The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 1 3. Shear in RC beams Shearforcesnormallyactincombinationwithflexure,axialloadandperhapstorsion.Itistherefore quite difficult to isolate the effect of shear forces acting alone. Shear transfer in RC beams relies on the tensileandcompressionstrengthofconcreteaswellastensilepropertiesoflongitudinaland,when provided,transversereinforcement.Inmostofthecases,failureduetoshearisbrittleinnatureand usually occurs with little warning. Basedontheseconsiderations,elementsaredesignedsothattheyfailinflexurebeforetheyfailin shear. 3.1. Elastic behaviour of uncracked beams Intheuncrackedstate,reinforcedconcretecanbeconsideredtobeahomogeneous,elasticmaterial. Withthisassumptioninmind,theaxialstressesduetobendingandtheshearstressescanbeeasily calculated according to Eqs. (1) and (2). MyI = (1) VSIb =(2) Where:M = bending moment V = shear force I = second moment of area of the section S = first moment of the hatched area Av (see Figure 1) = vA yy = distance from the neutral axis of the point at which the axial stress is calculated b = width of the section Once the above stresses at a point P in the beam are known, the principal stresses at this point can be derived as follows: 221,22 2 = + (3) The angle between the horizontal and f1 can be determined from 2tan 2= (4) As can be seen in Figure 2, the combination of shear stresses and bending stresses causes the principal stress trajectories to change directions along the depth of the beam. When the tensile stresses, 1, are higherthantheconcretetensilestrength,crackingwilloccurandreinforcementshouldbeprovided. The magnitude and the direction of 1is influenced by the shear stress v. Vertical cracks are expected to form at midspan, in the bottom of the beam. Away from midspan, a crack initiating at the bottom of the beam will progress upwards and change direction as the magnitude of shear stresses changes with consequentchangesof1.Attheneutralaxis,onlyshearstressesareactivecausingtheprincipal stresses to be inclined at 45. Thus, diagonal cracks initiating at the neutral axis will form at 45 to the horizontal. The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 2 Oncetheconcretehascracked,theassumptionofahomogeneousisotropicmaterialdoesnotapply andadifferentapproachwillhavetobeconsideredtopredict/estimatethedevelopmentofstress fields. P A A Section A-A V M M V n.a. P Shear force diagram Bending moment diagram axial stressesshear stresses 1stresses at point P 122section A-A y y A v Figure 1: Elastic behaviour of a beam Compression fieldTension field Figure 2: Principal stresses in an uncracked beam The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 3 3.2. Behaviour of cracked beams ConsiderthecrackedbeamshowninFigure3a.Considerpartcanddoffigure.Theaverageshear stress between the cracks of part c of the figure can be calculated as follows. The tensile forces in the rebar on each side of the crack can be written as MTz=and M dMT dTz++ =and therefore dMdTz= (5) where the lever arm z is assumed to be constant. For moment equilibrium of the element dM Vdx =and VdxdTz= (6) T+dTT T+dT T dx CC+dC V M+dMM V zz (a) cracked beam(b) free-body of cracked concrete below n.a. (c) equilibrium of a differential element (d) average shear stresses Figure 3: Shear stresses in a cracked concrete section IftheshadedportioninFigure3cisisolatedasshowninpartbofthefigure,theforcedTmustbe transferredbyhorizontalshearstressesonthetopoftheshadedelement.Theaveragevalueofthese stresses below the top of the crack is dTbdx =or Vbz = (7) The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 4 3.3. Basic element shear transfer mechanisms The internal shear resistance of a beam can be expressed by the following equation: ( )dM d dT dzV Tz z Tdx dx dx dx= = = + (8) From this equation it can be seen that shear is resisted by two combined effects: 1) beam action: the internal lever arm remains constant and the magnitude of the tensile force, T, changes along the length of the beam (first term on the right-hand side of Eq. (8)). 2) archaction:thetensileforce,T,remainsconstantandthevalueoftheinternalleverarm changes along the length of the beam (second term on the right-hand side of Eq. (8)). Thedevelopmentofbeamactionandarchactionsisaffectedbyvariousphenomena(e.g.cracking, bond behaviour) and these simple mechanisms are generally combined in a complex manner. Theinternalforceflowsinregionsofabeamwherestatic,geometricormaterialconditionsvary abruptly, can be represented appropriately by means of strut and tie models. Although such approaches represent a simplification to the problem, the basic load transfer mechanisms over an element can be easily determined and analysed. A strut and tie mechanism (Figure 4) develops when a force is transferred with a strut directly to the support. In this case, the tensile forces are carried by the flexural reinforcement, whilst the concrete is supposedtoactonlyincompression.Inastrutandtiemechanism,theshearisalways carried in the compressionzone,however,thereisevidencethatthistypeofmechanismdevelopsonlyinbeams with a shear ratio of less than 3 and that it becomes significant only for a shear ratio of 2 or less. strut struttie Figure 4: Strut and Tie mechanism An arching action (Figure 5) can be developed only when a predominantly distributed load is applied toastructure.Insuchcases,externalforcescausethecompressionforcetobediverted,whilethe tensionforceintheflexuralreinforcementmayormaynotremainconstant.Hence,theconcrete between the compression arc and the tensile tie is substantially redundant. If the tensile strength of the concreteismobilisedintheregionbetweenthecompressivearchandthetensiontie,however,an arching action can develop even without the presence of UD loading. In this case the arch mechanism becomes very similar to the strut and tie mechanism. The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 5 Figure 5: Arch mechanism Whentheforcecannotmigratedirectlytothesupport,atrussmechanismisactivatedbuta supplementary force is needed to divert it up and down the beam depth. This diverted force, whether carriedbystirrupsorconcrete,causesadecreaseinthetensionforceintheflexuralreinforcement closetothesupport,whichistypicalforatrussmechanism(Figure6).Inthecasethatinclinedor, mostcommonly,verticalreinforcementisprovided,tensilestressescanbetransmittedacrossshear cracks by the reinforcement and a truss-like mechanism is developed along the beam. In such a model, the shear is carried by the inclined compression struts and the tensile ties. Ft Figure 6: Truss mechanism Strut-and-tieorarchmechanismandsomekindofatrussmechanism,however,mayoccur simultaneouslyinaRCbeamandtheextenttowhateachmechanismcontributestothetotalshear resistancedependsontheexternalloading,geometryofthebeam,presenceoftransverse reinforcement and material properties of concrete as well as reinforcement. It can be observed that the contribution of a truss mechanism becomes relevant, in general, for shear-span to depth ratios greater than2while,forshear-spantodepthratioslessthan2,astrut-and-tieorarchmechanismcanbe considered to be the most relevant shear resisting mechanisms. 3.4. Shear modes of failure Failure of RC beams due to shear is always preceded by the formation of cracks inclined to the main axisofthebeam.Theformationofshearcrackschangestheinternalbehaviouroftheelementand failure can subsequently take place either simultaneously with the formation of new or extending shear cracksorafteranincreaseintheappliedload.KaniinvestigatedthevariationinthestrengthofRC beamswithoutshearreinforcementoverawiderangeofdifferentshear-spantodepthratiosand generalized the type of expected shear failure according to this ratio. By expressing the beam capacity as the ratio of the moment in the flexural region at failure, Mc, to the ultimate flexural capacity, Mf, the variation with the ratio a/d describes the so called shear valley (Figure 7). The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 6 Shear span to depth ratio (a/d)0 1 2 3 4 5 6 7 8Beam capacity (Mc/Mf)0.00.51.0III IIIIV Figure 7: Representation of Kani's shear valley Accordingtotheshearvalleyconcept,failurecanoccurinfourdifferentmodes.TypeIisaflexural typeoffailureandoccursinbeamswithalongshearspaninwhichflexuralcapacitycanbefully developed. Type II is a brittle failure caused by the propagation of a diagonal crack toward the point of loading, which initiates from the tip of a flexural crack, close to the support. In this case, the flexural capacityoftheelementreduceswiththeratioa/d.TypeIIIalsodescribesabrittlemodeoffailure, which is characterized by a diagonal crack, normally initiating in the web, that forms independently of theflexuralcracks.Whenthistypeoffailureoccurs,theflexuralcapacityofthememberincreases withthereductionintheratioa/d.TypeIVischaracterizedbythedevelopmentofadiagonalcrack that propagates from the support toward the point-load. This type of situation is typically encountered indeepbeamsandallowstheflexuralcapacitytobefullydeveloped.Typicalshearfailure mechanisms are described in detail in the following paragraphs. Type IType IIType IIIType IV Figure 8: Typical types of failure modes 3.4.1.Diagonal tension failure Diagonal tension failure occurs typically in beams with a shear-span to depth ratio (a/d) greater than 2, but could occur also for lower values of a/d. When this mechanism takes place a diagonal crack forms asacontinuationofanearlierflexuralcrack(1)andturnsgraduallyintoacrackmoreandmore inclinedundertheshearload(Figure9).Suchacrackdoesnotproceedimmediatelytofailurebut usuallyencountersresistanceasitmovesintothecompressionzoneandstops(1).Withafurther increaseintheappliedload,thetensioncrackextendsgraduallyataveryflatslopeuntilsudden failure occurs when the shear crack breaks through the compressive zone. Shortly before reaching the The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 7 critical failure point (2), the tension crack extends backward beyond and below the head of the original flexural crack (3). Usually, additional cracks develop at the level of the longitudinal reinforcement (4), and the concrete cover subsequently splits. 31124 Figure 9: Development of diagonal tension crack 3.4.2.Shear compression failure The development of a diagonal crack such as that described in the previous section is often contained bythepresenceofanexternallyappliedloadthatinducescompressivestressesinthenearbyregion thusreducingthepossibilityoffurthertensioncracking(Figure10).Inaddition,thecompressive stresses over the reaction support reduce the chance of bond splitting and diagonal cracking along the reinforcement. In such a case, failure of the structure will take place due to crushing of the concrete in the region adjacent to the load. This type of failure has been designated as a shear-compression failure because the failure zone carries most of the shear forces and the failure is caused by the combination of shear and compressive stresses. Such a failure can be expected to occur for values of the shear-span todepthratiolessthanfour,andforsmallshear-spantodepthratios(typicallylessthan2),the increase in the shear strength may be significant. Compression failureCompression stress Figure 10: Shear-compression failure 3.4.3.Anchorage failure Occasionally, if the main reinforcement is not adequately anchored beyond the crack, small diagonal cracks will cause the concrete to split along the reinforcing bars before compression failure can occur. Thebeamhereactslikeasimpletiearchmechanisminwhichtheexternalloadisresistedbya concretearchpassingabovethecrackandwithathrustlinefromtheloadtowardsthesupport.The mechanism fails if the full anchorage of the longitudinal bars is not provided beyond the crack. The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 8 3.4.4.Splitting or true shear failure Splitting failure can only occur in beams with a shear-span to depth ratio (a/d) less than one in which a directtransferoftheloadtothesupporttakesplace(Figure11).Alternativelytosplittingfailure, failureincompressionoftheregionadjacentthesupportmaytakeplace.Insuchmembers,usually referred to as deep beams, shear strength is much higher than in ordinary flexural beams. Figure 11: Shear failure in beams with shear-span to depth ratio less than one 3.5. Local shear carrying mechanisms When external loads act on a RC beam, a combination of internal mechanisms is mobilized in order to transfer the applied load to the support (Figure 12). These local mechanisms are: shear transfer in the compression zone, Vc; aggregate interlock, Va; dowel action, Vd, and, when provided, shear transfer via vertical or inclined shear reinforcement, Vs. VdVcVaTC FvTCVVs Figure 12: Shear resistance mechanisms in beams without shear reinforcement (left) and shear strength contribution of vertical stirrups (right) 3.5.1.Shear transfer in the compression zone Shearstressescandevelopintheconcretecompressionzoneasaresultofthevariationoftheaxial force along the beam (Figure 13) as well as the variation of the compressive-force-path depth (Figure 14). Thepresenceofashearforceactingonthecompressionzonealongwiththeaxialforceinducesa complexbiaxialstressdistributionthroughthecompressionzoneand,asaconsequence,thestrain-capacityoftheconcreteisaffected.Studiesrevealedthatthepresenceofatransversetensilestress affectsthestrain-capacityoftheconcretewhichreducesasthetensilestressincreases.Asa consequence,whenashearforcedevelopswithinthecompressionzone,thestraincapacityofthe The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design SMR: Professor K. Pilakoutas 9 concreteinthisregionreducesand,hence,sectionsintheshear-spanaremorelikelytofailbefore those subjected to pure flexure. c+ cneutral axiscc Figure 13: Shear stresses due to the variation of axial forces along the beam caxial forceshear forcecompressive force Figure 14: Shear stresses distribution due to variation of the compressive-force-path depth 3.5.2.Aggregate interlock Inthetensilezone,sheartransferacrossacrackbymechanicalinterlockisdevelopedwhenashear displacement parallel to the direction of the crack occurs (Figure 15). Figure 15: Transmission of shear stresses across cracks due to aggregate interlock 3.5.3.Dowel action of reinforcement Thetermdowelactionreferstothecombinationofthetensileresistanceoftheconcretealongthe splittingplaneandthebendingresistanceofthereinforcementbar.Dowelstrengthacrossashear planecanbedevelopedbythreemechanisms:theflexureofthereinforcingbars,theshearstrength across the bars and kinking of the reinforcement (Figure 16). As studies have shown, dowel action is a shearcarryingactionthatisofrelativelyminorimportanceincomparisontoothersheartransfer mechanisms. The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 10 VMMVVVVVflexure of the bar shear strength of the bar kinking of the bar Figure 16: The mechanism of dowel action across a shear interface 3.5.4.Contribution of shear reinforcement Tensilestressesarecarriedacrossshearcracksbyinclinedor,mostcommonly,byvertical reinforcementintheformofclosedlinks,bentupbarsorspiralreinforcement.Priortocracking, however,stirrupscarryessentiallynostress,possiblyonlyalittlecompressionduetovertical shrinkage of the concrete. When cracking occurs, the transverse reinforcement goes into tension as it is crossed by the diagonal crack and this tension controls and limits the progress of the crack, delaying the failure of the beam until higher loads are imposed. 3.6. Truss analogy ThistheoryidealisesaRCbeamasatrusscomprisingparallellongitudinalchordsandaweb composedofdiagonalconcretestrutsandtransversesteelties.Whenshearisappliedtothistruss, compressionforcesareresistedbytheconcretestruts,whiletensionisresistedinthetransverseties andinthelongitudinalchords(Figure17).Inmemberswithouttransversereinforcement,shearcan only be resisted by the concrete across the ties and consequently, shear resistance is lost when the first cracks in the concrete start to appear. If transverse reinforcement is provided, the member reaches its ultimateshearcapacitywhenthelinksstarttoyield.Theforcesdevelopedineachelementcanbe easily determined by static analysis. Beam cross-sectionLC Vbd Figure 17: Truss analogy3.6.1.Shear reinforcement contribution Figure18bshowsafreebodycutbysectionA-Aparalleltothediagonalcompressivestruts.Inthis section the shear force that is resisted by tension in the stirrups (Vs) is given by Eq. (9). s s sw ywV n A f = (9) Where:ns = numbers of shear links intersected by A-A Asw = cross-section area of shear links fyw = yield strength of shear reinforcement The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 11 The numbers of bars intersected by A-A can be easily calculated from the analysis of Figure 18c and is given in Eq.(10). 1cot= =sv vs zns s(10) Substituting Eq. (10) into Eq. (9) cots sw ywvzV A fs= (11) Considering the case of = 45 and assuming that z d, the equation can be written as s sw ywvdV A fs=(12) Beam cross-sectionV s A A V A A T C A s v s v A z s 1 b d (b) (a) (c) Figure 18: Truss analogy: shear reinforcement contribution 3.6.2.Stress in the concrete Considering now the vertical section B-B in Figure 19, the force V acting on the section is resisted by an inclined compressive force D, where sin=VD (13) The width of the diagonal concrete strut is zcos and the average compressive stress fcd is equal to 1tansin cos tan = = + cdV Vfbz bz(14) The University of Sheffield Department of Civil & Structural Engineering CIV6235 Advanced Concrete Design 12 Alimitforthisstressshouldbeimposedtoavoidcrushingoftheconcretestruts.Areasonablelimit depends on the angle but will range between 0.25fcu and 0.45fcu. Beam cross-sectionV B B V D s v b d (b) (a) (c) B B z cosD f cd NN/2N/2 Figure 19: Truss analogy: stress in the concrete Inordertoimprovetheaccuracyofthismodel,especiallywhendealingwithbeamswithoutweb reinforcement, it became accepted design practice to introduce an empirical term, commonly referred toastheconcretecontribution,alongwiththeshearresistancecalculatedaccordingtothetruss equation. This concrete contribution, which formulation may differ in the various codes of practice, takes into account, in an empirical way, the contribution to the total shear resistance provided by other shearcarryingactionsuchasincompression,aggregateinterlockanddoweloftheflexural reinforcement.Inaddition,inordertopreventtheconcretestrutsfromcrushing,limitshavetobe imposedtothemaximumvaluethatthedesignconcreteshearstresscanassumeandareusually expressed as a function of the design concrete strength. The simplicity of the model and its ability to yield an adequate level of safety has meant that the truss analogystillformsthebasisofparticularaspectsofdesigncodesinpracticetoday,suchasthe equations that are used to derive the required amount of transverse reinforcement. CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 1CIV 6235Shear Design of RC BeamsDepartment of Civil & Structural Engineering Slide No. 1Dr Maurizio GuadagniniCIV 6235 Aim: To examine shear in RC beams and learn how to design RC beams to sustain shear forces Overview of the lectureShear Design of RC BeamsDepartment of Civil & Structural Engineering Slide No. 2 Introduction Shear behaviour of RC beams Uncracked and cracked beams Shear resisting mechanisms & shear carrying mechanisms Shear Failures Shear: Theory and Design Truss analogy & Design codesCIV 6235Bending Moment and Shear ForceDepartment of Civil & Structural Engineering Slide No. 3 Shear force is the force in an element acting perpendicular to its longitudinal axis.CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 2CIV 6235Flexural or Shear Failure? We design beams to fail How come? Dont we want beams not to fail? Any beam will fail under increasing load. However, we want beams to fail exactly under the design load in a fully controlled manner by achieving the mostDepartment of Civil & Structural Engineering Slide No. 4fully controlled manner by achieving the most wanted failure mode.CIV 6235Is Shear Failure t d?Department of Civil & Structural Engineering Slide No. 8wanted?CIV 623513Rd,c200.(6.2) : V 0.12 1 100slck wwAEq f bdd bd (| || | (= + | | | (\ . \ . 1) Members not requiring shear reinforcementShear Code Equations: EC2Department of Civil & Structural Engineering Slide No. 9( ),.(6.8) : cot cot sinswRds ywdAEq V zfs = +2) Members requiring shear reinforcementCIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 3CIV 6235Un-cracked RC Beams In regions uncracked in bending the shear stress distribution is derived as follows:V SI bM y=Department of Civil & Structural Engineering Slide No. 10M yI=32VbhCIV 6235Un-cracked RC BeamsDepartment of Civil & Structural Engineering Slide No. 11CIV 6235Un-cracked RC BeamsDepartment of Civil & Structural Engineering Slide No. 12CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 4CIV 6235Cracked RC BeamsDepartment of Civil & Structural Engineering Slide No. 13CIV 6235Basic Shear Transfer MechanismsDepartment of Civil & Structural Engineering Slide No. 14CIV 6235Basic Shear Transfer Mechanismsstrut strutiDepartment of Civil & Structural Engineering Slide No. 15tieStrut and Tie MechanismCIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 5CIV 6235Basic Shear Transfer MechanismsDepartment of Civil & Structural Engineering Slide No. 16Arch MechanismCIV 6235Basic Shear Transfer MechanismsDepartment of Civil & Structural Engineering Slide No. 17Truss MechanismFtCIV 6235Types of FailuresDepartment of Civil & Structural Engineering Slide No. 18CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 6CIV 6235Shear Failures1124Diagonal Tension FailureDepartment of Civil & Structural Engineering Slide No. 193CIV 6235Shear FailuresCompression FailureDepartment of Civil & Structural Engineering Slide No. 20CIV 6235Shear FailuresSplitting / True ShearFailureDepartment of Civil & Structural Engineering Slide No. 21CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 7CIV 6235Beams with shear reinforcementDepartment of Civil & Structural Engineering Slide No. 22CIV 6235Truss analogyDepartment of Civil & Structural Engineering Slide No. 23CIV 6235Truss analogyDepartment of Civil & Structural Engineering Slide No. 24CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 8CIV 6235Beams without shear r/mentVdVcVaTCDepartment of Civil & Structural Engineering Slide No. 25VMMVVVflexure of the bar shear strength of the barVa= Aggregate interlock Vd= Dowel actionVc= Concrete in compressionCIV 6235ULS of shear Design equations Shear resistance - VRd Beam with no shear reinforcement Shear reinforcementEC2 i t l dDepartment of Civil & Structural Engineering Slide No. 26 EC2 compares internal and external shear forces to assess if shear reinforcement is required EC2 identifies four basic shear forces for design purposes:VEd, VRd,c, VRd,s and VRd,maxCIV 6235What are VEd, VRd,c , VRd,s and VRdmax? VEd Is the applied shear force i.e. the design shear force resulting from external loading VRd,c Is the shear resistance of a member without shear reinforcementDepartment of Civil & Structural Engineering Slide No. 27 VRd,s Is the shear resistance of a member governed by the yielding of shear reinforcement VRd,max Is the maximum shear resistance of a member limited by the crushing of compression strutsCIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 9CIV 6235 No shear reinforcement is necessary (minimum)1) VEd VRd,cwVRd,c2) VRd,c< VEd < VRd,maxVRd,maxVRd cBASICSDepartment of Civil & Structural Engineering Slide No. 28 Shear reinforcement should be provided in order that VEd VRdVRd,c Not allowed3) VEd> VRd,maxVRd,maxCIV 62351) Members not requiring shear reinforcement( )13Rd,c , 1 1Rd,c min 1Eq. (6.2.a): V 100with a minimum of V ( )where200k1 2.0, with in mm (size factor)dRdc ck cp wcp wC k f k bdv k bdd (= + ( = += + Department of Civil & Structural Engineering Slide No. 291, and must not be greater than 2%,where is slwEdcpcEdAbdNAN==the axial force due to loading or prestressing ( 0 compression) is the cross-sectional area of concrete is the smallest width of the cross section in the tensile area (see figure) is the effeEdcwNAbd>ctive depth of the cross sectionCIV 623513Rd,c3/ 21/ 2200V 0.12 1 100with a minimum of200V 0 035 1slck wwAf bdd bdbd f bd (| || | (= + | | | (\ . \ . | |+ |1) Members not requiring shear reinforcementDepartment of Civil & Structural Engineering Slide No. 301/ 2Rd,c minV 0.035 1w ck wv bd f bdd= = + | |\ . Advised value: CRd,c= 0.18/c= 0.12CIV 6235 - Advanced Concrete DesignDepartment of Civil and Structural Engineering 10CIV 62352) Members requiring shear reinforcement Variable angle truss analogywith strut inclination1 < cot < 2.521.8