Redistribution of force concentrations in reinforced concrete cantilever slab using 3D non-linear FE analyses MasterofScienceThesisintheMastersProgrammeStructuralEngineeringand Building Technology SONGLY LIM Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures CHALMERS UNIVERSITY OF TECHNOLOGY Gteborg, Sweden 2013 Masters Thesis 2013:100 x y my x my Linear elastic isotropic Linear elastic orthotropic Plastic MASTERS THESIS 2013:100 Redistribution of force concentrations in reinforced concrete cantilever slab using 3D non-linear FE analyses Master of Science Thesis in the Masters Programme Structural Engineering and Building TechnologySONGLY LIM Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures CHALMERS UNIVERSITY OF TECHNOLOGY Gteborg, Sweden 2013 Redistribution of force concentrations in reinforced concrete cantilever slab using 3D non-linear FE analyses MasterofScienceThesisintheMastersProgrammeStructuralEngineeringand Building TechnologySONGLY LIM SONGLY LIM, 2013 Examensarbete / Institutionen fr bygg- och miljteknik,Chalmers tekniska hgskola 2013:100 Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures Chalmers University of Technology SE-412 96 Gteborg SwedenTelephone: + 46 (0)31-772 1000 Cover: Cantilever slab subjected to a single point load and the resulting moment distribution along the support line for different types of structural response. Chalmers ReproserviceGteborg, Sweden 2013 I Redistribution of force concentrations in reinforced concrete cantilever slab using 3D non-linear FE analyses MasterofScienceThesisintheMastersProgrammeStructuralEngineeringand Building TechnologySONGLY LIM Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures Chalmers University of Technology ABSTRACT Finite element analyses (FE analyses) methods are nowadays commonly used for the analysisanddesignofcivilengineeringstructures.Whenmodelingreinforced concreteslabs,localforceconcentrationariseswhenusing3DlinearelasticFE analyses.Theseforceconcentrationswillinrealitybedistributedduetoconcrete cracking and yielding of tensile reinforcement. This master thesis uses 3D non-linear FE analyses and simplified methods to describe the structural response of a reinforced concretecantileverslabsubjectedtoasinglepointload.Thecantileverslabstudied was modeled as a homogeneous material with different material models such as linear elasticisotropicandorthotropicmaterial,andbilinearandmulti-linearelasto-plastic model in order to investigate the structural response of the slab in different states. For linear elastic isotropic and orthotropic analysis of the studied slab, a shell element modelwasusedtovalidatethebeamgrillagemodellaterusedinthemain investigation.ThecomparisonofFEresultsshowsthatshellelementisthemost appropriateelementtouseandthereisadivergenceuponmeshrefinementwhen usingbeamelementsduetotorsionaleffects.Nevertheless,thebeamgrillagemodel can be used in case that appropriate beam element mesh size is used.Fornon-linearanalysis,thestudiedcantileverslabwasanalyzedusingthebeam grillagemodelwithnon-linearmoment-curvaturerelationshipsinceshellelement models previously have proven to give incorrect results when used with elasto-plastic material response, due to unwanted biaxial effects. Concerning non-linear analysis in servicelimitstate(SLS),theorthotropicanalysisisshowntoprovideagood approximation of maximum moment. In the ultimate limit state (ULS), there is a small need for plastic rotation of the slab before the slab fails in non-linear analysis. Hence, there is a possibility to redistribute the moment along the entire length of the studied slabduetoyieldingoftensilereinforcement.Thecomparisonbetweenapresent guideline and the FE analyses shows that the recommendation provided in the former isconservativeintermsofdistributionwidthformomentincantileverslabs. However,theusershouldbecautiouswiththisconclusionsinceitisbasedonone particular load case. In order to provide a better understanding of the behaviour of the reinforcedconcretecantileverslabandexaminetherecommendationfurther,further studies should be carried out. Key words:Non-linear3DFEanalyses,redistributionofforces,reinforced concrete,cantileverslab,materialmodel,moment-curvature,beam element, shell element CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 IIContents ABSTRACTI CONTENTSII PREFACEVI NOTATIONSVII 1INTRODUCTION1 1.1Background1 1.2Purpose1 1.3Method1 1.4Limitations2 1.5Outline of the report2 2THEORY4 2.1Finite element4 2.2Structural response of reinforced concrete structures4 2.2.1Material response4 2.2.1.1Concrete4 2.2.1.2Reinforcing steel5 2.2.2Structural response of reinforced concrete members6 2.2.3Theory of plasticity and plastic hinges10 2.2.4Plastic rotation capacity11 2.2.5Non-linear response in service limit state, SLS14 2.2.6Non-linear response in ultimate limit state ULS15 2.3Beams16 2.3.1Introduction16 2.3.2Global methods for the response of a reinforced concrete beam16 2.3.3Simplified linear elastic analysis17 2.3.4Simplified linear elastic analysis with limited redistribution18 2.3.5Plastic analysis19 2.3.6Shear20 2.4Slabs22 2.4.1Introduction22 2.4.2Definition26 2.4.3Alternative methods for slab analysis26 2.4.4Design based on linear elastic analysis27 2.4.5Strip method28 2.4.6Yield line method30 2.4.7Distribution width for moment in ultimate limit states32 2.4.8Distribution width for moment in serviceability limit states34 2.4.9Distribution width for shear forces35 2.4.10Cantilever slabs39 2.4.10.1Distribution width for moment40 CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 III 2.4.10.2Distribution width for shear force42 3REINFORCED CONCRETE CANTILEVER BEAM45 3.1Introduction45 3.2Equivalent Youngs modulus45 3.3FE-elements46 3.4Geometry and loading46 3.5Elasto-plastic case46 3.5.12D beam element-bilinear plastic material model46 3.5.23D beam element-bilinear plastic material model48 3.5.33D beam element-moment curvature model52 3.5.4Alternative methods for plastic rotation of cantilever beam55 4CANTILEVER SLAB58 4.1Introduction58 4.2Geometry and loading58 4.3Isotropic case59 4.3.1Slab with shell elements59 4.3.1.1Moment60 4.3.1.2Shear61 4.3.1.3Vertical displacement62 4.3.2Slab with beam grillage model63 4.3.2.1Moment63 4.3.2.2Shear65 4.3.2.3Vertical displacement66 4.3.3Comparisons between shell elements and beam elements67 4.4Orthotropic case69 4.4.1Slab with shell elements69 4.4.1.1Convergence study69 4.4.2Structural response of orthotropic slab73 4.4.2.1Moment73 4.4.2.2Shear75 4.4.2.3Vertical displacement76 4.4.3Distribution width of orthotropic slab77 4.4.4Comparisons between shell elements and beam elements80 4.5Elastoplastic case-beam grillage model84 4.5.1Choice of model used for the analysis84 4.5.2Influence of mesh-isotropic case87 4.5.3Quadlinear M() - Mpl = 0.6Mel, EII = EII/589 4.5.3.1Moment89 4.5.3.2Curvature contour plot94 4.5.3.3Plastic rotation97 4.5.3.4Comparison with existing guideline98 4.5.3.5Shear99 4.5.4Influence of different state II stiffness in quadlinear M()102 CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 IV4.5.4.1Orientation102 4.5.4.2Load-displacement relation102 4.5.4.3Plastic rotation103 4.5.5Comparisons between trilinear and quadlinear M()104 4.5.5.1Load-displacement relation105 4.5.5.2Moment106 4.5.5.3Plastic rotation108 4.5.6Quadlinear M()-Mpl = 0.4Mel, EII = EII/5108 4.5.6.1Moment108 4.5.6.2Contour plot112 4.5.6.3Plastic rotation114 4.5.6.4Comparison with existing guideline114 4.5.7Trilinear M() with no yielding - EII = EII/5115 4.5.7.1Moment115 4.5.8Influence of state II stiffness in trilinear M() without yielding119 4.5.8.1Orientation119 4.5.8.2Load-displacement relation119 4.5.8.3Moment120 5CONCLUDING REMARKS123 5.1Conclusions123 5.2Further studies124 6REFERENCES125 APPENDIX AFE ANALYSIS-ADINA126 A.1Beam elements126 A.1.1Beam geometry126 A.1.2Elastic-plastic beam element126 A.1.3Bilinear plastic material model127 A.1.4Elastic-plastic moment curvature model127 A.2Shell elements128 APPENDIX BMODIFIEDALPHAFACTORFORTHEFICTITIOUSYIELD STRESS WHEN USING SEVEN INTEGRATION POINTS129 APPENDIX CMOMENT,SHEARFORCEANDDEFLECTIONFOR CANTILEVER SLAB131 C.1Isotropic case131 C.1.1Cantilever slab with shell elements131 C.1.2Cantilever slab with beam elements135 C.1.3Comparisons between shell model and beam grids model139 C.2Orthotropic case143 C.2.1Cantilever slab with shell elements143 C.2.2Cantilever slab with beam elements147 C.2.3Comparisons between shell model and beam grids model151 CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 V C.3Elasto-plastic case-cantilever slab with beam grillage model155 C.3.1Quadlinear M() with Mpl = 0.6 Mel155 C.3.1.1 EII = EI/5 = 6 GPa155 C.3.1.2 EII = EI/2 = 15 GPa160 C.3.1.3 EII = EI/10 = 3 GPa166 C.3.2Trilinear M() with Mpl = 0.6 Mel and EII = EI/5 = 6 GPa172 C.3.3Quadlinear M() with Mpl = 0.4 Mel and EII = EI/5 = 6 GPa177 C.3.4Trilinear M() without yielding180 C.3.4.1 EII = EI/5 = 6 GPa180 C.3.4.2 EII = EI/2 = 15 GPa184 C.3.4.3 EII = EI/10 = 3 GPa189 APPENDIX DSHEAR FORCES DISTRIBUTION194 D.1Isotropic 0.2 m shell element L = 3 m194 D.2Isotropic 0.2 m shell element L = 6 m195 APPENDIX EPLASTIC ROTATION OF A CANTILEVER BEAM196 E.1Geometry and loading196 E.2Alternative methods for plastic rotation of a cantilever beam196 E.3Bilinear moment-curvature input-2 different meshes199 E.4Bilinear moment-curvature input-single mesh201 E.5trilinear moment-curvature input202 APPENDIX FMODELLED SLAB VERSUS REAL SLAB205 F.1Geometry205 F.2Stiffness205 F.3Cracking moment207 F.4Maximum internal resistance207 F.5Plastic rotation capacity207 F.6Comparison between modelled slab and real reinforced concrete slab208 CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 VIPreface Thismastersthesistreatstheredistributionofforceconcentrationsin3Dfinite elementanalyses.TheworkofthestudyhasbeencarriedoutatReinertsensoffice fromJanuary2013toJune2013incooperationwiththeDivisionofStructural EngineeringatChalmersUniversityofTechnologyandReinertsenSverigeAB.Tools forthisstudyareacombinationofliteraturestudies,linearandnon-linearfinite element analyses in which present guidelines are compared. The aim of this masters thesisistodeterminethemostappropriatestripwidthtodistributetheforce concentrations over. Examiner has been Mario Plos.I would like to express my gratitude to my parents, Heng Leang Lim and Channy Sen, for their unconditional supports and encouragement throughout this academic journey. Very special thanks to my supervisor, Morgan Johansson, PhD Reinertsen and also to my supervisor and examiner, Mario Plos, Associate Professor at Chalmers University ofTechnologyfortheirvaluableadvice,supportandguidancethroughouttheentire working period. IwouldalsoliketothankM.Sc.GinkoGeorgievforhissupportandvaluable thoughts along the project. Finally,Iwouldliketothankmyopponents,DimosthenisFlorosandlafurgst Ingason for having assisting me in my work and providing me with their thoughts. SONGLY LIM Gteborg, June 2013 CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 VII Notations Roman upper case letters AArea AIEquivalent area state I AsArea of reinforcement DSlab stiffness EYoungs modulus EcYoungs modulus for concrete EIYoungs modulus for uncracked state EIIEquivalent Youngs modulus for cracked reinforced concrete EsYoungs modulus for steel FExternal force Fc Force in concrete Fs Force in steel GShear modulus IMoment of inertia IIMoment of inertia state I IIIMoment of inertia state II KvTorsional stiffness LLength LcCharacteristic span width MMoment McrCracking moment MEdDesign value of bending moment MRdUltimate moment capacity RReaction force VShear force VEdDesign value of shear Roman lower case letters b Width of cross-section c Concrete cover 1c Factor depending on the height and width of the rectangular cross section d Effective depth of cross-section CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 VIIIcf Concrete compressive strength cdf Design value of concrete compressive strength ctf Concrete tensile strength ctmf Mean value of concrete tensile strength yf Yield stress ydf Design value of yield stress modyf Fictitious yield stress h Height of cross-section h Slab thickness k Modification factor for rotation capacity l Length pll Development length of a plastic region mMoment crm Cracking moment rdm Ultimate moment capacity xym Linear torsional moment hingesn Number of plastic hinges of the structure r Radius of curvature s Reinforcement bar spacing t Height of a rectangular cross section t Thickness of the surfacing elu Elastic displacement plu Plastic displacement totu Total displacement zu Vertical displacement 0v Principal shear force wWidth of the cross section wDistribution width effw Effective width x Coordinate CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 IXux Height of compression zone y Coordinate csy Distance from the centre of the load to the critical cross section z Coordinate Greek lower case letters Angle that determine the direction of shear force vector Factor to transform fy to fymod Ratio between Youngs Modulus for reinforcement and concrete R Stress block factor R Stress block factor Curvature el Elastic curvature pl Plastic curvature y Yield curvature Ratio of the distributed chosen moment to the elastic bending moment Strain c Concrete compression strain cu Ultimate concrete strain s Steel strain y Yield strain el Elastic strain pl Plastic strain Bar diameter Different in percentage Shear slenderness Number of statically indeterminacy of the structure pl Plastic rotation rd Plastic rotation capacity min , rdMinimum plastic rotation capacity Stress CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 Xc Concrete compressive stress ct Concrete tensile stress Shear stress Poisson's ratio CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 1 1Introduction 1.1Background Finiteelementanalysesbasedonlinearelasticmaterialresponseisoftenusedto determine the force distribution in design of civil engineering structures. In the design ofreinforcedconcreteslabs,differenttypesandpositionsofloadsmayresultin various force concentrations. The knowledge of how these linearly determined forces canberedistributedfromonepointtoanotherisverycrucial.Therealforce distribution of a structure may be difficult to explain and therefore simplified approaches areoftenusedindesignofthestructure.Recentlyguidelineshavebeenpresentedof howtodesignreinforcedconcreteslabsbasedonlinearelasticanalysis,including Pacosteetal.(2012).Inthepast,structuralengineersusedtraditionaldesigntools such as tables and 2D analyses to design concrete slabs. In this case the distribution of forcesinthestructurewillbegivenbythechoiceofthestructuralmodel.However with the development of computer analysis capacity, there has been large increase in theuseofthree-dimensional(3D)finiteelementanalyses(FEanalyses)inanalyses anddesign.Fordesignofbridges,therehasbeenashiftfromtraditionalmethodsto 3D analyses in the last few years in Sweden. 3D FE analyses provide the possibility for a more accurate study of the structure than whatispossiblebyusingmoretraditionaldesigntools.Inliaisonwiththeincreased usageofsuchanalyses,however,therehavealsoappearedsomeproblemsthathave notbeenknownofwhenusingsimplified2Danalyses.Onesuchproblemishowto interpret local force concentrations that arise in linear elastic analysis, but that may be not so critical in reality. Normallyinconcretestructure,suchforceconcentrationsleadtocrackingand yielding of the reinforcement which result in the redistribution of forces. Hence, in a realstructure,therewillbeaconsiderablymorefavourableforcedistributionthan what is usually the case in a finite element analysis assuming linear elastic response. It wouldalsobeadvantageousfromaneconomicperspectivetobeabletodistribute such force concentrations over a larger region of the structure in the design process in order to optimise material. 1.2Purpose Theoverallaimofthismastersthesiswastoprovideamorecomprehensive understandingabouthowconcentratedforcesinlinearelasticFEanalysescanbe distributedoveralargerregioninthedesignofreinforcedconcreteslabsandto investigate technical and practical recommendations for such redistributions. The goal in this report was to determine a strip width in which concentrations of moments and shear forces acquired in a linear FE analysis may be distributed. Another objective of the study was to evaluate the recommendations given by Pacoste et al. (2012). 1.3Method Atheoreticalpartwascarriedoutthroughaliteraturestudyofbothtextbooksand researchpapersinordertogetagoodoverallpictureofhowthedistributionof moments and shear forces can be handled.CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 2Investigationswerecarriedoutbyperforminglinearandnon-linearFEanalysesofa cantileverslabinthestudentversionofthefiniteelementsoftwareADINA,Adina (2011). To begin with, simplified 3D linear elastic isotropic and orthotropic analyses were performed. The orthotropic analyses were conducted to simulate different stiffness in two directions; this is what will happen in reality when concrete cracks. At a later stage, more complex 3D non-linear elastoplastic FE analyses, used to simulate the effect of cracking and yielding, were carried out with the same structure as the one used in the linear elastic analyses. The result of the FE analyseswere verified by checking the procedure, assumptions and modelling made. Several analyses were made in order to get reliable results and to check the influence of different parameters.Finally,theresultsintermofmomentandshearforcedistributioninthestructurefrom 3D FE analyses were compared to those from the present guidelines found, Pacoste et al. (2012). 1.4Limitations TheFEanalysescarriedoutonlytreatreinforcedconcretecantileverslabssubjected topointloads.Asthereisalimitedtimeforthemasterthesisstudy,thisproject mainlyfocusesonbendingresponseofaslabbutalsobrieflytreatsthedistributionof shear forces. Thethesisinvestigatesthebehaviouroftheslabsindifferentstatesfromcracking untilthecollapseofthestructure.Thestudiedslabsweresubjectedtooutofplane load only. Other effects such as prestressing force, normal force, shrinkage, creep and thermal effects can be important, but were not taken into account in this master thesis study and these effects were not discussed in this report. Due to the complex material behaviourofreinforcedconcretestructures,theslabsstudiedwerenotmodelledin detailwithseparatematerialsforconcreteandreinforcement.Instead,theidealized behaviouroftheslabsweremodelledtobelinearelasticisotropic,linearelastic orthotropicandtohaveanon-linearmomentcurvatureresponse.Assimplified propertiesoftheslabswereused,therewillbeadifferencebetweentheresponseof the slabs obtained from FE analysis and the response of the real concrete slabs. 1.5Outline of the report Chapter2coversmaterialresponse,structuralresponseofreinforcedconcrete structures and design approaches for beams and slabs. This chapter also explains some recommendations on moment and shear force distribution in slabs based on Eurocode 2, CEN (2004), and other literature studied, including Pacoste et al. (2012). Chapter3examinesthebehaviourofareinforcedconcretecantileverbeamundera pointloadinelasto-plasticresponse.Italsodescribesdifferentapproachesthatare usedtoimitatetheresponseofthebeamintheelasto-plasticcaseincludingbilinear plasticmaterialandelasto-plasticmoment-curvaturerelationship.Comparisons between 2D beam and 3D beam using different material models were performed. This chapter also treats the plastic rotation of the cantilever beam studied. Chapter 4 describes the responses of the cantilever slabs studied loaded by one point load in different states from cracked state until the collapse of the structure is formed. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 3 Differenttypesofmodellingoftheslabwereinvestigated.Theresultsintermof supportreaction,momentdistribution,shearforcedistributionanddisplacement receivedfromFEanalysesusingbeamgrillagewithbeamelementsandslabwith shell elements were compared in order to see which is the most appropriate modelling in the design and in order to evaluate the accuracy of the modelling. This chapter also coveramethodonhowtodeterminetheplasticrotationoftheslab.studied.The recommendation on the distribution width of moments and shear forces from present guidelines were compared with the distribution width obtained from the FE analyses. Chapter5summarisesthediscussionandconclusionofresultsoftheprojects.The suggestions to further studies and investigation are presented. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 42Theory 2.1Finite element Theusageoffiniteelementprocedureshasincreasedsignificantlyoverthelast decades and these methods are nowadays commonly used for the analysis and design of engineering structures. In order to be able to rely on the solutions obtained from a FEanalysis,itisnecessarytochooseanappropriateandeffectivemodelthatwill predict the response of the structures which are being analysed and to be able to assess thelevelofaccuracyoftheanalysis.Someissuesrelatedtothismethodneedtobe mentionedandthecombinationoftheknowledgeofthestructuralbehaviourofthe structures studied and of FE method is needed. Forreinforcedconcreteslabs,linearfiniteanalysisisnormallyusedinthedesignin ordertosimplifyitsnon-linearresponse.Usingthismethod,wecansuperimpose different actions that act on the slab. This method is suitable as the design is based on moment and forces that fulfil equilibrium, i.e. and linear finite analysis is one of many possible solutions that fulfil equilibrium. 2.2Structural response of reinforced concrete structures 2.2.1Material response 2.2.1.1Concrete Concreteisaveryvariablematerialwhichhasawiderangeofstrengthsandstress-straincurve.Figure 2.1showsatypicalstress-strainforconcreteincompression according to Molsey et al.(2007). c c c1cu1 fcm Figure 2.1.Typical stress-strain relationship for concrete in compression AccordingtoEngstrm(2011a),fornormalstrengthconcrete,C12/16-C50/60,the ultimatestrainisthesame,i.e.cu1 = 0.0035,whilethestrainatthemaximumstress c1hasdifferentvaluesfordifferentconcretestrengthclasses.Forstrengthand deformation characteristics for concrete, reader can refer to Eurocode 2, CEN (2004). The stress-strain relationship for concrete is in fact non-linear from the beginning, but this non-linearity is approximated to be linear when the compressive stress in concrete issmallerthanthetensilestrengthofconcrete,i.e.c < fct.Thismeansthatthe material response of reinforced concrete structure is non-linear even if the small load isapplied.Thebehaviourofconcreteisweakintension.Asaruleofthumb,the CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 5 tensile strength of concrete is 10 times smaller than its compressive strength, but the actualproportiondependsonthestrengthoffcthough.Ifc > fct, stress-strain relationshipisnolongerlinearandtheconcretebehavesasaplasticmaterial.A permanent deformation remains when the applied load is removed. 2.2.1.2Reinforcing steel Figure 2.2 shows a typical stress-strain relationship for hot rolled reinforcing steel and cold worked reinforcing steel. (a)su s ft fy y (b)su ft s 0.2ft 0.002 s s Figure 2.2.Typicalstress-straincurveforreinforcingsteel(a)hotrolledsteel (Class B and Class C steel) (b) cold worked steel (Class A steel). FromFigure 2.2a,itisseenthathotrolledsteel(ClassBandClassCsteel)hasa pronouncedyieldstress,followedbyaplasticplateaubeforestrainhardeningstarts. After yield point, this becomes a plastic material and the strain increases rapidly up to theultimatevalue.Forcoldworkedsteel(ClassAsteel)hasnodistinctyieldstress and no plastic plateau, see Figure 2.2b. Since the ductility of reinforcing steel depends mainly on its elongation at maximum stress,Eurocode2,CEN(2004)definestheductilityclasseswithregardtoboththe ultimate strain and the ratio between tensile strength and yield strength. Table 2.1. Properties of reinforcement from Eurocode 2, CEN (2004). Product formBars and de-coiled rods ClassABC Characteristic yield strength fyk or f0.2k (MPa) 400 to 600 Minimum value of k = (ft/ fy)k 1.051.08 1.15 Ey and vice versa. There is a fluctuation of shear distribution at about x = 1.6 m to x = 3.6 m when Ex = 0.1 Ey. For all the cases, i.e. Ex = 0.2, 0.5, 1,2, 5, 10Ey in Figure 4.18, vy < 0 ataboutx = 3.6 mtox = 4 m.Thisbehaviouristhesameasthatobservedin Figure 4.5. This negative value of shear vy < 0 is due to the same reasons explained in Section 4.3.1.2. -380-340-300-260-220-180-140-100-60-20200.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x[m]Ex =0.01EyEx =0.2EyEx =0.5EyEx =EyEx =2EyEx =5EyEx =10EyCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 76 Figure 4.18.ShearforcedistributionvyalongL5 fromFE-linearelasticorthotropic analysis using 0.2 m shell elements with Poissons ratio = 0. 4.4.2.3Vertical displacement The relationship between the stiffnessand loaddistribution can also beexplained by the vertical displacement uz in Figure 4.19. When Ex = 0.1Ey much load is transferred towardtheelementiny-directionwhichmeansthatthetransversalelementalongL1 willresistmuchload.Therefore,alongthislinethedeflectionwhenEx = 0.1Eyis muchlargerthanthedeflectionwhenEx = 10Eywheretheloadtriestodistribute along the longitudinal direction. Figure 4.19.VerticaldisplacementuzfromFE-elasticorthotropicwithPoissons ratio = 0 using 0.2 m shell along L1. -60-40-200204060801001201401600.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Shear force, vy [kN/m] Coordinate, x[m]Ex = 0.1EyEx = 0.2EyEx = 0.5EyEx = EyEx = 2EyEx = 5EyEx = 10Ey-9-8-7-6-5-4-3-2-100.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6Vertical Displacement, uz[mm] Coordinate, y [m]Ex = 10EyEx = 5EyEx = 2EyEx = EyEx = 0.5EyEx = 0.2EyEx = 0.1EyCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 77 4.4.3Distribution width of orthotropic slab Due to the change of stiffness, moment and shear forces will distribute over a certain widthdenotedweffunderacertainloadlevel.Inordertoinvestigatetheeffectofthe effectivewidthintermofchangeofstiffnessdependingonfactor = Ex/Ey,itis necessary to determine the effective width for moments weff,m and the effective width for shear forces weff,v. The effective width for moments weff,m is computed as: max ,0,5yLiym effmmw== (4-10) 2) (, 1 ,105i y i yi iLiym mx x m =++=(4-11) Where=m effw,effective width for moment =ym moment distribution my along L5 =max , ym maximum moment my along L5 The effective width for shear forces weff,v is computed as: max ,0,5yLiyv effvvw== (4-12) 2) (, 1 ,105i y i yi iLiyv vx x v =++=(4-13) Where=v effw,effective width for shear =yv shear force distribution vy along L5 =max , yvmaximum shear force vy along L5 Themaximummomentandmomentdistributioninequation (4-10)aredetermined fromFigure 4.15andthemaximumshearandshearforcesdistributionin equation (4-12) are determined from Figure 4.18. Thedifferencebetweentheeffectivewidthincaseof 1andincaseof = 1for momentdiffernce,mandshearforcesdifference,varecomputedinequation (4-14)and equation (4-15), respectively. Ey Ex m effEy Ex m eff Ey Ex m effm differenceww w== = =, ,, , , ,,100 (4-14) CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 78Ey Ex v effEy Ex v eff Ey Ex v effv differenceww w== = =, ,, , , ,,100 (4-15) The ratio between the effective width in case of 1 and in case of = 1 for moment ratio,m and shear forces ratio,v are determined in equation (4-16) and equation (4-17). Ey Ex m effEy Ex m effm ratioww===, ,, ,, (4-16) Ey Ex v effEy Ex v effv ratioww===, ,, ,, (4-17) Theresultsofeffectivewidthweffofmomentsandshearforcesaresummarizedin Table 4.3 and Table 4.4. Table 4.3. Effectivewidthweff,m,differencedifference,mandratioratio,mbetweenthe effective width in case of 1 and in case of = 1 for moment. =50Liymmax , ymm effw, m difference ,m ratio,0.1-160-173.50.946.71.8 0.2-160-142.11.134.91.5 0.5-160-110.11.416.11.1 1-160-92.41.70.01.0 2-160-79.22.1-16.60.8 5-160-66.52.4-38.90.7 10-160-59.32.7-55.80.6 Table 4.4. Effectivewidthweff,v,differencedifference,vandratioratio,v,betweenthe effective width in case of 1 and in case of = 1 for shear. =50Liyvmax , yvv effw, v difference,v ratio,0.1100138.10.731.41.4 0.2100127.10.825.51.3 0.5100109.90.913.91.1 CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 79 110094.61.10.01.0 210078.21.2-20.80.8 510058.11.7-62.90.6 1010045.92.1-105.80.4 Figure 4.20.Effectivewidthintermof =Ex/Ey fromFE-elasticorthotropicwith Poissons ratio = 0 using 0.2 m shell element for moment and shear. 0.00.40.81.21.62.02.42.80.1 1 10Effective width, weff [m] Alpha factor, [-]momentshearCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 80 Figure 4.21.Ratiobetweeneffectivewidthorthotropicandisotropicintermof = Ex/Ey fromFE-elasticorthotropicwithPoissonsratio = 0using 0.2 m shell elements for moment and shear. FromFigure 4.20,itisseenthattheeffectivewidthformomentweff,mincreaseswith increasing factor. Figure 4.20 also shows that when Ey is greater than Ex, i.e. < 1, theeffectivewidthformomentweff,m issmallsincetheelementsiny-directionare stiffer than the elements in x-direction and a large portion of the load is carried by the elementsclosetothemiddleoftheslab.WhenEx isgreaterthanEy,i.e. > 1,the distribution width for moment weff,m in Figure 4.20 is large as the x-direction are stiffer thantheelementsalongthey-directionwhichleadstoamoreevenlydistributed moment my along L5 as shown in Figure 4.15b.Like the effective width for moment weff,m, the effective width for shear forces weff,y in Figure 4.20alsoincreaseswithregardtoincreasingoffactor.Figure 4.20also shows that the distribution width for shear force weff,v is also small when < 1. Based on Figure 4.21, it is observed that the change of effective width for shear ratio,v is not linear whereas the variation of the effective width for moment ratio,m increases more linearly when increases comparing to ratio,v. 4.4.4Comparisons between shell elements and beam elements It is not possible to combine orthotropic material with beam elements in ADINA. For orthotropicanalysis,stiffnessandshearmodulusindifferentdirectionsarerequired. Forbeamelements,onlyonestiffnessisneeded.Therefore,thiscombinationisnot available in FE program ADINA. As the beam grillage model consists of the beams in x-andy-directions,onewaytodealwithsuchaproblemistomodelthecantilever slab studied using beam elements with isotropic material, but the beam elements will have different stiffness for the beam located in x- and y-directions. It means that it is necessarytointroducethestiffnessExforthebeamslocatedinx-directionand 00.20.40.60.811.21.41.61.822.20.1 1 10Ratio of distribution width, ratio [-] Alpha factor, [-]momentshearCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 81 introducethestiffnessEyforthebeamlocatediny-directionsothatitsimulatesthe orthotropic behaviour of the slab. Inordertoevaluatethissimplifiedmodel,fourdifferentstiffnessinx-directionis chosen while the stiffness in y-direction is kept constant, i.e. Ex = 0.2, 0.5, 1, 5Ey.The geometry and orthotropic properties of the slab in Figure 4.2 using simplified model is depicted in Figure 4.22 and the result lines in Figure 4.2 are still valid in Figure 4.22. Intersection point L=4 m b=1.6 m Beam element along y-direction, Ey Beam element along x-direction,Ex = Ey ( = 0.2, 0.5, 1, 5.) symmetry line F=100 kN Figure 4.22.Simplified orthotropic slab model using beam elementsFrom the comparisons between shell elements and beam elements for isotropic case in Section 4.3.3,itisclearlyshownthat0.2 mbeamelementsprovidetheresultsmost similartothoseoftheshellelements.Therefore,itissufficienttouseonly0.2 m beam elements to compare with the results of 0.2 m shell elements in orthotropic case. The different in percentage of moment distribution my, shear forces distribution vy and verticaldisplacementuzbetween0.2 mshellelementsand0.2 mbeamelementsare computed using equation(4-18), equation (4-19) and equation (4-20) respectively. beam Ey Exshell Ey Ex beam Ey Exm orthomm m,, ,,100 == = = (4-18) beam Ey Exshell Ey Ex beam Ey Exv orthovv v,, ,,100 == = = (4-19) beam Ey Ex zshell Ey Ex z beam Ey Ex zu orthouu uz,, ,,,, ,100 == = = (4-20) Forthecomparisonsofshearforcesdistributionvyandverticaldisplacementuz, readers can refer to APPENDIX C. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 82 -160-140-120-100-80-60-40-20020400.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6Moment, my[kNm/m] Coordinate, y [m]Ex =0.2Ey, shellEx =0.2Ey, beamEx =0.5Ey, beamEx =0.5Ey, shellEx =Ey, beamEx =Ey, shellEx =5Ey, shellEx =5Ey, beam -20-10010200.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6Difference, ortho,m[%] Coordinate, y [m]Ex =0.2EyEx =0.5EyEx =EyEx =5Ey(a) (b) Figure 4.23.Comparisonsbetween0.2 mbeamelementsand0.2 mshellelements fromlinearelasticorthotropicFE-analysiswithPoissonsratio=0: (a)momentdistributionmyalongL1(b)differenceinpercentagemy along L1. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 83 -160-140-120-100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Ex =0.2Ey, shellEx =0.2Ey, beamEx =0.5Ey, beamEx =0.5Ey, shellEx =Ey, beamEx =Ey, shellEx =5Ey, beamEx =5Ey, shell -20-10010200.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Difference, ortho,m[%] Coordinate, x [m]Ex =0.2EyEx =0.5EyEx =EyEx =5Ey (a) (b) Figure 4.24.Comparisonsbetween0.2 mbeamelementsand0.2 mshellelements fromlinearelasticorthotropicFE-analysiswithPoissonsratio=0: (a)momentdistributionmyalongL1(b)differenceinpercentageofmy along L5. Whenobservingtheconvergenceanddivergenceofthe0.2 mbeamgrillagemodel and 0.2 m shell element model, it can be seen in Figure 4.23 and Figure 4.24 that there isthebestconvergenceincaseEx = 0.5EywhileEx =0.2Eyshowstheworst convergence among the four different cases that have been made. Ex =Ey and Ex = 5Ey casesgivearathergoodconvergencecomparingtoEx = 0.5Ey.Thereasonforthe fluctuation in these differences due to the change of stiffness is not clearly known and explained but it is believed to be due to the simplification that has been made in beam CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 84grillagemodelwhenanalysingofslabinorthotropiccase.Ascanbeseenin Figure 4.23,thedifferenceinmy alongL1is|| < 10 %fortheelementssituatedat about y < 0.8 m in the transversal direction from the fixed supports. For the elements situatedatabouty > 0.8 m,thedifferenceislarger.Thesedifferencesareconsidered acceptableinthestudysincethelargemomentthatisusedforthedesigngivesa small while a small absolute moment value give a large . This is not so critical in the design process. Similar to the difference in my along L1, the difference in my along L5 alsoshowsalargewherethereisasmallabsolutemomentvalue,i.e.in Figure 4.24, there is a huge divergence between 0.2 m beam elements and 0.2 m shell elementswherethemomentisclosetozero.Formy alongL5, thereisasmall difference around || = 10 % for a large moment value that is used for the design in the critical section. In actual design, the minimum reinforcement will be able to resist this moment value. Moreover for the moment at around x = 2 m, || < 5 %. This difference isacceptablesinceinelasto-plasticanalysis,thedistributionofmomentdueto yieldingwilltakeplaceonlyatthelocationwherethereisahugemomentwhichis largerthanyieldingcapacityofthesection.Tosumup,thebeamelementscanbe used for elasto-plastic analysis of the previous examined cantilever slab in Figure 4.2. Basedontheanalysisofthecantileverslabbothinlinearelasticisotropiccasein Section 4.3.3andorthotropiccaseinSection 4.4.4,itisconvincedthatshellelement isthemostsuitableandappropriatefiniteelementchoicetoexaminethecantilever slabstudiedinFigure 4.2.Therefore,itisrecommendedtouseshellelementswhen analysing a cantilever slab under point load in linear elastic isotropic and orthotropic case. 4.5Elastoplastic case-beam grillage model 4.5.1Choice of model used for the analysis There is a biaxial effect that increases the moment capacity of the slab when analysing a slab using shell elements with a combination of plastic material according to Augustsson and Hrenstam (2010). Augustsson and Hrenstam (2010) explained this effect by studying a one-way simply supported slab subjected to distributed loads over the whole surface as shown in Figure 4.25. The material model used is a bilinear-plastic material model depicted in Figure 4.26. b=1 m x y L=2.7 m Surface load Figure 4.25.One-way simply supported slab subjected to surface loads. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 85 EII = 0 1 fyd Figure 4.26.Bilinearplasticmaterialdataforsimplysupportedslabanalysisusing shell elements. Theplasticmomentcapacityoftheslabisdeterminebytheyieldstressprovidedin the material model. fyd Ultimate capacity due to biaxial effect Ultimate capacity due to uniaxial effect 3 2 1 Figure 4.27.Ultimatecapacityofthesimplysupportedone-wayslab:dotline representexpectedcapacity(uniaxialeffect),solidlinerepresent increased capacity from FE analysis (biaxial effect). In Figure 4.27, point 1 corresponds to the elastic state where the Poissons ratio = 0. Thestressesaredevelopedonlyinx-directionandtheresultobtainedfromtheFE analysis is equal to the expected solution. When continuing loading the slab, the stress reachestheyieldstresswhichisdenotedbypoint2inFigure 4.27.Point2 correspondstotheexpectedcapacityofthesectionwithregardtotheinputin Figure 4.26.Continuingloadingtheslabsfurther,itisobservedthatthestresskeeps increasingandpassestheyieldstressofthesection.AccordingtoAugustssonand Hrenstam (2010), this increase is due to the biaxial effect of plastic material used in ADINA.TheplasticmaterialusedinADINAisbasedontheVonMisesyield condition described, seeBathe (1996). When thematerial reaches plasticity,ADINA willsetPoissonsratio = 0.5nomatterwhatPoissonsratioinputdefinedbythe user. Consequently, the stress in the y-direction is developed due to restrain caused by the elements next to the element that yields when the stress in the x-direction reaches the yield stress defined in the material input. Due to the development of the stresses in the y-direction, a biaxial stress state is formed, which increases the yield stress in the x-direction. In order to avoid this effect, beam elements described in Section 4.3 and 4.4 are used instead.Further,sincetheresponseoftheslabindifferentstatesisofinterest,itis necessarytomodeltheslabinelasto-plasticcaseusing3Dbeamelementswith multilinear moment-curvature relationship model. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 86AsshowninSection 4.3andSection 4.4,whenanalysingtheslabusingbeam elementsinisotropicandorthotropiccase,0.2 mbeamelementsprovidesasimilar resulttothosewhenusingshellelements.Therefore,0.2 mbeamelementsizewas used in this case as well. From Section 3.5.3, it is seen that element mesh sizes have a large influence on the results when analysing the beam in elasto-plastic case using 3D beam elements. It is necessary to have sufficiently fine mesh in order to get acceptable results for the analysis of the slab. Hence based on Section 3.5.3, smaller element size than 0.2 m beam elements should be used. In order to assure the accuracy of the analysis of the slab in elasto-plastic case when usingbeamelements,thetorsionalstiffnessoftheslabreceivedfrom0.2 mbeam elementsneedtobekeptconstant.Itisknownthatinordertokeepthetorsional stiffness of the slab constant when using beam elements, the element mesh sizes at the intersectionpointsneedtobeunchanged.Itmeansthattheelementmeshsizesfor both x and y-direction have to have 0.2 m length at the intersection point.Onewaytodealwithsuchaproblemistosubdividethefirstelementofthe longitudinalbeam(beaminy-direction)into4smallerelements.Itmeansthat0.2 m elementmeshsize(firstelementofthebeaminy-direction)isdividedintofour smaller element mesh sizes equal to 0.05 m element size as depicted in Figure 4.28. It is enough to just sub-divide the first elementand not the other elements along they-directionsinceitisknownthatonlythefirstelementyieldsandotherelementswill notyield.Therefore,a0.05 melementmeshsizeischosenasitprovidesan acceptableresultaccordingtoSection 3.5.3.Itisalsonotedthatiftheelementisin elastic state, 0.2 m element size provides a good result. b=1.6 m Beam element along y-directionsymmetry lineIntersection point L=4 m Beam element along x-direction F=100 kN bel=0.2 m bel=0.05 m Lel=0.2 m bel=0.2 m Lel=0.2 m bel=0.2 m Intersection point (a) (b) Figure 4.28.Elementmeshsizesforcantileverslab(a)singlemesh0.2 melement size(b)2differentmeshes0.05 melementsizeforthefirstelementof beaminy-directionand0.2 melementsizesfortherestofthebeam elements. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 87 4.5.2Influence of mesh-isotropic case Since the cantilever slab in Figure 4.2 is now modelled with two different meshes, it is importanttocheckiftheslabinFigure 4.28bgivesthesameresultsastheslabin Figure 4.28ainelasticisotropiccase.ThematerialinputshowninFigure 4.3thatis usedforthecantileverslabstudiedwithsinglemeshsize0.2 misalsousedforthe cantilever slab studied with two different mesh sizes so that it is possible to compare the results. -100-80-60-40-2000.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6Moment, my [kNm/m] Coordinate, y [m]L1-single mesh-0.2mL1-two meshes-0.05 and 0.2 -100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]L5-single mesh-0.2mL5-two meshes-0.05m and 0.2m (a) (b) Figure 4.29.Comparisonsbetweenslabwithsinglemeshsizeandslabwithtwo different mesh sizes (a) my along L1 (b) my along L5. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 88 Figure 4.30.Comparisonsofshearforcesdistributionvybetweenslabwithsingle mesh size and slab with two different mesh sizes. Figure 4.31.Comparisons of vertical displacement uz between slab with single mesh size and slab with two different mesh sizes. From Figure 4.29a to Figure 4.31, it is observed that the results of the two models at the intersection points between theelements in x andy-direction coincides well. The resultsofthefirstelementoftwodifferentmeshesmodellocatedaty = 0.05 m, y = 0.1 m and y = 0.15 m away from the fixed support provides more accurate results -40-200204060801000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Shear force, vy [kN/m] Coordinate, x [m]L5-single mesh-0.2mL5-two meshes-0.05m and 0.2m-4-3.5-3-2.5-2-1.5-1-0.500.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6Vertical displacement, z [mm] Coordinate, y [m]L1-single mesh-0.2mL1-two meshes-0.05 and 0.2CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 89 as the result of single mesh model is computed based on linear interpolation between y = 0 m and y = 2 m. From Figure 4.29b and Figure 4.30, it is seen that the results of moment and shear forces along line L5 are identical as the elements along x-direction arenotsubdivided.ItisconcludedthatthecantileverslabinFigure 4.2canbe modeledwithbeamelementsusingtwodifferentmeshesfortheanalysisinelasto-plastic case. 4.5.3Quadlinear M() - Mpl = 0.6Mel, EII = EII/5 4.5.3.1Moment A cantilever slab inFigure 4.32awith the samegeometry and boundarycondition as the cantilever slab in Figure 4.2 is subjected to a point load until failure. Thereby, it is possible to see how far plastic hinges along the clamped boundary of the studied slab can be formed if the plastic rotation capacity of the slab is sufficient. b=1.6 m L=4 m F (failure load) bel=0.2 m bel=0.05 m Lel=0.2 m (a) (b) L=4 m b=1.6 m L1L2 L5 L4 L3 symmetry line 0.8 m 0.8 m 0.8 m F(failure load) symmetry line t=0.2 m (thickness) Figure 4.32.(a)Geometryandloadofacantileverslabwithtwodifferentelement mesh sizes (b) result lines for moment and shear force distribution. ThecantileverslabinFigure 4.32bwasanalyzedusingaquadlinearmoment curvatureinputillustratedinFigure 4.33.Astheanalysisdoesnotmodelthereal reinforcedconcreteslab,momentplasticandcrackingmomentischosenbasedon practical experience so that they are close to value of a certain type of real reinforced CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 90concrete slab, see APPENDIX F. According to Pacoste et al. (2012), moment plastic Mplasticcanbechosen0.6Melastic MplasticMelastictoredistributethereinforcement moment.HereMplastic = 0.6Melastic ischosenwhereMelasticistakenfromthelinear elasticisotropicanalysisofthecantileverslabinSection 4.3.2.Further,thecracked moment Mcr1 = Mplastic/3 and Mcr2 = Mplastic/2 is chosen as input for the analysis carried outhere.ThestiffnessinstateIiskeptthesameasthestiffnessinlinearelastic isotropiccaseEI = 30 GPawhilethestiffnessinstateIIischosenequalto EII = EI/5 = 6 GPa. These input data are chosen close to a reinforced concrete slab of 1 m width with B500B 516 and concrete strength C30/37 as shown in Figure F.1 in APPENDIX F. Itisimportanttonotethat0.2 mbeamelementsizeisusedtomodeltheslab.The moment-curvaturerelationshipinputforbeamelementinFEprogramADINAisin Nmunit.Inordertomakeiteasytocomparetheresultofmomentintheslab,the moment in Nm is distributed per unit width in order to have intensity in Nm/m. The cantilever slab studied in Figure 4.32 was analyzed using the following moment-curvature input: kNm 4 . 11 19 6 . 0 6 . 0 = = =elastic plasticM M (4-21) kNm/m 572 . 04 . 11= =plasticm (4-22) kNm 7 . 524 . 1122= = =plasticcrMM(4-23) kNm/m 5 . 282 . 07 . 52= =crm (4-24) kNm 8 . 334 . 1131= = =plasticcrMM(4-25) kNm/m 192 . 08 . 31= =crm (4-26) ThecurvatureI,II,IIIandIV,ofthemoment-curvaturediagraminFigure 4.33is determined from equation (4-27) to equation (4-30). [ ] m / 1 00095 . 0122 . 0 2 . 010 3010 8 . 33931= = =I EMIcrI (4-27) [ ] m / 1 007125 . 0122 . 0 2 . 010 610 7 . 53932= = =I EMIIcrII (4-28) CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 91 [ ] m / 1 01425 . 0122 . 0 2 . 010 610 4 . 11393= = =I EMIIplasticIII (4-29) [ ] m / 1 10 =IV (4-30) ) m / 1 ( M (kNm) EII Mpl Mcr1 00095 . 0 =I 01425 . 0 =III EI EI = 30 GPa EII = EI/5 = 6 GPa 10 =IV1 4 A C Mcr2 007125 . 0 =II B 2 3 Figure 4.33.Quadlinear moment-curvature relationship for the analysis of the studied cantilever slab in elasto-plastic case, EII = EI/5. TherearefourzonesinFigure 4.33thatwereusedtosimulatethebehaviourofthe cantileverslabstudied.Zone1determinedfrom0topointArepresentsuncracked state (state I). Zone 2 from point A to point B and zone 3 from point B to point C were usedtosimulatethetensionstiffeningeffectincrackedstate(stateII).Zone4 representsyieldingpartofthetensilereinforcement(stateIII).Formoredetailof structuralresponseofreinforcedconcretestructureinserviceabilitylimitstateand ultimate limit state, reader can refer to Section 2.2.5 and 2.2.6. Theratioofthemaximummomentmy,max receivedfromtheFEanalysistomoment distribution my,(x) along x-direction is determined in equation (4-31). x yyquadmm,max ,= (4-31) From Figure 4.34a, it is seen that when F = 20 kN, the maximum moment intensity at themiddleoftheslabismy.max = 19.2 kNm/m.Theremainingpartalongthefixed boundaryislessthan19 kNm/mwhichmeansthattheyareinelasticstate.When F = 40 kN,thereisanabruptchangeinstiffnessatx = 2 m.Thisindicatesthatparts of the slab along fixed boundary, 0 x 2 m, are cracked while the remaining parts located further away at x> 2 m is uncracked. For a load F = 60 kN, the cracks spread along L5 and there is a change in stiffness at x = 3 m. It means that the slab cracks at 0 x 3 mandisuncrackedatx > 3 m.Whentheloadkeepsincreasing,thecracks CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 92develop even further until the slab yields at a length Lyield = 0.6 m from the symmetry line when F = 100 kN. This shows that there is a plastic redistribution of moment due to yielding where the yielding part is equal to Lyield = 0.6 m. (a) (b) -100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=100kNPlastic, F=80kNPlastic, F=60kNPlastic, F=40kNPlastic, F=20kN 00.20.40.60.810.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Ratio, [ ] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=100kNPlastic, F=60kN Figure 4.34.Moment distribution my along L5 at different load level up to 100 kN (a) momentdistributionmy (b).ratioofmaximummomentmy,maxtomy,x moment along L5. FromFigure 4.34b,itisobservedthatthemomentdistributionshapeafterplastic redistribution is smoother than the moment distribution shape after redistribution due CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 93 tothechangeofstiffness.Comparedtotheothercurvesinthesameillustration,the distribution shape of moment from linear elastic FE analysis is very sharp. (a) (b) -100-90-80-70-60-50-40-30-20-1000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=143.8kNPlastic, F=140kNPlastic, F=130kNPlastic, F=120kNPlastic, F=100kN 00.20.40.60.810.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Ratio, [ ] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=143.8kNPlastic, F=120kN Figure 4.35.MomentdistributionmyalongL5 atdifferentloadlevelupto143.8 kN (a)momentdistributionmy (b).ratioofmaximummomentmy,maxtomy,x moment along L5. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 94Figure 4.35a shows that part of the slab that yields along the fixed boundary increases withincreasingloadF.Iftheloadisincreaseduntilfailure,afailureloadof F = 143.8 kNisobtainedasshowninFigure 4.35aandthewholepartoftheslab along L5 yields. In Figure 4.35b, it is seen that the moment redistribution shape of the studiedcantileverafteryieldingbecomesmoreevenwithincreasingappliedloadF andthemomentdistributionshapeforF = 143.8 kNisconstant,i.e.full redistribution. 4.5.3.2Curvature contour plot In order to investigate the structural response of the whole cantilever slab studied with regardtoincreasingofappliedloadF,thecontourplotofmyinFigure 4.36and Figure 4.37 are plotted. In Figure 4.33, there are four different zones that represent the structuralresponseofthecantileverslabstudiedfromuncrackedstateuntilfailure. Therefore, the contour plots in Figure 4.36 and Figure 4.37 of the same slab also have fourdifferentcoloursinwhicheachcolourcorrespondstothezonesinFigure 4.33. Green colour represent zone 1(uncracked state), orange colour represents zone 2 (first cracked state),yellow colour represents zone 3 (second cracked state) and red colour representszone4(yieldingoftensilereinforcement)ofthequadlinearmoment-curvature input in Figure 4.33. From Figure 4.36, it is seen that when F = 20 kN, only the part of the slab located at x = 0.2 mandy = 0.2misinzone2sincethemaximummomentintensityatthat positionmy.max = 19.2 kNm/mishigherthanMcr1 = 19 kNm/m.However,the remaining parts of the slab is in zone 1 as the moment intensity of those parts is less thanMcr1 = 19 kNm/m.WhenF = 40 kN,zone2propagatesfurther,butthereisa disturbance in part of the slab at x = 0.2 m and y = 1.4 m. This disturbance is believed to be due to the torsional effect at that position. Similar disturbances can also be seen in Figure 4.4a for 0.2 m shell element and Figure 4.8a for 0.2 m beam element. When F = 60 kN,cracksdevelopfurtheruntilsomepartsofthestudiedslabalongL5at about0 x 1 mareinzone3andalongL1 atabout0 y 1.4 mareinzone2. WhenF = 80 kN,zone3alongL5developsfurtheratabout0 x 1.8 mandit extends along L1 until 0 y 0.4 m. When the load increases up to F = 100 kN, some partsoftheslabalongfixedboundaryyieldsatabout0 x 0.8 mandtheyarein zone4whilethepartslocatedatabout0.8 x 3 mareinzone3andatabout 3 x 4 m are in zone 2. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 95 Figure 4.36.Contour plot at different load level up to 80 kN. 1.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.0F = 20kNF = 40kNF = 60kNF = 80kNF = 100kNCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 96 Figure 4.37.Contour plot at different load level up to 143.8 kN. InFigure 4.37,whenF = 120 kN,yieldingpartsoftheslabpropagatesfurtherat about0 x 1.8 manditisclearlyseenthattherearedisturbancesatposition (x = 0.2 m, y = 1.4 m) and position (x = 4 m, y = 0.2 m) of the slab. The development ofyieldingalongL5andcrackinginotherpartscontinuesuntilthecollapse mechanism is formed when F = 143.8 kN. When F = 143.8 kN, parts of the slab along L5 at about 0 x 3.8 m are in zone 4. There are disturbances at position (x = 0.2 m, y = 1.4 m) and position (x = 4 m, y = 0.2 m). 1.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.0F = 143.8kNF = 130kNF = 140kNF = 120kNCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 97 Itisnotedthattherearealwaysdisturbancesatposition(x = 0.2 m,y = 1.4m)and position (x = 4 m, y = 0.2 m) of the slab in each load level. These disturbances cannot be seen when these parts and their surrounding regions are located in the same zone, for example when F = 100 kN. However, these disturbances can be seen clearly when thesepositionsandtheregionsnexttothemsituatedindifferentzones,forinstance when F = 143.8 kN. 4.5.3.3Plastic rotation FromAPPENDIXF,thecantileverslabstudiedinFigure 4.32withinputdatain Figure 4.33issimilartotherealreinforcedconcretecantileverslabinFigure F.1. Therefore,fromAPPENDIXF,itisconcludedthatthestudiedcantileverslabin Figure 4.32hasaplasticrotationcapacityofaboutrd = 24.510-3 radwith xu/d = 0.179 and has a minimum plastic rotation capacity of rd,min = 10.610-3 rad. As shown in Section 3.5.4, method 2 presented in APPENDIX E is valid to determine the plastic rotation of the cantilever slab examined in Figure 4.32 in elastoplastic case. Therefore, method 2 was used to calculate the plastic rotation of the studied cantilever slab with quadlinear moment curvature relationship input depicted in Figure 4.33. Figure 4.38.PlasticrotationdistributionplalongL5 atdifferentloadlevelupto F = 143.8 kN. F = 100 kNresultsinthemaximumplasticrotationpl,max = 7.010-4 rad,see Figure 4.38.Sincepl,max = 7.010-4 radissmallerthantheminimumplasticrotation capacity of the cantilever slab studied rd,min = 10.610-3 rad, it means that it is possible toincreasetheloadfurther.Whencontinueloadingtheslabuntilafailureloadof F = 143.8 kN, all elements along L5 yields and the cantilever slab studied fails due to theglobalfailurewhenthemechanismisformed.AloadF = 143.8 kNgives pl,max = 8.010-3 rad. It is seen that even when the collapse of the structure takes place, 01234567890.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Plastic rotation, pl [10-3rad] Coordinate, x [m]Plastic, F=143.8kNPlastic, F=140kNPlastic, F=130kNPlastic, F=120kNPlastic, F=100kNPlastic, F = 80 kNCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 98pl,max remainssmallerthantheminimumplasticrotationcapacityofthecantilever slabstudiedpl,max = 8.010-3 rad < rd,min = 10.610-3 rad.Thisindicatesthatif Mplastic = 0.6Melastic, it is possible to redistribute the moment with a distribution width wx = 4 m.Thisalsoindicatesthatalargermomentcanbereducedfromthemoment receivedfromFEanalysis,i.e.Mplastic pl,max,EII = EII/5 = 6.410-3 rad > pl,max,EII =EII/2 = 6.310-3 rad .Whenthecantileverslabstudied reachesitscollapseloadforeachcases,itisseenthatpl,max,EII =EII/10 = 8.810-3 rad > pl,max,EII =EII/5 = 8.010-3 rad > pl,max,EII =EII/2 = 7.310-3 rad .Itisnoticedthatthe maximumplasticrotationpl,maxofthestudiedcantileverslabincase EII = EI/2 = 15 GPabecomessmallerthanpl,max inbothcaseEII = EI/5 = 5 GPaand EII = EI/10 = 3 GPa from F = 130 kN to F = 140 kN. The fluctuation of the maximum plastic rotation that took place between F = 130 kN and F = 140 kN has not yet been understood.Furtherinvestigationsareneededsothatreasonsforthisdisturbance might be provided. Figure 4.43.PlasticrotationdistributionplalongL5 atdifferentloadlevelupto failure load for EII = EI/2, EII = EI/5, EII = EI/10. From Figure 4.43, it is seen that for these three cases when the cantilever slab studied reached its collapse load, the yielding parts of the slab along L5 is Lyield = 4 m from the symmetry line. From Figure 4.43, it is also seen that the maximum plastic rotation for each case when the collapse mechanism was formed is less than the minimum plastic rotationcapacityofthestudiedcantileverslabrd,min = 10.610-3 rad.Thisindicates thatitispossibletoredistributemy = 0.6 my,elasticwithinw = 4 m.Therefore,itis concluded that the recommendation given by Pacoste et al. (2012) in equation (2-53) 0123456789100.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Plastic rotation, pl [10-3rad] Coordinate, x [m]Plastic, F=144kN, EII=EI/10Plastic, F=140kN, EII=EI/10Plastic, F=130kN, EII=EI/10Plastic, F=120kN, EII=EI/10Plastic, F=143.8kN, EII=EI/5Plastic, F=140kN, EII=EI/5Plastic, F=130kN, EII=EI/5Plastic, F=120kN, EII=EI/5Plastic, F=143.4kN, EII=EI/2Plastic, F=140 kN, EII=EI/2Plastic, F=130kN, EII=EI/2Plastic, F=120kN, EII=EI/2CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 104isconservativewithregardlessofthechangedstiffnessfromEII = EI/2 < EII < EII = EI/10whenanalysingthecantileverslabstudiedinFigure 4.32withelasto-plastic moment curvature input in Figure 4.33, Figure C.30 and Figure C.36. 4.5.5Comparisons between trilinear and quadlinear M() InordertoinvestigateofhowtrilinearmomentcurvaturerelationshipinFigure 4.44 affectsthestructuralresponseofthecantileverslabstudiedinFigure 4.32,a comparison of slab in Figure 4.32 with moment curvature input in Figure 4.44 and in Figure 4.33 was made.Mplastic = 0.6Melastic, and Mcr1 = Mplastic/3 in equation (4-21) and equation (4-25) in Section 4.5.3.1 were chosen for curvature I and II of the moment-curvature diagram in Figure 4.44 and I, II and III are determined as: [ ] m / 1 00095 . 0122 . 0 2 . 010 3010 8 . 33931= = =I EMIcrI (4-35) [ ] m / 1 01425 . 0122 . 0 2 . 010 610 4 . 11393= = =I EMIIplasticII (4-36) [ ] m / 1 10 =III (4-37) ) m / 1 ( M (kNm) EII Mpl Mcr 00095 . 0 =I 01425 . 0 =IIEI EI = 30 GPa EII = EI/5 = 30 GPa 10 =III1 2 3 A B Figure 4.44.Trilinearmoment-curvaturerelationshipfortheanalysisoftheslabin elasto-plastic case, EII = EI/5. Line number 1 in Figure 4.44 represents elastic part (state I). Point A is the boundary between the uncracked and cracked part of the section. Line number 2 in Figure 4.44 isusedtosimulatetheeffectoftensionstiffeningaftertheconcretecracks.Line number2endsatabreakpointBwheretheyieldingofthetensilereinforcement starts.Line number 3 isused to simplifyayielding part of the tensile reinforcement. Theratioofthemaximummomentmy,max receivedfromFEanalysistomoment distribution my,(x) along x-direction is determined in equation (4-38). CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 105 x yytrimm,max ,= (4-38) For the results of moment distribution, shear forces distribution and plastic rotation of thecantileverslabstudiedwiththetrilinearmomentcurvatureinputinFigure 4.44, Reader can refer to APPENDIX C. 4.5.5.1Load-displacement relation From Figure 4.45, it is seen that when F < 20 kN, the displacement due to trilinear and quadlinearmomentcurvaturerelationshipareidenticalsincetheslabstudiedhasthe same stiffness before the slab starts to crack. When F > 20 kN, the slab studied starts tocrackforbothcases.Thetrilinearproduceasmallerverticaldisplacement uz = 23.7 mm while a quadlinear produce a larger vertical displacement uz = 25.3 mm forthesamecollapsedloadF = 143.8 kN.Thereasonisthattheslabwithtrilinear moment curvature relationship in Figure 4.44 simulates a crack development which is slower than a crack development assumed by a quadlinear moment curvature input in Figure 4.33.TheresultsshowninFigure 4.45arereasonableasthestifferslab produceasmallerverticaldisplacement.Sincethesetwomodellinggivesthesame failureloadF = 143.8 kN,itisconcludedthattheassumptionaboutthetension stiffening effect of the slab does not affect the failure load of the slab. Figure 4.45.Comparisonsofverticaldisplacementuzatdifferentloadlevelupto collapseloadbetweentrilinearmomentcurvaturerelationshipand quadlinear moment curvature relationship, EII = EI/5. -160-140-120-100-80-60-40-200-30 -25 -20 -15 -10 -5 0Applied force, F [kN] Vertical displacement, uz [mm]trilinear, EII=EI/5, F=143.8kNquadlinear, EII=EI/5, F=143.8kNCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1064.5.5.2Moment ThemomentmydistributionalongL5forbothcasesfordifferentloadlevelfrom F = 60 kN to F = 143.8 kN are illustrated in Figure 4.46. FromFigure 4.46a,itisseenthatwhenF = 60 kNtheslabwithtrilinearmoment curvatureinputgiveslargermomentthanslabwithquadlinearmomentcurvature input,i.e.my,tri, = 38.9 kNm/m> my,quad = 36.8 kNm/mwhileF = 80 kNgives my,tri, = 51.2 kNm/m< my,quad = 52.8 kNm/mandwhenF = 100 kN, my,tri, = 58.3 kNm/m > my,quad = 58.29 kNm/m. From Figure 4.46b, it is seen that my,tri, < my,quadfrom F = 120 kN to F = 143.8 kN. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 107 (a) (b) -100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=100kN, quadlinearPlastic, F=80kN, quadlinearPlastic, F=60kN,quadlinearPlastic, F=100kN, trilinearPlastic, F=80kN, trilinearPlastic, F=60kN, trilinear -100-90-80-70-60-50-40-30-20-1000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=143.8kN, quadlinearPlastic, F=140kN, quadlinearPlastic, F=130kN, quadlinearPlastic, F=120kN, quadlinearPlastic, F=143.8kN, trilinearPlastic, F=140kN, trilinearPlastic, F=130kN, trilinearPlastic, F=120kN, trilinear Figure 4.46.Moment distribution my along L5 at different load level (a) moment distribution my up to 100 kN (b). moment distribution my up to 143.8 kN, EII = EI/5. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1084.5.5.3Plastic rotation The comparisons in Figure 4.47 between the cantilever slab studied with trilinear and quadlinear moment curvature input shows that plastic rotation due to trilinear moment curvatureinputissmallerthanplasticrotationwithquadlinearmomentcurvature input,i.e.pl,max,tri< pl,max,quad forF = 120 kN < F < F = 143.8 kN.Basedonthe comparisonsofplasticrotationbetweenslabwithtrilinearmomentcurvatureinput andslabwithquadlinearmomentcurvatureinput,itisconcludedthatthestifferthe stiffer the slab is, the smaller the plastic rotation is received. ForthefailureloadF = 143.8 kN,pl,max,tri = 7.9910-3 rad andpl,max,quad = 8.0310-3 rad .Thiscollapseloadleadstoyieldingalloverfixedboundaryconditionofthe slabs,i.e.Lyield = 4 m.Sincepl,max,tri = 7.9910-3 rad < pl,max,quad = 8.0310-3 rad< rd,min = 10.610-3 rad,itmeansthatitispossibletoredistributemy = 0.6 my,elastic withinwx = 4 m.Therefore,itisconcludedtherecommendationgivenbyPacosteet al. (2012) in equation (2-53) is conservative for these particular cases. Figure 4.47.Comparisons between trilinear and quadlinear moment curvature input for plastic rotation distribution pl along L5 at different load level up to failure load for EII = EI/5. 4.5.6Quadlinear M()-Mpl = 0.4Mel, EII = EII/5 4.5.6.1Moment As mentioned in Section 4.5.3.3, there is a possibility to choose Mplastic < 0.6Melastic to redistributethereinforcementmomentforthecantileverslabstudiedinFigure 4.32. This cantilever slab was in this case analysed with the moment-curvature relationship input in Figure 4.48. Here Mplastic = 0.4Melastic was chosen while Mcr1 and Mcr2 input in equation (4-23) to (4-26) were kept unchanged. 01234567890.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Plastic rotation, pl [10-3rad] Coordinate, x [m]Plastic, F=143.8kN, quadlinearPlastic, F=140kN, quadlinearPlastic, F=130kN, quadlinearPlastic, F=120kN, quadlinearPlastic, F=143.8kN, trilinearPlastic, F=140 kN, trilinearPlastic, F=130kN, trilinearPlastic, F=120kN, trilinearCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 109 kNm 6 . 7 19 6 . 0 4 . 0 = = =elastic plasticM M (4-39) kNm/m 382 . 06 . 7= =plasticm (4-40) ThecurvatureI,II,andIVinequation (4-27),equation (4-28)andequation (4-30) were also kept unchanged while III is determined as: [ ] m / 1 0095 . 0122 . 0 2 . 010 610 6 . 7393= = =I EMIIplasticIII (4-41) ) m / 1 ( M (kNm) EII Mpl Mcr1 10 =I 0095 . 0 =IIIEI EI = 30 GPa EII = EI/5 = 6 GPa 0095 . 0 =IV 1 4 A C Mcr2 007125 . 0 =IIB 2 3 Figure 4.48.Quadlinear moment-curvature relationship for the analysis of the slab in elasto-plastic case, Mpl = 0.4Mel, EII = EI/5. Figure 4.48 consists of four zones that were used to simulate the structural response of theslabstudiedinFigure 4.32.Theonlydifferencebetweenquadlinearmoment curvature input in Figure 4.33 and Figure 4.48 is that Mpl = 0.4Mel was used instead of Mpl = 0.6Mel. Theratioofthemaximummomentmy,max receivedfromFEanalysistomoment distribution my,(x) along x-direction can also be determined in equation (4-31). From Figure 4.49, it is seen that from F = 20 kN to F = 60 kN, the structural response oftheslabstudiedinthisSectionisthesameasthebehavioroftheslabstudied described in Section 4.5.3.1 since their geometry and moment-curvature input are the same for this load level. WhenF = 80 kN,themaximummomentintensityatthemiddleoftheslabis my,max = 38.9 kNm/mwhichislargerthantheyieldingmomentintensity my,yield = 38 kNm/m.TheslabstudiedyieldsatalengthLyield = 1.6 mfromthe symmetry line. When continuing loading the slab until collapse load F = 95.8 kN, it is CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 110seen that the yielding occurs all along fixed boundary, Lyield = 4 m and the slab failed when the mechanism was formed. FromFigure 4.49,itisobservedthattheslabstudiedyieldsandcollapseearlier comparedtoFigure 4.35.WhenMpl = 0.4Mel waschosen,itrequiresless reinforcementanditgiveslowermomentresistancethanthecaseofMpl = 0.6Mel. Therefore,theslabstudiedincaseofMpl = 0.4Mel collapsedbeforetheslabwith Mpl = 0.6Mel. ThedistributionshapeofmomentinFigure 4.49baftercrackingandplastic redistributionissmootherthandistributionshapeofmomentfromlinearelasticFE analysis. This indicates that the moment distribution from linear elastic FE analysis is critical in the design. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 111 (a) (b) -100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=95.8kNPlastic, F=80kNPlastic, F=60kNPlastic, F=40kNPlastic, F=20kN 00.20.40.60.810.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Ratio, [ ] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNPlastic, F=95.8kNPlastic, F=60kN Figure 4.49.Moment distribution my along L5 at different load level up to 95.8 kN (a) momentdistributionmy (b).ratioofmaximummomentmy,maxtomy,x moment along L5, Mpl = 0.4 Mel, EII = EI/5. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1124.5.6.2Contour plot In Figure 4.50, there are four different colors that represent the structural response of thecantileverslabfromuncrackedstateuntilfailure.Thesecolourshavethesame meaning as those described in Section 4.5.3.2. FromF = 20 kNtoF = 60 kN,thecontourplotsinFigure 4.50areidenticaltothe contour plots in Figure 4.36. When the load increases up to F = 80 kN, the slab examined starts to yield along fixed boundaryataboutLyield = 1.6 mfromthesymmetrylinewhiletheremainingparts located at about 1.6m x 2.4m m are in zone 3 and at about 2.4m x 4 m are in zone2.WhenF = 90 kN,yieldingpartsoftheslabpropagatesfurtheratabout Lyield = 3 m.WhenF = 95.8 kN,Lyield = 4 malongL5andthecrackingpartsofthe studiedslabdevelopfurther.Thedisturbancesatposition(x = 0.2 m,y = 1.4m)and position (x = 4 m, y = 0.2 m) can be seen clearly. Like Figure 4.36 and Figure 4.37, it is also noted that there are always disturbances at position(x = 0.2 m,y = 1.4m)andposition(x = 4 m,y = 0.2m),butthese disturbances cannot be seen when their surrounding regions located in the same zone. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 113 Figure 4.50.ContourplotatdifferentloadleveluptofailureloadF = 95.8 kN, Mpl = 0.4 Mel, EII = EI/5. 1.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.01.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00.0 1.0 2.0 3.0 4.0F = 95.8kNF = 20kNF = 40kNF = 60kNF = 80kNF = 90kNCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1144.5.6.3Plastic rotation Figure 4.38 shows thatwhen the slab collapsedat F = 95.8 kN, the maximum plastic rotationispl,max = 5.210-3 rad.Hence,thisrotationissmallerthantheminimum plasticrotationcapacityofthecantileverslabexaminedrd,min = 10.610-3 radas showninAPPENDIXF.Itmeansthatplasticrotationisnotaprobleminthiscase. Therefore,itispossibletoredistributethemomentMplastic=0.4Melastic witha distributionwidthw = 4 m(symmetricalcase)iftheappliedloadisnotgreaterthan F = 95.8 kN. Figure 4.51.PlasticrotationdistributionplalongL5 atdifferentloadlevelupto F = 143.8 kN, Mpl = 0.4 Mel, EII = EI/5. 4.5.6.4Comparison with existing guideline FromSection 4.5.3.4,themeanmomentdistributionmy,mean = 90.3 kNm/m, determined from the recommendation given by Pacoste et al. (2012), can be distribute within wx = 0.65 m for F = 100 kN. WhenchoosingMplastic = 0.6MelasticandMplastic = 0.4Melastic,afailureloadof F = 143.8 kNandF = 95.8 kNrespectively,wasobtained.Inordertodetermine MplasticforafailureloadF = 100 kN,linearinterpolationbetweenMplastic = 0.4Melastic and Mplastic = 0.6Melastic were performed as following: ( ) 42 . 0 4 . 0 6 . 08 . 95 8 . 1438 . 95 1004 . 0 = + = x (4-42) kNm 9 . 7 19 4175 . 0 4175 . 0 = = =elastic plasticM M (4-43) 01234560.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Plastic rotation, pl [10-3rad] Coordinate, x [m]Plastic, R=95.8kNPlastic, R = 90kNPlastic, R=80kNPlastic, R=70kNPlastic, R = 60 kNCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 115 kNm/m 7 . 392 . 093 . 7= =plasticm (4-44) Hence, in order to achieve a collapse load F = 100 kN, mplastic = 39.7 kNm/m needs to be provided. mplastic = 39.7 kNm/m resulted a plastic rotation as following: ( )3 3 3 3100 ,10 4 . 5 10 2 . 5 10 0 . 88 . 95 8 . 1438 . 95 10010 2 . 5 = = + =F pl (4-45) FromSection 4.5.3.4,my,mean = 0.6my,max = 57 kNm/mresultsinpl,max = 8.010-3 rad anditisredistributedwithinwx = 4 m.Therefore,itisconcludedthatforacollapse loadF = 100 kN,my,mean =0.42my,max = 39.7 kNm/mwithpl,max = 5.4210-3 radcan also be distributed within wx = 4 m. ForloadF = 100 kN,theratiobetweenthemeanmomentdistributionin equation (4-33)determinedfromtherecommendationgivenbyPacosteetal.(2012) and the mean moment distribution in equation (4-44) received from elasto-plastic FE analysis is calculated as: % 443 . 9065 . 39100 100, ,, ,= = =FE mean ytion recommenda mean ymm(4-46) = 44 %indicatesthatitispossibletoreduce44 %frommeanmomentobtained from the recommendation provided by Pacoste et al. (2012). For shear forces distribution, reader can refer to APPENDIX C. 4.5.7Trilinear M() with no yielding - EII = EII/5 4.5.7.1Moment Thiscasecorrespondstoacasewheretheyieldcapacityishighenoughnottobe reached.Thus,itcanbeusedtocomparewiththeresponseinservicelimitstate.To give a better understanding of the structural behavior of the studied cantilever slab in thiscase,threemoreparametricstudieswereperformed.Thecantileverslabin Figure 4.52wasexaminedwithatrilinearmomentcurvaturerelationshipwithno yieldinginputillustratedinFigure 4.53.Thiscantileverslabhasthesamegeometry and boundary condition as the slab in Figure 4.32, but was subject to the applied load F = 200 kN.Averylargevaluewasgivenforultimatemomentcapacityandthe correspondingcurvaturesothatyieldingwasnotreached.Fortheinputvaluein Figure 4.53,readerscanrefertoequation (4-21)to(4-29)inSection 4.5.3.1where Mcr1 = 19 kNm/m, Mcr2 = 28.5 kNm/m, Mpl = 57 kNm/m.CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 116L=4 m b=1.6 m L1L2 L5 L4 L3 symmetry line 0.8 m 0.8 m 0.8 m F = 200 kN Figure 4.52. Geometryandresultlinesofthestudiedslabusingtrilinearmoment-curvature relationship with no yielding. ) m / 1 ( M (kNm) EII Mpl Mcr1 00095 . 0 =I01425 . 0 =III EI EI = 30 GPa EII = EI/5 = 6 GPa 1 3 A C Mcr2 007125 . 0 =II B 2 3 100 Mpl D 425 . 1 =IV Figure 4.53.Trilinearmoment-curvaturerelationshipwithoutyieldingforthe analysis of the slab, EII = EI/5. InFigure 4.54a,F = 20 kN,F = 40 kN,F = 60 kNandF = 80 kNgivesidentical structuralresponsetothoseinFigure 4.34asincethesetwocaseshavethesame momentcurvatureinputformy,max = 52.5 kNm/m < mpl = 57 kNm/m.When F = 100 kN,itisseenthatmy,max = 69 kNm/m> mpl = 57 kNm/msincethereisno yieldinginthemomentcurvatureinputinFigure 4.53andthemomentkeeps increasingwithincreasingappliedload.WhenF = 100 kN,my,max = 69 kNm/mis closetothemaximummomentinelasticorthotropiccasemy,max,ortho = 65.1 kNm/m anddeviatesalotfromthemaximummomentintheelasticisotropiccasewhere my,max,iso = 95 kNm/m,seeFigure 4.54a.Themomentdistributionshapefor F = 100 kNisclosetothedistributionshapeoftheelasticisotropiccaseatabout CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 117 0 < x < 1 mandthesetwocurvesdeviaterathermuchfromeachotherataround 1 m< x < 4 m, see Figure 4.54b. (a) (b) -100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=100kNCracking, F=80kNCracking, F=60kNCracking, F=40kNCracking, F=20kN00.20.40.60.810.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Ratio,[ ] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=100kNCracking, F=60kN Figure 4.54.Moment distribution my along L5 at different load level up to 100 kN (a) momentdistributionmy (b).ratioofmaximummomentmy,maxtomy,x moment along L5, EII = EI/5. FromFigure 4.55a,itcanbeseenthatthemomentincreaseswithincreasingapplied load F. The moment distribution shape when F = 140 kN and F = 200 kN is similar to each other. The similarity in the response is because the stiffness in the cracked parts in the slab are similar for these high loads. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 118(a) (b) -160-140-120-100-80-60-40-2000.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=200kNCracking, F=180kNCracking, F=160kNCracking, F=140kNCracking, F=120kN 00.20.40.60.810.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0Ratio,[ ] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=200kNCracking, F=140kN Figure 4.55.Moment distribution my along L5 at different load level up to 200 kN (a) momentdistributionmy (b).ratioofmaximummomentmy,maxtomy,x moment along L5, EII = EI/5. SinceF = 100 kN,my,max = 69 kNm/misclosetothemaximummomentinelastic orthotropic case my,max,ortho = 65.1 kNm/m, it is concluded that orthotropic analysis can be used to rather well predict the maximum moment for design of the cantilever slab studiedinFigure 4.52.Consequently,themaximummomentreceivedfrom orthotropicanalysiscanbeuseddirectlyforthedesignanddonotneedtoreduceit likeinthecaseofisotropicelasticanalysis.Forshearforcesdistribution,thereader can refer to APPENDIX C. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 119 4.5.8Influence of state II stiffness in trilinear M() without yielding 4.5.8.1Orientation In order to investigate the behaviour of the studied cantilever slab in Figure 4.52 with the trilinear moment curvature relationship without yielding input, a parametric study of stateII stiffness, i.e. EII = EI/2 and EII = EI/10 was examined. The reader can refer to APPENDIX C for trilinear moment curvature relationsip without yielding input and for the moment distribution and shear forces distribution for these cases. 4.5.8.2Load-displacement relation In these cases, the cantilever slab studied was subjected to a load of F = 200 kN. The maximumverticaldisplacementuzatthemiddleoftheslabbecomeslargerwhen increasingtheappliedloadF,seeFigure 4.56.Thereisnoformationofthecollapse mechanism since there is no yield limit. The vertical displacement uz for these three cases are identical before the slab starts to crack.Afterthecracking,thestiffnessofthestudiedslabadaptedtothestiffnessof each case. It is seen that after the cracking, the stiffer slab provides a smaller vertical displacement, i.e. uz(EII=EI/2) = 13 mm < uz(EII=EI/5) = 25.3 mm < uz(EII=EI/10) = 43.6 mm. Figure 4.56.Verticaldisplacementuzatdifferentloadleveluptocollapseloadfor EII = EI/2, EII = EI/5, EII = EI/10. -200-160-120-80-400-50.0 -40.0 -30.0 -20.0 -10.0 0.0Applied force, F [kN] Vertical displacement, uz [mm]EII=EI/2, F=200kNEII=EI/5, F=200kNEII=EI/10, F=200kNCHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1204.5.8.3Moment Themomentdistributionsandtheratiosofmaximummomentmy,max tomy,x along fixed boundary for each case were plotted in Figure 4.57. Figure 4.57.Comparisonsofmomentdistributionfortrilinearmomentcurvature relationship without yielding at different load level up to F = 100 kN for EII = EI/2, EII = EI/5, EII = EI/10. (a)(b) (c)(d) (e)(f)-100-80-60-40-2000.0 1.0 2.0 3.0 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=2Ey, F=100kNCracking, F=100kNCracking, F=80kNCracking, F=60kNCracking, F=40kNCracking, F=20kN00.20.40.60.810.0 1.0 2.0 3.0 4.0Ratio, [-] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=2Ey, F=100kNCracking, F=100kNCracking, F=60kN-100-80-60-40-2000.0 1.0 2.0 3.0 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=100kNCracking, F=80kNCracking, F=60kNCracking, F=40kNCracking, F=20kN00.20.40.60.810.0 1.0 2.0 3.0 4.0Ratio, [-] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=5Ey, F=100kNCracking, F=100kNCracking, F=60kN-100-80-60-40-2000.0 1.0 2.0 3.0 4.0Moment, my [kNm/m] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=10Ey, F=100kNCracking, F=100kNCracking, F=80kNCracking, F=60kNCracking, F=40kNCracking, F=20kN00.20.40.60.810.0 1.0 2.0 3.0 4.0Ratio, [-] Coordinate, x [m]Elastic, Ex=Ey, F=100kNElastic, Ex=10Ey, F=100kNCracking, F=100kNCracking, F=60kNCHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 121 Theratioofthemaximummomentmy,max receivedfromtheFEanalysistomoment distribution my,(x) along x-direction is determined in equation (4-47). x yymm,max ,= (4-47) Thiscorrespondstothedistributionshapeforthemoment.Hence,y,m-kdefinesa distribution shape for the moment obtained using trilinear moment curvature without yieldinginputwhiley,isodefinesadistributionshapeforthemomentreceivedfrom the isotropic analysis. The difference for the maximum moment received from the FE analysis with trilinear momentcurvaturewithoutyieldinginputmy,m-k,max andmaximummomentobtained from orthotropic FE analysis my,ortho,max is calculated as: max , ,max , , max , ,100k m yortho y k m ymm m = (4-48) Where my,m-k,max = maximummomentreceivedfromtrilinearmomentcurvature without yielding input my,ortho,max = maximum moment obtained from orthotropic analysis Table 4.5. Differencebetweentrilinearmomentcurvaturerelationshipwithout yielding input and orthotropic analysis. Casemy,m-k,max [kNm]my,ortho,max [kNm] [%] EII = EII/2-81.6-80.61.3 EII = EII/5-69.0-65.15.9 EII = EII/10-62.5-56.98.9 Figure 4.57aandTable 4.5showsthatwhenF = 100 kN,thedifferencebetween my,m-k andmy,orthoforEII = EI/2is = 1.3 %.Thedistributionshapeofmomentfrom trilinear moment curvature relation without yieldin y,m-k when F = 100 kN is close to thatofelasticisotropiccasey,isoatabout0 < x < 1.4 m,seeFigure 4.57b.For EII = EI/5,Table 4.5showthat = 5.9 %andy,m-kisclosetoy,isoataround 0 < x < 1 m,seeFigure 4.57d.IncaseofEII = EI/10, = 8.9 %andy,m-kiscloseto y,iso at approximately 0 < x < 0.8 m as illustrated in Figure 4.57f. From the difference in percentage , it is seen that the orthotropic analysis provides a maximummomentwhichisclosetothemaximummomentobtainedfromthe analysisofthestudiedcantileverslabwithtrilinearmomentcurvaturewithout yielding input. Hence, it is concluded that an orthotropic analysis is a possible method topredictadesignmomentforthethreecasesmentionedinthisSection.Itis observedthatthestifferthestudiedslabis,thebettertheorthotropicFEanalysis predicts the design moment.Concerning the distribution shape for the moment, the distribution shape for moment from isotropic analysis can be used to predict the distribution shape for moment from CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 122the trilinear moment curvature without yielding input. Consequently, y,iso approaches y,m-k when the studied cantilever become stiffer and stiffer as shown in Figure 4.57b, Figure 4.57d, Figure 4.57f. CHALMERS Civil and Environmental Engineering, Masters Thesis 2013:100 123 5Concluding remarks 5.1Conclusions This thesis studied the distribution of forces in a cantilever slab subjected to a single pointload.Abeamgrillagemodelwasanalysedwithdifferentmaterialmodelsto capture the structural response of the cantilever slab. InthisMasterThesisproject,ashellelementmodelwasusedtovalidatethebeam grillageelementmodelforthecantileverslabstudiedinisotropicandorthotropic case. The result from the FE analysis shows that shell element is the most appropriate element when analysing the cantilever slab studied in elastic isotropic and orthotropic case.However,resultsreceivedfromtheFEanalysisalsoindicatesthatthereisa good agreement between shell element and beam grillage element if appropriate mesh sizes for the beam elements are provided. Therefore, beam grillage models should be usedwithcare,butcanstillbeusedtoanalysethecantileverslabstudiedifcorrect mesh sizes are used. Orthotropic analysis was used to simulate different stiffnesses of the slab in different directions. According to the comparisons made with nonlinear analysis it can be seen thatanorthotropicanalysisgivesagoodestimationofthemaximummoment obtainedbeforetheeffectofyieldingtakesplace.Carehastobetaken,though,of what different stiffnesses of the reinforced concrete structure that should be taken into account. Hence, such an orthotropic model taking this into account could successfully be used in service limit state (SLS) and is also as a conservative estimation in ultimate limit state (ULS). Inelasto-plasticanalysis,thecantileverslabstudiedwasexaminedusingabeam grillagemodelwithnon-linearmoment-curvaturerelationship.Inthiscase,the elementsizeaffectstheplasticmomentcapacityofthestructurewhenusingbeam elements.Thefinerthemeshis,theclosertheresultsobtainedfromtheFEanalysis convergestotheexpectedsolution.Basedontheanalysisofthecantileverslab studiedinelasto-plasticcase,theneedforplasticrotationissmallandhencethe possibilityofredistributionduetoyieldingisconsiderable.TheFEanalysesofthe cantileverslabstudiedshowsthatintermofdistributionwidthformoment,the recommendationgivenbyPacosteetal.(2012)isconservative.Theeffectofplastic redistribution, though, should be handled with care if there are moving load that might causeaccumulateddamagetothestructure,whichmightleadtoadditionalplastic rotation. CHALMERS, Civil and Environmental Engineering, Masters Thesis 2013:100 1245.2Further studies Some simplifications related to the modelling of the slab studied have been made.In thisreport,intermsofdistri