弹性力学课件三

2Theoryofplaneproblems2.1PlaneStressandPlaneStrain2.2DifferentialEquationsofEquilibrium2.3StressataPoint.PrincipalStresses2.4GeometricalEquations.Rigid-bodyDisplacements2.5StrainataPoint2.6PhysicalEquations2.7BoundaryConditions2.8Saint-Venant’sPrinciple2.9SolutionofPlaneProbleminTermsofDisplacements2.10SolutionofPlaneProbleminTermsofStresses

2.12Airy’sStressFunction.InverseMethodandSemi-InverseMethod2.11CaseofConstantBodyForces2.1.PLANESTRESSANDPLANESTRAINIfAbodyhasaparticularshape.

Theexternalforcesaredistributedinaparticularmanner.Thespatialproblem

PlanestressproblemPlanestrainproblemAplaneproblem

1)Planestress

a.Theshapeofthebody：Thinplate：thicknesstthelengthsoftheotherdimensionsxyozoytb.Mannerofexternalforces:Surfaceforces：loadontheedges,areparalleltothefacesand

distributeuniformlyoverthethickness.Bodyforces：areparalleltothefacesand

distributeuniformlyoverthethickness.xyozoytSincetheplateisverythinandtheexternalforcesareuniformlydistributedoverthethickness,thestresscomponentsz,

zxand

zy

maybeconsideredaszerowithinthewholeplate:z=0zx=0zy=0

zxz=+t/2=0zyz=+t/2=0zz=+t/2=0Notingtheabsenceofsurfaceforcesonthefacesoftheplate,wehave：Theremainingstresscomponentsx,

yand

xyforthesamereason,maybeconsideredtobefunctionsofx

and

y

only:x=f1(x,y)

y=f2(x,y)

xy=f3(x,y)Suchaproblemiscalledaplanestressproblemandthethinplateistobeinaplanestresscondition.yzxa.Theshapeofthebody:：Verylongcylindricalorprismaticalbodyb.Mannerofexternalforces:（1）beparalleltoacrosssection（2）don’tveryalongtheaxialdirection2)Planestrain

(２)Duetosymmetry(everycrosssectionbeingaplaneofsymmetry),theshearingstresseszxandzymustbezero:

zx=0zy=0

(1)Ifthebodywereofinfinitelength,allcomponentsofstress,strainanddisplacementwouldnotvaryintheaxialdirection.Withanycrosssectionofthebodyasxyplane,thecomponentswillbefunctionsofxandyonly.Forexample:

x=f1(x,y)

y=f2(x,y)

xy=f3(x,y)(3)

Therewillbenormalstress

z

onacrosssectionduetothemutualconstraintbetweenthematerialsontwosidesofthecrosssection.

(4)Alsoduetosymmetry,anypointofthebodywillnotmoveintheaxialdirection,andwehavew=0.Suchproblemshouldhavebeencalledaplanedisplacementproblem.Buttraditionally,ithasbeencalledaplanestrainproblem.xyOIsolateanelementintheformofarectangular.

t=1２.2DIFFERENTIALEQUATIONSOFEQUILIBRIUMyyxxyxcXYConsidertheequilibrium.yyxxyxXYxyoXYcMC=0（1）xy=yxX=0（2）Y=0（3）foraplaneproblemInthesetwoequations,therearethreeunknownfunctions.Hence,theelasticityproblemisstaticallyindeterminate.Tosolvefortheunknownstresses,wehavetoconsiderthestrainsanddisplacements.Inaplanestrainproblem,thenormalstresseszareinequilibriumthemselvesanddonotaffecttheformulationofaboveEqs..Whenthestresscomponentsx，y，xyatacertainpointPareknown,wecancomputethestressactingonanyplanepassingthroughthispoint,paralleltothezaxisbutinclinedtothexandyaxes.2.3STRESSATAPOINT.PRINCIPALSTRESSESyyxxyxXYxyoxyxyyxTakeaplaneABparalleltotheinclinedplaneconsideredandatasmalldistancefromP.t=1PA=dxPB=dyAB=dsyyxxyxxyoBAPsYNXNNNNThecosinesoftheanglesbetweentheoutwardnormaltotheplaneABare：cos(N,x)=lcos(N,y)=mSo:PA=m·dsPB=l·dsConsidertheequilibriumconditions：X=0yyxxyxxyoBAPsYNXNNNN-x·l·ds·1-xy·m·ds·1+XN·ds·1=0Whythebodyforcecomponentscanbeneglected?XN=lx+mxyY=0YN=my+lxyyyxxyxxyoBAPsYNXNNNNLetthenormalstressandshearingstressonABbeNandN,projectionofXNandYNonandperpendiculartothenormalNwillgiveN=lXN+mYNN=lYN-mXNThus,thenormalandshearingstressonanyplanedefinedbythedirectioncosinescaneasilybecalculated,providedthatthestresscomponentsx，yandxyatthepointPareknown.UsingtheexpressionsofXNandYN,weobtainN=l2x+m2y+2lmxyN=lm(y-x)+(l2-m2)xyxyoBAPYNXNNXN=lx+mxy=lYN=my+lxy=mIftheshearingstressonacertainplanethroughPvanishes,thatplaneiscalledaprincipalplaneatPandthenormalstressontheplaneiscalledaprincipalstressatP.SupposeABisaprincipalplane.2-（x+y）+（xy-xy2）=0Sincetheexpressionwithineachsignofsquarerootisalwayspositive,1and2mustberealquantities.ThisshowsthattwoprincipalstressesexistatthepointPAnd:1+2=x+yLet1betheanglebetween1andx.Then,Similarly,wecanobtaini.e.Because1+2=x+y2=x+y-1ThenWecanobtainThatexpressionshowsthatthetwoprincipalstressesareperpendiculartoeachother.xyOyyxxyxxyxyxy1122Whentheprincipalstressesatapointareknown,itisveryeasytodeterminethemaximumandminimumstressesatthepoint.Forsimplicityofanalysis,thexandyaxesareplacedinthedirectionsof1and2,respectively.xyxy1122Thus,wehave1=x2=yxy=0Thenormalstressonanyinclinedplanewillbe：xy1122N=l2x+m2y+2lmxy=l21+m22Therelationoflandmisl2+m2=1So,wecanobtainN=l21+（1-l2）2=l2（1-2）+2Whenl2=1，Nismaximum.Whenl2=0，Nisminimum.Whenl2=1，N=max=1Whenl2=0，N=min=2ThemaximumandminimumvaluesofNare1and2,respectively.TheshearingstressontheinclinedplaneisN=lm(y-x)+(l2-m2)xy=

lm(2-1)=

NWhen,theshearingstressNreachesitsextremevalues.Thisshowstheextremeshearingstressesareactingonplanesinclinedat450withthexzandyzplanes.xy1122NN２.4GEOMETRICALEQUATIONS.RIGID-BODYDISPLACEMENTSA’B’P’Taketwolineelements：PABxyoPA=dxPB=dyuvx：normalstrainofPAy：normalstrainofPBxyoryx：thechangeoftherightanglebetweenPAandPB.Thechangeis+.A’B’P’uvPABxyo

Then,wehavethefollowingthreegeometricalequations：Foragivensetofdisplacementcomponentsuandv,asdefinitefunctionsofxandy,thestraincomponentsaredefinedbytheserelations.whichexpresstherelationsbetweenthestraincomponentsandthedisplacementcomponentsinaplaneproblem.Foragivensetofstraincomponents,asdefinitefunctionsofxandy,however,thedisplacementcomponentsarenotwhollydeterminate.Letx=0，，y=0，xy=0u=f1（y）

v=f2（x）

u=f1（y）=u0-y

v=f2（x）=v0+x

Thesearethedisplacementcomponentscorrespondingtozerostrainsandbetherigid-bodydisplacements.u0andv0aretherigid-bodytranslationsinthexandydirections.istherigid-bodyrotation.Thiscanbeillustratedasfollows.TheresultantdisplacementisWhenonlyisdifferentfromzero,thedisplacementcomponentsofanypointofthebodyare：u=-yv=x

PxxyyrroLettheanglebetweenthedirectionofthisresultantdisplacementandtheyaxisbe,thenwehaveThisshowsthattheresultantdisplacementofPisperpendiculartotheradiallineOPandconsequentlyalongthetangentialdirectionatP.Theconstantistherigid-bodyrotationofthebodyaboutthezaxis.Pxxyyrru=yv=yoNowWeseethatanelasticbodycanhaveanyrigid-bodydisplacementsforzerostrains.Hence,atagivenstateofstrain,thebodymayhavedifferentrigid-bodydisplacementsunderdifferentconditionsofconstraint,reflectedbythedifferentvaluesoftheconstantsu0,v0and.Inordertodeterminetheactualdisplacementofthebody,theremustbethreeproperconditionsofconstraintforthedeterminationofthethreeconstants.2.5STRAINATAPOINTWhenthestraincomponentsx,y,xyatacertainpointPareknown,wecancalculatethenormalstrainofanylineelementPNparalleltoxyplaneandalsothechangeoftheanglebetweentwolineelementsPNandPN’paralleltoxyplane.xoPNN’uvyP1N1N1’111LetthecoordinatesP:(x,y)N:(x+dx,y+dy)thelengthofPN:drthedirectioncosinesofPN:landmSothatprojectionsofPNonthexandyaredx=ldr,dy=mdrxoPNN’uvyP1N1N1’111Displacementcomponents:P:uandvN:Hence,afterdeformation,whenPNisdisplacedtosomeposition,P1N1,theprojectionsofP1N1onthexandyaxesareIfwedenotethenormalstrainofPNbyN,thelengthofPNafterdeformationwillbeP1N1=dr+Ndrandwehavethegeometricalrelation(dr)2=(dx)2+(dy)2SinceNandthederivativetermsareverysmallincomparisonwithunit,theirsquaresandproductsmaybeneglectedandtheaboveequationreducestoSolvingitforN,andreplacingthederivativetermsbystraincomponents,weobtainNowweproceedtofindthechangeoftheanglebetweenPNandPN’.Afterdeformation,PNbecomesP1N1,whichhasdir