过程装备与控制工程专业英语上册

过程装备与控制工程专业英语翻译作业龙飞啸天ReadingMaterial1StaticAnalysisofBeamsAbarthatissubjectedtoforcesactingtransversetoitsaxisisaxisiscalledabeam.Inthissectionwewillconsideronlyafewofthesimplesttypesofbeams,suchasthoseshowninFig.1.2.Ineveryinstanceitisassumedthatthebeamhasaplaneofsymmetrythatisparalleltotheplaneofthefigureitself.Thus,thecrosssectionofthebeamhasaverticalaxisofsymmetry.Also,itisassumedthattheappliedloadsactintheplaneofsymmetry,andhencebendingofthebeamoccursinthatplane.Laterwewillconsideramoregeneralkindofbendinginwhichthebeammayhaveanunsymmetricalcrosssection.ThebeaminFig.1.2(a)，withapinsupportatoneendandarollersupportattheother,iscalledasimplysupportedbeam,orasimplybeam.Theessentialfeatureofasimplebeamisthatbothendsofthebeammayrotatefreelyduringbending,buttheycannottranslateinthelateraldirection.Also,oneendofthebeamcanmovefreelyintheaxialdirection(thatis,horizontally).Thesupportsofasimplebeammaysustainverticalreactionsactingeitherupwardanddownward.ThebeaminFig.1.2(b)whichisbuild-inorfixedatoneendandfreeattheotherend,iscalledacantileverbeam.Atthefixedsupportthebeamcanneitherrotatenortranslate,whileatthefreeenditmaydoboth.Thethirdexampleinthefigureshowsabeamwithanoverhang.ThebeamissimplysupportedatAandBandhasafreeendatC.Loadsonabeammaybeconcentratedforces,suchasP1andP2inFig.1.2(a)and(c).Distributedloads,suchastheloadqinFig.1.2(b).Distributedloadsarecharacterizedbytheirintensity,whichisexpressedinunitsofforceperunitdistancealongtheaxisofthebeam.Forauniformlydistributedload,illustratedinFig.1.2(b),theintensityisconstant;avaryingload,ontheotherhand,isoneinwhichtheintensityvariesasafunctionofdistancealongtheaxisofthebeam.ThebeansshowninFig.1.2arestaticallydeterminatebecausealltheirreactionscanbedeterminedfromequationsofstaticequilibrium.Forinstance,inthecaseofsimplebeamsupportingtheloadP1[Fig.1.2],bothreactionsarevertical,andtheirmagnitudescanbefoundbysummingmomentsabouttheends;thus,wefindRa=P1(L-a)/LRb=P1a/LThereactionsforthebeamwithanoverhang[Fig.1.2(c)]canbefoundinthesamemanner.Forthecantileverbeam[Fig.1.2(b)],theactionoftheappliedloadqisequilibratedbyaverticalforceRaandacoupleMaactingatthefixedsupportasshowninthefigure.Fromasummationofforcesintheverticaldirection,weconcludethatRa=qbAnd,fromasummationofmomentsaboutpointA,wefindMa=qb(a+b/2)ThereactivemomentMaactscounterclockwiseasshowninthefigure.Theprecedingexamplesillustratehowthereactions(forcesandmoments)ofstaticallydeterminatebeamsmanybecalculatedbystatics.Thedeterminationofthereactionsforstaticallyindeterminatebeamsrequiresaconsiderationofthebendingofthebeams,andhencethissubjectwillbepostponed.Theprecedingexamplesillustratehowthereaction(forcesandmoments)ofstaticallydeterminatebeamsmaybecalculatedbystatics.Thedeterminationofthereactionsforstaticallyindeterminatebeamsrequiresaconsiderationofthebendingofthebeams,andhencethissubjectwillbepostponed.TheidealizedsupportconditionsshowninFig.1.2areencounteredonlyoccasionallyinpractice.Asanexample,long-spanbeamsinbridgessometimesareconstructedwithpinandrollersupportsattheends.However,inbeamsofshorterspan,thereisusuallysomerestraintagainsthorizontalmovementofthesupports.Undermostconditionsthisrestrainthaslittleeffectonthebeamandcanbeneglected.However,ifthebeamisveryflexible,andifthehorizontalrestraintsattheendsareveryrigid,itmaybenecessarytoconsidertheireffects.ExampleFindthereactionsatthesupportsforasimplevbeamloadedasshowninFig.1.3(a).Neglecttheweightofthebeam.SolutionTheloadingofthebeamisalreadygivenindiagrammaticform.Thenatureofthesupportsisexaminednextandtheunknowncomponentsofthesereactionsareboldlyindicatedonthediagram.Thebeam,withtheunknownreactioncomponentsandalltheappliedforces,isredrawninFig.1.3(b)todeliberatelyemphasizethisimportantstepinconstructingafree-bodydiagram.AtA,twounknownreactioncomponentsmayexist,sincetheendispinned.ThereactionatBcanonlyactinaverticaldirectionsincetheendisonaroller.Thepointsofapplicationofallforcesarecarefullynoted.Afterafree-bodydiagramofthebeamismade,theequationsofstaticsappliedtoobtainthesolution.ΣFx=0,RAx=0ΣMA=0+,2000+100(10)+160(15)-RB(20)=0,RB=+2700lb↑ΣMB=0+,Σy=0↑+,-10-100-160+270=0NotethatΣFx=0usesuponeofthethreeindependentequationsofstatics,thusonlytwoadditionalreactioncomponentsmaybedeterminedfromstatics.Ifmoreunknownreactioncomponentsormomentsexistatthesupport,theproblemsbecomestaticallyindeterminate.NotethattheconcentratedmomentappliedatCentersonlyintotheexpressionsforthesummationofmoments.ThepositiveofRBindicatesthatthedirectionofRBhasbeencorrectlyassumedinfig.1.3(b).TheinverseisthecaseofRay,andtheverticalreactionatAisdownward.Notethatacheckonthearithmeticalworkisavailableifthecalculationsaremadeasshown.梁的静态分析梁是指一根在轴线上受到横向作用力的杆。在这篇课文中我们仅仅研究的只是一些简单常见的梁，比如图1－2中所示的那些。每个例子中梁都被假设有一个与其本身所在平面平行的对称平面。因此，梁在横断面有一个垂直轴向对称面。与此同时，假定负载只在对称的面中起作用，因此梁的弯曲部分仅在其中的一个平面中发生。以后我们可能会讨论更多具有普遍性的弯曲发生在不对称的横断面弯曲的梁。图1.2（a）示的梁，一端铰链支撑,另一方则以滚动轴承的支持,这就是所谓的简支梁支架顶梁,或梁。主要特点:梁弯曲的面内自由旋转的,但它们不可能水平移动。此外,梁的结束的轴向上可以自由移动。简支梁支架可以受到垂直方向的反作用力,垂直向上或者向下。如图1.2（b）中的梁，一端是嵌在固定端,另一端是自由端的。所以我们都叫悬臂梁。固定端梁也允许转动不允许移动,但这章反映到