Engineering Basic Mechanics I Statics 工程基础力学Ⅰ 静力学 课件 Chapter 13 Energy Method

StaticsStaticsofdeformablebodyChapter13

EnergyMethod

13.1Strainenergyofbars13.2Mohrtheorem13.3DiagrammultiplicationmethodforMohrintegration13.4Castigliano'stheoremContentsU=W(13-1)Principleofstrainenergymethod：theworkdonebytheexternalforceatthecorrespondingdisplacementisnumericallyequaltothestrainenergystoredinthedeformedbody.Thestrainenergymethod:Themethodsofsolvingproblemsusingtheoremsandprinciplesrelatedtotheconceptofstrainenergy.External:workWdonebyexternalforceInternal:potentialenergydeformationenergyU13.1Strainenergyofbars1.

StrainenergyforbasicdeformationThecalculationofthestrainenergyunderseveralbasicdeformationsisnowexamined.(1)AxialtensionorcompressionForaxialtensionorcompressionofstraightrodsofequalcross-section,theexternalforceislinearlyrelatedtotheaxialdeformationofthebarwithintheelasticrange.ABoFl

D

(a)FlD

l(b)FThestrainenergyofthebarcanthusbewrittenas

a.Iftheinternalforcevariescontinuouslyalongtheaxisofthebar,i.e.FN=FN(x),Thestrainenergyofthebarcanbe

b.Iftheinternalforcesvaryinsteps,

wheremisthenumberoftensionandcompressionbars.Thestrainenergy(strainenergydensityorspecificenergy)perunitvolumeofatension(compression)baris

ABoM

j(2)Torsionofcircularshafts(a)lMjjMWorkdonebytorque:(b)

AccordingtoequationU=W,thisworkisequaltothetorsionalstrainenergystoredinthecircularshaft.Whenthecircularshaftissubjectedtoexternalforcecouplesatbothendsonly,wehaveThus,thetorsionalstrainenergyofthecircularshaftcanbewrittenasDiscussion：Iftheinternalforcecouplemomentvariescontinuouslyalongtheaxisofthecircularshaft,i.e.Mn=Mn(x),thestrainenergyforthewholecircularshaftisIftheinternalcouplemomentvariesinstepsalongtheaxis,wehave

Thestrainenergyperunitvolumeofthecircularshaft,i.e.thestrainenergydensityinthepureshearstate,is

(3)PlanebendingPurebendingofstraightcantileverbeamsofequalsection.Astheconcentratedcouplemomentgraduallyincreasesfromzerotoitsfinalvaluem,theangleofrotationatthefreeendofthecantileverbeamalsograduallyincreasesfromzerotoitsfinalvalueθ(Fig.a).(b)ABoql(a)

workdonebyMcanbeexpressedintermsoftheareaofthetriangleOAB,i.e

Discussion:strainenergyofthepurelybendingbeamstrainenergyofthestraightbeamintransverseforcebending

Intherangeoflinearelasticityandunderstaticload,thestrainenergyofabarcanbeexpresseduniformlyas

F：generalizedforceδ：generalizeddisplacementF：force

δ：displacement；F：forcecouplemoment

δ：angular-displacement2.CharacteristicsofelasticstrainenergyThedeformationenergiescannotsimplybesuperimposedingeneral.Note：If

M1andM2denotethebendingmomentscausedbythetwoexternalforce(F1F2)actingalonerespectively,whentheyacttogether,thebendingmomentsofthebeamshouldbeM1+M2.Thestrainenergyofthebeamis(2)theelasticstrainenergyisindependentoftheorderofloadinganddependsentirelyonthefinalvalueoftheloadanddisplacement.(3)whenthecross-sectionalchangesorinternalforcesarerepresentedbydifferentfunctions,thedeformationenergyshouldbecalculatedinseparatesections.(4)therodisalinearelastomersatisfyingHooke'slaw,forthenon-linearelastomer,thedeformationenergywillbecome3、TheClapeyron’stheorem-------UniversalexpressionsfordeformationenergyδidenotesthegeneralizeddisplacementofthegeneralizedforceFiatthepointofactionalongitsdirectionofaction.δi

canbewrittenas

whereδi1representsthegeneralizeddisplacementatthepointofFialongitsactiondirection.ItiscausedbythegeneralizedforceF1.Therestaresimilar.β1⋯βm

areconstantsrelatedtothestructure.1F2F1d2dmFmd…..ThesumoftheworkdonebyeachloadisnumericallyequaltothestrainenergyofthestructureThisconclusioniscalledClapeyron’stheorem.Itcanbedescribedasthesumofthedeformationenergyofalinearelastomerequaltoone-halfoftheproductofeachexternalforceanditscorrespondingdisplacement.4.Strainenergyforcombineddeformation

Usingthegeneralexpressionforstrainenergy,thestrainenergyofabarsubjectedtothecombinedactionofbending,torsionandaxialtensioncanbeobtained.

Nowinterceptamicro-segmentoflengthdxinthebar,iftheaxialforce,bendingmomentandtorqueinthecrosssectionareFN(x),M(x)andMn(x)(forthemicro-sectiondx,FN(x),M(x)andMn(x)shouldberegardedasexternalforces).Therelativeaxialdisplacement,rotationangleandtorsionanglebetweenthetwoendcrosssectionsared(Δl),dθanddφ,respectivelySincethedeformationscausedbyeachofFN(x),M(x)andMn(x)areindependentofeachother,thestrainenergywithinthemicro-segmentdxshouldbeThen,thedeformationenergyoftheentirecombineddeformedbarcanbeobtainedbyintegratingtheaboveequation.Example1:TrytofindthestrainenergyofthesquaretrussstructureandfindtherelativedisplacementsatpointsAandC.ItisknownthateachbarhasthesametensileandcompressiverigidityEA.solution：Axialforces：Deformationenergy：BACDFFlWorkdonebyexternalforceFromU=W,

weobtainThen,wecangetBACDFFlExample1:TrytofindthestrainenergyofthesquaretrussstructureandfindtherelativedisplacementsatpointsAandC.ItisknownthateachbarhasthesametensileandcompressiverigidityEA.Example2:Rightfigureshowsaplanerigidframe.ThebendingrigidityandtensilerigidityoftheframeareknowntobeEIandEA,respectively.trytofindtheverticaldisplacementδAofendA.SolutionSectionAB：SectionBC：Deformationenergy：FBACaDeformationenergy：VerticaldisplacementofsectionA：Ifa=landcrosssectiondiameterisd(l=10d),thenExample2:Rightfigureshowsaplanerigidframe.ThebendingrigidityandtensilerigidityoftheframeareknowntobeEIandEA,respectively.trytofindtheverticaldisplacementδAofendA.FBACathen：Thesecondterminbracketsislessthan0.05%.So,theeffectofaxialforcescangenerallybeneglectedwhensolvingfordeformationsordisplacementsinbendingresistantbarstructures.Example2:Rightfigureshowsaplanerigidframe.ThebendingrigidityandtensilerigidityoftheframeareknowntobeEIandEA,respectively.trytofindtheverticaldisplacementδAofendA.FBACaExample3:Aplanecurvedbarwithasemi-circularaxisisshown.AconcentratedforceperpendiculartotheplaneinwhichtheaxisislocatedisactingatthefreeendA.TrytofindtheverticaldisplacementofsectionA.solution：Itcanbeseenfromfigure(b)thatthetorsionandbendingonthecrosssectionm-nare

AFROjdjFAmmndj(b)Deformationenergy:Deformationenergyofthewholerod:mndj(b)Example3:Aplanecurvedbarwithasemi-circularaxisisshown.AconcentratedforceperpendiculartotheplaneinwhichtheaxisislocatedisactingatthefreeendA.TrytofindtheverticaldisplacementofsectionA.LettheverticaldisplacementofAbe.Duringthedeformation,theworkdonebytheexternalforceisnumericallyequaltothestrainenergyofthecurvedbar,i.e.Therefore:mndj(b)Example3:Aplanecurvedbarwithasemi-circularaxisisshown.AconcentratedforceperpendiculartotheplaneinwhichtheaxisislocatedisactingatthefreeendA.TrytofindtheverticaldisplacementofsectionA.13.2Mohr’stheoremMohr’stheoremisaneffectivetoolfordeterminingdisplacementatanypointinanydirection.TheconceptandpropertiesofstrainenergyarenowusedtoderiveMohrtheorem,usingabeamasanexample.SupposethebeamisbentanddeformedundertheactionofanexternalforceF1,F2......,asshowninFigure(a).WecalculatethedeflectionδatanypointConthebeamundertheactionoftheaboveexternalforce.C1F2FABd(a)thestrainenergycausedbyM(x)canbefound….oneunitforceF0=1isappliedatpointCinthedirectionofdeflectionbendingmoment:thedeformationstoredinthebeam:

AddF1,F2

......backtothebeam.TheunitforceF0completestheworkwiththevalueF0δagain.InthecaseofFigure(c),thestrainenergyofthebeamcanbe(b)BAC0F0F2F1FCBAd(c)….SincethebendingmomentunderthejointactionofF0andF1,F2......isM(x)+M0(x),thestrainenergyofthebeamcanalsobeexpressedastwoequationsareequal,so：

ConsideringF0=1,weget：

ThisisMohrtheoremalsoknownastheMohrintegration.Forsmallcurvaturecurvedbar,theMohr'sintegralformulaforstraightbeamcanbeextendedtoobtaintheMohrintegralforthebendingdeformationofthecurvedbar

Theformulaforcalculatingthedisplacementofthenodeofthetrussstructure:

TheMohrformulaforcalculatingthedisplacementofacombineddeformedstructureis：PointstonotewhenusingMoore'stheorem：④ThecoordinatesystemofM0(x)andM(x)mustbethesame,andthecoordinatesystemofeachsegmentoftherodcanbeestablishedfreely.⑤TheMohrintegralmustcovertheentirestructure.②M0——Byremovingtheactiveforce,atthepointofthegeneralizeddisplacementwherethecalculationneedstobedone,alongtherequesteddirectionofthegeneralizeddisplacement,theinternalforcegeneratedbythestructurewhenthegeneralizedunitforceisadded.①M(x)：Internalforceofthestructureundertheoriginalload.③Theproductofthegeneralizedunitforceaddedandthegeneralizeddisplacementmusthavethesamedimensionaswork.ExampleThecantileverbeamsubjectedtouniformloadisshown.IfEIisaconstant,trytouseMohrtheoremtocalculatethedeflectionanddeflectionangleofsectionAatthefreeend.xlqA(a)x1(b)solution

Bendingmomentequation：Thebendingmomentcausedbyunitforceis：AccordingtoMohrtheorem,thedeflectionofsectionAis：

ExampleThecantileverbeamsubjectedtouniformloadisshown.IfEIisaconstant,trytouseMohr'stheoremtocalculatethedeflectionanddeflectionangleofsectionAatthefreeend.xlqA(a)x1(b)x1(c)Thebendingmomentcausedbytheunitcoupleis:FromtheMohrtheorem,ExampleThecantileverbeamsubjectedtouniformloadisshown.IfEIisaconstant,trytouseMohrtheoremtocalculatethedeflectionanddeflectionangleofsectionAatthefreeend.xlqA(a)x1(b)ExampleAsimpletrussstructureshownissubjectedtoforces.Letthetensile(compressive)rigidityEAofeachbarbethesame.TrytofindtherelativedisplacementbetweenthepointsBandD.31452llF2FDACB31452llDA11CBExampleAsteelframeofcircularsectionissubjectedtoforcesasshowninFigure(a).ThetorsionalrigidityofthewholeframeisGIpandEI,respectively.Iftheeffectofshearondeformationisexcluded,trytofindthedisplacementδCofsectionCalongtheverticaldirection.ABlq(a)ClThepositiveandnegativeinternalforcesineachsegmentcanstillfollowthesignregulationsfortheinternalforcesinthebarundervariousbasicdeformations.SectionBC:

SectionAB:

1x2x2x1x(b)ABC1ABl(a)ClqByusingcorrespondingformula,thenumericaldisplacementofsectionCcanbeobtainedasBC:AC:ExampleThesmallcurvaturebarisshown.TrytofindtheverticaldisplacementandtheangleofrotationofthefreeendA.TheEIisaconstant.FAdsdjjR(a)Solution:

bendingmomentcausedbyload:BendingmomentunderaconcentratedforceatpointA：A1(b)TheverticaldisplacementofpointAisExampleThesmallcurvaturebarisshown.TrytofindtheverticaldisplacementandtheangleofrotationofthefreeendA.TheEIisaconstant.FAdsdjjR(a)A1(b)A(c)AddaunitforcecoupleatpointA，wecanget:ExampleThesmallcurvaturebarisshown.TrytofindtheverticaldisplacementandtheangleofrotationofthefreeendA.TheEIisaconstant.FAdsdjjR(a)A1(b)13.3DiagrammultiplicationmethodforMohrintegrationForthebendingdeformationofstraightbeamswithequalsection,

（a）M0(x)istheinternalforcecausedbytheunitload.Itmustconsistofastraightlineorabrokenline.

LettheM(x)andM0(x)diagramstobethediagramsofmomentscausedbytheloadsandunitforce,respectively.AsectionofthegraphofM0

(x)isobliquestraightline.

Correspondingequation:CxxlSubstituteaboveequationintoequation(a),weget

(b)

M（x）dxxxCCx0CMx0()Mx()0Mxlsecondterm：ωistheareaoftheM(x)graphFirstterm：centroidM（x）()Mx()MxdxxcxCx0CMx0()MxlwhereisverticalcoordinateoftheM0(x)diagramcorrespondingtothecenterCoftheM(x)diagram.ThismethodofreducingtheMohrintegrationoperationtoanalgebraicoperationbetweengraphsisknownasthediagrammultiplicationmethod.（3）Thismethodcanbeusedtofindthedeformationordisplacementofallkindofstraightbarwithequalsection.note：（1）ωandMC0arebothgenerationalquantitieswiththesamepositiveandnegativesignsasM(x)andM0(x).（2）IfM(x)isasegmentedsmoothcurve,orifM0(x)isaline,thegraphicalmultiplicationformulashouldbeusedforthesegments,andthenfindthealgebraicsum.abh3l+a3l+blCh

n+2(n+1)llCn+2lh4

3llC4lh8

5llC8

3lToppointtriangle：

Quadraticparabola：Quadraticparabola：Nthdegreeparabola：AqBCMlaExampleAnexternallyoverhangingbeamshownisloaded.IfEIisaconstant,trytofindthedeflectionatthefreeendC.

ExampleAnexternallyoverhangingbeamshownisloaded.IfEIisaconstant,trytofindthedeflectionatthefreeendC.

28ql11Cw22wCM．C33w．．AqBCMlaTheparabolicpartwithareaω1iscausedbytheuniformload.Thefoldedpartwithareasω2andω3iscausedbytheconcentratedforcecouple.

28ql11Cw22wCM．．C33w．AqBCMlaThediagramM0(x)causedbytheunitforceisgiven.ThevalueofMC0correspondingtothecentroidsofthethreepartsoftheM(x)diagramcanbefoundusingtheproportionalrelationshipbetweenthelinesegments.thedeflectionoftheCsectioncanbefoundasABC101M02M03Ma28ql11Cw22wCM．．C33w．ExampleAsteelframeofconstantEIisshown,withbeamBCsubjectedtoauniformloadq.Iftheeffectofshearandaxialforcesondeformationisnotconsidered,trytofindtheverticaldisplacementofsectionA.BCAq2a2asolutionFirstdrawthebendingmomentdiagramofthesteelframeunderloadasshowninFigureblew.22qa22qaBCAq2a2aTocalculatetheverticaldisplacementofsectionA,aunitforceintheverticaldirectionisappliedonsectionAandthenthecorrespondingM0(x)diagramisdrawnasfollows.CBA12a2a02M01MBCAq2a2a22qa22qa2w2C．1C1w．Accordingtothecorrespondingformulain,theareaofthemomentdiagramofthetwobarsABandBCcanbefoundasBCAq2a2a22qa22qa1C1w．2w2C．

Mc0correspondingtothecentroidsofω1andω2inFigure(d)is2a2a02M01M22qa22qa1C1w．2w2C．fromtheequationtheverticaldisplacementofsectionAcanbefound.2a2a02M01M22qa22qa2w2C．1C1w．13.4Castigliano'stheorem1.Castigliano'stheoremLetthefreeendAofastraightcantileverbeamwithEIbesubjectedtoaconcentratedforceFA.ItisnotdifficulttofindthestrainenergystoredinthecantileverbeamThestrainenergyinthebeamisnumericallyequaltotheworkoftheexternalforceW,i.e.FAlxABThedeflectionofthefreeendofthecantileverbeamis

IfwetakethepartialderivativeofthestrainenergyUofthebeamwithrespecttotheconcentratedforceFatsectionA,wehaveThisisexactlyequaltothefreeenddeflection.Therefore,ThepartialderivativeofthestrainenergywithrespecttoFisequaltothedisplacementofthepointofFalongtheforcedirection,whichisknownasCastigliano'stheorem.TheCartesiantheoremcanbedescribedas:thepartialderivativeofthedeformationenergyoftheelasticbodytoanyloadisequaltothedisplacementoftheloadapplicationpointalongtheloadapplicationdirection.

Thebeamisnowusedtoprovethistheorem.LetasetofstaticloadsF1

、F2···actingonabeam.Thedisplacementsinresponsetotheseloadsareδ1

、δ2···.Duringthedeformationprocess,theworkdonebytheaboveloadisequaltothestrainenergystoredinthebeam.ThestrainenergyUisafunctionoftheloadF1、F2···andcanbeexpressedas(a)1F2FnF1d2dnd(a)…..IfFnisgivenanincrementdFn,thestrainenergyUwillalsohaveanincrement.Theelasticstrainenergyofthebeamcanbewrittenas

(b)1F2F…..Fn+

dFnChangetheloadingorderbyfirstaddingdFntothebeamandthenactingF1、F2···.WhendFnisfirstadded,itcausesadisplacementdδnatitspointalongthesamedirection.Thestrainenergyinthebeamatthistimeshouldbe1/2dFndδn.ndFndd1F2F1d2dnd…..dFn+FnnddBecausethestrainenergycausedbyF1

、F2···isstillU,thestrainenergystoredinthebeamshouldbe

(c)1F2F1d2dnd…..dFn+FnnddSincethestrainenergywithinthelinearelasticbodyisindependentoftheloadingorder,thestrainenergycausedbythetwodifferentloadingordersshouldbeequal,i.e.Neglectingthesecondordermicro-quantity,weget

ThisispreciselytheexpressionofCastigliano'stheoremofequation.TheCastigliano'stheoremonlyappliestolinearelasticstructures.2.SpecialformsofCastigliano'stheorem(1)Truss

Ifthewholetrussconsistsofmbars,thestrainenergyofthewholestructurecanbecalculatedbyequation(13-5),i.e.AccordingtotheCastigliano'stheoremthereis

(2)Straightbeam

Forstraightbeamswhereplanebendingoccurs,thestrainenergycanbecalculatedusingequation(13-14),i.e.ApplyingCastigliano'stheorem,wegetIntheaboveequation,onlythebendingmomentM(x)isrelatedtotheloadFn.TheintegralvariablexandFnarenotrelated.Sowecanfirsttakethepartialderivativeandthenintegrateit.(3)PlanecurvedbarsThestressdistributionofsmallcurvaturebarissimilartothatofastraightbeam.ThebendingstrainenergycanbewrittenasApplyingCastigliano'stheorem,weget

(4)Combineddeformationofbars

Forbarssubjectedtothecombinedactionoftension(compression),bendingandtorsion,thestrainenergycanbewrittenfromequation(13-19),i.e.ApplyingCastigliano'stheorem,weget

Solution

SectionAC：SectionBC:

Example

FindthedeflectionangleofsectionAandthedeflectionatthemid-pointC.FBACFBACExample

FindthedeflectionangleofsectionAandthedeflectionatthemid-pointC.FBACFBACdeflectionatthemid-pointC:Example

FindthedeflectionangleofsectionAandthedeflectionatthemid-pointC.FBACFBAC3.SpecialtreatmentofCastigliano'stheoremIfweusetheCastigliano'stheoremtocalculatethegeneralizeddisplacement,theremustbethegeneralizedexternalforcecorrespondingtotheformanddirectionoftherequestedgeneralizeddisplacement.Themethodofadditionalforces:firstly,appendageneralizedforcecorrespondingtotherequestedgeneralizeddisplacement,andthenCastigliano'stheoremisapplied