(1.5.3)-2.3 Changes in Lengths Under Non材料力學(xué)材料力學(xué)

Chapter2AxiallyLoadedMembers

MechanicsofMaterialsChangesinLengthsUnderNonuniformConditionsWelcometomechanicsofmaterials,intheprecedingsection,wediscussedthechangesinlengthsofaxiallyloadedmembers,andforaprismaticbar,thechangeinlengthscanbedescribedbytheforce-displacementrelation.Inthissection,wewillseehowthissameequationcanbeusedinmoregeneralsituations.BarswithIntermediateAxialLoads①②③RA＋δ:elongation－δ:shorteningAxialforcediagramThechangeinlengthforbarswithintermediateaxialloads:Inmoregeneralsituations,onecaseisthataprismaticbar*isloadedbyoneormoreaxialloadsactingatintermediatepointsalongtheaxis.Todeterminethelengthchangeforthisbar,first,dividethebarintosegmentsAB,BCandCDaccordingtothe*applicationpointsofthoesexternalforcesonthebar,includingpointA,pointB,pointCandpointDwherereactionforceRA,actionforcesPB,PCandPDact,respectively.Forconvenience,identifythesesegmentsas*1,2,and3,respectively.Next,usingthemethodofsections,determinetheinternalaxialforcesinsegments1,2,and3respectively.Makinganarbitratycuttinginsegment1,anddrawthis*freebodydiagram,bysummingforcesintheverticaldirection,*N1isobtained.Inasimilarway,*N2,and*N3canbefound.NoticethattheinternalaxialforcesaredenotedbytheletterNtodistinguishthem

fromtheexternalloadsP.Then,usingthe*force-displacementrelation,determinethe*changesinthelengthsofthesegments1,2and3respectively.Finally,thechangeinlengthoftheentirebarδ*isobtainedbyaddingδ1,δ2,andδ3.Note,thechangesinlengthsmustbeaddedalgebraically,with*elongationsbeingpositiveandshorteningsnegative.Inaddition,wecanalsosketch*theaxialforcediagramforthisbar,whichdisplaysthevariationofinternalaxialforceNoverthelengthofthisbar.Wecanreadtheinternalaxialforces*N1,N2andN3foreachsegmentfromthisdiagram.BarsConsistingofPrismaticSegmentsAxialforcediagramAxialdisplacementdiagramThechangeinlengthforbarsconsistingofprismaticsegments:Nowlet’slookatanothercase,whenabar*consistsofseveralprismaticsegments,eachmayhavedifferentaxialforces,differentdimensions,anddifferentmaterials.Thechangeinlengthisobtained*usingthesamegeneralapproach.Inthisexpressionmthesubscriptiisanumberingindexforthesegmentsandnisthetotalnumberofsegments.NiistheinternalaxialforceinSegmmenti.Forthisbar,thesketchoftheaxialforcediagramis*.Similarly,wecandrawaplot*ofdisplacementsforthisbar,whichisreferredtoasthe*axialdisplacementdiagram.Fromtheaxialdisplacementdiagram,thedisplacementatA,δA,isthechangeinlengthoftheentirebar,whichisequaltothesumofshorteningsofsegmentABandsegmentBC.BarswithContinuouslyVaryingLoadsorDimensionsThechangeinlengthofadifferentialelement:Thechangeinlengthoftheentirebar:Internalaxialforce:N(x)Cross-sectionalarea:A(x)Forbars*withtheaxialforceNandthecross-sectionalareaAvarycontinuouslyalongitsaxis,wecouldnotobtainthechangeinlengthoftheentirebardirectlyfromtheforce-displacementrelation,givenpreviously.Instead,thechangeinlengthoftheentirebarorpartofthebar,couldbedeterminedbyselecta*differentialelementatadistancexfromtheleft-handendofthebar，Theinternalaxialforce*N(x)actingatthiscrosssectionmaybedeterminedbythemethodofsections,fromequilibriumusingeithersegmentACorsegmentCBasafreebody.Ingeneral,thisforceisafunctionofx.Also,knowingthedimensionsofthebar,thecross-sectionalarea*A(x)isexpressedasafunctionofx.Then,theelongationδdofthedifferentialelementmaybeobtained*usingtheforce-displacementrelation,bysubstitutingN(x)forP,dxforL,andA(x)forA.Theelongationoftheentirebarisobtained*byintegratingoverthelength.Inaddition,theintegralcanbeevaluatedanalyticallyornumerically.Limitations①Thechangeinlengthforprismaticbars:②Thechangeinlengthforbarswithintermediateaxialloads,orconsistingofprismaticsegments:③ThechangeinlengthofContinuouslyVaryingLoadsorDimensions:Validonlyifanglebetweenthesidesofthebarissmall.Thoseequationsapplyonlytolinearlyelasticmaterials.Wehaveintroducedseveralequations*fordeterminingthechangsinlengthsfordifferenttypesofaxiallyloadedmembers.Alltheseequationsgivetherelationshipsbetweentheforcesactingonamemberanditschangesinlength,knownastheforce-displacementrelation.Theserelationshavevariousformsdependinguponthepropertiesofthemember.However,*thoseequationsapplyonlytobarsmadeoflinearlyelasticmaterials,asshownbythepresenceofthemodulusofelasticityEintheformulas.And,*thethirdequation,givessatisfactoryresultsforataperedbaronlyifanglebetweenthesidesofthebarissmall.AxialForceDiagram(AFD)-AgraphicaldisplayofthevariationininternalaxialforceN(x)overthelengthofabar.1N12N35kN3CABDq=20kN/m55kN5kNRARAx

RAq=20kN/mN2CABDq=20kN/m55kN5kN2m1m1.5mRAAscanbeseen,whendeterminingthechangesinlengthsofaxiallyloadedmembers,axialforcediagramsareusefulforfindingthecorrespondinginternalaxialforceNwhichappearintheforce-displacementrelaion.Axialforcediagrams*developagraphicaldisplayofthevariationininternalaxialforceN(x)overthelengthofabar.Fromsuchdiagrams,criticalregionsofthebar,suchasthelocationofmaximuminternalaxialforceNmaxcanbeindentified.Toconstructionofaxialforcediagrams,therelationshipsbetweenexternalforcesandinternalaxialforceN(x)mustbedeterminedusingthemethodofsections.Toshowhowtoconstructanaxialforcediagram,let’sseethisexample*.Therearedistributedloadqandconcentratedforcesactingonthisbar.First,determinethereactionforceRA*atthefixedendAusingtheequillibriumequationalongxdirection.Then,basedonthe*freebodydiagramofthebar,dividethebarinsegmentsatthelocationswhereexternalforcesact.Inthisexample,PointA,B,C,andD,willdividethebarintothreesegments,segmentAB,segmentBC,andsegmentCD.Next,applingthemethodofsectionsonceforeachsegment,wecandeterminetheaxialforceforeachsegment.Forexample,forthefirstsegmentAB,makean*imaginarycuttingatanarbitrarylocationwiththedistancexfromthelefthandendofthebar,select*theleft-handpart,drawthefreebodydiagramofit,N1*representstheaxialforceonthisarbitruarycrosssection.Setupthe*equilibriumequation,givestheaxialforceN1*forthesegmentAB,andN1isafunctionofx.Inasimilarway,wecanfindN2*forsegmentBCandN3*forsegmentCD.N50105+(kN)CABDq=20kN/m55kN5kN2m1m1.5mxBasedontheequations*ofaxialforceN1,N2andN3,wecanplottheaxialforcediagram*.Andfromthisdiagram,themaximum*axialforceis50KN,whichoccurinthesegmentBC.AxialDisplacementDiagram(ADD)-Adisplayofthevariationofaxialdisplacementd(x)overthelengthofthebar.①

TheslopeatanypointontheADDisequaltotheordinateontheAFD,N(x)dividedbyEAatthatsamepoint.②ThechangeinaxialdisplacementbetweenanytwopointsalongabarisequaltotheareaundertheAFDdividedbyEAbetweenthosesametwopoints.GuidelinesforconstructingtheADDfromtheAFDSimilartotheaxialforcediagram,anaxialdisplacementdiagramdisplays*thevariationofaxialdisplacementoverthelengthofabar.Usingthis*force-displacementrelation,wecabcreateaxialdisplacementdiagrams.Observingthisforce-displacementrelation,wecanobtaintworulesorguidelines*forconstructingtheaxialdisplacementdiagramfromtheaxialforcediagram.First,*theslopeatanypointontheaxialdisplacementdiagramisproportionaltoN(x),equaltoN(x)/EAatthesamepoint.Second,*thechangeinaxialdisplacementbetweenanytwopointsAandBisequaltotheareaundertheAFDdividedbyEAbetweenthosesametwopoints.SegmentAxialforce

AreaAB2PBC－PCDPUsingthesetworules,let’sconstructtheaxialdisplacementdiagramforthisbar.First,thefreebodydiagram*andtheaxialforcediagram*forthebarareplotted.Thenusetheaxialforcediagramtoconstructtheaxialdisplacementdiagram.Fromtheaxialforcediagram,itcanbeseen,theaixalforceisconstantforeachsegment.Theareaundertheaxialforcedigramforeachsegmenteuqaltotheaxialforcemultipliedbythelengthofeachsegment.Thesevaluesarepresentedinthistable*.Tocalculatethedispalcement,noticethatthedisplacementatthefixedsupportAδA*is0.And,thechangeofdisplacementbetweenpointBandpointAiseuqalto*,soδB*isobtained.Inasimilarway,wecandetermine*thedisplacementsatPointCandPointD.Therefore,theaxialdisplacementdiagram*canbeplotted.Eaxample1:

AverticalsteelbarABCispin-supportedatitsupperendandloadedbyaforceP1atitslowerend.AhorizontalbeamBDEispinnedtotheverticalbaratjointBandsupportedatpointD.

L1=20.0in.;A1=0.25in.2;L2=34.8in.;A2=0.15in.2;

a=28in.andb=25in..ThemodulusofelasticityofthesteelE=29.0×106psi.CalculatetheverticaldisplacementδCatpointCifPl=2100lbandP2=5600lb.(Disregardtheweightsofthebarandthebeam.)Nowlet’slookatthisexample,tofindtheverticaldisplacementatthepointContheverticalbar.Solution:

N1N2First,drawthefreebodydiagram*forbeamBDEandtheverticalbarABC.NoticethatP3areinoppositedirectionsonthebeamandtheverticalbar.Takingmoments*aboutpointDforthefreebodydiagramofthebeamBDE,P3isfound*.Then,fromtheequilibriumofthebar,Rais*.Thisbarisabarwithintermediateaxialloads,tofindthechangeinlengthforit,firstdeterminetheinternalaxialforcesonthetwosegments*ofthebarusingthemethodofsections*.Therefore,usingthesecondequa