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.)平衡梁的剪力和彎矩讓我們來共同探討像圖1.5(a)所示懸梁自由端在傾斜拉力P的作用下的問題。如果將平衡梁在截面mn處截斷且將其左邊部分作為隔離體（圖1.5（b）?？梢钥闯龈綦x體截面（右邊）的作用國必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)力就是應(yīng)力的合成力。比如靜止的固定梁合成力可由平衡方程得出，如圖1.5所示懸臂梁結(jié)構(gòu)。這樣就可以得到圖形另一部分中的圖示自由部分的三個平衡方程式。由水平合力和垂直合力的方向，可得：N=Pcos如果將平衡梁在截面mn處截斷且將其左邊部分作為隔離體（圖1.5（b）?？梢钥闯龈綦x體截面（右邊）的作用國必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)力就是應(yīng)力的合成力。比如靜止的固定梁合成力可由平衡方程得出，如圖1.5所示懸臂梁結(jié)構(gòu)。這樣就可以得到圖形另一部分中的圖示自由部分的三個平衡方程式。由水平合力和垂直合力的方向，可得：N=PcosV=Psin如果將平衡梁在截面mn處截斷且將其左邊部分作為隔離體（圖1.5（b）?？梢钥闯龈綦x體截面（右邊）的作用國必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)