外文翻译--平衡梁的剪力和弯矩.doc

ShearForceandBendingMomentinBeamsLetusnowconsider,asanexample,acantileverbeamacteduponbyaninclinedloadPatitsfreeendFig.1.5(a).Ifwecutthroughthebeamatacrosssectionmnandisolatetheleft-handpartofthebeamasfreebodyFig.1.5(b),weseethattheactionoftheremovedpartofthebeam(thatis,theright-handpart)upontheleft-handpartmustastoholdtheleft-handinequilibrium.Thedistributionofstressesoverthecrosssectionmnisnotknownatthisstageinourstudy,butweedoknowthattheresultantofthesestressesmustbesuchastoequilibratetheloadP.ItisconvenienttoresolvetotheresultantintoanaxialforceNactingnormaltothecrosssectionandpassingthroughthecentriodofthecrosssection,ashearforceVactingparalleltothecrosssection,andabendingmomentMactingintheplaneofthebeam.Theaxialforce,shearforce,andbendingmomentactingatacrosssectionofabeamareknownasstressresultants.Forastaticallydeterminatebeam,thestressresultantscanbedeterminedfromequationsofequilibrium.Thus,forthecantileverbeampicturedinFig.1.5,wemaywriterthreeequationsofstacticsforthefree-bodydiagramshowninthesecondpartofthefigure.Fromsummationsofforcesinthehorizontalandverticaldirectionswefind,respectively,N=PcosV=Psinand,fromasummationofmomentsaboutanaxisthroughthecentroidofcrosssectionmn,weobtainM=Pxsinwherexisthedistancefromthefreeendtosectionmn.Thus,throughtheuseofafree-bodydiagramandequationsofstaticequilibrium,weareabletocalculatethestressresultantswithoutdifficulty.ThestressinthebeamduetotheaxialforceNactingalonehavebeendiscussedinthetextofUnit.2；NowwewillseehowtoobtainthestressesassociatedwithbendingmomentMandtheshearforceV.ThestressresultantsN,VandMwillbeassumedtobepositivewhenthetheyactinthedirectionsshowninFig.1.5(b).Thissignconventionisonlyuseful,however,whenwearediscussingtheequilibriumoftheleft-handpartofthebeamisconsidered,wewillfindthatthestressresultantshavethesamemagnitudesbutoppositedirectionsseeFig.1.5(c).Therefore,wemustrecognizethatthealgebraicsignofastressresultantdoesnotdependuponitsdirectioninspace,suchastotheleftortotheright,butratheritdependsuponitsdirectionwithrespecttothematerialagainst,whichitacts.Toillustratethisfact,thesignconventionsforN,VandMarerepeatedinFig.1.6,wherethestressresultantsareshownactingonanelementofthebeam.Weseethatapositiveaxialforceisdirectedawayfromthesurfaceuponwhichisacts(tension),apositiveshearforceactsclockwiseaboutthesurfaceuponwhichitacts,andapositivebendingmomentisonethatcompressestheupperpartofthebeam.ExampleAsimplebeamABcarriestwoloads,aconcentratedforcePandacoupleMo,actingasshowninFig.1.7(a).Findtheshearforceandbendingmomentinthebeamatcrosssectionslocatedasfollows:(a)asmalldistancetotheleftofthemiddleofthebeamand(b)asmalldistancetotherightofthemiddleofthebeam.SolutionThefirststepintheanalysisofthisbeamistofindthereactionsRAandRB.TakingmomentsaboutendsAandBgivestwoequationsofequilibrium,fromwhichwefindRA=3P/4Mo/LRB=P/4+mo/LNext,thebeamiscutatacrosssectionjusttotheleftofthemiddle,andafree-bodydiagramisdrawnofeitherhalfofthebeam.Inthisexamplewechoosetheleft-handhalfofthebean,andthecorrespondingdiagramisshowninFig.1.7(b).TheforcepandthereactionRAappearinthisdiagram,asalsodotheunknownshearforceVandbendingmomentM,bothofwhichareshownintheirpositivedirections.ThecoupleModoesnotappearinthefigurebecausethebeamiscuttotheleftofthepointwhereMoisapplied.AsummationofforcesintheverticaldirectiongivesV=RP=-P/4-M0/LWhichshownthattheshearforceisnegative；hence,itactsintheoppositedirectiontothatassumedinFig.1.7(b).TakingmomentsaboutanaxisthroughthecrosssectionwherethebeamiscutFig.1.7(b)givesM=RAL/2-PL/4=PL/8-Mo/2Dependingupontherelativemagnitudesofthetermsinthisequation,weseethatthebendingmomentMmaybeeitherpositiveornegative.Toobtainthestressresultantsatacrosssectionjusttotherightofthemiddle,wecutthebeamatthatsectionandagaindrawanappropriatefree-bodydiagramFig.1.7(c).TheonlydifferencebetweenthisdiagramandtheformeroneisthatthecoupleMonowactsonthepartofthebeamtotheleftofthecutsection.Againsummingforceintheverticaldirection,andalsotakingmomentsaboutanaxisthroughthecutsection,weobtainV=-P/4-Mo/LM=PL/8+Mo/2WeseefromtheseresultsthattheshearforcedoesnotchangewhenthesectionisshiftedfromlefttorightofthecoupleMo,butthebendingmomentincreasesalgebraicallybyanamountequaltoMo.(Selectedfrom:StephenP.TimoshekoandJamesM.Gere,Mechanicsofmaterials,VanNostrandreinholdCompanyLtd.,1978.)平衡梁的剪力和弯矩让我们来共同探讨像图1.5(a)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：N=Pcos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：N=PcosV=Psin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：N=PcosV=Psin由通过截面mn质心的轴向总弯矩，可得M=Pxsin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1.5(b)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