外文翻译--平衡梁的剪力和弯矩.doc -- 5 元

ShearForceandBendingMomentinBeamsLetusnowconsider,asanexample,acantileverbeamacteduponbyaninclinedloadPatitsfreeendFig.1.5a.IfwecutthroughthebeamatacrosssectionmnandisolatethelefthandpartofthebeamasfreebodyFig.1.5b,weseethattheactionoftheremovedpartofthebeamthatis,therighthandpartuponthelefthandpartmustastoholdthelefthandinequilibrium.Thedistributionofstressesoverthecrosssectionmnisnotknownatthisstageinourstudy,butweedoknowthattheresultantofthesestressesmustbesuchastoequilibratetheloadP.ItisconvenienttoresolvetotheresultantintoanaxialforceNactingnormaltothecrosssectionandpassingthroughthecentriodofthecrosssection,ashearforceVactingparalleltothecrosssection,andabendingmomentMactingintheplaneofthebeam.Theaxialforce,shearforce,andbendingmomentactingatacrosssectionofabeamareknownasstressresultants.Forastaticallydeterminatebeam,thestressresultantscanbedeterminedfromequationsofequilibrium.Thus,forthecantileverbeampicturedinFig.1.5,wemaywriterthreeequationsofstacticsforthefreebodydiagramshowninthesecondpartofthefigure.Fromsummationsofforcesinthehorizontalandverticaldirectionswefind,respectively,NPcosβVPsinβand,fromasummationofmomentsaboutanaxisthroughthecentroidofcrosssectionmn,weobtainMPxsinβwherexisthedistancefromthefreeendtosectionmn.Thus,throughtheuseofafreebodydiagramandequationsofstaticequilibrium,weareabletocalculatethestressresultantswithoutdifficulty.ThestressinthebeamduetotheaxialforceNactingalonehavebeendiscussedinthetextofUnit.2NowwewillseehowtoobtainthestressesassociatedwithbendingmomentMandtheshearforceV.ThestressresultantsN,VandMwillbeassumedtobepositivewhenthetheyactinthedirectionsshowninFig.1.5b.Thissignconventionisonlyuseful,however,whenwearediscussingtheequilibriumofthelefthandpartofthebeamisconsidered,wewillfindthatthestressresultantshavethesamemagnitudesbutoppositedirectionsseeFig.1.5c.Therefore,wemustrecognizethatthealgebraicsignofastressresultantdoesnotdependuponitsdirectioninspace,suchastotheleftortotheright,butratheritdependsuponitsdirectionwithrespecttothematerialagainst,whichitacts.Toillustratethisfact,thesignconventionsforN,VandMarerepeatedinFig.1.6,wherethestressresultantsareshownactingonanelementofthebeam.Weseethatapositiveaxialforceisdirectedawayfromthesurfaceuponwhichisactstension,apositiveshearforceactsclockwiseaboutthesurfaceuponwhichitacts,andapositivebendingmomentisonethatcompressestheupperpartofthebeam.ExampleAsimplebeamABcarriestwoloads,aconcentratedforcePandacoupleMo,actingasshowninFig.1.7a.Findtheshearforceandbendingmomentinthebeamatcrosssectionslocatedasfollowsaasmalldistancetotheleftofthemiddleofthebeamandbasmalldistancetotherightofthemiddleofthebeam.SolutionThefirststepintheanalysisofthisbeamistofindthereactionsRAandRB.TakingmomentsaboutendsAandBgivestwoequationsofequilibrium,fromwhichwefindRA3P/4–Mo/LRBP/4mo/LNext,thebeamiscutatacrosssectionjusttotheleftofthemiddle,andafreebodydiagramisdrawnofeitherhalfofthebeam.Inthisexamplewechoosethelefthandhalfofthebean,andthecorrespondingdiagramisshowninFig.1.7b.TheforcepandthereactionRAappearinthisdiagram,asalsodotheunknownshearforceVandbendingmomentM,bothofwhichareshownintheirpositivedirections.ThecoupleModoesnotappearinthefigurebecausethebeamiscuttotheleftofthepointwhereMoisapplied.AsummationofforcesintheverticaldirectiongivesVR–PP/4M0/LWhichshownthattheshearforceisnegativehence,itactsintheoppositedirectiontothatassumedinFig.1.7b.TakingmomentsaboutanaxisthroughthecrosssectionwherethebeamiscutFig.1.7bgivesMRAL/2PL/4PL/8Mo/2Dependingupontherelativemagnitudesofthetermsinthisequation,weseethatthebendingmomentMmaybeeitherpositiveornegative.Toobtainthestressresultantsatacrosssectionjusttotherightofthemiddle,wecutthebeamatthatsectionandagaindrawanappropriatefreebodydiagramFig.1.7c.TheonlydifferencebetweenthisdiagramandtheformeroneisthatthecoupleMonowactsonthepartofthebeamtotheleftofthecutsection.Againsummingforceintheverticaldirection,andalsotakingmomentsaboutanaxisthroughthecutsection,weobtainVP/4Mo/LMPL/8Mo/2WeseefromtheseresultsthattheshearforcedoesnotchangewhenthesectionisshiftedfromlefttorightofthecoupleMo,butthebendingmomentincreasesalgebraicallybyanamountequaltoMo.SelectedfromStephenP.TimoshekoandJamesM.Gere,Mechanicsofmaterials,VanNostrandreinholdCompanyLtd.,1978.平衡梁的剪力和弯矩让我们来共同探讨像图1.5a所示悬梁自由端在倾斜拉力P的作用下的问题。如果将平衡梁在截面mn处截断且将其左边部分作为隔离体（图1.5（b）。可以看出隔离体截面（右边）的作用国必须和左边的作用力平衡，截面mn处应力的分布情况我们现阶段是不知道的，但我们知道这些应力的合力必须和拉力P平衡。按常规可将合力分解成为通过质点作用于横截面的轴向应力N、平行于截面的剪切力V和作用在平衡梁平面中的弯矩M。作用在截面上的轴向应力、剪切力和弯曲应力就是应力的合成力。比如静止的固定梁合成力可由平衡方程得出，如图1.5所示悬臂梁结构。这样就可以得到图形另一部分中的图示自由部分的三个平衡方程式。由水平合力和垂直合力的方向，可得NPcosβ如果将平衡梁在截面mn处截断且将其左边部分作为隔离体（图1.5（b）。可以看出隔离体截面（右边）的作用国必须和左边的作用力平衡，截面mn处应力的分布情况我们现阶段是不知道的，但我们知道这些应力的合力必须和拉力P平衡。按常规可将合力分解成为通过质点作用于横截面的轴向应力N、平行于截面的剪切力V和作用在平衡梁平面中的弯矩M。作用在截面上的轴向应力、剪切力和弯曲应力就是应力的合成力。比如静止的固定梁合成力可由平衡方程得出，如图1.5所示悬臂梁结构。这样就可以得到图形另一部分中的图示自由部分的三个平衡方程式。由水平合力和垂直合力的方向，可得NPcosβVPsinβ如果将平衡梁在截面mn处截断且将其左边部分作为隔离体（图1.5（b）。可以看出隔离体截面（右边）的作用国必须和左边的作用力平衡，截面mn处应力的分布情况我们现阶段是不知道的，但我们知道这些应力的合力必须和拉力P平衡。按常规可将合力分解成为通过质点作用于横截面的轴向应力N、平行于截面的剪切力V和作用在平衡梁平面中的弯矩M。作用在截面上的轴向应力、剪切力和弯曲应力就是应力的合成力。比如静止的固定梁合成力可由平衡方程得出，如图1.5所示悬臂梁结构。这样就可以得到图形另一部分中的图示自由部分的三个平衡方程式。由水平合力和垂直合力的方向，可得NPcosβVPsinβ由通过截面mn质心的轴向总弯矩，可得MPxsinβ其中力是自由端到截面mn的距离。因此，通过隔离体图解和静态平衡方程，可简单地计算出各合成力。属于单独作用的轴向应力N的应力已经在第二单元讨论过了，在这里我们将讨论怎样解出与这些应力有关的弯矩M和剪切力V。假设如图1.5b所示合成力N、V和弯矩M的作用方向为正，当我们在讨论梁