analysis and design of steel.pdf

Analysis and Design of Steel and Composite Structures

Analysis and Design of Steel and Composite Structures

Qing Quan Liang

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

© 2015 by Qing Quan LiangCRC Press is an imprint of Taylor & Francis Group, an Informa business

No claim to original U.S. Government worksVersion Date: 20140707

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This book is dedicated to the memory of my parents, Bo Fen Liang (1928–1981) and Xing Zi He (1936–1987), and to my wife, Xiao Dan Cai, and my sons, Samuel Zhi De Liang, Matthew Zhi Cheng Liang and John Zhi Guo Liang.

vii

Contents

Preface xviiAcknowledgements xix

1 Introduction 1

1.1 Steelandcompositestructures 11.2 Limitstatedesignphilosophy 3

1.2.1 Basicconceptsanddesigncriteria 31.2.2 Strengthlimitstate 31.2.3 Stabilitylimitstate 41.2.4 Serviceabilitylimitstate 5

1.3 Structuraldesignprocess 51.4 Materialproperties 7

1.4.1 Structuralsteel 71.4.2 Profiledsteel 81.4.3 Reinforcingsteel 81.4.4 Concrete 8

1.4.4.1 Short-termproperties 81.4.4.2 Time-dependentproperties 11

References 12

2 Design actions 15

2.1 Introduction 152.2 Permanentactions 152.3 Imposedactions 162.4 Windactions 17

2.4.1 Determinationofwindactions 172.4.2 Regionalwindspeeds 192.4.3 Siteexposuremultipliers 20

2.4.3.1 Terrain/heightmultiplier(Mz,cat) 202.4.3.2 Shieldingmultiplier(Ms) 202.4.3.3 Topographicmultiplier(Mt) 22

2.4.4 Aerodynamicshapefactor 222.4.4.1 Calculationofaerodynamicshapefactor 222.4.4.2 Internalpressurecoefficient 23

viii Contents

2.4.4.3 Externalpressurecoefficient 232.4.4.4 Areareductionfactor 242.4.4.5 Combinationfactor 242.4.4.6 Localpressurefactor 242.4.4.7 Permeablecladdingreductionfactor 242.4.4.8 Frictionaldragcoefficient 24

2.4.5 Dynamicresponsefactor 252.4.5.1 General 252.4.5.2 Along-windresponse 252.4.5.3 Crosswindresponse 272.4.5.4 Combinationoflong-windandcrosswindresponse 28

2.5 Combinationsofactions 282.5.1 Combinationsofactionsforstrengthlimitstate 282.5.2 Combinationsofactionsforstabilitylimitstate 282.5.3 Combinationsofactionsforserviceabilitylimitstate 29

References 35

3 Local buckling of thin steel plates 37

3.1 Introduction 373.2 Steelplatesunderuniformedgecompression 37

3.2.1 Elasticlocalbuckling 373.2.1.1 Simplysupportedsteelplates 373.2.1.2 Steelplatesfreeatoneunloadededge 41

3.2.2 Post-localbuckling 423.2.3 Designofslendersectionsaccountingforlocalbuckling 44

3.3 Steelplatesunderin-planebending 483.3.1 Elasticlocalbuckling 483.3.2 Ultimatestrength 493.3.3 Designofbeamsectionsaccountingforlocalbuckling 49

3.4 Steelplatesinshear 523.4.1 Elasticlocalbuckling 523.4.2 Ultimatestrength 54

3.5 Steelplatesinbendingandshear 553.5.1 Elasticlocalbuckling 553.5.2 Ultimatestrength 55

3.6 Steelplatesinbearing 563.6.1 Elasticlocalbuckling 563.6.2 Ultimatestrength 57

3.7 Steelplatesinconcrete-filledsteeltubularcolumns 573.7.1 Elasticlocalbuckling 573.7.2 Post-localbuckling 61

3.8 Doubleskincompositepanels 653.8.1 Localbucklingofplatesunderbiaxialcompression 653.8.2 Post-localbucklingofplatesunderbiaxialcompression 67

Contents ix

3.8.3 Localbucklingofplatesunderbiaxialcompressionandshear 673.8.4 Post-localbucklingofplatesunder

biaxialcompressionandshear 70References 70

4 Steel members under bending 73

4.1 Introduction 734.2 Behaviourofsteelmembersunderbending 734.3 Propertiesofthin-walledsections 75

4.3.1 Centroids 754.3.2 Secondmomentofarea 754.3.3 Torsionalandwarpingconstants 754.3.4 Elasticsectionmodulus 77

4.4 Sectionmomentcapacity 804.5 Membermomentcapacity 81

4.5.1 Restraints 814.5.2 Memberswithfulllateralrestraint 824.5.3 Memberswithoutfulllateralrestraint 84

4.5.3.1 Opensectionswithequalflanges 844.5.3.2 I-sectionswithunequalflanges 87

4.5.4 Designrequirementsformembersunderbending 884.6 Shearcapacityofwebs 92

4.6.1 Yieldcapacityofwebsinshear 924.6.2 Shearbucklingcapacityofwebs 944.6.3 Websincombinedshearandbending 954.6.4 Transversewebstiffeners 964.6.5 Longitudinalwebstiffeners 98

4.7 Bearingcapacityofwebs 1024.7.1 Yieldcapacityofwebsinbearing 1024.7.2 Bearingbucklingcapacityofwebs 1044.7.3 Websincombinedbearingandbending 1044.7.4 Load-bearingstiffeners 105

4.8 Designforserviceability 107References 108

5 Steel members under axial load and bending 109

5.1 Introduction 1095.2 Membersunderaxialcompression 109

5.2.1 Behaviourofmembersinaxialcompression 1095.2.2 Sectioncapacityinaxialcompression 1105.2.3 Elasticbucklingofcompressionmembers 1105.2.4 Membercapacityinaxialcompression 1165.2.5 Lacedandbattenedcompressionmembers 119

x Contents

5.3 Membersinaxialtension 1245.3.1 Behaviourofmembersinaxialtension 1245.3.2 Capacityofmembersinaxialtension 124

5.4 Membersunderaxialloadanduniaxialbending 1275.4.1 Behaviourofmembersundercombinedactions 1275.4.2 Sectionmomentcapacityreducedbyaxialforce 1275.4.3 In-planemembercapacity 1305.4.4 Out-of-planemembercapacity 131

5.5 Designofportalframeraftersandcolumns 1335.5.1 Rafters 1335.5.2 Portalframecolumns 134

5.6 Membersunderaxialloadandbiaxialbending 1395.6.1 Sectioncapacityunderbiaxialbending 1395.6.2 Membercapacityunderbiaxialbending 141

References 146

6 Steel connections 149

6.1 Introduction 1496.2 Typesofconnections 1496.3 Minimumdesignactions 1526.4 Boltedconnections 152

6.4.1 Typesofbolts 1526.4.2 Boltsinshear 1536.4.3 Boltsintension 1556.4.4 Boltsincombinedshearandtension 1566.4.5 Plyinbearing 1566.4.6 Designofboltgroups 157

6.4.6.1 Boltgroupsunderin-planeloading 1576.4.6.2 Boltgroupsunderout-of-planeloading 159

6.5 Weldedconnections 1616.5.1 Typesofwelds 1616.5.2 Buttwelds 1616.5.3 Filletwelds 1626.5.4 Weldgroups 163

6.5.4.1 Weldgroupunderin-planeactions 1636.5.4.2 Weldgroupunderout-of-planeactions 164

6.6 Boltedmomentendplateconnections 1676.6.1 Designactions 167

6.6.1.1 Designactionsforthedesignofbolts,endplatesandstiffeners 167

6.6.1.2 Designactionsforthedesignofflangeandwebwelds 1696.6.2 Designofbolts 1706.6.3 Designofendplate 1706.6.4 Designofbeam-to-end-platewelds 171

Contents xi

6.6.5 Designofcolumnstiffeners 1736.6.5.1 Tensionstiffeners 1736.6.5.2 Compressionstiffeners 1756.6.5.3 Shearstiffeners 1756.6.5.4 Stiffenedcolumnsintensionflangeregion 1766.6.5.5 Stiffenedcolumnsincompressionflangeregion 177

6.6.6 Geometricrequirements 1776.7 Pinnedcolumnbaseplateconnections 180

6.7.1 Connectionsundercompressionandshear 1816.7.1.1 Concretebearingstrength 1816.7.1.2 Baseplatesduetoaxialcompressionincolumns 1816.7.1.3 Columntobaseplatewelds 1836.7.1.4 Transferofshearforce 1836.7.1.5 Anchorboltsinshear 184

6.7.2 Connectionsundertensionandshear 1856.7.2.1 Baseplatesduetoaxialtensionincolumns 1856.7.2.2 Columntobaseplatewelds 1866.7.2.3 Anchorboltsunderaxialtension 1866.7.2.4 Anchorboltsundertensionandshear 187

References 192

7 Plastic analysis of steel beams and frames 195

7.1 Introduction 1957.2 Simpleplastictheory 195

7.2.1 Plastichinge 1957.2.2 Fullplasticmoment 1967.2.3 Effectofaxialforce 2007.2.4 Effectofshearforce 201

7.3 Plasticanalysisofsteelbeams 2027.3.1 Plasticcollapsemechanisms 2027.3.2 Workequation 2027.3.3 Plasticanalysisusingthemechanismmethod 204

7.4 Plasticanalysisofsteelframes 2087.4.1 Fundamentaltheorems 2087.4.2 Methodofcombinedmechanism 208

7.5 PlasticdesigntoAS4100 2137.5.1 Limitationsonplasticdesign 2137.5.2 Sectioncapacityunderaxialloadandbending 2147.5.3 Slendernesslimits 214

References 215

8 Composite slabs 217

8.1 Introduction 2178.2 Componentsofcompositeslabs 2178.3 Behaviourofcompositeslabs 219

xii Contents

8.4 Shearconnectionofcompositeslabs 2198.4.1 Basicconcepts 2198.4.2 Strengthofshearconnection 2198.4.3 Degreeofshearconnection 221

8.5 MomentcapacitybasedonEurocode4 2218.5.1 Completeshearconnectionwithneutralaxisabovesheeting 2218.5.2 Completeshearconnectionwithneutralaxiswithinsheeting 2228.5.3 Partialshearconnection 223

8.6 MomentcapacitybasedonAustralianpractice 2248.6.1 Positivemomentcapacitywithcompleteshearconnection 2248.6.2 Positivemomentcapacitywithpartialshearconnection 2268.6.3 Minimumbendingstrength 2288.6.4 Designfornegativemoments 230

8.7 Verticalshearcapacityofcompositeslabs 2328.7.1 Positiveverticalshearcapacity 2328.7.2 Negativeverticalshearcapacity 2338.7.3 VerticalshearcapacitybasedonEurocode4 234

8.8 Longitudinalshear 2348.9 Punchingshear 2358.10 Designconsiderations 235

8.10.1 Effectivespan 2358.10.2 Potentiallycriticalcrosssections 2358.10.3 Effectsofpropping 236

8.11 Designforserviceability 2408.11.1 Crackcontrolofcompositeslabs 2408.11.2 Short-termdeflectionsofcompositeslabs 2418.11.3 Long-termdeflectionsofcompositeslabs 2428.11.4 Span-to-depthratioforcompositeslabs 242

References 249

9 Composite beams 251

9.1 Introduction 2519.2 Componentsofcompositebeams 2519.3 Behaviourofcompositebeams 2539.4 Effectivesections 254

9.4.1 Effectivewidthofconcreteflange 2549.4.2 Effectiveportionofsteelbeamsection 256

9.5 Shearconnectionofcompositebeams 2569.5.1 Basicconcepts 2569.5.2 Load–slipbehaviourofshearconnectors 2589.5.3 Strengthofshearconnectors 2589.5.4 Degreeofshearconnection 2619.5.5 Detailingofshearconnectors 262

9.6 Verticalshearcapacityofcompositebeams 2629.6.1 Verticalshearcapacityignoringconcretecontribution 2629.6.2 Verticalshearcapacityconsideringconcretecontribution 263

Contents xiii

9.7 Designmomentcapacityforpositivebending 2669.7.1 Assumptions 2669.7.2 Crosssectionswithγ≤ 0.5andcompleteshearconnection 266

9.7.2.1 NominalmomentcapacityMbc 2669.7.2.2 Plasticneutralaxisdepth 268

9.7.3 Crosssectionswithγ≤ 0.5andpartialshearconnection 2709.7.3.1 NominalmomentcapacityMb 2709.7.3.2 Depthofthefirstplasticneutralaxis 2719.7.3.3 Depthofthesecondplasticneutralaxis 271

9.7.4 Crosssectionswithγ=1.0andcompleteshearconnection 2729.7.4.1 NominalmomentcapacityMbfc 2729.7.4.2 Plasticneutralaxisdepth 273

9.7.5 Crosssectionswithγ=1.0andpartialshearconnection 2739.7.5.1 NominalmomentcapacityMbf 2739.7.5.2 Depthofthefirstplasticneutralaxis 2749.7.5.3 Depthofthesecondplasticneutralaxis 275

9.7.6 Crosssectionswith0.5<γ≤ 1.0 2759.7.7 Minimumdegreeofshearconnection 276

9.8 Designmomentcapacityfornegativebending 2819.8.1 Designconcepts 2819.8.2 Keylevelsoflongitudinalreinforcement 282

9.8.2.1 Maximumareaofreinforcement 2829.8.2.2 PNAlocatedatthejunctionofthetopflangeandweb 2839.8.2.3 PNAlocatedintheweb 2839.8.2.4 PNAlocatedatthejunctionofthewebandbottomflange 2839.8.2.5 PNAlocatedatthejunctionofthe

bottomflangeandplate 2839.8.3 Plasticneutralaxisdepth 2839.8.4 Designnegativemomentcapacity 284

9.9 Transferoflongitudinalshearinconcreteslabs 2949.9.1 Longitudinalshearsurfaces 2949.9.2 Designlongitudinalshearforce 2959.9.3 Longitudinalshearcapacity 2969.9.4 Longitudinalshearreinforcement 296

9.10 Compositebeamswithprecasthollowcoreslabs 3049.11 Designforserviceability 305

9.11.1 Elasticsectionproperties 3059.11.2 Deflectioncomponentsofcompositebeams 3079.11.3 Deflectionsduetocreepandshrinkage 3089.11.4 Maximumstressinsteelbeam 309

References 313

10 Composite columns 317

10.1 Introduction 31710.2 Behaviouranddesignofshortcompositecolumns 318

10.2.1 Behaviourofshortcompositecolumns 318

xiv Contents

10.2.2 Shortcompositecolumnsunderaxialcompression 32010.2.3 Shortcompositecolumnsunderaxialloadanduniaxialbending 321

10.2.3.1 General 32110.2.3.2 Axialload–momentinteractiondiagram 322

10.3 Non-linearanalysisofshortcompositecolumns 33410.3.1 General 33410.3.2 Fibreelementmethod 33410.3.3 Fibrestraincalculations 33410.3.4 Materialconstitutivemodelsforstructuralsteels 33610.3.5 MaterialmodelsforconcreteinrectangularCFSTcolumns 33610.3.6 MaterialmodelsforconcreteincircularCFSTcolumns 33910.3.7 Modellingoflocalandpost-localbuckling 34010.3.8 Stressresultants 34210.3.9 Computationalalgorithmsbasedonthesecantmethod 342

10.3.9.1 Axialload–strainanalysis 34210.3.9.2 Moment–curvatureanalysis 34310.3.9.3 Axialload–momentinteractiondiagrams 344

10.4 Behaviouranddesignofslendercompositecolumns 34710.4.1 Behaviourofslendercompositecolumns 34710.4.2 Relativeslendernessandeffectiveflexuralstiffness 34710.4.3 Concentricallyloadedslendercompositecolumns 34810.4.4 Uniaxiallyloadedslendercompositecolumns 350

10.4.4.1 Second-ordereffects 35010.4.4.2 Designmomentcapacity 351

10.4.5 Biaxiallyloadedslendercompositebeam–columns 35710.5 Non-linearanalysisofslendercompositecolumns 357

10.5.1 General 35710.5.2 Modellingofload–deflectionbehaviour 35810.5.3 Modellingofaxialload–momentinteractiondiagrams 36010.5.4 NumericalsolutionschemebasedonMüller’smethod 36110.5.5 Compositecolumnswithpreloadeffects 364

10.5.5.1 General 36410.5.5.2 Non-linearanalysisofCFST

columnswithpreloadeffects 36410.5.5.3 AxiallyloadedCFSTcolumns 36410.5.5.4 BehaviourofCFSTbeam–columnswithpreloadeffects 365

10.5.6 Compositecolumnsundercyclicloading 36510.5.6.1 General 36510.5.6.2 Cyclicmaterialmodelsforconcrete 36610.5.6.3 Cyclicmaterialmodelsforstructuralsteels 36810.5.6.4 Modellingofcyclicload–deflectionresponses 369

References 371

11 Composite connections 377

11.1 Introduction 37711.2 Single-plateshearconnections 377

Contents xv

11.2.1 Behaviourofsingle-plateconnections 37811.2.2 Designrequirements 37911.2.3 Designofbolts 37911.2.4 Designofsingleplate 38011.2.5 Designofwelds 380

11.3 Teeshearconnections 38211.3.1 Behaviourofteeshearconnections 38311.3.2 Designofbolts 38311.3.3 Designofteestems 38411.3.4 Designofteeflanges 38411.3.5 Designofwelds 38411.3.6 Detailingrequirements 385

11.4 Beam-to-CECcolumnmomentconnections 38711.4.1 Behaviourofcompositemomentconnections 38811.4.2 Designactions 38911.4.3 Effectivewidthofconnection 39011.4.4 Verticalbearingcapacity 39111.4.5 Horizontalshearcapacity 39211.4.6 Detailingrequirements 394

11.4.6.1 Horizontalcolumnties 39411.4.6.2 Verticalcolumnties 39411.4.6.3 Face-bearingplates 39511.4.6.4 Steelbeamflanges 39511.4.6.5 Extendedface-bearingplatesandsteelcolumn 395

11.5 Beam-to-CFSTcolumnmomentconnections 40011.5.1 Resultantforcesinconnectionelements 40011.5.2 Neutralaxisdepth 40211.5.3 Shearcapacityofsteelbeamweb 40211.5.4 Shearcapacityofconcrete 403

11.6 Semi-rigidconnections 40511.6.1 Behaviourofsemi-rigidconnections 40611.6.2 Designmomentsatsupports 40611.6.3 Designofseatangle 40611.6.4 Designofslabreinforcement 40711.6.5 Designmomentcapacitiesofconnection 40711.6.6 Compatibilityconditions 40711.6.7 Designofwebangles 40811.6.8 Deflectionsofcompositebeams 40811.6.9 Designprocedure 409

References 409

Notations 411Index 431

xvii

Preface

Steelandcompositesteel–concretestructuresarewidelyusedinmodernbridges,buildings,sportstadia,towersandoffshorestructures.Theanalysisanddesignofsteelandcompos-itestructuresrequireasoundunderstandingofthebehaviourofstructuralmembersandsystems.Thisbookprovidesanintegratedandcomprehensiveintroductiontotheanalysisanddesignofsteelandcompositestructures.ItdescribesthefundamentalbehaviourofsteelandcompositemembersandstructuresandthelatestdesigncriteriaandproceduresgiveninAustralianStandardsAS/NZS1170,AS4100,AS2327.1,Eurocode4andAISC-LRFDspecifications.Thelatestresearchfindingsoncompositemembersbytheauthor’sresearchteamsarealsoincorporatedinthebook.Emphasisisplacedonasoundunderstandingofthefundamentalbehaviouranddesignprinciplesofsteelandcompositemembersandcon-nections.Numerousstep-by-stepexamplesareprovidedtoillustratethedetailedanalysisanddesignofsteelandcompositemembersandconnections.

Thisbookisanidealcoursetextbookonsteelandcompositestructuresforundergradu-ateandpostgraduatestudentsofstructuralandcivilengineering,and it isacomprehen-siveandindispensableresourceforpractisingstructuralandcivilengineersandacademicresearchers.

Chapter1introducesthelimitstatedesignphilosophy,thedesignprocessandmate-rial properties of steels and concrete. The estimation of design actions on steel andcomposite structures in accordance with AS/NZS 1170 is described in Chapter 2.Chapter 3presentsthelocalandpost-localbucklingbehaviourofthinsteelplatesunderin-plane actions, including compression, shear and bending of steel plates in contactwithconcrete.ThedesignofsteelmembersunderbendingistreatedinChapter4,whichincludesthedesignforbendingmomentsandtheshearandbearingofwebstoAS4100.Chapter 5isdevotedtosteelmembersunderaxialloadandbending.Theanalysisanddesignofsteelmembersunderaxialcompression,axialtensionandcombinedaxialloadandbendingtoAS4100arecovered.InChapter6,thedesignofboltedandweldedsteelconnections, includingboltedmomentendplateconnectionsandpinnedcolumnbaseplateconnections,ispresented.Chapter7introducestheplasticanalysisanddesignofsteelbeamsandframes.

ThebehaviouranddesignofcompositeslabsforstrengthandserviceabilitytoEurocode 4andAustralianpractice are treated in Chapter 8.Chapter 9presents the behaviour anddesignofsimplysupportedcompositebeamsforstrengthandserviceabilitytoAS2327.1.Thedesignmethod for continuous compositebeams is also covered.ThebehaviouranddesignofshortandslendercompositecolumnsunderaxialloadandbendinginaccordancewithEurocode4aregiveninChapter10.Thischapteralsopresentsthenonlinearinelasticanalysisofthin-walledconcrete-filledsteeltubularshortandslenderbeam-columnsunderaxialloadandbiaxialbending.Chapter11introducesthebehaviouranddesignofcomposite

xviii Preface

connectionsinaccordancewithAISC-LRFDspecifications,includingsingle-plateandteeshear connections, beam-to-composite columnmoment connections and semi-rigid com-positeconnections.

Qing Quan LiangAssociateProfessorVictoriaUniversity

Melbourne,Victoria,Australia

xix

Acknowledgements

Theauthor thanksProfessorYeong-BinYangatNationalTaiwanUniversity,Dr.AnneW.M.NgatVictoriaUniversity inMelbourne,BenjaminCheung, seniorprojectengi-neer in Melbourne, and Associate Professor Yanglin Gong at Lakehead University fortheirinvaluableandcontinuedsupport.Theauthoralsothanksallhisco-researchersfortheir contributions to the research work, particularly Associate Professor MuhammadN.S.HadiattheUniversityofWollongong,ProfessorBrianUyandProfessorMarkA.BradfordattheUniversityofNewSouthWales,ProfessorYi-MinXieatRMITUniversity,EmeritusProfessorGrantP.StevenattheUniversityofSydney,ProfessorJat-YuenRichardLiewattheNationalUniversityofSingapore,EmeritusProfessorHowardD.WrightattheUniversityofStrathclyde,Dr.HamidR.RonaghattheUniversityofQueenslandandDr.MostafaF. HassaneinandDr.OmniaF.KharoobatTantaUniversity.ThanksalsogotoProfessorJin-GuangTengatTheHongKongPolytechnicUniversity,ProfessorDennisLamattheUniversityofBradford,ProfessorBenYoungattheUniversityofHongKong,Professor Lin-Hai Han at Tsinghua University, Associate Professor Mario Attard andProfessorYong-LinPiandDr.SawekchaiTangaramvongattheUniversityofNewSouthWales, Dr. Zora Vrcelj at Victoria University and Professor N. E. Shanmugam at theNationalUniversityofMalaysiafortheirusefulcommunicationsandsupport.Gratefulacknowledgement ismade to theauthor’s formerPhDstudentDr.Vipulkumar I.Patelfor his contributions to the research work on composite columns and to ME studentsDr. Sukit Yindeesuk in the Department of Highways in Thailand and Hassan Nashidfortheir support.Finally,andmost importantly, theauthorthankshiswife,XiaoDanCai,andsons,Samuel,MatthewandJohn,fortheirgreatencouragement,supportandpatiencewhilehewaswritingthisbook.

1

Chapter 1

Introduction

1.1 Steel And comPoSIte StructureS

Steelandcompositesteel–concretestructuresarewidelyusedinmodernbridges,buildings,sportstadia,towersandoffshorestructures.Accordingtotheirintendedfunctions,build-ings canbe classified into industrial, residential, commercial and institutionalbuildings.A steelstructureiscomposedofsteelmembersjoinedtogetherbyboltedorweldedconnec-tions,whichmaybeintheformofapin-connectedtrussorarigidframe.Incomparisonwith reinforced concrete structures, steel structures have the advantages of lightweight,large-span,highductilityand rapid construction.The rapid steel constructionattributestothefactthatsteelmembersandconnectioncomponentscanbeprefabricatedinashop.As a result, significant savings in construction timeand costs canbe achieved. Perhaps,steelportalframesasdepictedinFigure1.1arethemostcommonlyusedsteelstructuresinindustrialbuildings.Theyareconstructedbycolumns,roofraftersandbracings,whicharejoinedtogetherbyknee,ridgeandcolumnbaseconnections.Thedesignofsteelportalframesistreatedinthisbook.

Theadvantagesoftherapidandeconomicalsteelconstructionofmultistoreybuildingscanonlybeutilisedbycomposite steel–concretestructures,whichareefficientandcost-effectivestructuralsystems.Compositestructuresareusuallyconstructedbycompositecol-umnsorsteelcolumnsandsteelbeamssupportingcompositeslabsorconcreteslabs.Itisnotedthatsteelisthemosteffectiveincarryingtensionandconcreteisthemosteffectiveinresistingcompression.Compositemembersmakethebestuseoftheeffectivematerialprop-ertiesofbothsteelandconcrete.AcompositebeamisformedbyattachingaconcreteslabtothetopflangeofasteelbeamasshowninFigure1.2.Bythecompositeactionachievedbyweldingshearconnectorstothetopflangeofthesteelbeam,thesteelbeamandthecon-creteslabworkstogetherasonestructuralmembertoresistdesignactions.Inacompositebeamunderbending,theconcreteslabissubjectedtocompression,whilethesteelbeamisintension,whichutilisestheeffectivematerialpropertiesofbothsteelandconcrete.Thecommontypesofcompositecolumnsincludeconcreteencasedcompositecolumns,rectan-gularconcrete-filledsteeltubularcolumnsandcircularconcrete-filledsteeltubularcolumnsaspresentedinFigure1.3.High-strengthcompositecolumnshaveincreasinglybeenusedinhigh-risecompositebuildingsduetotheirhighstructuralperformancesuchashighstrengthandhighstiffness.Thefundamentalbehaviourandthestate-of-the-artanalysisanddesignofcompositeslabs,compositebeams,compositecolumnsandcompositeconnectionsarecoveredinthisbook.

Thedesignofsteelandcompositestructuresisdrivenbythelimitedmaterialresources,environmental impacts and technological competition which demand lightweight, low-cost and high-performance structures. These demands require that structural designersmusthavea soundunderstandingof the fundamentalbehaviourof steelandcomposite

2 Analysis and design of steel and composite structures

structuresandthelatestdesignprinciplesandtechnologiesforthedesignofthesestructures.Theformsofsteelandcompositestructureshavebeenevolvinginthelastfewdecades,andmanyinnovativesteelandcompositestructureshavebeendesignedandconstructedaroundtheworld.Topologyoptimisationtechniquescanbeusedtofindtheoptimalandinnovativelayoutsofstructures(Liang2005).Itisrecognizedthattopologyoptimisation

(a) (b) (c)

Figure 1.3 Cross sections of composite columns: (a) concrete encased composite column, (b) rectangular concrete-filled steel tubular column, and (c) circular concrete-filled steel tubular column.

Steel beam

Composite slab Stud shear connector

Figure 1.2 Cross section of composite beam.

Figure 1.1 Steel portal frames.

Introduction 3

produces much more material savings and higher-performance optimal structures thanshapeandsizingoptimisation.

Thischapterintroducesthelimitstatedesignphilosophy,thestructuraldesignprocessandmaterialpropertiesofsteelsandconcreteusedintheconstructionofsteelandcompos-itestructures.

1.2 lImIt StAte deSIgn PhIloSoPhy

1.2.1 Basic concepts and design criteria

Thelimitstatedesignphilosophyhasbeenadoptedinthecurrentcodesofpracticeasthebasicdesignmethodforthedesignofsteelandcompositestructuresasitisbelievedthatthismethodiscapableofyieldingsaferandmoreeconomicaldesignsolutions.Thelimitstateisdefinedasthestatebeyondwhichthestructurewillnotsatisfythedesigncriteria.Thislimitstatemaybecausedbythefailureofoneormorestructuralmembers,theinstabilityofstructuralmembersorthewholestructure,orexcessivedeformationsofthestructure.Thelimitstatedesignistodesignastructureorstructuralcomponentthatcanperformtheintendedphysicalfunctionsinitsdesignlifetime.Inthelimitstatedesign,theperformanceofastructureisevaluatedbycomparisonofdesignactioneffectswithanumberoflimitingconditionsofusefulness.Thelimitstatesmayincludestrength,stability,serviceability,fire,fatigue,earthquakeandbrittlefracturelimitstates.

Structuraldesigncriteriaareexpressed in termsofachievingmultipledesignobjectives.Thereareusuallymultipledesignobjectivesthatmustbeconsideredbythestructuraldesignerwhendesigningastructure.Themainobjectivesarefunctionality,safety,economyandeaseofconstruction.Thesafetyisastructuraldesignobjectivewhichisrelatedtothestrengthandserviceability.Designcodesandstandardsimposelimitationsontheserviceabilityandstrengthofastructureorstructuralmemberstoensurethatthestructureorstructuralmem-bersdesignedwillperformnormalfunctions.Functionality,whichistheabilityofastructuretoperformitsintendednon-structuraluse,andeconomyarenon-structuraldesignobjectives.However,theycanbeusedtorankalternativedesignsthatsatisfystructuraldesigncriteria.

1.2.2 Strength limit state

Thestrengthdesigncriterionrequiresthatthestructuremustbedesignedsothatitwillnotfailinitsdesignlifetimeortheprobabilityofitsfailureisverylow.Thestrengthlimitstatedesignistodesignastructureincludingallofitsmembersandconnectionstohavedesigncapacitiesinexcessoftheirdesignactioneffects.Thiscanbeexpressedinthemathematicalformasfollows:

E Ra n≤ φ (1.1)

whereEaisthedesignactioneffectφisthecapacityreductionfactorRnisthenominalcapacityorresistanceofthestructuralmemberφRnisthedesigncapacityorthedesignresistanceofthestructuralmember

ThedesignactioneffectEarepresentsaninternalactionsuchasaxialforce,shearforceorbendingmoment,whichisdeterminedbystructuralanalysisusingfactoredcombinationsofdesignactionsappliedonthestructure.Inthestrengthlimitstatedesign,loadfactorsareusedtoincreasethenominalloadsonstructuralmembers,whilecapacityreductionfactorsareemployedtodecreasethecapacityofthestructuralmember.


Theuseofloadfactorsandcapacityreductionfactorsinthestrengthlimitstatedesignistoensurethattheprobabilityofthefailureofastructureunderthemostadversecombinationsofdesignactionsisverysmall.Thesefactorsareusedtoaccountfortheeffectsoferrorsanduncertaintiesencounteredintheestimationofdesignactionsonasteelorcompositestruc-tureandofitsbehaviour.Errorsmadebythedesignermaybecausedbysimplifiedassump-tionsandlackofprecisionintheestimationofdesignactions,instructuralanalysis,inthemanufactureandintheerectionofthestructure(TrahairandBradford1998).Thedesignactionsonastructurevarygreatly.Thismaybecausedbytheestimationofthemagnitudeofthepermanentactions(deadloads)owingtovariationsinthedensitiesofmaterials.Inaddition,imposedactions(liveloads)maychangecontinuallyduringthedesignlife.Windactionsvarysignificantlyandareusuallydeterminedbyprobabilisticmethods.Theuncer-taintiesaboutthestructureincludematerialproperties,residualstresslevels,cross-sectionaldimensionsofsteelsectionsandinitialgeometricimperfectionsofstructuralmembers.Theaforementionederrorsanduncertaintiesmayleadtotheunderestimateofthedesignactionsandtheoverestimateofthecapacityofthestructure.Loadfactorsandcapacityreductionfactorsareusedtocompensatetheseeffectsinthestrengthlimitstatedesign.

Probabilitymethodsareusuallyemployedtodetermineloadandcapacityfactorsonthebasisofstatisticaldistributionsofdesignactionsandcapacitiesofstructuralmembers.TheloadandcapacityfactorsgiveninAS4100werederivedbyusingtheconceptofsafetyindex.Thelimitstatedesigngenerallyyieldsslightlysaferdesignswithasafetyindexrangingfrom3.0to3.5incomparisonwiththetraditionalworkingstressdesign(Phametal.1986).Thecapacityreductionfactordependsonthemethodsemployedtodeterminethenominalcapaci-ties,nominaldesignactionsandthevaluesusedfortheloadfactors.Table1.1givesthecapac-ityreductionfactorsforsteelmembersandconnectionsforthestrengthlimitstatedesign.

1.2.3 Stability limit state

Thestabilitylimitstateisconcentredwiththelossofstaticequilibriumorofdeformationsofthestructureoritsmembersowingtosliding,upliftingoroverturning.Thestabilitylimitstaterequiresthatthefollowingconditionbesatisfied:

E E Ra dst a stb n. .− ≤ φ (1.2)

whereEa dst. isthedesignactioneffectofdestabilizingactionsEa stb. isthedesignactioneffectofstabilizingactions

Table 1.1 Capacity reduction factor (ϕ) for strength limit states

Structural component Capacity reduction factor (ϕ)

Steel member 0.9Connection component (excluding bolt, pin or weld) 0.9Bolted or pin connection 0.8Ply in bearing 0.9Welded connection SP category GP categoryComplete penetration butt weld 0.9 0.6Longitudinal fillet weld in RHS (t < 3 mm) 0.7 —Other welds 0.8 0.6

Source: AS 4100, Australian Standard for Steel Structures, Standards Australia, Sydney, New South Wales, Australia, 1998.

Introduction 5

1.2.4 Serviceability limit state

Theserviceabilitylimitstateisthestatebeyondwhichastructureorastructuralmemberwillnotsatisfythespecifiedservicedesigncriteria.Thismeansthatbeyondthelimitstate,thestructurewillnotfitfortheintendeduseunderserviceloadconditions.Serviceabilitylimitstatesmayincludedeformation,vibrationanddegradationlimitstates.Thedeforma-tionsofastructurearegovernedbythestiffnessdesignrequirementswhicharesystemper-formancecriteria.Forthestiffnesslimitstatedesign,thedeflectionsofthestructureundermostadverseserviceloadconditionsneedtobelimitedsothatthestructurecanperformthenormalfunctionwithoutimpairingitsappearance,safetyandpubliccomfort.Thiscanbeexpressedinthemathematicalformasfollows:

δ δj j≤ ∗ (1.3)

whereδ jisthejthdisplacementordeflectionofthestructureunderthemostadverseservice

loadcombinationsδ j∗isthelimitofthejthdisplacementordeflection

The deflections of a structure under service design actions are usually determined byperformingafirst-orderlinearelasticanalysisorasecond-ordernonlinearelasticanalysis.OnlythemostessentialdeflectionlimitsaregiveninAS4100(1998).Thestructuraldesignerneedstodeterminewhetherthestructuredesignedsatisfiestheserviceabilityrequirements.

1.3 StructurAl deSIgn ProceSS

Theoverallpurposeofthestructuraldesignistodevelopthebestfeasiblestructuralsys-temthatsatisfiesthedesignobjectives intermsofthefunctionality,safetyandeconomy.Structuraldesignisacomplex,iterative,trial-and-erroranddecision-makingprocess.Inthedesignprocess,aconceptualdesigniscreatedbythedesignerbasedonhisintuition,creativ-ityandpastexperience.Structuralanalysisisthenundertakentoevaluatetheperformanceofthedesign.Ifthedesigndoesnotsatisfythedesignobjectives,anewdesignisthendevel-oped.Thisprocessisrepeateduntilthedesignsatisfiesthemultipleperformanceobjectives.ThemainstepsoftheoverallstructuraldesignprocessareillustratedinFigure1.4.

Thefirststep in thestructuraldesignprocess is to investigate theoveralldesignprob-lem.Firstly,thedesignengineersdiscusstheneedsforthestructure,itsproposedfunction,requirementsandconstraintswiththeowner.Thefunctionalityistheabilityofastructuretoperformitsintendednon-structuraluse.Itisoneoftheimportantdesignobjectivesthatmustbeachievedforastructureandaffectsallstagesofthestructuraldesignprocess.Thesiteandgeotechnical investigationsarethenfollowed.Thestructuraldesignersalsoneedtostudysimilarstructuresandtoconsultauthoritiesfromwhompermissionsandapprov-alsmustbeobtained.Multipledesignobjectivesarethenidentifiedforthestructureandselectedbytheownerwhoconsultswiththestructuraldesignersbasedontheconsiderationofhis/herexpectations,economicanalysisandacceptablerisk.

In the conceptualdesign stage, the structuraldesignerdevelops thebest feasible struc-tural systems thatappear toachieve thedesignobjectivesdefined in thepreceding stage.Theselectionof structural systems isgenerally iterative innaturebasedon thedesigner’screation,intuitionandpastexperience.Inordertoobtainanoptimalstructure,anumberofalternativestructuralsystemsmustbeinventedandevaluated.Theinventionofstructural


systemsisthemostchallengingtaskinstructuraldesignsinceitinvolvesalargenumberofpossibilitiesforthestructurallayouts.Thetraditionaldesignprocessishighlytimeconsum-ingandexpensive.Sincethedevelopmentofstructuralsystemsisanoptimaltopologydesignproblem,automatedtopologyoptimisationtechniquesuchastheperformance-basedoptimi-sation(PBO)technique(Liang2005)canbeemployedintheconceptualdesignstagetogen-erateoptimalstructures.Theoptimalstructuralsystemisproducedbytopologyoptimisationtechniquesbasedonthedesigncriteriaandconstraintsbutnotonthepastexperience(Liangetal.2000a).Thedesigneralsoselectsthematerialsofconstructionforthestructure.

After the best feasible structure has been created, the preliminary design can be car-riedout.Thedesignloadsandloadcombinationsappliedtothestructureareestimatedinaccordancewiththeloadingcodes.Thestructuralanalysismethodormodernnumericaltechniquesuchasthefiniteelementmethod(ZienkiewiczandTaylor1989,1991)isthenemployedtoanalysethestructuretoevaluateitsstructuralperformance.Fromtheresultsofthestructuralanalysis,structuralmembersarepreliminarilysizedtosatisfythedesigncriteria.Thecostofthestructureisalsopreliminarilyestimated.Ifthestructuredoesnot

Start

Problem investigation

Conceptual design

Preliminary design

Final design

Documentation

Yes

No

Tendering

Inspection and certi�cation

End

Satisfy designobjectives?

Yes

NoSatisfy designobjectives?

Figure 1.4 The structural design process.

Introduction 7

satisfythefunction,structuralefficiencyandcostdesignobjectives,anewstructuralsystemmustbedevelopedandthedesignprocessisrepeated,asdepictedinFigure1.4.Itisobviousthatshapeandsizingoptimisationtechniquescanbeappliedinthepreliminarydesignstagetoachievecost-efficientdesigns.

Sincethestructureisapproximatelyproportionedinthepreliminarydesignstage,itmustbecheckedagainst thedesigncriteriaandobjectives in thefinaldesign stage.The loadsappliedtothestructurearerecalculatedandthestructureisreanalysed.Theperformanceofthestructureisthenevaluatedandcheckedwithperformancerequirements.Anychangeinthemembersizesmayrequireafurtherreanalysisandresizingofthestructure.Thedesignandredesignprocessisrepeateduntilnomoremodificationcanbemadetothestructure.Thestructureisevaluatedforthedesignobjectivessuchasfunction,serviceability,strengthandcost.Iftheseobjectivesarenotsatisfied,thestructuremaybemodifiedoranewcon-ceptualdesignmaybegenerated.ThedesignprocessisrepeatedasindicatedinFigure1.4.Inthefinaldesignstage,thesizingofthestructureisthemaintask.Therefore,sizingopti-misation techniques canbe employed toautomate thedesignprocess. It isworthnotingthattopologyoptimisationtechniquescanalsobeusedinthefinaldesignstage.Liangetal.(2000b,2001,2002)demonstratedthattheautomatedPBOtechniquecanbeemployedinthefinaldesignstagetogenerateoptimalstrut-and-tiemodelsforthedesignanddetailingofreinforcedandprestressedconcretestructures.

After the structure is finalised, the documentation such as the detailed drawings andspecificationscanbepreparedandtendersforconstructioncanbecalledfor.Atthefinalstage,thedesignerscarryoutinspectionandcertificationduringconstructiontoensurethatallperformanceobjectivesdefinedareachievedinthestructuraldesignprocess.

1.4 mAterIAl ProPertIeS

1.4.1 Structural steel

Structuralsteelisusuallyhotrolled,weldedfromflatplatesorcoldformedfromflatplatestoformstructuralsections,suchasI-sections,rectangularhollowsections(RHSs)andcir-cularhollowsections(CHSs).Figure1.5depictsanidealisedstress–straincurveformildstructuralsteel.Itcanbeseenthatthesteelinitiallyhasalinearstress–strainrelationshipuptotheelasticlimit,whichcanbeapproximatelydefinedbytheyieldstressfy.ThemostimportantpropertiesofmildstructuralsteelareitsYoung’smodulusofelasticityEsrang-ingfrom200to210GPaanditsyieldstressrangingfrom250to400MPa.Beyondthe

Strain ε 0

fy

fu

Elastic

Plastic

Fracture

Stre

ss σ Strain hardening

εstεy

Figure 1.5 Idealised stress–strain curve for mild structural steel.


elasticlimit,thesteelundergoeslargeplasticflowswithoutanyincreaseinthestressuntilreachingthehardeningstrainεst,whichisusually10or11timestheyieldstrainεy.Thisplasticplateauindicatestheductilityofthesteel.Afterreachingthehardeningstrainεst,thestressincreasesabovetheyieldstresswithanincreaseinthestrainuntiltheultimatetensilestrengthfuisattained.Thisisfollowedbytheneckingofthecrosssectionanddecreasinginstressuntilthetensilefractureoccurs.Thesteelnormallyfollowsthesamestress–straincurveintensionandcompression.Intheelasticrange,Poisson’sratioofsteelisabout0.3.InAS4100,Poisson’sratioistakenas0.25forAustralianstructuralsteels.

The yield stress is an important property of a structural steel, which depends on thechemicalcontentssuchascarbonandmanganese,theheattreatmentusedandtheamountofworkinginducedduringtherollingprocess.Coldworkingalsoincreasestheyieldstressofthesteel.Theyieldstressofastructuralsteelcanbedeterminedbystandardtensiontests.Theminimumyieldstressofthestructuralsteelgivenindesigncodesforuseinstructuraldesignisacharacteristicvaluethatisusuallylessthanthatdeterminedfromanystandardtensiontest.Thisimpliesthattheuseoftheyieldstressgivenindesigncodesusuallypro-videsconservativedesigns.

1.4.2 Profiled steel

Profiledsteelsheetingisusedincompositeslabsandbeamsasthepermanentformworkandpartofreinforcementfortheconcrete.Itismanufacturedbycoldrollingthinsteelplateintoshapewithwidesteelribs.Theyieldstressmaybeincreasedbythecold-rollingprocess.Thestress–straincurveforprofiledsteelisroundedwithoutawell-definedyieldstressasshowninFigure1.5.A0.2%proofstressof550MPaisusuallyusedforprofiledsteel,whileitselasticmodulusisabout200GPa.

Themajor typesofprofiled steel sheetingused incomposite construction inAustraliaareLYSAGHTBondekII,ComformandCondeckHP.Profiledsteelsheetingmighthaveanadverseinfluenceonthebehaviourofcompositebeams.AS2327.1imposesrestrictionson thegeometryofprofiled steel sheeting so that composite slabs incorporatingprofiledsteel sheetingcanbe treatedas solid slabswhencalculating thedesigncapacityof shearconnectors.

1.4.3 reinforcing steel

ThetypesofreinforcingsteelcommerciallyavailableinAustraliaarereinforcingbar,hard-drawnwireandweldedwire fabric.Reinforcingbarcanbeclassified intoseveralgrades,namely,400Yhighyieldwithaminimumguaranteedyieldstressof400MPa,plainbarwithaminimumguaranteedyieldstressof250and500MPasteelswithacharacteristicyieldstressof500MPa.The500MPasteelshavethreegradessuchas500L,500Nand500Egrades,wherethefinalletterdenotesthelevelofductility,withLindicatinglowductility,NdenotingnormalductilityandEstandingforspecialhigh-ductilitysteelforuseinearthquake-resistantdesign.Thestress–straincurveforreinforcingsteelisassumedtobeelastic–perfectlyplasticindesign.Theelasticmodulusofreinforcingsteelisusuallytakenas200GPa.

1.4.4 concrete

1.4.4.1 Short-term properties

Themainpropertiesofthehardenedconcreteareitscompressivestrength,elasticmodu-lusincompression,tensilestrengthanddurability.Thecharacteristiccompressivestrength

Introduction 9

′fc ofconcrete iscommonlyusedforconcretedueto the largevariationof theconcretestrength.Itisdeterminedasthestrengthattainedat28 daysby95%oftheconcreteasobtainedbystandardcompressiontests.Thenormal-strengthconcretehasacharacter-isticcompressivestrength ′fcupto50MPa;concretewithacompressivestrengthhigherthan50MPa is regardedashigh-strengthconcrete (Warneretal.1998),whichcanbemadebyusinghigh-qualityaggregatesandsuperplasticizers,andthestrengthmayexceed100MPa.

Figure1.6depictsthetypicalstress–straincurvesforconcreteinuniaxialcompressionwithvariouscompressivestrengths.Itappearsfromthefigurethatthestress–strainrela-tionship is linear for stressup to0 4. ′fc.However, at stresshigher than0 4. ′fc , the stress–strainrelationshipbecomesnonlinearduetotheeffectsoftheformationsanddevelopmentofmicrocracks at the interfaces between themortar and coarse aggregate.As shown inFigure 1.6,theshapeofthestress–straincurveforconcretevarieswiththeconcretecom-pressivestrengthsanditisaffectedbythetypeofaggregateusedandthestrainrateappliedinthecompressiontests.Thestress–straincurveforhigh-strengthconcreteissteeperthanfornormal-strengthconcrete.Thedescendingbranchinthepost-ultimaterangedecreasessharply with increasing the compressive strength of concrete. This indicates that high-strengthconcreteisverybrittle.

Empiricalequationshavebeenproposedbyvariousresearchersbasedonexperimentalresultstoexpressthestress–straincurvesfornormal-andhigh-strengthconcrete.Manderetal.(1988)presentedequationsformodellingthestress–strainbehaviourofunconfinedconcreteasfollows:

σλ ε ε

λ ε ε λcc c c

c c

f=

′ ′( )− + ′( )

/

/1 (1.4)

Strain

fc = 25 MPa

fc = 50 MPa

fc = 70 MPa

fc = 100 MPa

Stre

ss (M

Pa)

00

20

40

60

80

100

120

0.001 0.002 0.003 0.004 0.005 0.006 0.007

Figure 1.6 Stress–strain curves for normal- and high-strength concrete.


λ

ε=

− ′ ′( )E

E fc

c c c/ (1.5)

whereσcisthelongitudinalcompressivestressofconcreteεcisthelongitudinalcompressivestrainofconcrete′εcisthestrainat ′fcEcisYoung’smodulusofconcrete

Young’smodulusofconcretecanbedeterminedfromthemeasuredstress–straincurveasthesecantmodulusatastresslevelequalto0 45. ′fc .Young’smodulusofconcreteintensionisapproximatelythesameasthatofconcreteincompression.InAS3600(2001),Young’smodulusEcofnormal-strengthconcreteiscalculatedapproximatelyby

E fc cm= 0 043 1 5. .ρ MPa (1.6)

whereρisthedensityofconcreteinkg/m3

fcmisthemeancompressivestrengthofconcreteatanyparticularage

Fornormal-andhigh-strengthconcrete,thefollowingequationsuggestedbyACICommittee363(1992)canbeusedtoestimateYoung’smodulus:

E fc c= ′ +3320 6900MPa (1.7)

ItcanbeseenfromFigure1.6thatthestrain ′εcatthepeakstress ′fc ofconcretevarieswiththecompressivestrengthofconcrete.Thevalueofstrain ′εcisbetween0.002and0.003.Forthecompressivestrengthofconcretelessthan28MPa,thestrain ′εcis0.002,whileitcanbetakenas0.003forthecompressivestrengthofconcretehigherthan82MPa.Whenthecompressivestrengthofconcreteisbetween28and82MPa,thestrain ′εccanbedeterminedbylinearinterpolation.Poisson’sratio(ν)forconcreteisintherangeof0.15–0.22andcanbetakenas0.2intheanalysisanddesignofpracticalstructures.

Thetensilestrengthofconcreteappearstobemuchlowerthanitscompressivestrengthand it may be ignored in some design calculations. However, it needs to be taken intoaccount in the nonlinear inelastic analysis of composite beams and columns in order tocapturethetruebehaviours.Testssuchasdirecttensiontests,cylindersplittestsorflexuraltestscanbeconductedtodeterminethetensilestrengthofconcrete.However,thetensilestrengthofconcreteisoftenestimatedfromitscompressivestrength.InAS3600(2001),thecharacteristicflexuraltensilestrengthat28 daysisgivenby

′ = ′f fcf c0 6. MPa (1.8)

Indirecttension,thecharacteristicprincipaltensilestrengthofconcreteat28 daysmaybetakenas

′ = ′f fct c0 4. MPa (1.9)

Introduction 11

Anidealisedstress–straincurveisusuallyassumedforconcreteintensioninthenonlinearanalysis(Liang2009).Thetensionstressincreaseslinearlywithanincreaseintensilestrainuptoconcretecracking.Afterconcretecracking,thetensilestressdecreaseslinearlytozeroastheconcretesoftens.Theultimatetensilestrainistakenas10timesthestrainatcracking.

1.4.4.2 Time-dependent properties

Thestrainofaconcretememberunderasustainedloadisnotconstantandrather,itgradu-allyincreaseswithtime.Thistime-dependentbehaviourofconcreteiscausedbycreepandshrinkage.Creepstrainisinducedbythesustainedstressandisbothstressdependentandtimedependent.Shrinkagestrainismainlycausedbythelossofwaterinthedryingprocessof theconcreteand is stress independentand timedependent.Creepandshrinkagemayinduceaxialandlateraldeformationsofcompositesections,stressredistributionbetweentheconcreteandsteelcomponentsandlocalbucklingofsteelsectionsincompositemem-bers. More details on the time-dependent properties of concrete can be found in books(Gilbert1988;GilbertandMickleborough2004).

Consideraconcretememberunderaconstant sustainedaxial stressσofirstappliedattime τo.The total strainatany timegreater thanτo consistsof the instantaneous strainεel(τo),creepstrainεcr(t,τo)andshrinkagestrainεsh(t)asdemonstratedinFigure1.7andcanbeexpressedby

ε ε τ ε τ ε( ) ( ) ( , ) ( )t t tel o cr o sh= + + (1.10)

Theinstantaneousstrainεel(τo)oftheconcreteatserviceloadsisusuallylinearelasticandisgivenby

ε τ σel o

o

cE( ) = (1.11)

Thecreepfunctionorfactorisusuallyusedtoevaluatethecapacityofconcretetocreep,whichisdefinedastheratioofthecreepstraintotheinstantaneousstrainas

φ τ ε τ

ε τc ocr o

el o

tt

( , )( , )( )

= (1.12)

Time

εcr(t,τo)

εel(to)

εsh(t)

τo t

Strain

Figure 1.7 Time-dependent strain for concrete under constant stress.


Fromthisequation,thecreepstraincanbewrittenas

ε τ σ φ τcr o

o

cc ot

Et( , ) ( , )= (1.13)

At the time infinity, thecreepfunctionapproaches itsfinalmaximumvalueφc∗,which is

usuallyintherangeof1.5–4.0.Thestrainattimetcausedbyaconstantsustainedstressσoconsistsoftheelasticandcreepcomponentsasfollows(GilbertandMickleborough2004):

ε τ ε τ σ σ φ τ σ φ τ σel o cr o

o

c

o

cc o

o

cc o

o

ce

tE E

tE

tE

( ) ( , ) ( , ) ( , )+ = + = +[ ] =1(( , )t oτ

(1.14)

whereEce(t,τo)istheeffectivemodulusofconcreteandisexpressedby

E t

Et

ce oc

c o

( , )( , )

τφ τ

=+1

(1.15)

Thecompressivestressmaybegraduallyappliedtotheconcrete.Thisreducessignificantlythecreepstrainoftheconcreteduetotheagingoftheconcrete.ForastressincrementΔσ,thestress-dependentstrainisgivenby(Trost1967;Bažant1972)

ε τ ε τ σ χ φ τ σ

τel o cr o

ca c o

ce o

tE

tE t

( ) ( , ) ( , )( , )

+ = +[ ] = ∗∆ ∆

1 (1.16)

whereχa istheagingcoefficient(Trost1967;Bažant1972)E tce o∗( , )τ istheage-adjustedeffectivemodulusforconcrete,whichisexpressedby

E t

Et

ce oc

a c o

∗ =+

( , )( , )

τχ φ τ1

(1.17)

Theagingcoefficientχaisintherangeof0.6–1.0andisafunctionofthedurationofloadingandtheageatthefirstloading.

The shrinkage strain decreases with time. At the time infinity, the shrinkage strainapproachesitsfinalvalueεsh∗ .Theshrinkagedependsonallfactorsthatinfluencethedryingofconcrete,includingtherelativehumidity,themixdesignandthesizeandshapeoftheconcretemember.Thebasicshrinkagestrainofconcretecanbetakenas850×10−6assug-gestedinAS3600(2001).

referenceS

ACICommittee363(1992)StateoftheArtReportonHigh-StrengthConcrete,ACIPublication363R-92,Detroit,MI:AmericanConcreteInstitute.

AS3600 (2001)AustralianStandard forConcrete Structures,Sydney,NewSouthWales,Australia:StandardsAustralia.

Introduction 13

AS 4100 (1998) Australian Standard for Steel Structures, Sydney, New South Wales, Australia:StandardsAustralia.

Bažant,Z.P.(1972)Predictionofconcretecreepeffectsusingage-adjustedeffectivemodulusmethod,ACIJournal,69:212–217.

Gilbert,R.I.(1988)TimeEffectsinConcreteStructures,Amsterdam,theNetherlands:Elsevier.Gilbert,R.I. andMickleborough,N.C. (2004)DesignofPrestressedConcrete,London,U.K.:Spon

Press.Liang,Q.Q.(2005)Performance-BasedOptimizationofStructures:TheoryandApplications,London,

U.K.:SponPress.Liang,Q.Q.(2009)Performance-basedanalysisofconcrete-filledsteeltubularbeam-columns,partI:

Theoryandalgorithms,JournalofConstructionalSteelResearch,65(2):363–372.Liang,Q.Q.,Uy,B.andSteven,G.P.(2002)Performance-basedoptimizationforstrut-tiemodelingof

structuralconcrete,JournalofStructuralEngineering,ASCE,128(6):815–823.Liang,Q.Q.,Xie,Y.M.andSteven,G.P.(2000a)Optimaltopologydesignofbracingsystemsformulti-

storysteelframes,JournalofStructuralEngineering,ASCE,126(7):823–829.Liang, Q.Q., Xie,Y.M. and Steven, G.P. (2000b)Topology optimization of strut-and-tie models in

reinforcedconcretestructuresusinganevolutionaryprocedure,ACIStructuralJournal,97(2):322–330.

Liang,Q.Q.,Xie,Y.M.andSteven,G.P.(2001)Generatingoptimalstrut-and-tiemodelsinprestressedconcretebeamsbyperformance-basedoptimization,ACIStructuralJournal,98(2):226–232.

Mander,J.B.,Priestley,M.J.N.andPark,R.(1988)Theoreticalstress–strainmodelforconfinedcon-crete,JournalofStructuralEngineering,ASCE,114(8):1804–1826.

Pham,L.,Bridge,R.Q.andBradford,M.A.(1986)Calibrationoftheproposedlimitstateddesignrulesforsteelbeamsandcolumns,CivilEngineeringTransactions,InstitutionofEngineersAustralia,28(3):268–274.

Trahair,N.S.andBradford,M.A.(1998)TheBehaviourandDesignofSteelStructurestoAS4100,3rdedn.(Australian),London,U.K.:Taylor&FrancisGroup.

Trost,H.(1967)Auswirkungendessuperpositionsprinzipsaufkirech-undrelaxationsproblemebeibetonundspannbeton,Beton-undStahlbetonbau,62:230–238,261–269.

Warner, R.F., Rangan, B.V., Hall, A.S. and Faulkes, K.A. (1998) Concrete Structures, Melbourne,Victoria,Australia:AddisonWesleyLongman.

Zienkiewicz, O.C. and Taylor, R.L. (1989) The Finite Element Method, 4th edn., Vol. 1, BasicFormulationandLinearProblems,NewYork:McGraw-Hill.

Zienkiewicz,O.C.andTaylor,R.L.(1991)TheFiniteElementMethod,4thedn.,Vol.2,SolidandFluidMechanics,DynamicsandNonlinearity,NewYork:McGraw-Hill.

15

Chapter 2

design actions

2.1 IntroductIon

Inordertodesignasteelorcompositestructure,thestructuraldesignermustestimatethedesignactions(loads)actingonthestructure.Designactionsonsteelandcompositestruc-turesmaybedividedintopermanentactions,imposedactions,windactions,snowactions,earthquake actions and other indirect actions caused by temperature, foundation settle-mentandconcreteshrinkages.Thestructuraldesignermustdeterminenotonlythetypesandmagnitudesofdesignactionswhichwillbeappliedtothestructurebutalsothemostseverecombinationsofthesedesignactionsforwhichthestructuremustbedesigned.Thecombinationsofdesignactionsareundertakenbymultiplyingthenominaldesignactionsusingloadfactors.

Theaccurateestimationofdesignactionsonthestructureisveryimportantinstructuraldesignasitsignificantlyaffectsthefinaldesignandobjectives.Anyerrorintheestimationofdesignactionsmayleadtowrongresultsofstructuralanalysisonthestructureandleadtotheunrealisticsizingofitsstructuralmembersorevencollapseofthestructure.AS/NZS1170.0(2002)providesspecificationsontheestimationofdesignactionsbasedonstatisticalorprob-abilisticanalysesowingtouncertaintiesaboutdesignactionsonstructures.TheevaluationofpermanentandimposeddesignactionsisstraightforwardinaccordancewithAS/NZS1170.1(2002).However,theproceduregivenintheAS/NZS1170.2(2011)fordeterminingthewindactionsonbuildingsisquitecomplicated,particularlyforirregularandsensitivestructures.Thedetailedtreatmentofthecalculationofwindactionsisgiveninthischapter.

In this chapter, the estimation of design actions on steel and composite structures inaccordancewithAS/NZS1170.0,AS/NZS1170.1andAS/NZS1170.2ispresented.Thediscussiononpermanentactionsisgivenfirst.Thisisfollowedbythedescriptionofimposedactionsforvariousstructures.Thebasicprocedureandunderliningprincipalsfordetermin-ingwindactionsarethenprovided.Thecombinationsofactionsforultimatelimitstatesandserviceabilitylimitstatesarediscussed.Finally,aworkedexampleisprovidedtoillustratetheprocedureforcalculatingwindactionsonanindustrialbuilding.ThischaptershouldbereadwithAS/NZS1170.0(2002),AS/NZS1170.1(2002)andAS/NZS1170.2(2011).

2.2 PermAnent ActIonS

Permanentactionsareactionsactingcontinuouslyonastructurewithoutsignificantchangesinmagnitudeinitsdesignlife.Permanentactionsarecalculatedastheself-weightofthestructure including finishes, permanent construction materials, permanent equipments,fixedormovablepartitionsandstoredmaterials.Theself-weightofastructuralmemberiscalculatedfromitsdesignorknowndimensionsandtheunitweight,whichisgivenin


TablesA1andA2ofAS/NZS1170.1(2002).Itshouldbenotedthattheunitweightsofmaterialsgiveninthecodeareaveragevaluesforthespecificmaterials.

Thecalculatedself-weightofpermanentpartitionsmustbeappliedtotheactualpositionsinthestructure.Ifastructureisdesignedtoallowformovablepartitions,thecalculatedself-weightof themovablepartitionscanbeapplied toanyprobablepositionswhere thepartitionsmaybeplaced.Thestructuremustbedesignedforthedesignactions.AS/NZS1170.1requiresthataminimumuniformlydistributedpermanentloadof0.5kPashallbeusedtoconsidermovablepartitions.Thisistoensurethatthemassofthemovableparti-tionsistakenintoaccountindesigningthestructureunderanearthquake.Inaddition,theminimumloadof0.5kPaisadequatetocovertheself-weightofmostpartitionsmadeofstudssupportingglass,plywoodandplasterboard.

2.3 ImPoSed ActIonS

Imposedactions(orliveloads)areloadsonthestructurewhicharisefromtheintendeduseofthestructure,includinggraduallyappliedloads(staticloads)anddynamicloadssuchascyclic loadsand impact loads.The live loadsarecharacterisedby their time-dependencyandrandomdistributionsinspace.Themagnitudesanddistributionsofliveloadsvarysig-nificantlywiththeoccupancyandfunctionofthestructure.Imposedactionsonastructurevaryfromzerotothemaximumvalueswhichoccurrarelyandareregardedasthemaxi-mumloadsinthedesignlifeofthestructure.

TheimposedactionsgiveninAS/NZS1170.1arecharacteristicloads,whichrepresentthepeakloadsovera50-yeardesignlifehavinga5%probabilityofbeingexceeded.ImposedflooractionsaregiveninTable3.1ofAS/NZS1170.1(2002).Theuniformlydistributedloads(UDLs)andconcentratedloadsarelistedinthetable.Theconcentratedloadsareusedtorep-resentthelocalisedloadscausedbyheavyequipmentsorvehiclesthatmaynotbeadequatelycoveredbytheUDLs.However,itshouldbenotedthatthedistributedandconcentratedliveloadsshouldbeconsideredseparatelyandthestructuremustbedesignedforthemostadverseeffectofdesignactions.Theliveloadsgivenintheloadingcodeconsidertheimportanceanddesignworkinglifeofthestructure,whichareassumedtobepartoftheoccupancydescrip-tion.Thisimpliesthatoncetheoccupancyofthestructurehasbeendetermined,theimposedloadscanbeusedtodesignthestructureregardlessofitsimportanceanddesignworkinglife.

AS/NZS1170.1allowsforconsiderationofpatternloadingforliveloads.Thepurposeforthisistoaccountforthemostadverseeffectsofliveloadsonthestructure.Theconsider-ationofpatternloadingdependsontheratioofdeadtoliveloadandthetypeofstructuralmember.Fora structure subjected towind, earthquakeorfire loading,pattern imposedloadingoncontinuousbeamsorslabsneednotbeconsidered.

InAS/NZS1170.1,areductionfactorψaisusedtoreducetheuniformlydistributedliveloadsbasedontheresultsofloadsurveys.Thereductionfactorψaistakenas1.0forareasusedforoccupancytypesC3–C5specifiedinTable3.1ofAS/NZS1170.1,storageareassubjectedtoimposedloadsexceed5kPa,lightandmediumtrafficareasandone-wayslabs.Forotherareas,Clause3.4.2ofAS/NZS1170.1(2002)providesthefollowingformulafordeterminingthereductionfactorψa:

ψ ψa

ta

A= + ≤ ≤0 3

30 5 1 0. ( . . ) (2.1)

whereAt(m2)isthesumofthetributaryareassupportedbythestructuralmemberunderconsideration.Thereductionfactorψamustnotbegreaterthan1.0andnotlessthan0.5.

Design actions 17

Theroofsofindustrialbuildingsareusuallynon-trafficable.Forstructuralelementssuchaspurlinsandraftersandcladdingprovidingdirectsupport,theuniformlydistributedliveloadiscalculatedbythefollowingformulagiveninClause3.5.1ofAS/NZS1170.1(2002):

w

AQ

pa

= +

≥0 12

1 80 25.

.. kPa (2.2)

where Apa (m2) is the plan projection of the surface area of the roof supported by thestructuralmember.Theaforementionedformularepresentsanimposeddistributedloadof0.12kPaplusaconcentratedloadof1.8kNwhichisdistributedovertheareaApasup-portedbythestructuralmember.Theconcentratedloadof1.8kNistoaccountfortheweightofaheavyworkerstandingontheroof.AsshowninTable3.2ofAS/NZS1170.1,the structuralelementsof the roofmustbedesigned tosupportaconcentrated loadof1.4 kNatanypointandthecladdingmustsupportaconcentratedloadof1.1kN.

2.4 WInd ActIonS

WindactionsonstructuralmembersandstructuresorbuildingsarespecifiedinAS/NZS1170.2(2011).Thedesignofbuildings,particularlyindustrialbuildings,isinfluencedsig-nificantlybywindloads.Therefore,itisimportanttocarefullyestimatethewindloadsinaccordancewithloadingcodes.Windloadsarebothtimedependentandspacedependent.Theestimationofwindloadingisrelativelycomplicatedasitdependsonthelocationanddirectionofthebuildingbeingdesigned,siteconditionsrelatedtoterrain/height,shieldingandtopography,theshapeofthebuildingandthefundamentalfrequencyofthestructure(Holmesetal.1990;Holmes2001).TheestimationofwindactionsinaccordancewithAS/NZS1170.2isdescribedinthesubsequentsections.

2.4.1 determination of wind actions

Themainstepsfordeterminingwindactionsonstructuralmembersorstructuresaregivenasfollows:

1.Determinethesitewindspeeds. 2.Determinethedesignwindspeedfromthesitewindspeeds. 3.Calculatethedesignwindpressuresanddistributedforces. 4.Computewindactions.

InClause2.2ofAS/NZS1170.2(2011),thesitewindspeedsaredefinedfortheeightcardi-naldirectionsatthereferenceheightabovethegroundandarecalculatedby

V V M M M Msit R d z cat s t, ,( )β = (2.3)

whereVRistheregional3sgustwindspeed(m/s)forannualprobabilityofexceedanceof1/RMd isthewinddirectionalmultipliersfortheeightcardinaldirectionsMz cat, istheterrain/heightmultiplierMsistheshieldingmultiplierMtisthetopographicmultiplier


Thewind speed isgenerallydeterminedat theaverage roofheight (h)of thebuildingasshown inFigure2.1. If theorientationof thebuildingbeingdesigned isnotknown, theregionalwindshouldbeassumedtoactfromanycardinaldirectionsandMdcanbeconser-vativelytakenas1.0foralldirections.

Thebuildingorthogonaldesignwindspeed(Vdes,θ)isdeterminedasthemaximumcar-dinaldirectionsitewindspeed(Vsit,β)withinasectorof±45°totheorthogonaldirectionbeingconsidered.Thedesignwindspeed(Vdes,θ)mayvarywiththeorthogonaldirection.Itisrequiredthatfourorthogonaldirectionsmustbeconsideredinthedesignofabuilding.Thestructurecanbeconservativelydesignedbyusingthesitewindspeedandmultipliersfortheworstdirection.Theminimumdesignwindspeed(Vdes,θ)of30m/sissuggestedinAS/NZS1170.2fortheultimatelimitstatedesign.

Thedesignwindpressureactingnormaltothesurfaceofastructuralmemberorbuild-ingcanbecalculatedinaccordancewithClause2.4.1ofAS/NZS1170.2(2011)asfollows:

p V C Cair des fig dyn= 0 5 2. ,ρ θ (2.4)

wherepisthedesignwindpressure(Pa)ρairisthedensityofairtakenas1 2. kg/m2

CfigistheaerodynamicshapefactorCdynisthedynamicresponsefactor

Thedesignwindfrictionaldragforceperunitarea(f)onstructuralmembersandstructurescanalsobecalculatedusingEquation2.4.

Windactionsonastructureshouldbedeterminedbyconsideringthewindfromnofewerthanfourorthogonaldirections.TheClause2.5.3.1ofAS/NZS1170.2specifiesthattheforcesactingonstructuralmembersorsurfacesarecalculatedby

F p Az z=∑( ) (2.5)

whereFdenotestheforce(N)derivedfromwindactionspzstandsforthedesignwindpressure(Pa)normaltothesurfaceatheightzAzisthereferencearea( )m2 onwhichthewindpressurepzactsatheightz

h

Figure 2.1 Average roof height of structure.

Design actions 19

Forenclosedbuildings,externalpressuresaccountingfortheeffectsoflocalpressurefac-torsshouldbecombinedwithinternalpressuresandthestructuremustbedesignedforthemostseverecombinationsofwindactions.

2.4.2 regional wind speeds

Theregionalwindspeedsgiven inAS/NZS1170.2weredeterminedfromtheanalysesoflong-termrecordsofdailymaximumwindspeedsforeachparticularregioninAustralia.Theregionalwindspeedsareafunctionofthestandardsiteexposure,peakgust,annualproba-bilityofexceedanceandwinddirection.Thestandardsiteexposurerepresents10mheightinterraincategory2,whichisdefinedinSection2.4.3.1.Theannualprobabilityofexceedanceofthewindspeedistheinverseofthereturnperiodoraveragerecurrenceinterval,whichisrelatedtotheimportancelevelofthestructure.InAS/NZS1170.0,structuresareclassifiedintofiveimportancelevelsaccordingtotheirconsequenceoffailure,whicharegivenTableF1ofAS/NZS1170.0(2002).Oncethedesignworkinglifeandimportancelevelofthestruc-turehavebeendetermined,theannualprobabilityofexceedanceofthewindspeedcanbeobtainedfromTableF2ofAS/NZS1170.0.Theregionalwindspeeds(VR)foralldirectionsbasedon3sgustwinddataaregiveninTable3.1ofAS/NZS1170.2(2011).WindregionsinAustraliaareprovidedinAS/NZS1170.2.Theimportancelevelofnormalstructuresis2.Fornormalstructures,theannualprobabilityofexceedanceofthewindspeedis1/500.Theregionalwindspeeds(VR)fornormalstructuresareprovidedinTable2.1.

Thewinddirectionalmultipliers(Md)forregionsAandWareprovidedinTable3.2ofAS/NZS1170.2(2011).Thesesmultiplierscanbeusedforstrengthandserviceabilitylimitstatedesignsandwerederivedbasedontheassumptionthatonlythewindloadwithinthetwo45°directionalsectorsofthetypicalrectangularbuildingscontributetotheprobability(Melbourne1984).ItcanbeseenfromTable3.2ofAS/NZS1170.2thatthewinddirec-tionalmultiplier(Md)variesfrom0.8to1.0fortheeightcardinaldirections.However,itshouldbenotedthatiftheorientationofthebuildingbeingdesignedinregionsAandWisnotknown,thewindshouldbeassumedtoactinanydirectionsothatMdistakenas1.0.

ForbuildingsinregionsB–D,Clause3.3.2ofAS/NZS1170.2(2011)suggeststhatthewinddirectionalmultiplier(Md)foralldirectionsistakenasfollows:

1.0.95forcalculatingtheresultantforcesandoverturningmomentsonacompletebuild-ingandwindloadsonmajorstructuralframingmembers

2.1.0forallotherdesignsituations

TheregionsCandDarecyclonicregionswherethedirectionalmultipliersarenotused.Thisisbecausethemaximumwindspeedmayoccurinanydirection.However,thewinddirectionalmultiplierofMd=0.95canbeappliedtothewindspeedwhenitisusedtocalcu-latetheresultantforcesandovertrainingmomentsonacompletebuildingandwindactions

Table 2.1 Regional wind speeds for R500 for normal structures

Wind region Regional wind speed

(m/s)

Non-cyclonic A 45W 51B 57

Cyclonic C 66FC

D 80FD


onmajorstructuralmembersinregionsB–D.Thisfactorisusedtoaccountfortheaver-ageprobabilityofthedesignwindspeedbeingexceededforthebuilding(Davenport1977;Holmes1981).Forotherdesignsituations,suchasnon-majorstructuralmembersincludingcladdingandimmediatesupportingmembers,Mdistakenas1.0.

FactorsFCandFDappliedtowindspeedsinregionsCandDasgiveninTable2.1aretakenasFC=1.05andFD=1.1forR≥50 yearsandFC=FD=1.0forR<50 years.

2.4.3 Site exposure multipliers

Theexposuremultipliers(Mz,cat, Ms, Mt)areusedtoaccountfortheeffectsofthesitecondi-tionsofthebuildingonthesitewindspeeds(Vsit,β),whichincludeterrain/height,shieldingandtopography.Theterrainandsurroundingbuildingsprovidingshieldingmaychangeinthedesignworkinglifeofthebuildingduetonewdevelopmentinthearea.Therefore,itisimportanttoconsidertheknownfuturechangestotheterrainroughnessinevaluatingtheter-raincategoryandtothebuildingsthatprovideshieldinginestimatingtheshieldingmultiplier.

2.4.3.1 Terrain/height multiplier (Mz,cat)

The terrain/height multiplier (Mz,cat) varieswith the terrain roughness andheight of thebuilding.InClause4.2.1ofAS/NZS1170.2(2011),theterrainisdividedintofourcatego-riesasfollows:

1.Terrain category1 includes exposedopen terrainwith fewornoobstructions andwatersurfacesatserviceabilitywindspeeds.

2.Terraincategory2coversopenwatersurfaces,openterrain,grasslandandairfieldswithfew,well-scatteredobstructionswithheightsgenerallyfrom1.5to10m.

3.Terraincategory3includestheterrainwithnumerouscloselyspacedobstructionswith3–5mheight,forexample,intheareasofsuburbanhousingandlevelwoodedcountry.

4.Terraincategory4coverslargecitycentresandwell-developedindustrialareaswithnumerouslargeandcloselyspacedobstructionswithheightsfrom10to30m.

Theterrain/heightmultipliersforgustwindspeedsforfullydevelopedterrainsinallregionsforserviceability limitstatedesignandinregionsA1–A7,WandB forultimate limitstatedesignaregiveninTable4.1(A)ofAS/NZS1170.2.Itappearsfromthetablethattheterrain/heightmultiplier(Mz,cat)is1.0forbuildingheightof10minterraincategory2asthiscondi-tionisusedasareferenceforothercategoriesandbuildingheights.FortheultimatelimitstatedesignofbuildingsinregionsCandDwhicharecyclonicregions,theterrain/heightmultipli-ersareprovidedinTable4.1(B)ofAS/NZS1170.2.Theterrain/heightmultipliersforbuildingsinterraincategories1and2havingthesameheightarethesameandthisholdstrueforbuild-ingsinterraincategories3and4.ThedesigncodeallowsforMz,cattobetakenastheweightedaveragevalueovertheaveragingdistanceupwindofthebuildingwhentheterrainchanges.

2.4.3.2 Shielding multiplier (Ms)

Theshieldingmultiplier(Ms)isusedtoaccountfortheeffectsoftotalandlocalwindactionsonstructureswitharangeofshieldingconfigurations(HolmesandBest1979;HussainandLee1980). Itdependson theshielding factors including theaveragespacing, roofheightandbreadthofshieldingbuildingsnormaltothewinddirection,theaverageofroofheightofthebuildingbeingshieldedandthenumberofupwindshieldingbuildingswithina45°sectorofradius20h.Itshouldbenotedthatonlybuildingslocatedina45°sectorofradius

Design actions 21

20hsymmetricallypositionedaboutthedirectionandwhoseheightisgreaterthanorequaltotheaverageroofheightofthebuildingbeingshieldedcanprovideshieldingasdepictedinFigure2.2.Theshieldingmultiplier forbuildingswithvariousshieldingparameters isprovidedinTable4.3ofAS/NZS1170.2.Iftheaverageupwindgradientisgreaterthan0.2ornoshieldinginthewinddirection,theshieldingmultiplieristakenas1.0.

Clause4.3.3ofAS/ZNS1170.2(2011)providesequationsforcalculatingtheshieldingparametergiveninTable4.3ofAS/NZS1170.2asfollows:

s

lh bs

s s

= (2.6)

l h

ns

s

= +

105 (2.7)

wherels,hs andbs are the average spacing, roofheight andbreadthof shieldingbuildings,

respectivelyhistheaverageroofheightofthestructurebeingshieldednsisthetotalnumberofupwindshieldingbuildingswithina45°sectorofradius20h

Wind direction

Non-shielding building Shielding building

20h

45°

Building being designed

Figure 2.2 Shielding in complex urban areas.


2.4.3.3 Topographic multiplier (Mt)

Thetopographicmultiplier(Mt)considerstheeffectsoflocaltopographiczonesonthesitewindspeeds.InClause4.4.1ofAS/NZS1170.2(2011),forsitesinTasmaniaover500mabovethesealevel,MtistakenasMt=MhMlee(1+0.00015Esl),whereMhisthehill-shapemultiplier,theleemultiplierMlee=1andEslisthesiteelevationabovethemeansealevel(m).ForAustraliasites,MtistakenasthelargervalueofMhandMlee.Thehill-shapemultiplieris takenas1.0 except that for theparticular cardinaldirection in the local topographiczonesshowninFigures4.2through4.4ofAS/NZS1170.2.Forthelocaltopographiczonesdepictedinthesefigures,thehill-shapemultiplier(Mh)isgiveninClause4.4.2ofAS/NZS1170.2(2011)asfollows:

M

HL

Hz L

x

LHL

h

u

u

=

<

++

−

≤ <

1 02

0 05

13 5

1 0 0521 2

. .

. ( ).

for

for 0.45

11 0 71 122

+ −

>

.x

LHLu

for 0.45 (in separation zone)

(2.8)

whereHistheheightofthehill,ridgeorescarpmentLuisthehorizontaldistanceupwindfromthecrestofthehill,ridgeorescarpmentto

thelevelhalftheheightbelowthecrestzisthereferenceheightonthestructurefromtheaveragelocalgroundlevelxisthehorizontaldistancefromthestructuretothecrestofthehill,ridgeorescarpmentL1isthelengthscale(m)whichisthelargerof0 36. Luand0.4HL2isthelengthscale(m)whichistakenas4 1L upwindforalltypesanddownwindfor

hillsandridgesor10 1L downwindforescarpments

ItshouldbenotedthatforH/(2Lu)>0.45andinzonesotherthantheseparationzone,Mhistakenasthatfor0.05≤H/(2Lu)<0.45(Bowen1983;PatersonandHolmes1993).

Thehill-shapemultiplier(Mh)forthelocaltopographiczoneswithxandzarezeroisgiveninTable4.4ofAS/NZS1170.2.ForAustraliasites,theleemultiplier(Mlee)istakenas1.0.

2.4.4 Aerodynamic shape factor

2.4.4.1 Calculation of aerodynamic shape factor

Theaerodynamicshapefactor(Cfig)considerstheeffectsofthegeometryofastructureonthesurfacelocal,resultantoraveragewindpressure.Itisafunctionofthegeometryandshapeofthestructureandtherelativewinddirectionandspeed(ISO4354,1997).Thesignconven-tionoftheaerodynamicshapefactor(Cfig)assumesthatpositivevaluesindicatepressureact-ingtowardsthesurfaceandnegativevaluesindicatepressureactingawayfromthesurface.

Forenclosedbuildings,theaerodynamicshapefactorisgiveninClause5.2ofAS/NZS1170.2(2011)asfollows:

C C K K K Kfig p e a c e l p= , , for externalpressures (2.9)

Design actions 23

C C Kfig p i c i= , , for internalpressures (2.10)

C C K Kfig f a c= for frictionaldrag forces (2.11)

whereCp e, istheexternalpressurecoefficientKaistheareareductionfactorKcisthecombinationfactor(Kc e, forexternalpressuresandKc i, forinternalpressures)Kl isthelocalpressurefactorKpistheporouscladdingreductionfactorCp i, istheinternalpressurecoefficientCf isthefrictionaldragforcecoefficient

2.4.4.2 Internal pressure coefficient

Internalpressuredependsontherelativepermeabilityoftheexternalsurfacesofabuild-ing.Itmaybepositiveornegativethatdependsonthepositionandsizeoftheopening.Thepermeabilityofasurfaceiscalculatedasthesumoftheareasofopeningandleakageonthatsurfaceofthebuilding.Opendoorsandwindows,vents,ventilationsystemsandgapsincladdingare typicalopenings.Thedominantopeningmeans that itplaysadominanteffectontheinternalpressureinthebuilding.Ifthesumofallopeningsinthesurfaceisgreaterthanthesumofopeningsineachoftheothersurfacesinthebuilding,thesurfaceis regarded as containing dominant openings. Internal pressure coefficients for enclosedrectangularbuildingsaregiveninClause5.3ofAS/NZS1170.2.Table5.1(A)ofAS/NZS1170.2providestheinternalpressurecoefficientsforbuildingswithopeninteriorplanandhavingpermeablewallswithoutdominantopenings.Forbuildingswithopeninteriorplanandhavingdominantopeningsononesurface, internalpressurecoefficientsaregiven inTable5.1(B)ofAS/NZS1170.2.

Officesandhouseswithallwindowsclosedusuallyhavepermeabilitybetween0.01%and0.2%ofthewallarea,whichdependsonthedegreeofdraughtproofing.Thepermeabilityofindustrialandfarmbuildingscanbeupto0.5%ofthewallarea.Thewallsofindustrialbuildingsareusuallyconsideredtobepermeable,whileconcrete,concretemasonryorotherwallsdesignedtopreventairpassagemaybetreatedasnon-permeable.

2.4.4.3 External pressure coefficient

Forenclosedrectangularbuildings,externalpressurecoefficientsaregiveninTables5.2(A)to5.2(C)forwallsandTables5.3(A)to5.3(C)forroofsofAS/NZS1170.2(2011).Itcanbeobservedfromthesetablesthatinsomecases,twovaluesaregivenforthepressurecoef-ficient.Forthesecases,roofsurfacesmaybesubjectedtoeithervalueowingtoturbulencesothatroofsshouldbedesignedforbothvalues.Alternatively,externalpressuresarecombinedwithinternalpressurestoobtainthemostseverecombinationsofactionsforthedesignofthestructure.Discussionsonexternalpressuresonlow-risebuildingandmonosloperoofsaregivenbyHolmes(1985)andStathopoulosandMohammadian(1985).

Forcrosswindroofslopesandforallroofpitches(α),thevaluesgiveninTables5.3(A)and5.3(B)shouldbeusedtodeterminethemostsevereactioneffectsasfollows:

1.Applythemorenegativevalueofthetwogiveninthetabletobothhalvesoftheroof. 2.Applythemorepositivevalueofthetwogiveninthetabletobothhalvesoftheroof. 3.Applythemorenegativevaluetoonehalfandmorepositivevaluetotheotherhalf

oftheroof.


2.4.4.4 Area reduction factor

Theareareductionfactor(Ka)forroofsandsidewallsdependsonthetributaryarea(At),whichisdefinedastheareacontributingtotheforceunderconsideration.InClause5.4.2ofAS/NZS1170.2,theareareductionfactorKaistakenas1.0fortributaryareaAt ≤10 m2,0.9forAt =25 m2and0.8forAt ≥100 m2.Forintermediateareas,linearinterpolationcanbeusedtodeterminetheareareductionfactor.Forallothercases,theareareductionfactoristakenas1.0.Thevaluesgiveninthecodeweredeterminedbydirectmeasurementsoftotalroofloadsinwindtunneltests(Davenportetal.1977;KimandMehta1977).

2.4.4.5 Combination factor

Thecombinationfactor(Kc)accountsfortheeffectsofnon-coincidenceofpeakwindpres-suresondifferent surfacesof thebuilding.Forexamples,wallpressuresarewellcorre-latedwithroofpressures.Table5.5ofAS/NZS1170.2givescombinationfactorsKc,iandKc,eforwindpressuresonmajorstructuralelementsofanenclosedbuilding.However,itshouldbenotedthatthecombinationfactorsdonotapplytocladdingorpurlins.Whentheareareductionfactor(Ka)islessthan1.0,Kcforallsurfacesmustsatisfythefollowingcondition:

K

Kc

a

≥ 0 8. (2.12)

2.4.4.6 Local pressure factor

Thewindpressuresonsmallareasareevaluatedusingthelocalpressurefactor(Kl).ThepeakwindpressuresoftenoccuronsmallareasnearwindwardcornersandroofedgesofthebuildingasdepictedinFigure2.3,wherea=min(0.2b, 0.2d, h).Thelocalpressurefac-torisappliedtocladding,theirfixingsandthemembersthatdirectlysupportthecladdingandisgiveninTable5.6ofAS/NZS1170.2.ForareasSA1,RA1,RA3andWA1,Kl=1.5.ForareasSA2,RA2andRA4,Kl=2.ForothercaseswhicharenotspecifiedinthistableorFigure2.3,thelocalpressurefactoristakenas1.0.

2.4.4.7 Permeable cladding reduction factor

It has been found that negative surface pressures onpermeable cladding are lower thanthoseonasimilarnon-permeablecladdingowingtotheporoussurface.Thiseffectistakenintoaccountindeterminingtheaerodynamicshapefactorbythepermeablecladdingreduc-tionfactor(Kp),whichisgiveninClause5.4.5ofAS/NZS1170.2.Itshouldbenotedthatthisfactorisusedfornegativepressureonlywhenexternalsurfacesconsistingofpermeablecladdingandthesolidityratioislessthan0.999andgreaterthan0.99.Thesolidityratioofthesurfaceisdefinedastheratioofsolidareatothetotalareaofthesurface.

2.4.4.8 Frictional drag coefficient

Thefrictionaldragforcesonroofsandsidewallsofenclosedbuildingswitharatioofd/hord/bthatisgreaterthan4needtobeconsideredwhendesigningtheroofandwallbrac-ingsystems.Whendeterminingfrictionaldragforces,theaerodynamicshapefactor(Cfig)istakenasthefrictionaldragcoefficient(Cf)inthedirectionofthewind,whichisgiveninClause5.5ofAS/NZS1170.2(2011).

Design actions 25

2.4.5 dynamic response factor

2.4.5.1 General

Thedynamicresponsefactor(Cdyn)accountsforthedynamiceffectsofwindonflexible,lightweight,slenderorlightlydampedstructures.Itconsidersthecorrelationeffectsoffluc-tuatingalong-windforcesontallstructures,effectivepressuresduetoinertiaforces,reso-nantvibrationsandfluctuatingpressuresinthewakeofthestructure(ISO4354,1997).Thedynamicresponsefactordependsonthenaturalfirstmodefundamentalfrequenciesofthestructure.Moststructuresarenotflexible,lightweight,slenderorlightlydampedsothattheyarenotdynamicallywindsensitive.Thenaturalfirstmodefundamentalfrequenciesofmoststructuresaregreaterthan1.0 Hzandtheirdynamicresponsefactoristakenas1.0.Forstructureswithnaturalfirstmodefundamentalfrequenciesbetween0.2and1.0 Hz,thedynamicresponsefactorisdeterminedforalong-windresponseandcrosswindresponsediscussedinthefollowingsections.

2.4.5.2 Along-wind response

Thealong-windresponseofmost structures isdue to the incident turbulenceof thelongitudinalcomponentofthewindvelocity(Davenport1967;Vickery1971).Fortall

a

d

b

a

RA1

RA2

RA4RA3

SA1

WA1

d

a

b

SA2

SA1

RA2RA1

SA2WA1

Figure 2.3 Local pressure areas.


buildingsand towers, thedynamic response factor (Cdyn) isgiven inClause6.2.2ofAS/NZS1170.2(2011)asfollows:

C

I g B H g S

g Idyn

h v s s R s t

v h

=+ +

+1 2

1 2

2 2( )β ζ/ (2.13)

whereIhistheturbulenceintensitywhichisgiveninAS/NZS1170.2(2011)gvisthepeakfactorfortheupwindvelocityfluctuationsandistakenas3.7Hsistheheightfactorfortheresonantresponseandiscalculatedas[ ( ) ]1 2+ s h/sistheheightofthelevelatwhichwindloadsaredeterminedforthestructurehistheaverageroofheightofthestructureabovethegroundζistheratioofstructuraldampingtocriticaldampingofthestructure

ThebackgroundfactorBsinEquation2.13isusedtomeasuretheslowlyvaryingback-groundcomponentofthefluctuatingresponseinducedbylow-frequencywindspeedvaria-tions.Thisfactorcanbecalculatedby

Bh s b L

s

sh h

=+ − +( )

1

1 36 642 2( ) (2.14)

wherebshistheaveragebreadthofthestructurebetweenheightsandhLhistheintegralturbulencelengthscaleatheighthandistakenas85 10 0 25( ) .h/

ThepeakfactorgRforresonantresponsein10 minutesperiodisexpressedby

g fR e nc= ( )2 600log (2.15)

wherefncisthefirstmodenaturalfrequencyofthestructureinthecrosswinddirectioninHz.Thesizereductionfactor(βs)isexpressedas

β

θ θs

na v h des na h v h desf h g I V f b g I V=

+ + + +

11 3 5 1 1 4 10. ( ) ( ), ,

(2.16)

wherefnaisthefirstmodenaturalfrequencyofthestructureinthealong-winddirectioninHzb h0 istheaveragebreadthofthestructurebetweenheights0andh

Thespectrumoftheturbulenceofthestructureiscalculatedby

Sf

ft

nr

nr

=+( )π

1 70 8 256.

(2.17)

Design actions 27

wherefnristhereducedfrequency,whichisdeterminedby

f

f L g IV

nrna h v h

des

= +( )

,

1

θ (2.18)

2.4.5.3 Crosswind response

Crosswindexcitationofmoderntallbuildingsandstructurescanbeexpressedbywake,incidentturbulenceandcrosswinddisplacementmechanisms(Melbourne1975).Theequiv-alentcrosswindstaticwindforceperunitheight(N/m)fortallenclosedbuildingsandtow-ersofrectangularcrosssectionsisafunctionofz.ThiswindforceisgiveninClause6.3.2.1ofAS/NZS1170.2(2011)asfollows:

w z V d C Ceq air des fig dyn( ) . ( ),= 0 5 2ρ θ (2.19)

inwhichthedesignwindspeedVdes,θisestimatedatz=handdisthehorizontaldepthofthestructureparalleltothewinddirection.Theproductoftheaerodynamicshapefactorandtheaerodynamicresponsefactorisdeterminedby

C C gbd

K

g I

zh

Cfig dyn R

m

v h

kfs=

+( )

1 5

12.

πζ

(2.20)

whereKmisthemodeshapecorrectionfactorforcrosswindaccelerationandiscalculatedas

(0.76+0.24k)kisthemodeshapepowerexponentforthefundamentalmode

Thepowerexponentis1.5foruniformcantilever,0.5foraslenderframedstructure,1.0forabuildingwithcentralcoreandmoment resistingcurtainwallsand2.3 fora towerwhosestiffnessdecreaseswithheightorthevalueobtainedfromfittingϕ1(z)=(z/h)ktothecomputedmodeshapeofthestructure.ThecoefficientCfsisthecrosswindforcespectrumcoefficientforalinearmodeshape.

ThecrosswindbaseoverturningmomentMc(Nm)canbedeterminedbyintegratingthewindforceweq(z)from0toh.Clause6.3.2.2ofAS/NZS1170.2(2011)providesaformulaforcalculatingMcasfollows:

M g b

Vg I

hk

KC

c Rair des

v hm

fs=+

+

0 5

0 51

32

2

22.

.( )

,ρ πζ

θ (2.21)

Thecrosswindforcespectrumcoefficient(Cfs)generalizedforalinearmodeshapeisafunc-tionof theaspect ratioof thecrosssectionandheight, turbulence intensityandreducedvelocity (Vn).The reducedvelocity (Vn) isprovided inClause6.3.2.3ofAS/NZS1170.2(2011)asfollows:

V

Vf b g I

ndes

nc v h

=+,

( )θ

1 (2.22)

Thecrosswindforcespectrumcoefficient(Cfs)canbedeterminedfortheturbulenceinten-sityevaluatedat2h/3inaccordancewithClause6.3.2.3ofAS/NZS1170.2(2011).


2.4.5.4 Combination of long-wind and crosswind response

InClause6.4ofAS/NZS1170.2 (2011), the totalcombinedpeakscalardynamicactioneffectsuchasanaxialloadinacolumniscalculatedby

E E E Ea t a m a p c p, , , ,= + +2 2 (2.23)

whereEa p, istheactioneffectcausedbythepeakalong-windresponseEc p, istheactioneffectcausedbythepeakcrosswindresponseEa m, istheactioneffectcausedbythemeanalong-windresponseandisgivenby

E

EC g I

a ma p

dyn v h,

,

( )=

+1 2 (2.24)

2.5 comBInAtIonS of ActIonS

2.5.1 combinations of actions for strength limit state

Structuresmaybe subjected topermanent actionG (dead load), imposedactionQ (liveload),windactionW(windload),earthquakeEoracombinationofthem.ThefollowingbasiccombinationsofactionsforthestrengthlimitstatearesuggestedinClause4.4.2ofAS/NZS1170.0(2002):

1.1.35G 2.1.2G+1.5Q 3.1 2 1 5. .G Ql+ ψ 4.1 2. G Q Wc u+ +ψ 5.0 9. G Wu+ 6.G Q Ec u+ +ψ

Intheaforementionedloadcombinations,ψlandψcarethelong-termandcombinationfac-tors,respectively,andaregiveninTable4.1ofAS/NZS1170.0,EuistheearthquakeloadandWuistheultimatewindload.

2.5.2 combinations of actions for stability limit state

Thestabilitylimitstateisanultimatelimitstatewhichisconcernedwiththelossofthestaticequilibriumof structuralmembers or thewhole structure.TheClause 4.2.1 ofAS/NZS1170.0(2002)specifiesthatifthepermanentactionscausestabilizingeffects,thecombina-tionistakenas0.9G.However,ifthecombinationsofactionscausedestabilizingeffects,thecoderequiresthatcombinationsaretakenasfollows:

1.1.35G 2.1.2G+1.5Q 3.1 2 1 5. .G Ql+ ψ 4.1 2. G Q Wc u+ +ψ 5.0 9. G Wu+ 6.G Q Ec u+ +ψ

Design actions 29

2.5.3 combinations of actions for serviceability limit state

Fortheserviceabilitylimitstate,Clause4.3ofAS/NZS1170.0(2002)statesthatappropri-atecombinationsusingtheshort-termand long-termfactorsshouldbeusedfor theser-viceabilityconditionsconsidered.Thefollowingcombinationsofdeadload,liveloadandservicewindload(Ws)maybeconsidered:

1.G Qs+ ψ 2.G Ws+ 3.G Q Ws s+ +ψ 4.Ws

Theshort-termfactorψsisgiveninTable4.1ofAS/NZS1170.0.

Example 2.1: Calculation of wind actions on an industrial building

Figure2.4depictsaproposedsteelportalframedindustrialbuildingof28m×50m.Theheightoftheeaveofthebuildingis5m,whileitsridgeis8.75m.Oneofthewallscon-tainsaloadingdoorof4000×3600 mm,whichislocatedinthesecondbayoftheportalframes.TherearenoopeningsonotherwallsandroofsandtheridgeisnotventedasdepictedinFigure2.4.Internalsteelframesaretobespacedat5m.ThebuildingistobelocatedonaflatexposedsiteinregionA2,terraincategory2.Therearenosurroundingbuildingsandtheorientationofthebuildinghasnotbeenfinalised.Thedesignworkinglifeofthebuildingis50 years.Itisrequiredtodeterminetheinternalandexternalpres-suresonroofsandwallsandtheloadingonthefirstinternalframe.

1. Site wind speed

Thebuildingisanormalstructure,whichisdesignedforimportancelevel2.Thebuild-ingislocatedinregionA2whichisanon-cyclonicwindregionanditsdesignworkinglifeis50 years.Theannualprobabilityofexceedanceofthewindeventforthisnormalstructureis1/500.

TheregionalwindspeedcanbeobtainedfromTable2.1as

VR = 45m/s

Astheorientationofthebuildinghasnotbeenfinalised,thewinddirectionalmultiplierforregionA2isMd=1.0.

50 m

28 m

Figure 2.4 Steel-framed industrial building.


Theaverageroofheightofthebuildingis

h=5+3.75/2=6.875m

Theterrain/heightmultipliercanbeobtainedfromTable4.1(A)ofAS/NZS1170.2usinglinearinterpolationasfollows:

Mz cat,

( ).= +

− −−

=0.91(6.875 5)(1 0.91)

10 50 944

Thebuildingislocatedonaflatexposedsitewithoutupwindbuildingsothatthereisnoshielding.TheshieldingmultiplieristakenasMs=1.0.

ThetopographicmultiplierisMt=1.0.Thesitewindspeedcanbecalculatedas

V V M M M Msit R d z cat s t, , . . . . .β = ( ) = × × × × =45 1 0 0 944 1 0 1 0 42 48m/s

2. Design wind speed

Theorientationofthebuildinghasnotbeenfinalisedsothatthedesignwindspeedcanbetakenasthesitewindspeed:

Vdes, .θ = 42 48m/s

3. Aerodynamic shape factor

3.1. External pressure coefficients under crosswind

Windwardwall:Cp,e=0.7 Table5.2(A)(AS/NZS1170.2)Leewardwall:

Theroofpitch:α=15° >10°,d/b=28/50=0.56Therefore,Cp,e=−0.3 Table5.2(B)Roofs:α=15° >10°,h/d=6.875/28=0.246Forupwindslope:Cp,e=−0.5 Table5.3(B)Fordownwindslope:Cp,e=−0.5 Table5.3(C)

TheexternalpressurecoefficientsforsidewallsvarywiththehorizontaldistancefromthewindwardedgeandcanbeobtainedfromTable5.2(C)ofAS/NZS1170.2asfollows:

Horizontal distance from windward edge 0–6.875 m 6.875–13.75 m 13.75–20.625 m >20.625 m

Cp e, −0.65 −0.5 −0.3 −0.2

3.2. External pressure coefficients under longitudinal wind

Windwardwall:Cp,e=0.7 Table5.2(A)Leewardwall:

Theroofpitch:α=15° >10°,d/b=50/28=1.786Therefore,Cp,e=−0.3 Table5.2(B)

Theroofofthebuildingisagableroof.TheexternalpressurecoefficientsforthegableroofvarywiththehorizontaldistancefromthewindwardedgeandcanbeobtainedfromTable5.3(A)ofAS/NZS1170.2asfollows:

hd= =6 87528

0 246.

.

Design actions 31


Cp e, −0.9 −0.5 −0.3 −0.2

TheexternalpressurecoefficientsforsidewallsvarywiththehorizontaldistancefromthewindwardedgeandcanbeobtainedfromTable5.2(C)ofAS/NZS1170.2asfollows:


Cp e, −0.65 −0.5 −0.3 −0.2

3.3. Internal pressure coefficients under crosswind

Undercrosswind,thewindwardorleewardwallcontainsaloadingdoorwhichisadomi-nantopening.Itneedstocalculatethepermeabilityratioofthesurfacesofthebuildinginordertodeterminetheinternalpressurecoefficients.

Theareaofthedominantopeningis

Ado = × =4 3 6 14 4. . m2

Assumethebuildingleakageisat0.1%permeability.Thetotalbuildingleakageis0.1%of the area of all other surfaces excluding the one containing the dominant opening,whichiscalculatedasfollows:

Al = × × + ×

+ × ×( ) + ×

× =2 5 28 28

3 752

2 14 49 50 50 5 0 1 2 08.

. . % . 44m2

Thepermeabilityratiois

ξp

o

l

AA

= = = >14 42 084

6 9 6 0.

.. .

FromTable5.1(B), it canbe seen that the internalpressure coefficient is equal to theexternalpressurecoefficient:Cp,i=Cp,e.

Theworstcasefortheinternalpressureundercrosswindisthewindwardwalldooropensothattheinternalpressurecoefficientis

Cp i, .= +0 7

Theworstcasefortheinternalsuctionundercrosswindistheleewardwalldooropen.Forthiscase,theinternalpressurecoefficientis

Cp i, .= −0 3

3.4. Internal pressure coefficients under longitudinal wind

Theworst case for the internalpressureunder longitudinalwind is the sidewalldoorclosed.Theinternalpressurecoefficientis

Cp i, .= +0 0


Theworstcaseforthesuctionunderlongitudinalwindisthesidewalldooropen.Theinternalpressurecoefficientis

Cp i, .= −0 65

3.5. Area reduction factor

Thetributaryareaforrafterundercrosswindis

At = × × = >2 14 49 5 144 9 100. . m m2 2

Therefore,forrafter,Ka=0.8.Thetributaryareaforrafterandcolumnsunderlongitudinalwindis

At = × × + × × = >2 14 49 5 2 5 5 194 9 100. . m m2 2

Therefore,forrafterandcolumns,Ka=0.8.

3.6. Local pressure factor

h=6.875m,0.2b=0.2×50=10m,0.2d=0.2×28=5.6mTherefore,thedimensionofthelocalpressurezoneis

a . b . d h= ( ) =min , m0 2 0 2 5 6, .

Thelocalpressurefactorforlocalzonea×a=5.6×5.6m:Kl=1.5.Thelocalpressurefactorforlocalzone(0.5a×0.5a)=2.8×2.8m:Kl=2.0.

3.7. Combination factor

Fortheportalframeunderexternalandinternalwindloads,thecombinationfactoristakenasKc=1.0andsatisfiesthefollowingcondition:

K

Kc

a

≥ = =0 8 0 8

1 00 8

. ..

. , OK

3.8. Permeable cladding reduction factor

Thecladdingisnotpermeable,sothatKp=1.0.

3.9. Aerodynamic shape factors

Whencalculatingthewindpressuresonsurfacesratherthanontheportal frame,thefollowingfactorsaretakenas1.0:

K K K Ka c e c i l= = = =, , .1 0

Theaerodynamicshapefactorforexternalpressuresiscalculatedby

C C K K K K C Cfig p e a c l p p e p e= = × × × × =, , ,. . . .1 0 1 0 1 0 1 0

Theaerodynamicshapefactorforinternalpressuresiscalculatedby

C C K C Cfig p i c p e p i= = × =, , ,.1 0

Design actions 33

4. Design wind pressures on surfaces

Theindustrialbuildingisnotsensitivetowindanditsnaturalfrequencyisgreaterthan1.0 HzsothatitsdynamicresponsefactorcanbetakenasCdyn=1.0.

Thedesignwindpressureiscalculatedas

p V C C C Cair des fig dyn p e= ( ) = ×( )× × × =0 5 0 5 1 2 42 48 1 0 10832 2. . . . ., ,ρ θ pp e p eC, ,.= 1 083 kPa

Thedesignwindpressuresonsurfacesforvariouspressurecoefficientsarecalculatedasfollows:

Cp e, 0.7 −0.9 −0.65 −0.5 −0.3 −0.2

p (kPa) 0.758 −0.975 −0.704 −0.542 −0.325 −0.217

TheexternalwindpressuresonthesurfacesofthebuildingundercrossandlongitudinalwindsareshowninFigures2.5and2.6,respectively.

5. Loading on the first internal frame

a. Dead load (G)

Trimdeksheeting:4.28 kg/m2=0.0428kPa.

–0.704

–0.704

–0.542

–0.542

–0.542 kPa–0.542 kPa

–0.325

–0.325

–0.217

–0.217

–0.325 kPaWind

+0.758 kPa

7375687568756875

Figure 2.5 External wind pressures on surfaces of the industrial building under crosswind.


AssumingthatZ20019LYSAGHTpurlins(5.74 kg/m)at1200 mmspacingareused,theself-weightofthepurlinis

gp =

× ×=

−5 74 9 8 101 2

0 0473. .

.. kPa

Totalweightofsheetingandpurlin:g=0.0428+0.047=0.09≈0.1kPa.Thesheetingandpurlinloadonrafteris

G=0.1×5=0.5kN/m

b. Live load (Q)

w

AL = + = +

×= <0 12

1 8.

.0.12

1.85 28

0.133kPa 0.25kPa

Liveloadonrafter:Q=0.25×5=1.25kN/m.

c. Crosswind load

Theareareductionfactorforrafter:Ka=0.8.UDLonwindwardcolumn=0.758×5=3.79kN/m.UDLonleewardcolumn=0.325×5=1.63kN/m.UDLonrafter=0.8×0.542×5=2.17kN/m.

–0.704–0.704 –0.975–0.975

–0.325

–0.325 –0.325–0.325

–0.542–0.542–0.542–0.542

–0.325

–0.217–0.217–0.217 –0.217

Wind

+0.758 kPa

6875

6875

6875

Figure 2.6 External wind pressures on surfaces of the industrial building under longitudinal wind.

Design actions 35

d. Longitudinal wind on first internal frame

TheexternalwindpressuresonthefirstinternalframecolumnunderlongitudinalwindareshowninFigure2.7.Itisseenthatthecolumnsupportswindpressuresof0.704 kPaonanareaof4375×5000mm2andwindpressuresof0.542kPaonanareaof625×5000mm2becausethewindpressuresvarywiththehorizontaldistancefromthewind-wardedge.Theexternalwindpressuresontherafteralsovarywiththehorizontaldis-tancefromthewindwardedge.

TheareareductionfactorforroofandwallsisKa=0.8.Thelineloadsoncolumnsandraftersduetoexternalwindpressuresarecalculatedasfollows:

UDLoncolumns=0.8×(0.704×4.375+0.542×0.625)=2.74kN/m.UDLonrafter=0.8×(0.975×4.375+0.542×0.625)=3.68kN/m.

e. Internal pressure under crosswind

UDLonrafterandcolumns=0.758×5=3.79kN/m.

f. Internal pressure under longitudinal wind


g. Internal suction under crosswindUDLonrafterandcolumns=0.325×5=1.63kN/m.

h. Internal suction under longitudinal wind


referenceS

AS/NZS1170.0(2002)Australian/NewZealandStandardforStructuralDesignActions,Part0:GeneralPrinciples,Sydney,NewSouthWales,Australia:StandardsAustraliaandStandardsNewZealand.

AS/NZS 1170.1 (2002) Australian/New Zealand Standard for Structural Design Actions, Part 1:Permanent,ImposedandOtherActions,Sydney,NewSouthWales,Australia:StandardsAustraliaandStandardsNewZealand.

5000 5000

5000p=0.542(kPa)

p=0.704(kPa)

1875 6252500

4375

h=6875

25002500

Figure 2.7 External wind pressures on the first internal frame column under longitudinal wind.


AS/NZS1170.2(2011)Australian/NewZealandStandardforStructuralDesignActions,Part2:WindActions,Sydney,NewSouthWales,Australia:StandardsAustraliaandStandardsNewZealand.

Bowen,A.J.(1983)Thepredictionofmeanwindspeedsabovesimple2dhillshapes,JournalofWindEngineeringandIndustrialAerodynamics,15(1–3):259–270.

Davenport,A.G.(1967)Gustloadingfactors,JournaloftheStructuralDivision,ASCE,93:11–34.Davenport,A.G. (1977)Theprediction of risk underwind loading, Paper presented at the Second

InternationalConferenceonStructuralSafetyandReliability,Munich,Germany,pp.169–174.Davenport,A.G.,Surry,D.andStathopoulos,T.(1977)Windloadsonlow-risebuildings,Finalreport

ofphasesIandII,boundarylayerwindtunnelreport,BLWTSS8,UniversityofWesternOntario,London,Ontario,Canada.

Holmes,J.D.(1981)Reductionfactorsforwinddirectionforuseincodesandstandards,Paperpre-sentedattheColloque,DesignwiththeWind,Nantes,France,pp.VI.2.1–VI.2.15.

Holmes, J.D. (1985) Recent developments in the codification of wind loads on low-rise structures,Paper presented at the Asia-Pacific Symposium onWind Engineering, Roorkee, Uttarakhand,India,pp.iii–xvi.

Holmes,J.D.(2001)WindLoadingofStructures,London,U.K.:SponPress.Holmes,J.D.andBest,R.J.(1979)Awindtunnelstudyofwindpressuresongroupedtropicalhouses,

Windengineeringreport5/79,JamesCookUniversity,Townsville,Queensland,Australia.Holmes,J.D.,Melbourne,W.H.andWalker,G.R.(1990)ACommentaryontheAustralianStandardfor

WindLoads:AS1170Part2,1989,Melbourne,Victoria,Australia:AustralianWindEngineeringSociety.

Hussain,M.andLee,B.E.(1980)Awindtunnelstudyofthemeanpressureforcesactingonlargegroupsoflowrisebuilding,JournalofWindEngineeringandIndustrialAerodynamics,6:207–225.

ISO 4354 (1997) Wind Actions on Structures, International Organization for Standardization,Switzerland.

Kim,S.I.andMehta,K.C.(1977)Windloadsonflat-roofareathroughfull-scaleexperiment,InstituteforDisasterResearchReport,TexasTechnologyUniversity,Lubbock,TX.

Melbourne,W.H. (1975) Cross-wind response of structures to wind action, Paper presented at theFourth International Conference on Wind Effects on Buildings and Structures, CambridgeUniversityPress,London,U.K.

Melbourne,W.H.(1984)Designingfordirectionality,PaperpresentedattheFirstWorkshoponWindEngineeringandIndustrialAerodynamics,Highett,Victoria,Australia,pp.1–11.

Paterson,D.A.andHolmes,J.D.(1993)Computationofwindflowovertopography,JournalofWindEngineeringandIndustrialAerodynamics,6:207–225.

Stathopoulos,T.andMohammadian,A.R. (1985)Codeprovisions forwindpressuresonbuildingswith monoslope roofs, Paper presented at the Asia-Pacific Symposium onWind Engineering,Roorkee,Uttarakhand,India,pp.337–347.

Vickery,B.J.(1971)Onthereliabilityofgustloadingfactors,CivilEngineeringTransactions,InstituteofEngineersAustralia,13:1–9.

37

Chapter 3

local buckling of thin steel plates

3.1 IntroductIon

Steelandcompositemembersareusuallymadeofthin-walledsteelplateelementsbyhotrolling, welding or cold forming. Members composed of slender plate elements may failprematurelyowingtolocalbuckling.Localbucklingofthinsteelplatesremarkablyreducestheultimatestrengthandstiffnessofsteelandcompositemembers.Therefore,itisimpor-tanttounderstandthelocalbucklingbehaviourofthinsteelplatesundervariousloadingandboundaryconditionsandtoconsiderlocalbucklingeffectsinthedesignofsteelandcompositemembers.

Theelasticlocalbucklingbehaviourofathinsteelplatedependsonitswidth-to-thicknessratio(slendernessratio),materialproperties,geometricimperfections,loadingandbound-aryconditions.Aslenderthinsteelplatepossessessignificantpost-localbucklingreverseofstrength.Becauseofthis,slendersteelplateswillnotfailbyelasticlocalbuckling.Thepost-localbucklingstrengthofthinsteelplatesisinfluencedbytheiryieldstressandresid-ualstressesinducedbythehot-rolling,weldingorcold-formingprocess.Insteel–concretecompositememberssuchasconcrete-filledsteeltubular(CFST)columnsanddoubleskincompositepanels,steelplatesarerestrainedbyconcretesothattheycanonlybucklelocallyawayfromtheconcrete.Thelocalbucklingstressofthinsteelplatesincontactwithcon-creteismuchhigherthanthatoftheonesunrestrainedbyconcrete.

Thischapterdescribes thebehaviourofrectangularthinsteelplates thatformsteelorcompositemembers.Theplates considered are subjected to in-plane compression, shear,bending,bearingorcombinedstatesof stresses.Thedesignof steelandcompositecrosssectionscomposedofslendersteelplatesaccountingforlocalbucklingeffectsisdiscussed.

3.2 Steel PlAteS under unIform edge comPreSSIon

3.2.1 elastic local buckling

3.2.1.1 Simply supported steel plates

Steelcolumnscomposedofslenderplateelementsunderuniformcompression,suchashol-lowsteelboxcolumnsandI-sectioncolumns,mayundergolocalbuckling.Figure3.1showsthebuckledshapeofapin-endedhollowsteelboxshortcolumnunderuniformcompres-sion.Itcanbeseenfromthefigurethatthetwooppositesidesoftheboxbuckle locallyoutwardwhiletheothertwosidesbuckleinward.Itcanbeassumedthattheplateelementsarehingedalongtheircommonboundariesandcanrotatefreelyaboutthefouredges.Theflangesandwebsoftheboxcolumncanbeidealisedassimplysupportedontheirfouredges.


Similarly,thefouredgesofthewebinapin-endedsteelI-sectioncolumncanbetreatedassimplysupported.

AsimplysupportedthinflatsteelplateunderuniformedgecompressionontwooppositeedgesisschematicallydepictedinFigure3.2.ThelengthoftheplateisL,thewidthoftheplateisbanditsthicknessist.Whentheappliedcompressiveloadisequaltoitselasticbuck-lingload,thesteelplatebuckleslocallybydeflectingoutofitsplane.Theelasticbucklingloadofthethinplatecanbedeterminedbytheenergymethod(Bleich1952;TimoshenkoandGere1961;Bulson1970)orthefiniteelementmethod.Figure3.3showsthebuckledshapeofasimplysupportedlongsteelplateunderuniformedgecompression,whichwasmodelledbyfiniteelements.Thelocalbucklingdisplacementsoftheplatecanbedescribedbythefollowingdoubleseries:

u u

n xL

m yb

m=

sin sin

π π (3.1)

whereumistheundetermineddeflectionatthecentreoftheplatemisthenumberofhalfwavesacrossthewidthbnisthenumberofhalfwavesinthedirectionoftheappliedcompressiveload

The elastic buckling load can be calculated by the following equation (Bleich 1952;TimoshenkoandGere1961;Bulson1970):

P

L bDn

nL

mb

crr= +

π2 2

2

2

2

2

2

2

(3.2)

Figure 3.1 Buckled shape of a pin-ended hollow steel box short column under uniform compression.

Local buckling of thin steel plates 39

whereDristheplateflexuralrigidity,whichiswrittenas

D

E tr

s=−

3

212 1( )ν (3.3)

whereEsisYoung’smodulusofthesteelmaterialtisthethicknessofthesteelplateνisPoisson’sratio

ThelowestvalueofPcrcanbeobtainedbytakingm=1inEquation3.2.Thisimpliesthatthebuckledplatehasonlyonehalfwaveacrossitswidthbbutseveralhalfwavesinthedirectionoftheappliedloading.Theelasticbucklingstressoftheplateisexpressedbythefollowingequation(Bleich1952;TimoshenkoandGere1961;Bulson1970):

σ π

νcrb sk E

b t=

−

2

2 212 1( )( )/ (3.4)

wherekbistheelasticbucklingcoefficient,whichisgivenby

k

nbL

Lnb

b = +

2

(3.5)

ThisequationindicatesthattheelasticbucklingcoefficientofasimplysupportedflatplatedependsonitsaspectratioL/bandthenumberofhalfwavesnalongtheplateinthedirection

z

Figure 3.2 A simply supported steel plate under uniform edge compression.

Figure 3.3 Buckled shape of a long simply supported steel plate under uniform edge compression.


oftheappliedcompressiveload.ThebucklingcoefficientsofsimplysupportedsteelplatesunderuniformedgecompressionaregiveninFigure3.4.ItcanbeseenfromFigure3.4thattheminimumbucklingcoefficientkbis4.0regardlessofthenumberofhalfwaves.Themini-mumbucklingcoefficientoccurswhentheplateaspectratioL/bisanevennumbersuchas1,2,3,4and5.Thelargerthenumberofhalfwavesn,theflatterthebucklingcoefficientcurve.

Topreventtheelasticlocalbucklingfromoccurringbeforesteelyields,thelimitingwidth-to-thicknessratiocanbeobtainedfromEquation3.4bysettingthecriticalbucklingstresstoitsyieldstress.Thecalculatedwidth-to-thicknessratioforsimplysupportedplatesunderuniformcompressionisgreaterthantheslendernessyieldlimitgiveninAS4100astheyieldlimitgiveninthecodeconsiderstheeffectofresidualstresses.

ItcanbeobservedfromFigure3.3thatasimplysupportedlongsteelplateunderuniformcompressionwillbucklelocallyinseveralhalfwavesinthedirectionoftheloadingwithalengthaboutthewidthboftheplate.Asaresult,theuseoftransversestiffenerstorein-forcetheplatewillhavelittleeffectonthelocalbucklingstressunlessthespacingofthetransversestiffenersismuchlessthanthewidthoftheplate(TrahairandBradford1998).Aneconomicaldesigncanbeachievedbyweldingoneormorelongitudinalstiffenerstotheplate.Thelongitudinalstiffenersdividetheplateintosmallerpanels,remarkablyincreas-ingthebucklingstressoftheplateaccordingtoEquation3.4.Inaddition,thelongitudinalstiffenerscanwithstandaportionofthecompressiveload.

Topreventtheplatefromdeflectingatthestiffeners,intermediatelongitudinalstiffenersmusthaveadequateflexuralrigidities.Therequiredminimumsecondmomentofareaofanintermediatelongitudinalstiffenerplacedatthecentrelineofasimplysupportedsteelplate(TrahairandBradford1998)isgivenby

I b t

Ab t

Ab t

ss s= + +

4 5 1

2 31

21

3

1 1

..

(3.6)

whereb1istakenasb/2Asisthecross-sectionalareaofthestiffener

Plate aspect ratio L/b

Buck

ling

coe

cien

t kb

00

1

2

3

4

5

6

7

8

1 2 3 4 5 6

n= 4n= 3

n= 2

n= 1

Figure 3.4 Buckling coefficients of simply supported steel plates under uniform edge compression.


Stiffenersareusuallyattachedtoonesideoftheplateratherthantobothsides.Itshouldbenotedthatastiffenerisusuallymadeofsteelstrip,whichmaybucklelocallywhensubjectedtocompression.Therefore,stiffenersmustbeproportionedtopreventfromlocalbuckling.

Endstiffenersmaybeattachedtothesteelplatetoincreasethestiffnessoftheplateandtocarryaportionofthecompressiveload.TherequiredminimumsecondmomentofareaofanendlongitudinalstiffenercanbeobtainedbymodifyingEquation3.6asfollows(TrahairandBradford1998):

I bt

Abt

Abt

ss s= + +

2 25 1

4 61

23.

. (3.7)

3.2.1.2 Steel plates free at one unloaded edge

Thebuckledpatternsofapin-endedsteelI-sectionshortcolumnunderuniformcompressionarepresentedinFigure3.5.Localbucklingisinfluencedbytherelativestiffnessofthecon-nectedelementsinasteelsection.TheflangeoutstandofthesteelI-sectioncanbeassumedtobesimplysupportedbytheweb,whiletheoppositeedgeisfree.Asaresult,theflangeoutstandissimplysupportedattwoloadededgesandoneunloadededgeandfreeatoneunloadededgeasshowninFigure3.6.Theplateissubjectedtouniformcompressiveedgestressesontwooppositeedges.ThebuckledshapeofalongsteelplatefreeatoneunloadededgeandmodelledwithfiniteelementsispresentedinFigure3.7.Thefigureshowsthatthefreeunloadededgecausestheplatetobuckleinonehalfwaveinthedirectionofthecompressiveload.TheelasticbucklingstressforasteelplatefreeatoneunloadededgecanbeexpressedbyEquation 3.4.However,theelasticbucklingcoefficientkbisgivenby(Bulson1970)

k

bL

b = +

0 4252

. (3.8)

(a) (b) (c)

Figure 3.5 Buckled shapes of steel I-section short column under uniform compression: (a) mode 1, (b) mode 2 and (c) mode 3.


Equation3.8indicatesthatthebucklingcoefficientdependsontheplateaspectratioL/b.ThebucklingcoefficientsofthinsteelplateswithonefreeunloadededgeandunderuniformedgecompressionaredemonstratedinFigure3.8.Itappearsthatwhentheplateaspectratioislessthan2.0,thebucklingcoefficientdecreasessignificantlywithanincreaseinitsL/bratio.However,thisdecreasetendstobesmallwhentheL/bratioisgreaterthan2.0.ForlongsteelplateswithlargeL/bratiossuchastheflangeoutstandsofI-sectioncolumns,thebucklingcoefficientapproachestheminimumvalueof0.425asindicatedinEquation3.8.Therefore,thebucklingcoefficientkb=0.425canbeusedinthedesignofflangeoutstandsofI-sectionsinlongsteelcolumnsunderaxialcompression.

3.2.2 Post-local buckling

Afterinitiallocalbuckling,thinsteelplatescanstillcarryincreasedloadswithoutfailure.Thisbehaviourof thin steelplates is calledpost-localbuckling.Thepost-localbucklingbehaviourof a thin steel plateunder edge compression is characterisedby its transversedeflections and the in-plane stress redistribution within the buckled plate. The in-planestressredistributionisassociatedwiththein-planeboundaryconditionsoftheplate(TrahairandBradford1998).Theboundarylinesoftheloadededgesoftheplateundergoaconstantaxialshortening,whichiscausedbyboththetransversedeflectionsandtheaxialstrain.Theaxialshortening inducedbythetransversedeflectionsvariesacross theplate fromamaximumatthecentretoaminimumattheunloadededges.Thisvariationiscompensatedforbytheaxialshorteningcausedbytheaxialstrain,varyingfromaminimumatthecentre

L

F

S

S

S b

Figure 3.6 A steel plate with a free edge.

Figure 3.7 Buckled shape of a steel plate with a free unloaded edge.


toamaximumattheunloadededges.Thein-planestressdistributionwithinthebuckledplateintheloadingdirectionmustbethesameasthatoftheaxialstrain.Thisimpliesthatthe central portion of the buckled plate carry relatively lower stresses, while the loadededgestripswithstandhigherstresses.ThiswasconfirmedbytheresultsofthefiniteelementanalysiscarriedoutbyLiangandUy(1998).

Theeffectivewidthconceptisusuallyusedtodescribethepost-localbucklingbehaviourof thinsteelplates.Figure3.9adepicts the in-planeultimatestressdistribution inasim-plysupportedthinsteelplateunderuniformedgecompression.Thisactualultimatestressdistribution is transformed into an idealised stress distribution within the buckled plateasillustratedinFigure3.9b.Theeffectivewidthconceptassumesthatthecentralportionofthebuckledplatewithstandszerostresses,whiletheeffectivewidthbecarriestheyieldstress.Theeffectivewidthofathinsteelplatecanbeevaluatedby

bb fe u

y

= σ (3.9)

wherebeistheeffectivewidthoftheplateσuistheaverageultimatestressactingontheplate,whichcanbedeterminedbyexperi-

mentsornonlinearfiniteelementanalyses(LiangandUy2000;Lianget al.2007)

TheeffectivewidthofasimplysupportedthinsteelplateunderuniformedgecompressionwasdevelopedbyvonKarmanet al.(1932)as

bb fe cr

y

= σ (3.10)


Buck

ling

coe�

cien

t kb

00

1

2

3

4

5

6

1 2 3 4 5 6

Figure 3.8 Buckling coefficients of steel plates with a free unloaded edge.


For hot-rolled and welded plates with initial curvatures and residual stresses, AS 4100(1998)suggeststhattheeffectivewidthoftheplatesshouldbereducedbyareductionfactorasfollows:

bb fe cr

y

= α σ (3.11)

Thereductionfactorαaccountsfortheeffectofinitialcurvaturesandresidualstressesontheultimatestrengthoftheplate.Forhot-rolledplates,αistakenas0.65inAS4100(1998).Real steelplateshave small initial curvatureswhich reduce the stiffness and strengthofplates.Itisnotedthatinitialcurvatureshavelittleeffectonthestrengthofthickplatesbutsignificantly reduce the strength of plates with intermediate slenderness ratios. Residualstressespresentedinsteelplatesareusuallycausedbyunevencoolingafterrollingorweld-ing. Tensile stresses are presented at the junctions of plate elements, while compressivestressesactattheremainderoftheplate.Tensilestressesonasteelplatearebalancedbycompressivestressesactingonthesameplate.Residualstressescauseprematurebucklingandyieldingoftheplate.

Theeffectivewidthsofhot-rolledsteelplatescalculatedbyEquation3.11arepresentedinFigure3.10,wherethemodifiedplateslendernessisdefinedasλ σm y crf= / .ItappearsfromFigure3.10thatwhenλm≤0.65,theplateisfullyeffectiveinattainingitsyieldcapacity.Whenλm>0.65,theeffectivewidthoftheplatedecreaseswithincreasingitsslenderness.

Forcold-formedmembers,theeffectivewidthofplateelementswithinitialcurvaturescanbeexpressedbythefollowingequation(Winter1947):

bb f fe cr

y

cr

y

= −

σ σ1 0 22. (3.12)

3.2.3 design of slender sections accounting for local buckling

Asdiscussedintheprecedingsections,localandpost-localbucklingofsteelplatesreducestheultimatestrengthofthecrosssectionsofsteelmembersunderaxialcompression.Theeffectof localbuckling isconsidered inthedesignofaxially loadedsteelmembersmadeofslenderplateelementsinAS4100byusingtheeffectivewidthconcept(Bradford1985,1987;Bradfordet al.1987).TheeffectivewidthofaplateelementiscalculatedusingitsslendernessandyieldlimitgiveninClause6.2ofAS4100(1998).Theplateelementslender-nessisdefinedasfollows.

(a)

fy

b

fy

(b)

be2

be2

Figure 3.9 Effective width concept for simply support plates: (a) ultimate stress distribution and (b) effective width.


Theslendernessofaflatplateelementiscalculatedas

λe

ybt

f=

250 (3.13)

Theslendernessofacircularhollowsectionisexpressedby

λe

o ydt

f=

250

(3.14)

wheredoistheoutsidediameterofthecircularsectiontisthewallthicknessofthesection

Clause6.2.4ofAS4100(1998)givesasimplemethodfordeterminingtheeffectivewidthofflatplate elements and circularhollow section. In thismethod, the effectivewidthofaplateelementiscalculatedbyusingtheplateelementslendernessandtheelementyieldslendernesslimits(λey)(Bradford1985,1987;Bradfordet al.1987).Theelementyieldslen-dernesslimitsdependontheplatetype,supportcondition,stressdistributionandresidualstresslevelandaregiveninTable5.2ofAS4100.

Theeffectivewidthforaflatplateelementcanbecalculatedas

b b be

ey

e

=

≤

λλ

(3.15)

Theeffectiveoutsidediameterforacircularhollowsectionisdeterminedby

d d d de oey

eo

ey

eo=

≤min ,

λλ

λλ3

2

(3.16)

Modi�ed plate slenderness λm

0 0.5 1 1.5 2 2.5 3

Eec

tive w

idth

be/b

0

0.2

0.4

0.6

0.8

1

1.2

Figure 3.10 Effective widths of simply support plates under uniform edge compression.


Theplateelementunderuniformcompressionisslenderifλe>λey.Forasteelsectionmadeupofflatplateelements,thesectionslendernessλsistakenasthevalueoftheplateelementslendernessλewhichhasthegreatestvalueofλe/λey.

Theformfactorisusedtoaccountforlocalbucklingeffectsontheultimateaxialstrengthofslendersteelsectionsunderaxialcompression(Rasmussenet al.1989).Clause6.2.3ofAS4100(1998)definestheformfactoras

k

AA

fe

g

= ≤ 1 0. (3.17)

whereAeistheeffectiveareaofthesteelsectionAgisthegrossareaofthesection

TheeffectiveareaAe iscalculatedbysummingtheeffectiveareasofindividualelements.Itshouldbenotedthattheformfactorkfisastrengthreductionfactorwhichmustbelessthanorequalto1.0.Forasteelsectionwithoutlocalbucklingeffects,thesectionisfullyeffectiveandkf=1.0.

Thedesignsectionaxialcapacityofasteelmemberunderaxialcompressioncanbedeter-minedinaccordancewithClause6.2.1ofAS4100(1998)as

φ φN k A fs f n y= (3.18)

whereφ = 0 9. isthecapacityreductionfactorAnisthenetareaofthesectionwhichisusuallytakenasthegrossareaAgofthesectionfyistheminimumyieldstressforthesection

Thedesignrequirementforthesectionofasteelmemberunderaxialcompressionis

N Ns∗ ≤ φ (3.19)

whereN∗isdesignaxialloadactingonthesection.

Example 3.1: Section capacity of a steel column under compression

DeterminethedesignsectionaxialcapacityoftheheavilyweldedsteelI-sectionofasteelcolumnunderaxialcompression.ThecrosssectionofthecolumnisshowninFigure3.11.Theyieldstressofthesteelsectionfyis320MPa.

1. Plate element slenderness

ThedimensionsofthesteelI-sectionare

b t d tf f w= = = =420 12 450 10mm, mm, mm, mm

Theslendernessoftheflangeoutstandsis

λef

y f w

f

ybt

f b tt

f= =

−=

−=

2502

250420 10 2

12320250

19 33( ) ( )

./ /


Oneofthelongitudinaledgesoftheflangeoutstandissimplysupportedbythewebandtheopposite longitudinal edge is free.The topflangeof the section is underuniformcompression.FromTable5.2ofAS4100,theyieldslendernesslimitcanbeobtainedasλey=14.λ λef ey= > =19 33 14. ,theflangeisslender.Theslendernessofthewebis

λew

y f

w

ybt

f d tt

f= =

−=

− ×( )=

2502

250

450 2 12

10320250

48 2( )

.

Bothofthelongitudinaledgesofthewebaresimplysupportedbytheflangesandareunderuniformcompression.FromTable5.2ofAS4100,theyieldslendernesslimitcanbeobtainedasλey=35.λ λew ey= > =48 2 35. ,thewebisslender.

2. Effective area of steel section

Theeffectivewidthoftheflangeoutstandsiscomputedas

b b

b tef

ey

e

f w ey

ef

=

=

−

=

−

×

λλ

λλ2

420 102

11419 33

148 5.

.

= mm

Theeffectivewidthofthewebiscalculatedas

b b d tew

ey

ef

ey

ew

=

= −

= − ×( )×

λλ

λλ

( ).

2 450 2 123548 2

= 309 3. mm

Theeffectiveareaofthesectioncanbecalculatedas

A b t t b te ef w f ew w= + + = × × + × + × =2 2 2 2 148 5 10 12 309 3 10( ) ( . ) . 10,461mm2

TheeffectiveareaofthesteelI-sectionisillustratedinFigure3.12.Thegrossareaofthesteelsectionis

Ag = × × + − × × =2 420 12 450 2 12 10( ) 14,340mm2

12

12420

450 10

Figure 3.11 Section of compression member.


3. Design section axial capacity

Theformfactorcanbecalculatedas

k

AA

fe

g

= = =10,46114,340

0 73.

Thedesignsectionaxialcapacityis

φ φN k A fs f n y= = × × × =0 9 0 73 320 8. . .14,340 N 3,014 kN

3.3 Steel PlAteS under In-PlAne BendIng


Whenasteelbeamisunderbending,thewebofthebeamissubjectedtoin-planebend-ingstressesanditmaybucklelocally.ThebeamweboflengthL,widthdandthickness tis assumed to be simply supported on its four edges as schematically demonstrated inFigure 3.13.Theplateisunderin-planelinearlydistributedbendingstressesontwooppo-siteedges.Localbucklingoccurswhenthemaximumbendingstressactingontheplatereachestheelasticbucklingstressoftheplate.ThetypicalbuckledshapeofathinsteelplatewithanL/dratioof2andsubjectedtobendingstressesispresentedinFigure3.14.Thefigureshowsthattheportionoftheplateundercompressivestressesbucklesoutoftheplane,whiletheportionundertensilestressesdoesnotbuckle.Solutionstothelocalbucklingproblemofthinsteelplates inbendingcanbeobtainedbytheenergymethod

L

S

S

S

S d

Figure 3.13 A simply supported steel plate in bending.

12

12

10450E�ective area

Ine�ective area

Figure 3.12 Effective area of steel I-section.


(Bleich1952;TimoshenkoandGere1961;Bulson1970)orthefiniteelementmethod.Theelasticlocalbucklingstresscanbedeterminedby

σ π

νofb sk E

d t=

−

2

2 212 1( )( )/ (3.20)

inwhichtheelasticbucklingcoefficientkbisafunctionoftheplateaspectratioL/dandthenumberofbucklesinplate.Forlongsteelplates,thelengthofeachbuckleisabout2d/3andtheminimumbucklingcoefficientiskb=23.9.

Likethesimplysupportedsteelplates,transversestiffenersarenoteffectiveinpreventingthelocalbucklingoftheplatessubjectedtoin-planebendingstressesunlesstheirspacingislessthan2d/3.Longitudinalstiffenersattachedtotheplateunderin-planebendingareeffectiveinincreasingtheresistancetolocalbucklingastheyalterthebuckledpatternoftheplate.Thelongitudinalstiffenerismostefficientwhenitisplacedintheportionundercompressionatadistance0.2d2fromthecompressionedge.TherequiredminimumsecondmomentofareaforthelongitudinalstiffenerisspecifiedinAS4100.

3.3.2 ultimate strength

Theultimatestrengthofastockysteelplateunder in-planebending isdeterminedby itsplasticsectionmodulusandyieldstress.Foraslendersteelplatesubjectedtoin-planebend-ingstresses,theelasticlocalbucklingstressoftheplatewillbelessthanitsyieldstress.Thepost-localbucklingbehaviourofthinsteelplatesunderin-planebendingstressescanalsobedescribedbytheeffectivewidthconcept(Bulson1970;Usami1982;Shanmugamet al.1989;Lianget al.2007).Theeffectivewidthoftheplateislocatedwithintheportionundercompression,whiletheportionintensionisfullyeffectiveincarryingtensilestresses.

3.3.3 design of beam sections accounting for local buckling

OneoftheflangesofasteelbeamunderbendingsuchasahollowsteelboxorasteelI-beamissubjectedtocompressivestresses,whilethebeamwebisunderin-planebendingstresses.InAS4100,steelplateelementsinacrosssectionareclassifiedascompact,non-compactorslender

Figure 3.14 Buckled shape of a simply supported steel plate in bending.


basedontheirplateelementslendernessratio.Theeffectivesectionmodulusisusedtoaccountforlocalbucklingeffectsonthesectionmomentcapacityofasteelbeamunderbending.

Compactelementsundercompressionorin-planebendingdonotundergolocalbucklingandcanattaintheirfullplasticcapacities.Aplateelementiscompactifitsslenderness(λe)satisfies

λ λe ep≤ (3.21)

inwhichλepistheplasticityslendernesslimitgiveninTable5.2ofAS4100.Non-compactelementsundercompressionorin-planebendingcanattaintheirfirstyield

capacitiesbutundergolocalbucklingbeforetheirfullplasticcapacitiesarereached.Aplateelementisnon-compactifitsslenderness(λe)satisfies

λ λ λep e ey< ≤ (3.22)

Slender elements under compression or in-plane bending undergo elastic local bucklingbeforeyielding.Aplateelementisclassifiedasslenderifitsatisfies

λ λe ey> (3.23)

Thecross sectionsof steelbeamsarealsoclassifiedascompact,non-compactor slenderbasedontheclassificationoftheirelementsinAS4100.Allelementsmustbecompactinacompactsteelsection.Therearenoslenderelementsandatleastonenon-compactelementinanon-compactsteelsection.Thereisatleastoneslenderelementinaslendersteelsection.Thesectionslenderness(λs)ofasteelsectioncomposedofflatplateelementsistakenasthevalueoftheplateelementslenderness(λe)fortheelementofthesectionhavingthegreatestvalueofλe/λey.

InClause5.2.3ofAS4100(1998),theeffectivesectionmodulusZeforacompactsteelbeamsectionistakenas

Z Z S Ze c= = ≤ 1 5. (3.24)

whereZcistheeffectivesectionmodulusofacompactsectionSistheplasticsectionmodulusdefinedinSection7.2.2Zistheelasticsectionmodulus,whichisdefinedinSection4.3.4

However,foranon-compactsteelbeamsection,Clause5.2.4ofAS4100(1998)pro-vides an equation based on linear interpolation for determining the effective sectionmodulusas

Z Z Z Ze c

sy s

sy sp

= + −( ) −−

λ λλ λ

(3.25)

whereλs,λsyandλsparethevaluesofλe,λeyandλepfortheelementofthesectionhavingthegreatestvalueofλe/λey.

Clause5.2.5ofAS4100givesspecificationsfordeterminingtheeffectivesectionmodu-lusforslendersections,whicharedescribedherein.Forabeamwithaslenderflangeunder


uniformcompression, theeffective sectionmoduluscanbecalculatedusing theeffectivewidthorbythefollowingequation:

Z Ze

sy

s

=

λλ

(3.26)

Forabeamconsistingofaslenderweb,theeffectivesectionmoduluscanbedeterminedby

Z Ze

sy

s

=

λλ

2

(3.27)

Theeffectivesectionmodulusforaslendercircularhollowsteelsectionisgivenby

Z Z Zesy

s

sy

s

=

min ,λλ

λλ

2

2

(3.28)

Thenominalsectionmomentcapacityofasteelbeamiscalculatedby

M Z fs e y= (3.29)

MoredetailsonthemomentcapacityofsteelbeamsareprovidedinSection4.4.Thedesignrequirementforthesectionofasteelbeamunderbendingis

M Ms∗ ≤ φ (3.30)

inwhichϕ=0.9isthecapacityreductionfactor.

Example 3.2: Section moment capacity of a steel I-beam under bending

Determinethedesignsectionmomentcapacityofahot-rolled310UB32.0steelI-beambendingaboutitsprincipalx-axisasshowninFigure3.15.Thesectionpropertiesarefy=320MPa,Zx=424×103mm3andSx=475×103mm3.

1. Plate element slenderness


b t d tf f w= = = =149 8 298 5 5mm, mm, mm, mm.

Theslendernessoftheflangeoutstandsiscalculatedas

λef

y f w

f

ybt

f b tt

f= =

−=

−=

2502

250149 5 5 2

8320250

10 1( ) ( . )

./ /

Oneofthelongitudinaledgesoftheflangeoutstandissimplysupportedbythewebandtheoppositelongitudinaledgeisfree.Thetopflangeofthesectionisassumedtobeinuniformcompression.FromTable5.2ofAS4100, theplasticityandyieldslendernesslimitscanbeobtainedasλep=9 and λey=16.λ λ λep ef ey= < = < =9 10 1 16. ,theflangeisnon-compact.


Theslendernessofthewebiscomputedas

λew

y ef

w

ybt

f d tt

f= =

−=

− ×=

2502

250298 2 8

5 5320250

58( ) ( )

.

Bothofthelongitudinaledgesofthewebaresimplysupportedbytheflangesandareunderlinearbendingstresses.FromTable5.2ofAS4100,theplasticityslendernesslimitscanbeobtainedasλep=82.λ λew ep= < =58 82,thewebiscompact.

2. Effective section modulus

Thesectioncontainsanon-compactflangesothatthewholesectionisnon-compact.Forthenon-compactsection,theeffectivesectionmoduluscanbecalculatedby

Z Z Z Ze csy s

sy sp

= + −−−

= × + − × ×

−( ) ( )

λ λλ λ

424 10 475 424 10163 3 110 116 9

467 103.−

= × mm3

3. Design section moment capacity

Thedesignsectionmomentcapacityiscomputedas

φ φM Z fs e y= = × × × =0 9 467 10 320 134 53. .Nmm kNm

3.4 Steel PlAteS In SheAr


Thewebofasteelbeamnearthesupportsorzerobendingmomentmaybesubjectedtopureshearstressesalongitsedges.Figure3.16depictsasimplysupportedsteelplatewithlengthL,depthd and thickness t andunder shear stressesuniformlydistributedalong its fouredges.Localbucklingoccurswhentheshearstressesareequaltotheelasticbucklingstressoftheplate.Thislocalbucklingproblemofthinsteelplatesinshearcanbesolvedbynumer-icalmethodssuchasthefiniteelementmethod.ThebuckledshapeofasimplysupportedsteelplateundershearstressesonfouredgesisshowninFigure3.17,wheretheplateaspect

8

5.5

8

149

298

Figure 3.15 Hot-rolled steel I-section.


ratioL/dis2.FiniteelementanalysisresultsshowthatincreasingtheplateaspectratioL/dincreasesthenumberofbuckles.Theelasticlocalbucklingstresscanbeexpressedby

σ π

νovb sk E

d t=

−

2

2 212 1( )( )/ (3.31)

wherethebucklingcoefficientkbisafunctionoftheplateaspectratioL/d(TimoshenkoandGere1961)andcanbedeterminedby

k

dL

L d

dL

L d

b =

+ ≤

+

≥

5 35 4

5 35 4

2

2

.

.

for

for

(3.32)

Buckling coefficients calculated by Equation 3.32 are presented in Figure 3.18. The fig-ure demonstrates that when L ≤ d, the buckling coefficient decreases significantly withincreasingtheL/d ratio.However,whenL≥d, increasingplateL/d ratio leadstoonlya

L

dS S

S

S

Figure 3.16 A simply supported steel plate in shear.

Figure 3.17 Buckled shape of a simply supported steel plate in shear.


smalldecreaseinthebucklingcoefficient.Foraverylongsteelplateinshear,itsbucklingcoefficientapproachestotheminimumvalueof5.35.

AttachingintermediatetransversestiffenerstotheplateinpuresheartoreducetheaspectratioofL/d can significantly increase thebuckling coefficientandbuckling stressof theplate.Theelasticbucklingstressofaplateinshearcanalsobegreatlyincreasedbyusingthelongitudinalstiffenerstoreducethed/tratio.Toachieveefficientdesigns,theaspectratioofeachpaneldividedbystiffenersshouldbebetween0.5and2.


Astockyweb inan I-sectionbeamsubjected topure shearbehaves elasticallyuntilfirstyieldoccursatτy yf= / 3andundergoesincreasingplasticisationuntilitfullyyields.Theshearstressdistributioninthewebatfirstyieldisnearlyuniformandtheshearshapefactoriscloseto1.0.Becausestockywebsinsteelbeamsinshearyieldbeforebuckling,theyareusuallyunstiffenedandtheirultimatestrengthsaredeterminedbytheshearyieldstressasfollows:

V d tw w w y= τ (3.33)

wheredwisthecleardepthofthewebtwisthethicknessoftheweb

Slenderwebswithtransversestiffenerswillbuckleelasticallybeforeyieldingoccurs.Thereserveofthepost-localbucklingstrengthoftheslenderwebsisrelativelyhighcomparedtostockywebs.Theultimateshearstressofaslenderwebcanbeestimatedbyitselasticlocalbucklingstresswithlengthequaltothestiffenerspacingandthetensionfieldcontributionatyield(Basler1961;Evans1983).


30

25

20

15

10

5

00 654321

Buck

ing

coe�

cien

t kb

Figure 3.18 Buckling coefficients of simply supported steel plates in shear.


3.5 Steel PlAteS In BendIng And SheAr


Asimplysupportedthinflatsteelplateof lengthL,depthdandthicknesstunderbend-ingandshearisdepictedinFigure3.19.Theelasticbucklingstressofthethinplatecanbedetermined from the following interaction equation (Bleich1952;TimoshenkoandGere1961;Bulson1970):

σσ

ττ

f

of

v

ov

+

=

2 2

1 (3.34)

whereτovistheelasticbucklingstressoftheplateinpureshearσof istheelasticbucklingstressoftheplateinpurebendingτvandσf aretheelasticbucklingstressesoftheplateundercombinedbendingandshear

ItcanbefoundfromtheHencky–vonMisesyieldcriterionthatthemostsevereloadingcon-ditionforwhichelasticlocalbucklingandyieldingoccursimultaneouslyisthepureshear.


In steel beams, stocky unstiffened webs yield before elastic local buckling occurs. ThedesigncapacitiesofstockyunstiffenedwebscanbeestimatedbytheHencky–vonMisesyieldcriterionas

VV

MMu u

∗

+

∗

=

φ φ

2 2

1 (3.35)

whereV∗andM∗arethedesignshearforceandmomentinthewebVuisthenominalshearyieldcapacityoftheweb,whichiscalculatedas

V f d tu y w w= 0 6. (3.36)

L

d

S

S

S

S

Figure 3.19 A simply supported steel plate in bending and shear.


InEquation3.35,Muisthenominalfirstyieldmomentcapacityoftheweb,whichisdeter-minedby

M

d t fu

w w y=2

6 (3.37)

Forslenderunstiffenedwebsundercombinedbendingandshear,thereserveofpost-localbucklingissmallsothattheirultimatestrengthcanbeestimatedapproximatelybytheirelasticbucklingstressessatisfyingEquation3.34.Theultimatestrengthofastiffenedwebincombinedbendingandshear isgiveninClause5.12.3ofAS4100andisdiscussedinSection4.5.3.

3.6 Steel PlAteS In BeArIng


Steelplategirdersareoftensubjectedtoconcentratedorlocallydistributedloadsontheirtopflanges.ThelocalloadcauseslocalbearingstressesinthewebimmediatelybeneaththeloadasdepictedinFigure3.20.Thesebearingstressesareresistedbyverticalshearstressesatthetransversewebstiffenersofaslenderstiffenedplategirder.Plategirdersundertrans-verseloadsmaybesubjectedtocombinedbendingandshearorcombinedbending,shearandbearingataninteriorsupport.Forapanelofastiffedweb,theedgesofthepanelcanbeassumedtobesimplysupported.Theelasticbucklingstressofapanelunderpurebearingcanbecalculatedas

σ π

νobb sk E

d t=

−

2

2 212 1( )( )/ (3.38)

inwhichthebucklingcoefficientkbisafunctionofthepanelaspectratios/d(Bulson1970;TrahairandBradford1998).

bb

tf

2.52.511

11 1

1

Figure 3.20 Beam web in bearing.


Whenawebpanelissubjectedtocombinedbending,shearandbearing,theelasticbuck-lingstressesofthepanelcanbedeterminedbytheinteractionequation(Rockeyet al.1972;AllenandBulson1980)

σσ

στ

σσ

f

of

v

ov

b

ob

+

+ =

2 2

1 (3.39)

whereσof ,τovandσobareelasticbucklingstressesofaplateunderpurebending, shearor

bearingonlyσf ,τvandσbareelasticbucklingstressesoftheplateundercombinedbearing,shear

andbending


Thepointloadorlocallydistributedloadappliedonthetopoftheflangeisassumedtobedisperseduniformlythroughtheflangeataslopeof1:2.5andthroughthewebat a slope of 1:1 as depicted in Figure 3.20. The general yielding of a thick web inbearingoccurswhenthewebareadefinedbythedispersionoftheappliedloadyields.Theultimatestrengthofathickwebinbearingdependsonitsyieldstress.Whentheweb is subjected to combined bearing, shear and bending, its ultimate strength canbedeterminedfromtheHencky–vonMisesyieldcriterion.Thinstiffenedwebpanelsunderbearingstresseshaveaconsiderablereserveofpost-localbucklingstrength.Thisisattributedto itsability toredistribute in-planestresses fromthebuckledregiontothestiffeners.

3.7 Steel PlAteS In concrete-fIlled Steel tuBulAr columnS


Inathin-walledCFSTcolumnasdepictedinFigure3.21,thesteeltubewallsarerestrainedtobucklinglocallyoutwardbytheconcretecore.Figure3.22showsthebuckledshapeofthe tested square CFST columns under axial loading or eccentric loading. The restraintoftheconcretecoreconsiderably increasesthe localbucklingstressofthesteel tubeandthe ultimate strength of the CFST column (Ge and Usami 1992; Wright 1993; Uy andBradford1995;BridgeandO’Shear1998;LiangandUy2000;Uy2000;Lianget al.2007).SteelplatesinCFSTbeam–columnsmaybesubjectedtostressgradientscausedbyuniaxialbendingorbiaxialbending.Thisunilateral localbucklingproblemofsteelplatescanbesolvedbyusing thefinite elementmethod (LiangandUy2000;Liang et al. 2007).ThefouredgesofthesteelplaterestrainedbyconcreteareassumedtobeclampedasillustratedinFigure 3.23.ThebuckledshapeofsteelplatesrestrainedbyconcreteandunderuniformedgecompressionpredictedbythefiniteelementmethodisgiveninFigure3.24.Theelasticlocalbucklingstressof theclampedflatsteelplateundercompressivestressgradientsas


B

D

Concrete Steel tube

t d

b

Figure 3.21 Cross section of rectangular CFST column.

Figure 3.22 Local buckling of rectangular CFST short columns.

L

b

C

C

C

C

Figure 3.23 A clamped steel plate under uniform edge compression.


depictedinFigure 3.25canbedeterminedbyEquation3.4usingthebucklingcoefficientgivenbyLianget al.(2007)asfollows:

kb s s= − +18 89 14 38 5 3 2. . .α α (3.40)

whereαsisthestressgradientcoefficient,whichisdefinedastheratiooftheminimumedgestress(σ2)tothemaximumedgestress(σ1)actingontheplate.Figure3.26showsthebuck-lingcoefficientasafunctionofthestressgradientcoefficient.Itappearsthatincreasingthestressgradientcoefficientdecreasesthebucklingcoefficientkb.Whentheαs=1.0,theplateissubjectedtouniformcompressionandkb=9.81(LiangandUy2000).

Realsteelplateshaveinitialimperfectionsincludinginitialout-of-planedeflectionsandresidual stresses,which are induced in the process of construction and hot rolling, coldforming or welding. These imperfections will reduce the stiffness and strength of steelplates.Themaximummagnitudeof initialgeometric imperfectionsataplatecentrecanbetakenas0.1t.Figure3.27depictstheresidualstresspatterninweldedCFSTcolumns.Tensileresidualstressesthatreachthesteelyieldstressareinducedattheweldedcornersofthetubularcrosssection,whilecompressiveresidualstressesarepresentintheremainderofthetubewalls.Thetensileresidualstressesarebalancedbythecompressiveresidualstressesinatubewall.Thecompressiveresidualstressisusuallyabout25%–30%oftheyieldstressofthesteeltube(LiangandUy2000).

The initial local buckling stress of a steel plate with prescribed geometric imperfec-tionsandresidualstressesisafunctionofitsplatewidth-to-thicknessratio,stressgradient

Figure 3.24 Buckled shape of a clamped square steel plate under uniform edge compression.

L

b

Cσ2

σ1σ1

σ2

C

C

C

Figure 3.25 A clamped steel plate under compressive stress gradients.


coefficientandyieldstress.Forthinsteelplateswithb/tratiosrangingfrom30to100andunderlinearlyvaryingedgecompression,theirinitialbucklingstresscanbedeterminedby(Lianget al.2007)

σ1c

yfa a

bt

abt

abt

= +

+

+

1 2 3

2

4

3

(3.41)

whereσ1cistheinitiallocalbucklingstressofaplatewithimperfectionsa1,a2,a3anda4areconstantcoefficientswhichdependonthestressgradientcoefficient

αsandaregiveninTable3.1.

σr

σrσr

+ +

+

++

+

+ +

–

–

– –

σrfy

fy

fy

fy

Figure 3.27 Residual stress pattern in welded CFST columns.

00

2

4

6

8

10

12

14

16

18

20

0.2 0.4Stress gradient σ2/σ1

Buck

ling

coe�

cien

t kb

0.6 0.8 1 1.2

Figure 3.26 Buckling coefficients of clamped steel plates under compressive stress gradients.


3.7.2 Post-local buckling

Thepost-localbucklingstrengthofasteelplatewithprescribedgeometric imperfectionsandresidualstressesdependsonitsb/tratio,stressgradientcoefficient(αs)andyieldstress(fy)andcanbecalculatedby(Lianget al.2007)

σ1u

yfc c

bt

cbt

cbt

= +

+

+

1 2 3

2

4

3

(3.42)

whereσ1uistheultimatevalueofthemaximumedgestressσ1

c1,c2,c3andc4areconstantcoefficientswhicharegiveninTable3.2

Theultimatestrengthofsteelplateswithstressgradientcoefficientsgreaterthanzerocanbeapproximatelyestimatedby(Liang2009)

σ φ σ φ1u

ys

u

ys

f f= + ≤ <( . ) ( . )1 0 5 0 1 0 (3.43)

whereϕs=1−αsandσuistheultimatestressofsteelplatesunderuniformcompressionandcanbecalculatedusingEquation3.42withthestressgradientcoefficientofαs=1.0.

Table 3.1 Constant coefficients for determining the initial local buckling stresses of plate under stress gradients

αs a1 a2 a3 a4

0.0 0.6925 0.02394 − × −4 408 10 4. 1 718 10 6. × −

0.2 0.8293 0.01118 − × −2 427 10 4. 8 164 10 7. × −

0.4 0.6921 0.01223 − × −2 488 10 4. 8 676 10 7. × −

0.6 0.4028 0.02152 − × −3 742 10 4. 1 446 10 6. × −

0.8 0.5096 0.0112 − × −2 11 10 4. 7 092 10 7. × −

1.0 0.5507 0.005132 − × −9 869 10 5. 1 198 10 7. × −

Source: Adapted from Liang, Q.Q. et al., J. Constr. Steel Res., 63(3), 396, 2007.

Table 3.2 Constant coefficients for determining the ultimate strengths of plate under stress gradients

αs c1 c2 c3 c4

0.0 1.257 −0.006184 1 608 10 4. × − − × −1 407 10 6.

0.2 0.6855 0.02894 − × −4 89 10 4. 2 134 10 6. × −

0.4 0.6538 0.02888 − × −5 215 10 4. 2 424 10 6. × −

0.6 0.7468 0.01925 − × −3 689 10 4. 1 677 10 6. × −

0.8 0.6474 0.02088 − × −4 171 10 4. 2 058 10 6. × −

1.0 0.5554 0.02038 − × −3 944 10 4. 1 921 10 6. × −

−0.2 1.48 −0.01584 2 868 10 4. × − − × −1 742 10 6.

Source: Adapted from Liang, Q.Q. et al., J. Constr. Steel Res., 63(3), 396, 2007.


ForthesteelplatesinCFSTcolumnsunderuniformcompression,theireffectivewidthcanbeexpressedbythefollowingequationsgivenbyLiangandUy(2000):

bb f

fe cr

ycr y=

≤0 6751 3

./

σ σfor (3.44)

bb f

fe cr

cr ycr y=

+

>0 9151 3

./

σσ

σfor (3.45)

wherebeisthetotaleffectivewidthofthesteelplateσcr is theelasticcriticalbucklingstressof theperfect steelplateunderuniformedge

compression

TheeffectivewidthsofsteelplatesunderstressgradientsinCFSTcolumnsunderbiaxialbendingaredepictedinFigure3.28.EffectivewidthformulasofclampedsteelplatesundercompressivestressgradientsinCFSTbeam–columnswithb/tratiosrangingfrom30to100aregivenby(Lianget al.2007)

bb

bt

bt

e1 42

70 2777 0 01019 1 972 10 9 605 10= +

− ×

+ ×− −. . . .

bbt

s

>3

0 0for α . (3.46)

bb

bt

bt

e1 52

0 4186 0 002047 5 355 10 4 685 10= −

+ ×

− ×− −. . . . 77

3

0 0bt

s

=for α . (3.47)

bb

bb

es

e2 11= +( )φ (3.48)

wherebe1andbe2aretheeffectivewidthsasshowninFigure3.28.Fortheeffectivewidth(be1+be2)>b,thesteelplateisfullyeffectiveincarryingloadsandtheultimatestrengthofthesteelplatecanbedeterminedusingEquations3.42and3.43.

N.AB

D

t

σ2

σ1

σ1b be2 e1

σ2

Figure 3.28 Effective widths of steel tube walls under stress gradients.


Example 3.3: Effective area of steel section of a CFST column

Thecrosssection(600×700 mm)ofaCFSTcolumnunderbiaxialbendingisshowninFigure3.29.Oneofthesteelflangesissubjectedtocompressivestressgradientwithastressgradientcoefficientofαs=0.829,whileoneof thewebs isundercompressivestresses with a stress gradient coefficient of αs = 0.513. Calculate the effective cross-sectionalareaofthesteeltube.

1. Effective width of the flange under compressive stress gradient

Theclearwidthoftheflange:b=600−2×10=580 mm.Theeffectivewidthbe1oftheflangeundercompressivestressgradientiscalculatedas

bb

bt

bt

e1 42

70 2777 0 01019 1 972 10 9 605 10= +

− ×

+ ×− −. . . .

bbt

= +

− ×

−

3

40 2777 0 0101958010

1 972 1058010

. . . + ×

=−

27

3

9 605 1058010

0 393. .

be1 0 393 580 227 9= × =. . mm

Theeffectivewidthofbe2iscomputedasfollows:

φ αs s= − = − =1 1 0 829 0 171. .

bb

bb

es

e2 11= +( )φ

b be s e2 11 1 0 171 227 9 266 9= + = + × =( ) ( . ) . .φ mm

Thetotaleffectivewidthoftheflangeistherefore

b b b be e e= + = + = < =1 2 227 9 266 9 494 8 580. . . mm mm

N.A

700

600

10

αs=0.513

αs=0.829

Figure 3.29 Cross section of CFST column under biaxial bending.


2. Effective width of the web under compressive stress gradient

Theclearwidthoftheweb:b=700−2×10=680 mm.Theeffectivewidthbe1ofthewebundercompressivestressgradientiscalculatedas

bb

bt

bt

e1 42

70 2777 0 01019 1 972 10 9 605 10= +

− ×

+ ×− −. . . .

bbt

= +

− ×

−

3

40 2777 0 0101968010

1 972 1068010

. . . + ×

=−

27

3

9 605 1068010

0 361. .

be1 0 361 680 245 5= × =. . mm

Theeffectivewidthofbe2iscomputedasfollows:

φ αs s= − = − =1 1 0 513 0 487. .

b be s e2 11 1 0 487 245 5 365= + = + × =( ) ( . ) .φ mm

Thetotaleffectivewidthofthewebis

b b b be e e= + = + = < =1 2 245 5 610 5 680. .365 mm mm

3. Effective cross-sectional area of the steel tube

Assume thatonly theflangeandwebundercompressive stressgradientswillundergolocalbuckling.Theineffectivecross-sectionalareaoftheflangeundercompressivestressgradientisdeterminedas

A b b tnef e= − = − × =( ) ( . )580 494 8 10 852mm2

Theineffectivecross-sectionalareaofthewebundercompressivestressgradientis

A b b tnew e= − = − × =( ) ( . )680 610 5 10 695mm2

Thegrosscross-sectionalareaofthesteeltubeiscalculatedas

Ag = × − − × − × =600 700 600 2 10 700 2 10( )( ) 25,600mm2

Theeffectivecross-sectionalareaofthesteeltubeis

Ae = − − =25,600 24,053mm2852 695

TheeffectivesteelareasoftheCFSTcolumnunderbiaxialbendingareshowninFigure3.30.


3.8 douBle SkIn comPoSIte PAnelS

3.8.1 local buckling of plates under biaxial compression

Doubleskincomposite(DSC)panelsareformedbyfillingconcretebetweentwosteelplatesweldedwithstudorothertypeofshearconnectorsataregularspacingasschematicallydepictedinFigure3.31.Thesteelskinsareusedaspermanentformworkandbiaxialsteelreinforcementfortheconcretecore,providingsoundwaterproofinginmarineandfreshwa-terenvironment.Studshearconnectorscarrythelongitudinalshearbetweentheconcretecoreandthesteelskinsaswellasseparationattheinterface.Thiscompositesystemoffershighstrength,stiffnessandductilityandis increasinglyusedinsubmergedtubetunnels,militaryshelters,nuclearinstallations,shearwallsinbuildings,liquidandgascontainmentstructuresandoffshorestructures.DSCpanelsexhibittwoparticularfailuremodeswhichincludethelocalbucklingofsteelplatefieldsbetweenstudshearconnectorsandtheshearconnection failure between the steel skins and the concrete core (Oduyemi and Wright1989;Wrightet al.1991).

Steel plate

Concrete coreStud shear connector

Figure 3.31 Cross section of double skin composite panel.

600

365

245.5

227.9Ine�ective area

E�ective area266.9

10

700

Figure 3.30 Effective steel area of CFST column under biaxial bending.


Figure3.32depictsasingleplatefieldbetweenstudshearconnectors,whichisrestrainedatthecornersbystudshearconnectors.Itisassumedthattheedgesoftheplatefieldbetweenshearconnectorsarehingedandtherotationsatthecornersarerestrainedwhilethein-planetranslationsofstudshearconnectorsaredefinedbytheshear–slipmodel(Lianget al.2003).Whentheplatefieldis locatedattheedgeofthepanel,theedgeoftheplatefieldcanbeassumedtobeclampedastherotationsarerestrained.Theelasticlocalbucklingstressofasteelplateunderbiaxialcompressiondependsonitsaspectratio(spacingofshearconnectorsintwodirections),theplatethickness,compressivestressesintwodirections(αcs=σx/σy)andboundaryconditionsincludingtherestraintofshearconnectors.Theelasticbuckingstressinxdirectioncanbedeterminedby

σ π

νxcrxo sk E

b t=

−

2

2 212 1( )( )/ (3.49)

wherekxoistheelasticbucklingcoefficientinthexdirection.TheelasticbuckingstressσycrinydirectioncanbeobtainedbysubstitutingkyoandainEquation3.49.Elasticbucklingcoefficientsofplateswithvariousboundaryconditionsandloadingratiosofbiaxialcom-pressionsweregivenbyLianget al.(2003).

Elastic buckling coefficients can be used to determine the limiting width-to-thicknessratiosforsteelplatefieldsunderbiaxialcompressioninDSCpanels.Thelimitingwidth-to-thicknessratioofsteelplatefieldswithEs=200GPaandv=0.3canbeobtainedfromthevonMisesyieldcriterionasfollows(Lianget al.2003):

bt

fk

k k kyxo

xo yo yo

25026 89 2

2

2

4

1 4

= − +

.

/

ϕ ϕ (3.50)

whereφ=a/bistheplateaspectratio.Forasquaresteelplatefieldunderthesamecompres-sivestressesintwodirections(αcs=1.0),thelocalbucklingcoefficientiskxo=kyo=2.404(Lianget al.2003).Thelimitingwidth-to-thicknessratiois41.7.Ifthe16 mmthicksteelplateofGrade300withayieldstressof300MPaisused,themaximumspacingofstudshearconnectorsintwodirectionsis609 mm.

a

Stud shear connector

σy

σx b

Figure 3.32 Single plate element restrained by stud shear connector under biaxial compression.


3.8.2 Post-local buckling of plates under biaxial compression

SteelplatefieldsinDSCpanelsarerestrainedbystudshearconnectorswithafiniteshearstiffnesswhichconsiderablyincreasestheresistanceofplatefieldsagainstlocalbuckling.Slendersteelplatefieldsmaybucklelocallyinaunilateraldirectionbeforeshearconnec-torsfail.Inaddition,shearconnectorsmayfracturebeforestockysteelplatefieldsattaintheirfullplasticcapacities.Moreover,interactionmodesbetweenlocalbucklingandshearconnectionfailuremayexist.Theeffectofstudshearconnectorsontheplatebucklingcanbetakenintoaccountinthenonlinearanalysisbyusingtheshear–slipmodel(Lianget al.2003).

Thepost-localbucklingbehaviourofsteelplatefieldsinaDSCpanelcanbedescribedbybiaxialstrengthinteractionformulasderivedfromthevonMisesyieldellipseasfollows(Lianget al.2003):

σ ησ σ σ

γ γζ

xuo

ys

xuo yuo

y

yuo

yn n

f f f

c

+

+

= ≤2

2

1( ) (3.51)

whereσxuodenotestheultimatestrengthofaplateinxdirectionunderbiaxialcompressionσyuoistheultimatestrengthofaplateinydirectionunderbiaxialcompressionζcistheshapefactoroftheinteractioncurvedependingontheplateaspectratioand

slendernessηsisafunctionoftheplateslendernessγnistheuniaxialstrengthfactor

Theshapefactorηscanbeusedtodefineanyshapeofinteractioncurvesfromastraightline(ηs=2)tothevonMisesellipse(ηs=−1).Forsquareplates,theshapefactorζc=2andthevaluesofηsandγnaregiveninTable3.3.

3.8.3 local buckling of plates under biaxial compression and shear

WhenDSCpanelsareusedasslabsorshearwalls,steelplatefieldsbetweenstudshearcon-nectorsmaybesubjectedtobiaxialcompressionandin-planeshear.Figure3.33schemati-callydepictsaplatefieldundercombinedbiaxialcompressionandshear.Thislocalbuckling

Table 3.3 Parameters of strength interaction formulas for square plates in biaxial compression

b/t ζc ηs γn

100 2.0 1.4 0.1480 2.0 1.47 0.21160 2.0 1.45 0.35340 2.0 0.8 0.6520 2.0 0.0 0.846

Source: Adapted from Liang, Q.Q. et al., Proc. Inst. Civil Eng., Struct. Build., U.K., 156(2), 111, 2003.


problemofplatefieldscanbesolvedbyusingthefiniteelementmethod(Lianget al.2004).Theelasticbucklingcoefficientscanbecalculatedbythefollowingequations:

σ π

νxcrx sk E

b t=

−

2

2 212 1( )( )/ (3.52)

σ

πνycr

y sk Ea t

=−

2

2 212 1( )( )/ (3.53)

τ

πνxycr

xy sk Eb t

=−

2

2 212 1( )( )/ (3.54)

whereσxcrstandsfortheelasticbucklingstressinthexdirectionσycrrepresentstheelasticbucklingstressintheydirectionτxycrdenotestheelasticshearbucklingstresskxstandsfortheelasticbucklingcoefficientinthexdirectionky denotestheelasticbucklingcoefficientintheydirectionkxyistheelasticshearbucklingcoefficient

Thebucklingcoefficientofplatesundercombinedstatesofstressesaccountsfortheeffectsofplateaspectratio,boundaryconditionincludingrestraintsbyshearconnectorsandinter-actionbetweenbiaxialcompressionandshearonthecriticalbucklingstress.

Theinteractionformulafordeterminingtheelasticbucklingcoefficientsforsquareplatesunderbiaxialcompressionandshearisexpressedby(Lianget al.2004)

kk

kk

x

xo

xy

xyo

b

+

=

ζ 2

1 (3.55)

wherekxodenotesthebucklingcoefficientinthexdirectionintheabsenceofshearstresseskxyostandsfortheshearbucklingcoefficientintheabsenceofbiaxialcompressionζbisthebucklingshapefactordefiningtheshapeofabucklinginteractioncurve

a

b

σy

τxy

σx

Stud shear connector

Figure 3.33 Single plate element restrained by stud shear connector under biaxial compression and shear.


Thevaluesofbucklingcoefficientskxoandkxyoforsteelplateswithdifferentboundarycon-ditionsaregiveninTable3.4fordesign.

Bucklingcoefficientspresentedcanbeusedtodeterminethelimitingwidth-to-thicknessratiosforsteelplatesunderbiaxialcompressionandshearinDSCpanels.Thisensuresthattheelasticlocalbucklingofsteelplatesbetweenstudshearconnectorswillnotoccurbeforesteelyielding.TherelationshipbetweencriticalbucklingstresscomponentsatyieldcanbeexpressedbythevonMisesyieldcriterionas

σ σ σ σ τxcr xcr ycr ycr xycr yf

2 2 2− + + =3 2 (3.56)

IfthematerialpropertiesE=200GPaandν=0.3andtheplateaspectratioφ=a/bareused,thelimitingwidth-to-thicknessratiocanbederivedbysubstitutingEquations3.52through3.54intoEquation3.56as(Lianget al.2004)

bt

fk

k k kky

xx y y

xy250

26 89 32 4

1 4

= − + +

.

/

22

2

ϕ ϕ (3.57)

StressesactingattheedgesofaplatefieldinaDSCpanelcanbedeterminedbyundertakingaglobalstressanalysisontheDSCpanel.Itisassumedthatasquareplatefield(φ=1)withtheS-S-S-S+SCboundaryconditionisunderbiaxialcompressivestresses(αcs=1)andshearstressτxy=0.5σx.Thisgiveskx=kyandkxy=0.5kxaccordingtoEquations3.52through3.54.FromTable3.4,parametersforbucklinginteractionscanbeobtainedaskxo=2.404,kxyo=10.838andζb=1.1.BysubstitutingtheseparametersintoEquation3.55,bucklingcoefficientsareobtainedaskx=2.38andkxy=1.19.ByusingEquation3.57,thelimiting

Table 3.5 Parameters of strength interaction formulas for plates in biaxial compression and shear

b/t ζs σxuo yf/ τ τxyuo / 0

100 0.8 0.205 0.87580 1.1 0.248 0.98460 1.3 0.321 1.040 1.6 0.481 1.020 2.0 0.658 0.927

Source: Adapted from Liang, Q.Q. et al., J. Struct. Eng., ASCE, 130(3), 443, 2004.

Table 3.4 Parameters of buckling interaction formulas for plates in biaxial compression and shear

Boundary condition

kxo

kxyo ζb αcs =1 5. αcs =1 0. αcs = 0 5. αcs = 0 25. αcs = 0

C-C-S-S+SC 3.362 4.216 5.514 6.56 7.797 18.596 2C-S-S-S+SC 2.589 3.168 4.06 4.705 5.552 14.249 1.7S-S-S-S+SC 1.923 2.404 3.204 3.84 4.782 10.838 1.1

Source: Adapted from Liang, Q.Q. et al., J. Struct. Eng., ASCE, 130(3), 443, 2004.


width-to-thicknessratio for thisplatefieldwithayieldstressof300MPa is48. If thecompressionsteelskinwithathicknessof16 mmisused,themaximumstudspacingintwodirectionsinthisDSCpanelis700 mm.

3.8.4 Post-local buckling of plates under biaxial compression and shear

The shape of strength interaction curves strongly depends on the plate slenderness. Thepost-localbucklingstrengthofplatefieldsinDSCpanelscanbedescribedbythefollowingstrengthinteractionformulas(Lianget al.2004):

σσ

ττ

ζxu

xuo

xyu

xyuo

s

+

=2

1 (3.58)

whereσxudenotestheultimatestrengthofaplate inxdirectionunderbiaxialcompression

andshearσxuoistheultimatestrengthofaplateinxdirectionunderbiaxialcompressiononlyτxyurepresentstheultimateshearstrengthofaplateτxyuodenotestheultimatestrengthofaplateunderpureshearonlyζsisthestrengthshapefactoroftheultimatestrengthinteractioncurve

Table3.5givestheultimatestrengthofsquaresteelplatesundereitherbiaxialcompressionorshearaloneandthestrengthshapefactorsforplateswithvariousslendernessratios.

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Shanmugam,N.E.,Liew,J.Y.R.andLee,S.L.(1989)Thin-walledsteelboxcolumnsunderbiaxialload-ing,JournalofStructuralEngineering,ASCE,115(11):2706–2726.

Timoshenko,S.P.andGere,J.M.(1961)TheoryofElasticStability,2ndedn.,NewYork:McGraw-Hill.Trahair,N.S.andBradford,M.A.(1998)TheBehaviourandDesignofSteelStructurestoAS4100,3rd

edn.(Australian),London,U.K.:Taylor&FrancisGroup.Usami,T. (1982) Effective width of locally buckled plates in compression and bending, Journal of

StructuralEngineering,ASCE,119(5):1358–1373.Uy,B. (2000)Strengthof concrete-filled steelboxcolumns incorporating localbuckling, Journalof

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Experimentalandtheoreticalstudy,ProceedingsoftheInstitutionofCivilEngineers,StructuresandBuildings,U.K.,110:426–440.

von Karman, T., Sechler, E.E. and Donnel, L.H. (1932) Strength of thin plates in compression,TransactionsofASME,54:53–57.

Winter, G. (1947) Strength of Thin Steel Compression Flanges, Cornell University Eng. Exp. Stn.,ReprintNo.32.

Wright,H.D.(1993)Bucklingofplatesincontactwitharigidmedium,TheStructuralEngineer,71(12):209–215.

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73

Chapter 4

Steel members under bending

4.1 IntroductIon

Steelmembersunderbendingareflexuralmembers(beams)whichareusedtotransfertrans-verseloadstothesupports.Thetransverseloadsactingonabeammayinducetheactionsofbending,shearandbearinginthebeam.Therefore,steelbeamsneedtobedesignedforbending,shearandbearing.Steelbeamsareoftenmadeofthin-walledelementsbyhotroll-ing,weldingandcoldforming.TypicalsectionsforsteelbeamsaregiveninFigure4.1.Thebehaviourofasteelbeamdependsonitssectionslenderness,materialpropertiesandmem-berslenderness.Lateralandtorsionalrestraintsalongthesteelbeamsignificantlyincreaseitsmembermomentcapacity.Asaresult,theuseoflateralandtorsionalrestraintsleadstosignificanteconomies.Steelplategirdersareoftenmadeofslenderwebswhichmayundergoshearandbearingbuckling.Transversewebandload-bearingstiffenersareattachedtothewebsofsteelplategirderstoincreasetheirbucklingcapacities.Thedesignofasteelbeamfor strength includes the verification of its section and member moment capacities, webshearandbearingcapacitiesandthedesignofwebstiffenersandrestraints.

Thischapterpresents thebehaviouranddesignofsteelmembersunderbendingtoAS4100(1998).Thefundamentalbehaviourofsteelbeamsunderbendingisdiscussedfirst.Thebasicprinciplesfordeterminingtheelasticsectionpropertiesofthin-walledmembersaredescribed.Methodsforcalculatingthesectionmomentandmembermomentcapacitiesofsteelbeamsarepresented.Thedesignofsteelbeamwebswithorwithoutstiffenersforshearandbearingisalsogiven.

4.2 BehAvIour of Steel memBerS under BendIng

Thebehaviourofasteelmemberunderbendingisinfluencedbyitsmaterialproperties,sec-tionslenderness,memberslendernessandlateralandtorsionalrestraints.Forflexuralmem-berscomposedofslendersteelelements,localbucklingofthecompressionflangeorbendingwebmayoccurbeforesteelyields.AsdiscussedinChapter3,localplatebucklingremark-ablyreducestheultimatesectionmomentcapacityofsteelmembers inbending.Underahighshearforce,thewebofasteelbeammayfailbyshearbucklingoryielding.Thisresultsinafurtherreductioninthemomentcapacityofthesteelbeam.Underconcentratedloadsorreactionsatthesupports,thewebofasteelflexuralmemberissubjecttobearingstresses,whichmaycausethewebbearingbucklingoryielding.Theaforementionedlocalfailurespreventsteelmemberssubjectedtotransverseloadsfromattainingtheirfullplasticmomentcapacities.Steelbeamsmadeofcompactsteelsectionsrestrainedlaterallyandtorsionallywouldnotfailuntilwellafteryielding.Thesebeamsofcompactsectionscanattaintheirfullplasticmomentcapacitiesbeyondtheyieldmoments.


Ifasteelbeamunderin-planeloadingdoesnothavesufficientlateralstiffnessorlateralandtorsionalsupports,itmaybuckleoutofitsplaneoftheloadingbydeflectinglaterallyandtwistingasillustratedinFigure4.2.Thisbehaviouriscalledflexural–torsionalbuck-ling,whichsignificantlyreducesthein-planeload-carryingcapacityofthebeam(Trahair1993a).Whentheappliedmomentreachestheelasticbucklingmomentofthebeam,theelas-ticflexural–torsionalbucklingoccurs.LongandunrestrainedsteelI-beamshavesuchlowresistancestobendingandtorsionthattheircapacitiesaregovernedbytheelasticflexural–torsionalbuckling.Aperfectlystraightbeamwithan intermediateslendernessmayyieldbeforetheelasticflexural–torsionalbucklingoccurs.Stockysteelbeamsarenotaffectedby

(e) (f )

(d)(c)

(g)

(b)(a)

Figure 4.1 Typical steel sections for beams: (a) hot-rolled section, (b) welded section, (c) built-up section, (d) hot-rolled section with flange plates, (e) welded box section, (f) welded box section from channels and (g) cold-formed hollow section.

Figure 4.2 Flexural–torsional buckling of a simply supported steel I-beam modelled by finite elements.

Steel members under bending 75

lateralbuckling,andtheirinelasticbucklingmomentsarehigherthanthein-planeplasticcollapsemoments.Lateralandtorsionalrestraintsareoftenusedinsteelbeamsinpracticetopreventtheflexural–torsionalbuckling.

4.3 ProPertIeS of thIn-WAlled SectIonS

4.3.1 centroids

Thecentroidofacompoundthin-walledsectionisdefinedasthegeometriccentreofthecrosssection.Ifthesectioniscomposedofuniformorhomogeneousmaterial,thecentroidofthesectioncoincideswithitscentreofmassoritscentreofgravity.Forathin-walledsteelsectioncomposedofnelements,thecoordinatesofthecentroidposition(xc, yc)aboutthereferenceaxescanbedeterminedby

xA x

Ac

j jj

n

jj

n= =

=

∑∑

1

1

(4.1)

yA y

Ac

j jj

n

jj

n= =

=

∑∑

1

1

(4.2)

whereAjistheareaofelementjxjandyjarethecentroidalcoordinatesofelementjmeasuredfromthereferenceaxes

4.3.2 Second moment of area

Thesecondmomentofareaofacompoundthin-walledsteelsectionaboutitscentroidalaxescanbecalculatedusingtheparallelaxistheoremasfollows:

I I A y yx ox j j j c

j

n

= + −⋅

=∑[ ( ) ]2

1

(4.3)

I I A x xy oy j j j c

j

n

= + −⋅

=∑[ ( ) ]2

1

(4.4)

whereIox j⋅ isthesecondmomentofareaofthejthelementaboutitscentroidalaxisoxIoy j⋅ isthesecondmomentofareaofthejthelementaboutitscentroidalaxisoy

4.3.3 torsional and warping constants

Thetorsional loadingactingonasteelbeamisresistedbytwoshearstresscomponents.Whenasteelbeamissubjectedtouniformtorsion,therateofchangeintheangleoftwist


rotationandthelongitudinalwarpingdeflectionsisconstantalongthebeam(Kollbrunnerand Basler 1969; Trahair and Bradford 1998). A single set of shear stresses distributedaroundthecrosssectionresiststhetorqueactingatthecrosssection.ThestiffnessofthebeamassociatedwiththeseshearstressesisreferredtothetorsionalrigidityGJofthebeam,whereG istheshearmodulusandJ isthetorsionalconstant.Whenasteelbeamissub-jectedtonon-uniformtorsion,thelongitudinalwarpingdeflectionsvaryalongthebeam.Anadditionalsetofshearstressesmayacttogetherwiththoseinducedbyuniformtorsiontoresistthetorqueactingatthecrosssection.ThestiffnessofthebeamassociatedwiththeseadditionalshearstressesisreferredtothewarpingrigidityEIwofthebeam,whereIwisthewarpingconstant.Torsionalandwarpingconstantsareneededinthedeterminationoftheelasticbucklingmomentsofsteelbeams.

Thetorsionalconstant(J)ofasectionisthepolarmomentofinertiaofthecross-sectionalarea.Forcircularhollowsections,thetorsionalconstant(J)iscalculatedby

J d do i= −( )π

3244 (4.5)

wheredoanddiaretheouterandinnerdiametersofthecircularsection,respectively.Forthin-walledopensections,thetorsionalconstantcanbeapproximatelycomputedas

thesumofthetorsionalconstantofindividualrectangularelementbyneglectingthecontri-butionofthefilletregionwhereelementsarejoined:

J

bt≈∑3

3 (4.6)

wherebisthelengthtisthethicknessofeachrectangularelementthatformsthecrosssection

ForI-beamswithequalflanges,thewarpingconstantisgivenby

I

I dw

y fc=2

4 (4.7)

wheredfcisthedistancebetweenthecentroidsofthetwoflanges.FormonosymmetricI-sectionsasdepictedinFigure4.3,thewarpingconstantiscalcu-

latedby(KitipornchaiandTrahair1980;TrahairandBradford1998)

I

q b t dw

m fc= 13

12

12 (4.8)

whereqmisgivenby

q

b b t tm =

+1

1 1 23

1 2( ) ( )/ / (4.9)


4.3.4 elastic section modulus

Theelasticsectionmoduluscanbedeterminedfromthesecondmomentofareaasfollows:

Z

Iy

xx=

max

(4.10)

Z

Ix

yy=

max

(4.11)

whereZxandZyaretheelasticsectionmoduliaboutitscentroidalx-andy-axes,respectivelyxmaxandymaxarethemaximumdistancesfromthecentroidalx-andy-axesofthesec-

tiontoitsextremefibres,respectively

Theelasticsectionmodulusisusedinthecalculationofelasticstressesinsteelmembersunderbending.Itisnotedthattheeffectivesectionmodulus(Ze)isusedinthecalculationof thesectionmomentcapacitiesofsteelbeams.Asdiscussed inChapter3, theeffectivesectionmodulusofanon-compactorslendersteelsectionisdeterminedbyaccountingforlocalbucklingeffects.

t1

b1

tw dfc Ds

t2

b2

y

Figure 4.3 Dimensions of monosymmeric I-section.


Example 4.1: Calculation of elastic properties of a monosymmetric I-section

ThemonosymmetricsteelI-sectionisdepictedinFigure4.4.Thesectionissymmetricaboutitsminorprincipaly-axis.Calculatetheelasticpropertiesofthemonosymmetricsection.

1. Centroid of the section

Thecleardepthofthesectionisd1=400−16−14=370mm.Thecentroidpositionofthesectionmeasuredfromthetopfibreiscalculatedas

yA y

Ac

j jj

n

jj

n=

=× × + × × + + ×

=

=

∑∑

1

1

200 16 16 2 370 10 370 2 16 120 14( ) ( )/ / ×× −× + × + ×

=

( )

.

400 14 2200 16 370 10 120 14

166 6

/

mm

16

10

14

y

120

400

200

Figure 4.4 Dimensions of monosymmetric I-section.


2. Second moment of area

Thesecondmomentofareaaboutthemajorprincipalx-axisis

I I A y yx ox j j j c

j

n

= + −

=×

+ × × −

⋅

=∑ ( )

.

2

1

3200 1612

200 16 166 6162

×+ × × + −

2

3 210 370

1210 370

3702

16 166 6+ .

×+ × × − −

= ×+

120 1412

120 14 400142

166 6 213 29 103 2

. . 66 mm4

Thesecondmomentofareaabouttheminorprincipaly-axisis

I I A x xy oy j j j c

j

n

= + −

=×

+

×

⋅

=∑ ( )2

1

3 316 20012

370 1012

+×

= × mm414 120

1212 71 10

36.

3. Torsion and warping constants

Thetorsionconstantcanapproximatelybecalculatedas

J

bt≈ =

×+

×+

×= ×∑

3 3 3 33

3200 16

3370 10

3120 14

3506 16 10. mm4

Thewarpingconstantcanbecalculatedasfollows:

q

b b t tm =

+=

+ ×=

11

11 200 120 16 14

0 1591 2

31 2

3( ) ( ) ( ) ( ).

/ / / /

I

q b t dw

m cf= =× × × − −

= ×13

12 3 2

120 159 200 16 400 16 2 14 2

12251 39 1

. ( ).

/ /009 mm6

4. Section modulus

Thesectionmodulusaboutitsprincipalx-axisis

Z

Iy

xx= =

×−

= ×max

..

.213 29 10400 166 6

913 8 106

3 mm3

Thesectionmodulusaboutitsminorprincipaly-axisis

Z

Ix

yy= =

×= ×

max

..

12 71 10200 2

127 1 106

3

/mm3


4.4 SectIon moment cAPAcIty

ThesectionmomentcapacityofasteelsectioncanbederivedfromthestressdistributionshowninFigure4.5.Fortherectangularsection,thesecondmomentofareaaboutitssec-tionmajorprincipalx-axis is Ix=BD3/12.Theeffective sectionmodulusof this sectionwhichisassumedtobefullyeffectiveisdeterminedas

Z

Iy

BDD

BDex

x= = =max

3 2122 6//

(4.12)

TheextremefibreofthesectiondepictedinFigure4.5isassumedtoreachtheyieldstress(fy)ofthesteel.ThecompressionandtensionforcesinthesectionareC T BD f BDfy y= = =1

2142( ) ./

Thenominalmomentcapacityofthesectionforbendingaboutthesectionmajorprincipalx-axiscanbeobtainedbytakingmomentsaboutitscentroidas

M BDf D

BDfsx y y= ×

=

14

23 6

2

(4.13)

Theearlierequationcanberewrittenas

M Z fsx ex y= (4.14)

wherefyistakenastheminimumyieldstressforthesteelsection.Whenasteelbeamissubjectedtobendingaboutitssectionmajorprincipalx-axis,all

sectionsofthebeammustsatisfythefollowingdesignrequirement:

M Mx sx∗ ≤ φ (4.15)

whereMx∗isthefactoreddesignbendingmomentaboutthex-axis

φ = 0 9. isthecapacityreductionfactor

C

D

BCross section Stress distribution

D23

T

fy

fy

Figure 4.5 Stress distributions in rectangular steel section under bending.


Similarly,forasteelbeambendingaboutitssectionminorprincipaly-axis,allsectionsofthebeammustsatisfy

M My sy∗ ≤ φ (4.16)

whereMy∗isthefactoreddesignbendingmomentaboutthesectionminorprincipaly-axis

Msyisthenominalsectionmomentcapacityforbendingaboutthesectionminorprin-cipaly-axisandisdeterminedas

M Z fsy ey y= (4.17)

inwhichZeyistheeffectivesectionmodulusforbendingaboutthesectionminorprincipaly-axis.

4.5 memBer moment cAPAcIty

4.5.1 restraints

Themembermomentcapacityofasteelbeamunderbendingdependsonthelateralandtorsionalrestraintsatitsendsandalongthebeam.Therestraintsuchasanelement,supportorconnectionisusedtopreventabeamfromlateraldeflectionand/orlateralrotationabouttheminoraxisand/ortwistaboutthecentrelineofthebeam.VariousrestraintconditionsforcrosssectionsaredefinedinClause5.4ofAS4100(1998)andbrieflydescribedherein.Allsupportsareassumedtofullyorpartiallyrestrainthecrosssectionsagainstdeflectionsandtwistoutoftheplaneofloading.

Ifthelateraldeflectionofthecriticalflangeiseffectivelypreventedandthetwistrotationofthesectioniseithereffectivelypreventedorpartiallyprevented,thecrosssectionisconsideredtobefullyrestrained(F).Ifthelateraldeflectionofsomepointsinthecrosssectionratherthanthecriticalflangeandthetwistrotationofthesectioniseffectivelysuppressed,thecrosssectionisalsofullyrestrained.Thecriticalflangeistheflangethatwoulddeformfurtheriftherestraintisremoved.Thisisthecompressionflangeforasimplysupportedbeamandthetopflangeforacan-tileverundergravityloads.SomeofthefullyrestrainedcrosssectionsareillustratedinFigure4.6.

Apartiallyrestrained(P)crosssectionisthesectionwherethelateraldeflectionofsomepointsinthecrosssectionratherthanthecriticalflangeiseffectivelysuppressedwhilethetwistrotationofthesectionispartiallyprevented.Figure4.7schematicallydepictspartiallyrestrainedcrosssections.

Ifthelateraldeflectionofthecriticalflangeiseffectivelypreventedbytherestraintwhichineffectivelysuppressesthetwistrotationofthesection,thecrosssectionisconsideredtobelaterallyrestrained(L),asshowninFigure4.8.

Iftherotationofthecriticalflangeaboutthesection’sminoraxisinafullyorpartiallyrestrainedcrosssectionisprevented,thecrosssectionistreatedasrotationallyrestrainedasdemonstratedinFigure4.9.

Tobeeffectiveinrestrainingasegmentinasteelbeam,therestrainingelementsattheendsofthesegmentmustbeabletotransferatransverseforceactingatthecriticalflangeasspecifiedinClause5.4.3ofAS4100.Thenominaltransversedesignforce( )NR

∗ transferredbytherestraintagainstlateraldeflectionortwistrotationis

N NR f∗ = ∗0 025. (4.18)

whereNf∗isthemaximumforceinthecriticalflangesoftheadjacentsegments.


Whenparallelmembersarerestrainedbyacontinuousrestrainingelement,eachrestrain-ingelementshouldbedesignedtocarryatransverseforceequaltothesumof0 025. Nf

∗fromtheconnectedmemberand0.0125timesthesumofflangeforcesintheconnectedmembersbeyond.

4.5.2 members with full lateral restraint

Theflexural–torsionalbucklingofasteelbeamwithfulllateralrestraintiseffectivelypre-ventedbytherestraint.Thisimpliesthatthenominalmembermomentcapacity(Mb)ofa

To pinsupport

C

C

C

Concrete slab

To pinsupport

Fully weldedFully welded

C= critical ange

Figure 4.7 Partially restrained cross sections.

C = critical �ange

C

Or C here

Websti�ener

CC

C

Fly brace

Rafter

Purlin

To pinsupport

To pinsupport

Fully weldedFully welded

Figure 4.6 Fully restrained cross sections.


steelmemberwithfulllateralrestraintcanbetakenasthenominalsectionmomentcapacity(Ms)ofthecriticalsection.ThecriticalsectioninasegmentormemberisdefinedasthecrosssectionhavingthelargestratioofM*/Ms.

AsspecifiedinClause5.3.2ofAS4100(1998),asegmentfullyorpartiallyrestrainedatbothends isconsidered tohave full lateral restraint if it satisfiesoneof the followingrestraintconditions:

a.Thesegmenthascontinuousrestraintsatthecriticalflange. b.Thesegmenthasintermediatelateralrestraintsatthecriticalflangeandthelengthof

eachsub-segmentsatisfiestheslendernessrequirementsgivenin(c). c.Thesegmentsatisfiestheslenderness(l/ry)requirementsgiveninTable4.1,whereryis

theradiusofgyrationaboutthesectionminorprincipaly-axis.

ThemomentratioβmgiveninTable4.1istakenas−1.0or−0.8forsegmentssubjectedtotransverseloadsorβm M M= ± ∗ ∗

2 1/ forsegmentswithouttransverseloads,whereM1∗andM2

∗( )M M1 2

∗ ≥ ∗ aredesignbendingmomentsatthesegmentends.Themomentratioβmistakenaspositiveforbendinginreversecurvatureandnegativeforbendinginsinglecurvature.

To pinsupport

Purlin

Rafter

C

C

Pin connection

Figure 4.8 Laterally restrained cross sections.

Beam

Column

Sti�ener

Sti�enerHeavy end plate

C

Figure 4.9 Rotationally restrained cross sections.


4.5.3 members without full lateral restraint

Steelbeamswithoutfulllateralrestraintmayundergoflexural–torsionalbuckling,whichreducestheirmembermomentcapacities.Therefore,steelbeamswithoutfulllateralrestraintmustbedesignedagainstflexural–torsionalbuckling(Trahair1993a,b;Trahairetal.1993;TrahairandBradford1998).Theeffectofflexural–torsionalbucklingistakenintoaccountbyusingaslendernessreductionfactorαs.

4.5.3.1 Open sections with equal flanges

InClause5.6.1.1ofAS4100,thenominalmembermomentcapacity(Mb)foropensectionsegmentswithequalflangesandfullorpartialrestraintsatbothendsiscomputedby

M M Mb m s s s= ≤α α (4.19)

whereαm is the momentmodification factorwhich accounts for the effect of non-uniform

momentdistributionalongthesegmentαsistheslendernessreductionfactorwhichconsiderstheeffectofthesegmentslender-

nessonthemembermomentcapacityMsisthenominalsectionmomentcapacity

Itisnotedthatthemembermomentcapacityshouldnotbegreaterthanthesectionmomentcapacity.

Themomentmodificationfactor(αm),whichisusuallygreaterthan1.0,mayincreasethe member moment capacity. Economical designs can be achieved by using αm formemberswithhighmomentgradientsalongthesegments.Thisfactorcanbeobtained

Table 4.1 Slenderness requirements for full lateral restraint for segments fully or partially restrained at both ends

Segment section Slenderness limits

I-section with equal flangeslr fy

my

≤ +( )80 50250

β

Equal channellr fy

my

≤ +( )60 40250

β

I-section with unequal flanges lr

I Ad

I Z fym

cy fc

y ex y

≤ +

( )

.80 50

22 5

250β

RHS or square hollow section (SHS)lr

bd fy

mf

w y

≤ +

( )1800 1500

250β

Anglelr

bb fy

my

≤ +

( )210 175

2502

1

β



fromTable5.6.1ofAS4100orcalculatedfromthedesignbendingmomentdistributiondeterminedbystructuralanalysiswithinthesegmentasfollows:

αmmM

M M M

=∗

∗( ) + ∗( ) + ∗( )≤1 72 5

2

2

3

2

4

2

.. (4.20)

whereMm∗isthemaximumdesignbendingmomentwithinthesegmentconsidered

M2∗andM4

∗aredesignbendingmomentsatthequarterpointsofthesegmentM3∗isthedesignbendingmomentatthemidpointofthesegment

Thememberslendernessreductionfactor(αs),whichisusuallylessthan1.0,mayreducethemembermomentcapacity(Mb)belowthesectionmomentcapacity(Ms).Thisfactorisa functionof the sectionmomentcapacityand theelasticbucklingmoment (Moa)whichreflectstheslendernessofthememberandisdeterminedby

αss

oa

s

oa

MM

MM

=

+ −

≤0 6 3 1 0

2

. . (4.21)

whereMoacanbeeithertakenasthereferencebucklingmomentMoordeterminedfromanelasticbucklinganalysis.Figure4.10showstherelationshipbetweenαsandthemomentratioofMs/Moa.ItappearsthattheslendernessreductionfactordecreaseswithincreasingthemomentratioofMs/Moa.

Moment ratio Ms/Moa

00

0.2

0.4

Slen

dern

ess r

educ

tion

fact

or α

s

0.6

0.8

1

1.2

1 2 3 4 5 6

Figure 4.10 Slenderness reduction factor.


The reference buckling moment (Mo), which is the theoretical elastic lateral–torsionalbuckling strength of the beam under uniform bending moment (Timoshenko and Gere1961),isgiveninClause5.6.1.1ofAS4100(1998)asfollows:

ME IL

GJE IL

os y

e

s w

e

= +

π π2

2

2

2 (4.22)

whereEsisYoung’smodulusGistheshearmodulusofsteel(80,000MPa)JisthetorsionalconstantIwisthewarpingconstantLeistheeffectivelengthofthesegment

Figure4.11presentsthereferenceelasticbucklingmomentswithvariousslendernessratiosofLe/ry.Itcanbeseenthatincreasingthememberslendernessratiosignificantlyreducestheelasticbucklingmoment(Mo).Inotherwords,theelasticbucklingmomentcanbeincreasedbydecreasingLeandincreasingIyandIw.

Theeffectivelength(Le)ofasegmentdependsonitstwistrestraint,loadheightpositionandlateralrotationalrestraint(BradfordandTrahair1983).Clause5.6.3ofAS4100(1998)suggeststhattheeffectivelengthofasegmentshouldbedeterminedby

L k kk le t l r= (4.23)

wherelistheactuallengthofthesegmentktisthetwistrestraintfactorthataccountsfortheeffectofpartialtorsionalrestraint

00

0.2

0.4Buck

ling

mom

ent M

o/M

y

0.6

0.8

1

1.2

1.4

50 100 150 200 250 300

Slenderness ratio Le/ry

Figure 4.11 Elastic buckling moments of simply supported I-beams.


For segment endswith restraint conditionsofFP,PLorPU, the twist restraint factor isdeterminedby(BradfordandTrahair1983)

k

d l t tn

tf w

w

= +121

3( )( )/ / (4.24)

For segment ends with restraint conditions of PP, kt is calculated by (Bradford andTrahair1983)

k

dl

tt

nt

f

w

w

= +

12

21

3

(4.25)

wherenwisthenumberofwebinthesegmentsection.Forotherrestraintconditionsnotmentionedearlier,ktistakenas1.0.

Theloadheightfactorklisusedtoconsiderthedestabilizingeffectofgravityloadsatthetopflangeincomparisonwiththeloadingattheshearcentre.Theloadheightfactorklistakenas1.4forgravityloadswithinthesegmentandonthetopflangeofsegmentand2.0forgravityloadsonthetopflangeofcantilever.Forotherrestraintconditionsatsegmentendsandloadingatsegmentendsandforshearcentreloads,klistakenas1.0.

Thelateralrotationrestraintfactorkr is takenas0.85forsegmentendswithrestraintconditionsofFF,FPorPPandwithlateralrotationrestraintatoneendand0.7forsegmentswithlateralrotationrestraintsatbothends(TrahairandBradford1998).Forothercases,kristakenas1.0.

4.5.3.2 I-sections with unequal flanges

AsspecifiedinClause5.6.1.2ofAS4100(1998),thenominalmembermomentcapacitiesofsteelI-sectionswithunequalflangessymmetricalabouttheminoraxiscanalsobecalcu-latedusingEquation4.19andthereferencebucklingmoment(Mo)determinedeitherbyanelasticbucklinganalysisorbythefollowingequation:

ME IL

GJE IL

E IL

E IL

os y

e

s w

e

x s y

e

x s y

e

= + + +

π π β π β π2

2

2

2

2 2

2

2

24 2 (4.26)

whereβxisthemonosymmetricsectionconstant,whichcanbedeterminedby(KitipornchaiandTrahair1980)

βx fc

cy

y

dII

= −

0 8

21. (4.27)

Thenominalmembermomentcapacity(Mb)ofananglesectionmemberorarectangularhollowsection (RHS)membercanbedeterminedusingEquation4.19with thewarpingconstantofIw=0.


4.5.4 design requirements for members under bending

ForasteelmembersubjectedtoabendingmomentMx∗aboutitssectionmajorprincipalx-axis

whichisdeterminedbytheelasticmethodofstructuralanalysis,Clause5.1ofAS4100(1998)requiresthatboththesectionandmembermomentcapacitiesshallbecheckedasfollows:

M Mx sx∗ ≤ φ (4.28)

M Mx bx∗ ≤ φ (4.29)

inwhichMbxisthemembermomentcapacitybendingaboutthemajorprincipalx-axis.ForasteelmembersubjectedtoabendingmomentMy

∗aboutitssectionminorprincipaly-axiswhichisdeterminedbyelasticmethodofstructuralanalysis,thememberwillnotundergolateral–torsionalbucklingsothatonlyitsin-planesectionmomentcapacityneedstobecheckedasfollows:

M My sy∗ ≤ φ (4.30)

Example 4.2: Design of steel beam without intermediate lateral restraints

AsimplysupportedsteelI-beamisdepictedinFigure4.12.Thebeamissubjecttoauni-formlydistributeddeadloadof4.4kN/mandaliveloadof5.3kN/mandaconcentrateddeadloadof32kNandaconcentratedliveloadof37kN.Allloadsareappliedtothetopflangeofthebeam.Thebeamispartiallyrestrainedattheendswherethelateraldeflec-tionsareeffectivelypreventedandtwistrotationsarepartiallysuppressed.Therearenointermediate lateral restraintsbetween the supports.Check theadequacyof thebeamwitha610UB113sectionofGrade300steel.

Bending moment diagram (kN m)

7 m

3.5 m

PG= 32 kNPQ= 37 kN wQ= 5.3 kN/m

wG= 4.4 kN/m

3.5 m

206.9 206.9253.6

80.1

0.0 0.0

80.1

149.1 149.1

Figure 4.12 Steel beam without intermediate lateral restraints.


1. Design actions

Theuniformlydistributeddeadloadis4.4kN/m.Theself-weightofthesteelbeamis113×9.81×10−3=1.11kN/m.Theuniformlydistributedliveloadis5.3kN/m.Theuniformlydistributeddesignloadw*=1.2G+1.5Q=1.2×(4.4+1.11)+1.5×5.3=14.56kN/m.TheconcentrateddesignloadP*=1.2G+1.5Q=1.2×32+1.5×37=93.9kN.ThedesignbendingmomentdiagramofthebeamisshowninFigure4.12.

2. Section moment capacity

Thesectionpropertiesof610UB113ofGrade300steelare

d t t

I I

f w

y w

1

6 9

572 17 3 11 2

34 3 10 10

= = =

= × = ×

mm, mm, mm

mm 2980 mm4 6

. .

. , , JJ

Z G E fex s y

= ×

= × = × =

1140 mm

3290 mm MPa, 200,000MPa,

4

3

10

10 80 10

3

3 3, == 280MPa

Thenominalsectionmomentcapacitycanbecalculatedas

M Z fsx ex y= = × × =3290 10 280 921 23 Nmm kNm.

Thedesignsectionmomentcapacityis

φM Msx x= × = > ∗ =0 9 892 253 6. .921.2 kN m kN m

3. Moment modification factor

AsshowninFigure4.13,thedesignbendingmomentsare

M M M Mm∗ = ∗ = ∗ = ∗ =253 6 149 1 253 6 149 12 3 4. . . .kNm, kNm, kNm, kNm

Themomentmodificationfactoriscalculatedas

αmmM

M M M

=∗

∗( ) + ∗( ) + ∗( )=

×

( ) + ( ) +

1 7 1 7 253 6

149 1 253 6 12

2

3

2

4

2 2 2

. . .

. . 449 11 307 2 5

2.

. .( )

= ≤

4. Slenderness reduction factor

Thebeamispartiallyrestrainedatbothsupports(PP)sothatthetwistrestraintfactorcanbecalculatedas

k

d l t tn

tf w

w

= + = +× ×

=12 2

12 572 7000 17 3 2 11 2

11 071

3 3( )( ) ( )( . . ).

/ / / /55


TheloadsareappliedtothetopflangewithinthebeamwithPPrestraintsatthesupports.Theloadheightfactoristakenaskl=1.4.

Since none of the ends of the beam are restrained rotationally, the lateral rotationrestraintfactoriskr=1.0.

Theeffectivelengthofthebeamisdeterminedas

L k k k le t l r= = × × × =1 075 1 4 1 0. . . 7,000 10,535mm

Thereferencebucklingmomentiscalculatedasfollows:

ME IL

GJE IL

os y

e

s w

e

= +

=× × × ×

π π

π

2

2

2

2

2 3 6

2

200 10 34 3 108

.10,535

00 10 10200 10 10

296

3 32 3 9

× × × +× × × ×

=

1,1402,980

10,535Nmm2

π

.66kNm

Theslendernessreductionfactorisdeterminedby

αss

oa

s

oa

MM

MM

=

+ −

=

0 6 3 0 6921 2296 6

2

. ... + −

= <

2

3921 2296 6

0 27 1 0..

. .

5. Member moment capacity

Thenominalmembermomentcapacityofthebeamis

M Mbx m s sx= = × × =α α 1 307 0 27 921 2 325. . . kNm

Bending moment diagram (kN m)

7 m

3.5 m

PG= 32 kNPQ= 37 kN wQ= 5.3 kN/m

wG= 4.4 kN/m

3.5 m

206.9 206.9253.6

80.1

0.0 0.0

80.1

149.1 149.1

Figure 4.13 Steel beam with an intermediate lateral restraint.


Thedesignmembermomentcapacityofthebeamis

φM Mbx x= × = > ∗ =0 9 325 253 6. . 292.5kN m kN m, OK

Example 4.3: Design of steel beam with an intermediate lateral restraint

RedesignthesteelbeampresentedinExample4.2byincorporatingonelateralrestraintatthemid-spanofthebeamasdepictedinFigure4.13.

1. Design actions

ThedesignactionshavebeencalculatedinExample4.2andthebendingmomentdia-gramisshowninFigure4.13.

2. Section moment capacity

Trysection460UB67.1ofGrade300steel.Thepropertiesofthe460UB67.1are

d t t

I I J

f w

y w

1

6 9

428 12 7 8 5

14 5 10 708 10

= = =

= × = × =

mm, mm, mm

mm mm4 6

. .

. , , 3378 10

10 80 10 3

3

3 3

×

= × = × = =

mm

1,480 mm MPa, 200,000MPa,

4

3Z G E fex s y, 000MPa

Thenominalsectionmomentcapacitycanbecalculatedas

M Z fsx ex y= = × × =1480 10 300 4443 Nmm kNm

Thedesignsectionmomentcapacityis

φM Msx x= × = > ∗ =0 9 399 6 253 6. . .444 kN m kN m

3. Moment modification factor

AsshowninFigure4.13,thedesignbendingmomentsactingonthesegmentbetweenthemid-spanandthesupportare

M M M Mm∗ = ∗ = ∗ = ∗ =253 6 80 1 149 1 206 92 3 4. . . .kNm, kNm, kNm, kNm

Themomentmodificationfactoriscalculatedas

αmmM

M M M

=∗

∗( ) + ∗( ) + ∗( )=

×

+ +

1 7 1 7 253 6

80 1 149 1 22

2

3

2

4

2 2 2

. . .

( . ) ( . ) ( 006 91 613 2 5

2. ). .= ≤


Thesegmentbetweenthesupportandmid-spanispartiallyrestrainedatthesupport(P)andlaterallyrestrainedatthemid-span(L)sothatthetwistrestraintfactoriscalculatedas

k

d l t tn

tf w

w

= + = +×

=12

1428 3500 12 7 2 8 5

11 0511

3 3( ) ( ) ( )( . . ).

/ / / /


TheloadsareappliedtothetopflangewithinthesegmentwithPLrestraintsattheends.Theloadheightfactoristakenaskl=1.4.

Sincenoneoftheendsofthesegmentarerestrainedrotationally,thelateralrotationrestraintfactoriskr=1.0.

Theeffectivelengthofthesegmentisdeterminedas

L k k k le t l r= = × × × =1 051 1 4 1 0 3500 5150. . . mm

Thereferencebucklingmomentiscalculatedas

ME IL

GJE IL

os y

e

s w

e

= +

=× × × ×

×

π π

π

2

2

2

2

2 3 6

2

200 10 14 5 105150

80.

110 378 10200 10 708 10

5150

299

3 32 3 9

2× × +× × × ×

=

πNmm

kNm

Theslendernessreductionfactoriscomputedasfollows:

αss

oa

s

oa

MM

MM

=

+ −

=

+0 6 3 0 6

444299

2 2

. . 33444299

0 478 1 0−

= <. .

5. Member moment capacity

Thenominalmembermomentcapacityofthesegmentis

M Mbx m s sx= = × × =α α 1 613 0 478 444 342. . kNm

Thedesignmembermomentcapacityofthesegmentorthebeamis

φM Mbx x= × = > ∗ =0 9 342 253 6. . 308kN m kN m, OK

4.6 SheAr cAPAcIty of WeBS

4.6.1 yield capacity of webs in shear

Thewebofasteelbeamunderbendingissubjectedtoshear.Thecapacityofasteelwebinsheardependsonitsdepth-to-thicknessratioandthespacingoftransversewebstiffen-ers(Bradford1987;TrahairandBradford1998).Clause5.10ofAS4100(1998)providesrequirementsontheminimumthicknessofbeamwebsincludinganytransverseorlongitu-dinalstiffeners,whicharegiveninTable4.2.Forwebswiththestiffenerspacingtodepthratios/dpgreaterthan3.0,thewebsshouldbeconsideredtobeunstiffened.

Thedesignrequirementforasteelbeamwebunderadesignshearforce(V*)is

V Vv∗ ≤ φ (4.31)

whereVvisthenominalshearcapacityoftheweb.ThesheardistributioninwebsofmostI-sectionmembersisapproximatelyuniform.For

awebwithaapproximatelyuniformshearstressdistribution,Clause5.11.2ofAS4100


allowsVvbetakenasthenominalshearcapacityoftheweb(Vu)withauniformshearstressdistribution,whichisgivenby

VV

dt

f

Vdt

fu

wp

w

y

bp

w

y

=≤

>

for

for

25082

25082

(4.32)

wheredpisthecleardepthofthewebpaneltwisthethicknessofthewebVbistheshearbucklingcapacityofthewebVwisthenominalshearyieldcapacity,whichisdeterminedby

V f Aw y w= 0 6. (4.33)

Table 4.2 Minimum web thickness

Arrangement of webs Required thickness tw

Unstiffened web bounded by two flanges td f

wy≥ 1

180 250

Unstiffened web bounded by one free edge td f

wy≥ 1

90 250

Transversely stiffened webs

sd1

0 74≤ . td f

wy≥ 1

270 250

0 74 1 01

. .< ≤sd

ts f

wy≥

200 250

1 0 3 01

. .≤ ≤sd

td f

wy≥ 1

200 250

Webs with one longitudinal and transverse stiffeners

sd1

0 74< . td f

wy≥ 1

340 250

0 74 1 01

. .≤ ≤sd

ts f

wy≥

250 250

1 0 2 41

. .≤ ≤sd

td f

wy≥ 1

250 250

Webs with two longitudinal stiffeners and sd1

1 5< . td f

wy≥ 1

400 250

Webs containing plastic hinges td f

wy≥ 1

82 250



whereAwisthecross-sectionalareaoftheweb,whichistakenasAw=Dstwforhot-rolledsteel I-sectionswhereDs is the total depthof thehot-rolled I-section andAw =d1tw forweldedorbuilt-upsteelI-sections.

Itisnotedthatforastockywebwith( )d t fp w y/ /250 82≤ ,thewebyieldsbeforeelasticlocalbucklingsothatVuistakenastheshearyieldcapacityVw.Incontrast,foraslenderwebwith( )d t fp w y/ /250 82> ,thewebbuckleselasticallybeforeyieldingsothatVuistakenastheshearbucklingcapacityVb.

Theshearstressdistribution in thewebofasteelbeamwithunequalflanges,varyingwebthicknessorholesnotusedforfastenersisnon-uniform.InClause5.11.3ofAS4100,the shear capacity of the web with non-uniform shear stress distribution is determinedfrom the shearcapacityofthewebwithuniformshearstressdistribution(Vu)byconsider-ingtheeffectoftheshearstressratiointhewebas

VV

f fVv

u

vm va

u=+ ∗ ∗( ) ≤2

0 9. (4.34)

inwhichfvm∗ andfva

∗arethemaximumandaveragedesignshearstressesintheweb,respec-tively,andaredeterminedbyanelasticanalysis.

4.6.2 Shear buckling capacity of webs

AsspecifiedinClause5.11.5.1ofAS4100,thenominalshearbucklingcapacity(Vb)ofaslenderunstiffenedwebisbasedonitselasticlocalbucklingstressandiscalculatedby

Vd t f

V Vbp w y

w w=

≤82250

2

( )/ / (4.35)

If the design shear buckling capacity (ϕVb) of a slender unstiffened web is less than thedesign shear force (V*), intermediate transverse stiffeners may be welded to the web toincreasetheshearbucklingcapacityoftheweb.Thenominalshearbucklingcapacityofaslenderstiffenedwebwithaspacing-to-depthratioofs/dp≤3.0isgiveninClause5.11.5.2ofAS4100(1998)asfollows:

V V Vb v d f w w= ≤α α α (4.36)

whereαvisthestiffeningfactorwhichaccountsfortheeffectsoftheincreasedelasticbuck-lingresistanceduetotransversestiffenersandisgivenby

αvp w y pd t f s d

=

+

≤

82250

0 751 0 1 0

2

2( ).

( ). .

/ / /when 1..0 /≤ ≤s dp 3 0. (4.37)

αvp w y pd t f s d

s d=

+

≤

82250

10 75 1 0

2

2( ) ( ). .

/ / /when / pp ≤ 1 0. (4.38)

where s is the stiffener spacing.Theeffectof transverse stiffenerson the shearbucklingcapacityisincorporatedinfactorαv,anditdependsonthestiffenerspacingtodepthratioofthewebpanel.


InEquation4.36,αdisthetensionfieldcontributionfactorwhichconsidersthecontribu-tionofthetensionfieldtotheshearbucklingcapacityandisexpressedby

α α

αd

v

v ps d= + −

+1

1

1 15 1 2. ( )/ (4.39)

FactorαfinEquation4.36istheflangerestraintfactorreflectingtheincreaseintheshearbucklingcapacityofthewebduetotherestrainingeffectsprovidedbytheflangesandisgivenby

αf

fo f wb t d t= −

+ ( )( )1 6

0 6

1 40 212

..

(4.40)

wherebfoandtfarethewidthandthicknessoftheflangeoutstand,respectively.Forwebswithoutlongitudinalstiffeners,bfoistakenastheleastof( )12 250t ff y / ,thedistancefromthemid-planeofthewebtotheneareredgeoftheflangeorhalfthecleardistancebetweenthewebs.

4.6.3 Webs in combined shear and bending

WhenthebeamissubjectedtobothhighdesignmomentM*andshearforceV*,itmustbedesignedforcombinedbendingandshear.Clause5.12ofAS4100permitstwomethodsforthedesignofbeamwebsundercombinedbendingandshear:theproportioningandinterac-tionmethods.

Intheproportioningmethod,thebendingmomentisassumedtocarryonlybytheflangesandthewebresiststhewholeshearforce.Thedesignbendingmomentandshearforcemustsatisfy

M A d ffm fc y∗ ≤ φ (4.41)

V Vv∗ ≤ φ (4.42)

whereAfmisthelesseroftheflangeeffectiveareasforthecompressionflangeandthelesserof

thegrossareaoftheflangeand0 85. A f ffn u y/ forthetensionflange,inwhichAfnisthenetareaoftheflange

Vvisthenominalshearcapacityoftheweb

Theproportioningmethodisusedtodesignbeamswithslenderwebs.Theflangesofthesebeamsshouldbeatleastnon-compacttoachievebetterdesigns.

Inthe interactionmethod,thebendingmoment isassumedtobecarriedbythewholecrosssection.Thismethodisusedtodesignbeamswithlessslenderwebsandappliestobothstiffenedandunstiffenedwebs.Thebendingandshear interactiondiagramissche-maticallydepictedinFigure4.14,whichisexpressedbyequationsgiveninClause5.12.3ofAS4100(1998)asfollows:

V V M Mv s∗ ≤ ∗ ≤φ φ for 0 75. (4.43)

V V. MM

M M Mvs

s s∗ ≤ −

∗

≤ ∗ ≤φφ

φ φ2.2 for 1 6

0 75. (4.44)


4.6.4 transverse web stiffeners

Intermediatetransversewebstiffenerscanbeusedtopreventlocalbucklingofthewebinshear.Thesewebstiffenersmusthavenotonlyadequatestiffnesstoensurethattheelasticbucklingstressofapanelcanbeattainedbutalsoadequatestrengthtocarrythetensionfieldstiffenerforce.TransversewebstiffenersareusuallynotconnectedtothetensionflangeandcanbeattachedtoeitheronesideorbothsidesofthewebasillustratedinFigure4.15.Thespacingofthetransversewebstiffenersshouldbelessthan3d1inordertoeffectivelyresisttheshearforce.Forthestrengthdesign,Clause5.15.3ofAS4100(1998)givesthefol-lowingminimumareaofanintermediatewebstiffener:

AVV

sd

s d

s dAs w v

u p

p

p

w≥ −∗

−+

0 5 11

2

2. ( )

( )

( )γ α

φ/

/ (4.45)

End plate A

A

Longitudinal sti�ener

Intermediate transverse sti�ener

Elevation Section A-A

4tw

tw

0.2d2

Load–bearing sti�ener

Figure 4.15 Web stiffeners of a steel plate girder.

00

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4 0.6 0.8 1 1.2

M*/φMs

V*/φV

v

Figure 4.14 Strength interaction diagram for sections in bending and shear.


whereγwis1.0forapairofstiffeners,1.8forasingleanglestiffenerand2.4forasingleplatestiffener.Thisminimumareaistoensurethestiffenerhastheyieldcapacitythatissufficienttotransmittheforcecausedbythetensionfield.

Thedesignshearforce(V*)actingontheslenderstiffenedwebofasteelplategirderasillustratedinFigure4.15isresistedbythewebshearbucklingandthestiffener–webbuck-ling.Asaresult,anintermediatewebstiffenermustsatisfythebucklingstrengthrequire-mentgiveninClause5.15.4ofAS4100(1998)asfollows:

V R Vsb b∗ ≤ +φ( ) (4.46)

whereRsbisthebucklingcapacityofthestiffener–webcompressionmemberasawholeVbistheshearbucklingcapacityofthestiffenedwebgiveninEquation4.36

Thebucklingcapacityofthewebandtheintermediatewebstiffenerasawhole(Rsb)isdeterminedastheaxialloadcapacityofthestiffener–webcompressionmemberinaccor-dancewithClause6.3.3ofAS4100.Theeffectivecross-sectionalareaofthestiffener–webstrutistakenastheareaofthestiffenerplusthewebareahavinganeffectivewidthoneachsideofthecentrelineofthestiffenerconsideredasschematicallydepictedinFigure4.16.Theeffectivewidthoftheweb(bew)aspartofthestiffener–webcompressionmemberistakenas

bt

fs

eww

y

=

min.17 5250 2/

, (4.47)

wheretwisthethicknessofthewebsisthewebpanelwidthorspacingofthestiffeners

Transverse web stiffener

s/2 s/2

bew= min , S2

17.5twfy/250√

17.5twfy/250√

17.5twfy/250√

Figure 4.16 Effective width of the web as part of stiffener–web compression member.


Theeffectivelengthofthestiffener–webstrutistakenasd1.Thesecondmomentofareaofthestiffener–websectioniscalculatedabouttheaxisparalleltotheweb.Theslendernessreductionfactorαcisdeterminedbytakingαb=0.5andkf=1.0inaccordancewithClause6.3.3ofAS4100.

Thedesignbucklingcapacityoftheweb–stiffenercompressionmemberisexpressedby

φ φαR A fsb c ws y= (4.48)

whereAwsisthecross-sectionalareaofthestiffener–webcompressionmember.For an intermediate web stiffener that is not subjected to external loads ormoments,

theminimumsecondmomentofarea(Is)aboutthecentrelineofthewebisgiveninClause5.15.5ofAS4100as

Id t s d

d t s s ds

w

w

≥≤

>

0 75 2

1 5 2

13

1

13 3 2

1

.

.

for /

/ for / (4.49)

Attheendofaplategirder,theendstiffenermustresistthehorizontalcomponentofthetensionfieldintheendpanel.Toavoidthis,thelengthoftheendpanelcanbereducedsothatthecontributionofthetensionfieldtotheultimatestressisnotrequired.Thiscanbeachievedbydesigningtheendpanelwithαd=1.0inEquation4.36.Alternatively,anendpostconsistingofaload-bearingstiffenerandaparallelendplatecanbeusedtotransferthetensionfieldactionontheendofaplategirderasillustratedinFigure4.15.Clause5.15.9ofAS4100requiresthattheareaoftheendplatemustsatisfythefollowingcondition:

A

d V Vs f

epv w

ep y

≥∗ −1

8( )/φ α

(4.50)

wheresepisthedistancebetweentheendplateandload-bearingstiffener.

4.6.5 longitudinal web stiffeners

Longitudinalwebstiffenersarecontinuousandattached to the transversewebstiffenersto increase theeffectivenessof theweb in resisting shearandbending.When thedepth-to-thicknessratiooftheweb( )d t fw y1 250/ / isgreaterthan200,afirstlongitudinalstiff-ener isneededatadistance0.2d2 fromthecompressionflangeasschematicallydepictedinFigure 4.15.Thesecondmomentofarea(Is)ofthisstiffeneraboutthefaceoftheweb(TrahairandBradford1998)mustsatisfyClause5.16.2ofAS4100as

I d t

Ad t

Ad t

s ws

w

s

w

≥ + +

4 1

412

3

2 2

(4.51)

whered2istwicethecleardistancebetweentheneutralaxisandthecompressionflangeAsisthestiffenerarea(Bradford1987,1989)


Whenthedepth-to-thicknessratiooftheweb( )d t fw y1 250/ / isgreaterthan250,asecondlongitudinalstiffenerisrequiredattheneutralaxisofthesection,anditssecondmomentofareaaboutthefaceofthewebmustsatisfy

I d ts w≥ 23 (4.52)

Example 4.4: Design of a stiffened plate girder web for shear

ThecrosssectionofaplategirderofGrade300steelisshowninFigure4.17.Intermediatetransversewebstiffeners100×14 mmofGrade300steelarespacedat1500 mm.Thewidthoftheendpanelis1200 mm.Thereisnoendpost.Theflangesoftheplategirderarerestrainedbyotherstructuralmembersagainstrotation.Apairofload-bearingstiff-eners100×14 mmisusedabovethesupportoftheplategirder,whichissupportedbystiffbearing200 mmlong.Thedesignreactionis1200kN.Theyieldstressesofthewebandstiffenersarefy=310MPaandfys=300MPa,respectively:

a.Determinethedesignshearcapacityoftheweb. b.Determinethedesignshearcapacityoftheendpanel. c.Checktheadequacyoftheintermediatetransversewebstiffeners.

a. Shear capacity of the stiffened web

1. Slenderness of the web


b t d tf f w= = = =350 20 1200 10mm, mm, mm, mm

Theslendernessofthewebis

λew

p

w

y f

w

ydt

f d tt

f= =

−=

− ×= >

2502

2501200 2 20

10310250

129 2 82( ) ( )

.

1200

350

20

10

20

Figure 4.17 Section of plate girder.


Theweboftheplategirdersubjectedtouniformshearstresseswillundergoshearbuck-ling.Thus,Vu=Vb.

Since λew = <129 2 200. ,thelongitudinalwebstiffenerisnotrequired.

2. Partial factors

Thespacing-to-depthratiois

s dp/ /= − × = <1500 1200 2 20 1 29 3 0( ) . . ;thewebistreatedasastiffenedone.

Thestiffeningfactorαviscalculatedas

αvp w yd t f s d

=

+

=

82

2500 75

1 082

129 2

2

2( ).

( ).

./ / /

+

= <

2

2

0 751 29

1 0 0 584 1 0..

. . .

Thetensionfieldcontributionfactorαdisdeterminedas

αα

αd

v

v ps d= +

−

+= +

−

× × +=1

1

1 15 11

1 0 584

1 15 0 584 1 1 291 38

2 2. ( )

.

. . ..

/

Thewidthoftheflangeoutstandisbfo=(bf−tw)/2=(350−10)/2=170.Theflangerestraintfactorαfcanbecomputedas

αf

fo f wb t d t= −

+ ( )( )= −

+ × × ×1 6

0 6

1 401 6

0 6

1 40 170 20 1160 1212 2 2

..

..

( / 001 053

).

( )=

3. Shear capacity of the web

Theshearyieldcapacityofthewebis

V f Aw y w= = × × − ×( )× =0 6 0 6 310 1200 2 20 10 2157 6. . .N kN

Thenominalshearbucklingcapacityofthestiffenedwebiscomputedas

V V Vb v d f w w= = × × × = < =α α α 0 584 1 38 1 053 2157 6 1831 2157 6. . . . .kN kN kN

Thedesignshearcapacityofthewebisdeterminedas

φVu = × =0 9 1831 1648. kN

b. Shear capacity of the end panel

1. Slenderness of the end panel

Theslendernessoftheendpanelisthesameasthatoftheweb,λew=129.2>82.Theendpanelwillundergoshearbucklingbeforeyielding;takeVu=Vb.

2. Partial factors

Thewidthoftheendpanelsis1200 mm.Thespacing-to-depthratiooftheendpanelis

sdp

=− ×

= <1200

1200 2 201 034 3 0. .


Thestiffeningfactorαvfortheendpaneliscalculatedas

αvp w syd t f s d

=

( )

+

=

82250

0 751 0

82129 2

2

2( ).

../ / /

+

= <

2

2

0 751 034

1 0 0 685 1 0.

.. . .

Thewidthoftheendpanelhasbeenreducedfrom1500 mmstiffenerspacingto1200 mmandtheendpostisnotrequiredbytakingαd=1.0.Theflangerestraintfactorhasbeencalculatedasαf=1.053.

3. Shear capacity of the web

Thenominalshearbucklingcapacityoftheendpaneliscalculatedas

V V Vb v d f w w= = × × × = < =α α α 0 685 1 0 1 053 2157 6 1556 3 2157 6. . . . . .kN kN kN

Thedesignshearcapacityoftheendpanelcanbedeterminedby

φVu = × =0 9 1556 3 1401. . kN

c. Intermediate transverse stiffeners

1. Minimum area of transverse web stiffener

Thedimensionsandpropertiesofthetransversestiffenersare

b t fs s ys w= = = =100 14 300 1 0mm, mm, MPa, forapairof stiffenersγ . ( )

A s ds p= × × = = <2 100 14 2800 1 29 3 0mm /2, . .

TakingV*=ϕVu,theminimumareaoftheintermediatewebstiffeneriscalculatedas

AVV

sd

s d

s dAs w v

u p

p

p

w≥ −

−

+

≥

∗

0 5 11

0

2

2. ( )

( )

( )

.

γ αφ

/

/

55 1 0 1 0 584 1 0 1 291 29

1 1 291160 10

652

2

2× × −( )× × −

+

× ×

≥

. . . ..

.

mmm mm OK2 2< =As 2800 ,

2. Section properties of the stiffener–web compression member

Theeffectivelengthofthestiffenedweboneachsideofthestiffeneris

bt

f

sew

w

y

= =×

= < = =17 5

250 310 250157 2

21500

2750

..

/

17.5 10/

mm mm

Takebew=157.2mm.Thesectionpropertiesoftheweb–stiffenercompressionmemberare

Aws = × × + × × =2 157 2 10 5943. 2 100 14 mm2


Ib t t b t

wsew w s s w=

+

+=

×

+

× ×2

12212

2154 7 10

12

14 23 3 3( ) . 1100 10

1210 83 10

36+( )

= ×. mm6

rIA

wsws

ws

= =×

=10 83 10

594342 7

6..


Theeffectivelengthofthestiffener–webcompressionmemberis

L de = = − × =1 1200 2 20 1160mm

Themodifiedslendernessoftheweb–stiffenercanbecalculatedas

λn

e

wsf

yLr

kf

= = =250

116042 7

1 0300250

29 8.

. .

FromTable5.3ofChapter5withαb=0.5andkf=1.0,theslendernessreductionfactorcanbeobtainedasαc=0.918.

4. Buckling capacity of the stiffener–web compression member

Thenominalbucklingcapacityoftheweb–stiffenerstrutiscalculatedas

R A fsb c ws y= = × × =α 0 918 5943 300 1637. N kN

Thedesignbuckling capacityof the stiffener–web compressionmember canbedeter-minedas

φ φ( ) . ( ) .R V Vsb b u+ = × + = > =0 9 1637 1831 3121 2 1648kN kN, OK

For s d/ 1 1 29 2= <. , the requiredminimumsecondmomentofareaof the transversewebstiffeneris

I d t Is w ws≥ = × × = × < = ×0 75 0 75 1160 10 0 87 10 10 83 1013 3 6 6. . . . ,mm mm O6 6 KK

4.7 BeArIng cAPAcIty of WeBS

4.7.1 yield capacity of webs in bearing

ConcentratedloadsorlocallydistributedloadsonthetopflangeofasteelbeamandreactionsonthesupportsinducebearingstressesinthewebasschematicallydepictedinFigure 4.18.Thesebearingstressesmaycauseyieldingorbucklingof theweb.Therefore, theweb inbearingmustbedesignedforyieldingandbucklinglimitstates.Thewebofasteelbeaminbearingmustsatisfythefollowingstrengthrequirement:

R Rb∗ ≤ φ (4.53)

whereR∗isthedesignbearingforceonthewebφ = 0 9. isthecapacityreductionfactorRbisthenominalbearingcapacityoftheweb,whichistakenasthelesserofitsnominal

bearingyieldcapacity( )Rby andbearingbucklingcapacity( )Rbb


Stockywebscanattaintheirbearingyieldcapacitiesastheelasticbucklingofthewebswillnotoccur.Clause5.13.3ofAS4100(1998)suggeststhatthenominalbearingyieldcapacityofawebshouldbecalculatedby

R b t fby bf w y= 1 25. (4.54)

inwhichbbfisthebearingwidthoftheflangeoftheI-sectionbeamasdepictedinFigure4.18andisdeterminedasthelesserofthefollowingcalculatedvalues:

b b tbf s f= + 5 (4.55)

b b t bbf s f d= + +2 5. (4.56)

wherebdistheremainingdistancetotheendofthebeamasshowninFigure4.18.InClause5.13.3ofAS4100,thenominalbearingyieldcapacityofbothwebsinsquare

andRHSsisdeterminedby

R b t fby b p y= 2 α (4.57)

wherebbisthebearingwidthandistakenasb b r db s e= + +5 5,wherereistheoutsideradiusof

thesectionandd5istheflatwidthofthewebtisthethicknessofthehollowsectionαpisareductionfactorwhichisdifferentforinteriorbearingandendbearing(Zhao

et al.1996)andisgivenasfollows:

Forinteriorbearingwithbd≥1.5d5,αpisgivenby

α α αp

spm

s

vpm

vkkk k

= + −( ) + − −( )

12

1 1 1 114

2 22 (4.58)

End support bearing forcebd

2.511

2

2.5 2.51

11

N.A.

Interior bearing force

11

1

11

11

bs

d2

tf

bo bbf

bbf

bb

bb

bbw

bbw bbw

bs

2.5

Figure 4.18 Bearing force dispersions in the flanges and web of a steel I-beam.


inwhichd5istheflatwidthofwebandistakenasd5=d−2re,andthecoefficientαpmandratiosksandkvaregivenby

αpm

s vk k= +1 1

2 (4.59)

k

rt

se= −2

1 (4.60)

k

dt

v = 5 (4.61)

Forendbearingwithbd<1.5d5,αpisgivenby

αp s sk k= + −2 2 (4.62)

Thebearingwidthofendbearingiscalculatedas

b b r

db s e= + +2 5

25. (4.63)

4.7.2 Bearing buckling capacity of webs

Thebucklingcapacityofanunstiffenedwebunderbearingstresses isdeterminedas theaxialloadcapacityofanequivalentcompressionmemberwithanareatakenasAw=bbtwandaslendernessratio(Le/r)takenas(2.5d1)/tw.Thetotalbearingwidthoftheweb(bb)isobtainedbydispersionsataslopeof1:1frombbftotheneutralaxisasillustratedinFigure4.18.Forendbearing,thetotalbearingwidthisgivenby

b b b bb o bf bw= + + (4.64)

wherebo=bd − 2.5tf,bbf=bs+5tfandbbw=d2/2.Forinteriorbearing,boinEquation4.64isreplacedbybbwasshowninFigure4.18.

ForsquareandRHSs,however,theslendernessratio(Le/r)istakenas(3.5d5)/twforinte-riorbearingwithbd≥1.5d5andequalsto(3.8d5)/twforendbearingwithbd<1.5d5.

Theslendernessreductionfactorαcisdeterminedbytakingαb=0.5andkf=1.0inaccor-dancewithClause6.3.3ofAS4100,andthebearingbucklingcapacityofthewebcanbecalculatedby

φ φαR A fbb c w y= (4.65)

4.7.3 Webs in combined bearing and bending

Theultimate strengthsof squareandRHSbeamsunder combinedbearingandbendingare influenced by the interaction between bearing and bending. The presence of bend-ingmomentreducesthebearingstrength,whilethepresenceofbearingforcereducesthe


bending strengthof the sections. Interaction equations are given inClause5.13.5ofAS4100(1998)fordeterminingthecapacitiesofsquareandRHSsundercombinedbearingandbending(Zhaoetal.1996):

1 2 1 5 1 0 301. . .RR

MM

bb

dtb s

s

w

∗

+

∗

≤ ≥ ≤

φ φfor and (4.66)

0 8 1 0 1 0 301. . .RR

MM

bb

dtb s

s

w

∗

+

∗

≤ < >

φ φfor and (4.67)

whereφ = 0 9. andbisthetotalwidthofthesectionMsisthenominalsectionmomentcapacity

4.7.4 load-bearing stiffeners

Whenthewebhasinsufficientcapacitytowithstandtheimposedconcentratedloads,itmaybestrengthenedbyweldingbearingstiffenerstothewebadjacenttotheloadsasdepictedinFigure4.15.Thedesignrulesforload-bearingstiffenersareprovidedinClause5.14ofAS4100,whichspecifiesthattheoutstandsofthestiffenerfromthefaceofthewebmustsatisfythefollowingcondition:

bt

fes

s

ys

≤ 15250/

(4.68)

wheretsisthethicknessofthestiffenerfysistheyieldstressofthestiffener

Theload-bearingstiffenerandpartofthewebinthevicinityofthestiffenerconsideredaretreatedasacompressionmember.Theload-bearingstiffener–webcompressionmembermustbechecked for itsyieldandbucklingcapacitiesagainst thedesignbearing forceordesignreaction(R*)actingonthebearingstiffenerasfollows:

R Rsy∗ ≤ φ (4.69)

R Rsb∗ ≤ φ (4.70)

whereϕ=0.9andRsyistheyieldcapacityofthestiffener–webcompressionmember,whichisgiveninClause5.14.1ofAS4100(1998)as

R R A fsy by s ys= + (4.71)

whereRbyisthebearingyieldcapacityofthewebgiveninEquation4.54Asisthecross-sectionalareaofthestiffenerfysistheyieldstressofthestiffener

Thedesignbucklingcapacityofthewebandload-bearingstiffenerasawhole(ϕRsb)isgivenbyEquation4.48.


Example 4.5: Design of a stiffened plate girder web for bearing

ThesteelplategirderpresentedinExample4.4istobedesignedforbearingatthesup-portsas shown inFigure4.19.There isnoendpost.Apairof load-bearingstiffeners100 ×14 mm is tobeusedabove the supportof theplategirder,which is supportedbystiffbearing200 mmlong.Thedistancefromthesupporttotheendofthebeamis100 mmasillustratedinFigure4.19.Thedesignreactionis1200kN.Theyieldstressesofthewebandstiffenersarefy=310MPaandfys=300MPa,respectively:

a.Checkwhethertheload-bearingstiffenersarerequiredatthesupports. b.Checktheadequacyoftheload-bearingstiffenersatthesupports.

a. Bearing capacity of the web without load-bearing stiffeners

1. Bearing yield capacity of the web

Thebearingwidthoftheflangeiscalculatedas

b b tbf s f= + = + × =5 200 5 20 300mm

b b t bbf s f d= + + = + × + =2 5 200 2 5 20 100 350. . mm

Takebf=300mm.Thenominalbearingyieldcapacityisdeterminedby

R b t fby bf w y= = × × × =1 25 1 25 300 10 310 1162 5. . .N kN

Thedesignbearingyieldcapacityis

φR Rby = × = < ∗ =0 9 1162 5 1046 3 1200. . .kN kN kN, NOT OK!

Load-bearingstiffenersarerequiredinthewebatthesupportstotransferthebearingforce.

Elevation Section A-A100

116010

100A

A

Load–bearing stiffener

200100

Figure 4.19 Steel plate girder.


2. Bearing buckling capacity

Thetotalbearingwidthatthesupportcanbedeterminedas

b b b b b t b d t

.

b o bf bw d f bf f= + + = − + + −

= − × + + −

2 5 2 2

100 2 5 20 300 1200

. ( )

(

/

22 20 2 930× =)/ mm

Thecross-sectionalareaofthewebisAw=bbtw=930×20=18,600mm2.Theeffectivesectionofthewebistreatedascolumnsection.Themodifiedslenderness

ratiois

λn

ef

y

wf

yLr

kf d

tk

f= = =

×=

2502 5

2502 5 1160

201 0

310250

161 51. .. .

Takingαb=0.5,theslendernessreductionfactorcanbeobtainedasαc=0.242.Thebearingbucklingcapacityofthewebcompressionmemberis

φ φαR A f Rbb c w y= = × × × = > ∗ =0 9 0 242 310 8. . .18,600 1,255 kN 1,200kN, OK

b. Load-bearing stiffener

1. Bearing yield capacity of the stiffener–web compression member

Thebearinglengthatthejunctionofthewebandflangeis

b b tbf s f= + = + × =5 200 5 20 300mm

Thenominalyieldcapacityofthestiffener–webcompressionmembercanbecomputedas

R R A f b t f A fsy by s ys bf w y s ys= + = +

× × × + ×

1 25

1 25 300 10 310 2800 300

.

.= N == 2002 5. kN

Thedesignyieldcapacityofthestiffener–webmemberis

φR Rsy = × = > ∗ =0 9 2002 5 1802 1200. . kN kN, OK

2. Buckling capacity of the stiffener–web compression member

Thenominalbucklingcapacityofthestiffener–webcompressionmemberhasbeendeter-minedinExample4.4asRsb=1637kN.Thedesignbucklingcapacityofthememberis

φR Rsb = × = > ∗ =0 9 1637 1473 3 1200. . kN kN, OK

4.8 deSIgn for ServIceABIlIty

Thedesignofsteelbeamsforserviceabilityneedstocheckfordeflections,boltslipsorvibra-tions.Inserviceconditions,itisrequiredtocheckforthedeflectionsofthesteelbeamsunderserviceloadsdefinedinSection2.5.3.Underserviceloads,steelbeamsareusuallyassumedtobehaveelastically.Therefore,thefirst-orderlinearelasticanalysiscanbeperformedtodeterminethedeflectionsofsteelbeamsunderserviceloads.Forthispurpose,moderninter-activecomputersoftwaresuchasStrand7,MultiframeandSpaceGasscanbeused.


DeflectionlimitsonsteelbeamsaregiveninAS4100asfollows:

1.Forallbeams,thetotaldeflectionislimitedtoL/250forspansandL/125forcantilever. 2.Forbeamssupportingmasonrypartitions, the incrementaldeflection,whichoccurs

aftertheattachmentofpartitions,islimitedtoL/500forspansandL/250forcan-tileverwhereprovisionisprovidedtoreducetheeffectofmovement;otherwise,theincrementaldeflectionislimitedtoL/1000forspansandL/500forcantilever.

referenceS


Bradford,M.A.(1987)InelasticlocalbucklingoffabricatedI-beams,JournalofConstructionalSteelResearch,7:317–334.

Bradford, M.A. (1989) Buckling of longitudinally stiffened plates in bending and compression,CanadianJournalofCivilEngineering,16(5):607–614.

Bradford, M.A. and Trahair, N.S. (1983) Lateral stability of beam on seats, Journal of StructuralEngineering,ASCE,109(9):2212–2215.

Kitipornchai,S.andTrahair,N.S.(1980)BucklingpropertiesofmonosymmetricI-beams,JournaloftheStructuralDivision,ASCE,106(ST5):941–957.

Kollbrunner,C.F.andBasler,K.(1969)TorsioninStructures,2ndedn.,Berlin,Germany:Springer-Verlag.Timoshenko,S.P.andGere,J.M.(1961)TheoryofElasticStability,2ndedn.,NewYork:McGraw-Hill.Trahair,N.S.(1993a)Flexural-TorsionalBucklingofStructures,London,U.K.:SponPress.Trahair,N.S.(1993b)Designofunbracedcantilevers,SteelConstruction,AustralianInstituteofSteel

Construction,27(3):2–10.Trahair,N.S.andBradford,M.A.(1998)TheBehaviourandDesignofSteelStructurestoAS4100,3rd

edn.(Australian), London,U.K.:Taylor&FrancisGroup.Trahair,N.S.,Hogan,T.J.andSyam,A.(1993)Designofunbracedbeams,SteelConstruction,Australian

InstituteofSteelConstruction,27(1):2–26.Zhao,X.L.,Hancock,G.J. and Sully,R. (1996)Designof tubularmembers and connections using

amendmentnumber3toAS4100,SteelConstruction,AustralianInstituteofSteelConstruction,30(4):2–15.

109

Chapter 5

Steel members under axial load and bending

5.1 IntroductIon

Membersinsteeltrussesunderpointloadsatjointsaresubjectedtoeitheraxialcompres-sionoraxialtension.Incontrast,membersinsteelframesmaybesubjectedtothecombinedaxialloadandbending,whichmaybecausedbylateralloads,eccentricloadingorframeactions.Theaxial loadandbendingmayincludethecombinedactionsofaxial loadanduniaxialbendingandofaxialloadandbiaxialbending.Membersundercompressiveaxialloadandbendingareregardedasbeam–columns,whichcombinethefunctionsofbeamsandcolumns.

This chapter deals with the behaviour, analysis and design of steel members underaxialloadandbendinginaccordancewithAS4100.Thebehaviouranddesignofsteelmembers in axial compression are described first. This is followed by the discussionsofthedesignofmembers inaxialtension.Thebehaviouranddesignofsteelmembersundercombinedactionsofaxialloadanduniaxialbendingarethenpresented,includingmethods forcalculating thesectionmomentcapacityreducedbyaxial forces, in-planemember capacity and out-of-plane member capacity. In Section 5.6, the analysis anddesignofsteelmembersundercombinedactionsofaxial loadandbiaxialbendingaregivenindetail.

5.2 memBerS under AxIAl comPreSSIon

5.2.1 Behaviour of members in axial compression

Thebehaviourofasteelmemberinaxialcompressiondependsonitsmaterialproperties,sectionslendernessandmemberslenderness,initialgeometricimperfectionsandresidualstresses.Thedesignofaverystockymember isgovernedby itssectioncapacity,whichdependsontheyieldstress, slendernessandresidualstressesof thecrosssection.Foracompression member mode of slender steel elements, local buckling may occur beforesteelyields.Localbucklingmaysignificantlyreducetheultimateaxialsectioncapacityofsteelmembersandmustbetakenintoaccountindesign.Residualstressesinducedbyhotrollingorweldingmaycauseasignificantreductionintheaxialsectioncapacityduetoprematureyielding.

Theultimatestrengthofanaxiallyloadedsteelmemberdecreaseswithanincreaseinitslength.Thisiscausedbytheappliedaxialloadwhichinducesbendingactionsandlateraldeflectionsinthememberwithinitialgeometricimperfections.Thelateraldeflectionsandbending actions of the member increase with increasing the member slenderness, which


leadstoadecreaseinthestrengthofthemember.Anelasticsteelmemberwithoutinitialimperfectionswillnotdeflectuntil theappliedaxial compressive force reaches itselasticbucklingload,whichiscalledtheEulerbucklingload.Theelasticbucklingloadgivesanindicationoftheslendernessofanaxiallyloadedmember,whilethesquashloadreflectsonitsresistancetoyieldingandlocalbuckling.

Practicalcompressionmembersusuallyhaveinitialimperfectionswhichincludegeometricimperfections,residualstressesandinitialloadingeccentricity.Theseinitialimperfectionsreduce the strengthsof intermediate and slender compressionmembersbelow the elasticbuckling loads of the members. The load–deflection behaviour of practical compressionmembers isnonlinear inelastic.The strengthsof steelmembers inaxial compressionarefoundtodecreasewithanincreaseintheinitialimperfections.Geometricimperfectionsarealwayspresentinsteelmembersbecauseitisdifficulttomanufactureasteelcolumnwithan initialgeometric imperfection less thanL/1000at itsmid-length.Theeffectof initialgeometricimperfectionofL/1000atthemid-lengthofsteelcolumnshasbeentakenintoaccountinthedesigncodes.

5.2.2 Section capacity in axial compression

Axiallyloadedsteelmemberscomposedofslenderplateelementsmaybucklelocallybeforetheultimateaxialloadisattained.Theeffectoflocalbucklingonthesectioncapacityofcompressionmembersistakenintoaccountbythesectionformfactor(kf),whichwasdis-cussedbyRasmussenetal.(1989).InClause6.2ofAS4100(1998),thenominalsectioncapacityofasteelmembersubjecttoaxialcompressionisexpressedby

N k A fs f n y= (5.1)

inwhichAnisthenetareaofthecrosssectiontakingasA A d tn g h= −∑ ,whereAgisthegrosscross-sectionalarea,dhisthediameterofaholeandtisthethicknessofthememberat thehole.For sectionswithunfilledholesorpenetrations that reduce the sectionareabylessthan100{1−[fy/(0.85fu)]}%,Anistakenasthegrossarea(Ag).Theformfactorkfisexpressedby

k

AA

fe

g

= (5.2)

whereAeistheeffectivecross-sectionalareaofthesectionasgiveninChapter3.

5.2.3 elastic buckling of compression members

Theelasticbucklingloadofaperfectlystraightpin-endedmemberunderaxialcompressionasdepictedinFigure5.1canbedeterminedbyascertainingthedeflectedequilibriumposi-tion,whichisdefinedbythedisplacementfunctionasfollows:

u u

zL

m=

sin

π (5.3)

Steel members under axial load and bending 111

inwhichumisthedeflectionatthemid-lengthofthemember.TheelasticbucklingloadortheEulerbucklingload(TimoshenkoandGere1961;Bulson1970)canbeobtainedas

P

E IL

crs= π2

2 (5.4)

whereIisthesecondmomentofareaofthecolumncrosssectionabouttheprincipalaxisListhememberlength

Theelasticbucklingloadcanbeexpressedbythecolumnslendernessratio(L/r)as

P

E AL r

crs= π2

2( )/ (5.5)

whereAisthecross-sectionalarear I A= / istheradiusofgyration

Theelasticbucklingstresscanbedeterminedas

σ πcr

cr sPA

EL r

= =2

2( )/ (5.6)

ItshouldbenotedthatEquation5.5isvalidonlyforperfectlystraightpin-endedmem-berswithout residual stressesand loadedat thecentreofgravity. It canbe seen fromFigure5.2 thatEquation5.5overestimates the capacityof compression columnswith

y

P

L

L2

P

z

um

Figure 5.1 Pin-ended compression member.


aslendernessratio lessthan200sothat itcannotbeusedtocalculatethecapacityofintermediatelengthcolumns.

Ingeneral,theelasticbucklingloadofcompressionmemberswithendrestraintscanbeexpressedby

P

E Ik L

crs

e

= π2

2( ) (5.7)

wherekeisthemembereffectivelengthfactorL k Le e= is the effective length of a compression member, which is the unsupported

distancebetweenthezeromomentpoints

The member effective length factor (ke) depends on the translational and rotationalrestraints at the ends of the member. For members with idealised end restraints, thevaluesofke are given inFigure5.3 asprovided inAS4100.Forbraced compressionmembersinasteelframewithrigidconnections,theeffectivelengthfactor(ke)canbedeterminedfromthefollowingequation(DuanandChen1988;TrahairandBradford1998):

γ γ π γ γ π π π1 22

1 2

4 21

k k ke e e

+ +

−

+cot

tan( /222

1 0k

ke

e

)( )π/

− = (5.8)

00

0.2

0.1

0.3

0.4

0.5

P u/A

gfy

0.6

0.7

0.8

0.9

1

50 150Member slenderness ratio L/r

Column capacityElastic buckling load

100 200 250

Figure 5.2 Capacity of compression members.


whereγ1andγ2denotethestiffnessratiosofacompressionmemberatend1andend2,respectively.Thestiffnessratioofacompressionmemberinarectangularframewithnegli-gibleaxialforcesinbeamsisgiveninClause4.6.3.4ofAS4100as

γβ

j

c

e b

I L

I L= ∑∑

( )

( )

/

/ (5.9)

where( )I L c/∑ isthesumofthestiffnessintheplaneofbendingofallcompressionmembersrigidlyconnectedattheendofthememberconsidered,includingthememberitself( )I L b/∑ isthesumofthestiffnessintheplaneofbendingofallbeamsrigidlyconnectedattheendofthememberconsidered

Thestiffnessofanybeamspin-connectedtothememberisnotconsidered.Theγjvalueofacompressionmemberthatisnotrigidlyconnectedtoafootingshouldbetakenasgreaterthanorequalto10.Foracompressionmemberthatisrigidlyconnectedtoafooting,theγjvalueshouldbetakenasgreaterthanorequalto0.6.

Themodifyingfactor(βe)isusedtoaccountfortheconditionsatthefarendsofthebeam,whichisgiveninClause4.6.3.4ofAS4100(1998)asfollows:

1.Ifthefarendofthebeamispinned,βe=1.5whenthebeamrestrainsabracedmemberandβe=0.5whenthebeamrestrainsaswaymember.

2.Ifthefarendofthebeamisrigidlyconnectedtoacolumn,βe=1.0whenthebeamrestrainsabracedmemberoraswaymember.

3.Ifthefarendofthebeamisfixed,βe=2.0whenthebeamrestrainsabracedmemberandβe=0.67whenthebeamrestrainsaswaymember.

Fix

Fix

Fix

E�ective length factor (ke)

Fix Fix Fix

ke= 1.2ke= 2.2ke= 0.7ke= 0.85ke= 1.0

Pin

Pin Pin

Figure 5.3 Effective length factors for idealised columns. (Adapted from AS 4100, Australian standard for steel structures, Standards Australia, Sydney, New South Wales, Australia, 1998.)


Fora swaycompressionmember inanunbracedrigid-jointed frame, theeffective lengthfactorcanbedeterminedfromthefollowingequation(DuanandChen1989;TrahairandBradford1998):

γ γ πγ γ

π π1 22

1 2

366

0( )( )

cot/k

k ke

e e

−+

−

= (5.10)

TheeffectivelengthfactorsforbracedmembersandswaymembersinframesaregiveninFigure4.6.3.3inAS4100.Theeffectivelengthisalsousedinthecalculationofthemembercapacityofpracticalcompressionmemberswithimperfections.

Example 5.1: Calculation of effective length factors for columns in frame

Figure 5.4 shows a rigid-jointed plane steel frame whose base is fixed to the founda-tion.Theout-of-planebehaviouroftheframeisprevented.Allbeamsaresubjectedtonegligibleaxialforces.ThesectionsandtheirpropertiesusedintheanalysisaregiveninTable 5.1.Determinetheeffectivelengthfactorsforallcolumns.

1. Column 1-4

Thebaseofcolumn1-4isfixed;thestiffnessratioofthecolumnatend1canbetakenasγ1=0.6accordingtoClause4.6.4.4(a)ofAS4100.

Atcolumnend4:Thefarendofbeam4-5isrigidlyconnectedtoacolumn,andthebeam4-5restrains

aswaycolumn1-4;thus,βe=1.0.Thestiffnessratioofcolumn1-4atend4canbecalculatedas

γβ

461 3 4 0 61 3 3 6

1 0 142 6 51 481= =

+×

=∑∑

( )

( )

. . . .. .

.I L

I L

c

e b

/

/

/ //

6500 7500

21 3

654

87

3600

4000

Figure 5.4 Rigid-jointed plane steel frame.

Table 5.1 Section properties

Member Section Ix(mm4)

1-4, 4-7 200UC59.5 61.3 × 106

2-5, 5-8 250UC72.9 114 × 106

3-6 200UC46.2 45.9 × 106

Beams 360UB50.7 142 × 106


The effective length factor can be obtained by solving Equation 5.10 or from Figure4.6.3.3inAS4100aske=1.321.

2. Column 5-8

Atcolumnend7:Thefarendofbeam7-8isrigidlyconnectedtoacolumn,andthebeam7-8restrainsa

swaycolumn4-7;thus,βe=1.0.Thestiffnessratioofcolumn4-7attheend7canbecalculatedas

γβ

761 3 3 6

1 0 142 6 50 779= =

×=∑

∑( )

( )

. .. .

.I L

I L

c

e b

/

/

//

Theeffectivelengthfactorforcolumn4-7withγ4=1.481andγ7=0.779canbeobtainedbysolvingEquation5.10orfromFigure4.6.3.3inAS4100aske=1.35.

3. Column 2-5

Thebaseofcolumn2-5isfixed;itsstiffnessratioatend2isλ1=0.6accordingtoClause4.6.4.4(a)ofAS4100.

Atcolumnend5:Thefarendsofbeam4-5andbeam5-6arerigidlyconnectedtocolumns,andthese

twobeamsrestrainaswaycolumn2-5.Thisgivesβe=1.0.Thestiffnessratioofcolumn2-5atend5iscomputedas

γβ

5114 4 0 114 3 6

1 0 142 6 5 1 0 142 7 5= =

+× + ×

∑∑

( )

( )

. .. . . .

I L

I L

c

e b

/

/

/ // /

== 1 475.


4. Column 5-8

Atcolumnend8:Thefarendofbeam7-8isrigidlyconnectedtoacolumn,andbeam7-8restrainsa

swaycolumn5-8,whichgivesβe=1.0.Thestiffnessratioofcolumn5-8atend8is

γβ

8114 3 6

1 0 142 6 51 45= =

×=∑

∑( )

( )

.. .

.I L

I L

c

e b

/

/

//

Theeffectivelengthfactorforcolumn5-8withγ5=1.475andγ8=1.45canbeobtainedbysolvingEquation5.10orfromFigure4.6.3.3inAS4100aske=1.448.

5. Column 3-6

Thebaseofthecolumn3-6isfixedsothatitsstiffnessratioatend3istakenasλ1=0.6.

Atcolumnend6:Thefarendofbeam5-6isrigidlyconnectedtoacolumn,andbeam5-6restrainsa

swaycolumn3-6sothatβe=1.0.


Thestiffnessratioofcolumn3-6atend6canbecalculatedas

γβ

645 9 4 0

1 0 142 7 50 606= =

×=∑

∑( )

( )

. .. .

.I L

I L

c

e b

/

/

//


5.2.4 member capacity in axial compression

Alongcompressionsteelmemberwithresidualstressesandgeometricimperfectionshasaloweraxialstrengththanitssectioncapacity.Theeffectsofmemberslenderness,residualstresspatternandgeometricimperfectionsonthemembercapacityofcompressionmembersareaccountedforbythememberslendernessreductionfactor(Hancock1982;Rotter1982;Hancocketal.1987;Galambos1988;TrahairandBradford1998).InClause6.3.3ofAS4100(1998),thenominalmembercapacityofacompressionmemberwithconstantcrosssectioniscalculatedby

N N Nc c s s= ≤α (5.11)

whereαcisthememberslendernessreductionfactor.Asetofequationsforcalculatingthememberslendernessreductionfactor(αc)givenbyRotter(1982)isprovidedinClause6.3.3ofAS4100(1998)andisdescribedasfollows:

Themodifiedmemberslendernessisexpressedby

λn

ef

yLr

kf

=250

(5.12)

Theslendernessmodifieriscomputedas

α λ

λ λan

n n

= −− +2100 13 5

15 3 20502

( . ).

(5.13)

Thecombinedslendernessiswrittenas

λ λ α α= +n a b (5.14)

whereαbisthemembersectionconstantthataccountsfortheeffectsofresidualstresspat-ternonthecapacityofacolumnandisgiveninTable5.2.Thesectionconstantisinfluencedbythesectiontype,manufacturingandfabricatingmethodsthatinduceresidualstresses,thicknessofmainelementsandsectionformfactor(DavidsandHancock1985;Keyetal.1988;RasmussenandHancock1989).

Theimperfectionparameter(η)iscalculatedas

η λ= − ≥0 00326 13 5 0. ( . ) (5.15)


Thefactorξisafunctionofthecombinedslendernessandimperfectionparameter,whichisdeterminedas

ξ λ η

λ= + +( )

( )/

/90 12 90

2

2 (5.16)

Thememberslendernessreductionfactor(αc)isthereforecalculatedby

α ξξλc = − −

≤1 1

901 0

2

. (5.17)

Thememberslendernessreductionfactor(αc)canbecalculatedbyeithertheformulasgivenearlierorlinearinterpolationfromTable5.3,inwhichallαcvalueswerecalculatedusingtheearlierequations.ItcanbeseenfromTable5.3thattherearefivevaluesofthesectioncon-stant(αb),representingfiveresidualstresslevelsandimperfections.Thevalueofαb=−1.0representsthelowestimperfectionandresidualstress.Figure5.5demonstratestheeffectsofthemodifiedmemberslenderness(λn)andsectionconstant(αb)onthememberslendernessreductionfactor(αc).Thememberslendernessreductionfactorisshowntodecreasewithincreasingeitherthemodifiedslendernessratioortheresidualstresslevel.

Steelmembersofvaryingcrosssectionsaresometimesusedinportalframesastaperedcolumnsandrafters.Clause6.3.4ofAS4100statesthatthenominalsectioncapacityofacompressionmemberwithvaryingcrosssectionscanbetakenastheminimumsectioncapacityofallcrosssectionsalongthelengthofthemember.Themembercapacityiscalcu-latedusingthefollowingmodifiedmemberslenderness:

λn

s

om

NN

= 90 (5.18)

Table 5.2 Member section constant (αb)

Section

Section constant αb

kf = 1 0. kf < 1 0.

Hot-formed RHS and circular hollow section (CHS) −1.0 −0.5Cold-formed RHS and CHS (stress relieved) Cold-formed RHS and CHS (non-stress relieved) −0.5 −0.5Hot-rolled UB and UC sections (tf ≤ 40) 0 0Welded box sectionsWelded H- and I-sections 0 –Tees flame-cut from UB and UC, angles 0.5 –Hot-rolled channelsWelded H- and I-sections (tf ≤ 40) 0.5 0.5Hot-rolled UB and UC sections (tf > 40) 1.0 –Welded H- and I-sections (tf > 40)Other sections not listed in this table 0.5 1.0

Source: Adapted from AS 4100, Australian standard for steel structures, Standards Australia, Sydney, New South Wales, Australia, 1998.


Table 5.3 Member slenderness reduction factor (αc)

Modified member slenderness (λn)

Member slenderness reduction factor (αc)

αb = –1.0 αb = –0.5 αb = 0 αb = 0.5 αb = 1.0

≤10 1.000 1.000 1.000 1.000 1.00015 1.000 0.998 0.995 0.992 0.99020 1.000 0.989 0.978 0.967 0.95625 0.997 0.979 0.961 0.942 0.92330 0.991 0.968 0.943 0.917 0.88835 0.983 0.955 0.925 0.891 0.85340 0.973 0.940 0.905 0.865 0.81850 0.944 0.905 0.861 0.808 0.74760 0.907 0.862 0.809 0.746 0.67665 0.886 0.837 0.779 0.714 0.64270 0.861 0.809 0.748 0.680 0.60975 0.835 0.779 0.715 0.646 0.57680 0.805 0.746 0.681 0.612 0.54590 0.737 0.675 0.610 0.547 0.48795 0.700 0.638 0.575 0.515 0.461

100 0.661 0.600 0.541 0.485 0.435110 0.584 0.528 0.477 0.431 0.389115 0.546 0.495 0.448 0.406 0.368120 0.510 0.463 0.421 0.383 0.348125 0.476 0.434 0.395 0.361 0.330130 0.445 0.406 0.372 0.341 0.313140 0.389 0.357 0.330 0.304 0.282150 0.341 0.316 0.293 0.273 0.255155 0.320 0.298 0.277 0.259 0.242160 0.301 0.281 0.263 0.246 0.231170 0.267 0.251 0.236 0.222 0.210175 0.252 0.238 0.224 0.212 0.200180 0.239 0.225 0.213 0.202 0.192185 0.226 0.214 0.203 0.193 0.183190 0.214 0.203 0.193 0.184 0.175200 0.194 0.185 0.176 0.168 0.161205 0.184 0.176 0.168 0.161 0.154210 0.176 0.168 0.161 0.154 0.148215 0.167 0.161 0.154 0.148 0.142220 0.160 0.154 0.148 0.142 0.137225 0.153 0.147 0.142 0.137 0.132230 0.146 0.141 0.136 0.131 0.127240 0.134 0.130 0.126 0.122 0.118245 0.129 0.125 0.121 0.117 0.114250 0.124 0.120 0.116 0.113 0.110


inwhichNomistheelasticbucklingloadofthecompressionmemberpredictedbytheelasticbucklinganalysis(Galambos1988).

Formembersunderaxialcompression,thefollowingrequirementsmustbesatisfied:

N Ns∗≤ φ (5.19)

N Nc∗ ≤ φ (5.20)

whereN∗isthedesignaxialcompressionforceφ = 0 9. isthecapacityfactor

5.2.5 laced and battened compression members

Inpractice,twoormoreparallelsteelcomponentsmaybetiedtogetherbylacingorbatten-ingtoformasinglecompressionmembertocarryheavyaxialcompressionloads,includinglaced,battenedandback-to-backcompressionmembers.DesignrulesforthedesignofthesecompressionmembersaregiveninClauses6.4and6.5ofAS4100.Themaincomponentsandtheirconnectionsmustbedesignedtoresistadesigntransverseshearforcewhichis

0 50 100

Modified slenderness λn

0

0.2

0.4

0.6

α c

αb= –1.0

αb= –0.5

αb= 0

αb= 0.5

αb= 1.0

0.8

1

150 200 250

Figure 5.5 Member slenderness reduction factor αc.


appliedatanypointalongthelengthofthemember.Thedesigntransverseshearforce(V*)(McGuire1968)isdeterminedas

V

NN

Ns

c

n

∗ =−

∗π

λ

1 (5.21)

whereNsandNcarethenominalsectionandmembercapacityofthecompressionmember,

respectivelyN∗isthetotaldesignaxialforceappliedtothecompressionmemberλnisthemodifiedmemberslendernessgiveninClauses6.4and6.5ofAS4100

Thebattenanditsconnectionsmustbedesignedtoresistadesignlongitudinalshearforceandadesignbendingmoment.Thedesignlongitudinalshearforce( )Vl

∗ anddesignbendingmoment(M*)aregiveninClause6.4.3.7ofAS4100asfollows:

V

V sn d

lb

b b

∗ =∗

(5.22)

M

V sn

b

b

∗ =∗

2 (5.23)

whereV∗isthedesigntransverseshearforcesbisthelongitudinalcentre-to-centredistancebetweenthebattensnbisthenumberofparallelplanesofbattensdbisthelateraldistancebetweenthecentroidsoftheweldsoffasteners

Example 5.2: Checking the capacity of a compression steel column

A360UB44.7Grade300steelcolumnof8mlengthisfixedatitsbaseandpinnedatitstop.Thecolumnisbracedagainstbucklingaboutthey-axisbystrutsthatarepin-connectedtoitsmid-height.Thestrutspreventlateraldeflectionsintheminorprincipalplane.Thecolumnissubjectedtocompressiveforcesincludinganominaldeadloadof200kNandanominalliveloadof250kN.Thesectionpropertiesof360UB44.7Grade300 steel shown in Figure 5.6 are Ag = 5720 mm2, rx = 146 mm, ry = 37.6 mm andfy = 320MPa.Checkthecapacityofthecompressioncolumn.

1. Design axial load

Thedesignaxialloadis

N G Q∗ = + = × + × =1 2 1 5 1 2 200 1 5 250 615. . . . kN

2. Check the slenderness of elements

Theslendernessoftheflangeis

λ λef

yey

bt

f= =

−= < =

250171 6 9 2

9 7320250

9 57 16( . )

.. .

/Table 5 2 of AS41000

Hence,theflangeisnotslender.



λ λew

yey

bt

f= =

− ×= > =

250352 2 9 7

6 9320250

54 54 45( . )

.. .Table 5 2 of AS41100

Hence,thewebisslender.

3. Section capacity

Theeffectivewidthofthewebcanbecalculatedas

b bew

ey

e

=

= − ×

=

λλ

( . ).

.352 2 9 745

54 54274 4 mm

Theeffectiveareaofthesectionis

Ae = × × + × =2 171 9 7 274 4 6 9 5210 76( . ) . . . mm2

Thegrossareaofthesectioncanbecalculatedas

Ag = × × + − × × =2 171 9 7 352 2 9 7 6 9 5612 34( . ) ( . ) . . mm2

Theformfactorisdeterminedas

k

AA

fe

g

= = =5210 765612 34

0 928..

.

Thesectiondesigncapacityiscalculatedas

φ φN k A f Ns f n y= = × × × = > ∗ =0 9 0 928 5720 320 1529. . kN 615 kN, OK

4. Member capacity

Thecolumnisfixedatitsbaseandpinnedatitstopsothattheeffectivelengthfactorforbucklingaboutthex-axisiske=0.85.Theeffectivelengthis

L k Lex e= = × =0 85 8000 6800. mm

9.7

9.7

6.9352

171

Figure 5.6 Cross section of steel column.


Theendsoftheuppersegmentofthecolumnarepinned.Theeffectivelengthoftheuppersegmentbucklingabouty-axisis

L k Ley e= = × =1 0 4000 4000. mm

Themodifiedmemberslendernessiscalculatedasfollows:

λnx

ex

xf

yLr

kf

= = =250

6800146

0 928320250

50 76. .

λ λny

ey

yf

ynx

Lr

kf

= = = >250

400037 6

0 928320250

115 94.

. .

Hence,thecolumnwillbuckleaboutthey-axis,λn=115.94.Forhot-rolledI-sectionskf<0,thesectionconstantisobtainedfromTable5.2asαb=0.ThememberslendernessreductionfactorcanbeobtainedfromTable5.3as

αc = −

− −−

=0 4480 448 0 421 15 94 115

120 1150 443.

( . . )( . )( )

.

Thedesigncapacityofthecolumnis

φ φαN N Nc c s= = × = > ∗ =0 443 1529 677 615. kN kN, OK

Example 5.3: Checking the capacity of an RHS compression column

Thepin-endedrectangularhollowsection(RHS)column200×100×4.0RHSofGradeC350 steel as depicted in Figure 5.7 is 4 m length. The column is subjected to axialcompressionforcesincludinganominaldeadloadof100kNandanominalliveloadof120kN.ThesectionpropertiesoftheRHScolumnareAg=2280mm2,rx=72.1mm,ry = 42.3 mmandfy=350 MPa.Checkthecapacityofthecolumn.

1. Design axial load

Thedesignaxialloadis

N G Q∗ = + = × + × =1 2 1 5 1 2 100 1 5 120 300. . . . kN

200 4

100

Figure 5.7 Cross section of RHS steel column.


2. Check the slenderness of elements

Theslendernessoftheflangeisdeterminedas

λ λef

yey

bt

f= =

− ×= < =

250100 2 4

4350250

27 2 40 00( )

. .Table 5 2 of AS41

Hence,theflangeisnotslender.Theslendernessofthewebiscalculatedas

λ λew

yey

bt

f= =

− ×= > =

250200 2 4

4350250

56 79 40( )

.

Hence,thewebisslender.

3. Section capacity

Theeffectivewidthofthewebcanbecalculatedas

b bew

ey

e

=

= − ×

=

λλ

( ).

.200 2 440

56 79135 2 mm

Theeffectiveareaofthesectionisdeterminedas

Ae = × − × × + × × =2 100 2 4 4 2 135 2 4 1817 6( ) . . mm2

Thegrossareaofthesectioniscalculatedas

Ag = × − × × + − × × =2 100 2 4 4 200 2 4 4 2272( ) ( ) mm2

Theformfactoriskf=Ae/Ag=1817.6/2272=0.8.Thesectiondesigncapacityisdeterminedas

φ φN k A f Ns f n y= = × × × = > ∗ =0 9 0 8 2280 350 574 56. . . kN 300 kN, OK

4. Member capacity

Theeffectivelengthofcolumnbucklingabouttheminorprincipaly-axisis

L k Ley e= = × =1 0 4000 4000. mm

Themodifiedmemberslendernessiscomputedas

λny

ey

yf

yLr

kf

= = =250

400042 3

0 8350250

100.

.

Forcold-formedRHSwithkf<1.0,thesectionconstantcanbeobtainedfromTable5.2asαb =−0.5.ThememberslendernessreductionfactorcanbeobtainedfromTable5.3asαc=0.6.

Thedesigncapacityofthecolumniscalculatedas

φ φαN N Nc c s= = × = > ∗ =0 6 574 56 334 7 300. . . kN kN, OK


5.3 memBerS In AxIAl tenSIon

5.3.1 Behaviour of members in axial tension

Theload–extensionbehaviourofsteeltensionmemberswithoutholes,geometricimperfec-tionsandresidualstressesisfoundtofollowthematerialstress–strainrelationship.Theseidealisedtensionmembersdisplayductilebehaviourandcanattainthegrossyieldcapac-ity(Agfy)ofthesections.Residualstressesintensionmemberscauselocalearlyyieldingandstrainhardening.Holesintensionmembersleadtoearlyyieldingaroundtheholes.Consequently,theload–deflectionbehaviouroftensionmemberswithholesisnonlinear.Whentheholesare large, thememberunderaxial tensionmayfailbyfracturingattheholes,anditsstrengthisgovernedbythefacturecapacityofitscrosssection(DhallaandWinter1974a,b;Bennettsetal.1986).Thestrengthandbehaviourof steelmembers inaxialtensionmaybegovernedbyeitherthegrossyieldcapacityorthefracturecapacityoftheircrosssections.Therefore,thedesignoftensionmembersmustconsiderthesetwofailurecriteria.

5.3.2 capacity of members in axial tension

A steel member subject to a design axial tension force (N*) must satisfy the followingstrengthdesignrequirement:

N Nt∗ ≤ φ (5.24)

whereφ = 0 9. isthecapacityreductionfactorNtisthenominalsectioncapacityinaxialtension

Clause7.2ofAS4100(1998)specifiesthatNtistakenasthelesserofthegrossyieldcapacity(Nty)andfracturecapacity(Nta):

N A fty g y= (5.25)

N k A fta ct n u= 0 85. (5.26)

wherekct is the correction factor considering the effect of non-uniform force distributions

inducedbytheendconnectionsofthetensionmemberAnisthenetcross-sectionalareafuisthedesigntensilestrength

Thefactorof0.85intheearlierequationisusedtoaccountforthesuddenfailurebylocalbrittlefacturebehaviouratthenetsection.Thenominalmembercapacityofasteelmemberunderaxialtensionistakenasitsnominalsectioncapacity.

Guidelines for determining the correction factor are given in Clause 7.3 of AS 4100.The correction factor (kct) is taken as 1.0 for a tension member whose end connectionsaredesignedtoprovideuniformforcedistributioninthemember.Toachievethiscondi-tion, the endconnectionsmustbe symmetricallyplacedabout the centroidalaxisof themember,andeachpartoftheconnectionmustbecapableofresistingthedesignforcein


itsconnected part.Iftheendconnectionsofatensionmemberinducenon-uniformforcedistributionsinthemember,themembershouldbedesignedforcombinedactionsofaxialtensionandbending.However,foreccentricallyconnectedangles,channelsandtees,thecorrection factor (kct) given in Table 5.4 can be used in Equation 5.26 to determine itscapacity(Bennettsetal.1986).Thecorrectionfactor(kct)istakenas0.85forsymmetricalI-sectionsandchannelsectionsconnectedbybothflangesonlyifthelengthbetweenthefirstandlastlowsofboltsisgreaterthanthedepthofthemembersection.

TensionmemberswithstaggeringholesmayfailbyfracturealongazigzagpathABCDEFasdepictedinFigure5.8ratherthanalongthepathperpendiculartotheappliedaxialten-sileforce.ThefailurelinemaybealongthepathDGorthediagonalpathDEasshowninFigure5.8.Thecriticalfailurepathistheonethathastheminimumnetarea.Thediffer-encebetweenthepathDGandthepathDEisrepresentedbyalengthcorrection(Cochrane1922)as

l

ss

cp

g

=2

4 (5.27)

wherespstandsforthestaggeredpitch,whichisthecentre-to-centredistanceparalleltothe

directionofthetensileforceinthemembersgrepresentsthegaugethatisthecentre-to-centredistanceofholesmeasuredatright

angletothedirectionofthetensileforceinthemember

Table 5.4 Correction factor (kct)

Configuration kct

Single angle connection or twin angles on same sideUnequal angle connected by short leg 0.75Otherwise 0.85

Single channel (web connected to the plate) 0.85Single tee (flange connected to the plate) 0.9Back-to-back symmetric connection 1.0

Source: Adapted from AS 4100, Australian standard for steel structures, Standards Australia, Sydney, New South Wales, Australia, 1998.

G

Sp

Sg

F

E

D

N * N *C

B

A

Figure 5.8 Failure paths on net section.


Theareacorrectioniscalculatedbymultiplyingthelengthcorrectionandthethickness.Thenetcross-sectionalareaalongthezigzagpathcanbecalculatedby

A A d t

s ts

n g hp

g

= − +∑ ∑2

4 (5.28)

wheredhisthediameterofaholetisthethicknessoftheholedmaterial

Itisnotedthatthenetarea(An)mustbelessthanorequaltothegrossarea(Ag).Thisgivess s dp g h≤ 4 .Thismeansthatifthestaggerspacings s dp g h> 4 ,theholedoesnotreducetheareaofthemember(TrahairandBradford1998).Ifholesarenotstaggeredors s dp g h> 4 ,spistakenaszeroinEquation5.28.

Example 5.4: Capacity of a bolted steel member in axial tension

Bothflangesofa steel I-sectionmemberunderaxial tensionareboltedasdepicted inFigure 5.9. The depth of the I-section is 339 mm. The diameter of each bolt hole is24 mm.Thegrossareaofthemembersectionis29,300 mm2andthethicknessoftheflangeis32 mm.Theyieldstressofthesectionis280MPa,whileitstensilestrengthis430MPa.Determinethecapacityofthetensionmember.

1. Net area of the section

Theminimumstaggeriscalculatedas

s s d spm g h p= = × × = > =4 4 70 24 82 35mm mm

Thus, the failure path at each flange is staggered as indicated by the path ABCDEFshowninFigure5.9.Thisfailurepathateachflangeincludesfourholesandtwostaggers.Thenetareaofthesectioncanbecalculatedas

A A d ts ts

n g hp

g

= − + = − × × ×( ) + × ×××

∑ ∑2 2

42 4 24 32 2 2

35 324 70

29,300

= 23,716 mm2

N *N * 130

A

B

C

D

E

F

7070707070

70

70

70707070

Figure 5.9 Bolted flange of a welded I-section member.


2. Capacity of tension member

Thegrossyieldcapacitycanbecalculatedas

N A fty g y= = × × =−29,300 kN 8,204 kN280 10 3

ThisI-sectiontensionmemberisconnectedbybothflangesonly.Thelengthbetweenthefirstandlastlowsofboltsislcb=5×70=350 mm>D=339 mm.Thus,thecorrectionfactorofthisI-sectionmembersatisfyingtherequirementofClause7.3.2ofAS4100iskct=0.85.

Thefracturecapacitycanbecalculatedas

N k A fta ct n u= = × × × = <0 85 0 85 0 85 430. . . 23,716 N 7,368 kN 8,204 kN

Thus,Nt=7368 kN.Thedesigncapacitycanbedeterminedas

φNt = × =0 9 7368 6331. kN kN

5.4 memBerS under AxIAl loAd And unIAxIAl BendIng

5.4.1 Behaviour of members under combined actions

Steel members subject to combined actions of axial load and bending are called beam–columns.Thebendingmomentsare inducedbythe loadingeccentricity, the lateral loadsappliedtothecolumnsandtheoverallframeactions.Thebehaviourofsteelmembersundercombedactionsischaracterisedbythein-plane,out-of-planeandbiaxialbending.Whenasteelbeam–columnisconstrainedtobendaboutitsmajorprincipalaxisorwhenitisbentaboutitsminorprincipalaxis,itsdeformationsoccurintheplaneofbending.Thisisthein-planebehaviour,whichischaracterisedbythebendingofbeamsandbythebucklingofcompressionmembersintheplane.Whenasteelbeam–columnthatisnotrestrainedlater-allyisbentaboutitsprincipalaxis,itmayundergoflexural–torsionalbuckling.

Theultimatestrengthofasteelbeam–columnundercombinedaxialcompressionforceandbendingmomentisinfluencedbytheinteractionbetweentheaxialforceandbendingmoment.Theaxialcompressionforcereducesthemomentcapacityofthebeam–column,whilethebend-ingmomentreducesthememberaxialloadcapacity.Theinteractionbetweentheaxialcompres-sionforceanddeformationsleadstosecond-ordereffectswhichamplifythebendingmoments.

Forasteelmembersubjecttocombinedactionsofaxialtensionandbending,theaxialtensileforcereducesthesectionmomentcapacityofthememberbutincreases itsout-of-planemembercapacitywhenbendingaboutitsmajorprincipalx-axis.

5.4.2 Section moment capacity reduced by axial force

ThedesignrulesforsectionmomentcapacityreducedbyaxialforcearegiveninClause8.3ofAS4100.FurtherinformationcanbefoundinpublicationsbyWoolcockandKitipornchai(1986),Bradfordetal.(1987),BridgeandTrahair(1987)andTrahairandBradford(1998).Whenasteelmemberissubjectedtoanaxialforce(N*)andadesignbendingmoment( )Mx

∗ aboutitssectionmajorprincipalx-axis,thenominalsectionmomentcapacity(Mrx)reducedbytheaxialforceisgiveninClause8.3.2ofAS4100(1998)asfollows:

M M

NN

rx sxs

= −∗

1

φ (5.29)


whereNsstandsforthenominalsectionaxialloadcapacityforaxialcompressionoraxialtension.Equation5.29isbasedonthesimplestraight-lineinteractioncurveasdepictedinFigure5.10andisconservativeforcompactdoublysymmetricI-sectionsandrectangularandsquarehollowsections.

Forcompressionmemberswithkf=1.0andtensionmembersthatareofcompactdoublysymmetricI-sectionsandrectangularandsquarehollowsections,Clause8.3.2ofAS4100(1998)providesamoreaccurateexpressionforcalculatingMrxasfollows:

M M

NN

Mrx sxs

sx= −∗

≤1 18 1.

φ (5.30)

ThestrengthinteractioncurverepresentingEquation5.30forcompactdoublysymmetricI-sectionsandrectangularandsquarehollowsectionsisshowninFigure5.10.ItcanbeseenfromthefigurethattheearlierstrengthinteractionformulagiveshighersectioncapacitiesthanEquation5.29.

Ifcompressionmemberswithkf<1.0areofcompactdoublysymmetricI-sectionsandrectangularandsquarehollowsections,Clause8.3.2ofAS4100(1998)providesthefollow-ingmoreaccurateformulafordeterminingthereducedsectionmomentcapacity:

M MNN

Mrx sxs

w

wysx= −

∗

+ −

−

≤1 1 0 18

8282φ

λλ

. (5.31)

inwhichλwandλwyaretheslendernessandslendernessyieldlimitoftheweb,respectively.

00

0.2

0.4

N*/

φNs

0.6

0.8

1

1.2

0.2 0.6Mrx/Msx

0.4 0.8 1 1.2

Equation 5.29Equation 5.30

Figure 5.10 Strength interaction curves for compact doubly symmetric I-sections under axial force and uniaxial bending about principal y-axis.


AsgiveninClause8.3.3ofAS4100(1998),foramembersubjecttoaxialforce(N*)anddesignbendingmoment( )My

∗ aboutitssectionminorprincipaly-axis,thenominalsectionmomentcapacity(Mry)reducedbytheaxialtensionorcompressionforceisexpressedby

M M

NN

ry sys

= −∗

1

φ (5.32)

whereMsydenotesthenominalsectionmomentcapacityforbendingabouttheminorprinci-paly-axis.Equation5.32representsastraight-lineinteractioncurveasshowninFigure 5.11andisconservativeforcompactdoublysymmetric I-sectionsandrectangularandsquarehollowsections.

ForcompactdoublysymmetricI-sections,MrycanbemoreaccuratelycalculatedbythefollowingformulagiveninAS4100(1998)

M MNN

Mry sys

sy= −∗

≤1 19 1

2

.φ

(5.33)

ThestrengthinteractioncurvewhichrepresentsEquation5.33isalsoshowninFigure5.11.Itisshownthatthestraight-lineinteractioncurvebasedonEquation5.32isveryconserva-tive,andsignificanteconomycanbeachievedbyusingEquation5.33forcompactdoublysymmetricI-sections.

Forcompactrectangularandsquarehollowsections,thefollowingexpressiongiveninClause8.3.3ofAS4100(1998)providesmoreaccuratepredictionofMryasillustratedinFigure5.12:

M M

NN

Mry sys

sy= −∗

≤1 18 1.

φ (5.34)

0 0.2 0.4 0.6 0.8 1 1.2

Mry/Msy

N*/

φNs

0

0.2

0.4

0.6

0.8

1

1.2


Figure 5.11 Strength interaction curves for compact doubly symmetric I-sections under axial force and uniaxial bending about principal y-axis.


Forstrengthdesign,allsectionsofasteelmemberunderaxialforceanduniaxialbendingmustsatisfythefollowingconditions:

M Mx rx∗ ≤ φ (5.35)

M My ry∗ ≤ φ (5.36)

5.4.3 In-plane member capacity

Thedesignofmembersunderaxial loadanduniaxialbending for in-planebendingandbucklingisgiveninClause8.4.2ofAS4100formembersanalysedbytheelasticmethod.Thenominalin-planemembermomentcapacity(Mi)ofacompressionmemberisgiveninClause8.4.2.2ofAS4100(1998)as

M M

NN

i sc

= −∗

1

φ (5.37)

whereN∗isthedesignaxialcompressiveforceNcisthenominalmembercapacityinaxialcompressionforbucklingaboutthesame

principalaxisdeterminedusingtheeffectivelengthfactorofke = 1 0. forbothbracedandswaymembers,unlessalowervalueofkecanbedeterminedforbracedmembers

Thereasonfortakingkeas1.0foraswaymemberisthattheeffectsofendrestraintsonthememberbucklinghavebeenconsideredinthesecond-orderelasticanalysis.However, this

0 0.2 0.4 0.6 0.8 1 1.2

Mry/Msy

N*/

φNs

0

0.2

0.4

0.6

0.8

1

1.2


Figure 5.12 Strength interaction curves for compact rectangular and square sections under axial force and uniaxial bending about principal y-axis.


may result in unsafe designs for some sway compression members under small bendingmoments.Therefore, if the effective length factordetermined inaccordancewithClause4.6.3isusedinthecalculationofthecompressionmembercapacity,thedesignaxialcom-pressionforcealonemustbelessthanthesectionandmembercapacityinaxialcompres-sion.Figure5.13demonstratesthestrengthinteractioncurveforin-planebendingandaxialcompression.

For compact doubly symmetric I-sections and rectangular and square hollow sectionswithkf=1.0,Clause8.4.2ofAS4100providesthefollowingformulaforcalculatingthein-planemembermomentcapacity(Mi)(Trahair1986):

M MNN

i sm

c

m= − +

−∗

+

+

1

12

1 1 181

21

3 3βφ

β. −−

∗

≤N

NM M

crx ryφ

or (5.38)

whereβm=1.0foruniformbending.Thein-planemembermomentcapacityofmembersunderaxialtensionandbendingis

notreducedbyaxialtensionsothattheirdesignisgovernedbythesectioncapacities.Thedesignofasteelmemberforin-planebendingandaxialforcemustsatisfy

M M M Mx ix y iy∗ ≤ ∗ ≤φ φor (5.39)

whereMixandMiyarethenominalin-planemembermomentcapacitiesforbendingabouttheprincipalaxes,respectively.

5.4.4 out-of-plane member capacity

A steel member subject to axial load and uniaxial bending about its major principalx-axismayfailbybucklingoutoftheplaneofbending.Thedesignofthesemembersfor

0 0.2 0.4 0.6 0.8 1 1.2

Mi/Ms

N*/φN

c

0

0.2

0.4

0.6

0.8

1

1.2

Figure 5.13 Strength interaction curves for compression members under in-plane bending.


out-of-planebucklingisgiveninClause8.4.4ofAS4100.AsprovidedinClause8.4.4.1ofAS4100(1998),thenominalout-of-planemembermomentcapacity(Mox)ofacompressionmemberiscalculatedby

M M

NN

ox bxcy

= −∗

1

φ (5.40)

whereMbxisthenominalmembermomentcapacityofthememberwithoutfulllateralrestraintNcyisthenominalmembercapacityinaxialcompressionforbucklingabouttheminor

principaly-axis

Equation5.40isplottedasastraight-linestrengthinteractioncurveinFigure5.14.FormemberswithcompactdoublesymmetricI-sectionsfullyorpartiallyrestrainedat

bothendsandwithkf=1.0,amoreaccurateexpression(Cuketal.1986)isgiveninClause8.4.4.1ofAS4100as

M MNN

NN

Mox bc bxocy oz

rx= −∗

−

∗

≤α

φ φ1 1 (5.41)

wherethefactorαbcaccountsfortheeffectsofthemomentratio(βm)andtheaxialforce(N*)andcanbedeterminedby

αβ β

φ

bc

m m

cy

NN

=− + +

−

∗

1

12

12

0 4 0 233

. .

(5.42)

0 0.2 0.4 0.6 0.8 1 1.2

Mox/Mbx

N*/φN

cy

0

0.2

0.4

0.6

0.8

1

1.2

Figure 5.14 Strength interaction curves for compression members for out-of-plane buckling.


InEquation5.41,Mbxoisthenominalmembermomentcapacitywithoutfulllateralrestrainand under uniform design moment distribution, and Noz is the elastic torsional buckingcapacityofthemember,whichisgivenby

N

GJE Il

I I Aoz

s w

z

x y

=+

+

π2

2

( )/ (5.43)

inwhichlzisthedistancebetweenpartialorfulltorsionalrestraints.Forasteelmemberunder,anaxialtensileforceandadesignbendingmoment,thenominal

out-of-planemembermomentcapacity(Mox)isgiveninClause8.4.4.2ofAS4100(1998)as

M M

NN

Mox bxt

rx= +∗

≤1

φ (5.44)

whereNt isthenominalsectioncapacityinaxialtension.Itcanbeseenfromtheearlierequationthatthetensileforceincreasestheout-of-planemembermomentcapacity.

Thedesignof a steelmemberunder an axial compressive force and adesignbendingmomentaboutitsmajorprincipalx-axismustcheckforitsin-planeandout-of-planemem-bermomentcapacitiesasfollows:

M Mx i∗ ≤ φ (5.45)

M Mx ox∗ ≤ φ (5.46)

However,fortensionmembers,onlytheout-of-planemembermomentcapacityneedstobechecked.

5.5 deSIgn of PortAl frAme rAfterS And columnS

Rafters and columns inportal frames are subjected to combinedaxial force anduniax-ialbending.Thedesignofraftersandcolumnsinportalframesmaybegovernedbythestrengthcriteriaorbythedeflectioncriteria.Fordesignforthestrengthcriteria,theeco-nomicaldesignsofraftersandcolumnscanbeachievedbydesigningthemembercapacityascloseaspossibletothesectioncapacity.Thiscanbedonebyprovidingadequateflybracestolaterallyrestraintheinsidecompressionflangesoftheraftersandcolumns.Forlargespanportalframes,deflectionusuallygovernsthedesign.Forthiscase,haunchescanbeaddedtotherafterstoreducethedeflections.AtypicalportalframeisdepictedinFigure5.15.

5.5.1 rafters

Purlins,whichareboltedtothetopflangeofarafterinasteelportalframe,providelateralbutnotrotationalrestrainttothetopflangebecausetheboltedconnectionbetweenpur-linsandtheflangeallowsforrotation.Underdeadandliveloads,mostofthetopflangeofarafterissubjectedtocompression.Asaresult,theeffectivelengthcanbetakenasthedistancebetweenthepurlinswhencalculatingthemembermomentcapacity(Mbx)oftherafter.Underupwardwind loads,however,mostof thebottomflangeof the rafter is in


compression.Toincreasetheraftermembercapacity,flybracesintheformofsmallanglesectionmembersthatconnectthebottomflangetothepurlinsareusuallyusedtobracetherafterasdepictedinFigure5.16.Asflybracesprovidefullrestrainttothebottomflangeoftherafter,theeffectivelengthoftherafteristakenasthedistancebetweentheflybracesinthecalculationofitsmembermomentcapacity(Mbx).Itisrecommendedthatflybracesshouldbeprovidedwithinthefirstquarterofthetotalrafterspan,attheinsidecornerofthekneejointandneartheridge(Woolcocketal.2003).Whencalculatingthein-planemembercapacityofarafterundercombinedactions,thenominalmembercapacityinaxialcom-pression(Ncx)isrequired.Forthispurpose,theeffectivelength(Lex)istakenastheactualrafterlengthmeasuredfromthecentrelineofthecolumntotheridge.Forcolumnsundercombinedactions,however, thenominalmembercapacity (Ncy) inaxialcompression forbucklingaboutthey-axiscanbecomputedusingthedistancebetweenpurlinsastheeffec-tivelength(Ley).Thisisbecausepurlinsandroofsheetingactasarigiddiaphragmbetweenroofbracingnodes,whichforcetheraftertobucklebetweenthepurlins.

5.5.2 Portal frame columns

Atthebottomofaportalframecolumn,thebaseplateandboltsofferfulllateralandtor-sionalrestraintandnearlysomeminoraxisandwarpingrestraint.Atthetopofthecolumn,thewallbracingand theflybraceat the insidecornerof thehaunchprovide full lateralrestraint.Itisnotedthattherafterdoesnotofferminoraxisandwarpingrestrainttothecolumn.Whentheinsideflangeoftheportalcolumnwithoutflybracesisincompression,theeffectivelengthofthecolumncanbetakenasdistancefromthebaseplatetotheunder-sideofthehaunchfordeterminingitsmembermomentcapacity(Mbx).Forcolumnswithflybraces,theeffectivelengthcanbetakenasthedistancebetweenflybracesinthecalculationofMbx.Whentheoutsideflangeofaportalcolumnisincompression,itsmembermomentcapacity(Mbx)shouldbecalculatedusingthespacingofgirtsastheeffectivelength.

L

HSg

Sp

SfFly brace

Purlin

Figure 5.15 Portal frame.

Fly brace

Rafter

PurlinCleat

Figure 5.16 Double fly braces.


Thenominalmembercapacityinaxialcompression(Ncx)isusedtocalculatethein-planemembercapacityofaportalframecolumnundercombinedactions.Forthispurpose,theeffectivelength(Lex) istakenastheactualcolumnlength.Whendeterminingtheout-of-planemembercapacityofthecolumnundercombinedactions,theeffectivelength(Ley)istakenasthedistancebetweengirtsinthecalculationofthenominalmembercapacity(Ncy)inaxialcompressionforbucklingaboutthey-axis.Thereasonforthisisthatgirtsandwallsheetingactasarigiddiaphragmbetweenwallbracingnodeswhichshouldbeeffectiveinensuring thecolumnbucklebetweengirts.However,whendesigningheavily loadedcol-umns,theeffectivelengthshouldbetakenasthedistancebetweenflybraces.

Example 5.5: Design of a steel portal frame column under axial force and uniaxial bending

Asteelportalframecolumnof460UB74.6ofGrade300steelisschematicallydepictedinFigure5.17.Theheightoftheportalframecolumnmeasuredfromthefloortothecen-trelineofrafteris6000 mm,whiletheheightoftheundersideofthehaunchis5364 mm.Thecolumnispinnedatitsbaseandbracedbygirtswithaspacingof1400 mm.Thesecond-orderelasticanalysisoftheportalframeundervariousloadcombinationscalcu-latedinExample2.1hasbeenperformed.Checkthecapacitiesofthecolumnunderthefollowingdesignactionsobtainedfromthesecond-orderelasticanalysis:

a. M Nx c∗ = ∗ =330 kN m, 87 kN (compression)

b. M Nx t∗ = ∗ =420 kN m, 115 kN (tension)

1. Section properties

Thedimensionsandpropertiesof460UB74.6are

d b t t A

I

f f w g

x

= = = = =

= ×

457 190 14 5 9 1 9520

335 106

mm, mm, mm, mm, mm

mm

2. .

44 3 4mm, mm mm

mm,

, , .

.

r Z I

r Z

x ex y

y ey

= = × = ×

= =

188 1660 10 16 6 10

41 8 26

3 6

22 10 815 10 530 10

300 440

3 9 3× = × = ×

= =

mm mm mm

MPa, MPa,

3 6 3, ,I J

f f k

w

y u f == = × = ×0 948 200 10 80 103 3. , E Gs MPa, MPa

Compactnessaboutthex-axis=compact

60005364

1400

3000

19014.5

14.5

457 9.1

Figure 5.17 Steel portal frame column.


2. Axial section capacities

Thesectioncapacityinaxialcompressioniscalculatedas

N k A f

N

sc f n y

sc

= = × × =

= × =

0 948 9520 300 2707 5

0 9 2707 5 2436 7

. .

. . .

N kN

kφ NN kN, OK> ∗ =Nc 87

Thegrossyieldcapacityofthesectionis

N A fty g y= = × =9520 300 2856N kN

Thefracturecapacityofthesectionis

N k A fta t n u= = × × × = >0 85 0 85 1 0 8520 440 3186 5. . . .N kN 2856 kN

∴ =Nt 2856 kN

Thedesigncapacityinaxialtensionis

φN Nt t= × = > ∗ =0 9 2856 115. 2570.4 kN kN, OK

3. Section moment capacities

3.1. Section moment capacity without axial force

Thesectionmomentcapacityiscomputedas

M Z f

M

sx ex y

sx

= = × × × =

= × =

−1660 10 300 10 498

0 9 498 448 2

3 6 kNm

kNmφ . .

3.2. Reduced section moment capacity due to axial compression

460UB74.6isacompactdoublysymmetricI-sectionwithkf=0.948<1.0sothatitssectionmomentcapacityreducedbyaxialcompressioncanbecalculatedusingEquation 5.31.


λw

ybt

f= =

− ×=

250457 2 14 5

9 1300250

51 5.

..

TheslendernesslimitforthewebsupportedbytwoflangesunderuniformcompressioncanbeobtainedfromTable5.2ofAS4100asλwy=45.

Thesectionmomentcapacityreducedbyaxialcompressioniscalculatedas

M MNN

Mrx sxs

w

wysx= −

∗

+

−−

≤

=

1 1 0 188282φ

λλ

.

4498 187

1 0 1882 51 582 45

551 5× −

+ ×

−−

=2436.7

..

. kkNm kNm

kNm

> =

∴ =

M

M

sx

rx

498

498

φM Mrx x= × = > ∗ =0 9 498 448 2 330. . kN m kN m, OK


3.3. Reduced section moment capacity due to axial tension

ForcompactdoublysymmetricI-section,thesectionmomentcapacityreducedbyaxialtensioncanbecalculatedas

M MNN

Mrx sxs

sx= −∗

≤

= × × −

=

1 18 1

1152570 4

56

.

.

φ

1.18 498 1 11 3 498

498

. kNm kNm

kNm

> =

∴ =

M

M

sx

rx

φM Mrx x= × = > ∗ =0 9 498 448 2 420. . Nm kN m, OK

4. Axial member capacities

4.1. Axial member capacity Ncx

Theeffective length factorke is takenas1.0as required for combinedaction (Clause8.4.2.2).Theeffectivelengthis

L k Le e= = × =1 0 6000 6000. mm

Themodifiedmemberslendernesscanbecalculatedas

λnx

ex

xf

yLr

kf

= = =250

6000188

0 948300250

34.

Forhot-rolledUB(universalbeam)sectionwithkf<1.0,αb=0,theslendernessreductionfactorcanbeobtainedfromTable5.3usinglinearinterpolationasfollows:

αc = −

− −−

=0 9430 943 0 925 34 30

35 300 928.

( . . )( )( )

.

Thedesignaxialmembercapacityis

φ φαN Ncx c s= = × × =0 9 0 928 2707 5 2261 3. . . . kN

4.2. Axial member capacity Ncy

Theportalframecolumnisbracedbygirtswithaspacingof1400forbucklingaboutthe minor principal y-axis so that the effective length is taken as the girt spacingLey = 1400mm.

Themodifiedmemberslendernessis

λny

ey

yf

yLr

kf

= = =250

140041 8

0 948300250

35 7.

. .


Forhot-rolledUBsectionkf<1.0,αb=0,theslendernessreductionfactorcanbeobtainedfromTable5.3usinglinearinterpolationas

αc = −

− −−

=0 9250 925 0 905 35 7 35

40 350 922.

( . . )( . )( )

.

Thedesignaxialmembercapacityisdeterminedas

φ φαN Ncy c s= = × × =0 9 0 922 2707 5 2246 7. . . . kN

5. In-plane member moment capacity

Thein-planemembermomentcapacityofthecolumnofkf<1.0undercompressionandbendingcanbedeterminedas

M M

NN

i sc

= −∗

= × −

=1 498 1

872436 7

478 8φ .

. kNm

φM Mi x= × = > =∗0 9 478 8 431 330. . kN m kN m, OK

6. Out-of-plane member capacity

6.1. Member moment capacity without full lateral restraints

Theportalframecolumnistreatedastheonewithoutflybraces.Thelengthofthecol-umnforflexural–torsionalbucklingistakenasL= 5364 mm.

Asthecolumnisfullyrestrainedagainsttwistatbothends,thetwistrestraintfactoriskt= 1.0.

Thecolumnissubjectedtoonlymoments;theloadheightfactoriskl= 1.0.Asthecolumnisrestrainedbybaseplateagainstlateralrotation,thelateralrotational

restraintfactoriskr= 0.85.Theeffectivelengthfactoristherefore

L k k k Le t l r= = × × × = mm1 0 1 0 0 85 5364 4559 4. . . .

Theelasticbucklingmomentiscalculatedas

ME IL

GJE IL

os y

e

s w

e

= +

=× × × ×

π π

π

2

2

2

2

2 3 6

2

200 10 16 6 104559 4

8.

.00 10 530 10

200 10 815 104559 4

434 5

3 32 3 9

2× × × +× × × ×

=

π.

.

kN m

Thememberslendernessreductionfactoriscomputedas

αss

oa

s

oa

MM

MM

=

+ −

=

+ −0 6 3 0 6

498434 5

349

2 2

. ..

88434 5

0 558.

.

=


Thebendingmomentdistributionalongthecolumnis linearwithzeromomentatthebottomandmaximummomentatthetop.Themomentmodificationfactorisαm= 1.75.

Themembermomentcapacityis

M M M

M

M

bx m s sx sx

sx

bx

= ≤

= × × = < =

∴ =

α α

1 75 0 558 498 486 3 498

486

. . . kNm kNm

..3 kNm

6.2. Out-of-plane member capacity in axial compression and bending

Theout-of-planemembermomentcapacityofthecolumnunderaxialcompressionandbendingcanbecalculatedas

M MNN

Mox bxcy

rx= −∗

≤

= × −

=

1

486 3 187

2246 3467 3

φ

..

. kNmm kNm

467.5 kNm

< =

∴ =

M

M

rx

ox

498

φM Mox x= × = > ∗ =0 9 467 5 330. . 421 kN m kN m, OK

6.3. Out-of-plane member capacity in axial tension and bending

Foraxialtensionandbending,theout-of-planemembercapacityisdeterminedas

M MNN

Mox bxt

rx= +∗

≤

= × +

= >

1

486 3 1115

2570 4508

φ

..

kNm MM

M

rx

ox

=

∴ =

498

498

kNm

kNm

φM Mox x= × = > ∗ =0 9 498 420. 448.2 kN m kN m, OK

Therefore,thecapacityoftheportalframecolumnisadequate.

5.6 memBerS under AxIAl loAd And BIAxIAl BendIng

5.6.1 Section capacity under biaxial bending

Clause8.3.4ofAS4100providesasimplelinearexpressionforconservativelyestimat-ingtheaxialloadandbendingmomentinteractionsectioncapacitiesofsteelmembersunderbiaxialbending.Thedesignaxial tensileor compressive force (N*)anddesign


bendingmoments( )Mx∗ and( )My

∗ aboutthemajorandminorprincipalaxesmustsatisfythefollowingcondition:

NN

MM

MMs

x

sx

y

sy

∗+

∗+

∗≤

φ φ φ1 (5.47)

ForcompactdoublysymmetricI-sectionsandrectangularandsquarehollowsectionsunderbiaxialbending,thedesignbendingmomentsmustsatisfythefollowingpowerlawexpres-siongiveninClause8.3.4ofAS4100:

MM

MM

x

rx

y

ry

b b∗

+

∗

≤

φ φ

γ γ

1 (5.48)

whereγbisgivenas

γ

φbs

NN

= +∗≤1 4 2 0. . (5.49)

Whenthereisnoaxialforce(N*=0),thesectionissubjectedtobiaxialbendingmomentsandγb=1.4.Whentheaxialforce(N*)isgreaterthan0.6Ns,theexponentγbistakenas2.0. Figure 5.18 illustrates the strength interaction curves for sections under axial load

00

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4

M*x/φMrx

γb=1.4

γb=2.0

M* y/φM

ry

0.6 0.8 1 1.2

Figure 5.18 Strength interaction curves for compact symmetric I-sections under biaxial bending.


andbiaxialbending.ItisseenthatanysignificantdesignbendingmomentMy*remarkablyreducesthedesignbendingmomentcapacitiesϕMrxofthesection.

5.6.2 member capacity under biaxial bending

Themembercapacityofasteelbeam–columnunderaxialcompressionandbiaxialbendingmomentsdependsonitsin-planeandout-of-planemembermomentcapacities.Clause8.4.5ofAS4100specifiesthatasteelbeam–columnunderaxialcompressionandbiaxialbendingmustsatisfythefollowingstrengthinteractionformula:

MM

MM

x

cx

y

iy

∗

+

∗

≤

φ φ

1 4 1 4

1. .

(5.50)

whereMcx is taken as the lesser of the in-plane member moment capacity ( )Mix and the

nominalout-of-planemembermomentcapacity( )Mox forbendingaboutthemajorprincipalx-axis

Miystandsforthenominalin-planemembermomentcapacityabouttheminorprincipaly-axis

Figure5.19showsthestrengthinteractioncurveforsteelmemberunderaxialcompressionandbiaxialbending.

00

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4

M*x/φMcx

M* y/φM

iy

0.6 0.8 1 1.2

Figure 5.19 Strength interaction curves for compression members under biaxial bending.


Similarly,asnotedinClause8.4.5.2ofAS4100,asteeltensionmembersubjecttobiaxialbendingmustsatisfythefollowinginteractionexpression:

MM

MM

x

tx

y

ry

∗

+

∗

≤

φ φ

1 4 1 4

1. .

(5.51)

whereMtxistakenasthelesserofthenominalsectionmomentcapacity( )Mrx reducedbyaxial

tensionand thenominalout-of-planemembermomentcapacity( )Mox forbendingaboutthemajorprincipalx-axis

Mrydenotesthenominalsectionmomentcapacityreducedbyaxialtensionabouttheminorprincipaly-axis

Example 5.6: Design of a steel beam–column under axial compression and biaxial bending

An8mlengthsteelbeam–columnof200UC59.5Grade300isfixedatitsbaseandpinnedatitstop.Thecolumnisbracedaboutthey–y-axisbystrutsthatarepin-connectedtoitsmid-height.Thestrutsprevent lateraldeflections in theminorprincipalplane.ThecolumnissubjectedtoanaxialcompressiveforceofN*=450kNandbiaxialbendingmoments ofMx

∗ = 85 kNm andMy∗ = 15 kNm, which have been determined from the

second-orderelasticanalysis.Checkthecapacityofthebeam–column.

1. Section properties

Thedimensionsandpropertiesof200UC59.5are

d b t A I

r

f w g x

x

= = = = = ×

=

210 205 9 3 7620 61 3 10

89 7

6mm, mm, mm, mm mm2 4. , .

. mmm, mm mm mm

mm

3 4

3

Z I r

Z

ex y y

ey

= × = × =

= ×

656 10 20 4 10 51 7

299 10

3 6

3

, . , .

, II J f

E G

w y

s

= × = × =

= × = ×

195 10 477 10 300

200 10 80 10

9 3

3

mm mm MPa

MPa,

6 3, ,

33 MPa, = 1.0kf

Compactnessaboutthex-axis=compactCompactnessaboutthey-axis=compact

2. Axial section capacity

Thesectioncapacityinaxialcompressioniscalculatedasfollows:

N k A f

N N

sc f n y

sc

= = × × =

= × = > ∗ =

1 0 7620 300 2286

0 9 2286 2057 4 4

.

. .

N kN

kNφ 550 kN, OK

3. Section moment capacities

3.1. Section moment capacities without axial forces

Forbendingaboutthemajorprincipalx-axis,

M Z f

M

sx ex y

sx

= = × × × =

= × =

−656 10 300 196 8

0 9 196 8 177 12

3 610 kN m

kN

.

. . .φ mm kN m, OK> ∗ =Mx 85


Forbendingaboutthemajorprincipaly-axis,

M Z f

M M

sy ey y

sy y

= = × × × =

= × = >

−299 10 300 89 7

0 9 89 7 80 7

3 610 kN m

kN m

.

. . .φ ∗∗ = 15 kN m, OK

3.2. Reduced section moment capacities due to axial compression

For compact doubly symmetric I-section with kf = 1.0, the section moment capacityreducedbyaxialcompressionforbendingaboutx-axiscanbecalculatedasfollows:

M MNN

Mrx sxs

sx= −∗

≤

= × × −

=

1 18 1

1 18 196 8 1450

2057 4

.

. ..

φ

1181 4 196 8

181 4

. .

.

kNm kNm

kNm

< =

∴ =

M

M

sx

rx

φM Mrx x= × = > ∗ =0 9 181 4 163 3 85. . . kN m kN m, OK

Forbendingabouttheminorprincipaly-axis,thereducedsectionmomentcapacityis

M MNN

Mry sys

sy= −∗

≤

= × × −

1 19 1

1 19 89 7 1450

2057

2

.

. ..

φ

44101 6 89 7

89 7

2

= > =

∴ =

. .

.

kNm kNm

kNm

M

M

sy

ry

φM Mry y= × = > ∗ =0 9 89 7 80 73 15. . . kN m kN m, OK

3.3. Section capacities under biaxial bending

ForcompactdoublysymmetricI-sectionunderbiaxialbending,thesectioncapacityisdeterminedasfollows:

γ

φbs

NN

= +∗= + = <1 4 1 4

4502057 4

1 619 2. ..

.

MM

MM

x

rx

y

ry

b b∗

+

∗

≤

=

+

φ φ

γ γ

1

85163 29

1 619

.

.115

80 730 413 1 0

1 619

.. . ,

.

= < OK


4. Axial member capacities

4.1. Axial member capacity Ncx

Sincetheeffectofendrestraintshasbeentakenintoaccountinthesecond-orderelasticanalysis,theeffectivelengthfactorkeistakenas1.0forcombinedactions.TheeffectivelengthisLe=keL=1.0×8000=8000 mm.

Themodifiedmemberslendernessis

λnx

ex

xf

yLr

kf

= = =250

800089 7

1 0300250

97 7.

. .

Forhot-rolleduniversalcolumn(UC)sectionwithkf=1.0,αb=0.Theslendernessreduc-tionfactorcanbeobtainedfromTable5.3bylinearinterpolationas

αc = −

− −−

=0 5750 575 0 541 97 7 95

100 950 556.

( . . )( . )( )

.

Thedesignaxialcapacityistherefore

φ φαN Ncx c s= = × × =0 9 0 556 2286 1144. . kN

4.2. Axial member capacity Ncy

Theportalframecolumnisbracedbystrutsatitsmid-height,sothattheeffectivelengthistakenas1.0.TheeffectivelengthLe=keL=1.0×4000=4000 mm.

Themodifiedmemberslendernessiscalculatedas

λny

ey

yf

yLr

kf

= = =250

400051 7

1 0300250

84 75.

. .

Forhot-rolledUCsectionwithkf=1.0,αb=0.TheslendernessreductionfactorcanbeobtainedfromTable5.3bylinearinterpolationas

αc = −

− −−

=0 6810 681 0 645 84 75 80

85 800 647.

( . . )( . )( )

.

Thedesignaxialcapacityistherefore

φ φαN Ncy c s= = × × =0 9 0 647 2286 1331. . kN

5. In-plane member moment capacities

Forbendingaboutthemajorprincipalx-axis,thein-planemembermomentcapacitycanbecalculatedas

M M

NN

ix sxcx

= −∗

= × −

=1 196 8 1

4501144

119 4φ

. . kNm

φM Mix x= × = > =∗0 9 119 4 107 5 85. . . kN m kN m, OK


Forbendingabouttheminorprincipaly-axis,thein-planemembermomentcapacitycanbecalculatedas

M M

NN

iy sycy

= −∗

= × −

=1 89 7 1

4501331 5

59 4φ

..

. kNm

φM Miy y= × = > ∗ =0 9 59 4 53 5 15. . . kN m kN m, OK

6. Out-of-plane member capacity

6.1. Member moment capacity without full lateral restraints

Thebeam–columnisbracedbystrutsatitsmid-height.Thelengthoftheupperbeam–columnforflexural–torsionalbucklingistakenasL=4000 mm.

Asthesegmentisfullyrestrainedagainsttwistatbothends,thetwistrestraintfactoriskt= 1.0.

Thesegmentissubjectedtoonlymoments;theloadheightfactoriskl= 1.0.As the segment is restrained against lateral rotation, the lateral rotational restraint

factoristakenaskr= 1.0.Theeffectivelengthistherefore

L k k k Le t l r= = × × × = mm1 0 1 0 1 0 4000 4000. . .

Theelasticbucklingmomentiscalculatedas

ME IL

GJE IL

os y

e

s w

e

= +

=× × × ×

×

π π

π

2

2

2

2

2 3 6

2

200 10 20 4 104000

80.

110 477 10200 10 195 10

4000

396

3 32 3 9

2× × +× × × ×

=

π

kN m

Thememberslendernessreductionfactoriscomputedas

αss

oa

s

oa

MM

MM

=

+ −

=

+ −0 6 3 0 6

196 8396

319

2 2

. .. 66 8

3960 783

..

=

Assume that thebeam–columnabout thex-axis bendingundergoesdouble curvaturebending, having a contraflexure point at the mid-height lateral restraint with zeromoment.Themomentratiooftheupperbeam–columnsegmentis

βm

MM

=∗∗ = = 2

1

085

0

Themomentmodificationfactorisdeterminedas

α β βm m m= + + = 1.75 1.05 0 3 1 752. .


Themembermomentcapacityis

M M M

M

M

bx m s sx sx

sx

bx

= ≤

= × × = > =

∴

α α

1 75 0 783 196 8 269 7 196 8. . . . .kNm kNm

== 196 8. kNm

6.2. Out-of-plane member capacity

Theout-of-planemembermomentcapacityofthecolumnunderaxialcompressionandbendingcanbecalculatedas

M MNN

Mox bxcy

rx= −∗

≤

= × −

=

1

196 8 1450

1331130 3

φ

kN. . mm kN m

kN m

< =

∴ =

M

M

rx

ox

181 4

130 3

.

.

φM Mox x= × = > ∗ =0 9 130 3 85. . 117.3 kN m kN m, OK.

7. Member capacities under biaxial bending

Thein-planeandout-of-planemembermomentcapacitieshavebeencalculatedas

φ φ φM M Mix ox iy= = =107 5 53 5. , .kNm, 117.3 kNm kNm

Thecriticalmembermomentcapacityaboutthex-axisistakenas

φ φ φM M Mcx ix ox= = =min( , ) min( . , 117.3) 107.5 kN m107 5

Thebiaxialmembercapacitiesarecheckedasfollows:

MM

MM

x

cx

y

iy

∗

+

∗

≤

=

+

φ φ

1 4 1 4

1 4

1

85107 5

1

. .

.

.55

53 50 888 1 0

1 4

.. . ,

.

= < OK

Therefore, the capacity of the beam–column under axial load and biaxial bending isadequate.

referenceS

AS4100(1998)AustralianStandardforSteelStructures,Sydney,NewSouthWales,Australia:StandardsAustralia.

Bennetts,I.D.,Thomas,I.R.,andHogan,T.J.(1986)Designofstaticallyloadedtensionmembers,CivilEngineeringTransactions,InstitutionofEngineersAustralia,28(4),318–327.

Bradford, M.A., Bridge, R.Q., Hancock, G.J., Rotter, J.M., and Trahair, N.S. (1987) Australianlimitstatedesignrulesforthestabilityofsteelstructures,PaperpresentedattheInternationalConferenceonSteelandAluminiumStructures,Cardiff,UK,pp.11–23.


Bridge,R.Q.andTrahair,N.S.(1987)Limitstatedesignrulesforsteelbeam-columns,SteelConstruction,AustralianInstituteofSteelConstruction,21(2):2–11.

Bulson,P.S.(1970)TheStabilityofFlatPlates,London,U.K.:ChattoandWindus.Cochrane,V.H.(1922)Rulesforrivetholedeductionsintensionmembers,EngineeringNews-Record,

89(16):847–848.Cuk,P.E.,Bradford,M.A.,andTrahair,N.S.(1986)Inelasticlateralbucklingofsteelbeam-columns,

CanadianJournalofCivilEngineering,13(6):693–699.Davids,A.J.andHancock,G.J.(1985)Thestrengthoflong-lengthI-sectioncolumnsfabricatedfrom

slenderplates,CivilEngineeringTransactions,InstitutionofEngineersAustralia,27(4):347–352.Dhalla,A.K.andWinter,G.(1974a)Steelductilitymeasurements,JournaloftheStructuralDivision,

ASCE,100(ST2):427–444.Dhalla,A.K.andWinter,G.(1974b)Suggestedsteelductilityrequirements,JournaloftheStructural

Division,ASCE,100(ST2):445–462.Duan, L. and Chen, W.F. (1988) Effective length factor for columns in braced frames, Journal of

StructuralEngineering,ASCE,114(10):2357–2370.Duan, L. and Chen, W.F. (1989) Effective length factor for columns in unbraced frames, Journal

of StructuralEngineering,ASCE,115(1):149–165.Galambos,T.V.(ed.)(1988)GuidetoStabilityDesignCriteriaforMetalStructures,4thedn.,New York:

JohnWiley&Sons.Hancock,G.J. (1982)Designmethods for interactionbuckling inboxand I-section columns,Civil

EngineeringTransactions,InstitutionofEngineersAustralia,24(2):183–186.Hancock,G.J.,Davids,A.J.,Keys,P.W.,andRasmussen,K.(1987)Strengthtestsonthin-walledhigh

tensilesteelcolumns,PaperpresentedattheInternationalConferenceonSteelandAluminiumStructures,Cardiff,UK,pp.475–486.

Key,P.W.,Hasan,S.W.,andHancock,G.J.(1988)Columnbehaviourofcold-formedhollowsections,JournalofStructuralEngineering,ASCE,114(ST2):390–407.

McGuire,W.(1968)SteelStructures,EnglewoodCliffs,NJ:PrenticeHall.Rasmussen,K.J.R.andHancock,G.J.(1989)Compressiontestsofweldedchannelsectioncolumns,

JournalofStructuralEngineering,ASCE,115(ST4):789–808.Rasmussen,K.J.R.,Hancock,G.J.,andDavids,A.J.(1989)Limitstatedesignofcolumnsfabricated

fromslenderplates,CivilEngineeringTransactions, InstitutionofEngineersAustralia,27 (3):268–274.

Rotter,J.M.(1982)Multiplecolumncurvesbymodifyingfactors,JournaloftheStructuralDivision,ASCE,108(ST7):1665–1669.

Timoshenko,S.P.andGere,J.M.(1961)TheoryofElasticStability,2ndedn.,NewYork:McGraw-Hill.Trahair,N.S.(1986)Designstrengthsofsteelbeam-columns,CanadianJournalofCivilEngineering,

13(6):639–646.Trahair,N.S.andBradford,M.A.(1998)TheBehaviourandDesignofSteelStructurestoAS4100,3rd

edn.,London,U.K.:Taylor&FrancisGroup.Woolcock,S.T.andKitipornchai,S.(1986)Designofsingleanglewebstrutsintrusses,Journalofthe

StructuralDivision,ASCE,112(6):1665–1669.Woolcock, S.T., Kitipornchai, S., and Bradford, M.A. (2003) Limit State Design of Portal Frame

Buildings,Sydney,NewSouthWales,Australia:AustralianInstituteofSteelConstruction.

149

Chapter 6

Steel connections

6.1 IntroductIon

Structuralconnectionsareusedtoconnectastructuralmembertoanothermemberortothesupportsothattheforcescarriedbythestructuralmembercanbetransferredtotheothermemberortothesupport.Asteelconnectionconsistsofconnectioncomponentsandconnectors.Cleats,gussetplates,bracketsandconnectingplatesusedinsteelconnectionsarecalledconnectioncomponents,whilebolts,weldsandpinsareconnectors.Membersarejoinedtogetherinaconnectionthatconsistsofseveralelements,whichresultsincomplexstress distributions within the connection. Connections in a steel structure may becomepotentialweakspotsthatneedcarefulconsiderationsinthedesign.Structuralconnectionsare importantpartsofa steel structure that influence theoverallperformanceofa steelstructure.

Thischapterdealswiththebehaviouranddesignofstructuralsteelconnectionsinaccor-dancewithAS4100(1998).Thebehaviouranddesignofboltsandboltgroupsundershear,tensionandcombinedshearandtensionarediscussed.Thedesignsofweldsandweldgroupsunderin-planeandout-of-planedesignactionsarealsogiven.Oneoftheemphasesofthischapter isplacedon thedesignofboltedmoment endplate connections,which includesbeamnormaltocolumnconnections,kneeconnectionsandridgeconnectionsinrigidsteelconstruction.Thedesignprinciplespresentedforboltedmomentendplateconnectionscanbeextendedtothedesignofweldedbeam-to-columnmomentconnections.Anotherempha-sisisonthebehaviouranddesignofpinnedcolumnbaseplateconnections.Designproce-duresofstructuralsteelconnectionsareillustratedthroughworkedexamples.

6.2 tyPeS of connectIonS

Steelconnectionsmaybeclassifiedbytheamountofrotationalrestraintprovidedbytheconnections,whicharerelatedtothetypeofsteelframes.Steelconnectionsareusuallyclas-sifiedintorigid,simpleandsemi-rigidconnections.

Rigid connections provide full continuity at the connections which hold the anglesbetweenintersectingmembersunchangedafterdeformations.Thisrequiresthattherigidconnectionneedstohavetherotationalrestraintequaltoorgreaterthan90%ofthatneces-sarytopreventanyanglechangebetweentheintersectingmembers.Itisassumedthatthedeformationsofrigidconnectionshavenosignificanteffectsonthedistributionofdesignactionsorontheoveralldeformationoftheframe.Rigidconnectionsareusedtotransferthedesignactionsofbendingmoment, shear forceandaxial force fromonemember toanother in steel rigid frames.Typical examplesof rigid connections areweldedmoment


connectionsandboltedmomentendplateconnectionsdepicted inFigure6.1andboltedsplicesillustratedinFigure6.2.

Simpleconnectionsprovidelittlerotationalrestraintattheendsofamembersothattheendsofthemembercanrotateunderappliedloads.Insimpleconnections,thechangeintheoriginalanglebetweenintersectingmembersis80%ormoreofthatcausedbytheuse

(d)(c)

(b)(a)

Beam

Fillet weldStiffener

Stiffener

Column

End plate

Haunch

Fillet weld

RafterColumn

Stiffener

Beam

Fillet weld

Fillet weld

End plate

ColumnBeam

Fillet weld

End plate

Stiffener

Stiffener

Figure 6.1 Rigid connections: (a) welded moment connection, (b) bolted moment end plate connection, (c) knee joint and (d) ridge connection.

Plate

Plate

Plate

Figure 6.2 Splice connection.

Steel connections 151

offrictionlesshingedconnections.Simpleconnectionsaredesignedtotransfershearforceonlyfromonemembertoanotherinasimpleframingsystem.Somestandardsimplecon-nectionsaredepictedinFigures6.3and6.4,includingangleseat,bearingpad,flexibleendplate,anglecleat,beam-to-columnandbeam-to-beamwebsideplateconnections.Insimpleconstruction,simpleconnectionsmustbedesignedtonotonlywithstandthereactionsfrom

ColumnBeam

Angle

(d) (c)

End plate

Beam

Column

(a)

Angle seat

Cleat

CleatColumn Column

BeamPacker

Bearing pad

(b)

End plate

Figure 6.3 Flexible connections: (a) angle seat connection, (b) bearing pad connection, (c) flexible end plate connection and (d) angle cleat connection.

(a) (b)

PlatePlate

Column

Beam Beam Beam

Figure 6.4 Flexible web side plate connections: (a) beam-to-column connection and (b) beam-to-beam connection.


thesimplysupportedbeamsandthefactoredlateralloadsbutalsohavesufficientinelasticrotationcapacitytoallowanglechangesbetweenintersectingmembers.

Semi-rigidconnectionsprovidesomedegreesofrotationalrestraintattheendsofamem-berso that theconnectionscan transferbendingmomentsandshearandaxial forces insemi-rigidsteelframes.Therotationalstiffnessofsemi-rigidconnectionsisbetweenthatofrigidconnectionsandsimpleconnections.

6.3 mInImum deSIgn ActIonS

Clause9.1.4ofAS4100(1998)requiresthatsteelconnectionsbedesignedtotransmitthegreaterofthedesignactioninthememberortheminimumdesignactiongivenasfollows:

1.Theminimumdesignbendingmoment( )minM∗ forthedesignofarigidconnectionistakenas0.5ϕMb.

2.Theminimumdesignshearforce( )minV∗ forsimpleconnectionstoabeamistakenasthelesserof0.15ϕVvand40kN,whereϕVvisthememberdesignshearcapacity.

3.Theminimumdesignaxialforce( )minN∗ forconnectionsattheendsoftensileorcompres-sionmembersis0.3timesthememberdesigncapacity.

4.Theminimumtensileforceforthreadedrodbracingmemberwithturnbucklesistakenasthememberdesigncapacity.

Theminimumdesignactionsfordesigningspliceconnectionsintensionmembers,compres-sionmembers,flexuralmembersandmembersundercombinedactionsarealsospecifiedinClause9.1.4.ofAS4100asfollows:

1.Theminimumdesignforceforsplicesintensionmembersistakenas0.3ϕNt,whereϕNtisthememberdesigncapacityinaxialtension.

2.Splices in axial compression members prepared for full contact at their ends mustcarrythecompressiveactionsbybearingoncontactsurfaces.

3.Theminimumdesignforceforfasteners inthesplices is0.15ϕNc,whereϕNc is thememberdesigncapacityinaxialcompression.

4.Theminimumdesignforcesforspliceconnectionsincompressionmembersthatarenotpreparedforfullcontactis0.3ϕNc.

5.Spliceconnectionsbetweenpointsofeffectivelateralsupportsunderaxialcompres-sionmustbedesignedforcombinedactionsofaxialcompressionandbendingmomenttakingasM*=δmN*ls/1000,whereδmistheamplificationfactorandlsisthedistancebetweenpointsofeffectivelateralsupports.

6.TheminimumdesignbendingmomentforspliceconnectionsinflexuralmembersisM Mbmin . .∗ = 0 3φ

7.Thespliceconnectionsinmembersundercombinedactionsmustsatisfyallminimumdesignactionrequirementsformembersundersingleactionasdescribedearlier.

6.4 Bolted connectIonS

6.4.1 types of bolts

ThetypesofboltsusedinsteelconnectionsincludePropertyClass4.6commercialbolts,PropertyClass8.8high-strength structuralbolts andPropertyClass8.8,10.9and12.9precisionbolts.PropertyClass4.6commercialboltsconformingtoAS1111aremadeof


low-carbonsteel.Theyareusedonlyforsnug-tightinstallationdesignatedas4.6/Sbolts.PropertyClass8.8high-strengthstructuralboltsconformingtoAS/NZS1252aremadeofmediumcarbonsteel.Theirpropertiesareenhancedbyquenchingandtempering.Class8.8high-strengthstructuralboltscanbehighlytensionedandareusedforsnug-tightinstalla-tiondesignatedas8.8/S.Thesehigh-strengthstructuralboltsaredesignatedas8.8/TBwhenusedinbearingmodeconnectionsandas8.8/TFwhenusedinfrictionmodeconnections.PropertyClass8.8,10.9and12.9precisionboltsareusedformechanicalassembly.TheminimumyieldstressofPropertyClass4.6boltsis240MPa,whiletheirminimumtensilestrength is400MPa.PropertyClass8.8high-strength structuralboltshaveaminimumyieldstressof660MPaandaminimumtensilestrengthof830MPa.

6.4.2 Bolts in shear

BoltsinsteelconnectionsaresubjectedtoshearandbearingasdepictedinFigure6.5.Theshearstrengthsofboltscanbedeterminedbyexperimentsinwhichboltsaresubjectedtodoubleshearcausedbyplateseitherintensionorcompression.Testdataindicatedthattheaverageshearstrengthwasabout62%ofthetensilestrengthofthebolt(Kulaketal.1987).Inaddition,itwasfoundthatthelevelofinitialtensionappliedtotheboltdoesnothaveasignificanteffectontheultimateshearstrengthofthebolt.Theshearstrengthofaboltalsodependsontheshearareaofthebolt,thenumberofshearplanesandthelengthofthejoint.Thetotalstrengthofaboltedlapspliceconnectionwasfoundtodecreasewithanincreaseinthelengthoftheconnection.InAS4100,areductionfactorisusedtoaccountfortheeffectofthelengthoftheboltedlapconnectionsontheshearstrengthofthebolts.

ThenominalshearstrengthofaboltiscalculatedbythefollowingequationprovidedinClause9.3.2.1ofAS4100(1998)asfollows:

V f k n A n Af uf rc n c x o= +0 62. ( ) (6.1)

wherefuf standsfortheminimumtensilestrengthoftheboltkrcdenotesthereductionfactoraccountingfortheeffectofthelengthofaboltedlap

connection

Thefactorkrc is takenas1.0fortheconnectionlength(lj) lessthan300 mm,0.75forlj > 1300mmand(1.075− lj/4000)for300≤lj≤1300mm(McGuire1968;Kulaket al.1987). InEquation6.1,nn is thenumberofshearplaneswiththreads interceptingthe

Bearing stress

Shear stress

Figure 6.5 Bolt in shear and bearing.


shear plane,Ac is the core areaof the bolt,nx is the numberof shear planeswithoutthreads intercepting the shear plane, and Ao is the plain shank area of the bolt. Thenominaldiameters(df)ofcommonlyusedboltsvaryfrom12to36 mm.Thecore,shankand tensile stress areas of bolts are given inTable 6.1.BasedonAS1275 (1985), thetensilestressarea iscalculatedasAs=π(df − 0.9382p)2/4,wherep is thethreadpitch.ThecoreareaiscalculatedasAc = π(df − 1.0825p)2/4,andtheshankareaiscomputedasA do f= π 2 4/ .

Aboltunderadesignshearforce( )Vf∗ mustsatisfythefollowingstrengthrequirement:

V Vf f∗ ≤ φ (6.2)

wherethecapacityreductionfactorϕ=0.8.Thedesigncapacitiesof4.6/Sboltsandof8.8/Sand8.8/TBboltsinsinglesheararegiveninTables6.2and6.3,respectively.ThevalueinbracketinTable6.3forM208.8boltisthecurrentlyuseddesignvalue.

Table 6.1 Geometric properties of bolts

Nominal diameter df (mm) Thread pitch p (mm) Tensile stress area As (mm2) Core area Ac (mm2) Shank area Ao (mm2)

12 1.75 84.3 80.2 113.116 2 156.7 150.3 201.120 2.5 244.8 234.9 314.224 3 352.5 338.2 452.430 3.5 560.6 539.6 706.936 4 816.7 787.7 1017.9

Table 6.3 Design capacities of 8.8/S and 8.8/TB bolts

Nominal diameter df (mm) Axial tension ϕNtf (kN)

Single shear ϕVf (kN)

Threads included Threads excluded

16 104.0 61.9 82.820 162.5 96.7 (92.6) 129.324 234.1 139.2 186.230 372.2 222.1 291.036 542.3 324.3 419.0

Table 6.2 Design capacities of 4.6/S bolts

Nominal diameter df (mm) Axial tension ϕNtf (kN)

Single shear ϕVf (kN)

Threads included Threads excluded

12 27.0 15.9 22.416 50.1 29.8 39.920 78.3 46.6 62.324 112.8 67.1 89.830 179.4 107.1 140.236 261.3 156.3 202


Forfriction-typeconnectionssuchas8.8/TFcategorybolts,theslipneedstobelimitedunder the serviceability loads.Connectionswhere slip theoretically exceeds2–3 mmareclassifiedasslipcriticalandneedtobedesignedforserviceabilitylimitstate(Fisheretal.1978;Galambosetal.1982;Birkemoe1983).InClause9.3.3.1ofAS4100,thenominalshearcapacityofaboltunderserviceloadisgivenby

V n N ksf ei ti h= µ (6.3)

whereμistheslipfactorneidenotesthenumberofeffectiveinterfacesNtiistheminimumbolttensionatinstallationkhisthefactoraccountingfortheeffectofdifferentholetypesandistakenas1.0for

standardholes,0.85forshortslottedandoversizeholesand0.7forlongslottedholes

Ifsurfacesincontactarecleanasrolledsurfaces,theslipfactoristakenas0.35(Kulaket al.1987).

Thedesignrequirementofboltssubjectedtoadesignshearforcefortheserviceabilitylimitstateis

V Vsf sf∗ ≤ φ (6.4)

6.4.3 Bolts in tension

Thestrengthofaboltinaxialtensionisgovernedbythethreadedpartofthebolt.Beforesubjectedtotheappliedaxialtensileforce,theboltisusuallytightenedbyturningthenut.However,thisdoesnothaveasignificanteffectonthetensilestrengthofthebolt(Kulaket al.1987).Inaddition,itwasfoundthattensionedboltscanwithstanddirectaxialtensileforceswithoutanysignificantreductionintheirtensilestrength.

The nominal tensile capacity of a bolt can be determined in accordance with Clause9.3.2.2ofAS4100asfollows:

N A ftf s uf= (6.5)

whereAsisthetensilestressareaofaboltasgiveninAS1275andTable6.1.Thedesignofaboltinaxialtensionmustsatisfy

N Ntf tf∗ ≤ φ (6.6)

whereNtf∗isthedesigntensionforceandthecapacityreductionfactorϕ=0.8.Thedesign

capacities of 4.6/S and 8.8/S and 8.8/TB bolts in axial tension are given in Tables 6.2and 6.3,respectively.


6.4.4 Bolts in combined shear and tension

Foraboltsubjecttocombinedshearandtension,aninteractionrelationshipbasedonexper-imentalresults(Kulaketal.1987)isusedtodeterminetheultimatestrengthoftheboltasspecifiedinClause9.3.2.3ofAS4100:

VV

NN

f

f

tf

tf

∗

+

∗

≤

φ φ

2 2

1 0. (6.7)

whereφVf denotesthedesignshearcapacityoftheboltundershearforcealoneφNtf isthedesigntensilecapacityoftheboltsubjecttotensionforcealoneφ = 0 8.

Theslipoffriction-typeconnectionssubjectedtocombinedserviceloadsofshearandten-sionisrequiredtobelimitedfortheserviceabilitylimitstate.Forthispurpose,aboltundercombinedshearandtensionmustsatisfythefollowinglinearinteractionequation(ResearchCouncilonStructuralConnections1988)giveninClause9.3.3.3ofAS4100:

VV

NN

sf

sf

tf

tf

∗

+

∗

≤φ φ

1 0. (6.8)

whereVsf∗denotesthedesignshearforceactingontheboltintheplaneoftheinterfaceandstandsforthedesigntensionforceactingonthebolt

Vsf isthedesignshearcapacityoftheboltgiveninEquation6.1Ntf isthenominaltensilecapacityoftheboltandistakenastheminimumbolttension

atinstallation( )Nti

φ = 0 7.

6.4.5 Ply in bearing

Inaboltedconnectionundershearforce,theconnectionplate(ply)issubjectedtobearingduetoboltsinshearasillustratedinFigure6.6.Thelocalbearingfailureoftheplyoccursatabearingstressbetween4.5fypand4.9fyp(HoganandThomas1979a;Kulaketal.1987).

Bearing stress

Tear-out failure

ae

Figure 6.6 Bearing and tear-out of ply.


ThedesignequationgiveninClause9.3.2.4ofAS4100forcalculatingthenominalbearingcapacityofaplyduetoaboltinshearisbasedonthelowerbearingstressof4.5fypandisexpressedby

V d t fbp f p up= 3 2. (6.9)

wheretpisthethicknessoftheplyfupisthetensilestrengthoftheply

ForaplysubjectedtoaforceactingtowardsanedgeasshowninFigure6.6,thebearingortearingfailuremayoccur.Thestrengthofaplyinbearingmaybelimitedbythebearingortearingfailure.Thetearingfailureisusuallymorecriticalthanthebearingfailurewhentheenddistance(ae)measuredfromthecentreoftheboltholetotheedgeoftheplyinthedirectionoftheforceislessthan3.2df.AsspecifiedinClause9.3.2.4ofAS4100,thetear-outcapacityoftheplyisdeterminedas(Kulaketal.1987)

V a t ftp e p up= (6.10)

Thenominalbearingcapacity(Vfb)oftheplysubjectedtoaforcetowardsanedgeshouldbetakenasthelesserofVbpandVtp.Aplysubjectedtoadesignbearingforce( )Vb

∗ duetoaboltinshearmustsatisfythefollowingcondition:

V Vb fb∗ ≤ φ (6.11)

whereϕ=0.9isthecapacityreductionfactor.

6.4.6 design of bolt groups

6.4.6.1 Bolt groups under in-plane loading

Forboltgroupsubjectedtoin-planeloading,theelasticanalysiscanbeusedtodeterminethedesignactionsinaboltgroup,providedthattheassumptionsgiveninClause9.4.1ofAS4100aresatisfied.Theseassumptionsare:(a)theconnectionplatesmustberigid;(b)theconnectionplatesrotateabouttheinstantaneouscentreoftheboltgroup;(c) foraboltgroupsubjectedtoapurecouple,instantaneouscentreofrotationislocatedatthecentroidoftheboltgroup;(d)thesuperpositionmethodcanbeused;and(e)thedesignshearforceineachboltactsatrightangletotheradiusfromtheinstantaneouscentretothebolt.

Assumingthecross-sectionalareaofeachboltinagroupisunityandallboltshavethesamesize,thesecondmomentsofareaofaboltgroupcanbecomputedbythefollowing:

I yx n=∑ 2 (6.12)

I xy n=∑ 2 (6.13)


I I Ip x y= + (6.14)

whereIxandIyarethesecondmomentsofareaoftheboltgroupaboutitscentroidalaxisIprepresentsthepolarsecondmomentofareaoftheboltsinthegroupxnandynarethecoordinatesofabolt

ItisassumedthatthehorizontalforceVx∗andverticalforceVy

∗appliedtoaboltgroupaspresentedinFigure6.7aareequallysharedbyallboltsinthegroup.Theforcesonanyboltinthegroupcanbedeterminedby

V

Vn

xbx

b

∗ =∗

(6.15)

V

Vn

yby

b

∗ =∗

(6.16)

wherenbisthetotalnumberofboltsintheboltgroup.TheboltforceduetothedesignbendingmomentMz

∗aboutthecentroidoftheboltgroupis proportional to the distance from the centroid of the bolt group. The maximum boltforcesinxandydirectionsduetoMz

∗occuratthefarthestboltfromthecentroidoftheboltgroup(Thomasetal.1985;HoganandThomas1994)andaredeterminedby

V

M yI

xbmz

p

∗ =∗

max (6.17)

V

M xI

ybmz

p

∗ =∗

max (6.18)

inwhichxmaxandymaxarethedistancesfromtheboltgroupcentroidtothefarthestcornerbolt.

(a)

x

y

Vx*

Mx*Mz*

Nz*

Vy*

T1

T3

C

T2

(b)

Figure 6.7 Bolt groups: (a) in-plane actions and (b) out-of-plane actions.


Theresultantdesignshearforceontheboltlocatedfarthestawayfromthecentreoftheboltgroupcanbedeterminedas

V V V V Vres xb xbm yb ybm∗ = ∗ + ∗( ) + ∗ + ∗( )2 2

(6.19)

Theresultantdesignshearforce( )Vres∗ onthefarthestboltmustbelessthanthedesignshear

capacity(ϕVf)oftheboltandthebearingcapacity(ϕVfb)oftheply.

6.4.6.2 Bolt groups under out-of-plane loading

Foraboltgroupsubjectedtotheout-of-planeactionsasdepictedinFigure6.7b,theforcesintensionboltscanbedeterminedbyassumingalineardistributionofforcefromtheneu-tralaxistothefarthestbolts.Themethodsofanalysisforboltgroupsunderout-of-planeactionsaregivenbyMcGuire(1968),Kulaketal.(1987),AISC-LRFDManual(1994)andHoganandThomas(1994).Theneutralaxiscanbeassumedtobeplacedatthed/6fromthebottomoftheendplateofadepthd(Gorencetal.2005).Thetensionforceonaboltcanbecalculatedbythefollowingequation(TrahairandBradford1998):

TNn

M y

yi

z

b

x i

i

=∗+

∗

∑ 2 (6.20)

whereyiisthecoordinateoftheboltfromthecentroidoftheboltintheydirection.ThetensionforceineachofthecriticallyloadedboltisN T ntf b

∗ = 1 1/ ,wherenb1isthenum-berofboltsinthetoprow.Thedesignshearforce( )Vo

∗ ontheboltgroupisassumedtobeequallysharedbyallbolts.Therefore,thedesignshearforceoneachboltisV V nf o b

∗ = ∗ / .ThecapacityoftheboltundercombinedshearandtensioncanbecheckedusingEquation6.7.

Example 6.1: Capacity of bolted splice connection in tension

Figure6.8showsaboltedspliceconnectionindoublesheararrangementunderadesignaxialtensionforceofN*=850kN.Grade300steelandM208.8/Sboltsareused.Checkthedesigncapacityofthisboltedspliceconnection.

1. Design capacity of steel member

TheyieldstressandtensilestrengthofthememberandspliceplatesectionsareobtainedfromTable2.1ofAS4100asfy=300MPaandfu=430MPa,respectively.

2 × 200 × 18 splice plates

200

M20 8.8/S bolts35 3570

Steel member

220N* N*

Figure 6.8 Bolted splice connection.


Thegrosscross-sectionalareaofthesteelmemberis

Ag = × =220 20 4400 mm2

Thenetcross-sectionalareaofthespliceplateis

An = − × × =4400 2 24 20 3440 mm2

Theconnectionissymmetricsothatktc=1.0.Thefracturecapacityofthesteelmemberis

φ φN k A fta tc n u= = × × × × × =−0 85 0 9 0 85 1 0 3440 430 10 1131 63. . . . . kN

Thegrossyieldcapacityofthesteelmemberis

φ φN A fty g y= = × × × =−0 9 4400 300 10 11883. kN

Thus,ϕNt=min(1131.6;1188)=1131.6kN>N*=850kN, OKTheminimumdesignaxialtensionforceis

N N Ntmin . ( ) . . .∗ ∗= = × = < =0 3 0 3 1131 6 339 5 850φ kN kN

Therefore,thedesigntensionforceN*=850kNisusedinthedesignoftheconnection.

2. Design capacity of splice plate

Thegrosscross-sectionalareaofthesteelmemberis

Ag = × =200 18 3600 mm2

Thenetcross-sectionalareaofthespliceplateis

An = − × × =3600 2 24 18 2736 mm2

Theconnectionissymmetricsothatktc=1.0.Thefracturecapacityofthesteelspliceplateis

φ φN k A fta tc n u= = × × × × × =−0 85 0 9 0 85 1 0 2736 430 10 9003. . . . kN

Thegrossyieldcapacityofthesteelmemberis

φ φN A fty g y= = × × × =−0 9 3600 300 10 9723. kN

Thus,ϕNt=min(900;972)=900kN>N*=850kN, OK

3. Shear capacity of bolts

ThecoreandshankareasofaM20boltareobtainedfromTable6.1as

A Ac o= =234.9 mm 314.2 mm2 2,


Thedesigncapacityofaboltindoublesheariscomputedas

φ φV f k n A n Af uf rc n c x o= + = × × × × × + ×0 62 0 8 0 62 830 1 0 1 1. ( ) . . . ( 234.9 314..2 N kN) = 226

Thedesignshearcapacityof4bolts:4×226=904kN>N* =850kN, OK

4. Bearing capacity of connection plate

Thedesigntear-outcapacityofaplycanbecalculatedas

φ φV a t ftp e p up= = × × × × =−0 9 35 20 430 10 270 93. . kN

Thetotaldesigncapacityof4boltsinbearingis

4 4 270 9 1083 6 850φV Ntp = × = > ∗ =. . kN kN, OK

Thedesignbearingcapacityofthespliceplateduetoaboltinshearis

φ φV d t fbp f p up= = × × × × × =−3 2 0 9 3 2 20 20 430 10 495 43. . . . kN

Thetotaldesignbearingcapacityofthespliceplatedueto4boltsinshearis

4 4 495 4 1981 4 850φV Nbp = × = > ∗ =. . kN kN, OK

6.5 Welded connectIonS

6.5.1 types of welds

Welding isused inthefabricationofsteelsections,connectionsandmembersand intheattachmentofstiffeners.Thetypesofweldsusedinsteelconnectionsincludebutt,filletandcompoundwelds.Fromthestrengthconsideration,buttweldsarepreferable.Buttwelds,however,requirecarefulpreparationsoftheplatesforweldingandarehencecostly.Incon-trast,filletweldsrequireonlyminimalweldpreparationsinvolvingastraightforwardweld-ingprocess,whichmakesthemlesscostly.Compoundweldsconsistofbuttandfilletweldsandareusedtoprovideasmoothertransitionwhichreducesthestressconcentrations.Theweldqualitiesorcategories,whichareameasureofthepermittedlevelofdefectspresentondepositedwelds,areusuallyclassifiedintostructuralpurpose(SP)andgeneralpurpose(GP).SPweldcategoryisusedforhighlystressedwelds,whileGPweldcategoryisforlowlystressedweldsandnon-structuralwelds.

6.5.2 Butt welds

Buttweldscanbedividedintotwogroups,namely,completepenetrationbuttweldsandincompletepenetrationbuttwelds.AcompletepenetrationbuttweldhasfusionbetweentheweldandparentmetalthroughoutthecompletedepthofthejointasdepictedinFigure 6.9.Anincompletepenetrationbuttweldhasfusionbetweentheweldandparentmetaloverpartofthedepthofthejoint.

AsspecifiedinClause9.7.2.7ofAS4100,thedesigncapacityofacompletepenetrationbuttweldcanbetakenasthedesigncapacityoftheweakerpartofthepartsjoined,where


thecapacityreductionfactor(ϕ)istakenas0.9forSPcategoryweldsand0.6forGPcat-egorywelds.Thedesigncapacityofanincompletepenetrationbuttweldisdeterminedasforafilletweld.

6.5.3 fillet welds

Thefailureplaneofafilletweldmaybesubjectedtoresultantforcesincludingshearforceparallel to the longitudinal axis of theweld, shear forceperpendicular to the longitudi-nalaxisoftheweldandnormalcompressionortensileforcetothetheoreticalplane.Itisassumedthatthenormalorshearstressesonthefailureplaneareuniformlydistributed.Thecapacityofafilletweldisdeterminedbythenominalshearcapacityacrosstheweldthroator failureplane.Thenominal capacityof afilletweldperunit length is given inClause9.7.3.10ofAS4100asfollows:

v f t kw uw t rw= 0 6. (6.21)

inwhichfuwisthetensilestrengthofweldmetal,whichis410MPaforE41XXweldsand480MPaforE48XXwelds.Thedesignthroatthickness(tt)istakenas0.707Dw(Dwistheleglengthofthefilletweld).Thereductionfactorkrw,whichaccountsforeffectofthelength(lw)ofaweldedlapconnection,istakenasfollows:

• krw = 1 0. forlw ≤ 1 7. m• k lrw w= −1 10 0 06. . for1 7 8 0. .< ≤lw• krw = 0 62. forlw > 8m

Thefilletweldsubjectedtoadesignforceperunitlengthofweld( )vw∗ mustsatisfy

v vw w∗ ≤ φ (6.22)

wherethecapacityreductionfactorϕis0.8forSPcategorywelds,0.6forGPcategoryweldsand0.7forSPcategorylongitudinalweldstorectangularhollowsectionswithwallthick-nesslessthan3 mm.Thedesignforce( )vw∗ isthevectorresultantofallforcesactingonthefilletweld.Thedesigncapacitiesofequal-legfilletweldsaregiveninTable6.4.

Thedesigncapacityofanincompletebuttweldisdeterminedasthatofthefilletweldbytakingkrw=1.0.Thedesignofcompoundweldshouldsatisfythestrengthrequirementofabuttweld.

(a) (b) (c)

Figure 6.9 Butt and fillet welds: (a) complete penetration butt weld, (b) incomplete penetration butt weld and (c) fillet weld.


6.5.4 Weld groups

6.5.4.1 Weld group under in-plane actions

Weldedconnectionsmaybesubjectedtoin-planeactionsofforcesandbendingmoment.Weldgroupsintheconnectionneedtobedesignedtoresistthesein-planeactions.Tosim-plify theanalysisofweldgroups, the followingassumptionsaremade: (a) theweldsaretreatedashomogeneous,isotropicandelasticelements,(b)theplatebeingweldedisrigidintheplaneoftheweldgroupand(c)theeffectsofresidualstressesandstressconcentrationareignored(Swannell1979;HoganandThomas1994,1979b).

Figure6.10ashowsthein-planedesignforcesandbendingmomentactingontheweldgroup.Theforcesactatthecentroidoftheweldgroupwiththedesignbendingmomentaboutthecentroid.Theforcesperunitlengthintheweldsegmentfarthestfromthecentroidoftheweldgroupcanbecalculatedasfollows:

v

VL

M yI

xx

w

z

wp

∗ =∗−

∗max (6.23)

v

VL

M xI

yy

w

z

wp

∗ =∗+

∗max (6.24)

wherexmaxandymaxarethecoordinatesoftheweldsegmentlocatedfarthestfromthecentroid

oftheweldgroupLwisthetotallengthoftheweldintheweldgroup,andthepolarsecondmomentofarea

oftheweldgroupisgivenby

I x l y lwp i iw i iw= +( )∑ 2 2 (6.25)

wherexiandyiarethecoordinatesoftheithweldsegmentliwisthelengthoftheithweldsegment

table 6.4 Design capacities of fillet welds

Leg size Dw (mm)

φφvw (kN/mm)

Category SP Category GP

E41XX E48XX E41XX E48XX

4 0.557 0.652 0.417 0.4895 0.696 0.814 0.522 0.6116 0.835 0.977 0.626 0.7338 1.113 1.303 0.835 0.977

10 1.391 1.629 1.044 1.22212 1.670 1.955 1.252 1.466


The resultant forceperunit lengthactingon themost critically loadedpartof theweldgroupmustsatisfy

v v v vres x y w∗ = ∗( ) + ∗( ) ≤

2 2φ (6.26)

6.5.4.2 Weld group under out-of-plane actions

Theout-of-planedesignforcesandbendingmomentactingontheweldgrouparedepictedinFigure6.10b.Thesameassumptionsusedforweldgroupssubjectedtoin-planeactionsareadoptedfortheanalysisofweldgroupsunderout-of-planeactions.TheweldgroupissubjectedtoadesignbendingmomentMx

∗aboutthecentroid.Theforcesperunitlengthintheweldsegmentfarthestfromthecentroidoftheweldgroup

subjectedtotheout-of-planedesignactionsofNz∗andMx

∗canbecalculatedasfollows:

v

NL

M yI

zz

w

x

wx

∗ =∗+

∗max (6.27)

whereIwxisthesecondmomentofareaoftheweldgroupaboutthex-axisoftheweldgroupandisexpressedby

I y lwx i iw=∑ 2 (6.28)

Theweldgroupmayalsobesubjectedtoanin-planedesignshearforceVy∗.Theshearper

unitlengthintheweldsegmentfarthestfromthecentroidoftheweldgroupisgivenby

v

VL

yy

w

∗ =∗

(6.29)

The resultant forceperunit lengthactingon themost critically loadedpartof theweldgroupundercombinedin-planeandout-of-planedesignactionsmustsatisfy

v v v vres y z w∗ = ∗( ) + ∗( ) ≤

2 2φ (6.30)

(a) (b)

Mx*

Nz*

y

xVx*

Mz*

Vy*

Figure 6.10 Weld groups: (a) in-plane actions and (b) out-of-plane actions.


ForaweldgroupwithweldsaroundtheperimeterofasteelI-section,theweldgroupcanbedividedintosubgroupstosimplifytheanalysisoftheweldgroup.ItisassumedthattheweldsaroundtheflangesoftheI-sectionresistthebendingmomentandthetotalshearforceisresistedbytheweldsaroundtheweb(Gorencetal.2005).Theforceactingattheflangecausedbythebendingmomentis

N

Md t

fo

f

∗ =∗

− (6.31)

wheredisthedepthofthesteelI-sectiontf isthethicknessoftheflanges

Theflangefilletweldsmustsatisfy

N v Lf w w∗ ≤ φ (6.32)

wherethelengthoftheweldaroundeachflangeistakenasLw=2bfandbfisthewidthoftheflange.

Similarly,thefilletweldsaroundthewebundershearforceVz∗mustsatisfy

V v Lz w w∗ ≤ φ (6.33)

wherethelengthofthefilletweldsaroundwebistakenasLw=2d1andd1isthecleardepthoftheweb.

Example 6.2: Design of welded beam-to-column connection

A welded beam-to-column moment connection is subjected to a vertical design shearforceof35kNandanout-of-planedesignbendingmomentof142kNm.ThesteelbeamwithGrade300steel360UB50.7sectionshowninFigure6.11isfullyrestrainedfromlateralbuckling.Usethesimplemethodtodesignthisweldedconnection.

11.5

7.3356

171

11.5

Figure 6.11 Cross section of 360UB50.7.


1. Design actions

Thesectiondesigncapacityofthesteelbeamis

φ φM Z fsx ex y= = × × × × =−0 9 897 10 300 10 242 23 6. . kNm

Sincethesteelbeamisfullyrestrainedfromlateralbuckling,thememberdesignmomentcapacityofthebeamis

φ φM Mbx sx= = 242 2. kNm

Theminimumdesignbendingmomentfortheconnectionis

M M Mbxmin . . .∗ ∗= = × = < =0 5 0 5 242 2 142φ 121kN m kN m

Therefore,thedesignactionsforthedesignofconnectionsare

M V∗ ∗= =142 35kNm, kN

2. Design of flange welds

Theflangesandwebofthesteelbeamsectionarefilletweldedtotheflangeofthesteelcolumn. The flange forces due to the design bending moment M* are transmitted byflangeweldsaloneandarecalculatedas

N

Md t

ff

∗ ∗=

−=

×−

=142 10356 11 5

412 23

.. kN

ThetotallengthoffilletweldoneachflangeisLw=2bf=2×171=342 mm.Thedesignshearonfilletweldsonflangeistherefore

v

NL

ff

w

∗ ∗= = =

412 2342

1 2.

. kN/mm

Use8EXX48SPfilletweldstothebeamflanges;thedesignshearcapacityoftheweldperunitlengthisobtainedfromTable6.4as

φv vw f= > ∗ =1 303 1 2. .kN/mm kN/mm, OK

3. Design of web welds

Theshearforceisassumedtobetransmittedbythefilletweldsonbothsidesofthesteelbeamweb.Thetotallengthofthewebweldsis

L dw = = × =2 2 333 6661 mm

Thedesignshearonfilletweldsonthewebistherefore

v

VL

ww

∗ =∗= =

35666

0 053. kN/mm


Use6E48XXSPfilletweldstobothsidesoftheweb;thedesignshearcapacityoftheweldperunitlengthisobtainedfromTable6.4as

φv vw w= > ∗ =0 977 0 053. .kN/mm kN/mm, OK

6.6 Bolted moment end PlAte connectIonS

Boltedmomentendplateconnectionsareusedtotransferdesignbendingmoment,shearforceandaxialforcefrommemberstosupportingmembersinsteelportalframesormul-tistoreyrigidsteelframes.Thesteelbeamisusuallyshopweldedtotheendplatewhichisfieldboltedtothecolumnflangeorsupportingelement.Typicalboltedmomentendplateconnectionsarekneeandridgeconnectionsinportalframesandbeamnormaltocolumnconnectionsas shown inFigure6.1.Thebehaviourofboltedmomentendplate connec-tionsischaracterisedbytheirmoment–rotationcurves.Thebehaviouranddesignofboltedmomentendplateconnectionsareintroducedherein.FurtherinformationcanbefoundinthebookbyHoganandThomas(1994).

6.6.1 design actions

AboltedmomentendplateconnectionusedinarigidsteelframeissubjectedtoadesignbendingmomentM*,adesignshear forceV*andadesignaxial forceN*.Thesedesignaction effects can be determined by performing either a first-order elastic analysis withmomentamplificationorasecond-orderelasticanalysisoraplasticanalysis.Intheconnec-tiondesign,differentdesignactionsarecalculatedforthedesignofflangeandwebweldsandforthedesignofbolts,endplatesandstiffenersduetothefactthatdifferentassump-tionsareadoptedinthedesignmodels.

6.6.1.1 Design actions for the design of bolts, end plates and stiffeners

Whencalculatingthedesignactionsforthedesignofbolts,endplatesandstiffeners,itisassumedthatthedesignbendingmomentM*istransmittedbythetwoflanges,thedesignshearforceistransmittedbythewebandthedesignaxialforce(N*)istransmittedbythetwoflanges.Theaxialforcecarriedbyeachflangeisproportionaltoitscross-sectionalarea.TheforcecomponentsofdesignactionsactingontheconnectionaredepictedinFigure 6.12.Thedesignforceontheflangesduetothedesignbendingmomentisgivenby

N N

Md t

tm cmf

∗ = ∗ =∗

− (6.34)

wheredandtfarethedepthandthicknessoftheI-beamcrosssection,respectively.Designactionsonthecomponentsofaridgeconnectionwithasymmetriccrosssection

underbendingmoment(M*),axialforce(N*)andshearforce(V*)canbeobtainedfromFigure6.12asfollows:

N

Md t

N Vft

f

∗ =∗

−+

∗−

∗cos cos sinθ θ θ

2 2 (6.35)


N

Md t

N Vfc

f

∗ =∗

−−

∗+


2 2 (6.36)

V V Nvc∗ = ∗ + ∗cos sinθ θ (6.37)

whereNft∗istheresultanthorizontaldesignforceinthetensionflange

Nfc∗istheresultanthorizontaldesignforceinthecompressionflange

Vvc∗istheresultantverticaldesignshearforceattheendplateandcolumninterface

ItisnotedthatthesignsofdesignactionsarepositiveinthedirectionsshowninFigure6.12a.Equations6.35 through6.37canbeused tocalculate thedesignactionson thebeam

normaltocolumnconnectionsbytakingθ=0.ForkneeconnectionsinportalframesasillustratedinFigure6.12d,thedesignforcesintheflangesareexpressedby

N

Md t

N Vft

f

∗ =∗

−+

∗+


2 2 (6.38)

N

Md t

N Vfc

f

∗ =∗

−−

∗−


2 2 (6.39)

(a)

V *

V *V * V *

V *

M *

N *

M *N * N *

Ntm*

Ncm*

Ncm*

Ntm*

M *

θ

θ θθ

θ

θ

(c) (d)

(b)

Ntm* cos θ

Ncm* cos θNcm* sin θ

N *cos θ

V *sin θV *sin θ

V *cos θV *cos θ

N *sin θ

θ

θ

θ

N *

Ntm* sin θ

Figure 6.12 Design actions for the design of bolts, end plates and stiffeners: (a) design actions, (b) force components due to moment and axial force, (c) force components of shear force in ridge con-nection and (d) force components of shear force in knee joint.


6.6.1.2 Design actions for the design of flange and web welds

Forthedesignofflangeandwebwelds,itisassumedthatthedesignbendingmoment(M*)istransmittedbythewebandtheflanges.Theproportiontransmittedbyeachcomponentdependsonthesecondmomentsofareaofthewebandflanges.Thebendingmomentscar-riedbythewebandtwoflangesaredeterminedby(HoganandThomas1994)

M k Mw mw∗ = ∗ (6.40)

M k Mf mw∗ = − ∗( )1 (6.41)

whereMw∗andMf

∗are thedesignbendingmoments transmittedby thewebandflanges,respectively,andkmwiscalculatedby

k

II I

mwweb

web f

=+

(6.42)

whereIwebandIfarethesecondmomentsofareaofthewebandthetwoflangesabouttheprincipalx-axis,respectively.

Thedesignaxialforce(N*) isassumedtobecarriedbytheflangesandweb.Thepro-portionofthedesignaxialforcecarriedbyeachcomponentisproportionaltotheircross-sectional areas. The design axial forces transmitted by the web and each flange can bedeterminedasfollows(HoganandThomas1994):

N k Nw w∗ = ∗ (6.43)

N

k Nf

w∗ = − ∗( )12

(6.44)

wherethefactorkwisexpressedby

k

AA

ww

g

= (6.45)

whereAwisthecross-sectionalareaofthebeamwebAgisthegrosscross-sectionalareaofthebeamsection

Designactionsforthedesignofflangeweldsinridgeconnectionsunderbendingmoment(M*),axialforce(N*)andshearforce(V*)asillustratedinFigure6.12canbedeterminedby

N

Md t

NV

ftf

ff

∗ =∗

−+ ∗ −

∗

( )cos cos sinθ θ θ

2 (6.46)

N

Md t

NV

fcf

ff

∗ =∗

−− ∗ +

∗

( )cos cos sinθ θ θ

2 (6.47)


6.6.2 design of bolts

Theboltsinaboltedmomentendplateconnectionaresubjectedtodesigntensionforce( )Nft∗

intensionflangeandverticalshearforce( )Vvc∗ .Therefore,theboltsintheconnection(bolt

group)mustbecheckedfortheirdesigntensilecapacityϕNtbanddesignshearcapacityϕVfnasfollows:

N Nft tb∗ ≤ φ (6.48)

V Vvc fn∗ ≤ φ (6.49)

For a bolted end plate connection with four bolts placed symmetrically about the ten-sionflange,thedesigncapacityofboltsintension(ϕNtb)canbecalculatedby(HoganandThomas1994):

φ

φN

Nk

tbtf

pr

=+

41( )

(6.50)

wherethecapacityfactorϕ=0.8,ϕNtfisthedesigncapacityofaboltintensionandkpristhefactorthataccountsfortheeffectofadditionalboltforceduetoprying.Pryingoccursinboltedconnectionswhenboltsaresubjectedtotension.Theedgeoftheendplateunderbendingcausesbearingstressesonthematingsurface.Theresultingreactionactingontheendplatemustaddtothebolttension.Thepryingactionisfoundtoincreasethebolttensionforceby20%–33%(MannandMorris1979;Grundyetal.1980).Thefactorkprisbetween0.2and0.33.Atypicalvalueofkpr=0.25canbeusedinthedesignoftheconnections.

Becausetheboltsatthetensionflangehavebeenutilisedtocarrythetensionforce,onlythoseboltsalongthewebandatthecompressionflangeareassumedtobeeffectiveintrans-mittingthedesignshearforce.Thedesigncapacityofboltgroupinshearisdeterminedas

φ φV n Vfn cw fc= ( ) (6.51)

wherencwisthenumberofboltsalongthewebandatthecompressionflangeφVfcisthedesigncapacityofsingleboltinshear,whichistakenas

φ φ φ φV V V Vfc f fb bc= min( ; ; ) (6.52)

whereφVf isthedesignshearcapacityofaboltφVfbisthedesigncapacityoftheendplateduetothelocalbearingortear-outofthe

endplateφVbcisthedesigncapacityofthesupportingmemberduetolocalbearingortear-out

6.6.3 design of end plate

Theendplate is subjected tobending inducedby the tension forceat the tensionflange,verticalshearandhorizontalshearintheboltedendplateconnection.Theendplateundercombinedactionsmustsatisfy

N Nft pb∗ ≤ φ (6.53)


N N Vft fc ph∗ ∗ ≤and φ (6.54)

V Vvc pv∗ ≤ φ (6.55)

whereφNpbisthedesigncapacityoftheendplateunderbendingφVphisthedesigncapacityoftheendplateinhorizontalshearφVpvisthedesigncapacityoftheendplateinverticalshear

Assuming one dimensional yield line and double curvature bending (Sherbourne 1961;Grundy et al. 1980), the design capacity (ϕNpb) of the end plate under bending can beobtainedas

φN

f b ta

pbyp p p

fe

=0 9 2.

(6.56)

wherefypistheyieldstressoftheendplatebpandtparethewidthandthicknessoftheendplate,respectivelyafeeffectivedesignvalueofthedistanceaf showninFigure6.14

Thedesigncapacitiesoftheendplateunderhorizontalandverticalshearforcesaregivenby(HoganandThomas1994)

φV f b tph yp p p= 0 9 0 5. ( . ) (6.57)

φV f d tpv yp p p= 0 9 0 5. ( . ) (6.58)

wheredpisthedepthoftheendplate.

6.6.4 design of beam-to-end-plate welds

Inaboltedmomentendplateconnection,thebeamsectionisweldedtotheendplateasdepictedinFigure6.13.TheflangeweldstransferthetotalhorizontaldesignforcesNft

∗andNfc∗whicharecalculatedbyEquations6.46and6.47.Iffilletweldisusedalongtheflanges,

theweldmustsatisfythefollowingdesignrequirement:

φN N Nw ft fc≥ ∗ ∗and (6.59)

whereϕNwisthedesigncapacityoffilletweldaroundaflangeofthesteelI-section,whichisdeterminedas

φ φN L vw w w= 2 ( ) (6.60)

inwhichtheweldlengthLwacrosstheflangeistakenasthewidthofthebeamflangebfandϕvwisthedesigncapacityoffilletweldperunitlengthoftheweldgiveninTable6.4.


ThewebofthesteelbeamtransmitsthedesignactionsofMw∗,Nw

∗andV*asdepictedinFigure6.13.Itisassumedthateachsideofthewebofthesteelbeamisweldedtotheendplateusingfilletweld,whichtransmitsMw

∗,Nw∗andV*.Fromthestressdistributionshown

inFigure6.13,themomentequilibriumconditiongives

212 2

23

×∗

×

= ∗L v

L Mw zmw w (6.61)

wherevzm∗ is themaximum shear stress in thehorizontal direction causedby thedesignbendingmomentMw

∗.FromEquation6.61,vzm∗ canbeobtainedas

v

ML

zmw

w

∗ =∗3

2 (6.62)

inwhichLwistheweldlengthalongtheweb,whichistakenasLw=(d−2tf)/cosθforridgeconnection.

Thetotalhorizontaldesignforceactingononewebweldis

N

N Vwnv

w∗ =∗

−∗cos sinθ θ

2 2 (6.63)

TheshearinthezdirectioncausedbythedesignforceNwnv∗ isgivenby

v

NL

N VL

znvwnv

w

w

w

∗ =∗

=∗ − ∗cos sinθ θ

2 (6.64)

Thetotalshearinthezdirectioncanbedeterminedby

v

N VL

ML

zw

w

w

w

∗ =∗ − ∗

+∗cos sinθ θ

23

2 (6.65)

y

V *

θ

M *N *

Lw

v*zm

v*zm

Lw

zz

2Lw3

v*zmLw1 × ×2 2

v*zmLw1 × ×2 2

Figure 6.13 Shear in z direction caused by the bending moment.


TheshearononeweldcausedbytheverticaldesignshearforceVvc∗intheydirectionis

v

VL

VL

yvc

w

vc

w

∗ =∗

=∗/2

2 (6.66)

Theresultantshearontheweldperunitlengthis

v v vres z y∗ = ∗( ) + ∗( )2 2

(6.67)

Thedesignrequirementforthewebweldis

v v vz y w∗( ) + ∗( ) ≤

2 2φ (6.68)

6.6.5 design of column stiffeners

6.6.5.1 Tension stiffeners

Thetensionflangeofaboltedmomentendplateconnectionmaybesubjectedtoalargedesigntensionforce.Thisforcemaycauseexcessiveyieldinganddistortionofthecolumnflangewhichisboltedtotheendplate.Asaresult,thecolumnflangeorwebmayfail.Therefore,itisnecessarytochecktheneedforthecolumnstiffenersatthetensionflangeofthebeam.

Tensionstiffenersarerequiredifthefollowingconditionissatisfied:

N R R Rft t t t∗ > = ( )φ φ φmin ;1 2 (6.69)

whereϕRt1andϕRt2areexpressedby(PackerandMorris1977)

φR f t

a a s da

t ycf fcd c p h

d1

20 93 14 2

=+ + −

.

. ( ) (6.70)

φR f t

a a sa a

Na

at ycf fc

d c p

d ptf

p

d2

20 93 14 0 5

3 6=+ ++

+ ∗.

. ( ) .( )

.++

ap

(6.71)

wherefycf istheyieldstressofthecolumnflangetfcisthethicknessofthecolumnflangedepictedinFigure6.14spisthepitchofboltsNtf∗isthemaximumdesigntensionforceactingonabolt

distancesac,adandaparedeterminedas

a

b sc

fc g=−2

(6.72)

a

s t bd

g wc rc=− − 2

2 (6.73)


a

b sp

p g=−2

(6.74)

Ifcolumnstiffenersarerequired,columnstiffenersneedtobedesignedtocarrytheexcessofthedesigntensionforceasfollows(HoganandThomas1994):

NN R

ts

ft t∗ =∗ − φ (beam oon one side)

(beams on both sidesmax ;N R N Rft t ft t1 2∗ −( ) ∗ −( )

φ φ ))

(6.75)

Tensionstiffenersmustsatisfythefollowingdesignrequirement:

N Nts ts∗ ≤ φ (6.76)

wherethedesigncapacityofthetensionstiffenersisgivenby

φN f Ats ys s= 0 9. (6.77)

inwhichAsisthetotalcross-sectionalareaofthestiffeners,takenasAs=2bests,wherebesisthewidthofthestiffenerandtsisthethicknessofthestiffener.Thewidthofthestiffeneristakenasb t fes s ys≤ ( )15 250/ / asrequiredbytheClauseof5.14.3ofAS4100.Itiscommonpracticetodesignthestiffenerwithbes≥bf/3andts≥tf/2.

ae af tf

dp

sg

sp

tw

bf

dc

drcdwc

twc brc

bfc

ac

tfc

bf bp

drc

tp

bp

d

Figure 6.14 Beam-to-column connection details.


6.6.5.2 Compression stiffeners

Thecompressionflangeofthebeaminaboltedmomentendplateconnectionmaybesub-jectedtoalargedesigncompressionforce,whichmaycausethewebbucklingofthesteelcolumn.Therefore,itisnecessarytochecktheneedforthecolumnstiffenersatthecompres-sionflangeofthebeam.

Compressionstiffenersarerequiredifthefollowingconditionissatisfied:

N R R Rfc c c c∗ > = ( )φ φ φmin ;1 2 (6.78)

where ϕRc1 and ϕRc2 are the design bearing yield capacity and design bearing bucklingcapacityof thecolumnweb,respectively.ThedesignforceNfc

∗actingat thecompressionflangeisassumedtobedistributedona2.5:1slopetothelineatadistanceofdcrmeasuredfromthetopfaceofthecolumnflangeasdepictedinFigure6.14.ExpressionsforϕRc1andϕRc2derivedbasedontestresults(ChenandNewlin1973;Kulaketal.1987)aregivenby

φR f t t d tc ycw wc f rc p1 0 9 5 2= + +. ( ) (6.79)

φR

t f

dc

wc ycw

wc2

2

0 910 8

= ..

(6.80)

wherefycwistheyieldstressofthewebofthesteelcolumnandothersymbolsaredefinedinFigure6.14.

Alternatively,thedesignbearingyieldandbucklingcapacitiesofthecolumnwebcanbedeterminedusingthespecificationsgiveninAS4100.

Ifcompressionstiffenersarerequired,columnstiffenersneedtobedesignedtocarrytheexcessofthedesigncompressionforceasfollows:

NN R

ts

fc c∗ =∗ − φ (beam oon one side)

(beams on both sidesmax ;N R N Rfc c fc c1 2∗ −( ) ∗ −( )

φ φ ))

(6.81)

Thedesignofcompressionstiffenersissimilartothatofthetensionstiffeners.Ifcompres-sionstiffenersareprovided,thecapacityofthestiffenedcolumnwebneedstobechecked.

6.6.5.3 Shear stiffeners

Thecolumnwebintheconnectionregionissubjectedtoshearforcescomposedofahorizon-taldesignforceNft

∗orNfc∗ontheflangeandadesignshearforceVc

∗inthecolumnasshowninFigure6.15.Vc

∗istakenaspositiveifitactsinthesamedirectionasthedesignforceintheflangeofthebeam(HoganandThomas1994).Thecolumnwebundershearmayfailbyyield-ingorshearbuckling.Shearstiffenersarerequirediftheresultanthorizontalforce( )Vres

∗ actingontheflangeandcolumnwebisgreaterthanthedesigncapacity(ϕVc)ofthecolumnwebinshear.Thedesigncapacity(ϕVc)ofthecolumnwebinshearisdeterminedas

φ φ φV V Vc w b= min( ; ) (6.82)

whereφVwisthedesignshearyieldcapacityofthecolumnwebφVbisthedesignshearbucklingcapacityofthecolumnwebasgiveninChapter5


Whendiagonalstiffenersareusedas thewebstiffenersof thecolumnwithabeamcon-nectedononeside,thedesignforce( )Nvs

∗ carriedbythediagonalstiffenersistakenasthemaximumof ( )V Vres c

∗ − φ on the tensionand compressionflanges.Thediagonal stiffenersmustsatisfy

NNvs

vs

∗≤

cosθφ (6.83)

whereθistheanglebetweenthediagonalstiffenerandthehorizontalaxisandϕNvs = 0.9fysAs.

6.6.5.4 Stiffened columns in tension flange region

Whenconventionaltensionstiffenersareprovided,thestrengthofthestiffenedflangeofthecolumnneedstobechecked.Thisrequiresthatthedesigncapacityofthestiffenedcol-umnflange(ϕNts)mustbegreaterorequaltothedesigntensionforceatthetensionflangeNft∗.Thedesigncapacityofthecolumnflange(PackerandMorris1977)iscalculatedas

follows:

φN f t

w w da w w

a a dts ycf fch

dc d h= + − + +

+ −( )

0 9

2 2 1 12 22 1 2

1 2

. (6.84)

w a a a dd c d h1 0 5= + −( ). (6.85)

w s t t wp s w2 12 2= − − ≤( )/ (6.86)

IfφN Nts ft< ∗,alargersectionofthecolumnneedstobeusedorflangedoublerplatescanbeweldedtothecolumnflange.Thedesignrequirementforthestiffenedcolumnflangedoubler

(a)

V *c

V *c N*ft

N*fc

V *c

V *c

N*ft1 N*ft2

N*fc2N*fc1

(b)

Figure 6.15 Shear forces for the design of column stiffeners: (a) beam on one side of column and (b) beam on both sides of column.


platesisN Rft td∗ ≤ φ ,wherethedesigncapacityofstiffenedcolumnflangeϕRtdisestimated

by(Zoetemeijer1974)

φR t f t f

s a aa

td fc ycf d ydp d c

d

= +( ) + +

0 9 0 5

4 1 252 2. ..

(6.87)

wheretdisthethicknessofdoublerplatesfydistheyieldstressofdoublerplates

Whenconventionaltensionstiffenersareusedinadditiontodoublerplates,ϕRtdshouldbecalculatedusing(tfc+td)insteadoftfc.

6.6.5.5 Stiffened columns in compression flange region

Thestiffenedcolumnwebinthecompressionflangeregionmustwithstandthedesigncom-pressionforce( )Nfc

∗ actingatthecompressionflange.Thedesigncapacityofthestiffenedcolumnweb(MannandMorris1979)canbeestimatedby

φR f A f t b tcs ys s ycw fc fc wc= +( )0 9 1 63. . (6.88)

ThedesignrequirementisφR Ncs fc≥ ∗.

6.6.6 geometric requirements

Boltedmomentendplateconnectionsshallbedesignedtosatisfythegeometricrestrictions.ThesymbolsusedintheconnectiondesignsareshowninFigure6.14.Thegeometricrestric-tionsaregivenasfollows(HoganandThomas1994):

• b bp fc≤

• s b d s b d s sg f f fc f g g≤ − ≤ − ≥ ≥and but mm (M20 bolts) mm (M2g 2 5 80 120. , , 44 bolts)

• 30 36≤ ≤ ≤ ≤a d a de f e f2.5 mm (M20 bolts), 2.5 mm (M24 bolts)

• a a d L a d L af f f a f s s fas small as possible, but and≥ + ≥ + ≥cot , . cot ,φ φ0 5 00 5. d Lw w+

ThelengthLa istakenasLa=2.2df+grip(actualboltlength),andthedistanceds isthesocketdiametertakenasds=50mmforM20boltsandds=60mmforM24bolts.ThesocketlengthLsistakenasLs=65mmforM20boltsandLs=80mmforM24bolts.

Example 6.3: Design of bolted ridge connection

AboltedridgeconnectioninasteelportalframeissubjectedtoadesignbendingmomentM* = 160 kNm, a design axial tension force N* = 68 kN and a design shear forceV* = −7.5kN.TherafteroftheridgeconnectionisaGrade300steelsection360UB56.7.Therafterslopeis8°.Thedesignbendingmomentcapacityoftherafteris250kNm.Designthisboltedridgeconnection.

1. Design actions

a. Minimum design actions


Theminimumdesignbendingmomentistakenas

M M Mbmin . .∗ = = × = < ∗ =0 3 0 3 250 75 160kNm kNm

Thus,M*=160kNmisusedinthedesignoftheconnection.Thedesignshearforce|V*|<40kN;thus,V*istakenasV*=−40kNactinginthe

samedirectionoftheshearforce.

b. Design actions for the design of bolts and end plate

Thedimensionsof360UB56.7steelsectionare:d=359mm, bf=172mm, tf=13mmDesignforcesattheflangesandshearforcearecalculatedasfollows:

NMd t

N Vft

f

∗ =∗

−+

∗−

∗

=×−

° +

cos cos sin

cos cos

θ θ θ2 2

160 10359 13

8682

83

°° −−

° =402

8 494 4sin . kN

NMd t

N Vct

f

∗ =∗

−−

∗−

∗

=×−

° −

cos cos sin

cos cos

θ θ θ2 2

160 10359 13

8682

83

°° +−

° =402

8 421 5sin . kN

V V Nvc∗ = ∗ + ∗ = − × ° + × ° = −cos sin cos sinθ θ 40 8 68 8 30 kN

c. Design actions for the design of web and flange welds

Theloadsharingfactorsarecalculatedaskmw=0.155andkw=0.368.Thedesignbendingmomentstransmittedbythewebandflangesare

M k Mw mw∗ = ∗ = × =0 155 160 24 8. . kNm

M k Mf mw∗ ∗= − = − × =( ) ( . ) .1 1 0 155 160 135 2 kNm

Thedesignaxialforcestransmittedbythewebandflangeare

N k Nw w∗ = ∗ = × =0 368 68 25. kN

N

k Nf

w∗ =− ∗

=− ×

=( ) ( . )

.1

21 0 368 68

221 5 kN

Thedesignactionsforthedesignofflangeweldsarecalculatedas

NMd t

NV

ftf

ff

∗∗

∗∗

=−

+ −

= ×−

° +

( )cos cos sin

.cos

θ θ θ2

135 2 10359 13

8 213

.. cos sin5 8402

8 411° − − ° = kN


NMd t

NV

fcf

ff

∗ =∗

−− ∗ +

∗

=×−

° −

( )cos cos sin

.cos

θ θ θ2

135 2 10359 13

8 213

.. cos sin5 8402

8 363° +−

° = kN

2. Design of bolts

Use4M208.8/TBboltsateachflangeoftheraftersection;thecapacitiesofasingleboltareϕNtf=163 kN(tension) and ϕVf=92.6 kN(shear)(Table 6.3).

Takingkpr=0.25,thedesigncapacityofboltsatthetensionflangecanbecomputedas

φ

φN

Nk

Ntbtf

prft=

+=

×+

= > ∗ =41

4 1631 0 25

521 6 494 4( )

.. .kN kN, OK

Therearefourboltsatthecompressionflange,nw=4.Thedesigncapacityofboltsinshearisdeterminedas

φ φV n V Vfn cw f vc= = × = > ∗ =( ) . .4 92 6 370 4 30kN kN, OK

Adopttotal8M208.8/TBbolts.

3. Design of end plate

UseGrade250steelbp×tp=200×25mmendplate.Theyieldstressoftheendplateisfyp =250MPa.

Thepitchofboltsischosenas140 mm.Byplacingthetwoboltssymmetricallyaboutthecentroidofthetopflange,thedistanceafisaf=(140− 13)/2=63.5mm.

Theeffectivevalueofafisafe=af − dh/2=63.5− 24/2=51.5mmThedesigncapacityofendplateinflexurecanbecomputedas

φN

f b ta

Npbyp p p

feft= =

× × × ×= > ∗ =

−0 9 0 9 250 200 25 1051 5

546 492 2 3. .

.kN 44 4. kN, OK

Thedesigncapacityofendplateunderhorizontalshearwithdoubleshearplanesiscal-culatedas

φ φN f b t

N

ph yp p p

f

= = × × × × × ×

= >

−2 0 5 2 0 9 0 5 250 200 25 10

1125

3( . ) . ( . )

kN tt∗ = 494 4. kN, OK

Assuming35 mmedgedistance,thetotaldepthoftheendplateisdeterminedas

d a a t dp e f f= + − + = × + × − + =2 2 2 35 2 63 5 13 359 543. mm

Thedesigncapacityofendplateunderverticalshearwithdoubleshearplanesistherefore

φ φN f d t

V

ph yp p p= = × × × × × ×

= >

−2 0 5 2 0 9 0 5 250 543 25 10

3054 4

3( . ) . ( . )

. kN vvc∗ = 30 kN, OK

Thus,adopt200×25 mmendplate.


4. Design of flange welds

Use8E48XXSPfilletweldtoflanges.Thedesigncapacityoffilletweldperunitlengthisϕvw=1.303 kN/mTable6.4.Thedesigncapacityofthefilletweldtoeachflangeis

φ φN L v N

N

w w w ft

fc

= = × × = > ∗ =

> ∗ =

2 2 172 1 303 448 2 411

362

( ) . . kN kN, OK

kN,, OK

Adopt8E48XXSPfilletweldstotwoflanges.

5. Design of web welds

Thelengthoftheweldononesideofthewebis

L d tw f= − = − × ° =( ) cos ( ) cos .2 359 2 13 8 336 3/ / mmθ

Thehorizontalshearonwebweldiscomputedas

vN V

LML

zw

w

w

w

∗ =∗ − ∗

+∗=

× ° − − °×

+cos sin cos ( sin )

.θ θ2

3 25 8 40 82 336 3

32

×× ×=

24 8 10336 3

0 7013

2

..

. kN/mm

Theverticalshearonwebweldis

v

VL

yvc

w

∗ =∗=

×=

230

2 336 30 045

.. kN/mm

Theresultantshearisdeterminedas

v v vres z y∗ = ∗( ) + ∗( ) = + =

2 22 20 701 0 045 0 702. . . kN/mm

UsesixE48XXSPfilletweldstobothsidesoftheweb;fromTable6.4,weobtain

φv vw res= > ∗ =0 978 0 702. .kN/mm kN/mm, OK

Therefore,theboltedridgeconnectionisspecifiedasfollows:

8M208.8/TBbolts,90 mmgauge,140 mmpitch200×543×25 mmsteelendplate8E48XXSPfilletweldstoflanges6E48XXSPfilletweldstobothsidesofweb

6.7 PInned column BASe PlAte connectIonS

Pinnedcolumnbaseplateconnectionsareusedtotransmitthedesignactionsfromthesteelcolumns to the foundations. The components of a pinned column base plate connectionincludeconcretefoundation,steelbaseplate,filletweldsandanchorbolts.PinnedcolumnbaseplatesmaybesubjectedtoanaxialdesignforceN*(eithercompressionNc

∗ortension Nt∗)


and a designshearV*actinginthedirectionofprincipalaxisorboth(Vx∗,Vy

∗).Thedesignofpinnedcolumnbaseplateconnectionsmustcheckforthestrengthsoftheconnectioncompo-nentsunderaxialcompression/tensionandshearforces.Thebehaviouranddesignofpinnedcolumnbaseplateconnectionsareintroducedherein.FurtherinformationcanbefoundinthebookbyHoganandThomas(1994).

6.7.1 connections under compression and shear

6.7.1.1 Concrete bearing strength

Thelargeaxialcompressionforcetransmittedfromthesteelcolumntothebaseplateresultsinhighbearingstressesontheconcretefooting.Thisbearingstressmayreachthecompres-sivestrengthoftheconcrete,whichcausesthecrushingoftheconcrete.Thebearingstrengthoftheconcretedependsonthebearingareaofthebaseplate,thesupportingsurfaceareaofthefootingandthecompressivestrengthoftheconcrete.Clause12.3ofAS3600(2001)givesspecificationsonthedesignbearingstrengthoftheconcreteasfollows:

φ φ φN A f

AA

A fbc c c= ′ ≤ ′12

110 85 2. (6.89)

whereφ = 0 6. isthecapacityreductionfactor′fc isthecompressivestrengthofconcrete

A1isthebearingareaA2isthelargestareaofthesupportingsurfacethatisgeometricallysimilartoandcon-

centricwithA1

TheanchorboltholesofthebaseplateareignoredinthecalculationofthebearingareaA1.

6.7.1.2 Base plates due to axial compression in columns

Itisassumedthatthebaseplateisrigidandtheaxialcompressionforceisconcentratedoveranareaof0.8bfc×0.95dcforthesteelI-sectioncolumnbaseplateconnectionasshowninFigure6.16.Thebaseplateunderbearingstressescanbetreatedasacantileverplatebend-ingabouttheedgesofthisarea(Stockwell1975;DeWolf1978,1990).Themaximumvalue(amax)ofdistancesamandanisusedtocalculatethebendingmomentofthecantileverplateunderbearingstressofϕNsc1/A1.Thebendingmomentperunitwidthattheedgesofthisareaisequaltothemomentcapacityoftheplate:

φNA

a f tsc yp p1

1

2 2

20 9

4× =max .

(6.90)

wherefypistheyieldstressofthebaseplatetp is the thickness of thebaseplate and thedesign capacity of thebaseplate under

compressionφNsc1canbeobtainedas

φN

f t Aa

scyp p

1

21

2

0 92

=.

max

(6.91)


TheactualbearingstressdistributionunderthebaseplatemaynotbeuniformbutratherisconfinedtoanH-shapedareacharacterisedbythedimensionao(Stockwell1975;DeWolf1978;Murry1983)asdepictedinFigure6.17.Equation6.91canbemodifiedas

φN

f t Aa

scyp p H

o2

2

2

0 92

=.

(6.92)

wheretheH-shapedareaAHistakenasthelesserofthevaluescalculatedbythefollowingequations(Stockwell1975;DeWolf1978;Murry1983):

AN

f A b dH

c

c fc c

=∗

′=

φφ

0 85 2. ( )/( 0.6) (6.93)

Critical section for bending

N *c

bfc

dc dp

am

am

an 0.8bfc

0.95dc

an

amax

tp

bp

Figure 6.16 Critical section for bending of the cantilever plate.

bp

dc

bfc

0.5tfc

0.5tfc

ao

ao ao

ao

dp

Figure 6.17 H-shaped bearing area AH.


A

Nf

Hc

c

=∗

′=

φφ

2( 0.6) (6.94)

Thedimensionaocanbecalculatedas

a b d b d Ao fc c fc c H= + − + −

14

42( ) ( ) (6.95)

Thedesigncapacity(ϕNsc)ofthebaseplateundercompressionshouldbetakenasthelesserofϕNsc1andϕNsc2.

6.7.1.3 Column to base plate welds

Ifthecolumnendisnotpreparedforfullcontactwiththebaseplate,thefilletweldatthebaseofthecolumnunderaxialcompressionmustsatisfythefollowingrequirement:

N Nc w∗ ≤ φ (6.96)

whereϕNwisthedesigncapacityofthefilletweldatthebaseofcolumnandiscalculatedasϕNw=(ϕvw)Lw,whereLwisthetotallengthoffilletweld.

Thecolumnendisfilletweldedtothebaseplatetotransmittheaxialcompressionforceanddesignshearforces( , )V Vx y

∗ ∗ actinginbothprincipalaxes.Underthecombinedactionsofaxialcompressionandshear,thefilletweldmustsatisfy

v v v v vres x y z w∗ = ∗( ) + ∗( ) + ∗( ) ≤

2 2 2φ (6.97)

wherev V Lx x w∗ = ∗ / ,v V Ly y w

∗ = ∗ / ,v N Lz c w∗ = ∗ / andLwisthetotallengthoffilletweldaroundthe

columnsectionprofileφvwisthedesigncapacityoffilletweldperunitlength

6.7.1.4 Transfer of shear force

Inpinnedcolumnbaseplateconnections,thehorizontalshearforcemayberesistedby(a) theanchorbolts,(b)frictionbetweenthebaseplateandtheconcretefoundation,(c) shearkeyweldedtotheundersideofthebaseplateand(d)recessingthebaseplateintotheconcretefoundationoracombinationofthese(HoganandThomas1994).Itisnotrecommendedthatshearberesistedbytheanchorboltsalone.Thereasonforthisisthattheshearinducesbendingoftheanchorboltthathasalowbendingcapacity.Underaxialcompression,theshear shouldbe resistedby friction.However, if friction is not sufficient to resist shear,anchorboltscanbedesignedtoresistpartoftheshear,oracombinationoffrictionandashearkeymaybeused.Underaxialtension,theshearcanberesistedbyanchorboltsorashearkey.


Whenshearatthebaseplateisresistedbyfrictionalone,thedesignshearcapacity(ϕVd1)basedonfrictionmustbegreaterthantheresultantshear.Thiscanbeexpressedas

V V V Vres x y d∗ = ∗( ) + ∗( ) ≤

2 2

1φ (6.98)

wherethedesignshearcapacityisφ µV Nd c1 0 8= ∗. .Thecoefficientoffrictionμistakenas0.55forcontactplanebetweenthegroutandtherolledsteelcolumnabovetheconcretesur-face,0.7forcontactplaneattheconcretesurfaceand0.9forthecontactplaneofthebaseplatethicknessbelowtheconcretesurface(DeWolf1990).

6.7.1.5 Anchor bolts in shear

Shearforceonanchorboltistransferredbybearingonthesurroundingconcreteandbend-ing the bolt. The possible failure modes for anchor bolt under shear force (Ueda et al.1988)include(a)concretefailurewithwedgecone,(b)concretefailurewithoutwedgecone,(c) concretefailurewithpull-outconeand(d)shearfailureoftheanchorbolt.Thefailuremode(a)canbepreventedbysufficientedgedistance,whilefailuremode(c)canbepre-ventedbyprovidingsufficientembedmentoftheanchorbolt.

Thestrengthofboltinshearandthedistancebetweentheplaneoftheappliedshearforceandtheconcretesurfacehaveinfluencesontheshearcapacityofanchorbolt.Iftheshearactstowardsanedgeoftheconcretefooting,theedgedistancemaygoverntheshearcapac-ityoftheanchorbolt.Theconcretefailuresurfaceisassumedtobeasemi-coneofheightequaltotheedgedistanceandaninclinationof45°withrespecttotheconcreteedge.Thedesigncapacityoftheembeddedanchorboltundershearforcecanbeestimatedbyusingthetensilestrengthofconcreteovertheprojectedareaofthesemi-conesurface(ACI3491976)asfollows:

φ φV a fus e c= ′0 32 2. (6.99)

whereφ = 0 8. isthecapacityreductionfactoraeisthedistancemeasuredfromthecentreofananchorbolttotheconcreteedge

Theminimumdistanceaeistakenas

a df

fe f

uf

c

>′0 83.

(6.100)

Thedistanceaeshouldbegreaterthan12dfforGrade250rodorGrade4.6boltsand17dfforGrade8.8bolts.

Ananchorboltsubjectedtodesignshearforceinaprincipalaxisorinbothdirectionsmustsatisfy(HoganandThomas1994)

V Vf fe∗ ≤ φ (6.101)

whereV V nf x b∗ = ∗ / ,V V nf y b

∗ = ∗ / orV V nf res b∗ = ∗ / andthedesigncapacityoftheanchorboltin

shearϕVfe=min(ϕVf;ϕVus).Thedesigncapacityofasinglebolt inshear(ϕVf) isgiveninTables6.2and6.3.


6.7.2 connections under tension and shear

6.7.2.1 Base plates due to axial tension in columns

Thesteelbaseplateduetoaxialtensionincolumnissubjectedtoupliftforcebuthelddownbytheanchorbolts.ThefailuremechanismofthebaseplateweldedtoanI-sectioncolumnischaracterisedbythreeyield linesradiatingfromthecentreof thecolumnweb(Murry1983).Basedontheyieldlinetheory,thedesigncapacityofsteelbaseplateduetoaxialten-sioninthecolumncanbeestimatedby(Murry1983)

φ

φN

b f t

sn

bd

stfo yp p

g

bfo

c=

≤

4

2 2 2

2

for (6.102)

φ

φN

b d f t

s dn

bd

stfo c yp p

g c

bfo

c=+( )

>

2

2 2

2 2 2

for (6.103)

whereφ = 0 9. isthecapacityfactornbisthetotalnumberofboltsintheconnectionbfoislengthofyieldlinedefinedinFigure6.18sgisthegaugeofanchorbolts

twc

tc

t wc

tfc tfc

dc

dc do

do

tc dp

dcdp

dp

dp

am

an

an an anan

an an an

bp

bp bp

bc

bp

am

am am

amam

am

am

bfc bfcbfo

0.95dc

0.95dc 0.8do

0.8do

0.5bfo

0.5bfo 0.5bfo

0.95dc

0.8bfc

0.95bc

0.8bfc

Figure 6.18 Base plate connection details.


Theseequationscanbeusedforthedesignofbaseplateweldedtochannelsections,RHSandCHScolumnswithtwoparsofanchorbolts.However,thelengthofyieldlinesmustbesimilartothatforI-sectionsasdefinedinFigure6.18.

6.7.2.2 Column to base plate welds

Forthecolumntobaseplateweldssubjectedtoaxialtensionandshearforce,thedesigncapacityofthefilletweldinshearneedstobecheckedasfollows:

v v v v vres x y z w∗ = ∗( ) + ∗( ) + ∗( ) ≤

2 2 2φ (6.104)

wherev V Lx x w∗ = ∗ / ,v V Ly y w

∗ = ∗ / ,v N Lz t w∗ = ∗ / andLwisthetotallengthoffilletweldaroundthe

columnsectionprofileφvwisthedesigncapacityoffilletweldperunitlength

6.7.2.3 Anchor bolts under axial tension

Anchorboltsusedincolumnbaseplateconnectionsareclassifiedintocast-in-placeboltsanddrilled-inbolts.Cast-in-placeboltsincludehookedbolts,boltswithhead,boltswithnut,boltswithplateandU-bolts.Hookedboltsareoftenusedbutmayfailbystraighteningandpullingoutoftheconcretewhensubjectedtotension.Theyarerecommendedtobeusedincolumnbaseplateconnectionsunderaxialcompression(DeWolf1990).Boltswithhead,nutandplateor theU-boltsoffermorepositiveanchorage.Thefailuremodesofanchorboltsare(a)thefailureoftheboltgroupintensionand(b)thepull-outfailureofaconeofconcreteradiatingoutwardsat45°fromtheheadofthenutorboltasshowninFigure6.19.Topreventthesefailuresfromoccurring,anchorboltsmustsatisfy

N N N Nt t tb cc∗ ≤ =φ φ φmin( ; ) (6.105)

whereφNtisthedesigncapacityofembeddedboltsφNtbisthedesigncapacityoftheboltgroupcalculatedasφ φN n Ntb b tf= ( )φNcc is thepull-out resistanceofconcrete (MarshandBurdette1985;DeWolf1990)

givenby

φN f Acc c ps= ′( )0 8 0 33. . for all bolt types but hook bolts (6.106)

φN n f d Lcc b c f h= ′( )0 8 0 7. . for hook bolts (6.107)

wheredf isthediameterofthehookboltLhisthelengthofthehookApsistheprojectedareaoffailureconeofconcrete

Forisolatedsinglebolt,A Lps d= π 2 ,whereLdisthelengthembedment.TheprojectedareaoffailureconeofconcreteforboltgroupisgivenbyMarshandBurdette(1985).


Therequirementontheedgedistanceis

a df

fe f

uf

c

≥′6

(6.108)

Theedgedistance(ae)shouldbegreaterthan5dfforGrade250rodorGrade4.6boltsand7dfforGrade8.8boltsand100 mm.

6.7.2.4 Anchor bolts under tension and shear

Forananchorboltsubjecttocombinedtensionandshearforces,theboltmustsatisfythefollowingadditionalrequirement(HoganandThomas1994):

VV

NN

f

f

tf

tf

∗+

∗≤

φ φ1 0. (6.109)

whereN N ntf t b∗ = ∗ / .

Example 6.4: Design of column base plate connection

Designapinnedbaseplate connection for the steel columnof460UB74.6 subjectedtoaxial forcesandshearforces.Theendofthesteelcolumniscoldsawn.Thesteelcolumnissupportedona850 mmdiameterconcretepierfoundation.Thecompressivestrengthof concrete is ′ =fc 25MPa.Thecolumn is subjected to the followingdesignactions:(a) N Vc y

∗ = ∗ =87 30kN, kNand(b)N Vt y∗ = ∗ =105 70kN, kN.

a. Connection under axial compression and shear

1. Connection geometry

Basedonstandardbaseplateconnections,theinitialsizingofthebaseplateconnectionisselectedas200×490×20 mmendplate,4M204.6/Sboltswith300 mmpitchand100 mmgaugeasschematicallydepictedinFigure6.20.

Connectiongeometryandmaterialproperties:

d b t t

d b t

c fc fc wc

p p

= = = =

= =

457 190 14 5 9 1

490 200

mm, mm, mm, mm

mm, mm,

. .

pp yp

g p

f

s s

= =

= =

20 250

100 300

mm, MPa

mm, mm

45° 45°

Projected surface

LdLd

Figure 6.19 Failure cone of embedded bolt in tension.


2. Concrete bearing strength

Theareaofthebearingbaseplateis

A b dp p1 200 490= = × = 98,000 mm2

ThesupportingsurfaceareaA2thatisgeometricallysimilartoA1canbecalculatedas

A2 320 784= × = 250,880 mm2

Thedesignbearingstrengthofconcreteis

φ φN A f

AA

bc c= ′ = × × × × =12

1

0 85 0 6 0 85 25 1999. . . .98,000250,88098,000

22 kN

φ φN A fbc c= ′ = × × × × =−1

32 0 6 2 25 10. 98,000 2,940 kN

Thus, φN Nbc c= = > ∗ =min( . ; ) .1999 2 2940 1999 2 87kN kN, OK.

3. Base plate due to axial compression in column

Thedistanceamaxiscalculatedasfollows:

a d d

a b b

m p c

n p fc

= − = − × =

= − =

( . ) ( . )

( . ) (

0 95 2 490 0 95 457 2 28

0 8 2 20

/ / mm

/ 00 0 8 190 2 24

28

− × =

= =

. )

max( ;max

/ mm

24) 28mma

ThedesigncapacityofbaseplateϕNsc1is

φN

f t Aa

scyp p

1

21

2

2 3

2

0 92

0 9 250 20 102 28

= =× × × ×

×=

−. .

max

98,0005,625 kkN

TheH-shapedareaiscalculatedasfollows:

AN

f A b dH

c

c fc c

=∗

′=

×× × ×φ0 85

87 100 6 0 85 25 190 4572

3

. ( ) . . (/ 250,880/ ))= 4,014 mm2

490300 xx

y

y

100

200(a)

850490

(b)

200

A2

Figure 6.20 Base plate connection: (a) base plate and (b) area A2.


A

Nf

Hc

c

=∗

′=

×× ×

=φ2

87 100 6 2 25

3

.2900 mm2

Thus,AH=max(4014;2900)=4014mm2.ThedistanceaooftheH-shapedareais

a b d b d Ao fc c fc c H= + − + −

= + − + −

14

414

190 457 190 457 42 2( ) ( ) ( ) ( ) ××

=

4014

3.13 mm

ThedesigncapacityofbaseplateϕNsc2is

φN

f t Aa

scyp p H

o2

2

2

2 3

2

0 92

0 9 250 20 4014 102 3 13

18437 5= =× × × ×

×=

−. ..

. kkN

Thedesigncapacityofbaseplateistherefore

φ φ φN N N Nsc sc sc c= = = > ∗ =min( ; ) min( ; . )1 2 5625 18437 5 5625 87kN kN, OKK

4. Column to base plate welds

Thetotallengthoffilletweldaroundthecolumnsectionprofileis

L b b t d tw fc fc wc c fc= + − + − = × + × − + × − ×2 2 2 2 2 190 2 190 9 1 2 457 2 1( ) ( ) ( . ) ( 44 5

1598

. )

= mm

Theshearsperunitlengthundershearandaxialcompressionare

vVL

vNL

yy

w

zc

w

∗ =∗= =

∗ =∗= =

301598

0 019

871598

0 054

.

.

kN/mm

kN/mm

Theresultantshear v v vres y z∗ = ∗ + ∗ = + =( ) ( ) . .2 2 20 054 0 057 0.019 kN/mm.2

Use5EXX48GPfilletweld;φv vw res= > ∗ =0 522 0 057. .kN/mm kN/mm, OK.

5. Transfer of shear force

Theshearforceisassumedtoberesistedbyfrictionalone.Thebaseplateissupportedonagroutpadonthetopoftheconcretepierfoundationsothatthecoefficientoffrictionisμ=0.55.

Thedesignshearcapacityiscalculatedas

φ µV N Vd c y1 0 8 0 8 0 55 87 38 3 30= = × × = > =∗ ∗. . . . kN kN, OK


6. Anchor bolts in shear

Theminimumedgedistanceis

a df

f

d

e fuf

c

f

>′= × =

> = × =

0 8320

4000 83 25

196 4

12 20 240

. .. mm

12 mm

Adoptae=250mm.Thedesigncapacityofembeddedboltunderhorizontalshearis

φ φV a fus e c= ′ = × × × =0 32 0 8 0 32 250 25 802 2. . . N kN

Thedesign shear capacityof a singleboltwith threads included in the shearplane isobtainedfromTable6.2asϕVf=46.6kN.

Thus,ϕVfe=min(ϕVus;ϕVf)=min(80;46.6)=46.6kN.Thedesignshearforceonaboltis

V V Vf y fe∗ = ∗ = = < =/ / kN kN, OK4 30 4 7 5 46 6. .φ

b. Connection under axial tension and shear

1. Base plate due to axial tension on column

ForI-sectioncolumn,thelengthofyieldlineis

b

dfo

c= < = =1902

4572

323mm mm

Thedesigncapacityofthebaseplateis

φφ

Nb f t

s

nst

fo yp p

g

b=

=

× × × ××

4

2 20 9 4 190 250 20

2 10042

2 2. = > ∗ =N kN kN, OK967 3 105. Nt

2. Column to base plate welds

Thetotallengthoffilletweldaroundthecolumnsectionprofileis

Lw = 1598mm

Theshearsperunitlengthundershearandaxialtensionsare

vVL

vNL

yy

w

zt

w

∗ =∗= =

∗ =∗= =

701598

0 044

1051598

0 066

.

.

kN/mm

kN/mm

Theresultantshearv v vres y z∗ = ∗ + ∗ = + =( ) ( ) . .2 2 20 066 0 079 0.044 kN/mm.2

Use5EXX48GPfilletweld;φv vw res= > =∗0 522 0 079. .kN/mm kN/mm, OK.


3. Anchor bolts under axial tension

Theminimumlengthofboltembedmentis

L d Ld f d= = × = =12 12 20 240 250mm, adopt mm

Theprojectedareaoffailureconeforasingleboltis

A Lps d= = × =π 2 23 14 250. 196,250 mm2

ThecapacityofasingleboltintensionisϕNtf=78.3kN(Table6.2).Thepull-outresistanceofconcretecanbedeterminedas

φ φN f A Ncc c ps tf= ′( ) = × × × = > =−0 8 0 33 0 8 0 33 25 10 259 783. . . . 196,250 kN ..3 kN, OK

TheprojectedareaoffailureconesforboltgroupisillustratedinFigure6.21.TheareaAp1iscalculatedas

A L L s s s sp d d g p g p12 2

2 250 100 300 100 300

= + + +

= × + × × + + ×

π ( )

( )3.14 2502 == 426,250 mm2

TheshapedareaAp2asshowninFigure6.21iscalculatedasfollows(MarshandBurdette1985):

A L Ls s s L L

p d dp p p d d

22

2 1 2

24 2

2180

2 250 25

= − −

−°

= × −

−sin ( )/ π

003004

3002

300 2 250 250180

22 1 2

−

−× ×

°

= −

−sin ( ( ))/

45,000

π

440,219 4,781mm2=

A A Aps p p= − = − × =1 22 2426,250 4,781 416,688 mm2

850

100

300

Outline of area Ap1

Area Ap2

Figure 6.21 Projected area of failure cone.


Thepull-outresistanceofconcreteforboltgroupis

φN f Acc c ps= ′( ) = × × × =−0 8 0 33 0 8 0 33 25 10 5503. . . . 416,688 kN

Thecapacityoftheboltgroupintensionis

φ φ φN n N Ntb b tf cc= = × = < =( ) . .4 78 4 313 6 550kN kN

Thus,φN Nt t= > ∗ =313 6 105. kN kN, OK.Therequiredminimumedgedistancefortheboltis

a df

fde f

uf

cf>

′= × = < = × =

620

4006 25

73 5 20 100 00mm 5 mm adopt 1 mm,

4. Anchor bolts under tension and shear

Theforcesonasingleboltundercombinedtensionandshearforcesare

VVn

NNn

VV

NN

fy

b

tft

b

f

f

tf

tf

∗ =∗= =

∗ =∗= =

∗+

∗

704

17 5

1054

. kN

26.25 kN

φ φ== + = <17 544 6

26 2578 4

0 73 1 0..

..

. . , OK

referenceS

ACI349.(1976)Coderequirementsfornuclearsafetyrelatedstructures,ManualofConcretePractice,AmericanConcreteInstitute,Detroit,Michigan.

AISC-LRFDManual.(1994)LoadandResistanceFactorDesign,Vol.II,Connections,ManualofSteelConstruction,Chicago,IL:AmericanInstituteofSteelConstruction.

AS1275.(1985)Australianstandardformetricscrewthreadsforfasteners,Sydney,NewSouthWales,Australia:StandardsAustralia.

AS3600 (2001)AustralianStandard forConcrete Structures,Sydney,NewSouthWales,Australia:StandardsAustralia.


Birkemoe,P.C.(1983)High-strengthbolting:Recentresearchanddesignpractice,PaperpresentedattheW.H.MunseSymposiumonBehaviourofMetalStructures,ASCE,103–127,Philadelphia,PA,May1983.

Chen,W.F.andNewlin,D.E.(1973)Columnwebstrengthinbeam-to-columnconnections,technicalnotes,JournaloftheStructuralDivision,ASCE,99(ST9):1978–1984.

DeWolf,J.T.(1978)Axiallyloadedcolumnbaseplates,JournaloftheStructuralDivision,ASCE,104(ST5):781–794.

DeWolf,J.T.(1990)Columnbaseplates,Designguideseriesno.1,Chicago,IL:AmericanInstituteofSteelConstruction.


Fisher,J.W.,Galambos,T.V.,Kulak,G.L.,andRavindra,M.K.(1978)Loadandresistancefactordesigncriteriaforconnectors,JournaloftheStructuralDivision,ASCE,104(ST9):1427–1441.

Galambos,T.V.,Reinhold,T.A.,andEllingwood,B.(1982)Serviceabilitylimitstates:Connectionslip,JournaloftheStructuralDivision,ASCE,108(ST12):2668–2680.

Gorenc,B.E.,Tinyou,R.,andSyam,A.A.(2005)SteelDesigners’Handbook,7thedn.,Sydney,NewSouthWales,Australia:UNSWPress.

Grundy,P.,Thomas,I.R.,andBennetts,I.D.(1980)Beam-to-columnmomentconnections,JournaloftheStructuralDivision,ASCE,106(ST1):313–330.

Hogan,T.J.andThomas,I.R.(1979a)Bearingstressandedgedistancerequirementsforboltedsteel-workconnections,SteelConstruction,AustralianInstituteofSteelConstruction,13(3).

Hogan,T.J.andThomasI.R.(1979b)FilletwelddesignintheAISCStandardizedStructuralConnections,SteelConstruction,AustralianInstituteofSteelConstruction,13(1):16–29.

Hogan,T.J.andThomas,I.R.(1994)DesignofStructuralConnections,4thedn.,Sydney,NewSouthWales,Australia:AustralianInstitutionofSteelConstruction.

Kulak,G.L.,Fisher,J.W.,andStruik,J.H.A.(1987)GuidetoDesignCriteriaforBoltedandRivetedJoints,2ndedn.,NewYork:JohnWiley&Sons.

Mann,A.P. andMorris, J.L. (1979)Limitdesignof extended end-plate connections, Journalof theStructuralDivision,ASCE,105(ST3):511–526.

Marsh,M.L.andBurdette,E.G.(1985)Anchorageofsteelbuildingcomponentstoconcrete,EngineeringJournal,AmericanInstituteofSteelConstruction,22(1),33–39.

McGuire,W.(1968)SteelStructures,EnglewoodCliffs,NJ:PrenticeHall.Murry,T.M.(1983)Designoflightlyloadedsteelcolumnbaseplates,EngineeringJournal,American

InstituteofSteelConstruction,20(4),143–152.Packer,J.A.andMorris,L.J.(1977)Alimitstatedesignmethodforthetensionregionofboltedbeam-

columnconnections,TheStructuralEngineer,55(10):446–458.ResearchCouncilonStructuralConnections. (1988)Specification for structural jointsusingASTM

A325orA490bolts,AISC.Sherbourne,A.N.(1961)Boltedbeamtocolumnconnections,TheStructuralEngineer,39(6):203–210.Stockwell,F.W.(1975)Preliminarybaseplateselection,EngineeringJournal,AmericanInstituteofSteel

Construction,12(3),92–93.Swannell, P. (1979) Design of fillet weld groups, Steel Construction, Australian Institute of Steel

Construction,13(1):2–15.Thomas,I.R.,Bennetts,I.D.,andElward,S.J.(1985)Eccentricallyloadedboltedconnections,Paperpre-

sentedattheThirdConferenceonSteelDevelopments,AustralianInstituteofSteelConstruction,Melbourne,Victoria,Australia,May1985,pp.37–43.

Trahair,N.S.andBradford,M.A.(1998)TheBehaviourandDesignofSteelStructurestoAS4100,3rdedn.(Australian),London,U.K.:Taylor&FrancisGroup.

Ueda,T.,Kitipornchi,S.,andLink,K.(1988)Anexperimentalinvestigationofanchorboltsundershear,Research reportno.CE93,Brisbane,Queensland,Australia:DepartmentofCivilEngineering,UniversityofQueensland.

Zoetemeijer,P.(1974)Adesignmethodforthetensionsideofstaticallyloadedboltedbeam-to-columnconnections,Heron,STEVINLaboratoryandI.B.B.C.InstituteTNO,theNetherlands,20(1):1–59.

195

Chapter 7

Plastic analysis of steel beams and frames

7.1 IntroductIon

Theplasticanalysismethodsarewidelyusedinthedesignofsimplysupportedsteelbeams,continuoussteelbeams,steelportalframesandmultistoreyrectangularsteelframes.Thegoaloftheplasticanalysisistodeterminetheultimateloadsofasteelstructureatwhichthestructurewillfailduetothedevelopmentofexcessivedeflections(Neal1977).Theplasticmethodsofstructuralanalysisprovideeconomicaldesignsofsteelstructuresandhavetheadvantageofsimplicitycomparedtotheelasticmethodsofstructuralanalysis.Theplasticanalysisassumesthat(1)thebehaviourofthesteelstructurebeinganalysedisductile,(2)thedeflectionsofthestructurearenotthecriticaldesigncriteria,and(3)thelocalandoverallbucklingofthestructurewillnotoccurbeforethecollapseloadisreached.

Thischaptergivesanintroductiontotheplasticmethodsofstructuralanalysis.Thesimpleplastictheoryisdescribed,providinginsightintotheplastichinge,fullplasticmoment,plas-ticsectionmodulus,shapefactorandtheeffectsofaxialandshearforcesonthefullplasticmoment.Theplasticanalysisofsimplysupportedandcontinuoussteelbeamsispresentedbyintroducingthecollapsemechanism,theworkequationandthemechanismmethod.Themethodofcombinedmechanismsisprovidedtodealwiththeplasticanalysisofsteelframes.TheplasticdesigntoAS4100isalsodiscussed.

7.2 SImPle PlAStIc theory

7.2.1 Plastic hinge

Thebasicconceptsof thesimpleplastic theorycanbedemonstratedby investigatingtheactualbehaviourofasimplysupportedsteelbeamunderuniformlydistributedload.Thetypicalload–deflectioncurveofthesteelbeamisshowninFigure7.1a(BakerandHeyman1969).ThebehaviourfromOtoAontheload–deflectioncurveiselastic.WhentheloadisincreasedfromAtoB,thebeamdevelopssomepermanentdeformationswhichcannotberecoveredafterremovingtheload.Inadditiontothis,thedeflectionsincreasemorerapidlywithincreasingtheloading.ItcanbeobservedthatfurtherincreaseoftheloadingfromBtoCleadstorapidincreaseoflargedeflections.ThebeamisconsideredtohavecollapsedwhentheloadhasreachedtheloadingatpointB.Itshouldbenotedthatthestrainhardeningofthesteelmaterialresultsintheraisingcharacteristicsoftheload–deflectioncurvebeyondpointB.Theidealisedload–deflectioncurveisgiveninFigure7.1b,whichshowsthatundertheconstant load(Wc), thedeflectionincreaseswithout limit.This loadWc iscalledthecollapseloadofthesteelbeam.


Inthecollapsedstate,largedeflectionsoccuratthecentralkinkinthesimplysupportedsteelbeamduetotherotationofthehinge.Thishingeisknownasaplastichingethatformsatthesectionofmaximumbendingmomentinthebeam.Whenaplastichighformsinasteelmember,yieldingstartsatalocalsectionofthegreatestbendingmoment.ThegradualspreadofyieldingtowardstheneutralaxisandlocallyalongthemembertakesplacewhenthemomentcapacityisincreasedasdepictedinFigure7.2.Thisresultsintheplasticzoneattheplastichinge.Inthesimpleplastictheory,however,thespreadofplasticityalongthememberisusuallyignoredandtheplastichingeisassumedtobeconfinedatthecrosssec-tionofmaximumbendingmoment.

7.2.2 full plastic moment

The relationship between the bending moment and curvature can be derived from thestress–strainrelationbasedonthesimplebeamtheory.Figure7.3schematicallydepictsthestressdistributionsinarectangularcrosssectionofabeam.Thesectionisassumedtoremainplaneafterdeformation,whichresultsinalinearstraindistributionthroughthedepthofthesection.AsshowninFigure7.3d,theyieldstrain(εy)isattainedatadistancehfromtheneutralaxis.ThecompressionandtensionforcesshowninFigure7.3caredeter-minedasC1=T1=(1/2)bdfy.TheforcesshowninFigure7.3dareC2=T2=b(d−h)fyandC3=T3=(1/2)bhfy.ThebendingmomentcanbedeterminedfromthestressdistributiongiveninFigure7.3dasfollows:

M f bh h f b d h d h b d

hfy y y=

+ − + = −

12

43 3

22

( )( ) (7.1)

Thecurvatureisdeterminedasϕ=εy/h.Whenh=d,theyieldedzonesdisappearandtheextremefibreattainstheyieldstressasshowninFigure7.3c.Thecorrespondingmomentis

W

A

0(a)

CB

δ

A

0(b)

CBW

Wc

δ

Figure 7.1 Load–deflection curves for beam: (a) typical and (b) idealised.

Plastic zone

W

Figure 7.2 Plastic zone in simply supported beam under a concentrated load.

Plastic analysis of steel beams and frames 197

calledthefirstyieldmoment(My),whichisthegreatestmomentthatthesectioncanwith-standbeforeyielding.Thefirstyieldmoment(My)oftherectangularsectioncanbeobtainedfromEquation7.1as

M

bdfy y=

23

2

(7.2)

Thisequationcanbewrittenas

M Zfy y= (7.3)

whereZistheelasticsectionmodulus.ThecurvaturecorrespondingtoMyisϕy=εy/d.The bending moment–curvature relationship of the rectangular cross section can be

obtainedbycombiningEquations7.1and7.2asfollows(Neal1977):

MMy

y= −

1 5 0 52

. .φφ

(7.4)

Figure7.4showsthemoment–curvaturecurvefortherectangularsection.Itappearsthatwhenthecurvatureisverylarge,themomentMapproachesto1.5My.

Whenh=0,thestateoffullplasticityoftherectangularsteelcrosssectionisachievedasshowninFigure7.3e.FromthefullplasticstressdistributionillustratedinFigure7.3e,the full plastic moment can be calculated by taking moments about the plastic neutralaxis(PNA).ItisnotedthatthePNAisazerostressaxisthatdividesthesectionintotwoequalareas.ThefullplasticmomentoftherectangularsectioncanalsobeobtainedfromEquation7.1as

M bd fp y= ( )2 (7.5)

Equation7.5canberewrittenas

M Z fp p y= (7.6)

whereZp =bd2 is theplastic sectionmodulusof the rectangular cross section shown inFigure7.3.

(a)b

d

d

(b)

C1

h

C2

C3

TT3

T2T1

fy

fy

fy

fy

fy

fy

C

(c) (d) (e)

Figure 7.3 Stress distributions in rectangular section: (a) cross section, (b) strain, (c) at first yield, (d) par-tially plastic and (e) fully plastic.


Ingeneral,theplasticsectionmodulusofacrosssectioncomposedofelementscanbecomputedbysumming thefirstmomentofareaofeachelementabout thePNAof thesectionas

Z Ay A yp i i

i

m

j j

j

n

= += =∑ ∑

1 1

(7.7)

whereAiistheareaoftheithelementabovethePNAyiisthedistancefromthecentroidoftheithelementtothePNAAjistheareaofthejthelementbelowthePNAyjisthedistancefromthecentroidofthejthelementtothePNAmandnarethetotalnumberofelementsaboveandbelowthePNA,respectively

Theshapefactorisdefinedastheratiooftheplastictoelasticsectionmodulus(Neal1977):

ν =

ZZp (7.8)

Theshapefactorindicatestheadditionalmomentcapacitythatasectioncansupportbeyonditsfirstyieldmoment.

Example 7.1: Calculation of full plastic moment of T-section

Figure7.5showsaGrade300steelT-sectionbendingaboutitsprincipalx-axis.Theyieldstressofthesteelsectionis300MPa.Calculate(a)thefirstyieldmomentofthesection,(b)thefullplasticmomentand(c)theshapefactorofthesection.

a. First yield moment

Thecentroidlocationofthesectionmeasuredfromthetopfibreiscomputedas

yA y

Ac

n n

n

= =× × + × × +

× + ×=∑

∑200 20 20 2 18 250 250 2 20

200 20 18 2508

( ) ( )/ /11 5. mm

00

0.5

Mom

ent M

/My

1

1.5

2

2 4 6 8Curvature φ/φy

Figure 7.4 Typical moment–curvature curve of beam.


Thesecondmomentofareaofthesectionaboutthex-axisis

Ix =×

+ × × −

×+ ×

200 2012

200 20 81 5202

18 25012

18 2

3 2

3

.

+ 5502502

20 81 5 62 16 102

6× + −

= ×. . mm4

Theelasticsectionmodulusis

Z

Iy

x= =×

+ −=

max

..

62 16 10250 20 81 5

329 7366

, mm3

Thefirstyieldmomentiscalculatedas

M Zfy y= = × × =−329 736 300 10 98 96, kNm.

b. Full plastic moment

Thecross-sectionalareaoftheflangeis

Af = × =200 20 4000 mm2

Thecross-sectionalareaofthewebis

A Aw f= × = > =18 250 4500 4000mm mm2 2

ThePNAislocatedintheweb.ThedepthofthePNAcanbedeterminedas

200 20 18 20 18 250 20× + × − = × + −( ) ( )d dn n

Hence,dn=33.9mm.

18

y

250

20

200

Figure 7.5 Steel T-section.


Theplasticmodulusofthesectioniscomputedas

Zp = × × −

+ × − ×

−

+ × +

200 20 33 9202

18 33 9 2033 9 20

2. ( . )

( . )

18 (250 200 , mm3− ×+ −

=33 9250 20 33 9

2599 028. )

( . )

Thefullplasticmomentofthesectionistherefore

M Z fp p y= = × × =−599 028 300 10 179 716, kNm.

c. Shape factor

Theshapefactoris

ν = = =

ZZp 599 028

329 7361 82

,,

.

7.2.3 effect of axial force

Forasteelshortcolumnsubjecttoaxialloadandbending,thefullplasticmomentofthecolumnsectionisreducedbytheaxialload.Figure7.6depictstheplasticstressdistributionofarectangularcolumnsectionundercombinedaxialloadandbending.Inthefullplasticstate,theaxialforce(P)andfullplasticmoment(Mp)inthesectioncanbedeterminedasthestressresultants:

P b d f Py o= =( )( )α α2 (7.9)

M M P

dMp o o= −

= −α α

21 2( ) (7.10)

wherePoistheultimateaxialloadofthecrosssectionintheabsenceofbendingmomentMoisthefullplasticmomentintheabsenceoftheaxialload

(a)b

d

d

(b)

T0

P

2 fy

C0

C

T

αd

fy

fyfy

fy

(c)

Figure 7.6 Plastic stress distributions in a rectangular column section under axial load and bending: (a) cross-section; (b) actual plastic stress distribution; (c) equivalent plastic stress distribution.


Theaxialload–momentinteractionequationcanbeobtainedbycombiningEquations7.9and7.10as

PP

MMo

p

o

+

=

2

1 (7.11)

TheinteractioncurveforasteelshortcolumnunderaxialcompressionandbendingmomentisgiveninFigure7.7.Theinteractioncurverepresentsayieldsurfacewhichisanimportantconceptintheplastictheory(BakerandHeyman1969).Ifapointlieswithinthebound-aryoftheyieldsurface,thesectioncancarrythecombinationofaxialloadandbendingmoment.Apointontheboundaryoftheyieldsurfacejustcausesthesectiontobecomefullyplastic.Apointoutsidetheboundaryoftheyieldsurfacerepresentsanimpossiblestate.

7.2.4 effect of shear force

Thecrosssectionofasteelmemberundercombinedshearforceandbendingissubjectedtoa2Dstressstate.Thebendingstressesactinthelongitudinaldirection,whiletheshearstressesactinthetransversedirection.ItisassumedthattheflangesofasteelI-sectiondonotcarryshearstressesandshearstressesareuniformlydistributedovertheweb.Inthefullyplasticstate,thelongitudinalstress(σ)inthewebforresistingtheplasticmomentwillbe less thantheyieldstress (fy)duetothepresenceof theshearstresses(τ).ThiscanbeexpressedbythevonMisesyieldcriteriaasfollows:

σ τ2 2 23+ = fy (7.12)

ThelongitudinalbendingstressonthewebofthesteelI-sectioncanbeobtainedfromtheaforementionedequationas

σ τ= − ≤f fy y

2 23 (7.13)

00

0.2

Axi

al lo

ad P

/Po

0.4

0.6

0.8

1

1.2

0.2 0.4 0.6 0.8 1 1.2Moment Mp/Mo

Figure 7.7 Yield surface.


Figure7.8depictsthestressdistributionoverthecrosssectionsubjectedtocombinedbend-ingandshear.Itcanbeseenthatthecontributionfromwebtothefullplasticmomentisreducedbyshearstresses.FurtherdetailsontheeffectsofshearonthefullplasticmomentweregivenbyBakerandHeyman(1969).

7.3 PlAStIc AnAlySIS of Steel BeAmS

7.3.1 Plastic collapse mechanisms

Thefixedendedbeamdepicted inFigure7.9 isused todemonstrate thedevelopmentofplasticcollapsemechanism(BakerandHeyman1969).ThebeamofuniformcrosssectionissubjectedtoslowlyincreasingpointloadWuntilitcollapses.Theelasticbendingmomentdiagram is shown inFigure7.9b.ThebendingmomentsatpointsA,BandCareMA=6WL/27,MB=8WL/27andMC=12WL/27.AstheloadWisslowlyincreased,thebendingmomentatpointCapproachesthefullplasticmomentMpandthefirstplastichingeformsatpointCasillustratedinFigure7.9c.Theformationoftheplastichingecausesaredis-tributionofmoments.Astheloadiscontinuouslyincreased,asecondplastichingeformsatpoint B.Thetwoplastichingeshaveturnedtheredundantbeamintoastaticallydeter-minatestructure.Furtherincreaseintheloadingcausesthefinalplastichingetoformatpoint A.Theformationofthethirdplastichingeturnsthebeamintoamechanismofplasticcollapse.TheplasticmomentdistributionofthebeamisgiveninFigure7.9d.Itshouldbenotedthatthecollapseloaddoesnotdependontheorderofformationoftheplastichinges.Theplasticanalysisisconcernedwithonlythecollapsestateofastructure.

7.3.2 Work equation

Thevirtualworkequationcanbeusedtodescribetheenergybalanceforastructureinthecollapsestate(BakerandHeyman1969;Neal1977).Theformationofplastichingesinastructure turns the structure intoacollapse state.This implies thata smalldeformation

fy

fy

σ

σ

Figure 7.8 Effect of shear on the stress distribution in I-section.


of the collapse mechanism at constant values of the applied loads can occur. The workdonebytheappliedloadWunderasmalldeformationδisWδ.Thetotalworkdonebyallappliedloadsonthestructureis Wδ∑ .TheplastichingeswillabsorbtheworkdonebyexternalloadsbyrotatingcertainanglesθundertheconstantplasticmomentMp.Theworkabsorbedinallplastichingesis

Mp∑ θ.Theworkequationbasedonthesimpleenergybal-

ancetheoremisexpressedby

W Mpδ θ=∑∑ (7.14)

Theworkdissipatedataplastichingeisalwayspositive.Therefore,thesignsofhingerota-tions(θ)mustbetakenasthesameasthesignofthecorrespondingplasticmoment(Mp).Allcollapsemechanismsareusuallydrawnwithstraightmembersbetweenplastichinges.Theuseoftheworkequationiscalledthemechanismmethod.

B L

C

2L

(c)

A

W

B

L

C

2L

(b)

A

MA

MB

MC

B

L

C

2L

(d)

A

Mp

Mp

Mp

W

B

2L

(a)

L

A C

Figure 7.9 Development of plastic collapse mechanism: (a) steel beam with fixed ends, (b) elastic bending moment diagram, (c) plastic collapse mechanism and (d) plastic bending moment diagram.


7.3.3 Plastic analysis using the mechanism method

Theplasticdesignofbeamsistodetermineallpossiblecollapsemechanismsandthecorre-spondingvaluesoffullplasticmomentsandthendesignthebeamsbasedonthemechanismwhichprovidesthelargestfullplasticmoment.Intheplasticanalysisusingthemechanismmethod,thefollowingconsiderationsshouldbetakenintoaccount:

• Allpossiblemechanismsofcollapseshouldbeinvestigated.• Plastichingestendtoformattheendsofmembers,atpositionsofconcentratedloads

andatthepointofmaximumbendingmoment.• ThemechanismandMpof each span ina continuousbeamshouldbe investigated

individually.• Ateachsupportofacontinuousbeam,theplastichingeformsintheweakermember

withasmallervalueofMp.

Thepropped cantileverbeam shown inFigure7.10a isused to illustrate themechanismmethod(BakerandHeyman1969;HorneandMorris1981).TheproppedcantileverofspanLissubjectedtoslowlyincreaseuniformlydistributedloadw.ThecollapsemechanismisgiveninFigure7.10b,whichiscomposedoftworigidrinks.Thecentralhingeislocatedsomedistancexfromtheright-handsupport.Theangleofrotationattheleft-handendofthebeamisassumedtobeθ1.Otherrotationscanbedeterminedfromthegeometryintermsofθ1.Thisgivesθ2=(L−x)θ1/xandθ3=Lθ1/x.TheresultantforceactingoneachrigidrinkisshowninFigure7.10b.Undertheresultantforceoneachrigidrink,therigidrinkundergoesameandisplacementofδ/2,whereδisthedisplacementatthepointofcentralplastichinge.Theworkequationcanbewrittenas

[ ( ) ]

( )( )w L x wx

L xM M

Lx

p p− + × − = +

θ θ θ11 1

2 (7.15)

(b)

δ

L–x x

wxw(L–x)

(a)

w

L

θ2θ1

θ3

Figure 7.10 (a) Propped cantilever and (b) collapse mechanism.


Thefullplasticmomentcanbeobtainedfromtheprecedingequationas

M

wLx L xL x

p =−+

2

(7.16)

The maximum full plastic moment is Mp = wL2/11.66 when x = 0.414L (Horne andMorris1981).

Example 7.2: Largest plastic moment of two-span continuous beam

Atwo-spancontinuoussteelbeamwithdifferentuniformcrosssectionsunderfactoredconcentratedloadsisschematicallydepictedinFigure7.11a.Determinethelargestfullplasticmomentofthecontinuousbeam.

1. Mechanism 1

Mechanism 1 is shown in Figure 7.11b. At the support, the plastic hinge is correctlylocatedintheweakermember.Sincenomechanismhasbeenassumedinthesecondspan,

(a)

(b)

(c)

(d)

2θ

3θ

3θ

2θ

2θ

θ

θ

θ

δ

δ

δ

20 kN

20 kN

20 kN

20 kN 20 kN 20 kN

20 kN

20 kN

20 kN

20 kN

2.5 2.5

5 m 6 m

2 2 2

Mp 2Mp

20 kN 20 kN

θ

Figure 7.11 Mechanisms of two-span continuous beam: (a) continuous beam, (b) mechanism 1, (c) mecha-nism 2 and (d) mechanism 3.


thereisnodisplacementandtheworkdonebytheloadinginthatspaniszero.Theworkequationformechanism1canbewrittenas

W M

M M M

M

p

p p p

p

δ θ

θ θ θ θ

∑ ∑=× × = + +

∴ =

20 2 5 2

12 5

( . ) ( ) ( ) ( )

. kNm

2. Mechanism 2

Mechanism 2 is shown in Figure 7.11c. At the support, the plastic hinge is correctlylocatedintheweakermember.Theworkequationformechanism2canbewrittenas

W M

M M

M

p

p p

p

δ θ

θ θ θ θ

∑ ∑=× × + × × = +

∴ =

20 2 2 20 2 2 2 3

15

( ) ( ) ( ) ( )

kNm

3. Mechanism 3

Mechanism 3 is shown in Figure 7.11d. At the support, the plastic hinge is correctlylocatedintheweakermember.Theworkequationformechanism3canbewrittenas

W M

M M

M

p

p p

p

δ θ

θ θ θ θ

∑ ∑=× × + × × = +

∴ =

20 2 20 4 2 3

17 1

( ) ( ) ( ) ( )

. kNm

Therefore,thegreatestfullplasticmomentofthecontinuousbeamis17.1kNm.

Example 7.3: Collapse load of three-span continuous beam

A three-span continuous steel beam with a uniform cross section under concentratedloadsisshowninFigure7.12a.ThefullplasticmomentofthebeamcrosssectionisMp=450kNm.DeterminethecollapseloadWofthecontinuousbeam.

1. Mechanism 1

Mechanism1 is shown inFigure7.12b.Theworkequation formechanism1canbewrittenas

W M

W M M

WM

p

p p

p

δ θ

θ θ θ

∑ ∑=× × = +

∴ = =×

=

1 5 3 2

34 5

. ( ) ( ) ( )

.3 4504.5

300 kN


2. Mechanism 2

Mechanism2 is shown in Figure 7.12c.Thework equation formechanism2 canbewrittenas

W M

W M M M

WM

M

p

p p p

pp

δ θ

θ θ θ θ

∑ ∑=× × = + +

∴ = = =

( ) ( ) ( ) ( )4 2

44

450 kN

3. Mechanism 3

Mechanism 3 is shown in Figure 7.12d. The work equation for mechanism 3 can bewrittenas

W M

W M M

WM

p

p p

p

δ θ

θ θ θ

∑ ∑=× × = +

∴ = =×

=

2 2 5 2

35

3 4505

270

( . ) ( ) ( )

kN

Therefore,theminimumcollapseloadWofthecontinuousbeamis270kN.

(a)

(b)

(c)

(d)

δ

δ

δ

1.5W

1.5W

1.5W

1.5W

W

W

W

W

2W

2W

2W

2W

5 m8 m6 m3 3

2θ

2θ

2θ

4 4 2.5 2.5Mp Mp Mp

θ

θ θ

θθ

θ

Figure 7.12 Mechanisms of three-span continuous beam: (a) continuous beam, (b) mechanism 1, (c) mecha-nism 2 and (d) mechanism 3.


7.4 PlAStIc AnAlySIS of Steel frAmeS

7.4.1 fundamental theorems

Intheplasticdesign,onlyproportionalloadingisallowed.Thismeansthattheloadsappliedtoastructurewillnotvaryrandomlyandindependently.Itisconsideredthatthestructureisinitiallysubjectedtoworkingloadswhichcanbemultipliedbyacommonloadfactorλastheloadincreases.Thefundamentaltheoremsareconcernedwiththevalueoftheloadfactorλcatthecollapseofthestructure(BakerandHeyman1969).

Theuniquenesstheoremstatesthattheloadfactor(λc)atthecollapseofastructurehasadef-initevalue,whichisuniqueforthestructure.Astheloadsaregraduallyincreased,thestructurecollapsesatacertainvalueλc.Theunsafetheoremstatesthattheloadfactor(λ)determinedfromtheanalysisofanassumedcollapsemechanismwillbegreaterorequaltothetruecol-lapseloadfactorλc.Thistheoremmeansthatiftheassumedmechanismhappenstobecorrect,theloadfactorisequaltothecollapseloadfactorλc;otherwise,theloadfactorcalculatedfromtheassumedmechanismisgreaterthanλcandisoverestimated.Thesafetheoremisconcernedwiththeequilibriumstateofastructure.Thistheoremstatesthattheloadfactordeterminedfromtheequilibriumofbendingmomentdistributionwithexternalloadswillbelesstheorequaltothecollapseloadfactorλc.Thisimpliesthatifthebendingmomentdistributiondoesnotcauseacollapsemechanism,theloadfactordeterminedfromthatwillbelessthanλc.

Astructureatcollapsemustsatisfythreeconditions(BakerandHeyman1969).Thefirstiscalledthemechanismcondition,whichrequiresthatasufficientnumberofplastichingesmustbeformedtoturnthestructureintoamechanism.Thesecondiscalledtheequilibriumcondition,whichimpliesthatthebendingmomentdistributionmustbeinequilibriumwithexternalloadsatallloadingstages.Thethirdconditioniscalledtheyieldcondition,whichmeansthatthebendingmomentatanysectionmustnotexceedthefullplasticmomentMp.Ifthesethreeconditionsaresatisfiedsimultaneously,thestructureisatthestateofcollapseandtheloadfactordeterminedisequaltothecollapsefactorλcwhichisunique.

7.4.2 method of combined mechanism

Theworkequationcanbewrittenforanymechanismwhichsatisfiestheequilibriumcondi-tionofastructure.However,thereisalimittothenumberofindependentequationsofequi-libriumforastructure.Foramultistoreyandmulti-bayframe,thenumberofindependentmechanismscanbecalculatedas(HorneandMorris1981)

n k jm = +( )1 (7.17)

wherekisthetotalnumberofstoreysjisthetotalnumberofbays

BeamandswaymechanismsareindependentmechanismsasdepictedinFigure7.13bandcforaportalframe.Allothermechanismscanbededucedfromtheseindependentmecha-nisms.Thecombinedmechanismisobtainedbycombiningthebeamandswaymechanismsintoone.Someoftheplastichingesinthetwomechanismsarecancelledinordertolocktogether inanequilibriumstate.Figure7.13dshowsacombinedmechanism.TheplasticanalysisofframesusingthecombinedmechanismmethodisdemonstratedinExamples 7.4and 7.5. Further details on the plastic analysis of frames can be found in the book byHorne andMorris(1981).


Example 7.4: Collapse load factor of steel portal frame

Figure7.14ashowsa steelportal frameunderworking loads.Theportal framehasauniform fullplasticmomentof150kNm.Determine the collapse load factorof thisportalframe.

1. Beam mechanism

ThebeammechanismisshowninFigure7.14b.Theworkequationforthebeammecha-nismcanbewrittenas

W M

M M M

M M

p

p p p

p p

δ θ

λ θ θ θ θ

λ

∑ ∑=× × = + +

∴ = = =

100 4 2

4400

1501

( ) ( ) ( ) ( )

100

0001 5= .

2. Sway mechanism

TheswaymechanismisshowninFigure7.14c.Theworkequationfortheswaymecha-nismcanbewrittenas

W M

M M M M

M

p

p p p p

p

δ θ

λ θ θ θ θ θ

λ

∑ ∑=× × = + + +

∴ = =×

60 3 5

4210

4 150

( . ) ( ) ( ) ( ) ( )

2210= 2.86

(a)

(c) (d)

(b)

H

H H

V

VV

VH

h

h h

h

L

L L

LL/2 L/2L/2L/2

θ θ

θ

θ

θ

θ

2θ

θ

θ

θ

2θ

Figure 7.13 Beam, sway and combined mechanisms: (a) portal frame, (b) beam mechanism, (c) sway mecha-nism and (d) combined mechanism.


3. Combined mechanism

Thecombinedmechanism3isshowninFigure7.14d.Theworkequationforthecom-binedmechanismcanbewrittenas

W M

M M M M

p

p p p p

δ θ

λ θ λ θ θ θ θ θ

λ

∑ ∑=× × + × × = + + +

∴

100 4 60 3 5 2 2( ) ( . ) ( ) ( ) ( ) ( )

== =×

=6610

6 150610

Mp 1.48

Therefore,thecollapseloadfactorisλc=1.48.

Example 7.5: Collapse load factor of two-storey frame

Figure7.15showsatwo-storeysteelframeunderworkingloads.ThefullplasticmomentsMp(kNm)oftheframemembersareshowninthefigure.Determinethecollapseloadfactorofthistwo-storeyframe.

1. Number of independent mechanisms

Thisisatwo-storeyandone-bayframe;thus,k=2andj=1.Thenumberofindependentmechanismsis

n k jm = + = × + =( ) ( )1 2 1 1 4

There are four independent mechanisms, which include two beam and two swaymechanisms.

(a)

(c) (d)

(b)

100 kN

θ

θ

θθ

θ θ

θ

2θ3.5 m3.5 m

3.5 m

4 48 m

8 m 8 m

100 λ100 λ

θ θ

2θ3.5 m

448 m

100 λ60 λ

60 λ 60 λ

60 kN

Figure 7.14 Mechanisms of steel portal frame: (a) portal frame, (b) beam mechanism, (c) sway mechanism and (d) combined mechanism.


2. Mechanism 1

Mechanism1isabeammechanismshowninFigure7.16a.Theworkequationfortheswaymechanismcanbewrittenas

W Mpδ θ

λ θ θ θ θ

λ

∑ ∑=× × = × + × + ×

∴ = =

40 4 60 90 2 60

300160

1 875

( ) ( ) ( ) ( )

.

3. Mechanism 2

Mechanism2isalsoabeammechanismshowninFigure7.16b.Theworkequationforthismechanismcanbewrittenas

W Mpδ θ

λ θ θ θ θ

λ

∑ ∑=× × = × + × + ×

∴ = =

50 4 180 180 2 180

720200

3 6

( ) ( ) ( ) ( )

.

4. Mechanism 3

Mechanism3isaswaymechanismshowninFigure7.16c.Theworkequationforthismechanismcanbewrittenas

W Mpδ θ

λ θ θ θ θ θ

λ

∑ ∑=× × = × + × + × + ×

∴ = =

30 3 5 60 60 60 60

240200

2

( . ) ( ) ( ) ( ) ( )

..286

8 m

4 4

150

18060 kN

30 kN

60

50 kN

60

150

3.5 m

3.5 m

90

40 kN

Figure 7.15 Two-storey frame.


(a)

30 λ40 λ

50λ

180

150

4 48 m

150

2θ60

9060 3.5 m

3.5 m

θθ

θθ

60 λ

(b)

30 λ

40 λ

50 λ

180150

4 48 m

150

60

90

60 3.5 m

3.5 m

θ θ

2θ60 λ

(c)

30 λ

40 λ

50 λ

180

150

8 m

150

60

90

60 3.5 m

3.5 m

θ

θ θ

θ60 λ

(d)

30 λ

40 λ

50 λ

180

150

8 m

150

60

90

60 3.5 m

3.5 mθ θ

θθ

60 λ

(e)

30 λ

40 λ

50 λ

180

150

8 m

150

60

90

60 3.5 m

3.5 mθ

θ

θ θ

θ

θ

60 λ

(f )

30 λ 40 λ

50 λ

180

150

8 m

150

60 90 60 3.5 m

3.5 m

θ

θ θ

θ

θ

θ

2θ 2θ

60 λ

Figure 7.16 Mechanisms of two-storey frame: (a) mechanism 1, (b) mechanism 2, (c) mechanism 3, (d) mechanism 4, (e) mechanism 5 and (f) mechanism 6.


5. Mechanism 4

Mechanism4isalsoaswaymechanismshowninFigure7.16d.Theworkequationforthismechanismcanbewrittenas

W Mpδ θ

λ θ θ θ θ θ

λ

∑ ∑=+ × × = × + × + × + ×

∴

( ) ( . ) ( ) ( ) ( ) ( )60 30 3 5 150 150 150 150

== =600315

1 9.

6. Mechanism 5

Mechanism5isacombinedmechanismshowninFigure7.16e.Theworkequationforthismechanismcanbewrittenas

W Mpδ θ

λ θ λ θ θ θ θ

∑ ∑=× × + × × = × × + × × + × ×30 7 60 3 5 2 150 2 60 2 180( ) ( . ) ( ) ( ) ( )

∴∴ = =λ780420

1 857.

7. Mechanism 6

Mechanism6isacombinedmechanismshowninFigure7.16f.Theworkequationforthismechanismcanbewrittenas

W Mpδ θ

λ θ λ θ λ θ θ θ

∑ ∑=× × + × × + × × = × × + × +30 7 60 3 5 40 4 2 150 60 2( ) ( . ) ( ) ( ) ( ) 22 180

960580

1 655

× × + ×

∴ = =

( )

.

θ θ

λ

90 (2 )

Therefore,thecollapseloadfactorisλc=1.655.

7.5 PlAStIc deSIgn to AS 4100

7.5.1 limitations on plastic design

Clause4.5ofAS4100requiresthatiftheplasticmethodofstructuralanalysisisused,allofthefollowingconditionsshallbesatisfied:

• Themembersusedshallbehot-formed,doublysymmetric,compactI-sections.• Theminimumyieldstressofthesteelshallnotexceed450MPa.• Thestress–straincharacteristicsofthesteelshallnotbesignificantlydifferentfrom

thoseofAS/NZS3678orAS/NZS3679.1.• Thestress–straincurveofthesteelshallhaveayieldplateauextendingforatleastsix

timestheyieldstrain.• Theratiooffu/fyisnotlessthan1.2.• The elongation of the steel is not less than 15% and it exhibits strain-hardening

characteristics.


• Noimpactloadingorfluctuatingloadingthatrequiresafatigueassessmentisappliedtothemembers.

• The connections shall have the capacity to cope with the formation of the plastichingesanddonotsuppresstheformationofplastichinges.

7.5.2 Section capacity under axial load and bending

AS4100givesspecificationsontheplasticdesignof in-planebeams,beam–columnsandframes.However,thebiaxialbendingisnotconsideredinAS4100owingtothecomplexityofbiaxialinteractionbehaviour.

ThedesignmomentcapacityofthesectionreducedbyaxialforceforbendingaboutthemajorprincipalaxisisgiveninClause8.4.3.4ofAS4100(1998)asfollows:

φ φ

φφM M

NN

Mprx sxs

sx= −∗

≤1 18 1. (7.18)

whereϕ=0.9,thecapacityreductionfactorφMsxisthedesignsectionmomentcapacityforbendingaboutthemajorprincipalx-axisN∗isthedesignaxialforceφNsisthedesignaxialsectioncapacity

Forasectionbentabouttheminorprincipalaxis,AS4100providesthefollowingequationforcalculatingthereduceddesignmomentcapacityofthesection:

φ φφ

φM MNN

Mpry sys

sy= −∗

≤1 19 1

2

. (7.19)

whereϕ=0.9,thecapacityreductionfactorφMsyisthedesignsectionmomentcapacityforbendingabouttheminorprincipaly-axis

7.5.3 Slenderness limits

Clause8.4.3.2ofAS4100giveslimitsontheslendernessofmemberswhichcontainplastichingesintermsofthedesignaxialcompressiveforce.ThedesignaxialcompressiveforceN∗inamembercontainingaplastichingeshallsatisfythefollowingconditions:

NN N N

NNs

m

s cr s

∗≤ +

∗≤

φ φβ0 6 0 4

0 152

. ..

/when (7.20)

NN

N N

N N

NNs

m s cr

m s cr s

∗≤

+ −

+ +

∗>

φ φ

β

β

1

10 15

/

/when . (7.21)

whereNsisthenominalaxialsectioncapacityofthememberNcristheelasticbucklingloadofthememberβmistheratioofthesmallertothelargerendbendingmoments


Themember,whichdoesnothaveaplastichinge,shouldbedesignedbasedontheelasticmethodifthefollowingconditionissatisfied:

NN

N N

N N

NNs

m s cr

m s cr s

∗>

+ −

+ +

∗>

φ

β

β φ

1

10 15

/

/and . (7.22)

Clause 8.4.3.3 of AS 4100 also gives limits on the webs of members containing plastichingesintermsofthedesignaxialcompressionforce.Inmemberscontainingplastichinges,thedesignaxialcompressiveforcesshouldsatisfythefollowingconditions:

NNs

nn

∗≤ − ≤ ≤

φλ λ0.6 for 45137

82 (7.23)

NNs

nn

∗≤ − < <

φλ λ1.91 for 2524 7

45.

(7.24)

NNs

n

∗≤ ≤ ≤

φλ1 for 0.0 25 (7.25)

Whenλn>82, thewebof themember is slender so that itmustnotcontainanyplastichinge.Themembermustbedesignedbasedontheelasticmethodortheframeshouldberedesigned.

referenceS

AS4100(1998)Australianstandardforsteelstructures,Sydney,NewSouthWales,Australia:StandardsAustralia.

Baker,J.andHeyman,J.(1969)PlasticDesignofFrames,London,U.K.:CambridgeUniversityPress.Horne,M.R.andMorris,L.J.(1981)PlasticDesignofLow-RiseFrames,London,U.K.:Collins.Neal,B.G.(1977)ThePlasticMethodsofStructuralAnalysis,London,U.K.:ChapmanandHall.

217

Chapter 8

composite slabs

8.1 IntroductIon

Compositefloorsystemsareformedbyconnectingfloorslabstothetopflangesofstructuralsteelbeams,girdersortrussesusingmechanicalshearconnectors.Theconcretefloorslabcanbeaconventionalreinforcedconcreteslaboracompositeslabwithprofiledsteelsheet-ingsupportingtheconcrete.Compositeslabshavebeenwidelyusedinmultistoreycompos-itebuildingsinmanycountries.Thiscompositeslabsystemutilisesthebestload-resistingcharacteristicsof steelandconcretematerials.Structural steelhas thepropertiesofhighstrength,highductilityandhighspeedoferection,whilestructuralconcretehastheproper-tiesofexcellentfireresistance,inherentmassandlowmaterialcost.Compositeslabscanbedesignedaseithersimplysupportedone-wayslabsorcontinuousslabs.

Currently,therearenoAustralianStandardsavailableforthedesignofcompositeslabs.Thischapterpresentsthebehaviouranddesignofcompositeslabsforstrengthandservice-abilitytoEurocode4(2004)andtoAustralianpractice.Theconceptofshearconnectionisintroducedfirst.ThedesignofsimplysupportedcompositeslabswithcompleteandpartialshearconnectionstoEurocode4isthendescribed.ThisisfollowedbythepresentationsofthedesignofcontinuouscompositeslabsforpositivemomentandnegativemomentregionsintermsofflexuralandverticalshearstrengthsinaccordancewithAustralianpractice.Thelongitudinalshearandpunchingsheararealsocovered.Thedesignofcompositeslabsforserviceabilityisgiven.

8.2 comPonentS of comPoSIte SlABS

Thecomponentsofa composite slab include theprofiled steel sheeting, cast in situ con-creteandreinforcementintheformofwelded-wiremeshordeformedbarsasschematicallydepictedinFigure8.1.

Theprofiledsteelsheetingisverythinwithbasismetalthicknessbetween0.6and1.0 mmforAustralianproducts.Thesteelsheetingispressedorcoldrolledandisdesignedtospaninthelongitudinaldirectiononly.Intheconstructionstage,beforecastingtheconcrete,theprofiledsteelsheetingactsasaplatformforconstruction.Aftercastingtheslabconcrete,thesheetingsupportsthewetconcreteandactsaspermanentformworkfortheconcrete.Aftertheconcretehashardenedandcompositeactionbetweenthesheetingandthecon-cretehasbeendeveloped,thesteelsheetingactsasbottomfacetensilereinforcementfortheconcreteslab.


Theprofiledsteelsheetingusedincompositebeamconstructionmustsatisfythegeomet-ricrequirementsgivenintheClause1.2.4ofAS2327.1(2003)asillustratedinFigure8.2:

• Theheightofthesteelrib(hr)shouldnotbegreaterthan80 mm.• Theconcretecoverslabthickness(hc=Dc−hr)shouldnotbelessthan65 mm.• Theopeningwidthofthesteelribatitsbaseshouldnotbegreaterthan20 mm.• Theareaofthevoidsduetotheopeningoftheribshouldnotbegreaterthan20%of

theareaoftheconcretewithinthedepthoftheribs.• Thewidthofconcretebetweenthemid-heightsofadjacentribsshouldnotbelessthan

150 mm.

Theprofiledsteelsheetingusuallyprovidesmorethanadequatebottomreinforcementforthecompositeslabsothatitcanbedesignedassimplysupportedtoutilisethestrengthoftheprofiledsteelsheeting.However,toplongitudinalreinforcementatthesupportsisstillneededtocontrolcracksiftheslabsaretreatedassimplysupported.InAustralia,itiscom-monpracticetodesigncontinuouscompositeslabswithnegativetensilereinforcementoverthesupportsforbendingandcrackcontrol.Positivetensilereinforcementmaybeprovidedtoincreasethemomentcapacityofcompositeslabs.Transversereinforcementmustbepro-videdincompositeslabsforcrackcontrolduetoshrinkageandtemperatureeffects.

Negative tensile reinforcement

Uniformly distributed load

Shear connector

Steel beam

Pro�led steel sheetingPositive tensile reinforcement

Transverse reinforcementfor crack control

Longitudinal shearreinforcement

Figure 8.1 Components of composite slab.

Longitudinal stiffenerbb≤ 20

hr≤ 80

bsr

Dchc≥ 65

bcr ≥ 150

sr

Figure 8.2 Profiled steel sheeting geometric restrictions.

Composite slabs 219

8.3 BehAvIour of comPoSIte SlABS

Thebehaviourofcompositeslabscanbedeterminedbyeitherexperimentsornumericalanalysissuchasthefiniteelementanalysis.Therearethreepossiblefailuremodesassociatedwithasimplysupportedcompositeslabinatwo-pointloadtest(Johnson2004).Thefailuremodedependsontheratiooftheshearspantotheeffectivedepthoftheslab(Ls/De).WhentheLs/Deratioishigh,thecompositeslabfailsbyflexureintheregionofmaximumpositivebendingmoment.WhentheLs/Deratioislow,thecompositeslabfailsbytheverticalshearnearthesupports.AtintermediatevaluesofLs/De,longitudinalshearfailureoccursattheinterfaceofthesheetingribsandtheconcretecoverslab.Thelongitudinalshearfailureisinitialisedbythecrackintheconcreteunderoneoftheloadpoints,whichassociateswiththelossofbondalongtheshearspanandslipattheendoftheslab.Theshearconnectionbetweentheconcreteandsheetingisbrittleiflongitudinalshearfailureoccurs.Forcontinu-ouscompositeslabs,flexuralandverticalshearfailuresmayoccurinthenegativemomentregions.Thedesignofcompositeslabsistoensurethatthefailuremodesmentionedearlierwillnotoccur.Forthispurpose,continuouscompositeslabsneedtobedesignedforpositiveandnegativebendingmomentsandverticalshearforces.

8.4 SheAr connectIon of comPoSIte SlABS

8.4.1 Basic concepts

Theshearconnectionofacompositeslabistheinterconnectionbetweentheprofiledsteelsheetingandtheconcrete,whichenablesthetwocomponentstoactasasinglestructuralmember.Theshearconnectionresiststhelongitudinalslipattheinterfaceofthesteelsheet-ingandconcrete.Therearethreemechanismsthatcontributetotheshearconnectionofacompositeslab.Thefirstmechanismisthechemicalbondbetweenthetwocomponents.Thesecondmechanismisthemechanicalinterlockprovidedbythedimpleswhicharepressedintothesurfaceofthesteelsheeting.Thethirdmechanismistheendanchoragewhichmaybeprovidedbypins,weldingstudsthroughthesheetingtothetopflangeofthesteelbeamorfrictionbetweenthesheetingandthesupports.

Whenno shear connectionbetween the sheetingandconcrete isprovided, there isnobondbetweenthesetwocomponentssothattheyactseparately.Iftheslipandslipstraininacompositeslabareeverywherezero,thisconditioniscalledfullinteractionofacompositeslab.Thisimpliesthatplanesectionsremainplaneafterdeformation.Thefullinteractionofacompositeslabisastiffnesscriterion.Whentheslipattheinterfaceofthesheetingandconcreteoccursalongthelengthofacompositeslab,thisconditioniscalledpartialinter-action,whichisastiffnesscriterion.Complete/fullshearconnectionofacompositeslabistheconditionforwhichitssectionmomentcapacityisgovernedbythestrengthofthesteelsheetingorconcretecoverslababovethesteelribs.Incontrast,thepartialshearconnectionofcompositeslabistheconditionforwhichitssectionmomentcapacityisgovernedbythestrengthoftheshearconnection.Itisnotedthatthecompleteorpartialshearconnectionsisconcernedwiththestrengthofcompositeslabssothatitisastrengthcriterion.

8.4.2 Strength of shear connection

The shear connection strengthof a composite slabdependson themechanical resistancewhichincludesthecontributionsofchemicalbondandmechanicalinterlockalongtheslaband on the frictional resistance at its supports. The steel sheeting in a simply supportedcompositeslabunderbendingissubjectedtoaresultanttensileforce(Tp),whilethetoppart


oftheconcreteslabisincompressionasschematicallydepictedinFigure8.3.Theresultanttensileforce(Tp)atacriticalcrosssectionisresistedbythemechanicalresistance(Hm)andthefrictionalresistance(Hf).Itisassumedthatthemechanicalresistance(Hm)isdevelopeduniformlyacrossthefullwidthofthecompositeslabandisexpressedasforceperunitplanarea(kPa).Themechanicalresistanceofacompositeslab isusuallydeterminedbyeitherfull-scaleslabtestsorsmall-scaleslip-blocktests(Patrick1990;PatrickandBridge1994).Testresultsshowedthatthemechanicalresistance(Hm)dependsontheprofilegeometry,thesheetingthicknessandthecompressivestrengthofconcrete.Themechanicalresistance(Hm)isdeterminedexperimentallybyBridge(1998)as88 t fbm c′ ,235and210kPaforprofiledsteelsheetingBondekII,ComformandCondeckHP,respectively,andaregiveninTable8.1.

Atacrosssectionwithcompleteshearconnection,theresultanttensileforceinthesteelsheeting(Tpcs)canbedeterminedfromtheforceequilibriumconditionusingtherectangularstressblock theory.The strengthof complete shear connection isgovernedbyeither thestrengthofthesteelsheetingorthestrengthoftheconcretecoverslabincludingthecontri-butionoflongitudinaltensilereinforcementintheconcrete.Foracompositeslabreinforcedwithconventionaltensilereinforcementinthebottomface,thestrengthofthereinforcedconcretecoverslabcanbeexpressedby

F f b D h Tcst c c r yr= ′ − −0 85. ( ) (8.1)

whereT A fyr r yr= istheyieldcapacityofthesteelreinforcementinthebottomfaceofthecom-

positeslabAristhecross-sectionalareaofthereinforcementfyristheyieldstressofthereinforcement

Table 8.1 Properties of profiled steel sheeting

Profiled steel sheeting hr (mm) bcr (mm) sr (mm) Hm (kPa) Ap (mm2) Mup (kN m/m) ϕb

Bondek II 54 187 200 88 t fbm c′ 1678tbm 13 8. tbm 1 2−βsc

Comform 58 300 300 235 1563tbm 10 7. tbm 1 3−βsc

Condeck HP 55 300 300 210 1620tbm 11 6. tbm 1 3−βsc

Source: Adapted from Goh, C.C. et al., Design of composite slabs for strength, composite structures design manual – Design booklet DB3.1, BHP Integrated Steel, Melbourne, Victoria, Australia, 1998.

Support reactionFrictional resistance

Pro�led steel sheetingMechanical resistance

Uniformly distributed load

Cc

Tp

R*End slip

Figure 8.3 Mechanical and frictional resistance in composite slab.

Composite slabs 221

Theresultanttensileforceinthesteelsheeting(Tpcs)withcompleteshearconnectionistakenas

T F Tpcs cst yp= min( , ) (8.2)

whereT A fyp p yp= istheyieldcapacityofthesteelsheetingApisthecross-sectionalareaofthesheetingfypistheyieldstressofthesheeting

8.4.3 degree of shear connection

Thedegreeofshearconnectionatacrosssectioninacompositeslabisdefinedastheratiooftheresultanttensileforce(Tp)totheresultanttensileforce(Tpcs)inthesteelsheetingwithcompleteshearconnection(Gohetal.1998),whichisexpressedby

β βsc

p

pcssc

TT

= ≤ ≤0 1 0. (8.3)

Ifthedegreeofshearconnectionatacrosssectionisknown,thestrengthoftheshearcon-nectiongoverningthemomentcapacityofthecompositeslabwithpartialshearconnectionisobtainedfromEquation8.3asTp=βscTpcs.

8.5 moment cAPAcIty BASed on eurocode 4

Atacrosssectionofacompositeslabwithcompleteshearconnectionandunderbending,the plastic neutral axis of the cross section is usually located in the concrete cover slab(abovethesteelsheeting),exceptwherethesheetingisverydeepthattheplasticneutralaxismaylieinthesheeting.However,therearetwoneutralaxesinacrosssectionwithpartialshearconnection.Thefirstplasticneutralaxisliesintheconcretecoverslab,whilethesec-ondfallsinthesheeting.Theultimatemomentcapacityofacompositeslabwithanydegreeofshearconnectiondependsonthelocationoftheplasticneutralaxis.ThecalculationoftheultimatemomentcapacityofcompositeslabsbasedonEurocode4(2004)isgivenindetailinthefollowingsections.

8.5.1 complete shear connection with neutral axis above sheeting

The longitudinal bending stress distribution through the depthof the cross sectionof acompositeslabwithcompleteshearconnectionisschematicallydepictedinFigure8.4.Forclarity, onlypartof the cross sectionof the composite slab is shown inFigure8.4.Therectangularstressblocktheoryisassumedforconcreteincompression.Theplasticneutralaxisisassumedtobeabovethesheeting.Theeffectiveareaofwidth(b)ofsheetingandtheheight(hp)ofthecentreofareaabovethebottomofthesheetingaredeterminedbytests.Thecompressiveforceintheconcretecoverslabcanbecalculatedby

N f b dcc c n= ′0 85. γ (8.4)

wherednistheneutralaxisdepthγisgiveninAS3600(2001)as

γ γ= − ′ − ≤ ≤0 85 0 007 28 0 65 0 85. . ( ) . .fc (8.5)


Assumingbothreinforcingsteelandprofiledsteelsheetingareatyieldattheultimatelimitstate,thecompressiveforceinconcretewithcompleteshearconnectionis

N T Tcc yp yr= + (8.6)

Theneutralaxisdepthdncanbedeterminedfromtheforceequilibriumas

d

Nf b

ncc

c

=′0 85. γ

(8.7)

Thenominalultimatemomentcapacityofthecompositeslabcanbecalculatedbytakingmomentsaboutthetopfibreas

M T d T d N du yp p yr r cc n= + − ( . )0 5γ (8.8)

wheredpisthedistancefromthetopfibretotheelasticcentroidofthesheetingdristhedistancefromthetopfibretothecentroidofsteelreinforcement

8.5.2 complete shear connection with neutral axis within sheeting

When theplasticneutralaxis is locatedwithin the sheetingas shown inFigure8.5, thecompressiveforceintheconcretewithcompleteshearconnectionignoringthecompressiveconcreteintheribsisgivenby

N f bhcc c c= ′0 85. (8.9)

wherehc=(Dc−hr)istheheightoftheconcretecoverslababovetheribs.Asdepicted inFigure8.5, there is a compressive forceNac in the steel sheetingbelow

theplasticneutralaxis.Thereisnosimplemethodfordeterminingtheplasticneutralaxisdepth(dn)andNacduetothecomplexpropertiesofprofiledsteelsheeting.InEurocode4,theapproximatemethodisused(Johnson2004).Thetensileforceinsteelsheetingisdecom-posedintoaforceatthebottomequaltoNacandaforceNp=Ncc.ThemomentcapacityMpr

hc

hr

dp

hp

dr

ep

b

dn

fyr

fyp

Tyr

Ncc

Typ

γdn

0.85 f c

Dc

Figure 8.4 Stress distributions in section with complete shear connection: PNA above sheeting.

Composite slabs 223

duetothecoupleforces(Nac)isdeterminedasthemomentcapacityofthesteelsheeting(Mpa)reducedbytheaxialforceNcc.InEurocode4(2004),Mprisapproximatelydeterminedby

M M

NN

pr pacc

p

= −

1 25 1. (8.10)

Themomentcapacityofthecompositeslabis

M N z Mu cc pr= + (8.11)

wherethelevelarmzisgivenby

z D h e e h

NN

c c p p pcc

p

= − − + −0 5. ( ) (8.12)

whereepisthedistanceofplasticneutralaxisabovethebaseofsteelsheetinghpisthedistanceofelasticcentroidabovethebaseofsteelsheeting

8.5.3 Partial shear connection

ThestressdistributioninsectionwithpartialshearconnectionispresentedinFigure8.6.Whenthecrosssectionofacompositeslabisinpartialshearconnection,thecompressiveforceintheconcrete(Ncp)islessthanNccandisdeterminedbythestrengthoftheshearconnection.Thedepth(dn)oftheneutralaxisintheconcretecoverslabis

d

Nf b

ncp

c

=′0 85. γ

(8.13)

AsshowninFigure8.6,thesecondneutralaxisfallsinthesheetingandthestressdistribu-tionissimilartothatshowninFigure8.5.InEurocode4(2004),themomentcapacity(Mpr)duetocoupleforces(Nac)isapproximatelydeterminedby

M M

NN

Mpr pacp

ppa= −

≤1 25 1. (8.14)

hc

hr

dp

hp

dn

ep

b

hc

fyp

Tyr

Ncc

Nac

Nac

Np

0.85 f c

Dc

Figure 8.5 Stress distributions in section with complete shear connection: PNA in sheeting.


Themomentcapacityofthecompositeslabcanbecalculatedas

M N z Mu cp pr= + (8.15)

wherethelevelarmzisgivenby

z D d e e h

NN

c n p p pcp

p

= − − + −0 5. ( )γ (8.16)

8.6 moment cAPAcIty BASed on AuStrAlIAn PrActIce

8.6.1 Positive moment capacity with complete shear connection

InAustralianpracticeofcompositeslabdesign,thesimpleplasticrectangularstressblocktheoryisusedinthecalculationofthemomentcapacityofacompositeslab.Itisassumedthatconventionalreinforcementlocatedonthetensilesideoftheneutralaxisyieldsattheultimatemoment,otherwiseitisignored.Thesheetingislumpedattheheightofitscentroidabovethebottomofthecompositeslab.Theheight(yp)ofthesheetingcentroidvarieswiththedegreeofshearconnection,whichisgivenasfollows(Gohetal.1998):

a.BondekII: ypsc sc

sc sc

=< ≤

− < ≤

18 0 75

21 6 6 1 1 0

2β ββ β

for 0

for 0.75

.

. . .

b.Comform: ypsc sc

sc sc

=< ≤

− < ≤

18 0 75

23 1 9 7 1 0

3β ββ β

for 0

for 0.75

.

. . .

c.CondeckHP: ypsc sc

sc sc

=< ≤

− < ≤

16 0 75

24 1 11 3 1 0

2β ββ β

for 0

for 0.75

.

. . .

Figure8.7givesthestressdistributioninthesectionwithcompleteshearconnection.Foracrosssectionwithcompleteshearconnection,theneutralaxisdepththatliesabovetheconcretecoverslabcanbecalculatedusingEquation8.7,providingthatbothsteelreinforce-mentandsheetingareatyield.Thestraininthesteelreinforcementisgivenby

εr

r n

n

d dd

= × −0 003. (8.17)

hc

hr

dp

hp

dr

ep

b

dn

fyr

fyp

NacNp

Nac

Tyr

Ncpγdn

0.85 f c

Dc

Figure 8.6 Stress distributions in section with partial shear connection.

Composite slabs 225

Ifthestraininsteelreinforcementisgreaterthantheyieldstrainofthesteelreinforcement,itscontribution to themomentcapacityof thecomposite slab isconsidered,otherwise itis ignored.Thenominalmoment capacityof the composite slab canbe calculatedusingEquation8.8.

Example 8.1: Moment capacity of section with complete shear connection

ThecrosssectionofacompositeslabincorporatingBondekIIprofiledsteelsheetinghascompleteshearconnection.Theoveralldepthoftheslab(Dc)is150 mm.Thecompres-sivestrengthofconcrete( )′fc is32MPa.Thethicknessofthesheeting(tbm)is1.0 mm.Theyieldstressofthesheetingis550MPa.Thecross-sectionalareaofbottomfacetensilereinforcement (Ar) in thecompositeslab is393 mm2/m.Determine thedesignpositivemomentcapacityofthesectionwithcompleteshearconnection.

1. Resultant tensile force in sheeting

TheareaofBondekIIsheetingisobtainedfromTable8.1as

A tp bm= = × =1678 1678 1 0 1678. mm /m2

Forcompleteshearconnection,theresultanttensileforceinsheetingisequaltoitsyieldcapacity,whichiscomputedas

T A fyp p yp= = × × =−1678 550 10 922 93 . kN/m

2. Neutral axis depth

Assumetheplasticneutralaxisislocatedintheconcretecoverslabandthesteelrein-forcementisatyieldattheultimatemoment.Theyieldforceinreinforcementis

T A fyr r yr= = × × =−393 400 10 157 23 . kN/m

Thecompressiveforceintheconcretecoverslabis

C f b dc c n= ′0 85. γ

γ = − ′ − = − × − =0 85 0 007 28 0 85 0 007 32 28 0 822. . ( ) . . ( ) .fc

hc

hr

dp

yp

dr

b

dn

fyr

fyp

Typ

Tyr

Ccγdn

0.85 f c

Dc

Figure 8.7 Stress distributions in section with complete shear connection: PNA above sheeting.


FromtheforceequilibriumCc=Typ+Tyr,theneutralaxisdepthdniscomputedas

d

T Tf b

nyp yr

c

=+′

=+ ×

× × ×=

0 85922 9 157 2 10

0 85 32 1000 0 82248

3

.( . . ). .γ

..3 mm

h D h dc c r n= − = − = > =150 54 96 48 3mm mm OK. ,

3. Check strain in reinforcement

UsingY10barsinbothdirectionsinthecompositeslab,thedepthofthelongitudinalreinforcementfromthetopfibreoftheslabis

dr = − − − =150 54 10

102

81mm

Thestraininsteelreinforcementcanbecalculatedas

ε εr

r n

ny

d dd

= ×−

= ×−

= > = =0 003 0 00381 48 3

48 30 00203

400200 000

0. ..

..

,..002

Thesteelreinforcementyieldsattheultimatelimitstate.

4. Design moment capacity

TheheightypforwhichTypactsforsectionwithcompleteshearconnectionisdeterminedas

yp sc= − = × − =21 6 6 1 21 6 1 0 6 1 15 5. . . . . .β mm

dp = − =150 15 5 134 5. . mm

Thecompressiveforceintheconcretecoverslabiscomputedas

C f b dc c n= ′ = × × × × × =−0 85 0 85 32 1000 0 822 48 3 10 10803. . . .γ kN/m

Thenominalmomentcapacityofthesectionis

M T d T d C du yp p yr r c n= + −

= × + × − × ×

( . )

. . . ( .

0 5

922 9 134 5 157 2 81 1080 0 5 0

γ

.. . ) .822 48 3 115 4× =kNmm kNm/m

Thedesignmomentcapacityofthecompositeslabsectionistherefore

φMu = × =0 8 115 4 92 3. . . kNm/m

8.6.2 Positive moment capacity with partial shear connection

Theresultanttensileforce(Tp)developedinthesteelsheetingdependsonthedegreeofshearconnectionatthecrosssectionandisresistedbythemechanicalresistanceforceHmxandthefrictionalforceμR∗(Gohetal.1998).Theresultanttensileforceinsheetingatthecriti-calsectionwithadistancexfromoneendofthesheetinginthecompositeslabwithpartialshearconnectioncanbedeterminedby

T H x R Tp m pcs= + ∗ ≤( )µ (8.18)

Composite slabs 227

whereμisthefrictioncoefficient,takenas0.5.Itisnotedthatthetensileforceinthesheet-ingvarieswiththedistancefromtheendofthesteelsheetingandisaffectedbythesupportreaction.Ifthesteelsheetingdoesnotextendoverthefullwidthofthesupport,thefric-tionalresistanceistakenaszero.Theresultanttensileforce(Tp)insheetingshouldbetakenasthelesservaluesofTp⋅LandTp⋅RcalculatedusingEquation8.18forthecriticalsectionwiththedistancefromtheleftandrightendsofthesheeting.Byignoringthefrictionalresis-tanceforce,thedistancemeasuredfromtheendofthesheetingtothecrosssectionwherethecompleteshearconnectionisattainedcanbecomputedfromEquation8.18as

x

TH

csyp

m

= (8.19)

Crosssectionslocatedatadistancefromtheendofthesheetinglessthanxcsareinpar-tialshearconnectionandshallbedesignedbasedonthepartialshearconnectionstrengththeory.Forthecrosssectionwithpartialshearconnection,thefirstneutralaxisislocatedintheconcretecoverslabasshowninFigure8.8.Thecompressiveforceinconcreteisgivenby

C f b dc c n= ′0 85. γ (8.20)

Itisassumedthatthesteelreinforcementyieldsattheultimatemomentandtheresultanttensileforce(Tp)inthesheetingislessthanTpcs.Thisneutralaxisdepth(dn)intheconcretecoverslabcanbecalculatedby

d

T Tf b

np yr

c

=+′0 85. γ

(8.21)

Itshouldbenotedthatthestrainintheconventionalsteelreinforcementneedstobecheckedagainstitsyieldstrain.Ifthereinforcementisnotatyield,itcanbeignoredinthecalculation.

ThemomentcapacityduetothecoupleforcesNacisrepresentedbyMupφb,whichdependsontheaxialforceNacandthesectionpropertiesoftheprofiledsteelsheeting.Thenominalultimate moment capacity of the composite slab can be determined by taking momentsaboutthetopfibreofthesectionas

M T d T d C d Mu p p yr r c n up b= + − +( . )0 5γ ϕ (8.22)

whereMupisthenominalmomentcapacityofthesheetingaloneϕbisthebendingfactorofthesheetingwhichisafunctionofthedegreeofshearcon-

nectiongiveninTable8.1(Gohetal.1998)

hc

hr

dp

yp

dr

b

dn

fyr

fyp

Nac

TpNac

Tyr

Ccγdn

0.85 f c

Dc

Figure 8.8 Stress distributions in section with partial shear connection.


8.6.3 minimum bending strength

Topreventthesuddencollapseofcompositeslabsthatexhibitbrittlefailure,themomentcapacity at each cross section in thepositivemoment regionsmust satisfy the followingminimumbendingstrengthrequirement(AS36002001):

M M bD fu u c c≥ = ′,min .0 12 2 (8.23)

Inthepositivemomentregions,theminimumbendingstrengthrequirementissatisfiedifthemechanicalresistance(Hm)isgreaterthan100MPa,thecompositeslabsaresubjectedtouniformlydistributedloadsandtheslabsmeetthedeflectionlimitsandhaveaspantodepthratioofL/Dc≥15(Gohetal.1998).

Example 8.2: Moment capacity of section with partial shear connection

ThecrosssectionofacompositeslabincorporatingComformprofiledsteelsheetinghaspartialshearconnectionofβsc=0.6.Theoveralldepthoftheslab(Dc)is160 mm.Thecompressive strengthof concrete ( )′fc is40MPa.The thicknessof the sheeting (tbm) is1.0 mm.Theyieldstressofthesheetingis550MPa.Thecross-sectionalareaofbottomfacetensilereinforcement(Ast)inthecompositeslabis393 mm2/m.Determinethedesignpositivemomentcapacityofthesectionwithpartialshearconnection.

1. Resultant tensile force in sheeting

TheareaoftheComformsheetingiscalculatedas

A tp bm= = × =1563 1563 1 0 1563. mm /m2

Assumethereinforcementisatyieldattheultimatemoment.Theyieldcapacityofsteelreinforcementis

T A fyr r yr= = × × =−393 400 10 157 23 . kN/m

Thestrengthofreinforcedconcretecoverslabis

F f b D h Tcst c c r yr= ′ − −

= × × × − × − =−

0 85

0 85 40 1000 160 58 10 157 23

. ( )

. ( ) . 33310.8 kN/m

Theyieldcapacityofsheetingis

T A fyp p yp= = × × =−1563 550 10 859 653 . kN/m

Theresultanttensileforceinsheetingwithcompleteshearconnectionistakenas

T F Tpcs cst yp= = =min( , ) min( . , . ) .3310 8 859 65 859 65 kN/m

Theresultanttensileforceinsheetingwithpartialshearconnectionisgivenby

T Tp sc p cs= = × =⋅β 0 6 859 65 515 79. . . kN/m

Composite slabs 229


Forsectionwithpartialshearconnection,thefirstplasticneutralaxisislocatedintheconcretecoverslab.Thecompressiveforceintheconcretecoverslabis

C f b dc c n= ′0 85. γ

γ = − ′ − = − × − =0 85 0 007 28 0 85 0 007 40 28 0 766. . ( ) . . ( ) .fc

FromtheforceequilibriumCc=Tp+Tyr,theneutralaxisdepthdnis

d

T Tf b

np yr

c

=+′

=+ ×

× × ×=

0 85515 79 157 2 10

0 85 40 1000 0 76625

3

.( . . ). .γ

..8 mm

h D h dc c r n= − = − = > =160 58 102 25 8mm mm OK. ,

3. Check reinforcement strain

UsingY10barsinbothdirectionsinthecompositeslab,thedepthofthelongitudinalreinforcementfromthetopfibreoftheslabis

dr = − − − =160 58 10

102

87 mm

Thestraininsteelreinforcementcanbecalculatedas

ε εr

r n

ny

d dd

= ×−

= ×−

= > = =0 003 0 00387 25 8

25 8400

200 0000 0. .

..

.0.007,

002

Thesteelreinforcementisatyield.


Theheightofsheetingypforsectionwithβsc=0.6iscalculatedas

yp sc= = × =18 18 0 6 3 93 3β . . mm

dp = − =160 3 9 156 1. . mm

Thecompressiveforceinconcretecoverslabiscomputedas

C f b dc c n= ′ = × × × × × =−0 85 0 85 40 1000 0 766 25 8 10 617 93. . . . .γ kN/m

ThenominalmomentcapacityofthebaresheetingisobtainedfromTable8.1as

M tup bm= = × =10 7 10 7 1 0 10 7. . . . kNm/m

Thebendingfactorofthesheetingis

ϕ βb sc= − = − =1 1 0 6 0 7843 3. .


Thenominalmomentcapacityofthesectionis

M T d T d C d Mu p p yr r c n up b= + − +

= × + × −

( . )

. . . .

0 5

515 79 0 1561 157 2 0 087 6

γ ϕ

117 9 0 5 0 766 0 0258 10 7 0 784. ( . . . ) . .× × × + ×

= 96.5 kNm/m

Thedesignmomentcapacityofthecompositeslabsectionistherefore

φMu = × =0 8 96 5 77 2. . . kNm/m

8.6.4 design for negative moments

Continuous composite slabs at the interior supports are subjected to negative bendingmoments.Negativetensilereinforcementneedstobeprovidedatthetopfaceofthecon-tinuouscompositeslabover thesupports.Thecontributionofsteelsheetingtothenega-tivemomentcapacityofcompositeslabsisusuallyignored.Therefore,thedesignnegativemomentcapacityofthecompositeslabiscalculatedas

φ φ γ γM f b k d k du c u u= ′ −( . )( . )0 85 1 0 5 (8.24)

whereϕ=0.8isthecapacityreductionfactork d du n= /d istheeffectivedepthofthecompositeslabmeasuredfromthecentroidoftopface

reinforcementtotheextremefibreofcompression

Toachieveductiledesigns,theneutralaxisparameter(ku)mustnotexceed0.4asrequiredbyAS3600.TherequiredneutralaxisparametercorrespondingtotheminimumamountoftopfacereinforcementcanbedeterminedfromEquation8.24as

k

q q qu =

− −1 12

2

γ (8.25)

whereq1istakenas1.0q2isgivenby

q

Mf bdc

2 2

20 85

=∗

′−

φ . (8.26)

ThemajorAustralianproductsofprofiledsteelsheetinghaveahighyieldstressof550MPasothattheyprovidethecompositeslabwithalargepositivemomentcapacity.Toachieveeconomicaldesignsofcontinuouscompositeslabs,itisdesirabletoredistributethebendingmomentsfromthenegativemomentregionstothepositivemomentregions.ThemomentredistributionincontinuouscompositeslabsshouldbeinaccordancewiththeClause7.6.8ofAS3600(2001).Ifthemomentredistributionisusedinthedesign,ClassNconventional

Composite slabs 231

reinforcementmustbeusedasnegativetensilereinforcement.Thenegativedesignbendingmomentafterredistributionisgivenby

M MRm− −

∗ = − ∗( )1 ξ (8.27)

whereM−∗isthenegativedesignbendingmomentatthesupportobtainedbyelasticanalysis

ξmisthemomentredistributionparameter,whichistakenas0.3fortheneutralaxisparameterku ≤ 0 2. and( . . )0 3 0 75− ku for0 2 0 4. .< ≤ku (Gohetal.1998)

Fordesignincorporatingmomentredistributionfromnegativemomentregionstopositivemomentregions,theparametersq1andq2inEquation8.25aregivenby(Gohetal.1998)

q

k

Mf bd

k

u

cu

1

2

1 0 2

10 750 85

0 4=

≤

−∗

′< ≤

−

for

for 0.2

.

..

.φ γ

(8.28)

q

Mf bdc

2 2

1 40 85

=∗

′−.

.φ (8.29)

Theminimumcross-sectionalareaofnegativereinforcementcanbedeterminedfromtheforceequilibriumofthesectionasfollows:

A

f b k df

stc u

yr

=′0 85. γ

(8.30)

Example 8.3: Design of composite slab for negative moments

The interior supportofa continuouscomposite slab supportedon steelbeams is sub-jected to a negative design bending moment of 27 kNm. The depth of the compositeslabis140 mm.Thecompressivestrengthofconcreteis32MPa.Theconcretecoveris25 mm.Thedepthofnegativetensilereinforcementfromthetopfibreisassumedtobe30 mm.Theyieldstressofthereinforcementis400MPa.(a)Determinetheamountofnegativetensilereinforcementrequiredattheinteriorsupportfordesignnotincorporat-ing moment distribution. (b) Determine the amount of negative tensile reinforcementrequiredattheinteriorsupportfordesignincorporatingmomentdistribution.a. Design not incorporating moment redistributionTheeffectivedepthofthecompositeslabundernegativemomentis

d D dc ct= − = − =140 30 110 mm

Forthedesignofcompositeslabwithoutmomentredistribution,q1=1andq2iscalcu-latedasfollows:

q

Mf bdc

2 2

6

2

20 85

2 27 100 8 0 85 32 1000 110

0 205=∗

′=

× ×× × × ×

=−

φ . . ..

γ = − ′ − = − × − =0 85 0 007 28 0 85 0 007 32 28 0 822. . ( ) . . ( ) .fc


Theneutralaxisparameterkuiscomputedas

k

q q qu =

− −=

− −= <1 1

22

21 1 0 2050 822

0 132 0 4γ

..

. . ,OK

Therequiredminimumcross-sectionalareaofnegativetensilereinforcementis

A

f b k df

stc u

yr

=′

=× × × × ×

=0 85 0 85 32 1000 0 822 0 132 110

400812

. . . .γmm2 //m

b. Design incorporating moment redistribution

Assumetheneutralaxisparameterku≤0.2.Forthedesignofcompositeslabincorporat-ingmomentredistribution,q1=1andq2iscalculatedasfollows:

q

Mf bdc

2 2

6

2

1 40 85

1 4 27 100 8 0 85 32 1000 110

0 144=∗

′=

× ×× × × ×

=−..

.. .

.φ

Theneutralaxisparameterkuiscomputedas

k

q q qu =

− −=

− −= ≤1 1

22

21 1 0 1440 822

0 091 0 2γ

..

. . ,OK

Therequiredminimumcross-sectionalareaofnegativereinforcementis

A

f b k df

stc u

yr

=′

=× × × × ×

=0 85 0 85 32 1000 0 822 0 091 110

400560

. . . .γmm2 //m

8.7 vertIcAl SheAr cAPAcIty of comPoSIte SlABS

8.7.1 Positive vertical shear capacity

Experimentshavebeenconductedonsimplysupportedcompositeslabsincorporatingpro-filedsteelsheetingunderaverticallineloadplacedatadistanceof1.5Dcfromthesupport(Patrick1993).Testresultsindicatedthatthecompositeslabdidnotfailbyverticalshearbeforetheultimateloadcorrespondingtoitsmomentcapacitywasattained.Thisimpliesthat thepositive vertical shear capacity (ϕVuc) of a simply supported composite slab canbecalculatedbyitspositivemomentcapacity(ϕMu)atthecrosssectionwithadistanceof1.5Dcfromthesupport.Thesheetingandfullyanchoredreinforcementcontributetotheverticalshearcapacityofthecompositeslabinthepositivemomentregions.Ahypotheticallineloadisassumedtobeplacedatadistanceof1.5DcfromthefaceofthehypotheticalsupportasdepictedinFigure8.9.Forcontinuouscompositeslabsunderuniformlydistrib-utedloadonallspans,ahypotheticalsupportcanbeplacedateachpointofcontraflexure.Thedesignverticalshearcapacityofacompositeslabinthepositivemomentregionscanbecalculatedby(Gohetal.1998)

φ φV

MD y

uu

c p

=−1 5. ( )

(8.31)

whereϕ=0.8isthecapacityreductionfactorφMuisthedesignmomentcapacityofthecompositeslabypistheheightoftheprofiledsteelsheetingatwhichthetensileforceTpacts

ItshouldbenotedthatϕMuandyparecalculatedatthelocationofthehypotheticallineload.

Composite slabs 233

For simple spansand the edge support regionsof end spans, thedesignvertical shearcapacityconsideringthecontributionofsheetingbutignoringthecontributionofreinforce-mentcanbeapproximatelycomputedby(Gohetal.1998)

φ

φ

µV

H M bD bDu

m up c c=

+( )−

1 5

1 5

2. ( )

.

/ (8.32)

wherethecapacityreductionfactorϕ=0.8.

8.7.2 negative vertical shear capacity

Thedesignforverticalshearofcompositeslabsinnegativemomentregionsistreatedasthesameasthatofreinforcedconcreteslabs.Compositeslabsinnegativemomentregionsaretreatedassolidreinforcedconcreteslabs.AsspecifiedinAS3600(2001),thedesignnegativeverticalshearcapacityofacompositeslabiscalculatedby

φ φβ β βV b d

f Ab d

u v oc st

v o

=′

1 2 3

1 3/

(8.33)

wherebvistheeffectivewidthoftheslabforverticalshearβ2 1=β3 1=β1isgivenby

β1 1 1 1 6

10001 1= −

≥. . .

do (8.34)

P

(b)

Bending moment diagram

Shear force diagram

Point of contra�exure 1.5Dc

(a)

Figure 8.9 Model for positive vertical shear capacity: (a) actual continuous composite slab and (b) hypotheti-cal simply supported composite slab.


InEquation8.33,Astisthecross-sectionalareaoflongitudinalnegativetensilereinforce-mentwhichisfullyanchored.

8.7.3 vertical shear capacity based on eurocode 4

InEurocode4,theverticalshearcapacityofacompositeslabisassumedtobeprovidedbytheconcreteribs.Thereinforcementthatisfullyanchoredbeyondtheshearcriticalcrosssectionisconsideredtocontributetotheverticalshearcapacity.However,thecontributionofsteelsheetingisignored.Theresistanceofacompositeslabtoverticalshear(designverti-calshearcapacity)perunitwidthisgiveninEurocode4(2004)as

V

bs

d vucr

rp=

min (8.35)

wherebcristhewidthofconcreteribatthemid-heightofthesteelribsinthecompositeslabsristhespacingofsteelribsvministheshearstrengthoftheconcrete,whichisexpressedby

vd

fp

ckmin

/

.= +

0 035 1200

3 2

(8.36)

wheredp ≥ 200 mmvminandfckareinMPa

8.8 longItudInAl SheAr

AsdescribedinSection8.4.1,threemechanismscontributetothetransferoflongitudinalshearincompositeslabsincorporatingprofiledsteelsheeting.Shear-bondtestswereusuallyperformedtodeterminetheresistanceofcompositeslabstolongitudinalshear.Them–kmethod isused in thedesignof longitudinal shear incomposite slabs inEurocode4.AsspecifiedinEurocode4(2004),thedesignlongitudinalshearcapacityofacompositeslabmustsatisfy

φ φV bd

mAbL

k Vl pp

s

= +

≥

∗ (8.37)

whereϕ=0.8thecapacityreductionfactorbisthewidthofslabmandkareconstantsthataredeterminedbyexperimentsV∗istheverticalshearatanendsupportwherethelongitudinalshearfailureoccursin

ashearspanofLs(Johnson2004)

TheshearspanLsistakenasL/4foracompositeslabwithspanofLandunderuniformlydistributedload.

Composite slabs 235

Them–kmethodisshowntobeadequatefordesigningcompositeslabswithshortspans(Johnson2004).However,thismethodisnotbasedonamechanicalmodelanddoesnotaccountfortheeffectsofendanchorageandfrictionabovethesupports.

8.9 PunchIng SheAr

Punchingshear failuremayoccur in thincompositeslabsunderconcentrated loads.Thepunching shear capacity of thin composite slabs that support point loads needs to bechecked.Itisassumedthatpunchingshearoccuronacriticalperimeteroflengthups.Theloadedareaap×bpoftheconcentratedloadisassumedtospreadthroughascreedofthick-nesshfat45°.Theeffectivedepthofthecompositeslabistakenashc.Thecriticalperimeterlengthisdeterminedas(Johnson2004)

u h b h a h d hps c p r p f p c= + + + + + −2 2 2 2 2 2 2π ( ) ( ) (8.38)

ItisassumedthattheareasofreinforcingmeshperunitwidthabovethesteelsheetingribsareAsxandAsyinxandydirections,respectively.Thereinforcementratiosareρx=Asx/hcandρy=Asy/hc.TheeffectivereinforcementratioisgiveninEN1992-1-1asρ ρ ρs x y= ≤ 0 02. .Thedesignpunchingshearstressisgivenby(Eurocode42004)

v f

dvps s ck

om

= +

≥0 12 100 1

2001 3. ( ) /minρ (8.39)

wheredom≥200mmistheaverageeffectivedepthofthetwolayersofreinforcementandvminisgivenbyEquation8.36.

Thepunchingshearcapacityofthecompositeslabis

φV v u dps ps ps om= (8.40)

8.10 deSIgn conSIderAtIonS

8.10.1 effective span

Theeffectivespanofacompositeslabdependsonitssupportconditions.Whenacompos-iteslabissupportedonsteelbeams,itseffectivespanistakenasthedistancebetweenthecentrelinesofadjacentsteelbeams.Whenacompositeslabissupportedonmasonrywalls,itseffectivespanistakenasthelesserof[Ln+(bs1+bs2)/2]and(Ln+Dc),whereLndenotesthecleardistancebetweenthesupportfacesandbs1andbs2arethewidthsoftheadjacentmasonrysupports.Foracompositeslabwherethesteelribsarenotorientedperpendiculartothesupportlines,theslabshouldbedesignedasaseriesofparallelstrips.Theeffectivespanofeachdesignstripistakenasthedistancebetweenthecentrelinesofthestrip.

8.10.2 Potentially critical cross sections

Thepotentiallycriticalcrosssectionofacompositeslabisacrosssectionthatmaygoverntheflexuralandshearstrengthsoftheslab.Designcheckforstrengthsshouldbeundertaken


at thepotentially critical cross sectionsof a composite slab.Fordesign forbendingandshear,potentiallycriticalcrosssectionsareasfollows:

• Sectionssubjecttothemaximumdesignpositivebendingmoment• Sectionsubjecttomaximumnegativebendingmoment• Sectionssubjecttomaximumdesignshearforce• Sectionswithadistanceequaltothetensiledevelopmentlengthawayfromthetermi-

natedendofthereinforcement• Foracompositeslabunderuniformlydistributedload,sectionsatone-thirdandtwo-

thirdsofthedistancemeasuredfromthemaximumpositivemomenttotheendsofthespanoradjacentcontraflexurepoints

8.10.3 effects of propping

The construction of composite slabs is classified into unpropped and propped. Inunpropped construction, the profiled steel sheeting must support its self-weight, theweightofwetconcreteandreinforcementandanyconstructionloadsbeforethehard-eningof the concrete.The spanof composite slabswhichareunpropped in construc-tionisusually2–3m.Itisassumedthatthecompositeactionbetweentheinterfaceofthesteelsheetingandtheconcrete isachievedwhentheconcretecompressivestrengthreaches 15 MPa as specified in AS 2327.1 (2003). In propped construction, the steelsheetingspanscanbechosentoavoidlargedeflections.Thepositivemomentcapacityofacompositeslabisnotaffectedbytheconstructionmethod,namely,unproppedorproppedconstruction.Asaresultofthis,theconstructionsequenceisnotconsideredinthestrengthdesignofacompositeslab.

Example 8.4: Design of continuous composite slab for strength

Atwo-spancontinuouscompositeslabsupportedonsteelbeamsisshowninFigure8.10.Theslabissubjectedtoaliveloadof4kPaandasuperimposeddeadloadof1.0kPa.Theconcretecompressivestrength ( )′fc is25MPa.Thecross-sectionalareaofthebot-tomfacetensilereinforcementisAst=500 mm2/m.Thecentroidheightofthebottomfacereinforcementfromtheslabsoffitis60 mm.Theyieldstressofthereinforcementis400 MPa.TheCondeckHPprofiledsteelsheetingwithtbm=0.75 mmisused.Theyieldstressof the sheeting is550MPa.Calculate the amountofnegative tensile reinforce-mentatsupportBfordesignnotincorporatingmomentdistribution,checkthepositivemomentcapacityofthesectionwithadistancex=1401 mmmeasuredfromtheendofthesheetingasdepictedinFigure8.5andcheckthepositiveandnegativeverticalshearcapacitiesofthecompositeslab.

3200

89A B Pro�led steel sheeting 89

120C

x= 1401

3200

Figure 8.10 Two-span continuous composite slab.

Composite slabs 237

1. Design actions

Thedesignwidthoftheslabistakenas1mandtheunitweightofcompositeslabwithreinforcementistakenas25kN/m3.

Deadload:G=(0.12×25+1.0)×1=4kN/mLiveload:Q=4×1=4kN/mThedesignload:w∗=1.2G+1.5Q=1.2×4+1.5×4=10.8kN/m

Themaximumpositivedesignbendingmomentoccursatx=1401 mmfromtheendofthesheetingwhenliveloadisonthefirstspanonly:

M+∗ = 9 45. kNm/m

ThepositiveshearforceatsupportAisVA∗ = 14 3. kN/m.

Whenliveloadisonbothspans,themaximumdesignnegativebendingmomentatsup-portBisobtainedas

M−∗ = 13 3. kNm/m

ThenegativeshearforceatsupportBisVB∗ = 21 4. kN/m.

ThereactionatsupportBis

RA∗ = + × =14 3 15 3. .10.8 0.089 kN/m

2. Negative tensile reinforcement

Theeffectivedepthofthecompositeslabinthenegativemomentregionis

d D dc ct= − = − =120 30 90 mm

Themomentredistribution isnotconsidered inthedesignof thiscompositeslab.Theparametersq1=1andq2iscalculatedas

q

Mf bdc

2 2

6

2

20 85

2 13 3 100 8 0 85 25 1000 90

0 193=∗

′=

× ×× × × ×

=−

φ ..

. ..

γ = − ′ − = − × − = >0 85 0 007 28 0 85 0 007 25 28 0 871 0 85. . ( ) . . ( ) . .fc

∴ =γ 0 85.

Thenaturalaxisparameterkuiscomputedas

k

q q qu =

− −=

− −= <1 1

22

21 1 0 1930 85

0 12 0 4γ

..

. . ,OK

Therequiredminimumcross-sectionalareaofnegativereinforcementis

A

f b k df

stc u

sy

=′

=× × × × ×

=0 85 0 85 25 1000 0 85 0 12 90

400488

. . . .γmm /m2


3. Positive moment capacity

3.1. Resultant tensile force in sheeting

Thecross-sectionalareaofbaresteelsheetingis(Table8.1)

A tp bm= = × =1620 1620 0 75 1215. mm /m2

Theyieldcapacityofsteelsheetingiscomputedas

T A fyp p py= = × × =−1215 550 10 668 253 . kN/m

ThemechanicalresistanceofCondeckHPisHm=210 kPa.Thedistancexcsfromtheendofsheetingtothesectionwithcompleteshearconnection

isgivenby

x

TH

csyp

m

= = =668 25210

3 182.

. m

Sincex=1.401m<xcs=3.182m,thesectionatx=1.401misinpartialshearconnection.Theyieldcapacityofbottomreinforcementis

T A fyr r yr= = × × =−500 400 10 2003 kN

Thestrengthofthereinforcedconcretecoverslabiscomputedas


= × × × − × − =−

0 85

0 85 25 1000 120 55 10 200 113

. ( )

. ( ) 881 25. kN/m

Theresultanttensileforcedevelopedinsheetingwithcompleteshearconnectionis

T F Tpcs cst pcs= = =min( , ) min( . , . ) .1181 25 668 25 668 25 kN

Thetensileforceinsheetingatsectionwithdistancex=1.401mfromtheleftendofthesheetingisdeterminedas

T H x R Tp L m A pcs⋅ = + ∗ = × + × = < =µ 210 1 401 0 5 15 3 301 86 668 25. . . . .kN/m kN/mm

Hence,Tp=301.86 kN/m.

3.2. Neutral axis depth

Theneutralaxisdepthdnintheconcretecoverslabiscalculatedas

d

T Tf b

np yr

c

=+′

=+ ×

× × ×=

0 85301 86 200 10

0 85 25 1000 0 8527 78

3

.( . ). .

.γ

mmm

3.3. Check reinforcement strain

Thestraininthesteelreinforcementis

εr

r n

n

d dd

= ×−

= ×− −

=0 003 0 003120 60 27 78

27 780 0035. .

( . ).

.

Composite slabs 239

Theyieldstainofsteelreinforcementis

ε εsy

sy

sr

fE

= = = < =400

200 0000 002 0 0035

,. .

Hence,thesteelreinforcementyieldsatultimatemomentcapacity.

3.4. Design moment capacity

Thedegreeofshearconnectionatthesectionofx=1.401misgivenby

βsc

p

yp

TT

= = =301 86668 25

0 45..

.

Theheightofcentroidofsheetingfor0<βsc=0.45≤0.75isobtainedfromTable8.1as

yp sc= = × =16 16 0 45 1 463 3β . . mm

Hence,

d dp r= − = = − =120 1 46 118 54 120 60 60. . mm, mm


ϕ βb sc= − = − =1 1 0 45 0 9093 3. .


C f b dc c n= ′ = × × × × × =−0 85 0 85 25 1000 0 85 27 78 10 501 83. . . . .γ kN/m

Thenominalmomentcapacityofthebaresteelsheetingis

M tup bm= = × =11 6 11 6 0 75 8 7. . . . kNm/m

Theminimalpositivemomentcapacityofthecompositeslabatx=1.401miscalculatedas


= × + × −

( . )

. .

0 5

0 11854 200 0 060 50

γ ϕ

301.86 11 8 0 5 0 85 0 02778 8 7 0 909

49 8

. ( . . . ) . .

.

× × × + ×

= kN m/m

Thedesignpositivemomentcapacityis

φM Mu = × = > ∗ =+0 8 49 8 39 84 9 45. . . . ,kNm/m kNm/m OK

4. Positive vertical shear capacity

ThedesignpositiveshearforceatadistanceofDcfromthesupportAis

V∗ = − × =14 3 10 8 0 12 13. . . kN/m


Thedesignverticalshearcapacityofthecompositeslabiscalculatedas

φφ

µV

H M bD bDu

m up c c=

+( )−

=× × + ×(

1 5

1 5

0 8 1 5 210 8 7 1 0 12

2

2

. ( )

.

. . ( . . )

/

/ ))× ×

−= > ∗ =

1 0 12

1 5 0 588 24 13

.

. .. ,kN/m kN/m OKV

5. Negative vertical shear capacity

ThedesignnegativeshearforceatadistanceofDcfromthesupportBis

V∗ = − × =21 4 10 8 0 12 20 1. . . . kN/m

β1 1 1 1 6

10001 1 1 6

120 301000

1 66 1 1= −

= −

−

= >. . . . . .

do

Thedesignnegativeverticalshearcapacityofacompositeslabistherefore

φ φβ β βV b df Ab d

u v oc st

v o

=′

= × × × × ××

1 2 3

1 3

0 8 1 66 1 1 1000 9025

/

. .4488

1000 9047 4 20 1

1 3

×

= > ∗ =

/

. . ,kN/m kN/m OK.V


8.11.1 crack control of composite slabs

Crackcontrolisanimportantdesignconsiderationofcompositeslabs.Ifthecompositeslabiscontinuousovertheinternalsupport,crackingwilloccurinthetopfaceoftheslaboverthesupport.Eachspanoftheslabmaybedesignedassimplysupportedtousethebenefi-cial effectofhigh-strengthsteel sheetingmaterial.However, thiswill lead tomore severecrackinginthetopfaceoftheslaboverthesupport.Tocontrolcracking,longitudinalrein-forcementmustbeprovidedaboveinternalsupports.InEurocode4(2004),theminimumcross-sectionalareaofthisreinforcementistakenasfollows:0.2%ofthecross-sectionalareaoftheconcretecoverslababovetheribsshouldbeprovidedforunproppedconstructionand0.4%forproppedconstruction.

AsspecifiedinClause9.1.1ofAS3600,forreinforcedconcreteslabssupportedonbeamsorwalls,theminimumtensilereinforcementratioofAst/bdshouldnotbelessthan0.8/fsy.Forcompositeslabssupportedonbeamsorwalls,theminimumtensilesteelareaincludingtheareasofsteelsheetingandconventionalreinforcementshouldbetakenasnotlessthan0.002bhc.Tocontrolflexuralcrackingincompositeslabs,thecentre-to-centrespacingofbarsinprimarydirectionshouldnotexceedthelesserof2.5Dor500 mm.Theareaofsteelsrequiredtocontrolcrackingduetoshrinkageandtemperatureeffectsisinfluencedbytheflexureaction,thedegreeofrestraintagainstin-planemovementandexposureclassifica-tionandshouldbedetermined inaccordancewithClause9.4.3ofAS3600(2001).Thesteelsheetingisconsideredtocontributetothecontrolofcrackingduetoshrinkageandtemperatureeffects.

Composite slabs 241

8.11.2 Short-term deflections of composite slabs

InAS3600,thedeflectionofone-wayreinforcedconcreteslabsunderuniformlydistributedloadiscalculatedusingaprismaticbeamofunitwidth.AsimplifiedmethodisgiveninAS3600forcalculatingthedeflectionsofreinforcedconcretebeams.Thissimplifiedmethodisadoptedhereforcalculatingthedeflectionsofcompositeslabswithprofiledsteelsheet-ing. The immediate deflections of composite slab under short-term service loads can becalculatedusingYoung’smodulusofconcrete(Ecj)andtheeffectivesecondmomentofareaofthecompositeslab(Ief).Theeffectivesecondmomentofarea(Ief)ofasectionisbetweenthesecondmomentofareaofthecrackedsection(Icr)andthesecondmomentofareaoftheuncrackedgrosssection(Ig).Thesecondmomentofareaofthecrackedsection(Icr)inacompositeslabcanbecomputedusingthetransformedsectionmethodofelasticanalysis.Inthismethod,theareasofsteelsheetingandconventionalreinforcementsaretransformedtoequivalentconcreteareasusingthemodulusratio(n=Es/Ec)asdepictedinFigure8.11.Theneutralaxisdepthdncanbedeterminedbyequatingthefirstmomentsofareaofthecompressiveandtensileareasabouttheneutralaxisasfollows:

12

2bd nA d d nA d dn p p n r r n= − + −( ) ( ) (8.41)

Thesecondmomentofareaofthecrackedsectioncanbeobtainedbytakingthesecondmomentsofareasabouttheneutralaxisas

I bd nA d d nA d dcr n p p n r r n= + − + −1

33 2 2( ) ( ) (8.42)

The effective secondmomentof areaof the section considered is evaluatedby (Branson1963)

I I I I

MM

Ief cr g crcr

seg= + −

≤( )3

(8.43)

whereMseisthebendingmomentatthesectionundershort-termserviceloadMcristhecrackingmomentatthesection

hc

hr

dp

hp

drdndr

b b

dpnAr

nAp

(a) (b)

Dc

Figure 8.11 Transformed cracked section: (a) cross section and (b) transformed section.


Theconcretecrackswhenthetensilestressoftheconcretereachesitstensilestrength ′fct.Bysettingtheconcretetensilestressattheextremefibreofthecrosssectionequalto ′fct,thecrackingmomentatthesectioncanbedeterminedas

M f

Iy

cr ctg

t

= ′ (8.44)

where′ = ′f fct c0 6.

yt isthedistancefromthecentroidalaxisofthecrosssectiontotheextremetensilefibre

Foracompositeslabwithseveralregionsofpeakmoments,theshort-termdeflectioncanbecalculatedusingtheaveragevalue(Ief⋅av)oftheeffectivesecondmomentsofareaIefatnominatedcrosssectionsasfollows:

• Forsimplysupportedcompositeslab,Ief⋅av=Iefatthemid-span.• Foranendspanofacontinuouscompositeslab,Ief⋅av=0.5(Ief⋅M+Ief⋅S),whereIef⋅Mand

Ief⋅Saretheeffectivesecondmomentsofareaatmid-spanandatthecontinuoussup-port,respectively.

• Foraninteriorspanofacontinuouscompositeslab,Ief⋅av=0.5[Ief⋅M+0.5(Ief⋅L+Ief⋅R)],whereIef⋅LandIef⋅Raretheeffectivesecondmomentsofareaattheleftsupportandattherightsupport,respectively.

• Foracantilevercompositeslab,Ief⋅av=Ief⋅Satthesupport.

8.11.3 long-term deflections of composite slabs

Thelong-termdeflectionsofcompositeslabsunderlong-termserviceloadsareinducedbytheshrinkageandcreepofconcrete.Thedeflectionduetoshrinkageshouldbeestimatedusingtheshrinkagepropertiesoftheconcrete.Thedeflectionscausedbycreepofconcretecanbecalculatedbymultiplyingtheshort-termdeflectionsbythefinalcreepcoefficients.InAS3600(2001),asimplifiedmultipliermethodisusedtodeterminethelong-termdeflectionsinducedbyshrinkageandcreep.Inthismethod,theadditionallong-termdeflectioniscomputedbymultiplyingtheshort-termdeflectioncausedbythesustainedloadsbyamultipliergivenby

k

AA

cssc

r

= −

≥2 1 2 0 8. . (8.45)

whereAsc is the cross-sectional areaof compressive reinforcement in the top face,Ar is thecross-sectionalareaoftensilereinforcementinthebottom,thesteelratioAsc/Aristakenatthemid-spanforsimplysupportedcompositeslaboratthesupportforacantilevercompositeslab.

InEurocode4, the secondmomentofareaof thecomposite slab for internal spans istakenasthemeanvalueofthesecondmomentsofareaofthecrackedanduncrackedsec-tions.Deflectioncalculationcanbeomittediftheshearconnectionofthecompositeslabissostrongthattheendslipdoesnotoccurunderserviceloadsandthespantotheeffectivedepthratioislessthan20.

8.11.4 Span-to-depth ratio for composite slabs

TheClause9.3.4ofAS3600(2001)providesthespan-to-depthratiomethodasanalternativetocheckingthedeflectionsofreinforcedconcreteslabs.Iftheslabssatisfythespan-to-depth

Composite slabs 243

limits,thecalculationofdeflectionscanbeavoided.Thismethodisadoptedforcompositeslabswithuniformdepthandsubjectedtouniformlydistributedloadsandwheretheliveloaddoesnotexceedthedeadload.Thecompositeslabsatisfiesdeflectionlimitsifthespan-to-depthofthecompositeslabsatisfiesthefollowingcondition:

Ld

k kL EF

ef ef c

d ef

≤

3 4

( )

.

∆/ (8.46)

whereLef istheeffectivespandistheeffectivedepthofthecompositeslab∆/Lef isthedeflectionlimitk3 1=k4isthedeflectionconstantwhichis1.6forsimplysupportedslabs,2.0inanendspan

and2.4ininteriorspansofacontinuouscompositeslabwhereinadjoiningspans,ratiooflongerspantoshorterspandoesnotexceed1.2andwherenoendspanislongerthananinteriorspan

Theeffectivedesign loadperunit length(Fd⋅ef) inEquation8.46forcalculatingthetotaldeflectionistakenas

F k g k qd ef cs s cs l⋅ = + + +( . ) ( )1 0 ψ ψ (8.47)

Forcalculatingthedeflectionwhichoccursaftertheadditionorattachmentoftheparti-tions,Fd⋅efistakenas

F k g k qd ef cs s cs l⋅ = + +( )ψ ψ (8.48)

whereψsistheshort-termloadfactorψlisthelong-termloadfactor

Example 8.5: Design of simply supported composite slab

AsimplysupportedcompositeslabsupportedonsteelbeamsisshowninFigure8.12.The composite slab is tobe constructedunproppedand is subjected toa live loadof7.5kPaandasuperimposeddeadloadof1.0kPainadditiontoitsownweight.Intheconstructionstage1,theloadfromstackedmaterialsis4.0kPaandliveloadis1.0kPa.Theconcretecompressivestrength( )′fc is25MPa.TheBondekIIprofiledsteelsheetingwithtbm=0.75 mmisused.Theyieldstressofthesheetingis550MPa.(1)Checkthedeflectionandstrengthsofthesteelsheetingduringconstruction;(2)checktheflexuralandshearstrengthanddeflectionsofthecompositeslab.

a. Design of formwork

1. Design for serviceability

Self-weightofsheeting:Gp=0.1kPaSelf-weightofconcreteandreinforcement(0.1kPaforreinforcement):

Gc = × ×

+ =0 12 2400

9 81000

0 1 2 92..

. . kPa


Takingthedesignwidthoftheslabas1m,thedesignserviceloadis

w G Gp c= + × = + × = ( )( ) . ( . .1 0 0 1 2 92 0 ) 1.0 3.02 kN/m AS 36 1-1995

The second moment of area of the Bondek II profiled steel sheeting is Ip = 0.4798 ×106 mm4/mandEs=200×103MPa.Thedeflectionatthemid-spanofthesheetingis

δC

s p

wLE I

1 3

4 4

3 6

5384

5 3 02 1950384 200 10 0 4798 10

5 9..

..= =

× ×× × × ×

= mm

Thedeflectionlimitis

∆limit mm 5.9 mm OK= = = > =

LC

2501950250

7 8 1 3. ,.δ

1.2. Design for strength

Atstage1,beforeplacingconcrete,thedesignloadis

w G Q Qp uv m1 1 2 1 5 1 5 1 2 0 1 1 5 1 1 5 4 7 62∗ = + + = × + × + × =. . . . . . . . kN/m

Atstage2,afterplacingconcrete,thedesignloadis

w G G Q wp c uv2 11 2 1 2 1 5 1 2 0 1 1 2 2 92 1 5 1 5 12∗ = + + = × + × + × = < ∗. . . . . . . . . kN/m

Therefore,

w∗ = 7 62. kN/m

Thedesignmaximumbendingmomentatmid-spanoftheslabatstage1is

M

w L∗ =∗

=×

=2 2

87 62 1 95

83 62

. .. kNm/m

Thenominalmomentcapacityofthebaresteelsheetingis(Table8.1)

M tu bm= = × =13 8 13 8 0 75 10 35. . . . kNm/m

1950

Profiled steel sheetingA89

B89

120

Figure 8.12 Simply supported composite slab.

Composite slabs 245

Thedesignmomentcapacityofthesheetingis

φM Mu = × = > ∗ =0 8 10 35 8 28 3 62. . . . ,kNm/m kNm/m OK

b. Design of the composite slab

1. Design actions

Deadload:G=(0.12×25+1.0)×1=4kN/mLiveload:Q=7.5×1=7.5kN/mThedesignload:w∗=1.2G+1.5Q=1.2×4+1.5×7.5=16.05kN/m

Themaximumdesignbendingmomentiscalculatedas

M

w L∗ =∗

=×

=2 2

816 05 1 95

87 6

. .. kNm/m

Theverticalshearforce:Vw L∗ =∗

=×

=2

16 05 1 952

15 6. .

. kN/m

ThereactionatsupportA:RA∗ = + × =15 6 0 089 16 05 17 1. . . . kN


2.1. Resultant tensile force in sheeting

Thecross-sectionalareaandcapacityofbaresteelsheetingare(Table8.1)

A tp bm= = × =1678 1678 0 75 1258 5. . mm /m2

Theyieldcapacityofsteelsheetingiscomputedas

T A fyp p py= = × × =−1258 5 550 10 692 23. . kN/m

ThemechanicalresistanceofBondekIIis

H t fm bm c= ′ = × × =88 88 0 75 25 381. kPa

Thedistancexcsfromtheendofsheetingtothesectionwithcompleteshearconnectionisgivenby

x

TH

csyp

m

= = =692 2381

1 816.

. m

Thedistancefromtheendofsheetingtothemid-spanofthecompositeslabisx=1.95/2 +0.089=1.064 m<xcs=1.816 m;therefore,thesectionatisinpartialshearconnection.Thestrengthofthereinforcedconcretecoverslabiscomputedas


= × × × − × − =−

0 85

0 85 25 1000 120 54 10 0 14023

. ( )

. ( ) ..5 kN/m

Theresultanttensileforcedevelopedinsheetingwithcompleteshearconnectionis

T F Tpcs cst pcs= = =min( , ) min( . , ) .1402 5 692 2692.2 kN/m


Thetensileforceinsheetingatsectionwithdistancex=1.064mfromtheleftendofthesheetingisdeterminedas

T H x R Tp L m A pcs⋅ = + ∗ = × + × = < =µ 381 1 064 0 5 17 1 414 692 2. . . .kN/m kN/m

Hence,

Tp = 414 kN/m

2.2. Neutral axis depth

Theneutralaxisdepthdnintheconcretecoverslabis

d

T Tf b

np yr

c

=+′

=+ ×

× × ×=

0 85414 0 10

0 85 25 1000 0 8522 92

3

.( )

. ..

γmm

2.3. Design moment capacity

Thedegreeofshearconnectionatthesectionofx=1.064misgivenby

βsc

p

yp

TT

= = =414692 2

0 6.

.

TheheightofsheetingwhereTpactsfor0<βsc=0.6≤0.75iscalculatedas

yp sc= = × =18 18 0 6 6 482 2β . . mm

Hence,

dp = − =120 6 48 113 52. . mm


ϕ βb sc= − = − =1 1 0 6 0 642 2. .


C f b dc c n= ′ = × × × × × =−0 85 0 85 25 1000 0 85 22 92 10 4143. . . .γ kN

Thenominalpositivemomentcapacityofthecompositeslabatx=1.308miscalculatedas


= × + − × ×

( . )

. ( . .

0 5

0 11352 0 414 0 5 0 85

γ ϕ

414 ×× + ×

=

0 02292 10 35 0 64

49 6

. ) . .

. kNm/m

Thedesignpositivemomentcapacityis

φM Mu = × = > ∗ =0 8 49 6 39 7 7 6. . . . ,kNm/m kNm/m OK

Composite slabs 247

3. Positive vertical shear capacity

ThedesignpositiveshearforceatadistanceofDcfromthesupportAis

V ∗ = − × =15 6 16 05 0 12 13 7. . . . kN

Thedesignverticalshearcapacityofthecompositeslabiscalculatedas

φφ

µV

H M bD bDu

m up c c=

+( )−

=× × + ×

1 5

1 5

0 8 1 5 381 10 35 1 0 12

2

2

. ( )

.

. . ( . .

/

/ )) .

. .. . ,

( )× ×

−= > ∗ =

1 0 12

1 5 0 5123 9 13 7kN kN OKV

4. Deflection check

4.1. Second moment of area of cracked section

Young’smodulusofconcreteiscomputedas

E fc c= ′ + = × + =3 320 6 900 3 320 25 6 900 23 500, , , , , MPa

Themodulusratiois

n

EEs

c

= = =200 00023 500

8 5,,

.

The height of the elastic centroid from the sheeting bottom is 15.6 mm. Assume theneutralaxisisintheconcretecoverslab.Theneutralaxisdepthdncanbedeterminedbytakingthefirstmomentofareaabouttheneutralaxisas

12

12

1000 8 5 1258 5120 15 6

2

2

bd nA d d

d d

n p p n

n n

= −

× × = × − −

( )

. . ( . )

d hn c= < = − =37 8 120 54 66. mm mm;thus,theneutralaxisisintheconcretecoverslab.

Thesecondmomentofareaofthecrackedsectionis

Icr =×

+ × × − −

= ×

1000 37 83

8 5 1259 120 15 6 37 83

2.. ( . . )

65.53 10 mm /m6 4

4.2. Effective second moment of area

Thesecondmomentofareaofthecrosssectionignoringthesheetingis

Ig =

×= ×

1000 12012

3

144 10 mm /m6 4


Thecrackingmomentatthesectioniscalculatedas

M f

Iy

cr cg

t

= ′ = ××

=0 6 25144 10120 2

7 26

. ./

kNm/m

Theshort-termserviceloadis

w G Qs s= + = + + × =ψ ( . . ) . . .3 22 1 0 1 0 7 5 11 52 kN/m

Thebendingmomentatthemid-spanundershort-termserviceloadis

M

w Lse

s= =×

=2 2

811 52 1 95

85 5

. .. kNm/m

Theeffectivesecondmomentofareaatmid-spansectioniscomputedas

I I I IMM

Ief cr g crcr

seg= + −

≤

= × + × −

( )

. ( .

3

6 665 53 10 144 10 65 53××

= × >10

7 25 5

241 57 1063

6)..

. mm /m4 Ig

Hence,

I Ief g= = ×144 106 mm /m4

4.3. Short-term deflection

Theshort-termdeflectionduetoshort-termserviceloadis

δs

c ef

wLE I

= =× ×× × ×

=5

3845 11 52 1 950

384 23 500 144 100 64

4 4

6

..

,,

mm

4.4. Long-term deflection

Thelong-termserviceloadis

w G Ql= + = + × =ψ 4 0 6 7 5 8 5. . . kN/m

Thedeflectionduetothesustainedloadis

δsus = × =

8 511 5

0 64 0 47..

. . mm

Thereisnocompressivereinforcementinthecompositeslab,Asc=0:

k

AA

cssc

r

= −

=2 1 2 2.

Composite slabs 249

Thelong-termdeflectionduetoshrinkageandcreepistherefore

δ δl cs susk= = × =2 0 47 0 94. . mm

4.5. Total deflection

Thetotaldeflectionis

δ δ δ δ δtot C s l= + + = + + = < =1 3 5 9 0 64 0 94 7 5 7 8. . . . . . ,mm mm OKlimit

referenceS

AS 2327.1 (2003) Australian standard for composite structures, Part 1: Simply supported beams,Sydney,NewSouthWales,Australia:StandardsAustralia.

AS 3600 (2001)Australian standard for concrete structures, Sydney, New SouthWales,Australia:StandardsAustralia.

Branson,D.E.(1963)Instantaneousandtime-dependentdeflectionofsimpleandcontinuousreinforcedconcretebeams,HPRReportNo.7.Birmingham,AL:AlabamaHighwayDepartment,USBureauofPublicRoads.

Bridge,R.Q.(1998)ShearConnectionParametersforBondekII,ComformandCondeckHP,Sydney,NewSouthWales,Australia:UniversityofWesternSydney.

Eurocode4(2004)Designofcompositesteelandconcretestructures,Part1.1:Generalrulesandrulesforbuildings,Brussels,Belgium:EuropeanCommitteeforStandardization.

Goh,C.C.,Patrick,M.,Proe,D.,andWilkie,R.(1998)Designofcompositeslabsforstrength,com-positestructuresdesignmanual–DesignbookletDB3.1,Melbourne,Victoria,Australia:BHPIntegratedSteel.

Johnson,R.P.(2004)CompositeStructuresofSteelandConcrete:Beams,Slabs,Columns,andFramesforBuildings,Oxford,U.K.:BlackwellPublishing.

Liang,Q.Q.andPatrick,M. (2001)Designof the shear connectionof simply-supportedcompositebeamstoAustralianstandardsAS2327.1-1996,Compositestructuresdesignmanual–DesignbookletDB1.2,Sydney,NewSouthWales,Australia:OneSteelManufacturingLimited.

Patrick,M.(1990)Anewpartialshearconnectionstrengthmodelforcompositeslabs,SteelConstructionJournal,AustralianInstituteofSteelConstruction,24(3):2–17.

Patrick,M.(1993)TestinganddesignofBondekIIcompositeslabsforverticalshear,SteelConstructionJournal,AustralianInstituteofSteelConstruction,27(2):2–26.

Patrick,M.andBridge,R.Q.(1994)Partialshearconnectiondesignofcompositeslabs,EngineeringStructures,16(5):348–362.

251

Chapter 9

composite beams

9.1 IntroductIon

Asteel–concretecompositebeamisconstructedbyconnectingtheconcreteslabtothetopflangeofasteelbeambyshearconnectors.Inasimplysupportedcompositebeam,thecon-creteslabissubjectedtocompression,whilepartorwholeofthesteelbeamisintension.Thebestpropertiesofbothsteelandconcretematerialsareutilisedincompositebeamcon-struction.Shearconnectorsnotonlytransferthelongitudinalshearattheinterfaceoftheconcreteslabandthesteelbeambutalsoresistthelongitudinalslipandverticalseparationofthesetwocomponents.Thestrengthofcompositebeamsdependsonthedegreeofshearconnectionbetweentheconcreteslabandthesteelbeam.Continuouscompositebeamshavetheadvantagesofreducedsteelquantityandimprovedflexuralstiffnesscomparedtosimplysupported composite beams. However, additional slab reinforcement needs to be placedinthenegativemomentregions.Theuseofpartialshearconnectionleadstoeconomicaldesignsofsimplysupportedcompositebeamswhilecontinuouscompositebeamsareusu-allydesignedwithcompleteshearconnection.

This chapter presents thebehaviour anddesignof simply supported composite beamsfor strengthandserviceability toAS2327.1 (2003).Thedesignofcontinuouscompositebeams is also introduced.Themethod fordetermining the effective sectionsof concreteslabsandsteelbeamsisgivenfirst.Thebasicconceptsanddesignoftheshearconnectionofcompositebeamsisintroduced.Theveridicalshearcapacityofcompositebeamsisthendescribed.Thisisfollowedbytheintroductionofthedesignofcompositebeamsforpositiveandnegativemomentregions.Thedesignoflongitudinalshearreinforcementispresented.Thedesignofcompositebeamsforserviceabilityisdiscussed.

9.2 comPonentS of comPoSIte BeAmS

Themaincomponentsofacompositebeamconsistofthesteelbeam,concreteslabandshearconnectorsasschematicallydepictedinFigure9.1.Themostcommontypesofsteelbeamsinclude hot-rolled I-sections, welded I-sections, rectangular cold-formed hollow sections,fabricatedI-sectionsandanyofthesementionedsectionswithanadditionalplateweldedtothebottomflange,asshowninFigure9.2.Ingeneral,AS2327.1requiresthatthecrosssectionofthesteelbeammustbesymmetricalaboutitsverticalaxis.

Theconcreteslabcanbeeitherasolidslaboracompositeslabincorporatingprofiledsteelsheeting.Theconcreteslabmustbereinforcedwithdeformedbarsormeshtocarrytensileforcesandlongitudinalshearintheslabarisingfromdirectloading,shrinkageandtempera-tureeffectsorfire.Thedesignofsolidreinforcedconcreteslabsmustbeinaccordancewith


AS3600.ThedesignofcompositeslabsisgiveninChapter8.TheprofiledsteelsheetingincorporatedinacompositeslabmustsatisfythegeometricrequirementsgiveninClause1.2.4ofAS2327.1.ThemajorAustralianprofiledsteelsheetingproductssuchasBondekII,ComformandCondeckHPsatisfythesegeometricrequirements.

Theshearconnectorsareattachedtothetopflangeofthesteelbeamtoresistthelon-gitudinalslipatthe interfaceandtheverticalseparationbetweenthesteelbeamandtheconcreteslab.Thecommonlyusedshearconnectorsareheadedstuds,channelsandhigh-strengthstructuralboltsasshowninFigure9.3.Theheadedstudsarethemostwidelyusedshearconnectorsincompositebeamconstruction.

Steel reinforcement Stud shear connector

Pro led steel sheeting

Steel beam

Concrete

Figure 9.1 Components of a composite beam.

(a) (b) (c)

(d) (e) (f )

Figure 9.2 Typical composite beams incorporating profiled steel sheeting: (a) composite beam with hot-rolled steel I-section; (b) composite beam with welded steel I-section; (c) composite beam with hot-rolled steel I-section welded with bottom plate; (d) composite beam with cold-formed rect-angular hollow steel section; (e) composite beam with welded rectangular hollow steel section; (f) composite beam with steel T-section.

Composite beams 253

9.3 BehAvIour of comPoSIte BeAmS

Thebehaviourof compositebeams canbedeterminedby either experiments (ChapmanandBalakrishnan1964;Ansourian1981)ornumericalanalysissuchasthefiniteelementanalysis(Liangetal.2004,2005;Pietal.2006a,b;Ranzi2008;ZonaandRanzi2011).Itdependsontheshearconnectionbetweentheconcreteslabandthesteelbeam.Eitherpush-out testsor full-scalecompositebeamtestscanbeusedtodetermine the load–slipcharacteristicsandultimateshearcapacityofshearconnectors.Push-outtestsindicatethattheshearconnectionmayfailbycrushingoftheconcreteorbyshearingofftheshearcon-nectors (ChapmanandBalakrishnan1964).The extentof crackingand crushing in theconcrete slabdependson the typeanddiameterof thestuds.During the testsof simplysupported composite beams, a distinct bond-breaking soundmayoccur,which signifiesthatextensivesliphasoccurredbetweentheconcreteslabandthesteelbeam.However,insomecases,thebondmaybegraduallydestroyedsothatnobond-breakingsoundcanbeheard.Simplysupportedcompositebeamsunderaconcentratedloadappliedatthemid-spanmayfailbycrushingoftheconcreteinthetopfaceandyieldingofthesteelsectionatthemid-span.Theconnectionmayfailsuddenlybyshearingofftheconnectorsinonehalfofthebeam,whichsignificantlyreducestheload-carryingcapacityofthebeam.Thefailuremodeofshearconnectionisbrittle.Testresultsdemonstratethattheendslipsandslipsatthemid-spanoccur.Thepull-outfailureofshearconnectorsleadstoarapidincreaseinslipandupliftandindeflections.

(a)

(b)

(c)

Figure 9.3 Types of shear connectors: (a) headed studs, (b) channels and (c) high-strength structural bolts.


Testsontwo-spancontinuouscompositebeamsindicatethatthetopoftheconcreteslabatmid-spanmaycrushandspall,whiletheentiresteelsectionmayyieldintensionatthemid-span(Ansourian1981).Inaddition,thebottomflangeandwebintheinteriorsupportmaybucklelocally.Testresultsshowthattheconcreteslabandcompositeactioncontributesignificantlytotheverticalshearstrengthofcompositebeams(ClawsonandDarwin1982).ThiswasconfirmedbythefiniteelementanalysesundertakenbyLiangetal.(2004,2005)onsimplysupportedandcontinuouscompositebeams.

9.4 effectIve SectIonS

Thesectionmomentcapacityofacompositebeamiscalculatedusingitseffectivecrosssec-tion,whichiscomposedoftheeffectivewidthofconcreteflangeandtheeffectiveportionofthesteelbeamsection.

9.4.1 effective width of concrete flange

Thein-planeshearstrainintheconcreteslabofacompositesectionunderbendingcausesthelongitudinaldisplacementsinthepartsoftheslabremotefromthesteelwebtolagbehindthoseneartheweb.Thisphenomenoniscalledshearlag,whichaffectslongitudi-naldisplacementsandstressesinthecompositesection(MoffattandDowling1978).Thedistributionofelasticstrainsbetweentheconcreteslabandthesteelbeamisnotuniform.Thestrainsarelargeabovethesteelbeamanddecreasewiththedistancefromthebeam(Adekola1968;VallenillaandBjorhovde1985).Theeffectivewidthconceptisemployedas a simplifiedmethod fordetermining the strength and stiffnessof compositebeams,which indirectlyaccounts for shear lageffects.Thisconceptassumes that theeffectiveconcreteflange carries themaximumuniform stressover the steelbeam.The effectivewidth(bcf)ofconcreteflangeinacompositebeamdependsontheeffectivespan(Lef)ofthecompositebeam,centre-to-centre spacing (b1,b2)ofadjacentbeamsand theoverallthicknessoftheslab(Dc).

ForaninternalcompositebeamshowninFigure9.4a,theeffectivewidthsbe1andbe2oftheconcreteflangeinasolidslabaregiveninClause5.2.2ofAS2327.1(2003)asfollows:

b

L b bDe

ef fc1

1 1

8 2 28=

+

min , , (9.1)

b

L b bDe

ef fc2

2 1

8 2 28=

+

min , , (9.2)

wherebf1isthewidthofthetopflangeofthesteelsectioninthecompositebeam.Theeffec-tivespan(Lef)ofacompositebeamisthedistancebetweenpointsofzerobendingmoment.Forsimplysupportedbeams,itshouldbedeterminedinaccordancewithAppendixHofAS2327.1.InEurocode4(2004),forcontinuouscompositebeams,theeffectivespanforpositivebendingistakenas0.8Loforanendspanand0.7Loforaninteriorspan,whereLoisthecentre-to-centrespacingofthesupports.Fornegativebending,Lefistakenas(L1+L2)/4,whereL1andL2areadjacentspans.

Composite beams 255

ForanedgecompositebeamschematicallydepictedinFigure9.4b,Clause5.2.2ofAS2327.1(2003)suggeststhattheeffectivewidthsbe1andbe2oftheconcreteflangeinasolidslabarecalculatedby

b

Lb

bDe

ef fc1 1

1

8 26=

+

min , ,( ) (9.3)

b

L b bDe

ef fc2

2 1

8 2 28=

+

min , , (9.4)

TheeffectivewidthoftheconcreteflangeinacompositebeamwheretheslabisacompositeslabisillustratedinFigure9.5.Fortheportionoftheconcretecoverslababovetheribs,the

bcf

bcf

be2

be2

be1

be1

bf1

bf1

b2

b2

Dc

Dc

b1

b1

(a)

(b)

Figure 9.4 Effective width of concrete flange in composite beams with solid slabs: (a) internal beam and (b) edge beam.

(a)

bcf bcfbe2

hr λbcf

be2

bf1

Dc Dc

be1 be1

(b)

Figure 9.5 Effective width of concrete flange in composite beams with composite slabs: (a) ribs orientated parallel to steel beam and (b) ribs orientated with an angle to steel beam.


effectivewidthiscalculatedusingEquations9.1and9.2foraninternalcompositebeamandEquations9.3and9.4foranedgecompositebeam,respectively.Clause5.2.2ofAS2327.1specifiesthatfortheportionoftheslabwithinthedepthoftheribs,theeffectivewidthistakenasλbcf.Themultiplierλdependsontheorientationangle(θ)ofsheetingribswithrespecttothelongitudinalaxisofthesteelbeamandistakenas1.0for0<θ≤15°,(bcrcos2θ)/srfor15< θ≤60°and0.0forθ>60°.

9.4.2 effective portion of steel beam section

Whenpartoftheflangeorpartortheentirewebofthesteelbeamcrosssectionisincom-pression,localbucklingoftheseplateelementsmayoccur.AS2327.1doesnotallowsteelbeamswithslenderplateelementstobeusedincompositebeams.Ifasteelbeaminacom-positebeamhasacompactsection,theentiresteelsectionisassumedtobeeffective.Theeffectivewidthconceptcanbeusedtodeterminetheeffectiveportionofthesteelbeamwithnon-compactsection.

Figure9.6ashowstheeffectiveportionofsteelbeamwiththenon-compacttopflangeandweb.Theeffectiveportionofthenon-compactsteelwebcanbedeterminedbycalculatingtheineffectivelengthx,whichisgivenasx d t fw w y= − 30 250/ .TheClause5.2.3.3ofAS2327.1providesasimplifiedmethodfordeterminingtheeffectiveportionofanon-compactsteelwebasillustratedinFigure9.6b.Inthesimplifiedmethod,theeffectivethickness(tew)ofthesteelwebiscalculatedbyignoringtheineffectiveportionofthewebinthecompres-sionzoneastew=tw(dw−x)/dw.TheplateelementplasticityandyieldslendernesslimitsaregiveninAS2327.1.

9.5 SheAr connectIon of comPoSIte BeAmS

9.5.1 Basic concepts

Theshearconnectionofacompositebeamistheinterconnectionbetweentheconcreteslabandthesteelbeam,whichenablesthetwocomponentstoacttogetherasasinglestructuralmember.This isachievedbymechanical shearconnectorswhichareattached to the topflangeof the steelbeam.Theshearconnectionofacompositebeam iscomposedoffivecomponents, includingshearconnectors,concreteslab,topflangeofthesteelbeam,slab

(a)

tw

dn

dw

tew

tw

x

(b)

0.5dew

0.5dew

Figure 9.6 Effective portion of steel section in positive bending: (a) effective area of steel section and (b) simplified effective steel section.

Composite beams 257

reinforcementandprofiledsteelsheetingasschematicallydepictedinFigure9.1(LiangandPatrick2001).Thebehaviourofshearconnectionisinfluencedbythesecomponents.

Whennoshearconnectionisprovidedattheinterfacebetweentheconcreteslabandthesteelbeam, the twocomponentswillwork independently to resist the loadingas showninFigure9.7a.Theendoftheslabisfreetoslipandthereisaverticalseparationbetweenthesetwocomponents.Theultimatestrengthofthenon-compositebeamisconservativelydeterminedastheplasticcapacityof thesteelbeamaloneandthecontributionfromtheconcreteslabisignored.Perfectconnectionrequiresaconnectionwithinfiniteshear,bend-ingandaxialstiffness.Itisdifficulttoachieveperfectconnectionsincenomechanicalshearconnectorscanprovidethisdegreeofshearconnection.Inpractice,theshearconnectorsofasimplysupportedcompositebeamaredesignedtotransferthelongitudinalshearforce,whichisthesmallerofeitherthetensilecapacityofthesteelbeamortheeffectivecompres-sivecapacityoftheconcreteslab.Theconnectionsodesignediscalledcompleteshearcon-nectionorfullshearconnectionasdepictedinFigure9.7b,whichresultsinthemaximumpossiblecapacityofacompositesection(Liang2005).

The incomplete interaction or partial shear connection is between no connection andcompleteshearconnectionasillustratedinFigure9.7c.Inpartialshearconnection,thetotalsheartransferredbytheshearconnectorsinasimplysupportedcompositebeamislessthanthesmallerofthetensilecapacityofthesteelbeamandtheeffectivecompressivecapacityoftheconcreteslab.Thisimpliesthatthesectionmomentcapacityofthecompositebeamisgovernedbythestrengthofshearconnection.Thepartialshearconnectionofferseco-nomicaldesignsofsimplysupportedcompositebeams.ThepartialshearconnectiontheoryhasbeenadoptedinAS2327.1(2003),Eurocode4(2004)andAISC-LRFDSpecification(1994)forthedesignofsimplysupportedcompositebeams.Ontheotherhand,thecodesallowonlycompleteshearconnectiontobeconsideredinthedesignofcompositebeamsinnegativemomentregions.

(c)

Small slip

(b)

Strain distribution

Strain distribution

Strain distribution(a)

Vertical separation

Slip

No slip

Figure 9.7 Effect of shear connection on the behaviour of composite beams: (a) no shear connection, (b) complete shear connection and (c) partial shear connection.


9.5.2 load–slip behaviour of shear connectors

The behaviour of shear connectors embedded in the concrete slab of a composite beamunderappliedloadsischaracterisedbytheload–sliprelationship,whichcanbeobtainedfrompush-outtests(Ollgaardetal.1971;OehlersandCoughlan1986;LiangandPatrick2001;PatrickandLiang2002).Thestandardpush-outtestgiveninEurocode4(2004)isschematicallydepictedinFigure9.8.Theslipcapacityofthespecimenistakenasthelargerslipmeasuredatthecharacteristicloadandcanbedeterminedbystatisticalanalysisofthepush-outtestresults.Theload–sliprelationshipofstudshearconnectorsdevelopedbasedonexperimentalresults(Ollgaardetal.1971)isexpressedby

Q f en vs= − −( ) .1 18 0 4δ (9.5)

whereQnisthelongitudinalshearforceactingonashearconnectorfvsisthenominalshearcapacityofaweldedheadedstudδisthelongitudinalslip

Figure9.9showsatypicalload–slipcurvecalculatedusingEquation9.5fora19 mmdiam-eter stud shear connector embedded in 25 MPa concrete. It becomes apparent that thisheadedstudshearconnectorexhibitsaductilebehaviour.InAS2327.1,itisrequiredthattheshearconnectionofacompositebeammustbeductilebecausethedesignmethodsforcompositebeamsgiveninthecodesarebasedontheductilebehaviourofshearconnection.Shearconnectorswithaslipcapacityof6 mmareregardedasductileinEurocode4.

9.5.3 Strength of shear connectors

Clause8.2.2ofAS2327.1(2003)givessomegeometricrequirementsonheadedstuds,chan-nelsandhigh-strengthstructuralbolts.Standardheadedstudsare15.9or19 mmdiameterstuds.Theoverallheightofstudsafterweldingshouldnotbelessthanfourtimesthenomi-nalshankdiameter(dbs)and40 mmabovethetopofribsincompositeslabs.Thelengthofchannelshearconnectorsshouldbegreaterthan50 mmandlessthan60 mm.M20high-strengthstructuralboltsareusuallyusedincompositebeams.AS2327.1requiresthattheoverallheightoftheboltsmeasuredfromthetopfaceofthesteelflangetothetopofthebolt

P

Figure 9.8 Standard push-out test in Eurocode 4 (2004).

Composite beams 259

shouldnotbelessthan100 mm.Onlyautomaticallyweldedheadedstudsareallowedtobeattacheddirectlytothesteeltopflangethroughprofiledsteelsheeting.

Ashearconnectorinaconcreteslabundershearforcemayfailbyeithershearingofftheshearconnectorinstrongerconcreteorcrushingoftheconcretewhentheconcreteisweak.Theshearcapacityofashearconnectorembeddedinaconcreteslabisgovernedbyeitherthestudstrengthortheconcretestrength.InClause8.3.2.1ofAS2327.1(2003),thenomi-nalshearcapacity(fvs)ofaweldedheadedstudistakenasthelesservaluecalculatedbythefollowingequations:

f d fvs bs uc= 0 63 2. (9.6)

f d f Evs bs cj c= ′0 31 2. (9.7)

wheredbsdenotesthediameteroftheshankofthestudfucisthetensilestrengthofshearconnectormaterial(fuc ≤ 500 MPa)′fcjistheestimatedcharacteristiccompressivestrengthofconcreteatjdaysEccanbecalculatedasE fc c cj= ′0 043 1 5. .ρ fornormal-weightandlightweightconcrete

Thenominalshearcapacity(fvs)ofachannelshearconnectorembeddedinasolidconcreteslabisgiveninAISC-LRFDSpecification(1994)as

f t t L f Evs cf cw c cj c= + ′0 3 0 5. ( . ) (9.8)

wheretcf,tcwandLcaretheflangethickness,webthicknessandlengthofthechannelshearconnector.

Thenominal shear capacitiesof shear connectors innormal-weight concretearegiveninTable9.1.Thesevaluesarecalculatedusingtheconcretedensityofρc=2400kg/m3and

Slip δ (mm)0

0

Shea

r for

ce Q

n (kN

)

20

40

60

80

100

120

0.1 0.2 0.3 0.4 0.5

Figure 9.9 Typical load–slip curve for headed stud shear connector.


theminimumtensilesteelstrengthoffuc=410MPaforheadedstudsandfuc=500MPaforhigh-strengthstructuralboltshearconnectors.Forchannelsandhigh-strengthstructuralboltsinlightweightconcrete,fvsshallbetakenas80%ofthevaluesdeterminedfornormal-weightconcreteofthesamegrade.

InAS2327.1,thestrengthofshearconnectorslocatedintheribsofprofiledsteelsheet-ingthatsatisfiesthegeometryrequirementsspecifiedinClauseof1.2.4istakenasthesameasthatofshearconnectorsinsolidslabs.Profiledsteelsheetingthatdoesnotsatisfythesegeometry requirements may reduce the strength of shear connectors welded to the steelflangethroughthesheeting(Grantetal.1977;LiangandPatrick2001).Forribsorientedperpendiculartothesteelbeam,thestrengthreductionfactor(Grantetal.1977)forthestudisgivenby

ϕpe

x

cr

r

s

rnbh

hh

=

−

≤

0 851 1 0

.. (9.9)

wherehsistheheightofthestudafterweldingbcristhewidthofconcreteribatthemid-heightofsteelribsnxisthenumberofshearconnectorsatacrosssectionofthecompositebeam

Forribsorientedparalleltothesteelbeam,thestrengthreductionfactor(Grantetal.1977)forthestudisexpressedby

ϕpa

cr

r

s

r

bh

hh

=

−

≤0 6 1 1 0. . (9.10)

Inarealcompositebeam,shearconnectorsaredistributedalongthebeam.Thelongitudinalshearforceissharedbyshearconnectorsinthecompositebeam.Itisassumedthatallshearconnectorsareductileandhavethesamedesignshearcapacity,whichisinfluencedbythe

Table 9.1 Nominal shear capacity of shear connectors in normal-weight concrete

Type of shear connector

fvs (kN)

′ =fc 25 MPa ′ =fc 32 MPa ′ =fc 40 MPa

Headed studdbs = 19 mm 89 93 93dbs = 15.9 mm 62 65 65Channel (l = 50 mm)100TFC/100PFC 100 110 125High-strength structural boltM20/8.8 98 118 126

Source: Adapted from AS 2327.1, Australian standard for composite structures, Part 1: Simply supported beams, Standards Australia, Sydney, New South Wales, Australia, 2003.

Composite beams 261

numberofshearconnectorsinthegroup.InClause8.3.4ofAS2327.1(2003),thedesignshearcapacityofashearconnectorinagroupofshearconnectorsisgivenby

f k fds n vs= φ (9.11)

whereϕ=0.85isthecapacityreductionfactorknistheload-sharingfactor,whichisdeterminedas

k

nn

c

= −1 180 18

..

(9.12)

The number of shear connectors (nc) is taken as the lesser number of shear connectorsbetweeneachendofthebeamandthecrosssectionbeingconsidered.

9.5.4 degree of shear connection

Themomentcapacityofacompositebeamcrosssectionwithcompleteshearcompletionisgovernedbyeitherthetensilecapacity(Fst)ofthesteelbeamortheeffectivecompressivecapacity(Fcc)oftheconcreteslabasdepictedinFigure9.10.Thismeansthatthestrengthofshearconnection(Fsh)isgreaterthanFstandFcc.Incontrast,themomentcapacityofacompositebeamcrosssectionwithpartialshearconnectionisgovernedbythestrengthofshearconnection,whichimpliesthatFst>FshandFcc>Fsh.

The degree of shear connection of composite beams is defined in Clause 1.4.3 of AS2327.1as

β β= ≤ ≤

FFcp

cc

0 1.0 (9.13)

inwhichFccandFcparethecompressiveforcesintheconcreteslabwithcompleteshearcon-nectionandwithpartialshearconnection,respectively.Ifthedegreeofshearconnectionisknown,thecompressiveforceintheconcretewithpartialshearconnectioniscalculated

FccFsh

Fst

Figure 9.10 Strength of the components of a composite beam.


asFcp=βFcc.Ifthedistributionofshearconnectorsalongthecompositebeamisknown,thecompressiveforceintheconcreteslabatthepotentiallycriticalcrosssectionistakenas

F n f n f Fcp A ds A B ds B cc= ≤⋅ ⋅min[ ; ] (9.14)

wherenAandfds A⋅ arethenumberofshearconnectorsbetweentheleftendofthebeamtothe

sectionconsideredandtheircorrespondingdesignshearcapacity,respectivelynBandfds B⋅ arethenumberofshearconnectorsbetweentherightendofthebeamtothe

sectionconsideredandtheircorrespondingdesignshearcapacity,respectively

9.5.5 detailing of shear connectors

Clause8.4ofAS2327.1providesdetailingrequirementsforshearconnectordistributionsinlongitudinalandtransversedirections.Theshearconnectorsshouldbedetailedalongthelengthofthebeamasfollows:

• Shearconnectors shouldbeuniformlydistributedbetweenpotentially critical crosssectionsandtheendsofthebeam.

• Themaximumlongitudinalspacingofshearconnectorsistakenasthelesserof4Dcor600 mm.

• The minimum centre-to-centre spacing of headed studs or high-strength structuralboltsinsolidslabsandincompositeslabswithsheetingorientedparalleltothesteelbeamis5dbs.

• Forchannelshearconnectors,theminimumcleardistancebetweentheadjacentedgesis100 mm.

• Theminimumdistancebetweenadjacentfacesofaheadedstudandsheetingribmea-suredparalleltothelongitudinalaxisofthebeamis60 mm.

Theshearconnectorsshouldbedetailedalongthetransversecrosssectionofthebeamasfollows:

• Themaximumnumberofheadedstudshearconnectorspertransversecrosssectionisthreeforsolidslabsandtwoforcompositeslabs,whileitistwoforhigh-strengthstructuralboltsandheadedstudsincompositeslabs.

• Theminimumtransversespacingofheadedstudsandhigh-strengthstructuralboltsbetweentheirheadsis1.5dbs.

• Theminimumclearancebetweentheshearconnectorandthenearestpartofsheetingriborendofanopenedribprofiledis30 mm.

9.6 vertIcAl SheAr cAPAcIty of comPoSIte BeAmS

9.6.1 vertical shear capacity ignoring concrete contribution

InAS2327.1,theverticalshearcapacityofacompositebeamisassumedtoberesistedbythewebofthesteelbeamaloneandiscalculatedinaccordancewithAS4100.Thisimpliesthatthecontributionfromtheconcreteslabtotheverticalshearcapacityofcompositebeamisignored.ThedesignrequirementisexpressedbyV∗≤ϕVu,whereϕ=0.9isthecapacityreductionfactor,andthenominalshearcapacityofthesteelwebVuisgiveninSection4.5.

Composite beams 263

InAS2327.1,theshearratioisdefinedastheratioofthedesignverticalshearforcetothedesignverticalshearcapacityofthesteelweb,whichisexpressedby

γ

φ=

∗VVu

(9.15)

Thedesignsectionmomentcapacityofacompositebeammaybeinfluencedbythedesignshearforceactingonthesection.Thereisastrengthinteractionbetweenthemomentcapac-ityandtheverticalshearcapacity.Thedesignsectionmomentcapacityofacompositebeamdependsontheshearratio. Ifγ≤0.5, thedesignshear force is smallso that itdoesnotreducethemomentcapacityofthecompositebeam.However,if0.5<γ≤1.0,thedesignshearforcereducesthesectionmomentcapacityofthecompositebeamanditseffectmustbetakenintoaccountintheevaluationoftheflexuralstrength.

9.6.2 vertical shear capacity considering concrete contribution

Compositebeamsunderappliedloadsareoftensubjectedtocombinedactionsofbendingandverticalshear.Despiteexperimentalevidence,thecontributionsoftheconcreteslabandcompositeactiontotheverticalshearstrengthofcompositebeamsareignoredincurrentdesigncodes,suchasAS2327.1(2003),Eurocode4(2004)andAISC-LRFDSpecification(1994).Thedesigncodesassumethatthewebofthesteelsectionresiststheentireverticalshear.Thisassumptionobviously leads toconservativedesignsofcompositebeams.TheeffectsoftheconcreteslabandcompositeactionontheflexuralandverticalshearstrengthsofsimplysupportedandcontinuouscompositebeamshavebeeninvestigatedbyLiangetal.(2004,2005)usingthefiniteelementanalysismethod.Theirinvestigationsindicatethattheconcreteslabandcompositeactioncontributesignificantlytotheflexuralandverticalshearstrengthsofcompositebeams.

When no shear connection is provided between the steel beam and the concrete slab,theverticalshearcapacityofthenon-compositesectioncanbedeterminedby(Liangetal.2004,2005)

V V Vo c s= + (9.16)

whereVcisthecontributionoftheconcreteslabVsistheshearcapacityofthewebofthesteelbeam

Testsindicatedthatthepull-outfailureofstudshearconnectorsincompositebeamsmayoccur.Thisfailuremodelimitstheverticalshearcapacityoftheconcreteslab.Asaresult,thecontributionoftheconcreteslabVcshouldbetakenasthelesseroftheshearstrengthoftheconcreteslabVslabandthepull-outcapacityofstudshearconnectorsTpo.TheshearstrengthoftheconcreteslabproposedbyLiangetal.(2004,2005)isexpressedby

V f Aslab c ec= ′( )ϕ1

1 3/ (9.17)

whereϕ1isequalto1.16forsimplysupportedcompositebeamsand1.31forcontinuouscom-

positebeams′fc isthecompressivestrengthoftheconcrete(MPa)Aecistheeffectiveshearareaoftheconcreteslab


TheeffectiveshearareaofasolidslabcanbetakenasAec=(bf1+Dc)Dc,inwhichbf1isthewidthofthetopflangeofthesteelbeamandDcisthetotaldepthoftheconcreteslab.Foracompositeslabincorporatingprofiledsteelsheetingplacedperpendiculartothesteelbeam,Aeccanbetakenas(bf1+hr+Dc)(Dc−hr),inwhichhristheribheightoftheprofiledsteelsheeting.

Thepull-outcapacityofstudshearconnectorsinacompositebeamcomprisingasolidslabcanbecalculatedby

T d h n s h fpo s s x x s ct= + + −[ ( ) ( ) ]π 2 1 (9.18)

wheredsistheheaddiameteroftheheadedstudhsisthetotalheightofthestudnxisthenumberofstudspercrosssectionsxisthetransversespacingofstudsfctisthetensilestrengthofconcrete(MPa)

Thepull-out capacityof stud shearconnectors in composite slabs incorporatingprofiledsteelsheetingshouldbedeterminedusingtheeffectivepull-outfailuresurfacesintheafore-mentionedequations.Itshouldbenotedthatthetransversespacingofstudshearconnectorsshouldnotbegreaterthantwotimesthestudheight.

Theshearcapacityofthewebofthesteelbeamcanbecalculatedby

V f d ts w yw w w= 0 6. α (9.19)

wherefywistheyieldstrengthofthesteelweb(MPa)dwisthedepthofthesteelwebtwisthethicknessofthesteelwebαwisthereductionfactorforslenderwebsinshearbuckling

Forstockysteelwebswithoutshearbuckling,thereductionfactorαwisequalto1.0.Equation9.16canbeusedtodeterminetheverticalshearcapacityofnon-compositesec-

tions.Totakeadvantageofcompositeactions,designmodelsfortheverticalshearstrengthof compositebeamswith anydegreeof shear connectionwereproposedbyLiang et al.(2004,2005)as

V Vuo o= + ≤ ≤( ) 1 02ϕ β β 1 (9.20)

whereVuoistheultimateshearstrengthofthecompositesectioninpureshearϕ2is0.295forsimplysupportedcompositebeamsandsaggingmomentregionsincon-

tinuouscompositebeamsand0.092forhoggingmomentregionsincontinuouscom-positebeams

βisthedegreeofshearconnection

Composite beams 265

Itshouldalsobenotedthatthepull-outfailureofstudshearconnectorsleadstothedamageofcompositeaction.Ifthisoccurs,theultimateshearstrengthofthedamagedcompositebeam(Vuo)shouldbetakenasVoforsafety.

InteractionequationsareusedinAS2327.1andEurocode4toaccountfortheeffectofverticalshearontheultimatemomentcapacityofcompositebeams.However,thedesigncodes allow only the shear strength of the steel web to be considered in the interactionequations.StrengthinteractionequationsaccountingfortheeffectsoftheconcreteslabandcompositeactionweregivenbyLiangetal.(2004,2005)as

MM

VV

u

uo

eu

uo

em v

+

= 1 (9.21)

whereMuandVuaretheultimatemomentandshearcapacitiesofthecompositebeamincom-

binedbendingandshear,respectivelyMuoistheultimatemomentcapacityofthecompositesectioninpurebendingTheexponentsemandevareequalto6.0forsimplysupportedcompositebeamsand5.0

forsaggingmomentregionsincontinuouscompositebeams.Forhoggingmomentregionsincontinuouscompositebeams,emandevareequalto0.6and6.0,respectively

Themoment–shearinteractiondiagramsforcompositebeamsundersaggingandhoggingareshowninFigure9.11.Theultimatemomentcapacityofthecompositesection(Muo)canbedeterminedusingtherigidplasticanalysismethodinaccordancewiththecodesofprac-ticesuchasAS237.1andEurocode4.Itshouldbenotedthattheultimatemoment-to-shearratioisequaltotheappliedmoment-to-shearratio.Iftheappliedmomentandverticalshearareknown,theultimatestrengthsofacompositebeamincombinedactionsofbendingandshearcanbedeterminedusingEquation9.21.

00 0.2 0.4 0.6 0.8 1 1.2

0.2

0.4

0.6

0.8

1

1.2

Mu/M

uo

Vu/Vuo

Hogging

Sagging

Figure 9.11 Moment–shear interaction of composite beams.


9.7 deSIgn moment cAPAcIty for PoSItIve BendIng

9.7.1 Assumptions

Intheanalysisofthecrosssectionofacompositebeamfordeterminingitsultimatemomentcapacity,themainassumptionsareasfollows:

1.Eachoftheplanecrosssectionsofsteelbeamandconcreteflangeremainsplaneafterdefor-mation,resultinginlineardistributionofstrainonthecrosssectionofeachcomponent.

2.Theeffectiveportionofsteelsectionisstressedtoitsyieldstrengthincompressionorintension.

3.The rectangular stressblock from the extremecompressivefibreof concrete to theplasticneutralaxis(PNA)hasacompressivestressof0 85. ′fc .

4.Thetensilestrengthofconcreteisignored. 5.Shearconnectorsareductile.

9.7.2 cross sections with γ ≤ 0.5 and complete shear connection

9.7.2.1 Nominal moment capacity Mbc

Thedesignmomentcapacity(ϕMbv)ofthecrosssectionofacompositebeamunderpositivebendingisafunctionofthedegreeofshearconnection(β)andshearratio(γ)atthesec-tion.Figure9.12showsthedimensionlessmomentcapacitiesofatypicalcompositebeamwithvariousdegreesofshearconnectionandshearratios.Forcrosssectionswhereγ≤0.5,theverticalshearforcedoesnotaffectthedesignmomentcapacityofthecrosssections.Thedesignmomentcapacityofacompositebeamwithanydegreeshearconnectioncanbedeterminedfromtheplasticstressdistributionsinthecrosssection(AS2327.12003).Theequivalentplasticstressdistributioninthecompositebeamcrosssectionwithγ≤0.5andcompleteshearconnectionisschematicallypresentedinFigure9.13,wherethePNAisshowntolieinthewebofthesteelbeam.However,itshouldbenotedthatthePNAcanbelocatedintheconcretecoverslab,thesteelribs,thetopflangeorthewebofthesteelbeam.

FromtheequivalentplasticstressdistributiongiveninFigure9.13,thenominalmomentcapacity (Mbc) of the cross section with γ ≤ 0.5 and complete shear connection can be

Degree of shear connection (β)0

0

0.5

1

1.5

2

2.5

3

φMbv

/φM

sf

0.2 0.4 0.6

γ ≤ 0.5

γ = 0.75

γ = 1.0

0.8 1 1.2 1.4

Figure 9.12 Design moment capacity as a function of degree of shear connection.

Composite beams 267

obtainedbytakingmomentsaboutthelineofactionoftheresultantcompressiveforce(Fsc)inthesteelsectionasfollows:

M F d d F d dbc cc c sc st st sc= + + −( ) ( ) (9.22)

wheredcisthedistancefromthecentroidofthecompressiveforceFccintheconcreteslabto

thetopfaceofthesteelsectiondsc isthedistancefromthecentroidoftheresultantcompressiveforceFsc inthesteel

sectiontothetopfaceofthesteelsection

Forthecaseofnocompressioninthesteelsection,dsc=0.ThedesignmomentcapacityofthecompositecrosssectionisthereforeϕMbc,wherethecapacityreductionfactorϕ=0.9.

Thetensilecapacityofthesteelsectioniscalculatedas

F b t b t f d t fst f f f f yf w w yw= + +( )1 1 2 2 (9.23)

wheresubscript1referstothetopflangesubscript2referstothebottomflangesubscriptwreferstothewebofthesteelsectionfyf andfywaretheyieldstressoftheflangeandweb,respectively

ThedistancefromthelineofactionofFsttothetopfaceofthesteelsectionis

d

F t F d t F D tF

stf f w w f f s f

st

=+ + + −1 1 1 2 22 2 2( ) ( ) ( )/ / /

(9.24)

whereF b t ff f f yf1 1 1=F b t ff f f yf2 2 2=F d t fw w w yw=

Thecompressivecapacityoftheconcretecoverslab(Fc1)andconcretebetweentheribs(Fc2)arecalculatedby

F f b D hc c cf c r1 0 85= ′ −. ( ) (9.25)

F f b hc c cf r2 0 85= ′. λ (9.26)

bcf

dst

dn

Fstfyw

fyf

fyf

2fyf

0.85f c

2fyw

Fsc

dscFcc

dcDc hr

λbcfDs

Figure 9.13 Plastic stress distributions in composite section with γ ≤ 0.5 and β = 1.0.


9.7.2.2 Plastic neutral axis depth

Case 1: If the compressive capacityof the concrete cover slab is greater than the tensilecapacityofthesteelsection,suchasFc1≥Fst,thePNAfallsintheconcretecoverslababovethe steel ribs.This givesdn≤hc.The compressive force in concretewith complete shearconnectionisF f b dcc c cf n= ′0 85. ascanbeseenfromFigure9.13.FromtheforceequilibriumconditionofFcc=Fst,thedepthofthePNA(dn)canbeobtainedas

d

Ff b

nst

c cf

=′0 85.

(9.27)

Case 2: IfFc1<Fst≤(Fc1+Fc2), thePNAis locatedinthesteelribssothathc<dn≤Dc.The compressive force in concrete with complete shear connection is determined asF F f b d hcc c c cf n c= + ′ −1 0 85. ( )λ .FromtheforceequilibriumconditionofFcc=Fst,thedepthofthePNA(dn)canbedeterminedby

d h

F Ff b

n cst c

c cf

= + −′

1

0 85. λ (9.28)

Case 3:If(Fc1+Fc2)<Fst≤(Fc1+Fc2+2Ff1),thePNAliesinthetopflangeofthesteelsectionsothatDc<dn≤(Dc+tf1).Thecompressiveforceinconcretewithcompleteshearconnec-tionbecomesFcc=(Fc1+Fc2).ThecompressiveforceinthetopsteelflangecanbecalculatedasFsc=bf1(dn−Dc)(2fyf).Theforceequilibriumconditionrequiresthat(Fcc+Fsc)=Fst.ThedepthofthePNA(dn)isgivenby

d D

F Fb f

n cst cc

f yf

= + −

1 2( )

(9.29)

Case 4:If(Fc1+Fc2+2Ff1)<Fst,thePNAislocatedinthewebofthesteelsectionasillus-tratedinFigure9.13.Thisimpliesthat(Dc+tf1)<dn≤(Dc+dw).ThecompressiveforceinconcretewithcompleteshearconnectionisFcc=(Fc1+Fc2).ThecompressiveforceinthesteelsectioniscomputedasFsc=2Ff1+tw(dn−Dc−tf1)(2fyw).ThedepthofthePNA(dn)canbedeterminedfromtheforceequilibriumconditionof(Fcc+Fsc)=Fstas

d D t

F F Ft f

n c fst cc f

w yw

= + +− −

112

2( ) (9.30)

Example 9.1: Moment capacity of composite beam with complete shear connection

Figure9.14showsthecrosssectionofasimplysupportedcompositebeamwithcompleteshearconnection.Theprofiledsteelsheetingisorientatedθ=30°tothelongitudinalaxisofthesteelbeam.Thegeometricparametersofthesteelsheetingarehr=55mm,sr=bcr =300mm.ThesteelI-sectionis410UB53.7ofGrade300steelwithfyf=fyw=320MPa.Thedesignstrengthoftheconcreteflangeis ′ =fc 25MPa.Thedesignshearforceatthesectionconsideredis200kN.Determinethedesignmomentcapacityofthiscompositebeamcrosssection.

1. Vertical shear capacity

Theslendernessofthesteelwebis

λ λw

w

w

yyw

dt

f= =

− ×= < =

250403 2 10 9

7 6320250

..

56.7 82

Composite beams 269

Thewebisnotslender.Theshearyieldcapacityofthewebiscalculatedas

φ φV A f Vu w yw= = × × × × = > ∗ =( . ) . . . ,0 6 0 9 0 6 403 7 6 320 200N 529.3 kN kN OK

Theshearratiois

γ

φ=

∗= = <

VVu

200529 3

0 38 0 5.

. .

Therefore,thedesignmomentcapacityofthecompositebeamisnotaffectedbytheverti-calshear.

2. Plastic neutral axis depth

Thetensilecapacityofthesteelsectioniscomputedas

F b t b t f d t fst f f f f yf w w yw= + +

= × × × + × ×

( )1 1 2 2

(178 10.9 2) 320 381.2 7.6 3320 N 2168.8 kN=

Forcompleteshearconnection,Fcc=Fst=2168.8kN.Thecompressivecapacityoftheconcretecoverslabis

F f b D hc c cf c r130 85 0 85 25 1200 130 55 10 1912 5= ′ − = × × × − × =−. ( ) . ( ) . kN

SinceFc1<Fst,oneneedstocheckiftheneutralaxisisintheribs.Theanglebetweentheribsandthelongitudinalaxisofthesteelbeamisθ=30°.

Theparameterλiscalculatedas

λ

θ= =

× °=

bs

cr

r

cos cos.

2 2300 30300

0 75

Thecompressivecapacityofconcreteinthesteelribsis

F f b hc c cf r230 85 0 85 25 0 75 1200 55 10 1051 9= ′ = × × × × × =−. . . .λ kN

F F F Fc c st c1 2 11912 5 1051 9 2964 4 2168 8+ = + = > = >. . . .kN kN

Hence,thePNAislocatedwithintheribs.

178

403

10.9

7.6

130

1200

178 10.9

Figure 9.14 Cross section of composite beam under positive bending.


ThePNAdepthiscalculatedas

d h

F Ff b

n cst c

c cf

= +−′

= +− ×

× ×1

3

0 8575

2168 8 1912 5 100 85 25 0 7.( . . ). .λ 55 1200

88 4×

= . mm

3. Distances to centroid of forces

Thecompressiveforceintheconcretewithintheribsis

F f b d hcn c cf n c= ′ − = × × × × − =0 85 0 85 25 0 75 1200 88 4 75 256 3. ( ) . . ( . ) .λ N kNN

ThedistancefromthecentroidofFcntothetopfaceofthesteelsectionis

d h

d hcn r

n c= −−

= −−

=2

5588 4 75

248 3

.. mm

ThedistancefromthecentroidofFcctothetopfaceofthesteelsectionisdeterminedas

d

F d F dF

cc c cn cn

cc

=+

=× − + ×

=1 1 1912 5 130 75 2 256 3 48 32168 8

87. ( ) . .

..

/33 mm

410UB53.7isadoublysymmetricsection.ThedistancefromthecentroidofFsttothetopfaceofthesteelsectionisgivenas

d

Dst

s= = =2

4032

201 5. mm

ThecompressiveforceinthesteelsectionisFsc=0andthedistancefromthecentroidofFsctothetopfibreofthesteelsectionisdsc=0.


TakingmomentsaboutthelineofactionofthecompressiveforceinsteelsectionFsc,thenominalmomentcapacityiscalculatedas

M F d d F d dbc cc c sc st st sc= + + −

= × + + ×

( ) ( )

2168.8 (87.3 0) 2168.8 (201.55 0) kN mm kN m− = 626 3.

Thedesignmomentcapacityistherefore

φMbc = × =0 9 626 3. . 563.7 kNm

9.7.3 cross sections with γ ≤ 0.5 and partial shear connection

9.7.3.1 Nominal moment capacity Mb

Foracrosssectionwithpartialshearconnection(0<β<1.0),itsmomentcapacityisgov-ernedbythestrengthofshearconnection.Thecompressiveforceintheconcreteslabwithpartialshearconnectioncanbedeterminedbyoneofthefollowingexpressions:

F n f Fcp i ds cc= ≤ (9.31)

F F Fcp cc cc= ≤β (9.32)

whereniisthenumberofshearconnectorsbetweenthepotentiallycriticalcrosssectioniandtheendofthebeam.

Composite beams 271

Theequivalentplasticstressdistributioninthecrosssectionwithγ≤0.5andpartialshearconnection is illustrated inFigure9.15.Foracompositebeamcross sectionwithpartialshearconnection,therearetwoplasticneutralaxesinthesectionasshowninFigure9.15.ThefirstPNAislocatedintheconcreteslabeitherintheconcretecoverslaborinthesteelribs.Thesecondonefallsinthesteelsection,whichcanbelocatedinthetopflange,weborthebottomflange.

Thenominalmomentcapacity(Mb)ofthecrosssectionwithγ≤0.5andpartialshearconnectionisdeterminedbytakingmomentsaboutthelineofactionoftheresultantcom-pressiveforce(Fsc)inthesteelsectionasfollows:

M F d d F d db cp c sc st st sc= + + −( ) ( ) (9.33)

9.7.3.2 Depth of the first plastic neutral axis

Case 1:IfFcp≤Fc1,thefirstPNAliesintheconcretecoverslababovethesteelribssothatdn1≤hc.ThecompressiveforceinconcretewithpartialshearconnectionisdeterminedasF f b dcp c cf n= ′0 85 1. .TheforceFcpdependsonthedegreeofshearconnectionandistakenasFcp=βFcc.Consequently,thedepthofthefirstPNA(dn1)canbeexpressedby

d

Ff b

ncp

c cf1

0 85=

′. (9.34)

Case 2:IfFcp>Fc1,thefirstPNAislocatedinthesteelribs.Thedepthoftheneutralaxisisintherangeofhc<dn1≤Dc.Thecompressiveforceinconcretewithpartialshearconnec-tionisdeterminedasF F f b d hcp c c cf n c= + ′ −1 10 85. ( )λ .Thiscompressiveforce(Fcp)mustbeinequilibriumwiththestrength(Fsh)oftheshearconnection.ThedepthofthefirstPNA(dn1)isderivedasfollows:

d h

F Ff b

n ccp c

c cf1

1

0 85= +

−′. λ

(9.35)

9.7.3.3 Depth of the second plastic neutral axis

ItcanbeseenfromFigure9.15thatpartofthesteelsectionissubjectedtocompression.Theequilibriumconditionrequiresthattheresultantforceinthesteelsectionmustbeequaltothestrengthoftheshearconnection:Fst− Fsc=Fsh=Fcp.Theresultantcompressiveforce(Fsc)inthesteelsectionisdeterminedasFsc=Fst − Fcp.

Ds

Dc hr

λbcf

bcf

dn1

dn2

dstFst

fyw

fyf

fyf

2fyf2fyw

Fcpdc dscFsc

0.85 f c

Figure 9.15 Plastic stress distributions in composite section with γ ≤ 0.5 and 0 < β < 1.0.


Case 1:IfFsc≤2Ff1,thesecondPNAislocatedinthetopflangeofthesteelsectionsothatdn2≤tf1.ItisseenfromFigure9.15thatthecompressiveforceinthesteelsectionisdeterminedasFsc=bf1dn2(2fyf).ThedepthofthesecondPNA(dn2)canbecomputedas

d

F Fb f

nst cp

f yf2

1 2=

−( )

(9.36)

Case 2:If2Ff1<Fsc≤(2Ff1+2Fw),thesecondPNAliesinthesteelweb.Thisimpliesthattf1<dn2≤(tf1+dw).ItisseenfromFigure9.15thattheresultantcompressiveforceinthesteelsectionisobtainedasFsc=2Ff1+tw(dn2−tf1)(2fyw).ThedepthofthesecondPNAisgivenby

d t

F Ft f

n fsc f

w yw2 1

122

= +−( )

(9.37)

Case 3:IfFsc>(2Ff1+2Fw),thesecondPNAfallsinthebottomflangeofthesteelsection.Thisconditionleadsto(tf1+dw)<dn2≤Ds.Forthiscase,theresultantcompressiveforceinthesteelsectioniscalculatedasFsc=2Ff1+2Fw+bf2(dn2−dw−tf1)(2fyf).ThedepthofthesecondPNA(dn2)isderivedas

d t d

F F Fb f

n f wsc f w

f yf2 1

1

2

2 22

= + +− −

( ) (9.38)

9.7.4 cross sections with γ = 1.0 and complete shear connection

9.7.4.1 Nominal moment capacity Mbfc

Whentheshearratio(γ)atthecrosssectionofacompositebeamunderpositivebendingisequaltounity,thecontributionofthesteelwebtothemomentcapacityisignored.Thesteelwebisassumedtoresisttheentireverticaldesignshearforce.Theplasticstressdistributioninthecompositebeamcrosssectionwithγ=1.0andcompleteshearconnectionisschemati-callydepictedinFigure9.16,wherethestressesonthesteelwebarenotdrawnbecausethewebiscompletelyignoredinthecalculationofthemomentcapacity.ItisnotedthatthefigureshowsonlythetypicalcaseforwhichthePNAislocatedinthetopflangeofthesteelsection.

InAS2327.1,thedegreeofshearconnectionatthecrosssectionwithγ=1.0andcompleteshearconnectioniscalculatedas

ψ =

FFccf

cc

(9.39)

Ds

Dc

λbcf

bcf

dn

dstFstf

fyf

fyf

2fyf

FccfdcdscFsc

0.85 fc

hr

Figure 9.16 Plastic stress distributions in composite section with γ = 1.0 and complete shear connection.

Composite beams 273

inwhichFccfisthecompressiveforceintheconcreteslabwithβ=1.0whenthesteelwebisignored.ItisworthtonotingthatFccisthecompressiveforceintheconcreteslabwithβ=1.0whenthewholeeffectivesteelsectionistakenintoaccount.Forcrosssectionswithγ=1.0,thecompleteshearconnectionisdefinedastheconditionofψ≤β≤1.0.

Thenominalmomentcapacity(Mbfc)ofthecrosssectionwithγ=1.0andcompleteshearconnectionisdeterminedbytakingmomentsaboutthelineofactionoftheresultantcom-pressiveforce(Fsc)inthesteelsectionasfollows:

M F d d F d dbfc ccf c sc stf st sc= + + −( ) ( ) (9.40)

whereFstfisthetensilecapacityofthetwoflangesofthesteelsection.

9.7.4.2 Plastic neutral axis depth

Case 1: If the compressive capacityof the concrete cover slab is greater than the tensilecapacityofthesteeltwoflanges(Fstf),suchasFstf≤Fc1,thePNAliesintheconcretecoverslababovethesteelribssothatdn≤hc.ThecompressiveforceinconcretewithcompleteshearconnectionisF f b dccf c cf n= ′0 85. .TheforceequilibriumconditionofFccf=FstfgivesthedepthofthePNA(dn)asfollows:

d

Ff b

nstf

c cf

=′0 85.

(9.41)

Case 2:IfFc1<Fstf≤(Fc1+Fc2),thePNAislocatedinthesteelribs.Theneutralaxisdepthsatisfiestheconditionofhc<dn≤Dc.ThecompressiveforceinconcretewithcompleteshearconnectioniscomputedasF F f b d hccf c c cf n c= + ′ −1 0 85. ( )λ .Fromtheforceequilibriumcondi-tionofFccf=Fstf,thedepthofthePNA(dn)canbedeterminedas

d h

F Ff b

n cstf c

c cf

= +−′

1

0 85. λ (9.42)

Case 3: If (Fc1+Fc2)<Fstf≤ (Fc1+Fc2+2Ff1), thePNA lies in the topflangeof the steelsectionanditmeansthatDc<dn≤(Dc+tf1).ThecompressiveforceinconcretebecomesFccf=(Fc1+Fc2).ThecompressiveforceinthetopflangeisFsc=bf1(dn−Dc)(2fyf).Theforceequilibriumconditionisexpressedas(Fccf+Fsc)=Fstf.ThedepthofthePNA(dn)isderivedas

d D

F Fb f

n cstf ccf

f yf

= +−

1 2( ) (9.43)

9.7.5 cross sections with γ = 1.0 and partial shear connection

9.7.5.1 Nominal moment capacity Mbf

Foracrosssectionwithpartialshearconnection(0<β<ψ),itsmomentcapacityisgovernedbythestrengthofshearconnection.Thecompressiveforce(Fcpf)intheconcreteslabatacrosssectionwithγ=1.0andpartialshearconnectioncanbedeterminedbyoneofthefol-lowingexpressions:

F n f Fcpf i ds ccf= ≤ (9.44)


F F Fcpf cc ccf= ≤β (9.45)

whereFccfisthecompressiveforceintheconcreteslabatacrosssectionwithγ=1.0andcompleteshearconnection.Theequivalentplasticstressdistributioninthecrosssectionwithγ=1.0andpartialshearconnectionisillustratedinFigure9.17,whereshowsthattherearetwoplasticneutralaxesinthecrosssection.ThefirstPNAislocatedintheconcreteslabeitherintheconcretecoverslaborinthesteelribs.Thesecondonefalls inthetoporthebottomflangeofthesteelsection.

Thenominalmomentcapacity(Mbf)ofthecrosssectionwithγ=1.0andpartialshearconnectionisobtainedbytakingmomentsaboutthelineofactionofthecompressiveforce(Fsc)inthesteelflangeasfollows:

M F d d F d dbf cpf c sc stf st sc= + + −( ) ( ) (9.46)

9.7.5.2 Depth of the first plastic neutral axis

Case 1: IfFcpf ≤Fc1, thefirstPNA lies in the concrete cover slababove the steel ribs sothatdn1≤hc.ThecompressiveforceinconcretewithpartialshearconnectionisgivenasF f b dcpf c cf n= ′0 85 1. .TheforceFcpdependsonthedegreeofshearconnectionandistakenasFcpf=βFcc.ThedepthofthefirstPNA(dn1)isgivenby

d

Ff b

ncpf

c cf1

0 85=

′. (9.47)

Case 2:IfFcpf>Fc1,thefirstPNAliesinthesteelribs.Thedepthoftheneutralaxissatisfiestheconditionofhc<dn1≤Dc.Thecompressiveforceinconcretewithpartialshearconnec-tionisdeterminedas F F f b d hcpf c c cf n c= + ′ −1 10 85. ( )λ .Thiscompressiveforce(Fcp)mustbeinequilibriumwiththestrength(Fsh)oftheshearconnection.ThedepthofthefirstPNA(dn1)is

d h

F Ff b

n ccpf c

c cf1

1

0 85= +

−′. λ

(9.48)

Ds

Dc

λbcf

bcf

dn1

dn2

dstFstf

fyf

fyf

2fyf

Fcpfdc dscFsc

0.85 f c

hr

Figure 9.17 Plastic stress distributions in composite section with γ = 1.0 and partial shear connection.

Composite beams 275

9.7.5.3 Depth of the second plastic neutral axis

ItcanbeseenfromFigure9.17thatpartofthesteelsectionissubjectedtocompression.Theequilibriumconditionrequiresthattheresultantforceinthesteelsectionmustbeequaltothestrengthoftheshearconnection,suchasFstf− Fsc=Fsh=Fcpf.Theresultantcompressiveforce(Fsc)inthesteelsectioncanbeobtainedasFsc=Fstf− Fcpf.Case 1:IfFsc≤2Ff1,thesecondPNAislocatedinthetopflangeofthesteelsectionsothatdn2≤tf1.ItisseenfromFigure9.17thatthecompressiveforceinthesteelsectionisdeter-minedasFsc=bf1dn2(2fyf).ThedepthofthesecondPNA(dn2)canbecomputedas

d

F Fb f

nstf cpf

f yf2

1 2=

−( )

(9.49)

Case 2: If2Ff1<Fsc, thesecondPNAlies inthebottomflangeofthesteelsection.Thisimplies that (tf1 + dw) < dn2 ≤ Ds. The compressive force in the steel flange is obtainedasFsc=2Ff1+bf2(dn2−dw−tf1)(2fyf).ThedepthofthesecondPNAisderivedas

d t d

F Fb f

n f wsc f

f yf2 1

1

2

22

= + +−( )

(9.50)

9.7.6 cross sections with 0.5 < γ ≤ 1.0

Forbeamcrosssectionswith0.5<γ≤1.0,thedesignmomentcapacityϕMbvdependsontheshearratioγandthedegreeofshearconnectionβ.ThismeansthatthedesignverticalshearforceactingatthesectionreducesthedesignmomentcapacityϕMbv.AS2327.1allowsalinearinteractionequationtobeusedtodeterminethedesignmomentcapacityϕMbvofacompositebeamwith0.5<γ≤1.0.Themoment–shearinteractiondiagramispresentedinFigure9.18.Thedesignmomentcapacity(ϕMbv)ofthecrosssectionwith0.5<γ≤1.0iscalculatedbylinearinterpolationasfollows:

φ φ φ φ γ βM M M Mbv bf b bf= + − − <( )( )2 for 1.02 (9.51)

φ φ φ φ γ βM M M Mbv bfc bc bfc= + − − =( )( )2 for 1.02 (9.52)

0.0 0.5 1.0

φMbfc or φMbf

φMbv

φMbc or φMb

γ

Figure 9.18 Moment–shear interaction diagram for composite sections.


9.7.7 minimum degree of shear connection

InordertosatisfythestrengthrequirementM∗ ≤ϕMbvatapotentiallycriticalcrosssection,theminimumdegreeofshearconnectionatthatsectionneedstobedetermined.Forcrosssectionswithγ≤0.5,theminimumdegreeofshearconnectionβiisgiveninClause6.5.2ofAS2327.1(2003)asfollows:

β φ

φ φφ φi

s

b ss b

M MM M

M M M=∗ −

−≥ < ∗ ≤

⋅⋅

20

55

( )for (9.53)

β φ φ

φ φφ φi

bc b

bc bb bc

M M MM M

M M M=∗ + −

−≥ < ∗ ≤⋅

⋅⋅

22

05

55

( ),for (9.54)

whereφMsisthedesignmomentcapacityofthesteelsectionφMb⋅5isthedesignmomentcapacityofthecrosssectionbysettingβ=0.5

Forcrosssectionswith0.5<γ≤1.0,theminimumdegreeofshearconnectionβicanbecalculatedinaccordancewithClause6.5.3ofAS2327.1(2003)asfollows:For0<βi≤ψ,

β

γ φ γ φ ψγ φ γ φ γi

sf s

sf bfc

M M MM M

=∗ − − − −

− + − − −[ ( ) ( ) ]

( ) ( ) ( )2 1 2 1

1 2 2 1 2 1 φφ γ φ ψM Ms b+ −≥

⋅2 10

( ) (9.55)

Forψ<βi,

β ψ

ψ γ φ γ φγ φ φ

ψ

ψi

b bfc

bc b

M M MM M

= +− ∗ − − − −

− −≥⋅

⋅

( )[ ( ) ( ) ]( )( )

1 2 1 2 12 1

0 (9.56)

whereφMsf isthedesignmomentcapacityofthesteelsectionneglectingthecontribution

ofthewebφ ψMb⋅ isthedesignmomentcapacityofthecompositecrosssectionbysettingβ=ψ

Example 9.2: Design of simply supported composite beam with complete shear connection for strength

Figure9.19showsthecrosssectionofaninternalsecondarysimplysupportedcompositebeamwithcompleteshearconnection.Thespacingofthesecondarybeamsis3.2m.Theeffectivespanofthecompositebeamis8m.Theprofiledsteelsheetingisplacedperpen-diculartothesteelbeam.Thesteelsection360UB50.7ofGrade300steelisusedwithfyf=300MPaandfyw=320MPa.Thedesignstrengthoftheconcreteflangeis ′ =fc 32 MPa.Thecompositeslabissubjectedtoasuperimposeddeadloadof1.0kPaandaliveloadof4kPa.Checkthestrengthsofthecompositebeamandprovideadequatestudshearcon-nectorstothebeam.

Composite beams 277

1. Effective width of concrete flange

Theprofiledsteelsheetingisplacedperpendiculartothesteelbeam,λ=0.Theeffectivewidthoftheconcreteflangeiscalculatedasfollows:

be1

80008

32002

1712

8 120 1000=

+ ×

=min , , mmm

bcf = × =2 1000 2000 mm

2. Design action effects

Theself-weightofthesteelbeam:50.7×9.81×10−3=0.497kN/mTheself-weightoftheslab:0.12×25×3.2=9.6kN/m

Superimposeddeadload:1.0×3.2=3.2kN/mTotaldeadload:G=0.497+9.6+3.2=13.3kN/mLiveload:Q=4×3.2=12.8kN/mThedesignload:w∗=1.2G+1.5Q=1.2×13.3+1.5×12.8=35.16kN/m

Themaximumdesignbendingmomentatmid-spanofthecompositebeamis

M

w Lef∗ =∗

=×

=2

835.16 8

8281.3 kNm

2

Thedesignverticalshearatsupportis

V

w Lef∗ =∗

=×

=2

35.16 82

140.6 kN



λ λw

w

w

yyw

dt

f= = = < =

2503337 3

320250.

51.6 82

17111.5

11.5

55

bcf

7.3356

120

Figure 9.19 Cross section of composite beam under positive bending.



φ φV A f Vu w yw= = × × × × = > ∗ =( . ) . . . . ,0 6 0 9 0 6 356 7 3 320 140 6N 449 kN kN OK

0 5 0 5 140 6. . .φV Vu = × = > ∗ =449 kN 224.5 kN kN

Therefore,thedesignmomentcapacityofthecompositebeamisnotaffectedbytheverti-calshear.


Thetensilecapacityofthesteelsectioniscomputedas


= × × × + × ×

( )1 1 2 2

(171 11.5 2) 300 333 7.3 3200 N 1957.8 kN=

Thecompressivecapacityoftheconcretecoverslabiscomputedas

F f b D hc c cf c r130 85 0 85 32 2000 120 55 10 3536= ′ − = × × × − × =−. ( ) . ( ) kN

SinceFc1>Fst,thePNAislocatedintheconcretecoverslab.Forcompleteshearconnection,Fcc=Fst.

ThePNAdepthiscalculatedas

d

Ff b

hnst

c cfc=

′=

×× ×

= < = − =0 85

1957 8 100 85 32 2000

36 120 55 653

..

.mm mmm


ThedistancefromthecentroidofFcctothetopfaceofthesteelsectionis

d D

dc c

n= − = − =2

120362

102 mm


d

Dst

s= = =2

3562

178 mm

ThecompressiveforceinthesteelsectionisFsc=0,andthedistancefromthecentroidofFsctothetopfibreofthesteelsectionisdsc=0.



M F d d F d dbc cc c sc st st sc= + + −

= × + × + ×−

( ) ( )

1957.8 (102 0) 10 1957.8 (13 778 0) 10 kNm3− × =− 548 2.


φM Mbc = × = > ∗ =0 9 548 2 281 3. . . ,493.4 kNm kNm OK

Composite beams 279

7. Required number of shear connectors

Thenominalshearcapacityof19 mmdiameterheadedstudembeddedin32MPacon-crete isobtainedfromTable9.1asfvs=93kN.Takingfds=fvs=93kN,therequirednumberofshearconnectorsfromtheendofthecompositebeamtoitsmid-spancanbedeterminedas

n

Ff

ccc

ds

= = =1957 8

9321

.

Takingnc=22,theload-sharingfactoris

k

nn

c

= − = − =1 180 18

1 180 1822

1 14..

..

.

Thedesignshearcapacityofshearconnectorsinagroupiscomputedas

f k fds n vs= = × × =φ 0 85 1 14 93 90. . kN

Therequirednumberofstudshearconnectorsisfinalizedas

n

Ff

ccc

ds

= = = ≈1957 890

21 8 22.

.

Thedesignstrengthoftheshearconnectionisdeterminedas

F n f Fsh c ds cc= = × = > =22 90 1980 1957 8kN kN,OK.

Thetotalnumberofstudshearconnectorsinthewholecompositebeamis44.

Example 9.3: Design of simply supported composite beam with partial shear connection for strength

Asshown inExample9.1,only57%of thedesignmomentcapacityof thecompositebeam with complete shear connection is utilised. Redesign this composite beam withpartialshearconnectionofβ=0.6.


Thecompositebeamisdesignedwithβ=0.6atthemid-spansection.Thetensilecapacityofthesteelsectioniscomputedas


= × × × + × ×

( )1 1 2 2

(171 11.5 2) 300 333 7.3 3200 N 1957.8 kN=

Thecompressiveforceintheconcreteslabwithpartialshearconnectionis

F Fcp cc= = × =β 0 6 1957 8 1174 68. . . kN

Thecompressivecapacityoftheconcretecoverslabiscomputedas

F f b D hc c cf c r130 85 0 85 32 2000 120 55 10 3536= ′ − = × × × − × =−. ( ) . ( ) kN

SinceFcp<Fc1,thefirstPNAliesintheconcretecoverslab.


ThedepthofthefirstPNAintheconcreteslabiscalculatedas

d

Ff b

hncp

c cfc1

3

0 851174 68 100 85 32 2000

22 120 55 6=′

=×

× ×= < = − =

..

.mm 55 mm

Thecompressiveforceinsteelsectioniscomputedas

F F Fsc st cp= − = − =1957 8 1174 67 783. . kN

Theslendernessofthetopflangeincompressionis

λ λef

yey

bt

f= =

−= < =

250171 7 3 2

11 5300250

7 8 9 5 1 232( . )

.. .

/Table of AS 77 1.

Hence,thetopflangeofthesteelsectioniscompact.Thecapacityofthesteeltopflangeis

2 2 2 171 11 5 300 11801 1 1F b t ff f f yf= × = × × × =. N kN

IfFsc<2Ff1,thesecondneutralaxisliesinthetopflangeofthesteelsection.Thedepthofthesecondneutralaxisiscomputedas

d

Fb f

tnsc

f yff2

1

3

12

783 10171 2 300

7 6 11 5= =×

× ×= < =

( ) ( ). .mm mm


ThedistancefromthecentroidofFcptothetopfaceofthesteelsectionis

d D

dc c

n= − = − =1

2120

222

109 mm


d

Dst

s= = =2

3562

178 mm

ThedistancefromthecentroidofFsctothetopfibreofthesteelsectionis

d

dsc

n= = =2

27 62

3 8.

. mm.



M F d d F d db cp c sc st st sc= + + −

= × + × + ×−

( ) ( )

.1174 68 (109 3.8) 10 1957.83 ((178 3.8) 10 kNm3− × =− 473 6.

Composite beams 281


φM Mbc = × = > ∗ =0 9 473 6 281 3. . . ,426.24 kNm kNm OK


Takingfds=fvs=93kN,therequirednumberofshearconnectorsfromtheendofthecompositebeamtoitsmid-spancanbedeterminedas

n

Ff

ccp

ds

= = =1174 68

9312 63

..


k

nn

c

= − = − =1 180 18

1 180 1814

1 132..

..

.


f k fds n vs= = × × =φ 0 85 1 132 93 89 5. . . kN


n

Ff

ccp

ds

= = =1174 6789 5

13 13..

.

Thedesignstrengthoftheshearconnectionisdeterminedas

F n f Fsh c ds cp= = × = > =14 89 5 1253 1174 68. . ,kN kN OK

Thetotalnumberofstudshearconnectorsinthewholecompositebeamis28.

9.8 deSIgn moment cAPAcIty for negAtIve BendIng

9.8.1 design concepts

The cross sections of peak negative moments in a continuous composite beam must bedesignedforcompleteshearconnectiontopreventcatastrophicfailureinnegativemomentregions.Themaximumdesignshearforceusuallyoccursatthesupportsofacontinuouscompositebeam.Asaresult,itseffectonthedesignmomentcapacityofcrosssectionsinnegativebendingismorecriticalthanonthatofcrosssectionsinpositivebending.Whenγ ≤0.5,thedesignmomentcapacityofcrosssectionisnotaffectedbyverticalshearsothattheeffectiveportionofthesteelwebcontributestotheresistancetobending.Whenγ=1.0,thewebofthesteelsectionisusedtoresistverticalshearandisignoredinthecalculationofthedesignmomentcapacityofcrosssection.

Figure9.20presentstheplasticstressdistributioninageneralcompositecrosssectionwithγ≤0.5andinnegativebending.Themomentcapacityofthecrosssectiondependsontheareaoflongitudinaltensilereinforcement(Ar)intheconcreteslab.Anylevelofreinforcementleadstoatleastpartofthesteelsectionincompression.Themaximumareaoflongitudinalrein-forcementcorrespondstotheconditioninwhichtheentiresteelsectionisincompression.Thelocalbucklingoftheflangesandwebofthesteelsectioninthecompositebeaminnegative


momentregionsmayoccur.Thestrengthandductilityofthecompositesectioninnegativebendingmaybelimitedbylocalbucklingofthesteelsectionorfractureofthereinforcement.SlenderplateelementsarenotallowedtobeusedincompositecrosssectionsaccordingtoAS2327.1,whichisalsoappliedtothedesignfornegativemomentregions.Longitudinalwebstiffenersmaybeweldedtothewebtoreduceitsslenderness,andadditionalplatemaybeweldedtothebottomflangetolowerthePNAtoplacelessofthewebincompression.

9.8.2 key levels of longitudinal reinforcement

In the analysis of a composite cross section in negative bending to compute itsmomentcapacity,thelocationofthePNAneedstobedetermined.ThelocationofPNAdependsontheareaoflongitudinaltensilereinforcementintheconcreteslabandeffectivesteelcrosssection.ThecalculationofkeylevelsofreinforcementthatdefinesthekeylocationsofPNAgivesasimpledirectsolutiontotheproblem(Berryetal.2001a,b).Themethodforcalculat-ingthekeylevellongitudinaltensilereinforcementintheconcreteslabispresentedherein.FurtherinformationonthemethodisgivenbyBerryetal.(2001a,b).

9.8.2.1 Maximum area of reinforcement

Themaximumareaoflongitudinaltensilereinforcementintheconcreteslab,whichmakescontributionstothenegativemomentcapacityofacompositecrosssection,islimitedbythecompressivecapacityoftheeffectivesteelsection.ThePNAislocatedbetweenthetopfaceofthesteelsectionandthebottomofthereinforcement.Theforceequilibriumconditionisexpressedby

F F F F Frm ef ew ef efp= + + +1 2 (9.57)

whereF A frm rm yr=F b t fef ef f yf1 1 1 1=F d t few ew w yw=F b t fef ef f yf2 2 2 2=F b t fefp efp p yfp=

Thesubscripterepresentstheeffectivewidthofaplateelementoreffectiveforce.

Ds

Dc

tw

bf1

bef 2

befp

bcf

dndr

dst

fyr

fyf 1

fyw

fywdsc

Fsc

fyf 2fyfp

Fr

Fst

0.5dew

0.5dew

Figure 9.20 Plastic stress distribution in composite section under negative bending with γ ≤ 0.5.

Composite beams 283

Themaximumareaoflongitudinaltensilereinforcementintheconcreteslabistherefore

A

F F F Ff

rmef ew ef efp

yr

=+ + +1 2 (9.58)

9.8.2.2 PNA located at the junction of the top flange and web

WhenthePNAislocatedatthejunctionofthetopflangeandthewebofthesteelsection,thetopflangeisintension,whilethewebandthebottomflangeandplatearesubjectedtocompression.Fromtheforceequilibrium,therequiredareaofreinforcementiscalculatedby

A

F F F Ff

rfwew ef efp f

yr

=+ + −2 1 (9.59)

9.8.2.3 PNA located in the web

WhenthePNAliesintheweb,itdividesthesteelwebintotensionandcompressionzones.Ifthedepthofthewebincompressionisgreaterthand t few w yw= 30 250/ ,thelocalbucklingofthesteelweboccursandaholewilldevelopinthewebasshowninFigure9.20.Ifthedepthofthewebincompressionisequaltodew,thecompressiveforce(Fwc)inthewebisFwc=dewtwfyw.ThetensileforceinthewebiscomputedasFwt=Fw−Fwc.Therequiredareaofreinforcementisdeterminedfromtheforceequilibriumas

A

F F F F Ff

rhowc ef efp f wt

yr

=+ + − −2 1 (9.60)

9.8.2.4 PNA located at the junction of the web and bottom flange

WhenthePNAislocatedatthejunctionofthesteelwebandthebottomflange,thetopflangeandthewebareintension,whilethebottomflangeandadditionalflangeplateareincompression.Theforceequilibriumgives

A

F F F Ff

rwfef efp f w

yr

=+ − −2 1 (9.61)

9.8.2.5 PNA located at the junction of the bottom flange and plate

WhenthePNAislocatedatthejunctionofthesteelbottomflangeandtheadditionalflangeplate,theentiresteelI-sectionisintensionandtheadditionalflangeplateisincompres-sion.Forthiscase,theareaoflongitudinalreinforcementcanbecalculatedfromtheforceequilibriumas

A

F F F Ff

rfpefp f w f

yr

=− − −1 2 (9.62)

9.8.3 Plastic neutral axis depth

ThePNAofacompositecrosssectionundernegativebendingdependsontheareaoflongi-tudinaltensilereinforcementintheconcreteslab.Itmaybelocatedbetweenthebottomofthereinforcementandtopfaceofthetopflange,inthetopflange,web,bottomflangeandadditionalbottomflangeplate.


Case 1:IfArm≤Ar,thePNAislocatedbetweenthebottomofthelongitudinalreinforcementandthetopfaceofthesteeltopflange.Sincetheentiresteelsectionisincompression,theeffectiveportionofthesteelsectionshouldbeusedtocalculatethenegativemomentcapacity.Case 2:IfArfw≤Ar<Arm,thePNAliesinthetopflangeofthesteelsection.Forthiscase,theportionofthetopflangebelowthePNAisincompressionandtheeffectivewidthofthesteeltopflangeisusedincompressionandtension.ThedepthofthePNAisdeterminedusinglinearinterpolationas

d D

A AA A

tn crm r

rm rfwf= + −

−

1 (9.63)

Case 3:IfArho≤Ar<Arfw,thePNAfallsinthewebofthesteelsection.Aholeformsinthecompressiveportionoftheweb.Ontheonsetoflocalbucklingofthewebincompression,theeffectivedepthofthewebincompressionisdew,whilethedepthofthewebintensionisequaltodwt=dw−dew.ThePNAvarieswithinthedepthofdwt.ThedepthofthePNAisgivenby

d D t

A AA A

dn c frfw r

rfw rhowt= + +

−−

1 (9.64)

Case 4:IfArwf≤Ar<Arho,thePNAislocatedwithinthedepthdewofthewebmeasuredfromthejunctionofthewebandthebottomflange.ThedepthofthePNAisobtainedusinglinearinterpolationas

d D t d

A AA A

dn c f wtrho r

rho rwfew= + + + −

−

1 (9.65)

Case 5:IfArfp≤Ar<Arwf,thePNAliesinthebottomflange.Forthiscase,theportionofthebottomflangebelowthePNAisincompressionandtheeffectivewidthofthesteelbottomflangeisusedincompressionandtension.ThedepthofthePNAisexpressedby

d D t d

A AA A

tn c f wrwf r

rwf rfpf= + + +

−−

1 2 (9.66)

9.8.4 design negative moment capacity

OncethedepthofthePNAhasbeendetermined,thenominalnegativemomentcapacity(Mbc)ofthecompositecrosssectionwithγ≤0.5canbecalculatedbasedonthestressdistri-butionsdepictedinFigure9.20bytakingmomentsaboutthecentroidoftheresultanttensileforce(Fst)inthesteelsectionas

M F d d F d dbc r r st sc sc st= + + −( ) ( ) (9.67)

wheredr isthedistancefromthecentroidofthelongitudinalreinforcementintheconcrete

slabtothetopfaceofthesteelsectiondstisthedistancefromthecentroidoftheresultanttensileforceFstinthesteelsection

tothetopfaceofthesteelsection

Composite beams 285

Forthecaseofnotensioninthesteelsection,dst=0.ThedistancedscisthedistancefromthecentroidoftheresultantcompressiveforceFscinthesteelsectiontothetopfaceofthesteelsection.ItshouldbenotedthatFsciscalculatedusingtheeffectiveareasofsteelplateelementswhichliesbelowthePNA.

Foracrosssectionwithγ=1.0,thesteelwebisignoredinthedeterminationofitsnomi-nalnegativemomentcapacity(Mbfc).Theplasticstressdistributioninthecompositesectionwithγ=1.0ispresentedinFigure9.21.Forthissituation,theareasofkeylevellongitudinalreinforcementintheconcreteslabthatneedtobecalculatedareArm,Arfw,ArwfandArfp.ThedepthofthePNAcanbedeterminedusingtheequationsgivenintheprecedingsection.

For a cross section with 0.5 < γ ≤ 1.0, the design vertical shear reduces the negativemomentcapacityofthesection.Themoment–shearinteractiondiagramforcompositecrosssectionsundercombinednegativebendingandvertical shear ispresented inFigure9.18.Thedesignnegativemomentcapacity(ϕMbv)ofcrosssectionswith0.5<γ≤1.0canbecal-culatedusingEquation9.52.Figure9.22presentsthedesignnegativemomentcapacityofatypicalcompositesectionasafunctionoftheareaofreinforcementandshearratio.

Ds

Dc

bef 1

bef 2

befp

bcf

dn dr

dst

fyr

fyf1

dsc

Fsc

fyf1

fyf 2fyfp

Fr

Fst

Figure 9.21 Plastic stress distribution in composite section under negative bending with γ = 1.0.

Area of reinforcement Ar/Arm

00

φMbv

/φM

sf

0.5

1

1.5

2

2.5

0.2 0.4 0.6

γ = 0.75

γ ≤ 0.5

γ = 1.0

0.8 1 1.2

Figure 9.22 Design negative moment capacity as a function of area of reinforcement and shear ratio.


Therequirednumberofshearconnectorsbetweenthemaximumnegativemomentatthesupportandtheadjacentsectionofzeromomentcanbedeterminedby

n

Ff

cr

ds

= (9.68)

whereFr=Arfyr.

Example 9.4: Negative moment capacity of composite beam

Thecrosssectionofacompositebeamundernegativebendingandadesignverticalshearforceof320kNispresentedinFigure9.23.Theprofiledsteelsheetingisplacedperpen-diculartothesteelbeam.Thehot-rolledsteelsection460UB74.6ofGrade300steelwithfyf=300Mpaandfyw=320Mpaisused.Thecompressivestrengthoftheconcreteflangeis ′ =fc 32MPa.Thecross-sectionalareaoflongitudinaltensilereinforcementinthecon-creteflangeis1100 mm2andthedistancefromthecentroidofthereinforcementtothetopfaceoftheslabis35 mm.Theyieldstressofthereinforcementis500MPa.Calculatethedesignnegativemomentcapacityofthecompositebeamsection.


Theslendernessofthesteelwebunderverticalshearis

λ λew

yey

bt

f= =

− ×= < =

250457 2 14 5

9 1320250

..

53.2 82


φ φV A f Vu w yw= = × × × × = > =∗( . ) . . . ,0 6 0 9 0 6 457 9 1 320 320N 718.6 kN kN OK

Theshearratioatthesectionis

γ

φ=

∗= = <

VVu

320718 6

0 45 0 5.

. .

Therefore,thedesignnegativemomentcapacityofthecompositebeamisnotaffectedbytheverticalshear.

190

14.5

9.1457

140

1500

5414.5

Figure 9.23 Cross section of composite beam under negative bending.

Composite beams 287

2. Key levels of longitudinal reinforcement

Themaximumareaoflongitudinalreinforcementcanbecalculatedbytakingdn=Dc.Forthiscase,thewholesteelsectionisincompression.

Theslendernessoftheflangesis

λ λef

yey

bt

f= =

−= < =

250190 9 1 2

14 5300250

( . ).

/6.8 9 Table 5.1 of AS 2327.1

Thetwoflangesarecompact.Theeffectivewidthofthewebincompressioniscalculatedas

d t

fdew w

yw= = × × = < = − × =30

25030 9 1

250320

241 3 457 2 14 5 428. . .mm mm

Hence,localbucklingoccursandaholeformsinthesteelweb.Thecapacitiesoftheeffectivesteelflangesandwebarecalculatedasfollows:

F b t fef ef f yf1 1 1= = × × × =−190 14.5 300 10 826.5 kN3

F d t few ew w yw= = × × × =−241.3 9.1 320 10 702.7 kN3

F b t fef ef f yf2 2 2= = × × × =−190 14.5 300 10 826.5 kN3

Thecapacitiesofthewebintensionarecomputedas

F d t fw w w yw= = × × × =−428 9.1 320 10 1246.3 kN3

F F Fwt w ew= − = − =1246.3 702.7 543.6 kN

Theareasofkeylevellongitudinalreinforcementintheconcreteslabarecalculatedasfollows:

A

F F F Ff

rmef ew ef efp

yr

=+ + +

=+ + + ×

=1 23826 5 702 7 826 5 0 10

500471

( . . . )11mm2

A

F F F Ff

rfwew ef efp f

yr

=+ + −

=+ + − ×

=2 13702 7 826 5 0 826 5 10

500140

( . . . )55 mm2

AF F F F F

frho

ew ef efp f wt

yr

=+ + − −

=+ + − − ×2 1 702 7 826 5 0 826 5 543 6( . . . . ) 110

500318

3

= mm2

A A A Ar rho r rfw= < <1100mm hence2,

3. Depth of the plastic neutral axis

SinceArho<Ar<Arfw,thePNAislocatedinthewebofthesteelsection.ThedepthofthePNAiscalculatedas

d D tA AA A

dn c frfw r

rfw rhowt= + +

−−

= + +−

1

140 14 51405 110014

.005 318

428 241 3 206 9−

× − =( . ) . mm


4. Forces and distances to centroid of forces

Thetensileforceinreinforcementiscalculatedas

F A fr r yr= = × × =−1100 500 10 5503 kN

ThedistancefromthecentroidofFrtothetopofthesteelsectionis

d D dr c t= − = − =140 35 105 mm

Thetensileforceinthetopsteelflangeis

Fef1 = 826.5 kN

ThedistancefromthecentroidofFef1tothetopofsteelsectionis

d

tf

f1

1

214 52

7 25= = =.

. mm

Thetensileforceinthewebfordn=206.9mmiscomputedas

F d D t t fwt n c f w yw= − − = − − × × =( . ) .1 140 14 5 9 1 320) (206.9 N 152.6 kN

ThedistancefromthecentroidofFwttothetopofsteelsectionis

d

d D ttwt

n c ff=

− −+ =

− −+ =1

12

140 14 52

14 5 40 7 206.9

mm.

. .

Theresultanttensileforceinthesteelsectioniscomputedas

F F Fst ef wt= + = + =1 826 5 152 6 979 1. . . kN

ThedistancefromthecentroidofFsttothetopofthesteelsectioniscomputedas

d

F d F dF

stef f wt wt

st

=+

=× + ×

=1 1 826 5 7 25 152 6 40 7979 1

. . . ..

12.5 mm

Thecompressiveforceinthewebis

F Fwc ew= = 702.7 kN

ThedistancefromthecentroidofFwctothetopofsteelsectionis

d D tD t D d

wc s fs f c n= − −− + −

= − −− + −

=

22

2457 14 5

457 14 5 140 206 92

254

.. .

.77 mm

Theresultantcompressiveforceinthesteelsectionis

F F Fsc wc ef= + = + =2 702 7 826 5 1529 2. . . kN

Composite beams 289

ThedistancefromthecentroidofFsctothetopofthesteelsectioniscomputedas

dF d F D t

Fsc

ew wc ef s f

sc

=+ −

=× + × −

2 2 2

702 7 254 7 826 5 457 14 5 2

( )

. . . ( .

/

/ )).1529 2

= 360 mm

5. Design negative moment capacity

Thenominalnegativemomentcapacityofthecompositesectioniscomputedas

M F d d F d dbc r r st sc sc st= + + −

= × + + × −

( ) ( )

( . ) . ( .550 105 12 5 1529 2 360 12 55 596) kNmm kNm=

Thedesignnegativemomentcapacityofthecompositesectionis

φMbc = × =0 9 596 536 4. . kNm

Example 9.5: Design negative moment capacity of continuous composite beam

Figure9.24showsthecrosssectionofacontinuouscompositebeamundernegativebend-ing and a design vertical shear force of 350 kN. The profiled steel sheeting is placedparalleltothesteelbeam.Thehot-rolledsteelsection410UB53.7ofGrade300steelwithfyf = fyw=300Mpaisused.Thecompressivestrengthoftheconcreteflangeis ′ =fc 25MPa.The cross-sectional areaof longitudinal tensile reinforcement in the concreteflange is1600 mm2andthedistancefromthecentroidofthereinforcementtothetopfaceoftheslab is35 mm.Theyieldstressof thereinforcement is500MPa.Calculate thedesignnegativemomentcapacityofthecompositebeamsectionandtherequirednumberofstudshearconnectorsinthenegativemomentregiontoachievecompleteshearconnection.


Theslendernessofthesteelwebunderverticalshearis

λ λew

yey

bt

f= =

− ×= < =

250403 2 10 9

7 6320250

..

56.7 82

178

10.9

7.6403

178 10.9

1200

130

Figure 9.24 Cross section of continuous composite beam under negative bending.



φ φV A f Vu w yw= = × × × × = > ∗ =( . ) . . . ,0 6 0 9 0 6 403 7 6 320 350N 529.25 kN kN OK

Theshearratioatthesectionis

γ

φ=

∗= = >

VVu

350529 25

0 66 0 5.

. .

Therefore,thedesignnegativemomentcapacityofthecompositebeamisaffectedbytheverticalshear.ItneedstocalculateϕMbcwithγ=0.5andϕMbfcwithγ=1.0,respectively.

2. Design negative moment capacity with γ = 0.5

2.1. Key levels of longitudinal reinforcement

Themaximumareaoflongitudinalreinforcementcanbecalculatedbytakingdn=Dc.Forthiscase,thewholesteelsectionisincompression.

Theslendernessoftheflangesis

λ λef

yey

bt

f= =

−= < =

250178 7 6 2

10 9320250

( . ).

/8.8 9 Table 5.1 of AS 2327.1

Thetwoflangesarecompact.Theeffectivewidthofthewebincompressioniscalculatedas

d t

fdew w

yw= = × × = < =30

25030 7 6

250320

201 5 381. . mm mm

Hence,localbucklingoccursandaholeformsinthesteelweb.Thecapacitiesoftheeffectivesteelflangesandwebarecalculatedasfollows:

F b t fef ef f yf1 1 1= = × × × =−178 10.9 320 10 620.86 kN3

F d t few ew w yw= = × × × =−201.5 7.6 320 10 490 kN3

F b t fef ef f yf2 2 2= = × × × =−178 10.9 320 10 620.86 kN3

Theareasofkeylevellongitudinalreinforcementintheconcreteslabarecalculatedasfollows:

A

F F F Ff

rmef ew ef efp

yr

=+ + +

=+ + + ×

=1 23620 86 490 620 86 0 10

500346

( . . )33 mm2

A

F F F Ff

rfwew ef efp f

yr

=+ + −

=+ + − ×

=2 13490 620 86 0 620 86 10

500980

( . . )mmm2

A A A Ar rfw r rm= < <1600mm hence2, .

Composite beams 291

2.2. Depth of the plastic neutral axis

SinceArfw<Ar<Arm,thePNAliesinthetopflangeofthesteelsection.Thedepthoftheneutralaxisiscalculatedas

d D

A AA A

tn crm r

rm rfwf= +

−−

= +

−−

1 1303463 16003463 980 × =10 9 138 2. . mm

2.3. Forces and distances to centroid of forces

Thetensileforceinreinforcementis

F A fr r yr= = × × =−1600 500 10 8003 kN


d D dr c t= − = − =130 35 95 mm

Thecompressiveforceinthetopsteelflangeiscalculatedas

F b D t d fef c ef c f n yf1 1 13178 130 10 9 138 2 320 10 15⋅−= + − = × + − × × =( ) ( . . ) 33 8. kN

ThedistancefromthecentroidofFef1⋅ctothetopofsteelsectionis

d t

D t df c f

c f n1 1

1

210 9

130 10 9 138 22

9 55⋅ = −+ −

= −+ −

=.. .

. mm

Theresultantcompressiveforceinthesteelsectioniscomputedas

F F F Fsc ef c ew ef= + + = + + =⋅1 2 153 8 490 620 86 1264 7. . . kN

ThedistancefromthecentroidofFsctothetopofthesteelsectioniscomputedas

dF d F D F D t

Fsc

ef c f c ew s ef s f

sc

=+ + −

=× + ×

⋅1 1 2 22 2

153 8 9 55 490

( ) ( )

. .

/ /

(( ) . ( . ).

403 2 620 86 403 10 9 21264 7/ /

274.4 mm+ × −

=

Thedistance from the centroidof steel flange in tension to the topof steel section is

dd D

stn c=−

=−

=2

138 2 1302

4 1.

. mm

2.4. Design negative moment capacity


M F d d F d dbc r r st sc sc st= + + −

= × + + × −

( ) ( )

( . ) . ( . .800 95 4 1 1264 8 274 4 4 1)) .kNmm kNm= 421 13

Thedesignnegativemomentcapacityofthecompositesectionwithγ=0.5is

φMbc = × =0 9 421 13 379. . kNm



3.1. Key levels of longitudinal reinforcement

Atthecrosssectionwithγ=1.0,thesteelwebisignored.Theareasofkeylevellongitu-dinalreinforcementarecomputedas

A

F F F Ff

rmef ew ef efp

yr

=+ + +

=+ + + ×

=1 23620 86 0 620 86 0 10

5002483

( . . )mmm2

A

F F F Ff

rfwew ef efp f

yr

=+ + −

=+ + − ×

=2 130 620 86 0 620 86 10

5000

( . . )mm2

A A A Ar rfw r rm= < <1600mm hence2,

3.2. Depth of the plastic neutral axis

SinceArfw<Ar<Arm,thePNAliesinthetopflangeofthesteelsection.Thedepthoftheneutralaxisiscalculatedas

d D

A AA A

tn crm r

rm rfwf= +

−−

= +

−−

×1 130

2483 16002483 0

110 9 133 9. .= mm

3.3. Forces and distances to centroid of forces


d D dr c t= − = − =130 35 95 mm

Thecompressiveforceinthetopsteelflangeiscomputedas

F b D t d fef c ef c f n yf1 1 13178 130 10 9 133 9 320 10 39⋅−= + − = × + − × × =( ) ( . . ) 88 7. kN

ThedistancefromthecentroidofFef1⋅ctothetopofthesteelsectionis

d t

D t df c f

c f n1 1

1

210 9

130 10 9 133 92

7 4⋅ = −+ −

= −+ −

=.. .

. mm

Theresultantcompressiveforceinthesteelsectioniscomputedas

F F Fsc ef c ef= + = + =⋅1 2 398 7 620 86 1019 56. . . kN

ThedistancefromthecentroidofFsctothetopofthesteelsectionis

dF d F D t

Fsc

ef c f c ef s f

sc

=+ −

=× + × −

⋅ ⋅1 1 2 2 2

398 7 7 4 620 86 403 10

( )

. . . (

/

.. ).

9 21019 56

/245 mm=

Thedistancefromthecentroidofsteelflangeintensiontothetopofsteelsectionisdst=(dn−Dc)/2=(133.9−130)/2=1.95mm

Composite beams 293

3.4. Design negative moment capacity


M F d d F d dbfc r r st sc sc st= + + −

= × + × + ×−

( ) ( )

( . ) . (800 95 1 95 10 1019 56 23 445 1 95 10 325 43− × =−. ) . kNm

Thedesignnegativemomentcapacityofthecompositesectionwithγ=1.0istherefore

φMbfc = × =0 9 3 292 8. .25.4 kNm


Forthesectionwithγ=0.66,thedesignnegativemomentcapacityiscalculatedas

φ φ φ φ γM M M Mbv bfc bc bfc= + − −

= + − − × =

( )( )

. ( . )( . )

2 2

292 8 379 292 8 2 2 0 66 3351 4. kNm


FromTable9.1,fvs=89kN.Takingfds=fvs=89kN,therequirednumberofshearcon-nectorsbetweenthemaximumnegativemomentatthesupportandtheadjacentsectionofzeromomentcanbedeterminedas

n

Ff

cr

ds

= = =80089

8 99.


k

nn

c

= − = − =1 180 18

1 180 1810

1 123..

..

.


f k fds n vs= = × × =φ 0 85 1 123 89 85. . kN


n

Ff

cr

ds

= = =80085

9 4.

Adoptingnc=10,thedesignstrengthoftheshearconnectionisdeterminedas

F n f Fsh c ds r= = × = > =10 85 850 800kN kN OK,

The total number of stud shear connectors in the negative moment region which isassumedtobesymmetricaboutthesupportis20.


9.9 trAnSfer of longItudInAl SheAr In concrete SlABS

9.9.1 longitudinal shear surfaces

Shearconnectorstransferlongitudinalshearfromthesteelbeamtotheconcreteslabinacompositebeam.Thesheartransfermechanismintheconcreteslabcanbesimulatedbyeitherthestrut-and-tiemodel(Liangetal.2000;Liang2005)ortheshear–frictionmodel.Theshearconnectorsunderlongitudinalshear inducecompressiveforceontheconcrete,whichisdispersedthroughstrutsandinterconnectedbytensionties.Asaresult,longitu-dinalshearreinforcement(placedperpendiculartothesteelbeam)mustbeprovidedintheconcreteslabtoresistthetensileforces.

AS2327.1(2003)identifiesfourtypesoflongitudinalshearfailuresurfaces,whichareschematically illustrated in Figures 9.25 through 9.27. As shown in Figure 9.27, Type 4longitudinalshearfailuremayoccurincompositeedgebeamswithprofiledsteelsheetingplacedperpendiculartothesteelbeamwhentheoutstandofthecompositebeamislessthan600 mmandstudshearconnectorsareweldedthroughthesheeting.

AsshowninFigure9.25,thelongitudinalshearfailurecorrespondingtotheType1shearsurfacemayoccurattheoutsidefacesofshearconnectorgroups,atsectionswherelongi-tudinalshearreinforcementisterminatedoroverthesheetingribswhichareparalleltothesteelbeam.Theperimeterlength(up)ofType1shearsurfacesistakenasDcforsolidslabs,compositeslabswithsheetingribsperpendiculartothesteelbeamandforcompositeslabsbetweenribswhichareparalleltothesteelbeam.ForType2shearsurfaces,theperimeterlengthisdeterminedas(bx+2hs),wherebxistheoverallwidthacrossthetopofconnectorsin

Type 2

Type 1Type 1

Type 2

Type 1Type 1

Figure 9.25 Type 1 and 2 longitudinal shear failure surfaces.

(a)

≥30 ≥30

bx bx

hs

c1 c2 c1 c2

hs hs

bx

(b) (c)

Figure 9.26 Longitudinal shear surfaces: (a) shear surface 1, (b) shear surface 2 and (c) shear surface 3.

Composite beams 295

thecrosssectionandhsistheoverallheightoftheshearconnectorsabovethetopflangeofthesteelsection.TheType3shearsurfacesareassociatedwithlongitudinalshearfailurearoundtheshearconnectorgroupsincompositeslabs,asillustratedinFigure9.26.TheperimeterlengthofType3shearsurfacesistakenasup=min(u1,u2,u3),whicharedefinedinFigure9.26.

9.9.2 design longitudinal shear force

Thecompressiveforceintheconcreteslabofacompositebeamisassumedtobeuniformlydistributedacrosstheeffectivewidthoftheconcreteflange.Thisimpliesthatthelongitu-dinalshearflowintheconcreteslabisuniform.Thisuniformshearflowmodelisusedtodeterminethedesignlongitudinalshearforceperunitlength( )VL

∗ ofthecompositebeamforType1,2and3shearsurfacesatthebeamcrosssection.InAS2327.1(2003),VL

∗isassumedtovarylinearlyfromzeroattheextremitiesoftheeffectivewidthoftheconcreteslabtothemaximumoneachsideofthecentrelineofthesteelbeamasshowninFigure9.28.ForType1shearsurface,VL

∗iscalculatedby

V

xb

VLcf

L tot∗ =

∗

⋅ (9.69)

wherexisthedistancefromtheextremityoftheeffectivewidthtothecrosssectionwherethe

longitudinalshearforceiscalculatedVL tot⋅∗ isthetotaldesignlongitudinalshearforceperunitlength,givenby

V

n fs

L totx ds

c⋅∗ = (9.70)

wherenxisthenumberofconnectorsinacrosssectionfdsisthedesignshearcapacityofshearconnectorsinthebeamscisthelongitudinalspacingofshearconnectors

ForType2and3shearsurfaces,thecompressiveforceacrosstheconcreteslabistransferredbytheshearsurfaces.Therefore,thedesignlongitudinalshearforceactingonType2and3surfacesistakenasV VL L tot

∗ = ∗⋅ .

Type 4

Figure 9.27 Type 4 longitudinal shear surface in edge beam.


9.9.3 longitudinal shear capacity

Theshear–frictionmodelforreinforcedconcreteisadoptedinClauseof9.6ofAS2327.1(2003)tocalculatethenominal longitudinalshearcapacity(perunit length)ofType1,2 and 3 shear surfaces, which is taken as the lesser value calculated by the followingequations:

V u f A fL p c sv yr= ′( ) +0 36 0 9. . (9.71)

V f uL c p= ′0 32. (9.72)

whereAsvisthetotalcross-sectionalareaoflongitudinalshearreinforcementcrossingtheshearsurface(mm2).

9.9.4 longitudinal shear reinforcement

ItisnecessarytoensurethattheconcreteshearcapacityofType1,2and3shearsurfacesisnotlessthanthedesignlongitudinalshearforce,suchasφ0 32. ′ ≥ ∗f u Vc p L .Thetotalcross-sectionalareaoflongitudinalshearreinforcementforresistingType1,2and3shearsur-facescanbedeterminedbyusingthefollowingequation,respectively:

A

V u f

fsv

L p c

yr

=− ′∗ /φ 0 360 9..

(9.73)

x

V*L

be1

bcf

be2

Figure 9.28 Distribution of longitudinal shear force for Type 1 shear surface.

Composite beams 297

Theperimeterlengthupistakenasthelesseroftheperimeterlengthsu2andu3ofType2and3shearsurfacesinEquation9.73.However,thelargeroftheperimeterlengthsu2andu3shouldbeusedtocalculatetheminimumcross-sectionalareaofshearreinforcementforType2and3shearsurfaces.Anyexistingflexuralandshrinkagereinforcementplacedtrans-versetothesteelbeamintheconcreteslabcanbetreatedastheeffectivelongitudinalshearreinforcementiftheysatisfytheanchoragerequirementofClause9.7.3ofAS2327.1(2003).ItisnotedthattheadditionalreinforcementforType1shearsurfacedependsontheType2and3shearreinforcementaswellasexistingreinforcementintheconcreteslab.AS2327.1doesnotgivedesignrulesonthespacingoflongitudinalshearreinforcement.ItissuggestedthatthemaximumspacingoflongitudinalshearreinforcementforType1,2and3shearsurfacesshouldbetakenastheminimumof2sc,4Dcand600 mm(LiangandPatrick2001).

Thelongitudinalreinforcementmusthaveadequateanchoragelengthtodevelopitsyieldstress.ThestressdevelopmentlengthoflongitudinalreinforcementinconcreteslabsgiveninAS3600(2001)isadoptedhere,whichisexpressedby

L

k k f A

c d fdyst

yr b

b cb=

+ ′≥1 2

225

( ) (9.74)

wherek1 1 0= .k2 2 4= .dbisthediameterofthereinforcingbarAbisthecross-sectionalareaofthebarcisthecovertothereinforcingbars

Forbottomfacereinforcementincompositeslabs,cmaybetakenashr.TheType1longitudinalreinforcementshouldbeextended12dbfromthesectionwhere

longitudinalreinforcementisnotrequiredtoresistlongitudinalshear.SpecialsteelreinforcingproductshavebeendevelopedinAustraliaforuseincomposite

beamsaslongitudinalshearreinforcement(LiangandPatrick2001;Liangetal.2001).ThesenewreinforcingproductscomplementthenewdesignapproachtothelongitudinalshearincompositebeamsandhavebeenincorporatedinthecomputersoftwareCOMPSHEARforthedesignoftheshearconnectionofcompositebeams(Liangetal.2001).Waveformrein-forcingproductsDECKMESHcanbeusedincompositeedgebeamsincorporatingBondekII andCondeckHPprofiled steel sheeting toprevent rib shearing failure fromoccurringwhenthesheetingribsareplacedperpendiculartothesteelbeam(LiangandPatrick2001).

ThedesignprocedurefordeterminingType1,2and3longitudinalshearreinforcementintheconcreteslabofacompositebeamisgivenasfollows:

1.Calculatethedesignshearcapacityofshearconnectors,whichrequirestheminimumnumberofshearconnectorstobedetermined.

2.Calculatetotaldesignlongitudinalshearforceperunitlength. 3.CalculatetheperimeterlengthsofType1,2and3longitudinalshearsurfaces. 4.CheckfortheconcreteshearcapacityofType1,2and3longitudinalshearsurfaces,

suchthatφ0 32. ′ ≥ ∗f u Vc p L .Ifthisconditionisnotsatisfied,eithertheperimeterlengthsortheconcretecompressivestrengthshouldbeincreasedandthengobacktoStep1.

5.Calculatethecross-sectionalareasandlengthsofadditionallongitudinalshearrein-forcement for Type 2 and 3 shear surfaces. The cross-sectional area of any fullyanchoredbottomreinforcementintheconcreteslabplacedtransversetothelongitudi-nalaxisofthesteelbeamistakenintoaccount.


6.CalculatethedesignlongitudinalshearforceperunitlengthforType1surfaceatanydistancefromtheextremityoftheslabeffectivewidth.

7.Calculatethecross-sectionalareasandlengthsofadditionalreinforcementforType1shearsurfaceforeveryshear forceVL

∗computed.Thecross-sectionalareaofanyfullyanchoredtransversereinforcementandtheadditionalType2and3reinforcementshouldbetakenintoconsideration.

8.Determinethemaximumcross-sectionalareaandlengthsofadditionalreinforcementforType1shearsurface,which is treatedas the topreinforcement in theconcreteslab.

Example 9.6: Design of shear connection of internal composite beam

The cross section of an internal primary composite beam which is simply supportedis schematically depicted in Figure 9.29. The effective span of the composite beam is8.4 m. The profiled steel sheeting is placed parallel to the steel beam. The steel sec-tion410UB59.7ofGrade300steelisused.Thedesignstrengthoftheconcreteflangeis′ =fc 32 MPa.Twentyheadedstudshearconnectorsof19 mmdiameterareuniformlydis-

tributedbetweentheendandmid-spanofthecompositebeam.Theheightoftheheadedstudis95 mm.TheflexuralreinforcementofN10at240 mmisplacedatthetopfaceoftheconcreteslab.TheSL72mesh(Ast=179mm2/m)isplacedonthetopofthesheetingribstoprovidecrackcontrolforshrinkageandtemperatureeffects.Theexposureclas-sificationisA1.Designtheshearconnectionofthecompositebeam.

1. Design shear capacity of shear connectors

Thenominalshearcapacityof19 mmdiameterheadedstudembeddedin32MPacon-creteisobtainedfromTable9.1asfvs=93kN.

Theload-sharingfactoris

k

nn

c

= − = − =1 180 18

1 180 1820

1 14..

..

.

Thedesignshearcapacityofashearconnectorsinthecompositebeamiscomputedas

f k fds n vs= = × × =φ 0 85 1 14 93 90. . kN

2100

960

140

960

12.8SL72

54

410UB59.7

N10 at 240

Figure 9.29 Cross section of internal composite beam.

Composite beams 299

2. Total design longitudinal shear force

Thelongitudinalspacingofshearconnectorsisdeterminedassc=213mm.Thetotaldesignlongitudinalshearforceperunitlengthofthecompositebeamiscalcu-latedas

V

n fs

L totx ds

c⋅∗ = =

× ×=

1 90 1000213

423 N/mm

3. Perimeter lengths of shear surfaces

Theoverallwidthacrossthetopofconnectorinthecrosssectionis

b dx h= = 32 mm

TheperimeterlengthsofType1and2shearsurfacesarecomputedasfollows:

u D hc r1 140 54 86= − = − = mm

u b hx s2 2 32 2 95 222= + = + × = mm

Assumethatthestudisplacedatthecentreoftheadjacentribs,thedistancec1is

c

s br x1

2200 32

284=

−=

−= mm

u b h h cx s r32

12 2 22 32 2 95 54 84 219= + − + = + × − + =( ) ( ) mm

Assumec1=30mm:

u b h h c hx s r s32

12 2 232 95 54 30 95 178= + − + + = + − + + = <( ) ( ) mm 219 mm

Hence,

u3 219 178= =min( , ) 178 mm

4. Check for the concrete shear capacity

ThedesignlongitudinalshearforceperunitlengthofthebeamactingonType1shearsurfaceis

V

xb

VLcf

L tot∗ =

∗ =

× =⋅

10502100

423 211 5. N/mm

ThedesignshearcapacityoftheconcreteforType1shearsurfaceiscalculatedas

φ φV f u VL c p L= ′ = × × × = > ∗ =0 32 0 7 0 32 32 86 616 4 211 5. . . . . ,N/mm N/mm OK

TheminimumperimeterlengthofType2and3shearsurfacesis

u u up = = =min( , ) min( ,2 3 222 178) 178 mm

ThedesignshearcapacityoftheconcreteforType2and3shearsurfacesis

φ φV f u VL c p L tot= ′ = × × × = > ∗ =⋅0 32 0 7 0 32 32 178 1276 423. . . ,N/mm N/mm OK


5. Additional type 2 and 3 longitudinal shear reinforcement

Sinceu2>u3,upistakenasu3whichisusedtocalculatethetotalareaofshearreinforce-mentperunitlengthforType2and3shearsurfacesasfollows:

AV u f

fsv

L p c

yr

=∗ − ′

=− ×

×

=

/ /

m

φ 0 36

0 9423 0 7 0 36 178 32

0 9 500

0 537

.

.. .

.

. mm /mm 537 mm /m2 2=

Theminimumareaoflongitudinalshearreinforcementiscomputedas

A

uf

Asvp

yrsv⋅ = =

×= < =min

800 800 222500

5355 mm /m 37 mm /m2 2

TherequiredadditionalType3reinforcementis

Asb a⋅ = − =

5372

179 89 5. mm /m2

Thespacinglimitonlongitudinalshearreinforcementis

s s Db c c⋅ = = × × =max min( , , min( , ,2 2 213 4264 600) 4 140 600) mm

UseN10at400(Asb⋅a=196mm2/m).ThedevelopmentlengthofType2and3reinforcementistakenas

L dyst b= =25 250 mm

ThelengthofType3reinforcementiscomputedas

L L b c Lab ab x sy t⋅ ⋅ ⋅= = + + = × + + =1 2 10 5 0 5 32 30 250 296. . mm

TakeLab⋅1=Lab⋅2=300mm.

6. Additional type 1 longitudinal shear reinforcement

ThetotalareaoflongitudinalshearreinforcementperunitlengthforType1shearsur-faceiscalculatedasfollows:

AV u f

fsv

L p c

yr

=∗ − ′

=− ××

=

/ /φ 0 36

0 9211 5 0 7 0 36 86 32

0 9 500

0 282

.

.. . .

.

. mmm /mm 282 mm /mm2 2=

ExistingflexuralreinforcementatthetopfaceoftheconcreteslabN10at240:Ast=327mm2/m

TheSL72mesh:Asb=179mm2/m

TherequiredadditionalType1reinforcementiscalculatedas

Asb a⋅ = − − = −282 179 327 244 mm /m2

Therefore,noadditionalreinforcementisrequiredfortheType1shearsurfacesneartheshearconnector.

Composite beams 301

The design longitudinal shear force at section where Type 1 reinforcement is notrequiredcanbecalculatedas

V u fL p c∗ = ′ = × × × =φ0 36 0 7 0 36 86 32 122 6. . . . N/mm

ThedistancebetweentheextremityoftheeffectivewidthandtheType1shearplaneis

xV

VbL

L tot

cf=∗

∗

=

× =

⋅

122 6423

2100 609.

mm

Thedistancefromthisshearplanetotheverticalcentroidalaxisofthesteelbeamis

xc = − =1050 609 441mm

The length of the effective reinforcement measured from the centre line of the steelbeamis

x d L Lc b t t+ = + × = < = =⋅ ⋅12 441 12 10 561 9601 2mm mm

Therefore,theflexuralreinforcement(N10at240)placedatthetopfaceoftheconcreteslabisadequateforresistingType1shearfailure.

Example 9.7: Design of shear connection of edge composite beam

Thecrosssectionofasecondaryedgecompositebeamwhichissimplysupportedissche-maticallydepictedinFigure9.30.Theeffectivespanofthecompositebeamis6m.Theprofiledsteelsheetingisplacedparalleltothesteelbeam.Thesteelsection410UB59.7ofGrade300steel isused.Thedesignstrengthof theconcreteflange is ′ =fc 32 MPa .Twenty-twoheadedstudshearconnectorsof19 mmdiameterareuniformlydistributedinpairsbetweentheendandmid-spanofthecompositebeam.Theheightoftheheadedstudis95 mm.TheflexuralreinforcementofN10at300 mmisplacedatthetopfaceoftheconcreteslab.TheexposureclassificationisA1.Designtheshearconnectionofthecompositebeam.

410UB59.7

12.8

140

250 750

1000

N10 at 300

Figure 9.30 Cross section of edge composite beam.


1. Design shear capacity of shear connectors

Thenominalshearcapacityof19 mmdiameterheadedstudembeddedin25MPacon-creteisobtainedfromTable9.1asfvs=89kN.

Theload-sharingfactoris

k

nn

c

= − = − =1 180 18

1 180 1822

1 14..

..

.

Thedesignshearcapacityofashearconnectorsinthecompositebeamiscomputedas

f k fds n vs= = × × =φ 0 85 1 14 89 86 24. . . kN

2. Total design longitudinal shear force

Thetotaldesignlongitudinalshearforceperunitlengthofthecompositebeamiscalcu-latedas

V

n fs

L totx ds

c⋅∗ = =

× ×=

2 86 24 1000300

575.

N/mm

3. Perimeter lengths of shear surfaces

Theoverallwidthacrossthetopofconnectorinthecrosssectionis

b s dx x h= + = + =80 32 112 mm

TheperimeterlengthsofType1and2shearsurfacesarecomputedasfollows:

u Dc1 140= = mm

u b hx s2 2 112 2 95 302= + = + × = mm

4. Check for the concrete shear capacity

ForType1shearsurface,x=be2−sx/2−dh/2=750−80/2−32/2=694mm.ThedesignlongitudinalshearforceperunitlengthofthebeamactingonType1shear

surfaceis

V

xb

VLcf

L tot∗ =

∗ =

× =⋅

6941000

575 399N/mm

ThedesignshearcapacityoftheconcreteforType1shearsurfaceiscalculatedas

φ φV f u VL c p L= ′ = × × × = > ∗ =0 32 0 7 0 32 25 140 784 399. . . ,N/mm N/mm OK

ThedesignshearcapacityoftheconcreteforType2shearsurfaceis

φ φV f u VL c p L tot= ′ = × × × = > ∗ =⋅0 32 0 7 0 32 25 302 1691 2 575. . . . ,N/mm N/mm OOK

Composite beams 303

5. Type 2 longitudinal shear reinforcement

ThetotalareaofshearreinforcementperunitlengthforType2shearsurfaceiscalcu-latedas

AV u f

fsv

L p c

yr

=∗ − ′

=− × ×

×

=

/ /φ 0 36

0 9575 0 7 0 36 302 25

0 9 500

0 617

.

.. .

.

. mmm /mm 617 mm /m2 2=

Theminimumareaoflongitudinalshearreinforcementiscomputedas

A

uf

Asvp

yrsv⋅ = =

×= < =min

800 800 302500

483 mm /m 617 mm /m2 2

TherequiredadditionalType2reinforcementis

Asb a⋅ = =

6172

308 5. mm /m2

Thespacinglimitonlongitudinalshearreinforcementis

s s Db c c⋅ = = × × =max min( , , min( , ,2 2 300 5604 600) 4 140 600) mm

UseN10at250(Asb⋅a=314mm2/m).ThedevelopmentlengthforType2reinforcementistakenas25db=250mm.

ThelengthofType2reinforcementiscomputedas

L b Lab x sy t⋅ ⋅= + = × + =2 0 5 0 5 112 250 306. . mm andtakeLab⋅ =2 310 mm.

SinceLab⋅1=310mm>250−20=230mm,useU-bars.Hence,useN10at250U-bars,Lab⋅1=230mmandLab⋅2=310mm.

6. Additional type 1 longitudinal shear reinforcement

ThetotalareaoflongitudinalshearreinforcementperunitlengthforType1shearsur-faceiscalculatedasfollows:

AV u f

fsv

L p c

yr

=∗ − ′

=− × ×

×

=

/ /φ 0 36

0 9399 0 7 0 36 140 25

0 9 500

0 707

.

.. .

.

. mmm /mm 707 mm /mm2 2=

ExistingreinforcementatthetopfaceoftheconcreteslabN10at300:Ast=262mm2/mTherequiredadditionalType1reinforcementiscalculatedas

Asb a⋅ = − − =707 262 314 131mm /m2

UseN10at500(Ast⋅a=157mm2/m).


The design longitudinal shear force at section where Type 1 reinforcement is notrequiredcanbecalculatedas

V u fL p c∗ = ′ = × × × =φ0 36 0 7 0 36 140 25 176 4. . . . N/mm

ThedistancebetweentheextremityoftheeffectivewidthandtheType1shearplaneis

xV

VbL

L tot

cf=∗

∗

=

× =

⋅

176 4575

1000 307.

mm

Thedistancefromthisshearplanetotheverticalcentroidalaxisofthesteelbeamis

xc = − =750 307 443 mm

Thelengthoftheeffectivereinforcementmeasuredfromthecentrelineofthesteelbeamis

L x dat c b⋅ = + = + × =2 12 443 12 10 563 mm

Hence,useN10at500(Ast⋅a=157mm2/m)Lat⋅1=230,Lat⋅2=565mmastheadditionalType1reinforcement.

9.10 comPoSIte BeAmS WIth PrecASt holloW core SlABS

CompositebeamswithprecasthollowcoreconcreteslabsdepictedinFigure9.31arecom-monlyusedintheUnitedKingdomasalternativestocompositebeamsincorporatingpro-filedsteelsheeting.Themainadvantagesofthisformofcompositebeamconstructionare(a)precast concrete slabscan spanup to15mwithoutpropping, (b) theerectionof theprecastconcreteslabunitsaresimpleand(c)thepre-weldingofstudconnectorsonthesteelbeamsleadstorapidconstruction(Lam2002).Thedepthoftheprecasthollowcoreslabsisusuallybetween150and400 mm.

The design moment capacity of composite beams incorporating precast hollow coreslabs with complete or partial shear connection can be determined by the plastic stress

Precast hollow core slab In situ concrete in�ll

Transverse reinforcement

UB section

Headed stud

Figure 9.31 Composite beam with precast hollow core slab.

Composite beams 305

distributions,providingthatanappropriateeffectivewidthfortheconcreteflangeisused.Theeffectivewidthoftheprecasthollowcoreconcreteflangeisinfluencedbythestrengthofconcreteandthetransversereinforcement(Lametal.2000a).Basedontheresultsobtainedfromexperiments(Lametal.2000b),theeffectivewidthoftheprecasthollowcoreconcreteslabisgivenby(Lametal.2000a)

b

f fcf

cu t

=

′

+1000

25 0 4300

2.

(9.75)

wherefcuisthecompressiveconcretecubestrengthoftheinsituconcreteinfill(MPa)′ft istheeffectivetensilestrengthandisdeterminedas ′ =f A f At st yr c/ ,whereAstisthearea

oftransversereinforcementandAcisthecross-sectionalareaofconcrete

Forsimplicity,theeffectivewidthoftheprecasthollowcoreslab(bcf)canbetakenasspan/5.Push-outtests indicatethattheshearstrengthofshearconnectorsincompositebeams

withprecasthollowcoreslabsisinfluencedbytheinsituconcretegapwidth,thetransversejointsbetweenhollowcoreslabs,thestrengthofconcreteandtheamountoftransverserein-forcement(Lametal.2000a).ThenominalshearcapacityofheadedstudshearconnectorsincompositeprecasthollowcoreslabsistakenasthelesserofthevaluescalculatedusingthefollowingequationsbasedonEurocode4andpush-outtestresults(Lametal.2000a):

f d f Evs bs cj c= ′0 29 1 2 3

2. α α α ϖ (9.76)

f f

dvs u

bs=

0 8

4

2

.π

(9.77)

whereα1isthefactorwhichaccountsfortheeffectoftheheightofstudandisexpressedas

α1 0 2 1 1 0= + ≤. ( ) .h ds bs/α2isthefactorconsideringtheeffectoftheinsituinfillgap(g)betweenthehollowcore

slabsandisgivenbyα2 0 5 70 1 1 0= + ≤. ( ) .g / withg≥30α3isusedtotakeintoaccounttheeffectofthediameter( )db ofthetransversereinforce-

mentandisdeterminedbyα3 0 5 20 1 1 0= + ≤. ( ) .db /ϖisthetransversejointfactorandistakenasϖ = + ≤0 5 600 1 1 0. ( ) .bhcs / ,wherebhcsis

thewidthofthehollowcoreslab


9.11.1 elastic section properties

Theelasticsectionpropertiesofcompositebeamcrosssectionswithcompleteshearconnec-tionarecalculatedbyusingthetransformedsectionmethod.Forthispurpose,thefullinter-actionbetweentheconcreteslabandthesteelbeamisassumed.Theeffectivesectionofacompositebeamshouldbeusedinthecalculationofitselasticsectionproperties.Thetensilestrengthofconcreteisignored.Figure9.32showsthetransformedsectionofacompositebeam,whichisanequivalentsteelsection.Thetransformedeffectivewidthoftheconcrete


flangeisdeterminedasbtf=bcf/n.Themodularratio(n)iscalculatedasn=Es/Ece,whereEceistheeffectivemodulusofconcrete.Whencalculatingtheimmediatedeflectionsofacom-positebeamandthemaximumstressesinthesteelbeamusingthesecondmomentofarea(Iti),EceistakenasEc.Forthelong-termdeflectioncalculationsusingItl,Eceiscalculatedby

EE

cec

c

=+ ∗1 φ

(9.78)

wheretheconcretecreepfactorφc∗ = 2isusedinAS2327.1(2003).

Theelasticneutralaxisofthetransformedsectionislocatedeitherintheconcreteslaborinthesteelsection.Thedepthoftheelasticneutralaxisofthetransformedsectioncanbedeterminedbytakingthefirstmomentofareaabouttheelasticneutralaxis.IftheelasticneutralaxisliesintheconcretecoverslabofacompositeslaborasolidslabasdepictedinFigure9.32,thedepthoftheelasticneutralaxisofthecompositesectionwithcompleteshearconnectionisdeterminedas

( ) ( )b d

dA d dtf ne

nes sg ne× = −

2 (9.79)

whereAsisthetotaleffectiveareaofthesteelsectiondsgisthedistancefromthecentroidoftheeffectivesteelsectiontothetopofthecon-

creteslab

Theelasticneutralaxisdepth(dne)canbeobtainedfromtheaforementionedequationas

d c c d cne a a sg a= + −2 2 (9.80)

whereca=As/btf.The second moment of area of the transformed section can be calculated by taking

momentofareasabouttheelasticneutralaxisas

I

b dI A d dt

tf nes s sg ne= + + −

32

3( ) (9.81)

(a)

Dc

Ds

Dc

bcf btf

dne

dsghr

(b)

Figure 9.32 Transformed section of composite beam: (a) effective section and (b) transformed section.

Composite beams 307

Whentheelasticneutralaxisislocatedinthesteelribsofthecompositeslabwithλ=0orinthesteelsectionofthecompositebeamwithλ=0,thedepth(dne)oftheelasticneutralaxisandthesecondmomentofarea(It)aregivenasfollows:

d

b h h A db h A

netf c c s sg

tf c s

=++

( )/2 (9.82)

I

b hb h d

hI A d dt

tf ctf c ne

cs s sg ne= + −

+ + −

3 22

12 2( ) (9.83)

Foracompositebeamwithpartialshearconnectionatthecrosssectionofmaximumbend-ing,theeffectivesecondmomentsofareaaregiveninAS2327.1(2003)asfollows:

I I I Ieti ti mb ti s= − − −0 6 1. ( )( )β (9.84)

I I I Ietl tl mb tl s= − − −0 6 1. ( )( )β (9.85)

whereβmbisthedegreeofshearconnectionatthecrosssectionunderthemaximumbendingmoment.

9.11.2 deflection components of composite beams

Thedeflectionsofa compositebeam include the immediatedeflectionsof the compositebeamunder construction loadsduringvarious construction stagesandunder short-termin-serviceloadsanditslong-termdeflectionsduetocreepandshrinkageduringin-serviceconditions.Theexactcalculationofdeflectionsofcompositebeamsiscomplex.Thereasonsfor thisareas follows: (1) thechangeof loadsduring the lifeof thestructurecannotbepredictedinthedesignstage,(2)thestructuralmodelmaynotadequatelyaccountfor3Deffectsofthestructure,(3)thenon-linearload–slipbehaviourofshearconnectionisusuallyignoredand(4)themodulusofelasticityfortheconcretechangeswithtimeduetocreepandshrinkage(Viestetal.1997).Ifthespansarelarge,alargeportionofliveloadispresentoveralongperiodoftime,oriftheconcreteusedfortheslabissensitivetocreepandshrinkage,thelong-termdeflectionsduetocreepandshrinkageneedtobetakenintoaccount.

AsimplifiedmethodforcalculatingthedeflectionsofcompositebeamsissuggestedinAS2327.1.ThecomponentsofdeflectionofacompositebeamandthecorrespondingdesignloadsaredescribedinAS2327.1(2003)asfollows:

1.Immediatedeflection(δC1⋅3)ofsteelbeamduringconstructionstages1–3underdesignloads (GC1⋅3), which include the weight of the steel beam, formwork, concrete andreinforcement

2.Immediatedeflection(δC5⋅6)ofcompositebeamduringconstructionstages5–6underdesignloads,whichincludedeadloads(GC1⋅3)andsuperimposeddeadloads(Gsup)

3.Immediatedeflection(δQ)ofcompositebeamduringin-serviceconditionundershort-termliveload(ψsQ)

4.Long-termdeflection(δcr)ofcompositebeamduetoconcretecreepunderserviceloads,whichincludedeadloads(Gsup),long-termliveload(ψlQ)andforproppedconstruc-tion,(GC1⋅3)

5.Long-termdeflection(δsh)ofcompositebeamduetotheshrinkageofconcreteduringin-servicecondition


9.11.3 deflections due to creep and shrinkage

Thelong-termdeflectionofacompositebeamduetoconcretecreepcanbecalculatedusingthelong-termsectionpropertiesofitstransformedsection.However,thelong-termdeflec-tionsthuscalculatedincludetheimmediatedeflectionduetothesuperimposeddeadloadGsupandlong-termliveloadψlQand,ifpropped,GC1⋅3.Therefore,thedeflectioncomponentδcrduetocreephastobecomputedbysubtractingtheimmediatedeflectionduetotheseloadsfromthelong-termdeflectionasspecifiedinAS2327.1.

The final free shrinkage strain in unrestrained concrete given in AS 3600 (2001) isbetween300and1100microstrain.Unlike free shrinkage, the shrinkageof concrete incompositebeamsisrestrainedbythesteelbeamsthroughshearconnectors.Theshrink-ageofconcretecausescontraction,whichisresistedbyshearconnectors.Thecontractionoftheconcreteduetoshrinkageinducesdeflectionsandflexuralstresseswhichareinthesamedirectionas thosecausedbygravity loads (OehlersandBradford1999).Themid-spandeflectionsoftypicalsimplysupportedcompositebeamsarewithinthelimitofL/750(Alexander2003).

Figure9.33presentspartofacompositebeam.Thedeformationoftheconcreteduetoshrinkagestrainisrepresentedbyanexternalcompressiveforce(Nsh)actingatthecentroidoftheconcreteslab(Viestetal.1958;ChienandRitchie1984).Thisforceactingeccentri-cally to the elastic neutral axis of the transformed composite section induces a bendingmomentappliedattheendofthebeam.TheaxialforceNsh inducedbytheshrinkageofconcreteisexpressedby

N E Ash ce sh c= ε (9.86)

whereεshistherestrainedshrinkagestrainofconcreteinthecompositebeamAcistheeffectivecross-sectionalareaoftheconcreteslab

Therestrainedshrinkagestain(εsh)ofconcretemaybetakenas0 8. εcs∗ (Alexander2003),whereεcs∗ isthefinalfreeshrinkagestrainofconcreteestimatedinaccordancewithAS3600(2001).

ENA

ycn

Nshhc

hr

Dc

Ds

dne

Figure 9.33 Equivalent external force for shrinkage.

Composite beams 309

AsshowninFigure9.33,theeccentricityoftheaxialforceNsh isycn=(dne−hc/2).Thebendingmomentinducedbytheshrinkageofconcreteisdeterminedby

M N d

hsh sh ne

c= −

2

(9.87)

wheredneisthedepthoftheelasticneutralaxisofthetransformedsectiondeterminedusing

themodularratioofn E Es c= 3 /hcisthethicknessoftheconcreteslababovethesteelribs

TheshrinkageofconcreteproducesaconstantbendingmomentMshovertheentirelengthofthecompositebeam.Thedeflectionofthesimplysupportedcompositebeamwithcom-pleteshearconnectionduetoshrinkageiscalculatedby

δsh

sh

s tl

M LE I

=2

8 (9.88)

9.11.4 maximum stress in steel beam

WhenthesimplifiedmethodgiveninAS2327.1isusedtocalculatethedeflectionsofcom-positebeams, themaximumstress in thesteelbeamduringconstructionstages1–6andduringin-serviceconditionmustnotexceed0.9fy.Duringconstructionstages1–3,beforethedevelopmentofcompositeaction,themaximumstressinsteelbeamunderloadcombi-nationofG+Qiscalculatedseparatelyforeachconstructionstage.Duringconstructionstages5–6,themaximumstressinthesteelsectionofthecompositebeamiscalculatedbyconsideringthestresscausedbydesignloadsGC1⋅3duringconstructionstages1–3andthestressinducedbytheloadcombinationofGsup+Qactingonthecompositebeam.Duringin-servicecondition,theadditionalstressinsteelbeamofthecompositebeamundershort-termliveloadψsQiscalculatedbyassumingcompleteshearconnection.Thestressinthesteelbeamshouldbecomputedusingtheelasticsectionmoduliofthesteelbeamorcom-positebeamasappropriate.Atcrosssectionswithβ<0.4,thecompositeactionshouldbeignoredandthesectionmoduliofthesteelbeamshouldbeused.

Example 9.8: Deflection of simply supported composite beam

Checkforthedeflectionsofthesimplysupportedcompositebeamwithcompleteshearconnection presented in Example 9.1 and with partial shear connection presented inExample9.2,respectively.Thecompositebeamisproppedduringconstructionandthepropsareremovedattheendofconstructionstage5.Thepartitionsareinstalledafterthepropsareremoved.

1. Deflection of composite beam with complete shear connection

1.1. Short-term section properties

Young’smodulusofconcreteiscalculatedas

E fc c cj= ′ = × × =0 043 0 043 2 400 32 286001 5 1 5. .. .ρ , , MPa

Themodularratioforcalculatingshort-termsectionpropertiesisgivenby

n

EEs

c

= = =200 00028600

6 993,,

.


Thetransformedeffectivewidthoftheconcreteflangeis

b

bn

tfcf= = =

20006 993

286.

mm

Thegeometricparametersarecomputedasfollows:

h D hc c r= − = − =120 55 65 mm

A b t d ts f f w w= + = × × + − × × =2 2 171 11 5 356 2 11 5 7 3 63641 1 . ( . ) . mm2

Is = ×139 2 106. mm4

d D

Dsg c

s= + = + =2

1203562

298 mm

Assumetheelasticneutralaxisislocatedinthesteelrib.Thedepthoftheelasticneutralaxisiscomputedas

d

b h h A db h A

netf c c s sg

tf c s

=++

=× × + ×

×( ) ( )/ /2 286 65 65 2 6364 298

286 65++=

6364100 2. mm

Sincehc<dne<Dc,theelasticneutralaxisliesinthesteelribs.Thesecondmomentsofareaarecalculatedas

Ib h

b h dh

I A d dttf c

tf c nec

s s sg ne= + −

+ + −

=×

+

3 22

3

12 2

286 6512

( )

2286 65 100 2652

139 2 10 6364 298 100 22

6 2× × −

+ × + × −

= ×

. . ( . )

480 100 mm6 4

I Iti t= = ×480 10 mm6 4

1.2. Long-term section properties

Theeffectivemodulusofconcreteis

EE

cec

c

=+ ∗

=+

=1

286001 2

9 533φ

,, MPa

Themodularratiois

n

EE

s

ce

= = =200 000

9 53320 98

,,

.

Thetransformedeffectivewidthoftheconcreteflangeis

b

bn

tfcf= = =

200020 98

95 33.

. mm

Composite beams 311

Forthelong-termtransformedsection,theelasticneutralaxisislocatedinthesteelsec-tion.TheelasticneutralaxisdepthandItlareobtainedasfollows:

d Ine tl= = ×167 mm 363 10 mm6 4,

1.3. Deflection calculation

a. ImmediateDeflectionduringConstructionStages1–3 Sincethecompositebeamisproppedduringconstruction,δC1⋅3=0. b. ImmediateDeflectionduringConstructionStages5–6 Duringconstruction stages5–6, thepropsare removedand superimposeddead

loadisadded.

Theloading:w=GC1⋅3+Gsup=13.3kN/m

Theimmediatedeflectioniscalculatedas

δC

s ti

wLE I

5 6

4 4

3 6

5384

5384

13 3 8000200 10 480 10

7 39⋅ = = ××

× × ×=

.. mm

c. ImmediateDeflectionduringIn-ServiceCondition

Theshort-termliveload:w=ψsQ=0.7×12.8=8.96kN/m

Thedeflectionofcompositebeamundershort-termliveloadis

δQ

s ti

wLE I

= = ××

× × ×=

5384

5384

8 96 8000200 10 480 10

4 984 4

3 6

.. mm

d.Long-TermDeflectionduetoCreep Thelong-termserviceloadis

w G G QC l= + + = + × =⋅1 3 13 3 0 4 12 8 18 42sup . . . .ψ kN/m

Thelong-termdeflectionduetocreepiscalculatedas

δcr

s tl ti

wLE I I

= −

= ×

×× ×

5384

1 1 5384

18 42 8000200 10

1363 1

4 4

3

.00

1480 10

3 36 6−×

= . mm

e.Long-TermDeflectionduetoShrinkage Thefinalfreeshrinkagestrainofconcreteforthehypotheticalthicknessth=120mm

ofthecompositebeaminanear-coastalregioncanbeobtainedfromAS3600asεcs∗ = × −544 10 6.

Therestrainedshrinkagestrainofconcreteisestimatedas

ε εsh cs= ∗ = × × = ×− −0 8 0 8 544 10 435 106 6. .

Theaxialforceintheconcretecomponentduetoshrinkageis

N E Ash ce sh c= = × × × × × =− −ε 9533 435 10 2000 65 10 539 36 3 . kN

Themomentinducedbyshrinkageis

M N d

hsh sh ne

c= −

= × −

× =−

2539 3 167

652

10 72 53. . kNm


Thelong-termdeflectionduetoshrinkageiscomputedas

δsh

sh

s ti

M LE I

= =× ×

× × × ×=

2 6 2

3 6872 5 10 8000

8 200 10 480 107 99

.. mm

f.TotalandIncrementalDeflections Thetotaldeflectionofthecompositebeamis

δ δ δ δ δ δtot C C Q cr sh

L

= + + + + = + + + +

= <

⋅ ⋅1 3 5 6 0 7 39 4 98 3 3 7 99

23 72

. . . .

. mm550

32= mm OK,

Theincrementaldeflectionofthecompositebeamis

δ δ δ δinc Qi cr sh

L= + + = + + × = < =0 6 4 98 3 3 0 6 7 99 13 1

50016. . . . . . ,mm mm OK

Therefore,thecompositebeamwithcompleteshearconnectionsatisfiesthedeflec-tionlimits.

2. Deflection of composite beam with partial shear connection

2.1. Elastic section properties

ThecompositebeampresentedinExample9.2wasdesignedwithβ=0.6butwaspro-videdwith14headedstudshearconnectorsbetweentheendandmid-spanofthebeam.Asthisismorethanrequired,theactualdegreeofshearconnectionneedstobedeter-mined.FromExample9.2,weobtain

n f Fc ds cc= = =14 89 5 1957 8, . , .kN kN

Theactualcompressiveforceintheconcreteslabis

Fcp = × =14 89 5 1253. kN

Thedegreeofshearconnectionatmaximumbendingmomentis

βmb

cp

cc

FF

= = =12531957 8

0 64.

.

Theeffectivesecondmomentsofareaof thecompositebeamcrosssectionwithβmb=0.64arecalculatedasfollows:

I I I Ieti ti mb ti s= − − −

= × − × − × −

0 6 1

480 10 0 6 1 0 64 480 1396

. ( )( )

. ( . ) ( .

β

22 10 406 106 6)× = × mm4

I I I Ietl tl mb tl s= − − −

= × − × − × −

0 6 1

363 10 0 6 1 0 64 363 1396

. ( )( )

. ( . ) ( .

β

22 10 315 106 6)× = × mm4

Composite beams 313

2.2. Deflection calculation

Thedeflectioncomponentsarecalculatedusingthesameloadingcomponentsgivenintheprecedingsectionas

δ δ δ δ δC C Qi cr sh1 3 5 60 8 73 6 11 3 5 9 21⋅ ⋅= = = = =, . , . , . , .mm mm mm mm

Thetotaldeflectionofthecompositebeamis

δ δ δ δ δ δtot C C Q cr sh

L

= + + + + = + + + +

= <

⋅ ⋅1 3 5 6 0 8 73 6 11 3 5 9 21

27 56

. . . .

. mm2250

32= mm OK,

Theincrementaldeflectionofthecompositebeamis

δ δ δ δinc Q cr sh

L= + + = + + × = < =0 6 6 11 3 5 0 6 9 21 15 14

50016. . . . . . ,mm mm OK

Therefore,thecompositebeamwithβmb=0.64satisfiesthedeflectionlimits.

3. Maximum stress in steel beam

Sincethedeflectioniscalculatedusingthesimplifiedmethod,themaximumstressesinthesteelbeamneedtobechecked.Considerthebeamduringthein-servicecondition,theloadingis

w G G QC s= + + = + × =⋅1 3 11 7 0 7 12 8 20 66sup . . . .ψ kN/m

Themaximumbendingmomentunderthisserviceloadis

M

wL= =

×=

2 2

820 66 8

8165 28

.. kNm

Thesectionmodulusofthecompositesectioniscomputedusingitsshort-termsectionpropertyItiandassumingfullinteractionas

Z

Iy

bti= =

×+ −

= ×max .

.480 10

120 356 100 21 28 10

66 mm3

Themaximumstressatthebottomfibreofthesteelbeamisdeterminedas

σmax

..

. ,= =××

= < = × =MZ

fb

y165 28 101 28 10

129 0 9 300 2706

6 MPa 0.9 MPa OK

referenceS

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Grant, J.A., Fisher, J.W., and Slutter, R.G. (1977) Composite beams with formed steel deck, AISCEngineeringJournal,1stQuarter,14(1):24–43.

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Lam,D.,Elliott,K.S.,andNethercot,D.A.(2000a)Experimentsoncompositesteelbeamswithprecastconcretehollowcorefloor slabs,Proceedingsof InstitutionofCivilEngineers,StructuresandBuildings,U.K.,140:127–138.

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317

Chapter 10

composite columns

10.1 IntroductIon

Steel–concrete composite columns have been widely used in high-rise composite build-ings,bridgesandoffshorestructuresduetotheirhighstructuralperformance,suchashighstrength,high stiffness,highductilityand large strainenergyabsorptioncapacities.ThetypesofcompositecolumnsareshowninFigure10.1.Themostcommonlyusedcompositecolumnsareconcrete-encasedcomposite(CEC)columnsasshowninFigure10.1a,rectan-gularconcrete-filledsteeltubular(CFST)columnsasillustratedinFigure10.1candcircularCFSTcolumnsasdepictedinFigure10.1d.ACECcolumnisformedbyencasingastruc-turalsteelI-sectionintoareinforcedconcretecolumn.Studshearconnectorsmaybeweldedtothestructuralsteelsectiontotransferforcesbetweenthesteelsectionandtheconcrete.ACFSTcolumnisconstructedbyfillingconcreteintoacircularorrectangularhollowsteeltube.Longitudinalreinforcementmaybeplacedinsidethesteeltubestoincreasethecapaci-tiesofCFSTcolumns.

CECcolumnshaveahigherfireresistancethanCFSTcolumns.Inaddition,theconcreteeffectivelypreventsthesteelI-sectioninaCECcolumnfromlocalbuckling.ThesteeltubeofaCFSTcolumncompletelyencases theconcretecore,whichremarkably increases thestrengthandductilityoftheconcretecoreincircularCFSTcolumnsandthedualityoftheconcretecoreinrectangularCFSTcolumns.Ontheotherhand,theconcretecoreeffectivelypreventstheinwardlocalbucklingofthesteeltube,whichresultsinahigherlocalbucklingstrengthofthetubethanthehollowone.Moreover,thesteeltubeisutilizedaspermanentformworkandlongitudinalreinforcementfortheconcretecore,offeringsignificantreduc-tionsinconstructiontimeandcosts(Liang2009a).

Compositecolumnsareimportantstructuralmembers,whichareusedtosupportheavyaxial loads as compression members or gravity and lateral loads as beam–columns inmoment-resistingcompositeframes.Practicalcompositecolumnsareoftensubjectedtothecombinedactionsofaxialloadandbendingmoments.Thischapterpresentsthebehaviour,designandnon-linearanalysisofshortandslendercompositecolumns.ThedesignofshortandslendercompositecolumnsforstrengthtoEurocode4(2004)iscovered.Thenon-linearinelasticanalysisofshortandslenderCFSTbeam–columnsunderaxial loadandbiaxialbending,preloadsonthesteeltubesandcyclicloadingispresented.


10.2 BehAvIour And deSIgn of Short comPoSIte columnS

10.2.1 Behaviour of short composite columns

Experimentalstudieshavebeenconductedonthebehaviourofshortcompositecolumns(Furlong1967;KnowlesandPark1969;TomiiandSakino1979a,b;Shakir-KhalilandZeghiche1989;Shakir-KhalilandMouli1990;GeandUsami1992;BridgeandO’Shea1998;Schneider1998;Uy1998,2000,2001;Han2002;ZhaoandGrzebieta2002;GiakoumelisandLam2004;YoungandEllobody2006;ZhaoandPacker2009;Uyetal.2011).Thebehaviourofshortcompositecolumnsunderaxialcompressionischaracterisedbytheiraxialload–straincurveswhichindicatetheaxialstiffness,theultimateaxialstrength,thepost-peakbehaviourandtheaxialductilityofthecolumns.Figure10.2showsatypicalaxialload–straincurveforaCFSTshortcolumnpredictedbythecomputerprogramNACOMS(NonlinearAnalysisofCompositeColumns)developedbyLiang(2009a,b).CECshortcolumnsmayfailbyyieldingofthesteels

(a) (b) (c)

(d) (e) (f )

(g) (h) (i)

Figure 10.1 Types of composite columns: (a) concrete-encased composite column; (b) partially encased com-posite column; (c) rectangular concrete-filled steel tubular (CFST) column; (d) circular CFST col-umn; (e) rectangular CFST column with internal circular steel tube; (f) circular CFST column with internal circular steel tube; (g) circular CFST column with rectangular steel tube; (h) double skin rectangular CFST column with internal circular steel tube; (i) double skin circular CFST column.

Composite columns 319

andcrushingoftheconcrete.ThefailuremodesofCFSTshortcolumnsincludeyieldingorlocalbucklingofthesteelsectionandcrushingoftheinfillconcrete.Theultimateaxialstrengthofshortcompositecolumnsisgovernedbythesectionpropertiesandthematerialstrengthsofthesteelandconcrete.Thebehaviourofshortcompositecolumnsunderaxialloadandbendingischaracterisedbytheirmoment–curvaturecurveswhichindicatetheflexuralstiffness,ultimatemomentcapacity,post-peakbehaviourandcurvatureductilityofthecolumns.Themoment–curvaturecurveforatypicalCFSTshortcolumnunderaxialloadandbiaxialbendingpredictedbythecomputerprogramNACOMS(Liang2009a,b)isgiveninFigure10.3.

Strain0 0.01 0.02 0.03 0.04 0.05

0

0.2

0.4

0.6

0.8

1

1.2

Axi

al lo

ad P

/Po

Figure 10.2 Typical axial load–strain curve for a rectangular CFST column.

Curvature φ (× 10–5) (1/mm)0 2 4 6 8

0

Mom

ent M

/Mu

0.2

0.4

0.6

0.8

1

1.2

Figure 10.3 Typical moment–curvature curve for a rectangular CFST beam–column under axial load and biaxial bending.


NumericalstudiescarriedoutbyLiang(2009b,c)demonstratethatlocalbucklingofthesteeltuberemarkablyreducesthestiffness,axialstrengthandductilityofCFSTcolumns.In addition,increasingtheD/tratioofCFSTcolumnsectionsreducestheirsectionaxialper-formance,axialductility,flexuralstiffnessandstrengthandcurvatureductility.Moreover,increasingtheconcretecompressivestrengthincreasestheaxialloadandmomentcapacitiesofCFSTcolumnsbutdecreasestheirsectionaxialperformanceandductility.Furthermore,theaxialandflexuralstrengthsofCFSTcolumnsarefoundtoincreasebyincreasingtheyield strength of the steel tubes, but the axial ductility is generally shown to decrease.Numericalresultsalsoindicatethatincreasingtheaxialloadlevelsignificantlyreducestheflexuralstiffness,strengthandcurvatureductilityofcompositebeam–columns.

10.2.2 Short composite columns under axial compression

Theultimateaxialstrength(Po)ofaCECshortcolumnorarectangularCFSTshortcolumnunderaxialcompressionisthesumofthestrengthofsteelandconcretecomponentsofthecolumnandcanbeexpressedby

P f A f A f Ao c c c y se yr r= ′ + +γ (10.1)

whereAc isthecross-sectionalareaofconcreteAse istheeffectivestructuralsteelareaofthecrosssectionAr isthecross-sectionalareaoflongitudinalreinforcementγc is the reduction factorused toaccount for the effectof column sizeandconcrete

qualityonthecolumnstrength,proposedbyLiang(2009a)as

γ γc c cD= ≤ ≤−1 85 1 00 135. . ). (0.85 (10.2)

inwhichDc is thediameterof theconcretecoreandtakenas the largerof (B−2t)and(D − 2t)forarectangularcrosssection.

Circularsteeltubesprovideconfinementtotheconcretecore,whichincreasesthestrengthandductilityof theconcretecore incircularCFSTcolumns.Thesteel tubeofacircularCFSTcolumnisbiaxiallystressed.Thehooptensiondevelopedinthesteeltubereducesitsyieldstressinthelongitudinaldirection.TheultimateaxialstrengthofcircularCFSTshortcolumnsconsideringconfinementeffectsisgivenbyLiangandFragomeni(2009)as

P f f A f A f Ao c c rp c s y s yr r= ′ +( ) + +γ γ4 1. (10.3)

wherefrpdenotesthelateralconfiningpressureprovidedbythesteeltubeontheconcretecore.BasedontheworkofTangetal.(1996)andHuetal.(2003),aconfiningpressuremodelfornormalandhigh-strengthconcreteconfinedbyeitheranormalorhigh-strengthsteeltubewasproposedbyLiangandFragomeni(2009)as

f

tD t

fDt

Dt

rp

e s y

=

−−

≤

−

0 722

47

0 006241 0 0000357

. ( )

. .

ν ν for

< ≤

f

Dt

y for 47 150

(10.4)


inwhichνeandνsarePoisson’sratiosofthesteeltubewithorwithoutconcreteinfill,respec-tively.Poisson’sratioνsistakenas0.5atthemaximumstrengthpoint,andνeisgivenby(Tangetal.1996)

v v

ff

vff

e ec

ye

c

y

= + ′ −′

+ ′ ′

−0 2312 0 3582 0 1524 4 843. . . . 99 169

2

.′

ffc

y

(10.5)

′ = ×

− ×

+ ×

− − −vDt

Dt

Dt

e 0 881 10 2 58 10 1 953 1063

42

2. . . + 0 4011. (10.6)

Thefactorγsaccountsfortheeffectofhooptensilestressesandstrainhardeningontheyieldstressofthesteeltube.Forcarbonsteeltubes,γsisgivenbyLiang(2009a)as

γ γs s

Dt

=

≤ ≤( )−

1 458 1 10 1

. ..

0.9 (10.7)

InEurocode4(2004),theconfinementeffectthatincreasesthecompressivestrengthoftheconcretecoreincircularCFSTcolumnswitharelativeslendernessofλ ≤ 0 5. andasmallloadingeccentricratioofe/D<0.1istakenintoaccountinthecalculationoftheultimateaxialstrength(Npl,Rd)asfollows:

N A f A f

tDff

A fpl Rd s s y c c cy

cr yr, = + ′ +

′

+η η1 (10.8)

wherethefactorsηsandηcaregivenby

η λs = + ≤0 25 3 2 1 0. ( ) . (10.9)

η λ λc = − + ≥4 9 18 5 17 02

. . (10.10)

whereλistherelativeslendernessofthecolumngiveninSection10.4.2.Eurocode4(2004)provideslimitsonthewidth-to-thicknessratioforsteelelementsin

compositecolumnsasfollows:

• ForcircularCFSTcolumns,(D/t)≤90(235/fy).

• ForrectangularCFSTcolumns,( ) ( ).D t fy/ /≤ 52 235

• FortheflangesofpartiallyencasedI-sections,( ) ( ).b t ff f y/ /≤ 44 235

10.2.3 Short composite columns under axial load and uniaxial bending

10.2.3.1 General

Compositecolumnsincompositeframeswithrigidconnectionsareoftensubjectedtocom-binedactionsofaxialcompressionanduniaxialbending.Thecombinedactionsmayalsobecausedbytheeccentricityoftheappliedload.Thedesigncodesrequirethatallpractical


columnsshouldbedesignedasbeam–columns.Inthedesignofslendercompositebeam–col-umnsunderaxialloadanduniaxialbending,theaxialload–momentinteractionactiondia-gramsforthecolumnsneedtobedetermined.Thenon-linearinelasticanalysisofcompositebeam–columnsundereccentricloadingiscomplexwithouttheaidofcomputerprograms.

Inpractice,therigidplasticanalysisisusuallyusedtodeterminetheultimatestrengthsofcompositebeam–columnsundereccentricloading.Therigidplasticanalysisassumesthat(1)fullcompositeactionbetweensteelandconcretecomponentsuptofailureandplanesec-tionsremainplane,(2)allsteelyieldsincompressionandtensionattheultimatestrengthlimitstate,(3)arectangularconcretestressblockincompressionisstressedto0 85. ′fc ,(4)localbucklingisignoredforCECcolumns,(5)localbucklingofsteeltubesmaybeconsid-eredforCFSTcolumnsand(6)thetensilestrengthofconcreteisignored.

Eurocode4(2004)allowsasimplifieddesignmethoddevelopedbyRoikandBergmann(1989) tobeused fordevelopingaxial load–moment interactiondiagrams for compositeshortcolumns.Thissimplifiedmethodislimitedtomembersofdoublysymmetricalcrosssectionsincludingrolled,cold-formedorweldedsteelsections.Thelimitsonconcretethick-nesscovertothesteelsectionarecx≤0.4Bandcy≤0.3D.Thedepth-to-widthratio(D/B)ofthecompositecrosssectionshouldbewithinthelimitsof0.2and5.0.Theareaoflongitudi-nalreinforcementusedtocalculatetheultimateaxialandbendingstrengthsofacompositecolumnshouldnotexceed6%oftheconcreteareainthecompositesection.

10.2.3.2 Axial load–moment interaction diagram

Thetypicalaxialload–momentinteractiondiagramforacompositecolumnsectionissche-maticallydepictedinFigure10.4.Inthesimplifiedmethod,theaxialload–momentinter-actiondiagramfor thecolumnsection isapproximatedby thepolygonACDBasshowninFigure10.4(RoikandBergmann1989;OehlersandBradford1999;Eurocode42004;Johnson2004).Thesimplifiedmethodisintroducedhereinfordevelopingtheaxialload–momentinteractiondiagrams.

PointAinFigure10.4correspondstotheultimateaxialstrength(Po)ofthecolumnsec-tionunderaxialcompressiononly,whichcanbecalculatedusingEquation10.1.

0

2

Mu,maxMo

Mu

Pmo

C

D

Pmo

Po

Pu

A

B

Figure 10.4 Axial load–moment interaction diagram of a composite section.


PointBinFigure10.4correspondstotheultimatepurebendingmomentcapacity(Mo)ofthecolumnsectionunderbendingwithouttheaxialload.TheplasticstressdistributioninthecolumnsectionunderpurebendingisshowninFigure10.5.Theplasticneutralaxisislocatedatadistancehnabovethecentroidofthecolumncrosssection.ThecrosssectionisdividedintothreeregionsasillustratedinFigure10.5.Region1isabovethehndistancefromthecentroidofthesection,whileregion3isbelowthehndistancefromthecentroidofthesection.Region2iswithinthehndistanceaboveandbelowthecentroidofthesec-tion.Iftheneutralaxisislocatedinthewebofthesteelsection,thecompressiveforceintheconcreteiscalculatedby

F f B D h b t d h n t Ac c n f f w n w w r1 10 85 0 5 0 5= ′ − − − − −. [ ( . ) ( . )( ) ] (10.11)

wheredwisthecleardepthofthesteelwebnwisthetotalnumberofwebsinthesteelsectionAr1istheareaoflongitudinalreinforcementatthetopofthecrosssection

Forthecompositesectionunderpurebending,thesumofcompressionforcesmustequalthesumoftensionforcesinthesection:Fc1+Fs1+Fr1=Fs2+Fr2+Fs3+Fr3,whereFs1=Fs3andFr1=Fr3duetosymmetryofthesteelelementsaboutthecentroidofthesectionasshowninFigure10.5andFs2=2hn(nwtw)fy.Fromtheforceequilibriumcondition, thefollowingexpressioncanbeobtained

F h n t f Fc n w w y r1 22= +( ) (10.12)

Thedistancehncanbedeterminedfromthisequationas

h

f A Ff B n t n t f

dn

c cn r

c w w w w y

w=′ −

′ − +≤0 85

0 85 2 22.

. ( ) (10.13)

whereAcn=B(0.5D)− bftf− (0.5dw)(nwtw)− Ar1.

Cross section

Concrete Steel section Steel bars

0.85 f c

N.A.

B

D Ds

tw

dn

hn

hn

tf

bf

Fc1Fr1

Fs1

Fr2

Fr3

fy fyr

Fs2

Fs3

Stress distributions and forces

Figure 10.5 Plastic stress distributions in the cross section of a composite column: neutral axis above the centroid of section.


Iftheneutralaxisislocatedinthetopflangeofthesteelsection,thecompressiveforceintheconcreteiscomputedby

F f B D h b D h Ac c n f s n r1 10 85 0 5 0 5= ′ − − − − . ( . ) ( . ) (10.14)

Theforceequilibriumconditionyieldsthefollowingcondition:

F F F b h d fc r w f n w y1 2 2 0 5= + + −( . ) (10.15)

whereFw=dwtwfyandthedistancehnisgivenby

h

f A F F b d ff B b b f

nc cn r w f w y

c f f y

=′ − − +

′ − +0 85

0 85 22.

. ( ) (10.16)

whereAcn=B(0.5D)− bf(0.5Ds)− Ar1.Thenominalmomentcapacityofthecompositesectionunderpurebendingcanbecalcu-

latedbytakingmomentsaboutitscentroidas

M F d F d F do c c s s r r= + +1 1 1 1 1 12 2 (10.17)

inwhichdc1isthedistancefromthecentroidofFc1tothecentroidofthecrosssection,takenasdc1=0.5D− 0.5(0.5D− hn)forCECcolumnsanddc1=0.5dw− 0.5(0.5dw− hn)forCFSTcolumns.

SimpledesignformulasforcalculatingtheultimatemomentcapacitiesofcircularCFSTshortcolumnsunderpurebendingaregivenbyLiangandFragomeni(2010)asfollows:

M Z fo m fc y e y= λ α α (10.18)

λm

tD

tD

D t= +

−

≤ ≤0 0087 12 3 362

. . (10 / 120) (10.19)

αfc c cf f= ′( ) ≤ ′ ≤0 774

0 075.

.(30 120MPa) (10.20)

αy

y yy

f ff= + + ≤ ≤0 883

21 147 42026902.

.(250 MPa) (10.21)

whereλmisthefactoraccountingfortheeffectofD/tratioαfcisthefactoraccountingfortheeffectofconcretecompressivestrengthαyisthefactorusedtotakeintoaccounttheeffectoftheyieldstrengthofthesteeltubeZeistheelasticsectionmodulusofthecircularCFSTcolumn,calculatedasπD3 32/

Point C in Figure 10.4 corresponds to the point where the nominal moment capacityof the column sectionunderanaxial forceofPmo is equal to thepurebendingmoment


capacity (Mo).Forthiscase,theplasticneutralaxisislocatedatadistanceofhnbelowthecentroidofthecrosssectionasdepictedinFigure10.6,whichshowstheplasticstressdistri-butioninthe crosssection.Thevalueofhnhasbeendeterminedforthesectionunderpurebending.Thecompressiveforceintheconcreteinregion2iscalculatedby

F f B h h n tc c n n w w2 0 85 2 2= ′ −[ ]. ( ) ( )( ) (10.22)

TheresultantforceinthecompositesectioncanbeobtainedfromFigure10.6bysummingallforcesinthecrosssectionas

P F F Fmo s c r= + +2 2 2 2 (10.23)

Point D in Figure 10.4 corresponds to the point where the maximum moment capacity(Mu,max)ofthecolumnsectionunderanaxialforceofPmo/2occurs.Forthiscase,theplasticneutralaxisliesatthecentroidofthecrosssectionasshowninFigure10.7whichillustratestheplasticstressdistributioninthecrosssection.Theresultantaxialforceinthecompositesection isdeterminedasPu=Fc1+Fc2/2=Pmo/2.By takingmomentsabout thecentroidof thecross section, themaximummomentcapacity (Mu,max)of thecomposite section isobtainedas

M f A d M F du c cm cm s r r,max .= ′ + +0 85 2 1 1 (10.24)

whereAcm is the area of concrete above the plastic neutral axis and is calculated as

A BD A Acm s st= − −( )/ /2 2 1

dcmisthedistancefromthecentroidofAcmtothecentroidofthecompositesectionMsisthenominalmomentcapacityofthewholesteelsectionalone

Cross section

Concrete Steel section Steel bars

0.85 f c

N.A.

B

D Ds

tw hn

hn

tf

bf

Fc1Fr1

Fs1

Fs2 Fr2

Fr3

fy fyr

Fc2

Fs3


Figure 10.6 Plastic stress distributions in the cross section of a composite column: neutral axis below the centroid of section.


Example 10.1: Axial load–moment interaction diagram of CEC short column

Develop theaxial load–moment interactiondiagramfor theCECshortcolumnbend-ingabouttheprincipalx-axisasshowninFigure10.8.Theconcretedesignstrengthis′ =fc 32MPa.Theyieldstressofthesteelsectionis300MPa,whiletheyieldstressofthe

steelreinforcementis500MPa.

1. Point A: Ultimate axial strength

Thereductionfactorforconcreteγcis

γc cD= = × = <− −1 85 1 85 500 0 8 0 850 135 0 135. . . .. .

Hence,γc=0.85.Theareaofthestructuralsteelsectioniscomputedas

Ase = × × + − × × =350 16 2 350 2 16 12( ) 15,016mm2

Cross sectionConcrete Steel section Steel bars

0.85 f c

N.A.

B

D Ds

tw hn

hn

tf

bf

Fc1Fr1

Fs1

0.5Fs2

Fr3

fy fyr

0.5Fc2

0.5Fs2

Fs3


Figure 10.7 Plastic stress distributions in the cross section of a composite column: neutral axis at the centroid of section.

500

16

x

R10 Tie

4N20

50

350

12

75

500 350 x

30

75

Figure 10.8 Cross section of a CEC column.


Thetotalareaofreinforcementinthecrosssectionis

Ar = × ×

=4

204

2

π 1256.6mm2

Theareaofconcreteinthecrosssectioncanbecalculatedas

A BD A Ac se r= − − = × − − =500 500 15,016 1,256.6 233,727mm2

Theultimateaxialstrengthofthecompositesectionistherefore

P f A f A f Ao c c c y se yr r= ′ + +

= × × + × + ×

γ

0 85 32 300 500. 233,727 15,016 1,2566 N 11,490.5kN.6 =

2. Point B: Pure bending moment capacity

Assumetheplasticneutralaxisislocatedinthesteelweb.Thecleardepthofthewebisdw=Ds−2tf=350−2×16=318mm.TheareaofthetopreinforcementisAr1=Ar/2=628.3mm2.Thedistanceofthetopreinforcementtothecentroidofthecompositesectionis

dr1

5002

30 10202

200= − − − = mm

Thedistancehniscalculatedasfollows:

A B D b t d n t Acn f f w w w r= − − −

= × × − × − ×

( . ) ( . )( )

. .

0 5 0 5

500 0 5 500 350 16 0 5

1

3318 1 12 628 3 7× × − =( ) . .116,863 mm2

hf A F

f B n t n t fn

c cn r

c w w w w y

=′ −

′ − +

=× ×

0 850 85 2

0 85 32 7

2.. ( )

. .116,863 −−× × − × + × × ×

= < = =0

0 85 32 500 1 12 2 1 12 300155 3

23182

159. ( )

. mm mmdw

Hence,theplasticneutralaxisislocatedinthesteelweb.Thecompressiveforceintheconcreteinregion1iscomputedas

F f BD

h b td

h n t Ac c n f fw

n w w st1 10 852 2

= ′ −

− − −

−

.

= × × × −

− × − −

0 85 32 500

5002

155 3 350 163182

155 3. . . ×× × −

=

1 12 628 3. N

1117.3kN


ThedistanceofFc1tothecentroidofthecrosssectionis

d

D D hc

n1

222

5002

500 2 155 32

202 65= −−

= −−

=( ) ( . )

./ /

mm

The force in the steel topflange and itsdistance from the centroidof the sectionarecomputedas

F b t fs f f f y1 1 350 16 300⋅ = = × × =N 1680kN

d

D ts f

s f1

2 23502

162

167⋅ = − = − = mm

Theforceinthesteelwebanditsdistancefromthecentroidofthesectionarecalculatedas

F

dh n t fs w

wn w w y1

23182

155 3 1 12 300⋅ = −

= −

× × × =. N 13.32kNN

d

d d hs w

w w n1

222

3182

318 2 155 32

157 15⋅ = −−

= −−

=( ) ( . )

./ /

mm

Theresultantforceinthesteelcomponentsinregion1istherefore

Fs1 = + =1680 13.32 1693.32kN

ThedistancefromthecentroidofFs1tothecentroidofthesectionis

d

F d F dF

ss f s f s w s w

s1

1 1 1 1

1

1680 167 13 32 157 151693 32

=+

=× + ×

=⋅ ⋅ ⋅ ⋅ . ..

1166 92. mm

Theforceinthetopreinforcementis

F A fr r yr1 1 628 3 500= = × =. N 314.2kN

ThepurebendingmomentcapacityMoiscalculatedas

M F d F d F do c c s s st st= + +

= × + × ×

1 1 1 1 1 12 2

1117 3 202 65 2 1693 47 166 92. . . . ++ × × =2 314.2 200kN mm 917.4kN m

3. Point C: Mu=Mo

Theplasticneutralaxisislocatedatadistancehn=155.3mmbelowthecentroidofthesection.Theforceinthesteelcomponentinregion2iscomputedas

F h n t fs n w w y2 2 2 155 3 1 12 300= = × × × × =( ) . ( ) N 1118.2kN


Thecompressiveforceintheconcreteinregion2is

F f B h h n tc c n n w w2 0 85 2 2= ′ −

= × × × × − ×

. ( ) ( )

0.85 32 500 2 155.3 2 155.33 1 12 N 4122.8kN× × =

Theresultantaxialforceinthecompositesectionistherefore

F F Fmo s c= + = × + =2 2 1118 2 4122 82 2 . . 6359.2kN

4. Point D: Maximum moment capacity

Theplasticneutralaxisliesatthecentroidofthecrosssection.Theresultantforceinthecompositesectionisdeterminedas

P

Pu

mo= = =2

6359 22

3179 6.

. kN

Theareaofconcreteabovetheplasticneutralaxisanditsdistancetothecentroidofthesectionarecomputedas

A

BD AAcm

sst= − − = × − − =

2 2500

5002 2

628 3 7115,016

116,863 mm2. .

d

Dcm = = =

45004

125mm

ThemomentcapacityofthewholesteelI-sectioniscalculatedas

M b t f D td

n t fd

s f t y s fw

w w yw= − +

= × × ×

( ) ( )2 2

350 16 300 (3500 16) (1 12) 300 Nmm

652kNm

− +

× × × ×

=

3182

3182

Themaximummomentcapacityistherefore

M f A d M F du c cm cm s r r,max .= ′ + +

= × × × + +

0 85 2 1 1

0.85 32 11,683.7 125 652 2×× × =314.16 200Nmm 1,175kNm

Theaxialload–momentinteractiondiagramofthiscompositeshortcolumnisshowninFigure10.9.


Example 10.2: Axial load–moment interaction diagram of CFST short column

Develop the axial load–moment strength interaction diagram of the CFST short col-umnbendingabouttheprincipalx-axisasdepictedinFigure10.10.Theconcretedesignstrengthis ′ =fc 50MPa.Theyieldstressofthesteelsectionis300MPa.

1. Point A: Ultimate axial strength


Dt fy= = < =60020

30 52235

46

Thewebandthesectionarecompact.Thereductionfactorforconcrete(γc)is

γc cD= = × = <− −1 85 1 85 500 0 8 0 850 135 0 135. . . .. .

Hence,γc=0.85.

00

2,000

40,00

6,000

8,000

10,000

12,000

14,000

200 400 600 800

Moment Mu (kN m)

Axi

al lo

ad P

u (kN

)

1,000 1,200 1,400

Figure 10.9 Axial load–moment interaction diagram of the CEC short column.

500

600

20

xx

y

y

Figure 10.10 Cross section of a CFST column.


Theareaofconcreteinthecross-sectionis

A B t D tc = − − = − × − × =( )( ) ( )( )2 2 500 2 20 600 2 20 257600 mm2

Theareaofthestructuralsteelsectioniscomputedas

A BD As c= − = × − =500 600 257600 42400 mm2

Theultimateaxialstrengthofthecolumnsectionistherefore

P f A f A f Ao c c c y se yr r= ′ + +

= × × + × + =

γ

0 85 50 300 0. 257,600 42,400 N 23,6688kN

2. Point B: Pure bending moment capacity

Assumetheplasticneutralaxisislocatedinthesteelweb.Thecleardistanceofthewebisdw=Ds− 2tf=600− 2×20=560mm.Thedistancehniscalculatedasfollows:

A B D b t d n t Acn f f w w w r= − − −

= × × − × − ×

( . ) ( . )( )

. .

0 5 0 5

500 0 5 600 500 20 0 5

1

5560 2 20 0× × − =( ) 128,800mm2

hf A F

n t f f B n tn

c cn r

w w y c w w

=′ −

+ ′ −

=× × −

0 852 0 85

0 85 50 0

2.. ( )

. 128,80022 2 20 300 0 85 50 500 2 20

125 72

5602

280× × × + × × − ×

= < = =. ( )

. mm mmdw

Hence,theplasticneutralaxisislocatedinthesteelweb.Thecompressiveforceintheconcreteinregion1iscomputedas

F f B D h b t d h n t Ac c n f f w n w w st1 10 85 0 5 0 5

0 85 50

= ′ − − − − −

= ×

. ( . ) ( . )

. ×× × × − − × − × − × × −

=

500 0 5 600 125 7 500 20 0 5 560 125 7 2 20 0( . . ) ( . . ) N

30016.6kN

ThedistanceofFc1tothecentroidofthecrosssectionis

d

d d hc

w w n1

222

5602

560 2 125 72

202 85= −−

= −−

=( ) ( . )

./ /

mm

Theforceinthetopsteelflangeanditsdistancetothecentroidofthesectionare

F b t fs f f f y1 1 500 20 300⋅ = = × × =N 3000kN

d

D ts f

s f1

2 26002

202

290⋅ = − = − = mm


Theforceinthesteelwebabovetheplasticneutralaxisanditsdistancetothecentroidofthesectionarecalculatedas

F

dh n t fs w

wn w w y1

25602

125 7 2 20 300. .= −

= −

× × × =N 1851.6kkN

d

d d hs w

w w n1

222

5602

560 2 125 72

202 85⋅ = −−

= −−

=( ) ( . )

./ /

mm

Theresultantforceinthesteelcomponentsinregion1is

Fs1 = + =3000 1851.6 4851.6kN

ThedistancefromthecentroidofFs1tothecentroidofthesectionis

d

F d F dF

ss f s f s w s w

s1

1 1 1 1

1

3000 290 1851 6 202 854851 6

=+

=× + ×

=⋅ ⋅ ⋅ ⋅ . ..

2256 74. mm

ThepurebendingmomentcapacityMoiscalculatedas

M F d F d F do c c s s st st= + +

= × + × × +

1 1 1 1 1 12 2

3016 6 202 85 2 4851 6 256 74. . . . 00kNmm 3103kNm=

3. Point C: Mu=Mo

Theplasticneutralaxisislocatedatadistanceofhn=125.7mmbelowthecentroidofthesection.Theforceinthesteelcomponentinregion2iscalculatedas

F h n t fs n w w y2 2 2 125 7 2 20 300= = × × × × =( ) . ( ) N 3016.8kN

Thecompressiveforceintheconcreteinregion2is

F f B h h n tc c n n w w2 0 85 2 2= ′ −

= × × × × − ×

. ( ) ( )

0.85 50 500 2 125.7 2 125.77 2 20 N 4914.9kN× × =

Theresultantaxialforceinthecrosssectionistherefore

F F Fmo s c= + = × + =2 22 2 3016.8 4914.9 10948.5kN

4. Point D: Maximum moment capacity

Theplasticneutralaxisliesatthecentroidofthecrosssection.Theresultantforceinthesectionisdeterminedas

P

Pu

mo= = =2

52

2510,948

5474 kN.

.


Theareaofconcreteabovetheplasticneutralaxisanditsdistancetothecentroidofthesectionarecomputedas

A

BD AAcm

sst= − − = × − − =

2 2500

6002 2

0142,400

128,800mm2

d

dcm

w= = =4

5604

140mm

ThemomentcapacityofthewholesteelI-sectionis

M b t f D td

n t fd

s f t y s fw

w w yw= − +

= × × ×

( ) ( )2 2

500 20 300 (6000 20) (2 12) 300 Nmm

2680.8kNm

− +

× × × ×

=

5602

5602

Themaximummomentcapacityis

M f A d M F du c cm cm s st st,max .= ′ + +

= × × × +

0 85 2 1 1

0.85 50 128800 140 2680.88 0Nmm 3447.2kNm+ =

Theaxialload–momentinteractiondiagramofthiscompositeshortcolumnisshowninFigure10.11.

00

Axi

al lo

ad P

u (kN

)

5,000

10,000

15,000

20,000

25,000

Moment Mu (kN m)

1,000 2,000 3,000 4,000

Figure 10.11 Axial load–moment interaction diagram of the CFST short column.


10.3 non-lIneAr AnAlySIS of Short comPoSIte columnS

10.3.1 general

Thenon-linearmethodsofanalysis forcompositecolumnsandstructureswerereviewedbySpaconeandEl-Tawil(2004).Areviewonthestate-of-the-artdevelopmentofcompositecolumnswaspresentedbyShanmugamandLakshmi(2001).Analyticalandfibreelementmodelshavebeendevelopedbyvariousresearchersforthenon-linearinelasticanalysisofshortcompositecolumns(El-Tawiletal.1995;HajjarandGourley1996;MuñozandHsu1997;El-TawilandDeierlein1999;Chenetal.2001;LakshmiandShanmugam2002;Liangetal.2006,2007a;Liang2008,2009a,b,c;LiangandFragomeni2009,2010).Finiteele-mentanalysesofCFSTshortcolumnsandconcrete-filledstainless steel tubular (CFSST)columnswerealsoreportedintheliterature(Huetal.2003;EllobodyandYoung2006;Ellobodyetal.2006;Taoetal.2011;Hassaneinetal.2013a,b,c).ThenumericalmodelsdevelopedbyLiang(2008,2009a,b,c,2011a,b)forCFSTshortcolumnsunderaxialloadandbiaxialbendingaredescribedinthefollowingsections.

10.3.2 fibre element method

Thefibreelementmethod isanefficientandaccuratenumerical technique fordetermin-ingtheinelasticbehaviourofcompositecrosssections(El-Tawiletal.1995;Liang2009a).In this method, the cross section of a composite column is discretised into many smallfibreelementsasdepictedinFigure10.12.Eachelementrepresentsafibreofmaterialrun-ninglongitudinallyalongthememberandcanbeassignedeithersteelorconcretematerialproperties.Uniaxialstress–strainrelationshipsareusedtosimulatethematerialbehaviour.Stressresultantsareobtainedbynumericalintegrationofstressesthroughthecrosssection.Numericalmodelsbasedonthefibreelementmethodhavebeendevelopedforpredictingthenon-linearinelasticbehaviourofcompositeshortcolumnsunderaxialloadorcombinedaxialloadandbending.

10.3.3 fibre strain calculations

Thefibrestrainisafunctionofthecurvature(ϕ),orientation(θ)andthedepth(dn)oftheneutralaxisinthecrosssectionofarectangularCFSTcolumnunderaxialloadandbiaxialbendingasschematicallydepictedinFigure10.13.ThestraindistributioninacircularCFST

(b)D

x

t

y

(a)

Steel fibres

Concrete fibres

B

D

y

x

t

Figure 10.12 Fibre element discretization: (a) rectangular section and (b) circular section.


column section is illustrated in Figure 10.14. The plane sections are assumed to remainplaneafterdeformation,whichresultsinalinearstraindistributionthroughthedepthofthecrosssection.Thestrainattheextremefibre(εt)ofthesectionisequaltoϕdn.For0° ≤ θ <90°,thefibrestrainiscomputedasfollows(Liang2009a):

c

dn=cosθ

(10.25)

y x

B Dcn i i, tan= − + −

2 2

θ (10.26)

d y ye i i n i, , cos= − θ (10.27)

yn,i

yi

t

y

xc

Pa

dn

de,i

εt

εi

N.A.

D

B

α

θ

φ

θ

Figure 10.13 Strain distributions in rectangular CFST column section under axial load and biaxial bending.

εi

φ

εt

N.A.

x

y

D

dn

Figure 10.14 Strain distributions in a circular CFST column section.


ε

φφie i i n i

e i i n i

d y y

d y y=

≥− <

, ,

, ,

for

for (10.28)

wherede i, istheorthogonaldistancefromthecentroidofeachfibreelementtotheneutralaxisxiandyiarethecoordinatesofthefibreiεiisthestrainattheithfibre

Forθ=90°,thefibrestrainiscalculatedasfollows:

x

Bdn i n, = −

2 (10.29)

d x xe i i n i, ,= − (10.30)

ε

φφie i i n i

e i i n i

d x x

d x x=

≥− <

, ,

, ,

for

for (10.31)

10.3.4 material constitutive models for structural steels

Figure10.15showstheidealisedstress–straincurvesforstructuralsteels(Liang2009a).Atrilinearstress–strainrelationshipisassumedformildstructuralsteelsbothincompressionandtension.Thestress–stainbehaviourofhigh-strengthandcold-formedsteelsischaracter-isedbyaroundedstress–straincurve.Alinear-rounded-linearstress–straincurveisthereforeused for cold-formed steels,but forhigh-strength steels, the roundedpartof the curve isreplacedwithastraightlineasdepictedinFigure10.15.Theroundedpartofthestress–straincurveforcold-formedsteelsisdeterminedbythefollowingequationgivenbyLiang(2009a):

σ

ε εε ε

ε ε εs ys y

st yy s stf=

−−

< ≤

0 90 9

0 91 45

..

( . )/

(10.32)

whereσsdenotesthestressinasteelfibreεsrepresentsthestraininasteelfibreεystandsfortheyieldstrainofsteelεstisthesteelstrainatstrainhardeningasdepictedinFigure10.15

The hardening strain εst is taken as 10εy for mild structural steels and 0.005 for high-strengthandcold-formed steels.To reflect theductilityofdifferent structural steels, theultimatestrain(εsu)istakenas0.2formildstructuralsteels,whileitistakenas0.1forhigh-strengthandcold-formedsteels.

10.3.5 material models for concrete in rectangular cfSt columns

TheductilityoftheconcretecoreinarectangularCFSTcolumnisshowntoincreasedueto theconfinementprovidedbythesteel tube.However, theconfinementeffectdoesnotincrease the compressive strength of the concrete core. The idealised stress–strain curve


depictedinFigure10.16isusedinfibreelementmodelstosimulatethematerialbehaviourofconfinedconcreteinrectangularCFSTcolumns(Liang2009a).ThepartOAofthestress–straincurvegiveninFigure10.16ismodelledusingthefollowingequationssuggestedbyManderetal.(1988):

σλ ε ε

λ ε ε λcce c ce

c ce

f=

′ ′( )− + ′( )1

(10.33)

λ

ε=

− ′ ′( )E

E fc

c ce ce

(10.34)

E fc ce= ′ +3320 6900MPa (10.35)

′ =

′ ≤

+′ −

< ′ ≤εce

ce

cece

f

ff

0.002 for MPa

54,000for 28 MP

28

0 00228

82. aa

0.003 for MPa′ >

fce 82

(10.36)

whereσcstandsforthelongitudinalcompressiveconcretestress′fceistheeffectivecompressivestrengthofconcretewhichistakenas ′ = ′f fce c cγ

εcisthelongitudinalcompressiveconcretestrain′εceisthestrainat ′fceEcisYoung’smodulusofconcrete(ACI-3182011)

Thestrain ′εceisbetween0.002and0.003dependingontheeffectivecompressivestrengthofconcrete.Fortheeffectivecompressivestrengthofconcretebetween28and82MPa,thestrain ′εceisdeterminedbylinearinterpolation.

0.9 fy

fsu

fy

σs

0 0.9 εy εst εsu εs

Figure 10.15 Stress–strain curves for structural steels.


ThepartsAB,BCandCDofthestress–straincurveforconfinedconcretedepictedinFigure10.16areexpressedby(Liang2009a)

σ

ε ε ε

βε ε β

ε εc

ce ce c B

ce cecp c ce ce ce

cp B

f

ff f

=

′ ′ < ≤

′ +− ′ − ′( )

−

for

( )

( )ffor

for

ε ε ε

β ε ε

B c cp

ce ce c cpf

< ≤

′ >

(10.37)

whereεB = 0 005. and εcp = 0 015. are concrete compressive strains corresponding to points

B andCshowninFigure10.16

βce is the factoraccounting for the confinement effecton the strengthandductilityofconcreteinthepost-peakrange,dependingonthewidth-to-thicknessratio( )B ts / ofthesec-tion,whereBsistakenasthelargerofBandDforarectangularcrosssection.BasedontheexperimentalresultspresentedbyTomiiandSakino(1979a),βceisgivenbyLiang(2009a)as

βce

s

s s

s

Bt

Bt

Bt

Bt

=

≤

−

< ≤

>

1 0 24

1 5148

48

0 5 48

.

.

.

for

for 24

for

(10.38)

Thestress–straincurveforconcreteintensionisdepictedinFigure10.16.Itisassumedthatthetensilestressincreaseslinearlywithanincreaseintensilestrainuptoconcretecracking.Afterconcretecracking,thetensilestressdecreaseslinearlytozeroastheconcretesoftens.Thetensilestrengthofconcrete(fct)istakenas0 6. ′fce ,whileitsultimatetensilestrain(εtu)istakenas10timesofthestrainatcracking(εct).

εtu εB

A B

C D

E

F G

Concrete in circular tube

Concrete in rectangular tube

εcp εF εcuεcεct

βce f ce

f ce

f cc

βcc f cc

ε ce

fct

σc

ε cc0

Figure 10.16 Idealised stress–strain curves for concrete in CFST columns.


10.3.6 material models for concrete in circular cfSt columns

Theconcreteconfinementeffectincreasesboththestrengthandductilityofconcreteincir-cularCFSTcolumns.Anidealisedstress–straincurveaccountingfortheconfinementeffectisalsopresentedinFigure10.16(LiangandFragomeni2009;Liang2011a).ThepartOEofthestress–straincurveshowninFigure10.16isrepresentedusingtheequationssuggestedbyManderetal.(1988)as

σλ ε ε

λ ε ε λccc c cc

c cc

f=

′ ′( )− + ′( )1

(10.39)

λ

ε=

− ′ ′( )E

E fc

c cc cc

(10.40)

where′fccstandsforthecompressivestrengthoftheconfinedconcrete′εccdenotesthestrainat ′fcc

Whenconcreteissubjectedtoalaterallyconfiningpressure,theuniaxialcompressivestrength′fccandthecorrespondingstrain ′εccaremuchhigherthanthoseofunconfinedconcrete.The

equationsproposedbyManderetal.(1988)forthecompressivestrengthandstrainofcon-finedconcretearemodifiedusingthestrengthreductionfactorγc(Liang2011a)asfollows:

′ = ′ +f f k fcc c c rpγ 1 (10.41)

′ = ′ +

′

ε ε

γcc crp

c c

kff

1 2 (10.42)

wherefrpisthelateralconfiningpressureontheconcretecore,expressedbyEquation10.4.k1andk2aretakenas4.1and20.5,respectively,basedonexperimentalresultsreported

byRichartetal.(1928)

Thestrain ′εcisthestrainat ′fcoftheunconfinedconcrete,giveninEquation10.23.BasedontheworkofTangetal.(1996)andHuetal.(2003),LiangandFragomeni(2009)proposedanaccuratemodelforpredictingtheconfiningpressureonnormalorhigh-strengthconcretecon-finedbyeithernormalorhigh-strengthcircularsteeltubes,whichisgiveninEquation10.4.

ThepartsEFandFGofthestress–straincurveshowninFigure10.16areexpressedby

σβ

ε ε βε ε

ε ε ε

βc

cc ccF c cc cc cc

F cccc c F

cc

ff f

=′ +

− ′ − ′( )− ′( )

′ < ≤

′

( )for

ffcc c Fforε ε>

(10.43)

whereεFistakenas0.02basedonexperimentalresultsβcc is the factorusedtoconsider theeffectof theconfinementeffectprovidedbythe

circularsteeltubeonthepost-peakstrengthandductilityofconfinedconcrete,givenbyHuetal.(2003)as


βcc

Dt

Dt

Dt

=

≤

−

+

1 0 40

0 010085 1 34912

.

. .

for

0.0000339 foor 40 150< ≤

Dt

(10.44)

10.3.7 modelling of local and post-local buckling

Localbucklingof thinsteelplates is influencedbytheplateaspectratio,width-to-thick-nessratio,appliededgestressgradients,boundaryconditions,geometricimperfectionsandresidualstresses.Thelocalandpost-localbucklingbehaviourofthinsteelplatesinrectan-gularCFSTbeam–columnsunderstressgradientshasbeenstudiedbyLiangandUy(2000)andLiangetal.(2007b)usingthefiniteelementmethod.FormulashavebeendevelopedforpredictingtheinitiallocalbucklingstressesofsteeltubewallsinrectangularCFSTbeam–columns with initial geometric imperfections and residual stresses (Liang and Uy 2000;Liangetal.2007b).Theseformulascanbeincorporatedinnon-linearanalysistechniquestoaccountforthelocalbucklingeffectsofsteeltubesonthebehaviourofrectangularCFSTbeam–columns(Liang2009a).

Thinsteelplateshaveaveryhighreverseofpost-localbucklingstrengths(LiangandUy2000;Liangetal.2007b).Theeffectivestrengthconceptcanbeusedtodescribethepost-localbucklingstrengthsofsteelplatesinrectangularCFSTbeam–columnsunderaxialloadandbiaxialbending.TheeffectivestrengthformulasproposedbyLiangetal.(2007b)havebeenincorporatedinfibreelementmodelstoaccountfortheeffectsofpost-localbuckling(Liang2009a).Theultimatestrengthofthesteeltubewallsunderstressgradientsgreaterthanzerocanbeestimatedby

σ α σ1 1 5 0 5u

ys

u

yf f= −( ). . (10.45)

whereσ1urepresentstheultimatestresscorrespondingtothemaximumedgestressσ1atthe

ultimatestrengthlimitstateαsisthestressgradientwhichistheratiooftheminimumedgestressσ2tothemaxi-

mumedgestressσ1ontheplate

For intermediate stress gradients, the ultimate stress σ1u can be determined by linearinterpolation.

Theeffectivewidthconceptisusuallyusedtodeterminethepost-localbucklingstrengthofathinsteelplateunderstressgradientsasdepictedinFigure10.17.Effectivewidthformu-lascanbeincorporatedinnon-linearanalysismethodstoaccountforlocalbucklingeffectsonthebehaviourofrectangularCFSTcolumns(Liangetal.2006;Liang2009a).Theeffec-tivewidthformulasproposedbyLiangetal.(2007b)forsteelplatesinrectangularCFSTbeam–columnsundercompressivestressgradientsareexpressedby

bb

bt

bte1

42

70 2777 0 01019 1 972 10 9 605 10

=+

− ×

+ ×− −. . . .

bbt

bt

bt

s

>

−

+ ×

−

3

5

0 0

0 4186 0 002047 5 355 10

for α .

. . . − ×

=

−2

73

4 685 10 0 0. .bt

sfor

α

(10.46)


bb

bb

es

e2 12= −( )α (10.47)

wherebe1andbe2aretheeffectivewidthsasillustratedinFigure10.17.If(be1+be2)≥b,thesteelplateisfullyeffectiveincarryingloads.Forthiscase,theeffectivestrengthformulasshouldbeusedtovaluatetheultimatestrengthofthesteelplate.

Thepost-localbucklingbehaviourofthinsteelplatesunderincreasedcompressiveedgestressesischaracterisedbytheprogressivestressredistributionwithinthebuckledplates.Theheavilybuckledregioninasteelplatesustainsrelativelylowstresses,whileitstwoedgestrips carryhigh stresses (LiangandUy1998). For steel platesunderuniformcompres-sion,theeffectivewidthconceptassumesthateffectivesteelfibresarestressedtotheyieldstrengthofthesteelplates,whilethestressesinineffectivesteelfibresarezeroattheultimatestrengthlimitstate.Aftertheonsetoflocalbuckling,theineffectivewidthofasteeltubewallincreasesfromzerotothemaximumvalue(bne,max)whentheappliedloadisincreasedtoitsultimateload,wherebne,maxisgivenby

b b b bne e e,max = − +( )1 2 (10.48)

Theineffectivewidthofasteeltubewallbetweenzeroandbne,maxunderstressgradientsisapproximatelycalculatedusinglinearinterpolationbasedonitsstresslevelas

b

fbne

c

y cne= −

−

σ σσ

1 1

1,max (10.49)

σ1cistheinitiallocalbucklingstressofthesteeltubewallwithimperfections.Forasteeltubewallunderstressgradients, theeffectivewidthconceptassumesthatthesteel tubewall attains itsultimate strengthwhen itsmaximumedge stressσ1 is stressedtotheyieldstrengthofthesteelwall.Thesteelfibreswithintheineffectivewidth(bne)areassigned to zero stress, and their contributions to the strength of the CFST column areignoredasillustratedinFigure10.17.

be1

be2t

y

x

σ1

σ2

Pa

N.A.

D

B

α

θ

Figure 10.17 Effective steel areas of CFST beam–column under biaxial bending.


10.3.8 Stress resultants

In fibre element analysis, fibre stresses are calculated from fibre strains using materialstress–strainrelationships.Theaxialforceandbendingmomentsinthecompositesectionaredeterminedasstressresultants:

P A As i s i

i

ns

c j c j

j

nc

= += =∑ ∑σ σ, , , ,

1 1

(10.50)

M A y A yx s i s i i

i

ns

c j c j

j

nc

j= += =∑ ∑σ σ, , , ,

1 1

(10.51)

M A x A xy s i s i i

i

ns

c j c j

j

nc

j= += =∑ ∑σ σ, , , ,

1 1

(10.52)

wherePdenotestheaxialforceMxrepresentsthebendingmomentaboutthex-axisMyisthebendingmomentaboutthey-axisσs i, standsforthelongitudinalstressatthecentroidofsteelfibreiAs i, istheareaofsteelfibreiσc j, isthelongitudinalstressatthecentroidofconcretefibrejAc j, istheareaofconcretefibrejxiandyiarethecoordinatesofsteelfibreixjandyjarethecoordinatesofconcretefibrejnsisthetotalnumberofsteelfibreelementsncisthetotalnumberofconcretefibreelements

Compressivestressesaretakentobepositive.

10.3.9 computational algorithms based on the secant method

10.3.9.1 Axial load–strain analysis

Theultimateaxialstrengthofashortcompositecolumnunderaxialcompressionisdeterminedasthemaximumaxialloadfromitscompleteaxialload–straincurve.Theaxialload–straincurveforashortcompositecolumncanbeobtainedbygraduallyincreasingtheaxialstrainandcalculatingthecorrespondingstressresultantinthecrosssection.Theiterativeanalysisprocesscanbestoppedwhentheaxialloaddropsbelowaspecifiedpercentageofthemaximumaxialload(Pmax)suchas0.5Pmaxorwhentheaxialstraininconcreteexceedsthespecifiedultimatestrainεcu(Liang2009a).Theeffectsoflocalbucklingaretakenintoaccountintheultimateaxialloadofthin-walledCFSTcolumnsbyredistributingthenormalstressesonthesteeltubewalls.

Theaxialload–strainanalysisprocedureforCFSTshortcolumnsisgivenasfollows:

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.Initialiseaxialfibrestrainsε=Δε. 4.Computefibrestressesusingstress–strainrelationships.


5.Checklocalbucklingandupdatesteelfibrestressesaccordingly. 6.CalculatetheresultantaxialforceP. 7.Increaseaxialfibrestrainsbyε=ε+Δε. 8.RepeatSteps4–7untilP<0.5Pmaxorε>εcu.

10.3.9.2 Moment–curvature analysis

Theaxialload–moment–curvaturerelationshipsareestablishedtodeterminetheultimatemomentcapacitiesofshortcompositecolumnsundercombinedaxialloadandbiaxialbend-ing.Foragivenaxialload(Pu)appliedatafixedloadangle(α)asshowninFigure10.13,thecorrespondingultimatemomentcapacityofthecompositesectionisdeterminedasthemaximum moment from the moment–curvature curve, which is obtained by graduallyincreasingthecurvatureandsolvingforthecorrespondingmoment.Atypicalmovement–curvaturecurveforaCFSTshortcolumnunderbiaxial loadspredictedbythecomputerprogramNACOMS(Liang2009a,b)ispresentedinFigure10.3.Theequilibriumconditionsforthecompositesectionunderaxialloadandbiaxialbendingareexpressedby

P Pu − = 0 (10.53)

tanα − =

MM

y

x

0 (10.54)

Inthemoment–curvatureanalysis,thedepthoftheneutralaxis(dn)inthecompositesectionneedstobeiterativelyadjustedtosatisfytheforceequilibriumcondition.Aftertheforceequilib-riumhasbeenachieved,internalmomentsMxandMyarethencalculatedandtheorientationoftheneutralaxis(θ)isiterativelyadjustedtosatisfyboththeforceandmomentequilibriumcondi-tions.EfficientcomputationalalgorithmsbasedonthesecantmethodhavebeendevelopedandimplementedinthefibreelementanalysisprogramsbyLiang(2009a)toadjustthedepthandori-entationoftheneutralaxisinaCFSTbeam–columnsectiontosatisfyequilibriumconditions.

Thedepthoftheneutralaxis(dn)isadjustedbythefollowingequation(Liang2009a):

d d

d d rr r

n j n jn j n j p j

p j p j, ,

, , ,

, ,

( )

+ +

+ +

+= −

−−2 1

1 1

1

(10.55)

wherethesubscriptjistheiterationnumberr P Pp u= − istheresidualaxialforceinthecompositesectionatthecurrentiteration

Theconvergencecriterionfortheneutralaxisdepthdnisexpressedby|dn,j+1− dn|≤εk,whereεkistheconvergencetolerancewhichistakenas10−4.

Theorientationoftheneutralaxiswithrespecttothex-axisasshowninFigure10.13isadjustedbythefollowingequation(Liang2009a):

θ θ

θ θk k

k k m k

m k m k

r

r r+ +

+ +

+= −

−( )−2 1

1 1

1

,

, ,

(10.56)

wherethesubscriptkistheiterationnumberr M Mm y x= −tanα / is the residual moment in the composite section at the current

iteration


Theconvergencecriterionfortheorientationoftheneutralaxisθisgivenby|θk+1− θk|≤εk.Thesecantmethodneedstwoinitialvaluestostarttheiterativeprocess.Initialvaluesfor

theneutralaxisdepthdn,1anddn,2canbesettoDandD/2,respectively,whileinitialvaluesfortheorientationoftheneutralaxisθ1andθ2canbesettoαandα/2(Liang2009a).Inordertoadjustdn,theforceresidualsrp,1andrp,2arecalculatedusingdn,1anddn,2,respectively.Similarly,themomentresidualsrm,1andrm,2arecomputedinordertoadjusttheorientationoftheneutralaxis.Itshouldbenotedthatforshortcompositecolumnsunderaxialloadanduniaxialbending,onlythedepthoftheneutralaxisneedstobeadjusted(Liang2011a).

Themoment–curvatureanalysisprocedureforCFSTshortbeam–columnsincorporatinglocalbucklingeffectsisgivenasfollows:

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.Initialisecurvatureϕ=Δϕ. 4.Initialiseθ1=α, θ2=α/2, dn, 1=D, dn, 2=D/2. 5.Computefibrestressesusingstress–strainrelationships. 6.Checklocalbucklingandupdatesteelfibrestressesaccordingly. 7.Calculateresidualforcesandmomentsrp,1, rp,2, rm,1andrm,2. 8.Computefibrestressesusingstress–strainrelationships. 9.Checklocalbucklingandupdatesteelfibrestressesaccordingly. 10.CalculatetheresultantaxialforceP. 11.Adjusttheneutralaxisdepth(dn)usingthesecantmethod. 12.RepeatSteps8–11until|rp| <εk. 13.ComputebendingmomentsMx and My. 14.Adjusttheneutralaxisorientation(θ)usingthesecantmethod. 15.RepeatSteps8–14until|rm| <εk. 16.ComputetheresultantmomentM M Mx y= +2 2 . 17.Increasethecurvaturebyϕ=ϕ+Δϕ. 18.RepeatSteps4–17untilM<0.5Mmaxorεc >εcu.

10.3.9.3 Axial load–moment interaction diagrams

Inordertodeveloptheaxialload–momentinteractiondiagramforashortcompositecolumnunderaxialloadandbiaxialbending,theultimateaxialstrength(Po)ofthecompositecol-umnunderaxialcompressioniscalculatedfirstbyconductinganaxialload–strainanalysisofthecompositesection.Theaxialload(Pu)isincreasedfromzerotoamaximumvalueof0.9Po,andeachloadstepistakenas0.1Po.Foragivenloadincrement(Pu)appliedatafixedloadangle(α),themoment–curvatureanalysisofthecompositesectionisperformedtoobtainthecorrespondingmomentcapacityMu.Bygraduallyincreasingtheappliedloadandsolvingforthecorrespondingmomentcapacity,asetofaxialloadsandmomentcapaci-ties can be obtained and used to plot the axial load–moment interaction diagram. ThecomputerprogramNACOMSdevelopedbyLiang(2009a,b)cangenerateaxialload–straincurves,moment–curvaturecurvesandaxialload–momentinteractiondiagramsforbiaxi-allyloadedthin-walledCFSTshortbeam–columnswithlocalbucklingeffects.

The computational procedure for determining the axial load–moment interaction dia-gramsforcompositecolumnsunderaxialloadandbiaxialbendingisgivenasfollows:

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.ComputePousingtheaxialload–strainanalysisprocedure.


4.SettheaxialloadPu=0. 5.CalculateMuusingthemoment–curvatureanalysisprocedure. 6.IncreasetheaxialloadbyPu=Pu+ΔPu,whereΔPu=Po/10. 7.RepeatSteps5–6untilPu>0.9Po. 8.Plottheaxialload–momentinteractiondiagram.

Example 10.3: Computer analysis of CFST short column under axial load and biaxial bending

The computerprogramNACOMS is employed to analyse a square thin-walledCFSTshortcolumnunderaxialloadandbiaxialbending.Thedimensionsofthecrosssectionofthecolumnare500×500 mm.Thethicknessofthesteeltubeis8 mm.Thecompressivestrengthoftheconcreteinfillis40MPa.Theyieldstressofthesteelsectionis320 MPa,whileitstensilestrengthis430MPa.Young’smodulusofthesteeltubeis200 GPa.Inthemoment–curvatureanalysis,theaxialloadappliedtothecolumnistakenas0.6Po.The angleoftheappliedaxialloadisfixedat45°withrespecttoy-axisofthecolumncrosssection.Itisrequiredtodeterminetheaxialload–straincurve,moment–curvaturecurve andaxial load–moment interactiondiagram for thisCFST short columnunderaxialloadandbiaxialbending.

Computer solution

Theslendernessofthesteeltubewallis

Dt fy= = > = =5008

62 5 52235

52235320

44 6. .

Hence,thesteeltubeisnon-compact.ThemethodgiveninEurocode4cannotbeusedtodeterminetheaxialload–momentinteractiondiagramofthisCFSTshortcolumn.TheeffectoflocalbucklingistakenintoaccountinthecomputeranalysisoftheCFSTcolumn.

Inthefibreelementanalysis,thesteeltubewallisdividedintofivelayersthroughitsthickness,andtheconcretecoreisdividedinto80×80fibreelements.Theaxialload–straincurveforthiscolumnispresentedinFigure10.18.Itappearsthattheultimateaxial

Strain

Axi

al lo

ad P

(kN

)

00

2,000

4,000

6,000

8,000

10,000

12,000

14,000

0.01 0.02 0.03 0.04 0.05

Figure 10.18 Axial load–strain curve for the thin-walled CFST short column.


loadoftheCFSTshortcolumnis11,863kN.Thepredictedmoment–curvaturecurveforthecolumnisshowninFigure10.19.Thepredictedultimatemomentofthecompositesectionundertheaxial loadlevelof0.6Po is983kNm.Figure10.20showstheaxialload–momentinteractiondiagramofthecompositesectionunderaxialloadandbiaxialbending.Itcanbeseenfromthefigurethattheultimatepurebendingmomentis1104.3kNm,whilethemaximumultimatemomentoftheCFSTcolumnsectionis1299kNm.

Curvature φ(×10–5) (1/mm)

Mom

ent M

(kN

m)

00

200

400

600

800

1000

1200

1 2 3 4 65

Figure 10.19 Moment–curvature curve for the thin-walled CFST short column under axial load and biaxial bending.

Moment Mu (kN m)

Axi

al lo

ad P

u (kN

)

00

2,000

4,000

6,000

8,000

10,000

12,000

14,000

200 400 600 800 1,000 1,200 1,400

Figure 10.20 Axial load–moment interaction diagram of the thin-walled CFST short column under axial load and biaxial bending.


10.4 BehAvIour And deSIgn of Slender comPoSIte columnS

10.4.1 Behaviour of slender composite columns

Eccentricallyloadedbeam–columnswithaslendernessratio(L/r)greaterthan22areusu-allytreatedasslenderbeam–columnsindesigncodes.Thebehaviourofslendercompositecolumnshasbeeninvestigatedexperimentallybyresearchers(Furlong1967;KnowlesandPark 1969; Neogi et al. 1969; Bridge 1976; Shakir-Khalil and Zeghiche 1989; Shakir-KhalilandMouli1990;RanganandJoyce1992;MuñozandHsu1997;VrceljandUy2002a;MursiandUy2006).Thebehaviourofslendercompositecolumnsundereccentricloadingischaracterisedbytheiraxialload–deflectioncurveswhichindicatetheflexuralstiffness,theultimateaxialstrength,thepost-peakbehaviourandtheflexuralductilityofthecolumns.CECslendercolumnsmayfailbyinelasticglobalbucklingassociatedwithyieldingofthesteelsectionandreinforcementandcrushingoftheconcrete.CFSTslendercolumnsmayfailbytheinteractionofinelasticlocalandglobalbucklingassociatedwithyieldingofthesteelsectionandcrushingoftheinfillconcrete.Veryslendercompositecol-umnsfailbyelasticglobalbuckling.Theultimatestrengthsofslendercompositecolumnsareusuallygovernedbytheflexuralstiffnessratherthanthematerialstrengthsofthesteelandconcrete.

Thefundamentalbehaviourofslendercompositebeam–columnsisinfluencedbymanyparameters,includingcolumnslendernessratio,depth-to-thicknessratio,loadingeccen-tricity,concretecompressivestrength,steelyieldstrength,initialgeometricimperfectionsandsecond-ordereffects.NumericalstudiesconductedbyLiang(2011b)andPateletal.(2012a,b,c;2014c)demonstratethatincreasingthecolumnslendernessratio(L/r)ortheloading eccentric ratio (e/D) significantly reduces the initial flexural stiffness and ulti-mateaxialstrengthof theCFSTbeam–columnundereccentric loadingbutremarkablyincreases its lateral deflection and displacement ductility. For a given axial load level,thecorrespondingultimatemomentcapacityoftheCFSTcolumnisfoundtoreducebyincreasingthecolumnslendernessratio.Inaddition,localbucklingisfoundtoreducetheflexural stiffness and strength of rectangular CFST slender beam–columns. Moreover,increasingtheconcretecompressivestrengthslightlyincreasestheinitialflexuralstiffnessbutsignificantly increases theultimateaxialstrengthof theCFSTbeam–columnundereccentric loading. Furthermore, the initial flexural stiffness of the eccentrically loadedCFST beam–column is shown to be not affected by the yield stress of the steel tube.However, increasing theyield stressof the steel tube remarkably increases theultimateaxialstrengthoftheCFSTbeam–column.ThestudiesonconfinementeffectsconductedbyLiang(2011b)showthat incircularCFSTbeam–columns, theconcreteconfinementeffectdecreaseswithanincreaseinthecolumnslendernessratioortheloadingeccentric-ity.ForveryslendercircularCFSTbeam–columnswithanL/rratiogreaterthan70orforslendercircularCFSTbeam–columnswithane/Dratiogreaterthan2,theconfinementeffectcanbeignoredinthedesign.

10.4.2 relative slenderness and effective flexural stiffness

InEurocode(2004),theslendernessofacompositecolumnismeasuredbyitsrelativeslen-dernessforthebendingplanebeingconsideredasfollows:

λ = P

Po

cr

(10.57)


wherePoistheultimateaxialstrengthofthecompositecolumnsectionunderaxialcompres-

sionignoringtheconfinementeffect,givenbyEquation10.1,inwhichγcistakenas0.85forCECcolumnsand1.0forCFSTcolumns

Pcristheelasticcriticalbucklingloadofthecompositecolumnunderaxialcompression,givenby

P

EI

Lcr

eff

e

=π2

2

( ) (10.58)

inwhich(EI)effrepresentstheeffectiveflexuralstiffnessofthecrosssectionofacompositecolumn,whichisexpressedby

( ) .EI E I E I E Ieff s s cm c r r= + +0 6 (10.59)

whereEs,Ecm andEr are the elasticmoduliof structural steel, concrete and reinforcement,

respectivelyIs,IcandIrare thesecondmomentsofareaofstructural steel section,concreteand

reinforcement,respectively

Theeffectiveflexuralstiffness(EI)effshouldaccountforthelong-termeffectsduetocon-cretecreepontheelasticmodulusofconcrete(Ecm)byusingtheeffectiveelasticmodulusofconcreteconsideringthelong-termeffectofconcretecreep,whichisexpressedby

EE

P Pc eff

cm

G c

, =+ ∗ ∗( ) ∗1 / φ

(10.60)

whereφc∗isthefinalconcretecreepfactorPG∗isthepermanentpartofthedesignaxialforceP∗

Theelasticmodulusofconcrete(Ecm)canbecalculatedby

E

fcm

c=′ +

22000

810

0 3.

MPa (10.61)

Fordetermininginternaldesignactionsonaslendercompositecolumn,theeffectiveflex-uralstiffnessconsideringlong-termeffectsisgiveninEurocode4as

( ) . ( . ), ,EI E I E I E Ieff II s s c eff c r r= + +0 9 0 5 (10.62)

10.4.3 concentrically loaded slender composite columns

AsimplemethodisgiveninEurocode4(2004)fordeterminingtheultimateaxialstrengthofslendercompositecolumnsunderaxialcompressionasfollows:

P Pu o= χ (10.63)


whereχisthereductionfactorwhichisafunctionoftherelativeslendernessλandimperfec-tionsgiveninEurocode3(2005)andisexpressedby

χϕ ϕ λ

=+ −

≤11 0

2 2. (10.64)

ϕ α λ λ= + −( ) +

0 5 1 0 22

. .g (10.65)

whereαgistheimperfectionfactorgiveninTable10.1.The column buckling curves for slender composite columns under axial compression

determined using Equation 10.64 are presented in Figure 10.21. The buckling curvesandimperfectionsfordifferentcompositecolumnsaregiveninTable10.1asspecifiedinEurocode4(2004).

0 0.5 1

a

bc

1.5 2 2.5

Relative slenderness λ–3

0

Redu

ctio

n fa

ctor

χ

0.2

0.4

0.6

0.8

1

12

Figure 10.21 Buckling curves for composite columns under axial compression.

Table 10.1 Buckling curves and member imperfections for composite columns

Composite column Buckling curve Imperfection factor αg Member imperfection

CEC or partially encased composite columnsBending about the x-axis b 0.34 L/200Bending about the y-axis c 0.49 L/150

Circular/rectangular CFST columns

ρs ≤ 3% a 0.21 L/300

3 6% %< ≤ρs b 0.34 L/200

Circular CFST columns with I-sections

b 0.34 L/200

Source: Eurocode 4 (2004) Design of Composite Steel and Concrete Structures, Part 1-1: General Rules and Rules for Buildings, European Committee for Standardization, CEN.


10.4.4 uniaxially loaded slender composite columns

10.4.4.1 Second-order effects

Eurocode4(2004)suggeststhatdesignactionsonslendercompositecolumnsmaybecalcu-latedbyanelasticglobalanalysisofthecompositeframeincorporatingtheglobalsecond-ordereffectsandglobalimperfections.Equivalentgeometricimperfectionscanbeusedtoaccountfortheeffectsofinitialgeometricimperfectionsandresidualstressesonthestrengthandbehaviourofslendercompositecolumns.Theanalysisofindividualslendercompositebeam–columnunderaxialforceandendmomentsdeterminedfromtheglobalanalysismustconsiderthesecond-ordereffectsinthecolumnandthecolumnimperfections.Thesecond-ordereffectsandequivalentgeometricimperfectionsamplifythedesignbendingmomentsontheslendercompositecolumn.Thedesignmethodistodeterminetheamplifieddesignbendingmomentontheslendercompositecolumnduetosecond-ordereffectsaswellasequivalentgeometric imperfections.Foragivendesignaxial load, iftheamplifieddesignmomentisstilllessthanorequaltothedesignmomentcapacityofthecolumncrosssection,theslendercompositecolumnsatisfiesthestrengthrequirement.

A pin-ended slender composite beam–column subjected to an axial load and bendingmomentsM1

∗andM2∗showninFigure10.22isconsideredheretoexplainthesecond-order

effects.TheendmomentsM1∗andM2

∗causethebeam–columntobendintoasinglecur-vature. This results in an additional deflection u along the beam–column and an addi-tional moment P*u, which is called the secondary moment. The maximum moment onthebeam–columnduetosecond-ordereffects isusedtodesignthebeam–columnand isdeterminedbyamplifyingthemaximumendmomentM1

∗usingthemomentamplificationfactor.Thismeansthattheendmomentamplifiedbythesecond-ordereffectsisdeterminedasM Mend m

∗ = ∗δ 1 .InEurocode4(2004),themomentamplificationfactorisexpressedby

δmb

cr eff

c

P P=

− ∗( )1 / ,

(10.66)

wherePcr,effistheelasticbucklingloadatthecompositecolumncalculatedusing(EI)eff,IIandcbaccountsfortheeffectsofdifferentmomentsatcolumnends,givenby

cb m= − ≥0 66 0 44 0 44. . .β (10.67)

P*

P*

uL

P*u

M*2 M*2

M*end

M*1M*1

Figure 10.22 Second-order effects on a slender composite beam–column.


inwhichthemomentratioβm M M= ± ∗ ∗2 1/ ,whichistakenasnegativeforsinglecurvature

bendingandpositivefordoublecurvaturebending.Thesecond-ordereffectsduetotheequivalentgeometricimperfection(uo)atthemid-height

ofaslendercompositebeam–columnalsocauseanadditionalmomentP*uoatitsmid-height.Themomentatthemid-heightofthecompositecolumninducedbygeometricimperfectionsisdeterminedasM P uimp m o

∗ = ∗δ ,whereδmiscalculatedusingcb=1.0inEquation10.66.Thedesignbendingmoment for the slender composite columnaccounting for second-

ordereffectsiscalculatedas

M M Mend imp∗ = ∗ + ∗ (10.68)

10.4.4.2 Design moment capacity

Thedesignmomentcapacityofaslendercompositebeam–columndependsonthedesignaxialloadlevel.Theloadratioiscalculatedasχd=P*/Po,whichisdrawnonthedimension-lessaxialload–momentinteractiondiagramofthecompositecolumnsection.ThemomentcapacityfactorμdcorrespondingtoχdcanbedeterminedfromtheinteractiondiagramforthecompositesectionasillustratedinFigure10.23.Theslendercompositecolumnunderaxialloadanduniaxialbendingmustsatisfythefollowingdesignrequirement:

M Mu∗ ≤ φ (10.69)

whereφ = 0 8. isthecapacityreductionfactorM Mu M d o= α µ isthenominalmomentcapacityoftheslendercompositecolumn

ThereductionfactorαMaccountsfortheeffectofunconservativeassumptionoftherect-angularstressblockwhichisextendedtotheplasticneutralaxis.ThefactorαMistakenas

Pu Po

Mu Mo0

χd

1.0

B

D

C

A

μd 1.0

Figure 10.23 Dimensionless axial load–moment interaction diagram of a composite short column.


0.9forsteelgradeswithyieldstressbetween235and355MPaand0.8forsteelgradeswithyieldstressbetween420and460MPa.

Themainstepsforcheckingthestrengthofaslendercompositecolumnunderaxialloadanduniaxialbendingaregivenasfollows:

1.Determinetheaxialload–momentinteractiondiagramforthecolumnsection. 2.Calculatetheeffectiveflexuralstiffness(EI)eff,IIforthecompositecolumn. 3.ComputethecriticalbucklingloadPcr,effusing(EI)eff,II. 4.CalculateM*accountingforsecond-ordereffectsandgeometricimperfections. 5.Determineμdcorrespondingtoχdontheinteractiondiagram. 6.Checkthedesignmomentcapacity:M* ≤ϕMu.

Example 10.4: Strength of CEC slender column under axial load and uniaxial bending

ThecrosssectionofaCECslendercolumnisshowninFigure10.8.Thecolumnof4mlengthissubjectedtoadesignaxialcompressiveforceP*=7469kNofwhich4855kNispermanentanddesignbendingmomentsM1 300∗ = kNmandM2 150∗ = kNmattheends.Thecolumnisbentintosinglecurvatureaboutthex–x-axis.Thedesigndataareasfol-lows: ′ =fc 32MPa,fy=300MPa,fyr=500MPaandEs=Er=200,000MPa.Theaxialload–moment interaction diagram for the composite column section has been deter-mined inExample10.1.Thefinalconcretecreepfactor isφc

∗ = 3 0. .Checkthedesignmomentcapacityofthisslendercompositecolumn.

1. Second moments of area of uncracked section

Thesecondmomentofareaofsteelsectionis

I

b D b t ds

f s f w w= −−

=×

−− ×

= ×3 3 3 3

12 12350 350

12350 12 318

12344 75

( ) ( ). 1106 mm4

Thesecondmomentofareaofreinforcementis

Ir

r dr r= +

= +

×

=4

44

20 24

202

200 04

24 2

ππ

ππ

( ).

/22827 106× mm4

Thesecondmomentofareaofconcreteinthesectioniscomputedas

I

BDI Ic s r= − − =

×− × − × = ×

3 36 6 6

12500 500

12344 75 10 0 2827 10 4860 10. . mm44

2. Effective flexural stiffness

Theeffectivemodulusofconcreteiscalculatedas

E

fcm

c=′ +

=22,000 33,346MPa

810

0 3.

EE

P Pc eff

cm

G c

, =+ ∗ ∗( ) ∗ =

+ ×=

1 /

33,3461 (4,855/7,469) 3

11,304MPaφ


Theeffectiveflexuralstiffnessofthecompositecolumniscomputedas

( ) . ( . )

.

, ,EI E I E I E Ieff II s s c eff c r r= + +

= × × ×

0 9 0 5

344 75 0.9 (200 103 ++ × × + × × ×

= ×

0 5 200 10 0 2827 103 6. . )11,304 4,860

868 10 N mm11 2

3. Amplified design bending moment

Thecriticalbucklingloadofthecompositecolumnisdeterminedas

P

EIL

cr effeff II

e,

,( )= =

× ×=

π π2

2

2 11

2

868 104,000

N 53,543kN

Thecompositecolumnisbentintoasinglecurvaturesothatitsmomentratiois

βm

MM

= −∗∗ = − = −2

1

150300

0 5.

Themomentamplificationfactoriscalculatedasfollows:

cb m= − = − × − =0 66 0 44 0 66 0 44 0 5 0 88. . . . ( . ) .β

δm b

cr eff

cP P

=− ∗ =

−=

10 88

11 023

( ).

( ).

,/ 7,469/53,543

Theamplifieddesignbendingmomentatthecolumnendduetosecond-ordereffectsiscomputedas

M Mend m∗ = ∗ = × =δ 1 1 023 300 306 8. . kNm

Theequivalentgeometricimperfectionatthemid-heightofthecolumnis

u

Lo = = =

2004000200

20mm

Formomentcausedbygeometricimperfection,cb=1.0.Themomentamplificationfactorforgeometricimperfectionsiscalculatedas

δm b

cr eff

cP P

=− ∗ =

−=

11 0

11 162

( ).

( ).

,/ 7,469/53,543

Theamplifieddesignbendingmomentduetogeometricimperfectionsis

M P uimp m o∗ = ∗ = × × =δ ( ) . ( . ) .1 162 7469 0 02 173 5kNm

Thedesignmomentonthecompositecolumnconsideringsecond-ordereffectsis

M M M Mend imp∗ = ∗ + ∗ = + = > ∗ =306 8 173 5 3001. . 480.3kNm kNm



Theaxialloadratioiscomputedas

χd

o

PP

=∗= =

7,46911,490

0.65.5

From the axial load–moment interaction diagram of the composite section shown inFigure10.24,themomentcapacityfactorisobtainedas

µd = 0.78

Thedesignmomentcapacityoftheslendercompositecolumnisdeterminedas

φ φα µM M Mu M d o= = × × × = > ∗ =0 8 0 9 0 78 917 5 515 2 480 3. . . . . .kN m kN m, OK

Example 10.5: Strength of CFST slender column under axial load and uniaxial bending

ThecrosssectionofaCFSTslendercolumnisshowninFigure10.10.Thecolumnof8 mlengthissubjectedtoadesignaxialcompressiveforceP*=1420kNofwhich9230 kNispermanentanddesignbendingmomentsM1 1200∗ = kNmandM2 0∗ = kNmattheends.Thecolumnisbentintosinglecurvatureaboutthex–x-axis.Thedesigndataareasfol-lows: ′ =fc 50MPa ,fy=300MPa,Es=Er=200,000MPa.Theaxialload–momentinter-actiondiagramforthecompositecolumnsectionhasbeendeterminedinExample 10.2.Thefinalconcretecreep factor isφc

∗ = 3 0. .Check thedesignmomentcapacityof thisslendercompositecolumn.

1. Second moments of area of uncracked section

Thesecondmomentofareaoftheconcretecoreiscomputedas

IB t D t

c =− −

=− × − ×

= ×( )( ) ( )( )2 2

12500 2 20 600 2 20

126732 10

3 36 mm4

Mu/Mo

P u/P

o

00

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4 0.6 0.8 1.2 1.41

Figure 10.24 Dimensionless axial load–moment interaction diagram of the CEC short column.


Thesecondmomentofareaofsteelsectionis

I

BDIs c= − =

×− × = ×

3 36 6

12500 600

126732 10 2268 10 mm4

2. Effective flexural stiffness

Theeffectivemodulusofconcreteiscalculatedas

E

fcm

c=′ +

=

+

=22,000 22,000 37,278MPa

810

50 810

0 3 0 3. .

EE

P Pc eff

cm

G c

, =+ ∗ ∗( ) ∗ =

+ ×=

1 /

37,2781 (9,230/14,200) 3

12,637MPaφ

Theeffectiveflexuralstiffnessofthecompositecolumniscomputedas

( ) . ( . ), ,EI E I E I E Ieff II s s c eff c r r= + +

= × × × +

0 9 0 5

00.9 (200 10 2,2683 .. )5 0 106× × + ×

= ×

12,637 6,732

446.5 10 Nmm12 2

3. Amplified design bending moment

Thecriticalbucklingloadofthecompositecolumnisdeterminedas

P

EIL

cr effeff II

e,

,( ) .= =

× ×=

π π2

2

2 12

2

446 5 108,000

N 68,856kN

Thecompositecolumnisbentintoasinglecurvaturesothatitsmomentratiois

βm

MM

= −∗∗ = − =2

1

01200

0

Themomentamplificationfactoriscalculatedasfollows:

cb m= − = − × =0 66 0 44 0 66 0 44 0 0 66. . . . .β

δm b

cr eff

cP P

=− ∗ =

−=

10 66

10 83

( ).

( ).

,/ 14,200/68,856

Theamplifieddesignbendingmomentatthecolumnendduetosecond-ordereffectsiscomputedas

M Mend m∗ = ∗ = × =δ 1 0 83 1200 996. kNm

Theequivalentgeometricimperfectionatthemid-heightofthecolumnis

u

Lo = = =

3008000300

27mm


Formomentcausedbygeometricimperfection,cb=1.0.Themomentamplificationfactorforgeometricimperfectionsiscalculatedas

δm b

cr eff

cP P

=− ∗ =

−=

11 0

11 26

( ).

( ).

,/ 14,200/68,856

Theamplifieddesignbendingmomentduetogeometricimperfectionsis

M P uimp m o∗ = ∗ = × × =δ ( ) . ( . )1 26 0 027 48314,200 kNm

Thedesignmomentonthecompositecolumnconsideringsecond-ordereffectsis

M M M Mend imp∗ = ∗ + ∗ = + = > ∗ =996 483 120011479kNm kNm


Theaxialloadratioiscomputedas

χd

o

PP

=∗= =14,20023,668

0.6

From the axial load–moment interaction diagram of the composite section shown inFigure10.25,themomentcapacityfactorisobtainedas

µd = 0.67

Thedesignmomentcapacityoftheslendercompositecolumnisdeterminedas

φ φα µM M Mu M d o= = × × × = > ∗ =0 8 0 9 0 67 3103 1497 1479. . . kN m kN m, OK

Mu/Mo

P u/P

o

00

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4 0.6 0.8 1.21

Figure 10.25 Dimensionless axial load–moment interaction diagram of the CFST short column.


10.4.5 Biaxially loaded slender composite beam–columns

Forslendercompositecolumnsunderaxialloadandbiaxialbending,thedesignmomentcapacitiesneedtobedeterminedseparatelyforeachprincipalaxis.Eurocode4suggeststhatimperfectionsshouldbeconsideredonlyintheplaneinwhichfailureisexpectedtooccur.Ifthecriticalplaneisnotknown,checksshouldbeundertakenforbothbendingplanes.Eurocode4(2004)requiresthattheslendercompositecolumnunderaxialcompressionandbiaxialbendingmustsatisfythefollowingconditions:

M Mx ux∗ ≤ φ (10.70)

M My uy∗ ≤ φ (10.71)

MM

MM

x

ux

y

uy

∗+

∗≤

φ φ1 0. (10.72)

whereMx∗andMy

∗aretheamplifieddesignbendingmomentsabouttheprincipalx-andy-axes,respectively

MuxandMuyarethenominalmomentcapacityoftheslendercompositecolumnbendingabouttheprincipalx-andy-axes,respectively,andaregivenby

M Mux M dx uox= α µ (10.73)

M Muy M dy uoy= α µ (10.74)

whereµdx andµdy are themoment capacity factors for bending about theprincipalx- and

y-axes,respectivelyMuoxandMuoyarethepuremomentcapacitiesofthecolumnsectionforbendingabout

theprincipalx-andy-axes,respectively

10.5 non-lIneAr AnAlySIS of Slender comPoSIte columnS

10.5.1 general

Fibreelementmodelsweredevelopedforthenon-linearanalysisofCECbeam–columnsunderaxialloadandbiaxialbending(El-Tawilletal.1995;MuñozandHsu1997).Analyticalandnumericalmodelswerealsodevelopedforpredictingthebehaviourofcircularandrectan-gularCFSTslenderbeam–columns(Neogietal.1969;Bradford1996;Hajjaretal.1998;LakshmiandShanmugam2002;Shanmugametal.2002;VrceljandUy2002b;MursiandUy2003;ValipourandFoster2010;Liang2011a,b;Portolésetal.2011;Liangetal.2012;Pateletal.2012a,b,c;2014c).ThefibreelementmodelsdevelopedbyLiang(2011a),Liangetal.(2012)andPateletal.(2012a)forCFSTslenderbeam–columnsunderaxialloadandbendingareintroducedinthefollowingsections.


10.5.2 modelling of load–deflection behaviour

Theload–deflectionresponsesofslendercompositebeam–columnsunderincreasedloadingareinfluencedbytheinelasticcross-sectionalbehaviour,columnslenderness,loadingeccen-tricity, imperfections and second-order effects. The inelastic stability analysis of slendercompositebeam–columnsmusttakeintoaccountthegeometricandmaterialnonlinearitiesof the beam–columns. Numerical models have been developed for the inelastic stabilityanalysisofcircularandrectangularCFSTslenderbeam–columns,whichincorporatestheeffectsofbothgeometricandmaterialnonlinearities(Liang2011a;Liangetal.2012;Pateletal.2012a).Thebeam–columnconsideredispinendedandsubjectedtosinglecurvaturebendingasschematicallydepictedinFigure10.26.ItisassumedthatthedeflectedshapeofCFSTbeam–columnsisapartofasinewaveandisexpressedby

u u

zL

m=

sin

π (10.75)

whereumrepresentsthedeflectionatthemid-heightofthebeam–column.Theinitialgeo-metricimperfectionofthebeam–columnmaybedescribedbythesameformofthedisplace-mentfunctionas

u u

zL

oy o=

sin

π (10.76)

inwhichuoistheinitialgeometricimperfectionatthemid-heightofthebeam–column.Thecurvature(ϕ)ofthebeam–columncanbeobtainedfromEquation10.75as

φ π π= ∂

∂=

2

2

2uz L

uzL

m sin (10.77)

Thecurvatureatthemid-heightofthebeam–columncanbederivedas

φ πm m

Lu=

2

(10.78)

z

y

eP

L

L

Pe

2

um

Figure 10.26 Pin-ended beam–column model.


Theexternalbendingmomentatthemid-heightofthebeam–columnwithaninitialgeo-metricimperfectionuoandundereccentricloadingcanbecalculatedby

M P e u ume m o= + +( ) (10.79)

wherePistheappliedloadeistheeccentricityoftheappliedloadasshowninFigure10.26

Thedeflectioncontrolmethodisemployedinnumericalmodelstopredictthecompleteload–deflectioncurvesforslendercompositebeam–columnsunderuniaxialorbiaxialloads(Liang2011a;Liangetal.2012;Pateletal.2012a).Thedeflectionatthemid-heightumofthe slenderbeam–column is gradually increased, and the corresponding curvatureϕm atthemid-heightofthebeam–columniscalculated.ThedepthandorientationoftheneutralaxiscanbeadjustedbythesecantmethodorMüller’smethod(Müller1956)toachievethemomentequilibriumatthemid-heightofthebeam–column.Theequilibriumconditionsfortheslenderbeam–columnunderbiaxialbendingareexpressedby

P e u u Mm o mi( )+ + − = 0 (10.80)

tanα − =

MM

y

x

0 (10.81)

whereM M Mmi x y= +2 2 istheresultantmomentinthecompositesection.Intheiterativenumericalanalysis,residualmomentsinthecompositesectionarecalcu-

latedby

r P e u u Mmc

m o mi= + + −( ) (10.82)

r

MM

mb y

x

= −tanα (10.83)

If rmc

k< ε and rmb

k< ε ,theequilibriumconditionsaresatisfied.Theconvergencetoleranceεkcanbetakenas10−4inthenumericalanalysis.

Thecomputationalprocedureforpredictingtheload–deflectioncurvesforslendercom-positebeam–columnsunderbiaxialloadsisdescribedasfollows:

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.Initialisethemid-heightdeflectionum=Δum. 4.Calculatethecurvatureϕmatthemid-heightofthebeam–column. 5.Adjusttheneutralaxisdepth(dn)usingMüller’smethod. 6.ComputestressresultantsPandMmi. 7.RepeatSteps5–6until rm

ck< ε .

8.ComputebendingmomentsMxandMy. 9.Adjusttheneutralaxisorientation(θ)usingMüller’smethod. 10.RepeatSteps5–9until rm

bk< ε .


11.Increasethemid-heightdeflectionbyum=um+Δum. 12.RepeatSteps4–11untilP≤0.5Pmaxor u um m> ∗.

10.5.3 modelling of axial load–moment interaction diagrams

Theaxial load–moment interactiondiagrams for slendercompositebeam–columnunderaxialloadandbiaxialbendingaregeneratedbyanincrementalanditerativeanalysispro-cedure.Foragivenaxialload(Pu)appliedatafixedloadangle(α),theultimatebendingstrength(Mu)ofaslenderbeam–columnisdeterminedasthemaximummoment(Me,max)thatcanbeappliedtothecolumnends.Themomentequilibriumismaintainedatthemid-heightofthebeam–column.Theexternalmomentatthemid-heightoftheslenderbeam–columnisgivenby

M M P u ume e u m o= + +( ) (10.84)

whereMeisthemomentatthecolumnends.Thedeflectionatthemid-heightoftheslenderbeam–columncanbecalculatedfromthe

curvatureby

u

Lm m=

πφ

2

(10.85)

Togeneratetheinteractiondiagram,thecurvature(ϕm)atthemid-heightofthebeam–col-umnisgraduallyincreasedandthecorrespondinginternalmoment(Mmi)iscomputedbythemoment–curvatureanalysisprocedure.Thecurvatureatthecolumnends(ϕe)isadjusted,andthecorrespondingmomentatthecolumnends(Me) iscalculateduntilthemaximummomentatthecolumnends(Me,max)isobtained.Theaxialloadisincreasedandtheaxialload–moment interaction diagram of the slender composite column can be generated byrepeatingtheprecedingprocess.Forbiaxialbending,equilibriumequationsareexpressedby

P Pu − = 0 (10.86)

tanα − =

MM

y

x

0 (10.87)

M P u u Me u m o mi+ + − =( ) 0 (10.88)

Inthenumericalanalysis,theresidualforceandmomentsateachiterationarecalculatedasγma uP P= − ,γ αm

by xM M= −tan / andγmc e u m o miM P u u M= + + −( ) .Iftheabsolutevaluesofthe

residualforceandmomentsarelessthanthespecifiedtoleranceεk(εk=10−4),theequilibriumstatesareattained.

The computational procedure for determining the axial load–moment interaction dia-gramsofslendercompositecolumnsunderbiaxialloadsisdescribedasfollows:

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.ComputetheultimateaxialloadPoaoftheslendercolumnunderaxialcompression

usingtheload–deflectionanalysisproduce.


4.InitialisetheaxialloadasPu=0. 5.Initialisethecurvatureatthemid-heightofthecolumnasϕm=Δϕm. 6.Computethemid-heightdeflectionum. 7.Adjusttheneutralaxisdepth(dn)usingMüller’smethod. 8.CalculatetheresultantaxialforcePinthecompositesection. 9.RepeatSteps7–8until rm

ak< ε .

10.ComputebendingmomentsMxandMyinthecompositesection. 11.Adjusttheneutralaxisorientation(θ)usingMüller’smethod. 12.RepeatSteps7–11until rm

bk< ε .

13.ComputetheresultantmomentMmi. 14.AdjustthecurvatureatthecolumnendϕeusingMüller’smethod. 15.ComputeMeusingthemoment–curvatureanalysisprocedure. 16.RepeatSteps13–15until rm

ck< ε .

17.Increasethecurvatureatthemid-heightofthecolumnbyϕm=ϕm+Δϕm. 18.RepeatSteps6–17untilMe,maxatthecolumnendsisobtained. 19.IncreasetheaxialloadbyPu=Pu+ΔPu,whereΔPu=Poa/10. 20.RepeatSteps5–19untilPu>0.9Poa.

10.5.4 numerical solution scheme based on müller’s method

In thenon-linear analysisof a slender compositebeam–columnunderbiaxial loads, thedepthandorientationoftheneutralaxisandthecurvatureatthecolumnendsneedtobeiterativelyadjustedinordertosatisfytheforceandmomentequilibriumconditions.Forthispurpose, computational algorithmsbasedon the secantmethodhavebeendevelopedbyLiang(2009a,2011a).Forslenderbeam–columnsunderuniaxialbending,thecurvatureatthecolumnends(ϕe)isadjustedbythefollowingequation(Liang2011a):

φ φ

φ φe k e k

e k e k m k

m k m k

r

r r, ,

, , ,

, ,+ +

+ +

+= −

−( )−2 1

1 1

1

(10.89)

wherethesubscriptkistheiterationnumberr M P u u Mm e u m o mi= + + −( )

Itappearsthatcomputationalalgorithmsbasedonthesecantmethodareefficientandreli-ableforobtainingconvergedsolutions(Liang2009a,2011a).ThegeneralizeddisplacementcontrolmethodproposedbyYangandShieh (1990)canbeused tosolve the incrementalequilibriumequations(YangandKuo1994).Müller’smethod(1956)isageneralizationofthesecantmethod,whichcanalsobeusedtosolvenon-linearequations.Pateletal.(2012a)andLiangetal.(2012)havedevelopedcomputationalalgorithmsbasedonMüller’smethodtoadjustthedepthandorientationoftheneutralaxisandthecurvatureatthecolumnends.Thedepth(dn)andorientation(θ)oftheneutralaxisandthecurvature(ϕe)aretreatedasvari-ableswhicharedenotedbyω.Threeinitialvaluesofthevariablesω1,ω2andω3arerequiredbyMüller’smethodtostarttheiterativecomputationalprocess.Thecorrespondingresidualforcesormomentsrm,1,rm,2andrm,3arecalculated.Thenewvariableω4thatapproachesthetruevalueiscomputedbythefollowingequations(Pateletal.2012a;Liangetal.2012):

ω ω4 32

2

4= + −

± −

c

b b a cm

m m m m

(10.90)


a

r r r rm

m m m m=− − − − −

− − −( )( ) ( )( )

( )( )(, , , ,ω ω ω ω

ω ω ω ω ω2 3 1 3 1 3 2 3

1 2 1 3 2 ωω3) (10.91)

b

r r r rm

m m m m=− − − − −

− −( ) ( ) ( ) ( )

( )( )(, , , ,ω ω ω ω

ω ω ω ω ω1 3

22 3 2 3

21 3

1 2 1 3 22 3− ω ) (10.92)

c rm m= ,3 (10.93)

ThesignofthesquarerootterminthedenominatorofEquation10.90istakentobethesameasthatofbmwhentheequationisusedtoadjustthedepthandorientationoftheneutral axis. However, this sign is taken as positive when the equation is employed toadjustthecurvatureatthecolumnends.Inordertoobtainconvergedsolutions,theval-uesofω1,ω2andω3andcorrespondingresidualforcesormomentsrm,1,rm,2andrm,3needtobeswapped(Pateletal.2012a).Equation10.90andtheexchangeofdesignvariablesandforceormomentfunctionsareexecutediterativelyuntiltheconvergencecriterionof|rm| < εkissatisfied.

Theinitialvaluesofthedepthandorientationoftheneutralaxisandthecurvatureatthecolumnendscanbetakenasfollows:dn,1=D/4,dn,3=D,dn,2=(dn,1+dn,2)/2;θ1=α/4,θ3 = α,θ2=(θ1+θ3)/2;ϕe,1=10−10,ϕe,3=10−6,ϕe,2=(ϕe,1+ϕe,3)/2.

ComputationalalgorithmsusingthemixedsecantandMüller’smethodhavebeendevel-opedandimplementedinthecomputerprogramNACOMSbytheauthorforthenon-linearinelasticanalysisofthin-walledCFSTslenderbeam–columnsunderaxialloadandbiaxialbending.Inthecomputationalalgorithms,theultimateaxialstrengthofCFSTslendercol-umnsunderaxialcompression iscomputedusingMüller’smethod,while theanalysisofCFSTslenderbeam–columnsundercombinedaxialloadandbiaxialbendingisperformedusingthesecantmethod.

Example 10.6: Computer analysis of CFST slender beam–column under axial load and biaxial bending

The computerprogramNACOMS is employed to analyse a square thin-walledCFSTslenderbeam–columnunderaxialloadandbiaxialbending.Thedimensionsandmate-rialpropertiesofthebeam–columncrosssectionaregiveninExample10.3.Thelengthof thebeam–column is5m.The eccentricity ratio (e/D)of theaxial load is takenas0.2.Theinitialgeometricimperfectionatthemid-heightofthebeam–columnistakenasL/1000.Theangleoftheappliedaxialloadisfixedat45°withrespecttothey-axisof the columncross section. It is required todetermine the load–deflectionandaxialload–momentinteractioncurvesforthisCFSTslenderbeam–columnunderaxial loadandbiaxialbending.

Computer solution

Thesteel tube section isnon-compactas shown inExample10.3.Hence, themethodgiven in Eurocode 4 cannot be used to determine the axial load–moment interactiondiagramofthisCFSTcolumn.TheeffectoflocalbucklingistakenintoaccountinthecomputeranalysisoftheCFSTcolumn.Inthefibreelementanalysis,thesteeltubewallisdividedintofivelayersthroughitsthicknessandtheconcretecoreisdividedinto80× 80fibreelements.Theload–deflectioncurveforthiscolumnispresentedinFigure10.27.ItappearsthattheultimateaxialloadoftheCFSTslenderbeam–columnundereccentricloadingis7171kN.Figure10.28showstheaxialload–momentinteractiondiagramfortheCFSTslenderbeam–columnunderaxial loadandbiaxialbending. It canbe seen


fromthefigurethattheultimateaxialloadoftheslendercolumnwithoutthepresenceofbendingmoment is 11,264kN.Thepurebendingmoment is 1104.3kNm,whilethe maximum ultimate moment of the CFST column section is 1142.7 kN m. It canbeobservedthattheslendernessandloadingeccentricityreducetheultimateaxialandbendingstrengthsoftheCFSTcolumn.However,thepurebendingmomentcapacityisnotaffectedbythelengthoftheCFSTcolumn.

Moment Mu (kN m)

Axi

al lo

ad P

u (kN

)

00

2,000

4,000

6,000

8,000

10,000

12,000

200 400 600 800 1,000 1,200 1,400

Figure 10.28 Axial load–moment interaction diagram of the thin-walled CFST slender beam–column under axial load and biaxial bending.

Mid-height deflection um (mm)

Axi

al lo

ad (

kN)

00

1000

2000

3000

4000

5000

6000

7000

8000

50 100 150 200

Figure 10.27 Load–deflection curve for the thin-walled CFST slender beam–column under axial load and biaxial bending.


10.5.5 composite columns with preload effects

10.5.5.1 General

Thecommonconstructionpracticeofhigh-risecompositebuildingsistoerectthehollowsteeltubesandcompositefloorsseveralstoreysbeforefillingthewetconcreteintothesteeltubes.Thisconstructionpracticeimposespreloadsarisingfromtheconstrictionloadsandpermanentloadsoftheupperfloorsonthesteeltubes.Thepreloadscauseinitialstressesanddeformationsinthesteeltubes,whichmaysignificantlyreducethestiffnessandulti-matestrengthofCFSTslenderbeam–columns.Therefore,itisofpracticalimportancetoaccountfortheeffectsofpreloadsonthesteeltubesinthenon-linearanalysisanddesignofCFSTslenderbeam–columnsinmultistoreycompositeframes.

No experiments have been conducted on biaxially loaded rectangular CFST slenderbeam–columnsconsideringpreloadeffects.Onlylimitedtestsonthebehaviourofuniaxi-ally loaded CFST columns with preload effects have been undertaken in the past (Hanand Yao 2003; Xiong and Zha 2007; Liew and Xiong 2009). Test results indicate thatthepreloadonthesteeltubemightreducetheultimateaxialstrengthoftheCFSTslenderbeam–columnby15%ifthepreloadwasgreaterthan60%oftheultimateaxialstrengthofthehollowsteeltube.ThestrengthandbehaviourofshortCFSTcolumnsarenotaffectedbypreloads.FiniteelementanalysesofcircularCFSTcolumnswithpreloadeffectswereperformedbyXiongandZha (2007)andLiewandXiong (2009).FibreelementmodelsweredevelopedbyPateletal.(2013,2014a)forsimulatingtheload–deflectionbehaviourofcircularandrectangularCFSTslenderbeam–columnsunderuniaxialandbiaxialbendingaccountingfortheeffectsofpreloads.

10.5.5.2 Non-linear analysis of CFST columns with preload effects

Thepreloadsonthesteeltubeinduceinitialstressesanddeflections inthesteeltube.Themid-heightdeflectionofahollowsteeltubeunderthepreloadcanbedeterminedbyperformingaload–deflectionanalysisbasedontheloadcontrolmethod(Patelet al.2013,2014a).Thedeflectionatthemid-height(umo)ofthesteeltubecausedbythepre-loadistreatedasanadditionalgeometricimperfectioninthenon-linearanalysisoftheCFSTslenderbeam–columnusingthedeflectioncontrolmethod.The load–deflectionresponsesofCFST slenderbeam–columnswithpreload effects canbedeterminedbyusing the load–deflection analysis procedure given in Section10.4 (Patel et al. 2013,2014a).

10.5.5.3 Axially loaded CFST columns

TheultimateaxialstrengthofCFSTcolumnsunderaxialcompressionisafunctionofthepreloadratio(βa),relativeslenderness( )λ andgeometricimperfections.BasedontheresultsoffibreelementanalysesconsideringgeometricimperfectionsofL/1000atthemid-heightofrectangularCFSTcolumnsandEurocode4(2004),adesignmodelfordeterminingtheultimateaxialstrengthsofconcentricallyloadedCFSTslendercolumnswithpreloadeffectsisgivenbyPateletal.(2014a)asfollows:

P Pup prg o= χ (10.94)

wherePoistheultimateaxialstrengthofthecolumnsectionunderaxialcompression,takenasP f A f Ao c c y s= ′ +0 85. .


Thecolumnstrengthreductionfactorχprgaccountsfortheeffectsofpreloadratio,relativeslendernessandgeometricimperfectionsontheultimatestrengthofCFSTslendercolumnsunderaxialcompressionandisgivenbyPateletal.(2014a)asfollows:

χφ φ λ

prg

prg prg

=+ −

12 2

(10.95)

φ

α β λ ζ λprg

prg a=

+ +( ) − + +( ) 1 1 1 1 0 05 1

2

2. .

(10.96)

α β

β β ζprg

a

a a

= −

−( ) − −( ) − +

1 2

11 5 1 2 4 3 1 2 1 5 602

.

. . . . . (10.97)

ζβ

β β=

≤− < ≤

0 0 4

0 44

0 8

for

for 0.4

a

aa

.

..

(10.98)

wheretherelativeslendernessλiscalculatedusingPoinEquation10.57.

10.5.5.4 Behaviour of CFST beam–columns with preload effects

NumericalstudiesperformedbyPateletal.(2013,2014a)demonstratethatincreasingthepreloadratiodecreasestheultimateaxial load,bendingstrengthandflexuralstiffnessofCFSTslenderbeam–columns.ThereductionintheultimatestrengthsofCFSTcolumnsduetopreloadeffectsisfoundtoincreasewithanincreaseinthecolumnslendernessratio(L/r).Thepreloadwitharatioof0.6mayreducetheultimateaxialstrengthoftheCFSTcolumnwithanL/rratioof100by15.8%.However,thepreloadonlyhasaminoreffectonCFSTshortbeam–columnswithanL/rratiooflessthan22orCFSTslenderbeam–columnswithsmallpreloadratios,andthusitseffectcanbeignoredinthedesign.Thestrengthreductionduetopreloadeffectsisshowntoincreasewithanincreaseintheloadingeccentricityratio(e/D)from0.0to0.4.Whene/D>0.4,however,thestrengthreductiontendstodecreasewithanincreaseinthee/Dratio.ItisinterestingtonotethatthereductionintheultimateaxialstrengthofCFSTcolumnsduetopreloadeffectsismaximizedwhenthee/Dratioisequalto0.4.Itwouldappearthatthepreloadhasmorepronounceeffectsonhigh-strengthCFSTslenderbeam–columnsthanonnormalstrengthones.Thepreloadhavingapreloadratioof0.8mayreducetheultimateaxialstrengthofthehigh-strengthcircularCFSTslen-derbeam–columnwithyieldsteelstrengthof690MPaby17.3%.

10.5.6 composite columns under cyclic loading

10.5.6.1 General

Inseismicregions,thin-walledrectangularCFSTslenderbeam–columnsmaybesubjectedtoaconstantaxialloadfromupperfloorsandcyclicallyvaryinglateralloadingduetotheearthquake.TheseCFSTbeam–columnsmayundergocyclic localandglobal interactionbuckling,whichreducestheirstrength,flexuralstiffnessandductility.Experimentsonnor-malandhigh-strengthrectangularCFSTbeam–columnsunderaxialloadandcycliclateralloadinghavebeenundertakenbyresearchers(Varmaetal.2002,2004;Hanetal.2003).High-strengthconcreteupto110MPaandhigh-strengthsteeltubeswithyieldstressupto


660MPawereusedtoconstructCFSTcolumns.ThefailuremodesassociatedwiththeseCFSTcolumnswerecrackingoftheconcretecoreandlocalbucklingofthesteeltubes.TheoutwardlocalbucklingofsomeCFSTbeam–columnswasobservedaftersteelyielding.

Numerical models have been developed to predict the cyclic responses of rectangularCFSTbeam–columns consideringor ignoring localbuckling effects (Varma et al. 2002;Gayathrietal.2004a,b;Chungetal.2007;ZubydanandElSabbagh2011).Someofthesemodelsapproximatelyaccountforlocalbuckleeffectsbymodifyingthestress–straincurveforsteelincompression.However,thismethodcannotsimulatetheprogressivecycliclocalbucklingofthesteeltubefromtheonsettothepost-localbuckling.Ithasbeenfoundthatthe modified stress–strain curve method might overestimate or underestimate the cycliclocalbucklingstrengthsofsteeltubesunderstressgradients(Pateletal.2014b).Pateletal.(2014b)developedafibreelementmodelforsimulatingthecycliclocalandglobalinterac-tionbucklingbehaviourofrectangularCFSTslenderbeam–columnsunderconstantaxialloadandcyclicallyvaryinglateralloading,whichisintroducedinthefollowingsections.

10.5.6.2 Cyclic material models for concrete

Thecyclicstress–straincurves forconcrete inCFSTcolumnsareshowninFigure10.29(Pateletal.2014b).Thestiffnessdegradationandcrackopeningandclosingcharacteristicsoftheconcreteundercyclicloadingaretakenintoaccountinthecyclicmaterialconstitu-tivemodel.Theenvelopecurveforconcreteundercyclicaxialcompressionisdefinedbythemonotonicstress–straincurvegiveninSection10.3.

Theconcreteundercompression is initially loadeduptoanunloadingstrainandthenunloadedtoazerostresslevel.Alinearstress–strainrelationshipisassumedfortheconcretereloadingfromthezerostressuptotheenvelopecurve.Theparabolicstress–straincurvefortheconcreteunloadingasdepictedinFigure10.29isgivenbyManderetal.(1988)as

σ σσ λ ε ε

ε ε

λ ε εε ε

λc un

un uc un

pl un

uc un

pl un

u= −

−−

− + −−

1

ε ε εpl c un< <( ) (10.99)

εtu

A B

C D

εB

(σun, εun)

(σre, εre)

εcp εcuεc

εctεun εp1

βce f ce

f ce

fct

fro

σc

σun

ε ce0

1

23

4

5

6

Figure 10.29 Cyclic stress–strain curves for concrete in rectangular CFST columns.


λσ

ε ε

uun

unun

un pl

E

E

=−

−

(10.100)

whereσundenotesthecompressivestressofconcreteattheunloadingεunrepresentsthestrainatσun

εplistheplasticstrainwhichiscalculatedby(Manderetal.1988)

ε ε σ ε σ ε

σ εpl unun un un a

un c aE= − +

+ (10.101)

whereε ε εa c un cea= ′ andacistakenasthelargerof ′ ′ +ε ε εce ce un/( )and0.09εun/εccforarect-angularcrosssection.

InEquation10.100,Eunistheinitialmodulusofelasticityofconcreteattheunloadingandiswrittenas

E

fEun

un

ce

ce

unc=

′

′

σ εε

(10.102)

where( ) .σun cef/ ′ ≥ 1 0and ′ ≤ε εce un/ 1 0. .Thelinearstress–strainrelationshipforconcreteatreloadingisdefinedby

σ σ

ε εε ε ε ε εc

ro re

ro rec ro ro pl c ro

ff= −

−

− + < <( ) ( ) (10.103)

wherefroistheconcretestressatthereloadingεroisthestrainatfroεreandσrearethereturnstrainandstressonthemonotoniccurveasshowninFigure10.29

Thestress–straincurveforconcreteintensionisalsogiveninFigure10.29.Itisassumedthattheconcretetensilestressincreaseslinearlyuptocrackingandthendecreaseslinearlytozeroattheultimatetensilestrain.Thetensilestrengthofconcrete istakenas0 6. ,′fce whiletheultimatetensilestrain isassumedtobe10timesofthestrainatcracking.Thetensilestressintheconcreteforunloadingfromthecompressiveenvelopeisdeterminedby

σ

ε εε ε

ε ε ε

ε εε

t

ct c pl

ct tutu c ct

ct c pl

ct

f

f=

′ −′ − ′( )

′ < ≤ ′

′ −′

( )

( )

for

foor ′ < <

ε ε εct c pl

(10.104)

′ = −

′

f fct ct

pl1εεce

(10.105)

′ = +ε ε εct ct pl (10.106)


Thetensilestrengthofconcreteisassumedtodecreasewithanincreaseinthecycles.Thisimpliesthatiftheprevioustensionloadingwentalongthepath1–2–3–4–5,thecurrentten-sionloadingwillfollowthe5–6pathasillustratedinFigure10.29.

10.5.6.3 Cyclic material models for structural steels

Figure10.30depictsthecyclicstress–strainmodelforthestructuralsteels.Itisnotedthatthestress–straincurveatunloadingfollowsastraightlinewiththesameslopeastheinitialstiffness,whichisexpressedby

σ ε ε ε ε εs s s mo o s moE= − < ≤( ) ( ) (10.107)

whereεmo=εso−fso/Es,εsoisthestrainattheunloadingandfsoisthestressattheunloading.Thestress–straincurveforstructuralsteelsatreloadingisgivenbyShietal.(2012)as

follows:

σ ε ε η ε ε ε ε εs s s mo s k s so mo s bE E E= − − − − < ≤( ) ( )( ) ( ) (10.108)

Ek

b

b mo

=−σ

ε ε (10.109)

ηε ε ε ε

ε ε=

−− − +

− ≥

−

1 0480 05

0 050 04

1 0740 08

..

( ) / ( ) ..

..

s so b sob sofor

(( ) / ( ) ..

ε ε ε εε ε

s so b soso− − +

− <

0 08

0 04for b

(10.110)

εst

εmo–εmo

–εsu –εy –0.9εy0 0.9 εy

0.9 fyb

0.9 fy

σs

εsεsu

fy

fy

fsu

fsu

Figure 10.30 Cyclic stress–strain curves for structural steels.


TheinitialvalueofthestrainεbatreloadingasindicatedbypointBinFigure10.30istakenas0.9εy.Thestressσbatthestrainεbisdeterminedfromthemonotonicstress–straincurves.Ifthestrainisgreaterthanεb,thesteelstressisdeterminedfromthecyclicskeletoncurve.Afterinitialreloading,thereloadingisdirectedtowardsthepreviousunloading.

10.5.6.4 Modelling of cyclic load–deflection responses

Afibreelementmodelwasdevelopedforcantilevercolumnsunderconstantaxialload(Pa)andcyclicallyvaryinglateralloading(F)asillustratedinFigure10.31.Theeffectivelengthof the cantilever column is takenas2L.Thedeflected shapeof the cantilever column isassumedtobepartofasinewave.Thecurvatureatthebaseofthecantilevercolumncanbedeterminedfromthedisplacementfunctionas

φ πb t

Lu=

2

2

(10.111)

whereutisthelateraldeflectionatthetipofthecolumn.Theexternalmomentatthebaseofthecantilevercolumniscalculatedby

M FL P e u ume a t to= + + +( ) (10.112)

whereeistheeccentricityoftheaxialloadandistakenaszeroforthecolumnunderconcen-

tricaxialloadutoistheinitialgeometricimperfectionatthetipofthecantilevercolumn

Pa

F F

L

ut

Pa

Figure 10.31 Cantilever column under constant axial load and cyclically varying lateral loading.


Theequilibriumconditionsforthecantilevercolumnareexpressedby

P Pa − = 0 (10.113)

FL P e u u Ma t to+ + + − =( ) 0 (10.114)

wherePandMaretheinternalforceandbendingmomentinthecompositesection.ThelateralloadcanbeobtainedfromEquation10.114as

F

M P e u uL

a t to= − + +( ) (10.115)

Inthecyclicload–deflectionanalysis,thelateraldeflectionatthetipofthecantilevercol-umnisgraduallyincreaseduptothepredefinedunloadingdeflectionandthendecreasedtothereloadinglevel.ThecomputationalalgorithmsbasedonMüller’smethod(Liangetal.2012;Pateletal.2012a)areusedtoadjusttheneutralaxisdepthtomaintaintheforceequilibriuminthecompositesection.ThelateralloadFatthetipofthecantilevercolumniscomputedfromthemomentequilibriumstate.Thestress–strainhistoriesofthecom-positesectionunderpreviouscyclicloadingarestoredinordertodeterminethecurrentstatesof stresses.Byrepeating theaforementionedanalysisprocess, thecompletecyclicload–deflectioncurvescanbeobtained.

Thecomputationalprocedureforsimulatingthecyclicload–deflectionresponsesofCFSTbeam–columnsisgivenasfollows(Pateletal.2014b):

1.Inputdata. 2.Discretisethecompositesectionintofibreelements. 3.Initialisethefirstunloadingdeflectionuut. 4.Initialisethelateraldeflectionatthetipofthecolumnut=Δut. 5.Calculatethecurvatureϕbatthebaseofthecolumn. 6.Ifut>(uut−Δut)orut<(−uut−Δut),thenΔut=−Δut. 7.If(ut−ulast)(ulast−uold)<0andut>ulast,setthenextunloadingdeflectionuut. 8.Recalltheunloadingstrainsandstressesattheunloadingdeflection. 9.Adjusttheneutralaxisdepth(dn)usingMüller’smethod. 10.ComputetheresultantforcePconsideringlocalbucklingeffects. 11.RepeatSteps9–10until|rp|<εk. 12.CalculatethecycliclateralforceFfromthemomentequilibrium. 13.Recordthedeflectionsuold=ulastandulast=ut. 14.Storethefibrestrainsandfibrestressesunderthecurrentdeflection. 15.Increasethedeflectionatthetipofthecantilevercolumnbyut=ut+Δut. 16.RepeatSteps5–16untilF<0.85Fmaxorut>u*.

Thetypicalcycliclateralload–deflectioncurvesforarectangularCFSTcantilevercolumnpredictedusingtheprecedingcomputationalprocedureareshowninFigure10.32.


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oad F

(kN

)

–120–1000

–800

–600

–400

0

–200

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1000

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377

Chapter 11

composite connections

11.1 IntroductIon

Composite connections are used to transfer forces between composite members and tomaintain the integrityofa composite structureunderapplied loads.Thebehaviourofacompositeconnectionischaracterisedbyitsmoment–rotationcurve,whichexpressesthemomentasafunctionoftheanglebetweenthebeamandthecolumn.Therotationalstiff-nessofacompositeconnectionisdeterminedbytheslopeof itsmoment–rotationcurve.Compositeconnectionsareclassifiedassimple,rigidandsemi-rigidconnectionsbasedonthestiffnesscriteriaandasfull-strengthandpartial-strengthconnectionsbasedonthestrengthcriteria.Thetypesofcompositeconnectionsincludethebaseplateconnectionsofcompositecolumns,compositecolumnsplices,beam-to-columnshearconnections,beam-to-columnmomentconnectionsandsemi-rigidconnections(Viestetal.1997).Double-angleconnec-tions,single-plateshearconnectionsandteeshearconnectionsarebeam-to-columnshearconnections.Thebeam-to-columnmomentconnectionsincludecompositeconnectionswithsteelbeampassingthroughconcrete-encasedcomposite(CEC)columns,reinforcedconcretecolumns,concrete-filledsteeltubular(CFST)columnsorsteelcolumns.

Composite connections in a composite frame are potential weak spots that must bedesignedforalargermarginofsafetythantheconnectingmembers.Ingeneral,compositeconnectionsmustsatisfythestrength,serviceabilityandconstructioncriteria.Thestrengthcriteriarequirethatcompositeconnectionsmustbedesignedtoresistaxialforce,bendingmoment,shearandtorsionarisingfromtheappliedloads.Theserviceabilitycriteriarequirethatthedesignofcompositeconnectionsmustensurethatthejointrotationinmomentcon-nectionsunderserviceloadsdoesnotleadtoexcessivedeflections,crackingordistressinothermembersinthecompositestructure.Theconstructioncriteriaforcompositeconnec-tionsrequiresimpleandrapidfabricationandconstruction.

ThischapterpresentsthebehaviouranddesignofcompositeconnectionsinaccordancewiththeAISC-LRFDManual(1994).Thedesignofsingle-plateshearconnections,teeshearconnections,beam-to-CECcolumnmomentconnections,beam-to-CFSTcolumnmomentconnectionsandsemi-rigidconnectionsisintroduced.Thedesignofsingle-angleanddou-ble-angleshearconnectionsasshowninFigure11.1isgivenintheAISC-LRFDManual(1994)andbyGong(2008,2009,2013).

11.2 SIngle-PlAte SheAr connectIonS

Single-plateshearconnectionsasdepictedinFigure11.2areusedtotransfertheendreaction(shear)ofsimplysupportedsteelorcompositebeamstothesteelorcompositecolumns.Thesinglesteelplateisusuallyshopweldedtothecolumnandfiledboltedtothewebofthesteelbeam.


Thefabricationanderectionofsingle-plateconnectionsareeasyandsimple.Thistypeofconnec-tionsisusedinbothsteelandcompositeconstruction.Theeffectsofthecompositeslabortheslabreinforcementontheperformanceofsingle-plateconnectionsarenotconsideredinthedesign.

11.2.1 Behaviour of single-plate connections

The behaviour of single-plate shear connections is characterised by their shear–rotationcurveswhichexpress theshear forceasa functionof theendrotationof thebeam.The

Column

Angle

Steel beam

Concrete slab

Figure 11.1 Bolted double-angle shear connection.

Steel beam

Single plate

Concrete slab

Column

Figure 11.2 Single-plate shear connection.

Composite connections 379

shear–rotationcurvesforsingle-plateconnectionscanbedeterminedbyexperiments.TestresultspresentedbyAstanehetal. (1989)andAstaneh-Asletal. (1993) indicatethatthesingleplateyieldedinshearsothatinelasticsheardeformationsdevelopedintheconnec-tions.Theconnectionstestedfailedbyshearfractureoftheboltswhichconnectedthesingleplatestothewebofthesteelbeams.Otherfailuremodesassociatedwithsingle-platecon-nectionswereobservedfromexperiments,includingbearingfailureofboltholes,fractureofthenetsectionofthesingleplate,fractureoftheplateedgeandfractureofthewelds.Theplateshearyieldingandbearingyieldingofboltholeswerefoundtobeductile,whileotherfailuremodeswerebrittle.

11.2.2 design requirements

Single-plate shear connections must be designed to satisfy the following requirements:(1)havingsufficientstrengthtotransfertheshearforcefromthebeamreaction,(2)havingsufficientrotationcapacitytomeetthedemandofasimplysupportedbeamand(3)thecon-nectionshouldbeflexiblesothatthebeamendmomentsarenegligible.

Thedesignofsingle-plateshearconnectionsshouldsatisfythefollowinggeometricandmaterialrequirements(Astaneh-Asletal.1993;Viestetal.1997):

1.Theconnectionshouldhaveonlyoneverticalrowofbolts,havingthenumberofboltswithintherangeof2and9.

2.Theboltspacingis76 mm. 3.Theedgedistanceisae≥1.5df. 4.Thedistancefromtheboltcentrelinetoweldlineisabw≥76mm. 5.Thesingleplateshouldbemadeofmildsteel. 6.E41XXorE48XXfilletweldsshouldbeused. 7.Thethicknessofthesingleplateistp≥0.5df+1.6mm. 8.Theratiodp/abw≥2,wheredpisthedepthofthesingleplate. 9.M20orM24high-strengthstructuralboltsshouldbeused.


Boltsinsingle-plateconnectionsaresubjectedtoshearforceandbendingmomentwhichiscausedbytheeccentricityofthebeamendreactionfromtheboltline.Therefore,boltsaredesignedforcombinedshearandbending.Theeccentricity(eb)ofthereactionfortheplateweldedtoarigidsupportingelementiscomputedby

e n ab b bw= − −25 4 1. ( ) (11.1)

Forthesingleplateweldedtoaflexiblesupportingelement,theeccentricity(eb)ofthereac-tiontotheboltcentrelineistakenasthelargervalueobtainedfromEquation11.1andabw.

ThemomentcanbedeterminedasM V eb b∗ = ∗ .Thedesignofboltsforcombinedactions

ofshearandmomentisgivenintheAISC-LRFDManual(1994)andinSection6.4.6.IntheAISC-LRFDManual,thedesignshearcapacityofaboltgroupundereccentricloadingisdeterminedby

φ φV C Vfb f= ( ) (11.2)

whereCisthecoefficientaccountingfortheeffectofeccentricloadingonthedesignshearcapacityoftheboltgroup,whichisgiveninTable8.18intheAISC-LRFDManual(1994).


Itisafunctionoftheeccentricityoftheloading,spacingofboltsandnumberofboltsinoneverticalrowintheconnection.

11.2.4 design of single plate

Therealsteelplateinasingle-plateshearconnectionisactuallysubjectedtoalargeshearforceandasmallbendingmoment.However,thesteelplateisdesignedtoyieldundertheshearforceonly.Thisistofacilitatetheearlyyieldingoftheplatemadeofmildsteel.Theshearyieldcapacityoftheplateisgivenby

φ φV f Au y p= 0 6. (11.3)

whereφ = 0 9.Apisthecross-sectionalareaofthesingleplate,takenasA d tp p p=

Experimentsshowthattheshearfractureofthenetsectionoccursalongaverticalplaneclosetotheedgeoftheboltholesratherthanalongthecentrelineoftheboltholes(AstanehandNader1990).Thedesignshearfracturecapacityofthenetsectionisdeterminedby

φ φV f Ans u n= 0 6. (11.4)

whereφ = 0 75. isthecapacityreductionfactorfuisthetensilestrengthoftheplateAnisthenetcross-sectionalareaoftheshearplanepassingthroughthecentrelineofthe

bolts,whichisgivenby(AISC-LRFDManual1994)

A A n d tn g b f p= − +( . )1 5 (11.5)

wherenbisthenumberofboltsdf isthediameterofthebolt

Thebearingcapacityoftheplateinshearisdeterminedby(AISC-LRFDManual1994)

φ φR C f d tb u f p= ( . )2 4 (11.6)

Topreventtheedgefailure,theverticalandhorizontaledgedistances(ae)mustnotbelessthan1.5dfandtheverticaldistanceaemustnotbelessthan38 mmregardlessoftheboltdiameters.Localbucklingofthebottomportionofthesingleplatemayoccur.Thedepth-to-thicknesslimitontheplateistakenasdp/tp≤64topreventlocalbucklingfromoccurring.

11.2.5 design of welds

Theweldsinsingle-plateshearconnectionsaredesignedforthecombinedactionsofshearandbendingmoment.Thebendingmomentiscausedbytheeccentricity(ew)ofthebeam


reactionandiscalculatedasM V ew w∗ = ∗ ,whereewistakenasthelargervalueof25.4nband

abw.Theweldsaredesignedtoyieldafteryieldingoftheplatetopreventthebrittlefailureofthewelds.Thisimpliesthattheweldisstrongerthantheplateinasingle-plateshearcon-nection.Toensurethis,theshear–momentinteractioncurvefortheplateshouldlieinsidetheshear–momentcurveforthewelds.Theweldsizecanbederivedfromthisconditionas(Astaneh-Asletal.1993)

D

ff

twy

uwp>

1 41. (11.7)

wherefuwisthetensilestrengthoftheweldmetal.Theweldsizesatisfyingtheaforemen-tionedconditionwillensurethattheplatefailurewilloccurbeforetheweldfails.

Example 11.1: Design of single-plate shear connection

Designasingle-plateshearconnectionwhichconnectsacompositebeamtotheflangeofasteelcolumn.Thereactionofthecompositebeamiscomposedofanominaldeadloadof200kNandanominalliveloadof180kN.Thesteelbeamsectionis610UB125ofGrade300steel(tw=11.9mm).M208.8/Shigh-strengthstructuralboltsareusedwithaspacingof76 mm.

1. Design of bolts

ThedesignshearforceisV∗=1.2G+1.5Q=1.2×200+1.5×180=510kN.TheshearcapacityofanM20boltisϕVf=92.6 kN.Therequirednumberofboltsis

n

VV

bf

=∗= =

φ51092 6

5 51.

. Try 6 bolts

Theflangeofthesupportingsteelcolumnisconsideredasrigid.Theeccentricityofthereactionis

e n a eb b bw x= − − = × − − = =25 4 1 25 4 6 1 76 51. ( ) . ( ) mm

FromTable8.18inVol.IIoftheAISC-LRFDManual,thecoefficientCisobtainedasC=5.45.

Thedesignshearstrengthoftheboltgroupis

φ φV C V Vfb f= = × = < ∗ =( ) . . .5 45 92 6 504 67 510kN kN, N.G.

Try7bolts;thedesignshearstrengthoftheboltgroupisdeterminedasfollows:

e n a eb b bw x= − − = × − − = =25 4 1 25 4 7 1 76 76 4. ( ) . ( ) . mm

C=6.06

φ φV C V Vfb f= = × = > ∗ =( ) . . .6 06 92 6 561 2 510kN kN, OK


2. Design of the single plate

Thedepthofthesingleplateis

d s np b b= = × =76 7 532mm

Theshearfracturecapacityofthenetsectioniscalculatedby

φ φV f d n d tns u p b b p= − +( . )[ ( . )]0 6 1 6

Therequiredthicknessoftheplateistherefore

t

Vf d n d

pns

u p b b

≥− × +

=×

× × × −φ

φ( . )[ ( . )] . . [0 6 1 60510 10

0 75 0 6 430 532

3

77 20 1 66 9

× +=

( . )]. mm

Trytp=8.5mm;checktheplatethicknesslimitasfollows:

t dp b≤ + = × + =0 5 1 6 0 5 20 1 6 11 6. . . . . mm, OK

t

dp

p> = =64

52564

8 2. mm, OK

Theshearyieldcapacityoftheplateis

φ φV f A Vu y p= = × × × × = > ∗ =( . ) . . . .0 6 0 9 0 6 300 532 8 5 732 6 510N kN kN, OK

Thebearingcapacityoftheplateiscomputedas

φ φR C f d t

V

b u f p= = × × × × ×

= > ∗ =

( . ) . . . .

.

2 4 6 06 0 75 2 4 430 20 8 5

797 4 510

N

kN kkN, OK

Sincethebeamwebisthickerthanthesingleplate,itisnotrequiredtocheckthebearingstrengthofthebeamweb.

3. Design of fillet welds

Thesizeofthefilletweldisdeterminedas

D

ff

twy

uwp=

= ×

× =1 41 1 41

300480

8 5 7 5. . . . mm

Use8 mmE48XXfilletweldsonbothsidesoftheplate.

11.3 tee SheAr connectIonS

Teeshearconnectionsareusedtotransfertheendshearreactionofsimplysupportedsteelorcompositebeamstothesteelorcompositecolumns.Ateeconnectionisconstructedbyconnectingittoasteelbeamwebandtoacolumn.Theteecanbecutfromawideflangeorfabricatedbyweldingtwoplates.Eitherboltsorweldscanbeusedasfastenersinteecon-nections.Therearefourcommontypesofteeshearconnectionswhichareusedinbothsteelandcompositestructuresdependingontheuseoffasteners.Theteeshearconnectionshown


inFigure11.3,wheretheteestemisboltedtothesteelbeamwebandtheteeflangeisweldedtothecolumnflange,isconsideredhere.Theeffectsofcompositeslabsortheslabreinforce-mentonthestrengthandbehaviourofteeshearconnectionsarenotconsideredinthedesign.

11.3.1 Behaviour of tee shear connections

The behaviour of tee shear connections is characterised by their shear–rotation curveswhichexpress theshear forceasa functionof theendrotationof thebeam.Theshear–rotationcurvesforteeshearconnectionscanbedeterminedbyexperiments.Testsconductedby AstanehandNader (1989,1990)demonstrate thatall specimensunder themoment–rotationtestsexperiencedminoryielding.Whentherotationreached0.06rad,theweldsfractured.Themoment–rotationresponsesindicatethatteeshearconnectionswereflexibleandductilesothattheycouldberegardedassimpleconnections.Sixfailuremodeswereobservedfromtheshear–rotationtests(AstanehandNader1989,1990), includingshearyieldingoftheteestem,yieldingoftheteeflange,bearingfailureofthesteelbeamwebandtheteestem,shearfractureofthenetsectionoftheteestem,shearfractureoftheboltsandfractureofwelds.Thefailuremodesofyieldingofthesteelstemandflangeareductile.However,theboltandweldfractureresultsinbrittlefailuremodeoftheteeconnections.


Bolts in tee shearconnections shouldbedesigned fordirect shear.When the supportingelementisrigid,theeccentricity(eb)ofthereactiontotheboltlineissosmallthatitcanbeignored.Theflangeofacolumnoranembeddedsteelplateisconsideredasrigidsupport-ingelement.Forthiscase,ebistakenaszero.Whenthesupportingelementisrotationallyflexible,theinflectionpointisassumedtobelocatedattheweldline.Asaresult,theboltsaresubjectedtoshearforce(V∗)andabendingmomentwhichisequaltoV∗eb,wheretheeccentricityebistakenasabw.Theboltsarethereforedesignedforcombinedshearandbend-ing.Forthispurpose,Table9.10intheAISC-LRFDManual(1994)canbeused.

Column

Concrete slab

Tee section

Steel beam

Figure 11.3 Tee shear connection.


11.3.3 design of tee stems

Theteesteminactualteeshearconnectionsissubjectedtoashearforceandasmallbendingmoment.Thesmallbendingmomentisnotconsideredinthedesignoftheteesteminteeconnectionsanditisdesignedforshearforceonly.Thenominalshearyieldcapacityoftheteestemisexpressedby

V f d tts y t ts= 0 6. (11.8)

wheredtisthedepthoftheteesectionttsisthethicknessoftheteestem

Theshearfracturefailureofthenetsectionoftheteesteminteeshearconnectionsissimilar to thatof thesingleplate insingle-plateshearconnections.Experiments indicatethatthefracturefailureoftheteesteminshearoccursatthenetsectionalongtheedgeofboltholesratherthanalongtheircentreline.TheshearfracturecapacityoftheteestemcanbecalculatedusingEquation11.4.However,theeffectivenetareainsheariscalculatedusingtheaverageofthenetareaalongtheboltcentrelineandthegrossareaoftheteestemasfollows(AstanehandNader1990):

A A n d tn g b f ts= − +0 5 1 5. ( . ) (11.9)

Thedesignbearingcapacityoftheteesteminshearisgivenby(AISC-LRFDManual1994)

φ φR n f d tb b u f ts= ( . )2 4 (11.10)

whereϕ=0.75isthecapacityreductionfactor.Forthesteelbeamweb,theearlierequationcanbeusedtocalculateitsdesignbearingcapacitybysubstitutingttsbytw.

11.3.4 design of tee flanges

Ifthethicknessoftheteeflangeislessthanthethicknessoftheteesteminateeshearcon-nection,theteeflangewillyieldbeforetheteestem.Thenominalshearyieldcapacityoftheteeflangeisdeterminedby

V f d ttf y t f= 2 0 6( . ) (11.11)

wheretfisthethicknessoftheteeflange.

11.3.5 design of welds

Inteeshearconnections,filletweldsareusedtoconnecttheteeflangetothesupportingelementsuchastheflangeofacolumnasdepictedinFigure11.3.Theweldsaresubjectedtoshearforceandbendingmomentcausedbytheeccentricityewofthebeamreactionfromtheweldline.Theeccentricityewcanbeconservativelytakenasthedistancebetweenthe


boltandweldlines,suchasew=abw.UsingTable8.38intheAISC-LRFDManual(1994),thedesignstrengthofeccentricallyloadedweldgroupundershearforceV∗andbendingmomentofV∗ewcanbedeterminedby

φR CC D Lw w= 1 16 (11.12)

whereC is thecoefficient includingthecapacityfactorϕ=0.75,given inTable8.38 inthe

AISC-LRFDManual(1994)C1istheelectrodestrengthcoefficientgiveninTable8.37D16isthenumberof16thofaninchintheweldsizeLwistheweldlength

11.3.6 detailing requirements

Thedesignmethodforteeshearconnectionsdescribedintheprecedingsectionswasdevel-opedbasedonlimitedtestresults.Theteeshearconnectionsdesignedusingthismethodarerestrictedtosomegeometricandmaterialrequirementsdescribedinthissection(AstanehandNader1990).

Topreventthelocalbucklingofthelowerhalfoftheteestemincompression,theratioofdt/abwoftheteestemshouldbegreaterthan2.Thewidth-to-thicknessratio(0.5bf /tf)oftheteeflangeoutstandshouldbegreaterthan6.5toensuretheflexibilityoftheconnection.Thedepth-to-widthratio(dt/bf)oftheteesectionshouldnotexceed3.5topreventlargeinelastictensilestrainfromdevelopinginthewelds.Toincreaseconnectionductility,theratioof(tts/df)/(tf/tts)shouldbelessthan0.25.

Theteesectionshouldbemadeofmildsteeltoensuregoodshearandrotationalductility.M20orM24high-strengthstructuralboltsmaybeusedinonlyoneverticalrow.Snug-tightboltsarepreferred.Theverticalspacingofboltsshouldbeequalto76 mm.Thenumberofboltsshouldnotbelessthan2andmorethan9.Filletweldsshouldbeusedtoweldtheteeflangetothesupportingelement.Thetopofthefilletweldsshouldbereturnedadistanceof2Dw.Iftheteeflangeisweldedtotheflangeofasteelcolumn,thethicknessofthecolumnflangeshouldbegreaterthanthatoftheteeflange.

Example 11.2: Design of tee shear connection

Designateeshearconnectionwhichconnectsacompositebeamtotheflangeofasteelcolumn.Thereactionofthecompositebeamunderfactoreddesignloadsis300kN.Thesteelbeamsectionis410UB59.7ofGrade300steelandthesteelcolumnis250UCofGrade300steel.TheM208.8/Sboltsareusedwithaspacingof76 mm.TheE48XXfilletweldsareused.

1. Design of bolts

TheshearcapacityofanM20boltisϕVf=92.6kN.Therequirednumberofboltsis

n

VV

bf

=∗= =

φ30092 6

3 24.

.

Adoptfourbolts.


2. Check geometric requirements of the tee section

Therequiredgrossareasoftheteestemcanbedeterminedfromitsshearyieldcapacityas

A

Vf

tsy

=∗

=××

=0 6

300 100 6 300

16673

. .mm2

Thedimensionsoftheteesectionareselectedasfollows:

b t t a sf f ts e g= = = = =170 13 8 35 76mm mm mm mm mm, , , ,

Thewidth-to-thicknessratiooftheteestemiscalculatedas

btf

f2170

2 13=

×= >6.54 6.5, OK

Theratioofdf/ttsis

dtf

ts

= = >208

2 5 2 0. . , OK

Theedgedistanceoftheteestemis

a de f= > = × =35 1 5 1 5 20 30mm mm, OK. .

Thedepthoftheteestemisdeterminedas

d n s at b g e= − + = − × + × =( ) ( )1 2 4 1 76 2 35 298mm

Thecross-sectionalareaoftheteestemiscomputedas

d t At ts ts= × = > =298 8 2384 1667mm mm OK2 2,

Thedepth-to-widthratiooftheteesectionis

dbt

f

= = <298170

1 75 3 5. . , OK

Thethicknessofthecolumnistfc=14.2mm>tf=13mm, OK.Thethicknessratiooftheteesectionis

t dt tts f

f ts

//

//

OK= = <8 2013 8

0 246 0 25. . ,

Thecleardepthofthesteelbeamwebisdw=406− 2×12.8=380.4mm>dt=298mm, OK.

3. Design strengths of the tee stem

Thenominalshearyieldcapacityoftheteestemiscomputedas

φ φV f d t Vts y t ts= = × × × × = > ∗ =( . ) . . .0 6 0 9 0 6 300 298 8 386 2 300N kN kN, OK

Thenetareaoftheteesteminsheariscalculatedas

A A n d tn g b f ts= − + = − × × + × =0 5 1 5 2384 0 5 4 20 1 5 8 2040. ( . ) . ( . ) mm2


Theshearfracturecapacityofthenetsectionoftheteestemis

φ φ φV f A Vns u ns ts= = × × × = > =( . ) . . . .0 6 0 75 0 6 430 2040 394 7 386 2N kN kN, OOK

Thebearingcapacityoftheteestemiscalculatedas

φ φ

φ

R n f d t

V

b b u f ts

ts

= = × × × × × =

=

( . ) . .2 4 4 0 75 2 4 430 20 8

386

N 495.36kN

> ..2kN, OK

Sincetw=8.5mm>tts=8mm,theteestemwillgovernthebearingfailure.Theshearcapacityoftheteeflangeis

φ φ

φ

V f d t

V

tf y t f

ts

= = × × × × × =

> =

2 0 6 0 9 2 0 6 300 298 13

386

( . ) . . N 1255.2kN

..2kN, OK

4. Design of fillet welds

Thefilletweldsaredesignedforcombinedshearandout-of-planebendingmoment.Theeccentricityisabw=76mm.Theeccentricityratiois

a

eLw

w

= = =76298

0 252.

Withk=0forout-of-planebending,thecoefficientCisobtainedfromTable8.38intheAISC-LRFDManualasC=2.48.

UsingC1=1.0forE48XXfilletwelds,therequiredweldsizein16thofaninchis

D

VCC L

ts

w16

1

429 4 44802 48 1 0 298 25 4

3 32= =× ×

=( / . )

. . ( / . ).

ThesizeoftheweldsisDw=(D16×25.4)/16=(3.32×25.4)/16=5.3mm.Use6 mmE48XXfilletweldsonbothsidesoftheteeflange.

11.4 BeAm-to-cec column moment connectIonS

Beam-to-column moment connections between steel or composite beams and reinforcedconcreteorcompositecolumnsareemployedinmoderate-tohigh-risecompositebuildings.Beam-to-columnmomentconnectionsareusedtotransfertheaxialforce,bendingmomentandshearforcearisingfromappliedloadsfromthebeamstothecompositecolumns.Asteelbeam-to-CECcolumnmomentconnectionisconstructedbypassingthesteelbeamthroughaCECcolumnas illustrated inFigure11.4.Face-bearingplates (FBPs)andverticalrein-forcementmaybeattachedtothesteelbeamtoresistbearingforces.Horizontalreinforcingtiesareprovidedinthecolumnwithinthebeamdepthandaboveandbelowthebeamtocarrytensionforcesdevelopedintheconnection.

Thedesignmethodforsteelbeam-to-CECcolumnmomentconnectionspresentedhereinisbasedon theworkofSheikhetal. (1989),Deierlein etal. (1989)and theASCETaskCommittee(1994).Itisapplicableonlytointeriorandexteriormomentconnectionsbetweensteelbeamsandreinforcedconcreteorcompositecolumns.Theeffectsofcompositeslabsortheslabreinforcementonthestrengthandbehaviourofcompositeconnectionsarenot


consideredinthedesign.Theaspectratiooftheconnectionislimitedto0.75≤D/Ds≤2.0.Themethod is limited tonormalweightconcretewith ′ ≤fc 40MPa, structural steelwithyieldstressoffy≤345MPaandreinforcingbarswithyieldstressoffyr≤410MPa.

11.4.1 Behaviour of composite moment connections

Thebehaviourof steelbeam-to-encased composite columnmoment connections is char-acterisedby twoprimary failuremodes,namely, thepanel shear failureand theverticalbearingfailure(Sheikhetal.1989).Inthecompositeconnection,bothstructuralsteelandreinforcedconcreteareinvolvedinthepanelshearfailure,whichissimilartothestructuralsteelorreinforcedconcreteconnection.Bearingfailureoccursattheupperandlowercor-nersoftheconnectionssubjectedtohighcompressivestresses.Theforcesintheconnectionare transferred by three shear mechanisms, which are the steel web panel, the concretecompressionstrutandconcretecompressionfield.Thesteelwebissubjectedtopureshearstressoveraneffectivepanellength.Theverticalstiffenerplatesattachedtothesteelbeam

CEC column

Steel beam

Concrete slab

CEC column

Face-bearing plate

Concrete slab

Steel column

Steel beam

Steel column

B

bp

bp

D

do

do

Ds

Face-bearing plate

Figure 11.4 Beam-to-CEC column moment connection.


mobilisetheconcretecompressionstrut.Theconcretecompressionfieldcomposingofsev-eralcompressionstrutswithhorizontalreinforcementformsastrut-and-tiesystemtocarrytheforcesintheconnection.

11.4.2 design actions

TheforcesactingonthecompositeconnectionareshowninFigure11.5.Theaxialforcesinthesteelbeamareusuallysmallsothattheyarenotconsideredinthestrengthcalculationoftheconnection.Thecompressiveaxialforcesinthecolumnarealsonottakenintoaccountasexperimentsindicatethatitisconservativetoneglecttheireffects.Ifthenettensionforcesexistintheconnection,theconcretecontributiontotheshearstrengthoftheconcretecom-pressionfieldshouldbeignored.Theforcesonacompositeconnectioncanbeexpressedby(ASCETaskCommittee1994)

M M V D V Dc b b c s∗ = ∗ + ∗ − ∗∑∑ (11.13)

where

M M Mc c c∗ = ∗ + ∗∑ 1 2 (11.14)

M M Mb b b∗ = ∗ + ∗∑ 1 2 (11.15)

V

V Vb

b b∗ =∗ + ∗( )1 2

2 (11.16)

V

V Vc

c c∗ =∗ + ∗( )1 2

2 (11.17)

M*c2

M*c1

M*b1 M*b2

V*c2

V*c1

V*b1

V*b2

Figure 11.5 Design actions on interior beam-to-CEC column moment connection.


11.4.3 effective width of connection

The effectivewidthof the composite connectionwithin a composite column is givenby(ASCETaskCommittee1994)

b b bj i o= + (11.18)

wherebi istheinnerpanelwidthwhichistakenasthelargerofthewidthoftheFBP( )bp and

thewidthofthebeamflange( )bfboistheouterpanelwidthasdepictedinFigure11.6

FortheextendedFBPsorsteelcolumns,bo isdeterminedbasedontheoverallcross-sec-tionalgeometryasfollows:

b h b b do xy i o= − <( )max 2 (11.19)

b

b Bb D bff fmax

( ).=

+< + <

21 75 (11.20)

B

CEC column

Steel beam

Concrete slab

Steel column

2bo

bi

bj

do

do

DsFace-bearing plate

Figure 11.6 Effective width of beam-to-CEC column moment connection.


h

xD

yb

xyf

=

(11.21)

whered Do s= 0 25. whenthecolumnisasteelcolumnordoistakenasthelesserof0 25. Dsand

theheightoftheextendedFBPswhentheseplatesareusedBisthecolumnwidthorientedperpendiculartothebeamDisthedepthofthecolumnyisthelargerofthesteelcolumnorextendedFBPwidthx=DwheretheextendedFBPsareusedorx=D/2+dc/2whenonlythesteelcolumn

isusedasillustratedinFigure11.7

11.4.4 vertical bearing capacity

The vertical bearing forces on the connection are the results of combined shears andmomentstransferredbetweenthebeamandcolumnasdepictedinFigure11.8,wherethemomentsMc1

∗andMc2∗ onthecolumnarerepresentedbythebearingforcesCcandtheforces

intheverticalreinforcementTvr(tension)andCvr(compression).Thelengthofthebearingzone(ac)aboveandbelowthebeamisassumedtobe0.3D.Thenominalconcretebearingstrengthisdeterminedby

C f b Dcb c j= ′2 0 3( . ) (11.22)

wherethebearingstress is takenas2 ′fc duetotheconcreteconfinementprovidedbythereinforcementandsurroundingconcrete.

Reinforcingbars,rodsorsteelanglescanbeattachedtothesteelbeamasverticalrein-forcementtocarryverticalbearingforcesintheconnection.However,itshouldbenotedthat providing a large amount of vertical reinforcement may induce high bearing stress

D

By

dc

x

bf

Face-bearingplate

Steel beam

Tie

Strut

Figure 11.7 Strut-and-tie model for horizontal force transfer.


ontheconcretebetweenthetwoflangesofthesteelbeam.Toavoidthis,thestrengthsoftheverticalreinforcementintension(Tvr)andincompression(Cvr)arelimited(ASCETaskCommittee1994)by

T C f b Dvr vr c j+ ≤ ′0 3. (11.23)

Replacing thevertical forceswith their respectivenominal strengthvalues and from themoment equilibrium, the following expression is obtained for the composite connectionsubjectedtoverticalbearing(ASCETaskCommittee1994):

M D V C D T C dc b cb vr vr vr∗ + ∗ ≤ + +[ ]∑ 0 35 0 7. ( . ) ( )∆ φ (11.24)

where∆V V Vb b b

∗ = ∗ − ∗2 1

dvristhedistancebetweenthebars

Theverticalreinforcementisassumedtocarrybothtensionandcompressionforcesorcom-pressiononly(Tvr=0).

11.4.5 horizontal shear capacity

Thehorizontalshearinasteelbeam-to-encasedcompositecolumnconnectionisresistedby three shear mechanisms, which consist of the steel web panel, the inner concretecompressionstrutandtheouterconcretecompressionfieldasillustratedinFigure11.8.

D

Strut

Rebar

V*c2

V*b1

V*b1 M*b2

V*c1

M*b1

Tvr

ac

Ds

Cvr

Cc do

do

Vcs

Vcs

Tvr

dvr

Cvr

Cc

Figure 11.8 Forces in beam-to-CEC column moment connection.


Thesteelwebpanelissubjectedtopureshearanditsstrengthisgovernedbyitsshearyieldcapacityasfollows:

V f L twp yw p w= 0 6. (11.25)

whereLpisthepanelwidth.Itwouldappearthattheconcretecompressionstrutisadiagonalcompressionmember

that forms within the inner panel width (bi) as shown in Figure 11.8. The compressionforceinthediagonalconcretestrutprovidesbearingstressontheFBPswithinthedepthofthesteelbeam.Thenominalstrengthofthecompressionstrutisgivenby(ASCETaskCommittee1994)

V f b D f b dcs c p c p w= ′ ≤ ′1 7 0 5. . (11.26)

where1 7. ′fc is the average limiting horizontal shear stress for concrete, the concretestrength ′fc isinMPaandtheeffectivewidthoftheFBPistakenasb b t bp f p f≤ + ≤5 1 5. .Thebearingfailureofconcreteattheendsofthestrutmayoccur.Topreventthis,thehorizontalshearislimitedbyamaximumbearingstressof2 ′fc actingonanareaofbp(0.25dw)atthetopandbottomoftheFBPs.

Thecompressionfieldsdevelopintheouterpanelwidth(bo).Thecompressionfieldsaremobilisedby thehorizontal strutsandcolumntieswhich formastrut-and-tie systembybearingagainstthesteelcolumnand/orextendedFBPsasshowninFigure11.7.Thenomi-nalstrengthoftheconcretecompressionfieldisgovernedbythestrengthoftheconcreteandthehorizontalcolumntiesandcanbecomputedby(ASCETaskCommittee1994)

V V V f b Dcf c s c o= ′ + ′ ≤ ′1 7. (11.27)

where ′Vc isthestrengthprovidedbytheconcreteincompression,whichisgivenby

′ = ′V f b Dc c o0 4. (11.28)

Ifthecolumnisintension, ′ =Vc 0.Thestrengthprovidedbythehorizontaltiesisdeterminedas

′ =V A f D ss sr yr sr0 9. / (11.29)

whereAsristhecross-sectionalareaofreinforcingbarsineachlayeroftiesspacedatssrinthedepthofthebeamwebandAsr≥0.004bssr.

Thehorizontalshearstrengthoftheconnectionisthesumoftheshearstrengthofthesteelwebpanel,theinnerconcretecompressionstrutandtheouterconcretecompressionfield.Theverticalshearintheconnectioncausedbyappliedloadsisequaltothetotalshearstrengthoftheconnection.Thehorizontalshearstrengthoftheconnectionmustsatisfythefollowingcondition(ASCETaskCommittee1994):

M V L V d V d V D dc b p wp fc cs w cf s o∗ − ∗ ≤ + + + ∑ φ ( . ) ( )0 75 (11.30)


wheredfcisthedistancebetweenthecentroidsofthebeamflange,andthepanelwidthLpiscalculatedasfollows:

LM

C T C VDp

c

c vr vr b

=∗

+ + − ∗ ≥∑φ( ) .

.0 5

0 7∆

(11.31)

C f b ac c j c= ′2 (11.32)

a

D DK Dc = − − ≤

2 40 3

2

. (11.33)

K

M V D T C d

f b

c b vr vr vr

c j

=∗ + ∗ − +

′∑ ∆ ( ) ( )

( )

/2

2

φ

φ (11.34)

11.4.6 detailing requirements

The detailing requirements on the steel beam-to-CEC column moment connections aregivenbytheASCETaskCommittee(1994)andarediscussedinthissection.

11.4.6.1 Horizontal column ties

HorizontalreinforcingtiesshouldbeprovidedinthecolumnwithinthedepthofthesteelbeamandaboveandbelowthebeamtosustaintensionforcesdevelopedintheconnectionasshowninFigure11.7.Horizontalreinforcingtieswithinthebeamdepthareusedtocarrythetensionforcesassociatedwiththecompressionfields.Onepairoftiesineachlayerinthebeamdepthshouldpassthroughholesinthebeamwebtoprovidecontinuousconfinementtotheconcrete.

Reinforcingtiesaboveandbelowthebeamarepartofthehorizontalstrut-and-tiesystem.Threelayersoftiesshouldbeprovidedaboveandbelowthesteelbeamwithinadistanceof0.4Dsfromthebeamflangeasfollows:(1)forB≤500 mm,10 mmbarswithfourlegsineachlayer;(2)for500<B≤750 mm,12 mmbarswithfourlegsineachlayerand(3)forB>750 mm,16 mmbarswithfourlegsineachlayer.TheminimumamountoftiesaboveandbelowthebeammaybegovernedbytheforceinthecompressionfieldVcff(≤Vcf).Theminimumtotalcross-sectionalareaoftieswithinthedepthof0.4Dsshouldsatisfy

A

Vf

tiecff

yr

≥ (11.35)

11.4.6.2 Vertical column ties

Thelargechangesinreinforcingbarforcesowingtothetransferofmomentsintheconnec-tionmayoccur,whichleadstotheslipofverticalbars.Tolimitthebarslip,thesizeoftheverticalcolumnbarsshouldbetakenasfollows:

d

D db

o< +( )220

(11.36)

wheredb is thediameter of the vertical bar or thediameter of a bar equivalent to thebundlebars.


Ifthechangeinforceinverticalbarssatisfiesthefollowingrequirement,largersizethanthelimitbyEquation11.36canbeused:

∆F D d fb o c< + ′80 2( ) (11.37)

11.4.6.3 Face-bearing plates

FBPswithinthebeamdepthareusedtocarrythehorizontalforcesintheconcretestrut.IfsplitFBPsareemployed,theplateheight(dp)shouldnotbelessthan0.45dw.Therequiredthickness of the FBP is influenced by the distribution of the concrete bearing stress, itsgeometry,supportconditionsandyieldstress.ThethicknessoftheFBPshouldsatisfythefollowingrequirements:

t

V b t fb f

pcs f w yw

f up

≥−3( )

(11.38)

t

Vb f

pcs

f up

≥ 32

(11.39)

t

V bf d

pcs p

yp w

≥ 0 2. (11.40)

t

bp

p≥22

(11.41)

t

b bp

p f≥−( )5

(11.42)

wherefypandfuparetheyieldandtensilestrengthsofthebearingplate,respectively.

11.4.6.4 Steel beam flanges

Theflangesofthesteelbeamunderverticalbearingforcesinthecompositeconnectionaresubjectedtotransversebending.Theflangesofthesteelbeammusthavesufficientflexuralstiffnesstoresistthetransversebending.Forthispurpose,thethicknessofthebeamflangesmustsatisfythefollowingrequirement:

t

b t D fDf

ff w s yw

yf

≥ 0 3. (11.43)

11.4.6.5 Extended face-bearing plates and steel column

TheextendedFBPsand/orsteelcolumnsaresubjectedtocompressivebearingforcesinthehorizontalstruts.ThenetbearingforceisequaltotheshearforceVcff (≤Vcf)carriedbythecompressionstrut.Whenasteelcolumnisused,onlyoneofthecolumnflangesissubjectedtobearingforceasdepictedinFigure11.7.Thedesignoftheseelementsisgovernedbythe


transversebendingoftheplate,shearstrengthofthesupportingelementandtheconnectiontothesteelbeam.ThethicknessoftheextendedFBPsorthecolumnflangesislimitedby

t

V bd f

fcff p

o y

≥′

0 12. (11.44)

where′bpisthewidthoftheextendedFBPorthewidthoftheflangeofthesteelcolumnVcff canbetakenasVcf

ThethicknessoftheextendedbearingplateshouldbegreaterthanthatoftheFBPbetweentheflangesofthebeam.

Example 11.3: Design of steel beam-to-CEC column moment connection

Check the capacity and design the details of the steel beam-to-CEC column momentconnectionshowninFigure11.9.Theconnectionissubjectedtothefollowingfactoreddesignactions:

M M V V M Mb b b b c c1 2 1 2 1 2300∗ = ∗ = ∗ = ∗ = ∗ = ∗ =635,750kNmm, kN, 600,000kNmmm

V Vc c1 2 500∗ = ∗ = kN

DesigndatashowninFigure11.9are

Composite column: mm

Steel beam: mm, m

B D

b tf f

= =

= =

650

209 15 6. mm, mm, mm, mm

D t d

d

s w fc= = =533 10 2 517 4. .

ww yf yw

c

f f

d

= = =

=

502 300 320

203

mm, MPa, MPa,

Steel column: mmm, mm

Face-bearing plates: mm, mm,

t

b b t

cf

p p

=

= ′ =

11

209 203 pp yp

up

f

f

= =

=

16 300

4

mm, MPa

330

0

MPa

Vertical reinforcement: T Cvr vr= =

650

209

Steel beam

Tie

650203

203

Face-bearingplate

Figure 11.9 Steel beam-to-CEC column moment connection.


1. Design actions

Thedesignactionsarecalculatedasfollows:

M M Mc c c∗ = ∗ + ∗ = + =∑ 1 2 600,000 600,000 1,200,000kNmm

V

V Vb

b b∗ =∗ + ∗( )

=+

=1 2

2300 300

2300

( )kN

∆V V Vb b b∗ = ∗ − ∗ = − =1 2 300 300 0kN

2. Effective width of the connection

Themaximumwidthoftheconnectionis

bb B

b D bff fmax .

.

=+

≤ + ≤

=+

= ≤ + =

>

21 75

2429 5 209 650 859

209 650mm mm

1.75×× =209 mm365 75.

Hence,bmax=365.75mm.Theeffectivewidthoftheconnectioniscomputedas

x

D dc= + = + =2 2

6502

2032

426 5. mm

h

xD

yb

xyf

=

=

=

426 5650

203209

0 637.

.

b h b bo xy i= − = × − =( ) . ( . )max 0 637 365 75 209 100mm

b b bj i o= + = = =209 100 309mm

3. Vertical bearing capacity

Thenominalconcretebearingstrengthiscalculatedas

C f b Dcb c j= ′ = × × × × =2 0 3 2 40 309 0 3 650 4820 4( . ) ( . ) .N kN

Thedesignactionsontheconnectionarecomputedas

M D Vc b∗ + = + × × =∑ 0 35 0 35 650 0. .∆ 1,200,000 1,200,000kNmm

Theverticalbearingcapacityoftheconnectioniscalculatedas

φ C D T C dcb vr vr vr( . ) ( ) . . .0 7 0 7 4820 4 0 7 650 0+ + = × × × +

= 1,535,,297 kN m 1,200,000kN mm, OK>


4. Horizontal shear capacity

Thewidthoftheshearpaneliscalculatedasfollows:

K

M V D T C d

f b

c b vr vr vr

c j

=∗ + ∗ − +

′=

× + −∑ ∆ ( ) ( )

( )

/ 1,200,0002

210 0 0

0

3φ

φ ..7 2 40 309× × ×= 69,348mm2

aD D

K Dc = − − ≤

= − − = < × =

=

2 40 3

6504

134 5 195

2

2

.

.6502

69,348 mm 0.3 650 mm

1334.5mm

C f b ac c j c= ′ = × × × =2 2 40 309 134 5 3325. N kN

LM

C T C VDp

c

c vr vr b

=∗

+ + − ∗≥

=× + +

∑φ( ) .

.

. ( )

0 50 7

0 7 0 0

∆

1,200,0003,325 −− ×

= > × =

=

0 5 0515 6 455

.. mm 0.7 650 mm

515.6 mm

Thenominalshearyieldcapacityofthesteelwebpanelis

V f L twp yw p w= = × × × =0 6 0 6 320 515 6 10 2 1009 8. . . . .N kN

Thenominalstrengthofthecompressionstrutiscalculatedas

V f b Dcs c p= ′ = × × × =1 7 1 7 40 209 650 1460 6. . .N kN

0 5 0 5 40 209 502 2098 4 1460 6. . . .′ = × × × = > =f b d Vc p w csN kN kN

Hence,Vcs=1460.6kN.Assumingthetiesareadequate,thenominalstrengthofthecompressionfieldisdeter-

minedas

V f b Dcf c o= ′ = × × × =1 7 1 7 40 100 650 698 9. . .N kN

Thedesignactionsontheconnectionarecomputedas

M V Lc b p∗ − = − × =∑ 1,200,000 1,045,320kNmm300 515 6.

Thehorizontalshearcapacityoftheconnectioniscalculatedas

φ V d V d V D dwp fc cs w cf s o+ + +

= × × +

( . ) ( )

. . . .

0 75

0 7 1009 8 517 4 1460 6×× × + × + ×

= >

0 75 502 698 9 533 0 25 533. . ( . )

1,076,602kN mm 1,045,3200kN mm, OK


5. Detailing

5.1. Column ties within beam depth

Thestrengthprovidedbyconcreteincompressionis

′ = ′ = × × × =V f b Dc c o0 4 0 4 40 100 650 164 4. . .N kN

Thestrengthprovidedbythehorizontaltiesisdeterminedas

′ = − ′ = − =V V Vs cf c 698 9 164 5 534 5. . . kN

Therequiredcross-sectionalareaofcolumntiesperunitlengthis

As

VDf

sr

sr

s

yr

=′

=×

× ×=

0 9534 5 10000 9 650 400

2 28.

..

. mm /mm2

As

Bsr

sr

= = × =min

. . .0 004 0 004 650 2 6mm /mm2

Use4-legsY12tiesforeachlayer,Asr=4×110=440mm2;thespacingofthetiesis

ssr = =

4402 6

169.

mm

Use4-legsY12at160 mm.

5.2. Column ties adjacent to connection

Therequiredareaofcolumntiesis

A

Vf

tiecf

yr

= =×

=698 9 1000

4001747

.mm2

Thedepthinwhichthetiesareplacedis0.4Ds=0.4×533=213mm.Use4-layersY12at70 mm(Atie=1810mm2).

5.3. Thickness of face-bearing plates

ThethicknessoftheFBPsiscalculatedasfollows:

t

V b t fb f

pcs f w yw

f up

≥−

=× × − × ×

×=

3 3 1460 6 10 209 10 2 320209 430

3( ) ( . . )115mm

t

Vb f

pcs

f up

≥ =× ×× ×

=3

23 1460 6 102 209 430

143.

mm

t

V bf d

pcs p

yp w

≥ =× ××

=0 2 0 21460 6 10 209

430 5028 7

3

. ..

. mm

t

bp

p≥ = =22

20922

9 5. mm

Hence,tp=16mm>15mm, OK.


5.4. Steel beam flanges

Therequiredthicknessofthesteelbeamflangesis

t

b t D fDf

ff w s yw

yf

≥ =× × ×

×=0 3 0 3

209 10 2 533 320650 300

15 2. ..

. mm

tf = >15 6 15 2. .mm mm, OK

5.5. Flange thickness of the steel column

Therequiredthicknessofthesteelcolumnflangesis

t

V bd f

cfcff p

o y

≥′=

× ×× ×

=0 12 0 3698 2 10 2030 25 533 300

6 923

. ..

.. mm

tcf = >11 6 92mm mm, OK.

11.5 BeAm-to-cfSt column moment connectIonS

High-strength thin-walled CFST columns with concrete compressive strengths above70 MPaareincreasinglyusedinhigh-risecompositebuildingstocarrylargeaxialandlat-eralloads(Liang2009,2011a,b).ThetubewallsofCFSTcolumnsarerelativelythin,whichprohibitsdirectweldingofthesteelbeamstothetubes.Consequently,anchorboltsareusedtoconnectaT-sectiontothetubeandthesteelbeamisboltedtotheT-section.Alternatively,theconnectingelementscanbeembeddedintheconcretecoreviaslotscutinthesteeltube(AzizinaminiandPrakash1993).Thecapacityofthesecompositeconnectionsmaybelim-itedbythepull-outcapacityoftheanchorboltsortheconnectionelements.Asteelbeam-to-CFSTcolumnmomentconnection isconstructedbypassingthesteelbeamthroughaCFSTcolumn.Thebeam-to-CFSTcolumnmomentconnectioncanbeshopfabricatedbyweldingashortbeampassingthroughacertainheightsteeltube.Theshortsteelbeamoftheconnectioncanbefieldboltedtothegirder.ThedesignofsteelbeamtocircularCFSTcolumnmomentconnectionsispresentedherein,whichisbasedontheworkofAzizinaminiandPrakash(1993).Theeffectsoftheconcreteslabofslabreinforcementonthestrengthofthecompositeconnectionarenotconsideredinthedesign.

11.5.1 resultant forces in connection elements

Thedesignactionsontheconnectionareassumedtoberelatedasfollows:

M l Vc c c∗ = ∗ (11.45)

V Vc cb b∗ = ∗α (11.46)

M l Vb b b∗ = ∗ (11.47)

TheweboftheconnectionisdepictedinFigure11.10,whiletheuppercolumnisshowninFigure11.11. It isassumedthat (1) thedistributionofconcretestress is linear; (2) the


widthoftheconcretestressblockisequaltothewidthofthesteelbeamflangesand(3)thestraindistributionovertheuppercolumnislinear.AsillustratedinFigure11.11,theuppercolumnshearcarriedbythesteelbeamistakenasμCc,whereCcistheresultantconcretecompressiveforceonthebeamflangeandμisthefrictioncoefficient.FromFigure11.11,themaximumstraininconcreteisobtainedas

ε εc

n

nt

dD d

=−

(11.48)

D

Strut

CsCc

Ds

Cc

Cs Ts

dnDs

Ds

Ds

Ds

Ts

V*c

V*c

V*b V*b

M*b

M*b

M*b

M*b

Figure 11.10 Force transfer mechanism in beam-to-CFST column moment connection.

Cs

Cc

εc

lc

D

V*c

dnεt

Ts

V*c

Figure 11.11 Stress distributions in the upper column.


Themaximumstressinconcrete(σc),stressinsteeltubeincompression(σsc)andstressinsteeltubeintension(σst)aredeterminedasfollows:

σ εc c cE= (11.49)

σ εsc s cE= (11.50)

σ εst s tE= (11.51)

Theareaoftheconcreteincompressionistakenasbfdn,wherednistheneutralaxisdepth.Itisnotedthatonlypartofthesteeltubethatsupportsthesteelbeamiseffectiveincarry-ingtheforcetransferredfromthesteelbeam.Theeffectiveareaofthesteeltubeincarryingcompressionortensionforcesisassumedtobe2bft.Theresultantforcesinconnectionele-mentscanbedeterminedasfollows(AzizinaminiandPrakash1993):

C

b dn

dD d

fcf n n

nt y=

−

2( )φ (11.52)

C b t

dD d

fs fn

nt y=

−

2 ( )φ (11.53)

T b t fs f t y= 2 ( )φ (11.54)

wheren E Es c= / isthemodulusratioφt yf isthestresslevelinthesteeltubeattheultimatestrengthlimitstateandφt = 0 75.fyistheyieldstressofthesteeltube

11.5.2 neutral axis depth

TheverticalforceequilibriumoftheuppercolumnasshowninFigure11.11isexpressedbyCc+Cs=Ts.Fromthiscondition,therequiredthicknessofthesteeltubecanbeobtainedas

t

nd

D dn

n

=−

14 2

2

(11.55)

Fromthemomentequilibriumoftheuppercolumn,thefollowingequationcanbederivedfordeterminingthedepthoftheneutralaxis(AzizinaminiandPrakash1993):

dD d

DdD d

d Dd n

fl Vb

n

n

n

nn

n

t y

cb c b

f− −+ −

−

∗

2

2 32φ

α = 0. (11.56)

11.5.3 Shear capacity of steel beam web

Thehorizontal shear in the connection is resistedby thewebof the steelbeamand theconcretebetweenthebeamflanges.Theshearforceinthesteelbeamwebattheultimate


conditioncanbeobtainedfromthehorizontalforceequilibriuminthefreebodydiagramshowninFigure11.10asfollows(AzizinaminiandPrakash1993):

V

MD

C Cwb

sc cs

∗ =∗− −2 µ θcos (11.57)

whereCcsistheresultantforceinthecompressionstrutandθ=arctan(Ds/D).ItisassumedthatthesteelbeamwebunderthefactoreddesignshearforceVw

∗startstoyield.Theshearyieldcapacityofthesteelbeamwebinhorizontalshearisgivenby

V f Dtw yw w= 0 6. (11.58)

11.5.4 Shear capacity of concrete

ThedesignshearcapacityofconcreteininteriorreinforcedconcreteconnectionsisgivenbytheACI-ASCECommittee352(1985)as

φ φV f b Dcc c f= ′1 7 2. ( ) (11.59)

whereϕ=0.85isthecapacityreductionfactor.Theeffectivewidthoftheconcretecompres-sionstrutintheconnectionistakenas2bf.

Example 11.4: Design of steel beam-to-CFST column moment connection

CheckthecapacityanddesignthedetailsofthesteelbeamtothecircularCFSTcolumnmomentconnection.Theconnectionissubjectedtothefollowingfactoreddesignactions:

M V

V

Vl

M

Vb b cb

c

b

cc

c

∗ = ∗ = =∗

∗= =

∗

∗ =280 400 0 85kNmm kN 850mm, , . ,α

Designdataare

Steel tube: mm, MPa

Steel beam: mm,

D f

b D

y

f s

= =

= =

600 300

178 4406 7 8 320

200 000 70

mm, mm, MPa,

MPa, MP

t f

E f

w yw

s c

= =

= ′ =

.

, aa


Young’smodulusofconcreteiscomputedas

E fc c= ′ + = + =3,320 6,900 3,320 6,900 34,677MPa70

Themodulusratioisn=Es/Ec=200,000/34,677=5.768.Theneutralaxisdepthiscalculatedasfollows:

dD d

DdD d

d Dd n

fl Vb

n

n

n

nn

n

t y

cb c b

f− −+ −

−

∗

2

2 32φ

α = 0


dd

dd

ddn

n

n

nn

n

600600600 2

6003

2 5 7680 75 3

2

−×−

+ −

−

××.

. 0000 85 850 400 10

1780

3. × × ×

=

The neutral axis depth dn can be solved by using the Goal Seek function in What-IfAnalysisinExcel.Forthiscase,dn=184.42mm.

2. Required thickness of the steel tube

Therequiredthicknessofthesteeltubeiscomputedas

t

nd

D dn

n

=−

= × − ×

=

14 2

14 5 768

184 42600 2 184 42

6 42 2

..

.. mmm

Uset=7 mmforthesteeltube.

3. Check stresses in connection elements

Thestrainsinthesteeltubeandconcretearecalculatedasfollows:

ε

φt

t y

s

fE

= =×

=0 75 300

0 001125.

.200,000

ε εc

n

nt

dD d

=−

=×−

=184 4 0 001125

600 184 40 00049924

. ..

.

Thestressesinconcreteandinsteeltubearecomputedas

σ εc c c cE f= = × = < ′ =34,677 MPa MPa, OK0 00049924 17 3 70. .

σ εsc s c yE f= = × = < =200,000 MPa MPa, OK0 00049921 99 8 300. .

σ εst s t yE f= = × = < =200,000 MPa MPa, OK0 001125 225 300.

4. Forces in concrete compression strut

Theforceinthecompressionstrutiscalculatedasfollows:

θ =

=

= °arctan arctan .

DDs 406

60034 08

C

b dn

dD d

fcf n n

nt y=

−

=

××

×−2

178 184 422 5 768

184 42600 18

( ).

..

φ44 42

0 75 300 284.

.

× × =N kN

C

Ccs

c= =°=

sin sin ..

θ28434 08

506 8kN

5. Shear capacity of steel beam web

Thedesignshearforceinthesteelbeamwebiscalculatedas

V

MD

C Cwb

sc cs

∗ =∗− − =

× ×− × − °

2 2 280 10406

0 5 284 506 8 34 083

µ θcos . . cos . == 817 6. kN


Theshearyieldcapacityofthesteelbeamwebis

V f Dt Vw yw w w= = × × × = > ∗ =0 6 0 6 320 600 7 8 898 6 817 6. . . . .N kN kN, OK

6. Shear capacity of the concrete in connection

Theshearforcecarriedbytheconcretewithinthebeamflangesis

V Cc cs∗ = = × ° =cos . cos .θ 506 8 34 08 420kN

Theshearcapacityoftheconcreteintheconnectioniscomputedas

φ φV f b D

V

cc c f

c

= ′ = × × × × ×

= > ∗ =

1 7 2 0 85 1 7 70 2 178 600

2582 4 4

. ( ) . . ( )

.

N

kN 220kN, OK

11.6 SemI-rIgId connectIonS

Semi-rigidcompositeconnectionscanbeusedtotransmitmomentsandshearforcescausedbystaticloadsaswellasseismicloadsinlow-andmoderate-heightcompositeframes.Thiscomposite connection utilises the strength and stiffness offered by the floor slab whichisprovidedwithadditional studshearconnectorsandslabreinforcement in the negativemoment regions adjacent to the columns. Figure 11.12 schematically depicted a typicalsemi-rigidcompositeconnection,whichconnectsacompositebeamtoasteelcolumn.Themoment is transmittedbytheslabreinforcementandthebottomseatangle,whiletheverti-calshearistransmittedbythewebangles.Semi-rigidcompositeconnectionsarefoundtoprovideaneconomicalsolutiontocompositeconstruction.Therestraintprovidedbysemi-rigidcompositeconnectionstocompositebeamsreducesdeflections,crackingandvibrationsassociatedwithcompositefloors.Therestraintalsoreducestheeffectivelengthofcolumns.Theuseofsemi-rigidcompositeconnectionsleadstosignificantreductionsintheoverallstructuralsteelcosts.Thedesignmethodforsemi-rigidcompositeconnectionspresentedinthissectionisbasedontheworkofAmmermanandLeon(1990),LeonandAmmerman(1990)andtheASCETaskCommittee(1998).Itshouldbenotedthatthemethodshouldnot

Seat angle

Steel beam

Web angle

Slab reinforcementColumnConcrete slab

Stud shearconnector

Figure 11.12 Semi-rigid composite connection.


beusedforbeamswithspanslongerthan15m,forbeamsdeeperthanW27andforbeamswithflangethicknesslargerthan20 mm.

11.6.1 Behaviour of semi-rigid connections

The behaviour of semi-rigid composite connections is characterised by their moment–rotationcurves.Semi-rigidcompositeconnectionsarepartiallyrestrainedastheyusuallyhavea ratioof the secant stiffnessKser to the stiffnessof the framing steelbeamEIb/Lbbetween0.5and20(Gerstle1985).Thedesignactionsandload–deflectionbehaviourofcomposite frames with semi-rigid composite connections depend on the rotational stiff-nessoftheconnections.Semi-rigidcompositeconnectionsarepartial-strengthconnections.Thismeansthatthemomentcapacityoftheconnectionissmallerthanthatoftheframingsteelbeam.A semi-rigidcompositeconnectionsubjectedtonegativebendinghasahighermomentcapacitythanatypicaltopandseatangleconnectionduetothehigherstrengthofreinforcementandlargermomentarm.Thecapacityofthesemi-rigidcompositeconnectionmaybelimitedbytheshearfailureoftheboltsthatconnecttheseatangletothebottombeamflange.Underloadreversalsandpositivebending,thebottomanglemaypulloutatrelativelylowloads.Thefailuremodesassociatedwithsemi-rigidcompositeconnectionsincludetheshearfailureofboltsattachingtheseatangletothebeambottomflange,bear-ingfailureofboltholes,yieldingandfractureoftheseatangle,tensionfailureofboltsatthebeamweb,shearfailureofthewebangles,yieldingoftheslabreinforcementandshearfailureofstuds(ASCETaskCommittee1998).

11.6.2 design moments at supports

Forthedesignofthetypicalsemi-rigidcompositeconnectionshowninFigure11.12,itisassumedthat(1)theconnectionsframeintothemajoraxisofthesteelcolumn;(2)unproppedconstructionisused;(3)thetotalnumberofboltsattachedtothebottomflangeofthesteelbeamislimitedtosix(threeateachseatangle)and(4)completeshearconnectionisusedinthenegativemomentregions.

Thedesignof semi-rigidcomposite connections requires the selectionof thedegreeoffixityat thecolumns.This isachievedbyassuming theamountof the factored live loadmomentMqe

∗ atthesupports.Thelowerandupperboundsonthemomentsatthesupportsfor common load cases are given by Leon and Ammerman (1990). The factored designmomentatthemid-spanMqm

∗ canbeobtainedfromMqe∗ .Acompositebeamcanbeselected

tocarryMqm∗ wherethesteelbeamcancarrythedesignmomentMcm

∗ causedbyfactoredconstruction loads without reaching its plastic capacity andMdm

∗ induced by dead loadswithoutyielding.

11.6.3 design of seat angle

TheseatangleisusedtoresistthehorizontaldesignforceinducedbythedesignmomentMqe∗ .

TherequiredhorizontaldesignforceFh∗onthebottomangleisdeterminedby

F

MD d

hqe

s cf

∗ =∗

+ (11.60)

wheredcfisthedistancefromthetopofthesteelbeamtothecentroidoftheslabforce.Fornegativebending,dcfisthedistancefromthecentroidoflongitudinaltensilereinforcementintheconcreteslabtothetopofthesteelbeam.


Therequiredareaoftheseatangleleg(Asa)iscomputedby(LeonandAmmerman1990)

A

Ff

sah

ysa

=∗1 33.

(11.61)

wherefysaistheyieldstressoftheseatangleandthefactor1.33isusedtoensurethattheslabreinforcementwillyieldbeforetheseatangleunderthesamehorizontalforceFh∗.Bytakingthewidthoftheseatanglebsaatleastequaltothewidthofthebeamflange(bf),thethicknessoftheanglecanbedeterminedas

t

Ab

sasa

sa

= (11.62)

11.6.4 design of slab reinforcement

Theeffectivewidthof theconcrete slab in thenegativemomentregion isassumedtobeseventimesofthecolumnwidth.Longitudinalreinforcementintheconcreteslabisplacedwithintheeffectivewidthoftheconcreteslab.Thecross-sectionalareaofslabreinforce-mentiscalculatedby

A

Ff

rh

yr

=∗

(11.63)

11.6.5 design moment capacities of connection

Thedesignmomentcapacitiesofthesemi-rigidcompositeconnectionunderserviceandulti-mateloadscanbeestimatedbythefollowingequationsgivenbyLeonandForcier(1992)as

φ φM A f A f D dser r yr a ya s cf= + +0 17 4. ( )( ) (11.64)

φ φM A f A f D du r yr a ya s cf= + +0 245 4. ( )( ) (11.65)

whereϕMseristhedesignmomentcapacityoftheconnectionunderserviceloadsandϕ=0.85.

11.6.6 compatibility conditions

Thecompletemoment–rotationcurveforsemi-rigidcompositeconnectionsisexpressedbythefollowingequations(AmmermanandLeon1990):

M C e CC= −( ) +−

2 41 3θ θ (11.66)

C A f D dr yr s cf2 = +( ) (11.67)

C

AA

D da

rs cf3

0 15

32 9=

+. ( )

.

(11.68)

C A f D da ya s cf4 24= +( )

(11.69)

whereθistherotationinradians.


Therotationofthesemi-rigidcompositeconnectionislimitedto2.5mradfordesignforserviceabilitycriteriaandto10mradfordesignforstrengthcriteria.Thecompatibilitycon-ditionrequiresthatthedesignmomentcapacitiesoftheconnectioncalculatedusingtheselimitsmustsatisfy

φ φM C e C MC

sqe2 5 20 0025

41 0 00253.

. .= −( ) +

≥

∗− (11.70)

φ φM C e C MC

qe10 20 01

41 0 013= −( ) +

≥

∗− . . (11.71)

whereMsqe∗ isthemomentatthesupportunderserviceliveloads.

11.6.7 design of web angles

Thewebanglesareusedtotransmittheverticalshearforceinthecompositeconnection.Theshearandbearingcapacitiesoftheanglesandbeamwebneedtobechecked.Thenum-berofboltsrequiredcanbedeterminedfromtheshearandbearingcapacitiesofthebeamwebandistakenasthelargerofthevaluescalculatedby

n

VV

bf

=∗

φ (11.72)

n

Vf d t

buwa f w

=∗

φ2 4. (11.73)

wherefuwisthetensilestrengthofthesteelbeamweb.Thelargervalueoftheaforemen-tionednumbersofboltsisusedinthedesign.Thethicknessofthewebanglecanbedeter-minedfromitsbearingcapacityas

t

Vf n d

auwa b f

=∗

φ2 4. (11.74)

wherefuwaisthetensilestrengthofthewebangle.

11.6.8 deflections of composite beams

Thedeflectioncalculationsofcompositebeamsunderserviceliveloadsshouldaccountfortheeffectofthedifferentsectionpropertiesinpositiveandnegativebendingandtheflexibil-ityofsemi-rigidcompositeconnections.Thesecondmomentsofareaofcompositebeamsare highly different for positive and negative bending. The use of either positive secondmomentofareaornegativesecondmomentofareawillresultinsignificanterrorsinthecalculationsofthecompositebeamdeflections.Theeffectivesecondmomentofareaforacompositesectionisdeterminedby(AmmermanandLeon1990)

I I Ics p n= +0 6 0 4. . (11.75)


whereIpandInarethesecondmomentsofareaofthecompositesectionunderpositiveandnegativebending,respectively.ThelowerboundvaluesofmomentofinertiaforpositiveandnegativebendingaregivenintheAISC-LRFDManual(1994).

Thedeflectionsofcompositebeamswithsemi-rigidcompositeconnectionsmaybecalcu-latedas(Hoffman1994;Leonetal.1996)

δ δ θsr FF

sL= +4

(11.76)

whereδFFisthedeflectionofthefixed-endcompositebeamunderthesameloadingθsistherotationoftheconnectionunderserviceloadsListhebeamlength

11.6.9 design procedure

Design examples for semi-rigid composite connections were given elsewhere (Leon andAmmerman1990;Viestetal.1997;ASCETaskCommittee1998).Thedesignprocedureissummarizedasfollows:

1.Computethedesignmomentsofthesimplysupportedcompositebeamunderfactoredconstructionloadsandthedesignmomentunderdeadloads.

2.Calculatethefactoredliveloadmomentatthesupportandmid-span. 3.Select thesteelbeamsectiontocarry theconstruction loadmomentanddead load

moment. 4.Computetheareaandthicknessoftheseatangle. 5.Calculatetheareaofslabreinforcement. 6.Calculatethemomentcapacitiesoftheconnectionunderserviceandultimateloads. 7.Checkthecompatibilityconditionusingthemoment–rotationrelationships. 8.Designthewebangleandbolts. 9.Determinetherequirednumberofshearconnectorsinthecompositebeams. 10.Calculatethedeflectionsofcompositebeamsunderserviceloads. 11.Checkthestressesinthesteelbeamunderserviceloads.

referenceS

ACI-ASCE Committee 352 (March 1985) Recommendations for design of beam column joints inmonolithicreinforcedconcretestructures,ACIStructuralJournal,82(3),266–283.

AISC-LRFDManual(1994)LoadandResistanceFactorDesign,Vol.II,Connections,ManualofSteelConstruction,Chicago,IL:AmericanInstituteofSteelConstruction.

Ammerman,D.J.andLeon,R.T.(1990)Unbracedframeswithsemi-rigidcompositeconnections,AISCEngineeringJournal,27(1):12–21.

ASCETaskCommittee (onDesignCriteria forCompositeStructures inSteelandConcrete) (1994)Guidelinesfordesignofjointsbetweensteelbeamsandreinforcedconcretecolumns,JournalofStructuralEngineering,ASCE,120(8):2330–2357.

ASCETaskCommittee (onDesignCriteria forCompositeStructures inSteelandConcrete) (1998)Designguideforpartiallyrestrainedcompositeconnections,JournalofStructuralEngineering,ASCE,124(10):1099–1114.

Astaneh,A.,Call,S.M.andMcMullin,K.M.(1989)Designofsingleplateshearconnections,AISCEngineeringJournal,26(1),21–32.


Astaneh, A. and Nader, M.N. (1989) Design of tee framing shear connections, AISC EngineeringJournal,26(1):1–20.

Astaneh,A.andNader,M.N.(1990)Experimentalstudiesanddesignofsteelteeshearconnections,JournalofStructuralEngineering,ASCE,116(10):2882–2902.

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Viest, I.M.,Colaco,J.P.,Furlong,R.W.,GriffisL.G.,Leon,R.T.,andWyllie,L.A.(1997)CompositeConstructionDesignforBuildings,NewYork:ASCEandMcGraw-Hill.

411

Notations

a Lengthofaplatefieldbetweenstudshearconnectorsabw Distancefromtheboltcentrelinetotheweldlineac Distance from thebolt centre line to the edgeof the columnflangeor

strainratioorconcretebearingwidthad Distancefromtheboltcentrelinetothefilletedgeofthewebae Edgedistancefromtheboltcentrelinetotheedgeofaplyaf,afe Distancefromtheboltcentrelinetothetopflangeofthebeamandits

designvalueam Distancefromthecentroidofthecolumnflangetotheedgeofthebase

plateamax =max(am,an)an Distancefromtheedgeofthecolumnbearingareatotheedgeofthebase

plateao DimensionoftheH-shapebearingareainbaseplateconnectionap Distancefromboltcentrelinetotheedgeoftheendplate;widthofloaded

areaA1 BearingareaA2 Largestareaofthesupportingsurface,whichisgeometricallysimilartoA1

Ab Cross-sectionalareaofareinforcingbarAc Cross-sectionalareaofaboltcoreorcross-sectionalareaofconcreteAcm Areaofconcreteabovetheplasticneutralaxis(PNA)inthecrosssection

ofacompositecolumnAcn Areaofconcreteabovehndistancefromthecentroidofacompositecol-

umnsectionAH H-shapebearingareaAe Effectivecross-sectionalareaofaplateorsectionAec EffectiveshearareaofconcreteslabAfm FlangeeffectiveareaAfn NetareaofaflangeAg Grosscross-sectionalareaofasectionAn Netcross-sectionalareaofasteelsectionorplateAo Cross-sectionalareaoftheplainshankofaboltAp Cross-sectionalareaofaplateApa PlanprojectionofthesurfaceareaofroofAps ProjectedareaoffailureconeofconcreteAr,Ar1 Cross-sectionalareasofreinforcementArfw RequiredareaoftensilereinforcementintheconcreteslabwhenthePNA

islocatedatthejunctionofthetopflangeandthewebofthesteelsection

412 Notations

Arfp RequiredareaoftensilereinforcementintheconcreteslabwhenthePNAis locatedat the junctionof the steelbottomflangeand theadditionalflangeplate

Arho RequiredareaoftensilereinforcementintheconcreteslabwhenPNAislocatedinthesteelwebwhereaholeforms

Arwf RequiredareaoftensilereinforcementintheconcreteslabwhenthePNAislocatedatthejunctionofthesteelwebandthebottomflange

As Cross-sectionalareaofastiffenerorsectionortensilestressareaofabolt

Asa RequiredareaoftheseatanglelegAse Totaleffectivecross-sectionalareaofstructuralsteelAsr Cross-sectionalareaofreinforcingbarsineachlayeroftiesinthedepth

ofthebeamwebAst Totalcross-sectionalareaofreinforcementAsv Total cross-sectional area of longitudinal shear reinforcement crossing

theshearsurfaceAsv⋅min MinimumareaoflongitudinalshearreinforcementAt TributaryareaAtie Minimumtotalcross-sectionalareaoftieswithinthedepthof0.4Ds

Aw Cross-sectionalareaofasteelwebAws Cross-sectionalareaofthestiffener-webcompressionmemberAsc Cross-sectionalareaofcompressivereinforcementintheslabAsx,Asy Cross-sectionalareasofreinforcementinxandydirectionsAz Areaofwindpressureb Widthofaplateorplatefiledorsectionb1,b2 WidthsofflangesofamonosymmetricI-section;centre-to-centrespacing

adjacentbeamsbb Bearingwidthbbf,bbw Bearingwidthsintheflangeandwebofasteelsectionbcf Effectivewidthoftheconcreteflangeofacompositebeambcr Widthofconcreteribsatthemid-heightofsteelsheetingribsbd Distancefromthebearingplatetotheendofthebeambe Effectivewidthofasteelplatebe1,be2 Effectivewidthsofasteelplateorconcreteflangeofacompositebeambef1,bef2,befp Effectivewidthsofthetopandbottomflangesofasteelsectionandaddi-

tionalbottomplatebes Widthofthewebtransversestiffenerbew Effectivewidthofasteelwebbf Widthofthetopflangeofasteelbeambf1,bf2 Widthsofthetopandbottomflangesofasteelsectionbfc Widthofasteelcolumnflangebfo WidthoftheflangeoutstandofasteelI-sectionorlengthofyieldlinein

baseplateconnectionbhcs Widthofahollowcoreslabbi Innerpanelwidthofacompositeconnectionbj Effectivewidthofabeam-to-CECcolumnmomentconnectionbne,bne,max Ineffectivewidthsofasteeltubewallanditsmaximumvaluebm Coefficientbo Bearingwidthinthewebmeasuredfromtheedgeofthebearingsupport

tothebeamendorouterpanelwidthofacompositeconnection

Notations 413

bp Widthofloadedareaunderconcentratedloadoreffectivewidthoftheface-bearingplate

bs Averagebreadthofshieldingbuildingsorwidthofthebearingstressbsa Widthofseatanglebtf Transformedeffectivewidthoftheconcreteflangeinacompositebeamb0h Averagebreadthofastructurebetweenheight0andhbsh Averagebreadthofastructurebetweenheightsandhbv Effectivewidthofaslabbx OverallwidthacrossthetopofconnectorsinacrosssectionB WidthofacolumncrosssectionBs BackgroundfactororwidthtakenasthelargervalueofBandDfora

rectangularcrosssectionc Covertoreinforcingbarsordistancecb Coefficientaccountsfortheeffectsofdifferentmomentsatthecolumn

endscm CoefficientC CoefficientorcompressionforceC1,C2,C3,C4 CompressionforcesorcoefficientsorC1=electrodestrengthcoefficientCc CompressiveforceinconcreteCcb NominalconcretebearingstrengthCcs ResultantforceinaconcretecompressionstrutCdyn DynamicshapefactorCf FrictionaldragforcecoefficientCfs CrosswindforcespectrumcoefficientCfig AerodynamicshapefactorCp,e,Cp,i ExternalandinternalpressurecoefficientsCs CompressiveforceinthesteeltubeCvr Forceintheverticalreinforcementincompressiond DepthofasteelI-sectionorstructureorsection;effectivedepthcompos-

iteslabd1 Cleardepthofthewebofasteelsectiond2 Twice thecleardistancebetween theneutralaxisand thecompression

flanged5 Flatwidthofthewebofahollowsteelsectiondb Lateraldistancebetweenthecentroidsoftheweldsoffastenersordiam-

eterofareinforcingbardbs Diameteroftheshankofastuddc DistancefromthecentroidofFccintheconcreteslabtothetopfaceofthe

steelsectiondc1 DistancefromthecentroidofFc1tothecentroidofacolumncrosssectiondcf Distancefromthetopofthesteelbeamtothecentroidoftheslabforcedcm DistancefromthecentroidofAcmtothecentroidofacolumncrosssectionde Effectiveoutsidediameterofacircularsteelsectionde,i Orthogonaldistancefromthecentroidofeachfibreelementtotheneutral

axisinacompositecolumncrosssectiondew Effectivedepthofthewebofasteelsectiondfc Distancebetweenthecentroidsofthetwoflangesofasteelsectiondf Nominaldiameterofaboltdh Diameterofafastenerholedi Innerdiameterofacircularsteelsection

414 Notations

dn,dne Depthsofplasticandelasticneutralaxis,respectivelydn1,dn2 DepthsofthefirstandsecondPNAintheconcreteslab,respectivelydo Outsidediameterofacircularsteelsectionordistancefromtheouter-

mostlayeroftensilereinforcementtotheextremecompressivefibreoftheslabordoistakenasthelesserof0.25Dsandtheheightoftheextendedface-bearingplates(dp)

dom Averageeffectivedepthofthetwolayersofreinforcementdp Depthofaplateorpanel;ordistancefromthetopfibretotheelasticcen-

troidofthesheetingdr Distancefromthetopfibretotheelasticcentroidofsteelreinforcement

or distance from the centroid of the longitudinal reinforcement in theconcreteslabtothetopfaceofthesteelsection

drc Distancefromthetopfaceofthecolumnflangetothefilletofthewebds Headdiameterofaheadedstudorasocketdsc DistancefromthecentroidofFscinthesteelsectiontothetopfaceofthe

steelsectiondsg Distancefromthecentroidoftheeffectivesteelsectiontothetopofthe

concreteslabdst DistancefromthecentroidofFstinthesteelsectiontothetopfaceofthe

steelsectiondt Depthofasteelteesectiondvr Distancebetweenthebarsinacompositeconnectiondw CleardepthofthewebofasteelI-sectionorpaneldwc Cleardepthofasteelcolumnwebdwt DepthofthesteelwebintensionD DepthofacolumncrosssectionD16 NumberofsixteenthofaninchintheweldsizeDc OveralldepthofaconcreteslabDr PlateflexuralrigidityDs DepthofasteelsectionDw Leglengthoffilletwelde Eccentricityofloadingeb Eccentricityofthereactiontotheboltcentrelineeh Distanceofelasticcentroidabovethebaseofsheetingep DistanceofthePNAabovethebaseofsheetingoreccentricityofthereac-

tionfortheplateweldedtoarigidsupportingelementew EccentricityofthebeamreactiontotheweldE Young’smodulusofmaterialEa DesignactioneffectEa⋅dst DesignactioneffectofdestabilizingactionEa⋅m,Ea⋅p Action effects caused by the mean and peak along-wind response,

respectivelyEa⋅t TotalcombinedpeakscaledynamicactioneffectEa⋅stb DesignactioneffectofstabilizingactionEc,Ecm Young’smoduliofconcreteEce EffectivemodulusofconcreteEc,eff Effectiveelasticmodulusofconcreteaccountingforlong-termeffectEce(t,τo) EffectivemodulusofconcreteE tce o∗( , )τ Age-adjustedeffectivemodulusofconcrete

Ec,p Actioneffectcausedbythepeakcrosswindresponse

Notations 415

(EI)eff Effectiveflexuralstiffnessofacompositecolumn(EI)eff,II Effectiveflexural stiffnessof a composite columnaccounting for long-

termeffectEs Young’smodulusofsteelmaterialEsl SiteelevationabovethemeansealevelEu EarthquakeactionEun Initialmodulusofelasticityofconcreteattheunloading′fc Compressivecylinderstrengthofconcreteat28 days′fcc Compressivestrengthofconfinedconcrete′fce Effectivecompressivestrengthofconcrete′fcf Characteristicflexuraltensilestrengthofconcreteat28 days′fcj Characteristiccompressivestrengthofconcreteatjdays

fck Characteristiccompressivestrengthofconcretefcm Meancompressivestrengthofconcreteatanyagefct Tensilestrengthofconcrete′fct Characteristicprincipaltensilestrengthofconcreteat28 days

fcu Compressiveconcretecubestrengthoftheinsituconcreteinfillfds Designshearcapacityofashearconnectorfna,fnc Firstmodenaturalfrequenciesofastructureinthealong-windandcross-

winddirections,respectivelyfnr Reducedfrequencyofastructurefro Concretestressatthereloadingfrp Lateralconfiningpressureprovidedbyacircularsteeltubeonconcretefso Steelstressattheunloadingfu Tensilestrengthofsteelfuc Tensilestrengthofshearconnectormaterialfuf Minimumtensilestrengthofaboltfup Tensilestrengthofaplyorplatefuw Tensilestrengthofweldmetalfuwa Tensilestrengthofwebanglefva∗ Averagedesignshearstressinthewebofasteelsection

fvm∗ Maximumdesignshearstressinthewebofasteelsectionfvs

Nominalshearcapacityofaweldedheadedstudfy Yieldstrengthofstructuralsteelfycf,fycw Yieldstrengthsofthesteelcolumnflangeandweb,respectivelyfyd Yieldstrengthofthedoublerplatefyf,fyw Yieldstrengthsoftheflangeandwebofasteelbeam,respectivelyfyf1,fyf2 Yieldstrengthsofthetopandbottomflangesofasteelbeam,respectivelyfyfp Yieldstrengthoftheadditionalbottomflangeplatefyp Yieldstrengthofsteelsheetingorbearingplatefyr Yieldstrengthofsteelreinforcementfys Yieldstrengthofstiffenerfysa YieldstressofseatangleF ForcederivedfromwindactionorhorizontalcyclicforceFC FactorappliedtowindspeedsinregionCFc1 Compressivecapacityoftheconcretecoverslabwithintheslabeffective

widthFc2 Compressivecapacityoftheconcretebetweensteelribswithintheslab

effectivewidth

416 Notations

Fcc Compressive force in theconcrete slabwithcomplete shearconnectionandγ≤0.5

Fccf Compressiveforceintheconcreteslabwithβ=1.0whenthesteelwebisignored

Fcp Compressiveforceintheconcreteslabwithpartialshearconnectionandγ≤0.5

Fcpf Compressiveforceintheconcreteslabofacompositebeamcrosssectionwithγ=1.0andpartialshearconnection

Fcst StrengthofreinforcedconcretecoverslabFd⋅ef EffectivedesignloadperunitlengthFD FactorappliedtowindspeedsinregionDFef1,Fef2,Fefp Effectivecapacitiesofthetopflange,bottomflangeandadditionalplate,

respectivelyFew EffectivecapacityofthewebofasteelsectionFf1,Ff2 Capacitiesofthetopandbottomflangesofasteelsection,respectivelyFh∗ Required horizontal design force on the bottom angle in a semi-rigid

compositeconnectionFr YieldcapacityofreinforcementintheconcreteslabFr1,Fr2,Fr3 Yieldcapacitiesofreinforcementinregions1,2and3,respectivelyFrm Maximumcapacityoflongitudinaltensilereinforcementintheconcrete

slabusedtocalculatethemomentcapacityofacompositebeamFs1,Fs2,Fs3 Tensionforcesinsteelcomponentsinregions1,2and3,respectivelyFsc ResultantcompressiveforceinthesteelsectionFsh StrengthofshearconnectionFst TensilecapacityofasteelbeamsectionFstf TensilecapacityofthetwoflangesofasteelsectionFw CapacityofthewebofasteelsectionFwc,Fwt Compressiveandtensileforcesinthewebofasteelsection,respectivelygv,gR Peak factors for upwind velocity fluctuations and resonant response,

respectivelyG PermanentactionordeadloadGsup Superimposeddeadloadsh Averageroofheightofabuildinghc Heightoftheconcretecoverslabinacompositeslabhn DistancebetweenthePNAandthecentroidofthecrosssectionofacom-

positecolumnhr Ribheightofprofiledsteelsheetinghs Average roof height of shielding buildings; height of a stud after

weldingH HeightoftheportalframeorhillHi ImpulseresponsematrixHm MechanicalresistanceforceHs HeightfactorforresonantresponseIc SecondmomentofareaofconcreteinacompositecolumnsectionIcr SecondmomentofareaofthecrackedsectionIcs EffectivesecondmomentofareaofacompositesectionIcy ModifiedmomentofinertiaofacompositesectionIef Effectivesecondmomentofareaofacrosssection

Notations 417

Ieti,Ietl Secondmomentsofareaofatransformedcompositebeamsectionwithpartialshearconnectionforshort-termandlong-termdeflectioncalcula-tions,respectively

If Secondmomentofareaofthetwoflangesofasectionaboutthecentroidofthesection

Ig SecondmomentofareaofgrosscrosssectionIh TurbulenceintensityIn MomentofinertiaofcompositesectioninnegativebendingIp Polarmomentofareaofboltsormomentofinertiaofcompositesection

inpositivebendingIr SecondmomentofareaofreinforcementIs SecondmomentofareaofastiffenerorasteelsectionIt Secondmomentofareaofatransformedcompositebeamsectionwith

completeshearconnectionIti,Itl Secondmomentsofareaofatransformedcompositebeamsectionwith

completeshearconnectionforshort-termandlong-termdeflectioncalcu-lations,respectively

Iox⋅j,Ioy⋅j Secondmomentsofareaofthejthelementaboutitscentroidalx-axisandy-axis,respectively

Ix,Iy Secondmomentsofareaofacrosssectionaboutitscentroidalx-axisandy-axis,respectively

Iw WarpingconstantIweb Second moment of area of the web of an I-section about the section

centroidIwp PolarsecondmomentofareaofaweldgroupIwx Secondmomentofareaofaweldgroupaboutthex-axisJ Torsionalconstantk1,k2 Coefficientsk3,k4 Deflectionconstantskb Elasticbucklingcoefficientke Membereffectivelengthfactorkct Correction factor considering the effect of non-uniform force distribu-

tionsinducedbyendconnectionskf Formfactoraccountingfortheeffectofplatelocalbucklingkh Factoraccountingfortheeffectholetypekl Load height factor accounting for the destabilizing effect of gravity

loadskmw Ratio of the second moment of area of the web to that of the whole

I-sectionkn Load-sharingfactorkpr Factoraccountingfortheeffectofadditionalboltforceduetopryingkr Lateralrotationalrestraintfactorkrc,krw Lengthreductionfactorsforboltedlapconnectionsandweld,respectivelykt Twistrestraintfactorku Neutralaxisparameterkv Flatwidthtothicknessratioofthewebkw Ratioofthecross-sectionalareaofthewebtothegrossareaofthesectionkx,ky Elasticlocalbucklingcoefficientsinthexandydirections,respectively

418 Notations

kxo Elastic local buckling coefficient in the x direction under biaxialcompression

kxy Elasticshearbucklingcoefficientkxyo CriticalshearbucklingcoefficientunderpureshearKa AreareductionfactorKc CombinationfactorappliedtowindpressuresKc,e,Kc,i Combination factors applied to external and internal wind pressures,

respectivelyKl LocalpressurefactorKm ModeshapecorrectionfactorforcrosswindaccelerationKp Porouscladdingreductionfactorl Lengthofasegmentlb = ∗ ∗M Vb b/lc Lengthcorrectionfactororl M Vc c c= ∗ ∗/liw Lengthoftheithweldsegmentlj Connectionlengthls Averagespacingofshieldingbuildingslw Lengthofaweldedlapconnectionl SpanofabeamorlengthofaplateL1,L2 Lengthscalesforhills,ridgesandescarpmentsLa LengthofaboltLc LengthofchannelshearconnectorLd LengthembedmentLe EffectivelengthofamemberLef EffectivespanLex,Ley Effectivelengthsofamemberbendingaboutitssectionmajorandminor

principalaxes,respectivelyLh InternalturbulencelengthscaleatheighthorlengthofthehookofaboltLp PanelwidthLs Socketlength;shearspanLu HorizontaldistanceupwindfromthecrestofahillLw LengthofaweldLyst StressdevelopmentlengthoflongitudinalreinforcementinconcreteslabsM∗ DesignbendingmomentM1∗ Largerdesignbendingmomentattheendofacolumn

M2∗ Smallerdesignbendingmomentattheendofacolumnordesignmoment

atthequarterpointofasegmentM3∗,M4

∗,Mm∗ Designmomentsatthemidpoint,quarterpointandmaximummomentof

asegment,respectivelyM_∗,M R

_∗ Negativedesignmomentsatthesupportbeforeandafterredistribution

Mb Nominalmembermomentcapacityofasteelmemberornominalmomentcapacityofacompositebeamcrosssectionwithγ≤0.5andpartialshearconnection

Mb∗ Bendingmoment causedbyeccentricityof shear forceor sumofMb1

∗ andMb2

∗

Mb1∗ ,Mb2

∗ Design bending moments on the left and right beams of a compositeconnection

Mb⋅5 Nominalmomentcapacityofacompositebeamcrosssectionwithγ≤0.5andβ=0.5

Notations 419

ϕMb⋅ ψ Nominalmomentcapacityofacompositebeamcrosssectionwith0.5<γ≤1.0andβ=ψ

Mbc Nominalmomentcapacityofacompositebeamcrosssectionwithγ≤0.5andcompleteshearconnection

Mbf Nominalmomentcapacityofacompositebeamcrosssectionwithγ=1.0andpartialshearconnection

Mbfc Nominalmomentcapacityofacompositebeamcrosssectionwithγ=1.0andcompleteshearconnection

Mbv Nominalmomentcapacityofacompositebeamcrosssectionwith0.5<γ≤1.0

Mbx Member moment capacity bending about its section major principalx-axis

Mbxo Nominal member moment capacity without full lateral restraint andunderuniformbendingmoment

Mc CrosswindbaseoverturningmomentMc∗ SumofMc1

∗andMc2∗

Mc1∗ ,Mc2

∗ Designbendingmomentsattheupperandlowercolumns,respectivelyMcm∗ ,Mdm

∗ Design moments at the mid-span of a composite beam under factoredconstructionloadsanddeadloads,respectively

Mcr CrackingmomentMcx =min(Mix;Mox)Md WinddirectionalmultiplierMe,Me,max MomentattheendsofacolumnanditsmaximumvalueMend∗ Designbendingmomentatthecolumnendamplifiedbythesecond-order

effectMf∗ DesignbendingmomentcarriedbythetwoflangesofanI-section

Mh Hill-shapemultiplierMi Nominalin-planemembermomentcapacityMimp∗ Design bending moment at the mid-height of the composite column

inducedbygeometricimperfectionsMlee TheleemultiplierMme Externalbendingmomentatthemid-heightofabeam–columnMmi Resultantbendingmomentatthemid-heightofabeam–columnMmin∗ Minimumdesignbendingmoment

Mo Referencebucklingmomentofasteelmemberunderbendingorultimatepurebendingmomentcapacityofacolumn

Moa ElasticbucklingmomentofasteelmemberunderbendingMox Nominalout-of-planemembermomentcapacityofamemberunderaxial

compressionandbendingMp FullplasticmomentMpa NominalsectionmomentcapacityofsteelsheetingMpr NominalmomentcapacityduetocoupleforcesincompositeslabMprx,Mpry Nominalplastic sectionmomentcapacitiesabout themajorandminor

principalx-andy-axesreducedbyaxialforce,respectivelyMqe∗ Factoredliveloadmomentatthesupports

Mqm∗ Factoreddesignmomentatthemid-span

Mrx,Mry Nominalsectionmomentcapacitiesaboutthemajorandminorprincipalx-andy-axesreducedbyaxialforce,respectively

Ms Shieldingmultiplierorsectionmomentcapacity

420 Notations

Msf Nominalmomentcapacityofthesteelsectionneglectingthecontributionoftheweb

Msh BendingmomentinducedbyconcreteshrinkageMse Bendingmomentatthesectionundershort-termserviceloadMser NominalmomentcapacityofaconnectionunderserviceloadsMsqe∗ Momentatthesupportunderserviceliveloads

Msx,Msy Nominalsectionmomentcapacitiesaboutthemajorandminorprincipalx-andy-axes,respectively

Mt TopographicmultiplierMtx =min(Mrx;Mox)Mu Ultimatemomentcapacityofacompositebeamincombinedbendingand

shearMu,min MinimumbendingstrengthofcompositeslabinpositivemomentregionMu,max Maximummomentcapacityofacompositecolumnunderaxialloadand

bendingMuox,Muoy Puremomentcapacitiesofthecolumnsectionforbendingaboutthesec-

tionmajorandminorprincipalaxes,respectivelyMup NominalsectionmomentcapacityofthesteelsheetingaloneMx∗,My

∗ Designbendingmoments about the sectionmajor andminorprincipalx-andy-axes,respectively

My SectionfirstyieldmomentcapacityMuo UltimatemomentcapacityofacompositesectioninpurebendingMux,Muy Nominalmomentcapacitiesofaslendercompositecolumnbendingabout

thesectionmajorandminorprincipalaxes,respectivelyMw∗ Designbendingmomentcarriedbytheweborbendingmomentcausedby

eccentricitytotheweldMz∗ Designbendingmomentaboutthecentroidofaboltgroup

Mz,cat Terrain/heightmultipliern Numberofhalfwavesinthedirectionoftheappliedload;modulusrationb Numberofparallelplanesofbattensornumberofboltsinaboltgroupnc Numberofshearconnectorsbetweentheendofthebeamandthecross

sectionbeingconsideredncw Numberofboltsalongthewebandatthecompressionflangeni Numberofshearconnectorsbetweenthepotentiallycriticalcrosssection

iandtheendofthebeamnn Numberofshearplaneswiththreadsinterceptingtheshearplanesns Totalnumberofupwindshieldingbuildingswithina45°sectionofradius

20hnw Numberofwebsinasegmentnx Numberof shearplaneswithout threads intercepting the shearplanes;

numberofshearconnectorsatacrosssectionofacompositebeamN∗ DesignaxialloadNbc NominalbearingstrengthofconcreteNc NominalmembercapacityofacompressionmemberNcc Pull-outresistanceofconcreteorcompressiveforceinconcretecoverslabNc∗ Designaxialcompressionforce

Ncm∗ ,Ntm

∗ Designforcesincompressionandtensionflangesduetobendingmoment,respectively

Ncp CompressiveforceinconcreteofcompositeslabwithpartialshearconnectionNcr Elasticbucklingloadofamember

Notations 421

Ncy Nominalmembercapacity inaxialcompressionforbucklingaboutthesectionminorprincipaly-axis

Nf∗ Maximumforceinthecriticalflangesofadjacentsegments

Nfc∗ Resultanthorizontaldesignforceincompressionflange

Nfc1∗ ,Nfc2

∗ Designcompressionforcesinflangesontheleftandrightsidesofthesteelcolumn,respectively

Nft∗ Resultanthorizontaldesignforceintensionflange

Nft1∗ ,Nft2

∗ Tensionforcesinthebeamflangeontheleftandrightsidesofthesteelcolumn,respectively

NR∗ Nominaltransversedesignforcecarriedbyrestraint

Nom Elasticbucklingloadofacompressionmemberdeterminedbytheelasticbucklinganalysis

Noz ElastictorsionalbucklingcapacityofamemberNp TensileforceinsheetingNpb NominalcapacityoftheendplateinbendingNpl,Rd UltimateaxialstrengthofcompositecolumnsectionNs NominalsectionaxialcapacityofasteelmemberNsc =min(Nsc1,Nsc2)Nsc1,Nsc2 NominalcapacitiesofbaseplateundercompressionNsh AxialforceinducedbytheshrinkageofconcreteNst NominalcapacityofsteelbaseplateduetoaxialtensioninthecolumnNt Nominalsectioncapacityinaxialtensionorcapacityofanchorboltin

tensionNt∗ Designaxialtensionforce

Ntb NominaltensilecapacityofaboltgroupNtf NominaltensilecapacityofaboltNtf∗ Designtensionforceonabolt

Nti MinimumbolttensionforceatinstallationNts NominalcapacityofatensionstiffenerorcolumnflangeNts∗ Resultanttensionforceinthebeamflangesofthebeam–columnconnection

Nty,Nta Nominal gross yield and fracture capacities of a steel section in axialtension,respectively

Nvs CapacityofdiagonalstiffenerNvs∗ Designforceonthediagonalstiffener

Nw NominalcapacityoffilletweldaroundasteelelementNwnv∗ Totalhorizontaldesignforceononeweldontheweb

Nz∗ Out-of-planetensionforceonaboltgroupinthezdirection

p Windpressurepz DesignwindpressureonsurfaceatheightzP PointloadoraxialforceP∗ DesignaxialforcePa AppliedaxialloadPG∗ PermanentpartofthedesignaxialforceP*Pcr ElasticbucklingloadPcr,eff Elasticbucklingloadofacompositecolumncalculatedusing(EI)eff,II

Pmax MaximumaxialloadofashortcompositecolumnPmo Ultimateaxialloadofashortcolumnwhenitsmomentcapacityisequal

toMo

Po,Poa Ultimateaxialloadsofshortandslendercolumnsunderaxialcompres-sion,respectively

422 Notations

Pup UltimateaxialstrengthsofconcentricallyloadedCFSTslendercolumnswithpreloadeffects

Pu UltimateaxialloadofacompositeshortcolumnΔPu AxialloadincrementQn LongitudinalshearforceonashearconnectorQ Imposedactionorliveloadr Radiusofgyrationofasectionre Outsideradiusofhollowcrosssectionrm,rm

b,rmc Residualmomentsinacompositecolumnsection

rma ,rp Residualforcesinacompositecolumnsection

rx,ry Radiiofgyrationofasectionaboutitsmajorandminorprincipalx-andy-axes,respectively

R∗ DesignbearingforceorreactionforceRb NominalbearingcapacityofthewebofasteelsectionRbb,Rby Nominal bearing buckling and yield capacities of a steel web,

respectivelyRc NominalbearingcapacityofthecolumncompressionflangeRc1,Rc2 Nominalbearingbucklingandyieldcapacitiesofthecolumncompres-

sionflange,respectivelyRcs NominalcapacityofstiffenedcolumnwebRn NominalcapacityorresistanceofastructuralmemberRsb,Rsy Nominalbucklingandyieldcapacitiesofthestiffener-webcompression

member,respectivelyRt =min(Rt1,Rt2)Rt1,Rt2 NominalresistancesofcolumnflangeundertensionRtd NominalcapacityofstiffenedcolumnflangeRw Nominalstrengthofeccentrically loadedweldgroupundershearforce

andbendingmoments Spacingoftransversewebstiffenerssb Longitudinalcentre-to-centredistancebetweenbattenssep Distancebetweentheendplateandloadbearingstiffenersg,sp Gaugeandpitchofbolts,respectivelysr Centre-to-centrespacingofsteelribsssr Spacingoftiesinthedepthofthebeamwebsx TransversespacingofstudsinthecrosssectionofacompositebeamSt Spectrumoftheturbulenceofastructuret Thicknessofaplatet1,t2 ThicknessoftheflangesofamonosymmetricsteelI-sectiontcf,tcw Thicknessoftheflangeandwebofachannel,respectivelyta Thicknessofwebangletd Thicknessofdoublerplatetew Effectivethicknessofasteelwebtf Thicknessofasteelflangetf1,tf2 Thicknessofthetopandbottomflangesofasteelsection,respectivelytfc,twc Thicknessoftheflangeandwebofasteelcolumn,respectivelytp Thicknessofaplatetts Thicknessoftheteestemofasteelteesectiontw ThicknessofasteelwebTp Resultanttensileforceinthesteelsheetingofacompositeslabwithpar-

tialshearconnection

Notations 423

Tpcs Resultanttensileforceinthesteelsheetingofacompositeslabwithcom-pleteshearconnection

Ti TensionforceontheithboltTs TensileforceinthesteeltubeTvr ForceintheverticalreinforcementintensionTyp YieldcapacityofsteelsheetingTyr Yieldcapacityofsteelreinforcementu Displacementu1,u2,u3 Perimeterlengthsoflongitudinalshearsurfacesuo Initialgeometric imperfectionat themid-heightofa slendercomposite

beam–columnumo Deflectionatthemid-heightofthesteeltubecausedbythepreloadulast Deflectionatthelastiterationum Displacement/deflectionatthemid-heightofcolumnorcentreofaplateuold Deflectionatthepreviousiterationup PerimeterlengthofType1shearsurfacesups Criticalperimeterlengthut LateraldeflectionatthetipofacantilevercolumnΔut Deflectionincrementatthetipofacantilevercolumnuto Initialgeometricimperfectionatthetipofacantilevercolumnν Poisson’sratioνe Poisson’sratioofthesteeltubewithconcreteinfillνs Poisson’sratioofthesteeltubewithoutconcreteinfillvres∗ Resultantforceperunitlengthontheweldsegmentvmin Shearstrengthofconcretevps Designpunchingshearstressvw Nominalcapacityofafilletweldperunitlengthvw∗ Designforceperunitlengthofweldvx∗,vy∗,vz∗ Designforcesperunitlengthinweldsegmentinthex,yandzdirections,

respectivelyvzm∗ Maximum shear stress in the horizontal direction caused by bending

momentvznv∗ ShearinthezdirectioncausedbyNwnv

∗Vb NominalshearbucklingcapacityofthewebofasteelsectionVb∗ DesignbearingforceorV V Vb b b

∗ = ∗ + ∗( )1 2 2/Vb1∗,Vb2

∗ Designshearforcesintheleftandrightbeamsofabeam–columnconnec-tion,respectively

Vbc Nominalbearingortear-outcapacityofthesupportingplateVbp NominalbearingcapacityoftheplyduetoaboltinshearVc Nominalshearcapacityofthewebofasteelcolumnorcontributionof

theconcreteslabtotheverticalshearcapacityVcc Nominal shear capacity of concrete in interior reinforced concrete

connections′Vc Strengthprovidedbytheconcreteincompression

Vc∗ = ∗ + ∗( )V Vc c1 2 2/

Vc1∗,Vc2

∗ Designshearforcesintheupperandlowercolumnsofabeam–columnconnection,respectively

Vcf,Vcff Nominalstrengthandforceoftheconcretecompressionfield,respectivelyVcs NominalstrengthofthecompressionstrutVf Nominalshearcapacityofabolt

424 Notations

Vfe EffectiveshearcapacityofanchorboltVfn NominalshearcapacityofaboltgroupV∗ DesignshearforceVL∗ DesignlongitudinalshearforceperunitlengthonType1,2and3shear

surfacesVL tot⋅∗ Totaldesignlongitudinalshearforceperunitlengthofcompositebeam

Vd1 NominalshearcapacityofbaseplatebasedonfrictionVdes,θ DesignwindspeedVf∗ Designshearforceonabolt

Vfb NominalbearingcapacityofaplyVfn NominalshearcapacityofboltgroupVl NominallongitudinalshearcapacityofacompositeslabVl∗ Designlongitudinalshearforce

Vmin∗ Minimumdesignshearforce

Vn ReducedvelocityVns ShearfracturecapacityofthenetsectionofasteelplateVo Verticalshearcapacityofnon-compositesectionVo∗ Designshearforceonaboltgroup

Vph NominalcapacityoftheendplateinhorizontalshearVps NominalpunchingshearcapacityofcompositeslabVpv NominalcapacityoftheendplateinverticalshearVR Regional3sgustwindspeedVres∗ Resultantdesignshearforceonabolt

Vs Shearcapacityofthewebofasteelbeam′Vs Strengthprovidedbythehorizontalties

Vsf NominalshearcapacityofaboltunderserviceloadVsf∗ Designshearforceinservicecondition

Vslab VerticalshearstrengthoftheconcreteslabVsit,β SitewindspeedVtp Tear-outcapacityofaplyVtf,Vts Nominalshearyieldcapacitiesoftheteeflangeandstemofasteelteesec-

tion,respectivelyVu NominalshearcapacityofasectionorwebVuo UltimateshearstrengthofcompositesectioninpureshearVus NominalshearcapacityofembeddedanchorboltinshearVv NominalshearcapacityofasteelwebVvc∗ Resultantverticaldesignshearforceontheendplate

Vw NominalshearyieldcapacityofasteelwebVw∗ Design shear force in the steel beam web in a beam-to-CFST column

connectionVxb∗ ,Vyb

∗ Designshearforcesonaboltinthexandydirections,respectivelyVxbm∗ ,Vybm

∗ MaximumboltforcesduetoMz∗inthexandydirections,respectively

weq(z) WindforceW AppliedloadWu UltimatewindloadWs Servicewindloadx Horizontaldistancefromastructuretothecrestofthehillorineffective

lengthofthewebofasteelsectionxc Coordinateofthecentroidofasection

Notations 425

xcs Distancefromtheendofthesteelsheetingtothecrosssectionwithcom-pleteshearconnection

xj Centroidalcoordinateofelementjxmax Maximumdistancefromcentroidalx-axisofasectiontoitsextremefibrexn Coordinateoftheboltnxn,i Distancefromthecentroidoftheithfibreelementyi,yj Coordinatesofanelementjymax Maximumdistancefromcentroidaly-axisofasectiontoitsextremefibreyn Coordinateoftheboltnyn,i Distancefromthecentroidoftheithfibreelementyp TheheightofthetensileforceTpactsyt Distancefromthecentroidalaxisofthecrosssectiontotheextremeten-

silefibrez LevelarmZ ElasticsectionmodulusZc EffectivesectionmodulusofacompactsteelsectionZe EffectivesectionmodulusofasteelsectionZex,Zey Effectivesectionmoduliforbendingaboutthesectionmajorandminor

principalaxes,respectivelyZp PlasticsectionmodulusZx,Zy Elasticsectionmoduliaboutitscentroidalx-andy-axes,respectivelyα Coefficientorloadanglewithrespecttothey-axisofacompositecolumn

sectionαa Slendernessmodifierαb Membersectionconstantaccountingfortheeffectofresidualstresspatternsαbc Factoraccountingfortheeffectsofmomentratioandaxialforceonthe

out-of-planemembermomentcapacityαc Memberslendernessreductionfactorαcb = ∗ ∗V Vc b/αcs Ratioofcompressivestressesintwodirections,αcs=σx/σy

αd Tensilefieldcontributionfactoraccountingforthecontributionoftensilefieldtoshearbucklingcapacityofasteelweb

αf Flangerestraintfactoraccountingfortherestrainingeffectofflangesontheshearbucklingcapacityofasteelweb

αfc FactoraccountingfortheeffectofconcretecompressivestrengthonthemomentcapacityofacircularCFSTcolumnsection

αg Imperfectionfactorαm MomentmodificationfactorαM Reductionfactoraccountingfortheeffectofunconservativeassumption

oftherectangularstressblockthatisextendedtothePNAαp Reductionfactorforplateinbearingαs Stressgradientcoefficientorslendernessreductionfactorαv Stiffeningfactoraccountingfortheeffectsoftransversestiffenersonthe

shearbucklingcapacityofasteelwebαw Reductionfactorduetoshearbucklingαy Factoraccountingfortheeffectoftheyieldstrengthofthesteeltubeon

themomentcapacityofacircularCFSTcolumnsectionβ Degreeofshearconnectionβa Preloadratio

426 Notations

βce Factor used to consider the confinement effect provided by the rect-angularsteeltubeonthepost-peakstrengthandductilityofconfinedconcrete

βcc Factorusedtoconsidertheconfinementeffectprovidedbythecircularsteeltubeonthepost-peakstrengthandductilityofconfinedconcrete

βe Modifyingfactoraccountingfortheconditionatthefarendsofabeamβi Minimumdegreeofshearconnectionβm Momentratioβm M M= ± ∗ ∗

2 1/βmb Degreeofshearconnectionatthecrosssectionunderthemaximumbend-

ingmomentβs Sizereductionfactorβsc Degreeofshearconnectionofcompositeslabβx Monosymmetricsectionconstantχ Reductionfactoraccountingfortheeffectofrelativeslendernessλand

imperfectionsonthestrengthofcolumnχa Ageingcoefficientχd Loadratio,χd=P∗/Po

χprg Strengthreductionfactoraccountingfortheeffectsofpreloadratio,rela-tiveslendernessandgeometricimperfectionsontheultimatestrengthofCFSTslendercolumnunderaxialcompression

δ LongitudinalslipδC1⋅3 Deflectioncausedconstructionloadsatstages1–3δC5.6 Immediatedeflectionofcompositebeamduringconstructionstages5–6δcr Long-termdeflectionofcompositebeamduetoconcretecreepδFF Deflectionofthefixedendcompositebeamδj,δ j

∗ Thejthdisplacementordeflectionofastructureanditslimitδl Long-termdeflectionδm AmplificationfactorδQ Immediatedeflectionofcompositebeamundershort-termliveload(ψsQ)δs Short-termdeflectionδsh Long-termdeflectionofcompositebeamduetoconcreteshrinkageδsr Deflectionofcompositebeamswithsemi-rigidcompositeconnectionδsus Deflectionduetosustainedloadδtot Totaldeflectionεa Concretestrainεb InitialvalueofthesteelstrainatreloadingεB ConcretestrainatpointB,εB=0.005εc Longitudinalcompressivestrainofconcrete′εc Concretestraincorrespondingto ′fc′εcc Compressiveconcretestrainat ′fcc′εce Strainat ′fce

εcp Concretestrain=0.015εcr(t,τo) Concretecreepstrainεcs∗ Finalfreeshrinkagestrainofconcreteεct Concretestrainatcracking′εct Concretestrainintension

εel(τo) InstantaneousstrainofconcreteεF Concretecompressivestrain,takenas0.02εi Strainattheithfibres

Notations 427

εk Convergencetoleranceεmo Steelstrainεpl Plasticstrainofconcreteεr Straininreinforcementεre Returnstrainonthemonotoniccurveεro Concretestrainatfro

εs Straininasteelfibreε(t) Totalstrainofconcreteεsh(t) Shrinkagestrainofconcreteεsh Restrainedshrinkagestrainofconcreteinacompositebeamεso Steelstrainattheunloadingεst Steelstrainatstrainhardeningεsu Ultimatestrainofsteelεtu Ultimatetensilestrainofconcreteεy YieldstrainofsteelmaterialΔε Axialfibrestrainincrementεun Concretestrainatunloadingcorrespondingtoσun

ϕ CapacityreductionfactororcurvatureΔϕ Curvatureincrementϕb Curvatureatthebaseofacantilevercolumnϕc(t,τo) Creepfunctionorfactorofconcreteφc∗ Finalcreepfactorofconcrete

ϕe Curvatureatthecolumnendsϕm Curvatureatthemid-heightofabeam–columnΔϕm Curvatureincrementϕprg Coefficientϕs Factorϕs=1−αs

ϕt Strengthreductionfactorsteeltubeϕt=0.75ϕy Yieldcurvatureφ Plateaspectratioorcoefficientφ1 Coefficientfordeterminingtheverticalshearcapacityofconcreteslabφ2 Coefficientfordeterminingtheverticalshearcapacityofcompositebeamφb Bendingfactorofprofiledsteelsheetingφpa,φpe Strengthreductionfactorsforstudsincompositeslabwithribsoriented

parallelandperpendiculartothesteelbeam,respectivelyγ Reductionfactorforconcretestrengthorshearratioγ1,γ2,γj Stiffnessratiosofacompressionmemberatend1andend2γb Exponentofthestrengthinteractionactioncurveγn Uniaxialstrengthfactorγs Strengthfactoraccountingfortheeffectofhooptensilestressesandstrain

hardeningontheyieldstressofthesteeltubeγw Factoraccountingfortheeffectofstiffenertypesη Imperfectionparameterλ Combined slenderness of a member or load factor or multiplier or

coefficientλ Relativeslendernessofacolumnλc Collapseloadfactorλe Slendernessofaplateλep Elementslendernessplasticitylimit

428 Notations

λey Elementyieldslendernesslimitλm FactoraccountingfortheeffectofD/tratioonthemomentcapacityof

CFSTcolumnsectionλn Modifiedmemberslendernessλs,λsp,λsy Theslendernessplasticitylimitandyieldlimitofanelementhavingthe

greatestvalueofλe/λsyinthesectionμ Slipfactororfrictioncoefficientμd Momentcapacityfactorcorrespondingtoχd

μdx,μdy Momentcapacityfactorforbendingaboutthesectionmajorandminorprincipalaxes,respectively

θ Pitchofroof/rafterorangleorrotationororientationoftheneutralaxiswithrespecttothex-axisinacompositecolumnsection

ρ Densityofmaterial;effectivereinforcementratioρair Densityofairρs Effectivereinforcementratioρx,ρy Reinforcementratiosinxandydirections,respectivelyσ1,σ2 Maximumandminimumedgestressesonaplate,respectivelyσ1c Initialbucklingstressofaplateσ1u Ultimatevalueofthemaximumedgestressσ1onaplateσb Elasticbearingbucklingstressofaplateundercombinedactionsorsteel

stressatthestrainεb

σc Longitudinalcompressivestressofconcreteσcr Criticallocalbucklingstressσf Elasticbendingbucklingstressofaplateundercombinedbendingand

shearσob Elasticbucklingstressofaplateinpurebearingσo Axialstressappliedattimeτo

σv Elasticshearbucklingstressofaplateundercombinedbendingandshearσof Elasticlocalbucklingstressofaplateunderin-planebendingσov Elasticlocalbucklingstressofaplateinshearσre Returnstressonthemonotoniccurveσs Stressinasteelfibreσsc Stressinsteeltubeincompressionσst Stressinsteeltubeintensionσt Tensilestressintheconcreteforunloadingfromthecompressiveenvelopeσu Averageultimatestressactingonaplateσun Compressivestressofconcreteattheunloadingσx,σy Normalstressesinxandydirections,respectivelyσxcr,σycr Elasticbucklingstressesinxandydirections,respectivelyσxu Ultimatestrengthofasteelplateinthexdirectionσxuo Ultimatestrengthofasteelplateunderbiaxialcompressiononlyinthex

directionσyu Ultimatestrengthofasteelplateintheydirectionτ Shearstressτo Initialtimewhenaxialstressσoappliedtoconcreteτov Elasticshearbucklingstressofaplateinpureshearτv Elasticshearbucklingstressofaplateundercombinedactionsτxy Shearstressτxyu Ultimateshearstrengthofsteelplateτxyuo Ultimateshearstrengthofsteelplateunderpureshear

Notations 429

τy Shearyieldstressν Poisson’sratioorshapefactorξ Factor that is a function of combined slenderness and imperfection

parameterξm Momentredistributionparameterψ Degreeofshearconnectionatthecrosssectionwithγ=1.0andcomplete

shearconnectionψa Reductionfactorusedtoreducetheuniformlydistributedliveloadsψc,ψs,ψl Combination,short-termandlong-termfactors,respectivelyω,ω1,ω2,ω3 Variableandinitialvaluesofthevariablesζ Ratio of structural damping to critical damping of a structure or

coefficient

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