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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
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