statics

Statics of Bending: Shear and Bending Moment Diagrams David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 November 15, 2000 Introduction Beamsarelongandslenderstructuralelements,dieringfromtrusselementsinthattheyare calledontosupporttransverseaswellasaxialloads. Theirattachmentpointscanalsobemorecomplicatedthanthoseoftrusselements:theymaybeboltedorweldedtogether,sotheattachments cantransmitbendingmomentsortransverseforcesintothebeam. Beamsareamongthemostcommonofallstructuralelements,beingthesupportingframesofairplanes, buildings,cars,people,andmuchelse. Thenomenclatureofbeamsisratherstandard:asshowninFig.1, Listhelength,orspan; bisthewidth,and histheheight(alsocalledthe depth). Thecross-sectionalshapeneednot berectangular,andoftenconsistsofavertical webseparatinghorizontal flangesatthetopand bottomofthebeam 1. Figure1:Beamnomenclature. AswillbeseeninModules13and14,thestressesanddeflectionsinducedinabeamunder bendingloadsvaryalongthebeam’slengthandheight.Therststepincalculatingthesequan-titiesandtheirspatialvariationconsistsofconstructing shearandbendingmoment diagrams, V(x)andM(x),whicharetheinternalshearingforcesandbendingmomentsinducedinthe beam,plottedalongthebeam’slength.Thefollowingsectionswilldescribehowthesediagramsaremade. 1 Figure2:Acantileveredbeam. Free-body diagrams Asasimplestartingexample,considerabeamclamped(\cantilevered")atoneendandsub- jectedtoaloadPatthefreeendasshowninFig.2. Afreebodydiagramofasectioncut transverselyatposition xshowsthatashearforce VandamomentMmustexistonthecut sectiontomaintainequilibrium.WewillshowinModule13thatthesearetheresultantsofshearandnormalstressesthataresetuponinternalplanesbythebendingloads. Asusual,wewillconsidersectionareaswhosenormalspointinthe+ xdirectiontobepositive;thenshearforces pointinginthe+ ydirectionon+xfaceswillbeconsideredpositive. Momentswhosevector directionasgivenbytheright-handruleareinthe+ zdirection(vectoroutoftheplaneofthe paper,ortendingtocausecounterclockwiserotationintheplaneofthepaper)willbepositivewhenactingon+ xfaces.Anotherwaytorecognizepositivebendingmomentsisthattheycause thebendingshapetobeconcaveupward.Forthisexamplebeam,thestaticsequationsgive: X Fy=0=V+P)V=constant=−P (1) X M0=0=−M+Px)M=M(x)=Px (2) Notethatthemomentincreaseswithdistancefromtheloadedend,sothemagnitudeofthe maximumvalueof McomparedwithVincreasesasthebeambecomeslonger. Thisistrueof mostbeams,sosheareectsareusuallymoreimportantinbeamswithsmalllength-to-height ratios. Figure3:Shearandbendingmomentdiagrams. 1Thereisastandardizedprotocolfordenotingstructuralsteelbeams; forinstanceW8 40indicatesa wide-flangebeamwithanominaldepthof800andweighing40lb/ftoflength 2 Asstatedearlier,thestressesanddeflectionswillbeshowntobefunctionsof VandM,soit isimportanttobeabletocomputehowthesequantitiesvaryalongthebeam’slength.PlotsofV(x)andM(x)areknownasshearandbendingmomentdiagrams, anditisnecessarytoobtain thembeforethestressescanbedetermined.Fortheend-loadedcantilever,thediagramsshown inFig.3areobviousfromEqns.1and2. Figure4:Wallreactionsforthecantileveredbeam. Itwaseasiesttoanalyzethecantileveredbeambybeginningatthefreeend,butthechoice oforiginisarbitrary. Itisnotalwayspossibletoguesstheeasiestwaytoproceed,soconsider whatwouldhavehappenediftheoriginwereplacedatthewallasinFig.4. Nowwhenafreebodydiagramisconstructed,forcesmustbeplacedattheorigintoreplacethereactionsthatwereimposedbythewalltokeepthebeaminequilibriumwiththeappliedload.Thesereactionscanbedeterminedfromfree-bodydiagramsofthebeamasawhole(ifthebeamisstaticallydeterminate),andmustbefoundbeforetheproblemcanproceed.ForthebeamofFig.4: X Fy=0=−VR+P)VR=P X Mo=0=MR−PL)MR=PL Theshearandbendingmomentat xarethen V(x)=VR=P=constant M(x)=MR−VRx=PL−Px Thischoiceoforiginproducessomeextraalgebra,butthe V(x)andM(x)diagramsshownin Fig.5arethesameasbefore(exceptforchangesofsign): Visconstantandequalto P,andM varieslinearlyfromzeroatthefreeendto PLatthewall. Distributed loads Transverseloadsmaybeappliedtobeamsinadistributedratherthanat-a-pointmanneras depictedinFig.6,whichmightbevisualizedassandpiledonthebeam. Itisconvenienttodescribethesedistributedloadsintermsof forceperunitlength, sothatq(x)dxwouldbethe loadappliedtoasmallsectionoflength dxbyadistributedload q(x).Theshearforce V(x)set upinreactiontosuchaloadis V(x)=−Zx x0q()d (3) 3 Figure5:Alternativeshearandbendingmomentdiagramsforthecantileveredbeam. Figure6:Adistributedloadandafree-bodysection. wherex0isthevalueofxatwhichq(x)begins,andisadummylengthvariablethatlooks backwardfromx.HenceV(x)istheareaunderthe q(x)diagramuptoposition x.Themoment balanceisobtainedconsideringtheincrementofload q()dappliedtoasmallwidth dofbeam, adistancefrompointx.Theincrementalmomentofthisloadaroundpoint xisq()d,so themomentM(x)is M=Zx x0q()d (4) Thiscanberelatedtothecentroidoftheareaunderthe q(x)curveuptox,whosedistance fromxis =Rq()dRq()d HenceEqn.4canbewritten M=Q (5) whereQ=Rq()disthearea. Therefore,thedistributedload q(x)isstaticallyequivalentto aconcentratedloadofmagnitude Qplacedatthecentroidoftheareaunderthe q(x)diagram. Example 1 Considerasimply-supportedbeamcarryingatriangularandaconcentratedloadasshowninFig.7.For 4 Figure7:Distributedandconcentratedloads. thepurposeofdeterminingthesupportreactionforces R1andR2,thedistributedtriangularloadcanbe replacedbyitsstaticequivalent.Themagnitudeofthisequivalentforceis Q=Z2 0(−600x)dx=−1200 Theequivalentforceactsthroughthecentroidofthetriangulararea,whichisis2/3ofthedistancefrom itsnarrowend(seeProb.1).Thereaction R2cannowbefoundbytakingmomentsaroundtheleftend: X MA=0=−500(1)−(1200)(2=3)+R2(2)!R2=650 Theotherreactioncanthenbefoundfromverticalequilibrium: X Fy=0=R1−500−1200+650=1050 Successive integration method Figure8:Relationsbetweendistributedloadsandinternalshearforcesandbendingmoments. WehavealreadynotedinEqn.3thattheshearcurveisthenegativeintegraloftheloading curve. Anotherwayofdevelopingthisistoconsiderafreebodybalanceonasmallincrement 5 oflengthdxoverwhichtheshearandmomentchangesfrom VandMtoV+dVandM+dM (seeFig.8).Thedistributedload q(x)canbetakenasconstantoverthesmallinterval,sothe forcebalanceis: X Fy=0=V+dV+qdx−V=0 dV dx=−q (6) or V(x)=−Z q(x)dx (7) whichisequivalenttoEqn.3.Amomentbalancearoundthecenteroftheincrementgives X Mo=(M+dM)+(V+dV)dx 2+Vdx 2−M Astheincrement dxisreducedtothelimit,thetermcontainingthehigher-orderdierential dVdxvanishesincomparisonwiththeothers,leaving dM dx=−V (8) or M(x)=−Z V(x)dx (9) Hencethevalueoftheshearcurveatanyaxiallocationalongthebeamisequaltothenegative oftheslopeofthemomentcurveatthatpoint,andthevalueofthemomentcurveatanypointisequaltothenegativeoftheareaundertheshearcurveuptothatpoint. Theshearandmomentcurvescanbeobtainedbysuccessiveintegrationofthe q(x)distri- bution,asillustratedinthefollowingexample. Example 2 Consideracantileveredbeamsubjectedtoanegativedistributedload q(x)=−q0=constantasshown inFig.9;then V(x)=−Z q(x)dx=q0x+c1 wherec1isaconstantofintegration. Afreebodydiagramofasmallsliveroflengthnear x=0shows thatV(0)=0,sothec1mustbezeroaswell.Themomentfunctionisobtainedbyintegratingagain: M(x)=−Z V(x)dx=−1 2q0x2+c2 wherec2isanotherconstantofintegrationthatisalsozero,since M(0)=0. Admittedly,thisproblemwaseasybecausewepickedonewithnullboundaryconditions,and withonlyoneloadingsegment. Whenconcentratedordistributedloadsarefoundatdierent 6 Figure9:Shearandmomentdistributionsinacantileveredbeam. positionsalongthebeam,itisnecessarytointegrateovereachsectionbetweenloadsseparately. Eachintegrationwillproduceanunknownconstant,andthesemustbedeterminedbyinvokingthecontinuityofslopesanddeflectionsfromsectiontosection.Thisisalaboriousprocess,butonethatcanbemademucheasierusing singularityfunctions thatwillbeintroducedshortly. Itisoftenpossibletosketch VandMdiagramswithoutactuallydrawingfreebodydia- gramsorwritingequilibriumequations.Thisismadeeasierbecausethecurvesareintegralsorderivativesofoneanother,sographicalsketchingcantakeadvantageofrelationsamongslopes andareas. Theserulescanbeusedtoworkgraduallyfromthe q(x)curvetoV(x)andthentoM(x). Whereveraconcentratedloadappearsonthebeam,the V(x)curvemustjumpbythatvalue, butintheoppositedirection;similarly,the M(x)curvemustjumpdiscontinuouslywherevera coupleisappliedtothebeam. Example 3 Figure10:Asimplysupportedbeam. Toillustratethisprocess,considerasimply-supportedbeamoflength LasshowninFig.10,loaded 7 overhalfitslengthbyanegativedistributedload q=−q0. Thesolutionfor V(x)andM(x)takesthe followingsteps: 1. Thereactionsatthesupportsarefoundfromstaticequilibrium. Replacingthedistributedload byaconcentratedload Q=−q0(L=2)atthemidpointofthe qdistribution(Fig.10(b))andtaking momentsaround A: RBL=q0L 23L 4 )RB=3q0L 8 Thereactionattherightendisthenfoundfromaverticalforcebalance: RA=q0L 2−RB=q0L 8 Notethatonlytwoequilibriumequationswereavailable,sinceahorizontalforcebalancewould providenorelevantinformation.Hencethebeamwillbestaticallyindeterminateifmorethantwosupportsarepresent. Theq(x)diagramisthenjustthebeamwiththeendreactionsshowninFig.10(c). 2. Beginningthesheardiagramattheleft, Vimmediatelyjumpsdowntoavalueof −q 0L=8in oppositiontothediscontinuouslyappliedreactionforceat A;itremainsatthisvalueuntil x=L=2 asshowninFig.10(d). 3. Atx=L=2,theV(x)curvestartstorisewithaconstantslopeof+ q0astheareaunderthe q(x) distributionbeginstoaccumulate. When x=L,theshearcurvewillhaverisenbyanamount q0L=2,thetotalareaunderthe q(x)curve;itsvalueisthen( −q0l=8)+(q0L=2)=(3q0L=8).The shearcurvethendropstozeroinoppositiontothereactionforce RB=(3q0L=8).(TheVandM diagramsshouldalwaysclose,andthisprovidesacheckonthework.) 4. ThemomentdiagramstartsfromzeroasshowninFig.10(e),sincethereisnodiscontinuously appliedmomentattheleftend.Itmovesupwardataconstantslopeof+ q0L=8,thevalueofthe sheardiagraminthersthalfofthebeam.When x=L=2,itwillhaverisentoavalueof q0L2=16. 5. Afterx=L=2,theslopeofthemomentdiagramstartstofallasthevalueofthesheardiagram rises.Themomentdiagramisnowparabolic,alwaysbeingoneorderhigherthanthesheardiagram. Thesheardiagramcrossesthe V=0axisatx=5L=8,andatthispointtheslopeofthemoment diagramwillhavedroppedtozero. Themaximumvalueof Mis9q0L2=32,thetotalareaunder theVcurveuptothispoint. 6. Afterx=5L=8,themomentdiagramfallsparabolically,reachingzeroat x=L. Singularity functions Thisspecialfamilyoffunctionsprovidesanautomaticwayofhandlingtheirregularitiesof loadingthatusuallyoccurinbeamproblems. Theyaremuchlikeconventional polynomialfactors,butwiththepropertyofbeingzerountil\activated"atdesiredpointsalongthebeam.Theformaldenitionis f n(x)=hx−ain=( 0;x < a (x−a)n;x > a(10) wheren=−2;−1;0;1;2;


. Thefunctionhx−ai0isaunitstepfunction, hx−ai−1isa concentratedload,and hx−ai−2isaconcentratedcouple.Therstveofthesefunctionsare sketchedinFig.11. 8 Figure11:Singularityfunctions. Thesingularityfunctionsareintegratedmuchlikeconventionalpolynomials: Zx −1hx−aindx=hx−ain+1 n+1n0 (11) However,therearespecialintegrationrulesforthe n=−1andn=−2members,andthis specialhandlingisemphasizedbyusingsubscriptsforthe nindex: Zx −1hx−ai−2dx=hx−ai−1 (12) Zx −1hx−ai−1dx=hx−ai0(13) Example 4 ApplyingsingularityfunctionstothebeamofExample4.3,theloadingfunctionwouldbewritten q(x)=+q0L 8hx−0i−1−q0hx−L 2i0 Thereactionforceattherightendcouldalsobeincluded,butitbecomesactivatedonlyastheproblem isover.Integratingonce: V(x)=−Z q(x)dx=−q0L 8hxi0+q0hx−L 2i1 Theconstantofintegrationisincludedautomaticallyhere,sincetheinfluenceofthereactionat Ahas beenincludedexplicitly.Integratingagain: M(x)=−Z V(x)dx=q0L 8hxi1−q0 2hx−L 2i2 Examinationofthisresultwillshowthatitisthesameasthatdevelopedpreviously. MapleTMsymbolicmanipulationsoftwareprovidesanecientmeansofplottingthesefunctions.The followingshowshowthemomentequationofthisexamplemightbeplotted,usingtheHeavisidefunction toprovidethesingularity. 9 # Define function sfn in terms of a and n >sfn:=proc(a,n) (x-a)^n*Heaviside(x-a) end; sfn := proc(a, n) (x - a)^n*Heaviside(x - a) end proc # Input moment equation using singularity functions >M(x):=(q*L/8)*sfn(0,1)-(q/2)*sfn(L/2,2); M(x) := 1/ 8qLx Heaviside(x) 2 - 1/2 q (x - 1/2 L) Heaviside(x - 1/2 L) # Provide numerical values for q and L: >q:=1: L:=10: # Plot function >plot(M(x),x=0..10); Figure12:Maplesingularityplot Problems 1. (a){(c)Locatethemagnitudeandpositionoftheforceequivalenttotheloadingdistribu- tionsshownhere. 2. (a){(c)DeterminethereactionforcesatthesupportsofthecasesinProb.1.3. (a){(h)Sketchtheshearandbendingmomentdiagramsfortheloadcasesshownhere.4. (a){(h)Writesingularity-functionexpressionsfortheshearandbendingmomentdistribu- tionsforthecasesinProb.3. 5. (a){(h)UseMaple(orother)softwaretoplottheshearandbendingmomentdistributions forthecasesinProb.3,usingthevalues(asneeded) L=25in;a=5in;w=10lb=in;P= 150lb. 10 Prob.1 Prob.3 6. Thetransversedeflectionofabeamunderanaxialload Pistakentobe(y)=0sin(y=L), asshownhere.Determinethebendingmoment M(y)alongthebeam. 7. Determinethebendingmoment M()alongthecircularcurvedbeamshown. 11 Prob.6 Prob.7 12