英汉双语弹性力学8.ppt

1,SpatialProblem,Chapter8,2,第八章空间问题,3,SpaceProblem,Chapter8SpaceProblem,8-4TheSphericalSymmetricProblemofSpace,8-3TheAxiallySymmetricProblemofSpace,8-2TheBasicEquationundeRectangularCoordinate,8-1Introduction,4,空间问题,第八章空间问题,8-4空间球对称问题,8-3空间轴对称问题,8-2直角坐标下的基本方程,8-1概述,5,Inthischapterwefirstgiveouttheequationsofequilibrium,thegeometricequationsandthephysicalequationsunderrectangularcoordinateforspatialproblems.Fortheanalyticsolutionsofspatialproblemscanonlybeobtainedunderpeculiarboundaryconditions,wediscusstheaxialsymmetricproblemsandtheballsymmetricproblemsofspaceemphatically.,8-1Introduction,SpaceProblem,6,本章首先给出空间问题直角坐标下的平衡方程、几何方程和物理方程。针对空间问题的解析解一般只能在特殊边界条件下才可以得到，我们着重讨论空间轴对称问题和空间球对称问题。,8-1概述,空间问题,7,8-2BasicEquationsunderRectangularCoordinate,One.DifferentialEquationsofEquilibrium,Consideranarbitrarypointinsidethebodyandfetchasmallparallelhexahedron,whichstresscomponentsoneachsideareshownasfigure.,Ifabdenotesthelinewhichjoinsthecentersoftwofacesofthehexahedron,thenfromweget,Cancelingtermsandneglectinghigherordersmallvariables,weget,SpaceProblem,8,8-2直角坐标下的基本方程,空间问题,一平衡微分方程,在物体内任意一点P，取图示微小平行六面体。微小平行六面体各面上的应力分量如图所示。,若以连接六面体前后两面中心的直线为ab，则由得,化简并略去高阶微量，得,9,Similarly,weget,Hereweprovetherelationoftheequalityofcrossshearsagain,from,Listtheequations,cancelterms,weget,Thesearedifferentialequationsofequilibriumunderrectangularcoordinateofspace,Two.GeometricEquations,Forspatialproblems,deformationcomponentsanddisplacementcomponentsshouldsatisfyfollowinggeometricequations,Ofwhichthefirsttwoandthelasthavebeenobtainedamongplaneproblems,theotherthreecanbeledoutwiththesamemethod.,SpaceProblem,10,空间问题,同理可得,这只是又一次证明了剪应力的互等关系。,由,立出方程，经约简后得,这就是空间直角坐标下的平衡微分方程。,二几何方程,在空间问题中，形变分量与位移分量应当满足下列6个几何方程,其中的第一式、第二式和第六式已在平面问题中导出，其余三式可用相同的方法导出。,11,Three.PhysicalEquations,Foranisotropicbody,therelationsbetweendeformationcomponentsandstresscomponentsareasfollows:,Thesearephysicalequationsforspatialproblems.,Ifstresscomponentsaredenotedbystraincomponents,physicalequationscanbewrittenas:,where：,SpaceProblem,12,空间问题,三物理方程,对于各向同性体，形变分量与应力分量之间的关系如下：,这就是空间问题的物理方程。,将应力分量用应变分量表示，物理方程又可表示为：,其中：,13,FourEquationsofCompatibility,Differentiatethesecondandthethirdformulaofgeometricequationsattheleft.Addingthesetwo,weget,Substitutethefourthformulaofgeometricequationsintotheaboveequation,weget,（a）,SpaceProblem,14,空间问题,四相容方程,将几何方程第二式左边对z的二阶导数与第三式左边对y的二阶导数相加，得,将几何方程第四式代入，得,（a）,15,DifferentiatethelatethreeformulasofgeometricequationsseparatelyforX,Y,Z,weget,Fromtheaboveequations,weget,SpaceProblem,16,空间问题,将几何方程中的后三式分别对x、y、z求导，得,并由此而得,17,Theequationsof(a),(b),(c),(d)arecalledcompatibilityconditionsofdeformation,alsoknownasequationsofcompatibility.,Substitutingphysicalequationsintotheaboveequations,andcancelingtermsaccordingtodifferentiateequationsofequilibrium,wegetthecompatibilityequationswhichareexpressedwithstresscomponents:,SpaceProblem,18,空间问题,方程(a)、(b)、(c)、(d)称为变形协调条件，也称相容方程。,将物理方程代入上述相容方程，并利用平衡微分方程简化后，得用应力分量表示的相容方程：,19,WecallthemMichelcompatibilityequations.,SpaceProblem,20,空间问题,称其为密切尔相容方程。,21,Amongspatialproblems,iftheelasticitybodysgeometricshape,restraintconditionandanyexternalfactorsaresymmetricalinacertainaxis(anyplanewhichpassesthisaxisisallsymmetricalone),thenallstresses,deformationsanddisplacementsaresymmetricalinthisaxis.Thiskindofproblemiscalledaxialsymmetryproblemofspace.,Theformsofelastomersofaxialsymmetryproblemaregenerallydividedintotwokinds:cylinderorhalfspacebody.,Accordingtothecharacteristicofaxialsymmetry,weshouldadoptthecylindricalcoordinates.ifwetakezaxisastheaxisofsymmetry,thenallthestresscomponents,straincomponentsanddisplacementcomponentswillbeonlythefunctionofrandz,withthecoordinatehavenothingtodowith.,8-3AxiallySymmetricProblemsforSpace,SpaceProblem,22,空间问题,8-3空间轴对称问题,23,One.DifferentialEquationsofEquilibrium,Considerasmallelementasshowninfigure.Foraxialsymmetry,theelementstwocylindricalplanesexistonlynormalstressesandaxialshearstresses;itstwohorizontalplanesexistonlynormalstressesandradialshearstresses;itstwoperpendicularplanesexistonlyroundnormalstresses,whichareshowninfigure.,Accordingtotheassumptionofcontinuity,stresscomponentsofthesmallelementspositiveplaneshaveasmallincreasecomparedwiththenegativeones.Attention:theincreaseofroundnormalstressesarezeroatthismoment.,Forequilibriumatradialdirectionandaxialdirectionandfrom,cancelingtermsandignoringthehighordersmallvalues,weget,SpaceProblem,24,空间问题,25,Thesearethedifferentialequationsofequilibriumforaxialsymmetryproblemsintermsofcylindricalcoordinates.,Two.GeometricEquations,Similartotheanalysisofplaneproblemintermofpolarcoordinates,weget,thestraincomponentscausedbyradialdisplacementare:,Thestraincomponentscausedbyaxialdisplacementare:,Fromtheprincipleofsuperposing,namelywegetthegeometricequationsforspatialaxialsymmetryproblems:,SpaceProblem,26,空间问题,这就是轴对称问题的柱坐标平衡微分方程。,二几何方程,通过与平面问题及极坐标中同样的分析，可见，由径向位移引起的形变分量为：,由轴向位移引起的形变分量为：,由叠加原理，即得空间轴对称问题的几何方程：,27,Three.PhysicalEquations,Becausethecylindricalcoordinatesareorthogonalcoordinatesastherectangularones,wecangetthephysicalequationsdirectlyfromHookeslaw:,Ifstresscomponentsareexpressedwithstraincomponents,theaboveequationscanbewrittenas:,Where:,SpaceProblem,28,空间问题,三物理方程,由于圆柱坐标，是和直角坐标一样的正交坐标，所以可直接根据虎克定律得物理方程：,应力分量用形变分量表示的物理方程：,其中：,29,Four.SolutionofAxialSymmetryProblems,Substitutethegeometricequationsintothephysicalequationswhichstresscomponentsareexpressedwithstraincomponents,wegettheelasticequations:,Where:,Substitutetheaboveequationsintothedifferentialequationsofequilibrium,andusethenotation:,Weget,Theseareknownasbasicdifferentialequationsforsolvingthespatialaxialsymmetryproblemsintermsofdisplacementcomponents.,Obviously,thedisplacementcomponentsinaboveequationsarefunctionscoordinatesrandz,theycantbesolveddirectly.Soweintroducethefollowingmethod:,SpaceProblem,30,空间问题,四轴对称问题的求解,将几何方程代入应力分量用应变分量表示的物理方程，得弹性方程：,其中：,再将弹性方程代入平衡微分方程，并记：,得到,这就是按位移求解空间轴对称问题所需要的基本微分方程。,显然，上述基本微分方程中的位移分量是坐标r、z的函数，不可能直接求解，为此介绍下列方法：,31,Five.DisplacementTendencyFunction,Forsimplicity,ignoringthebodyforce,thebasicdifferentialequationsintermofdisplacementcomponentscanbesimplifiedas:,Supposingnowthedisplacementhastendency,weusedisplacementtendencyfunctiontodenotethedisplacementcomponents:,Thusweget:,Substitutewiththebasicdifferentialequationswhichignoringthebodyforce,weget:,Namely,SpaceProblem,32,空间问题,五位移势函数,为简单起见，不计体力。位移分量的基本微分方程简化为：,现在假设位移是有势的，把位移分量用位移势函数表示为：,从而有,代入不计体力的基本微分方程，得,即,33,Soforanaxialsymmetryproblem,ifwefindasuitablemediationfunction,fromwhichthedisplacementcomponentsandstresscomponentssatisfytheboundaryconditions,thenwegetthecorrectsolutionoftheproblem.,Inordertosolveaxialsymmetryproblems,Lameintroducesadisplacementfunction,Attention:notallthedisplacementfunctionsofspatialproblemshavetendency.Butiftheyhave,thevolumetricstrain.,SixLameDisplacementFunction,Define,Where,SpaceProblem,34,空间问题,取，则。即,为调和函数，由位移势函数求应力分量的表达式为：,为求解轴对称问题，拉甫引用一个位移函数,六拉甫位移函数,令,其中,35,Substitutetheabovefunctionsintothebasicdifferentialfunctionswhichintheabsenceofbodyforce,weget:,Namelyisarepeatedmediationfunction,wecallitLamedisplacementfunction.Therepresentationsofstresscomponentsfromthisfunctionare:,Soforanaxialsymmetryproblem,ifwefindasuitablerepeatedmediationfunction,fromwhichthedisplacementcomponentsandstresscomponentssatisfytheboundaryconditions,thenwegetthecorrectsolutionoftheproblem.,SpaceProblem,36,空间问题,将上式代入不计体力位移分量的基本微分方程，可见：,即是重调和函数，称为拉甫位移函数。由拉甫位移函数求应力分量的表达式为：,可见，对于一个轴对称问题，只须找到恰当的重调和的拉甫位移函数，使得该位移函数给出的位移分量和应力分量能够满足边界条件，就得到该问题的正确解答。,37,SevenExample:halfspacebodywhichisundertheactionofoutwarddrawnconcentratedforcesintheboundary,Considerahalfspacebody,whichbodyforcesareignored.Itreceivesoutwarddrawnconcentratedforcesintheboundary,asshowninfigure.Pleasesolveitsstressesanddisplacements.,Solution:choosethecoordinatesystemasfig.Throughthedimensionalanalysis,LamesdisplacementfunctionispositiveoneorderpoweroflengthcoordinateofwhichFmultipliesR、z、.Afterpreliminarycalculation,wesetdisplacementfunctionas:,Accordingtotherelationsofdisplacementcomponentsandstresscomponentsanddisplacementfunction:,SpaceProblem,38,空间问题,七举例：半空间体在边界上受法向集中力,设有半空间体，体力不计，在其边界上受有法向集中力，如图所示。试求其应力与位移。,解：取坐标系如图。通过量纲分析，拉甫位移函数应是F乘以R、z、等长度坐标的正一次幂，试算后，设位移函数为,根据位移分量和应力分量与位移函数的关系：,39,Wecanobtainthedisplacementcomponentsandthestresscomponents,SpaceProblem,40,空间问题,可以求得位移分量和应力分量,41,Theboundaryconditionsare,AccordingtotheSaint-VenantsPrinciple,wehave,(c),Theboundarycondition(a)issatisfied.Fromboundarycondition(b),weget,(d),Fromcondition(c),weget,(e),Solvingintermsof(d)and(e),weget,SpaceProblem,42,空间问题,由(d)及(e)二式的联立求解，得,43,SubstitutetheobtainedA1andA2intotheforgoingrepresentations,weget,SpaceProblem,44,空间问题,将得出的A1及A2回代，得,45,Amongspatialproblems,iftheelasticitybodysgeometricshape,restraintconditionandanyexternalfactorsaresymmetricalinacertainpoint(anyplanewhichpassesthispointisallsymmetricalone),thenallstresses,strainsanddisplacementsaresymmetricalinthispoint.Thiskindofproblemiscalledsphericallysymmetryproblemofspace.,Accordingtothecharacteristicofsphericallysymmetry,weshouldadoptthesphericalcoordinates.ifwetakeelasticitybodyssymmetricalpointasthecoordinatesorigin,thenallthestresscomponents,straincomponentsanddisplacementcomponentswillbeonlythefunctionofradialcoordinater,withtheothertwocoordinateshavenothingtodowith.,Obviously,sphericallysymmetricproblemscanonlyexistinholloworsolidroundspheroid.,8-4SphericallySymmetricProblemForSpace,SpaceProblem,46,空间问题,在空间问题中，如果弹性体的几何形状、约束情况以及所受的外来因素，都对称于某一点（通过这一点的任意平面都是对称面），则所有的应力、形变和位移也对称于这一点。这种问题称为空间球对称问题。,根据球对称的特点，应采用球坐标表示。若以弹性体的对称点为坐标原点，则球对称问题的应力分量、形变分量和位移分量都将只是径向坐标r的函数，而与其余两个坐标无关。,显然，球对称问题只可能发生于空心或实心的圆球体中。,8-4空间球对称问题,47,One.DifferentialEquationsofEquilibrium,Forsymmetry,thesmallelementonlyhasradialvolumeforce.Fromradialequilibrium,andconsidering,Neglectingthehigherordersmallvariables,wegetthedifferentialequationsofequilibriumforsphericallysymmetricproblems:,Fetchasmallelement.Fetchasmallhexahedronfromtheelastomer.Itisformedbytwopelletfaces,whichdistanceis,andtwopairsofradialplanes,whichangleisrespectively.Forsphericalsymmetry,eachplaneonlyhasnormalstress.Itsstresssituationsareshowninfig.,SpaceProblem,48,空间问题,49,TwoGeometricEquations,ThephysicalequationsforsphericallysymmetricproblemscandirectlybeledoutfromHookeslaw,Ifstresscomponentsareexpressedwithstraincomponents,weget,SpaceProblem,50,空间问题,二几何方程,由于对称，只可能发生径向位移；又由于对称，只可能发生径向正应变及切向正应变，不可能发生坐标方向的剪应变。球对称问题的几何方程为：,三物理方程,球对称问题的物理方程可直接根据虎克定律得来：,将应力用应变表示为：,51,Four.TheBasicDifferentialEquationinTermsofDisplacement,Substitutethegeometricequationsintothephysicalequations,wegettheelasticequations:,Substitutetheaboveequationsintothedifferentialequationsofequilibrium,weget,Thisisknownasthebasicdifferentialequationsforsolvingthesphericallysymmetricproblemsintermsofdisplacement.,SpaceProblem,52,空间问题,四位移法求解的基本微分方程,将几何方程代入物理方程，得弹性方程,再代入平衡微分方程，得,这就是按位移求解球对称问题时所需要用的基本微分方程。,53,Example:ahollowpelletwhichisunderactionoftheevendistributedpressure,considerahollowpellet.Itsinteriorradiusisa,theexteriorisb,theinnerpressureisqa,outerpressureisqb.Attheabsenceofbodyforce,pleasefinditsstressesanddisplacements.,Itssolutionis,Andthestresscomponentsare:,Solution:forignoringthebodyforce,thedifferentialequationforsphericallysymmetricproblemscanbesimplifiedas,SpaceProblem,Five,54,空间问题,五举例：空心圆球受均布压力,设有空心圆球，内半径为a，外半径为b，内压为qa，外压为qb，体力不计，试求其应力及位移。,其解为,得应力分量,解：由于体力不计，球对称问题的微分方程简化为,55,Substitutetheboundaryconditions,intotheaboveformulas,weget,Andthenwegettheradialdisplacementoftheproblem:,Thestressexpressionsare:,SpaceProblem,56,空间问题,于是得问题的径向位移,应力表达式,57,Exercise8.1supposethereisaequalsectionpolewitharbitraryshape.itsdensityis,withitsupperendhungandlowerendfree,whichisshownasfig.Trytoprovethestresscomponents,besuitableforanycondition.,Solution:thestresscomponentsare:,Thebodyforcecomponentsare:,SpaceProblem,58,空间问题,练习8.1设有任意形状的等截面杆，密度为，上端悬挂,下端自由，如图所示。试证明应力分量,能满足所有一切条件。,z,y,解：已知应力分量为,体力分量为,59,OneTheInspectionofDifferentialEquationsofEquilibrium,Obviouslytheyaresatisfied.,Two.TheInspectionofCompatibility,Becausethebodyforceisaconstant,thecompatibilityequ