the behaviour and design of steel structures to ec3
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The Behaviour and Design ofSteel Structures to EC3
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The Behaviour and Designof Steel Structures to EC3
Fourth edition
NS Trahair MA BradfordDA Nethercot and L Gardner
First edition published 1977Second edition 1988 revised second edition published 1988Third edition ndash Australian (AS4100) published 1998Third edition ndash British (BS5950) published 2001
Fourth edition published 2008byTaylor amp Francis2 Park Square Milton Park Abingdon Oxon OX14 4RN
Simultaneously published in the USA and CanadabyTaylor amp Francis270 Madison Ave NewYork NY 10016
Taylor amp Francis is an imprint of theTaylor amp Francis Groupan informa business
copy 1977 1988 2001 2008 NS Trahair MA BradfordDA Nethercot and L Gardner
All rights reserved No part of this book may be reprinted orreproduced or utilised in any form or by any electronicmechanical or other means now known or hereafterinvented including photocopying and recording or in anyinformation storage or retrieval system without permission inwriting from the publishers
The publisher makes no representation express or impliedwith regard to the accuracy of the information contained in this book andcannot accept any legal responsibility or liability for any efforts oromissions that may be made
British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library
Library of Congress Cataloging in Publication DataThe behaviour and design of steel structures to EC3 NS Trahair
[et al] ndash 4th edp cm
Includes bibliographical references and index1 Building Iron and steel I Trahair NS
TA684T7 20086241prime821ndashdc22 2007016421
ISBN10 0ndash415ndash41865ndash8 (hbk)ISBN10 0ndash415ndash41866ndash6 (pbk)ISBN10 0ndash203ndash93593ndash4 (ebk)
ISBN13 978ndash0ndash415ndash41865ndash2 (hbk)ISBN13 978ndash0ndash415ndash41866ndash9 (pbk)ISBN13 978ndash0ndash203ndash93593ndash4 (ebk)
This edition published in the Taylor amp Francis e-Library 2007
ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo
ISBN 0-203-93593-4 Master e-book ISBN
Contents
Preface ixUnits and conversion factors xGlossary of terms xiNotations xv
1 Introduction 1
11 Steel structures 112 Design 313 Material behaviour 814 Member and structure behaviour 1415 Loads 1716 Analysis of steel structures 2117 Design of steel structures 24
References 30
2 Tension members 33
21 Introduction 3322 Concentrically loaded tension members 3323 Eccentrically and locally connected tension members 3724 Bending of tension members 3925 Stress concentrations 4026 Design of tension members 4227 Worked examples 4528 Unworked examples 48
References 49
3 Compression members 50
31 Introduction 5032 Elastic compression members 51
vi Contents
33 Inelastic compression members 5534 Real compression members 6135 Design of compression members 6236 Restrained compression members 6537 Other compression members 7438 Appendix ndash elastic compression members 7839 Appendix ndash inelastic compression members 81310 Appendix ndash effective lengths of compression members 83311 Appendix ndash torsional buckling 88312 Worked examples 89313 Unworked examples 96
References 98
4 Local buckling of thin-plate elements 100
41 Introduction 10042 Plate elements in compression 10243 Plate elements in shear 11344 Plate elements in bending 11845 Plate elements in bending and shear 12146 Plate elements in bearing 12447 Design against local buckling 12648 Appendix ndash elastic buckling of plate elements
in compression 13949 Worked examples 141410 Unworked examples 151
References 152
5 In-plane bending of beams 154
51 Introduction 15452 Elastic analysis of beams 15653 Bending stresses in elastic beams 15754 Shear stresses in elastic beams 16355 Plastic analysis of beams 17556 Strength design of beams 18357 Serviceability design of beams 18958 Appendix ndash bending stresses in elastic beams 19059 Appendix ndash thin-walled section properties 191510 Appendix ndash shear stresses in elastic beams 195511 Appendix ndash plastic analysis of beams 197
Contents vii
512 Worked examples 205513 Unworked examples 224
References 225
6 Lateral buckling of beams 227
61 Introduction 22762 Elastic beams 22863 Inelastic beams 23764 Real beams 23965 Design against lateral buckling 24066 Restrained beams 24567 Cantilevers and overhanging beams 25368 Braced and continuous beams 25669 Rigid frames 261610 Monosymmetric beams 263611 Non-uniform beams 266612 Appendix ndash elastic beams 268613 Appendix ndash effective lengths of beams 273614 Appendix ndash monosymmetric beams 274615 Worked examples 275616 Unworked examples 290
References 291
7 Beam-columns 295
71 Introduction 29572 In-plane behaviour of isolated beam-columns 29673 Flexuralndashtorsional buckling of isolated
beam-columns 31174 Biaxial bending of isolated beam-columns 31975 Appendix ndash in-plane behaviour of elastic beam-columns 32376 Appendix ndash flexuralndashtorsional buckling of
elastic beam-columns 32677 Worked examples 32978 Unworked examples 343
References 345
8 Frames 347
81 Introduction 34782 Triangulated frames 348
viii Contents
83 Two-dimensional flexural frames 35084 Three-dimensional flexural frames 37285 Worked examples 37386 Unworked examples 386
References 388
9 Joints 392
91 Introduction 39292 Joint components 39293 Arrangement of joints 39694 Behaviour of joints 39895 Common joints 40696 Design of bolts 41097 Design of bolted plates 41498 Design of welds 41799 Appendix ndash elastic analysis of joints 420910 Worked examples 423911 Unworked examples 431
References 432
10 Torsion members 433
101 Introduction 433102 Uniform torsion 436103 Non-uniform torsion 449104 Torsion design 461105 Torsion and bending 466106 Distortion 469107 Appendix ndash uniform torsion 471108 Appendix ndash non-uniform torsion 473109 Worked examples 4781010 Unworked examples 484
References 485
Index 487
Preface
This fourth British edition has been directed specifically to the design of steel structures inaccordance with Eurocode 3 Design of Steel Structures The principal part of this is Part1-1General Rules and Rules for Buildings and this is referred to generally in the text asEC3 Also referred to in the text are Part 1-5 Plated Structural Elements and Part 1-8Design Of Joints which are referred to as EC3-1-5 and EC3-1-8 EC3 will be accompaniedby NationalAnnexes which will contain any National Determined Parameters for the UnitedKingdom which differ from the recommendations given in EC3
Designers who have previously used BS5950 (which is discussed in the third Britishedition of this book) will see a number of significant differences in EC3 One of the moreobvious is the notation The notation in this book has been changed generally so that it isconsistent with EC3
Another significant difference is the general absence of tables of values computed fromthe basic design equations which might be used to facilitate manual design Some design-ers will want to prepare their own tables but in some cases the complexities of the basicequations are such that computer programs are required for efficient design This is espe-cially the case for members under combined compression and bending which are discussedin Chapter 7 However the examples in this book are worked in full and do not rely on suchdesign aids
EC3 does not provide approximations for calculating the lateral buckling resistancesof beams but instead expects the designer to be able to determine the elastic bucklingmoment to be used in the design equations Additional information to assist designers inthis determination has been given in Chapter 6 of this book EC3 also expects the designerto be able to determine the elastic buckling loads of compression members The additionalinformation given in Chapter 3 has been retained to assist designers in the calculation ofthe elastic buckling loads
EC3 provides elementary rules for the design of members in torsion These are gener-alised and extended in Chapter 10 which contains a general treatment of torsion togetherwith a number of design aids
The preparation of this fourth British edition has provided an opportunity to revise thetext generally to incorporate the results of recent findings and research This is in accordancewith the principal objective of the book to provide students and practising engineers withan understanding of the relationships between structural behaviour and the design criteriaimplied by the rules of design codes such as EC3
NS Trahair MA Bradford DA Nethercot and L GardnerApril 2007
Units and conversion factors
Units
While most expressions and equations used in this book are arranged so that theyare non-dimensional there are a number of exceptions In all of these SI units areused which are derived from the basic units of kilogram (kg) for mass metre (m)for length and second (s) for time
The SI unit of force is the newton (N) which is the force which causes a mass of1 kg to have an acceleration of 1 ms2 The acceleration due to gravity is 9807 ms2
approximately and so the weight of a mass of 1 kg is 9807 NThe SI unit of stress is the pascal (Pa) which is the average stress exerted by a
force of 1 N on an area of 1 m2 The pascal is too small to be convenient in structuralengineering and it is common practice to use either the megapascal (1 MPa =106 Pa) or the identical newton per square millimetre (1 Nmm2 = 106 Pa) Thenewton per square millimetre (Nmm2) is used generally in this book
Table of conversion factors
To Imperial (British) units To SI units
1 kg = 0068 53 slug 1 slug = 1459 kg1 m = 3281 ft 1 ft = 0304 8 m
= 3937 in 1 in = 0025 4 m1 mm = 0003 281 ft 1 ft = 3048 mm
= 0039 37 in 1 in = 254 mm1 N = 0224 8 lb 1 lb = 4448 N1 kN = 0224 8 kip 1 kip = 4448 kN
= 0100 36 ton 1 ton = 9964 kN1 Nmm2lowastdagger = 0145 0 kipin2 (ksi) 1 kipin2 = 6895 Nmm2
= 0064 75 tonin2 1 tonin2 = 1544 Nmm2
1 kNm = 0737 6 kip ft 1 kip ft = 1356 kNm= 0329 3 ton ft 1 ton ft = 3037 kNm
Noteslowast 1 Nmm2 = 1 MPadagger There are some dimensionally inconsistent equations used in this book which arise
because a numerical value (in Nmm2) is substituted for the Youngrsquos modulus of elasticityE while the yield stress fy remains algebraic The value of the yield stress fy used in theseequations should therefore be expressed in Nmm2 Care should be used in convertingthese equations from SI to Imperial units
xii Glossary of terms
Distortion A mode of deformation in which the cross-section of a memberchanges shape
Effective length The length of an equivalent simply supported member whichhas the same elastic buckling load as the actual member Referred to in EC3 asthe buckling length
Effective width That portion of the width of a flat plate which has a non-uniformstress distribution (caused by local buckling or shear lag) which may be con-sidered as fully effective when the non-uniformity of the stress distribution isignored
Elastic buckling analysis An analysis of the elastic buckling of the member orframe out of the plane of loading
Elastic buckling load The load at elastic buckling Referred to in EC3 as theelastic critical buckling load
Elastic buckling stress The maximum stress at elastic buckling Referred to inEC3 as the elastic critical buckling stress
Factor of safety The factor by which the strength is divided to obtain the workingload capacity and the maximum permissible stress
Fastener A bolt pin rivet or weld used in a connectionFatigue A mode of failure in which a member fractures after many applications
of loadFirst-order analysis An analysis in which equilibrium is formulated for the
undeformed position of the structure so that the moments caused by productsof the loads and deflections are ignored
Flexural buckling A mode of buckling in which a member deflectsFlexuralndashtorsional buckling A mode of buckling in which a member deflects
and twists Referred to in EC3 as torsionalndashflexural buckling or lateralndashtorsionalbuckling
Friction-grip joint A joint in which forces are transferred by friction forces gen-erated between plates by clamping them together with preloaded high-strengthbolts
Geometrical imperfection Initial crookedness or twist of a memberGirt A horizontal member between columns which supports wall sheetingGusset A short-plate element used in a connectionImposed load The load assumed to act as a result of the use of the structure but
excluding wind loadInelastic behaviour Deformations accompanied by yieldingIn-plane behaviour The behaviour of a member which deforms only in the plane
of the applied loadsJoint The means by which members are connected together and through which
forces and moments are transmittedLateral buckling Flexuralndashtorsional buckling of a beam Referred to in EC3 as
lateralndashtorsional bucklingLimit states design Amethod of design in which the performance of the structure
is assessed by comparison with a number of limiting conditions of usefulness
Glossary of terms xiii
The most common conditions are the strength limit state and the serviceabilitylimit state
Load effects Internal forces and moments induced by the loadsLoad factor A factor used to multiply a nominal load to obtain part of the design
loadLoads Forces acting on a structureLocal buckling A mode of buckling which occurs locally (rather than generally)
in a thin-plate element of a memberMechanism A structural system with a sufficient number of frictionless and
plastic hinges to allow it to deform indefinitely under constant loadMember A one-dimensional structural element which supports transverse or
longitudinal loads or momentsNominal load The load magnitude determined from a loading code or
specificationNon-uniform torsion The general state of torsion in which the twist of the
member varies non-uniformlyPlastic analysis Amethod of analysis in which the ultimate strength of a structure
is computed by considering the conditions for which there are sufficient plastichinges to transform the structure into a mechanism
Plastic hinge A fully yielded cross-section of a member which allows themember portions on either side to rotate under constant moment (the plasticmoment)
Plastic section A section capable of reaching and maintaining the full plasticmoment until a plastic collapse mechanism is formed Referred to in EC3 as aClass 1 section
Post-buckling strength A reserve of strength after buckling which is possessedby some thin-plate elements
Preloaded bolts High-strength bolts used in friction-grip jointsPurlin A horizontal member between main beams which supports roof sheetingReduced modulus The modulus of elasticity used to predict the buckling of
inelastic members under the so called constant applied load because it isreduced below the elastic modulus
Residual stresses The stresses in an unloaded member caused by non-uniformplastic deformation or by uneven cooling after rolling flame cutting or welding
Rigid frame A frame with rigid connections between members Referred to inEC3 as a continuous frame
Second-order analysis An analysis in which equilibrium is formulated for thedeformed position of the structure so that the moments caused by the productsof the loads and deflections are included
Semi-compact section A section which can reach the yield stress but whichdoes not have sufficient resistance to inelastic local buckling to allow it to reachor to maintain the full plastic moment while a plastic mechanism is formingReferred to in EC3 as a Class 3 section
Semi-rigid frame A frame with semi-rigid connections between membersReferred to in EC3 as a semi-continuous frame
xiv Glossary of terms
Service loads The design loads appropriate for the serviceability limit stateShear centre The point in the cross-section of a beam through which the resultant
transverse force must act if the beam is not to twistShear lag A phenomenon which occurs in thin wide flanges of beams in which
shear straining causes the distribution of bending normal stresses to becomesensibly non-uniform
Simple frame A frame for which the joints may be assumed not to transmitmoments
Slender section Asection which does not have sufficient resistance to local buck-ling to allow it to reach the yield stress Referred to in EC3 as a Class 4 section
Splice A connection between two similar collinear membersSquash load The value of the compressive axial load which will cause yielding
throughout a short memberStiffener A plate or section attached to a web to strengthen a memberStrain-hardening Astressndashstrain state which occurs at stresses which are greater
than the yield stressStrength limit state The state of collapse or loss of structural integritySystem length Length between adjacent lateral brace points or between brace
point and an adjacent end of the memberTangent modulus The slope of the inelastic stressndashstrain curve which is used to
predict buckling of inelastic members under increasing loadTensile strength The maximum nominal stress which can be reached in tensionTension field A mode of shear transfer in the thin web of a stiffened plate girder
which occurs after elastic local buckling takes place In this mode the tensiondiagonal of each stiffened panel behaves in the same way as does the diagonaltension member of a parallel chord truss
Tension member A member which supports axial tension loadsTorsional buckling A mode of buckling in which a member twistsUltimate load design A method of design in which the ultimate load capacity
of the structure is compared with factored loadsUniform torque That part of the total torque which is associated with the rate
of change of the angle of twist of the member Referred to in EC3 as St Venanttorque
Uniform torsion The special state of torsion in which the angle of twist of themember varies linearly Referred to in EC3 as St Venant torsion
Warping A mode of deformation in which plane cross-sections do not remainin plane
Warping torque The other part of the total torque (than the uniform torque)This only occurs during non-uniform torsion and is associated with changes inthe warping of the cross-sections
Working load design A method of design in which the stresses caused by theservice loads are compared with maximum permissible stresses
Yield strength The average stress during yielding when significant strainingtakes place Usually the minimum yield strength in tension specified for theparticular steel
Glossary of terms
Actions The loads to which a structure is subjectedAdvanced analysis An analysis which takes account of second-order effects
inelastic behaviour residual stresses and geometrical imperfectionsBeam A member which supports transverse loads or moments onlyBeam-column A member which supports transverse loads or moments which
cause bending and axial loads which cause compressionBiaxial bending The general state of a member which is subjected to bending
actions in both principal planes together with axial compression and torsionactions
Brittle fracture Amode of failure under a tension action in which fracture occurswithout yielding
Buckling A mode of failure in which there is a sudden deformation in a directionor plane normal to that of the loads or moments acting
Buckling length The length of an equivalent simply supported member whichhas the same elastic buckling load as the actual member
Cleat A short-length component (often of angle cross-section) used in aconnection
Column A member which supports axial compression loadsCompact section Asection capable of reaching the full plastic moment Referred
to in EC3 as a Class 2 sectionComponent method of design A method of joint design in which the behaviour
of the joint is synthesised from the characteristics of its componentsConnection A jointDead load The weight of all permanent construction Referred to in EC3 as
permanent loadDeformation capacity A measure of the ability of a structure to deform as a
plastic collapse mechanism develops without otherwise failingDesign load A combination of factored nominal loads which the structure is
required to resistDesign resistance The capacity of the structure or element to resist the design
load
Notations
The following notation is used in this book Usually only one meaning is assignedto each symbol but in those cases where more meanings than one are possiblethen the correct one will be evident from the context in which it is used
Main symbols
A AreaB Bimomentb WidthC Coefficientc Width of part of sectiond Depth or DiameterE Youngrsquos modulus of elasticitye Eccentricity or ExtensionF Force or Force per unit lengthf Stress property of steelG Dead load or Shear modulus of elasticityH Horizontal forceh Height or Overall depth of sectionI Second moment of areai Integer or Radius of gyrationk Buckling coefficient or Factor or Relative stiffness ratioL LengthM Momentm IntegerN Axial force or Number of load cyclesn Integerp Distance between holes or rows of holesQ Loadq Intensity of distributed loadR Radius or Reaction or Resistancer Radius
xvi Notations
s SpacingT Torquet ThicknessU Strain energyu Deflection in x directionV Shear or Vertical loadv Deflection in y directionW Section modulus or Work donew Deflection in z directionx Longitudinal axisy Principal axis of cross-sectionz Principal axis of cross-sectionα Angle or Factor or Load factor at failure or Stiffnessχ Reduction factor∆ Deflection∆σ Stress rangeδ Amplification factor or Deflectionε Normal strain or Yield stress coefficient = radic
(235fy)φ Angle of twist rotation or Global sway imperfectionγ Partial factor or Shear strainκ Curvatureλ Plate slenderness = (ct)ελ Generalised slendernessmicro Slip factorν Poissonrsquos ratioθ Angleσ Normal stressτ Shear stress
Subscripts
as AntisymmetricB Bottomb Beam or Bearing or Bending or Bolt or Bracedc Centroid or Column or Compressioncr Elastic (critical) bucklingd DesignEd Design load effecteff Effectiveel ElasticF Forcef FlangeG Dead loadI Imposed load
Notations xvii
i Initial or Integerj Jointk Characteristic valueL LeftLT Lateral (or lateralndashtorsional) bucklingM Materialm Momentmax Maximummin MinimumN Axial forcen Integer or Nominal valuenet Netop Out-of-planep Bearing or Platep pl PlasticQ Variable loadR Resistance or Rightr Rafter or ReducedRd Design resistanceRk Characteristic resistances Slip or Storey or Sway or Symmetricser Servicest Stiffener or Strain hardeningT Top or Torsional bucklingt St Venant or uniform torsion or TensionTF Flexuralndashtorsional (or torsionalndashflexural) bucklingtf Tension fieldult UltimateV v ShearW Wind loadw Warping or Web or Weldx x axisy y axis or Yieldz z axisσ Normal stressτ Shear stress0 Initial value1ndash4 Cross-section class
Additional notationsAe Area enclosed by hollow sectionAf max Flange area at maximum sectionAf min Flange area at minimum section
xviii Notations
Ah Area of hole reduced for staggerAnt Anv Net areas subjected to tension or shearAs Tensile stress area of a boltAv Shear area of sectionC Index for portal frame bucklingCm Equivalent uniform moment factorD Plate rigidity Et312(1 minus v2)
D Vector of nodal deformationsEr Reduced modulusEt Tangent modulusF Buckling factor for beam-columns with unequal end momentsFpC Bolt preloadFL FT Weld longitudinal and transverse forces per unit lengthFT Rd Design resistance of a T-stub flange[G] Stability matrixIcz Second moment of area of compression flangeIm Second moment of area of memberIn = b3
ntn12Ir Second moment of area of restraining member or rafterIt Uniform torsion section constantIyz Product second moment of areaIw Warping torsion section constantIzm Value of Iz for critical segmentIzr Value of Iz for restraining segmentK Beam or torsion constant = radic
(π2EIwGJL2) or Fatigue lifeconstant
[K] Elastic stiffness matrixKm = radic
(π2EIyd2f 4GItL2)
Lc Distance between restraints or Length of column which failsunder N alone
Lj Length between end bolts in a long jointLm Length of critical segment or Member lengthLr Length of restraining segment or rafterLstable Stable length for member with plastic hingesLF Load factorMA MB End momentsMb0yRd Design member moment resistance when N = 0 and Mz = 0Mc0zRd Design member moment resistance when N = 0 and My = 0McrMN Elastic buckling moment reduced for axial compressionMN yRd MN zRd Major and minor axis beam section moment resistancesME = (πL)
radic(EIyGIt)
Mf First-order end moment of frame memberMfb Braced component of Mf
Notations xix
Mfp Major axis moment resisted by plastic flangesMfs Sway component of Mf
MI Inelastic beam buckling momentMIu Value of MI for uniform bendingML Limiting end moment on a crooked and twisted beam at first
yieldMmax0 Value of Mmax when N = 0MN Plastic moment reduced for axial forceMb Out-of-plane member moment resistance for bending aloneMbt Out-of-plane member moment resistance for bending and
tensionMry Mrz Section moment capacities reduced for axial loadMS Simple beam momentMty Lesser of Mry and Mbt
Mzx Value of Mcr for simply supported beam in uniform bendingMzxr Value of Mzx reduced for incomplete torsional end restraintNi Vector of initial axial forcesNbRd Design member axial force resistance when My = 0 and Mz = 0NcrMN Elastic buckling load reduced for bending momentNcrL = π2EIL2
Ncrr Reduced modulus buckling loadNim Constant amplitude fatigue life for ith stress rangeNcrt Tangent modulus buckling loadQD Concentrated dead loadQI Concentrated imposed loadQm Upper-bound mechanism estimate of Qult
Qms Value of Qs for the critical segmentQrs Value of Qs for an adjacent restraining segmentQs Buckling load for an unrestrained segment or Lower bound
static estimate of Qult
R Radius of circular cross-section or Ratio of column and rafterstiffnesses or Ratio of minimum to maximum stress
RH Ratio of rafter rise to column heightR R1minus4 Restraint parametersSF Factor of safetySj Joint stiffnessTM Torque exerted by bending momentTP Torque exerted by axial loadVR Resultant shear forceVTy VTz Transverse shear forces in a fillet weldVvi Shear force in ith fastenera = radic
(EIwGJ ) or Distance along member or Distance fromweb to shear centre or Effective throat size of a weld or Ratioof web to total section area or Spacing of transverse stiffeners
xx Notations
a0 Distance from shear centreb cf or cw
c Factor for flange contribution to shear resistancecm Bending coefficient for beam-columns with unequal end
momentsde Depth of elastic coredf Distance between flange centroidsd0 Hole diametere1 End distance in a platee2 Edge distance in a plateeNy Shift of effective compression force from centroidf Factor used to modify χLT
hw Clear distance between flangesif z Radius of gyration of equivalent compression flangeip Polar radius of gyration
i0 =radic(i2
p + y20 + z2
0)
k Deflection coefficient or Modulus of foundation reactionkc Slenderness correction factor or Correction factor for moment
distributionkij Interaction factors for bending and compressionks Factor for hole shape and sizekt Axial stiffness of connectorkv Shear stiffness of connectorkσ Plate buckling coefficientk1 Factor for plate tension fracturek1 k2 Stiffness factorseff Effective length of a fillet weld or Effective length of an
unstiffened column flangey Effective loaded lengthm Fatigue life index or Torque per unit lengthn Axial compression ratio or Number of shear planespF Probability of failurep(x) Particular integralp1 Pitch of bolt holesp2 Spacing of bolt hole liness Distance around thin-walled sectionss Stiff bearing lengths Staggered pitch of holessm Minimum staggered pitch for no reduction in effective areas1 s2 Side widths of a fillet weldw = WplWel
wAB Settlement of B relative to Awc Mid-span deflectionz Distance to centroid
Notations xxi
yp zp Distances to plastic neutral axesyr zr Coordinates of centre of rotationy0 z0 Coordinates of shear centrezc Distance to buckling centre of rotation or Distance to centroidzn Distance below centroid to neutral axiszQ Distance below centroid to loadzt Distance below centroid to translational restraintα Coefficient used to determine effective width or Unit warping
(see equation 1035)αbc Buckling coefficient for beam columns with unequal end
momentsαbcI Inelastic moment modification factor for bending and com-
pressionαbcu Value of αbc for ultimate strengthαd Factor for plate tear outαi In-plane load factorαL Limiting value of α for second mode bucklingαLT Imperfection factor for lateral bucklingαLα0 Indices in interaction equations for biaxial bendingαm Moment modification factor for beam lateral buckling
αn = (1A)int E
0 αtdsαr αt Rotational and translational stiffnessesαxαy Stiffnesses of rotational restraints acting about the x y axesαst Buckling moment factor for stepped and tapered beamsα β Indices in section interaction equations for biaxial bendingβ Correction factor for the lateral buckling of rolled sections or
Safety indexβe Stiffness factor for far end restraint conditionsβLf Reduction factor for bolts in long jointsβm End moment factor or Ratio of end momentsβw Fillet weld correlation factorβy Monosymmetry section constant for I-beamβ23 Effective net area factors for eccentrically connected tension
membersσC Reference value for fatigue siteσL Fatigue endurance limitε Load height parameter = (Kπ)2zQdf
Φ Cumulative frequency distribution of a standard normalvariate or Value used to determine χ
φCd Design rotation capacity of a jointφj Joint rotationγF γG γQ Load partial factorsγm γn γs Factors used in moment amplification
xxii Notations
γ12 Relative stiffnessesη Crookedness or imperfection parameter or Web shear resis-
tance factor (= 12 for steels up to S460)λc0 Slenderness limit of equivalent compression flangeλp Generalised plate slenderness = radic
(fyσcr)
λ1 = πradic(Efy)
micro = radic(NEI)
θ Central twist or Slope change at plastic hinge or Torsion stressfunction
ρ Perpendicular distance from centroid or Reduction factorρm Monosymmetric section parameter = IzcIz
ρc ρr Column and rafter factors for portal frame bucklingρ0 Perpendicular distance from shear centreσac σat Stresses due to axial compression and tensionσbcy Compression stress due to bending about y axisσbty σbtz Tension stresses due to bending about y z axesσcrl Bending stress at local bucklingσcrp Bearing stress at local bucklingσL Limiting major axis stress in a crooked and twisted beam at
first yieldτh τv Shear stresses due to Vy Vz
τhc τvc Shear stresses due to a circulating shear flowτho τvo Shear stresses in an open sectionψ End moment ratio or Stress ratioψ0 Load combination factor
Chapter 1
Introduction
11 Steel structures
Engineering structures are required to support loads and resist forces and totransfer these loads and forces to the foundations of the structures The loads andforces may arise from the masses of the structure or from manrsquos use of the struc-tures or from the forces of nature The uses of structures include the enclosure ofspace (buildings) the provision of access (bridges) the storage of materials (tanksand silos) transportation (vehicles) or the processing of materials (machines)Structures may be made from a number of different materials including steelconcrete wood aluminium stone plastic etc or from combinations of these
Structures are usually three-dimensional in their extent but sometimes they areessentially two-dimensional (plates and shells) or even one-dimensional (linesand cables) Solid steel structures invariably include comparatively high volumesof high-cost structural steel which are understressed and uneconomic except invery small-scale components Because of this steel structures are usually formedfrom one-dimensional members (as in rectangular and triangulated frames) orfrom two-dimensional members (as in box girders) or from both (as in stressedskin industrial buildings) Three-dimensional steel structures are often arranged sothat they act as if composed of a number of independent two-dimensional framesor one-dimensional members (Figure 11)
Structural steel members may be one-dimensional as for beams and columns(whose lengths are much greater than their transverse dimensions) or two-dimensional as for plates (whose lengths and widths are much greater than theirthicknesses) as shown in Figure 12c While one-dimensional steel members maybe solid they are usually thin-walled in that their thicknesses are much less thantheir other transverse dimensions Thin-walled steel members are rolled in a num-ber of cross-sectional shapes [1] or are built up by connecting together a numberof rolled sections or plates as shown in Figure 12b Structural members canbe classified as tension or compression members beams beam-columns torsionmembers or plates (Figure 13) according to the method by which they transmitthe forces in the structure The behaviour and design of these structural membersare discussed in this book
2 Introduction
[2-D]rigid frames
+
[1-D]beams(purlins and girts)
+
[2-D] bracingtrusses
[3-D] structure
equiv
Figure 11 Reduction of a [3-D] structure to simpler forms
(a) Solid [1-D] member (b) Thin-walled [1-D] members
(c) [2-D] member
t d ltlt L~~ t ltlt d ltlt L
t ltlt d L~~
Figure 12 Types of structural steel members
Structural steel members may be connected together at joints in a number ofways and by using a variety of connectors These include pins rivets bolts andwelds of various types Steel plate gussets or angle cleats or other elements mayalso be used in the connections The behaviour and design of these connectors andjoints are also discussed in this book
Introduction 3
(a) Tension member(b) Compression
member (c) Beam
(d) Beam-column (e) Torsion member (f) Plate
Figure 13 Load transmission by structural members
This book deals chiefly with steel frame structures composed of one-dimensionalmembers but much of the information given is also relevant to plate structuresThe members are generally assumed to be hot-rolled or fabricated from hot-rolledelements while the frames considered are those used in buildings However muchof the material presented is also relevant to bridge structures [2 3] and to structuralmembers cold-formed from light-gauge steel plates [4ndash7]
The purposes of this chapter are first to consider the complete design process andthe relationships between the behaviour and analysis of steel structures and theirstructural design and second to present information of a general nature (includinginformation on material properties and structural loads) which is required for usein the later chapters The nature of the design process is discussed first and thenbrief summaries are made of the relevant material properties of structural steeland of the structural behaviour of members and frames The loads acting on thestructures are considered and the choice of appropriate methods of analysing thesteel structures is discussed Finally the considerations governing the synthesis ofan understanding of the structural behaviour with the results of analysis to formthe design processes of EC3 [8] are treated
12 Design
121 Design requirements
The principal design requirement of a structure is that it should be effective thatis it should fulfil the objectives and satisfy the needs for which it was created The
4 Introduction
structure may provide shelter and protection against the environment by enclosingspace as in buildings or it may provide access for people and materials as inbridges or it may store materials as in tanks and silos or it may form part of amachine for transporting people or materials as in vehicles or for operating onmaterials The design requirement of effectiveness is paramount as there is littlepoint in considering a structure which will not fulfil its purpose
The satisfaction of the effectiveness requirement depends on whether the struc-ture satisfies the structural and other requirements The structural requirementsrelate to the way in which the structure resists and transfers the forces and loadsacting on it The primary structural requirement is that of safety and the first con-sideration of the structural engineer is to produce a structure which will not fail inits design lifetime or which has an acceptably low risk of failure The other impor-tant structural requirement is usually concerned with the stiffness of the structurewhich must be sufficient to ensure that the serviceability of the structure is notimpaired by excessive deflections vibrations and the like
The other design requirements include those of economy and of harmony Thecost of the structure which includes both the initial cost and the cost of mainte-nance is usually of great importance to the owner and the requirement of economyusually has a significant influence on the design of the structure The cost of thestructure is affected not only by the type and quantity of the materials used butalso by the methods of fabricating and erecting it The designer must thereforegive careful consideration to the methods of construction as well as to the sizes ofthe members of the structure
The requirements of harmony within the structure are affected by the relation-ships between the different systems of the structure including the load resistanceand transfer system (the structural system) the architectural system the mechan-ical and electrical systems and the functional systems required by the use of thestructure The serviceability of the structure is usually directly affected by the har-mony or lack of it between the systems The structure should also be in harmonywith its environment and should not react unfavourably with either the communityor its physical surroundings
122 The design process
The overall purpose of design is to invent a structure which will satisfy the designrequirements outlined in Section 121 Thus the structural engineer seeks to inventa structural system which will resist and transfer the forces and loads acting onit with adequate safety while making due allowance for the requirements of ser-viceability economy and harmony The process by which this may be achieved issummarised in Figure 14
The first step is to define the overall problem by determining the effectivenessrequirements and the constraints imposed by the social and physical environmentsand by the ownerrsquos time and money The structural engineer will need to consult theowner the architect the site construction mechanical and electrical engineers
Introduction 5
Definition of the problem
Invention of alternatives
Preliminary design
Preliminary evaluation
Selection
Modification
Final design
Final evaluation
Documentation
Execution
Use
Needs
Constraints
Objectives
Overall system
Structural system
Other systems
Structural designLoadsAnalysisProportioningOther design
Effectiveness
Safety and serviceability
Economy
Harmony
Structural design
Other design
DrawingsSpecification
TenderingConstruction andsupervisionCertification
Figure 14 The overall design process
and any authorities from whom permissions and approvals must be obtainedA set of objectives can then be specified which if met will ensure the successfulsolution of the overall design problem
The second step is to invent a number of alternative overall systems and theirassociated structural systems which appear to meet the objectives In doing sothe designer may use personal knowledge and experience or that which can be
6 Introduction
Invention or modificationof structural system
Preliminary analysis
Proportioning membersand joints
KnowledgeExperienceImaginationIntuitionCreativity
Approximations
Loads
Behaviour
Design criteria
Design codes
larrlarrlarrlarrlarr
---
larrlarr
AnalysisLoads
Behaviour
--
EvaluationDesign criteria
Design codes
larrlarr
Figure 15 The structural design process
gathered from others [9ndash12] or the designer may use his or her own imaginationintuition and creativity [13] or a combination of all of these
Following these first two steps of definition and invention come a series of stepswhich include the structural design evaluation selection and modification of thestructural system These may be repeated a number of times before the structuralrequirements are met and the structural design is finalised A typical structuraldesign process is summarised in Figure 15
After the structural system has been invented it must be analysed to obtain theinformation required for determining the member sizes First the loads supportedby and the forces acting on the structure must be determined For this purpose load-ing codes [14 15] are usually consulted but sometimes the designer determinesthe loading conditions or commissions experts to do this Anumber of approximateassumptions are made about the behaviour of the structure which is then analysedand the forces and moments acting on the members and joints of the structureare determined These are used to proportion the structure so that it satisfies thestructural requirements usually by referring to a design code such as EC3 [8]
At this stage a preliminary design of the structure will have been completed how-ever because of the approximate assumptions made about the structural behaviourit is necessary to check the design The first steps are to recalculate the loads and to
Introduction 7
reanalyse the structure designed and these are carried out with more precision thanwas either possible or appropriate for the preliminary analysis The performanceof the structure is then evaluated in relation to the structural requirements and anychanges in the member and joint sizes are decided on These changes may requirea further reanalysis and re-proportioning of the structure and this cycle may berepeated until no further change is required Alternatively it may be necessary tomodify the original structural system and repeat the structural design process untila satisfactory structure is achieved
The alternative overall systems are then evaluated in terms of their service-ability economy and harmony and a final system is selected as indicatedin Figure 14 This final overall system may be modified before the design isfinalised The detailed drawings and specifications can then be prepared andtenders for the construction can be called for and let and the structure canbe constructed Further modifications may have to be made as a consequenceof the tenders submitted or due to unforeseen circumstances discovered duringconstruction
This book is concerned with the structural behaviour of steel structures andthe relationships between their behaviour and the methods of proportioning themparticularly in relation to the structural requirements of the European steel struc-tures code EC3 and the modifications of these are given in the National AnnexesThis code consists of six parts with basic design using the conventional membersbeing treated in Part 1 This part is divided into 12 sub-parts with those likely tobe required most frequently being
Part 11 General Rules and Rules for Buildings [8]Part 15 Plated Structural Elements [16]Part 18 Design of Joints [17] andPart 110 Selection of Steel for Fracture Toughness and Through-Thickness
Properties [18]
Other parts that may be required from time to time include Part 12 that coversresistance to fire Part 13 dealing with cold-formed steel Part 19 dealing withfatigue and Part 2 [19] that covers bridges Composite construction is covered byEC4 [20] Since Part 11 of EC3 is the document most relevant to much of thecontent of this text (with the exception of Chapter 9 on joints) all references toEC3 made herein should be taken to mean Part 11 [8] including any modificationsgiven in the National Annex unless otherwise indicated
Detailed discussions of the overall design process are beyond the scope of thisbook but further information is given in [13] on the definition of the designproblem the invention of solutions and their evaluation and in [21ndash24] on theexecution of design Further the conventional methods of structural analysis areadequately treated in many textbooks [25ndash27] and are discussed in only a fewisolated cases in this book
8 Introduction
13 Material behaviour
131 Mechanical properties under static load
The important mechanical properties of most structural steels under static loadare indicated in the idealised tensile stressndashstrain diagram shown in Figure 16Initially the steel has a linear stressndashstrain curve whose slope is the Youngrsquos mod-ulus of elasticity E The values of E vary in the range 200 000ndash210 000 Nmm2and the approximate value of 205 000 Nmm2 is often assumed (EC3 uses210 000 Nmm2) The steel remains elastic while in this linear range and recoversperfectly on unloading The limit of the linear elastic behaviour is often closelyapproximated by the yield stress fy and the corresponding yield strain εy = fyEBeyond this limit the steel flows plastically without any increase in stress untilthe strain-hardening strain εst is reached This plastic range is usually consider-able and accounts for the ductility of the steel The stress increases above theyield stress fy when the strain-hardening strain εst is exceeded and this contin-ues until the ultimate tensile stress fu is reached After this large local reductionsin the cross-section occur and the load capacity decreases until tensile fracturetakes place
The yield stress fy is perhaps the most important strength characteristic of a struc-tural steel This varies significantly with the chemical constituents of the steel themost important of which are carbon and manganese both of which increase theyield stress The yield stress also varies with the heat treatment used andwith the amount of working which occurs during the rolling process Thus thinnerplates which are more worked have higher yield stresses than thicker plates of thesame constituency The yield stress is also increased by cold working The rateof straining affects the yield stress and high rates of strain increase the upper orfirst yield stress (see the broken line in Figure 16) as well as the lower yieldstress fy The strain rates used in tests to determine the yield stress of a particular
Upper yield stressStress
Strain
Strain-hardening E
Tensile rupture Plastic
Elastic E
f
f
0
u
y
y st
st
Not to scale
Figure 16 Idealised stressndashstrain relationship for structural steel
Introduction 9
steel type are significantly higher than the nearly static rates often encountered inactual structures
For design purposes a lsquominimumrsquo yield stress is identified for each differ-ent steel classification For EC3 these classifications are made on the basis of thechemical composition and the heat treatment and so the yield stresses in each clas-sification decrease as the greatest thickness of the rolled section or plate increasesThe minimum yield stress of a particular steel is determined from the results ofa number of standard tension tests There is a significant scatter in these resultsbecause of small variations in the local composition heat treatment amount ofworking thickness and the rate of testing and this scatter closely follows a nor-mal distribution curve Because of this the minimum yield stress fy quoted fora particular steel and used in design is usually a characteristic value which hasa particular chance (often 95) of being exceeded in any standard tension testConsequently it is likely that an isolated test result will be significantly higherthan the quoted yield stress This difference will of course be accentuated if thetest is made for any but the thickest portion of the cross-section In EC3 [8] theyield stress to be used in design is listed in Table 31 for hot-rolled structural steeland for structural hollow sections for each of the structural grades
The yield stress fy determined for uniaxial tension is usually accepted as beingvalid for uniaxial compression However the general state of stress at a point ina thin-walled member is one of biaxial tension andor compression and yieldingunder these conditions is not so simply determined Perhaps the most generallyaccepted theory of two-dimensional yielding under biaxial stresses acting in the1prime2prime plane is the maximum distortion-energy theory (often associated with namesof Huber von Mises or Hencky) and the stresses at yield according to this theorysatisfy the condition
σ 21prime minus σ1primeσ2prime + σ 2
2prime + 3σ 21prime2prime = f 2
y (11)
in which σ1prime σ2prime are the normal stresses and σ1prime2prime is the shear stress at the pointFor the case where 1prime and 2prime are the principal stress directions 1 and 2 equation 11takes the form of the ellipse shown in Figure 17 while for the case of pure shear(σ1prime = σ2prime = 0 so that σ1 = minusσ2 = σ1prime2prime) equation 11 reduces to
σ1prime2prime = fyradic
3 = τy (12)
which defines the shear yield stress τy
132 Fatigue failure under repeated loads
Structural steel may fracture at low average tensile stresses after a large numberof cycles of fluctuating load This high-cycle fatigue failure is initiated by localdamage caused by the repeated loads which leads to the formation of a small localcrack The extent of the fatigue crack is gradually increased by the subsequent loadrepetitions until finally the effective cross-section is so reduced that catastrophic
10 Introduction
Pri
ncip
al s
tres
s ra
tio
2f
y
100 10
10
ndash
ndash
10
Un i a x i a l t e n si o n
Pu r e s h e a r
Un i a x i a lco m p r e s s i o n
P r i n c i p a l s t r es s
Ma x i m u m d i s t o r t i o n - e n er g ycr i t e r i o n 1
2 ndash 12 + 12 = fy
2
rati o 1fy 2
Figure 17 Yielding under biaxial stresses
failure may occur High-cycle fatigue is only a design consideration when a largenumber of loading cycles involving tensile stresses is likely to occur during thedesign life of the structure (compressive stresses do not cause fatigue) This isoften the case for bridges cranes and structures which support machinery windand wave loading may also lead to fatigue problems
Factors which significantly influence the resistance to fatigue failure includethe number of load cycles N the range of stress
σ = σmax minus σmin (13)
during a load cycle and the magnitudes of local stress concentrations An indi-cation of the effect of the number of load cycles is given in Figure 18 whichshows that the maximum tensile stress decreases from its ultimate static value fuin an approximately linear fashion as the logarithm of the number of cycles Nincreases As the number of cycles increases further the curve may flatten out andthe maximum tensile stress may approach the endurance limit σL
The effects of the stress magnitude and stress ratio on the fatigue life are demon-strated in Figure 19 It can be seen that the fatigue life N decreases with increasingstress magnitude σmax and with decreasing stress ratio R = σminσmax
The effect of stress concentration is to increase the stress locally leading to localdamage and crack initiation Stress concentrations arise from sudden changes inthe general geometry and loading of a member and from local changes due tobolt and rivet holes and welds Stress concentrations also occur at defects in themember or its connectors and welds These may be due to the original rolling of thesteel or due to subsequent fabrication processes including punching shearing
Introduction 11
Max
imum
tens
ile
stre
ngh
Stress range = max ndash min
Stress ratio R = min max
max
1cycle
Ultimate tensile strength
Endurance limit
Number of cycles N (log scale)
(b) Fatigue strength(a) Stress cycle
Stress
min
f
Time1
u
L
Figure 18 Variation of fatigue strength with number of load cycles
Alternating
N =
N = 1
Max
imum
tens
ile
stre
ss
max
(Compression) Minimum stress min (Tension)
Staticloading
loadingTypical ofexperiments
Approximateequation 14
fufu
fu
R = ndash10 ndash05 0 05 10
infin
Figure 19 Variation of fatigue life with stress magnitudes
and welding or due to damage such as that caused by stray arc fusions duringwelding
It is generally accepted for design purposes that the fatigue life N varies withthe stress range σ according to equations of the type
(σ
σC
)m (N
2 times 106
)= K (14)
in which the reference value σC depends on the details of the fatigue site andthe constants m and K may change with the number of cycles N This assumeddependence of the fatigue life on the stress range produces the approximatingstraight lines shown in Figure 19
12 Introduction
Reference values ∆C
Str
ess
rang
e ∆
=
max
ndash m
in(N
mm
2 )
Constant amplitudefatigue limit
m = 5 Cut-off limit1
Number of stress cycles N10 10 10 10 104 6 7 85 5 5 5 52
m = 3
m =3
m
10
50
100
500
1000
1401129071564536
16012510080635040
∆C =∆ when N = 2 times 106
Reference values ∆C
Figure 110 Variation of the EC3-1-9 fatigue life with stress range
EC3-1-1 [8] does not provide a treatment of fatigue since it is usually the casethat either the stress range σ or the number of high amplitude stress cycles Nis comparatively small However for structures supporting vibrating machineryand plant reference should be made to EC3-1-9 [28] The general relationshipsbetween the fatigue life N and the service stress rangeσ for constant amplitudestress cycles are shown in Figure 110 for reference valuesσC which correspondto different detail categories For N le 5 times 106 m = 3 and K = 1 so thatthe reference value σC corresponds to the value of σ at N = 2 times 106 For5 times 106 le N le 108 m = 5 and K = 0423 asymp 0543
Fatigue failure under variable amplitude stress cycles is normally assessed usingMinerrsquos rule [29]sum
NiNim le 1 (15)
in which Ni is the number of cycles of a particular stress range σi and Nim theconstant amplitude fatigue life for that stress range If any of the stress rangesexceeds the constant amplitude fatigue limit (at N = 5 times 106) then the effects ofstress ranges below this limit are included in equation 15
Designing against fatigue involves a consideration of joint arrangement as wellas of permissible stress Joints should generally be so arranged as to minimisestress concentrations and produce as smooth a lsquostress flowrsquo through the joint as ispracticable This may be done by giving proper consideration to the layout of ajoint by making gradual changes in section and by increasing the amount of mate-rial used at points of concentrated load Weld details should also be determined
Introduction 13
with this in mind and unnecessary lsquostress-raisersrsquo should be avoided It willalso be advantageous to restrict where practicable the locations of joints to lowstress regions such as at points of contraflexure or near the neutral axis Furtherinformation and guidance on fatigue design are given in [30ndash33]
133 Brittle fracture under impact load
Structural steel does not always exhibit a ductile behaviour and under some cir-cumstances a sudden and catastrophic fracture may occur even though the nominaltensile stresses are low Brittle fracture is initiated by the existence or formation ofa small crack in a region of high local stress Once initiated the crack may prop-agate in a ductile (or stable) fashion for which the external forces must supply theenergy required to tear the steel More serious are cracks which propagate at highspeed in a brittle (or unstable) fashion for which some of the internal elastic strainenergy stored in steel is released and used to fracture the steel Such a crack isself-propagating while there is sufficient internal strain energy and will continueuntil arrested by ductile elements in its path which have sufficient deformationcapacity to absorb the internal energy released
The resistance of a structure to brittle fracture depends on the magnitude of localstress concentrations on the ductility of the steel and on the three-dimensionalgeometrical constraints High local stresses facilitate crack initiation and so stressconcentrations due to poor geometry and loading arrangements (including impactloading) are dangerous Also of great importance are flaws and defects in thematerial which not only increase the local stresses but also provide potential sitesfor crack initiation
The ductility of a structural steel depends on its composition heat treatmentand thickness and varies with temperature and strain rate Figure 111 shows theincrease with temperature of the capacity of the steel to absorb energy during
Crack initiationand propagation easy
Crack initiationmore difficult
Crack initiationdifficult
Temperature
Ene
rgy
abso
rpti
on
Figure 111 Effect of temperature on resistance to brittle fracture
14 Introduction
impact At low temperatures the energy absorption is low and initiation andpropagation of brittle fractures are comparatively easy while at high temperaturesthe energy absorption is high because of ductile yielding and the propagation ofcracks can be arrested Between these two extremes is a transitional range in whichcrack initiation becomes increasingly difficult The likelihood of brittle fracture isalso increased by high strain rates due to dynamic loading since the consequentincrease in the yield stress reduces the possibility of energy absorption by ductileyielding The chemical composition of steel has a marked influence on its ductilitybrittleness is increased by the presence of excessive amounts of most non-metallicelements while ductility is increased by the presence of some metallic elementsSteel with large grain size tends to be more brittle and this is significantly influ-enced by heat treatment of the steel and by its thickness (the grain size tendsto be larger in thicker sections) EC3-1-10 [18] provides values of the maximumthickness t1 for different steel grades and minimum service temperatures as wellas advice on using a more advanced fracture mechanics [34] based approach andguidance on safeguarding against lamellar tearing
Three-dimensional geometrical constraints such as those occurring in thickeror more massive elements also encourage brittleness because of the higher localstresses and because of the greater release of energy during cracking and theconsequent increase in the ease of propagation of the crack
The risk of brittle fracture can be reduced by selecting steel types which haveductilities appropriate to the service temperatures and by designing joints with aview to minimising stress concentrations and geometrical constraints Fabricationtechniques should be such that they will avoid introducing potentially dangerousflaws or defects Critical details in important structures may be subjected to inspec-tion procedures aimed at detecting significant flaws Of course the designer mustgive proper consideration to the extra cost of special steels fabrication techniquesand inspection and correction procedures Further information on brittle fractureis given in [31 32 34]
14 Member and structure behaviour
141 Member behaviour
Structural steel members are required to transmit axial and transverse forces andmoments and torques as shown in Figure 13 The response of a member tothese actions can be described by the load-deformation characteristics shown inFigure 112
A member may have the linear response shown by curve 1 in Figure 112at least until the material reaches the yield stress The magnitudes of the defor-mations depend on the elastic moduli E and G Theoretically a member can onlybehave linearly while the maximum stress does not exceed the yield stress fyand so the presence of residual stresses or stress concentrations will cause earlynon-linearity However the high ductility of steel causes a local redistributionafter this premature yielding and it can often be assumed without serious error
Introduction 15
Load Buckling
Linear
Geometric non-linearity
Materialnon-linearity
Material and geometricnon-linearity
Deformation
1
2
3
4
5
Brittle fracture8Local buckling7
Fully plastic6
Figure 112 Member behaviour
that the member response remains linear until more general yielding occurs Themember behaviour then becomes non-linear (curve 2) and approaches the condi-tion associated with full plasticity (curve 6) This condition depends on the yieldstress fy
The member may also exhibit geometric non-linearity in that the bendingmoments and torques acting at any section may be influenced by the deformationsas well as by the applied forces This non-linearity which depends on the elasticmoduli E and G may cause the deformations to become very large (curve 3) as thecondition of elastic buckling is approached (curve 4) This behaviour is modifiedwhen the material becomes non-linear after first yield and the load may approacha maximum value and then decrease (curve 5)
The member may also behave in a brittle fashion because of local bucklingin a thin plate element of the member (curve 7) or because of material fracture(curve 8)
The actual behaviour of an individual member will depend on the forces actingon it Thus tension members laterally supported beams and torsion membersremain linear until their material non-linearity becomes important and then theyapproach the fully plastic condition However compression members and laterallyunsupported beams show geometric non-linearity as they approach their bucklingloads Beam-columns are members which transmit both transverse and axial loadsand so they display both material and geometric non-linearities
142 Structure behaviour
The behaviour of a structure depends on the load-transferring action of its mem-bers and joints This may be almost entirely by axial tension or compressionas in the triangulated structures with joint loading as shown in Figure 113a
16 Introduction
(a) Axial force (b) Bending (c) Shear(d) Axial force and bending
Figure 113 Structural load-transfer actions
Alternatively the members may support transverse loads which are transferred bybending and shear actions Usually the bending action dominates in structures com-posed of one-dimensional members such as beams and many single-storey rigidframes (Figure 113b) while shear becomes more important in two-dimensionalplate structures (Figure 113c) The members of many structures are subjectedto both axial forces and transverse loads such as those in multistorey buildings(Figure 113d) The load-transferring action of the members of a structure dependson the arrangement of the structure including the geometrical layout and the jointdetails and on the loading arrangement
In some structures the loading and joints are such that the members are effec-tively independent For example in triangulated structures with joint loads anyflexural effects are secondary and the members can be assumed to act as ifpin-jointed while in rectangular frames with simple flexible joints the momenttransfers between beams and columns may be ignored In such cases the responseof the structure is obtained directly from the individual member responses
More generally however there will be interactions between the members andthe structure behaviour is not unlike the general behaviour of a member as canbe seen by comparing Figures 114 and 112 Thus it has been traditional toassume that a steel structure behaves elastically under the service loads Thisassumption ignores local premature yielding due to residual stresses and stressconcentrations but these are not usually serious Purely flexural structures andpurely axial structures with lightly loaded compression members behave as iflinear (curve 1 in Figure 114) However structures with both flexural and axialactions behave non-linearly even near the service loads (curve 3 in Figure 114)This is a result of the geometrically non-linear behaviour of its members (seeFigure 112)
Most steel structures behave non-linearly near their ultimate loads unless theyfail prematurely due to brittle fracture fatigue or local buckling This non-linearbehaviour is due either to material yielding (curve 2 in Figure 114) or memberor frame buckling (curve 4) or both (curve 5) In axial structures failure may
Introduction 17
Load Buckling
Linear
Geometric non-linearity
Material non-linearity
Material and geometricnon-linearity
Deformation
1
2
3
4
5
Figure 114 Structure behaviour
involve yielding of some tension members or buckling either of some compressionmembers or of the frame or both In flexural structures failure is associated withfull plasticity occurring at a sufficient number of locations that the structure canform a collapse mechanism In structures with both axial and flexural actions thereis an interaction between yielding and buckling (curve 5 in Figure 114) and thefailure load is often difficult to determine The transitions shown in Figure 114between the elastic and ultimate behaviour often take place in a series of non-linearsteps as individual elements become fully plastic or buckle
15 Loads
151 General
The loads acting on steel structures may be classified as dead loads as imposedloads including both gradually applied and dynamic loads as wind loads as earthor ground-water loads or as indirect forces including those due to temperaturechanges foundation settlement and the like The more general collective termActions is used throughout the Eurocodes The structural engineer must evaluatethe magnitudes of any of these loads which will act and must determine thosewhich are the most severe combinations of loads for which the structure must bedesigned These loads are discussed in the following subsections both individuallyand in combinations
152 Dead loads
The dead loads acting on a structure arise from the weight of the structure includingthe finishes and from any other permanent construction or equipment The dead
18 Introduction
loads will vary during construction but thereafter will remain constant unlesssignificant modifications are made to the structure or its permanent equipment
The dead load may be assessed from the knowledge of the dimensions and spe-cific weights or from the total weights of all the permanent items which contributeto the total dead load Guidance on specific weights is given in [14] the valuesin which are average values representative of the particular materials The dimen-sions used to estimate dead loads should also be average and representative inorder that consistent estimates of the dead loads can be made By making theseassumptions the statistical distribution of dead loads is often taken as being of aWeibull type [35] The practice sometimes used of consistently overestimatingdimensions and specific weights is often wasteful and may also be danger-ous in cases where the dead load component acts in the opposite sense to theresultant load
153 Imposed loads
The imposed loads acting on a structure are gravity loads other than the dead loadsand arise from the weights of materials added to the structure as a result of its usesuch as materials stored people and snow Imposed loads usually vary both inspace and time Imposed loads may be sub-divided into two groups dependingon whether they are gradually applied in which case static load equivalents canbe used or whether they are dynamic including repeated loads and impact orimpulsive loads
Gradually applied imposed loads may be sustained over long periods of timeor may vary slowly with time [36] The past practice however was to consideronly the total imposed load and so only extreme values (which occur rarely andmay be regarded as lifetime maximum loads) were specified The present imposedloads specified in loading codes [14] often represent peak loads which have 95probability of not being exceeded over a 50-year period based on a Weibull typedistribution [35]
It is usual to consider the most severe spatial distribution of the imposed loadsand this can only be determined by using both the maximum and minimum valuesof the imposed loads In the absence of definite knowledge it is often assumed thatthe minimum values are zero When the distribution of imposed load over largeareas is being considered the maximum imposed loads specified which representrare events are often reduced in order to make some allowance for the decreasedprobability that the maximum imposed loads will act on all areas at the same time
Dynamic loads which act on structures include both repeated loads and impactand blast loads Repeated loads are of significance in fatigue problems (see Section132) in which case the designer is concerned with both the magnitudes rangesand the number of repetitions of loads which are very frequently applied At theother extreme impact loads (which are particularly important in the brittle fractureproblems discussed in Section 133) are usually specified by values of extrememagnitude which represent rare events In structures for which the static loads
Introduction 19
dominate it is common to replace the dynamic loads by static force equivalents[14] However such a procedure is likely to be inappropriate when the dynamicloads form a significant proportion of the total load in which case a proper dynamicanalysis [37 38] of the structure and its response should be made
154 Wind loads
The wind loads which act on structures have traditionally been allowed for byusing static force equivalents The first step is usually to determine a basic windspeed for the general region in which the structure is to be built by using infor-mation derived from meteorological studies This basic wind speed may representan extreme velocity measured at a height of 10 m and averaged over a period of3 seconds which has a return period of 50 years (ie a velocity which will onaverage be reached or exceeded once in 50 years or have a probability of beingexceeded of 150) The basic wind speed may be adjusted to account for the topog-raphy of the site for the ground roughness structure size and height above groundand for the degree of safety required and the period of exposure The resultingdesign wind speed may then be converted into the static pressure which will beexerted by the wind on a plane surface area (this is often referred to as the dynamicwind pressure because it is produced by decelerating the approaching wind veloc-ity to zero at the surface area) The wind force acting on the structure may thenbe calculated by using pressure coefficients appropriate to each individual surfaceof the structure or by using force coefficients appropriate to the total structureMany values of these coefficients are tabulated in [15] but in special cases wherethese are inappropriate the results of wind tunnel tests on model structures maybe used
In some cases it is not sufficient to treat wind loads as static forces For examplewhen fatigue is a problem both the magnitudes and the number of wind fluctuationsmust be estimated In other cases the dynamic response of a structure to wind loadsmay have to be evaluated (this is often the case with very flexible structures whoselong natural periods of vibration are close to those of some of the wind gusts) andthis may be done analytically [37 38] or by specialists using wind tunnel testsIn these cases special care must be taken to model correctly those properties ofthe structure which affect its response including its mass stiffness and dampingas well as the wind characteristics and any interactions between wind and structure
155 Earth or ground-water loads
Earth or ground-water loads act as pressure loads normal to the contact surface ofthe structure Such loads are usually considered to be essentially static
However earthquake loads are dynamic in nature and their effects on the struc-ture must be allowed for Very flexible structures with long natural periods ofvibration respond in an equivalent static manner to the high frequencies of earth-quake movements and so can be designed as if loaded by static force equivalents
20 Introduction
On the other hand stiff structures with short natural periods of vibration respondsignificantly and so in such a case a proper dynamic analysis [37 38] shouldbe made The intensities of earthquake loads vary with the region in which thestructure is to be built but they are not considered to be significant in the UK
156 Indirect forces
Indirect forces may be described as those forces which result from the strainingof a structure or its components and may be distinguished from the direct forcescaused by the dead and applied loads and pressures The straining may arise fromtemperature changes from foundation settlement from shrinkage creep or crack-ing of structural or other materials and from the manufacturing process as in thecase of residual stresses The values of indirect forces are not usually specifiedand so it is common for the designer to determine which of these forces should beallowed for and what force magnitudes should be adopted
157 Combinations of loads
The different loads discussed in the preceding subsections do not occur alone but incombinations and so the designer must determine which combination is the mostcritical for the structure However if the individual loads which have differentprobabilities of occurrence and degrees of variability were combined directlythe resulting load combination would have a greatly reduced probability Thusit is logical to reduce the magnitudes of the various components of a combinationaccording to their probabilities of occurrence This is similar to the procedure usedin reducing the imposed load intensities used over large areas
The past design practice was to use the worst combination of dead load withimposed load andor wind load and to allow increased stresses whenever thewind load was included (which is equivalent to reducing the load magnitudes)These increases seem to be logical when imposed and wind loads act togetherbecause the probability that both of these loads will attain their maximum valuessimultaneously is greatly reduced However they are unjustified when applied inthe case of dead and wind load for which the probability of occurrence is virtuallyunchanged from that of the wind load
A different and more logical method of combining loads is used in the EC3 limitstates design method [8] which is based on statistical analyses of the loads andthe structure capacities (see Section 1734) Strength design is usually carriedout for the most severe combination of actions for normal (termed persistent) ortemporary (termed transient) conditions usingsum
jge1
γG jGk j + γQ1Qk 1 +sumigt1
γQiψ0iQk i (16)
where implies lsquothe combined effect ofrsquo γG and γQ are partial factors for thepersistent G and variable Q actions andψ0 is a combination factor The concept is
Introduction 21
Table 11 Partial load factors for common situations
Ultimate limit state Permanent actions γG Variable actions γQ
Unfavourable Favourable Unfavourable Favourable
EQU 11 09 15 0STRGEO 135 10 15 0
thus to use all the persistent actions Gk j such as self-weight and fixed equipmentwith a leading variable action Qk 1 such as imposed snow or wind load andreduced values of the other variable actions Qk i More information on the Eurocodeapproach to loading for steel structures is given in [39ndash41]
This approach is applied to the following forms of ultimate limit state
EQU = loss of static equilibrium of the structure on any part of itSTR = failure by excessive deformation transformation of the structure or
any part of it into a mechanism rupture or loss of stability of thestructure or of any part of it
GEO = failure or excessive deformation of the groundFAT = fatigue failure
For the most common set of design situations use of the appropriate values from[41] gives the load factors of Table 11
Using the combination factors of ψ0 = 07 and 05 of [41] for most variableactions and wind actions respectively leads to the following common STR loadcombinations for buildings
135 G + 15 QI + 075 QW
135 G + 105 QI + 15 QW
minus 10 G + 15 QI and
minus 10 G + 15 QW
in which the minus signs indicate that the permanent action is favourable
16 Analysis of steel structures
161 General
In the design process the assessment of whether the structural design require-ments will be met or not requires the knowledge of the stiffness and strength ofthe structure under load and of its local stresses and deformations The term struc-tural analysis is used to denote the analytical process by which this knowledge
22 Introduction
of the response of the structure can be obtained The basis for this process is theknowledge of the material behaviour and this is used first to analyse the behaviourof the individual members and joints of the structure The behaviour of the completestructure is then synthesised from these individual behaviours
The methods of structural analysis are fully treated in many textbooks[eg 25ndash27] and so details of these are not within the scope of this book Howeversome discussion of the concepts and assumptions of structural analysis is neces-sary so that the designer can make appropriate assumptions about the structure andmake a suitable choice of the method of analysis
In most methods of structural analysis the distribution of forces and momentsthroughout the structure is determined by using the conditions of static equilib-rium and of geometric compatibility between the members at the joints The wayin which this is done depends on whether a structure is statically determinate(in which case the complete distribution of forces and moments can be determinedby statics alone) or is statically indeterminate (in which case the compatibilityconditions for the deformed structure must also be used before the analysis can becompleted)
An important feature of the methods of structural analysis is the constitutiverelationships between the forces and moments acting on a member or connectionand its deformations These play the same role for the structural element as do thestressndashstrain relationships for an infinitesimal element of a structural material Theconstitutive relationship may be linear (force proportional to deflection) and elastic(perfect recovery on unloading) or they may be non-linear because of material non-linearities such as yielding (inelastic) or because of geometrical non-linearities(elastic) such as when the deformations themselves induce additional momentsas in stability problems
It is common for the designer to idealise the structure and its behaviour so asto simplify the analysis A three-dimensional frame structure may be analysed asthe group of a number of independent two-dimensional frames while individualmembers are usually considered as one-dimensional and the joints as points Thejoints may be assumed to be frictionless hinges or to be semi-rigid or rigid Insome cases the analysis may be replaced or supplemented by tests made on anidealised model which approximates part or all of the structure
162 Analysis of statically determinate membersand structures
For an isolated statically determinate member the forces and moments acting onthe member are already known and the structural analysis is only used to determinethe stiffness and strength of the member A linear elastic (or first-order elastic)analysis is usually made of the stiffness of the member when the material non-linearities are generally unimportant and the geometrical non-linearities are oftensmall The strength of the member however is not so easily determined as oneor both of the material and geometric non-linearities are most important Instead
Introduction 23
the designer usually relies on a design code or specification for this informationThe strength of the isolated statically determinate members is fully discussed inChapters 2ndash7 and 10
For a statically determinate structure the principles of static equilibrium areused in the structural analysis to determine the member forces and moments andthe stiffness and strength of each member are then determined in the same way asfor statically determinate members
163 Analysis of statically indeterminate structures
A statically indeterminate structure can be approximately analysed if a sufficientnumber of assumptions are made about its behaviour to allow it to be treated asif determinate One method of doing this is to guess the locations of points ofzero bending moment and to assume there are frictionless hinges at a sufficientnumber of these locations that the member forces and moments can be determinedby statics alone Such a procedure is commonly used in the preliminary analysisof a structure and followed at a later stage by a more precise analysis Howevera structure designed only on the basis of an approximate analysis can still be safeprovided the structure has sufficient ductility to redistribute any excess forces andmoments Indeed the method is often conservative and its economy increaseswith the accuracy of the estimated locations of the points of zero bending momentMore commonly a preliminary analysis is made of the structure based on thelinear elastic computer methods of analysis [42 43] using approximate memberstiffnesses
The accurate analysis of statically indeterminate structures is complicated by theinteraction between members the equilibrium and compatibility conditions andthe constitutive relationships must all be used in determining the member forcesand moments There are a number of different types of analysis which might bemade and some indication of the relevance of these is given in Figure 115 andin the following discussion Many of these can only be used for two-dimensionalframes
For many structures it is common to use a first-order elastic analysis which isbased on linear elastic constitutive relationships and which ignores any geometricalnon-linearities and associated instability problems The deformations determinedby such an analysis are proportional to the applied loads and so the principle ofsuperposition can be used to simplify the analysis It is often assumed that axialand shear deformations can be ignored in structures whose action is predominantlyflexural and that flexural and shear deformations can be ignored in structureswhose member forces are predominantly axial These assumptions further simplifythe analysis which can then be carried out by any of the well-known methods[25ndash27] for which many computer programs are available [44 45] Some of theseprograms can be used for three-dimensional frames
However a first-order elastic analysis will underestimate the forces andmoments in and the deformations of a structure when instability effects are present
24 Introduction
Deformation
Load
First-order elastic analysis
Elastic stability analysis
Second-orderelastic analysis
First-order plasticanalysis
Advancedanalysis
Strength load
Service load
1
4
3
2
5
Figure 115 Predictions of structural analyses
Some estimate of the importance of these in the absence of flexural effects can beobtained by making an elastic stability analysis A second-order elastic analysisaccounts for both flexure and instability but this is difficult to carry out althoughcomputer programs are now generally available [44 45] EC3 permits the useof the results of an elastic stability analysis in the amplification of the first-ordermoments as an alternative to second-order analysis
The analysis of statically indeterminate structures near the ultimate load isfurther complicated by the decisive influence of the material and geometricalnon-linearities In structures without material non-linearities an elastic stabilityanalysis is appropriate when there are no flexural effects but this is a rare occur-rence On the other hand many flexural structures have very small axial forcesand instability effects in which case it is comparatively easy to use a first-orderplastic analysis according to which a sufficient number of plastic hinges mustform to transform the structure into a collapse mechanism
More generally the effects of instability must be allowed for and as a firstapproximation the nominal first yield load determined from a second-order elasticanalysis may be used as a conservative estimate of the ultimate load A much moreaccurate estimate may be obtained for structures where local and lateral bucklingis prevented by using an advanced analysis [46] in which the actual behaviour isclosely analysed by allowing for instability yielding residual stresses and initialcrookedness However this method is not yet in general use
17 Design of steel structures
171 Structural requirements and design criteria
The designerrsquos task of assessing whether or not a structure will satisfy the structuralrequirements of serviceability and strength is complicated by the existence of errors
Introduction 25
and uncertainties in his or her analysis of the structural behaviour and estimation ofthe loads acting and even in the structural requirements themselves The designerusually simplifies this task by using a number of design criteria which allow him orher to relate the structural behaviour predicted by his or her analysis to the structuralrequirements Thus the designer equates the satisfaction of these criteria by thepredicted structural behaviour with satisfaction of the structural requirements bythe actual structure
In general the various structural design requirements relate to correspondinglimit states and so the design of a structure to satisfy all the appropriate require-ments is often referred to as a limit states design The requirements are commonlypresented in a deterministic fashion by requiring that the structure shall not failor that its deflections shall not exceed prescribed limits However it is not possibleto be completely certain about the structure and its loading and so the structuralrequirements may also be presented in probabilistic forms or in deterministicforms derived from probabilistic considerations This may be done by defining anacceptably low risk of failure within the design life of the structure after reachingsome sort of balance between the initial cost of the structure and the economic andhuman losses resulting from failure In many cases there will be a number of struc-tural requirements which operate at different load levels and it is not unusual torequire a structure to suffer no damage at one load level but to permit some minordamage to occur at a higher load level provided there is no catastrophic failure
The structural design criteria may be determined by the designer or he or shemay use those stated or implied in design codes The stiffness design criteriaadopted are usually related to the serviceability limit state of the structure underthe service loads and are concerned with ensuring that the structure has sufficientstiffness to prevent excessive deflections such as sagging distortion and settle-ment and excessive motions under dynamic load including sway bounce andvibration
The strength limit state design criteria are related to the possible methods offailure of the structure under overload and understrength conditions and so thesedesign criteria are concerned with yielding buckling brittle fracture and fatigueAlso of importance is the ductility of the structure at and near failure ductilestructures give a warning of the impending failure and often redistribute the loadeffects away from the critical regions while ductility provides a method of energydissipation which will reduce the damage due to earthquake and blast loading Onthe other hand a brittle failure is more serious as it occurs with no warning offailure and in a catastrophic fashion with a consequent release of stored energyand increase in damage Other design criteria may also be adopted such as thoserelated to corrosion and fire
172 Errors and uncertainties
In determining the limitations prescribed by design criteria account must be takenof the deliberate and accidental errors made by the designer and of the uncertainties
26 Introduction
in his or her knowledge of the structure and its loads Deliberate errors includethose resulting from the assumptions made to simplify the analysis of the loadingand of the structural behaviour These assumptions are often made so that anyerrors involved are on the safe side but in many cases the nature of the errorsinvolved is not precisely known and some possibility of danger exists
Accidental errors include those due to a general lack of precision either inthe estimation of the loads and the analysis of the structural behaviour or in themanufacture and erection of the structure The designer usually attempts to controlthe magnitudes of these by limiting them to what he or she judges to be suitablysmall values Other accidental errors include what are usually termed blundersThese may be of a gross magnitude leading to failure or to uneconomic structuresor they may be less important Attempts are usually made to eliminate blunders byusing checking procedures but often these are unreliable and the logic of such aprocess is open to debate
As well as the errors described above there exist a number of uncertaintiesabout the structure itself and its loads The material properties of steel fluctuateespecially the yield stress and the residual stresses The practice of using a min-imum or characteristic yield stress for design purposes usually leads to oversafedesigns especially for redundant structures of reasonable size for which an aver-age yield stress would be more appropriate because of the redistribution of loadwhich takes place after early yielding Variations in the residual stress levels arenot often accounted for in design codes but there is a growing tendency to adjustdesign criteria in accordance with the method of manufacture so as to make someallowance for gross variations in the residual stresses This is undertaken to someextent in EC3
The cross-sectional dimensions of rolled-steel sections vary and the valuesgiven in section handbooks are only nominal especially for the thicknesses ofuniversal sections The fabricated lengths of a structural member will vary slightlyfrom the nominal length but this is usually of little importance except wherethe variation induces additional stresses because of lack-of-fit problems or wherethere is a cumulative geometrical effect Of some significance to members subjectto instability problems are the variations in their straightness which arise duringmanufacture fabrication and erection Some allowances for these are usuallymade in design codes while fabrication and erection tolerances are specified inEN1090 [47] to prevent excessive crookedness
The loads acting on a structure vary significantly Uncertainty exists in thedesignerrsquos estimate of the magnitude of the dead load because of the variations inthe densities of materials and because of the minor modifications to the structureduring or subsequent to its erection Usually these variations are not very signif-icant and a common practice is to err on the safe side by making conservativeassumptions Imposed loadings fluctuate significantly during the design usage ofthe structure and may change dramatically with changes in usage These fluctua-tions are usually accounted for by specifying what appear to be extreme values inloading codes but there is often a finite chance that these values will be exceeded
Introduction 27
Wind loads vary greatly and the magnitudes specified in loading codes are usuallyobtained by probabilistic methods
173 Strength design
1731 Load and capacity factors and factors of safety
The errors and uncertainties involved in the estimation of the loads on and thebehaviour of a structure may be allowed for in strength design by using load fac-tors to increase the nominal loads and capacity factors to decrease the structuralstrength In the previous codes that employed the traditional working stress designthis was achieved by using factors of safety to reduce the failure stresses to permis-sible working stress values The purpose of using various factors is to ensure thatthe probability of failure under the most adverse conditions of structural overloadand understrength remains very small The use of these factors is discussed in thefollowing subsections
1732 Working stress design
The working stress methods of design given in previous codes and specificationsrequired that the stresses calculated from the most adverse combination of loadsmust not exceed the specified permissible stresses These specified stresses wereobtained after making some allowances for the non-linear stability and materialeffects on the strength of isolated members and in effect were expressions of theirultimate strengths divided by the factors of safety SF Thus
Working stress le Permissible stress asymp Ultimate stress
SF(16)
It was traditional to use factors of safety of 17 approximatelyThe working stress method of a previous steel design code [48] has been replaced
by the limit states design method of EC3 Detailed discussions of the working stressmethod are available in the first edition of this book [49]
1733 Ultimate load design
The ultimate load methods of designing steel structures required that the calculatedultimate load-carrying capacity of the complete structure must not exceed the mostadverse combination of the loads obtained by multiplying the working loads bythe appropriate load factors LF Thussum
(Working load times LF) le Ultimate load (17)
These load factors allowed some margins for any deliberate and accidentalerrors and for the uncertainties in the structure and its loads and also provided the
28 Introduction
structure with a reserve of strength The values of the factors should depend on theload type and combination and also on the risk of failure that could be expectedand the consequences of failure A simplified approach often employed (perhapsillogically) was to use a single load factor on the most adverse combination of theworking loads
A previous code [48] allowed the use of the plastic method of ultimate loaddesign when stability effects were unimportant These have used load factors of170 approximately However this ultimate load method has also been replacedby the limit states design method in EC3 and will not be discussed further
1734 Limit states design
It was pointed out in Section 156 that different types of load have different proba-bilities of occurrence and different degrees of variability and that the probabilitiesassociated with these loads change in different ways as the degree of overloadconsidered increases Because of this different load factors should be used for thedifferent load types
Thus for limit states design the structure is deemed to be satisfactory if itsdesign load effect does not exceed its design resistance The design load effectis an appropriate bending moment torque axial force or shear force and iscalculated from the sum of the effects of the specified (or characteristic) loadsFk multiplied by partial factors γGQ which allow for the variabilities of the loadsand the structural behaviour The design resistance RkγM is calculated from thespecified (or characteristic) resistance Rk divided by the partial factor γM whichallows for the variability of the resistance Thus
Design load effect le Design resistance (18a)
or sumγgQ times (effect of specified loads) le (specified resistanceγM ) (18b)
Although the limit states design method is presented in deterministic formin equations 18 the partial factors involved are usually obtained by usingprobabilistic models based on statistical distributions of the loads and thecapacities Typical statistical distributions of the total load and the structuralcapacity are shown in Figure 116 The probability of failure pF is indicatedby the region for which the load distribution exceeds that for the structuralcapacity
In the development of limit state codes the probability of failure pF is usuallyrelated to a parameter β called the safety index by the transformation
Φ(minusβ) = pF (19)
where the functionΦ is the cumulative frequency distribution of a standard normalvariate [35] The relationship between β and pF shown in Figure 117 indicates
Introduction 29
Total load effect or resistance
Pro
babi
lity
dens
ity
Total loadeffect
Resistance
Shaded area illustratesprobability of failure
Sum
of
effe
cts
of s
peci
fied
load
sD
esig
n lo
ad e
ffec
t
Des
ign
resi
stan
ce
Spe
cifi
ed r
esis
tanc
e
Figure 116 Limit states design
ndash8 ndash6 ndash4 ndash2
Probability of failure pF
0
1
2
3
4
5
6
Saf
ety
inde
x
10 10 10 10
Figure 117 Relationship between safety index and probability of failure
that an increase in β of 05 implies a decrease in the probability of failure byapproximately an order of magnitude
The concept of the safety index was used to derive the partial factors for EC3This was done with reference to previous national codes such as BS 5950 [50]to obtain comparable values of the probability of failure pF although much ofthe detailed calibration treated the load and resistance sides of equations 18separately
30 Introduction
174 Stiffness design
In the stiffness design of steel structures the designer seeks to make the structuresufficiently stiff so that its deflections under the most adverse working load con-ditions will not impair its strength or serviceability These deflections are usuallycalculated by a first-order linear elastic analysis although the effects of geometri-cal non-linearities should be included when these are significant as in structureswhich are susceptible to instability problems The design criteria used in the stiff-ness design relate principally to the serviceability of the structure in that theflexibility of the structure should not lead to damage of any non-structural compo-nents while the deflections should not be unsightly and the structure should notvibrate excessively It is usually left to the designer to choose limiting values foruse in these criteria which are appropriate to the structure although a few valuesare suggested in some design codes The stiffness design criteria which relate tothe strength of the structure itself are automatically satisfied when the appropriatestrength design criteria are satisfied
References
1 British Standards Institution (2005) BS4-1 Structural Steel Sections-Part 1 Specifica-tion for hot-rolled sections BSI London
2 OrsquoConnor C (1971) Design of Bridge Superstructures John Wiley New York3 Chatterjee S (1991) The Design of Modern Steel Bridges BSP Professional Books
Oxford4 Rhodes J (editor) (1991) Design of Cold Formed Members Elsevier Applied Science
London5 Rhodes J and Lawson RM (1992) Design of Steel Structures Using Cold Formed
Steel Sections Steel Construction Institute Ascot6 Yu W-W (2000) Cold-Formed Steel Design 3rd edition John Wiley New York7 Hancock GJ (1998) Design of Cold-Formed Steel Structures 3rd edition Australian
Institute of Steel Construction Sydney8 British Standards Institution (2005) Eurocode 3 Design of Steel Structures Part 11
General Rules and Rules for Buildings BS EN 1993-1-1 BSI London9 Nervi PL (1956) Structures McGraw-Hill New York
10 Salvadori MG and Heller R (1963) Structure in Architecture Prentice-HallEnglewood Cliffs New Jersey
11 Torroja E (1958) The Structures of Educardo Torroja FW Dodge Corp New York12 Fraser DJ (1981) Conceptual Design and Preliminary Analysis of Structures Pitman
Publishing Inc Marshfield Massachusetts13 Krick EV (1969) An Introduction to Engineering and Engineering Design 2nd
edition John Wiley New York14 British Standards Institution (2002) Eurocode 1 Actions on Structures Part 11
General Actions ndash Densities Self-weight Imposed Loads for Buildings BS EN1991-1-1 BSI London
15 British Standards Institution (2005) Eurocode 1 Actions on Structures Part 14General Actions ndash Wind Actions BS EN 1991-1-4 BSI London
Introduction 31
16 British Standards Institution (2006) Eurocode 3 Design of Steel Structures Part 15Plated Structural Elements BS EN 1993-1-5 BSI London
17 British Standards Institution (2006) Eurocode 3 Design of Steel Structures Part 18Design of Joints BS EN 1993-1-8 BSI London
18 British Standards Institution (2006) Eurocode 3 Design of Steel Structures Part 110Selection of Steel for Fracture Toughness and Through-Thickness Properties BS EN1993-1-10 BSI London
19 British Standards Institution (2006) Eurocode 3 Design of Steel Structures Part 2Steel Bridges BS EN 1993-2 BSI London
20 British Standards Institution (2005) Eurocode 4 Design of Composite Steel and Con-crete Structures Part 11 General Rules and Rules for Buildings BS EN 1994-1-1BSI London
21 Davison B and Owens GW (eds) (2003) Steel Designersrsquo Manual 6th editionBlackwell Publishing Oxford
22 CIMsteel (1997) Design for Construction Steel Construction Institute Ascot23 Antill JM and Ryan PWS (1982) Civil Engineering Construction 5th edition
McGraw-Hill Sydney24 Australian Institute of Steel Construction (1991) Economical Structural Steelwork
AISC Sydney25 Norris CH Wilbur JB and Utku S (1976) Elementary Structural Analysis 3rd
edition McGraw-Hill New York26 Coates RC Coutie MG and Kong FK (1990) Structural Analysis 3rd edition
Van Nostrand Reinhold (UK) Wokingham27 Ghali A Neville AM and Brown TG (1997) Structural Analysis ndash A Unified
Classical and Matrix Approach 5th edition Routledge Oxford28 British Standards Institution (2006) Eurocode 3 Design of Steel Structures Part 19
Fatigue Strength of Steel Structures BS EN 1993-1-9 BSI London Rules for BuildingsECS Brussels
29 Miner MA (1945) Cumulative damage in fatigue Journal of Applied MechanicsASME 12 No 3 September pp A-159ndashA-164
30 Gurney TR (1979) Fatigue of Welded Structures 2nd edition Cambridge UniversityPress
31 Lay MG (1982) Structural Steel Fundamentals Australian Road Research BoardMelbourne
32 Rolfe ST and Barsoum JM (1977) Fracture and Fatigue Control in Steel StructuresPrentice-Hall Englewood Cliffs New Jersey
33 Grundy P (1985) Fatigue limit state for steel structures Civil Engineering Transac-tions Institution of Engineers Australia CE27 No 1 February pp 143ndash8
34 Navy Department Advisory Committee on Structural Steels (1970) Brittle Fracture inSteel Structures (ed GM Boyd) Butterworth London
35 Walpole RE Myers RH Myers SL and Ye K (2007) Probability and Statisticsfor Engineers and Scientists 8th edition Prentice-Hall Englewood Cliffs
36 Ravindra MK and Galambos TV (1978) Load and resistance factor design for steelJournal of the Structural Division ASCE 104 No ST9 September pp 1337ndash54
37 Clough RW and Penzien J (2004) Dynamics of Structures 2nd edition CSI BooksBerkeley California
38 Irvine HM (1986) Structural Dynamics for the Practising Engineer Allen andUnwin London
32 Introduction
39 Gulvanesian H Calgaro J-A and Holicky M (2002) DesignersrsquoGuide to EN 1990Eurocode Basis of Structural Design Thomas Telford Publishing London
40 Gardner L and Nethercot DA (2005) Designersrsquo Guide to EN 1993-1-1 ThomasTelford Publishing London
41 Gardner L and Grubb PJ (2008) Guide to Eurocode Load Combinations for SteelStructures British Constructional Steelwork Association
42 Harrison HB (1973) Computer Methods in Structural Analysis Prentice-HallEnglewood Cliffs New Jersey
43 Harrison HB (1990) Structural Analysis and Design Parts 1 and 2 Pergamon PressOxford
44 Computer Service Consultants (2006) S-Frame 3D Structural Analysis Suite CSC(UK) Limited Leeds
45 Research Engineers Europe Limited (2006) STAADPro Space Frame Analysis REEBristol
46 Clarke MJ (1994) Plastic-zone analysis of frames Chapter 6 of Advanced Analysis ofSteel Frames Theory Software and Applications (eds WF Chen and S Toma) CRCPress Inc Boca Raton Florida pp 259ndash319
47 British Standards Institution (1998) Execution of Steel Structures ndash General Rules andRules for Buildings BS DD ENV 1090-1 BSI London
48 British Standards Institution (1969) BS449 Part 2 Specification for the Use ofStructural Steel in Building BSI London
49 Trahair NS (1977) The Behaviour and Design of Steel Structures 1st editionChapman and Hall London
50 British Standards Institution (2000) BS5950 The Structural Use of Steelwork inBuilding Part 1 Code of Practice for Design Rolled and Welded Sections BSILondon
Chapter 2
Tension members
21 Introduction
Concentrically loaded uniform tension members are perhaps the simplest structuralelements as they are nominally in a state of uniform axial stress Because of thistheir load-deformation behaviour very closely parallels the stressndashstrain behaviourof structural steel obtained from the results of tensile tests (see Section 131)Thus a member remains essentially linear and elastic until the general yield loadis approached even if it has residual stresses and initial crookedness
However in many cases a tension member is not loaded or connected concen-trically or it has transverse loads acting resulting in bending actions as well asan axial tension action Simple design procedures are available which enable thebending actions in some members with eccentric connections to be ignored butmore generally special account must be taken of the bending action in design
Tension members often have comparatively high average stresses and in somecases the effects of local stress concentrations may be significant especially whenthere is a possibility that the steel material may not act in a ductile fashion In suchcases the causes of stress concentrations should be minimised and the maximumlocal stresses should be estimated and accounted for
In this chapter the behaviour and design of steel tension members are discussedThe case of concentrically loaded members is dealt with first and then a procedurewhich allows the simple design of some eccentrically connected tension membersis presented The design of tension members with eccentric or transverse loads isthen considered the effects of stress concentrations are discussed and finally thedesign of tension members according to EC3 is dealt with
22 Concentrically loaded tension members
221 Members without holes
The straight concentrically loaded steel tension member of length L and constantcross-sectional area A which is shown in Figure 21a has no holes and is free fromresidual stress The axial extension e of the member varies with the load N in the
34 Tension members
Load N
Npl
Nu
Elastic
Plastic
Strain-hardeningFracture
Not to scale
Axial extension e
(b) Axial extension e
ey est
E AN N
L e
(a) Tension member
Figure 21 Load-extension behaviour of a perfect tension member
same way as does the average strain ε= eL with the average stress σ = NA andso the load-extension relationship for the member shown in Figure 21b is similarto the material stressndashstrain relationship shown in Figure 16 Thus the extensionat first increases linearly with the load and is equal to
e = NL
EA (21)
where E is the Youngrsquos modulus of elasticity This linear increase continues untilthe yield stress fy of the steel is reached at the general yield load
Npl = Afy (22)
when the extension increases with little or no increase in load until strain-hardeningcommences After this the load increases slowly until the maximum value
Nu = Afu (23)
is reached in which fu is the ultimate tensile strength of the steel Beyond thisa local cross-section of the member necks down and the load N decreases untilfracture occurs
The behaviour of the tension member is described as ductile in that it can reachand sustain the general yield load while significant extensions occur before itfractures The general yield load Npl is often taken as the load capacity of themember
If the tension member is not initially stress free but has a set of residual stressesinduced during its manufacture such as that shown in Figure 22b then localyielding commences before the general yield load Npl is reached (Figure 22c) andthe range over which the load-extension behaviour is linear decreases Howeverthe general yield load Npl at which the whole cross-section is yielded can stillbe reached because the early yielding causes a redistribution of the stresses The
Tension members 35
1 Residual stress(N = 0)
First yield
b
Fully yielded(N = Npl)
2
3
4
fy
t
(a) Section
(b) Stressdistributions
tc
c
Load N
Npl
est
Extension e
ey1
23 4
(c) Load-extension behaviour
Figure 22 Effect of residual stresses on load-extension behaviour
N = 0 N = 0 N Nww0
(a) Initial crookedness (b) Deflection under load
Figure 23 Tension member with initial crookedness
residual stresses also cause local early strain-hardening and while the plastic rangeis shortened (Figure 22c) the member behaviour is still regarded as ductile
If however the tension member has an initial crookedness w0 (Figure 23a)the axial load causes it to bend and deflect laterally w (Figure 23b) These lateraldeflections partially straighten the member so that the bending action in the centralregion is reduced The bending induces additional axial stresses (see Section 53)which cause local early yielding and strain-hardening in much the same way asdo residual stresses (see Figure 22c) The resulting reductions in the linear andplastic ranges are comparatively small when the initial crookedness is small as isnormally the case and the member behaviour is ductile
222 Members with small holes
The presence of small local holes in a tension member (such as small bolt holesused for the connections of the member) causes early yielding around the holesso that the load-deflection behaviour becomes non-linear When the holes aresmall the member may reach the gross yield load (see equation 22) calculated
36 Tension members
No holesHoles not significantSignificant holes
Extension e
Load N
Afu
Afy
Anet fu
Figure 24 Effect of holes on load-extension behaviour
on the gross area A as shown in Figure 24 because of strain-hardening effectsaround the holes In this case the member behaviour is ductile and the non-linearbehaviour can be ignored because the axial extension of the member under load isnot significantly increased except when there are so many holes along the lengthof the member that the average cross-sectional area is significantly reduced Thusthe extension e can normally be calculated by using the gross cross-sectional areaA in equation 21
223 Members with significant holes
When the holes are large the member may fail before the gross yield load Npl isreached by fracturing at a hole as shown in Figure 24 The local fracture load
Nu = Anet fu (24)
is calculated on the net area of the cross-section Anet measured perpendicular tothe line of action of the load and is given by
Anet = A minussum
d0t (25)
where d0 is the diameter of a hole t the thickness of the member at the hole andthe summation is carried out for all holes in the cross-section under considerationThe fracture load Nu is determined by the weakest cross-section and therefore bythe minimum net area Anet A member which fails by fracture before the grossyield load can be reached is not ductile and there is little warning of failure
In many practical tension members with more than one row of holes the reduc-tion in the cross-sectional area may be reduced by staggering the rows of holes(Figure 25) In this case the possibility must be considered of failure along a zigndashzag path such as ABCDE in Figure 25 instead of across the section perpendicular
Tension members 37
A
B
C
D
E
p
p
s s
N N
Figure 25 Possible failure path with staggered holes
to the load The minimum amount of stagger sm for which a hole no longer reducesthe area of the member depends on the diameter d0 of the hole and the inclinationps of the failure path where p is the gauge distance between the rows of holesAn approximate expression for this minimum stagger is
sm asymp (4pd0)12 (26)
When the actual stagger s is less than sm some reduced part of the hole areaAh must be deducted from the gross area A and this can be approximated byAh = d0t(1 minus s2s2
m) whence
Ah = d0t(1 minus s24pd0) (27)
Anet = A minussum
d0t +sum
s2t4p (28)
where the summations are made for all the holes on the zigndashzag path consideredand for all the staggers in the path The use of equation 28 is allowed for inClause 6222 of EC3 and is illustrated in Section 271
Holes in tension members also cause local stress increases at the hole boundariesas well as the increased average stresses NAnet discussed above These localstress concentrations and their influence on member strength are discussed laterin Section 25
23 Eccentrically and locally connectedtension members
In many cases the fabrication of tension members is simplified by making theirend connections eccentric (ie the centroid of the connection does not coincidewith the centroidal axis of the member) and by connecting to some but not all ofthe elements in the cross-section It is common for example to make connections
38 Tension members
(a) (b) (c) (d)
Figure 26 Eccentrically and locally connected tension members
to an angle section through one leg only to a tee-section through the flange (ortable) or to a channel section through the web (Figure 26) The effect of eccentricconnections is to induce bending moments in the member whilst the effect ofconnecting to some but not all elements in the cross-section is to cause thoseregions most remote from the connection point(s) to carry less load The latter isessentially a shear lag effect (see Section 545) Both of these effects are localto the connections decreasing along the member length and are reduced furtherby ductile stress redistribution after the onset of yielding Members connected bysome but not all of the elements in the cross-section (including those connectedsymmetrically as in Figure 26d) are also discussed in Section 25
While tension members in bending can be designed rationally by using the pro-cedure described in Section 24 simpler methods [1ndash3] also produce satisfactoryresults In these simpler methods the effects described above are approximatedby reducing the cross-sectional area of the member (to an effective net area)and by designing it as if concentrically loaded For a single angle in tension con-nected by a single row of bolts in one leg the effective net section Aneteff to beused in place of the net area Anet in equation 24 is defined in EC3-1-8 [4] It isdependent on the number of bolts and the pitch p1 and for one bolt is given by
Aneteff = 20(e2 minus 05 d0)t (29)
for two bolts by
Aneteff = β2Anet (210)
and for three or more bolts by
Aneteff = β3Anet (211)
The symbols in equations 29ndash211 are defined below and in Figure 27 and Anet
is the net area of the angle For an unequal angle connected by its smaller leg Anet
Tension members 39
e1 p1 p1
e2 d0
Figure 27 Definition of symbols for tension connections in single angles
should be taken as the net section of an equivalent equal angle of leg length equalto the smaller leg of the unequal angle In the case of welded end connections foran equal angle or an unequal angle connected by its larger leg the eccentricitymay be neglected and the effective area Anet may be taken as equal to the grossarea A (Clause 413(2) of EC3-1-8 [4])
For a pair of bolts in a single row in an angle leg β2 in equation 210 is takenas 04 for closely spaced bolts (p1 le 25 d0) as 07 if they are widely spaced(p1 ge 5 d0) or as a linear interpolation for intermediate spacings For three ormore bolts in a single row in an angle leg β3 in equation 211 is taken as 05 forclosely spaced bolts (p1 le 25 d0) as 07 if they are widely spaced (p1 ge 5 d0)or again as a linear interpolation for intermediate spacings
24 Bending of tension members
Tension members often have bending actions caused by eccentric connectionseccentric loads or transverse loads including self-weight as shown in Figure 28These bending actions which interact with the tensile loads reduce the ultimatestrengths of tension members and must therefore be accounted for in design
The axial tension N and transverse deflections w of a tension member causedby any bending action induce restoring moments Nw which oppose the bend-ing action It is a common (and conservative) practice to ignore these restoringmoments which are small when either the bending deflections w or the tensile loadN are small In this case the maximum stress σmax in the member can be safelyapproximated by
σmax = σat + σbty + σbtz (212)
where σat = NA is the average tensile stress and σbty and σbtz are the maximumtensile bending stresses caused by the major and minor axis bending actions My
and Mz alone (see Section 53) The nominal first yield of the member thereforeoccurs when σmax = fy whence
σat + σbty + σbtz = fy (213)
When either the tensile load N or the moments My and Mz are not small the firstyield prediction of equation 213 is inaccurate A suggested interaction equation
40 Tension members
N Nw
(a) Transverse load only (b) Transverse load and tension
Bending moment diagrams
Nw
Figure 28 Bending of a tension member
for failure is the linear inequality
MyEd
MtyRd+ MzEd
MrzRdle 1 (214)
where MrzRd is the cross-section resistance for tension and bending about theminor principal (z) axis given by
MrzRd = MczRd(1 minus NtEdNtRd) (215)
and MtyRd is the lesser of the cross-section resistance MryRd for tension andbending about the major principal (y) axis given by
MryRd = McyRd(1 minus NtEdNtRd) (216)
and the out-of-plane member buckling resistance MbtRd for tension and bendingabout the major principal axis given by
MbtRd = MbRd(1 + NtEdNtRd) le McyRd (217)
in which MbRd is the lateral buckling resistance when N = 0 (see Chapter 6)In these equations NtRd is the tensile resistance in the absence of bending
(taken as the lesser of Npl and Nu) while McyRd and MczRd are the cross-sectionresistances for bending alone about the y and z axes (see Sections 472 and 5613)Equation 214 is similar to the first yield condition of equation 213 but includesa simple approximation for the possibility of lateral buckling under large valuesof MyEd through the use of equation 217
25 Stress concentrations
High local stress concentrations are not usually important in ductile materialsunder static loading situations because the local yielding resulting from these
Tension members 41
concentrations causes a favourable redistribution of stress However in situationsfor which it is doubtful whether the steel will behave in a ductile manner aswhen there is a possibility of brittle fracture of a tension member under dynamicloads (see Section 133) or when repeated load applications may lead to fatiguefailure (see Section 132) stress concentrations become very significant It isusual to try to avoid or minimise stress concentrations by providing suitable jointand member details but this is not always possible in which case some estimateof the magnitudes of the local stresses must be made
Stress concentrations in tension members occur at holes in the member andwhere there are changes or very local reductions in the cross-section and at pointswhere concentrated forces act The effects of a hole in a tension member (ofnet width bnet and thickness t) are shown by the uppermost curve in Figure 29which is a plot of the variation of the stress concentration factor (the ratio ofthe maximum tensile stress to the nominal stress averaged over the reduced cross-section bnet t)with the ratio of the hole radius r to the net width bnet The maximumstress approaches three times the nominal stress when the plate width is very large(rbnet rarr 0)
This curve may also be used for plates with a series of holes spaced equallyacross the plate width provided bnet is taken as the minimum width betweenadjacent holes The effects of changes in the cross-section of a plate in tension areshown by the lower curves in Figure 29 while the effects of some notches or verylocal reductions in the cross-section are given in [5] Methods of analysing theseand other stress concentrations are discussed in [6]
Large concentrated forces are usually transmitted to structural members by localbearing and while this produces high local compressive stresses any subsequent
Str
ess
conc
entr
atio
n fa
ctor
0 0 2 0 4 0 6 0 8 1 01 0
1 5
2 0
2 5
3 0
3 5
N o m i n al s t r e s s i sc a l cu la t e d f o r n e t area bnet times t
b
2 r bnet 2
b r bnet
Ratio rbnet
15 225
(b-bnet )2r=1
Figure 29 Stress concentrations in tension members
42 Tension members
plastic yielding is not usually serious Of more importance are the increased tensilestresses which result when only some of the plate elements of a tension memberare loaded (see Figure 26) as discussed in Section 23 In this case conservativeestimates can be made of these stresses by assuming that the resulting bending andaxial actions are resisted solely by the loaded elements
26 Design of tension members
261 General
In order to design or to check a tension member the design tensile force NtEd ateach cross-section in the member is obtained by a frame analysis (see Chapter 8)or by statics if the member is statically determinate using the appropriate loadsand partial load factors γG γQ (see Section 17) If there is substantial bendingpresent the design moments MyEd and MzEd about the major and minor principalaxes respectively are also obtained
The process of checking a specified tension member or of designing an unknownmember is summarised in Figure 210 for the case where bending actions can beignored If a specified member is to be checked then both the strength limit statesof gross yielding and the net fracture are considered so that both the yield strengthfy and the ultimate strength fu are required When the member size is not knownan approximate target area A can be established The trial member selected maybe checked for the fracture limit state once its holes connection eccentricities andnet area Anet have been established and modified if necessary
The following sub-sections describe each of the EC3 check and design processesfor a statically loaded tension member Worked examples of their application aregiven in Sections 271ndash275
262 Concentrically loaded tension members
The EC3 method of strength design of tension members which are loaded concen-trically follows the philosophy of Section 22 with the two separate limit statesof yield of the gross section and fracture of the net section represented by a singleequation
NtEd le NtRd (218)
where NtRd is the design tension resistance which is taken as the lesser of theyield (or plastic) resistance of the cross-section NplRd and the ultimate (or fracture)resistance of the cross-section containing holes NuRd
The yield limit state of equation 22 is given in EC3 as
NplRd = AfyγM0 (219)
where A is the gross area of the cross-section and γM0 is the partial factor forcross-section resistance with a value of 10
Tension members 43
Checking Designing
Section
Loading
Analysis for NEd
Find fy fu Anet
Assume fy
Find trial section A ge
ge
NEd fy
NtRd is the lesser of NplRd and NuRd
Check NEd Nt Rd
Check complete
Design complete
Find new section
Yes
No
NplRd = Afy γM0 NuRd = 09Anet fu γM2
Figure 210 Flow chart for the design of tension members
The fracture limit state of equation 24 is given in EC3 as
NuRd = 09Anet fuγM2 (220)
where Anet is the net area of the cross-section and γM2 is the partial factor forresistance in tension to fracture with a value of 11 given in the National Annex
44 Tension members
to EC3 The factor 09 in equation 220 ensures that the effective partial factorγM209(asymp122) for the limit state of material fracture (NuRd) is suitably higherthan the value of γM0(=10) for the limit state of yielding (NplRd) reflecting theinfluence of greater variability in fu and the reduced ductility of members whichfail by fracture at bolt holes
263 Eccentrically connected tension members
The EC3 method of strength design of simple angle tension members which areconnected eccentrically as shown in Figure 26 is similar to the method discussedin Section 262 for concentrically loaded members with the lsquo09Anetrsquo used inequation 220 being replaced by an effective net area Aneteff as described inSection 23
264 Tension members with bending
2641 Cross-section resistance
EC3 provides the conservative inequality
NtEd
NtRd+ MyEd
MyRd+ MzEd
MzRdle 1 (221)
for the cross-section resistance of tension members with design bending momentsMyEd and MzEd where MyRd and MzRd are the cross-section moment resistances(Sections 472 and 5613)
Equation 221 is rather conservative and so EC3 allows I-section tensionmembers with Class 1 or Class 2 cross-sections (see Section 472) with bendingabout the major (y) axis to satisfy
MyEd le MN yRd = MplyRd
(1 minus NtEdNplRd
1 minus 05a
)le MplyRd (222)
in which MN yRd is the reduced plastic design moment resistance about themajor (y) axis (reduced from the full plastic design moment resistance MplyRd toaccount for the axial force (see Section 7241) and
a = (A minus 2btf )A le 05 (223)
in which b and tf are the flange width and thickness respectively For I-sectiontension members with Class 1 or Class 2 cross-sections with bending about the
Tension members 45
minor (z) axis EC3 allows
MzEd le MN zRd = MplzRd
1 minus
(NtEdNplRd minus a
1 minus a
)2
le MplzRd (224)
where MN zRd is the reduced plastic design moment resistance about the minor (z)axis
EC3 also provides an alternative to equation 221 for the section resistance toaxial tension and bending about both principal axes of Class 1 or Class 2 cross-sections This is the more accurate interaction equation (see Section 742)(
MyEd
MN yRd
)α+
(MzEd
MN z Rd
)βle 1 (225)
where the values of α and β depend on the cross-section type (eg α= 20 andβ = 5NtEdNplRd for I-sections and α=β = 2 for circular hollow sections)
A worked example of a tension member with bending actions is given inSection 275
2642 Member resistance
EC3 does not provide any specific rules for determining the influence of lateralbuckling on the member resistance of tension members with bending This maybe accounted for conservatively by using
MyEd le MbRd (226)
in equation 221 in which MbRd is the design buckling moment resistance of themember (see Section 65) This equation conservatively omits the strengtheningeffect of tension on lateral buckling A less conservative method is provided byequations 214ndash217
27 Worked examples
271 Example 1 ndash net area of a bolted universal columnsection member
Problem Both flanges of a universal column section member have 22 mm diam-eter holes arranged as shown in Figure 211a If the gross area of the section is201 times 102 mm2 and the flange thickness is 25 mm determine the net area Anet ofthe member which is effective in tensionSolution Using equation 26 the minimum stagger is sm = radic
(4 times 60 times 22)=727 mmgt 30 mm = s The failure path through each flange is therefore stag-gered and by inspection it includes four holes and two staggers The net area can
46 Tension members
6 0 6 0 6 0
6 0
1 2 0
6 0
6 0 6 0
Gross area of member = 201times102 mm2
Flange is 3112 mm times 25 mm Holes are 22 mm diameter
(a) Flange of universal column section
1 0 0
1 0 0
1 0 2 9 1
2 6 1 2
1 2 y
z
(b) Universal beam section
(c) Double angle section
Dimensions in mm
A = 159 cm2 A = 2 times 227 cm2
Figure 211 Examples 1ndash6
therefore be calculated from equation 28 (or Clause 6222(4) of EC3) as
Anet = [201 times 102] minus [2 times 4 times (22 times 25)] + [2 times 2 times (302 times 25)(4 times 60)]= 161 times 102 mm2
272 Example 2 ndash checking a bolted universal columnsection member
Problem Determine the tension resistance of the tension member of example 1assuming it is of S355 steel
Solution
tf = 25 mm fy = 345 Nmm2 fu = 490 Nmm2 EN 10025-2
NplRd = AfyγM0 = 201 times 102 times 34510 = 6935 kN 623(2)a
From Section 271
Anet = 161 times 102 mm2
NuRd = 09Anet fuγM2 = 09 times 161 times 102 times 49011 = 6445 kN623(2)b
NtRd = 6445 kN (the lesser of NplRd and NuRd) 623(2)
Tension members 47
273 Example 3 ndash checking a bolted universal beamsection member
Problem A 610times229 UB 125 tension member of S355 steel is connected throughboth flanges by 20 mm bolts (in 22 mm diameter bolt holes) in four lines twoin each flange as shown in Figure 211b Check the member for a design tensionforce of NtEd = 4000 kN
Solution
tf = 196 mm fy = 345 Nmm2 fu = 490 Nmm2 EN 10025-2
A = 15 900 mm2
NplRd = AfyγM0 = 15900 times 34510 = 5486 kN 623(2)a
Anet = 15 900 minus (4 times 22 times 196) = 14175 mm2 6222(3)
NuRd = 09Anet fuγM2 = 09 times 14175 times 49011 = 5683 kN 623(2)b
NtRd = 5486 kN (the lesser of NplRd and NuRd) gt 4000 kN = NtEd
623(2)
and so the member is satisfactory
274 Example 4 ndash checking an eccentrically connectedsingle (unequal) angle
Problem A tension member consists of a 150 times 75 times 10 single unequal anglewhose ends are connected to gusset plates through the larger leg by a singlerow of four 22 mm bolts in 24 mm holes at 60 mm centres Use the method ofSection 23 to check the member for a design tension force of NtEd = 340 kN if theangle is of S355 steel and has a gross area of 217 cm2
Solution
t = 10 mm fy = 355 Nmm2 fu = 490 Nmm2 EN 10025-2
Gross area of cross-section A = 2170 mm2
NplRd = AfyγM0 = 2170 times 35510 = 7704 kN 623(2)a
Anet = 2170 minus (24 times 10) = 1930 mm2 6222(3)
β3 = 05 (since the pitch p1 = 60 mm = 25d0) EC3-1-8 3103(2)
NuRd = β3Anet fuγM2 = 05 times 1930 times 49011 = 4299 kN 623(2)b
NtRd = 4299 kN (the lesser of NplRd and NuRd) gt 340 kN = NtEd
623(2)
and so the member is satisfactory
48 Tension members
275 Example 5 ndash checking a member under combinedtension and bending
Problem A tension member consists of two equal angles of S355 steel whose endsare connected to gusset plates as shown in Figure 211c If the load eccentricityfor the tension member is 391 mm and the tension and bending resistances are1438 kN and 377 kNm determine the resistance of the member by treating it asa member under combined tension and bending
SolutionSubstituting into equation 221 leads to
NtEd1438 + NtEd times (3911000)377 le 1 621(7)
so that NtEd le 5772 kNBecause the load eccentricity causes the member to bend about its minor axis
there is no need to check for lateral buckling which only occurs when there ismajor axis bending
276 Example 6 ndash estimating the stress concentration factor
Problem Estimate the maximum stress concentration factor for the tensionmember of Section 271
Solution For the inner line of holes the net width is
bnet = 60 + 30 minus 22 = 68 mm and so
rbnet = (222)68 = 016
and so using Figure 29 the stress concentration factor is approximately 25However the actual maximum stress is likely to be greater than 25 times the
nominal average stress calculated from the effective area Anet = 161 times 102 mm2
determined in Section 271 because the unconnected web is not completelyeffective A safe estimate of the maximum stress can be determined on the basisof the flange areas only of
Anet = [2 times 3106 times 25] minus [2 times 4 times 22 times 25] + [2 times 2 times (302 times 25)(4 times 60)]= 1151 times 102 mm2
28 Unworked examples
281 Example 7 ndash bolting arrangement
A channel section tension member has an overall depth of 381 mm width of1016 mm flange and web thicknesses of 163 and 104 mm respectively and an
Tension members 49
yield strength of 345 Nmm2 Determine an arrangement for bolting the web of thechannel to a gusset plate so as to minimise the gusset plate length and maximisethe tension member resistance
282 Example 8 ndash checking a channel section memberconnected by bolts
Determine the design tensile resistance NtRd of the tension member of Section281
283 Example 9 ndash checking a channel section memberconnected by welds
If the tension member of Section 281 is fillet welded to the gusset plate insteadof bolted determine its design tensile resistance NtRd
284 Example 10 ndash checking a member under combinedtension and bending
The beam of example 1 in Section 6151 has a concentric axial tensile load 5Q inaddition to the central transverse load Q Determine the maximum value of Q
References
1 Regan PE and Salter PR (1984) Tests on welded-angle tension members StructuralEngineer 62B(2) pp 25ndash30
2 Nelson HM (1953) Angles in Tension Publication No 7 British Constructional SteelAssociation pp 9ndash18
3 Bennetts ID Thomas IR and Hogan TJ (1986) Design of statically loaded tensionmembers Civil Engineering Transactions Institution of Engineers Australia CE28No 4 November pp 318ndash27
4 British Standards Institute (2006) Eurocode 3 Design of Steel Structures ndash Part 1ndash8Design of Joints BSI London
5 Roark RJ (1965) Formulas for Stress and Strain 4th edition McGraw-HillNew York
6 Timoshenko SP and Goodier JN (1970) Theory of Elasticity 3rd edition McGraw-Hill New York
Chapter 3
Compression members
31 Introduction
The compression member is the second type of axially loaded structural elementthe first type being the tension member discussed in Chapter 2 Very stocky com-pression members behave in the same way as do tension members until after thematerial begins to flow plastically at the squash load Ny = Afy However the resis-tance of a compression member decreases as its length increases in contrast tothe axially loaded tension member whose resistance is independent of its lengthThus the compressive resistance of a very slender member may be much less thanits tensile resistance as shown in Figure 31
This decrease in resistance is caused by the action of the applied compressiveload N which causes bending in a member with initial curvature (see Figure 32a)In a tension member the corresponding action decreases the initial curvature andso this effect is usually ignored However the curvature and the lateral deflectionof a compression member increase with the load as shown in Figure 32b Thecompressive stresses on the concave side of the member also increase until themember fails due to excessive yielding This bending action is accentuated bythe slenderness of the member and so the resistance of a compression memberdecreases as its length increases
For the hypothetical limiting case of a perfectly straight elastic member thereis no bending until the applied load reaches the elastic buckling value Ncr (EC3refers to Ncr as the elastic critical force) At this load the compression memberbegins to deflect laterally as shown in Figure 32b and these deflections grow untilfailure occurs at the beginning of compressive yielding This action of suddenlydeflecting laterally is called flexural buckling
The elastic buckling load Ncr provides a measure of the slenderness of acompression member while the squash load Ny gives an indication of itsresistance to yielding In this chapter the influences of the elastic bucklingload and the squash load on the behaviour of concentrically loaded compres-sion members are discussed and related to their design according to EC3Local buckling of thin-plate elements in compression members is treated inChapter 4 while the behaviour and design of eccentrically loaded compression
Compression members 51
Tension memberand very stocky compression member
Slender compression member
Change in length
Axi
al lo
ad
Ny = Afy
fy LE
Figure 31 Resistances of axially loaded members
Straight member 0 = 0 equation 32
Member with initial curvature0 =0 sinπxLequation 39
Straight member 0 = 0
0 5 10 15 20
02
04
06
08
10
Dim
ensi
onle
ss lo
ad N
Ncr
Dimensionless deflection 0
(b) Central deflection(a) Compression member
N
N
L2
L2x
Initial position(N = 0)
δ 0
δ
0
Figure 32 Elastic behaviour of a compression member
members are discussed in Chapter 7 and compression members in frames inChapter 8
32 Elastic compression members
321 Buckling of straight members
Aperfectly straight member of a linear elastic material is shown in Figure 33a Themember has a frictionless hinge at each end its lower end being fixed in positionwhile its upper end is free to move vertically but is prevented from deflectinghorizontally It is assumed that the deflections of the member remain small
52 Compression members
N
N N
N
v
x
L
(a) Unloaded member (b) Undeflected loaded member
(c) Deflected member
Figure 33 Straight compression member
The unloaded position of the member is shown in Figure 33a A concentricaxial load N is applied to the upper end of the member which remains straight(Figure 33b) The member is then deflected laterally by a small amount v asshown in Figure 33c and is held in this position If the original straight positionof Figure 33b is one of stable equilibrium the member will return to it whenreleased from the deflected position while if the original position is one of unstableequilibrium the member will collapse away from it when released When theequilibrium of the original position is neither stable nor unstable the member isin a condition described as one of neutral equilibrium For this case the deflectedposition is one of equilibrium and the member will remain in this position whenreleased Thus when the load N reaches the elastic buckling value Ncr at which theoriginal straight position of the member is one of neutral equilibrium the membermay deflect laterally without any change in the load as shown in Figure 32b
The load Ncr at which a straight compression member buckles laterally can bedetermined by finding a deflected position which is one of equilibrium It is shownin Section 381 that this position is given by
v = δ sin πxL (31)
in which δ is the undetermined magnitude of the central deflection and that theelastic buckling load is
Ncr = π2EIL2 (32)
in which EI is the flexural rigidity of the compression member
Compression members 53
The elastic buckling load Ncr and the elastic buckling stress
σcr = NcrA (33)
can be expressed in terms of the geometrical slenderness ratio Li by
Ncr = σcrA = π2EA
(Li)2(34)
in which i = radic(IA) is the radius of gyration (which can be determined for a
number of sections by using Figure 56) The buckling load varies inversely asthe square of the slenderness ratio Li as shown in Figure 34 in which thedimensionless buckling load NcrNy is plotted against the generalised slendernessratio
λ =radic
Ny
Ncr=
radicfyσcr
= L
i
radicfyπ2E
(35)
in which
Ny = Afy (36)
is the squash load If the material ceases to be linear elastic at the yield stress fythen the above analysis is only valid for λ = radic
(NyNcr) = radic(fyσcr) ge 1 This
limit is equivalent to a slenderness ratio Li of approximately 85 for a materialwith a yield stress fy of 275 Nmm2
0 05 10 15 20 25
Generalised slenderness cry NN =
0
02
04
06
08
10
12
Dim
ensi
onle
ss lo
ad N
crN
y or
NLN
y Complete
yielding
First yield NLNydue to initial curvature (equations 311ndash314)
Elastic buckling NcrNy
(equations 34 and 35)
Figure 34 Buckling and yielding of compression members
54 Compression members
322 Bending of members with initial curvature
Real structural members are not perfectly straight but have small initial curva-tures as shown in Figure 32a The buckling behaviour of the hypothetical straightmembers discussed in Section 321 must therefore be interpreted as the limit-ing behaviour of real members with infinitesimally small initial curvatures Theinitial curvature of the real member causes it to bend from the commencementof application of the axial load and this increases the maximum stress in themember
If the initial curvature is such that
v0 = δ0 sin πxL (37)
then the deflection of member is given by
v = δ sin πxL (38)
where
δ
δ0= NNcr
1 minus NNcr (39)
as shown in Section 382 The variation of the dimensionless central deflectionδδ0 is shown in Figure 32b and it can be seen that deflection begins at thecommencement of loading and increases rapidly as the elastic buckling load Ncr
is approachedThe simple load-deflection relationship of equation 39 is the basis of the
Southwell plot technique for extrapolating the elastic buckling load from experi-mental measurements If equation 39 is rearranged as
δ
N= 1
Ncrδ + δ0
Ncr (310)
then the linear relation between δN and δ shown in Figure 35 is obtained Thusif a straight line is drawn which best fits the points determined from experimen-tal measurements of N and δ the reciprocal of the slope of this line gives anexperimental estimate of the buckling load Ncr An estimate of the magnitudeδ0 of the initial crookedness can also be determined from the intercept on thehorizontal axis
As the deflections v increase with the load N so also do the bending momentsand the stresses It is shown in Section 382 that the limiting axial load NL at whichthe compression member first yields (due to a combination of axial plus bending
Compression members 55
Central deflection
0
NStraight line of best fit (see equation 310)
Slope = crN
1
0
Figure 35 Southwell plot
stresses) is given by
NL
Ny= 1
Φ +radicΦ2 minus λ
2 (311)
in which
Φ = 1 + η + λ2
2 (312)
η = δ0b
2i2 (313)
and b is the width of the member The variation of the dimensionless limiting axialload NLNy with the generalised slenderness ratio λ is shown in Figure 34 for thecase when
η = 1
4
Ny
Ncr (314)
For stocky members the limiting load NL approaches the squash load Ny whilefor slender members the limiting load approaches the elastic buckling load Ncr Equations 311 and 312 are the basis for the member buckling resistance in EC3
33 Inelastic compression members
331 Tangent modulus theory of buckling
The analysis of a perfectly straight elastic compression member given inSection 321 applies only to a material whose stressndashstrain relationship remains
56 Compression members
Strain Tangent modulus Et Slenderness Li
(a) Elastic stressndashstrainrelationship
(b) Tangent modulus (c) Buckling stress
E
E
Et
Et
Ncr
Ncrt(equation 317)B
uckl
ing
load
Ncr
t
Str
ess
NA
Str
ess
NA
(equation 34)
Figure 36 Tangent modulus theory of buckling
linear However the buckling of an elastic member of a non-linear material suchas that whose stressndashstrain relationship is shown in Figure 36a can be analysedby a simple modification of the linear elastic treatment It is only necessary tonote that the small bending stresses and strains which occur during buckling arerelated by the tangent modulus of elasticity Et corresponding to the average com-pressive stress NA (Figure 36a and b) instead of the initial modulus E Thus theflexural rigidity is reduced from EI to EtI and the tangent modulus buckling loadNcrt is obtained from equation 34 by substituting Et for E whence
Ncrt = π2EtA
(Li)2 (315)
The deviation of this tangent modulus buckling load Ncrt from the elastic bucklingload Ncr is shown in Figure 36c It can be seen that the deviation of Ncrt fromNcr increases as the slenderness ratio Li decreases
332 Reduced modulus theory of buckling
The tangent modulus theory of buckling is only valid for elastic materials Forinelastic nonlinear materials the changes in the stresses and strains are related bythe initial modulus E when the total strain is decreasing and the tangent modulusEt only applies when the total strain is increasing as shown in Figure 37 Theflexural rigidity ErI of an inelastic member during buckling therefore depends onboth E and Et As a direct consequence of this the effective (or reduced) modulusof the section Er depends on the geometry of the cross-section as well It is shownin Section 391 that the reduced modulus of a rectangular section is given by
Er = 4EEt(radicE + radic
Et
)2 (316)
Compression members 57
E
E
Et
Strain
Str
ess
NA
Figure 37 Inelastic stressndashstrain relationship
Lateral deflection
Load
Rate of increase in load required to prevent strain reversal
Ncr
Ncrr
Ncrt
Nmax
Figure 38 Shanleyrsquos theory of inelastic buckling
The reduced modulus buckling load Ncrr can be obtained by substituting Er for Ein equation 34 whence
Ncrr = π2ErA
(Li)2 (317)
Since the tangent modulus Et is less than the initial modulus E it follows thatEt lt Er lt E and so the reduced modulus buckling load Ncrr lies between thetangent modulus buckling load Ncrt and the elastic buckling load Ncr as indicatedin Figure 38
58 Compression members
333 Shanleyrsquos theory of inelastic buckling
Although the tangent modulus theory appears to be invalid for inelastic materialscareful experiments have shown that it leads to more accurate predictions thanthe apparently rigorous reduced modulus theory This paradox was resolved byShanley [1] who reasoned that the tangent modulus theory is valid when bucklingis accompanied by a simultaneous increase in the applied load (see Figure 38) ofsufficient magnitude to prevent strain reversal in the member When this happensall the bending stresses and strains are related by the tangent modulus of elasticityEt the initial modulus E does not feature and so the buckling load is equal to thetangent modulus value Ncrt
As the lateral deflection of the member increases as shown in Figure 38 thetangent modulus Et decreases (see Figure 36b) because of the increased axial andbending strains and the post-buckling curve approaches a maximum load Nmax
which defines the ultimate resistance of the member Also shown in Figure 38is a post-buckling curve which commences at the reduced modulus load Ncrr
(at which buckling can take place without any increase in the load) The tangentmodulus load Ncrt is the lowest load at which buckling can begin and the reducedmodulus load Ncrr is the highest load for which the member can remain straightIt is theoretically possible for buckling to begin at any load between Ncrt and Ncrr
It can be seen that not only is the tangent modulus load more easily calculatedbut it also provides a conservative estimate of the member resistance and is incloser agreement with experimental results than the reduced modulus load Forthese reasons the tangent modulus theory of inelastic buckling has gained wideacceptance
334 Buckling of members with residual stresses
The presence of residual stresses in an intermediate length steel compressionmember may cause a significant reduction in its buckling resistance Resid-ual stresses are established during the cooling of a hot-rolled or welded steelmember (and during plastic deformation such as cold-rolling) The shrinking ofthe late-cooling regions of the member induces residual compressive stresses inthe early-cooling regions and these are balanced by equilibrating tensile stressesin the late-cooling regions In hot-rolled I-section members the flange ndash webjunctions are the least exposed to cooling influences and so these are the regionsof residual tensile stress as shown in Figure 39 while the more exposed flangetips are regions of residual compressive stress In a straight intermediate lengthcompression member the residual compressive stresses cause premature yield-ing under reduced axial loads as shown in Figure 310 and the member bucklesinelastically at a load which is less than the elastic buckling load Ncr
In applying the tangent modulus concept of inelastic buckling to a steel whichhas the stressndashstrain relationships shown in Figure 16 (see Chapter 1) the strain-hardening modulus Est is sometimes used in the yielded regions as well as in
Compression members 59
05fy compression
y
z
03fy tension
03fy compression
Figure 39 Idealised residual stress pattern
N = 0 025Ny N Ny
de
de
0
d
bz
y
(EI)t = EI EIYielded
areas
(a)Load
c ct
fyResidual stress pattern
05Ny
05fy
05fy
(b)Stress
distribution
(c)Cross-section
(d)Effective flexural
rigidity
Fully yielded
EI (EI)t (EI)t = 0
Shaded areas due to axial load
First yield
Elastic
Elasticcore
Elastic
Figure 310 Effective section of a member with residual stresses
the strain-hardened regions This use is based on the slip theory of dislocationin which yielding is represented by a series of dynamic jumps instead of by asmooth quasi-static flow Thus the material in the yielded region is either elasticor strain-hardened and its subsequent behaviour may be estimated conservativelyby using the strain-hardening modulus However the even more conservativeassumption that the tangent modulus Et is zero in both the yielded and strain-hardened regions is frequently used because of its simplicity According to thisassumption these regions of the member are ineffective during buckling and themoment of resistance is entirely due to the elastic core of the section
60 Compression members
Thus for a rectangular section member which has the simplified residual stressdistribution shown in Figure 310 the effective flexural rigidity (EI)t about the zaxis is (see Section 392)
(EI)t = EI [2(1 minus NNy)]12 (318)
when the axial load N is greater than 05Ny at which first yield occurs and theaxial load at buckling Ncrt is given by
Ncrt
Ny=
(Ncr
Ny
)2
[1 + 2(NyNcr)2]12 minus 1 (319)
The variation of this dimensionless tangent modulus buckling load Ncrt Ny withthe generalised slenderness ratio λ = radic
(NyNcr) is shown in Figure 311For stocky members the buckling load Ncrt approaches the squash load Nywhile for intermediate length members it approaches the elastic buckling loadNcr as λ = radic
(NyNcr) approachesradic
2 For more slender members prema-ture yielding does not occur and these members buckle at the elastic bucklingload Ncr
Also shown in Figure 311 is the dimensionless tangent modulus buckling loadNcrtNy given by
Ncrt
Ny= 1 minus 1
4
Ny
Ncr (320)
0 05 10 15 20 25
Generalised slenderness cry NN =
0
02
04
06
08
10
12
Dim
ensi
onle
ss b
uckl
ing
load
Ncr
tN
y Completeyield
Elastic buckling NcrNy
Inelastic buckling of rectangular section shown in Fig 310 (equation 319)
NN
crt
y---- = 1ndash 1
4----N
Ny
cr----
(equation 320)
radic2
Figure 311 Inelastic buckling of compression members with residual stresses
Compression members 61
which was developed as a compromise between major and minor axis bucklingof hot-rolled I-section members It can be seen that this simple compromise isvery similar to the relationship given by equation 319 for the rectangular sectionmember
34 Real compression members
The conditions under which real members act differ in many ways from theidealised conditions assumed in Section 321 for the analysis of the elastic buck-ling of a perfect member Real members are not perfectly straight and theirloads are applied eccentrically while accidental transverse loads may act Theeffects of small imperfections of these types are qualitatively the same as thoseof initial curvature which were described in Section 322 These imperfectionscan therefore be represented by an increased equivalent initial curvature whichhas a similar effect on the behaviour of the member as the combined effectof all of these imperfections The resulting behaviour is shown by curve A inFigure 312
A real member also has residual stresses and its elastic modulus E and yieldstress fy may vary throughout the member The effects of these material vari-ations are qualitatively the same as those of the residual stresses which weredescribed in Section 334 and so they can be represented by an equivalent set ofthe residual stresses The resulting behaviour of the member is shown by curve B inFigure 312
Since real members have both kinds of imperfections their behaviour is asshown by curve C in Figure 312 which is a combination of curves A and B Thusthe real member behaves as a member with equivalent initial curvature (curve A)until the elastic limit is reached It then follows a path which is similar to andapproaches that of a member with equivalent residual stresses (curve B)
Elastic bending
Curve A ndash equivalentinitial curvature
Curve B ndash equivalent residual stresses
Curve C ndash real members
Elastic limit
Nmax
Elastic buckling
Ncr
NL
Ncrt
Lateral deflection
Load
Figure 312 Behaviour of real compression members
62 Compression members
35 Design of compression members
351 EC3 design buckling resistance
Real compression members may be analysed using a model in which the initialcrookedness and load eccentricity are specified Residual stresses may be includedand the realistic stressndashstrain behaviour may be incorporated in the prediction of theload versus deflection relationship Thus curve C in Figure 312 may be generatedand the maximum load Nmax ascertained
Rational computer analyses based on the above modelling are rarely used exceptin research and are inappropriate for the routine design of real compression mem-bers because of the uncertainties and variations that exist in the initial crookednessand residual stresses Instead simplified design predictions for the maximumload (or resistance) are used in EC3 that have been developed from the results ofcomputer analyses and correlations with available test data
A close prediction of numerical solutions and test results may be obtained byusing equations 311 and 312 which are based on the first yield of a geometricallyimperfect member This is achieved by writing the imperfection parameter η ofequation 312 as
η = α(λminus 02) ge 0 (321)
in which α is a constant (imperfection factor) which shifts the resistance curve asshown in Figure 313 for different cross-section types proportions thicknessesbuckling axes and material strengths (Table 62 of EC3) The advantage of thisapproach is that the resistances (referred to as buckling resistances in EC3) of aparticular group of sections can be determined by assigning an appropriate value
0
02
04
06
08
1
12
0 05 1 15 2 25
Generalised slenderness cry NN =
Dim
ensi
onle
ss b
uckl
ing
resi
stan
ce χ
= N
bR
dN
y
Curve d
Curve c
Curve b
Curve a
Curve a0
Figure 313 Compression resistances of EC3
Compression members 63
of α to them The design buckling resistance NbRd is then given by
NbRd = χNyγM1 (322)
in which γM1 is the partial factor for member instability which has a recommendedvalue of 10 in EC3
The dimensionless compressive design buckling resistances NbRdNy for α =013 021 034 049 and 076 which represent the EC3 compression membercurves (a0) (a) (b) (c) and (d) respectively are shown in Figure 313 Memberswith higher initial crookedness have lower strengths due to premature yieldingand these are associated with higher values of α On the other hand members withlower initial crookedness are not as greatly affected by premature yielding andhave lower values of α assigned to them
352 Elastic buckling load
Although the design buckling resistance NbRd given by equation 322 was derivedby using the expression for the elastic buckling load Ncr given by equation 32 for apin-ended compression member equation 322 is also used for other compressionmembers by generalising equation 32 to
Ncr = π2EIL2cr (323)
in which Lcr is the effective length (referred to as the buckling length in EC3)The elastic buckling load Ncr varies with the member geometry loading and
restraints but there is no guidance given in EC3 for determining either Ncr or Lcr Section 36 following summarises methods of determining the effects of restraintson the effective length Lcr while Section 37 discusses the design of compressionmembers with variable sections or loading
353 Effects of local buckling
Compression members containing thin-plate elements are likely to be affected bylocal buckling of the cross-section (see Chapter 4) Local buckling reduces theresistances of short compression members below their squash loads Ny and theresistances of longer members which fail by flexural buckling
Local buckling of a short compression member is accounted for by using areduced effective area Aeff instead of the gross area A as discussed in Section 471Local buckling of a longer compression member is accounted for by using Aeff
instead of A and by reducing the generalised slenderness to
λ =radic
Aeff fyNcr (324)
but using the gross cross-sectional properties to determine Ncr
64 Compression members
354 Design procedures
For the design of a compression member the design axial force NEd is determinedby a rational frame analysis as in Chapter 8 or by statics for a statically determinatestructure The design loads (factored loads) FEd are for the ultimate limit state andare determined by summing up the specified loads multiplied by the appropriatepartial load factors γF (see Section 156)
To check the compression resistance of a member both the cross-sectionresistance NcRd and the member buckling resistance NbRd should generally beconsidered though it is usually the case for practical members that the bucklingresistance will govern The cross-section resistance of a compression memberNcRd must satisfy
NEd le NcRd (325)
in which
NcRd = AfyγM0 (326)
for a fully effective section in which γM0 is the partial factor for cross-sectionresistance (with a recommended value of 10 in EC3) or
NcRd = Aeff fyγM0 (327)
for the cross-sections that are susceptible to local buckling prior to yieldingThe member buckling resistance NbRd must satisfy
NEd le NbRd (328)
where the member buckling resistance NbRd is given by equation 322The procedure for checking a specified compression member is summarised
in Figure 314 The cross-section is checked to determine if it is fully effective(Aeff = A where A is the gross area) or slender (in which case the reducedeffective area Aeff is used) and the design buckling resistance NbRd is then foundand compared with the design compression force NEd
An iterative series of calculations is required when designing a compressionmember as indicated in Figure 314 An initial trial section is first selected eitherby using tabulations formulations or graphs of design buckling resistance NbRd
versus effective length Lcr or by making initial guesses for fy Aeff A and χ (say05) and calculating a target area A A trial section is then selected and checkedIf the section is not satisfactory then a new section is selected using the latestvalues of fy Aeff A and χ and the checking process is repeated The iterationsusually converge within a few cycles but convergence can be hastened by using themean of the previous and current values of χ in the calculation of the target area A
A worked example of checking the resistance of a compression member is givenin Section 3121 while worked examples of the design of compression membersare given in Sections 3122 and 3123
Compression members 65
No
Yes
Yes
Checking Designing
Section Length and loading
Analysis for NEd
Find fy Aeff
Find
Find
Compression resistance NbRd = Aeff fy M1
Check if NEd NbRd
Check complete
Design complete
Find section A NEd( times fy)
No
Table or graph
Trial section
Find Lcr
Initial values fy (= nom) Aeff (= A)
Find section NbRd NEd
Test convergence
Modify to hasten
convergence
(= 05)
ge
ge
le
Figure 314 Flow chart for the design of compression members
36 Restrained compression members
361 Simple supports and rigid restraints
In the previous sections it was assumed that the compression member was sup-ported only at its ends as shown in Figure 315a If the member has an additionallateral support which prevents it from deflecting at its centre so that (v)L2 = 0
66 Compression members
Ncr Ncr Ncr NcrNcr
Lcr
Lcr
Lcr
Lcr
Lcr LcrL L L
L
L
22 LEINcr π=Lcr = L
(a)
22 4 LEINcr π=Lcr = L2
(b)
22 2 LEINcr πasympasympLcr 07L
(c)
22 4 LEINcr π=Lcr = L2
(d)
22 4 LEINcr π=Lcr = 2L
(e)
Figure 315 Effective lengths of columns
as shown in Figure 315b then its buckled shape v is given by
v = δ sin 2πxL (329)
and its elastic buckling load Ncr is given by
Ncr = 4π2EI
L2 (330)
The end supports of a compression member may also differ from the simple sup-ports shown in Figure 315a which allow the member ends to rotate but preventthem from deflecting laterally For example one or both ends may be rigidly built-in so as to prevent end rotation (Figure 315c and d) or one end may be completelyfree (Figure 315e) In each case the elastic buckling load of the member may beobtained by finding the solution of the differential equilibrium equation whichsatisfies the boundary conditions
All of these buckling loads can be expressed by the generalisation ofequation 323 which replaces the member length L of equation 32 by the effectivelength Lcr Expressions for Lcr are shown in Figure 315 and in each case it canbe seen that the effective length of the member is equal to the distance betweenthe inflexion points of its buckled shape
The effects of the variations in the support and restraint conditions on the com-pression member buckling resistance NbRd may be accounted for by replacingthe actual length L used in equation 32 by the effective length Lcr in the calcula-tion of the elastic buckling load Ncr and hence the generalised slenderness λ andusing this modified generalised slenderness throughout the resistance equationsIt should be noted that it is often necessary to consider the member behaviour ineach principal plane since the effective lengths Lcry and Lcrz may also differ aswell as the radii of gyration iy and iz
Compression members 67
0 10 20 30 40Stiffness (16EIL3 )
0
2
4
6
8
Second mode(equation 330)
L = 16 2EIL3 (equation 331)
L2
L2
Ncr
2
x
v
22 4 LEI
(a) Braced firstmode
(b) Second mode
(c) Buckling loadB
uckl
ing
load
Ncr
2
EI
L2
Braced first mode (equations 365 and 366)
Llt L
2
ge
Figure 316 Compression member with an elastic intermediate restraint
362 Intermediate restraints
In Section 361 it was shown that the elastic buckling load of a simply supportedmember is increased by a factor of 4 to Ncr = 4π2EIL2 when an additionallateral restraint is provided which prevents it from deflecting at its centre asshown in Figure 315b This restraint need not be completely rigid but may beelastic (Figure 316) provided its stiffness exceeds a certain minimum value If thestiffness α of the restraint is defined by the force αδ acting on the restraint whichcauses its length to change by δ then the minimum stiffness αL is determined inSection 3101 as
αL = 16π2EIL3 (331)
The limiting stiffness αL can be expressed in terms of the buckling load Ncr =4π2EIL2 as
αL = 4NcrL (332)
Compression member restraints are generally required to be able to transmit 1(Clause 533(3)) of EC3) of the force in the member restrained This is a littleless than the value of 15 suggested [2] as leading to the braces which aresufficiently stiff
363 Elastic end restraints
When the ends of a compression member are rigidly connected to its adjacentelastic members as shown in Figure 317a then at buckling the adjacent members
68 Compression members
N
N
EI L
1
2
(a) Frame (b) Member
2
1
x
v
N
NM1
M2 M1 + M2 + N
L
M1 + M2 + N
L
∆
∆
∆
13
13
Figure 317 Buckling of a member of a rigid frame
exert total elastic restraining moments M1 M2 which oppose buckling and whichare proportional to the end rotations θ1 θ2 of the compression member Thus
M1 = minusα1θ1
M2 = minusα2θ2
(333)
in which
α1 =sum
1
α (334)
where α is the stiffness of any adjacent member connected to the end 1 of thecompression member and α2 is similarly defined The stiffness α of an adjacentmember depends not only on its length L and flexural rigidity EI but also on itssupport conditions and on the magnitude of any axial load transmitted by it
The particular case of a braced restraining member which acts as if simplysupported at both ends as shown in Figure 318a and which provides equal andopposite end moments M and has an axial force N is analysed in Section 3102where it is shown that when the axial load N is compressive the stiffness α isgiven by
α = 2EI
L
(π2)radic
NNcrL
tan(π2)radic
NNcrL(335)
where
NcrL = π2EIL2 (336)
Compression members 69
01 2 3123 23
ndash2
ndash4
ndash6
2
4
Tension Compression
Dimensionless axial load
(b) Stiffness
Exact equation 335
NNcrLL
N
N
v
x
M
M
(a) Braced member
Exact equation 337
Approx equation 338
(2E
IL)
Dim
ensi
onle
ss s
tiff
ness
Figure 318 Stiffness of a braced member
and by
α = 2EI
L
(π2)radic
NNcrL
tanh(π2)radic
NNcrL(337)
when the axial load N is tensile These relationships are shown in Figure 318band it can be seen that the stiffness decreases almost linearly from 2EI L to zeroas the compressive axial load increases from zero to NcrL and that the stiffness isnegative when the axial load exceeds NcrL In this case the adjacent member nolonger restrains the buckling member but disturbs it When the axial load causestension the stiffness is increased above the value 2EI L
Also shown in Figure 318b is the simple approximation
α = 2EI
L
(1 minus N
NcrL
) (338)
The term (1 minus NNcrL) in this equation is the reciprocal of the amplification fac-tor which expresses the fact that the first-order rotations ML2EI associated withthe end moments M are amplified by the compressive axial load N to (ML2EI )(1 minus NNcrL) The approximation provided by equation 338 is close and con-servative in the range 0 lt NNcrL lt 1 but errs on the unsafe side and withincreasing error as the axial load N increases away from this range
70 Compression members
N N N N N N
N N N N N N
(a) Braced members
M M M
M M M M M M
(b) Unbraced members
)1(2
LcrN
N
L
EI minus )1(3
LcrN
N
L
EI minus
)(while
)41(6
Lcr
Lcr
NN
N
N
L
EI
lt
minus ) ) )21( ( (4
LcrN
N
L
EI minus
minusminus
Lcr
Lcr
NN
NN
L
EI
2
1
42 π
minus
minusLcr
Lcr
NN
NN
L
EI
2
41
π
Figure 319 Approximate stiffnesses of restraining members
Similar analyses may be made of braced restraining members under other endconditions and some approximate solutions for the stiffnesses of these are sum-marised in Figure 319 These approximations have accuracies comparable withthat of equation 338
When a restraining member is unbraced its ends sway ∆ as shown inFigure 320a and restoring end moments M are required to maintain equilibriumSuch a member which receives equal end moments is analysed in Section 3103where it is shown that its stiffness is given by
α = minus(
2EI
L
)π
2
radicN
NcrLtan
π
2
radicN
NcrL(339)
when the axial load N is compressive and by
α =(
2EI
L
)π
2
radicN
NcrLtanh
π
2
radicN
NcrL(340)
when the axial load N is tensile These relationships are shown in Figure 320band it can be seen that the stiffness α decreases as the tensile load N decreases andbecomes negative as the load changes to compressive Also shown in Figure 320bis the approximation
α = minus2EI
L
(π24)(NNcrL)
(1 minus NNcrL) (341)
Compression members 71
4 3 2 10
1
ndash2
ndash4
ndash6
2
4
Tension
(b) Stiffness
L
N
N
x
M
M
(a) Unbraced member
Exact equation 340
Approx equation 341
Compression
Exact equation 339
Dimensionless axial load NNcrL
(2E
IL
)D
imen
sion
less
sti
ffne
ss
Figure 320 Stiffness of an unbraced member
which is close and conservative when the axial load N is compressive and whicherrs on the safe side when the axial load N is tensile It can be seen from Figure 320bthat the stiffness of the member is negative when it is subjected to compressionso that it disturbs the buckling member rather than restraining it
Similar analyses may be made of unbraced restraining members with other endconditions and another approximate solution is given in Figure 319 This has anaccuracy comparable with that of equation 341
364 Buckling of braced members with end restraints
The buckling of an end-restrained compression member 1ndash2 is analysed inSection 3104 where it is shown that if the member is braced so that it cannotsway (∆ = 0) then its elastic buckling load Ncr can be expressed in the gen-eral form of equation 323 when the effective length ratio kcr = LcrL is thesolution of
γ1γ2
4
(π
kcr
)2
+(γ1 + γ2
2
)(1 minus π
kcrcot
π
kcr
)+ tan(π2kcr)
π2kcr= 1 (342)
where the relative stiffness of the braced member at its end 1 is
γ1 = (2EIL)12sum1α
(343)
72 Compression members
k
0
02
04
06
08
10
2
0
02
04
06
08
10
0 02 04 06 08 10 0 02 04 06 08 10
k2
k1 k1
10 infin
050525
0550575
060625
0650675
07
07508
08509
095
10105
11115
12125
1314
1516
1718
1920
2224
2628
34
5
(a) Bracedmembers (equations 342 344)
(b) Unbraced members (equations 346 348)
Figure 321 Effective length ratios LcrL for members in rigid-jointed frames
in which the summationsum
1 α is for all the other members at end 1 γ2 is similarlydefined and 0 le γ le infin The relative stiffnesses may also be expressed as
k1 = (2EIL)12
05sum1α + (2EIL)12
= 2γ1
1 + 2γ1(344)
and a similar definition of k2 (0 le k le 1) Values of the effective lengthratio LcrL which satisfy equations 342ndash344 are presented in chart form inFigure 321a
In using the chart of Figure 321a the stiffness factors k may be calculatedfrom equation 344 by using the stiffness approximations of Figure 319 Thus forbraced restraining members of the type shown in Figure 318a the factor k1 canbe approximated by
k1 = (IL)12
05sum1(IL)(1 minus NNcrL)+ (IL)12
(345)
in which N is the compression force in the restraining member and NcrL its elasticbuckling load
365 Buckling of unbraced members with end restraints
The buckling of an end-restrained compression member 1ndash2 is analysed inSection 3105 where it is shown that if the member is unbraced against sidesway
Compression members 73
then its elastic buckling load Ncr can be expressed in the general form ofequation 323 when the effective length ratio kcr = LcrL is the solution of
γ1γ2(πkcr)2 minus 36
6(γ1 + γ2)= π
kcrcot
π
kcr (346)
where the relative stiffness of the unbraced member at its end 1 is
γ1 = (6EIL)12sum1α
(347)
in which the summationsum
1 α is for all the other members at end 1 and γ2 issimilarly defined The relative stiffnesses may also be expressed as
k1 = (6EIL)12
15sum1α + (6EIL)12
= γ1
15 + γ1(348)
and a similar definition of k2 Values of the effective length ratio Lcr L whichsatisfy equation 346 are presented in chart form in Figure 321b
In using the chart of Figure 321b the stiffness factors k may be calculatedfrom equation 348 by using the stiffness approximations of Figure 319 Thus forbraced restraining members of the third type shown in Figure 319a
k1 = (IL)12
15sum1(IL)(1 minus N4NcrL)+ (IL)12
(349)
366 Bracing stiffness required for a braced member
The elastic buckling load of an unbraced compression member may be increasedsubstantially by providing a translational bracing system which effectively pre-vents sway The bracing system need not be completely rigid but may be elasticas shown in Figure 322 provided its stiffness α exceeds a certain minimumvalue αL It is shown in Section 3106 that the minimum value for a pin-endedcompression member is
αL = π2EIL3 (350)
This conclusion can be extended to compression members with rotational endrestraints and it can be shown that if the sway bracing stiffness α is greater than
αL = NcrL (351)
where Ncr is the elastic buckling load for the braced mode then the member iseffectively braced against sway
74 Compression members
0 10 20 30 400
05
10
15
20Ncr L
EIL22
(a) Mode 2
L
(b) Mode 1 (c) Buckling load
Mode 1
Mode 2
L = 2EIL3
(equation 350)
(EIL3)
Buc
klin
g lo
ad N
cr
2 EI
L2
lt L
Stiffness
ge
Figure 322 Compression member with an elastic sway brace
37 Other compression members
371 General
The design method outlined in Section 35 together with the effective length con-cept developed in Section 36 are examples of a general approach to the analysisand design of the compression members whose ultimate resistances are governedby the interaction between yielding and buckling This approach often termeddesign by buckling analysis originates from the dependence of the design com-pression resistance NbRd of a simply supported uniform compression member onits squash load Ny and its elastic buckling load Ncr as shown in Figure 323which is adapted from Figure 313 The generalisation of this relationship to othercompression members allows the design buckling resistance NbRd of any memberto be determined from its yield load Ny and its elastic buckling load Ncr by usingFigure 323
In the following subsections this design by buckling analysis method isextended to the in-plane behaviour of rigid-jointed frames with joint loadingand also to the design of compression members which either are non-uniformor have intermediate loads or twist during buckling This method has alsobeen used for compression members with oblique restraints [3] Similar andrelated methods may be used for the flexuralndashtorsional buckling of beams(Sections 66ndash69)
372 Rigid-jointed frames with joint loads only
The application of the method of design by buckling analysis to rigid-jointedframes which only have joint loads is a simple extrapolation of the design
Compression members 75
0
02
04
06
08
1
12
050 1 15 2 25
Generalised slenderness cry NN =
Dim
ensi
onle
ss b
uckl
ing
resi
stan
ce N
bR
dN
y Elastic buckling NcrNy
Design buckling resistance NbRdNy
Figure 323 Compression resistance of structures designed by buckling analysis
method for simply supported compression members For this extrapolation itis assumed that the resistance NbRd of a compression member in a frame isrelated to its squash load Ny and to the axial force Ncr carried by it whenthe frame buckles elastically and that this relationship (Figure 323) is thesame as that used for simply supported compression members of the same type(Figure 313)
This method of design by buckling analysis is virtually the same as the effectivelength method which uses the design charts of Figure 321 because these chartswere obtained from the elastic buckling analyses of restrained members The onlydifference is that the elastic buckling load Ncr may be calculated directly for themethod of design by buckling analysis as well as by determining the approximateeffective length from the design charts
The elastic buckling load of the member may often be determined approximatelyby the effective length method discussed in Section 36 In cases where this is notsatisfactory a more accurate method must be used Many such methods have beendeveloped and some of these are discussed in Sections 8353 and 8354 and in[4ndash13] while other methods are referred to in the literature [14ndash16] The bucklingloads of many frames have already been determined and extensive tabulations ofapproximations for some of these are available [14 16ndash20]
While the method of design by buckling analysis might also be applied torigid frames whose loads act between joints the bending actions present in thoseframes make this less rational Because of this consideration of the effects ofbuckling on the design of frames with bending actions will be deferred untilChapter 8
76 Compression members
373 Non-uniform members
It was assumed in the preceding discussions that the members are uniform butsome practical compression members are of variable cross-section Non-uniformmembers may be stepped or tapered but in either case the elastic buckling loadNcr can be determined by solving a differential equilibrium equation similar to thatgoverning the buckling of uniform members (see equation 358) but which hasvariable values of EI In general this can best be done using numerical techniques[4ndash6] and the tedium of these can be relieved by making use of a suitable computerprogram Many particular cases have been solved and tabulations and graphs ofsolutions are available [5 6 16 21ndash23]
Once the elastic buckling load Ncr of the member has been determined themethod of design by buckling analysis can be used For this the squash load Ny
is calculated for the most highly stressed cross-section which is the section ofminimum area The design buckling resistance NbRd = χNyγM1 can then beobtained from Figure 323
374 Members with intermediate axial loads
The elastic buckling of members with intermediate as well as end loads is alsobest analysed by numerical methods [4ndash6] while solutions of many particularcases are available [5 16 21] Once again the method of design by bucklinganalysis can be used and for this the squash load will be determined by themost heavily stressed section If the elastic buckling load Ncr is calculated for thesame member then the design buckling resistance NbRd can be determined fromFigure 323
375 Flexuralndashtorsional buckling
In the previous sections attention was confined to compression members whichbuckle by deflecting laterally either perpendicular to the section minor axis at anelastic buckling load
Ncrz = π2EIzL2crz (352)
or perpendicular to the major axis at
Ncry = π2EIyL2cry (353)
However thin-walled open section compression members may also buckle bytwisting about a longitudinal axis as shown in Figure 324 for a cruciform sectionor by combined bending and twisting
Compression members 77
L
f
y
x
z
Figure 324 Torsional buckling of a cruciform section
A compression member of doubly symmetric cross-section may buckle elasti-cally by twisting at a torsional buckling load (see Section 311) given by
NcrT = 1
i20
(GIt + π2EIw
L2crT
)(354)
in which GIt and EIw are the torsional and warping rigidities (see Chapter 10)LcrT is the distance between inflexion points of the twisted shape and
i20 = i2
p + y20 + z2
0 (355)
in which y0 z0 are the shear centre coordinates (which are zero for doublysymmetric sections see Section 543) and
ip = radic(Iy + Iz)A (356)
is the polar radius of gyration For most rolled steel sections the minor axisbuckling load Ncrz is less than NcrT and the possibility of torsional bucklingcan be ignored However short members which have low torsional and warpingrigidities (such as thin-walled cruciforms) should be checked Such members canbe designed by using Figure 323 with the value of NcrT substituted for the elasticbuckling load Ncr
Monosymmetric and asymmetric section members (such as thin-walled teesand angles) may buckle in a combined mode by twisting and deflecting Thisaction takes place because the axis of twist through the shear centre does not
78 Compression members
0 1 2 3 4 5Length L (m)
0
250
500
750
1000
NcrzNcry
NcrT
Ncr
75
100
6
(a) Cross-section
y
z
(b) Elastic buckling loadsL
oads
Ncr
y N
crz
Ncr
T a
nd N
cr
Figure 325 Elastic buckling of a simply supported angle section column
coincide with the loading axis through the centroid and any twisting which occurscauses the centroidal axis to deflect For simply supported members it is shown in[5 24 25] that the (lowest) elastic buckling load Ncr is the lowest root of the cubicequation
N 3cr
i20 minus y2
0 minus z20
minus N 2cr
(Ncry + Ncrz + NcrT )i
20 minus Ncrzy2
0 minus Ncryz20
+ Ncri2
0NcryNcrz + NcrzNcrT + NcrT Ncry minus NcryNcrzNcrT i20 = 0
(357)
For example the elastic buckling load Ncr for a pin-ended unequal angle is shownin Figure 325 where it can be seen that Ncr is less than any of Ncry Ncrz orNcrT
These and other cases of flexuralndashtorsional buckling (referred to as torsionalndashflexural buckling in EC3) are treated in a number of textbooks and papers [4ndash624ndash26] and tabulations of solutions are also available [16 27] Once the bucklingload Ncr has been determined the compression resistance NbRd can be found fromFigure 323
38 Appendix ndash elastic compression members
381 Buckling of straight members
The elastic buckling load Ncr of the compression member shown in Figure 33 canbe determined by finding a deflected position which is one of equilibrium Thedifferential equilibrium equation of bending of the member is
EId2v
dx2= minusNcrv (358)
Compression members 79
This equation states that for equilibrium the internal moment of resistanceEI (d2vdx2) must exactly balance the external disturbing moment minusNcrv at anypoint along the length of the member When this equation is satisfied at all pointsthe displaced position is one of equilibrium
The solution of equation 358 which satisfies the boundary condition at the lowerend that (v)0 = 0 is
v = δ sinπx
kcrL
where
1
k2cr
= Ncr
π2EIL2
and δ is an undetermined constant The boundary condition at the upper end that(v)L = 0 is satisfied when either
δ = 0v = 0
(359)
or kcr = 1n in which n is an integer so that
Ncr = n2π2EIL2 (360)
v = δ sin nπxL (361)
The first solution (equations 359) defines the straight stable equilibrium positionwhich is valid for all loads N less than the lowest value of Ncr as shown inFigure 32b The second solution (equations 360 and 361) defines the bucklingloads Ncr at which displaced equilibrium positions can exist This solution does notdetermine the magnitude δ of the central deflection as indicated in Figure 32bThe lowest buckling load is the most important and this occurs when n = 1so that
Ncr = π2EIL2 (32)
v = δ sin πxL (31)
382 Bending of members with initial curvature
The bending of the compression member with initial curvature shown inFigure 32a can be analysed by considering the differential equilibrium equation
EId2v
dx2= minusN (v + v0) (362)
which is obtained from equation 358 for a straight member by adding the additionalbending moment minusNv0 induced by the initial curvature
80 Compression members
If the initial curvature of the member is such that
v0 = δ0 sin πxL (37)
then the solution of equation 362 which satisfies the boundary conditions(v)0L = 0 is the deflected shape
v = δ sin πxL (38)
where
δ
δ0= NNcr
1 minus NNcr (39)
The maximum moment in the compression member is N (δ+ δ0) and so the max-imum bending stress is N (δ + δ0)Wel where Wel is the elastic section modulusThus the maximum total stress is
σmax = N
A+ N (δ + δ0)
Wel
If the elastic limit is taken as the yield stress fy then the limiting axial load NL forwhich the above elastic analysis is valid is given by
NL = Ny minus NL(δ + δ0)A
Wel (363)
where Ny = Afy is the squash load By writing Wel = 2Ib in which b is themember width equation 363 becomes
NL = Ny minus δ0b
2i2
NL
(1 minus NLNcr)
which can be solved for the dimensionless limiting load NLNy as
NL
Ny=
[1 + (1 + η)NcrNy
2
]minus
[1 + (1 + η)Ncr Ny
2
]2
minus Ncr
Ny
12
(364)
where
η = δ0b
2i2 (313)
Alternatively equation 364 can be rearranged to give the dimensionless limitingload NLNy as
NL
Ny= 1
Φ +radicΦ2 minus λ
2(311)
Compression members 81
in which
Φ = 1 + η + λ2
2(312)
and
λ = radicNyNcr (35)
39 Appendix ndash inelastic compression members
391 Reduced modulus theory of buckling
The reduced modulus buckling load Ncrr of a rectangular section compressionmember which buckles in the y direction (see Figure 326a) can be determinedfrom the bending strain and stress distributions which are related to the curvatureκ(= minusd2vdx2) and the moduli E and Et as shown in Figure 326b and c Theposition of the line of zero bending stress can be found by using the condition thatthe axial force remains constant during buckling from which it follows that theforce resultant of the bending stresses must be zero so that
1
2dbcEtbcκ = 1
2dbtEbtκ
or
bc
b=
radicEradic
E + radicEt
Axial strain b c
b t
bt
bc
(b) Straindistribution
bt
bc
(a) Rectangular cross-section
d
y
z
Et
E
b
(c) Stressdistribution
Axial stress NA Et bc
Eb t
Figure 326 Reduced modulus buckling of a rectangular section member
82 Compression members
The moment of resistance of the section ErI(d2vdx2) is equal to the momentresultant of the bending stresses so that
ErId2v
dx2= minusdbc
2Etbcκ
2bc
3minus dbt
2Ebtκ
2bt
3
or
ErId2v
dx2= minus4b2
cEt
b2
db3
12κ
The reduced modulus of elasticity Er is therefore given by
Er = 4EEt
(radic
E + radicEt)
2 (316)
392 Buckling of members with residual stresses
Arectangular section member with a simplified residual stress distribution is shownin Figure 310 The section first yields at the edges z = plusmnd2 at a load N = 05Nyand yielding then spreads through the section as the load approaches the squashload Ny If the depth of the elastic core is de then the flexural rigidity for bendingabout the z axis is
(EI)t = EI
(de
d
)
where I = b3d12 and the axial force is
N = Ny
(1 minus 1
2
d2e
d2
)
By combining these two relationships the effective flexural rigidity of the partiallyyielded section can be written as
(EI)t = EI [2(1 minus NNy)]12 (318)
Thus the axial load at buckling Nt is given by
Ncrt
Ncr= [2(1 minus NcrtNy)]12
which can be rearranged as
Ncrt
Ny=
(Ncr
Ny
)2
[1 + 2(NyNcr)2]12 minus 1 (319)
Compression members 83
310 Appendix ndash effective lengths of compressionmembers
3101 Intermediate restraints
A straight pin-ended compression member with a central elastic restraint is shownin Figure 316a It is assumed that when the buckling load Ncr is applied to themember it buckles symmetrically as shown in Figure 316a with a central deflec-tion δ and the restraint exerts a restoring force αδ The equilibrium equation forthis buckled position is
EId2v
dx2= minusNcrv + αδ
2x
for 0 le x le L2
The solution of this equation which satisfies the boundary conditions (v)0 =(dvdx)L2 = 0 is given by
v = αδL
2Ncr
(x
Lminus sin πxkcrL
2(π2kcr) cosπ2kcr
)
where kcr = LcrL and Lcr is given by equation 323 Since δ = (v)L2 itfollows that(
π
2kcr
)3
cotπ
2kcr(π
2kcrcot
π
2kcrminus 1
) = αL3
16EI (365)
The variation with the dimensionless restraint stiffness αL316EI of the dimen-sionless buckling load
Ncr
π2EIL2= 4
π2
(π
2kcr
)2
(366)
which satisfies equation 365 is shown in Figure 316c It can be seen that thebuckling load for this symmetrical mode varies from π2EIL2 when the restraintis of zero stiffness to approximately 8π2EIL2 when the restraint is rigid Whenthe restraint stiffness exceeds
αL = 16π2EIL3 (331)
the buckling load obtained from equations 365 and 366 exceeds the value of4π2EIL2 for which the member buckles in the antisymmetrical second modeshown in Figure 316b Since buckling always takes place at the lowest possibleload it follows that the member buckles at 4π2EIL2 in the second mode shownin Figure 316b for all restraint stiffnesses α which exceed αL
84 Compression members
3102 Stiffness of a braced member
When a structural member is braced so that its ends act as if simply supportedas shown in Figure 318a then its response to equal and opposite disturbing endmoments M can be obtained by considering the differential equilibrium equation
EId2v
dx2= minusNv minus M (367)
The solution of this which satisfies the boundary conditions (v)0 = (v)L = 0 whenN is compressive is
v = M
N
1 minus cosπ
radicNNcrL
sin πradic
NNcrLsin
[π
radicN
NcrL
x
L
]+ cos
[π
radicN
NcrL
x
L
]minus 1
where
NcrL = π2EIL2 (336)
The end rotation θ = (dvdx)0 is
θ = 2M
NL
π
2
radicN
NcrLtan
π
2
radicN
NcrL
whence
α = M
θ= 2EI
L
(π2)radic
NNcrL
tan(π2)radic
NNcrL (335)
When the axial load N is tensile the solution of equation 367 is
v = M
N
1 minus cosh π
radicNNcrL
sinh πradic
NNcrLsinh
[π
radicN
NcrL
x
L
]
+ cosh
[π
radicN
NcrL
x
L
]minus 1
whence
α = 2EI
L
(π2)radic
NNcrL
tanh(π2)radic
NNcrL (337)
Compression members 85
3103 Stiffness of an unbraced member
When a structural member is unbraced so that its ends sway ∆ as shown inFigure 320a restoring end moments M are required to maintain equilibriumThe differential equation for equilibrium of a member with equal end momentsM = N∆2 is
EId2v
dx2= minusNv + M (368)
and its solution which satisfies the boundary conditions (v)0 = 0 (v)L = ∆ is
v = M
N
1 + cosπ
radicNNcrL
sin πradic
NNcrLsin
[π
radicN
NcrL
x
L
]minus cos
[π
radicN
NcrL
x
L
]+ 1
The end rotation θ = (dvdx)0 is
θ = 2M
NL
π
2
radicN
NcrLcot
π
2
radicN
NcrL
whence
α = minus(
2EI
L
)π
2
radicN
NcrLtan
π
2
radicN
NcrL (339)
where the negative sign occurs because the end moments M are restoring insteadof disturbing moments
When the axial load N is tensile the end moments M are disturbing momentsand the solution of equation 368 is
v = M
N
1 + cosh π
radicNNcrL
sinh πradic
NNcrLsinh
[π
radicN
NcrL
x
L
]
minus cosh
[π
radicN
NcrL
x
L
]+ 1
whence
α =(
2EI
L
)π
2
radicN
NcrLtanh
π
2
radicN
NcrL (340)
3104 Buckling of a braced member
A typical compression member 1ndash2 in a rigid-jointed frame is shown inFigure 317b When the frame buckles the member sways∆ and its ends rotate by
86 Compression members
θ1 θ2 as shown Because of these actions there are end shears (M1 +M2 +N∆)Land end moments M1 M2 The equilibrium equation for the member is
EId2v
dx2= minusNv minus M1 + (M1 + M2)
x
L+ N∆
x
L
and the solution of this is
v = minus(
M1 cosπkcr + M2
Ncr sin πkcr
)sin
πx
kcrL+ M1
Ncr
(cos
πx
kcrLminus 1
)
+(
M1 + M2
Ncr
)x
L+
x
L
where Ncr is the elastic buckling load transmitted by the member (seeequation 323) and kcr = LcrL By using this to obtain expressions for the endrotations θ1 θ2 and by substituting equation 333 the following equations can beobtained
M1
(NcrL
α1+ 1 minus π
kcrcot
π
kcr
)+ M2
(1 minus π
kcrcosec
π
kcr
)+ Ncr∆ = 0
M1
(1 minus π
kcrcosec
π
kcr
)+ M2
(NcrL
α2+ 1 minus π
kcrcot
π
kcr
)+ Ncr∆ = 0
(369)
If the compression member is braced so that joint translation is effectively pre-vented the sway terms Ncr∆ disappear from equations 369 The moments M1M2 can then be eliminated whence
(EIL)2
α1α2
(π
kcr
)2
+ EI
L
(1
α1+ 1
α2
)(1 minus π
kcrcot
π
kcr
)+ tan π2kcr
π2kcr= 1
(370)
By writing the relative stiffness of the braced member at the end 1 as
γ1 = 2EIL
α1= (2EIL)12sum
1α
(343)
with a similar definition for γ2 equation 370 becomes
γ1γ2
4
(π
kcr
)2
+(γ1 + γ2
2
)(1 minus π
kcrcot
π
kcr
)+ tan(π2kcr)
π2kcr= 1 (342)
3105 Buckling of an unbraced member
If there are no reaction points or braces to supply the member end shears (M1 +M2 + N∆)L the member will sway as shown in Figure 317b and the sway
Compression members 87
moment N∆ will be completely resisted by the end moments M1 and M2 so that
N∆ = minus(M1 + M2)
If this is substituted into equation 369 and if the moments M1 and M2 areeliminated then
(EIL)2
α1α2
(π
kcr
)2
minus 1 = EI
L
(1
α1+ 1
α2
)π
kcrcot
π
kcr(371)
where kcr = LcrL By writing the relative stiffnesses of the unbraced member atend 1 as
γ1 = 6EIL
α1= (6EIL)12sum
1α
(347)
with a similar definition of γ 2 equation 371 becomes
γ1γ2(πkcr)2 minus 36
6(γ1 + γ2)= π
kcrcot
π
kcr (346)
3106 Stiffness of bracing for a rigid frame
If the simply supported member shown in Figure 322 has a sufficiently stiff swaybrace acting at one end then it will buckle as if rigidly braced as shown inFigure 322a at a load π2EIL2 If the stiffness α of the brace is reduced below aminimum value αL the member will buckle in the rigid body sway mode shownin Figure 322b The elastic buckling load Ncr for this mode can be obtained fromthe equilibrium condition that
Ncr∆ = α∆L
where α∆ is the restraining force in the brace so that
Ncr = αL
The variation of Ncr with α is compared in Figure 322c with the braced modebuckling load of π2EIL2 It can be seen that when the brace stiffness α isless than
αL = π2EIL3 (350)
the member buckles in the sway mode and that when α is greater than αL itbuckles in the braced mode
88 Compression members
311 Appendix ndash torsional buckling
Thin-walled open-section compression members may buckle by twisting as shownin Figure 324 for a cruciform section or by combined bending and twisting Whenthis type of buckling takes place the twisting of the member causes the axialcompressive stresses NA to exert a disturbing torque which is opposed by thetorsional resistance of the section For the typical longitudinal element of cross-sectional area δA shown in Figure 327 which rotates a0(dφdx) (where a0 is thedistance to the axis of twist) when the section rotates φ the axial force (NA)δAhas a component (NA)δAa0(dφdx) which exerts a torque (NA)δAa0(dφdx)a0
about the axis of twist (which passes through the shear centre y0 z0 as shown inChapter 5) The total disturbing torque TP exerted is therefore
TP = N
A
dφ
dx
intA
a20dA
where
a20 = (y minus y0)
2 + (z minus z0)2
This torque can also be written as
TP = Ni20
dφ
dx
where
i20 = i2
p + y20 + z2
0 (355)
ip = radic (Iy + Iz)A
(356)
For members of doubly symmetric cross-section (y0 = z0 = 0) a twistedequilibrium position is possible when the disturbing torque TP exactly balances
xast
AN
d
d 0
xa
d
d0
δ
δ
δδ
xAxis of twist
stAN st
A
N
f
f
Figure 327 Torque exerted by axial load during twisting
Compression members 89
the internal resisting torque
Mx = GItdφ
dxminus EIw
d3φ
dx3
in which GIt and EIw are the torsional and warping rigidities (see Chapter 10)Thus at torsional buckling
N
Ai20
dφ
dx= GIt
dφ
dxminus EIw
d3φ
dx3 (372)
The solution of this which satisfies the boundary conditions of end twistingprevented ((φ)0L = 0) and ends free to warp ((d2φdx2)0L = 0) (see Chapter 10)is φ = (φ)L2 sin πxL in which (φ)L2 is the undetermined magnitude of theangle of twist rotation at the centre of the member and the buckling load NcrT is
NcrT = 1
i20
(GIt + π2EIw
L2
)
This solution may be generalised for compression members with other endconditions by writing it in the form
NcrT = 1
i20
(GIt + π2EIw
L2crT
)(354)
in which the torsional buckling effective length LcrT is the distance betweeninflexion points in the twisted shape
312 Worked examples
3121 Example 1 ndash checking a UB compression member
Problem The 457 times 191 UB 82 compression member of S275 steel of Figure 328ais simply supported about both principal axes at each end (Lcry = 120 m) andhas a central brace which prevents lateral deflections in the minor principal plane(Lcrz = 60 m) Check the adequacy of the member for a factored axial compres-sive load corresponding to a nominal dead load of 160 kN and a nominal imposedload of 230 kN
Factored axial load NEd = (135 times 160)+ (15 times 230) = 561 kN
Classifying the sectionFor S275 steel with tf = 16 mm fy = 275 Nmm2 EN 10025-2
ε = (235275)05 = 0924
cf (tf ε) = [(1913 minus 99 minus 2 times 102)2](160 times 0924) = 544 lt 14T52
90 Compression members
125
125
75
12
10
y
z
y
z z
y
1913
4600
99
160
RHS
(a) Example 1
457 times 191 UB 82 Iz = 1871 cm4
A = 104 cm2
iy = 188 cm iz = 423 cm r = 102 mm
(b) Example 3
2 ndash 125 times 75 times 10 UA Iy = 1495 cm4
Iz = 1642 cm4
(c) Example 4
Figure 328 Examples 1 3 and 4
cw = (4600 minus 2 times 160 minus 2 times 102) = 4076 mm T52
cw(twε) = 4076(99 times 0924) = 445 gt 42 T52
and so the web is Class 4 (slender) T52
Effective area
λp =radic
fyσcr
= bt
284εradic
kσ= 407699
284 times 0924 times radic40
= 0784 EC3-1-5 44(2)
ρ= λp minus 0055(3 +ψ)λ
2p
= 0784 minus 0055(3 + 1)
07842= 0918 EC3-1-5 44(2)
d minus deff = (1 minus 0918)times 4076 = 336 mm
Aeff = 104 times 102 minus 336 times 99 = 10 067 mm2
Cross-section compression resistance
NcRd = Aeff fyγM0
= 10 067 times 275
10= 2768 kN gt 561 kN = NEd 624(2)
Compression members 91
Member buckling resistance
λy =radic
Aeff fyNcry
= Lcry
iy
radicAeff A
λ1= 12 000
(188 times 10)
radic10 06710 400
939 times 0924= 0724
6313(1)
λz =radic
Aeff fyNcrz
= Lcrz
iz
radicAeff A
λ1= 6000
(423 times 10)
radic10 06710 400
939 times 0924
= 1608 gt 0724 6313(1)
Buckling will occur about the minor (z) axis For a rolled UB section (with hb gt12 and tf le 40 mm) buckling about the z-axis use buckling curve (b) withα = 034 T62 T61
Φz = 05[1 + 034(1608 minus 02)+ 16082] = 2032 6312(1)
χz = 1
2032 + radic20322 minus 16082
= 0305 6312(1)
NbzRd = χAeff fyγM1
= 0305 times 10 067 times 275
10= 844 kN gt 561 kN = NEd
6311(3)
and so the member is satisfactory
3122 Example 2 ndash designing a UC compression member
Problem Design a suitable UC of S355 steel to resist the loading of example 1 inSection 3121
Design axial load NEd = 561 kN as in Section 3121
Target area and first section choiceAssume fy = 355 Nmm2 and χ = 05
A ge 561 times 103(05 times 355) = 3161 mm2
Try a 152 times 152 UC 30 with A = 383 cm2 iy = 676 cm iz = 383 cmtf = 94 mm
ε = (235355)05 = 0814 T52
λy =radic
AfyNcry
= Lcry
iy
1
λ1= 12 000
(676 times 10)
1
939 times 0814= 2322 6313(1)
λz =radic
AfyNcrz
= Lcrz
iz
1
λ1= 6000
(383 times 10)
1
939 times 0814= 2050 lt 2322
6313(1)
92 Compression members
Buckling will occur about the major (y) axis For a rolled UC section (with hb le12 and tf le 100 mm) buckling about the y-axis use buckling curve (b) withα = 034 T62 T61
Φy = 05[1 + 034(2322 minus 02)+ 23222] = 3558 6312(1)
χy = 1
3558 + radic35582 minus 23222
= 0160 6312(1)
which is much less than the guessed value of 05
Second section choice
Guess χ = (05 + 0160)2 = 033
A ge 561 times 103(033 times 355) = 4789 mm2
Try a 203 times 203 UC 52 with A = 663 cm2 iy = 891 cm tf = 125 mmFor S355 steel with tf = 125 mm fy = 355 Nmm2 EN 10025-2
ε = (235355)05 = 0814 T52
cf (tf ε) = [(2043 minus 79 minus 2 times 102)2](125 times 0814) = 865 lt 14 T52
cw(twε) = (2062 minus 2 times 125 minus 2 times 102)(79 times 0814) = 250 lt 42T52
and so the cross-section is fully effective
λy =radic
AfyNcry
= Lcry
iy
1
λ1= 12 000
(891 times 10)
1
939 times 0814= 1763 6313(1)
For a rolled UC section (with hb le 12 and tf le 100 mm) buckling about they-axis use buckling curve (b) with α = 034 T62 T61
Φy = 05[1 + 034(1763 minus 02)+ 17632] = 2320 6312(1)
χy = 1
2320 + radic23202 minus 17632
= 0261 6312(1)
NbyRd = χAfyγM1
= 0261 times 663 times 102 times 355
10= 615 kN gt 561 kN = NEd
6311(3)
and so the 203 times 203 UC 52 is satisfactory
3123 Example 3 ndash designing an RHS compression member
Problem Design a suitable hot-finished RHS of S355 steel to resist the loading ofexample 1 in Section 3121
Compression members 93
Design axial load NEd = 561 kN as in Section 3121SolutionGuess χ = 03
A ge 561 times 103(03 times 355) = 5268 mm2
Try a 250 times 150 times 8 RHS with A = 608 cm2 iy = 917 cm iz = 615 cmt = 80 mmFor S355 steel with t = 8 mm fy = 355 Nmm2 EN 10025-2
ε = (235355)05 = 0814
cw(tε) = (2500 minus 2 times 80 minus 2 times 40)(80 times 0814) = 347 lt 42 T52
and so the cross-section is fully effective
λy =radic
AfyNcry
= Lcry
iy
1
λ1= 12 000
(918 times 10)
1
939 times 0814= 1710 6313(1)
λz =radic
AfyNcrz
= Lcrz
iz
1
λ1= 6000
(615 times 10)
1
939 times 0814= 1276 lt 1710
6313(1)
Buckling will occur about the major (y) axis For a hot-finished RHS use bucklingcurve (a) with α = 021 T62 T61
Φy = 05[1 + 021(1710 minus 02)+ 17102] = 2121 6312(1)
χy = 1
2121 + radic21212 minus 17102
= 0296 6312(1)
NbyRd = χAfyγM1
= 0296 times 608 times 102 times 355
10= 640 kN gt 561 kN = NEd
6311(3)
and so the 250 times 150 times 8 RHS is satisfactory
3124 Example 4 ndash buckling of double angles
Problem Two steel 125 times 75 times 10 UA are connected together at 15 m intervalsto form the long compression member whose properties are given in Figure 328cThe minimum second moment of area of each angle is 499 cm4 The member issimply supported about its major axis at 45 m intervals and about its minor axisat 15 m intervals Determine the elastic buckling load of the member
94 Compression members
Member buckling about the major axis
Ncry = π2 times 210 000 times 1495 times 10445002 N = 1530 kN
Member buckling about the minor axis
Ncrz = π2 times 210 000 times 1642 times 10415002 N = 1513 kN
Single angle buckling
Ncrmin = π2 times 210 000 times 499 times 10415002 N = 4597 kN
and so for both angles 2Ncrmin = 2 times 4597 = 919 kN lt 1520 kNIt can be seen that the lowest buckling load of 919 kN corresponds to the case
where each unequal angle buckles about its own minimum axis
3125 Example 5 ndash effective length factor inan unbraced frame
Problem Determine the effective length factor of member 1ndash2 of the unbracedframe shown in Figure 329a
Solution Using equation 348
k1 = 6EIL
15 times 0 + 6EIL= 10
k2 = 6EIL
15 times 6(2EI)(2L)+ 6EIL= 04
Using Figure 321b Lcr L = 23
2058
2096
142
94y
z
EI
2EI
2L
L
5N
1
2 3
(a) Example 5
5N
EI
203 times 203 UC 60Iy = 6125 cm4
Iz = 2065 cm4
A = 764 cm2
iz = 520 cmr = 102 mm
(b) Example 7
Figure 329 Examples 5 and 7
Compression members 95
3126 Example 6 ndash effective length factor in a braced frame
Problem Determine the effective length factor of member 1ndash2 of the frame shownin Figure 329a if bracing is provided which prevents sway buckling
SolutionUsing equation 344
k1 = 2EIL
05 times 0 + 2EIL= 10
k2 = 2EIL
05 times 2(2EI)(2L)+ 2EIL= 0667
Using Figure 321a Lcr L = 087
3127 Example 7 ndash checking a non-uniform member
Problem The 50 m long simply supported 203 times 203 UC 60 member shown inFigure 329b has two steel plates 250 mm times 12 mm times 2 m long welded to itone to the central length of each flange This increases the elastic buckling loadabout the minor axis by 70 If the yield strengths of the plates and the UC are355 Nmm2 determine the compression resistance
Solution
ε = (235355)05 = 0814 T52
cf (tf ε) = [(2058 minus 93 minus 2 times 102)2](142 times 0814) = 762 lt 14T52
cw(twε) = (2096 minus 2 times 142 minus 2 times 102)(93 times 0814) = 2124 lt 42T52
and so the cross-section is fully effective
Ncrz = 17 times π2 times 210 000 times 2065 times 10450002 = 2910 kN
λz =radic
AfyNcrz
=radic
764 times 102 times 355
2910 times 103= 0965
For a rolled UC section with welded flanges (tf le 40 mm) buckling about thez-axis use buckling curve (c) with α = 049 T62 T61
Φz = 05[1 + 049(0965 minus 02)+ 09652] = 1153 6312(1)
χz = 1
1153 + radic11532 minus 09652
= 0560 6312(1)
NbzRd = χAfyγM1
= 0560 times 764 times 102 times 355
10= 1520 kN 6311(3)
96 Compression members
3128 Example 8 ndash checking a member with intermediateaxial loads
Problem A 5 m long simply supported 457 times 191 UB 82 compression member ofS355 steel whose properties are given in Figure 328a has concentric axial loads ofNEd and 2NEd at its ends and a concentric axial load NEd at its midpoint At elasticbuckling the maximum compression force in the member is 2000 kN Determinethe compression resistance
SolutionFrom Section 3121 the cross-section is Class 4 slender and Aeff =10 067 mm2
Ncrz = 2000 kN
λz =radic
Aeff fyNcrz
=radic
10 067 times 355
2000 times 103= 1337
For a rolled UB section (with hb gt 12 and tf le 40 mm) buckling about thez-axis use buckling curve (b) with α = 034 T62 T61
Φz = 05[1 + 034(1337 minus 02)+ 13372] = 1587 6312(1)
χz = 1
1587 + radic15872 minus 13372
= 0410 6312(1)
NbzRd = χAeff fyγM1
= 0410 times 10 067 times 355
10= 1465 kN 6311(3)
313 Unworked examples
3131 Example 9 ndash checking a welded column section
The 140 m long welded column section compression member of S355 steel shownin Figure 330(a) is simply supported about both principal axes at each end (Lcry =140 m) and has a central brace that prevents lateral deflections in the minorprincipal plane (Lcrz = 70 m) Check the adequacy of the member for a factoredaxial compressive force corresponding to a nominal dead load of 420 kN and anominal imposed load of 640 kN
3132 Example 10 ndash designing a UC section
Design a suitable UC of S275 steel to resist the loading of example 9 inSection 3131
Compression members 97
300
40
28
355
100 100
175
100
380
25
25090
90
8
Cy y
z z
u
v
Welded column section
Iy = 747 times 10 6 mm4
Iz = 286 times 106 mm4
iy = 145 mm iz = 896 mm A = 35 700 mm2
(a) Example 9
2 ndash 380 times 100 channels
For each channel
Iy = 15034 times 106 mm4
Iz = 643 times 106 mm4
A = 6870 mm2
(b) Example 11
90 times 90 times 8 EA
Iu = 166 times 106 mm4
Iv = 0431 times 106 mm4
iu = 345 mm iv = 176 mm A = 1390 mm2
It = 328 times 103 mm4
y0 = 250 mm
(c) Example 14
Figure 330 Examples 9 11 and 14
L2
N
EA EA
E I L
L2
Figure 331 Example 12
3133 Example 11 ndash checking a compound section
Determine the minimum elastic buckling load of the two laced 380 times 100 channelsshown in Figure 330(b) if the effective length about each axis is Lcr = 30 m
3134 Example 12 ndash elastic buckling
Determine the elastic buckling load of the guyed column shown in Figure 331 Itshould be assumed that the guys have negligible flexural stiffness that (EA)guy =2(EI )columnL2 and that the column is built-in at its base
98 Compression members
3135 Example 13 ndash checking a non-uniform compressionmember
Determine the maximum design load NEd of a lightly welded box section can-tilever compression member 1ndash2 made from plates of S355 steel which taper from300 mm times 300 mm times 12 mm at end 1 (at the cantilever tip) to 500 mm times 500 mmtimes 12 mm at end 2 (at the cantilever support) The cantilever has a length of 100 m
3136 Example 14 ndash flexuralndashtorsional buckling
A simply supported 90 times 90 times 8 EA compression member is shown inFigure 330(c) Determine the variation of the elastic buckling load with memberlength
References
1 Shanley FR (1947) Inelastic column theory Journal of the Aeronautical Sciences 14No 5 May pp 261ndash7
2 Trahair NS (2000) Column bracing forces Australian Journal of StructuralEngineering Institution of Engineers Australia 3 Nos 2 amp 3 pp 163ndash8
3 Trahair NS and Rasmussen KJR (2005) Finite-element analysis of the flexuralbuckling of columns with oblique restraints Journal of Structural Engineering ASCE131 No 3 pp 481ndash7
4 Structural Stability Research Council (1998) Guide to Stability Design Criteria forMetal Structures 5th edition (ed TV Galambos) John Wiley New York
5 Timoshenko SP and Gere JM (1961) Theory of Elastic Stability 2nd editionMcGraw-Hill New York
6 Bleich F (1952) Buckling Strength of Metal Structures McGraw-Hill New York7 Livesley RK (1964) Matrix Methods of Structural Analysis Pergamon Press Oxford8 Horne MR and Merchant W (1965) The Stability of Frames Pergamon Press
Oxford9 Gregory MS (1967) Elastic Instability E amp FN Spon London
10 McMinn SJ (1961) The determination of the critical loads of plane frames TheStructural Engineer 39 No 7 July pp 221ndash7
11 Stevens LK and Schmidt LC (1963) Determination of elastic critical loads Journalof the Structural Division ASCE 89 No ST6 pp 137ndash58
12 Harrison HB (1967) The analysis of triangulated plane and space structures account-ing for temperature and geometrical changes Space Structures (ed RM Davies)Blackwell Scientific Publications Oxford pp 231ndash43
13 Harrison HB (1973) Computer Methods in Structural Analysis Prentice-HallEnglewood Cliffs New Jersey
14 Lu LW (1962) A survey of literature on the stability of frames Welding ResearchCouncil Bulletin No 81 pp 1ndash11
15 Archer JS et al (1963) Bibliography on the use of digital computers in structuralengineering Journal of the Structural Division ASCE 89 No ST6 pp 461ndash91
16 Column Research Committee of Japan (1971) Handbook of Structural StabilityCorona Tokyo
Compression members 99
17 Brotton DM (1960) Elastic critical loads of multibay pitched roof portal frames withrigid external stanchions The Structural Engineer 38 No 3 March pp 88ndash99
18 Switzky H and Wang PC (1969) Design and analysis of frames for stability Journalof the Structural Division ASCE 95 No ST4 pp 695ndash713
19 Davies JM (1990) In-plane stability of portal frames The Structural Engineer 68No 8 April pp 141ndash7
20 Davies JM (1991) The stability of multi-bay portal frames The Structural Engineer69 No 12 June pp 223ndash9
21 Bresler B Lin TY and Scalzi JB (1968) Design of Steel Structures 2nd editionJohn Wiley New York
22 Gere JM and Carter WO (1962) Critical buckling loads for tapered columns Journalof the Structural Division ASCE 88 No ST1 pp 1ndash11
23 Bradford MA andAbdoli Yazdi N (1999)ANewmark-based method for the stabilityof columns Computers and Structures 71 No 6 pp 689ndash700
24 Trahair NS (1993) FlexuralndashTorsional Buckling of Structures E amp FN SponLondon
25 Galambos TV (1968) Structural Members and Frames Prentice-Hall EnglewoodCliffs New Jersey
26 Trahair NS and Rasmussen KJR (2005) Flexuralndashtorsional buckling of columnswith oblique eccentric restraints Journal of Structural Engineering ASCE 131 No 11pp 1731ndash7
27 Chajes A and Winter G (1965) Torsionalndashflexural buckling of thin-walled membersJournal of the Structural Division ASCE 91 No ST4 pp 103ndash24
Chapter 4
Local buckling ofthin-plate elements
41 Introduction
The behaviour of compression members was discussed in Chapter 3 where itwas assumed that no local distortion of the cross-section took place so thatfailure was only due to overall buckling and yielding This treatment is appro-priate for solid section members and for members whose cross-sections arecomposed of comparatively thick-plate elements including many hot-rolled steelsections
However in some cases the member cross-section is composed of more slender-plate elements as for example in some built-up members and in most light-gaugecold-formed members These slender-plate elements may buckle locally as shownin Figure 41 and the member may fail prematurely as indicated by the reduc-tion from the full line to the dashed line in Figure 42 A slender-plate elementdoes not fail by elastic buckling but exhibits significant post-buckling behaviouras indicated in Figure 43 Because of this the platersquos resistance to local fail-ure depends not only on its slenderness but also on its yield strength andresidual stresses as shown in Figure 44 The resistance of a plate elementof intermediate slenderness is also influenced significantly by its geometricalimperfections while the resistance of a stocky-plate element depends primar-ily on its yield stress and strain-hardening moduli of elasticity as indicated inFigure 44
In this chapter the behaviour of thin rectangular plates subjected to in-planecompression shear bending or bearing is discussed The behaviour under com-pression is applied to the design of plate elements in compression members andcompression flanges in beams The design of beam webs is also discussed andthe influence of the behaviour of thin plates on the design of plate girders to EC3is treated in detail
Figure 41 Local buckling of an I-section column
0 05 10 15 20 25 30 350
02
04
06
08
10
Complete yielding
Overall elasticbuckling
Ultimate strengthwithout local buckling
Ultimate strengthreduced by local buckling
Modified slenderness (Ny Ncr)
Ult
imat
e lo
ad r
atio
Nul
t N
y
NcrNy
Nult Ny
Nult Ny
Figure 42 Effects of local buckling on the resistances of compression members
102 Local buckling of thin-plate elements
0 05 10 15 20 25Dimensionless transverse deflection t
0
05
10
15
20First yield
Failure
Elasticpost-bucklingbehaviour
Buckling
Flat plate
Plate with initialcurvatureD
imen
sion
less
load
NN
cr
Figure 43 Post-buckling behaviour of thin plates
0 05 10 15 20 25 30
Modified plate slenderness
0
02
04
06
08
10
12
14
Strain-hardening bucklingplates free along one edge
simply supported plates
Yielding
Ultimate strengths of simplysupported plates ultfy
Flat plates withoutresidual stresses
Plates withinitialcurvature
Plates with residualstresses and initial
curvature
Elastic buckling crfy
Str
ess
rati
o c
rf y
ul
tfy
k
f
Et
bf
cr
y2
2)1(12
) )( (=
ndash
Figure 44 Ultimate strengths of plates in compression
42 Plate elements in compression
421 Elastic buckling
4211 Simply supported plates
The thin flat plate element of length L width b and thickness t shown in Figure 45bis simply supported along all four edges The applied compressive loads N are
Local buckling of thin-plate elements 103
L
L
EI
t
t
N
N
b
b= Ebt
12-----------
3
Ebt
12(1ndash2)--------------------------
3bD
Nbt
Nbt
SimplysupportedSimply
supported
(a) Column buckling (b) Plate buckling
SS
2
2
4
4
22
4
4
4
2
2
4
4
2partx
partN
z
zx
x
bDdx
N
dx
EI ndash=++ndash=
xx
yy
z
z
( )
=
partpartpartpartpartdd partpart
Figure 45 Comparison of column and plate buckling
uniformly distributed over each end of the plate When the applied loads are equalto the elastic buckling loads the plate can buckle by deflecting v laterally out ofits original plane into an adjacent position [1ndash4]
It is shown in Section 481 that this equilibrium position is given by
v = δ sinmπx
Lsin
nπz
b(41)
with n = 1 where δ is the undetermined magnitude of the central deflection Theelastic buckling load Ncr at which the plate buckles corresponds to the bucklingstress
σcr = Ncrbt (42)
which is given by
σcr = π2E
12(1 minus ν2)
kσ(bt)2
(43)
The lowest value of the buckling coefficient kσ (see Figure 46) is
kσ = 4 (44)
which is appropriate for the high aspect ratios Lb of most structural steel members(see Figure 47)
104 Local buckling of thin-plate elements
0 1 2 3 4 5 6Plate aspect ratio Lb
0
2
4
6
8
10
L
b2 3 4 5
2
34
m = 1
n = 1
Buc
klin
g co
effi
cien
t k
Figure 46 Buckling coefficients of simply supported plates in compression
b
b b b
L = bm
cr cr
Figure 47 Buckled pattern of a long simply supported plate in compression
The buckling stress σcr varies inversely as the square of the plate slendernessor widthndashthickness ratio bt as shown in Figure 44 in which the dimensionlessbuckling stress σcrfy is plotted against a modified plate slenderness ratio
radicfyσcr
= b
t
radicfyE
12(1 minus ν2)
π2kσ (45)
If the material ceases to be linear elastic at the yield stress fy the above analysisis only valid for
radic(fyσcr) ge 1 This limit is equivalent to a widthndashthickness ratio
bt given by
b
t
radicfy
235= 568 (46)
Local buckling of thin-plate elements 105
for a steel with E = 210 000 Nmm2 and ν = 03 (The yield stress fy inequation 46 and in all of the similar equations which appear later in this chaptermust be expressed in Nmm2)
The values of the buckling coefficient kσ shown in Figure 46 indicate that theuse of intermediate transverse stiffeners to increase the elastic buckling stress ofa plate in compression is not effective except when their spacing is significantlyless than the plate width Because of this it is more economical to use interme-diate longitudinal stiffeners which cause the plate to buckle in a number of halfwaves across its width When such stiffeners are used equation 43 still holds pro-vided b is taken as the stiffener spacing Longitudinal stiffeners have an additionaladvantage when they are able to transfer compressive stresses in which case theircross-sectional areas may be added to that of the plate
An intermediate longitudinal stiffener must be sufficiently stiff flexurally toprevent the plate from deflecting at the stiffener An approximate value for theminimum second moment of area Ist of a longitudinal stiffener at the centre lineof a simply supported plate is given by
Ist = 45bt3[
1 + 23Ast
bt
(1 + 05
Ast
bt
)](47)
in which b is now the half width of the plate and Ast is the area of the stiffenerThis minimum increases with the stiffener area because the load transmitted bythe stiffener also increases with its area and the additional second moment of areais required to resist the buckling action of the stiffener load
Stiffeners should also be proportioned to resist local buckling A stiffener isusually fixed to one side of the plate rather than placed symmetrically about themid-plane and in this case its effective second moment of area is greater than thevalue calculated for its centroid It is often suggested [2 3] that the value of Ist
can be approximated by the value calculated for the mid-plane of the plate butthis may provide an overestimate in some cases [4ndash6]
The behaviour of an edge stiffener is different from that of an intermediatestiffener in that theoretically it must be of infinite stiffness before it can providean effective simple support to the plate However if a minimum value of thesecond moment of area of an edge stiffener of
Ist = 225bt3[
1 + 46Ast
bt
(1 + Ast
bt
)](48)
is provided the resulting reduction in the plate buckling stress is only a few percent[5] Equation 48 is derived from equation 47 on the basis that an edge stiffeneronly has to support a plate on one side of the stiffener while an intermediatestiffener has to support plates on both sides
The effectiveness of longitudinal stiffeners decreases as their number increasesand the minimum stiffness required for them to provide effective lateral supportincreases Some guidance on this is given in [3 5]
106 Local buckling of thin-plate elements
b
L
Simply supported Simply supported
Free edge
cr cr
Figure 48 Buckled pattern of a plate free along one edge
4212 Plates free along one longitudinal edge
The thin flat plate shown in Figure 48 is simply supported along both transverseedges and one longitudinal edge and is free along the other The differen-tial equation of equilibrium of the plate in a buckled position is the sameas equation 4105 (see Section 481) The buckled shape which satisfies thisequation differs however from the approximately square buckles of the simplysupported plate shown in Figure 47 The different boundary conditions along thefree edge cause the plate to buckle with a single half wave along its length asshown in Figure 48 Despite this the solution for the elastic buckling stress σcr
can still be expressed in the general form of equation 43 in which the bucklingcoefficient kσ is now approximated by
kσ = 0425 +(
b
L
)2
(49)
as shown in Figure 49For the long-plate elements which are used as flange outstands in many structural
steel members the buckling coefficient kσ is close to the minimum value of 0425In this case the elastic buckling stress (for a steel for which E = 210 000 Nmm2
and ν = 03) is equal to the yield stress fy when
b
t
radicfy
235= 185 (410)
Once again it is more economical to use longitudinal stiffeners to increase theelastic buckling stress than transverse stiffeners The stiffness requirements ofintermediate and edge longitudinal stiffeners are discussed in Section 4211
4213 Plates with other support conditions
The edges of flat-plate elements may be fixed or elastically restrained insteadof being simply supported or free The elastic buckling loads of flat plates with
Local buckling of thin-plate elements 107
0 1 2 3 4 5Plate aspect ratio Lb
0
1
2
3
4
5
L
b
Free
Exact= 0425 + (bL)2
0425
k
Buc
klin
g co
effi
cien
t k
Figure 49 Buckling coefficients of plate free along one edge
various support conditions have been determined and many values of the bucklingcoefficient kσ to be used in equation 43 are given in [2ndash6]
4214 Plate assemblies
Many structural steel compression members are assemblies of flat-plate elementswhich are rigidly connected together along their common boundaries The localbuckling of such an assembly can be analysed approximately by assuming that theplate elements are hinged along their common boundaries so that each plate actsas if simply supported along its connected boundary or boundaries and free alongany unconnected boundary The buckling stress of each plate element can then bedetermined from equation 43 with kσ = 4 or 0425 as appropriate and the lowestof these can be used as an approximation for determining the buckling load of themember
This approximation is conservative because the rigidity of the joints between theplate elements causes all plates to buckle simultaneously at a stress intermediatebetween the lowest and the highest of the buckling stresses of the individual plateelements Anumber of analyses have been made of the stress at which simultaneousbuckling takes place [4ndash6]
For example values of the elastic buckling coefficient kσ for an I-section inuniform compression are shown in Figure 410 and for a box section in uniformcompression in Figure 411 The buckling stress can be obtained from these figuresby using equation 43 with the plate thickness t replaced by the flange thicknesstf The use of these stresses and other results [4ndash6] leads to economic thin-walledcompression members
108 Local buckling of thin-plate elements
0 01 02 03 04 05 06 07 08 09 10Outstand width to depth ratio bfdf
0
02
04
06
08
10
= E--------------------------
2
12(1ndash 2)k ----------------------
( )b tf f2
bf
df
tw
tf
t twf =100
150
200
075
Fla
nge
loca
l buc
klin
g co
effi
cien
t k
cr
Figure 410 Local buckling coefficients for I-section compression members
0 01 02 03 04 05 06 07 08 09 10Width to depth ratio bfdf
0
1
2
3
4
5
6
bf
dftw
tf
t twf = 075
Fla
nge
loca
l buc
klin
g co
effi
cien
t k
100
125
150
200
= E--------------------------
2
12(1ndash 2)k----------------------
( )b tf f2cr
Figure 411 Local buckling coefficients for box section compression members
422 Ultimate strength
4221 Inelastic buckling of thick plates
The discussion given in Section 421 on the elastic buckling of rectangular platesapplies only to materials whose stressndashstrain relationships remain linear Thusfor stocky steel plates for which the calculated elastic buckling stress exceeds theyield stress fy the elastic analysis must be modified accordingly
One particularly simple modification which can be applied to strain-hardenedsteel plates is to use the strain-hardening modulus Est and
radic(EEst) instead of E
Local buckling of thin-plate elements 109
in the terms of equation 4105 (Section 481) which represent the longitudinalbending and the twisting resistances to buckling while still using E in the termfor the lateral bending resistance With these modifications the strain-hardeningbuckling stress of a long simply supported plate is equal to the yield stress fy when
b
t
radicfy
235= 237 (411)
More accurate investigations [7] of the strain-hardening buckling of steel plateshave shown that this simple approach is too conservative and that the strain-hardening buckling stress is equal to the yield stress when
b
t
radicfy
235= 321 (412)
for simply supported plates and when
b
t
radicfy
235= 82 (413)
for plates which are free along one longitudinal edge If these are compared withequations 46 and 410 then it can be seen that these limits can be expressed interms of the elastic buckling stress σcr byradic
fyσcr
= 057 (414)
for simply supported plates and byradicfyσcr
= 046 (415)
for plates which are free along one longitudinal edge These limits are shown inFigure 44
4222 Post-buckling strength of thin plates
A thin elastic plate does not fail soon after it buckles but can support loads signifi-cantly greater than its elastic buckling load without deflecting excessively This isin contrast to the behaviour of an elastic compression member which can only carryvery slightly increased loads before its deflections become excessive as indicatedin Figure 412 This post-buckling behaviour of a thin plate is due to a number ofcauses but the main reason is that the deflected shape of the buckled plate cannotbe developed from the pre-buckled configuration without some redistribution ofthe in-plane stresses within the plate This redistribution which is ignored in the
110 Local buckling of thin-plate elements
Central deflection
Simply supported thinplates with constantdisplacement of loadededges
Constant displacementalong unloaded edges
Unloaded edgesfree to deflect
Thin column
Small deflection theoryCom
pres
sive
load
on
plat
e N
Ncr
Figure 412 Post-buckling behaviour of thin elastic plates
small deflection theory of elastic buckling usually favours the less stiff portionsof the plate and causes an increase in the efficiency of the plate
One of the most common causes of this redistribution is associated with thein-plane boundary conditions at the loaded edges of the plate In long structuralmembers the continuity conditions along the transverse lines dividing consecu-tive buckled panels require that each of these boundary lines deflects a constantamount longitudinally However the longitudinal shortening of the panel due toits transverse deflections varies across the panel from a maximum at the centreto a minimum at the supported edges This variation must therefore be compen-sated for by a corresponding variation in the longitudinal shortening due to axialstrain from a minimum at the centre to a maximum at the edges The longitudinalstress distribution must be similar to the shortening due to strain and so the stressat the centre of the panel is reduced below the average stress while the stress atthe supported edges is increased above the average This redistribution whichis equivalent to a transfer of stress from the more flexible central region of thepanel to the regions near the supported edges leads to a reduction in the transversedeflection as indicated in Figure 412
A further redistribution of the in-plane stresses takes places when the longitu-dinal edges of the panel are supported by very stiff elements which ensure thatthese edges deflect a constant amount laterally in the plane of the panel In thiscase the variation along the panel of the lateral shortening due to the transversedeflections induces a self-equilibrating set of lateral in-plane stresses which aretensile at the centre of the panel and compressive at the loaded edges The tensilestresses help support the less stiff central region of the panel and lead to furtherreductions in the transverse deflections as indicated in Figure 412 However this
Local buckling of thin-plate elements 111
action is not usually fully developed because the elements supporting the panelsof most structural members are not very stiff
The post-buckling effect is greater in plates supported along both longitudinaledges than it is in plates which are free along one longitudinal edge This is becausethe deflected shapes of the latter have much less curvature than the former and theredistributions of the in-plane stresses are not as pronounced In addition it is notpossible to develop any lateral in-plane stresses along free edges and so it is notuncommon to ignore any post-buckling reserves of slender flange outstands
The redistribution of the in-plane stresses after buckling continues with increas-ing load until the yield stress fy is reached at the supported edges Yielding thenspreads rapidly and the plate fails soon after as indicated in Figure 43 The occur-rence of the first yield in an initially flat plate depends on its slenderness and athick plate yields before its elastic buckling stress σcr is reached As the slender-ness increases and the elastic buckling stress decreases below the yield stress theratio of the ultimate stress fult to the elastic buckling stress increases as shown inFigure 44
Although the analytical determination of the ultimate strength of a thin flat plateis difficult it has been found that the use of an effective width concept can lead tosatisfactory approximations According to this concept the actual ultimate stressdistribution in a simply supported plate (see Figure 413) is replaced by a simplifieddistribution for which the central portion of the plate is ignored and the remainingeffective width beff carries the yield stress fy It was proposed that this effectivewidth should be approximated by
beff
b=
radicσcr
fy (416)
Actual ultimatestress distribution
Averageultimatestress ult
Centralexcess widthignored
(a) Ultimate stress distribution (b) Effective width concept
asymp fyfy
b
equiv
beff 2 beff 2
y
ulteff
fb
b=
Figure 413 Effective width concept for simply supported plates
112 Local buckling of thin-plate elements
which is equivalent to supposing that the ultimate load carrying capacity of the platefybeff t is equal to the elastic buckling load of a plate of width beff Alternativelythis proposal can be regarded as determining an effective average ultimate stressfult which acts on the full width b of the plate This average ultimate stress whichis given by
fult
fy=
radicσcr
fy(417)
is shown in Figure 44Experiments on real plates with initial curvatures and residual stresses have
confirmed the qualitative validity of this effective width approach but suggestthat the quantitative values of the effective width for hot-rolled and welded platesshould be obtained from equations of the type
beff
b= α
radicσcr
fy (418)
where α reflects the influence of the initial curvatures and residual stresses Forexample test results for the ultimate stresses fult(= fybeff b) of hot-rolled sim-ply supported plates with residual stresses and initial curvatures are shown inFigure 414 and suggest a value of α equal to 065 Other values of α are given in[8] Tests on cold-formed members also support the effective width concept withquantitative values of the effective width being obtained from
beff
b=
radicσcr
fy
(1 minus 022
radicσcr
fy
) (419)
4223 Effects of initial curvature and residual stresses
Real plates are not perfectly flat but have small initial curvatures similar to thosein real columns (see Section 322) and real beams (see Section 622) The initialcurvature of a plate causes it to deflect transversely as soon as it is loaded as shownin Figure 43 These deflections increase rapidly as the elastic buckling stress isapproached but slow down beyond the buckling stress and approach those of aninitially flat plate
In a thin plate with initial curvature the first yield and failure occur only slightlybefore they do in a flat plate as indicated in Figure 43 while the initial curvaturehas little effect on the resistance of a thick plate (see Figure 44) It is only in a plateof intermediate slenderness that the initial curvature causes a significant reductionin the resistance as indicated in Figure 44
Real plates usually have residual stresses induced by uneven cooling after rollingor welding These stresses are generally tensile at the junctions between plateelements and compressive in the regions away from the junctions Residual com-pressive stresses in the central region of a simply supported thin plate cause it
Local buckling of thin-plate elements 113
0 02 04 06 08 10 12 14 16
Modified plate slenderness
02
04
06
08
10
12
(f y )
Test results
Yielding
Elastic buckling
equation 418 ( = 065)
cr
Eff
ecti
ve w
idth
rat
io b
eff
b
radic
Figure 414 Effective widths of simply supported plates
to buckle prematurely and reduce its ultimate strength as shown in Figure 44Residual stresses also cause premature yielding in plates of intermediate slender-ness as indicated in Figure 44 but have a negligible effect on the strain-hardeningbuckling of stocky plates
Some typical test results for thin supported plates with initial curvatures andresidual stresses are shown in Figure 414
43 Plate elements in shear
431 Elastic buckling
The thin flat plate of length L depth d and thickness t shown in Figure 415is simply supported along all four edges The plate is loaded by shear stressesdistributed uniformly along its edges When these stresses are equal to the elasticbuckling value τcr then the plate can buckle by deflecting v laterally out of itsoriginal plane into an adjacent position For this adjacent position to be one ofequilibrium the differential equilibrium equation [1ndash4](
part4v
partx4+ 2
part4v
partx2partz2+ part4v
partz4
)= minus2τcrt
D
part2v
partxpartz(420)
must be satisfied (this may be compared with the corresponding equation 4105 ofSection 481 for a plate in compression)
Closed form solutions of this equation are not available but numerical solutionshave been obtained These indicate that the plate tends to buckle along compressiondiagonals as shown in Figure 415 The shape of the buckle is influenced bythe tensile forces acting along the other diagonal while the number of buckles
114 Local buckling of thin-plate elements
L
d
Line of zero deflection
Compression diagonals
cr
cr
Figure 415 Buckling pattern of a simply supported plate in shear
increases with the aspect ratio Ld The numerical solutions for the elastic bucklingstress τcr can be expressed in the form
τcr = π2E
12(1 minus ν2)
kτ(dt)2
(421)
in which the buckling coefficient kτ is approximated by
kτ = 534 + 4
(d
L
)2
(422)
when L ge d and by
kτ = 534
(d
L
)2
+ 4 (423)
when L le d as shown in Figure 416The shear stresses in many structural members are transmitted by unstiffened
webs for which the aspect ratio Ld is large In this case the buckling coefficientkτ approaches 534 and the buckling stress can be closely approximated by
τcr = 534π2E12(1 minus ν2)
(dt)2 (424)
This elastic buckling stress is equal to the yield stress in shear τy = fyradic
3 (seeSection 131) of a steel for which E = 210 000 Nmm2 and ν = 03 when
d
t
radicfy
235= 864 (425)
Local buckling of thin-plate elements 115
Plate aspect ratio Lb
0
5
10
15
20
25
L
b
5341 buckle 2 buckles 3 buckles
Approximations ofequations 422 and 423
Buc
klin
g co
effi
cien
t k
0 1 2 3 4 5 6
Figure 416 Buckling coefficients of plates in shear
The values of the buckling coefficient kτ shown in Figure 416 and the form ofequation 421 indicate that the elastic buckling stress may be significantly increasedeither by using intermediate transverse stiffeners to decrease the aspect ratio Ldand to increase the buckling coefficient kτ or by using longitudinal stiffeners todecrease the depthndashthickness ratio dt It is apparent from Figure 416 that suchstiffeners are likely to be most efficient when the stiffener spacing a is such thatthe aspect ratio of each panel lies between 05 and 2 so that only one buckle canform in each panel
Any intermediate stiffeners used must be sufficiently stiff to ensure that theelastic buckling stress τcr is increased to the value calculated for the stiffenedpanel Some values of the required stiffener second moment of area Ist are givenin [3 5] for various panel aspect ratios ad These can be conservatively approx-imated by using
Ist
at312(1 minus ν2)= 6
ad(426)
when ad ge 1 and by using
Ist
at312(1 minus ν2)= 6
(ad)4(427)
when ad le 1
116 Local buckling of thin-plate elements
432 Ultimate strength
4321 Unstiffened plates
A stocky unstiffened web in an I-section beam in pure shear is shown inFigure 417a The elastic shear stress distribution in such a section is analysedin Section 5102 The web behaves elastically in shear until first yield occurs atτy = fy
radic3 and then undergoes increasing plastification until the web is fully
yielded in shear (Figure 417b) Because the shear stress distribution at first yieldis nearly uniform the nominal first yield and fully plastic loads are nearly equaland the shear shape factor is usually very close to 10 Stocky unstiffened webs insteel beams reach first yield before they buckle elastically so that their resistancesare determined by the shear stress τy as indicated in Figure 418
(a) I-section (b) Shear stress distributions
y y
d
Elastic First yield Fully plastic
Figure 417 Plastification of an I-section web in shear
0 05 10 15 20 25 300
02
04
06
08
10
Yieldingad = 04
0610
15
Ultimate strengthsof stiffened plates
Elastic bucklingUltimate strength ofunstiffened plates
Modified plate slenderness
cr fy
ult fy
kE
f
t
df y
cr
y
2
2)1(12
3
3
3( () )
ult fy33
=ndash
She
ar r
atio
s 3
crf
y 3
ult
fy
Figure 418 Ultimate strengths of plates in shear
Local buckling of thin-plate elements 117
Thus the resistance of a stocky web in a flanged section for which the shearshape factor is close to unity is closely approximated by the web plastic shearresistance
Vpl = dtτy (428)
When the depthndashthickness ratio dt of a long unstiffened steel web exceeds864
radic(fy235) its elastic buckling shear stress τcr is less than the shear yield
stress (see equation 425) The post-buckling reserve of shear strength of such anunstiffened web is not great and its ultimate stress τult can be approximated withreasonable accuracy by the elastic buckling stress τcr given by equation 424 andshown by the dashed line in Figure 418
4322 Stiffened plates
Stocky stiffened webs in steel beams yield in shear before they buckle and so thestiffeners do not contribute to the ultimate strength Because of this stocky websare usually unstiffened
In a slender web with transverse stiffeners which buckles elastically before ityields there is a significant reserve of strength after buckling This is causedby a redistribution of stress in which the diagonal tension stresses in the webpanel continue to increase with the applied shear while the diagonal compressivestresses remain substantially unchanged The increased diagonal tension stressesform a tension field which combines with the transverse stiffeners and the flangesto transfer the additional shear in a truss-type action [7 9] as shown in Figure 419The ultimate shear stresses τult of the web is reached soon after yield occurs in thetension field and this can be approximated by
τult = τcr + τtf (429)
in which τcr is the elastic buckling stress given by equations 421ndash423 with Lequal to the stiffener spacing a and τtf is the tension field contribution at yieldwhich can be approximated by
τtf = fy2
(1 minus radic3τcrfy)radic
1 + (ad)2 (430)
Thus the approximate ultimate strength can be obtained from
radic3τult
fy=
radic3τcr
fy+
radic3
2
(1 minus radic3τcrfy)radic
1 + (ad)2 (431)
which is shown in Figure 418 This equation is conservative as it ignores anycontributions made by the bending resistance of the flanges to the ultimate strengthof the web [9 10]
118 Local buckling of thin-plate elements
Chord Web strut Tension member
(a) Equivalent truss
(b)Tension fields
Tension fieldWeb stiffenerFlange
a a a
d
Figure 419 Tension fields in a stiffened web
Not only must the intermediate transverse stiffeners be sufficiently stiff to ensurethat the elastic buckling stress τcr is reached as explained in Section 431 butthey must also be strong enough to transmit the stiffener force
Nst = fydt
2
(1 minus
radic3τcr
fy
)[a
dminus (ad)2radic
1 + (ad)2
](432)
required by the tension field action Intermediate transverse stiffeners are often sostocky that their ultimate strengths can be approximated by their squash loads inwhich case the required area Ast of a symmetrical pair of stiffeners is given by
Ast = Nstfy (433)
However the stability of very slender stiffeners should be checked in whichcase the second moment of area Ist required to resist the stiffener force can beconservatively estimated from
Ist = Nstd2
π2E(434)
which is based on the elastic buckling of a pin-ended column of length d
44 Plate elements in bending
441 Elastic buckling
The thin flat plate of length L width d and thickness t shown in Figure 420is simply supported along all four edges The plate is loaded by bending stress
Local buckling of thin-plate elements 119
Ld
All edges simply supported
2d3
2d3
2d3
crl
crl
crl
crl
Figure 420 Buckled pattern of a plate in bending
distributions which vary linearly across its width When the maximum stressreaches the elastic buckling value σcrl the plate can buckle out of its originalplane as shown in Figure 420 The elastic buckling stress can be expressed inthe form
σcrl = π2E
12(1 minus ν2)
kσ(dt)2
(435)
where the buckling coefficient kσ depends on the aspect ratio Ld of the plate andthe number of buckles along the plate For long plates the value of kσ is close toits minimum value of
kσ = 239 (436)
for which the length of each buckle is approximately 2d3 By using thisvalue of kσ it can be shown that the elastic buckling stress for a steel for whichE = 210 000 Nmm2 and ν = 03 is equal to the yield stress fy when
d
t
radicfy
235= 1389 (437)
The buckling coefficient kσ is not significantly greater than 239 except when thebuckle length is reduced below 2d3 For this reason transverse stiffeners areineffective unless more closely spaced than 2d3 On the other hand longitudinalstiffeners may be quite effective in changing the buckled shape and therefore thevalue of kσ Such a stiffener is most efficient when it is placed in the compression
120 Local buckling of thin-plate elements
region at about d5 from the compression edge This is close to the position ofthe crests in the buckles of an unstiffened plate Such a stiffener may increase thebuckling coefficient kσ significantly the maximum value of 1294 being achievedwhen the stiffener acts as if rigid The values given in [5 6] indicate that a hypo-thetical stiffener of zero area Ast will produce this effect if its second moment ofarea Ist is equal to 4dt3 This value should be increased to allow for the compres-sive load in the real stiffener and it is suggested that a suitable value might begiven by
Ist = 4dt3[
1 + 4Ast
dt
(1 + Ast
dt
)] (438)
which is of a similar form to that given by equation 47 for the longitudinal stiffenersof plates in uniform compression
442 Plate assemblies
The local buckling of a plate assembly in bending such as an I-beam bent about itsmajor axis can be analysed approximately by assuming that the plate elements arehinged along their common boundaries in much the same fashion as that describedin Section 4214 for compression members The elastic buckling moment canthen be approximated by using kσ = 0425 or 4 for the flanges as appropriateand kσ = 239 for the web and determining the element which is closest tobuckling The results of more accurate analyses of the simultaneous buckling ofthe flange and web elements in I-beams and box section beams in bending aregiven in Figures 421 and 422 respectively
0 01 02 03 04 05 06 07 08 09 10Outstand width to depth ratio bf df
0
02
04
06
08
10
t twf
bf
df
tw
t f
= E
--------------------------2
12(1ndash )2----------------------( )b tf f
2 = 10
15
20
40
kcrl
Fla
nge
loca
l buc
klin
g co
effi
cien
t k
Figure 421 Local buckling coefficients for I-beams
Local buckling of thin-plate elements 121
0 01 02 03 04 05 06 07 08 09 10
Width to depth ratio bfdf
0
1
2
3
4
5
6
7
bf
dftw
tf
= E--------------------------
2----------------------( )b tf f
2= 075
100125
150
200
t twf
Fla
nge
loca
l buc
klin
g co
effi
cien
t k crl
k
12(1ndash 2)
Figure 422 Local buckling coefficients for box section beams
443 Ultimate strength
The ultimate strength of a thick plate in bending is governed by its yield stress fyand by its plastic shape factor which is equal to 15 for a constant thickness plateas discussed in Section 552
When the width to thickness ratio dt exceeds 1389radic(fy235) the elastic
buckling stress σcrl of a simply supported plate is less than the yield stress fy(see equation 437) A long slender plate such as this has a significant reserve ofstrength after buckling because it is able to redistribute the compressive stressesfrom the buckled region to the area close to the supported compression edge in thesame way as a plate in uniform compression (see Section 4222) Solutions forthe post-buckling reserve of strength of a plate in bending given in [6 11ndash13] showthat an effective width treatment can be used with the effective width being takenover part of the compressive portion of the plate In the more general case of theslender plate shown in Figure 420 being loaded by combined bending and axialforce the post-buckling strength reserve of the plate can be substantial For suchplates the effective width is taken over a part of the compressive portion of the plateEffective widths can be obtained in this way from the local buckling stressσcr undercombined bending and axial force using the modified plate slenderness
radic(f yσcr)
in a similar fashion to that depicted in Figure 414 although the procedure is morecomplex
45 Plate elements in bending and shear
451 Elastic buckling
The elastic buckling of the simply supported thin flat plate shown in Figure 423which is subjected to combined shear and bending can be predicted by using the
122 Local buckling of thin-plate elements
L
d
cr
cr
crcr
crl
crl
crl
crl
Figure 423 Simply supported plate under shear and bending
approximate interaction equation [2 4 5 13]
(τcr
τcro
)2
+(σcrl
σcrlo
)2
= 1 (439)
in which τcro is the elastic buckling stress when the plate is in pure shear(see equations 421ndash424) and σcrlo is the elastic buckling stress for pure bending(see equations 435 and 436) If the elastic buckling stresses τcr σcrl are used inthe HenckyndashVon Mises yield criterion (see Section 131)
3τ 2cr + σ 2
crl = f 2y (440)
it can shown from equations 439 and 440 that the most severe loading conditionfor which elastic buckling and yielding occur simultaneously is that of pure shearThus in an unstiffened web of a steel for which E = 210 000 Nmm2 and ν = 03yielding will occur before buckling while (see equation 425)
d
t
radicfy
235le 864 (441)
452 Ultimate strength
4521 Unstiffened plates
Stocky unstiffened webs in steel beams yield before they buckle and their designresistances can be estimated approximately by using the HenckyndashVon Mises yieldcriterion (see Section 131) so that the shear force Vw and moment Mw in theweb satisfy
3
(Vw
dt
)2
+(
6Mw
d2t
)2
= f 2y (442)
Local buckling of thin-plate elements 123
This approximation is conservative as it assumes that the plastic shape factor forweb bending is 10 instead of 15
While the elastic buckling of slender unstiffened webs under combined shearand bending has been studied the ultimate strengths of such webs have not beenfully investigated However it seems possible that there is only a small reserveof strength after elastic buckling so that the ultimate strength can be approxi-mated conservatively by the elastic buckling stress combinations which satisfyequation 439
4522 Stiffened plates
The ultimate strength of a stiffened web under combined shear and bending can bediscussed in terms of the interaction diagram shown in Figure 424 for the ultimateshear force V and bending moment M acting on a plate girder When there isno shear the ultimate moment capacity is equal to the full plastic moment Mpproviding the flanges are Class 1 or Class 2 (see Section 472) while the ultimateshear capacity in the absence of bending moment is
Vult = dtτult (443)
where the ultimate stress τult is given approximately by equation 431 (whichignores any contributions made by the bending resistance of the flanges) Thisshear capacity remains unchanged as the bending moment increases to the valueMfp which is sufficient to fully yield the flanges if they alone resist the momentThis fact forms the basis for the widely used proportioning procedure for which itis assumed that the web resists only the shear and that the flanges are required to
10
08
06
04
02
0
MMp
VV
ult
Equation 444
Typical interaction curve
0 02 04 06 08 10
Figure 424 Ultimate strengths of stiffened webs in shear and bending
124 Local buckling of thin-plate elements
resist the full moment As the bending moment increases beyond Mfp the ultimateshear capacity falls off rapidly as shown A simple approximation for the reducedshear capacity is given by
V
Vult= 05
(1 +
radic1 minus MMp
1 minus MfpMp
)(444)
while MfpMp le MMp le 1 This approximation ignores the influence of thebending stresses on the tension field and the bending resistance of the flanges Amore accurate method of analysis is discussed in [14]
46 Plate elements in bearing
461 Elastic buckling
Plate elements are subjected to bearing stresses by concentrated or locally dis-tributed edge loads For example a concentrated load applied to the top flangeof a plate girder induces local bearing stresses in the web immediately beneaththe load In a slender girder with transverse web stiffeners the load is resisted byvertical shear stresses acting at the stiffeners as shown in Figure 425a Bearingcommonly occurs in conjunction with shear (as at an end support see Figure 425b)bending (as at mid-span see Figure 425c) and with combined shear and bending(as at an interior support see Figure 425d)
In the case of a panel of a stiffened web the edges of the panel may be regardedas simply supported When the bearing load is distributed along the full length a ofthe panel and there is no shear or bending the elastic bearing buckling stress σcrp
VR
VR
VLVL
M M
M M
(a) Bearing (c) Bearing and bending
(b) Bearing and shear (d) Bearing shear and bending
p
p p
p
Figure 425 Plate girder webs in bearing
Local buckling of thin-plate elements 125
0 05 10 15 20 25 30
Panel aspect ratio ad
0
5
10
15
20
25
= E
--------------------------2
12(1ndash )2
k---------------( )dt 2
a
a
d
06 =1
=1
0ad = infin
[4]
[15]
Ela
stic
buc
klin
g co
effi
cien
t k
crp
crp
Figure 426 Buckling coefficients of plates in bearing
can be expressed as
σcrp = π2E
12(1 minus ν2)
kσ(dt)2
(445)
in which the buckling coefficient kσ varies with the panel aspect ratio ad as shownin Figure 426
When a patch-bearing load acts along a reduced length αa of the top edge ofthe panel the value of αkσ is decreased This effect has been investigated [15]and some values of αkσ are shown in Figure 426 These values suggest a limitingvalue of αkσ = 08 approximately for the case of unstiffened webs (ad rarr infin)
with concentrated loads (α rarr 0)For the case of a panel in combined bearing shear and bending it has been
suggested [6 15] that the elastic buckling stresses can be determined from theinteraction equation
σcrp
σcrpo+
(τcr
τcro
)2
+(σcrl
σcrlo
)2
= 1 (446)
in which the final subscripts o indicate the appropriate elastic buckling stress whenonly that type of loading is applied Some specific interaction diagrams are givenin [15]
462 Ultimate strength
The ultimate strength of a thick web in bearing depends chiefly on its yield stressfy Although yielding first occurs under the centre of the bearing plate general
126 Local buckling of thin-plate elements
yielding does not take place until the applied load is large enough to yield a webarea defined by a dispersion of the applied stress through the flange Even at thisload the web does not collapse catastrophically and some further yielding andredistribution is possible When the web is also subjected to shear and bendinggeneral yielding in bearing can be approximated by using the HenckyndashVon Misesyield criterion (see Section 131)
σ 2p + σ 2
b minus σpσb + 3τ 2 = f 2y (447)
in which σp is the bearing stress σb the bending stress and τ the shear stressThin stiffened web panels in bearing have a reserve of strength after elastic buck-
ling which is due to a redistribution of stress from the more flexible central regionto the stiffeners Studies of this effect suggest that the reserve of strength decreasesas the bearing loads become more concentrated [15] The collapse behaviour ofstiffened panels is described in [16 17]
47 Design against local buckling
471 Compression members
The cross-sections of compression members must be designed so that the designcompression force NEd does not exceed the design compressive resistance NcRd whence
NEd le NcRd (448)
This ignores overall member buckling as discussed in Chapter 3 for whichthe non-dimensional slenderness λ le 02 The EC3 design expression forcross-section resistance under uniform compression is
NcRd = Aeff fyγM0
(449)
in which Aeff is the effective area of the cross-section fy is the nominalyield strength for the section and γM0(=1) is the partial section resistancefactor
The cross-section of a compression member may buckle locally before it reachesits yield stress fy in which case it is defined in EC3 as a Class 4 cross-sectionCross-sections of compression members which yield prior to local buckling arelsquoeffectiversquo and are defined in EC3 as being either of Class 1 Class 2 or Class 3For these cross-sections which are unaffected by local bucking the effective areaAeff in equation 449 is taken as the gross area A A cross-section composed offlat-plate elements has no local buckling effects when the widthndashthickness ratio
Local buckling of thin-plate elements 127
bt of every element of the cross-section satisfies
b
tle λ3ε (450)
where λ3ε is the appropriate Class 3 slenderness limit of Table 52 of EC3 andε is a constant given by
ε =radic
235fy (451)
in which fy is in Nmm2 units Some values of λ3ε are given in Figures 427and 428
Class 4 cross-sections are those containing at least one slender element for whichequation 450 is not satisfied For these the effective area is reduced to
Aeff =sum
Aceff (452)
in which Aceff is the effective area of a flat compression element comprising thecross-section which is obtained from its gross area Ac by
Aceff = ρAc (453)
where ρ is a reduction factor given by
ρ = (λp minus 022)λ2p le 10 (454)
along both edges
---- -
Section andelement widths
---Diameter d0
b1 b1 b2 b1
d1
d0
d1 d1
b2
Flange outstand b1
Flange b2 supportedalong both edges
Web d1 supported
Section description UB UC CHS Box RHS
14 14
42 42
4242 42
902
Figure 427 Yield limits for some compression elements
128 Local buckling of thin-plate elements
for element supported on both edges and
ρ = (λp minus 0188)λ2p le 10 (455)
for outstands and where
λp =radic
fyσcr
= b
t
1
284εradic
kσ(456)
is the modified plate slenderness (see equation 45) Values of the local buck-ling coefficient kσ are given in Table 42 of EC3-1-5 and are similar to those inFigures 410 and 411
A circular hollow section member has no local buckling effects when itsdiameterndashthickness ratio dt satisfies
d
tle 90ε2 (457)
If the diameterndashthickness ratio does not satisfy equation 457 then it has a Class4 cross-section whose effective area is reduced to
Aeff =radic
90
d(tε2)A (458)
Worked examples for checking the section capacities of compression members aregiven in Sections 491 and 3121ndash3123
472 Beam flanges and webs in compression
Compression elements in a beam cross-section are classified in EC3 as Class 1Class 2 Class 3 or Class 4 depending on their local buckling resistance
Class 1 elements are unaffected by local buckling and are able to developand maintain their fully plastic capacities (Section 55) while inelastic momentredistribution takes place in the beam Class 1 elements satisfy
bt le λ1ε (459)
in which λ1ε is the appropriate plasticity slenderness limit given in Table 52 ofEC3 (some values of λ1ε are shown in Figures 428 and 533) These limits areclosely related to equations 412 and 413 for the strain-hardening buckling stressesof inelastic plates
Class 2 elements are unaffected by local buckling in the development of theirfully plastic capacities (Section 55) but may be unable to maintain these capacities
Local buckling of thin-plate elements 129
Beams Mp
Not to scale
Columns
Mp
0 1 2 3 = bt
Beams
NRd MRd prop1
Wel fy
Ny My
Ny
Sec
tion
res
ista
nces
NR
d M
Rd
1 2 3 4 Section classification
Flange outstands 9 10 14
33 38 42Simplysupportedflanges
Figure 428 EC3 section resistances
while inelastic moment redistribution takes place in the beam Class 2 sectionssatisfy
λ1ε le bt le λ2ε (460)
in which λ2ε is the appropriate Class 2 slenderness limit given in Table 52 of EC3(some values of λ2ε are shown in Figures 428 and 533) These limits are slightlygreater than those implied in equations 412 and 413 for the strain-hardeningbuckling of inelastic plates
Class 3 elements are able to reach first yield but buckle locally before theybecome fully plastic Class 3 elements satisfy
λ2ε le bt le λ3ε (461)
in which λ3ε is the appropriate yield slenderness limit given in Table 52 of EC3(some values of λ3ε are shown in Figures 428 and 533) These limits are closelyrelated to equations 418 used to define the effective widths of flange plates inuniform compression or modified from equation 437 for web plates in bendingClass 4 elements buckle locally before they reach first yield and satisfy
λ3ε lt bt (462)
Beam cross-sections are classified as being Class 1 Class 2 Class 3 or Class 4depending on the classification of their elements A Class 1 cross-section has allof its elements being Class 1 A Class 2 cross-section has no Class 3 or Class 4elements and has at least one Class 2 element while a Class 3 cross-section hasno Class 4 elements and at least one Class 3 element A Class 4 cross-section hasat least one Class 4 element
130 Local buckling of thin-plate elements
A beam cross-section must be designed so that its factored design moment MEd
does not exceed the design section moment resistance so that
MEd le McRd (463)
Beams must also be designed for shear (Section 473) and against lateral buckling(Section 69) The section resistance McRd is given by
McRd = WfyγM0
(464)
in which W is the appropriate section modulus fy the yield strength and γM0(=1)the partial section resistance factor For Class 1 and 2 beam cross-sections
W = Wpl (465)
where Wpl is the plastic section modulus and which allows many beams to bedesigned for the full plastic moment
Mp = Wplfy (466)
as indicated in Figure 428For Class 3 beam cross-sections the effective section modulus is taken as the
minimum elastic section modulus Welmin which is that based on the extremefibre that reaches yield first Some I-sections have Class 1 or 2 flanges but con-tain a Class 3 web and so based on the EC3 classification they would be Class3 cross-sections and unsuitable for plastic design However the EC3 permitsthese sections to be classified as effective Class 2 cross-sections by neglectingpart of the compression portion of the web This simple procedure for hot-rolled or welded sections conveniently replaces the compressed portion of theweb by a part of width 20εtw adjacent to the compression flange and withanother part of width 20εtw adjacent to the plastic neutral axis of the effectivecross-section
For a beam with a slender compression flange supported along both edgesthe effective section modulus Wel may be determined by calculating the elasticsection modulus of an effective cross-section obtained by using a compres-sion flange effective width beff obtained using equation 453 The calculationof Wel must incorporate the possibility that the effective section is not sym-metric and that the centroids of the gross and effective sections do notcoincide
Worked examples of classifying the section and checking the section momentcapacity are given in Sections 492ndash494 51215 and 51217
Local buckling of thin-plate elements 131
473 Members in compression and bending
For cross-sections subjected to a design compressive force NEd and a design majoraxis bending moment MyEd EC3 requires the interaction equation
NEd
NRd+ MyEd
MyRdle 1 (467)
to be satisfied where NRd is the compression resistance of the cross-section deter-mined from equation 449 and MyRd is the bending resistance of the cross-sectiondetermined according to equation 463 Generally the provision of equation 467is overly conservative and is of use only for preliminary member sizing and soEC3 provides less conservative and more detailed member checks depending onthe section classification
If the cross-section is Class 1 or 2 EC3 reduces the design bending resistanceMyRd to a value MyN Rd dependent on the coincident axial force NEd Providedthat
NEd le 025NplRd and (468)
NEd le 05hwtwfyγM0
(469)
where NplRd is the resistance based on yielding given by equation 449 andγM0(= 1) is the partial section resistance factor no reduction in the plastic momentcapacity MyRd = MplyRd is needed since for small axial loads the theoreticalreduction in the plastic moment is offset by strain hardening If either of equations468 or 469 is not satisfied EC3 requires that
MN yRd = MplyRd
(1 minus n
1 minus 05a
)(470)
where
n = NEdNplRd (471)
is the ratio of the applied compression load to the plastic compression resistanceof the cross-section and
a = (A minus 2bf tf )A le 05 (472)
is the ratio of the area of the web to the total area of the cross-sectionFor Class 3 cross-sections EC3 allows only a linear interaction of the stresses
arising from the combined bending moment and axial force so that the maximumlongitudinal stress is limited to the yield stress that is
σxEd le fyγM0 (473)
As for Class 3 cross-sections the longitudinal stress in Class 4 sections sub-jected to combined compression and bending is calculated based on the effective
132 Local buckling of thin-plate elements
properties of the cross-section so that equation 473 is satisfied The resultingexpression that satisfies equation 473 is
NEd
Aeff fyγM0+ MyEd + NEdeNy
Weff ymin fyγM0le 1 (474)
Equation 474 accounts for the additional bending moment caused by a shift eNy
in the compression force NEd from the geometric centroid of the net cross-sectionto the centroid of the effective cross-section
474 Longitudinal stiffeners
A logical basis for the design of a web in pure bending is to limit its proportions sothat its maximum elastic bending strength can be used so that the web is a Class 3element When this is done the section resistance MyRd of the beam is governedby the slenderness of the flanges Thus EC3 requires a fully effective unstiffenedweb to satisfy
hwtw le 124ε (475)
This limit is close to the value of 1269 at which the elastic buckling stress is equalto the yield stress (see equation 437)
Unstiffened webs whose slenderness exceeds this limit are Class 4 elementsbut the use of one or more longitudinal stiffeners can delay local buckling so thatthey become Class 3 elements and the cross-section can be designed as a Class 2section as discussed in Section 472 EC3 does not specify the minimum stiffnessof longitudinal stiffeners to prevent the web plate from deflecting at the stiffenerlocation during local buckling such as in equation 438 Rather it allows theplate slenderness λp in equation 456 to be determined from the local bucklingcoefficient kσ p which incorporates both the area and second moment of the areaof the stiffener If the stiffened web plate is proportioned such that λp le 0874 thereduction factor in equation 453 is unity and the longitudinal stress in the stiffenedweb can reach its yield strength prior to local buckling
The Australian standard AS4100 [18] gives a simpler method of design inwhich a first longitudinal stiffener whose second moment of area is at least thatof equation 428 is placed one-fifth of the web depth from the compression whenthe overall depthndashthickness ratio of the web exceeds the limit of
hwtw = 194ε (476)
which is greater than that of equation 475 because it considers the effect of theflanges on the local buckling stress of the web An additional stiffener whosesecond moment of area exceeds
Ist = hwt3w (477)
is required at the neutral axis when hwtw exceeds 242ε
Local buckling of thin-plate elements 133
475 Beam webs in shear
4751 Stocky webs
The factored design shear force VEd on a cross-section must satisfy
VEd le VcRd (478)
in which VcRd is the design uniform shear resistance which may be calculatedbased on a plastic (VplRd) or elastic distribution of shear stress
I-sections have distributions of shear stress through their webs that are approx-imately uniform (Section 542) and for stocky webs with hwtw lt 72ε the yieldstress in shear τy = fy
radic3 is reached before local buckling Hence EC3 requires
that
VcRd = VplRd = Av(fyradic
3)
γM0(479)
where γM0 = 1 and Av is the shear area of the web which is defined in Clause626 of EC3 This resistance is close to but more conservative than the resistancegiven in equation 428 since the shear area Av is reduced for hot-rolled sectionsby including the root radius in its calculation and because equation 479 ignoresthe effects of strain hardening Some sections such as a monosymmetric I-sectionhave non-uniform distributions of the shear stress τEd in their webs Based on anelastic stress distribution the maximum value of τEd may be determined as inSection 542 and EC3 then requires that
τEd le fyradic
3
γM0 (480)
Cross-sections for which hwtw le 72ε are stocky and the webs of all UBrsquos andUCrsquos in Grade 275 steel satisfy hwtw le 72ε
4752 Slender webs
The shear resistance of slender unstiffened webs for which hwtw gt 72ε decreasesrapidly from the value in equation 479 as the slenderness hwtw increases Forthese
VcRd = VbwRd le ηfyw
radic3
γM1hwtw (481)
where VbwRd is the design resistance governed by buckling of the web in shearfyw is the yield strength of the web η is a factor for the shear area that can betaken as 12 for steels up to S460 and 10 otherwise and γM1(= 1) is a partialresistance factor based on buckling The buckling resistance of the web is given
134 Local buckling of thin-plate elements
in EC3 as
VbwRd = χwfyw
radic3
γM1hwtw (482)
in which the web reduction factor on the yield strength hwtwfywradic
3 due tobuckling is
χw =
1 λw lt 083083λw λw ge 083
(483)
where the modified web plate slenderness is
λw =radic
fywradic
3
τcrasymp 076
radicfyw
τcr(484)
and τcr is the elastic local buckling stress in shear given in equation 421equation 484 is of the same form as equation 45 For an unstiffened webλw = (hwtw)864ε and so it can be seen from equation 482 that the resis-tance of an unstiffened web decreases with an increase in its depthndashthickness ratiohwtw
The shear resistance of a slender web may be increased by providing transversestiffeners which increase the resistance to elastic buckling (Section 431) andalso permit the development of tension field action (Section 4322) Thus
VcRd = VbwRd + Vbf Rd le ηfyw
radic3
γM1hwtw (485)
in which VbwRd is the web resistance given in equation 481 and Vbf Rd isthe flange shear contribution provided by the understressed flanges during thedevelopment of the tension field in the web which further enhances the shearresistance In determining the web resistance reduction factor due to buckling χw
in equation 482 the modified plate slenderness is given by
λw = hw
tw
1
374εradic
kτ(486)
where kτ is the local buckling coefficient given in equations 422 or 423 with Lequal to the stiffener spacing a and d equal to the web depth hw The additionalpost-buckling reserve of capacity due to tension field action for web plates withλw gt 108 is achieved by the use of an enhanced reduction factor of χw = 137(07+λw) instead of the more conservativeχw = 083λw The presence of tensionfield action is also accounted for implicitly in the additional flange resistance
Vbf Rd = bf t2f
c
fyf
γM1
[1 minus
(MEd
Mf Rd
)2]
(487)
where MEd is the design bending moment Mf Rd is the moment of resistance ofthe cross-section consisting of the area of the effective flanges only fyf is the
Local buckling of thin-plate elements 135
yield strength of the flanges and where
c = a
(025 + 16
bf t2f
twh2w
fyf
fyw
) (488)
Worked examples of checking the shear capacity of a slender web are given inSections 495 and 496
4753 Transverse stiffeners
Provisions are made in EC3 for a minimum value for the second moment of areaIst of intermediate transverse stiffeners These must be stiff enough to ensure thatthe elastic buckling stress τcr of a panel can be reached and strong enough totransmit the tension field stiffener force The stiffness requirement of EC3 is thatthe second moment of area of a transverse stiffener must satisfy
Ist ge 075hwt3w (489)
when ahw gtradic
2 in which hw is the depth of the web and
Ist ge 15h3wt3
wa2 (490)
when ahw le radic2 These limits are shown in Figure 429 and are close to those of
equations 427 and 426The strength requirement of EC3 for transverse stiffeners is that they should
be designed as compression members of length hw with an initial imperfection
Value of ad
00 1 2 3
1
2
equation 427
equation 489
equation 490
equation 426
Val
ue o
f I s
at w
3
Figure 429 Stiffness requirements for web stiffeners
136 Local buckling of thin-plate elements
w0 = hw300 using second-order elastic analysis for which the maximum stressis limited by
σzRd = fyγM1 (491)
The effective stiffener cross-section for this design check consists of the cross-section of the stiffener itself plus a width of the web 15εtw each side of the stiffenerand the force to be resisted by the effective stiffener cross-section is
NzEd = VEd minus hwtw
λ2w
fywradic
3
γM1(492)
where λw is given by equation 485 Using equation 38 for the bending of aninitially crooked compression member this procedure implies that the maximumstress in the effective stiffener cross-section is
σzRd = NzEd
Ast
[1 + hwAst
300Wxstmin
(NzEdNcr
1 minus NzEdNcr
)](493)
where
Ncr = π2EIsth2w (494)
is the elastic flexural buckling load of the effective stiffener area and Ast and Ist
are the area and second moment of area of the effective stiffener area and Wxstmin
is its minimum elastic section modulusA worked example of checking a pair of transverse stiffeners is given in
Section 498
4754 End panels
At the end of a plate girder there is no adjacent panel to absorb the horizontalcomponent of the tension field in the last panel and so this load may have to beresisted by the end stiffener A rigid end post is required if tension field action is tobe utilised in design Special treatment of this end stiffener may be avoided if thelength of the last (anchor) panel is reduced so that the tension field contributionτtf to the ultimate stress τult is not required (see equation 429) This is achievedby designing the anchor panel so that its shear buckling resistance VbwRd givenby equation 482 is not less than the design shear force VEd A worked example ofchecking an end panel is given in Section 497
Alternatively a rigid end post consisting of two double-sided load-bearingtransverse stiffeners (see Section 4762) may be used to anchor the tension fieldat the end of a plate girder The end post region consisting of the web and twinpairs of stiffeners can be thought of as a short beam of length hw and EC3 requiresthis beam to be designed to resist the in-plane stresses in the web resulting fromthe shear VEd
Local buckling of thin-plate elements 137
476 Beam webs in shear and bending
Where the design moment MEd and shear VEd are both high the beam must bedesigned against combined shear and bending For stocky webs when the shearforce is less than half the plastic resistance VplRd EC3 allows the effect of theshear on the moment resistance to be neglected When VEd gt 05VplRd EC3requires the bending resistance to be determined using a reduced yield strength
fyr = (1 minus ρ)fyw (495)
for the shear area where
ρ =(
2VEd
VplRdminus 1
)2
(496)
The interaction curve between the design shear force VEd and design moment MEd
is shown in Figure 430 These equations are less conservative than the von-Misesyield condition (see Section 131)
f 2yr + 3τ 2
Ed = f 2y (497)
Equations 495 and 496 may be used conservatively for the well-known propor-tioning method of design for which only the flanges are used to resist the momentand the web to resist the shear force and which is equivalent to using a reducedyield strength fyr = 0 in equation 495 for the web
10
08
06
04
02
00 02 04 06 08 10
MvEd MplRd
VE
dV
plR
d
07
05
Figure 430 Section resistances of beams under bending and shear
138 Local buckling of thin-plate elements
EC3 also uses a simpler reduced design plastic resistance MyV Rd for I-sectionswith a stocky web in lieu of the more tedious use of equations 495 and 496 Hence
MyV Rd = (Wply minus ρA2w4tw)fy
γM0 (498)
where ρ is given in equation 496 Aw = hwtw is the area of the web and Wply isthe major axis plastic section modulus If the web is slender so that its resistanceis governed by shear buckling EC3 allows the effect of shear on the bendingresistance to be neglected when VEd lt 05VbwRd When this is not the case thereduced bending resistance is
MV Rd = MplRd(1 minus ρ2)+ Mf Rdρ2 (499)
where ρ is given in equation 496 using the shear buckling resistance VbwRd
instead of the plastic resistance VplRd Mf Rd is the plastic moment of resistanceconsisting of the effective area of the flanges and MplRd is the plastic resistanceof the cross-section consisting of the effective area of the flanges and the fullyeffective web irrespective of the section class The reduced bending strength inequation 499 is similar to that implied in equation 444 If the contribution ofthe web is conservatively neglected in determining MplRd for the cross-sectionequation 499 leads to MV Rd = Mf Rd which is the basis of the proportioningmethod of design
477 Beam webs in bearing
4771 Unstiffened webs
For the design of webs in bearing according to EC3 the bearing resistance FRd
based on yielding and local buckling is taken as
FRd = χFfywtwy
γM1(4100)
in which y is the effective loaded length of the web and
χF = 05
λF(4101)
is a reduction factor for the yield strength Fy = fywtwy due to local buckling inbearing in which
λF =radic
Fy
Fcr(4102)
Local buckling of thin-plate elements 139
is a modified plate slenderness for bearing buckling (which is of the same form asequation 45) and
Fcr = kFπ2E
12(1 minus ν2)
(twhw
)2
hwtw (4103)
is the elastic buckling load for a web of area hwtw in bearing The buck-ling coefficients for plates in bearing are similar to those in Figure 426 withkF = 6 + 2(ahw)
2 being used for a web of the type shown in Figure 425a Theeffective loaded length used in determining the yield resistance Fy in bearing is
y = ss + 2ntf (4104)
in which ss is the stiff bearing length and ntf is the additional length assuming adispersion of the bearing at 1 n through the flange thickness More convenientlythe dispersion can be thought of as being at a slope 11 through a depth of (nminus1)tfinto the web For a thick web (for which λF lt 05) n = 1 + radic
(bf tw) while fora slender web (for which λF gt 05) n = 1 + radic[bf tw + 002(hwtf )2]
4772 Stiffened webs
When a web alone has insufficient bearing capacity it may be strengthened byadding one or more pairs of load-bearing stiffeners These stiffeners increase theyield and buckling resistances by increasing the effective section to that of thestiffeners together with the web lengths 15twε on either side of the stiffeners ifavailable The effective length of the compression member is taken as the stiffenerlength hw or as 075hw if flange restraints act to reduce the stiffener end rotationsduring buckling and curve c of Figure 313 for compression members should beused
A worked example of checking load-bearing stiffeners is given in Section 499
48 Appendix ndash elastic buckling of plateelements in compression
481 Simply supported plates
A simply supported rectangular plate element of length L width b and thicknesst is shown in Figure 45b Applied compressive loads N are uniformly distributedover both edges b of the plate The elastic buckling load Ncr can be determinedby finding a deflected position such as that shown in Figure 45b which is one ofequilibrium The differential equation for this equilibrium position is [1ndash4]
bD
(part4v
partx4+ 2
part4v
partx2partz2+ part4v
partz4
)= minusNcr
part2v
partxpartz(4105)
140 Local buckling of thin-plate elements
where
bD = Ebt3
12(1 minus ν2)(4106)
is the flexural rigidity of the plateEquation 4105 is compared in Figure 45 with the corresponding differential
equilibrium equation for a simply supported rectangular section column (obtainedby differentiating equation 356 twice) It can be seen that bD for the plate corre-sponds to the flexural rigidity EI of the column except for the (1minusν2) term whichis due to the Poissonrsquos ratio effect in wide plates Thus the term bDpart4vpartx4 repre-sents the resistance generated by longitudinal flexure of the plate to the disturbingeffect minusNpart2vpartx2 of the applied load The additional terms 2bDpart4vpartx2partz2 andbDpart4vpartz4 in the plate equation represent the additional resistances generated bytwisting and lateral bending of the plate
A solution of equation 4105 which satisfies the boundary conditions along thesimply supported edges [1ndash4] is
v = δ sinmπx
Lsin
nπz
b (41)
where δ is the undetermined magnitude of the deflected shape When this is sub-stituted into equation 4105 an expression for the elastic buckling load Ncr isobtained as
Ncr = n2π2bD
L2
[1 + 2
(nL
mb
)2
+(
nL
mb
)4]
which has its lowest values when n = 1 and the buckled shape has one half waveacross the width of the plate b Thus the elastic buckling stress
σcr = Ncr
bt(42)
can be expressed as
σcr = π2E
12(1 minus ν2)
kσ(bt)2
(43)
in which the buckling coefficient kσ is given by
kσ =[(
mb
L
)2
+ 2 +(
L
mb
)2]
(4107)
The variation of the buckling coefficient kσ with the aspect ratio Lb of the plateand the number of half waves m along the plate is shown in Figure 46 It can beseen that the minimum value of kσ is 4 and that this occurs whenever the buckles
Local buckling of thin-plate elements 141
are square (mbL = 1) as shown in Figure 47 In most structural steel membersthe aspect ratio Lb of the plate elements is large so that the value of the bucklingcoefficient kσ is always close to the minimum value of 4 The elastic bucklingstress can therefore be closely approximated by
σcr = π2E3(1 minus ν2)
(bt)2 (4108)
49 Worked examples
491 Example 1 ndash compression resistance ofa Class 4 compression member
Problem Determine the compression resistance of the cross-section of the membershown in Figure 431a The weld size is 8 mm
Classifying the section plate elements
tf = 10 mm tw = 10 mm fy = 355 Nmm2 EN10025-2
ε = radic(235355) = 0814 T52
cf (tf ε) = (4002 minus 102 minus 8)(10 times 0814) T52
= 230 gt 14 and so the flange is Class 4 T52
cw(twε) = (420 minus 2 times 10 minus 2 times 8)(10 times 0814) T52
= 472 gt 42 and so the web is Class 4 T52
400
400
10
10
8 8 1070
1711 97
= 102
10
10
410
20
430
450
420
S355 steel
r
z
S355 steel S355 steel z
351
4
S355 steel z
1540
(a) Section of (b) Section of (c) Section of (d) Section of 356 times 171 UB 45 beam
A = 573 cm2
Wpl = 775 cm3
compression member box beam plate girder
Figure 431 Worked examples
142 Local buckling of thin-plate elements
Effective flange area
kσ f = 043 EC3-1-5 T42
λp f = (4002 minus 102 minus 8)10
284 times 0814 times radic043
= 123 gt 0748 EC3-1-5 44(2)
ρf = (123 minus 0188)1232 = 0687 EC3-1-5 44(2)
Aeff f = 0687 times 4 times (4002 minus 102 minus 8)times 10
+ (10 + 2 times 8)times 10 times 2 EC3-1-5 44(1)
= 5658 mm2
Effective web area
kσ w = 40 EC3-1-5 T41
λpw = (420 minus 2 times 10 minus 2 times 8)10
284 times 0814 times radic40
= 0831 gt 0673 EC3-1-5 44(2)
ρw = 0831 minus 0055 times (3 + 1)08312 = 0885 EC3-1-5 44(2)
Aeff w = 0885times (420minus2times10minus2 times 8)times (10+8times10times2)
= 3558 mm2 EC3-1-5 44(1)
Compression resistance
Aeff = 5658 + 3558 = 9216 mm2
NcRd = 9216 times 35510 N = 3272 kN
492 Example 2 ndash section moment resistance of aClass 3 I-beam
Problem Determine the section moment resistance and examine the suitability forplastic design of the 356 times 171 UB 45 of S355 steel shown in Figure 431b
Classifying the section-plate elements
tf = 97 mm tw = 70 mm fy = 355 Nmm2 EN10025-2
ε = radic(235355) = 0814 T52
cf (tf ε) = (17112 minus 702 minus 102)(97 times 0814) T52
= 91 gt 9 but lt 10 and so the flange is Class 2 T52
cw(twε) = (3514 minus 2 times 97 minus 2 times 102)(70 times 0814) T52
= 547 lt 72 and so the web is Class 1 T52
Local buckling of thin-plate elements 143
Section moment resistanceThe cross-section is Class 2 and therefore unsuitable for plastic design
McRd = 775 times 103 times 35510 N mm = 2751 kNm 625(2)
493 Example 3 ndash section moment resistance ofa Class 4 box beam
Problem Determine the section moment resistance of the welded-box sectionbeam of S355 steel shown in Figure 431c The weld size is 6 mm
Classifying the section-plate elements
tf = 10 mm tw = 8 mm fy = 355 Nmm2 EN10025-2
ε = radic(235355) = 0814 T52
cf (tf ε) = 410(10 times 0814) T52
= 504 gt 42 and so the flange is Class 4 T52
cw(twε) = (430 minus 2 times 10 minus 2 times 6)(8 times 0814) T52
= 611 lt 72 and so the web is Class 1 T52
The cross-section is therefore Class 4 since the flange is Class 4
Effective cross-section
kσ = 40 EC3-1-5 T42
λp = 410
10 times 284 times 0814 times radic40
= 0887 gt 0673
EC3-1-5 44(2)
ψ = 1 EC3-1-5 T41
ρ = (0887 minus 0055 times (3 + 1))08872 = 0848 EC3-1-5 44(2)
bceff = 0848 times 410 = 3475 mm
Aeff = (450 minus 410 + 3475)times 10 + (450 times 10)
+ 2 times (430 minus 2 times 10)times 8
= 14 935 mm2
14 935 times zc = (450 minus 410 + 3475)times 10 times (430 minus 102)
+ 450 times 10 times 102 + 2 times (430 minus 2 times 10)times 8 times 4302
zc = 2062 mm
Ieff = (450 minus 410 + 3475)times 10 times (430 minus 102 minus 2062)2
+ 450 times 10 times (2062 minus 102)2 + 2 times (430 minus 2 times 10)3 times 812
+ 2 times (430 minus 2 times 10)times 8 times (4302 minus 2062)2 mm4
= 4601 times 106 mm4
144 Local buckling of thin-plate elements
Section moment resistance
Weff min = 4601 times 106(430 minus 2062) = 2056 times 106 mm3
McRd = 2056 times 106 times 35510 Nmm = 7299 kNm 625(2)
494 Example 4 ndash section moment resistance of aslender plate girder
Problem Determine the section moment resistance of the welded plate girder ofS355 steel shown in Figure 431d The weld size is 6 mm
Solution
tf = 20 mm tw = 10 mm fyf = 345 Nmm2 EN10025-2
ε = radic(235345) = 0825 T52
cf (tf ε) = (4002 minus 102 minus 6)(20 times 0825) T52
= 115 gt 10 but lt 14 and so the flange is Class 3 T52
cw(twε) = (1540 minus 2 times 20 minus 2 times 6)(10 times 0825) T52
= 1803 gt 124 and so the web is Class 4 T52
A conservative approximation for the cross-section moment resistance may beobtained by ignoring the web completely so that
McRd = Mf = (400 times 20)times (1540 minus 20)times 34510 Nmm = 4195 kNm
A higher resistance may be calculated by determining the effective width of theweb
ψ = minus1 EC3-1-5 T41
kσ = 239 EC3-1-5 T41
λp = (1540 minus 2 times 20 minus 2 times 6)10
284 times 0825 times radic239
= 1299 EC3-1-5 44(2)
ρ = (1299 minus 0055 times (3 minus 1))12992 = 0705 EC3-1-5 44(2)
bc = (1540 minus 2 times 20 minus 2 times 6)1 minus (minus1) = 7440 mm EC3-1-5 T41
beff = 0705 times 7440 = 5244 mm EC3-1-5 T41
be1 = 04 times 5244 = 2098 mm EC3-1-5 T41
be2 = 06 times 5244 = 3146 mm EC3-1-5 T41
Local buckling of thin-plate elements 145
and the ineffective width of the web is
bc minus be1 minus be2 = 7440 minus 2098 minus 3146 = 2196 mm EC3-1-5 T41
Aeff = (1540 minus 2 times 20 minus 2196)times 10 + 2 times 400 times 20
= 28 804 mm2
28 804 times zc = (2 times 400 times 20 + (1540 minus 2 times 20)times 10)times (15402)
minus 2196 times 10 times (1540 minus 20 minus 6 minus 2098 minus 21962)
zc = 7376 mm
Ieff = (400 times 20)times (1540 minus 10 minus 7376)2
+ (400 times 20)times (7376 minus 10)2
+ (1540 minus 2 times 20)3 times 1012 + (1540 minus 2 times 20)
times 10 times (15402 minus 7376)2
minus 21963 times 1012 minus 2196
times 10 times (1540 minus 20 minus 6 minus 2098 minus 21962 minus 7376)2 mm4
= 1162 times 109 mm4
Weff = 1162 times 109(1540 minus 7376) = 1448 times 106 mm3
McRd = 1448 times 106 times 34510 Nmm = 4996 kNm
495 Example 5 ndash shear buckling resistance of anunstiffened plate girder web
Problem Determine the shear buckling resistance of the unstiffened plate girderweb of S355 steel shown in Figure 431d
Solution
tw = 10 mm fyw = 355 Nmm2 EN10025-2
ε = radic(235355) = 0814 T52
η = 12 EC3-1-5 51(2)
hw = 1540 minus 2 times 20 = 1500 mm EC3-1-5 F51
ηhw(twε) = 12 times 1500(10 times 0814)
= 2212 gt 72 and so the web is slender EC3-1-5 51(2)
ahw = infinhw = infin kτ st = 0 EC3-1-5 A3(1)
kτ = 534 EC3-1-5 A3(1)
146 Local buckling of thin-plate elements
τcr = 534 times 190 000 times (101500)2 = 451 Nmm2
EC3-1-5 A1(2)
λw = 076 times radic(355451) = 2132 gt 108 EC3-1-5 53(3)
Assuming that there is a non-rigid end post then
χw = 0832132 = 0389 EC3-1-5 T51
Neglecting any contribution from the flanges
VbRd = VbwRd = 0389 times 355 times 1500 times 10radic3 times 10
N = 1196 kN EC3-1-5 2(1)
496 Example 6 ndash shear buckling resistance of astiffened plate girder web
Problem Determine the shear buckling resistance of the plate girder web of S355steel shown in Figure 431d if intermediate transverse stiffeners are placed at1800 mm spacing
Solution
tw = 10 mm fyw = 355 Nmm2 EN10025-2
ε = radic(235355) = 0814 T52
hw = 1540 minus 2 times 20 = 1500 mm
ahw = 18001500 = 120 kτ st = 0 EC3-1-5 A3(1)
kτ = 534 + 4001202 = 812 EC3-1-5 A3(1)
τcr = 812 times 190 000 times (101500)2 = 686 Nmm2 EC3-1-5 A1(2)
Assuming that there is a rigid end post then
λw = 076 times radic(355686) = 173 gt 108 EC3-1-5 53(3)
χw = 137(07 + 173) = 0564 EC3-1-5 T51
Neglecting any contribution from the flanges
VbRd = VbwRd = 0564 times 355 times 1500 times 10radic3 times 10
N = 1734 kN
EC3-1-5 52(1)
Local buckling of thin-plate elements 147
If the flange contribution is considered and MEd = 0
2 times (15εtf )+ tw = 2 times (15 times 0814 times 20)+ 10 EC3-1-5 54(1)
= 498 mm gt 400 mm and so bf = 400 mm EC3-1-5 54(1)
c = 1800 times(
025 + 16 times 400 times 202 times 345
10 times 15002 times 355
)= 4699 mm
EC3-1-5 54(1)
Vbf Rd = (400 times 202 times 345 times 10)(4699 times 10) N = 117 kNEC3-1-5 54(1)
VbRd = 1734 + 117 = 1851 kN EC3-1-5 52(1)
497 Anchor panel in a stiffened plate girder web
Problem Determine the shear elastic buckling resistance of the end anchor panelof the welded stiffened plate girder of Section 496 if the width of the panel is1800 mm
Solution
tw = 10 mm fyw = 355 Nmm2 EN10025-2
Using λw = 173 (Section 496) the shear elastic buckling resistance of the endpanel is
VpRd = (11732)times 355 times 1500 times 10(103 times radic3 times 10) N = 1027 kN
EC3-1-5 933(3)
498 Example 8 ndash intermediate transverse stiffener
Problem Check the adequacy of a pair of intermediate transverse web stiffeners100 times 16 of S460 steel for the plate girder of Section 496
SolutionUsing VEd = 1734 kN (Section 496) and VpRd = 1027 kN (Section 497)
the stiffener section must resist an axial force of
NEds = 1734 minus 1027 = 7064 kN EC3-1-5 933(3)
The stiffener section consists of the two stiffener plates and a length of the webdefined in Figure 91 of EC3-1-5 For the stiffener plates
tp = 16 mm fyp = 460 Nmm2 EN10025-2
ε = radic(235460) = 0715 T52
cp(tpε) = 100(16 times 0715) = 874 lt 9 T51
and so the plates are fully effective and the stiffener section is Class 1
148 Local buckling of thin-plate elements
Using the lower of the web and plate yield stresses of fyw = 355 Nmm2 andε = 0814
Aeff s = 2 times (16 times 100)+ (2 times 15 times 0814 times 10 + 16)times 10
= 5801 mm2 EC3-1-5 F91
Ieff s = (2 times 100 + 10)3 times 1612 = 1235 times 106 mm4 EC3-1-5 F91
ieff s = radic (1235 times 1065801
) = 461 mm
λ1 = 939 times 0814 = 764 6313(1)
Lcr = 10 times 1500 = 1500 mm EC3-1-5 94(2)
λ = 1500(461 times 764) = 0426 6313(1)
For buckling curve c α = 049 EC3-1-5 94(2) T61
Φ = 05 times [1 + 049 times (0426 minus 02)+ 04262] = 06466312(1)
χ = 1[0646 + radic(06462 minus 04262)] = 0884 6312(1)
NbRd = 0884 times 5801 times 35510 N 6311(3)
= 1820 kN gt 7064 kN = NEds OK
ahw = 18001500 = 120 ltradic
2 EC3-1-5 933(3)
15h3wt3
wa2 = 15 times 15003 times 10318002 EC3-1-5 933(3)
= 1563 times 106 mm4 lt 1235 times 106 mm4 = Is OK
499 Example 9 ndash load-bearing stiffener
Problem Check the adequacy of a pair of load-bearing stiffeners 100 times 16 of S460steel which are above the support of the plate girder of Section 496 The flangesof the girder are not restrained by other structural elements against rotation Thegirder is supported on a stiff bearing ss = 300 mm long the end panel width is1000 mm and the design reaction is 1400 kN
Stiffener yield resistance
ts = 16 mm fy = 460 Nmm2 EN10025-2
ε = radic(235460) = 0715 T52
cs(tsε) = 100(16 times 0715) T51
= 874 lt 9 and so the stiffeners are fully effective T51
FsRd = 2 times (100 times 16)times 46010 N = 1472 kN
Local buckling of thin-plate elements 149
Unstiffened web resistance
kF = 2 + 6 times (300 + 0)1500 = 32 EC3-1-5 F61
Fcr = 09 times 32 times 210 000 times 1031500 N = 4032 kN EC3-1-5 64(1)
e = 32 times 210 000 times 102
2 times 355 times 1500= 631 mm lt 300 mm
= ss + c EC3-1-5 65(3)
m1 = (355 times 400)(355 times 10) = 40 EC3-1-5 65(1)
m2 = 002 times (150020)2 = 1125 EC3-1-5 65(1)
ly611 = 631 + 20 timesradic
402 + (63120)2 + 1125
= 3018 mm EC3-1-5 65(3)
ly612 = 631 + 20 times radic40 + 1125
= 3101 mm gt 3018 mm = ly611 EC3-1-5 65(3)
ly = 301 8 mm EC3-1-5 65(3)
λF =radic
3018 times 10 times 355
4032 times 103= 1630(gt05) EC3-1-5 64(1)
χF = 051630 = 0307 EC3-1-5 64(1)
Leff = 0307 times 3018 = 926 mm EC3-1-5 64(1)
FRd = 355 times 926 times 1010 N EC3-1-5 62(1)
= 3286 kN lt 1400 kN and so load bearing stiffeners are required
Stiffener buckling resistance
Aeff s = 2 times (16 times 100)+ (2 times 15 times 0814 times 10 + 16)times 10
= 5801 mm2 EC3-1-5 F91
Ieff s = (2 times 100 + 10)3 times 1612 = 1235 times 106 mm4 EC3-1-5 F91
ieff s = radic (1235 times 1065801
) = 461 mm
λ1 = 939 times 0814 = 764 6313(1)
Lcr = 10 times 1500 mm = 1500 mm EC3-1-5 94(2)
λ = 1500(461 times 764) = 0426 6313(1)
For buckling curve c α = 049 EC3-1-5 94(2) T61
Φ = 05 times [1 + 049 times (0426 minus 02)+ 04262] = 0646 6312(1)
χ = 1[0646 + radic(06462 minus 04262)] = 0884 6312(1)
NbRd = 0884 times 5801 times 35510 N = 1820 kN gt 1400 kN OK6311(3)
150 Local buckling of thin-plate elements
4910 Example 10 ndash shear and bending of a Class 2 beam
Problem Determine the design resistance of the 356 times 171 UB 45 of S355steel shown in Figure 431b at a point where the design moment is MEd =230 kNm
SolutionAs in Section 492 fy = 355 Nmm2 ε = 0814 η = 12 the flange is Class 2
and the web is Class 1 the section is Class 2 and Wpl = 775 cm3
Av = 573 times 102 minus 2 times 1711 times 97 + (70 + 2 times 102)times 97
= 2676 mm2 626(3)
η = 12 EC3-1-5 51(2)
ηhwtw = 12 times (3514 minus 2 times 97)times 70 = 2789 mm2 gt 2676 mm2
626(3)
VplRd = 2789 times (355radic
3)10 N = 5716 kN 626(2)
ηhw(twε) = 12 times (3514 minus 2 times 97)(70 times 0814) = 700 lt 72EC3-1-5 51(2)
so that shear buckling need not be consideredUsing a reduced bending resistance corresponding to MV Rd = MEd = 230
kNm then
(1 minus ρ)times 355 times 775 times 103 = 230 times 106 so that 628(3)
ρ = 0164
Now ρ = (2VEdVplRd minus 1)2 which leads to 628(3)
VcM RdVplRd = [radic(0164)+ 1]2 = 0702 so that
VcM Rd = 0702 times 5716 = 4015 kN
4911 Example 11 ndash shear and bending of a stiffenedplate girder
Problem Determine the shear resistance of the stiffened plate web girder of Section496 shown and shown in Figure 431d at the point where the design moment isMEd = 4000 kNm
SolutionAs previously fyf = 345 Nmm2 εf = 0825 and the flanges are Class 3
(Section 494) fyw = 345 Nmm2 and εw = 0814 (Section 495) and VbwRd =1733 kNm (Section 496)
Local buckling of thin-plate elements 151
The resistance of the flanges to bending is
Mf Rd = (400 times 20)times (1540 minus 20)times 34510 Nmm
= 4195 kNm gt 4000 kNm = MEd EC3-1-5 54(1)
and so the flange resistance is not completely utilised in resisting the bendingmoment
2 times (15εtf )+ tw = 2 times (15 times 0814 times 20)+ 10 = 498 mm gt 400 mm
EC3-1-5 54(1)
bf = 400 mm lt 498 mm EC3-1-5 54(1)
c = 1800 times(
025 + 16 times 400 times 202 times 345
10 times 15002 times 355
)= 4699 mm
EC3-1-5 54(1)
Vbf Rd = 400 times 202 times 355
4699 times 10times
[1 minus
(4000
4195
)2]
N = 11 kN
EC3-1-5 54(1)
VbwRd + Vbf Rd = 1733 + 11 = 1744 kN EC3-1-5 52(1)
410 Unworked examples
4101 Example 12 ndash elastic local buckling of a beam
Determine the elastic local buckling moment for the beam shown in Figure 432a
4102 Example 13 ndash elastic local buckling of a beam-column
Determine the elastic local buckling load for the beam-column shown inFigure 432b when MN = 20 mm
4103 Example 14 ndash section capacity of a welded box beam
Determine the section moment resistance for the welded box beam of S275 steelshown in Figure 432c
4104 Example 15 ndash designing a plate web girder
The overall depth of the laterally supported plate girder of S275 steel shown inFigure 432d must not exceed 1800 mm Design a constant section girder for the
152 Local buckling of thin-plate elements
150
10 mm6 mm
150 150600
(a) Beam section
Z = 139 times 10 mm
300
300
6 mm
(b) Beam-column section
300 times 6 mm
200 times 6 mm
(c) Box beam section
A B C
500 kN
15 000 mm 000 mm
(d) Plate girder
95 Nmm
10 mm
max6 3
Z = 091 times10 mmmin6 3
5
Figure 432 Unworked examples
factored loads shown and determine
(a) The flange proportions(b) The web thickness(c) The distribution of any intermediate stiffeners(d) The stiffener proportions and(e) The proportions and arrangements of any load-bearing stiffeners
References
1 Timoshenko SP and Woinowsky-Krieger S (1959) Theory of Plates and Shells2nd edition McGraw-Hill New York
2 Timoshenko SP and Gere JM (1961) Theory of Elastic Stability 2nd editionMcGraw-Hill New York
3 Bleich F (1952) Buckling Strength of Metal Structures McGraw-Hill New York4 Bulson PS (1970) The Stability of Flat Plates Chatto and Windus London5 Column Research Committee of Japan (1971) Handbook of Structural Stability
Corona Tokyo6 Allen HG and Bulson PS (1980) Background to Buckling McGraw-Hill (UK)7 Basler K (1961) Strength of plate girders in shear Journal of the Structural Division
ASCE 87 No ST7 pp 151ndash808 Bradford MA Bridge RQ Hancock GJ Rotter JM and Trahair NS (1987)
Australian limit state design rules for the stability of steel structures Proceedings FirstStructural Engineering Conference Institution of Engineers Australia Melbournepp 209ndash16
Local buckling of thin-plate elements 153
9 Evans HR (1983) Longitudinally and transversely reinforced plate girders Chapter1 in Plated Structures Stability and Strength (ed R Narayanan) Applied SciencePublishers London pp 1ndash37
10 Rockey KC and Skaloud M (1972) The ultimate load behaviour of plate girdersloaded in shear The Structural Engineer 50 No 1 pp 29ndash47
11 Usami T (1982) Postbuckling of plates in compression and bending Journal of theStructural Division ASCE 108 No ST3 pp 591ndash609
12 Kalyanaraman V and Ramakrishna P (1984) Non-uniformly compressed stiffenedelements Proceedings Seventh International Specialty Conference Cold-FormedStructures St Louis Department of Civil Engineering University of Missouri-Rollapp 75ndash92
13 Merrison Committee of the Department of Environment (1973) Inquiry into theBasis of Design and Method of Erection of Steel Box Girder Bridges Her MajestyrsquosStationery Office London
14 Rockey KC (1971)An ultimate load method of design for plate girders Developmentsin Bridge Design and Construction (eds KC Rockey JL Bannister and HR Evans)Crosby Lockwood and Son London pp 487ndash504
15 Rockey KC El-Gaaly MA and Bagchi DK (1972) Failure of thin-walled mem-bers under patch loading Journal of the Structural Division ASCE 98 No ST12pp 2739ndash52
16 Roberts TM (1981) Slender plate girders subjected to edge loading ProceedingsInstitution of Civil Engineers 71 Part 2 September pp 805ndash19
17 Roberts TM (1983) Patch loading on plate girders Chapter 3 in Plated Struc-tures Stability and Strength (ed R Narayanan) Applied Science Publishers Londonpp 77ndash102
18 SA (1998) AS4100ndash1998 Steel Structures Standards Australia Sydney Australia
Chapter 5
In-plane bending of beams
51 Introduction
Beams are structural members which transfer the transverse loads they carry to thesupports by bending and shear actions Beams generally develop higher stressesthan axially loaded members with similar loads while the bending deflections aremuch higher These bending deflections of a beam are often therefore a primarydesign consideration On the other hand most beams have small shear deflectionsand these are usually neglected
Beam cross-sections may take many different forms as shown in Figure 51 andthese represent various methods of obtaining an efficient and economical mem-ber Thus most steel beams are not of solid cross-section but have their materialdistributed more efficiently in thin walls Thin-walled sections may be open andwhile these tend to be weak in torsion they are often cheaper to manufacture thanthe stiffer closed sections Perhaps the most economic method of manufacturingsteel beams is by hot-rolling but only a limited number of open cross-sections isavailable When a suitable hot-rolled beam cannot be found a substitute may befabricated by connecting together a series of rolled plates and this has becomeincreasingly common Fabricating techniques also allow the production of beamscompounded from hot-rolled members and plates and of hybrid members in whichthe flange material is of a higher yield stress than the web Tapered and castellatedbeams can also be fabricated from hot-rolled beams In many cases a steel beamis required to support a reinforced concrete slab and in this case its strength maybe increased by connecting the steel and concrete together so that they act com-positely The fire resistance of a steel beam may also be increased by encasing itin concrete The final member cross-section chosen will depend on its suitabilityfor the use intended and on the overall economy
The strength of a steel beam in the plane of loading depends on its sectionproperties and on its yield stress fy When bending predominates in a determinatebeam the effective ultimate strength is reached when the most highly stressedcross-section becomes fully yielded so that it forms a plastic hinge The momentMp at which this occurs is somewhat higher than the first yield moment My at whichelastic behaviour nominally ceases as shown in Figure 52 and for hot-rolled
In-plane bending of beams 155
Solid section Rolled section
Cut
WeldTapered I-beam
Composite beam
Encased beam
Castellated I-beam
WeldWeld
Compoundsection
Fabricatedsection
Cut
Reverse
Shearconnector
Thin-walledopen section
Thin-walledclosed section
Weld
Displace
Figure 51 Beam types
M
2
2
d
d
x
w
Rigid plastic assumption
MpMy 4
5
1
Effect of residual stresses
(d) Momentndashcurvature relationships
3
2
2
d
d
x
w
1 2 3 4 5
fy fy fy fy
(c) Stress distributions
First yield 2
Elastic Firstyield
Strain-hardened
Fullyplastic
Elastic-plastic
Figure 52 Momentndashcurvature relationships for steel beams
I-beams this margin varies between 10 and 20 approximately The attainmentof the first plastic hinge forms the basis for the traditional method of design ofbeams in which the bending moment distribution is calculated from an elasticanalysis
156 In-plane bending of beams
However in an indeterminate beam a substantial redistribution of bendingmoment may occur after the first hinge forms and failure does not occur untilsufficient plastic hinges have formed to cause the beam to become a mechanismThe load which causes this mechanism to form provides the basis for the morerational method of plastic design
When shear predominates as in some heavily loaded deep beams of short spanthe ultimate strength is controlled by the shear force which causes complete plas-tification of the web In the more common sections this is close to the shear forcewhich causes the nominal first yield in shear and so the shear design is carriedout for the shear forces determined from an elastic analysis In this chapter thein-plane behaviour and design of beams are discussed It is assumed that neitherlocal buckling (which is treated in Chapter 4) nor lateral buckling (which is treatedin Chapter 6) occurs Beams with axial loads are discussed in Chapter 7 while thetorsion of beams is treated in Chapter 10
52 Elastic analysis of beams
The design of a steel beam is often preceded by an elastic analysis of the bendingof the beam One purpose of such an analysis is to determine the bending momentand shear force distributions throughout the beam so that the maximum bend-ing moments and shear forces can be found and compared with the moment andshear capacities of the beam An elastic analysis is also required to determine thedeflections of the beam so that these can be compared with the desirable limitingvalues
The data required for an elastic analysis include both the distribution and themagnitudes of the applied loads and the geometry of the beam In particular thevariation along the beam of the effective second moment of area I of the cross-section is needed to determine the deflections of the beam and to determine themoments and shears when the beam is statically indeterminate For this purposelocal variations in the cross-section such as those due to bolt holes may be ignoredbut more general variations including any general reductions arising from theuse of the effective width concept for excessively thin compression flanges (seeSection 4222) should be allowed for
The bending moments and shear forces in statically determinate beams can bedetermined by making use of the principles of static equilibrium These are fullydiscussed in standard textbooks on structural analysis [1 2] as are various methodsof analysing the deflections of such beams On the other hand the conditions ofstatics are not sufficient to determine the bending moments and shear forces instatically indeterminate beams and the conditions of compatibility between thevarious elements of the beam or between the beam and its supports must alsobe used This is done by analysing the deflections of the statically indeterminatebeam Many methods are available for this analysis both manual and computerand these are fully described in standard textbooks [3ndash8] These methods can also
In-plane bending of beams 157
MS = MA ndash MB MS = qL2 8 MS = QL 4 MS = QL 3 MS = QL 4
k = 298 k = k = k = k = 496 397 507 546
L2
MA
L L L2
qQ Q Q QQ
(a) Mid-span deflection wc
(b) Moments MS and deflection coefficients k
and units wc = kMS L
2I wc ndash mm
MS ndash kNm L ndash m I ndash cm4
L3 L3 L3 L4 L2 L4
MB
Figure 53 Mid-span deflections of steel beams
be used to analyse the behaviour of structural frames in which the member axialforces are small
While the deflections of beams can be determined accurately by using themethods referred to above it is often sufficient to use approximate estimatesfor comparison with the desirable maximum values Thus in simply supported orcontinuous beams it is usually accurate enough to use the mid-span deflection(see Figure 53) The mid-span deflection wc (measured relative to the level of theleft-hand support) depends on the distribution of the applied load the end momentsMA and MB (taken as clockwise positive) and the sinking wAB of the right-handsupport below the left-hand support and can be expressed as
wc = kMS + 298(MA minus MB)L2
I+ wAB
2 (51)
In this equation the deflections wc and wAB are expressed in mm the momentsMS MA MB are in kNm the span L is in m and the second moment of area I is incm4 Expressions for the simple beam moments MS and values of the coefficientsk are given in Figure 53 for a number of loading distributions These can becombined together to find the central deflections caused by many other loadingdistributions
53 Bending stresses in elastic beams
531 Bending in a principal plane
The distribution of the longitudinal bending stresses σ in an elastic beam bent in aprincipal plane xz can be deduced from the experimentally confirmed assumptionthat plane sections remain substantially plane during bending provided any shearlag effects which may occur in beams with very wide flanges (see Section 545)
158 In-plane bending of beams
are negligible Thus the longitudinal strains ε vary linearly through the depth ofthe beam as shown in Figure 54 as do the longitudinal stresses σ It is shown inSection 581 that the moment resultant My of these stresses is
My = minusEIyd2w
dx2(52)
where the sign conventions for the moment My and the deflection w are as shownin Figure 55 and that the stress at any point in the section is
σ = Myz
Iy(53)
in which tensile stresses are positive In particular the maximum stresses occur atthe extreme fibres of the cross-section and are given by
in compression and
σmax = MyzT
Iy= minus My
WelyT
σmax = MyzB
Iy= My
WelyB
(54)
in tension in which WelyT = minus IyzT and WelyB = IyzB are the elastic sectionmoduli for the top and bottom fibres respectively for bending about the y axis
The corresponding equations for bending in the principal plane xy are
Mz = EIzd2v
dx2(55)
where the sign conventions for the moment Mz and the deflection v are also shownin Figure 55
σ = minusMzyIz (56)
(c) Strains
δx δx
δx
(b) Bending (a) Cross-section (d) Stresses
= E
=
= ndash
zdAA
(e) Moment resultant
y
ndashzT
z
My
z2
2
dd
xwEI
M
y
y
x
w
d
d2
2
My
zB
z
C
b
Figure 54 Elastic bending of beams
In-plane bending of beams 159
z w
xC
y v
xCMy
My
Mz
Mz
C
y
z
x
+ve My
ndashve d2wdx2
+ve Mz
+ve d2vdx
(b) Curvature in xz plane
(c) Curvature in xy plane(a) Positive bending actions
Figure 55 Bending sign conventions
andσmax = MzWelzL
σmax = minusMzWelzR
(57)
in which WelzL = minus IzyL and WelzR = IzyR are the elastic section moduli forbending about the z axis
Values of Iy Iz Welz Welz for hot-rolled steel sections are given in [9] whilevalues for other sections can be calculated as indicated in Section 59 or in standardtextbooks [1 2] Expressions for the properties of some thin-walled sections aregiven in Figure 56 [10] When a section has local holes or excessive widths (seeSection 4222) these properties may need to be reduced accordingly
Worked examples of the calculation of cross-section properties are given inSections 5121ndash5124
532 Biaxial bending
When a beam deflects only in a plane xz1 which is not a principal plane (seeFigure 57a) so that its curvature d2v1d x2 in the perpendicular xy1 plane is zerothen the bending stresses
σ = minusEz1d2w1dx2 (58)
have moment resultants
andMy1 = minusEIy1d2w1dx2
Mz1 = minusEIy1z1d2w1dx2
(59)
160 In-plane bending of beams
b1
2b
b33t
2t
1t
y
z
1b
1t
2t3b3t
3t
z
y
1b = b2
dt
A
I
I
y
z
y
z
0
0
A + A + A
A z + I
I + I
I b12
12
z
A z + z t
A bnsum2
n 4
0
b
3
2
3
3
321
nnsum3
1
2
sum2
n pn pn 2)(
I I12( )Iz2
A ndashndash ndash
ndash ndash
ndash
ndash
ndash A12( )A2
A + A + A321
A z +1 A z +21 2
22
2
2
(
]
2
2 2
SectionGeometry
Wply12
Wplz12
Wely12
Welz12
Iy z
bn3tn 12
A z +323 2I3
I + I + A b 21 21 2 3
I zny
I b1z2
A z + z t3sum2
n pn pn )(
A bnsum2
n 4 + A b13
0
b3I I12( )
Iz2
A A12( )
A2
d
d
d
d
d
d
d
t
t3 8
t3 8
t2 4
t2 4
t2
d t2
0
0
An = b tnn In=
z1 2 = b A + A3 2 1 2) A3
z3 = ( )z z2 1 2
zpn=[0 A An2( )2t3le le b3
Figure 56a Thin-walled section properties
where Iy1z1 is the product second moment of area (see Section 59) Thus theresultant bending moment
M = radic(M 2
y1 + M 2z1
) (510)
is inclined to the xz1 plane of the bending deflections as shown in Figure 57aThe simplest method of analysing this or any other biaxial bending situation is
to replace the moments My1 Mz1 by their principal plane static equivalents My
In-plane bending of beams 161
b1
1b
2b
2b
2b
2b
1b
3b
3b
b3b3
3t
1t
1t t t
y
y
y
z
z
z
SectionGeometry
A 2A +
+ + +++
A3
A2
1 2 A 2sumA+2 A + A321 22 n
sumA2 n
I
I
y
z
y
y1
y2
y3
yp
z
0
0
Wply
Wplz
Wely
Welz
A b 2 2 22
31
A b 23 b
3b
22
1I3
I3 I2 ndash( ) A2 b3
b2
2ndash( 2)
4 +
sumI2
n+
A3 b3
2
2A1y12 +A3y3
2 +2I1 2A1y12 +2A2y2
2 + A3y32+2I1 A2b3
2 4 + A3b224
2Iyb3
Iz(b1 ndash y3) and Izy3
A1b3 + A3b34 A1b3 + A2(b3 ndash b2) + A3b34
(b1 ndash yp)2t1 + yp
2t1+A3yp
A1 b32b1
b1A32A
b1A1A
(A ndash 2 A3)4t1 ge 0 (A ndash 2 A3)4t ge 0
An= bntn In = bn3 tn12
(b1 + 2b2)A1A
b12 ndash y3
b1 ndash y3
4Iy y3
+ (b3 yp)t
ndash 8b233
(b1 ndash yp)2 + yp2 +2b2(b1 ndash yp)
Izy2 and Izy3
2Iyb3
+
0 0 0
ndash
ndash
ndash
ndash
ndash
y3 +b32(b1 + 2b2) A1
4Iy
radic2Iyb3
2radic2Iz(b2+b3)
(b32+2b2 b3 ndash b2
2)t2radic2
(b2+b3)2t2radic2
3radic2Iy
(b2+b3) 2radic2+ (3b3ndash2b2)b2
2b3t
Figure 56b Thin-walled section properties
Mz calculated from
My = My1 cosα + Mz1 sin αMz = minusMy1 sin α + Mz1 cosα
(511)
162 In-plane bending of beams
M
M
y
y1 (v1 = 0)
Plane of moment
My
Mz
Mz1
y
z1
zz1
z
zM 1
My1
My1y1
Principalplanes
(a) Deflection in a non-principal plane (b) Principal plane moments My Mz
Plane ofdeflection
Figure 57 Bending in a non-principal plane
zw
x
y v
x
y
z
q
C
qy ( ) +
ndash~
+
+
+
qz ( )
My
Mz ve
(b) Bending My in xz plane qy ( )
qz ( )
(a) Section and loading (c) Bending Mz in xy plane
+ ve
Figure 58 Biaxial bending of a zed beam
in which α is the angle between the y1 z1 axes and the principal y z axes as shownin Figure 57b The bending stresses can then be determined from
σ = Myz
Iyminus Mzy
Iz (512)
in which Iy Iz are the principal second moments of area If the values of α Iy Iz
are unknown they can be determined from Iy1 Iz1 Iy1z1 as shown in Section 59Care needs to be taken to ensure that the correct signs are used for My and Mz
in equation 512 as well as for y and z For example if the zed section shown inFigure 58a is simply supported at both ends with a uniformly distributed load qacting in the plane of the web then its principal plane components qy qz causepositive bending My and negative bending Mz as indicated in Figure 58b and c
In-plane bending of beams 163
The deflections of the beam can also be determined from the principal planemoments by solving equations 52 and 55 in the usual way [1ndash8] for the principalplane deflections w and v and by adding these vectorially
Worked examples of the calculation of the elastic stresses and deflections in anangle section cantilever are given in Section 5125
54 Shear stresses in elastic beams
541 Solid cross-sections
A vertical shear force Vz acting parallel to the minor principal axis z of a section ofa beam (see Figure 59a) induces shear stresses τxy τxz in the plane of the sectionIn solid section beams these are usually assumed to act parallel to the shear force(ie τxy = 0) and to be uniformly distributed across the width of the section asshown in Figure 59a The distribution of the vertical shear stresses τxz can bedetermined by considering the horizontal equilibrium of an element of the beamas shown in Figure 59b Because the bending normal stresses σ vary with x theycreate an imbalance of force in the x direction which can only be compensatedfor by the horizontal shear stresses τzx = τv which are equal to the vertical shearstresses τxz It is shown in Section 5101 that the stress τv at a distance z2 fromthe centroid where the section width is b2 is given by
τv = minus Vz
Iyb2
int z2
zT
bzdz (513)
y
z
x
V
ndash z
z
bx2
z
b
z
T
2
bδz
x+ )(
+z
)(
(b) Horizontal equilibrium(a) Assumed shear stress distribution
Shear stresses vparallel to shearforce Vz and constantacross width of section
vbδx
δx
bδzδx
(vb)vb
δz
δzpart
part
partpart
Figure 59 Shear stresses in a solid section
164 In-plane bending of beams
This can also be expressed as
τv = minusVzA2z2
Iyb2(514)
in which A2 is the area above b2 and z2 is the height of its centroidFor the particular example of the rectangular section beam of width b and depth
d shown in Figure 510a the width is constant and so equation 513 becomes
τv = Vz
bd
(15 minus 6z2
d2
) (515)
This shear stress distribution is parabolic as shown in Figure 510b and has amaximum value at the y axis of 15 times of average shear stress Vzbd
The shear stresses calculated from equations 513 or 514 are reasonably accuratefor solid cross-sections except near the unstressed edges where the shear stress isparallel to the edge rather than to the applied shear force
542 Thin-walled open cross-sections
The shear stress distributions in thin-walled open-section beams differ from thosegiven by equations 513 or 514 for solid section beams in that the shear stressesare parallel to the wall of the section as shown in Figure 511a instead of parallelto the applied shear force Because of the thinness of the walls it is quite accurateto assume that the shear stresses are uniformly distributed across the thickness t ofthe thin-walled section Their distribution can be determined by considering thehorizontal equilibrium of an element of the beam as shown in Figure 511b It isshown in Section 5102 that the stress τv at a distance s from the end of the section
b
y
z z
d
Vz
15Vz bd
(a) Cross-section (b) Shear stress distribution
v
Figure 510 Shear stress distribution in a rectangular section
In-plane bending of beams 165
t
yy
zV
0
z
O
C
E
xy
z
x
+
st
)(δx δsδxv
tδx
δx
v
t
tδs
δs δx t δs
δs
tv
v
(a) Shear stress distribution (b) Horizontal equilibrium
partpart
partxpart
Figure 511 Shear stresses in a thin-walled open section
caused by a vertical shear force Vz can be obtained from the shear flow
τvt = minusVz
Iy
int s
0ztds (516)
This can also be expressed as
τv = minusVzAszs
Iyt(517)
in which As is the area from the free end to the point s and zs is the height of thecentroid of this area above the point s
At a junction in the cross-section wall such as that of the top flange and theweb of the I-section shown in Figures 512 and 513 horizontal equilibrium of thezero area junction element requires
(τvtf )21δx + (τvtf )23δx = (τvtw)24δx (518)
This can be thought of as an analogous flow continuity condition for the corre-sponding shear flows in each cross-section element at the junction as shown inFigure 513c so that
(τvtf )21 + (τvtf )23 = (τvtw)24 (519)
which is a particular example of the general junction conditionsum(τvt) = 0 (520)
in which each shear flow is now taken as positive if it acts towards the junction
166 In-plane bending of beams
b
dy
sf
f
f
t sw
wt
z
1 2 3
45 6
42
btdIVvtw
ff
y
z
4==
42
2btdIVvtw
ff
y
zIV
y
z8td wf
= +
btdIV
vtfff
y
z
(b) Shear flow distribution(a) I-section
Figure 512 Shear flow distribution in an I-section
0
12
3
4
y
z
x
V
2
2
2z
4
1 2 3
0
0
My
My My+
(c) Flow analogy(b) Flange-web junction
(a) Sign of shear flow
δx
δ
δA
(+δ)δA
(tf)21δx
(tw)24δx
(tf)23δx
(tw)24
(tf)23(tf)21
(tw)24δx
tf δx
Figure 513 Horizontal equilibrium considerations for shear flow
Afree end of a cross-section element is the special case of a one-element junctionfor which equation 520 reduces to
τvt = 0 (521)
This condition has already been used (at s = 0) in deriving equation 516
In-plane bending of beams 167
An example of the analysis of the shear flow distribution in the I-section beamshown in Figure 512a is given in Section 5126 The shear flow distributionis shown in Figure 512b It can be seen that the shear stress in the web is nearlyconstant and equal to its average value
τv(av) = Vz
df tw (522)
which provides the basis for the commonly used assumption that the applied shearis resisted only by the web A similar result is obtained for the shear stress in theweb of the channel section shown in Figure 514 and analysed in Section 5127
When a horizontal shear force Vy acts parallel to the y axis additional shearstresses τh parallel to the walls of the section are induced These can be obtainedfrom the shear flow
τht = minusVy
Iz
int s
0ytds (523)
which is similar to equation 517 for the shear flow due to a vertical shear forceA worked example of the calculation of the shear flow distribution caused by ahorizontal shear force is given in Section 5128
y
z+
CS
ya
t
b
w
0
df
tf
tf
vf
Vf
Vw
btfIy Iydf tf
df tf b df twVz Vzb
2
2
82
2
+=wv t
Iy
df tf bVz
2=fv t
Iy
df tf bVz
2=wv t
(a) Shear flow due to shear force Vz (b) Shear centre and flange and web shears
Figure 514 Shear flow distribution in a channel section
168 In-plane bending of beams
543 Shear centre
The shear stresses τv induced by the vertical shear force Vz exert a torque equalto intE
0 τvtρds about the centroid C of the thin-walled cross-section as shown inFigure 515 and are therefore statically equivalent to a vertical shear force Vz
which acts at a distance y0 from the centroid equal to
y0 = 1
Vz
int E
0τvtρds (524)
Similarly the shear stresses τh induced by the horizontal shear force Vy arestatically equivalent to a horizontal shear force Vy which acts at a distance z0 fromthe centroid equal to
z0 = minus 1
Vy
int E
0τhtρds (525)
The coordinates y0 z0 define the position of the shear centre S of the cross-sectionthrough which the resultant of the bending shear stresses must act When anyapplied load does not act through the shear centre as shown in Figure 516then it induces another set of shear stresses in the section which are additionalto those caused by the changes in the bending normal stresses described above(see equations 516 and 523) These additional shear stresses are statically equiv-alent to the torque exerted by the eccentric applied load about the shear centreThey can be calculated as shown in Sections 10214 and 10312
s
t
y
z
V
y0
CE
O
z
vtδs
δs
Figure 515 Moment of shear stress τv about centroid
In-plane bending of beams 169
e
y
z
Vz
Vz
VzeS
C y
z
S
C y
z
S
C+
+Pure bendingBending shear stresses
Off shear centreloading
Pure torsionTorsion shear stresses
Figure 516 Off shear centre loading
Centroid C Shear centre S
Figure 517 Centroids and shear centres
The position of the shear centre of the channel section shown in Figure 514 isdetermined in Section 5128 Its y0 coordinate is given by
y0 = d2f tf b2
4Iy+ b2tf
2btf + df tw(526)
while its z0 coordinate is zero because the section is symmetrical about the y axisIn Figure 517 the shear centres of a number of thin-walled open sections are
compared with their centroids It can be seen that
(a) if the section has an axis of symmetry then the shear centre and centroid lieon it
170 In-plane bending of beams
(b) if the section is of a channel type then the shear centre lies outside the weband the centroid inside it and
(c) if the section consists of a set of concurrent rectangular elements (tees anglesand cruciforms) then the shear centre lies at the common point
A general matrix method for analysing the shear stress distributions and fordetermining the shear centres of thin-walled open-section beams has been prepared[11 12]
Worked examples of the determination of the shear centre position are given inSections 5128 and 51210
544 Thin-walled closed cross-sections
The shear stress distribution in a thin-walled closed-section beam is similar tothat in an open-section beam except that there is an additional constant shearflow τvct around the section This additional shear flow is required to prevent anydiscontinuity in the longitudinal warping displacements u which arise from theshear straining of the walls of the closed sections To show this consider the slitrectangular box whose shear flow distribution τvot due to a vertical shear force Vz isas shown in Figure 518a Because the beam is not twisted the longitudinal fibresremain parallel to the centroidal axis so that the transverse fibres rotate throughangles τvoG equal to the shear strain in the wall as shown in Figure 519 Theserotations lead to the longitudinal warping displacements u shown in Figure 518bthe relative warping displacement at the slit being
int E0 (τvoG)ds
b
df
Vz
y
yz
z
x
E
E
y0
C
C
S
S
O
O
Slit
EvoG
0
ds
(b) Warping displacements due to shear stress
Relative warpingdisplacement
(a) Shear flow vo t
Figure 518 Warping of a slit box
In-plane bending of beams 171
v
v v
v
x
s
G
Projection of s x axes onto plane ofelementδs times δx
δsδu
δs
δx=
Change in warpingdisplacement due toshear strain
Element δs times δx times t
Figure 519 Warping displacements due to shear
b
y
z
= +
Thickness tw
Thickness tf
df Vz
(a) Circulating shear flow vc t(slit box shear flow vo t
shown in Fig 518a)
v t vc t vo t (b) Total shear flow
Figure 520 Shear stress distribution in a rectangular box
However when the box is not slit as shown in Figure 520 this relative warpingdisplacement must be zero Thus
∮τv
Gds = 0 (527)
in which total shear stress τv can be considered as the sum
τv = τvc + τvo (528)
of the slit box shear stress τvo (see Figure 518a) and the shear stress τvc due to aconstant shear flow τvct circulating around the closed section (see Figure 520a)Alternatively equation 516 for the shear flow in an open section can be modified
172 In-plane bending of beams
for the closed section to
τvt = τvct minus Vz
Iy
int s
0ztds (529)
since it can no longer be said that τv is zero at s = 0 this not being a free end (theclosed section has no free ends) Mathematically τvct is a constant of integrationSubstituting equation 528 or 529 into equation 527 leads to
τvct = minus∮τvods∮(1t)ds
(530)
which allows the circulating shear flow τvct to be determinedThe shear stress distribution in any single-cell closed section can be obtained
by using equations 529 and 530 The shear centre of the section can then bedetermined by using equations 524 and 525 as for open sections For the particularcase of the rectangular box shown in Figure 520
τvct = Vz
Iy
(d2
f tw
8+ df tf b
4
) (531)
and the resultant shear flow shown in Figure 520b is symmetrical because of thesymmetry of the cross-section while the shear centre coincides with the centroid
A worked example of the calculation of the shear stresses in a thin-walled closedsection is given in Section 51211
The shear stress distributions in multi-cell closed sections can be determinedby extending this method as indicated in Section 5103 A general matrix methodof analysing the shear flows in thin-walled closed sections has been described[11 12] This can be used for both open and closed sections including compositeand asymmetric sections and sections with open and closed parts It can also beused to determine the centroid principal axes section constants and the bendingnormal stress distribution
545 Shear lag
In the conventional theory of bending shear strains are neglected so that it can beassumed that plane sections remain plane after loading From this assumptionfollow the simple linear distributions of the bending strains and stresses dis-cussed in Section 53 and from these the shear stress distributions discussed inSections 541ndash544 The term shear lag [13] is related to some of the discrep-ancies between this approximate theory of the bending of beams and their realbehaviour and in particular refers to the increases of the bending stresses near theflange-to-web junctions and the corresponding decreases in the flange stressesaway from these junctions
The shear lag effects near the mid-point of the simply supported centrally loadedI-section beam shown in Figure 521a are illustrated in Figure 522 The shear
In-plane bending of beams 173
Q
Q2
Q2
(+)
(ndash)
Warping displacementsdue to positive shear
Warping displacementsdue to negative shear
(b) Warping displacements calculated from conventional theory
(a) Beam and shear for diagram
Q2
Q2
Figure 521 Incompatible warping displacements at a shear discontinuity
Real behaviour
Approximateconventionaltheory
Real behaviour
(b) Shear flow distribution(a) Bending stress distribution
Approximateconventionaltheory
Figure 522 Shear lag effects in an I-section beam
stresses calculated by the conventional theory are as shown in Figure 523a andthese induce the warping (longitudinal) displacements shown in Figure 523b Thewarping displacements of the web are almost linear and these are responsible forthe shear deflections [13] which are usually neglected when calculating the beamrsquosdeflections The warping displacements of the flanges vary parabolically and itis these which are responsible for most of the shear lag effect The applicationof the conventional theory of bending at either side of a point in a beam where
174 In-plane bending of beams
y
z
x
C
(b) Warping displacements due to shear
(a) Shear flow due to a vertical shear Vz
Approximate shear flow givenby conventional theory
Figure 523 Warping of an I-section beam
there is a sudden change in the shear force such as at the mid-point of the beamof Figure 521a leads to two different distributions of warping displacement asshown in Figure 521b These are clearly incompatible and so changes are requiredin the bending stress distribution (and consequently in the shear stress distribution)to remove the incompatibility These changes which are illustrated in Figure 522constitute the shear lag effect
Shear lag effects are usually very small except near points of high concentratedload or at reaction points in short-span beams with thin wide flanges In particularshear lag effects may be significant in light-gauge cold-formed sections [14]and in stiffened box girders [15ndash17] Shear lag has no serious consequences ina ductile structure in which any premature local yielding leads to a favourableredistribution of stress However the increased stresses due to shear lag may beof consequence in a tension flange which is liable to brittle fracture or fatiguedamage or in a compression flange whose strength is controlled by its resistanceto local buckling
An approximate method of dealing with shear lag is to use an effective widthconcept in which the actual width b of a flange is replaced by a reduced widthbeff given by
beff
b= nominal bending stress
maximum bending stress (532)
This is equivalent to replacing the actual flange bending stresses by constantstresses which are equal to the actual maximum stress and distributed over theeffective flange area beff times t Some values of effective widths are given in [14ndash16]
In-plane bending of beams 175
This approach is similar to that used to allow for the redistribution of stress whichtakes place in a thin compression flange after local buckling (see Chapter 4) How-ever the two effects of shear lag and local buckling are quite distinct and shouldnot be confused
55 Plastic analysis of beams
551 General
As the load on a ductile steel beam is increased the stresses in the beam alsoincrease until the yield stress is reached With further increases in the load yieldingspreads through the most highly strained cross-section of the beam until it becomesfully plastic at a moment Mp At this stage the section forms a plastic hinge whichallows the beam segments on either side to rotate freely under the moment MpIf the beam was originally statically determinate this plastic hinge reduces it to amechanism and prevents it from supporting any additional load
However if the beam was statically indeterminate the plastic hinge does notreduce it to a mechanism and it can support additional load This additional loadcauses a redistribution of the bending moment during which the moment at theplastic hinge remains fixed at Mp while the moment at another highly strainedcross-section increases until it forms a plastic hinge This process is repeated untilenough plastic hinges have formed to reduce the beam to a mechanism The beamis then unable to support any further increase in load and its ultimate strength isreached
In the plastic analysis of beams this mechanism condition is investigated todetermine the ultimate strength The principles and methods of plastic analysis arefully described in many textbooks [18ndash24] and so only a brief summary is givenin the following sub-sections
552 The plastic hinge
The bending stresses in an elastic beam are distributed linearly across any section ofthe beam as shown in Figure 54 and the bending moment M is proportional to thecurvature minusd2wdx2 (see equation 52) However once the yield strain εy = fyE(see Figure 524) of a steel beam is exceeded the stress distribution is no longerlinear as indicated in Figure 52c Nevertheless the strain distribution remainslinear and so the inelastic bending stress distributions are similar to the basicstressndashstrain relationship shown in Figure 524 provided the influence of shearon yielding can be ignored (which is a reasonable assumption for many I-sectionbeams) The moment resultant M of the bending stresses is no longer proportionalto the curvature minusd2wdx2 but varies as shown in Figure 52d Thus the sectionbecomes elastic-plastic when the yield moment My = fyWel is exceeded and thecurvature increases rapidly as yielding progresses through the section At highcurvatures the limiting situation is approached for which the section is completely
176 In-plane bending of beams
0 sty
Stress
Plastic Strain-hardened
Est
Strain
Rigid plasticassumption
Elastic E
fy
Figure 524 Idealised stressndashstrain relationships for structural steel
yielded at the fully plastic moment
Mp = fyWpl (533)
where Wpl is the plastic section modulus Methods of calculating the fully plasticmoment Mp are discussed in Section 5111 and worked examples are given inSections 51212 and 51213 For solid rectangular sections the shape factorWplWel is 15 but for rolled I-sections WplWel varies between 11 and 12approximately Values of Wpl for hot-rolled I-section members are given in [9]
In real beams strain-hardening commences just before Mp is reached and thereal momentndashcurvature relationship rises above the fully plastic limit of Mp asshown in Figure 52d On the other hand high shear forces cause small reductionsin Mp below fyWpl due principally to reductions in the plastic bending capacityof the web This effect is discussed in Section 452
The approximate approach of the momentndashcurvature relationship shown inFigure 52d to the fully plastic limit forms the basis of the simple rigid-plasticassumption for which the basic stressndashstrain relationship is replaced by the rect-angular block shown by the dashed line in Figure 524 Thus elastic strainsare completely ignored as are the increased stresses due to strain-hardeningThis assumption ignores the curvature of any elastic and elasticndashplastic regions(M ltMp) and assumes that the curvature becomes infinite at any point whereM = Mp
The consequences of the rigid-plastic assumption on the theoretical behaviourof a simply supported beam with a central concentrated load are shown inFigures 525 and 526 In the real beam shown in Figure 525a there is a finitelength of the beam which is elastic-plastic and in which the curvatures are largewhile the remaining portions are elastic and have small curvatures Howeveraccording to the rigid-plastic assumption the curvature becomes infinite at mid-span when this section becomes fully plastic (M = Mp) while the two halves ofthe beam have zero curvature and remain straight as shown in Figure 525b The
In-plane bending of beams 177
Qult = 4Mp L Qult = 4Mp L
Plastichinge
(b) Rigid-plastic beam(a) Real beam
Qult Qult
13
Rigid Rigid
Curvature
Deflectedshape
Beam
Elastic Elastic
Elastic-plasticinfin
Figure 525 Beam with full plastic moment Mp at mid-span
Q
Q
1
2
3 4
5
0
Mp
My 12
34
5
Real
(b) Bending moment
Elastic
Strain-hardened Fully plastic
Elasticndashplastic First yield
Rigidndashplastic
(c) Deflection
(a) Beam
L2L2
Figure 526 Behaviour of a simply supported beam
infinite curvature at mid-span causes a finite change θ in the beam slope and sothe deflected shape of the rigid-plastic beam closely approximates that of the realbeam despite the very different curvature distributions The successive stages inthe development of the bending moment diagram and the central deflection of thebeam as the load is increased are summarised in Figure 526b and c
178 In-plane bending of beams
M M
MM
Mp
(b) Frictionless hinge(a ) Pl a s tic h i n g e s
Behaviour
1313
1313
00
Hinges
Figure 527 Plastic hinge behaviour
The infinite curvature at the point of full plasticity and the finite slope changeθ predicted by the rigid-plastic assumption lead to the plastic hinge concept illus-trated in Figure 527a The plastic hinge can assume any slope change θ once thefull plastic moment Mp has been reached This behaviour is contrasted with thatshown in Figure 527b for a frictionless hinge which can assume any slope changeθ at zero moment
It should be noted that when the full plastic moment Mp of the simply supportedbeam shown in Figure 526a is reached the rigid-plastic assumption predicts thata two-bar mechanism will be formed by the plastic hinge and the two frictionlesssupport hinges as shown in Figure 525b and that the beam will deform freelywithout any further increase in load as shown in Figure 526c Thus the ultimateload of the beam is
Qult = 4MpL (534)
which reduces the beam to a plastic collapse mechanism
553 Moment redistribution in indeterminate beams
The increase in the full plastic moment Mp of a beam over its nominal first yieldmoment My is accounted for in elastic design (ie design based on an elasticanalysis of the bending of the beam) by the use of Mp for the moment capacityof the cross-section Thus elastic design might be considered to be lsquofirst hingersquodesign However the feature of plastic design which distinguishes it from elasticdesign is that it takes into account the favourable redistribution of the bendingmoment which takes place in an indeterminate structure after the first hinge formsThis redistribution may be considerable and the final load at which the collapse
In-plane bending of beams 179
mechanism forms may be significantly higher than that at which the first hinge isdeveloped Thus a first hinge design based on an elastic analysis of the bendingmoment may significantly underestimate the ultimate strength
The redistribution of bending moment is illustrated in Figure 528 for a built-inbeam with a concentrated load at a third point This beam has two redundan-cies and so three plastic hinges must form before it can be reduced to a collapsemechanism According to the rigid-plastic assumption all the bending momentsremain proportional to the load until the first hinge forms at the left-hand sup-port A at Q = 675MpL As the load increases further the moment at this hingeremains constant at Mp while the moments at the load point and the right-handsupport increase until the second hinge forms at the load point B at Q = 868MpLThe moment at this hinge then remains constant at Mp while the moment at theright-hand support C increases until the third and final hinge forms at this point atQ = 900MpL At this load the beam becomes a mechanism and so the ultimateload is Qult = 900MpL which is 33 higher than the first hinge load of 675MpLThe redistribution of bending moment is shown in Figure 528b and d while thedeflection of the load point (derived from an elasticndashplastic assumption) is shownin Figure 528c
554 Plastic collapse mechanisms
A number of examples of plastic collapse mechanisms in cantilevers and single-and multi-span beams is shown in Figure 529 Cantilevers and overhanging beams
Q Q Q
Mp
A B C
Q
M------
p90
AB
C
B
MM----
p10
MpMp
Mp Mp
MpMp
MA
MB
MC
QLM------
p
QL
Mp
Mp
QL
(d ) Ben d ing moment diagrams
( b ) Moment (a ) B u i l t - i n b ea m (c) Deflection
= 868 Mp
QL= 90Mp
QL= 675
redistribution
2L3 L3
675868
90
675868
Hinges
Figure 528 Moment redistribution in a built-in beam
180 In-plane bending of beams
(a ) Can til e v e rs and over- hanging beams
(b) Single-span beams
(c ) Multi-span beams
Figure 529 Beam plastic collapse mechanisms
generally collapse as single-bar mechanisms with a plastic hinge at the supportWhen there is a reduction in section capacity then the plastic hinge may form inthe weaker section
Single-span beams generally collapse as two-bar mechanisms with a hinge(plastic or frictionless) at each support and a plastic hinge within the spanSometimes general plasticity may occur along a uniform moment region
Multi-span beams generally collapse in one span only as a local two-bar mech-anism with a hinge (plastic or frictionless) at each support and a plastic hingewithin the span Sometimes two adjacent spans may combine to form a three-barmechanism with the common support acting as a frictionless pivot and one plastichinge forming within each span Similar mechanisms may form in over-hangingbeams
Potential locations for plastic hinges include supports points of concentratedload and points of cross-section change The location of a plastic hinge in a beamwith distributed load is often not well defined
555 Methods of plastic analysis
The purpose of the methods of plastic analysis is to determine the ultimate load atwhich a collapse mechanism first forms Thus it is only this final mechanism con-dition which must be found and any intermediate load conditions can be ignoredA further important simplification arises from the fact that in its collapse condi-tion the beam is a mechanism and can be analysed by statics without any of thedifficulties associated with the elastic analysis of a statically indeterminate beam
The basic method of plastic analysis is to assume the locations of a series ofplastic hinges and to investigate whether the three conditions of equilibrium mech-anism and plasticity are satisfied The equilibrium condition is that the bending
In-plane bending of beams 181
moment distribution defined by the assumed plastic hinges must be in staticequilibrium with the applied loads and reactions This condition applies to allbeams elastic or plastic The mechanism condition is that there must be a suffi-cient number of plastic and frictionless hinges for the beam to form a mechanismThis condition is usually satisfied directly by the choice of hinges The plastic-ity condition is that the full plastic moment of every cross-section must not beexceeded so that
minusMp le M le Mp (535)
If the assumed plastic hinges satisfy these three conditions they are the correctones and they define the collapse mechanism of the beam so that
(Collapse mechanism) satisfies
Equilibrium
MechanismPlasticity
(536)
The ultimate load can then be determined directly from the equilibrium conditionsHowever it is usually possible to assume more than one series of plastic hinges
and so while the assumed plastic hinges may satisfy the equilibrium and mechanismconditions the plasticity condition (equation 535) may be violated In this casethe load calculated from the equilibrium condition is greater than the true ultimateload (QmgtQult) This forms the basis of the mechanism method of plastic analysis(
Mechanism methodQm ge Qult
)satisfies
(EquilibriumMechanism
) (537)
which provides an upper-bound solution for the true ultimate loadA lower bound solution (Qs le Qult) for the true ultimate load can be obtained
by reducing the loads and bending moments obtained by the mechanism methodproportionally (which ensures that the equilibrium condition remains satisfied)until the plasticity condition is satisfied everywhere These reductions decreasethe number of plastic hinges and so the mechanism condition is not satisfied Thisis the statical method of plastic analysis(
Statical MethodQs le Qult
)satisfies
(EquilibriumPlasticity
)(538)
If the upper and lower bound solutions obtained by the mechanism and staticalmethods coincide or are sufficiently close the beam may be designed immediatelyIf however these bounds are not precise enough the original series of hingesmust be modified (and the bending moments determined in the statical analysiswill provide some indication of how to do this) and the analysis repeated
The use of the methods of plastic analysis is demonstrated in Sections 5112and 5113 for the examples of the built-in beam and the propped cantilever shown
182 In-plane bending of beams
in Figures 530a and 531a A worked example of the plastic collapse analysis ofa non-uniform beam is given in Section 51214
The limitations on the use of the method of plastic analysis in the EC3 strengthdesign of beams are discussed in Section 562
The application of the methods of plastic analysis to rigid-jointed frames isdiscussed in Section 8355 This application can only be made provided the effectsof any axial forces in the members are small enough to preclude any instabilityeffects and provided any significant decreases in the full plastic moments due toaxial forces are accounted for (see Section 722)
321 4L 3 L 3 L 3
Q
Q1
2
4
Q
Q 2 Q
2 Q 2 Q
2δ13
2δ13
δ13
δ13
2 Q
2 Q1
3
4
Mp
Mp
MpMp
Mp Mp
M1 M2 M3 M4
Mp
Mp
Q
M4 M3ndashL3
1 2 3 42 3
(b) Beam segments
(c ) Fi r st mec han i sm and virtual displacements
(a ) Bu i lt- in b e a m
(d ) Co r r ec t mech a n i sm and virtual displacements
M3 M2ndashL3
M2 M1ndashL3
Figure 530 Plastic analysis of a built-in beam
Mp
MpMp
q
L05L
0586L
(a) Propped cantilever (c) First mechanism
(d) Second mechanism(b) Correct mechanism
0583L
Figure 531 Plastic analysis of a propped cantilever
In-plane bending of beams 183
56 Strength design of beams
561 Elastic design of beams
5611 General
For the elastic method of designing a beam for the strength limit state the strengthdesign loads are first obtained by multiplying the nominal loads (dead imposedor wind) by the appropriate load combination factors (see Section 156) The dis-tributions of the design bending moments and shear forces in the beam under thestrength design load combinations are determined by an elastic bending analysis(when the structure is statically indeterminate) or by statics (when it is staticallydeterminate) EC3 also permits a moment redistribution to be made in each ade-quately braced Class 1 or Class 2 span of a continuous beam of up to 15 of thespanrsquos peak elastic moment provided equilibrium is maintained The beam mustbe designed to be able to resist the design moments and shears as well as anyconcentrated forces arising from the factored loads and their reactions
The processes of checking a specified member or of designing an unknownmember for the design actions are summarised in Figure 532 When a specified
Check serviceability
Guess fy and sectionclassification
Check fy andclassification
Select trial section(usually for moment)
Check bracingCheck shearCheck shear and bendingCheck bearing
Analysis for MEd VEd REd
DesigningChecking
Section Length and loading
Find fy Classify section
Check moment
Figure 532 Flow chart for the elastic design of beams
184 In-plane bending of beams
member is to be checked then the cross-section should first be classified asClass 1 2 3 or 4 This allows the effective section modulus to be determinedand the design section moment resistance to be compared with the maximumdesign moment Following this the lateral bracing should be checked then theweb shear resistance and then combined bending and shear should be checkedat any cross-sections where both the design moment and shear are high Finallythe web bearing resistance should be checked at reactions and concentrated loadpoints
When the member size is not known then a target value for the effective sectionmodulus may be determined from the maximum design moment and a trial sectionchosen whose effective section modulus exceeds this target The remainder of thedesign process is then the same as that for checking a specified member Usuallythe moment resistance governs the design but when it doesnrsquot then an iterativeprocess may need to be followed as indicated in Figure 532
The following sub-sections describe each of the EC3 checking processes sum-marised in Figure 532 A worked example of their application is given inSection 51215
5612 Section classification
The moment shear and concentrated load bearing resistances of beams whoseplate elements are slender may be significantly influenced by local buckling con-siderations (Chapter 4) Because of this beam cross-sections are classified asClass 1 2 3 or 4 depending on the ability of the elements to resist local buckling(Section 472)
Class 1 sections are unaffected by local buckling and are able to develop andmaintain their fully plastic resistances until a collapse mechanism forms Class 2sections are able to form a first plastic hinge but local buckling prevents subsequentmoment redistribution Class 3 sections are able to reach the yield stress but localbuckling prevents full plastification of the cross-section Class 4 sections have theirresistances reduced below their first yield resistances by local buckling effects
Sections are classified by comparing the slendernessλ= (ct)radic(fy235)of eachcompression element with the appropriate limits of Table 52 of EC3 These limitsdepend on the way in which the longitudinal edges of the element are supported(either one edge supported as for the flange outstand of an I-section or two edgessupported as for the internal flange element of a box section) the bending stressdistribution (uniform compression as in a flange or varying stresses as in a web)and the type of section as indicated in Figures 533 and 429
The section classification is that of the lowest classification of its elements withClass 1 being the highest possible and Class 4 the lowest All UBrsquos in S275 steelare Class 1 all UCrsquos are Class 1 or 2 but welded I-section members may also beClass 3 or 4
Worked examples of section classification are given in Sections 51215 and492ndash494
In-plane bending of beams 185
cw
cf1 cf1
cf1cf1 cf 2 cf1
cf 2
cw cw cw
Section description
Section andelement widths
Welded box RHS Hot-rolled UBUC
Welded PWG
Flangeoutstandcf1
Class 1Class 2Class 3
Class 1Class 2Class 3
ndashndashndash
ndashndashndash
91014
7283
124
910143338427283
124
3338427283
124
Class 1Class 2Class 3
Flange supportedalong bothedges cf 2
Web cw
Figure 533 Local buckling limits λ = (ct)ε for some beam sections
5613 Checking the moment resistance
For the moment resistance check of a beam that has adequate bracing against lateralbuckling the inequality
MEd le McRd (539)
must be satisfied in which MEd is the maximum design moment and
McRd = WfyγM0 (540)
is the design section moment resistance in which W is the appropriate sectionmodulus fy the nominal yield strength for the section and γM0(=1) the partialfactor for section resistance
For Class 1 and Class 2 sections the appropriate section modulus is the plasticsection modulus Wpl so that
W = Wpl (541)
For Class 3 sections the appropriate section modulus may simply be taken as theminimum elastic section modulus Welmin For Class 4 sections the appropriatesection modulus is reduced below the elastic section modulus Welmin as discussedin Section 472
186 In-plane bending of beams
Worked examples of checking the moment resistance are given inSections 51215 and 492ndash494 and a worked example of using the momentresistance check to design a suitable section is given in Section 51216
5614 Checking the lateral bracing
Beams with insufficient lateral bracing must also be designed to resist lateralbuckling The design of beams against lateral buckling is discussed in Section 69It is assumed in this chapter that there is sufficient bracing to prevent lateralbuckling While the adequacy of bracing for this purpose may be checked bydetermining the lateral buckling moment resistance as in Section 69 EC3 alsogives simpler (and often conservative) rules for this purpose For example lateralbuckling may be assumed to be prevented if the geometrical slenderness of eachsegment between restraints satisfies
Lc
if zle 3756
kc
McRd
MyEd
radic235
fy(542)
in which if z is the radius of gyration about the z axis of an equivalent compressionflange kc is a correction factor which allows for the moment distribution McRd isthe design resistance of a fully braced segment and MyEd is the maximum designmoment in the segment
A worked example of checking the lateral bracing is given in Section 51215
5615 Checking the shear resistance
For the shear resistance check the inequality
VEd le VcRd (543)
must be satisfied in which VEd is the maximum design shear force and VcRd thedesign shear resistance
For a stocky web with dwtw le 72radic(235fy) and for which the elastic shear
stress distribution is approximately uniform (as in the case of an equal flangedI-section) the uniform shear resistance VcRd is usually given by
VcRd = Av(fyradic
3) (544)
in which fyradic
3 is the shear yield stress τy (Section 131) and Av is the shear areaof the web defined in Clause 626(2) of EC3 All the webs of UBrsquos and UCrsquosin S275 steel satisfy dwtw le 72
radic(235fy) A worked example of checking the
uniform shear capacity of a compact web is given in Section 51215
In-plane bending of beams 187
For a stocky web with dwtw le 72radic(235fy) and with a non-uniform elastic
shear stress distribution (such as an unequal flanged I-section) the inequality
τEd le fyradic
3 (545)
must be satisfied in which τEd is the maximum design shear stress induced in theweb by VEd as determined by an elastic shear stress analysis (Section 54)
Webs for which dwtw ge 72radic(235fy) have their shear resistances reduced by
local buckling effects as discussed in Section 474 The shear design of thin websis governed by EC3-1-5 [25] Worked examples of checking the shear capacity ofsuch webs are given in Sections 495 and 496
5616 Checking for bending and shear
High shear forces may reduce the ability of a web to resist bending moment asdiscussed in Section 45 To allow for this Class 1 and Class 2 beams undercombined bending and shear in addition to satisfying equations 539 and 543must also satisfy
MEd le McRd
1 minus
(2VEd
VplRdminus 1
)2
(546)
in which VEd is the design shear at the cross-section being checked and VplRd isthe design plastic shear resistance This condition need only be considered whenVEd gt 05VplRd as indicated in Figure 432 It therefore need only be checked atcross-sections under high bending and shear such as at the internal supports ofcontinuous beams Even so many such beams have VEd le 05VplRd and so thischeck for combined bending and shear rarely governs
5617 Checking the bearing resistance
For the bearing check the inequality
FEd le FRd (547)
must be satisfied at all supports and points of concentrated load In this equationFEd is the design-concentrated load or reaction and FRd the bearing resistance ofthe web and its load-bearing stiffeners if any The bearing design of webs andload-bearing stiffeners is governed by EC3-1-5 [25] and discussed in Sections 46and 476 Stocky webs often have sufficient bearing resistance and do not requirethe addition of load-bearing stiffeners
A worked example of checking the bearing resistance of a stocky unstiffenedweb is given in Section 51215 while a worked example of checking the bearingresistance of a slender web with load-bearing stiffeners is given in Section 499
188 In-plane bending of beams
562 Plastic design of beams
In the plastic method of designing for the strength limit state the distributions ofbending moment and shear force in a beam under factored loads are determinedby a plastic bending analysis (Section 555) Generally the material propertiesfor plastic design should be limited to ensure that the stressndashstrain curve has anadequate plastic plateau and exhibits strain-hardening so that plastic hinges canbe formed and maintained until the plastic collapse mechanism is fully developedThe use of the plastic method is restricted to beams which satisfy certain limitationson cross-section form and lateral bracing (additional limitations are imposed onmembers with axial forces as discussed in Section 7242)
EC3 permits only Class 1 sections which are symmetrical about the axis per-pendicular to the axis of bending to be used at plastic hinge locations in order toensure that local buckling effects (Sections 42 and 43) do not reduce the ability ofthe section to maintain the full plastic moment until the plastic collapse mechanismis fully developed while Class 1 or Class 2 sections should be used elsewhereThe slenderness limits for Class 1 and Class 2 beam flanges and webs are shownin Figure 533
The spacing of lateral braces is limited in plastic design to ensure that inelasticlateral buckling (see Section 63) does not prevent the complete development ofthe plastic collapse mechanism EC3 requires the spacing of braces at both endsof a plastic hinge segment to be limited A conservative approximation for thespacing limit for sections with htf le 40
radic(235fy) is given by
Lstable le 35izradic(235fy) (548)
in which iz is the radius of gyration about the minor z axis although higher valuesmay be calculated by allowing for the effects of moment gradient
The resistance of a beam designed plastically is based principally on its plasticcollapse load Qult calculated by plastic analysis using the full plastic moment Mp
of the beam The plastic collapse load must satisfy QEd le Qult in which QEd is thecorresponding design load
The shear capacity of a beam analysed plastically is based on the fully plasticshear capacity of the web (see Section 4321) Thus the maximum shear VEd
determined by plastic analysis under the design loads is limited to
VEd le VplRd = Avfyradic
3 (549)
EC3 also requires web stiffeners to be provided near all plastic hinge locationswhere an applied load or reaction exceeds 10 of the above limit
In-plane bending of beams 189
The ability of a beam web to transmit concentrated loads and reactions isdiscussed in Section 462 The web should be designed as in Section 476 usingthe design loads and reactions calculated by plastic analysis
A worked example of the plastic design of a beam is given in Section 51217
57 Serviceability design of beams
The design of a beam is often governed by the serviceability limit state forwhich the behaviour of the beam should be so limited as to give a high proba-bility that the beam will provide the serviceability necessary for it to carry out itsintended function The most common serviceability criteria are associated withthe stiffness of the beam which governs its deflections under load These mayneed to be limited so as to avoid a number of undesirable situations an unsightlyappearance cracking or distortion of elements fixed to the beam such as claddinglinings and partitions interference with other elements such as crane girders orvibration under dynamic loads such as traffic wind or machinery loads
Aserviceability design should be carried out by making appropriate assumptionsfor the load types combinations and levels for the structural response to loadand for the serviceability criteria Because of the wide range of serviceability limitstates design rules are usually of limited extent The rules given are often advisoryrather than mandatory because of the uncertainties as to the appropriate values tobe used These uncertainties result from the variable behaviour of real structuresand what is often seen to be the non-catastrophic (in terms of human life) natureof serviceability failures
Serviceability design against unsightly appearance should be based on the totalsustained load and so should include the effects of dead and long-term imposedloads The design against cracking or distortion of elements fixed to a beam shouldbe based on the loads imposed after their installation and will often include theunfactored imposed load or wind load but exclude the dead load The designagainst interference may need to include the effects of dead load as well as ofimposed and wind loads Where imposed and wind loads act simultaneously itmay be appropriate to multiply the unfactored loads by combination factors suchas 08 The design against vibration will require an assessment to be made of thedynamic nature of the loads
Most stiffness serviceability limit states are assessed by using an elastic anal-ysis to predict the static deflections of the beam For vibration design it may besufficient in some cases to represent the dynamic loads by static equivalents andcarry out an analysis of the static deflections More generally however a dynamicelastic analysis will need to be made
The serviceability deflection limits depend on the criterion being used To avoidunsightly appearance a value of L200 (L180 for cantilevers) might be used aftermaking allowances for any pre-camber Values as low as L360 have been usedfor design against the cracking of plaster finishes while a limit of L600 has been
190 In-plane bending of beams
used for crane gantry girders Avalue of H300 has been used for horizontal storeydrift in buildings under wind load
A worked example of the serviceability design is given in Section 51218
58 Appendix ndash bending stresses in elastic beams
581 Bending in a principal plane
When a beam bends in the xz principal plane plane cross-sections rotate asshown in Figure 54 so that sections δx apart become inclined to each other atminus(d2wdx2)δx where w is the deflection in the z principal direction as shownin Figure 55 and δx the length along the centroidal axis between the two cross-sections The length between the two cross-sections at a distance z from the axisis greater than δx by z(minusd2wd x2)δx so that the longitudinal strain is
ε = minuszd2w
d x2
The corresponding tensile stress σ = Eε is
σ = minusEzd2w
d x2(550)
which has the stress resultants
N =int
AσdA = 0
sinceint
A zdA = 0 for centroidal axes (see Section 59) and
My =int
Aσ zdA
whence
My = minusEIyd2w
dx2(52)
sinceint
A z2dA = Iy (see Section 59) If this is substituted into equation 550 thenthe bending tensile stress σ is obtained as
σ = Myz
Iy (53)
In-plane bending of beams 191
582 Alternative formulation for biaxial bending
When bending moments My1 Mz1 acting about non-principal axes y1 z1 causecurvatures d2w1dx2 d2v1dx2 in the non-principal planes xz1 xy1 the stress σat a point y1 sz1 is given by
σ = minusEz1d2w1dx2 minus Ey1d2v1dx2 (551)
and the moment resultants of these stresses are
andMy1 = minusEIy1d2w1dx2 minus EIy1z1d2v1dx2
Mz1 = EIy1z1d2w1dx2 + EIz1d2v1dx2
(552)
These can be rearranged as
and
minusEd2w1
dx2= My1Iz1 + Mz1Iy1z1
Iy1Iz1 minus I2y1z1
Ed2v1
dx2= My1Iy1z1 + Mz1Iy1
Iy1Iz1 minus I2y1z1
(553)
which are much more complicated than the principal plane formulations ofequations 59
If equations 553 are substituted then equation 551 can be expressed as
σ =(
My1Iz1 + Mz1Iy1z1
Iy1Iz1 minus I2y1z1
)z1 minus
(My1Iy1z1 + Mz1Iy1
Iy1Iz1 minus I2y1z1
)y1 (554)
which is much more complicated than the principal plane formulation ofequation 512
59 Appendix ndash thin-walled section properties
The cross-section properties of many steel beams can be analysed with suffi-cient accuracy by using the thin-walled assumption demonstrated in Figure 534in which the cross-section is replaced by its mid-thickness line While the thin-walled assumption may lead to small errors in some properties due to junction andthickness effects its consistent use will avoid anomalies that arise when a moreaccurate theory is combined with thin-walled theory as might be the case in thedetermination of shear flows The small errors are typically of the same order asthose introduced by the common practice of ignoring the fillets or corner radii ofhot-rolled sections
192 In-plane bending of beams
t2
t1
b1 times t1
b2 times t2
b3 times t3
b1
b 2
b3
t 3
aa
(a) Actual cross-section (b) Thin-walled approximation
Figure 534 Thin-walled cross-section assumption
The area A of a thin-walled section composed of a series of thin rectangles bn
wide and tn thick (bn tn) is given by
A =int
AdA =
sumn
An (555)
in which An = bntn This approximation may make small errors of the order of(tb)2 at some junctions
The centroid of a cross-section is defined by
intA
ydA =int
AzdA = 0
and for a thin-walled section these conditions become
sumn
Anyn =sum
n
Anzn = 0 (556)
in which yn zn are the centroidal distances of the centre of the rectangular ele-ment as demonstrated in Figure 535a The position of the centroid can be foundby adopting any convenient initial set of axes yi zi as shown in Figure 535aso that
yn = yin minus yic
zn = zin minus zic
In-plane bending of beams 193
A2 = b2 times t2 A2 = b2 times t2
z1z1
y1
zi
y12y12 yic
yi2
yi
z12
zi2
zic
132
z12
y1C 0
2
C
2
(b) Second moments of area(a) Centroid
Figure 535 Calculation of cross-section properties
When these are substituted into equation 556 the centroid position is found from
yic =sum
AnyinA
zic =sum
AnzinA
(557)
The second moments of area about the centroidal axes are defined by
Iy =int
Az2 dA =
sum(Anz2
n + In sin2 θn)
Iz =int
Ay2 dA =
sum(Any2
n + In cos2 θn)
Iyz =int
Ayz dA =
sum(Anynzn + In sin θn cos θn)
(558)
in which θn is always the angle from the y axis to the rectangular element (positivefrom the y axis to the z axis) as shown in Figure 535b and
In = b3ntn12 (559)
These thin-walled approximations ignore small terms bnt3n12 while the use of
the angle θn adjusts for the inclination of the element to the y z axes
194 In-plane bending of beams
The principal axis directions y z are defined by the condition that
Iyz = 0 (560)
These directions can be found by considering a rotation of the centroidal axes fromy1 z1 to y2 z2 through an angle α as shown in Figure 536a The y2 z2 coordinatesof any point are related to its y1 z1 coordinates as demonstrated in Figure 536a by
y2 = y1 cosα + z1 sinαz2 = minusy1 sinα + z1 cosα
(561)
The second moments of area about the y2 z2 axes are
Iy2 =int
Az2
2 dA = Iy1 cos2 α minus 2Iy1z1 sinα cosα + Iz1 sin2 α
Iz2 =int
Ay2
2 dA = Iy1 sin2α + 2Iy1z1 sinα cosα + Iz1 cos2 α
Iy2z2 =int
Ay2z2 dA = 1
2(Iy1 minus Iz1) sin 2α + Iy1z1 cos 2α
(562)
The principal axis condition of equation 560 requires Iy2z2 = 0 so that
tan 2α = minus2Iy1z1(Iy1 minus Iz1) (563)
In this book y and z are reserved exclusively for the principal centroidal axes
C
(y1 z1)
z1z2
y2
z2
z1
z1 cos
Iy
1z
IyIz
(Iy1 ndashIy1z1)
(Iy1Iy1z1)
Iyz
y1 sin
2
z1 sin y1 cos
y2
y1
(y2 z2)
(a) Rotation of axes (b) Mohrrsquos circle construction
Figure 536 Calculation of principal axis properties
In-plane bending of beams 195
The principal axis second moments of area can be obtained by using this valueof α in equation 562 whence
Iy = 1
2(Iy1 + Iz1)+ (Iy1 minus Iz1)2 cos 2α
Iz = 1
2(Iy1 + Iz1)minus (Iy1 minus Iz1)2 cos 2α
(564)
These results are summarised by the Mohrrsquos circle constructed in Figure 536b onthe diameter whose ends are defined by (Iy1 Iy1z1) and (Iz1 minusIy1z1) The anglesbetween this diameter and the horizontal axis are equal to 2α and (2α+ 180)while the intersections of the circle with the horizontal axis define the principalaxis second moments of area Iy Iz
Worked examples of the calculation of cross-section properties are given inSections 5121ndash5124
510 Appendix ndash shear stresses in elastic beams
5101 Solid cross-sections
A solid cross-section beam with a shear force Vz parallel to the z principal axis isshown in Figure 59a It is assumed that the resulting shear stresses also act parallelto the z axis and are constant across the width b of the section The correspondinghorizontal shear stresses τzx = τv are in equilibrium with the horizontal bendingstresses σ For the element b times δz times δx shown in Figure 59b this equilibriumcondition reduces to
bpartσ
partxδxδz + part(τvb)
partzδzδx = 0
whence
τv = minus 1
b2
int z2
zT
bpartσ
partxdz (565)
which satisfies the condition that the shear stress is zero at the unstressed boundaryzT The bending normal stresses σ are given by
σ = Myz
Iy
and the shear force Vz is
Vz = dMy
dx
196 In-plane bending of beams
and so
partσ
partx= Vzz
Iy (566)
Substituting this into equation 565 leads to
τv = minus Vz
Iyb2
int z2
zT
bz dz (513)
5102 Thin-walled open cross-sections
A thin-walled open cross-section beam with a shear force Vz parallel to the z prin-cipal axis is shown in Figure 511a It is assumed that the resulting shear stressesact parallel to the walls of the section and are constant across the wall thicknessThe corresponding horizontal shear stresses τv are in equilibrium with the hori-zontal bending stresses σ For the element t times δs times δx shown in Figure 511b thisequilibrium condition reduces to
partσ
partxδxtδs + part(τvt)
partxδsδx = 0
and substitution of equation 566 and integration leads to
τvt = minusVz
Iy
int s
0zt ds (516)
which satisfies the condition that the shear stress is zero at the unstressed boundarys = 0
The direction of the shear stress can be determined from the sign of the right-hand side of equation 516 If this is positive then the direction of the shear stressis the same as the positive direction of s and vice versa The direction of the shearstress may also be determined from the direction of the corresponding horizontalshear force required to keep the out-of-balance bending force in equilibrium asshown for example in Figure 513a Here a horizontal force τvtf δx in the positive xdirection is required to balance the resultant bending force δσ δA (due to the shearforce Vz = dMydx) and so the direction of the corresponding flange shear stressis from the tip of the flange towards its centre
5103 Multi-cell thin-walled closed sections
The shear stress distributions in multi-cell closed sections can be determined byextending the method discussed in Section 544 for single-cell closed sections Ifthe section consists of n junctions connected by m walls then there are (m minus n + 1)independent cells In each wall there is an unknown circulating shear flow τvct
In-plane bending of beams 197
There are (m minus n + 1) independent cell warping continuity conditions of the type(see equation 527)
sumcell
intwall(τG) ds = 0
and (n minus 1) junction equilibrium equations of the typesumjunction
(τ t) = 0
These equations can be solved simultaneously for the m circulating shearflows τvct
511 Appendix ndash plastic analysis of beams
5111 Full plastic moment
The fully plastic stress distribution in a section bent about a y axis of symmetryis antisymmetric about that axis as shown in Figure 537a for a channel sectionexample The plastic neutral axis which divides the cross-section into two equalareas so that there is no axial force resultant of the stress distribution coincides withthe axis of symmetry The moment resultant of the fully plastic stress distribution is
Mpy = fysum
T
Anzn minus fysum
C
Anzn (567)
in which the summationssum
T sum
C are carried out for the tension and com-pression areas of the cross-section respectively Note that zn is negative forthe compression areas of the channel section so that the two summations in
yy
z
z
T
C
T
CT
C
(a) Bending about an axis of symmetry
(b) Bending of a monosymmetric section in a plane of symmetry
(c) Bending of an asymmetric section
b2 times t
b times tf
df times twb times tf
b times tf
df times tw b1 times t
y1 Mpy1
Mpz1
Mp
zp
zp
z1
fyfyfy fy fy
fy
Figure 537 Full plastic moment
198 In-plane bending of beams
equation 567 are additive The plastic section modulus Wply = Mpyfy is obtainedfrom equation 567 as
Wply =sum
T
Anzn minussum
C
Anzn (568)
A worked example of the determination of the full plastic moment of a doublysymmetric I-section is given in Section 51212
The fully plastic stress distribution in a monosymmetric section bent in a planeof symmetry is not antisymmetric as can be seen in Figure 537b for a channelsection example The plastic neutral axis again divides the cross-section into twoequal areas so that there is no axial force resultant but the neutral axis no longerpasses through the centroid of the cross-section The fully plastic moment Mpy
and the plastic section modulus Wply are given by equations 567 and 568 Forconvenience any suitable origin may be used for computing zn Aworked exampleof the determination of the full plastic moment of a monosymmetric tee-section isgiven in Section 51213
The fully plastic stress distribution in an asymmetric section is also not antisym-metric as can be seen in Figure 537c for an angle section example The plasticneutral axis again divides the cross-section into two equal areas The direction ofthe plastic neutral axis is perpendicular to the plane in which the moment resul-tant acts The full plastic moment is given by the vector addition of the momentresultants about any convenient y1 z1 axes so that
Mp = radic(M 2
py1 + M 2pz1) (569)
where Mpy1 is given by an equation similar to equation 567 and Mpz1 isgiven by
Mpz1 = minusfysum
T
Any1n + fysum
C
Any1n (570)
The inclination α of the plastic neutral axis to the y1 axis (and of the plane of thefull plastic moment to the zx plane) can be obtained from
tan α = Mpz1Mpy1 (571)
5112 Built-in beam
51121 General
The built-in beam shown in Figure 530a has two redundancies and so three plastichinges must form before it can be reduced to a collapse mechanism In this casethe only possible plastic hinge locations are at the support points 1 and 4 and at theload points 2 and 3 and so there are two possible mechanisms one with a hinge at
In-plane bending of beams 199
point 2 (Figure 530c) and the other with a hinge at point 3 (Figure 530d) Threedifferent methods of analysing these mechanisms are presented in the followingsub-sections
51122 Equilibrium equation solution
The equilibrium conditions for the beam express the relationships between theapplied loads and the moments at these points These relationships can be obtainedby dividing the beam into the segments shown in Figure 530b and expressing theend shears of each segment in terms of its end moments so that
V12 = (M2 minus M1)(L3)
V23 = (M3 minus M2)(L3)
V34 = (M4 minus M3)(L3)
For equilibrium each applied load must be equal to the algebraic sum of the shearsat the load point in question and so
Q = V12 minus V23 = 1
L(minus3M1 + 6M2 minus 3M3)
2Q = V23 minus V34 = 1
L(minus3M2 + 6M3 minus 3M4)
which can be rearranged as
4QL = minus 6M1 + 9M2 minus 3M4
5QL = minus 3M1 + 9M3 minus 6M4
(572)
If the first mechanism chosen (incorrectly as will be shown later) is that withplastic hinges at points 1 2 and 4 (see Figure 530c) so that
minusM1 = M2 = minusM4 = Mp
then substitution of this into the first of equations 572 leads to 4QL = 18Mp
and so
Qult le 45MpL
If this result is substituted into the second of equations 572 then
M3 = 15 Mp gt Mp
200 In-plane bending of beams
and the plasticity condition is violated The statical method can now be used toobtain a lower bound by reducing Q until M3 = Mp whence
Q = 45Mp
L
M3
Mp= 3Mp
L
and so
3MpL lt Qult lt 45MpL
The true collapse load can be obtained by assuming a collapse mechanismwith plastic hinges at point 3 (the point of maximum moment for the previousmechanism) and at points 1 and 4 For this mechanism
minusM1 = M3 = minusM4 = Mp
and so from the second of equations 572
Qult le 36MpL
If this result is substituted into the first of equations 572 then
M2 = 06Mp lt Mp
and so the plasticity condition is also satisfied Thus
Qult = 36MpL
51123 Graphical solution
A collapse mechanism can be analysed graphically by first plotting the known freemoment diagram (for the applied loading on a statically determinate beam) andthen the reactant bending moment diagram (for the redundant reactions) whichcorresponds to the collapse mechanism
For the built-in beam (Figure 538a) the free moment diagram for a simplysupported beam is shown in Figure 538c It is obtained by first calculating theleft-hand reaction as
RL = (Q times 2L3 + 2Q times L3)L = 4Q3
In-plane bending of beams 201
(+)
L3 L3 L3
4QL9 5QL9
5QL9
321 4
(+)
(-) (-)
(-)Mp MpMp
Mp
(d) Reactant moment diagram(b) Collapse moment diagram
(c) Free moment diagram(a) Built-in beam
Q 2Q
Figure 538 Graphical solution for the plastic collapse of a built-in beam
and then using this to calculate the moments
M2 = (4Q3)times L3 = 4QL9
and
M3 = (4Q3)times 2L3 minus QL3 = 5QL9
Both of the possible collapse mechanisms (Figure 529c and d) require plastichinges at the supports and so the reactant moment diagram is one of uniformnegative bending as shown in Figure 538d When this is combined with the freemoment diagram as shown in Figure 538b it becomes obvious that there must bea plastic hinge at point 3 and not at point 2 and that at collapse
2Mp = 5QultL9
so that
Qult = 36MpL
as before
51124 Virtual work solution
The virtual work principle [4 7 8] can also be used to analyse each mechanismFor the first mechanism shown in Figure 530c the virtual external work done
202 In-plane bending of beams
by the forces Q 2Q during the incremental virtual displacements defined by theincremental virtual rotations δθ and 2δθ shown is given by
δW = Q times (δθ times 2L3)+ 2Q times (δθ times L3) = 4QLδθ3
while the internal work absorbed at the plastic hinge locations during theincremental virtual rotations is given by
δU = Mp times 2δθ + Mp times (2δθ + δθ)+ Mp times δθ = 6Mpδθ
For equilibrium of the mechanism the virtual work principle requires
δW = δU
so that the upper-bound estimate of the collapse load is
Qult le 45MpL
as in Section 51122Similarly for the second mechanism shown in Figure 530d
δW = 5QLδθ3
δU = 6Mpδθ
so that
Qult le 36MpL
once more
5113 Propped cantilever
51131 General
The exact collapse mechanisms for beams with distributed loads are not always aseasily obtained as those for beams with concentrated loads only but sufficientlyaccurate solutions can usually be found without great difficulty Three differentmethods of determining these mechanisms are presented in the following sub-sections for the propped cantilever shown in Figure 531
In-plane bending of beams 203
51132 Equilibrium equation solution
If the plastic hinge is assumed to be at the mid-span of the propped cantilever asshown in Figure 531c (the other plastic hinge must obviously be located at thebuilt-in support) then equilibrium considerations of the left and right segments ofthe beam allow the left and right reactions to be determined from
RLL2 = 2Mp + (qL2)L4
and
RRL2 = Mp + (qL2)L4
while for overall equilibrium
qL = RL + RR
Thus
qL = (4MpL + qL4)+ (2MpL + qL4)
so that
qult le 12MpL2
The left-hand reaction is therefore given by
RL = 4MpL + (12MpL2)L4 = 7MpL
The bending moment variation along the propped cantilever is therefore given by
M = minusMp + (7MpL)x minus (12MpL2)x22
which has a maximum value of 25Mp24gtMp at x = 7L12 Thus the lowerbound is given by
qult ge (12MpL2)(2524) asymp 1152MpL
2
and so
1152Mp
L2lt qult lt
12Mp
L2
If desired a more accurate solution can be obtained by assuming that the hinge islocated at a distance 7L12(= 0583L) from the built-in support (ie at the point
204 In-plane bending of beams
of maximum moment for the previous mechanism) whence
116567Mp
L2lt qult lt
116575Mp
L2
These are very close to the exact solution of
qult = (6 + 4radic
2)MpL2 asymp 116569MpL
2
for which the hinge is located at a distance (2 minus radic2)L asymp 05858L from the built-in
support
51133 Graphical solution
For the propped cantilever of Figure 531 the parabolic free moment diagramfor a simply supported beam is shown in Figure 539c All the possible collapsemechanisms have a plastic hinge at the left-hand support and a frictionless hingeat the right-hand support so that the reactant moment diagram is triangular asshown in Figure 539d A first attempt at combining this with the free momentdiagram so as to produce a trial collapse mechanism is shown in Figure 539b Forthis mechanism the interior plastic hinge occurs at mid-span so that
Mp + Mp2 = qL28
which leads to
q = 12MpL2
as before
q
LRL RR
(+)
Mp ( + ) qL 8
L 2 L 2
Mp
Mp 22
qL 82
Mp
MpMmax
(-)( - )
(d) Reactant moment diagram
(a) Propped cantilever
(b) T ri a l c ol l ap s e m o m e n t d ia gram
(c) Free moment diagram
Figure 539 Graphical analysis of the plastic collapse of a propped cantilever
In-plane bending of beams 205
Figure 539b indicates that in this case the maximum moment occurs to the rightof mid-span and is greater than Mp so that the solution above is an upper boundA more accurate solution can be obtained by trial and error by increasing theslope of the reactant moment diagram until the moment of the left-hand supportis equal to the interior maximum moment as indicated by the dashed line inFigure 539b
51134 Virtual work solution
For the first mechanism for the propped cantilever shown in Figure 531c forvirtual rotations δθ
δW = qL2δθ4
δU = 3Mpδθ
and
qult le 12MpL2
as before
512 Worked examples
5121 Example 1 ndash properties of a plated UB section
Problem The 610 times 229 UB 125 section shown in Figure 540a is strengthened bywelding a 300 mm times 20 mm plate to each flange Determine the section propertiesIy and Wely
Solution Because of the symmetry of the section the centroid of the plated UBis at the web centre Adapting equation 558
Iy = 98 610 + 2 times 300 times 20 times [(6122 + 20)2]2104 = 218 500 cm4
Wely = 218 500[(6122 + 2 times 20)(2 times 10)] = 6701 cm3
5122 Example 2 ndash properties of a tee-section
Problem Determine the section properties Iy and Wely of the welded tee-sectionshown in Figure 540b
206 In-plane bending of beams
80
5
8010
y
zz
y
zpzi
(a) 610 times 229 UB 125
Iy = 98 610 cm4
Wply = 3676 cm3
(b) Tee-section
6122
1 1 9
196229
Figure 540 Worked examples 1 2 and 18
Solution Using the thin-walled assumption and equations 557 and 558
zi = (80 + 52)times 10 times (80 + 52)2
(80 times 5)+ (80 + 52)times 10= 278 mm
Iy = (80 times 5)times 2782 + (8253 times 1012)+ (825 times 10)
times (8252 minus 278)2 mm4
= 9263 cm4
Wely = 9263(825 minus 278)10 = 1693 cm3
Alternatively the equations of Figure 56 may be used
5123 Example 3 ndash properties of an angle section
Problem Determine the properties A Iy1 Iz1 Iy1z1 Iy Iz and α of the thin-walledangle section shown in Figure 541c
Solution Using the thin-walled assumption the angle section is replaced by tworectangles 1425(= 150 minus 152)times 15 and 825(= 90 minus 152)times 15 as shown in
In-plane bending of beams 207
37 mm
967 mm
188 mm
Deflection of tip centroid
(c) Actual cross-section
90 mm
15 mm
15 mm
6 kN15 m
150 90 15unequal angle
A
C
S
6 kN
z
y
93 mm
100 mm
y1
z1
150
mm
Iy = 8426 cm4
Iz = 1205 cm4
=1982o
(a) Cantilever
(b) Section properties(d) Thin-walled section (deflections to different scale)
825 times 15
times times 1425 times 15
Figure 541 Worked examples 3 and 5
Figure 541d If the initial origin is taken at the intersection of the legs and theinitial yi zi axes parallel to the legs then using equations 555 and 557
A = (1425 times 15)+ (825 times 15) = 3375 mm2
yic = (1425 times 15)times 0 + (825 times 15)times (minus8252)3375
= minus151 mm
zic = (1425 times 15)times (minus14252)+ (825 times 15)times 03375
= minus451 mm
Transferring the origin to the centroid and using equations 558
Iy1 = (1425 times 15)times (minus14252 + 451)2 + (14253 times 1512)times sin2 90
+ (825 times 15)times (451)2 + (8253 times 1512)times sin2 180
= 7596 times 106 mm4 = 7596 cm4
208 In-plane bending of beams
Iz1 = (1425 times 15)times (151)2 + (14253 times 1512)times cos2 90
+ (825 times 15)times (minus8252 + 151)2 + (8253 times 1512)times cos2 180
= 2035 times 106 mm4 = 2035 cm4
Iy1z1 = (1425 times 15)times (minus14252 + 451)times 151 + (14253 times 1512)
times sin 90 cos 90 + (825 times 15)times (451)times (minus8252 + 151)
+ (8253 times 1512)times sin 180 cos 180
= minus2303 times 106 mm4 = minus2303 cm4
Using equation 563
tan 2α = minus2 times (minus2303)(7596 minus 2035) = 08283
whence 2α= 3963 and α= 1982Using Figure 530b the Mohrrsquos circle centre is at
(Iy1 + Iz1)2 = (7596 + 2035)2 = 4816 cm4
and the Mohrrsquos circle diameter is
radic(Iy1 minus Iz1)2 + (2Iy1z1)
2 = radic(7596 minus 2035)2 + (2 times 2303)2= 7221 cm4
and so
Iy = 4816 + 72212 = 8426 cm4
Iz = 4816 minus 72212 = 1205 cm4
5124 Example 4 ndash properties of a zed section
Problem Determine the properties Iyl Izl Iylzl Iy Iz and α of the thin-walledZ-section shown in Figure 542aSolution
Iy1 = 2 times (75 times 10 times 752)+ (1503 times 512) = 9844 times 103 mm4
= 9844 cm4
In-plane bending of beams 209
150
50
875
62
5
6 6
6
125
75 100 100505075
1 0
1 0
yy
z
z1
6 6
1
y
z
121 2 3 4
6 5
1 25
4 3
= = 1400 cm4 600 cm4I Iz z
z
yCC
C
150
(b) Pi-section (c) Rectangular hollow section (All dimensions in mm)
(a) Z-section
Figure 542 Worked examples 4 9 10 and 11
Iz1 = 2 times (75 times 10 times 3752)+ 2 times (753 times 1012)
= 2813 times 103 mm4 = 2813 cm4
Iy1z1 = 75 times 10 times 375 times (minus75)+ 75 times 10 times (minus375)times 75
= minus4219 times 103 mm4 = minus4219 cm4
The centre of the Mohrrsquos circle (see Figure 537b) is at
Iy1 + Iz1
2=
[9844 + 2813
2
]= 6328 cm4
The diameter of the circle isradic[(9844 minus 2813)2 + (2 times 4219)2] = 10983 cm4
and so
Iy = (6328 + 109832) = 1182 cm4
Iz = (6328 minus 109832) = 837 cm4
tan 2α = 2 times 4219
(9844 minus 2813)= 1200
α = 251
5125 Example 5 ndash biaxial bending of an anglesection cantilever
Problem Determine the maximum elastic stress and deflection of the angle sectioncantilever shown in Figure 541 whose section properties were determined inSection 5123
210 In-plane bending of beams
Solution for maximum stress The shear centre (see Section 543) load of 6 kNacting in the plane of the long leg of the angle has principal plane components of
6 cos 1982 = 5645 kN parallel to the z axis and
6 sin 1982 = 2034 kN parallel to the y axis
as indicated in Figure 541d These load components cause principal axis momentcomponents at the support of
My = minus5645 times 15 = minus8467 kNm about the y axis and
Mz = +2034 times 15 = +3052 kNm about the z axis
The maximum elastic bending stress occurs at the point A (y1A = 151 mmz1A = minus974 mm) shown in Figure 541d The coordinates of this point can beobtained by using equation 561 as
y = 151 cos 1982 minus 974 sin 1982 = minus188 mm and
z = minus151 sin 1982 minus 974 cos 1982 = minus967 mm
The maximum stress at A can be obtained by using equation 512 as
σmax = (minus8467 times 106)times (minus967)
8426 times 104minus (3052 times 106)times (minus188)
1205 times 104
= 1469 Nmm2
The stresses at the other leg ends should be checked to confirm that the maximumstress is at A
Solution for maximum deflection The maximum deflection occurs at the tip of thecantilever Its components v and w can be calculated using QL33EI Thus
w = (5645 times 103)times (1500)3(3 times 210 000 times 8426 times 104) = 36 mm
and
v = (2034 times 103)times (1500)3(3 times 210 000 times 1205 times 104) = 90 mm
The resultant deflection can be obtained by vector addition as
δ = radic(362 + 902) = 97 mm
Using non-principal plane properties Problems of this type are sometimes incor-rectly analysed on the basis that the plane of loading can be assumed to be a
In-plane bending of beams 211
principal plane When this approach is used the maximum stress calculated forthe cantilever is
σmax = (6 times 103)times 1500 times (974)7596 times 104 = 115 Nmm2
which seriously underestimates the correct value by more than 20 The correctnon-principal plane formulations which are needed to solve the example in thisway are given in Section 582 The use of these formulations should be avoidedsince their excessive complication makes them very error-prone The much simplerprincipal plane formulations of equations 59ndash512 should be used instead
It should be noted that in this book y and z are reserved exclusively for theprincipal axis directions instead of also being used for the directions parallel tothe legs
5126 Example 6 ndash shear stresses in an I-section
Problem Determine the shear flow distribution and the maximum shear stress inthe I-section shown in Figure 512a
Solution Applying equation 516 to the left-hand half of the top flange
τvtf = minus(VzIy)
int sf
0(minusdf 2)tf dsf = (VzIy)df tf sf 2
which is linear as shown in Figure 512bAt the flange web junction (τvtf )12 = (VzIy)df tf b4 which is positive and
so the shear flow acts into the flangendashweb junction as shown in Figure 512aSimilarly for the right-hand half of the top flange (τvtf )23 = (VzIy)df tf b4 andagain the shear flow acts in to the flangendashweb junction
For horizontal equilibrium at the flangendashweb junction the shear flow from thejunction into the web is obtained from equation 520 as
(τvtw)24 = (VzIy)df tf b4 + (VzIy)df tf b4 = (VzIy)df tf b2
Adapting equation 516 for the web
τvtw = (VzIy)(df tf b2)minus (VzIy)
int sw
0(minusdf 2 + sw)twdsw
= (VzIy)(df tf b2)+ (VzIy)(df sw2 minus s2w2)tw
which is parabolic as shown in Figure 512b At the web centre (τvtw)df 2 =(VzIy)(df tf b2 + d2
f tw8) which is the maximum shear flowAt the bottom of the web
(τvtw)df = (VzIy)(df tf b2 + d2f tw2 minus d2
f tw2)
= (VzIy)(df tf b2) = (τvtw)24
which provides a symmetry check
212 In-plane bending of beams
The maximum shear stress is
(τv)df 2 = (VzIy)(df tf b2 + d2f tw8)tw
A vertical equilibrium check is provided by finding the resultant of the web shearflow as
Vw =int df
0(τvtw)dsw
= (VzIy)[df tf bsw2 + df tws2w4 minus s3
wtw6]df
0
= (VzIy)(d2f tf b2 + d3
f tw12)
= Vz
when Iy = d2f tf b2 + d3
f tw12 is substituted
5127 Example 7 ndash shear stress in a channel section
Problem Determine the shear flow distribution in the channel section shown inFigure 514
Solution Applying equation 515 to the top flange
τvtf = minus(VzIy)
int sf
0(minusdf 2)tf dsf = (VzIy)df tf sf 2
(τvtf )b = (VzIy)df tf b2
τvtw = (VzIy)(df tf b2 minusint sw
0(minusdf 2 + sw)twdsw)
= (VzIy)(df tf b2 + df swtw2 minus s2wtw2)
(τvtw)df = (VzIy)(df tf b2 + d2f tw2 minus d2
f tw2)
= (VzIy)(df tf b2) = (τvtf )b
which provides a symmetry checkA vertical equilibrium check is provided by finding the resultant of the web
flows as
Vw =int df
0(τvtw)dsw
= (VzIy)[df tf bsw2 + df s2wtw4 minus s3
wtw6]df
0
= (VzIy)(d2f tf b2 + d3
f tw12)
= Vz
when Iy = d2f tf b2 + d3
f tw12 is substituted
In-plane bending of beams 213
5128 Example 8 ndash shear centre of a channel section
Problem Determine the position of the shear centre of the channel section shownin Figure 514
Solution The shear flow distribution in the channel section was analysed inSection 5127 and is shown in Figure 514a The resultant flange shear forcesshown in Figure 514b are obtained from
Vf =int b
0(τvtf )dsf
= (VzIy)[df tf s2f 4]b
0
= (VzIy)(df tf b24)
while the resultant web shear force is Vw = Vz (see Section 5127)These resultant shear forces are statically equivalent to a shear force Vz acting
through the point S which is a distance
a = Vf df
Vw= d2
f tf b2
4Iyfrom the web and so
y0 = d2f tf b2
4Iy+ b2tf
2btf + df tw
5129 Example 9 ndash shear stresses in a pi-section
Problem The pi shaped section shown in Figure 542b has a shear force of 100kN acting parallel to the y axis Determine the shear stress distribution
Solution Using equation 523 the shear flow in the segment 12 is
(τh times 12)12 = minus100 times 103
1400 times 104
int s1
0(minus100 + s1)times 12 times ds1
whence (τh)12 = 0007143 times (100s1 minus s212) Nmm2
Similarly the shear flow in the segment 62 is
(τh times 6)62 = minus100 times 103
1400 times 104
int s6
0(minus50)times 6 times ds1
whence (τh)62 = 0007143 times (50s6) Nmm2
214 In-plane bending of beams
For horizontal equilibrium of the longitudinal shear forces at the junction 2 theshear flows in the segments 12 62 and 23 must balance (equation 520) and so
[(τh times 12)23]s3=0 = [(τh times 12)12]s1=50 + [(τh times 6)62]s6=200
= 0007143
[(100 times 50 minus 502
2
)times 12 + 50 times 200 times 6
]
= 0007143 times 8750
The shear flow in the segment 23 is
(τh times 12)23 = 0007143 times 12 times 8750 minus 100 times 103
1400 times 104
timesint s2
0(minus50 + s2)times 12 times ds2
whence (τh)23 = 0007143(8750 + 50s2 minus s222) Nmm2
As a check the resultant of the shear stresses in the segments 12 23 and 34 is
intτhtds = 2 times 0007143 times 12
int 50
0(100s1 minus s2
12)ds1
+ 0007143 times 12 timesint 100
0(8750 + 50s2 minus s2
22)ds2
= 0007143 times 122 times (100 times 5022 minus 5036)
+ (8750 times 100 + 50 times 10022 minus 10036)N= 100 kN
which is equal to the shear force
51210 Example 10 ndash shear centre of a pi-section
Problem Determine the position of the shear centre of the pi section shown inFigure 542b
Solution The resultant of the shear stresses in the segment 62 is
intτhtds = 0007143 times 6
int 200
0(50s6)times ds6
= 0007143 times 6 times 50 times 20022 N
= 4286 kN
The resultant of the shear stresses in the segment 53 is equal and opposite to this
In-plane bending of beams 215
The torque about the centroid exerted by the shear stresses is equal to the sumof the torques of the force in the flange 1234 and the forces in the webs 62 and 53Thus
intAτhtρds = (100 times 50)+ (2 times 4286 times 50) = 9286 kNmm
For the applied shear to be statically equivalent to the shear stresses it must exertan equal torque about the centroid and so
minus100 times z0 = 9286
whence z0 = minus929 mmBecause of symmetry the shear centre lies on the z axis and so y0 = 0
51211 Example 11 ndash shear stresses in a box section
Problem The rectangular box section shown in Figure 542c has a shear force of100 kN acting parallel to the y axis Determine the maximum shear stress
Solution If initially a slit is assumed at point 1 on the axis of symmetry so that(τ hot)1 = 0 then
(τhot)12 = minus100 times 103
6 times 106
int s1
0s1 times 12 times ds1 = minus01s2
1
(τhot)2 = minus01 times 502 = minus01 times 2500 Nmm
(τhot)23 = minus01 times 2500 minus 100 times 103
6 times 106
int s2
050 times 6 times ds2
= minus01(2500 + 50s2)
(τhot)3 = minus01(2500 + 50 times 150) = minus01 times 10 000 Nmm
(τhot)34 = minus01 times 10 000 minus 100 times 103
6 times 106
int s3
0(50 minus s3)times 6 times ds3
= minus01[10 000 + (50s3 minus s232)]
(τhot)4 = minus01[10 000 + (50 times 100 minus 10022)] = minus01 times 10 000 Nmm
= (τhot)3 which is a symmetry check
The remainder of the shear stress distribution can therefore be obtained bysymmetry
216 In-plane bending of beams
The circulating shear flow τ hct can be determined by adapting equation 530For this∮
τhods = 2 times (01 times 5033)12
+ 2 times (minus01)times (2500 times 150 + 50 times 15022)6
+ (minus01)times (10 000 times 100 + 50 times 10022 minus 10036)6
= (minus01)times (6 000 000)12 Nmm∮1
tds = 100
12+ 150
6+ 100
6+ 150
6
= 90012
Thus
τhct = (minus01)times (6 000 000)
12times 12
900
= 66667 Nmm
The maximum shear stress occurs at the centre of the segment 34 and adaptingequation 529
(τh)max = minus01 times (10 000 + 50 times 50 minus 5022)+ 66676 Nmm2
= minus764 Nmm2
51212 Example 12 ndash fully plastic moment of a plated UB
Problem A 610 times 229 UB 125 beam (Figure 540a) of S275 steel is strengthenedby welding a 300 mm times 20 mm plate of S275 steel to each flange Compare thefull plastic moments of the unplated and the plated sections
Unplated section
For tf = 196 mm fy = 265 Nmm2 EN10025-2
Mplu = fyWply = 265 times 3676 times 103 Nmm = 974 kNm
Plated section
For tp = 20 mm fy = 265 Nmm2 EN10025-2
Because of the symmetry of the cross-section the plastic neutral axis is at thecentroid Adapting equation 567
Mplp = 974 + 2 times 265 times (300 times 20)times (61222 + 202)106 kNm
= 1979 kNm asymp 2 times Mplu
In-plane bending of beams 217
51213 Example 13 ndash fully plastic moment of a tee-section
Problem The welded tee-section shown in Figure 540b is fabricated from twoS275 steel plates Compare the first yield and fully plastic moments
First yield moment The minimum elastic section modulus was determined inSection 5122 as
Wely = 1693 cm3
For tmax = 10 mm fy = 275 Nmm2 EN10025-2
Mely = fyWely = 275 times 1693 times 103 Nmm = 4656 kNm
Full plastic moment For the plastic neutral axis to divide the cross-section intotwo equal areas then using the thin-walled assumption leads to
(80 times 5)+ (zp times 10) = (825 minus zp)times 10
so that zp = (825 times 10 minus 80 times 5)(2 times 10) = 213 mmUsing equation 567
Mply = 275 times [(80 times 5)times 213 + (213 times 10)times 2132
+ (825 minus 213)2 times 102] Nmm
= 8117 kNm
Alternatively the section properties Wely and Wply can be calculated usingFigure 56 and used to calculate Mely and Mply
51214 Example 14 ndash plastic collapse of a non-uniform beam
Problem The two-span continuous beam shown in Figure 543a is a 610 times 229UB 125 of S275 steel with 2 flange plates 300 mm times 20 mm of S275 steelextending over a central length of 12 m Determine the value of the applied loadsQ at plastic collapse
Solution The full plastic moments of the beam (unplated and plated) weredetermined in Section 51212 as Mplu = 974 kNm and Mplp = 1979 kNm
The bending moment diagram is shown in Figure 543b Two plastic collapsemechanisms are possible both with a plastic hinge (minusMplp) at the central supportFor the first mechanism (Figure 543c) plastic hinges (Mplp) occur in the platedbeam at the load points but for the second mechanism (Figure 543d) plastichinges (Mplu) occur in the unplated beam at the changes of cross-section
218 In-plane bending of beams
Q Q
M M Mpu pp pu
Q 9 6 4
(a) C o n t in u ou s b ea m (le n g t h s
(b) Bending moment
Mpp Mpp
Mpu Mpu
96
48
96
36Mpp
(c) First me chanism
(d) Second mechanism
Mpp
times
36 12 48 3 61248
(+ ) ( + ) ( )ndash
m )i n
Figure 543 Worked example 14
For the first mechanism shown in Figure 543c inspection of the bendingmoment diagram shows that
Mplp + Mplp2 = Q1 times 964
so that
Q1 = (3 times 19792)times 496 = 1237 kN
For the second mechanism shown in Figure 543d inspection of the bendingmoment diagram shows that
Mplu + 36Mplp96 = (3648)times Q2 times 964
so that
Q2 = [974 + (3696)times 1979] times (4836)times 496 = 954 kN
which is less than Q1 = 1237 kNThus plastic collapse occurs at Q = 954 kN by the second mechanism
51215 Example 15 ndash checking a simply supported beam
Problem The simply supported 610 times 229 UB 125 of S275 steel shown inFigure 544a has a span of 60 m and is laterally braced at 15 m intervals Checkthe adequacy of the beam for a nominal uniformly distributed dead load of 60 kNmtogether with a nominal uniformly distributed imposed load of 70 kNm
In-plane bending of beams 219
4 1 5 = 6 0 m h
pl
bftf
tw
3 0 m
qD = 60 kNm qI = 70 kNm
QD = 40 kNQI
qD = 20 kNm
qI = 10 kNm
(a) Examples 15 17 and 18
(c) Example 16
bf = 229 mm A = 159 cm2
h = 6122 mm Iz = 3932 cm4
tf = 196 mm i z = 497 mmtw = 119 mm W y = 3676 cm3
r = 127 mm
(b) 610 229 UB 125
= 120 kN
times
Figure 544 Worked examples 15ndash18
Classifying the section
tf = 196 mm fy = 265 Nmm2 EN10025-2
ε =radic(235265) = 0942 T52
cf (tf ε) = (2292 minus 1192 minus 127)(196 times 0942) T52
= 519 lt 9 and the flange is Class 1 T52
cw(twε) = (6122 minus 2 times 196 minus 2 times 127)(119 times 0942) T52
= 489 lt 72 and the web is Class 1 T52
(Note the general use of the minimum fy obtained for the flange)
Checking for moment
qEd = (135 times 60)+ (15 times 70) = 186 kNm
MEd = 186 times 628 = 837 kNm
McRd = 3676 times 103 times 26510 Nmm 625
= 974 kNm gt 837 kNm = MEd
which is satisfactory
220 In-plane bending of beams
Checking for lateral bracing
if z =radic
2293 times 19612
229 times 196 + (6122 minus 2 times 196)times 1196= 591 mm 6324
kcLc(if zλ1) = 10 times 1500(591 times 939 times 0942) = 0287 6324
λc0McRdMEd = 04 times 974837 = 0465 gt 0287 6324
and the bracing is satisfactory
Checking for shear
VEd = 186 times 62 = 558 kN
Av = 159 times 102 minus 2 times 229 times 196 + (119 + 2 times 127)times 196
= 7654 mm2 626(3)
VcRd = 7654 times (265radic
3)10 N = 1171 kN gt 558 kN = VEd 626(2)
which is satisfactory
Checking for bending and shear The maximum MEd occurs at mid-span whereVEd = 0 and the maximum VEd occurs at the support where MEd = 0 and so thereis no need to check for combined bending and shear (Note that in any case 05VcRd = 05 times 1171 = 5855 kN gt 558 kN = VEd and so the combined bendingand shear condition does not operate)
Checking for bearing
REd = 186 times 62 = 558 kN
tw = 119 fyw = 275 Nmm2 EN10025-2
hw = 6122 minus 2 times 196 = 5730 mm
Assume (ss + c) = 100 mm which is not difficult to achieve
kF = 2 + 6 times 1005730 = 3047 EC3-1-5 61(4)(c)
Fcr = 09 times 3047 times 210 000 times 11935730 N = 1694 kNEC3-1-5 64(1)
m1 = (265 times 229)(275 times 119) = 1854 EC3-1-5 65(1)
m2 = 002 times (5727196)2 = 1709 if λF gt 05 EC3-1-5 65(1)
In-plane bending of beams 221
e = 3047 times 210 000 times 1192(2 times 275 times 5730) = 2875 gt 100EC3-1-5 65(3)
y1 = 100 + 196 times radic(18542 + (100196)2 + 1709) = 2419
EC3-1-5 65(3)
y2 = 100 + 196 times radic(1854 + 1709) = 2170 lt 2419 EC3-1-5 65(3)
y = 2170 mm EC3-1-5 65(3)
λF =radic
2170 times 119 times 275
1694 times 103= 0648 gt 05 EC3-1-5 64(1)
χF = 050648 = 0772 EC3-1-5 62
Leff = 0772 times 2170 = 1676 mm EC3-1-5 62
FRd = 275 times 1676 times 11910 N = 548 kN EC3-1-5 62
lt 558 kN = REd and so the assumed value of (ss + c) = 100 mm
is inadequate
51216 Example 16 ndash designing a cantilever
Problem A 30 m long cantilever has the nominal imposed and dead loads (whichinclude allowances for self weight) shown in Figure 544c Design a suitable UBsection in S275 steel if the cantilever has sufficient bracing to prevent lateralbuckling
Selecting a trial section
MEd = 135 times (40 times 3 + 20 times 322)+ 15 times (120 times 3 + 10 times 322)
= 891 kNm
Assume that fy = 265 MPa and that the section is Class 2
Wply ge MEdfy = 891 times 106265 mm3 = 3362 cm3 625
Choose a 610 times 229 UB 125 with Wply = 3676 cm3 gt 3362 cm3
Checking the trial section The trial section must now be checked by classifyingthe section and checking for moment shear moment and shear and bearing Thisprocess is similar to that used in Section 51215 for checking a simply supportedbeam If the section chosen fails any of the checks then a new section must bechosen
222 In-plane bending of beams
51217 Example 17 ndash checking a plastically analysed beam
Problem Check the adequacy of the non-uniform beam shown in Figure 543and analysed plastically in Section 51214 The beam and its flange plates areof S275 steel and the concentrated loads have nominal dead load components of250 kN (which include allowances for self weight) and imposed load componentsof 400 kN
Classifying the sections The 610 times 229 UB 125 was found to be Class 1 inSection 51215For the 300 times 20 plate tp = 20 mm fy = 265 Nmm2 EN10025-2
ε= radic(235265)= 0942 T52
For the 300 times 20 plate as two outstands
cp(tpε) = (3002)(196 times 0942) T52
= 813 lt 9 and the outstands are Class 1 T52
For the internal element of the 300 times 20 plate
cp(tpε) = 229(20 times 0942) T52
= 122 lt 33 and the internal element is Class 1 T52
Checking for plastic collapse The design loads are QEd = (135 times 250)+ (15 times400)= 9375 kN
The plastic collapse load based on the nominal full plastic moments of thebeam was calculated in Section 51214 as
Q = 954 kN gt 9375 kN = QEd
and so the resistance appears to be adequate
Checking for lateral bracing For the unplated segment
Lstable = (60 minus 40 times 0)times 0924 times 497 times 10 = 2808 mm 6353
Thus a brace should be provided at the change of section (a plastic hinge location)and a second brace between this and the support
For the plated segment
iz =radic
(3932 times 104)+ (2 times 3003 times 2012)
(159 times 102)+ (2 times 300 times 20)
= 681 mm
and so
Lstable = 60 minus 40 times 974(minus1979) times 0924 times 681 = 5109 mm 6353
Thus a single brace should be provided in the plated segment A satisfactorylocation is at the load point