(1.7.2)--ch4-2 Types of Beams, loads and材料力学材料力学.ppt
Chapter 4 Shear forces andd bending moments Mechanics of MaterialsTypes of Beams,loads and reactionsWelcometoMechanicsofMaterias,lastvideogaveanintrodutiontothemaintopicsinthischapter,thisvideodiscussestypesofbeams,loadsandreactions.Simplification of beamsAsaforementioned,beamsareslenderstructuralcomponentswhichrestonsuppportsandcarryexternalloads.Ingeneral,supportsandexternalforcesofbeamsareverycomplex.*Weshoulddosomenecessary*simplificationforthemtohelpwithbeamanalysis.*Therefore,apropersketchforbeamsneedstobedonebeforeanyfurthercalculation.Simplification of beams:representedbyitsaxis.Simplification of loads:Loads(includingthereaction)actingonthebeammaybereducedintothreetypes：Simplification of supportsSimplification of beamsMPqPin supportRoller supportFixed supportAAAAAAAXAYConcentratedforceConcentratedcoupleDistributedforce.Ingenral,*beamsarerepresentedbyitsaxis,*asshowninthefigure.*Externalloads,includingreactionforces,actingonthebeammaybereducedinto*threetypes:*concentratedforce,*concentratedcoupleand*distributedforce.Distributedloadsaremeasuredbytheirintensity,whichisexpressedinunitsofforceperunitdistance(forexample,newtonspermeter).Auniformlydistributedload,oruniformload,hasaconstantintensityqperunitdistance.*Avaryingloadhasanintensitythatchangeswithdistancealongtheaxis;forinstance,thelinearlyvaryingloadhasanintensitythatvarieslinearlyfromq1toq2.*Supportsofbeamscanbeclassifiedintothreemaintypes,*pinsupport,*rollersupportand*fixedsupport.Forthepinsurpport,theessentialfeatureofitisthatitpreventstranslationattheendofabeambutdoesnotpreventrotation.Thus,endAofthebeam,asshowninthefigure,cannotmovehorizontallyorvertically,buttheaxisofthebeamcanrotateintheplaneofthefigure.*Consequently,apinsupportiscapableofdevelopingaforcereactionwithbothhorizontalandverticalcomponents(AXandAY),butitcannotdevelopamomentreaction.While,arollersupportprovidesoneverticalforcereaction,andthefixedsupportmayproducebothforceandmomentreactions,asshowninthefirgure.Types of beams Simple beam Cantilever beam Overhaning beamSimply supportedOverhangSketchPinRollerFixed endBeamsareusuallydescribedbythemannerinwhichtheyaresupported.*Therearethreemaintypesofbeams,simplebeam,cantileverbeamandoverhangingbeam.*Asimplysupportedbeamorsimplebeamisabeamwith*apinsupportatoneendand*arollersupportattheother.*Acantileverbeamisfixedatoneendandfreeattheother.*Thethirdexampleisabeamwithanoverhang.*Thisbeamissimplysupportedattwopoints,thatis,ithasapinsupportandarollersupport,*butitprojectsbeyondoneofortwosupports.Whendrawingsketchesofbeams,thesupportsareidentifiedbyconventionalsymbols,suchasthoseshownintheFigures.Thesesymbolsindicatethemannerinwhichthebeamisrestrained,andthereforetheyalsoindicatethenatureofthereactiveforcesandmoments.However,thesymbolsdonotrepresenttheactualphysicalconstruction.Symbols of supports Move freely in horizontal direction Constraint agaist rotation is small Symbol of roller supportForinstance,hereisawide-flangebeamsupportedonaconcretewallandhelddownbyanchorboltsthatpassthroughslottedholesinthelowerflangeofthebeam.Thisconnectionrestrainsthebeamagainstverticalmovementbutdoesnotpreventhorizontalmovement.Also,anyrestraintagainstrotationofthebeamissmallandordinarilymaybedisregarded.Consequently,thistypeofsupportisusuallyrepresentedbyaroller.ExternalreactionsInternalreactionsshearforceVbendingmomentMBeamanalysisstressdeformationReactionsFindingthe reactionsisusuallythefirststepintheanalysisofabeam.StaticanalysisFindingthereactionsisusuallythefirststepintheanalysisofabeam.*Ifabeamissupportedinastaticallydeterminatemanner,allreactionscanbefoundfromfree-bodydiagramsandequationsofequilibrium.Thisisstaticanalysis.*Oncethereactionsareknown,*theshearforcesandbendingmomentscanbefound,andbeamanalysiscouldbedonesubsequently,asdescribedlaterinthiscourse.Internal ReleasesMoment release M=0,N,V0Torque release T=0,N,V,M0Axial release N=0,V,M0Shear release V=0,N,M0Insomeinstances,itmaybenecessarytoaddinternalreleasesintothebeamorframemodeltobetterrepresentactualconditionsofconstructionthatmayhaveanimportanteffectonoverallstructurebehavior.Thefirstisanaxialreleaseillustratedbytheinsertionofpipe2intopipe1.ForcePappliedtopipe1cannotbetransmittedtopipe2becausetheinternalaxialforceN,intheabsenceoffriction,mustbezeroatthisconnection.HowevershearforceVandbendingmomentMcanbecarriedbythisconnection.Thesymbolforanaxialreleaseisshownatthebottom.Similarly,therearethreeothertypesofinternalreleases,torquerelease,shearreleaseandmomentrelease.Andifreleasesarepresentinthestructuremodel,thestructuremustbebrokenintoseparatefree-bodydiagramsbycuttingthroughtherelease;anadditionalequationofequilibriumisthenavailableforuseinsolvingfortheunknownsupportreactionsincludedinthatfree-bodydiagram.Letslookthefollowingexampletoseehowtocalculatereactionsforbeamswitharelease.Example 1:FindthesupportreactionsforthetwobeamsshowninFig.aandb.Beam(a)isacantileverbeamsubjectedtoaninclinedconcentratedloadP4andalinearlyvaryingdistributedload.Beam(b)hasthesameloadingbuthasapinconnection/momentrelease(enlargedinFigb)totherightofsupportAandarollersupportatjointB.ThisexampleasksyoutofindthesupportreactionsforthetwobeamsasshowninFigureaandb.Beam(a)isacantileverbeamsubjectedtoaninclinedconcentratedloadP4andalinearlyvaryingdistributedload.Beam(b)hasthesameloadingbuthasapinconnection(ormomentrelease)totherightofsupportAandarollersupportatjointB.ThemomentreleaseisenlargedinFig.4-14bforemphasis.Useafour-step problem-solving approachforthesebeams.1.Conceptualize:Overallfree-bodydiagramsofeachbeamareshowninthefollowingfigures.Solution:*Useafour-stepproblem-solvingapproachforthesebeams.*Stepone,Conceptualize.*Firstlydrawtheoverallfree-bodydiagramsofeachbeam.*Bothbeam(a)and*beam(b)havethreereactionsatA(twoforcecomponentsandamoment)and,*beam(b)hasanadditionalforcereactionatrollersupportB.2.Categorize:Beam(b)has4unknowreactionforces,needstobecutintoseparatefree-bodydiagramstofindoneadditionalequation ofequilibrium:Beam(a)isstaticallydeterminate;M=0*step2Categorize:*Beam(a)isreadilyidentifiedasstaticallydeterminatebecausetherearethreereactionforceunknownsandthreestaticequilibriumequations.*Beam(b)hasfourreactionforceunknowns,needstobecutintoseparate*left-and*right-handfree-bodydiagramstofindoneadditionalequationofequilibrium:*m=0becuseatthemomentreleasethemomentisassumedtobeequaltozero.Thisadditionalequilibriumequationisneededtosolveforallfourreactioncomponents.3.Analyze:External reactions-Beam(a):2/3bSaticssignconventionusedtowritetheseequilibriumequations.5/13P4FdStep3,analyze.letsfindexternalreactionsforbeamafirstly.Basedonthefreebodydiagram,wewritethreeequilibriumequationsonebyone.firstlysummingalltheforcesin*xdirection,asshowninthefreebodydiagram,therearetwoforcesinxdirection,oneisAX,theotheroneis*xcomponentofforceP4,equalsto5/13P4.So,thisequationcanbewrittenas*.*AXcanbeobtaineddirectlyfromthisequation.Note,astaticssignconventionisusedtowritetheseequilibriumequations.Thensummingalltheforcesin*ydirection,therearethreeverticalforces,AY,YcomponentofP4,anddistributedload.Beforewritingtheequation,itisnecessarytodividethedistributedloadtrapezoidintotwotriangles,asshownbythe*dashedlineintheFigure.Eachloadtrianglecanbereplacedbyitsresultantwithmagnitudeequaltotheareaofthetriangle,*halfq1band*halfq2brespectively,and*lineofactionforeachresultant,isthroughthecentroidofthetriangle.Therefore,theequationinydirectionis*,sametoreactionAX,*ForceAycanbefounddirectlybythisequation.Finally,*summingmomentsatpointA,theequationis*,whichleadstothe*solutionofMAdirectly.Nowletslookatthetermsinthemomentequation.ThefirsttermisMA,withacouter-clockwisedirection,sopositivesignaccodingtothestaticsignconvention.ThesecondtermisthemomentproducedbyforceP4,equaltoitsverticalcomponent12/13P4multipliedbythedistancea,frompointAtotheverticalcomponent,directionofthismomentisclockwise,sonegative.*Thelasttwotermsrepresentthemomentproducedbythedistributedload,equaltothesumofthemomentscausedbythetwoloadtriangle.Thethirdtermrepresentsthemomentproducedbyoneoftheforcetriangles,equalstoitsresultanthalfq1bmultipliedbythedistancefrompointAtoitsactionline,(L-2/3b).Thelasttermisthemomentcausedbytheotherforcetriangle,equaltotheresultant*halfq2bmultipliedbythe*distancetopointA,(L-b/3).Becausethedirectionofthesetwomomentsisclockwise,theyarebothnegativeaccordingtothestaticsignconvention.3.Analyze:External reactions-Beam(b):Overall free body diagram Right-hand free body diagramNowletsfindexternalreactionsfor*beamb.*First,sumalltheforcesinxdirectionfortheoverallfreebodydiagramofbeamb,solvingtheequation,reactionAXcanbeobtained,equalto5/13P4.ItcanbefoundthatreactionAxisthesameforbeams(a)and(b),Itcanalsobeseenthat*theequationinxdirectionisthesameforbothbeams.*Thencutthroughthemomentreleaseandusetheright-handfree-bodydiagramasshown.*SummomentsatthecuttofindreactionBy.Theequationis*.Letlookatthetermsinthisequation,thefirsttermrepresentsthemomentproducedbyBy,equaltoforceBymultipliedbyitsdistance,fromthecuttoBy,(L-a),directionofitiscounter-clockwise,sopositiveaccoringtothestaticsignconvention.Thesecondtermrepresentsthemomentproducedbyoneoftheforcetriangles,equalstoitsresultanthalfq1bmultipliedbythedistancefromthecuttoitsactionline,(L-a-2/3b),directionofitisclockwise,sonegative.Thelasttermisthemomentcausedbytheotherforcetriangle,equaltotheresultanthalfq2bmultipliedbythedistance(L-a-b/3),clockwiseandnegative.SolvetheequationandBycanbegetthereby.Next,*sumforcesintheydirectionfortheentirefree-bodydiagramtofindreactionforceAy.Thisequationis*Solvetheequationand*ReactionAycanbefound.Finally,*summomentsaboutAfortheentirefree-bodydiagramtogetreactionmoment*MA.4.Finalize:Overall FBD for Beam(a)Beam(b)Overall FBD for Beam(b)Additionalequationatthemomentrelease:Tosummarizethisexample,firstly,forbeam(b),itwasnecessarytocutitintotwopartsthroughthemomentreleasetofind*anadditionalequilibriumequationthatresultedinasolutionforreactionBy.Secondly,symbolicexamplesratherthannumericalexampleswereusedinthisExampleinordertoshowhowtheindividualstepsarecarriedout.Ineachcase,free-bodydiagramswereessentialtoanefficientsolutionforallreactions.EndThatistheofthisvideo,nextvideoisgoingtodiscussshearforcesandbendingmomentsinbeams.
