Structural Steelwork: Design to Limit State Theory Third edition Dennis Lam School of Civil Engineering The University of Leeds, Leeds, UK Thien-Cheong Ang School of Civil and Environmental Engineering Nanyang Technological University, Singapore Sing-Ping Chiew School of Civil and Environmental Engineering Nanyang Technological University, Singapore AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DEIGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Elsevier Butterworth-Heinemann Linacre House, Jordan Hill, Oxford OX2 8DP 200 Wheeler Road, Burlington, MA 01803 First published 1987 Reprinted 1988 (with corrections), 1990, 1991 Second edition 1992 Reprinted 1993 (twice), 1994, 1995, 1997, 1998, 1999, 2001, 2002 Third edition 2004 Copyright © 1987, 1992, 2004, Elsevier Ltd. 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You may also complete your request on-line via the Elsevier homepage (http://www.elsevier.com), by selecting ‘Customer Support’ and then ‘Obtaining Permissions’. Cover Image: Swiss Re Building – 30 St Mary Axe – The Gherkin. Photo with kind permission from Grant Smith Photographer British Library Cataloguing in Publication Data Lam, Dennis Structural steelwork : design to limit state theory. – 3rd ed. 1. Steel, Structural 2. Building, Iron and steel I. Title II. Ang, Paul III. Chiew, Sing-Ping 624.1′ 821 ISBN 0 7506 59122 Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 0 7506 59122 For information on all Butterworth-Heinemann publications visit our website at www.bh.com Typeset by Newgen Imaging Systems (P) Ltd, Chennai, India Printed and bound in Great Britain Contents Preface to the third edition Preface to the second edition Preface to the first edition vi vii viii Chapter 1 INTRODUCTION 1.1 1.2 1.3 1.4 1.5 1.6 1 Steel structures 1 Structural elements 1 Structural design 4 Design methods 4 Design calculations and computing Detailing 7 Chapter 2 LIMIT STATE DESIGN 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 3.1 3.2 3.3 3.4 12 15 Structural steel properties 15 Design considerations 15 Steel sections 18 Section properties 21 Chapter 4 BEAMS 4.1 4.2 4.3 4.4 4.5 8 Limit state design principles 8 Limit states for steel design 8 Working and factored loads 9 Stability limit states 11 Structural integrity 11 Serviceability limit state deflection Design strength of materials 13 Design methods for buildings 14 Chapter 3 MATERIALS 6 24 Types and uses 24 Beam loads 25 Classification of beam cross-sections Bending stresses and moment capacity Lateral torsional buckling 34 26 28 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 Shear in beams 40 Deflection of beams 42 Beam connections 42 Examples of beam design Compound beams 56 Crane beams 62 Purlins 77 Sheeting rails 85 Chapter 5 PLATE GIRDERS 5.1 5.2 5.3 5.4 5.5 94 Design considerations 94 Behaviour of a plate girder 97 Design to BS 5950: Part 1 101 Design of a plate girder 112 Design utilizing tension field action Chapter 6 TENSION MEMBERS 6.1 6.2 6.3 6.4 6.5 47 131 Uses, types and design considerations 131 End connections 132 Structural behaviour of tension members 134 Design of tension members 139 Design examples 141 Chapter 7 COMPRESSION MEMBERS 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 144 Types and uses 144 Loads on compression members 146 Classification of cross-sections 148 Axially loaded compression members 148 Beam columns 165 Eccentrically loaded columns in buildings 173 Cased columns subjected to axial load and moment 182 Side column for a single-storey industrial building 184 Crane columns 194 Column bases 204 Chapter 8 TRUSSES AND BRACING 8.1 8.2 8.3 8.4 8.5 8.6 8.7 120 210 Trusses—types, uses and truss members 210 Loads on trusses 210 Analysis of trusses 212 Design of truss members 214 Truss connections 218 Design of a roof truss for an industrial building Bracing 233 220 Chapter 9 PORTAL FRAMES 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 Design and construction 246 Elastic design 248 Plastic design 259 In-plane stability 264 Restraints and member stability 267 Serviceability check for eaves deflection Design of joints 271 Design example of a portal frame 274 Further reading for portal design 282 Chapter 10 CONNECTIONS 10.1 10.2 10.3 10.4 10.5 246 270 284 Types of connections 284 Non-preloaded bolts 284 Preloaded bolts 301 Welded connections 306 Further considerations in design of connections 318 Chapter 11 WORKSHOP STEELWORK DESIGN EXAMPLE 11.1 11.2 11.3 11.4 11.5 11.6 Introduction 325 Basic design loads 325 Computer analysis data 327 Results of computer analysis 330 Structural design of members 334 Steelwork detailing 337 Chapter 12 STEELWORK DETAILING 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 338 Drawings 338 General recommendations 339 Steel sections 340 Grids and marking plans 341 Bolts 343 Welds 344 Beams 347 Plate girders 347 Columns and bases 348 Trusses and lattice girders 349 Computer-aided drafting 350 References 352 Index 353 325 Preface to the third edition This is the third edition of the Structural Steelwork: Design to Limit State Theory by T.J. MacGinley and T.C. Ang, which proved to be very popular with both students and practising engineers. The change of authorship was forced upon by the deceased of Mr T.J. MacGinley. All the chapters have been updated and rearranged to comply with the latest revision of the BS 5950-1:2000 Structural use of steelwork in building - Part 1: Code of practice for design rolled and welded sections, it may be used as a stand-alone text or in conjunction with BS 5950. The book contains detailed explanation of the principles underlying steel design and is intended for students reading for civil and/or structural engineering degrees in universities. It should be useful to final year students involved in design projects and also sufficiently practical for practising engineers and architects who require an introduction to the latest revision of BS 5950. Every topic is illustrated with fully worked examples and problems are also provided for practice. D.L. Preface to the second edition The book has been updated to comply with BS 5950: Part I: 1990 Structural Use of Steelwork in Building Code of Practice for Design in Simple and Continuous Construction: Hot Rolled Sections A new chapter on portal design has been added to round out its contents. This type of structure is in constant demand for warehouses, factories and for many other purposes and is the most common single-storey building in use. The inclusion of this material introduces the reader to elastic and plastic rigid frame design, member stability problems and design of moment-transmitting joints. T.J.M. T.C.A. Preface to the first edition The purpose of this book is to show basic steel design to the new limit state code BS 5950. It has been written primarily for undergraduates who will now start learning steel design to the new code, and will also be of use to recent graduates and designers wishing to update their knowledge. The book covers design of elements and joints in steel construction to the simple design method; its scheme is the same as that used in the previous book by the principal author, Structural Steelwork Calculations and Detailing, Butterworths, 1973. Design theory with some of the background to the code procedures is given and separate elements and a complete building frame are designed to show the use of the code. The application of microcomputers in the design process is discussed and the listings for some programs are given. Recommendations for detailing are included with a mention of computer-aided drafting (CAD). T.J.M. T.C.A. 1 Introduction 1.1 Steel structures Steel frame buildings consist of a skeletal framework which carries all the loads to which the building is subjected. The sections through three common types of buildings are shown in Figure 1.1. These are: (1) single-storey lattice roof building; (2) single-storey portal frame building; (3) medium-rise braced multi-storey building. These three types cover many of the uses of steel frame buildings such as factories, warehouses, offices, flats, schools, etc. A design for the lattice roof building (Figure 1.1(a)) is given and the design of the elements for the braced multi-storey building (Figure 1.1(c)) is also included. Design of portal frame is described separately in Chapter 9. The building frame is made up of separate elements—the beams, columns, trusses and bracing—listed beside each section in Figure 1.1. These must be joined together and the building attached to the foundations. Elements are discussed more fully in Section 1.2. Buildings are three dimensional and only the sectional frame has been shown in Figure 1.1. These frames must be propped and braced laterally so that they remain in position and carry the loads without buckling out of the plane of the section. Structural framing plans are shown in Figures 1.2 and 1.3 for the building types illustrated in Figures 1.1(a) and 1.1(c). Various methods for analysis and design have been developed over the years. In Figure 1.1, the single-storey structure in (a) and the multi-storey building in (c) are designed by the simple design method, while the portal frame in (b) is designed by the continuous design method. All design is based on the newly revised limit state design code BS 5950-1: 2000: Part 1. Design theories are discussed briefly in Section 1.4 and design methods are set out in detail in Chapter 2. 1.2 Structural elements As mentioned above, steel buildings are composed of distinct elements: (1) Beams and girders—members carrying lateral loads in bending and shear; (2) Ties—members carrying axial loads in tension; 1 2 Introduction (a) Single-storey lattice roof building with crane 1 Elements 1 Lattice girder 2 Crane column 3 Crane girder 3 2 Fixed base (b) Single-storey rigid pinned base portal Elements 1 Portal rafter 2 Portal column 1 Haunched joint 2 Pinned base (c) Multi-storey building 3 Elements 1 Floor beam 2 Plate girder 3 Column 4 Bracing 1 3 2 4 Figure 1.1 Three common types of steel buildings (3) Struts, columns or stanchions—members carrying axial loads in compression. These members are often subjected to bending as well as compression; (4) Trusses and lattice girders—framed members carrying lateral loads. These are composed of struts and ties; (5) Purlins—beam members carrying roof sheeting; (6) Sheeting rails—beam members supporting wall cladding; (7) Bracing—diagonal struts and ties that, with columns and roof trusses, form vertical and horizontal trusses to resist wind loads and hence provided the stability of the building. Joints connect members together such as the joints in trusses, joints between floor beams and columns or other floor beams. Bases transmit the loads from the columns to the foundations. Structural elements 3 3 5 Roof plan 9 6 Lower chord bracing 8 3 4 7 3 Side elevation 2 1 10 4 Section Gable framing Building elements 1 2 3 4 5 Lattice girder Column Purlins and sheeting rails Crane girder Roof bracing 6 7 8 9 10 Lower chord bracing Wall bracing Eaves tie Ties Gable column Figure 1.2 Factory building The structural elements are listed in Figures 1.1–1.3, and the types of members making up the various elements are discussed in Chapter 3. Some details for a factory and a multi-storey building are shown in Figure 1.4. 4 Introduction 2 3 1 2 4 4 Front elevation End elevation 4 I I I I I I I I I I I Building elements I I 1 2 3 4 4 2 I I 2 I I 3 I I Column Floor beams Plate girder Bracing I Plan first floor level Figure 1.3 Multi-storey office building 1.3 Structural design Building design nowadays usually carried out by a multi-discipline design team. An architect draws up plans for a building to meet the client’s requirements. The structural engineer examines various alternative framing arrangements and may carry out preliminary designs to determine which is the most economical. This is termed the ‘conceptual design stage’. For a given framing arrangement, the problem in structural design consists of: (1) estimation of loading; (2) analysis of main frames, trusses or lattice girders, floor systems, bracing and connections to determine axial loads, shears and moments at critical points in all members; (3) design of the elements and connections using design data from step (2); (4) production of arrangement and detail drawings from the designer’s sketches. This book covers the design of elements first. Then, to show various elements in their true context in a building, the design for the basic single-storey structure with lattice roof shown in Figure 1.2 is given. 1.4 Design methods Steel design may be based on three design theories: (1) elastic design; (2) plastic design; (3) limit state design. Elastic design is the traditional method and is still commonly used in the United States. Steel is almost perfectly elastic up to the yield point and elastic theory is a very good method on which the method is based. Structures are analysed by elastic theory and sections are sized so that the permissible stresses are not Design methods 5 (a) Factory building C + + + + truss to column joint column base crane girder (b) Multi-storey building + + + + + + + + column base beams column joints Figure 1.4 Factory and multi-storey building exceeded. Design in accordance with BS 449-2: 1967: The Use of Structural Steel in Building is still acceptable in the United Kingdom. Plastic theory developed to take account of behaviour past the yield point is based on finding the load that causes the structure to collapse. Then the working load is the collapse load divided by a load factor. This too is permitted under BS 449. Finally, limit state design has been developed to take account of all conditions that can make the structure become unfit for use. The design is based on the actual behaviour of materials and structures in use and is in accordance 6 Introduction with BS 5950: The Structural Use of Steelwork in Building; Part 1—Code of Practice for Design—Rolled and Welded Sections. The code requirements relevant to the worked problems are noted and discussed. The complete code should be obtained and read in conjunction with this book. The aim of structural design is to produce a safe and economical structure that fulfils its required purpose. Theoretical knowledge of structural analysis must be combined with knowledge of design principles and theory and the constraints given in the standard to give a safe design. A thorough knowledge of properties of materials, methods of fabrication and erection is essential for the experienced designer. The learner must start with the basics and gradually build up experience through doing coursework exercises in conjunction with a study of design principles and theory. British Standards are drawn up by panels of experts from the professional institutions, and include engineers from educational and research institutions, consulting engineers, government authorities and the fabrication and construction industries. The standards give the design methods, factors of safety, design loads, design strengths, deflection limits and safe construction practices. As well as the main design standard for steelwork in buildings, BS 5950-1: 2000: Part 1, reference must be made to other relevant standards, including: (1) BS EN 10020: 2000. This gives definition and classification of grades of steel. (2) BS EN 10029: 1991 (plates); BS EN 10025: 1993 (sections); BS EN 102101: 1994 (hot finished hollow sections); BS EN 10219-1: 1997 (cold formed hollow sections). This gives the mechanical properties for the various types of steel sections. (3) BS 6399-1: 1996 Part 1, Code of Practice for Dead and Imposed Loads. (4) BS 6399-2: 1997 Part 2, Code of Practice for Wind Loads. (5) BS 6399-3: 1998 Part 1, Code of Practice for Imposed Roof Loads. Representative loading may be taken for element design. Wind loading depends on the complete building and must be estimated using the wind code. 1.5 Design calculations and computing Calculations are needed in the design process to determine the loading on the structure, carry out the analysis and design the elements and joints, and must be set out clearly in a standard form. Design sketches to illustrate and amplify the calculations are an integral part of the procedure and are used to produce the detail drawings. Computing now forms an increasingly larger part of design work, and all routine calculations can be readily carried out on a PC. The use of the computer speeds up calculation and enables alternative sections to be checked, giving the designer a wider choice than would be possible with manual working. However, it is most important that students understand the design principles involved before using computer programs. It is through doing exercises that the student consolidates the design theory given in lectures. Problems are given at the end of most chapters. Detailing 7 1.6 Detailing Chapter 12 deals with the detailing of structural steelwork. In the earlier chapters, sketches are made in design problems to show building arrangements, loading on frames, trusses, members, connections and other features pertinent to the design. It is often necessary to make a sketch showing the arrangement of a joint before the design can be carried out. At the end of the problem, sketches are made to show basic design information such as section size, span, plate sizes, drilling, welding, etc. These sketches are used to produce the working drawings. The general arrangement drawing and marking plans give the information for erection. The detailed drawings show all the particulars for fabrication of the elements. The designer must know the conventions for making steelwork drawings, such as the scales to be used, the methods for specifying members, plates, bolts, welding, etc. He/she must be able to draw standard joint details and must also have a knowledge of methods of fabrication and erection. AutoCAD is becoming generally available and the student should be given an appreciation of their use. 2 Limit state design 2.1 Limit state design principles The central concepts of limit state design are as follows: (1) All the separate conditions that make the structure unfit for use are taken into account. These are the separate limit states. (2) The design is based on the actual behaviour of materials and performance of structures and members in service. (3) Ideally, design should be based on statistical methods with a small probability of the structure reaching a limit state. The three concepts are examined in more detail below. Requirement (1) means that the structure should not overturn under applied loads and its members and joints should be strong enough to carry the forces to which they are subjected. In addition, other conditions such as excessive deflection of beams or unacceptable vibration, though not in fact causing collapse, should not make the structure unfit for use. In concept (2), the strengths are calculated using plastic theory and postbuckling behaviour is taken into account. The effect of imperfections on design strength is also included. It is recognized that calculations cannot be made in all cases to ensure that limit states are not reached. In cases such as brittle fracture, good practice must be followed to ensure that damage or failure does not occur. Concept (3) implies recognition of the fact that loads and material strengths vary, approximations are used in design and imperfections in fabrication and erection affect the strength in service. All these factors can only be realistically assessed in statistical terms. However, it is not yet possible to adopt a complete probability basis for design, and the method adopted is to ensure safety by using suitable factors. Partial factors of safety are introduced to take account of all the uncertainties in loads, materials strengths, etc. mentioned above. These are discussed more fully below. 2.2 Limit states for steel design The limit states for which steelwork is to be designed are set out in Section 2 of BS 5950-1: 2000. These are as follows. 8 Working and factored loads 9 2.2.1 Ultimate limit states The ultimate limit states include the following: (1) strength (including general yielding, rupture, buckling and transformation into a mechanism); (2) stability against overturning and sway; (3) fracture due to fatigue; (4) brittle fracture. When the ultimate limit states are exceeded, the whole structure or part of it collapses. 2.2.2 Serviceability limit states The serviceability limit states consist of the following: (5) (6) (7) (8) deflection; vibration (for example, wind-induced oscillation); repairable damage due to fatigue; corrosion and durability. The serviceability limit states, when exceeded, make the structure or part of it unfit for normal use but do not indicate that collapse has occurred. All relevant limit states should be considered, but usually it will be appropriate to design on the basis of strength and stability at ultimate loading and then check that deflection is not excessive under serviceability loading. Some recommendations regarding the other limit states will be noted when appropriate, but detailed treatment of these topics is outside the scope of this book. 2.3 Working and factored loads 2.3.1 Working loads The working loads (also known as the specified, characteristic or nominal loads) are the actual loads the structure is designed to carry. These are normally thought of as the maximum loads which will not be exceeded during the life of the structure. In statistical terms, characteristic loads have a 95 per cent probability of not being exceeded. The main loads on buildings may be classified as: (1) Dead loads: These are due to the weights of floor slabs, roofs, walls, ceilings, partitions, finishes, services and self-weight of steel. When sizes are known, dead loads can be calculated from weights of materials or from the manufacturer’s literature. However, at the start of a design, sizes are not known accurately and dead loads must often be estimated from experience. The values used should be checked when the final design is complete. For examples on element design, representative loading has been chosen, but for the building design examples actual loads from BS 6399: Part 1 are used. (2) Imposed loads: These take account of the loads caused by people, furniture, equipment, stock, etc. on the floors of buildings and snow on roofs. The values of the floor loads used depend on the use of the building. Imposed loads are given in BS 6399: Part 1 and snow load is given in BS 6399: Part 3. 10 Limit state design (3) Wind loads: These loads depend on the location and building size. Wind ∗ loads are given in BS 6399: Part 2. Calculation of wind loads is given in the examples on building design. (4) Dynamic loads: These are caused mainly by cranes. An allowance is made for impact by increasing the static vertical loads and the inertia effects are taken into account by applying a proportion of the vertical loads as horizontal loads. Dynamic loads from cranes are given in BS 6399: Part 1. Design examples show how these loads are calculated and applied to crane girders and columns. Other loads on the structures are caused by waves, ice, seismic effects, etc. and these are outside the scope of this book. 2.3.2 Factored loads for the ultimate limit states In accordance with Section 2.4.1 of BS 5950-1: 2000, factored loads are used in design calculations for strength and stability. Factored load = working or nominal load × relevant partial load factor, γf The partial load factor takes account of: (1) the unfavourable deviation of loads from their nominal values; and (2) the reduced probability that various loads will all be at their nominal value simultaneously. It also allows for the uncertainties in the behaviour of materials and of the structure as opposed to those assumed in design. The partial load factors, γf are given in Table 2 of BS 5950-1: 2000 and some of the factors are given in Table 2.1. Clause 2.4.1.1 of BS 5950-1: 2000 states that the factored loads should be applied in the most unfavourable manner and members and connections should Table 2.1 Partial factors for load, γf Loading Factors γf Dead load Dead load restraining uplift or overturning Dead load, wind load and imposed load Imposed load Wind load 1.4 1.0 1.2 1.6 1.4 Crane loads Vertical load Vertical and horizontal load Horizontal load Crane loads and wind load 1.6 1.4 1.6 1.2 ∗ Note: In countries other than United Kingdom, loads can be determined in accordance with this clause, or in accordance with local or national provisions as appropriate. Structural integrity 11 not fail under these load conditions. Brief comments are given on some of the load combinations: (1) The main load for design of most members and structures is dead plus imposed load. (2) In light roof structures uplift and load reversal occurs and tall structures must be checked for overturning. The load combination of dead plus wind load is used in these cases with a load factor of 1.0 for dead and 1.4 for wind load. (3) It is improbable that wind and imposed loads will simultaneously reach their maximum values and load factors are reduced accordingly. (4) It is also unlikely that the impact and surge load from cranes will reach maximum values together and so the load factors are reduced. Again, when wind is considered with crane loads the factors are further reduced. 2.4 Stability limit states To ensure stability, Clause 2.4.2 of BS 5950 states that structures must be checked using factored loads for the following two conditions: (1) Overturning: The structure must not overturn or lift off its seat. (2) Sway: To ensure adequate resistance, two design checks are required: (a) Design to resist the applied horizontal loads. (b) A separate design for notional horizontal loads. These are to be taken as 0.5 per cent of the factored dead plus imposed load, and are to be applied at the roof and each floor level. They are to act with 1.4 times the dead and 1.6 times the imposed load. Sway resistance may be provided by bracing rigid-construction shear walls, stair wells or lift shafts. The designer should clearly indicate the system he is using. In examples in this book, stability against sway will be ensured by bracing and rigid portal action. 2.5 Structural integrity The provisions of Section 2.4.5 of BS 5950 ensure that the structure complies with the Building Regulations and has the ability to resist progressive collapse following accidental damage. The main parts of the clause are summarized below: (1) All structures must be effectively tied at all floors and roofs. Columns must be anchored in two directions approximately at right angles. The ties may be steel beams or reinforcement in slabs. End connections must be able to resist a factored tensile load of 75 kN for floors and for roofs except where the steelwork only supports cladding that weighs not more than 0.7 kN/m2 and that carries only imposed roof loads and wind loads. (2) Additional requirements are set out for certain multi-storey buildings where the extent of accidental damage must be limited. In general, tied buildings will be satisfactory if the following five conditions are met: (a) sway resistance is distributed throughout the building; (b) extra tying is to be provided as specified; 12 Limit state design (c) column splices are designed to resist a specified tensile force; (d) any beam carrying a column is checked as set out in (3) below; and (e) precast floor units are tied and anchored. (3) Where required in (2) the above damage must be localized by checking to see if at any storey any single column or beam carrying a column may be removed without causing more than a limited amount of damage. If the removal of a member causes more than the permissible limit, it must be designed as a key element. These critical members are designed for accidental loads set out in the Building Regulations. The complete section in the code and the Building Regulations should be consulted. 2.6 Serviceability limit state deflection Deflection is the main serviceability limit state that must be considered in design. The limit state of vibration is outside the scope of this book and fatigue was briefly discussed in Section 2.2.1 and, again, is not covered in detail. The protection for steel to prevent the limit state of corrosion being reached was mentioned in Section 2.2.4. BS 5950-1: 2000 states in Clause 2.5.1 that deflection 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 finishings. The serviceability loads used are the unfactored imposed loads except in the following cases: (1) Dead + imposed + wind. Apply 80 per cent of the imposed and wind load. (2) Crane surge + wind. The greater effect of either only is considered. The structure is considered to be elastic and the most adverse combination of loads is assumed. Deflection limitations are given in Table 8 of BS 5950-1: 2000. These are given here in Table 2.2. These limitations cover beams and structures other than pitched-roof portal frames. It should be noted that calculated deflections are seldom realized in the finished structure. The deflection is based on the beam or frame steel section only and composite action with slabs or sheeting is ignored. Again, the full value of the imposed load used in the calculations is rarely achieved in practice. Table 2.2 Deflection limits Deflection of beams due to unfactored imposed loads Cantilevers Beams carrying plaster All other beams Horizontal deflection of columns due to unfactored imposed and wind loads Tops of columns in single-storey buildings In each storey of a building with more than one storey Crane gantry girders Vertical deflection due to static wheel loads Horizontal deflection (calculated on top flange properties alone) due to crane surge Length/180 Span/360 Span/200 Height/300 Storey height/300 Span/600 Span/500 Design methods for buildings 13 2.7 Design strength of materials The design strengths for steel complying with BS 5950-2 are given in Section 3.1.1 of BS 5950-1: 2000. Note that the material strength factor γm , part of the overall safety factor in limit state design, is taken as 1.0 in the code. The design strength may be taken as py = 1.0 Ys but not greater than Us /1.2 where Ys is the minimum yield strength, ReH and Us the minimum ultimate tensile strength, Rm . For the common types of steel values of py are given in Table 9 of the code and reproduced in Table 2.3. Table 2.3 Design strengths py for steel Steel grade Thickness (mm) less than or equal to Sections, plates and hollow sections, py (N/mm2 ) S275 16 40 63 80 100 150 275 265 255 245 235 225 S355 16 40 63 80 100 150 355 345 335 325 315 295 S460 16 40 63 80 100 460 440 430 410 400 The code states that the following values for the elastic properties are to be used: Modulus of elasticity, E = 205 000 N/mm2 Shear modulus, G = E/[2(1 + v)] Poisson’s ratio, v = 0.30 Coefficient of linear thermal expansion (in the ambient temperature range), α = 12 × 10−6 /◦ C 14 Limit state design 2.8 Design methods for buildings The design of buildings must be carried out in accordance with one of the methods given in Clause 2.1.2 of BS 5950-1: 2000. The design methods are as follows: (1) Simple design: In this method, the connections between members are assumed not to develop moments adversely affecting either the members or structure as a whole. The structure is assumed to be pin jointed for analysis. Bracing or shear walls are necessary to provide resistance to horizontal loading. (2) Continuous design: The connections are assumed to be capable of developing the strength and/or stiffness required by an analysis assuming full continuity. The analysis may be made using either elastic or plastic methods. (3) Semi-continuous design: This method may be used where the joints have some degree of strength and stiffness, but insufficient to develop full continuity. Either elastic or plastic analysis may be used. The moment capacity, rotational stiffness and rotation capacity of the joints should be based on experimental evidence. This may permit some limited plasticity, provided that the capacity of the bolts or welds is not the failure criterion. On this basis, the design should satisfy the strength, stiffness and in-plane stability requirements of all parts of the structure when partial continuity at the joints is taken into account in determining the moments and forces in the members. (4) Experimental verification: The code states that where the design of a structure or element by calculation in accordance with any of the above methods is not practicable, the strength and stiffness may be confirmed by loading tests. The test procedure is set out in Section 7 of the code. In practice, structures are designed to either the simple or continuous methods of design. Semi-continuous design has never found general favour with designers. Examples in this book are generally of the simple method of design. 3 Materials 3.1 Structural steel properties Structural steel products are manufactured to conform to new specifications given in BS 5950 Part 2: 2001. The previously used specification for weldable structural steels, BS 4360: 1990 has been replaced by a series of Euronorm specifications for technical delivery requirements, dimensions and tolerances such as BS EN10025, BS EN10029, BS EN10051, BS EN10113, BS EN10137, BS EN10155, BS EN10163, BS EN10210, BS EN10219 and others. Steel is composed of about 98 per cent of iron with the main alloying elements carbon, silicon and manganese. Copper and chromium are added to produce the weather-resistant steels that do not require corrosion protection. Structural steel is basically produced in three strength grades S275, S355 and S460. The important design properties are strength, ductility, impact resistance and weldability. The stress–strain curves for the three grades of steel are shown in Figure 3.1(a) and these are the basis for the design methods used for steel. Elastic design is kept within the elastic region and because steel is almost perfectly elastic, design based on elastic theory is a very good method to use. The stress–strain curves show a small plateau beyond the elastic limit and then an increase in strength due to strain hardening. Plastic design is based on the horizontal part of the stress–strain shown in Figure 3.1(b). The mechanical properties for steels are set out in the respective specifications mentioned earlier. The yield strengths for the various grades vary with the thickness and other important design properties are given in Section 2.7 of this book. 3.2 Design considerations Special problems occur with steelwork and good practice must be followed to ensure satisfactory performance in service. These factors are discussed briefly below in order to bring them to the attention of students and designers, although they are not generally of great importance in the design problems covered in this book. However, it is worth noting that the material safety factor γm is set to unity in BS 5950 which implies a certain level of quality and testing in steel usage. Weld procedures are qualified by maximum carbon equivalent values. 15 16 Materials Stress N/mm2 (a) Stress–strain diagrams for structural steels S460 S355 460 355 275 S275 Thickness ⭐16 mm 0 0.1 0.2 Strain 0.3 0.4 Stress N/mm2 (b) Stress–strain diagram for plastic design Yield stress Plastic range Elastic range 0 0.1 0.2 Strain Figure 3.1 Stress–strain diagrams for structural steels Attention to weldability should be given when dealing with special, thick and higher grade steel to avoid hydrogen induced cracking. Reader can refer to BS EN10229: 1998 for more information if necessary. 3.2.1 Fatigue Fatigue failure can occur in members or structures subjected to fluctuating loads such as crane girders, bridges and offshore structures. Failure occurs through initiation and propagation of a crack that starts at a fault or structural discontinuity and the failure load may be well below its static value. Welded connections have the greatest effect on the fatigue strength of steel structures. Tests show that butt welds give the best performance in service while continuous fillet welds are much superior to intermittent fillet welds. Bolted connections do not reduce the strength under fatigue loading. To help avoid fatigue failure, detail should be such that stress concentrations and abrupt changes of section are avoided in regions of tensile stress. Cases where fatigue could occur are noted in this book, and for further information the reader should consult reference (1). Design considerations 17 3.2.2 Brittle fracture Structural steel is ductile at temperatures above 10◦ C but it becomes more brittle as the temperature falls, and fracture can occur at low stresses below 0◦ C. The Charpy impact test is used to determine the resistance of steel to brittle fracture. In this test, a small specimen is broken by a hammer and the energy or toughness to cause failure at a given test temperature is measured. In design, brittle fracture should be avoided by using steel quality grade with adequate impact toughness. Quality steels are designated JR, J0, J2, K2 and so forth in order of increasing resistance to brittle fracture. The Charpy impact fracture toughness is specified for the various steel quality grades: for example, Grade S275 J0 steel is to have a minimum fracture toughness of 27 J at a test temperature of 0◦ C. In addition to taking care in the selection of steel grade to be used, it is also necessary to pay special attention to the design details to reduce the likelihood of brittle fracture. Thin plates are more resistant than thick ones. Abrupt changes of section and stress concentration should be avoided. Fillets welds should not be laid down across tension flanges and intermittent welding should not be used. Cases where brittle fracture may occur in design of structural elements are noted in this book. For further information, the reader should consult reference (2). 3.2.3 Fire protection Structural steelwork performs badly in fires, with the strength decreasing with increase in temperature. At 550◦ C, the yield stress has fallen to approximately 0.7 of its value at normal temperatures; that is, it has reached its working stress and failure occurs under working loads. The statutory requirements for fire protection are usually set out clearly in the approved documents from the local Building Regulations (3) or Fire Safety Authority. These lay down the fire-resistance period that any load-bearing element in a given building must have, and also give the fire-resistance periods for different types of fire protection. Fire protection can be provided by encasing the member in concrete, fire board or cementitious fibre materials. The main types of fire protection for columns and beams are shown in Figure 3.2. More recently, intumescent paint is being used especially for exposed steelwork. Solid casing Hollow casing Profile casing Figure 3.2 Fire protections for columns and beams 18 Materials All multi-storey steel buildings require fire protection. Single-storey factory buildings normally do not require fire protection for the steel frame. Further information is given in reference (4). 3.2.4 Corrosion protection Exposed steelwork can be severely affected by corrosion in the atmosphere, particularly if pollutants are present, and it is necessary to provide surface protection in all cases. The type of protection depends on the surface conditions and length of life required. The main types of protective coatings are: (1) Metallic coatings: Either a sprayed-on in line coating of aluminium or zinc is used or the member is coated by hot-dipping it in a bath of molten zinc in the galvanizing process. (2) Painting: where various systems are used. One common system consists of using a primer of zinc chromate followed by finishing coats of micaceous iron oxide. Plastic and bituminous paints are used in special cases. The single most important factor in achieving a sound corrosion-protection coating is surface preparation. Steel is covered with mill scale when it cools after rolling, and this must be removed before the protection is applied, otherwise the scale can subsequently loosen and break the film. Blast cleaning makes the best preparation prior to painting. Acid pickling is used in the galvanizing process. Other methods of corrosion protection which can also be considered are sacrificial allowance, sherardizing, concrete encasement and cathodic protection. Careful attention to design detail is also required (for example, upturned channels that form a cavity where water can collect should be avoided) and access for future maintenance should also be provided. For further information the reader should consult BS EN ISO12944 Parts 1-8: 1998—Corrosion Protection of Steel Structures by Protective Paint Systems and BS EN ISO14713: 1999—Protection against Corrosion of Iron and Steel in Structures, Zinc and Aluminium Coatings. 3.3 Steel sections 3.3.1 Rolled and formed sections Rolled and formed sections are produced in steel mills from steel blooms, beam blanks or coils by passing them through a series of rollers. The main sections are shown on Figure 3.3 and their principal properties and uses are discussed briefly below: (1) Universal beams: These are very efficient sections for resisting bending moment about the major axis. (2) Universal columns: These are sections produced primarily to resist axial load with a high radius of gyration about the minor axis to prevent buckling in that plane. (3) Channels: These are used for beams, bracing members, truss members and in compound members. Steel sections B B D D D B D × B 152 × 152 to D × B 100 × 50 to 356 × 406 400 × 100 D × B 203 × 133 to 914 × 419 Universal beam Universal column Parallel flange channel B A A A 19 A A × A 25 × 25 to 250 × 250 Equal angle B A × B 40 × 25 to 200 × 150 Unequal angle D B × A 133 × 102 to 305 × 457 Structural tee cut from UB B D 21.3 to 508 Circular hollow section D D D D × D 40 × 40 to 400 × 400 D × B 50 × 30 to 500 × 300 Square hollow section Rectangular hollow section Figure 3.3 Rolled and formed sections (4) Equal and unequal angles: These are used for bracing members, truss members and for purlins, side and sheeting rails. (5) Structural tees: The sections shown are produced by cutting a universal beam or column into two parts. Tees are used for truss members, ties and light beams. (6) Circular, square and rectangular hollow sections: These are mostly produced from hot-rolled coils, and may be hot-finished or cold-formed. A welded mother tube is first formed and then it is rolled to its final square or rectangular shape. In the hot process, the final shaping is done at the steel normalising temperature whereas in the cold process, it is done at ambient room temperature. Both types of hollow sections are now permitted in BS 5950. These sections make very efficient compression members, and are used in a wide range of applications as members in roof trusses, lattice girders, in building frames, for purlins, sheeting rails, etc. Note that the range in serial sizes is given for the members shown in Figure 3.3. A number of different members are produced in each serial size by varying the flange, web, leg or wall thicknesses. The material properties, tolerances and dimensions of the various sections can be found in the following standards as shown in Table 3.1. 20 Materials Table 3.1 Material properties, dimensions and tolerances of various sections Sections Materials Dimensions and tolerances Universal beams, columns, tees, bearing piles Channels (hot-finished) Purlins (cold-formed) Angles Flats (strips) Plates Hot-finished hollows Cold-formed hollows EN 10025 EN 10113 EN 10025 EN 10149 EN 10025 EN 10025 EN 10025 EN 10210-1 EN 10219-1 EN 10034 BS 4-1 BS 4-1 BS 5950-7 EN 10056 EN 10048 EN 10029 EN 10210-2 EN 10219-2 (a) Compound beam (b) (c) Crane girder (d) Battened member Laced member Figure 3.4 Compound sections 3.3.2 Compound sections Compound sections are formed by the following means (Figure 3.4): (1) strengthening a rolled section such as a universal beam by welding on cover plates, as shown in Figure 3.4(a); (2) combining two separate rolled sections, as in the case of the crane girder in Figure 3.4(b). The two members carry loads from separate directions. (3) connecting two members together to form a strong combined member. Examples are the laced and battened members shown in Figures 3.4(c) and (d). Section properties Plate girder Built-up section Box girder 21 Box column Figure 3.5 Built-up sections Zed section Sigma section Lipped channel Figure 3.6 Cold-rolled sections 3.3.3 Built-up sections Built-up sections are made by welding plates together to form I, H or box members which are termed plate girders, built-up columns, box girders or columns, respectively. These members are used where heavy loads have to be carried and in the case of plate and box girders where long spans may be required. Examples of built-up sections are shown in Figure 3.5. 3.3.4 Cold-rolled open sections Thin steel plates can be formed into a wide range of sections by cold rolling. The most important uses for cold-rolled open sections in steel structures are for purlins, side and sheeting rails. Three common sections-the zed, sigma and lipped channel-are shown in Figure 3.6. Reference should be made to manufacturer’s specialised literature for the full range of sizes available and the section properties. Some members and their properties are given in Sections 4.12.6 and 4.13.5 in design of purlins and sheeting rails. 3.4 Section properties For a given member serial size, the section properties are: (1) the exact section dimensions; (2) the location of the centroid if the section is asymmetrical about one or both axes; (3) area of cross-section; (4) moments of inertia about various axes; 22 Materials (5) radii of gyration about various axes; (6) moduli of section for various axes, both elastic and plastic. The section properties for hot rolled and formed sections are also listed in SCI Publication 202: Steelwork Design Guide to BS 5950: Part 1: 2000, Volume 1 Section Properties and Member Capacities, 6th edition with amendments, The Steel Construction Institute, United Kingdom. For compound and built-up sections, the properties must be calculated from first principles. The section properties for the symmetrical I-section with dimensions as shown in Figure 3.7(a) are as follows: (1) Elastic properties: Area A = 2BT + dt Moment of inertia x–x axis Ix = BD 3 /12 − (B − t)d 3 /12 Moment of inertia y–y axis Iy = 2TB 3 /12 + dt 3 /12 Radius of gyration x–x axis rx = (Ix /A)0.5 Radius of gyration y–y axis ry = (Iy /A)0.5 Modulus of section x–x axis Zx = 2Ix /D Modulus of section y–y axis Zy = 2Iy /B (a) x x t d D B y y Symmetrical I-section (b) y x1 x1 x x x x xx Centroidal axis x1x1 Equal area axis Asymmetrical I-section y Figure 3.7 Beam section Section properties 23 (2) Plastic moduli of section: The plastic modulus of section is equal to the algebraic sum of the first moments of area about the equal area axis. For the I-section shown: Sx = 2BT (D − T )/2 + td 2 /4 Sy = 2TB 2 /4 + dt 2 /4 For asymmetrical sections such as those shown in Figure 3.7(b), the neutral axis must be located first. In elastic analysis, the neutral axis is the centroidal axis while in plastic analysis it is the equal area axis. The other properties may then be calculated using procedures from strength of materials (5). Calculations of properties for unsymmetrical sections are given in various parts of this book. Other properties of universal beams, columns, joists and channels, used for determining the buckling resistance moment are: buckling parameter, u; torsional index, x; warping constant, H ; torsional constant, J . These properties may be calculated from formulae given in Appendix B of BS 5950: Part I or from Section A of the SCI Publication 202: Steelwork Design Guide to BS 5950: Part 1: 2000, Volume 1 Section Properties and Member Capacities, 6th edition with amendments, The Steel Construction Institute, United Kingdom. 4 Beams 4.1 Types and uses Beams span between supports to carry lateral loads which are resisted by bending and shear. However, deflections and local stresses are also important. Beams may be cantilevered, simply supported, fixed ended or continuous, as shown in Figure 4.1(a). The main uses of beams are to support floors and columns, carry roof sheeting as purlins and side cladding as sheeting rails. (a) Cantilever Simply supported Fixed ended Continuous Types of beams (b) Universal beam Compound beam Crane beam Purlins and sheeting rails Beam sections Figure 4.1 Types of beams and beam sections 24 Channel Beam loads 25 Bending moment diagram Cover plates Simply supported beam Bending moment diagram Haunched ends Fixed ended beam Figure 4.2 Non-uniform beam Any section may serve as a beam, and common beam sections are shown in Figure 4.1(b). Some comments on the different sections are given: (1) The universal beam where the material is concentrated in the flanges is the most efficient section to resist uniaxial bending. (2) The universal column may be used where the depth is limited, but it is less efficient. (3) The compound beam consisting of a universal beam and flange plates is used where the depth is limited and the universal beam itself is not strong enough to carry the load. (4) The crane beam consists of a universal beam and channel. It is because the beam needs to resist bending in both horizontal and vertical directions. Beams may be of uniform or non-uniform section. Sections may be strengthened in regions of maximum moment by adding cover plates or haunches. Some examples are shown in Figure 4.2. 4.2 Beam loads Types of beam loads are: (1) concentrated loads from secondary beams and columns; (2) distributed loads from self-weight and floor slabs. The loads are further classified into: (1) dead loads from self weight, slabs, finishes, etc. (2) imposed loads from people, fittings, snow on roofs, etc. (3) wind loads, mainly on purlins and sheeting rails. Loads on floor beams in a steel frame building are shown in Figure 4.3(a). The figure shows loads from a two-way spanning slab which gives trapezoidal 26 Beams (a) A B C 1 2 3 Beam B1 Beams 2A and 2B Slab loads on floor beams (b) Floor slab Column Secondary beam Support Actual beam Load diagram Shear force diagram Bending moment diagram Actual loads on a beam Figure 4.3 Beam loads and triangular loads on the beams. One-way spanning floor slabs give uniform loads. An actual beam with the floor slab and members it supports is shown in Figure 4.3(b). The load diagram and shear force and bending moment diagrams constructed from it are also shown. 4.3 Classification of beam cross-sections The projecting flange of an I-beam will buckle prematurely if it is too thin. Webs will also buckle under compressive stress from bending and from shear. Classification of beam cross-sections 27 This problem is discussed in more detail in Section 5.2 in Chapter 5 (see also reference (6)). To prevent local buckling from occurring, limiting outstand/thickness ratios for flanges and depth/thickness ratios for webs are given in BS 5950-1: 2000 in Section 3.5. Beam cross-sections are classified as follows in accordance with their behaviour in bending: Class 1 Plastic cross-section: This can develop a plastic hinge with sufficient rotation capacity to permit redistribution of moments in the structure. Only class I sections can be used for plastic design. Class 2 Compact cross-section: This can develop the plastic moment capacity, but local buckling prevents rotation at constant moment. Class 3 Semi-compact cross-section: The stress in the extreme fibres should be limited to the yield stress because local buckling prevents development of the plastic moment capacity. Class 4 Slender cross-section: Premature buckling occurs before yield is reached. Flat elements in a cross section are classified as: (1) Internal elements supported on both longitudinal edges. (2) Outstand elements attached on one edge with the other free. Elements are generally of uniform thickness, but, if tapered, the average thickness is used. Elements are classified as plastic, compact or semi-compact if they meet limits given in Tables 11 and 12 in association with Figures 5 and 6 of the code. An example for the limiting proportions for elements of universal beams and channels are shown in Figure 4.4. b d t n n t I-section Compression element Outstand element of Rolled compression flange section Web with neutral axis at middepth T d T b Channel Ratio Class 1 plastic Limiting value Class 2 compact Class 3 semi-compact b/ T 9ε 10ε 15ε d/t 80ε 100ε 120ε The parameter, ε = (275/py)0.5 Figure 4.4 Limiting proportions for rolled sections 28 Beams 4.4 Bending stresses and moment capacity Both elastic and plastic theories are discussed here. Short or restrained beams are considered in this section. Plastic properties are used for plastic and compact sections and elastic properties for semi-compact sections to determine moment capacities. For slender sections, only effective elastic properties are used. 4.4.1 Elastic theory (1) Uniaxial bending The bending stress distributions for an I-section beam subjected to uniaxial moment are shown in Figure 4.5(a). We define following terms for the I-section: M = applied bending moment; Ix = moment of inertia about x–x axis; Zx = 2Ix /D = modulus of section for x–x axis; and D = overall depth of beam. The maximum stress in the extreme fibres top and bottom is: fbc = fbt = Mx /Zx The moment capacity, Mc = σb Zx where σb is the allowable stress. The moment capacity for a semi-compact section subjected to a moment due to factored loads is given in Clause 4.2.5.2 of BS 5950-1: 2000 as Mc = py Z where py is the design strength. (a) fbc (b) fbc Y y1 Y X X y2 D X X Y Section fbt Stress T-section with two axes of symmetry Figure 4.5 Beams in uniaxial bending Y Section Crane beam with one axis of symmetry fbt Stress Bending stresses and moment capacity 29 For the asymmetrical crane beam section shown in Figure 4.5(b), the additional terms require definition: Z1 = Ix /y1 = modulus of section for top flange, Z2 = Ix /y2 = modulus of section for bottom flange, y1 , y2 = distance from centroid to top and bottom fibres. The bending stresses are: Top fibre in compression fbc = Mx /Z1 Bottom fibre in tension fbt = Mx /Z2 The moment capacity controlled by the stress in the bottom flange is Mc = py Z2 (2) Biaxial bending Consider that I-section in Figure 4.6(a) which is subject to bending about both axes. We define the following terms: Mx = moment about the x–x axis, My = moment about the y–y axis, Zx = modulus of section for the x–x axis, Zy = modulus of section for the y–y axis. The maximum stress at A or B is: fA = fB = Mx /Zx + My /Zy If the allowable stress is σb , the moment capacities with respect to x–x and y–y axes are: Mcx = σb Zx Mcy = σb Zy Taking the maximum stress as σb and substituting for Zx and Zy in the expression above gives the interaction relationship My Mx + =1 Mcx Mcy This is shown graphically in Figure 4.6(b). (3) Asymmetrical sections Note that with the channel section shown in Figure 4.7(a), the vertical load must be applied through the shear centre for bending in the free member to 30 Beams (a) Mx Zx Vertical load My Mx X Horizontal load Compression Y B Mx X X X Mx Zx Y Tension Zy Y My My Zy A My Vertical bending stresses Y Compression Tension Horizontal bending stresses Bending stresses (b) 1.0 My Mcy Mx Mcx 1.0 Interaction diagram Figure 4.6 Biaxial bending take place about the x–x axis, otherwise twisting and biaxial bending occurs. However, a horizontal load applied through the centroid causes bending about the y–y axis only. For an asymmetrical section such as the unequal angle shown in Figure 4.7(b), bending takes place about the principal axes u–u and v–v in the free member when the load is applied through the shear centre. When the angle is used as a purlin, the cladding restrains the member so that it bends about the x–x axis. 4.4.2 Plastic theory (1) Uniaxial bending The stress–strain curve for steel on which plastic theory is based is shown in Figure 4.8(a). In the plastic region after yield, the strain increases without Bending stresses and moment capacity (a) Y Vertical load Horizontal load 31 X X Shear centre Vertical bending stress Y Horizontal bending stress Channel section (b) Vertical load Y V Shear centre U X X U V U V U V Y Bending stresses U - U axis Bending stresses V - V axis Unequal angle Figure 4.7 Bending of asymmetrical sections increase in stress. Consider the I-section shown in Figure 4.8(b). Under moment, the stress first follows an elastic distribution. As the moment increases, the stress at the extreme fibre reaches the yield stress and the plastic region proceeds inwards as shown, until the full plastic moment is reached and a plastic hinge is formed. For single axis bending, the following terms are defined: Mc = plastic moment capacity, S = plastic modulus of section, Z = elastic modulus of section, py = design strength. The moment capacity given in Clause 4.2.5.2 of BS 5950-1: 2000 for class 1 and 2 sections with low shear load is: Mc = py S, ≤ 1.2 py Z. 32 Beams Stress (a) Yield stress, Ys Plastic Elastic design region design region Strain Simplified stress-strain curve (b) ⭐py X py py X ⭐py Section py py Elastic Partly plastic Fully plastic Behaviour in bending Figure 4.8 Behaviour in bending py ⭐py Centroidal axis X1 Compression X X X X1 X1 X1 py b X Yield stress Tension Equal area axis Section Elastic stresses Plastic stress distribution Figure 4.9 Section with one axis of symmetry The first expression is the plastic moment capacity, the second ensures that yield does not occur at working loads in I-sections bent about the y–y axis. For single-axis bending for a section with one axis of symmetry, consider the T-section shown in Figure 4.9. In the elastic range, bending takes place about the centroidal axis and there are two values for the elastic modulus of section. Bending stresses and moment capacity 33 In the plastic range, bending takes place about the equal area axis and there is one value for the plastic modulus of section: S = Mc /py = Ab/2 where A is the area of cross section and b is the lever arm between the tension and compression forces. (2) Biaxial bending When a beam section is bent about both axes, the neutral axis will lie at an angle to the rectangular axes which depends on the section properties and values of the moments. Solutions have been obtained for various cases and a relationship established between the ratios of the applied moments and the moment capacities about each axis. The relationship expressed in Sections 4.9 of BS 5950-1: 2000 for plastic or compact cross sections is given in the following form:     My Z2 Mx Z1 + ≤1 Mcx Mcy where Mx = factored moment about the x–x axis, My = factored moment about the y–y axis, Mcx = moment capacity about the x–x axis, Mcy = moment capacity about the y–y axis, Z1 = 2 for I- and H-sections and 1 for other open sections Z2 = 1 for all open sections. A conservative result is given if Z1 = Z2 = 1. The interaction diagram is shown in Figure 4.10. (3) Unsymmetrical sections For sections with no axis of symmetry, plastic analysis for bending is complicated, but solutions have been obtained. In many cases where such sections are used, the member is constrained to bend about the rectangular axis (see Section 4.4.1(3)). Such cases can also be treated by elastic theory using factored loads with the maximum stress limited to the design strength. (a) (b) Y X X Y Section Figure 4.10 Biaxial bending 1.0 My Mcy I-section z1 = 2.0 z2 = 1.0 z1–z2 = 1 Mx Mcx Interaction diagram 1.0 34 Beams 4.5 Lateral torsional buckling 4.5.1 General considerations The compression flange of an I-beam acts like a column, and will buckle sideways if the beam is not sufficiently stiff or the flange is not restrained laterally. The load at which the beam buckles can be much less than that causing the full moment capacity to develop. Only a general description of the phenomenon and factors affecting it are set out here. The reader should consult references (6) and (7) for further information. Consider the simply supported beam with ends free to rotate in plan but restrained against torsion and subjected to end moments, as shown in Figure 4.11. Initially, the beam deflects in the vertical plane due to bending, but as the moment increases, it reaches a critical value ME less than the moment capacity, where it buckles sideways, twists and collapses. Elastic theory is used to set up equilibrium equations to equate the disturbing effect to the lateral bending and torsional resistances of the beam. The solution of this equation gives the elastic critical moment:  π 2 EH π EIy GJ 1 + 2 ME = L L GJ (a) M M Elevation Rotation Section at centre Plan Y Buckled position of beam (b) X X Elastic critical moment M/Me 10 Y I-section M Mc Intermediate Short 0 50 Slender 100 150 Slenderness LE /ry Behaviour curve Figure 4.11 Lateral torsional buckling 200 250 Lateral torsional buckling 35 where E = Young’s modulus, G = shear modulus, J = torsion constant for the section, H = warping constant for the section, L = span Iy = moment of inertia about the y–y axis. The theoretical solution applies to a beam subjected to a uniform moment. In other cases where the moment varies, the tendency to buckling is reduced. If the load is applied to the top flange and can move sideways, it is destabilizing, and buckling occurs at lower loads than if the load were applied at the centroid, or to the bottom flange. In the theoretical analysis, the beam was assumed to be straight. Practical beams have initial curvature and twisting, residual stresses, and the loads are applied eccentrically. The theory set out above requires modification to cover actual behaviour. Theoretical studies and tests show that slender beams fail at the elastic critical moment, ME and short or restrained beams fail at the plastic moment capacity Mc . A lower bound curve running between the two extremes can be drawn to contain the behaviour of intermediate beams. Beam behaviour as a function of slenderness is shown in Figure 4.11(b). To summarize, factors influencing lateral torsional buckling are: (1) The unrestrained length of compression flange: The longer this is, the weaker the beam. Lateral buckling is prevented by providing props at intermediate points. (2) The end conditions: Rotational restraint in plan helps to prevent buckling. (3) Section shape: Sections with greater lateral bending and torsional stiffness have greater resistance to buckling. (4) Note that lateral restraint to the tension flange also helps to resist buckling (see Figure 4.11). (5) The application of the loads and shape of the bending moment diagram between restraints. A practical design procedure must take into account the effects noted above. Terms used in the curve are defined as follows: M = moment causing failure, Mc = moment capacity for a restrained beam, ME = elastic critical moment, LE /ry = slenderness with respect to the y–y axis (see the next section). 4.5.2 Lateral restraints and effective length The code states in Clause 4.2.2 that full lateral restraint is provided by a floor slab if the friction or shear connection is capable of resisting a lateral force of 2.5 per cent of the maximum factored force in the compression flange. Other suitable construction can also be used. Members not provided with full lateral restraint must be checked for buckling. The following two types of restraints are defined in Sections 4.3.2 and 4.3.3 of the code: (1) Intermediate lateral restraint, which prevents sideways movement of the compression flange; and 36 Beams (a) Floor slab provides full lateral restraint Secondary beam provides lateral restraint Torsional restraint free to rotate in plan Lateral and torsional restraint (b) l1 Slab l3 LE = l 3 l4 LE = 0.714 Open Torsional restraint free to rotate in plan LE = l2 Fully restrained l2 LE = l1 Fixed ends Effective lengths Figure 4.12 Restraints and effective lengths (2) Torsional restraint, which prevents movement of one flange relative to the other. Restraints are provided by floor slabs, end joints, secondary beams, stays, sheeting, etc., and some restraints are shown in Figure 4.12(a). The effective length LE for a beam is defined in Section 1 of the code as the length between points of effective restraints multiplied by a factor to take account of the end conditions and loading. Note that a destabilizing load (where the load is applied to the top flange and can move with it) is taken account of by increasing the effective length of member under consideration. Lateral torsional buckling 37 Table 4.1 Effective length LE –Beams Support conditions Loading conditions Beam partial torsionally unrestrained Compression flange laterally unrestrained Both flanges free to rotate on plan Beam torsionally restrained Compression flange laterally restrained Compression flange only free to rotate on plan Beam torsionally restrained Both flanges NOT free to rotate on plan Normal Destabilizing 1.2LLT + 2D 1.4LLT + 2D 1.0LLT 1.2LLT 0.7LLT 0.85LLT LLT = length of beam between restraints. D = depth of beam. The effective length for beams is discussed in Section 4.3.5 of BS 5950-1: 2000. When the beam is restrained at the ends only, that is, without intermediate restraint, the effective length should be obtained from Table 13 in the code. Some values from this table are given in Table 4.1. Where the beam is restrained at intervals by other members the effective length LE may be taken as L, the distance between restraints. Some effective lengths for floor beams are shown in Figure 4.12(b). 4.5.3 Code design procedure (1) General procedure The general procedure for checking the resistance to lateral torsional buckling is out lined in Section 4.3.6 of BS 5950-1: 2000: (1) Resistance to lateral-torsional buckling need not be checked separately (and the buckling resistance moment Mb may be taken as equal to the relevant moment capacity Mc ) in the following cases: bending about the minor axis; CHS, square RHS or circular or square solid bars; RHS, unless LE /ry exceeds the limiting value given in Table 15 of BS 5950-1: 2000 for the relevant value of D/B; I, H, channel or box sections, if λLT does not exceed λL0 , Otherwise, for members subject to bending about their major axis, reference should be made as follows: Mx ≤ Mb /mLT and Mx ≤ Mcx (2) Calculate the equivalent uniform moment factor mLT : The value of the equivalent uniform moment factor mLT which depend on the ratio and direction of the major axis moment. For the normal loading condition, the equivalent uniform moment factor for lateral-torsional buckling should be obtained from Table 18 of BS 5950-1: 2000. For destabilizing loading condition, mLT should be taken as 1.0. Values for some common load cases are shown in Figure 4.13. 38 Beams M βM β = 1 mLT = 1.0 M M β=0 mLT = 0.60 βM β = 0.5 mLT = 0.80 βM β = –1.0 mLT = 0.44 Figure 4.13 Equivalent uniform moment factor mLT for lateral-torsional buckling (3) Estimate the effective length LE of the unrestrained compression flange using the rules from Section 4.5.2. Minor axis slenderness, λ = LE /ry , where ry = radius of gyration for the y–y axis. (4) Calculate the equivalent slenderness, λLT √ λLT = uvλ βw where u = buckling parameter allowing for torsional resistance. This may be calculated from the formulae in Appendix B or taken from the published table in the Guide to BS 5950: Part 1: 2000, Vol. 1, Section properties, Member Capacities, SCI. It can also conservatively taken as 0.9 for an uniform rolled I-section, v = slenderness factor which depends on values of η and λ/x. where η= Iyc Iyc + Iyt Iyc = second moment of area of the compression flange about the minor axis of the section; Iyt = second moment of area of the tension flange about the minor axis of the section, η = 0.5 for a symmetrical section, x = torsional index. This can be calculated from the formula in Appendix B or obtained from the published table in the Guide to BS 5950: Part 1: 2000. The torsional index can be taken conservatively approximately equal to D/T where D is the overall depth of beam and T the thickness of the compression flange. Values of v are given in Table 19 of the BS 5950-1: 2000. Alternatively, v can be determined by the formulae in Appendix B or Clause 4.3.6.7. in the code. (5) Ratio βw should be determined in accordance to Clause 4.3.6.9. The ratio is dependent on the classification of the sections. For class 1 or class 2 sections, βw is taken as 1.0. (6) Read the bending strength, pb from Table 16 for rolled sections and Table 17 for welded sections in the BS 5950-1: 2000. Values of pb depend on the equivalent slenderness λLT and design strength py . Lateral torsional buckling 39 (7) Calculate the buckling resistance moment. for class 1 plastic or class 2 compact cross-sections: Mb = pb Sx . for class 3 semi-compact cross-sections: Mb = pb Zx ; or alternatively, Mb = pb Sx,eff for class 4 slender cross-sections: Mb = pb Zx,eff . where pb is the bending strength; Sx is the plastic modulus about the major axis; Sx,eff is the effective plastic modulus about the major axis; Zx is the section modulus about the major axis; Zx,eff is the effective section modulus about the major axis. (2) Conservative approach for equal flanged rolled sections The code gives a conservative approach for equal flanged rolled sections in Section 4.3.7. The buckling resistance moment Mb of a plain rolled I, H or channel section with equal flanges may be determined using the bending strength, pb obtained from Table 20 for the relevant values of (βw )0.5 LE /ry and D/T as follows: for a class 1 plastic or class 2 compact cross-section: Mb = pb Sx for a class 3 semi-compact cross-section: Mb = pb Zx 4.5.4 Biaxial bending Lateral torsional buckling affects the moment capacity with respect to the major axis only of I-section beams. When the section is bent about only the minor axis, it will reach the moment capacity given in Section 4.4.2(1). Where biaxial bending occurs, BS 5950-1: 2000 specifies in Section 4.9 that the following simplified interaction expressions must be satisfied for plastic or compact sections: (1) Cross-section capacity check at point of maximum combined moments: My Mx + ≤1 Mcx Mcy This design check was discussed in Section 4.4.2(2) above. 40 Beams (2) Member buckling check at the centre of the beam: my My mx M x + ≤1 py Zx py Zy my My mLT MLT + ≤1 Mb py Zy where Mb is the buckling resistance moment, MLT is the maximum major axis moment in the segment length L governing Mb ; Mx is the maximum major axis moment in the segment length Lx ; My is the maximum minor axis moment in the segment length Ly ; Zx is the section modulus about the major axis; Zy is the section modulus about the minor axis. The equivalent uniform moment factors mLT , mx and my should be obtained from Clause 4.8.3.3.4 of BS 5950-1: 2000. More exact expressions are given in the code. Biaxial bending is discussed more fully in Chapter 7 of this book. 4.6 Shear in beams 4.6.1 Elastic theory The value of shear stress at any point in a beam section is given by the following expression (see Figure 4.14(a)): fs = V Ay Ix t where V = shear force at the section A = area between the point where the shear stress is required and a free edge y = distance from the centroid of the area A to the centroid of the section Ix = second moment of area about the x–x axis t = thickness of the section at the point where the shear stress is required. Using this formula, the shear stresses at various points in the beam section can be found. Thus, the maximum shear stress at the centroid in terms of the beam dimensions shown in the figure is: fmax = V Ix t  BT (d + T ) td 2 + 2 8  Shear in beams (a) Maximum stress X D d T B X 41 T t Elastic shear stress distribution D t d (b) t T-section Rolled beam Shear areas Yeild stress (c) X X Section Shear stress Yield stress Net bending stress Plastic theory-shear and moment Figure 4.14 Shear in beams Note that the distribution shows that the web carries the bulk of the shear. It has been customary in design to check the average shear stress in the web given by: fav = V /Dt which should not exceed an allowable value. 4.6.2 Plastic theory Shear is considered in BS 5950-1: 2000 in Section 4.2.3. For a rolled member subjected to shear only, the shear force is assumed to be resisted by the web area Av shown in Figure 4.14(b), where: Av = web thickness × overall depth = tD 42 Beams For the T section shown in the figure: Av = 0.9Ao where Ao is the area of the rectilinear element which has the largest dimension in the direction parallel to the shear force and equal to td. √ The shear area may be stressed to the yield stress in shear, that is, to 1/ 3 of the yield stress in tension. The capacity is given in the code as: Pv = 0.6py Av If the ratio d/t exceeds 70ε for a rolled section, or 62ε for a welded section, the web should be checked for shear buckling in accordance with Clause 4.4.5 in BS 5950-1: 2000. If moment as well as shear occurs at the section, the web is assumed to resist all the shear while the flanges are stressed to yield by bending. The section analysis is based on the shear stress and bending stress distributions shown in Figure 4.14(c). The web is at yield under the combined bending and shear stresses and von Mises’ criterion is adopted for failure in the web. The shear reduces the moment capacity, but the reduction is small for all but high values of shear force. The analysis for shear and bending is given in reference (8). BS 5950-1: 2000 gives the following expression in Section 4.2.5.3 for the moment capacity for plastic or compact sections in the presence of high shear load. When the average shear force F , is less than 0.6 of the shear capacity Pv , no reduction in moment capacity is required. When Fv is greater than 0.6 Pv , the reduced moment capacity for class 1 or class 2 cross-sections is given by: Mc = py (S − ρSv ) where Sv = tD 2 /4 for a rolled section with equal flanges ρ = [2(Fv /Pv ) − 1]2 . 4.7 Deflection of beams The deflection limits for beams specified in Section 2.5.1 of BS 5950-1: 2000 were set in Section 2.6 of this book. The serviceability loads are the unfactored imposed loads. Deflection formulae are given in design manuals. Deflections for some common load cases for simply supported beams together with the maximum moments are given in Figure 4.15. For general load cases deflections can be calculated by the moment area method. (see references (9) and (10).) 4.8 Beam connections End connections to columns and other beams form an essential part of beam design. Checks for local failure are required at supports and points where concentrated loads are applied. Beam connections Maximum moment Beam and load 43 Deflection at centre W L/2 WL3 WL /4 L/2 48 EI W/2 W/2 W 5 WL3 WL /8 384 EI L W/2 W/2 W a Wb L b Wab/L Wa L L WL3 48 EI 3a – 4 a L L 3 W a b L W/2 W/2 W(a/2 + b/8) c W/2 Wa/3 b L W/2 WL/6 W/2 [16a2 – 20ab + 5b2] W/2 L W/2 Wa 120 EI a 2W/L W/2 [8L3 – 4Lb2 + b3] W/2 W/a a W 384 EI W/2 W/2 W/2 L/2 WL/8 L/2 WL3 60 EI W/2 WL3 73.14 EI Figure 4.15 Simply supported beam maximum moments and deflections 4.8.1 Bearing resistance of beam webs The local bearing capacity of the web at its junction with the flange must be checked at supports and at points where loads are applied. The bearing capacity is given in Section 4.5.2 of BS 5950-1: 2000. An end bearing and an intermediate bearing are shown in Figures 4.16(a) and (b), respectively: Pbw = (b1 + nk)tpyw in which: n = 5 for intermediate bearing, n = 2 + 0.6be /k but n ≤ 5 for end bearing, and k is obtained as follows: for a rolled I- or H-section: k = T + r, for a welded I- or H-section: k = T , 44 Beams (a) Clearance b1 nk 1 2.5 z z Check bearing r 45° End bearing on a angle bracket (b) b1 – 5 1 2.5 Check bearing 45° Intermediate bearing Figure 4.16 Web and bracket bearing where b1 is the stiff bearing length, be is the distance to the nearer end of the member from the end of the stiff bearing; pyw is the design strength of the web; r is the root radius; T is the flange thickness: t is the web thickness. The stiff bearing length b1 is defined in Section 4.5.1.3 as the length which cannot deform appreciably in bending. The dispersion of the load is taken as 45◦ through solid material. Stiff bearing lengths b1 are shown in Figure 4.16(a). For the unstiffened angle, a tangent is drawn at 45◦ to the fillet and the length b1 in terms of dimensions shown is: b1 = t + t + 0.8r − clearance where r = radius of the fillet t = thickness of the angle leg Beam connections 45 For the beam supported on the angle bracket as shown in Figure 4.16(a) the bracket is checked in bearing at Section ZZ and the weld or bolts to connect it to the column are designed for direct shear only. If the bearing capacity of the beam web is exceeded, stiffeners must be provided to carry the load (see Section 5.3.7). 4.8.2 Buckling resistance of beam webs Types of buckling caused by a load applied to the top flange are shown in Figure 4.17. The web buckles at the centre if the flanges are restrained, otherwise sideways movement or rotation of one flange relative to the other occurs. The buckling resistance of a web to loads applied through the flange is given in Section 4.5.3 of BS 5950-1: 2000. If the flange through which the load or reaction is applied is effectively restrained against both: (a) rotation relative to the web; Figure 4.17(c) and (b) lateral movement relative to the other flange; Figure 4.17(b), then, provided that the distance αe from the load or reaction to the nearer end of the member is at least 0.7d, the buckling resistance of the unstiffened web should be taken as Px given by: Px = √ 25εt Pbw (b1 + nk)d where d = depth of the web Pbw = the bearing capacity of the unstiffened web at the web-to-flange connection. If the distance αe from the load or reaction to the nearer end of the member is less than 0.7d, the buckling resistance Px of the web should be taken as: Px = 25εt αe + 0.7d Pbw √ 1.4d (b1 + nk)d and b1 , k, n and t are as defined in Section 4.8.1. above. This applies when the flange where the load is applied is effectively restrained against (a) rotation relative to the web and (b) lateral movement (a) (b) (c) Sway Rotation Web buckles Restrained flanges Figure 4.17 Types of buckling Sway between flanges Rotation of flanges 46 Beams relative to the other flange. Where (a) or (b) is not met, the buckling resistance of the web should be reduced to Pxr , given that: Pxr = 0.7d Px LE in which LE is the effective length of the web acting as a compression member. If the load exceeds the buckling resistance of the web, stiffeners should be provided (see Section 5.3.7). 4.8.3 Beam-end shear connections Design procedures for flexible end shear connections for simply supported beams are set out here. The recommendations are from the SCI publication (11). Two types of shear connections, beam to column and beam to beam, are shown in Figures 4.18(a) and (b), respectively. Design recommendations for the end plate are: (1) Length—maximum = clear depth of web, minimum = 0.6 of the beam depth. (2) Thickness—8 mm for beams up to 457 × 191 serial sizes, 10 mm for larger beams. (3) Positioning—the upper edge should be near the compression flange. (b) (a) Beam to column joint (c) t Beam to beam joint (d) θ z L End plate flexes g a θ A Rotation at beam support Figure 4.18 Flexible shear connection z Notched end Examples of beam design 47 Flexure of the end plate permits the beam end to rotate about its bottom edge, as shown in Figure 4.18(c). The end plate is arranged so that the beam flange at A does not bear on the column flange. The end rotation is taken as 0.03 radians, which represents the maximum slope likely to occur at the end of the beam. If the bottom flange just touches the column at A then t/a = 0.03 or a/t should be made < 33 to prevent contact. The joint is subjected to shear only. The steps in the design are: (1) (2) (3) (4) Design the bolts for shear and bearing. Check the end plate in shear and bearing. Check for block shear. Design the weld between the end plate and beam web. If the beam is notched as shown in Figure 4.18(d), the beam web should be checked for shear and bending at section z–z. To ensure that the web at the top of the notch does not buckle, the BCSA manual limits the maximum length of notch g to 24t for Grade S275 steel and 20t for Grade S355 steel, where t is the web thickness. 4.9 Examples of beam design 4.9.1 Floor beams for an office building The steel beams for part of the floor of a library with book storage are shown in Figure 4.19(a). The floor is a reinforced concrete slab supported on universal beams. The design loading has been estimated as: Dead load—slab, self weight of steel, finishes, ceiling, partitions, services and fire protection: = 6.0 kN/m2 Imposed load from Table I of BS 6399: Part 1 = 4.0 kN/m2 Determine the section required for beams 2A and 1B and design the end connections. Use Grade 275 steel. The distribution of the floor loads to the two beams assuming two-way spanning slabs is shown in Figure 4.19: (1) Beam 2A Service dead load =6×3 Service imposed load = 4 × 3 Factored shear = (1.4 × 31.5) + (1.6 × 21) Factored moment = 1.4[(31.5 × 2.5) − (13.5 × 1.5) −(18 × 0.5)] + 1.6[(21 × 2.5) −(9 × 1.5) − (12 × 0.5)] = 18 kN/m, = 12 kN/m, = 77.7 kN, = 122.1 kN m. Beams A (a) B 1A C 1B A1 B1 1 3m 48 2A 2B 3A 3B 3m 2 3 5m 5m Part floor plan and load distribution (b) 9 24 9 13.5 36 13.5 Imposed 21 21 Dead 1.5 m 2.0 m 5.0 m 31.5 1.5 m 31.5 Working loads on beam 2A-kN (c) 18 39 27 42 63 3m 58.5 18 27 Imposed 39 Dead 3m 6m 58.5 Working loads on beam B1-kN Figure 4.19 Library: part floor plan and beam loads Design strength, Grade 275—steel, thickness ≤ 16 mm, py = 275 N/mm2 , M 122.1 × 103 Plastic modulus S = = = 444 cm3 , py 275 Try 356 × 127 UB33 Sx = 539.8 cm3 , Zx = 470.6 cm3 , Ix = 8200 cm4 . The dimensions for the section are shown in Figure 4.20(a). The classification checks from Tables 11 BS 5950-1: 2000 are: ε = (275/py )0.5 = 1.0 b/T = 62.7/8.5 = 7.37 < 9 d/T = 311.1/5.9 = 52.7 < 80 This is a plastic section. The moment capacity is py S ≤ 1.2py Z py Sx = 275 × 539.8 × 10−3 = 148.4 kN m, 1.2py Zx = 1.2 × 275 × 470.6 × 10−3 = 155.3 kN m. The section is satisfactory for the moment. Examples of beam design 125.4 b = 62.7 (a) 49 (b) 82 4 no. 20 mm  bolts t = 5.9 Beam 2A 356 × 127 × 33 UB T = 8.5 d = 311.1 D = 348.5 r = 10.2 Beam B1 457 × 152 × 60 UB Connection of beam 2A to B1 Section dimensions (c) ⬎24t = 141.6 (d) 6 mm fillet weld End notch and plate 213.7 Equal area axis Centroidal axis 104.8 318.5 245.2 215 77.7 kN 73.3 a = 103.5 40 135 40 30 82 8 Section of notch Figure 4.20 Section and end-connection beam 2A The deflection due to the unfactored imposed load using formulae from Figure 4.15 is: δ= 18 × 103 × 1500 120 × 205 × 103 × 8200 × 104 × [16 × 15002 + 20 × 1500 × 2000 + 5 × 20002 ] 24 × 103 384 × 205 × 103 × 8200 × 104 × [8 × 50003 − 4 × 5000 × 20002 + 20003 ] = 1.553 + 3.45 = 5.00 mm. δ/span = 5.00/5000 = 1/1000 < 1/360. + The beam is satisfactory for deflection. The end connection is shown in Figure 4.20(b) and the end shear is 77.7 kN. The notch required to clear the flange and fillet on beam B1 is shown in Figure 4.20(c). The end plate conforms to recommendations given 50 Beams in Section 4.8.3 above. To ensure end rotation: a/t = 103.5/8 = 12.93 < 33 The bolts are 20 mm diameter, Grade 8.8: Single shear value on threads Capacity of four bolts = 91.9 kN, = 4 × 91.9 = 367.6 kN, Bearing capacity of a bolt on 8 mm thick end plate = 73.6 kN, Bearing on the end plate = 73.6 kN. Bolts and end plate are satisfactory in bearing. The web of beam B1 is checked for bearing below: Shear capacity of end plate in shear on both sides Pv = 2 × 0.9 × 0.6 × 275 × 8(215 − 44) × 10−3 = 406.2 kN Provide 6 mm fillet weld in two lengths of 215 mm each. The strength at 0.92 kN/mm = 2(215 − 12) × 0.92 = 374 kN Check the beam end in shear at the notch (see Figure 4.20(d)) Pv = 318 × 5.9 × 0.9 × 0.6 × 275 × 10−3 = 279.0 kN Check the beam end in bending at the notch. The locations of the centroid and equal area axes of the T section are shown in Figure 4.20(d). The elastic and plastic properties may be calculated from first principles. The properties are: Elastic modulus top, Z = 148.5 cm3 , Plastic modulus, S = 263.3 cm3 . Moment capacity assuming a semi-compact section with the maximum stress limited to the design strength: Mc = 148.5 × 275 × 10−3 = 40.8 kN m Factored moment at the end of the notch: M = 77.7 × 70 × 10−3 = 5.44 kN m The beam end is satisfactory. Note that the notch length 70 mm is taken from the Guide to BS 5950-1: 2000, Vol.1, SCI. Examples of beam design 51 (2) Beam B1 The beam loads are shown in Figure 4.21(c). The point load at the centre is twice the reaction of Beam 2A. The triangular loads are: Dead Imposed Factored shear Factored moment Plastic modulus, S Try 457 × 152 UB 60 : = 2 × 1.52 × 6 = 27 kN, = 2 × 1.52 × 4 = 18 kN, = (1.4 × 58.5) + (1.6 × 39) = 144.3 kN, = 1.4[(58.5 × 3) − (27 × 1.5)] + 1.6[(39 × 3) −(18 × 1.5)] = 333 kN m, = 333 × 103 /275 = 1210.9 cm3 . Sx = 1284 cm3 , Zx =1120 cm3 , Ix = 25464 cm4 The section is checked and found to be plastic. The moment capacity is py S ≤ 1.2py Z py Sx = 275 × 1284 × 10−3 = 353.1 kN m 1.2py Zx = 1.2 × 275 × 1120 × 10−3 = 369.6 kN m The section is satisfactory for the moment. Shear capacity Pv = 0.6 × 275 × 8.0 × 454.7 × 10−2 = 600.2 kN (satisfactory). The deflection due to the unfactored imposed loads using formula from Figure 4.15 is: 36 × 103 × 60003 42 × 103 × 60003 + 48 × 205 × 103 × 25464 × 104 73.14 × 205 × 103 × 25464 × 104 = 5.65 mm δ= δ/span = 5.65/6000 = 1/1062 < 1/360 (satisfactory). The end connection is shown in Figure 4.21(a) with the beam supported on an angle bracket 150 × 75 × 10 RSA. Details for the various checks are shown below: (1) Bearing check (see Figure 4.21(c)): Bearing capacity, Pbw = (b1 + nk)tpyw = (23.4 + 60.84) × 8.0 ×275 × 10−3 = 185.3 kN Satisfactory, Reaction = 144.3 kN (2) Buckling check (see Figure 4.21(b)): Px = 25εt αe + 0.7d Pbw √ 1.4d (b1 + nk)d 25(1.0)(8.0) 23.4 + 0.7(407.7) (185) √ 1.4(407.7) (23.4 + 60.84)(407.7) = 107.5 kN. Px = Web stiffener required, (see Chapter 5 for the design of stiffener) Beams (a) Clearance 3 mm 23.5 (b) b1 m = 227.35 23.4 55 23.5 227.35 d = 407.7 d = 454.7 45° r = 11 55 457 × 152 × 60 UB 203 × 203 × 46 UC 150 × 75 × 10 L Connection Web bucking (c) (d) 3 23.4 n2 = 58.75 8.0 45° 1 13.3 10 52 X 45° X 23.5 11 2.5 X X 4 no. 20 mm ø bolts Web bearing 52.5 46.55 52.5 155 Angle bracket in bearing Figure 4.21 End-connection beam B1 (3) Check bracket angle for bearing at Section x-x (see Figures 4.21(d)): Stiff bearing bl = 46.55 mm Length in bearing = 46.55 + 5 × (10 + 11) = 151.6 mm, Bearing capacity = 151.6 × 10 × 275 × 10−3 = 417 kN. (4) Bracket bolts: Provide four No. 20 mm diameter Grade 8.8 bolts. Shear capacity = 4 × 91.9 = 367.6 kN. Satisfactory, the bolts are adequate. (5) Check beam B1 for bolts from 2 No. beams 2A bearing on web: For 8.0 mm thick: Reactions = 2 × 77.7 = 155.4 kN, Bearing capacity of bolts = 4 × 460 × 20 × 8.0 × 10−3 = 294.4 kN. The joint is satisfactory. 4.9.2 Beam with unrestrained compression flange Design the simply supported beam for the loading shown in Figure 4.22. The loads P are normal loads. The beam ends are restrained against torsion with the compression flange free to rotate in plan. The compression flange is unrestrained between supports. Use Grade S275 steel. Examples of beam design P P 1.0 w P 1.5 1.5 53 1.0 5.0 m P = 25 kN dead load 12 kN imposed load W = 2.0 kN/m dead load Figure 4.22 Beam with unrestrained compression flange Factored load = (1.4 × 37.5) + (1.4 × 5) + (1.6 × 18) = 88.3 kN, Factored moment = 1.4(37.5 × 2.5 − 25 × 1.5) + 1.4 × 2 × 52 /8 +1.6(18 × 2.5 − 12 × 1.5) = 130.7 kN m. Try 457 × 152 UB 60. The properties are: ry = 3.23 cm; x = 37.5, u = 0.869, Sx = 1280 cm3 . Note that a check will confirm this is a plastic section. Design strength py = 275 N/mm2 (Table 9, BS 5950) The effective length, LE from Table 13 of BS 5950-1: 2000: LE = 1.0LLT = 5000 mm. √ Equivalent slenderness λLT = uvλ βw λ = 5000/32.3 = 154.8, η = 0.5 and x = 37.5, λ/x = 154.8/37.5 = 4.13. v = 0.855 from Table 19 of BS 5950-1: 2000, λLT = 0.869 × 0.855 × 154.8 × 1.0 = 115. Bending strength, pb = 102 N/mm2 (Table 17, BS 5950) Buckling resistance moment: Mb = 102 × 1280 × 103 = 130.6 kN m mLT = 0.925 Mb /mLT = 130.6/0.925 = 141.2 kN m Shear capacity: Overall depth, D = 454.7 mm, Web thickness, t = 8.0 mm, Pv = 0.6 × 275 × 454.7 × 8 × 10−3 = 600.2 kN, The section is satisfactory. 54 Beams The conservative approach in Section 4.3.7. of BS 5950-1: 2000 gives: LE /ry = 154.8 √ βw = 1.0 2 D/T = 454.7/13.3 = 34.2, pb = 100.0 N/mm —(Table 20, BS 5950), Mb = 100.0 × 1280 × 10−3 = 128.0 kN m, Mb /mLT = 128.0/0.925 = 138.4 kN m. The section is satisfactory. 4.9.3 Beam subjected to bending about two axes A beam of span 5 m with simply supported ends not restrained against torsion has its major principal axis inclined at 30◦ to the horizontal, as shown in Figure 4.23. The beam is supported at its ends on sloping roof girders. The unrestrained length of the compression flange is 5 m. If the beam is 457 × 152 UB 52, find the maximum factored load that can be carried at the centre. The load is applied by slings to the top flange. Let the centre factored load = W kN. The beam self weight is unfactored. Moments Mx = [W × 5/4 + (1.4 × 52 × 9.81 × 52 × 10−3 )/8] cos 30◦ = 1.083 W + 1.933 My = Mx tan 30◦ = 0.625 W + 1.116. Properties for 457 × 152 UB 52: Sx = 1090 cm3 , Zy = 84.6 cm3 , x = 43.9, u = 0.859. ry = 3.11 cm, The section is a plastic section. The design strength py = 275 N/mm2 (Table 9, BS 5950). (1) Moment capacity for x–x axis Effective length: the ends are torsionally unrestrained and free to rotate in plan and the load is destabilizing. (Refer to Table 13 of BS 5950.) Y X 2.5 m X 30° Y Figure 4.23 Beam in biaxial bending 2.5 m Examples of beam design 55 LE = 1.4LLT + 2D, LE = 1.4(5000) + 2 × 449.8 = 7899.6 mm. Slenderness λ = 7899.6/31.1 = 254.0. The load is destabilizing, mLT = 1.0 (Clause 4.3.6.6 BS 5950). η = 0.5, uniform I section. λ/x = 254.0/43.9 = 5.79. v = 0.778 (Table 19 of BS 5950). Equivalent slenderness: λLT = 0.859 × 0.768 × 254 × 1.0 = 167.6. Bending strength, pb = 55.5 N/mm2 (Table 16 of BS 5950). Buckling resistance moment, Mb = 55.5 × 1090 × 10−3 = 60.5 kN m. (2) Biaxial bending The capacity in biaxial bending is determined by the buckling capacity at the centre of the beam (see Section 4.5.4). The interaction relationships to be satisfied are: my My mx Mx + ≤1 py Zx py Zy my My mLT MLT + ≤1 Mb py Zy The moment capacity for the x–x axis: py Zx = 275 × 950 × 10−3 = 261.3 kN m. The moment capacity for the y–y axis: py Zy = 275 × 84.6 × 10−3 = 23.3 kN m. Factor mx and my = 0.9 (Table 26 BS 5950) 0.9(1.083W + 1.933) 0.9(0.625W + 1.116) + = 1, 261.3 23.3 (4.1) W = 34.1 kN 1.0(1.083W + 1.933) 0.9(0.625W + 1.116) + = 1. 60.5 23.3 W = 22.0 kN (4.2) 56 Beams Clearly, Equation (2) is more critical, therefore the maximum that the beam can be carried at the centre is 22.0 kN. 4.10 Compound beams 4.10.1 Design considerations A compound beam consisting of two equal flange plates welded to a universal beam is shown in Figure 4.24. (1) Section classification Compound sections are classified into plastic, compact, semi-compact and slender in the same way as discussed for universal beams in Section 4.3. However, the compound beam is treated as a section built up by welding. The limiting proportions from Table 11 of BS 5950-1: 2000 for such sections are shown in Figure 6 of the code. The manner in which the checks are to be applied set out in Section 3.5.3 of the code is as follows: (1) Whole flange consisting of flange plate and universal beam flange is checked using b1 /T , where b1 is the total outstand of the compound beam flange and T the thickness of the original universal beam flange. (2) The outstand b2 of the flange plate from the universal beam flange is checked using b2 /Tf , where Tf is the thickness of the flange plate. (3) The width/thickness ratio of the flange plate between welds b3 /Tf is checked, where b3 is the internal width of the universal beam flange. (4) The universal beam flange itself and the web must also be checked as set out in Section 4.3. (2) Moment capacity The area of flange plates to be added to a given universal beam to increase the strength by a required amount may be determined as follows. This applies to a restrained beam (see Figure 4.25(a)): Total plastic modulus required: Sx = M/py where M is the applied factored moment. b2 D T T = Tf Tf b1 b3 Compound beam Figure 4.24 Compound beams fabricated by welding Compound beams (a) 57 2 Tf D (D + Tf ) Tf B Compound beam and flange plates (b) Theoretical cut-off Actual cut-off w A X B P wL 2 wL 2 L Curtailment of flange plates B 2 Fillet welds (D + Tf) D/2 Tf (c) Flange weld Figure 4.25 Compound beam design If SUB is the plastic modulus for the universal beam, the additional plastic modulus required is: Sax = Sx − SUB = 2BTf (D + Tf )/2 where B × Tf is the flange area and D is the depth of universal beam. Suitable dimensions for the flange plates can be quickly established. If the beam is unrestrained, successive trials will be required. (3) Curtailment of flange plates For a restrained beam with a uniform load the theoretical cut-off points for the flange plates can be determined as follows (see Figure 4.25(b)): The moment capacity of the universal beam: MUB = py SUB ≤ 1.2py ZUB where ZUB is the elastic modulus for the universal beam. Equate MUB to the moment at P a distance x from the support: wLx/2 − wx 2 /2 = MUB where w is the factored uniform load and L the span of the beam. 58 Beams Solve the equation for x. The flange plate should be carried beyond the theoretical cut-off point so that the weld on the extension can develop the load in the plate at the theoretical cut-off. (4) Web The universal beam web must be checked for shear. It must also be checked for buckling and crushing if the beam is supported on a bracket or column or if a point load is applied to the top flange. (5) Flange plates to universal beam welds The fillet welds between the flange plates and universal beam are designed to resist horizontal shear using elastic theory (see Figure 4.25(c)). Horizontal shear in each fillet weld: Fv BTf (D − Tf ) 4Ix where Fv is the factored shear and Ix is the moment of inertia about x–x axis. The other terms have been defined above. The leg length can be selected from Table 10.5. In some cases a very small fillet weld is required, but the minimum recommended size of 6 mm should be used. Intermittent welds may be specified, but continuous welds made automatically are to be preferred. These welds considerably reduce the likelihood of failure due to fatigue or brittle fracture. 4.10.2 Design of a compound beam A compound beam is to carry a uniformly distributed dead load of 400 kN and an imposed load of 600 kN. The beam is simply supported and has a span of 11 m. Allow 30 kN for the weight of the beam. The overall depth must not exceed 700 mm. The length of stiff bearing at the ends is 215.9 mm where the beam is supported on 203 × 203 UC 71 columns. Full lateral support is provided for the compression flange. Use Grade S275 steel. (1) Design the beam section and check deflection, assuming a uniform section throughout. (2) Determine the theoretical and actual cut-off points for the flange plates and the possible saving in weight that would result if the flange plates were curtailed. (3) Check the web for shear, buckling and bearing, assuming that plates are not curtailed. (4) Design the flange plate to universal beam welds. (1) Design of the beam section The total factored load carried by the beam = 1.4(400 + 30) + (1.6 × 600) = 1562 kN (i.e. 142 kN/m). Maximum moment = 1562 × 11/8 = 2147.8 kN m. Compound beams (a) 59 142 kN/m = 1562 kN A X P 781 kN B 11 m Loading 781 kN 781 kN Shear force diagram 781 kN 2147.8 kNm Bending moment diagram Loading, shear force and bending moment diagrams b1 = 150.0 b2 = 34.95 Tf = 25 (b) 667 617 d = 547.2 T = 22.1 59.9 b = 115.05 Tf = 25 b3 = 230.1 300 59.9 t = 13.1 Beam section Figure 4.26 Compound beam The loading, shear force and bending moment diagrams are shown in Figure 4.26(a). Assume that the flanges of the universal beam are thicker than 16 mm: py = 265 N/mm2 (from Table 9, BS 5950) Plastic modulus required, Sx = 2147.8 × 103 /265 = 8104.9 cm3 . Try 610 × 229 UB 140, where Sx = 4146 cm3 . The beam section is shown in Figure 4.26(b). The additional plastic modulus required: = 8104.9 − 4146 = 3958.9 cm3 = 2 × 300 × Tf (617 + Tf )/(2 × 103 ), where the flange plate thickness Tf is to be determined for a width of 300 mm. This reduces to: Tf2 + 617Tf − 13196 = 0. 60 Beams Solving gives Tf = 20.69 mm. Provide plates 300 mm × 25 mm. The total depth is 667 mm (satisfactory). Check the beam dimensions for local buckling: ε = (275/265)0.5 = 1.02. Universal beam (see Figure 4.4): Flange: b/T = 115.1/22.1 = 5.21 < 9.0 × 1.02 = 9.18, Web: d/t = 547.2/13.1 = 41.7 < 80 × 1.02 = 81.6. Compound beam flange (see Figure 4.25): Flange b1 /T = 150/22.1 = 6.79 < 1.02 × 8.0 = 8.16 b2 /Tf = 34.95/25 = 1.40 < 8.16 b3 /Tf = 230.1/25 = 9.2 < 28 × 1.02 = 28.56. The section meets the requirements for a plastic section. The moment of inertia about the x–x axis for the compound section is calculated. Note for the universal beam: Ix = 111844 cm3 , Ix = 111844 + 2 × 30 × 2.5 × 32.12 + 2 × 30 × 2.52 /12 = 266483 cm4 . The deflection due to the unfactored imposed load is 5 × 600 × 103 × 110003 = 19.03 mm 384 × 205 × 103 × 266483 × 104 δ/span = 19.03/11000 = 1/578 < 1/360 (Satisfactory) δ= (2) Curtailment of flange plates Moment capacity of the universal beam: Mc = 4146 × 265 × 10−3 = 1098.7 kN m. Referring to Figure 4.26(a), determine the position of P where the bending moment in the beam is 1098.7 kN m from the following equation: 781x − 142x 2 /2 = 1098.7 This reduces to x 2 − 11x + 15.47 = 0 x = 1.656 m from each end. Compound beams 61 The compound section will be the elastic range at this point with an average stress in the plate for the factored loads 1098.7 × 106 × 321 = 132.4 N/mm2 , 266483 × 104 Force in the flange plate = 132.4 × 300 × 25 × 10−3 = 993 kN. Assume 6 mm fillet weld, strength 0.92 kN/mm from Table 4.5 (see (4) below). Length of weld to develop the force in the plate = [993/(2 × 0.92)] + 6 = 546 mm, Actual cut-off length = 1656 − 546 = 1110 mm. Cut plates off at 1000 mm from each end. Saving in material from curtailment: Area of universal beam = 178.4 cm2 , Area of flange plates = 150 cm2 . Volume of the compound beam with no curtailment of plates = 328.4 × 1100 = 36.12 × 104 cm3 . Volume of material saved = 200 × 150 = 3.0 × 104 cm3 . Saving in material = 8.3% (3) Web in shear, buckling and bearing (1) Shear capacity (see Figure 4.26(b)). This is checked on the web of the universal beam. Pv = 0.6 × 265 × 617 × 13.1 × 10−3 = 1285 kN, Factored shear, Fv = 781 kN. (2) Web bearing (see Figures 4.26(b) and 4.27(a)): Pbw = (b1 + nk)tpyw Pbw = (215.9 + 149.75) × 13.1 × 265 × 10−3 = 1269.3 kN (satisfactory). (3) Web buckling: 25εt αe + 0.7d Pbw √ 1.4d (b1 + nk)d 25 × 1.0 × 13.1 108 + 0.7 × (547.2) Px = 1269.3 √ 1.4 × (547.2) (215.9 + 149.75)547.2 = 595 kN. Px = Web stiffeners required. (see Chapter 5 for the design of stiffener) 62 Beams (a) b1 = 215.9 n1 = 333.5 L Beam 333.5 45° 203 × 203 × 71 UC support Web buckling (b) b1 = 215.9 n2 = 149.75 1 2.5 59.7 End of fillet (c) 321 300 25 Web crushing Flange plate to universal beam weld design Figure 4.27 Bearing and buckling check and flange weld design (4) Flange plate to universal beam weld (see Figure 4.27(b)): Factored shear at support Fv = 781 kN 781 × 300 × 25 × 321 Horizontal shear on two fillet welds = 2 × 266483 × 104 = 0.353 kN/mm. Provide 6 mm fillet welds, strength 0.92 kN/mm. This is the minimum size weld to be used. 4.11 Crane beams 4.11.1 Types and uses Crane beams carry hand-operated or electric overhead cranes in industrial buildings such as factories, workshops, steelworks, etc. Types of beams used are shown in Figures 4.28(a) and (b). These beams are subjected to vertical and horizontal loads due to the weight of the crane, the hook load and dynamic loads. Because the beams are subjected to horizontal loading, a larger flange or horizontal beam is provided at the top on all but beams for very light cranes. Light crane beams consist of a universal beam only or of a universal beam and channel, as shown in Figure 4.28(a). Heavy cranes require a plate girder Crane beams (a) 63 (b) Channel Surge girder and walkway Universal beam Universal beam Plate girder Heavy crane girders Light crane beams (c) Light rail Standard rail Heavy rail Parker Clip Crane rails and fixing crane rail to beam (d) Connection-crane beam to column Figure 4.28 Type of crane beams and rails and connection to column with surge girder, as shown in Figure 4.28(b). Only light crane beams are considered in this book. Some typical crane rails and the fixing of a rail to the top flange are shown in Figure 4.28(c). The connection of a crane girder to the bracket and column is shown in Figure 4.28(d). The size of crane rails depends on the capacity and use of the crane. 4.11.2 Crane data Crane data can be obtained from the manufacturer’s literature. The data required for crane beam design are: Crane capacity Span Weight of crane Weight of the crab End carriage wheel centres Minimum hook approach Maximum static wheel load The data are shown in Figure 4.29. 64 Beams End carriage wheel centres Crane bridge Hook load Minimum hook approach Span Wheel loads Figure 4.29 Crane design data (1) Loads on crane beams Crane beams are subjected to: (1) Vertical loads from self weight, the weight of the crane, the hook load and impact; and (2) horizontal loads from crane surge. Cranes are classified into four classes in BS 2573: Rules for Design of Cranes, Part 1: Specification for Classification, Stress Calculations and Design Criteria for Structures. The classes are: Class 1—light. The safe working load is rarely hoisted; Class 2—moderate. The safe working load is hoisted fairly frequently; Classes 3 and 4 are heavy and very heavy cranes. Only beams for cranes of classes 1 and 2 are considered in this book. The dynamic loads caused by these classes of cranes are given in BS 6399: Part 1, Section 7. The loading specified in the code is set out below. The following allowances shall be deemed to cover all forces set up by vibration, shock from slipping of slings, kinetic action of acceleration and retardation and impact of wheel loads: (1) For loads acting vertically, the maximum static wheel loads shall be increased by the following percentages: For electric overhead cranes: 25% For hand-operated cranes: 10% (2) The horizontal force acting transverse to the rails shall be taken as a percentage of the combined weight of the crab and the load lifted as follows: For electric overhead cranes: 10% For hand-operated cranes: 5% (3) The horizontal force acting along the rails shall be taken as 5 per cent of the static wheel loads for either electric or hand-operated cranes. The forces specified in (2) or (3) may be considered as acting at rail level. Either of these forces may act at the same time as the vertical load. The load factors to be used with crane loads given in Table 2 in the code are: Vertical or horizontal crane loads considered separately: γf = 1.6 Vertical and horizontal crane loads acting together: γf = 1.4 The application of these clauses will be shown in an example. Crane beams (a) (b) 65 Maximum moment Wheel loads Wheel loads = Spacing Span Maximum shear Center of gravity of loads = Beam E Maximum moment Figure 4.30 Rolling loads: maximum shear and moment (2) Maximum shear and moment The wheel loads are rolling loads, and must be placed in position to give maximum shear and moment. For two equal wheel loads: (1) The maximum shear occurs when one load is nearly over a support; (2) The maximum moment occurs when the centre of gravity of the loads and one load are placed equidistant about the centre line of the girder. The maximum moment occurs under the wheel load nearest the centre of the girder. The load cases are shown in Figure 4.30. Note that if the spacing between the loads is greater than 0.586 of the span of the beam, the maximum moment will be given by placing one wheel load at the centre of the beam. 4.11.3 Crane beam design (1) Buckling resistance moment for x–x axis Section properties A crane girder section consisting of a universal beam and channel is shown in Figure 4.31(a) and the elastic properties for a range of sections are given in the Structural Steelwork Handbook. The plastic properties are calculated as follows for the plastic stress distribution with bending about the equal area axis shown in Figure 4.31(b). First locate the equal area axis by trial and error and then calculate the positions of the centroids of the tension and compression areas. If z is the lever arm between these centroids, the plastic modulus: Sx = Az/2 where A is the total area of cross-section. The plastic modulus may also be calculated from the definition. This is the algebraic sum of the first moments of area about the equal area axis. Lateral torsional buckling The code specifies in Section 4.11.3 that no reduction is to be made for the equivalent uniform moment factor mLT = 1.0. The effective length LE = span for a simply supported beam with the ends torsionally restrained and the compression flange laterally restrained but free to rotate on plan. The slenderness λ = LE /ry , where ry = radius of gyration for the whole section about the y–y axis. 66 Beams Centroid compression Y area (a) pb (b) Compression X1 X1 Equal area axis Centroidal axis Lever arm, z X X Tension Centroid tension area pb Y Section Plastic stress distribution (d) Y (e) Area connected A . . hs Approx value y DL (c) Ict X1 Y Ict Section residing horizontal moment X1 Weld-channel to universal beam Values for determining V XR T HR (f) Local compression under wheels Figure 4.31 Column beam design The factors modifying the slenderness are set out in Appendix B.2.4 of the code. The buckling parameter u = 1.0. This may also be calculated from a formula in Appendix B. The torsional index x for a flanged section symmetrical about the minor axis is: x = 0.566 hs (A/J )0.5 where hs is the distance between the shear centres of the flanges. As a conservative approximation, hs may be taken as the distance from the centre of the bottom flange to the centroid of the channel web and universal beam flange, as shown in Figure 4.31(c): A = area of cross-section, J = torsion constant = 1/3( bt 3 + hw tw3 ), b = flange width, t = flange thickness, hw = web depth, tw = web thickness. Crane beams 67 Note that the top flange of the universal beam and channel web act together, so t is the sum of the thicknesses. The width may be taken as the average of the widths of the universal beam flange and the depth of web of the channel: η= Icf Icf + Itf where Icf is the moment of inertia of the top flange about the y–y axis = Ix (channel) +(1/2) Iy (universal beam), Itf is the moment of inertia of the bottom flange about the y–y axis = (1/2) Iy (universal beam). The monosymmetry index ψ for an I- or T-section with lipped flange is: ψ = 0.8(2η − 1)(1 + 0.5DL /D) where D denote overall depth of section, DL the depth of lip and the breadth of channel flange. The slenderness factor: v= 1 [(4η(1 − η) + 0.05(λ/x)2 + ψ 2 )0.5 + ψ]0.5 The modified slenderness: √ λLT = uv · λ · βw The bending strength pb is obtained from Table 17 for welded sections. The buckling resistance moment: Mb = Sx pb This must exceed the factored moment for the vertical loads only including impact with load factor 1.6. (2) Moment capacity for the y–y axis (see Figure 4.31(d)) The horizontal bending moment is assumed to be taken by the channel and top flange of the universal beam. The elastic modulus Zy for this section is given in the Structural Steelwork Handbook. The moment capacity: Mcy = Zy py (3) Biaxial bending check The overall buckling check using the simplified approach is given in Section 4.8.3 of BS 5950-1: 2000. This is: My Mx + ≤1 py Zx py Zy My Mx + ≤1 Mb py Zy 68 Beams Two checks are required: (1) Vertical crane loads with no impact and horizontal loads only with load factor 1.6; and (2) Vertical crane loads with impact and horizontal loads both with load factor 1.4. (4) Shear capacity The vertical shear capacity is checked as for a normal beam (see Section 4.6.2). The horizontal shear load is small and is usually not checked. (5) Weld between channel and universal beam (see Figure 4.31(e)) The horizontal shear force in each weld: F Ay 2Ix where F = factored shear, A = area connected by the weld = area of the channel, y = distance from the centroid of the channel to the centroid of the crane beam Ix = moment of inertia of the crane beam about the x–x axis. The elastic properties are given in the Structural Steelwork Handbook. (6) Web buckling and bearing The web is to be checked for buckling and bearing as set out in Section 4.8. The length to be taken for stiff bearing depends on the bracket construction or other support for the crane beam (for example, if it is carried on a crane column). (7) Local compression under wheels (see Figure 4.31(f)) BS 5950-1: 2000 specifies in Section 4.11.4 that the local compression on the web may be obtained by distributing the crane wheel load over a length: xR = 2(HR + T ) but xR ≤ sw where HR = rail height; Sw = the minimum distance between centres of adjacent wheels; T = flange thickness. Bearing stress = p/(txR ) where p = crane wheel load t = web thickness. This stress should not exceed the design strength of the web pyw . Crane beams P 69 P Self weight w a b c L Deflection at centre PL3 48 EI 3(a+c) L – 4(a3+c3) + 5 wL3 384 EI L3 Figure 4.32 Crane beam deflection 4.11.4 Crane beam deflection The deflection limitations for crane beams given in Table 8 of BS 5950-1: 2000 are quoted in Table 2.2 in this book. These are: (1) Vertical deflection due to static wheel loads = span/600. (2) Horizontal deflection due to crane surge, calculated using the top flange properties alone = span/500. The formula for deflection at the centre of the beam is given in Figure 4.32 for crane wheel loads placed in the position to give the maximum moment. The deflection should also be checked with the loads placed equidistant about the centre of the beam, when a = c in the formula given. 4.11.5 Design of a crane beam Design a simply supported beam to carry an electric overhead crane. The design data are as follows: Crane capacity = 100 kN, Span between crane rails = 20 m, Weight of crane = 90 kN, Weight of crab = 20 kN, Minimum hook approach = 1.1 m, End carriage wheel centres = 2.5 m, Span of crane girder = 5.5 m, Self weight of crane girder = 8 kN. Use Grade S275 steel. (1) Maximum wheel loads, moments and shear The crane loads are shown in Figure 4.33(a). The maximum static wheel loads at A 90 120 × 18.9 = + = 79.2 kN. 4 20 × 2 The vertical wheel load, including impact = 79.2 + 25% = 99 kN. The horizontal surge load transmitted by friction to the rail through four wheels: = 10%(100 + 20)/4 = 3 kN. 70 Beams (a) 20 kN A 79.2 kN/wheel B 100 kN 18.9 m 20 m Crane girder centres 1.1 m Crane loads 99 kN (b) CG loads A 0.875 125.5 kN 2.5 £ 99 kN Self weight 8 kN beam B 0.625 0.625 c 2.125 5.5 m Vertical loads – maximum moment 3 kN 80.5 kN 3 kN A B C 3.62 kN 2.38 kN Horizontal loads – maximun moment 99 kN 99 kN Self weight 8 kN A B 2.5 m 3.0 m 45.25 kN 160.75 kN Loads causing maximum vertical shear Crane beam loads Figure 4.33 Crane and crane beam loads Load factors from Table 2.1: Dead load − self weight γf = 1.4 Vertical and horizontal crane loads considered separately γf = 1.6, Vertical and horizontal crane loads acting together γf = 1.4. The crane loads in a position to give maximum vertical and horizontal moments and maximum vertical shear are shown in Figure 4.33(b). The maximum vertical moments due to dead load and crane loads are calculated separately: Dead load RB = 4 kN, Mc = (4 × 2.125) − (8 × 2.1252 /(5.5 × 2)) = 5.22 kN m. Crane beams 71 Crane load, including impact RB = 99(0.875 + 3.375)/5.5 = 76.5 kN, Mc = 76.5 × 2.125 = 162.6 kN m. Crane loads with no impact Mc = 162.6 × 79.2/99 = 130.1 kN m. The maximum horizontal moment due to crane surge RB = 3(0.875 + 3.375)/5.5 = 2.32 kN, Mc = 2.32 × 2.125 = 4.93 kN m. The maximum vertical shear Dead load RA = 4 kN. Crane loads, including impact RA = 99 + 99 × 3.0/5.5 = 153.0 kN. The load factors are introduced to calculate the design moments and shear for the various load combinations: (1) Vertical crane loads with impact and no horizontal crane load. Maximum moment Mc = (1.4 × 5.22) + (1.6 × 162.6) = 267.5 kN m, Maximum shear FA = (1.4 × 4) + (1.6 × 153.0) = 250.4 kN. (2) Horizontal crane loads and vertical crane loads with no impact. Maximum horizontal moment Mc = 1.6 × 4.93 = 7.89 kN m, Maximum vertical moment Mc = (1.4 × 5.22) + (1.6 × 130.1) = 215.47 kN m. (3) Vertical crane loads with impact and horizontal crane loads acting together. Maximum vertical moment Mc = (1.4 × 5.22) + (1.4 × 162.6) = 234.95 kN m, Maximum horizontal moment Mc = 1.4 × 4.93 = 6.9 kN m. (2) Buckling resistance moment for the x–x axis The following trial section is selected: 457 × 191 UB 74 + 254 × 76 Channel. Referring to Figure 4.34, the equal area axis x–x and centroids of the tension and compression areas are located and the plastic modulus calculated. Computations are shown in the figure. The elastic properties for this crane beam are taken from the Structural Steelwork Handbook, and these are shown in Figure 4.35. Beams z 312.89 465.3 1 3 2 4 X X 9.1 10.9 125.6 X X 457 × 191 × 74 UB 254 × 76 C 1 399.09 9.99 y–1 Compression area 2 Section 190.5 Tension area Simplified section Figure 4.34 Crane beam: plastic properties X 8.1 X X 407.9 X 457.2 254 10.9 20.0 Y 24.65 190.5 Y 74.5 76.2 9.1 24.65 Y Y 254 × 76 C A = 36.03 cm2 IX = 3367 cm4 457 × 191 × 74 UB X Y Y 7.25 Section resisting horizontal moment Crane beam IX = 45983 cm4 IY = 6.2 cm Figure 4.35 Crane beam: elastic properties ZY = 331cm3 IY = 4202cm4 10.52 Y Distance to centroid X 465.3 176.7 Y hZ = 447.53 10.52 IY = 1671 cm4 b Area A Thickness of top flange T = A/b D/T = 24.5 66.21 X X 254 8.1 17.81 –y 1 (b) –y (a) 43.61 145 72 Crane beams 73 1. Locate equal area axis Total area = (25.4 × 0.81) + (6.81 × 2 × 1.09) + (2 × 19.05 × 1.45) + (42.82 × 0.91) = 129.64 cm2 . 64.82 = (25.4 × 0.82) + (19.05 × 1.45) + 2 × 1.09(y − 0.81) + 0.91(y − 1.45 − 0.81) ȳ = 6.618 cm 2. Locate centroids of compression and tension areas Compression area area moments about top No 1 2 3 4 Area 20.54 27.62 12.67 3.97 Sum 64.83 y 0.405 1.535 3.175 4.441 Ay 8.32 42.39 47.07 17.63 115.41 Tension area area moments about bottom No 1 2 3 Area 27.62 35.00 2.18 y 0.725 20.27 38.59 64.80 y1 = 1.78 Ay 20.02 709.5 84.12 813.59 y2 = 12.56 3. Lever arm, Z = 46.53 − 1.78 − 12.56 = 31.19 cm 4. Plastic modulus, Sx = 64.83 × 32.19 = 2086.8 cm3 The bending strength, pb taking lateral torsional buckling into account, is determined: Effective length LE = span = 5500, Slenderness λ = LE /ry = 5500/62 = 88.7, Factors modifying slenderness: Buckling parameter, u = 1.0. This is conservative: the value of 0.81 is calculated from the formula in Appendix B of the code. The slenderness factor v is calculated from the formulae in Appendix B: Icf = Ix (channel) + 21 Iy (UB) = 3367 + 835.5 = 4202.5 cm4 , Itf = 21 Iy (UB) = 835.5 cm4 . η= 4202.5 = 0.834. 4202.5 + 835.5 74 Beams The distance between the shear centres of the flanges: hs = distance from centre of bottom flange to centroid of channel web and universal beam flange = 447.53 mm approximately (see Figure 4.35). Torsion constant: J = 13 [(14.52 × 190.5) + (9.12 × 428.2) + (22.62 × 211.35) + (2 × 10.93 × 76.2)] = 1.18 × 106 mm4 . AreaA = 12964 mm2 . The torsional index x = 0.566 × 447.53(13103/1.18 × 106 )0.5 = 26.5. This compares with D/T = 24.5 from the section table. The monosymmetry index ψ for a T-section with lipped flanges, where DL = depth of the lip = 76.2 mm, D = overall depth = 465.3 mm. ψ = 0.8[(2 × 0.834) − 1][1 + (76.2/2 × 465.3)] = 0.578. The slenderness factor 1 v=  0.5 (4(0.834)(1 − 0.834) + 0.05(88.7/26.5)2 + 0.5782 )0.5 + 0.578 = 0.75 Table 14 gives v = 0.769 for η = 0.834 and λ/x = 88.7/26.5 = 3.34. The equivalent slenderness λLT = 1.0 × 0.769 × 88.7 × 1.0 = 68.2. From Table 17 for welded sections for py = 265 N/mm2 . Top flange thickness = 23.6 mm total pb = 154.3 N/mm2 . The buckling resistance Mb = 2086.8 × 154.3 × 10−3 = 321.9 kN m. Crane beams 75 (3) Moment capacity for the top section for the y–y axis Mcy = 265 × 331 × 10−3 = 87.7 kN m, Zy = 331 cm3 (from the section table). (4) Check beam in bending (1) Vertical moment, no horizontal moment: Mx = 267.5 kN m < Mb = 321.9 kN m. (2) Vertical moment no impact + horizontal moment: My Mx 215.5 7.89 + = 0.76 < 1. + = Mb py Zy 321.9 87.7 (3) Vertical moment with impact + horizontal moment: My Mx 234.9 6.9 + = 0.81 < 1. + = Mb py Zy 321.9 87.7 The crane girder is satisfactory in bending. (5) Shear Capacity (see Section 4.62) Pv = 0.6 × 457.2 × 9.1 × 265 × 10−3 = 611.5 kN, Maximum factored shear = 250.4 kN. (6) Weld between channel and universal beam The dimensions for determining the horizontal shear are shown in Figure 4.36(a). The location of the centroidal axis is taken from the Structural Steelwork Handbook. Horizontal shear force in each weld = 250.4 × 3603 × 158.1 = 0.31 kN/mm. 45983 × 104 Provide 6 mm continuous fillet (weld strength = 0.92 kN/mm). Beams 24.65 76 (b) Channel 407.9 18.6 302.1 45° 158.1 y 9.1 X Centroidal axis X 24.65 176.7 Area = 36.03 cm2 457.2 (a) 73.5 Ixx = 45.983 cm4 Stiff bearing Web buckling Channel to universal beam weld (c) (d) 73.5 Web bearing 65 1 2.5 Rail 175.2 22.6 24.65 135.1 Local compression under wheel Figure 4.36 Diagram for crane beam design (7) Web bearing and buckling Assume a stiffened bracket 200 mm wide. The stiff bearing length allowing 3 mm clearance between the beams is 73.5 mm (see Section 4.8 and Figure 4. 36(c)). Factored reaction = 250.4 kN Bearing capacity = 135.1 × 9.1 × 265 × 10−3 = 325.8 kN ∴ Satisfactory Buckling resistance:  25(1.0)(9.1) 200 + 0.7(407.9) Px = (325.8) √ 1.4(407.9) (135.1)(407.9) = 268.4 kN ∴ Satisfactory (8) Local compression under wheels A 25 kg/m crane rail is used, and the height HR is 65 mm. The length in bearing is shown in Figure 4.36(d): Bearing capacity = 265 × 9.1 × 175.2 × 10−3 = 422.4 kN Factored crane wheel load = 99 × 1.6 = 158.4 kN ∴ Satisfactory Purlins 77 (9) Deflection The vertical deflection due to the static wheel load must not exceed: Span/600 = 5500/600 = 9.17 mm. See Figures 4.32 and 4.33. The horizontal deflection due to crane surge must not exceed: Span/500 = 5500/500 = 11 mm. The vertical deflection at the centre with the loads in position for maximum moment is   3(875 + 2125) 4(8753 + 21253 ) 79.2 × 103 × 55003 − δ= 5500 48 × 205000 × 45983 × 104 55003 5 × 8000 × 55003 = 4.03 + 0.18 = 4.21 mm + 384 × 205000 × 45983 × 104 If the loads are placed equidistant about the centre line of beam, a = c is 1500 mm: δ = 4.29 + 0.18 = 4.47 mm. This gives the maximum deflection. The horizontal deflection due to the surge loads = 4.29 × 3 × 45983 = 1.77 mm. 79.2 × 4202.5 The crane girder is satisfactory with respect to deflection. 4.12 Purlins 4.12.1 Types and uses The purlin is a beam and it supports roof decking on flat roofs or cladding on sloping roofs on industrial buildings. Members used for purlins are shown in Figure 4.37. These are cold-rolled sections, angles, channels joists and structural hollow sections. Cold-rolled sections are now used on most industrial buildings. 78 Beams Cold rolled sections Angle Channel Structural hollow sections Joist Figure 4.37 Section used for purlins and sheeting rails Corrugated sheeting Roof decking Belt Insulation board Sheeting Purlins Joist Decking Top chord of roof turss Ceiling Flat roof construction Cladding for sloping roof Figure 4.38 Roof material and constructions 4.12.2 Loading Roof loads are due to the weight of the roof material and the imposed load. The sheeting may be steel or aluminium corrugated or profile sheets or decking. On sloping roofs, sheeting is placed over insulation board or glass wool. On flat roofs, insulation board, felt and bitumen are laid over the steel decking. Typical roof cladding and roof construction for flat and sloping roofs are shown in Figure 4.38. The weight of roofing varies from 0.3 to 1.0 kN/m2 , including the weight of purlins or joists, and the manufacturer’s literature should be consulted. Purlins carrying sheeting are usually spaced at from 1.4–2.0 m centres. Joists carrying roof decking can be spaced at larger centres up to 6 m or more, depending on thickness of decking sheet and depth of profile. Imposed loading for roofs is specified in BS 6399: Part 3 in Section 4. (1) Flat roofs: On flat roofs and sloping roofs up to and including 10◦ , where access in addition to that necessary for cleaning and repair is provided to the roof, allowance shall be made for an imposed load, including snow of 1.5 kN/m2 measured on plan or a load of 1.8 kN concentrated. On flat roofs and sloping roofs up to and including 10◦ , where no access is provided to the roof other than that necessary for cleaning and repair, allowance shall be made for an imposed load, including snow of 0.6 kN/m2 measured on plan or a load of 0.9 kN concentrated. Purlins 79 (2) Sloping roofs: On roofs with a slope greater than 10◦ and with no access provided to the roof other than that necessary for cleaning and repair the following imposed loads, including snow, shall be allowed for: (a) For a roof slope of 30◦ or less, 0.6 kN/m2 measured on plan or a vertical load of 0.9 kN concentrated; (b) For a roof slope of 60◦ or more, no allowance is necessary. For roof slopes between 30◦ and 60◦ , a uniformly distributed load of 0.6[(60 − α)/30] kN/m2 measured on plan where α is the roof slope. Wind loads are generally upward, or cause suction on all but steeply sloping roofs. In some instances, the design may be controlled by the dead-wind load cases. Wind loads are estimated in accordance with BS 6399 Part 2. The calculation of wind loads on a roof is given in Chapter 8 of this book. 4.12.3 Purlins for a flat roof These members are designed as beams with the decking providing full lateral restraint to the top flange. If the ceiling is directly connected to the bottom flange the deflection due to imposed load may need to be limited to span/360, in accordance with Table 8 of BS 5950-1: 2000. In other cases the code states in Section 4.12.2 that the deflection should be limited to suit the characteristics of the cladding system. 4.12.4 Purlins for a sloping roof Consider a purlin on a sloping roof as shown in Figure 4.39(a). The load on an interior purlin is from a width of roof equal to the purlin spacing S. The load is made up of dead and imposed load acting vertically downwards. A conservative method of design is to neglect the in-plane strength of the roof, resolve the load normal and tangential to the roof surface and design the purlin for moments about the x–x and y–y axes (see Figure 4.39(c)). If a section such as a channel is used where the strength about the y–y axis is much less than that about the x–x axis, a system of sag rods to support the purlin about the weak axis may be introduced, as shown in Figure 4.39(b). The purlin is then designed as a simply supported beam for bending about the x–x axis and a continuous beam for bending about the y–y axis. A more realistic and economic design results if the in-plane strength of the cladding is taken into account. The purlin is designed for bending about the x–x axis with the whole vertical load assumed to cause moment. An angle purlin bent at the full plastic moment about the x–x axis is shown in Figure 4.39(d). Note that the internal resultant forces act at the centroids of the tension and compression areas. These forces cause a secondary moment about the y–y axis. It is assumed in design that the sheeting absorbs this moment. BS 5950-1: 2000 gives the classification for angles in Table 11, where limiting width/thickness ratios are given for legs. The sheeting restrains the angle member so that bending take place about the x–x axis. The unsupported downward leg is in tension in simply supported purlins, but it would be in compression under uplift from wind load or near the supports in continuous purlins. 80 Beams (a) Roof carried by purlin S (b) Sog rods Purlin spacing S Roof trusses Side elevation Section through roof W (c) Y S X W X Normal Tangential Y Conservative design method py Centroid of compression area (d) Equal area axis X X Centroid of tension area py Eccentricity section Stress distribution Angle at full plasticity Figure 4.39 Design of purlins for a sloping roof The moment capacity for semi-compact outstand elements and a conservative value for plastic and compact sections is: Mc = py Zx , Zx = elastic modulus for the x–x axis. 4.12.5 Design of purlins to BS 5950-1: 2000, Section 4.12 The code states that the cladding may be assumed to provide restraint to an angle section or to the face against which it is connected in the case of other sections. Deflections as mentioned above are to be limited to suit the characteristics of the cladding used. Purlins 81 The empirical design method is set out in Section 4.12.4 of the code, and the general requirements are: (1) (2) (3) (4) The member should be of steel to a minimum of grade S275. Unfactored loads are used in the design; The span is not to exceed 6.5 m centre to centre of main supports; If the purlin spans one bay it must be connected by two fasteners at each end; (5) If the purlins are continuous over two or more bays with staggered joints in adjacent lines, at least one end of any single-bay member should be connected by not less than two fasteners. The rules for empirical design of angle purlins are: (1) The roof slope should not exceed 30◦ . (2) The load should be substantially uniformly distributed. Not more than 10 per cent of the total load should be due to other types of load; (3) The elastic modulus about the axis parallel to the plane of cladding should not be less than the larger value of Wp /1800 cm3 or Wq L/2250 cm3 , where Wp is the total unfactored load on one span (kN) due to dead and imposed load and Wq is the total unfactored load on one span (kN) due to dead minus wind load and L is the span (mm). (4) Dimension D perpendicular to the plane of the cladding is not to be less than L/45. Dimension B parallel to the plane of the cladding is not to be less than L/60. The code notes that where sag rods are provided the sag rod spacing may be used to determine B only. 4.12.6 Cold-rolled purlins Cold-rolled purlins are almost exclusively adopted for industrial buildings. The design is to conform to BS 5950: Part 5, Code of Practice for Design of Cold Formed Sections. Detailed design of these sections is outside the scope of this book. The purlin section for a given roof may be selected from manufacturer’s data. Ward Building Components Ltd has kindly given permission for some of their design data to be reproduced in this book. This firm produces complete systems for purlins and cladding rails based on their cold-formed multibeam section. Full information including fixing methods and accessories is given in their Technical Handbook (12). In addition, they have produced the multibeam design software system for optimum design for their purlins and side rails. The multibeam cold-formed section and ultimate loads for double-span purlins for a limited range of purlins are shown in Table 4.2. Notes for use of the table are: (1) The loads tables show the ultimate loads that can be applied. The section self-weight has not been deducted. Loadings have also been tabulated that will produce the noted deflection ratio. (2) The loads given are based on lateral restraint being provided to the top flange by the cladding, (3) The values given are also the ultimate uplift load due to wind uplift. 82 Beams Table 4.2 Ward building components cold-formed purlin—design data Y 20 13 30 Section depth X X Units 14 nominal Y 30 mm Example: Section P175150 Depth = 175 mm;thickness = 1.5 mm Double Span Loads Span m Section Depth D Self wt Kg/m Ult gravity kN Ult uplift kN Def limit L/180 kN 4.5 P145130 P145145 P145155 P145170 P175140 P175150 P175160 P175170 145 145 145 145 175 175 175 175 3.03 3.38 3.62 3.97 3.59 3.85 4.05 4.36 12.90 15.58 17.38 19.93 18.36 20.62 22.52 25.01 10.32 12.47 13.90 15.94 14.69 16.50 18.02 20.01 12.96 14.43 15.46 16.89 21.71 23.26 24.47 26.32 5.0 P145130 P145145 P145155 P145170 P175140 P175150 P175160 P175170 145 145 145 145 175 175 175 175 3.03 3.38 3.62 3.97 3.59 3.85 4.05 4.36 11.76 14.18 15.80 18.10 16.77 18.81 20.51 22.77 9.41 11.34 12.64 14.48 13.42 15.05 16.41 18.22 10.50 11.69 12.52 13.68 17.59 18.84 19.84 21.32 4.12.7 Purlin design examples Example 1. Design of a purlin for a flat roof The roof consists of steel decking with insulation board, felt and rolled-steel joist purlins with a ceiling on the underside. The total dead load is 0.9 kN/m2 and the imposed load is 1.5 kN/m2 . The purlins span 4 m and are at 2.5 m centres. The roof arrangement and loading are shown in Figure 4.40. Use Grade S275 steel. Dead load = 0.9 × 4 × 2.5 = 9 kN, Imposed load = 1.5 × 4 × 2.5 = 15 kN, Design load = (1.4 × 9) + (1.6 × 15) = 36.6 kN, Moment = 36.6 × 4/8 = 18.3 kN, Design strength, py = 275 N/mm2 , Purlins 4m 83 4m 2.5 m Dead load = 9 kN Imposed load = 15 kN Purlin 2.5 m 4m Lattice girder Loading Load on one purlin Part roof plan Figure 4.40 Purlin for a flat roof Modulus required, Zreq′ d = 18.3 × l03 /275 = 66.54 cm3 . Try 127 × 76 joist 13.36 kg/m, Zx = 74.94 cm3 , Ix = 475.9 cm4 . Deflection due to imposed load: 5 × 15 × 103 × 40003 = 12.81 mm, 384 × 205 × 103 × 475.9 × 104 δ/span = 12.81/4000 = 1/312 > 1/360, δ= Increase section to 127 × 76 joist 16.37 kg/m, Ix = 569.4 cm4 . δ/span = 1/373 (satisfactory), Purlin 127 × 76 joist 16.37 kg/m. Example 2. Design of an angle purlin for a sloping roof Design an angle purlin for a roof with slope 1 in 2.5. The purlins are simply supported and span 5.0 m between roof trusses at a spacing of 1.6 m. The total dead load, including purlin weight, is 0.32 kN/m2 on the slope and the imposed load is 0.6 kN/m2 on plan. Use Grade S275 steel. The arrangement of purlins on the roof slope and loading are shown in Figure 4. 41. Dead load on slope = 0.32 × 5 × 1.6 = 2.56 kN, Imposed load on plan = 0.6 × 5 × 1.6 × 2.5/2.69 = 4.46 kN, Design load = (1.4 × 2.56) + (1.6 × 4.46) = 10.72 kN, Moment = 10.72 × 5/8 = 6.7 kN m. Assume that the angle bending about the x–x axis resists the vertical load. The horizontal component is taken by the sheeting. Design strength, py = 275 N/mm2 , Applied moment = moment capacity of a single angle 6.7 × 103 = 275×Zx 84 Beams Cladding Purlin Top chord .6 m 1 1.6 m Dead load = 2.56 kN Imposed load = 5.58 kN 5m Loading Figure 4.41 Angle purlin for a sloping roof Elastic modulus Zx = 24.4 cm3 , Provide 125 × 75 × 8 L × 12.2 kg/m, Zx = 29.6 cm3 . Deflection need not be checked in this case. Example 3. Design using empirical method from BS 5950-1: 2000 Redesign the angle purlin above using the empirical method from Section 4.12.4.The purlin specified meets the requirements for the design rules. Wp = total unfactored dead + imposed load = 7.02 kN, Wq = total unfactored wind + dead load = 3.07 kN 7.02 × 5000 = 19.5 cm3 , Zp = 1800 3.07 × 5000 Zq = = 6.82 cm3 , 2250 Elastic modulus, Z = 19.5 cm3 . Leg length perpendicular to plane of cladding, D = 5000/45 = 111.1 mm, Leg length parallel to plane of cladding, B = 5000/60 = 83.3 mm, Provide 120 × 120 × 8 L × 14.7 kg/m, Zx = 29.5 cm3 . Example 4. Select a cold-formed purlin to meet the above requirements Try purlin section P145130 from Table 4.2. Dead load on slope = 0.32 × 5 × 1.6 = 2.56 kN, Imposed load on plan = 0.6 × 5 × 1.6 × 2.5/2.69 = 4.46 kN, Wind load = 0.7 × 5 × 1.6 = 5.6 kN, Design load (gravity) = (1.4 × 2.56) + (1.6 × 4.46) = 10.72 kN, Design load (uplift) = (1.0 × 2.56) − (1.4 × 5.6) = −5.28 kN. The section is satisfactory and is much lighter than angle section. Sheeting rails 85 4.13 Sheeting rails 4.13.1 Types of uses Sheeting rails support cladding on walls and the sections used are the same as those for the purlins shown in Figure 4.37. 4.13.2 Loading Sheeting rails carry a horizontal load from the wind and a vertical one from self-weight and the weight of the cladding. The cladding materials are the same as used for sloping roofs (metal sheeting on insulation board). Wind loads are estimated using BS 6399: Part 2. For design examples in this section suitable values for wind loads will be assumed. The arrangement of sheeting rails on the side of a building is shown in Figure 4.42(a) and the loading on the rails is shown in Figure 4.42(b). The wind may act in either direction due to pressure or suction on the building walls. (a) Sheeting rail (b) S Cladding and self weight S Wind load S Column Cladding Sheeting rails on side of building (c) Loads on sheeting rail Y X X Cladding Y Angle sheeting rail axes for bending Figure 4.42 Sheeting rails: arrangement and loading 86 Beams 4.13.3 Design of angle sheeting rail Sheeting rails may be designed as beams bending about two axes. It is assumed for angle sheeting rails that the sheeting restrains the member and bending takes place about the vertical and horizontal axes. Eccentricity of the vertical loading (shown in Figure 4.42(b)) is not taken into account. The sheeting rail is fully supported on the downward leg. The outstand leg for simply supported sheeting rails is in compression from dead load and tension or compression from wind load. The moment capacity is (see Section 4.12.4): Mc = py Z where Z is the elastic modulus for the appropriate axis. For biaxial bending: my My mx M x + ≤ 1, py Zx py Zy my My mLT MLT + ≤ 1. Mb py Zy 4.13.4 Design of angle sheeting rails to BS 5950-1: 2000 The general requirements from Section 4.12.4.1 of the code set out for purlins in Section 4.12.4 above must be satisfied. Empirical rules for design of sheeting rails are given in Section 4.12.4.3 of the BS 5950. These state that: (1) The loading should generally be due to wind load and weight of cladding. Not more than 10 per cent should be due to other loads or due to loads not uniformly distributed. (2) The elastic moduli for the two axes of the sheeting rail from Table 28 in the code should not be less than the following values for an angle (see Figure 4.42(c): (a) y–y axis—parallel to plane of the cladding: Z1 > W1 L1 /2250 cm3 , where W1 = unfactored load on one rail acting perpendicular to the plane of the cladding in kN. (This is the wind load.) L1 = span in millimetres, centre to centre of columns. (b) x–x axis—perpendicular to the plane of the cladding: Z2 > W2 L2 /1200 cm3 , where W2 = unfactored load on one railing acting parallel to the plane of the cladding in kN. (This is the weight of the cladding and rail.) L2 = span centre to centre of columns or spacing of sag rods where these are provided and properly supported. (3) The dimensions of the angle should not be less than the following: D—perpendicular to the cladding < L1 /45, B—parallel to the cladding < L2 /60. L1 and L2 were defined above. Sheeting rails (a) 87 Eaves member Maker up Columns 11 Tubular strut Rail 11 X Rail Rail 6.1 m maximum 6.1 m maximum High tensile steel rope diagonal 30° minimum angle Side elevation of building Maximum height X-metal bonded insulation-16.0 m -metal single skin -18.0 m (b) Cold rolled sheeting rail Fixing piece 14 nominal Y 30 mm 30 Section depth X 20 Y 13 X Units Rail fixing Figure 4.43 Cold-formed sheeting rail system 4.13.5 Cold-formed sheeting rails The system using cold-formed sheeting rails designed and marketed by Ward Building Components is described briefly with their kind permission. The rail member is the Multibeam section placed with the major axis vertical. For bay widths up to 6.1 m, a single tubular steel strut is provided to support the rails at mid-span. The strut is supported by diagonal wire rope ties and the cladding system can be levelled before sheeting by adjusting the ties. The system is shown in Figure 4.43. For larger width bays, two struts are provided. The allowable applied wind loads for a limited selection of sheeting rail spans and their ultimate loads are given in Table 4.3. Notes regarding use of the table are given below. The manufacturer’s Technical Handbook should be consulted for full particulars regarding safe wind loads and fixing details for rails, support system and cladding. Notes relating to Table 4.3 are: (1) The loads shown are valid only when the rails and cladding are fixed exactly as indicated by the manufacturer. (2) The loads shown are for positive external wind loads (ultimate pressure) and negative suction loads (ultimate suction). (3) Interpolation of the ultimate loads shown is permissible on a linear basis. 4.13.6 Sheeting rail design examples Example 1. Design of an angle sheering rail A simply supported sheeting rail spans 5 m. The rails are at 1.5 m centres. The total weight of cladding and self weight of rail is 0.32 kN/m2 . The wind 88 Beams 14 nominal Y 30 mm X Units 30 Section depth X 20 Y 13 Table 4.3 Ward Multibeam Cladding Rails Example: Section P145170 Depth = 145 mm;thickness = 1.7 mm Double Span Loads Span m Section Depth D Self wt Kg/m Ult pressure kN Ult suction kN Def limit L/150 kN 4.5 P145130 P145145 P145155 P145170 P175140 P175150 145 145 145 145 175 175 3.03 3.38 3.62 3.97 3.59 3.85 12.898 15.583 17.377 19.928 18.932 20.624 10.318 12.456 13.901 15.942 14.690 16.499 ***** ***** ***** ***** ***** ***** 5.0 P145130 P145145 P145155 P145170 P175140 P175150 P145160 145 145 145 145 175 175 175 3.03 3.98 3.62 3.97 3.59 3.85 4.05 11.763 14.184 15.800 18.098 16.722 18.811 20.513 9.410 11.347 12.640 14.478 19.418 15.049 16.410 ***** 14.029 15.023 16.420 ***** ***** ***** 5.5 P145130 P145145 P145155 P145170 P175140 P175150 145 145 145 145 175 175 3.03 3.38 3.62 3.97 3.59 3.85 10.908 13.012 14.483 16.573 15.430 17.268 8.674 10.410 11.587 13.259 12.344 13.829 10.409 11.594 12.416 13.570 ***** ***** ∗∗∗∗∗ Indicates the load to produce a deflection of Span/150 exceeds ultimate UDL capacity loading on the wall is ± 0.5 kN/m2 . The wind load would have to be carefully estimated for the particular building and the maximum suction and pressure may be different. The sheeting rail arrangement is shown in Figure 4.44(a). Use Grade S275 steel. Vertical load = 0.32 × 1.5 × 5 = 2.4 kN, Horizontal load = 0.5 × 1.5 × 5 = 3.75 kN. The loading is shown in Figure 4.44(b). The load factor γf = 1.4 for a wind load acting with dead load only. (Table 2 of BS 5950-1: 2000). Factored vertical moment, Mcx = 1.4 × 2.4 × 5/8 = 2.10 kN m, Factored horizontal moment, Mcy = 1.4 × 3.75 × 5/8 = 3.28 kN m. Sheeting rails (b) (a) 89 24 kN 1.5 m 5m Vertical loading 3.75 kN Horizontal loading Unfactored loading Sheeting rail (c) Y X X Y Angle rail Figure 4.44 Angle sheeting rail Design strength, py = 275 N/mm2 . Try 100 × 100 × 10 L where Z = 24.6 cm3 . The moment capacity: Mb = Mcy = 0.8 × 275 × 24.6 × 10−3 = 5.41 kN m. The biaxial bending interaction relationship: My 3.28 2.1 Mx + = + = 0.99 < 1.0. Mb Mcy 5.41 5.41 Provide 100 × 100 × 10 L × 15 kg/m. For the outstand leg, blt = 10 compact (Table 11). Example 2. Design using empirical method from BS 5950-1: 2000 Redesign the angle sheeting rail above using the empirical method from BS 5950. Unfactored wind load W1 = 3.75 kN. Elastic modulus Z1 = Zy = 3.75 × 5000/2250 = 8.33 cm3 . Unfactored dead load W2 = 2.4 kN. 90 Beams Elastic modulus Z2 = Zx = 2.4 × 5000/1200 = 10.0 cm3 . Dimensions specified are to be D—perpendicular to cladding < 5000/45 = 111.1 mm, B—parallel to cladding < 5000/60 = 83.3 mm. 120 × 120 × 8 L is the smallest angle to meet all the requirements. Example 3. Select a cold-rolled sheeting rail to meet the following requirements Wind load = ±0.5 kN/m2 , Span = 5.0 m, Spacing = 1.5 m. Try cladding rail section P145130 from Table 4.3. Horizontal load = 0.5 × 1.5 × 5 = 3.75 kN, Design load (pressure or suction) = 1.4 × 3.75 = 5.25 kN. This section is satisfactory. (See Figure 4.43 for the rail support system.) Problems 4.1 A simply supported steel beam of 6.0 m span is required to carry a uniform dead load of 40 kN/m and an imposed load of 20 kN/m. The floor slab system provides full lateral restraint to the beam. If a 457 × 191 UB 67 of Grade S275 steel is available for this purpose, check its adequacy in terms of bending, shear and deflection. 4.2 The beam carries the same loads as in Problem 4.1, but no lateral restraint is provided along the span of the beam. Determine the new size of universal beam required. 4.3 A steel beam of 8.0 m span carries the loading as shown in Figure 4.45. Lateral restraint is provided at the supports and the point of concentrated load (by cross beams). Using Grade S275 steel, select a suitable universal beam section to satisfy bending, shear and the code’s serviceability requirements. DL = 90 kN IL = 70 kN DL = 10 kN/m 80 kNm 4.0 m 4.0 m 8.0 m Figure 4.45 45 kNm Problems 91 4.4 It is required to design a beam with an overhanging end. The dimension and loading are shown in Figure 4.46. The beam has torsional restraints at the supports but no intermediate lateral support. Select a suitable universal beam using Grade S275 steel. P 2m P 3m 2m 7.0 m DL = 5 kN/m IL = 10 kN/m 3.0 m For P, DL = 50 kN IL = 40 kN Figure 4.46 4.5 A 610 × 229 UB 125 is used as a roof beam. The arrangement is shown in Figure 4.47 and the beam is of Grade S275 steel and fully restrained by the roof decking. Check the adequacy of the section in bending and shear and the web in buckling and crushing. PU = 300 kN 610 × 229 UB 125 WU = 150 kN Cap plate B A 3.0 m 3.0 m 254 × 254 × 73 UC 3.0 m Cap plates 270 × 270 × 20 mm thk All grade 43 steel Figure 4.47 4.6 The part floor plan for the internal panel of an office building is shown in Figure 4.48. The floor is precast concrete slabs 125 mm thick supported on 6.0 m All column 203 × 203 UC 60 Similar panels surround bands shown = = 6.0 m = = = = 6.0 m Part office floor plan – internal panel Figure 4.48 92 Beams steel beams. The following loading data may be used: 125 mm concrete slab = 3.0 kN/m2 , Screed finishes = 1.0 kN/m, Partition = 1.0 kN/m2 , Imposed load = 3.0 kN/m2 . Design the floor beams, assuming that the self weight of main beams and secondary beams may be taken as 0.5 and 1.0 kN/m run, respectively. 4.7 A simply supported girder is required to span 7.0 m. The total load including self-weight of girder is 130 kN/m uniformly distributed. The overall depth of the girder must not exceed 500 mm and a compound girder is proposed. If the compression flange has adequate lateral restraint and the two flange plates are not curtailed, carry out the following work: (a) Check that a section consisting of 457×191 UB 98 and two No. 15×250 flange plates is satisfactory; (b) Determine the weld size required for the plate-to-flange weld at the point of maximum shear; (c) If the girder is supported on brackets at each end with a stiff bearing length of 80 mm, check the web shear, buckling and crushing. 4.8 A simply supported crane girder for a 200 kN (working load) capacity electric overhead crane spans 7 m. The maximum static wheel loads from the end carriage are shown in Figure 4.49. It is proposed to use a crane girder consisting of 533 × 210 UB 122 and 305 × 89 × 42 kg/m Channel. The weight of the crab is 40 kN and the self-weight of the girder may be taken as 15 kN. Check the adequacy of the girder section. 160 kN 160 kN 2.8 m 305 × 89 × 42 kg/m channel 533 × 210 UB 122 Crane girder Figure 4.49 4.9 A factory building has combined roof and crane columns at 8 m centres. It is required to install an electric overhead travelling crane. Design the crane girder using simply supported spans between columns. The crane data are as follows: Hook load = 150 kN, Span of crane = 15 m, Weight of crane bridge = 180 kN, Weight of crab = 40 kN, No. of wheels in end carriage = 2, Problems Wheels centres in end carriage = 3 m, Minimum hook approach = 1 m. 93 4.10 Select a suitable size for a simply supported cold-rolled purlin. The purlin span is 5.0 m and the spacing is 1.8 m. The total dead load and imposed load on plan are 0.22 and 0.6 kN/m2 , respectively. Use Table 4.2 in the design. Redesign the purlin using the rules from BS 5950-1: 2000. 5 Plate girders 5.1 Design considerations 5.1.1 Uses and construction Plate girders are used to carry larger loads over longer spans than are possible with rolled universal or compound beams. They are used in buildings and industrial structures for long-span floor girders, heavy crane girders and in bridges. Plate girders are constructed by welding steel plates together to form I-sections. A closed section is termed a ‘box girder’. Typical sections, including a heavy fabricated crane girder, are shown in Figure 5.1(a). (a) Weld Flange Web Stiffener Weld Section Stiffeners Box grider Heavy crane girder Plate girder Sections for fabricated girders (b) End post Stiffener Flange Web Side elevation of a plate girder Figure 5.1 Plate girder construction 94 Design considerations 95 To be competitive and cost effective, the web of a plate girder is made relatively thin compared to rolled section, and stiffeners are introduced to prevent buckling either due to compression from bending or shear. Tension field action is utilized to increase the shear buckling resistance of the thin web. Stiffeners are also required at load points and supports. Thus the side elevation of a plate girder has an array of stiffeners as shown in Figure 5.1(b). 5.1.2 Depth and breadth of flange The depth of a plate girder may be fixed by headroom requirements but it can often be selected by the designer. The depth is usually made from one-tenth to one-twelfth of the span. The breadth of flange plate is made about one-third of the depth. The deeper the girder is made, the smaller are the flange plates required. However, the web plate must then be made thicker or additional stiffeners provided to meet particular design requirements. A method to obtain the optimum depth is given in Section 5.3.4. A shallow girder can be very much heavier than a deeper girder in carrying the same loads. 5.1.3 Variation in girder sections Flange cover plates can be curtailed or single flange plates can be reduced in thickness when reduction in bending moment permits. This is shown in Figure 5.2(a). In the second case mentioned, the girder depth is kept constant throughout. For simply supported girders, where the bending moment is maximum at the centre, the depth may be varied, as shown in Figure 5.2(b). In the past, hog-back or fish-belly girders were commonly used. In modem practice with (a) Flange plate tapered at splice Flange plate Cover plate Curtailed covered plates Web splice Constant depth girder (b) Tapered ends Variable depth girders (c) Haunched ends continuous girder Figure 5.2 Variation in plate girder sections Fish belly Hog-back 96 Plate girders automatic methods of fabrication, it is more economical to make girders of uniform depth and section throughout. In rigid frame construction and in continuous girders, the maximum moment occurs at the supports. The girders may be haunched to resist these moments, as shown in Figure 5.2(c). (a) Beam to girder connections (b) Hanger Loads from columns, beams and hangers (c) Full strength welds Welded and bolted splices (d) Plate girder end connections Figure 5.3 Plate girder connections and splices Behaviour of a plate girder 97 5.1.4 Plate girder loads Loads are applied to plate girders through floor slabs, floor beams framing into the girder, columns carried on the girder or loads suspended from it through hangers. Some examples of loads applied to plate girders through secondary beams, a column and hanger are shown in Figure 5.3. 5.1.5 Plate girder connections and splices Typical connections of beams and columns to plate girders are shown in Figures 5.3(a) and (b). Splices are necessary in long girders. Bolted and welded splices are shown in Figure 5.3(c) and end supports in Figure 5.3(d). 5.2 Behaviour of a plate girder 5.2.1 Girder stresses The stresses from moment and shear for a plate and box girder in the elastic state are shown in Figure 5.4. The flanges have uniform direct stresses and the web shear and varying direct stress. Plate and box girders are composed of flat plate elements supported on one or both edges and loaded in plane by bending and shear. The way in which the girder acts is determined by the behaviour of the individual plates. 5.2.2 Elastic buckling of plates The components of the plate and box girder under stress can be represented by the four plates loaded as shown in Figure 5.5. The way in which the plates buckle and their critical buckling stresses depend on the edge conditions, dimensions and loading. The buckled plate patterns are also shown in the figure. In all cases, the critical buckling stress can be expressed by the equation: π 2E pcr = K 12 1 − ν 2  2 t b (a) Flange bending stresses Figure 5.4 Stresses in plate and box girders (b) Web shear and bending stresses Plate girders a b b (a) b b 98 Square Long Uniform compression a b (b) Free Uniform compression-one edge free a b (c) Shear stress a b (d) In plane bending stress All edges simply supported except as noted in (b) Figure 5.5 Elastic buckling of plates where K is the buckling coefficient that depends on the ratio of plate length to width a/b (the edge conditions and loading case), E the Young’s modulus, ν the Poisson’s ratio and t the plate thickness. Some values of K for the four plates are shown in Figure 5.6. Note that the plate length a shown is also the stiffener spacing on a plate girder. The critical stress depends on the width/thickness ratio b/t. Limiting values of b/t, where the critical stress equals the yield stress, are also shown in Figure 5.6. These values are for Grade S275 steel for plate up to 16 mm thick, where the yield stress py = 275 N/mm2 . The values form the basis for Class 3 semi-compact section classification given in Table 11 in BS 5950. The web plates of girders are subjected to combined stresses caused by direct bending stress and shear. An interaction formula is used to obtain critical stress combinations. Discussion of this topic is outside the scope of this book, where simplified design procedures given in the code are used. The reader should consult references (13) and Annex H of BS5950-1: 2000. Behaviour of a plate girder Length a = Width b b Plate and load 99 Limiting value Buckling coeffcient, K of b/t for pcr = yield stress 1.0 5.0 4.0 4.0 52.9 51.9 1.0 1.425 0.425 10.3 16.0 9.35 5.35 78.1 60.0 25.6 minimum 23 131.3 124.4 a Free ∞ 1.0 ∞ 1.0 All edges simply supported except as noted Figure 5.6 Buckling coefficients and limiting values of width/thickness ratios (a) (b) Yield stress Actual stress distribution Effective width simplification Figure 5.7 Post-buckling strength: plate in compression 5.2.3 Post-buckling strength of plates (1) Plates in compression The plate supported on two long edges shown in Figure 5.7(a) can support more load on the outer parts following buckling of the centre portion. The behavior can be approximated by assuming that the load is carried by strips at the edge, as shown in Figure 5.7(b). The load is considered to be carried on an effective width of plate. This effective section principle is now used in the design of thin plate members which are classified as Class 4 slender sections (see Section 3.6 of the code). 100 Plate girders A plate supported on one long edge buckles more readily than the plate above and the strength gain is not as great. Stiffeners increase the load that can be carried (see Figure 5.6). (2) Plates in edge bending These plates can sustain load in excess of that causing buckling. Longitudinal stiffeners in the compression region are very effective in increasing the load that can be carried. Such stiffeners are commonly provided on deep plate girders used in bridges (see Figure 5.5). However, it is not so commonly found in building steelworks and hence, longitudinal stiffeners are outside the scope of BS 5950. (3) Plates in shear A strength gain is possible with plates in shear where tension field action is considered. Thin unstiffened plates cannot carry much load after buckling. Referring to Figure 5.6, the critical buckling stress is increased if stiffeners are added. However, the stiffened plate can carry more loads after buckling in the diagonal tension field, as shown in Figure 5.8. The flanges, stiffeners and tension field now act like a truss (14). If the bending strength of the flanges is ignored, the tension field develops between the stiffeners, as shown in Figure 5.8(a). If the flange contribution is included, the tension field spreads as shown in Figure 5.8(b). Failure in the girder panel occurs when plastic hinges form in the flanges and a yield zone in the web, as shown in Figure 5.8(c). Design formulae based on theoretical and experimental work have been developed to take tension field action into account. The design method in the code also includes the flange contribution. The resistance of the web is thus the sum of the elastic buckling strength, the tension field and the flange strength. (a) (b) Tension field in web only (c) Flange contribution included Plastic hinges Yield zone in web Failure mechanism Figure 5.8 Tension field action and failure mechanism Design to BS 5950: Part 1 101 Note that in the internal panels tension fields in adjacent panels support each other. In the end panels, the end post must be designed as a vertical beam supported by the flanges to carry the tension field (see Figures 5.8(a) and (b)). Expressions have been derived for loads on the end post. 5.3 Design to BS 5950: Part 1 5.3.1 Design strength The strength of the thin web may be higher than that of the thicker flange due to the thickness requirements, for example, S275 steel has a design strength of 275 N/mm2 when less than 16 mm thick and 265 N/mm2 when greater than 16 mm thick. The code requires that if the web strength is greater than the flange strength (pyw ≥ pyf ), the flange strength should be used in all calculations, including section classification, except those for shear or forces transverse to the web where the web strength may be used. If the web strength is less than the flange strength (pyw ≤ pyf ), both strengths may be used when considering moment or axial force, but the web strength should be used in all calculations involving shear or forces transverse to the web. 5.3.2 Classification of girder cross-sections The classification of cross-sections from Section 3.5 of BS 5950: Part 1 was given in Section 4.3 (beams). The limiting proportions for flanges and webs for built-up sections from Table 11 in the code are given in Figure 5.9. The limits for welded sections are lower than those for rolled sections because welded sections have more severe residual stresses and fabrication errors can also adversely affect behaviour. (The reader should refer to the code for treatment of Class 4 slender cross-sections.) 5.3.3 Moment capacity If the depth/thickness ratio d/t for the web is less than or equal to 62ε, the web is not susceptible to shear buckling and the moment capacity is determined in the same way as for restrained beams given in Section 4.4.2. The stress distribution is shown in Figure 5.10(a). If the depth/thickness ratio d/t for the web is greater than 62ε, the web is susceptible to shear buckling. The post-buckled shear resistance of the web is defined as the simple shear buckling resistance, Vw = dtqw . The shear buckling strength qw is given in Table 21 or Annex H.1 of the code and depends on the d/t of the web and a/d of the web panel. When the applied shear reaches this level, the web will already buckle. Although the web is still capable of carrying further shear in its buckled state, its ability to take part in resisting bending moment or longitudinal compression will be reduced. Therefore, the moment capacity of the section will depend on the level of applied shear and may be obtained using one of the following methods: (1) Low shear: If the applied shear is less than or equal to 60 per cent of the simple shear buckling resistance Vw , then it will not cause shear buckling Plate girders b d T T b d 102 t t T T t Class of section Type of element Class 1 Plastic Class 2 Compact Class 3 Semi-compact Outstand element of compression flange b T 8 9 13  Internal element of compression flange b T 28  32  40  Web with neutral axis at mid depth b T 80  100  120  ε = (275/py)0.5 Figure 5.9 Classification of girder cross-sections and the moment capacity is determined in the usual way as for restrained beams, (2) High shear—flange only method: If the applied shear is greater than 60 per cent Vw , the web is designed for shear only and the flanges are not Class 4 slender, then the moment capacity may be obtained by assuming that the moment is resisted by the flanges alone with each flange subject to a uniform stress not exceeding pyf . (3) High shear—general method: If the applied shear is greater than 60 per cent Vw and the moment does not exceed the low shear moment capacity given in (a), then the moment capacity may be based on the capacity of the flanges plus the capacity of the web. Checks on the web contribution should be carried out to Annex H of the code. Only method (2), i.e. flange-only method will be considered further in this book. The stress distribution in bending for this case is shown in Figure 5.10(b). The moment capacity for a girder with laterally restrained compression flange is: Mc = BT (d + T )pyf where B is the flange, T the flange thickness and d the web depth. For cases where the compression flange is not restrained, lateral torsional buckling may occur. This is treated in the same way that was set out for beams in Section 4.5. The bending strength Pb for welded sections is taken from Table 17 in the code. Design to BS 5950: Part 1 py B T (a) 103 d d py t py T d/t ≤ 62  Whole section resists moment (c) Flanges only resist moment 250 R = 200 150 1500 R = do /t 1000 do Optimum depth do mm 2000 py d/t ≥ 62  t 500 Section 0 10 20 30 40 50 60 Modulus of section Sx × 103 cm3 Optimal depth design chart Figure 5.10 Plate girder stresses and optimum depth 5.3.4 Optimum depth The optimum depth based on minimum area of cross-section may be derived as follows. This treatment applies to a girder with restrained compression flange for a given web depth/thickness ratio. Define terms: d0 = distance between centres of flanges, = d, clear depth of web approximately, R = ratio of web depth/thickness = d0 lt, Af = area of flange, S = plastic modulus based on the flanges only, = A f d0 , Af = S/d0 , Aw = area of web = d0 t = d02 /R, A = total area = 2S/d0 + d02 /R. Differentiate with respect to d0 and equate to zero to give d0 = (RS)1/3 104 Plate girders Curves drawn for depth d0 against plastic modulus S for values of R of 150, 200 and 250 are shown in Figure 5.10(b). For the required value of S = M/pyf the optimum depth d0 can be read from the chart for a given value of R, where M is the applied moment. 5.3.5 Shear buckling resistance and web design (1) Minimum thickness of web This is given in Section 4.4.3 of BS 5950: Part 1. The following two conditions must be satisfied for webs with intermediate transverse stiffeners: (i) Serviceability to prevent damage in handling: Stiffener spacing a > d: t ≥ d/250 Stiffener spacing a ≤ d: t ≥ (d/250)(a/d)0.5 (ii) To avoid the flange buckling into the web. This type of failure has been observed in girders with thin webs:13 Stiffener spacing a > 1.5d: t ≥ (d/250)(pyf /345) Stiffener spacing a ≤ 1.5d: t ≥ (d/250)(pyf /455)0.5 where d is the depth of web, t the thickness of web and pyf the design strength of compression flange. (2) Design for shear buckling resistance The shear buckling resistance of the thin webs with d/t > 62ε is covered in Section 4.4.5 of BS 5950: Part 1 and applies to webs carrying shear only. Those that are used to carry bending moment and/or axial load in addition to shear should be designed to Annex H of the code. Thin webs with intermediate stiffeners may be designed either using the simplified or more exact method. In the simplified method, it is assumed that the flanges play no part in resisting the shear. The shear buckling resistance Vb of the thin web with intermediate transverse stiffeners should be based on the simple shear buckling resistance Vw as given in Section 4.4.5.2 of the code as: Vb = Vw = dtqw where qw is the post-buckled shear buckling strength assuming tension field action and is given in Tables 21 or Annex H.1 of the code. Unlike the old code, the critical shear buckling resistance before utilizing tension field action Vcr is now expressed as equation in Cl.4.4.5.4 or Annex H.2 of the code (see equations given in Section 5.3.7(4)). The more exact method assumes that the flanges can play a part in resisting the shear. The stress in the flange due to axial load and/or bending moment as well as the strength of the flange must therefore be considered. If the flange is fully stressed (ff = pyf ) then the shear buckling resistance is the same as for the simplified method. Design to BS 5950: Part 1 105 If the flanges are not fully stressed (ff ≤ pyf ), the shear buckling resistance may be increased to: Vb = Vw + Vf but Vb ≤ Pv and Vf , the flange-dependent shear buckling resistance is given by: Vf = Pv (d/a) 1 − ff /pyf 2 1 + 0.15 Mpw /Mpf where, ff is the mean longitudinal stress in the flange due to moment, M/(d0 BT ), Mpf the plastic moment capacity of the flange, pyf BT 2 /4, Mpw the plastic moment capacity of the web, pyw td 2 /4, pyf the design strength of the flange and M the applied moment. The dimensions B, T , t, d and d0 defined above are shown in Figure 5.10. 5.3.6 Stiffener design Two main types of stiffeners used in plate girders are: (1) Intermediate transverse web stiffeners: These divide the web into panels and prevent the web from buckling due to shear. They also have to resist direct forces from tension field action and possibly external loads acting as well. (2) Load carrying and bearing stiffeners: These are required at all points where substantial external loads are applied through the flange and at supports to prevent local buckling and crushing of the web. The stiffeners at the supports are also termed ‘end posts’. The design of the end posts to provide end anchorage for tension field action to develop in the end panel is dealt with in Section 5.3.9. Note that other special-purpose web stiffeners are defined in BS 5950: Part 1 in Section 4.5.1.1. Only the types mentioned above will be discussed in this book. 5.3.7 Intermediate transverse web stiffeners Transverse stiffeners may be placed on either one or both sides of the web, as shown in Figure 5.11. Flats are the most common stiffener section used. The requirements and design procedure are set out in Section 4.4.6 of BS 5950: Part 1. Only stiffeners not subjected to any external loads or moments are considered here. The code should be consulted for design of stiffeners subjected to external loads or moments. The design process is: (1) Spacing This depends on: (i) minimum web thickness (see Section 5.3.5(1)); (ii) web shear buckling resistance required. The closer the spacing is, the greater the shear buckling resistance (see Section 5.3.5(2)). 106 Plate girders 4t t Elevation Section Outstand bs Sectional plan ts Stiffened detail Figure 5.11 intermediate transverse web stiffeners (2) Outstand This is given in Section 4.5.1.2 of the code. The outstand should not exceed 19ts ε (see Figure 5.11), where ts is the thickness of stiffener and ε equals (275/py )0.5 . When the outstand is between 13ts ε and 19ts ε, the design is to be based on a core effective section with an outstand of 13ts ε. (3) Minimum stiffness Transverse stiffeners not subjected to any external loads or moments should have a second moment of inertia Is about the centerline of the web not less than Is given by: √ 2: √ for a/d < 2 : for a/d ≥ 3 Is = 0.75dtmin 3 Is = 1.5(d/a)2 dtmin where a is the actual stiffener spacing, d the depth of web and tmin the minimum required web thickness for actual stiffener spacing a. Note that additional stiffness for external loading are stipulated in Section 4.4.6.5 of the code is required where stiffeners are subject to lateral loads or to moments due to eccentricity of transverse loads relative to the web. No increase is needed where transverse loads are in line with the web. Design to BS 5950: Part 1 107 (4) Buckling resistance This check is only required for intermediate stiffeners in webs when tension field action is utilized. The stiffener should be checked for buckling for a force: Fq = V − Vcr < Pq where V is the maximum shear in the web panel adjacent to the stiffener, Vcr the critical shear buckling resistance of the same web panel given by the following: if Vw = Pv if Pv > Vw > 0.72Pv if Vw ≤ 0.72Pv Vcr = Pv Vcr = (9Vw − 2Pv )/7 Vcr = (Vw /0.9)2 /Pv where Pv is the shear capacity of the web panel = 0.6Py dt and Pq the buckling resistance of the intermediate web stiffener (see Section 6.3.8). (5) Connection to web of intermediate stiffeners The connection between each plate and the web is to be designed for a shear of not less than: t 2 /(5bs ) (kN/mm) where t is the web thickness (mm) and bs the outstand of the stiffener (mm). The code states that intermediate stiffeners that are not subject to external forces or moments may be cut-off at about 4t above the tension flange. The stiffeners should extend to the compression flange but need not be connected to it (see Figure 5.11). 5.3.8 Load carrying and bearing stiffeners Load carrying and bearing stiffeners are required to prevent local buckling and crushing of the web due to concentrated loads applied through the flange when the web itself cannot support the load. The capacity of the web alone in buckling and bearing was discussed in the earlier chapter in Sections 5.8.1 and 5.8.2, respectively. The design procedure for these stiffeners is set out in Section 4.5 of BS 5950: Part 1. The process is as follows: (1) Outstand This is the same as set out for intermediate stiffeners in Section 5.3.7 (2). (2) Buckling resistance of stiffeners This is set out in Section 4.5.3.3 of BS 5950: Part 1 (see Figure 5.12(a)). The stiffener is designed as a ‘cruciform’ strut of cross-sectional area As at the centre of the girder where As is the area of stiffener plus 15 times the web thickness on either side of the centre line of the stiffener (= 2bs ts + 30t 2 where bs the density stiffener outstand, ts the stiffener thickness and t the web thickness). 108 Plate girders The radius of gyration is taken about the centroidal axis of the strut area parallel to the web. The effective length Le to be used in calculating the slenderness ratio of the stiffener acting as the strut is: (a) Intermediate transverse stiffeners: Le = 0.7L. (b) Load carrying stiffeners where the flange through which the load is applied is restrained against lateral movement is: (i) where the flange is restrained against rotation in the plane of the stiffener by other elements: Le = 0.7L. (ii) where the flange is not so restrained: Le = 1.0L. where L is the length of stiffener. Note that the code states that if no effective lateral restraint is provided the stiffener should be designed as part of the compression member applying the load. The design strength py from Table 9 of the code is the minimum for the web or stiffener. The reduction of 20 N/mm2 referred to in the code for welded construction should not be applied unless the stiffeners themselves are welded sections (see Clause 4.5.3.3 in the code). The compressive strength pc is taken from Table 24(c) of the code. The buckling resistance is: for intermediate stiffener for load carrying stiffener Pq = pc As > Fq , Px = pc As > Fx where Fq is the intermediate stiffener force (see Section 6.3.7(4) above) and Fx the external load or reaction. If the load carrying stiffener also acts as an intermediate web stiffener the code states that it should be checked for the effect of combined loads due to Fq and Fx in accordance with Clause 4.4.6.6 of the code. (3) Bearing resistance This is set out in Section 4.5.2.2 of BS 5950: Part 1. Bearing stiffeners should be designed for the applied force Fx minus the bearing capacity of the unstiffened web. The area of stiffener As.net in contact with the flange is the net crosssectional area after allowing for cope holes for welding. The bearing capacity Ps of the stiffener is given by: Ps = As.net py . The area A is shown in Figure 5.12(c). Note that the stiffener has been coped at the top to clear the web/flange weld. Design to BS 5950: Part 1 (a) (b) 15t 109 15t t bs Load, Fx Stiffener cut back ts Area acting as a strut (c) Girder section Bearing area at top of stiffener Figure 5.12 Load-bearing stiffeners (4) Web check between stiffeners It may be necessary to check the compression edge of the web if loads are applied to it direct or through a flange between web stiffeners. A procedure to make this check is set out in Section 4.5.3.2 of BS 5950: Part 1. (The reader is referred to the code.) 5.3.9 End-post design End anchorage should be provided to carry the longitudinal anchor force Hq representing the longitudinal component of the tension field at the end panel of the web with intermediate transverse stiffeners. The end post of a plate girder is provided for this purpose, and may consist of a single or twin stiffeners, as shown on Figure 5.13. The design procedure is set out in Sections 4.4.5.4 and Annex H.4 of BS 5950: Part 1. This is summarized as follows: (1) Sufficient shear buckling resistance is available without having to utilize tension field action. Design the end post as a load carrying and bearing stiffener as set out in Section 5.3.7. (2) End panel and internal panels are designed utilizing tension field action. In addition to carrying the reaction, the end post must be designed as a beam spanning between the flanges. The two cases shown in the figure are discussed below. Further references should be made to the code for the case where the interior panels are designed utilizing tension field action but the end panel is not. 110 Plate girders (a) Load carrying stiffener and end post (b) End post Full strength welds Load carrying stiffener Reaction Reaction 15 t 15 t t t 15 t Strut area for buckling and area resisting moment Strut area for buckling Single stiffener S T B t T Area resisting moment and shear Twin stiffeners Figure 5.13 End-post design (1) Single stiffener end-post (see Figure 5.13(a)) The single stiffener end-post also acts as both load carrying and bearing stiffener. It must be connected by full-strength welds to the flanges. The design is made for: (i) compression due to the vertical reaction, and (ii) in-plane bending moment Mtf due to the anchor force Hq . (2) Twin stiffener end-post (see Figure 5.13(b)) The inner stiffener carries the vertical reaction from the girder. It is checked for bearing at the end and for buckling at the centre (see Section 5.3.7). The end-post is checked as a vertical beam spanning between the flanges of the girder, with the stiffeners forming the flanges of the beam. It is designed to resist a shear force Rtf and a moment Mtf due to the longitudinal component of the tension field anchor force Hq . The moment Mtf induces a tension in the inner stiffener and a compression in the outer end stiffener as these form the lower and upper flanges of the vertical beam. This force Ftf is equal to the Design to BS 5950: Part 1 111 moment divided by the ‘depth S’ of the vertical beam. Thus, the stiffener must also be designed to resist this force, plus any force arising from the reaction of the plate girder. The expressions to derive the shear, moment and compressive force given in the code are: Shear Rtf = 0.75Hq Moment Mtf = 0.15Hq d Force Ftf = Mtf /S The anchor force Hq from the tension field: (i) if the web is fully loaded in shear (Fv ≥ Vw ) Hq = 0.5dtpy (1 − Vcr /Pv )0.5 (ii) if the web is not fully loaded in shear (Fv < Vw )  Fv − Vcr (1 − Vcr /Pv )0.5 Hq = 0.5dtpy Vw − Vcr where, d is the depth of the web, Fv the maximum shear force, Pv the shear capacity, t the web thickness, Vcr the critical shear buckling resistance and Vw the simple shear buckling resistance. The shear capacity of the end-post is: Pv = 0.6py St where S is the length of web between stiffeners and t the web thickness. The shear capacity Pv must exceed the shear from the tension field Rtf . The moment capacity of the end-post at the centre of the girder, assuming that the flanges resist the whole moment, is: Mcx = py BT (S + T ) where B is the stiffener width and T the stiffener thickness. Note that proportions should be selected so that the plates selected are Class 3 semi-compact as a minimum requirement. The moment capacity Mcx must exceed the moment due to tension field action Mtf . The welds between the stiffener and web must be designed to carry the reaction and the shear from the end-post beam action. The application of the design procedure is given in the example in Section 6.4. 5.3.10 Flange to web welds Fillet welds are used for the flange to web welds (see Figure 5.14). The welds are designed for the horizontal shear per weld: = F Ay/2Ix where F is the applied shear, A the area of flange, y the distance of the centroid of A from the centroid of the girder and Ix the moment of inertia of the girder about the x–x axis. 112 Plate girders A y Fillet welds x x Figure 5.14 Flange-to-web weld The fillet weld can be intermittent or continuous, but continuous welds made by automatic welding are generally used. 5.4 Design of a plate girder A simply supported plate girder has a span of 12 m and carries two concentrated loads on the top flange at the third points consisting of 450 kN dead load and 300 kN imposed load. In addition, it carries a uniformly distributed dead load of 20 kN/m, which includes an allowance for self-weight and an imposed load of 10 kN/m. The compression flange is fully restrained laterally. The girder is supported on a heavy stiffened bracket at each end. The material is Grade S275 steel. Design the girder using the simplified method for web first. 5.4.1 Loads, shears and moments The factored loads are: Concentrated loads = (1.4 × 450) + (1.6 × 300) = 1110 kN Distributed load = (1.4 × 20) + (1.6 × 10) = 44 kN/m The loads and reactions are shown in Figure 5.15(a) and the shear force diagram in Figure 5.15(b). The moments are: MC = (1374 × 4)−(44 × 4 × 2) = 5144 kNm ME = (1374 × 6) − (1110 × 2) − (44 × 6 × 3) = 5232 kNm The bending moment diagram is shown in Figure 5.15(c). 5.4.2 Girder section for moment (1) Design for girder depth span/10 Take the overall depth of the girder as 1200 mm and assume that the flange plates are over 40 mm thick. Then the design strength from BS 5950: Part 1, Table 9 for plates is py = 255 N/mm2 . The flanges resist all the moment by a couple with lever arm of, say, 1140 mm, as shown in Figure 5.16(a). The flange area is: = 5232 × 106 = 17 998 mm2 1140 × 255 Make the flange plates 450 × 45 mm2 , giving an area of 20 250 mm2 . The girder section with web plate 10 mm thick is shown in Figure 5.16(b). Design of a plate girder 113 (a) 1110 kN 1110 kN E C 44 kN/m D A B 4m 4m 4m 1374 kN 1374 kN Loading (b) 1 330 1242 1 374 1198 88 1m 88 3.0 m 1198 1374 Shear force diagram shears-kN (c) 1m C E D 5144 5232 5144 A B 1352 Bending moment diagram moments-kNm Figure 5.15 Load, shear and moment diagrams The flange projection b is 220 mm and the ratio: b/T = 220/45 = 4.89. Referring to Table 11 of the code, the ratio:   275 0.5 b = 8.31 = 4.89 ≤ 8ε = 8 T 255 The flanges are Class 1 plastic and the area of cross section is 51 600 mm2 . (2) Design using the optimum depth chart Redesign the girder using the optimum depth chart shown in Figure 5.10. Assume that the flange plates are between 16 and 40 mm thick. Then the Plate girders (b) 1 110 Lever arm 1140 Flange force 450 b= 220 1200 (a) T= 45 114 T t = 10 Flange forces b= 245 500 1440 1500 t = 10 T d0 = 1470 T = 30 (c) Section for depth 1200 mm Section at optimum depth Figure 5.16 Plate girder sections design strength from Table 9 of the code is: py = 265 N/mm2 Plastic modulus Sx = 5232 × 101/265 = 19.74 × 103 cm3 Using curve d0 /t = 150, the optimum depth d0 = 1450 mm Make the depth 1500 mm: 5232 × 106 Flange area = = 13 162 mm2 1500 × 265 Provide flanges 500×30 mm2 giving an area of 15 000 mm2 . The girder section with web plate 10 mm thick is shown in Figure 5.16(c). Note that the actual d0 /t ratio is 144. The flange projection b is 245 mm and the ratio b/T = 245/30 = 8.17. Referring to the limits in Table 11 in the code, the flanges are Class 2 compact. The area of cross-section is 44 400 mm2 . The saving in material compared with the first design is 13.9 per cent. The design will be based on a depth of 1200 mm because of headroom restriction. Design of a plate girder End post Girder Load bearing stiffener Intermediate stiffeners a = 1000 0.9d 115 a = 1333.3 Web = 10 mm 1.2 d 4@1000 = 4000 3@1333.3 = 4000 Figure 5.17 Stiffener arrangement 5.4.3 Design of web (no tension field action, Vcr > Fv ) (1) Minimum thickness of web (Section 4.4.2 of BS 5950) An arrangement for the stiffeners is set out in Figure 5.17. The design strength of the web py = 275 kN/mm2 from Table 9 of BS 5950: Part 1 for plate less than 16 mm thick. The minimum thickness is the greater of: (1) Serviceability. Stiffener spacing a > depth d in the centre of the girder. Web thickness t > 1110/250 = 4.4 mm. (2) To prevent the flange buckling into the web: Stiffener spacing a < 1.5 depth d: Web thickness t ≥ 1110 250  275 455 0.5 = 3.45 mm (2) Buckling resistance of web (Section 4.4.5.2 of BS 5950) Try a 10 mm thick web plate. The buckling resistance is checked for the maximum shear in the end panel: Web depth/thickness ratio d/t = 111 Stiffener spacing/web depth ratio a/d = 100/1110 = 0.9 From Table 21 in the code the shear buckling strength: qw = 143 N/mm2 Shear buckling resistance: Vb = Vw = 143 × 10 × 1110/103 = 1587.3 kN Critical shear buckling resistance: Pv = 0.6Py Av = 0.6 × 275 × 1110 × 10 × 10−3 = 1831 kN 116 Plate girders Since Pv > Vw > 0.72Pv Vcr = (9Vw − 2Pv )/7 = 1517.2 kN Factored applied shear Fv = 1374 kN < 1517.2 kN The stiffener arrangement and web thickness are satisfactory. Since the critical shear buckling resistance Vcr of the stiffened web is sufficient to resist the applied shear force Fv , tension field action is not developed in the web. The design of the intermediate, load carrying and bearing stiffeners, and end-post is therefore greatly simplified as given below. 5.4.4 Intermediate stiffeners (1) Trial size and outstand (Section 4.5.1.2 of BS 5950) Try stiffeners composed of 2 No. 60 × 8 mm2 flats: Design strength py = 275 kN/mm2 (Table 9) Factor ε = 1.0 Outstand 60 < 13 × 8 = 104 mm. (2) Minimum stiffness (Section 4.4.6.4 of BS 5950) The intermediate stiffener is shown in Figure 5.18. The moment of inertia about the centre of the web is: Is = 8 × 1303 /12 = 1.464 × 106 mm4 > 1.5 × 11103 × 83 = 1.05 × 106 mm4 . 10002 When the spacing a = 1 000 mm < √ 2(1100) = 1569.5 mm. (a) (b) 8 130 32 1078 Stiffeners 60 × 8 Stiffener Figure 5.18 Intermediate stiffener 4 no.6 mm fillet welds Section Design of a plate girder 117 Note that t, the minimum required web thickness for spacing a = 1000 mm using tension field action, is 8 mm (see Section 5.5 below). The stiffener is satisfactory with respect to stiffness. In a conservative design, t = 10 mm. Is ≥ 1.5 × 1110 × 103 /10002 = 2.05 × 106 mm4 Stiffeners 70 × 8 mm2 are then required. (3) Connection to web (Section 4.4.6.7 of BS 5950) Shear between each flat and web = 102 /8 × 60 = 0.208 kN/mm on two welds Use 6 mm fillet weld, strength 0.924 kN/mm. Four continuous fillet welds are provided. 5.4.5 Load carrying and bearing stiffener (1) Trial size and outstand Try stiffeners composed of 2 No. 150×15 mm2 plates as shown in Figure 5.19: Outstand 150 < 13 × 15 = 195 mm The stiffener is fully effective in resisting load. (2) Bearing check (Section 4.5.2.2 of BS 5950) The area in bearing at the top of the stiffener is shown in Figure 5.19(b). The stiffeners have been cut back 15 mm to clear the web to flange welds: Design strength of stiffener Pys = 275 N/mm2 As.net = 2 × 15 × 135 = 4050 mm2 Ps = 4050 × 275 × 10−3 = 1113.8 kN > 1110 kN The bearing capacity Ps from the stiffener itself is already sufficient, no needs to include the bearing capacity Pbw from the unstiffened web. (3) Buckling check (Section 4.5.3.3 of BS 5950) The stiffener area at the centre of the girder acting as a strut is shown in Figure 5.19(c). The stiffener properties are calculated from the dimensions shown: A = (2 × 150 × 15) + (300 × 10) = 7500 mm2 Ix = 15 × 3103 /12 = 37.23 × 106 mm4 Rx = (37.23 × 106/8500)0.5 = 66.1 mm 118 (a) Plate girders 150 135 15 (b) (c) 4 no. 6 mm fillet weld 10 310 1 110 15 15 135 135 Stiffeners 150 × 15 X X 150 150 Section Bearing area at top Strut area at centre Figure 5.19 Load carrying and bearing stiffener Assume that the flange is restrained against lateral movement and against rotation in the plane of the stiffeners: Slenderness λ = 0.7 × 1110/66.1 = 11.8 Design strength = 275 N/mm2 (No reduction necessary for welded stiffener) Compressive strength pc = 275 N/mm2 (Table 24 for strut curve c) Buckling resistance: Px = 275 × 7500/103 = 2062.5 kN The size selected is satisfactory. (4) Connection to web Shear between each flat and web: = 1110 102 + = 0.583 kN/mm on two welds. 8 × 150 2 × 1110 Use 6-mm continuous fillet weld, strength is 0.924 kN/mm. Four fillet welds are provided. Note that the bearing area required controls the stiffener size. 5.4.6 End-post (1) Trial size and outstand The trial size for the end-post consisting of a single plate 450 × 15 mm2 is shown in Figure 5.20(a). The end-post is also designed as a load carrying and Design of a plate girder (b) (c) (d) 150 7.5 Core 400 450 1 110 15 192.5 X X 10 (a) 119 2 no. 6 mm fillet weld Section 1374 kN Bearing area at bottom Strut area at centre End reaction Figure 5.20 End-post bearing stiffener because no tension field action is necessary in the end panel, no anchorage and hence no anchor force is developed. Outstand = 220 mm > 13 × 15 = 195 mm < 19 × 15 = 285 mm Base design on a stiffener core 400 mm × 15 mm Design strength = 275 N/mm2 (Table 9) (2) Bearing check The bearing area is shown in Figure 5.20(c): As.net = 15 × 400 = 6000 mm2 Ps = 6000 × 275 × 10−3 = 1650 kN > 1374 kN (satisfactory) (3) Buckling check The area at the centre line acting as a strut is shown in Figure 5.20(d): A = (400 × 15) + (142.5 × 10) = 7425 mm2 Ix = 15 × 4003 /12 = 80 × 106 mm4 rx = (80.0 × 106 /7925)0.5 = 100.4 λ = 0.7 × 1110/100.4 = 7.7 pc = 275 N/mm2 (Table 24 for strut curve c) Px = 275 × 7425/103 = 2041.9 kN Load carried = 1374 kN The size is satisfactory. 120 Plate girders (4) Connection to web Shear between end plate and web: = 1374 = 0.62 kN per weld 2 × 1110 Provide 6-mm continuous fillet, weld strength 0.924 kN/mm. Two lengths of weld are provided. 5.4.7 Flange to web weld See Figure 5.16(b) for the girder dimension: Ix = (450 × 12003 − 440 × 11103 )/12 = 14.65 × 109 mm4 Horizontal shear per weld (see Section 5.3.9): = 1374 × 450 × 45 × 577.5 = 0.548 kN/mm 14.65 × 109 × 2 Provide 6-mm continuous fillet, weld strength 0.924 kN/mm. 5.4.8 Design drawing A design drawing of the girder is shown in Figure 5.21. 5.5 Design utilizing tension field action (Vb = Vw + Vf ) Redesign the web, stiffeners and end post for the girder in Section 5.4 using the more exact method for web (i.e. utilizing tension field action in the web). 5.5.1 Design of the web Try an 8-mm thick web with the stiffeners spaced at 1000 mm in the end 4 m of the girder, as shown in Figure 5.22. The web design is set out in Section 4.4.5.3 of BS 5950: d/t = 1110/8 = 138.75 a/d = 1000/1110 = 0.9 The shear buckling strength from Table 21 in the code: qw = 118 N/mm2 Design utilizing tension field action (Vb = Vw + Vf ) 121 Girder B 4 @ 1000 = 4000 3 @ 1333.3 = 4000 Load A Flanges 450 × 45 End Post Web A Reaction B 1110 × 10 All fillet weld 6mm continuous weld 450 45 Full strength weld Chamfer 15 mm weld 1110 1200 Chamfer 15 mm to clear web to flange weld 32 Stiffener 2 no. 150 × 15 plates Stiffeners 2 no. 60 × 8 plates End plate 450 × 15 45 Do not weld Section A – A Intermediate stiffener Section B – B Load carrying stiffener Section C – C end post Figure 5.21 Design without utilizing tension field action End post Load carryimg stiffener Girder d = 1110 Intermediate stiffeners Web 8 mm a = 1000 0.9d 4 @ 1000 = 4000 3 @ 1333.3 = 4000 Figure 5.22 Stiffener arrangement The shear buckling resistance of the stiffened panel is: Vb = Vw = 118 × 1110 × 8/103 = 1047.8 kN This is less than the applied shear of 1374 kN. The contribution to shear buckling resistance from the flanges is now necessary; hence use the more exact method. 122 Plate girders Vb = Vw + Vf but Vb ≤ Pv The plastic moment capacity of the flange where the design strength of the flange pyf = 255 N/mm2 : Mpf = 255 × 450 × 452 = 58.1 kNm 4 × 106 The plastic moment capacity of the web where the design strength of the web pyw = 275 N/mm2 : Mpw = 275 × 8 × 11102 = 677.6 kNm 4 × 106 The maximum moment in the end panel is 1352 kNm (see Figure 5.15(c)). The mean longitudinal stress in the flange due to moment: 1352 × 106 = 57.8 N/mm2 1155 × 45 × 450 Pv = 0.6Py Av = 0.6 × 275 × 8 × 1110 × 10−3 = 1465.2 kN ff = Since the flanges are not fully stressed (ff < Pyf ), Vb = Vw + Vf but Vb ≤ Pv . The flange-dependent shear buckling resistance: Vf =   2 Pv (d/a) 1 − ff /pyf 1 + 0.15 Mpw /Mpf   1465.2(1100/1000) 1 − (57.8/255)2 = 556.1 kN = 1 + 0.15(677.6/58.1) The total shear buckling resistance: Vb = 1047.8 + 556.1 = 1603.9 ≤ 1465.2 kN This exceeds the applied shear of 1374 kN (hence, satisfactory). Design utilizing tension field action (Vb = Vw + Vf ) 123 Check the web in the panel at 3.0 m from the support: Applied shear = 1242 kN ff = Vf = 3924 × 106 = 167.8 N/mm2 1155 × 45 × 450   2 Pv (d/a) 1 − ff /pyf 1 + 0.15 Mpw /Mpf   1465.2(1100/1000) 1 − (167.8/255)2 = = 330.9 kN 1 + 0.15(677.6/58.1) Total shear buckling resistance: Vb = 1047.8 + 330.9 = 1378.7 ≤ 1465.2 kN The girder is satisfactory for the stiffener arrangement assumed. 5.5.2 Intermediate stiffeners (1) Minimum stiffness Try stiffeners composed of two No. 80×8 mm2 flats (see Section 5.4.4 above). The outstand is satisfactory. The stiffener is shown in Figure 5.23: Is = 8 × 1683 /12 = 3.161 × 106 mm4 > 1.05 × 106 mm4 (2) Buckling check Maximum shear adjacent to the stiffener at 1.0 m from support (see Figure 5.15(b)): V = 1330 kN (a) (b) 8 120 120 Section Figure 5.23 Intermediate stiffener 4 no. 6 mm fillet welds 8 168 Stiffeners 80 × 8 Strut area 124 Plate girders The critical shear buckling resistance of web: Vw = 1047.8 kN Pv = 0.6Py Av = 0.6 × 275 × 1110 × 8 × 10−3 = 1465.2 kN Since Vw ≤ 0.72Pv Vcr = (Vw /0.9)2 /Pv = 924.9 kN Stiffener force Fq = V − Vcr = 1330 − 924.9 = 405.1 kN The stiffener properties are: A = (160 × 8) + (240 × 8) = 3200 mm2 rx = (3.161 × 106 /3200)0.5 = 31.43 mm λ = 0.7 × 1110/31.43 = 24.7 pc = 254 N/mm2 from Table 24(curve c) for py = 255 N/mm2 Buckling resistance: Pq = 254 × 3200/103 = 812.8 kN > Fq The stiffener is satisfactory. (Note: these stiffeners are to extend from flange to flange, not permitted to terminate clear of the tension flange in this case, see clause 4.4.6.7 of the code). (3) Connection to web Provide 6-mm continuous fillet weld. 5.5.3 Load carrying and bearing stiffener Try stiffeners composed of 2 Nos. 150 × 20 mm2 plates as shown in Figure 5.24. The stiffener outstand will be satisfactory and the bearing check will also be adequate (refer to Section 5.4.5 earlier). (a) 15 chamfer 20 Stiffeners 150 × 20 308 8 (b) 120 120 4 no. 6 mm fillet weld Section Figure 5.24 Load carrying and bearing stiffener Strut area Design utilizing tension field action (Vb = Vw + Vf ) 125 Only the buckling check is carried out here: A = (2 × 150 × 20) + (240 × 8) = 7920 mm2 Ix = 20 × 3083 /12 = 48.69 × 106 mm4 rx = (48.69 × 106 /7920)0.5 = 78.4 mm λ = 0.7 × 1110/78.4 = 9.91 py = 275 N/mm2 (Table 9) pc = 275 N/mm2 (Table 24 for curve c) Px = 275 × 7920/103 = 2178 kN Combined external transverse shear force Fx = 1110 + 1198 − 924.9 = 1383.1 kN < Px The size selected for the stiffener is satisfactory. Provide 6-mm continuous fillet weld between stiffeners and web. 5.5.4 End-post The design will be made using twin stiffener end post as shown in Figure 5.25. (a) (b) 470 1000 15 chamfer 1110 End post Bearing stiffener A Reaction A End past (c) Section A – A (d) (e) 20 20 20 450 X Figure 5.25 End-post 2 no. 6 mm fillet welds 8 120 120 Bearing area 20 221 8 206 150 206 X Strut area End post 4 no. 6 mm fillet welds 126 Plate girders (1) Bearing check The reaction is carried on the inner stiffener. The stiffener ends are chamfered to clear the web to flange welds and the bearing area is shown in Figure 5.25(c). pys = 265 N/mm2 (Table 9) As.net = 2 × 20 × 206 = 8240 mm2 Ps = 265 × 8240 = 2183.6 kN > 1374 kN  275 0.5 = 264.9 mm Note that outstand = 221 mm < 13 × 20 265 The full area of the stiffener is effective and the stiffener is satisfactory for bearing.  (2) Buckling check The area at the centre line of the bearing stiffener acting as a strut is shown in Figure 5.25(d): A = (2 × 20 × 221) + (2 × 120 × 8)= 10760 mm2 Ix = 20 × 4503 /12= 151.87 × 106 mm4 rx = (151.87 × 106 /10760)0.5 = 118.8 mm λ = 0.7 × 1110/118.8= 6.54 pc = 265 N/mm2 (Table 27 curve c) Px = 265 × 10760/103 = 2851 kN This exceeds the reaction of 1374 kN. Therefore satisfactory. (3) Shear and moment from the tension field anchor force Refer to Section 4.4.5.4 and Annex H.4 of BS 5950. d/t = 138.75 and a/d = 0.9 From Table 21, shear buckling strength qw = 118 N/mm2 . The simple shear buckling resistance Vw = 118 × 1110 × 8/103 = 1047.8 kN Applied shear force in the end panel: Fv = 1374 kN Since Fv > Vw , the web is fully loaded in shear and the tension field longitudinal anchor force Hq cannot be reduced. This anchor force is: Hq = 0.5 dtpy (1 − Vcr /Pv )0.5 = 0.5 × 1100 × 8 × 275 × 10−3 {1 − (924.9/1465.2)}0.5 = 734.8 kN Design utilizing tension field action (Vb = Vw + Vf ) 127 Shear from the tension field force: Rtf = 0.75 Hq = 551.1 kN Moment in the end-post: Mtf = 0.15 × 734.8 × 1110 = 121.2 kNm 103 (4) Shear capacity of end-post The end-post is shown in Figure 5.25(e). The web 450 × 8 mm2 resists shear (see Table 11 of code for limiting proportions for webs): d/t = 450/8 = 56.25 < 80ε The section is Class 1 Plastic Shear capacity, Pv = 0.6 × 275 × 450 × 8/103 = 594 kN The end-post is satisfactory with respect to shear. (5) Moment capacity Referring to Figure 5.25(e), the flange proportions are: b/T = 221/20 = 11.05 Design strength py = 265 N/mm2 (Table 9 in the code). From Table 11, b/T < 13ε = 13.2. The flanges are Class 3 semi-compact. Moment capacity check at the centre of the girder: Mcx = 265 × 450 × 20 × 470/106 = 1120.9 kNm. This exceeds the moment from tension field action of 121.1 kNm. (6) Additional stiffener force due to moment The moment Mtf induces additional compressive force Ftf in the inner stiffener which must be checked for (see Section H.4.3 of the code). Ftf = Mtf /ae where ae is the centre-to-centre of the twin stiffeners. Ftf = 121.1 × 103 /470 = 257.7 kN The inner twin stiffener is still satisfactory for bearing and buckling with this additional force (see Section 6.5.4 (1) and (2)). 128 Plate girders Load 470 4 @ 1000 = 4000 3 @ 1333.3 = 4000 Girder C B A Flanges 450 × 45 1110 × 8 Web D D C B A All fillet weld 6 mm continuous weld Reaction 45 Chamfer 15 mm to clear web to flange weld 1110 1200 Stiffener 2 no. 80 × 8 plate Full strength weld 20 Stiffener 2 no. 150 × 20 plates 450 20 450 × 20 plate 2 no. 221 × 20 plates Do not weld 45 Section AA load carrying stiffener Section BB intermediate stiffener Section CC end post Section DD end post Figure 5.26 Design utilizing tension field action (7) Weld sizes The four fillet welds shown in Figure 5.25(e) to connect the bearing stiffener to the web is designed first. The welds must support the reaction and the beam shear from the end-post: End-post Ix = (450 × 4903 − 442 × 4503 )/12 = 1055 × 106 mm4 551.1 × 2 × 221 × 20 × 234 1374 + Weld force = 4 × 1110 4 × 1055 × 106 = 0.31 + 0.27 = 0.58 kN/mm Provide 6 mm continuous fillet weld; strength 0.924 kN/mm. This size of weld will be satisfactory for the welds between the end-plate and web. 5.5.5 Design drawing A drawing of the girder designed using the more exact method for the web and utilizing tension field action is shown in Figure 5.26. Problems 5.1 A welded plate girder fabricated from Grade S275 steel is proportioned as shown in Figure 5.27. It spans 15.0 m between centres of brackets and supports a 254 × 254 UC 107 column at mid-span. The loading is shown Problems 129 in the figure. The compression flange is effectively restrained over the span and intermediate stiffeners are provided at 1.875 m centres between supports and the centre load. Assuming that the plate girder and fire-protection casing weigh 20 kN/m, carry out the following design work: (1) Check the adequacy of the section with respect to bending, shear and deflection. (2) Design a suitable load carrying and bearing stiffener for the supports and concentrated load positions. (3) Determine the weld size required at the point of maximum shear. 420 P DL = 900 kN IL = 550 kN 40 7.5m 12 1920 2000 DL = 20 kN/m 40 15 m Figure 5.27 5.2 A welded plate girder of Grade S275 steel carries two concentrated loads transmitted from 254×254 UB 107 columns at the third points. The columns rest on the top flange and the loads are each 400 kN dead load and 250 kN imposed load, respectively. The plate girder is 12 m span and is simply supported at its ends. The compression flange has adequate lateral restraint at the points of concentrated loads and at the supports. Assume that the weight of the girder is 4 kN/m and that the girder is supported on brackets at each end. (1) Assuming that the depth limit is 1400 mm for the plate girder, design the girder section. (2) Design the intermediate, load carrying and bearing stiffeners. (3) Design the web-to-flange weld. (4) Sketch the arrangement and details of the plate girder. 5.3 The framing plans for a four-storey building are shown in Figure 5.28. The front elevation is to have a plate girder at first-floor level to carry wall and floors and give clear access between columns B and C. The plate girder is simply supported with a shear connection between the girder end plates and the column flanges. Columns B and C are 305 × 305 UC 158. The loading from floor, roof and wall is as follows: Dead loads: Front wall between B and C (includes glazing and columns) = 0.7 kN/m2 Floors of r.c. slab: (includes screed, finish, ceilings, etc.) = 6.0 kN/m2 Roof of r.c. slab: (includes screed, finish, ceilings, etc.) = 4.0 kN/m2 130 Plate girders 6@ 4 m = 24 m 1st 5m Plate grider A B Ground floor Stairs, lifts, services Ground-floor plan = 12 m 2nd 3@ 4 m 3rd =12 m 3@ 4 m Roof Bracing C D Front elevation Plan first floor to roof Figure 5.28 Framing plan for a three-storey building Imposed loads: Roof = 1.5 kN/m2 Floors = 2.5 kN/m2 (1) Calculate the loads on the girder. (2) Design the plate girder and show all design information on a sketch. 6 Tension members 6.1 Uses, types and design considerations 6.1.1 Uses and types A tension member transmits a direct axial pull between two points in a structural frame. A rope supporting a load or cables in a suspension bridge are obvious examples. In building frames, tension members occur as: (1) tension chords and internal ties in trusses; (2) tension bracing members; (3) hangers supporting floor beams. Examples of these members are shown in Figure 6.1. The main sections used for tension members are: (1) open sections such as angles, channels, tees, joists, universal beams and columns; (2) closed sections. Circular, square and rectangular hollow sections; (3) compound and built-up sections. Double angles and double channels are common compound sections used in trusses. Built-up sections are used in bridge trusses. Round bars, flats and cables can also be used for tension members where there is no reversal of load. These elements as well as single angles are used in cross bracing, where the tension diagonal only is effective in carrying a load, as shown in Figure 6.1(d). Common tension member sections are shown in Figure 6.2. 6.1.2 Design considerations Theoretically, the tension member is the most efficient structural element, but its efficiency may be seriously affected by the following factors: (1) The end connections. For example, bolt holes reduce the member section. (2) The member may be subject to reversal of load, in which case it is liable to buckle because a tension member is more slender than a compression member. 131 132 Tension members (a) Ties Roof truss (b) Ties Lattice girder (c) (d) Tie Member ineffective Ties Multi-storey building Industrial building (e) Ties Hangers Floor beam Hangers supporting floor beam Figure 6.1 Tension members in buildings (3) Many tension members must also resist moment as well as axial load. The moment is due to eccentricity in the end connections or to lateral load on the member. 6.2 End connections Some common end connections for tension members are shown in Figures 6.3(a) and (b). Comments on the various types are: (1) Bolt or threaded bar. The strength is determined by the tensile area at the threads. End connections 133 (a) Rolled and formed sections (b) Compound and built-up sections Figure 6.2 Tension member sections (a) (b) Angle connections Threaded bar (c) Bolted splice (d) Backing strip Welded splice Figure 6.3 End connections and splices (2) Single angle connected through one leg. The outstanding leg is not fully effective, and if bolts are used the connected leg is also weakened by the bolt hole. Full-strength joints can be made by welding. Examples occur in lattice girders made from hollow sections. However, for ease of erection, most site joints are bolted, and welding is normally confined to shop joints. Site splices are needed to connect together large trusses that have been fabricated in sections for convenience in transport. Shop splices are needed in long members or where the member section changes. Examples of bolted and welded splices in tension members are shown in Figures 6.2(c) and (d). 134 Tension members 6.3 Structural behaviour of tension members 6.3.1 Direct tension The tension member behaves in the same way as a tensile test specimen. In the elastic region: Tensile stress ft = P /A, Elongation δ = P L/AE where P is the load on the member, A the area of cross section and L the length. 6.3.2 Tension and moment: elastic analysis (1) Moment about one axis Consider the I-section shown in Figure 6.4(b), which has two axes of symmetry. If the section is subjected to an axial tension P and moment Mx about the x–x axis, the stresses are: Direct tensile stress ft = P /A, Tensile bending stress fbx = Mx /Zx , Maximum tensile stress fmax = ft + fbx , where Zx is the elastic modulus for the x–x axis. The stress diagrams ace shown in Figure 6.4(b). Define the allowable stresses: pt —direct tension, pbt —tension due to bending. Then the interaction expression ft fbt ≤1 + Pt Pbt gives permissible combinations of stresses. This is shown graphically in Figure 6.4(c). A section with one axis of symmetry may be treated similarly. (2) Moment about two axes If the section is subjected to axial tension P and moments Mx and Mx about the x–x and y–y axes, respectively, the individual stresses and maximum stress are: Direct tensile stress ft = P /A, Tensile bending stress x–x axis fbx = Mx /Zx , Tensile bending stress y–y axis fby = My /Zy , Maximum stress fmax = ft + fbx + fby Zy = elastic modulus for the y–y axis. Structural behaviour of tension members (a) (b) X ft – fbx – fbx ft 135 X P Section ft fbx ft + fbx Direct Bending Combined Stresses (c) Direct tension 1.0 A ft Pt fbx Pt 1.0 Tension due to bending X–X axis Interaction diagram Figure 6.4 Elastic analysis: tension and moment about one axis These stresses are shown in Figures 6.5(b)–(d). The interaction expression to give permissible combinations of stresses is: fby ft fbx + ≤1 + Pt Pbt Pbt This may be represented graphically by the plane in Figure 6.5(e). Sections with one axis of symmetry or with no axis of symmetry which are free to bend about the principal axes can be treated similarly. 6.3.3 Tension and moment: plastic analysis (1) Moment about one axis For a section with two axes of symmetry (as shown in Figure 6.6(a)), the moment is resisted by two equal areas extending inwards from the extreme fibres. The central core resists the axial tension. The stress distribution is shown in Figure 6.6(b) for the case where the tension area lies in the web. At higher loads, the area needed to support tension spreads to the flanges, as shown in Figure 6.6(c). 136 Tension members (a) (b) Y X (c) ft –fbx X P fbx Y Eccentricities Section fby (d) Direct stress Bending stress X–X axis –fby Maximum stress fmax = ft + fbx + fby Bending stress Y–Y axis (e) Direct tension 1.0 ft /Pt fby /Pbt A fbx/Pbt 1.0 1.0 Tension due to bending Y–Y axis Tension due to bending X–X axis Interaction diagram Figure 6.5 Elastic analysis: tension and moments about two axes For design strength py , the maximum tension the section can support is: Pt = py A. If moment only is applied, the section can resist: Plastic moment Mcx = py Sx Elastic moment MEx = py Zx where Sx denotes plastic modulus and Zx the elastic modulus. For values F of tension less than Pt if the tension area is in the web (as shown in Figure 6.6(a), the length a of web supporting F is: a = F /(py t). where t is the web thickness. Structural behaviour of tension members (a) (b) py py py 137 Compression from moment X X Direct tension Tension from moment t py Section Axial tension Tension area in web py Moment Resultant stress Stress distribution (c) py py Compression X X Direct tension Tension Section Resultant stress Tension area spread to flanges (d) Tension 1.0 Elastic stress distribution Linear interaction, M/Mcx Plastic stress distribution, Mrx /Mcx F/Pt Mrx /Mcx1, M/Mcx 1.0 Moment Interaction diagram Figure 6.6 Plastic analysis: tension and moment about one axis The reduced moment capacity in the presence of axial load is: Mrx = (Sx − ta 2 /4)py . A more complicated formula is needed for the case where the tension area enters the flanges, as shown in Figure 6.6(b). The curve of F /Pt against Mrx /Mcx , for an I-section bent about the x–x axis is convex (see Figure 6.6(d)), but a conservative design results if the straight line joining the end points is 138 Tension members adopted. This gives the linear interaction expression: F Mx + =1 Pt Mcx where Mx is the applied moment. The elastic curve is also shown. The strength gain due to plasticity is the area between the two curves. In calculating the reduced moment capacity, it is convenient to use a reduced plastic modulus. This was given for the case above by: Srx = Sx − ta 2 /4. If the average stress on the whole section of area A: f = F /A, then the formula for reduced plastic modulus can be written after substituting for a as: Srx = Sx − n2 A2 /4t where n = f/py . The expression is more complicated when the tension area spreads into the flanges. These are the formulae given in Steelwork Design, Guide to BS 5980: Part 1: Volume 1, to calculate the reduced plastic modulus. The change value of n is given to indicate when the tension area enters the flanges. Note that the reduced plastic modulus is not required if the linear interaction expression is adopted. The analysis for sections with one axis of symmetry is more complicated. (2) Moment about two axes Solutions can be found for sections subject to axial tension and moments about both axes at full plasticity. I-sections with two axes of symmetry have been found to give a convex failure surface, as shown in Figure 6.7. This interaction surface is constructed in terms of: F /Pt , Mrx /Mcx , Mry /Mcy Where F is the axial tension, Pt the tension capacity, Mcx the moment capacity for the x–x axis in the absence of axial load, Mrx the reduced moment capacity for the x–x axis in the presence of axial load, Mcy the moment capacity for the y–y axis in the absence of axial load, and Mry the reduced moment capacity for the y–y axis in the presence of axial load. In practice, Mcy is restricted with some sections. Any point A on the failure surface gives the permissible combination of axial load and moments the section can support. Design of tension members 139 Tension 1.0 F Pt Mrx Mcx Mry Mcy A My Mcy Mx Mcx 1.0 Moment Y–Y axis 1.0 Moment X–X axis Plane for linear interaction expression Interaction diagram Figure 6.7 Plastic analysis: tension and moment about two axes A plane may be drawn through the terminal points on the failure surface. This can be used to give a simplified and conservative linear interaction expression: My Ft Mx + + =1 Pt Mcx Mcy where Mx is the applied moment about the x–x axis and My the applied moment about the y–y axis. 6.4 Design of tension members 6.4.1 Axially loaded tension members The tension capacity is given in Section 4.6.1 of BS 5950: Part 1. This is: Pt = Ae py where Ae is the effective area of the section defined in Sections 3.4.3, 4.6.2 and 4.6.3 of the code. From Section 3.4.3, the effective area of each element of a member is given by: Ae = Ke × net area where holes occur ≤ gross area Ke = 1.2 for Grade S275 and 1.1 for S355 steel (Net area = gross area less holes.) Tests show that holes do not reduce the capacity of a member in tension provided that the ratio of net area to gross area is greater than the ratio of yield strength to ultimate strength. 140 Tension members 6.4.2 Simple tension members (1) Single angles, channels or T-section members connected through one leg These may be designed in accordance with Section 4.6.3 of the code as axially loaded members with an effective area (see Figure 6.3(b)): For bolted connection: Pt = py (Ae − 0.5a2 ) For welded connection: Pt = py (Ag − 0.3a2 ) where a2 equals (Ag − a1 ), where Ag is the gross cross-sectional area and a1 the gross sectional area of the connected leg. (2) Double angles, channels or T-section members connected through one side of a gusset For bolted connection: Pt = py (Ae − 0.25a2 ), For welded connection: Pt = py (Ag − 0.15a2 ). (3) Double angles, channels or T-section members connected to both sides of gusset plates If these members are connected together as specified in the code, they can be designed as axially loaded members using the net area specified in Section 3.3.2 of the code. This is the gross area minus the deduction for holes. 6.4.3 Tension members with moments The code states in Sections 4.6.2 and 4.8.1 that moments from eccentric end connections and other causes must be taken into account in design. Single angles, double angles and T-sections carrying direct tension only may be designed as axially loaded members, as set out in Section 4.6.3 of the code. Design of tension members with moments is covered in Section 4.8.2 of the code. This states that the member should be checked for capacity at points of greatest moment using the simplified interaction expression: My F Mx + + ≤1 Ae py Mcx Mcy where F is the applied axial load, Ae the effective area, Mx the applied moment about the x–x axis, Mcx the moment capacity about the x–x axis in the absence of axial load, and My the applied moment about the y–y axis. Mcy the moment capacity about the y–y axis in the absence of axial load. The interaction expression was discussed in Section 6.3.3(2) above. (See Section 5.4.2 for calculation of Mcx and Mcy .) For bending about one axis, the terms for the other axis are deleted. An alternative expression given in the code takes account of convexity of the failure surface. This leads to greater economy in the design of plastic and compact sections. Design examples 141 6.5 Design examples 6.5.1 Angle connected through one leg Design a single angle to carry a dead load of 70 kN and an imposed load of 35 kN. (1) Bolted connection Factored load = (1.4 × 70) + (1.6 × 35) = 154 kN. Try 80 × 60 × 7 angle connected through the long leg, as shown in Figure 6.8(a). The bolt hole is 22 mm diameter for 20 mm diameter bolts. Design strength from Table 6 in the code py = 275 N/mm2 a1 = net area of connected leg = (76.5 − 22)7 = 381.5 mm2 , a2 = area of unconnected leg = 56.5 × 7 = 395.5 mm2 , Effective area Ae = a1 + a2 = 777 mm2 . Tension capacity: Pt = py (Ae − 0.5a2 ) = 275(777 − 0.5 × 395.5)/103 = 159 kN. The angle is satisfactory. Note that the connection would require either 3 No. Grade 8.8 or 3 No. friction-grip 20 mm diameter bolts to support the load. (2) Welded connection Try 75 × 50 × 6 L connected through the long leg (see Figure 6.8(b)): a1 = 72 × 6 = 432 mm2 , a2 = 47 × 6 = 282 mm2 , Ag = a1 + a2 = 714 mm2 . Tension capacity: Pt = py (Ag − 0.3a2 ) = 275(714 − 0.3 × 282)/103 = 173 kN. The angle is satisfactory. (b) 76.5 3 50 3.5 47 22 φ hole 56.5 3.5 60 (a) 3 80 Bolted connection Figure 6.8 Single angle connected through one leg 72 75 Welded connection 142 Tension members 6.5.2 Hanger supporting floor beams A high-strength Grade S460 steel hanger consisting of a 203 × 203 UC 46 carries the factored loads from beams framing into it and from the floor below, as shown in Figure 6.9(a). Check the hanger at the main floor beam connection. The design strength from Table 9 of BS 5950: Part 1 for sections less than 16 mm thick is: py = 460 N/mm2 . The net section is shown in Figure 6.9(b). For S460 steel the effective section is equal to the net section. The factor Ke from Section 3.4.3 of the code is 1.0. The connection plates are not considered. Check the limiting proportions of the flanges using Table 6a in the code: ε = (275/460)0.5 = 0.773, b/t = 101.6/11 = 9.23 < 15ε = 11.72. Values of b and t are shown in Figure 6.9(b). The section is semi-compact. The moment capacity is calculated using the elastic properties. This can be calculated using first principles, and the properties are: Location of the centroidal axis is shown. Effective area = 53.1 cm2 . Minimum value of elastic modulus Z = 363 cm3 . (a) (b) 2 at 120 kN 120 kN 590 kN X1 Hanger 203 × 203 × 46 UC 590 kN Grade 55 steel X b = 101.6 320 kN 8.76 X All holes 22φ X1 92.84 T = 11 120 kN 305 × 165 × 54 UB 457 × 152 × 74 UB Connection Figure 6.9 High strength hanger Eccentricity 201.6 Hanger-net section and bolt holes 320 kN Problems 143 The moment capacity for the major axis: Mcx = 363 × 460/103 = 166.9 kN m. The applied axial load: F = (2 × 120) + 590 + 320 = 1150 kN m. The applied moment about the x1 –x1 axis: Mx = (320 × 0.21) + (2 × 120 + 590)0.0088 = 74.5 kN m. Substitute into the interaction expression: Mx 1150 × 10 74.5 F + = 0.92 < 1. + = Ae py Mcx 53.1 × 450 166.9 The hanger is satisfactory. Problems 6.1 A tie member in a roof truss is subjected to an ultimate tension of 1000 kN. Design this member using Grade S275 steel and an equal angle section. 6.2 A tension member in Grade S275 steel consists of 2 No. 150×100×8 mm unequal angles placed back to back. At the connection, two rows of 2 No. 22 mm diameter holes are drilled through the longer legs of the angles. Determine the ultimate tensile load that can be carried by the member. 6.3 A tension member from a heavy truss is subjected to an ultimate axial load and bending moment of 2000 kN and 500 kN m, respectively. Design a suitable universal beam section in Grade S275 steel. Assume that the gross section will resist the load and moment. 6.4 A tie member in a certain steel structure is subjected to tension and biaxial bending. The ultimate tensile load was found to be 3000 kN while the ultimate moments about the major and minor axes were 160 kN m and 90 kN m, respectively. Check whether a 305 × 305 UC 158 in Grade S275 steel is adequate. Assume that the gross section resists the loads and moments. 7 Compression members 7.1 Types and uses 7.1.1 Types of compression members Compression members are one of the basic structural elements, and are described by the terms ‘columns’, ‘stanchions’ or ‘struts’, all of which primarily resist axial load. Columns are vertical members supporting floors, roofs and cranes in buildings. Though internal columns in buildings are essentially axially loaded and are designed as such, most columns are subjected to axial load and moment. The term ‘strut’ is often used to describe other compression members such as those in trusses, lattice girders or bracing. Some types of compression members are shown in Figure 7.1. Building columns will be discussed in this chapter and trusses and lattice girders are dealt with in Chapter 8. 7.1.2 Compression member sections Compression members must resist buckling, so they tend to be stocky with square sections. The tube is the ideal shape, as will be shown below. These are in contrast to the slender and more compact tension members and deep beam sections. Bracing strut Struts in truss (b) Wind (a) Crane column Building columns Multi-storey building Industrial building Figure 7.1 Types of compression members 144 Types and uses 145 Fillet welds Universal column Built-up H-section Battened column Box column Crane and building column Strut sections for trusses, lattices, girders and bracing Figure 7.2 Compression member sections Rolled, compound and built-up sections are used for columns. Universal columns are used in buildings where axial load predominates, and universal beams are often used to resist heavy moments that occur in columns in industrial buildings. Single angles, double angles, tees, channels and structural hollow sections are the common sections used for struts in trusses, lattice girders and bracing. Compression member sections are shown in Figure 7.2. 7.1.3 Construction details Construction details for columns in buildings are: (1) (2) (3) (4) beam-to-column connections; column cap connections; column splices; column bases. (1) Beam-to-column cap connections Typical beam-to-column connections and column cap connections are shown in Figures 7.3(a) and (b), respectively. (2) Column splices Splices in compression members are discussed in Section 6.1.8.2 of BS 5950: Part 1. The code states that where the members are not prepared for full contact in bearing, the splice should be designed to transmit all the moments and forces to which the member is subjected. Where the members are prepared for full contact, the splice should provide continuity of stiffness about both axes and resist any tension caused by bending. 146 Compression members (a) Flexible beam to column connections (b) Column cap connections (c) Column splices Figure 7.3 Column construction details In multi-storey buildings, splices are usually located just above floor level. If butted directly together, the ends are usually machined for bearing. Fully bolted splices and combined bolted and welded splices are used. If the axial load is high and the moment does not cause tension the splice holds the columns’ lengths in position. Where high moments have to be resisted, high strength or friction-grip bolts or a full-strength welded splice may be required. Some typical column splices are shown in Figure 7.3(c). (3) Column bases Column bases are discussed in Section 7.10. 7.2 Loads on compression members Axial loading on columns in buildings is due to loads from roofs, floors and walls transmitted to the column through beams and to self weight (see Figure 7.4(a)). Floor beam reactions are eccentric to the column axis, as shown, Loads on compression members 147 Roof (a) Wind Floor Beam reactions Wall Elevation A Axial load C A B B C Unsymmetrical loads Plan Column in multi-storey buildings (b) Eccentricity Roof Roof Wheels Surge Wind Crane Roof load Column Crane loads Column in an industrial building (c) Roof Wind Floors Roof Column Moments Wall Wind Wall Multi-storey frame Portal Rigid frame buildings Figure 7.4 Loads and moments on compression members and if the beam arrangement or loading is asymmetrical, moments are transmitted to the column. Wind loads on multi-storey buildings designed to the simple design method are usually taken to be applied at floor levels and to be resisted by the bracing, and so do not cause moments. In industrial buildings, loads from cranes and wind cause moments in columns, as shown in Figure 7.4(b). In this case, the wind is applied as a distributed load to the column through the sheeting rails. In rigid frame construction moments are transmitted through the joints from beams to column, as shown in Figure 7.4(c). Rigid frame design is outside the scope of this book. 148 Compression members 7.3 Classification of cross-sections The same classification that was set out for beams in Section 5.3 is used for compression members. That is, to prevent local buckling, limiting proportions for flanges and webs in axial compression are given in Table 11, BS 5950: Part 1. The proportions for rolled and welded column sections are shown in Figure 7.5. 7.4 Axially loaded compression members 7.4.1 General behaviour Compression members may be classified by length. A short column, post or pedestal fails by crushing or squashing, as shown in Figure 7.6(a). The squash load Py in terms of the design strength is: Py = py A where A is the area of cross-section. A long or slender column fails by buckling, as shown in Figure 7.6(b). The failure load is less than the squash load and depends on the degree of slenderness. Most practical columns fail by buckling. For example, a universal column under axial load fails in flexural buckling about the weaker y–y axis (see Figure 7.6(c)). b T T b t t Universal column d d d T b t H-section column Box column Limiting proportions Element Section Type Outstand element of compression flange Rolled b/T ≤ Welded b/T ≤ Internal element of Welded b/T ≤ compression flange Web subject to compression throughout Rolled d/t ≤ Class 1- Plastic Section Class 2- Compact Section Class 3- Semi Compact Section 9.0 ε 10.0 ε 15.0 ε 8.0 ε 8.0 ε 8.0 ε 8.0 ε 8.0 ε 8.0 ε _ _ Welded d/t ≤ All elements in compression due to axial load: ε = (275/py)0.5; r2 =Fc /(Agpyw) Figure 7.5 Limiting proportions for rolled and welded column sections 120 ε / (1+ 2r2) but ε 40 ε Axially loaded compression members (a) (b) Crushing (c) Direction of buckling Buckling x Slender column Short column 149 x Universal column (d) Bar and tube of same area Figure 7.6 Behaviour of members in axial compression The strength of a column depends on its resistance to buckling. Thus the column of tubular section shown in Figure 7.6(d) will carry a much higher load than the bar of the same cross-sectional area. This is easily demonstrated with a sheet of A4 paper. Open or flat, the paper cannot be stand on edge to carry its own weight; but rolled into a tube, it will carry a considerable load. The tubular section is the optimum column section having equal resistance to buckling in all directions. 7.4.2 Basic strut theory (1) Euler load Consider a pin-ended straight column. The critical value of axial load P is found by equating disturbing and restoring moments when the strut has been given a small deflection y, as shown in Figure 7.7(a). The equilibrium equation is: EIy d2 y = −Py dx 2 This is solved to give the Euler or lowest critical load: PE = π 2 EIy /L2 In terms of stress, the equation is: PE = π 2E π 2E = (L/ry )2 λ2 150 Compression members (a) (b) PE (c) P P A y x x y L y0 C y r C L Initial position Final position P PE Initially straight strut euler load (d) x Strut with initial curvature Initial position Final position P Strut with end eccentricity h Y X X Y Column section Figure 7.7 Load cases for struts where Iy is the moment of inertia about the minor axis y–y, L the length of the strut, PP the axial load, ry the radius of gyration for the minor axis y–y = (Iy /A)0.5 , pE = PE /A = Euler critical stress and λ = slenderness ratio = L/ry . The slenderness λ is the only variable affecting the critical stress. At the critical load, the strut is in neutral equilibrium. The central deflection is not defined and may be of unlimited extent. The curve of Euler stress against slenderness for a universal column section is shown in Figure 7.9. (2) Strut with initial curvature In practice, columns are generally not straight, and the effect of out of straightness on strength is studied in this section. Consider a strut with an initial curvature bent in a half sine wave, as shown in Figure 7.7(b). If the initial deflection at x from A is y0 and the strut deflects y further under load P , the equilibrium equation is: EIy d2 y = P (y + y0 ) dx 2 Axially loaded compression members 151 where deflection y = sin(π x/L). If δ0 is the initial deflection at the centre and δ the additional deflection caused by P , then it can be shown by solving the equilibrium equation that: δ= δ0 (PE /P ) − 1 The maximum stress at the centre of the strut is given by: Pmax = P P (δ0 + δ)h + A IY where h is shown in Figure 7.7(d). In the above equation, pmax = py = design strength, pc = P /A = average stress, pE = PE /A = Euler stress, Iy = Ary2 = moment of inertia about the y–y axis, A = area of cross-section, ry = radius of gyration for the y–y axis, h = half the flange breath. The equation for maximum stress can be written:   δ0 h 1 py = pc + pc 1 + (pE /pc ) − 1 ry2 Put η = δ0 h/ry2 and rearrange to give: (pE − pc )(py − pc ) = ηpE pc The value of pc the limiting strength at which the maximum stress equals the design strength, can be found by solving this equation and η is the Perry factor. This is to redefined in terms of slenderness. (See Section 7.4.3 (2) below. The design strength curve is also discussed in that section.) (3) Eccentrically loaded strut Most struts are eccentrically loaded, and the effect of this on strut strength is examined here. A strut with end eccentricities e is shown in Figure 7.7(c). If y is deflection from the initially straight strut the equilibrium equation is: EIy d2 y = −p(e + y) dx 2 This can be solved to give the secant formula for limiting stress. 152 Compression members Theoretical studies and tests show that the behaviour of a strut with end eccentricity is similar to that of one with initial curvature. Thus the two cases can be combined with the Perry factor, taking account of both imperfections. 7.4.3 Practical strut behaviour and design strength (1) Residual stresses As noted above, in general, practical struts are not straight and the load is not applied concentrically. In addition, rolled and welded strut sections have residual stresses which are locked in when the section cools. A typical pattern of residual stress for a hot-rolled H-section is shown in Figure 7.8. If the section is subjected to a uniform load, the presence of these stresses causes yielding to occur first at the ends of the flanges. This reduces the flexural rigidity of the section, which is now based on the elastic core, as shown in Figure 7.8(b). The effect on buckling about the y–y axis is more severe than for the x–x axis. Theoretical studies and tests show that the effect of residual stresses can be taken into account by adjusting the Perry factor η. (2) Column tests and design strengths An extensive column-testing programme has been carried out, and this has shown that different design curves are required for: (a) Tension Compression Residual stress pattern (b) Compression Elastic core Residual stress Applied stress Py Plastic ends Spread of yield Figure 7.8 Residual stresses Combined stress Axially loaded compression members 153 (1) different column sections; (2) the same section buckling about different axes; (3) sections with different thicknesses of metal. For example, H-sections have high residual compressive stresses at the ends of the flanges, and these affect the column strength if buckling takes place about the minor axis. The total effect of the imperfections discussed above (initial curvature, end eccentricity and residual stresses on strength) are combined into the Perry constant η. This is adjusted to make the equation for limiting stress pc a lower bound to the test results. The constant η is defined by: η = 0.001 a(λ − λ0 ) λ = 0.2 (π 2 E/py )0.5 The value λ0 gives the limit to the plateau over which the design strength py controls the strut load. The Robertson constant a is assigned different values to give the different design curves. For H-sections buckling about the minor axis, a has the value 5.5 to give design curve (c) (Table 24(c)). A strut table selection is given in Table 23 in BS 4950: Part 1. For example, for rolled and welded H sections with metal thicknesses up to 40 mm, the following design curves are used: (1) buckling about the major axis x–x curve (b) (Table 24(b)); (2) buckling about the minor axis y–y curve (c) (Table 24(c)). The compressive strength is given by the smaller root of the equation that was derived above for a strut with initial curvature. This is: (pE − pc )(py − pc ) = ηpE pc pE py pc = (φ + φ 2 − pE py )0.5 φ = [py + (η + l)pE ]/2 The curves for Euler stress pE and limiting stress or compressive strength pc for a rolled H-section column buckling about the minor axis are shown in 154 Compression members 300 Design strength 240 200 Euler curve Code strut curve C 0 λ = 85.7 100 λo = 17.15 Euler stress Compressive strength –pE (N/mm2) –pC (N/mm2) 275 100 200 Slenderness λ 300 350 Figure 7.9 Strut strength curves Figure 7.9. It can be noted that short struts fail at the design strength while slender ones approach the Euler critical stress. For intermediate struts, the compressive strength is a lower bound to the test results, as noted above. Compressive strengths for struts for curves a, b, c and d are given in Tables 24(a)–(d) in BS 5950: Part 1. 7.4.4 Effective lengths (1) Theoretical considerations The actual length of a compression member on any plane is the distance between effective positional or directional restraints in that plane. A positional restraint should be connected to a bracing system which should be capable of resisting 1% of the axial force in the restrained member. See Clause 4.7.1 of BS 5950. The actual column is replaced by an equivalent pin-ended column of the same strength that has an effective length: LE = KL where L is the actual length, and K the effective length ratio and K is to be determined from the end conditions. An alternative method is to determine the distance between points of contraflexure in the deflected strut. These points may lie within the strut length or they may be imaginary points on the extended elastic curve. The distance so defined is the effective length. Pinned end Fixed end Pin fixed Free fixed 155 LE = L LE = 2 L L E = 0.7 L L E = 0.5 L LE = L Axially loaded compression members Fixed ends sway Figure 7.10 Figure effective lengths The theoretical effective lengths for standard cases are shown in Figure 7.10. Note that for the cantilever and sway case the point of contraflexure is outside the strut length. (2) Code definitions and rules The effective length is defined in Section 1.2.14 of BS 5950: Part 1 as the length between points of effective restraint of a member multiplied by a factor to take account of the end conditions and loading. Effective lengths for compression members are set out in Section 4.7.2 of the code. This states that for members other than angles, channels and T-sections, the effective length should be determined from the actual length and conditions of restraint in the relevant plane. The code specifies: (1) That restraining members which carry more than 90 per cent of their moment capacity after reduction for axial load shall be taken as incapable of providing directional restraint. (2) Table 22 is used for standard conditions of restraint. (3) Appendix D1 is used for stanchions in single-storey buildings of simple construction (see Section 7.6). (4) Appendix E is used for members forming part of a frame with rigid joints. The normal effective lengths LE are given in Table 22 of the code. Some values from this table for various end conditions where L is the actual length are: (1) Effectively held in position at both ends (a) Restrained in direction at both ends, LE = 0.7L (b) Partially restrained in direction. at both ends, LE = 0.85L (c) Not restrained in direction at either end, LE = L (2) One end effectively held in position and restrained in direction. Other end not held in position (a) Partially restrained in direction, LE = 1.5L (b) Not restrained in direction, LE = 2.0L The reader should consult the table in the code for other cases. 156 Compression members Note the case for the fixed end strut, where the effective length is given as 0.7 L, is to allow for practical ends where true fixity is rarely achieved. The theoretical value shown in Figure 7.10 is 0.5 L. 7.4.5 Slenderness The slenderness λ is defined in Section 4.7.3 of the code as: λ= Effective length LE = Radius of gyration about relevant axis r The code states that, for members resisting loads other than wind load, λ must not exceed 180. Wind load cases are dealt with in Chapter 8 of this book. 7.4.6 Compression resistance The compression resistance of a strut is defined in Section 4.7.4 of BS 5950: Part 1 as: (1) Plastic, compact or semi-compact sections: Pc = Ag pc (2) Slender sections: Pc = Aeff pcs where Ag is the gross sectional area defined in Section 3.4.1 of the code, Aeff the effective sectional area defined in Section 3.6.2 of the code, pc the compressive strength from Section 4.7.5 and Tables 27(a)–(d) of the code and pcs the value pc from clause 4.75 for a reduced slenderness of λ(Aeff /Ag )0.5 in which λ is based on the radius of gyration r of the gross cross-sections. 7.4.7 Column design Column design is indirect, and the process is as follows (the tables referred to are in the code): (1) The steel grade and section is selected. (2) The design strength py , is taken from Table 9. (3) The effective length LE is estimated using Table 22 for the appropriate end conditions. (4) The slenderness λ is calculated for the relevant axis. (5) The strut curve is selected from Table 23. (6) The compressive strength is read from the appropriate part of Tables 24(a)–(d). (7) The compression resistance Pc is calculated (see Section 7.4.6 above). For a safe design, Pc should just exceed the applied load, and successive trials are needed to obtain an economical design. Load tables can be formed to give the compression resistance for various sections for different values of effective length. Table 7.1 gives compression resistances for some universal column sections. Column sizes may be selected from tables in the Guide to BS 5950: Part 1, Volume 1, Section Properties, Member Capacities, Steel Construction Institute. Axially loaded compression members 157 Table 7.1 Compression resistance of S275 steel U.C. sections Serial size (mm) Mass per metre (kg) Compression resistances (kn) for effective lengths (m) 2 2.5 3 3.5 4 5 6 8 10 254 × 254 universal column 167 132 107 89 73 5230 4160 3360 2790 2350 4990 3960 3200 2650 2230 4730 3750 3020 2510 2110 4460 3530 2840 2360 1970 4180 3300 2650 2200 1830 3590 2820 2260 1860 1550 3010 2360 1880 1550 1270 2080 1620 1280 1060 860 3580 1140 909 742 602 203 × 203 universal column 86 71 60 52 46 2570 2130 1820 1600 1410 2400 1980 1700 1480 1310 2220 1830 1560 1360 1200 2030 1670 1410 1230 1080 1830 1510 1270 1100 968 1460 1200 995 865 757 1150 943 778 676 590 740 605 494 1429 374 – – – – – 152 × 152 universal column 37 30 23 1030 825 632 910 727 552 787 627 472 671 533 397 568 450 334 411 325 239 306 241 177 – – – – – – 7.4.8 Example: universal column A part plan of an office floor and the elevation of internal column stack A are shown in Figures 7.11(a) and (b). The roof and floor loads are as follows: Roof: Dead load (total) = 5 kN/m2 ; Imposed load = 1.5 kN/m2 . Floors: Dead load (total) = 7 kN/m2 ; Imposed load = 3 kN/m2 Design column A for axial load only. The self-weight of the column, including fire protection, may be taken as 1 kN/m. The roof and floor steel have the same layout. Use Grade S275 steel. When calculating the loads on the column lengths, the imposed loads may be reduced in accordance with Table 2 of BS 6399: Part 1. This is permitted because it is unlikely that all floors will be fully loaded simultaneously. Values from the table are: Number of floor carried by member Reduction in imposed load (%) 1 2 3 0 10 20 The roof is regarded as a floor for reckoning purposes. The slabs for the floor and roof are precast one-way spanning slabs. The dead and imposed loads are calculated separately. 158 Compression members (a) 7.6 m (b) 7.6 m B2 B1 B1 slab 4m 4m B1 2nd floor 1st floor 5m 6m B1 of B1 Span B1 B2 6m Roof Base Part floor plan Column stack ‘A’ (c) 22.8 w 22.8 w 7.6 m Beam B1 11.4 w 6m 11.4 w Beam B2 Beam loads Figure 7.11 Column design example (1) Loading Four floor beams are supported at column A. These are designated as B1 and B2 in Figure 7.11 (a). The reactions from these beams in terms of a uniformly distributed load are shown in Figure 7.11(c): Load on beam B1 = 7.6 × 3 × 10 = 22.8 w (kN) where w is the uniformly distributed load. The dead and imposed loads must be calculated separately in order to introduce the different load factors. The self weight of beam B2 is included in the reaction from beam B1. The design loading on the column can be set out as shown in Figure 7.12. The design loads are required just above the first floor, the second floor and the base. (2) Column design (1) Top length = Roof to second floor Design load = 434.2 kN Try 152 × 152 UC 30 A = 38.2 cm2 ; ry = 3.82 cm Design strength py = 275 N/mm2 (Table 9) where section thickness is less than 16 mm. Axially loaded compression members Roof 159 B1 Dead load kN B2 Self weight 4 kN Nil 2nd floor B1 Self weight 4 kN 114 114 4 34.2 34.2 Total 232 68.4 30.8 30.8 27.4 27.4 Design load = (1.4 × 232) + (1.6 × 68.4) 2 w = 3.0 kN/m 434.2 2 2 No B1 2 No B2 Self weight 159.6 159.6 4.0 61.6 61.6 Total 555.2 184.8 54.7 54.7 Design load = (1.4 × 555.2) + (1.6 × 184.8) B1 w = 7 kN/m B2 Self weight 5 kN Toatal design load w = 1.5 kN/m2 2 No B1 2 No B2 Self weight w = 7 kN/m B2 10% 1st floor w = 5 kN/m2 Imposed load (kN) Reduction 0% 10% 20% 2 w = 3.0 kN/m 1073 2 2 No B1 2 No B2 Self weight 159.6 159.6 5.0 54.7 54.7 Total 879.6 273.6 Design load = (1.4 × 879.6) + (1.6 × 273.6) 1669 20% Base Figure 7.12 Column design loads If the beam connections are the shear type discussed in Section 5.8.3, where end rotation is permitted, the effective length, from Table 22: LE = 0.85 × 4000 = 3400 mm Slenderness, λ = 3400/38.2 = 89 For a rolled H section thickness less than 40 mm buckling about the minor y–y axis, use Table 24, curve (c): Compressive strength pc = 144 N/mm2 Compressive resistance pc = 144 × 38.2/10 = 550.1 kN The column splice and floor beam connections at second-floor level are shown in Figure 7.13(a). The net section at the splice is shown in Figure 7.13(b) with 4 No. 22 mm diameter holes. The section is satisfactory. (2) Intermediate length—first floor to second floor. Design load = 1073 kN. Try 203 × 203 UC 46. 160 Compression members (a) (b) 9.4 4 no. 22 ø holes 152 × 152 × 30 UC 152 × 152 × 30 UC Net section at splice 2nd floor (c) B1 B2 B2 B2 B2 B1 Section at floor beam connections Beam B1 not shown Spice and floor beam connections Figure 7.13 Column connection details A = 58.8 cm2 ; ry = 5.11 cm py = 275 N/mm2 λ = 3400/51.1 = 66.5 pc = 188 N/mm2 Pc = 188 × 58.8/10 = 1105.4 kN The section is satisfactory. (3) Bottom length—base to first floor. Design load = 1669 kN. Try 254 × 254 UC 73: A = 92.9 cm2 ; ry = 6.46 cm. The flange is 14.2 mm thick. The design strength from Table 9 in the code is py = 275 N/mm2 The beam connections do not restrain the column in direction at the first floor level. The base can be considered fixed. The effective length is taken as: LE = 0.85 × 5000 = 4250 mm λ = 4250/64.6 = 65.8 pc = 189.4 N/mm2 (Table 24(c)) Pc = 92.9 × 189.4/10 = 1759.5 kN Axially loaded compression members 161 The section selected is satisfactory. The same sections could have been selected from Table 7.1. 7.4.9 Built-up column: design The two main types of columns built up from steel plates are the H and box sections shown in Figure 7.2. The classification for cross-sections is given in Figure 7.5. For plastic, compact or semi-compact cross-sections, the local compression capacity is based on the gross section. The code states in Section 4.7.5 that the design strength py for sections fabricated by welding is to be the value from Table 9 reduced by 20 N/mm2 . This takes account of the severe residual stresses and possible distortion due to welding. Slender cross-sections are dealt with in Section 3.6 of the code. The capacity of these sections is limited by local buckling and the design should be based on the effectively cross-sectional area. The compressive strength pcs should be evaluated from Clause 4.75 with a reduced slenderness of λ(Aeff /Ag )0.5 . 7.4.10 Example: built-up column Determine the compression resistance of the column section shown in Figure 7.14. The effective length of the column is 8 m and the steel is S275. (1) Flanges The design strength from Table 9 for plate 30mm thick py = 265 N/mm2 . Reducing by 20 N/mm2 for a welded section gives: py = 245 N/mm2 ε = (275/245)0.5 = 1.059 Flange outstand b = 442.5 = 14.75 T > 13 εT = 313.77 T Referring to Table 11, the flange is slender. The effective area per flange is: 13εT × 2 × T = 13 × 1.059 × 30 × 2 × 30 = 24 780 mm2 . 900 840 900 30 Y X X 30 15 Figure 7.14 Built-up H column Y 162 Compression members (2) Web This is an internal element in axial compression. py = 275 − 20 = 255 N/mm2 ε = (275/255)0.5 = 1.038 The effective area of the web is taken as 20εt from each end, hence for the web effective area is 20 × 1.038 × 15 × 2 × 15 = 9342 mm2 (3) Properties of the gross section and effective section Gross area = (2 × 30 × 900) + (840 × 15) = 66 600 mm2 Iy = (60 × 9003 /12) + (neglect web) = 3.645 × 109 mm4 ; ry = [3.645 × 109 /6.66 × 104 ]0.5 = 233.9 mm; λ = 8000/233.9 = 34.2. Effective sectional area = 2 × 24 780 + 9342 = 58 902 mm2 . (4) Compressive resistance of the column The compressive strength of the column pc is obtained with py of 245 N/mm2 and reduced slenderness of λ(Aeff /Ag )0.5 = 34.2 × (58 902/66 600)0.5 = 32.2. From Table 24c, pc is 225 N/mm2 , Pc = pc Aeff = 225 × 58 902/1000 = 13 253 kN. This is compared with the strength of 12 300 kN calculated from the previous version of the code. Note that the new procedure is much easier. 7.4.11 Cased columns: design (1) General requirements Solid concrete casing acts as fire protection for steel columns and the casing assists in carrying the load and preventing the column from buckling about the weak axis. Regulations governing design are set out in Section 4.14 of BS 5950: Part 1. The column must meet the following general requirements: (1) The steel section is either a single-rolled or fabricated I- or H-section with equal flanges. Channels and compound sections can also be used. (Refer to the code for requirements.) (2) The steel section is not to exceed 1000 × 500 mm2 . The dimension 1000 mm is in the direction of the web. (3) Primary structural connections should be made to the steel section. Axially loaded compression members 163 (4) The steel section is unpainted and free from dirt, grease, rust, scale, etc. (5) The steel section is encased in concrete of at least Grade 25, to BS 8110. (6) The cover on the steel is to be not less than 50 mm. The corners may be chamfered. (7) The concrete extends the full length of the member and is thoroughly compacted. (8) The casing is reinforced with steel fabric mesh #D98 per BS 4483 or alternatively with rebars not less than 5 mm diameter at a maximum spacing of 200 mm to form a cage of closed links and longitudinal bars. The reinforcement is to pass through the centre of the cover, as shown in Figure 7.15(a). (9) The effective length is not to exceed 40bc , 100bc2 /dc or 250r, whichever is the least, where bc is the minimum width of solid casing, dc the minimum depth of solid casing and r the minimum radius of gyration of the steel section. (2) Compression resistance The design basis set out in Section 4.14.2 of the code is as follows: (1) The radius of gyration about the y–y axis shown in Figure 7.15, ry should be taken as 0.2bc but not more than 0.2(B + 150), where B is the overall width of the steel flange. The radius of gyration for the x–x axis rx should be taken as that of the steel section. (2) The compression resistance Pc is   fcu Pc = Ag + 0.45 Ac pc py Cover ⬍ 50 (b) B 203.9 206.2 Cover ⬍ 50 Y X X 203 × 203 × 52 UC dc X X Y 12.5 (a) Y Steel core Bars and links (c) 310 Y Y bc ⬍ b + 100 310 Cased section X X Y Cased column example Figure 7.15 Cased column 164 Compression members but not more than the short strut capacity,   fcu Pcs = Ag + 0.25 A c py py where Ac is the gross sectional area of the concrete. Casing in excess of 75 mm from the steel section is neglected. Finish is neglected. Ag the gross area of the steel section, fcu the characteristic strength of the concrete at 28 days. This is not to exceed 40 N/mm2 , pc the compress strength of the steel section determined using rx and ry , in the determination of which py < 355 N/mm2 , and py the design strength of the steel. 7.4.12 Example: cased column An internal column in a multi-storey building has an actual length of 4.2 m centre-to-centre of floor beams. The steel section is a 203 × 203 UC 52. Calculate the compression resistance of the column if it is cased in accordance with Section 4.14 of BS 5950: Part 1. The steel is Grade S275 and the concrete Grade 25. The steel core and cased section are shown in Figure 7.15(b). The casing has been made 310 mm2 . The properties of the steel section are: A = 66.4 cm2 , rx = 8.9 cm, ry = 5.16 cm For the cased section: ry = 0.2 × 310 = 62 mm ≥ 0.2(203.9 + 150) = 70.78 mm. Because the column is cased throughout, the effective length is taken from Table 22 as 0.7 of the actual length: Effective length LE = 0.7 × 4200 = 2940 mm. The effective length LE is not to exceed: 40bc = 40 × 310 = 12 400 mm 100bc2 /dc = 100 × 310 = 3100 mm 250r = 250 × 51.6 = 12 900 mm Slenderness, λ = 2940/62 = 47.4 The design strength from Table 9, py = 275 N/mm2 . For buckling about y–y, select curve (c) from Table 23. Compressive strength from Table 24(c): pc = 225.2 N/mm2 The gross sectional area of the concrete: Ac = 310 × 310 = 96 100 mm2 Beam columns 165 Compressive resistance of the cased section:   25 × 96 100 225.2 Pc = 66.4 × 102 + 0.45 275 103 = 1495.3 + 885.3 = 2380.6 kN This is not to exceed the short strut capacity:   25 × 96 100 275 2 Pcs = 66.4 × 10 + 0.25 275 103 = 1826 + 600.6 = 2426.6 kN The compression resistance is 2380.6 kN. 7.5 Beam columns 7.5.1 General behaviour (1) Behaviour classification As already stated at the beginning of this chapter, most columns are subjected to bending moment in addition to axial load. These members, termed ‘beamcolumns’, represent the general load case of an element in a structural frame. The beam and axially loaded column are limiting cases. Consider a plastic or compact H-section column as shown in Figure 7.16(a). The behaviour depends on the column length, how the moments are applied and the lateral support, if any, provided. The behaviour can be classified into the following five cases: Case 1: A short column subjected to axial load and uniaxial bending about either axis or biaxial bending. Failure generally occurs when the plastic capacity of the section is reached. Note limitations set in (2) below. Case 2: A slender column subjected to axial load and uniaxial bending about the major axis x–x. If the column is supported laterally against buckling about the minor axis y–y out of the plane of bending, the column fails by buckling about the x–x axis. This is not a common case (see Figure 7.17(a)). At low axial loads or if the column is not very slender, a plastic hinge forms at the end or point of maximum moment Case 3: A slender column subjected to axial load and uniaxial bending about the minor axis y–y. The column does not require lateral support and there is no buckling out-of-the-plane of bending. The column fails by buckling about the y–y axis. At very low axial loads, it will reach the bending capacity for y–y axis (see Figure 7.17(b)). Case 4: A slender column subjected to axial load and uniaxial axial bending about the major axis x–x. This time the column has no lateral support. The column fails due to a combination of column buckling aboutthe y–y axis 166 Compression members (a) (b) Y Bending–tension Axial load–compression X X Bending–compression Y Universal column Plastic stress distribution bending about XX axis (c) Mry 1.0 1.0 F/Pc Mcy Mry Mrx F/Pc Mrx Mcx Mcy Mcx Mx My Mcx Mcy Mrx Mry Mx My Mx Mcx A My Mcy 1.0 1.0 1.0 Mcx Mcy Mcx Mcy Interaction curves for universal bending XX and YY axes Interaction surface biaxial bending full plasticity Figure 7.16 Short-column behaviour and lateral torsional buckling where the column section twists as well as deflecting in the x–x and y–y planes (see Figure 7.17(c)). Case 5: A slender column subject to axial load and biaxial bending. The column has no lateral support. The failure is the same as in Case 4 above but minor axis buckling will usually have the greatest effect. This is the general loading case (see Figure 7.17(d)). Some of these cases are discussed in more detail below. (2) Short-column failure The behaviour of short columns subjected to axial load and moment is the same as for tension members subjected to identical loads. This was discussed in Section 7.3.3. The plastic stress distribution for uniaxial bending is shown in Figure 7.16(b). The moment capacity for plastic or compact sections in the Beam columns (a) (b) X Y X X Plastic hinge may form Y M1 Y n io ct le ef 167 X n io ct le ef D D Lateral Restraints M1 > M2 X X X Y Y X Moments about XX axis buckling restrained about YY axis Moments about YY axis no restraint Y (c) (d) X X X ct le tat n io ct le ef Ro Y io Ro n tat io n ion X D ef D Deflection Deflection Y X X Moments about XX axis no restraint X X Y Moments about XX axis and YY axis no restraint Figure 7.17 Slender columns subjected to axial load and moment absence of axial load is given by: Mc = Spy ≤ 1.2 Zpy (see Section 4.2.5 of BS 5950: Part 1) where S is the plastic modulus for the relevant axis and Z the elastic-modulus for the relevant axis. 168 Compression members The interaction curves for axial load and bending about the two principal axes separately are shown in Figure 7.18(a). Note the effect of the limitation of bending capacity for the y–y axis. These curves are in terms of F /Pc against Mrx /Mcx and Mry /Mcy , where F is the applied axial load, Pc the py A, the squash load, Mrx the reduced moment capacity about the x–x axis in the presence of axial load, Mcx the moment capacity about the x–x axis in the absence of axial load Mry the reduced moment capacity about the y–y axis in the presence of axial load and Mcy the moment capacity about the y–y axis in the absence of axial load. Values for Mrx and Mry are calculated using equations for reduced plastic modulus given in the Guide to BS 5950: Part 1: Volume 1, Section Properties, Member Capacities, S.C.I. Linear interaction expressions can be adopted. These are: F /Pc + Mx /Mcx = 1 and F /Pc + My /Mcy = 1 where Mx is the applied moment about the x–x axis and My the applied moment about the y–y axis. This simplification gives a conservative design. Plastic and compact H-sections subjected to axial load and biaxial bending are found to give a convex failure surface, as shown in Figure 7.18(a). At any point A on the surface the combination of axial load and moments about the x–x and y–y axes Mx and My respectively, that the section can support can be read off. A plane drawn through the terminal points of the surface gives a linear interaction expression. My F Mx + + =1 Pc Mcx Mcy This results in a conservative design. (3) Failure of slender columns With slender columns, buckling effects must be taken into account. These are minor axis buckling from axial load and lateral torsional buckling from moments applied about the major axis. The effect of moment gradient must also be considered. All the imperfections, initial curvature, eccentricity of application of load and residual stresses affect the behaviour. The end conditions have to be taken into account in estimating the effective length. Theoretical solutions have been derived and compared with test results. Failure surfaces for H-section columns plotted from the more exact approach given in the code are shown in Figure 7.18(a) for various values of slenderness. Failure contours are shown in Figure 7.18(b). These represent lower bounds to exact behaviour. Beam columns (a) 169 Axial load, F/Pc 10 Mry Mcy Mrx Mcx 0 λ Mox 0 10 50 10 0 0 50 λ Mcx Moy Mcy 100 0 5 0 1.0 1.0 XX axis λ YY axis Failure surfaces 0. 5 0 YY axis 0.1 27 0. 0 YY axis 4 0 1.0 3 53 0. 6 0.3 0.6 YY axis F/Pc 1.0 XX axis 0. 1.0 45 XX axis 8 1.0 0. XX axis 1. 0 (b) F/Pc 1.0 λ=0 F/Pc 1.0 λ = 50 λ = 100 Failure contours Figure 7.18 Failure surface for slender beam-column The failure surfaces are presented in the following terms: Slenderness λ = 0 Fc /Pc ; Mx /Mcx ; My /Mcy λ = 50, 100 Fc /Pc ; Max /Mcx ; May /Mcy Some of the terms were defined in Section 7.5.1(2) above. New terms used are: Max = maximum buckling moment about the x–x axis in the presence of axial load, May = maximum buckling moment about the y–y axis in the presence of axial load. The following points are to be noted. (1) Mcy , the moment capacity about the y–y axis, is not subjected to the restriction 1.2py Zy . 170 Compression members (2) At zero axial load, slenderness does not affect the bending strength of an H section about the y–y axis. (3) At high values of slenderness the buckling resistance moment Mb about the x–x axis controls the moment capacity for bending about that axis. (4) As the slenderness increases, the failure curves in the F /Pc , y–y-axis plane change from convex to concave, showing the increasing dominance of minor axis buckling. For design purposes, the results are presented in the form of an interaction expression, and this is discussed in the next section. 7.5.2 Code design procedure The code design procedure for compression members with moments is set out in Section 4.8.3 of BS 5950: Part 1. This requires the following two checks to be carried out: (1) cross-section capacity check and (2) member buckling check. In each case, two procedures are given. These are a simplified approach and a more exact one. Only the simplified approach will be used in design examples in this book. (1) Cross-section capacity check The member should be checked at the point of greatest bending moment and axial load. This is usually at the end, but it could be within the column height if lateral loads are also applied. The capacity is controlled by yielding or local buckling. (Local buckling was considered in Section 7.3.) Except for Class 4 members, with the simplified approach, the interaction relationship for Classes 1, 2 and 3 members given in Section 4.8.3.2 of the code is: My Fc Mx + + ≤1 Ag py Mcx Mcy where F is the applied axial load, Ag the gross cross-sectional area, Mx the applied moment about the major axis x–x, Mcx the moment capacity about the major axis x–x in the absence of axial load, My the applied moment about the minor axis y–y, and Mcy the moment capacity about the minor axis y–y in the absence of axial load. Alternatively, a more rigorous interaction relationship for plastic and compact sections given in the code can also be used. This is based on the convex failure surface discussed above and gives greater economy in design. For Class 4 members, the interaction relationship is: My Fc Mx + + ≤1 Aeff py Mcx Mcy where the additional term Aeff is the effective cross-sectional area defined by the code under Clause 3.6. Beam columns 171 (2) Member buckling resistance Under Clause 4.8.3.3.1 of the code, for simplified method, the buckling resistance of a member may be verified by checking the following relationships so that both are satisfied: my My mx Mx Fc + + ≤1 Ag pc Mb py Zy and my My mLT MLT Fc + + ≤1 Ag pcy Mb py Zy where Fc m = axial compressive load = equivalent uniform moment factor (x or y axis) from Table 18 of the code, Mb = buckling resistance moment capacity about the major axis x–x, MLT = the maximum major axis moment in the segment length L governing Mb , Mx = the maximum major axis moment in the segment length Lx govering pcx , My = the maximum minor axis moment in the segment length Ly govering pcy , pc = the smaller of pcx and pcy , pcy = the compression resistance, considering buckling about the minor axis only, Zy = elastic modulus of section for the minor axis y–y, Zy = elastic modulus of section for the minor axis y–y. The value for Mb is determined using the methods set out in Section 5.5 of this book (dealing with lateral torsional buckling of beams). A more exact approach is also given in the code. This uses the convex failure surfaces discussed above. 7.5.3 Example of beam column design A braced column 4.5 m long is subjected to the factored end loads and moments about the x–x axis, as shown in Figure 7.19(a). The column is held in position but only partially restrained in direction at the ends. Check that a 203 × 203 UC 52 in Grade 43 steel is adequate. (1) Column-section classification Design strength from Table 9 py = 275 N/mm2 Factor ε = (275/py )0.5 = 1.0 (see Figure 7.19(b)) Flange b/T = 101.95/12.5 = 8.156 < 9.0 Web d/t = 160.8/8.0 = 20.1 < 40 Compression members 880 kN (b) d = 160.8 35 kNm t = 8.0 X r = 10.2 6.5 m Moments applied about XX axis T = 12.5 203.9 (a) b = 101.95 172 X Trial section 12 kNm 88.6 kN Column length and loads Figure 7.19 Beam column design example Referring to Table 11, the flanges are plastic and the web semi-compact. (2) Cross-section capacity check Section properties for 203 × 203 UC 52 are: A = 66.4 cm2 ; x = 15.8; Zx = 510 cm3 ; u = 0.848; ry = 516 cm Sx = 568 cm3 Moment capacity about the x–x axis: Mcx = 275 × 568/193 = 156.2 kN m < 1.2 × 275 × 510/103 = 168.4 kN m Interaction expression: 35 880 × 10 + = 0.48 + 0.22 = 0.7 < 1 66.4 × 275 156.2 The section is satisfactory with respect to local capacity. = 0.48 + 0.22 = 0.7 < 1 (3) Member buckling check The effective length from Table 22: LE = 0.85 × 4500 = 3825 Slenderness λ = 3825/51.6 = 74.1 Eccentrically loaded columns in buildings 173 From Table 23, select Table 24(c) for buckling about the y–y axis. From Table 24(c), compressive strength py = 172.8 N/mm2 . Referring to Table 13, the support conditions for the beam column are that it is laterally restrained and restrained against torsion but partially free to rotate in plan: Effective length LE = 0.85 × 4500 = 3825 mm Slenderness λ = 74.1 The ratio of end moments: β = 12/35 = 0.342 From Table 18 the equivalent uniform moment factor mx = 0.697. λLT = uvλ where u = 0.848 and denotes buckling parameter for H-section, N = 0.5 for uniform section with equal flanges, x = 15.8, the torsional index, λ/x = 74.1/15.8 = 4.69, v = 0.832, the slendemess factor from Table 19, λLT = 0.848 × 0.832 × 74.1 = 52.2 From Table 16, the bending strength: pb = 232.7 N/mm2 Buckling resistance moment: Mb = 232.7 × 568/103 = 132.1 kN m Interaction expression: my My mx Mx Fc + + ≤1 Ag pc Mb py Zy 0.697 × 35 880 × 10 + + 0 = 0.77 + 0.18 = 0.95 < 1.0 132.1 172.8 × 66.4 The section is also satisfactory with respect to overall buckling. 7.6 Eccentrically loaded columns in buildings 7.6.1 Eccentricities from connections The eccentricities to be used in column design in simple construction for beam and truss reactions are given in Clause 4.7.6 of BS 5950: Part 1. These are as follows: (1) For a beam supported on a cap plate, the load should be taken as acting at the face of the column or edge of the packing. (2) For a roof truss on a cap plate, the eccentricity may be neglected provided that simple connections are used. 174 Compression members (3) In all other cases, the load should be taken as acting at a distance from the face of the column equal to 100 mm or at the centre of the stiff bearing, whichever gives the greater eccentricity. The eccentricities for the various connections are shown in Figure 7.20. 7.6.2 Moments in columns of simple construction The design of columns is set out in Section 4.7.7 of the code. The moments are calculated using eccentricities given in Section 7.6.1 above. For multi-storey columns effectively continuous at splices, the net moment applied at any one level may be divided between lengths above and below in proportion to the stiffness I /L of each length. When the ratio of stiffness does not exceed 1.5, the moments may be divided equally. These moments have no effect at levels above or below that at which they are applied. The following interaction equation should be satisfied for the overall buckling check: My Fc Mx + + ≤1 Ag pc Mbs py Zy (a) (b) Reaction Reaction Eccentricity Beam to column connection Truss to column connection (d) Reaction 100 Reaction Eccentricity (c) 100 Eccentricity Eccentricities for beam-column connections Stiff bearing Beam supported on bracket Figure 7.20 Eccentricies for end reactions Eccentrically loaded columns in buildings 175 where Mbs is the buckling resistance moment for a simple column calculated using an equivalent slenderness, λLT = 0.5L/ry , I the moment of inertia of the column about the relevant axis, L the distance between levels at which both axes are restrained, ry the radius of gyration about the minor axis and Fc the compressive force in the column. Other terms are defined in Section 7.5.2. 7.6.3 Example: corner column in a building The part plan of the floor and roof steel for an office building is shown in Figure 7.21(a) and an elevation of the corner column is shown in Figure 7.21(b). The roof and floor loading is as follows: Roof: Total dead load = 5 kN/m2 Imposed load = 1.5 kN/m2 Floors: Total dead load = 7 kN/m2 Imposed load = 3 kN/m2 The self-weight of the column, including fire protection, is 1.5 kN/m. The external beams carry the following loads due to brick walls and concrete casing (they include self-weight): Roof beams—parapet and casing = 2 kN/m Floor beams—walls and casing = 6 kN/m The reinforced concrete slabs for the roof and floors are one-way slabs spanning in the direction shown in the figure. Design the corner column of the building using S275 steel. In accordance with Table 2 of BS 6399: Part 1, the imposed loads may be reduced as follows: One floor carried by member—no reduction Two floors carried by member—10% reduction Three floors carried by member—20% reduction (a) 7.6 m (b) 4m 2nd floor 1st floor 5m Span of Floor Slab 6m 4m Floor Base Proof and floor plan Figure 7.21 Corner-column design example Column stack 176 Compression members The roof is counted as a floor. Note that the reduction is only taken into account in the axial load on the column. The full imposed load at that section is taken in calculating the moments due to eccentric beam reactions. (1) Loading and reactions floor beams Mark numbers for the floor beams are shown in Figure 7.22(a). The end reactions are calculated below: Roof B1 Dead load = (5 × 3.8 × 1.5) + (2 × 3.8) = 36.1 kN Imposed load = 1.5 × 3.8 × 1.5 = 8.55 kN B2 Dead load = 5 × 3.8 × 3 = 57.0 kN Imposed load = 1.5 × 3.8 × 3 = 17.1 kN B3 Dead load = (0.5 × 57.0) + (2 × 3) = 34.5 kN Imposed load = 0.5 × 17.1 = 8.55 kN Floors B1 Dead load = (7 × 3.8 × 1.5) + (6 × 3.8) = 62.7 kN Imposed load = 3 × 3.8 × 1.5 = 17.1 kN B2 Dead load = 7 × 3.8 × 3 = 79.8 kN Imposed load = 3 × 3.8 × 3 = 34.2 kN B3 Dead load = (0.5 × 79.8) + (6 × 3) = 57.9 kN Imposed load = 0.5 × 34.2 = 17.1 kN The roof and floor beam reactions are shown in Figure 7.22(b). (2) Loads and moments at roof and floor levels The loading at the roof, second floor, first floor and base is calculated from values shown in Figure 7.22(b). The values for imposed load are calculated separately, so that reductions permitted can be made and the appropriate load factors for dead and imposed load introduced to give the design loads and moments. The moments due to the eccentricities of the roof and floor beam reactions are based on the following assumed sizes for the column lengths: Roof to second floor: 152 × 152 UC where the inertia I is proportional to 1.0; Second floor to first floor: 203 × 203 UC where the inertia I is proportional to 2.5; First floor to base: 203 × 203 UC where the inertia I is proportional to 2.5. Further, it will be assumed initially that the moments at the floor levels can be divided between the upper and lower column lengths in proportion to the stiffnesses which are based on the inertia ratios given above. The actual values are not required. The division of moments is made as follows: (1) Joint at second floor level Upper column length—stiffness I /L = 1/4 = 0.25 Lower column length—stiffness I /L = 2.5/4 = 0.625 Eccentrically loaded columns in buildings (a) 177 B1 B3 B2 Reactions in kN Beam mark numbers (b) 36.1 8.55 B1 57.0 17.1 36.1 8.55 62.7 Dead Imposed 17.1 57.0 17.1 Dead 79.8 Imposed 34.2 B2 57.0 17.1 34.5 8.35 Dead Imposed B1 B2 79.8 34.2 79.8 34.2 Dead Imposed 34.5 8.35 62.7 17.1 57.9 17.1 Roof beam 57.9 17.1 Floor beams Roof and floor beam reactions Figure 7.22 Floor–beam reactions If M is the moment due to the eccentric floor beam reaction then the moment in the upper column length is: Mu = [0.25/(0.25 + 0.625)]M = 0.286 M Moment in the lower column length is: Ml = (1 − 0.286)M = 0.714 M (2) Joint at first level It will be assumed that the same column section will be used for the two lower lengths. Hence the moments of inertia are the same and the stiffnesses are inversely proportional to the column lengths. Upper column length—stiffness = 1/4 = 0.25 Lower column length—stiffness = 1/5 = 0.20 The stiffness of the upper column length does not exceed 1.5 times the stiffness of the lower length. Thus the moments may be divided equally between the upper and lower lengths. The eccentricities of the beam reactions and the column loads and moments from dead and imposed loads are shown in Figure 7.23. 178 Column sections Y 176 Roof 6 kN 176 × Nil I = 1.0 X × 10% I = 2.5 6 kN 202 2nd floor 1st floor Y 34.5 D 8.55 I 202 Y 202 7.5 kN Imposed load Reduced imp. load Dead Mx Imposee Mx Dead My Imposed My 70.6 17.1 17.1 6.07 1.57 6.35 1.51 76.6 17.1 17.1 3.34 0.99 3.52 0.99 197.2 31.3 46.2 8.35 2.47 9.01 2.47 203.2 31.3 46.2 5.84 1.79 6.33 1.73 2nd floor 62.7 D 17.1 I 323.8 85.5 68.4 5.84 1.79 6.33 1.73 Base 331.3 68.4 – – – Roof 8.55 I 36.1 D Above 2nd floor Below 2nd floor X X 62.7 D 17.1 I Above Y 57.9 D 1st Floor 17.1 I Y 202 Below X × 20% I = 2.5 X Dead load Position X Y 57.9 D 17.1 I × 10% permitted values for reduction in imposed loads loads are in kN. moments in kNm Figure 7.23 Loads and moments from actual and imposed loads – Compression members Column stack Eccentrically loaded columns in buildings 179 (3) Column design Roof to second floor: Referring to Figure 7.23, the design load and moments at roof level are: Axial load F = (1.4 × 70.6) + (1.6 × 17.1) = 126.2 kN Moment Mx = (1.4 × 6.07) + (1.6 × 1.51) = 10.92 kN m My = (1.4 × 6.35) + (1.6 × 1.51) = 11.32 kN m Try 152 × 152 UC 30, the properties of which are: A = 38.2 cm2 ; ry = 3.82 cm; Zy = 73.06 cm3 ; Zx = 221.2 cm3 ; Sx = 247.1 cm3 ; Sy = 111.2 cm3 . The roof beam connections and column section dimensions are shown in Figure 7.24(a). The design strength from Table 9, py = 275 N/mm2 , Flange, b/T = 76.45/9.4 = 8.13 < 9.0—plastic, Web, d/T = 123.4/6.6 = 18.7 < 40— semi-compact. The limiting proportions are from Table 11 of the code. Local capacity check: Moment capacities for the x–x and y–y axes are: Mcx = 247.1 × 275/103 = 67.95 kN m, < 1.2 × 221.2 × 275/103 = 73.0 kN m. Mcy = 111.2 × 275/103 = 30.58 kN m, < 1.2 × 275 × 73.06/103 = 24.10 kN m. Interaction expression: 126.2 × 10 10.92 11.31 + + = 0.75 < 1.0 38.2 × 275 67.95 24.10 The section is satisfactory. Overall buckling check: The column is effectively held in position and partially restrained in direction at both ends. From Table 22, the effective length is: LE = 0.85 × 4000 = 3400 mm λ = 3400/38.2 = 89 From Table 24(c) pc = 144 N/mm2 The axial load at the centre of the column is = 126.2 + (3 × 1.4) = 130.4 kN Compression members 152.9 b = 76.45 8 mm plate X X 6.6 Y X 157.5 d=123.4 Y X 9.4 152 × 152 × 30 UC Y 180 20 mm φ bolts 90 Column section Roof beam connections a) Column – roof to second floor X Y 8.0 203.2 X 160.8 203 × 2.3 × 48 UC Y 11.0 203.2 20mm φ bolts 90 Floor beam connections b) Column – second floor to base Figure 7.24 Column connections and section dimensions The buckling resistance moment Mb is calculated using Section 4.77 of the code: λLT = 0.5 × 4000/38.2 = 52.35 pb = 232.2 N/mm2 (Table 16) Mb = 232.2 × 247.1/101 = 57.4 kN m Interaction expression: 130.4 × 10 10.92 11.32 × 103 + + = 0.98 < 1.0. 38.2 × 144 57.4 275 × 73.06 The section is satisfactory. Eccentrically loaded columns in buildings 181 Second floor to base: The same column section will be used from the second floor to the base. The lower column length between first floor and base will be designed. Referring to Figure 7.23, the design load and moments just below first floor level are: F = (1.4 × 323.8) + (1.6 × 68.4) = 562.76 kN, Mx = (1.4 × 5.84) + (1.6 × 1.73) = 10.94 kN m, My = (1.4 × 6.33) + (1.6 × 1.73) = 11.63 kN m. Try 203 × 203 UC 46, the properties of which are: A = 58.8 cm2 ; 3 Zy = 151.5 cm ; ry = 5.11 cm; Sx = 479.4 cm3 . Local capacity check: The floor beam connections and column section dimensions are shown in Figure 7.24(b). The section is plastic and py = 275 N/mm2 The moment capacities are: Mcy = 275 × 497.4/103 = 136.8 kN m, Mcy = 1.2 × 151.5 × 275/103 = 50.0 kN m. Interaction expression: 562.76 × 10 10.94 11.63 + + = 0.66 < 1.0. 58.8 × 275 36.8 50.0 Overall buckling check: λ = 0.85 × 5000/51 = 83.2 pc = 155.2 N/mm2 (Table 24(c)) Axial load at centre of column: λLT = 562.76 + (1.4 × 3.75) = 568.01 kN, = 0.5 × 5000/51.1 = 48.9. pb = 240.6 N/mm2 —Table 11. Mb = 240.6 × 497.4/103 = 119.6 kN m. 182 Compression members Interaction expression: 568.01 × 10 10.94 11.63 × 103 + + = 0.992 < 1.0. 58.8 × 155.1 119.6 275 × 151.5 The section is satisfactory. 7.7 Cased columns subjected to axial load and moment 7.7.1 Code design requirements The design of cased members subjected to axial load and moment is set out in Section 4.14.4 of BS 5950: Part 1. The member must satisfy two conditions. (1) Capacity check My Fc Mx + + ≤1 Pcs Mcx Mcy where Fc is the compressive force due to axial load, Pcs the compressive resistance of a cased strut with zero slenderness(see Section 7.4.1 1), Mx the applied moment about the x–x axis and My the applied moment about the y–y axis, Mcx the moment capacity of the steel section about the x–x axis, and Mcy the moment capacity of the steel section about the y–y axis. (2) Buckling resistance my My m x Mx Fc + + ≤1 Pc Mb Mcy where Pc is the compression resistance (see Section 7.4.11), m the equivalent uniform moment factor and Mb the buckling resistance moment calculated using the radius of gyration ry for a cased section. 7.7.2 Example A column of length 7 m is subjected to the factored loads and moments as shown in Figure 7.25. Design the column using S275 steel and Grade 30 concrete. Try 203 × 203 UC 60, the properties of which are: A = 75.8 cm2 , Sx = 652 cm3 , rx = 8.96 cm, ry = 5.19 cm, u = 0.847, x = 14.1 The section is plastic and design strength py = 275 N/mm2 . The cased section 320 × 320 mm2 is shown in Figure 7.25(b). Cased columns subjected to axial load and moment (a) 1200 kN (b) 183 320 7m 320 215.9 85 kN 51 kNm 206.2 203 × 203 × 60 UC 610 kN Column and loads Cased section Figure 7.25 Cased column: design example (1) Capacity check The terms for the interaction expression in Section 7.7.1(1) above are calculated:   30 × 322 275 = 2852 kN. Pcs = 75.8 + 0.25 × 275 10 Mcx = 652 × 275/103 = 179.3 kN m. Interaction expression: 85 1200 + = 0.42 + 0.474 = 0.894 < 1.0. 2852 179.3 This is satisfactory. (2) Buckling resistance For the cased section ry = 0.2 × 320 = 64 mm. The strut is taken to be held in position and partially restrained in direction at the ends: LE = 7000 × 0.85 = 5950 mm (Table 22). λ = 5950/64 = 93.0. Pc = 137 N/mm2 (Table 24(c)).   0.45 × 30 × 322 137 = 1727 kN. Pc = 75.8 + 275 10 184 Compression members From Table 18, for β = −51/85 = −0.6: mLT = 0.44, λ = 93.0 (same as above), λ/x = 93/14.1 = 6.59, v = 0.746 (Table 19), λLT = 0.847 × 0.746 × 93 = 58.7, Pb = 216.3 N/mm2 (Table 16), Mb = 216.3 × 652/103 = 141 kN m. Interaction expression: 1200 0.44 × 85 + = 0.694 + 0.265 = 0.959 < 1.0. 1727 141 The section is satisfactory. 7.8 Side column for a single-storey industrial building 7.8.1 Arrangement and loading The cross-section and side elevation of a single-storey industrial building are shown in Figures 7.26(a) and (b). The columns are assumed to be fixed at the base and pinned at the top, and act as partially propped cantilevers in resisting lateral loads. The top of the column is held in the longitudinal direction by the caves member and bracing, as shown on the side elevation. (a) (c) Eaves tie X Section through building L Y (b) Side elevation Side column Figure 7.26 Side column in a single-storey industrial building Y X Side column for a single-storey industrial building 185 The loading on the column is due to: (1) dead and imposed load from the roof and dead load from the walls and column; and (2) wind loading on roof and walls. The load on the roof consists of: (1) dead load due to sheeting, insulation board, purlins and weight of truss and bracing. This is approximately 0.3–0.5 kN/m2 on the slope length of the roof; and (2) Imposed load due to snow, erection and maintenance loads. This is given in BS 6399: Part 1 as 0.75 kN/m2 on plan area. The loading on the walls is due to sheeting, insulation board, sheeting rails and the weight of the column and bracing. The weight is approximately the same as for the roof. The wind load depends on the location and dimensions of the building. The method of calculating the wind load is taken from CP3: Chapter V: Part 2. This is shown in the following example. The breakdown and diagrams for the calculation of the loading and moments on the column are shown in Figure 7.27, and the following comments are made on these figures. (1) The dead and imposed loads give an axial reaction R at the base of the column (see Figure 7.27(a)). (2) The wind on the roof and walls is shown in Figure 7.27(b). There may be a pressure or suction on the windward slope, depending on the angle of the slope. The reactions from wind on the roof only are shown in Figure 7.27(c). The uplift results in vertical reactions R1 , and R2 . The net horizontal reaction is assumed to be divided equally between the two columns. This is 0.5(H2 − H1 ), where H2 and H1 are the horizontal components of the wind loads on the roof slopes. (3) The wind on the walls causes the frame to deflect, as shown in Figure 7.27(d). The top of each column moves by the same amount S. The wind p, and P2 on each wall, taken as uniformly distributed, will have different values, and this results in a force P in the bottom chord of the truss, as shown in Figure 7.27(e). The value of P may be found by equating deflections at the top of each column. For the case where p1 , is greater than P2 there is a compression P in the bottom chord: P L3 p2 L4 P L3 p 1 L4 − = + 8EI 3EI 8EI 3EI This gives P = 3L(p1 − p2 )/16. where I denotes the moment of inertia of the column about the x–x axis (same for each column), E the Young’s modulus and L the column height. 186 Compression members (a) Roof cladding, purlins truss, imposed load Wind Sheeting rails cladding Column R R Dead and imposed loads (b) V1 M1 N1 Resultant loads N 2 Wind loads V2  Wind  H2 H2 – H1 2 R1 H2 – H1 2 R2 Wind on roof Deflected frame Wind (c) From Wind Roof P p1 p2 p1 > p2 Wind Column and wall Wind on walts Column loads Figure 7.27 Loads on side column of an industrial building (4) The resultant loading on the column is shown in Figure 7.27(f), where the horizontal point load at the top is: H = P + (H2 − H1 )/2 The column moments are due entirely to wind load. 7.8.2 Column design procedure (1) Section classification Universal beams are often used for these columns where the axial load is small, but the moment due to wind is large. Referring to Figure 7.28(a), the Side column for a single-storey industrial building 187 classification is checked as follows: (1) Flanges are checked using Table 11 of the code where limits for b/T are given, where b is the flange outstand as shown in the figure and T is the flange thickness. (2) Webs are in combined axial and flexural compression. The classification can be checked using Table 11 and Section 3.5.4 of the code. For example, from Table 11 for webs generally a plastic section has the limit: d 80ε ≤ but ≥ 40ε t 1 + r1 where d is the clear depth of web, t the thickness of web and r1 the stress ratio as defined in Clause 3.55 of the code. (2) Effective length for axial compression Effective lengths for cantilever columns connected by roof trusses are given in Appendix D of BS 5950: Part 1. The tops must be held in position longitudinally by eaves members connected to a braced bay. Two cases are shown in Figure 7.28(b): (1) Column with no restraints: x–x axis LE = 1.5L y–y axis LE = 0.85L. If the base is not effectively fixed about the y–y axis: LE = 1.0L (b) T X Y L1 t L X Eaves member L d Y b L2 (a) Column sector Fixed base No restraint Restraint near centre Column conditions (c) Laced member Lateral support for column Figure 7.28 Side column design features Stays from sheeting rail 188 Compression members (2) Column with restraints: The restraint provides lateral support against buckling about the weak axis: x–x axis LE = 1.5L y–y axis LE = 0.85L1 or L2 , whichever is the greater. The restraint is often provided by a laced member or stays from a sheeting rail, as shown in Figure 8.28(c). (3) Effective length for calculating the buckling resistance moment The effective length of compression flange is estimated using Sections 4.3.5, 4.3.6 and Tables 13 and 14 of BS 5950: Part 1, and the effective lengths for the two cases shown in Figure 7.28(b) are: (1) Column with no restraints (Table 14): The column is fixed at the base and restrained laterally and torsionally at the top. For normal loading LE = 0.5L. Note that the code specifies in this case that the uniform moment factor m is taken as 1.0. (2) Column with restraints: This is to be treated as a beam and the effective length taken from Table 13: LE = 0.85L1 or 1.0L2 in the case shown. (4) Column design The column moment is due to wind and controls the design. The load combination is then dead plus imposed plus wind load. The load factor from Table 2 of the code is γf = 1.2. The following two checks are required in design: (1) Local capacity check at base; and (2) Overall buckling check. The design procedure is shown in the example that follows. (5) Deflection at the column cap The deflection at the column cap must not exceed the limit given in Table 8 of the code for a single-storey building. The limit is height/300. 7.8.3 Example: design of a side column A section through a single-storey building is shown in Figure 7.29. The frames are at 5 m centres and the length of the building is 30 m. The columns are pinned at the top and fixed at the base. The loading is as follows: Roof Dead load–measured on slope Sheeting, insulation board, purlins and truss = 0.45 kN/m2 Imposed load–measured on plan = 0.75 kN/m2 Walls Sheeting, insulation board, sheeting rails = 0.35 kN/m2 Column Estimate = 3.0 kN Wind load: The new code of practice for wind loading is BS 6399, Part 2. Side column for a single-storey industrial building 189 10.7 6m 4m 7m 20 m Figure 7.29 Section through building Determine the loads and moments on the side column and design the member using S275 steel. Note that the column is taken as not being supported laterally between the top and base. (1) Column loads and moments Dead and imposed load Roof Dead load = 10 × 5 × 0.45 × 10.77/10 = 24.23 kN Imposed load = 10 × 5 × 0.75 = 37.5 kN Walls = 6 × 5 × 0.35 = 10.5 kN Column = 3.0 kN Total load at base = 75.23 kN Wind load Location: north-east England. The site wind speed Vs is 26 m/s and the wind load factor Sb is 1.73 taken from Table 4, the effective wind speed, Ve = Vs × Sb = 45 m/s. The wind pressure coefficients and wind loads for the building are shown in Figure 7.30(b). The wind load normal to the walls and roof slope is given by: W = 5 qL (Cpe − Cpi ), L = height of wall or length of roof slope, q = dynamic pressure for walls or roof slopes. The resultant normal loads on the roof and the horizontal and vertical resolved parts are shown in Figure 7.30(c). The horizontal reaction is divided equally between each support and the vertical reactions are found by taking moments about supports. The reactions at the top of the columns for the two wind-load cases are shown in the figure. The wind loading on the walls requires the analysis set out above in Section 7.8.1, where the column tops deflect by an equal amount and a force P is transmitted through the bottom chord of the truss. For the internal pressure case, see Figure 7.30(d): P = 3(8.31 − 6.64)/16 = 0.313 kN The loads and moments on the columns are summarized in Figure 7.31. E G F H α=0 21° 8⬘ h (a) A B 6m Compression members l = 30 m 190 w = 20 m h/w = 6/20 < 1/2 1 < l/w = 20/30 < 1/2 Plan Roof Walls EF A Cpe = – 0.32 Cpe = + 0.7 GH B Cpe = –0.4 Cpe = –0.2 – 0.32 – 0.4 External pressure coefficients Cpe – Wind angle α = 0 (b) – 0.4 – 0.32 – 0.2 0.7 – 0.7 0.7 – 0.3 – 0.2 Pressure coefficients 19.04 kN + 0.2 8.31 kN 3.66 kN 0.732 kN 21.97 kN 1.66 kN –0.3 6.64 kN 16.62 kN Wind loads Pressure coefficients and wind loads (c) 17.68 19.04 21.97 0.55 7.07 20.4 8.16 18.36 0.55 0.68 0.732 0.55 0.27 3.66 3.4 1.36 1.36 19.72 0.55 2.72 Roof loads and reactions –0.3 1.66 +0.2 P = 3.42 16.62 P = 0.313 6.64 8.31 (d) Wind on walls Figure 7.30 Wind-pressure coefficients and loads (2) Notional horizontal loads To ensure stability, the structure is checked for a notional horizontal load in accordance with Clause 2.4.2.3 of BS 5950: Part 1. The notional force from the roof loads is taken as the greater of. One per cent of the factored dead loads = 0.01 × 1.4 × 24.23 = 0.34 kN or 0.5 per cent of the factored dead load plus vertical imposed load = 0.005 1(1.4 × 24.23) + (1.6 × 37.5)1 = 0.5 kN. This load is applied at the top of each column and is taken to act simultaneously with 1.4 times the dead and 1.3 times the imposed vertical loads. Side column for a single-storey industrial building Wind case Column Internal pressure Windward Leeward Internal suction Windward Leeward Dead 24.23 24.23 24.23 24.23 Imposed 39.5 37.5 37.5 37.5 Wind 18.36 19.72 1.36 .237 Wind wall column 8.31 .863 13.5 2.87 6.64 13.5 191 2.72 3.97 16.6 13.5 13.5 1.66 Dead 37.73 37.73 37.73 37.73 Imposed 37.54 37.5 37.5 37.5 Wind 18.36 19.72 1.36 2.72 Wind moment 26.35 25.1 32.64 18.84 Loads are in kN, Moments are in kNm Figure 7.31 Summary of loads and moments The design load at the base is P = (1.4 × 37.73) + (1.3 × 37.5) = 101.57 kN. The moment is: M = 0.5 × 6 = 3.0 kN m. The design conditions for this case are less severe than those for the combination dead + imposed + wind loads. (3) Column design The maximum design condition is for the wind-load case of internal suction. For the windward column, the load combination is dead plus imposed plus wind loads. Local capacity check (see Section 7.52) Design load = 1.2(37.73 + 37.5 − 1.36) = 88.64 kN Design moment = 1.2 × 32.64 = 39.17 kN m Try 406 × 140 UB 39, the properties of which are: A = 49.4 cm2 ; rx = 15.88 cm; Sx = 721 cm3 ; ry = 2.89 cm; Zx = 627 cm3 , x = 47.4, I = 12 452 cm4 . 192 Compression members Check the section classification using Table 7. The section dimensions are shown in Figure 7.32: Design strength py = 275 N/mm2 (Table 9), Factor ε = 1.0, Flanges b/T = 70.9/8.6 = 8.2 < 9.0 (plastic), Web. This is in combined axial and flexural compression. Design axial load = 88.64 kN, Length of web supporting the load at the design strength: = 88.64 × 103 = 51.1 mm, 275 × 6.3 For the web: 359.6 d = = 57.1 t 6.3 Limiting value for plastic web: d 80ε < but ≥ 40ε t 1 + r1 where the web stress ratio r1 is 88.64 × 103 Fc = 0.1422, = dtPyw 359.6 × 6.3 × 275 80 × 1.0 d < = 70 > 57.1 (plastic web) t 1 + 0.1422 r1 = The moment capacity about the x–x axis: Mcx = 275 × 721/103 = 198.22 kN m < 1.2 × 275 × 627/103 = 206.9 kN m X 6.3 Figure 7.32 Column section Supports axial load 205.3 X 51.1 359.6 8.6 141.8 b = 70.9 X1 X1 Plastic neutral axis Side column for a single-storey industrial building 193 Interaction expression: 88.64 × 10 39.17 + = 0.27 < 1.0 49.4 × 275 198.22 The sect