B cos μx +

μ2 0 cos (πx/L) π2 /L2 − μ2

when x = ±L/2, y = 0 and hence A = B = 0 and y=

μ2 0 cos (πx/L) π2/L2 − μ2

If the Euler buckling load NE = π2 EI/L2 , then y=

N0 cos (πx/L) NE − N

(6.7)

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If N/A = f and NE /A = fE then the deflection y=

πx f 0 cos fE − f L

and the total deflection at any point πx fE + 1 0 cos y0 + y = L fE − f The maximum deflection at x = 0 fE ymax = 0 fE − f and the maximum bending moment is fE Mmax = 0 N fE − f If dexf is the distance of the extreme fibre from the neutral axis then the maximum compressive stress 0 dexf N f fexf = +f I fE − f If (0 dexf N/I) = (0 dexf Af/I) = (0 dexf /i2 )f = ηf where i is the radius of gyration of the column section then η fE +1 fexf = f fE − f Assuming that the critical buckling load is reached when yielding commences in the extreme fibres of the strut, that is, when fexf = fy and f = fPR then rearranging the Perry–Robertson buckling stress fPR = 0,5[ fy + (η + 1) fE ] − {0,52 [ fy + (η + 1) fE ]2 − fy fE }1/2

(6.8)

where the Euler critical buckling stress fE = π2 E/λ2 . Equation (6.8) is known as the Perry–Robertson formula and its adoption is explained by Dwight (1975). The value of the function η has varied over the years and the value originally obtained experimentally by Robertson in 1925 related to the slenderness ratio (λ) for circular sections was η = 0,003λ

(6.9)

The value suggested later by Godfrey (1962) to give more economical designs, based on experimental work by Duthiel in France, was η = 0,3

λ 100

2 (6.10)

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•

Chapter 6 / Axially Loaded Members Equation (6.8) has been rearranged in the European Code (Eq. (6.49), EN 1993-1-1 (2005)) to express the buckling stress ( fPR ) in terms of the stress ratio ( fE /fy = λ) and a reduction factor (χ) related to column imperfections. If ζ = 0,5[ fy + (η + 1) fE ] then fPR = ς − (ς 2 − fy fE )1/2 = = ζ fE

+

0

ζ fE

fy 2

−

f y fE

[ς + ς 2 − fy fE

1/2

]

= χ fy f 11/2 y

(6.11)

fE

In the European Code ζ/fE = , fy /fE = λ and the reduction factor (χ) is then expressed in terms of λ =

[fy /fE + (η + 1)] ζ = fE 2

η = 0,001a (λ − λ0 ) but not less than zero, where a varies from 2 to 8 depending on the shape of the section and the limiting slenderness ratio π2 E λ0 = 0,2 >0 fy Combining these equations (cl 6.3.1.2, EN 1993-1-1 (2005)) 2

= 0,5[1 + α(λ − 0,2) + λ ] where α = 0,001a (π2 E/fy )1/2 is an imperfection factor. For uniform members the buckling stress reduction factor (Eq. (6.49), EN 1993-1-1 (2005)) 2

χ = 1/[ + (2 − λ )1/2 ] ≤ 1

(6.12)

The buckling curves are related to the shape of the sections, axis of buckling, and thickness of material as shown in Tables 6.1 and 6.2, and Fig. 6.4, EN 1993-1-1 (2005). The curves are based on experimental results for real columns as described in the ECCS (1976) report, and expressed theoretically by Beer and Schultz (1970). As the lowest value of the central deflection () is related to the initial curvature, bending moments are generated as soon as the axial load is applied and the buckling process starts immediately. Therefore there is no condition of elastic instability as defined by Euler (1759) and the average compressive stress may never reach the Euler critical stress for a strut of finite length. Nevertheless the failure of a strut is sudden when compared with the ductile failure of a tension member. For a fuller development of the buckling theory see Trahair and Bradford (1988).

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6.3.3 Local Buckling (cl 5.5.2(2) and Table 5.2, EN 1993-1-1 (2005)) Slender elements of a section (e.g. flanges), which are primarily in compression, may buckle locally before overall buckling of the member occurs. This is likely to occur with Class 4 sections for high values of the width (c) to thickness (t) ratios of elements of a cross-section. To allow for local buckling the European Code reduces the crosssectional area to an effective area and consequently the member supports less load.

6.3.4 Column Cross-section Studies (Bleich, 1952) over the year have shown that the strength of a column is influenced by the cross-section. Initial theoretical investigations showed that material concentrated at the centre of gravity was more effective in resisting buckling. Later research identified other factors (Table 6.2, EN 1993-1-1 (2005)) which affect buckling, for example, hot finished or cold formed, yield strength, buckling axis, welding and shape of cross-section. For design purposes these factors are related to an appropriate buckling curve (Fig. 6.4, EN 1993-1-1 (2005)).

6.3.5 Buckling Length of a Column The buckling theory, developed previously, is based on the assumption that the ends of the column are pinned, that is, frictionless joints which prevent linear movement but which can rotate freely about any axis of the section. Pin-ended columns are rare in practice but the pin-ended condition is a useful theoretical concept which can be shown to relate to other end conditions. Alternative end support conditions can be simulated by replacing the actual length of the column by an effective length. Consider the theoretical end conditions shown in Fig. 6.5 which vary from complete end fixity to complete freedom at the end of a cantilever. The theoretical effective length (l) is expressed as a proportion of the actual length (L) and is the distance between real or theoretical pins (i.e. points of contraflexure). The effective length is important in design calculations because the Euler buckling load is inversely proportional to the square of the effective length. In practice the full rigidity of a fixed built-in end is never achieved and some rotation occurs due to the flexibility of the connection or the support. Only a small rotation is necessary to transform a built-in end to a pinned end and thus reduce the buckling resistance of a column. In comparison small translational movements at supports are not so critical and may be limited by supporting members. Practical end conditions therefore allow for some rotation and translation at the ends of a real column. For the column in Fig. 6.6 the theoretical effective length is 0,7 L but in practice the built-in end can rotate and the real effective length is 0,85 L, which

l = 2L

l=L

l = 0,5 L

l = 0,7 L

Chapter 6 / Axially Loaded Members

Rotation at support

L

0,85 L

FIGURE 6.5 Theoretical effective lengths

0,7 L

•

L

186

FIGURE 6.6 Rotation of the end of a strut

reduces the axial load at failure. Because perfect rigidity and completely free ends do not occur in practice the range of common practical values is between 0,7 L and 1,5 L. Methods exist (Williams and Sharp, 1990; Wood, 1974) for determining the effective lengths for columns in rigid frames which are based on the relative total column stiffness at a joint to the total stiffness of all members at the joint. For situations where sway of the column does not occur, for example, where cross bracing is present in simple construction, the effective length is less than the real length (0,5 L to 1,0 L). Where sway occurs, for example, an unpropped portal frame, the effective length is greater than the real length (1,0 L to 2,0 L). For single angles used as struts eccentricity of axial load may generally be ignored. As a rough guide the effective length is approximately 0,85 L for two fasteners (or weld)

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and 1,0 L for one fastener. Buckling failure about the minor axis must be considered for which a single equal angle, is at 45◦ to the major axis of bending. An alternative method of determining a modified slenderness ratio for angles is given in cl BB 1.2, EN 1993-1-1 (2005). For double angle struts the effective lengths are similar to those for a single angle but failure about the 45◦ axis may not be possible because of restraints.

6.3.6 Maximum Slenderness Ratios At high values of the slenderness ratio struts become so flexible that deflections under their own weight are sufficient to introduce stresses in excess of the Perry–Robertson buckling formula. The following empirical limits have been used in the past. The slenderness ratio (λ) should not generally exceed the following: (a) for members resisting loads other than wind loads (b) for members resisting self-weight and wind loads only (c) for any member normally acting as a tie but subject to reversal of stress resulting from the action of wind

180 250 350

Members whose slenderness exceeds 180 should be checked for self-weight deflection. If this exceeds (length/1000) the effect of the bending should be taken into account in design.

6.3.7 Intermediate Restraints A member that provides an intermediate restraint and prevents buckling of a strut, reduces the effective length and increases the strength of the strut. The restraint need not be rigid and may be elastic provided that its stiffness exceeds a certain value (Trahair and Bradford, 1988). Often restraining members associated with built-up members are required to resist not less than 1% of the axial force in the restrained member.

6.3.8 Combined Bending, Shear and Axial Force (cl 6.2.10, EN 1993-1-1 (2005)) Th effect of a shear force on the Euler buckling load for solid sections is small (Bleich, 1952). However as explained in Chapter 4 the plastic moment of resistance is reduced when the design shear force is greater than 50% of the plastic shear resistance. Also for large values of shear shear buckling may reduce the resistance of the section (cl 5, EN 1993-1-5 (2005)).

6.3.9 Design Buckling Resistance (cl 6.3.1, EN 1993-1-1 (2005)) Equation (6.8) expresses failure as a buckling stress but the European Code expresses failure as a design load where local buckling and the buckling stress factor (χ) reduce the buckling load.

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•

Chapter 6 / Axially Loaded Members The design buckling resistance of a compression member for Classes 1, 2 and 3 crosssections Nb,Rd = χA

fy γM1

(6.13a)

and for Class 4 sections Nb,Rd = χAeff

fy γM1

(6.13b)

where χ is the reduction factor for the relevant buckling mode which is generally ‘flexural buckling’. In other cases ‘torsional’ or ‘flexural–torsional’ modes may govern. 2

χ = 1[ + (2 − λ )1/2 ] ≤ 1 2

= 0,5[1 + α(λ − 0,2) + λ ] fy 1/2 (Lcr /i) λ= A = Ncr λ1 1/2 E λ1 = π = 93,9ε fy 235 1/2 ε= fy The imperfection factor (α) corresponding to the appropriate buckling curve is obtained from Table 6.1, EN 1993-1-1 (2005). The value of the cross sectional area is A for Classes 1, 2 and 3 sections with no reductions for local buckling provided that the maximum width-to-thickness ratios are within the limits of Table 5.2, EN 1993-1-1 (2005). For Class 4 sections the effective area, reduced because of local buckling, is calculated from cl 4, EN 1993-1-5 (2003).

EXAMPLE 6.2 Angle strut in a roof truss. A steel roof truss is composed of angles and tee sections and has been analysed assuming pin joints. A particular member is subject to the following axial forces. Permanent action (FG = −7,2 kN), variable snow load (Fs = −10,7 kN), variable wind downwards (Fw = −2,0 kN) and variable wind upwards (Fw = +19,1 kN). If an angle, welded at the ends, is chosen to resist the loads, determine the size if the actual length is 2,1 m. Design load cases: (a) no wind 1,35FG + 1,5 Fs = 1,35(−7,2) + 1,5(−10,7) = −25,77 kN (compression)

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(b) wind down 1,35 FG + 1,5(Fs + Fw ) = 1,35(−7,2) + 1,5(−10,7 − 2) = −28,77 kN (compression) (c) wind up 1,0 FG + 1,5 Fw = −7,2 + 1,5(+19,1) = +21,45 kN (tension) Design value Nb,Ed = 28,77 kN assuming uniform compressive stress across the section. Try a 65 × 50 × 6 mm angle grade S275 steel, long leg attached (A = 659 mm2 , iv = 10,7 mm from Section Tables). Design as an axially loaded Class 3 compression member, ignoring eccentric loads (cl BB 1.2, EN 1993-1-1 (2005)). The effective area need not be reduced because of end holes (cl 6.3.1.1(4), EN 1993-1-1 (2005)). Check maximum width-to-thickness ratios (Table 5.2, EN 1993-1-1 2005) h 65 235 1/2 = = 10,8 < 15 × = 13,9 satisfactory t 6 275 b+h 50 + 65 235 1/2 = = 9,58 < 11,5 × = 10,6 2t 2×6 275 satisfactory, no reduction of area for local buckling. Non-dimensional slenderness about the v–v axis (Eq. (6.50), EN 1993-1-1 (2005)) Afy 1/2 (Lcr /iv ) (Lcr /iv ) = = λv = Ncr λ1 93,9ε =

2100/10,7 = 2,26 [93,9 × (235/275)1/2 ]

Largest effective slenderness ratio is about the v–v axis for an angle (cl BB 1.2, EN 1993-1-1 (2005)) λeff,v = 0,35 + 0,7λv = 0,35 + 0,7 × 2,26 = 1,93 For λeff,v = 1,93 and curve b (Table 6.2, EN 1993-1-1 (2005)) the buckling reduction factor (Fig. 6.4, EN 1993-1-1 (2005)) χ = 0,23 Or by calculation (Eq. (6.49), EN 1993-1-1 (2005)) the buckling reduction factor χ=

[ + (2

1 1 = 2 1/2 [2,66 + (2,662 − 1,932 )1/2 ] − λeff,v ) ]

= 0,223 < (graph value 0,23)

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•

Chapter 6 / Axially Loaded Members where = 0,5[1 + α(λeff,v − 0,2) + λeff,v 2 ] = 0,5 × [1 + 0,34 × (1,93 − 0,2) + 1,932 ] = 2,66 which includes the imperfection factor for buckling curve b (Table 6.1, EN 1993-1-1 (2005)) α = 0,34 Design buckling resistance (Eq. (6.47), EN 1993-1-1 (2005)) Nb,Rd = χA

fy 275 = 0,223 × 659 × γM1 1,1 × 1E3

= 36,7 > (Nb,Fd = 28,77) kN satisfactory.

6.4 COMBINED BENDING AND AXIAL FORCE – WITH BUCKLING (cl 6.3.3, EN 1993-1-1 (2005)) 6.4.1 Introduction In practical situations axial forces in columns are accompanied by bending moments acting about the major and minor axes of bending. The axial force and bending moment vary along the length of the member and an exact analysis is complicated (Culver, 1966). In some situations overall buckling, lateral torsional buckling and local buckling can occur together. This is more likely to be a problem with Class 4 sections where outstand/thickness ratios are large. Exact theoretical solutions for buckling of columns are not available and in any case would be too complicated for design. Alternatively design interaction equations (Eqs (6.61) and (6.62), EN 1993-1-1 (2005)) are used based on elastic and plastic limits. A steel member subject to bending moments and axial forces fails when a stress is greater than first yield but less than full plasticity of the section. First yield theories are conservative while full plasticity theories are unsafe. Joint rotation reduces the buckling load of a column and this can now be incorporated into the analysis of frames (cl 2, EN 1993-1-8 (2005)). This can be important because even small rotations can significantly reduce load capacity. In design it is often assumed that parts of frames can be analysed separately and the compression members can be isolated and analysed accordingly. For traditional simple design methods beam and column structures are assumed to be connected together with pin joints and braced to prevent sidesway collapse. The pin joints are assumed eccentric to the column axis and thus introduce bending moments to the column which reduces the failure load.

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191

0

3m

18

0

3m

D 12

18

0

12

C 18

4,6 m

B 18

y

z A

z

y

FIGURE 6.7 Example: three storey corner column

EXAMPLE 6.3 Bending moments in a three storey corner column. The first three storeys of a corner stanchion are sketched in Fig. 6.7. The beam support reactions are indicated in kN on each beam. Assuming simple design, eccentric pin joints and cross bracing to reduce sway, determine the bending moments in the columns AB and BC. Try the following column sizes from Section Tables. Columns AB and BC: 203 × 203 × 60 UC h = 209,6 mm, Iy = 6103 cm4 , Iz = 2047 cm4 . Column CD: 203 × 203 × 46 UC h = 203,2 mm, Iy = 4565 cm4 , Iz = 1539 cm4 . Consider a corner column buckling with bending about the major and minor axes. Bending about the major y–y axis Stiffness of columns (cm units) Column AB, KAB = Iy /LAB = 6103/460 = 13,27 Column BC, KBC = Iy /LBC = 6103/300 = 20,34 Column CD, KCD = Iy /LCD = 4564/300 = 15,21

Chapter 6 / Axially Loaded Members (a) First floor level (at joint B): Moment about the y–y axis from eccentricity of the joint assuming MBy = (beam reaction)(100 + h/2) = 180 ×

(100 + 209,6/2) = 36,86 kNm. 1E3

For simple design moments are distributed in proportion to stiffness MBy KAB 36,86 × 13,27 = 14,55 kNm = KAB + KBC 13,27 + 20,34 MBy KBC 36,86 × 20,34 = = = 22,31 kNm KAB + KBC 13,27 + 20,34

MBA = MBC

(b) Second floor level (at joint C): Assuming that the splice between columns BC and CD lies above first floor level, then the design moment is the same at the first floor (MCy = 36,86 kNm). If moments are distributed equally for simple multi-storey construction MCB = MCD =

MCy 36,86 = = 18,43 kNm. 2 2

It should be appreciated that the simple method does not incorporate joint stiffness which reduces moments at mid-span for the beams. This is beneficial, but moments in the column are inaccurate. However, the method has been used extensively in the past.

EXAMPLE 6.4 Two storey corner column. Determine the size of a column section required for a two storey corner column shown in Fig. 6.8. The design loads (kN) shown are the end reactions from the beams. The beams are connected to the column using cleats and bolts and pin joints are assumed. The columns and beams are encased in concrete. C

3,7 m

•

4,6 m

192

11

7

18

0

12

B 18

y

z A

z

y

FIGURE 6.8 Example: two storey corner column

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Axial design loads on the column: (a) First floor Floor beam (self-weight included) Wall beam (self-weight included) Self-weight of column BC (estimated) Total

kN 117 12 10 139

(b) Ground floor Floor beam (self-weight included) Wall beam (self-weight included) Upper storey Total (excluding self-weight of column AB)

kN 180 18 139 337

Try 152 × 152 × 37 kg UC grade S275 steel for both storeys. From Section Tables h = 161,8 mm, b = 154,4 mm, tw = 8,1 mm tf = 11,5 mm, Agross = 4730 mm2 , iz = 38,7 mm, iy = 68,4 mm, Wel,y = 274E3 mm3 , Wel,z = 91,5E3 mm3 , Wpl,y = 309E3 mm3 , Wpl,z = 140E3 mm3 , Iz = 706E4 mm4 , Iy = 2213E4 mm4 , Iw = 0,0399 dm6 , It = 19,3E4 mm4 . This is a Class 1 section, that is, c/t outstand ratios are within limits (Table 5.2, EN 1993-1-1 (2005)) and local buckling does not occur. However, the section must be checked for overall buckling and lateral torsional buckling. Assuming simple nominal bending moments from eccentricity of the beam reactions (R) at join B: R(100 + h/2) 180(100 + 161,8/2) = = 32,56 kNm 1E3 1E3 R(100 + tw /2) 18(100 + 8,1/2) = = = 1,87 kNm 1E3 1E3

MBy = MBz

Design bending moments applied to the column (same section throughout) My,Ed = =

MBy (IBC /LBC ) [(IBC /LBC ) + (IBA /LBA )] 32,56(1/3,7) = 32,56 × 0,554 = 18,03 kNm [(1/3,7) + (1/4,6)]

Mz,Ed = MBz × 0,554 = 1,87 × 0,554 = 1,036 kNm Check combined axial load, bending and lateral torsional buckling about the stronger y–y axis using the interaction formula (Eq. (6.61), EN 1993-1-1 (2005)) kyy (My,Ed + My,Ed ) kyz (Mz,Ed + Mz,Ed ) NEd + + ≤1 χy NRk /γM1 χLT My,Rk /γM1 Mz,Rk /γM1 337 0,465 × (18,03 + 0) 0,453 × (1,036 + 0) + + 946 0,751 × 77,3 35 = 0,356 + 0,144 + 0,013 = 0,513 < 1 satisfactory. =

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•

Chapter 6 / Axially Loaded Members Check combined axial load, bending and lateral torsional buckling about the weaker z–z axis using the interaction formula (Eq. (6.62), EN 1993-1-1 (2005)) kzy (My,Ed + My,Ed ) kzz (Mz,Ed + Mz,Ed ) NEd + + ≤1 χz NRk /γM1 χLT My,Rk /γM1 Mz,Rk /γM1 337 0,453 × (18,03 + 0) 0,755 × (1,036 + 0) + + 532 0,751 × 77,3 35 = 0,633 + 0,141 + 0,022 = 0,796 < 1 satisfactory. =

The two previous equations include the following values. For buckling load calculations, consider column AB just below B, assuming the buckling length about the y–y and z–z axes, Lcr = 0,85 L. Use buckling curve b for buckling about the y–y axis and curve c for the z–z axis (Table 6.2, EN 1993-1-1 (2005)) based on h/b < 1,2; tf < 100 mm and steel grade S275. The non-dimensional slenderness ratio (Eq. (6.50), EN 1993-1-1 (2005)) λy = =

Afy Ncr

1/2

=

Lcr iy

×

1 λ1

(0,85 × 4600) 1 × = 0,659 68,4 [93,9 × (235/275)1/2 ]

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve b), λy = 0,659 and a reduction factor χy = 0,80 the design buckling resistance for a Class 1 section (Eq. (6.47), EN 1993-1-1 (2005)) Nb,Rd = χy A

fy 275 = 0, 80 × 4730 × = 946 kN. γM1 1,1 × 1E3

My,Rd = Wpl,y

fy 275 = 309E3 × = 77, 3 kNm. γM1 1,1 × 1E6

1 Lcr × iz λ1 0,85 × 4600 1 = = 1,16 × 38,7 [93,9 × (235/275)1/2 ]

λz =

Afy Ncr

1/2

=

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) and λz = 1,16 the reduction factor χz = 0,45 and the design buckling resistance for a Class 1 section (Eq. (6.48), EN 1993-1-1 (2005)) Nb,Rd = χz A

fy 275 = 0,45 × 4730 × = 532 kN. γM1 1,1 × 1E3

Mz,Rd = Wpl,z

fy 275 = 35 kNm. = 140E3 × 1,1 × 1E6 γM1

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For a Class 1 section (Table 6.7, EN 1993-1-1 (2005)) My,Ed = Mz,Ed = 0 Interaction factor for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesser value of (λy − 0,2)NEd kyy = cmy 1 + χy NRk /γM1 (0,659 − 0,2) × 337E3 = 0,4 × 1 + = 0,465 0,80 × 4730 × 275/1,1 or

kyy = cmy 1 +

0,8NEd χy NRK /γM1

= 0,4 × 1 +

0,8 × 337E3 0,80 × 4730 × 275/1,1

= 0,514 > 0,465 use 0,465 where (Table B3, EN 1993-1-1 (2005)) cmy = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4 Interaction factors for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesser value of NEd kzz = cmz 1 + (2λz − 0,6) χz NRk /γM1 337E3 = 0,4 × 1 + (2 × 1,16 − 0,6) × = 0,836 0,45 × 4730 × 275/1,1 or

kzz = cmz 1 +

1,4NEd χz NRk /γM1

1,4 × 337E3 = 0,4 × 1 + 0,45 × 4730 × 275/1,1

= 0,755 < 0,836 use 0,755 where (Table B3, EN 1993-1-1 (2005)) cmz = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4 and kyz = 0,6kzz = 0,6 × 0,755 = 0,453

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•

Chapter 6 / Axially Loaded Members Reduction factor for lateral torsional buckling (Eq. (6.57), EN 1993-1-1 (2005)) χLT = =

1 [LT + (2LT

2

− βλLT )1/2 ]

[0,881 + (0,8812

1 = 0,751 < 1 − 0,75 × 0,9052 )1/2 ]

or 1 2 λLT

=

1 = 1,22 > 1, use 0,751 0,9052

which includes the factor 2

LT = 0,5[1 + αLT (λLT − λLT,0 ) + βλLT ] = 0,5 × [1 + 0,21 × (0,905 − 0,2) + 0,75 × 0,9052 ] = 0,881 where the imperfection factor αLT = 0,21 (Table 6.3, EN 1993-1-1 (2005)). λLT =

Wply fy Mcr

1/2

=

309E3 × 275 103,85E6

1/2 = 0,905

The elastic critical bending moment Eq. (5.5) 1/2 Iw L2 GIt Mcr = + 2 Iz π EIz 1/2 0,0399E-6 4,62 × 15,59 π2 × 1483 = + 2 706E-8 4,62 π × 1483

π2 EIz L2

= 103,85 kNm where L = 4,6 m EIz = 210E6 × 706E-8 = 1483 kNm2 GIt = 80,77E6 × 19,3E-8 = 15,59 kNm2 EIw = 210E6 × 0,0399E-6 = 8,379 kNm4 For further information on elastic critical bending moments see Chapter 5. Check the self-weight of column BC. Minimum overall dimensions of cased column H = 161,8 + 100 = 261,8, say 270 mm B = 154,4 + 100 = 254,4, say 260 mm Ac = Ag − As = 270 × 260 − 4740 = 65460 mm2 .

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Total weight = steel column + concrete casing = Lρg + LAc ρg = 3,7 × 37 × 9,81/1E3 + 3,7 × 65460 × 2400 × 9,81/1E9 = 7 kN < 10 kN (assumed) satisfactory. These calculations show that a 152 × 152 × 37 kg UC section grade 275 steel is satisfactory, however a lesser weight of 152 × 152 × 30 kg UC might be suitable. Calculations are more extensive for Class 4 sections which are reduced in area to allow for local buckling (cl 5.5.2(2) EN 1993-1-1; and cl 4.3 EN 1993-1-5). Section properties are based on the effective cross-sections.

REFERENCES Beer, H. and Schultz, G. (1970). Theoretical basis for the European column curves. Construction Metallique, No. 3. Bleich, F. (1952). Buckling strength of metal structures. McGraw-Hill. Culver, G.C. (1966). Exact solution for biaxial bending equations. American Society of Civil Engineers, Str. Div., 92(ST2). Dwight, J.B. (1975). Adaption of the Perry formula to represent the new European steel column curves, Steel Construction, AISC, 9(1). The background to British Standards for structural steel work. Imperial College, London, and Constrado. ECCS (1976). Manual on stability, introductory report, second international colloquium on stability. European Convention for Structural Steelwork. Liege. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-5 (2003). Plated structural elements. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Euler, L. (1759). Sur la force de collones. Memoires de l’Acadamie de Berlin. Godfrey, G.B. (1962). The allowable stress in axially loaded struts, Structural Engineer, March 1962. Nelson, H.M. (1953). Angles in tension. British Constructional Steelwork Association publication No. 7. Regan, P.E. and Salter, P.R. (1984). Tests on welded angle tension members, Structural Engineer, 62B(2). Robertson, A. (1925). The strength of struts, Selected Engineering Paper No. 28, Institution of Civil Engineers. Trahair, N.S. and Bradford, M.S. (1988). The Behaviour and Design of Steel Structures. Chapman and Hall. Williams, F.W. and Sharp, G. (1990). Simple elastic critical load and effective length calculations for multi-storey rigid sway frames. Proceedings of the ICE, 90. Wood, R.H. (1974). Effective lengths for columns in multi-storey buildings. Structural Engineer, 52.

Chapter

7 / Structural Joints

(EN 1993-1-8, 2005)

7.1 INTRODUCTION Structural steel connections, referred to as joints in the Code, are required to ensure continuity at the intersection members and foundations. They are also used to form splices and to construct brackets to support loads. Generally structural steel joints are composed of plates, or parts of sections, shop welded in controlled conditions and bolted together on site. Welding can be carried out on site but it needs to be carefully supervised and is limited because of the expense. The physical appearance of some joints is shown in Figs 7.22 and 7.23. Structural joints transmit internal forces and moments in a structure and strength is of major importance. However, the rigidity of joints also needs to be considered. All joints are semi-rigid with associated small linear and larger rotational movements. The linear movements at a joint are generally small and generally need not be considered, but the rotational movements affect the distribution of forces and moments which must be taken into account in structural analysis. For theoretical purposes in the analysis of structures, joints are classified (Table 5.1, EN 1993-1-8 (2005)) by strength and rigidity as: (a) Pinned – low moment of resistance. (b) Rigid – full strength and all deformations are insignificant. (c) Semi-rigid – characteristics of the connection lie between (a) and (b). These theoretical and practical descriptions are important to recognize when analyzing a structure to determine the distribution of forces and moments using global analysis (cl 5.1, EN 1993-1-8 (2005)). There are three methods of global analysis: (1) Elastic – joints are classified by rotational stiffness. (2) Rigid-plastic – joints are classified by strength. (3) Elastic plastic – joints are classified by stiffness and strength.

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7.2 THE IDEAL STRUCTURAL JOINT The types of joints in common use have been developed and modified to suit the manufacturing and assembly processes and the ideal requirements are: (a) Simple to manufacture and assemble. (b) Standardized for situations where the dimensions and loads are similar thus avoiding a multiplicity of dimensions, plate thicknesses, weld sizes and bolts. (c) Manufactured from materials and components that are readily available. (d) Designed and detailed so that work is from the top of the joint not from below where the workman’s arms will be above his head. There should also be sufficient room to locate a spanner, or space to weld if required. (e) Designed so that the welding is confined to the workshops to ensure a good quality and to reduce costs. (f) Detailed to allow sufficient clearance and adjustment to accommodate the lack of accuracy in site dimensions. (g) Designed to withstand normal working loads and also erection forces. (h) Designed to avoid the use of temporary supports during erection. (i) Designed to develop the required load–deformation characteristics at service and ultimate loads. (j) Detailed to resist corrosion and to be of reasonable appearance. (k) Low in cost and cheap to maintain.

7.3 WELDED JOINTS Welding is a method of connecting components by heating the materials to a suitable temperature so that fusion occurs. The most common method for heating steelwork is by means of an electric are between a coated wire electrode and the materials being joined. The electrical circuit is shown in Fig. 7.1(a). During the process, which is illustrated in Fig. 7.1(b), the coated electrode is consumed, the wire becomes the filler material and the coating is converted partly into a shielding gas, partly into slag, and some part is absorbed by the weld metal. This method, known as the manual metal arc welding process, is still the most common for structural joints because of low capital cost and flexibility. However, for long continuous welds automatic processes are preferred because of consistency. Generally the electrode is stronger than the parent metal. For manual metal arc welding the electrodes should be compatible with the steel being welded (BSEN 499, 1995; Gourd, 1980). The main reason for the flux covering to the electrode in the manual metal arc welding process is to provide an inert gas which shields the molten metal from atmospheric contamination. In addition the flux forms a slag to protect the weld until it is cooled to room temperature, when the slag should be easily detachable. Other functions of the flux include: arc stabilization, control of surface profile, control of weld metal composition, alloying and deoxidization. However, it should be noted that

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Electrode wire Welding machine Coating on electrode

Molten arc pool Electrode Slag

Arc

Gaseous shield Arc stream

Weld Base metal Components being welded

Circuit

(a) Arc welding circuit

(b) Shielded arc welding

FIGURE 7.1 Shielded metal are welding

the flux can be a source of hydrogen contamination from absorbed and chemically combined moisture. The absorbed moisture can be removed by drying. The particular advantage of welding is that it forms a rigid joint, however the manufacture of welded joints requires more skill and supervision than bolted joints. Most welded structural joints are affected using the manual metal arc process but long continuous welds, which occur in built-up girders, are laid down with automatic welding equipment. The automatic processes achieve the exclusion of atmospheric pollution by gas shielding, flux core or submerged arc. Types of welds used in structural engineering and allowed in the European Code are fillet, slot, butt, plug and flare groove. Some common types are shown in Fig. 7.2. The two types in most common use are butt and fillet welds. Butt welds, often used to lengthen plates in the end on position, may be considered as strong as the parent plate as long as full penetration for the weld is achieved. For thin plates penetration is achieved without preparing the plate, but on thicker plates V or double J preparation is required. Butt welds are also used to connect plates at right angles but the plates require edge preparation. Partial penetration butt welds are not favoured in design and should not be used intermittently or in fatigue situations (BSEN 1011-1 (1998)). Fillet welds are generally formed with equal leg lengths. They do not require special edge preparation of the plates and are therefore cheaper than butt welds. Generally in connections, plates intersect at right angles but intersection angles of between 60◦ and 120◦ can be used provided that the correct throat size is used in design calculations. In order to accommodate lack of fit the minimum leg length of fillet weld in structural engineering is 5 mm although 6 mm is often preferred. The maximum size of fillet weld from a single run metal arc process is 8 mm, but 6 mm is

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Butt weld made from one side

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201

Butt weld made from two side T-butt weld made from two sides

T-butt weld made from two sides

t

b d t and c 3 mm or c t / 5

Slope should not exceed 1:1 in butt joints

b

d c

T-butt weld The weld is regarded to be of equal or higher strength than the vertical member

(a) Examples of butt welds

T-fillet weld

Lap joint

(b) Examples of fillet welds

FIGURE 7.2 Types of welds in structural joints

preferred to guarantee quality. When larger fillet welds are required they are formed from multiple runs. The use of intermittent butt and fillet welds is permitted (cl 4.3.2.2, EN 1993-1-8 (2005)). Intermittent welds are not favoured in structural engineering because they introduce stress discontinuities, act as stress raisers, may introduce fatigue cracks, may act as corrosion pockets and are difficult to produce with an automatic welding machine. The spacing of intermittent fillet welds is shown in Fig. 4.1, EN 1993-1-8 (2005).

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Deep penetration a

a

a

a

a

FIGURE 7.3 Throat thickness of fillet welds (Figs 4.3 and 4.4, EN 1993-1-8)

7.3.1 Throat Thickness of a Weld (cl 4.5.2, EN 1993-1-8 (2005)) The size of a weld is often described by the leg length but the strength is calculated using the effective throat thickness (a) as defined in Fig. 7.3. The effective throat thickness should not be less than 3 mm.

7.3.2 Effective Length of a Weld (cl 4.5.1, EN 1993-1-8 (2005)) The effective length of a fillet weld should be taken as the length over which the fillet is full size. The minimum length allowed to transmit loading is six times the throat thickness, and not less than 30 mm. In practice, fillet welds terminating at the ends, or sides, of parts are returned continuously round the corners for a distance of not less than twice the leg length, unless impracticable. The continuation round the corner is to reduce stress concentrations and its strength is generally ignored in strength calculations. The effective length of a weld is reduced if a component distorts under load in situations similar to that shown In Fig. 7.4, where the deformations in the weld adjacent to the web are greater than those at the end of the flange. The larger deformations at the web initiate failure in the weld at this point with consequent loss of strength for the total length of the weld (Elzen, 1966; Rolloos, 1969). For design the effective breadth (beff ) of a weld (cl 4.10, EN 1993-1-8 (2005)) is: For a rolled ‘I’ or ‘H’ section (Eqs (4.6a) and (4.6b), EN 1993-1-8 (2005)) fy,f tf beff = tw + 2r + 7ktf ≤ tw + 2r + 7 tf (7.1a) tp fy,p For box or channel sections where the widths of the connected plate and the flange are similar fy,f tf beff = 2tw + 5 tf ≤ 2tw + 5 tf (7.1b) tp fy,p

7.3.3 Long Welded Joints (cl 4.11, EN 1993-1-8 (2005)) The stress distribution along the length of a long lap joint is not uniform being greatest at the ends. To allow for this the length of the weld is reduced. For joints longer

Structural Design of Steelwork to EN 1993 and EN 1994 Shear force V

203

V

Bending moment M

Column I section

•

Deformation of welds due to shear force V determining flexible beam flanges

Bracket I section

Side elevation

End elevation Deformation of weld due to bending moment M deforming the flexible column flange

Plan

FIGURE 7.4 Reduction in strength of welds associated with flange deformations

than 150 times the throat thickness (a), the reduction factor (Eq. (4.9), EN 1993-1-8 (2005)) βLW ,1 = 1, 2 − 0, 2Lj /(150a) ≤ 1

(7.2)

where Lj is the overall length of the lap in the direction of the force transfer.

7.3.4 Design Resistance of Fillet Welds (cl 4.5.3, EN 1993-1-8 (2005)) The real external forces acting on a 90◦ fillet weld are probably those shown in Fig. 7.5 (a) (Clarke, 1971). Experiments (Biggs et al., 1981) on 90◦ fillet welds of equal leg length loaded to failure show that the fracture plane varies between 10◦ and 80◦ depending on the combination of external forces. The actual distribution of stress on the failure plane is uncertain but a theoretical distribution (Kato and Morita, 1974) shows peak stresses at the root of the weld which reduce towards the face of the weld. This distribution appears to be confirmed by experimental observations of cracks initiating at the root. The situation is complicated further by residual stresses and variables such as the type of electrode type of steel, ratio of the size of weld to the plate thickness, the quality of weld and whether the loading is static or dynamic. If stresses on the failure plane are assumed to be uniform then the relationship between the average shear stress and tensile stress on the failure plane has been shown (Biggs et al., 1981) to approximate to an ellipse. An ellipse of failure stresses combined with a variable

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) z x

Mwx

τf

Fwx

Mwy Fwy y

f

f

sf

Mwz Fwz (a) Complex system of forces

gth

en

it l Un

Fwx

τ1 Fwx

Fwz

Fwy f = 45° Fwz

Fwy

τ2 s

f = 45° th Th ick ro es at s a

•

Leg length s

204

(b) Simple system of forces

FIGURE 7.5 Forces acting on a 90◦ fillet weld

fracture angle can be used theoretically to predict the magnitude of the external forces, but the method is unnecessarily complicated for design purposes. For design purposes a complex system of external forces acting on a fillet weld is reduced forces acting in three perpendicular directions on a unit length of weld as shown in Fig. 7.5 (b). The vector sum of all the design forces should not exceed the design resistance of a fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) and may be expressed as 2 2 2 2 Fwx + Fwy + Fwz ≤ Fw,Rd

(7.3)

The term Fw,Rd = a fvw,d = a( fu /31/2 )/(βw γM2 ) is the design strength of a fillet weld per unit length. fu = nominal ultimate tensile strength of the weaker part joined (Table 3.1, EN 1993-1-1 (2005)) βw = correlation factor dependent on steel grade (Table 4.1, EN 1993-1-8 (2005)) γM2 = partial safety factors for joints, recommended value 1,25 (Table 2.1, EN 1993-1-8 (2005)).

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The design strength of a fillet weld has been shown (Ligtenberg, 1968) to be related to the strength of the parent material. The correct type and strength of electrode must be used for each grade of steel. An alternative directional method of design (cl 4.5.3.2, EN 1993-1-8 (2005)) involves calculating the normal and shear stresses on the throat section of the weld and combining them using the yield criteria developed in Chapter 2. This method is more laborious and introduces the possibility of further errors when resolving the forces onto the critical plane. The relationship between the forces Fx , Fy and Fz for this method can be expressed I the same form as Eq. (7.3) for a 45◦ plane (Holmes and Martin, 1983). The size of the fillet weld obtained using this method is slightly less than using the vector addition method.

7.3.5 Load–Deformation Relationships for Fillet Welds The strength of the weld in a connection is of primary importance but the load– deformation characteristics of the weld should also be considered. The deformation at the maximum load varies from approximately 0.6 to 1.4 mm depending on the orientation of the weld in relation to the applied load (Clarke, 1970) as shown in Fig. 7.6. The maximum deformation for the side fillet weld which is parallel to the applied load and the minimum is for an end fillet weld. To allow for this effect design stresses are based on the weaker side fillet welds. At the stress the disparity in deformations for end and side fillet welds is less than at failure.

7.3.6 Conditions Affecting the Strength of Welded Joints The following conditions affect the strength of welded connections: (a) Use of an incorrect steel (BS 7668, 1994). (b) Use of an incorrect electrode (BSEN 499, 1995). (c) Cavities and slag inclusions in the weld. These may be detected by non-destructive testing.

Average throat stress (MPa)

End fillet weld θ = 0° 500

θ = 30° θ = 60° θ = 90° P

Side fillet weld

θ varies

Test weld Design stress P 0

1 2 Weld deformation (mm)

FIGURE 7.6 Load-deformation relationships for an 8 mm fillet weld (Clarke 1970)

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) (d) (e) (f) (g)

(h)

(i) (j) (k)

(l)

Excessive lack of fit between components. Stress concentrations combined with oscillating loads producing fatigue. Residual stresses introduced from differential heating during welding. Hydrogen cracks associated with welding occur when the cooling rate is too rapid (Fig. 7.7(a)). Excessive hardening occurs in the heat affected zone which cracks under the action of residual stresses if sufficient hydrogen is present in the weld. This defect can be avoided by controlling the cooling and the hydrogen input to the weld (BSEN 1011-1 (1998)). Lamellar tearing may occur when welding plate connections of the type shown in Fig. 7.7 (b). the cracks are produced by a combination of low ductility in the plate in the transverse direction and high point restraint in the weld which induces tensile forces adjacent to the connection. The low ductility in the plate is produced by inclusions of non-metallic substances formed in the steel making process. When the ingot is rolled to make steel these inclusions form as plates parallel to the direction of rolling. Only a small percentage of plates are susceptible to lamellar tearing, and where it occurs joint details can be changed to reduce the chances of if affecting the strength of the connection (Farrar and Dolby, 1972). Brittle fracture. Corrosion which reduces the size of components or causes pitting which may initiate fatigue cracks. Insufficient penetration of the parent metal which leads to a reduction in strength of the weld. The welder uses a voltage and arc length which produces a stable arc and a satisfactory weld profile. The current then becomes the main factor in controlling penetration. Another important factor in depth of penetration is edge preparation. Plates of 6 mm with square edges can be butt welded from one side, but the edges of thicker material must be bevelled to provide access for the arc. Lack of side wall fusion occurs if there is poor bond between the parent and weld metal. Good bonding can only occur when the surface of the parent metal has been melted before the weld metal is allowed to flow into the joint.

Further information on faults in welds can be found in Gourd (1980).

Fillet weld

Hydrogen cracks (a) Hydrogen cracks

FIGURE 7.7 Faults associated with welding

Fillet weld

Lamellar tearing (b) Lamellar tearing

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7.3.7 Design Strength of Fillet Welds (cl 4.5.3.3 and Table 4.1, EN 1993-1-8 (2005)) The design strength of a unit length of fillet weld (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = (throat thickness) × (unit length) × (design shear stress)/factors fu a 31/2 = βw γM2

(7.4)

The factor (βw ) is related to steel grade, and varies between 0,8 and 1,0 (Table 4.1, EN 1993-1-8 (2005)). The ultimate stress ( fu ) is that of the weaker material joined (Table 3.1, EN 1993-1-1 (2005)). Typical values are fu = 430 MPa for S275 grade steel, and fu = 510 MPa for S355 grade steel. The value ( fu /31/2 ) is from the shear distortion strain energy theory as explained in Chapter 2. It should be noted that although the strength of the weld is calculated using the throat thickness (a) (Figs 4.3 and 4.4, EN 1993-1-8 (2005)) the weld is often specified by the leg length. A table of the strength of fillet welds is given in Annex Al.

7.4 BOLTED JOINTS (cl 3, EN 1993-1-8 (2005)) The advantage of bolted joints is that they require less supervision than welding, and therefore are ideal for site conditions. Other advantages are that the connection can be fastened quickly, supports load as soon as the bolts are tightened and accommodates minor discrepancies in dimensions. Disadvantages of bolted connections are that for large forces the space required for the joint is extensive, and the connection is not as rigid as a welded connection even when friction grip bolts are used. Steel bolts are identified by their gross diameter, strength and use. The preferred sizes of bolts in general use are 16, 20, 24, 30 and 36 mm diameter. The most common size use in structural connections is 20 mm. The types of bolt in common use (Table 3.1, EN 1993-1-8 (2005)) are: (a) Ordinary bolts Classes 4.6–6.8 and includes foundation bolts. (b) Pre-loaded bolts Classes 8.8 and 10.9. A Class 4.6 bolt is low in cost, can be installed with the use of simple tools and requires little supervision during the erection. At fracture the bolt has a relatively large extension of 25%, a property which is preferred at plastic collapse. Where forces are large, or where space for the connection is limited, or where erection costs can be reduced by using fewer bolts, then the higher grade bolts are used. The percentage elongation of 12% at failure is less, but is still acceptable for design purposes.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Where a more rigid bolted connection is required, for example in plastic methods of design, pre-loaded bolts are used. The strength of these bolts is greater with an increased cost for the additional site supervision which is necessary to ensure that the bolts are axially pre-loaded in tension to the design values. The object of the pre-load is to ensure that the friction between the ‘faying’ surfaces prevents slip when subject to external shear forces and thus produce a more rigid joint. The nuts of pre-loaded bolts are tightened with a torque wrench which is calibrated in relation to the required axial pre-load. A simpler method of measuring the axial force in the bolts is to use a load indicating washer, under the head of the nut, which reduces in thickness to a specified value for a specified pre-load. The washer is less accurate than the torque spanner (Bahia and Martin, 1981). A further alternative method of ensuring that the bolt is pre-loaded is to specify ‘turns of the nut’. Investigations (Fisher and Struck, 1974) showed that in general the pre-load produced by the torque wrench and ‘turns of the nut’ method on site exceeded the specified value. Close tolerance turned bolts are used only where accurate alignment of components or structural elements is required. The shank of the bolt is at least 2 mm greater in diameter than the threaded portion of the bolt and the hole is only 0,15 mm greater than the shank diameter. This small tolerance necessitates the use of special methods to ensure that the holes align correctly. Foundation bolts, or holding down bolts, are used for connection structural elements to concrete pads or concrete foundations. Generally the bolts are cast into the concrete before erection of the steel work and thus require accurate setting out. Where uplift forces occur the bolts must be anchored by a washer plate. Most bolts used are Class 4.6 but higher strengths are available. Sometimes bolts are grouted in the holes during erection using epoxy resin. Rivets were used extensively in the past in the fabrication shop and on site. They were difficult and expensive to place but they resulted in a rigid connection because the hot rivet, after driving, expanded to fill the hole. Rivets have now been superseded by welding and bolts. There are five categories of bolted joints (Table 3.2, EN 1993-1-8 (2005)) related to the type of connection or bolt and pre-loading. Shear connections Category A: bearing type where there is no pre-loading nor special provision for contact surfaces. Design for shear and bearing resistance. This is the cheapest type of connection where complete rigidity and plasticity are not important. Category B: slip resistant at serviceability limit state. Design for slip resistance at the serviceability limit state and shear and bearing resistance at the ultimate limit state. Connection used to provide full rigidity in the elastic stage of behaviour when deflections are critical.

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Category C: slip resistant at ultimate limit state. Design for slip resistance and bearing resistance at the ultimate limit state. Connection used to limit movements at the ultimate plastic limit state. Tension connections Category D: No pre-loading of bolts. Category E: Pre-loaded bolts.

7.4.1 Washers In British practice most bolts have steel washers under the head and under the nut. The washer distributes the bolt force and prevents the nut, or bolt head, from damaging the component or member. However washers are not essential in all cases (ECSS, 1981), and they are now being omitted in British practice. Ordinary washers (BSEN 4320, (1998)) are in general use but hardened washers (BSEN 14399-1 to 5 (2005)) are used for pre-loaded bolts. The outside diameter of a washer is an important dimension when detailing, for example to avoid overlapping an adjacent weld.

7.4.2 Bolt Holes Bolt holes are usually drilled, but may be punched full size, or punched under size and reamed. Holes should never be formed by gas cutting because of the inaccuracy and the effect on the local properties of the steel. Punched holes are preferred by steel fabricators because it saves time and reduces cost. However research (Owens et al., 1981) shows that distortion in the vicinity of a hole reduces toughness and ductility and can lead to brittle fracture. Punched holes should not be used in locations where plastic tensile straining can occur. Bolt holes are made larger than the bolt diameter to facilitate erection and to allow for inaccuracies. The clearance is 2 mm for bolts not exceeding 24 mm diameter and 3 mm for bolts exceeding 24 mm diameter. Oversize and slotted holes are allowed but not often used. Slotted holes are sometimes used for pre-loaded bolts to facilitate erection with unusual shaped structures, or alternatively they can be used to accommodate movement in a structure. The clearance for a close tolerance turned bolt is 0,15 mm. Bolt holes reduce the gross cross-sectional area of a plate to the net cross-sectional area. The net value is used for calculations where the structural element, or parts of an element, are in tension. Bolt holes also produce stress concentrations, but it is argued that these are offset by the fact that at yield the highly stressed cross-section will work harder before fracture and yield will by then have occurred at adjacent cross-sections. The gross cross-section of a member is used in compression because at yield the bolt

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) hole deforms and the shank of the bolt resists part of the load in bearing. Consideration should be given to corrosion and local buckling when deciding the position of holes.

7.4.3 Spacing of Bolt Holes (Table 3.3, EN 1993-1-8 (2005)) The longitudinal spacing between the centre line of bolts in the direction of the axial stress in a member is called the pitch. The minimum spacing in the direction of the load of 2,2 × (diameter of the hole) is specified to prevent excessive reduction in the cross-sectional area of a member, to provide sufficient space to tighten the bolts and to prevent overlapping of the washers. Other critical distances are given in Table 3.3 and Fig. 3.1, EN 1993-1-8 (2005). These values are specified to prevent buckling of plates in compression between bolts, to ensure that bolts act together as a group to resist forces, and to minimize corrosion.

7.4.4 Edge and End Distances for Bolt Holes (Table 3.3, EN 1993-1-8 (2005)) Edge and end distances are specified to resist the load, to prevent local buckling, to limit corrosion and to provide space for the bolt head, washer and nut. The edge distance is from the centre of a hole to the nearest edge measured at right angles to the direction of the load. The minimum edge distance specified is 1,2 × (diameter of the hole) and the maximum should not exceed 4t + 40 mm. The end distance is from the centre of a hole to the adjacent edge in the direction of the load transfer. The minimum end distance in the direction of the load is 1,2 × (diameter of the hole) and the maximum should not exceed 4t + 40 mm. The end distance should also be sufficient for bearing capacity. There are recommended positions, spacing and diameter of holes in Section Tables. These distances are based on providing sufficent clearances to the web and adequate edge distances (Annexes A4 to A6).

7.4.5 Deductions for Holes in Tension Members (cl 3.10, EN 1993-1-8 (2005)) Holes are drilled in tension members to accommodate fasteners at connections. A hole reduces the gross cross-sectional area and weakens a tension member because the fastener in the hole does not transmit the axial force. In contrast a hole in a compression member has little effect on the buckling strength because as the member compresses the axial force is transmitted by bearing on the shank of the bolt. When designing a tie the net cross-sectional area is used in calculations to determine the design axial force. The net cross-sectional area is the gross area reduced by the maximum sum of the sectional areas of the holes. These holes may be in line at right

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211

C

110 95 A

95 B

95

95

430

110

P = 110

A

•

Direction of stress

95

C

s = 95

FIGURE 7.8 Example: net area of a plate

angles to the axial stress in the member (line AA in Fig. 7.8), or staggered (lines BB and CC in Fig. 7.8). Typical areas to be deducted for bolt holes are: Bolt diameter + 2 mm clearance for bolts not exceeding 24 mm in diameter or Bolt diameter + 3 mm clearance for bolts exceeding 24 mm in diameter. For staggered fastener holes the area to be deducted shall be the greater of: (a) the deduction for non-staggered holes; (b) the sum of the sectional areas of all holes in any diagonal or zig-zag line extending progressively across the member, or part of the member, less s2 t/4p for each gauge space in the chain of holes. where (Fig. 7.8) s is the staggered pitch p is the spacing of the holes t is the thickness of the holed material. For sections such as angles with holes in both legs the spacing is approximately the sum of the back marks to each hole, less the leg thickness. The arrangement and spacing of holes in a member should not significantly weaken a member at a section.

EXAMPLE 7.1 Net area of a plate with holes. Calculate the net cross-sectional area for the plate shown in Fig. 7.8 which is subject to a tensile force. The plate is 20 mm thick and contains four lines of staggered holes drilled for 24 mm diameter bolts. From Fig. 7.8, s = 95 mm and p = 110 mm. Diameter of hole do = d + 2 = 24 + 2 = 26 mm.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Gross cross-sectional area perpendicular to the direction of stress = 20 × 430 = 8600 mm2 Areas to be deducted at possible failure lines are: nh t do − ngs

s2 t 4p

where ngs is the number of gauge spaces in the chain of holes. Line AA: 2 × 20 × 26 = 1040 mm2 Line BB: 3 × 20 × 26 − 1 ×952 × 20/(4 × 110) = 1150 mm2 Line CC: 4 × 20 × 26 − 3 × 952 × 20/(4 × 110) = 849 mm2 . Minimum net area for line BB = 8600 − 1150 = 7450 mm2 .

7.4.6 Design Resistance of Single Bolts (Tables 3.1 and 3.4, EN 1993-1-8 (2005)) Bolted connections consist of two or more bolts and each bolt may be subject to any combination of tension, shear or bearing forces. The design resistance of a single bolt when subject to these forces is now considered in detail. A bolt in tension fails at the smallest cross-section, that is, the root of the threads where the net area is approximately 80% of the gross area. Where the bolt fails across the reduced cross-section at the root of the thread the tension resistance As Ft,Rd = k2 fub (7.5) γM2 where the ultimate tensile strength of the bolt (fub ) is obtained from Table 3.1. EN 1993-1-8 (2005) and k2 = 0,9 except for a countersunk bolt where k2 = 0,63. Where the bolt assembly fails by the bolt head, or nut, shear punching through the plate then the tension resistance of the plate (Table 3.4, EN 1993-1-8 (2005)) Bp,Rd = 0,6 π dm tp

fu γM2

(7.6)

where fu is the ultimate tensile stress of the plate. tp is the thickness of the plate under the head of the bolt. dm is the mean of across points and across flats dimensions of the bolt head or the nut, whichever is smaller. The shear resistance per shear plane for a single bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = αv fub

A γM2

(7.7)

Structural Design of Steelwork to EN 1993 and EN 1994

•

213

Where the shear plane passes through the unthreaded portion of the bolt αv = 0,6 and A is the gross area. Where the shear plane passes through the threads αv = 0,5 or 0,6, depending on the grade of the bolt, and A = As the reduced area. The values of the reduced bolt areas used in this chapter (BS 3692, 2001; BS 4190, 2001) are: Bolt diameter (mm) 12 16 Reduced area (mm2 ) 84,3 157

20 245

22 303

24 353

27 459

30 561

36 817

Values may also be obtained from Section Tables. Shearing of a bolt occurs on the shank, that is, the gross area of the bolt, if the thread length on a bolt is carefully specified, but it is safer to assume that it occurs on the reduced area. It also simplifies calculations and avoids confusion. Experiments (Bahia and Martin, 1980) and other investigators have found shear values that vary between 0,62fu and 0,71fu . The value from the Huber–Von–Mises–Hencky shear distortion strain energy theory is fu /31/2 as shown in Chapter 2. Ordinary bolts deform when subject to shear stresses but it is important to realize that the shear deformation of the connection is increased by the bearing stresses on the plate. The higher the bearing stresses the greater the deformation as shown in Fig. 7.9.

7.4.7 Design Resistance of a Bolt Subject to Shear and Tension Forces (Table 3.4, EN 1993-1-8 (2005)) Shear and tensile resistances are related by the linear interaction formula Fv,Ed Ft,Ed + ≤ 1, 0 Fv,Rd 1,4Ft,Rd

(7.8)

Generally it is assumed that the failure plane passes through the threaded portion of the bolt. This equation is to be compared with a non-linear experimental relationship (Chesson et al., 1965) based on the net cross-sectional area of the bolt (Fig. 7.10). Alternative elliptical relationships are given in ECSS (1981) and BS 5400 (2000).

7.4.8 Design Bearing Resistance for a Bolt (Table 3.4, EN 1993-1-8 (2005)) A bolt subject to a shear force, such as shown in Fig. 7.11, comes in contact with the plate when the shear load is applied and slip occurs. The bearing stresses between the bolt and plate need to be controlled to limit deformations.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

tp

tp

=2

160

0, = 1 15 m 7 m , tp = 1 10 m m 4, tp = 1 10 m 2,0 m 7m m

High-strength bolts 20 mm diameter

0m

m

140

tp

=2

120

Shear load (kN)

•

100

Black bolt 20 mm diameter

80

60

40

20

0

4 0

2

6 2

8 4

10 High-strength bolts 6 black bolts

Total deformation (mm) Notes : Permanent deformation of holes shown as a broken line; bolts failed in single shear across the threads; tp is the plate thickness.

FIGURE 7.9 Relation between shear load and deformation for single bolt tests (Bahia and Martin, 1980)

Experimental results by Chesson et al., 1965 for high strength bolts with shear plane on shank and on threads

1,0 0,8

2 ⎤ ft ⎤ ⎥ ⎥ + ⎦ ftu ⎦

fv ⎤ 2 ⎤ ⎥ ⎥ =1 ⎦ 0,62 ftu ⎦

0,6 f v /f tu

214

0,4 0,2

0

0,2

0,4

0,6 f t /f tu

0,8

1,0

1,2

FIGURE 7.10 Relationship between shear and tensile stresses for bolts

Structural Design of Steelwork to EN 1993 and EN 1994

•

215

Bolts

A Side elevation

dh

Thickness tp

A

cp

Plan section AA (a) End bearing failure for a small end distance

(b) Enclosed bearing failure for a large end distance

Plan

Sectional elevation (c) Actual distribution of bearing stress

(d) Theoretical distribution of bearing stress

FIGURE 7.11 Bearing stresses for bolts

The bolt or plate may deform because of high local bearing stresses between bolt and plate, or a bolt may shear through the end of the plate. The bearing resistance Fb,Rd = k1 αb fu

dt γM2

(7.9)

Values of k1 control end distances and values of αb control edge distances (Table 3.4, EN 1993-1-8 (2005)). The value of fu is the weaker of the bolt grade or the adjacent plate or section. The values of the ultimate tensile strength for bolts (Table 3.1 EN 1993-1-8 (2005)) used in this chapter are: Bolt class 4.6 fub MPa 400 αv 0,6

4.8 400 0,5

5.6 5.8 500 500 0,6 0,5

6.8 600 0,5

8.8 800 0,6

10.9 1000 0,5

Values of steel grades used in examples in this chapter are S275 ( fu = 430 MPa) and S355 ( fu = 510 MPa). Equation (7.9) assumes uniform bearing stresses as shown in Fig. 7.11(d) whereas in reality they are closer to those shown in Fig. 7.11(c). Design bearing stresses are high

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Permanent deformation of hole shown as a broken line

700

tp = 12,07 mm

600 Average bearing stress (MPa)

216

tp = 14,10 mm

500

tp = 17,10 mm

400

tp = 20,15 mm

300 Yield stress of plates 221,8 ± 8,03 MPa

200 100

0

2 4 6 8 10 Total shear deformation (mm)

Note: Bolts failed on the threads in single shear at a shear stress of 600 MPa

FIGURE 7.12 Relationships between bearing stress and deformation for an M20 high stress single bolt (Bahia and Martin, 1980)

in relation to the yield stress because material subject to bearing stresses is generally confined by other parts which restricts deformation. High bearing stresses are not disastrous but lead to excessive deformation of a connection as shown in the experimental results (Fig. 7.12). Equations (7.5)–(7.9) express the design tensile and shear strengths of a bolt and can be presented in the form of tables to reduce calculations (Annex A2).

7.4.9 Bolts Through Packings (cl 3.6.1(12), EN 1993-1-8 (2005)) Where the total thickness of the packing (tp ) is greater than three times the nominal diameter (d) of the bolts the design shear resistance is multiplied by a reduction factor βp =

9d ≤1 8d + 3tp

(7.10)

The reason for the reduction in strength is because as the grip length increases the bolt is subject to greater bending moments from shear forces which move further apart.

7.4.10 Long Bolt Joints (cl 3.8, EN 1993-1-8(2005)) For long joints the load is not shared equally by the bolts or rivets. The fasteners on the end resist the greatest force and the resistance gradually reduces to the centre line of the joint. The reduction is due to friction, errors in marking out and deformations

Structural Design of Steelwork to EN 1993 and EN 1994

•

217

Lj Cover plate

FIGURE 7.13 Length (Lj ) for long bolted joints in the materials. Where the length of the joint Lj > 15d (Fig. 7.13) measured in the direction off the transfer force then the design shear resistance reduction factor βLf = 1 −

Lj − 15d 200d

(7.11)

but 0,75 ≤ βLf ≤ 1,0

7.4.11 Design of Slip Resistant Joints (cl 3.9, EN 1993-1-8 (2005)) High strength bolts can be pre-loaded when installed and designed to be slip resistant at working load or ultimate load. The design slip resistance of a pre-loaded bolt when subject to external tensile and shear forces is now considered in detail. The design slip resistance of pre-loaded high strength bolt of Classes 8.8 or 10.9 (cl 3.9.1, EN 1993-1-8 (2005)) is Fs,Rd =

ks n μFp,C γM3

(7.12)

where Fp,C = 0,7fub As is the design preloading force. μ = slip factor as listed in Table 3.7, EN 1993-1-8 (2005). ks = 1 for a bolt in a clearance hole and reduced for slotted holes (Table 3.6, EN 1993-1-8 (2005)) n = number of friction interfaces. T ypical Surface treatment

Slip factor (μ)

A – blasted with shot or grit, loose rust removed, no pitting B – ditto, painted with zinc C – cleaned and loose rust removed D – surfaces not treated

0,50 0,40 0,30 0,20

Table 7.2 Slip factors(μ) for pre-loaded bolts (Table 3.7, EN 1993-1-8 (2005)).

218

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) The effect of a tensile force acting on a pre-loaded bolt is to reduce the frictional resistance (cl 3.9.2, EN 1993-1-8 (2005)). The design slip resistance per bolt subject to a combination of shear and tension forces is Category B connection (Eq. (3.8a), EN 1993-1-8 (2005)) at service load Fs,Rd,ser =

ks nμ(Fp,C − 0,8Ft,Ed,ser ) γM3,serv

(7.13)

Category C connection (Eq. (3.8b), EN 1993-1-8 (2005)) at ultimate load Fs,Rd =

ks nμ(Fp,C − 0,8Ft,Ed ) γM3

(7.14)

The factor of 0,8 is introduced to allow for the fact that the minimum shank tension may not be achieved. The design tensile, shear and bearing strength of a parallel shank pre-loaded bolt and can be presented in the form of a table to reduce calculations (Annex A3).

7.5 PLATE THICKNESSES FOR JOINT COMPONENTS Plates, or parts of sections acting as flange plates, often form part of structural connections. The length and breadth of plates are generally determined from the geometry of the connection but the thickness is calculated from the elastic or plastic theory of bending. Backing plates are used to strengthen flanges of columns as shown in Fig. 6.3, EN 1993-1-8 (2005). In particular, they are used to strengthen T-stubs and methods are given in Table 6.2, EN 1993-1-8 (2005). The limits on dimensions of the plates are given in cl 6.2.4.3, EN 1993-1-8 (2005).

7.5.1 Plastic Methods for Plates (cl 6.2.4, EN 1993-1-8 (2005)) Parts of connections may be idealized as T-stubs as shown in Fig. 7.14 where the external force (Ft ) is balanced by bolt force (Bt ) and prying force (Q). A prying force is an additional axial tensile force that is induced in a bolt due to the flexing and reaction of components. There are three possible conditions of equilibrium for a T-stub at the ultimate limit state (Fig. 7.15). Case 1: Bolt failure as shown in Fig. 7.15(a) Resolving forces vertically Ft + Bt = 0

(7.15)

•

Tensile force in the bolt (Bt )

Structural Design of Steelwork to EN 1993 and EN 1994

219

Ft Prying force Q at bolt failure

Pre-loaded

Bt = Ft Bt

Q

Bt

Q

Prying force Q at the elastic stage of behaviour Bolt with no preload

m

e

Applied external force (Ft )

Ft (a) Relationship between external force and bolt force for a T-stub

(b)

FIGURE 7.14 Prying forces for T-stubs

Thick plates

Elastic plates

Plastic hinges

Prying force

Bolt failure

Bolt failure

(a) No Prying force

Elastic bolt

(b) Forms of prying force failure

(c)

F t = 2Mpl + e ΣB t (m + e)

F t = ΣB t

Ft = 4Mpl m

A ΣB t /2

ΣBt /2

C Q

ΣBt /2

ΣBt /2

Q

Q Ft

Mpl

Q

B

2

Ft

+Q Mpl

2

+Q

Ft = ΣB t Idealized failure modes

FIGURE 7.15 Failure modes of T-stub flanges

Mpl

Mpl

220

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) In this case the prying force (Q) is zero, the thickness of plate is a maximum and the force in the bolt is a minimum (Table 6.2, EN 1993-1-8 (2005), mode 3). Case 2: Bolt failure with partial yielding of the flange as shown in Fig. 7.15(b) Taking moments of forces about C −Mpl +

Bt Ft + =0 2(m + e) 2e

Rearranging, the design tension resistance Ft =

2Mpl + Bt e m+e

(7.16)

Prying force Q=

Bt /2m − Mpl m+e

(7.17)

This is the case that is most likely to occur in practice because plate thicknesses are limited to those available (Table 6.2, EN 1993-1-8 (2005), mode 2). Case 3: Complete yielding of the flange as shown in Fig. 7.15(c) Taking moments of forces about A −Mpl + (Ft /2 + Q)m − Q(m + e) = 0

(i)

Taking moments about B Mpl − Qe = 0

(ii)

Combining (i) and (ii) to eliminate Q Ft =

4Mpl m

(7.18)

They prying force Q=

Mpl e

(7.19)

This case results in the smallest thickness of plate but the largest bolt force (Table 6.2, EN 1993-1-8 (2005), mode 1). If needed the prying force can be calculated more accurately (Holmes and Martin, 1983) and the theory has been shown to agree with the experimental results (Bahia et al., 1981). Prying forces can be avoided by using non-flexible components or by the use of stiffeners.

7.5.2 Plastic Method for the Thickness of Flange Plates When considering T-stubs the length of the flange is known, but when analysing column flanges in similar situations (Fig. 7.16(a)) an effective length needs to be determined.

Structural Design of Steelwork to EN 1993 and EN 1994

•

221

The method of determining the effective length is illustrated by a simple example for the plastic yield lines shown in Fig. 7.16(b). For a single bolt force then from virtual work 2 tp fyp Ft m 4(m + e) 2x = + m+e 4 x m+e

(i)

Differentiating with respect to x to determine the value of x for which Ft is a minimum ∂Ft 2 (m + e) + = −4 =0 2 ∂x m+e x

√ hence x = (m + e) 2

(ii)

Combining Eqs (i) and (ii) and rearranging, the thickness of the plate Ft mfyp tp = √ 2(m + e) From Eq. (i) the effective length √ 2x 4m + e + (m + e) = 4 2(m + e) leff,b = x m+e

N

Δ x

f

M

m e

(a)

m e

(c) leff,b 4 m 1,25e (inner and end bolts)

p

p

(b) Simple theory

m e

m e

(d) leff,b p (inner bolts) 0,5p 2 m 0,625 e (outer bolts)

(e) leff,b 2 πm (inner and end bolts)

FIGURE 7.16 Effective lengths for column flanges

(7.20)

222

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) The work Eq. (i) can therefore be written as an equilibrium equation 2 tp fyp Ft m = leff,b 4

(7.21)

The above theory assumes simple idealized conditions and ignores, welds, washers, limited rotation at the plastic hinge and strain hardening. However the simple example shows the basic theoretical approach. Other solutions are related to research work (Stark and Bijlaard, 1988). In practice the yield lines are more complicated (Figs 7.16(c), (d) and (e)) and more difficult to analyse. Analysis using yield lines is avoided in the European Code by giving the effective length of plate (leff,b ) (Fig. 7.16) and applying the T-stub equations (Table 6.4, EN 1993-1-8 (2005)).

7.5.3 Elastic Methods for Plates (a) Base plates are used to distribute the load from column sections as shown by the area enclosed by the dotted line in Fig. 7.17. Elastic stress distribution with the maximum stress at the yield strength is used at the ultimate limit state to ensure that large displacements do not occur. If the load is axial the pressure (fj ) beneath the base plate is uniform and the projection (c) of the steel plate beyond the edge of the column, then for a cantilever from the simple theory of elastic bending at first yield per unit width M = fy W f j c2 fy t 2 = 2 6 rearranging, the projection of the plate (Eq. (6.5), EN 1993-1-8 (2005)) c=t

fy (3fj γM0 )

1/2 (7.22)

Justification for this theory is given in Holmes and Martin (1983).

t

fj per unit area c

FIGURE 7.17 Bending strength of a column base plate

Structural Design of Steelwork to EN 1993 and EN 1994

Gusset plate

Hg Lg

223

Gusset plate

(b) Bracket

(a) Column base

Shape of deflected free edge observed in experiments for

•

Approximate buckling stress distribution where Lg/ig 185

Edge stress depends on slenderness ratio

Hg Lg

fgy o δbg

sg

Hg

(e) bg Buckling stress distribution

Bg Theoretical hinge

Fu

Edge stress depends on slenderness ratio

Element strut

o

fgy sg

Fu Lg

Theoretical hinge

(c)

o

sg

Fu

(d) Theoretical model (Martin)

FIGURE 7.18 Gusset plates (b) Gusset plates are used to stiffen base plates and brackets as shown in Fig. 7.18. The following theory for the buckling of a gusset plate is based on experimental work (Martin, 1979; Martin and Robinson, 1981). No advice is given in the European Code. The basic structural unit is a triangular plate with loading applied to one edge as shown in Fig. 7.18(c). For theoretical purposes the plate is assumed to be composed of a series of fixed ended struts parallel to the free edge. The distribution of direct stress across the width bg is shown on an element of the gusset plate in Fig. 7.18(d). The buckling stress varies depending on the slenderness ratio of the elemental strut. At the hinge the stress is for a very short strut (i.e. yield). At the free edge the value is for the slenderness ratio of the elemental strut at the free edge. For simplicity the buckling stress distribution shown in Fig. 7.18(d) can be replaced by a linear distribution as shown in Fig. 7.18(e), provided that the slenderness ratio of the free edge lg /ig < 185. This restraint is acceptable because slenderness ratios of gusset plates in structural engineering do not often exceed this value.

224

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Taking moments of forces about the theoretical hinge at O (Fig. 7.18(e)) and ignoring the moment of resistance of the base plate as justified (Martin and Robinson, 1981). Bg F u sg = (bg fg tg )δbg (7.23) 0

For each strip the buckling stress ( fg ) is linearly related to the slenderness ratio (lg /ig ). The effective length lg = bg when Lg = Hg , and from experiments (Martin, 1979) this is approximately correct when Lg = Hg . The buckling stress for each strip can therefore be expressed as bg provided that lg /ig < 185 (7.24) fg = fgy 1 − 185ig Combining Eqs (7.23) and (7.24), integrating, expressing the radius of gyration as ig = tg /(2 × 31/2 ), and rearranging tg =

2Fu sg Bg + 2 80 fgy Bg

(7.25)

where from the geometry of the plate Bg =

Lg [(Lg /Hg )2 + 1]1/2

(7.26)

The slenderness ratio of the gusset plate may be defined as the slenderness ratio of a strip of unit width parallel to the free edge. From this definition and Eq. (7.26) lg (2 × 31/2 )Bg 2 × 31/2 (Lg /tg ) = = ig tg [(Lg /Hg )2 + 1]1/2

(7.27)

This theory is for non-slender gusset plates, that is, for lg /ig < 185. The theory for slender gusset plates is given elsewhere (Martin, 1979).

7.6 JOINTS SUBJECT TO SHEAR FORCES Two simple connections subject to shear forces are shown in Fig. 7.19. The forces in the members are assumed to be axial and to act through the centroidal axes of the members. This is correct in some situations, for example the bolted joint shown in Fig. 7.19(a). However it is not correct for the welded lap connection shown in Fig. 7.19(b) because the eccentricity of the force produces a moment which results in distortion at ultimate load. It is not correct for a roof truss joint as shown in Fig. 7.24 because although the centroidal axes intersect and there are axial forces in the members there are also secondary moments. A further assumption for simple joints is that the external forces are distributed evenly to the bolts or welds. This is not correct for long bolted and welded joints and allowance must be made for this.

Structural Design of Steelwork to EN 1993 and EN 1994

x

F

F

x

F

Flat F

e

Tie F

x x x x x

b

x

225

Angle

l

F

•

F

F

(a) Bolted connection F

F Distortion at ultimate load (b) Welded connection

FIGURE 7.19 Joints subject to shear forces

The overlap distance (l) is important for simple joints. For bolted joints the minimum of two bolts and the required end distances generally ensure that the lap is sufficient. However for welds (Fig. 7.19) the greater strength may indicate that the lap distance can be small, but it must be appreciated that there must be room for stop and start lengths and that stress concentrations can occur. The minimum lap length is generally not be less than four times the thickness of the thinner part joined where the weld is continuous. The minimum length of weld is 40 mm or six times the throat thickness. For a joint with side welds only the lap is generally not less than the width of the member and there should be end returns of twice the leg length of the weld (cl 4.3.2.1(4), EN 1993-1-8 (2005)) to reduce stress concentrations.

7.7 JOINTS SUBJECT TO ECCENTRIC SHEAR FORCES Joints, such as shown in Fig. 7.20(a), are subject to eccentric shear forces which tend to rotate the joint. This produces a resultant shear force on a fastener (bolt, or unit length of weld) from the direct shear force and the moment. The forces acting on a group of fasteners can be idealized as shown in Fig. 7.20(b). The bolt group rotates about the theoretical instantaneous centre of rotation which varies in position depending on the magnitudes of the external forces V and H and the eccentricity e. In the linear elastic stage of behaviour it is reasonable to assume that the force acting on a fastener is proportional to the distance from the centre of rotation. At ultimate load this assumption is not strictly correct but the error involved is not great. For a rigorous solution to the theory and accuracy at ultimate load see Bahia and Martin (1980).

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

(a)

N

o θ n

z Fmax Fastener r1 Fmax rn n rn V H y Zn

y

r2 F rn max r2 θ2

V

r1 θ1 y

θ

H

rG n θ

M Ve

Vertical component of force from torsional moment zG

•

V

226

G

yn

Centroid of fastener group

H yG

e

z (b) Diagram for general theory

Horizontal component of force from torsional moment

(c) Vector digram for simple vector addition theory for a typical fastener

FIGURE 7.20 Joints subject to eccentric shear forces Although this is the correct approach to the theory there is a simpler more practical method in common use which gives the same values. It is assumed that rotation occurs about the centroid of the fastener group and for convenience the forces acting on a fastener are parallel to the z–z and y–y axes as shown in Fig. 7.20(c) There forces are combined vectorially and the resultant force on a fastener furthest from the centre of rotation is 1/2 MyG 2 MzG 2 V H 2 2 1/2 FR = [Fy + Fz ] = + + + (7.28) n Ix n Ix where n is the number of fasteners in the group yG and zG are coordinates of a fastener related to the centroid of the fastener group IX = IY + IZ IX , IY and IZ are second moments of area of unit size fasteners about the X–X, Y–Y and Z–Z axes The method is used in practice for ordinary bolts, high strength friction grip bolts and welds.

7.8 JOINTS WITH END BEARING Some joints involve end bearing between components (Fig. 7.21). End bearing can occur in beam-to-column (Fig. 6.15, EN 1993-1-8 (2005)), bracket-to-column, beamto-beam and column-to-base joints.

Structural Design of Steelwork to EN 1993 and EN 1994

•

227

End bearing

End bearing (a)

(b)

V

End bearing

Ro

H M O

dp

dr

R

μRo Centre of rotation at stiff bearing O

(c)

(d)

FIGURE 7.21 Joints with end bearing

Where end bearing occurs, rotation takes place about a stiff axis of rotation, axis O–O shown in Fig. 7.21(d). The reaction force Ro is generally large and the bearing may have to be reinforced if it is not to distort under the load. The balancing tensile force R is resisted by bolts or welds. If there is slip at the stiff bearing a frictional force μRo develops parallel to the stiff bearing surface. Consider an end bearing joint subject to external forces V , H and moment M as shown in Fig. 7.21(d). If slip occurs resolving forces vertically V − μRo < 0

(7.29)

Taking moments of forces about force R M + −H(dr − dp ) − Ro dr = 0

(7.30)

where R is the resultant force of the fasteners acting at a distance dr from the axis of rotation O–O. Combining Eqs (7.29) and (7.30) to eliminate Ro , then slip will not occur if μ[M ± H(dr − dp )] >1 Vdr

(7.31)

Most joints with end bearing do not slip and therefore the fasteners are not subject to the external shear force. One exception is a bracket supporting a load with a small eccentricity.

228

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) In the elastic stage of behaviour it is assumed that the forces acting on a fastener are proportional to the distance from the axis of rotation O–O. If, conservatively, this is also assumed to occur at ultimate load then: Taking moment of forces about axis O–O the maximum tensile force resisted by a fastener Ft(max) =

[M ± Hdp ]zmax IO

(7.32)

where IO = z2 is the second moment of area of unit size fasteners about axis O–O Resolving forces vertically the shear force on a fastener 2 " #3 M V − μ ± H(1 − d /d ) p r (V − μRo ) dr Fs = = (7.33) n n This equation is formed assuming that bolts of the same size and design strength resist equal shear forces and also that slip has taken place. If the fasteners are welds then Ft(max) and Fs are combined vectorially. If the fasteners are ordinary bolts the forces are combined using Eq. (7.8) and related to design strengths. If friction grip bolts are used the forces are combined using Eqs (7.13) and (7.14) and related to design strengths. Traditional design methods ignore the existence of the frictional force which errs on the side of safety. However, research (Bahia et al., 1981) has shown that the frictional force does resist part of the shear force. If pre-loaded bolts are used then slip may occur at service load or at ultimate load.

7.9 ‘PINNED’ JOINTS (CL 3.13, EN 1993-1-8 (2005)) Some simple joints (e.g. a tie bar) are connected by real pins as shown in Fig. 7.22(a). Provided that the pins are not corroded, or blocked with debris, they will act as pin joints, that is, they will resist forces but not moments. Tie bars are rarely used now because of the cost of manufacture, risk of seizure from corrosion or debris, and because safety depends on a single pin. Other connections shown in Fig. 7.22 are designated as ‘pins’ because the rotational restraint is small. In the past these joints have been designed assuming that the rotational resistance is zero and the connection resists direct forces only. The general approach to design of these ‘pinned’ connections follows.

7.9.1 Pinned Beam-to-Column Joints (Figs 7.22(b), (c), (e) and (f)) In design calculations for the joint shown in Fig. 7.22 (b) it is assumed that the shear force is resisted by the four bolts connecting the bottom cleat to the column flange.

Structural Design of Steelwork to EN 1993 and EN 1994

•

229

Shear force Angle section

Pin joint Axial force

Tie force (a) Tie rod Angle section (b) Beam-to-column Shear force

Shear force

Secondary beam Tie force

Tie force

Angle section

Angle section

Main beam (d) Beam-to-beam

(c) Beam-to-column

Bolted ‘pin’ connections Shear force

Shear force

Plate welded to beam

Plate welded to column Tie force

Tie force

Weld

(f) Beam-to-column

(e) Beam-to-column

Axial force Shear force

Base plate welded to column Tie force Reinforced concrete foundation

Plate welded to end of beam (g) Beam-to-beam

(h) Column-to-foundation

Welded-bolted ‘pin’ connections

FIGURE 7.22 Examples of ‘pinned’ joints During erection the bottom cleat, which is bolted or welded to the column, is used as a marker by the crane operator when the beam is placed. The top cleat is assumed to resist no vertical load but it does provide torsional resistance which is important for lateral stability. The top and bottom cleats also resist the tie force. The resistance of

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) the web of the beam to shear, bearing and buckling must be checked. At ultimate load the rotation at the end of the beam often introduces end moments. In practice, these are assumed to be small and are ignored. Other types of ‘pinned’ beam-to-column joints are shown in Figs. 7.22(c), (e) and (f) where the depth of the joint is kept to a minimum. For example, the end plate depth for Fig. 7.22(e) is kept to a minimum to reduce end moments and the empirical thickness is 8 mm for UB sizes up to 457 × 191 kg, and 10 mm for sizes greater than 533 × 210 kg. Examples of ‘pin’ joints used in practice are given by Pillinger (1988). Where the depth of the connection for Fig. 7.22(e) is greater and is also slip resistant then advice on the distribution of forces to the bolts is given in Fig. 6.15, EN 1993-1-8 (2005).

7.9.2 ‘Pinned’ Beam-to-Beam Joints (Figs 7.22 (d) and (g)) The transverse secondary beam is connected to the main beam through angle cleats as shown in Fig. 7.22(d). It is assumed in design that the shear force is transferred to the main beam via the bolts in the web of the main beam. These bolts are therefore designed for single shear and bearing on the web of the main beam and on the angle cleats. It follows therefore that the shear force is eccentric to the bolts in the web of the secondary beam. These bolts are double shear and bearing on the web of the secondary beam and the angle cleats. Steel fabricators and erectors often prefer a welded and plate (Fig. 7.22 (g)) as an alternative to the angle cleats shown in Fig. 7.22 (d). This results in a more rigid joint and an end moment is introduced to the end of the secondary beam which is dependant on the torsional stiffness of the main beam. If there are secondary beams on both sides of the main beam the secondary moment can be large.

7.9.3 ‘Pinned’ Column-to-Foundation Joints (Fig. 7.22 (h)) The column is fastened to the base plate which is connected to the foundation by foundation bolts. This type of joint is used where the predominant force in the column is axial but there is generally a small shear force. The size of the foundation bolt is based on resisting the forces, but with a minimum size of M16. The thickness of the base plate is related to pressure beneath the base and the cantilever effect the plate.

7.10 ‘RIGID’ JOINTS ‘Rigid’ (or ‘fixed’) joints (Fig. 7.23) exhibit small rotational displacements in the elastic stage of behaviour. They are now more common as connections to members are welded in the workshops and bolted on site. They are useful to limit deflections of members,

Structural Design of Steelwork to EN 1993 and EN 1994

(a)

231

(c)

(b)

(d)

•

(e)

(f) RHS sections

(h) (g) Strap

d

M

Haunch

Ro Web stiffener

(i)

End plate

(j)

FIGURE 7.23 Examples of ‘rigid’ joints resist fatigue and resist impact loading. However, generally, the design procedure for the frame requires global analysis and the components are more highly stressed.

7.10.1 ‘Rigid’ Column Bracket Joints (Figs 7.23 (a), (b) and (c)) For the brackets shown in Figs 7.23 (a) and (c) no end bearing is involved whereas the bracket shown in Fig. 7.23 (b) involves end bearing.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

7.10.2 ‘Rigid’ Beam-to-Column Joints (Figs 7.23 (d) and (e)) The most rigid type of joint is where the beam is welded directly to the column on site, as shown in Fig. 7.23(d), but this is expensive and it is difficult to control the quality of the weld. Alternatively stub cantilever beams can be welded to the column in the workshops and a suspended beam site bolted between the ends of the cantilever. The bolted connection is positioned as close to the point of contraflexure as possible. The method commonly used in Britain is to weld end plates to the beam in the workshops and to bolt these to the columns on site as shown in Fig. 7.23(e). The numbers of bolts is usually six, as shown, because of the limited depth available. If the moment of resistance needs to be increased then it is necessary to increase the lever arm by haunching the beam at the end. This connection is less rigid than welding the beam directly to the column but it is easier to manufacture and erect. The amount of rotation depends on the thickness of the end plate, thickness of the column flanges and the extensibility of the bolts. Advice on details is given in cl 6.2.7.2, EN 1993-1-8 (2005). If the connection is close to a plastic hinge then it must be decided whether the plastic hinge should forming the beam, or the column, or the connection. Recent research favours the formation of the hinge in the beam and therefore the column and the connection must be overdesigned. End bearing occurs between beam and column and the first step in design is to check whether slip occurs using Eq. (7.31). Generally, because the bending moment is large, slip does not occur and the size of the top four bolts required can be determined approximately by taking moments of forces about the compression flange of the beam. The tensile force in a bolt Ft =

M 4(h − tf )

(7.34)

The tensile force in a bolt is increased by a prying force. Equation (7.34) assumes that the two bolts close to the compression flange of the beam do not resist any part of the bending moment. Lever arm recommendations are given in Fig. 6.15 (EN 19931-8 (2005)), The thicknesses of the end plate is determined by assuming equivalent T-stubs and associated yield lines. This method can be compared with a survey of existing literature on end plates by Mann and Morris (1979) who recommended ⎡ ⎢ ⎣

⎤1/2 M

dbf fy

4w sv

p

⎥ ⎦ dbf

+ s h

Mbp < tp < 2wp dbf fy

1/2

provided that Bp = 9db , sh = 6db , sv = 6db and eb > 2,5db .

(7.35)

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233

The thickness of the column flange can also be determined from equivalent T-stubs with yield line patterns. For comparison a further survey by Mann and Morris (1979) recommended: For unstiffened column flanges the thickness of the flange M 1/2 M 1/2 0,28 < tcf < 0,39 dbf fy dbf fy For stiffened column flanges the thickness of the flange M 1/2 M 1/2 0,23 < tcf < 0,32 dbf fy dbf fy

(7.36)

(7.37)

Equations (7.36) and (7.37) are valid provided that bc = 2,5db and cc = tst + 5db . Where the column flange thickness is inadequate backing plates can be used. The force Ro at the axis of rotation may produce failure by bearing or buckling of the web of the column. If failure is likely to occur then stiffeners can be welded into the web of the column. However, stiffeners increase costs and reduce the room available for bolts. The tensile force balancing Ro (Fig. 7.21) which acts on the web of the column is often not critical but the strength of the column web must be checked. The size of the fillet weld connecting the end plate to the beam is determined by assuming rotation about the axis O–O at the bottom flange of the beam, but an alternative more conservative method is to assume rotation about the centroidal axis of the weld.

7.10.3 ‘Rigid’ Beam Splices (Fig. 7.23(g)) Beam splices are introduced to extend standard bar lengths, or to facilitate construction and transport. Splices are generally located at sections where the forces are minimum, and to avoid local geometrical deformations of the structure pre-loaded bolts are used. The connection is usually made on site and therefore bolts are used. A simple design method is to assume that the entire shear force is resisted by the web splice and the flanges resist the entire bending moment. These assumptions are not correct but it simplifies the design and the errors are generally not large. Alternatively part of the moment is assumed to be resisted by the web. The bolts in the web, in double shear, are subject to an eccentric shear load. The bolts in the flange are also in double shear for two flange plates and the force on a bolt is obtained a simple moment equation F = M/(nd).

7.10.4 ‘Rigid’ Column-to-Column Joints (Fig. 7.23(f)) Column splices are used to extend standard bar lengths, to facilitate erection and transport, and for economy by reducing the section size. Where bending moments are large then, as with beam splices, pre-loaded bolts are used to maintain the axial line of

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) the column. Where sections change size steel packings and an end plate are required to ensure that a good fit is obtained. Generally the ends of the columns are in contact, or in contact with the end plate, and therefore end bearing occurs. Any shear force will be resisted by the friction at the end bearing and the web plates (or cleats). Where forces are small sizes of components are decided from experience, practicality and corrosion resistance.

7.10.5 ‘Rigid’ Hollow Section Joints (Fig. 7.23(h)) This is a typical ‘T’ joint for a Vierendeel girder where members of different width intersect. If forces and moments are not too large then the connection can be made without using stiffeners. Rectangular hollow sections are also used in braced triangulated trusses but it is difficult to arrange for the centre lines of the members to intersect at a point. The offsets introduce moments which should be taken into account in the analysis of the structure (Purkiss and Croxton, 1981). Further information on analysis and design of connections is given by Davies (1981). Circular hollow sections are also used for trusses but the geometry of the connection is more difficult and connections are more expensive to form. A review of methods of analysis of the strength of these connections is given by Stamenkovic and Sparrow (1981). The failure modes for hollow section joints are (cl 7.2.2, EN 1993-1-8 (2005)): (a) (b) (c) (d) (e) (f)

chord face failure, chord side wall failure, chord shear failure, punching shear failure, brace failure, local buckling.

The design resistance of numerous types of joints are shown in Tables 7.2–7.24, EN 1993-1-8 (2005). These take into account in plane and out of plane forces. Guidance on strength of welds for connecting members is given in cl 7.5.1, EN 1993-1-8 (2005). These are only applicable if they meet the joint parameters of Table 7.8, EN 1993-1-8 (2005).

7.10.6 ‘Rigid’ Column-to-Foundation Joints (Fig. 7.23(i)) Where bending moments are small a simple slab base is satisfactory with the bolts in line with the column axis. As the bending moment increases the bolts are off-set from the column axis. A built-up base is used where moments are large. The basic design method is based on ‘T’ stub theory (cl 6.2.8, EN 1993-1-8 (2005)).

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For a gussetted slab base subject to an axial load and bending moment it is assumed that at ultimate load the distribution of stresses beneath the steel base plate are as shown in Fig. 7.34(c). The depth of the compression zone is determined approximately by taking moments of forces about the tensile bolts " # M N + 2 d x= (7.38) bc fij where bc is based on the cantilever length (c) and fij is the allowable bearing pressure. Alternatively it can be assumed that the centroid of the compression zone is located under the column flange (Table 6.7, EN 1993-1-8 (2005)). From taking moments of forces about the compressive force the tensile force in a bolt is # " M N − 2 d Ft = (7.39) n The thickness of the base plate is found based on elastic bending of the cantilever length (c). For the built-up base the base plate thickness is determined by the same method. The gusset plate size is determined by the method shown in Section 7.5.3. The size of the welds connecting the gusset plate to the base and to the column can be determined assuming end bearing. Alternatively end bearing can be ignored and a larger size determined by assuming rotation about the centroid of the weld group.

7.10.7 ‘Rigid’ Knee Joint for a Portal Frame (Fig. 7.23(j)) This type of joint is similar to the beam-to-column connection shown in Fig. 7.23(e). However because portal frames are often designed using plastic analysis the magnitude of the reaction Ro = M/d is generally high and consequently the shear stresses in the web are close to the limit. The magnitude of Ro is reduced by haunching the beam and consequently increasing the distance d, but web stiffeners are generally required for the column. To reduce the shear deformation in the column web stiffeners can be introduced as described and tested by Morris and Newsome (1981). The tensile force which balances Ro can be resisted by a strap, or by a group of bolts through the flange of the column. The strap may interfere with the placing of purlins in the roof and the alternative group of bolts may cause distortion of the column flanges if the column flange thickness is small. The distortion can be controlled by the use of flange stiffeners, or increase the size of the column section, but both increase the cost.

EXAMPLE 7.2 Design of a ‘pin’ joint for a roof truss (Fig. 7.24). The forces, size of angles and tees have been obtained from an analysis at ultimate load assuming pin joints and axial forces in the members (cl 5.1.5(2), EN 1993-1-8 (2005)).

30 k N

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

310

310

bg

125 75 8 angles

60 60 6 angles

30

Gusset plate tg 10

kN

N

0k

10 28

16

330

15

24

t 6,1

215 kN

6 mm fillet welds

Strap

60

80

40

10,2

165 152 20 kg tee

10

165 152 20 kg tee 151,9

•

347

236

31 125 kN

Grade S275 steel M20 class 4.6 bolts

FIGURE 7.24 ‘Pinned’ roof truss joint The centroidal axes of the members intersect so there is no eccentricity to be taken into account (cls 2.7 and 3.10.3, EN 1993-1-8 (2005)). The thickness of the gusset plate is at least 6 mm to resist corrosion, and at least equal to the minimum thickness of the angle or tee (6,1 mm). Use a 10 mm thick plate grade S275 steel. A rectangular plate is simple to mark and cut, and low in fabrication cost. Alternatively a more complicated shape can be used which is aesthetically more acceptable but the fabrication cost is greater. Member 24, structural tee cut from a UB (165 × 152 × 20 kg) welded to a gusset plate, design force NEd = 215 kN. Tensile resistance of the tee Grade S275 steel (Eqs (6.6) and (6.7), EN 1993-1-1 (2005)) Npl,Rd =

Afy 2580 × 275 = γM0 1,00 × 1E3

= 709,5 > 215 kN satisfactory. Nu,Rd = 0,9Anet

fu 430 = 0,9 × (2580 − 2 × 22) × γM2 (1,25 × 1E3)

= 766,9 > 215 kN satisfactory. Assuming a 6 mm fillet weld for Grade S275 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = =

fu a 1/2 (3 βw γM2 ) 6 430 × 0,7 × 1E3

(31/2 × 0,85 × 1,25)

= 0,981 kN/mm

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237

Effective length of weld required to resist the tensile force using the simplified method (cl 4.5.3.3, EN 1993-1-8 (2005)) NEd 215 = 219,2 mm = Fw,Rd 0,981 Two side fillet welds of 110 mm length would be satisfactory but in practice the lengths would probably be the full overlap (i.e. 2 × 310 = 620 mm). Member 31, structural tee cut from a UB (165 × 152 × 20 kg) bolted to the gusset plate and strap, design force NEd = 125 kN. Resistance of 4-M20 class 4.6 bolts in single shear (Table 3.4, EN 1993-1-8 (2005)) fub Fv,Rd = nb 0,6As γM2 400 = 4 × 0,6 × 245 × = 188 > (NEd = 125) kN satisfactory. 1,25 × 1E3 Resistance of 2-M20 class 4.6 bolts in bearing on the web of the tee section (t = 6,1 mm) (Table 3.4, EN 1993-1-8 (2005)) k1 αb fub dt Fb,Rd = nb γM2 2,5 × 0,909 × 400 × 20 × 6,1 =2× 1,25 × 1E3 = 177,4 > (NEd = 125) kN satisfactory where for an end bolt αb = e1 /(3do ) = 60/(3 × 22) = 0,909 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 55/22 − 1,7 = 5,3 > 2, 5 The strap increases the out of plane stiffness of the truss. Connections for other members can be designed by the same method.

EXAMPLE 7.3

‘Rigid’ column bracket. Determine the size of the components required to connect the bracket to the column shown in Fig. 7.25 using Grade S355 steel. The forces shown are applied to one gusset plate at ultimate load. For the 10 bolts (Fig. 7.25(a)) of unit cross-sectional area the properties of the bolt group are: Second moment of area of the bolt group about the centroidal y–y axis Iy = (∂A)z2 = 4(802 + 1602 ) = 128E3 mm4

Second moment of area of the bolt group about the centroidal z–z axis Iz = (∂A)y2 = 10(70)2 = 49E3 mm4

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

eh 250

H 45 kN

y G

Gusset plate tg 10

Bg 196,1

z H g dw 400

e v 350

4 at 80 320

Bg 231

Gusset plate y

G y z

140 bf 255,9

z

ev 350

sg 150

sg

y

V 210 kN H 45 kN

z 40

tf 17,3

V 210 kN

hg 150

eh 250

254 254 89 kg UC

40

•

Hg 400

238

bw 200

tg 10 Lg 225

Lg 282,95 2 229 76 channels

Grade S 355 steel

(a) Bracket bolted to a UC

h 228,6 (b) Bracket welded to a compound column

FIGURE 7.25 ‘Rigid’ column brackets

Second moment of area of the bolt group about the centroidal polar x–x axis Ix = Iy + Iz = (128 + 49)E3 = 177E3 mm4 From Eq. (7.28) the maximum vector force in the direction of the z–z axis on a bolt furthest from the centroid of the bolt group V (Veh + Hev )yn + nb Ix 210 (210 × 250 + 45 × 350)70 = + = 48 kN 10 177E3

Fz =

The maximum vector force in the direction of the y–y axis on the same bolt H (Veh + Hev )zn + nb Ix 45 (210 × 250 + 45 × 350)160 = + = 66,2 kN 10 177E3

Fy =

Resultant vector design force on this bolt Fr = (Fz2 + Fy2 )1/2 = (482 + 66, 22 )1/2 = 81, 77 kN Solution (a) using class 4.6 bolts. Shear resistance of a M30 class 4.6 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub γM2

= 0,6 × 561 ×

400 = 107,7 > (Fr,Ed = 81, 77) kN satisfactory. 1,25 × 1E3

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239

However the recommended maximum bolt diameter for a column flange width of 254 mm is 24 mm (Annex A4). Use a higher class of bolt. Solution (b) using M20 class 8.8 bolts not pre-loaded. Shear resistance of an M20 class 8.8 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0, 6As

fub γM2

= 0,6 × 245 ×

800 = 94,1 > (Fr,Ed = 81,77) kN satisfactory. 1,25 × 1E3

M20 class 8.8 bolt in bearing on the gusset plate (t = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Bearing strength k1 αb fup dt γM2

Fb,Rd =

2,5 × 0,606 × 510 × 20 × 10 = 123,6 > (Fr,Ed = 81, 77) kN 1,25 × 1E3

= where

for an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 58/22 − 1,7 = 5,68 > 2,5 Solution (c) using pre-loaded M22 class 10.9 bolts (cl 3.9, Eqs (3.6) and (3.7), EN 1993-1-8 (2005)) Fs,Rd = =

ks nμb Fp,C γM3 303 1,0 × 1,0 × 0,5 × 0,7 × 1E3 × 1E3

1,25

= 84,8 > (Fr,Ed = 81,77) kN

To determine the thickness of the gusset plate for the bolted joint Fig. 7.25(a) (255,9 − 140) = 282,95 mm 2 (255,9 − 140) sg = 150 + = 207,95 mm 2

Lg = 225 +

Width of the gusset plate perpendicular to the free edge (Eq. (7.26)) Bg =

Lg 282,95 = 1/2 = 231,0 mm 1/2 Lg 2 282,95 2 +1 +1 400 H g

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) From Eq. (7.25), replacing the term (Pu sg ) with (Vsg + Hhg ), the thickness of the gusset plate Grade S355 steel tg =

2(VSg + Hhg ) Bg 2 + fgy Bg 80 γM1

=

2 × (210 × 207,95 + 45 × 150)E3 231 + 355×2312 80 1,0

= 8, 21 mm; use a 10 mm thick plate of Grade S355 steel Check the slenderness ratio of the gusset plate (Eq. (7.27)). lg Bg 231 = 2 × 31/2 = 2 × 31/2 × ig tg 10 = 80,02 < 185 the limit of the slenderness ratio for the application of the theory, satisfactory. Solution (d) using welds (Fig. 7.25(b)). Where it is not possible to bolt to a column, for example the compound channel column shown in Fig. 7.25(b), then welds are used. The connection is rigid and for welds of unit size the properties of the weld group are: Total length of weld Lw = 2(dw + bw ) = 2(400 + 200) = 1200 mm Second moment of area of the weld group about the centroidal y–y axis 2 dw d 3 w Iy = (∂A) z2 = 2 + bw 12 2 4003 400 2 =2 = 26, 67E6 mm4 + 200 12 2

Second moment of area of the weld group about the centroidal z–z axis 2 3 bw b Iz = (∂A)y2 = 2 w + dw 12 2 2003 200 2 =2 = 9, 33E6 mm4 + 400 12 2

Second moment of area of the weld group about the centroidal polar x–x axis Ix = Iy + Iz = (26, 67 + 9, 33)E6 = 36E6 mm4

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241

Maximum vector force in the direction of the z–z axis on a weld element furthest from the centroid of the weld group from Eq. (7.28). V (Veh + Hev )yn + Lw Ix 210 (210 × 250 + 45 × 350)100 = + = 0,365 kN/mm 1200 36E6

Fz =

Maximum vector force in the direction of the y–y axis on the same weld element H (Veh + Hev )zn + Lw Ix 45 (210 × 250 + 45 × 350)200 = + = 0,417 kN/mm 1200 36E6

Fy =

Resultant vector design force on this weld element Fr,Ed = (Fz2 + Fy2 )1/2 = (0, 3652 + 0, 4172 )1/2 = 0, 554 kN/mm Assuming a 6 mm fillet weld for Grade S275 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a 1

(3 2 βw γM2 )

=

6 430 × 0,7 × 1E3

(31/2 × 0,85 × 1,25)

= 0,981 < (Fr,Ed = 0, 554) kN/mm satisfactory. To determine the thickness of the gusset plate for the welded joint Fig. 7.25(b). From Eq. (7.26) the width of the gusset plate perpendicular to the free edge Bg =

Lg 225 = 1/2 = 196, 1 mm 1/2 Lg 2 225 2 +1 +1 H 400 g

From Eq. (7.25) replacing the term Pu sg by (Vsg + Hhg ) tg =

2(Vsg + Hhg ) Bg 2 + fgy Bg 80 γM1

=

2 × (210 × 150 + 45 × 150)E3 196, 1 + 355 80 2 × 196,1 1,0

= 8, 05 mm, use an 10 mm thick plate of Grade S355 steel. Check the slenderness ratio of the gusset plate from Eq. (7.27) lg 2 × 31/2 Bg 2 × 31/2 × 196, 1 = = 10 ig tg = 67, 93 < 185 the limit of the slenderness ratio for the application of this theory.

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) 203 203 86 UC

End shear force on beam V 225 kN Top angle cleat 80 80 10 mm angle supports no vertical load

Clearance 5 mm b 192,8 tw 11,4 r 10,2 tf 20,5 tw 13

h 467,4

242

tf 19,6

457 191 98 kg UB Tie force 75 kN Grade S 355 steel M20 class 8,8 bolts

Bottom cleat 125 75 10 mm angle support all vertical load

bf 208,8

FIGURE 7.26 ‘Pinned’ beam-to-column joint

EXAMPLE 7.4 ‘Pinned’ beam-to-column connection. Check the size of components for the connection shown in Fig. 7.26 at ultimate load. (a) If the design shear force of Fv,Ed = 225 kN is resisted by 4-M20 grade 8.8 bolts in the bottom cleat (125 × 75 × 10 mm angle), then the design shear force per bolt Fv,Ed =

225 = 56,25 kN 4

Shear resistance of an M20 grade 8.8 bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub γM2

= 0,6 × 245 ×

800 = 94,1 > (Fv,Ed = 56,25) kN satisfactory. 1,25 × 1E3

Bearing resistance of an M20 class 8.8 bolt bearing on the leg of the angle (ta = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd =

k1 αb fua dta 2,12 × 0,682 × 510 × 20 × 10 = γM2 1,25 × 1E3

= 118 > (Fv,Ed = 56,25) kN satisfactory. where for an end bolt αb = e1 /(3do ) = 45/(3 × 22) = 0,682 and for an edge bolt k1 = 2,8 (e2 /do ) − 1,7 = 2,8 × (30/22) − 1,7 = 2,12 < 2,5 Assume the 4-M20 class 8.8 bolts connecting the bottom cleat to the column which resist the shear force of 225 kN also resist the tensile force of 75 kN. Design tensile force per bolt from the 75 kN tie force is Ft,Ed =

75 = 18,75 kN 4

Tensile resistance of an M-20 class 8.8 bolt Ft,Rd = 0,9As

fu 800 = 0,9 × 245 × = 141,1 kN γM2 1,25 × 1E3

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243

Combined shear and tension for a bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Ed Ft,Ed + Fv,Rd (1,4Ft,Rd ) =

56,25 18,75 + = 0,693 < 1 satisfactory. 94,1 (1,4 × 141,1)

(b) Alternatively the four bolts in the vertical leg of the bottom angle could be replaced by fillet welds along the two vertical edges of the angle. Resistance of two 6 mm fillet welds for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) 2lw Fw,Rd = =

2lw fu a (31/2 βw γM2 ) 6 2 × 125 × 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 275 > (VEd = 225) kN.

The top cleat is used to provide torsional resistance against lateral buckling of the beam and to resist tie and erection forces. The angle must be at least 6 mm thick to resist corrosion and the leg of sufficient length to accommodate M20 bolts. From Section Tables a 80 × 80 × 10 mm angle is chosen. For resistance to transverse shear forces for the beam see Example 4.12

EXAMPLE 7.5 ‘Pinned’ beam-to-beam connection. Determine the size of the components required for the connection shown in Fig. 7.27. The beam sizes have been determined from bending calculations at ultimate load. Assuming that the M20 class 4.6 bolts through the web of the main beam B (Grade S275 steel) are subject to single shear forces. Shear resistance of a M20 class 4.6 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub 400 = 0,6 × 245 × = 47,0 kN γM2 1,25 × 1E3

M20 class 4.6 bolt in bearing on the web of the transverse beam A (tw = 6,9 mm) (Table 3.4 EN 1993-1-8 (2005)). For an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 For an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 30/22 − 1,7 = 2,12 < 2,5 k1 αb fub dtw γM2 2,12 × 0,606 × 400 × 20 × 6,9 = = 56,7 > (Fv,Rd = 47,0) kN. 1,25 × 1E3 VEd 150 nb = = = 3,19 use 4-M20 class 4.6 bolts. Fv,Rd 47,0

Fb,Rd =

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Transverse beam A 356 171 45 kg UB

z

y

a

Grade S275 steel M20 class 4,6 bolts

tw 6,9

40

y

h 352,0

b

3 70 210

160

65

40

bf 171,0

End shear force V 150 kN 38

158

tw 6,9

40 8 mm Clearance z tw 11,9

Side elevation

tf 9,7

tf 19,7

bf 304,8

65

•

h 609,6

244

70 70 10 angles

Main beam B 610 305 149 kg UB

End elevation

FIGURE 7.27 ‘Pinned’ beam-to-beam joint

Assuming that the bolts connecting the angle cleats to the web of the transverse beam A are in double shear and subject to an eccentric load. Second moments of area of the bolt group about the centroidal axis for bolts of unit area are Iy = (∂A)z2 = 2(352 + 1052 ) = 24,5E3 mm4 Iz = 0 Ix = Iy + Iz = 24,5E3 mm4 Maximum shear force on a bolt in the y direction from Eq. (7.28) Fy =

VEd ezmax 150 × 40 × 105 = = 25,71 kN Ix 24,5E3

Average shear force on a bolt in the z direction Fz =

VEd 150 = 37,5 kN = nb 4

Maximum resultant design shear force on a bolt Fr,Ed = (Fy2 + Fz2 )1/2 = (25,712 + 37,52 )1/2 = 45,47 kN Double shear strength of an M20 class 4.6 bolt 2Fv,Rd = 2 × 47,0 = 94,0 > (Fr,Ed = 45,47) kN, satisfactory.

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Check ‘block shear tearing’ (cl 3.10.2, EN 1993-1-8 (2005)) for line of holes in end of transverse beam A 1 Ant Anv + 1/2 fy γM2 γMO 3 1 1194 = 0 + 1/2 × 275 × 1,0 × 1E3 3 = 189,6 > (VEd = 150) kN satisfactory.

Veff,2,Rd = 0,5fu

which includes Anv = (3 × hole spacing + end distance − 3,5 × hole diameter)tw = (3 × 70 + 40 − 3,5 × 22) × 6,9 = 1194 mm2

EXAMPLE 7.6 ‘Pinned’ column-to-foundation connection. Determine the size of the components for the axially loaded base shown in Fig. 7.28 at the ultimate limit state. Concrete cylinder crushing strength fck = 20 MPa. N = 1500 kN

120

254 254 107 UC

6 mm fillet weld

M 20 class 4,6 bolts

Lp 450 C

C

C

t w 13 80

bf 258,3 C

Bp 42,5

h 266,7

tf 20,5

Grade S355 steel

FIGURE 7.28 ‘Pinned’ column to foundation joint

246

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Assuming that the bearing area is bounded by the dotted line shown previously in Fig. 7.17 and the pressure beneath base plate (cl 6.2.5(7), EN 1993-1-8 (2005)), N 2 = fj = fck A 3 Rearranging and inserting known numerical values, bearing area required A=

NEd 1500E3 NEd = = 112,5E3 mm2 = 2 2×20 fj f 3 ck 3

(i)

To determine the minimum thickness of the steel base plate the bearing area enclosed by the dotted line (Fig. 7.17) A = (bf + 2c)(h + 2c) − (h − 2tf − 2c)(bf − tw ) = (258,3 + 2c)(266,7 + 2c) − (266,7 − 2 × 20,5 − 2c)(258,3 − 13) = 13524,4 + 1540,6c + 4c2

(ii)

Equating Eqs (i) and (ii) the projection of the dotted area c = (192,572 + 24744)1/2 − 192,57 = 56,1 mm Check if areas overlap along the bolt line between flanges (h − 2tf ) (266,7 − 2 × 20,5) = 2 2 = 112,85 > (c = 56,1) mm satisfactory. Thickness of base plate Grade S355 steel (cl 6.2.5(4), EN 1993-1-8(2005))

3fj γM0 tp = c fy

1/2 = 56,1

3×

2 3

× 20 × 1 355

1/2

= 18,8 mm, use 20 mm thick base plate. Minimum length of base plate Dp = h + 2c = 266,7 + 2 × 56,1 = 322,8 mm Minimum breadth of base plate Bp = bf + 2c = 258,3 + 2 × 56,1 = 370,5 mm Use 450 × 425 × 20 mm base plate Grade S355 steel. If the end of the column is machined then the load is assumed to be transferred directly to the base plate and a minimum size of fillet weld of 6 mm is used to connect the base plate to the column.

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Alternatively if the end of the column is not machined then the force per unit length of weld is approximately Fw,Ed =

NEd 1500 = = 0,957 kN/mm (4bf +2h) (4 × 258,3 + 2 × 266,7)

Assuming a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a 1/2 (3 βw γM2 )

=

510 × 0,7 × 6/1E3 31/2 × 0,9 × 1,25)

= 1,1 > (Fw,Ed = 0,957) kN/mm satisfactor. The base plate is subject to a compressive force which is not transferred to the holding down bolts. The bolts are therefore subject only to erection forces and if these are not known then experience has shown that a bolt size approximately equal to the plate thickness is suitable. Use 2M20 class 4.6 holding down bolts. If there is a bending moment applied to the column and hence to the base plate then the bearing area is located beneath the column flange as shown in Fig. 6.4, EN 1993-1-8 (2005).

EXAMPLE 7.7

‘Rigid’ column bracket. Determine the size of fillet welds for the bracket shown in Fig. 7.29 at the ultimate limit-state.

N 500 kN z

B tw 19,2 r 15,2

g

52 6

tw 16,1 r 17,8

838 292 226 kg UB Ro

bf 293,8

dt 824,1

tf 31,4

100 tf 26,8

h 850,9

V 405 kN

y

Hg 797,3

305 305 198 kg UC e 600 bf 314,1

G

G

o

Lg 700

o

Grade S 355 steel d 246,7 h 339,9 Side elevation

FIGURE 7.29 ‘Rigid’ column bracket

End elevation

248

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) There are two possible solutions based on failure mechanisms (a) assuming rotation about axis G–G which is the simple traditional conservative method and (b) assuming rotation about axis O–O which is more correct but the calculations are more extensive. (a) Rotation about axis G–G The fillet weld is continuous round the bracket section as shown in Fig. 7.29. If there are no stiffeners in the web of the column then the strength of the weld around the flanges of the bracket is reduced because of the flexibility of the column flange. Effective length of the column flange weld (Eq. (4.6a), EN 1993-1-8(2005)) beff = tw + 2r + 7ktf = 19,2 + 2 × 15,2 + 7 × 1 × 31,4 = 269,4 mm which includes fy,f tf 31,4 355 k= = = 1,17 > 1 use 1. tp fy,p 26,8 355 Check if stiffeners required for column web (Eq. (4.7), EN 1993-1-8(2005)) fyp 355 bp = × 293,8 beff = fup 510 = 204,5 < (beff = 269,4) mm therefore no stiffeners required. Rotation about axis G–G The effective second moment of area of the weld group about axis G–G IG =

2dw3 + 4beff (df /2)2 12

2(850,9 − 2 × 26,8)3 (850,9 − 26,8) 2 = + 4 × 269,4 × 12 2 = (84,5 + 183,0)E6 = 267,5E6 mm4 Maximum force per unit length on weld in the y direction (Eq. (7.28)) d (Ve) 2f Fy = IG 405 × 600 × (850,9 − 26,8) = = 0,374 kN/mm (2 × 267,5E6) Maximum force per unit length of weld in the z direction Fz = V/Lw = =

V [4beff + 2(h − 2tf )] 405 = 0,152 kN/mm 4 × 269,4 + 2 × (850,9 − 2 × 26,8)

Maximum resultant design force per unit length of weld Fr,Ed = (Fy2 + Fz2 )1/2 = (0,3742 + 0,1522 )1/2 = 0,404 kN/mm

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For a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a (3

1/2

βw γM2 )

=

510 × 0,7 × 6/1E3 (31/2 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0,404) kN/mm satisfactory. Rotation about axis O–O The second moment of area of the weld group about axis O–O 2 Io = d3 + 2beff df2 3 w 2 = (850,9 − 2 × 26,8)3 + 2 × 269,4 × (850,9 − 26,8)2 3 = (337,9 + 365,9)E6 = 703,8E6 mm4 Maximum force per unit length of weld in the y direction Fy = (Ve)df /Io = (405 × 600) × (850,9 − 26,8)/703,8E6 = 0,285 kN/mm Effective length of weld resisting shear Leff = 4beff + 2dw = 4 × 269,4 + 2(850,9 − 2 × 26,8) = 2672 mm Distance (dr ) from the axis O–O to the resultant force in the weld is determined from equating the moments of the forces in the weld group about the axis O–O moment of the parts = moments of the whole (2beff + 0,5 × 2df )Fx dr = Fx Io /df Rearranging and putting Io = (2/3)df3 + 2beff df2 2b 2 + deff dr 3 f = 2beff df 1+ df

"

= "

2 3

2×269,4

+ 850,9−26,8 2×269,4

1 + 850,9−26,8

#

# = 0,798

Check whether slip occurs by substituting in Eq. (7.31) μs M 0,45 × 600 μs e = = (Vdr ) dr [0,798(850,9 − 26,8)] = 0,411 < 1, therefore slip occurs. Maximum force per unit length of weld in the z direction V μs R V μs (Ve/dr ) − = − Leff Leff Leff Leff # " 405×600 0,45 405 0,798(850,9−26,8) = − = 0,089 kN/mm 2672 2672

Fz =

250

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Maximum resultant design force per unit length of weld Fr,Ed = (Fy2 + Fz2 )1/2 = (0,2852 + 0,0892 )1/2 = 0,299 kN/mm For a 6 mm fillet weld and Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a 510 × 0,7 × 6/1E3 = 1/2 (31/2 βw γM2 ) (3 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0, 299) kN/mm satisfactory. An alternative method related to the European Code: Assume the applied vertical shear force is resisted by the two 6 mm web welds VR,Ed = 2df Fw,Rd = 2 × (850,9 − 2 × 26,8) × 1,1 = 1450 > (VEd = 405) kN satisfactory, and the applied bending moment is resisted by two 6 mm effective flange welds with rotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8(2005)) MR,Ed = 2beff df Fw,Rd = 2 × 269,4 × (850,9 − 26,8) × 1,1/1E3 = 488,4 > (MEd = 243) kNm satisfactory. Gusset plate design Check the thickness of the web of the 838 × 292 × 226 kg UB acting as a gusset plate. From Eq. (7.26) Bg =

Lg 700 = = 526,0 mm [(Lg /Hg )2 + 1]1/2 [(700/797,3)2 + 1]1/2

Required thickness of the web of the UB acting as a gusset plate (Eq. (7.25)) Bg 2Pu sg tg = 2 + fgy Bg 80 γM1

=

2 × 405E3 × 600 526 + 80 (355/1,0 × 526)2

= 11,52 < 16,1 mm (thickness of web of UB), satisfaactory. Check the slenderness ratio of the web of UB acting as a gusset plate (Eq. (7.27)) lg √ Bg √ 526 =2 3 =2 3× ig tg 16,1 = 113,2 < 185 limit of application of the theory, acceptable. Column web in transverse compression Reaction Ro may buckle or crush the web of the 305 × 305 × 198 kg UC.

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Reaction (Eq. (7.30)) Ro,Ed =

Ve 405 × 600 = = 369,5 kN dr [0,798(850,9 − 26,8)]

The design resistance of the unstiffened column web (Eq. (6.9), EN 1993-1-8(2005)) Fc,wc,Rd =

ωkwc beff,c,wc twe fywe γM0

0,764 × 1 × 271,7 × 19,2 × 355 1,0 × 1E3 = 1415 > (Ro,Ed = 369, 5) kN satisfactory. =

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,764 × 1 × 1 × 271,7 × 19,2 × 355 1,0 × 1E3 = 1415 > (Ro,Ed = 369,5) kN satisfactory. =

From (Eq. (6.10), EN 1993-1-8(2005)) for a welded connection beff,c,wc = tf b + 2 × 21/2 ab + 5(tfc + s) = 26,8 + 2 × 21/2 × 0,7 × 6 + 5 × (31,4 + 15,2) = 271,7 mm The maximum longitudinal stress in the flange of the column from the axial and eccentric loads (Eq. (6.14), EN 1993-1-8(2005)) (N + V ) V (e + h/2) + A Wel (500 + 405)E3 405E3(600 + 339, 9/2) = + 252E2 2993E3 = 140,1 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

σcom,Ed =

and (Table (6.3), EN 1993-1-8(2005)) β = 1(Table 5.4, EN 1993-1-8(2005)) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

2 1/2

Avc

1

=

1 + 1,3 × 271,7 ×

1/2 19,2 2 7032

= 0,764

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) where the shear area of the column (cl 6.2.6(3), EN 1993-1-1(2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 252E2 − 2 × 314,1 × 31,4 + (19,2 + 2 × 15,2) × 31,4 = 7032 mm2 ηhw tw = 1,0 × (339,9 − 2 × 31,4) × 19,2 = 5320 < (Avc = 7032) mm2 Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8(2005))

beff,c,wc dwc fy,wc 1/2 λp = 0,932 2 ) (Etwc 271,7 × 246,7 × 355 1/2 = 0,932 × (210E3 × 19,22 ) = 0,516 < 0,72 use ρ = 1,0 Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8(2005)) Vwp,Rd = 0,9

Avc fy,wc 7032 × 355 = 0,9 × 1/2 1/2 (3 γM0 ) 3 × 1 × 1E3

= 1297 > (Ro,Ed = 369,5) kN satisfactory.

EXAMPLE 7.8 ‘Rigid’ beam-to-column connection. Determine the size of the components for the connection shown in Fig. 7.30 at the ultimate limit state.

bf 208,5

20 45 35

Qbe Fbt

O

d 160,8

M 97,5 kNm

Ro

(a)

FIGURE 7.30 ‘Rigid’ beam-to-column joint

90

85 45

End plate 200 400 20 mm h 222,3

o

bf 166,8

9,1

r 10,2

200

V tw 7,7 60 kN

70

tw 13,0

h 310,9

bf 166,8

End plate o

35

30516554 UB

tf 20,5

O

d3 54,05 d2 254,05 d1 339,05

203 203 86 UC

tw 7,7

Grade S 355 steel M20 pre-loaded bolts

(b)

O

O

O

(c)

df 297,2

N 100 kN

d 265,6

•

t f 13,7

252

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253

Check for slip assuming rotation about axis O–O (Eq. (7.31)) μM 0,45 × 97,5E6 = = 2,46 > 1 [V (h − tf b ) [60E3 × (310,9 − 13,7)] therefore rotation about the compression flange of the beam at O without slip. Assuming rotation about axis O–O the design tensile force acting on a single bolt Ft,Ed =

M 97,5E6 = = 82 kN [4(h − tf b )] [4 × (310,9 − 13,7) × 1E3]

Design tensile resistance of an M20 pre-loaded class 8.8 bolt (Table 3.4, EN 1993-1-8 (2005)) assuming not subject to a shear force Ft,Rd = k2 As

fub 800 = 0,9 × 245 × γM2 1,25 × 1E3

= 141 > (Ft,Ed = 82) kN satisfactory. Shear resistance of an M20 class 8.8 pre-loaded bolt in single shear and subject to a tensile force of 82 kN (Eq. (3.8a), EN 1993-1-8 (2005)) Fs,Rd =

ks nμ (Fp,C − 0,8Ft,Ed ) γM3

1,0 × 1,0 × 0,5 (0,7 × 800 × 245 − 0,8 × 82E3) × 1,25 1E3 60 = 28,6 > Fs,Ed = = 10 kN satisfactory. 6

=

M20 class 8.8 bolt in bearing on end plate (t = 20 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dt γM2 2,5 × 0,682 × 510 × 20 × 20 = 278,2 > (Fs,Ed = 10) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 45/(3 × 22) = 0, 682 for an edge bold k1 = 2,8 (e2 /do ) − 1,7 = 2,8 × (55/22) − 1,7 = 5,3 > 2,5 Thickness of end plate related to a single bolt (Table 6.2, EN 1993-1-8 (2005), Method 1) ⎡ ⎤1/2 ⎢ 4Ft,Ed m ⎥ tf = ⎣ l f ⎦ eff y

γM0

4 × 82E3 × 35 = 100 × 355/1,0

1/2 = 18,0 mm; use 20 mm plate.

254

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8 (2005)) Fc,wc,Rd = =

ωkwc beff,c,wc twe fywe γM0 0,685 × 1 × 224,1 × 13 × 355 = 708,4 > 328 kN satisfactory. 1,0 × 1E3

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,685 × 1 × 1 × 224,1 × 13 × 355 1,0 × 1E3 = 708,4 > 328 kN satisfactory. =

where for a bolted end plate connection (Eq. (6.12), EN 1993-1-8 (2005)) beff,c,wc = tfb + 2 × 21/2 ap + 5(tfc + s) + sp = 13,7 + 2 × 21/2 × 0,7 × 6 + 5 × (20,5 + 15,2) + 20 = 224,1 mm The maximum longitudinal stress in the flange of the column from the axial load and eccentric load (Eq. (6.14), EN 1993-1-8 (2005))

σcom,Ed

60E3×223,2 Vh/2 (N + V ) (100 + 60)E3 2 + + = = A 110E2 851E3 Wel = 22,4 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

Table 5.4, EN 1993-1-8 (2005) β=1 Table 6.3, EN 1993-1-8 (2005) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

Avc

1

=

2 1/2

1 + 1,3 × 224,1 ×

1/2 13 2 3124

= 0,685

where (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 110E2 − 2 × 208,8 × 20,5 + (13 + 2 × 10,2) × 20,5 = 3124 mm2 ηhw tw = 1,0 × (222,3 − 2 × 20,5) × 13 = 2357 < (Avc = 3124) mm2

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Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

beff,c,wc dwe fy,wc 1/2 λp = 0,932 2 ) (Etwc 224,1 × 161,8 × 355 1/2 = 0,932 = 0,561 < 0,72 use ρ = 1,0 210E3 × 132 Reaction Ro = 4Ft = 4 × 82 = 328 kN Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8 (2005)) Vwp,Rd = 0,9

Avc fy,wc 0,9 × 3124 × 355 = 1/2 (31/2 γM0 ) (3 × 1 × 1E3)

= 576 > (Ro = 328) kN satisfactory. Alternative calculations for comparison Previous calculations assume the top four bolts resist all of the applied moment with a single lever arm. Alternatively assume a linear variation of forces from axis O–O to the bolts furthest from the axis. Maximum tensile force acting on the bolt furthest from axis O–O M Ft = 2 d1 + d2 + d3 =

97,5E3 = 75,3 kN [2(339,05 + 254,05 + 54,05)]

Prying force for each bolt assuming ∂b = 0 for a pre-loaded bolt and elastic behavior (Holmes and Martin, 1983). 2EI∂b Fbe − a b2 p p Qbe = a 2 2ap p 1 + 3 b b p

=

p

[90,62 − 0] 2 = 29,02 kN 2×45 1 × 45 + 35 3 35

and the maximum tensile force on a bolt Fbt + Qbe = 75,3 + 29,0 = 104,3 < (Ft,Rd = 141) kN satisfactory for no slip. For the welded connection between the end plate and the beam assume the applied vertical shear force is resisted by two 6 mm web welds VR,Ed = 2df Fw,Rd = 2 × (310,9 − 2 × 13,7) × 1,1 = 623,7 > (VEd = 60) kN satisfactory.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Assume that the applied bending moment is resisted by two 10 mm effective flange welds with rotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8 (2005)) MR,Ed = 2beff df Fw,Rd = 2 × 91,2 × (310,9 − 13,7) ×

1,83 1E3

= 99,2 > (MEd = 97,5) kNm satisfactory (cl 6.2.3(4), EN 1993-1-8 (2005)). For strength of welds (Fw,Rd ) see Annex 1 and for beff see Eq. (4.6a), EN 1993-1-8 (2005).

EXAMPLE 7.9 ‘Rigid’ beam-to-beam joint. Determine the size of the components for the rigid beam-to-beam joint shown in Fig. 7.31 at the ultimate limit state. Assuming rotation about axis O–O and the applied moment of 97,5 kNm is resisted entirely by the cover plate then the design tensile force in the flange cover plate Ff =

M 97,5E6 = = 320,9 kN. h 303,8 × 1E3

Thickness of flange connection plate, bp = 165 and 20 mm pre-loaded bolts Ff

tp = "

(bp − 2dh )

fy γM1

320,9E3

# = "

(165 − 2 × 22) ×

355 1,0

#

= 7, 47 mm, use 8 mm thick Grade S355 steel plate.

V 45 kN

40 60 60 40

75 6 mm fillet weld

tf 10,2

tp 10

R0 O

tw 10,7 10 mm thick end plate

tf 18,9

30516540 kg UB bf 165,1 tw 6,1

h 303,8

Connecting plate

tp 8

M 97,5 kNm

Grade S 355 steel M20 pre-loaded bolts

30

•

h 465,1

256

Alternative connections bf 153,5 457 152 82 kg UB

(a) Welded end plate connection

FIGURE 7.31 ‘Rigid’ beam-to-beam joint

(b) With bottom cleats

(c) With web cleats

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257

Single shear resistance of M20 class 8.8 pre-loaded bolt (Table 3.4, EN 1993-1-8 (2005)) Fs,Rd = =

ks nμ0,7Fp,C As γM2 245 1,0 × 1, 0 × 0,5 × 0,7 × 800 × 1E3

1,25

= 54,9 kN

M20 class 8.8 bolt in bearing on plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dt γM2 2,34 × 0,606 × 510 × 20 × 8 = 92,6 > (Fs,Rd = 54,9) kN 1,25 × 1E3

for an end bold αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8(e2 /do ) − 1,7 = 2,8 × 31,75/22 − 1,7 = 2,34 < 2,5 Number of bolts required for the connecting plate nb =

Fs,Ed 320,9 = = 5,85 use 6-M20 class 8.8 pre-loaded bolts. Fs,Rd 54,9

Reaction at the hinge is equal to the force in the flange, Ro = Ff = 320,9 kN and the frictional resistance at the hinge μRo = 0,45 × 320,9 = 144,4 > (VEd = 45) kN therefore no slip occurs. Use a 10 mm thick end plate welded to the end of the 305 × 165 × 40 kg UB and bolted to the 457 × 152 × 82 kg UB as shown in Fig. 7.31(a). Shear resistance of 4-M20 pre-loaded bolts in double shear in the end plate = 2 × 4 × 54,9 = 439,2 > (VEd = 90) kN satisfactory. Shear resistance of two 6 mm fillet welds connecting the end plate to the 305 × 165 × 40 kg UB using grade S355 steel Fw,Rd = =

fu a (3

1/2

βw γM2 )

6 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,10 kN/mm

2lw Fw,Rd = 2 × (303,8 − 30 − 10,2) × 1,10 = 580 > (VEd = 45) kN satisfactory.

EXAMPLE 7.10

‘Rigid’ beam splice. Determine the size of components for the beam splice shown in Fig. 7.32 at the ultimate limit state.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) V 600 kN M 450 kNm

bf 209,3

40 80 80 40

140

2

Web plate 10 mm thick

533 210 92 kg UB tw 10,2

h 533,1

tf 15,6 8 8

•

35 35

z (a) Elevation 40 80 80

y

40 80 80 80 80 40

258

Grade S355 steel

Section

V 600 kN Mw 87,4 kN y

M20 pre-loaded bolts e 121 z (b) Web connection resisting shear force and web bending moment

FIGURE 7.32 ‘Rigid’ beam splice

Check if the beam is in the elastic stage of behaviour f =

M 450E6 = = 216, 8 < ( fy = 355) MPa, therefore elastic behaviour. We 2076E3

Second moment of area of the web of the beam Iweb =

tw (h − 2tf )3 10,2 × (533,1 − 2 × 15,6)3 = = 107, 5E6 mm4 12 12

From Section Tables the gross second moment of area of the beam section Igross = 553,5E6 mm4 Assumed proportion of the applied bending moment taken by the web 107,5E6 Iweb Mweb = × 450 = 87,39 kNm M= Igross 553,5E6 Check the strength of the arrangement of bolts in shear in the web plate (Fig. 7.32(b))

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259

Second moment of area of bolts of unit area about the centroidal y–y axis Iy = (∂A)z2 = 6 × (802 + 1602 ) = 192E3 mm4 Second moment of area of bolts of unit area about the centroidal z–z axis Iz = (∂A)y2 = 10 × 802 = 64E3 mm4 Second moment of area of bolts of unit area about the centroidal polar x–x axis Ix = Iy + Iz = (192 + 64)E3 = 256E3 mm4 Eccentricity of the applied shear force relative to the centroid of half the bolt group is 121 mm (Fig. 7.32(b)). This eccentricity produces a moment which is increased by the bending moment resisted by the web. The equivalent eccentricity e = e +

Mweb 87,4E3 = 121 + = 266,6 mm V 600

Maximum vector shear force in the y–y direction acting on a bolt furthest from the centroid of the web bolt group Fy = Ve

zn 160 = 600 × 266,6 × = 99,99 kN Ix 256E3

Maximum vector shear force in the z–z direction acting on the same bolt Fz =

V yn 600 80 + Ve = + 600 × 266,6 × = 90,0 kN n Ix 15 256E3

Resultant maximum vector force acting on the same bolt Fr = [Fy2 + Fz2 ]1/2 = [99,992 + 90,02 ]1/2 = 134,5 kN Double shear strength of an M20 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5) in the web (cl 3.9.1, EN 1993-1-8(2005)) Fv,Rd =

1, 0 × 2, 0 × 0,5 × 245 × 0,7 × 1000 ks nμFp,C 1E3 = 1,25 γM3

= 137,2 > (Fr = 134,5) kN, satisfactory. M20 class 8.8 bolt in bearing on the web (t = 10,2 mm) (Table 3.4, EN 1993-1-8(2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,606 × 510 × 20 × 10,2 = 126,1 < (Fv,Rd = 137,2) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 106,5/22 − 1,7 = 11,9 > 2,5

260

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Required number of M20 pre-loaded bolts in bearing for the flange splice # " (M−Mweb ) Ff (h−tf ) nb = = Fb,Rd Fb,Rd " # =

(450−87,4)E3 (533,1−15,6)

126,1

= 5,56 use 6 bolts.

Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005)) (Lj − 15d) (2 × 80 − 15 × 20) =1− 200d 200 × 20 = 1,035 use 1,0

βLf = 1 −

Thickness of the outer and inner flange cover plates Ff tp = (b −2d +2w −2d )f f

p

h

h y

γM1

=

700,7E3

= 7,55 mm (209,3 − 2 × 22 + 2 × 70 − 2 × 22) × 355 1,0

Use 8 mm plates Grade S355 steel as shown in Fig. 7.32.

EXAMPLE 7.11 ‘Rigid’ column splice. Determine the size of the components for the rigid column splice shown in Fig. 7.33 at the ultimate limit state. Where column sections are of the same serial size it is possible to connect them directly with web and flange plates. The ends of the column are machined and will be in contact. Rotation will take place about an axis near the outer edge of the flange of the upper column. Thickness of the flange plate, from moments of forces about the axis of rotation Nh M − 2u tp = " f # y (bp − 2dh ) γ hu M1 # " 712,5E3×355,6 480E6 − 2 # = " 355 (365 − 2 × 24) 1,0 × 355,6 = 8,83 mm, use 10 mm Grade S355 steel plate. Single shear strength of an M22 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5) in the flange (cl 3.9.1, EN 1993-1-8 (2005)) Fv,Rd = =

ks nμFp,C γM2 1,0 × 1,0 × 0,5 × 303 × 0,7 × 1000 1E3 1,25

= 84,8 kN.

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261

N 712,5 kN M 480 kNm 356 368 129 kg UC bf 368,3

Packing

hu 355,6 Packing tf 17,5

Machined plate

tw 10,7 Angle

Flange plate

960 365 10 Flange plate 40

140 80

(b) Column sections of different serial size

140

40 80 80 80 80 80

V 150 kN

Web plate thickness 10 mm

40 150 40 Packing tw 16,8

Welded end plates

tf 27 Grade S355 steel M22 pre-loaded bolts

hL 374,7 356 368 202 kg UC bf 374,4

(a) Column sections of the same serial size

(c) Welded and plate column connection

FIGURE 7.33 ‘Rigid’ splices in steel columns

M22 class 10,9 bolt in bearing on plate (t = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,555 × 510 × 22 × 10 = 124,5 > (Fv,Rd = 84,8)kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 40/(3 × 24) = 0,555 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 108/24 − 1,7 = 10,9 > 2,5

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Number of bolts required # " M − Nh2 u nb = (Fv,Rd hu ) " # 712,5E3×355,6 480E6 − 2 = = 11,7 use 12-M22 bolts. 84,8E3 × 355,6 Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005)) (Lj − 15d) (5 × 80 − 15 × 22) =1− 200d 200 × 22 = 0, 984. This factor does not affect the number of bolts required.

βLf = 1 −

Where the ends of the column are machined and in contact the horizontal shear force on the column is resisted by the friction force, in part or whole, at the point of contact, that is, at the axis of rotation. Assuming machined surfaces μ = 0,15 the frictional resistance M N =μ + hu 2 480E3 712,5 = 0,15 × + = 255,9 > 150 kN (applied shear force). 355,6 2 Theoretically no shear connection required but in practice a web plate is generally provided to align the webs. If the frictional resistance is ignored then the web splice is designed to resist the entire shear force as follows: Second moments of area of two bolts of unit area on one side of the web connection about the centroidal axes are: Iy = 0 Iz = 2 × 752 = 11,25E3 mm4 Ix = Iy + Iz = 11,25E3 mm4 Vector force on a bolt in the y–y direction Fx =

V 150 = = 75 kN nb 2

Vector force on a bolt furthest from the centre of rotation in the z–z direction Fz = (Ve)

yn 75 = 150 × 40 × = 40 kN lx 11,25E3

Maximum vector shear force on the same bolt Fr = (Fy2 + Fz2 )1/2 = (752 + 402 )1/2 = 85 < (Fb,Rd = 124,5) kN, acceptable.

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263

N 198 kN M 264 kNm

35

fc = 20

Bp 460

H 49,5 kN M30 class 4,6 bolts

eb 70

113,1 C 104,4

300

Gusset plate tg 12,5

70

203 203 86 UC bf 208,8; h 222,3

Grade S355 steel

Dp 800

(a)

(b)

Lq 278,9

Bg

y

y G

Dp 800 G

Dw 300

Hg 300

h 222,3 z

G

z

2/3 20

xp 83,6

o (d) Column-to-gusset plate welds

o

o

(e) Base plate-to-gusset plate welds

Fg Sg 247,1 (c)

FIGURE 7.34 ‘Rigid’ column-to-foundation joint

EXAMPLE 7.12 ‘Rigid’ built-up column base connection. Determine the size of the components for the connection shown in Fig. 7.34 at the ultimate limit state assuming fc = 20 MPa. Tensile strength of an M30 class 4.6 holding down bolt (Table 3.4, EN 1993-1-8 (2005)) Ft,Rd =

0,9As fub 0,9 × 561 × 400 = = 161,6 kN γM2 1,25 × 1E3

Distance required between holding down bolts (Eq. (7.39)) dp = " N 2

M + nFt,Rd

#=

264E3 198 2

+ 2 × 161,6

= 625,3 mm

The size of the base plate is determined as follows: Assume a bolt edge distance of approximately 2d = 2 × 30 = 60 mm use 70 mm.

264

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Total length of base plate Dp = dp + 2eb = 625,3 + 2 × 70 = 765,3 mm, use length of 800 mm Minimum width of base plate Bp = bf + 2tg + washer + 2 × welds + 2eb = 208,8 + 2 × 12,5 + 66 + 2 × 10 + 2 × 70 = 459,8 mm. Use width of 460 mm. The thickness of gusset plate(tg ) and size of welds are assumed at this stage. Assume the projection length for the base plate is half the width of the column (Fig. 7.34(b)) c=

bf 208,8 = = 104,4 mm 2 2

Thickness of base plate Grade S355 steel (Eq. (6.5), EN 1993-1-8 (2005)) 1/2 3 × 23 × 20 × 1,0 3fjd γM0 1/2 tp = c = 104,4 = 35 mm fy 355 Length of concrete compression zone beneath the steel baseplate (Eq. (7.38)) assuming lever arm la = 660 mm M N + la 2 xp = (bf + 2tg + 2c)fjd 264E6 198E3 + 660 2 = [(208,8 + 2 × 15 + 2 × 104,4) × 23 × 20] = 83,6 mm Lever arm for resistance of concrete in bending at ultimate limit state la = Dp − eb −

xp 83,6 = 800 − 70 − = 688,2 mm 2 2

Tensile force in a holding down bolt (Eq. (7.39)) D Xp p M −N 2 − 2 Fbt = n la b 83,6 264E3 − 198 × 800 − 2 2 = = 140,3 < (Ft,Rd = 161,6) kN 2 × 688,2 Use 800 × 460 × 35 mm base plate Grade S355 steel. Force from bearing pressure applied to each gusset plate (Fig. 7.34(c)) Bj 2 460 83,6 Fg = fjd xp = × 20 × × = 256,4 kN 2 3 2 1E3

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265

Length of gusset plate allowing for 10 mm for weld Lg =

(Dp − 2 × 10 − h) (800 − 20 − 222,3) = = 278,9 mm 2 2

Assume height of gusset plate Hg = 300 mm Eccentricity of force Fg in relation to the inner corner of the gusset plate (Fig. 7.34(c)) sg =

(Dp − h) (800 − 222,3) − xp /2 = − 83,6/2 = 247,1 mm 2 2

Width of gusset plate (Eq. (7.26)) Lg L 2 1/2 =

Bg =

g

1+ H g

1+

278,9 2 1/2 = 204,2 mm 278,9 300

Thickness of gusset plate Grade S355 steel (Eq. (7.25)) 2Fg sg Bg tg = f + yg 80 B2 γM1 g =

2 × 256,4E3 × 247,1 204,2 + = 11,1 mm 355 80 2 × 204,2 1,0

Use 12,5 mm thick Grade S355 steel gusset plate. Check slenderness ratio of gusset plate (Eq. (7.27)) √ √ lg 2 3Bg 2 3 × 204,2 = = = 56,6 < 185, satisfactory. ig tg 12,5 Minimum length of foundation bolt (Holmes and Martin, 1983) (cl 6.2.6.12, EN 1993-1-8 (2005)) Fbt 161,6E3 Lb = = = 186,8 mm 2/3 πftc π × 0,3 × 20 1,5

Use 4-M30 class 4.6 holding down bolts, 300 mm long anchored by washer plates. The size of the fillet weld connecting the base plate to the gusset assuming no friction is obtained as follows: Length of weld (Fig. 7.34(e)) Lw = 2Dp = 2 × 800 = 1,6E3 mm Second moment of area about centroid of the weld group for unit size weld IwG =

2D3p 12

=

2 × 8003 = 85,33E6 mm4 12

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Maximum vertical force compressive per unit length weld in the z direction D p M N 2 Fwz = + Iwg Lw 264E3 × 800 198 2 = + = 1,36 kN/mm 1,6E3 85,33E6 Horizontal force per unit length of weld in the x direction Fwx =

H 49,5 = = 0,0309 KN/mm Lw 1,6E3

Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwz + Fwx ) = (1,362 + 0,03092 )1/2 = 1,362 kN/mm

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (31/2 βw γM2 )

=

8 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,362) kN/mm Alternatively if surface between the base plate and edge of the gusset plate are machined and bearing is assumed at the axis of rotation O–O (Fig. 7.34(e)). From Eq. (7.30) ⎛ Ro =

⎞

NDp +M ⎝ 6 ⎠ 2Dp 3

198×800 =

6

+ 264E3

2×800 3

= 544,5 kN Frictional resistance at Ro = μRo = 0,15 × 544,5 = 81,65 > (H = 49,5) kN, satisfactory. Second moment of area about axis O–O for unit size weld (Fig. 7.34(e)) IwO =

2D3p 3

=

2 × 8003 = 341,3E6 mm4 3

Maximum vertical force per unit length of weld in the z direction ND

Fwz =

M − 2w 264E3 − 198×800 2 = = 0, 542 kN/mm IwO 341,3E3

Assuming a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

6 510 × 0,7 × 1E3 fu a = (31/2 βw γM2 ) (31/2 × 0,9 × 1, 25)

= 1,10 > (Fwz = 0,542) kN/mm

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267

If the end of the column is not machined then rotation is assumed to be about axis G–G and the size of the weld connecting the column to the gusset plate is obtained as follows. Length of weld (Fig. 7.34(d)) Lw = 4Dw = 4 × 300 = 1,2E3 mm Second moments of area about the centroid of the weld group (axis G–G) fo unit size welds 4D3w 4 × 3003 IwGy = = = 9E6 mm4 12 12 2 h 222,3 2 IwGz = 4Dw = 4 × 300 = 14,83E6 mm4 2 2 Polar second moment of area IwGx = IwGy + IwGz = (9,0 + 14,83)E6 = 23,83E6 mm4 Maximum vertical force per unit length of weld in the z direction on an element furthest from the axis of rotation h N HDw 2 Fwz = + M− Lw 2 IwGx 222,3 198 49,5 × 300 2 = + 264E3 − × 1,2E3 2 23,83E6 = 1, 362 kN/mm Horizontal force per unit length of weld in the y direction on the same element Dw H HDw 2 Fwy = + M− Lw 2 IwGx 300 49,5 49,5 × 300 2 = + 264E3 − × 1,2E3 2 23,83E6 = 1,656 kN/mm Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwy + Fwz ) = (1,6562 + 1,3622 )1/2 = 2,144 kN/mm

Assuming a 12 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (301/2 βw γM2 )

=

12 510 × 0, 7 × 1E3

(31/2 × 0, 9 × 1, 25)

= 2, 20 > (Fwr = 2, 144) kN/mm Alternatively if the contact between the base plate and the end of the column is machined then the axis of rotation is O–O (Fig. 7.34(d)).

268

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Second moments of area about centroid of the weld group for unit size weld about axis O–O IwOy =

4 × 3003 4D3w = = 36E6 mm4 3 3

Iwbz = 2Dw h2 = 2 × 300 × 222,32 = 29, 65E6 mm4 Polar second moment of area IwOx = IwOy + IwOz = (36 + 29, 65)E6 = 65,65E6 mm4 Force per unit length of weld in the z direction on an element furthest from the axis of rotation h M − Nh 2 Fwz = IwOx 222,3 222,3 = 264E3 − 198 × × = 0, 819 kN/mm 2 65,65E6 Force per unit length of weld in the y direction on the same element M − Nh Dw 2 Fwy = IwOx 222,3 300 = 264E3 − 198 × × = 1,106 kN/mm 2 65,65E6 Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwy + Fwz ) = (1,1062 + 0,8192 )1/2 = 1,376 kN/mm

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd

8 510 × 0,7 × 1E3 fu a = 1/2 = 1/2 (3 βw γM2 ) (3 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,376) kN/mm Check shear resistance beneath the base plate (cl 6.2.2(8), Eq. (6.3), EN 1993-18(2005)) F1,v,Rd = 107,7 kN (see Annex A2) αb = 0,44 − 0,0003fyb = 044 − 0,0003 × 240 = 0,368 As 561 = 66,1 < 107,7 kN = 0,368 × 400 × γMb 1,25 = 0,2N + nb F1,v,Rd = 0,2 × 198 + 4 × 66,1

F2,v,Rd = αb fub Fv,Rd

= 304 > (H = 49,5) kN satisfactory.

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269

b1 150

h1 250 250 150 10 RHS

N 66,75 kN M 85 kNm

V 40 kN

Chord face

•

t1 10

8 mm fillet weld

o h0 300

R0 300 200 10 RHS Side wall

t0 10

b0 200

Side elevation

End elevation

150

o Plan of weld o 250

R

Assumed stress distribution

dwr

FIGURE 7.35 ‘Rigid’ RHS joint

EXAMPLE 7.13 ‘Rigid’ RHS connection. Check the strength of the rigid rectangular hollow section connection at the ultimate limit state assuming Grade S355 steel (Fig. 7.35). Check validity of the joint (Table 7.8, EN 1993-1-8 (2005)) b1 150 = = 15 < 35 satisfactory. t1 10 h1 250 = = 25 < 35 satisfactory. t1 10 Check for chord face failure of the horizontal member (Table 7.14, EN 1993-1-8 (2005)). " # η 1 2 + + 1/2 (2η) (1−β) (1−β) Mip,1,Rd = kn fyo to2 h1 γM5 5/3 1 2 (2×5/3) + (1−0,75)1/2 + (1−0,75) = 1,0 × 355 × 102 × 250 × (1,0 × 1E6) = 97,3 > (MEd = 85) kNm satisfactory.

270

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Included in the previous calculations b1 150 = = 0,75 < 0,85 b0 200 kn = 1,3 − 0,4n/β 0,282 = 1,3 − 0,4 × 0,75 = 1,15 > 1 use 1 h1 5 250 η= = = b0 150 3 β=

If the weld group is assumed to rotate about the axis O–O (Fig. 7.35) then the effective width of the weld (Table 7.13, EN 1993-1-8 (2005)) furthest from the axis O–O fy0 t0 b1 # beff = 10 " b0 fy1 t1 t 0

355 × 10 × 150 # = 75 mm = 10 × " 200 × 355 × 10 10 beff 75 = = 0,5 b1 150 Distance(dwr ) from the axis O–O from the resultant force in the weld is obtained as follows moment of the parts = moment of the whole 1 Fz Ioe beff + × 2dw Fz dwr = 2 dwr rearranging and substituting Ioe = 2dw3 /3 + beff dw2 beff 2 + dwr 3 dw = beff dw 1+ d w 2 75 + 3 250 = 0,744 = 75 1 + 250 Resultant reaction from Eq. (7.30) Ro = =

M + N(dwr − dw /2) dwr # " 85E3 + 66,75 × 0,744 × 250 − 250 2 0,744 × 250

= 478,9 kN

Frictional force at the stiff bearing if the end of the vertical RHS is machined μRo = 0, 15 × 478, 9 = 71, 8 > 40 kN (applied shear force). The weld group is subject to the actions from N and M acting about axis O–O.

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271

Second moment of area about axis O–O of the weld group (Fig. 7.35) for unit size welds IO =

2dw3 2 × 2503 + beff dw2 = + 75 × 2502 = 15,1E6 mm4 3 3

Moment applied about axis O–O M = M −

Ndw 250 = 85 − 66,75 × = 76,65 kNm 2 2 × 1E3

Maximum tensile force per unit length of weld furthest from the axis of rotation

Fw =

Md 76,65E3 × 250 = = 1,27 kN/mm IO 15,1E6

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (31/2 βw γM2 )

=

8 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,47 > (Fwy = 1,27) kN/mm An alternative more conservative calculation to determine the size of weld is to assume rotation about the centroid of the weld group. This results in a larger weld size. For interest check for side wall crushing in horizontal member at O–O (Table 7.14, EN 1993-1-8 (2005)) Mip,1,Rd = =

0,5fyk t0 (h1 + 5t0 )2 γM5 0,5 × 355 × 10 × (300 + 5 × 10)2 = 217,4 > Mip,1,Rd = 155 kNm. (1,0 × 1E6)

This calculation is not necessary because the ratio β = b1 /b0 = 150/200 = 0,75 < 0,85 indicates that it is not critical.

EXAMPLE 7.14 ‘Rigid’ knee connection for a portal frame. Check the strength of the knee joint components for the frame (Fig. 7.36) at the ultimate limit state. Moment acting about axis O–O MO = 250,2 − 116,8 × 0,5379 = 187,4 kNm Thickness of strap assuming 190 mm wide and 22 mm diameter hole MO tst = l b f a st y,st γM1

187,4E6

= "

717,2 × cos 10◦ × (190 − 2 × 22) × 355 1,0

= 5,12 mm. Use a 8 mm thick strap.

#

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) V 174,4 kN Strap welded to column t 8

100 x

45719167 kg UB bf 189,9, h 453,6 d 407,9

tw 8,5 r 10,2

Grade S 355 steel M20 pre-loaded bolts 6 mm fillet welds

O

50

13

H 116,8 kN tw 8

10°

Separate bolted strap

M 250,2 kNm 717,2

•

537,9

272

Cap plate t 12,5

R0 end plate t 12,5 tf 12,7

Web stiffener ts 8 (a)

(b)

FIGURE 7.36 ‘Rigid’ knee joint for a portal frame

Strap welded to the top of the column and bolted to the rafter (Fig. 7.36(a)). For a single M20 pre-loaded bolt slip resistance (Eq. (3.6), EN 1993-1-8 (2005)) Fs,Rd = ks n μfub

As γM3

= 1,0 × 1,0 × 0,4 × 0,7 × 800 ×

245 = 43,9 kN 1,25 × 1E3

M20 class 8.8 bolt in bearing on the plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,758 × 510 × 20 × 8 = 123,7 > (Fs,Rd = 43,9) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 50/(3 × 22) = 0,758 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 50/22 − 1,7 = 4,66 > 2,5 Number of M20 pre-loaded bolts in single shear for the strap nb =

MO 187,4E3 = = 6,04 use 6 bolts. (la Fs,Rd ) 717,2 × cos 10◦ × 43,9

Force per unit length acting on the two welds connecting strap to head of column Fw,Ed = =

MO la n w l w 187,4E3 717,2 × cos 10◦ × 2 × cos 10◦ 407,9

= 0,320 kN/mm

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273

Shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) Fw,Rd = =

fu a (3

1/2

(31/2

βw γM2 ) 510 × 0,7 × 6 = 1,1 > (Fw,Ed = 0,320) kN/mm × 0,9 × 1,25 × 1E3)

These calculations assume, conservatively, that the force is resisted only by the two welds along the web of the column, but the strap is also welded to the flanges of the column which increases the strength. Moments of forces about X to determine the reaction Ro at O −717, 2Ro + 250, 2E3 + 116, 8 × (717, 2 − 537, 9) = 0;

hence Ro = 378, 1 kN

Vertical frictional force at O = μRo = 0, 4 × 378, 1 = 151, 2 < (VEd = 174, 4) kN, therefore slip occurs. Bolt end plate (tep = 12,5 mm) to the flange of the column using M20 pre-loaded bolts (fpu = 800 MPa, μ = 0,4). For a single bolt slip resistance (Eq. (3.6), EN 1993-1-8 (2005)). Fs,Rd = 43,9 kN as calculated previously. Number of bolts required nb =

VEd 174,4 = = 3,97 use 4 bolts. Fv,Rd 43,9

Shear strength of the web of the column (cl 6.2.6.1 (2), EN 1993-1-8 (2005)) f y,wc

Vpl,Rd = 0,9Avc

γM0

31/2

= 0,9 × 4094 ×

355 1,0

(31/2 × 1E3)

= 755,2 > (Ro = 378,1) kN, satisfactory. Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8(2005)) Fc,wc,Rd =

ωkwc beff,c,wc twe fywe γM0

0,932 × 1 × 0,731 × 164,4 × 8,5 × 355 1,0 × 1E3 = 338,0 < (Ro = 378,1) kN not satisfactory. =

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,932 × 1 × 0,731 × 164,4 × 8,5 × 355 1,00 × 1E3 = 338,0 < (Ro = 378,1) kN not satisfactory stiffeners required. =

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Previous calculations include the effective width (Eq. (6.11), EN 1993-1-8 (2005)) beff,c,wc = tfb + 2 × 21/2 ap + 5(tfc + s) + sp = 13 + 2 × 21/2 × 0,7 × 6 + 5 × (12,7 + 10,2) + 2 × 12,5 = 164,4 mm Maximum longitudinal stress in the flange of the column from the axial load and eccentric load (cl 6.2.6.2(2), EN 1993-1-8 (2005)) σcom,Ed =

M+

V + A

Vh 2

Wel 453,6 250,2E6 + 174,4E3 × 174,4E3 2 = + 85,5E2 1297E3 = 244 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

β = 1 (Table 5.4, EN 1993-1-8 (2005)) and from Table 6.3, EN 1993-1-8 (2005) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

Avc

1

= 1 + 1,3

2 1/2

1/2 164,4×8,5 2 4094

= 0,932

which includes (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 85, 5E2 − 2 × 189, 9 × 12, 7 + (8, 5 + 2 × 10, 2) × 12, 7 = 4094 mm2 ηhwc twc = 1,0 × (453,6 − 2 × 12,7) × 8,5 = 3640 < (Avc = 4094) mm2 satisfactory. Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

beff,c,wc dwe fy,wc 1/2 λp = 0,932 2 ) (E twc 164,4 × 407,9 × 355 1/2 = 0,932 = 1,167 > 0,72 therefore (210E3 × 8,52 ) ρ=

λp − 0,2 2 λp

=

1,196 − 0,2 = 0,731 1,1962

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Thickness of the load bearing web stiffeners in compression in the column (cl 6.2.6.2(4), EN 1993-1-8 (2005)) Ro

ts = "

fyw

(bcf − tcw − 2rc ) γ

#

M1

378, 1E3

= "

(189,9 − 8,5 − 2 × 10,2) 355 1,0

#

= 6,62 mm. Use 8 mm thick stiffeners either side of the web of the column. Force per unit length of weld connecting the web stiffener to the column web Ro 378,1 = = 0,463 kN/mm 2 × 407,9 n w lw

Fw =

As previously the shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) Fw,Rd = 1, 1 > (Fw = 0,463) kN/mm It is assumed conservatively that the force is resisted by the welds along the web. To avoid damage to the strap in transit an alternative arrangement is shown in Fig. 7.36(b).

7.11 JOINT ROTATIONAL STIFFNESS (CL 6.3, EN 1993-1-8 (2005)) Previous sections in this chapter show calculations for joint strength. Another aspect of joint behaviour that needs consideration is the relationship between joint rotational stiffness and member stiffness. This relationship is important in the analysis of structures (cl 5, EN 1993-1-8 (2005)). A simple example of joint rotational stiffness (Fig. 7.37(a)) is as follows. The theoretical rotational stiffness related to the elastic and plastic behaviour of the top cleat can be developed for the joint. Initially assuming elastic behaviour of the steel top cleat with deformations as shown El 2 6 ma3 z2 M Ft z ba ta3 Sj = z2 = = = 12E φ 12m3 z

= "

E

z2

1 (ba ta3 /m3 )

#

(7.40)

This equation considers only deformation of the top cleat and can be compared with the theoretical expression (Eq. (6.27), EN 1993-1-8 (2005)) which replaces ba with leff and includes factors for other deformations (Table 6.11, EN 1993-1-8 (2005)). E z2 # μ k1

Sj = "

1

(7.41)

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

6E1/m 2

m

M

Z

276

(a)

0,4 2,0 Hu

f A

HI

Q

Mj

0,3

0,5

kc /kb

Mj 0,2 Q L/8 u Mj/Sj

L

(b)

0,1

0

0,2

0,4

0,6

0,8

1,0

Sj /kb (c)

FIGURE 7.37 Stiffness of joints

Other factors that affect the rotational stiffness of a joint are: extension and shear deformation of bolts, and deformation of column flanges, end plates and column webs. Each factor (Tables 6.10 and 6.11, EN 1993-1-8 (2005)) produces values of 1/k1 which are added and inserted in Eq. (7.41) as shown in Example 7.15. Experimental results (Maxwell et al., 1981) which show that the rotational stiffness of a joint is non-linear as ultimate resistance approaches. The stiffness of the joint reduces and to allow for this the value of μ increases (cl 6.3.1(6), EN 1993-1-8 (2005)). It is also necessary to check for other forms of failure. For this joint (Fig. 7.37(a)) the ultimate moment of resistance based on yielding of the top angle fy b ta2 Mjya,Rd = 2 (7.42a) 4 mz

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Alternatively for this joint if the bolts reach ultimate load Mjub,Rd = nb Ab fu z

(7.42b)

7.12 FRAME-TO-JOINT STIFFNESS Ideally the analysis of a structure should incorporate the stiffness of the joints and the members. An example of the theoretical relationship between the stiffness of joint and members for a simple frame follows. Consider the elastic behaviour of the simple frame in Fig. 7.37(b). From the application of the area moment method (Croxton and Martin, 1987 and 1989), the moment of resistance of the joint (Mj ) for a point load at the mid-span of the beam For the beam at A Mj M j Lb 1 Lb QLb EIb θ + = × × − Sj 2 2 4 2

(i)

For the upper column at A EIcu θ =

1 × Mju Hu 4

(ii)

For the lower column at A EIcl θ =

1 × Mjl Hl 4

(iii)

and Mj = Mju + Mjl

(iv)

Combining (i) to (iv) to eliminate θ and expressing stiffness as k = El/L Mj =

1+

QL 8 1 2(kcU /kb +kcL /kb )

2k

+ Sjb

(7.43)

The stiffness of the joint (Sj ), beam (kb ) and columns (kc ) can be seen to affect the moment of resistance at the joint (Mj ).

EXAMPLE 7.15 Stiffness of a ‘pin’ joint. Determine the stiffners of the ‘Pin’ shown in Fig. 7.26. Joint stiffness (Eq. (6.27), EN 1993-1-8(2005)) assuming μ = 1 E z2 # μ k1

Sj = "

=

210E3 × 517,42 = 22668 kNm/radian [(1 × 2,48 × 1E6)]

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) where (Fig. 6.15(b), EN 1993-1-8 (2005)) z = h + ea +

ta 10 = 467,4 + 45 + = 517,4 mm 2 2

and for a column web panel in shear (Table 6.11, EN 1993-1-8 (2005)) k1 =

0,38 × 3130 0,38Avc = = 2,3 βz 1 × 517,4

and for a column web panel in compression (Table 6.11, EN 1993-1-8 (2005)) k2 =

0,7beff,c,wc twc 0,7 × 179,5 × 13 = = 10,2 dc 160,8

where (Table 6.12, EN 1993-1-8 (2005)) beff,c,we = 2ta + 0,6ra + 5(tfc + s) = 2 × 10 + 0,6 × 10 + 5 × (20,5 + 10,2) = 179,5 mm and for a column web in tension (Table 6.11, EN 1993-1-8 (2005)) k3 =

0,7beff,c,wc tfc 0,7 × 210 × 13 = 11,9 = dc 160,8

where (Table 6.4, EN 1993-1-8 (2005)) beff,c,wc is the lesser value of 2πm = 2π × 45 = 283 mm and (208,8 − 140) πm + 2e1 = π × 45 + 2 × = 210 mm 2 and for a column flange in bending (Table 6.11, EN 1993-1-8 (2005)) k4 =

3 0,9leff tfe 0,9 × 146,5 × 20,53 = = 12,5 m3 453

where (Table 6.4, EN 1993-1-8 (2005)) leff is the lesser value of leff = 4m + 1,25e = 4 × 45 + 1,25 ×

(208,8 − 140) = 223 mm or 2

leff = 2m + 0,625e + e1 = 2 × 45 + 0,625 +

(208,8 − 140) + 35 = 146,5 mm 2

and for a cleat in bending (Table 6.11, EN 1993-1-8 (2005)) k6 =

0,9leff t3 0,9(208,8) 103 = ×

3 = 1,03 2 m3 45 − 10/2

and for bolts in tension (Table 6.11, EN 1993-1-8 (2005)) k10 =

1,6As 1,6 × 245 = 6,91 = Lb 20,5 + 10 + 3,7 + 13 + 16 2

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and for bolts in shear (Table 6.11, EN 1993-1-8 (2005)) k11 =

16 nb d2 fub 16 × 2 × 202 × 800 = = 3,45 E dM16 210E3 × 14,1

and for bolts in bearing (Table 6.11, EN 1993-1-8 (2005)) k12 =

24 nb kb kt d fu 24 × 2 × 1,06 × 1,06 × 20 × 510 = = 2,62 E 210E3

where 0,25eb 0,25 × 45 + 0,5 = + 0,5 = 1,06 d 20 1,5tj 1,5 × 10 = kt = = 1,06 dM16 14,1

kb =

The total value of 1 1 1 1 1 1 1 1 1 μ =1× + + + + + + + = 2,48. k 2,3 10,2 11,9 12,5 1,03 6,91 3,45 2,62

EXAMPLE 7.16

Effect of ‘pin’ joint stiffness on a simple frame shown in Fig. 7.37(b). Assume Q = 52 kN, Lb = 10 m, kb = 9612 kNm, kb /kc = 1 and Sj = 22668 kNm obtained in Example 7.15. The joint design moment of resistance at A (Eq. (7.43)) QL 8

Mj,Rd = ⎛ ⎝1 +

1 kcu kcl 2 + kb k b

=

1+

⎞ + Sb ⎠ j 2k

52E3×10E3 (8×1E6) 1 2×9612 + 2(1+1) 22668

= 31 kNm The effect of introducing the joint stiffness to the above equation is to reduce the end moment on the beam (Mj,Rd ) from 51,8 to 31 kNm. If Mj,Ed = 20 kNm then Mj,Ed /Mj,Rd = 20/31 = 0,65 < 2/3 and the adoption of μ = 1 is satisfactory (cl 6.3.1(6), EN 1993-1-8 (2005)). Check the ultimate moment of resistance of the joint based on yielding of the top angle (Eq. (7.42a)) fy b ta2 Mjya,Rd = 2 4 mz 355 2 × 517,4 10 40 = 2 × 208,8 × × = 47,94 > (Mj,Rd = 31) kNm 4 1E6

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) which is the maximum moment of resistance for the joint. Check the ultimate moment of resistance of the two top bolts (2M-20 class 8.8) failing in tension (Eq. (7.42b)) 517,4 Mjub,Rd = 2Ab fu z = 2 × 245 × 800 × 1E6 = 202,8 > (Mjya,Rd = 47,94) kNm The effect of varying joint to beam stiffness can be shown for the simple structure (Fig. 7.37(b)) using Eq. (7.43). The ratio (Mj /(QL/8)), which includes the moment of resistance at the end of the beam, varies with the ratios of joint-to-beam stiffness (Sj /kb ) and column-to-beam stiffness(kc /kb ) as shown in Fig. 7.37(c). This relationship is related to the structure in Fig. 7.37(b) for a beam stiffness of 4E3 kNm. Elastic behaviour is assumed but plastic failure may limit the value of the joint resistance. For example the joint moment (Mj ) reduces the moment at mid-span for the beam, which is beneficial, but introduces a moment to the columns which if slender can reduce their capacity. However for more complicated structures an increase in connection stiffness reduces slenderness ratios and column deflections, which increase the load capacity of a column. Numerical investigations (Jones et al., 1981) show that including joint stiffnesses in the analysis of frames can reduce the weight of steel.

REFERENCES Bahia, C.S. and Martin, L.H. (1980). Bolt groups subject to torsion and shear, Proceedings of the I.C.E. Pt2, V69. Bahia, C.S. and Martin, L.H. (1981). Experiments on stressed and unstressed bolt groups subject to torsion and shear, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Bahia, C.S., Graham, J. and Martin, L.H. (1981). Experiments on rigid beam-to-column connections subject to shear and bending forces, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-1 to 3 (1993) and BSEN 10210-1 (1994). Specification for weldable structural steels. BSI. BS 5400 (2000). Steel concrete and composite bridges, Pt3 Code of practice for the design of steel bridges. BSI. BS 3692 (2001). ISO Metric precision hexagon bolts, screws and nuts. BSI. BS 4190 (2001). ISO Metric black hexagon bolts, screws and nuts. BSI. BSEN 499 (1995). Covered electrodes for the manual metal arc welding of carbon and carbon manganese steels. BSI. BSEN ISO 4320 (1998). Metal washers for general engineering purposes. BSI. BSEN 1011-1 (1998) and 2 (2001). Specification for the process of arc welding of carbon and carbon manganese steels. BSI. BSEN 14399-1 to 5 (2005). High strength friction grip bolts and associated nuts and washers for structural engineering: Pt1 General grade, Pt2 Higher grade bolts and nuts and general grade washers. BSI. Biggs, M.S.A.B., Crofts, M.R., Higgs, J.D., Martin, L.H. and Tzogius, A. (1981). Failure of fillet welded connections subject to static loading, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Chesson, Jr., E. Faustino, N.L. and Munse, W.H. (1965). High strength bolts subject to torsion and shear, A.S.C.E. (Structural Division), V91, ST5.

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Clarke, A. (1970). The strength of fillet welded connections. MSc Thesis, Imperial College, University of London. Clarke, P.J. (1971). Basis for the design of fillet welded joints under static loading, Conference Proceedings. Welding Institution, Improving Welding Design Paper 10, V1. Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving Problems in Structures Vols. 1 and 2. Longman Scientific and Technical. Davies, G. (1981). Estimating the strength of some welded lap joints formed from rectangular hollow section members, Conference Proceedings, Joints in Structural Steelwork, Teeside Polytechnic. Elzen, L.W.A. (1966). Welding beams in beam-to-column connections without the use of stiffening plates. Report 6-66-2. I.I.W. document XV-213-66. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. European Convention for Structural Steelwork (1981). European recommendations for steel construction. Construction Press. Farrar, J.C.M. and Dolby, R.E. (1972). Lamellar tearing in welded steel fabrication. Welding Institute, Cambridge, England. Fisher, J.W. and Struik, J.H.A. (1974). Guide to Design Criteria for Bolted and Riveted Joints. John Wiley and Sons. Gourd, L.M. (1980) Principles of Welding Technology. Edward Arnold. Holmes, M. and Martin, L.H. (1983). Analysis and Design of Structural Connections. Ellis Horwood Ltd. Jones, S.W., Kirby, P.A. and Nethercot, D.A. (1981). Modelling of semi-rigid connection behaviour and its influence on steel column behaviour, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Kato, B. and Morita, K. (1974). Strength of transverse fillet welded joints. Welding Research. Ligtenberg, F.K. (1968). International test series, final report, Stevin Laboratory, Technological University of Delft, Doc XV-242-68. Mann, A.P. and Morris, L.J. (1979). Limit state design of extended plate connections, ASCE Journal of Structural Engineering, 105(ST3): 511–526. Martin, L.H. (1979). Methods for limit state design of triangular steel gusset plates, Building and Environment. Martin, L.H. and Robinson, S. (1981). Experiments to investigate parameters associated with the failure of gusset plates, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Maxwell, S.M., Jenkins, W.M. and Howlett, J.H. (1981). A theoretical approach to the analysis of connection behaviour. Conference Proceedings, Joints in Steelwork. Teeside Polytechnic. Morris, L.J. and Newsome, C.P. (1981). Bolted corner connection subjected to an out of balance moment – the behaviour of the web panel, Conference Proceedings, Joints in Structural Steelwork, Teeside Polytechnic. Owens, G.W., Driver, P.J. and Kriege, G.J. (1981). Punched holes in structural steelwork, Constructional Steel Research, 1(3). Pillinger, A.H. (1988). Structural steel work: a flexible approach to the design of simple construction, Structural Eng 66(19/4), October. Purkiss, J.A. and Croxton, P.C.L. (1981). Design of eccentric welded connections in rolled hollow sections, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Rolloos, A. (1969). The effective weld length of beam-to-column connections with stiffening plates, Final report, I.I.W. Document XV-276-69. Stamenkovic, A. and Sparrow, K.D. (1981). A review of existing methods for the determination of the static axial strength of welded T, Y, N, K, and X joints in circular hollow steel sections, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Stark, J.W.B. and Bijlaard, F.S.K. (1988). Design rules for beam-to-column connections in Europe, Steel Beam-to Column Building Connections. pub. Elsevier Applied Science.

Chapter

8 / Frames and Framing

The previous chapters have dealt with design of beams, columns and connections. This chapter, and the next, deal with the way the individual components are assembled together and also with design problems associated with the whole structure. In the first place it is necessary to give some consideration to how the choice is made of the structural form employed to carry the primary loading. For convenience this survey is divided into single and multi-storey structures.

8.1 SINGLE STOREY STRUCTURES T ypical examples of such structures include sports complexes, exhibition halls, factory units or assembly buildings. Unless architectural considerations prevail, the most economic solution will be obtained using one-way spanning structural systems rather than space frame structures. Systems which at first sight appear two-way spanning often comprise a number of overlaid one-way systems. Roof systems can be conveniently divided into flat and pitched roof systems.

8.1.1 Flat Roof Systems For spans up to around 15 m rolled sections form the most economic solution. However, it should be noted that the potential extra cost of beams over 12 m long needs to be taken into consideration, and the use of 12–15 m long beams should only be contemplated if a large number are required. At around 14 m and up to 20 m the use of castellated or beams with circular openings in the web become economic. Although this type of beam incurs high fabrication costs and requires a higher construction depth than ordinary rolled sections, the holes in the web allow services to be contained within the beam depth. Above 20 m it is usual to use parallel chord lattice trusses fabricated either from rolled hollow sections or from lightweight cold formed sections (top and bottom chords) and bar members for the web. The roof decking is then generally lightweight steel sheeting with suitable finishes and insulation, although timber decking with asphaltic waterproofing, woodwool slabs or pre-cast concrete decking units may be used. One problem with flat roofs is drainage and thus sufficient cross-fall must be provided to give adequate run-off and avoid local ponding. The cross-fall is

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generally provided by variable depth purlins and joists or by adjusting screed depths. To some extent these problems are alleviated by employing a sloping top chord to the roof system.

8.1.2 Pitched Roof Systems These fall into two main divisions: trusses with a sloping top chord and pitched roof portal frames. Modern trusses have a relatively low slope to the top chord. A pitch of between 4◦ and 10◦ is adequate to allow run off and to allow the joints in the roof sheeting to remain watertight. This type of truss is generally fabricated from square or rectangular rolled hollow sections with fully welded nodal connections, with circular hollow sections sometimes being used for the web members in Grade S355 steel, owing to availability, Pitches greater than 10◦ are only seen where an existing building with traditional large pitch trusses designed for tiles or slates as roofing materials is being extended, or where a large pitch is required for architectural reasons in, for example, shopping malls. The most common method of single storey construction is the pitched roof portal frame whether for factory units, small sports complexes or warehouses. This is basically due to the high speed and simplicity of construction. The internal bays are designed as rigid jointed frames. The end frames, unless there is likely to be an extension to the structure, are much lighter and have the rafters designed as spanning across the gable end posts which are also used to support the sheeting rails for the cladding. The most economic frame spacing is generally 7.5 m or 9.0 m for much higher frame spans. For spans below 20 m a frame spacing of 6.0 m may be adopted (Horridge, 1985; Horridge and Morris, 1986). Both truss systems and portal frames can be used for spans up to 60 m, although for spans over 30 m multi-bay structures become an option. Multi-bay construction requires internal columns which may reduce the flexibility of usage. However, this situation may be mitigated by the use of internal lattice girder support systems. Note that where the roof system comprises trusses the internal support system can be within the depth of the truss. This is not possible with portal frames, so unless reduced headroom is acceptable internal columns must be used. It should also be remembered that where multi-bay construction is used, there will be the need to supply valley guttering and associated drainage. Such valleys cause build-up in snow loading and can be potential areas of leakage and cause problems with access for maintenance.

8.2 MULTI-STOREY CONSTRUCTION 8.2.1 Multi-storey Steel Skeleton In UK practice the steel skeleton is generally designed to carry vertical loading due to the permanent and variable actions only, with the horizontal loading from wind and

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Chapter 8 / Frames and Framing the notional horizontal loading taken by a bracing system, or more commonly the lift shaft(s) and stair well(s). The economics of various types of multi-storey construction is discussed in Gibbons (1995). When using lift shafts or stairwells as bracing care must be taken in their layout as torsional effects from lateral loading on an asymmetric layout must be avoided.

8.2.2 Flooring Systems The flooring system is generally required to act as a horizontal diaphragm to carry horizontal forces from their point of application to the bracing or lift core. Thus it is essential that adequate lateral stiffness exists in the plane of the floor and the flooring system is adequately tied to the frame. Historically the floors of multi-storey steel frames have comprised cast in-situ normal weight concrete slabs. This system is now little used partly owing to excess weight of the concrete and partly owing to the slow construction time and the need for propping the deck often over two storeys. The problems with propping can be reduced by the use of proprietary falsework systems involving lightweight trusses which allow the actual formwork to be struck before the props are removed. The use of in-situ concrete floor systems except for small areas not otherwise able to be handled has been superseded by pre-cast pre-stressed concrete units or by composite steel–concrete decking. Pre-cast pre-stressed concrete units may be placed on the top flange of the steel beam, supported on shelf angles welded to the web of the beam, or placed between the flanges of a column section used as a beam. In all cases the flange of the supporting beam system may be considered to be restrained against lateral torsional buckling, although for shelf angle floors consideration may need to be given to the torsional load on the beam during construction. An alternative for ease of construction is to use asymmetric sections with the bottom flange larger than the top (Mullett, 1992; Lawson et al., 1997). These alternatives are illustrated in Fig. 8.1. For pre-cast units placed on the top flange small discrete vertical plates are welded to the top flange to give shear anchorage. For floor systems where the pre-cast unit sits on a bottom flange or shelf angle, the top corners of the units will need chamfering to ensure the units can be placed in position. Such floors also have the advantage of increased fire performance as the top flange is shielded by the concrete section from the effects of a fire on the underside. Equally the bottom flange of the steel section will be either partially or totally adjacent to a heat sink formed by the concrete slab, thus also inducing lower temperatures than would otherwise exist. For most types of flooring a screed is required to give a proper finish and it is recommended that a light structural mesh reinforcement is placed over the support beams to help control cracking in the screed. One potential disadvantage of using pre-cast units is the large amount of hook time required to place the units which could otherwise have been used to hoist materials required for finishes, etc.

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Pre-cast units

Shelf angles (a) Shelf angle floors (screed omitted)

(b) Pre-cast units resting on bottom flange of column section (screed omitted)

in-situ concrete

Deep trough decking (c) Asymmetric section with deep trough decking

FIGURE 8.1 Types of Flooring

Composite steel–concrete floors use thin gauge trough steel sheeting which initially acts as permanent shuttering to the concrete before acting compositely after the concrete has set to provide both tensile and shear ‘reinforcement’ to the finished concrete slab. The concrete used is lightweight concrete with a specific weight of around 18–20 kN/m3 , and designed to be placed using concrete pumps rather than skips. The only cranage required is that to lift the bundled steel sheets to the correct level, as the sheets are individually light enough to be manhandled. To provide shear continuity between the deck and the steel support beams in-situ through deck shear stud welding is employed. Reinforcement will be required in the form of mesh in the top face of the slab in areas of hogging moments if only to control the effects of cracking. This mesh will also contribute to the fire performance of the deck. Any propping can be avoided by limiting the spans of the decking whilst acting non-compositely by additional support beams. This option is more cost effective than propping. A recent development is to use deep decking over the whole depth of a UC used as a beam with the decking supported on flange plates welded to the lower flange of the UC or specially rolled asymmetric sections. This system is known as ‘Slimflor’ construction (Mullett and Lawson, 1993), in which the outer beams are downstand UB’s or rolled hollow sections with plates welded to the soffit (Mullett, 1997). A very useful practical

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Chapter 8 / Frames and Framing guide for an overview of design and construction is published by the SCI (Couchman et al., 2000).

8.3 INFLUENCE OF CONNECTION DESIGN AND DETAILING From the previous chapter it will be noticed that connections are either designed to take the effect of beam reactions in the form of shear (with nominal moments) or to resist the effects of moments, and the coincident shear and axial force. The first type is generally designated a ‘pin’ joint which allows large relative rotations between the members in the connection (and leads to the concept of simple construction). The second is generally designated a ‘fixed’ or ‘rigid’ joint in which rotational compatibility exists between the members framing into the connection. This type of connection allows full moment transfer and should ideally be welded, although in UK practice a heavily bolted stiff connection is taken as rigid. Obviously the type of connection in the structure will markedly influence the behaviour of the structure under whatever actions are applied to it. A fuller discussion of the implications of the above paragraph is given in Section 8.12, but it is first necessary to consider the actions applied to a structure.

8.4 STRUCTURAL ACTIONS These may either be physical loading or imposed deformations due, for example, to differential settlement.

8.4.1 Physical Loading For convenience this is divided into two categories: gravity and non-gravity loading.

8.4.1.1 Gravity Loading This covers the self-weight of the structure, the finishes on the structure, the actions due to the usage of the structure (variable actions) and roof loading whether as a nominal load or as snow loading (uniform or drifting as appropriate). Values of loads are given in EN 1991-1-1. It should however be noted that snow drift loading given in EN 1991-1-3 constitutes an accidental load case and therefore takes lower partial safety factors, and thus may not be critical. It is advisable to design the structure under uniform roof loading and then check the structure, if appropriate, under non-uniform snow drift loads.

8.4.1.2 Non-gravity Loading This can be considered under a series of sub-headings: • Wind loading: This is covered by EN 1991-1-4. • Inertia and impact loads: These have to be considered where dynamic loading is considered, for example cranes and supporting systems (EN 1991-3). Impact loading

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needs considering in the case of car parks where columns can be damaged due to collisions or for bridge piers where vehicular impact may occur (EN 1991-1-7). • Seismic loads: These are not of general importance in the UK except for nuclear installations and other similar structures, and are covered in EN 1998. • Accidental loads: These can be due to snow drift loading (EN 1991-1-3), fire (EN 1991-1-2) or explosions (EN 1991-1-7). In the case of explosions either the implications of progressive collapse needs to be considered or the structure must be tied together and the resultant tying forces considered.

8.4.2 Deformations Structural deformations can be either due to differential settlement (Section 8.11.1) or thermal movements (EN 1991-1-5) (Section 8.11.2).

8.4.3 Load Combinations Load combinations for single variable loads have been covered in Chapter 3, and it therefore remains to cover load combinations for multiple variable loads. The classic example of this is where there are variable loads due to structure usage (e.g. office loading) and variable loads due to wind. The basis behind the combination rules in EN 1990 is that it is deemed statistically unlikely that all the variable loads will be acting at their maximum intensity at the same time. It should be noted that roof loading and floor loading taken for the structure as a whole are considered as separate variable actions. The rules take account of the non-simultaneity of maximum effects by introducing ψ factors on the non-principal variable actions. EN 1990 allows a number of combination rules – this is reflected by the UK National Annexe to EN 1990. At ultimate limit state, the following combination rules are available: 1,35Gk + 1,5Qk,1 +

n

1,5ψ0, j Qk, j

(8.1)

2

where Gk is the permanent load, Qk,1 is the principal (or leading) variable load and Qk,2 to Qk,n are the accompanying variable actions. Note however that where the permanent load is unfavourable (i.e. is of opposite sign to the permanent load, the partial safety factor γG is set equal to 1,0, and where the variable loading is unfavourable its partial safety factor γQ is taken as zero. Equation (8.1) must be applied taking each variable load as the principal variable load in turn and the remainder as accompanying actions. Experience, however, may be used in reducing the calculations when it is clear which leading action is critical. Note it is possible for different types of actions to have different values of ψ0, j . 1,35Gk + 1,5ψ0,1 Qk,1 +

n 2

1,5ψ0, j Qk, j

(8.2)

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Chapter 8 / Frames and Framing It is necessary with this combination to examine the effects of different ψ0, j values, although it is likely to give lower loads than those determined from Eq. (8.1). 1,35ξGk + 1,5Qk,1 +

n

1,5ψ0, j Qk, j

(8.3)

2

The additional factor on the permanent load ξ is subject to the limit 0.85 ≤ ξ ≤ 1.0. The UK National Annexe specifies a value of 0,925. Equation (8.3) will give the lowest total load provided the permanent load is unfavourable. It would appear permissible to use the ξ factor where there is only a single variable load. For serviceability loading the combination rules are similar except the maximum values of γG and γQ are set equal to one as is ξ. The other change is that ψ0 may be replaced by ψ1 or ψ2 . The factor ψ1 is used to determine effects under reversible limits states (known as the frequent combination) and ψ2 is used to determine long-term effects or the appearance of the structure (known as quasi-permanent). The use of ψ0 is only required for irreversible limit states. Deflection is generally checked under the quasi-permanent combination although it would be advisable to check it under full variable load together with permanent load in order to assess any maximum instantaneous deflection. It is now necessary to consider how structures resist the forces due to actions applied to the structure. The major concern is with the transfer of horizontally applied forces, although some consideration is given to the distribution to supporting members of vertically applied forces. Single storey structures will be considered in detail, before continuing by looking at multi-storey structures.

8.5 SINGLE STOREY STRUCTURES UNDER HORIZONTAL LOADING Consider initially a basic structure comprising a single bay flat roof portal type frame with encastré feet under vertical loading only (Fig. 8.2). It does not at first sight matter whether the connections at B and C are rigid or pinned, as the structure can be analysed and the members designed under the resultant forces. However if the joints at B and C are pinned the rafter will be a heavier section as no ‘fixing’ moment will exist at the connections, and there will be an increased deflection at the centre of the rafter. The column section will be lighter and the connection detail simplified. However, the bottom ends of the column are fixed with respect to the bases. This condition will cause problems for the foundation design (Section 8.11.1) as the ratio of moment to axial load will be too high to give an economic design. The general custom on such a frame is therefore to use non-moment resisting connections between the feet of the stanchion and foundations, thus if non-moment resisting connections are used between the rafter and the columns the frame is inherently unstable. This

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Vertical loading B

C Rigid joint

A

D

(a) Basic frame with rigid joints at B and C B

C Pins

A

D

(b) Basic frame with pinned joints at B and C

FIGURE 8.2 Simple single bay flat roof portal

statement is clearly true for horizontal loading, but is also true for vertical loading due to inherent imperfections in both the members and through construction. Two solutions, not considering moment resistant connections at the feet of the columns, are possible. The first is to restore the moment connections at B and C; the second is to find an alternative method to resist any horizontal forces. The first method has the drawback that for single bay frames the total moments at the connections may reverse in sign due to the change in direction of the application of wind forces. The moment due to the wind force alone must change sign (Fig. 8.3). Allowing the connections to resist wind forces is possible in multi-bay structures where the wind force is adsorbed through a large number of connections, thus reducing the moment on each individual connection (see Section 8.12). The second method is to adopt a bracing system within the frame. So for the wind force P (Fig. 8.4(a)), a diagonal member is placed such that the diagonal member is in tension (i.e. capable of taking full design load). This has the effect of triangulating the structure. When the wind blows from the reverse direction (force P in Fig. 8.4(b)), the tie AC now becomes a strut with much reduced load capacity owing to buckling. Thus a further tie BD is now inserted giving rise to cross-bracing. The cross-bracing is designed by only considering the forces in tension members (i.e. the compression members are assumed to have zero load capacity, a conservative assumption). It should also be noted that since the structure is triangulated the frame analysis (and design) is much simplified.

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P (Wind force)

A

D (a) Wind from left to right

B

C P′ (Wind force)

A

D

FIGURE 8.3 Effect of wind reversal

(b) Wind from right to left B

C

P (Wind force)

Tie to take wind force P

A

D

(a) Triangulation to take effect of wind force P B

C P′ (Wind force)

Tie to take wind force P ′

A

D

Note: AC now acts as a strut, and for calculating the force in BD is assumed to have zero strength (b) Triangulation to take effect of wind force P ′

FIGURE 8.4 Elementary wind bracing

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Eaves beam

(a) Single storey multi-bay frame – each frame braced

(b) Single storey multi-bay frame with wind girder

FIGURE 8.5 Wind girder bracing

No structure exists as a single frame. Where a set of such frames described above, complete with cross-bracing, to be assembled as multi-bay single span structure, it would be unusable (Fig. 8.5(a)). This state of affairs can be made acceptable by retaining the cross-bracing in each of the end frames only and by supplying full diagonal crossbracing between each of the frames at rafter level (Fig. 8.5(b)). Such a bracing system is known as a wind girder, and transmits any horizontal forces through the girder to

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Beam

(a) Section

Intermittent steel flats welded to top flange

(b) Elevation (pre-cast units omitted)

FIGURE 8.6 Shear key detail

the bracing in the end frames. A wind girder is always needed where lightweight roofing systems are used in conjunction with light gauge purlin systems. Where stressed skin construction is used the wind girder can be omitted as the sheeting is designed to resist the shear due to the horizontal loadings (Davies and Bryan, 1982). If pre-cast concrete units are used then these may replace the wind girder provided adequate shear connection between the rafters and the units is provided (Fig. 8.6). Composite decking with through deck shear studs will not generally need additional bracing (but see Section 8.9.3). For multi-bay multi-span structures a wind girder is needed around the periphery of the structure and full diagonal bracing in each of the corners of the structure (Fig. 8.7). For portal frame systems, the wind forces are taken by moment resisting connections in the plane of the frame, but for wind applied along the structure, i.e., normal to the plane of the portals, wind bracing must be provided. It may also be necessary to brace the end or gable frame for wind in the plane of the frame if it is designed as rafters spanning over intermediate gable end columns. Although this section has referred to discrete bracing it is possible to replace the bracing by alternatives. The most common replacement for vertical bracing is masonry shear wall construction. For single storey construction this need be no more than conventional masonry cladding provided such cladding is fully ties to the steel frame using, for example, half wall ties spot welded to the stanchions or approved proprietary tying systems. The design of such cladding is to EN 1996-1-1. It should however be remembered that in the case of masonry cladding

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Key : Main members; Wind bracing Note : All corners must be fully braced

FIGURE 8.7 Schematic layout of wind bracing for a multi-bay structure the frame is erected first and that the frame will need temporary bracing during this stage. It is the designer’s responsibility to ensure this is present (Construction ‘Design and Management’ Regulations) (HMSO, 1994).

8.6 MULTI-STOREY CONSTRUCTION The situation with respect to vertical and horizontal planes can be discussed separately.

8.6.1 Vertical Plane The effect of horizontal forces in the vertical plane can be handled in a number of ways: • Bracing The structure can be braced either with the bracing left as an external feature and hidden behind lightweight cladding. Such bracing can be at each corner of the structure, or provided the floors act as stiff diaphragms, in the centre of each face of the structure. The latter layout is only really suitable for structures symmetric about their centre lines as otherwise torsional effects are introduced which cannot easily be resisted. • Shear walls This solution is generally adopted for the smaller end panels of structures which have a high aspect ratio with few columns in the smaller direction. The shear walls in order to act properly must be relatively unpierced, i.e., free from dominant or significant openings. This can produce architectural constraints. Irwin (1984) provides

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Chapter 8 / Frames and Framing information on design of shear walls. In the longitudinal direction the wind force is lower (as it acts on the smaller face) and there are a large number of bays to take the wind forces on the columns and the column–beam connections. • Cores In the UK, the most common solution is to allow the wind forces to be adsorbed by the core(s) provided for the lift shaft(s) and/or stairwell(s). Such cores are generally of reinforced concrete construction, although the walls surrounding stairwells may be masonry. The walls surrounding lifts or stairwells do not have large openings as they generally provide access to fire escape routes and protected access areas for fire fighting, and thus stiff enough to provide the lateral restraint to the structure necessary. It should be noted that tolerance problems can arise when marrying up steelwork and reinforced concrete construction, and that shrinkage and creep effects in the concrete should not be ignored (Irwin, 1984). • Rigid/semi-rigid construction The case where the frame is taken as rigid for both the vertical loading and horizontal loading is rare in the UK. However, a hybrid method of design is becoming popular in which the frame is consider pin jointed (simple construction) for the vertical loading but as rigid jointed for the horizontal loading. There are restrictions on this method mostly related to the size and shape of the structure (Salter et al., 1999).

8.6.2 Horizontal Plane Composite steel–concrete flooring systems will, when the concrete has hardened, prove a very stiff diaphragm to transmit horizontal forces provided adequate shear connection between the deck and the steel skeleton is available. The shear stud requirement to provide flexural composite action will generally be adequate. When pre-cast concrete units are used a shear key detail such as that illustrated in Fig. 8.6 should be used, and structural mesh provided in the screed in order to give full diaphragm action and prevent cracking over the beam.

8.7 BEHAVIOUR UNDER ACCIDENTAL EFFECTS Accidental actions should be considered in a number of possible circumstances: • Explosions whether due to gas or terrorism. • Impact due to vehicles or aircraft. Should the risk of such an incident be high and the effects be catastrophic, or in certain circumstances the need to check be mandatory (e.g. vehicular impact on bridge piers), then the designer must ensure that the structure is designed and detailed to ensure that should an accidental situation occur, the structure does not suffer complete or partial collapse from either the accidental situation itself or subsequent events, such as spread of fire.

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Four now classic cases where collapse occurred due to accidental or terrorist action are: (1) Ronan Point block of flats where a gas explosion blew out a wall panel causing progressive collapse of a corner of the structure (Wearne, 1999). (2) The Alfred P. Murrah Building in Oklahoma City where collapse was caused by a terrorist bomb blast (Wearne, 1999). (3) World Trade Centre Towers in New York on 11 September 2001 due to impact from deliberate low flying aircraft and subsequent fire spread. The towers were steel framed structures and it has since been demonstrated that the prime cause of collapse was due to loss of fire protection on the floor members, as it was considered that the fire load from the impacting aircraft was not unduly high (Dowling, 2005). Information on considerations of the design of high rise construction is given in ISE (2002). (4) Pentagon Building on 11 September 2001 following the deliberate crash of a commercial airline. The concrete structure suffered partial but not extensive collapse due to the impact and subsequent fire (Mlakar et al., 2003).

8.7.1 Progressive Collapse EN 1990 identifies the need to consider that in the event of an accident such as explosion or fire the structure should not exhibit disproportionate damage. This is reinforced by the requirements of the relevant Building Regulations within the UK, for example the recently revised Approved Document A of the England and Wales Building Regulations. Such damage can be mitigated by: • Attempting to reduce or limit the hazard In the case of fire this could be done by full consideration being given to the use of non-flammable materials within the structure and by the provisional of relevant active fire protection measures such as sprinklers. In the case of industrial processes where explosions are a risk then such potential risks need to be taken into account by, say, enclosing the process in blast proof enclosures. • Maintenance The recent Pipers Row Car Park collapse of a lift slab concrete structure (Wood, 2003) indicated the need for adequate design especially where punching shear in a concrete slab may be critical as this is a quasi-brittle failure. The need for adequate inspection and maintenance where it is known that environmental effects will be severe was also highlighted. Although the failure was due to poor maintenance, it was exacerbated by uneven reaction distribution at the slab–column interface and no tying through the lower face of the slab. Admittedly the construction technique used would have made the latter extremely difficult although it contributed to the former.

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Chapter 8 / Frames and Framing • Consideration of the structural form A case where this is relevant is when the structure could be subject to externally provoked explosions. One reason why the damage was extensive following the Oklahoma City explosion was that the blast was amplified by an overhanging portion of the structure above ground floor level. A solution for structures known to be at potential risk is to provide an external curtain which is not structural and is easily blown out whilst ensuring the structure is stabilized by a core which is in the centre of the building. Additionally the risk can be mitigated by ensuring vehicles cannot come within certain limiting distances of the structure. More information on this, including assessment of blast forces is given in a SCI publication (Yandzio and Gough, 1999). • Ensuring the structure is adequately tied to resist collapse This needs considering on two levels. The first is to determine the magnitudes of the likely levels of tying force required. The second is to provide properly detailed connections between elements of the structure. This is needed for both connections in the horizontal plane: beam-to beam or beam to column and in the vertical plane: column splices. • Removal of key elements to establish stability In this approach a risk analysis is carried out to identify the key elements in a structure which are vital for its stability. A structural analysis is carried out with any one of these elements removed to establish the stability of the remaining structure and its ability to carry the loading so induced. As this is an accidental limit state lower partial safety factors are used on both the loading and the material strengths determining the member strengths. It is not necessary to check any serviceability limit states. Approved Document A divides structures into four categories: 1, 2A, 2B and 3, with Category 1 being the least onerous. Category 1 constitutes low-rise housing, agricultural buildings and buildings with restricted access. These require no specific checks. For Category 2A which are medium consequence risk structures, the consideration of tying forces is adequate in most cases. For Category 2B defined as high consequence risk structures, then the requirement must be satisfied by using removal of key elements. For Category 3 structures which are very high consequence risk structures such as large capacity grandstands, hospitals, structures over 15 storeys or structures in which hazardous materials are involved require special consideration (Way, 2005). Alexander (2004) gives examples on risk assessment with the hazards grouped into two categories: those that must be considered for any Class 3 structure and those that only need considering for specific location dependent structure such as flooding.

8.7.2 Structure Stability Any structure with high lateral loading or where vertical loading can be applied outside the frame envelope must be checked for overturning. Also any continuous member with a cantilever must be checked for the possibility of uplift on any support.

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6 m GK = 1500 kN

wK = 6 kN/m

35 m gK = 32 kN/m

1.5 m

A

FIGURE 8.8 Design data for Example 8.1

14 m

In all these cases the loading which causes overturning is deemed to be unfavourable, and the loading tending to restore the situation is known as favourable. This is defined in EN 1990 as verification of static equilibrium EQU. For this situation the partial safety factors corresponding to EQU in EN 1990 (Table A1.2 A) must be used.

EXAMPLE 8.1 High wind load. Consider the possibility of overturning for the water tower shown in Fig. 8.8. Taking moments about point A: Unfavourable effects due to wind:

35 1,5 × 120(3 + 35 + 1,5) + 1,5 × 6 × 35 + 1,5 = 13095 kNm 2

Favourable effects due to self-weight: 0,9 × 1500

14 14 + 0,9 × 32 × 35 = 16506 kNm 2 2

The moment due to the favourable effects exceeds that due to the unfavourable effect, thus the structure will not overturn.

EXAMPLE 8.2 External loading. Consider the cantilevering frame shown in Fig. 8.9. Here the check is more complex, as the permanent load of 220 kN per storey is both favourable and unfavourable, in that the load acting on the cantilever is unfavourable as it contributes to the overturning effect. The permanent load per storey may be expressed as 27,5 kN/m run. The favourable part of the permanent loading on the internal span takes a factor of 0,9, and that on the cantilever of 1,1.

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[email protected] 3.5 m GK = 220 kN/storey

wk = 13 kN/m

7m

B 1m

FIGURE 8.9 Design data for Example 8.2

Taking moments about B: Effect due to the wind loading (unfavourable): 1,5 × 13

17,52 = 2986 kNm 2

Effect due to permanent action (per storey): 0,9 × 27,5

72 12 − 1,1 × 27,5 = 591 kNm 2 2

Total restoring (or favourable) moment: 5 × 591 = 2955 kNm The structure will just overturn as the disturbing moment is marginally greater than the restoring moment. Thus either the structure needs tying down using, say, tension piles or ground anchors, or the permanent loading could be marginally increased. The latter is the cheaper option owing to the small margin involved.

8.8 TRANSMISSION OF LOADING 8.8.1 Transmission of Loading from Flooring Systems When the loading is considered as a UDL, then the distribution of such loading to any supporting beam system will depend whether the decking system may be considered as one or two way spanning. Deck systems comprising lightweight proprietary systems (roofing only), pre-cast concrete slabs, composite steel–concrete decks and timber are all one way spanning. Thus half the load is taken to either end of the pre-cast unit, purlin system, joists or composite deck (in the direction of the troughs) (Fig. 8.10). Where a decking system is continuous over the supporting beam system then the reactions should be determined using a suitable analysis, although for approximate design continuity may be ignored. Where in-situ concrete is used then loading is distributed to

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Concrete slab with profiled sheet steel decking Main beam Decking Load on main Reaction (UDL) beam

Reaction from decking (UDL) Joist or purlin

Intermediate beam reactions

Reaction from joist

Reaction from joist

Main beams Note: The loading on the main beam will be appiled through web cleats (not shown) (a) Transfer of load from one way spanning concrete deck

(b) Load transfer from decking through a joist or purlin system

FIGURE 8.10 Load transfer for one way spanning system

A

B

45°

E

D

F

C

Note: (a) Load on area AED is supported by beam AD (b) Load on area AEFB is supported by beam AB

FIGURE 8.11 Approximate load transfer from two way spanning slabs

all the supporting beams. The loads on individual beams should be determined using the same assumptions as in the slab design, such as Johansen’s yield line method or the Hillerborg strip approach. However, for approximate calculations a 45◦ dispersion from each corner of the slab may be used to partition the loading to the beams (Fig. 8.11).

8.8.2 Lintels In the absence of other data a 60◦ dispersion of the load above the lintel may be used, that is, the lintel only carries the loading within the complete 60◦ equilateral triangle. Where the triangle is incomplete due to openings then the whole load above should be taken.

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8.9 DESIGN OF BRACING 8.9.1 Permanent Bracing The forces to be used to design such bracing are given in cl 5.3: (a) any horizontal loads applied to the frame being braced, (b) any horizontal or vertical loads applied directly to the bracing system (c) the effects of initial imperfections, or the equivalent horizontal forces, from the bracing system itself and from all the frames it braces. For wind bracing the forces in (a) and (b) will need considering, whereas for sway bracing in the vertical plane it is a combination of all three. The equivalent horizontal forces for sway are covered in Section 8.12.2. There is an additional requirement for stability forces at the splice in a column or beam which is covered in the next section. The design of diagonal bracing is generally simplified by ignoring diagonal members in compression and assuming all joints are pinned. In the case of wind bracing the member should not be slender enough to induce an acceptable degree of sag under its own self-weight.

8.9.2 Restraint Bracing to Compression Flanges and Column Splices • Compression flange bracing The problem can either be handled by adding an additional bow in the members to be restrained and designing for the resultant additional moments, or using an equivalent stabilizing force. In the first case the initial bow imperfection e0 is given by L 500

e0 = αm

(8.4)

where L is the span of the bracing system and αm is given by αm =

1 0,5 1 + m

(8.5)

where m is the number of members to be restrained. For convenience the effective bow imperfection in the members to be restrained by a bracing system may be replaced by the equivalent stabilizing force qd is given by qd =

NEd 8

e0 + δ q L2

(8.6)

where δq is the in-plane deflection due to the load q and any external loads calculated from a first order analysis. If second order analysis is used then δq = 0.

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The notional force NEd in the flange is defined as MEd /h where MEd is the maximum moment in the flange and h is the overall depth of the beam. • Restraint to splices An additional requirement at splices is that the bracing should be able to resist a local force of αm NEd /100 together with any other applied forces. Note that the application of this local force is not co-existent with the force q defined above.

8.9.3 Temporary or Erection Bracing Under the Temporary Workplaces Directive enforced in the UK by the Construction (Design and Management) Regulations, the client must appoint a planning supervisor at the Design stage on all health and safety matters throughout the execution of the project. It thus falls to the supervisor to ensure the structure is capable of being erected safely and that the requisite safety measures are in force. This means that the specification and design of temporary bracing should not be left to the steelwork erector without approval by the planning supervisor. The necessity for such bracing is often due to the intended construction sequence where construction starts at the centre of a structure in order to reduce cumulative tolerance errors whereas the permanent bracing designed to give stability to the completed structure is in the end bays. Further reasons are that support conditions or the effect of temporary loadings may be such that tension members are in compression or the signs of bending moments are reduced. There is also the requirement to ensure that any forces applied to the structure from safety equipment used during erection are considered. Where in-situ concrete is being used the compression flange will not receive full restraint until the concrete is hardened, and thus during construction the compression flange of the support beam system is unrestrained and therefore prone to lateral torsional buckling. Where profiled sheet steel decking is being used provided through deck welded shear studs are used, the beams running normal to the deck profiles can be considered as fully restrained, but those running parallel with the profiles will need checking for buckling though the deck will provide some degree of restraint (Lawson and Nethercot, 1985). An additional point to be noted is that large members may not be stable when being lifted into position partly due to the change in support conditions and partly due to the changes in restraints. Such members may need to be braced together at the correct spacing and lifted in pairs. Such bracing should if possible form part of the permanent bracing.

8.10 FIRE PERFORMANCE It is recommended that the reader is referred to Ham et al. (1999) for an overview of the basic procedures and of methods of protection. Whilst it is not intended to cover detail fire design of steel structures, a brief overview is necessary. For full details reference should be made, for example, to Purkiss (2007).

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8.10.1 Single Storey Structures For most single storey structures the situation with respect to fire performance is relatively straightforward as there is generally no specific requirement within the England and Wales Building Regulations (Approved Document B) (DTLR, 2000). The only problem tends to be caused by outward collapse of walls and columns of single storey frames. This tends to be worse with pitched roof portals (see Simms and Newman, 2002). In addition there is the problem of ensuring that fire fighters have complete safe access to fight the fire during its complete duration.

8.10.2 Multi-storey Structures Traditionally design has been by considering single elements with no interaction between such elements. However, fire tests on the steel frame structure at Cardington have demonstrated that with structural interaction between beams and columns it is possible to leave all beams unprotected (except at connections) and achieve a very good fire performance. Interim design data allowing some beams within floor systems being left unprotected has been published by the SCI (Newman et al., 2006). It is also possible to encase columns in brickwork or blockwork or to fill rolled hollow section columns with concrete to achieve fire performance without additional measures (Bailey et al., 1999). Although a complete discussion of the Cardington tests is given elsewhere (Purkiss, 2007), an overview of the tests will be presented. The tests at Cardington demonstrated that temperatures in the unprotected steel beams which were composite with the floor slab reached temperatures on the bottom flange in excess of 1000◦ C. This is around 400◦ C higher than the limiting temperatures allowed after correcting for load level in design codes. It should be noted that the loading applied on the floors was around onethird of the ambient design value of 2,5 kPa giving a load ratio lower than that of most office type structures. Although the structure retained its load carrying capacity the deflections were extremely high with values of up to 640 mm being recorded. Although the tests showed far better performance than would be expected from the standard furnace test on individual elements, a number of points need to be raised: (1) The high deflections reached would probably mean part at least of the structure would need replacing. (2) Early tests in the series were performed with unprotected columns which suffered severe buckling just below the beam columns connections. (3) Deflections of some beams were more than sufficient to cause internal lightweight compartment walls to fail leading to loss of compartmentation and thus increased fire spread (Bailey, 2004). (4) Buckling in the lower flanges of the beams occurred at the ends owing to highinduced compressive forces as the moments were redistributed away from midspan. This has led to the recommendation that design allowing for increased

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moment capacity at the connection (Lawson, 1990a, b) should not be used where the beam is restrained against lateral movement during a fire (Bailey et al., 1999). (5) During the cooling phase the beams do not recover plastic deformations and potentially thus place the connections in tension. This in fact resulted in the partial failure of some connections due to shear in the bolts, excessive flexure of end plates or failure in the welds connecting the end plates to the beams (Bailey, 2004). Design methods have, however, been introduced which allow for some secondary beams in composite construction to be left unprotected (Newman et al., 2006) and which allow a better analysis of composite floor slab action under the compressive membrane force system induced in the fire limit state (Bailey (2001, 2003)). Even with the traditional approach shelf angle beams will give 30 min standard fire performance and may give 60 min (Newman, 1993). Slim floor construction should give 60 min with no additional fire protection. Under no circumstances should columns be left unprotected. For fire performance periods of 30 min infilling the web space of a UC with non-structural blockwork can be sufficient (Newman, 1992).

8.11 ADDITIONAL DESIGN CONSTRAINTS 8.11.1 Ground Conditions and Foundations This section is intended to be neither a comprehensive survey of foundation design or construction nor of problems associated with ground conditions. It is offered as an aidememoir and covers the implications of any problematic areas. For a full consideration, the reader is referred to Henry (1986). Also the design of concrete foundations is covered in Martin and Purkiss (2006). The selection of foundation type generally depends on the potential bearing capacity of the ground and its susceptibility to absolute or differential settlement. The most economic form of foundation is the simple pad foundation which in order to be of a reasonable size requires a reasonably high bearing capacity. If bearing capacities are low, and the loads applied to the foundations are high, the individual pad foundations will start to overlap and will become either combined foundations or in the extreme case raft foundations. Where the bearing capacity is low then either piled foundations, ground stabilization, or the use of replacement imported fill should be considered. The decision will be primarily based on cost, although environmental considerations such as noise, dust and extra traffic must be taken into account. Ideally pad or raft foundations should be designed for uniform bearing pressure at ultimate limit state under imposed loading other than wind or accidental actions. This will generally be possible for raft foundations, but may not be for pad foundations. It is recommended that for pad foundations the ratio of the maximum to minimum bearing pressures should not exceed around 1,5. In order to achieve this it is possible to offset any stanchion away from the centre line of the foundation except where severe

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Chapter 8 / Frames and Framing moment reversals may occur. It also leads to the adoption of nominally pinned feet where the axial load in single storey structures is low as the required ratio cannot be satisfied with fixed feet. Uplift should never be allowed below a foundation under normal conditions. It should be noted that it may be necessary, in practice, to supply a degree of fixity to bases of columns in single bay portal frames when considering such frames at site boundaries in the case of fire and thereby to accept a high ratio of maximum to minimum bearing pressures, even uplift, as this is an accidental situation. Any horizontal reaction at the base of a column will also apply a moment to the underside of the foundation and must be included. If the structure is tied together by ground beams then only the net horizontal force need be considered, otherwise each reaction must be considered independently. For multi-storey structures it is general practice to employ tower cranes to facilitate erection. These need substantial foundations which are often incorporated into the structure as the foundations to the lift shafts. Differential settlement may either be caused by non-uniform loading imposed by the structure on its foundations or by variable ground conditions beneath the foundations. The effect of non-uniform loading can be mitigated by placing the whole structure on a single raft foundation, and thus producing sensibly uniform bearing pressures. Also to avoid differential movement between any masonry cladding and the frame, the masonry can be supported on ground beams tied to the pad foundations below the columns. Alternatively the components of the structure can be allowed to settle independently by the provision of movement joints, in which case the cladding may not be considered to act as bracing and full wind bracing will need to be designed. Where the structure is such that differential settlement cannot be avoided then either the structure must be designed to accept these movements (note, differential settlement does not affect moments when determined using plastic analysis; they do when an elastic analysis is used), or the structure can be articulated such that the movements do not affect the internal forces by the provision of hinges, often making the structure iso-static. Typically such hinges should be at the points of contraflexure, and must be checked for rotational capacity expected through such movements (e.g. Fig. 8.12). This type of articulation is frequently used on bridge decks where there may be substantial differential settlement due often to mining subsidence. Often in this case where such subsidence could be large, provision is built in to jack up the structure, in which case a realistic estimate will be needed of such movements.

8.11.2 Expansion and Contraction Joints It can be a matter of some debate on whether these should be provided and if so at what spacing. It is suggested that masonry should have movement joints at around 7 m centres (or the frame spacing), and concrete floor slabs at around 20–30 m (Deacon, 1986; Bussell and Cather, 1995; Concrete Society, 2003). This suggests that continuous multi-bay frames should have movement joints at around the same spacing. Thus either

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UDL

Note : BMD is drawn on the tension face on the beam

FIGURE 8.12 Cantilever-suspended span structure

movement joints should be provided through the whole height of the structure, in which case full wind bracing needs providing in each portion of the structure, or the structure will need designing to take the actions caused by such movements. For information on movement joints see Alexander and Lawson (1981).

8.11.3 Stability Stability can be considered at two levels: member stability and frame stability. • Member stability This effectively ensures that there will be sufficient ductility in the member to ensure that plastic rotational capacity is available when required by the design/analysis synthesis. The ductility check is made by considering limiting values of flange and web slenderness. It is essential that where plastic analysis is used Class 1 sections are mandatory, although where only a single hinge is needed for collapse, as in a simply supported beam, Class 2 sections may be used. It should be noted that virtually all UKBs are Class 1 whether Grade S255 or S335. The assumption is made in plastic analysis that the member can achieve its full plastic moment capacity before the onset of elasto-plastic buckling. For members in simple construction or isolated beam elements, this condition is not necessary as the load carrying capacity of the member can be checked using reduced strengths which allow for buckling. For rigid jointed frames premature elasto-plastic buckling is not permissible and must be counteracted either by bracing which reduces effective, or system, lengths below critical values or by increasing member sizes.

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Chapter 8 / Frames and Framing • Frame stability For portal frame systems this takes the form of two checks. The first is the determination of vertical deflections at the ridges and horizontal deflections at the top of the stanchions at the eaves. The second check is only for multi-span portals and is designed to avoid ‘snap through’ of the rafters. For multi-storey structures both the relative lateral deflections on each storey and the overall lateral deflection of the whole structure is subject to limits.

8.12 DESIGN PHILOSOPHIES The basic analysis methods allowed are elastic or plastic (the use of elasto-plastic methods may be needed where sway is important). Traditionally the UK has adopted simple framing where the frame carries the vertical loading and, the bracing system, the horizontal loading. Continuous framing (except for portal frames) and semi-continuous framing are rarer.

8.12.1 Frame Classification A frame may be classified as sway or non-sway. A non-sway frame is one where horizontal deformations are limited by the provision of substantial bracing members often in the form of lift shafts or stairwells often as part of a central core. A frame with bracing may still be classified as a sway frame. The criteria for neglecting global second order effects are for • Elastic analysis αcr =

Fcr ≥ 10,0 FEd

(8.7)

• Plastic analysis αcr =

Fcr ≥ 15,0 FEd

(8.8)

where αcr is the ratio by which the design loading would have to be increased to cause global elastic instability, FEd is the design load on the structure and Fcr is the elastic critical global buckling load. The higher limit for plastic analysis is due to the issue that imperfections become more important in plastic analysis due to P– δ effects. For shallow pitch portal frames (roof slope less than 26◦ ) or beam and column type structures αcr is given by αcr =

HEd h VEd δh,Ed

(8.9)

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where HEd is the total horizontal reaction at the bottom of a storey (including notional loads), VEd is total vertical loading at the bottom of a storey, δh,Ed is the horizontal displacement due to horizontal loading including notional horizontal loads and h is the storey height. The value of αcr may also be determined by applying nominal geometric rotations, or equivalent horizontal static forces, to the nodes in a frame and determining the resultant horizontal deformations. The disturbing forces are small so that any sway is low and will not cause large deflections. The method is based on the theory proposed by Horne (1975) for the determination of the elastic critical load factor. A braced frame cannot a priori be considered as a non-sway frame. Determination of sway classification is either carried out by imposing a rotation of ϕ at the foot of each stanchion or by imposing horizontal forces of ϕ W at each level of the frame where W is the total factored vertical loading at that level. For each level the sawy index /h is determined. From Horne and Morris (1981) the value of αcr is given by the minimum value of 0,009 αcr =

(8.10)

h max

For multi-storey frames with diagonal bracing Zalka (1999) provides a very quick way of estimating αcr . The critical frame buckling load Ncr for a frame loaded with uniformly distributed loading is given by Ncr = λK

(8.11)

where the parameter λ is tabulated in terms of a parameter βs defined by βs =

K Ng

(8.12)

where Ng is the global bending critical load given by Ng =

7,837Ec Ig n n + 1,588 H2

(8.13)

Where n is the number of storeys, H is the height of the building and Ec Ig is the global flexural rigidity of the columns determined using the parallel axis theorem with the second moment of the columns themselves being neglected. The shear stiffness K is dependent upon the type of bracing. For single direction bracing, −1 d3 1 K= − Ah Eh h Ad Ed hl2

(8.14)

and for complete diagonal cross-bracing, K = Ad Ed

hl2 d3

(8.15)

•

Chapter 8 / Frames and Framing 1.0 0.9 0.8 0.7 0.6 λ

308

0.5 0.4 0.3 0.2 0.1 0 0

2

4

6

8

10

12

14

16

18

20

βs

FIGURE 8.13 Critical load parameter λ (Zalka, 1999) by permission where Ad Ed is the axial rigidity of the bracing member, Ah Eh is the axial rigidity of the horizontal member, d is the length of the diagonal member, l is the width of the braced bay and h is the storey height. The formulations of K for other bracing layouts are given in Zalka (1999). Figure 8.13 from Zalka (1999) gives the relationship between βs and λ (the data are also given in tabular form in Zalka).

8.12.2 Frame Imperfections For frames sensitive to sway buckling, an allowance in the frame analysis should be made by introducing and initial sway imperfection and an initial bow.

8.12.2.1 Initial Sway Imperfection The imperfection factor φ is determined from φ = αh αm φ0

(8.16)

where the reduction factor for the building height αh is given by 2 αh = √ h

(8.17)

where h is the height of the building (in m) and 2/3 ≤ αh ≤ 1,0. The factor αm depends on the number of columns m in a row, 1 αm = 0,5 1 + m The basic imperfection factor φ0 is equal to 1/200.

(8.18)

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Note, the effects of this may be ignored if HEd ≥ 0,15VEd , provided the normalized slenderness ratio λ complies with Afy λ ≥ 0,3 (8.19) NEd

8.12.2.2 Initial Bow This is expressed as the initial bow e0 divided by the length l. It is a function of the strut buckling curve and whether elastic or plastic methods of analysis are used. The values are given in Table 5.1 of EN 1993-1-1. These effects should however be ignored if the members are checked for buckling.

8.12.3 Simple Framing In simple framing the joints between the members may be assumed to offer negligible moment transfer, and the beams should be designed as simply supported. This means any horizontal loading must be taken by bracing or any equivalent substitute. Simple framing can be used to design conventional multi-storey beam and column structures or triangulated truss systems where there will be no significant secondary moments due to joint fixity.

8.12.4 Continuous Framing In this case the connections are designed to give full rigidity between members at a joint. Note on sway and non-sway frames. Elastic, elasto-plastic or full rigid plastic analyses may be used. It is suggested that reference should be made to Ghali and Neville (1989), Coates et al. (1988), Horne and Morris (1981), Neale (1977) or Moy (1996).

8.12.5 Semi-continuous Framing There are two basic approaches to semi-continuous framing: the first is to employ moment capacity/rotation characteristics and the second is a hybrid between simple and continuous. The basic approach is to use an analysis which incorporates the connection characteristics in the analysis (Roberts, 1981; Taylor, 1981). In an elastic analysis the actual moment–rotation characteristic is input as an equivalent spring at the joint, in a plastic analysis the moment capacity is used as a local plastic moment capacity, provided the connection has sufficient ductility. In the hybrid method, the frame is designed as simple for the purpose of determining the member sizes required to carry the imposed vertical loading, but as continuous when determining the effects of lateral, or horizontal, loading on the structure (Salter

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•

Chapter 8 / Frames and Framing et al., 1999). There are limitations on the method based on number of storeys and layout, but it can be clearly advantageous as it will avoid the need to provide any wind bracing, but at the expense of providing connections capable of transmitting the moments due to wind. An alternative approach is to use an elastic analysis and then redistribute the moments to simulate plastic action. EN 1993-1-1, cl 5.4 4(B) places restrictions by specifying that the maximum redistribution is limited to 15%, the section classifications of members where moments are reduced must be no worse than Class 2, and that lateral torsional buckling must be prevented.

8.13 DESIGN ISSUES FOR MULTI-STOREY STRUCTURES Brown et al. (2004) provide a useful overview of constructional details for multi-storey frames. A distinction needs to be drawn between frames which are rigid jointed under all applied loading and frames in which the connections are either semi-rigid or pinned. With pinned connections it should be reiterated that the frame requires bracing to resist the effects of any horizontal actions on the frame, and that even if a frame is braced it cannot be assumed that the frame is a no-sway frame (Section 8.12).

8.13.1 Simple Construction Frames In simple construction frames, the beams are designed as simply supported and the columns designed to take the beam reactions together with nominal fixing moments. The nominal moments should be calculated assuming the beam reactions are applied at a minimum of 100 mm from the face of the column for major axis loading. For minor axis loading, which should be much smaller, the load is applied at either 100 mm from the web or the edges of the flanges depending upon the connection detail. If the reaction at the connection acts at a fixed point then this distance should be taken if it exceeds 100 mm. The nominal fixing moments may be distributed to the lengths of the columns immediately above and below the level being designed. There is no carry over to other storeys outside those above and below. The distribution should be in proportion to the stiffness of the column segments.

8.13.2 Semi-rigid Frames Two methods are here available, either the frame is analysed using actual or design connection moment–rotation characteristics and the frame designed in the same manner as a rigid frame, or a hybrid method is used combining simple and continuous construction. The latter method has already been briefly outlined (Section 8.11.3), but further comment is necessary. The method has the effect of reducing the beam crosssection and the beam deflection. This will ensure that the combined effects of wind, permanent and variable loads will not be critical on the beam itself. The columns need

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to be designed to resist the worst effects of either the permanent and variable loads or the combination of wind, permanent and variable. In both cases the nominal fixing moments from the beams together with the moments induced by the nominal frame imperfection loads must be considered, and may well produce a heavier column cross-section than for simple construction.

8.13.3 Continuous Construction For continuous construction either elastic or plastic methods may be used to carry out the analysis. There is a restriction that the compression flange at the hinge must be Class 1, and where a transverse force exceeding 10% of the shear resistance is applied to the web at the position of the plastic hinge then web stiffeners need to be supplied at a distance of h/2 from the hinge where h is the beam depth (cl 5.6 2b). For varying cross-section members, the web thickness should not be reduced for a distance 2d either side of the hinge (where d is the depth between fillets), the flanges must be Class 1 for the greater distance of 2d or the point where the moment drops to 80% of the plastic moment at the hinge. Restraints must be provided at rotated plastic hinges (i.e. all but the last one to form) (cl. 6.3.5.2). This may be provided by diagonal bracing from the lower flange to beams or purlins fixed to the top flange. Where members carry moment only, or moment with axial tension, and are connected to a concrete slab by shear connectors, then this is deemed sufficient for rolled ‘I’ or ‘H’ sections. At each hinge the restraint and connections should be capable of resisting a force equal to 0,0025Nf,Ed where Nf,Ed is the axial force in the compression flange. In addition to the imperfection force given by Eq. (8.6), the bracing system should be able to carry a force Qm is given by Qm = 1,5αm

Nf,Ed 100

(8.20)

The lateral torsional buckling check of segments between restraints need not be carried out if the length between restraints is less than Lstable and the moment gradient is linear, where Lstable is given as for 0,625 ≤ ψ ≤ 1,0 Lstable = 35iz

235 fy

(8.21)

and for −1,0 ≤ ψ ≤ 0,625 Lstable = (60 − 40ψ)iz

235 fy

(8.22)

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•

Chapter 8 / Frames and Framing where the moment ratio ψ is given by ψ=

MEd,mim Mpl,Rd

(8.23)

Note that where a rotated plastic hinge location occurs immediately adjacent to the end of a haunch, the tapered section need not be checked for stability if the restraint is placed at h/2 along the tapered section (not the uniform section) and the haunch remains elastic along its length.

8.13.4 Effective Length Factors for Continuous Construction For either elastic or plastic analyses, the design of the columns takes account of the relative joint stiffnesses to determine the effective length factor for the column being designed. The two distribution factors k1 and k2 at the top and bottom of the column are defined as k1 =

K c + K1 Kc + K1 + K11 + K12

(8.24)

k2 =

K c + K2 Kc + K2 + K21 + K22

(8.25)

where Kc is the stiffness of the column, K1 and K2 the stiffnesses of the columns at ends 1 and 2, K11 and K12 the stiffnesses of the beams framing in at end 1 and K21 and K22 the stiffnesses of the beams framing in at end 2 (Fig. 8.14). • Non-sway frames The available information to calculate the effective length l is given in BS 5950 Part 1: 2000. The ratio of the effective length l to the actual length L is given by either l = 0,5 + 0,14(k1 + k2 ) + 0,055(k1 + k2 )2 L

(8.26)

Note that the moments due to the nominal frame imperfections must be included in the design moments for the frame in addition to those caused by any imposed actions. • Sway frames The sway mode effective lengths are calculated from l 1 − 0,2(k1 + k2 ) − 0,12k1 k2 1/2 = L 1 − 0,8(k1 + k2 ) + 0,6k1 k2

(8.27)

Williams and Sharp (1990) provide alternative formulations: l π =! k +k2 −k1 k2 L 12 − 36 1 4−k k 1 2

(8.28)

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K1 K12

Kc

K12

Column being designed

K21

K22

K2

Note : The K values are the stiffnesses of the upper and lower beams, and the upper and lower columns in consistent units.

FIGURE 8.14 Data for determination of column effective lengths or π(1,05 − 0,05k1 k2 ) l =! k +k2 −k1 k2 L 12 − 36 1 4−k 1 k2

(8.29)

Note that for a sway frame l/L may exceed 1. Sway moments should be increased using an amplification factor given by 1/(1 − 1/αcr ), provided αcr ≥ 3,0.

8.13.5 Column Loads in Multi-storey Structures In a multi-storey structure, the frame analysis will provide the reactions in the columns. However, such analyses assume that all floors are fully loaded at all times. Statistically this is conservative, thus the axial loads due to variable loading may be reduced in the lower columns of a multi-storey structure. Note, there is clearly no reduction in loads due to permanent (or quasi-permanent) loading, and each individual floor must be designed under full variable and permanent actions. The variable loading may also be adjusted for loaded area, as the loading will be more concentrated over smaller floor areas (and less concentrated over large areas). This means that column reactions may be reduced for this reason also. However, reductions may be made either for the number of floors or loaded area. These reductions are only for loading categories

314

•

Chapter 8 / Frames and Framing A to D of Table 6.1 of EN 1991-1-1. This prohibits any reduction on situations where the loads are predominantly storage or industrial. The reduction factor αn for the number of stories n (n > 2) is given by αn =

2 + (n − 2) ψ0 n

(8.30)

The reduction factor αA for the loaded area is given by αA =

5 A0 ψ0 + ≤ 1,0 7 A

(8.31)

where A is the loaded area, A0 the reference value of 10 m2 . The reduction factor is also subject to the restriction that for load categories C and D it may not be less than 0,6.

8.14 PORTAL FRAME DESIGN 8.14.1 Single Bay Portal To illustrate the design procedure an internal frame with full rigid connections is considered (EXAMPLE 8.3). • A heavier section will be used for the stanchion than the rafter. Although this requires haunches at both the eaves and the ridge to accommodate the connections, the solution is still more economic than the use of a uniform section throughout. • To avoid stability problems in the eaves haunch the stresses in the eaves are kept elastic by extending the haunch for a distance of span/10 from the face of the stanchion. The haunch is taken at its maximum depth of twice that of the rafter in order to ease fabrication. • Plastic hinges are assumed to occur in the stanchion at the base of the haunch, that is, 1,5 times the rafter depth below the intersection of the rafter–stanchion centre lines, and at the second purlin point below the ridge. It is then usual to check the moment at the first purlin point. • The frame will be designed under variable and permanent actions. The frame will not be checked for wind, as this case is rarely critical on a portal frame. • Stability will be checked by applying notional forces of ϕN at the top of the stanchion. The approach given by Davies (1990) will also be checked. • The connections at the eaves, ridge and base will not be designed (see Chapter 7). • The background to the stanchion and rafter stability checks is given in Horne and Ajmani (1971a, b), Horne et al. (1979).

EXAMPLE 8.3 Portal frame design. Prepare a design in grade S355 steel for the frame whose basic geometry is given in Fig. 8.15.

•

315

Spacing 9000 ctrs. Purlin spacing 1250 ctrs (plan)

5000

2200

Structural Design of Steelwork to EN 1993 and EN 1994

25000

FIGURE 8.15 Basic frame geometry for Example 8.3 Permanent loading (kPa):

Sheeting Purlins Frame Total

0,18 0,06 0,18 0,42

UDL due to permanent loading: 0,42 × 9,0 = 3,78 kN/m Nominal variable load (EN 1991-1-1): 0,60 kPa (or 5,40 kN/m) Total factored design load: q = 1,35 × 3,78 + 1,5 × 5,40 = 13,20 kN/m Since the purlins are at 1,25 m centres, it will be accurate enough to apply the roof loading as a UDL. Preliminary design (geometry in Fig. 8.16): Assume hh = 0,65 m; h2 = hc − hh = 5,00 − 0,65 = 4,35 m Distance from ridge to second purlin point, b = 2,5 m l = L/2 − b = 12,5 − 2,5 = 10,0 m Gradient of rafter, sr = 2,2/12,5 = 0,176 Height to second purlin point, r: r = hc + lsr = 5,0 + 10 × 0,176 = 6,76 m Moment at base of haunch, MB1 : MB1 = Hh2 = 4,35H

(8.32)

Moment at second purlin point, M2 : The vertical reaction at the base of the stanchion V is given by V =

qL 25 = 13,2 = 165 kN 2 2

M2 = Vl − Hr −

13,2 × 102 qL2 = 10 × 165 − 6,76H − = 990 − 6,76H 2 2

(8.33)

Chapter 8 / Frames and Framing 1st 2nd purlin purlin X Y

cL

hh

hr

r

B b

h2

•

h

316

Note : All distances to member cL’s unless otherwise indicated

A

H

V

I

b L/2

FIGURE 8.16 Detail frame geometry

By setting M2 equal to Mb2 (i.e. equal strength rafter and column) Eqs (8.32) and (8.33) may be solved to give H = Hbal = 89,1 kN This will give equal plastic moment capacities in the column and rafter. This is not an optimal solution as it is general practice to use a larger section column than rafter. An optimal solution can be obtained by increasing Hbal by around 15–20%. Increase Hbal to 104 kN (i.e. an increase of 16,7%). Thus from Eq. (8.32), MB1 (=Mpl,stanchion ) is given as MB1 = Hh2 = 4,35H = 4,35 × 104 = 452,4 kNm γM0 1,0 = 452,4 × 103 Wpl = Mpl = 1274 cm3 fy 355 Select a 457 × 191 × 67 UKB (section classification Class 1: Mpl = 522 kNm) and from Eq. (8.33) M2 (=Mpl,rafter ): M2 = 990 − 6,76H = 990 − 6,76 × 104 = 287 kNm γM0 1,0 = 287 × 103 Wpl = Mpl = 808 cm3 fy 355 Select a 356 × 171 × 57 UKB (section classification is Class 1: Mpl = 359 kNm)

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Note both the rafter and stanchion have been overdesigned to allow for any reduction in carrying capacity due to sway classification and any slight changes in frame geometry over those assumed in the preliminary assessment of dimensions. Check haunch depth: h = 358 mm, so depth to base of haunch, hh = 1,5 × 358 = 537 mm Hinge at B1 occurs at 5,00 − 0,537 = 4,463 m above base. Thus H = Mpl,stanchion /h2 = 475/4,463 = 106,4 kN The slight change in H can be ignored as it is small (2,3%). Check load required to give collapse (q1 ) at the second purlin point, as the first hinge to occur is at the eaves (from an elastic distribution of bending moments). Mpl,rafter = 10 × 12,5q1 − 106,4 × 6,76 −

q1 102 = 75q1 − 719 2

(8.34)

Equate the value of Mpl,rafter for Eq. (8.34) with the actual value to give 75q1 − 719 = 326 or q1 = 13,9 kN/m This is greater than the design load of 13,2 kN/m, thus collapse does not occur at the second purlin point. Check moment at first purlin point from the apex: l = 11,25 m; r = 5,00 + 0,176 × 11,25 = 6,98 m M1 = Vl − Hr −

ql2 13,2 × 11,252 = 165 × 11,25 − 106,4 × 6,98 − = 278 kNm 2 2

This is less than Mpl,rafter , and is therefore satisfactory. Check moment at end of haunch: b = 2,5 + 0,5hstanchion = 2,5 + 0,5 × 0,4534 = 2,727 m MH = Vb − (5,00 + 0,176b)H −

qb2 2

= 165 × 2,727 − 106,4(5,00 + 0,176 × 2,727) −

13,2 × 2,7272 = −182 kNm 2

Morris and Randall (1979) suggested this moment should be limited numerically to 0,87Mpl,rafter /γ, where γ is the load factor between the total ultimate load and the characteristic loads. The 0,87 factor accounts for the approximate ratio between the moment to first yield and the plastic moment capacity (i.e. the shape factor).

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Chapter 8 / Frames and Framing Total load factor, γ: Load at ULS = 13,2 kN/m and total characteristic load = 3,78 + 5,40 = 9,18 kN/m, so γ = 13,2/9,18 = 1,44. 0,87Mpl,rafter 0,87 × 359 = = 217 kNm γ 1,44 An alternative is to calculate the moment capacity to first yield Myield divided by the total load factor: Myield fy 1 355 1 = Wel 896 = × 10−3 = 221 kNm γ γ γM0 1,44 1,0 Both methods give similar results, are less than the applied moment, and are therefore satisfactory. The maximum load the frame can carry qmax assuming hinges occur simultaneously at the ridge and the base of the haunch is given by (Fig. 8.16): ⎡ ⎤ ⎞ ⎛ hr 1 + 8 h ⎠ qmax = 2 ⎣Mpl,R + ⎝ (8.35) Mpl,c ⎦ hh L 1− h

or

⎡

qmax

⎞ ⎤ ⎡ ⎛ ⎞ ⎤ 2,2 hr 1 + 1 + 8 8 h ⎠ 5 ⎠ Mpl,c ⎦ = 2 ⎣359 + ⎝ 522⎦ = 2 ⎣Mpl,R + ⎝ 0,537 hh L 25 1− 5 1− h ⎛

= 15,37 kN/m Thus the actual load factor γ max on the applied loading is γmax =

15,37 = 1,67 9,18

The required load factor γ from above is 1,44. Horizontal sway stability The deflection δh under a force ϕN applied at the top of the (unhaunched) frame is given by ⎤ ⎡ ⎛ ⎛ ⎞ h1 Ir h Ir h 2I L + 6 + 3 h ⎥ ⎢ Ic Lr + cos θ ⎜ c r ⎥ ⎢⎝ ⎠ ⎜1 − 2 ⎥ ⎢ ⎝ Ir h h1 h1 Ir h ⎥ ⎢ 2 I L +3+3 h + h Ic Lr ⎥ ⎢ 3 c r φNh ⎢ ⎥ ⎛ ⎞2 ⎞ δh = ⎥ ⎢ 2 ⎥ ⎢ 3EIc h1 h1 Ir h ⎥ ⎢ 0,25 − 2I L + 6 + 3 h ⎜ ⎟ ⎟ ⎥ ⎢ h c r ⎟ ⎟− ⎥ ⎢ +2⎜ cos θ 2 ⎠ ⎠ ⎝ I h ⎦ ⎣ r h h Ih 4 I rL + 3 + 3 h1 h1 Ic L r c r

(8.36) The notation is given in Fig. 8.17

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Lr

Ir

Ir

P wN Ic

Ic

h

P wN

FIGURE 8.17 Idealized frame for stanchion deflection

Calculation of ϕ from Eq. (8.16): Determine αm from Eq. (8.18): αm =

1 0,5 1 + m

=

1 0,5 1 + 2

= 0,866

Determine αh from Eq. (8.17) taking h as the height to eaves (as this is more conservative) 2 2 αh = √ = √ = 0,894 h 5 The value αh is within the limits of 2/3 and 1,0, thus φ = αm αh φ0 = 0,866 × 0,894

1 1 = 200 258

If N is taken as the axial load in the column, then it may be taken equal to V (=165 kN). The length of the rafter Lr is given by Lr =

L 25 = = 12,69 m 2 cos θ 2 × cos 10

Equation (8.36) is best evaluated in sections: Ir h 16040 × 5 = = 0,215 Ic L r 29380 × 12,69 Ir h 2,2 h1 = 7,75 + 6 + 3 = 2 × 0,215 + 6 + 3 Ic L r h 5 2 Ir h h1 2,2 2 h1 2,2 + +3+3 + = 0,215 + 3 + 3 = 4,729 Ic L r h h 5 5

2

320

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Chapter 8 / Frames and Framing

δh =

1 165 × 53 258

3 × 210 × 106 × 29380 × 10−8 2 0,215 + cos 10 7,75 7,75 × 1− +2 0,215 2 × 4,729 4 × 4,729 ⎞ 2 2,2 0,25 − 5 ⎟ − cos 10⎠ 0,215

= 1,133 × 10−3 m Horne and Morris (1981) suggested that under a horizontal load of 1% of the total applied load the deflection should be limited to 0,0009hc . This is equivalent to 0,00045hc under 0.5% (1/200) of the load applied to a single column. If the initial imperfection ϕ0 is modified by αm αh then the limiting deflection δlim is reduced to 0,0045αm αh hc . δlim = 0,00045αm αh hc = 0,00045 × 0,866 × 0,894 × 5 = 1,74 × 10−3 m The actual deflection is lower and has been determined on an unhaunched frame. An elastic analysis on the haunched frame gives a horizontal deflection of 0,823 mm, or h/6075 (=164,6 × 10−3 h). The deflection determined using the unhaunched frame is approximately 40% higher. This is acceptable. Horne (1975) originally suggested that for multi-storey frames the elastic critical load factor αcr,H was given by αcr,H =

Hnom hc δnom VULS

(8.37)

where δnom is the horizontal deflection due to a nominal horizontal force Hnom , and VULS is the axial force in the column. Thus using the results determined above for Hnom (as ϕN) and δnom as δh , αcr is given as αcr,H =

1 165 258

5 = 17,1 1,133 × 10−3 165

The actual value of the notional horizontal load is not critical as only the load-deflection ratio is required, but clearly the applied load should not cause any type of failure. However, Lim et al. (2005) have suggested that Horne’s approach to calculating αcr is unconservative for portal frames, that is, it overestimates αcr . Lim et al. (2005) suggest a better estimate of αcr,est is given by NR,ULS αcr,est = 0.8 1 − αcr,H NR,cr max

(8.38)

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where NR,ULS is the axial load in the rafter and NR,cr is the Euler buckling load of the rafter based on the span of the frame, and that the ratio should be taken at its maximum. The maximum axial load in the rafter is at the haunch, and equals 133,5 kN. Owing to the purlin restraints at 1,25 m centres means the rafter as a whole can only buckle about the major axis. The nominal buckling load Ncr assuming pinned ends and neglecting the effect of the haunches is given by Ncr =

π2 EIr π2 210 × 106 × 29380 × 10−8 = = 974,3 kN L2 252

Determine αcr,est from Eq. (8.38) NR,ULS 133,5 αcr,est = 0,8 1 − αcr,H = 0,8 1 − 17,1 = 11,8 NR,cr max 974,3 From Lim et al. (2005), the second order plastic collapse load factor for a Category A type frames, the second order plastic collapse load factor αp2 may be calculated using the Merchant–Rankine equation and the first order plastic collapse load factor αp1 as αp2 = αpl

αcr − 1 αcr

(8.39)

From earlier the load factor αp1 (determined as γ max ) is 1,67, thus αp2 = αp1

αcr − 1 11,8 − 1 = 1,67 = 1,53 αcr 11,8

The required load factor on the frame is 1.44, thus the frame is satisfactory. Check the method proposed by Davies (1990) for determining λcr : λcr =

3EIr Lr (hc Nc,char + 0.3Lr Nr,char )

(8.40)

where Nr,char and Nc,char are the forces in the rafter and column under characteristic frame loading, Lr is the rafter length, hc is the height to eaves, and Ic and Ir are the major axis second moments of area of the column and rafter, respectively. L = 25,0 mm; Lr = 12,69 m; hc = 5,0 m; Ic = 29380 cm4 ; Ir = 16040 cm4 (as only relative values are required). Characteristic load/per unit run = 3,78 + 5,40 = 9,18 kN/m, Nc,char =

25 × 9,18 qL = = 114,8 kN 2 2

To estimate Nr,char the following equation may be used Nr,char =

qL2 (3 + 5m) cos θ + 0,25qL sin θ 16Nhc

(8.41)

322

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Chapter 8 / Frames and Framing where θ is the roof slope and the parameters N and m are given by I r hc N = 2 1 + m + m2 + Lr Ic

(8.42)

and h1 hc

m=1+

(8.43)

where h1 is the rise from eaves to ridge. h1 2,2 =1+ = 1,44 hc 5 I r hc 5 × 16040 N = 2 1 + m + m2 + = 2 1 + 1,44 + 1,442 + = 9,457 L r Ic 12,69 × 29380

m=1+

Nr,char = = λcr =

qL2 (3 + 5m) cos θ + 0,25qL sin θ 16Nhc 9,18 × 252 (3 + 5 × 1.44) cos 10 + 0,25 × 9,18 × 25 sin 10 = 86,0 kN 16 × 9,475 × 5

3EIr 3 × 210 × 106 × 16040 × 10−8 = 8,83 = Lr (hc Nc,char + 0,3Lr Nr,char ) 12,69(5 × 114,8 + 0,3 × 12,69 × 86,0)

This is a lesser value than that derived from the imposition of a 0,5% sway load. Adopt the criterion in Horne and Merchant (1965) for the plastic load factor derived from a modified Rankine type interaction equation with the lesser value of λcr . As 4,6 < λcr < 10,0: λp = 0,9

λcr 8,83 = 0,9 = 1,015 λcr − 1 8,83 − 1

Design UDL at ULS = 13,2 kN/m: actual carrying capacity = 15,37 kN/m (from above); surplus factor = 15,37/13,2 = 1,16. This is greater than that required by the Horne and Merchant check, so frame is satisfactory. An alternative approach to checking stability is given in Horne and Morris (1981), where a span/depth ratio check is carried out. The notation has been slightly changed to give Ic Lr

L−b 240 50W0 L Ir h cos θ ≤ h λW hc 2 + Ic Lr fy Ih

(8.44)

r

W0 is the load required to cause collapse of the rafter assuming it is a straight member of exactly the same section as the original rafter. Any stiffening caused by the ridge haunch can be ignored as it will have little effect. The eaves haunches cannot be ignored, however. There are two possible collapse mechanisms to give W0 whereby a hinge always forms at the centre of the rafter beam, and the remaining two hinges

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can occur either at the ends of the haunch in the parent rafter section, or at the end of the haunch in the enlarged section. For the general case W0 is given by W0 =

8 (Mpl,end + Mpl,centre ) (L − 2b)2

(8.45)

Where L is the span, b is the lengthy of the haunch (taken as zero for hinges forming adjacent to the stanchions), Mpl,end is the numerical value of the plastic moment capacity at the end of rafter and Mpl,centre that of the numerical value of that at the eaves. Consider first the hinges at the haunch–rafter intersection: In this case, b = 2,5 m and Mpl,end = Mpl,centre = 359 kNm, thus W0 is given by 8 8 (Mpl,end + Mpl,centre ) = (359 + 359) = 14, 36 kN/m (L − 2b)2 (25 − 2 × 2,5)2 Now, consider hinges at the end of the haunch and the centre: W0 =

In this case, the hinge at the end of the haunch must occur in the weaker member, that is, the column with b = 0; W0 =

8 8 (Mpl,end + Mpl,centre ) = 2 (359 + 522) = 11,28 kN/m (L − 2b)2 25

Take the lesser value of W0 , that is, 11,28 kN/m. Ic L r

50W0 L Ir h 240 cos θ λW hc 2+ Ir Lr fy Ic h ⎞ ⎛ 29380×12,69 50 × 11,28 25 ⎝ 5×16040 ⎠ 240 cos 10 = 99,5 = 12,69×29380 13,2 5 355 2+ 16040×5

L−b 25 − 2,5 = = 62,9 h 0,358 The criterion is therefore satisfied, and the frame is stable. Deflection check: For an unhaunched frame with pinned feet under a UDL the vertical deflection δv at the ridge is given by ⎡ h h 8 + 5 h1 3 + 2 h1 qL4 ⎢ ⎢ c c δv = ⎢10 − 2 768EIr ⎣ 2 h h Ih 3+ h1 + 3 h1 + I r Lc c c c r ⎤ 2 Ir h cos θ − 1 h1 h1 I r hc ⎥ ⎥ × (8.46) +3+ +3 + ⎥ Ic Lr cos θ hc hc Ic L r ⎦

324

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Chapter 8 / Frames and Framing Ih

θ−1 In Eq. (8.46) the term I rL cos is negligible compared with the remainder of the c r cos θ terms in the last set of round parentheses, and thus Eq. (8.46) reduces to ⎡ ⎤ h

h

8 + 5 h1 3 + 2 h1 ⎥ qL4 ⎢ c c ⎢10 − ⎥ δv = ⎣ ⎦ 2 768EIr h h Ih 3 + h1 + 3 h1 + I r Lc c

c

c r

The horizontal deflection at the eaves δh is given by δh = δv tan θ

(8.47)

Determine δv and δh in terms of q: ⎡ ⎤ h1 h1 4 8 + 5 3 + 2 qL ⎢ hc hc ⎥ δv = ⎣10 − ⎦ 2 768EIr h1 h1 Ir hc 3+ h + 3h + I L c r c c ⎡

⎤ 2,2 2,2 3+2 5 8+5 5 ⎢ ⎥ = ⎣10 − ⎦ 2 768 × 210 × 106 × 16040 × 10−8 2,2 2,2 16040×5 3+ 5 + 3 5 + 29380×12,69 254 q

= 0,0246q δh = δv tan θ = 0,0246q tan 10 = 4,34 × 10−3 q Vertical deflections: Variable load of 5,4 kN/m: δv = 0,0246q = 0,0246 × 5,4 = 0,133 m This is equivalent to span/188. Permanent load of 3,78 kN/m δv = 0,0246q = 0,0246 × 3,78 = 0.093 m This is equivalent to span/270. Deflection under variable and permanent loading: δv = 0,093 + 0,133 = 0,226 m This is equivalent to span/110. Horizontal deflections: Variable load of 5,4 kN/m: δh = 4,34 × 10−3 q = 4,34 × 10−3 × 5,4 = 0.023 m This is equivalent to height/217.

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Permanent load of 3,78 kN/m δh = 4,34 × 10−3 q = 4,34 × 10−3 × 3,78 = 0,016 m This is equivalent to height/313 Deflection under variable and permanent loading: δh = 0,016 + 0,023 = 0,039 m This is equivalent to height/128. Woolcock and Kitipornchai (1986) recommended limits under service permanent loading of span/360, variable loading span/240 for vertical deflection and height/150 for horizontal deflection. It appears unclear which loads are to be used for horizontal deflections. The horizontal deflection under variable and permanent loading marginally exceeds their recommendation but the vertical deflection well exceeds them. However, it should be noted that the frame is stiffer than assumed in the calculations owing to the presence of haunches and the stiffness of the roof sheeting has been ignored. Thus the vertical deflection calculated above will be accepted as satisfactory. A computer analysis allowing for the haunches gives a vertical deflection at the eaves under Variable load: 79,2 mm (or span/316) Permanent plus variable load: 134,7 mm (or span/186) The analysis without haunches overpredicts the vertical deflections by almost 70%. Thus the estimates with no allowance for haunches are extremely conservative. The deflections determined with haunches will be reduced even further if the effects of cladding are considered. Column stability: There are two possible approaches, the first is to supply a torsional restraint at a distance Lm below the hinge position and then to check the remainder of the column below this restraint for the combined effects of strut buckling and lateral torsional buckling. The second is to check the whole length of the column below the hinge making use of restraint to the tension flange by the sheeting rails. Method 1 The maximum distance to point of restraint in a column Lm is given by 38iz

Lm = 1 NEd 57.4 A

+

2 1 Wpl,y 2 756C1 AIt

f 2 y 235

(8.48)

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Chapter 8 / Frames and Framing Note, Eq. (8.48) must be solved iteratively as the moment gradient factor C1 is dependent upon the value of the bending moment at a distance Lm below the hinge. A conservative solution may be obtained by setting C1 = 1.0 (uniform moment). As an alternative, the length to the restraint may be taken equal to Ls where Ls is given by Ls = Lk Cm

Mpl,y,Rk MN,y,Rk + aNEd

(8.49)

where Cm is the modification factor for a linear moment gradient, MN,y,Rd is the moment capacity of the column reduced due to axial load and Lk is given h 600fy 5,4 + E iz t Lk = f (8.50) fy 5,4 E th

2

f

−1

Use Eq. (8.49): Force in the column (NEd ) = 165 kN; NEd 165 × 103 = = 19,3 MPa A 8550 Suitable sheeting rails for this frame are 202 mm deep, thus a = 0,5(453,4 + 202) = 327,7 mm Determine MN,y,Rk : Determine the ratio n between the applied force NEd and the plastic axial resistance Npl,Rd n=

NEd 165 × 103 = = 0,054 Npl,Rd 8550 355 1,0

Determine the ratio a between the total web area (excluding just the flanges) and the cross-section area: a=

A − 2btf 8550 − 2 × 189,9 × 12,7 = = 0,436 A 8550

MN,y,Rd = Mpl,y,Rd

1−n 1 − 0,054 = Mpl,y,Rd = 1,21 Mpl,y,Rd 1 − 0,5a 1 − 0,5 × 0,436

Thus MN,y,Rd = Mpl,y,Rd = 522 kNm. 38iz

Lm = 1 NEd 57,4 A

=

2 Wpl,y

1 + 756C 2 AI t 1

f 2 y 235

38 × 41,2 (1471×103 )2 1 19,3 + C12 756×8550×37,1×104 57,4 1

355 235

2 =

1566 2,059

0,336 + C2 1

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522,4 kNm 1137

Underside of haunch

Position of restraint 4467

(354 kNm)

FIGURE 8.18 BMD in stanchion AB

A

Use Eq. (5.40) to determine C1 . Under a constant moment gradient, C1 = 1,0, thus Lm = 1012 mm. After three iterations, ψ = 0,678, C1 = 1,148 and Lm = 1137 mm. Check column buckling below this level under axial load and moment (Fig. 8.18): Mz,Ed = 0: NSd = 165 kN: My,Ed = 106,4 × (4,463 − 1,137) = 354 kNm. Section classification: Flanges: Class 1. Web: Actual slenderness: c d 407,6 = = = 48,0 tw tw 8,5 Length of web xw required to carry axial force: xweb =

NEd fy

tw γ

=

M0

165 × 103 8,5 355 1,0

= 54,7 mm

Depth of web in compression αc: αc =

d xweb 407,6 + 54,7 + = = 231,2 mm 2 2 2

or α=

(αc) 231,2 = = 0,567 c 407,6

Limit for Class 1, with α > 0,5: 396 235 f

! 396 235 355

y d = = = 50,6 tw 13α − 1 13 × 0,567 − 1

328

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Chapter 8 / Frames and Framing The actual web slenderness is below this, therefore section is Class 1. As the compression flange is unrestrained, the lateral torsional buckling must be checked. As the section is Class 1, Eqs (6.61) and (6.62) of EN 1993-1-1 may be simplified as My,Ed and My,Ed are both zero, as is Mz,Ed to give NEd fy χy A γ M1

NEd fy

χz A γ

My,Ed

+ kyy

fy

χLT Wpl,y γ

M1

My,Ed

+ kzy

fy

χLT Wpl,y γ

M1

≤ 1,0

≤ 1,0

M1

Buckling curves: h 453,4 = = 2.39 > 1.2 b 189,9 y–y axis curve ‘a’ (α = 0.21) and z–z axis curve ‘b’ (α = 0,34). Calculation of χy : For yy axis buckling, the system length is taken as base of the haunch and the bottom of the column as restraints are present at both these points, that is, a length of 4,463 m π2 EI y π2 × 210 × 106 × 29380 × 10−8 = = 30572 kN L2y 4,4632 Af y 8550 × 355 λ¯ y = = = 0,315 Ncr,y 30572 × 103

Ncr,y =

y = 0,5[1 + α(λ¯ y − 0,2) + (λ¯ y )2 ] = 0,5[1 + 0.21(0,315 − 0,2) + 0,3152 ] = 0,562 χy =

y +

!

1

1

2 = 0,562 + 0,5622 − 0,3152 = 0,973 2y − λ¯ y

Calculation of χz : For yy axis buckling, the system length is taken between the restraint 1,137 m below the haunch and the bottom of the column as restraints are present at both these points, that is, a length of 3,362 m π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN L2z 3,3622 Af y 8550 × 355 λ¯ z = = = 1,064 Ncr,y 2663 × 103

Ncr,z =

z = 0,5[1 + α(λ¯ z − 0,2) + (λ¯ z )2 ] = 0,5[1 + 0,34(1,064 − 0,2) + 1,0642 ] = 1,213 χz =

z +

!

1

1

2 = 1,213 + 1,2132 − 1,0642 = 0,567 2z − λ¯ z

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Calculation of χLT : The system length for lateral torsional buckling is taken as the distance between the restraint 1,137 m below the haunch and the bottom of the column, that is, a length of 3,362 m Determine Mcr from Eq. (5.5) Iw 0,705 × 10−6 = = 0,0486 m2 Iz 1452 × 10−8 π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN L2 3,3622 L2 GI t 81 × 106 × 37,1 × 10−8 = = 0,0113 m2 2663 π2 EI z 1/2 π2 EI z Iw L2 GI t Mcr = + 2 = 2663[0,0486 + 0,0113]1/2 = 652 kNm Iz L2 π EI z Determine C1 from Eq. (5.40): ψ = 0, 1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0 = 0,6 C1 or, C1 = 1,667. From Eq. (5.12) with Mcr modified by C1 , Wpl fy 1471 × 103 × 355 × 10−6 λ¯ LT = = 0,693 = 1,667 × 652 C1 Mcr Use the general case for lateral torsion buckling, as h/b > 2, αLT = 0,34 (from Table 5.1) LT = 0,5[1 + αLT (λ¯ LT − 0,2) + (λ¯ LT )2 ] = 0,5[1 + 0,34(0,693 − 0,2) + 0,6932 ] = 0,824 χLT =

LT +

!

1

1 = = 0,788 2 0,824 + 0,8242 − 0,6932 2LT − λ¯ LT

Use Annex B of EN 1993-1-1 to determine kyy and kzy : As the moment gradient is linear with the least value of moment equal to zero, ψ = 0, so from Table B.3, all the Cm values are 0,6. ⎛ ⎞ ⎛ ⎞ NEd ⎠ NEd ⎠ ≤ Cmy ⎝1 + 0,8 kyy = Cmy ⎝1 + (λ¯ y − 0,2) fy fy χy A γ χy A γ M1 M1 ⎛ ⎞ ⎛ ⎞ 3 3 165 × 10 165 × 10 ⎠ ≤ 0,6 ⎝1 + 0,8 ⎠ = 0,6 ⎝1 + (0,315 − 0,2) 355 0,973 × 8550 355 0,973 × 8550 1,0 1,0 = 0,604 ≤ 0,627

330

•

Chapter 8 / Frames and Framing Thus, kyy = 0,604. As torsional deformations can occur (although they will be in practice slightly limited by the sheeting rails) kzy = 1 −

0,1λ¯ z 0,1 NEd NEd ≥1− f y CmLT − 0,25 χ A CmLT − 0,25 χ A fy z

= 1−

z

γM1

γM1

0,1 × 1,064 165 × 103 165 × 103 0,1 ≥ 1 − 0,6 − 0,25 0,567 × 8550 355 0,6 − 0,25 0,567 × 8550 355 1,0

1,0

= 0,971 ≥ 0,973 kxy = 0,973 NEd fy

χy A γ

My,Ed

+ kyy

fy

χLT Wpl,y γ

M1

=

M1

165 × 103

354 × 106 + 0,604 0,973 × 8550 355 0,788 × 1471 × 103 355 1,0 1,0

= 0,576 ≤ 1,0 NEd fy

χz A γ

My,Ed

+ kzy

fy

χLT Wpl,y γ

M1

M1

=

165 × 103

354 × 106 + 0,973 0,567 × 8550 355 0,788 × 1471 × 103 355 1,0 1,0

= 0,942 ≤ 1,0 Thus the column is satisfactory below the torsional restraint. Method 2 Determine the elastic critical torsional buckling load. This is given by Eq. (5.100) together with an additional term π2 EIz a2 /L2t to allow for the restraint from the sheeting rails. a = 327,7 mm (from Method 1) NcrT

1 = 2 is

π2 EIz a2 π2 EIw + + GIt L2t L2t

where Lt is the distance between restraints to both flanges. Try Lt as the height from the underside of the haunch to the base, that is, 4,463 m, and is from (Eq. (5.101)) modified by an additional term a2 to give i2s = i2z + i2y + a2 or i2s = i2z + i2y + a2 = (41,22 + 1852 + 327,72 ) × 10−6 = 0,1433 m2

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331

or, is = 379 mm. 1 π2 EIz a2 π2 EIw NcrT = 2 + + GIt is L2t L2t 1 π2 × 210 × 106 × 1452 × 10−8 = 0,1433 4,4632 π2 × 210 × 106 × 0,705 × 10−6 + + 81 × 106 × 37,1 × 10−8 4,4632

= 11265 kN NcrE =

π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 1511 kN 4,4632 L2t

NcrE 1511 = = 0,134 NcrT 11265 1 + 10η 1 + 10 × 0,134 B0 = = = 0,636 1 + 20η 1 + 20 × 0,134 √ √ 5 η 5 0,134 B1 = = 0,269 √ √ = π + 10 η π + 10 0,134 η=

B2 =

0,5 0,5 0,5 0,5 = = 0,097 − √ √ − 1 + π η 1 + 20η 1 + π 0,134 1 + 20 × 0,134

Determine the ratio βt defined as the ratio of the algebraically smaller end moment to the larger end moment. As the smaller end moment (at the base) is zero, βt = 0. Cm =

1 B0 + B1 βt + B2 βt2

(8.51)

As βt = 0, Eq. (8.51) reduces to Cm =

1 1 = = 1,572 B0 0,636

Lk is given by Eq. (8.50): 600fy h 5,4 + E tf Lk = iz = 41,2 fy 5,4 E

h 2 tf

−1

453,4 12,7

5,4 + 600×355 210×103

355 5,4 210×10 3

453,4 2 12,7

= 2893 mm −1

The reduced moment capacity MN,y,Rk due to the axial load equals the plastic moment capacity Mpl,Rk. Mpl,y,Rk 522 = 3286 mm = 2893 1,572 Ls = C m Lk 522 + 165 × 0,3277 Mn,y,Rk + aNEd This is less than the height to the underside of the haunch, and slightly less than the height to the restraint of 3362 mm.

Chapter 8 / Frames and Framing Recheck using Lt = 3362 mm. NcrT

π2 EIz a2 π2 EIw + + GIt L2t L2t 1 π2 × 210 × 106 × 1452 × 10−8 = 0,1433 3,3622

1 = 2 is

π2 × 210 × 106 × 0,705 × 10−6 + + 81 × 106 × 37,1 × 10−8 3,3622 = 19692 kN NcrE =

π2 EIz π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN 3,3622 L2t

NcrE 2663 = 0,135 = NcrT 19692 1 + 10η 1 + 10 × 0,135 B0 = = = 0,635 1 + 20η 1 + 20 × 0,135 η=

Hence, Lk = 3290 mm. This is probably adequate. Rafter (a) Haunch between connection and bracing at first purlin point. Relevant dimensions are given in Fig. 8.19.

‘zed’ purlins at 1250 ctrs

179

•

227

538

332

227

FIGURE 8.19 Haunch geometry

2500

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At any point along the rafter the value of the applied moment MSd is given by 13,2x2 2 At the connection, x = 0,227 m (ignoring end plate) and MEd,x = 165x − 106,4(5 + 0,176x) −

MSd = −499 kNm. At the restraint x = 1,25 m, MSd = −359,5 kNm. Section classification (with axial force) Flanges are Class 1. Web: Determine the axial force in the member: NSd = H cos 10◦ + V sin 10◦ = 106,4 cos 10 + 165 sin 10 = 133,4 kN. Web classification check is not needed as the web is restrained from buckling due to the end plate welded to the web and flanges forming the end plate for the connection. Determine section properties ignoring both the central flange of the haunch and the fillets. A = 2btf + dtw = 2 × 0,1722 × 0,013 + 0,0081(0,716 − 2 × 0,013) = 0,010066 m2 # 1" 3 bh − (b − tw ) (h − 2tf )3 12 1 = [0,1722 × 0,7163 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)3 ] 12

Iy =

= 0,775 × 10−3 m4 # 1" 2tf b3 + (h − 2tf )tw3 12 # 1" = 2 × 0,013 × 0,17223 + (0,716 − 2 × 0,013)0,00813 = 11,09 × 10−6 m4 12

Iz =

Plastic section modulus, Wpl,y : # 1" 2 Wpl,y = bh − (b − tw )(h − 2tf )2 4 # 1" = 0,1722 × 0,7162 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)2 4 = 2,538 × 10−3 m3 Warping constant, Iw : 11,09 × 10−6 (0,716 − 0,013)2 Iz h2s = = 1,370 × 10−6 m6 4 4 where hs is the depth between the centroids of the flanges. Iw =

334

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Chapter 8 / Frames and Framing To determine the torsional constant, It , it will be sufficiently accurate to treat the flanges and webs as thin, thus, # 1 3 1" It = bt = 2 × 0,1722 × 0,0133 + (0,716 − 2 × 0,013) 0,00813 3 3 = 0,374 × 10−6 m4 The distance between restraints, L, is the slope distance between the end of the rafter and the first purlin, and is given by 1,250 − 0,277 = 0,988 m cos 10◦ Determine Mcr from Eq. (5.5) L=

Iw 1,370 × 10−6 = = 0,1235 m2 Iz 11,09 × 10−6 π2 EIz π2 × 210 × 106 × 11,09 × 10−6 = = 23547 kN 2 L 0,9882 L2 GIt 81 × 106 × 0,374 × 10−6 = = 1,287 × 10−3 m2 2 23547 π EIz 1/2 π2 EIz Iw L2 GIt 1/2 Mcr = + 2 = 23547[0,1235 + 1,287 × 10−3 ] 2 Iz L π EIz = 8318 kNm Determine C1 from Eq. (5.40), ψ=

359,5 = 0,731 492

1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0,731 = 0,892 C1 or, C1 = 1,121. Wpl fy 2,538 × 10−3 × 355 λLT = = = 0,311 C1 Mcr 1,121 × 8318 × 10−3 As λLT < 0,4, lateral torsional buckling cannot occur. Haunch instability An alternative approach is to calculate the limiting length between lateral torsion restraints (i.e. between the haunch and the purlin immediately after the end of the haunch) (cl BB.3). √ Lk C n Ls = c

(8.52)

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where Lk is the basic critical length, and c is a factor allowing for the shape of the haunch given by 2 3 hh 3 Lh c=1+ h (8.53) Ly − 9 hs tf

where h/tf is the ratio of the depth to flange thickness tf for the basic section, Lh Ly is the proportion of the length between the restraints taken up by the taper, and hh /hs is the ratio of additional depth of the beam at the haunch to that of the basic section. The basic length Lk is given by Eq. (8.50). Determine Lh /Ly : Length of taper 2500 mm, and horizontal distance from inside of stanchion to next purlin beyond end of taper is 3 × 1250 − 227 = 3523 mm, so Lh 2500 = = 0,710 Ly 3523 As the haunch is fabricated using the same section size as the parent section minus one flange, hh = 358 − 13 = 345 mm, so hh 345 = = 0,964 hs 358 2 2 3 hh 3 Lh 3 c=1+ h = 1 + 358 0,964 3 0,710 = 1,133 hs Ly −9 tf − 9 13 Determine Lk from Eq. (8.50): Lk = iz

f

355 y 5,4 + 600 210×10 5,4 + 600 E 3 358 = 39,1 1/2 1/2 = 2838 mm 13 2 2 fy h 355 358 5,4 E t 5,4 210×103 13 −1 −1

h tf

f

Determination of Cn : To determine Cn values of the parameter R are required which is defined by R=

My,Ed + aNEd Wpl,y, fy

From safe load tables a suitable depth purlin is 232 mm, thus the depth a between the centroid of the purlins and the centroid of the member is given by a = 0,5h + 0,5 × 232 Take the value of NEd as at the end of the member, that is, 134,4 kN. The values of R are determined in Table 8.1 From Table 8.1 the maximum value of R, RS = 0,633: RE is max (R1 , R5 ) = 0,627.

336

•

Chapter 8 / Frames and Framing Cn =

12 R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE )

RS − RE = 0,633 − 0,627 = 0,006 > 0 So, Cn =

12 R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE )

12 = 1,584 0,625 + 3 × 0,629 + 4 × 0,632 + 3 × 0,633 + 0,627 + 2 × 0,006 √ √ 2838 1,584 Lk C n Ls = = = 3158 mm c 1,13 =

This exceeds the length of the haunch so there need only be restraints at haunch connection and the end of the haunch. Check between haunch and first purlin point beyond point of contraflexure. Point of contraflexure occurs when M = 0, or 165x − 106,4(5 + 0,176x) − 6,6x2 = 0, or x = 4,584 m The distance from the end of the haunch to the point of contraflexure is 4,584 − 2,727 = 1,857 m. The distance to the next purlin point is 5,00 − 2,727 = 2,273 m. This is less than the value of Lk and therefore no additional restraint is required. (c) Ridge As the moment gradient is non-linear, the stable length is given by Lk = 2833 mm (calculated above). This is greater than the slope length between purlins.

8.14.2 Notes on the Design of a Gable End Frame • The rafter is usually designed as a continuous beam spanning over the gable end columns, with simple, non-moment transferring connections to the frame. This will TABLE 8.1 Calculation of the values of R for the haunch.

Distance x (mm) Depth h (mm) A (mm2 ) I y (×108 mm4 ) W pl (×106 mm3 ) M x (kNm) aN Ed (kNm) R

1

2

3

4

5

227 716 10156 7,854 2,538 499,1 63,7 0,625

852 626,5 9431 5,556 2,104 412,2 57,7 0,629

1477 537 8706 4,031 1,702 330,4 51,7 0,632

2102 447,5 7981 2,658 1,332 253,7 45,7 0,633

2727 358,0 7256 1,603 0,996 182,2 39,6 0,627

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mean that if the spans are equal, the first bay will be critical since it will only be continuous at one end. • The centre lines of the gable end columns should be coincident with the spacing of the roof purlins in order to resist the reaction from the wind loading on the gable end columns without subjecting the rafters to bi-axial bending. It is also preferable if the spacing of the gable end columns can be similar to the frame spacing in order to keep the same section for the sheeting rails, although this is not essential. • Often the loading on the gable end will require the use of only relatively small sections, but detailing requirements and compatibility with the rest of the frame may make it necessary to use larger sections.

8.14.3 Multi-bay portal frames It is not intended to present a design example of such a frame since the principles and methods are similar to those for single bay frames. It is intended, rather, to point out additional factors that need to be taken into account. • The critical bay, assuming as in normal practice, that all the spans are equal, is the end bay. However, it must be noted that it is no longer possible to assume uniform snow loading. The frame should be first checked under the nominal roof loading of 0,6 kPa (modified for slope if necessary) as a variable load together with the permanent loading. It should then be checked for the accidental snow drift loading together with the permanent loading considered as an accidental load combination with reduced partial safety factors.

(a) Interior bays before snap through

(b) Exterior bays after snap through

FIGURE 8.20 Snap through failure of multi-bay portal frames

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Chapter 8 / Frames and Framing • Although with balanced loading the interior stanchions carry axial load only, and the internal rafters have a higher reserve of strength, it is normal practice to detail all bays identically. This is partly to achieve fabrication economies, and to avoid problems on site with varying rafter sizes, and partly to avoid deflection and ‘snap through’ failure. • Sway stability is either handled by the imposition of a nominal sway force as for single bay frames, or if the empirical method of Horne and Morris (1981) is used as in Eq. (8.44), then the ratio of the column stiffness (Ic /hc ) to the rafter stiffness (Ir /L) should be halved. • In frames with more than two bays it is possible for snap through to occur when instability happens as the top of the stanchions spread followed by inversion of the rafter (Fig. 8.20) This can be checked using the following empirical equation (Horne and Morris, 1981) Ic L 1 + 25 4 + L−b 240 h I c r tan 2θ (8.54) = λW λW h fy −1 W0

W0

where symbols are defined in Eq. (8.44). If the arching ratio is less than one, snap through cannot occur, so there is no limit on the effective length to depth ratio.

REFERENCES Alexander, S.W. (2004). New approach to disproportionate collapse, Structural Engineer, 82(23/24), 14–18. Alexander, S.J. and Lawson, R.M. (1981). Movement design in buildings. Technical Note 107. CIRIA. Bailey, C.G. (2001). Steel structures supporting composite floor slabs: design for fire. Digest 462. BRE. Bailey, C.G. (2003). New fire design method for steel frames with composite floor slabs. FBE Report 5. BRE. Bailey, C.G. (2004). Structural fire design: core or specialist subject? Structural Engineer, 82(9), 32–38. Bailey, C.G., Newman, G.M. and Simms, W.I. (1999). Design of steel framed buildings without applied fire protection. P186. SCI. Brown, D.G., King, C.M., Rackham, J.W. and Way, A.G.J. (2004). Design of multi-storey braced frames. P334. SCI. BS 5950-1. Structural use of steelwork in building – Part 1: Code of practice for design – rolled and welded sections. BSI. Bussell, M.N. and Cather, R. (1995). Design and construction of joints in concrete structures. Report 146. CIRIA. Coates, R.C., Coutie, M.G. and Kong, F.K. (1988). Structural Analysis (3rd edition). Van Nostrand Reinhold. Concrete Society (2003). Concrete industrial ground floors (3rd edition). Report TR 43. Concrete Society. Couchman, G.H., Mullett, D.L. and Rackham, J.W. (2000). Composite slabs and beams using steel decking: Best practice for design and construction. Technical Paper No.13 (SCI P300). Metal Cladding and Roofing Manufacturers Association/SCU. Davies, J.M. (1990). In plane stability in portal frames, Structural Engineer, 68(8), 141–147. Davies, J.M. and Bryan, E.R. (1982). Manual of stressed skin diaphragm construction. Granada Irwin.

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Deacon, R.C. (1986). Concrete ground floors: Their design, construction and finish. Cement and Concrete Association. Dowling, J. (2005). United States can learn from worldwide fire engineering expertise, New Steel Construction, 13(9), 26–26. DTLR (2000). The Building Regulations: Approved Document B: Fire safety. DTLR. EN 1990. Eurocode – basis of structural design. EN 1991-1-1. Eurocode 1: Actions on structures: Part 1–1: General actions – densities, self-weight, imposed loads for buildings. CEN/BSI. EN 1991-1-2. Eurocode 1: Actions on structures: Part 1–2: Actions on structures exposed to fire. CEN/BSI. EN 1991-1-3. Eurocode 1: Actions on structures: Part 1–3: Snow loads. CEN/BSI. EN 1991-1-4. Eurocode 1: Actions on structures: Part 1–4: Wind loads. CEN/BSI. EN 1991-1-5. Eurocode 1: Actions on structures: Part 1–5: General actions – thermal actions. CEN/BSI. EN 1991-1-7. Eurocode 1: Actions on structures: Part 1–7: Accidental actions due to impact and explosions. CEN/BSI. EN 1991-3. Eurocode 1: Actions on structures: Part 3: Actions induced by cranes and machinery. CEN/BSI. EN 1996-1-1. Eurocode 6: Design of masonry structures: Part 1–1: Rules for reinforced and unreinforced masonry. CEN/BSI. EN 1998. Eurocode 8: Design of structures for earthquake resistance. CEN/BSI. Ghali, A. and Neville, A.M. (1989). Structural analysis – a unified classical and matrix approach (3rd edition). Chapman and Hall. Gibbons, C. (1995). Economic steelwork design, Structural Engineer, 73(15), 250–253. Ham, S.J., Newman, G.M., Smith, C.I. and Newman, L.C. (1999). Structural fire safety: A handbook for architects and engineers. P197. SCI. Henry, F.D.C. (ed) (1986). Design and construction of engineering foundations. Chapman and Hall. HMSO (1994). Construction (Design and Management) Regulations. HMSO. Horne, M.R. (1975). An approximate method for calculating the elastic critical loads of multi-storey plane frames, The Structural Engineer, 53, 242–8. Horne, M.R. and Ajmani, J.L. (1971a). Design of columns restrained by side-rails, Structural Engineer, 49(8), 339–345. Horne, M.R. and Ajmani, J.L. (1971b). The post-buckling behaviour of laterally restrained columns, Structural Engineer, 49(8), 346–352. Horne, M.R. and Merchant, W. (1965). The stability of frames. Pergammon Press. Horne, M.R. and Morris, L.J. (1981). Plastic design of low rise frames. Constrado Monographs (republished Collins, 1985). Granada publishing. Horne, M.R., Shakir-Khalil, H. and Akhtar, S. (1979). The stability of tapered and haunched beams, Proceedings of the Institution of Civil Engineers, 67, 677–694. Horridge, J.F. (1985). The design of industrial buildings, Civil Engineering Steel Supplement, 13–16. Horridge, J.F. and Morris, L.J. (1986). Comparative costs of a single-storey steel framed structures, Structural Engineer, 64A(7), 177–181. Irwin, A.W. (1984). Design of shear wall buildings. Report 112. CIRIA. ISE (2002). Safety in tall buildings and other buildings with large occupancy. Institution of Structural Engineers. Lawson, R.M. (1990a). Behaviour of steel beam to column connections in fire, Structural Engineer, 68, 263–271. Lawson, R.M. (1990b). Enhancement of fire resistance of beams by beam to column connections. TR-086. SCI. Lawson, R.M. and Nethercot, D.J. (1985). Lateral stability of I-beams restrained by profiled sheeting, Structural Engineer, 63(B), 1–13. Lawson, R.M., Mullett, D.L. and Rackham, J.W. (1997). Design of Asymmetric Slimflor ® Beams using deep composite decking. Publication 175. SCI.

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Chapter 8 / Frames and Framing Lim, J.B.P., King, C.M., Rathbone, A.J., Davies, J.M. and Edmonson, V. (2005). Eurocode 3 and the in-plane stability of portal frames, Structural Engineer, 83(21), 43–49. Martin, L.H. and Purkiss, J.A. (2006). Concrete design to EN 1992 (2nd edition). ButterworthHeinemann. Mlakar, P.F., Dusenberry, D.O., Harris, J.R., Haynes, G., Phan, L.T. and Sozen, M.A. (2003). Pentagon building performance report, Civil Engineering, 73(2), 43–55. Morris, L.J. and Randall, A.L. (1979). Plastic Design. Constrado. Moy, S.S.J. (1996). Plastic methods for steel and concrete structures (2nd edition). MacMillan. Mullett, D.L. (1992). Slim floor design and construction. Publication 110. SCI. Mullett, D.L. (1997). Design of RHS Slimflor ® Edge Beams. Publication 169. SCI. Mullett, D.L. and Lawson, R.M. (1993). Slim floor construction using deep decking. Technical Report 120. SCI. Neale, B.G. (1977). The plastic methods of structural analysis (3rd edition). Chapman and Hall. Newman, G.M. (1992). The fire resistance of web-infilled steel columns. Publication 124. SCI. Newman, G.M. (1993). The fire resistance of shelf angle floor beams to BS 5950: Part 8. Publication 126. SCI. Newman, G.M., Robinson, J.T. and Bailey, C.G. (2006). Fire safe design: A new approach to multi-storey steel-framed buildings, P. 288. Steel Construction Institute. Purkiss, J.A. (2007). Fire safety engineering design of structures (2nd edition). Butterworth-Heinemann. Roberts, E.H. (1981). Semi-rigid design using the variable stiffness method of column design, In Joints in Structural Steelwork (eds J.H. Howlett, W.M. Jenkins, and R. Stainsby); Proceedings of the International Conference. Teesside Polytechnic (now Teesside University), 6–9 April, 1981, Pentech Press, 5.36–5.49. Salter, P.R., Couchman, G.H. and Anderson, D. (1999). Wind-moment design of low-rise frames. P-263. SCI. Simms, W.I. and Newman, G.M. (2002). Single storey steel framed buildings in fire boundary conditions. P313. SCI. Taylor, J.C. (1981). Semi-rigid beam connections: Effects on column design: B20 Code Method, In Joints in structural steelwork (eds J.H. Howlett, W.M. Jenkins, and R. Stainsby); Proceedings of the International Conference. Teesside Polytechnic (now Teesside University), 6–9 April, 1981, Pentech Press, 5.50–5.57. Way, A.G.J. (2005) Guidance on meeting the robustness requirements in Approved Document A (2004 Edition), Publication P341. SCI. Wearne, P. (1999). Collapse – Why buildings fall down. Channel 4 Books. Williams, F.W. and Sharp, G. (1990). Simple elastic critical load and effective length calculations for multi-storey rigid sway frames, Proceedings of the Institution of Civil Engineers, 90, 279–287. Wood, J.G.M. (2003). Pipers row car park collapse: Identifying risk. Concrete, 37(9), 29–31. Woolcock, S.T. and Kitipornchai, S. (1986). Deflection limits for portal frames, Steel Construction, 20(3), 2–10. Yandzio, E. and Gough, M. (1999) Protection of buildings against explosions. Publication 244. SCI. Zalka, K.A. (1999). Full-height buckling of frameworks with cross-bracing, Structures and Buildings, Proceedings of the Institution of Civil Engineers, 134, 181–191.

Chapter

9 / Trusses

This chapter is concerned with the design of triangulated and non-triangulated trusses. With the advent of rolled hollow sections the design of both types of truss has been revolutionized. The use of rolled hollow sections for trusses provides a far more efficient use of material as the buckling strengths are higher as radii of gyration are larger and lateral torsional buckling is either non-existent in the case of square or circular sections, or the effects are much reduced for rectangular sections. Rolled section trusses are much easier to fabricate as the welding is generally fillet welds or full strength butt welds. However, it needs remembering that room must be provided to allow the welds to be properly formed. This can be achieved by limiting the number of members at a node to the main chord member and at most two subsidiary web members and limiting the minimum angle between members to 30◦ . A problem however with rolled sections is forming joints to enable large trusses to be transported. If circular sections are used as main chord members fabrication problems and resultant high costs may ensue. Maintenance, and painting, is much easier with rolled sections as spray techniques can be employed. Note, rolled hollow sections are commonly available in Grade S355, but it may be difficult to obtain such sections in S275.

9.1 TRIANGULATED TRUSSES It is generally acceptable to analyze triangulated trusses on the basis of pinned joints at the nodes together with loading from the roof system applied at nodes. However, where purlins exist between the nodes then the relevant members need to designed under the effects of flexure form the loads concerned. It is conservative to consider the main chord members as simply supported between nodes, although use of continuity can be made. Where it is necessary to consider the existence of services within the roof space, such loading can be considered as nodal loads on the bottom chord of the truss. The effect of service loads must be neglected when considering the effects of wind uplift. It is not generally necessary to consider the effects of secondary moments due to joint fixity as these are usually low. Note, however if the original pin-jointed analysis is performed using a plane frame computer analysis package then it is relatively easy to

Chapter 9 / Trusses check the effect of secondary moments by using the sections determined as suitable to carry the pinned forces and simply altering the end conditions of the members. Also, provided the secondary moments are negligible, the deflections can be determined from a pin-joint analysis. It should be noted that the actual deflections in a truss will be less than the calculated deflections as any stiffening effect of the roof cladding and purlins has been ignored. Note that even if the applied actions due to wind uplift and permanent loads are lower than those due to wind downthrust, roof loading and permanent loads, consideration still needs to be given to bracing on the lower boom of the truss as this is now in compression. This will almost certainly mean the provision of longitudinal bracing to ensure system lengths are kept reasonably small to ensure buckling does not occur out of plane.

EXAMPLE 9.1 Triangulated truss design. Prepare a design using square and rectangular rolled hollow sections in Grade S355 steel for the truss whose geometry is given in Fig. 9.1. The load due to initial imperfections has been omitted partly because the value of the load is small and partly because the truss is triangulated and thus will exhibit virtually no sway effects. Actions (kPa): Permanent:

Sheeting Purlins Total

0,22 0,08 0,30

UDL for each truss = 0,30 × 9,0 = 2,7 kN/m. Assume self-weight is 1,5 kN/m, thus total permanent action is 4,2 kN/m. Since the purlins are at nodal points, nodal permanent action = 3,0 × 4, 2 = 12,60 kN (on an internal node). Variable action: From EN 1991-1-1 variable action (with no access, snow drift ignored) is 0,6 kPa, so nodal load = 0,6 × 9,0 × 3,0 = 16,2 kN J

H

L

F

0,8 m

•

3,2 m

342

B A

C

N

P R

D

E

G

I 8 bays @ 3 m

FIGURE 9.1 Truss geometry for EXAMPLE 9.1

K

M

O

Q

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Although wind loading will need considering as it may cause uplift due to suction, in this example only permanent and variable actions due to roof loading will be considered. Thus the only combination of actions needed is 1,35gk + 1,5qk . So the total nodal action is 1,35 × 12,6 + 1,5 × 16,2, that is, 41,31 kN per internal node. The truss will be designed using a pin-joint analysis, and then checked using a fixedjoint analysis. The effect of notional horizontal loading will be ignored for this design example as its effect will be negligible. Table 9.1 shows the results from a pin-joint analysis under factored variable and permanent actions for half the truss. Member design: The top and bottom chord together with the end posts will be fabricated from the same section. Maximum force (compressive) NSd is in member CF and is 379 kN. (The maximum tensile force is 320 kN and will not be critical.) From cl BB.1.3 the buckling length may be taken as 0,9L in both planes where L is the system length. Effective length: LFC = 3,105 m (slope length).

TABLE 9.1 Member forces for EXAMPLE 9.1. Fixed-joint analysis

Member

Pin-joint analysis Axial force kN

Axial force kN

Shear force kN

Moment End 1 kNm

End 2 kNm

AB CD EF GH IJ AD DE EG GI BC CF FH HJ BD CE FG HI

−165,24 −82,62 −28,92 9,54 82,62 0 309,83 371,79 357,50 −315,98 −379,14 −364,56 −315,98 320,64 68,37 −17,18 −63,08

−163,38 −82,12 −29,33 9,17 81,63 7,19 307,14 371,29 356,79 −311,97 −378,60 −364,02 −316,14 308,46 73,28 −17,27 −61,76

−7,19 −1,57 0,04 0,12 0 1,86 −0,50 0,04 −0,37 1,82 −0,37 0,12 −0,54 0,17 −0,08 0 0

2,85 −1,07 0,04 −0,12 0 −2,85 1,90 0,50 0,83 0,29 −1,61 −0,45 0,66 0,29 0,04 0,04 0,04

−2,93 1,16 −0,04 0,17 0 2,77 0,87 0,62 0,29 −0,29 −0,45 −0,78 −0,99 −0,29 0,25 0,08 0

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Chapter 9 / Trusses Out-of-plane buckling: Buckling effective length is the length of the member, that is, 3,105 m. In-plane buckling Buckling length is 0,9LFC = 2,795 m, as there is a large degree of restraint owing to the welded joints (and continuity of the chord member). Try a 100 × 100 × 8 HFRHS. A = 2910 mm2 , i = 37,4 mm, I = 4,08 × 10−6 m4 . Section classification: c b − 2r − 2t 100 − 2 × 8 − 2 × 8 = = = 8,5 t t 8 Limit for Class 1: 235 235 33ε = 33 = 33 = 26,8 fy 355 Section is Class 1. Use the greatest system length of 3,105 to calculate Ncr : π2 EI π2 × 210 × 106 × 4,08 × 10−6 = = 877 kN 2 L 3,1052 Afy 2910 × 355 × 10−3 λ= = = 1,085 Ncr 877

Ncr =

For a rolled hollow section, buckling curve ‘a’ is used, α = 0,21: = 0,5 [1 + α(λ − 0,2) + (λ)2 ] = 0,5 [1 + 0,21(1,085 − 0,2) + 1,0852 ] = 1,182 χ=

+

!

Nb,Rd = χA

1 2 − (λ)2

=

1,182 +

1 1,1822 − 1,0852

= 0,606

fy 355 × 10−3 = 626 kN = 0,606 × 2910 γM1 1,0

All the remaining members will be the same section size in order to aid fabrication. Maximum force NSd is in member BC and is 320 kN (tension). Try a 60 × 60 × 6,3 HFRHS. A = 1330 mm2 , i = 21,7 mm, I = 63,4 × 108 mm4 . Section classification: c b − 2r − 2t 60 − 2 × 6,3 − 2 × 6,3 = = = 5,52 t t 6,3

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Limiting value for Class 1: c 235 235 = 33 = 33 = 26,8. t fy 355 Section is therefore Class 1. Nt,Rd = A

fy 355 = 1330 × 10−3 = 472 kN γM0 1,0

However the compression members also need checking. Member DC carries the largest compression force of −82,6 kN. Length of DC is 1,4 m. Critical effective length is that for out-of-plane buckling, that is, 1,4 m. π2 EI π2 × 210 × 106 × 63,4 × 10−8 = = 670 kN 2 L 1,42 Afy 1330 × 355 λ= = = 0,839 Ncr 670 × 103

Ncr =

For a hot rolled RHS, α = 0,21, so = 0,5 [1 + α(λ − 0,2) + (λ)2 ] = 0,5 [1 + 0,21(0,839 − 0,2) + 0, 8392 ] = 0,987 χ=

+

!

Nb,Rd = χA

1 2 − (λ)2

=

0,987 +

1 0,9872 − 0,8392

= 0,664

fy 355 = 0,664 × 1330 × 10−3 = 314 kN > NSd γM1 1,0

The deflections were also assessed on a pin-jointed analysis and gave 0,018 m under variable load (span/1333) and 0,033 m (span/727) under variable together with permanent loading. Both these deflection ratios are acceptable. Check on self-weight of the truss: From the member sizes determined on the basis of the force distribution for the pin-joint analysis, the total mass of the truss was 1,6 ton compared with the estimate of 3,6 ton. This means that both the total forces and the total deflection have been overestimated by around 10%. The check on self-weight is carried out in Table 9.2. Web capacity in the bottom chord at A (Section 4.8). The reaction is 165,2 kN. Note as a rolled hollow section is being used, fyw = fyf = 355 MPa. Since the length of stiff bearing ss is not known, set ss = 0 and determine the value required should the check fail. For an end support with c = ss = 0, kF = 2 (type c)

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Chapter 9 / Trusses TABLE 9.2 Check on self-weight estimate for EXAMPLE 9.1. Member

Total length (m)

AQ AB and QR BJ and JR BC and OR DE and PH FG and NK HI and LI DC and PO EF and NH HG and LK JI Total mass

24 1,6 24,48 6,21 6,62 7,21 7,68 2,8 4,0 5,2 3,2

Sub total (m)

Mass/unit length (kg/m)

Mass (kg)

50,08

22,9

1147

42,92

10,5

451 1598

Determine m1 : m1 =

fyf bf bf 50 = = = 6,25 tw 8 fyw tw

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 6,25 = 8,0 = 14,1 mm 2 2

The depth of the web hw has been taken as the clear depth of the web, hw = h − 2r − 2t = h − 4t = 100 − 4 × 8 = 68 mm tw3 8,03 = 2846 kN = 0,9 × 2 × 210 hw 68 ly tw fyw 14,1 × 8,0 × 355 λF = = = 0, 119 FCR 2846 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0 χF =

0,5 0,5 = = 4,2 0,119 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 14,1 = 14,1 mm FRd = Leff tw

fyw 355 = 14,1 × 8,0 × 10−3 = 40 kN γM1 1,0

The total resistance therefore is 80 kN (from both webs). A stiff bearing is therefore required to supply a further 165,2 − 80 = 85,2 kN.

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Try a 25 mm stiff bearing. 25 + 0 ss + c kF = 2 + 6 = 4,21 =2+6 hw 68 lc =

kF Etw2 3,33 × 210 × 103 × 82 ≤ 25 + 0 = 927 ≤ 25 ≤ ss + c = 2fyw hw 2 × 355 × 68

The limiting value of lc is greater than ss , so lc = 25 mm. 2 2 m1 lc 25 6, 25 ly = lc + tf = 25 + 8,0 = 53,7 mm + + 2 tf 2 8 √ ly = lc + tf m1 = 25 + 8,0 6,25 = 45 mm

(a) (b)

The value of ly using the equation for cases (a) and (b) is clearly larger, thus the value of lc is the lesser of those above, that is, 45 mm tw3 8,03 = 0,9 × 4,21 × 210 = 5991 kN hw 68 ly tw fyw 45,0 × 8,0 × 355 λF = = = 0,146 FCR 5991 × 103

FCR = 0,9kF E

χF =

0,5 0,5 = = 3,42 0,146 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 45 = 45 mm FRd = Leff tw

fyw 355 = 45 × 8,0 × 10−3 = 128 kN γM1 1,0

The total resistance therefore is 256 kN (from both webs). This exceeds the reaction of 165,2 kN. Thus the length of stiff bearing could be reduced to around 20 mm. Note also that the end of the bottom member will be sealed by a thin plate welded over its end to prevent corrosive matter reaching the inside of the tubular section. This will have a slight stiffening effect. The results from a fixed-joint analysis are also given in Table 9.1 where it will be observed that with the exception of the members close to the supports, the axial force resultants show negligible difference. The exception is member AD which now carries an axial force of 7 kN and a moment of 2,85 kNm (Mpl,Rd = 35,5 kNm). Thus it may be concluded the secondary forces induced by joint fixity are negligible, and may be ignored. In the fixed-joint analysis the deflections are reduced by around 1–2%.

9.2 NON-TRIANGULATED TRUSSES Another common form of truss is the Vierendeel girder. This is non-triangulated, even though the top and bottom cords may not be parallel. As the loading is carried by sway,

•

Chapter 9 / Trusses there are substantial moments at the nodes, although the shears and axial forces of reasonable size. As a result, the Vierendeel girder is automatically classified as a sway frame and is generally analyzed elastically with the system or effective lengths of the members determined for the sway case. The notional horizontal load is applied at the end of the truss in the direction of the top chord.

EXAMPLE 9.2 Design of a parallel chord Vierendeel girder. Prepare a design in Grade S355 steel using rectangular hollow section for the girder detailed in Fig. 9.2. Loading: Permanent (kPa):

Sheeting Purlins Total

0,22 0,10 0,32

UDL per truss = 0,32 × 8,5 = 2,72 kN/m Self-weight (estimated) = 2,00 kN/m Total permanent = 4,72 kN/m or a nodal load of 3,0 × 4,72 = 14,16 kN (internal node) Variable action = 0,6 kPa, or 8,5 × 3,0 × 0,6 = 15,3 kN per node Total factored ultimate load per internal node = 1,35 × 14,16 + 1,5 × 15,3 = 42,07 kN For this example any load combination involving wind will not be considered. Note, however, where the load cases including wind to be considered then the moments due to wind would need multiplying by the sway amplification factor. Determination of notional horizontal loading: Determine αh from Eq. (8.17): 2 2 αh = √ = √ = 1,414 2 h The maximum value allowed for αh is 1,0,

A

C

E

G

J

K

M

O

Q

B

D

F

H

K

L

N

P

R

2,0 m

348

8 bays @ 3 m

FIGURE 9.2 Truss geometry for EXAMPLE 9.2

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thus αh = 1,0. Determine αm from Eq. (8.18): 1 1 αm = 0,5 1 + = 0,5 1 + = 0,745 m 9 Determine ϕ from Eq. (8.16): φ = φ0 αh αm = 1,0 × 0,745

1 = 3,725 × 10−3 200

Total factored vertical loading is 8 × 42,07 = 336,6 kN. Thus the notional horizontal load is 336,6 × 3,725 × 10−3 = 1,25 kN. Determination of αcr : A notional load of 1% of the total factored load is applied at the top chord level. The load is 0,01 × 336,6 = 3,37 kN. The actual analysis for both these cases was carried out using a computer package with a 1 kN load, and the results were obtained pro-rata. The results for the notional horizontal load of 1,25 kN is given in Table 9.3, and the deflections (absolute and relative) for the notional load of 3,37 kN are given in Table 9.4. As only a single storey is being considered, then from Table 9.4 the maximum relative deflection is 0,058 mm. Thus 0,058 = = 29 × 10−6 h 2000

TABLE 9.3 Internal stress resultants due to the notional horizontal load. Moment Member

Axial force kN

Shear force kN

End 1 kNm

End 2 kNm

AB CD EF GH IJ AC CE EG GI BD DF FH HJ

−0,52 0 0 0 0 −1,16 −1,00 −0,86 −0,70 1,16 1,0 0,85 0,70

0,093 0,15 0,15 0,16 0,16 −0,05 −0,05 −0,05 −0,05 0,05 −0,05 −0,05 −0,05

−0,0845 −0,1475 −0,154 −0,155 −0,155 −0,085 −0,077 −0,075 −0,074 0,085 0,077 0,075 0,074

0,0845 0,1475 0,154 0,155 0,155 0,072 0,079 0,081 0,083 −0,072 −0,079 −0,081 0,083

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Chapter 9 / Trusses TABLE 9.4 Determination of αcr . Deflection Member

Top mm

Bottom mm

Net deflection mm

AB CD EF GH IJ KL MN OP QR

0,058 0,053 0,049 0,045 0,042 0,040 0,038 0,037 0,037

0 −0,005 −0,009 −0,012 −0,016 −0,018 −0,020 −0,021 −0,021

0,058 0,058 0,058 0,057 0,058 0,058 0,058 0,058 0,058

From Eq. (8.10), the value of αcr is given by αcr =

0,009 h

=

0,009 = 310 29 × 10−6

The minimum value for elastic analysis of αcr is 10, thus second order effects may be ignored. The magnification factor to be applied to moments causing sway (i.e. for this particular example only the moments due to the notional horizontal loads) is (cl 5.2.2 (5)B, EN 1993-1-1) 1 1−

1 αcr

1

=

= 1,003 1 1 − 310

The stress resultants due to the vertical loads are given in Table 9.5. The bending moment, shear force and axial force diagrams for the total loading are plotted in Fig. 9.3(a) to (c), respectively for half the frame Try a 250 × 150 × 12,5 S355J2H Member checks: Section classification: Web: The web is subject to compression: Maximum axial compressive force NEd is 479 kN. Length of web χw to resist the compressive force is given by χw =

NEd fy

2tw γ

M0

=

479 × 103 2 × 12,5 355 1,0

= 54 mm

Structural Design of Steelwork to EN 1993 and EN 1994

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TABLE 9.5 Internal stress resultants due to the factored vertical load. Moment Member

Axial force kN

Shear force kN

End 1 kNm

End 2 kNm

AB CD EF GH IJ AC CE EG GI BD DF FH HJ

−94,53 −21,20 −20,99 −21,04 −21,04 −116,11 −291,67 −415,78 −478,29 116,11 291,67 415,78 478,29

−116,11 −175,56 −62,52 −62,52 0 73,50 52,63 31,55 10,52 73,75 52,55 31,55 −10,52

−166,11 −175,56 −124,11 −62,52 0 115,95 −70,93 −35,04 −5,01 116,24 70,72 37,15 5,01

116,24 175,56 123,90 62,52 0 104,59 86,96 57,51 26,55 −104,93 −86,92 −57,51 −26,55

479 −117

−

− −21

−95 118

416

293

−

−

−21

−21

+

+

293

−21 + 479

417 (a) AFD (kN)

+73

+55

−116

−175

+73

+53

+32

+10

−123

−62

+32

+10

(b) SFD (kN)

71

175

124

37 63

176 87

104 116

71

58

26 0

5

37 175

5

124

63

87 104 (c) BMD (kNm) contension force

FIGURE 9.3 Final AF, SF and BM diagrams for EXAMPLE 9.2

26

351

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Chapter 9 / Trusses c = h − 2tf − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm αc = 0,5c − 0,5ax = 100 + 27 = 127 mm 127 α= = 0,635 200 c 200 = = 16 tw 12,5 Class 1 limit for α > 0,5: 235 fy

!

235

c 355 = 396 = 396 = 44,4 tw 13α − 1 13 × 0,635 − 1 Web is Class 1. Flange: c = b − 2tf − 2r = 150 − 2 × 12,5 − 2 × 12,5 = 100 mm 100 c = =8 tf 12,5 Limit for Class 1 (Table 5.2 Sheet 1): c 235 235 = 72 = 72 = 58,6 tf fy 355 Flange is Class 1. The section is therefore Class 1. Shear: The shear area Av for a load parallel to the depth is given in cl 6.2.6 (EN 1993-1-1) as h 250 = 9300 = 5813 mm2 h+b 250 + 150 fy 355 1 1 = √ Av = √ 5813 × 10−3 = 1191 kN γM0 1,0 3 3

Av = A Vpl,Rd

The maximum applied shear force is 175,4 kN. Thus the section is satisfactory and there is no moment capacity reduction due to shear (VEd /Vpl,Rd = 0,15). Members BD, DF, FH and HJ need checking under combined tension and bending. This check reduces to determining whether the reduced flexural capacity due to axial load exceeds the applied moment. For box sections: MN,y,Rd = aw =

1−n Mpl,y,Rd 1 − 0,5aw A − 2bt 9300 − 2 × 150 × 12,5 = = 0,597 A 9300

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Limiting maximum value of aw = 0,5, therefore aw = 0,5. 1 − 0,5aw = 1 − 0,5 × 0,5 = 0,75 fy 355 Mpl,y,Rd = Wpl,y = 751000 × 10−6 = 267 kNm γM0 1,0 fy 355 = 9300 × 10−3 = 3302 kN γM0 1,0 NEd n= Npl,Rd

Npl,Rd = A

The checks are carried out in Table 9.6, where it should be noted that in no case is the plastic moment capacity reduced, and that the values quoted for My,Ed are the larger absolute values within the member. Members AB, CD, DF, GH, AC, CE, EG, GI need checking under combined bending and compressive axial force. The check for reduction in moment capacity for members AC to GI is exactly the same as for BD to HJ. Members AB and CD carry lower axial forces and will not therefore be critical for moment capacity reductions. Lateral torsional buckling: The system length may be taken as the worst case of member length, that is, 3 m. Lateral torsional buckling is checked using Section 5.1.8.1. Determine φb from Eq. (5.63):

φb =

=

" #" # W 2 1 − Iz 1 − GIt pl,y Iy EIy AIt " #" # 751 × 103 2 1 − 3310 1 − 81×7317 7518 210×7518 9300 × 7317 × 104

= 0,538

The moment gradient factor C1 should be determined for the greatest ratio of end moments which occurs in member GI. ψ=

−5,08 = −0,191 26,63 TABLE 9.6 Member capacity checks for members in tension. Member

NEd kN

n

MN,y,Rd kNm

Mpl,y,Rd kNm

My,Ed kNm

BD DF FH HJ

117,3 292,7 416,6 479,0

0,036 0,089 0,126 0,145

343 324 311 304

267 267 267 267

116,3 87,0 57,6 26,6

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Chapter 9 / Trusses Determine C1 from Eq. (5.40): 1 = 0,6 + 0,4ψ = 0,6 + 0,4(−0,191) = 0,524 > 0,4 C1 Thus C1 = 1,908 The slenderness λ is given by λ=

L 3000 = = 50,3 iz 59,7

Determine the lateral torsional buckling slenderness ratio λLT from Eq. (5.62): 1/2 1/2 E 1 210 1 λLT = 1/2 π (φb λ)1/2 = √ (0,538 × 50,3)1/2 = 8,47 π G 81 1,908 C 1

Determine the normalized slenderness ratio λLT from Eq. (5.60): λLT =

λLT π E f y

=

!

8,47

3 π 210×10 355

= 0,111

As λLT < 0,4, lateral torsional buckling cannot occur. Check the critical length for lateral torsional buckling lcrit from Eq. (5.64a): b−t 2 3 + b−t 113400 (h − t) h−t h−t lcrit = b−t fy 1+3 1 + b−t h−t

h−t

150−12,5 2 3 + 113400 (250 − 12,5) 250−12,5 = 150−12,5 355 1 + 3 250−12,5 1 +

150−12,5 250−12,5 150−12,5 250−12,5

= 14000 mm This is well in excess of the system length of 3 m. This also confirms lateral torsional buckling will not occur. Check Eq. (5.64b) from Kaim (2006), 25 235 λz,lim = h fy b 235 25 235 55167 55167 λz,lim = λz,lim λ1 = h 93,9 = h = 250 = 93,2 fy fy 355 f y 150 b b L = λz,lim iz = 93,2 × 59,7 = 5560 mm The actual length of 3 m is below the critical value, therefore lateral torsional buckling will not occur. Thus as lateral torsional buckling is not critical Table B.1 can be used to determine the values of kyy and kzy .

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355

TABLE 9.7 Member capacity checks for compression members. Member Calculation

AC

CE

EG

GI

KC K1 K 11 K 12 k1 K2 K 21 K 22 k2 l cr,y /L (Eq. (8.22)) l cr,y /L (Eq. (8.23)) l cr,y /L (Eq. (8.24)) M Ed,max M Ed,min N Ed N cr,y λy χy N cr,z λz χz kyy kzy NEd χy Afy kyy MEd,max Wpl,y fy Total NEd χz Afy kzy MEd,max Wpl,y fy Total

0,3333 0 0 0,5 0,4 0,3333 0 0,5 0,571 1,470 1,418 1,473 116,03 −104,66 117,27 7979 0,643 0,873 7623 0,658 0,867 0,407 0,244

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 87,04 −71,01 292,68 6565 0,709 0,843 7623 0,658 0,867 0,421 0,253

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 57,59 −35,12 416,63 6565 0,709 0,843 7623 0,658 0,867 0,430 0,258

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 26,63 kNm −5,08 kNm 478,99 kN 6565 kN 0,709 0,843 7623 kN 0,658 0,867 0,570 0,342

0,041

0,105

0,150

0,172

0,177

0,138

0,093

0,057

0,218

0,243

0,243

0,229

0,041

0,102

0,146

0,168

0,106

0,083

0,056

0,034

0,147

0,185

0,201

0,202

The calculations for member capacity are carried out in Table 9.7 where it will be seen that the member capacities are overgenerous. The 2 m verticals (AB, CD, EF and GH) have not been checked as they are shorter than the top chord compression members (therefore have a higher buckling load) and carry lower axial forces. The strut buckling lengths have been determined assuming the frame can sway. This is conservative. The buckling length co-efficients have been determined using Eqs (8.22)–(8.24) for comparison. As will be noted in Table 9.7 There is little difference between the results. In this example those from Eq. (8.24) have been used for buckling in-plane. For out-of-plane buckling, the actual length has been used, although this is

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•

Chapter 9 / Trusses conservative. To determine the effective lengths, the EI value of the section has been taken as unity as all the members are fabricated from the same section. Thus EI/L reduces to 1/L. The axial forces although compression are given as positive, as is the larger end moment. The smaller one is given the appropriate sign. Deflection: Permanent load deflection:

22,4 mm (span/1070)

Variable load deflection:

24,2 mm (span/992)

Total deflection:

44,6 mm (span/538)

These are satisfactory. Web check (Section 4.8): The web needs checking at B with the capacity calculated for a single web and then doubled. Reaction = 168,3 kN, moment = 116 kNm. For an end support with c = ss = 0, kF = 2 (type c) Determine m1 : m1 =

bf fyf bf 75 = 6,0 = = fyw tw tw 12,5

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 6,0 = 12,5 = 21,7 mm 2 2

The depth of the web hw has been taken as hw = d − 2t − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm tw3 12,53 = 0,9 × 2 × 210 = 3691 kN hw 200 ly tw fyw 21,7 × 12,5 × 355 λF = = = 0,162 FCR 3691 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. χF =

0,5 0,5 = 3,1 = 0,162 λF

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357

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 21,7 = 21,7 mm FRd = Leff tw

fyw 355 = 21,7 × 12,5 × 10−3 = 96,3 kN γM1 1,0

The total for both webs is 192,6 kN. This exceeds the reaction of 168,3 kN. η2 =

FEd fyw Leff tw γ M1

=

FEd 168,3 = = 0,874 FRd 192,6

η2 = 7,17 kNm/m 2 2 Hogging: Design as reinforced concrete ignoring sheeting: The depth allowing for the trapezoidal indents is given by 105 − 0,5 × 51(38 + 12) = 96 mm. Assume 25 mm cover and 10 mm bars, d = 96 − 25 − 5 = 66 mm. Assuming αcc = 0,85, then using Eq. (6.22) in Martin and Purkiss (2006), MSd 8,72 × 106 = 0,080 = bd2 fck 1000 × 662 × 25 As fyk MSd = 0,652 − 0,425 − 1,5 2 = 0,652 − 0,452 − 1,5 × 0,080 = 0,0997 bdfck bd fck As =

0,0997 × 1000 × 66 × 25 = 329 mm2/m. 500

Fix B385 mesh [385 mm2 /m] Flexural shear: Maximum shear is at A1 on span AA1: Co-efficient for complete UDL is 1,418 + 0,124 − 0,019 + 0,019 − 0,008 + 0,002 = 1,537 Shear due to permanent loading = 1,537 × 6,08 = 9,345 kN/m

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369

Co-efficient for UDL on AB and BC is 1,418 + 0,124 = 1,542 Shear due to variable loading = 1,542 × 6,0 = 9,252 kN/m. Total shear = 9,345 + 9,252 = 18,6 kN/m. b0 is the distance between rib centres, that is, 150 mm. Applied shear per rib = 18,6(150/1000) = 2,79 kN/m Using Eq. (10.6) to calculate Vv,Rd , 385 = 5,83 × 10−3 1000 × 66 200 200 k =1+ =1+ = 2,74 > 2,0 d 66 ρ1 =

1 0,15 3 Vv,Rd = CRd,c k 100ρ1 fck 3 b0 d = (2 × 150 × 66) 100 × 5,83 × 10−3 × 25 1,25 = 5,8 kN/rib This is greater than the applied shear. Shear bond: For an end span Ls = 0,9 L = 0,9 × 2,5 = 2,25 m, as the loading is a UDL Ls /4 is used, that is, 2250/4 = 562.5 mm. The values of m (=166,6) and k (=0,005) were from British standard tests, thus whilst the value of m remains unchanged the value of k has to be amended. 1/2

k = 0,005 × 1,12 × fck = 0,005 × 1,12 × 251/2 = 0,028 Use Eq. (10.3) to calculate Vl,Rd : bdp mAp 1000 × 88,3 166,6 × 1507 Vl,Rd = + 0,028 +k = γvs 1, 25 1000 × 625 bLs = 32,0 kN/m > VSd Deflection: Initially assume there will be no problems with end slip. From Table 7.4 EN 1992-1-1, basic span depth ratio for the end span of a continuous beam (or slab) is 26 (lightly stressed). Actual span 2500 = = 28,3 effective depth ratio 105 − 16,7 This is higher than the allowable of 26, but since the mid-span capacity is much higher than the applied moment, the section will be relatively uncracked thus reducing deflections.

370

•

Chapter 10 / Composite Construction End anchorage: Use 20 mm shear studs. Tensile force required = Np = 434000 N/m. Bearing resistance check Eq. (10.4): For a single shear stud, take a at its minimum distance, that is, 2d0 , thus from Eq. (10.5), kφ = 1 +

a 2 × dd0 =1+ =3 0,5 or 36ε/α when α < 0,5. For Class 2 sections the constants are 456 and 41,5. The ratio α is the normalized depth of the compression zone of the web. Where partial interaction is used, the web classification must be carried out using full interaction.

10.3.2 Design Criteria In addition to the expected checks for flexure, vertical shear (carried by the web only), web capacity under in plane forces and deflection, the transverse shear and the shear connector capacity also need checking.

10.3.3 Flexural Design For stresses during construction the difference between propped and unpropped methods need to be noted. • Propped construction The load due to the concrete is not transferred to the composite section until the concrete has hardened. However, additional forces due to the release of the props may also need considering. • Unpropped construction The effect of the wet concrete is taken on the steel beam alone before the onset of composite action. In both cases the effects of finishes and variable loads are taken on the composite section. It should be noted that in unpropped construction the steel beam may still need checking for lateral torsional buckling during the construction stage. Where the profile decking runs parallel to the span of the beam, the sheeting does not provide full restraint. In the case of decking perpendicular to the span, full restraint is likely to be available (Lawson and Nethercot, 1985). At ultimate limit state, it is not relevant whether the construction was propped or unpropped.

372

•

Chapter 10 / Composite Construction Owing to the relatively high flange widths of composite beams, the whole flange cannot be taken in calculating the moment capacity of the beam owing to shear lag. The effective width either side of the web centre line should be considered separately. The partial effective width be,i should be taken as Le /8, where Le is the span for simply supported beams. The total effective width beff is then given as 2be,i + b0 , where b0 is the transverse distance between the shear connectors. In practice the contribution of b0 may be neglected. The total effective width beff for continuous beams, see cl 5.4.1 (EN 1994-1-1). The partial effective width should not exceed either the distance to the free edge or half the spacing to the adjacent beam centre line. The following assumptions are made when determining the flexural capacity at ULS: • The concrete in between the ribs is ignored, and the remainder is stressed with a uniform strength of 0,85fck /γc . • The profile decking may be included if it is in tension and is then stressed to fy /γap , otherwise it is ignored. • Reinforcement is stressed to fsk /γs where fsk is the characteristic strength of the reinforcement (compression reinforcement may be ignored). • The steel beam is stressed to a uniform stress of fy /γa .

10.3.3.1 Sagging All symbols are defined in Fig. 10.6(a). The force in the concrete Nc is given by

0,85fck Nc = beff hf − hp γc

(10.8)

The force in the steel section Na is given by Na = Aa

fy γa

(10.9)

where Aa is the cross-sectional area of the steel beam. There are three possibilities for the position of the plastic neutral axis, in the concrete slab, the top flange of the steel beam or the web of the steel beam. If the neutral axis lies in the concrete slab, then Nc > Na , otherwise the neutral axis is in the steel beam. (a) Neutral axis in concrete slab (Fig. 10.6(b)). Depth of neutral axis in the slab x is given by Na x = 0,85f ck beff γ

(10.10)

c

The moment capacity Mpl,Rd is then given by h x Mpl,Rd = Na + hf − 2 2

(10.11)

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373

0,85 fck gc

beff Concrete slab

hp

hf

Nc

Neutral axis

h

Area

x

Aa

tw

tf bf

(a) Basic data

(b) Neutral axis in slab

Nc

N af Naw

x

Na

Neutral axis

ht f x 2

h 2

ht f 2 (c) Neutral axis in web

Nc Neutral axis

N afx

x

Na

h x 2

(d) Neutral axis in flange

FIGURE 10.6 Determination of plastic moment capacity (sagg mg). (b) Neutral axis in the steel beam (Na > Nc ) Define an out-of-balance steel compression force Nac as Nac = Na − Nc

(10.12)

and the capacity of the flange Naf (ignoring fillets) as Naf = bf tf

fy γa

(10.13)

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•

Chapter 10 / Composite Construction The out-of-balance force is introduced to enable the force in the beam Na to be retained with its point of application of h/2 in the moment capacity calculations. The neutral axis is • in the web if Nac > 2Naf or • in the flange if Nac < 2Naf . Case 1: Neutral axis in the web (Fig. 10.6(c)) Depth of web x in compression is given by x=

Nac − Naf

(10.14)

fy

2tw γ

a

The compression force in the web Naw is given by Naw = xtw

fy γa

(10.15)

and the design moment of resistance Mpl,Rd is given by

h p + hs h+ 2

Mpl,Rd = Nc

+ 2Naf

tf h− 2

h x + 2Naw h − tf − − Na (10.16) 2 2

Case 2: Neutral axis in the flange (Fig. 10.6(d)) Depth of flange x in compression is given by x=

Nac

(10.17)

fy

2b γ

a

The force in the flange Nafx is given by Nafx = xb

fy γa

(10.18)

and the design moment of resistance Mpl,Rd is given by h p + hf h x Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2

(10.19)

10.3.3.2 Hogging (negative moment) Conservatively the contribution of the steel profile decking will be ignored. All symbols are defined in Fig. 10.7(a). The force in the reinforcement Nr is given by Nr = As

fsk γs

(10.20)

•

Structural Design of Steelwork to EN 1993 and EN 1994 Reinforcement As

As

375

c

h

Nr

Naw

Neutral axis

Fs

Na

tf 2

tw

h

h 2

h

Steel beam (area As)

Naf

tf x

tf

h tf

x 2

bf

tf

(a) Definition of symbols

(b) Neutral axis in web

As

Nr Nafx

h

x 2

x

Neutral axis

Na

(c) Neutral axis in top flange

FIGURE 10.7 Calculation of the plastic moment capacity (hogging). The force in the steel beam Na is given by Eq. (10.9), and the force in a flange Naf is given by Eq. (10.13). The neutral axis is in • the web, if Na > Nr + 2Naf , • the flange, if Na < Nr + 2Naf . Case 1: Neutral axis in web (Fig. 10.7(b)) The depth of the web x in tension is given by x=

Na − Nr − 2Naf fy

2tw γ

(10.21)

a

The capacity of the tension part of the web Naw is given by fy

Naw = xtw γ

a

The moment capacity of the section Mpl,Rd is given by

tf h x Mpl,Rd = Nr h + hs − c + Naf h − + Naw h − tf − − Na 2 2 2

(10.22)

(10.23)

Case 2: Neutral axis in web (Fig. 10.7(c)) The depth of the flange x in tension is given by x=

Na − Nr fy

2b γ

a

(10.24)

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•

Chapter 10 / Composite Construction The capacity of the tension part of the flange Nafx is given by Nafx = xbf

fy γa

(10.25)

The moment capacity of the section Mpl,Rd is given by

x h Mpl,Rd = Nr h + hf − c + Nafx h − − Na 2 2

(10.26)

10.3.3.3 Reduction in Plastic Moment Capacity Due to Shear In a similar fashion to plain steel beams, the moment capacity is reduced when the vertical shear VEd exceeds half the plastic shear resistance VRd . A reduced design strength (1 − ρ)fyd to determine the capacity of the steel section is used where ρ is defined as 2 2VEd −1 (10.27) ρ= VRd For calculation of plastic shear capacity see Section 10.3.4.

10.3.3.4 Reduction in Sagging Moment Capacity with Partial Shear Connection (cl 6.2.1.3) The section capacity should be calculated using a force in the concrete section of Nc,f where Nc,f is given by Nc,f = ηNc

(10.28)

where η is the degree of shear connection. The moment capacity Mpl,Rd is then conservatively given by MRd = Mpl,a,Rd +

Nc Mpl,Rd − Mpl,a,Rd Nc,f

(10.29)

where Mpl,Rd is the moment capacity of the composite section with full shear connection and Mpl,a,Rd is the plastic moment capacity of the steel section alone.

10.3.3.5 Elastic Capacity For elastic resistance conventional elastic theory is used with the slab taking its effective width beff and where appropriate account is taken of creep (cl 6.2.1.5). The limiting stresses for Mel,Rd are 0,85 fck /γc for concrete, fy /γa for structural steel (Class 1,2 or 3 cross-sections) and fsk /γs for reinforcing steel. Account should be taken of creep and shrinkage in determining the elastic capacity. However, for beams with only one flange composite this may be achieved by using an appropriate modular ratio nL (cl 5.4.2.2.2). The modular ratio nL is defined as nL = n0 (1 + ψL φt )

(10.30)

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377

where n0 is the short-term modular ratio defined as Ea /Ecm , ϕt is the creep co-efficient defined as ϕ(t,t0 ) in EN 1992-1-1, and ψL is creep multiplier depending on the type of loading (1,1 for permanent loads, 0,55 for shrinkage and 1,5 for prestressing by imposed deformations). For cases where any amplification of internal forces is less than 10% due to deformations, the structure is not mainly intended for storage nor prestressed by imposed deformations, the effect of creep can be taken into account for both short- and long-term loading by using a nominal modular ratio n determined using an effective concrete elastic modulus Ec,eff given by 0,5Ecm . Note, there appears to be no explicit requirement to check serviceability elastic stresses if the plastic moment capacity is used to determine the strength of the beam.

10.3.4 Flexural Shear The flexural shear capacity Vpl,Rd is calculated exactly as for a normal steel beam. Additionally for an unstiffened and uncased web d/tw should not exceed 69ε, for a cased web the constant is 124.

10.3.5 Design of Shear Connectors The strength of shear connectors is dependant upon both the strength and elasticity of the concrete and the ultimate tensile strength of the connector itself. Following a large series of tests Olgaard et al. (1971) proposed the following equation for the strength of shear stud connectors kd2 fcu Ec PRd = (10.31) 1,25 where d is the diameter of the stud, Ec Young’s Modulus of the concrete, fcu concrete cube strength and k is an empirical constant allowing for the height to diameter ratio of the stud. Using the 80% utilization factor proposed by Yam and Chapman (1968), and converting the cube strength to cylinder strength by a factor of 0,8 gives the following strength formula, 0,29αd2 fck Ecm PRd = (10.32) γv where γv takes a value of 1,25 and for 3 < hsc /d < 4 hsc α = 0,2 +1 d

(10.33)

for h/d > 4 α = 1,0

(10.34)

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Chapter 10 / Composite Construction Oehlers and Johnson (1987) carried out further examination of shear stud capacity and proposed an upper bound which was dependant upon the ultimate strength of the shear stud. Their results have been simplified in the code and are given as PRd =

0,8fu πd2 γv 4

(10.35)

where fu is the ultimate tensile strength of the stud material (taken as not greater than 500 MPa). For decks with ribs parallel to the beam the strength of the shear studs needs to be reduced by the parameter k1 as the containment by the concrete is incomplete (Mottram and Johnson, 1990), kl = 0,6

b0 hp

hsc − 1 ≤ 1,0 hp

(10.36)

where hp is the height of the profile and hsc is the overall height of the stud ( sf cot θf

(10.45)

where θf is the angle of inclination of the concrete truss member such that 1,0 ≤ cot θf ≤ 2,0. For normal weight concrete the shear resistance vEd is given by vEd = 0,5vfcd

(10.46)

where v is given by 0,6 (1 − fck /250). For lightweight concrete the shear resistance vEd is given by vEd = 0,5η1 v1 fcd

(10.47)

where v1 is given by 0,5η1 (1 − fck /250), and η1 is given by 0,4 + 0,6ρ/2200 where ρ is the density of the concrete. A limit is placed on vEd such that vEd < vfcd sin θf cos θf

(10.48)

It would appear acceptable to take the traditional approach and use an angle of 45◦ √ in the truss analogy giving cot θf = 1,0 and sin θf = cos θf = 1/ 2 (Johnson, 2004).

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1

At 2

2

Ab

1

Plane 11 22

Steel area Ab At 2Ab

(a) Solid slab 3

At

Sheeting

3

Plane 33

Steel area 2At

(b) Composite slab (sheeting perpendicular to span) 4

5

4

5

At

Lab joint in sheeting

Plane 44 55

Steel area

(c) Composite slab (sheeting parallel to span)

2At

FIGURE 10.8 Critical transverse shear planes.

Where the profile sheet decking is normal to the span and is continuous over the beam then Eq. (10.45) applied to vertical shear planes may be enhanced by the effect of the decking to give Asf fyd vEd hf + Ape fyp,d > sf cot θf

(10.49)

where Ape is the effective cross-sectional area of the decking and fyp,d is its design strength.

Chapter 10 / Composite Construction beff

Concrete slab hp

hs

•

h

382

Steel beam (second moment of area,Ia area, Aa, modular ratio n)

FIGURE 10.9 Second moment of area calculation for a composite beam.

10.3.11 Deflection (cl 5.2.2) For the deflection of the steel member alone the principles of EN 1993-1-1 are applied. No account need be taken of partial interaction if η ≥ 0,5 or the force in the shear connector does not exceed PRd at the serviceability limit state, and for ribbed slabs transverse to the beam the rib height does not exceed 80 mm. The composite second moment of area may be determined as follows (Fig. 10.9)

3 2 2

beff hs − hp h s + hp h Ic = nIa + + nAa − x + beff hs − hp h + −x 12 2 2 (10.50) where x is given by

h +h nAa + hs − hp beff h + s 2 p

x= nAa + hs − hp beff

(10.51)

10.3.12 Vibration This is only likely to be critical on lightly loaded, long span beams. It is suggested that the following formula is used (Wyatt, 1989) 18 f =√ δsw

(10.52)

where f is the frequency (Hz) and δsw is the instantaneous deflection (mm) due to selfweight and permanent loading. The suggested limits for f are 4 Hz for most buildings, 3 for car parks and 5 for sports halls (Lawson and Chung, 1994).

10.3.13 Detailing 10.3.13.1 Cover This should be the greater of 20 mm or the values specified in EN 1992-1-1 (Table 4.4) less 5 mm (cl 6.6.5.2).

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10.3.13.2 Spacing (cl 6.6.5.5) This should be not less than 22tf ε for a solid slab or 15tf ε for a ribbed slab where the flange becomes a Class 1 or Class 2 by virtue of restraint due to shear connection with the edge distance not exceeding 9tf ε. The maximum spacing should not exceed the lesser of 800 mm or six times the slab thickness. The edge distance should also not exceed 20 mm. The overall height of a stud connector should be not less than 3d (where d is the shank diameter), the spacing in the direction of the shear force should be not less than 5d, and transverse to the shear force 2,5d for solid slabs and 4d for other cases. Unless the stud is directly over the web the diameter of the stud should not exceed 2,5tf (cl 6.6.5.7). For through-deck welding, the profile sheet steel decking should not exceed 1,25 mm thick if galvanized and 1,5 if not. The studs should extend 2d above the deck after welding. Johnson (2005) expresses concern over the detailing rules (and stud capacities) when applied to open trapezoidal decks.

EXAMPLE 10.2 Composite beam (decking transverse to span) Design a composite beam (ref C1/23 (Fig. 10.3)) in Grade S355 steel to carry the composite slab designed in Example 10.1. The concrete is lightweight LC25/30 and a dry specific weight of 19 kN/m3 . The span is 4 m and carries the two 2,5 m deck spans. Loading due to wet concrete: 0,105 × 19 = 2,0 kPa Finishes: 2,5 kPa Variable loading: 4,0 kPa The loading co-efficients for the reaction at A1 (from SFDs in Fig. 10.4) are given in Table 10.2 for all spans loaded. For the variable loading, spans AA1, A1B, CC1 and C1D are loaded. This gives an overall co-efficient of 3,062

TABLE 10.2 Loading co-efficients for reactions A1. Span

Individual components

Total

AA1 A1B BC CC1 C1D Overall total

1,418 + 0,218 0,124 + 1,254 −0,019 − 0,096 0,039 + 0,008 0,009 + 0,003

1,636 1,369 −0,115 0,047 0,012 2,949

Chapter 10 / Composite Construction Load per unit run on the beam: Permanent: 2,949(2,0 + 2,5) = 13,27 kN/m Variable: 3,062 × 4 = 12,25 kN/m Ultimate limit state design: MSd = (1,35 × 13,27 + 1,5 × 12,25) × 42 /8 = 72,6 kNm Effective width of slab, beff (ignoring spacing between connectors as it is likely only a single row will be needed) be,i = Le /8 = 4000/8 = 500 mm beff = 2be,i = 1000 mm. Try a 203 × 133 × 25 UKB (Grade S355) Since the flange is restrained by through deck studs, the beam flanges are automatically Class 1. The web slenderness satisfies the shear buckling check. Actual width between beam centre lines = 2500 mm > beff Dimensions to calculate Mpl,Rd assuming full shear connection are given in Fig. 10.10. From Eq. (10.8), the force in the concrete flange Nc is given by Nc =

0,85 × 25 0,85fck beff hf − hp = 1000 × (105 − 51) × 10−3 = 765 kN γc 1,5

From Eq. (10.9) the force in the steel beam Na is given by Na = Aa

fy 355 = 3200 × 103 = 1136 kN γa 1,0

From Eq. (10.12) the out-of-balance force Nac is given by Nac = Na − Nc = 1136 − 765 = 371 kN

1000

Concrete slab

51

Steel beam 20313325 UKB S 375

7,8 7,8

105

•

203,2

384

5,7

133,2

FIGURE 10.10 Design dimensions for EXAMPLE 10.2

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From Eq. (10.13) the capacity of the flange Naf is given by Naf = btf

fy 355 × 10−3 = 369 kN = 133,2 × 7,8 1,0 γa

Since Nac < 2Naf , the neutral axis is in the flange. From Eq. (10.17) the depth of flange x is given by x=

371 × 103 = = 3,92 mm fy 2 × 133,2 × 355 2b 1,0 Nac

γa

From Eq. (10.18) the compression force in the flange Nafx is given by Nafx = xb

fy 355 = 3,92 × 133,2 × 10−3 = 185 kN γa 1,0

From Eq. (10.19) Mpl,Rd is given by h p + hf x h Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2 105 + 51 3,78 203,2 = 0,765 203,2 + + 2 × 0,185 203,2 − − 1,136 2 2 2 = 174,2 kNm Note, for convenience all the forces have been expressed in MN. Then if the dimensions are left in mm, the moment is in kNm. Since the neutral axis for full shear connection is in the flange, the only web check is for shear buckling (see above). As Mpl,Rd exceeds MEd by a substantial margin, partial interaction may be used. For the beam alone the section is Class 2 (which is satisfactory). Check shear capacity: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 262 kN = √ 1280 1,0 γ 3 3 a VEd =

(1,35 × 13,27 + 1,5 × 12,25) × 4 = 72,6 kN 2

Since VSd < 0, 5Vpl,Rd , there is no reduction in moment capacity. Shear connectors: Use 100 mm long by 19 mm diameter stud connectors. The load capacity of a single stud, Eq. (10.32): h/d = 100/19 = 5,3, so from Eq. (10.34), α = 1,0 fck = 25 MPa

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Chapter 10 / Composite Construction For normal weight concrete, Ecm = 22

fck + 8 10

1/3 = 22

25 + 8 10

1/3 = 31,5 GPa

Modification factor for lightweight concrete ηE (from cl 11.3.2 of EN 1992-1-1) ρ 2 1900 2 ηE = = = 0,746 2200 2200 So, Ecm = 0,746 × 31,5 = 23,5 GPa From Eq. (10.32) PRd

0,29αd2 fck Ecm 0,29 × 1 × 192 × 25 × 23,5 × 103 = = × 10−3 = 64,2 kN γv 1,25

Limiting capacity: For ribs transverse to beam the ultimate connector strength (fu = 450 MPa) needs to be multiplied by the factor kt . Determine kt from Eq. (10.37) In view of the large overdesign on moment capacity, number of connectors per rib will be 1, that is, nr = 1. For a dovetail deck b0 is the distance between the top of the dovetails, that is b0 = 150 − 38 = 112 mm Profile height hp = 55 mm, height of connector, h = 100 mm, 0,7 b0 kt = √ nr hp

hsc 0,7 112 100 −1 = √ − 1 = 1,17 hp 55 1 55

However the maximum value allowed for kt is 1,0. Limiting capacity of studs from Eq. (10.35) PRd =

0,8fu πd2 0,8 × 450 π × 192 = × 10−3 = 81,7 kN γv 4 1,25 4

Shear stud capacity is the lower of the two values, that is 64,2 kN. No. of connectors for full shear connection: Nc = 765 kN, PRd = 64,2 kN, thus nf for the half beam span is given by nf = Nc /PRd = 765/64,2 = 11,9 Use 1 connector per rib, thus n for the whole beam is given by span over distance between centre lines of ribs = 4000/300 = 13,3.

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The ratio between η between n and nr is given by η=

n 13,3 = = 0,56 nr 2 × 11,9

The limiting value of η is given by Eq. (10.39) with Le = 4,0 m, thus η≥1−

355 355 0,75 − 0,03Le ≥ 0,4 = 1 − 0,75 − 0,03 × 4 = 0,37 fy 355

The limiting value is 0,4, as the actual value is 0,56, it is therefore satisfactory. The moment capacity MRd is given by Eq. (10.29). The moment capacity of the steel section Mpl,a,Rd alone is given by 355 Mpl,a,Rd = 258 × × 10−3 = 91,6 kNm 1,0 MRd = Mpl,a,Rd +

nc Mpl,Rd − Mpl,a,Rd = 91,6 + 0,56 (174,2 − 91,6) nc,f

= 137,9 kNm This is greater than MEd . Longitudinal shear: The only plane requiring checking is 4–4 (or 5–5) of Fig. 10.8 Determine VSd from Eq. (10.44), VSd =

nr PRd 1,0 × 64,2 = = 214 kN/m sL 0,3

The shear stress VSd is determined using the thickness of the concrete above the ribs, that is, 105 − 51 = 54 mm. Also the shear may be equally divided between the two shear planes, thus VEd is given by 214 = 1,98 MPa 54 As the sheeting is continuous across the beam with the ribs running normal to the beam, the contribution of the sheeting may be mobilized. √ Using a 45◦ angle for the truss analogy, cos θf = sin θf = 1/ 2 and cot θf = 1,0. vEd = 0,5 ×

For lightweight concrete vEd is given by Eq. (10.47). Determine v1 : The reduction factor for lightweight concrete η1 is given by ρ 1900 η1 = 0,4 + 0,6 = 0,4 + 0,6 = 0,918 2200 2200 fck 25 v1 = 0,5η1 1 − = 0,5 × 0,918 1 − = 0,413 250 250

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Chapter 10 / Composite Construction VEd = 0, 5v1

fck 25 = 0,5 × 0,413 × = 3,44 MPa γc 1,5

From Eq. (10.48) the maximum value of VEd is given by 25 25 1 1 VEd = vfcd sin θr cos θr = 0,6 1 − √ √ = 4,5 MPa 250 1,5 2 2 Use Eq. (10.49) to determine the requirement for Asf sf as the sheeting runs normal to the span. Evaluate the right hand side of Eq. (10.49), vEd hf 3,44 × 55 = = 1892,2 N/mm cot θf 1,0 Evaluate the contribution of the sheeting: Ape fyp,d =

1597 280 = 447,2 N/mm 1000 1,0

The sheeting overprovides the required resistance, therefore only minimum reinforcement is necessary. Deflections and service stresses: a) Wet concrete: This is taken on the steel beam alone. The uniformly distributed load due to the wet concrete q is given by q = 20 × 0,105 × 2,5 = 5,25 kN/m Deflection, δconc is given by: δconc =

5 qL4 5 5,25 × 44 = = 3,6 × 10−3 m 384 EI 384 210 × 106 × 2340 × 10−4

MEd =

qL2 5,25 × 42 = = 10,5 kNm 8 8

The numerical value of the stress σconc is given by σconc =

MEd 10,5 × 103 = = 46 MPa Wel 230

b) Variable and permanent loads: As the ratio n/nf of the number of shear studs provided to that required for full connection is 0,57 (and is therefore greater than the critical ratio of 0,5), no account need be taken of partial shear connection in determining the deflection. Assume Ec,eff = Ecm /2 = 23,5/2 = 11,75 GPa The value of n from cl 5.4.2.2 (11) is Es /Ec,eff (=210/11,75 = 17,9)

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Use Eq. (10.51) to determine the elastic centroidal axis x,

hs +hp nAa + hs − hp beff h + 2

x= nAa + hs − hp beff

203,2 17,9 × 3200 2 + 105 − 51 2500 203,2 + 105+51 2

= 227, 7 mm = 17,9 × 3200 + 105 − 51 2500 From Eq. (10.50)

3 2 2

beff hs − hp hs + hp h Ic = nIa + + nAa − x + beff hs − hp h + −x 12 2 2

3 2 2500 105 − 51 203,2 6 = 17,9 × 23,4 × 10 + + 17,9 × 3200 − 227,7 12 2 2

105 + 51 + 2500 105 − 51 203,2 + − 227,7 = 1,75 × 109 mm4 2 Ic Ec,eff = 1,75 × 109 × 11,75 = 20,45 × 109 kNmm2 = 20,56 × 103 kNm2 Variable load is 13,27 kN/m, and the moment MEd is 26,54 kNm. Thus the deflection δv is given by δv =

5 qL4 5 13,27 × 44 = = 2,15 × 10−3 m 384 EI 384 20,56 × 103

This is equivalent to span/1860 which is acceptable. Although not strictly necessary, determine the stresses due to the variable load: Top of the concrete slab σv,top,c :

MEd h + hs − x 26,54 × 106 203,2 + 105 − 227,7 σv,top,c = − = −1,22 MPa =− Ic 1,75 × 109 Stress at the soffit of the steel beam, σv,soffit : σv,soffit =

nMEd x 17,9 × 26,54 × 106 × 227,7 = = 61,8 MPa Ic 1,75 × 109

Stress at the top of the steel beam, σv,top,A : σv,top,c = −

17,9 × 26,54 × 106 (203,2 − 227,7) nMEd (h − x) =− = 6,65 MPa Ic 1,75 × 109

Permanent load is 12,07 kN/m, and the moment MEd = 24,14 kNm. δv =

5 12,07 × 44 5 qL4 = = 2,0 × 10−3 m 384 EI 384 20,45 × 103

Total deflection: δtotal = δp + δv + δconc = 2,2 + 2,0 + 3,6 = 7,8 mm

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Chapter 10 / Composite Construction This is equivalent to span/513 which is acceptable. Although not strictly necessary determine the stresses due to the variable load: Top of the concrete slab σp,top,c :

MEd h + hs − x 24,14 × 106 203,2 + 105 − 227,7 σp,top,c = − =− Ic 1,75 × 109 = −1,11 MPa Stress at the soffit of the steel beam, σp,soffit : σp,soffit =

nMEd x 17,9 × 24,14 × 106 × 227,7 = = 56,2 MPa Ic 1,75 × 109

Stress at the top of the steel beam, σp,top,A :

nMEd h − x 17,9 × 24,14 × 106 (203,2 − 227,7) σp,top,c = − = 6,05 MPa =− Ic 1,75 × 109 Final total stresses: Top of concrete slab: −1,11 − 1,22 = −2,33 MPa Top of steel beam: −46 + 6,65 + 6,05 = −33,3 MPa Soffit of steel beam: 46 + 61,8 + 56,2 = 164 MPa All these stresses are acceptable. Vibration: Use ψ2 = 0,3, so quasi-permanent load is 12,07 = 0,3 × 13,27 = 16,05 kN/m Deflection under this load δsw is given by δsw =

5 qL4 5 16,05 × 44 = = 0,0026 m = 2,6 mm 384 EI 384 20,45 × 103

Determine the frequency f from Eq. (10.52) 18 18 = 11,2 Hz =√ f =√ δsw 2,6 This is well above the recommended limit of 3 Hz. If the full variable load is taken, then f = 8,7 Hz (and is still acceptable).

EXAMPLE 10.3 Composite beam design (decking parallel to the span). Prepare a design in Grade S355 steel for the beam Mark 3AD of Fig. 10.3 where there are no intermediate columns. The actions on the beam are given in Fig. 10.11. The composite deck is that of EXAMPLE 10.1.

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Due to the long span, the beam is designed as propped to eliminate the high deflections under the permanent loading due to the wet concrete. The resultant shear force and bending moment diagrams for the applied loading are given in Fig. 10.12. The effective width, either side of the beam centre line, be,i = Le /8 = 12000/8 = 1500 mm. Thus total effective width beff = 2be,i = 3000 mm. (The actual beam spacing is 4 m.) To determine the capacity of the beam, the effect of the voids in the deck will be ignored as these run in the direction of the beam and are small. Try a 838 × 292 × 194 Grade S355 UKB. 94

214

214

94

Permanent (kN)

70

240

240

70

Variable (kN)

2,5

2,5

2,5

2,0

2,5

FIGURE 10.11 Loading for EXAMPLE 10.3 232

2500

649

2500

649

2000

232

2500

All loads in kN

2500

BMD (kNm)

2203

3825 881 649

SFD (kN)

649 881

FIGURE 10.12 BM and SF EXAMPLE 10.3

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Chapter 10 / Composite Construction Moment capacity with full shear connection: Determine the force in the concrete Nc from Eq. (10.8), Nc =

0,85 × 25 0,85fck beff hf − hp = 3000 × 105 × 10−6 = 4,463 MN γc 1,5

Determine the force in the steel section Na from Eq. (10.9): Na = Aa

fy 355 = 24700 × 10−6 = 8,769 MN γa 1,0

Determine the maximum force in a flange Naf from Eq. (10.13): Naf = bf tf

fy 355 = 292,4 × 21,7 × 10−6 = 2,253 MN γa 1,0

Determine the out-of-balance force Nac from Eq. (10.12): Nac = Na − Nc = 8,769 − 4,463 = 4,306 MN As Nac < 2Naf , therefore neutral axis lies in the flange. Determine the position of the neutral axis x from Eq. (10.17): x=

4,306 × 106 = = 20,74 mm fy 2 × 292,4 355 2b 1,0 Nac

γa

Determine the compression force Nafx in the flange from Eq. (10.19): Nafx = xb

fy 355 = 20,74 × 292,4 × 10−6 = 2,153 MN γa 1,0

Determine the plastic moment capacity Mpl,Rd from Eq. (10.19): h p + hf x h Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2 105 20,74 840,7 = 4,463 840,7 + + 2 × 2,153 840,7 − − 8,769 2 2 2 = 3876 kNm This is greater than MEd (=3825 kNm). Check the ratio of Mpl,Rd to Mpl,a,Rd (the moment capacity of bare steel section): Mpl,Rd Mpl,Rd 3876 = = = 1,43 355 fy Mpl,a,Rd 7640 1,0 × 10−3 Wpl γ a This is less than the critical value of 2,5. Since the neutral axis is in the flange, the web check is unnecessary and thus the section is Class 1.

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Flexural shear: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 2685 kN = √ 13100 1,0 γ 3 3 a VEd = 881 kN ( = = 0,985 mm sf fyd cot θf 500 1,0 This can be divided between top and bottom, so Ast /Sr on each face is 0,493 mm2 /mm. Use B503 Mesh [503 mm2 /m] Deflection: The loading for determining variable and total deflections is given in Fig. 10.11. Use the formula δ = (WL3 /48EI)(3(a/L) − 4(a/L)3 ) To simplify calculations determine L3 /48EI as a constant for the composite and steel sections, and then determine W (3(a/l) − 4(a/L)3 ) for each load. Determination of Ic : Neglect effect of profiles and take hp = 0, beff = 3000 mm and αe = 17,9 (as EXAMPLE 10.2). Summarizing results: x = 396,3 mm, Ic = 0,162 × 1012 mm4 . Ec,eff Ic = 11,75 × 106 × 0,162 = 1,904 × 106 kNm2 L3/48Ec,eff Ic = 123/48 × 1,904 × 106 = 18,9 × 10−6 m/kN Loads at A1: a/L = 2,5/12 = 0,208; 3(a/L) − 4(a/L)3 = 0,589 Total load: W = 94 + 70 = 164 kN, δ = 164 × 0,589 × 18,9 × 10−6 = 0,0018 m

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Loads at B: a/L = 5/12 = 0,417; 3(a/L) − 4(a/L)3 = 0,961 Total load: W = 214 + 240 = 454 kN, δ = 454 × 0,961 × 18,9 × 10−6 = 0,0082 m Final deflection under total loads: δ = 2(0,0018 + 0,0082) = 0, 0202 m Span/deflection ratio is 12/0,0202 = 594. This is acceptable. Vibration: Total variable deflection = 10,2 × 10−3 m Total permanent deflection = 9,87 × 10−3 m Variable deflection due to a value of ψ2 = 0,3: 0,3 × 10,2 × 10−3 = 0,0031 m δsw = 0,0031 + 0,0102 = 0,0133 m = 13,3 mm √ From Eq. (10.52), f = 18/ 13,3 = 4,94 Hz This is higher than the minimum recommended value of 3 Hz. If the total variable load is taken then f = 4,0 Hz, which is still acceptable.

10.4 COMPOSITE COLUMNS These can take a variety of forms but fall essentially into two categories; partially or totally encased Universal Columns (or H sections) and filled rolled hollow sections, with or without additional reinforcement. Typical configurations are given in Fig. 10.13. This text only considers composite columns which are symmetric about both axes. The methods given in EC 1994-1-1 only hold if (a) the steel contribution ratio δ defined as δ=

Aa fyd Npl,Rd

(10.53)

satisfies the limits 0,2 ≤ δ ≤ 0,9. For δ < 0,2 the column should be designed as reinforced concrete, and for δ > 0,9 designed as non-composite steel. (b) the normalized slenderness ratio λ is less than 2,0.

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Traditional encased column

Web in filled column

Concrete unfilled hollow section

FIGURE 10.13 Types of composite columns.

10.4.1 Axial Compression The design condition for axial compression is that the design resistance χNpl,Rd should exceed the applied load NEd . The buckling co-efficient χ is determined using a nondimensionalized slenderness ratio λ (defined in Eq. (10.73)) in combination with buckling curve ‘a’ for concrete filled hollow sections with ρs ≤ 3% or ‘b’ 3% < ρs < 6% (where ρs is the percentage of reinforcement), ‘b’ for partially or fully encased I sections bending about the major axis, and ‘c’ for partially or fully encased sections bending about the minor axis (the buckling curve designations are those used in EN 1993-1-1).

10.4.2 Uniaxial Bending and Axial Compression Initially, the resistance of the cross-section is determined using an interaction diagram between the axial load resistance and bending moment resistance in a similar fashion to reinforced concrete columns. The diagram is shown schematically in Fig. 10.14, where Npl,Rd is the axial squash capacity (Point A), and Mpl,Rd is the plastic moment capacity (Point B). Although the interaction diagram which is of a similar shape to that for reinforced concrete columns is strictly curved, it may be approximated to a series of straight lines. Point C is established by the application of the moment Mpl,Rd and a resultant axial capacity of the concrete alone Npm,Rd , Point D by the moment capacity Mmax,Rd under 0,5Npm,Rd .

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N

N pl,Rd

A

C

N pm,Rd ½N pm,Rd

D B N pl,Rd

M mav,Rd

M

FIGURE 10.14 Interaction diagram for composite columns

The moment capacity μdy Mpl,y,Rd (or Mpl,N,Rd ) is determined from the interaction diagram under an axial load of NEd . The member is deemed to have sufficient capacity when MEd MEd = = αM Mpl,N,Rd μd Mpl,Rd

(10.54)

The co-efficient αM takes a value of 0,9 for steel grades of S235 and S355 and 0,8 for grades S420 and S460. The factor is partially needed to compensate for the assumption over the depth of the rectangular stress block between EN 1992-1-1 where it is taken as extending over 0,8x but over x in EN 1994-1-1, since the same stress level of 0,85fck /γc is used in both, and partially to allow for the adverse effect of the higher yield strain on the crushing on the concrete. Should the bending moment be entirely due to the eccentricity of the axial load, then μd can be greater than unity.

10.4.3 Biaxial Bending The procedure is similar to uniaxial bending except that two parameters μdz and μdy now need to be determined. However, the effect of imperfections needs only to be considered for the likely failure axis (usually the minor or zz axis). The column is then satisfactory if the following conditions are satisfied, My,Ed ≤ αM,y μdy Mpl,y,Rd

(10.55)

Mz,Ed ≤ αM,z μdz Mpl,z,Rd

(10.56)

My,Ed Mz,Ed + ≤ 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd

(10.57)

and

398

•

Chapter 10 / Composite Construction The co-efficients αM,y and αM,z are to be taken as αM . The increase of the interaction co-efficient to 1 in the combined equation is due to the higher crushing strength of the concrete under biaxial bending.

10.4.4 Determination of Member Capacities The formulae for the flexural capacity of concrete filled rectangular tubes (with reinforcing steel ignored) are taken from Johnson and Anderson (2004).

10.4.4.1 Axial Squash Capacity, Npl,Rd [Point A] This is given by the sum of the individual components due to the steel section, the concrete and the reinforcement (Fig. 10.15(a)). So in general Npl,Rd is given by Npl,Rd = Aa

fy 0,85fck fsk + Ac + As γa γc γs

(10.58)

where Aa is the area of the steel section and fy is the yield strength, Ac is the concrete area and fck is the characteristic cylinder strength, As is the area of the reinforcement and fsk is the characteristic strength. For concrete filled hollow sections the full cylinder strength fck may be used owing to the containment of the concrete. For concrete filled circular tubes Npl,Rd is modified to take account of the triaxial stresses in the concrete due to its containment, and of the reduction in the allowable strength of the steel cross-section owing to the induced hoop tension from the concrete triaxial stresses. The modified value of Npl,Rd is subject to two conditions: (1) λ < 0,5 and (2) e = MEd /NEd ≤ d/10 (where d is the diameter) The equation for Npl,Rd becomes fy t fy fck fsk Npl,Rd = Aa ηa + Ac 1 + ηc + As γa d fck γc γs

(10.59)

where t is the thickness of the tube and ηa and ηc are co-efficients determined as follows, e ηc = ηc0 1 − 10 (10.60) d where ηc0 is given by ηc0 = 4,9 − 18,5λ + 17(λ)2 ≥ 0

(10.61)

and ηa = ηa0 + (1 − ηa0 )10

e d

(10.62)

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Ac

•

399

As

Ax (a) Point A

hn

hn

(b) Point B and C

Neutral axis centroidal axis

Concrete below axis ignored (c) Point D

FIGURE 10.15 Calculation of section capacities

where ηa0 is given by ηa0 = 0,25(3 + 2λ) ≤ 1,0

(10.63)

For e > d/10, ηc = 0 and ηa = 1,0.

10.4.4.2 Calculation of Mpl,Rd [Point B] For the typical section given in Fig. 10.15(b), the forces in the flanges and the reinforcement cancel out, and thus the tension force in the web must balance the compression block in the concrete. It is therefore straightforward to determine the height of the web hn above the centroidal axis. Equating compressive and tensile forces gives hn =

2b fγckc

Npm,Rd f + 4tw 2 γya −

fck γc

where Npm,Rd is given by Eq. (10.67).

(10.64)

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Chapter 10 / Composite Construction The plastic moment capacity is determined by taking moments about the centroidal axis of the section. The moment capacity Mpl,Rd is thus given by Mpl,Rd = Mmax,Rd − Mn,Rd

(10.65)

The values of Mmax,Rd and Mn,Rd are given by Eqs (10.71) and (10.68), respectively.

10.4.4.3 Determination of Npm,Rd [Point C] This can be done by noting that the neutral axis shifts from hn above the centroidal axis to the same distance below it (Fig 10.15(b)). Npm,Rd is determined from horizontal force equilibrium recognizing that the forces in the steel flanges and reinforcement cancel out, that is, the axial load is carried by the concrete alone. The area of the concrete Ac is given by Ac = (b − 2t)(h − 2t) − (4 − π)r 2

(10.66)

where r is the corner radius taken equal to t, and Npm,Rd is given by Npm,Rd = Ac

fck γc

(10.67)

The moment capacity Mn,Rd is given by Mn,Rd = Wp,a,n

fy fck + 0, 5Wp,c,n γa γc

(10.68)

where Wp,a,n and Wp,c,n are the plastic section moduli for the portions of the steel tube and concrete contained within ± hn , and are given by Wp,c,n = (b − 2t)h2n

(10.69)

Wp,a,n = bh2n − Wp,c,n

(10.70)

and

10.4.4.4 Determination of Mmax,Rd [Point D] For this case when an axial force of 0,5Npm,Rd acts the neutral axis coincides with the centroidal axis (Fig. 10.15(c)), and thus Mmax,Rd is simply given by the sum of the plastic moment capacities of the reinforcement, the steel section and the concrete above the centroidal axis. Mmax,Rd is given by Mmax,Rd = Wpa

fy fck + 0,5Wpc γa γc

(10.71)

where Wpa is the plastic section modulus for the steel section (taken from tables) and Wpc is calculated from Wpc =

2 (b − 2t)(h − 2t)2 − r 3 − (4 − π)(0,5h − t − r)r 2 4 3

(10.72)

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Note, the equations have been derived for bending about the major (yy) axis. For minor axis (zz) bending, h and b are simply interchanged.

10.4.5 Buckling The non-dimensionalized slenderness ratio λ is defined by Npl,Rk λ= Ncr

(10.73)

where Npl,Rk is the characteristic plastic axial load capacity from Eq. (10.58) or Eq. (10.59) with all the materials’ partial safety factors set equal to unity and the effective buckling load Ncr is given by Ncr =

π2 (EI)eff l2

(10.74)

where l is the buckling length with the effective flexural stiffness is given by (EI)eff = Ea Ia + Ke Ecm Ic + Es Is

(10.75)

where Ea Ia is the flexural rigidity of the steel section alone, Ecm Ic is the flexural rigidity of the concrete, and Es Is is the flexural rigidity of the reinforcement and Ke is a correction factor taken as 0,6. Ecm is given in Table 3.1 of EN 1992-1-1. Where appropriate, account should be taken of the influence of long-term loading by using an effective concrete modulus Ec,eff determined from Ec,eff =

Ecm N

1 + φt NG,Ed Ed

(10.76)

where NEd is the total design normal force and NG,Ed is that portion which is permanent and ϕt is the creep co-efficient determined from EN 1992-1-1.

10.4.6 Design Moments Second order effects within the column length may be allowed for by increasing the larger design bending moment determined from a first order analysis by a factor k given by k=

β N

1 − N Ed cr,eff

≥ 1,0

(10.77)

where the moment ratio factor β for end moments is given by β = 0,66 + 0,44r ≥ 0,44

(10.78)

and Ncr,eff is an effective buckling load determined using the actual column length and an effective stiffness (EI)eff,II given by (EI)eff,II = K0 (Ea Ia + Ke,II Ecm Ic + Es Is )

(10.79)

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Chapter 10 / Composite Construction where K0 is a calibration factor (=0,9) and Ke,II is a correction factor representing the effects of cracking in the concrete (=0,5). Johnson (2004) indicates that where Ncr,eff ≥ 10NEd , then the second order effects need considering and there is an additional moment induced by the additional imperfection. Thus the design moment MEd is given by MEd = kend M + kimp NEd e0

(10.80)

where kend is magnification factor due to the moment gradient, M is the larger end moment, e0 is the eccentricity due to imperfections, kimp is the magnification factor determined from Eq. (10.78) with β = 1,0.

10.4.7 Other Checks and Detailing 10.4.7.1 Local Buckling For circular hollow sections: d/t ≤ 90ε2 ; Rectangular hollow sections: h/t ≤ 52ε; Partially encased sections: b/tf ≤ 44ε.

10.4.7.2 Cover For reinforcement, this is governed by the requirements in EN 1992-1-1. For encased steel sections, this should be a maximum of 40 mm or b/6, where b is the flange width.

10.4.7.3 Shear The shear bond between the steel section and the concrete should be checked using an elastic distribution of forces on the uncracked section with a transmission length not exceeding twice the relevant transverse direction. The values of shear bond should not exceed 0,3 MPa for fully encased sections, 0,55 MPa for circular concrete filled sections, 0,40 MPa for circular concrete filled sections and 0,2 MPa for the flanges only in partially encased sections. Where necessary shear studs should be used on encased I sections to resist shear.

10.4.7.4 Fire For concrete filled hollow sections it is essential that two vent holes of 20 mm diameter should be drilled through the steel section at the top and bottom of each storey subject to a maximum spacing of 5 m (Newman and Simms, 2000). These holes must not be within the depth of the floor construction. The purpose of these holes is to allow the build up of water vapour to escape whilst the moisture within the concrete is driven off in the early stages of heating in the fire.

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It is clear that the method of designing composite columns under uniaxial or biaxial bending is complex, and leads itself readily to the use of spreadsheets or design charts. The analysis of given sections to determine their carrying capacity is much more straightforward. The first example illustrates the determination of axial carrying capacity. To avoid duplication of calculations the second and third both use the same section, one under uniaxial bending about the major axis, and the other with biaxial bending. In each case the loading is considered totally short term, that is, Ec,eff is taken as Ecm . Also any reinforcement is considered negligible and is neglected.

EXAMPLE 10.4 Determination of axial load capacity of a composite column. Determine the axial load carrying capacity of 4 m effective length 150 × 150 × 8 Grade S355 rolled hollow section filled with Grade C25/30 concrete. Check h/t: Actual: h 150 = = 18,75 t 8 Allowable: 52ε = 52 × (235/355)1/2 = 42,3, therefore satisfactory Determination of Npl,Rd : Use Eq. (10.58) with the concrete taken at full strength, # 25 fy 355 " 0,85fck fsk Npl,Rd = Aa + Ac + As = 4510 + (150 − 16)2 − (4 − π)82 γa γc γs 1,0 1,5 = 1,899 MN Determine the load contribution ratio δ from Eq. (10.53): fy

δ=

Aa γ

a

Npl,Rd

=

4510 355 × 10−6 1,0 1,899

= 0,84

As δ lies between 0,2 and 0,9, the column may be designed as composite. Use Eq. (10.75) to determine the effective stiffness (Es Is = 0) fck + 8 1/3 25 + 8 1/3 Ecm = 22 = 22 = 31,5 GPa 10 10 (EI)eff = Ea Ia + 0,6Ecm Ic = 210 × 106 × 1510 × 10−8 + 0, 6 × 31, 5 × 106 ×

(150 − 16)4 = 3678 kNm2 12

Determine the Euler critical load Ncr from Eq. (10.74) Ncr =

π2 (EI)eff π2 × 3678 = = 2269 kN l2 42

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•

Chapter 10 / Composite Construction Determine Npl,Rk using Eq. (10.58) with the materials’ partial safety factors set equal to 1, # 25 fy 0,85fck fsk 355 " Npl,Rk = Aa + Ac + As = 4510 + (150 − 16)2 − (4 − π)82 γa γc γs 1,0 1,0 = 2,048 MN Determine the normalized slenderness ratio λ from Eq. (10.73), Npl,Rk 2048 λ= = = 0,95 Ncr 2269 This is less than the critical value of 2,0. The strength reduction factor is determined using buckling curve ‘a’ with α = 0,21, φ = 0, 5(1 + α(λ − 0, 2) + (λ)2 ) = 0, 5(1 + 0,21(0,95 − 0,2) + 0,952 ) = 1,03 χ=

φ+

!

1 φ2 − (λ)2

=

1,03 +

1 1,032 − 0,952

= 0,70

Determine the axial capacity NRd : NRd = χNpl,Rd = 0,70 × 1899 = 1329 kN

EXAMPLE 10.5 Axial load and uniaxial bending about the major axis. Determine whether a column having a system length of 4 m fabricated from 150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete can carry an axial load at ULS of 400 kN and a moment at ULS about the major axis of 18 kNm. Aa = 3710 mm2 ; Wpl,y = 183 × 103 mm3 . Determine Wpc from Eq. (10.72) 2 (b − 2t)(h − 2t)2 − r 3 − (4 − π)(0,5h − t − r)r 2 4 3 (100 − 2 × 8)(150 − 2 × 8)2 150 2 = − 83 − (4 − π) −8−8 4 3 2

Wpc =

= 373500 mm3 The area of the concrete Ac is given by Eq. (10.66) Ac = (b − 2t)(h − 2t) − (4 − π)r 2 = (150 − 2 × 8)(100 − 2 × 8) − (4 − π)82 = 11200 mm2 Axial squash capacity Npl,Rd is given by Eq. (10.58): Npl,Rd = Aa

fy 0,85fck fsk 355 25 + Ac + As = 3710 + 11200 = 1504 kN γa γc γs 1,0 1,5

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405

Determine the contribution ratio, δ from Eq. (10.53): 3710 355 × 10−3 As fyd 1,0 δ= = = 0,876 Npl,Rd 1504 This is within the limits for design as a composite column. Maximum moment capacity Mmax,Rd from Eq. (10.71): Mmax,Rd = Wpa

fy fck 355 25 + 0,5Wpc = 183000 × 10−6 + 0,5 × 373500 × 10−6 γa γc 1,0 1,5

= 68,1 kNm Determine Npm,Rd from Eq. (10.67): Npm,Rd = Ac

fck 25 × 10−3 = 187 kN = 11200 γc 1,5

Npm,Rd = 93,5 kN 2 Determine hn from Eq. (10.64) determine hn : hn =

f 2b γck c

Npm,Rd 18700 = 7,33 mm f = fck 25 25 y 2 × 100 × 1,5 + 4 × 8 2 355 − + 4tw 2 γ − γ 1,0 1,5 a c

Determine Wp,c,n from Eq. (10.69): Wp,c,n = (b − 2t)h2n = (100 − 2 × 8)7,332 = 4513 mm3 Determine Wp,a,n from Eq. (10.70): Wp,a,n = bh2n − Wp,c,n = 100 × 7,332 − 4513 = 860 mm3 Determine Mn,Rd from Eq. (10.68): Mn,Rd = Wp,a,n

fy fck 355 25 + 0,5Wp,c,n = 860 × 10−6 + 0,5 × 4513 × 10−6 γa γc 1,0 1,5

= 0,3 kNm Determine Mpl,Rd from Eq. (10.65): Mpl,Rd = Mmax,Rd − Mn,Rd = 68,1 − 0,3 = 67,8 kNm The values required to plot the interaction diagram are given in Table 10.3.

TABLE 10.3 Values required for major axis interaction diagram for EXAMPLE 10.5. Point A B C D

(0, N pl,Rd ) (M pl,Rd , 0) (M pl,Rd , N pm,Rd ) (M max,Rd , 0,5N pm,Rd )

Moment capacity (kNm)

Axial capacity (kN)

0 67,8 67,8 69,1

1504 0 187 93,5

406

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Chapter 10 / Composite Construction These values are plotted in Fig. 10.15. Determine the resistance to axial buckling about the major axis: Ea Ia = 210 × 106 × 1106 × 10−8 = 2323 kNm2 ( fck + 8) 1/3 Ecm = 22 = 31,5 GPa 10 Ic =

(0,150 − 0,016)3 (0,100 − 0,016) = 16,84 × 10−6 m4 12

Ecd Ic = 31,5 × 106 × 16,84 × 10−6 = 530 kNm2 Determine (EI)eff,II from Eq. (10.79): (EI)eff,II = 0,9(Ea Ia + 0,5Ecm Ic ) = 0,9(2323 + 0,5 × 530) = 2329 kNm2 Determine Ncr,eff from Eq. (10.74) with (EI)eff replaced by (EI)eff,II : Ncr,eff =

π2 (EI)eff,II π2 × 2329 = = 1437 kN l2 42

NEd = 400 kN > Ncr,eff /10, therefore second order effects need to be considered. Npl,Rk = (3710 × 355 + 11200 × 25) × 10−3 = 1597 kN. Determine λ from Eq. (10.73): λ=

Npl,Rk = Ncr,eff

1597 = 1,054 > 2,0 1437

Thus the column satisfies the limits for composite design. Second order effects: (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend =

β 1−

NEd Ncr,eff

=

1,0

= 1,386 400 1 − 1437

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp = 1,386 The design moment MEd is given by Eq. (10.80): MEd = kend M + kimp NEd e0 = 1,386 × 18 + 1,386 × 400

4 = 32,3 kNm 300

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From the interaction diagram in Fig. 10.15, the moment M400 corresponding to an axial load of 400 kN is given by M400 = Mpl,Rd μd =

Npl,Rd − NEd 1504 − 400 = 56,5 kNm = 67,8 Npl,Rd − Npm,Rd 1504 − 187

M400 56,5 = = 0,833 Mpl,Rd 67,4

MEd MEd 32,3 = = = 0,572 < 0, 9 μd Mpl,Rd M400 56,5 The limiting value of αM is 0,9 as Grade S355 steel is being used. Thus the column is therefore satisfactory.

EXAMPLE 10.6 Axial load and biaxial bending. Determine whether a column having a system length of 4 m fabricated from 150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete can carry an axial load at ULS of 350 kN and a moment at ULS about the major axis of 15 and 10 kNm about the minor axis. Two interaction diagrams are required, one for each axis. For the major axis the interaction diagram is as EXAMPLE 10.5. For the minor axis, the formulae in Eqs (10.64)–(10.73) are used but with b and h interchanged. Aa = 3710 mm2 ; Wpl,y = 183 × 103 mm3 ; Wpl,z = 133 × 103 mm3 . Determine Wpc from Eq. (10.72): (h − 2t)(b − 2t)2 2 − r 3 − (4 − π)(0, 5b − t − r) r 2 4 3 (150 − 2 × 8)(100 − 2 × 8)2 2 3 100 = − 8 − (4 − π) − 8 − 8 = 234200 mm3 4 3 2

Wpc =

From EXAMPLE 10.5, Ac = 11200 mm2 , Npl,Rd = 1504 kN and δ = 0,876. Determine the maximum moment capacity Mmax,Rd from Eq. (10.71): Mmax,Rd = Wpa

fy fck 355 25 × 10−6 + 0,5 × 234200 × 10−6 + 0,5Wpc = 133000 γa γc 1,0 1,5

= 49,2 kNm From EXAMPLE 10.5, Npm,Rd = 187 kN, and 0,5Npm,Rd = 93,5 kN Determine hn from Eq. (10.64): hn =

Npm,Rd 187000 = 6,41 mm f = f f 25 25 y 2 × 150 1,5 + 4 × 8 2 355 − 2h γck + 4tw 2 γ − γck 1,0 1,5 c a c

•

Chapter 10 / Composite Construction Determine Wpc,n from Eq. (10.69): Wpc,n = (h − 2t)h2n = (150 − 2 × 8)6,412 = 5506 mm3 Determine Wpa,n from Eq. (10.70): Wpa,n = hh2n − Wpc,n = 150 × 6,412 − 5506 = 657 mm3 Determine Mn,Rd from Eq. (10.68): Mn,Rd = Wpa,n

fy 355 fck 25 + 0,5Wpc,n = 657 × 10−6 + 0,5 × 5506 × 10−6 γa γc 1,0 1,5

= 0,3 kNm Determine Mpl,Rd from Eq. (10.65): Mpl,Rd = Mmax,Rd − Mn,Rd = 49,2 − 0,3 = 48,9 kNm The values required to plot the interaction diagram are given in Table 10.4. These values are plotted in Fig. 10.16.

TABLE 10.4 Values required for minor axis interaction diagram for EXAMPLE 10.6. Point A B C D

(0, N pl,Rd ) (M pl,Rd ,0) (M pl,Rd , N pm,Rd ) (M max,Rd , 0,5N pm,Rd )

Moment capacity (kNm)

Axial capacity (kN)

0 48,9 48,9 49,12

1504 0 187 93,5

1600 1400 1200 Axial capacity (kN)

408

1000 800 600 400 200 0 0

20

40

60

Moment capacity (kNm)

FIGURE 10.16 Major axis M–N interaction diagram (EXAMPLE 10.5)

80

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409

Determine the resistance to axial buckling about the minor axis: Ea Ia = 210 × 106 × 577 × 10−8 = 1212 kNm2 Ecm = 31,5 GPa (as EXAMPLE 10.5) Ic =

(0,100 − 0,016)3 (0,150 − 0,016) = 7,11 × 10−6 m4 12

Ecd Ic = 31,5 × 106 × 7,11 × 10−6 = 224 kNm2 (EI)eff,II = 0,9(Ea Ia + 0,5Ecm Ic ) = 0,9(1212 + 0,5 × 224) = 1192 kNm2 Ncr,eff =

π2 (EI)eff,II π2 × 1192 = = 735 kN l2 42

NEd = 350 kN > Ncr,eff /10, therefore second order effects need to be considered. Npl,Rk = Aa fy + Ac fck = 3710 × 355 × 10−3 + 11200 × 25 × 10−3 = 1597 kN Npl,Rk 1597 λ= = = 1,474 < 2,0 Ncr 735 Second order effects (major yy axis) using the appropriate critical buckling load: (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend,y =

β 1−

NEd Ncr,eff

=

1,0

= 1,322 350 1 − 1437

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp,z = 1,322. Second order effects (minor zz axis): (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend,z =

β 1−

NEd Ncr,eff

=

1,0

= 1,921 1 − 350 730

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp,z = 1,921.

•

Chapter 10 / Composite Construction 1600 1400 1200 Axial capacity (kN)

410

1000 800 600 400 200 0 0

10

20

30 40 Moment capacity (kNm)

50

60

FIGURE 10.17 Minor axis M–N interaction diagram (EXAMPLE 10.6)

From the interaction diagram for the major axis in Fig. 10.16, the moment M350 corresponding to an axial load of 350 kN is given by M350 = Mpl,Rd μdy =

Npl,Rd − NEd 1504 − 350 = 67,8 = 59,4 kNm Npl,Rd − Npm,Rd 1504 − 187

M350 59,4 = = 0,881 Mpl,Rd 67,4

From the interaction diagram in Fig. 10.17, for the minor axis the moment M350 corresponding to an axial load of 350 kN is given by M350 = Mpl,Rd μdy =

Npl,Rd − NEd 1504 − 350 = 48,9 = 42,8 kNm Npl,Rd − Npm,Rd 1504 − 187

M350 42,8 = = 0,875 Mpl,Rd 48,9

The second order effect due the bow may only be applied on one axis. Use Eq. (10.80) to calculate the design moments. Two cases, therefore, need considering: (a) Bow on major axis. MEd, y = kend, y MSd, y + kimp, y NEd e0 = 1,322 × 15 + 1,322 × 350 MEd, z = kend, z MSd, z = 1,921 × 10 = 19,2 kNm

4 = 26,0 kNm 300

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MEd,y 26,0 = 0,438 < 0,9 = μdy Mpl,y,Rd 59,4 MEd,z 19,2 = = 0,45 < 0,9 μdz Mpl,z,Rd 42,8 MEd,y MEd,z 26,0 19,2 + = + = 0,887 < 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd 59,4 42,8 (b) Bow on minor axis. MEd,y = kend,y MSd,y = 1,322 × 15 = 19,8 kNm MEd,z = kend,z MSd,z + kimp,z NEd e0 = 1,921 × 10 + 1,921 × 350

4 300

= 28,2 kNm MEd,y 19,8 = = 0,333 < 0,9 μdy Mpl,y,Rd 59,4 MEd,z 28,2 = = 0,66 < 0,9 μsdz Mpl,z,Rd 42,8 MEd,y MEd,z 19,8 28,2 + = + = 0,992 < 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd 59,4 42,8 It will be noted that the application of the initial bow to the minor axis is the critical case. This is due to both the lower moment capacity and the lower buckling load.

REFERENCES Bunn, R. and Heywood, M. (2004) Supporting services from structure. Co-Construct (BSRIA). EN 1991-1-1 Eurocode 1: Actions on structures – Part 1–1: General actions – Densities, self-weight, imposed loads for buildings. CEN/BSI. EN 1991-1-6 Eurocode 1: Actions on structures – Part 1–6: Actions during execution. CEN/BSI. EN 1992-1-1 Eurocode 2: Design of concrete structures – Part 1–1: General rules and rules for buildings. CEN/BSI. EN 1993-1-1 Eurocode 3: Design of steel structures – Part 1.1: General rules and rules for buildings. CEN/BSI. EN 1994-1-1 Eurocode 4: Design of composite steel and concrete structures – Part 1.1: General rules and rules for buildings. CEN/BSI. Evans, H.R. and Wright, H.D. (1988). Steel–concrete composite flooring deck structures, In Steel– concrete composite structures (ed R. Narayanan). Elsevier. Johnson, R.P. (2004). Composite structures of steel and Concrete (3rd edition). Blackwell Publishing. Johnson, R.P. (2005). Shear connection in beams that support composite slabs – BS 5950 and EN 1994-1-1, Structural Engineer, 83(22), 21–24. Johnson, R.P. and Anderson, D. (2004). Designers’ guide to EN 1994-1-1 Eurocode 4: Design of composite steel and concrete structures – Part 1.1: General rules and rules for buildings. Thomas Telford. Johnson, R.P. and May, I.M. (1975). Partial-interaction design of composite beams, The Structural Engineer, 53(8), 305–311. Lawson, R.M. and Chung, K.F. (1994). Composite beam design to eurocode 4. Publication 121. SCI.

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Chapter 10 / Composite Construction Lawson, R.M. and Nethercot, D.A. (1985). Lateral stability of I-beams restrained by profiled sheeting, The Structural Engineer, 63B(1), 1–7, 13. Lawson, R.M., Mullett, D.L. and Rackham, J.W. (1997). Design of asymmetric Slimflor ® beams using deep composite decking. Publication 175. SCI. Martin, L.H. and Purkiss, J.A. (2006). Concrete design to EN 1992. Butterworth-Heinemann. Mottram, J.T. and Johnson, R.P. (1990). Push tests on studs welded through profiled steel sheeting, The Structural Engineer, 68(10), 187–193. Mullett, D.L. (1992). Slim floor design and construction. Publication 110. SCI. Mullett, D.L. (1997). Design of RHS Slimflor ® Edge Beams. Publication 169. SCI. Mullett, D.L. and Lawson, R.M. (1992). Slim floor construction using deep decking. Technical Report 120. SCI. Newman, G.M. and Simms, W.I. (2000). The fire resistance of concrete filled tubes to Eurocode 4. Technical Report 259. SCI. Oehlers, D.J. and Johnson, R.P. (1987). The strength of shear stud connections in composite beams, The Structural Engineer, 65B(2), 44–48. Olgaard, J.G., Slutter, R.G. and Fisher, J.W. (1971). Shear strength of stud connectors in lightweight and normal-weight concrete, Engineering Journal, American Institute of Steel Construction, 8(2), 55–64. Wyatt, T.A. (1989). Design guide on the vibration of floors. Publication 076. SCI. Yam, L.C.P. and Chapman, J.C. (1968). The inelastic behaviour of simply supported composite beams of steel and concrete, Proceedings of the Institution of Civil Engineers, 41, 651–683.

Chapter

11 / Cold-formed Steel Sections

Thin-walled, cold-formed steel sections are widely used as purlins and rails, the intermediate members between the main structural frame and the corrugated roof or wall sheeting in buildings for farming and industrial use (see Fig. 11.1). Trapezoidal sheeting is usually fixed to these members in order to enclose the building. The most common sections are the zed, channel and sigma shapes, which may be plain or have lips. The lips are small additional elements provided to a section to improve its efficiency under compressive loads by enhancing the section ability against local buckling. Cold-formed steel sections are fabricated by means of folding, press-braking of plates or cold-rolling of coils made from carbon steel. Sheet steel used in cold-formed sections is typically 0.9–8 mm thick. It is usually supplied pre-galvanized in accordance with European Standard EN 10142. Galvanizing gives adequate protection for internal members or those adjacent to the boundaries of the building envelope. Cold working of the steel increases its yield strength but also lowers its ductility (see Fig. 11.2). For example, a 20% reduction in thickness can increase yield strength by 50% but reduces elongation to as little as 7%, which probably represents the limit of formability for simple shapes. The main benefits of using a cold-formed section are not only its high strengthto-weight ratio but also its lightness, which can save costs on transport, erection and the construction of foundation, and flexibility that the members can be produced in a wide variety of sectional profiles, which can result in more cost effective designs. Examples of the structural use of cold-formed sections include roof and wall members, steel framing, wall partitions, large panels for housing, lintels, floor joists, modular frames for commercial buildings, trusses, space frames, curtain walling, prefabricated buildings, frameless steel buildings, storage racking, lighting and transmission towers, motorway crash barriers, etc. The prime difference between the behaviour of cold-formed sections and hot rolled structural sections is that cold-formed members involve thin plate elements which tend to buckle locally under compression. Cold-formed cross-sections are therefore

This chapter is contributed by Long-yuan Li (Aston University) and Xiao-ting Chu (Auckland University of Technology, New Zealand)

Side and gable rails

Rail extensions

Corner detail Inwardly lipped channel

Side rail supports

Purlins

Purlin extensions

Diagonal tie wires and fixing cleats

Rail sleeves Eaves beam

Chapter 11 / Cold-formed Steel Sections

Apex anti-sag bars Speed fix anti-sag bars Purlin sleeves

•

Cleader angle

414

FIGURE 11.1 The building using zed and channel sections as purlins and rails (Copy from Albion Sections Ltd design manual by permission)

usually classified as slender because they cannot generally reach their full strength based on the amount of material in the cross-section (Rhodes and Lawson, 1992). The secondary difference is that cold-formed members have low lateral stiffness and low torsional stiffness because of their open, thin, cross-sectional geometry, which gives great flexural rigidity about one axis at the expense of low torsional rigidity and low flexural rigidity about a perpendicular axis. This leads to cold-formed members being susceptible to distortional buckling and lateral–torsional buckling.

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s Increase of yield strength due to strain hardening

Ultimate strength

Yield point after cold working Fracture

Yield strength

Proportional limit

Further loading after cold working

e Perfect plasticity Linear region

Strain hardening

Necking

FIGURE 11.2 The influence of cold forming on the stress–strain curve of steel

11.1 ANALYTICAL MODEL Most purlins and rails are laterally restrained by their supported cladding or sheeting either partially or completely. Hence, it is necessary to consider the influence of the lateral restraints when establishing an analytical model. Also, it is well known that, when a thin-walled beam has one or more cross-sections that are constrained against warping, a complex distribution of longitudinal warping stresses can be developed. These warping stresses together with the longitudinal stresses generated by bending moments may cause the beam to have local, distortional and lateral–torsional buckling. Consider a zed section beam that is partially restrained by the sheeting on its upper flange. Without loss of generality, the restraints of the sheeting can be simplified by using a translational spring and a rotational spring, both of which are uniformly distributed along the longitudinal direction of the beam (see Fig. 11.3). Let the origin of the coordinate system (x, y, z) be the centroid of the cross-section, with x-axis being along the longitudinal direction of the beam, and y- and z-axes taken in the plane of the cross-section, as shown in Fig. 11.3. According to the bending theory of asymmetric beams (Vlasov, 1961; Oden, 1967) and noticing that for a zed section y- and z-axes used in Fig. 11.3 are not the principal axes, the constitutive relationships between moments and generalized strains can be expressed as d2 w d2 v − EI yz dx2 dx2 2 d w d2 v Mz = −EIyz 2 − EIz 2 dx dx My = −EIy

(11.1)

416

•

Chapter 11 / Cold-formed Steel Sections kf

qz qy

ky

Pk

Pq

φ

y, v Sheeting

My w Mz

Purlin

v

z,w (a)

(b)

FIGURE 11.3 (a) Purlin-sheeting system and (b) a simplified analysis model d2 φ dx2 dφ MT = GIT dx

Mω = EIw

where My and Mz are the bending moments about y- and z-axes, Mω is the warping moment, MT is the twisting moment, E is the modulus of elasticity, G is the shear modulus, Iy and Iz are the second moments of the cross-sectional area about y- and z-axes, Iyz is the product moment of the cross-sectional area, Iw is the warping constant, IT is the torsion constant, v and w are the y- and z-components of displacement of the centroid of the cross-section, φ is the angle of twisting. The above four equations together with three equilibrium equations can be used to determine seven unknowns (four moments, My , Mz , Mω and MT and three displacements, v, w and φ). For the present problem it is convenient to derive the equilibrium equations by using the principle of minimum potential energy. For the partially restrained beam the total potential energy involves the strain energy of the beam, the strain energy of the two springs and the potential of the applied loads, that is = Ub + Us + W

(11.2)

in which, 1 Ub = 2

l

My

d2 w − 2 dx

+ Mz

0

= strain energy of the beam

d2 v − 2 dx

d2 φ dφ + M ω 2 + MT dx dx dx

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1 Us = 2

•

417

l "

# ky (v + zk φ)2 + kφ φ2 dx = strain energy of the springs

0

l W =− (v + zq φ)qy + (w − yq φ)qz dx = potential of the applied loads 0

where l is the length of the beam, ky and kφ are the stiffness constants per unit length of the translational and rotational springs, qy and qz are the uniformly distributed loads in y- and z-directions, (−yk , −zk ) and (−yq , −zq ) are the coordinates of the spring and loading points Pk and Pq (see Fig. 11.3), respectively. Substituting Eq. (11.1) into Eq. (11.2) yields 1 = 2

⎡

l

⎣EIy

d2 w dx2

⎡

0

1 + 2

l

⎣EIw

2

d2 φ dx2

d2 v d2 w + 2EIyz 2 2 + EIz dx dx 2

+ GIT

dφ dx

2

0

d2 v dx2

⎤ ⎦dx + 1 2

2 ⎤ ⎦dx

l "

# ky (v + zk φ)2 + kφ φ2 dx

0

l −

[(v + zq φ)qy + (w − yq φ)qz ]dx

(11.3)

0

The following three equilibrium equations can be obtained by the variation of the total potential energy with respect to the displacement components, v and w, and the angle of twisting, φ d4 v d4 w + ky (v + zk φ) + EIyz 4 = qy 4 dx dx 4 4 d v d w EIyz 4 + EIy 4 = qz dx dx 4 d φ d2 φ EIw 4 − GIT 2 + (zk2 ky + kφ )φ + zk ky v = qy zq − qz yq dx dx EIz

(11.4)

For beams that have no restraints, that is, ky = kφ = 0, Eq. (11.4) are simplified to d4 v d4 w + EI = qy yz dx4 dx4 d4 v d4 w EIyz 4 + EIy 4 = qz dx dx 4 d φ d2 φ EIw 4 − GIT 2 = qy zq − qz yq dx dx EIz

(11.5)

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•

Chapter 11 / Cold-formed Steel Sections For beams that are fully restrained, that is, v = φ = 0, Eq. (11.4) are simplified to Rk + EIyz

d4 w = qy dx4

d4 w = qz dx4 Rk zk + Mk = qy zk − qz yk EIy

(11.6)

where Rk and Mk are the reaction force and reaction moment at the restrained point Pk . Most cold-formed sections are supported by cleats bolted to the web of the section as shown in Fig. 11.4. The boundary conditions thus can be assumed as v = 0 Mz ≈ 0 w = 0 My ≈ 0 φ = 0 Mω ≈ 0

(11.7)

The cleats are designed so that the lower flange of the section does not bear directly on the rafter, and web crippling problems are avoided. However, the shear or bearing strength of the connecting bolts is critical to the design. Governing Eqs (11.4), (11.5) or (11.6) together with boundary conditions (11.7) can be used to determine the displacements and angle of twisting of the beam under the action of external loads, qy and qz . The bending moments at any place can be calculated using Eq. (11.1). The bending and shear stresses thus can be calculated from the moments

Purlin Upslope

Gap

Detail of Butted joint

FIGURE 11.4 Purlin butted to rafter beam by a cleat

Cleat

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419

and shear force as follows: σx = τmax =

Mz Iy − My Iyz My Iz − Mz Iyz d2 φ y + z + E(ω − ω) 2 2 Iy Iz − Iyz Iy Iz − Iyz dx2 3MT V + Av Ls t 2

(11.8)

where ω is the sectorial coordinate with respect to the shear centre, ω is the average value of ω, Ls is the total length of the middle line section, t is the thickness, V is the shear force, Av is the shear area. The sectorial coordinates are properties of the cross-section and are calculated as follows (Chu et al., 2004a,b) s ω=

hs ds and 0

1 ω= Ls

Ls ωds

(11.9)

0

where hs is the perpendicular distance from a tangent at the point under consideration to the shear centre, and s is the distance from any chosen origin to the same point measured along the middle line of the section. Equation (11.8) indicates that when warping torsion is involved twisting produces not only the shear stress but also axial stress. More about warping torsion can be found in the books of Oden (1967) and Walker (1975). Ye et al. (2004) investigated the influences of restraints on the magnitude and distribution of the axial stress within the cross-section through varying the stiffness constants of two springs. They also used the stress pattern obtained from Eq. (11.8) as an input to the finite strip analysis program and investigated the influence of restraints on the behaviour of local, distortional and lateral–torsional buckling of channel and zed section beams (Ye et al., 2002). It is interesting to notice from Eq. (11.6) that, if it is fully restrained the zed section beam bends only in the plane of the web and the bending stress and deflection can be calculated simply based on the bending rigidity of the beam in the plane of the web although the section itself is point symmetric, that is, σx =

My z Iy

d4 w qz = 4 EIy dx

(11.10)

11.2 LOCAL BUCKLING Cold-formed members are usually very thin, and thus the thin plate elements tend to buckle locally under compression. The local buckling mode of a cold-formed member normally involves plate flexure along, with no transverse deformation of a line or lines of intersection of adjoining plates, and can be characterized by a relatively short

420

•

Chapter 11 / Cold-formed Steel Sections

(a)

(b)

FIGURE 11.5 Local buckling modes of a zed section (h = 120 mm, b = 75 mm, c = 20 mm, t = 2,5 mm). (a) Under a pure compression (web buckle) and (b) under a pure bending (flange buckle)

half-wavelength of the order of magnitude of individual plate elements, as illustrated in Fig. 11.5.

11.2.1 Elastic Local Buckling Stress It is known from the stability of plates that, a simply supported rectangular plate may buckle if it is subjected to compressive loads in the plane of its middle surface. The elastic critical compressive stress when the plate buckles is expressed as (Bulson, 1970)

σcr,p =

kσ π 2 E 12(1 − ν2 )

t bp

2 (11.11)

where bp is the width of the plate, ν is Poisson’s ratio and kσ is the buckling coefficient determined from kσ =

l mbp

2

+2+

mbp l

2 (11.12)

where l is the length of a plate and m is the number of half waves of the buckling mode in the longitudinal direction in which the plate is compressed. Note that, kσ varies with m. When m = l/bp , kσ has a minimum value of 4, which makes the compressive stress σcr,p critical. This indicates that the buckles approximate to square wave forms, as demonstrated in Fig. 11.5. Equations (11.11) and (11.12) are only for a plate simply supported on the two long sides and subjected to uniform compressive stresses. For a compression element of

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421

width bp with different support conditions and/or subjected to non-uniform compressive stresses the critical compressive stress can still be calculated using Eq. (11.11) but the buckling coefficient needs to take account the influence of both boundary conditions and stress pattern. When these factors have been taken into account kσ is expressed as follows: For doubly supported compression elements (Table 4.1 in EN 1993-1-5 (2006))

kσ =

⎧ 8,2 ⎪ ⎪ ⎨ 1,05 + ψ

0≤ψ≤1

7,81 − 6,29ψ + 9,78ψ2 ⎪ ⎪ ⎩ 5,98(1 − ψ)2

−1 < ψ < 0 −3 < ψ ≤ −1

(11.13a)

For outstand compression elements (support at σ1 ) (Table 4.2 in EN 1993-1-5 (2006)) ⎧ ⎨ 0,578 kσ = 0,34 + ψ ⎩ 1,7 − 5ψ + 17,1ψ2

0≤ψ≤1

(11.13b)

−1 ≤ ψ < 0

For outstand compression elements (support at σ2 ) (Table 4.2 in EN 1993-1-5 (2006)) kσ = 0,57 − 0,21ψ + 0,07ψ2

−3 < ψ ≤ 1

(11.13c)

For single edge compression stiffener elements (lips) (cl 5.5.3.2.5 in EN 1993-1-3 (2006)) 5 kσ =

0,5

cp /bp ≤ 0,35

0,5 + 0,83(cp /bp − 0,35)2/3

0,35 < cp /bp ≤ 0,6

(11.13d)

where ψ = σ2 /σ1 is the ratio of stresses at the two ends of the element (σ1 is the larger compressive stress, σ2 is the tensile stress or smaller compressive stress, and the compressive stress is assumed to be positive), cp and bp are the lengths of the middle lines of the lip and flange, respectively. Equations (11.11) and (11.13) are used to calculate the critical stress of local buckling of a compression element. For channel and zed sections, the web and lipped flange may be treated as the doubly supported elements if the lip satisfies the requirement specified in Section 5.2 in EN 1993-1-3 (2006). Flanges that have no intermediate stiffeners and no edge lips are treated as the outstand elements. When a web or a flange has an intermediate stiffener, the actual width of the element should be taken as the width of the individual part separated by the stiffener. More details for dealing with elements with intermediate stiffeners can be found in EN 1993-1-3 (2006) and EN 1993-1-5 (2006).

422

•

Chapter 11 / Cold-formed Steel Sections

11.2.2 Post-Buckling Behaviour and the Calculation of Effective Width When an element buckles locally it does not necessarily mean that this element will collapse or loss its ability of carrying loads. In fact, a plate can be allowed to take a considerably increased load beyond initial buckling before any danger of collapse occurs. This is because the deflections due to buckling are accompanied by stretching of the middle surface of the plate. It is not always possible for practical reasons to allow some elements of a structure to buckle, but if stable buckles can be tolerated, a considerable gain follows in structural efficiency. For a uniformly compressed rectangular plate, up to the buckling load, the stress distribution is uniform. With increase in load, the central unconstrained portion of the plate will start to deflect laterally and will therefore not support much additional load, whereas the portions close to the supported edges will be constrained to remain straight and will continue to carry increasing stresses. Figure 11.6 shows the typical variation of the stress distribution in a plate in pre- and post-buckling stages. The ultimate strength of the plate is when the maximum stress at the edges reaches the compressive yield strength of the material. Thus the ultimate load of the plate should be calculated based on the stress distribution at failure through the width of the plate. The problem, however, is that analysis of the post-buckled plate is a complicated process and no exact closed form results have been obtained for compressed plates. Therefore, instead of using the stress distribution in the postbuckling range, an alternative approach to assessing the ultimate load of the plate is to use an effective width concept.

0,5beff

0,5beff

bp

fy b

(a)

(b)

fy b

(c)

FIGURE 11.6 The concept of effective width. (a) Stress distribution up to buckling, (b) stress distribution at failure and (c) stress distribution in effective width

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423

The concept of effective width was originally developed by Von Karman et al. (1932) and calibrated for cold-formed members by Winter (1968). The method assumes that when the ultimate stress is reached, the total load is carried by two fictitious strips adjacent to the edges of the plate (see Fig. 11.6c), which carry a uniform stress equal to the yield strength of the material, and the central region is unstressed. Obviously, the calculation of the effective width is dependent on the stress distribution at the time when the plate fails, which is influenced by a number of factors including the pattern of applied compressive stresses and the boundary conditions, relative slenderness and geometrical imperfections of the plate. Based on large numbers of tests, empirical functions have been developed. In EN 1993-1-5 (2006) the following equations have been recommended for calculating the effective width of a compression element. For doubly supported compression elements (cl 4.4.2 in EN 1993-1-5 (2006)) ⎧ 1,0 ⎪ ⎪ ⎨ ρ = λp,red − 0,055(3 + ψ) ⎪ ⎪ ⎩ 2 λp,red

λp,red ≤ 0,673 0,673 < λp,red

(11.14a)

For outstand compression elements or single edge compression stiffener elements (cl 4.4.2 in EN 1993-1-5 (2006))

ρ=

⎧ 1,0 ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λp,red ≤ 0,748

λp,red − 0,188 2 λp,red

0,748 < λp,red

(11.14b)

in which, λp,red = λp λp =

σcom,Ed = reduced slenderness fyb /γM0

fyb = relative slenderness for local buckling σcr,p

where ρ is the reduction factor to determine the effective width of a compression element defined in Tables 11.1 and 11.2, fyb is the basic yield strength of the material, σcom,Ed (σcom,Ed ≤ fyb /γM0 ) is the largest compressive stress in the compression element, and γM0 is the partial safety factor for resistance of the cross-section. After the effective widths of individual compression elements have been determined, the effective area, second moments of the effective area and effective section modulus can be calculated, from which the design values of the resistance to bending moments can be determined.

424

•

Chapter 11 / Cold-formed Steel Sections TABLE 11.1 Doubly supported compression elements (ψ = σ2 /σ1 ). Stress distribution (compression positive)

Effective width, beff

s1

ψ=1 beff = ρb be1 = be2 = 0,5beff

s2

be1

be2

b

s1

s2

be1

bc

be1

b

be1 = 52b−effψ be2 = beff − be1

be2

b

s1

1>ψ≥0 beff = ρb

0>ψ ρb beff = ρbc = 1−ψ be1 = 0,4beff be2 = 0,6beff

bt

be2

s2

11.3 DISTORTIONAL BUCKLING Distortional buckling involves both rotation and translation at the corners of the crosssection. Distortional buckling of flexural members such as channel and zed sections involves rotation of only the compression flange and lip about the flange–web junction as shown in Fig. 11.7. The web undergoes flexure at the same half-wavelength as the flange buckle, and the compression flange may translate in a direction normal to the web, also at the same half-wavelength as the flange and web buckling deformations. The elastic distortional buckling stress of cold-formed flexural members can be determined using either analytical methods, such as those suggested in AS/NZS 4600 (1996) and EN 1993-1-3 (2006) or numerical methods, such as the finite strip method (FSM) (Schafer, 1997) and the generalized beam theory (GBT) method (Davies et al., 1993).

11.3.1 The Calculation Method in EN 1993-1-3 (2006) In EN 1993-1-3 (2006), the design of compression elements with intermediate or edge stiffeners is based on the assumption that the stiffener behaves as a compression

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425

TABLE 11.2 Outstand compression element (ψ = σ2 /σ1 ). Stress distribution (compression positive)

Effective width, beff

s1

1>ψ≥0 beff = ρb s2

beff

b

s1

bc

0>ψ beff = ρbc = 1 ρb −ψ

bt

beff

s2

b

1>ψ≥0 beff = ρb

s1 s2

beff

b

bc

s1

0>ψ beff = ρbc = 1 ρb −ψ

bt

s2 b

beff

(a)

(b)

FIGURE 11.7 Distortional buckling modes of a zed section (h = 120 mm, b = 75 mm, c = 20 mm, t = 2,5 mm) (a) under a pure compression and (b) under a pure bending

Chapter 11 / Cold-formed Steel Sections member with continuous partial restraint, with a spring stiffness that depends on the boundary conditions and the flexural stiffness of the adjacent plane elements of the cross-section. The spring stiffness of the stiffener is determined by applying a unit load per unit length to the cross-section at the location of the stiffener, as illustrated in Fig. 11.8, and is determined from K=

1 1 Et3 = δ 4(1 − ν2 ) b21 (b1 + hp )

(11.15)

where δ is the deflection of the centroid of the stiffener due to a unit load, b1 is the horizontal distance from the web line to the centroid of the effective area of the edge stiffener, and hp is the depth of the web. The elastic critical buckling stress for a long strut on an elastic foundation of a spring stiffness coefficient K is given by Timoshenko and Gere (1961) as follows: σcr,d =

π2 EIs Kλ2 + 2 As λ As π 2

(11.16)

where As and Is are the area and second moment of the effective section of the stiffener, as illustrated in Fig. 11.9 for an edge stiffener, and λ = l/m is the half-wavelength of the distortional buckling (l is the member length and m is the number of half waves). For a sufficiently long strut, the critical half-wavelength can be obtained by minimizing the critical stress as defined by Eq. (11.6) with respect to λ, to give, EIs 1/4 λcr = π (11.17) K

bp Cθ

U

b1 U δ

•

δ

426

θ (a) be2

K (b)

(c)

FIGURE 11.8 (a) Distortional buckling model used in EN 1993-1-3, (2006) (b) edge stiffener on an elastic foundation of a spring stiffness coefficient and (c) model used to determine spring stiffness coefficient (copy from EN 1993-1-3 (2006) by permission)

Structural Design of Steelwork to EN 1993 and EN 1994

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427

b bP be1

be2 b a

b1

C

Ceff Cp

a

b

FIGURE 11.9 Effective cross-sectional area of an edge stiffener in EN 1993-1-3 (2006) (copy from EN 1993-1-3 (2006) by permission)

K

As, ls

Substituting Eq. (11.17) into Eq. (11.16) yields, √ 2 KEIs σcr,d = As

(11.18)

Equation (11.18) is given in EN 1993-1-3 (2006) for calculating the critical stress of distortional buckling of the edge stiffener. The design strength in EN 1993-1-3 (2006) (Section 5.5.3.1) for distortional buckling is considered by using a reduced thickness of the edge stiffener. The reduction factor is calculated in terms of the relative slenderness as follows, ⎧ ⎪ λd ≤ 0,65 1,0 ⎪ ⎨ χd = 1,47 − 0,723λd 0,65 < λd < 1,38 (11.19) ⎪ ⎪ ⎩ 0,66/λ 1,38 ≤ λ d

d

in which, λd =

fyb = relative slenderness for distortional buckling σcr,d

The procedure for calculating χd can be summarized as follows (Section 5.5.3.2 in EN 1993-1-3 (2006)). • Step 1 Obtain an initial effective cross-section for the stiffener using effective widths determined by assuming that the stiffener gives full restraint and that σcom,Ed = fyb /γM0 . • Step 2 Use the initial effective cross-section of the stiffener to determine the reduction factor for distortional buckling (flexural buckling of a stiffener), allowing for the effects of the continuous spring restraint. • Step 3 Optionally iterate to refine the value of the reduction factor for buckling of the stiffener; that is, re-calculate the effective widths of the lip and the part of the flange near the lip based on the compressive stress σcom,Ed = χd fyb /γM0 and calculate the reduction factor again based on the newly calculated effective widths.

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•

Chapter 11 / Cold-formed Steel Sections The design value of the resistance to bending moment about the y-axis due to both local and distortional buckling is determined based on the elastic section modulus of the effective section, Mc,Rd =

fyb Weff,y γM0

(11.20)

where Weff,y is the section modulus of the effective section for bending about y-axis, in which, apart from the effective widths of the web and the part of the flange near to web are calculated using the local buckling formulae, the effective widths and thicknesses of the lip and the part of the flange near the lip are calculated using both local and distortional buckling formulae, based on the reduced compressive stress, σcom,Ed = χd fyb /γM0 .

11.3.2 The Calculation Methods in AS/NZS 4600 (1996) In EN 1993-1-3 (2006) the critical stress of distortional buckling of a section is calculated based on the model of an edge stiffener on an elastic foundation and the effect of the distortional buckling on the section properties is taken into account by reducing the thickness of the stiffener. An alternative to determine the critical stress of distortional buckling of a cold-formed steel section is to use AS/NZS 4600 (1996) design code. The elastic distortional buckling formulae for channel and zed sections in AS/NZS 4600 (1996) are based on a simple flange buckling model where the flange is treated as a thin-walled compression member, as shown in Fig. 11.10, undergoing flexural–torsional buckling (Lau and Hancock, 1987; Hancock, 1997). The rotational spring stiffness kθ represents the flexural restraint provided by the web which is in flexure, and the translational spring stiffness kx represents the resistance to translational movement of the section in the buckling mode. The model includes a reduction in the flexural restraint provided by the web as a result of the compressive stress in the web. Lau and Hancock (1987) showed that the translational spring stiffness has no significant influence on results and thus is assumed to be zero. The rotational spring stiffness and the critical stress at distortional buckling are given as h4p λ2 1,11σcr,d 2Et3 kθ = (11.21) 1− 5,46(hp + 0,06λ) Et2 12,56λ4 + 2,192h4p + 13,39λ2 h2p

x Kx

Shear centre

Kθ y

x

y

FIGURE 11.10 Flange elastically restrained along flange-web junction in AS/NZS 4600 (1996)

Structural Design of Steelwork to EN 1993 and EN 1994 σcr,d =

! E (α1 + α2 ) − (α1 + α2 )2 − 4α3 2 Af

•

429

(11.22)

in which, η kθ Ixf b2p + 0,039Jf λ2 + β1 β1 ηE 2 α2 = η Iyf + ybp Ixyf β1 η 2 2 α3 = η α1 Iyf − Ixyf bp β1 Ixf + Iyf β1 = x2 + Af 1/4 Ixf b2p hp λ = 4,80 2t3 α1 =

η=

π 2 λ

where Af is the full cross-sectional area of the compression flange and lip, Ixf and Iyf are the second moments of the area Af about x - and y -axes, respectively, where the x - and y -axes are located at the centroid of area Af with x -axis parallel with flange, Ixyf is the product moment of the area Af about x - and y -axes, Jf is the St Venant torsion constant of the area Af , x and y are the distances from the flange-web junction to the centroid of area Af in the x - and y -directions, respectively. Due to the coupling of σcr,d and kθ in Eqs (11.21) and (11.22), Hancock (1997) suggested that kθ can be calculated based on an initial σcr,d obtained by assuming kθ = 0 and after then σcr,d can be calculated based on the obtained kθ value. In the iteration, if kθ < 0, kθ should be calculated using σcr,d = 0. The elastic critical moment for distortional buckling is calculated based on the critical buckling stress as follows, Mcr,d = σcr,d Wy

(11.23)

where Wy is the elastic section modulus of the gross cross-section for the extreme compression fibre. The design value of the resistance to bending moment about y-axis due to distortional buckling which involves rotation of the compression flange and lip about the flange-web junction is calculated as follows:

Mc,Rd =

⎧ ⎪ ⎨ Mc

0 ≤ kθ

Weff,y ⎪ ⎩ Mc Wy

kθ < 0

(11.24)

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Chapter 11 / Cold-formed Steel Sections in which, ⎧ Myield Weff,y ⎪ ⎪ 1 − M ⎪ yield ⎪ 4Mcr,d Wy ⎪ ⎨ ⎡ ⎤ 2 Mc = ⎪ M Weff,y ⎪ yield ⎪ ⎣0,055 ⎪ M − 3,6 + 0,237⎦ ⎪ ⎩ yield Mcr,d Wy

0,5Myield < Mcr,d Mcr,d ≤ 0,5Myield

where Myield = fyb Wy is the moment causing initial yield at the extreme compression fibre of the gross section. A comparison of the critical stresses of distortional buckling by using EN 1993-1-3 (draft version of 2001) and AS/NZS 4600 (1996) with experimental data was made by Kesti and Davies (1999). It was found that Lau and Hancock’s analytical expressions give a good prediction of the distortional buckling stress. The method given in EN 1993-1-3 does not correlate as well as Lau and Hancock’s method. The error in the distortional buckling stress could lead to a consequential error in the effective cross-sectional area depending on the distortional buckling stress level.

11.4 LATERAL–TORSIONAL BUCKLING In practice, purlins and rails are usually used together with their supported cladding or sheeting, and thus they are generally considered to be restrained against lateral deflections perpendicular to the line of action of the loading. If a beam is fully restrained on its translational and rotational degrees neither will the beam rotate nor deflect laterally. However, if the cladding or sheeting is not strong enough then it is possible for the beam to become unstable and for very large lateral deflections to occur at a critical value of the applied load. This type of behaviour is called lateral–torsional buckling. Chapter 5 has discussed the lateral–torsional buckling of unrestrained beams. In this section it is to deal with the lateral–torsional buckling of zed section beams with partial restraints from the sheeting. For the lateral–torsional buckling of partially restrained channel section beams readers can see the work of Chu et al. (2004).

11.4.1 Critical Moment of Lateral–Torsional Buckling The model presented here for analysing the lateral–torsional buckling of partially restrained beams was originally developed by Li (2004) and lately expanded by Chu et al. (2004b), which is similar to that described in Section 11.1. Assume that the displacements and moments in a state of equilibrium are (v,w,φ) and (My , Mz , Mω , MT ). Now let vb and wb be the y- and z-components of the buckling displacement of the centroid of the cross-section and φb be the buckling angle of twisting of the section. Assuming that the displacements in pre-buckling are very small, the increase

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of the strain energy of the system due to the lateral–torsional buckling thus can be expressed as ⎡ 2 2 ⎤ l d 2 wb d 2 vb ⎦ d 2 v b d 2 wb 1 ⎣ U2 = EIy dx + 2EIyz 2 + EIz 2 dx2 dx dx2 dx2 0

1 + 2

l

⎡ ⎣EIw

d 2 φb dx2

2

+ GIT

dφb dx

2

⎤ ⎦dx

0

1 + 2

l [ky (vb + zk φb )2 + kφ φb2 ]dx

(11.25)

0

On the other hand, the lateral–torsional buckling leads to a decrease of the potential of the pre-buckling moments and loads, which can be expressed as (Chu, 2004 Li, 2004; Chu et al., 2004b) l

d 2 vb 1 d 2 wb (Mz φb ) − M φ + Mω y b 2 dx2 dx2

W2 =

dφb dx

2 dx

0

l +

1 [(qy yq + qz zq )φb2 ]dx 2

(11.26)

0

If the net change of the total potential is positive for any of possible buckling displacements, that is, U2 > W2 , then the equilibrium of the pre-buckling state is said to be stable because the generation of the buckling displacements requires an energy input into the system. On the other hand, if the net change of the total potential is negative, that is, U2 < W2 , then the equilibrium of the pre-buckling state is said to be unstable because the buckling displacements can be generated without any input of energy. A critical state between stable and unstable equilibria from which the critical load of the lateral–torsional buckling can be determined is U2 = W2

(11.27)

Equation (11.27) is an eigenvalue type equation. For given reference loads the smallest eigenvalue and corresponding eigenvector calculated from Eq. (11.27) represent the critical loading factor and corresponding buckling mode. The above model can be applied directly to the purlins with intermediate lateral restraints such as provided by anti-sag bars if the pre-buckling moments are calculated based on the same model and the buckling displacements satisfy the displacement restraint conditions at the places where the anti-sag bars are placed. It is difficult to achieve closed form solutions of Eq. (11.27). In most cases only numerical solutions can be obtained. A general numerical computation procedure has been described by Li (2004) and Chu et al. (2004b) to obtain the critical buckling load, in which the following

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•

Chapter 11 / Cold-formed Steel Sections piecewise cubic spline functions are used to construct the displacement fields before and during buckling ⎧ ⎫ ⎡ ⎪ ⎨ v(x) ⎪ ⎬ Ni (x) ⎢ w(x) = ⎣ 0 ⎪ ⎪ ⎩ φ(x) ⎭ 0

⎤⎧ ⎫ ⎪ 0 ⎬ ⎨ δvi ⎪ ⎥ 0 ⎦ δwi ⎪ ⎪ Ni (x) ⎩δφi ⎭ ⎧ ⎫ ⎡ ⎤⎧ b ⎫ ⎪ δ ⎪ ⎪ ⎪ ⎪ 0 0 ⎨ vb (x) ⎬ Ni (x) ⎬ ⎨ vi ⎪ ⎢ ⎥ b wb (x) = Ni (x) 0 ⎦ δwi ⎣ 0 ⎪ ⎪ ⎪ ⎩ φ (x) ⎪ ⎭ ⎭ ⎩ δb ⎪ 0 0 Ni (x) ⎪ b 0 Ni (x) 0

(11.28)

(11.29)

φi

where Ni (x) is the spline interpolation function at node i and {δvi , δwi , δφi } and {δbvi , δbwi , δbφi } are the nodal displacement vectors before and during buckling. By using the displacement expressions (11.28) and (11.29), the equilibrium Eq. (11.4) and buckling Eq. (11.27) can be simplified into the following algebraic matrix equations ⎡

K vv ⎢ ⎣Kvw Kvφ ⎡ Kvv ⎢ ⎣Kvw Kvφ

Kvw Kww 0 Kvw Kww 0

⎤⎧ ⎫ ⎧ ⎫ Kvφ ⎪ ⎨ δv ⎪ ⎬ ⎪ ⎨ Fv ⎪ ⎬ ⎥ 0 ⎦ δw = Fw ⎪ ⎪ ⎪ ⎪ Kφφ ⎩δφ ⎭ ⎩Fφ ⎭ ⎤⎧ ⎫ ⎡ b 0 Kvφ ⎪ ⎨ δv ⎪ ⎬ qcr ⎢ ⎥ b 0 ⎦ δw = ⎣ 0 ⎪ ⎪ qref 0 Kφφ ⎩δbφ ⎭ Kvφ

(11.30)

0 0 0 Kwφ

⎤⎧ ⎫ 0 b ⎪ Kvφ ⎬ ⎨ δv ⎪ ⎥ 0 Kwφ ⎦ δbw ⎪ ⎭ ⎩ δb ⎪ 0 Kφφ φ

(11.31)

where K ij (i, j = v,w,φ) is the stiffness matrix, Kij0 (i, j = v,w,φ) is the geometric stiffness matrix, δj ( j = v,w,φ) is the nodal displacement vector, F j ( j = v,w,φ) is the nodal force vector, δbj ( j = v,w,φ) is the nodal vector of buckling displacements, (qcr /qref ) represents the scale factor between the critical and referenced loads. As an example, Fig. 11.11 shows the critical load factors of simply supported zed purlin beams (web depth h = 202 mm, flange width b = 75 mm, lip length c = 20 mm and thickness t = 2,3 mm), with zero, one and two anti-sag bars, subjected to a uniformly distributed uplift load. It can be seen from the figure that the lateral restraint has remarkable influence on the lateral–torsional buckling of the beam with no antisag bars. The influence is found to decrease with the increase of the beam length. Interestingly, when the beam has one or two anti-sag bars, the influence of the lateral restraint on the lateral–torsional buckling becomes almost negligible. This implies that the anti-sag bar not only has the ability of increasing the critical load but also can reduce the influence of the lateral restraint provided by cladding. Practically, purlins are often used as continuous beams over two or more spans, in which case the boundary conditions of the beam can be regarded as simply supported at one end and fixed at the other end. Fig. 11.12 shows the critical load factors of the zed purlin beam with this kind of boundary conditions. The results show that there is a significant increase in critical load when the purlin has a fixed boundary condition.

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5 ky0, no bars

4.5

ky, no bars

4

ky0, 1 bar

Critical loading factor

3.5

ky, 1 bar ky0, 2 bars

3

ky, 2 bars 2.5 2 1.5 1 0.5 0 3000

4000

5000

6000

7000

8000

9000

10 000

Beam length, mm

FIGURE 11.11 Lateral–torsional buckling of simply supported zed section beams (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformly distributed uplift loading, with various different restraints 5 ky0, no bars

Critical loading factor

4.5

ky, no bars

4

ky0, 1 bar

3.5

ky, 1 bar ky0, 2 bars

3

ky, 2 bars

2.5 2 1.5 1 0.5 0 3000

4000

5000

6000

7000

8000

9000

10 000

Beam length, mm

FIGURE 11.12 Lateral–torsional buckling of pinned-fixed zed section beams (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformly distributed uplift loading, with various different restraints

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11.4.2 Buckling Resistance Moment of Beams Subject to Bending The critical load calculated from the lateral–torsional buckling is based on an idealized model in which the beam has no geometrical imperfections, the cross-section of the beam does not deform, and the beam does not buckle either locally or distortionally before the lateral–torsional buckling occurs. When designing a real member, however, these factors should be taken into account. The method for determining the design buckling resistance moment for the lateral–torsional buckling of cold-formed section beams is the same as that used for other steel section members (Section 6.3.2.2 in EN 1993-1-1 (2005)), that is, Mb,Rd = χLT

Weff,y fyb γM1

(11.32)

in which, χLT =

1 2 1/2

2 − λ φLT + φLT LT

= reduction factor for lateral–torsional buckling (χLT ≤ 1,0)

2 φLT = 0, 5[1 + αLT λLT − 0,2 + λLT ] = factor used to calculate the reduction factor Mc,Rd 1/2 λLT = = relative slenderness for lateral–torsional buckling Mcr,LT where αLT = 0,34 is the imperfection factor, γM1 = 1,0 is the partial factor for resistance of members to instability and Mcr,LT is the elastic critical moment of the gross crosssection for lateral–torsional buckling about the main axis.

11.5 CALCULATION OF DEFLECTIONS The deflections of channel and zed section beams under uniformly distributed transverse loads can be evaluated using the analytical model presented in Section 11.1, provided that the load does not exceed the critical loads of local and distortional buckling. For the evaluation of deflections at loads greater than any critical load the influence of the local buckling and/or distortional buckling must be taken into account. The precise analysis of the post-buckling behaviour of a cold-formed section beam, however, is very difficult. A simple approach is to assume that the relationships between the load and deflection are linear for both pre- and post-buckling analyses as illustrated in Fig. 11.13. In pre-buckling region, the deflections can be calculated using simple beam theory and the gross section properties of the beam, since the beam is fully effective before buckling. In post-buckling region, the deflections can be evaluated using simple beam theory and the reduced section properties of the beam, since the beam is not fully effective after buckling. This assumption is only approximate since, in reality, the line in post-buckling region is not a straight line, but the errors introduced in the approximation are acceptable and conservative when fully reduced

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M Rd

cr

MRd Mcr

D

FIGURE 11.13 Simplified model used for the calculation of deflections section properties are used. Note that, in Fig. 11.13 in the pre-buckling region and ( − cr ) in the post-buckling region are linearly proportional to M and (M − Mcr ), respectively. Therefore, the deflection of the beam, , at an applied moment, M, can be expressed as ⎧ M ⎪ M ≤ Mcr ⎪ ⎨ cr M cr = ⎪ M − Mcr I ⎪ ⎩ cr 1 + Mcr < M ≤ MRd Mcr Ieff

(11.33)

where Mcr and MRd are the critical moment and moment resistance of the beam, cr is the deflection of beam when it buckles, I is the second moment of the gross cross-sectional area, and Ieff is the second moment of the effective cross-sectional area.

11.6 FINITE STRIP METHODS The FSM was originally developed by Cheung (1976) and it can be considered as a specialization of the finite element method (Zienkiewicz and Taylor, 2000). The method is mainly applied to structures whose geometries do not vary with at least one of the coordinate axes. The approach of the FSM is considered particularly favourable when dealing with the initial buckling or natural frequency characteristics of thinwalled prismatic structures. In the FSM, the prismatic structure is discretized into a number of longitudinal strips and the displacement fields associated with each strip

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Chapter 11 / Cold-formed Steel Sections

(a)

(b)

FIGURE 11.14 (a) The finite element analysis and (b) the finite strip analysis

vary sinusoidally along the strip length and algebraically across the strip width. Similar to the finite element method, shape functions are also used to define the variation of displacement fields along the strip width but they are only the functions of the cross-section coordinates of the strip (see Fig. 11.14). The use of the FSM for understanding and predicting the behaviour of cold-formed steel members was pioneered by Lau and Hancock in Australia (1986, 1989). They modified the stiffness matrices derived in Cheung’s book and created a commercial computer program for solution of the elastic buckling problem of open thin-walled members via finite strip called Thin-Wall. Similar programs were developed by Loughlan (1993) in the UK and by Schafer (1997) in USA. The program developed by Schafer is available in internet (http://www.ce.jhu.edu/bschafer/cufsm) and is particularly friendly to use. The codes were written in Matlab language and thus can easily be modified by users.

11.6.1 Element Stiffness Matrix of the Strip Consider a strip shown in Fig. 11.15, in which the local coordinate system is defined as that, the x- and y-axes are the two axes within the plane of the strip and the z-axis is normal to the plane of the strip. The three components of buckling displacement of the strip at a point (x, y) can be expressed in terms of the nodal displacements as follows ⎫ ⎧ ⎡ ⎤⎡ ⎤ ⎪ v1m ⎪ mπx y y ⎪ ⎪ 5 6 ⎪ sin 0 0 0 ⎪ 1− ⎬ ⎨ v(x, y) a b b ⎢ ⎥⎢ ⎥ u1m =⎣ (11.34) ⎦ ⎣ ⎦ mπx y y ⎪ u(x, y) v2m ⎪ ⎪ ⎪ ⎪ ⎪ 0 cos 0 1− 0 a b b ⎩ u2m ⎭

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⎧ ⎫ w1m ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ mπx 3y2 y3 3y2 2y3 2y2 2y3 y3 y2 ⎨ θ1m ⎬ w(x, y) = sin 1− 2 + 3 , y− + 2, 2 − 3 , 2 − (11.35) a b b ⎪ b b b b b b w2m ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ θ2m

where u(x,y) and v(x,y) are the plane displacements, w(x,y) is the deflection, (u1m , v1m , w1m , θ1m ) and (u2m , v2m , w2m , θ2m ) are the nodal displacements associated with wave number m, a and b are the length and width of the strip, respectively. The assumed displacement functions satisfy only the simply supported conditions at the two end sides of the strip. The strain energy of the strip is given as follows Et U2 = 2(1 − ν2 )

a b

εx + ε y

o

Et3 + 24(1 − ν2 )

2

− 2(1 − ν) εx εy −

o

a b

κx + κ y

o

2

− 2(1 − ν)

2 γxy

4

2 κx κy − κxy

dx dy dx dy

(11.36)

o

The membrane and bending strains in Eq. (11.36) are defined as follows, εx =

∂u ∂x

∂2 w κx = − 2 ∂x

εy =

∂v ∂y

∂2 w κy = − 2 ∂y

γxy =

∂u ∂v + ∂y ∂x

∂2 w κxy = − ∂x ∂y

(11.37)

The element stiffness matrix can be obtained by substituting Eqs (11.34) and (11.35) into Eq. (11.37) and then into Eq. (11.36), that is, U2 =

1 T {δ} [K]m {δ}m 2 m

in which, {δ}m = u1m , v1m , u2m , v2m , w1m , θ1m , w2m , θ2m T = element nodal vector ⎤ ⎡ uv uv K uv uv K14 K11 K12 13 ⎥ ⎢ uv −K uv K uv K22 ⎥ ⎢ 14 24 [0] ⎥ ⎢ 4x4 uv uv K11 −K12 ⎥ ⎢ ⎥ ⎢ uv ⎥ ⎢ K 22 ⎥ ⎢ [K]m = ⎢ wθ wθ wθ wθ K11 K12 K13 K14 ⎥ ⎥ ⎢ ⎥ ⎢ wθ −K wθ K wθ K22 ⎥ ⎢ 14 24 ⎥ ⎢ symmetric wθ wθ ⎣ K11 −K12 ⎦ wθ K22 = element stiffness matrix uv K11 uv K12

12D 1 a 1 − ν b 2 = 2 + (mπ) 2b 12 a t 12D 3ν − 1 = 2 (mπ) 8 t

(11.38)

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Chapter 11 / Cold-formed Steel Sections 1a 1−νb 12D 2 − + (mπ) 2b 24 a t2 12D 1 + ν (mπ) = 2 8 t 12D 1 − ν a 1 b = 2 + (mπ)2 4 b 6a t 1−νa 12D 1 b 2 = 2 − + (mπ) 4 b 12 a t

uv K13 = uv K14 uv K22 uv K24

wθ K11

=

wθ K12 =

wθ K13 =

wθ K14 =

wθ K22 =

wθ = K24

D=

a 3 D 13 b 6a 4 2 (mπ) + (mπ) + 6 5b b a2 70 a 2 a 2 D 11 b 1 + 5ν 4 2 (mπ) + (mπ) + 3 a 420 a 10 b D 9 b 6a a 3 (mπ)4 − (mπ)2 − 6 2 5b b a 140 a 2 a 2 D 13 b 1 4 2 − (mπ) + (mπ) + 3 a 840 a 10 b 3 1 b 2 b a D (mπ)4 + (mπ)2 + 2 210 a 15 a b 3 3 b 1 b a 4 2 D − (mπ) − (mπ) + 840 a 30 a b Et3 12(1 − ν2 )

11.6.2 Element Geometric Stiffness Matrix of the Strip The geometric stiffness matrix of a strip subjected to linearly varying edge traction can be derived by considering the change of the potential of the in-plane forces during buckling. Similar to the approach described in Section 11.4, the change of the potential of the in-plane forces can be expressed as, t W2 = 2

a b " o

o

y# σ1 − (σ1 − σ2 ) b

∂u ∂x

2 +

∂v ∂x

2 +

∂w ∂x

2 dx dy

(11.39)

where σ1 and σ2 are the compressive stresses at nodes 1 and 2 (see Fig. 11.15). Substituting Eqs (11.34) and (11.35) into Eq. (11.39) yields W2 =

1 T {δ} Kg m {δ}m 2 m

(11.40)

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b

a

a x y u1 θ1

v1

u2

θ2

w1

z

v2

s1

w2

s2

FIGURE 11.15 Local coordinates, degrees of freedom and stress distribution in a strip

in which,

b(mπ)2 [Kguv ] [0]4x4 [Kg ]m = = element geometric stiffness matrix 1680a [0]4x4 [Kgwθ ] ⎡ ⎤ 70(T1 + T2 ) 0 70(3T1 + T2 ) 0 ⎢ 70(T1 + T2 ) ⎥ 70(3T1 + T2 ) 0 ⎢ ⎥ [Kguv ] = ⎢ ⎥ ⎣ ⎦ 70(T1 + 3T2 ) 0 symmetric 70(T1 + 3T2 ) ⎡ ⎤ −2b(7T1 + 6T2 ) 8(30T1 + 9T2 ) 2b(15T1 + 7T2 ) 54(T1 + T2 ) ⎢ b2 (5T1 + 3T2 ) 2b(6T1 + 7T2 ) −3b2 (T1 + T2 ) ⎥ ⎢ ⎥ [Kgwθ ] = ⎢ ⎥ ⎣ 24(3T1 + 10T2 ) −2b(7T1 + 15T2 ) ⎦ symmetric b2 (3T1 + 5T2 )

For a member composed of multiple strips the global stiffness matrix and global geometric stiffness matrix can be obtained by the assembly of element stiffness matrices and element geometric stiffness matrices, that is [Km ] = [K]m [Kgm ] = [Kg ]m (11.41) k

k

where the index k denotes the k-th element. The summation implies proper coordinate transformations and correct addition of the stiffness terms in the global coordinates and degrees of freedom. The elastic buckling problem is a standard eigenvalue problem of the following form: [Km ]{δm } = λm [Kgm ]{δm }

(11.42)

where eigenvalues, λm , and eigenvectors, {δm }, are the buckling loads and corresponding buckling modes.

•

Chapter 11 / Cold-formed Steel Sections

11.6.3 Finite Strip Solution Methods The finite strip analysis employs single wave functions and thus can only be applied to the case where the longitudinal stresses do not vary along the longitudinal axis. Note that both [K m ] and [K gm ] are the functions of the strip length, a, and wave number, m. The buckling loads and buckling modes solved from Eq. (11.42) are also the functions of them. Since only the critical buckling load, which is the smallest eigenvalue for any of possible wave numbers, is of interest, the strip length and wave number are not independent each other. There are two quick ways to find the critical buckling load. One is to let a equal to the real length of the strip and solve the problem for several numbers of m. The other is to let m = 1 and solve the problem for a number of lengths. The former provides the results only for the given length of the strip, whereas the latter provides a complete picture of the critical buckling loads and modes at various different half-wavelengths, which has clear physical meanings. For this reason the latter method is often used. Figure 11.16 shows the buckling curves of a laterally restrained (both displacement and rotation) zed section beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) subjected to pure bending, in which the three local minima represent the local, distortional and secondary distortional buckling. The secondary distortional buckling is sometimes called lateral–distortional buckling to distinguish it from flange distortional buckling. The secondary distortional buckling exists only when the section is restrained

3 Critical load curve Buckling curve m = 1 Buckling curve m = 2–8

2.5

Critical loading factor

440

2 110 mm

1.5 565 mm

3000mm

1

0.5

0

102

103 Beam length, mm

FIGURE 11.16 Buckling curves of a simply supported zed section beam restrained at the junction between web and tension flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) (copy from Chu et al. (2006) by permission)

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laterally. Note that the buckling curves only provide information about how and when the beam buckles. The actual critical load of the beam for a given length should be taken as the lowest value from all of the buckling curves of the same length, and this critical load curve is the solid line plotted in Fig. 11.16. The original FSM can only deal with the buckling problem of members subjected to either pure compression or pure bending. Recently, Chu et al. (2005, 2006) modified the geometric stiffness matrix in the FSM by allowing for the variation of the prebuckling stress along the longitudinal axis, thus leading to a semi-analytical FSM which is able to deal with the buckling problem of cold-formed section beams subjected to uniformly distributed transverse loading. As an example, Fig. 11.17 shows the critical load curve of the restrained zed section beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) under uniformly distributed loading. In order to compare the difference in critical load between pure bending and uniformly distributed loading, the results of the beam under pure bending is also superimposed in the figure. It is evident that the critical load associated with uniformly distributed loading is significantly higher than that arising from pure bending although, for both local and distortional buckling, the differences between the two loading cases decrease with the beam length. For example, for a 5 m long beam, the critical load of a uniformly distributed loading beam is 12% higher than that of the pure bending beam compared to 20% higher for a 2 m long beam.

4 Uniformly distributed load 3.5

Pure bending 110 mm

Critical loading factor

3 2.5 2 565 mm 1.5 3000mm 1 0.5 0

102

103 Beam length, mm

FIGURE 11.17 Critical load curves of a simply supported zed section beam restrained at the junction between web and tension flange subjected to a uniformly distributed uplift load and a pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) (copy from Chu et al. (2006) by permission)

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11.7 DESIGN METHODS FOR BEAMS PARTIALLY RESTRAINED BY SHEETING Purlins and rails are usually used together with their supported trapetzoidal sheeting. Thus it is generally assumed that the sheeting takes the load in the plane of the sheeting and the purlin takes the load normal to the plane of the sheeting, that is, in the plane of web. Also, the purlin may be regarded as being laterally restrained in the plane of the sheeting and partially restrained in twisting if the trapetzoidal sheeting is connected to a purlin and the connection meets the following condition (cl 10.1.1.6 in EN 1993-1-3 (2006)) 70 π2 EIw π2 h2 EIz S≥ 2 (11.43) + GIT + h l2 4l2 where S is the portion of the shear stiffness provided by the sheeting for the examined member connected to the sheeting at each rib (if the sheeting is connected to a purlin every second rib only, then S should be substituted by 0,2S), h is the web depth and l is the span length. The partial torsional restraint may be represented by a rotational spring with a spring stiffness CD , which can be calculated based on the stiffness of the sheeting and the connection between the sheeting and the purlin, as follows, 1 1 1 = + CD CD,A CD,C

(11.44)

where CD,A is the rotational stiffness of the connection between the sheeting and the purlin and CD,C is the rotational stiffness corresponding to the flexural stiffness of the sheeting. Both CD,A and CD,C are specified in Section 10.1.5.2 in EN 1993-1-3 (2006). The restraints of the sheeting to the purlin have important influence on the buckling behaviour of the purlin. Fig. 11.18 shows the buckling curves of a simply supported zed purlin beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) with various different lateral restraints applied at the junction between the web and the compression flange when subjected to a pure bending. The figure shows that, when the translational displacement of the compression flange is restrained the purlin does not buckle lateral torsionally. On the other hand, when the rotation of the compression flange is restrained the critical stresses of local buckling and distortional buckling are increased quite significantly. However, when the restraints are applied at the junction between the web and the tension flange, it is only the rotational restraint that influences the lateral–torsional buckling of the purlin (see Fig. 11.19). Similar to the buckling behaviour, the lateral restraints also have considerable influence on the bending behaviour of the purlin. It has been found from both finite element analyses and experiments that, the bending behaviours are different when the restraints are applied at tension and compression flanges, and the free flange and the restrained flange are bent differently, particular when the free flange is under

Structural Design of Steelwork to EN 1993 and EN 1994

Ratio of critical buckling stress to yield strength

5 4.5 4

•

443

No restraint Translational restraint Rotational restraint Full restraints

3.5 3 2.5 2 1.5 1 0.5 0

102

103 Half-wavelength, mm

FIGURE 11.18 Buckling curves of a simply supported zed section beam with different restraint applied at the junction between web and compression flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) Ratio of critical buckling stress to yield strength

5 4.5 4

No restraint Translational restraint Rotational restraint Full restraints

3.5 3 2.5 2 1.5 1 0.5 0

102

103 Half-wavelength, mm

FIGURE 11.19 Buckling curves of a simply supported zed section beam with different restraint applied at the junction between web and tension flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) compression (Lucas et al., 1997a, b; Vrany, 2002). This led to development of different treatments for sections when subjected to gravity loading and uplift loading, since, for instance, for a simply supported beam, the free flange is in tension for gravity loading but in compression for uplift loading. In EN 1993-1-3 (2006), the stress in the

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Chapter 11 / Cold-formed Steel Sections restrained flange is calculated based on the bending moment in the plane of web as follows (assuming there is no axial force), σmax,Ed =

My,Ed fy ≤ Weff,y γM0

(11.45)

where fy is the yield strength. While the stress in the free flange is calculated based on not only the bending moment in the plane of web but also the bending moment in the free flange due to the equivalent lateral load acting on the free flange caused by torsion and lateral bending as follows, σmax,Ed =

My,Ed fy Mfz,Ed + ≤ Weff,y Wfz γM0

(11.46)

where Mfz,Ed is the bending moment in the free flange due to the equivalent lateral load as defined in Fig. 10.3 in EN 1993-1-3 (2006), and Wfz is the gross elastic section modulus of the free flange plus 1/5 of the web height for the point of web–flange intersection, for bending about the z–z axis (see Fig. 11.20). The determination of Mfz,Ed is dependent on the section dimensions, loading position, span length, number of anti-sag bars, and spring stiffness CD , the detail of which is specified in Section 10.1.4.1 in EN 1993-1-3 (2006). It should be pointed out that Eq. (11.46) applies to only the case where the free flange is under compression. For the free flange under tension, where due to positive influence of flange curling and second order effect moment Mfz,Ed may be taken equal to zero.

qEd

NEd

NEd

1/5 h

h kh qEd b

z

y

y

1/5 h

ez2

z My,Ed Weff,y

NEd Aeff

ez1

Mfz,Ed Wfz

FIGURE 11.20 Superposition of stresses in free flange (copy from EN 1993-1-3 (2006) by permission)

1/6 h

Structural Design of Steelwork to EN 1993 and EN 1994

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Similar to any compression members, the flexural buckling of the free flange under compression also need to be considered. The buckling resistance of the free flange is verified by using the following formula fy Mfz,Ed 1 My,Ed + ≤ χLT Weff,y Wfz γM1

(11.47)

2 − 0, 75(λ )2 )1/2 ] = reduction factor for flexural buckin which, χLT = 1/[φLT + (φLT fz

2 ling of the free flange (χLT ≤ 1,0 and χLT ≤ 1/ λLT )

2 φLT = 0,5 1 + αLT λfz − 0,4 + 0,75 λfz = factor used to calculate the reduction factor

where αLT = 0, 34 is the imperfection factor and λfz is the relative slenderness for flexural buckling of the free flange and is determined from lfz fyb λfz = (11.48) ifz π2 E where ifz is the radius of gyration of the gross cross-section of the free flange plus the contributing part of the web for bending about the z–z axis and lfz is the buckling length for the free flange which is specified in Section 10.1.4.2 in EN 1993-1-3 (2006).

11.8 WORKING EXAMPLES In order to assist designers, manufactures of cold-formed steel sections often provide design manuals for their products, which, in general, include gross section properties, effective section properties, and load tables for a number of specified span lengths under various different sets and/or types of connection. The gross section properties are calculated based on the geometric dimensions of the section, whereas the effective section properties are calculated based on the effective widths by taking into account the effects of local and distortion buckling. The load tables can be obtained from either tests or calculations by considering the following modes of failure: • • • • • •

Flexural failure involving local and distortional buckling in compression Lateral–torsional buckling due to insufficient lateral restraints Excessive deflection Shear failure Web crushing under direct loads or reactions Combined effects between bending and web crushing, and bending and shear.

The load tables design their sections for specific uses and give the performance of sections under various given circumstances. The following is an example of calculation of gross and effective section properties for a cold-formed lipped zed section subjected to bending in the plane of web.

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•

Chapter 11 / Cold-formed Steel Sections The calculation is based on the method recommended in EN 1993-1-3 (2006). The dimensions of the cross-section are (where the influence of rounding of the corners is neglected): Depth of web Width of flange in compression Width of flange in tension Length of lip Thickness

h = 202 mm bc = 75 mm bt = 75 mm c = 20 mm t = 2 mm

The material properties of the section are: Modulus of elasticity Poisson’s ratio Basic yield strength Partial factor

E = 210000 N/mm2 ν = 0,3 fyb = 390 N/mm2 γM0 = 1,00

Checking of geometrical proportions The design method of EN 1993-1-3 (2006) can be applied if the following conditions are satisfied (Section 5.2): b/t ≤ 60

bc /t = 75/2 = 37,5 < 60 bt /t = 75/2 = 37,5 < 60 c/t ≤ 50 c/t = 20/2 = 10 < 50 h/t ≤ 500 h/t = 202/2 = 101 < 500

→ ok → ok → ok → ok

In order to provide sufficient stiffness and avoid primary buckling of the stiffener itself, the size of stiffener should be within the following range (Section 5.2 in EN 1993-1-3 (2006)): 0,2 ≤ c/b ≤ 0,6 c/bc = 20/75 = 0,27 c/bt = 20/75 = 0,27

→ ok → ok

For cold-formed steel sections the section properties are usually calculated based on the dimensions of the section middle line as follows (see Fig. 11.21), Depth of web Width of flange in compression Width of flange in tension Length of lip

hp = h − t = 202 − 2 = 200 mm bp1 = bc − t = 75 − 2 = 73 mm bp2 = bt − t = 75 − 2 = 73 mm cp = c − t/2 = 20 − 2/2 = 19 mm

Calculation of gross section properties Gross cross-section area: A = t(2cp + bp1 + bp2 + hp ) = 2 × (2 × 19 + 73 + 73 + 200) = 768 mm2

Structural Design of Steelwork to EN 1993 and EN 1994

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447

hp

cp

bp1

cp

t

FIGURE 11.21 Symbols used for representing the dimensions of middle lines of a zed section

bp2

Position of the neutral axis with regard to the flange in compression: zc =

hp 200 = = 100 mm 2 2

Position of the neutral axis with regard to the flange in tension: zt = hp − zc = 200 − 100 = 100 mm Second moment of the gross cross-sectional area: Iy =

(h3p + 2c3p )t

+

(bp1 + bp2 )t3

c p 2 + zc2 bp1 t + zt2 bp2 t + cp t zc − 2

12 12 2 cp + cp t zt − = 4880000 mm4 2

Gross section modulus with regard to the flange in compression: Wy,c =

Iy 4880000 = = 48800 mm3 zc 100

Gross section modulus with regard to the flange in tension: Wy,t =

Iy 4880000 = = 48800 mm3 zt 100

Calculation of effective section properties The general (iterative) procedure is applied to calculate the effective properties of the compression flange and the lip (plane element with edge stiffener). The calculation should be carried out in three steps: Step 1 Obtain an initial effective cross-section for the stiffener using effective widths of the flange and lip determined by assuming that the compression flange is doubly supported, the stiffener gives full restraint (K = ∞) and that the design strength is not reduced, that is, σcom,Ed = fyb /γM0 . • Effective width of the compressed flange

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•

Chapter 11 / Cold-formed Steel Sections For the internal compression flange the stress ratio ψ = 1 (uniform compression), so the buckling coefficient is taken as kσ = 4. The relative slenderness thus is: fyb 390 λb,red = λb = = 0,827 = σcr,b 570 where σcr,b =

kσ π2 E 3,142 × 210000 4 = = 570 N/mm2 2 2 2 12(1 − ν ) (bp1 /t) 12(1 − 0,3 ) (73/2)2

Since λb,red = 0,827 > 0,673, the width reduction factor for the doubly supported compression element is calculated by (Eq. (11.14a)) ρ=

λb,red − 0,055(3 + ψ) 2 λb,red

=

0,827 − 0,055(3 + 1) = 0,887 0,8272

The effective width of the compressed flange thus is: beff = ρbp1 = 0,887 × 73 = 64,8 mm be1 = be2 = 0,5beff = 0,5 × 64,8 = 32,4 mm • Effective length of the lip For the compression lip, the buckling coefficient should be taken as follows (Eq. (11.13d)) kσ = 0,5 kσ = 0,5 + 0,83(cp /bp1 − 0,35)2/3

if cp /bp1 ≤ 0,35 if 0,35 ≤ cp /bp1 ≤ 0,6

For cp /bp1 = 19/73 = 0,260 < 0,35 kσ = 0,5. The relative slenderness is:

fyb 390 = = 0,609 σcr,c 1051 π2 E kσ 3,142 × 210000 0,5 where σcr,c = = = 1050 N/mm2 12(1 − ν2 ) (cp /t)2 12(1 − 0,32 ) (19/2)2 Since λc,red = 0, 609 < 0, 673, the width reduction factor for the outstand compression element thus is given by (Eq. (11.14b)): ρ = 1,0 The effective length of the compression lip thus is: ceff = ρcp = 1,0 × 19 = 19 mm The corresponding effective area of the edge stiffener is: As = t(be2 + ceff ) = 2 × (32,4 + 19) = 103 mm2 λc,red = λc =

Step 2 Use the initial effective cross-section of the stiffener to determine the reduction factor, allowing for the effects of the distortional buckling. The elastic critical stress of the distortional buckling for the edge stiffener is (Eq. (11.18)) √ 2 KEIs σcr,d = = 343 N/mm2 As

Structural Design of Steelwork to EN 1993 and EN 1994 where K =

•

449

Et3 = 0,445 N/mm2 4(1 − ν2 )(b21 hp + b31 )

b1 = bp1 −

b2e2 = 62,8 mm 2(be2 + ceff )

2 2 c3eff t c2eff c2eff be2 t3 ceff + + be2 t Is = + ceff t − 12 12 2(be2 + ceff ) 2 2(be2 + ceff ) = 3330 mm4 Thickness reduction factor for the edge stiffener is calculated based on the relative slenderness of the edge stiffener as follows (Eq. (11.19)): λd =

fyb = σcr,d

χd = 1,0 χd = 1,47 − 0,723λd χd = 0,66/λd

390 = 1,066 343 if λd ≤ 0,65 if 0,65 < λd < 1,38 if λd ≥ 1,38

0,65 < λd < 1,38 so χd = 1,47 − 0,723λd = 0,699 Step 3 As the reduction factor for the buckling of the stiffener is χd = 0,699 < 1, iterations are required to refine the value of the reduction factor. The iterations are carried out based on the reduced design strength, σcom,Ed,i = χd,i−1 fyb /γM0 to obtain new effective widths of the lip and flange in the stiffener and recalculate the critical stress of distortional buckling of the stiffener and thus to obtain new reduction factor. The iteration stops when the reduction factor χd converges. The final values obtained after iterations are be2 = 36,1, ceff = 19,0 and χd = 0,689. • Effective width of the web The position of the initial neutral axis (web is assumed as fully effective) with regard to the flange in compression is given by h2p c2 χ d cp cp hp − 2 + bp2 hp + 2 + eff2 hc = = 106 mm cp + bp2 + hp + be1 + (be2 + ceff )χd The stress ratio thus is: ψ=−

hp − hc 200 − 106 =− = −0,89 hc 106

The corresponding buckling coefficient is calculated by (Eq. (11.13a)) kσ = 7,81 − 6,29ψ + 9,78ψ2 = 21,2

450

•

Chapter 11 / Cold-formed Steel Sections The relative slenderness thus is: fyb 390 λh,red = λh = = = 0,986 σcr,h 1051 where σcr,h =

kσ π2 E 3,142 × 210000 21,2 = = 402 N/mm2 12(1 − ν2 ) (hp /t)2 12(1 − 0,32 ) (200/2)2

The width reduction factor thus is (Eq. (11.14a)) ρ=

λh,red − 0,055(3 + ψ) 2 λh,red

=

0,986 − 0,055(3 − 0,89) = 0,895 0,9862

The effective width of the zone in compression of the web is: heff = ρhc = 0,895 × 106 = 94,9 mm Part of the effective width near the flange is (see Table 11.1): he1 = 0,4heff = 0,4 × 94,9 = 37,9 mm Part of the effective width near the neutral axis is: he2 = 0,6heff = 0,6 × 94,9 = 56,8 mm Thus, h1 = he1 = 37,9 mm h2 = (hp − hc ) + he2 = (200 − 106) + 56,8 = 151 mm

The effective widths of the web obtained above are based on the position of the initial neutral axis (web is assumed as fully effective). To refine the result iterations are required which is based on the newly obtained effective widths, he1 and he2 , to determine the new position of the neutral axis. The stress ratio, buckling coefficient, relative slenderness, width reduction factor and effective widths of the web thus are recalculated according to the new position of the neutral axis. Iteration continues until it converges. The final values obtained after iterations are he1 = 37,9 mm, he2 = 56,8 mm and h2 = 149 mm (see Fig. 11.22). • Effective properties of the section (see Fig. 11.23) Effective cross-section area: Aeff = t[cp + bp2 + h2 + h1 + be1 + (be2 + ceff )χd ] = 698 mm2

Structural Design of Steelwork to EN 1993 and EN 1994

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451

bp1 be2

hp

he2

hc

he1

ceff

be1

n.a.

cp

t

FIGURE 11.22 Symbols used for representing the dimensions of the effective cross-section

bp2

bp1

hp

he2

zc

tχd

ceff

be2

he1

h1

be1

t

zt

h2

n.a

cp

FIGURE 11.23 Symbols used for representing the properties of the effective cross-section

bp2

Position of the neutral axis with regard to the flange in compression: h2 c2 χd cp h t cp hp − 2 + bp2 hp + h2 hp − 22 + 21 + eff2 zc = = 108 mm Aeff Position of the neutral axis with regard to the flange in tension: zt = hp − zc = 92 mm Second moment of the effective sectional area: Ieff,y =

t(h31 + h32 + c3p + χd c3eff )

t3 (bp2 + be1 + be2 χd3 ) 12 12 cp 2 h2 2 h1 2 2 + cp t zt − + bp2 tzt + h2 t zt − + h1 t zc − 2 2 2 2 c eff + be1 tzc2 + be2 (χd t)zc2 + ceff (χd t) zc − 2

= 4340000 mm4

+

452

•

Chapter 11 / Cold-formed Steel Sections Effective section modulus with regard to the flange in compression: Weff,y,c =

Ieff,y = 40100 mm3 zc

Effective section modulus with regard to the flange in tension: Weff,y,t =

Ieff,y = 47200 mm3 zt

The design value of the resistance of the section to bending moment about the y-axis due to local and distortional buckling is Mc,Rd =

fyb Weff,y,c 390 × 40100 × 10−6 = = 15,6 kNm γM0 1,0

The load table of a section is defined based on the specified type of connection and given span lengths. Assume that the section is used as a single span of 5 m long, both ends are butted to rafter beam using cleats, the section is subjected to gravity loading and is fully restrained laterally. The design value of the resistance to gravity load on the span due to local and distortional buckling is: Pc,Rd =

8Mc,Rd 8 × 15,6 = = 25,0 kN l 5

The design value of the plastic shear resistance is (Section 6.2.6 in EN 1993-1-1 (2005)): √ 200×2×390 √ Av fyb / 3 × 10−3 3 Vpl,Rd = = = 90,1 kN γM0 1,0 The design value of the shear buckling resistance is (Section 6.1.5 in EN 1993-1-3 (2006)): Vb,Rd =

hp tf sin φ bv

γM0

=

200 sin 90◦

×2×

0,48×390 1,49

1,0

× 10−3

= 50,3 kN

in which, fbv = 0,58fyb

if λw ≤ 0, 83

fbv = 0,48fyb /λw

if 0,83 < λw

λw = 0, 346

hp t

fyb 200 = 0,346 × × E 2

390 = 1,49 210000

So, the design shear resistance is Vc,Rd = min(Vpl,Rd , Vb,Rd ) = 50,3 kN Since the maximum shear force for the simply supported beam occurs at the support, the value of which is only half of the span load, it is obvious that for the present case

Structural Design of Steelwork to EN 1993 and EN 1994

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453

the design load is controlled by the design value of the resistance due to local and distortional buckling. Therefore the ultimate design load for the section is 25 kN. Note that, the load factors used for permanent actions and for variable actions are different. The design load for any combinations of dead and imposed loads thus should satisfy γG PG + γQ PQ ≤ Pc,Rd = 25 kN where γG = 1,35 and γQ = 1,5 are the load factors for dead and imposed loads, PG and PQ are the dead and imposed loads, respectively. The deflection check is usually done by using computer software which is normally provided by the manufacturer. The allowable deflection is determined based on individual cases and is not specified in most design standards.

11.9 CHAPTER CONCLUSIONS This chapter has described the mechanism of failures of the cold-formed steel sections and the corresponding methods for analyses and principles for design. The focuses have been placed on the analyses of local, distortional and lateral–torsional buckling of the sections for these are the main differences between the cold-formed steel section and the hot rolled steel section. For the cold-formed steel sections the most important properties are the effective section properties. For summarizing the calculation procedure a flowchart of calculation of effective section properties is given in Figs. 11.24 and 11.25. The design of a cold-formed steel section is much more complicated than that of a hot rolled section. This is partly because the section involves elements which have large width-to-thickness ratios and thus are easier to buckle locally and distortionally, and partly because the section is locally restrained by its supported trapetzoidal sheeting, which complicates the loading system and generates different stresses in restrained and free flanges. Simplified design methods may be used for channel, zed and sigma purlin systems that have no anti-sag bars, no use of sleeves and overlapping between two adjacent sections (see Annex E in EN 1993-1-3 (2006)). It is well known that for a short beam the design is normally controlled by the bending moment and/or shear force, while for a long beam it is usually controlled by the deflection and lateral–torsional buckling. Thus, purlins are usually designed to be continuous by using sleeves or overlapping in order to satisfy deflection limits and anti-sag bars to prevent twisting during erection and to stabilize the lower flange against wind uplift. For simply supported members, it is the sagging (positive) moment conditions that determine the capacity of the member. For continuous members over one or more internal supports, moments are to be determined elastically. Plastic hinge analysis is not permitted because the slender sections are not able to maintain their full

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•

Chapter 11 / Cold-formed Steel Sections

Start

Dimensions of cross-section; Material data

Determine geometrical proportions of cross-section

EN 1993-1-3 §5.1(3)

b/t 60 c/t 50 h/t 500 0,2 c/b 0,6 ?

The influence of rounding of the corners may neglected if. r/t 5; r/bp 0,1

No

Stop

EN 1993-1-3 §5.2 Yes The generai (iterative) procedure is applied to calculate the effective properties of the flanges and lips in compression. The calculation should be carried out in three steps.

EN 1993-1-3 §5.5.3.2

EN 1993-1-3 §5.5.2

Axial compression

Determine effective widths for the flanges and lips in compression

Determine the effective section properties of the web

Compute the gross cross-section properties

χd1, be12, ceff1, be11, χd2, be22, ceff2, be21

Major axis bending

EN 1993-1-3 §5.5.3.2

EN 1993-1-3 §5.5.2

he1, he2

EN 1993-1-5 §4.4

The general (iterative) procedure is applied to calculate the effective properties of the flange and lip in compression. The calculation should be carried out in three steps.

Determine effective widths for the flange and lip in comperssion

χd1,be2, ceff,be1

Determine the effective section properties of the web

he1, he2

Determine the properties of effective cross-section: leff, Weff

leff, Weff

EN 1993-1-5 §4.4

Determine effective area of the crosssection: Aeff

Stop

Aeff

Recalculate the position of the neutral axis taking into account the effective properties of compression flange and lip

Stop

FIGURE 11.24 The flowchart of calculation of effective section properties for cold-formed steel sections under compression or bending (copy from www.access-steel.com by permission) moment capacity when rotations exceed the point at which the section reaches yield. By utilizing the flexibility of sleeved or overlapping purlins at the supports, some elastic redistribution of moment may be achieved, and hence lead to more efficient design of the member. However, when the purlins are sleeved or overlapped, the design of the

For the initial evaluation, it is assumed that the flange in compression is double supported (K∞) and that the design strength is not reduced (σcom,Ed fyb / γM0)

Start

EN 1993-1-3 §5.5.3.2(3)

A

EN 1993-1-3 §5.5.3.2(5a)

Step 1 Determine the effective cross-section for the stiffener using effective width of flange

be1,be2,ceff

Determine the effective area of the edge stiffener: Ae

As

EN 1993-1-5 §4.4

EN 1993-1-3 §5.5.3.2(6)

Allow for the effects of continuous spring restraint

EN 1993-1-3 §5.5.3.2(3)

Step 2 Use the intial effective cross-section of the stiffener to determine the reduction factor

Where: K is the spring stiffness per unit length and Is is the effective second moment of area of the stiffener

EN 1993-1-3 §5.5.3.2(7) Determine the elastic buckling stress for the edge stiffener.

σσ,s

Determine the reduction factor for the distortional buckling resistance of the edge stiffener

χd

EN 1993-1-3 §5.5.3.1(5)

EN 1993-1-3 §5.5.3.1(7)

EN 1993-1-5 §4.4(2)

Xd < 1 ?

No

Return

Yes EN 1993-1-3 §5.5.3.2 (3), (10), (12)

Step 3 iterate to refine the value of the reduction factor for buckling of the stiffener.

Return

To A

The iterations are carried out based on modified values of ρ, obtained using: σcom,Edfyb / γM0 The iteration stops when the reduction factor χ converges.

FIGURE 11.25 The flowchart of iterative procedure for calculation of reduction factor due to distortion buckling (copy from www.access-steel.com by permission)

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•

Chapter 11 / Cold-formed Steel Sections system is normally developed based on testing or the combination of analysis and testing to achieve economic solutions. Standardized testing and evaluation procedures have been given in EN 1993-1-3 (2006) (Annex A) for both the cold-formed members and sheeting.

REFERENCES AS/NZS 4600 (1996). Australia/New Zealand Standard for Cold-formed Steel Structures. Standards Australia, Sydney. Bulson, P.S. (1970). The stability of flat plates. Chatto & Windus, London. Cheung, Y.K. (1976). Finite strip methods in structural analysis. Pergramon Press, New York. Chu, X.T. (2004). Failure Analysis of Cold-formed Steel Sections. Ph.D. Thesis, Aston University, Birmingham. UK. Chu, X.T., Kettle, R. and Li, L.Y. (2004a). Lateral–torsion buckling analysis of partial-laterally restrained thin-walled channel-sections beams, Journal of Constructional Steel Research, 60(8), 1159–1175. Chu, X.T., Li, L.Y. and Kettle, R. (2004b). The effect of warping stress on the lateral torsion buckling of cold-formed zed-purlins, Journal of Applied Mechanics, 71(5), 742–744. Chu, X.T., Rickard, J. and Li, L.Y. (2005a). Influence of lateral restraint on lateral–torsional buckling of cold-formed steel purlins. Thin-Walled Structures, 43, 800–810. Chu, X.T., Ye, Z.M, Kettle, R. and Li, L.Y. (2005b). Buckling behaviour of cold-formed channel sections under uniformly distributed loads. Thin-Walled Structures, 43, 531–542. Chu, X.T., Ye, Z.M, Li, L.Y. and Kettle, R. (2006). Local and distortional buckling of cold-formed zedsection beams under uniformly distributed transverse loads, International Journal of Mechanical Sciences, 48(8), 378–388. Davies, J.M., Leach, P. and Heinz, D. (1993). Second-order generalized beam theory, Journal of Constructional Steel Research, 31, 221–241. EN 1993-1-1 (2005). Eurocode 3: Design of steel structures – Part 1–1: General rules and rules for buildings. BSI. EN 1993-1-3 (2006). Eurocode 3: Design of steel structures – Part 1–3: General – Cold formed thin gauge members and sheeting. BSI. EN 1993-1-5 (2006). Eurocode 3: Design of steel structures – Part 1–5: General – Strength and stability of planar plated structures without transverse loading. BSI. Hancock, G.J. (1997). Design for distortional buckling of flexural members, Thin-Walled Structures, 27(1), 3–12. Kesti, J. and Davies, J.M. (1999). Local and distortional buckling of thin-walled short columns, Thin-Walled Structures, 34, 115–134. Lau, S.C.W. and Hancock, G.J. (1986). Buckling of thin flat-walled structures by a spline finite strip method, Thin-Walled Structures, 4, 269–294. Lau, S.C.W. and Hancock, G.J. (1987). Distortional buckling formulas for channel columns, Journal of Structural Engineering, ASCE, 113(5), 1063–1078. Lau, S.C.W. and Hancock, G.J. (1989). Inelastic buckling analyses of beams, columns and plates using the spline finite strip method, Thin-Walled Structures, 7, 213–238. Li, L.Y. (2004). Lateral–torsional buckling of cold-formed zed-purlins partial-laterally restrained by metal sheeting, Thin-Walled Structures, 42(7), 995–1011. Loughlan, J. (1993). Thin-walled cold-formed sections subjected to compressive loading, Thin-Walled Structures, 16(1–4), 65–109. Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997a). Modelling of cold-formed purlinsheeting systems – Part 1. Full model, Thin-Walled Structures, 27(3), 223–243.

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Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997b). Modelling of cold-formed purlinsheeting systems – Part 2. Simplified model, Thin-Walled Structures, 27(4), 263–286. Oden, J.T. (1967). Mechanics of elastic structures. McGraw-Hill Book Company, New York. Rhodes, J. and Lawson, R.M. (1992). Design of structures using cold-formed steel sections. SCI Publication, 089. SCI. Schafer, B.W. (1997). Cold-formed Steel Behaviour and Design: Analytical and Numerical Modelling of Elements and Members with Longitudinal Stiffeners. Ph.D. Thesis. Cornell University. Timoshenko, S.P. and Gere, J.M. (1961). Theory of elastic stability. McGraw-Hill Book Company, New York. Vlasov, V.Z. (1961). Thin-walled elastic beams. Israel Program for Scientific Translations, Jerusalem, Israel. Von Karman, T., Sechler, E.E. and Donnell, L.H. (1932). The strength design of thin plates in compression, Transactions ASME, 54, 53–55. Vrany, T. (2002). Torsional restraint of cold-formed beams provided by corrugated sheeting for arbitrary input variables. Proceedings of Eurosteel. Coimbra, cmm, 2002, 734–742. Walker, A.C. (1975). Design and analysis of cold-formed sections. International Textbook Company Ltd, London. Winter, G. (1968). Thin walled structures – theoretical solutions and test results, Preliminary Publications 8th Congress IABSE, 101–112. Ye, Z.M., Kettle, R., Li, L.Y. and Schafer, B. (2002). Buckling behaviour of cold-formed zed-purlins partially restrained by steel sheeting, Thin-Walled Structures, 40, 853–864. Ye, Z.M., Kettle, R. and Li, L.Y. (2004). Analysis of cold-formed zed-purlins partially restrained by steel sheeting, Computer and Structures, 82, 731–739. Zienkiewicz, O.C. and Taylor, R.L. (2000). The finite element method, Butterworth Heinemann, Oxford.

Annex

A1 /

Design Strengths for Fillet Welds

Design strength per unit length (Fw,Rd ) in KN/mm

Leg length (mm)

Steel grade S275 fu = 430 MPa βw = 0,85

Steel grade S355 fu = 510 MPa βw = 0,90

4 5 6 8 10 12 15 18 20 22 25

0,654 0,818 0,981 1,308 1,636 1,963 2,453 2,944 3,271 3,598 4,089

0,733 0,916 1,099 1,466 1,832 2,199 2,748 3,298 3,664 4,031 4,580

Notes: (1) F w,Rd = f u /(31/2 βw γM2 ) × 0,7 × (leg length) for equal leg lengths and γM2 = 1,25. (2) Steel to Standard 10025-2 with appropriate electrodes (BSEN 499).

•

Annex

A2 / Design Strengths for Class 4.6 Ordinary Bolts

Bolt diameter d (mm)

Reduced area of bolt As (mm2 )

Tension stress using reduced area Ft, Rd (kN)

Single shear stress using reduced area Fv,Rd (kN)

Single shear stress using gross area Fv,Rd (kN)

(M12) M16 M20 (M22) M24 (M27) M30 (M33) M36

84,3 157 245 303 353 459 561 694 817

24,3 45,2 70,6 87,3 101,7 132,2 161,6 199,9 235,3

16,2 30,1 47,0 58,2 67,8 88,1 107,7 133,2 156,9

21,7 38,6 60,3 73,0 86,9 109,9 135,7 164,2 195,4

Notes: (1) F v,Rd = 0,6 (A or As ) f ub /γM2 (2) F t,Rd = 0,9 As f ub /γM2 (3) f ub = 400 MPa; γM2 = 1,25 (4) Bolt sizes in brackets are not preferred. (5) Shear values for other classes of bolt are multiplied by a factor Class 4.6 4.8 5.6 5.8 6.8 8.8 10.9 Factor 1,0 0,83 1,25 1,04 1,25 2,0 2,08 (6) Tension values for other classes of bolt are multiplied by a factor Class 4.6 4.8 5.6 5.8 6.8 8.8 10.9 Factor 1,0 1,0 1,25 1,25 1,5 2,0 2,5

459

• 460 Annex

A3 /

Design Strengths for Preloaded Class 8.8 and 10.9 Bolts

Bolt size d (mm)

Reduced area of bolt As (mm2 )

Preload force class 8.8 F p,C (kN)

Preload force class 10.9 F p,C (kN)

Slip resistance class 8.8 F s,Rd (kN)

Slip resistance class 10.9 F s,Rd (kN)

(M12) M16 M20 M22 M24 M27 M30 M36

84,3 157 245 303 353 459 561 817

47,2 87,9 137,2 169,7 197,7 257,0 314,2 457,5

59,1 109,9 171,5 212,1 247,1 321,3 392,7 571,9

18,9 35,2 54,9 67,9 79,1 102,8 125,7 183,0

23,6 44,0 68,6 84,8 98,8 128,5 157,1 228,8

Notes: (1) Preload F p,C = 0,7 f ub As where f ub = 800 (Class 8.8) and 1000 (Class 10.9) MPa. (2) Design slip resistance F s,Rd = ks nμF p,C /γM3 where ks = 1, n = 1, μ = 0,5 and γM3 = 1,25 (3) Bolt sizes in brackets are not preferred.

•

Annex

A4 / Space of Holes in Beams, Columns, Joists and Tees 461

Beams, columns, joists and tees Nominal flange widths (mm)

Spacings in millimetres S4

Recommended diameter of bolt (mm)

Actual bmin (mm)

S1

S2

S3

419–368 330–305 330–305 292–203 190–165 152 146–114 102 89 76 64 51

140 140 140 140 90 90 70 54 50 40 34 30

140 120 120 – – – – – – – – –

75 60 60 – – – – – – – – –

290 240 240 – – – – – – – – –

24 24 20 24 24 20 20 12 – – – –

362 312 300 212 162 150 130 98 – – – –

S1

S1 S1

S3 S2 S3 S4

462

•

Annex

A5 / Spacing of Holes in Angles

Angles Spacing of holes

Maximum diameter of bolt

Nominal leg length S 1 S2 S3 S4 S5 S6 S1 S 2 and S 3 (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm)

S 4 ,S 5 and S 6 (mm)

200 150 125 120 100 90 80 75 70 65 60 50 45 40 30 25

20 – – – – – – – – – – – – – – –

– – – – 55 50 45 45 40 35 35 28 25 23 20 15

75 55 45 45 – – – – – – – – – – – –

75 55 50 50 – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

– – – – 24 24 20 20 20 20 16 12 – – – –

30 20 20 20 – – – – – – – – – – – –

Leg

S3 S2 S1 Leg

Leg

S6 S5 S4

Annex

A6 / Spacing of Holes in Channels

•

463

Channels Nominal flange width (mm)

S1 (mm)

Recommended diameter of bolt (mm)

102 89 76 64 51 38

55 55 45 35 30 22

24 20 20 16 10 –

S1

464

•

A7 /

Annex 7 / From BS5950: 1: 1990 by Permission

Annex

From BS5950: 1: 1990 by Permission

TABLE 15 Slenderness correction factor, n, for members with applied loading substantially concentrated within the middle fifth of the unrestrained length. Note 1. All hogging moments are +ve. Note 2. β is defined in Table 18. Note 3. M0 is the mid-length moment on a simply supported span equal to the unrestrained length (see Table 17).

L 10 maximum M0

M

BM

Unrestrained length L

β positive

β negative

γ = M/M 0

1.0

0.8

0.6

0.4

0.2

0.0

−0.2

−0.4

−0.6

−0.8

−1.0

+50.00 +10.00 +5.00 +2.00 +1.50 +1.00 +0.50 0.00 −0.10 −0.20 −0.30 −0.40 −0.50 −0.60 −0.70 −0.80 −0.90 −1.00

1.00 0.99 0.98 0.96 0.95 0.93 0.90 0.86 0.85 0.83 0.81 0.79 0.77 0.62 0.56 0.56 0.59 0.62

0.96 0.99 0.98 0.95 0.95 0.92 0.90 0.86 0.85 0.83 0.82 0.80 0.78 0.66 0.56 0.53 0.57 0.58

0.92 0.94 0.97 0.95 0.94 0.92 0.90 0.86 0.85 0.83 0.82 0.81 0.79 0.72 0.61 0.54 0.54 0.54

0.87 0.90 0.93 0.95 0.94 0.92 0.89 0.86 0.85 0.84 0.83 0.81 0.80 0.77 0.67 0.59 0.53 0.52

0.82 0.85 0.89 0.94 0.93 0.92 0.89 0.86 0.85 0.84 0.83 0.82 0.82 0.80 0.73 0.65 0.57 0.54

0.77 0.80 0.84 0.94 0.93 0.91 0.89 0.86 0.86 0.85 0.84 0.83 0.83 0.82 0.79 0.71 0.64 0.59

0.72 0.75 0.79 0.89 0.92 0.91 0.89 0.86 0.86 0.85 0.85 0.84 0.85 0.84 0.83 0.77 0.71 0.66

0.67 0.69 0.73 0.84 0.90 0.91 0.89 0.86 0.86 0.85 0.85 0.85 0.86 0.85 0.85 0.83 0.77 0.72

0.66 0.68 0.71 0.79 0.85 0.91 0.88 0.86 0.86 0.86 0.86 0.85 0.86 0.86 0.87 0.89 0.84 0.80

0.66 0.68 0.70 0.77 0.80 0.92 0.88 0.86 0.86 0.86 0.86 0.86 0.87 0.87 0.88 0.90 0.88 0.85

0.65 0.67 0.70 0.76 0.80 0.92 0.88 0.86 0.86 0.86 0.87 0.87 0.88 0.88 0.89 0.90 0.91 0.92

(Continued)

Structural Design of Steelwork to EN 1993 and EN 1994

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465

TABLE 15 Continued −1.10 −1.20 −1.30 −1.40 −1.50 −1.60 −1.70 −1.80 −1.90 −2.00 −5.00 −50.00 Infinity

0.66 0.70 0.73 0.74 0.75 0.76 0.77 0.79 0.80 0.81 0.93 0.99 1.00

0.62 0.66 0.69 0.70 0.70 0.72 0.74 0.77 0.79 0.81 0.89 0.95 0.96

0.57 0.60 0.63 0.64 0.64 0.65 0.66 0.68 0.69 0.70 0.83 0.90 0.91

0.54 0.55 0.57 0.58 0.59 0.60 0.61 0.63 0.64 0.65 0.77 0.86 0.86

0.54 0.54 0.55 0.56 0.56 0.57 0.58 0.59 0.60 0.61 0.72 0.79 0.82

0.57 0.55 0.54 0.54 0.54 0.55 0.56 0.56 0.57 0.58 0.67 0.74 0.77

0.63 0.60 0.57 0.55 0.55 0.55 0.55 0.56 0.56 0.56 0.64 0.70 0.72

0.68 0.65 0.61 0.60 0.59 0.58 0.58 0.57 0.57 0.56 0.61 0.67 0.68

0.76 0.73 0.69 0.66 0.65 0.64 0.63 0.62 0.61 0.60 0.60 0.64 0.65

0.83 0.80 0.77 0.74 0.73 0.72 0.70 0.69 0.67 0.66 0.62 0.63 0.65

0.89 0.87 0.83 0.81 0.80 0.80 0.78 0.76 0.75 0.74 0.65 0.65 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section. Note 5. Values for intermediate values of β and γ may be interpolated.

TABLE 16 Slenderness correction factor, n, for members with applied loading other than as for Table 15.

M

BM

M0

Note 1. All hogging moments are +ve. Note 2. β is defined in Table 18. Note 3. M0 is the mid-length moment on a simply supported span equal to the unrestrained length (see Table 17).

Unrestrained length L

β positive

β negative

γ = M/M 0

1.0

0.8

0.6

0.4

0.2

0.0

−0.2 −0.4 −0.6 −0.8 −1.0

+50.00 +10.00 +5.00 +2.00 +1.50 +1.00 +0.50 0.00

1.00 0.99 0.99 0.98 0.97 0.97 0.96 0.94

0.96 0.98 0.98 0.98 0.97 0.97 0.96 0.94

0.92 0.95 0.97 0.97 0.97 0.97 0.96 0.94

0.87 0.91 0.94 0.96 0.96 0.96 0.96 0.94

0.83 0.86 0.90 0.94 0.95 0.96 0.96 0.94

0.77 0.81 0.85 0.92 0.93 0.95 0.95 0.94

0.72 0.76 0.80 0.90 0.92 0.94 0.94 0.94

0.67 0.70 0.75 0.86 0.89 0.93 0.94 0.94

0.66 0.68 0.71 0.82 0.86 0.93 0.94 0.94

0.66 0.68 0.70 0.78 0.83 0.91 0.93 0.94

0.65 0.67 0.70 0.76 0.79 0.89 0.92 0.94

(Continued)

466

•

Annex 7 / From BS5950: 1: 1990 by Permission

TABLE 16 Continued −0.10 −0.20 −0.30 −0.40 −0.50 −0.60 −0.70 −0.80 −0.90 −1.00 −1.10 −1.20 −1.30 −1.40 −1.50 −1.60 −1.70 −1.80 −1.90 −2.00 −5.00 −50.00 Infinity

0.93 0.92 0.91 0.90 0.89 0.71 0.57 0.47 0.47 0.50 0.54 0.57 0.61 0.64 0.67 0.69 0.71 0.74 0.76 0.78 0.91 0.99 1.00

0.93 0.92 0.91 0.90 0.90 0.77 0.64 0.52 0.46 0.48 0.51 0.54 0.56 0.59 0.62 0.64 0.66 0.69 0.71 0.73 0.86 0.95 0.96

0.93 0.92 0.92 0.91 0.91 0.84 0.70 0.59 0.50 0.46 0.48 0.50 0.52 0.55 0.57 0.59 0.60 0.62 0.63 0.65 0.80 0.89 0.91

0.93 0.92 0.92 0.91 0.91 0.87 0.77 0.67 0.58 0.51 0.49 0.47 0.47 0.49 0.51 0.52 0.54 0.55 0.57 0.58 0.74 0.84 0.86

0.94 0.93 0.93 0.92 0.92 0.89 0.82 0.73 0.65 0.58 0.54 0.51 0.49 0.48 0.47 0.48 0.50 0.51 0.53 0.54 0.70 0.79 0.82

0.94 0.93 0.93 0.92 0.92 0.91 0.87 0.80 0.73 0.66 0.61 0.56 0.53 0.51 0.49 0.50 0.51 0.51 0.52 0.53 0.65 0.74 0.77

0.94 0.93 0.93 0.92 0.92 0.92 0.89 0.86 0.80 0.73 0.69 0.64 0.61 0.58 0.56 0.55 0.55 0.54 0.54 0.53 0.62 0.70 0.72

0.94 0.93 0.93 0.92 0.92 0.92 0.91 0.90 0.87 0.81 0.77 0.73 0.70 0.67 0.64 0.63 0.61 0.60 0.58 0.57 0.59 0.66 0.68

0.94 0.94 0.94 0.93 0.92 0.92 0.92 0.92 0.90 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.68 0.66 0.65 0.63 0.58 0.63 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section. Note 5. Values for intermediate values of β and γ may be interpolated.

0.94 0.94 0.94 0.93 0.92 0.92 0.92 0.92 0.90 0.89 0.87 0.84 0.82 0.79 0.77 0.76 0.74 0.73 0.71 0.70 0.61 0.62 0.65

0.94 0.93 0.94 0.93 0.92 0.92 0.91 0.92 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 0.82 0.81 0.80 0.79 0.67 0.65 0.65

Structural Design of Steelwork to EN 1993 and EN 1994 TABLE 17 Moment diagram between adjacent points of lateral restraint. M0

M

BM b ve

g ve M0

M

BM b ve

g ve M0

M

BM

b ve

g ve M0

M BM bve

gve

M0 M BM bve

gve

M0 M BM bve

gve

•

467

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Index Actions accidental, 29, 287, 294 characteristic, 27 classification, 27 combinations, 30, 287 design values, 29 destabilizing, 98, 104 envelopes, 32 erection, 29, 301, 359 fundamental, 30 gravity, 286 pattern loading, 31, 103 permanent, 28 progressive collapse, 295 seismic, 30, 287 snow, 28, 286 stabilising, 104 thermal, 29 transmission, 298 variable, 28 wind, 29, 286 analysis of frames elastic, 309 pitched roof portal, 314 plastic, 309 angles and tees as beams, 42 as tension members, 175 Beams deflection, 49 elastic analysis, 38 laterally restrained, 35, 37 principal axes, 39 section axes, 39 shear stress, 54 span/depth ratios, 53 symmetrical, 41 transverse forces, 78 types, 36 unsymmetrical, 42 bolts bearing strength, 213 clearance, 209 close tolerance, 208 design tables ordinary, Annex A2 design tables preloaded, Annex A3 edge and end distances, 210 foundation, 208 holes, 209

large grip lengths, 216 long joints, 216 net area, 212 ordinary bolts, packings, 216 preferred sizes, 207 prying forces, 218 shear strength and deformation, 214 shear strength and tension, 213 sizes, 213 slip resistant, 217 spacing, 210 tension and shear, 213 types, 207 washers, 209 bracing, 288, 300 brittle fracture, 22 buckling flange, 424 gusset plates, 223 lateral torsional buckling, 35, 91, 430 lip, 424 local, 35, 419 Perry-robertson, 183 struts, 180 theory, 181 webs, 138 characteristic load, 27 characteristic strength, 18 Charpy V-notch test, 22 classification of sections, 4, 371 cold formed sections analytical model, 414 deflections, 434 distortional buckling, 424 effective width, 422 effective section properties, 454 elastic local buckling stress, 420 finite strip methods, 435 lateral torsional buckling, 430 local buckling, 419 post buckling behaviour, 422 columns, 181, 300, 312, 325 composite construction composite beams deflection, 382 elastic capacity, 376 flexural design, 371 flexural shear, 377

470

•

Index composite beams (continued) longitudinal shear, 379 partial shear connection, 376, 378 shear connector design, 377 transverse shear, 380 vibration, 382 composite columns axial compression, 396 biaxial bending, 397 buckling, 401 design moments, 401 determination of member capacity, 398 fire, 402 uniaxial bending, 396 composite slabs deflection, 360, 362 flexural shear, 361, 362 flexure, 360, 361 longitudinal shear, 362, 379 punching shear, 362 shear bond, 360 transverse shear, 380 compression members axially loaded, 175 buckling length, 186, 312 buckling theory, 181 compression members, 180 effective length, 185 local buckling, 185 slenderness ratio, 185 with moments, 190, 401 connections (see joints) corrosion of steel, 20 Dead loads (see actions) deflections limits, 53 beams, 49 composite slabs, 360 portal frame, 325 span/depth ratios, 53 deformations, 287 design envelope, 19 methods for frames, 198 rigid, 294, 39, 311 sem-rigid, 294, 309, 310 simple, 309, 310 strength, 19 ductility, 20 durability, 20 drawings, 12 Effective area of an angle tie, 175 effective length beam, 97 cantilever, 100, 105 column, 185, 312 errors in design, 11 expansion and contraction joints, 304

Fabrication of steelwork, 12 failure criteria bolts, 213 steel, 24 welds, 203 fatigue, 23 fire single storey structures, 302 multi-storey structures, 302 composite columns, 402 flooring systems, 284 frames classification, 306 imperfections, 308 single storey, 282, 288 multi-storey, 283, 293 forces bracing, 290, 300 erection, 301 seismic, 287 tying, 296 foundation design 303 frame types, 2 Gable end frame, 336 gusset plates, 223, 250, 265 Hardness, 20, 317 haunches, 314, 317, 332 holes in members, Annexes A4, A5, A6 hollow sections, 234, 269 lateral torsional buckling, 123 hydrogen cracks, 205 Intermediate vertical stiffener, 145 Joints beam splice, 233, 257 beam-to-beam, 243, 256 beam-to-column, 232, 242, 252 column bracket, 231, 237, 247 column-to-column, 233, 260 column-to-foundation, 234, 245, 263 construction movement, 304 deformation, 205 eccentric shear, 225 effective length, 202 end bearing, 226 effective length, 202 expansion and contraction, 304 fillet weld, 203 ideal, 199 knee for portal frame, 235, 271 long 202 ‘pinned’, 228 RHS-to-RHS, 234, 269 ‘rigid’, 230 roof truss, 235 rotational stiffness, 275 washers, 209 welded, 199

Index Lateral restraint, 77 torsional buckling, 36, 91, 123, 125 limit state design, 9 lintels, 299 loads (see actions) Manufacture of steel, 5 material variation, 17 multi-storey frames, 283, 293 Overhanging beams, 107 Partial safety factors, 18 plastic analysis effect of axial load, 177 effect of shear force, 73 hinges, 37 lateral restraint, 77 methods of analysis, 71 reduction of modulus due to shear, 73, 376 section properties, 67 theorems, 72 plate girder fabrication, 132 intermediate vertical stiffeners, 145 lateral torsional buckling, 134 load bearing stiffeners, 142 minimum web thickness, 133 minimum weight, 134 web buckling, 138 welding, 132, 146 plate thickness elastic method, 222 plastic method, 218 standard thicknesses, 3 Poisson’s ratio, 19 portal frame single bay, 314 multi-bay, 337 principal axes, 38 progressive collapse, 295 prying force, 218 Rafter stability, 332 residual stresses, 22, 94 restraints, 99, 300, 325, 332 Section axes, 39 cold formed, 413 classification, 35 elastic properties, 38 modulus, 41 second moment of area, 40 types, 3 unsymmetrical, 42 plastic properties, 67 seismic actions, 287 settlement, 304

shear centre, 57 modulus, 19 closed sections, 54 lag, 57 open sections, 56 plate grider, 138 stress, 54 sign conventions, 39 single storey frames, 282, 288 slenderness ratio, 185 splices column, 300 plate, 146 stability of structures, 296, 305 steel cold formed, 3 first use, 5 manufacture, 5 production, 6 standard sections, 3 stock sizes, 3 types of, 3 stiffeners load bearing, 142 plate girder, 142 transverse, 144 strength characteristic, 18 design, 19 stress concentrations, 24 residual, 22 strain relationship, 38 yield, 38 structural design, 7 structural framing, 10 T-stubs, 218 torsional buckling interaction, 131 tension members angles, 175 axially loaded, 175 with moments, 177 net area, 210 effective area, 176 thermal coefficient of expansion, 19 testing of steel, 14 tolerances, 12 torsional buckling, 127 torsional restraint, 62, 97, 100 toughness, 22 transverse forces, 78 trusses non-triangulated trusses, 347 triangulated, 341 Vibration, 382 vierendeel girder, 348

•

471

472

•

Index Washers, 209 web buckling of plate girders, 138 welds brittle fracture, 206 butt welds, 201 defects, 205 design table, Annex A1 effective length, 202 electrodes, 199 factors affecting strength, 205 failure criteria, 204 hydrogen cracks, 206

lamellar tearing, 206 load-deformation, 205 stress concentrations, 206 throat thickness, 202 types, 201 wind bracing, 288 girder, 291 loading, 286 Yield stress, 18 Young’s modulus, 19

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Structural Design of Steelwork to EN 1993 and EN 1994 Third edition

L.H. Martin Bsc, PhD, CEng FICE J.A. Purkiss Bsc (Eng), PhD

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Butterworth-Heinemann is an imprint of Elsevier

Butterworth-Heinemann is an imprint of Elsevier Linacre House, Jordan Hill, Oxford OX2 8DP, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA First edition published by Edward Arnold 1984 Second edition 1992 Third edition 2008 Copyright © 2008, L.H. Martin and J.A. Purkiss. All rights reserved The right of L.H. Martin and J.A. Purkiss to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [email protected] Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN13: 978-0-7506-5060-1 For information on all Butterworth-Heinemann publications visit our web site at books.elsevier.com Typeset by Charon Tec Ltd (A Macmillan Company), Chennai, India www.charontec.com Printed and bound in Great Britan 07 08 09 10

10 9 8 7 6 5 4 3 2 1

Contents Preface Acknowledgements Principal Symbols

ix xi xiii

CHAPTER

1

General 1.1 1.2 1.3 1.4

Description of steel structures Development, manufacture and types of steel Structural design Fabrication of steelwork

1 1 5 7 12

CHAPTER

2

Mechanical Properties of Structural Steel 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Variation of material properties Characteristic strength Design strength Other design values for steel Corrosion and durability of steelwork Brittle fracture Residual stresses Fatigue Stress concentrations Failure criteria for steel

17 17 18 19 19 20 22 22 23 24 24

CHAPTER

3

Actions 3.1 3.2 3.3 3.4 3.5

Description Classification of actions Actions varying in time Design values of actions Actions with spatial variation

27 27 27 28 29 31

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Contents

CHAPTER

4

Laterally Restrained Beams 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Structural classification of sections Elastic section properties and analysis in bending Elastic shear stresses Elastic torsional shear stresses Plastic section properties and analysis Effect of shear force on the plastic moment of resistance Lateral restraint Resistance of beams to transverse forces

35 35 38 54 62 67 73 77 78

CHAPTER

5

Laterally Unrestrained Beams

91

5.1 5.2 5.3

91 127 132

Lateral torsional buckling of rolled sections symmetric about both axes Pure torsional buckling Plate girders

CHAPTER

6

Axially Loaded Members 6.1 6.2 6.3 6.4

Axially loaded tension members Combined bending and axial force – excluding buckling Buckling of axially loaded compression members Combined bending and axial force – with buckling

175 175 177 180 190

CHAPTER

7

Structural Joints 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

Introduction The ideal structural joint Welded joints Bolted joints Plate thicknesses for joint components Joints subject to shear forces Joints subject to eccentric shear forces Joints with end bearing ‘Pinned’ joints ‘Rigid’ joints

198 198 199 199 207 218 224 225 226 228 230

Contents 7.11 7.12

Joint rotational stiffness Frame-to-joint stiffness

•

vii 275 277

CHAPTER

8

Frames and Framing 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14

Single storey structures Multi-storey construction Influence of connection design and detailing Structural actions Single storey structures under horizontal loading Multi-storey construction Behaviour under accidental effects Transmission of loading Design of bracing Fire performance Additional design constraints Design philosophies Design issues for multi-storey structures Portal frame design

282 282 283 286 286 288 293 294 298 300 301 303 306 310 314

CHAPTER

9

Trusses 9.1 9.2

Triangulated trusses Non-triangulated trusses

341 341 347

CHAPTER

10

Composite Construction 10.1 10.2 10.3 10.4

Composite slabs Design of decking Composite beams Composite columns

358 358 359 370 395

CHAPTER

11

Cold-formed Steel Sections 11.1 11.2

Analytical model Local buckling

413 414 419

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Contents 11.3 11.4 11.5 11.6 11.7 11.8 11.9

Distortional buckling Lateral–torsional buckling Calculation of deflections Finite strip methods Design methods for beams partially restrained by sheeting Working examples Chapter conclusions

424 430 434 435 442 445 453

ANNEX

458

INDEX

469

Preface This book conforms to the latest recommendations for the design of steel and composite steel–concrete structures as described in Eurocode 3: Design of steel structures and Eurocode 4: Design of composite steel–concrete structures. References to relevant clauses of the Codes are given where appropriate. Note that for normal steelwork design, including joints, three sections of EN 1993 are required: • Part 1–1 General rules and rules for buildings • Part 1–5 Plated structural elements • Part 1–8 Design of joints Additionally if design for cold formed sections is carried out from first principles then Part 1–3 Cold formed thin gauge members and sheeting is also required. Whilst it has not been assumed that the reader has a knowledge of structural design, a knowledge of structural mechanics and stress analysis is a prerequisite. However, as noted below certain specialist areas of analysis have been covered in detail since the Codes do not provide the requisite information. Thus the book contains detailed explanations of the principles underlying steelwork design and provides appropriate references and suggestions for further reading. The text should prove useful to students reading for engineering degrees at University, especially for design projects. It will also aid designers who require an introduction to the new Eurocodes. For those familiar with current practice, the major changes are: (1) There is need to refer to more than one part of the various codes with calculations generally becoming more extensive and complex. (2) Steelwork design stresses are increased as the gamma values on steel are taken as 1,0, and the strength of high yield reinforcement is 500 MPa albeit with a gamma factor of 1,15. (3) A deeper understanding of buckling phenomena is required as the Codes do not supply the relevant formulae. (4) Flexure and axial force interaction equations are more complex, thus increasing the calculations for column design. (5) The checking of webs for in-plane forces is more complex. (6) Although tension field theory (or its equivalent) may be used for plate girders, the calculations are simplified compared to earlier versions of the Code. (7) Joints are required to be designed for both strength and stiffness. (8) More comprehensive information is given on thin-walled sections.

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Acknowledgements Lawrence Martin and John Purkiss would like to thank Long-yuan Li and Xiao-ting Chu for writing Chapter 11 on Cold-formed steel sections. The second author would like to thank Andrew Orton (Corus) for help with problems over limiting critical lengths for lateral–torsional buckling of rolled sections. The authors further wish to thank the following for permission to reproduce material: • • • •

Albion Sections Ltd for Fig. 11.1 www.access-steel.com for Figs 11.24 and 11.25 Karoly Zalka and the Institution of Civil Engineers for Fig. 8.13 BSI BSI Ref.

Book Ref.

BS 5950: Part 1: 1990

Tables 15–17

Annex A7

BS 5950: Part 1: 2000

Table 14

Table 5.3

EN 1993-1-3

Fig. 5.6 Fig. 5.7a Fig. 10.2

Fig. 11.8 Fig. 11.9 Fig. 11.20

EN 1994-1-1

Fig. 9.8

Fig. 10.3

British Standards may be obtained from BSI Customer Services, 389 Chiswick High Road, London W4 4AL. Tel: +44 (0)20 8996 9001. e-mail: [email protected]

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Principal Symbols Listed below are the symbols and suffixes common to European Codes

LATIN UPPER AND LOWER CASE A a B b c d E e F f G H h i I k L l M N n p Q q R r S s T t uu vv

accidental action; area distance; throat thickness of a weld bolt force breadth outstand depth of web; diameter modulus of elasticity edge distance; end distance action; force strength of a material permanent action; shear modulus of steel total horizontal load or reaction; warping constant of section height radius of gyration second moment of area stiffness length; span; buckling length effective buckling length; torsion constant; warping constant bending moment axial force number pitch; spacing variable action; prying force uniformly distributed action resistance; reaction radius; root radius; number of redundancies stiffness staggered pitch; distance; bearing length torsional moment thickness principal major axis principal minor axis

xiv

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Principal Symbols V v W w

shear force; total vertical load or reaction shear stress section modulus deflection

GREEK LOWER CASE α β γ δ ε η θ λ μ ν ρ σ τ φ χ ψ

coefficient of linear thermal expansion; angle; ratio; factor angle; ratio; factor partial safety factor deflection; deformation strain; coefficient (235/fy )1/2 where fy is in MPa distribution factor; shear area factor; critical buckling mode; buckling imperfection coefficient angle; slope slenderness ratio; ratio slip factor Poisson’s ratio unit mass; factor normal stress; standard deviation shear stress rotation; slope; ratio reduction factor for buckling stress ratio

SUFFIXES Ed design strength el elastic f flange j joint o initial; hole p plate pl plastic Rd resistance strength t torsion u ultimate strength v shear w web; warping x x-x axis y y-y axis; yield strength z z-z axis

Chapter

1 / General

1.1 DESCRIPTION OF STEEL STRUCTURES 1.1.1 Shapes of Steel Structures The introduction of structural steel, circa 1856, provided an additional building material to stone, brick, timber, wrought iron and cast iron. The advantages of steel are high strength, high stiffness and good ductility combined with relative ease of fabrication and competitive cost. Steel is most often used for structures where loads and spans are large and therefore is not often used for domestic architecture. Steel structures include low-rise and high-rise buildings, bridges, towers, pylons, floors, oil rigs, etc. and are essentially composed of frames which support the self-weight, dead loads and external imposed loads (wind, snow, traffic, etc.). For convenience load bearing frames may be classified as: (a) Miscellaneous isolated simple structural elements (e.g. beams and columns) or simple groups of elements (e.g. floors). (b) Bridgeworks. (c) Single storey factory units (e.g. portal frames). (d) Multi-storey units (e.g. tower blocks). (e) Oil rigs. A real structure consists of a load bearing frame, cladding and services as shown in Fig. 1.1(a). A load bearing frame is an assemblage of members (structural elements) arranged in a regular geometrical pattern in such a way that they interact through structural connections to support loads and maintain them in equilibrium without excessive deformation. Large deflections and distortions in structures are controlled by the use of bracing which stiffens the structure and can be in the form of diagonal structural elements, masonry walls, reinforced concrete lift shafts, etc. A load bearing steel frame is idealized, for the purposes of structural design, as center lines representing structural elements which intersect at joints, as shown in Fig. 1.1(b). Other shapes of load bearing frames are shown in Figs 1.1 (c) to (e). Structural elements are required to resist forces and displacements in a variety of ways, and may act in tension, compression, flexure, shear, torsion or in any combination of

2

•

Chapter 1 / General Dead and snow loading

Connection Wind loading

Services Cladding

Load bearing frame Connection (a) Real structure

(b) Idealized load bearing frame

Pinned connection

Bracing (c)

(d) Rigid connection

(e)

FIGURE 1.1 Typical load bearing frames

these forces. The structural behaviour of a steel element depends on the nature of the forces, the length and shape of the cross section of the member, the elastic properties and the magnitude of the yield stress. For example a tie behaves in a linear elastic manner until yield is reached. A slender strut behaves in a non-linear elastic manner until first yield is attained, provided that local buckling does not occur first. A laterally supported beam behaves elastically until a plastic hinge forms, while an unbraced beam fails by elastic torsional buckling. These modes of behaviour are considered in detail in the following chapters. The structural elements are made to act as a frame by connections. These are composed of plates, welds and bolts which are arranged to resist the forces involved. The connections are described for structural design purposes as pinned, semi-rigid and rigid, depending on the amount of rotation, and are described, analysed and designed in detail in Chapter 7.

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1.1.2 Standard Steel Sections The optimization of costs in steel construction favours the use of structural steel elements with standard cross-sections and common bar lengths of 12 or 15 m. The billets of steel are hot rolled to form bars, flats, plates, angles, tees, channels, I sections and hollow sections as shown in Fig. 1.2. The detailed dimensions of these sections are given in BS 4, Pt 1 (2005), BSEN 10056-1 (1990), and BSEN 10210-2 (1997). Where thickness varies, for example, Universal beams, columns and channels, sections are identified by the nominal size, that is, ‘depth × breadth × mass per unit length × shape’. Where thickness is constant, for example, tees and angle sections, the identification is ‘breadth × depth × thickness × shape’. In addition a section is identified by the grade of steel. To optimize on costs steel plates should be selected from available stock sizes. Thicknesses are in the range of 6, 8, 10, 12,5, 15 mm and then in 5 mm increments. Thicknesses of less than 6 mm are available but because of lower strength and poorer corrosion resistance their use is limited to cold formed sections. Stock plate widths are in the range 1, 1,25, 1,5, 2, 2,5 and 3 m, but narrow plate widths are also available. Stock plate lengths are in the range 2, 2,5, 3, 4, 5, 6, 10 and 12 m. The adoption of stock widths and lengths avoids work in cutting to size and also reduces waste. The application of some types of section is obvious, for example, when a member is in tension a round or flat bar is the obvious choice. However, a member in tension may

Universal beam (UB)

Universal column (UC)

Circular hollow section

Retangular hollow section

FIGURE 1.2 Standard steel sections

Channel

Angle

Bars

Structural tee from UB

Plate

4

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Chapter 1 / General be in compression under alternative loading and an angle, tee, or tube is often more appropriate. The connection at the end of a bar or tube, however, is more difficult to make. If a structural element is in bending about one axis then the ‘I’ section is the most efficient because a large proportion of the material is in the flanges, that is, at the extreme fibres. Alternatively, if a member is in bending about two axes at right angles and also supports an axial load then a tube, or rectangular hollow section, is more appropriate. Other steel sections available are cold formed from steel plate into a variety of cross sections for use as lightweight lattice beams, glazing bars, shelf racks, etc. Not all these sections are standardized because of the large variety of possible shapes and uses, however, there is a wide range of sections listed in BSEN 10162 (2003). Local buckling can be a problem and edges are stiffened using lips. Also when used as beams the relative thinness of the material may lead to web crushing, shear buckling and lateral torsional buckling. Although the thickness of the material (1–3 mm) is less than that of the standard sections the resistance to corrosion is good because of the surface finish obtained by pickling and oiling. After degreasing this surface can be protected by galvanizing, or painting, or plastic coating. The use in building of cold formed sections in light gauge plate, sheet and strip steel 6 mm thick and under is dealt with in BSEN 5950 (2001) and EN 1993-1-1 (2005).

1.1.3 Structural Classification of Steel Sections (cl 5.5. EN 1993-1-1 (2005)) A section, or element of a member, in compression due to an axial load may fail by local buckling. Local buckling can be avoided by limiting the width to thickness ratios (b/tf or d/tw ) of each element of a cross-section. The use of the limiting values given in Table 5.2, EN 1993-1-1 (2005) avoids tedious and complicated calculations. Depending on the b/tf or d/tw ratios standard or built-up sections are classified for structural purposes as: • Class 1: Low values of b/tf or d/tw where a plastic hinge can be developed with sufficient rotation capacity to allow redistribution of moments within the structure. • Class 2: Full plastic moment capacity can be developed but local buckling may prevent development of a plastic hinge with sufficient rotation capacity to permit plastic design. • Class 3: High values of b/tf and d/tw , where stress at the extreme fibres can reach design strength but local buckling may prevent the development of the full plastic moment. • Class 4: Local buckling may prevent the stress from reaching the design strength. Effective widths are used to allow for local buckling (cl 5.5.2(2), EN 1993-1-1 (2005)).

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1.1.4 Structural Joints (EN 1993-1-8 (2005)) Structural elements are connected together at joints which are not necessarily at the ends of members. A structural connection is an assembly of components (plates, bolts, welds, etc.) arranged to transmit forces from one member to another. A connection may be subject to any combination of axial force, shear force and bending moment in relation to three perpendicular axes, but for simplicity, where appropriate, the situation is reduced to forces in one plane. There are other types of joints in structures which are not structural connections. For example a movement joint is introduced into a structure to take up the free expansion and contraction that may occur on either side of the joint due to temperature, shrinkage, expansion, creep, settlement, etc. These joints may be detailed to be watertight but do not generally transmit forces. Detailed recommendations are given by Alexander and Lawson (1981). Another example is a construction joint which is introduced because components are manufactured to a convenient size for transportation and need to be connected together on site. In some cases these joints transmit forces but in other situations may only need to be waterproof.

1.2 DEVELOPMENT, MANUFACTURE AND TYPES OF STEEL 1.2.1 Outline of Developments in Design Using Ferrous Metals Prior to 1779, when the Iron Bridge at Coalbrookdale on the Severn was completed, the most important materials used for load bearing structures were masonry and timber. Ferrous materials were only used for fastenings, armaments and chains. The earliest use of cast iron columns in factory buildings (circa 1780) enabled relatively large span floors to be constructed. Due to a large number of disastrous fires around 1795, timber beams were replaced by cast iron with the floors carried on brick jack arches between the beams. This mode of construction was pioneered by Strutt in an effort to attain a fire proof construction technique. Cast iron, however, is weak in tension and necessitates a tension flange larger than the compression flange and consequently cast iron was used mainly for compression members. Large span cast iron beams were impractical, and on occasions disastrous as in the collapse of the Dee bridge designed by Robert Stephenson in 1874. The last probable use of cast iron in bridge works was in the piers of the Tay bridge in 1879 when the bridge collapsed in high winds due to poor design and unsatisfactory supervision during construction. In an effort to overcome the tensile weakness of cast iron, wrought iron was introduced in 1784 by Henry Cort. Wrought iron enabled the Victorian engineers to produce the following classic structures. Robert Stephenson’s Brittania Bridge was the first box girder bridge and represented the first major collaboration between engineer, fabricator (Fairburn) and scientist (Hodgkinson). I.K. Brunel’s Royal Albert Bridge

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Chapter 1 / General at Saltash combined an arch and suspension bridge. Telford’s Menai suspension bridge used wrought iron chains which have sine been replaced by steel chains. Telford’s Pont Cysyllte is a canal aqueduct near Llangollen. The first of the four structures was replaced after a fire in 1970. The introduction of wrought iron revolutionized ship building and enabled Brunel to produce the S.S. Great Britain. Steel was first produced in 1740, but was not available in large quantities until Bessemer invented the converter in 1856. The first major structure to use the new steel exclusively was Fowler and Baker’s railway bridge at the Firth of Forth. The first steel rail was rolled in 1857 and installed at Derby where it was still in use 10 years later. Cast iron rails in the same position lasted about 3 months. Steel rails were in regular production at Crewe under Ramsbottom from 1866. By 1840 standard shapes in wrought iron, mainly rolled flats, tees and angles, were in regular production and were appearing in structures about 10 years later. Compound girders were fabricated by riveting together the standard sections. Wrought iron remained in use until around the end of the nineteenth century. By 1880 the rolling of steel ‘I’ sections had become widespread under the influence of companies such as Dorman Long. Riveting continued in use as a fastening method until around 1950 when it was superseded by welding. Bessemer steel production in Britain ended in 1974 and the last open hearth furnace closed in 1980. Further information on the history of steel making can be found in Buchanan (1972), Cossons (1975), Derry and Williams (1960), Pannel (1964) and Rolt (1970).

1.2.2 Manufacture of Steel Sections The manufacture of standard steel sections, although now a continuous process, can be conveniently divided into three stages: (1) Iron production (2) Steel production (3) Rolling. Iron production is a continuous process and consists of chemically reducing iron ore in a blast furnace using coke and crushed limestone. The resulting material, called cast iron, is high in carbon, sulphur and phosphorus. Steel production is a batch process and consists in reducing the carbon, sulphur and phosphorus levels and adding, where necessary, manganese, chromium, nickel, vanadium, etc. This process is now carried out using a Basic Oxygen Converter, which consists of a vessel charged with molten cast iron, scrap steel and limestone through which oxygen is passed under pressure to reduce the carbon content by oxidation. This is a batch process which typically produces about 250–300 tons every 40 min. The alternative electric arc furnace is in limited use (approximately 5% of the UK steel production), and is generally used for special steels such as stainless steel.

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From the converter the steel is ‘teemed’ into ingots which are then passed to the rolling mills for successive reduction in size until the finished standard section is produced. The greater the reduction in size the greater the work hardening, which produces varying properties in a section. The variation in cooling rates of different thicknesses introduces residual stresses which may be relieved by the subsequent straightening process. Steel plate is now produced using a continuous casting procedure which eliminates, ingot casting, mould stripping, heating in soaking pits and primary rolling. Continuous casting permits, tighter control, improved quality, reduced wastage and lower costs.

1.2.3 Types of Steel The steel used in structural engineering is a compound of approximately 98% iron and small percentages of carbon, silicon, manganese, phosphorus, sulphur, niobium and vanadium as specified in BS 4360 (1990). Increasing the carbon content increases strength and hardness but reduces ductility and toughness. Carbon content therefore is restricted to between 0,25% and 0,2% to produce a steel that is weldable and not brittle. The niobium and vanadium are introduced to raise the yield strength of the steel; the manganese improves corrosion resistance; and the phosphorus and sulphur are impurities. BS 4360 (1990) also specifies tolerances, testing procedure and specific requirements for weldable structural steel. Steels used in practice are identified by letters and number, for example, S235 is steel with a tensile yield strength of 235 MPa (Table 3.1, EN 1993-1-1 (2005)).

1.3 STRUCTURAL DESIGN 1.3.1 Initiation of a Design The demand for a structure originates with the client. The client may be a private person, private or public firm, local or national government, or a nationalized industry. In the first stage preliminary drawings and estimates of costs are produced, followed by consideration of which structural materials to use, that is, reinforced concrete, steel, timber, brickwork, etc. If the structure is a building, an architect only may be involved at this stage, but if the structure is a bridge or industrial building then a civil or structural engineer prepares the documents. If the client is satisfied with the layout and estimated costs then detailed design calculations, drawings and costs are prepared and incorporated in a legal contract document. The design documents should be adequate to detail, fabricate and erect the structure. The contract document is usually prepared by the consultant engineer and work is carried out by a contractor who is supervised by the consultant engineer. However, larger firms, local and national government, and nationalized industries, generally employ their own consultant engineer.

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Chapter 1 / General The work is generally carried out by a contractor, but alternatively direct labour may be used. A further alternative is for the contractor to produce a design and construct package, where the contractor is responsible for all parts and stages of the work.

1.3.2 The Object of Structural Design The object of structural design is to produce a structure that will not become unserviceable or collapse in its lifetime, and which fulfils the requirements of the client and user at reasonable cost. The requirements of the client and user may include any or all of the following: (a) The structure should not collapse locally or overall. (b) It should not be so flexible that deformations under load are unsightly or alarming, or cause damage to the internal partitions and fixtures; neither should any movement due to live loads, such as wind, cause discomfort or alarm to the occupants/users. (c) It should not require excessive repair or maintenance due to accidental overload, or because of the action of weather. (d) In the case of a building, the structure should be sufficiently fire resistant to, give the occupants time to escape, enable the fire brigade to fight the fire in safety and to restrict the spread of fire to adjacent structures. The designer should be conscious of the costs involved which include: (a) The initial cost which includes fees, site preparation, cost of materials and construction. (b) Maintenace costs (e.g. decoration and structural repair). (c) Insurance chiefly against fire damage. (d) Eventual demolition. It is the responsibility of the structural engineer to design a structure that is safe and which conforms to the requirements of the local bye-laws and building regulations. Information and methods of design are obtained from Standards and Codes of Practice and these are ‘deemed to satisfy’ the local bye-laws and building regulations. In exceptional circumstances, for example, the use of methods validated by research or testing, an alternative design may be accepted. A structural engineer is expected to keep up to date with the latest research information. In the event of a collapse or malfunction where it can be shown that the engineer has failed to reasonably anticipate the cause or action leading to collapse, or has failed to apply properly the information at his disposal, that is, Codes of Practice, British Standards, Building Regulations, research or information supplied by the manufacturers, then he may be sued for professional negligence. Consultants and contractors carry liability insurance to mitigate the effects of such legal action.

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1.3.3 Limit State Design (cl 2.2, EN 1993-1-1 (2005)) It is self-evident that a structure should be ‘safe’ during its lifetime, that is, free from the risk of collapse. There are, however, other risks associated with a structure and the term safe is now replaced by the term ‘serviceable’. A structure should not during its lifetime become ‘unserviceable’, that is, it should be free from risk of collapse, rapid deterioration, fire, cracking, excessive deflection, etc. Ideally it should be possible to calculate mathematically the risk involved in structural safety based on the variation in strengths of the material and variation in the loads. Reports, such as the CIRIA Report 63 (1977), have introduced the designer to elegant and powerful concept of ‘structural reliability’. Methods have been devised whereby engineering judgement and experience can be combined with statistical analysis for the rational computation of partial safety factors in codes of practice. However, in the absence of complete understanding and data concerning aspects of structural behaviour, absolute values of reliability cannot be determined. It is not practical, nor is it economically possible, to design a structure that will never fail. It is always possible that the structure will contain material that is less than the required strength or that it will be subject to loads greater than the design loads. If actions (forces) and resistance (strength of materials) are determined statistically then the relationship can be represented as shown in Fig. 1.3. The design value of resistance (Rd ) must be greater than the design value of the actions (Ad ). It is therefore accepted that 5% of the material in a structure is below the design strength, and that 5% of the applied loads are greater than the design loads. This does not mean therefore that collapse is inevitable, because it is extremely unlikely that the weak material and overloading will combine simultaneously to produce collapse. The philosophy and objectives must be translated into a tangible form using calculations. A structure should be designed to be safe under all conditions of its useful life Frequency

Resistance R

Actions A

Resistance or actions

Ad Rd Design value of actions

Design value of resistance

Rd − Ad > 0

FIGURE 1.3 Statistical relationship between actions and resistance

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Chapter 1 / General and to ensure that this is accomplished certain distinct performance requirements, called ‘limit states’, have been identified. The method of limit state design recognizes the variability of loads, materials, construction methods and approximations in the theory and calculations. Limit states may be at any stage of the life of a structure, or at any stage of loading and are important for the design of steelwork. To reduce the number of load cases to be considered only serviceability and ultimate limit states are specified. Each of these sections is subdivided although some may not be critical in every design. Calculations for limit states involve loads and load factors (Chapter 3), and material factors and strengths (Chapter 2). Stability, an ultimate limit state, is the ability of a structure, or part of a structure, to resist overturning, overall failure and sway. Calculations should consider the worst realistic combination of loads at all stages of construction. All structures, and parts of structures, should be capable of resisting sway forces, for example, by the use of bracing, ‘rigid’ joints, or shear walls. Sway forces arise from horizontal loads, for example, winds, and also from practical imperfections, for example, lack of verticality. The sway forces from practical imperfections are difficult to quantify and advice is given in cl 5.3.3, EN 1993-1-1 (2005). Also involved in limit state design is the concept of structural integrity. Essentially this means that the structure should be tied together as a whole, but if damage occurs, it should be localized. Deflection is a serviceability limit state. Deflections should not impair the efficiency of a structure, or its components, nor cause damage to the finishes. Generally the worst realistic combination of unfactored imposed loads is used to calculate elastic deflections. These values are compared with empirical values related to the length of a member or height. Dynamic effects to be considered at the serviceability limit state are vibrations caused by machines, and oscillations caused by harmonic resonance, for example, wind gusts on buildings. The natural frequency of the building should be different from the exciting source to avoid resonance. Fortunately there are few structural failures and when they do occur they are often associated with human error involved in design calculations, or construction, or in the use of the structure.

1.3.4 Structural Systems Structural frame systems may be described as: (a) simple frames, (b) continuous frames, (c) semi-continuous frames.

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These titles refer to the types of joints and whether bracing is included. Simple design assumes that ‘pin joints’ connect the members and joint rotations are prevented by bracing. Historically this method was popular because parts of the structure could be designed in isolation and calculations could be done by hand. With the advent of the computer calculations are less onerous but the method is still in use. Continuous frames assume that the connections between members are rigid and therefore the angles between members can be maintained without the use of bracing. Calculations for the design of members and connections are more complicated and a computer is generally used. Global analysis of the frame is based on elastic, plastic, or elastic–plastic analysis assuming full continuity. Semi-continuous frames acknowledges that in reality, end moments and rotations exist at the connections. Global analysis using the computer is based on the moment– rotation and force displacement characteristics of the connections. Bracing is often necessary for this type of frame to reduce sway.

1.3.5 Errors The consequences of an error in structural design can lead to loss of life and damage to property, and it is necessary to appreciate where errors can occur. Small errors in design calculations can occur in the rounding off of figures but these generally do not lead to failures. The common sense advice is that the accuracy of the calculation should match the accuracy of the values given in the European Code. Errors that occur in structural design calculations and which affect structural safety are: (1) Ignorance of the physical behaviour of the structure under load and which consequently introduces errors in the basic assumptions used in the theoretical analysis. (2) Errors in estimating the loads, especially the erection forces. (3) Numerical errors in the calculations. These should be eliminated by checking, but when speed is paramount checks are often ignored. (4) Ignorance of the significance of certain effects (e.g. residual stresses, fatigue, etc). (5) Introduction of new materials, or methods, which have not been proved by tests. (6) Insufficient allowance for tolerances or temperature strains. (7) Insufficient information (e.g. in erection procedures). Errors that can occur in workshops or on construction sites are: (1) Using the wrong grade of steel, and when welding using the wrong type of electrode. (2) Using the wrong weight of section. A number of sections are the same nominal size but differ in web or flange thickness. (3) Errors in manufacture (e.g. holes in the wrong position).

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Chapter 1 / General Errors that occur in the life of a structure and also affect safety are: (1) Overloading (2) Removal of structural material (e.g. to insert service ducts) (3) Poor maintenance.

1.4 FABRICATION OF STEELWORK 1.4.1 Drawings Detailed design calculations are essential for any steel work design but the sizes of the members, dimensions and geometrical arrangement are usually presented as drawings. Initially the drawings are used by the fabricator and eventually by the contractor on site. General arrangement drawings are often drawn to scale of 1:100, while details are drawn to a scale of 1:20 or 1:10. Special details are drawn to larger scales where necessary. Drawings should be easy to read and should not include superfluous detail. Some important notes are: (a) Members and components should be identified by logically related mark numbers, for example, related to the grid system used in the drawings. (b) The main members should be presented by a bold outline (0,4 mm wide) and dimension lines should be unobtrusive (0,1 mm wide). (c) Dimensions should be related to centre lines, or from one end; strings of dimensions should be avoided. Dimensions should appear once only so that ambiguity cannot arise when revisions occur. Fabricators should not be put in the position of having to do arithmetic in order to obtain an essential dimension. (d) Tolerances for erection purposes should be clearly shown. (e) The grade of steel to be used should be clearly indicated. (f) The size, weight and type of section to be used should be clearly stated. (g) Detailing should take account of possible variations due to rolling margins and fabrication variations. (h) Keep the design and construction as simple as possible. Where possible use simple connections, avoid stiffeners, use the minimum number of sections and avoid changes in section along the length of a member. (i) Site access, transport and use of cranes should be considered.

1.4.2 Tolerances (cl 3.2.5, EN 1993-1-1 (2005)) Tolerances are limits places on unintentional inaccuracies that occur in dimensions which must be allowed for in design if structural elements and components are to fit together. In steelwork variations occur in the rolling process, marking out, cutting and drilling during fabrication, and in setting out during erection.

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In the rolling process the allowable tolerances for length, width, thickness and flatness for plates are given in BS 4360 (1990). Length and width tolerances are positive while those for thickness and flatness are negative and positive. The dimensional and weight tolerances for sections are given in BS 4 Pt 1 (2005), or BS 10056-1 (1990), as appropriate. During fabrication there is a tendency for members and components to increase rather than reduce, and the tolerance is therefore often specified as negative; it is often cheaper and simpler to insert packing rather than shorten a member, provided that the packing is not excessive. Where concrete work is associated with steelwork variations in dimensions are likely to be greater. When casting concrete, for example, errors in dimensions may arise from shrinkage or from warping of the shuttering, especially when it is re-used. Therefore, by virtue of the construction method, larger tolerances are specified for work involving concrete. To facilitate erection all members and connections should be provided with the maximum tolerance that is acceptable from structural and architectural considerations. A typical example is a connection between a steel column and a reinforced concrete base. It would produce great difficulties if the base were set too high and a tolerance of approximately 50 mm is often included in the design, with provision for grouting under the base. Tolerances are also provided to allow lateral adjustment of the foundation bolts. Tolerances between concrete and steelwork are also important because two different contractors are involved.

1.4.3 Fabrication, Assembly and Erection of Steelwork The drawings produced by the structural designer are used first by the steel fabricator and later by the contractor on site. The steel fabricator obtains the steel either direct from the rolling mills or from the steel stockiest, and then cuts, drills and welds the steel components to form the structural elements as shown on the drawings. In general, for British practice, the welding is confined to the workshop and the connections on site are made using bolts. In American, however, site welding is common practice. When marking out, the measurements of length for overall size, position of holes, etc. can be done by hand, but if there are several identical components then wooden or cardboard templates are made and repeated measurements avoided. Now automatic machines, controlled by a computer, or punched paper tape, are used to cut and drill standard sections. When completed, the steel work should be marked clearly and manufactured to the accepted tolerances. When fabricated, parts of the structure are delivered to the site in the largest pieces that can be transported and erected. For example a lattice girder may be sent fully assembled to a site in this country, but sent in pieces to fit a standard transport container

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Chapter 1 / General for erection abroad. All components should be assembled within the tolerances and cambers specified, and should not be bent or twisted or otherwise damaged. On site the general contractor may be responsible for the assembly, erection, connections, alignment and leveling of the complete structure. Alternatively the erection work may be done by the steel fabricator, or sublet to a specialist steel erector. The objective of the erection process is to assemble the steelwork in the most cost-effective method whilst maintaining the stability of individual members, and/or part or complete structure. To do this it may be necessary to introduce cranes and temporary bracing which must also be designed to resist the loads involved. During assembly on site it is inevitable that some components will not fit, despite the tolerances that have been allowed. A typical example is that the faying surfaces for a friction grip joint are not in contact when the bolts are stressed. Other examples are given by Mann and Morris (1981). The correction of some faults and the consequent litigation can be expensive.

1.4.4 Testing of Steelwork (cl 2.5, EN 1993-1-1 (2005)) Steel is routinely sampled and tested during production to maintain quality. However, occasionally new methods of construction are suggested and there may be some doubt as to the validity of the assumptions of behaviour of the structure. Alternatively if the structure collapses there may be some dispute as to the strength of a component, or member, of the structure. In such cases testing of components, or part of the structure, may be necessary. However it is generally expensive because of, the accuracy required, cost of material, cost of fabrication, necessity to repeat tests to allow for variations and to report accurately. Tests may be classified as: (a) acceptance tests – non-destructive for confirming structural performance, (b) strength tests – used to confirm the calculated capacity of a component or structure, (c) tests to failure – to determine the real mode of failure and the true capacity of a specimen, (d) check tests – where the component assembly is designed on the basis of tests. The size, shape, position of the gauges, and method of testing of small sample pieces of steel is given in BS 4360 (1990) and BSEN 10002-1 (2001). The tensile test is most frequently employed, and gives values of, Young’s modulus, limit of proportionality, yield stress or proof stress, percentage elongation and ultimate stress. Methods of destructive testing fusion welded joints and weld metal in steel are given in numerous Standards. The Charpy V-notch test for impact resistance is used to measure toughness, that is, the total energy, elastic and plastic, which can be absorbed by a specimen before

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fracture. The test specimen is a small beam of rectangular cross section with a ‘V’ notch at mid-length. The beam is fractured by a blow from a swinging pendulum, and the amount of energy absorbed is calculated from the loss of height of the pendulum swing after fracture. Details of the test specimen and procedure are given in BS 4360 (1990) and BSEN 10045-1 (1990). The Charpy V-notch test is often used to determine the temperature at which transition from brittle to ductile behaviour occurs. Structures which are unconventional, and/or method of design which are unusual or not fully validated by research, should be subject to acceptance tests. Essentially these consist of loading the structure to ensure that it has adequate strength to support, for example, 1 (test dead load) + 1,15 (remainder of dead load) + 1,25 (imposed load). Where welds are of vital importance, for example, in pressure vessels, they should be subject to non-destructive tests. The defects that can occur in welds are: slag inclusions, porosity, lack of penetration and sidewall fusion, liquation, solidification, hydrogen cracking, lamellar tearing and brittle fracture. A surface crack in a weld may be detected visibly but alternatively a dye can be sprayed onto the joint which seeps into the cracks. After removing any surplus dye the weld is resprayed with a fine chalk suspension and the crack then shows as a coloured line on the white chalk background. A variant of this technique is to use fluorescent dye and a crack then shows as a bright green line in ultra violet light. A surface crack may also be detected if the weld joint area is magnetized and sprayed with iron powder. The powder congregates along a crack, which shows as a black line. Other weld defects cannot be detected on the surface and alternative methods must be used. Radiographic methods use an X-ray, or gamma-ray, source on one side of the weld and a photographic film on the other. Rays are absorbed by the weld metal, but if there is a hole or crack there is less absorbtion which shows as a dark area on the film. Not all defects are detected by radiography since the method is sensitive to the orientation of the flaw, for example, cracks at right angles to the X-ray beam are not detected. Radiography also requires access to both sides of the joint. The method is therefore most suitable for in-line butt weld for plates. An alternative method to detect hidden defects in welds uses ultrasonics. If a weld contains a flaw then high frequency vibrations are reflected. The presence of a flaw can therefore be indicated by monitoring the reduction of transmission of ultrasonic vibrations, or by monitoring the reflections. The reflection method is extremely useful for welds where access is only possible from one side. Further details can be obtained from Gourd (1980).

REFERENCES Alexander S.J. and Lawson R.M. (1981). Movement design in buildings. Technical Note 107. Construction Industry Research and Information Association Publication.

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Chapter 1 / General BS 4 (2005). Specification for hot rolled sections Pt1. BSI. BSEN 10002-1 (2001). Methods for tensile testing of metals. BSI. BSEN 10045-1 (1990). The Charpy V-notch impact test on metals. BSI. BSEN 10162 (2003). Cold rolled sections. BSI. BS 4360 (1990) Specification for weldable structural steels. BSI. BSEN 10056-1 (1990). Hot rolled structural steel sections. BSI. BSEN 10210-2 (1997). Hot rolled structural steel hollow sections. BSI. BSEN 5950 (2001 and 1998). Structural use of steel work in buildings. BSI. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Buchanan R.A. (1972). Industrial archeology in Britain. Penguin. CIRIA (1977). Rationalisation of safety and serviceability factors in structural codes. Report 63. Construction Industry Research and Information Association Publication. Cossons N. (1975). The BP book of industrial archeology. David and Charles. Derry T.K. and Williams T.I. (1960). A short history of technology. Oxford University Press. Gourd L.M. (1980). Principles of welding technology. Edward Arnold. Mann A.P. and Morris L.J. (1981). Lack of fit in steel structures, Technical Report 87. Construction Industry Research and Information Association Publication. Pannel J.P.M. (1964). An illustrated history of civil engineering. Thames and Hudson. Rolt, L.T.C. (1970). Victorian engineering. Penguin.

Chapter

2 / Mechanical Properties of Structural Steel

2.1 VARIATION OF MATERIAL PROPERTIES All manufactured material properties vary because the molecular structure of the material is not uniform and because of inconsistencies in the manufacturing process. The variations that occur in the manufacturing process are dependent on the degree of control. Variations in material properties must be recognized and incorporated into the design process. The material properties that are of most importance for structural design using steel are strength and Young’s modulus. Other properties which are of lesser importance are hardness, impact resistance and melting point. If a number of samples are tested for a particular property, for example, strength, and the number of specimens with the same strength (frequency) plotted against the strength, then the results approximately fit a normal distribution curve as shown in Fig. 2.1. This curve can be expressed mathematically by the equation shown in Fig. 2.1 which can be used to define ‘safe’ values for design purposes.

y

Frequency

⎛

y=

⎛2

x−x 1 e−½ ⎜⎝ s ⎜⎝ s √2 p

5% of results

x Characteristic strength

1,64s

Strength Mean strength

x

FIGURE 2.1 Variation in material properties

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Chapter 2 / Mechanical Properties of Structural Steel

2.2 CHARACTERISTIC STRENGTH (cl 3.1, EN 1993-1-1 (2005)) A strength to be used as a basis for design must be selected from the variation in values shown in Fig. 2.1. This strength, when defined, is called the characteristic strength. If the characteristic strength is defined as the mean strength, then clearly from Fig. 2.1, 50% of the material is below this value. This is not acceptable. Ideally the characteristic strength should include 100% of the samples, but this is also impractical because it is a low value and results in heavy and costly structures. A risk is therefore accepted and it is therefore recognized that 5% of the samples fall below the characteristic strength. The characteristic strength is calculated from the equation fy = fmean − 1,64σ where for n samples the standard deviation σ=

( fmean − f )2 (n − 1)

1/2

Nominal values of the characteristic yield strengths and ultimate tensile strengths are given in Table 3.1, EN 1993-1-1 (2005) and some examples are given in Table 2.1.

TABLE 2.1 Some nominal values of yield strength for hot rolled steel (Table 3.1, EN 1993-1-1 (2005)). Standard EN 10025-2: Grade S235 S275 S355

Nominal thickness of material t ≤ 40 mm

40 < t ≤ 80 mm

fy (MPa)

fu (MPa)

fy (MPa)

fu (MPa)

235 275 355

360 430 510

215 255 335

360 410 470

TABLE 2.2 Some partial safety factors (cl 6.1, EN 1993-1-1 (2005) and Table 2.1, EN 1993-1-8 (2005)). Situation

Symbol

Value

Buildings

γM0 γM1 γM2 γM3 γM4 γM5 γM7

1,00 1,00 1,25 1,25 1,00 1,00 1,10

Joints

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2.3 DESIGN STRENGTH (cl 6.1, EN 1993-1-1 (2005)) The characteristic strength of steel is the value obtained from tests at the rolling mills, but by the time the steel becomes part of a finished structure this strength may be reduced (e.g. by corrosion or accidental damage). The strength to be used in design calculations is therefore the characteristic strength divided by a partial safety factor (γM ) (Table 2.2).

2.4 OTHER DESIGN VALUES FOR STEEL (cl 3.2.6, EN 1993-1-1 (2005)) The elastic modulus for steel (E) is obtained from the relationship between stress and strain as shown in Fig. 2.2. This is a material property and therefore values from a set of samples vary. However, the variation for steel is very small and the European Code assumes E = 210 GPa. The elastic shear modulus (G) is related to Young’s modulus by the expression E = 2G(1 + ν) where Poissons ratio ν = 0,3 in the elastic range and is used in calculations involving plates. The thermal coefficient of expansion for steel is given as α = 12E-6/K for T ≤ 100◦ C and is used in calculations involving temperature changes.

Fracture Fracture

Yield

Stress

Stress

Yield

Strain Plastic

Strain hardening

0,002

Strain

Elastic (a) Low-strength grade S 235

FIGURE 2.2 Tensile stress–strain relationships for steel

(b) High-strength grade S 450

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Chapter 2 / Mechanical Properties of Structural Steel Hardness is material property that is occasionally of importance in structural steel design. It is measured by the resistance the surface of the steel offers to, the indentation of a hardened steel ball (Brinell test), a square-based diamond pyramid (Vickers test) or a diamond cone (Rockwell test). Higher strength often correlates with greater hardness but this relationship is not infallible. Ductility may be described as the ability of a material to change its shape without fracture. This is measured by the percentage elongation, that is, 100 × (change in length)/(original length). Values of 20% can be obtained for mild steel but it is less for high-strength steel. A high value is advantageous because it allows the redistribution of stresses at ultimate load and the formation of plastic hinges.

2.5 CORROSION AND DURABILITY OF STEELWORK (cl 4, EN 1993-1-1 (2005)) Durability is a service limit state and the following factors should be considered at the design stage: (a) (b) (c) (d) (e) (f)

Environment, Degree of exposure, Shape of the members and details, Quality of workmanship and control, Protective measures, Maintenance.

Methods of protecting steel work are given in BSEN ISO 12944 (1998) and the specification for weather resistant steel is given is BS 7668 (1994). Corrosion of steel work reduces the cross-section of members and thus affects safety. Corrosion, which occurs on the surface of steel, is a chemical reaction between iron, water and oxygen, which produces a hydrated iron oxide called rust. Electrons are liberated in the reaction and a small electrical current flows from the corroded area to the uncorroded area. The elimination of water, oxygen or the electrical current, reduces the rate of corrosion. In contrast pollutants in the air, for example, sulphur dioxides from industrial atmospheres and salt from marine atmospheres, increase the electrical conductivity of water and accelerate the corrosion reaction. Steel is particularly susceptible to atmospheric corrosion which is often severe in coastal or industrial environments and the corrosion may reduce the section size due to pitting or flaking of the surface. Modern rolling techniques and higher-strength steels result in less material being used, for example, the web of an ‘I’ section may be only 6 mm thick. Generally in structural engineering 8 mm is the minimum thickness

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used for exposed steel, and 6 mm for unexposed steel. For sealed hollow sections these limits are reduced to 4 and 3 mm, respectively. Corrosion of steel usually takes the form of rust which is a complex oxide of iron. The rust builds up a deposit on the surface and may eventually flake off. The coating of rust does not inhibit corrosion, except in special steels, and corrosion progresses beneath the rust forming conical pits and the thickness of the metal is reduced. The conical pits can act as stress raisers, that is, centres of high local stress, and in cases where there are cyclic reversals of stress, may become the initiating points of fatigue cracks or brittle fracture. The corrosion resistance of unprotected steel is dependent on its chemical composition, the degree of pollution in the atmosphere, and the frequency of wetting and drying. Low-strength carbon steels are inexpensive but are particularly susceptible to atmospheric corrosion which is often greatest in industrial or coastal environments. High-strength low-alloy steels (Cr–Si–Cu–P) do not pit as severely as carbon steels and the rust that forms becomes a protective coating against further deterioration. These steels therefore have several times the corrosion resistance of carbon steels. The longer steel remains wet the greater the corrosion and therefore the detailing of steelwork should include drainage holes, avoid pockets and allow the free flow of air for rapid drying. The most common, and cheapest form of protection process is to clean the surface by sand or shot blasting, and then to paint with a lead primer, generally in the workshop prior to delivery on site. Joint contact surfaces need not be protected unless specified. On site the steel is erected and protection is completed with an undercoat and finishing coat, or coats, of paint. In the case of surfaces to be welded steel should not be painted, nor metal coated, within a suitable distance of any edges to be welded, if the paint specified or the metal coating is likely to be harmful to welders or impair the quality of the welds. Welds and adjacent parent metal should not be painted prior to de-slagging, inspection and approval. Encasing steel in concrete provides an alkaline environment and no corrosion will take place unless water diffuses through the concrete carrying with it SO2 and CO2 gases from the air in the form of weak acids. The resulting corrosion of the steel and the increase in pressure spalls the concrete. Parts to be encased in concrete should not be painted nor oiled, and where friction grip fasteners are used protective treatment should not be applied to the faying surfaces. A more expensive protection is zinc, or aluminium spray coating which is sometimes specified in corrosive atmospheres. Further improvements are hot dip zinc galvanizing, or the use of stainless steels. These and other forms of protection are described in BSEN ISO 12944 (1998). Recently zinc coated highly stressed steel has been shown to be susceptible to hydrogen cracking.

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2.6 BRITTLE FRACTURE (cl 3.2.3, EN 1993-1-1 (2005)) Brittle fracture is critical at the ultimate limit state. Evidence of brittle fracture is a small crack, which may or may not be visible, and which extends rapidly to produce a sudden failure with few signs of plastic deformation. This type of fracture is more likely to occur in welded structures (Stout et al., 2000). The essential conditions leading to brittle fracture are: (a) There must be a tensile stress in the material but it need not be very high, and may be a residual stress from welding. (b) There must be a notch, or defect, or hole in the material which produces a stress concentration. (c) The temperature of the material must be below the transition temperature (generally below room temperature). At low temperatures crack initiation and propagation is more likely because of lower ductility. The mechanism of failure is that the notch, defect or hole raises the local tensile stresses to values as high as three times the average tensile stress. The material which generally fails by a shearing mechanism now tends to fail by a brittle fracture cleavage mechanism which exhibits considerably less plastic deformation. A drop in the temperature encourages the cleavage failure. A ductile material which has an extensive plastic range is more likely to resist brittle fracture and a test used as a guide to resistance to brittle fracture is the Charpy V-notch impact test (BS 7668 (1994)). The importance of brittle fracture was shown by the failure of the welded ‘liberty’ cargo ships mass produced by the USA during the Second World War. The ships broke apart in harbour and at sea during the cold weather. Brittle fracture is considered only where tensile stresses exist. The mode of failure is mainly dependent on the following: (a) (b) (c) (d) (e) (f)

Steel strength grade Thickness of material Loading speed Lowest service temperature Material toughness T ype of structural element.

No further check for brittle fracture need to be made if the conditions given in EN 1993-1-10 (2005) are satisfied for lowest temperature. For further information see NDAC (1970).

2.7 RESIDUAL STRESSES Residual stresses are present in steel due to uneven heating and cooling. The stresses are induced in steel during, rolling, welding which constrains the structure to a

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particular geometry, force fitting of individual components, lifting and transportation. These stresses may be relieved by subsequent reheating and slow cooling but the process is expensive. The presence of residual stresses adversely affects the buckling of columns, introduces premature yielding, fatigue resistance and brittle fracture. Welding raises the local temperature of the steel which expands relative to the surrounding metal. When it cools it contracts inducing tensile stresses in the weld and the immediately adjacent metal. These tensile stresses are balanced by compressive stresses in the metal on either side. During rolling the whole of the steel section is initially at a uniform temperature, but as the rolling progresses some parts of the cross-section become thinner than others and consequently cool more quickly. Thus, as in the welded joint, the parts which cool last have a residual tensile stress and the parts which cool first may be in compression. Since the cooling rate also affects the yield strength of the steel, the thinner sections tend to have a higher yield stress than the thicker sections. A tensile test piece cut from the thin web of a Universal Beam will probably have a higher yield stress than one cut from the thicker flange. The residual stress and yield stress in rolled sections are also affected by the cold straightening which is necessary for many rolled sections before leaving the mills. Residual stresses are not considered directly in the European Code but are allowed for in material factors. For further information see Ogle (1982).

2.8 FATIGUE This is an ultimate limit state. The term fatigue is generally associated with metals and is the reduction in strength that occurs due to progressive development of existing small pits, grooves or cracks when subject to fluctuating loads. The rate of development of these cracks depends on the size of the crack and on the magnitude of the stress variation in the material and also the metallurgical properties. The number of stress variations, or cycles of stress, that a material will sustain before failure is called fatigue life and there is a linear experimental relationship between the log of the stress range and the log of the number of cycles. Welds are susceptible to a reduction in strength due to fatigue because of the presence of small cracks, local stress concentrations and abrupt changes of geometry. Research into the fatigue strength of welded structures is described by Munse (1984). Other references are BS 5400 (1980), Grundy (1985) and ECSS. All structures are subject to varying loads but the variation may not be significant. Stress changes due to fluctuations in wind loading need not be considered, but windinduced oscillations must not be ignored. The variation in stress depends on the ratio of dead load to imposed load, or whether the load is cyclic in nature, for example, where machinery is involved. For bridges and cranes fatigue effects are

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Chapter 2 / Mechanical Properties of Structural Steel more likely to occur because of the cyclic nature of the loading which causes reversals of stress. Generally calculations are only required for: (a) (b) (c) (d)

Lifting appliances or rolling loads, Vibrating machinery, Wind-induced oscillations, Crowd-induced oscillations.

The design stress range spectrum must be determined, but simplified design calculations for loading may be based on equivalent fatigue loading if more accurate data is not available. The design strength of the steel is then related to the number and range of stress cycles. For further information see EN 1993-1-9 (2005).

2.9 STRESS CONCENTRATIONS Structural elements and connections often have abrupt changes in geometry and also contain holes for bolts. These features produce stress concentrations, which are localized stresses greater than the average stress in the element, for example, tensile stresses adjacent to a hole are approximately three times the average tensile stress. If the average stress in a component is low then the stress concentration may be ignored, but if high then appropriate methods of structural analysis must be used to cater for this effect. The effect of stress concentrations has been shown to be critical in plate web girders in recent history. Stress concentrations are also associated with fatigue and can affect brittle fracture. Formulae for stress concentrations are given in Roark and Young (1975).

2.10 FAILURE CRITERIA FOR STEEL The structural behaviour of a metal at or close to failure may be described as ductile or brittle. A typical brittle metal is cast iron which exhibits a linear load–displacement relationship until fracture occurs suddenly with little or no plastic deformation. In contrast mild steel is a ductile material which also exhibits a linear load–displacement relationship, but at yield large plastic deformations occur before fracture. The nominal yield strength is a characteristic strength in the European Code and is therefore an important failure criterion for steel. The tensile yield condition can be related to various stress situations, for example, tension, compression, shear or various combinations of stresses. There are four generally acceptable theoretical yield criteria as follows: (1) The maximum stress theory, which states that yield occurs when the maximum principal stress reaches the uniaxial tensile stress.

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(2) The maximum strain theory, which states that yield occurs when the maximum principal tensile strain reaches the uniaxial tensile strain at yield. (3) The maximum shear stress theory, which states that yield occurs when the maximum shear stress reaches half of the yield stress in uniaxial tension. (4) The distortion strain energy theory, or shear strain energy theory, which states that yielding occurs when the shear strain energy reaches the shear strain energy in simple tension. For a material subject to principal stresses σ1 , σ2 and σ3 it is shown (Timoshenko, 1946) that this occurs when (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2fy2

(2.1)

This theory was originally developed by Huber, Von-Mises and Hencky. Alternatively Eq. (2.1) can be expressed in terms of direct stresses σb , σbc and σbt , and shear stress τ on two mutually perpendicular planes. It can be shown from Mohr’s circle of stress that the principal stresses 1/2 (σb + σbc ) (σb − σbc )2 2 σ1 = (2.2) − +τ 2 4 and

1/2 (σb − σbc ) (σb − σbc )2 σ2 = + + τ2 2 4

(2.3)

If Eqs (2.2) and (2.3) are inserted in Eq. (2.1) with σ3 = 0 and fy is equal to the design stress fy /γM then 2 ( fy /γM )2 = σbc + σb2 − σbc σb + 3τ 2

(2.4)

If σbc is replaced by σbt with a change in sign then 2 ( fy /γM )2 = σbc + 3τ 2

(2.5)

This equation is a yield criteria which is applicable in some design situations, for example the design of welds (EN 1993-1-8 (2005)).

REFERENCES BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-2 (1993), BSEN 10113-3 (1993) and BS 10210-1 (1994). Specification for weldable structural steels. BSI. BS 5400 (1980). Code of practice for fatigue, Pt 10. BSI. BSEN ISO 12944-1 to 8 (1998) and BSEN ISO 14713 (1999). Code of practice for protection of iron and steel structures. BSI. ECCS (1981). Recommendations for fatigue design of steel structures. European Convention for Structural Steelwork. Construction Press. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. EN 1993-1-9 (2005). Design of steel structures: Fatigue strength of steel structures. BSI.

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Chapter 2 / Mechanical Properties of Structural Steel EN 1993-1-10 (2005). Design of steel structures: Selection of steel for fracture toughness and through thickness properties. BSI. Grundy, P. (1985). Fatigue limit design for steel structures. Civil Engineering Transactions, Institution of Civil Engineers, Australia, CE27, No. 1. Munse, W.H. (1984). Fatigue of welded structures, Welding Research Council. NDAC (1970). Brittle fracture in steel structures, Navy Department Advisory Committee on Structural Steels (ed G.M. Boyd). Butterworth. Ogle, M.H. (1982). Residual stresses in a steel box-girder bridge. Tech Note 110. Construction Industry Research and Information Association Publication. Roark, J.R. and Young, W.C. (1975). Formula for Stress and Strain (5th edition). McGraw-Hill. Stout, R.D., Tor, S.S. and Ruzek, J.M. (1951). The effect of fabrication procedures on steels used in pressure vessels, Welding Journal 30. Timoshenko, S. (1946). Strength of Materials, Pt II. D. Van Nostrand.

Chapter

3 / Actions

3.1 DESCRIPTION Actions are a set of forces (loads) applied to a structure, or/and deformations produced by temperature, settlement or earthquakes (EN 1990 (2002)). Values of actions are obtained by determining characteristic or representative values of loads or forces. Ideally, loads applied to a structure during its working life, should be analysed statistically and a characteristic load is determined. The characteristic load might then be defined as the load above which no more than 5% of the loads exceed, as shown in Fig. 3.1. However, data is not available and the characteristic value of an action is given as a mean value, an upper value or a nominal value.

3.2 CLASSIFICATION OF ACTIONS Actions are classified as: Permanent Variable Accidental Seismic

Frequency

(1) (2) (3) (4)

5% of results

Loads Mean load

1,64s

Characteristic load

FIGURE 3.1 Variation in loads

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Chapter 3 / Actions In addition actions can be classified by: (1) Variation in time: (a) Permanent actions (G), for example, self-weight and fixed equipment. (b) Variable actions (Q), for example, imposed loads, wind actions or snow loads. (2) Spacial variation: (a) Fixed actions, for example, structures sensitive to self-weight. (b) Free actions which result in different arrangements of actions, for example, movable imposed loads, wind actions and snow loads.

3.3 ACTIONS VARYING IN TIME 3.3.1 Permanent Actions (G) Permanent actions are due to the weight of the structure, that is, walls, permanent partitions, floors, roofs, finishes and services. The actual weights of the materials (Gk ) should be used in the design calculations, but if these are unknown values of density in kN/m3 may be obtained from EN 1991-1-1 (2002). Also included in this group are water and soil pressures, prestressing force and the indirect actions such as settlement of supports.

3.3.2 Variable Actions (Q) (a) Imposed floor loads are variable actions and for various dwellings are given in EN 1991-1-1 (2002). These loads include a small allowance for impact and other dynamic effects that may occur in normal occupancy. They do not include forces resulting from the acceleration and braking of vehicles or movement of crowds. The loads are usually given in the form of a distributed load or an alternative concentrated load. The one that gives the most severe effect is used in design calculations. When designing a floor it is not necessary to consider the concentrated load if the floor is capable of distributing the load and for the design of the supporting beams the distributed load is always used. When it is known that mechanical stacking of materials is intended, or other abnormal loads are to be applied to the floor, then actual values of the loads should be used, not those obtained from EN 1991-1-1 (2002). In multi-storey buildings the probability that all the floors will simultaneously be required to support the maximum loads is remote and reductions to column loads are therefore allowed. (b) Snow roof loads are variable actions and are related to access for maintenance. They are specified in EN 1991-1-3 (2002) and, as with floor loads, they are expressed as a uniformly distributed load on plan, or as an alternative concentrated load. The magnitude of the loads decrease as the roof slope increases and in special situations, where roof shapes are likely to result in drifting snow, then loads are increased.

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29

(c) Wind actions are variable but for convenience they are expressed as static pressures in EN 1991-1-4 (2002). The pressure at any point on a structure is related to the shape of the building, the basic wind speed, topography and ground roughness. The effects of vibration, such as resonance in tall buildings must be considered separately. (d) Thermal effects need to be considered for chimneys, cooling towers, tanks, hot and cold storage, and services. They are classed as indirect variable actions. Elements of structures which are restrained or highly redundant introduce stresses which need to be determined (EN 1991-1-5 (2003)). (e) EN 1991-1-2 (2002) covers the actions to be taken into account in the structural design of buildings which are required to give adequate performance in fire.

3.3.3 Accidental Actions (A) (a) Accidental actions during execution include scaffolding, props and bracing (EN 1991-1-6 (2002)). These may involve consideration of construction loads, instability and collapse prior to the completion of the project. Erection forces are of great importance in steelwork construction because prefabrication is normal practice. Compression members which will be restrained in a completed structure may buckle during erection when subject to relatively minor forces. Joints which are rigid when fully bolted may, during erection, act as a pin and induce collapse of the structure. Suspension points for members or parts of structures may have to be specified to avoid damage to components. It is extremely difficult to anticipate all possible erection forces and the contractor is responsible for erection which should be carried out with due care and attention. Nevertheless a designer should have knowledge of the most likely method of erection and design accordingly. If necessary temporary stiffening or supports should be specified, and/or instructions given. (b) Accidental actions include impact and explosions which are covered in EN 1991-17 (2004). No structure can be expected to resist all actions but it must be designed so that it does not suffer extreme damage from probable actions, for example, vehicle collisions in a multi-storey car park. Local damage from accidental actions is acceptable. (c) When designing for earthquakes the inertial forces must be calculated as described in EN 1998-8 (2004). This is not of major importance in the UK. Actions induced by cranes and machinery are dealt with in EN 1991-3 (2004).

3.4 DESIGN VALUES OF ACTIONS Partial safety factors allow for the probability that there will be a variation in the effect of the action, for example, a variable action is more likely to vary than a permanent action. The values also allow for inaccurate modelling of the actions, uncertainties

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Chapter 3 / Actions in the assessment of the effects of actions, and uncertainties in the assessment of the limit state considered. The design value of an action is obtained by multiplying the characteristic value by a partial safety factor, for example, for a permanent action the design value Gd = γG Gk . For a variable action the design value Qd = ψ0 Qk or ψ1 Qk or ψ2 Qk . These represent combination, frequency and quasi-permanent values. The combination value (ψ0 Qk ) allows for the reduced probability that unfavourable independent actions occur simultaneously at the ultimate limit state. The frequency value (ψ1 Qk ) involves accidental actions and reversible ultimate limit states. The quasi-permanent value (ψ2 Qk ) also involves accidental actions and reversible serviceability limit states. Recommended values of ψ0 , ψ1 and ψ2 are given in EN 1990 (2002).

3.4.1 Combination of Design Actions For the ultimate limit state three alternative combinations of actions, modified by appropriate partial safety factors (γ), must be investigated: (a) Fundamental: a combination of all permanent actions including self-weight (Gk ), the prestressing action (P), the dominant variable action (Qk ) and combination values of all other variable actions (ψ0 Qk ). (b) Accidental: a combination value of the dominant variable actions (ψ0 Qk ). This combination assumes that accidents (explosions, fire or vehicular impact) of short duration have a low probability of occurrence. (c) Seismic: reduces the permanent action partial safety factors (γG ) with a reduction factor (ξ) between 0,85 and 1. For the serviceability limit state three alternative types of combination of actions must be investigated: (a) The characteristic rare combination occurring in cases when exceeding a limit state causes permanent local damage or deformation. (b) The combination which produces large deformations or vibrations which are temporary. (c) Quasi-permanent combinations used mainly when long-term effects are important. A combination of actions can be symbolically represented for design purposes, for example, for one of three conditions at the ultimate limit state:

γG Gk + γP P +

γ Q ψ 0 Qk

Similar equations can be formed for the other two conditions at ultimate limit state and for the three conditions at the serviceability limit state (EN 1990 (2002)).

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31

3.5 ACTIONS WITH SPACIAL VARIATION 3.5.1 Pattern Loading All possible actions relevant to a structure should be considered in design calculations. The actions should be considered separately and in realistic combinations to determine which is most critical for strength and stability of the structure. For continuous structures, connected by rigid joints or continuous over the supports, vertical actions should be arranged in the most unfavourable but realistic pattern for each element. Permanent actions need not be varied when considering such pattern loading, but should be varied when considering stability against overturning. Where horizontal actions are being considered pattern loading of vertical actions need not be considered. For the design of a simply supported beam it is obvious that the critical condition for strength is when the beam supports the maximum permanent action and maximum variable action at the ultimate limit state. The size of the beam is then determined from this condition and checked for deflection at the serviceability limit state. A more complicated structure is a simply supported beam with a cantilever as shown in Fig. 3.2(a). Assuming that the beam is of uniform section and that the permanent actions are uniformly applied over the full length of the beam, it is necessary to consider various combinations of the variable actions as shown in Figs. 3.2(b)–(d). Although partial

A

B

C

(a) Imposed design load Dead design load (b) Imposed design load Dead design load (c) Imposed design load Dead design load (d)

FIGURE 3.2 Pattern loading

3m

Chapter 3 / Actions

3m

Imposed loads

b

3m

a

3m

•

3m

32

5m

5m

5m

FIGURE 3.3 Multi-storey frame

loading of spans is possible this is not generally considered except in special cases of rolling actions (e.g. a train on a bridge span). For a particular section it is not immediately apparent which combination of actions is most critical because it depends on the relative span dimensions and magnitude of the actions. Therefore calculations are necessary to determine the condition and section for maximum bending moment and shear force at the ultimate limit state. Analysis of a multi-storey building is more complicated as shown by Holicky (1996). Where loads on other storeys affect a particular span they may be considered as uniformly distributed (EN 1991-1-1 (2002)). However, the critical positioning of the loads may be different, for example, Fig. 3.3 shows the load positions for verification of the bending resistance at points a and b. In other situations, for example, when checking the overturning of a structure, the critical combination of actions may be the minimum permanent action, minimum imposed action and maximum wind action.

3.5.2 Design Envelopes The effect of pattern loading can be seen by constructing a design envelope. This is a graph showing, at any point on a structural member, the most critical effect that results from various realistic combinations of actions. Generally the most useful envelopes

Structural Design of Steelwork to EN 1993 and EN 1994 A

B

(2) 91,0 81,7

(1) (3)66,4

11,7

20,0

•

33

C

Case 1 Case 2 Case 3 Envelope 38,3 99,0 108,3(1)

Shear force (kN) 66,4 (1)(3)

2,34 m

20 F

E 6,9

0,21 m

87,9 109(2) 2,40 m Bending moment (kNm)

FIGURE 3.4 Example: Design envelope

are for shear force and bending moment at the ultimate limit state. The formation and use of a design envelope is demonstrated by the following example.

EXAMPLE 3.1 Example of a design envelope. The beam ABC in Fig. 3.4-carries the following characteristic loads: Dead load Gk = 10 kN/m on both spans; Imposed loads Qk = 15 kN/m on span AB, 12 kN/m on span BC. Sketch the design envelope for the bending moment and shear force at the ultimate limit state. Indicate all the maximum values and positions of zero bending moment (points of contraflexure). The maximum and minimum design loads on the spans are: Maximum on AB = γG Gk + γQ Qk = 1,4 × 10 + 1,6 × 15 = 38 kN/m. Maximum on BC = γG Gk + γQ Qk = 1,4 × 10 + 1,6 × 12 = 33, 2 kN/m. Minimum on AB or BC = γG Gk = 1,0 × 10 = 10 kN/m.

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Chapter 3 / Actions Consider the following design load cases: (1) Maximum on AB and BC. (2) Maximum on AB, minimum on BC. (3) Minimum on AB, maximum on BC. The bending moment and shear force diagrams are shown in Fig. 3.4.

Comments (a) Only the numerical value of the shear force is required in design, the sign however, may be important in the analysis of the structure. (b) Positive (sagging) bending moments indicate that the bottom of the beam will be in tension; negative (hogging) moments indicate that the top of the beam will be in tension. (c) The envelope, shown as a heavy line, indicates the maximum values produced by any of the load cases. Note that on AB the envelope for shear force changes from case (2) to case (1) at the point where the numerical values of the shear force are equal. This process is tedious and an experienced designer knows the critical action combinations and the positions of the critical values and avoids some of the work involved. Alternatively the diagrams can be generated from input data using computer graphics. In more complicated structures and loading situations, envelopes are useful in determining where a change of member size could occur and where splices could be inserted. In other situations wind action is a further alternative to combinations of permanent and variable actions.

REFERENCES EN 1990 (2002). Basis of structural design. BSI. EN 1991-1-1 (2002). Densities, self weight and imposed loads. BSI. EN 1991-1-2 (2002). Actions on structures exposed to fire. BSI. EN 1991-1-3 (2002). Snow loads. BSI. EN 1991-1-4 (2002). Wind loads. BSI. EN 1991-1-5 (2003). Thermal actions. BSI. EN 1991-1-6 (2002). Actions during execution. BSI. EN 1991-1-7 (2004). Accidental actions due to impact and explosions. BSI. EN 1998-8 (2004). Design of structures for earthquake resistance. BSI. EN 1991-3(2004). Actions induced by cranes and machinery. BSI. Holicky, M. (1996). Densities, self weight and imposed loads on buildings. Czech Technical University, Prague.

Chapter

4 / Laterally Restrained Beams

4.1 STRUCTURAL CLASSIFICATION OF SECTIONS (CL 5.5, EN 1993-1-1 (2005)) Chapters 4 and 5 are concerned with the design of members which are predominantly in bending, that is, where axial loads, if any, are small and transverse shear forces are not excessive. Chapter 4 contains basic theoretical work on section properties and the design of laterally restrained beams using Class 1 standard sections and plastic methods of analysis. Sections of steel beams in common use are shown in Fig. 4.1. The rolled sections shown at (a) are used most often and of these the ‘I’ section is used widely. Some sections are of uniform thickness while others are of different thickness for the web and flange. The rolled sections are generally in stock, are lowest in cost, require less design and connections are straightforward. Hollow sections are not as efficient in bending but corrosion resistance is better and aesthetically they may be more acceptable. Cold formed sections are thinner and are therefore more susceptible to corrosion unless protected, however they are very economical for use as puriins. Fabricated sections are used when a suitable rolled section is not available, but costs are higher and delivery times are longer. Castellated sections are used for large spans with relatively low loads and where transverse shear forces are not excessive. Tapered beams are efficient in resisting bending moments but must be checked for shear forces. Composite steel–concrete sections are used for floors. The four classes of cross-section of steel ‘I’ beams are described in cl 5.5.2, EN 1993-1-1 (2005). To allow for flange buckling sections are reduced to effective sections. All members subject to bending should be checked for the following at critical sections: (a) (b) (c) (d) (e)

A combination of bending and shear force Deflection Lateral restraint Local buckling Web bearing and buckling.

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Chapter 4 / Laterally Restrained Beams

(a) Hot rolled

(b) Hollow

(c) Cold formed, light gauge

Crane rail

Channel welded to an ‘I’ beam (d) Fabricated

Weld

Cut

Standard ‘I’ beam (e) Tapered Weld

Cut

Standard ‘I’ beam

60°

(f) Castellated Shear connectors

Reinforced concrete

Reinforced concrete (g) Composite

Steel beam encased in concrete

FIGURE 4.1 Types of steel beams

This chapter is concerned with members which are predominantly subject to bending and where lateral torsional buckling and local buckling of the compression flange are prevented. It is important to recognize the characteristics of these two forms of buckling shown in Fig. 4.2. Lateral torsional buckling exhibits vertical movement (bending about about the y–y axis), lateral displacement (bending about the z–z axis),

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37

Compression flange

Local buckling

Moment

z x y

y x

Load

Moment

z (a) Lateral torsional buckling of a cantilever

(b) Local buckling of a flange

FIGURE 4.2 Buckling of beams

and torsional rotation (rotation about the x–x axis). Local buckling exhibits local deformation of an outstand, for example, a flange of an ‘I’ beam. Lateral torsional buckling occurs when the buckling resistance about the z–z axis and the torsional resistance about the x–x axis are low. The buckling resistance about the z–z axis can be improved by lateral restraints, for example, transverse members which prevent lateral movement of the compression flange. Local buckling occurs when the flange outstand to thickness ratio (b/tf ) is high and is avoided by choosing Class 1 sections (cl 5.5.2, EN 1993-1-1 (2005)). Where both types of buckling are prevented, as for Class 1 and Class 2 sections, then the section can be stressed to the maximum design stresses in bending, i.e. plastic methods of analysis and design can be used. If Class 3 or Class 4 sections are used the plastic moment capacity is reduced to an elastic moment capacity. If a steel ‘I’ section is used as a simply supported beam and loaded with a uniformly distributed load then the bending moment distribution varies parabolically. If the section is bent about a major axis then the stress distribution at centre span at various stages of loading is shown in Fig. 4.3(c). In the early stages of loading the stress distribution is elastic, then elastic–plastic and finally fully plastic. The corresponding moment curvature relationship is shown in Fig. 4.3(b). The fully plastic stage corresponds to the condition for the tensile stress–strain relationship for the steel shown in Fig. 4.3(a). Theoretically the load cannot be increased beyond this plastic condition but strain hardening occurs and this increases the resistance. Note that for full plasticity large strains occur, of the order of 20%, which makes mild steel ideal, while other steels with less plastic strain behave in a more brittle fashion. Although bending is the predominant design criteria checks must be made for the magnitude of the shear stresses. Shear stresses are introduced from vertical shear forces, or torsion moments (cls 6.2.6 and 6.2.7, EN 1993-1-1 (2005)).

Chapter 4 / Laterally Restrained Beams

Moment

•

Stress

38

Plastic Elastic

Elastic–plastic

Plastic Strain hardening

Elastic Elastic–plastic

Strain (a) Tensile stress–strain relationship

Strain hardening

Curvature (1/R ) (b) Moment–curvature relationship

sy

Neutral axis

(c) Change of bending stress distribution as a plastic hinge forms

(d) Development of a plastic hinge

FIGURE 4.3 Development of a plastic hinge

For the design of beams calculations are required in the elastic stage of behaviour, for example, stresses and deflections, and also at the fully plastic stage (e.g. collapse load). The calculations involve certain basic section properties.

4.2 ELASTIC SECTION PROPERTIES AND ANALYSIS IN BENDING 4.2.1 Sectional Axes and Sign Conventions For all standard sections rectangular centroidal axes y–y and z–z are defined parallel to the main faces of the section, as shown in Fig. 4.4. The position of these axes is given in Section Tables. For angles, and other sections where the rectangular and principal axes do not coincide, the principal axes are denoted by u–u and v–v. The major axis u–u is conventionally inclined to the y–y axis by an angle α, as shown in Figs 4.4(e) and (f). For equal angles, α = 45◦ . For problems involving simple uniaxial or biaxial bending of symmetrical sections a strict sign convention is not necessary, but for the solution of complex problems it is desirable. In this chapter the positive conventions of sagging curvature and downward deflections are adopted; and the direction of the angle α is anti-clockwise, consistent with the Section Tables. Fig. 4.5(a) shows the coordinates of a point P in the positive

•

Structural Design of Steelwork to EN 1993 and EN 1994 z

z

z

y

y

y

y

y

y

z

z (a) ‘I’ section

z

(b) Tee

v

z

(c) Hollow section

z

y

a y

y

z

(d) Channel

u

v

(e) Unequal angle

u a 45° y

y

u z

z

v

u y

39

z

v

(f) Equal angle

FIGURE 4.4 Sectional axes

Fv

Fz Mz

Mv

α

z

y v

Mu My

O

Pu

u

O

Fu

y

Mz Vy

Vy

O Fy

My y

y

z

Vz

z

x

Vz z

v

(a) Coordinates and external loads

x (b) Stress resultants

FIGURE 4.5 Sign conventions

quadrant of a section and the positive directions for the externally applied forces and couples. The positive directions of the corresponding stress resultants (shear forces and bending moments) are shown for the horizontal and vertical planes in Fig. 4.5(b). Positive directions relative to the u–u and v–v axes can be inferred. The convention for the moments has been chosen so that positive moments give tensile stresses in the positive quadrant of the section.

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Chapter 4 / Laterally Restrained Beams Normally the coordinates of points in a section relative to the rectangular axes are known, or can easily be obtained. The coordinates relative to the principal axes are given by u = y cos α + z sin α v = z cos α − y sin α

(4.1)

External forces and shear forces transform in exactly the same way, thus Fu = Fy cos α + Fz sin α Fv = Fz cos α − Fy sin α

(4.2)

However the directions chosen for the moments are consistent with the rules for a right hand set of axes, which gives rise to changes in sign, thus Mu = My cos α − Mz sin α Mv = Mz cos α + My sin α

(4.3)

4.2.2 Elastic Second Moments of Area This property is derived from the simple theory of elastic bending (Croxton and Martin Vol 1 (1987 and 1989)). In design it is used to calculate stresses and deflections in the elastic stage of behaviour, that is, at service loads. Second moments of area for all standard sections are given in Section Tables but for fabricated sections they must be calculated. The procedure involves application of the theorems of parallel axes which, for the single element of area A in Fig. 4.6, can be stated as follows: Iy = Ia + Az2 Iz = Ib + Ay2

(4.4)

Iyz = Iab + Ayz where Iy = elemental second moment of area about y–y Iz = elemental second moment of area about z–z Iyz = elemental product moment of area about y–y and z–z a–a and b–b are centroidal axes through the element, parallel to y–y and z–z, respectively. For the determination of Iyz , which can be either positive or negative, the correct signs must be allocated to the coordinates y and z. The positive directions are indicated by arrows in Fig. 4.6.

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41

O

y b z c

a

a y

Area A b

z

FIGURE 4.6 Parallel axes for an element

When second moments of area about the rectangular axes have been computed, the direction of the principal axes can be obtained from tan 2α = 2Iyz /(Iz − Iy )

(4.5)

The principal second moments of area are then give by Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α

(4.6)

If Iy is arranged to be greater than Iz , then α will be less than 45◦ and Iu will be the major principal second moment of area. A negative result for Eq. (4.5) indicates that α is to be measured clockwise from y–y.

4.2.3 Elastic Section Moduli These values are derived from the second moments of area by dividing by the distance to the extreme fibres (i.e. Wel = I/z). Values of section moduli are given in Section Tables. For structural tees two values of Wel are given, referring to the extreme fibres in the table and the stalk.

4.2.4 Elastic Bending of Symmetrical Sections When either of the rectangular axes is an axis of symmetry the normal bending stress at any point in the section is given by σ=

Mx z Mz y + Iy Iz

(4.7)

If the directions of the bending moments and the coordinates are in accordance with the sign convention of Fig. 4.5, a positive result indicates that the stress is tensile.

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Chapter 4 / Laterally Restrained Beams For simple bending about the y–y axis, which is the most common event, the stress in the extreme fibres σmax =

My zmax My = Iy Wely

(4.8)

Similarly for bending about the z–z axis σmax =

My zmax My = Iy Wely

(4.9)

4.2.5 Elastic Bending of Unsymmetrical Sections When a section is subject to bending about an axis which is not a principal axis the effect is the same as if the section were subject to the components of the bending moment acting about the principal axis. In other words the bending is biaxial. For standard rolled angles the principal second moments of area and the directions of the principal axes are given in Section Tables. Transforming bending moments and coordinates to the principal axes by means of Eqs (4.3) and (4.1). Bending stress σ=

Mu v Mv u + Iu Iv

(4.10)

This is the same as Eq. (4.7) but with all the terms treated to the principal axes. If the sign convention of Fig. 4.5 is observed, a positive result indicates tension. In other cases the additional calculations required for the solution of problems by principal axes can be avoided by the use of ‘effective bending moments’. These are modified bending moments which can be considered to act about the rectangular axes of the section. The bending stress is then given by an expression having exactly the same from as Eq. (4.7) σ=

Mey z Mez y + Iy Iz

(4.11)

where Mey and Mez are effective bending moments about y–y and z–z axes, respectively and are given by Mey =

My − Mz Iyz /Iz 1 − Iyz2 /(Iy Iz )

Mez =

Mz − My Iyz /Iy 1 − Iyz2 /(Iy Iz )

(4.12)

These expressions are derived from the application of conventional elastic bending theory to curvature in both the yx and zx planes. One such derivation is given by Megson (1980).

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By successive differentiation with respect to x, the longitudinal dimension, similar expressions for the effective shear force and effective load intensity can be obtained, thus Vey =

Vy − Vz Iyz /Iy 1 − Iyz2 /(Iy Iz )

Vez =

Vz − Vy Iyz /Iz 1 − Iyz2 /(Iy Iz )

(4.13)

and fey =

( fy − fz Iyz /Iy ) 1 − Iyz2 /(Iy Iz )

fez =

( fz − fy Iyz /Iz ) 1 − Iyz2 /(Iy Iz )

(4.14)

It should be noted that the quantities Iy and Iz are interchanged in Eqs (4.13) and (4.14). This is because the expressions for the shear force and the load intensity in the y direction are obtained by successive differentiation of bending moments along the z–z axis and vice versa. All bending moment problems with unsymmetrical sections can be solved simply by replacing ordinary loads, shears, and bending moments by their effective counterparts. Note however that these effective counterparts have values related to both the y–y and z–z axes, even if the section is only loaded in the direction of one of the rectangular axes.

EXAMPLE 4.1 Principal axes for an unequal angle section. Find the directions of the principal axes and the values of the principal second moments of area for the angle section in Fig. 4.7(a). For the calculation of section properties the work is simplified considerably, with insignificant loss of accuracy, by using the dimensions of the section profile, that is, the shape formed by the centre line of the elements, as shown in Fig. 4.7(b). The position of the centroid O is found by taking moments of the area about the centre lines of each leg in turn

Areas

mm2

A B A C

140 × 20 = 2800 290 × 20 = 5800 Total = 8600

Chapter 4 / Laterally Restrained Beams 150

140 c z

cz

A

c y

B

cy

O a 290

y

O 300

•

20

44

u

20

z (a) Actual section

c

v

(b) Section profile

FIGURE 4.7 Example: principal axes for an unequal angle Taking moments about A B 8600cy =

5800 × 290 2

hence cy = 97,8 mm

Taking moments about A C 8600cz =

2800 × 140 2

hence cz = 22,8 mm

Hence for the full section (Fig. 4.7(a)) cy = 107,8 mm

and

cz = 32,8 mm

Coordinates of the centroids of the legs AB and AC are therefore given by Leg A B

140 − cz = 47,2 mm 2 z = −cy = −97,8 mm

Leg A C

y = −cz = −22,8 mm

y=

z=

290 − cy = 47,2 mm 2

The second moments of area about the rectangular axes are obtained in the usual way be applying the parallel axes formula to each leg. bh3 + A(leg)z2 12 where b and h are dimensions of the leg parallel to the y–y and z–z axes, respectively. Iy (leg) =

Structural Design of Steelwork to EN 1993 and EN 1994

Leg A B

•

45

203 + 2800 × (−97,8)2 = 26,87E6 mm4 12 2903 + 5800 × 47,22 = 53,57E6 mm4 20 × 12

140 ×

Leg A C

Iy = 80,44E6 mm4 Iz (leg) =

hb3 + A(leg)y2 12

Leg A B Leg A C

20 ×

290 ×

1403 + 2800 × 47,22 = 10,81E6 mm4 12

203 + 5800 × (−22,8)2 = 3,21E6 mm4 12 Iz = 14,02E6 mm4

The product moment of area Iyz is obtained by applying the parallel axis formula to each leg Iyz (leg) = Iab + A(leg)yz For each leg the term Iab is equal to zero, because the parallel axes through the centroid of the leg are principal axes. Leg A B

2800 × 42,7 × (−97,8) = −12,93E6 mm4

Leg A C

5800 × (−22,8) × 47,2 = −6,24E6 mm4 Iyz = −19,17E6 mm4

Direction of the principal axes from Eq. (4.5) tan 2α =

2Iyz Iz − I y

2 × (−19, 17) 2α = arctan 14,02 − 80,44

hence α = 15◦

Principal second moments of area from Eq. (4.6) Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α Substituting values Iu = 85,58E6 mm4

and

Iv = 8,88E6 mm4 .

Chapter 4 / Laterally Restrained Beams bf tf

z

P cy

•

R y

y

h

46

100 kN m tw Q 50 kN m z

FIGURE 4.8 Example: structural tee in biaxial bending

As a check on the transformation Iy + Iz = Iu + Iv which is correct.

EXAMPLE 4.2 Structural tee in biaxial bending. Calculate the maximum extreme fibre stresses in a standard 292 × 419 × 113 kg structural tee cut from a Universal Beam. The tee is loaded by two moments as shown in Fig. 4.8. From Section Tables bf = 293,8 mm, h = 425,5 mm, tw = 16,1 mm, tf = 26,8 mm, Cy = 108 mm, Iy = 246,6E6 mm4 , Iz = 56,76E6 mm4 , Wely (flange) = 2,277E6 mm3 , Wely (toe) = 0,7776E6 mm3 , Welz = 0,3865E6 mm3 . By inspection the maximum compressive stress occurs at a point P because the stresses from both moments are compressive. σ= =

My Mz + Wely (flange) Welz 50E6 100E6 + = 173 MPa 2,277E6 0,3865E6

The maximum tensile stress can occur at point Q or point R, depending on the relative magnitude of the bending moments. It is necessary to check both points. Using the sign convention of Fig. 4.5 both bending moments are positive and the coordinates of the points are given by Point Q, Point R,

tw 16,1 = = 8,05 mm, z = h − cy = 425,5 − 108 = 317,5 mm. 2 2 bf 293,8 y= = = 146,9 mm, z = tf − cy = 26,8 − 108 = −81,2 mm. 2 2

y=

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From Eq. (4.7) the stresses My z Mz y + Iy Iz

σ=

100 × 317,5 50 × 8,05 + = 135,8 MPa 246,6 56,76 100 × (−81,2) 50 × 146,9 σR = + = 96,5 MPa 246,6 56,76

σQ =

Summarizing, the maximum stresses are: At Q, σQ = +135,8 MPa (tension) At P, σP = −173 MPa (compression)

EXAMPLE 4.3 Bending stresses in an unequal angle section. Calculate the bending stresses in the angle section shown in Fig. 4.9 where Iy = 80,44E6 Iz = 14,02E6

Iyz = −19,17E6 mm4

Effective moments from Eq. (4.12) Mey =

(My − Mz Iyz /Iz ) = 74,92 kNm (1 − Iyz2 /Iy Iz )

Mez =

(Mz − My Iyz /Iy ) = 32,86 kNm (1 − Iyz2 /Iy Iz )

Maximum compressive stress at A σA =

Mey zA Mez yA + Iy Iz

74,92E6 × (−107,8) 32,86E6 × (−32,8) + 80,44E6 14,02E6 = −177,3 MPa (compression)

=

Check for maximum tensile stress at B, and at C σB = =

Mey zB Mez yB + Iy Iz 74,92E6 × (−87,8) 32,86E6 × 117,2 + 80,44E6 14,02E6

= +192,9 MPa (tension) σC = =

Mey zC Mez yC + Iz Iy 74,92E6 × 192,2 32,86E6 × (−12,8) + 80,44E6 14,02E6

= +149,0 MPa (tension)

Chapter 4 / Laterally Restrained Beams 150 117,2

107,8

32,8 A

B 30 kN m

300

y

15 kN m

192,2 C 12,8

(a)

z

Mz

183,6 B 78,4

Mz y

My z

144,5 A

155

•

87,8

48

My Neutral axis x

(b) Positive bending moments

(c)

C125,6

FIGURE 4.9 Example: bending stresses in an unequal angle

Summarizing, the maximum stresses are: At B, σB = +192,9 MPa (tension) At A, σA = −177,3 MPa (compression) Note that although the position of the centroid and the values of the second moment of area can be calculated without significant error from the profile dimensions of the section, the same is not true of the stresses.

EXAMPLE 4.4 Bending about principal axes of an angle section. Recalculate the stresses at points A, B and C in the previous example considering bending about the principal axes.

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The coordinates of the points are transformed in accordance with Eq. (4.1), u = y cos α + z sin α sin α = 0, 2588 and Point

y/z axes y −32,8 117,2 −12,8

A B C

and v = z cos α − y sin α. cos α = 0, 9659, u/v axes

z u −107,8 −59,6 −87,8 90,5 192,2 37,5

v −95,6 −115,1 189,0

Bending moments transform in accordance with Eq. (4.3) Mu = My cos α − Mz sin α = 25,10 kNm Mv = Mz cos α + My sin α = 22,25 kNm Bending stresses from Eq. (4.10) σ=

Mu v Mv u + , Iu Iv

σA = −177,3

σB = 192,9

σC = 149,0 MPa

which are the same as obtained previously.

4.2.6 Elastic Analysis of Beams The elastic analysis of simply supported beams with examples of shear force and bending moment diagrams and deflection calculations are given in many introductory books on structural analysis (e.g. Croxton and Martin Vol. 1 (1987 and 1989)). The analysis of continuous beams is more complicated but there is a choice of methods such as area– moment, moment distribution, slope deflection, and matrix methods. These methods are also covered by Croxton and Martin Vol. 2 (1987 and 1989). with a computer program for analysis using matrix methods.

4.2.7 Elastic Deflections of Beams (cl 7.2.1, EN 1993-1-1 (2005)) The deflections under serviceability loads of a building or part should not impair the strength or efficiency of the structure or its components or cause damage to the finishes. When checking for deflections the most adverse realistic combination and arrangement of serviceability loads should be assumed, and the structure may assumed to be elastic. The theory and the methods of calculating deflections for static and hyperstatic structures are given in Croxton and Martin Vols. 1 and 2 (1987 and 1989). For simple beams standard cases can be superimposed and some useful cases are shown in Figs 4.10 and 4.11.

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Chapter 4 / Laterally Restrained Beams Maximum deflection (at free end) Q a 2(3 a)QL3/(6EI ) aL L (a) q

a 3(4 a)qL4/(24EI )

aL (b) q a 3(5 a)qL4/(120EI ) aL

FIGURE 4.10 Deflections of cantilevers

(c)

For simply supported beams the central deflection rather than the maximum is given, so that deflections from individual cases can be added. For most loading cases the central deflection only differs by a small percentage from the maximum. In case (a) of Fig. 4.11, for example, the difference is always within 2,5%. A notable exception is the case of equal end moments acting in the same direction, when the central deflection is zero. However in such a case the deflection at other points along the beam are likely to be small. A more accurate analysis can be formed if it is suspected that the deflection is likely to exceed the limit. Recommendations for limiting values for deflections are given in cl 7.2.1, EN 1993-1-1 (2005).

EXAMPLE 4.5 Deflections for a hyperstatic structure. The size of the members for the symmetrical structure shown in Fig. 4.12 have been determined and the structure requires to be checked for deflections in the elastic range of behaviour. The imposed variable characteristic loads are shown in Fig. 4.12 and the second moment of area is Iy = 127,56E6 mm4 . The bending moments (positive clockwise) at the joints are given in the following table. Joint

Span

Moment (kNm)

B B C C C G

AB BC BC CG CD CG

+60 −60 −6 +72 −66 +36

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Central deflection, and rotation at supports Q End 1

End 2 For a 0.5 d (3a 4a3)QL3/(48EI) u1 (2a 3a2 a3)QL2/(6EI ) u2 (a3 a)QL2/(6EI )

aL

(a)

L

q

d 5qL4/(384EI ) u1 qL3/(24EI ) u2 u1

(b)

q

For a 0.5 d (3a2 2a4)qL4/(96EI) u1 (a4 4a3 4a2)qL3/(24EI ) u2 (a4 2a2)qL3/(24EI)

(c) aL

d qL4/(120EI) u1 5qL3/(192EI ) u2 u1

q (d)

M1

M2 d (M1M2)L2/(16EI ) u1 (2M1M2)L /(6EI ) u2 (2M2M1)L /(6EI )

(e)

24 A

30 B

24 36

C

FIGURE 4.11 Displacements of simply supported beams

E 30

D 8

G 4

18

H 3

6

3

18

4

Dimensions in m; loads in kN.

FIGURE 4.12 Example: deflections of a symmetrical continuous structure

F

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Chapter 4 / Laterally Restrained Beams For all beams EIy = 210E3 × 127,56E6 = 26,79E12 N mm2

Span CD 1. Uniform load (Fig. 4.11(b)) ∂ul = =

5qL4 5QL3 = 384EIy 384EIy 5 × 36E3 × 12E33 = 30,2 mm 384 × 26,79E12

2. Concentrated loads (Fig. 4.11(a)) ∂cl = =

2[3a − 4a3 ]QL3 48EIy 2[3 × (3/12) − 4 × (3/12)3 ] × 24E3 × 12E33 = 44,3 mm. 48 × 26,79E12

3. End moments (Fig. 4.11(e)) ∂M = L2

(M1 − M2 ) 16EIy

= 12E32 ×

(−66 − 66)E6 = −44,3 mm (upwards) 16 × 26,79E12

Total deflection = ∂ul + ∂cl + ∂M = 30,2 + 44,3 + (−44,3) = 30,2 mm A limit for beams with plaster finish = L/350 = 12E3/350 = 34,3 mm

Span BC End moments (Fig. 4.11(e)) ∂ 3 = L2

(M1 − M2 ) 16EIy

= 18E32 ×

(−60 + 6)E6 = −40,8 mm (upwards) 16 × 26,79E12

An accurate analysis gives −42,4 mm at 7,35 m from B, A limit for beams with plaster finish = L/350 = 18E3/350 = 51,43 mm

Cantilever span AB For this span the deflection is due to the flexure of the cantilever, assuming the beam is horizontal at B, plus the effect of the anti-clockwise rotation of the beam at B, that is, −θ1 for span BC.

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End moments for span BC (Fig. 4.11(e)) θ1 = L ×

(2M1 − M2 ) 6EIy

= 18E3 ×

(−2 × 60 + 6)E6 = −0,01277 rad 6 × 26,79E12

1. Deflection at A due to rotation ∂R = −Lθ1 = −4E3 × (−0,01277) = 51,07 mm 2. Deflection due to load (Fig. 4.10(b)) ∂ul = =

a3 (4 − a)qL4 QL3 = 24EI 8EIy 30E3 × 4E33 = 8,96 mm 8 × 26,79E12

Total deflection = ∂R + ∂ul = 51,1 + 9,0 = 60,1 mm A limit for a cantilever beam with plaster finish 2L 2 × 4E3 = = 22,9 mm 350 350 If the deflection of the cantilever exceeds the limit and stiffening is required then increase the size of the section, or add flange plates. It is also necessary to stiffen spans AB and BC because the deflection is dependent on both.

4.2.8 Span/Depth Ratios for Simply Supported Beams An initial estimate for the depth of a simply supported ‘I’ beam carrying a uniformly distributed load can be obtained by using the deflection limit. If σ max is the maximum elastic bending stress at service load then from elastic bending theory σmax =

Mz QL h/2 = × Iy 8 Iy

rearranging Q=

16σmax Iy Lh

(i)

Assuming a deflection limit for beams with plaster finish of L 5QL3 = 350 384EIy

(ii)

Eliminating Q by combining Eqs (i) and (ii) and putting E = 210 GPa, the span depth ratio L 2880 = h σmax

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Chapter 4 / Laterally Restrained Beams If a beam is laterally restrained so that lateral torsional buckling does not occur, then if the maximum bending stress is approximately fy /1,5, the span/depth ratios for different grades of steel and different limits are: TABLE 4.1 Span/depth ratios for beams. Span/depth ratio Grade

Characteristic stress (fy )

L/350

L/250

S235 S275 S355

235 275 355

18,4 15,7 12,2

25,7 22,0 17,0

Note that since Youngs modulus (E) is a constant for all grades of steel, the stiffness of a beam does not increase with a higher grade of steel. If a design is governed by deflection then there is no advantage in using a higher grade of steel.

4.3 ELASTIC SHEAR STRESSES 4.3.1 Elastic Shear Stress Distribution for a Symmetrical Section When a beam is bent elastically by a system of transverse loads, plane sections no longer remain plane after bending, but are warped by shear strains. In most cases the effect is small and the errors introduced in the use of conventional bending theory are negligible. Important exceptions are discussed briefly in Section 4.3.2. Formulae for the calculation of shear stresses in an elastic beam are derived by considering the variation in bending stresses along a short length of beam. Consider the very short length of beam in Fig. 4.13(a). At a point S in the web the shear stresses on the vertical and the horizontal section are complementary and are given by the established formula (Eq. (6.20), EN 1993-1-1 (2005)) vs =

VAzc It

(4.15)

where V = the vertical shear force on the section A = the hatched area, that is, the part of the section between point S and the extreme fibres zc = the distance from the centroid of the area A to the neutral axis I = the second moment of area of the whole section about the neutral axis t = the thickness of the section at the point S.

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b

P

tf

tw

zc

NA

vs S

vs bf

Area A

vp tf

P

(a) Shear flow h

tw S

(b) Stress distribution

vs

FIGURE 4.13 Shear stresses in an ‘I’ beam

The formula cannot be used to obtain the vertical shear stress in the outstanding parts of the flange. However as this must be equal to zero at the top and bottom faces, it must be very small. In fact the resistance of the section to vertical shear is provided almost entirely by the web. The resultant of the longitudinal shear stress in the web is in equilibrium with the change in the normal tensile force on the area A due to the variation in bending moment along the beam. Similar longitudinal stresses exist in the flanges and give rise to horizontal complementary stresses in the directions shown. For example, at point P in the top flange. A = bf tf ,

tf = t,

zc =

(h − tf ) 2

and Eq. (4.15) becomes vp =

Vb(h − tf ) 2I

(4.16)

This expression is linear with respect to the variable b, and vp has a maximum value at the centre of the flange where b = bf /2, that is vp(max) =

Vbf (h − tf ) 4I

(4.17)

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Chapter 4 / Laterally Restrained Beams The complete distribution of shear stress on the cross-section is shown in Fig. 4.13(b). Equation (4.15) can be expressed in terms of the shear flow, which is the product of the shear stress and the thickness of the section, thus v = tvs =

VAzc I

(4.18)

In the longitudinal sense the shear flow is equal to the shear force per unit length of beam, and is a convenient quantity for the calculation of the shear force to be resisted by bolts or welds in a fabricated section.

4.3.2 Elastic Shear Stresses in Thin Walled Open Sections The shear stress distribution for a rectangular cross-section subject to a transverse shear force is shown in Fig. 4.14(a) and can be calculated using Eq. (4.15). The shear stress distribution for the open cross-section is different as shown in Fig. 4.14(b). the distribution for an ‘I’ section is shown in Fig. 4.14(c). Steel sections are usually composed of relatively thin elements, for which the analysis can be simplified by: (a) referring all dimensions to the profile of the section; (b) assuming that the shear stress does not vary across the thickness; (c) ignoring any shear stresses acting at right angles to the section profile. As these are equal to zero at each outside surface they must always be very small in a thin walled section. If it is further assumed that the load is applied in such a way that no twisting of the beam occurs, the shear flow at a point S on the profile of the section is given by s s Vey Vez vs = vo − tz ds − ty ds (4.19) Iy Iz o o here Vey and Vez are the effective shear forces obtained from Eq. (4.13) or by applying the effective loads obtained from Eq. (4.14). The variable s is the distance around the profile to the point of interest, starting from any point at which the shear flow

(a)

(b)

(c)

FIGURE 4.14 Distribution of shear stresses

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vo is known. At any open end, such as the end of a flange, the value of vo is zero. The direction of s can be chosen arbitrarily and, provided that the sign convention of Fig. 4.5 is adopted, a positive sign for vs indicates that the shear flow is in the direction chosen for s.

4.3.3 Elastic Shear Stresses in Thin Walled Closed Sections The shear stress and shear flow in a symmetrical closed section can be obtained directly from Eqs (4.15) and (4.18), respectively. For an unsymmetrical section Eq. (4.19) can be used, but analysis is complicated by the fact that that vo is not known at any point. The problem can be solved by first cutting the section at some point and finding the position of the shear centre in the resulting open section. The shear flow in the closed cross-section then results from the combined action of the applied shear loads transferred to the shear centre of the cut (open) section, and the torque on the closed section due to the transference of loads. Examples of shear stress distribution are given in Fig. 4.14. A similar approach is used to find the position of the shear centre of the closed section.

4.3.4 Elastic Shear Lag (cl 6.2.2.3, EN 1993-1-1 (2005)) The simple theory of bending is based on the assumption that plane sections remain plane after bending. In reality shear strains cause the section to warp. The effect in the flanges is to modify the bending stresses obtained by the simple theory, producing higher stresses near the junction of a web and lower stresses at points remote from it as shown in Fig. 4.15. This effect is described as ‘shear lag’. The discrepancies produced by shear lag are minimal in rolled sections, which have relatively narrow and thick flanges. However in plate girders, or box sections, having wide thin flanges the effects can be significant when they are subjected to high shear forces, especially in the vicinity of concentrated loads where the sudden change in shear force produces highly incompatible warping distortions. Shear lag effects can be allowed for by using effective widths (cl 3, EN 1993-1-5 (2003)).

Shear force

Bending stress non-uniform Warping displacements

FIGURE 4.15 Shear lag effects for an ‘I’ section

58

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Chapter 4 / Laterally Restrained Beams

(a)

(b) Shear centre

(c)

(d)

Centroid

FIGURE 4.16 Position of shear centre

Vz y′ Vy N B

C D

A

Shear centre Centroid

FIGURE 4.17 Shear centre–unsymmetrical section

4.3.5 Elastic Shear Centre Equation (4.19) is only valid if no twisting of the beam occurs at the section considered. Torsion in a section can be generated by a transverse load if the resultants of the shear stresses in the elements of the section produce a torque. To counteract this the line of action of the applied load must pass through the shear centre. In a symmetrical section the shear centre lies on an axis of symmetry, and loads applied along such an axis do not cause twisting. In some sections the position of the shear centre can be inferred from the direction of the shear flow (Fig. 4.16). In (a) the shear center lies at the intersection of the two axes of symmetry and is coincident with the centroid; in (b) and (c) it lies at the intersection of lines of shear flow; in (d), if the flanges are of the same size, the shear stresses in them set up opposing torques about the centroid which is therefore the shear centre. For the general case of an unsymmetrical thin walled open section subject to biaxial bending, with shear forces Vy and Vz , the position of the shear centre can be found by determination of the shear flow from Eqs (4.18) or (4.19), applying Vy and Vz in turn, assuming that they pass through the shear centre. Consider, for example, the section profile in Fig. 4.17. If point B is chosen as the fulcrum it is only necessary to find the resultant shear forces in the leg CD due to Vy and Vz in turn. These forces produce torques equal to Vy z and Vz y , respectively. By taking moments about B the values of

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59

z and y can be obtained. There is no need to calculate the shear stresses in AB or BC because their lines of action pass through the point B, and generate no moment. The resultant shear forces in CD are obtained by integrating the shear stresses obtained by Eq. (4.19) along the leg. The above process is tedious since, for each value of Vy and Vz , the corresponding effective shear forces Vey and Vez must be calculated and applied. If there is an axis of symmetry Eq. (4.18) can be used and the analysis is simplified.

EXAMPLE 4.6

Distribution of shear stresses for an angle section. Calculate the shear stresses in the simply supported angle section shown in Fig. 4.18(a). Iy = 80,44E6, Iz = 14,02E6, Iyz = −19,17E6 mm4 . To the left of mid-span the shear forces are Vy = 5

and

Vz = 10 kN

5

140 20 kN 117,2

s1

97,8

10

10 kN

22,8

y

s2

sp

192,2

6m

290

5

10

an

(a)

(b)

z

78,4 Vy y

2,4 (max)

20,3 0,9

z

155,2

Vy Vz

2,8 (max)

Vz (c)

(d)

FIGURE 4.18 Example: distribution of shear stresses for an unsymmetrical section

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Chapter 4 / Laterally Restrained Beams Effective shear forces from Eq. (4.13) Vey =

(Vy − Vz Iyz /Iy ) = 10,95 kN 1 − Iyz2 /Iy Iz

Vez =

(Vz − Vy Iyz /Iz ) = 24,97 kN 1 − Iyz2 /Iy Iz

Shear flow from Eq. (4.19) s s Vey Vez vs = vo − tz ds − ty ds Iy Iz o o

(i)

For the horizontal leg starting from the left hand end vo = 0,

s = s1 ,

y = (117,2 − s1 )

and

z = −97,8 mm

Substituting these values in (i) and integrating vs1 = 0,00781s21 − 1,224s1

(ii)

This equation shows that vs1 = 0 only when s1 = 0. Differentiating with respect to s1 and equating to zero gives a turning point at s1 = 78,4 mm. Hence from (ii) vs1 (max) = −47,96 N/mm, and s1 = 140 mm, vs1 = −18,28 N/mm. The negative signs indicate that the shear flow is in the opposite direction to s1 . For the vertical leg s = s2 ,

y = −22,8 mm,

z = (s2 − 97,8) mm,

vo = vA = −18,28 N/mm.

Substitution of these values into (i) gives vs2 = −18,28 + 0,9633s2 − 0,003104s22

(iii)

Solving (iii) for s2 shows that when vs2 = 0, s2 = 20,3 mm. There is also a turning point at s2 = 155,2 mm. Hence from (iii) vs2 (max) = 56,46 N/mm. The positive sign indicates that the shear flow is in the direction of s2 . As a check, putting s2 = 290 mm gives vs2 = 0, which is correct. The shear stresses in MPa are obtained by dividing the shear flows by the thickness, that is, 20 mm, and are plotted for the whole section in Fig. 4.18(d). This example demonstrates the method of analysis but the shear stresses are very low and do not justify such a detailed treatment. As a rough guide as to whether detailed analysis is required the shear forces are divided by the area of the appropriate leg. If the shear stress is only required at particular points in the section Eq. (4.15) can be used with the effective shear forces, taking each axis in turn and superimposing the results. Integration is avoided but the directions of the stresses have to be found by inspection. It can be seen from the position of the neutral axis in Example 4.3 that the maximum shear stresses occur where the neutral axis intersects the profile of the

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61

vz

b

A

y

d

y

t Shear centre

FIGURE 4.19 Example: shear centre of a channel

section, as in a symmetrical section. If these points have previously been found, then the maximum shear stress can be calculated directly as above, using Eq. (4.15).

EXAMPLE 4.7 Shear centre for a channel section. Find the position for the shear centre for the channel shown in Fig. 4.19 which has a uniform thickness. As the shear centre lies on the axis of symmetry y–y, there is no need to consider Vy . If point A at the intersection of axis y–y with the centre line of the web is taken as the fulcrum, then only the shear force in the flanges need to be considered since the resultant shear force in the web produces no moment about A. Equation (4.18) is vs =

VAzc I

The distribution of shear flow in the flanges in linear with zero at the ends and maximum at the web centre line. Hence, for maximum shear flow A = bt,

V = Vz ,

vs(max) =

Vz btd 2Iy

zc =

d 2

and

I = Iy , which gives

The resultant shear force is equal to half the maximum shear flow multiplied by the flange width, that is Force =

Vz b2 td 4Iy

The torque about point A from both flanges is equivalent to the torque produced by the applied shear force when it passes through the shear centre, thus V z y =

Vz b2 td2 4Iy

from which y =

b2 d2 t 4Iy

(4.20)

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4.4 ELASTIC TORSIONAL SHEAR STRESSES (CL 6.2.7, EN 1993-1-1 (2005)) 4.4.1 Elastic Torsion Generally torsion is not a major problem in the design of beams except for special cases. Torsional shear stresses arise from a variety of causes including, beams cranked or curved in plan, and loads whose line of action does not pass through the shear centre of the section. It is important to realize that in torsional problems the lever arm for the torque is measured from the shear centre, not the centroid. In particular, the distributed loads on unsymmetrical sections, such as angles or channels, usually act through the centroid of the section, as in the case of self-weight, although generally self-weight does not produce large torsional stresses. However, if other loads are offset from the shear centre the effect can be considerable, for example, for a 381 × 102 channel section the lever arm for loads acting through the centroid is approximately 50 mm. The total torsional moment at any section is the sum of the elastic torsion (St. Venant) plus the internal warping torsion. The warping torsion consists of a bi-moment plus warping.

4.4.2 Uniform and Non-uniform Elastic Torsion In general the cross-sections of members subject to torsion do not remain plane, but tend to warp. Warping is the change in geometric shape of the member so that crosssections do not remain plane as shown in Fig. 4.20. The degree of warping that takes place depends on the shape of the section, and is most pronounced in thin walled channels. In some sections, such as angles and tees, solid and hollow circular sections and square box sections of uniform thickness, warping is virtually non existent, while in others, such as closed box sections of general shape, its effect is small. In ‘I’ sections most of the warping takes place in the flanges and its effect on the web is very small and can be ignored. If torque is applied only at the ends of the member and warping is not restrained, the flanges remain virtually straight and maintain their original shape as shown in Fig. 4.20(a). The result is that the sectional planes of the flanges rotate in opposite directions, producing warping displacements which are constant along the whole length of the member. Under these circumstances the member is said to be in the state of uniform, or St. Venant, torsion. If warping is prevented, for example, by rigid supports at the ends of the member, the torsional stiffness is increased and longitudinal tensile and compressive stresses are induced. In practice warping is not so obvious but may arise from the action of structural connections, or from the incompatibility of warping displacements that occur when the torque is not uniform along the length of the member. Warping restraint increases the torsional stiffness of a member, and at any point along its length the applied torque is resisted by the two components, one due to the St. Venant torsion,

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Mb

(a) tf

H

h Mx H

Mb

(c)

T

(b)

T Stress (d)

Stress (e)

FIGURE 4.20 Torsion of thin walled sections

and the other to warping torsion from the effects of the restraints. The proportions of the two components depend on the type of loading and the distance from a restraint. Both components of torsion produce shear stresses parallel to the walls of the section, and their combined effects can locally be greater than the effect of St. Venant alone. However in members other than channels with very thin walls the increase in shear stresses can usually be ignored in design. In beams the maximum shear stress can be obtained approximately by combining the effects of transverse shears, using Eq. (4.15), with the shear stresses from torsion, assuming that the whole of the applied torque results in St. Venant torsion. A more significant effect of warping restraint in the design of beams is the introduction of longitudinal stresses. The effect is illustrated for an ‘I’ beam in Fig. 4.20(b). In this case the warping displacements are confirmed to the flanges, whose positions, if warping were allowed to occur freely, are shown by dotted outlines. Bi-moments Mb are induced in the planes of the flanges when warping is restrained, and these give rise to tensile and compressive stresses as shown. A full treatment of the analysis of members subject to warping torsion is beyond the scope of this book and the reader is referred to Zbirohowski-Koscia (1967). The analysis is tedious, but for ‘I’ beams a conservative estimate of the longitudinal stresses due to warping torsion can be obtained by assuming that each flange acts independently and is bent in its own plane by an analogous system of lateral loads which replace the applied torques, as in Fig. 4.20(c). The value of the equal and opposite lateral loads H, analogous to the applied torque Mx is H=

Mx h − tf

(4.21)

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Chapter 4 / Laterally Restrained Beams The ends of the flanges can be assumed to be either fixed or simply supported, depending on whether or not warping is restrained by the structural connections. The results obtained by this method are conservative because in reality the warping stresses are produced only by the warping component of the applied torque, not the whole torque as assumed. However, Eq. (4.21) can be useful in preliminary designs where it is necessary to assess whether the effects of torsion are likely to be significant.

4.4.3 Elastic Torsion of Circular Sections The elastic (St. Venant) theory of torsion of prismatic members with solid or hollow sections can be expressed by the well established formula T v Gθ = = It r L

(4.22)

where T is the torsional moment It is the torsion constant which, for a circular section, is equal to the polar second moment of area v is the shear stress at radius r G is the shear modulus θ is the angle of twist L is the length of the member The polar second moment of area for a solid section of radius R is It =

πR4 2

(4.23)

For a thin walled tube of mean radius Rm and wall thickness t, an approximate formula is It = 2πR3m t

(4.24)

The error is below 3% for t/Rm ratios of 1/3 or less, and is on the safe side. The polar second moment of area (It ) is twice the second moment of area (I) about a diameter, values for which are given in Section Tables. The elastic distribution of shear stress along the radius of a solid circular section is linear, with zero at the centre and a maximum at the outside surface. For a thin walled tube, the stress varies linearly across the wall thickness and in the range of standard structural tubes unsafe errors in the shear stress of up to about 18% are introduced by the use of Rm in Eq. (4.22) instead of the outside radius.

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4.4.4 Elastic Torsion of Thin Walled Open Sections The torsional constant It for a thin rectangle of width b and thickness t is It =

bt3 3

(4.25)

This formula is accurate when b/t is infinite and gives unsafe errors of 6% when b/t = 10, and 10% when b/t = 6. The b/t ratios are typical of the flanges of Universal Column and Beam sections. Most sections in steelwork design are composed of thin rectangles, and for a complete section the torsional constant can be obtained by summing the torsional constants for each rectangular element, that is

3 bt It = (4.26) 3 Values of It are given in Section Tables. In standard rolled sections the root fillets at the junctions of the web and the flanges give additional torsional stiffness. The shear stresses in an open section under St. Venant torsion vary from zero on the centre line of the wall to a maximum on the outside surface (Fig. 4.20(d)), and their direction is reversed on each side of the centre line. The shear flow constitutes a closed loop. The maximum stress in any element of thickness t is v=

Tt It

(4.27)

The maximum shear stress in a section therefore occurs in the thickest element. At re-entrant corners the flow lines are crowded together, giving rise to very high stress concentrations. The effect is reduced by fillet radii. The shear stress is zero at the outside corners. The angle of twist θ=

TL GIt

(4.28)

where It is calculated from Eq. (4.26).

4.4.5 Elastic Torsion of Thin Walled Closed Sections The shear stress distribution for closed sections is shown in Fig. 4.20(e). The flow is unidirectional with respect to the profile, contrasting with open sections. Variations in stress across the thickness of the section are ignored. The shear flow is constant at all points on the profile and is given by v=

T 2A

where A is the cross-sectional area of the profile.

(4.29)

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Chapter 4 / Laterally Restrained Beams The shear stress is a maximum in the thinnest part of the section and is obtained by dividing the shear flow by the thickness. This is in direct contrast to the open section where maximum shear stress occurs in the thickest part. Angle of twist TL θ= 4A2 G

ds t

(4.30)

As in open sections additional stresses are introduced when warping is restrained. Equations (4.29) and (4.30) are derived from the Bredt–Batho hypothesis in which it is assumed that the shape of the section remains unchanged. To ensure that this assumption remains valid it may be necessary to stiffen the section with internal diaphragms at intervals along its length, and at points where concentrated loads are applied.

EXAMPLE 4.7 Torsion and transverse shear in a box section. Find the maximum shear stress in the box section shown in Fig. 4.21 which is subject to a torque T = 200 kNm and a shear force V = 500 kN. Assume that the section is adequately stiffened to prevent distortion of the profile and ignore the effects of warping restraints. Calculate the angle of twist per metre length. The torsion and shear may be considered separately and the resulting shear flows are shown in Figs 4.21(a) and (b). For both calculations the profile dimensions shown in Fig. 4.21(c) can be used. The shear modulus is 80 GPa. First consider torsion. The area enclosed by the profile A = 790 × 380 = 0,3002E6 mm2 . From Eq. (4.29) the shear flow v=

T 200E6 = = 333 N/mm 2A 2 × 0,3002E6

The shear stress is maximum in the web where the section is thinnest vt =

333 v = = 33,3 MPa t 10

500 kN

800

790

20 10

T

380

400

T 200 kNm (a) Torsion

(b) Shear

(c) Profile

FIGURE 4.21 Example: torsion and transverse shear in a box section

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From Eq. (4.30) the angle of twist TL ds θ= 2 t 4A G 200E6 × 1E3 790 380 = × 2 + 20 10 (4 × 0,30022 × 1E12 × 80 × 1E3) = 0,00107 rad/m This is small which shows that the box is very stiff torsionally. Now consider the direct shear force. The maximum shear stress is in the webs at the neutral axis, and the first moment of areas 380 = 3,002E6 mm3 2 380 380 = 2 × 10 × × = 0,361E6 mm3 2 4

Azc(flange) = 20 × 790 × Azc(webs)

Total = 3,363E6 mm3 Second moment of area 3803 2 I = 2 10 × + 20 × 790 × 190 = 1232E6 mm4 12 Direct shear stress from Eq. (4.15) vs =

VAzc 500E3 × 3,363E6 = = 68,2 MPa It 1232E6 × 2 × 10

Combing torsional and direct shear stresses v(combined) = vt + vs = 33,3 + 68,2 = 101,5 MPa Information on combining shear and torsion in design calculations is given in cl 6.2.7, EN 1993-1-1 (2005).

4.5 PLASTIC SECTION PROPERTIES AND ANALYSIS 4.5.1 Plastic Section Modulus Plastic global analysis may be used in the design of steel structures for class 1 crosssections (cl 5.6, EN 1993-1-1 (2005)). For the general case of a steel section symmetrical about the plane of bending, the stress distributions in the elastic and fully plastic state are shown in Fig. 4.3(c). For equilibrium of normal forces, the tensile and compressive forces must be equal. In the elastic state, when the bending stress varies from zero at the neutral axis to a maximum at the extreme fibres, this condition is achieved when the neutral axis passes through the centroid of the section. In the fully plastic state,

Chapter 4 / Laterally Restrained Beams b f 420,5

A4/2

z A2 tw 21,5 z2

y

A3 A1

z1

y

tf 36,6

•

h 920,5

68

y3 tw/4

A4/2 y4

z (a)

(b)

(c)

FIGURE 4.22 Example: plastic section modulus for an ‘I’ section

because the stress is uniformly equal to the yield stress, equilibrium is obtained when the neutral axis divides the section into two equal areas. Mpl = (first moment of area about the plastic NA)fy

(4.31)

EXAMPLE 4.8 Plastic section moduli for an ‘I’ section. Determine the plastic section moduli about the y–y and z–z axes for the ‘I’ section shown in Fig. 4.22(a). The section is for a 914 × 419 × 388 kg Universal Beam with the root radius omitted. To determine the plastic section modulus about the y–y axis divide the section into A1 and A2 as shown in Fig. 4.22(b) where h 920,5 A1 = tw = 21,5 = 9895,375 mm2 2 2 A2 = (bf − tw )tf = (420,5 − 21,5)36,6 = 14603,5 mm2 and h 920,5 = = 230,125 mm 4 4 h tf 920,5 − 36,6 z2 = − = = 441,95 mm 2 2 2 z1 =

Plastic section modulus Wply = 2(A1 z1 + A2 z2 ) = 2(9895,375 × 230,125 + 14603,4 × 441,95) = 17,462E6 mm3 The value obtained from Section Tables is 17,657E6 mm3 which is slightly greater because of the additional material at the root radius.

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Similarly for the plastic section modulus about the z–z axis divide the section into areas A3 and A4 as shown is Fig. 4.22(c) where (h − 2tf )tw (920,5 − 2 × 36,6)21,5 = = 9108,475 mm2 2 2 bf 420,5 A4 = 2 tf = 2 36,6 = 15390,3 mm2 2 2 A3 =

and tw 21,5 = = 5,375 mm 4 4 bf 420,5 y4 = = = 105,125 mm 4 4

y3 =

Plastic section modulus Wplz = 2(A3 y3 + A4 y4 ) = 2(9108,475 × 5,375 + 15390,3 × 105,125) = 3,334E6 mm3 . The value obtained from Section Tables is 3,339E6 mm3 which is slightly greater because of the additional material at the root radius. From Section Tables the ratio of plastic/elastic section modulus (shape factor) for this section about the x–x axis is Wply /Wely = 17,657E6/15,616E6 = 1,1307 The value of the shape factor for bending about the y–y axis generally quoted for ‘I’ sections in current use in design is 1,15. The corresponding value of the shape factor for bending about the z–z axis is Wplz /Welz = 3,339E6/2,160E6 = 1,55 which is typical for an ‘I’ section.

EXAMPLE 4.9 Plastic section moduli for a channel section. Determine the plastic section moduli about the y–y and z–z axes for the channel section shown in Fig. 4.23(a). The section is for a 432 × 102 × 65,54 kg channel with the root radius omitted. To determine the plastic section modulus about the y–y axis divide the section into A1 and A2 and shown in Fig. 4.23(b) where h 431,8 A1 = tw = 12,2 = 2633,98 mm2 2 2 A2 = (bf − tw )tf = (101,6 − 12,2)16,8 = 1501,92 mm2 and h 431,8 = = 107,95 mm 4 4 431,8 − 16,8 h tf z2 = − = = 207,5 mm 2 2 2 z1 =

Chapter 4 / Laterally Restrained Beams bf 101,6

A2

A5

yn

y

y

A1

A3

z4

z1

z2

tw 12,2

z5

A4

tf 16,8

•

h 431,8

70

(a)

(b)

(c)

FIGURE 4.23 Example: plastic section modulus for a channel section

Plastic section modulus Wply = 2(A1 z1 + A2 z2 ) = 2(2633,98 × 107,95 + 1501,92 × 207,5) = 1,192E6 mm3 . The value obtained from Section Tables is 1,207E6 mm3 which is slightly different because of the additional material at the root radius and the fact that the flanges taper. Similarly for the plastic section modulus about the z–z axis divide the section into areas A3 , A4 and A5 and shown in Fig. 4.23(c). It is first necessary to determine the position of the neutral axis z–z. Since the axis z–z divides the total area into two equal parts yn =

A1 + A 2 2633,98 + 1501,92 = = 9,578 mm h 431,8

Areas

h 431,8 A3 = yn = 9,578 = 2067,89 mm2 2 2 h 431,8 A4 = − tf (tw − yn ) = − 16,8 (12,2 − 9,578) = 522,04 mm2 2 2 A5 = (bf − yn )tf = (101,6 − 9,578)16,8 = 1545,97 mm2

and yn 9,578 = = 4,789 mm 2 2 tw − yn 12,2 − 9,578 y4 = = = 1,311 mm 2 2 bf − y n 101,6 − 9,578 y5 = = = 46,011 mm 2 2

y3 =

•

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Plastic section modulus Wpz = 2(A3 y3 + A4 y4 + A5 y5 ) = 2(2067,89 × 4,789 + 522,04 × 1,311 + 1545,97 × 46,011) = 0,1634E6 mm3 The value obtained from Section Tables is 0,1531E6 mm3 which is slightly different because of the additional material at the root radius and the taper on the flanges.

4.5.2 Plastic Methods of Analysis A plastic collapse mechanism depends on the formation of a plastic hinge(s) and this will now be considered in detail. The tensile stress–strain curve for mild steel is shown in Fig. 4.3(a). The curve is idealized into three stages namely elastic, plastic, and strain hardening stages of deformation. The moment–curvature for a beam made of the same material is shown in Fig. 4.3(b) with the corresponding distribution of stress at various loading stages shown in Fig. 4.3(c). The spread of the plastic hinge along the length of the beam is shown in Fig. 4.3(d). The amount of rotation that can take place at a plastic hinge is determined by the length of the yield plateau shown in Fig. 4.3(b). For mild steel the length is considerable and the work hardening stage is ignored. For higher grade steels work hardening occurs immediately after yielding and there is no plateau. A plastic hinge does form but with an increasing moment of resistance. In design this increase in resistance due to work hardening is ignored which errs on the side of safety. Plastic methods of analysis and design consider a structure at collapse when sufficient plastic hinges have formed to produce a mechanism. Examples of collapse mechanisms are shown in Fig. 4.24. In simple situations, are shown in Fig. 4.24(a), the position of the plastic hinge is obvious and it is simple to calculate the collapse load using the method of virtual work. q/unit length

L

Pin

θ

q/unit length

L

Pin

Plastic hinge

Plastic hinge (a)

(b)

FIGURE 4.24 Examples of plastic collapse mechanisms

θ1 x

θ2 (c)

Pin

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Chapter 4 / Laterally Restrained Beams The method must only be applied to structures where the material becomes plastic at the yield stress and is capable of accommodating large plastic deformations. It must therefore not be applied to brittle materials such as cast iron. However it can be applied to reinforced concrete because the steel reinforcement behaves plastically at collapse but care must taken to check the rotation at the hinge for heavily reinforced sections. The plastic method can be seen as a more rational method for design because all parts of the structure can be given the same safety factor against collapse. In contrast for elastic methods the safety factor varies. Intrinsically the plastic method of analysis is simpler than the elastic method because there is no need to satisfy elastic strain compatibilty conditions. However calculations for instability and elastic deflections require careful consideration when using the plastic method, but nevertheless it is very popular for the design of some structures (e.g. beams and portal frames). The method of analysis demonstrated in this chapter is based on the principle of virtual work. This states that if a structure, which is equilibrium, is given a set of small displacements then the work done by the external loads on the external displacements is equal to the work done by the internal forces on the internal displacements. More concisely, external work equals internal work. The displacements need not be real, they can be arbitrary, which explains the use of the word ‘virtual’. However the external and internal geometry must be compatible. It is tacitly assumed that collapse is due to the formation of plastic hinges at certain locations and that other possible causes of failure, for example, local or general instability, axial or shear forces, are prevented from occurring. It is also important to understand that at collapse: (a) the structure is in equilibrium, that is, the forces and moments, externally and internally, balance, (b) no bending moment exceeds the plastic moment of resistance of a member, (c) there are sufficient hinges to form a collapse mechanism. These three conditions lead to three theorems for plastic analysis. (1) Lower bound theorem: if only conditions (a) and (b) are satisfied then the solution is less than or equal to the collapse load. (2) Upper bound theorem: if only conditions (a) and (c) are satisfied then solution is greater than or equal to the collapse load. (3) Uniqueness theorem: if conditions (a), (b) and (c) are satisfied then the solution is equal to the collapse load. Settlement of the supports has no effect on the solution at collapse because the only effect is to change the amount of rotation required. This is in contrast to elastic methods of analysis where settlement calculations must be included. Plastic hinges form in a member at the maximum bending moment. However at the intersection of two members, where the bending moment is the same, the hinge forms

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in the weaker member. Generally the locations of hinges are at restrained ends, intersection of members and at point loads. The hinges may not form simultaneously as loading increases but this is not important for calculating the final collapse load. Generally the number of plastic hinges n=r+1 where r is the number of redundancies. However there are exceptions, for example, partial collapse of a beam in a structure. Equating external to internal work for the member shown in Fig. 4.24(a) L wL θ = Mpl (2θ) hence Mpl = wL2/8 4 In other conditions, as shown in Fig. 4.24(c), the position of the plastic hinge for minimum collapse load is not obvious and the calculations are more complicated. Equating external to internal work for the member shown in Fig. 4.24(c). xθ1 θ2 qx (i) + q(L − x)2 = Mpl (2θ1 + θ2 ) 2 2 From geometry xθ1 = (L − x)θ2 Combining (i) and (ii) 2 1 qL = 2Mpl − x L−x

(ii)

(iii)

Differentiating (iii) to determine the value of x for which q is a minimum √ x2 − 4Lx + 2L2 = 0 hence x = L(2 − 2)

(iv)

Combining (iii) and (iv) √ Mpl = (1,5 − 2)qL2

(v)

The method can be applied to a variety of structures. Further explanation and examples are given in Moy (1981), and Croxton and Martin Vol. 2 (1987 and 1989).

4.6 EFFECT OF A SHEAR FORCE ON THE PLASTIC MOMENT OF RESISTANCE (CLS 5.6 AND 6.2.6, EN 1993-1-1 (2005)) In general the effect of a transverse shear force is to reduce the plastic moment of resistance but the reduction for an ‘I’ section is small and may be ignored (cl 6.2.8, EN 1993-1-1 (2005)) if VEd ≤ 0,5 Vpl,Rd

(4.32)

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Chapter 4 / Laterally Restrained Beams where VEd is the design shear force, and the plastic shear resistance and (Eq. (6.18), EN 1993-1-1 (2005))

Av fy /31/2 Vpl,Rd = (4.33) γM0 The areas resisting shear (Av ) for various sections are given in cl 6.2.6(3), EN 1993-1-1 (2005) and in Section Tables. The area resisting shear for an ‘I’ section is Av = A − 2btf + (tw + 2r)tf

(4.34)

This is a slight increase on the web area (htw ) which has been used in the past. The maximum shear stress of fy /31/2 is based on the failure criterion expressed in Eq. (2.5). There is no reduction in the plastic moment of resistance for plastic or compact sections provided that the design shear force does not exceed 50% of the plastic shear resistance. This recommendation is related to the work of Morris and Randall (1979) who stated that shear can be ignored unless the average shear stress in the web exceeds fy /3, or fy /4 when the ratio of the overall depth to flange width (h/bf ratio) exceeds 2.5. Where the design shear force exceeds 50% of the plastic shear resistance the European Code recommends that the yield strength is reduced to (1 − ρ)fy where for shear (Eq. (6.29), EN 1993-1-1 (2005)) 2 VEd ρ= 2 −1 (4.35) Vpl,Rd and for torsion 2 VEd ρ= 2 −1 Vpl,T,Rd

(4.35a)

These recommendations may be compared with theory by Horne (1971). If a shear force is applied to an ‘I’ section most of the shear force is resisted by the web. If it is assumed that all of the shear force is resisted by the web the effect on the plastic moment of resistance is shown in Fig. 4.25(a). The outer fibres in bending are at yield while the inner fibres are linear elastic. It is not possible to resist shear forces on the outer fibres and thus the inner fibres resist all of the shear force. Plastic moment of resistance of the web dp2 dp2 dp2 d2 d2 Mplw = tw − tw + tw fy = tw − tw fy 4 4 6 4 12

(i)

From the parabolic distribution of shear stress V =

2 tw dp τy 3

(ii)

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fy

tw

d

h

Parabolic distribution of shear stress τy

dp

(a) 0 V/Vmax 2/3 fy τ σ

τy

(b) 2/3 V/Vmax 1 1,0

Factor

0,8

Eq. (4.36)

0,6 Eq. (4.37) 0,4

0,2

0

0,2

0,4 0,6 V /Vmax

0,8

1,0

(c)

FIGURE 4.25 Effect of a shear force on the plastic section modulus of the web of an ‘I’ section

Defining Vmax = tw hτy Combining (i) to (iii) 2 tw d2 fy 3 V Mplw = 1− 4 4 Vmax This expression is valid for 0 ≤ V/Vmax ≤ 2/3.

(iii)

(4.36)

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Chapter 4 / Laterally Restrained Beams When 2/3 ≤ V/Vmax ≤ 1 then assuming the stress distributions shown in Fig. 4.25(b) the plastic moment of resistance of the web tw d2 σ 6 Applied shear force Mplw =

(i)

2 V = dtw τ + dtw (τy − τ) 3 If Vmax = dtw τy , ratio 1 τ V/Vmax = 2+ 3 τy

(ii)

(iii)

Adopting a failure criterion of the form shown Eq. (2.5) 2 2 σ τ + =1 fy τy

(iv)

Combining (i) to (iv) Mplw

⎤ ⎡ 2 tw d2 fy 2 3V ×⎣ = 1− −2 ⎦ 4 3 Vmax

(4.37)

The expressions in the square brackets in Eqs (4.36) and (4.37) are plotted in Fig. 4.25(c) expressed as Mplw = factor(tw d2 fy /4). This theory is approximate and conservative but it does give a general appreciation of the effect of a shear force on the plastic moment of resistance. A fuller description and less conservative theories are given in Horne (1958). In addition when the ratio hw /tw > 72ε/η (cl 6.2.6, EN 1993-1-1 (2005)) the web should be checked for shear buckling. The limiting value of hw /tw is related to experimental work by Longbottom and Heyman (1956), and later work by Horne (1958).

EXAMPLE 4.10

Plastic section modulus reduced by shear. Determine the plastic modulus for a 762 × 267 × 197 kg UB grade S275, using (a) Horne’s method, (b) European Code method, when subject to a design value of shear force VEd = 1145 kN. Ignore the material factor for this example. (a) Horne’s method: VEd VEd 1145E3 = = = 0,6 1/2 Vmax htw fy /3 769,6 × 15,6 × 275/31/2 According to Horne, dividing Eq. (4.36) by fy Wpl(web)

td2 3 VEd 2 = 1− 4 4 Vmax = 0,73(td2/4),

that is 27% loss in the web.

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Reduced plastic section modulus of the complete section Wpl(reduced) = Wpl(whole) − Wpl(web) (1 − factor) 15,6 × 685,82 = 7,164E6 − (1 − 0,73) = 6,672E6 mm3 4 Percentage reduction in plastic modulus for the whole section =

100(7,167E6 − 6,672E6) = 6,91% 7,167E6

(b) EN 1993-1-1 (2005) method: VEd VEd = Vpl,Rd (Av fy /31/2 ) =

1145E3 = 0,568 (12,7E3 × 275/31/2 )

where Av = 12,7E3 is obtained from Section Tables. According to cl 6.29, EN 1993-1-1 (2005) there is a reduction in the plastic modulus if VEd > 0,5 Vpl,Rd From Eq. (4.35a) 2 VEd ρ= 2 − 1 = (2 × 0,568 − 1)2 = 0,0185 Vpl,Rd and Wpl(reduced) =

Mv,Rd ρA2v = Wpl − fy 4tw

= 7,164E6 − 0,0185 ×

12,7E32 4 × 15,6

= 7,116E6 mm3 (7,164E6 − 7,116E6) × 100 Percentage reduction = = 0,67%. 7,164E6

4.7 LATERAL RESTRAINT (CL 6.3.5, EN 1993-1-1 (2005)) The full rotation required at a plastic hinge in a beam may not be realized unless lateral support is provided at the hinge position. It may also be necessary to provide lateral support at other points along the span to ensure that lateral torsional buckling does not occur. Lateral torsional buckling is considered to be prevented if the compression flange is prevented from moving laterally, either by an intersecting member, or by frictional restraint from intersecting floor units.

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4.8 RESISTANCE OF BEAMS TO TRANSVERSE FORCES The first check for transverse forces is the shear stress in the web at the neutral axis in the elastic stage of behaviour (cl 6.2.6(4), EN 1993-1-1 (2005)). In the plastic stage of behaviour the strength of the web is determined using values given in cl 6.2.6(2), EN 1993-1-1 (2005). Also it is necessary to consider web plate shear buckling at the ultimate limit state by checking if 72ε/η > hw /tw (cl 5.1(2), EN 1993-1-5(2003)). Design calculations are also required for concentrated transverse forces applied to girders from supports, cross beams, columns, etc. (Fig. 4.26). The concentrated loads are dispersed through plates, angles and flanges to the web of the supporting girder. The deformations that occur to the supporting beam are shown in Fig. 4.27 and include yielding of the flange and local buckling of the web as shown in experiments (Astill et al. (1980)). The design resistance is expressed simply as (cl 6.2, Eq. (6.1), EN 1993-1-5(2003)) FRd =

fyw Leff tw γM1

The effective bearing length (Leff ) is an extension of the stiff bearing length (ss ) which assumes a 45◦ dispersion through plates, flanges and angles as shown in Fig. 4.26. The root radius of a section increases the length of the stiff bearing by (2 − 21/2 )r.

Clearance

ss

ss = h + 2t p

h tp

Ss

‘I’ beam

45° 45°

ss = 2ta + (2 − √2 )ra − Clearance

ra

‘I’ beam

Angle support ta (a)

(b)

‘I’ beam

tf

ss 45°

ss = 2t f + t w + 2(2 − √2 )rb

rb ‘I’ beam tw (c)

FIGURE 4.26 Stiff bearing lengths

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tf

Yielding of flange

ss

Buckling of web

Leff

tw

FIGURE 4.27 Transverse concentrated load The extension of the stiff bearing length is based on theory of flange yielding and buckling of the web as shown in Examples 4.11 and 4.12. If the transverse resistance of an unstiffened web is insufficient stiffeners are required (cl 9.4(1), EN 1993-1-5 (2003)) designed according to cl 6.3.3 or 6.3.4, EN 1993-1-1 (2005) with a buckling length of not less than 0,75hw and using buckling curve c (Fig. 6.4, EN 1993-1-1 (2005)).

EXAMPLE 4.11 Simply supported beam carrying a uniformly distributed load and laterally restrained. The floor of an office building consists of 125 mm precast concrete units, with a mass of 205 kg/m2 , topped with a 40 mm concrete screed and 20 mm wood blocks. Lightweight partitions supported by the floor are equivalent to a superficial load of 1,0 kN/m2 and the suspended ceiling has a mass of 40 kg/m2 . The floor rests on the top flanges of simply supported steel beams spanning 8 m and at a pitch of 3,75 m. Characteristic loads Dead load Self-weight of steel beam (assumed) 125 mm precast units 40 mm concrete screed 2400 × 0,04 20 mm wood blocks 900 × 0,02 Suspended ceiling

kg/m2

20 205 96 18 40 379 × 9,81/1E3 =

kN/m2

Lightweight partitions

3,72 1,00

Total dead load Imposed load for an office (EN 1991-1-1 (2002))

4,72 2,50

Maximum design bending moment at mid-span

kNm

Permanent load BM = γG QL/8 = 1,35 × 4,72 × 82 × 3,75/8 = Variable load BM = γQ QL/8 = 1,5 × 2,5 × 82 × 3,75/8 =

191,2 112,5

Total BM

303,7

80

•

Chapter 4 / Laterally Restrained Beams Using grade S275 steel, fy = 275 MPa, the plastic section modulus required Wply =

M 303,7E6 = 1,215E6 mm3 . = fy /γM 275/1,1

From Table 4.1 for L/250, L/h = 22, hence h = 8E3/22 = 364 mm. From Section Tables, try 457 × 152 × 60 kg UB, Wply = 1,284E6 mm3 , tf = 13,3 mm, tw = 8 mm, d = 407,7 mm, Av = 3890 mm2 , bf = 152,9 mm, h = 454,7 mm, hw = (h − 2tf ) = (454,7 − 2 × 13,3) = 428,1 mm, r = 10,2 mm. Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) d 407,8 = = 50,96 tw 8 235 1/2 235 1/2 72ε = 72 = 72 = 66,6 > 50,96 satisfactory. fy 275 Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(152,9 − 8)/2 − 10,2] = = = 4,67 tf tf 13,3 235 1/2 235 1/2 9ε = 9 × =9 = 8,32 > 4,67 satisfactory. fy 275 This is a Class 1 section and calculations can be reduced by using Section Tables. Check deflection at the service limit state Permanent and variable load on span = 2,5 × 8 × 3,75 = 75 kN 5QL3 384EI 5 × 75 × (8E3)3 = 9,35 mm = 384 × 210 × 254,64E6

Maximum deflection = δ1 + δ2 =

Deflection limit from Table 4.1 δmax =

L 8E3 = = 32 > 9,35 mm satisfactory. 250 250

Check shear resistance of web at the ultimate limit state. Design shear force at the support VEd =

QL (1,35 × 4,72 + 1,5 × 2,5)8 × 3,75 = = 151,8 kN. 2 2

Area of web (cl 6.2.6(3), EN 1993-1-1 (2005)) Av = A − 2btf + (tw + 2r)tf = 7580 − 2 × 152,9 × 13,3 + (8 + 2 × 10,2)13,3 = 3890 mm2 , or obtain value from Section Tables.

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Design plastic shear resistance (cl 6.2.6(1), EN 1993-1-1 (2005)) Vpl,Rd =

Av τy Av (fy /31/2 ) = γM0 γM0

= 3890 ×

(275/31/2 ) = 561,5 kN 1,1 × 1E3

151,8 VEd = = 0,27 < 1,0 satisfactory. Vpl,Rd 561,5 Check web plate buckling from shear at the ultimate limit state (cl 5.1 (2), EN 1993-1-5 (2003)) 72ε 72(235/275)1/2 = = 55,5 η 1, 2 hw 428,1 = 53,5 < 55,5 satisfactory. = tw 8 Assuming 150 × 75 × 10 mm angle supports at the ends of the beam (Fig. 4.26) the transverse shear buckling load (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)) FRd =

fyw Leff tw 275 × 77,53 × 8 = γM1 1,0 × 1E3

= 170,6 > VEd = 151,8 kN satisfactory no stiffener required. The effective length (Leff ) in the previous equation is obtained as follows: The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), for a 150 × 75 × 10 mm angle support (Fig. 4.26) ss = 2ta + (2 − 21/2 )ra − clearance = 2 × 10 + (2 − 21/2 ) × 11 − 3 = 23,44 < hw = 428,1 mm Buckling co-efficient for the load application (cl 6.1(4), Type(c), EN 1993-1-5 (2003)) kF = 2 +

6(ss + c) 6(23,44 + 0) =2+ = 2,33 < 6 satisfactory. hw 428,1

Effective load length (cl 6.5, Eq. (6.13), EN 1993-1-5 (2003)) le =

kF Etw2 2fyw hw

2,33 × 210E3 × 82 2 × 275 × 428,1 = 133,0 > (ss + c = 23,44 + 0) use 23,44 mm. =

Force (cl 6.4 Eq. (6.5) EN 1993-1-5 (2003)) Fcr =

0,9 kF Etw3 0,9 × 2,33 × 210E3 × 83 = 526,7 kN. = hw (428,1 × 1E3)

82

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Chapter 4 / Laterally Restrained Beams The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003)) m1 =

fyf bf 275 × 152,9 = 19,11 = fyw tw 275 × 8

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003)) m2 = 0,02

hw tf

2

= 0,02

428,1 13,3

2 = 20,72

Yield length (cl 6.5, Eq. (6.12), EN 1993-1-5 (2003)) ly = le + tf (m1 + m2 )1/2 = 23,44 + 13,3(19,11 + 20,72)1/2 = 107,4 mm An alternative yield length (cl 6.5, Eq. (6.11), EN 1993-1-5 (2003)) 1/2 2 m1 le ly = le + tf + + m2 2 tf 1/2 19,11 23,44 2 = 23,44 + 13,3 + + 20,72 2 13,3 = 100,3 mm use minimum. An alternative yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003)) ly = ss + 2tf [1 + (m1 + m2 )1/2 ] = 23,44 + 2 × 13,3[1 + (19,11 + 20,72)1/2 ] = 217,9 mm. Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003)) ly tw fyw 1/2 100,3 × 8 × 275 1/2 λF = = = 0,647 Fcr 526,7E3 Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003)) χF =

0,5 0,5 = = 0,773 0,647 λF

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003)) Leff = χF ly = 0,773 × 100,3 = 77,53 < hw = 428,1 mm. Check self-weight of steel beam = 60/3,75 = 16 < 20 kg/m2 assumed in loading calculations, acceptable.

EXAMPLE 4.12 Support for a conveyor. Part of the support for a conveyor consists of a pair of identical beams as shown in Fig. 4.28. Each beam is connected to a stanchion at end A by a cleat and is supported on a cross beam at D by bolting through the connecting flanges. Lateral restraint is provided by transverse beams at A, B and E connected to rigid supports. The loads shown are at the ultimate limit state.

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•

83

450 kN

15

203 × 203 × 46 kg UC 225 kN

A

B

C

D

150 × 75 × 10 mm angle

2m

150 kN

533 × 210 × 101 kg UB

E

610 × 229 × 140 kg UB

3m

3m

2m 300

Bending moment (kNm) 600 825 300 75

150

375

Shear force (kN)

FIGURE 4.28 Example: support for a conveyor

Assuming pin joints at A and D. Reactions are determined by taking moments about A, then about D. 2 × 225 + 5 × 450 + 10 × 150 = 525 kN 8 6 × 225 + 3 × 450 − 2 × 150 RA = = 300 kN 8 RD =

Important bending moments are MB = RA × 2 = 300 × 2 = 600 kNm MC = RA × 5 − 225 × 3 = 300 × 5 − 225 × 3 = 825 kNm MD = −150 × 2 = −300 kNm The shear force and bending moment diagrams at the ultimate limit state are shown in Fig. 4.28.

84

•

Chapter 4 / Laterally Restrained Beams Using Grade S355 steel with a characteristic strength fy = 355 MPa. The plastic section modulus required Wply =

Mmax 825E6 = 2,56E6 mm3 . = fy /γM0 355/1,1

Deflection limits (Table 4.1), L/250, L/h = 17, hence h = 8E3/17 = 470 mm. From Section Tables try 533 × 210 × 101 kg UB, a Class 1 section, Wply = 2,619E6 mm3 , tf = 17,4 mm, tw = 10,9 mm, Av = 6250 mm2 , d = 476,5 mm. Check deflections at service load Using area and area moment methods the deflections due to the imposed loads at service loads are: mid-span in ABCD 11,3 mm (downwards) end of cantilever DE −6,5 mm (upwards) Deflection limits (Table 4.1) L 8E3 = = 32 > 11,3 mm satisfactory. 250 250 2L 2 × 2E3 span DE = = = 16 > 6,5 mm satisfactory. 250 250

span ABCD =

Check web shear resistance between C and D at the ultimate limit state. Design shear force VEd = 375 kN. Plastic shear resistance Vpl,Rd =

Av ( fy /31/2 ) γM

6240 × (355/31/2 ) 1,1 × 1E3 = 1163 > 375 kN satisfactory. =

Shear area (cl 6.2.6, EN 1993-1-1 (2005)) Av = A − 2btf + (tw + 2r)tf = 129,3 × 100 − 2 × 210,1 × 17,4 + (10,9 + 2 × 12,7) × 17,4 = 6250 mm2 or obtain a value from Section Tables. VEd /Vpl,Rd = 375/1163 = 0,322 < 0,5 therefore no reduction in the plastic section modulus (cl 6.2.8, EN 1993-1-1 (2005)). Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) c 476,5 = = 43,7 tw 10,9 235 1/2 235 1/2 72ε = 72 = 72 = 58,6 > 43,7 satisfactory. fy 355

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Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(210,1 − 10,9)/2 − 12,7] = = = 4,99 tf tf 17,4 235 1/2 235 1/2 9ε = 9 × =9× = 7,32 > 4,99 satisfactory. fy 355 This is a Class 1 section and calculations can be reduced by using Section Tables. Assuming a 150 × 75 × 10 mm angle support at A at the end of the beam (Fig. 4.26) the transverse shear buckling strength (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)) is determined as shown previously in Example 4.11. At D (Fig. 4.28) two UBs intersect which have the following dimensions. From Section Tables the 533 × 210 × 101 kg upper load carrying beam ABCDE. h = 536,7 mm, bf = 210,1 mm, hw = (h − 2tf ) = (536,7 − 2 × 17,4) = 501,9 mm, tf =17,4 mm, tw = 10,9 mm, r = 12,7 mm. From Section Tables the 610 × 229 × 140 kg lower support beam at D. h = 617 mm, bf = 230,1 mm, hw = (h − 2tf ) = (617 − 2 × 22,1) = 572,8 mm, 22,1 mm, tw =13,1 mm, r = 12,7 mm.

tf =

The transverse shear buckling strength of the upper 533 × 210 × 101 kg load carrying beam (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)). FRd =

fyw Leff tw 355 × 501,9 × 10,9 = γM1 1,0 × 1E3

= 1942 > RD = 525 kN satisfactory no stiffener required. The previous calculation for FRd includes the effective bearing length (Leff ) which is calculated as follows: The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), provided by the flange and web of the lower 610 × 229 × 140 UB ss = 2tf + tw + (2 − 21/2 )rb = 2 × 22,1 + 13,1 + (2 − 21/2 ) × 12,7 = 64,74 < hw = 501,9 mm Buckling co-efficient for the load application (cl 6.1(4), Type (a), EN 1993-1-5 (2003)) assuming a is large kF = 6 + 2

hw a

2 =6

Force (cl 6.4, Eq. (6.5), EN 1993-1-5 (2003)) Fcr =

0,9 kF Etw3 0,9 × 6 × 210E3 × 10,93 = = 2926 kN hw 501,9 × 1E3

86

•

Chapter 4 / Laterally Restrained Beams The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003)) m1 =

fyf bf 355 × 210,1 = 19,28 = fyw tw 355 × 10,9

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003)) m2 = 0,02

hw tf

2

= 0,02

501,9 17,4

2 = 16,64

Yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003)) ly = ss + 2tf [1 + (m1 + m2 )1/2 ] = 64,74 + 2 × 17,4[1 + (19,28 + 16,64)1/2 ] = 308,1 mm Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003)) λF =

ly tw fyw Fcr

1/2

=

308,1 × 10,9 × 355 2926E3

1/2 = 0,638

Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003)) χF =

0,5 0,5 = = 0,784 < 1 0,638 λF

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003)) Leff = χF ly = 0,784 × 308,1 = 241,6 < hw = 501,9 mm Similar calculations are required for the web buckling strength of the upper 533 × 210 × 101 kg UB at B, C and E. Stiffeners are not required for the 533 × 210 × 101 kg UB but in the past many designers have inserted them where large point loads are applied. Assuming that one is required at D(RD = 525 kN) the following calculations are required. Assume a symmetrical stiffener of 10 mm thickness welded to the web and the flanges. The non-dimensional slenderness ratio of the stiffener (Eq. (6.50), EN 1993-1-1 (2005)) λz = =

Afy Ncr

1/2 =

(Lcr /iy ) × 1 λ1

(0,75 × 572,8/15E3) × 1 = 0,375E-3 [93,9 × (235/355)1/2 ]

assuming a stiffener on both sides of the web, breadth of stiffener bs = bf − tw − 2 (weld leg) = 210,1 − 10,9 − 2 × 6 = 187,2 (use 180) mm iz = bs ts3 /12 = 180 × 103/12 = 15E3 mm4

Structural Design of Steelwork to EN 1993 and EN 1994 Q 85 kN

q = 42,5 kN/m A

B

C

E 3m

3m

478,1

478,1

Bending moment (kNm)

441

441

3m

87

Q 85 kN

D

11 m

•

Collapse of span AB

Collapse of span BC 329,4 201,9

190,3 116,9

Shear force (kN) 10,6 95,6 223,1

277,2

FIGURE 4.29 Example: two span beam

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) with λz = 0,375E-3 the reduction factor χz = 1,0 and the design buckling resistance χz Afy 1,0 × 180 × 10 × 355 = λM 1,1 × 1E3 = 581 > RD = 525 kN satisfactory.

Nb,Rd =

The design resistance of four vertical 6 mm fillet welds connecting the stiffener to both sides of the web (cl 4.5.3.3(3), Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = 4hw fu /(31/2 βγM2 ) × 0,7a = 4 × 572,8 × 510/(31/2 × 0,9 × 1,25) × 0,7 × 6/1E3 = 2519 > RD = 525 kN satisfactory.

EXAMPLE 4.13 Two span beam. Determine the size of Universal Beam required to support the design loads at the ultimate limit state as shown in Fig. 4.29. Assume that the compression flange is fully restrained and that lateral torsional buckling does not occur.

88

•

Chapter 4 / Laterally Restrained Beams Plastic analysis of the beam produces the following. Collapse of span AB considered as a propped cantilever √ √ Mpl = 1,5 − 2 qL2 = 1,5 − 2 42,5 × 112 = 441 kNm. Collapse of span BC with plastic hinges assumed to be at E and B (Fig. 4.29) External work = internal work 2L 2L θ1 2L L L L Q θ1 + Q +q θ1 + q θ2 3 3 2 3 3 3 6 = Mpl (θ1 + θ1 + θ2 ) and from geometry 2L L θ1 = θ2 3 3

hence θ2 = 2θ1

(i)

(ii)

Combining eqs (i) and (ii) and rearranging QL qL2 85 × 9 42,5 × 92 + = + = 478,1 kNm 4 12 4 12

Mpl =

This value of Mpl is the greater than the value for span AB and therefore is used to determine the size of a section which is continuous for two spans. The assumption that a plastic hinge is at E is not correct, it actually occurs at 3,25 m from C and Mpl = 479,1 kNm. However the error is small and is ignored. Used Grade S355 steel and assuming a characteristic strength fy = 355 MPa the plastic section modulus required Wply =

Mmax 478,1E6 = 1,347E6 mm3 = fy /γM 355/1,0

From Section Tables try 457 × 191 × 67 kg UB which is a Class 1 section bending about the y–y axis, Wply = 1,471E6 mm3 , tf = 12,7 mm, tw = 8,5 mm, Av = 4100 mm2 , d = 407,8 mm, bf = 189,9 mm. Check deflections at service load. Using area and area moment methods the deflections due to the imposed load are 20,7 mm for AB and 25 mm for BC. Deflection limit (Table 4.1) AB is

L 11E3 = = 44 > 20,7 mm, satisfactory 250 250

BC is

L 9E3 = = 36 > 25 mm, satisfactory. 250 250

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Sketch the shear force diagram (Fig. 4.29) at the ultimate limit state for the collapse of span BC. For EC

E + 478,1 + 42,5 × 3 × 1,5 − 3RC = 0

hence RC = 223,1 kN

For AB

B + 478,1 −

42,5 × 11 × 11 + 11RA = 0 2

hence RA = 190,3 kN

For ABC ↑ + RA + RB + RC −

qL −

Q=0

+190,3 + RB + 223,1 − 42,5 × 20 − 2 × 85 = 0

hence RB = 606,6 kN

Check web shear resistance at B at the ultimate limit state. Design shear force VEd = 329,4 kN (Fig. 4.29) Design plastic shear resistance Vpl,Rd =

Av τy Av (fy /31/2 ) 4100 × (355/31/2 ) = = γM γM 1,0 × 1E3

= 840 > 329,4 kN satisfactory. VEd /Vpl,Rd = 329,4/840 = 0,392 < 0,5 therefore no reduction in the plastic section modulus (cl 6.2.8, EN 1993-1-1 (2005)). Check for buckling of the web (Table 5.2 (sheet 1), EN 1993-1-1 (2005)) c 407,9 = 48,0 = tw 8,5 235 1/2 235 1/2 72ε = 72 = 72 = 58,6 > 48,0 satisfactory. fy 355 Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005)) c [(B − tw )/2 − r] [(189,9 − 8,5)/2 − 10,2] = = = 6,33 tf tf 12,7 235 1/2 235 1/2 9ε = 9 × =9× = 7,32 > 6,33 satisfactory. fy 355 This is a Class 1 section and calculations can be reduced by using Section Tables.

90

•

Chapter 4 / Laterally Restrained Beams A continuous 457 × 191 × 67 kg UB can be used for both spans. Alternatively if a minimum weight design is required then: (a) A smaller section could be used for span AB but there would have to a splice which would increase fabrication costs. (b) A smaller section could be used for span AB, with flange plates added to increase the moment of resistance for span BC, but again this would increase fabrication costs. Calculations, similar to Example 4.12, are required for the web buckling strength of the 457 × 191 × 67 kg UB at A, B, and C.

REFERENCES Astill, A.W., Holmes, M. and Martin, L.H. (1980). Web buckling of steel ‘I’ beams. CIRIA Tech. note 102. Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving problems in structures Vols. 1 and 2. Longman Scientific and Technical. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-5 (2003). Plated structural elements. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Horne, M.R. (1971). Plastic theory of structures. Nelson. Horne, M.R. (1958). The full plastic moment of sections subjected to shear force and axial loads, British Welding Journal, 5, 170. Longbottom, E. and Heyman, J. (1956). Experimental verification of the strength of plate girders designed in accordance with the revised BS153: tests on full size and on model plate girders, Proceedings of the ICE, 5(III), 462. Megson, T.H.G. (1980). Strength of materials for civil engineers. Nelson. Morris L.J. and Randall A.L. (1979) Plastic Design, Constrado. Moy, S.S. (1981). Plastic methods for steel and concrete structures. Macmillan Press. Zbirohowski-Koscia, K. (1967). Thin walled beams. crosby Lockwood.

Chapter

5 / Laterally Unrestrained Beams

In the previous chapter it was assumed that the compression flange of a beam was fully restrained, that is, the flange is unable to move laterally under the effect of loads or actions. In practice, this is rarely the case and although compression flanges may be restrained at discrete points the flange is still capable of buckling between restraints. Such buckling reduces the moment capacity of the member.

5.1 LATERAL TORSIONAL BUCKLING OF ROLLED SECTIONS SYMMETRIC ABOUT BOTH AXES 5.1.1 Basic Theory A beam under the action of flexure alone due to the application of point moments producing single curvature is considered as the basic case. The supports allow rotation about a vertical axis but do not allow relative displacement of the top and bottom flanges, that is, twisting is not allowed (Fig. 5.1). The governing equation for flexure about the minor axis is EIz

d2 v = −Mφ dx2

(5.1)

where EI z is the flexural rigidity about the minor axis. The governing equation for torsion is given by GIt

dφ d3 φ dv − EIw 3 = M dx dx dx

(5.2)

where GI t is the torsional rigidity, EI w the warping rigidity and the final term is the disturbing torque. Eliminate the term dv/dx between Eqs (5.1) and (5.2) to give GIt

d2 φ d4 φ M2 − EI = − φ w EIz dx2 dx4

(5.3)

•

Chapter 5 / Laterally Unrestrained Beams M Pins

Pins M

L (a) Elevation of beam loaded in single curvature

Unbuckled position

Buckled position (b) Support arrangement with f =

d2f dx 2

=0

L dv dx

x

V

92

cL of unbuckled

Beam cL after buckling (c) Plan view of buckled beam

V

Original position Buckled position

f

(d) Section

FIGURE 5.1 Lateral torsional buckling of beams

beam

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93

The solution to Eq. (5.3) is given by the first term of a sine series φ = φ0 sin

πx L

(5.4)

where φ0 is the angle of twist at mid-span. Substituting Eq. (5.4) into Eq. (5.3) with the boundary condition that φ = 0 at x = L gives the elastic critical moment Mcr as π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 (5.5)

or 1/2 √ π EIz GIt π2 EIw Mcr = 1+ 2 L L GIt

(5.6)

The full derivation of the elastic critical moment is given in Kirby and Nethercot (1979) or Trahair and Bradford (1988). In practice, real beams do not achieve full elastic buckling except at very high slenderness ratios and are also subject to the limit imposed by the section plastic capacity. Reductions in elastic buckling capacity are caused by the existence of residual stresses and any lack of initial straightness. The residual stresses within the section arise in the case of rolled sections due to differential rates of cooling in the web and flanges after hot rolling, and of plate girders due to both the preparation of the web and flange plates and the welding procedure (Nethercot, 1974a). These residual stresses do not affect the plastic capacity (Nethercot, 1974b; Kirby and Nethercot, 1979). The value of the critical moment given by Eq. (5.5) or (5.6) takes no account of major axis flexure. Trahair and Bradford (1988) indicate the critical moment obtained from Eq. (5.5) or (5.6) should be divided by a factor K where K is given by K=

1−

EIz EIy

1−

GIt π2 EIw 1+ EIy GIt L2

(5.7)

In practice for I or H sections GIt /EIy 1 + π2 EIw /GIt L2 is negligible compared to unity and thus K may be taken as K=

EIz 1− = EIy

1−

Iz Iy

(5.8)

For British Universal Beams the value of K is between 0,94 and 0,97 thus increasing Mcr by around 3% to 6%. This enhancement may therefore be neglected for Universal Beams. For Universal Columns, however, K is between 0,80 and 0,83, thus increasing Mcr by around 17% to 20%, and should therefore possibly be taken into account.

Chapter 5 / Laterally Unrestrained Beams Full plastic action

Elastic buckling curve (Eq.(5.5) or Eq.(5.6))

1.0

Perfect beam behaviour MpI

•

Mb,Rd

94

Experimental results 0.5

Design curve (Lower bound to experimental results)

0

0.4

0.8

1.2

1.6

⎛ M ⎛½ pl Effective slenderness ⎜ ⎜ ⎝ M cr ⎝

FIGURE 5.2 Beam behaviour with lateral torsional buckling

5.1.2 Interaction between Plastic Moment Capacity, Elastic Critical Moment and Allowable Bending Moment If the results for beam strength against effective slenderness are plotted in nondimensional format the results appear as in Fig. 5.2. The performance of beams is reduced below the perfect conditions due to the residual stresses and initial imperfections. A convenient lower bound to test data is given by a Perry–Robertson type approach to modelling the interaction. The imperfection co-efficient makes allowance both for geometric imperfections and residual stress levels. It should be noted that unlike columns there is no theoretical justification for such an approach.

5.1.2.1 Design Interaction between Buckling and Plasticity For any type of beam, the interaction equation is given by (Mb,Rd − Mcr )(Mb,Rd − Mpl,Rd ) = ηLT Mcr Mb,Rd

(5.9)

where ηLT is the imperfection factor for lateral torsional buckling. Equation (5.9) may be written in a normalized form to give Mb,Rd Mb,Rd Mcr Mb,Rd Mcr − − 1 = ηLT Mpl,Rd Mpl,Rd Mpl,Rd Mpl,Rd Mpl,Rd

(5.10)

It is convenient to define two parameters χLT and λLT given as χLT =

Mb,Rd Mpl,Rd

(5.11)

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TABLE 5.1 Values of αLT for the generalized lateral torsional buckling case. Cross section

Size limits (h/b)

αLT

Rolled I sections

≤2 >2 ≤2 >2

0,21 0,34 0,49 0,76 0,76

Welded I sections Other cross-sections

λLT

Mpl,Rd = Mcr

1/2 (5.12)

Equation (5.10) then becomes (χLT − 1)(χLT − (λLT )2 ) = ηLT

χLT (λLT )2

(5.13)

The solution to which is given by χLT =

1 − (λLT )2 ]1/2

LT + [2LT

(5.14)

where LT is given by LT = 0,5[1 + ηLT + (λLT )2 ]

(5.15)

with ηLT given as ηLT = αLT (λLT − 0,2)

(5.16)

The values of αLT are dependant upon the type of section and whether rolled or welded. The appropriate values are given in Table 5.1 (cl 6.3. 2.1. EN 1993-1-1) In design, full plastic moment capacity may be mobilized when λLT is less than λLT,0 which may be taken as 0,4 or when MEd /Mcr ≤ (λLT,0 )2 .

5.1.2.2 Rolled Sections and Equivalent Welded Sections (cl 6.3.2.3 EN 1993-1-1) For rolled sections and welded sections which are by implication symmetric, an alternative formulation of the interaction formula may be used, χLT =

LT + [2LT

1 − β(λLT )2 ]1/2

(5.17)

where LT is given by LT = 0,5[1 + ηLT + β(λLT )2 ]

(5.18)

with ηLT as ηLT = αLT (λLT − λLT,0 )

(5.19)

96

•

Chapter 5 / Laterally Unrestrained Beams TABLE 5.2 Values of αLT for lateral torsional buckling of rolled and welded sections. Cross section

Size limits (h/b)

αLT

Rolled I sections

≤2 >2 ≤2 >2

0,34 0,49 0,49 0,76

Welded I sections

The parameter χLT is subject to the limits χLT ≤ 1,0 and ≤ (λLT )2 . The code recommends that the maximum value of λLT,0 should be 0,4 and the minimum value of β as 0,75. The UK National Annex will adopt these values but will effectively specify the approach should be limited to rolled sections, and that, therefore, the generalized method must be used for welded sections. Again the values of αLT are dependant upon the type of section and whether rolled or welded. The appropriate values are given in Table 5.2. Additionally, if this method is used, then the calculation of Mb,Rd needs modifying by using a factor χLT,mod instead of χLT where χLT,mod =

χLT f

(5.20)

The factor f is given by f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] ≤ 1,0

(5.21)

The factor kc is determined from the type of loading. For loading only at points of restraint, kc =

1 1,33 − 0,33ψ

(5.22)

where ψ is the ratio between the moments at restraint points subject to the condition that −1 ≤ ψ ≤ 1. For loading between restraints values of kc are given in Table 6.6 of EN 1993-1-1. The additional factor is due to the critical moment being determined elastically but failure will be by generation of plasticity not necessarily at the point of maximum deflection for buckling.

5.1.2.3 Simplified Assessment Methods for Building Structures (cl 6.3.2.4 EN 1993-1-1) If there are discrete restraints to the compression flange, then lateral torsional buckling will not occur if the length Lc between restraints or the resultant slenderness λf satisfies the following equation λf =

Mc,Rd kc Lc = λc,0 ifz λ1 My,Ed

(5.23)

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where Mc,Rd is the flexural capacity of the section, kc is the factor to correct for moment gradient, if,z is the radius of gyration of the effective compression flange which comprises the actual compression flange together with one-third of the part of the web which is in compression and λ1 is given by 93,9(235/fy )1/2 and λC,0 is the normalized slenderness ratio of the effective compression flange.

5.1.2.4 Modifications Dependant upon Section Classification (cl 6.3.2.1 (3) EN 1993-1-1) For sections other than Classes 1 and 2, a further modification is required as for Class 3 sections the capacity is based on the elastic section modulus, and Class 4 an effective section modulus is used. The modification is made by defining the flexural capacity as Wel,y fy for Class 3 and Weff,y fy for Class 4. Also the buckling capacity Mb,Rd is defined as χLT Wpl,y fy/γM1 for Class 1 or Class 2, χLT Wel,y fy/γM1 for Class 3, χLT Weff,y fy/γM1 for Class 4. Having established the basic determination of the allowable moment capacity, Mb,Rd , two additional modifications due to varying support conditions and non-uniform flexural loading need to be examined.

5.1.3 Effect of Support Conditions If the beam has a rotational end restraint of R (where R takes a value of 0 for no restraint and 1 for full restraint), then the effective length kL is given by Eq. (5.24) (Trahair and Bradford, 1988). R π π = − cot 1−R 2k 2k

(5.24)

With a little loss in accuracy Eq. (5.24) can be written as k = 1 − 0,5R

(5.25)

Both Eqs (5.24) and (5.25) are plotted in Fig. 5.3. The results from Eq. (5.25) represent ideal conditions therefore in practice the values of kL used are higher at more complete fixity as absolute rigid joints do not exist. Pillinger (1988) gave some indications of the relationship between practical end conditions and effective lengths. Figure 5.4 gives typical end conditions in terms of connection type and degree of restraint. For beams the effective length L between restraints can be taken as lying between 0,7 and 1,0 times the actual length with the lower factor implying full rotational restraint to both flanges and the higher factor with both flanges free to rotate in plan. If the rotational restraint parameter R can be assessed, then Eq. (5.25) can be modified to give k = 1 − 0,3R

(5.26)

•

Chapter 5 / Laterally Unrestrained Beams 1.2 1 Effective length factor (k)

98

0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

1.2

Degree or restraint (R) Exact solution

Linear appproximation

Practical values

FIGURE 5.3 Values of effective length factor (k)

The values of kL for beams should be increased by 20% for destabilizing loads (BS 5950-1: Part 1 namely 1990 and 2000). For cantilevers the situation can be more complex as the effective length will depend on both the restraint at the encastré end and at the tip. Table 5.3 (from BS 5950-1 (2000)) gives suitable values. It should be noted that when the loading is destabilizing the effective length factors can be extremely high. Guidance is also given in Nethercot and Lawson (1992). Although it was suggested that for beams it is generally conservative to take the actual length, this may not be appropriate where the beam is supported solely on its bottom flange with no web or top flange restraint (Fig. 5.5). This type of situation will produce low lateral torsion buckling resistance, and Bradford (1989) suggests that in this case the system length L should be taken as, ⎛ ⎞ b hs tw 3 ⎝ 1 + hs ⎠ L=1+α (5.27) 6 tf 2 where the beam is under a moment gradient, or ⎛ ⎞ 3/2 1 + b h s tw hs ⎠ ⎝ L = 1 + 10 6 tf 2 under a central point load.

(5.28)

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Stability cleat Beam compression flange free to rotate on plan

Full torsional end restraint to beam

(a) Thin end plate to web only

(b) Bottom flange cleat

Beam compression flange free to rotate on plan Compression flange free to rotate. No torsional restraint

(c) Web cleats bolted to beam

Full torsional end restraint to beam

(d) Flange bolted to padstone

Compression flange rotation reduced. Torsional end restraint

(e) Full depth end plate welded to web and flange

(f) Flange bolted to padstone, web stiffener Full torsional restraint. No rotation of compression flange (g) Full end stiffener

FIGURE 5.4 Typical connection and support detail In both cases l is the span, hs the distance between the centroids of the flanges, b the flange width and tf and tw the thickness of the flange and web respectively, and α is given by α = 4 + 7ψ + 4ψ2

(5.29)

where ψ is the ratio of the end moments.

5.1.4 Intermediate Restraints Consider the beam in Fig. 5.6 but with a spring restraints giving a horizontal restraint stiffness of αt (force/per unit displacement) and rotational restraint stiffness of αr

TABLE 5.3 Effective length LE for cantilevers without intermediate restraint. Restraint conditions

Loading conditions

At support

At tip

Normal

Destabilizing

(a) Continuous, with lateral restraint to top flange

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

3.0L 2.7L

7.5L 7.5L

2.4L 2.1L

4.5L 3.6L

(b) Continuous, with partial (1) Free torsional restraint (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

2.0L 1.8L

5.0L 5.0L

1.6L 1.4L

3.0L 2.4L

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

1.0L 0.9L

2.5L 2.5L

0.8L 0.7L

1.5L 1.2L

(1) Free (2) Lateral restraint to top flange (3) Torsional restraint (4) Lateral and torsional restraint

0.8L 0.7L

1.4L 1.4L

0.6L 0.5L

0.6L 0.5L

L

L

(c) Continuous, with lateral and torsional restraint

L

(d) Restrained laterally, torsionally and against rotation on plan

L

(1) Free

(not braced on plan) Source: BS 5950 Part 1:2000

Tip restraint conditions (2) Lateral restraint to (3) Torsional restraint top flange

(4) Lateral and torsional restraint

(braced on plan in at least one bay)

(braced on plan in at least one bay)

(not braced on plan)

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A

A

cL of bolts

Section AA

FIGURE 5.5 Beam supported on bottom flange with no restraint to top flange

at mid-span (Fig 5.6(a)) at a height of bs above the centroidal axis. It can be shown (Mutton and Trahair, 1973) that the elastic critical moment is given by Eq. (5.6) or (5.7) with kL (the effective, or system, length) substituted for L, with k related to αt and αr by ⎛ L3

αt ⎝1 + 16EIz αr L3 16EIw 2b 2b 1 − h s h0

2bs h 2b0 h

⎞ ⎠=

π 3 π cot 2k 2k π π cot 2k −1 2k

(5.30)

=

π 3 π cot 2k 2k π π cot 2k −1 2k

(5.31)

where the distance of the centre of rotation below the shear centre b0 is given by b0 =

Mcr EI

π2 (kL)z 2

2 h k = 1+ 2 K

(5.32)

and K is defined K=

π2 EIw L2 GIt

1/2 (5.33)

•

Chapter 5 / Laterally Unrestrained Beams x

φL

Axis of buckled beam

2 ⎛

vL

⎛

α1 ⎜VL bs φL ⎜

2

⎝ 2

2⎝

bs

αeφL 2

y

z

FIGURE 5.6(a) Beam with elastic intermediate restraints

1.2

1 Effective length factor (k)

102

Braced mode 1 buckling 0.8

0.6 Mode 2 buckling

0,5 0.4

0.2

0 0

5

π2 10

15 20 Normalized restraint factor

25

30

FIGURE 5.6(b) Effect of spring restraint on effective length factor

The relationships from Eqs (5.30) and (5.31) are plotted in Fig 5.6(b). The value of k changes from 1 (when the beam buckles as an entity) to k = 0,5 when the beam buckles as two half waves when the right hand sides of Eq. (5.30) or (5.31) reach a value of π2 . From Eq. (5.32) the minimum value of b0 is h/2. From Eq. (5.30) the maximum value of αt applied at the top flange (bs = h/2) to force a change into second mode is given by αt =

4Mcr 4Pf = Lh L

where Pf is the force in the flange.

(5.34)

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5.1.5 Loading The effect of the applied loading on the beam system needs considering under two headings. The first is concerned with how the load is applied whether at restraint points or between restraints. The second is concerned with any possible destabilizing effects of the load.

5.1.5.1 Load Pattern The background to any amendments to the value of Mcr to allow for load pattern is given in Nethercot and Rockey (1971), or Trahair and Bradford (1988). For loading at points of lateral torsional restraint, the critical moment Mcr is modified by a factor C1 which is dependant solely on the moment gradient within the section of the beam being considered. The moment gradient is defined by the ratio of the applied moments at either end of the beam segment. The small moment within the segment caused by self-weight is not taken into account in this calculation. From Kirby and Nethercot (1979) C1 is given by 1 = 0,57 + 0,33β1 + 0,1β12 ≥ 0,43 C1

(5.35)

where β1 = 1,0 for single curvature (or −1 for double curvature). And from Trahair and Bradford (1988) as either, C1 = 1,75 + 1,05β2 + 0,3β22 ≤ 2,56

(5.36)

1 = 0,6 − 0,4β2 ≥ 0,4 C1

(5.37)

or,

where β2 = −1 for single curvature (and +1 for double curvature). Rewrite Eqs (5.35) to (5.37) using ψ rather than β1 or β2 , defining ψ = 1 for single curvature to give 1 = 0,57 + 0,33ψ + 0,1ψ2 ≤ 0,43 C1

(5.38)

C1 = 1,75 − 1,05ψ + 0,3ψ2 ≤ 2,56

(5.39)

1 = 0,6 + 0,4ψ ≥ 0,4 C1

(5.40)

A comparison between the values of C1 from Eqs (5.38) to (5.39) is given in Table 5.4 and Fig. 5.7 where it is observed that there is little difference between the values, and that this difference is reduced as the value of the normalized slenderness ratio requires 1/2 the value of C1 . For the examples herein Eq. (5.40) will be used.

Chapter 5 / Laterally Unrestrained Beams TABLE 5.4 Comparison of values of C 1 . ψ

Eq. (5.36)

Eq. (5.37)

Eq. (5.38)

1,0 0,75 0,5 0,25 0 −0,25 −0,5 −0,75 −1,0

1,00 1,14 1,32 1,52 1,75 2,03 2,33 2,33 2,33

1,00 1,13 1,30 1,51 1,75 2,03 2,35 2,56 2,56

1,00 1,11 1,25 1,43 1,67 2,00 2,50 2,50 2,50

3

2.5

2

1.5

1

0.5

2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1

1

0.

0

0.

1

2

0.

3

0.

4

0.

5

0.

6

0.

7

0.

8

0.

9

0.

1

0

0.

•

C1

104

Moment ratio Eq. (5.36)

Eq. (5.37)

Eq. (5.38)

FIGURE 5.7 Values of C 1 For loads between restraints use the n factor method from Tables 15 to 17 of BS 5950-1 (1990) should be used. In this case C1 is defined by 1 C1 = √ n

(5.41)

Tables 15 to 17 of BS 5950: Part 1: 1990 are reproduced in Annexe A7.

5.1.5.2 Destabilizing and Stabilizing Loads A destabilizing load is one which is applied to the compression flange and is free to move as the flange buckles laterally. Such a load has the effect of reducing the elastic critical moment as an additional disturbing torque is introduced (Anderson and Trahair, 1972; Trahair and Bradford, 1988). A stabilizing load, therefore, is one

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105

Additional torque

No additional torque

Decreased torque

FIGURE 5.8 Destabilizing loads applied in such a way that the effect counterbalances the buckling effect and thus enhances the elastic critical moment. Thus for simply supported beams if a load free to move is applied above the shear centre it is destablizing, and below the shear centre it is stabilizing (Fig. 5.8). The reverse is true for a cantilever! For loading applied at other than the centroid of the section it is only possible to determine the critical moment under a given load pattern. Timoshenko and Gere (1961) give the value of the critical load for a simply supported I beam under either a uniformly distributed load (UDL) or central point load as √ EIz GIt (5.42) Pcr = γ2 L2 where γ2 is dependant upon the position of the load (top flange, centroid or bottom flange) and the factor L2 GI t /EI w . Values of γ2 are plotted in Fig. 5.9 for values of beam stiffness 1/K 2 (where K is given by Eq. (5.33)).

5.1.5.3 Cantilevers and Beams Cantilevering Over Supports • Cantilevers Trahair (1983) demonstrated that for cantilevers built in at the support that the elastic critical moment Mcr could be given with little loss of accuracy as √ EIz GIt Mcr = (1,6 + 0,8K) (5.43) L √ where K is given by Eq. (5.33) as (π2 EI w /(L2 GI t )). It will be noted that this differs from the more usual solution given in Eq. (5.6) where the elastic critical moment is proportional to (1 + K 2 )1/2 .

Chapter 5 / Laterally Unrestrained Beams 160 140 120 100 Gamma 2

•

80 60 40 20 0 0

50

100

150

200

250

300

350

400

450

400

450

Beam stiffness (1/K 2) Load at top

Load at centre

Load at bottom

FIGURE 5.9(a) Factor for point load not applied at shear centre 250

200

Gamma 2

106

150

100

50

0 0

50

100

150

200

250

300

350

Beam stiffness (1/K 2) Load at top

Load at centre

Load at bottom

FIGURE 5.9(b) Factor for UDL not applied at shear centre For a point load Pcr at the free end of the cantilever Trahair gives the following expression, √ 1,2ε 1,2(ε − 0,1) EIz GIt Pcr = 11 1 + + 4(K − 2) 1 + L2 1 + 1,22 (ε − 0,1)2 1 + 1,22 ε2 (5.44)

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where the parameter ε defines the position relative to the centroidal axis of the point of application of the load, and is given by a EIz 2a K ε= = (5.45) L GIt h π where a is the distance of the point of application of the load For a UDL of qcr applied to whole length of the cantilever, √ 2 EIz GIt 1,4 (ε − 0,1) qcr = 27 1 + L3 1 + 1,42 (ε − 0,1)2 1,3(ε − 0,1) + 10(K − 2) 1 + 1 + 1,32 (ε − 0,1)2 For loading applied at the centroid, ε = 0, and Eq. (5.44) reduces to √ EIz GIt [3,96 + 3,52K] Pcr = L2 The moment Mcr,P due to this load is given by √ EIz GIt [3,96 + 3,52K] Mcr,P = Pcr L = L Equation (5.46) becomes when loading is applied at the centroid √ 2 EIz GIt [5,83 + 8,71K] qcr = L3 The moment Mcr,q due to this load is given by √ qcr L2 EIz GIt [5,83 + 8,71K] Mcr,q = = 2 L

(5.46)

(5.47)

(5.48)

(5.49)

(5.50)

Equations (5.43), (5.48) and (5.50) are plotted in Fig. 5.10, where it is observed that as the loading progresses from uniform moment through a point load at the end to a UDL, the equivalent elastic moment increases. This is due to the fact that for the UDL the moment quickly drops off from the support and thus has less of an effect. The moment due to the point load also drops off, but more slowly. It is therefore to be noted that the general UK practice of ignoring the beneficial effects of moment gradients on a cantilever (i.e. C1 = 1,0) is extremely conservative. • Overhanging beams For the situation shown in Fig. 5.11(a), where a beam overhangs the supports symmetrically and is loaded by end point loads, then the lateral torsional buckling of such a system is controlled by the restraint at the internal supports which may often be of the type shown in Fig. 5.11(b), where such restraints can be considered elastic. Such restraints increase the torsional flexibility of the beam and thereby decrease the buckling moment. The decrease P/P0 is dependant upon the stiffness

•

Chapter 5 / Laterally Unrestrained Beams 35 Normalized elastic critical moment

108

30 25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

Stiffness parameter K Pure moment

Point load

UDL

FIGURE 5.10 Normalized elastic critical moment for a cantilever

P

P

L

γL

L

(a) Basic geometry Over hanging beam

Bolts

Supporting beam

(b) Type support detail

FIGURE 5.11 Overhanging beams

at the support αR , and Trahair (1983) gives the following equation for beams loaded centroidally, αR L P 2a GIt = 1 − Kβ2 (1 − β) (5.51) αR L 4K 2 P0 h 5 + 1+K 2 + GI t

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109

20 kN Permanent 30 kN Variable

A

D B 3.0 m

C 3.0 m

3.0 m

Notes: (1) All loads are characteristic loads (2) Lateral torsional restraints exists at A, B, C and D

FIGURE 5.12 Design data for Example 5.1

where β is defined by β=

αR L GIt αR L 1 + GI t

(5.52)

It should be noted for situations where the internal span is greater than the sum of the overhangs, then the internal span dominates the behaviour.

EXAMPLE 5.1 Beam with loading applied at restraints. Prepare a design in Grade S355 steel for the beam for which the data are given in Fig. 5.12. Factored actions at ULS: at B: 1,35 × 40 + 1,5 × 70 = 159 kN at C: 1,35 × 20 + 1,5 × 30 = 72 kN Total load = 231 kN The BM and SF diagrams are drawn in Fig. 5.13. The critical section for design is the central section BC as the moment gradient is the least. Try a 406 × 178 × 74 UKB Mpl,Rd = Wpl,y

fy 355 = 1501000 × 10−6 = 533 kNm γM0 1,0

MSd = 390 kNm, beam satisfies the plastic capacity criterion. Section classification: Compression flange: c = 0,5[b − 2r − tw ] = 0,5[179,5 − 2 × 10,2 − 9,5] = 74,8 mm c/tf = 74,8/16 = 4,68

110

•

Chapter 5 / Laterally Unrestrained Beams A

B

C

D

303 390 (a) BMD (kNm) 130

B A

C

D

−29 −101 (b) SFD (kN)

FIGURE 5.13 BMD and SFD for Example 5.1

Maximum value for a Class 1 flange is 9ε = 9(235/355)1/2 = 7,32. Web: c/t = d/tw = 37,9. Maximum value for a Class 1 web is 9ε = 72(235/355)1/2 = 58,6. Thus a 406 × 178 × 74 UKB Grade S355 is Class 1. Shear check: fy 1 1 355 Vpl,Rd = √ Av = √ 4260 × 10−3 = 873 kN 1,0 3 γM0 3 By inspection, the moment capacity is not reduced due to shear. Calculation of Mcr : E = 210 GPa; G = 81 GPa. The most foolproof method to determine Mcr given the use of the cm and dm in section property tables is to work in kN and m. Use Eq. (5.5) to determine Mcr . System length L = 3 m. Iw 0,608 × 10−6 = = 0,03924 m2 Iz 1545 × 10−8 π2 EIz π2 × 210 × 106 × 1545 × 10−8 = = 3558 kN L2 32

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L2 GIt GIt 81 × 106 × 62,8 × 10−8 = 0,0143 m2 = = 2 π EIz 3558 π2 EIz 2 L

π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 = 3558 [0,03924 + 0,0143]1/2 = 823 kNm

Determine the value of C1 using Eq. (5.40): The moment ratio ψ is given by ψ=

MEd,C 303 = 0,777 = 390 MEd,B

1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0,777 = 0,911 C1 or C1 =

1 = 1,098 0,911

The value of Mcr to be used in Eq. (5.12) in the calculation of the normalized slenderness ratio is C1 Mcr , thus λLT is given as Wy fy 1501000 × 355 λLT = = = 0,768 Mcr 1,098 × 823 × 106 Both methods available to determine the strength reduction factors due to lateral torsional buckling will be used in this example to demonstrate any differences. (a) General case The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.1, αLT = 0,34 (curve b). Use Eqs (5.15) and (5.16) to determine LT : LT = 0,5[1 + ηLT + (λLT )2 ] = 0,5[1 + 0,34(0,768 − 0,2) + 0,7682 ] = 0,891 Use Eq. (5.14) to determine χLT : χLT =

1 1 = = 0,745 2 − 0,7682 ]1/2 2 1/2 0,891 + [0,891 − (λLT ) ]

LT + [2LT

Mb,Rd = χLT Wpl,y

fy 355 = 0,745 × 1501000 × 10−6 = 397 kNm γM1 1,0

This is greater than the moment at B (390 kNm). Clearly the end section AB does not need checking as the system length is the same as BC (therefore the basic value of Mcr does not change and since ψ = 0, C1 now becomes 1,75 with the effect of reducing the value of λLT and of increasing the value of χLT and hence Mb,Rd . (b) Method for rolled sections The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.2, αLT = 0,49 (curve c).

112

•

Chapter 5 / Laterally Unrestrained Beams Determine ηLT from Eq. (5.19) ηLT = αLT (λLT − λLT,0 ) = 0,49(0,768 − 0,4) = 0,180 Determine LT from Eq. (5.18) LT = 0,5[1 + ηLT + β(λLT )2 ] = 0,5[1 + 0,180 + 0,75 × 0,7682 ] = 0,811 Determine χLT from Eq. (5.17) χLT =

1 1 = = 0,784 0,811 + [0,8112 − 0,75 × 0,7682 ] LT + [2LT − β(λLT )2 ]1/2

Initially ignore the correction factor f , to give Mb,Rd as Mb,Rd = χLT Wpl,y

fy 355 = 0,784 × 1501000 × 10−6 = 418 kNm γM1 1,0

Determine f : The moment ratio ψ = 303/390 = 0,777, so from Eq. (5.33), kc =

1 1 = = 0,931 1,33 − 0,33ψ 1,33 − 0,33 × 0,777

From Eq. (5.32) f is given by f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] ≤ 1,0 = 1 − 0,5(1 − 0,931)[1 − 2,0(0,768 − 0,8)2 ] = 0,966 Using Eq. (5.31) Mb,Rd now becomes 418/0,966 = 433 kNm. The effect of the factor f is not terribly significant. Even without the factor, f lateral torsional buckling clauses for rolled sections produces a marginally higher value of Mb,Rd by around 20 kNm (or around 5%). Deflection (under service loading): EI = 210 × (27430 × 104 ) × 10−6 = 57603 kNm2 Mid-span deflection due to an asymmetric point load is given by 3 WL3 b b δ= 3 −4 48EI L L Variable action deflection: Load at B: W = 70 kN, b = 3 m, l = 9 m, δ = 0,016 m Load at C: W = 30 kN, b = 3 m, L = 9 m, δ = 0,007 m Deflection under service variable actions = 0,023 m. This deflection is equivalent to span/390, and is therefore acceptable.

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113

Deflection under total actions: Load at B: W = 110 kN, b = 3 m, l = 9 m, δ = 0,025 m Load at C: W = 50 kN, b = 3 m, L = 9 m, δ = 0,011 m Total deflection = 0,036 m. This deflection is equivalent to span/250, and is therefore acceptable. Web check (refer to Section 4.8 and cl 6 EN 1993-1-5): The web capacity under transverse forces needs checking at A and B. Note as a rolled section is being used, fyw = fyf = 355 MPa. Since at either point the length of stiff bearing ss is not known, set ss = 0 and determine the value required should the check fail At A: FSd = RA = 130 kN. For an end support with c = ss = 0, kF = 2 (type c) Determine m1 : m1 =

fyf bf bf 179,5 = = = 18,9 fyw tw tw 9,5

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 18,9 = 16,0 = 49,2 mm 2 2

The depth of the web hw has been taken as d, the depth between fillets. tw3 9,53 = 0,9 × 2 × 210 = 889 kN hw 360,4 ly tw fyw 49,2 × 9,5 × 355 λF = = = 0,432 FCR 889 × 103

FCR = 0,9kF E

As λF < 0.5, m2 = 0. χF =

0,5 0,5 = = 1,16 0,432 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 49,2 = 49,2 mm FRd = Leff tw

fyw 355 × 10−3 = 166 kN = 49,2 × 9,5 γM1 1,0

As FRd is greater than FEd (=RA = 130 kN), η2 < 1,0, therefore the web resistance at A is satisfactory without a stiff bearing.

114

•

Chapter 5 / Laterally Unrestrained Beams Check at B: The applied load is 159 kN, and the applied moment is 390 kNm. For the situation where the load is applied through the top flange, kF = 6 (type a, with the stiffener spacing a effectively taken as infinity) Determine m1 : m1 =

fyf bf bf 179,5 = = = 18,9 fyw tw tw 9,5

As m2 is dependant upon λF initially assume m2 = 0. As ss has been assumed to be zero, then the value of ly is given by √ ly = 2tf (1 + m1 ) = 2 × 16,0(1 + 18,9) = 171 mm The depth of the web hw has been taken as d, the depth between fillets. tw3 9,53 = 0,9 × 6 × 210 = 2698 kN hw 360,4 ly tw fyw 171 × 9,5 × 355 λF = = = 0,462 FCR 2698 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. 0,5 0,5 χF = = = 1,08 0,462 λF The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 171 = 171 mm FRd = Leff tw η2 =

fyw 355 = 171 × 9,5 × 10−3 = 577 kN γM1 1,0

FEd fyw Leff tw γM1

=

FEd 159 = = 0,276 FRd 577

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing. However an interaction equation needs checking owing to the co-existence of shear and bending moment: η2 + 0,8η1 ≤ 1,4 As there is no axial force and no shift in the neutral axis as the section is Class 1, the equation for η1 reduces to η1 =

MEd f

y Wpl γM1

=

390 × 106 1501 × 103 355 1,0

= 0,732

η2 + 0,8η1 = 0,276 + 0,8 × 0,732 = 0,862 ≤ 1,4 The web at B is therefore satisfactory.

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EXAMPLE 5.2 Beam design with loads applied between lateral torsional restraints. Prepare a design in Grade S355 steel for the beam whose data are given in Fig. 5.14 Factored actions: Factored UDL: 1,35 × 10 + 1,5 × 20 = 43,5 kN/m Factored point load: 1,5 × 15 = 22,5 kN Figures 5.15 (b) and (c) show the resultant BM and SF diagrams. Try a 457 × 191 × 98 UKB Grade S355 Section classification: Flanges: c = 0,5(b − 2r − tw ) = 0,5(192,8 − 2 × 10,2 − 11,4) = 80,5 mm c 80,5 = = 4,11 tf 19,6 Class 1 limit: 9ε = 9

235 = 7,32 355

Flanges are Class 1. Web: c d 407,6 = = = 35,8 tw tw 11,4 Class 1 limit: 72ε = 72

235 = 58,56 355

Web is Class 1, therefore the section classification is Class 1.

A

20 kN/m Variable 10 kN/m Permanent

15 kN Variable B C

7.0 m

Notes: (1) All loading is characteristic loading (2) Lateral torsional restraints exist at A and B (3) The TOP flange is restrained at C

FIGURE 5.14 Design data for Example 5.2

3.5 m

116

•

Chapter 5 / Laterally Unrestrained Beams 43,5 kN/m

22,5 kN

A

B C

(a) Beam loading at ULS 2,37 174,7 102,9

D

22,5

B

C

−201,6 (b) SFD (kN) −192,3

A

D

B C

345,2 (c) BMD (kNm)

FIGURE 5.15 BMD and SFD for Example 5.2 Plastic moment capacity: Mpl,Rd = Wpl,y

fy 355 = 2232 × 103 × 10−6 = 793 kNm γM0 1,0

This exceeds the maximum applied moment of 345,2 kNm. Shear capacity: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 1146 kN = √ 5590 γ 1,0 3 3 M0

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By inspection, there is no reduction in moment capacity for the effect of shear at the cantilever support. System length AB: The system length is taken as the span AB, not the distance between the points of contraflexure calculated from the applied loading as these will not form nodal points in the post-buckled shape of the beam. Such nodal points may only form at points of lateral restraint since at a nodal point the lateral deflection must be zero (Kirby and Nethercot, 1979). Determination of C1 : As the loading is between restraints at A and C, the ‘n’ factor method from BS 5950: Part 1: 1990 must be used (Annexe A7). From Table 17, the BMD is that of the fifth diagram with β = 0 (as the BM at A is zero), and γ is negative. The bending moment M0 assuming the beam to be simply supported between A and C is given by qult L2 43,5 × 72 = = 266,4 kNm 8 8 −345,2 M = γ= = −1,3 M0 266,4

M0 =

From Table 16 with β = 0 and γ = −1,3, n = 0,53 1 1 C1 = √ = √ = 1,374 n 0,53 Determine Mcr from Eq. (5.5): Iw 1,18 × 10−6 = = 0,0503 m2 Iz 2347 × 10−8 L = 7,0 m. π2 EIz π2 × 210 × 106 × 2347 × 10−8 = = 993 kN L2 72 L2 GIt GIt 81 × 106 × 121 × 10−8 = 2 = = 0,0987 m2 2 π EIz 993 π EIz 2 L

Mcr =

π2 EI L2

z

Iw L2 GIt + 2 Iz π EIz

1/2 = 993[0,0503 + 0,0987]1/2 = 383 kNm

Determine λLT using the moment gradient modified value of Mcr in Eq. (5.12), 1/2 Wpl,y fy 1/2 2232 × 103 × 355 × 10−6 λLT = = = 1,227 1,374 × 383 C1 Mcr

118

•

Chapter 5 / Laterally Unrestrained Beams As in the previous example calculate the strength reduction factor using both methods: • General method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.1, αLT = 0,34 Determine LT from Eqs (5.15) and (5.16) LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,34(1,227 − 0,2) + 1,2272 ] = 1,427 Determine χLT from Eq. (5.14) χLT =

LT +

1

2LT

Mb,Rd = χLT Wpl,y

− (λLT

1/2 )2

=

1 = 0,464 1,427 + [1,4272 − 1,2272 ]1/2

fy 355 = 0,464 × 2232 × 103 × 10−6 = 368 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. • Rolled section method: h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.2, αLT = 0,49 Determine LT from Eqs (5.18) and (5.19) LT = 0,5[1 + αLT (λLT − λLT,0 ) + β(λLT )2 ] = 0,5[1 + 0,49(1,227 − 0,4) + 0,75 × 1,2272 ] = 1,267 Determine χLT from Eq. (5.17) χLT =

LT + [2LT

1 1 = = 0,511 2 2 1/2 1,267 + [1,267 − 0,75 × 1,2272]1/2 − β(λLT) ]

Without the correction factor f : Mb,Rd = χLT Wpl,y

fy 355 = 0,511 × 2232 × 103 × 10−6 = 405 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. From Table 6.6 of EN 1993-1-1, kc = 0,91 Determine f from Eq. (5.32): f = 1 − 0,5(1 − kc )[1 − 2,0(λLT − 0,8)2 ] = 1 − 0,5(1 − 0,91)[1 − 2,0(1,227 − 0,8)2 ] = 0,971 Thus the corrected value of Mb,Rd given by Eq. (5.31) is Mb,Rd =

405 = 417 kNm 0,971

It is again noted that the rolled section approach even without the factor f gives slightly higher values of Mb,Rd than the general case.

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System length BC. Take the conventional approach and adopt C1 = 1,0 Determine Mcr from Eq. (5.5): Iw 1,18 × 10−6 = = 0,0503 m2 Iz 2347 × 10−8 L = 3,5 m π2 EIz π2 × 210 × 106 × 2347 × 10−8 = = 3971 kN L2 3,52 L2 GIt GIt 81 × 106 × 121 × 10−8 = 2 = = 0,0247 m2 2 π EIz 3971 π EIz L2

π2 EIz Mcr = L2

Iw L2 GIt + 2 Iz π EIz

1/2 = 3971[0,0503 + 0,0247]1/2 = 1088 kNm

Determine λLT from Eq. (5.12), 1/2 Wpl,y fy 1/2 2232 × 103 × 355 × 10−6 λLT = = = 0,853 C1 Mcr 1088 As in the previous example calculate the strength reduction factor using both methods: • General method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.1, αLT = 0,34 Determine LT from Eqs (5.15) and (5.16) LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,34(0,853 − 0,2) + 0,8532 ] = 0,975 Determine χLT from Eq. (5.14) χLT =

1

LT + 2LT − (λLT )2

Mb,Rd = χLT Wpl,y

1/2 =

1 = 0,691 0,975 + [0,9752 − 0,8532 ]1/2

fy 355 = 0,691 × 2232 × 103 × 10−6 = 548 kNm γM1 1,0

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. • Rolled section method h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.2, αLT = 0,49 Determine LT from Eqs (5.18) and (5.19) LT = 0,5[1 + αLT (λLT − λLT,0 ) + β(λLT )2 ] = 0,5[1 + 0,49(0,853 − 0,4) + 0,75 × 0,8532 ] = 0,884

120

•

Chapter 5 / Laterally Unrestrained Beams A

B

C

El constant a Deflection: At mid-span (span AB) At C

b (qa 2/384 EI)(5a 2 − 12b 2) (q b/24 EI)(3b 3 + 4 ab2 − a 3) (a) UDL P

A

B C

El constant a

b

Deflection: At mid-span (span AB) −Pba 2/16EI At C Pb 2(a + b)/3EI (b) Point load

FIGURE 5.16 Deflection formulae for Example 5.2

Determine χLT from Eq. (5.17) χLT = =

LT + [2LT

1 − β(λLT )2 ]1/2

1 = 0,730 0,884 + [0,8842 − 0,75 × 0,8532 ]1/2

As there appears to be no consideration given to cantilevers in Table 6.6, f will be taken as 1,0. Mb,Rd = χLT Wpl,y

fy 355 = 0,730 × 2232 × 103 × 10−6 = 578 kNm 1,0 γM1

This exceeds the absolute value of the maximum moment at C of 345,2 kNm. It is again noted that the rolled section approach even without the factor f gives slightly higher values of Mb,Rd than the general case. For this particular design case, the span AC is critical. Deflection check: EI = 210 × 10−6 × 45730 × 104 = 96033 kNm2 . The relevant formulae are given in Fig. 5.16. (a) Variable action check: UDL: 20 kN/m

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121

Span AB, central deflection δ=

qa2 20 × 72 (5a2 − 12b2 ) = (5 × 72 − 12 × 3,52 ) = 0,0026 m 384EI 384 × 96033

Span BC, at C qb 20 × 3,5 (3b3 + 4ab2 − a3 ) = (3 × 3,53 + 4 × 7 × 3,52 − 73 ) 24EI 24 × 96033 = 0,0039 m

δ=

Point load at C Span AB, mid-span δ=−

Pba2 15 × 3,5 × 72 =− = −0,0017 m 16EI 16 × 96033

Span BC, at C δ=

P(a + b)b2 15(7 + 3,5)3,52 = = 0,0067 m 3EI 3 × 96033

Net deflections: at mid-span = 0,0026 − 0,0017 = 0,0009 m Span deflection ratio is 7/0,0009 = 7780. This is more than acceptable. at C = 0,0039 + 0,0067 = 0,0106 m Span deflection ratio (based on twice the span) is 2 × 3,5/0,0106 = 660. This is acceptable. (b) Check under total actions. Deflection due to point load as above. Total UDL of 30 kN/m: Mid-span, δ = 0,0039 m; at C, δ = 0,0059 m. Total deflection at mid-span = 0,0039 − 0,0017 = 0,0022 m Span deflection ratio: 7/0,0022 = 3180. This is satisfactory. Total deflection at C = 0,0059 + 0,0067 = 0,0126 m Span deflection ratio (based on twice the span) is 2 × 3,5/0,0126 = 556. This is satisfactory. Web check at B (refer to Section 4.8 and cl 6 EN 1993-1-5): This is the only point that needs checking, as the other reaction point has a much lower force (even allowing for reduced dispersion length) and no coincident moment. RSd = 124,7 kN (Reaction at B); M = 114,5 kNm Ignore any stiff bearing (ss = 0). For the situation where the load is applied through the top flange, kF = 6 (type a, with the stiffener spacing a effectively taken as infinity)

122

•

Chapter 5 / Laterally Unrestrained Beams Determine m1 : m1 =

fyf bf bf 192,8 = = = 16,9 fyw tw tw 11,4

As m2 is dependant upon λF initially assume m2 = 0. As ss has been assumed to be zero, then the value of ly is given by √ ly = 2tf (1 + m1 ) = 2 × 19,6(1 + 16,9) = 200 mm The depth of the web hw has been taken as d, the depth between fillets. tw3 11,43 = 0,9 × 6 × 210 = 4122 kN hw 407,6 ly tw fyw 200 × 11,4 × 355 λF = = = 0,443 FCR 4122 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. χF =

0.5 0,5 = 1,13 = 0,443 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 200 = 200 mm FRd = Leff tw η2 =

fyw 355 = 200 × 11,4 × 10−3 = 809 kN γM1 1,0

FEd fyw Leff tw γ M1

=

FEd 124,7 = = 0,154 FRd 809

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing. However an interaction equation needs checking owing to the co-existence of shear and bending moment: η2 + 0,8η1 ≤ 1,4 As there is no axial force and no shift in the neutral axis as the section is Class 1, the equation for η1 reduces to η1 =

MEd fy

Wpl γ

M1

=

114,5 × 106 2232 × 103 355 1,0

= 0,145

η2 + 0,8η1 = 0,154 + 0,8 × 0,145 = 0,27 ≤ 1,4 The web at B is therefore satisfactory.

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123

5.1.6 Other Section Profiles A number of special cases need considering, hollow sections as the earlier equations for critical moment are not applicable, rectangular sections as the warping stiffness is zero, and T sections.

5.1.6.1 Rolled Hollow Sections Rees (1990) indicates that thin wall tubes of circular and triangular cross with uniform thickness cannot warp. For rectangular thin walled tubes only those whose wall thicknesses are in a constant ratio the length of the sides do not warp. These tubes are known as Neuber tubes. If there is no warping then lateral torsional buckling can only be resisted by torsion. For conventional hollow sections where the wall thickness is constant, then lateral torsional buckling is in part resisted by warping. It will be conservative to neglect the warping stiffness, and, therefore as a result, Eq. (5.3) can be reduced to GIt

d2 φ M2 = − φ EIz dx2

(5.53)

The resultant value of Mcr (with no allowance for major axis bending) is given from Eq. (5.6) as √ π EIz GIt (5.54) Mcr = L The factor K from Eq. (5.7) with the warping constant Iw set equal to zero becomes, EIz GIt K= 1− 1− (5.55) EIy EIy Combining Eqs (5.54) and (5.55) gives the elastic critical moment as √ π EIz GIt Mcr = EI GI L 1 − EIz 1 − EI t y

(5.56)

y

or the normalized slenderness ratio, λLT , is given from Eq. (5.12) with the introduction of the moment gradient factor C1 as t Wpl,y fy L 1 − IIz 1 − GI EI y y Wpl,y fy λLT = = (5.57) √ C1 Mcr C1 π EIz GIt The normalized lateral torsional buckling slenderness ratio λLT is also given by λLT =

λLT λ1

(5.58)

124

•

Chapter 5 / Laterally Unrestrained Beams or λLT = λLT λ1 = π

E λLT fy

t πEWpl,y L 1 − IIz 1 − GI EIy y = √ C1 EIz GIt

(5.59)

Rewrite Eq. (5.59) as

λLT =

1

t 1/2 Wpl,y L 1 − IIz 1 − GI EI y y E π √ G Iz It

1/2

C1

(5.60)

Define the slenderness ratio λ as λ=

L L =! iz Iz

(5.61)

A

when Eq. (5.62) becomes

λLT

t 1/2 Wpl,y λ 1 − IIz 1 − GI EIy y 1 E = 1/2 π √ G AIt C1

(5.62)

or λLT is given as λLT =

1 1/2

C1

E π G

1/2 (φb λ)1/2

(5.63)

where φb is defined as by " #" # ⎞1/2 I GI 2 1 − Iz 1 − EI t Wpl,y y y ⎠ φb = ⎝ AIt ⎛

(5.64)

This is the equation given in Section B2.6.1 of BS 5950-1. The term in rectangular parentheses in Eq. (5.63) has a value of 2,25. An alternative approach avoiding the calculation of lateral torsion buckling is to determine the critical length lcrit (in mm) corresponding to a value of λLT equalling 0,4 (below which buckling will not occur) (Rondal, et al., 1992). This value is given by

lcrit =

113400(h − t) fy

b−t 2 h−t b−t 1 + 3 h−t

3 + b−t h−t 1+

b−t h−t

(5.64a)

It should be noted that there is an apparent anomaly in Eq. (5.64) in that for a square section (b = h), lcrit remains finite, whereas Eq. (5.57) indicates φb (and hence λLT ) = 0.

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125

The reason is that the values from Eq. (5.64) in Rondal, et al. are also given in tabular form for discrete values of (h − t)/(b − t), and that interpolation would not be possible if the value for a square section were given as infinity. It should be noted that there is an error in the formula quoted in Rondal, et al., but that the values in Table 15 in the same publication are correct. Eq. (5.168) has been corrected (Kaim, 2006). It should be noted that the values of lcrit are extremely safe. Kaim (2006) suggests the normalized slenderness limit λz,lim is given by 25

λz,lim = h b

235 fy

(5.64b)

5.1.6.2 Rectangular Sections For rectangular sections of width b and depth h the equation for critical moment Mcr given in Eq. (5.6) reduces to Mcr =

π EIz GIt L

(5.65)

as the warping constant Iw is zero. (a) Thin sections For thin sections, It = hb3 /3 and Iz = hb3 /12, so Eq. (5.65) reduces to Mcr =

hb3 πE L 72(1 + ν)

(5.66)

(b) Thick sections In this case It is no longer given by ht3 /3. The following approximate formula can be used hb3 b 1 b 4 It = (5.67) 1 − 0,63 1− 3 h 12 h If h/b > 2, Eq. (5.67) can with little loss in accuracy be reduced to hb3 b It = 1 − 0,63 3 h

(5.68)

However the major axis bending is now important and to ignore it would be too conservative, thus the parameter K from Eq. (5.8) must be introduced to give Mcr as Mcr =

πhb3 √ 6L

1 − 0,63 b h EG 2 b 1− h

(5.69)

Chapter 5 / Laterally Unrestrained Beams

t1

b1

z0

zc

Axis through shear centre

hs

Elastic neutral axis

tw

t2

•

d

126

FIGURE 5.17 Calculation of monosymmetry index

b2

5.1.6.3 Monosymmetric Beams For beams with only one axis of symmetry (usually the minor axis), the centroid and the shear centre do not coincide, thus an additional disturbing torque occurs due to the longitudinal flexural stresses. From Trahair and Bradford (1988) the elastic critical moment Mcr for the monosymmetric beam in Fig. 5.17 is given by ⎧ ⎫ ⎨ 2 2 π π EIw πγM ⎬ πγM Mcr = EIz GIt 1+ + + (5.70) ⎩ L 2 ⎭ 2 GIt L2 where γM is given by βy EIz γM = L GIt

(5.71)

The monosymmetry parameter for the section βy is given by βy = where βy1

βy2

+ 1* βy1 − βy2 + βy3 − 2z0 Iy

(5.72)

b3 t2 = (hs − zc ) 2 + b2 t2 (hs − zc ) 12 b31 t1 = zc + b1 t1 zc2 12

(5.73)

(5.74)

Structural Design of Steelwork to EN 1993 and EN 1994

βy3

tw = 4

t2 hs − zc − 2

4

t1 − zc − 2

•

127

4 (5.75)

where hs = d −

t 1 + t2 2

(5.76)

t

b2 t2 hs + 2w (d − t1 − t2 )(d − t2 ) zc = b1 t1 + b2 t2 + (d − t1 − t2 )tw

(5.77)

z0 = αhs − zc

(5.78)

α=

1 b

1 + b1 2

3

t1 t2

(5.79)

The warping constant Iw is now given by Iw = α(1 − α)Iz h2s

(5.80)

5.1.6.4 ‘T’ Beams For ‘T’ beams, it should be first checked that the value of K from Eq. (5.8) is real as a large number of commercial ‘T’ beams have Iy > Iz in which case lateral torsional buckling cannot occur. Where Mcr needs calculating, it should be noted Iw is zero as α = 0 (Eq. (5.80)) and z0 = −zc . The position of the centroid zc and Iy are tabulated in section property tables.

5.1.6.5 Parallel Flange Channels The procedure for calculating Mcr follows that for ‘I’ beams except that Iw is calculated as Iw =

tf b3f h2 3bf tf + 2htw 12 6bf tf + htw

(5.81)

where bf and tf are the width and thickness of the flange and h and tw are the height and thickness of the web. For channels with tapered flanges tf may be taken as the mean thickness of the flange (Kirby and Nethercot, 1979).

5.2 PURE TORSIONAL BUCKLING This form of failure can only occur in open sections and is most likely to only where the sections are thin walled. There can then be an interaction between strut buckling or buckling in pure torsion.

128

•

Chapter 5 / Laterally Unrestrained Beams σ x Thickness (t ) b

b

b

b

x (a) Basic layout

cL

(b) Cross-section

s

dv

φ φ

dH

dθ dd

(c) Deflected geometry

FIGURE 5.18 Cruciform strut

5.2.1 Interaction between Torsional Buckling and Strut Buckling This is best illustrated by considering the case of a cruciform strut loaded by a uniform stress σ over its cross-section. To demonstrate the principles a thin section strut is considered (Fig. 5.18). The load induced by the stress σ remains parallel to the x axis, and therefore induces a lateral force in the yz plane. If the vertical component in Fig 5.18(c) is dV then the horizontal component dH is given by +dH = dV tan φ = φdV = s

dθ dV dx

(5.82)

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129

as φ is small, and where dθ/dx is the angle of twist per unit length and s is a distance measured from the x axis. The force dV is given by dV = σdA = σtds

(5.83)

where t is the thickness. The incremental torque dT is given by dT = sdH

(5.84)

Substitute Eqs (5.81) and (5.83) into Eq. (5.84) to give dT = σ

dθ 2 ts ds dx

(5.85)

Integrate Eq. (5.85) over the four arms of the strut to give h dθ dθ 4 3 T = 4σ t s2 ds = σ b t dx 0 dx 3

(5.86)

where 2b is the width of the strut. Equation (5.86) may be rewritten as T =σ

dθ Ix dx

(5.87)

where Ix is the polar second moment of area about the x axis, and is given by Ix =

4 3 b t 3

(5.88)

The second moment of area of the strut about either the z or y axis is given by Iz = Iy =

1 2 (2b)3 t = b3 t 12 3

(5.89)

as the other arm of the strut has negligible second moment of area (bt3 /6) compared with the other direction. Note that

2 3 4 Ix = Iy + Iz = 2 b t = b3 t 3 3

(5.90)

From torsion theory the torque that may be carried by the section is given by T = GIt

dθ dx

(5.91)

where G is the shear modulus and Tt is the torsional second moment of area. For the cruciform thin walled section It is given by 4 1 3 bt = bt3 It = 4 (5.92) 3 3

•

Chapter 5 / Laterally Unrestrained Beams

L b Strut buckling

π2E 6s0

130

Yield

Torsional buckling b t

E 2(1v)σ0

FIGURE 5.19 Interaction diagram

Substitute Eq. (5.92) into Eq. (5.91) to give T=

4 3 dθ bt G 3 dx

(5.93)

For the section to be able to sustain its twisted shape the two values of the torques from Eqs (5.87) and (5.93) must be equal, so σ=G

2 t b

(5.94)

There also exists the possibility that the strut can undergo normal Euler buckling under a stress σcr given by σcr =

π2 E 23 b3 t π2 EIz π2 E = = 6 AL2 4btL2

2 b L

(5.95)

Assuming no interaction, there are three possibilities of behaviour; the stress produces yield; torsional buckling occurs, or strut buckling occurs (Fig. 5.19). The stress which causes torsional buckling and strut buckling to occur simultaneously is given when the stresses from Eqs (5.94) and (5.95) are equal, or 2 t π2 E b 2 G = b 6 L

(5.96)

Structural Design of Steelwork to EN 1993 and EN 1994 or

L b = b t

π2 E b = 6G t

π2 (1 + v) 3

•

131

(5.97)

as G = E/2(1 + v). The transition from yield to torsional buckling occurs when the stress in Eq. (5.94) equals the yield stress σ0 , or b G E = (5.98) = t σ0 2(1 + ν)σ0 and from yield to Euler buckling when the stress in Eq. (5.95) equals σ0 , or L π2 E = b 6σ0

(5.99)

Note, other sections such as thin walled angles may also suffer similar behaviour, but the analysis is more complex as buckling is about the principal axes.

5.2.2 Torsional Buckling Interaction The critical axial load for torsional buckling Ncr is given in Chapman and Buhagiar (1993) as 1 π2 EIw Ncr,T = 2 GIt + (5.100) is lT2 where lT is the buckling length, and is the polar radius of gyration given by i2s = i2z + i2y + y20 + z02

(5.101)

where iz and iy are the flexural radii of gyration and z0 and y0 are distances from the shear centre to the geometric centroid. For a section whose centroid and shear centre co-incide y0 and z0 are zero. Alternatively the critical stress σcr,T is given by 1 π2 EIw σcr,T = GIt + (5.102) I0 lT2 where I0 is the polar moment of area. Timoshenko and Gere (1961) give the following interaction equation between strut buckling and torsion buckling i2s (N − Ncr,z )(N − Ncr,y )(N − Ncr,T ) − N 2 z02 (N − Ncr,y ) − N 2 y20 (N − Ncr,z ) = 0 (5.103) where N is the critical value of the axial load and Ncr,z and Ncr,y are the Euler buckling loads about the zz and xx axes.

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Chapter 5 / Laterally Unrestrained Beams For a section whose centroid and shear centre co-incide, Eq. (5.103) reduces to (N − Ncr,z )(N − Ncr,y )(N − Ncr,T ) = 0

(5.104)

That is, N is therefore the least of Ncr,x , Ncr,y and Ncr,T . Chapman and Buhagiar (1993) also indicate that where buckling can occur about both axes (i.e. where the buckling lengths and second moments of area are approximately equal, or where the critical loads are similar), then the imperfection factor η in the standard strut buckling interaction equation should be taken as twice its normal value to allow for torsional buckling.

5.3 PLATE GIRDERS Plate girders are used either on long spans where a rolled section would need to be spliced and as a result may be inefficient, or to support heavy loads such as on a bridge structure. It is important to note that although plate girders may be lighter than other forms of compound beams, fabrication costs are likely to be much higher. Also as Corus now roll UKB’s with a depth of 1016 mm, the use of plate girders in building structures unless spans are extremely high is less likely. Plate girders are built up from two flange plates and a web plate, generally from the same grade of steel. Continuous automatic electric arc or submerged gas welding is used to form the fillet welds between the flange and web (Fig. 5.20). Such welding is generally performed as a double pass one on either side of the girder. It should be noted that this process may cause very high residual stresses to exist in the flanges and web. The exact magnitudes will depend also on whether the flange plates were sheared or flame cut to size (Nethercot, 1974a). The stiffeners are then welded in place often manually. Such stiffeners are needed either to help combat the effects of web buckling or to provide support to any concentrated load or reaction. Only straight girders with equal flanges and vertical stiffeners are considered in this text. Flange plate

Web plate

Welds on 2nd pass

Fillet welds

1st pass

FIGURE 5.20 Plate girder fabrication

Welds from 1st pass

2nd pass

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5.3.1 Minimum Web Thickness With no web stiffeners the web should be sized to avoid the flange undergoing local buckling due to the web being unable to support the flange. This is known as flange induced buckling (cl. 8 EN 1993-1-5). In pure bending the flanges are subjected to equal and opposite forces. The force per unit length is given by σv tw =

fyf Afc R

(5.105)

where σv is the vertical stress in the web, Afc is the area of the compression flange and R is the radius of curvature. The curvature is dependant upon the variation of strain εf occurring at the mid-depth of the flanges. The residual strain due to fabrication εf is assumed to have a value of 0,5εy , and εf is the sum of the yield strain εy and the residual strain εf , that is, a total of 1,5εy . The radius of curvature R is then given by R=

0,5hw εf

(5.106)

where for simplicity hw has been taken as the depth between the centroids of the flanges rather than the clear depth of the web. It should be noted any error will be small. From Eqs (5.105) and (5.106), σv = 3

2 Afc fyf Aw E

(5.107)

The stress σv cannot exceed the elastic critical buckling stress for a simply supported thin plate which is given by (Bulson, 1970) as π2 E σv = 12(1 − ν2 )

tw hw

2 (5.108)

Equating Eqs (5.107) and (5.108) gives with a slight change in notation with Ac replacing Afc hw E Aw ≤ 0,55 (5.109) tw fyf Ac EN 1993-1-5 modifies the co-efficient of 0,55 to allow for situations where higher strains are required. Thus the critical hw /tw ratio is given by hw E Aw ≤k (5.110) tw fyf Ac where fyf is the yield strength of the compression flange, Afc is the effective area of the compression flange and Aw is the area of the web. The parameter k takes values of 0,3 where plastic hinge rotation is utilized, 0,4 if the plastic resistance is utilized

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Chapter 5 / Laterally Unrestrained Beams and 0,55 if the elastic resistance is utilized. Thus for rigid (continuous) design k = 0,3 unless the analysis is elastic with no redistribution. For simply supported beams k may be taken as 0,4.

5.3.2 Bending Resistance The section classification is determined in the same manner as rolled sections.

5.3.2.1 Compression Flange Restrained (i.e. Lateral Torsional Buckling cannot Occur) There are two methods that can be used for girder design: (1) The flanges carrying the bending moment and the web the shear force This is probably best used where the maximum bending moment and maximum shear force are not coincident, as the ability of the flange to contribute towards shear capacity may be utilized. Thus for a restrained beam under a UDL, this method may be advantageous. If the maximum bending moment and maximum shear are coincident in either a simply supported beam under point loading or in a continuous beam at the internal support, then the flange capacity will not be able to be utilized to resist shear, thus probably necessitating a thicker web. With this method, the moment capacity is only dependant on the section classification of the flanges as the web does not carry compression. (2) The girder carrying the forces as an entity This method of design is more complex as the beam is likely to be Class 4 and may not show any resultant economies over the first method, but should be utilized where maximum moment and maximum shear are co-incident.

5.3.2.2 Lateral Buckling May Occur In this case the second method must be used. The design then follows that for rolled beams, except that the general method for calculating χLT should be used with the value of αLT appropriate for welded sections. Section properties will need to be calculated from first principles.

5.3.3 Basic Dimensioning One method of dimensioning is to consider a minimum weight solution. It must be noted that a minimum weight solution is not synonymous with a minimum cost solution (Gibbons, 1995). Method 1: Assuming the moment to be resisted by the flanges alone, then MRd = fyd bf tf hw

(5.111)

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where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. Equation (5.111) is slightly conservative for beams of Classes 1 to 3. The cross-sectional area A is given by A = 2bf tf + hw t

(5.112)

Eliminate bf tf between Eqs (5.111) and (5.112) to give A=

2MRd + hw t hw fyd

(5.113)

Define the web slenderness ratio hw /t as λ, then Eq. (5.113) becomes A=

2MRd + λt2 λtfyd

(5.114)

For an optimum solution, dA/dt = 0, so Eq. (5.114) becomes, dA 2MRd + 2λt = 0 =− dt λfyd t2 or,

t=

3

and

MRd λ2 fyd

hw =

3

(5.115)

(5.116)

λMRd fyd

(5.117)

The area of the web, Aw is then given by 2 MRd 3 Aw = 2 λfyd Using Eq. (5.112), the flange area, Af is given by 2 MRd MRd 3 Af = bf tf = = 2 λfyd λM fyd 3 f Rd

(5.118)

(5.119)

yd

Thus the area of a single flange is equal to that of the web. Method 2 (Classes 1 and 2): It is recognized that this is not a likely case but is included for completeness. The moment is resisted by the complete section, when the moment capacity is given by that due to the flanges (Eq. (5.111)) and the additional plastic capacity of the web MRd = fyd bf tf hw + fyd

th2w 4

(5.120)

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•

Chapter 5 / Laterally Unrestrained Beams where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. The cross-sectional area A is given by Eqs (5.112), thus from Eqs (5.112) and (5.120), bf tf is given by bf tf =

MRd hw t − fyd hw 4

(5.121)

Eliminate bf tf between Eqs (5.112) and (5.121) to give A=

2MRd hw t hw t 2MRd + hw t − 2 + = hw fyd 4 hw fyd 2

(5.122)

Define the web slenderness ratio hw /t as λ, then Eq. (5.122) becomes A=

2MRd λt2 + λtfyd 2

(5.123)

For an optimum solution, dA/dt = 0, so Eq. (5.123) becomes, dA 2MRd + λt = 0 =− dt λfyd t2

(5.124)

or, t=

3

and

2MRd λ2 fyd

hw =

3

(5.125)

2λMRd fyd

(5.126)

The area of the web, Aw is then given by 2 4MRd 3 Aw = 2 λfyd Using Eq. (5.121), the flange area, Af is given by 2 MRd 3 1 1 3 3 Af = bf tf = − 2 2 16 λfyd or, 3

Af = Aw

2 MRd 2 λfyd

! 3

3

1 2

−

2 4MRd 2 λfyd

! 3

1 16

(5.127)

(5.128)

=

1 4

Thus the area of the web is equal four times that of a single flange.

(5.129)

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Method 2 (Class 3): The moment is resisted by the complete section, when the moment capacity is given by that due to the flanges (Eq. (5.111)) and the additional elastic capacity of the web MRd = fyd bf tf hw + fyd

th2w 6

(5.130)

where fyd is the design strength of the flanges, tf and bf the thickness and width of the flange plates and hw the distance between the internal faces of the flanges. The cross-sectional area A is given by Eq. (5.112), thus from Eqs (5.112) and (5.130), bf tf is given by bf tf =

MRd hw t − fyd hw 6

(5.131)

Eliminate bf tf between Eqs (5.112) and (5.131) to give A=

2MRd hw t 2hw t 2MRd + hw t − 2 + = hw fyd 6 hw fyd 3

(5.132)

Define the web slenderness ratio hw /t as λ, then Eq. (5.132) becomes A=

2MRd 2λt2 + λtfyd 3

(5.133)

For an optimum solution, dA/dt = 0, so Eq. (5.133) becomes, dA 2MRd 4 =− + λt = 0 2 dt 3 λfyd t

(5.134)

or, t=

3

3MRd 2λ2 fyd

(5.135)

and hw =

3

3λMRd 2fyd

(5.136)

The area of the web, Aw is then given by 2 9MRd 3 Aw = 2 4λfyd

(5.137)

Using Eq. (5.131), the flange area, Af is given by 2 MRd 3 2 1 3 Af = bf tf = − 2 3 6 λfyd

3

9 4

(5.138)

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•

Chapter 5 / Laterally Unrestrained Beams Thus 3

Af = Aw

2 MRd 2 λfyd

! 3

2 3

− 16

! 3

2 3 9MRd

9 4

=

1 2

(5.139)

4λfyd

Thus the area of the web is equal twice that of a single flange. It should be noted that a plate girder may well be Class 4 in which case an effective section needs calculating, and thus minimum weight optimization is not directly possible, although it can be simulated by increasing the applied moment (see Example 5.4).

5.3.4 Web Design Experimental work (Basler, 1961; Porter et al., 1975; Rockey et al., 1978; Davies and Griffith, 1999) showed that after web buckling occurred there was still a reserve of strength in the web. This additional reserve of strength in the web is due to a tension field forming in the central diagonal portion of the web (Fig. 5.21). The shear capacity in the web is determined using a shear buckling slenderness λw which is dependant upon the critical shear strength τcr (cl. A.1 EN 1993-1-5) The critical shear strength τcr is given by τcr = kτ σE

(5.140)

where kτ is a shear buckling co-efficient dependant upon the aspect ratio of a web panel and σE is the elastic critical stress, From classical plate buckling theory (Bulson, 1970) σE =

π2 E 12(1 − ν2 )

t hw

2

= 19000

t hw

2 (5.141)

where t is the thickness and hw the depth. The parameter kτ is given as (Bulson, 1970), for a/hw ≥ 1,0 kτ = 5,34 + 4,00

hw a

2 (5.142)

for a/hw < 1,0 kτ = 4,00 + 5,34

hw a

2 (5.143)

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τcr

hw

a

τcr

(a) Buckled shape

τ Buckled panel

Stiffener

Tension panel

Buckled panel

(b) Panel showing tension field

FIGURE 5.21 Shear failure of a plate girder

The non-dimensionalized web slenderness ratio λw is defined as (cl. 5.3 EN 1993-1-5) ⎤1/2 ⎡ f √yw

⎥ ⎢ λw = ⎣ 3 ⎦ τcr

= 0,76

fyw τcr

(5.144)

For webs with transverse stiffeners at the supports and either intermediate transverse or longitudinal stiffeners, the normalized web slenderness ratio λw is obtained by

140

•

Chapter 5 / Laterally Unrestrained Beams substituting Eqs (5.140) and (5.141) into (5.144) λw =

hw √ 37,4εt kτ

(5.145)

For webs with transverse stiffeners only at the supports, a/hw is large, hence with little loss in accuracy kτ = 5,34 from Eq. (5.142). Thus λw is given by λw =

hw hw = √ 86,4εt 37,4εt 5,34

(5.146)

Clearly the upper limit to the value the shear force that may be carried by the web is the plastic shear capacity. The limiting value of web slenderness beyond which buckling need to be considered may be derived as follows. From Eq. (5.142) for an infinitely long web, kI = 5,34, thus from Eq. (5.140), the limiting value of hw /t is given when √ τcr = τyw = fyk / 3, or hw = t

5,34π2 E 235 × 12(1 − ν2 )

235 = fyk

5,34π2 × 210 × 103 ε = 65,7ε 12(1 − 0,32 ) × 235

(5.147)

√ The code uses a lower limit of 72ε/η for unstiffened webs and 31ε kτ /η (cl 5.1.(2), EN 1993-1-5). The recommended value of η for steel grades of S460 or lower is 1,2. For steel grades higher than S460, η = 1,0. With η = 1,2, the lower limit for steel grades up to and including S460 is 72ε/1,2 = 60ε which is slightly more conservative than the figure derived in Eq. (5.147). Basler (1961) and Rockey and Škaloud (1971) recognized that actual web behaviour could be categorized by three regimes: pure shear, elastic buckling at the extremes and a transition phase between the two limits. (1) Non-rigid end post In the case of a non-rigid end post (which cannot generate post-buckling strength as it is incapable of resisting the additional horizontal forces) EN 1993-1-5 only defines two zones for the contribution of the web to shear buckling resistance χw where a tension field cannot be generated as the anchorage force is unable to be sustained, λw ≤ 0,83/η χw = η λw > 0,83/η 0,83 χw = λw

(5.148)

(5.149)

(2) Rigid end post Where there is a rigid end stiffener, the anchorage force from tension field theory can be sustained. The original theory behind tension field theory was outlined by Porter et al. (1975) and Rockey et al. (1978). EN 1993-1-5 has adopted Höglund’s rotating stress–field theory (Davies and Griffith, 1999) which is easier to apply

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than the original tension field theory. Höglund’s rotating stress–field theory also mobilizes post-buckling behaviour, but only if there is a rigid end post. In this case the relationships between normalized web slenderness λw and χw are given by λw ≤ 0, 83/η χw = η

(5.150)

0,83/η ≤ λw < 1,08 χw =

0,83 λw

(5.151)

λw ≥ 1,08 χw =

1,37 0,7 + λw

(5.152)

It will be observed that the difference between the two methods is that the web shear parameter χw is enhanced for λw ≥ 1,08. The design resistance of a web Vb,Rd whether stiffened or unstiffened (cl 5.2 EN 1993-1-5) is given by Vb,Rd = χv hw t

f √yw 3

γM1

(5.153)

where shear co-efficient χv is given by χv = χw + χf ≤ η

(5.154)

The parameter χf represents the contribution to shear resistance from the flanges for MEd < Mf,Rd and is given by √ bf tf2 fyf 3 MEd 2 χf = 1− (5.155) cthw fyw Mf,Rd where bf is the width of the flange taken as not greater than 15εtf on each side of the web, tf is the web thickness, Mf,Rd is design moment of resistance of the cross-section determined using the effective flanges only and c is the width of the portion of the web between the plastic hinges (see Fig. 5.22) and is given by Mpl,f 1,6bf tf2 fyf c = a 0,25 + 1,6 (5.156) = a 0,25 + Mpl,w th2w fyw Equation (5.155) is derived by determining the shear that may be carried by the portion of both flanges of length c with the yield strength fyf reduced by considering the effect of induced axial forces in the flanges. The background to the simplified method adopted in the Code is given in Davies and Griffith (1999).

•

Chapter 5 / Laterally Unrestrained Beams Position of hinges

Flange

Web

d

142

c a (a) Basic geometry of web panel

(b) Failure mechanism used to derive post buckling strength of web

FIGURE 5.22 Overall behaviour of web at failure The calculations for web shear capacity are iterative as both the shear factors and the applied moment are dependant upon the stiffener spacing. The calculations for shear capacity in the examples following were performed on a spreadsheet.

5.3.5 Stiffeners 5.3.5.1 Rigid End Post (cl. 9.3.1 EN 1993-1-5) This may either be a set of flats welded at the end and above the support with the centroids a distance e apart, or the end post may comprise a rolled section in which case e is the distance between the flange centroids. The horizontal stress σh for large slenderness ratios can be given as 0.43 σh = fy (5.157) λw Substitute the value of λw from Eq. (5.145) into Eq. (5.157) to give an upper bound for the UDL qh as √ fy t 2 ε k τ qh = σh t = 16.1 (5.158) hw

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Taking the maximum value of ε as 1,0 and the maximum value of kτ as 9,34, the value of qh becomes qh = 49

t2 fy hw

(5.159)

The distributed load is not uniform over the depth of the girder and the theoretical value of σh is high, the co-efficient of 49 in Eq. (5.159) is replaced by 32. Assuming the end post is simply supported, the maximum moment Mmax is given by Mmax =

qh h2w 8

(5.160)

The section modulus, ignoring any contribution from the web, W is given by Amin e, thus assuming the maximum stress is given by fy , then 32t2 fy h2w

σmax

Mmax h 8 = fy = = w W Amin e

(5.161)

Thus the minimum cross-sectional area of each pair Amin is given by Amin =

4h2w t e

(5.162)

with e > 0,1hw . The restriction on e would appear to be a detailing requirement. The end panel may be designed as non-rigid shear panel carrying the whole of the applied shear. This was originally proposed by Basler (1961). From Eq. (5.153) with χf = 0, the required value of χw is given by √ VEd 3γM1 χw = fyw hw t

(5.163)

From Eq. (5.149), the normalized web slenderness λw is given by λw =

0,83 χw

(5.164)

The buckling parameter kτ is given from Eq. (5.145) as kτ =

hw 37,4tελw

2 (5.165)

As a/hw < 1, kτ is given by Eq. (5.143), or the required panel width a is given as 5,35 a = hw (5.166) kτ − 4,00 Note, this will give an upper bound to the value of a.

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5.3.5.2 Transverse Stiffeners (cl 9.2.1 EN 1993-1-5) These should be checked as a simply supported beam with an initial sinusoidal imperfection w0 given by Eq. (5.169) together with any eccentricities. The transverse stiffener should carry the deviation forces from the adjacent panels assuming the adjacent transverse stiffeners are rigid and straight. A second order analysis should be used to determine that the maximum stress does not exceed fyd nor any additional deflection b/300. This will be more critical for single sided stiffeners. As double sided plate stiffeners have been used in the ensuing examples, only the stress criterion has been checked. In the absence of transverse loads or axial forces in the stiffeners, then the strength and deflection criteria are satisfied if they have a second moment of area given by σM b 4 300 Ist = 1 + w0 u (5.167) E π b where σM is given by σM

σcr,c NEd = σcr,p b

1 1 + a1 a2

(5.168)

where a1 and a2 are the panel lengths either side of the stiffener under consideration, NEd is the larger compressive force in the adjacent panels, b is the height of the stiffener. The initial imperfection w0 is given as w0 =

1 LEAST(a1 , a2 , b) 300

(5.169)

The parameter u is given by u=

π2 Eemax 300bfy γM1

≥ 1,0

(5.170)

The distance emax is taken from the extreme fibre of the stiffener to the centroid of the stiffener. The critical stress for plate between vertical stiffeners σcr,c is given by σcr,c =

π2 Et2 12(1 − ν2 )a2

(5.171)

The critical stress σcr,p is given by kσ,p σE . The value of σE is given by Eq. (5.141). The value for plates with longitudinal stiffeners in Annex A of EN 1993-1-5. For unstiffened plates σcr,p = σcr,c To avoid lateral torsional buckling of the stiffener, fy IT ≥ 5,3 Ip E

(5.172)

where IT and Ip are St. Venant torsional constant for the stiffener alone and Ip is the polar second moment of area about the edge fixed to the plate. Eq. (5.172) can be

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derived as follows. The critical buckling stress σcr for open section stiffeners (with negligible warping stiffness) is given by Eq. (5.102) with Iw = 0, σcr = G

IT Ip

(5.173)

From cl 9.2.1(8) of EN 1993-1-5, the critical stress σcr is limited by σcr ≥ θfy

(5.174)

For plate stiffeners, θ is taken as 2,0, thus Eq. (5.173) becomes fy IT ≥ 5,2 Ip E

(5.175)

The code replaces the co-efficient 5,2 by 5,3.

5.3.5.3 Intermediate Transverse Stiffeners (cl. 9.3.3 EN 1993-1-5) The force Ns,Rd to be resisted by a stiffener is given by Ns,Rd = VEd − χw hw t

f √yw 3

γM1

(5.176)

Note, χw is calculated for the web panel between adjacent stiffeners assuming the stiffener under consideration is removed. In the case of variable shear, then the check is performed at a distance 0,5hw from the edge of the pane with the larger shear force. To determine the buckling resistance of the stiffener a portion of the web may taken into account (Rockey et al., 1981). A section of the web in length equal to 15εt either side of the stiffener may be considered (cl 9.1, EN 1933-1-3) (Fig. 5.23) For a symmetric stiffener, the effective area Ae is given by Aequiv = Ast + 30εt2

(5.177)

and the effective second moment of area Iequiv by Iequiv = Ist +

1 30εt4 12

(5.178)

where Ast is the area of the stiffener and Ist is the second moment of area of the stiffener. For end stiffeners the co-efficient of 30 in Eqs (5.177) and (5.178) should be replaced by 15. The effective length of the stiffener may be taken as 0,75hw and buckling curve ‘c’ used to determine the strength reduction factor (cl 9.4 EN 1993-1-5). In order to provide adequate restraint against buckling it was found that the stiffeners need to possess a minimum second moment of area (Rockey et al., 1981).

146

•

Chapter 5 / Laterally Unrestrained Beams

A

A

Elevation

t

15 εt

15 εt

Section AA

FIGURE 5.23 Stiffener geometry The minimum second moment of area Is is given by √ for a/hw < 2 Ist ≥ 1,5 for a/hw ≥

√

h3w t3 a2

(5.179)

2

Ist ≥ 0,75hw t3

(5.180)

It can be demonstrated that for compression buckling a change from 1 to 2 half sine √ waves occurs at a/hw = 2, and that thereafter the buckling co-efficient is sensibly independent of the aspect ratio of the panel. Thus Eq. (5.180) is determined from √ Eq. (5.179) by substituting a = hw 2 (cl 9.3.3 (3) EN 1993-1-5).

5.3.5.4 Plate Splices (cl 9.2.3 EN 1993-1-5) The splice whether in the web or flanges, should ideally occur at a transverse stiffener. If not then the stiffener should be at a distance no greater than b0 /2 along the thinner plate where b0 is the depth of the web (or the least spacing of longitudinal stiffeners).

5.3.5.5 Longitudinal Welds (Web to Flange) (cl 9.3.5 EN 1993-1-5) The weld between the web and flange(s) should be designed for a shear flow of VEd /hw , provided VEd < χw hw t

f √yw 3

γM1

(5.181)

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If the condition in Eq. (5.181) is not satisfied, the welds should be designed under a √ shear flow of ηt(fyw / 3)/γM1 .

EXAMPLE 5.3 Design of a laterally restrained plate girder Design a plate girder in Grade S355 steel to carry a characteristic variable load of 150 kN/m over a span of 20 m. The compression flange is fully restrained against lateral torsional buckling. For a plate girder span/depth ratios are generally around 8 or 10 to 1. The higher this ratio is the lower the flange size but at the probable expense of a thicker web to overcome buckling. To calculate the imposed bending moment, assume the total weight of the beam is 100 kN. Total loading on the beam = 1,5 × 150 + 1,35 × (100/20) = 232 kN/m MSd = 232 × 202 /8 = 11,6 MNm The critical slenderness ratio for the web can be controlled by flange-induced buckling with k = 0,4 as plastic rotation is not utilized. If the flanges resist the bending moment, Aw = Af (from Eq. (5.119)), thus the critical hw /t ratio is given by Eq. (5.110) hw E ≤k t fyf

Aw 210 × 103 √ = 0,4 1 = 237 Ac 355

From Eq. (5.117) calculate hw hw =

3

λMRd 237 × 11,6 × 109 = = 1978 mm 355 fyd 1,0

The thickness t is given as t=

hw 1978 = = 8,34 λ 237

Use t = 9 mm Maximum ratio flange outstand to flange thickness beyond the weld for a Class 1 section is 9ε where ε = (235/355)1/2 (=0,814). So maximum flange outstand is 9 × tf × (235/355)1/2 = 7,33tf So flange area is 14,66tf2 (ignoring the effect of weld width). Mpl,Rd ≈ Af

fy hw γM0

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•

Chapter 5 / Laterally Unrestrained Beams or Af =

MEd fy hw γ M0

=

11,6 × 109 1978 355 1,0

= 16, 520 mm2

or 14,66tf2 = 16,520 or Inset tf = 33,6 mm Use 35 mm thick plate with a width of 500 mm. Overall depth, h: h = 2 × 35 + 1978 = 2048 mm. Use h = 2000 mm. Check actual hw /t ratio: Actual web slenderness: hw 1930 = = 214 t 9 Web is Class 4. However as the webs do not carry any compression, then the section may be treated dependant upon the classification of the flanges (cl 5.5.2 (12) EN 1993-1-1). Maximum hw /t ratio: hw E Aw 210 × 103 =k = 0,4 t fyf Af 355

1930 × 9 = 236 500 × 35

The actual value is below the allowable, and is therefore satisfactory. Plastic moment of resistance of the flanges, Mpl,Rd : Mpl,Rd = Af

fy 355 (2000 − 35) = 12,2 MNm (h − tf ) = 35 × 500 γM0 1,0

Vpl,Rd = 232 × 10 = 2320 kN (at the support) hw = h − 2tf = 2000 − 2 × 35 = 1930 mm The web will be designed both ways, non-rigid and rigid end post: The determined flange and web contributions take the maximum shear and maximum moment in a web panel, even though they are at opposite ends of the panel. This will be conservative. Determination of χf : To determine χf , the flange width is limited to 15εtf on either side of the web:

235 15εtf = 15 × 35 355

1/2 = 427 mm

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Actual flange width = 0,5(500 − 9) = 245 mm. This is less, therefore use actual width, so Mf,Rd = 12,2 MNm (from above). Both the web and flanges have a yield strength of 355 MPa. Non-rigid end post: First panel from the support: Intermediate stiffener is 1,2 m from the support. Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

MEd = 2320 × 1,2 − 232

1,6 × 500 × 352 × 355 = 1200 0,25 + 9 × 19302 × 355

= 335 mm

1,22 = 2617 kNm 2

Determine the flange contribution factor χf from Eq. (5.155) √ √ bf tf2 fyf 3 MEd 2 500 × 352 × 355 × 3 2617 2 χf = 1− = 1− cthw fyw Mf,Rd 335 × 9 × 1930 × 355 12200 = 0,174 hw 1930 = = 1,608 a 1200 a = 0,622 hw For a/hw < 1,0, kτ is given by Eq. (5.143), or

hw kτ = 4 + 5, 34 a

2 = 4 + 5,34 × 1,6082 = 17,81

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 1,670 ! √ = √ 235 37,4εt kτ 37,4 355 × 9 17,81

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,497 1,670 λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,174 + 0,497 = 0,671

150

•

Chapter 5 / Laterally Unrestrained Beams TABLE 5.5 Shear capacity calculations with non-rigid end post (Example 5.3). Distance from Support (m) 1,2 2,55 4,15 6,25 10 Panel width (m) 1,2 1,35 1,6 2,1 3,75 2617 5162 7630 9969 11600 Moment M Ed (kNm) 2320 2042 1728 1357 870 Shear V Ed (kN) c (mm) 335 377 447 586 1047 0,174 0,133 0,083 0,034 0,006 χf 0,622 0,699 0,829 1,088 1,943 a/hw 17,81 14,91 11,77 8,719 6,400 kτ λw 1,670 1,825 2,054 2,387 2,786 0,497 0,455 0,404 0,378 0,298 χw 0,671 0,588 0,487 0,382 0,304 χv 2389 2093 1735 1361 1081 V Rd (kN)

The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t

f √yw 3

γM1

355 0,671 × 1930 × 9 √3 = = 2389 kN 1000 1,0

The calculations for subsequent panels are summarized in Table 5.5 (together with the first panel) Intermediate stiffeners: Check strength: First stiffener: The axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5 hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(1,2 − 0,5 × 1,930) = 2265 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1200 + 1350 = 2550 mm hw 1930 = = 0,757 a 2550 a = 1,321 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or 2 hw kτ = 5,34 + 4 = 5,34 + 4 × 0,7572 = 7,63 a

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Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,55 ! √ √ = 37,4εt kτ 37,4 × 9 235 7,63 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = 0,325 = 2,55 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 2265 − 0,325 × 1930 × 9 × 10−3

355 √ 3

1,0

= 1108 kN

Second stiffener: Axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(2,55 − 0,5 × 1,930) = 1952 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1350 + 1600 = 2950 mm hw 1930 = = 0,654 a 2950 a = 1,528 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or kτ = 5,34 + 4

hw a

2 = 5,34 + 4 × 0,6542 = 7,05

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,65 ! √ √ = 37,4εt kτ 37,4 × 9 235 7,05 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,313 2,65 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1952 − 0,313 × 1930 × 9 × 10

−3

355 √ 3

1,0

= 838 kN

152

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Chapter 5 / Laterally Unrestrained Beams Third stiffener: Axial force NEd is given by Eq. (5.176) as NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(4,15 − 0,5 × 1,930) = 1581 kN The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 1600 + 2100 = 3700 mm hw 1930 = = 0,522 a 3700 a = 1,917 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or

hw kτ = 5,34 + 4 a

2 = 5,34 + 4 × 0,5222 = 6,43

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,78 ! √ √ = 37,4εt kτ 37,4 × 9 235 6,43 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,299 2,78 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1581 − 0,299 × 1930 × 9 × 10−3

355 √ 3

1,0

= 517 kN

Fourth stiffener: Axial force NEd is given by Eq. (5.176) NRd = VEd − χw hw t

f √yw 3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel with the higher shear: VEd = 2320 − 232(6,25 − 0,5 × 1,930) = 1094 kN

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The web contribution parameter χw is calculated assuming the stiffener is removed, thus a = 2100 + 3750 = 5850 mm hw 1930 = = 0,330 a 5850 a = 3,03 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or kτ = 5,34 + 4

hw a

2 = 5,34 + 4 × 0,332 = 5,78

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 2,93 ! √ √ = 37,4εt kτ 37,4 × 9 235 5,78 355

As λw > 1,08, χw is given by Eq. (5.145) as χw =

0,83 0,83 = = 0,283 2,93 λw

NRd = VEd − χw hw t

f √yw 3

γM1

= 1094 − 0,283 × 1930 × 9 × 10

−3

355 √ 3

1,0

= 86 kN

This value is small, thus there is no need to check the centre stiffener. The minimum stiffness requirement for all but the two panels either side of the central √ stiffener is given by the case a < 2hw , so design on the least value of a: Is = 1,5

h3w t3 19303 × 93 = 1,5 = 5,46 × 106 mm4 2 a 12002

Use 9 mm thick plate, then the total breadth of the stiffener b is given by b=

3

12 × 5,46 × 106 = 194 mm 9

Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as NRd = 2 × 200 × 9

355 × 10−3 = 1278 kN. 1,0

Buckling check: Effective length = 0,75 × 1930 = 1448 mm From Eq. (5.177), the effective area Aequiv is given as Aequiv = Ast + 30εt2 = 200 × 9 + 30 × 9

235 = 2020 mm2 355

154

•

Chapter 5 / Laterally Unrestrained Beams From Eq. (5.178), the effective area Iequiv is given as Iequiv = Ist + 30εt4 =

235 1 1 2003 × 9 + 30 × 94 = 6,013 × 106 mm4 12 12 355

π2 EIequiv π2 × 210 × 6,013 × 106 = = 5944 kN L2 14482 Aequiv fy 2020 × 355 × 10−3 λ= = 0,121 = Ncr 5944

Ncr =

As λ ≤ 0,2 strut buckling need not be checked (cl 6.3.1.2 (4) (EN 1993-1-1)). Thus the minimum stiffener size will be adequate for all the intermediate stiffeners. End stiffener: Try a 500 wide by 15 thick plate: The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 15εt = 15 × 9 (235/355) = 110 mm As,eff = As + Aweb = 500 × 15 + 110 × 9 = 8490 mm2 15 × 5003 110 × 93 Is,eff = Is + Iweb = + = 0,156 × 109 mm4 12 12 Is,eff 0,156 × 109 i= = = 135,6 mm As,eff 8490 Use an effective length of 0,75hw = 0,75 × 1930 = 1448 mm π2 EI 210 × 0,156 × 109 π2 = = 154210 kN l2 14482 Afy 8490 × 355 λ= = = 0,14 Ncr 154210 × 103

Ncr =

From cl 6.3.1.2 (4) (EN 1993-1-1), there is no reduction for strut buckling as λ < 0,2. So, fy 355 × 10−3 = 3014 kN = 8490 γM0 1.0 This exceeds the reaction of 2320 kN. NRd = A

Check cl 9.2.1 (7) Determine Ip : The second moment of area about the web centre line, Iy : Iy =

15 × 5003 bh3 = = 0,156 × 109 mm4 12 12

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The second moment of area normal to the web centre line about one edge, Iy : Iy =

500 × 153 bh3 = = 0,563 × 106 mm4 3 3

Ip = Ix + Iy = 0,156 × 109 + 0,563 × 106 = 0,157 × 109 mm4 bh3 = 0,563 × 106 mm4 3 IT 0,563 × 106 = = 3,59 × 10−3 Ip 0,157 × 109 IT =

Limiting value: 5,3

fy 355 = 5,3 = 8,96 × 10−3 E 210 × 103

The actual value is less than the limiting value thus the stiffener size must be increased. A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT /IP = 0,01) (and will clearly satisfy the strength and buckling criteria). Flange to web welds: √ From Table 5.5 VEd > hw t( fyw / 3)/γM1 as the flange contribution χf has been mobilized. Thus welds should be designed for a shear flow of

ηt

f √yw 3

γM1

= 1,2 × 9

355 √ 3

1,0

= 2214 N/mm

The final layout of the girder whose self-weight is 91,1 kN is given in Fig 5.24(a). Rigid end post: First panel from the support: Intermediate stiffener is 1,45 m from the support. Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

1,6 × 500 × 352 × 355 = 1450 0,25 + 9 × 19302 × 355

= 405 mm MEd = 2320 × 1,45 − 232

1,452 = 3120 kNm 2

Determine the flange contribution factor χf from Eq. (5.155) √ √ bf tf2 fyf 3 MEd 2 500 × 352 × 355 × 3 3120 2 χf = 1− = 1− cthw fyw Mf,Rd 405 × 9 × 1930 × 355 12200 = 0,141

Chapter 5 / Laterally Unrestrained Beams cL

35 500 flanges 9 mm web

2,0

•

End stiffeners 500 25

1,2 2,55 4,15

Intermediate stiffeners 2 200 9

6,25 (a) Non-rigid end post

cL

Intermediate 2 200 9

2,0

156

End post 2 500 15

1,45 0,3

3,25 6,0 (b) Rigid end post

FIGURE 5.24 Final layout for the beam in Example 5.3

1930 hw = = 1,331 a 1450 a = 0,751 hw For a/hw < 1,0, kτ is given by Eq. (5.142), or 2 hw kτ = 4 + 5,34 = 4 + 5,34 × 1,3312 = 13,46 a Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 1930 = 1,920 ! √ = √ 37,4εt kτ 37,4 235 × 9 13,46 355

As λw > 1,08, χw is given by Eq. (5.152) as χw =

1,37 1,37 = = 0,523 0,7 + 1,92 0,7 + λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,141 + 0,523 = 0,664

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The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t

f √yw 3

γM1

355 0,664 × 1930 × 9 √3 = 2363 kN = 1000 1,0

The resultant spacing of stiffeners and a summary of the remaining calculations is given in Table 5.6 The final result for the calculation of stiffener forces are also given in Table 5.6. The minimum stiffness requirement for all but the two panels either side of the central √ stiffener is given by the case a < 2hw , so design on the least value of a: Is = 1,5

h3w t3 19303 × 93 = 1,5 = 3,74 × 106 mm4 a2 14502

Use 9 mm thick plate, then the total breadth of the stiffener b is given by 6 3 12 × 3,74 × 10 b= = 171 mm 9 Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as NRd = 2 × 200 × 9

355 × 10−3 = 1278 kN. 1,0

This exceeds the values of NEd in the last line of Table 5.6. Rigid end post: Use Eqs (5.163)–(5.166) to determine the maximum panel size in accordance with cl 9.3.1(4) (EN 1993-1-5) Determine χw from Eq. (5.163) √ √ VEd 3γM1 2320 × 103 3 × 1,0 χw = = = 0,652 fyw hw t 355 × 1930 × 9

TABLE 5.6 Shear capacity calculations with rigid end post (Example 5.3). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) Stiffener force N Rd (kN)

1,45 1,45 3120 2320 405 0,141 0,751 13,46 1,920 0,523 0,664 2363 1118

3,25 1,80 6315 1984 503 0,089 0,933 10,14 2,213 0,470 0,559 1991 758

6,0 2,75 9744 1566 768 0,029 1,425 7,310 2,607 0,414 0,443 1578 154

10 4,00 11 600 928 1117 0,005 2,073 6,271 2,814 0,390 0,395 1407 −826

158

•

Chapter 5 / Laterally Unrestrained Beams The required normalized web slenderness λw from Eq. (5.164) is given by λw =

0,83 0,83 = = 1,273 χw 0,652

The buckling parameter kτ is given by Eq. (5.165) as kτ =

hw 37,4tελw

2

⎛ ⎜ =⎝

⎞2 1930 ⎟ ! ⎠ = 30,65 235 37,4 × 9 × 355 × 1,273

As a/hw < 1, kτ is given by Eq. (5.166), or the required panel width a is given as a = hw

5,35 5,35 = 1930 = 865 mm kτ − 4,00 30,65 − 4

Centre to centre distance of the pair of double stiffeners should be greater than 0,1 hw (=193 mm). Use a centre to centre distance of 300 mm (The shear resistance will be greater than 2320 kN). Minimum cross-sectional area of each plate is 4hw t2 4 × 1930 × 92 = = 2084 mm2 e 300 As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is 2084/500 (=4 mm). As for the non-rigid end post, a stiffener 500 mm by 15 mm suffices for resisting the reaction, use the same here. The layout of the beam whose self-weight is 93 kN is given in Fig 5.24 (b). The selfweight per unit length of the beam is 4,56 kN/m for the non-rigid end post and 4,51 for the rigid end post (i.e. almost identical). It would be possible to further optimize the weight of the girder by decreasing the web thickness towards the centre and the flange width and/or thickness towards the ends of the beam. Note, however, that so doing may also increase the stiffener requirements.

EXAMPLE 5.4 Design of a plate girder with lateral torsional buckling. Design a plate girder in Grade S355 steel to carry a characteristic variable load of 1500 kN at the centre of a span of 20 m. Lateral torsional restraints exist at the supports and the load. To calculate the imposed bending moment, assume the total weight of the beam is 150 kN. MEd =

1,35 × 150 × 20 1,5 × 1500 × 20 + = 11760 kNm 8 4

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To size the beam allowing for the fact that its classification will be Class 4 (needing the determination of an effective section) and lateral torsion buckling will occur, increase the moment by 50% and use the optimization equations for Class 3. MEd =

1,5 × 1500 × 20 1,35 × 200 × 20 + = 11925 kNm 4 8

Approximate design: As the moment resistance is calculated on an elastic resistance utilization, the factor k in Eq. (5.110) may be taken as 0,55, with the flange area equal to half the web area for an optimal design under elastic stress distribution (Eq. (5.139)), so hw E Aw 210 × 103 √ λ= =k = 0,55 2 = 460 t fyf Af 355 To determine the optimal solution increase MEd by 50% to 1,5 × 11925 (=17888 kNm). From Eq. (5.135), t is given by 6 3M 3 3 × 17888 × 10 Ed t= 3 2 = = 7,1 mm 2 2λ fyd 2 × 460 × 355 Use 8 mm plate. Determine hw from Eq. (5.136) 3λMEd 3 × 460 × 17888 × 106 hw = = = 3264 mm 2fyd 2 × 355 As the plate thickness has been rounded up from 7,1 to 8 mm, then the web height may be rounded down to 3240 mm. The area of the flange Af is given by Eq. (5.139) Af =

8 × 3240 Aw = = 12960 mm2 2 2

Use the limit of the flange outstand as a Class 1 section (i.e. 7,33tf ) so the area of the flange is 14,66tf2 . Thus 14,66tf2 = 12960, or tf = 29,7 mm Use a flange plate 30 mm thick. bf ≈ 14,66tf = 14,66 × 30 = 440 mm Use a flange plate 500 mm wide (Af = 15000 mm2 ). The reason for keeping the flange classification as low as possible is in order to mobilize as much of the flange capacity as possible to resist shear.

160

•

Chapter 5 / Laterally Unrestrained Beams Actual web slenderness hw 3240 = = 405 t 8 Allowable web slenderness: hw 210 × 103 E Aw = 0,55 =k t fyf Af 355

3240 × 8 = 427 500 × 30

Thus flange-induced buckling will not occur. From the actual web slenderness the web (and therefore the complete section) is Class 4. Determination of effective section properties (cl 4.4 EN 1993-1-5): The process is iterative as the amount of web not considered is a function of the stresses at the top and bottom of the web. Note, the calculation takes compressive stresses positive rather than the normal stress analysis convention of compressive stresses negative. First Iteration: Determine the stresses at the top and bottom of the web with no loss of section: Gross Area, A A = 2bf tf + hw t = 2 × 500 × 30 + 3240 × 8 = 55920 mm2 Gross Iy : Iy =

1 1 [b(hw + 2tf )3 − (b − t)h3w ] = [500 × 33003 − 492 × 32403 ] 12 12

= 102, 88 × 109 mm4 Stress at the top of the web, σ1 : 11 925 × 106 × 1620 = 188 MPa 102,88 × 109 Stress at the top of the web, σ2 : σ1 =

11 925 × 106 × 1620 = −188 MPa 102,88 × 109 Determine the stress ratio, ψ: −188 σ2 = −1,0 = ψ= σ1 188 From Table 4.1 of EN 1993-1-5, the buckling factor kσ = 23,9 for ψ = −1,0. σ2 = −

The normalized slenderness ratio λp is given by λp =

b 3240 = 3,585 ! √ √ = 28,4tε kσ 28,4 × 8 235 23,9 355

where b is the depth of the web.

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The reduction factor ρ for an internal compression member is given by ρ=

λp − 0,055(3 + ψ) (λp

)2

=

3,585 − 0,055(3 + (−1)) = 0,271 3,5852

The effective depth beff is given by beff =

ρhw 0,271 × 3240 = = 439 mm 1−ψ 1 − (−1)

The depth of web left at the top be1 : be1 = 0,4beff = 0,4 × 439 = 176 mm The depth of web left at the bottom (above the centroidal axis), be2 : be1 = 0,6beff = 0,6 × 439 = 263 mm The ineffective portion of web has a length lw = 1620 − 176 − 263 = 1181 mm The net loss of area of the web, Aw is given by Aw = 1181 × 8 = 9448 mm2 Effective area of section, Aeff : Aeff = A − Aw = 59920 − 9448 = 50472 mm2 Position of effective centroid, zeff : l A h2 − Aw zeff,prev + be2 + 2w zeff = Aeff 3300 1181 59920 3300 − 9448 + 263 + 2 2 2 = = 1490 mm 50472 Note, zeff,prev is the neutral axis position at the previous iteration. For the first iteration, zeff,prev = h/2. Effective second moment of area, Iy,eff : 2 tlw3 lw − Aw zeff,prev + be2 + − zeff 12 2 2 3300 8 × 11813 = 102,88 × 109 + 59920 − 1490 − 2 12 2 1181 3300 + 263 + − 1490 − 9448 2 2

Iy,eff = Iy + A

h − zeff 2

2

−

= 102,88 × 109 + 1,53 × 109 − 1,10 × 109 − 9,70 × 109 = 93,61 × 109 mm4

162

•

Chapter 5 / Laterally Unrestrained Beams Stress at the top of the web, σ1 : 11925 × 106 × (3270 − 1490) = 227 MPa 93,61 × 109

σ1 =

Stress at the bottom of the web, σ2 : σ2 = −

11925 × 106 × (1490 − 30) = −186 MPa 93,61 × 109

Determine the stress ratio, ψ: σ2 −186 = = −0,819 σ1 227

ψ=

From Table 4.1 of EN 1993-1-5, the buckling factor kσ for ψ = −0,794 is given by kσ = 7,81 − 6,29ψ + 9,78ψ2 = 7,81 − 6,29(−0,819) + 9,78(−0,819)2 = 19,5 The normalized slenderness ratio λp is given by λp =

b

√

28,4tε kσ

=

3240 = 3,97 ! √ 28,4 × 8 235 19,5 355

where b is the depth of the web. The reduction factor ρ for an internal compression member is given by ρ=

λp − 0,055(3 + ψ) (λp

)2

=

3,97 − 0,055(3 + (−0,819)) = 0,244 3,972

The effective depth beff is given by beff =

ρhw 0,244 × 3240 = = 435 mm 1−ψ 1 − (−0,819)

The depth of web left at the top be1 : be1 = 0,4beff = 0,4 × 435 = 174 mm The depth of web left at the bottom (above the centroidal axis), be2 : be1 = 0,6beff = 0,6 × 435 = 261 mm The ineffective portion of web has a length lw = (3240 − 1490) − 174 − 261 = 1315 mm The net loss of area of the web, Aw is given by Aw = 1315 × 8 = 10520 mm2 Effective area of section, Aeff : Aeff = A − Aw = 59920 − 10520 = 49400 mm2

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Position of effective centroid, zeff : l A h2 − Aw zeff,prev + be2 + 2w zeff = Aeff 1315 59920 3300 − 10520 1490 + 261 + 2 2 = = 1488 mm 49400 Effective second moment of area, Iy,eff :

2 tlw3 lw − Aw zeff,prev + be2 + − zeff 12 2 2 3300 8 × 13153 = 102,88 × 109 + 59920 − 1488 − 2 12 2 1315 − 1488 − 10520 1490 + 261 + 2

Iy,eff = Iy + A

h − zeff 2

2

−

= 102,88 × 109 + 1,57 × 109 − 1,52 × 109 − 8,91 × 109 = 94,02 × 109 mm4 Stress at the top of the web, σ1 : σ1 =

11925 × 106 × (3270 − 1488) = 226 MPa 94,02 × 109

Stress at the bottom of the web, σ2 : σ2 = −

11925 × 106 × (1488 − 30) = −185 MPa 94,02 × 109

As these stresses are virtually identical to those on the first iteration, there is no need to continue. Note, that as the changes in be1 and be2 were small the iteration could have stopped without further calculations of σ1 and σ2 (ψ = −0,819) The lesser elastic section modulus Weff,y is given as Ieff 94,02 × 109 = 64,49 × 106 mm3 = zeff − tf 1488 − 30 fy 355 × 10−6 = 22900 kNm = Weff,y = 64,49 × 106 1,0 γM0

Weff,y = MRd

Lateral torsional buckling check Moment gradient factor, C1 from Eq. (5.40), with ψ = 0: 1 = 0,4ψ + 0,6 = 0,6 C1

164

•

Chapter 5 / Laterally Unrestrained Beams The gross section is used to calculate the section properties required for Mcr . 1 1 [2tf b3f + hw t3 ] = [2 × 0,030 × 0,53 + 3,24 × 0,0083 ] 12 12

Iz =

= 0,625 × 10−3 m4 Iw = Iz

h2s 3,2702 = 0,625 × 10−3 = 1.67 × 10−3 m6 4 4

Note: hs is the distance between the centroids of the flanges. The torsional second moment of area may be calculated on the thin plate assumption. It =

1 1 [2bf tf3 + hw t3 ] = [2 × 0,5 × 0,0303 + 3,24 × 0,0083 ] = 9,55 × 10−6 m4 3 3

Use Eq. (5.5) to determine Mcr 1/2 π2 EIz Iw L2 GIt Mcr = + 2 Iz L2 π EIz L = 10 m. π2 EIz π2 × 210 × 106 × 0,625 × 10−3 = = 12954 kN L2 102 GIt = 81 × 106 × 9,55 × 10−6 = 774 kNm2 GIt π2 EI

z

L2

=

774 = 0,060 m2 12954

Iw 1,67 × 10−3 = = 2,672 m2 Iz 0,625 × 10−3 Mcr = 12954[2,672 + 0,060]1/2 = 21410 kNm From Eq. (5.12), λLT is given by fy Weff,y 355 × 64,49 × 106 × 10−6 λLT = = 0,6 = 0,801 C1 Mcr 21410 h/b > 2, so αLT = 0,76 LT = 0,5[1 + αLT (λLT − 0,2) + (λLT )2 ] = 0,5[1 + 0,76(0,801 − 0,2) + 0,8012 ] = 1,049 χLT =

1 1 = = 0,579 2 − 0,8012 )1/2 2 1/2 1,049 + (1,049 − (λLT ) )

LT + (2LT

Mb,Rd = χLT Weff,y

fy 355 = 0,579 × 64,49 × 106 × 10−6 = 13260 kNm γM1 1,0

This exceeds the applied moment of 11925 kN. The beam is therefore satisfactory for flexure.

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Deflection check: Igross = 102,88 × 109 mm4 For a worst case scenario assume that all the variable load contributes to the deflection, δ=

1 WL3 1 1500 × 203 = 0,012 m = 48 EI 48 210 × 106 × 102,88 × 10−3

This is equivalent to span/1667, which is satisfactory. Web design: As the web carries compression due to flexure, then the following interaction equation must be satisfied (cl 7.1(1), EN 1993-1-5), Mf,Rd η1 + 1 − (2η3 − 1)2 ≤ 1,0 (5.182) Mpl,Rd where Mf,Rd is the plastic moment resistance of the flanges and Mpl,Rd is the plastic moment of resistance of the section (both calculations are irrespective of classification of the section). The shear contribution factor η3 is defined as (cl. 5.5 (1), EN 1993-1-5) η3 =

VEd ≤ 1, 0 √ χv hw t[(fyw / 3)/γM1 ]

(5.183)

as there is no axial force the equation for η1 from cl 4.6 (1) (EN 1993-1-5) reduces to Med η1 = f W ≤ 1,0 y eff

(5.184)

γM0

If η3 < 0,5 there is no reduction in moment capacity (cl 7.1 (1), EN 1993-1-5) Mpl,Rd = Wpl

fy 500 × 33002 − 492 × 32402 355 × 10−6 = 24870 kNm = γM0 4 1,0

Mf,Rd = Af hs 1−

Mf,Rd Mpl,Rd

= 1−

fy 355 = 500 × 30 × (3300 − 30) × 10−6 = 17410 kNm γM0 1,0

17410 = 0,3 24870

Flexible end post: Three intermediate stiffeners are required at 2,8, 5,6 and 7,8 m from the support, together with a load bearing stiffener at the centre. Stiffener at 2,8 m: MEd = 1260 × 2,8 − 6,75 × 2,82 = 3475 kNm VEd = 1260 kN (at support)

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Chapter 5 / Laterally Unrestrained Beams Determine c from Eq. (5.156) 1,6bf tf2 fyf 1,6 × 500 × 302 × 355 c = a 0,25 + = 2800 0,25 + = 724 mm th2w fyw 8 × 32402 × 355 Determine the flange contribution factor χf from Eq. (5.155) √ bf tf2 fyf 3 MEd 2 χf = 1− cthw fyw Mf,Rd √ 500 × 302 × 355 × 3 3475 2 = = 0,040 1− 724 × 8 × 3240 × 355 17410 hw 3240 = = 1,157 a 2800 a = 0,864 hw For a/hw < 1,0, kτ is given by Eq. (5.143), or 2 hw kτ = 4 + 5,34 = 4 + 5,34 × 1,1572 = 11,15 a Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 3240 = 3.985 ! √ = √ 235 37.4εt kτ 37.4 355 × 8 11.15

As λw > 1,08, χw is given by Eq. (5.149) as χw =

0,83 0,83 = = 0,208 3,985 λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,040 + 0,208 = 0,248 The shear capacity Vb,Rd is determined from Eq. (5.153) Vb,Rd = χv hw t η3 =

f √yw 3

γM1

355 0,248 × 3240 × 8 √3 = = 1318 kN 1000 1,0

1260 VEd = = 0,956 √ 1318 χv hw t[(fyw / 3)/γM1 ]

As η3 > 0,5 the interaction equation between moment and shear must be considered (Eq. (5.152)) Med 3475 η1 = f W = = 0,152 y eff 22900 γM0

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TABLE 5.7 Shear capacity calculations with non-rigid end post (Example 5.4). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) η1 η3 Interaction equation Stiffener force (kN)

2,8 2,8 3475 1260 724 0,040 0,864 11,15 3,990 0,208 0,248 1318 0,152 0,956 0,401 1079

5,6 2,8 6844 1241 724 0,035 0,864 11,15 3,990 0,208 0,243 1293 0,209 0,960 0,553 210

7,8 2,2 9417 1222 568 0,038 0,679 15,58 3,371 0,246 0,284 1506 0,411 0,811 0,527 −134

10 2,2 11 925 1207 568 0,028 0,679 15,58 3,371 0,246 0,274 1457 0,521 0,829 0,650 −932

Mf,Rd (2η3 − 1)2 = 0,152 + 0,3(2 × 0,956 − 1)2 = 0,402 ≤ 1,0 η1 + 1 − Mpl,Rd The calculations for the remaining stiffeners are carried out in Table 5.7, together with the calculation of the forces on the stiffeners. Intermediate stiffeners: √ √ For all the panels, a ≤ 2hw (=3,24 2 = 4,58 m), thus based on the lesser value of a (=2,2 m) the stiffener requirement, Ist is given by Ist = 1,5

h3w t3 32403 × 83 = 1,5 = 5,4 × 106 mm4 2 a 22002

Use 8 mm plate, so 6 3 12 × 5,4 × 10 b= = 200 mm 8 Use double intermediate stiffeners of 200 × 8 mm plates either side of the web. Load capacity of stiffeners NRd (with no buckling): NRd = Ast

fy 355 × 10−3 = 1136 kN = 2 × 200 × 8 1,0 γM0

This exceeds the maximum stiffener force of 1079 kN. As in earlier examples strut buckling is not critical, it will not be checked in this example. End post: VEd = 1260 kN Use a 500 wide by 10 mm thick plate.

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Chapter 5 / Laterally Unrestrained Beams The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 15εt = 15 × 8 (235/355) = 98 mm From Eq. (5.177) Aequiv is given by Aequiv = Ast + 15εt2 = 500 × 10 + 98 × 8 = 5784 mm2 and Iequiv by 1 10 × 5003 98 × 83 15εt4 = + = 0,104 × 109 mm4 12 12 12 Iequiv 0,104 × 109 i= = = 134,1 mm Aequiv 5784

Iequiv = Ist +

Use an effective length of 0,75hw = 0,75 × 3240 = 2430 mm π2 EI 210 × 0,104 × 109 π2 = = 36500 kN l2 24302 Afy 5784 × 355 λ= = = 0,237 Ncr 36500 × 103

Ncr =

Use buckling curve ‘c’ (α = 0,49) = 0,5[1 + α(λ − 0,2) + (λ)2 ] = 0,5[1 + 0,49(0,237 − 0,2) + 0,2372 ] = 0,537 χ=

+

NRd = χA

!

1 2 − (λ)2

=

0,537 +

1 0,5372 − 0,2372

= 0,981

fy 355 × 10−3 = 2014 kN = 0,981 × 5784 1,0 γM0

This exceeds the reaction of 1260 kN. Check cl 9.2.1 (7) Determine Ip : The second moment of area about the web centre line, Iy : bh3 10 × 5003 = = 0,104 × 109 mm4 12 12 The second moment of area normal to the web centre line about one edge, Iy : Iy =

Iy =

bh3 500 × 103 = = 0,042 × 106 mm4 3 3

Ip = Ix + Iy = 0,104 × 109 + 0,042 × 106 = 0,104 × 109 mm4 bh3 = 0,042 × 106 mm4 3 IT 0,042 × 106 = = 0,4 × 10−3 Ip 0,104 × 109 IT =

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Limiting value: 5,3

fy 355 = 8,96 × 10−3 = 5,3 E 210 × 103

The actual value is less than the limiting value thus the stiffener size must be increased. A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT /IP = 0,01) (and will clearly satisfy the strength and buckling criteria). Central stiffener: NEd = 1,5 × 1500 = 2250 kN Use two 240 wide by 15 mm thick plates. The section must be checked for buckling, but a proportion of the web may be taken into account. √ Length of web = 30εt = 30 × 8 (235/355) = 195 mm Aequiv = Ast + 30εt2 = 2 × 240 × 15 + 195 × 8 = 8760 mm2 15 × (240 + 8 + 240)3 195 × 83 Isequiv = Ist + 30εt4 = + = 0,145 × 109 mm4 12 12 Iequiv 0,145 × 109 i= = = 128,7 mm Aequiv 8760 Use an effective length of 0,75 hw = 0,75 × 3240 = 2430 mm π2 EI 210 × 0,145 × 109 π2 = = 50895 kN l2 24302 Afy 8760 × 355 λ= = = 0,247 Ncr 50895 × 103

Ncr =

Use buckling curve ‘c’ (α = 0,49): = 0,5[1 + α(λ − 0,2) + (λ)2 ] = 0,5[1 + 0,49(0,247 − 0,2) + 0,2472 ] = 0,542 χ=

+

NRd = χA

!

1 2 − (λ)2

=

0,542 +

1 0,5422 − 0,2472

fy 355 × 10−3 = 3035 kN = 0,976 × 8760 γM0 1,0

This exceeds the applied load of 2250 kN.

= 0,976

•

Chapter 5 / Laterally Unrestrained Beams

cL

500 30 flanges 8 mm web 3,3

170

Intermediate 2 200 8 End 500 25 Central 2 240 15

2,8 5,6 7,8 (a) Non-rigid end post

cL

End post 500 25 Central 2 240 15

0,4

Intermediate 2 150 8

5,0

(b) Rigid end post

FIGURE 5.25 Final layout for the girder of Example 5.4

Flange to web welds: √ From Table 5.7, VEd > hw t[(fyw / 3)/γM1 ] as the flange contribution χf has been mobilized. Thus welds should be designed for a shear flow of ηt

f √yw 3

γM1

= 1.2 × 8

355 √ 3

1.0

= 1968 N/mm

The final layout is given in Fig 5.25 (a). The self-weight of the beam is 101 kN (or 5,05 kN/m). (b) Rigid end post First panel from the support: Intermediate stiffener is 5 m from the support. MEd = 1260 × 5 − 6,75 × 52 = 6131 kNm VEd = 1260 kN (at the support)

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Determine c from Eq. (5.156)

1,6bf tf2 fyf c = a 0,25 + th2w fyw

1,6 × 500 × 302 × 355 = 5000 0,25 + 8 × 32402 × 355

Determine the flange contribution factor χf from Eq. (5.155) √ bf tf2 fyf 3 MEd 2 χf = 1− cthw fyw Mf,Rd √ 500 × 302 × 355 × 3 6131 2 = = 0,020 1− 1293 × 8 × 3240 × 355 17410 hw 3240 = = 0,648 a 5000 a = 1,543 hw For a/hw > 1,0, kτ is given by Eq. (5.142), or

hw kτ = 5,34 + 4 a

2 = 5,34 + 4 × 0,6482 = 7,02

Determine the normalized web slenderness ratio λw from Eq. (5.145) λw =

hw 3240 = 5,02 ! √ = √ 37,4εt kτ 37,4 235 × 8 7,02 355

As λw > 1,08, χw is given by Eq. (5.149) as χw =

1,37 1,37 = = 0,240 0,7 + 5,02 0,7 + λw

From Eq. (5.154), the total shear co-efficient χv is given by χv = χf + χw = 0,020 + 0,240 = 0,260 The shear capacity Vb,Rd is determined from Eq. (5.153) f √yw 3

355

0,260 × 3240 × 8 √3 = 1380 kN γM1 1000 1,0 VEd 1260 η3 = = 0,913 = √ 1380 χv hw t[(fyw / 3)/γM1 ]

Vb,Rd = χv hw t

=

= 1293 mm

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•

Chapter 5 / Laterally Unrestrained Beams As η3 > 0,5 an interaction equation between moment and shear must be considered, Mf,Rd η1 + 1 − (2η3 − 1)2 ≤ 1,0 Mpl,Rd Med 6131 η1 = f W = = 0,268 y eff 22900 γM0

Mf,Rd η1 + 1 − (2η3 − 1)2 = 0,268 + 0,3(2 × 0,913 − 1)2 = 0,473 ≤ 1,0 Mpl,Rd The resultant spacing of stiffeners and a summary of the remaining calculations is given in Table 5.8. The final result for the calculation of stiffener forces are also given in Table 5.8. Note, it is possible to eliminate the intermediate stiffener by increasing the web thickness to 9 mm. This may be more economic when considering the overall materials and fabrication costs. The minimum stiffness requirement for both panels either side of the central stiffener √ is given by the case a > 2hw , so design on the least value of a: Ist = 0,75hw t2 = 0,75 × 3240 × 92 = 0,197 × 106 mm4 Use 8 mm thick plate, then the total breadth of the stiffener b is given by 6 3 12 × 0,197 × 10 b= = 67 mm 8 Minimum area to carry a stiffener force of 757 kN is given as A=

757 × 103 = 2132 mm2 355

TABLE 5.8 Shear capacity calculations with rigid end post (Example 5,4). Distance from support (m) Panel width (m) Moment M Ed (kNm) Shear V Ed (kN) c (mm) χf a/hw kτ λw χw χv V Rd (kN) η1 η3 Interaction equation Stiffener force N Rd (kN)

5 5 6131 1260 1293 0,020 1,543 7,02 5,02 0,239 0,259 1380 0,268 0,913 0,473 757

10 5 11 925 1226 1293 0,012 1,543 7,02 5,02 0,239 0,261 1337 0,521 0,917 0,729 −419

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This would require a total width of 2132/8 = 267 mm. Use two plates 150 mm by 8 mm as intermediate stiffeners. Design of rigid end post. Use Eqs (5.163) to (5.166) to determine the maximum panel size in accordance with cl 9.3.1(4) (EN 1993-1-5) Determine χw from Eq. (5.163) χw =

√ √ VEd 3γM1 1260 × 103 3 × 1.0 = 0,237 = fyw hw t 355 × 3240 × 8

The required normalized web slenderness λw from Eq. (5.164) is given by λw =

0,83 0,83 = = 3,502 χw 0,237

The buckling parameter kτ is given by Eq. (5.165) as kτ =

hw 37,4tελw

2

⎛ ⎜ =⎝

⎞2 3240 ⎟ ! ⎠ = 11,44 37,4 × 8 × 235 × 3,502 355

As a/hw < 1, kτ is given by Eq. (5.143), or the required panel width a is given as a = hw

5,35 5,35 = 3240 = 2750 mm kτ − 4,00 11,44 − 4

Centre to centre distance of the pair of double stiffeners should be greater than 0.1 hw (=324 mm). Use a centre to centre distance of 400 mm (the shear resistance will be greater than 1260 kN). Minimum cross-sectional area of each plate is 4hw t2 4 × 3240 × 82 = = 2074 mm2 e 400 As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is 2074/500 (=4 mm). As for the non-rigid end post, a stiffener 500 mm by 25 mm is needed. Stiffener under the point load: The calculations are the same as those for the non-rigid end post. The final layout is given in Fig 5.25 (b). The self-weight is 105 kN (5,06 kN/m)

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REFERENCES Anderson, J.M. and Trahair, N.S. (1972). Stability of mono-symmetric beams and cantilevers, Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, 98, 269–286. Basler, K. (1961). Strength of plate girders in shear, Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, 87, 151–180. Bradford, M.A. (1989). Buckling of beams supported on seats, Structural Engineer, 69(23), 411–414. Bulson, P.S. (1970). The stability of flat plates. Chatto and Windus. Chapman, J.C. and Buhagiar, D. (1993). Application of Young’s buckling equation to design against torsional buckling. Proceedings of the Institution of Civil Engineers, Structures and Buildings, 99, 359–369. Davies, A.W. and Griffiths, D.S.C. (1999). Shear strength of steel plate girders, Proceedings of the Institution of Civil Engineers, 134, 147–157. Gibbons, C. (1995). Economic steelwork design, Structural Engineer, 73(15), 250–253. Kaim, P. (2006). Buckling of members with rectangular hollow sections, In Tubular structures XI (eds Packer and Willibald). Taylor and Francis Group, 443–449. Kirby, P.A. and Nethercot, D.A. (1979). Design for structural stability. Granada. Mutton, B.R. and Trahair, N.S. (1973). Stiffness requirements for lateral bracing, Journal of the Structural Division, ASCE, 99, 2167. Nethercot, D.A. (1974a). Residual stresses and their influence upon the lateral buckling of rolled steel beams, Structural Engineer, 52(3), 89–96. Nethercot, D.A. (1974b). Buckling of welded beams and girders, IABSE, 34, 163–182. Nethercot, D.A.and Lawson, R.M. (1992). Lateral stability of steel beams and columns – common cases of restraint, Publication 093. Steel Construction Institute. Nethercot, D.A. and Rockey, K.C. (1971). A unified approach to the elastic critical buckling of beams, Structural Engineer, 49(7), 321–330. Pillinger, A.H. (1988). Structural steelwork: a flexible approach to the design of joints in simple construction, Structural Engineer, 66(19), 316–321. Porter, D.M., Rockey, K.C. and Evans, H.R. (1975). The collapse of plate girders loaded in shear, Structural Engineer, 53(8), 313–325. Rees, D.W.A. (1990). Mechanics of solids and structures. McGraw Hill. Rockey, K.C. and Škaloud, M. (1971). The ultimate load behaviour of plate girders loaded in shear, IABSE Colloquium – design of plate girders for ultimate strength. London, 1–19. Rockey, K.C., Evans, H.R. and Porter, D.M. (1978). A design method for predicting the collapse behaviour of plate girders, Proceedings of The Institution of Civil Engineers, 65, 85–112. Rockey, K.C., Valtinat, G. and Tang, K.H. (1981). The design of transverse stiffeners on webs loaded in shear – an ultimate load approach, Proceedings of The Institution of Civil Engineers, 71(2), 1069–1099. Rondal, J., Würker, K-G., Dutta, D., Wardenier, J. and Yeomans, N. (1992). Structural stability of hollow sections. Verlag TÜV Rheinland. Timoshenko, S.T. and Gere, J.M. (1961). Theory of elastic stability. McGraw Hill. Trahair, N.S. (1983). Lateral buckling of overhanging beams, In Instability and plastic collapse of steel structures (ed L.J. Morris). Granada, 503–518. Trahair, N.S. and Bradford, M.A. (1988). The behaviour and design of steel structures (2nd Edition). Chapman and Hall. BS 5950-1: Structural use of steelwork in building – Part 1: Code of practice for design – Rolled and welded sections. BSI. EN 1993-1-1. Eurocode 3: Design of steel structures, Part 1-1: General rules and rules for buildings. CEN/BSI. EN 1993-1-5. Eurocode 3: Design of steel structures, Part 1-5: Plated structural elements. CEN/BSI.

Chapter

6 / Axially Loaded Members

6.1 AXIALLY LOADED TENSION MEMBERS A member subject to axial tension extends and tends to remain straight or, if there is a small initial curvature, to straighten out as the axial load is increased. Tension members (ties) occur in trusses, bracing and hangers for floor beams. A flat can be used as a tie, but this is generally impractical because it buckles if it goes into compression. Tie sections are therefore angles and tees for small loads and ‘I’ sections for larger loads. In situations where the load is not applied axially then the member is designed to resist an axial force plus a bending moment. A tension member extends when subject to an axial load and is deemed to have failed when the yield or ultimate stress is reached. The failure load is independent of the length of the member which is in contrast to an axially loaded compression member which fails by buckling. A member which is purely in tension does not buckle locally or overall and is therefore not affected by the classification of sections. The characteristic stress is not reduced, for design purposes, except by the material factor.

6.1.1 Angles as Tension Members (cl 4.13, EN 1993-1-8 (2005)) Generally angles are connected by one leg at the end of the member and this introduces an eccentric load. Where an angle is connected to one side of a gusset plate (as in a truss) bending moments are introduced in addition to the direct axial force. For a tension member these moments produce lateral deflections which reduce the eccentricity of the load near the middle of the member. Thus under increasing load the bending stresses become concentrated more towards the ends of the member. For angles connected by one leg the principal sectional axes are inclined to the plane containing the bending moment. Secondary deflections therefore occur normal to the plane of bending and, because of the restraints provided by the gusset plates, twisting also takes place. Generally eccentrically loaded members are designed to resist an axial load and bending moment. However angle and tee experiments (Nelson (1953); Regan and Salter

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•

Chapter 6 / Axially Loaded Members (1984)) demonstrated that the above effects could be compensated for in design by reducing the cross-sectional area of the member. If there are holes then these also reduce the area of the cross-section. Angles may be treated as axially loaded members provided that the net area is reduced to the effective area (cl 4.13, EN 1993-1-8 (2005)). For an equal angle, or an unequal angle connected by the larger leg the effective area is the gross area. For an unequal angle connected by the smaller leg the effective area is twice that of the smaller leg.

6.1.2 Design Value of a Tension Member (cl 6.2.3, EN 1993-1-1 (2005)) The design value of the tensile force NEd at each cross-section should satisfy (Eq. (6.5), EN 1993-1-1 (2005)) NEd ≤1 Nt,Rd

(6.1)

where Nt,Rd is the design tension resistance taken as the smaller of: (a) The design plastic resistance of the gross cross-section (Eq. (6.6), EN 1993-1-1 (2005)) Npl,Rd =

Afy γM0

(6.2a)

(b) The design ultimate resistance of the net cross-section at holes for fasteners (Eq. (6.7), EN 1993-1-1 (2005)) Nu,Rd = 0,9

Anet fu γM2

(6.2b)

(c) The design ultimate resistance of the net cross-section at holes for fasteners which are preloaded or non-preloaded (Eq. (6.8), EN 1993-1-1 (2005) and cl 3.4.2.1(1), EN 1993-1-8 (2005)) Nnet,Rd =

Anet fy γM0

(6.2c)

EXAMPLE 6.1 An angle in tension connected by one leg. A 100 × 65 × 6 mm single angle tie is connected through the smaller leg by two 20 mm diameter bolts in line with a pitch of 2,5 d0 . Determine the design ultimate resistance of the angle assuming S275 steel and material factors of γM2 = 1,25 and γM0 = 1. Net area of angle connected by the smaller leg and allowing for holes Anet = (b − do )t + bt = (65 − 22) × 6 + 65 × 6 = 648 mm2 Compare this value with the gross area = 1120 mm2 (Section Tables).

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Design ultimate tensile resistance of the net cross-section for a two bolt angle connection (Eq. (6.7), EN 1993-1-1 (2005)) Nu,Rd = 0,9

Anet fu 0,9 × 648 × 430 = = 200,6 kN. γM2 1,25 × 1E3

or Npl,Rd =

Afy (2 × 65 × 6) × 275 = = 214,5 > 200,6 kN. γM0 1,0 × 1E3

A check must be made for the strength of the bolts as shown in Chapter 7.

6.2 COMBINED BENDING AND AXIAL FORCE – EXCLUDING BUCKLING (cl 6.2.9, EN 1993-1-1 (2005)) Combining an axial force (tension or compression) and bending moment induces stresses which vary across a ductile steel section. At a certain point the stresses may combine to produce a yield stress but this does not produce collapse of the member because collapse only occurs if the entire section is at yield stress (i.e. the section is plastic). If the axial force is compressive it is assumed that buckling does not occur.

6.2.1 Rectangular Sections (cl 6.2.9.1(3), EN 1993-1-1 (2005)) A rectangular section subject to an axial force and bending moment produces a stress diagram as shown in Fig. 6.1. The neutral axis is displaced from the equal area position and the stress diagram can be represented in two parts, one for the axial load and one for the reduced plastic modulus. Plastic moment of resistance about y–y axis My = fy bh1 (h − h1 )

(6.3)

Axial load N = fy bh(h − 2h1 )

(6.4) fy

fy

fy

h1

b

y

h

y

fy

fy

FIGURE 6.1 Effect of axial force on the plastic moment of resistance of a rectangular section

Chapter 6 / Axially Loaded Members bf

fy

fy

y

y

N

2c

tf

•

h

178

A

tw

(a) Neutral axis (NA) in web

y

y N

A

(b) Neutral axis in flange

FIGURE 6.2 Effect of axial force on the plastic moment of resistance of an ‘I’ section Combining Eqs (6.3) and (6.4) and eliminating h1 My /(bh2 fy /4) + [N/(bhfy )]2 = 1

(6.5a)

This relationship is plotted on Fig. 6.3. The term (bh2 fy /4) is the plastic moment of resistance for a rectangular section. This form of the equation is given in Eq. (6.32), EN 1993-1-1 (2005) and is applicable to rectangular solid sections without holes. NEd 2 MN,Rd = Mpl,Rd 1 − (6.5b) Npl,Rd

6.2.2 ‘I’ Sections The relationship between bending moment and axial force for an ‘I’ section is more complicated. An ‘I’ section subject to an axial load and bending moment produces a stress diagram as shown in Fig. 6.2. The neutral axis is displaced from the equal area position. This stress diagram can be represented in two parts, one for the axial load and one for the reduced plastic modulus. A convenient ratio to determine the reduced plastic section modulus is σ n= fy where σ is the mean axial stress fy is the specified minimum yield strength of steel. If the axial force is small the neutral axis is in the web. Alternatively if the axial force is large the neutral axis is in the flange as shown in Fig. 6.2. For bending about the y–y axis, the neutral axis moves from the web into the flange when n > nc where h − 2tf nc = tw A A is the total area of the section.

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Subtracting the plastic modulus of the hatched area from the whole section. Wyr = Wy − W(shaded area) W(shaded area) =

tw (2c)2 4

From equilibrium σA = (2c)tw fy and since by definition n = σ/fy When n ≤ nc Wyr = Wy − an2 When n > nc Wyr = b(1 − n)(c + n) where a=

A2 4tw

A2 4bf 2bf h c= −1 A

b=

For bending about the y–y axis the change point for n is nc =

tw h A

and the values of A2 4h A2 b= 8tf 4tf bf c= −1 A

a=

Values of a, b, c and nc for ‘I’ sections are given in Section Tables. The plastic section modulus determined in this way is applicable in tension and for short columns. For longer columns instability effects due to deflection of the column reduce this value. The relationship between N/Np and M/Mp for a typical ‘I’ section is plotted in Fig. 6.3.

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•

Chapter 6 / Axially Loaded Members

N/N p

1.0

0

M/Mp

1.0

FIGURE 6.3 Relationship between N/Np and M/Mp at collapse

The above theory for ‘I’ and ‘H’ sections is simplified in the European Code by the following recommendations (cl 6.2.9.1(4), EN 1993-1-1 (2005)). (a) For Class 1 and Class 2 cross-sections the plastic moment of resistance is not reduced if axial loads are limited. Bending about the y–y axis (Eqs (6.33) and (6.44), EN 1993-1-1 (2005)) NEd ≤ 0,25Npl,Rd NEd ≤ 0,5

h w t w fy γM0

Bending about the z–z axis NEd ≤

hw tw fy γM0

(b) For Class 3 and Class 4 cross-sections the maximum longitudinal stress is limited to fy /γM0 . In the case of Class 4 sections the effective area of the section is used.

6.3 BUCKLING OF AXIALLY LOADED COMPRESSION MEMBERS Compression members are present in many structures, for example, trusses, bracing and columns. They are generally greater in cross-sectional area than tension members because they may fail in buckling (Fig. 6.4). If buckling is likely to occur then sections must be capable of resisting bending moments.

6.3.1 Compression Members (cl 6.3.1, EN 1993-1-1 (2005)) An efficient cross-section for a strut is a hot finished tube because residual stresses are a minimum and the buckling resistance is the same for all axes of bending. However the use of a tube is not always practical because of the difficulties of making connections.

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N

L /2

L /2

NE

x y

NE (a) Euler strut

L /2

a L /2

•

O

Δ0

y

y0

x

N (b) Practical strut

FIGURE 6.4 Buckling behaviour based on Euler strut and practical strut

Connections to Universal Column ‘I’ sections are simpler but the section is not as efficient because of the weaker z–z axis of bending. Where loads are relatively small angles and ‘T’ sections are used as struts (e.g. in roof trusses). The strength of a compression member is not reduced significantly by welding or holes with fasteners but they must be arranged sensibly. Short steel compression members fail by squashing at the yield stress, while long, or more accurately slender members, fail by buckling. Buckling may occur at an axial stress which is less than the yield stress and is related to slenderness ratio, lack of straightness and non-axial loads. As buckling progresses the load becomes progressively more eccentric to the longitudinal axis of the member and a bending moment is introduced as shown in Fig. 6.4(a).

6.3.2 Buckling Theory The Euler theory was the first attempt to produce a rational explanation of buckling behaviour of a strut. It was based on the differential equation, related to Fig. 6.4(a), which shows the final deflected form of a pin-ended strut EI

d2 y = M = NE (a − y) dx2

(6.6)

The solution of Eq. (6.6) shows that an axially loaded pin-ended strut becomes elastically unstable and buckles at the Euler critical stress fE =

π2 E λ2

where E Young’s elastic modulus

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•

Chapter 6 / Axially Loaded Members L length of a pin-ended strut i radius of gyration λ slenderness ratio = L/i The Euler buckling theory makes no allowance for: (a) (b) (c) (d) (e) (f) (g) (h) (i)

homogeneity of column material, isotropy of column material, variation of E value and elastic–plastic behaviour of column material, loading not axial, residual stresses, lack of straightness of a column, cross-section of a column not rectangular, local buckling, alternative end conditions to a column.

These factors are present in practice and reduce the Euler buckling load. The Euler differential equation can be modified in a number of ways (Bleich (1952)) to take account of these factors. One method produces the Perry–Robertson formula which is used in the European Code and is related to the Euler buckling stress. Small unavoidable eccentricities of loading and lack of initial straightness can be simulated mathematically by assuming an initial curvature which produces a small central deflection 0 (Fig. 6.4(b)). When a load N is applied the deflection at x is increased by y and the differential equation of bending similar to Eq. (6.6) is EI

d2 y = M = −N[ y + y0 ] dx2

Adopting a sinusoidal function for the initial curvature y0 = 0 cos (πx/L) and putting μ2 = N/EI then πx # " d2 y 2 + μ cos y + =0 0 L dx2 The solution to this equation is y = A sin μx + B cos μx +

μ2 0 cos (πx/L) π2 /L2 − μ2

when x = ±L/2, y = 0 and hence A = B = 0 and y=

μ2 0 cos (πx/L) π2/L2 − μ2

If the Euler buckling load NE = π2 EI/L2 , then y=

N0 cos (πx/L) NE − N

(6.7)

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If N/A = f and NE /A = fE then the deflection y=

πx f 0 cos fE − f L

and the total deflection at any point πx fE + 1 0 cos y0 + y = L fE − f The maximum deflection at x = 0 fE ymax = 0 fE − f and the maximum bending moment is fE Mmax = 0 N fE − f If dexf is the distance of the extreme fibre from the neutral axis then the maximum compressive stress 0 dexf N f fexf = +f I fE − f If (0 dexf N/I) = (0 dexf Af/I) = (0 dexf /i2 )f = ηf where i is the radius of gyration of the column section then η fE +1 fexf = f fE − f Assuming that the critical buckling load is reached when yielding commences in the extreme fibres of the strut, that is, when fexf = fy and f = fPR then rearranging the Perry–Robertson buckling stress fPR = 0,5[ fy + (η + 1) fE ] − {0,52 [ fy + (η + 1) fE ]2 − fy fE }1/2

(6.8)

where the Euler critical buckling stress fE = π2 E/λ2 . Equation (6.8) is known as the Perry–Robertson formula and its adoption is explained by Dwight (1975). The value of the function η has varied over the years and the value originally obtained experimentally by Robertson in 1925 related to the slenderness ratio (λ) for circular sections was η = 0,003λ

(6.9)

The value suggested later by Godfrey (1962) to give more economical designs, based on experimental work by Duthiel in France, was η = 0,3

λ 100

2 (6.10)

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•

Chapter 6 / Axially Loaded Members Equation (6.8) has been rearranged in the European Code (Eq. (6.49), EN 1993-1-1 (2005)) to express the buckling stress ( fPR ) in terms of the stress ratio ( fE /fy = λ) and a reduction factor (χ) related to column imperfections. If ζ = 0,5[ fy + (η + 1) fE ] then fPR = ς − (ς 2 − fy fE )1/2 = = ζ fE

+

0

ζ fE

fy 2

−

f y fE

[ς + ς 2 − fy fE

1/2

]

= χ fy f 11/2 y

(6.11)

fE

In the European Code ζ/fE = , fy /fE = λ and the reduction factor (χ) is then expressed in terms of λ =

[fy /fE + (η + 1)] ζ = fE 2

η = 0,001a (λ − λ0 ) but not less than zero, where a varies from 2 to 8 depending on the shape of the section and the limiting slenderness ratio π2 E λ0 = 0,2 >0 fy Combining these equations (cl 6.3.1.2, EN 1993-1-1 (2005)) 2

= 0,5[1 + α(λ − 0,2) + λ ] where α = 0,001a (π2 E/fy )1/2 is an imperfection factor. For uniform members the buckling stress reduction factor (Eq. (6.49), EN 1993-1-1 (2005)) 2

χ = 1/[ + (2 − λ )1/2 ] ≤ 1

(6.12)

The buckling curves are related to the shape of the sections, axis of buckling, and thickness of material as shown in Tables 6.1 and 6.2, and Fig. 6.4, EN 1993-1-1 (2005). The curves are based on experimental results for real columns as described in the ECCS (1976) report, and expressed theoretically by Beer and Schultz (1970). As the lowest value of the central deflection () is related to the initial curvature, bending moments are generated as soon as the axial load is applied and the buckling process starts immediately. Therefore there is no condition of elastic instability as defined by Euler (1759) and the average compressive stress may never reach the Euler critical stress for a strut of finite length. Nevertheless the failure of a strut is sudden when compared with the ductile failure of a tension member. For a fuller development of the buckling theory see Trahair and Bradford (1988).

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6.3.3 Local Buckling (cl 5.5.2(2) and Table 5.2, EN 1993-1-1 (2005)) Slender elements of a section (e.g. flanges), which are primarily in compression, may buckle locally before overall buckling of the member occurs. This is likely to occur with Class 4 sections for high values of the width (c) to thickness (t) ratios of elements of a cross-section. To allow for local buckling the European Code reduces the crosssectional area to an effective area and consequently the member supports less load.

6.3.4 Column Cross-section Studies (Bleich, 1952) over the year have shown that the strength of a column is influenced by the cross-section. Initial theoretical investigations showed that material concentrated at the centre of gravity was more effective in resisting buckling. Later research identified other factors (Table 6.2, EN 1993-1-1 (2005)) which affect buckling, for example, hot finished or cold formed, yield strength, buckling axis, welding and shape of cross-section. For design purposes these factors are related to an appropriate buckling curve (Fig. 6.4, EN 1993-1-1 (2005)).

6.3.5 Buckling Length of a Column The buckling theory, developed previously, is based on the assumption that the ends of the column are pinned, that is, frictionless joints which prevent linear movement but which can rotate freely about any axis of the section. Pin-ended columns are rare in practice but the pin-ended condition is a useful theoretical concept which can be shown to relate to other end conditions. Alternative end support conditions can be simulated by replacing the actual length of the column by an effective length. Consider the theoretical end conditions shown in Fig. 6.5 which vary from complete end fixity to complete freedom at the end of a cantilever. The theoretical effective length (l) is expressed as a proportion of the actual length (L) and is the distance between real or theoretical pins (i.e. points of contraflexure). The effective length is important in design calculations because the Euler buckling load is inversely proportional to the square of the effective length. In practice the full rigidity of a fixed built-in end is never achieved and some rotation occurs due to the flexibility of the connection or the support. Only a small rotation is necessary to transform a built-in end to a pinned end and thus reduce the buckling resistance of a column. In comparison small translational movements at supports are not so critical and may be limited by supporting members. Practical end conditions therefore allow for some rotation and translation at the ends of a real column. For the column in Fig. 6.6 the theoretical effective length is 0,7 L but in practice the built-in end can rotate and the real effective length is 0,85 L, which

l = 2L

l=L

l = 0,5 L

l = 0,7 L

Chapter 6 / Axially Loaded Members

Rotation at support

L

0,85 L

FIGURE 6.5 Theoretical effective lengths

0,7 L

•

L

186

FIGURE 6.6 Rotation of the end of a strut

reduces the axial load at failure. Because perfect rigidity and completely free ends do not occur in practice the range of common practical values is between 0,7 L and 1,5 L. Methods exist (Williams and Sharp, 1990; Wood, 1974) for determining the effective lengths for columns in rigid frames which are based on the relative total column stiffness at a joint to the total stiffness of all members at the joint. For situations where sway of the column does not occur, for example, where cross bracing is present in simple construction, the effective length is less than the real length (0,5 L to 1,0 L). Where sway occurs, for example, an unpropped portal frame, the effective length is greater than the real length (1,0 L to 2,0 L). For single angles used as struts eccentricity of axial load may generally be ignored. As a rough guide the effective length is approximately 0,85 L for two fasteners (or weld)

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and 1,0 L for one fastener. Buckling failure about the minor axis must be considered for which a single equal angle, is at 45◦ to the major axis of bending. An alternative method of determining a modified slenderness ratio for angles is given in cl BB 1.2, EN 1993-1-1 (2005). For double angle struts the effective lengths are similar to those for a single angle but failure about the 45◦ axis may not be possible because of restraints.

6.3.6 Maximum Slenderness Ratios At high values of the slenderness ratio struts become so flexible that deflections under their own weight are sufficient to introduce stresses in excess of the Perry–Robertson buckling formula. The following empirical limits have been used in the past. The slenderness ratio (λ) should not generally exceed the following: (a) for members resisting loads other than wind loads (b) for members resisting self-weight and wind loads only (c) for any member normally acting as a tie but subject to reversal of stress resulting from the action of wind

180 250 350

Members whose slenderness exceeds 180 should be checked for self-weight deflection. If this exceeds (length/1000) the effect of the bending should be taken into account in design.

6.3.7 Intermediate Restraints A member that provides an intermediate restraint and prevents buckling of a strut, reduces the effective length and increases the strength of the strut. The restraint need not be rigid and may be elastic provided that its stiffness exceeds a certain value (Trahair and Bradford, 1988). Often restraining members associated with built-up members are required to resist not less than 1% of the axial force in the restrained member.

6.3.8 Combined Bending, Shear and Axial Force (cl 6.2.10, EN 1993-1-1 (2005)) Th effect of a shear force on the Euler buckling load for solid sections is small (Bleich, 1952). However as explained in Chapter 4 the plastic moment of resistance is reduced when the design shear force is greater than 50% of the plastic shear resistance. Also for large values of shear shear buckling may reduce the resistance of the section (cl 5, EN 1993-1-5 (2005)).

6.3.9 Design Buckling Resistance (cl 6.3.1, EN 1993-1-1 (2005)) Equation (6.8) expresses failure as a buckling stress but the European Code expresses failure as a design load where local buckling and the buckling stress factor (χ) reduce the buckling load.

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•

Chapter 6 / Axially Loaded Members The design buckling resistance of a compression member for Classes 1, 2 and 3 crosssections Nb,Rd = χA

fy γM1

(6.13a)

and for Class 4 sections Nb,Rd = χAeff

fy γM1

(6.13b)

where χ is the reduction factor for the relevant buckling mode which is generally ‘flexural buckling’. In other cases ‘torsional’ or ‘flexural–torsional’ modes may govern. 2

χ = 1[ + (2 − λ )1/2 ] ≤ 1 2

= 0,5[1 + α(λ − 0,2) + λ ] fy 1/2 (Lcr /i) λ= A = Ncr λ1 1/2 E λ1 = π = 93,9ε fy 235 1/2 ε= fy The imperfection factor (α) corresponding to the appropriate buckling curve is obtained from Table 6.1, EN 1993-1-1 (2005). The value of the cross sectional area is A for Classes 1, 2 and 3 sections with no reductions for local buckling provided that the maximum width-to-thickness ratios are within the limits of Table 5.2, EN 1993-1-1 (2005). For Class 4 sections the effective area, reduced because of local buckling, is calculated from cl 4, EN 1993-1-5 (2003).

EXAMPLE 6.2 Angle strut in a roof truss. A steel roof truss is composed of angles and tee sections and has been analysed assuming pin joints. A particular member is subject to the following axial forces. Permanent action (FG = −7,2 kN), variable snow load (Fs = −10,7 kN), variable wind downwards (Fw = −2,0 kN) and variable wind upwards (Fw = +19,1 kN). If an angle, welded at the ends, is chosen to resist the loads, determine the size if the actual length is 2,1 m. Design load cases: (a) no wind 1,35FG + 1,5 Fs = 1,35(−7,2) + 1,5(−10,7) = −25,77 kN (compression)

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(b) wind down 1,35 FG + 1,5(Fs + Fw ) = 1,35(−7,2) + 1,5(−10,7 − 2) = −28,77 kN (compression) (c) wind up 1,0 FG + 1,5 Fw = −7,2 + 1,5(+19,1) = +21,45 kN (tension) Design value Nb,Ed = 28,77 kN assuming uniform compressive stress across the section. Try a 65 × 50 × 6 mm angle grade S275 steel, long leg attached (A = 659 mm2 , iv = 10,7 mm from Section Tables). Design as an axially loaded Class 3 compression member, ignoring eccentric loads (cl BB 1.2, EN 1993-1-1 (2005)). The effective area need not be reduced because of end holes (cl 6.3.1.1(4), EN 1993-1-1 (2005)). Check maximum width-to-thickness ratios (Table 5.2, EN 1993-1-1 2005) h 65 235 1/2 = = 10,8 < 15 × = 13,9 satisfactory t 6 275 b+h 50 + 65 235 1/2 = = 9,58 < 11,5 × = 10,6 2t 2×6 275 satisfactory, no reduction of area for local buckling. Non-dimensional slenderness about the v–v axis (Eq. (6.50), EN 1993-1-1 (2005)) Afy 1/2 (Lcr /iv ) (Lcr /iv ) = = λv = Ncr λ1 93,9ε =

2100/10,7 = 2,26 [93,9 × (235/275)1/2 ]

Largest effective slenderness ratio is about the v–v axis for an angle (cl BB 1.2, EN 1993-1-1 (2005)) λeff,v = 0,35 + 0,7λv = 0,35 + 0,7 × 2,26 = 1,93 For λeff,v = 1,93 and curve b (Table 6.2, EN 1993-1-1 (2005)) the buckling reduction factor (Fig. 6.4, EN 1993-1-1 (2005)) χ = 0,23 Or by calculation (Eq. (6.49), EN 1993-1-1 (2005)) the buckling reduction factor χ=

[ + (2

1 1 = 2 1/2 [2,66 + (2,662 − 1,932 )1/2 ] − λeff,v ) ]

= 0,223 < (graph value 0,23)

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•

Chapter 6 / Axially Loaded Members where = 0,5[1 + α(λeff,v − 0,2) + λeff,v 2 ] = 0,5 × [1 + 0,34 × (1,93 − 0,2) + 1,932 ] = 2,66 which includes the imperfection factor for buckling curve b (Table 6.1, EN 1993-1-1 (2005)) α = 0,34 Design buckling resistance (Eq. (6.47), EN 1993-1-1 (2005)) Nb,Rd = χA

fy 275 = 0,223 × 659 × γM1 1,1 × 1E3

= 36,7 > (Nb,Fd = 28,77) kN satisfactory.

6.4 COMBINED BENDING AND AXIAL FORCE – WITH BUCKLING (cl 6.3.3, EN 1993-1-1 (2005)) 6.4.1 Introduction In practical situations axial forces in columns are accompanied by bending moments acting about the major and minor axes of bending. The axial force and bending moment vary along the length of the member and an exact analysis is complicated (Culver, 1966). In some situations overall buckling, lateral torsional buckling and local buckling can occur together. This is more likely to be a problem with Class 4 sections where outstand/thickness ratios are large. Exact theoretical solutions for buckling of columns are not available and in any case would be too complicated for design. Alternatively design interaction equations (Eqs (6.61) and (6.62), EN 1993-1-1 (2005)) are used based on elastic and plastic limits. A steel member subject to bending moments and axial forces fails when a stress is greater than first yield but less than full plasticity of the section. First yield theories are conservative while full plasticity theories are unsafe. Joint rotation reduces the buckling load of a column and this can now be incorporated into the analysis of frames (cl 2, EN 1993-1-8 (2005)). This can be important because even small rotations can significantly reduce load capacity. In design it is often assumed that parts of frames can be analysed separately and the compression members can be isolated and analysed accordingly. For traditional simple design methods beam and column structures are assumed to be connected together with pin joints and braced to prevent sidesway collapse. The pin joints are assumed eccentric to the column axis and thus introduce bending moments to the column which reduces the failure load.

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191

0

3m

18

0

3m

D 12

18

0

12

C 18

4,6 m

B 18

y

z A

z

y

FIGURE 6.7 Example: three storey corner column

EXAMPLE 6.3 Bending moments in a three storey corner column. The first three storeys of a corner stanchion are sketched in Fig. 6.7. The beam support reactions are indicated in kN on each beam. Assuming simple design, eccentric pin joints and cross bracing to reduce sway, determine the bending moments in the columns AB and BC. Try the following column sizes from Section Tables. Columns AB and BC: 203 × 203 × 60 UC h = 209,6 mm, Iy = 6103 cm4 , Iz = 2047 cm4 . Column CD: 203 × 203 × 46 UC h = 203,2 mm, Iy = 4565 cm4 , Iz = 1539 cm4 . Consider a corner column buckling with bending about the major and minor axes. Bending about the major y–y axis Stiffness of columns (cm units) Column AB, KAB = Iy /LAB = 6103/460 = 13,27 Column BC, KBC = Iy /LBC = 6103/300 = 20,34 Column CD, KCD = Iy /LCD = 4564/300 = 15,21

Chapter 6 / Axially Loaded Members (a) First floor level (at joint B): Moment about the y–y axis from eccentricity of the joint assuming MBy = (beam reaction)(100 + h/2) = 180 ×

(100 + 209,6/2) = 36,86 kNm. 1E3

For simple design moments are distributed in proportion to stiffness MBy KAB 36,86 × 13,27 = 14,55 kNm = KAB + KBC 13,27 + 20,34 MBy KBC 36,86 × 20,34 = = = 22,31 kNm KAB + KBC 13,27 + 20,34

MBA = MBC

(b) Second floor level (at joint C): Assuming that the splice between columns BC and CD lies above first floor level, then the design moment is the same at the first floor (MCy = 36,86 kNm). If moments are distributed equally for simple multi-storey construction MCB = MCD =

MCy 36,86 = = 18,43 kNm. 2 2

It should be appreciated that the simple method does not incorporate joint stiffness which reduces moments at mid-span for the beams. This is beneficial, but moments in the column are inaccurate. However, the method has been used extensively in the past.

EXAMPLE 6.4 Two storey corner column. Determine the size of a column section required for a two storey corner column shown in Fig. 6.8. The design loads (kN) shown are the end reactions from the beams. The beams are connected to the column using cleats and bolts and pin joints are assumed. The columns and beams are encased in concrete. C

3,7 m

•

4,6 m

192

11

7

18

0

12

B 18

y

z A

z

y

FIGURE 6.8 Example: two storey corner column

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Axial design loads on the column: (a) First floor Floor beam (self-weight included) Wall beam (self-weight included) Self-weight of column BC (estimated) Total

kN 117 12 10 139

(b) Ground floor Floor beam (self-weight included) Wall beam (self-weight included) Upper storey Total (excluding self-weight of column AB)

kN 180 18 139 337

Try 152 × 152 × 37 kg UC grade S275 steel for both storeys. From Section Tables h = 161,8 mm, b = 154,4 mm, tw = 8,1 mm tf = 11,5 mm, Agross = 4730 mm2 , iz = 38,7 mm, iy = 68,4 mm, Wel,y = 274E3 mm3 , Wel,z = 91,5E3 mm3 , Wpl,y = 309E3 mm3 , Wpl,z = 140E3 mm3 , Iz = 706E4 mm4 , Iy = 2213E4 mm4 , Iw = 0,0399 dm6 , It = 19,3E4 mm4 . This is a Class 1 section, that is, c/t outstand ratios are within limits (Table 5.2, EN 1993-1-1 (2005)) and local buckling does not occur. However, the section must be checked for overall buckling and lateral torsional buckling. Assuming simple nominal bending moments from eccentricity of the beam reactions (R) at join B: R(100 + h/2) 180(100 + 161,8/2) = = 32,56 kNm 1E3 1E3 R(100 + tw /2) 18(100 + 8,1/2) = = = 1,87 kNm 1E3 1E3

MBy = MBz

Design bending moments applied to the column (same section throughout) My,Ed = =

MBy (IBC /LBC ) [(IBC /LBC ) + (IBA /LBA )] 32,56(1/3,7) = 32,56 × 0,554 = 18,03 kNm [(1/3,7) + (1/4,6)]

Mz,Ed = MBz × 0,554 = 1,87 × 0,554 = 1,036 kNm Check combined axial load, bending and lateral torsional buckling about the stronger y–y axis using the interaction formula (Eq. (6.61), EN 1993-1-1 (2005)) kyy (My,Ed + My,Ed ) kyz (Mz,Ed + Mz,Ed ) NEd + + ≤1 χy NRk /γM1 χLT My,Rk /γM1 Mz,Rk /γM1 337 0,465 × (18,03 + 0) 0,453 × (1,036 + 0) + + 946 0,751 × 77,3 35 = 0,356 + 0,144 + 0,013 = 0,513 < 1 satisfactory. =

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•

Chapter 6 / Axially Loaded Members Check combined axial load, bending and lateral torsional buckling about the weaker z–z axis using the interaction formula (Eq. (6.62), EN 1993-1-1 (2005)) kzy (My,Ed + My,Ed ) kzz (Mz,Ed + Mz,Ed ) NEd + + ≤1 χz NRk /γM1 χLT My,Rk /γM1 Mz,Rk /γM1 337 0,453 × (18,03 + 0) 0,755 × (1,036 + 0) + + 532 0,751 × 77,3 35 = 0,633 + 0,141 + 0,022 = 0,796 < 1 satisfactory. =

The two previous equations include the following values. For buckling load calculations, consider column AB just below B, assuming the buckling length about the y–y and z–z axes, Lcr = 0,85 L. Use buckling curve b for buckling about the y–y axis and curve c for the z–z axis (Table 6.2, EN 1993-1-1 (2005)) based on h/b < 1,2; tf < 100 mm and steel grade S275. The non-dimensional slenderness ratio (Eq. (6.50), EN 1993-1-1 (2005)) λy = =

Afy Ncr

1/2

=

Lcr iy

×

1 λ1

(0,85 × 4600) 1 × = 0,659 68,4 [93,9 × (235/275)1/2 ]

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve b), λy = 0,659 and a reduction factor χy = 0,80 the design buckling resistance for a Class 1 section (Eq. (6.47), EN 1993-1-1 (2005)) Nb,Rd = χy A

fy 275 = 0, 80 × 4730 × = 946 kN. γM1 1,1 × 1E3

My,Rd = Wpl,y

fy 275 = 309E3 × = 77, 3 kNm. γM1 1,1 × 1E6

1 Lcr × iz λ1 0,85 × 4600 1 = = 1,16 × 38,7 [93,9 × (235/275)1/2 ]

λz =

Afy Ncr

1/2

=

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) and λz = 1,16 the reduction factor χz = 0,45 and the design buckling resistance for a Class 1 section (Eq. (6.48), EN 1993-1-1 (2005)) Nb,Rd = χz A

fy 275 = 0,45 × 4730 × = 532 kN. γM1 1,1 × 1E3

Mz,Rd = Wpl,z

fy 275 = 35 kNm. = 140E3 × 1,1 × 1E6 γM1

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For a Class 1 section (Table 6.7, EN 1993-1-1 (2005)) My,Ed = Mz,Ed = 0 Interaction factor for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesser value of (λy − 0,2)NEd kyy = cmy 1 + χy NRk /γM1 (0,659 − 0,2) × 337E3 = 0,4 × 1 + = 0,465 0,80 × 4730 × 275/1,1 or

kyy = cmy 1 +

0,8NEd χy NRK /γM1

= 0,4 × 1 +

0,8 × 337E3 0,80 × 4730 × 275/1,1

= 0,514 > 0,465 use 0,465 where (Table B3, EN 1993-1-1 (2005)) cmy = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4 Interaction factors for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesser value of NEd kzz = cmz 1 + (2λz − 0,6) χz NRk /γM1 337E3 = 0,4 × 1 + (2 × 1,16 − 0,6) × = 0,836 0,45 × 4730 × 275/1,1 or

kzz = cmz 1 +

1,4NEd χz NRk /γM1

1,4 × 337E3 = 0,4 × 1 + 0,45 × 4730 × 275/1,1

= 0,755 < 0,836 use 0,755 where (Table B3, EN 1993-1-1 (2005)) cmz = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4 and kyz = 0,6kzz = 0,6 × 0,755 = 0,453

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•

Chapter 6 / Axially Loaded Members Reduction factor for lateral torsional buckling (Eq. (6.57), EN 1993-1-1 (2005)) χLT = =

1 [LT + (2LT

2

− βλLT )1/2 ]

[0,881 + (0,8812

1 = 0,751 < 1 − 0,75 × 0,9052 )1/2 ]

or 1 2 λLT

=

1 = 1,22 > 1, use 0,751 0,9052

which includes the factor 2

LT = 0,5[1 + αLT (λLT − λLT,0 ) + βλLT ] = 0,5 × [1 + 0,21 × (0,905 − 0,2) + 0,75 × 0,9052 ] = 0,881 where the imperfection factor αLT = 0,21 (Table 6.3, EN 1993-1-1 (2005)). λLT =

Wply fy Mcr

1/2

=

309E3 × 275 103,85E6

1/2 = 0,905

The elastic critical bending moment Eq. (5.5) 1/2 Iw L2 GIt Mcr = + 2 Iz π EIz 1/2 0,0399E-6 4,62 × 15,59 π2 × 1483 = + 2 706E-8 4,62 π × 1483

π2 EIz L2

= 103,85 kNm where L = 4,6 m EIz = 210E6 × 706E-8 = 1483 kNm2 GIt = 80,77E6 × 19,3E-8 = 15,59 kNm2 EIw = 210E6 × 0,0399E-6 = 8,379 kNm4 For further information on elastic critical bending moments see Chapter 5. Check the self-weight of column BC. Minimum overall dimensions of cased column H = 161,8 + 100 = 261,8, say 270 mm B = 154,4 + 100 = 254,4, say 260 mm Ac = Ag − As = 270 × 260 − 4740 = 65460 mm2 .

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Total weight = steel column + concrete casing = Lρg + LAc ρg = 3,7 × 37 × 9,81/1E3 + 3,7 × 65460 × 2400 × 9,81/1E9 = 7 kN < 10 kN (assumed) satisfactory. These calculations show that a 152 × 152 × 37 kg UC section grade 275 steel is satisfactory, however a lesser weight of 152 × 152 × 30 kg UC might be suitable. Calculations are more extensive for Class 4 sections which are reduced in area to allow for local buckling (cl 5.5.2(2) EN 1993-1-1; and cl 4.3 EN 1993-1-5). Section properties are based on the effective cross-sections.

REFERENCES Beer, H. and Schultz, G. (1970). Theoretical basis for the European column curves. Construction Metallique, No. 3. Bleich, F. (1952). Buckling strength of metal structures. McGraw-Hill. Culver, G.C. (1966). Exact solution for biaxial bending equations. American Society of Civil Engineers, Str. Div., 92(ST2). Dwight, J.B. (1975). Adaption of the Perry formula to represent the new European steel column curves, Steel Construction, AISC, 9(1). The background to British Standards for structural steel work. Imperial College, London, and Constrado. ECCS (1976). Manual on stability, introductory report, second international colloquium on stability. European Convention for Structural Steelwork. Liege. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-5 (2003). Plated structural elements. BSI. EN 1993-1-8 (2005). Design of joints. BSI. Euler, L. (1759). Sur la force de collones. Memoires de l’Acadamie de Berlin. Godfrey, G.B. (1962). The allowable stress in axially loaded struts, Structural Engineer, March 1962. Nelson, H.M. (1953). Angles in tension. British Constructional Steelwork Association publication No. 7. Regan, P.E. and Salter, P.R. (1984). Tests on welded angle tension members, Structural Engineer, 62B(2). Robertson, A. (1925). The strength of struts, Selected Engineering Paper No. 28, Institution of Civil Engineers. Trahair, N.S. and Bradford, M.S. (1988). The Behaviour and Design of Steel Structures. Chapman and Hall. Williams, F.W. and Sharp, G. (1990). Simple elastic critical load and effective length calculations for multi-storey rigid sway frames. Proceedings of the ICE, 90. Wood, R.H. (1974). Effective lengths for columns in multi-storey buildings. Structural Engineer, 52.

Chapter

7 / Structural Joints

(EN 1993-1-8, 2005)

7.1 INTRODUCTION Structural steel connections, referred to as joints in the Code, are required to ensure continuity at the intersection members and foundations. They are also used to form splices and to construct brackets to support loads. Generally structural steel joints are composed of plates, or parts of sections, shop welded in controlled conditions and bolted together on site. Welding can be carried out on site but it needs to be carefully supervised and is limited because of the expense. The physical appearance of some joints is shown in Figs 7.22 and 7.23. Structural joints transmit internal forces and moments in a structure and strength is of major importance. However, the rigidity of joints also needs to be considered. All joints are semi-rigid with associated small linear and larger rotational movements. The linear movements at a joint are generally small and generally need not be considered, but the rotational movements affect the distribution of forces and moments which must be taken into account in structural analysis. For theoretical purposes in the analysis of structures, joints are classified (Table 5.1, EN 1993-1-8 (2005)) by strength and rigidity as: (a) Pinned – low moment of resistance. (b) Rigid – full strength and all deformations are insignificant. (c) Semi-rigid – characteristics of the connection lie between (a) and (b). These theoretical and practical descriptions are important to recognize when analyzing a structure to determine the distribution of forces and moments using global analysis (cl 5.1, EN 1993-1-8 (2005)). There are three methods of global analysis: (1) Elastic – joints are classified by rotational stiffness. (2) Rigid-plastic – joints are classified by strength. (3) Elastic plastic – joints are classified by stiffness and strength.

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7.2 THE IDEAL STRUCTURAL JOINT The types of joints in common use have been developed and modified to suit the manufacturing and assembly processes and the ideal requirements are: (a) Simple to manufacture and assemble. (b) Standardized for situations where the dimensions and loads are similar thus avoiding a multiplicity of dimensions, plate thicknesses, weld sizes and bolts. (c) Manufactured from materials and components that are readily available. (d) Designed and detailed so that work is from the top of the joint not from below where the workman’s arms will be above his head. There should also be sufficient room to locate a spanner, or space to weld if required. (e) Designed so that the welding is confined to the workshops to ensure a good quality and to reduce costs. (f) Detailed to allow sufficient clearance and adjustment to accommodate the lack of accuracy in site dimensions. (g) Designed to withstand normal working loads and also erection forces. (h) Designed to avoid the use of temporary supports during erection. (i) Designed to develop the required load–deformation characteristics at service and ultimate loads. (j) Detailed to resist corrosion and to be of reasonable appearance. (k) Low in cost and cheap to maintain.

7.3 WELDED JOINTS Welding is a method of connecting components by heating the materials to a suitable temperature so that fusion occurs. The most common method for heating steelwork is by means of an electric are between a coated wire electrode and the materials being joined. The electrical circuit is shown in Fig. 7.1(a). During the process, which is illustrated in Fig. 7.1(b), the coated electrode is consumed, the wire becomes the filler material and the coating is converted partly into a shielding gas, partly into slag, and some part is absorbed by the weld metal. This method, known as the manual metal arc welding process, is still the most common for structural joints because of low capital cost and flexibility. However, for long continuous welds automatic processes are preferred because of consistency. Generally the electrode is stronger than the parent metal. For manual metal arc welding the electrodes should be compatible with the steel being welded (BSEN 499, 1995; Gourd, 1980). The main reason for the flux covering to the electrode in the manual metal arc welding process is to provide an inert gas which shields the molten metal from atmospheric contamination. In addition the flux forms a slag to protect the weld until it is cooled to room temperature, when the slag should be easily detachable. Other functions of the flux include: arc stabilization, control of surface profile, control of weld metal composition, alloying and deoxidization. However, it should be noted that

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Electrode wire Welding machine Coating on electrode

Molten arc pool Electrode Slag

Arc

Gaseous shield Arc stream

Weld Base metal Components being welded

Circuit

(a) Arc welding circuit

(b) Shielded arc welding

FIGURE 7.1 Shielded metal are welding

the flux can be a source of hydrogen contamination from absorbed and chemically combined moisture. The absorbed moisture can be removed by drying. The particular advantage of welding is that it forms a rigid joint, however the manufacture of welded joints requires more skill and supervision than bolted joints. Most welded structural joints are affected using the manual metal arc process but long continuous welds, which occur in built-up girders, are laid down with automatic welding equipment. The automatic processes achieve the exclusion of atmospheric pollution by gas shielding, flux core or submerged arc. Types of welds used in structural engineering and allowed in the European Code are fillet, slot, butt, plug and flare groove. Some common types are shown in Fig. 7.2. The two types in most common use are butt and fillet welds. Butt welds, often used to lengthen plates in the end on position, may be considered as strong as the parent plate as long as full penetration for the weld is achieved. For thin plates penetration is achieved without preparing the plate, but on thicker plates V or double J preparation is required. Butt welds are also used to connect plates at right angles but the plates require edge preparation. Partial penetration butt welds are not favoured in design and should not be used intermittently or in fatigue situations (BSEN 1011-1 (1998)). Fillet welds are generally formed with equal leg lengths. They do not require special edge preparation of the plates and are therefore cheaper than butt welds. Generally in connections, plates intersect at right angles but intersection angles of between 60◦ and 120◦ can be used provided that the correct throat size is used in design calculations. In order to accommodate lack of fit the minimum leg length of fillet weld in structural engineering is 5 mm although 6 mm is often preferred. The maximum size of fillet weld from a single run metal arc process is 8 mm, but 6 mm is

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Butt weld made from one side

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201

Butt weld made from two side T-butt weld made from two sides

T-butt weld made from two sides

t

b d t and c 3 mm or c t / 5

Slope should not exceed 1:1 in butt joints

b

d c

T-butt weld The weld is regarded to be of equal or higher strength than the vertical member

(a) Examples of butt welds

T-fillet weld

Lap joint

(b) Examples of fillet welds

FIGURE 7.2 Types of welds in structural joints

preferred to guarantee quality. When larger fillet welds are required they are formed from multiple runs. The use of intermittent butt and fillet welds is permitted (cl 4.3.2.2, EN 1993-1-8 (2005)). Intermittent welds are not favoured in structural engineering because they introduce stress discontinuities, act as stress raisers, may introduce fatigue cracks, may act as corrosion pockets and are difficult to produce with an automatic welding machine. The spacing of intermittent fillet welds is shown in Fig. 4.1, EN 1993-1-8 (2005).

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Deep penetration a

a

a

a

a

FIGURE 7.3 Throat thickness of fillet welds (Figs 4.3 and 4.4, EN 1993-1-8)

7.3.1 Throat Thickness of a Weld (cl 4.5.2, EN 1993-1-8 (2005)) The size of a weld is often described by the leg length but the strength is calculated using the effective throat thickness (a) as defined in Fig. 7.3. The effective throat thickness should not be less than 3 mm.

7.3.2 Effective Length of a Weld (cl 4.5.1, EN 1993-1-8 (2005)) The effective length of a fillet weld should be taken as the length over which the fillet is full size. The minimum length allowed to transmit loading is six times the throat thickness, and not less than 30 mm. In practice, fillet welds terminating at the ends, or sides, of parts are returned continuously round the corners for a distance of not less than twice the leg length, unless impracticable. The continuation round the corner is to reduce stress concentrations and its strength is generally ignored in strength calculations. The effective length of a weld is reduced if a component distorts under load in situations similar to that shown In Fig. 7.4, where the deformations in the weld adjacent to the web are greater than those at the end of the flange. The larger deformations at the web initiate failure in the weld at this point with consequent loss of strength for the total length of the weld (Elzen, 1966; Rolloos, 1969). For design the effective breadth (beff ) of a weld (cl 4.10, EN 1993-1-8 (2005)) is: For a rolled ‘I’ or ‘H’ section (Eqs (4.6a) and (4.6b), EN 1993-1-8 (2005)) fy,f tf beff = tw + 2r + 7ktf ≤ tw + 2r + 7 tf (7.1a) tp fy,p For box or channel sections where the widths of the connected plate and the flange are similar fy,f tf beff = 2tw + 5 tf ≤ 2tw + 5 tf (7.1b) tp fy,p

7.3.3 Long Welded Joints (cl 4.11, EN 1993-1-8 (2005)) The stress distribution along the length of a long lap joint is not uniform being greatest at the ends. To allow for this the length of the weld is reduced. For joints longer

Structural Design of Steelwork to EN 1993 and EN 1994 Shear force V

203

V

Bending moment M

Column I section

•

Deformation of welds due to shear force V determining flexible beam flanges

Bracket I section

Side elevation

End elevation Deformation of weld due to bending moment M deforming the flexible column flange

Plan

FIGURE 7.4 Reduction in strength of welds associated with flange deformations

than 150 times the throat thickness (a), the reduction factor (Eq. (4.9), EN 1993-1-8 (2005)) βLW ,1 = 1, 2 − 0, 2Lj /(150a) ≤ 1

(7.2)

where Lj is the overall length of the lap in the direction of the force transfer.

7.3.4 Design Resistance of Fillet Welds (cl 4.5.3, EN 1993-1-8 (2005)) The real external forces acting on a 90◦ fillet weld are probably those shown in Fig. 7.5 (a) (Clarke, 1971). Experiments (Biggs et al., 1981) on 90◦ fillet welds of equal leg length loaded to failure show that the fracture plane varies between 10◦ and 80◦ depending on the combination of external forces. The actual distribution of stress on the failure plane is uncertain but a theoretical distribution (Kato and Morita, 1974) shows peak stresses at the root of the weld which reduce towards the face of the weld. This distribution appears to be confirmed by experimental observations of cracks initiating at the root. The situation is complicated further by residual stresses and variables such as the type of electrode type of steel, ratio of the size of weld to the plate thickness, the quality of weld and whether the loading is static or dynamic. If stresses on the failure plane are assumed to be uniform then the relationship between the average shear stress and tensile stress on the failure plane has been shown (Biggs et al., 1981) to approximate to an ellipse. An ellipse of failure stresses combined with a variable

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) z x

Mwx

τf

Fwx

Mwy Fwy y

f

f

sf

Mwz Fwz (a) Complex system of forces

gth

en

it l Un

Fwx

τ1 Fwx

Fwz

Fwy f = 45° Fwz

Fwy

τ2 s

f = 45° th Th ick ro es at s a

•

Leg length s

204

(b) Simple system of forces

FIGURE 7.5 Forces acting on a 90◦ fillet weld

fracture angle can be used theoretically to predict the magnitude of the external forces, but the method is unnecessarily complicated for design purposes. For design purposes a complex system of external forces acting on a fillet weld is reduced forces acting in three perpendicular directions on a unit length of weld as shown in Fig. 7.5 (b). The vector sum of all the design forces should not exceed the design resistance of a fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) and may be expressed as 2 2 2 2 Fwx + Fwy + Fwz ≤ Fw,Rd

(7.3)

The term Fw,Rd = a fvw,d = a( fu /31/2 )/(βw γM2 ) is the design strength of a fillet weld per unit length. fu = nominal ultimate tensile strength of the weaker part joined (Table 3.1, EN 1993-1-1 (2005)) βw = correlation factor dependent on steel grade (Table 4.1, EN 1993-1-8 (2005)) γM2 = partial safety factors for joints, recommended value 1,25 (Table 2.1, EN 1993-1-8 (2005)).

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The design strength of a fillet weld has been shown (Ligtenberg, 1968) to be related to the strength of the parent material. The correct type and strength of electrode must be used for each grade of steel. An alternative directional method of design (cl 4.5.3.2, EN 1993-1-8 (2005)) involves calculating the normal and shear stresses on the throat section of the weld and combining them using the yield criteria developed in Chapter 2. This method is more laborious and introduces the possibility of further errors when resolving the forces onto the critical plane. The relationship between the forces Fx , Fy and Fz for this method can be expressed I the same form as Eq. (7.3) for a 45◦ plane (Holmes and Martin, 1983). The size of the fillet weld obtained using this method is slightly less than using the vector addition method.

7.3.5 Load–Deformation Relationships for Fillet Welds The strength of the weld in a connection is of primary importance but the load– deformation characteristics of the weld should also be considered. The deformation at the maximum load varies from approximately 0.6 to 1.4 mm depending on the orientation of the weld in relation to the applied load (Clarke, 1970) as shown in Fig. 7.6. The maximum deformation for the side fillet weld which is parallel to the applied load and the minimum is for an end fillet weld. To allow for this effect design stresses are based on the weaker side fillet welds. At the stress the disparity in deformations for end and side fillet welds is less than at failure.

7.3.6 Conditions Affecting the Strength of Welded Joints The following conditions affect the strength of welded connections: (a) Use of an incorrect steel (BS 7668, 1994). (b) Use of an incorrect electrode (BSEN 499, 1995). (c) Cavities and slag inclusions in the weld. These may be detected by non-destructive testing.

Average throat stress (MPa)

End fillet weld θ = 0° 500

θ = 30° θ = 60° θ = 90° P

Side fillet weld

θ varies

Test weld Design stress P 0

1 2 Weld deformation (mm)

FIGURE 7.6 Load-deformation relationships for an 8 mm fillet weld (Clarke 1970)

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) (d) (e) (f) (g)

(h)

(i) (j) (k)

(l)

Excessive lack of fit between components. Stress concentrations combined with oscillating loads producing fatigue. Residual stresses introduced from differential heating during welding. Hydrogen cracks associated with welding occur when the cooling rate is too rapid (Fig. 7.7(a)). Excessive hardening occurs in the heat affected zone which cracks under the action of residual stresses if sufficient hydrogen is present in the weld. This defect can be avoided by controlling the cooling and the hydrogen input to the weld (BSEN 1011-1 (1998)). Lamellar tearing may occur when welding plate connections of the type shown in Fig. 7.7 (b). the cracks are produced by a combination of low ductility in the plate in the transverse direction and high point restraint in the weld which induces tensile forces adjacent to the connection. The low ductility in the plate is produced by inclusions of non-metallic substances formed in the steel making process. When the ingot is rolled to make steel these inclusions form as plates parallel to the direction of rolling. Only a small percentage of plates are susceptible to lamellar tearing, and where it occurs joint details can be changed to reduce the chances of if affecting the strength of the connection (Farrar and Dolby, 1972). Brittle fracture. Corrosion which reduces the size of components or causes pitting which may initiate fatigue cracks. Insufficient penetration of the parent metal which leads to a reduction in strength of the weld. The welder uses a voltage and arc length which produces a stable arc and a satisfactory weld profile. The current then becomes the main factor in controlling penetration. Another important factor in depth of penetration is edge preparation. Plates of 6 mm with square edges can be butt welded from one side, but the edges of thicker material must be bevelled to provide access for the arc. Lack of side wall fusion occurs if there is poor bond between the parent and weld metal. Good bonding can only occur when the surface of the parent metal has been melted before the weld metal is allowed to flow into the joint.

Further information on faults in welds can be found in Gourd (1980).

Fillet weld

Hydrogen cracks (a) Hydrogen cracks

FIGURE 7.7 Faults associated with welding

Fillet weld

Lamellar tearing (b) Lamellar tearing

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7.3.7 Design Strength of Fillet Welds (cl 4.5.3.3 and Table 4.1, EN 1993-1-8 (2005)) The design strength of a unit length of fillet weld (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = (throat thickness) × (unit length) × (design shear stress)/factors fu a 31/2 = βw γM2

(7.4)

The factor (βw ) is related to steel grade, and varies between 0,8 and 1,0 (Table 4.1, EN 1993-1-8 (2005)). The ultimate stress ( fu ) is that of the weaker material joined (Table 3.1, EN 1993-1-1 (2005)). Typical values are fu = 430 MPa for S275 grade steel, and fu = 510 MPa for S355 grade steel. The value ( fu /31/2 ) is from the shear distortion strain energy theory as explained in Chapter 2. It should be noted that although the strength of the weld is calculated using the throat thickness (a) (Figs 4.3 and 4.4, EN 1993-1-8 (2005)) the weld is often specified by the leg length. A table of the strength of fillet welds is given in Annex Al.

7.4 BOLTED JOINTS (cl 3, EN 1993-1-8 (2005)) The advantage of bolted joints is that they require less supervision than welding, and therefore are ideal for site conditions. Other advantages are that the connection can be fastened quickly, supports load as soon as the bolts are tightened and accommodates minor discrepancies in dimensions. Disadvantages of bolted connections are that for large forces the space required for the joint is extensive, and the connection is not as rigid as a welded connection even when friction grip bolts are used. Steel bolts are identified by their gross diameter, strength and use. The preferred sizes of bolts in general use are 16, 20, 24, 30 and 36 mm diameter. The most common size use in structural connections is 20 mm. The types of bolt in common use (Table 3.1, EN 1993-1-8 (2005)) are: (a) Ordinary bolts Classes 4.6–6.8 and includes foundation bolts. (b) Pre-loaded bolts Classes 8.8 and 10.9. A Class 4.6 bolt is low in cost, can be installed with the use of simple tools and requires little supervision during the erection. At fracture the bolt has a relatively large extension of 25%, a property which is preferred at plastic collapse. Where forces are large, or where space for the connection is limited, or where erection costs can be reduced by using fewer bolts, then the higher grade bolts are used. The percentage elongation of 12% at failure is less, but is still acceptable for design purposes.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Where a more rigid bolted connection is required, for example in plastic methods of design, pre-loaded bolts are used. The strength of these bolts is greater with an increased cost for the additional site supervision which is necessary to ensure that the bolts are axially pre-loaded in tension to the design values. The object of the pre-load is to ensure that the friction between the ‘faying’ surfaces prevents slip when subject to external shear forces and thus produce a more rigid joint. The nuts of pre-loaded bolts are tightened with a torque wrench which is calibrated in relation to the required axial pre-load. A simpler method of measuring the axial force in the bolts is to use a load indicating washer, under the head of the nut, which reduces in thickness to a specified value for a specified pre-load. The washer is less accurate than the torque spanner (Bahia and Martin, 1981). A further alternative method of ensuring that the bolt is pre-loaded is to specify ‘turns of the nut’. Investigations (Fisher and Struck, 1974) showed that in general the pre-load produced by the torque wrench and ‘turns of the nut’ method on site exceeded the specified value. Close tolerance turned bolts are used only where accurate alignment of components or structural elements is required. The shank of the bolt is at least 2 mm greater in diameter than the threaded portion of the bolt and the hole is only 0,15 mm greater than the shank diameter. This small tolerance necessitates the use of special methods to ensure that the holes align correctly. Foundation bolts, or holding down bolts, are used for connection structural elements to concrete pads or concrete foundations. Generally the bolts are cast into the concrete before erection of the steel work and thus require accurate setting out. Where uplift forces occur the bolts must be anchored by a washer plate. Most bolts used are Class 4.6 but higher strengths are available. Sometimes bolts are grouted in the holes during erection using epoxy resin. Rivets were used extensively in the past in the fabrication shop and on site. They were difficult and expensive to place but they resulted in a rigid connection because the hot rivet, after driving, expanded to fill the hole. Rivets have now been superseded by welding and bolts. There are five categories of bolted joints (Table 3.2, EN 1993-1-8 (2005)) related to the type of connection or bolt and pre-loading. Shear connections Category A: bearing type where there is no pre-loading nor special provision for contact surfaces. Design for shear and bearing resistance. This is the cheapest type of connection where complete rigidity and plasticity are not important. Category B: slip resistant at serviceability limit state. Design for slip resistance at the serviceability limit state and shear and bearing resistance at the ultimate limit state. Connection used to provide full rigidity in the elastic stage of behaviour when deflections are critical.

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Category C: slip resistant at ultimate limit state. Design for slip resistance and bearing resistance at the ultimate limit state. Connection used to limit movements at the ultimate plastic limit state. Tension connections Category D: No pre-loading of bolts. Category E: Pre-loaded bolts.

7.4.1 Washers In British practice most bolts have steel washers under the head and under the nut. The washer distributes the bolt force and prevents the nut, or bolt head, from damaging the component or member. However washers are not essential in all cases (ECSS, 1981), and they are now being omitted in British practice. Ordinary washers (BSEN 4320, (1998)) are in general use but hardened washers (BSEN 14399-1 to 5 (2005)) are used for pre-loaded bolts. The outside diameter of a washer is an important dimension when detailing, for example to avoid overlapping an adjacent weld.

7.4.2 Bolt Holes Bolt holes are usually drilled, but may be punched full size, or punched under size and reamed. Holes should never be formed by gas cutting because of the inaccuracy and the effect on the local properties of the steel. Punched holes are preferred by steel fabricators because it saves time and reduces cost. However research (Owens et al., 1981) shows that distortion in the vicinity of a hole reduces toughness and ductility and can lead to brittle fracture. Punched holes should not be used in locations where plastic tensile straining can occur. Bolt holes are made larger than the bolt diameter to facilitate erection and to allow for inaccuracies. The clearance is 2 mm for bolts not exceeding 24 mm diameter and 3 mm for bolts exceeding 24 mm diameter. Oversize and slotted holes are allowed but not often used. Slotted holes are sometimes used for pre-loaded bolts to facilitate erection with unusual shaped structures, or alternatively they can be used to accommodate movement in a structure. The clearance for a close tolerance turned bolt is 0,15 mm. Bolt holes reduce the gross cross-sectional area of a plate to the net cross-sectional area. The net value is used for calculations where the structural element, or parts of an element, are in tension. Bolt holes also produce stress concentrations, but it is argued that these are offset by the fact that at yield the highly stressed cross-section will work harder before fracture and yield will by then have occurred at adjacent cross-sections. The gross cross-section of a member is used in compression because at yield the bolt

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) hole deforms and the shank of the bolt resists part of the load in bearing. Consideration should be given to corrosion and local buckling when deciding the position of holes.

7.4.3 Spacing of Bolt Holes (Table 3.3, EN 1993-1-8 (2005)) The longitudinal spacing between the centre line of bolts in the direction of the axial stress in a member is called the pitch. The minimum spacing in the direction of the load of 2,2 × (diameter of the hole) is specified to prevent excessive reduction in the cross-sectional area of a member, to provide sufficient space to tighten the bolts and to prevent overlapping of the washers. Other critical distances are given in Table 3.3 and Fig. 3.1, EN 1993-1-8 (2005). These values are specified to prevent buckling of plates in compression between bolts, to ensure that bolts act together as a group to resist forces, and to minimize corrosion.

7.4.4 Edge and End Distances for Bolt Holes (Table 3.3, EN 1993-1-8 (2005)) Edge and end distances are specified to resist the load, to prevent local buckling, to limit corrosion and to provide space for the bolt head, washer and nut. The edge distance is from the centre of a hole to the nearest edge measured at right angles to the direction of the load. The minimum edge distance specified is 1,2 × (diameter of the hole) and the maximum should not exceed 4t + 40 mm. The end distance is from the centre of a hole to the adjacent edge in the direction of the load transfer. The minimum end distance in the direction of the load is 1,2 × (diameter of the hole) and the maximum should not exceed 4t + 40 mm. The end distance should also be sufficient for bearing capacity. There are recommended positions, spacing and diameter of holes in Section Tables. These distances are based on providing sufficent clearances to the web and adequate edge distances (Annexes A4 to A6).

7.4.5 Deductions for Holes in Tension Members (cl 3.10, EN 1993-1-8 (2005)) Holes are drilled in tension members to accommodate fasteners at connections. A hole reduces the gross cross-sectional area and weakens a tension member because the fastener in the hole does not transmit the axial force. In contrast a hole in a compression member has little effect on the buckling strength because as the member compresses the axial force is transmitted by bearing on the shank of the bolt. When designing a tie the net cross-sectional area is used in calculations to determine the design axial force. The net cross-sectional area is the gross area reduced by the maximum sum of the sectional areas of the holes. These holes may be in line at right

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211

C

110 95 A

95 B

95

95

430

110

P = 110

A

•

Direction of stress

95

C

s = 95

FIGURE 7.8 Example: net area of a plate

angles to the axial stress in the member (line AA in Fig. 7.8), or staggered (lines BB and CC in Fig. 7.8). Typical areas to be deducted for bolt holes are: Bolt diameter + 2 mm clearance for bolts not exceeding 24 mm in diameter or Bolt diameter + 3 mm clearance for bolts exceeding 24 mm in diameter. For staggered fastener holes the area to be deducted shall be the greater of: (a) the deduction for non-staggered holes; (b) the sum of the sectional areas of all holes in any diagonal or zig-zag line extending progressively across the member, or part of the member, less s2 t/4p for each gauge space in the chain of holes. where (Fig. 7.8) s is the staggered pitch p is the spacing of the holes t is the thickness of the holed material. For sections such as angles with holes in both legs the spacing is approximately the sum of the back marks to each hole, less the leg thickness. The arrangement and spacing of holes in a member should not significantly weaken a member at a section.

EXAMPLE 7.1 Net area of a plate with holes. Calculate the net cross-sectional area for the plate shown in Fig. 7.8 which is subject to a tensile force. The plate is 20 mm thick and contains four lines of staggered holes drilled for 24 mm diameter bolts. From Fig. 7.8, s = 95 mm and p = 110 mm. Diameter of hole do = d + 2 = 24 + 2 = 26 mm.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Gross cross-sectional area perpendicular to the direction of stress = 20 × 430 = 8600 mm2 Areas to be deducted at possible failure lines are: nh t do − ngs

s2 t 4p

where ngs is the number of gauge spaces in the chain of holes. Line AA: 2 × 20 × 26 = 1040 mm2 Line BB: 3 × 20 × 26 − 1 ×952 × 20/(4 × 110) = 1150 mm2 Line CC: 4 × 20 × 26 − 3 × 952 × 20/(4 × 110) = 849 mm2 . Minimum net area for line BB = 8600 − 1150 = 7450 mm2 .

7.4.6 Design Resistance of Single Bolts (Tables 3.1 and 3.4, EN 1993-1-8 (2005)) Bolted connections consist of two or more bolts and each bolt may be subject to any combination of tension, shear or bearing forces. The design resistance of a single bolt when subject to these forces is now considered in detail. A bolt in tension fails at the smallest cross-section, that is, the root of the threads where the net area is approximately 80% of the gross area. Where the bolt fails across the reduced cross-section at the root of the thread the tension resistance As Ft,Rd = k2 fub (7.5) γM2 where the ultimate tensile strength of the bolt (fub ) is obtained from Table 3.1. EN 1993-1-8 (2005) and k2 = 0,9 except for a countersunk bolt where k2 = 0,63. Where the bolt assembly fails by the bolt head, or nut, shear punching through the plate then the tension resistance of the plate (Table 3.4, EN 1993-1-8 (2005)) Bp,Rd = 0,6 π dm tp

fu γM2

(7.6)

where fu is the ultimate tensile stress of the plate. tp is the thickness of the plate under the head of the bolt. dm is the mean of across points and across flats dimensions of the bolt head or the nut, whichever is smaller. The shear resistance per shear plane for a single bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = αv fub

A γM2

(7.7)

Structural Design of Steelwork to EN 1993 and EN 1994

•

213

Where the shear plane passes through the unthreaded portion of the bolt αv = 0,6 and A is the gross area. Where the shear plane passes through the threads αv = 0,5 or 0,6, depending on the grade of the bolt, and A = As the reduced area. The values of the reduced bolt areas used in this chapter (BS 3692, 2001; BS 4190, 2001) are: Bolt diameter (mm) 12 16 Reduced area (mm2 ) 84,3 157

20 245

22 303

24 353

27 459

30 561

36 817

Values may also be obtained from Section Tables. Shearing of a bolt occurs on the shank, that is, the gross area of the bolt, if the thread length on a bolt is carefully specified, but it is safer to assume that it occurs on the reduced area. It also simplifies calculations and avoids confusion. Experiments (Bahia and Martin, 1980) and other investigators have found shear values that vary between 0,62fu and 0,71fu . The value from the Huber–Von–Mises–Hencky shear distortion strain energy theory is fu /31/2 as shown in Chapter 2. Ordinary bolts deform when subject to shear stresses but it is important to realize that the shear deformation of the connection is increased by the bearing stresses on the plate. The higher the bearing stresses the greater the deformation as shown in Fig. 7.9.

7.4.7 Design Resistance of a Bolt Subject to Shear and Tension Forces (Table 3.4, EN 1993-1-8 (2005)) Shear and tensile resistances are related by the linear interaction formula Fv,Ed Ft,Ed + ≤ 1, 0 Fv,Rd 1,4Ft,Rd

(7.8)

Generally it is assumed that the failure plane passes through the threaded portion of the bolt. This equation is to be compared with a non-linear experimental relationship (Chesson et al., 1965) based on the net cross-sectional area of the bolt (Fig. 7.10). Alternative elliptical relationships are given in ECSS (1981) and BS 5400 (2000).

7.4.8 Design Bearing Resistance for a Bolt (Table 3.4, EN 1993-1-8 (2005)) A bolt subject to a shear force, such as shown in Fig. 7.11, comes in contact with the plate when the shear load is applied and slip occurs. The bearing stresses between the bolt and plate need to be controlled to limit deformations.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

tp

tp

=2

160

0, = 1 15 m 7 m , tp = 1 10 m m 4, tp = 1 10 m 2,0 m 7m m

High-strength bolts 20 mm diameter

0m

m

140

tp

=2

120

Shear load (kN)

•

100

Black bolt 20 mm diameter

80

60

40

20

0

4 0

2

6 2

8 4

10 High-strength bolts 6 black bolts

Total deformation (mm) Notes : Permanent deformation of holes shown as a broken line; bolts failed in single shear across the threads; tp is the plate thickness.

FIGURE 7.9 Relation between shear load and deformation for single bolt tests (Bahia and Martin, 1980)

Experimental results by Chesson et al., 1965 for high strength bolts with shear plane on shank and on threads

1,0 0,8

2 ⎤ ft ⎤ ⎥ ⎥ + ⎦ ftu ⎦

fv ⎤ 2 ⎤ ⎥ ⎥ =1 ⎦ 0,62 ftu ⎦

0,6 f v /f tu

214

0,4 0,2

0

0,2

0,4

0,6 f t /f tu

0,8

1,0

1,2

FIGURE 7.10 Relationship between shear and tensile stresses for bolts

Structural Design of Steelwork to EN 1993 and EN 1994

•

215

Bolts

A Side elevation

dh

Thickness tp

A

cp

Plan section AA (a) End bearing failure for a small end distance

(b) Enclosed bearing failure for a large end distance

Plan

Sectional elevation (c) Actual distribution of bearing stress

(d) Theoretical distribution of bearing stress

FIGURE 7.11 Bearing stresses for bolts

The bolt or plate may deform because of high local bearing stresses between bolt and plate, or a bolt may shear through the end of the plate. The bearing resistance Fb,Rd = k1 αb fu

dt γM2

(7.9)

Values of k1 control end distances and values of αb control edge distances (Table 3.4, EN 1993-1-8 (2005)). The value of fu is the weaker of the bolt grade or the adjacent plate or section. The values of the ultimate tensile strength for bolts (Table 3.1 EN 1993-1-8 (2005)) used in this chapter are: Bolt class 4.6 fub MPa 400 αv 0,6

4.8 400 0,5

5.6 5.8 500 500 0,6 0,5

6.8 600 0,5

8.8 800 0,6

10.9 1000 0,5

Values of steel grades used in examples in this chapter are S275 ( fu = 430 MPa) and S355 ( fu = 510 MPa). Equation (7.9) assumes uniform bearing stresses as shown in Fig. 7.11(d) whereas in reality they are closer to those shown in Fig. 7.11(c). Design bearing stresses are high

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Permanent deformation of hole shown as a broken line

700

tp = 12,07 mm

600 Average bearing stress (MPa)

216

tp = 14,10 mm

500

tp = 17,10 mm

400

tp = 20,15 mm

300 Yield stress of plates 221,8 ± 8,03 MPa

200 100

0

2 4 6 8 10 Total shear deformation (mm)

Note: Bolts failed on the threads in single shear at a shear stress of 600 MPa

FIGURE 7.12 Relationships between bearing stress and deformation for an M20 high stress single bolt (Bahia and Martin, 1980)

in relation to the yield stress because material subject to bearing stresses is generally confined by other parts which restricts deformation. High bearing stresses are not disastrous but lead to excessive deformation of a connection as shown in the experimental results (Fig. 7.12). Equations (7.5)–(7.9) express the design tensile and shear strengths of a bolt and can be presented in the form of tables to reduce calculations (Annex A2).

7.4.9 Bolts Through Packings (cl 3.6.1(12), EN 1993-1-8 (2005)) Where the total thickness of the packing (tp ) is greater than three times the nominal diameter (d) of the bolts the design shear resistance is multiplied by a reduction factor βp =

9d ≤1 8d + 3tp

(7.10)

The reason for the reduction in strength is because as the grip length increases the bolt is subject to greater bending moments from shear forces which move further apart.

7.4.10 Long Bolt Joints (cl 3.8, EN 1993-1-8(2005)) For long joints the load is not shared equally by the bolts or rivets. The fasteners on the end resist the greatest force and the resistance gradually reduces to the centre line of the joint. The reduction is due to friction, errors in marking out and deformations

Structural Design of Steelwork to EN 1993 and EN 1994

•

217

Lj Cover plate

FIGURE 7.13 Length (Lj ) for long bolted joints in the materials. Where the length of the joint Lj > 15d (Fig. 7.13) measured in the direction off the transfer force then the design shear resistance reduction factor βLf = 1 −

Lj − 15d 200d

(7.11)

but 0,75 ≤ βLf ≤ 1,0

7.4.11 Design of Slip Resistant Joints (cl 3.9, EN 1993-1-8 (2005)) High strength bolts can be pre-loaded when installed and designed to be slip resistant at working load or ultimate load. The design slip resistance of a pre-loaded bolt when subject to external tensile and shear forces is now considered in detail. The design slip resistance of pre-loaded high strength bolt of Classes 8.8 or 10.9 (cl 3.9.1, EN 1993-1-8 (2005)) is Fs,Rd =

ks n μFp,C γM3

(7.12)

where Fp,C = 0,7fub As is the design preloading force. μ = slip factor as listed in Table 3.7, EN 1993-1-8 (2005). ks = 1 for a bolt in a clearance hole and reduced for slotted holes (Table 3.6, EN 1993-1-8 (2005)) n = number of friction interfaces. T ypical Surface treatment

Slip factor (μ)

A – blasted with shot or grit, loose rust removed, no pitting B – ditto, painted with zinc C – cleaned and loose rust removed D – surfaces not treated

0,50 0,40 0,30 0,20

Table 7.2 Slip factors(μ) for pre-loaded bolts (Table 3.7, EN 1993-1-8 (2005)).

218

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) The effect of a tensile force acting on a pre-loaded bolt is to reduce the frictional resistance (cl 3.9.2, EN 1993-1-8 (2005)). The design slip resistance per bolt subject to a combination of shear and tension forces is Category B connection (Eq. (3.8a), EN 1993-1-8 (2005)) at service load Fs,Rd,ser =

ks nμ(Fp,C − 0,8Ft,Ed,ser ) γM3,serv

(7.13)

Category C connection (Eq. (3.8b), EN 1993-1-8 (2005)) at ultimate load Fs,Rd =

ks nμ(Fp,C − 0,8Ft,Ed ) γM3

(7.14)

The factor of 0,8 is introduced to allow for the fact that the minimum shank tension may not be achieved. The design tensile, shear and bearing strength of a parallel shank pre-loaded bolt and can be presented in the form of a table to reduce calculations (Annex A3).

7.5 PLATE THICKNESSES FOR JOINT COMPONENTS Plates, or parts of sections acting as flange plates, often form part of structural connections. The length and breadth of plates are generally determined from the geometry of the connection but the thickness is calculated from the elastic or plastic theory of bending. Backing plates are used to strengthen flanges of columns as shown in Fig. 6.3, EN 1993-1-8 (2005). In particular, they are used to strengthen T-stubs and methods are given in Table 6.2, EN 1993-1-8 (2005). The limits on dimensions of the plates are given in cl 6.2.4.3, EN 1993-1-8 (2005).

7.5.1 Plastic Methods for Plates (cl 6.2.4, EN 1993-1-8 (2005)) Parts of connections may be idealized as T-stubs as shown in Fig. 7.14 where the external force (Ft ) is balanced by bolt force (Bt ) and prying force (Q). A prying force is an additional axial tensile force that is induced in a bolt due to the flexing and reaction of components. There are three possible conditions of equilibrium for a T-stub at the ultimate limit state (Fig. 7.15). Case 1: Bolt failure as shown in Fig. 7.15(a) Resolving forces vertically Ft + Bt = 0

(7.15)

•

Tensile force in the bolt (Bt )

Structural Design of Steelwork to EN 1993 and EN 1994

219

Ft Prying force Q at bolt failure

Pre-loaded

Bt = Ft Bt

Q

Bt

Q

Prying force Q at the elastic stage of behaviour Bolt with no preload

m

e

Applied external force (Ft )

Ft (a) Relationship between external force and bolt force for a T-stub

(b)

FIGURE 7.14 Prying forces for T-stubs

Thick plates

Elastic plates

Plastic hinges

Prying force

Bolt failure

Bolt failure

(a) No Prying force

Elastic bolt

(b) Forms of prying force failure

(c)

F t = 2Mpl + e ΣB t (m + e)

F t = ΣB t

Ft = 4Mpl m

A ΣB t /2

ΣBt /2

C Q

ΣBt /2

ΣBt /2

Q

Q Ft

Mpl

Q

B

2

Ft

+Q Mpl

2

+Q

Ft = ΣB t Idealized failure modes

FIGURE 7.15 Failure modes of T-stub flanges

Mpl

Mpl

220

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) In this case the prying force (Q) is zero, the thickness of plate is a maximum and the force in the bolt is a minimum (Table 6.2, EN 1993-1-8 (2005), mode 3). Case 2: Bolt failure with partial yielding of the flange as shown in Fig. 7.15(b) Taking moments of forces about C −Mpl +

Bt Ft + =0 2(m + e) 2e

Rearranging, the design tension resistance Ft =

2Mpl + Bt e m+e

(7.16)

Prying force Q=

Bt /2m − Mpl m+e

(7.17)

This is the case that is most likely to occur in practice because plate thicknesses are limited to those available (Table 6.2, EN 1993-1-8 (2005), mode 2). Case 3: Complete yielding of the flange as shown in Fig. 7.15(c) Taking moments of forces about A −Mpl + (Ft /2 + Q)m − Q(m + e) = 0

(i)

Taking moments about B Mpl − Qe = 0

(ii)

Combining (i) and (ii) to eliminate Q Ft =

4Mpl m

(7.18)

They prying force Q=

Mpl e

(7.19)

This case results in the smallest thickness of plate but the largest bolt force (Table 6.2, EN 1993-1-8 (2005), mode 1). If needed the prying force can be calculated more accurately (Holmes and Martin, 1983) and the theory has been shown to agree with the experimental results (Bahia et al., 1981). Prying forces can be avoided by using non-flexible components or by the use of stiffeners.

7.5.2 Plastic Method for the Thickness of Flange Plates When considering T-stubs the length of the flange is known, but when analysing column flanges in similar situations (Fig. 7.16(a)) an effective length needs to be determined.

Structural Design of Steelwork to EN 1993 and EN 1994

•

221

The method of determining the effective length is illustrated by a simple example for the plastic yield lines shown in Fig. 7.16(b). For a single bolt force then from virtual work 2 tp fyp Ft m 4(m + e) 2x = + m+e 4 x m+e

(i)

Differentiating with respect to x to determine the value of x for which Ft is a minimum ∂Ft 2 (m + e) + = −4 =0 2 ∂x m+e x

√ hence x = (m + e) 2

(ii)

Combining Eqs (i) and (ii) and rearranging, the thickness of the plate Ft mfyp tp = √ 2(m + e) From Eq. (i) the effective length √ 2x 4m + e + (m + e) = 4 2(m + e) leff,b = x m+e

N

Δ x

f

M

m e

(a)

m e

(c) leff,b 4 m 1,25e (inner and end bolts)

p

p

(b) Simple theory

m e

m e

(d) leff,b p (inner bolts) 0,5p 2 m 0,625 e (outer bolts)

(e) leff,b 2 πm (inner and end bolts)

FIGURE 7.16 Effective lengths for column flanges

(7.20)

222

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) The work Eq. (i) can therefore be written as an equilibrium equation 2 tp fyp Ft m = leff,b 4

(7.21)

The above theory assumes simple idealized conditions and ignores, welds, washers, limited rotation at the plastic hinge and strain hardening. However the simple example shows the basic theoretical approach. Other solutions are related to research work (Stark and Bijlaard, 1988). In practice the yield lines are more complicated (Figs 7.16(c), (d) and (e)) and more difficult to analyse. Analysis using yield lines is avoided in the European Code by giving the effective length of plate (leff,b ) (Fig. 7.16) and applying the T-stub equations (Table 6.4, EN 1993-1-8 (2005)).

7.5.3 Elastic Methods for Plates (a) Base plates are used to distribute the load from column sections as shown by the area enclosed by the dotted line in Fig. 7.17. Elastic stress distribution with the maximum stress at the yield strength is used at the ultimate limit state to ensure that large displacements do not occur. If the load is axial the pressure (fj ) beneath the base plate is uniform and the projection (c) of the steel plate beyond the edge of the column, then for a cantilever from the simple theory of elastic bending at first yield per unit width M = fy W f j c2 fy t 2 = 2 6 rearranging, the projection of the plate (Eq. (6.5), EN 1993-1-8 (2005)) c=t

fy (3fj γM0 )

1/2 (7.22)

Justification for this theory is given in Holmes and Martin (1983).

t

fj per unit area c

FIGURE 7.17 Bending strength of a column base plate

Structural Design of Steelwork to EN 1993 and EN 1994

Gusset plate

Hg Lg

223

Gusset plate

(b) Bracket

(a) Column base

Shape of deflected free edge observed in experiments for

•

Approximate buckling stress distribution where Lg/ig 185

Edge stress depends on slenderness ratio

Hg Lg

fgy o δbg

sg

Hg

(e) bg Buckling stress distribution

Bg Theoretical hinge

Fu

Edge stress depends on slenderness ratio

Element strut

o

fgy sg

Fu Lg

Theoretical hinge

(c)

o

sg

Fu

(d) Theoretical model (Martin)

FIGURE 7.18 Gusset plates (b) Gusset plates are used to stiffen base plates and brackets as shown in Fig. 7.18. The following theory for the buckling of a gusset plate is based on experimental work (Martin, 1979; Martin and Robinson, 1981). No advice is given in the European Code. The basic structural unit is a triangular plate with loading applied to one edge as shown in Fig. 7.18(c). For theoretical purposes the plate is assumed to be composed of a series of fixed ended struts parallel to the free edge. The distribution of direct stress across the width bg is shown on an element of the gusset plate in Fig. 7.18(d). The buckling stress varies depending on the slenderness ratio of the elemental strut. At the hinge the stress is for a very short strut (i.e. yield). At the free edge the value is for the slenderness ratio of the elemental strut at the free edge. For simplicity the buckling stress distribution shown in Fig. 7.18(d) can be replaced by a linear distribution as shown in Fig. 7.18(e), provided that the slenderness ratio of the free edge lg /ig < 185. This restraint is acceptable because slenderness ratios of gusset plates in structural engineering do not often exceed this value.

224

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Taking moments of forces about the theoretical hinge at O (Fig. 7.18(e)) and ignoring the moment of resistance of the base plate as justified (Martin and Robinson, 1981). Bg F u sg = (bg fg tg )δbg (7.23) 0

For each strip the buckling stress ( fg ) is linearly related to the slenderness ratio (lg /ig ). The effective length lg = bg when Lg = Hg , and from experiments (Martin, 1979) this is approximately correct when Lg = Hg . The buckling stress for each strip can therefore be expressed as bg provided that lg /ig < 185 (7.24) fg = fgy 1 − 185ig Combining Eqs (7.23) and (7.24), integrating, expressing the radius of gyration as ig = tg /(2 × 31/2 ), and rearranging tg =

2Fu sg Bg + 2 80 fgy Bg

(7.25)

where from the geometry of the plate Bg =

Lg [(Lg /Hg )2 + 1]1/2

(7.26)

The slenderness ratio of the gusset plate may be defined as the slenderness ratio of a strip of unit width parallel to the free edge. From this definition and Eq. (7.26) lg (2 × 31/2 )Bg 2 × 31/2 (Lg /tg ) = = ig tg [(Lg /Hg )2 + 1]1/2

(7.27)

This theory is for non-slender gusset plates, that is, for lg /ig < 185. The theory for slender gusset plates is given elsewhere (Martin, 1979).

7.6 JOINTS SUBJECT TO SHEAR FORCES Two simple connections subject to shear forces are shown in Fig. 7.19. The forces in the members are assumed to be axial and to act through the centroidal axes of the members. This is correct in some situations, for example the bolted joint shown in Fig. 7.19(a). However it is not correct for the welded lap connection shown in Fig. 7.19(b) because the eccentricity of the force produces a moment which results in distortion at ultimate load. It is not correct for a roof truss joint as shown in Fig. 7.24 because although the centroidal axes intersect and there are axial forces in the members there are also secondary moments. A further assumption for simple joints is that the external forces are distributed evenly to the bolts or welds. This is not correct for long bolted and welded joints and allowance must be made for this.

Structural Design of Steelwork to EN 1993 and EN 1994

x

F

F

x

F

Flat F

e

Tie F

x x x x x

b

x

225

Angle

l

F

•

F

F

(a) Bolted connection F

F Distortion at ultimate load (b) Welded connection

FIGURE 7.19 Joints subject to shear forces

The overlap distance (l) is important for simple joints. For bolted joints the minimum of two bolts and the required end distances generally ensure that the lap is sufficient. However for welds (Fig. 7.19) the greater strength may indicate that the lap distance can be small, but it must be appreciated that there must be room for stop and start lengths and that stress concentrations can occur. The minimum lap length is generally not be less than four times the thickness of the thinner part joined where the weld is continuous. The minimum length of weld is 40 mm or six times the throat thickness. For a joint with side welds only the lap is generally not less than the width of the member and there should be end returns of twice the leg length of the weld (cl 4.3.2.1(4), EN 1993-1-8 (2005)) to reduce stress concentrations.

7.7 JOINTS SUBJECT TO ECCENTRIC SHEAR FORCES Joints, such as shown in Fig. 7.20(a), are subject to eccentric shear forces which tend to rotate the joint. This produces a resultant shear force on a fastener (bolt, or unit length of weld) from the direct shear force and the moment. The forces acting on a group of fasteners can be idealized as shown in Fig. 7.20(b). The bolt group rotates about the theoretical instantaneous centre of rotation which varies in position depending on the magnitudes of the external forces V and H and the eccentricity e. In the linear elastic stage of behaviour it is reasonable to assume that the force acting on a fastener is proportional to the distance from the centre of rotation. At ultimate load this assumption is not strictly correct but the error involved is not great. For a rigorous solution to the theory and accuracy at ultimate load see Bahia and Martin (1980).

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

(a)

N

o θ n

z Fmax Fastener r1 Fmax rn n rn V H y Zn

y

r2 F rn max r2 θ2

V

r1 θ1 y

θ

H

rG n θ

M Ve

Vertical component of force from torsional moment zG

•

V

226

G

yn

Centroid of fastener group

H yG

e

z (b) Diagram for general theory

Horizontal component of force from torsional moment

(c) Vector digram for simple vector addition theory for a typical fastener

FIGURE 7.20 Joints subject to eccentric shear forces Although this is the correct approach to the theory there is a simpler more practical method in common use which gives the same values. It is assumed that rotation occurs about the centroid of the fastener group and for convenience the forces acting on a fastener are parallel to the z–z and y–y axes as shown in Fig. 7.20(c) There forces are combined vectorially and the resultant force on a fastener furthest from the centre of rotation is 1/2 MyG 2 MzG 2 V H 2 2 1/2 FR = [Fy + Fz ] = + + + (7.28) n Ix n Ix where n is the number of fasteners in the group yG and zG are coordinates of a fastener related to the centroid of the fastener group IX = IY + IZ IX , IY and IZ are second moments of area of unit size fasteners about the X–X, Y–Y and Z–Z axes The method is used in practice for ordinary bolts, high strength friction grip bolts and welds.

7.8 JOINTS WITH END BEARING Some joints involve end bearing between components (Fig. 7.21). End bearing can occur in beam-to-column (Fig. 6.15, EN 1993-1-8 (2005)), bracket-to-column, beamto-beam and column-to-base joints.

Structural Design of Steelwork to EN 1993 and EN 1994

•

227

End bearing

End bearing (a)

(b)

V

End bearing

Ro

H M O

dp

dr

R

μRo Centre of rotation at stiff bearing O

(c)

(d)

FIGURE 7.21 Joints with end bearing

Where end bearing occurs, rotation takes place about a stiff axis of rotation, axis O–O shown in Fig. 7.21(d). The reaction force Ro is generally large and the bearing may have to be reinforced if it is not to distort under the load. The balancing tensile force R is resisted by bolts or welds. If there is slip at the stiff bearing a frictional force μRo develops parallel to the stiff bearing surface. Consider an end bearing joint subject to external forces V , H and moment M as shown in Fig. 7.21(d). If slip occurs resolving forces vertically V − μRo < 0

(7.29)

Taking moments of forces about force R M + −H(dr − dp ) − Ro dr = 0

(7.30)

where R is the resultant force of the fasteners acting at a distance dr from the axis of rotation O–O. Combining Eqs (7.29) and (7.30) to eliminate Ro , then slip will not occur if μ[M ± H(dr − dp )] >1 Vdr

(7.31)

Most joints with end bearing do not slip and therefore the fasteners are not subject to the external shear force. One exception is a bracket supporting a load with a small eccentricity.

228

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) In the elastic stage of behaviour it is assumed that the forces acting on a fastener are proportional to the distance from the axis of rotation O–O. If, conservatively, this is also assumed to occur at ultimate load then: Taking moment of forces about axis O–O the maximum tensile force resisted by a fastener Ft(max) =

[M ± Hdp ]zmax IO

(7.32)

where IO = z2 is the second moment of area of unit size fasteners about axis O–O Resolving forces vertically the shear force on a fastener 2 " #3 M V − μ ± H(1 − d /d ) p r (V − μRo ) dr Fs = = (7.33) n n This equation is formed assuming that bolts of the same size and design strength resist equal shear forces and also that slip has taken place. If the fasteners are welds then Ft(max) and Fs are combined vectorially. If the fasteners are ordinary bolts the forces are combined using Eq. (7.8) and related to design strengths. If friction grip bolts are used the forces are combined using Eqs (7.13) and (7.14) and related to design strengths. Traditional design methods ignore the existence of the frictional force which errs on the side of safety. However, research (Bahia et al., 1981) has shown that the frictional force does resist part of the shear force. If pre-loaded bolts are used then slip may occur at service load or at ultimate load.

7.9 ‘PINNED’ JOINTS (CL 3.13, EN 1993-1-8 (2005)) Some simple joints (e.g. a tie bar) are connected by real pins as shown in Fig. 7.22(a). Provided that the pins are not corroded, or blocked with debris, they will act as pin joints, that is, they will resist forces but not moments. Tie bars are rarely used now because of the cost of manufacture, risk of seizure from corrosion or debris, and because safety depends on a single pin. Other connections shown in Fig. 7.22 are designated as ‘pins’ because the rotational restraint is small. In the past these joints have been designed assuming that the rotational resistance is zero and the connection resists direct forces only. The general approach to design of these ‘pinned’ connections follows.

7.9.1 Pinned Beam-to-Column Joints (Figs 7.22(b), (c), (e) and (f)) In design calculations for the joint shown in Fig. 7.22 (b) it is assumed that the shear force is resisted by the four bolts connecting the bottom cleat to the column flange.

Structural Design of Steelwork to EN 1993 and EN 1994

•

229

Shear force Angle section

Pin joint Axial force

Tie force (a) Tie rod Angle section (b) Beam-to-column Shear force

Shear force

Secondary beam Tie force

Tie force

Angle section

Angle section

Main beam (d) Beam-to-beam

(c) Beam-to-column

Bolted ‘pin’ connections Shear force

Shear force

Plate welded to beam

Plate welded to column Tie force

Tie force

Weld

(f) Beam-to-column

(e) Beam-to-column

Axial force Shear force

Base plate welded to column Tie force Reinforced concrete foundation

Plate welded to end of beam (g) Beam-to-beam

(h) Column-to-foundation

Welded-bolted ‘pin’ connections

FIGURE 7.22 Examples of ‘pinned’ joints During erection the bottom cleat, which is bolted or welded to the column, is used as a marker by the crane operator when the beam is placed. The top cleat is assumed to resist no vertical load but it does provide torsional resistance which is important for lateral stability. The top and bottom cleats also resist the tie force. The resistance of

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) the web of the beam to shear, bearing and buckling must be checked. At ultimate load the rotation at the end of the beam often introduces end moments. In practice, these are assumed to be small and are ignored. Other types of ‘pinned’ beam-to-column joints are shown in Figs. 7.22(c), (e) and (f) where the depth of the joint is kept to a minimum. For example, the end plate depth for Fig. 7.22(e) is kept to a minimum to reduce end moments and the empirical thickness is 8 mm for UB sizes up to 457 × 191 kg, and 10 mm for sizes greater than 533 × 210 kg. Examples of ‘pin’ joints used in practice are given by Pillinger (1988). Where the depth of the connection for Fig. 7.22(e) is greater and is also slip resistant then advice on the distribution of forces to the bolts is given in Fig. 6.15, EN 1993-1-8 (2005).

7.9.2 ‘Pinned’ Beam-to-Beam Joints (Figs 7.22 (d) and (g)) The transverse secondary beam is connected to the main beam through angle cleats as shown in Fig. 7.22(d). It is assumed in design that the shear force is transferred to the main beam via the bolts in the web of the main beam. These bolts are therefore designed for single shear and bearing on the web of the main beam and on the angle cleats. It follows therefore that the shear force is eccentric to the bolts in the web of the secondary beam. These bolts are double shear and bearing on the web of the secondary beam and the angle cleats. Steel fabricators and erectors often prefer a welded and plate (Fig. 7.22 (g)) as an alternative to the angle cleats shown in Fig. 7.22 (d). This results in a more rigid joint and an end moment is introduced to the end of the secondary beam which is dependant on the torsional stiffness of the main beam. If there are secondary beams on both sides of the main beam the secondary moment can be large.

7.9.3 ‘Pinned’ Column-to-Foundation Joints (Fig. 7.22 (h)) The column is fastened to the base plate which is connected to the foundation by foundation bolts. This type of joint is used where the predominant force in the column is axial but there is generally a small shear force. The size of the foundation bolt is based on resisting the forces, but with a minimum size of M16. The thickness of the base plate is related to pressure beneath the base and the cantilever effect the plate.

7.10 ‘RIGID’ JOINTS ‘Rigid’ (or ‘fixed’) joints (Fig. 7.23) exhibit small rotational displacements in the elastic stage of behaviour. They are now more common as connections to members are welded in the workshops and bolted on site. They are useful to limit deflections of members,

Structural Design of Steelwork to EN 1993 and EN 1994

(a)

231

(c)

(b)

(d)

•

(e)

(f) RHS sections

(h) (g) Strap

d

M

Haunch

Ro Web stiffener

(i)

End plate

(j)

FIGURE 7.23 Examples of ‘rigid’ joints resist fatigue and resist impact loading. However, generally, the design procedure for the frame requires global analysis and the components are more highly stressed.

7.10.1 ‘Rigid’ Column Bracket Joints (Figs 7.23 (a), (b) and (c)) For the brackets shown in Figs 7.23 (a) and (c) no end bearing is involved whereas the bracket shown in Fig. 7.23 (b) involves end bearing.

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

7.10.2 ‘Rigid’ Beam-to-Column Joints (Figs 7.23 (d) and (e)) The most rigid type of joint is where the beam is welded directly to the column on site, as shown in Fig. 7.23(d), but this is expensive and it is difficult to control the quality of the weld. Alternatively stub cantilever beams can be welded to the column in the workshops and a suspended beam site bolted between the ends of the cantilever. The bolted connection is positioned as close to the point of contraflexure as possible. The method commonly used in Britain is to weld end plates to the beam in the workshops and to bolt these to the columns on site as shown in Fig. 7.23(e). The numbers of bolts is usually six, as shown, because of the limited depth available. If the moment of resistance needs to be increased then it is necessary to increase the lever arm by haunching the beam at the end. This connection is less rigid than welding the beam directly to the column but it is easier to manufacture and erect. The amount of rotation depends on the thickness of the end plate, thickness of the column flanges and the extensibility of the bolts. Advice on details is given in cl 6.2.7.2, EN 1993-1-8 (2005). If the connection is close to a plastic hinge then it must be decided whether the plastic hinge should forming the beam, or the column, or the connection. Recent research favours the formation of the hinge in the beam and therefore the column and the connection must be overdesigned. End bearing occurs between beam and column and the first step in design is to check whether slip occurs using Eq. (7.31). Generally, because the bending moment is large, slip does not occur and the size of the top four bolts required can be determined approximately by taking moments of forces about the compression flange of the beam. The tensile force in a bolt Ft =

M 4(h − tf )

(7.34)

The tensile force in a bolt is increased by a prying force. Equation (7.34) assumes that the two bolts close to the compression flange of the beam do not resist any part of the bending moment. Lever arm recommendations are given in Fig. 6.15 (EN 19931-8 (2005)), The thicknesses of the end plate is determined by assuming equivalent T-stubs and associated yield lines. This method can be compared with a survey of existing literature on end plates by Mann and Morris (1979) who recommended ⎡ ⎢ ⎣

⎤1/2 M

dbf fy

4w sv

p

⎥ ⎦ dbf

+ s h

Mbp < tp < 2wp dbf fy

1/2

provided that Bp = 9db , sh = 6db , sv = 6db and eb > 2,5db .

(7.35)

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233

The thickness of the column flange can also be determined from equivalent T-stubs with yield line patterns. For comparison a further survey by Mann and Morris (1979) recommended: For unstiffened column flanges the thickness of the flange M 1/2 M 1/2 0,28 < tcf < 0,39 dbf fy dbf fy For stiffened column flanges the thickness of the flange M 1/2 M 1/2 0,23 < tcf < 0,32 dbf fy dbf fy

(7.36)

(7.37)

Equations (7.36) and (7.37) are valid provided that bc = 2,5db and cc = tst + 5db . Where the column flange thickness is inadequate backing plates can be used. The force Ro at the axis of rotation may produce failure by bearing or buckling of the web of the column. If failure is likely to occur then stiffeners can be welded into the web of the column. However, stiffeners increase costs and reduce the room available for bolts. The tensile force balancing Ro (Fig. 7.21) which acts on the web of the column is often not critical but the strength of the column web must be checked. The size of the fillet weld connecting the end plate to the beam is determined by assuming rotation about the axis O–O at the bottom flange of the beam, but an alternative more conservative method is to assume rotation about the centroidal axis of the weld.

7.10.3 ‘Rigid’ Beam Splices (Fig. 7.23(g)) Beam splices are introduced to extend standard bar lengths, or to facilitate construction and transport. Splices are generally located at sections where the forces are minimum, and to avoid local geometrical deformations of the structure pre-loaded bolts are used. The connection is usually made on site and therefore bolts are used. A simple design method is to assume that the entire shear force is resisted by the web splice and the flanges resist the entire bending moment. These assumptions are not correct but it simplifies the design and the errors are generally not large. Alternatively part of the moment is assumed to be resisted by the web. The bolts in the web, in double shear, are subject to an eccentric shear load. The bolts in the flange are also in double shear for two flange plates and the force on a bolt is obtained a simple moment equation F = M/(nd).

7.10.4 ‘Rigid’ Column-to-Column Joints (Fig. 7.23(f)) Column splices are used to extend standard bar lengths, to facilitate erection and transport, and for economy by reducing the section size. Where bending moments are large then, as with beam splices, pre-loaded bolts are used to maintain the axial line of

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) the column. Where sections change size steel packings and an end plate are required to ensure that a good fit is obtained. Generally the ends of the columns are in contact, or in contact with the end plate, and therefore end bearing occurs. Any shear force will be resisted by the friction at the end bearing and the web plates (or cleats). Where forces are small sizes of components are decided from experience, practicality and corrosion resistance.

7.10.5 ‘Rigid’ Hollow Section Joints (Fig. 7.23(h)) This is a typical ‘T’ joint for a Vierendeel girder where members of different width intersect. If forces and moments are not too large then the connection can be made without using stiffeners. Rectangular hollow sections are also used in braced triangulated trusses but it is difficult to arrange for the centre lines of the members to intersect at a point. The offsets introduce moments which should be taken into account in the analysis of the structure (Purkiss and Croxton, 1981). Further information on analysis and design of connections is given by Davies (1981). Circular hollow sections are also used for trusses but the geometry of the connection is more difficult and connections are more expensive to form. A review of methods of analysis of the strength of these connections is given by Stamenkovic and Sparrow (1981). The failure modes for hollow section joints are (cl 7.2.2, EN 1993-1-8 (2005)): (a) (b) (c) (d) (e) (f)

chord face failure, chord side wall failure, chord shear failure, punching shear failure, brace failure, local buckling.

The design resistance of numerous types of joints are shown in Tables 7.2–7.24, EN 1993-1-8 (2005). These take into account in plane and out of plane forces. Guidance on strength of welds for connecting members is given in cl 7.5.1, EN 1993-1-8 (2005). These are only applicable if they meet the joint parameters of Table 7.8, EN 1993-1-8 (2005).

7.10.6 ‘Rigid’ Column-to-Foundation Joints (Fig. 7.23(i)) Where bending moments are small a simple slab base is satisfactory with the bolts in line with the column axis. As the bending moment increases the bolts are off-set from the column axis. A built-up base is used where moments are large. The basic design method is based on ‘T’ stub theory (cl 6.2.8, EN 1993-1-8 (2005)).

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For a gussetted slab base subject to an axial load and bending moment it is assumed that at ultimate load the distribution of stresses beneath the steel base plate are as shown in Fig. 7.34(c). The depth of the compression zone is determined approximately by taking moments of forces about the tensile bolts " # M N + 2 d x= (7.38) bc fij where bc is based on the cantilever length (c) and fij is the allowable bearing pressure. Alternatively it can be assumed that the centroid of the compression zone is located under the column flange (Table 6.7, EN 1993-1-8 (2005)). From taking moments of forces about the compressive force the tensile force in a bolt is # " M N − 2 d Ft = (7.39) n The thickness of the base plate is found based on elastic bending of the cantilever length (c). For the built-up base the base plate thickness is determined by the same method. The gusset plate size is determined by the method shown in Section 7.5.3. The size of the welds connecting the gusset plate to the base and to the column can be determined assuming end bearing. Alternatively end bearing can be ignored and a larger size determined by assuming rotation about the centroid of the weld group.

7.10.7 ‘Rigid’ Knee Joint for a Portal Frame (Fig. 7.23(j)) This type of joint is similar to the beam-to-column connection shown in Fig. 7.23(e). However because portal frames are often designed using plastic analysis the magnitude of the reaction Ro = M/d is generally high and consequently the shear stresses in the web are close to the limit. The magnitude of Ro is reduced by haunching the beam and consequently increasing the distance d, but web stiffeners are generally required for the column. To reduce the shear deformation in the column web stiffeners can be introduced as described and tested by Morris and Newsome (1981). The tensile force which balances Ro can be resisted by a strap, or by a group of bolts through the flange of the column. The strap may interfere with the placing of purlins in the roof and the alternative group of bolts may cause distortion of the column flanges if the column flange thickness is small. The distortion can be controlled by the use of flange stiffeners, or increase the size of the column section, but both increase the cost.

EXAMPLE 7.2 Design of a ‘pin’ joint for a roof truss (Fig. 7.24). The forces, size of angles and tees have been obtained from an analysis at ultimate load assuming pin joints and axial forces in the members (cl 5.1.5(2), EN 1993-1-8 (2005)).

30 k N

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

310

310

bg

125 75 8 angles

60 60 6 angles

30

Gusset plate tg 10

kN

N

0k

10 28

16

330

15

24

t 6,1

215 kN

6 mm fillet welds

Strap

60

80

40

10,2

165 152 20 kg tee

10

165 152 20 kg tee 151,9

•

347

236

31 125 kN

Grade S275 steel M20 class 4.6 bolts

FIGURE 7.24 ‘Pinned’ roof truss joint The centroidal axes of the members intersect so there is no eccentricity to be taken into account (cls 2.7 and 3.10.3, EN 1993-1-8 (2005)). The thickness of the gusset plate is at least 6 mm to resist corrosion, and at least equal to the minimum thickness of the angle or tee (6,1 mm). Use a 10 mm thick plate grade S275 steel. A rectangular plate is simple to mark and cut, and low in fabrication cost. Alternatively a more complicated shape can be used which is aesthetically more acceptable but the fabrication cost is greater. Member 24, structural tee cut from a UB (165 × 152 × 20 kg) welded to a gusset plate, design force NEd = 215 kN. Tensile resistance of the tee Grade S275 steel (Eqs (6.6) and (6.7), EN 1993-1-1 (2005)) Npl,Rd =

Afy 2580 × 275 = γM0 1,00 × 1E3

= 709,5 > 215 kN satisfactory. Nu,Rd = 0,9Anet

fu 430 = 0,9 × (2580 − 2 × 22) × γM2 (1,25 × 1E3)

= 766,9 > 215 kN satisfactory. Assuming a 6 mm fillet weld for Grade S275 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd = =

fu a 1/2 (3 βw γM2 ) 6 430 × 0,7 × 1E3

(31/2 × 0,85 × 1,25)

= 0,981 kN/mm

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237

Effective length of weld required to resist the tensile force using the simplified method (cl 4.5.3.3, EN 1993-1-8 (2005)) NEd 215 = 219,2 mm = Fw,Rd 0,981 Two side fillet welds of 110 mm length would be satisfactory but in practice the lengths would probably be the full overlap (i.e. 2 × 310 = 620 mm). Member 31, structural tee cut from a UB (165 × 152 × 20 kg) bolted to the gusset plate and strap, design force NEd = 125 kN. Resistance of 4-M20 class 4.6 bolts in single shear (Table 3.4, EN 1993-1-8 (2005)) fub Fv,Rd = nb 0,6As γM2 400 = 4 × 0,6 × 245 × = 188 > (NEd = 125) kN satisfactory. 1,25 × 1E3 Resistance of 2-M20 class 4.6 bolts in bearing on the web of the tee section (t = 6,1 mm) (Table 3.4, EN 1993-1-8 (2005)) k1 αb fub dt Fb,Rd = nb γM2 2,5 × 0,909 × 400 × 20 × 6,1 =2× 1,25 × 1E3 = 177,4 > (NEd = 125) kN satisfactory where for an end bolt αb = e1 /(3do ) = 60/(3 × 22) = 0,909 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 55/22 − 1,7 = 5,3 > 2, 5 The strap increases the out of plane stiffness of the truss. Connections for other members can be designed by the same method.

EXAMPLE 7.3

‘Rigid’ column bracket. Determine the size of the components required to connect the bracket to the column shown in Fig. 7.25 using Grade S355 steel. The forces shown are applied to one gusset plate at ultimate load. For the 10 bolts (Fig. 7.25(a)) of unit cross-sectional area the properties of the bolt group are: Second moment of area of the bolt group about the centroidal y–y axis Iy = (∂A)z2 = 4(802 + 1602 ) = 128E3 mm4

Second moment of area of the bolt group about the centroidal z–z axis Iz = (∂A)y2 = 10(70)2 = 49E3 mm4

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

eh 250

H 45 kN

y G

Gusset plate tg 10

Bg 196,1

z H g dw 400

e v 350

4 at 80 320

Bg 231

Gusset plate y

G y z

140 bf 255,9

z

ev 350

sg 150

sg

y

V 210 kN H 45 kN

z 40

tf 17,3

V 210 kN

hg 150

eh 250

254 254 89 kg UC

40

•

Hg 400

238

bw 200

tg 10 Lg 225

Lg 282,95 2 229 76 channels

Grade S 355 steel

(a) Bracket bolted to a UC

h 228,6 (b) Bracket welded to a compound column

FIGURE 7.25 ‘Rigid’ column brackets

Second moment of area of the bolt group about the centroidal polar x–x axis Ix = Iy + Iz = (128 + 49)E3 = 177E3 mm4 From Eq. (7.28) the maximum vector force in the direction of the z–z axis on a bolt furthest from the centroid of the bolt group V (Veh + Hev )yn + nb Ix 210 (210 × 250 + 45 × 350)70 = + = 48 kN 10 177E3

Fz =

The maximum vector force in the direction of the y–y axis on the same bolt H (Veh + Hev )zn + nb Ix 45 (210 × 250 + 45 × 350)160 = + = 66,2 kN 10 177E3

Fy =

Resultant vector design force on this bolt Fr = (Fz2 + Fy2 )1/2 = (482 + 66, 22 )1/2 = 81, 77 kN Solution (a) using class 4.6 bolts. Shear resistance of a M30 class 4.6 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub γM2

= 0,6 × 561 ×

400 = 107,7 > (Fr,Ed = 81, 77) kN satisfactory. 1,25 × 1E3

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239

However the recommended maximum bolt diameter for a column flange width of 254 mm is 24 mm (Annex A4). Use a higher class of bolt. Solution (b) using M20 class 8.8 bolts not pre-loaded. Shear resistance of an M20 class 8.8 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0, 6As

fub γM2

= 0,6 × 245 ×

800 = 94,1 > (Fr,Ed = 81,77) kN satisfactory. 1,25 × 1E3

M20 class 8.8 bolt in bearing on the gusset plate (t = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Bearing strength k1 αb fup dt γM2

Fb,Rd =

2,5 × 0,606 × 510 × 20 × 10 = 123,6 > (Fr,Ed = 81, 77) kN 1,25 × 1E3

= where

for an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 58/22 − 1,7 = 5,68 > 2,5 Solution (c) using pre-loaded M22 class 10.9 bolts (cl 3.9, Eqs (3.6) and (3.7), EN 1993-1-8 (2005)) Fs,Rd = =

ks nμb Fp,C γM3 303 1,0 × 1,0 × 0,5 × 0,7 × 1E3 × 1E3

1,25

= 84,8 > (Fr,Ed = 81,77) kN

To determine the thickness of the gusset plate for the bolted joint Fig. 7.25(a) (255,9 − 140) = 282,95 mm 2 (255,9 − 140) sg = 150 + = 207,95 mm 2

Lg = 225 +

Width of the gusset plate perpendicular to the free edge (Eq. (7.26)) Bg =

Lg 282,95 = 1/2 = 231,0 mm 1/2 Lg 2 282,95 2 +1 +1 400 H g

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) From Eq. (7.25), replacing the term (Pu sg ) with (Vsg + Hhg ), the thickness of the gusset plate Grade S355 steel tg =

2(VSg + Hhg ) Bg 2 + fgy Bg 80 γM1

=

2 × (210 × 207,95 + 45 × 150)E3 231 + 355×2312 80 1,0

= 8, 21 mm; use a 10 mm thick plate of Grade S355 steel Check the slenderness ratio of the gusset plate (Eq. (7.27)). lg Bg 231 = 2 × 31/2 = 2 × 31/2 × ig tg 10 = 80,02 < 185 the limit of the slenderness ratio for the application of the theory, satisfactory. Solution (d) using welds (Fig. 7.25(b)). Where it is not possible to bolt to a column, for example the compound channel column shown in Fig. 7.25(b), then welds are used. The connection is rigid and for welds of unit size the properties of the weld group are: Total length of weld Lw = 2(dw + bw ) = 2(400 + 200) = 1200 mm Second moment of area of the weld group about the centroidal y–y axis 2 dw d 3 w Iy = (∂A) z2 = 2 + bw 12 2 4003 400 2 =2 = 26, 67E6 mm4 + 200 12 2

Second moment of area of the weld group about the centroidal z–z axis 2 3 bw b Iz = (∂A)y2 = 2 w + dw 12 2 2003 200 2 =2 = 9, 33E6 mm4 + 400 12 2

Second moment of area of the weld group about the centroidal polar x–x axis Ix = Iy + Iz = (26, 67 + 9, 33)E6 = 36E6 mm4

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241

Maximum vector force in the direction of the z–z axis on a weld element furthest from the centroid of the weld group from Eq. (7.28). V (Veh + Hev )yn + Lw Ix 210 (210 × 250 + 45 × 350)100 = + = 0,365 kN/mm 1200 36E6

Fz =

Maximum vector force in the direction of the y–y axis on the same weld element H (Veh + Hev )zn + Lw Ix 45 (210 × 250 + 45 × 350)200 = + = 0,417 kN/mm 1200 36E6

Fy =

Resultant vector design force on this weld element Fr,Ed = (Fz2 + Fy2 )1/2 = (0, 3652 + 0, 4172 )1/2 = 0, 554 kN/mm Assuming a 6 mm fillet weld for Grade S275 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a 1

(3 2 βw γM2 )

=

6 430 × 0,7 × 1E3

(31/2 × 0,85 × 1,25)

= 0,981 < (Fr,Ed = 0, 554) kN/mm satisfactory. To determine the thickness of the gusset plate for the welded joint Fig. 7.25(b). From Eq. (7.26) the width of the gusset plate perpendicular to the free edge Bg =

Lg 225 = 1/2 = 196, 1 mm 1/2 Lg 2 225 2 +1 +1 H 400 g

From Eq. (7.25) replacing the term Pu sg by (Vsg + Hhg ) tg =

2(Vsg + Hhg ) Bg 2 + fgy Bg 80 γM1

=

2 × (210 × 150 + 45 × 150)E3 196, 1 + 355 80 2 × 196,1 1,0

= 8, 05 mm, use an 10 mm thick plate of Grade S355 steel. Check the slenderness ratio of the gusset plate from Eq. (7.27) lg 2 × 31/2 Bg 2 × 31/2 × 196, 1 = = 10 ig tg = 67, 93 < 185 the limit of the slenderness ratio for the application of this theory.

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) 203 203 86 UC

End shear force on beam V 225 kN Top angle cleat 80 80 10 mm angle supports no vertical load

Clearance 5 mm b 192,8 tw 11,4 r 10,2 tf 20,5 tw 13

h 467,4

242

tf 19,6

457 191 98 kg UB Tie force 75 kN Grade S 355 steel M20 class 8,8 bolts

Bottom cleat 125 75 10 mm angle support all vertical load

bf 208,8

FIGURE 7.26 ‘Pinned’ beam-to-column joint

EXAMPLE 7.4 ‘Pinned’ beam-to-column connection. Check the size of components for the connection shown in Fig. 7.26 at ultimate load. (a) If the design shear force of Fv,Ed = 225 kN is resisted by 4-M20 grade 8.8 bolts in the bottom cleat (125 × 75 × 10 mm angle), then the design shear force per bolt Fv,Ed =

225 = 56,25 kN 4

Shear resistance of an M20 grade 8.8 bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub γM2

= 0,6 × 245 ×

800 = 94,1 > (Fv,Ed = 56,25) kN satisfactory. 1,25 × 1E3

Bearing resistance of an M20 class 8.8 bolt bearing on the leg of the angle (ta = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd =

k1 αb fua dta 2,12 × 0,682 × 510 × 20 × 10 = γM2 1,25 × 1E3

= 118 > (Fv,Ed = 56,25) kN satisfactory. where for an end bolt αb = e1 /(3do ) = 45/(3 × 22) = 0,682 and for an edge bolt k1 = 2,8 (e2 /do ) − 1,7 = 2,8 × (30/22) − 1,7 = 2,12 < 2,5 Assume the 4-M20 class 8.8 bolts connecting the bottom cleat to the column which resist the shear force of 225 kN also resist the tensile force of 75 kN. Design tensile force per bolt from the 75 kN tie force is Ft,Ed =

75 = 18,75 kN 4

Tensile resistance of an M-20 class 8.8 bolt Ft,Rd = 0,9As

fu 800 = 0,9 × 245 × = 141,1 kN γM2 1,25 × 1E3

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243

Combined shear and tension for a bolt (Table 3.4, EN 1993-1-8 (2005)) Fv,Ed Ft,Ed + Fv,Rd (1,4Ft,Rd ) =

56,25 18,75 + = 0,693 < 1 satisfactory. 94,1 (1,4 × 141,1)

(b) Alternatively the four bolts in the vertical leg of the bottom angle could be replaced by fillet welds along the two vertical edges of the angle. Resistance of two 6 mm fillet welds for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) 2lw Fw,Rd = =

2lw fu a (31/2 βw γM2 ) 6 2 × 125 × 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 275 > (VEd = 225) kN.

The top cleat is used to provide torsional resistance against lateral buckling of the beam and to resist tie and erection forces. The angle must be at least 6 mm thick to resist corrosion and the leg of sufficient length to accommodate M20 bolts. From Section Tables a 80 × 80 × 10 mm angle is chosen. For resistance to transverse shear forces for the beam see Example 4.12

EXAMPLE 7.5 ‘Pinned’ beam-to-beam connection. Determine the size of the components required for the connection shown in Fig. 7.27. The beam sizes have been determined from bending calculations at ultimate load. Assuming that the M20 class 4.6 bolts through the web of the main beam B (Grade S275 steel) are subject to single shear forces. Shear resistance of a M20 class 4.6 bolt in single shear (Table 3.4, EN 1993-1-8 (2005)) Fv,Rd = 0,6As

fub 400 = 0,6 × 245 × = 47,0 kN γM2 1,25 × 1E3

M20 class 4.6 bolt in bearing on the web of the transverse beam A (tw = 6,9 mm) (Table 3.4 EN 1993-1-8 (2005)). For an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 For an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 30/22 − 1,7 = 2,12 < 2,5 k1 αb fub dtw γM2 2,12 × 0,606 × 400 × 20 × 6,9 = = 56,7 > (Fv,Rd = 47,0) kN. 1,25 × 1E3 VEd 150 nb = = = 3,19 use 4-M20 class 4.6 bolts. Fv,Rd 47,0

Fb,Rd =

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Transverse beam A 356 171 45 kg UB

z

y

a

Grade S275 steel M20 class 4,6 bolts

tw 6,9

40

y

h 352,0

b

3 70 210

160

65

40

bf 171,0

End shear force V 150 kN 38

158

tw 6,9

40 8 mm Clearance z tw 11,9

Side elevation

tf 9,7

tf 19,7

bf 304,8

65

•

h 609,6

244

70 70 10 angles

Main beam B 610 305 149 kg UB

End elevation

FIGURE 7.27 ‘Pinned’ beam-to-beam joint

Assuming that the bolts connecting the angle cleats to the web of the transverse beam A are in double shear and subject to an eccentric load. Second moments of area of the bolt group about the centroidal axis for bolts of unit area are Iy = (∂A)z2 = 2(352 + 1052 ) = 24,5E3 mm4 Iz = 0 Ix = Iy + Iz = 24,5E3 mm4 Maximum shear force on a bolt in the y direction from Eq. (7.28) Fy =

VEd ezmax 150 × 40 × 105 = = 25,71 kN Ix 24,5E3

Average shear force on a bolt in the z direction Fz =

VEd 150 = 37,5 kN = nb 4

Maximum resultant design shear force on a bolt Fr,Ed = (Fy2 + Fz2 )1/2 = (25,712 + 37,52 )1/2 = 45,47 kN Double shear strength of an M20 class 4.6 bolt 2Fv,Rd = 2 × 47,0 = 94,0 > (Fr,Ed = 45,47) kN, satisfactory.

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Check ‘block shear tearing’ (cl 3.10.2, EN 1993-1-8 (2005)) for line of holes in end of transverse beam A 1 Ant Anv + 1/2 fy γM2 γMO 3 1 1194 = 0 + 1/2 × 275 × 1,0 × 1E3 3 = 189,6 > (VEd = 150) kN satisfactory.

Veff,2,Rd = 0,5fu

which includes Anv = (3 × hole spacing + end distance − 3,5 × hole diameter)tw = (3 × 70 + 40 − 3,5 × 22) × 6,9 = 1194 mm2

EXAMPLE 7.6 ‘Pinned’ column-to-foundation connection. Determine the size of the components for the axially loaded base shown in Fig. 7.28 at the ultimate limit state. Concrete cylinder crushing strength fck = 20 MPa. N = 1500 kN

120

254 254 107 UC

6 mm fillet weld

M 20 class 4,6 bolts

Lp 450 C

C

C

t w 13 80

bf 258,3 C

Bp 42,5

h 266,7

tf 20,5

Grade S355 steel

FIGURE 7.28 ‘Pinned’ column to foundation joint

246

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Assuming that the bearing area is bounded by the dotted line shown previously in Fig. 7.17 and the pressure beneath base plate (cl 6.2.5(7), EN 1993-1-8 (2005)), N 2 = fj = fck A 3 Rearranging and inserting known numerical values, bearing area required A=

NEd 1500E3 NEd = = 112,5E3 mm2 = 2 2×20 fj f 3 ck 3

(i)

To determine the minimum thickness of the steel base plate the bearing area enclosed by the dotted line (Fig. 7.17) A = (bf + 2c)(h + 2c) − (h − 2tf − 2c)(bf − tw ) = (258,3 + 2c)(266,7 + 2c) − (266,7 − 2 × 20,5 − 2c)(258,3 − 13) = 13524,4 + 1540,6c + 4c2

(ii)

Equating Eqs (i) and (ii) the projection of the dotted area c = (192,572 + 24744)1/2 − 192,57 = 56,1 mm Check if areas overlap along the bolt line between flanges (h − 2tf ) (266,7 − 2 × 20,5) = 2 2 = 112,85 > (c = 56,1) mm satisfactory. Thickness of base plate Grade S355 steel (cl 6.2.5(4), EN 1993-1-8(2005))

3fj γM0 tp = c fy

1/2 = 56,1

3×

2 3

× 20 × 1 355

1/2

= 18,8 mm, use 20 mm thick base plate. Minimum length of base plate Dp = h + 2c = 266,7 + 2 × 56,1 = 322,8 mm Minimum breadth of base plate Bp = bf + 2c = 258,3 + 2 × 56,1 = 370,5 mm Use 450 × 425 × 20 mm base plate Grade S355 steel. If the end of the column is machined then the load is assumed to be transferred directly to the base plate and a minimum size of fillet weld of 6 mm is used to connect the base plate to the column.

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Alternatively if the end of the column is not machined then the force per unit length of weld is approximately Fw,Ed =

NEd 1500 = = 0,957 kN/mm (4bf +2h) (4 × 258,3 + 2 × 266,7)

Assuming a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a 1/2 (3 βw γM2 )

=

510 × 0,7 × 6/1E3 31/2 × 0,9 × 1,25)

= 1,1 > (Fw,Ed = 0,957) kN/mm satisfactor. The base plate is subject to a compressive force which is not transferred to the holding down bolts. The bolts are therefore subject only to erection forces and if these are not known then experience has shown that a bolt size approximately equal to the plate thickness is suitable. Use 2M20 class 4.6 holding down bolts. If there is a bending moment applied to the column and hence to the base plate then the bearing area is located beneath the column flange as shown in Fig. 6.4, EN 1993-1-8 (2005).

EXAMPLE 7.7

‘Rigid’ column bracket. Determine the size of fillet welds for the bracket shown in Fig. 7.29 at the ultimate limit-state.

N 500 kN z

B tw 19,2 r 15,2

g

52 6

tw 16,1 r 17,8

838 292 226 kg UB Ro

bf 293,8

dt 824,1

tf 31,4

100 tf 26,8

h 850,9

V 405 kN

y

Hg 797,3

305 305 198 kg UC e 600 bf 314,1

G

G

o

Lg 700

o

Grade S 355 steel d 246,7 h 339,9 Side elevation

FIGURE 7.29 ‘Rigid’ column bracket

End elevation

248

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) There are two possible solutions based on failure mechanisms (a) assuming rotation about axis G–G which is the simple traditional conservative method and (b) assuming rotation about axis O–O which is more correct but the calculations are more extensive. (a) Rotation about axis G–G The fillet weld is continuous round the bracket section as shown in Fig. 7.29. If there are no stiffeners in the web of the column then the strength of the weld around the flanges of the bracket is reduced because of the flexibility of the column flange. Effective length of the column flange weld (Eq. (4.6a), EN 1993-1-8(2005)) beff = tw + 2r + 7ktf = 19,2 + 2 × 15,2 + 7 × 1 × 31,4 = 269,4 mm which includes fy,f tf 31,4 355 k= = = 1,17 > 1 use 1. tp fy,p 26,8 355 Check if stiffeners required for column web (Eq. (4.7), EN 1993-1-8(2005)) fyp 355 bp = × 293,8 beff = fup 510 = 204,5 < (beff = 269,4) mm therefore no stiffeners required. Rotation about axis G–G The effective second moment of area of the weld group about axis G–G IG =

2dw3 + 4beff (df /2)2 12

2(850,9 − 2 × 26,8)3 (850,9 − 26,8) 2 = + 4 × 269,4 × 12 2 = (84,5 + 183,0)E6 = 267,5E6 mm4 Maximum force per unit length on weld in the y direction (Eq. (7.28)) d (Ve) 2f Fy = IG 405 × 600 × (850,9 − 26,8) = = 0,374 kN/mm (2 × 267,5E6) Maximum force per unit length of weld in the z direction Fz = V/Lw = =

V [4beff + 2(h − 2tf )] 405 = 0,152 kN/mm 4 × 269,4 + 2 × (850,9 − 2 × 26,8)

Maximum resultant design force per unit length of weld Fr,Ed = (Fy2 + Fz2 )1/2 = (0,3742 + 0,1522 )1/2 = 0,404 kN/mm

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For a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a (3

1/2

βw γM2 )

=

510 × 0,7 × 6/1E3 (31/2 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0,404) kN/mm satisfactory. Rotation about axis O–O The second moment of area of the weld group about axis O–O 2 Io = d3 + 2beff df2 3 w 2 = (850,9 − 2 × 26,8)3 + 2 × 269,4 × (850,9 − 26,8)2 3 = (337,9 + 365,9)E6 = 703,8E6 mm4 Maximum force per unit length of weld in the y direction Fy = (Ve)df /Io = (405 × 600) × (850,9 − 26,8)/703,8E6 = 0,285 kN/mm Effective length of weld resisting shear Leff = 4beff + 2dw = 4 × 269,4 + 2(850,9 − 2 × 26,8) = 2672 mm Distance (dr ) from the axis O–O to the resultant force in the weld is determined from equating the moments of the forces in the weld group about the axis O–O moment of the parts = moments of the whole (2beff + 0,5 × 2df )Fx dr = Fx Io /df Rearranging and putting Io = (2/3)df3 + 2beff df2 2b 2 + deff dr 3 f = 2beff df 1+ df

"

= "

2 3

2×269,4

+ 850,9−26,8 2×269,4

1 + 850,9−26,8

#

# = 0,798

Check whether slip occurs by substituting in Eq. (7.31) μs M 0,45 × 600 μs e = = (Vdr ) dr [0,798(850,9 − 26,8)] = 0,411 < 1, therefore slip occurs. Maximum force per unit length of weld in the z direction V μs R V μs (Ve/dr ) − = − Leff Leff Leff Leff # " 405×600 0,45 405 0,798(850,9−26,8) = − = 0,089 kN/mm 2672 2672

Fz =

250

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Maximum resultant design force per unit length of weld Fr,Ed = (Fy2 + Fz2 )1/2 = (0,2852 + 0,0892 )1/2 = 0,299 kN/mm For a 6 mm fillet weld and Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005)) Fw,Rd =

fu a 510 × 0,7 × 6/1E3 = 1/2 (31/2 βw γM2 ) (3 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0, 299) kN/mm satisfactory. An alternative method related to the European Code: Assume the applied vertical shear force is resisted by the two 6 mm web welds VR,Ed = 2df Fw,Rd = 2 × (850,9 − 2 × 26,8) × 1,1 = 1450 > (VEd = 405) kN satisfactory, and the applied bending moment is resisted by two 6 mm effective flange welds with rotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8(2005)) MR,Ed = 2beff df Fw,Rd = 2 × 269,4 × (850,9 − 26,8) × 1,1/1E3 = 488,4 > (MEd = 243) kNm satisfactory. Gusset plate design Check the thickness of the web of the 838 × 292 × 226 kg UB acting as a gusset plate. From Eq. (7.26) Bg =

Lg 700 = = 526,0 mm [(Lg /Hg )2 + 1]1/2 [(700/797,3)2 + 1]1/2

Required thickness of the web of the UB acting as a gusset plate (Eq. (7.25)) Bg 2Pu sg tg = 2 + fgy Bg 80 γM1

=

2 × 405E3 × 600 526 + 80 (355/1,0 × 526)2

= 11,52 < 16,1 mm (thickness of web of UB), satisfaactory. Check the slenderness ratio of the web of UB acting as a gusset plate (Eq. (7.27)) lg √ Bg √ 526 =2 3 =2 3× ig tg 16,1 = 113,2 < 185 limit of application of the theory, acceptable. Column web in transverse compression Reaction Ro may buckle or crush the web of the 305 × 305 × 198 kg UC.

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Reaction (Eq. (7.30)) Ro,Ed =

Ve 405 × 600 = = 369,5 kN dr [0,798(850,9 − 26,8)]

The design resistance of the unstiffened column web (Eq. (6.9), EN 1993-1-8(2005)) Fc,wc,Rd =

ωkwc beff,c,wc twe fywe γM0

0,764 × 1 × 271,7 × 19,2 × 355 1,0 × 1E3 = 1415 > (Ro,Ed = 369, 5) kN satisfactory. =

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,764 × 1 × 1 × 271,7 × 19,2 × 355 1,0 × 1E3 = 1415 > (Ro,Ed = 369,5) kN satisfactory. =

From (Eq. (6.10), EN 1993-1-8(2005)) for a welded connection beff,c,wc = tf b + 2 × 21/2 ab + 5(tfc + s) = 26,8 + 2 × 21/2 × 0,7 × 6 + 5 × (31,4 + 15,2) = 271,7 mm The maximum longitudinal stress in the flange of the column from the axial and eccentric loads (Eq. (6.14), EN 1993-1-8(2005)) (N + V ) V (e + h/2) + A Wel (500 + 405)E3 405E3(600 + 339, 9/2) = + 252E2 2993E3 = 140,1 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

σcom,Ed =

and (Table (6.3), EN 1993-1-8(2005)) β = 1(Table 5.4, EN 1993-1-8(2005)) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

2 1/2

Avc

1

=

1 + 1,3 × 271,7 ×

1/2 19,2 2 7032

= 0,764

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) where the shear area of the column (cl 6.2.6(3), EN 1993-1-1(2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 252E2 − 2 × 314,1 × 31,4 + (19,2 + 2 × 15,2) × 31,4 = 7032 mm2 ηhw tw = 1,0 × (339,9 − 2 × 31,4) × 19,2 = 5320 < (Avc = 7032) mm2 Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8(2005))

beff,c,wc dwc fy,wc 1/2 λp = 0,932 2 ) (Etwc 271,7 × 246,7 × 355 1/2 = 0,932 × (210E3 × 19,22 ) = 0,516 < 0,72 use ρ = 1,0 Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8(2005)) Vwp,Rd = 0,9

Avc fy,wc 7032 × 355 = 0,9 × 1/2 1/2 (3 γM0 ) 3 × 1 × 1E3

= 1297 > (Ro,Ed = 369,5) kN satisfactory.

EXAMPLE 7.8 ‘Rigid’ beam-to-column connection. Determine the size of the components for the connection shown in Fig. 7.30 at the ultimate limit state.

bf 208,5

20 45 35

Qbe Fbt

O

d 160,8

M 97,5 kNm

Ro

(a)

FIGURE 7.30 ‘Rigid’ beam-to-column joint

90

85 45

End plate 200 400 20 mm h 222,3

o

bf 166,8

9,1

r 10,2

200

V tw 7,7 60 kN

70

tw 13,0

h 310,9

bf 166,8

End plate o

35

30516554 UB

tf 20,5

O

d3 54,05 d2 254,05 d1 339,05

203 203 86 UC

tw 7,7

Grade S 355 steel M20 pre-loaded bolts

(b)

O

O

O

(c)

df 297,2

N 100 kN

d 265,6

•

t f 13,7

252

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253

Check for slip assuming rotation about axis O–O (Eq. (7.31)) μM 0,45 × 97,5E6 = = 2,46 > 1 [V (h − tf b ) [60E3 × (310,9 − 13,7)] therefore rotation about the compression flange of the beam at O without slip. Assuming rotation about axis O–O the design tensile force acting on a single bolt Ft,Ed =

M 97,5E6 = = 82 kN [4(h − tf b )] [4 × (310,9 − 13,7) × 1E3]

Design tensile resistance of an M20 pre-loaded class 8.8 bolt (Table 3.4, EN 1993-1-8 (2005)) assuming not subject to a shear force Ft,Rd = k2 As

fub 800 = 0,9 × 245 × γM2 1,25 × 1E3

= 141 > (Ft,Ed = 82) kN satisfactory. Shear resistance of an M20 class 8.8 pre-loaded bolt in single shear and subject to a tensile force of 82 kN (Eq. (3.8a), EN 1993-1-8 (2005)) Fs,Rd =

ks nμ (Fp,C − 0,8Ft,Ed ) γM3

1,0 × 1,0 × 0,5 (0,7 × 800 × 245 − 0,8 × 82E3) × 1,25 1E3 60 = 28,6 > Fs,Ed = = 10 kN satisfactory. 6

=

M20 class 8.8 bolt in bearing on end plate (t = 20 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dt γM2 2,5 × 0,682 × 510 × 20 × 20 = 278,2 > (Fs,Ed = 10) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 45/(3 × 22) = 0, 682 for an edge bold k1 = 2,8 (e2 /do ) − 1,7 = 2,8 × (55/22) − 1,7 = 5,3 > 2,5 Thickness of end plate related to a single bolt (Table 6.2, EN 1993-1-8 (2005), Method 1) ⎡ ⎤1/2 ⎢ 4Ft,Ed m ⎥ tf = ⎣ l f ⎦ eff y

γM0

4 × 82E3 × 35 = 100 × 355/1,0

1/2 = 18,0 mm; use 20 mm plate.

254

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8 (2005)) Fc,wc,Rd = =

ωkwc beff,c,wc twe fywe γM0 0,685 × 1 × 224,1 × 13 × 355 = 708,4 > 328 kN satisfactory. 1,0 × 1E3

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,685 × 1 × 1 × 224,1 × 13 × 355 1,0 × 1E3 = 708,4 > 328 kN satisfactory. =

where for a bolted end plate connection (Eq. (6.12), EN 1993-1-8 (2005)) beff,c,wc = tfb + 2 × 21/2 ap + 5(tfc + s) + sp = 13,7 + 2 × 21/2 × 0,7 × 6 + 5 × (20,5 + 15,2) + 20 = 224,1 mm The maximum longitudinal stress in the flange of the column from the axial load and eccentric load (Eq. (6.14), EN 1993-1-8 (2005))

σcom,Ed

60E3×223,2 Vh/2 (N + V ) (100 + 60)E3 2 + + = = A 110E2 851E3 Wel = 22,4 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

Table 5.4, EN 1993-1-8 (2005) β=1 Table 6.3, EN 1993-1-8 (2005) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

Avc

1

=

2 1/2

1 + 1,3 × 224,1 ×

1/2 13 2 3124

= 0,685

where (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 110E2 − 2 × 208,8 × 20,5 + (13 + 2 × 10,2) × 20,5 = 3124 mm2 ηhw tw = 1,0 × (222,3 − 2 × 20,5) × 13 = 2357 < (Avc = 3124) mm2

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Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

beff,c,wc dwe fy,wc 1/2 λp = 0,932 2 ) (Etwc 224,1 × 161,8 × 355 1/2 = 0,932 = 0,561 < 0,72 use ρ = 1,0 210E3 × 132 Reaction Ro = 4Ft = 4 × 82 = 328 kN Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8 (2005)) Vwp,Rd = 0,9

Avc fy,wc 0,9 × 3124 × 355 = 1/2 (31/2 γM0 ) (3 × 1 × 1E3)

= 576 > (Ro = 328) kN satisfactory. Alternative calculations for comparison Previous calculations assume the top four bolts resist all of the applied moment with a single lever arm. Alternatively assume a linear variation of forces from axis O–O to the bolts furthest from the axis. Maximum tensile force acting on the bolt furthest from axis O–O M Ft = 2 d1 + d2 + d3 =

97,5E3 = 75,3 kN [2(339,05 + 254,05 + 54,05)]

Prying force for each bolt assuming ∂b = 0 for a pre-loaded bolt and elastic behavior (Holmes and Martin, 1983). 2EI∂b Fbe − a b2 p p Qbe = a 2 2ap p 1 + 3 b b p

=

p

[90,62 − 0] 2 = 29,02 kN 2×45 1 × 45 + 35 3 35

and the maximum tensile force on a bolt Fbt + Qbe = 75,3 + 29,0 = 104,3 < (Ft,Rd = 141) kN satisfactory for no slip. For the welded connection between the end plate and the beam assume the applied vertical shear force is resisted by two 6 mm web welds VR,Ed = 2df Fw,Rd = 2 × (310,9 − 2 × 13,7) × 1,1 = 623,7 > (VEd = 60) kN satisfactory.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Assume that the applied bending moment is resisted by two 10 mm effective flange welds with rotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8 (2005)) MR,Ed = 2beff df Fw,Rd = 2 × 91,2 × (310,9 − 13,7) ×

1,83 1E3

= 99,2 > (MEd = 97,5) kNm satisfactory (cl 6.2.3(4), EN 1993-1-8 (2005)). For strength of welds (Fw,Rd ) see Annex 1 and for beff see Eq. (4.6a), EN 1993-1-8 (2005).

EXAMPLE 7.9 ‘Rigid’ beam-to-beam joint. Determine the size of the components for the rigid beam-to-beam joint shown in Fig. 7.31 at the ultimate limit state. Assuming rotation about axis O–O and the applied moment of 97,5 kNm is resisted entirely by the cover plate then the design tensile force in the flange cover plate Ff =

M 97,5E6 = = 320,9 kN. h 303,8 × 1E3

Thickness of flange connection plate, bp = 165 and 20 mm pre-loaded bolts Ff

tp = "

(bp − 2dh )

fy γM1

320,9E3

# = "

(165 − 2 × 22) ×

355 1,0

#

= 7, 47 mm, use 8 mm thick Grade S355 steel plate.

V 45 kN

40 60 60 40

75 6 mm fillet weld

tf 10,2

tp 10

R0 O

tw 10,7 10 mm thick end plate

tf 18,9

30516540 kg UB bf 165,1 tw 6,1

h 303,8

Connecting plate

tp 8

M 97,5 kNm

Grade S 355 steel M20 pre-loaded bolts

30

•

h 465,1

256

Alternative connections bf 153,5 457 152 82 kg UB

(a) Welded end plate connection

FIGURE 7.31 ‘Rigid’ beam-to-beam joint

(b) With bottom cleats

(c) With web cleats

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257

Single shear resistance of M20 class 8.8 pre-loaded bolt (Table 3.4, EN 1993-1-8 (2005)) Fs,Rd = =

ks nμ0,7Fp,C As γM2 245 1,0 × 1, 0 × 0,5 × 0,7 × 800 × 1E3

1,25

= 54,9 kN

M20 class 8.8 bolt in bearing on plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dt γM2 2,34 × 0,606 × 510 × 20 × 8 = 92,6 > (Fs,Rd = 54,9) kN 1,25 × 1E3

for an end bold αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8(e2 /do ) − 1,7 = 2,8 × 31,75/22 − 1,7 = 2,34 < 2,5 Number of bolts required for the connecting plate nb =

Fs,Ed 320,9 = = 5,85 use 6-M20 class 8.8 pre-loaded bolts. Fs,Rd 54,9

Reaction at the hinge is equal to the force in the flange, Ro = Ff = 320,9 kN and the frictional resistance at the hinge μRo = 0,45 × 320,9 = 144,4 > (VEd = 45) kN therefore no slip occurs. Use a 10 mm thick end plate welded to the end of the 305 × 165 × 40 kg UB and bolted to the 457 × 152 × 82 kg UB as shown in Fig. 7.31(a). Shear resistance of 4-M20 pre-loaded bolts in double shear in the end plate = 2 × 4 × 54,9 = 439,2 > (VEd = 90) kN satisfactory. Shear resistance of two 6 mm fillet welds connecting the end plate to the 305 × 165 × 40 kg UB using grade S355 steel Fw,Rd = =

fu a (3

1/2

βw γM2 )

6 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,10 kN/mm

2lw Fw,Rd = 2 × (303,8 − 30 − 10,2) × 1,10 = 580 > (VEd = 45) kN satisfactory.

EXAMPLE 7.10

‘Rigid’ beam splice. Determine the size of components for the beam splice shown in Fig. 7.32 at the ultimate limit state.

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) V 600 kN M 450 kNm

bf 209,3

40 80 80 40

140

2

Web plate 10 mm thick

533 210 92 kg UB tw 10,2

h 533,1

tf 15,6 8 8

•

35 35

z (a) Elevation 40 80 80

y

40 80 80 80 80 40

258

Grade S355 steel

Section

V 600 kN Mw 87,4 kN y

M20 pre-loaded bolts e 121 z (b) Web connection resisting shear force and web bending moment

FIGURE 7.32 ‘Rigid’ beam splice

Check if the beam is in the elastic stage of behaviour f =

M 450E6 = = 216, 8 < ( fy = 355) MPa, therefore elastic behaviour. We 2076E3

Second moment of area of the web of the beam Iweb =

tw (h − 2tf )3 10,2 × (533,1 − 2 × 15,6)3 = = 107, 5E6 mm4 12 12

From Section Tables the gross second moment of area of the beam section Igross = 553,5E6 mm4 Assumed proportion of the applied bending moment taken by the web 107,5E6 Iweb Mweb = × 450 = 87,39 kNm M= Igross 553,5E6 Check the strength of the arrangement of bolts in shear in the web plate (Fig. 7.32(b))

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259

Second moment of area of bolts of unit area about the centroidal y–y axis Iy = (∂A)z2 = 6 × (802 + 1602 ) = 192E3 mm4 Second moment of area of bolts of unit area about the centroidal z–z axis Iz = (∂A)y2 = 10 × 802 = 64E3 mm4 Second moment of area of bolts of unit area about the centroidal polar x–x axis Ix = Iy + Iz = (192 + 64)E3 = 256E3 mm4 Eccentricity of the applied shear force relative to the centroid of half the bolt group is 121 mm (Fig. 7.32(b)). This eccentricity produces a moment which is increased by the bending moment resisted by the web. The equivalent eccentricity e = e +

Mweb 87,4E3 = 121 + = 266,6 mm V 600

Maximum vector shear force in the y–y direction acting on a bolt furthest from the centroid of the web bolt group Fy = Ve

zn 160 = 600 × 266,6 × = 99,99 kN Ix 256E3

Maximum vector shear force in the z–z direction acting on the same bolt Fz =

V yn 600 80 + Ve = + 600 × 266,6 × = 90,0 kN n Ix 15 256E3

Resultant maximum vector force acting on the same bolt Fr = [Fy2 + Fz2 ]1/2 = [99,992 + 90,02 ]1/2 = 134,5 kN Double shear strength of an M20 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5) in the web (cl 3.9.1, EN 1993-1-8(2005)) Fv,Rd =

1, 0 × 2, 0 × 0,5 × 245 × 0,7 × 1000 ks nμFp,C 1E3 = 1,25 γM3

= 137,2 > (Fr = 134,5) kN, satisfactory. M20 class 8.8 bolt in bearing on the web (t = 10,2 mm) (Table 3.4, EN 1993-1-8(2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,606 × 510 × 20 × 10,2 = 126,1 < (Fv,Rd = 137,2) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 40/(3 × 22) = 0,606 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 106,5/22 − 1,7 = 11,9 > 2,5

260

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Required number of M20 pre-loaded bolts in bearing for the flange splice # " (M−Mweb ) Ff (h−tf ) nb = = Fb,Rd Fb,Rd " # =

(450−87,4)E3 (533,1−15,6)

126,1

= 5,56 use 6 bolts.

Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005)) (Lj − 15d) (2 × 80 − 15 × 20) =1− 200d 200 × 20 = 1,035 use 1,0

βLf = 1 −

Thickness of the outer and inner flange cover plates Ff tp = (b −2d +2w −2d )f f

p

h

h y

γM1

=

700,7E3

= 7,55 mm (209,3 − 2 × 22 + 2 × 70 − 2 × 22) × 355 1,0

Use 8 mm plates Grade S355 steel as shown in Fig. 7.32.

EXAMPLE 7.11 ‘Rigid’ column splice. Determine the size of the components for the rigid column splice shown in Fig. 7.33 at the ultimate limit state. Where column sections are of the same serial size it is possible to connect them directly with web and flange plates. The ends of the column are machined and will be in contact. Rotation will take place about an axis near the outer edge of the flange of the upper column. Thickness of the flange plate, from moments of forces about the axis of rotation Nh M − 2u tp = " f # y (bp − 2dh ) γ hu M1 # " 712,5E3×355,6 480E6 − 2 # = " 355 (365 − 2 × 24) 1,0 × 355,6 = 8,83 mm, use 10 mm Grade S355 steel plate. Single shear strength of an M22 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5) in the flange (cl 3.9.1, EN 1993-1-8 (2005)) Fv,Rd = =

ks nμFp,C γM2 1,0 × 1,0 × 0,5 × 303 × 0,7 × 1000 1E3 1,25

= 84,8 kN.

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261

N 712,5 kN M 480 kNm 356 368 129 kg UC bf 368,3

Packing

hu 355,6 Packing tf 17,5

Machined plate

tw 10,7 Angle

Flange plate

960 365 10 Flange plate 40

140 80

(b) Column sections of different serial size

140

40 80 80 80 80 80

V 150 kN

Web plate thickness 10 mm

40 150 40 Packing tw 16,8

Welded end plates

tf 27 Grade S355 steel M22 pre-loaded bolts

hL 374,7 356 368 202 kg UC bf 374,4

(a) Column sections of the same serial size

(c) Welded and plate column connection

FIGURE 7.33 ‘Rigid’ splices in steel columns

M22 class 10,9 bolt in bearing on plate (t = 10 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,555 × 510 × 22 × 10 = 124,5 > (Fv,Rd = 84,8)kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 40/(3 × 24) = 0,555 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 108/24 − 1,7 = 10,9 > 2,5

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Number of bolts required # " M − Nh2 u nb = (Fv,Rd hu ) " # 712,5E3×355,6 480E6 − 2 = = 11,7 use 12-M22 bolts. 84,8E3 × 355,6 Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005)) (Lj − 15d) (5 × 80 − 15 × 22) =1− 200d 200 × 22 = 0, 984. This factor does not affect the number of bolts required.

βLf = 1 −

Where the ends of the column are machined and in contact the horizontal shear force on the column is resisted by the friction force, in part or whole, at the point of contact, that is, at the axis of rotation. Assuming machined surfaces μ = 0,15 the frictional resistance M N =μ + hu 2 480E3 712,5 = 0,15 × + = 255,9 > 150 kN (applied shear force). 355,6 2 Theoretically no shear connection required but in practice a web plate is generally provided to align the webs. If the frictional resistance is ignored then the web splice is designed to resist the entire shear force as follows: Second moments of area of two bolts of unit area on one side of the web connection about the centroidal axes are: Iy = 0 Iz = 2 × 752 = 11,25E3 mm4 Ix = Iy + Iz = 11,25E3 mm4 Vector force on a bolt in the y–y direction Fx =

V 150 = = 75 kN nb 2

Vector force on a bolt furthest from the centre of rotation in the z–z direction Fz = (Ve)

yn 75 = 150 × 40 × = 40 kN lx 11,25E3

Maximum vector shear force on the same bolt Fr = (Fy2 + Fz2 )1/2 = (752 + 402 )1/2 = 85 < (Fb,Rd = 124,5) kN, acceptable.

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263

N 198 kN M 264 kNm

35

fc = 20

Bp 460

H 49,5 kN M30 class 4,6 bolts

eb 70

113,1 C 104,4

300

Gusset plate tg 12,5

70

203 203 86 UC bf 208,8; h 222,3

Grade S355 steel

Dp 800

(a)

(b)

Lq 278,9

Bg

y

y G

Dp 800 G

Dw 300

Hg 300

h 222,3 z

G

z

2/3 20

xp 83,6

o (d) Column-to-gusset plate welds

o

o

(e) Base plate-to-gusset plate welds

Fg Sg 247,1 (c)

FIGURE 7.34 ‘Rigid’ column-to-foundation joint

EXAMPLE 7.12 ‘Rigid’ built-up column base connection. Determine the size of the components for the connection shown in Fig. 7.34 at the ultimate limit state assuming fc = 20 MPa. Tensile strength of an M30 class 4.6 holding down bolt (Table 3.4, EN 1993-1-8 (2005)) Ft,Rd =

0,9As fub 0,9 × 561 × 400 = = 161,6 kN γM2 1,25 × 1E3

Distance required between holding down bolts (Eq. (7.39)) dp = " N 2

M + nFt,Rd

#=

264E3 198 2

+ 2 × 161,6

= 625,3 mm

The size of the base plate is determined as follows: Assume a bolt edge distance of approximately 2d = 2 × 30 = 60 mm use 70 mm.

264

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Total length of base plate Dp = dp + 2eb = 625,3 + 2 × 70 = 765,3 mm, use length of 800 mm Minimum width of base plate Bp = bf + 2tg + washer + 2 × welds + 2eb = 208,8 + 2 × 12,5 + 66 + 2 × 10 + 2 × 70 = 459,8 mm. Use width of 460 mm. The thickness of gusset plate(tg ) and size of welds are assumed at this stage. Assume the projection length for the base plate is half the width of the column (Fig. 7.34(b)) c=

bf 208,8 = = 104,4 mm 2 2

Thickness of base plate Grade S355 steel (Eq. (6.5), EN 1993-1-8 (2005)) 1/2 3 × 23 × 20 × 1,0 3fjd γM0 1/2 tp = c = 104,4 = 35 mm fy 355 Length of concrete compression zone beneath the steel baseplate (Eq. (7.38)) assuming lever arm la = 660 mm M N + la 2 xp = (bf + 2tg + 2c)fjd 264E6 198E3 + 660 2 = [(208,8 + 2 × 15 + 2 × 104,4) × 23 × 20] = 83,6 mm Lever arm for resistance of concrete in bending at ultimate limit state la = Dp − eb −

xp 83,6 = 800 − 70 − = 688,2 mm 2 2

Tensile force in a holding down bolt (Eq. (7.39)) D Xp p M −N 2 − 2 Fbt = n la b 83,6 264E3 − 198 × 800 − 2 2 = = 140,3 < (Ft,Rd = 161,6) kN 2 × 688,2 Use 800 × 460 × 35 mm base plate Grade S355 steel. Force from bearing pressure applied to each gusset plate (Fig. 7.34(c)) Bj 2 460 83,6 Fg = fjd xp = × 20 × × = 256,4 kN 2 3 2 1E3

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265

Length of gusset plate allowing for 10 mm for weld Lg =

(Dp − 2 × 10 − h) (800 − 20 − 222,3) = = 278,9 mm 2 2

Assume height of gusset plate Hg = 300 mm Eccentricity of force Fg in relation to the inner corner of the gusset plate (Fig. 7.34(c)) sg =

(Dp − h) (800 − 222,3) − xp /2 = − 83,6/2 = 247,1 mm 2 2

Width of gusset plate (Eq. (7.26)) Lg L 2 1/2 =

Bg =

g

1+ H g

1+

278,9 2 1/2 = 204,2 mm 278,9 300

Thickness of gusset plate Grade S355 steel (Eq. (7.25)) 2Fg sg Bg tg = f + yg 80 B2 γM1 g =

2 × 256,4E3 × 247,1 204,2 + = 11,1 mm 355 80 2 × 204,2 1,0

Use 12,5 mm thick Grade S355 steel gusset plate. Check slenderness ratio of gusset plate (Eq. (7.27)) √ √ lg 2 3Bg 2 3 × 204,2 = = = 56,6 < 185, satisfactory. ig tg 12,5 Minimum length of foundation bolt (Holmes and Martin, 1983) (cl 6.2.6.12, EN 1993-1-8 (2005)) Fbt 161,6E3 Lb = = = 186,8 mm 2/3 πftc π × 0,3 × 20 1,5

Use 4-M30 class 4.6 holding down bolts, 300 mm long anchored by washer plates. The size of the fillet weld connecting the base plate to the gusset assuming no friction is obtained as follows: Length of weld (Fig. 7.34(e)) Lw = 2Dp = 2 × 800 = 1,6E3 mm Second moment of area about centroid of the weld group for unit size weld IwG =

2D3p 12

=

2 × 8003 = 85,33E6 mm4 12

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Maximum vertical force compressive per unit length weld in the z direction D p M N 2 Fwz = + Iwg Lw 264E3 × 800 198 2 = + = 1,36 kN/mm 1,6E3 85,33E6 Horizontal force per unit length of weld in the x direction Fwx =

H 49,5 = = 0,0309 KN/mm Lw 1,6E3

Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwz + Fwx ) = (1,362 + 0,03092 )1/2 = 1,362 kN/mm

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (31/2 βw γM2 )

=

8 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,362) kN/mm Alternatively if surface between the base plate and edge of the gusset plate are machined and bearing is assumed at the axis of rotation O–O (Fig. 7.34(e)). From Eq. (7.30) ⎛ Ro =

⎞

NDp +M ⎝ 6 ⎠ 2Dp 3

198×800 =

6

+ 264E3

2×800 3

= 544,5 kN Frictional resistance at Ro = μRo = 0,15 × 544,5 = 81,65 > (H = 49,5) kN, satisfactory. Second moment of area about axis O–O for unit size weld (Fig. 7.34(e)) IwO =

2D3p 3

=

2 × 8003 = 341,3E6 mm4 3

Maximum vertical force per unit length of weld in the z direction ND

Fwz =

M − 2w 264E3 − 198×800 2 = = 0, 542 kN/mm IwO 341,3E3

Assuming a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

6 510 × 0,7 × 1E3 fu a = (31/2 βw γM2 ) (31/2 × 0,9 × 1, 25)

= 1,10 > (Fwz = 0,542) kN/mm

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267

If the end of the column is not machined then rotation is assumed to be about axis G–G and the size of the weld connecting the column to the gusset plate is obtained as follows. Length of weld (Fig. 7.34(d)) Lw = 4Dw = 4 × 300 = 1,2E3 mm Second moments of area about the centroid of the weld group (axis G–G) fo unit size welds 4D3w 4 × 3003 IwGy = = = 9E6 mm4 12 12 2 h 222,3 2 IwGz = 4Dw = 4 × 300 = 14,83E6 mm4 2 2 Polar second moment of area IwGx = IwGy + IwGz = (9,0 + 14,83)E6 = 23,83E6 mm4 Maximum vertical force per unit length of weld in the z direction on an element furthest from the axis of rotation h N HDw 2 Fwz = + M− Lw 2 IwGx 222,3 198 49,5 × 300 2 = + 264E3 − × 1,2E3 2 23,83E6 = 1, 362 kN/mm Horizontal force per unit length of weld in the y direction on the same element Dw H HDw 2 Fwy = + M− Lw 2 IwGx 300 49,5 49,5 × 300 2 = + 264E3 − × 1,2E3 2 23,83E6 = 1,656 kN/mm Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwy + Fwz ) = (1,6562 + 1,3622 )1/2 = 2,144 kN/mm

Assuming a 12 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (301/2 βw γM2 )

=

12 510 × 0, 7 × 1E3

(31/2 × 0, 9 × 1, 25)

= 2, 20 > (Fwr = 2, 144) kN/mm Alternatively if the contact between the base plate and the end of the column is machined then the axis of rotation is O–O (Fig. 7.34(d)).

268

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Second moments of area about centroid of the weld group for unit size weld about axis O–O IwOy =

4 × 3003 4D3w = = 36E6 mm4 3 3

Iwbz = 2Dw h2 = 2 × 300 × 222,32 = 29, 65E6 mm4 Polar second moment of area IwOx = IwOy + IwOz = (36 + 29, 65)E6 = 65,65E6 mm4 Force per unit length of weld in the z direction on an element furthest from the axis of rotation h M − Nh 2 Fwz = IwOx 222,3 222,3 = 264E3 − 198 × × = 0, 819 kN/mm 2 65,65E6 Force per unit length of weld in the y direction on the same element M − Nh Dw 2 Fwy = IwOx 222,3 300 = 264E3 − 198 × × = 1,106 kN/mm 2 65,65E6 Resultant vector force per unit length of weld 2 2 1/2 Fwr = (Fwy + Fwz ) = (1,1062 + 0,8192 )1/2 = 1,376 kN/mm

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd

8 510 × 0,7 × 1E3 fu a = 1/2 = 1/2 (3 βw γM2 ) (3 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,376) kN/mm Check shear resistance beneath the base plate (cl 6.2.2(8), Eq. (6.3), EN 1993-18(2005)) F1,v,Rd = 107,7 kN (see Annex A2) αb = 0,44 − 0,0003fyb = 044 − 0,0003 × 240 = 0,368 As 561 = 66,1 < 107,7 kN = 0,368 × 400 × γMb 1,25 = 0,2N + nb F1,v,Rd = 0,2 × 198 + 4 × 66,1

F2,v,Rd = αb fub Fv,Rd

= 304 > (H = 49,5) kN satisfactory.

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269

b1 150

h1 250 250 150 10 RHS

N 66,75 kN M 85 kNm

V 40 kN

Chord face

•

t1 10

8 mm fillet weld

o h0 300

R0 300 200 10 RHS Side wall

t0 10

b0 200

Side elevation

End elevation

150

o Plan of weld o 250

R

Assumed stress distribution

dwr

FIGURE 7.35 ‘Rigid’ RHS joint

EXAMPLE 7.13 ‘Rigid’ RHS connection. Check the strength of the rigid rectangular hollow section connection at the ultimate limit state assuming Grade S355 steel (Fig. 7.35). Check validity of the joint (Table 7.8, EN 1993-1-8 (2005)) b1 150 = = 15 < 35 satisfactory. t1 10 h1 250 = = 25 < 35 satisfactory. t1 10 Check for chord face failure of the horizontal member (Table 7.14, EN 1993-1-8 (2005)). " # η 1 2 + + 1/2 (2η) (1−β) (1−β) Mip,1,Rd = kn fyo to2 h1 γM5 5/3 1 2 (2×5/3) + (1−0,75)1/2 + (1−0,75) = 1,0 × 355 × 102 × 250 × (1,0 × 1E6) = 97,3 > (MEd = 85) kNm satisfactory.

270

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Included in the previous calculations b1 150 = = 0,75 < 0,85 b0 200 kn = 1,3 − 0,4n/β 0,282 = 1,3 − 0,4 × 0,75 = 1,15 > 1 use 1 h1 5 250 η= = = b0 150 3 β=

If the weld group is assumed to rotate about the axis O–O (Fig. 7.35) then the effective width of the weld (Table 7.13, EN 1993-1-8 (2005)) furthest from the axis O–O fy0 t0 b1 # beff = 10 " b0 fy1 t1 t 0

355 × 10 × 150 # = 75 mm = 10 × " 200 × 355 × 10 10 beff 75 = = 0,5 b1 150 Distance(dwr ) from the axis O–O from the resultant force in the weld is obtained as follows moment of the parts = moment of the whole 1 Fz Ioe beff + × 2dw Fz dwr = 2 dwr rearranging and substituting Ioe = 2dw3 /3 + beff dw2 beff 2 + dwr 3 dw = beff dw 1+ d w 2 75 + 3 250 = 0,744 = 75 1 + 250 Resultant reaction from Eq. (7.30) Ro = =

M + N(dwr − dw /2) dwr # " 85E3 + 66,75 × 0,744 × 250 − 250 2 0,744 × 250

= 478,9 kN

Frictional force at the stiff bearing if the end of the vertical RHS is machined μRo = 0, 15 × 478, 9 = 71, 8 > 40 kN (applied shear force). The weld group is subject to the actions from N and M acting about axis O–O.

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271

Second moment of area about axis O–O of the weld group (Fig. 7.35) for unit size welds IO =

2dw3 2 × 2503 + beff dw2 = + 75 × 2502 = 15,1E6 mm4 3 3

Moment applied about axis O–O M = M −

Ndw 250 = 85 − 66,75 × = 76,65 kNm 2 2 × 1E3

Maximum tensile force per unit length of weld furthest from the axis of rotation

Fw =

Md 76,65E3 × 250 = = 1,27 kN/mm IO 15,1E6

Assuming a 8 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8 (2005)) Fw,Rd =

fu a (31/2 βw γM2 )

=

8 510 × 0,7 × 1E3

(31/2 × 0,9 × 1,25)

= 1,47 > (Fwy = 1,27) kN/mm An alternative more conservative calculation to determine the size of weld is to assume rotation about the centroid of the weld group. This results in a larger weld size. For interest check for side wall crushing in horizontal member at O–O (Table 7.14, EN 1993-1-8 (2005)) Mip,1,Rd = =

0,5fyk t0 (h1 + 5t0 )2 γM5 0,5 × 355 × 10 × (300 + 5 × 10)2 = 217,4 > Mip,1,Rd = 155 kNm. (1,0 × 1E6)

This calculation is not necessary because the ratio β = b1 /b0 = 150/200 = 0,75 < 0,85 indicates that it is not critical.

EXAMPLE 7.14 ‘Rigid’ knee connection for a portal frame. Check the strength of the knee joint components for the frame (Fig. 7.36) at the ultimate limit state. Moment acting about axis O–O MO = 250,2 − 116,8 × 0,5379 = 187,4 kNm Thickness of strap assuming 190 mm wide and 22 mm diameter hole MO tst = l b f a st y,st γM1

187,4E6

= "

717,2 × cos 10◦ × (190 − 2 × 22) × 355 1,0

= 5,12 mm. Use a 8 mm thick strap.

#

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) V 174,4 kN Strap welded to column t 8

100 x

45719167 kg UB bf 189,9, h 453,6 d 407,9

tw 8,5 r 10,2

Grade S 355 steel M20 pre-loaded bolts 6 mm fillet welds

O

50

13

H 116,8 kN tw 8

10°

Separate bolted strap

M 250,2 kNm 717,2

•

537,9

272

Cap plate t 12,5

R0 end plate t 12,5 tf 12,7

Web stiffener ts 8 (a)

(b)

FIGURE 7.36 ‘Rigid’ knee joint for a portal frame

Strap welded to the top of the column and bolted to the rafter (Fig. 7.36(a)). For a single M20 pre-loaded bolt slip resistance (Eq. (3.6), EN 1993-1-8 (2005)) Fs,Rd = ks n μfub

As γM3

= 1,0 × 1,0 × 0,4 × 0,7 × 800 ×

245 = 43,9 kN 1,25 × 1E3

M20 class 8.8 bolt in bearing on the plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)). Fb,Rd = =

k1 αb fup dtw γM2 2,5 × 0,758 × 510 × 20 × 8 = 123,7 > (Fs,Rd = 43,9) kN 1,25 × 1E3

for an end bolt αb = e1 /(3do ) = 50/(3 × 22) = 0,758 for an edge bolt k1 = 2,8e2 /do − 1,7 = 2,8 × 50/22 − 1,7 = 4,66 > 2,5 Number of M20 pre-loaded bolts in single shear for the strap nb =

MO 187,4E3 = = 6,04 use 6 bolts. (la Fs,Rd ) 717,2 × cos 10◦ × 43,9

Force per unit length acting on the two welds connecting strap to head of column Fw,Ed = =

MO la n w l w 187,4E3 717,2 × cos 10◦ × 2 × cos 10◦ 407,9

= 0,320 kN/mm

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273

Shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) Fw,Rd = =

fu a (3

1/2

(31/2

βw γM2 ) 510 × 0,7 × 6 = 1,1 > (Fw,Ed = 0,320) kN/mm × 0,9 × 1,25 × 1E3)

These calculations assume, conservatively, that the force is resisted only by the two welds along the web of the column, but the strap is also welded to the flanges of the column which increases the strength. Moments of forces about X to determine the reaction Ro at O −717, 2Ro + 250, 2E3 + 116, 8 × (717, 2 − 537, 9) = 0;

hence Ro = 378, 1 kN

Vertical frictional force at O = μRo = 0, 4 × 378, 1 = 151, 2 < (VEd = 174, 4) kN, therefore slip occurs. Bolt end plate (tep = 12,5 mm) to the flange of the column using M20 pre-loaded bolts (fpu = 800 MPa, μ = 0,4). For a single bolt slip resistance (Eq. (3.6), EN 1993-1-8 (2005)). Fs,Rd = 43,9 kN as calculated previously. Number of bolts required nb =

VEd 174,4 = = 3,97 use 4 bolts. Fv,Rd 43,9

Shear strength of the web of the column (cl 6.2.6.1 (2), EN 1993-1-8 (2005)) f y,wc

Vpl,Rd = 0,9Avc

γM0

31/2

= 0,9 × 4094 ×

355 1,0

(31/2 × 1E3)

= 755,2 > (Ro = 378,1) kN, satisfactory. Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8(2005)) Fc,wc,Rd =

ωkwc beff,c,wc twe fywe γM0

0,932 × 1 × 0,731 × 164,4 × 8,5 × 355 1,0 × 1E3 = 338,0 < (Ro = 378,1) kN not satisfactory. =

or Fc,wc,Rd =

ωkwc ρbeff,c,wc twe fywe γM1

0,932 × 1 × 0,731 × 164,4 × 8,5 × 355 1,00 × 1E3 = 338,0 < (Ro = 378,1) kN not satisfactory stiffeners required. =

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•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005) Previous calculations include the effective width (Eq. (6.11), EN 1993-1-8 (2005)) beff,c,wc = tfb + 2 × 21/2 ap + 5(tfc + s) + sp = 13 + 2 × 21/2 × 0,7 × 6 + 5 × (12,7 + 10,2) + 2 × 12,5 = 164,4 mm Maximum longitudinal stress in the flange of the column from the axial load and eccentric load (cl 6.2.6.2(2), EN 1993-1-8 (2005)) σcom,Ed =

M+

V + A

Vh 2

Wel 453,6 250,2E6 + 174,4E3 × 174,4E3 2 = + 85,5E2 1297E3 = 244 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

β = 1 (Table 5.4, EN 1993-1-8 (2005)) and from Table 6.3, EN 1993-1-8 (2005) 1

ω = ω1 = 1 + 1,3

b

eff,c,wc twc

Avc

1

= 1 + 1,3

2 1/2

1/2 164,4×8,5 2 4094

= 0,932

which includes (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables Avc = A − 2btf + (tw + 2r)tf = 85, 5E2 − 2 × 189, 9 × 12, 7 + (8, 5 + 2 × 10, 2) × 12, 7 = 4094 mm2 ηhwc twc = 1,0 × (453,6 − 2 × 12,7) × 8,5 = 3640 < (Avc = 4094) mm2 satisfactory. Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

beff,c,wc dwe fy,wc 1/2 λp = 0,932 2 ) (E twc 164,4 × 407,9 × 355 1/2 = 0,932 = 1,167 > 0,72 therefore (210E3 × 8,52 ) ρ=

λp − 0,2 2 λp

=

1,196 − 0,2 = 0,731 1,1962

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Thickness of the load bearing web stiffeners in compression in the column (cl 6.2.6.2(4), EN 1993-1-8 (2005)) Ro

ts = "

fyw

(bcf − tcw − 2rc ) γ

#

M1

378, 1E3

= "

(189,9 − 8,5 − 2 × 10,2) 355 1,0

#

= 6,62 mm. Use 8 mm thick stiffeners either side of the web of the column. Force per unit length of weld connecting the web stiffener to the column web Ro 378,1 = = 0,463 kN/mm 2 × 407,9 n w lw

Fw =

As previously the shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) Fw,Rd = 1, 1 > (Fw = 0,463) kN/mm It is assumed conservatively that the force is resisted by the welds along the web. To avoid damage to the strap in transit an alternative arrangement is shown in Fig. 7.36(b).

7.11 JOINT ROTATIONAL STIFFNESS (CL 6.3, EN 1993-1-8 (2005)) Previous sections in this chapter show calculations for joint strength. Another aspect of joint behaviour that needs consideration is the relationship between joint rotational stiffness and member stiffness. This relationship is important in the analysis of structures (cl 5, EN 1993-1-8 (2005)). A simple example of joint rotational stiffness (Fig. 7.37(a)) is as follows. The theoretical rotational stiffness related to the elastic and plastic behaviour of the top cleat can be developed for the joint. Initially assuming elastic behaviour of the steel top cleat with deformations as shown El 2 6 ma3 z2 M Ft z ba ta3 Sj = z2 = = = 12E φ 12m3 z

= "

E

z2

1 (ba ta3 /m3 )

#

(7.40)

This equation considers only deformation of the top cleat and can be compared with the theoretical expression (Eq. (6.27), EN 1993-1-8 (2005)) which replaces ba with leff and includes factors for other deformations (Table 6.11, EN 1993-1-8 (2005)). E z2 # μ k1

Sj = "

1

(7.41)

•

Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

6E1/m 2

m

M

Z

276

(a)

0,4 2,0 Hu

f A

HI

Q

Mj

0,3

0,5

kc /kb

Mj 0,2 Q L/8 u Mj/Sj

L

(b)

0,1

0

0,2

0,4

0,6

0,8

1,0

Sj /kb (c)

FIGURE 7.37 Stiffness of joints

Other factors that affect the rotational stiffness of a joint are: extension and shear deformation of bolts, and deformation of column flanges, end plates and column webs. Each factor (Tables 6.10 and 6.11, EN 1993-1-8 (2005)) produces values of 1/k1 which are added and inserted in Eq. (7.41) as shown in Example 7.15. Experimental results (Maxwell et al., 1981) which show that the rotational stiffness of a joint is non-linear as ultimate resistance approaches. The stiffness of the joint reduces and to allow for this the value of μ increases (cl 6.3.1(6), EN 1993-1-8 (2005)). It is also necessary to check for other forms of failure. For this joint (Fig. 7.37(a)) the ultimate moment of resistance based on yielding of the top angle fy b ta2 Mjya,Rd = 2 (7.42a) 4 mz

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Alternatively for this joint if the bolts reach ultimate load Mjub,Rd = nb Ab fu z

(7.42b)

7.12 FRAME-TO-JOINT STIFFNESS Ideally the analysis of a structure should incorporate the stiffness of the joints and the members. An example of the theoretical relationship between the stiffness of joint and members for a simple frame follows. Consider the elastic behaviour of the simple frame in Fig. 7.37(b). From the application of the area moment method (Croxton and Martin, 1987 and 1989), the moment of resistance of the joint (Mj ) for a point load at the mid-span of the beam For the beam at A Mj M j Lb 1 Lb QLb EIb θ + = × × − Sj 2 2 4 2

(i)

For the upper column at A EIcu θ =

1 × Mju Hu 4

(ii)

For the lower column at A EIcl θ =

1 × Mjl Hl 4

(iii)

and Mj = Mju + Mjl

(iv)

Combining (i) to (iv) to eliminate θ and expressing stiffness as k = El/L Mj =

1+

QL 8 1 2(kcU /kb +kcL /kb )

2k

+ Sjb

(7.43)

The stiffness of the joint (Sj ), beam (kb ) and columns (kc ) can be seen to affect the moment of resistance at the joint (Mj ).

EXAMPLE 7.15 Stiffness of a ‘pin’ joint. Determine the stiffners of the ‘Pin’ shown in Fig. 7.26. Joint stiffness (Eq. (6.27), EN 1993-1-8(2005)) assuming μ = 1 E z2 # μ k1

Sj = "

=

210E3 × 517,42 = 22668 kNm/radian [(1 × 2,48 × 1E6)]

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) where (Fig. 6.15(b), EN 1993-1-8 (2005)) z = h + ea +

ta 10 = 467,4 + 45 + = 517,4 mm 2 2

and for a column web panel in shear (Table 6.11, EN 1993-1-8 (2005)) k1 =

0,38 × 3130 0,38Avc = = 2,3 βz 1 × 517,4

and for a column web panel in compression (Table 6.11, EN 1993-1-8 (2005)) k2 =

0,7beff,c,wc twc 0,7 × 179,5 × 13 = = 10,2 dc 160,8

where (Table 6.12, EN 1993-1-8 (2005)) beff,c,we = 2ta + 0,6ra + 5(tfc + s) = 2 × 10 + 0,6 × 10 + 5 × (20,5 + 10,2) = 179,5 mm and for a column web in tension (Table 6.11, EN 1993-1-8 (2005)) k3 =

0,7beff,c,wc tfc 0,7 × 210 × 13 = 11,9 = dc 160,8

where (Table 6.4, EN 1993-1-8 (2005)) beff,c,wc is the lesser value of 2πm = 2π × 45 = 283 mm and (208,8 − 140) πm + 2e1 = π × 45 + 2 × = 210 mm 2 and for a column flange in bending (Table 6.11, EN 1993-1-8 (2005)) k4 =

3 0,9leff tfe 0,9 × 146,5 × 20,53 = = 12,5 m3 453

where (Table 6.4, EN 1993-1-8 (2005)) leff is the lesser value of leff = 4m + 1,25e = 4 × 45 + 1,25 ×

(208,8 − 140) = 223 mm or 2

leff = 2m + 0,625e + e1 = 2 × 45 + 0,625 +

(208,8 − 140) + 35 = 146,5 mm 2

and for a cleat in bending (Table 6.11, EN 1993-1-8 (2005)) k6 =

0,9leff t3 0,9(208,8) 103 = ×

3 = 1,03 2 m3 45 − 10/2

and for bolts in tension (Table 6.11, EN 1993-1-8 (2005)) k10 =

1,6As 1,6 × 245 = 6,91 = Lb 20,5 + 10 + 3,7 + 13 + 16 2

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and for bolts in shear (Table 6.11, EN 1993-1-8 (2005)) k11 =

16 nb d2 fub 16 × 2 × 202 × 800 = = 3,45 E dM16 210E3 × 14,1

and for bolts in bearing (Table 6.11, EN 1993-1-8 (2005)) k12 =

24 nb kb kt d fu 24 × 2 × 1,06 × 1,06 × 20 × 510 = = 2,62 E 210E3

where 0,25eb 0,25 × 45 + 0,5 = + 0,5 = 1,06 d 20 1,5tj 1,5 × 10 = kt = = 1,06 dM16 14,1

kb =

The total value of 1 1 1 1 1 1 1 1 1 μ =1× + + + + + + + = 2,48. k 2,3 10,2 11,9 12,5 1,03 6,91 3,45 2,62

EXAMPLE 7.16

Effect of ‘pin’ joint stiffness on a simple frame shown in Fig. 7.37(b). Assume Q = 52 kN, Lb = 10 m, kb = 9612 kNm, kb /kc = 1 and Sj = 22668 kNm obtained in Example 7.15. The joint design moment of resistance at A (Eq. (7.43)) QL 8

Mj,Rd = ⎛ ⎝1 +

1 kcu kcl 2 + kb k b

=

1+

⎞ + Sb ⎠ j 2k

52E3×10E3 (8×1E6) 1 2×9612 + 2(1+1) 22668

= 31 kNm The effect of introducing the joint stiffness to the above equation is to reduce the end moment on the beam (Mj,Rd ) from 51,8 to 31 kNm. If Mj,Ed = 20 kNm then Mj,Ed /Mj,Rd = 20/31 = 0,65 < 2/3 and the adoption of μ = 1 is satisfactory (cl 6.3.1(6), EN 1993-1-8 (2005)). Check the ultimate moment of resistance of the joint based on yielding of the top angle (Eq. (7.42a)) fy b ta2 Mjya,Rd = 2 4 mz 355 2 × 517,4 10 40 = 2 × 208,8 × × = 47,94 > (Mj,Rd = 31) kNm 4 1E6

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Chapter 7 / Structural Joints (EN 1993-1-8, 2005) which is the maximum moment of resistance for the joint. Check the ultimate moment of resistance of the two top bolts (2M-20 class 8.8) failing in tension (Eq. (7.42b)) 517,4 Mjub,Rd = 2Ab fu z = 2 × 245 × 800 × 1E6 = 202,8 > (Mjya,Rd = 47,94) kNm The effect of varying joint to beam stiffness can be shown for the simple structure (Fig. 7.37(b)) using Eq. (7.43). The ratio (Mj /(QL/8)), which includes the moment of resistance at the end of the beam, varies with the ratios of joint-to-beam stiffness (Sj /kb ) and column-to-beam stiffness(kc /kb ) as shown in Fig. 7.37(c). This relationship is related to the structure in Fig. 7.37(b) for a beam stiffness of 4E3 kNm. Elastic behaviour is assumed but plastic failure may limit the value of the joint resistance. For example the joint moment (Mj ) reduces the moment at mid-span for the beam, which is beneficial, but introduces a moment to the columns which if slender can reduce their capacity. However for more complicated structures an increase in connection stiffness reduces slenderness ratios and column deflections, which increase the load capacity of a column. Numerical investigations (Jones et al., 1981) show that including joint stiffnesses in the analysis of frames can reduce the weight of steel.

REFERENCES Bahia, C.S. and Martin, L.H. (1980). Bolt groups subject to torsion and shear, Proceedings of the I.C.E. Pt2, V69. Bahia, C.S. and Martin, L.H. (1981). Experiments on stressed and unstressed bolt groups subject to torsion and shear, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Bahia, C.S., Graham, J. and Martin, L.H. (1981). Experiments on rigid beam-to-column connections subject to shear and bending forces, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-1 to 3 (1993) and BSEN 10210-1 (1994). Specification for weldable structural steels. BSI. BS 5400 (2000). Steel concrete and composite bridges, Pt3 Code of practice for the design of steel bridges. BSI. BS 3692 (2001). ISO Metric precision hexagon bolts, screws and nuts. BSI. BS 4190 (2001). ISO Metric black hexagon bolts, screws and nuts. BSI. BSEN 499 (1995). Covered electrodes for the manual metal arc welding of carbon and carbon manganese steels. BSI. BSEN ISO 4320 (1998). Metal washers for general engineering purposes. BSI. BSEN 1011-1 (1998) and 2 (2001). Specification for the process of arc welding of carbon and carbon manganese steels. BSI. BSEN 14399-1 to 5 (2005). High strength friction grip bolts and associated nuts and washers for structural engineering: Pt1 General grade, Pt2 Higher grade bolts and nuts and general grade washers. BSI. Biggs, M.S.A.B., Crofts, M.R., Higgs, J.D., Martin, L.H. and Tzogius, A. (1981). Failure of fillet welded connections subject to static loading, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Chesson, Jr., E. Faustino, N.L. and Munse, W.H. (1965). High strength bolts subject to torsion and shear, A.S.C.E. (Structural Division), V91, ST5.

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Clarke, A. (1970). The strength of fillet welded connections. MSc Thesis, Imperial College, University of London. Clarke, P.J. (1971). Basis for the design of fillet welded joints under static loading, Conference Proceedings. Welding Institution, Improving Welding Design Paper 10, V1. Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving Problems in Structures Vols. 1 and 2. Longman Scientific and Technical. Davies, G. (1981). Estimating the strength of some welded lap joints formed from rectangular hollow section members, Conference Proceedings, Joints in Structural Steelwork, Teeside Polytechnic. Elzen, L.W.A. (1966). Welding beams in beam-to-column connections without the use of stiffening plates. Report 6-66-2. I.I.W. document XV-213-66. EN 1993-1-1 (2005). General rules and rules for buildings. BSI. EN 1993-1-8 (2005). Design of joints. BSI. European Convention for Structural Steelwork (1981). European recommendations for steel construction. Construction Press. Farrar, J.C.M. and Dolby, R.E. (1972). Lamellar tearing in welded steel fabrication. Welding Institute, Cambridge, England. Fisher, J.W. and Struik, J.H.A. (1974). Guide to Design Criteria for Bolted and Riveted Joints. John Wiley and Sons. Gourd, L.M. (1980) Principles of Welding Technology. Edward Arnold. Holmes, M. and Martin, L.H. (1983). Analysis and Design of Structural Connections. Ellis Horwood Ltd. Jones, S.W., Kirby, P.A. and Nethercot, D.A. (1981). Modelling of semi-rigid connection behaviour and its influence on steel column behaviour, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Kato, B. and Morita, K. (1974). Strength of transverse fillet welded joints. Welding Research. Ligtenberg, F.K. (1968). International test series, final report, Stevin Laboratory, Technological University of Delft, Doc XV-242-68. Mann, A.P. and Morris, L.J. (1979). Limit state design of extended plate connections, ASCE Journal of Structural Engineering, 105(ST3): 511–526. Martin, L.H. (1979). Methods for limit state design of triangular steel gusset plates, Building and Environment. Martin, L.H. and Robinson, S. (1981). Experiments to investigate parameters associated with the failure of gusset plates, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic. Maxwell, S.M., Jenkins, W.M. and Howlett, J.H. (1981). A theoretical approach to the analysis of connection behaviour. Conference Proceedings, Joints in Steelwork. Teeside Polytechnic. Morris, L.J. and Newsome, C.P. (1981). Bolted corner connection subjected to an out of balance moment – the behaviour of the web panel, Conference Proceedings, Joints in Structural Steelwork, Teeside Polytechnic. Owens, G.W., Driver, P.J. and Kriege, G.J. (1981). Punched holes in structural steelwork, Constructional Steel Research, 1(3). Pillinger, A.H. (1988). Structural steel work: a flexible approach to the design of simple construction, Structural Eng 66(19/4), October. Purkiss, J.A. and Croxton, P.C.L. (1981). Design of eccentric welded connections in rolled hollow sections, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Rolloos, A. (1969). The effective weld length of beam-to-column connections with stiffening plates, Final report, I.I.W. Document XV-276-69. Stamenkovic, A. and Sparrow, K.D. (1981). A review of existing methods for the determination of the static axial strength of welded T, Y, N, K, and X joints in circular hollow steel sections, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic. Stark, J.W.B. and Bijlaard, F.S.K. (1988). Design rules for beam-to-column connections in Europe, Steel Beam-to Column Building Connections. pub. Elsevier Applied Science.

Chapter

8 / Frames and Framing

The previous chapters have dealt with design of beams, columns and connections. This chapter, and the next, deal with the way the individual components are assembled together and also with design problems associated with the whole structure. In the first place it is necessary to give some consideration to how the choice is made of the structural form employed to carry the primary loading. For convenience this survey is divided into single and multi-storey structures.

8.1 SINGLE STOREY STRUCTURES T ypical examples of such structures include sports complexes, exhibition halls, factory units or assembly buildings. Unless architectural considerations prevail, the most economic solution will be obtained using one-way spanning structural systems rather than space frame structures. Systems which at first sight appear two-way spanning often comprise a number of overlaid one-way systems. Roof systems can be conveniently divided into flat and pitched roof systems.

8.1.1 Flat Roof Systems For spans up to around 15 m rolled sections form the most economic solution. However, it should be noted that the potential extra cost of beams over 12 m long needs to be taken into consideration, and the use of 12–15 m long beams should only be contemplated if a large number are required. At around 14 m and up to 20 m the use of castellated or beams with circular openings in the web become economic. Although this type of beam incurs high fabrication costs and requires a higher construction depth than ordinary rolled sections, the holes in the web allow services to be contained within the beam depth. Above 20 m it is usual to use parallel chord lattice trusses fabricated either from rolled hollow sections or from lightweight cold formed sections (top and bottom chords) and bar members for the web. The roof decking is then generally lightweight steel sheeting with suitable finishes and insulation, although timber decking with asphaltic waterproofing, woodwool slabs or pre-cast concrete decking units may be used. One problem with flat roofs is drainage and thus sufficient cross-fall must be provided to give adequate run-off and avoid local ponding. The cross-fall is

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generally provided by variable depth purlins and joists or by adjusting screed depths. To some extent these problems are alleviated by employing a sloping top chord to the roof system.

8.1.2 Pitched Roof Systems These fall into two main divisions: trusses with a sloping top chord and pitched roof portal frames. Modern trusses have a relatively low slope to the top chord. A pitch of between 4◦ and 10◦ is adequate to allow run off and to allow the joints in the roof sheeting to remain watertight. This type of truss is generally fabricated from square or rectangular rolled hollow sections with fully welded nodal connections, with circular hollow sections sometimes being used for the web members in Grade S355 steel, owing to availability, Pitches greater than 10◦ are only seen where an existing building with traditional large pitch trusses designed for tiles or slates as roofing materials is being extended, or where a large pitch is required for architectural reasons in, for example, shopping malls. The most common method of single storey construction is the pitched roof portal frame whether for factory units, small sports complexes or warehouses. This is basically due to the high speed and simplicity of construction. The internal bays are designed as rigid jointed frames. The end frames, unless there is likely to be an extension to the structure, are much lighter and have the rafters designed as spanning across the gable end posts which are also used to support the sheeting rails for the cladding. The most economic frame spacing is generally 7.5 m or 9.0 m for much higher frame spans. For spans below 20 m a frame spacing of 6.0 m may be adopted (Horridge, 1985; Horridge and Morris, 1986). Both truss systems and portal frames can be used for spans up to 60 m, although for spans over 30 m multi-bay structures become an option. Multi-bay construction requires internal columns which may reduce the flexibility of usage. However, this situation may be mitigated by the use of internal lattice girder support systems. Note that where the roof system comprises trusses the internal support system can be within the depth of the truss. This is not possible with portal frames, so unless reduced headroom is acceptable internal columns must be used. It should also be remembered that where multi-bay construction is used, there will be the need to supply valley guttering and associated drainage. Such valleys cause build-up in snow loading and can be potential areas of leakage and cause problems with access for maintenance.

8.2 MULTI-STOREY CONSTRUCTION 8.2.1 Multi-storey Steel Skeleton In UK practice the steel skeleton is generally designed to carry vertical loading due to the permanent and variable actions only, with the horizontal loading from wind and

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Chapter 8 / Frames and Framing the notional horizontal loading taken by a bracing system, or more commonly the lift shaft(s) and stair well(s). The economics of various types of multi-storey construction is discussed in Gibbons (1995). When using lift shafts or stairwells as bracing care must be taken in their layout as torsional effects from lateral loading on an asymmetric layout must be avoided.

8.2.2 Flooring Systems The flooring system is generally required to act as a horizontal diaphragm to carry horizontal forces from their point of application to the bracing or lift core. Thus it is essential that adequate lateral stiffness exists in the plane of the floor and the flooring system is adequately tied to the frame. Historically the floors of multi-storey steel frames have comprised cast in-situ normal weight concrete slabs. This system is now little used partly owing to excess weight of the concrete and partly owing to the slow construction time and the need for propping the deck often over two storeys. The problems with propping can be reduced by the use of proprietary falsework systems involving lightweight trusses which allow the actual formwork to be struck before the props are removed. The use of in-situ concrete floor systems except for small areas not otherwise able to be handled has been superseded by pre-cast pre-stressed concrete units or by composite steel–concrete decking. Pre-cast pre-stressed concrete units may be placed on the top flange of the steel beam, supported on shelf angles welded to the web of the beam, or placed between the flanges of a column section used as a beam. In all cases the flange of the supporting beam system may be considered to be restrained against lateral torsional buckling, although for shelf angle floors consideration may need to be given to the torsional load on the beam during construction. An alternative for ease of construction is to use asymmetric sections with the bottom flange larger than the top (Mullett, 1992; Lawson et al., 1997). These alternatives are illustrated in Fig. 8.1. For pre-cast units placed on the top flange small discrete vertical plates are welded to the top flange to give shear anchorage. For floor systems where the pre-cast unit sits on a bottom flange or shelf angle, the top corners of the units will need chamfering to ensure the units can be placed in position. Such floors also have the advantage of increased fire performance as the top flange is shielded by the concrete section from the effects of a fire on the underside. Equally the bottom flange of the steel section will be either partially or totally adjacent to a heat sink formed by the concrete slab, thus also inducing lower temperatures than would otherwise exist. For most types of flooring a screed is required to give a proper finish and it is recommended that a light structural mesh reinforcement is placed over the support beams to help control cracking in the screed. One potential disadvantage of using pre-cast units is the large amount of hook time required to place the units which could otherwise have been used to hoist materials required for finishes, etc.

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Pre-cast units

Shelf angles (a) Shelf angle floors (screed omitted)

(b) Pre-cast units resting on bottom flange of column section (screed omitted)

in-situ concrete

Deep trough decking (c) Asymmetric section with deep trough decking

FIGURE 8.1 Types of Flooring

Composite steel–concrete floors use thin gauge trough steel sheeting which initially acts as permanent shuttering to the concrete before acting compositely after the concrete has set to provide both tensile and shear ‘reinforcement’ to the finished concrete slab. The concrete used is lightweight concrete with a specific weight of around 18–20 kN/m3 , and designed to be placed using concrete pumps rather than skips. The only cranage required is that to lift the bundled steel sheets to the correct level, as the sheets are individually light enough to be manhandled. To provide shear continuity between the deck and the steel support beams in-situ through deck shear stud welding is employed. Reinforcement will be required in the form of mesh in the top face of the slab in areas of hogging moments if only to control the effects of cracking. This mesh will also contribute to the fire performance of the deck. Any propping can be avoided by limiting the spans of the decking whilst acting non-compositely by additional support beams. This option is more cost effective than propping. A recent development is to use deep decking over the whole depth of a UC used as a beam with the decking supported on flange plates welded to the lower flange of the UC or specially rolled asymmetric sections. This system is known as ‘Slimflor’ construction (Mullett and Lawson, 1993), in which the outer beams are downstand UB’s or rolled hollow sections with plates welded to the soffit (Mullett, 1997). A very useful practical

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Chapter 8 / Frames and Framing guide for an overview of design and construction is published by the SCI (Couchman et al., 2000).

8.3 INFLUENCE OF CONNECTION DESIGN AND DETAILING From the previous chapter it will be noticed that connections are either designed to take the effect of beam reactions in the form of shear (with nominal moments) or to resist the effects of moments, and the coincident shear and axial force. The first type is generally designated a ‘pin’ joint which allows large relative rotations between the members in the connection (and leads to the concept of simple construction). The second is generally designated a ‘fixed’ or ‘rigid’ joint in which rotational compatibility exists between the members framing into the connection. This type of connection allows full moment transfer and should ideally be welded, although in UK practice a heavily bolted stiff connection is taken as rigid. Obviously the type of connection in the structure will markedly influence the behaviour of the structure under whatever actions are applied to it. A fuller discussion of the implications of the above paragraph is given in Section 8.12, but it is first necessary to consider the actions applied to a structure.

8.4 STRUCTURAL ACTIONS These may either be physical loading or imposed deformations due, for example, to differential settlement.

8.4.1 Physical Loading For convenience this is divided into two categories: gravity and non-gravity loading.

8.4.1.1 Gravity Loading This covers the self-weight of the structure, the finishes on the structure, the actions due to the usage of the structure (variable actions) and roof loading whether as a nominal load or as snow loading (uniform or drifting as appropriate). Values of loads are given in EN 1991-1-1. It should however be noted that snow drift loading given in EN 1991-1-3 constitutes an accidental load case and therefore takes lower partial safety factors, and thus may not be critical. It is advisable to design the structure under uniform roof loading and then check the structure, if appropriate, under non-uniform snow drift loads.

8.4.1.2 Non-gravity Loading This can be considered under a series of sub-headings: • Wind loading: This is covered by EN 1991-1-4. • Inertia and impact loads: These have to be considered where dynamic loading is considered, for example cranes and supporting systems (EN 1991-3). Impact loading

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needs considering in the case of car parks where columns can be damaged due to collisions or for bridge piers where vehicular impact may occur (EN 1991-1-7). • Seismic loads: These are not of general importance in the UK except for nuclear installations and other similar structures, and are covered in EN 1998. • Accidental loads: These can be due to snow drift loading (EN 1991-1-3), fire (EN 1991-1-2) or explosions (EN 1991-1-7). In the case of explosions either the implications of progressive collapse needs to be considered or the structure must be tied together and the resultant tying forces considered.

8.4.2 Deformations Structural deformations can be either due to differential settlement (Section 8.11.1) or thermal movements (EN 1991-1-5) (Section 8.11.2).

8.4.3 Load Combinations Load combinations for single variable loads have been covered in Chapter 3, and it therefore remains to cover load combinations for multiple variable loads. The classic example of this is where there are variable loads due to structure usage (e.g. office loading) and variable loads due to wind. The basis behind the combination rules in EN 1990 is that it is deemed statistically unlikely that all the variable loads will be acting at their maximum intensity at the same time. It should be noted that roof loading and floor loading taken for the structure as a whole are considered as separate variable actions. The rules take account of the non-simultaneity of maximum effects by introducing ψ factors on the non-principal variable actions. EN 1990 allows a number of combination rules – this is reflected by the UK National Annexe to EN 1990. At ultimate limit state, the following combination rules are available: 1,35Gk + 1,5Qk,1 +

n

1,5ψ0, j Qk, j

(8.1)

2

where Gk is the permanent load, Qk,1 is the principal (or leading) variable load and Qk,2 to Qk,n are the accompanying variable actions. Note however that where the permanent load is unfavourable (i.e. is of opposite sign to the permanent load, the partial safety factor γG is set equal to 1,0, and where the variable loading is unfavourable its partial safety factor γQ is taken as zero. Equation (8.1) must be applied taking each variable load as the principal variable load in turn and the remainder as accompanying actions. Experience, however, may be used in reducing the calculations when it is clear which leading action is critical. Note it is possible for different types of actions to have different values of ψ0, j . 1,35Gk + 1,5ψ0,1 Qk,1 +

n 2

1,5ψ0, j Qk, j

(8.2)

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Chapter 8 / Frames and Framing It is necessary with this combination to examine the effects of different ψ0, j values, although it is likely to give lower loads than those determined from Eq. (8.1). 1,35ξGk + 1,5Qk,1 +

n

1,5ψ0, j Qk, j

(8.3)

2

The additional factor on the permanent load ξ is subject to the limit 0.85 ≤ ξ ≤ 1.0. The UK National Annexe specifies a value of 0,925. Equation (8.3) will give the lowest total load provided the permanent load is unfavourable. It would appear permissible to use the ξ factor where there is only a single variable load. For serviceability loading the combination rules are similar except the maximum values of γG and γQ are set equal to one as is ξ. The other change is that ψ0 may be replaced by ψ1 or ψ2 . The factor ψ1 is used to determine effects under reversible limits states (known as the frequent combination) and ψ2 is used to determine long-term effects or the appearance of the structure (known as quasi-permanent). The use of ψ0 is only required for irreversible limit states. Deflection is generally checked under the quasi-permanent combination although it would be advisable to check it under full variable load together with permanent load in order to assess any maximum instantaneous deflection. It is now necessary to consider how structures resist the forces due to actions applied to the structure. The major concern is with the transfer of horizontally applied forces, although some consideration is given to the distribution to supporting members of vertically applied forces. Single storey structures will be considered in detail, before continuing by looking at multi-storey structures.

8.5 SINGLE STOREY STRUCTURES UNDER HORIZONTAL LOADING Consider initially a basic structure comprising a single bay flat roof portal type frame with encastré feet under vertical loading only (Fig. 8.2). It does not at first sight matter whether the connections at B and C are rigid or pinned, as the structure can be analysed and the members designed under the resultant forces. However if the joints at B and C are pinned the rafter will be a heavier section as no ‘fixing’ moment will exist at the connections, and there will be an increased deflection at the centre of the rafter. The column section will be lighter and the connection detail simplified. However, the bottom ends of the column are fixed with respect to the bases. This condition will cause problems for the foundation design (Section 8.11.1) as the ratio of moment to axial load will be too high to give an economic design. The general custom on such a frame is therefore to use non-moment resisting connections between the feet of the stanchion and foundations, thus if non-moment resisting connections are used between the rafter and the columns the frame is inherently unstable. This

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Vertical loading B

C Rigid joint

A

D

(a) Basic frame with rigid joints at B and C B

C Pins

A

D

(b) Basic frame with pinned joints at B and C

FIGURE 8.2 Simple single bay flat roof portal

statement is clearly true for horizontal loading, but is also true for vertical loading due to inherent imperfections in both the members and through construction. Two solutions, not considering moment resistant connections at the feet of the columns, are possible. The first is to restore the moment connections at B and C; the second is to find an alternative method to resist any horizontal forces. The first method has the drawback that for single bay frames the total moments at the connections may reverse in sign due to the change in direction of the application of wind forces. The moment due to the wind force alone must change sign (Fig. 8.3). Allowing the connections to resist wind forces is possible in multi-bay structures where the wind force is adsorbed through a large number of connections, thus reducing the moment on each individual connection (see Section 8.12). The second method is to adopt a bracing system within the frame. So for the wind force P (Fig. 8.4(a)), a diagonal member is placed such that the diagonal member is in tension (i.e. capable of taking full design load). This has the effect of triangulating the structure. When the wind blows from the reverse direction (force P in Fig. 8.4(b)), the tie AC now becomes a strut with much reduced load capacity owing to buckling. Thus a further tie BD is now inserted giving rise to cross-bracing. The cross-bracing is designed by only considering the forces in tension members (i.e. the compression members are assumed to have zero load capacity, a conservative assumption). It should also be noted that since the structure is triangulated the frame analysis (and design) is much simplified.

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P (Wind force)

A

D (a) Wind from left to right

B

C P′ (Wind force)

A

D

FIGURE 8.3 Effect of wind reversal

(b) Wind from right to left B

C

P (Wind force)

Tie to take wind force P

A

D

(a) Triangulation to take effect of wind force P B

C P′ (Wind force)

Tie to take wind force P ′

A

D

Note: AC now acts as a strut, and for calculating the force in BD is assumed to have zero strength (b) Triangulation to take effect of wind force P ′

FIGURE 8.4 Elementary wind bracing

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Eaves beam

(a) Single storey multi-bay frame – each frame braced

(b) Single storey multi-bay frame with wind girder

FIGURE 8.5 Wind girder bracing

No structure exists as a single frame. Where a set of such frames described above, complete with cross-bracing, to be assembled as multi-bay single span structure, it would be unusable (Fig. 8.5(a)). This state of affairs can be made acceptable by retaining the cross-bracing in each of the end frames only and by supplying full diagonal crossbracing between each of the frames at rafter level (Fig. 8.5(b)). Such a bracing system is known as a wind girder, and transmits any horizontal forces through the girder to

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Beam

(a) Section

Intermittent steel flats welded to top flange

(b) Elevation (pre-cast units omitted)

FIGURE 8.6 Shear key detail

the bracing in the end frames. A wind girder is always needed where lightweight roofing systems are used in conjunction with light gauge purlin systems. Where stressed skin construction is used the wind girder can be omitted as the sheeting is designed to resist the shear due to the horizontal loadings (Davies and Bryan, 1982). If pre-cast concrete units are used then these may replace the wind girder provided adequate shear connection between the rafters and the units is provided (Fig. 8.6). Composite decking with through deck shear studs will not generally need additional bracing (but see Section 8.9.3). For multi-bay multi-span structures a wind girder is needed around the periphery of the structure and full diagonal bracing in each of the corners of the structure (Fig. 8.7). For portal frame systems, the wind forces are taken by moment resisting connections in the plane of the frame, but for wind applied along the structure, i.e., normal to the plane of the portals, wind bracing must be provided. It may also be necessary to brace the end or gable frame for wind in the plane of the frame if it is designed as rafters spanning over intermediate gable end columns. Although this section has referred to discrete bracing it is possible to replace the bracing by alternatives. The most common replacement for vertical bracing is masonry shear wall construction. For single storey construction this need be no more than conventional masonry cladding provided such cladding is fully ties to the steel frame using, for example, half wall ties spot welded to the stanchions or approved proprietary tying systems. The design of such cladding is to EN 1996-1-1. It should however be remembered that in the case of masonry cladding

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Key : Main members; Wind bracing Note : All corners must be fully braced

FIGURE 8.7 Schematic layout of wind bracing for a multi-bay structure the frame is erected first and that the frame will need temporary bracing during this stage. It is the designer’s responsibility to ensure this is present (Construction ‘Design and Management’ Regulations) (HMSO, 1994).

8.6 MULTI-STOREY CONSTRUCTION The situation with respect to vertical and horizontal planes can be discussed separately.

8.6.1 Vertical Plane The effect of horizontal forces in the vertical plane can be handled in a number of ways: • Bracing The structure can be braced either with the bracing left as an external feature and hidden behind lightweight cladding. Such bracing can be at each corner of the structure, or provided the floors act as stiff diaphragms, in the centre of each face of the structure. The latter layout is only really suitable for structures symmetric about their centre lines as otherwise torsional effects are introduced which cannot easily be resisted. • Shear walls This solution is generally adopted for the smaller end panels of structures which have a high aspect ratio with few columns in the smaller direction. The shear walls in order to act properly must be relatively unpierced, i.e., free from dominant or significant openings. This can produce architectural constraints. Irwin (1984) provides

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Chapter 8 / Frames and Framing information on design of shear walls. In the longitudinal direction the wind force is lower (as it acts on the smaller face) and there are a large number of bays to take the wind forces on the columns and the column–beam connections. • Cores In the UK, the most common solution is to allow the wind forces to be adsorbed by the core(s) provided for the lift shaft(s) and/or stairwell(s). Such cores are generally of reinforced concrete construction, although the walls surrounding stairwells may be masonry. The walls surrounding lifts or stairwells do not have large openings as they generally provide access to fire escape routes and protected access areas for fire fighting, and thus stiff enough to provide the lateral restraint to the structure necessary. It should be noted that tolerance problems can arise when marrying up steelwork and reinforced concrete construction, and that shrinkage and creep effects in the concrete should not be ignored (Irwin, 1984). • Rigid/semi-rigid construction The case where the frame is taken as rigid for both the vertical loading and horizontal loading is rare in the UK. However, a hybrid method of design is becoming popular in which the frame is consider pin jointed (simple construction) for the vertical loading but as rigid jointed for the horizontal loading. There are restrictions on this method mostly related to the size and shape of the structure (Salter et al., 1999).

8.6.2 Horizontal Plane Composite steel–concrete flooring systems will, when the concrete has hardened, prove a very stiff diaphragm to transmit horizontal forces provided adequate shear connection between the deck and the steel skeleton is available. The shear stud requirement to provide flexural composite action will generally be adequate. When pre-cast concrete units are used a shear key detail such as that illustrated in Fig. 8.6 should be used, and structural mesh provided in the screed in order to give full diaphragm action and prevent cracking over the beam.

8.7 BEHAVIOUR UNDER ACCIDENTAL EFFECTS Accidental actions should be considered in a number of possible circumstances: • Explosions whether due to gas or terrorism. • Impact due to vehicles or aircraft. Should the risk of such an incident be high and the effects be catastrophic, or in certain circumstances the need to check be mandatory (e.g. vehicular impact on bridge piers), then the designer must ensure that the structure is designed and detailed to ensure that should an accidental situation occur, the structure does not suffer complete or partial collapse from either the accidental situation itself or subsequent events, such as spread of fire.

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Four now classic cases where collapse occurred due to accidental or terrorist action are: (1) Ronan Point block of flats where a gas explosion blew out a wall panel causing progressive collapse of a corner of the structure (Wearne, 1999). (2) The Alfred P. Murrah Building in Oklahoma City where collapse was caused by a terrorist bomb blast (Wearne, 1999). (3) World Trade Centre Towers in New York on 11 September 2001 due to impact from deliberate low flying aircraft and subsequent fire spread. The towers were steel framed structures and it has since been demonstrated that the prime cause of collapse was due to loss of fire protection on the floor members, as it was considered that the fire load from the impacting aircraft was not unduly high (Dowling, 2005). Information on considerations of the design of high rise construction is given in ISE (2002). (4) Pentagon Building on 11 September 2001 following the deliberate crash of a commercial airline. The concrete structure suffered partial but not extensive collapse due to the impact and subsequent fire (Mlakar et al., 2003).

8.7.1 Progressive Collapse EN 1990 identifies the need to consider that in the event of an accident such as explosion or fire the structure should not exhibit disproportionate damage. This is reinforced by the requirements of the relevant Building Regulations within the UK, for example the recently revised Approved Document A of the England and Wales Building Regulations. Such damage can be mitigated by: • Attempting to reduce or limit the hazard In the case of fire this could be done by full consideration being given to the use of non-flammable materials within the structure and by the provisional of relevant active fire protection measures such as sprinklers. In the case of industrial processes where explosions are a risk then such potential risks need to be taken into account by, say, enclosing the process in blast proof enclosures. • Maintenance The recent Pipers Row Car Park collapse of a lift slab concrete structure (Wood, 2003) indicated the need for adequate design especially where punching shear in a concrete slab may be critical as this is a quasi-brittle failure. The need for adequate inspection and maintenance where it is known that environmental effects will be severe was also highlighted. Although the failure was due to poor maintenance, it was exacerbated by uneven reaction distribution at the slab–column interface and no tying through the lower face of the slab. Admittedly the construction technique used would have made the latter extremely difficult although it contributed to the former.

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Chapter 8 / Frames and Framing • Consideration of the structural form A case where this is relevant is when the structure could be subject to externally provoked explosions. One reason why the damage was extensive following the Oklahoma City explosion was that the blast was amplified by an overhanging portion of the structure above ground floor level. A solution for structures known to be at potential risk is to provide an external curtain which is not structural and is easily blown out whilst ensuring the structure is stabilized by a core which is in the centre of the building. Additionally the risk can be mitigated by ensuring vehicles cannot come within certain limiting distances of the structure. More information on this, including assessment of blast forces is given in a SCI publication (Yandzio and Gough, 1999). • Ensuring the structure is adequately tied to resist collapse This needs considering on two levels. The first is to determine the magnitudes of the likely levels of tying force required. The second is to provide properly detailed connections between elements of the structure. This is needed for both connections in the horizontal plane: beam-to beam or beam to column and in the vertical plane: column splices. • Removal of key elements to establish stability In this approach a risk analysis is carried out to identify the key elements in a structure which are vital for its stability. A structural analysis is carried out with any one of these elements removed to establish the stability of the remaining structure and its ability to carry the loading so induced. As this is an accidental limit state lower partial safety factors are used on both the loading and the material strengths determining the member strengths. It is not necessary to check any serviceability limit states. Approved Document A divides structures into four categories: 1, 2A, 2B and 3, with Category 1 being the least onerous. Category 1 constitutes low-rise housing, agricultural buildings and buildings with restricted access. These require no specific checks. For Category 2A which are medium consequence risk structures, the consideration of tying forces is adequate in most cases. For Category 2B defined as high consequence risk structures, then the requirement must be satisfied by using removal of key elements. For Category 3 structures which are very high consequence risk structures such as large capacity grandstands, hospitals, structures over 15 storeys or structures in which hazardous materials are involved require special consideration (Way, 2005). Alexander (2004) gives examples on risk assessment with the hazards grouped into two categories: those that must be considered for any Class 3 structure and those that only need considering for specific location dependent structure such as flooding.

8.7.2 Structure Stability Any structure with high lateral loading or where vertical loading can be applied outside the frame envelope must be checked for overturning. Also any continuous member with a cantilever must be checked for the possibility of uplift on any support.

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6 m GK = 1500 kN

wK = 6 kN/m

35 m gK = 32 kN/m

1.5 m

A

FIGURE 8.8 Design data for Example 8.1

14 m

In all these cases the loading which causes overturning is deemed to be unfavourable, and the loading tending to restore the situation is known as favourable. This is defined in EN 1990 as verification of static equilibrium EQU. For this situation the partial safety factors corresponding to EQU in EN 1990 (Table A1.2 A) must be used.

EXAMPLE 8.1 High wind load. Consider the possibility of overturning for the water tower shown in Fig. 8.8. Taking moments about point A: Unfavourable effects due to wind:

35 1,5 × 120(3 + 35 + 1,5) + 1,5 × 6 × 35 + 1,5 = 13095 kNm 2

Favourable effects due to self-weight: 0,9 × 1500

14 14 + 0,9 × 32 × 35 = 16506 kNm 2 2

The moment due to the favourable effects exceeds that due to the unfavourable effect, thus the structure will not overturn.

EXAMPLE 8.2 External loading. Consider the cantilevering frame shown in Fig. 8.9. Here the check is more complex, as the permanent load of 220 kN per storey is both favourable and unfavourable, in that the load acting on the cantilever is unfavourable as it contributes to the overturning effect. The permanent load per storey may be expressed as 27,5 kN/m run. The favourable part of the permanent loading on the internal span takes a factor of 0,9, and that on the cantilever of 1,1.

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[email protected] 3.5 m GK = 220 kN/storey

wk = 13 kN/m

7m

B 1m

FIGURE 8.9 Design data for Example 8.2

Taking moments about B: Effect due to the wind loading (unfavourable): 1,5 × 13

17,52 = 2986 kNm 2

Effect due to permanent action (per storey): 0,9 × 27,5

72 12 − 1,1 × 27,5 = 591 kNm 2 2

Total restoring (or favourable) moment: 5 × 591 = 2955 kNm The structure will just overturn as the disturbing moment is marginally greater than the restoring moment. Thus either the structure needs tying down using, say, tension piles or ground anchors, or the permanent loading could be marginally increased. The latter is the cheaper option owing to the small margin involved.

8.8 TRANSMISSION OF LOADING 8.8.1 Transmission of Loading from Flooring Systems When the loading is considered as a UDL, then the distribution of such loading to any supporting beam system will depend whether the decking system may be considered as one or two way spanning. Deck systems comprising lightweight proprietary systems (roofing only), pre-cast concrete slabs, composite steel–concrete decks and timber are all one way spanning. Thus half the load is taken to either end of the pre-cast unit, purlin system, joists or composite deck (in the direction of the troughs) (Fig. 8.10). Where a decking system is continuous over the supporting beam system then the reactions should be determined using a suitable analysis, although for approximate design continuity may be ignored. Where in-situ concrete is used then loading is distributed to

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Concrete slab with profiled sheet steel decking Main beam Decking Load on main Reaction (UDL) beam

Reaction from decking (UDL) Joist or purlin

Intermediate beam reactions

Reaction from joist

Reaction from joist

Main beams Note: The loading on the main beam will be appiled through web cleats (not shown) (a) Transfer of load from one way spanning concrete deck

(b) Load transfer from decking through a joist or purlin system

FIGURE 8.10 Load transfer for one way spanning system

A

B

45°

E

D

F

C

Note: (a) Load on area AED is supported by beam AD (b) Load on area AEFB is supported by beam AB

FIGURE 8.11 Approximate load transfer from two way spanning slabs

all the supporting beams. The loads on individual beams should be determined using the same assumptions as in the slab design, such as Johansen’s yield line method or the Hillerborg strip approach. However, for approximate calculations a 45◦ dispersion from each corner of the slab may be used to partition the loading to the beams (Fig. 8.11).

8.8.2 Lintels In the absence of other data a 60◦ dispersion of the load above the lintel may be used, that is, the lintel only carries the loading within the complete 60◦ equilateral triangle. Where the triangle is incomplete due to openings then the whole load above should be taken.

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8.9 DESIGN OF BRACING 8.9.1 Permanent Bracing The forces to be used to design such bracing are given in cl 5.3: (a) any horizontal loads applied to the frame being braced, (b) any horizontal or vertical loads applied directly to the bracing system (c) the effects of initial imperfections, or the equivalent horizontal forces, from the bracing system itself and from all the frames it braces. For wind bracing the forces in (a) and (b) will need considering, whereas for sway bracing in the vertical plane it is a combination of all three. The equivalent horizontal forces for sway are covered in Section 8.12.2. There is an additional requirement for stability forces at the splice in a column or beam which is covered in the next section. The design of diagonal bracing is generally simplified by ignoring diagonal members in compression and assuming all joints are pinned. In the case of wind bracing the member should not be slender enough to induce an acceptable degree of sag under its own self-weight.

8.9.2 Restraint Bracing to Compression Flanges and Column Splices • Compression flange bracing The problem can either be handled by adding an additional bow in the members to be restrained and designing for the resultant additional moments, or using an equivalent stabilizing force. In the first case the initial bow imperfection e0 is given by L 500

e0 = αm

(8.4)

where L is the span of the bracing system and αm is given by αm =

1 0,5 1 + m

(8.5)

where m is the number of members to be restrained. For convenience the effective bow imperfection in the members to be restrained by a bracing system may be replaced by the equivalent stabilizing force qd is given by qd =

NEd 8

e0 + δ q L2

(8.6)

where δq is the in-plane deflection due to the load q and any external loads calculated from a first order analysis. If second order analysis is used then δq = 0.

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The notional force NEd in the flange is defined as MEd /h where MEd is the maximum moment in the flange and h is the overall depth of the beam. • Restraint to splices An additional requirement at splices is that the bracing should be able to resist a local force of αm NEd /100 together with any other applied forces. Note that the application of this local force is not co-existent with the force q defined above.

8.9.3 Temporary or Erection Bracing Under the Temporary Workplaces Directive enforced in the UK by the Construction (Design and Management) Regulations, the client must appoint a planning supervisor at the Design stage on all health and safety matters throughout the execution of the project. It thus falls to the supervisor to ensure the structure is capable of being erected safely and that the requisite safety measures are in force. This means that the specification and design of temporary bracing should not be left to the steelwork erector without approval by the planning supervisor. The necessity for such bracing is often due to the intended construction sequence where construction starts at the centre of a structure in order to reduce cumulative tolerance errors whereas the permanent bracing designed to give stability to the completed structure is in the end bays. Further reasons are that support conditions or the effect of temporary loadings may be such that tension members are in compression or the signs of bending moments are reduced. There is also the requirement to ensure that any forces applied to the structure from safety equipment used during erection are considered. Where in-situ concrete is being used the compression flange will not receive full restraint until the concrete is hardened, and thus during construction the compression flange of the support beam system is unrestrained and therefore prone to lateral torsional buckling. Where profiled sheet steel decking is being used provided through deck welded shear studs are used, the beams running normal to the deck profiles can be considered as fully restrained, but those running parallel with the profiles will need checking for buckling though the deck will provide some degree of restraint (Lawson and Nethercot, 1985). An additional point to be noted is that large members may not be stable when being lifted into position partly due to the change in support conditions and partly due to the changes in restraints. Such members may need to be braced together at the correct spacing and lifted in pairs. Such bracing should if possible form part of the permanent bracing.

8.10 FIRE PERFORMANCE It is recommended that the reader is referred to Ham et al. (1999) for an overview of the basic procedures and of methods of protection. Whilst it is not intended to cover detail fire design of steel structures, a brief overview is necessary. For full details reference should be made, for example, to Purkiss (2007).

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8.10.1 Single Storey Structures For most single storey structures the situation with respect to fire performance is relatively straightforward as there is generally no specific requirement within the England and Wales Building Regulations (Approved Document B) (DTLR, 2000). The only problem tends to be caused by outward collapse of walls and columns of single storey frames. This tends to be worse with pitched roof portals (see Simms and Newman, 2002). In addition there is the problem of ensuring that fire fighters have complete safe access to fight the fire during its complete duration.

8.10.2 Multi-storey Structures Traditionally design has been by considering single elements with no interaction between such elements. However, fire tests on the steel frame structure at Cardington have demonstrated that with structural interaction between beams and columns it is possible to leave all beams unprotected (except at connections) and achieve a very good fire performance. Interim design data allowing some beams within floor systems being left unprotected has been published by the SCI (Newman et al., 2006). It is also possible to encase columns in brickwork or blockwork or to fill rolled hollow section columns with concrete to achieve fire performance without additional measures (Bailey et al., 1999). Although a complete discussion of the Cardington tests is given elsewhere (Purkiss, 2007), an overview of the tests will be presented. The tests at Cardington demonstrated that temperatures in the unprotected steel beams which were composite with the floor slab reached temperatures on the bottom flange in excess of 1000◦ C. This is around 400◦ C higher than the limiting temperatures allowed after correcting for load level in design codes. It should be noted that the loading applied on the floors was around onethird of the ambient design value of 2,5 kPa giving a load ratio lower than that of most office type structures. Although the structure retained its load carrying capacity the deflections were extremely high with values of up to 640 mm being recorded. Although the tests showed far better performance than would be expected from the standard furnace test on individual elements, a number of points need to be raised: (1) The high deflections reached would probably mean part at least of the structure would need replacing. (2) Early tests in the series were performed with unprotected columns which suffered severe buckling just below the beam columns connections. (3) Deflections of some beams were more than sufficient to cause internal lightweight compartment walls to fail leading to loss of compartmentation and thus increased fire spread (Bailey, 2004). (4) Buckling in the lower flanges of the beams occurred at the ends owing to highinduced compressive forces as the moments were redistributed away from midspan. This has led to the recommendation that design allowing for increased

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moment capacity at the connection (Lawson, 1990a, b) should not be used where the beam is restrained against lateral movement during a fire (Bailey et al., 1999). (5) During the cooling phase the beams do not recover plastic deformations and potentially thus place the connections in tension. This in fact resulted in the partial failure of some connections due to shear in the bolts, excessive flexure of end plates or failure in the welds connecting the end plates to the beams (Bailey, 2004). Design methods have, however, been introduced which allow for some secondary beams in composite construction to be left unprotected (Newman et al., 2006) and which allow a better analysis of composite floor slab action under the compressive membrane force system induced in the fire limit state (Bailey (2001, 2003)). Even with the traditional approach shelf angle beams will give 30 min standard fire performance and may give 60 min (Newman, 1993). Slim floor construction should give 60 min with no additional fire protection. Under no circumstances should columns be left unprotected. For fire performance periods of 30 min infilling the web space of a UC with non-structural blockwork can be sufficient (Newman, 1992).

8.11 ADDITIONAL DESIGN CONSTRAINTS 8.11.1 Ground Conditions and Foundations This section is intended to be neither a comprehensive survey of foundation design or construction nor of problems associated with ground conditions. It is offered as an aidememoir and covers the implications of any problematic areas. For a full consideration, the reader is referred to Henry (1986). Also the design of concrete foundations is covered in Martin and Purkiss (2006). The selection of foundation type generally depends on the potential bearing capacity of the ground and its susceptibility to absolute or differential settlement. The most economic form of foundation is the simple pad foundation which in order to be of a reasonable size requires a reasonably high bearing capacity. If bearing capacities are low, and the loads applied to the foundations are high, the individual pad foundations will start to overlap and will become either combined foundations or in the extreme case raft foundations. Where the bearing capacity is low then either piled foundations, ground stabilization, or the use of replacement imported fill should be considered. The decision will be primarily based on cost, although environmental considerations such as noise, dust and extra traffic must be taken into account. Ideally pad or raft foundations should be designed for uniform bearing pressure at ultimate limit state under imposed loading other than wind or accidental actions. This will generally be possible for raft foundations, but may not be for pad foundations. It is recommended that for pad foundations the ratio of the maximum to minimum bearing pressures should not exceed around 1,5. In order to achieve this it is possible to offset any stanchion away from the centre line of the foundation except where severe

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Chapter 8 / Frames and Framing moment reversals may occur. It also leads to the adoption of nominally pinned feet where the axial load in single storey structures is low as the required ratio cannot be satisfied with fixed feet. Uplift should never be allowed below a foundation under normal conditions. It should be noted that it may be necessary, in practice, to supply a degree of fixity to bases of columns in single bay portal frames when considering such frames at site boundaries in the case of fire and thereby to accept a high ratio of maximum to minimum bearing pressures, even uplift, as this is an accidental situation. Any horizontal reaction at the base of a column will also apply a moment to the underside of the foundation and must be included. If the structure is tied together by ground beams then only the net horizontal force need be considered, otherwise each reaction must be considered independently. For multi-storey structures it is general practice to employ tower cranes to facilitate erection. These need substantial foundations which are often incorporated into the structure as the foundations to the lift shafts. Differential settlement may either be caused by non-uniform loading imposed by the structure on its foundations or by variable ground conditions beneath the foundations. The effect of non-uniform loading can be mitigated by placing the whole structure on a single raft foundation, and thus producing sensibly uniform bearing pressures. Also to avoid differential movement between any masonry cladding and the frame, the masonry can be supported on ground beams tied to the pad foundations below the columns. Alternatively the components of the structure can be allowed to settle independently by the provision of movement joints, in which case the cladding may not be considered to act as bracing and full wind bracing will need to be designed. Where the structure is such that differential settlement cannot be avoided then either the structure must be designed to accept these movements (note, differential settlement does not affect moments when determined using plastic analysis; they do when an elastic analysis is used), or the structure can be articulated such that the movements do not affect the internal forces by the provision of hinges, often making the structure iso-static. Typically such hinges should be at the points of contraflexure, and must be checked for rotational capacity expected through such movements (e.g. Fig. 8.12). This type of articulation is frequently used on bridge decks where there may be substantial differential settlement due often to mining subsidence. Often in this case where such subsidence could be large, provision is built in to jack up the structure, in which case a realistic estimate will be needed of such movements.

8.11.2 Expansion and Contraction Joints It can be a matter of some debate on whether these should be provided and if so at what spacing. It is suggested that masonry should have movement joints at around 7 m centres (or the frame spacing), and concrete floor slabs at around 20–30 m (Deacon, 1986; Bussell and Cather, 1995; Concrete Society, 2003). This suggests that continuous multi-bay frames should have movement joints at around the same spacing. Thus either

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UDL

Note : BMD is drawn on the tension face on the beam

FIGURE 8.12 Cantilever-suspended span structure

movement joints should be provided through the whole height of the structure, in which case full wind bracing needs providing in each portion of the structure, or the structure will need designing to take the actions caused by such movements. For information on movement joints see Alexander and Lawson (1981).

8.11.3 Stability Stability can be considered at two levels: member stability and frame stability. • Member stability This effectively ensures that there will be sufficient ductility in the member to ensure that plastic rotational capacity is available when required by the design/analysis synthesis. The ductility check is made by considering limiting values of flange and web slenderness. It is essential that where plastic analysis is used Class 1 sections are mandatory, although where only a single hinge is needed for collapse, as in a simply supported beam, Class 2 sections may be used. It should be noted that virtually all UKBs are Class 1 whether Grade S255 or S335. The assumption is made in plastic analysis that the member can achieve its full plastic moment capacity before the onset of elasto-plastic buckling. For members in simple construction or isolated beam elements, this condition is not necessary as the load carrying capacity of the member can be checked using reduced strengths which allow for buckling. For rigid jointed frames premature elasto-plastic buckling is not permissible and must be counteracted either by bracing which reduces effective, or system, lengths below critical values or by increasing member sizes.

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Chapter 8 / Frames and Framing • Frame stability For portal frame systems this takes the form of two checks. The first is the determination of vertical deflections at the ridges and horizontal deflections at the top of the stanchions at the eaves. The second check is only for multi-span portals and is designed to avoid ‘snap through’ of the rafters. For multi-storey structures both the relative lateral deflections on each storey and the overall lateral deflection of the whole structure is subject to limits.

8.12 DESIGN PHILOSOPHIES The basic analysis methods allowed are elastic or plastic (the use of elasto-plastic methods may be needed where sway is important). Traditionally the UK has adopted simple framing where the frame carries the vertical loading and, the bracing system, the horizontal loading. Continuous framing (except for portal frames) and semi-continuous framing are rarer.

8.12.1 Frame Classification A frame may be classified as sway or non-sway. A non-sway frame is one where horizontal deformations are limited by the provision of substantial bracing members often in the form of lift shafts or stairwells often as part of a central core. A frame with bracing may still be classified as a sway frame. The criteria for neglecting global second order effects are for • Elastic analysis αcr =

Fcr ≥ 10,0 FEd

(8.7)

• Plastic analysis αcr =

Fcr ≥ 15,0 FEd

(8.8)

where αcr is the ratio by which the design loading would have to be increased to cause global elastic instability, FEd is the design load on the structure and Fcr is the elastic critical global buckling load. The higher limit for plastic analysis is due to the issue that imperfections become more important in plastic analysis due to P– δ effects. For shallow pitch portal frames (roof slope less than 26◦ ) or beam and column type structures αcr is given by αcr =

HEd h VEd δh,Ed

(8.9)

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where HEd is the total horizontal reaction at the bottom of a storey (including notional loads), VEd is total vertical loading at the bottom of a storey, δh,Ed is the horizontal displacement due to horizontal loading including notional horizontal loads and h is the storey height. The value of αcr may also be determined by applying nominal geometric rotations, or equivalent horizontal static forces, to the nodes in a frame and determining the resultant horizontal deformations. The disturbing forces are small so that any sway is low and will not cause large deflections. The method is based on the theory proposed by Horne (1975) for the determination of the elastic critical load factor. A braced frame cannot a priori be considered as a non-sway frame. Determination of sway classification is either carried out by imposing a rotation of ϕ at the foot of each stanchion or by imposing horizontal forces of ϕ W at each level of the frame where W is the total factored vertical loading at that level. For each level the sawy index /h is determined. From Horne and Morris (1981) the value of αcr is given by the minimum value of 0,009 αcr =

(8.10)

h max

For multi-storey frames with diagonal bracing Zalka (1999) provides a very quick way of estimating αcr . The critical frame buckling load Ncr for a frame loaded with uniformly distributed loading is given by Ncr = λK

(8.11)

where the parameter λ is tabulated in terms of a parameter βs defined by βs =

K Ng

(8.12)

where Ng is the global bending critical load given by Ng =

7,837Ec Ig n n + 1,588 H2

(8.13)

Where n is the number of storeys, H is the height of the building and Ec Ig is the global flexural rigidity of the columns determined using the parallel axis theorem with the second moment of the columns themselves being neglected. The shear stiffness K is dependent upon the type of bracing. For single direction bracing, −1 d3 1 K= − Ah Eh h Ad Ed hl2

(8.14)

and for complete diagonal cross-bracing, K = Ad Ed

hl2 d3

(8.15)

•

Chapter 8 / Frames and Framing 1.0 0.9 0.8 0.7 0.6 λ

308

0.5 0.4 0.3 0.2 0.1 0 0

2

4

6

8

10

12

14

16

18

20

βs

FIGURE 8.13 Critical load parameter λ (Zalka, 1999) by permission where Ad Ed is the axial rigidity of the bracing member, Ah Eh is the axial rigidity of the horizontal member, d is the length of the diagonal member, l is the width of the braced bay and h is the storey height. The formulations of K for other bracing layouts are given in Zalka (1999). Figure 8.13 from Zalka (1999) gives the relationship between βs and λ (the data are also given in tabular form in Zalka).

8.12.2 Frame Imperfections For frames sensitive to sway buckling, an allowance in the frame analysis should be made by introducing and initial sway imperfection and an initial bow.

8.12.2.1 Initial Sway Imperfection The imperfection factor φ is determined from φ = αh αm φ0

(8.16)

where the reduction factor for the building height αh is given by 2 αh = √ h

(8.17)

where h is the height of the building (in m) and 2/3 ≤ αh ≤ 1,0. The factor αm depends on the number of columns m in a row, 1 αm = 0,5 1 + m The basic imperfection factor φ0 is equal to 1/200.

(8.18)

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Note, the effects of this may be ignored if HEd ≥ 0,15VEd , provided the normalized slenderness ratio λ complies with Afy λ ≥ 0,3 (8.19) NEd

8.12.2.2 Initial Bow This is expressed as the initial bow e0 divided by the length l. It is a function of the strut buckling curve and whether elastic or plastic methods of analysis are used. The values are given in Table 5.1 of EN 1993-1-1. These effects should however be ignored if the members are checked for buckling.

8.12.3 Simple Framing In simple framing the joints between the members may be assumed to offer negligible moment transfer, and the beams should be designed as simply supported. This means any horizontal loading must be taken by bracing or any equivalent substitute. Simple framing can be used to design conventional multi-storey beam and column structures or triangulated truss systems where there will be no significant secondary moments due to joint fixity.

8.12.4 Continuous Framing In this case the connections are designed to give full rigidity between members at a joint. Note on sway and non-sway frames. Elastic, elasto-plastic or full rigid plastic analyses may be used. It is suggested that reference should be made to Ghali and Neville (1989), Coates et al. (1988), Horne and Morris (1981), Neale (1977) or Moy (1996).

8.12.5 Semi-continuous Framing There are two basic approaches to semi-continuous framing: the first is to employ moment capacity/rotation characteristics and the second is a hybrid between simple and continuous. The basic approach is to use an analysis which incorporates the connection characteristics in the analysis (Roberts, 1981; Taylor, 1981). In an elastic analysis the actual moment–rotation characteristic is input as an equivalent spring at the joint, in a plastic analysis the moment capacity is used as a local plastic moment capacity, provided the connection has sufficient ductility. In the hybrid method, the frame is designed as simple for the purpose of determining the member sizes required to carry the imposed vertical loading, but as continuous when determining the effects of lateral, or horizontal, loading on the structure (Salter

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•

Chapter 8 / Frames and Framing et al., 1999). There are limitations on the method based on number of storeys and layout, but it can be clearly advantageous as it will avoid the need to provide any wind bracing, but at the expense of providing connections capable of transmitting the moments due to wind. An alternative approach is to use an elastic analysis and then redistribute the moments to simulate plastic action. EN 1993-1-1, cl 5.4 4(B) places restrictions by specifying that the maximum redistribution is limited to 15%, the section classifications of members where moments are reduced must be no worse than Class 2, and that lateral torsional buckling must be prevented.

8.13 DESIGN ISSUES FOR MULTI-STOREY STRUCTURES Brown et al. (2004) provide a useful overview of constructional details for multi-storey frames. A distinction needs to be drawn between frames which are rigid jointed under all applied loading and frames in which the connections are either semi-rigid or pinned. With pinned connections it should be reiterated that the frame requires bracing to resist the effects of any horizontal actions on the frame, and that even if a frame is braced it cannot be assumed that the frame is a no-sway frame (Section 8.12).

8.13.1 Simple Construction Frames In simple construction frames, the beams are designed as simply supported and the columns designed to take the beam reactions together with nominal fixing moments. The nominal moments should be calculated assuming the beam reactions are applied at a minimum of 100 mm from the face of the column for major axis loading. For minor axis loading, which should be much smaller, the load is applied at either 100 mm from the web or the edges of the flanges depending upon the connection detail. If the reaction at the connection acts at a fixed point then this distance should be taken if it exceeds 100 mm. The nominal fixing moments may be distributed to the lengths of the columns immediately above and below the level being designed. There is no carry over to other storeys outside those above and below. The distribution should be in proportion to the stiffness of the column segments.

8.13.2 Semi-rigid Frames Two methods are here available, either the frame is analysed using actual or design connection moment–rotation characteristics and the frame designed in the same manner as a rigid frame, or a hybrid method is used combining simple and continuous construction. The latter method has already been briefly outlined (Section 8.11.3), but further comment is necessary. The method has the effect of reducing the beam crosssection and the beam deflection. This will ensure that the combined effects of wind, permanent and variable loads will not be critical on the beam itself. The columns need

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to be designed to resist the worst effects of either the permanent and variable loads or the combination of wind, permanent and variable. In both cases the nominal fixing moments from the beams together with the moments induced by the nominal frame imperfection loads must be considered, and may well produce a heavier column cross-section than for simple construction.

8.13.3 Continuous Construction For continuous construction either elastic or plastic methods may be used to carry out the analysis. There is a restriction that the compression flange at the hinge must be Class 1, and where a transverse force exceeding 10% of the shear resistance is applied to the web at the position of the plastic hinge then web stiffeners need to be supplied at a distance of h/2 from the hinge where h is the beam depth (cl 5.6 2b). For varying cross-section members, the web thickness should not be reduced for a distance 2d either side of the hinge (where d is the depth between fillets), the flanges must be Class 1 for the greater distance of 2d or the point where the moment drops to 80% of the plastic moment at the hinge. Restraints must be provided at rotated plastic hinges (i.e. all but the last one to form) (cl. 6.3.5.2). This may be provided by diagonal bracing from the lower flange to beams or purlins fixed to the top flange. Where members carry moment only, or moment with axial tension, and are connected to a concrete slab by shear connectors, then this is deemed sufficient for rolled ‘I’ or ‘H’ sections. At each hinge the restraint and connections should be capable of resisting a force equal to 0,0025Nf,Ed where Nf,Ed is the axial force in the compression flange. In addition to the imperfection force given by Eq. (8.6), the bracing system should be able to carry a force Qm is given by Qm = 1,5αm

Nf,Ed 100

(8.20)

The lateral torsional buckling check of segments between restraints need not be carried out if the length between restraints is less than Lstable and the moment gradient is linear, where Lstable is given as for 0,625 ≤ ψ ≤ 1,0 Lstable = 35iz

235 fy

(8.21)

and for −1,0 ≤ ψ ≤ 0,625 Lstable = (60 − 40ψ)iz

235 fy

(8.22)

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•

Chapter 8 / Frames and Framing where the moment ratio ψ is given by ψ=

MEd,mim Mpl,Rd

(8.23)

Note that where a rotated plastic hinge location occurs immediately adjacent to the end of a haunch, the tapered section need not be checked for stability if the restraint is placed at h/2 along the tapered section (not the uniform section) and the haunch remains elastic along its length.

8.13.4 Effective Length Factors for Continuous Construction For either elastic or plastic analyses, the design of the columns takes account of the relative joint stiffnesses to determine the effective length factor for the column being designed. The two distribution factors k1 and k2 at the top and bottom of the column are defined as k1 =

K c + K1 Kc + K1 + K11 + K12

(8.24)

k2 =

K c + K2 Kc + K2 + K21 + K22

(8.25)

where Kc is the stiffness of the column, K1 and K2 the stiffnesses of the columns at ends 1 and 2, K11 and K12 the stiffnesses of the beams framing in at end 1 and K21 and K22 the stiffnesses of the beams framing in at end 2 (Fig. 8.14). • Non-sway frames The available information to calculate the effective length l is given in BS 5950 Part 1: 2000. The ratio of the effective length l to the actual length L is given by either l = 0,5 + 0,14(k1 + k2 ) + 0,055(k1 + k2 )2 L

(8.26)

Note that the moments due to the nominal frame imperfections must be included in the design moments for the frame in addition to those caused by any imposed actions. • Sway frames The sway mode effective lengths are calculated from l 1 − 0,2(k1 + k2 ) − 0,12k1 k2 1/2 = L 1 − 0,8(k1 + k2 ) + 0,6k1 k2

(8.27)

Williams and Sharp (1990) provide alternative formulations: l π =! k +k2 −k1 k2 L 12 − 36 1 4−k k 1 2

(8.28)

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K1 K12

Kc

K12

Column being designed

K21

K22

K2

Note : The K values are the stiffnesses of the upper and lower beams, and the upper and lower columns in consistent units.

FIGURE 8.14 Data for determination of column effective lengths or π(1,05 − 0,05k1 k2 ) l =! k +k2 −k1 k2 L 12 − 36 1 4−k 1 k2

(8.29)

Note that for a sway frame l/L may exceed 1. Sway moments should be increased using an amplification factor given by 1/(1 − 1/αcr ), provided αcr ≥ 3,0.

8.13.5 Column Loads in Multi-storey Structures In a multi-storey structure, the frame analysis will provide the reactions in the columns. However, such analyses assume that all floors are fully loaded at all times. Statistically this is conservative, thus the axial loads due to variable loading may be reduced in the lower columns of a multi-storey structure. Note, there is clearly no reduction in loads due to permanent (or quasi-permanent) loading, and each individual floor must be designed under full variable and permanent actions. The variable loading may also be adjusted for loaded area, as the loading will be more concentrated over smaller floor areas (and less concentrated over large areas). This means that column reactions may be reduced for this reason also. However, reductions may be made either for the number of floors or loaded area. These reductions are only for loading categories

314

•

Chapter 8 / Frames and Framing A to D of Table 6.1 of EN 1991-1-1. This prohibits any reduction on situations where the loads are predominantly storage or industrial. The reduction factor αn for the number of stories n (n > 2) is given by αn =

2 + (n − 2) ψ0 n

(8.30)

The reduction factor αA for the loaded area is given by αA =

5 A0 ψ0 + ≤ 1,0 7 A

(8.31)

where A is the loaded area, A0 the reference value of 10 m2 . The reduction factor is also subject to the restriction that for load categories C and D it may not be less than 0,6.

8.14 PORTAL FRAME DESIGN 8.14.1 Single Bay Portal To illustrate the design procedure an internal frame with full rigid connections is considered (EXAMPLE 8.3). • A heavier section will be used for the stanchion than the rafter. Although this requires haunches at both the eaves and the ridge to accommodate the connections, the solution is still more economic than the use of a uniform section throughout. • To avoid stability problems in the eaves haunch the stresses in the eaves are kept elastic by extending the haunch for a distance of span/10 from the face of the stanchion. The haunch is taken at its maximum depth of twice that of the rafter in order to ease fabrication. • Plastic hinges are assumed to occur in the stanchion at the base of the haunch, that is, 1,5 times the rafter depth below the intersection of the rafter–stanchion centre lines, and at the second purlin point below the ridge. It is then usual to check the moment at the first purlin point. • The frame will be designed under variable and permanent actions. The frame will not be checked for wind, as this case is rarely critical on a portal frame. • Stability will be checked by applying notional forces of ϕN at the top of the stanchion. The approach given by Davies (1990) will also be checked. • The connections at the eaves, ridge and base will not be designed (see Chapter 7). • The background to the stanchion and rafter stability checks is given in Horne and Ajmani (1971a, b), Horne et al. (1979).

EXAMPLE 8.3 Portal frame design. Prepare a design in grade S355 steel for the frame whose basic geometry is given in Fig. 8.15.

•

315

Spacing 9000 ctrs. Purlin spacing 1250 ctrs (plan)

5000

2200

Structural Design of Steelwork to EN 1993 and EN 1994

25000

FIGURE 8.15 Basic frame geometry for Example 8.3 Permanent loading (kPa):

Sheeting Purlins Frame Total

0,18 0,06 0,18 0,42

UDL due to permanent loading: 0,42 × 9,0 = 3,78 kN/m Nominal variable load (EN 1991-1-1): 0,60 kPa (or 5,40 kN/m) Total factored design load: q = 1,35 × 3,78 + 1,5 × 5,40 = 13,20 kN/m Since the purlins are at 1,25 m centres, it will be accurate enough to apply the roof loading as a UDL. Preliminary design (geometry in Fig. 8.16): Assume hh = 0,65 m; h2 = hc − hh = 5,00 − 0,65 = 4,35 m Distance from ridge to second purlin point, b = 2,5 m l = L/2 − b = 12,5 − 2,5 = 10,0 m Gradient of rafter, sr = 2,2/12,5 = 0,176 Height to second purlin point, r: r = hc + lsr = 5,0 + 10 × 0,176 = 6,76 m Moment at base of haunch, MB1 : MB1 = Hh2 = 4,35H

(8.32)

Moment at second purlin point, M2 : The vertical reaction at the base of the stanchion V is given by V =

qL 25 = 13,2 = 165 kN 2 2

M2 = Vl − Hr −

13,2 × 102 qL2 = 10 × 165 − 6,76H − = 990 − 6,76H 2 2

(8.33)

Chapter 8 / Frames and Framing 1st 2nd purlin purlin X Y

cL

hh

hr

r

B b

h2

•

h

316

Note : All distances to member cL’s unless otherwise indicated

A

H

V

I

b L/2

FIGURE 8.16 Detail frame geometry

By setting M2 equal to Mb2 (i.e. equal strength rafter and column) Eqs (8.32) and (8.33) may be solved to give H = Hbal = 89,1 kN This will give equal plastic moment capacities in the column and rafter. This is not an optimal solution as it is general practice to use a larger section column than rafter. An optimal solution can be obtained by increasing Hbal by around 15–20%. Increase Hbal to 104 kN (i.e. an increase of 16,7%). Thus from Eq. (8.32), MB1 (=Mpl,stanchion ) is given as MB1 = Hh2 = 4,35H = 4,35 × 104 = 452,4 kNm γM0 1,0 = 452,4 × 103 Wpl = Mpl = 1274 cm3 fy 355 Select a 457 × 191 × 67 UKB (section classification Class 1: Mpl = 522 kNm) and from Eq. (8.33) M2 (=Mpl,rafter ): M2 = 990 − 6,76H = 990 − 6,76 × 104 = 287 kNm γM0 1,0 = 287 × 103 Wpl = Mpl = 808 cm3 fy 355 Select a 356 × 171 × 57 UKB (section classification is Class 1: Mpl = 359 kNm)

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Note both the rafter and stanchion have been overdesigned to allow for any reduction in carrying capacity due to sway classification and any slight changes in frame geometry over those assumed in the preliminary assessment of dimensions. Check haunch depth: h = 358 mm, so depth to base of haunch, hh = 1,5 × 358 = 537 mm Hinge at B1 occurs at 5,00 − 0,537 = 4,463 m above base. Thus H = Mpl,stanchion /h2 = 475/4,463 = 106,4 kN The slight change in H can be ignored as it is small (2,3%). Check load required to give collapse (q1 ) at the second purlin point, as the first hinge to occur is at the eaves (from an elastic distribution of bending moments). Mpl,rafter = 10 × 12,5q1 − 106,4 × 6,76 −

q1 102 = 75q1 − 719 2

(8.34)

Equate the value of Mpl,rafter for Eq. (8.34) with the actual value to give 75q1 − 719 = 326 or q1 = 13,9 kN/m This is greater than the design load of 13,2 kN/m, thus collapse does not occur at the second purlin point. Check moment at first purlin point from the apex: l = 11,25 m; r = 5,00 + 0,176 × 11,25 = 6,98 m M1 = Vl − Hr −

ql2 13,2 × 11,252 = 165 × 11,25 − 106,4 × 6,98 − = 278 kNm 2 2

This is less than Mpl,rafter , and is therefore satisfactory. Check moment at end of haunch: b = 2,5 + 0,5hstanchion = 2,5 + 0,5 × 0,4534 = 2,727 m MH = Vb − (5,00 + 0,176b)H −

qb2 2

= 165 × 2,727 − 106,4(5,00 + 0,176 × 2,727) −

13,2 × 2,7272 = −182 kNm 2

Morris and Randall (1979) suggested this moment should be limited numerically to 0,87Mpl,rafter /γ, where γ is the load factor between the total ultimate load and the characteristic loads. The 0,87 factor accounts for the approximate ratio between the moment to first yield and the plastic moment capacity (i.e. the shape factor).

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Chapter 8 / Frames and Framing Total load factor, γ: Load at ULS = 13,2 kN/m and total characteristic load = 3,78 + 5,40 = 9,18 kN/m, so γ = 13,2/9,18 = 1,44. 0,87Mpl,rafter 0,87 × 359 = = 217 kNm γ 1,44 An alternative is to calculate the moment capacity to first yield Myield divided by the total load factor: Myield fy 1 355 1 = Wel 896 = × 10−3 = 221 kNm γ γ γM0 1,44 1,0 Both methods give similar results, are less than the applied moment, and are therefore satisfactory. The maximum load the frame can carry qmax assuming hinges occur simultaneously at the ridge and the base of the haunch is given by (Fig. 8.16): ⎡ ⎤ ⎞ ⎛ hr 1 + 8 h ⎠ qmax = 2 ⎣Mpl,R + ⎝ (8.35) Mpl,c ⎦ hh L 1− h

or

⎡

qmax

⎞ ⎤ ⎡ ⎛ ⎞ ⎤ 2,2 hr 1 + 1 + 8 8 h ⎠ 5 ⎠ Mpl,c ⎦ = 2 ⎣359 + ⎝ 522⎦ = 2 ⎣Mpl,R + ⎝ 0,537 hh L 25 1− 5 1− h ⎛

= 15,37 kN/m Thus the actual load factor γ max on the applied loading is γmax =

15,37 = 1,67 9,18

The required load factor γ from above is 1,44. Horizontal sway stability The deflection δh under a force ϕN applied at the top of the (unhaunched) frame is given by ⎤ ⎡ ⎛ ⎛ ⎞ h1 Ir h Ir h 2I L + 6 + 3 h ⎥ ⎢ Ic Lr + cos θ ⎜ c r ⎥ ⎢⎝ ⎠ ⎜1 − 2 ⎥ ⎢ ⎝ Ir h h1 h1 Ir h ⎥ ⎢ 2 I L +3+3 h + h Ic Lr ⎥ ⎢ 3 c r φNh ⎢ ⎥ ⎛ ⎞2 ⎞ δh = ⎥ ⎢ 2 ⎥ ⎢ 3EIc h1 h1 Ir h ⎥ ⎢ 0,25 − 2I L + 6 + 3 h ⎜ ⎟ ⎟ ⎥ ⎢ h c r ⎟ ⎟− ⎥ ⎢ +2⎜ cos θ 2 ⎠ ⎠ ⎝ I h ⎦ ⎣ r h h Ih 4 I rL + 3 + 3 h1 h1 Ic L r c r

(8.36) The notation is given in Fig. 8.17

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Lr

Ir

Ir

P wN Ic

Ic

h

P wN

FIGURE 8.17 Idealized frame for stanchion deflection

Calculation of ϕ from Eq. (8.16): Determine αm from Eq. (8.18): αm =

1 0,5 1 + m

=

1 0,5 1 + 2

= 0,866

Determine αh from Eq. (8.17) taking h as the height to eaves (as this is more conservative) 2 2 αh = √ = √ = 0,894 h 5 The value αh is within the limits of 2/3 and 1,0, thus φ = αm αh φ0 = 0,866 × 0,894

1 1 = 200 258

If N is taken as the axial load in the column, then it may be taken equal to V (=165 kN). The length of the rafter Lr is given by Lr =

L 25 = = 12,69 m 2 cos θ 2 × cos 10

Equation (8.36) is best evaluated in sections: Ir h 16040 × 5 = = 0,215 Ic L r 29380 × 12,69 Ir h 2,2 h1 = 7,75 + 6 + 3 = 2 × 0,215 + 6 + 3 Ic L r h 5 2 Ir h h1 2,2 2 h1 2,2 + +3+3 + = 0,215 + 3 + 3 = 4,729 Ic L r h h 5 5

2

320

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Chapter 8 / Frames and Framing

δh =

1 165 × 53 258

3 × 210 × 106 × 29380 × 10−8 2 0,215 + cos 10 7,75 7,75 × 1− +2 0,215 2 × 4,729 4 × 4,729 ⎞ 2 2,2 0,25 − 5 ⎟ − cos 10⎠ 0,215

= 1,133 × 10−3 m Horne and Morris (1981) suggested that under a horizontal load of 1% of the total applied load the deflection should be limited to 0,0009hc . This is equivalent to 0,00045hc under 0.5% (1/200) of the load applied to a single column. If the initial imperfection ϕ0 is modified by αm αh then the limiting deflection δlim is reduced to 0,0045αm αh hc . δlim = 0,00045αm αh hc = 0,00045 × 0,866 × 0,894 × 5 = 1,74 × 10−3 m The actual deflection is lower and has been determined on an unhaunched frame. An elastic analysis on the haunched frame gives a horizontal deflection of 0,823 mm, or h/6075 (=164,6 × 10−3 h). The deflection determined using the unhaunched frame is approximately 40% higher. This is acceptable. Horne (1975) originally suggested that for multi-storey frames the elastic critical load factor αcr,H was given by αcr,H =

Hnom hc δnom VULS

(8.37)

where δnom is the horizontal deflection due to a nominal horizontal force Hnom , and VULS is the axial force in the column. Thus using the results determined above for Hnom (as ϕN) and δnom as δh , αcr is given as αcr,H =

1 165 258

5 = 17,1 1,133 × 10−3 165

The actual value of the notional horizontal load is not critical as only the load-deflection ratio is required, but clearly the applied load should not cause any type of failure. However, Lim et al. (2005) have suggested that Horne’s approach to calculating αcr is unconservative for portal frames, that is, it overestimates αcr . Lim et al. (2005) suggest a better estimate of αcr,est is given by NR,ULS αcr,est = 0.8 1 − αcr,H NR,cr max

(8.38)

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where NR,ULS is the axial load in the rafter and NR,cr is the Euler buckling load of the rafter based on the span of the frame, and that the ratio should be taken at its maximum. The maximum axial load in the rafter is at the haunch, and equals 133,5 kN. Owing to the purlin restraints at 1,25 m centres means the rafter as a whole can only buckle about the major axis. The nominal buckling load Ncr assuming pinned ends and neglecting the effect of the haunches is given by Ncr =

π2 EIr π2 210 × 106 × 29380 × 10−8 = = 974,3 kN L2 252

Determine αcr,est from Eq. (8.38) NR,ULS 133,5 αcr,est = 0,8 1 − αcr,H = 0,8 1 − 17,1 = 11,8 NR,cr max 974,3 From Lim et al. (2005), the second order plastic collapse load factor for a Category A type frames, the second order plastic collapse load factor αp2 may be calculated using the Merchant–Rankine equation and the first order plastic collapse load factor αp1 as αp2 = αpl

αcr − 1 αcr

(8.39)

From earlier the load factor αp1 (determined as γ max ) is 1,67, thus αp2 = αp1

αcr − 1 11,8 − 1 = 1,67 = 1,53 αcr 11,8

The required load factor on the frame is 1.44, thus the frame is satisfactory. Check the method proposed by Davies (1990) for determining λcr : λcr =

3EIr Lr (hc Nc,char + 0.3Lr Nr,char )

(8.40)

where Nr,char and Nc,char are the forces in the rafter and column under characteristic frame loading, Lr is the rafter length, hc is the height to eaves, and Ic and Ir are the major axis second moments of area of the column and rafter, respectively. L = 25,0 mm; Lr = 12,69 m; hc = 5,0 m; Ic = 29380 cm4 ; Ir = 16040 cm4 (as only relative values are required). Characteristic load/per unit run = 3,78 + 5,40 = 9,18 kN/m, Nc,char =

25 × 9,18 qL = = 114,8 kN 2 2

To estimate Nr,char the following equation may be used Nr,char =

qL2 (3 + 5m) cos θ + 0,25qL sin θ 16Nhc

(8.41)

322

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Chapter 8 / Frames and Framing where θ is the roof slope and the parameters N and m are given by I r hc N = 2 1 + m + m2 + Lr Ic

(8.42)

and h1 hc

m=1+

(8.43)

where h1 is the rise from eaves to ridge. h1 2,2 =1+ = 1,44 hc 5 I r hc 5 × 16040 N = 2 1 + m + m2 + = 2 1 + 1,44 + 1,442 + = 9,457 L r Ic 12,69 × 29380

m=1+

Nr,char = = λcr =

qL2 (3 + 5m) cos θ + 0,25qL sin θ 16Nhc 9,18 × 252 (3 + 5 × 1.44) cos 10 + 0,25 × 9,18 × 25 sin 10 = 86,0 kN 16 × 9,475 × 5

3EIr 3 × 210 × 106 × 16040 × 10−8 = 8,83 = Lr (hc Nc,char + 0,3Lr Nr,char ) 12,69(5 × 114,8 + 0,3 × 12,69 × 86,0)

This is a lesser value than that derived from the imposition of a 0,5% sway load. Adopt the criterion in Horne and Merchant (1965) for the plastic load factor derived from a modified Rankine type interaction equation with the lesser value of λcr . As 4,6 < λcr < 10,0: λp = 0,9

λcr 8,83 = 0,9 = 1,015 λcr − 1 8,83 − 1

Design UDL at ULS = 13,2 kN/m: actual carrying capacity = 15,37 kN/m (from above); surplus factor = 15,37/13,2 = 1,16. This is greater than that required by the Horne and Merchant check, so frame is satisfactory. An alternative approach to checking stability is given in Horne and Morris (1981), where a span/depth ratio check is carried out. The notation has been slightly changed to give Ic Lr

L−b 240 50W0 L Ir h cos θ ≤ h λW hc 2 + Ic Lr fy Ih

(8.44)

r

W0 is the load required to cause collapse of the rafter assuming it is a straight member of exactly the same section as the original rafter. Any stiffening caused by the ridge haunch can be ignored as it will have little effect. The eaves haunches cannot be ignored, however. There are two possible collapse mechanisms to give W0 whereby a hinge always forms at the centre of the rafter beam, and the remaining two hinges

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can occur either at the ends of the haunch in the parent rafter section, or at the end of the haunch in the enlarged section. For the general case W0 is given by W0 =

8 (Mpl,end + Mpl,centre ) (L − 2b)2

(8.45)

Where L is the span, b is the lengthy of the haunch (taken as zero for hinges forming adjacent to the stanchions), Mpl,end is the numerical value of the plastic moment capacity at the end of rafter and Mpl,centre that of the numerical value of that at the eaves. Consider first the hinges at the haunch–rafter intersection: In this case, b = 2,5 m and Mpl,end = Mpl,centre = 359 kNm, thus W0 is given by 8 8 (Mpl,end + Mpl,centre ) = (359 + 359) = 14, 36 kN/m (L − 2b)2 (25 − 2 × 2,5)2 Now, consider hinges at the end of the haunch and the centre: W0 =

In this case, the hinge at the end of the haunch must occur in the weaker member, that is, the column with b = 0; W0 =

8 8 (Mpl,end + Mpl,centre ) = 2 (359 + 522) = 11,28 kN/m (L − 2b)2 25

Take the lesser value of W0 , that is, 11,28 kN/m. Ic L r

50W0 L Ir h 240 cos θ λW hc 2+ Ir Lr fy Ic h ⎞ ⎛ 29380×12,69 50 × 11,28 25 ⎝ 5×16040 ⎠ 240 cos 10 = 99,5 = 12,69×29380 13,2 5 355 2+ 16040×5

L−b 25 − 2,5 = = 62,9 h 0,358 The criterion is therefore satisfied, and the frame is stable. Deflection check: For an unhaunched frame with pinned feet under a UDL the vertical deflection δv at the ridge is given by ⎡ h h 8 + 5 h1 3 + 2 h1 qL4 ⎢ ⎢ c c δv = ⎢10 − 2 768EIr ⎣ 2 h h Ih 3+ h1 + 3 h1 + I r Lc c c c r ⎤ 2 Ir h cos θ − 1 h1 h1 I r hc ⎥ ⎥ × (8.46) +3+ +3 + ⎥ Ic Lr cos θ hc hc Ic L r ⎦

324

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Chapter 8 / Frames and Framing Ih

θ−1 In Eq. (8.46) the term I rL cos is negligible compared with the remainder of the c r cos θ terms in the last set of round parentheses, and thus Eq. (8.46) reduces to ⎡ ⎤ h

h

8 + 5 h1 3 + 2 h1 ⎥ qL4 ⎢ c c ⎢10 − ⎥ δv = ⎣ ⎦ 2 768EIr h h Ih 3 + h1 + 3 h1 + I r Lc c

c

c r

The horizontal deflection at the eaves δh is given by δh = δv tan θ

(8.47)

Determine δv and δh in terms of q: ⎡ ⎤ h1 h1 4 8 + 5 3 + 2 qL ⎢ hc hc ⎥ δv = ⎣10 − ⎦ 2 768EIr h1 h1 Ir hc 3+ h + 3h + I L c r c c ⎡

⎤ 2,2 2,2 3+2 5 8+5 5 ⎢ ⎥ = ⎣10 − ⎦ 2 768 × 210 × 106 × 16040 × 10−8 2,2 2,2 16040×5 3+ 5 + 3 5 + 29380×12,69 254 q

= 0,0246q δh = δv tan θ = 0,0246q tan 10 = 4,34 × 10−3 q Vertical deflections: Variable load of 5,4 kN/m: δv = 0,0246q = 0,0246 × 5,4 = 0,133 m This is equivalent to span/188. Permanent load of 3,78 kN/m δv = 0,0246q = 0,0246 × 3,78 = 0.093 m This is equivalent to span/270. Deflection under variable and permanent loading: δv = 0,093 + 0,133 = 0,226 m This is equivalent to span/110. Horizontal deflections: Variable load of 5,4 kN/m: δh = 4,34 × 10−3 q = 4,34 × 10−3 × 5,4 = 0.023 m This is equivalent to height/217.

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Permanent load of 3,78 kN/m δh = 4,34 × 10−3 q = 4,34 × 10−3 × 3,78 = 0,016 m This is equivalent to height/313 Deflection under variable and permanent loading: δh = 0,016 + 0,023 = 0,039 m This is equivalent to height/128. Woolcock and Kitipornchai (1986) recommended limits under service permanent loading of span/360, variable loading span/240 for vertical deflection and height/150 for horizontal deflection. It appears unclear which loads are to be used for horizontal deflections. The horizontal deflection under variable and permanent loading marginally exceeds their recommendation but the vertical deflection well exceeds them. However, it should be noted that the frame is stiffer than assumed in the calculations owing to the presence of haunches and the stiffness of the roof sheeting has been ignored. Thus the vertical deflection calculated above will be accepted as satisfactory. A computer analysis allowing for the haunches gives a vertical deflection at the eaves under Variable load: 79,2 mm (or span/316) Permanent plus variable load: 134,7 mm (or span/186) The analysis without haunches overpredicts the vertical deflections by almost 70%. Thus the estimates with no allowance for haunches are extremely conservative. The deflections determined with haunches will be reduced even further if the effects of cladding are considered. Column stability: There are two possible approaches, the first is to supply a torsional restraint at a distance Lm below the hinge position and then to check the remainder of the column below this restraint for the combined effects of strut buckling and lateral torsional buckling. The second is to check the whole length of the column below the hinge making use of restraint to the tension flange by the sheeting rails. Method 1 The maximum distance to point of restraint in a column Lm is given by 38iz

Lm = 1 NEd 57.4 A

+

2 1 Wpl,y 2 756C1 AIt

f 2 y 235

(8.48)

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Chapter 8 / Frames and Framing Note, Eq. (8.48) must be solved iteratively as the moment gradient factor C1 is dependent upon the value of the bending moment at a distance Lm below the hinge. A conservative solution may be obtained by setting C1 = 1.0 (uniform moment). As an alternative, the length to the restraint may be taken equal to Ls where Ls is given by Ls = Lk Cm

Mpl,y,Rk MN,y,Rk + aNEd

(8.49)

where Cm is the modification factor for a linear moment gradient, MN,y,Rd is the moment capacity of the column reduced due to axial load and Lk is given h 600fy 5,4 + E iz t Lk = f (8.50) fy 5,4 E th

2

f

−1

Use Eq. (8.49): Force in the column (NEd ) = 165 kN; NEd 165 × 103 = = 19,3 MPa A 8550 Suitable sheeting rails for this frame are 202 mm deep, thus a = 0,5(453,4 + 202) = 327,7 mm Determine MN,y,Rk : Determine the ratio n between the applied force NEd and the plastic axial resistance Npl,Rd n=

NEd 165 × 103 = = 0,054 Npl,Rd 8550 355 1,0

Determine the ratio a between the total web area (excluding just the flanges) and the cross-section area: a=

A − 2btf 8550 − 2 × 189,9 × 12,7 = = 0,436 A 8550

MN,y,Rd = Mpl,y,Rd

1−n 1 − 0,054 = Mpl,y,Rd = 1,21 Mpl,y,Rd 1 − 0,5a 1 − 0,5 × 0,436

Thus MN,y,Rd = Mpl,y,Rd = 522 kNm. 38iz

Lm = 1 NEd 57,4 A

=

2 Wpl,y

1 + 756C 2 AI t 1

f 2 y 235

38 × 41,2 (1471×103 )2 1 19,3 + C12 756×8550×37,1×104 57,4 1

355 235

2 =

1566 2,059

0,336 + C2 1

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522,4 kNm 1137

Underside of haunch

Position of restraint 4467

(354 kNm)

FIGURE 8.18 BMD in stanchion AB

A

Use Eq. (5.40) to determine C1 . Under a constant moment gradient, C1 = 1,0, thus Lm = 1012 mm. After three iterations, ψ = 0,678, C1 = 1,148 and Lm = 1137 mm. Check column buckling below this level under axial load and moment (Fig. 8.18): Mz,Ed = 0: NSd = 165 kN: My,Ed = 106,4 × (4,463 − 1,137) = 354 kNm. Section classification: Flanges: Class 1. Web: Actual slenderness: c d 407,6 = = = 48,0 tw tw 8,5 Length of web xw required to carry axial force: xweb =

NEd fy

tw γ

=

M0

165 × 103 8,5 355 1,0

= 54,7 mm

Depth of web in compression αc: αc =

d xweb 407,6 + 54,7 + = = 231,2 mm 2 2 2

or α=

(αc) 231,2 = = 0,567 c 407,6

Limit for Class 1, with α > 0,5: 396 235 f

! 396 235 355

y d = = = 50,6 tw 13α − 1 13 × 0,567 − 1

328

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Chapter 8 / Frames and Framing The actual web slenderness is below this, therefore section is Class 1. As the compression flange is unrestrained, the lateral torsional buckling must be checked. As the section is Class 1, Eqs (6.61) and (6.62) of EN 1993-1-1 may be simplified as My,Ed and My,Ed are both zero, as is Mz,Ed to give NEd fy χy A γ M1

NEd fy

χz A γ

My,Ed

+ kyy

fy

χLT Wpl,y γ

M1

My,Ed

+ kzy

fy

χLT Wpl,y γ

M1

≤ 1,0

≤ 1,0

M1

Buckling curves: h 453,4 = = 2.39 > 1.2 b 189,9 y–y axis curve ‘a’ (α = 0.21) and z–z axis curve ‘b’ (α = 0,34). Calculation of χy : For yy axis buckling, the system length is taken as base of the haunch and the bottom of the column as restraints are present at both these points, that is, a length of 4,463 m π2 EI y π2 × 210 × 106 × 29380 × 10−8 = = 30572 kN L2y 4,4632 Af y 8550 × 355 λ¯ y = = = 0,315 Ncr,y 30572 × 103

Ncr,y =

y = 0,5[1 + α(λ¯ y − 0,2) + (λ¯ y )2 ] = 0,5[1 + 0.21(0,315 − 0,2) + 0,3152 ] = 0,562 χy =

y +

!

1

1

2 = 0,562 + 0,5622 − 0,3152 = 0,973 2y − λ¯ y

Calculation of χz : For yy axis buckling, the system length is taken between the restraint 1,137 m below the haunch and the bottom of the column as restraints are present at both these points, that is, a length of 3,362 m π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN L2z 3,3622 Af y 8550 × 355 λ¯ z = = = 1,064 Ncr,y 2663 × 103

Ncr,z =

z = 0,5[1 + α(λ¯ z − 0,2) + (λ¯ z )2 ] = 0,5[1 + 0,34(1,064 − 0,2) + 1,0642 ] = 1,213 χz =

z +

!

1

1

2 = 1,213 + 1,2132 − 1,0642 = 0,567 2z − λ¯ z

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Calculation of χLT : The system length for lateral torsional buckling is taken as the distance between the restraint 1,137 m below the haunch and the bottom of the column, that is, a length of 3,362 m Determine Mcr from Eq. (5.5) Iw 0,705 × 10−6 = = 0,0486 m2 Iz 1452 × 10−8 π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN L2 3,3622 L2 GI t 81 × 106 × 37,1 × 10−8 = = 0,0113 m2 2663 π2 EI z 1/2 π2 EI z Iw L2 GI t Mcr = + 2 = 2663[0,0486 + 0,0113]1/2 = 652 kNm Iz L2 π EI z Determine C1 from Eq. (5.40): ψ = 0, 1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0 = 0,6 C1 or, C1 = 1,667. From Eq. (5.12) with Mcr modified by C1 , Wpl fy 1471 × 103 × 355 × 10−6 λ¯ LT = = 0,693 = 1,667 × 652 C1 Mcr Use the general case for lateral torsion buckling, as h/b > 2, αLT = 0,34 (from Table 5.1) LT = 0,5[1 + αLT (λ¯ LT − 0,2) + (λ¯ LT )2 ] = 0,5[1 + 0,34(0,693 − 0,2) + 0,6932 ] = 0,824 χLT =

LT +

!

1

1 = = 0,788 2 0,824 + 0,8242 − 0,6932 2LT − λ¯ LT

Use Annex B of EN 1993-1-1 to determine kyy and kzy : As the moment gradient is linear with the least value of moment equal to zero, ψ = 0, so from Table B.3, all the Cm values are 0,6. ⎛ ⎞ ⎛ ⎞ NEd ⎠ NEd ⎠ ≤ Cmy ⎝1 + 0,8 kyy = Cmy ⎝1 + (λ¯ y − 0,2) fy fy χy A γ χy A γ M1 M1 ⎛ ⎞ ⎛ ⎞ 3 3 165 × 10 165 × 10 ⎠ ≤ 0,6 ⎝1 + 0,8 ⎠ = 0,6 ⎝1 + (0,315 − 0,2) 355 0,973 × 8550 355 0,973 × 8550 1,0 1,0 = 0,604 ≤ 0,627

330

•

Chapter 8 / Frames and Framing Thus, kyy = 0,604. As torsional deformations can occur (although they will be in practice slightly limited by the sheeting rails) kzy = 1 −

0,1λ¯ z 0,1 NEd NEd ≥1− f y CmLT − 0,25 χ A CmLT − 0,25 χ A fy z

= 1−

z

γM1

γM1

0,1 × 1,064 165 × 103 165 × 103 0,1 ≥ 1 − 0,6 − 0,25 0,567 × 8550 355 0,6 − 0,25 0,567 × 8550 355 1,0

1,0

= 0,971 ≥ 0,973 kxy = 0,973 NEd fy

χy A γ

My,Ed

+ kyy

fy

χLT Wpl,y γ

M1

=

M1

165 × 103

354 × 106 + 0,604 0,973 × 8550 355 0,788 × 1471 × 103 355 1,0 1,0

= 0,576 ≤ 1,0 NEd fy

χz A γ

My,Ed

+ kzy

fy

χLT Wpl,y γ

M1

M1

=

165 × 103

354 × 106 + 0,973 0,567 × 8550 355 0,788 × 1471 × 103 355 1,0 1,0

= 0,942 ≤ 1,0 Thus the column is satisfactory below the torsional restraint. Method 2 Determine the elastic critical torsional buckling load. This is given by Eq. (5.100) together with an additional term π2 EIz a2 /L2t to allow for the restraint from the sheeting rails. a = 327,7 mm (from Method 1) NcrT

1 = 2 is

π2 EIz a2 π2 EIw + + GIt L2t L2t

where Lt is the distance between restraints to both flanges. Try Lt as the height from the underside of the haunch to the base, that is, 4,463 m, and is from (Eq. (5.101)) modified by an additional term a2 to give i2s = i2z + i2y + a2 or i2s = i2z + i2y + a2 = (41,22 + 1852 + 327,72 ) × 10−6 = 0,1433 m2

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331

or, is = 379 mm. 1 π2 EIz a2 π2 EIw NcrT = 2 + + GIt is L2t L2t 1 π2 × 210 × 106 × 1452 × 10−8 = 0,1433 4,4632 π2 × 210 × 106 × 0,705 × 10−6 + + 81 × 106 × 37,1 × 10−8 4,4632

= 11265 kN NcrE =

π2 EI z π2 × 210 × 106 × 1452 × 10−8 = = 1511 kN 4,4632 L2t

NcrE 1511 = = 0,134 NcrT 11265 1 + 10η 1 + 10 × 0,134 B0 = = = 0,636 1 + 20η 1 + 20 × 0,134 √ √ 5 η 5 0,134 B1 = = 0,269 √ √ = π + 10 η π + 10 0,134 η=

B2 =

0,5 0,5 0,5 0,5 = = 0,097 − √ √ − 1 + π η 1 + 20η 1 + π 0,134 1 + 20 × 0,134

Determine the ratio βt defined as the ratio of the algebraically smaller end moment to the larger end moment. As the smaller end moment (at the base) is zero, βt = 0. Cm =

1 B0 + B1 βt + B2 βt2

(8.51)

As βt = 0, Eq. (8.51) reduces to Cm =

1 1 = = 1,572 B0 0,636

Lk is given by Eq. (8.50): 600fy h 5,4 + E tf Lk = iz = 41,2 fy 5,4 E

h 2 tf

−1

453,4 12,7

5,4 + 600×355 210×103

355 5,4 210×10 3

453,4 2 12,7

= 2893 mm −1

The reduced moment capacity MN,y,Rk due to the axial load equals the plastic moment capacity Mpl,Rk. Mpl,y,Rk 522 = 3286 mm = 2893 1,572 Ls = C m Lk 522 + 165 × 0,3277 Mn,y,Rk + aNEd This is less than the height to the underside of the haunch, and slightly less than the height to the restraint of 3362 mm.

Chapter 8 / Frames and Framing Recheck using Lt = 3362 mm. NcrT

π2 EIz a2 π2 EIw + + GIt L2t L2t 1 π2 × 210 × 106 × 1452 × 10−8 = 0,1433 3,3622

1 = 2 is

π2 × 210 × 106 × 0,705 × 10−6 + + 81 × 106 × 37,1 × 10−8 3,3622 = 19692 kN NcrE =

π2 EIz π2 × 210 × 106 × 1452 × 10−8 = = 2663 kN 3,3622 L2t

NcrE 2663 = 0,135 = NcrT 19692 1 + 10η 1 + 10 × 0,135 B0 = = = 0,635 1 + 20η 1 + 20 × 0,135 η=

Hence, Lk = 3290 mm. This is probably adequate. Rafter (a) Haunch between connection and bracing at first purlin point. Relevant dimensions are given in Fig. 8.19.

‘zed’ purlins at 1250 ctrs

179

•

227

538

332

227

FIGURE 8.19 Haunch geometry

2500

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At any point along the rafter the value of the applied moment MSd is given by 13,2x2 2 At the connection, x = 0,227 m (ignoring end plate) and MEd,x = 165x − 106,4(5 + 0,176x) −

MSd = −499 kNm. At the restraint x = 1,25 m, MSd = −359,5 kNm. Section classification (with axial force) Flanges are Class 1. Web: Determine the axial force in the member: NSd = H cos 10◦ + V sin 10◦ = 106,4 cos 10 + 165 sin 10 = 133,4 kN. Web classification check is not needed as the web is restrained from buckling due to the end plate welded to the web and flanges forming the end plate for the connection. Determine section properties ignoring both the central flange of the haunch and the fillets. A = 2btf + dtw = 2 × 0,1722 × 0,013 + 0,0081(0,716 − 2 × 0,013) = 0,010066 m2 # 1" 3 bh − (b − tw ) (h − 2tf )3 12 1 = [0,1722 × 0,7163 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)3 ] 12

Iy =

= 0,775 × 10−3 m4 # 1" 2tf b3 + (h − 2tf )tw3 12 # 1" = 2 × 0,013 × 0,17223 + (0,716 − 2 × 0,013)0,00813 = 11,09 × 10−6 m4 12

Iz =

Plastic section modulus, Wpl,y : # 1" 2 Wpl,y = bh − (b − tw )(h − 2tf )2 4 # 1" = 0,1722 × 0,7162 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)2 4 = 2,538 × 10−3 m3 Warping constant, Iw : 11,09 × 10−6 (0,716 − 0,013)2 Iz h2s = = 1,370 × 10−6 m6 4 4 where hs is the depth between the centroids of the flanges. Iw =

334

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Chapter 8 / Frames and Framing To determine the torsional constant, It , it will be sufficiently accurate to treat the flanges and webs as thin, thus, # 1 3 1" It = bt = 2 × 0,1722 × 0,0133 + (0,716 − 2 × 0,013) 0,00813 3 3 = 0,374 × 10−6 m4 The distance between restraints, L, is the slope distance between the end of the rafter and the first purlin, and is given by 1,250 − 0,277 = 0,988 m cos 10◦ Determine Mcr from Eq. (5.5) L=

Iw 1,370 × 10−6 = = 0,1235 m2 Iz 11,09 × 10−6 π2 EIz π2 × 210 × 106 × 11,09 × 10−6 = = 23547 kN 2 L 0,9882 L2 GIt 81 × 106 × 0,374 × 10−6 = = 1,287 × 10−3 m2 2 23547 π EIz 1/2 π2 EIz Iw L2 GIt 1/2 Mcr = + 2 = 23547[0,1235 + 1,287 × 10−3 ] 2 Iz L π EIz = 8318 kNm Determine C1 from Eq. (5.40), ψ=

359,5 = 0,731 492

1 = 0,6 + 0,4ψ = 0,6 + 0,4 × 0,731 = 0,892 C1 or, C1 = 1,121. Wpl fy 2,538 × 10−3 × 355 λLT = = = 0,311 C1 Mcr 1,121 × 8318 × 10−3 As λLT < 0,4, lateral torsional buckling cannot occur. Haunch instability An alternative approach is to calculate the limiting length between lateral torsion restraints (i.e. between the haunch and the purlin immediately after the end of the haunch) (cl BB.3). √ Lk C n Ls = c

(8.52)

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where Lk is the basic critical length, and c is a factor allowing for the shape of the haunch given by 2 3 hh 3 Lh c=1+ h (8.53) Ly − 9 hs tf

where h/tf is the ratio of the depth to flange thickness tf for the basic section, Lh Ly is the proportion of the length between the restraints taken up by the taper, and hh /hs is the ratio of additional depth of the beam at the haunch to that of the basic section. The basic length Lk is given by Eq. (8.50). Determine Lh /Ly : Length of taper 2500 mm, and horizontal distance from inside of stanchion to next purlin beyond end of taper is 3 × 1250 − 227 = 3523 mm, so Lh 2500 = = 0,710 Ly 3523 As the haunch is fabricated using the same section size as the parent section minus one flange, hh = 358 − 13 = 345 mm, so hh 345 = = 0,964 hs 358 2 2 3 hh 3 Lh 3 c=1+ h = 1 + 358 0,964 3 0,710 = 1,133 hs Ly −9 tf − 9 13 Determine Lk from Eq. (8.50): Lk = iz

f

355 y 5,4 + 600 210×10 5,4 + 600 E 3 358 = 39,1 1/2 1/2 = 2838 mm 13 2 2 fy h 355 358 5,4 E t 5,4 210×103 13 −1 −1

h tf

f

Determination of Cn : To determine Cn values of the parameter R are required which is defined by R=

My,Ed + aNEd Wpl,y, fy

From safe load tables a suitable depth purlin is 232 mm, thus the depth a between the centroid of the purlins and the centroid of the member is given by a = 0,5h + 0,5 × 232 Take the value of NEd as at the end of the member, that is, 134,4 kN. The values of R are determined in Table 8.1 From Table 8.1 the maximum value of R, RS = 0,633: RE is max (R1 , R5 ) = 0,627.

336

•

Chapter 8 / Frames and Framing Cn =

12 R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE )

RS − RE = 0,633 − 0,627 = 0,006 > 0 So, Cn =

12 R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE )

12 = 1,584 0,625 + 3 × 0,629 + 4 × 0,632 + 3 × 0,633 + 0,627 + 2 × 0,006 √ √ 2838 1,584 Lk C n Ls = = = 3158 mm c 1,13 =

This exceeds the length of the haunch so there need only be restraints at haunch connection and the end of the haunch. Check between haunch and first purlin point beyond point of contraflexure. Point of contraflexure occurs when M = 0, or 165x − 106,4(5 + 0,176x) − 6,6x2 = 0, or x = 4,584 m The distance from the end of the haunch to the point of contraflexure is 4,584 − 2,727 = 1,857 m. The distance to the next purlin point is 5,00 − 2,727 = 2,273 m. This is less than the value of Lk and therefore no additional restraint is required. (c) Ridge As the moment gradient is non-linear, the stable length is given by Lk = 2833 mm (calculated above). This is greater than the slope length between purlins.

8.14.2 Notes on the Design of a Gable End Frame • The rafter is usually designed as a continuous beam spanning over the gable end columns, with simple, non-moment transferring connections to the frame. This will TABLE 8.1 Calculation of the values of R for the haunch.

Distance x (mm) Depth h (mm) A (mm2 ) I y (×108 mm4 ) W pl (×106 mm3 ) M x (kNm) aN Ed (kNm) R

1

2

3

4

5

227 716 10156 7,854 2,538 499,1 63,7 0,625

852 626,5 9431 5,556 2,104 412,2 57,7 0,629

1477 537 8706 4,031 1,702 330,4 51,7 0,632

2102 447,5 7981 2,658 1,332 253,7 45,7 0,633

2727 358,0 7256 1,603 0,996 182,2 39,6 0,627

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mean that if the spans are equal, the first bay will be critical since it will only be continuous at one end. • The centre lines of the gable end columns should be coincident with the spacing of the roof purlins in order to resist the reaction from the wind loading on the gable end columns without subjecting the rafters to bi-axial bending. It is also preferable if the spacing of the gable end columns can be similar to the frame spacing in order to keep the same section for the sheeting rails, although this is not essential. • Often the loading on the gable end will require the use of only relatively small sections, but detailing requirements and compatibility with the rest of the frame may make it necessary to use larger sections.

8.14.3 Multi-bay portal frames It is not intended to present a design example of such a frame since the principles and methods are similar to those for single bay frames. It is intended, rather, to point out additional factors that need to be taken into account. • The critical bay, assuming as in normal practice, that all the spans are equal, is the end bay. However, it must be noted that it is no longer possible to assume uniform snow loading. The frame should be first checked under the nominal roof loading of 0,6 kPa (modified for slope if necessary) as a variable load together with the permanent loading. It should then be checked for the accidental snow drift loading together with the permanent loading considered as an accidental load combination with reduced partial safety factors.

(a) Interior bays before snap through

(b) Exterior bays after snap through

FIGURE 8.20 Snap through failure of multi-bay portal frames

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Chapter 8 / Frames and Framing • Although with balanced loading the interior stanchions carry axial load only, and the internal rafters have a higher reserve of strength, it is normal practice to detail all bays identically. This is partly to achieve fabrication economies, and to avoid problems on site with varying rafter sizes, and partly to avoid deflection and ‘snap through’ failure. • Sway stability is either handled by the imposition of a nominal sway force as for single bay frames, or if the empirical method of Horne and Morris (1981) is used as in Eq. (8.44), then the ratio of the column stiffness (Ic /hc ) to the rafter stiffness (Ir /L) should be halved. • In frames with more than two bays it is possible for snap through to occur when instability happens as the top of the stanchions spread followed by inversion of the rafter (Fig. 8.20) This can be checked using the following empirical equation (Horne and Morris, 1981) Ic L 1 + 25 4 + L−b 240 h I c r tan 2θ (8.54) = λW λW h fy −1 W0

W0

where symbols are defined in Eq. (8.44). If the arching ratio is less than one, snap through cannot occur, so there is no limit on the effective length to depth ratio.

REFERENCES Alexander, S.W. (2004). New approach to disproportionate collapse, Structural Engineer, 82(23/24), 14–18. Alexander, S.J. and Lawson, R.M. (1981). Movement design in buildings. Technical Note 107. CIRIA. Bailey, C.G. (2001). Steel structures supporting composite floor slabs: design for fire. Digest 462. BRE. Bailey, C.G. (2003). New fire design method for steel frames with composite floor slabs. FBE Report 5. BRE. Bailey, C.G. (2004). Structural fire design: core or specialist subject? Structural Engineer, 82(9), 32–38. Bailey, C.G., Newman, G.M. and Simms, W.I. (1999). Design of steel framed buildings without applied fire protection. P186. SCI. Brown, D.G., King, C.M., Rackham, J.W. and Way, A.G.J. (2004). Design of multi-storey braced frames. P334. SCI. BS 5950-1. Structural use of steelwork in building – Part 1: Code of practice for design – rolled and welded sections. BSI. Bussell, M.N. and Cather, R. (1995). Design and construction of joints in concrete structures. Report 146. CIRIA. Coates, R.C., Coutie, M.G. and Kong, F.K. (1988). Structural Analysis (3rd edition). Van Nostrand Reinhold. Concrete Society (2003). Concrete industrial ground floors (3rd edition). Report TR 43. Concrete Society. Couchman, G.H., Mullett, D.L. and Rackham, J.W. (2000). Composite slabs and beams using steel decking: Best practice for design and construction. Technical Paper No.13 (SCI P300). Metal Cladding and Roofing Manufacturers Association/SCU. Davies, J.M. (1990). In plane stability in portal frames, Structural Engineer, 68(8), 141–147. Davies, J.M. and Bryan, E.R. (1982). Manual of stressed skin diaphragm construction. Granada Irwin.

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Deacon, R.C. (1986). Concrete ground floors: Their design, construction and finish. Cement and Concrete Association. Dowling, J. (2005). United States can learn from worldwide fire engineering expertise, New Steel Construction, 13(9), 26–26. DTLR (2000). The Building Regulations: Approved Document B: Fire safety. DTLR. EN 1990. Eurocode – basis of structural design. EN 1991-1-1. Eurocode 1: Actions on structures: Part 1–1: General actions – densities, self-weight, imposed loads for buildings. CEN/BSI. EN 1991-1-2. Eurocode 1: Actions on structures: Part 1–2: Actions on structures exposed to fire. CEN/BSI. EN 1991-1-3. Eurocode 1: Actions on structures: Part 1–3: Snow loads. CEN/BSI. EN 1991-1-4. Eurocode 1: Actions on structures: Part 1–4: Wind loads. CEN/BSI. EN 1991-1-5. Eurocode 1: Actions on structures: Part 1–5: General actions – thermal actions. CEN/BSI. EN 1991-1-7. Eurocode 1: Actions on structures: Part 1–7: Accidental actions due to impact and explosions. CEN/BSI. EN 1991-3. Eurocode 1: Actions on structures: Part 3: Actions induced by cranes and machinery. CEN/BSI. EN 1996-1-1. Eurocode 6: Design of masonry structures: Part 1–1: Rules for reinforced and unreinforced masonry. CEN/BSI. EN 1998. Eurocode 8: Design of structures for earthquake resistance. CEN/BSI. Ghali, A. and Neville, A.M. (1989). Structural analysis – a unified classical and matrix approach (3rd edition). Chapman and Hall. Gibbons, C. (1995). Economic steelwork design, Structural Engineer, 73(15), 250–253. Ham, S.J., Newman, G.M., Smith, C.I. and Newman, L.C. (1999). Structural fire safety: A handbook for architects and engineers. P197. SCI. Henry, F.D.C. (ed) (1986). Design and construction of engineering foundations. Chapman and Hall. HMSO (1994). Construction (Design and Management) Regulations. HMSO. Horne, M.R. (1975). An approximate method for calculating the elastic critical loads of multi-storey plane frames, The Structural Engineer, 53, 242–8. Horne, M.R. and Ajmani, J.L. (1971a). Design of columns restrained by side-rails, Structural Engineer, 49(8), 339–345. Horne, M.R. and Ajmani, J.L. (1971b). The post-buckling behaviour of laterally restrained columns, Structural Engineer, 49(8), 346–352. Horne, M.R. and Merchant, W. (1965). The stability of frames. Pergammon Press. Horne, M.R. and Morris, L.J. (1981). Plastic design of low rise frames. Constrado Monographs (republished Collins, 1985). Granada publishing. Horne, M.R., Shakir-Khalil, H. and Akhtar, S. (1979). The stability of tapered and haunched beams, Proceedings of the Institution of Civil Engineers, 67, 677–694. Horridge, J.F. (1985). The design of industrial buildings, Civil Engineering Steel Supplement, 13–16. Horridge, J.F. and Morris, L.J. (1986). Comparative costs of a single-storey steel framed structures, Structural Engineer, 64A(7), 177–181. Irwin, A.W. (1984). Design of shear wall buildings. Report 112. CIRIA. ISE (2002). Safety in tall buildings and other buildings with large occupancy. Institution of Structural Engineers. Lawson, R.M. (1990a). Behaviour of steel beam to column connections in fire, Structural Engineer, 68, 263–271. Lawson, R.M. (1990b). Enhancement of fire resistance of beams by beam to column connections. TR-086. SCI. Lawson, R.M. and Nethercot, D.J. (1985). Lateral stability of I-beams restrained by profiled sheeting, Structural Engineer, 63(B), 1–13. Lawson, R.M., Mullett, D.L. and Rackham, J.W. (1997). Design of Asymmetric Slimflor ® Beams using deep composite decking. Publication 175. SCI.

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Chapter 8 / Frames and Framing Lim, J.B.P., King, C.M., Rathbone, A.J., Davies, J.M. and Edmonson, V. (2005). Eurocode 3 and the in-plane stability of portal frames, Structural Engineer, 83(21), 43–49. Martin, L.H. and Purkiss, J.A. (2006). Concrete design to EN 1992 (2nd edition). ButterworthHeinemann. Mlakar, P.F., Dusenberry, D.O., Harris, J.R., Haynes, G., Phan, L.T. and Sozen, M.A. (2003). Pentagon building performance report, Civil Engineering, 73(2), 43–55. Morris, L.J. and Randall, A.L. (1979). Plastic Design. Constrado. Moy, S.S.J. (1996). Plastic methods for steel and concrete structures (2nd edition). MacMillan. Mullett, D.L. (1992). Slim floor design and construction. Publication 110. SCI. Mullett, D.L. (1997). Design of RHS Slimflor ® Edge Beams. Publication 169. SCI. Mullett, D.L. and Lawson, R.M. (1993). Slim floor construction using deep decking. Technical Report 120. SCI. Neale, B.G. (1977). The plastic methods of structural analysis (3rd edition). Chapman and Hall. Newman, G.M. (1992). The fire resistance of web-infilled steel columns. Publication 124. SCI. Newman, G.M. (1993). The fire resistance of shelf angle floor beams to BS 5950: Part 8. Publication 126. SCI. Newman, G.M., Robinson, J.T. and Bailey, C.G. (2006). Fire safe design: A new approach to multi-storey steel-framed buildings, P. 288. Steel Construction Institute. Purkiss, J.A. (2007). Fire safety engineering design of structures (2nd edition). Butterworth-Heinemann. Roberts, E.H. (1981). Semi-rigid design using the variable stiffness method of column design, In Joints in Structural Steelwork (eds J.H. Howlett, W.M. Jenkins, and R. Stainsby); Proceedings of the International Conference. Teesside Polytechnic (now Teesside University), 6–9 April, 1981, Pentech Press, 5.36–5.49. Salter, P.R., Couchman, G.H. and Anderson, D. (1999). Wind-moment design of low-rise frames. P-263. SCI. Simms, W.I. and Newman, G.M. (2002). Single storey steel framed buildings in fire boundary conditions. P313. SCI. Taylor, J.C. (1981). Semi-rigid beam connections: Effects on column design: B20 Code Method, In Joints in structural steelwork (eds J.H. Howlett, W.M. Jenkins, and R. Stainsby); Proceedings of the International Conference. Teesside Polytechnic (now Teesside University), 6–9 April, 1981, Pentech Press, 5.50–5.57. Way, A.G.J. (2005) Guidance on meeting the robustness requirements in Approved Document A (2004 Edition), Publication P341. SCI. Wearne, P. (1999). Collapse – Why buildings fall down. Channel 4 Books. Williams, F.W. and Sharp, G. (1990). Simple elastic critical load and effective length calculations for multi-storey rigid sway frames, Proceedings of the Institution of Civil Engineers, 90, 279–287. Wood, J.G.M. (2003). Pipers row car park collapse: Identifying risk. Concrete, 37(9), 29–31. Woolcock, S.T. and Kitipornchai, S. (1986). Deflection limits for portal frames, Steel Construction, 20(3), 2–10. Yandzio, E. and Gough, M. (1999) Protection of buildings against explosions. Publication 244. SCI. Zalka, K.A. (1999). Full-height buckling of frameworks with cross-bracing, Structures and Buildings, Proceedings of the Institution of Civil Engineers, 134, 181–191.

Chapter

9 / Trusses

This chapter is concerned with the design of triangulated and non-triangulated trusses. With the advent of rolled hollow sections the design of both types of truss has been revolutionized. The use of rolled hollow sections for trusses provides a far more efficient use of material as the buckling strengths are higher as radii of gyration are larger and lateral torsional buckling is either non-existent in the case of square or circular sections, or the effects are much reduced for rectangular sections. Rolled section trusses are much easier to fabricate as the welding is generally fillet welds or full strength butt welds. However, it needs remembering that room must be provided to allow the welds to be properly formed. This can be achieved by limiting the number of members at a node to the main chord member and at most two subsidiary web members and limiting the minimum angle between members to 30◦ . A problem however with rolled sections is forming joints to enable large trusses to be transported. If circular sections are used as main chord members fabrication problems and resultant high costs may ensue. Maintenance, and painting, is much easier with rolled sections as spray techniques can be employed. Note, rolled hollow sections are commonly available in Grade S355, but it may be difficult to obtain such sections in S275.

9.1 TRIANGULATED TRUSSES It is generally acceptable to analyze triangulated trusses on the basis of pinned joints at the nodes together with loading from the roof system applied at nodes. However, where purlins exist between the nodes then the relevant members need to designed under the effects of flexure form the loads concerned. It is conservative to consider the main chord members as simply supported between nodes, although use of continuity can be made. Where it is necessary to consider the existence of services within the roof space, such loading can be considered as nodal loads on the bottom chord of the truss. The effect of service loads must be neglected when considering the effects of wind uplift. It is not generally necessary to consider the effects of secondary moments due to joint fixity as these are usually low. Note, however if the original pin-jointed analysis is performed using a plane frame computer analysis package then it is relatively easy to

Chapter 9 / Trusses check the effect of secondary moments by using the sections determined as suitable to carry the pinned forces and simply altering the end conditions of the members. Also, provided the secondary moments are negligible, the deflections can be determined from a pin-joint analysis. It should be noted that the actual deflections in a truss will be less than the calculated deflections as any stiffening effect of the roof cladding and purlins has been ignored. Note that even if the applied actions due to wind uplift and permanent loads are lower than those due to wind downthrust, roof loading and permanent loads, consideration still needs to be given to bracing on the lower boom of the truss as this is now in compression. This will almost certainly mean the provision of longitudinal bracing to ensure system lengths are kept reasonably small to ensure buckling does not occur out of plane.

EXAMPLE 9.1 Triangulated truss design. Prepare a design using square and rectangular rolled hollow sections in Grade S355 steel for the truss whose geometry is given in Fig. 9.1. The load due to initial imperfections has been omitted partly because the value of the load is small and partly because the truss is triangulated and thus will exhibit virtually no sway effects. Actions (kPa): Permanent:

Sheeting Purlins Total

0,22 0,08 0,30

UDL for each truss = 0,30 × 9,0 = 2,7 kN/m. Assume self-weight is 1,5 kN/m, thus total permanent action is 4,2 kN/m. Since the purlins are at nodal points, nodal permanent action = 3,0 × 4, 2 = 12,60 kN (on an internal node). Variable action: From EN 1991-1-1 variable action (with no access, snow drift ignored) is 0,6 kPa, so nodal load = 0,6 × 9,0 × 3,0 = 16,2 kN J

H

L

F

0,8 m

•

3,2 m

342

B A

C

N

P R

D

E

G

I 8 bays @ 3 m

FIGURE 9.1 Truss geometry for EXAMPLE 9.1

K

M

O

Q

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Although wind loading will need considering as it may cause uplift due to suction, in this example only permanent and variable actions due to roof loading will be considered. Thus the only combination of actions needed is 1,35gk + 1,5qk . So the total nodal action is 1,35 × 12,6 + 1,5 × 16,2, that is, 41,31 kN per internal node. The truss will be designed using a pin-joint analysis, and then checked using a fixedjoint analysis. The effect of notional horizontal loading will be ignored for this design example as its effect will be negligible. Table 9.1 shows the results from a pin-joint analysis under factored variable and permanent actions for half the truss. Member design: The top and bottom chord together with the end posts will be fabricated from the same section. Maximum force (compressive) NSd is in member CF and is 379 kN. (The maximum tensile force is 320 kN and will not be critical.) From cl BB.1.3 the buckling length may be taken as 0,9L in both planes where L is the system length. Effective length: LFC = 3,105 m (slope length).

TABLE 9.1 Member forces for EXAMPLE 9.1. Fixed-joint analysis

Member

Pin-joint analysis Axial force kN

Axial force kN

Shear force kN

Moment End 1 kNm

End 2 kNm

AB CD EF GH IJ AD DE EG GI BC CF FH HJ BD CE FG HI

−165,24 −82,62 −28,92 9,54 82,62 0 309,83 371,79 357,50 −315,98 −379,14 −364,56 −315,98 320,64 68,37 −17,18 −63,08

−163,38 −82,12 −29,33 9,17 81,63 7,19 307,14 371,29 356,79 −311,97 −378,60 −364,02 −316,14 308,46 73,28 −17,27 −61,76

−7,19 −1,57 0,04 0,12 0 1,86 −0,50 0,04 −0,37 1,82 −0,37 0,12 −0,54 0,17 −0,08 0 0

2,85 −1,07 0,04 −0,12 0 −2,85 1,90 0,50 0,83 0,29 −1,61 −0,45 0,66 0,29 0,04 0,04 0,04

−2,93 1,16 −0,04 0,17 0 2,77 0,87 0,62 0,29 −0,29 −0,45 −0,78 −0,99 −0,29 0,25 0,08 0

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Chapter 9 / Trusses Out-of-plane buckling: Buckling effective length is the length of the member, that is, 3,105 m. In-plane buckling Buckling length is 0,9LFC = 2,795 m, as there is a large degree of restraint owing to the welded joints (and continuity of the chord member). Try a 100 × 100 × 8 HFRHS. A = 2910 mm2 , i = 37,4 mm, I = 4,08 × 10−6 m4 . Section classification: c b − 2r − 2t 100 − 2 × 8 − 2 × 8 = = = 8,5 t t 8 Limit for Class 1: 235 235 33ε = 33 = 33 = 26,8 fy 355 Section is Class 1. Use the greatest system length of 3,105 to calculate Ncr : π2 EI π2 × 210 × 106 × 4,08 × 10−6 = = 877 kN 2 L 3,1052 Afy 2910 × 355 × 10−3 λ= = = 1,085 Ncr 877

Ncr =

For a rolled hollow section, buckling curve ‘a’ is used, α = 0,21: = 0,5 [1 + α(λ − 0,2) + (λ)2 ] = 0,5 [1 + 0,21(1,085 − 0,2) + 1,0852 ] = 1,182 χ=

+

!

Nb,Rd = χA

1 2 − (λ)2

=

1,182 +

1 1,1822 − 1,0852

= 0,606

fy 355 × 10−3 = 626 kN = 0,606 × 2910 γM1 1,0

All the remaining members will be the same section size in order to aid fabrication. Maximum force NSd is in member BC and is 320 kN (tension). Try a 60 × 60 × 6,3 HFRHS. A = 1330 mm2 , i = 21,7 mm, I = 63,4 × 108 mm4 . Section classification: c b − 2r − 2t 60 − 2 × 6,3 − 2 × 6,3 = = = 5,52 t t 6,3

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Limiting value for Class 1: c 235 235 = 33 = 33 = 26,8. t fy 355 Section is therefore Class 1. Nt,Rd = A

fy 355 = 1330 × 10−3 = 472 kN γM0 1,0

However the compression members also need checking. Member DC carries the largest compression force of −82,6 kN. Length of DC is 1,4 m. Critical effective length is that for out-of-plane buckling, that is, 1,4 m. π2 EI π2 × 210 × 106 × 63,4 × 10−8 = = 670 kN 2 L 1,42 Afy 1330 × 355 λ= = = 0,839 Ncr 670 × 103

Ncr =

For a hot rolled RHS, α = 0,21, so = 0,5 [1 + α(λ − 0,2) + (λ)2 ] = 0,5 [1 + 0,21(0,839 − 0,2) + 0, 8392 ] = 0,987 χ=

+

!

Nb,Rd = χA

1 2 − (λ)2

=

0,987 +

1 0,9872 − 0,8392

= 0,664

fy 355 = 0,664 × 1330 × 10−3 = 314 kN > NSd γM1 1,0

The deflections were also assessed on a pin-jointed analysis and gave 0,018 m under variable load (span/1333) and 0,033 m (span/727) under variable together with permanent loading. Both these deflection ratios are acceptable. Check on self-weight of the truss: From the member sizes determined on the basis of the force distribution for the pin-joint analysis, the total mass of the truss was 1,6 ton compared with the estimate of 3,6 ton. This means that both the total forces and the total deflection have been overestimated by around 10%. The check on self-weight is carried out in Table 9.2. Web capacity in the bottom chord at A (Section 4.8). The reaction is 165,2 kN. Note as a rolled hollow section is being used, fyw = fyf = 355 MPa. Since the length of stiff bearing ss is not known, set ss = 0 and determine the value required should the check fail. For an end support with c = ss = 0, kF = 2 (type c)

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Chapter 9 / Trusses TABLE 9.2 Check on self-weight estimate for EXAMPLE 9.1. Member

Total length (m)

AQ AB and QR BJ and JR BC and OR DE and PH FG and NK HI and LI DC and PO EF and NH HG and LK JI Total mass

24 1,6 24,48 6,21 6,62 7,21 7,68 2,8 4,0 5,2 3,2

Sub total (m)

Mass/unit length (kg/m)

Mass (kg)

50,08

22,9

1147

42,92

10,5

451 1598

Determine m1 : m1 =

fyf bf bf 50 = = = 6,25 tw 8 fyw tw

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 6,25 = 8,0 = 14,1 mm 2 2

The depth of the web hw has been taken as the clear depth of the web, hw = h − 2r − 2t = h − 4t = 100 − 4 × 8 = 68 mm tw3 8,03 = 2846 kN = 0,9 × 2 × 210 hw 68 ly tw fyw 14,1 × 8,0 × 355 λF = = = 0, 119 FCR 2846 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0 χF =

0,5 0,5 = = 4,2 0,119 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 14,1 = 14,1 mm FRd = Leff tw

fyw 355 = 14,1 × 8,0 × 10−3 = 40 kN γM1 1,0

The total resistance therefore is 80 kN (from both webs). A stiff bearing is therefore required to supply a further 165,2 − 80 = 85,2 kN.

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Try a 25 mm stiff bearing. 25 + 0 ss + c kF = 2 + 6 = 4,21 =2+6 hw 68 lc =

kF Etw2 3,33 × 210 × 103 × 82 ≤ 25 + 0 = 927 ≤ 25 ≤ ss + c = 2fyw hw 2 × 355 × 68

The limiting value of lc is greater than ss , so lc = 25 mm. 2 2 m1 lc 25 6, 25 ly = lc + tf = 25 + 8,0 = 53,7 mm + + 2 tf 2 8 √ ly = lc + tf m1 = 25 + 8,0 6,25 = 45 mm

(a) (b)

The value of ly using the equation for cases (a) and (b) is clearly larger, thus the value of lc is the lesser of those above, that is, 45 mm tw3 8,03 = 0,9 × 4,21 × 210 = 5991 kN hw 68 ly tw fyw 45,0 × 8,0 × 355 λF = = = 0,146 FCR 5991 × 103

FCR = 0,9kF E

χF =

0,5 0,5 = = 3,42 0,146 λF

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 45 = 45 mm FRd = Leff tw

fyw 355 = 45 × 8,0 × 10−3 = 128 kN γM1 1,0

The total resistance therefore is 256 kN (from both webs). This exceeds the reaction of 165,2 kN. Thus the length of stiff bearing could be reduced to around 20 mm. Note also that the end of the bottom member will be sealed by a thin plate welded over its end to prevent corrosive matter reaching the inside of the tubular section. This will have a slight stiffening effect. The results from a fixed-joint analysis are also given in Table 9.1 where it will be observed that with the exception of the members close to the supports, the axial force resultants show negligible difference. The exception is member AD which now carries an axial force of 7 kN and a moment of 2,85 kNm (Mpl,Rd = 35,5 kNm). Thus it may be concluded the secondary forces induced by joint fixity are negligible, and may be ignored. In the fixed-joint analysis the deflections are reduced by around 1–2%.

9.2 NON-TRIANGULATED TRUSSES Another common form of truss is the Vierendeel girder. This is non-triangulated, even though the top and bottom cords may not be parallel. As the loading is carried by sway,

•

Chapter 9 / Trusses there are substantial moments at the nodes, although the shears and axial forces of reasonable size. As a result, the Vierendeel girder is automatically classified as a sway frame and is generally analyzed elastically with the system or effective lengths of the members determined for the sway case. The notional horizontal load is applied at the end of the truss in the direction of the top chord.

EXAMPLE 9.2 Design of a parallel chord Vierendeel girder. Prepare a design in Grade S355 steel using rectangular hollow section for the girder detailed in Fig. 9.2. Loading: Permanent (kPa):

Sheeting Purlins Total

0,22 0,10 0,32

UDL per truss = 0,32 × 8,5 = 2,72 kN/m Self-weight (estimated) = 2,00 kN/m Total permanent = 4,72 kN/m or a nodal load of 3,0 × 4,72 = 14,16 kN (internal node) Variable action = 0,6 kPa, or 8,5 × 3,0 × 0,6 = 15,3 kN per node Total factored ultimate load per internal node = 1,35 × 14,16 + 1,5 × 15,3 = 42,07 kN For this example any load combination involving wind will not be considered. Note, however, where the load cases including wind to be considered then the moments due to wind would need multiplying by the sway amplification factor. Determination of notional horizontal loading: Determine αh from Eq. (8.17): 2 2 αh = √ = √ = 1,414 2 h The maximum value allowed for αh is 1,0,

A

C

E

G

J

K

M

O

Q

B

D

F

H

K

L

N

P

R

2,0 m

348

8 bays @ 3 m

FIGURE 9.2 Truss geometry for EXAMPLE 9.2

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thus αh = 1,0. Determine αm from Eq. (8.18): 1 1 αm = 0,5 1 + = 0,5 1 + = 0,745 m 9 Determine ϕ from Eq. (8.16): φ = φ0 αh αm = 1,0 × 0,745

1 = 3,725 × 10−3 200

Total factored vertical loading is 8 × 42,07 = 336,6 kN. Thus the notional horizontal load is 336,6 × 3,725 × 10−3 = 1,25 kN. Determination of αcr : A notional load of 1% of the total factored load is applied at the top chord level. The load is 0,01 × 336,6 = 3,37 kN. The actual analysis for both these cases was carried out using a computer package with a 1 kN load, and the results were obtained pro-rata. The results for the notional horizontal load of 1,25 kN is given in Table 9.3, and the deflections (absolute and relative) for the notional load of 3,37 kN are given in Table 9.4. As only a single storey is being considered, then from Table 9.4 the maximum relative deflection is 0,058 mm. Thus 0,058 = = 29 × 10−6 h 2000

TABLE 9.3 Internal stress resultants due to the notional horizontal load. Moment Member

Axial force kN

Shear force kN

End 1 kNm

End 2 kNm

AB CD EF GH IJ AC CE EG GI BD DF FH HJ

−0,52 0 0 0 0 −1,16 −1,00 −0,86 −0,70 1,16 1,0 0,85 0,70

0,093 0,15 0,15 0,16 0,16 −0,05 −0,05 −0,05 −0,05 0,05 −0,05 −0,05 −0,05

−0,0845 −0,1475 −0,154 −0,155 −0,155 −0,085 −0,077 −0,075 −0,074 0,085 0,077 0,075 0,074

0,0845 0,1475 0,154 0,155 0,155 0,072 0,079 0,081 0,083 −0,072 −0,079 −0,081 0,083

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Chapter 9 / Trusses TABLE 9.4 Determination of αcr . Deflection Member

Top mm

Bottom mm

Net deflection mm

AB CD EF GH IJ KL MN OP QR

0,058 0,053 0,049 0,045 0,042 0,040 0,038 0,037 0,037

0 −0,005 −0,009 −0,012 −0,016 −0,018 −0,020 −0,021 −0,021

0,058 0,058 0,058 0,057 0,058 0,058 0,058 0,058 0,058

From Eq. (8.10), the value of αcr is given by αcr =

0,009 h

=

0,009 = 310 29 × 10−6

The minimum value for elastic analysis of αcr is 10, thus second order effects may be ignored. The magnification factor to be applied to moments causing sway (i.e. for this particular example only the moments due to the notional horizontal loads) is (cl 5.2.2 (5)B, EN 1993-1-1) 1 1−

1 αcr

1

=

= 1,003 1 1 − 310

The stress resultants due to the vertical loads are given in Table 9.5. The bending moment, shear force and axial force diagrams for the total loading are plotted in Fig. 9.3(a) to (c), respectively for half the frame Try a 250 × 150 × 12,5 S355J2H Member checks: Section classification: Web: The web is subject to compression: Maximum axial compressive force NEd is 479 kN. Length of web χw to resist the compressive force is given by χw =

NEd fy

2tw γ

M0

=

479 × 103 2 × 12,5 355 1,0

= 54 mm

Structural Design of Steelwork to EN 1993 and EN 1994

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TABLE 9.5 Internal stress resultants due to the factored vertical load. Moment Member

Axial force kN

Shear force kN

End 1 kNm

End 2 kNm

AB CD EF GH IJ AC CE EG GI BD DF FH HJ

−94,53 −21,20 −20,99 −21,04 −21,04 −116,11 −291,67 −415,78 −478,29 116,11 291,67 415,78 478,29

−116,11 −175,56 −62,52 −62,52 0 73,50 52,63 31,55 10,52 73,75 52,55 31,55 −10,52

−166,11 −175,56 −124,11 −62,52 0 115,95 −70,93 −35,04 −5,01 116,24 70,72 37,15 5,01

116,24 175,56 123,90 62,52 0 104,59 86,96 57,51 26,55 −104,93 −86,92 −57,51 −26,55

479 −117

−

− −21

−95 118

416

293

−

−

−21

−21

+

+

293

−21 + 479

417 (a) AFD (kN)

+73

+55

−116

−175

+73

+53

+32

+10

−123

−62

+32

+10

(b) SFD (kN)

71

175

124

37 63

176 87

104 116

71

58

26 0

5

37 175

5

124

63

87 104 (c) BMD (kNm) contension force

FIGURE 9.3 Final AF, SF and BM diagrams for EXAMPLE 9.2

26

351

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Chapter 9 / Trusses c = h − 2tf − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm αc = 0,5c − 0,5ax = 100 + 27 = 127 mm 127 α= = 0,635 200 c 200 = = 16 tw 12,5 Class 1 limit for α > 0,5: 235 fy

!

235

c 355 = 396 = 396 = 44,4 tw 13α − 1 13 × 0,635 − 1 Web is Class 1. Flange: c = b − 2tf − 2r = 150 − 2 × 12,5 − 2 × 12,5 = 100 mm 100 c = =8 tf 12,5 Limit for Class 1 (Table 5.2 Sheet 1): c 235 235 = 72 = 72 = 58,6 tf fy 355 Flange is Class 1. The section is therefore Class 1. Shear: The shear area Av for a load parallel to the depth is given in cl 6.2.6 (EN 1993-1-1) as h 250 = 9300 = 5813 mm2 h+b 250 + 150 fy 355 1 1 = √ Av = √ 5813 × 10−3 = 1191 kN γM0 1,0 3 3

Av = A Vpl,Rd

The maximum applied shear force is 175,4 kN. Thus the section is satisfactory and there is no moment capacity reduction due to shear (VEd /Vpl,Rd = 0,15). Members BD, DF, FH and HJ need checking under combined tension and bending. This check reduces to determining whether the reduced flexural capacity due to axial load exceeds the applied moment. For box sections: MN,y,Rd = aw =

1−n Mpl,y,Rd 1 − 0,5aw A − 2bt 9300 − 2 × 150 × 12,5 = = 0,597 A 9300

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Limiting maximum value of aw = 0,5, therefore aw = 0,5. 1 − 0,5aw = 1 − 0,5 × 0,5 = 0,75 fy 355 Mpl,y,Rd = Wpl,y = 751000 × 10−6 = 267 kNm γM0 1,0 fy 355 = 9300 × 10−3 = 3302 kN γM0 1,0 NEd n= Npl,Rd

Npl,Rd = A

The checks are carried out in Table 9.6, where it should be noted that in no case is the plastic moment capacity reduced, and that the values quoted for My,Ed are the larger absolute values within the member. Members AB, CD, DF, GH, AC, CE, EG, GI need checking under combined bending and compressive axial force. The check for reduction in moment capacity for members AC to GI is exactly the same as for BD to HJ. Members AB and CD carry lower axial forces and will not therefore be critical for moment capacity reductions. Lateral torsional buckling: The system length may be taken as the worst case of member length, that is, 3 m. Lateral torsional buckling is checked using Section 5.1.8.1. Determine φb from Eq. (5.63):

φb =

=

" #" # W 2 1 − Iz 1 − GIt pl,y Iy EIy AIt " #" # 751 × 103 2 1 − 3310 1 − 81×7317 7518 210×7518 9300 × 7317 × 104

= 0,538

The moment gradient factor C1 should be determined for the greatest ratio of end moments which occurs in member GI. ψ=

−5,08 = −0,191 26,63 TABLE 9.6 Member capacity checks for members in tension. Member

NEd kN

n

MN,y,Rd kNm

Mpl,y,Rd kNm

My,Ed kNm

BD DF FH HJ

117,3 292,7 416,6 479,0

0,036 0,089 0,126 0,145

343 324 311 304

267 267 267 267

116,3 87,0 57,6 26,6

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Chapter 9 / Trusses Determine C1 from Eq. (5.40): 1 = 0,6 + 0,4ψ = 0,6 + 0,4(−0,191) = 0,524 > 0,4 C1 Thus C1 = 1,908 The slenderness λ is given by λ=

L 3000 = = 50,3 iz 59,7

Determine the lateral torsional buckling slenderness ratio λLT from Eq. (5.62): 1/2 1/2 E 1 210 1 λLT = 1/2 π (φb λ)1/2 = √ (0,538 × 50,3)1/2 = 8,47 π G 81 1,908 C 1

Determine the normalized slenderness ratio λLT from Eq. (5.60): λLT =

λLT π E f y

=

!

8,47

3 π 210×10 355

= 0,111

As λLT < 0,4, lateral torsional buckling cannot occur. Check the critical length for lateral torsional buckling lcrit from Eq. (5.64a): b−t 2 3 + b−t 113400 (h − t) h−t h−t lcrit = b−t fy 1+3 1 + b−t h−t

h−t

150−12,5 2 3 + 113400 (250 − 12,5) 250−12,5 = 150−12,5 355 1 + 3 250−12,5 1 +

150−12,5 250−12,5 150−12,5 250−12,5

= 14000 mm This is well in excess of the system length of 3 m. This also confirms lateral torsional buckling will not occur. Check Eq. (5.64b) from Kaim (2006), 25 235 λz,lim = h fy b 235 25 235 55167 55167 λz,lim = λz,lim λ1 = h 93,9 = h = 250 = 93,2 fy fy 355 f y 150 b b L = λz,lim iz = 93,2 × 59,7 = 5560 mm The actual length of 3 m is below the critical value, therefore lateral torsional buckling will not occur. Thus as lateral torsional buckling is not critical Table B.1 can be used to determine the values of kyy and kzy .

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355

TABLE 9.7 Member capacity checks for compression members. Member Calculation

AC

CE

EG

GI

KC K1 K 11 K 12 k1 K2 K 21 K 22 k2 l cr,y /L (Eq. (8.22)) l cr,y /L (Eq. (8.23)) l cr,y /L (Eq. (8.24)) M Ed,max M Ed,min N Ed N cr,y λy χy N cr,z λz χz kyy kzy NEd χy Afy kyy MEd,max Wpl,y fy Total NEd χz Afy kzy MEd,max Wpl,y fy Total

0,3333 0 0 0,5 0,4 0,3333 0 0,5 0,571 1,470 1,418 1,473 116,03 −104,66 117,27 7979 0,643 0,873 7623 0,658 0,867 0,407 0,244

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 87,04 −71,01 292,68 6565 0,709 0,843 7623 0,658 0,867 0,421 0,253

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 57,59 −35,12 416,63 6565 0,709 0,843 7623 0,658 0,867 0,430 0,258

0,3333 0,3333 0 0,5 0,571 0,3333 0 0,5 0,571 1,612 1,571 1,624 26,63 kNm −5,08 kNm 478,99 kN 6565 kN 0,709 0,843 7623 kN 0,658 0,867 0,570 0,342

0,041

0,105

0,150

0,172

0,177

0,138

0,093

0,057

0,218

0,243

0,243

0,229

0,041

0,102

0,146

0,168

0,106

0,083

0,056

0,034

0,147

0,185

0,201

0,202

The calculations for member capacity are carried out in Table 9.7 where it will be seen that the member capacities are overgenerous. The 2 m verticals (AB, CD, EF and GH) have not been checked as they are shorter than the top chord compression members (therefore have a higher buckling load) and carry lower axial forces. The strut buckling lengths have been determined assuming the frame can sway. This is conservative. The buckling length co-efficients have been determined using Eqs (8.22)–(8.24) for comparison. As will be noted in Table 9.7 There is little difference between the results. In this example those from Eq. (8.24) have been used for buckling in-plane. For out-of-plane buckling, the actual length has been used, although this is

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•

Chapter 9 / Trusses conservative. To determine the effective lengths, the EI value of the section has been taken as unity as all the members are fabricated from the same section. Thus EI/L reduces to 1/L. The axial forces although compression are given as positive, as is the larger end moment. The smaller one is given the appropriate sign. Deflection: Permanent load deflection:

22,4 mm (span/1070)

Variable load deflection:

24,2 mm (span/992)

Total deflection:

44,6 mm (span/538)

These are satisfactory. Web check (Section 4.8): The web needs checking at B with the capacity calculated for a single web and then doubled. Reaction = 168,3 kN, moment = 116 kNm. For an end support with c = ss = 0, kF = 2 (type c) Determine m1 : m1 =

bf fyf bf 75 = 6,0 = = fyw tw tw 12,5

As m2 is dependant upon λF initially assume m2 = 0. As ss and c have been assumed to be zero, then lc = 0, then the least value of ly is given by ly = tf

m1 6,0 = 12,5 = 21,7 mm 2 2

The depth of the web hw has been taken as hw = d − 2t − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm tw3 12,53 = 0,9 × 2 × 210 = 3691 kN hw 200 ly tw fyw 21,7 × 12,5 × 355 λF = = = 0,162 FCR 3691 × 103

FCR = 0,9kF E

As λF < 0,5, m2 = 0. χF =

0,5 0,5 = 3,1 = 0,162 λF

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357

The maximum value of χF is 1,0, thus Leff = χF ly = 1,0 × 21,7 = 21,7 mm FRd = Leff tw

fyw 355 = 21,7 × 12,5 × 10−3 = 96,3 kN γM1 1,0

The total for both webs is 192,6 kN. This exceeds the reaction of 168,3 kN. η2 =

FEd fyw Leff tw γ M1

=

FEd 168,3 = = 0,874 FRd 192,6

η2 = 7,17 kNm/m 2 2 Hogging: Design as reinforced concrete ignoring sheeting: The depth allowing for the trapezoidal indents is given by 105 − 0,5 × 51(38 + 12) = 96 mm. Assume 25 mm cover and 10 mm bars, d = 96 − 25 − 5 = 66 mm. Assuming αcc = 0,85, then using Eq. (6.22) in Martin and Purkiss (2006), MSd 8,72 × 106 = 0,080 = bd2 fck 1000 × 662 × 25 As fyk MSd = 0,652 − 0,425 − 1,5 2 = 0,652 − 0,452 − 1,5 × 0,080 = 0,0997 bdfck bd fck As =

0,0997 × 1000 × 66 × 25 = 329 mm2/m. 500

Fix B385 mesh [385 mm2 /m] Flexural shear: Maximum shear is at A1 on span AA1: Co-efficient for complete UDL is 1,418 + 0,124 − 0,019 + 0,019 − 0,008 + 0,002 = 1,537 Shear due to permanent loading = 1,537 × 6,08 = 9,345 kN/m

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369

Co-efficient for UDL on AB and BC is 1,418 + 0,124 = 1,542 Shear due to variable loading = 1,542 × 6,0 = 9,252 kN/m. Total shear = 9,345 + 9,252 = 18,6 kN/m. b0 is the distance between rib centres, that is, 150 mm. Applied shear per rib = 18,6(150/1000) = 2,79 kN/m Using Eq. (10.6) to calculate Vv,Rd , 385 = 5,83 × 10−3 1000 × 66 200 200 k =1+ =1+ = 2,74 > 2,0 d 66 ρ1 =

1 0,15 3 Vv,Rd = CRd,c k 100ρ1 fck 3 b0 d = (2 × 150 × 66) 100 × 5,83 × 10−3 × 25 1,25 = 5,8 kN/rib This is greater than the applied shear. Shear bond: For an end span Ls = 0,9 L = 0,9 × 2,5 = 2,25 m, as the loading is a UDL Ls /4 is used, that is, 2250/4 = 562.5 mm. The values of m (=166,6) and k (=0,005) were from British standard tests, thus whilst the value of m remains unchanged the value of k has to be amended. 1/2

k = 0,005 × 1,12 × fck = 0,005 × 1,12 × 251/2 = 0,028 Use Eq. (10.3) to calculate Vl,Rd : bdp mAp 1000 × 88,3 166,6 × 1507 Vl,Rd = + 0,028 +k = γvs 1, 25 1000 × 625 bLs = 32,0 kN/m > VSd Deflection: Initially assume there will be no problems with end slip. From Table 7.4 EN 1992-1-1, basic span depth ratio for the end span of a continuous beam (or slab) is 26 (lightly stressed). Actual span 2500 = = 28,3 effective depth ratio 105 − 16,7 This is higher than the allowable of 26, but since the mid-span capacity is much higher than the applied moment, the section will be relatively uncracked thus reducing deflections.

370

•

Chapter 10 / Composite Construction End anchorage: Use 20 mm shear studs. Tensile force required = Np = 434000 N/m. Bearing resistance check Eq. (10.4): For a single shear stud, take a at its minimum distance, that is, 2d0 , thus from Eq. (10.5), kφ = 1 +

a 2 × dd0 =1+ =3 0,5 or 36ε/α when α < 0,5. For Class 2 sections the constants are 456 and 41,5. The ratio α is the normalized depth of the compression zone of the web. Where partial interaction is used, the web classification must be carried out using full interaction.

10.3.2 Design Criteria In addition to the expected checks for flexure, vertical shear (carried by the web only), web capacity under in plane forces and deflection, the transverse shear and the shear connector capacity also need checking.

10.3.3 Flexural Design For stresses during construction the difference between propped and unpropped methods need to be noted. • Propped construction The load due to the concrete is not transferred to the composite section until the concrete has hardened. However, additional forces due to the release of the props may also need considering. • Unpropped construction The effect of the wet concrete is taken on the steel beam alone before the onset of composite action. In both cases the effects of finishes and variable loads are taken on the composite section. It should be noted that in unpropped construction the steel beam may still need checking for lateral torsional buckling during the construction stage. Where the profile decking runs parallel to the span of the beam, the sheeting does not provide full restraint. In the case of decking perpendicular to the span, full restraint is likely to be available (Lawson and Nethercot, 1985). At ultimate limit state, it is not relevant whether the construction was propped or unpropped.

372

•

Chapter 10 / Composite Construction Owing to the relatively high flange widths of composite beams, the whole flange cannot be taken in calculating the moment capacity of the beam owing to shear lag. The effective width either side of the web centre line should be considered separately. The partial effective width be,i should be taken as Le /8, where Le is the span for simply supported beams. The total effective width beff is then given as 2be,i + b0 , where b0 is the transverse distance between the shear connectors. In practice the contribution of b0 may be neglected. The total effective width beff for continuous beams, see cl 5.4.1 (EN 1994-1-1). The partial effective width should not exceed either the distance to the free edge or half the spacing to the adjacent beam centre line. The following assumptions are made when determining the flexural capacity at ULS: • The concrete in between the ribs is ignored, and the remainder is stressed with a uniform strength of 0,85fck /γc . • The profile decking may be included if it is in tension and is then stressed to fy /γap , otherwise it is ignored. • Reinforcement is stressed to fsk /γs where fsk is the characteristic strength of the reinforcement (compression reinforcement may be ignored). • The steel beam is stressed to a uniform stress of fy /γa .

10.3.3.1 Sagging All symbols are defined in Fig. 10.6(a). The force in the concrete Nc is given by

0,85fck Nc = beff hf − hp γc

(10.8)

The force in the steel section Na is given by Na = Aa

fy γa

(10.9)

where Aa is the cross-sectional area of the steel beam. There are three possibilities for the position of the plastic neutral axis, in the concrete slab, the top flange of the steel beam or the web of the steel beam. If the neutral axis lies in the concrete slab, then Nc > Na , otherwise the neutral axis is in the steel beam. (a) Neutral axis in concrete slab (Fig. 10.6(b)). Depth of neutral axis in the slab x is given by Na x = 0,85f ck beff γ

(10.10)

c

The moment capacity Mpl,Rd is then given by h x Mpl,Rd = Na + hf − 2 2

(10.11)

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373

0,85 fck gc

beff Concrete slab

hp

hf

Nc

Neutral axis

h

Area

x

Aa

tw

tf bf

(a) Basic data

(b) Neutral axis in slab

Nc

N af Naw

x

Na

Neutral axis

ht f x 2

h 2

ht f 2 (c) Neutral axis in web

Nc Neutral axis

N afx

x

Na

h x 2

(d) Neutral axis in flange

FIGURE 10.6 Determination of plastic moment capacity (sagg mg). (b) Neutral axis in the steel beam (Na > Nc ) Define an out-of-balance steel compression force Nac as Nac = Na − Nc

(10.12)

and the capacity of the flange Naf (ignoring fillets) as Naf = bf tf

fy γa

(10.13)

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•

Chapter 10 / Composite Construction The out-of-balance force is introduced to enable the force in the beam Na to be retained with its point of application of h/2 in the moment capacity calculations. The neutral axis is • in the web if Nac > 2Naf or • in the flange if Nac < 2Naf . Case 1: Neutral axis in the web (Fig. 10.6(c)) Depth of web x in compression is given by x=

Nac − Naf

(10.14)

fy

2tw γ

a

The compression force in the web Naw is given by Naw = xtw

fy γa

(10.15)

and the design moment of resistance Mpl,Rd is given by

h p + hs h+ 2

Mpl,Rd = Nc

+ 2Naf

tf h− 2

h x + 2Naw h − tf − − Na (10.16) 2 2

Case 2: Neutral axis in the flange (Fig. 10.6(d)) Depth of flange x in compression is given by x=

Nac

(10.17)

fy

2b γ

a

The force in the flange Nafx is given by Nafx = xb

fy γa

(10.18)

and the design moment of resistance Mpl,Rd is given by h p + hf h x Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2

(10.19)

10.3.3.2 Hogging (negative moment) Conservatively the contribution of the steel profile decking will be ignored. All symbols are defined in Fig. 10.7(a). The force in the reinforcement Nr is given by Nr = As

fsk γs

(10.20)

•

Structural Design of Steelwork to EN 1993 and EN 1994 Reinforcement As

As

375

c

h

Nr

Naw

Neutral axis

Fs

Na

tf 2

tw

h

h 2

h

Steel beam (area As)

Naf

tf x

tf

h tf

x 2

bf

tf

(a) Definition of symbols

(b) Neutral axis in web

As

Nr Nafx

h

x 2

x

Neutral axis

Na

(c) Neutral axis in top flange

FIGURE 10.7 Calculation of the plastic moment capacity (hogging). The force in the steel beam Na is given by Eq. (10.9), and the force in a flange Naf is given by Eq. (10.13). The neutral axis is in • the web, if Na > Nr + 2Naf , • the flange, if Na < Nr + 2Naf . Case 1: Neutral axis in web (Fig. 10.7(b)) The depth of the web x in tension is given by x=

Na − Nr − 2Naf fy

2tw γ

(10.21)

a

The capacity of the tension part of the web Naw is given by fy

Naw = xtw γ

a

The moment capacity of the section Mpl,Rd is given by

tf h x Mpl,Rd = Nr h + hs − c + Naf h − + Naw h − tf − − Na 2 2 2

(10.22)

(10.23)

Case 2: Neutral axis in web (Fig. 10.7(c)) The depth of the flange x in tension is given by x=

Na − Nr fy

2b γ

a

(10.24)

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•

Chapter 10 / Composite Construction The capacity of the tension part of the flange Nafx is given by Nafx = xbf

fy γa

(10.25)

The moment capacity of the section Mpl,Rd is given by

x h Mpl,Rd = Nr h + hf − c + Nafx h − − Na 2 2

(10.26)

10.3.3.3 Reduction in Plastic Moment Capacity Due to Shear In a similar fashion to plain steel beams, the moment capacity is reduced when the vertical shear VEd exceeds half the plastic shear resistance VRd . A reduced design strength (1 − ρ)fyd to determine the capacity of the steel section is used where ρ is defined as 2 2VEd −1 (10.27) ρ= VRd For calculation of plastic shear capacity see Section 10.3.4.

10.3.3.4 Reduction in Sagging Moment Capacity with Partial Shear Connection (cl 6.2.1.3) The section capacity should be calculated using a force in the concrete section of Nc,f where Nc,f is given by Nc,f = ηNc

(10.28)

where η is the degree of shear connection. The moment capacity Mpl,Rd is then conservatively given by MRd = Mpl,a,Rd +

Nc Mpl,Rd − Mpl,a,Rd Nc,f

(10.29)

where Mpl,Rd is the moment capacity of the composite section with full shear connection and Mpl,a,Rd is the plastic moment capacity of the steel section alone.

10.3.3.5 Elastic Capacity For elastic resistance conventional elastic theory is used with the slab taking its effective width beff and where appropriate account is taken of creep (cl 6.2.1.5). The limiting stresses for Mel,Rd are 0,85 fck /γc for concrete, fy /γa for structural steel (Class 1,2 or 3 cross-sections) and fsk /γs for reinforcing steel. Account should be taken of creep and shrinkage in determining the elastic capacity. However, for beams with only one flange composite this may be achieved by using an appropriate modular ratio nL (cl 5.4.2.2.2). The modular ratio nL is defined as nL = n0 (1 + ψL φt )

(10.30)

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377

where n0 is the short-term modular ratio defined as Ea /Ecm , ϕt is the creep co-efficient defined as ϕ(t,t0 ) in EN 1992-1-1, and ψL is creep multiplier depending on the type of loading (1,1 for permanent loads, 0,55 for shrinkage and 1,5 for prestressing by imposed deformations). For cases where any amplification of internal forces is less than 10% due to deformations, the structure is not mainly intended for storage nor prestressed by imposed deformations, the effect of creep can be taken into account for both short- and long-term loading by using a nominal modular ratio n determined using an effective concrete elastic modulus Ec,eff given by 0,5Ecm . Note, there appears to be no explicit requirement to check serviceability elastic stresses if the plastic moment capacity is used to determine the strength of the beam.

10.3.4 Flexural Shear The flexural shear capacity Vpl,Rd is calculated exactly as for a normal steel beam. Additionally for an unstiffened and uncased web d/tw should not exceed 69ε, for a cased web the constant is 124.

10.3.5 Design of Shear Connectors The strength of shear connectors is dependant upon both the strength and elasticity of the concrete and the ultimate tensile strength of the connector itself. Following a large series of tests Olgaard et al. (1971) proposed the following equation for the strength of shear stud connectors kd2 fcu Ec PRd = (10.31) 1,25 where d is the diameter of the stud, Ec Young’s Modulus of the concrete, fcu concrete cube strength and k is an empirical constant allowing for the height to diameter ratio of the stud. Using the 80% utilization factor proposed by Yam and Chapman (1968), and converting the cube strength to cylinder strength by a factor of 0,8 gives the following strength formula, 0,29αd2 fck Ecm PRd = (10.32) γv where γv takes a value of 1,25 and for 3 < hsc /d < 4 hsc α = 0,2 +1 d

(10.33)

for h/d > 4 α = 1,0

(10.34)

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Chapter 10 / Composite Construction Oehlers and Johnson (1987) carried out further examination of shear stud capacity and proposed an upper bound which was dependant upon the ultimate strength of the shear stud. Their results have been simplified in the code and are given as PRd =

0,8fu πd2 γv 4

(10.35)

where fu is the ultimate tensile strength of the stud material (taken as not greater than 500 MPa). For decks with ribs parallel to the beam the strength of the shear studs needs to be reduced by the parameter k1 as the containment by the concrete is incomplete (Mottram and Johnson, 1990), kl = 0,6

b0 hp

hsc − 1 ≤ 1,0 hp

(10.36)

where hp is the height of the profile and hsc is the overall height of the stud ( sf cot θf

(10.45)

where θf is the angle of inclination of the concrete truss member such that 1,0 ≤ cot θf ≤ 2,0. For normal weight concrete the shear resistance vEd is given by vEd = 0,5vfcd

(10.46)

where v is given by 0,6 (1 − fck /250). For lightweight concrete the shear resistance vEd is given by vEd = 0,5η1 v1 fcd

(10.47)

where v1 is given by 0,5η1 (1 − fck /250), and η1 is given by 0,4 + 0,6ρ/2200 where ρ is the density of the concrete. A limit is placed on vEd such that vEd < vfcd sin θf cos θf

(10.48)

It would appear acceptable to take the traditional approach and use an angle of 45◦ √ in the truss analogy giving cot θf = 1,0 and sin θf = cos θf = 1/ 2 (Johnson, 2004).

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1

At 2

2

Ab

1

Plane 11 22

Steel area Ab At 2Ab

(a) Solid slab 3

At

Sheeting

3

Plane 33

Steel area 2At

(b) Composite slab (sheeting perpendicular to span) 4

5

4

5

At

Lab joint in sheeting

Plane 44 55

Steel area

(c) Composite slab (sheeting parallel to span)

2At

FIGURE 10.8 Critical transverse shear planes.

Where the profile sheet decking is normal to the span and is continuous over the beam then Eq. (10.45) applied to vertical shear planes may be enhanced by the effect of the decking to give Asf fyd vEd hf + Ape fyp,d > sf cot θf

(10.49)

where Ape is the effective cross-sectional area of the decking and fyp,d is its design strength.

Chapter 10 / Composite Construction beff

Concrete slab hp

hs

•

h

382

Steel beam (second moment of area,Ia area, Aa, modular ratio n)

FIGURE 10.9 Second moment of area calculation for a composite beam.

10.3.11 Deflection (cl 5.2.2) For the deflection of the steel member alone the principles of EN 1993-1-1 are applied. No account need be taken of partial interaction if η ≥ 0,5 or the force in the shear connector does not exceed PRd at the serviceability limit state, and for ribbed slabs transverse to the beam the rib height does not exceed 80 mm. The composite second moment of area may be determined as follows (Fig. 10.9)

3 2 2

beff hs − hp h s + hp h Ic = nIa + + nAa − x + beff hs − hp h + −x 12 2 2 (10.50) where x is given by

h +h nAa + hs − hp beff h + s 2 p

x= nAa + hs − hp beff

(10.51)

10.3.12 Vibration This is only likely to be critical on lightly loaded, long span beams. It is suggested that the following formula is used (Wyatt, 1989) 18 f =√ δsw

(10.52)

where f is the frequency (Hz) and δsw is the instantaneous deflection (mm) due to selfweight and permanent loading. The suggested limits for f are 4 Hz for most buildings, 3 for car parks and 5 for sports halls (Lawson and Chung, 1994).

10.3.13 Detailing 10.3.13.1 Cover This should be the greater of 20 mm or the values specified in EN 1992-1-1 (Table 4.4) less 5 mm (cl 6.6.5.2).

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10.3.13.2 Spacing (cl 6.6.5.5) This should be not less than 22tf ε for a solid slab or 15tf ε for a ribbed slab where the flange becomes a Class 1 or Class 2 by virtue of restraint due to shear connection with the edge distance not exceeding 9tf ε. The maximum spacing should not exceed the lesser of 800 mm or six times the slab thickness. The edge distance should also not exceed 20 mm. The overall height of a stud connector should be not less than 3d (where d is the shank diameter), the spacing in the direction of the shear force should be not less than 5d, and transverse to the shear force 2,5d for solid slabs and 4d for other cases. Unless the stud is directly over the web the diameter of the stud should not exceed 2,5tf (cl 6.6.5.7). For through-deck welding, the profile sheet steel decking should not exceed 1,25 mm thick if galvanized and 1,5 if not. The studs should extend 2d above the deck after welding. Johnson (2005) expresses concern over the detailing rules (and stud capacities) when applied to open trapezoidal decks.

EXAMPLE 10.2 Composite beam (decking transverse to span) Design a composite beam (ref C1/23 (Fig. 10.3)) in Grade S355 steel to carry the composite slab designed in Example 10.1. The concrete is lightweight LC25/30 and a dry specific weight of 19 kN/m3 . The span is 4 m and carries the two 2,5 m deck spans. Loading due to wet concrete: 0,105 × 19 = 2,0 kPa Finishes: 2,5 kPa Variable loading: 4,0 kPa The loading co-efficients for the reaction at A1 (from SFDs in Fig. 10.4) are given in Table 10.2 for all spans loaded. For the variable loading, spans AA1, A1B, CC1 and C1D are loaded. This gives an overall co-efficient of 3,062

TABLE 10.2 Loading co-efficients for reactions A1. Span

Individual components

Total

AA1 A1B BC CC1 C1D Overall total

1,418 + 0,218 0,124 + 1,254 −0,019 − 0,096 0,039 + 0,008 0,009 + 0,003

1,636 1,369 −0,115 0,047 0,012 2,949

Chapter 10 / Composite Construction Load per unit run on the beam: Permanent: 2,949(2,0 + 2,5) = 13,27 kN/m Variable: 3,062 × 4 = 12,25 kN/m Ultimate limit state design: MSd = (1,35 × 13,27 + 1,5 × 12,25) × 42 /8 = 72,6 kNm Effective width of slab, beff (ignoring spacing between connectors as it is likely only a single row will be needed) be,i = Le /8 = 4000/8 = 500 mm beff = 2be,i = 1000 mm. Try a 203 × 133 × 25 UKB (Grade S355) Since the flange is restrained by through deck studs, the beam flanges are automatically Class 1. The web slenderness satisfies the shear buckling check. Actual width between beam centre lines = 2500 mm > beff Dimensions to calculate Mpl,Rd assuming full shear connection are given in Fig. 10.10. From Eq. (10.8), the force in the concrete flange Nc is given by Nc =

0,85 × 25 0,85fck beff hf − hp = 1000 × (105 − 51) × 10−3 = 765 kN γc 1,5

From Eq. (10.9) the force in the steel beam Na is given by Na = Aa

fy 355 = 3200 × 103 = 1136 kN γa 1,0

From Eq. (10.12) the out-of-balance force Nac is given by Nac = Na − Nc = 1136 − 765 = 371 kN

1000

Concrete slab

51

Steel beam 20313325 UKB S 375

7,8 7,8

105

•

203,2

384

5,7

133,2

FIGURE 10.10 Design dimensions for EXAMPLE 10.2

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From Eq. (10.13) the capacity of the flange Naf is given by Naf = btf

fy 355 × 10−3 = 369 kN = 133,2 × 7,8 1,0 γa

Since Nac < 2Naf , the neutral axis is in the flange. From Eq. (10.17) the depth of flange x is given by x=

371 × 103 = = 3,92 mm fy 2 × 133,2 × 355 2b 1,0 Nac

γa

From Eq. (10.18) the compression force in the flange Nafx is given by Nafx = xb

fy 355 = 3,92 × 133,2 × 10−3 = 185 kN γa 1,0

From Eq. (10.19) Mpl,Rd is given by h p + hf x h Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2 105 + 51 3,78 203,2 = 0,765 203,2 + + 2 × 0,185 203,2 − − 1,136 2 2 2 = 174,2 kNm Note, for convenience all the forces have been expressed in MN. Then if the dimensions are left in mm, the moment is in kNm. Since the neutral axis for full shear connection is in the flange, the only web check is for shear buckling (see above). As Mpl,Rd exceeds MEd by a substantial margin, partial interaction may be used. For the beam alone the section is Class 2 (which is satisfactory). Check shear capacity: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 262 kN = √ 1280 1,0 γ 3 3 a VEd =

(1,35 × 13,27 + 1,5 × 12,25) × 4 = 72,6 kN 2

Since VSd < 0, 5Vpl,Rd , there is no reduction in moment capacity. Shear connectors: Use 100 mm long by 19 mm diameter stud connectors. The load capacity of a single stud, Eq. (10.32): h/d = 100/19 = 5,3, so from Eq. (10.34), α = 1,0 fck = 25 MPa

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Chapter 10 / Composite Construction For normal weight concrete, Ecm = 22

fck + 8 10

1/3 = 22

25 + 8 10

1/3 = 31,5 GPa

Modification factor for lightweight concrete ηE (from cl 11.3.2 of EN 1992-1-1) ρ 2 1900 2 ηE = = = 0,746 2200 2200 So, Ecm = 0,746 × 31,5 = 23,5 GPa From Eq. (10.32) PRd

0,29αd2 fck Ecm 0,29 × 1 × 192 × 25 × 23,5 × 103 = = × 10−3 = 64,2 kN γv 1,25

Limiting capacity: For ribs transverse to beam the ultimate connector strength (fu = 450 MPa) needs to be multiplied by the factor kt . Determine kt from Eq. (10.37) In view of the large overdesign on moment capacity, number of connectors per rib will be 1, that is, nr = 1. For a dovetail deck b0 is the distance between the top of the dovetails, that is b0 = 150 − 38 = 112 mm Profile height hp = 55 mm, height of connector, h = 100 mm, 0,7 b0 kt = √ nr hp

hsc 0,7 112 100 −1 = √ − 1 = 1,17 hp 55 1 55

However the maximum value allowed for kt is 1,0. Limiting capacity of studs from Eq. (10.35) PRd =

0,8fu πd2 0,8 × 450 π × 192 = × 10−3 = 81,7 kN γv 4 1,25 4

Shear stud capacity is the lower of the two values, that is 64,2 kN. No. of connectors for full shear connection: Nc = 765 kN, PRd = 64,2 kN, thus nf for the half beam span is given by nf = Nc /PRd = 765/64,2 = 11,9 Use 1 connector per rib, thus n for the whole beam is given by span over distance between centre lines of ribs = 4000/300 = 13,3.

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The ratio between η between n and nr is given by η=

n 13,3 = = 0,56 nr 2 × 11,9

The limiting value of η is given by Eq. (10.39) with Le = 4,0 m, thus η≥1−

355 355 0,75 − 0,03Le ≥ 0,4 = 1 − 0,75 − 0,03 × 4 = 0,37 fy 355

The limiting value is 0,4, as the actual value is 0,56, it is therefore satisfactory. The moment capacity MRd is given by Eq. (10.29). The moment capacity of the steel section Mpl,a,Rd alone is given by 355 Mpl,a,Rd = 258 × × 10−3 = 91,6 kNm 1,0 MRd = Mpl,a,Rd +

nc Mpl,Rd − Mpl,a,Rd = 91,6 + 0,56 (174,2 − 91,6) nc,f

= 137,9 kNm This is greater than MEd . Longitudinal shear: The only plane requiring checking is 4–4 (or 5–5) of Fig. 10.8 Determine VSd from Eq. (10.44), VSd =

nr PRd 1,0 × 64,2 = = 214 kN/m sL 0,3

The shear stress VSd is determined using the thickness of the concrete above the ribs, that is, 105 − 51 = 54 mm. Also the shear may be equally divided between the two shear planes, thus VEd is given by 214 = 1,98 MPa 54 As the sheeting is continuous across the beam with the ribs running normal to the beam, the contribution of the sheeting may be mobilized. √ Using a 45◦ angle for the truss analogy, cos θf = sin θf = 1/ 2 and cot θf = 1,0. vEd = 0,5 ×

For lightweight concrete vEd is given by Eq. (10.47). Determine v1 : The reduction factor for lightweight concrete η1 is given by ρ 1900 η1 = 0,4 + 0,6 = 0,4 + 0,6 = 0,918 2200 2200 fck 25 v1 = 0,5η1 1 − = 0,5 × 0,918 1 − = 0,413 250 250

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Chapter 10 / Composite Construction VEd = 0, 5v1

fck 25 = 0,5 × 0,413 × = 3,44 MPa γc 1,5

From Eq. (10.48) the maximum value of VEd is given by 25 25 1 1 VEd = vfcd sin θr cos θr = 0,6 1 − √ √ = 4,5 MPa 250 1,5 2 2 Use Eq. (10.49) to determine the requirement for Asf sf as the sheeting runs normal to the span. Evaluate the right hand side of Eq. (10.49), vEd hf 3,44 × 55 = = 1892,2 N/mm cot θf 1,0 Evaluate the contribution of the sheeting: Ape fyp,d =

1597 280 = 447,2 N/mm 1000 1,0

The sheeting overprovides the required resistance, therefore only minimum reinforcement is necessary. Deflections and service stresses: a) Wet concrete: This is taken on the steel beam alone. The uniformly distributed load due to the wet concrete q is given by q = 20 × 0,105 × 2,5 = 5,25 kN/m Deflection, δconc is given by: δconc =

5 qL4 5 5,25 × 44 = = 3,6 × 10−3 m 384 EI 384 210 × 106 × 2340 × 10−4

MEd =

qL2 5,25 × 42 = = 10,5 kNm 8 8

The numerical value of the stress σconc is given by σconc =

MEd 10,5 × 103 = = 46 MPa Wel 230

b) Variable and permanent loads: As the ratio n/nf of the number of shear studs provided to that required for full connection is 0,57 (and is therefore greater than the critical ratio of 0,5), no account need be taken of partial shear connection in determining the deflection. Assume Ec,eff = Ecm /2 = 23,5/2 = 11,75 GPa The value of n from cl 5.4.2.2 (11) is Es /Ec,eff (=210/11,75 = 17,9)

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Use Eq. (10.51) to determine the elastic centroidal axis x,

hs +hp nAa + hs − hp beff h + 2

x= nAa + hs − hp beff

203,2 17,9 × 3200 2 + 105 − 51 2500 203,2 + 105+51 2

= 227, 7 mm = 17,9 × 3200 + 105 − 51 2500 From Eq. (10.50)

3 2 2

beff hs − hp hs + hp h Ic = nIa + + nAa − x + beff hs − hp h + −x 12 2 2

3 2 2500 105 − 51 203,2 6 = 17,9 × 23,4 × 10 + + 17,9 × 3200 − 227,7 12 2 2

105 + 51 + 2500 105 − 51 203,2 + − 227,7 = 1,75 × 109 mm4 2 Ic Ec,eff = 1,75 × 109 × 11,75 = 20,45 × 109 kNmm2 = 20,56 × 103 kNm2 Variable load is 13,27 kN/m, and the moment MEd is 26,54 kNm. Thus the deflection δv is given by δv =

5 qL4 5 13,27 × 44 = = 2,15 × 10−3 m 384 EI 384 20,56 × 103

This is equivalent to span/1860 which is acceptable. Although not strictly necessary, determine the stresses due to the variable load: Top of the concrete slab σv,top,c :

MEd h + hs − x 26,54 × 106 203,2 + 105 − 227,7 σv,top,c = − = −1,22 MPa =− Ic 1,75 × 109 Stress at the soffit of the steel beam, σv,soffit : σv,soffit =

nMEd x 17,9 × 26,54 × 106 × 227,7 = = 61,8 MPa Ic 1,75 × 109

Stress at the top of the steel beam, σv,top,A : σv,top,c = −

17,9 × 26,54 × 106 (203,2 − 227,7) nMEd (h − x) =− = 6,65 MPa Ic 1,75 × 109

Permanent load is 12,07 kN/m, and the moment MEd = 24,14 kNm. δv =

5 12,07 × 44 5 qL4 = = 2,0 × 10−3 m 384 EI 384 20,45 × 103

Total deflection: δtotal = δp + δv + δconc = 2,2 + 2,0 + 3,6 = 7,8 mm

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Chapter 10 / Composite Construction This is equivalent to span/513 which is acceptable. Although not strictly necessary determine the stresses due to the variable load: Top of the concrete slab σp,top,c :

MEd h + hs − x 24,14 × 106 203,2 + 105 − 227,7 σp,top,c = − =− Ic 1,75 × 109 = −1,11 MPa Stress at the soffit of the steel beam, σp,soffit : σp,soffit =

nMEd x 17,9 × 24,14 × 106 × 227,7 = = 56,2 MPa Ic 1,75 × 109

Stress at the top of the steel beam, σp,top,A :

nMEd h − x 17,9 × 24,14 × 106 (203,2 − 227,7) σp,top,c = − = 6,05 MPa =− Ic 1,75 × 109 Final total stresses: Top of concrete slab: −1,11 − 1,22 = −2,33 MPa Top of steel beam: −46 + 6,65 + 6,05 = −33,3 MPa Soffit of steel beam: 46 + 61,8 + 56,2 = 164 MPa All these stresses are acceptable. Vibration: Use ψ2 = 0,3, so quasi-permanent load is 12,07 = 0,3 × 13,27 = 16,05 kN/m Deflection under this load δsw is given by δsw =

5 qL4 5 16,05 × 44 = = 0,0026 m = 2,6 mm 384 EI 384 20,45 × 103

Determine the frequency f from Eq. (10.52) 18 18 = 11,2 Hz =√ f =√ δsw 2,6 This is well above the recommended limit of 3 Hz. If the full variable load is taken, then f = 8,7 Hz (and is still acceptable).

EXAMPLE 10.3 Composite beam design (decking parallel to the span). Prepare a design in Grade S355 steel for the beam Mark 3AD of Fig. 10.3 where there are no intermediate columns. The actions on the beam are given in Fig. 10.11. The composite deck is that of EXAMPLE 10.1.

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Due to the long span, the beam is designed as propped to eliminate the high deflections under the permanent loading due to the wet concrete. The resultant shear force and bending moment diagrams for the applied loading are given in Fig. 10.12. The effective width, either side of the beam centre line, be,i = Le /8 = 12000/8 = 1500 mm. Thus total effective width beff = 2be,i = 3000 mm. (The actual beam spacing is 4 m.) To determine the capacity of the beam, the effect of the voids in the deck will be ignored as these run in the direction of the beam and are small. Try a 838 × 292 × 194 Grade S355 UKB. 94

214

214

94

Permanent (kN)

70

240

240

70

Variable (kN)

2,5

2,5

2,5

2,0

2,5

FIGURE 10.11 Loading for EXAMPLE 10.3 232

2500

649

2500

649

2000

232

2500

All loads in kN

2500

BMD (kNm)

2203

3825 881 649

SFD (kN)

649 881

FIGURE 10.12 BM and SF EXAMPLE 10.3

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Chapter 10 / Composite Construction Moment capacity with full shear connection: Determine the force in the concrete Nc from Eq. (10.8), Nc =

0,85 × 25 0,85fck beff hf − hp = 3000 × 105 × 10−6 = 4,463 MN γc 1,5

Determine the force in the steel section Na from Eq. (10.9): Na = Aa

fy 355 = 24700 × 10−6 = 8,769 MN γa 1,0

Determine the maximum force in a flange Naf from Eq. (10.13): Naf = bf tf

fy 355 = 292,4 × 21,7 × 10−6 = 2,253 MN γa 1,0

Determine the out-of-balance force Nac from Eq. (10.12): Nac = Na − Nc = 8,769 − 4,463 = 4,306 MN As Nac < 2Naf , therefore neutral axis lies in the flange. Determine the position of the neutral axis x from Eq. (10.17): x=

4,306 × 106 = = 20,74 mm fy 2 × 292,4 355 2b 1,0 Nac

γa

Determine the compression force Nafx in the flange from Eq. (10.19): Nafx = xb

fy 355 = 20,74 × 292,4 × 10−6 = 2,153 MN γa 1,0

Determine the plastic moment capacity Mpl,Rd from Eq. (10.19): h p + hf x h Mpl,Rd = Nc h + + 2Nafx h − − Na 2 2 2 105 20,74 840,7 = 4,463 840,7 + + 2 × 2,153 840,7 − − 8,769 2 2 2 = 3876 kNm This is greater than MEd (=3825 kNm). Check the ratio of Mpl,Rd to Mpl,a,Rd (the moment capacity of bare steel section): Mpl,Rd Mpl,Rd 3876 = = = 1,43 355 fy Mpl,a,Rd 7640 1,0 × 10−3 Wpl γ a This is less than the critical value of 2,5. Since the neutral axis is in the flange, the web check is unnecessary and thus the section is Class 1.

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Flexural shear: fy 1 1 355 Vpl,Rd = √ Av × 10−3 = 2685 kN = √ 13100 1,0 γ 3 3 a VEd = 881 kN ( = = 0,985 mm sf fyd cot θf 500 1,0 This can be divided between top and bottom, so Ast /Sr on each face is 0,493 mm2 /mm. Use B503 Mesh [503 mm2 /m] Deflection: The loading for determining variable and total deflections is given in Fig. 10.11. Use the formula δ = (WL3 /48EI)(3(a/L) − 4(a/L)3 ) To simplify calculations determine L3 /48EI as a constant for the composite and steel sections, and then determine W (3(a/l) − 4(a/L)3 ) for each load. Determination of Ic : Neglect effect of profiles and take hp = 0, beff = 3000 mm and αe = 17,9 (as EXAMPLE 10.2). Summarizing results: x = 396,3 mm, Ic = 0,162 × 1012 mm4 . Ec,eff Ic = 11,75 × 106 × 0,162 = 1,904 × 106 kNm2 L3/48Ec,eff Ic = 123/48 × 1,904 × 106 = 18,9 × 10−6 m/kN Loads at A1: a/L = 2,5/12 = 0,208; 3(a/L) − 4(a/L)3 = 0,589 Total load: W = 94 + 70 = 164 kN, δ = 164 × 0,589 × 18,9 × 10−6 = 0,0018 m

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Loads at B: a/L = 5/12 = 0,417; 3(a/L) − 4(a/L)3 = 0,961 Total load: W = 214 + 240 = 454 kN, δ = 454 × 0,961 × 18,9 × 10−6 = 0,0082 m Final deflection under total loads: δ = 2(0,0018 + 0,0082) = 0, 0202 m Span/deflection ratio is 12/0,0202 = 594. This is acceptable. Vibration: Total variable deflection = 10,2 × 10−3 m Total permanent deflection = 9,87 × 10−3 m Variable deflection due to a value of ψ2 = 0,3: 0,3 × 10,2 × 10−3 = 0,0031 m δsw = 0,0031 + 0,0102 = 0,0133 m = 13,3 mm √ From Eq. (10.52), f = 18/ 13,3 = 4,94 Hz This is higher than the minimum recommended value of 3 Hz. If the total variable load is taken then f = 4,0 Hz, which is still acceptable.

10.4 COMPOSITE COLUMNS These can take a variety of forms but fall essentially into two categories; partially or totally encased Universal Columns (or H sections) and filled rolled hollow sections, with or without additional reinforcement. Typical configurations are given in Fig. 10.13. This text only considers composite columns which are symmetric about both axes. The methods given in EC 1994-1-1 only hold if (a) the steel contribution ratio δ defined as δ=

Aa fyd Npl,Rd

(10.53)

satisfies the limits 0,2 ≤ δ ≤ 0,9. For δ < 0,2 the column should be designed as reinforced concrete, and for δ > 0,9 designed as non-composite steel. (b) the normalized slenderness ratio λ is less than 2,0.

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Traditional encased column

Web in filled column

Concrete unfilled hollow section

FIGURE 10.13 Types of composite columns.

10.4.1 Axial Compression The design condition for axial compression is that the design resistance χNpl,Rd should exceed the applied load NEd . The buckling co-efficient χ is determined using a nondimensionalized slenderness ratio λ (defined in Eq. (10.73)) in combination with buckling curve ‘a’ for concrete filled hollow sections with ρs ≤ 3% or ‘b’ 3% < ρs < 6% (where ρs is the percentage of reinforcement), ‘b’ for partially or fully encased I sections bending about the major axis, and ‘c’ for partially or fully encased sections bending about the minor axis (the buckling curve designations are those used in EN 1993-1-1).

10.4.2 Uniaxial Bending and Axial Compression Initially, the resistance of the cross-section is determined using an interaction diagram between the axial load resistance and bending moment resistance in a similar fashion to reinforced concrete columns. The diagram is shown schematically in Fig. 10.14, where Npl,Rd is the axial squash capacity (Point A), and Mpl,Rd is the plastic moment capacity (Point B). Although the interaction diagram which is of a similar shape to that for reinforced concrete columns is strictly curved, it may be approximated to a series of straight lines. Point C is established by the application of the moment Mpl,Rd and a resultant axial capacity of the concrete alone Npm,Rd , Point D by the moment capacity Mmax,Rd under 0,5Npm,Rd .

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N

N pl,Rd

A

C

N pm,Rd ½N pm,Rd

D B N pl,Rd

M mav,Rd

M

FIGURE 10.14 Interaction diagram for composite columns

The moment capacity μdy Mpl,y,Rd (or Mpl,N,Rd ) is determined from the interaction diagram under an axial load of NEd . The member is deemed to have sufficient capacity when MEd MEd = = αM Mpl,N,Rd μd Mpl,Rd

(10.54)

The co-efficient αM takes a value of 0,9 for steel grades of S235 and S355 and 0,8 for grades S420 and S460. The factor is partially needed to compensate for the assumption over the depth of the rectangular stress block between EN 1992-1-1 where it is taken as extending over 0,8x but over x in EN 1994-1-1, since the same stress level of 0,85fck /γc is used in both, and partially to allow for the adverse effect of the higher yield strain on the crushing on the concrete. Should the bending moment be entirely due to the eccentricity of the axial load, then μd can be greater than unity.

10.4.3 Biaxial Bending The procedure is similar to uniaxial bending except that two parameters μdz and μdy now need to be determined. However, the effect of imperfections needs only to be considered for the likely failure axis (usually the minor or zz axis). The column is then satisfactory if the following conditions are satisfied, My,Ed ≤ αM,y μdy Mpl,y,Rd

(10.55)

Mz,Ed ≤ αM,z μdz Mpl,z,Rd

(10.56)

My,Ed Mz,Ed + ≤ 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd

(10.57)

and

398

•

Chapter 10 / Composite Construction The co-efficients αM,y and αM,z are to be taken as αM . The increase of the interaction co-efficient to 1 in the combined equation is due to the higher crushing strength of the concrete under biaxial bending.

10.4.4 Determination of Member Capacities The formulae for the flexural capacity of concrete filled rectangular tubes (with reinforcing steel ignored) are taken from Johnson and Anderson (2004).

10.4.4.1 Axial Squash Capacity, Npl,Rd [Point A] This is given by the sum of the individual components due to the steel section, the concrete and the reinforcement (Fig. 10.15(a)). So in general Npl,Rd is given by Npl,Rd = Aa

fy 0,85fck fsk + Ac + As γa γc γs

(10.58)

where Aa is the area of the steel section and fy is the yield strength, Ac is the concrete area and fck is the characteristic cylinder strength, As is the area of the reinforcement and fsk is the characteristic strength. For concrete filled hollow sections the full cylinder strength fck may be used owing to the containment of the concrete. For concrete filled circular tubes Npl,Rd is modified to take account of the triaxial stresses in the concrete due to its containment, and of the reduction in the allowable strength of the steel cross-section owing to the induced hoop tension from the concrete triaxial stresses. The modified value of Npl,Rd is subject to two conditions: (1) λ < 0,5 and (2) e = MEd /NEd ≤ d/10 (where d is the diameter) The equation for Npl,Rd becomes fy t fy fck fsk Npl,Rd = Aa ηa + Ac 1 + ηc + As γa d fck γc γs

(10.59)

where t is the thickness of the tube and ηa and ηc are co-efficients determined as follows, e ηc = ηc0 1 − 10 (10.60) d where ηc0 is given by ηc0 = 4,9 − 18,5λ + 17(λ)2 ≥ 0

(10.61)

and ηa = ηa0 + (1 − ηa0 )10

e d

(10.62)

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Ac

•

399

As

Ax (a) Point A

hn

hn

(b) Point B and C

Neutral axis centroidal axis

Concrete below axis ignored (c) Point D

FIGURE 10.15 Calculation of section capacities

where ηa0 is given by ηa0 = 0,25(3 + 2λ) ≤ 1,0

(10.63)

For e > d/10, ηc = 0 and ηa = 1,0.

10.4.4.2 Calculation of Mpl,Rd [Point B] For the typical section given in Fig. 10.15(b), the forces in the flanges and the reinforcement cancel out, and thus the tension force in the web must balance the compression block in the concrete. It is therefore straightforward to determine the height of the web hn above the centroidal axis. Equating compressive and tensile forces gives hn =

2b fγckc

Npm,Rd f + 4tw 2 γya −

fck γc

where Npm,Rd is given by Eq. (10.67).

(10.64)

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Chapter 10 / Composite Construction The plastic moment capacity is determined by taking moments about the centroidal axis of the section. The moment capacity Mpl,Rd is thus given by Mpl,Rd = Mmax,Rd − Mn,Rd

(10.65)

The values of Mmax,Rd and Mn,Rd are given by Eqs (10.71) and (10.68), respectively.

10.4.4.3 Determination of Npm,Rd [Point C] This can be done by noting that the neutral axis shifts from hn above the centroidal axis to the same distance below it (Fig 10.15(b)). Npm,Rd is determined from horizontal force equilibrium recognizing that the forces in the steel flanges and reinforcement cancel out, that is, the axial load is carried by the concrete alone. The area of the concrete Ac is given by Ac = (b − 2t)(h − 2t) − (4 − π)r 2

(10.66)

where r is the corner radius taken equal to t, and Npm,Rd is given by Npm,Rd = Ac

fck γc

(10.67)

The moment capacity Mn,Rd is given by Mn,Rd = Wp,a,n

fy fck + 0, 5Wp,c,n γa γc

(10.68)

where Wp,a,n and Wp,c,n are the plastic section moduli for the portions of the steel tube and concrete contained within ± hn , and are given by Wp,c,n = (b − 2t)h2n

(10.69)

Wp,a,n = bh2n − Wp,c,n

(10.70)

and

10.4.4.4 Determination of Mmax,Rd [Point D] For this case when an axial force of 0,5Npm,Rd acts the neutral axis coincides with the centroidal axis (Fig. 10.15(c)), and thus Mmax,Rd is simply given by the sum of the plastic moment capacities of the reinforcement, the steel section and the concrete above the centroidal axis. Mmax,Rd is given by Mmax,Rd = Wpa

fy fck + 0,5Wpc γa γc

(10.71)

where Wpa is the plastic section modulus for the steel section (taken from tables) and Wpc is calculated from Wpc =

2 (b − 2t)(h − 2t)2 − r 3 − (4 − π)(0,5h − t − r)r 2 4 3

(10.72)

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Note, the equations have been derived for bending about the major (yy) axis. For minor axis (zz) bending, h and b are simply interchanged.

10.4.5 Buckling The non-dimensionalized slenderness ratio λ is defined by Npl,Rk λ= Ncr

(10.73)

where Npl,Rk is the characteristic plastic axial load capacity from Eq. (10.58) or Eq. (10.59) with all the materials’ partial safety factors set equal to unity and the effective buckling load Ncr is given by Ncr =

π2 (EI)eff l2

(10.74)

where l is the buckling length with the effective flexural stiffness is given by (EI)eff = Ea Ia + Ke Ecm Ic + Es Is

(10.75)

where Ea Ia is the flexural rigidity of the steel section alone, Ecm Ic is the flexural rigidity of the concrete, and Es Is is the flexural rigidity of the reinforcement and Ke is a correction factor taken as 0,6. Ecm is given in Table 3.1 of EN 1992-1-1. Where appropriate, account should be taken of the influence of long-term loading by using an effective concrete modulus Ec,eff determined from Ec,eff =

Ecm N

1 + φt NG,Ed Ed

(10.76)

where NEd is the total design normal force and NG,Ed is that portion which is permanent and ϕt is the creep co-efficient determined from EN 1992-1-1.

10.4.6 Design Moments Second order effects within the column length may be allowed for by increasing the larger design bending moment determined from a first order analysis by a factor k given by k=

β N

1 − N Ed cr,eff

≥ 1,0

(10.77)

where the moment ratio factor β for end moments is given by β = 0,66 + 0,44r ≥ 0,44

(10.78)

and Ncr,eff is an effective buckling load determined using the actual column length and an effective stiffness (EI)eff,II given by (EI)eff,II = K0 (Ea Ia + Ke,II Ecm Ic + Es Is )

(10.79)

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Chapter 10 / Composite Construction where K0 is a calibration factor (=0,9) and Ke,II is a correction factor representing the effects of cracking in the concrete (=0,5). Johnson (2004) indicates that where Ncr,eff ≥ 10NEd , then the second order effects need considering and there is an additional moment induced by the additional imperfection. Thus the design moment MEd is given by MEd = kend M + kimp NEd e0

(10.80)

where kend is magnification factor due to the moment gradient, M is the larger end moment, e0 is the eccentricity due to imperfections, kimp is the magnification factor determined from Eq. (10.78) with β = 1,0.

10.4.7 Other Checks and Detailing 10.4.7.1 Local Buckling For circular hollow sections: d/t ≤ 90ε2 ; Rectangular hollow sections: h/t ≤ 52ε; Partially encased sections: b/tf ≤ 44ε.

10.4.7.2 Cover For reinforcement, this is governed by the requirements in EN 1992-1-1. For encased steel sections, this should be a maximum of 40 mm or b/6, where b is the flange width.

10.4.7.3 Shear The shear bond between the steel section and the concrete should be checked using an elastic distribution of forces on the uncracked section with a transmission length not exceeding twice the relevant transverse direction. The values of shear bond should not exceed 0,3 MPa for fully encased sections, 0,55 MPa for circular concrete filled sections, 0,40 MPa for circular concrete filled sections and 0,2 MPa for the flanges only in partially encased sections. Where necessary shear studs should be used on encased I sections to resist shear.

10.4.7.4 Fire For concrete filled hollow sections it is essential that two vent holes of 20 mm diameter should be drilled through the steel section at the top and bottom of each storey subject to a maximum spacing of 5 m (Newman and Simms, 2000). These holes must not be within the depth of the floor construction. The purpose of these holes is to allow the build up of water vapour to escape whilst the moisture within the concrete is driven off in the early stages of heating in the fire.

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It is clear that the method of designing composite columns under uniaxial or biaxial bending is complex, and leads itself readily to the use of spreadsheets or design charts. The analysis of given sections to determine their carrying capacity is much more straightforward. The first example illustrates the determination of axial carrying capacity. To avoid duplication of calculations the second and third both use the same section, one under uniaxial bending about the major axis, and the other with biaxial bending. In each case the loading is considered totally short term, that is, Ec,eff is taken as Ecm . Also any reinforcement is considered negligible and is neglected.

EXAMPLE 10.4 Determination of axial load capacity of a composite column. Determine the axial load carrying capacity of 4 m effective length 150 × 150 × 8 Grade S355 rolled hollow section filled with Grade C25/30 concrete. Check h/t: Actual: h 150 = = 18,75 t 8 Allowable: 52ε = 52 × (235/355)1/2 = 42,3, therefore satisfactory Determination of Npl,Rd : Use Eq. (10.58) with the concrete taken at full strength, # 25 fy 355 " 0,85fck fsk Npl,Rd = Aa + Ac + As = 4510 + (150 − 16)2 − (4 − π)82 γa γc γs 1,0 1,5 = 1,899 MN Determine the load contribution ratio δ from Eq. (10.53): fy

δ=

Aa γ

a

Npl,Rd

=

4510 355 × 10−6 1,0 1,899

= 0,84

As δ lies between 0,2 and 0,9, the column may be designed as composite. Use Eq. (10.75) to determine the effective stiffness (Es Is = 0) fck + 8 1/3 25 + 8 1/3 Ecm = 22 = 22 = 31,5 GPa 10 10 (EI)eff = Ea Ia + 0,6Ecm Ic = 210 × 106 × 1510 × 10−8 + 0, 6 × 31, 5 × 106 ×

(150 − 16)4 = 3678 kNm2 12

Determine the Euler critical load Ncr from Eq. (10.74) Ncr =

π2 (EI)eff π2 × 3678 = = 2269 kN l2 42

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•

Chapter 10 / Composite Construction Determine Npl,Rk using Eq. (10.58) with the materials’ partial safety factors set equal to 1, # 25 fy 0,85fck fsk 355 " Npl,Rk = Aa + Ac + As = 4510 + (150 − 16)2 − (4 − π)82 γa γc γs 1,0 1,0 = 2,048 MN Determine the normalized slenderness ratio λ from Eq. (10.73), Npl,Rk 2048 λ= = = 0,95 Ncr 2269 This is less than the critical value of 2,0. The strength reduction factor is determined using buckling curve ‘a’ with α = 0,21, φ = 0, 5(1 + α(λ − 0, 2) + (λ)2 ) = 0, 5(1 + 0,21(0,95 − 0,2) + 0,952 ) = 1,03 χ=

φ+

!

1 φ2 − (λ)2

=

1,03 +

1 1,032 − 0,952

= 0,70

Determine the axial capacity NRd : NRd = χNpl,Rd = 0,70 × 1899 = 1329 kN

EXAMPLE 10.5 Axial load and uniaxial bending about the major axis. Determine whether a column having a system length of 4 m fabricated from 150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete can carry an axial load at ULS of 400 kN and a moment at ULS about the major axis of 18 kNm. Aa = 3710 mm2 ; Wpl,y = 183 × 103 mm3 . Determine Wpc from Eq. (10.72) 2 (b − 2t)(h − 2t)2 − r 3 − (4 − π)(0,5h − t − r)r 2 4 3 (100 − 2 × 8)(150 − 2 × 8)2 150 2 = − 83 − (4 − π) −8−8 4 3 2

Wpc =

= 373500 mm3 The area of the concrete Ac is given by Eq. (10.66) Ac = (b − 2t)(h − 2t) − (4 − π)r 2 = (150 − 2 × 8)(100 − 2 × 8) − (4 − π)82 = 11200 mm2 Axial squash capacity Npl,Rd is given by Eq. (10.58): Npl,Rd = Aa

fy 0,85fck fsk 355 25 + Ac + As = 3710 + 11200 = 1504 kN γa γc γs 1,0 1,5

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405

Determine the contribution ratio, δ from Eq. (10.53): 3710 355 × 10−3 As fyd 1,0 δ= = = 0,876 Npl,Rd 1504 This is within the limits for design as a composite column. Maximum moment capacity Mmax,Rd from Eq. (10.71): Mmax,Rd = Wpa

fy fck 355 25 + 0,5Wpc = 183000 × 10−6 + 0,5 × 373500 × 10−6 γa γc 1,0 1,5

= 68,1 kNm Determine Npm,Rd from Eq. (10.67): Npm,Rd = Ac

fck 25 × 10−3 = 187 kN = 11200 γc 1,5

Npm,Rd = 93,5 kN 2 Determine hn from Eq. (10.64) determine hn : hn =

f 2b γck c

Npm,Rd 18700 = 7,33 mm f = fck 25 25 y 2 × 100 × 1,5 + 4 × 8 2 355 − + 4tw 2 γ − γ 1,0 1,5 a c

Determine Wp,c,n from Eq. (10.69): Wp,c,n = (b − 2t)h2n = (100 − 2 × 8)7,332 = 4513 mm3 Determine Wp,a,n from Eq. (10.70): Wp,a,n = bh2n − Wp,c,n = 100 × 7,332 − 4513 = 860 mm3 Determine Mn,Rd from Eq. (10.68): Mn,Rd = Wp,a,n

fy fck 355 25 + 0,5Wp,c,n = 860 × 10−6 + 0,5 × 4513 × 10−6 γa γc 1,0 1,5

= 0,3 kNm Determine Mpl,Rd from Eq. (10.65): Mpl,Rd = Mmax,Rd − Mn,Rd = 68,1 − 0,3 = 67,8 kNm The values required to plot the interaction diagram are given in Table 10.3.

TABLE 10.3 Values required for major axis interaction diagram for EXAMPLE 10.5. Point A B C D

(0, N pl,Rd ) (M pl,Rd , 0) (M pl,Rd , N pm,Rd ) (M max,Rd , 0,5N pm,Rd )

Moment capacity (kNm)

Axial capacity (kN)

0 67,8 67,8 69,1

1504 0 187 93,5

406

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Chapter 10 / Composite Construction These values are plotted in Fig. 10.15. Determine the resistance to axial buckling about the major axis: Ea Ia = 210 × 106 × 1106 × 10−8 = 2323 kNm2 ( fck + 8) 1/3 Ecm = 22 = 31,5 GPa 10 Ic =

(0,150 − 0,016)3 (0,100 − 0,016) = 16,84 × 10−6 m4 12

Ecd Ic = 31,5 × 106 × 16,84 × 10−6 = 530 kNm2 Determine (EI)eff,II from Eq. (10.79): (EI)eff,II = 0,9(Ea Ia + 0,5Ecm Ic ) = 0,9(2323 + 0,5 × 530) = 2329 kNm2 Determine Ncr,eff from Eq. (10.74) with (EI)eff replaced by (EI)eff,II : Ncr,eff =

π2 (EI)eff,II π2 × 2329 = = 1437 kN l2 42

NEd = 400 kN > Ncr,eff /10, therefore second order effects need to be considered. Npl,Rk = (3710 × 355 + 11200 × 25) × 10−3 = 1597 kN. Determine λ from Eq. (10.73): λ=

Npl,Rk = Ncr,eff

1597 = 1,054 > 2,0 1437

Thus the column satisfies the limits for composite design. Second order effects: (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend =

β 1−

NEd Ncr,eff

=

1,0

= 1,386 400 1 − 1437

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp = 1,386 The design moment MEd is given by Eq. (10.80): MEd = kend M + kimp NEd e0 = 1,386 × 18 + 1,386 × 400

4 = 32,3 kNm 300

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From the interaction diagram in Fig. 10.15, the moment M400 corresponding to an axial load of 400 kN is given by M400 = Mpl,Rd μd =

Npl,Rd − NEd 1504 − 400 = 56,5 kNm = 67,8 Npl,Rd − Npm,Rd 1504 − 187

M400 56,5 = = 0,833 Mpl,Rd 67,4

MEd MEd 32,3 = = = 0,572 < 0, 9 μd Mpl,Rd M400 56,5 The limiting value of αM is 0,9 as Grade S355 steel is being used. Thus the column is therefore satisfactory.

EXAMPLE 10.6 Axial load and biaxial bending. Determine whether a column having a system length of 4 m fabricated from 150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete can carry an axial load at ULS of 350 kN and a moment at ULS about the major axis of 15 and 10 kNm about the minor axis. Two interaction diagrams are required, one for each axis. For the major axis the interaction diagram is as EXAMPLE 10.5. For the minor axis, the formulae in Eqs (10.64)–(10.73) are used but with b and h interchanged. Aa = 3710 mm2 ; Wpl,y = 183 × 103 mm3 ; Wpl,z = 133 × 103 mm3 . Determine Wpc from Eq. (10.72): (h − 2t)(b − 2t)2 2 − r 3 − (4 − π)(0, 5b − t − r) r 2 4 3 (150 − 2 × 8)(100 − 2 × 8)2 2 3 100 = − 8 − (4 − π) − 8 − 8 = 234200 mm3 4 3 2

Wpc =

From EXAMPLE 10.5, Ac = 11200 mm2 , Npl,Rd = 1504 kN and δ = 0,876. Determine the maximum moment capacity Mmax,Rd from Eq. (10.71): Mmax,Rd = Wpa

fy fck 355 25 × 10−6 + 0,5 × 234200 × 10−6 + 0,5Wpc = 133000 γa γc 1,0 1,5

= 49,2 kNm From EXAMPLE 10.5, Npm,Rd = 187 kN, and 0,5Npm,Rd = 93,5 kN Determine hn from Eq. (10.64): hn =

Npm,Rd 187000 = 6,41 mm f = f f 25 25 y 2 × 150 1,5 + 4 × 8 2 355 − 2h γck + 4tw 2 γ − γck 1,0 1,5 c a c

•

Chapter 10 / Composite Construction Determine Wpc,n from Eq. (10.69): Wpc,n = (h − 2t)h2n = (150 − 2 × 8)6,412 = 5506 mm3 Determine Wpa,n from Eq. (10.70): Wpa,n = hh2n − Wpc,n = 150 × 6,412 − 5506 = 657 mm3 Determine Mn,Rd from Eq. (10.68): Mn,Rd = Wpa,n

fy 355 fck 25 + 0,5Wpc,n = 657 × 10−6 + 0,5 × 5506 × 10−6 γa γc 1,0 1,5

= 0,3 kNm Determine Mpl,Rd from Eq. (10.65): Mpl,Rd = Mmax,Rd − Mn,Rd = 49,2 − 0,3 = 48,9 kNm The values required to plot the interaction diagram are given in Table 10.4. These values are plotted in Fig. 10.16.

TABLE 10.4 Values required for minor axis interaction diagram for EXAMPLE 10.6. Point A B C D

(0, N pl,Rd ) (M pl,Rd ,0) (M pl,Rd , N pm,Rd ) (M max,Rd , 0,5N pm,Rd )

Moment capacity (kNm)

Axial capacity (kN)

0 48,9 48,9 49,12

1504 0 187 93,5

1600 1400 1200 Axial capacity (kN)

408

1000 800 600 400 200 0 0

20

40

60

Moment capacity (kNm)

FIGURE 10.16 Major axis M–N interaction diagram (EXAMPLE 10.5)

80

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409

Determine the resistance to axial buckling about the minor axis: Ea Ia = 210 × 106 × 577 × 10−8 = 1212 kNm2 Ecm = 31,5 GPa (as EXAMPLE 10.5) Ic =

(0,100 − 0,016)3 (0,150 − 0,016) = 7,11 × 10−6 m4 12

Ecd Ic = 31,5 × 106 × 7,11 × 10−6 = 224 kNm2 (EI)eff,II = 0,9(Ea Ia + 0,5Ecm Ic ) = 0,9(1212 + 0,5 × 224) = 1192 kNm2 Ncr,eff =

π2 (EI)eff,II π2 × 1192 = = 735 kN l2 42

NEd = 350 kN > Ncr,eff /10, therefore second order effects need to be considered. Npl,Rk = Aa fy + Ac fck = 3710 × 355 × 10−3 + 11200 × 25 × 10−3 = 1597 kN Npl,Rk 1597 λ= = = 1,474 < 2,0 Ncr 735 Second order effects (major yy axis) using the appropriate critical buckling load: (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend,y =

β 1−

NEd Ncr,eff

=

1,0

= 1,322 350 1 − 1437

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp,z = 1,322. Second order effects (minor zz axis): (a) Within the column length: Assuming the column is in single curvature, r = 1,0, so β = 1,0. kend,z =

β 1−

NEd Ncr,eff

=

1,0

= 1,921 1 − 350 730

(b) Due to initial bow: From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300 As β = 1,0 for initial bow, kimp,z = 1,921.

•

Chapter 10 / Composite Construction 1600 1400 1200 Axial capacity (kN)

410

1000 800 600 400 200 0 0

10

20

30 40 Moment capacity (kNm)

50

60

FIGURE 10.17 Minor axis M–N interaction diagram (EXAMPLE 10.6)

From the interaction diagram for the major axis in Fig. 10.16, the moment M350 corresponding to an axial load of 350 kN is given by M350 = Mpl,Rd μdy =

Npl,Rd − NEd 1504 − 350 = 67,8 = 59,4 kNm Npl,Rd − Npm,Rd 1504 − 187

M350 59,4 = = 0,881 Mpl,Rd 67,4

From the interaction diagram in Fig. 10.17, for the minor axis the moment M350 corresponding to an axial load of 350 kN is given by M350 = Mpl,Rd μdy =

Npl,Rd − NEd 1504 − 350 = 48,9 = 42,8 kNm Npl,Rd − Npm,Rd 1504 − 187

M350 42,8 = = 0,875 Mpl,Rd 48,9

The second order effect due the bow may only be applied on one axis. Use Eq. (10.80) to calculate the design moments. Two cases, therefore, need considering: (a) Bow on major axis. MEd, y = kend, y MSd, y + kimp, y NEd e0 = 1,322 × 15 + 1,322 × 350 MEd, z = kend, z MSd, z = 1,921 × 10 = 19,2 kNm

4 = 26,0 kNm 300

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MEd,y 26,0 = 0,438 < 0,9 = μdy Mpl,y,Rd 59,4 MEd,z 19,2 = = 0,45 < 0,9 μdz Mpl,z,Rd 42,8 MEd,y MEd,z 26,0 19,2 + = + = 0,887 < 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd 59,4 42,8 (b) Bow on minor axis. MEd,y = kend,y MSd,y = 1,322 × 15 = 19,8 kNm MEd,z = kend,z MSd,z + kimp,z NEd e0 = 1,921 × 10 + 1,921 × 350

4 300

= 28,2 kNm MEd,y 19,8 = = 0,333 < 0,9 μdy Mpl,y,Rd 59,4 MEd,z 28,2 = = 0,66 < 0,9 μsdz Mpl,z,Rd 42,8 MEd,y MEd,z 19,8 28,2 + = + = 0,992 < 1,0 μdy Mpl,y,Rd μdz Mpl,z,Rd 59,4 42,8 It will be noted that the application of the initial bow to the minor axis is the critical case. This is due to both the lower moment capacity and the lower buckling load.

REFERENCES Bunn, R. and Heywood, M. (2004) Supporting services from structure. Co-Construct (BSRIA). EN 1991-1-1 Eurocode 1: Actions on structures – Part 1–1: General actions – Densities, self-weight, imposed loads for buildings. CEN/BSI. EN 1991-1-6 Eurocode 1: Actions on structures – Part 1–6: Actions during execution. CEN/BSI. EN 1992-1-1 Eurocode 2: Design of concrete structures – Part 1–1: General rules and rules for buildings. CEN/BSI. EN 1993-1-1 Eurocode 3: Design of steel structures – Part 1.1: General rules and rules for buildings. CEN/BSI. EN 1994-1-1 Eurocode 4: Design of composite steel and concrete structures – Part 1.1: General rules and rules for buildings. CEN/BSI. Evans, H.R. and Wright, H.D. (1988). Steel–concrete composite flooring deck structures, In Steel– concrete composite structures (ed R. Narayanan). Elsevier. Johnson, R.P. (2004). Composite structures of steel and Concrete (3rd edition). Blackwell Publishing. Johnson, R.P. (2005). Shear connection in beams that support composite slabs – BS 5950 and EN 1994-1-1, Structural Engineer, 83(22), 21–24. Johnson, R.P. and Anderson, D. (2004). Designers’ guide to EN 1994-1-1 Eurocode 4: Design of composite steel and concrete structures – Part 1.1: General rules and rules for buildings. Thomas Telford. Johnson, R.P. and May, I.M. (1975). Partial-interaction design of composite beams, The Structural Engineer, 53(8), 305–311. Lawson, R.M. and Chung, K.F. (1994). Composite beam design to eurocode 4. Publication 121. SCI.

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Chapter 10 / Composite Construction Lawson, R.M. and Nethercot, D.A. (1985). Lateral stability of I-beams restrained by profiled sheeting, The Structural Engineer, 63B(1), 1–7, 13. Lawson, R.M., Mullett, D.L. and Rackham, J.W. (1997). Design of asymmetric Slimflor ® beams using deep composite decking. Publication 175. SCI. Martin, L.H. and Purkiss, J.A. (2006). Concrete design to EN 1992. Butterworth-Heinemann. Mottram, J.T. and Johnson, R.P. (1990). Push tests on studs welded through profiled steel sheeting, The Structural Engineer, 68(10), 187–193. Mullett, D.L. (1992). Slim floor design and construction. Publication 110. SCI. Mullett, D.L. (1997). Design of RHS Slimflor ® Edge Beams. Publication 169. SCI. Mullett, D.L. and Lawson, R.M. (1992). Slim floor construction using deep decking. Technical Report 120. SCI. Newman, G.M. and Simms, W.I. (2000). The fire resistance of concrete filled tubes to Eurocode 4. Technical Report 259. SCI. Oehlers, D.J. and Johnson, R.P. (1987). The strength of shear stud connections in composite beams, The Structural Engineer, 65B(2), 44–48. Olgaard, J.G., Slutter, R.G. and Fisher, J.W. (1971). Shear strength of stud connectors in lightweight and normal-weight concrete, Engineering Journal, American Institute of Steel Construction, 8(2), 55–64. Wyatt, T.A. (1989). Design guide on the vibration of floors. Publication 076. SCI. Yam, L.C.P. and Chapman, J.C. (1968). The inelastic behaviour of simply supported composite beams of steel and concrete, Proceedings of the Institution of Civil Engineers, 41, 651–683.

Chapter

11 / Cold-formed Steel Sections

Thin-walled, cold-formed steel sections are widely used as purlins and rails, the intermediate members between the main structural frame and the corrugated roof or wall sheeting in buildings for farming and industrial use (see Fig. 11.1). Trapezoidal sheeting is usually fixed to these members in order to enclose the building. The most common sections are the zed, channel and sigma shapes, which may be plain or have lips. The lips are small additional elements provided to a section to improve its efficiency under compressive loads by enhancing the section ability against local buckling. Cold-formed steel sections are fabricated by means of folding, press-braking of plates or cold-rolling of coils made from carbon steel. Sheet steel used in cold-formed sections is typically 0.9–8 mm thick. It is usually supplied pre-galvanized in accordance with European Standard EN 10142. Galvanizing gives adequate protection for internal members or those adjacent to the boundaries of the building envelope. Cold working of the steel increases its yield strength but also lowers its ductility (see Fig. 11.2). For example, a 20% reduction in thickness can increase yield strength by 50% but reduces elongation to as little as 7%, which probably represents the limit of formability for simple shapes. The main benefits of using a cold-formed section are not only its high strengthto-weight ratio but also its lightness, which can save costs on transport, erection and the construction of foundation, and flexibility that the members can be produced in a wide variety of sectional profiles, which can result in more cost effective designs. Examples of the structural use of cold-formed sections include roof and wall members, steel framing, wall partitions, large panels for housing, lintels, floor joists, modular frames for commercial buildings, trusses, space frames, curtain walling, prefabricated buildings, frameless steel buildings, storage racking, lighting and transmission towers, motorway crash barriers, etc. The prime difference between the behaviour of cold-formed sections and hot rolled structural sections is that cold-formed members involve thin plate elements which tend to buckle locally under compression. Cold-formed cross-sections are therefore

This chapter is contributed by Long-yuan Li (Aston University) and Xiao-ting Chu (Auckland University of Technology, New Zealand)

Side and gable rails

Rail extensions

Corner detail Inwardly lipped channel

Side rail supports

Purlins

Purlin extensions

Diagonal tie wires and fixing cleats

Rail sleeves Eaves beam

Chapter 11 / Cold-formed Steel Sections

Apex anti-sag bars Speed fix anti-sag bars Purlin sleeves

•

Cleader angle

414

FIGURE 11.1 The building using zed and channel sections as purlins and rails (Copy from Albion Sections Ltd design manual by permission)

usually classified as slender because they cannot generally reach their full strength based on the amount of material in the cross-section (Rhodes and Lawson, 1992). The secondary difference is that cold-formed members have low lateral stiffness and low torsional stiffness because of their open, thin, cross-sectional geometry, which gives great flexural rigidity about one axis at the expense of low torsional rigidity and low flexural rigidity about a perpendicular axis. This leads to cold-formed members being susceptible to distortional buckling and lateral–torsional buckling.

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s Increase of yield strength due to strain hardening

Ultimate strength

Yield point after cold working Fracture

Yield strength

Proportional limit

Further loading after cold working

e Perfect plasticity Linear region

Strain hardening

Necking

FIGURE 11.2 The influence of cold forming on the stress–strain curve of steel

11.1 ANALYTICAL MODEL Most purlins and rails are laterally restrained by their supported cladding or sheeting either partially or completely. Hence, it is necessary to consider the influence of the lateral restraints when establishing an analytical model. Also, it is well known that, when a thin-walled beam has one or more cross-sections that are constrained against warping, a complex distribution of longitudinal warping stresses can be developed. These warping stresses together with the longitudinal stresses generated by bending moments may cause the beam to have local, distortional and lateral–torsional buckling. Consider a zed section beam that is partially restrained by the sheeting on its upper flange. Without loss of generality, the restraints of the sheeting can be simplified by using a translational spring and a rotational spring, both of which are uniformly distributed along the longitudinal direction of the beam (see Fig. 11.3). Let the origin of the coordinate system (x, y, z) be the centroid of the cross-section, with x-axis being along the longitudinal direction of the beam, and y- and z-axes taken in the plane of the cross-section, as shown in Fig. 11.3. According to the bending theory of asymmetric beams (Vlasov, 1961; Oden, 1967) and noticing that for a zed section y- and z-axes used in Fig. 11.3 are not the principal axes, the constitutive relationships between moments and generalized strains can be expressed as d2 w d2 v − EI yz dx2 dx2 2 d w d2 v Mz = −EIyz 2 − EIz 2 dx dx My = −EIy

(11.1)

416

•

Chapter 11 / Cold-formed Steel Sections kf

qz qy

ky

Pk

Pq

φ

y, v Sheeting

My w Mz

Purlin

v

z,w (a)

(b)

FIGURE 11.3 (a) Purlin-sheeting system and (b) a simplified analysis model d2 φ dx2 dφ MT = GIT dx

Mω = EIw

where My and Mz are the bending moments about y- and z-axes, Mω is the warping moment, MT is the twisting moment, E is the modulus of elasticity, G is the shear modulus, Iy and Iz are the second moments of the cross-sectional area about y- and z-axes, Iyz is the product moment of the cross-sectional area, Iw is the warping constant, IT is the torsion constant, v and w are the y- and z-components of displacement of the centroid of the cross-section, φ is the angle of twisting. The above four equations together with three equilibrium equations can be used to determine seven unknowns (four moments, My , Mz , Mω and MT and three displacements, v, w and φ). For the present problem it is convenient to derive the equilibrium equations by using the principle of minimum potential energy. For the partially restrained beam the total potential energy involves the strain energy of the beam, the strain energy of the two springs and the potential of the applied loads, that is = Ub + Us + W

(11.2)

in which, 1 Ub = 2

l

My

d2 w − 2 dx

+ Mz

0

= strain energy of the beam

d2 v − 2 dx

d2 φ dφ + M ω 2 + MT dx dx dx

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1 Us = 2

•

417

l "

# ky (v + zk φ)2 + kφ φ2 dx = strain energy of the springs

0

l W =− (v + zq φ)qy + (w − yq φ)qz dx = potential of the applied loads 0

where l is the length of the beam, ky and kφ are the stiffness constants per unit length of the translational and rotational springs, qy and qz are the uniformly distributed loads in y- and z-directions, (−yk , −zk ) and (−yq , −zq ) are the coordinates of the spring and loading points Pk and Pq (see Fig. 11.3), respectively. Substituting Eq. (11.1) into Eq. (11.2) yields 1 = 2

⎡

l

⎣EIy

d2 w dx2

⎡

0

1 + 2

l

⎣EIw

2

d2 φ dx2

d2 v d2 w + 2EIyz 2 2 + EIz dx dx 2

+ GIT

dφ dx

2

0

d2 v dx2

⎤ ⎦dx + 1 2

2 ⎤ ⎦dx

l "

# ky (v + zk φ)2 + kφ φ2 dx

0

l −

[(v + zq φ)qy + (w − yq φ)qz ]dx

(11.3)

0

The following three equilibrium equations can be obtained by the variation of the total potential energy with respect to the displacement components, v and w, and the angle of twisting, φ d4 v d4 w + ky (v + zk φ) + EIyz 4 = qy 4 dx dx 4 4 d v d w EIyz 4 + EIy 4 = qz dx dx 4 d φ d2 φ EIw 4 − GIT 2 + (zk2 ky + kφ )φ + zk ky v = qy zq − qz yq dx dx EIz

(11.4)

For beams that have no restraints, that is, ky = kφ = 0, Eq. (11.4) are simplified to d4 v d4 w + EI = qy yz dx4 dx4 d4 v d4 w EIyz 4 + EIy 4 = qz dx dx 4 d φ d2 φ EIw 4 − GIT 2 = qy zq − qz yq dx dx EIz

(11.5)

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•

Chapter 11 / Cold-formed Steel Sections For beams that are fully restrained, that is, v = φ = 0, Eq. (11.4) are simplified to Rk + EIyz

d4 w = qy dx4

d4 w = qz dx4 Rk zk + Mk = qy zk − qz yk EIy

(11.6)

where Rk and Mk are the reaction force and reaction moment at the restrained point Pk . Most cold-formed sections are supported by cleats bolted to the web of the section as shown in Fig. 11.4. The boundary conditions thus can be assumed as v = 0 Mz ≈ 0 w = 0 My ≈ 0 φ = 0 Mω ≈ 0

(11.7)

The cleats are designed so that the lower flange of the section does not bear directly on the rafter, and web crippling problems are avoided. However, the shear or bearing strength of the connecting bolts is critical to the design. Governing Eqs (11.4), (11.5) or (11.6) together with boundary conditions (11.7) can be used to determine the displacements and angle of twisting of the beam under the action of external loads, qy and qz . The bending moments at any place can be calculated using Eq. (11.1). The bending and shear stresses thus can be calculated from the moments

Purlin Upslope

Gap

Detail of Butted joint

FIGURE 11.4 Purlin butted to rafter beam by a cleat

Cleat

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419

and shear force as follows: σx = τmax =

Mz Iy − My Iyz My Iz − Mz Iyz d2 φ y + z + E(ω − ω) 2 2 Iy Iz − Iyz Iy Iz − Iyz dx2 3MT V + Av Ls t 2

(11.8)

where ω is the sectorial coordinate with respect to the shear centre, ω is the average value of ω, Ls is the total length of the middle line section, t is the thickness, V is the shear force, Av is the shear area. The sectorial coordinates are properties of the cross-section and are calculated as follows (Chu et al., 2004a,b) s ω=

hs ds and 0

1 ω= Ls

Ls ωds

(11.9)

0

where hs is the perpendicular distance from a tangent at the point under consideration to the shear centre, and s is the distance from any chosen origin to the same point measured along the middle line of the section. Equation (11.8) indicates that when warping torsion is involved twisting produces not only the shear stress but also axial stress. More about warping torsion can be found in the books of Oden (1967) and Walker (1975). Ye et al. (2004) investigated the influences of restraints on the magnitude and distribution of the axial stress within the cross-section through varying the stiffness constants of two springs. They also used the stress pattern obtained from Eq. (11.8) as an input to the finite strip analysis program and investigated the influence of restraints on the behaviour of local, distortional and lateral–torsional buckling of channel and zed section beams (Ye et al., 2002). It is interesting to notice from Eq. (11.6) that, if it is fully restrained the zed section beam bends only in the plane of the web and the bending stress and deflection can be calculated simply based on the bending rigidity of the beam in the plane of the web although the section itself is point symmetric, that is, σx =

My z Iy

d4 w qz = 4 EIy dx

(11.10)

11.2 LOCAL BUCKLING Cold-formed members are usually very thin, and thus the thin plate elements tend to buckle locally under compression. The local buckling mode of a cold-formed member normally involves plate flexure along, with no transverse deformation of a line or lines of intersection of adjoining plates, and can be characterized by a relatively short

420

•

Chapter 11 / Cold-formed Steel Sections

(a)

(b)

FIGURE 11.5 Local buckling modes of a zed section (h = 120 mm, b = 75 mm, c = 20 mm, t = 2,5 mm). (a) Under a pure compression (web buckle) and (b) under a pure bending (flange buckle)

half-wavelength of the order of magnitude of individual plate elements, as illustrated in Fig. 11.5.

11.2.1 Elastic Local Buckling Stress It is known from the stability of plates that, a simply supported rectangular plate may buckle if it is subjected to compressive loads in the plane of its middle surface. The elastic critical compressive stress when the plate buckles is expressed as (Bulson, 1970)

σcr,p =

kσ π 2 E 12(1 − ν2 )

t bp

2 (11.11)

where bp is the width of the plate, ν is Poisson’s ratio and kσ is the buckling coefficient determined from kσ =

l mbp

2

+2+

mbp l

2 (11.12)

where l is the length of a plate and m is the number of half waves of the buckling mode in the longitudinal direction in which the plate is compressed. Note that, kσ varies with m. When m = l/bp , kσ has a minimum value of 4, which makes the compressive stress σcr,p critical. This indicates that the buckles approximate to square wave forms, as demonstrated in Fig. 11.5. Equations (11.11) and (11.12) are only for a plate simply supported on the two long sides and subjected to uniform compressive stresses. For a compression element of

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421

width bp with different support conditions and/or subjected to non-uniform compressive stresses the critical compressive stress can still be calculated using Eq. (11.11) but the buckling coefficient needs to take account the influence of both boundary conditions and stress pattern. When these factors have been taken into account kσ is expressed as follows: For doubly supported compression elements (Table 4.1 in EN 1993-1-5 (2006))

kσ =

⎧ 8,2 ⎪ ⎪ ⎨ 1,05 + ψ

0≤ψ≤1

7,81 − 6,29ψ + 9,78ψ2 ⎪ ⎪ ⎩ 5,98(1 − ψ)2

−1 < ψ < 0 −3 < ψ ≤ −1

(11.13a)

For outstand compression elements (support at σ1 ) (Table 4.2 in EN 1993-1-5 (2006)) ⎧ ⎨ 0,578 kσ = 0,34 + ψ ⎩ 1,7 − 5ψ + 17,1ψ2

0≤ψ≤1

(11.13b)

−1 ≤ ψ < 0

For outstand compression elements (support at σ2 ) (Table 4.2 in EN 1993-1-5 (2006)) kσ = 0,57 − 0,21ψ + 0,07ψ2

−3 < ψ ≤ 1

(11.13c)

For single edge compression stiffener elements (lips) (cl 5.5.3.2.5 in EN 1993-1-3 (2006)) 5 kσ =

0,5

cp /bp ≤ 0,35

0,5 + 0,83(cp /bp − 0,35)2/3

0,35 < cp /bp ≤ 0,6

(11.13d)

where ψ = σ2 /σ1 is the ratio of stresses at the two ends of the element (σ1 is the larger compressive stress, σ2 is the tensile stress or smaller compressive stress, and the compressive stress is assumed to be positive), cp and bp are the lengths of the middle lines of the lip and flange, respectively. Equations (11.11) and (11.13) are used to calculate the critical stress of local buckling of a compression element. For channel and zed sections, the web and lipped flange may be treated as the doubly supported elements if the lip satisfies the requirement specified in Section 5.2 in EN 1993-1-3 (2006). Flanges that have no intermediate stiffeners and no edge lips are treated as the outstand elements. When a web or a flange has an intermediate stiffener, the actual width of the element should be taken as the width of the individual part separated by the stiffener. More details for dealing with elements with intermediate stiffeners can be found in EN 1993-1-3 (2006) and EN 1993-1-5 (2006).

422

•

Chapter 11 / Cold-formed Steel Sections

11.2.2 Post-Buckling Behaviour and the Calculation of Effective Width When an element buckles locally it does not necessarily mean that this element will collapse or loss its ability of carrying loads. In fact, a plate can be allowed to take a considerably increased load beyond initial buckling before any danger of collapse occurs. This is because the deflections due to buckling are accompanied by stretching of the middle surface of the plate. It is not always possible for practical reasons to allow some elements of a structure to buckle, but if stable buckles can be tolerated, a considerable gain follows in structural efficiency. For a uniformly compressed rectangular plate, up to the buckling load, the stress distribution is uniform. With increase in load, the central unconstrained portion of the plate will start to deflect laterally and will therefore not support much additional load, whereas the portions close to the supported edges will be constrained to remain straight and will continue to carry increasing stresses. Figure 11.6 shows the typical variation of the stress distribution in a plate in pre- and post-buckling stages. The ultimate strength of the plate is when the maximum stress at the edges reaches the compressive yield strength of the material. Thus the ultimate load of the plate should be calculated based on the stress distribution at failure through the width of the plate. The problem, however, is that analysis of the post-buckled plate is a complicated process and no exact closed form results have been obtained for compressed plates. Therefore, instead of using the stress distribution in the postbuckling range, an alternative approach to assessing the ultimate load of the plate is to use an effective width concept.

0,5beff

0,5beff

bp

fy b

(a)

(b)

fy b

(c)

FIGURE 11.6 The concept of effective width. (a) Stress distribution up to buckling, (b) stress distribution at failure and (c) stress distribution in effective width

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423

The concept of effective width was originally developed by Von Karman et al. (1932) and calibrated for cold-formed members by Winter (1968). The method assumes that when the ultimate stress is reached, the total load is carried by two fictitious strips adjacent to the edges of the plate (see Fig. 11.6c), which carry a uniform stress equal to the yield strength of the material, and the central region is unstressed. Obviously, the calculation of the effective width is dependent on the stress distribution at the time when the plate fails, which is influenced by a number of factors including the pattern of applied compressive stresses and the boundary conditions, relative slenderness and geometrical imperfections of the plate. Based on large numbers of tests, empirical functions have been developed. In EN 1993-1-5 (2006) the following equations have been recommended for calculating the effective width of a compression element. For doubly supported compression elements (cl 4.4.2 in EN 1993-1-5 (2006)) ⎧ 1,0 ⎪ ⎪ ⎨ ρ = λp,red − 0,055(3 + ψ) ⎪ ⎪ ⎩ 2 λp,red

λp,red ≤ 0,673 0,673 < λp,red

(11.14a)

For outstand compression elements or single edge compression stiffener elements (cl 4.4.2 in EN 1993-1-5 (2006))

ρ=

⎧ 1,0 ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λp,red ≤ 0,748

λp,red − 0,188 2 λp,red

0,748 < λp,red

(11.14b)

in which, λp,red = λp λp =

σcom,Ed = reduced slenderness fyb /γM0

fyb = relative slenderness for local buckling σcr,p

where ρ is the reduction factor to determine the effective width of a compression element defined in Tables 11.1 and 11.2, fyb is the basic yield strength of the material, σcom,Ed (σcom,Ed ≤ fyb /γM0 ) is the largest compressive stress in the compression element, and γM0 is the partial safety factor for resistance of the cross-section. After the effective widths of individual compression elements have been determined, the effective area, second moments of the effective area and effective section modulus can be calculated, from which the design values of the resistance to bending moments can be determined.

424

•

Chapter 11 / Cold-formed Steel Sections TABLE 11.1 Doubly supported compression elements (ψ = σ2 /σ1 ). Stress distribution (compression positive)

Effective width, beff

s1

ψ=1 beff = ρb be1 = be2 = 0,5beff

s2

be1

be2

b

s1

s2

be1

bc

be1

b

be1 = 52b−effψ be2 = beff − be1

be2

b

s1

1>ψ≥0 beff = ρb

0>ψ ρb beff = ρbc = 1−ψ be1 = 0,4beff be2 = 0,6beff

bt

be2

s2

11.3 DISTORTIONAL BUCKLING Distortional buckling involves both rotation and translation at the corners of the crosssection. Distortional buckling of flexural members such as channel and zed sections involves rotation of only the compression flange and lip about the flange–web junction as shown in Fig. 11.7. The web undergoes flexure at the same half-wavelength as the flange buckle, and the compression flange may translate in a direction normal to the web, also at the same half-wavelength as the flange and web buckling deformations. The elastic distortional buckling stress of cold-formed flexural members can be determined using either analytical methods, such as those suggested in AS/NZS 4600 (1996) and EN 1993-1-3 (2006) or numerical methods, such as the finite strip method (FSM) (Schafer, 1997) and the generalized beam theory (GBT) method (Davies et al., 1993).

11.3.1 The Calculation Method in EN 1993-1-3 (2006) In EN 1993-1-3 (2006), the design of compression elements with intermediate or edge stiffeners is based on the assumption that the stiffener behaves as a compression

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425

TABLE 11.2 Outstand compression element (ψ = σ2 /σ1 ). Stress distribution (compression positive)

Effective width, beff

s1

1>ψ≥0 beff = ρb s2

beff

b

s1

bc

0>ψ beff = ρbc = 1 ρb −ψ

bt

beff

s2

b

1>ψ≥0 beff = ρb

s1 s2

beff

b

bc

s1

0>ψ beff = ρbc = 1 ρb −ψ

bt

s2 b

beff

(a)

(b)

FIGURE 11.7 Distortional buckling modes of a zed section (h = 120 mm, b = 75 mm, c = 20 mm, t = 2,5 mm) (a) under a pure compression and (b) under a pure bending

Chapter 11 / Cold-formed Steel Sections member with continuous partial restraint, with a spring stiffness that depends on the boundary conditions and the flexural stiffness of the adjacent plane elements of the cross-section. The spring stiffness of the stiffener is determined by applying a unit load per unit length to the cross-section at the location of the stiffener, as illustrated in Fig. 11.8, and is determined from K=

1 1 Et3 = δ 4(1 − ν2 ) b21 (b1 + hp )

(11.15)

where δ is the deflection of the centroid of the stiffener due to a unit load, b1 is the horizontal distance from the web line to the centroid of the effective area of the edge stiffener, and hp is the depth of the web. The elastic critical buckling stress for a long strut on an elastic foundation of a spring stiffness coefficient K is given by Timoshenko and Gere (1961) as follows: σcr,d =

π2 EIs Kλ2 + 2 As λ As π 2

(11.16)

where As and Is are the area and second moment of the effective section of the stiffener, as illustrated in Fig. 11.9 for an edge stiffener, and λ = l/m is the half-wavelength of the distortional buckling (l is the member length and m is the number of half waves). For a sufficiently long strut, the critical half-wavelength can be obtained by minimizing the critical stress as defined by Eq. (11.6) with respect to λ, to give, EIs 1/4 λcr = π (11.17) K

bp Cθ

U

b1 U δ

•

δ

426

θ (a) be2

K (b)

(c)

FIGURE 11.8 (a) Distortional buckling model used in EN 1993-1-3, (2006) (b) edge stiffener on an elastic foundation of a spring stiffness coefficient and (c) model used to determine spring stiffness coefficient (copy from EN 1993-1-3 (2006) by permission)

Structural Design of Steelwork to EN 1993 and EN 1994

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427

b bP be1

be2 b a

b1

C

Ceff Cp

a

b

FIGURE 11.9 Effective cross-sectional area of an edge stiffener in EN 1993-1-3 (2006) (copy from EN 1993-1-3 (2006) by permission)

K

As, ls

Substituting Eq. (11.17) into Eq. (11.16) yields, √ 2 KEIs σcr,d = As

(11.18)

Equation (11.18) is given in EN 1993-1-3 (2006) for calculating the critical stress of distortional buckling of the edge stiffener. The design strength in EN 1993-1-3 (2006) (Section 5.5.3.1) for distortional buckling is considered by using a reduced thickness of the edge stiffener. The reduction factor is calculated in terms of the relative slenderness as follows, ⎧ ⎪ λd ≤ 0,65 1,0 ⎪ ⎨ χd = 1,47 − 0,723λd 0,65 < λd < 1,38 (11.19) ⎪ ⎪ ⎩ 0,66/λ 1,38 ≤ λ d

d

in which, λd =

fyb = relative slenderness for distortional buckling σcr,d

The procedure for calculating χd can be summarized as follows (Section 5.5.3.2 in EN 1993-1-3 (2006)). • Step 1 Obtain an initial effective cross-section for the stiffener using effective widths determined by assuming that the stiffener gives full restraint and that σcom,Ed = fyb /γM0 . • Step 2 Use the initial effective cross-section of the stiffener to determine the reduction factor for distortional buckling (flexural buckling of a stiffener), allowing for the effects of the continuous spring restraint. • Step 3 Optionally iterate to refine the value of the reduction factor for buckling of the stiffener; that is, re-calculate the effective widths of the lip and the part of the flange near the lip based on the compressive stress σcom,Ed = χd fyb /γM0 and calculate the reduction factor again based on the newly calculated effective widths.

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•

Chapter 11 / Cold-formed Steel Sections The design value of the resistance to bending moment about the y-axis due to both local and distortional buckling is determined based on the elastic section modulus of the effective section, Mc,Rd =

fyb Weff,y γM0

(11.20)

where Weff,y is the section modulus of the effective section for bending about y-axis, in which, apart from the effective widths of the web and the part of the flange near to web are calculated using the local buckling formulae, the effective widths and thicknesses of the lip and the part of the flange near the lip are calculated using both local and distortional buckling formulae, based on the reduced compressive stress, σcom,Ed = χd fyb /γM0 .

11.3.2 The Calculation Methods in AS/NZS 4600 (1996) In EN 1993-1-3 (2006) the critical stress of distortional buckling of a section is calculated based on the model of an edge stiffener on an elastic foundation and the effect of the distortional buckling on the section properties is taken into account by reducing the thickness of the stiffener. An alternative to determine the critical stress of distortional buckling of a cold-formed steel section is to use AS/NZS 4600 (1996) design code. The elastic distortional buckling formulae for channel and zed sections in AS/NZS 4600 (1996) are based on a simple flange buckling model where the flange is treated as a thin-walled compression member, as shown in Fig. 11.10, undergoing flexural–torsional buckling (Lau and Hancock, 1987; Hancock, 1997). The rotational spring stiffness kθ represents the flexural restraint provided by the web which is in flexure, and the translational spring stiffness kx represents the resistance to translational movement of the section in the buckling mode. The model includes a reduction in the flexural restraint provided by the web as a result of the compressive stress in the web. Lau and Hancock (1987) showed that the translational spring stiffness has no significant influence on results and thus is assumed to be zero. The rotational spring stiffness and the critical stress at distortional buckling are given as h4p λ2 1,11σcr,d 2Et3 kθ = (11.21) 1− 5,46(hp + 0,06λ) Et2 12,56λ4 + 2,192h4p + 13,39λ2 h2p

x Kx

Shear centre

Kθ y

x

y

FIGURE 11.10 Flange elastically restrained along flange-web junction in AS/NZS 4600 (1996)

Structural Design of Steelwork to EN 1993 and EN 1994 σcr,d =

! E (α1 + α2 ) − (α1 + α2 )2 − 4α3 2 Af

•

429

(11.22)

in which, η kθ Ixf b2p + 0,039Jf λ2 + β1 β1 ηE 2 α2 = η Iyf + ybp Ixyf β1 η 2 2 α3 = η α1 Iyf − Ixyf bp β1 Ixf + Iyf β1 = x2 + Af 1/4 Ixf b2p hp λ = 4,80 2t3 α1 =

η=

π 2 λ

where Af is the full cross-sectional area of the compression flange and lip, Ixf and Iyf are the second moments of the area Af about x - and y -axes, respectively, where the x - and y -axes are located at the centroid of area Af with x -axis parallel with flange, Ixyf is the product moment of the area Af about x - and y -axes, Jf is the St Venant torsion constant of the area Af , x and y are the distances from the flange-web junction to the centroid of area Af in the x - and y -directions, respectively. Due to the coupling of σcr,d and kθ in Eqs (11.21) and (11.22), Hancock (1997) suggested that kθ can be calculated based on an initial σcr,d obtained by assuming kθ = 0 and after then σcr,d can be calculated based on the obtained kθ value. In the iteration, if kθ < 0, kθ should be calculated using σcr,d = 0. The elastic critical moment for distortional buckling is calculated based on the critical buckling stress as follows, Mcr,d = σcr,d Wy

(11.23)

where Wy is the elastic section modulus of the gross cross-section for the extreme compression fibre. The design value of the resistance to bending moment about y-axis due to distortional buckling which involves rotation of the compression flange and lip about the flange-web junction is calculated as follows:

Mc,Rd =

⎧ ⎪ ⎨ Mc

0 ≤ kθ

Weff,y ⎪ ⎩ Mc Wy

kθ < 0

(11.24)

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Chapter 11 / Cold-formed Steel Sections in which, ⎧ Myield Weff,y ⎪ ⎪ 1 − M ⎪ yield ⎪ 4Mcr,d Wy ⎪ ⎨ ⎡ ⎤ 2 Mc = ⎪ M Weff,y ⎪ yield ⎪ ⎣0,055 ⎪ M − 3,6 + 0,237⎦ ⎪ ⎩ yield Mcr,d Wy

0,5Myield < Mcr,d Mcr,d ≤ 0,5Myield

where Myield = fyb Wy is the moment causing initial yield at the extreme compression fibre of the gross section. A comparison of the critical stresses of distortional buckling by using EN 1993-1-3 (draft version of 2001) and AS/NZS 4600 (1996) with experimental data was made by Kesti and Davies (1999). It was found that Lau and Hancock’s analytical expressions give a good prediction of the distortional buckling stress. The method given in EN 1993-1-3 does not correlate as well as Lau and Hancock’s method. The error in the distortional buckling stress could lead to a consequential error in the effective cross-sectional area depending on the distortional buckling stress level.

11.4 LATERAL–TORSIONAL BUCKLING In practice, purlins and rails are usually used together with their supported cladding or sheeting, and thus they are generally considered to be restrained against lateral deflections perpendicular to the line of action of the loading. If a beam is fully restrained on its translational and rotational degrees neither will the beam rotate nor deflect laterally. However, if the cladding or sheeting is not strong enough then it is possible for the beam to become unstable and for very large lateral deflections to occur at a critical value of the applied load. This type of behaviour is called lateral–torsional buckling. Chapter 5 has discussed the lateral–torsional buckling of unrestrained beams. In this section it is to deal with the lateral–torsional buckling of zed section beams with partial restraints from the sheeting. For the lateral–torsional buckling of partially restrained channel section beams readers can see the work of Chu et al. (2004).

11.4.1 Critical Moment of Lateral–Torsional Buckling The model presented here for analysing the lateral–torsional buckling of partially restrained beams was originally developed by Li (2004) and lately expanded by Chu et al. (2004b), which is similar to that described in Section 11.1. Assume that the displacements and moments in a state of equilibrium are (v,w,φ) and (My , Mz , Mω , MT ). Now let vb and wb be the y- and z-components of the buckling displacement of the centroid of the cross-section and φb be the buckling angle of twisting of the section. Assuming that the displacements in pre-buckling are very small, the increase

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of the strain energy of the system due to the lateral–torsional buckling thus can be expressed as ⎡ 2 2 ⎤ l d 2 wb d 2 vb ⎦ d 2 v b d 2 wb 1 ⎣ U2 = EIy dx + 2EIyz 2 + EIz 2 dx2 dx dx2 dx2 0

1 + 2

l

⎡ ⎣EIw

d 2 φb dx2

2

+ GIT

dφb dx

2

⎤ ⎦dx

0

1 + 2

l [ky (vb + zk φb )2 + kφ φb2 ]dx

(11.25)

0

On the other hand, the lateral–torsional buckling leads to a decrease of the potential of the pre-buckling moments and loads, which can be expressed as (Chu, 2004 Li, 2004; Chu et al., 2004b) l

d 2 vb 1 d 2 wb (Mz φb ) − M φ + Mω y b 2 dx2 dx2

W2 =

dφb dx

2 dx

0

l +

1 [(qy yq + qz zq )φb2 ]dx 2

(11.26)

0

If the net change of the total potential is positive for any of possible buckling displacements, that is, U2 > W2 , then the equilibrium of the pre-buckling state is said to be stable because the generation of the buckling displacements requires an energy input into the system. On the other hand, if the net change of the total potential is negative, that is, U2 < W2 , then the equilibrium of the pre-buckling state is said to be unstable because the buckling displacements can be generated without any input of energy. A critical state between stable and unstable equilibria from which the critical load of the lateral–torsional buckling can be determined is U2 = W2

(11.27)

Equation (11.27) is an eigenvalue type equation. For given reference loads the smallest eigenvalue and corresponding eigenvector calculated from Eq. (11.27) represent the critical loading factor and corresponding buckling mode. The above model can be applied directly to the purlins with intermediate lateral restraints such as provided by anti-sag bars if the pre-buckling moments are calculated based on the same model and the buckling displacements satisfy the displacement restraint conditions at the places where the anti-sag bars are placed. It is difficult to achieve closed form solutions of Eq. (11.27). In most cases only numerical solutions can be obtained. A general numerical computation procedure has been described by Li (2004) and Chu et al. (2004b) to obtain the critical buckling load, in which the following

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•

Chapter 11 / Cold-formed Steel Sections piecewise cubic spline functions are used to construct the displacement fields before and during buckling ⎧ ⎫ ⎡ ⎪ ⎨ v(x) ⎪ ⎬ Ni (x) ⎢ w(x) = ⎣ 0 ⎪ ⎪ ⎩ φ(x) ⎭ 0

⎤⎧ ⎫ ⎪ 0 ⎬ ⎨ δvi ⎪ ⎥ 0 ⎦ δwi ⎪ ⎪ Ni (x) ⎩δφi ⎭ ⎧ ⎫ ⎡ ⎤⎧ b ⎫ ⎪ δ ⎪ ⎪ ⎪ ⎪ 0 0 ⎨ vb (x) ⎬ Ni (x) ⎬ ⎨ vi ⎪ ⎢ ⎥ b wb (x) = Ni (x) 0 ⎦ δwi ⎣ 0 ⎪ ⎪ ⎪ ⎩ φ (x) ⎪ ⎭ ⎭ ⎩ δb ⎪ 0 0 Ni (x) ⎪ b 0 Ni (x) 0

(11.28)

(11.29)

φi

where Ni (x) is the spline interpolation function at node i and {δvi , δwi , δφi } and {δbvi , δbwi , δbφi } are the nodal displacement vectors before and during buckling. By using the displacement expressions (11.28) and (11.29), the equilibrium Eq. (11.4) and buckling Eq. (11.27) can be simplified into the following algebraic matrix equations ⎡

K vv ⎢ ⎣Kvw Kvφ ⎡ Kvv ⎢ ⎣Kvw Kvφ

Kvw Kww 0 Kvw Kww 0

⎤⎧ ⎫ ⎧ ⎫ Kvφ ⎪ ⎨ δv ⎪ ⎬ ⎪ ⎨ Fv ⎪ ⎬ ⎥ 0 ⎦ δw = Fw ⎪ ⎪ ⎪ ⎪ Kφφ ⎩δφ ⎭ ⎩Fφ ⎭ ⎤⎧ ⎫ ⎡ b 0 Kvφ ⎪ ⎨ δv ⎪ ⎬ qcr ⎢ ⎥ b 0 ⎦ δw = ⎣ 0 ⎪ ⎪ qref 0 Kφφ ⎩δbφ ⎭ Kvφ

(11.30)

0 0 0 Kwφ

⎤⎧ ⎫ 0 b ⎪ Kvφ ⎬ ⎨ δv ⎪ ⎥ 0 Kwφ ⎦ δbw ⎪ ⎭ ⎩ δb ⎪ 0 Kφφ φ

(11.31)

where K ij (i, j = v,w,φ) is the stiffness matrix, Kij0 (i, j = v,w,φ) is the geometric stiffness matrix, δj ( j = v,w,φ) is the nodal displacement vector, F j ( j = v,w,φ) is the nodal force vector, δbj ( j = v,w,φ) is the nodal vector of buckling displacements, (qcr /qref ) represents the scale factor between the critical and referenced loads. As an example, Fig. 11.11 shows the critical load factors of simply supported zed purlin beams (web depth h = 202 mm, flange width b = 75 mm, lip length c = 20 mm and thickness t = 2,3 mm), with zero, one and two anti-sag bars, subjected to a uniformly distributed uplift load. It can be seen from the figure that the lateral restraint has remarkable influence on the lateral–torsional buckling of the beam with no antisag bars. The influence is found to decrease with the increase of the beam length. Interestingly, when the beam has one or two anti-sag bars, the influence of the lateral restraint on the lateral–torsional buckling becomes almost negligible. This implies that the anti-sag bar not only has the ability of increasing the critical load but also can reduce the influence of the lateral restraint provided by cladding. Practically, purlins are often used as continuous beams over two or more spans, in which case the boundary conditions of the beam can be regarded as simply supported at one end and fixed at the other end. Fig. 11.12 shows the critical load factors of the zed purlin beam with this kind of boundary conditions. The results show that there is a significant increase in critical load when the purlin has a fixed boundary condition.

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5 ky0, no bars

4.5

ky, no bars

4

ky0, 1 bar

Critical loading factor

3.5

ky, 1 bar ky0, 2 bars

3

ky, 2 bars 2.5 2 1.5 1 0.5 0 3000

4000

5000

6000

7000

8000

9000

10 000

Beam length, mm

FIGURE 11.11 Lateral–torsional buckling of simply supported zed section beams (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformly distributed uplift loading, with various different restraints 5 ky0, no bars

Critical loading factor

4.5

ky, no bars

4

ky0, 1 bar

3.5

ky, 1 bar ky0, 2 bars

3

ky, 2 bars

2.5 2 1.5 1 0.5 0 3000

4000

5000

6000

7000

8000

9000

10 000

Beam length, mm

FIGURE 11.12 Lateral–torsional buckling of pinned-fixed zed section beams (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformly distributed uplift loading, with various different restraints

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11.4.2 Buckling Resistance Moment of Beams Subject to Bending The critical load calculated from the lateral–torsional buckling is based on an idealized model in which the beam has no geometrical imperfections, the cross-section of the beam does not deform, and the beam does not buckle either locally or distortionally before the lateral–torsional buckling occurs. When designing a real member, however, these factors should be taken into account. The method for determining the design buckling resistance moment for the lateral–torsional buckling of cold-formed section beams is the same as that used for other steel section members (Section 6.3.2.2 in EN 1993-1-1 (2005)), that is, Mb,Rd = χLT

Weff,y fyb γM1

(11.32)

in which, χLT =

1 2 1/2

2 − λ φLT + φLT LT

= reduction factor for lateral–torsional buckling (χLT ≤ 1,0)

2 φLT = 0, 5[1 + αLT λLT − 0,2 + λLT ] = factor used to calculate the reduction factor Mc,Rd 1/2 λLT = = relative slenderness for lateral–torsional buckling Mcr,LT where αLT = 0,34 is the imperfection factor, γM1 = 1,0 is the partial factor for resistance of members to instability and Mcr,LT is the elastic critical moment of the gross crosssection for lateral–torsional buckling about the main axis.

11.5 CALCULATION OF DEFLECTIONS The deflections of channel and zed section beams under uniformly distributed transverse loads can be evaluated using the analytical model presented in Section 11.1, provided that the load does not exceed the critical loads of local and distortional buckling. For the evaluation of deflections at loads greater than any critical load the influence of the local buckling and/or distortional buckling must be taken into account. The precise analysis of the post-buckling behaviour of a cold-formed section beam, however, is very difficult. A simple approach is to assume that the relationships between the load and deflection are linear for both pre- and post-buckling analyses as illustrated in Fig. 11.13. In pre-buckling region, the deflections can be calculated using simple beam theory and the gross section properties of the beam, since the beam is fully effective before buckling. In post-buckling region, the deflections can be evaluated using simple beam theory and the reduced section properties of the beam, since the beam is not fully effective after buckling. This assumption is only approximate since, in reality, the line in post-buckling region is not a straight line, but the errors introduced in the approximation are acceptable and conservative when fully reduced

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M Rd

cr

MRd Mcr

D

FIGURE 11.13 Simplified model used for the calculation of deflections section properties are used. Note that, in Fig. 11.13 in the pre-buckling region and ( − cr ) in the post-buckling region are linearly proportional to M and (M − Mcr ), respectively. Therefore, the deflection of the beam, , at an applied moment, M, can be expressed as ⎧ M ⎪ M ≤ Mcr ⎪ ⎨ cr M cr = ⎪ M − Mcr I ⎪ ⎩ cr 1 + Mcr < M ≤ MRd Mcr Ieff

(11.33)

where Mcr and MRd are the critical moment and moment resistance of the beam, cr is the deflection of beam when it buckles, I is the second moment of the gross cross-sectional area, and Ieff is the second moment of the effective cross-sectional area.

11.6 FINITE STRIP METHODS The FSM was originally developed by Cheung (1976) and it can be considered as a specialization of the finite element method (Zienkiewicz and Taylor, 2000). The method is mainly applied to structures whose geometries do not vary with at least one of the coordinate axes. The approach of the FSM is considered particularly favourable when dealing with the initial buckling or natural frequency characteristics of thinwalled prismatic structures. In the FSM, the prismatic structure is discretized into a number of longitudinal strips and the displacement fields associated with each strip

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Chapter 11 / Cold-formed Steel Sections

(a)

(b)

FIGURE 11.14 (a) The finite element analysis and (b) the finite strip analysis

vary sinusoidally along the strip length and algebraically across the strip width. Similar to the finite element method, shape functions are also used to define the variation of displacement fields along the strip width but they are only the functions of the cross-section coordinates of the strip (see Fig. 11.14). The use of the FSM for understanding and predicting the behaviour of cold-formed steel members was pioneered by Lau and Hancock in Australia (1986, 1989). They modified the stiffness matrices derived in Cheung’s book and created a commercial computer program for solution of the elastic buckling problem of open thin-walled members via finite strip called Thin-Wall. Similar programs were developed by Loughlan (1993) in the UK and by Schafer (1997) in USA. The program developed by Schafer is available in internet (http://www.ce.jhu.edu/bschafer/cufsm) and is particularly friendly to use. The codes were written in Matlab language and thus can easily be modified by users.

11.6.1 Element Stiffness Matrix of the Strip Consider a strip shown in Fig. 11.15, in which the local coordinate system is defined as that, the x- and y-axes are the two axes within the plane of the strip and the z-axis is normal to the plane of the strip. The three components of buckling displacement of the strip at a point (x, y) can be expressed in terms of the nodal displacements as follows ⎫ ⎧ ⎡ ⎤⎡ ⎤ ⎪ v1m ⎪ mπx y y ⎪ ⎪ 5 6 ⎪ sin 0 0 0 ⎪ 1− ⎬ ⎨ v(x, y) a b b ⎢ ⎥⎢ ⎥ u1m =⎣ (11.34) ⎦ ⎣ ⎦ mπx y y ⎪ u(x, y) v2m ⎪ ⎪ ⎪ ⎪ ⎪ 0 cos 0 1− 0 a b b ⎩ u2m ⎭

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⎧ ⎫ w1m ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ mπx 3y2 y3 3y2 2y3 2y2 2y3 y3 y2 ⎨ θ1m ⎬ w(x, y) = sin 1− 2 + 3 , y− + 2, 2 − 3 , 2 − (11.35) a b b ⎪ b b b b b b w2m ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ θ2m

where u(x,y) and v(x,y) are the plane displacements, w(x,y) is the deflection, (u1m , v1m , w1m , θ1m ) and (u2m , v2m , w2m , θ2m ) are the nodal displacements associated with wave number m, a and b are the length and width of the strip, respectively. The assumed displacement functions satisfy only the simply supported conditions at the two end sides of the strip. The strain energy of the strip is given as follows Et U2 = 2(1 − ν2 )

a b

εx + ε y

o

Et3 + 24(1 − ν2 )

2

− 2(1 − ν) εx εy −

o

a b

κx + κ y

o

2

− 2(1 − ν)

2 γxy

4

2 κx κy − κxy

dx dy dx dy

(11.36)

o

The membrane and bending strains in Eq. (11.36) are defined as follows, εx =

∂u ∂x

∂2 w κx = − 2 ∂x

εy =

∂v ∂y

∂2 w κy = − 2 ∂y

γxy =

∂u ∂v + ∂y ∂x

∂2 w κxy = − ∂x ∂y

(11.37)

The element stiffness matrix can be obtained by substituting Eqs (11.34) and (11.35) into Eq. (11.37) and then into Eq. (11.36), that is, U2 =

1 T {δ} [K]m {δ}m 2 m

in which, {δ}m = u1m , v1m , u2m , v2m , w1m , θ1m , w2m , θ2m T = element nodal vector ⎤ ⎡ uv uv K uv uv K14 K11 K12 13 ⎥ ⎢ uv −K uv K uv K22 ⎥ ⎢ 14 24 [0] ⎥ ⎢ 4x4 uv uv K11 −K12 ⎥ ⎢ ⎥ ⎢ uv ⎥ ⎢ K 22 ⎥ ⎢ [K]m = ⎢ wθ wθ wθ wθ K11 K12 K13 K14 ⎥ ⎥ ⎢ ⎥ ⎢ wθ −K wθ K wθ K22 ⎥ ⎢ 14 24 ⎥ ⎢ symmetric wθ wθ ⎣ K11 −K12 ⎦ wθ K22 = element stiffness matrix uv K11 uv K12

12D 1 a 1 − ν b 2 = 2 + (mπ) 2b 12 a t 12D 3ν − 1 = 2 (mπ) 8 t

(11.38)

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Chapter 11 / Cold-formed Steel Sections 1a 1−νb 12D 2 − + (mπ) 2b 24 a t2 12D 1 + ν (mπ) = 2 8 t 12D 1 − ν a 1 b = 2 + (mπ)2 4 b 6a t 1−νa 12D 1 b 2 = 2 − + (mπ) 4 b 12 a t

uv K13 = uv K14 uv K22 uv K24

wθ K11

=

wθ K12 =

wθ K13 =

wθ K14 =

wθ K22 =

wθ = K24

D=

a 3 D 13 b 6a 4 2 (mπ) + (mπ) + 6 5b b a2 70 a 2 a 2 D 11 b 1 + 5ν 4 2 (mπ) + (mπ) + 3 a 420 a 10 b D 9 b 6a a 3 (mπ)4 − (mπ)2 − 6 2 5b b a 140 a 2 a 2 D 13 b 1 4 2 − (mπ) + (mπ) + 3 a 840 a 10 b 3 1 b 2 b a D (mπ)4 + (mπ)2 + 2 210 a 15 a b 3 3 b 1 b a 4 2 D − (mπ) − (mπ) + 840 a 30 a b Et3 12(1 − ν2 )

11.6.2 Element Geometric Stiffness Matrix of the Strip The geometric stiffness matrix of a strip subjected to linearly varying edge traction can be derived by considering the change of the potential of the in-plane forces during buckling. Similar to the approach described in Section 11.4, the change of the potential of the in-plane forces can be expressed as, t W2 = 2

a b " o

o

y# σ1 − (σ1 − σ2 ) b

∂u ∂x

2 +

∂v ∂x

2 +

∂w ∂x

2 dx dy

(11.39)

where σ1 and σ2 are the compressive stresses at nodes 1 and 2 (see Fig. 11.15). Substituting Eqs (11.34) and (11.35) into Eq. (11.39) yields W2 =

1 T {δ} Kg m {δ}m 2 m

(11.40)

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b

a

a x y u1 θ1

v1

u2

θ2

w1

z

v2

s1

w2

s2

FIGURE 11.15 Local coordinates, degrees of freedom and stress distribution in a strip

in which,

b(mπ)2 [Kguv ] [0]4x4 [Kg ]m = = element geometric stiffness matrix 1680a [0]4x4 [Kgwθ ] ⎡ ⎤ 70(T1 + T2 ) 0 70(3T1 + T2 ) 0 ⎢ 70(T1 + T2 ) ⎥ 70(3T1 + T2 ) 0 ⎢ ⎥ [Kguv ] = ⎢ ⎥ ⎣ ⎦ 70(T1 + 3T2 ) 0 symmetric 70(T1 + 3T2 ) ⎡ ⎤ −2b(7T1 + 6T2 ) 8(30T1 + 9T2 ) 2b(15T1 + 7T2 ) 54(T1 + T2 ) ⎢ b2 (5T1 + 3T2 ) 2b(6T1 + 7T2 ) −3b2 (T1 + T2 ) ⎥ ⎢ ⎥ [Kgwθ ] = ⎢ ⎥ ⎣ 24(3T1 + 10T2 ) −2b(7T1 + 15T2 ) ⎦ symmetric b2 (3T1 + 5T2 )

For a member composed of multiple strips the global stiffness matrix and global geometric stiffness matrix can be obtained by the assembly of element stiffness matrices and element geometric stiffness matrices, that is [Km ] = [K]m [Kgm ] = [Kg ]m (11.41) k

k

where the index k denotes the k-th element. The summation implies proper coordinate transformations and correct addition of the stiffness terms in the global coordinates and degrees of freedom. The elastic buckling problem is a standard eigenvalue problem of the following form: [Km ]{δm } = λm [Kgm ]{δm }

(11.42)

where eigenvalues, λm , and eigenvectors, {δm }, are the buckling loads and corresponding buckling modes.

•

Chapter 11 / Cold-formed Steel Sections

11.6.3 Finite Strip Solution Methods The finite strip analysis employs single wave functions and thus can only be applied to the case where the longitudinal stresses do not vary along the longitudinal axis. Note that both [K m ] and [K gm ] are the functions of the strip length, a, and wave number, m. The buckling loads and buckling modes solved from Eq. (11.42) are also the functions of them. Since only the critical buckling load, which is the smallest eigenvalue for any of possible wave numbers, is of interest, the strip length and wave number are not independent each other. There are two quick ways to find the critical buckling load. One is to let a equal to the real length of the strip and solve the problem for several numbers of m. The other is to let m = 1 and solve the problem for a number of lengths. The former provides the results only for the given length of the strip, whereas the latter provides a complete picture of the critical buckling loads and modes at various different half-wavelengths, which has clear physical meanings. For this reason the latter method is often used. Figure 11.16 shows the buckling curves of a laterally restrained (both displacement and rotation) zed section beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) subjected to pure bending, in which the three local minima represent the local, distortional and secondary distortional buckling. The secondary distortional buckling is sometimes called lateral–distortional buckling to distinguish it from flange distortional buckling. The secondary distortional buckling exists only when the section is restrained

3 Critical load curve Buckling curve m = 1 Buckling curve m = 2–8

2.5

Critical loading factor

440

2 110 mm

1.5 565 mm

3000mm

1

0.5

0

102

103 Beam length, mm

FIGURE 11.16 Buckling curves of a simply supported zed section beam restrained at the junction between web and tension flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) (copy from Chu et al. (2006) by permission)

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laterally. Note that the buckling curves only provide information about how and when the beam buckles. The actual critical load of the beam for a given length should be taken as the lowest value from all of the buckling curves of the same length, and this critical load curve is the solid line plotted in Fig. 11.16. The original FSM can only deal with the buckling problem of members subjected to either pure compression or pure bending. Recently, Chu et al. (2005, 2006) modified the geometric stiffness matrix in the FSM by allowing for the variation of the prebuckling stress along the longitudinal axis, thus leading to a semi-analytical FSM which is able to deal with the buckling problem of cold-formed section beams subjected to uniformly distributed transverse loading. As an example, Fig. 11.17 shows the critical load curve of the restrained zed section beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) under uniformly distributed loading. In order to compare the difference in critical load between pure bending and uniformly distributed loading, the results of the beam under pure bending is also superimposed in the figure. It is evident that the critical load associated with uniformly distributed loading is significantly higher than that arising from pure bending although, for both local and distortional buckling, the differences between the two loading cases decrease with the beam length. For example, for a 5 m long beam, the critical load of a uniformly distributed loading beam is 12% higher than that of the pure bending beam compared to 20% higher for a 2 m long beam.

4 Uniformly distributed load 3.5

Pure bending 110 mm

Critical loading factor

3 2.5 2 565 mm 1.5 3000mm 1 0.5 0

102

103 Beam length, mm

FIGURE 11.17 Critical load curves of a simply supported zed section beam restrained at the junction between web and tension flange subjected to a uniformly distributed uplift load and a pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) (copy from Chu et al. (2006) by permission)

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11.7 DESIGN METHODS FOR BEAMS PARTIALLY RESTRAINED BY SHEETING Purlins and rails are usually used together with their supported trapetzoidal sheeting. Thus it is generally assumed that the sheeting takes the load in the plane of the sheeting and the purlin takes the load normal to the plane of the sheeting, that is, in the plane of web. Also, the purlin may be regarded as being laterally restrained in the plane of the sheeting and partially restrained in twisting if the trapetzoidal sheeting is connected to a purlin and the connection meets the following condition (cl 10.1.1.6 in EN 1993-1-3 (2006)) 70 π2 EIw π2 h2 EIz S≥ 2 (11.43) + GIT + h l2 4l2 where S is the portion of the shear stiffness provided by the sheeting for the examined member connected to the sheeting at each rib (if the sheeting is connected to a purlin every second rib only, then S should be substituted by 0,2S), h is the web depth and l is the span length. The partial torsional restraint may be represented by a rotational spring with a spring stiffness CD , which can be calculated based on the stiffness of the sheeting and the connection between the sheeting and the purlin, as follows, 1 1 1 = + CD CD,A CD,C

(11.44)

where CD,A is the rotational stiffness of the connection between the sheeting and the purlin and CD,C is the rotational stiffness corresponding to the flexural stiffness of the sheeting. Both CD,A and CD,C are specified in Section 10.1.5.2 in EN 1993-1-3 (2006). The restraints of the sheeting to the purlin have important influence on the buckling behaviour of the purlin. Fig. 11.18 shows the buckling curves of a simply supported zed purlin beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) with various different lateral restraints applied at the junction between the web and the compression flange when subjected to a pure bending. The figure shows that, when the translational displacement of the compression flange is restrained the purlin does not buckle lateral torsionally. On the other hand, when the rotation of the compression flange is restrained the critical stresses of local buckling and distortional buckling are increased quite significantly. However, when the restraints are applied at the junction between the web and the tension flange, it is only the rotational restraint that influences the lateral–torsional buckling of the purlin (see Fig. 11.19). Similar to the buckling behaviour, the lateral restraints also have considerable influence on the bending behaviour of the purlin. It has been found from both finite element analyses and experiments that, the bending behaviours are different when the restraints are applied at tension and compression flanges, and the free flange and the restrained flange are bent differently, particular when the free flange is under

Structural Design of Steelwork to EN 1993 and EN 1994

Ratio of critical buckling stress to yield strength

5 4.5 4

•

443

No restraint Translational restraint Rotational restraint Full restraints

3.5 3 2.5 2 1.5 1 0.5 0

102

103 Half-wavelength, mm

FIGURE 11.18 Buckling curves of a simply supported zed section beam with different restraint applied at the junction between web and compression flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) Ratio of critical buckling stress to yield strength

5 4.5 4

No restraint Translational restraint Rotational restraint Full restraints

3.5 3 2.5 2 1.5 1 0.5 0

102

103 Half-wavelength, mm

FIGURE 11.19 Buckling curves of a simply supported zed section beam with different restraint applied at the junction between web and tension flange subjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) compression (Lucas et al., 1997a, b; Vrany, 2002). This led to development of different treatments for sections when subjected to gravity loading and uplift loading, since, for instance, for a simply supported beam, the free flange is in tension for gravity loading but in compression for uplift loading. In EN 1993-1-3 (2006), the stress in the

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Chapter 11 / Cold-formed Steel Sections restrained flange is calculated based on the bending moment in the plane of web as follows (assuming there is no axial force), σmax,Ed =

My,Ed fy ≤ Weff,y γM0

(11.45)

where fy is the yield strength. While the stress in the free flange is calculated based on not only the bending moment in the plane of web but also the bending moment in the free flange due to the equivalent lateral load acting on the free flange caused by torsion and lateral bending as follows, σmax,Ed =

My,Ed fy Mfz,Ed + ≤ Weff,y Wfz γM0

(11.46)

where Mfz,Ed is the bending moment in the free flange due to the equivalent lateral load as defined in Fig. 10.3 in EN 1993-1-3 (2006), and Wfz is the gross elastic section modulus of the free flange plus 1/5 of the web height for the point of web–flange intersection, for bending about the z–z axis (see Fig. 11.20). The determination of Mfz,Ed is dependent on the section dimensions, loading position, span length, number of anti-sag bars, and spring stiffness CD , the detail of which is specified in Section 10.1.4.1 in EN 1993-1-3 (2006). It should be pointed out that Eq. (11.46) applies to only the case where the free flange is under compression. For the free flange under tension, where due to positive influence of flange curling and second order effect moment Mfz,Ed may be taken equal to zero.

qEd

NEd

NEd

1/5 h

h kh qEd b

z

y

y

1/5 h

ez2

z My,Ed Weff,y

NEd Aeff

ez1

Mfz,Ed Wfz

FIGURE 11.20 Superposition of stresses in free flange (copy from EN 1993-1-3 (2006) by permission)

1/6 h

Structural Design of Steelwork to EN 1993 and EN 1994

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Similar to any compression members, the flexural buckling of the free flange under compression also need to be considered. The buckling resistance of the free flange is verified by using the following formula fy Mfz,Ed 1 My,Ed + ≤ χLT Weff,y Wfz γM1

(11.47)

2 − 0, 75(λ )2 )1/2 ] = reduction factor for flexural buckin which, χLT = 1/[φLT + (φLT fz

2 ling of the free flange (χLT ≤ 1,0 and χLT ≤ 1/ λLT )

2 φLT = 0,5 1 + αLT λfz − 0,4 + 0,75 λfz = factor used to calculate the reduction factor

where αLT = 0, 34 is the imperfection factor and λfz is the relative slenderness for flexural buckling of the free flange and is determined from lfz fyb λfz = (11.48) ifz π2 E where ifz is the radius of gyration of the gross cross-section of the free flange plus the contributing part of the web for bending about the z–z axis and lfz is the buckling length for the free flange which is specified in Section 10.1.4.2 in EN 1993-1-3 (2006).

11.8 WORKING EXAMPLES In order to assist designers, manufactures of cold-formed steel sections often provide design manuals for their products, which, in general, include gross section properties, effective section properties, and load tables for a number of specified span lengths under various different sets and/or types of connection. The gross section properties are calculated based on the geometric dimensions of the section, whereas the effective section properties are calculated based on the effective widths by taking into account the effects of local and distortion buckling. The load tables can be obtained from either tests or calculations by considering the following modes of failure: • • • • • •

Flexural failure involving local and distortional buckling in compression Lateral–torsional buckling due to insufficient lateral restraints Excessive deflection Shear failure Web crushing under direct loads or reactions Combined effects between bending and web crushing, and bending and shear.

The load tables design their sections for specific uses and give the performance of sections under various given circumstances. The following is an example of calculation of gross and effective section properties for a cold-formed lipped zed section subjected to bending in the plane of web.

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•

Chapter 11 / Cold-formed Steel Sections The calculation is based on the method recommended in EN 1993-1-3 (2006). The dimensions of the cross-section are (where the influence of rounding of the corners is neglected): Depth of web Width of flange in compression Width of flange in tension Length of lip Thickness

h = 202 mm bc = 75 mm bt = 75 mm c = 20 mm t = 2 mm

The material properties of the section are: Modulus of elasticity Poisson’s ratio Basic yield strength Partial factor

E = 210000 N/mm2 ν = 0,3 fyb = 390 N/mm2 γM0 = 1,00

Checking of geometrical proportions The design method of EN 1993-1-3 (2006) can be applied if the following conditions are satisfied (Section 5.2): b/t ≤ 60

bc /t = 75/2 = 37,5 < 60 bt /t = 75/2 = 37,5 < 60 c/t ≤ 50 c/t = 20/2 = 10 < 50 h/t ≤ 500 h/t = 202/2 = 101 < 500

→ ok → ok → ok → ok

In order to provide sufficient stiffness and avoid primary buckling of the stiffener itself, the size of stiffener should be within the following range (Section 5.2 in EN 1993-1-3 (2006)): 0,2 ≤ c/b ≤ 0,6 c/bc = 20/75 = 0,27 c/bt = 20/75 = 0,27

→ ok → ok

For cold-formed steel sections the section properties are usually calculated based on the dimensions of the section middle line as follows (see Fig. 11.21), Depth of web Width of flange in compression Width of flange in tension Length of lip

hp = h − t = 202 − 2 = 200 mm bp1 = bc − t = 75 − 2 = 73 mm bp2 = bt − t = 75 − 2 = 73 mm cp = c − t/2 = 20 − 2/2 = 19 mm

Calculation of gross section properties Gross cross-section area: A = t(2cp + bp1 + bp2 + hp ) = 2 × (2 × 19 + 73 + 73 + 200) = 768 mm2

Structural Design of Steelwork to EN 1993 and EN 1994

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447

hp

cp

bp1

cp

t

FIGURE 11.21 Symbols used for representing the dimensions of middle lines of a zed section

bp2

Position of the neutral axis with regard to the flange in compression: zc =

hp 200 = = 100 mm 2 2

Position of the neutral axis with regard to the flange in tension: zt = hp − zc = 200 − 100 = 100 mm Second moment of the gross cross-sectional area: Iy =

(h3p + 2c3p )t

+

(bp1 + bp2 )t3

c p 2 + zc2 bp1 t + zt2 bp2 t + cp t zc − 2

12 12 2 cp + cp t zt − = 4880000 mm4 2

Gross section modulus with regard to the flange in compression: Wy,c =

Iy 4880000 = = 48800 mm3 zc 100

Gross section modulus with regard to the flange in tension: Wy,t =

Iy 4880000 = = 48800 mm3 zt 100

Calculation of effective section properties The general (iterative) procedure is applied to calculate the effective properties of the compression flange and the lip (plane element with edge stiffener). The calculation should be carried out in three steps: Step 1 Obtain an initial effective cross-section for the stiffener using effective widths of the flange and lip determined by assuming that the compression flange is doubly supported, the stiffener gives full restraint (K = ∞) and that the design strength is not reduced, that is, σcom,Ed = fyb /γM0 . • Effective width of the compressed flange

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•

Chapter 11 / Cold-formed Steel Sections For the internal compression flange the stress ratio ψ = 1 (uniform compression), so the buckling coefficient is taken as kσ = 4. The relative slenderness thus is: fyb 390 λb,red = λb = = 0,827 = σcr,b 570 where σcr,b =

kσ π2 E 3,142 × 210000 4 = = 570 N/mm2 2 2 2 12(1 − ν ) (bp1 /t) 12(1 − 0,3 ) (73/2)2

Since λb,red = 0,827 > 0,673, the width reduction factor for the doubly supported compression element is calculated by (Eq. (11.14a)) ρ=

λb,red − 0,055(3 + ψ) 2 λb,red

=

0,827 − 0,055(3 + 1) = 0,887 0,8272

The effective width of the compressed flange thus is: beff = ρbp1 = 0,887 × 73 = 64,8 mm be1 = be2 = 0,5beff = 0,5 × 64,8 = 32,4 mm • Effective length of the lip For the compression lip, the buckling coefficient should be taken as follows (Eq. (11.13d)) kσ = 0,5 kσ = 0,5 + 0,83(cp /bp1 − 0,35)2/3

if cp /bp1 ≤ 0,35 if 0,35 ≤ cp /bp1 ≤ 0,6

For cp /bp1 = 19/73 = 0,260 < 0,35 kσ = 0,5. The relative slenderness is:

fyb 390 = = 0,609 σcr,c 1051 π2 E kσ 3,142 × 210000 0,5 where σcr,c = = = 1050 N/mm2 12(1 − ν2 ) (cp /t)2 12(1 − 0,32 ) (19/2)2 Since λc,red = 0, 609 < 0, 673, the width reduction factor for the outstand compression element thus is given by (Eq. (11.14b)): ρ = 1,0 The effective length of the compression lip thus is: ceff = ρcp = 1,0 × 19 = 19 mm The corresponding effective area of the edge stiffener is: As = t(be2 + ceff ) = 2 × (32,4 + 19) = 103 mm2 λc,red = λc =

Step 2 Use the initial effective cross-section of the stiffener to determine the reduction factor, allowing for the effects of the distortional buckling. The elastic critical stress of the distortional buckling for the edge stiffener is (Eq. (11.18)) √ 2 KEIs σcr,d = = 343 N/mm2 As

Structural Design of Steelwork to EN 1993 and EN 1994 where K =

•

449

Et3 = 0,445 N/mm2 4(1 − ν2 )(b21 hp + b31 )

b1 = bp1 −

b2e2 = 62,8 mm 2(be2 + ceff )

2 2 c3eff t c2eff c2eff be2 t3 ceff + + be2 t Is = + ceff t − 12 12 2(be2 + ceff ) 2 2(be2 + ceff ) = 3330 mm4 Thickness reduction factor for the edge stiffener is calculated based on the relative slenderness of the edge stiffener as follows (Eq. (11.19)): λd =

fyb = σcr,d

χd = 1,0 χd = 1,47 − 0,723λd χd = 0,66/λd

390 = 1,066 343 if λd ≤ 0,65 if 0,65 < λd < 1,38 if λd ≥ 1,38

0,65 < λd < 1,38 so χd = 1,47 − 0,723λd = 0,699 Step 3 As the reduction factor for the buckling of the stiffener is χd = 0,699 < 1, iterations are required to refine the value of the reduction factor. The iterations are carried out based on the reduced design strength, σcom,Ed,i = χd,i−1 fyb /γM0 to obtain new effective widths of the lip and flange in the stiffener and recalculate the critical stress of distortional buckling of the stiffener and thus to obtain new reduction factor. The iteration stops when the reduction factor χd converges. The final values obtained after iterations are be2 = 36,1, ceff = 19,0 and χd = 0,689. • Effective width of the web The position of the initial neutral axis (web is assumed as fully effective) with regard to the flange in compression is given by h2p c2 χ d cp cp hp − 2 + bp2 hp + 2 + eff2 hc = = 106 mm cp + bp2 + hp + be1 + (be2 + ceff )χd The stress ratio thus is: ψ=−

hp − hc 200 − 106 =− = −0,89 hc 106

The corresponding buckling coefficient is calculated by (Eq. (11.13a)) kσ = 7,81 − 6,29ψ + 9,78ψ2 = 21,2

450

•

Chapter 11 / Cold-formed Steel Sections The relative slenderness thus is: fyb 390 λh,red = λh = = = 0,986 σcr,h 1051 where σcr,h =

kσ π2 E 3,142 × 210000 21,2 = = 402 N/mm2 12(1 − ν2 ) (hp /t)2 12(1 − 0,32 ) (200/2)2

The width reduction factor thus is (Eq. (11.14a)) ρ=

λh,red − 0,055(3 + ψ) 2 λh,red

=

0,986 − 0,055(3 − 0,89) = 0,895 0,9862

The effective width of the zone in compression of the web is: heff = ρhc = 0,895 × 106 = 94,9 mm Part of the effective width near the flange is (see Table 11.1): he1 = 0,4heff = 0,4 × 94,9 = 37,9 mm Part of the effective width near the neutral axis is: he2 = 0,6heff = 0,6 × 94,9 = 56,8 mm Thus, h1 = he1 = 37,9 mm h2 = (hp − hc ) + he2 = (200 − 106) + 56,8 = 151 mm

The effective widths of the web obtained above are based on the position of the initial neutral axis (web is assumed as fully effective). To refine the result iterations are required which is based on the newly obtained effective widths, he1 and he2 , to determine the new position of the neutral axis. The stress ratio, buckling coefficient, relative slenderness, width reduction factor and effective widths of the web thus are recalculated according to the new position of the neutral axis. Iteration continues until it converges. The final values obtained after iterations are he1 = 37,9 mm, he2 = 56,8 mm and h2 = 149 mm (see Fig. 11.22). • Effective properties of the section (see Fig. 11.23) Effective cross-section area: Aeff = t[cp + bp2 + h2 + h1 + be1 + (be2 + ceff )χd ] = 698 mm2

Structural Design of Steelwork to EN 1993 and EN 1994

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451

bp1 be2

hp

he2

hc

he1

ceff

be1

n.a.

cp

t

FIGURE 11.22 Symbols used for representing the dimensions of the effective cross-section

bp2

bp1

hp

he2

zc

tχd

ceff

be2

he1

h1

be1

t

zt

h2

n.a

cp

FIGURE 11.23 Symbols used for representing the properties of the effective cross-section

bp2

Position of the neutral axis with regard to the flange in compression: h2 c2 χd cp h t cp hp − 2 + bp2 hp + h2 hp − 22 + 21 + eff2 zc = = 108 mm Aeff Position of the neutral axis with regard to the flange in tension: zt = hp − zc = 92 mm Second moment of the effective sectional area: Ieff,y =

t(h31 + h32 + c3p + χd c3eff )

t3 (bp2 + be1 + be2 χd3 ) 12 12 cp 2 h2 2 h1 2 2 + cp t zt − + bp2 tzt + h2 t zt − + h1 t zc − 2 2 2 2 c eff + be1 tzc2 + be2 (χd t)zc2 + ceff (χd t) zc − 2

= 4340000 mm4

+

452

•

Chapter 11 / Cold-formed Steel Sections Effective section modulus with regard to the flange in compression: Weff,y,c =

Ieff,y = 40100 mm3 zc

Effective section modulus with regard to the flange in tension: Weff,y,t =

Ieff,y = 47200 mm3 zt

The design value of the resistance of the section to bending moment about the y-axis due to local and distortional buckling is Mc,Rd =

fyb Weff,y,c 390 × 40100 × 10−6 = = 15,6 kNm γM0 1,0

The load table of a section is defined based on the specified type of connection and given span lengths. Assume that the section is used as a single span of 5 m long, both ends are butted to rafter beam using cleats, the section is subjected to gravity loading and is fully restrained laterally. The design value of the resistance to gravity load on the span due to local and distortional buckling is: Pc,Rd =

8Mc,Rd 8 × 15,6 = = 25,0 kN l 5

The design value of the plastic shear resistance is (Section 6.2.6 in EN 1993-1-1 (2005)): √ 200×2×390 √ Av fyb / 3 × 10−3 3 Vpl,Rd = = = 90,1 kN γM0 1,0 The design value of the shear buckling resistance is (Section 6.1.5 in EN 1993-1-3 (2006)): Vb,Rd =

hp tf sin φ bv

γM0

=

200 sin 90◦

×2×

0,48×390 1,49

1,0

× 10−3

= 50,3 kN

in which, fbv = 0,58fyb

if λw ≤ 0, 83

fbv = 0,48fyb /λw

if 0,83 < λw

λw = 0, 346

hp t

fyb 200 = 0,346 × × E 2

390 = 1,49 210000

So, the design shear resistance is Vc,Rd = min(Vpl,Rd , Vb,Rd ) = 50,3 kN Since the maximum shear force for the simply supported beam occurs at the support, the value of which is only half of the span load, it is obvious that for the present case

Structural Design of Steelwork to EN 1993 and EN 1994

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453

the design load is controlled by the design value of the resistance due to local and distortional buckling. Therefore the ultimate design load for the section is 25 kN. Note that, the load factors used for permanent actions and for variable actions are different. The design load for any combinations of dead and imposed loads thus should satisfy γG PG + γQ PQ ≤ Pc,Rd = 25 kN where γG = 1,35 and γQ = 1,5 are the load factors for dead and imposed loads, PG and PQ are the dead and imposed loads, respectively. The deflection check is usually done by using computer software which is normally provided by the manufacturer. The allowable deflection is determined based on individual cases and is not specified in most design standards.

11.9 CHAPTER CONCLUSIONS This chapter has described the mechanism of failures of the cold-formed steel sections and the corresponding methods for analyses and principles for design. The focuses have been placed on the analyses of local, distortional and lateral–torsional buckling of the sections for these are the main differences between the cold-formed steel section and the hot rolled steel section. For the cold-formed steel sections the most important properties are the effective section properties. For summarizing the calculation procedure a flowchart of calculation of effective section properties is given in Figs. 11.24 and 11.25. The design of a cold-formed steel section is much more complicated than that of a hot rolled section. This is partly because the section involves elements which have large width-to-thickness ratios and thus are easier to buckle locally and distortionally, and partly because the section is locally restrained by its supported trapetzoidal sheeting, which complicates the loading system and generates different stresses in restrained and free flanges. Simplified design methods may be used for channel, zed and sigma purlin systems that have no anti-sag bars, no use of sleeves and overlapping between two adjacent sections (see Annex E in EN 1993-1-3 (2006)). It is well known that for a short beam the design is normally controlled by the bending moment and/or shear force, while for a long beam it is usually controlled by the deflection and lateral–torsional buckling. Thus, purlins are usually designed to be continuous by using sleeves or overlapping in order to satisfy deflection limits and anti-sag bars to prevent twisting during erection and to stabilize the lower flange against wind uplift. For simply supported members, it is the sagging (positive) moment conditions that determine the capacity of the member. For continuous members over one or more internal supports, moments are to be determined elastically. Plastic hinge analysis is not permitted because the slender sections are not able to maintain their full

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•

Chapter 11 / Cold-formed Steel Sections

Start

Dimensions of cross-section; Material data

Determine geometrical proportions of cross-section

EN 1993-1-3 §5.1(3)

b/t 60 c/t 50 h/t 500 0,2 c/b 0,6 ?

The influence of rounding of the corners may neglected if. r/t 5; r/bp 0,1

No

Stop

EN 1993-1-3 §5.2 Yes The generai (iterative) procedure is applied to calculate the effective properties of the flanges and lips in compression. The calculation should be carried out in three steps.

EN 1993-1-3 §5.5.3.2

EN 1993-1-3 §5.5.2

Axial compression

Determine effective widths for the flanges and lips in compression

Determine the effective section properties of the web

Compute the gross cross-section properties

χd1, be12, ceff1, be11, χd2, be22, ceff2, be21

Major axis bending

EN 1993-1-3 §5.5.3.2

EN 1993-1-3 §5.5.2

he1, he2

EN 1993-1-5 §4.4

The general (iterative) procedure is applied to calculate the effective properties of the flange and lip in compression. The calculation should be carried out in three steps.

Determine effective widths for the flange and lip in comperssion

χd1,be2, ceff,be1

Determine the effective section properties of the web

he1, he2

Determine the properties of effective cross-section: leff, Weff

leff, Weff

EN 1993-1-5 §4.4

Determine effective area of the crosssection: Aeff

Stop

Aeff

Recalculate the position of the neutral axis taking into account the effective properties of compression flange and lip

Stop

FIGURE 11.24 The flowchart of calculation of effective section properties for cold-formed steel sections under compression or bending (copy from www.access-steel.com by permission) moment capacity when rotations exceed the point at which the section reaches yield. By utilizing the flexibility of sleeved or overlapping purlins at the supports, some elastic redistribution of moment may be achieved, and hence lead to more efficient design of the member. However, when the purlins are sleeved or overlapped, the design of the

For the initial evaluation, it is assumed that the flange in compression is double supported (K∞) and that the design strength is not reduced (σcom,Ed fyb / γM0)

Start

EN 1993-1-3 §5.5.3.2(3)

A

EN 1993-1-3 §5.5.3.2(5a)

Step 1 Determine the effective cross-section for the stiffener using effective width of flange

be1,be2,ceff

Determine the effective area of the edge stiffener: Ae

As

EN 1993-1-5 §4.4

EN 1993-1-3 §5.5.3.2(6)

Allow for the effects of continuous spring restraint

EN 1993-1-3 §5.5.3.2(3)

Step 2 Use the intial effective cross-section of the stiffener to determine the reduction factor

Where: K is the spring stiffness per unit length and Is is the effective second moment of area of the stiffener

EN 1993-1-3 §5.5.3.2(7) Determine the elastic buckling stress for the edge stiffener.

σσ,s

Determine the reduction factor for the distortional buckling resistance of the edge stiffener

χd

EN 1993-1-3 §5.5.3.1(5)

EN 1993-1-3 §5.5.3.1(7)

EN 1993-1-5 §4.4(2)

Xd < 1 ?

No

Return

Yes EN 1993-1-3 §5.5.3.2 (3), (10), (12)

Step 3 iterate to refine the value of the reduction factor for buckling of the stiffener.

Return

To A

The iterations are carried out based on modified values of ρ, obtained using: σcom,Edfyb / γM0 The iteration stops when the reduction factor χ converges.

FIGURE 11.25 The flowchart of iterative procedure for calculation of reduction factor due to distortion buckling (copy from www.access-steel.com by permission)

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•

Chapter 11 / Cold-formed Steel Sections system is normally developed based on testing or the combination of analysis and testing to achieve economic solutions. Standardized testing and evaluation procedures have been given in EN 1993-1-3 (2006) (Annex A) for both the cold-formed members and sheeting.

REFERENCES AS/NZS 4600 (1996). Australia/New Zealand Standard for Cold-formed Steel Structures. Standards Australia, Sydney. Bulson, P.S. (1970). The stability of flat plates. Chatto & Windus, London. Cheung, Y.K. (1976). Finite strip methods in structural analysis. Pergramon Press, New York. Chu, X.T. (2004). Failure Analysis of Cold-formed Steel Sections. Ph.D. Thesis, Aston University, Birmingham. UK. Chu, X.T., Kettle, R. and Li, L.Y. (2004a). Lateral–torsion buckling analysis of partial-laterally restrained thin-walled channel-sections beams, Journal of Constructional Steel Research, 60(8), 1159–1175. Chu, X.T., Li, L.Y. and Kettle, R. (2004b). The effect of warping stress on the lateral torsion buckling of cold-formed zed-purlins, Journal of Applied Mechanics, 71(5), 742–744. Chu, X.T., Rickard, J. and Li, L.Y. (2005a). Influence of lateral restraint on lateral–torsional buckling of cold-formed steel purlins. Thin-Walled Structures, 43, 800–810. Chu, X.T., Ye, Z.M, Kettle, R. and Li, L.Y. (2005b). Buckling behaviour of cold-formed channel sections under uniformly distributed loads. Thin-Walled Structures, 43, 531–542. Chu, X.T., Ye, Z.M, Li, L.Y. and Kettle, R. (2006). Local and distortional buckling of cold-formed zedsection beams under uniformly distributed transverse loads, International Journal of Mechanical Sciences, 48(8), 378–388. Davies, J.M., Leach, P. and Heinz, D. (1993). Second-order generalized beam theory, Journal of Constructional Steel Research, 31, 221–241. EN 1993-1-1 (2005). Eurocode 3: Design of steel structures – Part 1–1: General rules and rules for buildings. BSI. EN 1993-1-3 (2006). Eurocode 3: Design of steel structures – Part 1–3: General – Cold formed thin gauge members and sheeting. BSI. EN 1993-1-5 (2006). Eurocode 3: Design of steel structures – Part 1–5: General – Strength and stability of planar plated structures without transverse loading. BSI. Hancock, G.J. (1997). Design for distortional buckling of flexural members, Thin-Walled Structures, 27(1), 3–12. Kesti, J. and Davies, J.M. (1999). Local and distortional buckling of thin-walled short columns, Thin-Walled Structures, 34, 115–134. Lau, S.C.W. and Hancock, G.J. (1986). Buckling of thin flat-walled structures by a spline finite strip method, Thin-Walled Structures, 4, 269–294. Lau, S.C.W. and Hancock, G.J. (1987). Distortional buckling formulas for channel columns, Journal of Structural Engineering, ASCE, 113(5), 1063–1078. Lau, S.C.W. and Hancock, G.J. (1989). Inelastic buckling analyses of beams, columns and plates using the spline finite strip method, Thin-Walled Structures, 7, 213–238. Li, L.Y. (2004). Lateral–torsional buckling of cold-formed zed-purlins partial-laterally restrained by metal sheeting, Thin-Walled Structures, 42(7), 995–1011. Loughlan, J. (1993). Thin-walled cold-formed sections subjected to compressive loading, Thin-Walled Structures, 16(1–4), 65–109. Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997a). Modelling of cold-formed purlinsheeting systems – Part 1. Full model, Thin-Walled Structures, 27(3), 223–243.

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Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997b). Modelling of cold-formed purlinsheeting systems – Part 2. Simplified model, Thin-Walled Structures, 27(4), 263–286. Oden, J.T. (1967). Mechanics of elastic structures. McGraw-Hill Book Company, New York. Rhodes, J. and Lawson, R.M. (1992). Design of structures using cold-formed steel sections. SCI Publication, 089. SCI. Schafer, B.W. (1997). Cold-formed Steel Behaviour and Design: Analytical and Numerical Modelling of Elements and Members with Longitudinal Stiffeners. Ph.D. Thesis. Cornell University. Timoshenko, S.P. and Gere, J.M. (1961). Theory of elastic stability. McGraw-Hill Book Company, New York. Vlasov, V.Z. (1961). Thin-walled elastic beams. Israel Program for Scientific Translations, Jerusalem, Israel. Von Karman, T., Sechler, E.E. and Donnell, L.H. (1932). The strength design of thin plates in compression, Transactions ASME, 54, 53–55. Vrany, T. (2002). Torsional restraint of cold-formed beams provided by corrugated sheeting for arbitrary input variables. Proceedings of Eurosteel. Coimbra, cmm, 2002, 734–742. Walker, A.C. (1975). Design and analysis of cold-formed sections. International Textbook Company Ltd, London. Winter, G. (1968). Thin walled structures – theoretical solutions and test results, Preliminary Publications 8th Congress IABSE, 101–112. Ye, Z.M., Kettle, R., Li, L.Y. and Schafer, B. (2002). Buckling behaviour of cold-formed zed-purlins partially restrained by steel sheeting, Thin-Walled Structures, 40, 853–864. Ye, Z.M., Kettle, R. and Li, L.Y. (2004). Analysis of cold-formed zed-purlins partially restrained by steel sheeting, Computer and Structures, 82, 731–739. Zienkiewicz, O.C. and Taylor, R.L. (2000). The finite element method, Butterworth Heinemann, Oxford.

Annex

A1 /

Design Strengths for Fillet Welds

Design strength per unit length (Fw,Rd ) in KN/mm

Leg length (mm)

Steel grade S275 fu = 430 MPa βw = 0,85

Steel grade S355 fu = 510 MPa βw = 0,90

4 5 6 8 10 12 15 18 20 22 25

0,654 0,818 0,981 1,308 1,636 1,963 2,453 2,944 3,271 3,598 4,089

0,733 0,916 1,099 1,466 1,832 2,199 2,748 3,298 3,664 4,031 4,580

Notes: (1) F w,Rd = f u /(31/2 βw γM2 ) × 0,7 × (leg length) for equal leg lengths and γM2 = 1,25. (2) Steel to Standard 10025-2 with appropriate electrodes (BSEN 499).

•

Annex

A2 / Design Strengths for Class 4.6 Ordinary Bolts

Bolt diameter d (mm)

Reduced area of bolt As (mm2 )

Tension stress using reduced area Ft, Rd (kN)

Single shear stress using reduced area Fv,Rd (kN)

Single shear stress using gross area Fv,Rd (kN)

(M12) M16 M20 (M22) M24 (M27) M30 (M33) M36

84,3 157 245 303 353 459 561 694 817

24,3 45,2 70,6 87,3 101,7 132,2 161,6 199,9 235,3

16,2 30,1 47,0 58,2 67,8 88,1 107,7 133,2 156,9

21,7 38,6 60,3 73,0 86,9 109,9 135,7 164,2 195,4

Notes: (1) F v,Rd = 0,6 (A or As ) f ub /γM2 (2) F t,Rd = 0,9 As f ub /γM2 (3) f ub = 400 MPa; γM2 = 1,25 (4) Bolt sizes in brackets are not preferred. (5) Shear values for other classes of bolt are multiplied by a factor Class 4.6 4.8 5.6 5.8 6.8 8.8 10.9 Factor 1,0 0,83 1,25 1,04 1,25 2,0 2,08 (6) Tension values for other classes of bolt are multiplied by a factor Class 4.6 4.8 5.6 5.8 6.8 8.8 10.9 Factor 1,0 1,0 1,25 1,25 1,5 2,0 2,5

459

• 460 Annex

A3 /

Design Strengths for Preloaded Class 8.8 and 10.9 Bolts

Bolt size d (mm)

Reduced area of bolt As (mm2 )

Preload force class 8.8 F p,C (kN)

Preload force class 10.9 F p,C (kN)

Slip resistance class 8.8 F s,Rd (kN)

Slip resistance class 10.9 F s,Rd (kN)

(M12) M16 M20 M22 M24 M27 M30 M36

84,3 157 245 303 353 459 561 817

47,2 87,9 137,2 169,7 197,7 257,0 314,2 457,5

59,1 109,9 171,5 212,1 247,1 321,3 392,7 571,9

18,9 35,2 54,9 67,9 79,1 102,8 125,7 183,0

23,6 44,0 68,6 84,8 98,8 128,5 157,1 228,8

Notes: (1) Preload F p,C = 0,7 f ub As where f ub = 800 (Class 8.8) and 1000 (Class 10.9) MPa. (2) Design slip resistance F s,Rd = ks nμF p,C /γM3 where ks = 1, n = 1, μ = 0,5 and γM3 = 1,25 (3) Bolt sizes in brackets are not preferred.

•

Annex

A4 / Space of Holes in Beams, Columns, Joists and Tees 461

Beams, columns, joists and tees Nominal flange widths (mm)

Spacings in millimetres S4

Recommended diameter of bolt (mm)

Actual bmin (mm)

S1

S2

S3

419–368 330–305 330–305 292–203 190–165 152 146–114 102 89 76 64 51

140 140 140 140 90 90 70 54 50 40 34 30

140 120 120 – – – – – – – – –

75 60 60 – – – – – – – – –

290 240 240 – – – – – – – – –

24 24 20 24 24 20 20 12 – – – –

362 312 300 212 162 150 130 98 – – – –

S1

S1 S1

S3 S2 S3 S4

462

•

Annex

A5 / Spacing of Holes in Angles

Angles Spacing of holes

Maximum diameter of bolt

Nominal leg length S 1 S2 S3 S4 S5 S6 S1 S 2 and S 3 (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm)

S 4 ,S 5 and S 6 (mm)

200 150 125 120 100 90 80 75 70 65 60 50 45 40 30 25

20 – – – – – – – – – – – – – – –

– – – – 55 50 45 45 40 35 35 28 25 23 20 15

75 55 45 45 – – – – – – – – – – – –

75 55 50 50 – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

55 – – – – – – – – – – – – – – –

– – – – 24 24 20 20 20 20 16 12 – – – –

30 20 20 20 – – – – – – – – – – – –

Leg

S3 S2 S1 Leg

Leg

S6 S5 S4

Annex

A6 / Spacing of Holes in Channels

•

463

Channels Nominal flange width (mm)

S1 (mm)

Recommended diameter of bolt (mm)

102 89 76 64 51 38

55 55 45 35 30 22

24 20 20 16 10 –

S1

464

•

A7 /

Annex 7 / From BS5950: 1: 1990 by Permission

Annex

From BS5950: 1: 1990 by Permission

TABLE 15 Slenderness correction factor, n, for members with applied loading substantially concentrated within the middle fifth of the unrestrained length. Note 1. All hogging moments are +ve. Note 2. β is defined in Table 18. Note 3. M0 is the mid-length moment on a simply supported span equal to the unrestrained length (see Table 17).

L 10 maximum M0

M

BM

Unrestrained length L

β positive

β negative

γ = M/M 0

1.0

0.8

0.6

0.4

0.2

0.0

−0.2

−0.4

−0.6

−0.8

−1.0

+50.00 +10.00 +5.00 +2.00 +1.50 +1.00 +0.50 0.00 −0.10 −0.20 −0.30 −0.40 −0.50 −0.60 −0.70 −0.80 −0.90 −1.00

1.00 0.99 0.98 0.96 0.95 0.93 0.90 0.86 0.85 0.83 0.81 0.79 0.77 0.62 0.56 0.56 0.59 0.62

0.96 0.99 0.98 0.95 0.95 0.92 0.90 0.86 0.85 0.83 0.82 0.80 0.78 0.66 0.56 0.53 0.57 0.58

0.92 0.94 0.97 0.95 0.94 0.92 0.90 0.86 0.85 0.83 0.82 0.81 0.79 0.72 0.61 0.54 0.54 0.54

0.87 0.90 0.93 0.95 0.94 0.92 0.89 0.86 0.85 0.84 0.83 0.81 0.80 0.77 0.67 0.59 0.53 0.52

0.82 0.85 0.89 0.94 0.93 0.92 0.89 0.86 0.85 0.84 0.83 0.82 0.82 0.80 0.73 0.65 0.57 0.54

0.77 0.80 0.84 0.94 0.93 0.91 0.89 0.86 0.86 0.85 0.84 0.83 0.83 0.82 0.79 0.71 0.64 0.59

0.72 0.75 0.79 0.89 0.92 0.91 0.89 0.86 0.86 0.85 0.85 0.84 0.85 0.84 0.83 0.77 0.71 0.66

0.67 0.69 0.73 0.84 0.90 0.91 0.89 0.86 0.86 0.85 0.85 0.85 0.86 0.85 0.85 0.83 0.77 0.72

0.66 0.68 0.71 0.79 0.85 0.91 0.88 0.86 0.86 0.86 0.86 0.85 0.86 0.86 0.87 0.89 0.84 0.80

0.66 0.68 0.70 0.77 0.80 0.92 0.88 0.86 0.86 0.86 0.86 0.86 0.87 0.87 0.88 0.90 0.88 0.85

0.65 0.67 0.70 0.76 0.80 0.92 0.88 0.86 0.86 0.86 0.87 0.87 0.88 0.88 0.89 0.90 0.91 0.92

(Continued)

Structural Design of Steelwork to EN 1993 and EN 1994

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465

TABLE 15 Continued −1.10 −1.20 −1.30 −1.40 −1.50 −1.60 −1.70 −1.80 −1.90 −2.00 −5.00 −50.00 Infinity

0.66 0.70 0.73 0.74 0.75 0.76 0.77 0.79 0.80 0.81 0.93 0.99 1.00

0.62 0.66 0.69 0.70 0.70 0.72 0.74 0.77 0.79 0.81 0.89 0.95 0.96

0.57 0.60 0.63 0.64 0.64 0.65 0.66 0.68 0.69 0.70 0.83 0.90 0.91

0.54 0.55 0.57 0.58 0.59 0.60 0.61 0.63 0.64 0.65 0.77 0.86 0.86

0.54 0.54 0.55 0.56 0.56 0.57 0.58 0.59 0.60 0.61 0.72 0.79 0.82

0.57 0.55 0.54 0.54 0.54 0.55 0.56 0.56 0.57 0.58 0.67 0.74 0.77

0.63 0.60 0.57 0.55 0.55 0.55 0.55 0.56 0.56 0.56 0.64 0.70 0.72

0.68 0.65 0.61 0.60 0.59 0.58 0.58 0.57 0.57 0.56 0.61 0.67 0.68

0.76 0.73 0.69 0.66 0.65 0.64 0.63 0.62 0.61 0.60 0.60 0.64 0.65

0.83 0.80 0.77 0.74 0.73 0.72 0.70 0.69 0.67 0.66 0.62 0.63 0.65

0.89 0.87 0.83 0.81 0.80 0.80 0.78 0.76 0.75 0.74 0.65 0.65 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section. Note 5. Values for intermediate values of β and γ may be interpolated.

TABLE 16 Slenderness correction factor, n, for members with applied loading other than as for Table 15.

M

BM

M0

Note 1. All hogging moments are +ve. Note 2. β is defined in Table 18. Note 3. M0 is the mid-length moment on a simply supported span equal to the unrestrained length (see Table 17).

Unrestrained length L

β positive

β negative

γ = M/M 0

1.0

0.8

0.6

0.4

0.2

0.0

−0.2 −0.4 −0.6 −0.8 −1.0

+50.00 +10.00 +5.00 +2.00 +1.50 +1.00 +0.50 0.00

1.00 0.99 0.99 0.98 0.97 0.97 0.96 0.94

0.96 0.98 0.98 0.98 0.97 0.97 0.96 0.94

0.92 0.95 0.97 0.97 0.97 0.97 0.96 0.94

0.87 0.91 0.94 0.96 0.96 0.96 0.96 0.94

0.83 0.86 0.90 0.94 0.95 0.96 0.96 0.94

0.77 0.81 0.85 0.92 0.93 0.95 0.95 0.94

0.72 0.76 0.80 0.90 0.92 0.94 0.94 0.94

0.67 0.70 0.75 0.86 0.89 0.93 0.94 0.94

0.66 0.68 0.71 0.82 0.86 0.93 0.94 0.94

0.66 0.68 0.70 0.78 0.83 0.91 0.93 0.94

0.65 0.67 0.70 0.76 0.79 0.89 0.92 0.94

(Continued)

466

•

Annex 7 / From BS5950: 1: 1990 by Permission

TABLE 16 Continued −0.10 −0.20 −0.30 −0.40 −0.50 −0.60 −0.70 −0.80 −0.90 −1.00 −1.10 −1.20 −1.30 −1.40 −1.50 −1.60 −1.70 −1.80 −1.90 −2.00 −5.00 −50.00 Infinity

0.93 0.92 0.91 0.90 0.89 0.71 0.57 0.47 0.47 0.50 0.54 0.57 0.61 0.64 0.67 0.69 0.71 0.74 0.76 0.78 0.91 0.99 1.00

0.93 0.92 0.91 0.90 0.90 0.77 0.64 0.52 0.46 0.48 0.51 0.54 0.56 0.59 0.62 0.64 0.66 0.69 0.71 0.73 0.86 0.95 0.96

0.93 0.92 0.92 0.91 0.91 0.84 0.70 0.59 0.50 0.46 0.48 0.50 0.52 0.55 0.57 0.59 0.60 0.62 0.63 0.65 0.80 0.89 0.91

0.93 0.92 0.92 0.91 0.91 0.87 0.77 0.67 0.58 0.51 0.49 0.47 0.47 0.49 0.51 0.52 0.54 0.55 0.57 0.58 0.74 0.84 0.86

0.94 0.93 0.93 0.92 0.92 0.89 0.82 0.73 0.65 0.58 0.54 0.51 0.49 0.48 0.47 0.48 0.50 0.51 0.53 0.54 0.70 0.79 0.82

0.94 0.93 0.93 0.92 0.92 0.91 0.87 0.80 0.73 0.66 0.61 0.56 0.53 0.51 0.49 0.50 0.51 0.51 0.52 0.53 0.65 0.74 0.77

0.94 0.93 0.93 0.92 0.92 0.92 0.89 0.86 0.80 0.73 0.69 0.64 0.61 0.58 0.56 0.55 0.55 0.54 0.54 0.53 0.62 0.70 0.72

0.94 0.93 0.93 0.92 0.92 0.92 0.91 0.90 0.87 0.81 0.77 0.73 0.70 0.67 0.64 0.63 0.61 0.60 0.58 0.57 0.59 0.66 0.68

0.94 0.94 0.94 0.93 0.92 0.92 0.92 0.92 0.90 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.68 0.66 0.65 0.63 0.58 0.63 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section. Note 5. Values for intermediate values of β and γ may be interpolated.

0.94 0.94 0.94 0.93 0.92 0.92 0.92 0.92 0.90 0.89 0.87 0.84 0.82 0.79 0.77 0.76 0.74 0.73 0.71 0.70 0.61 0.62 0.65

0.94 0.93 0.94 0.93 0.92 0.92 0.91 0.92 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 0.82 0.81 0.80 0.79 0.67 0.65 0.65

Structural Design of Steelwork to EN 1993 and EN 1994 TABLE 17 Moment diagram between adjacent points of lateral restraint. M0

M

BM b ve

g ve M0

M

BM b ve

g ve M0

M

BM

b ve

g ve M0

M BM bve

gve

M0 M BM bve

gve

M0 M BM bve

gve

•

467

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Index Actions accidental, 29, 287, 294 characteristic, 27 classification, 27 combinations, 30, 287 design values, 29 destabilizing, 98, 104 envelopes, 32 erection, 29, 301, 359 fundamental, 30 gravity, 286 pattern loading, 31, 103 permanent, 28 progressive collapse, 295 seismic, 30, 287 snow, 28, 286 stabilising, 104 thermal, 29 transmission, 298 variable, 28 wind, 29, 286 analysis of frames elastic, 309 pitched roof portal, 314 plastic, 309 angles and tees as beams, 42 as tension members, 175 Beams deflection, 49 elastic analysis, 38 laterally restrained, 35, 37 principal axes, 39 section axes, 39 shear stress, 54 span/depth ratios, 53 symmetrical, 41 transverse forces, 78 types, 36 unsymmetrical, 42 bolts bearing strength, 213 clearance, 209 close tolerance, 208 design tables ordinary, Annex A2 design tables preloaded, Annex A3 edge and end distances, 210 foundation, 208 holes, 209

large grip lengths, 216 long joints, 216 net area, 212 ordinary bolts, packings, 216 preferred sizes, 207 prying forces, 218 shear strength and deformation, 214 shear strength and tension, 213 sizes, 213 slip resistant, 217 spacing, 210 tension and shear, 213 types, 207 washers, 209 bracing, 288, 300 brittle fracture, 22 buckling flange, 424 gusset plates, 223 lateral torsional buckling, 35, 91, 430 lip, 424 local, 35, 419 Perry-robertson, 183 struts, 180 theory, 181 webs, 138 characteristic load, 27 characteristic strength, 18 Charpy V-notch test, 22 classification of sections, 4, 371 cold formed sections analytical model, 414 deflections, 434 distortional buckling, 424 effective width, 422 effective section properties, 454 elastic local buckling stress, 420 finite strip methods, 435 lateral torsional buckling, 430 local buckling, 419 post buckling behaviour, 422 columns, 181, 300, 312, 325 composite construction composite beams deflection, 382 elastic capacity, 376 flexural design, 371 flexural shear, 377

470

•

Index composite beams (continued) longitudinal shear, 379 partial shear connection, 376, 378 shear connector design, 377 transverse shear, 380 vibration, 382 composite columns axial compression, 396 biaxial bending, 397 buckling, 401 design moments, 401 determination of member capacity, 398 fire, 402 uniaxial bending, 396 composite slabs deflection, 360, 362 flexural shear, 361, 362 flexure, 360, 361 longitudinal shear, 362, 379 punching shear, 362 shear bond, 360 transverse shear, 380 compression members axially loaded, 175 buckling length, 186, 312 buckling theory, 181 compression members, 180 effective length, 185 local buckling, 185 slenderness ratio, 185 with moments, 190, 401 connections (see joints) corrosion of steel, 20 Dead loads (see actions) deflections limits, 53 beams, 49 composite slabs, 360 portal frame, 325 span/depth ratios, 53 deformations, 287 design envelope, 19 methods for frames, 198 rigid, 294, 39, 311 sem-rigid, 294, 309, 310 simple, 309, 310 strength, 19 ductility, 20 durability, 20 drawings, 12 Effective area of an angle tie, 175 effective length beam, 97 cantilever, 100, 105 column, 185, 312 errors in design, 11 expansion and contraction joints, 304

Fabrication of steelwork, 12 failure criteria bolts, 213 steel, 24 welds, 203 fatigue, 23 fire single storey structures, 302 multi-storey structures, 302 composite columns, 402 flooring systems, 284 frames classification, 306 imperfections, 308 single storey, 282, 288 multi-storey, 283, 293 forces bracing, 290, 300 erection, 301 seismic, 287 tying, 296 foundation design 303 frame types, 2 Gable end frame, 336 gusset plates, 223, 250, 265 Hardness, 20, 317 haunches, 314, 317, 332 holes in members, Annexes A4, A5, A6 hollow sections, 234, 269 lateral torsional buckling, 123 hydrogen cracks, 205 Intermediate vertical stiffener, 145 Joints beam splice, 233, 257 beam-to-beam, 243, 256 beam-to-column, 232, 242, 252 column bracket, 231, 237, 247 column-to-column, 233, 260 column-to-foundation, 234, 245, 263 construction movement, 304 deformation, 205 eccentric shear, 225 effective length, 202 end bearing, 226 effective length, 202 expansion and contraction, 304 fillet weld, 203 ideal, 199 knee for portal frame, 235, 271 long 202 ‘pinned’, 228 RHS-to-RHS, 234, 269 ‘rigid’, 230 roof truss, 235 rotational stiffness, 275 washers, 209 welded, 199

Index Lateral restraint, 77 torsional buckling, 36, 91, 123, 125 limit state design, 9 lintels, 299 loads (see actions) Manufacture of steel, 5 material variation, 17 multi-storey frames, 283, 293 Overhanging beams, 107 Partial safety factors, 18 plastic analysis effect of axial load, 177 effect of shear force, 73 hinges, 37 lateral restraint, 77 methods of analysis, 71 reduction of modulus due to shear, 73, 376 section properties, 67 theorems, 72 plate girder fabrication, 132 intermediate vertical stiffeners, 145 lateral torsional buckling, 134 load bearing stiffeners, 142 minimum web thickness, 133 minimum weight, 134 web buckling, 138 welding, 132, 146 plate thickness elastic method, 222 plastic method, 218 standard thicknesses, 3 Poisson’s ratio, 19 portal frame single bay, 314 multi-bay, 337 principal axes, 38 progressive collapse, 295 prying force, 218 Rafter stability, 332 residual stresses, 22, 94 restraints, 99, 300, 325, 332 Section axes, 39 cold formed, 413 classification, 35 elastic properties, 38 modulus, 41 second moment of area, 40 types, 3 unsymmetrical, 42 plastic properties, 67 seismic actions, 287 settlement, 304

shear centre, 57 modulus, 19 closed sections, 54 lag, 57 open sections, 56 plate grider, 138 stress, 54 sign conventions, 39 single storey frames, 282, 288 slenderness ratio, 185 splices column, 300 plate, 146 stability of structures, 296, 305 steel cold formed, 3 first use, 5 manufacture, 5 production, 6 standard sections, 3 stock sizes, 3 types of, 3 stiffeners load bearing, 142 plate girder, 142 transverse, 144 strength characteristic, 18 design, 19 stress concentrations, 24 residual, 22 strain relationship, 38 yield, 38 structural design, 7 structural framing, 10 T-stubs, 218 torsional buckling interaction, 131 tension members angles, 175 axially loaded, 175 with moments, 177 net area, 210 effective area, 176 thermal coefficient of expansion, 19 testing of steel, 14 tolerances, 12 torsional buckling, 127 torsional restraint, 62, 97, 100 toughness, 22 transverse forces, 78 trusses non-triangulated trusses, 347 triangulated, 341 Vibration, 382 vierendeel girder, 348

•

471

472

•

Index Washers, 209 web buckling of plate girders, 138 welds brittle fracture, 206 butt welds, 201 defects, 205 design table, Annex A1 effective length, 202 electrodes, 199 factors affecting strength, 205 failure criteria, 204 hydrogen cracks, 206

lamellar tearing, 206 load-deformation, 205 stress concentrations, 206 throat thickness, 202 types, 201 wind bracing, 288 girder, 291 loading, 286 Yield stress, 18 Young’s modulus, 19

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