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StaticsStaticsofdeformablebodyChapter10
PlaneBending10.1Introduction 10.2Internalforceinbeams 10.3Shearforceandbendingmomentdiagrams 10.4Relationsamongshearforce,bendingmomentandloads 10.5Normalstressoncrosssectioninpurebending10.6Shearstressinbending 10.7Calculationofstrength 10.8Displacementandrigidityconditionofbeams10.9Deflectionofbeams
Contents10.1Introduction1.PlanebendingofbeamsBendingisabasicformofdeformationinengineeringpractice.examples:
FFDiagramofthecranecrossbeam(b)AxleSchematicdiagramoftheaxle:1F2FFFFFDiagramofthecranecrossbeam(b)AxleThecommonfeatureofthesemembersisthattheycanallbesimplifiedtoabar,whichintheplanepassingthroughtheaxisissubjectedtoanexternalforce(transverseforce)perpendiculartotheaxisofthebaroranexternalcouple.Thisformofdeformationiscalledbending.Inengineering,themembersofbendingdeformationareusuallycalledbeams.FFDiagramofthecranecrossbeam(b)AxleLongitudinalplaneofsymmetry——theplanepassingthroughtheaxisofthebeamandtheaxisofsymmetryofthesectionPlanebending
——theexternalloadonthebeamislocatedinthelongitudinalplaneofsymmetry,theaxisofthebeambendintoaplanecurvelocatedinthesameplane.ABFRAFRB1F2FAxisofsymmetryLongitudinalplaneofsymmetryAxisofthebeamafterdeformation2.SupportsandloadsTheexternalforcesactingonabeam,includingloadsandsupportreactions,canbesimplifiedtothefollowingthreetypes:(1)Concentratedforces;unit:Newton(N)(2)Distributedload;unit:Newton/meter(N/m)q—Distributedload—ConcentratedforceF(3)
Concentratedcouple;unit:Newton·meter(N·m)M—ConcentratedcoupleThewayinwhichthebeamsaresupportedcanbesimplifiedtothefollowingthreeforms:Fixedhingesupport:2constraints,1degreeoffreedom.Supports:Rollersupport:
1constraints,2degreeoffreedom.Fixedendsupport:3
constraints,0degreeoffreedom.Staticallydeterminatebeam
——thebearingreactionforcesofabeamcanbefullyderivedfromthestaticequilibriumequations.Superstaticbeam——thebearingreactionforcesofabeamcannotbe(orfully)derivedfromthestaticequilibriumequations.Thecommonstaticbeamsinengineeringpracticehavethefollowingthreeforms:(1)Simplysupportedbeam:
(2)Cantileverbeam:
(3)Overhangingbeam:Theportionofthebeambetweentwosupportsiscalledthespananditslengthiscalledthespanlengthorspanofthebeam.Commonly,staticbeamsaremostlysinglespan.10.2Internalforceinbeams1.SectionmethodforcalculatinginternalforcesinbeamsWhenallexternalforcesonthebeamareknown,thesectionmethodcanbeusedtodeterminetheinternalforcesinanycrosssectionofthebeam.ThesimplysupportedbeamABshowninFig.(a)isanalyzedforinternalforcesatcrosssectionm-matadistancexfromendA.yamAB2F1F3FxnnfindthesupportreactionforceRA,RB(2)findtheinternalforce–sectionmethod2F3FMFQMFQ1FmyaxOyaxmnnAB2F1F3FWherethepoint
O
isthecentroidofthesectionm-m(a)(b)AccordingtoWeget:similarlyyaxmnnAB2F1F3Fya1FmQMx02F3FMsimilarly:
andMofsectionm-mcanbeobtainedfromtherightbeamaccordingtoitsequilibriumconditions,whichareequalinmagnitudebutoppositeindirectiontothoseobtainedfromtheleftbeam.FQFQThesignsforshearforceandbendingmomentsarespecifiedasfollows:①shearforce:IftheshearforceFQtendstorotateclockwisearoundthemicro-section,itispositive.Intheoppositedirection,itisnegative.FQ(+)FQ(–)FQ(–)FQ(+)M(+)M(+)M(–)M(–)②
bendingmoment:IfthebendingmomentMcausesadownwardconvexdeformationofthemicro-section,thebendingmomentMispositive.Intheoppositedirection,itisnegative.MyaF1MeFQMxFRA0TheshearforceFQ
andthebendingmomentMinthiscrosssectionmustbeofequalmagnitudeandthesamesign.FQ(+)FQ(+)M(+)M(+)FRBFQF2F3(1)Theshearforceinacrosssectionisnumericallyequaltothealgebraicsumofalllateralforcesinthebeamtotheleft(orright)ofthesection.Anupwardforceontheleft-handbeam(oradownwardforceontheright-handbeam)producesapositiveshearforce;theoppositeproducesanegativeshearforce.leftupper,rightlowerFQ(+)FQ(+)yaF1MeFQMxFRA0MFRBFQF2F3(2)Thebendingmomentinasectionisnumericallyequaltothealgebraicsumofthemomentsofallexternalforcestotheleft(orright)ofthesectionwithrespecttothecentroidofthesection.1.Theclockwiseexternalforcecoupleattheleftendofthecross-sectionortheclockwiseexternalmomentgeneratedbytheexternalforceproducesapositivebendingmomentonthecross-section:leftclockwiseyaF1MeFQMxFRA0MFRBFQF2F32.Thecounterclockwiseexternalforcecoupleattherightendofthesectionorthecounterclockwiseexternalmomentgeneratedbytheexternalforceproducesapositivebendingmomentonthesection:rightcounterclockwiseyaF1MeFQMxFRA0MFRBFQF2F3
Example1Findtheinternalforcesinsections1-1and2-2showninfollowingpicture.ABq11222M=qacDCFDFaaaqasolution:(1)supportreactionforces(2)externalforcesSection1-1:Takingtheleftsegmentalbeamastheobjectofstudy,Section2-2:Takingtherightsegmentalbeamastheobjectofstudy,ABq11222M=qacDCFDFaaaqa10.3Shearforceandbendingmomentdiagrams1.ShearforceandbendingmomentequationsTheshearforceandbendingmomentineachcrosssectionofthebeamcanbeexpressedasafunctionofx,2.Shearforceandbendingmoment
diagramsThegraphsofFQ(x)andM(x)areplottedwithxasthehorizontalcoordinateandFQ
orMastheverticalcoordinate,respectively,whicharecalledshearforceandbendingmomentdiagrams.Inthisway,themaximumvaluesofshearforceandbendingmomentsinthebeamcanbedetermined,aswellasthelocationofthecorrespondingcrosssection.Example2ThecantileverbeamshowninFig.(a)issubjectedtoaconcentratedforceFatitsfreeend.Trytolisttheshearequationandbendingmomentequationofthebeamandmakeashearforcediagramandbendingmomentdiagram.AyBl(a)(b)(c)xxFQMFlx(2)Makeshearforceandbendingmomentdiagrams.xFsolution:(1)Findtheequationofshearandtheequationofbendingmoment.+-(a)(b)FAyBl(a)(b)(c)xxFQMFlxxF+-Eq.(a)showsthattheshearforceineachcrosssectionofthebeamisthesameF.Theresultingshearforcediagramisastraightlinelocatedontheuppersideofthexaxisandparalleltothexaxis
.Fromequation(b)weknowthatthebendingmomentisalinearfunctionofx,andthereforethebendingmomentdiagramisaninclinedstraightline.ThebendingmomentdiagramasshowninFig.(c)canbedrawnbydeterminingthetwopoint,suchasM=-Flatx=0andM=0atx=l.(a)(b)FAsseeninFig.(b)(c),theshearforceisthesameinallcrosssectionsofthebeam,andthebendingmomentismaximumatthefixedendsection.AyBl(a)(b)(c)xxFQMFlxxF+-FExample3AsimplysupportedbeamshowninFig.(a),issubjectedtoauniformloadq.Findtheshearforceandbendingmomentequations,andthenmakeashearforcediagramandbendingmomentdiagram.Solution:(1)Findthesupportreactionforce.(2)Findtheequationsofshearforceandbendingmoment.qyAxB(a)xlBFAF(a)(b)qyAFAxxBFlB(a)+FQ(b)(c)2qlx2qlMx4l34l2232ql2232ql(3)Makeshearforceandbendingmomentdiagrams.theshearforcediagramisaninclinedstraightlineandcanbedrawnbytwopointsonit.themomentdiagramisaquadraticparabola.+-themomentdiagramisaquadraticparabola,qyAFAxxBFlB(a)+FQ(b)(c)2qlx2qlMx4l34l2232ql2232ql+-Fromtheshearforceandbendingmomentdiagramswecanget:qyAFAxxBFlB(a)+FQ(b)(c)2qlx2qlMx4l34l2232ql2232ql+-Example4AsimplysupportedbeamshowninFig.(a),issubjectedtoaconcentratedforceF.Findtheshearforceandbendingmomentequations,andthenmakeashearforcediagramandbendingmomentdiagram.solution:(1)Findthesupportreactionforce.derivedfromtheequilibriumconditions∑MB=0and∑MA=0,weget
yxABlPBFC(↑)(↑)AFab(2)Findtheshearforceequationandbendingmomentequation.
Takeanarbitrarysectionatx1fromtheorigininthesectionAC
Takeanarbitrarysectionatx2fromtheorigininthesectionCByxABlPBFCAFab+Mx+-x(3)
Makeshearforceandbendingmomentdiagrams.yxABlPBFCAFab2x1xFQThebendingmomentatthesectionCwheretheconcentratedforceFactsismaximumAtthepointwheretheconcentratedforceisapplied,theshearforceundergoesasuddenchange,thevalueofwhichisequaltothevalueoftheconcentratedforce,+Mx+-xyxABlPBFCAFab2x1xFQThedistributedloadsetq(x)isacontinuousfunctionandisspecifiedtobepositiveupwards.Oyxxdx(a)dxcM(x)FQ(x)M(x)+dM(x)FQ(x)+dFQ(x)Accordingtotheconditionsofstaticequilibrium,dxcM(x)FQ(x)M(x)+dM(x)FQ(x)+dFQ(x)
wecanget
bendingmomentshearforceintensityofdistributedload(descendingpower)(descendingpower)Itisnotingthattheloadintensityispositiveintheupwarddirectionandthexaxisispositiveintherightwarddirection.TheshearforceFQ(x)
andthebendingmomentM(x)arebothcontinuousfunctionsofxintheinterval.Thereisneitheraconcentratedforcenoraconcentratedcoupleintheinterval.(1)Attheconcentrationofforce:theshearstresschangesabruptly.Thebendingmomentdiagramformsaturningpoint.(2)Attheconcentrationoftheforcecouple:thebendingmomentdiagramchangesabruptly.Fa/4F/2Fa/2a/2aqqa2/
8qa/
2M/
2Ma/2a/2M/
a(3)FQ(x)=0,i.e.dM/dx=0,theslopeofthemomentdiagramiszero.Themomentinthatsectionisanextremevalue.(4)q(x)=0,i.e.FQ(x)=c,
theshearforceonthesectionisconstant.Andthebendingmomentisalinearfunctionofx.Fa/4F/2Fa/2a/2aqqa2/
8qa/
2M/
2Ma/2a/2M/
a(5)q(x)=c.i.e.FQ(x)=cx+b.Theshearforceonthissectionisalinearfunctionofx,thebendingmomentisaquadraticfunctionofx.Theshearforcediagramisaslopingstraightlineandthebendingmomentdiagramisaquadraticparabola.Fa/4F/2Fa/2a/2aqqa2/
8qa/
2m/
2Ma/2a/2m/
aFa/4F/2Fa/2a/2aqqa2/
8qa/
2M/
2Ma/2a/2M/
aInfact,theconcentratedforcesaredistributedoveramicro-segmentofthebeam,andtheshearforceandbendingmomentdiagramsshouldvarycontinuouslyoverthismicro-segment.Forthesakeofsimplicity,thedistributedforcesonthemicro-segment∆lareconsideredasconcentratedforces(i.e.
∆l→0).+-pl(a)+(b)(c)Asimilarexplanationcanbegivenfortheabruptchangesinthemomentdiagramatthecouple.Example6Usingtherelationshipsbetweenbendingmoment,shearforceanddistributedloadintensity,drawtheshearforceandbendingmomentdiagramsoftheoverhangingbeamshowninFig.(a)andcheck.Aqacq2qaB3x1x2xCFDFDaa4aFQ(a)solution:(1)Findthesupportreactionforce.from
,wehave
(↑)
from,wehave(↑)Aqacq2qaB3x1x2xCFDFDaa4aFQ(a)(2)Drawshearforcediagram.SectionACFQAright=-qaSectionDBFQBleft=0SectionCDFQDleft=-2qaFQCright=2qa3a2qa2qaqaAqacq2qaB3x1x2xCFDFDaa4axFQ(a)(b)(3)DrawbendingmomentdiagramSectionAC2qa22qaqax3a2qa2qaqaAqacq2qaB3x1x2xCFDFDaa4aFQ(a)(b)xM(c)SectionCDSectionDB10.5Normalstressinpurebending1.PurebendingFQABaaCDFFxTheshearforceandbendingmomentdiagrams+-FQFFx+xMFaAC、DB:bothshearforceandbendingmomentarenotzero-----------thetransverseloadbendingCD:theshearforceisequaltozeroandthebendingmomentisconstant-----------purebendingpurebendingtransverseloadbending2.Normalstressinpurebending(1)InferenceideasGeometricrelationPhysicalrelationStaticrelationRelationshipbetweennormalstrainandneutralsurfacecurvatureRelationshipbetweennormalstressandneutralsurfacecurvatureDeterminingtheneutralaxisNeutralsurfacecurvatureexpressionsandnormalstressexpressions(planesectionhypothesis
)(Hook'sLaw)(Relationshipbetweenaxialforce,bendingmomentandnormalstressinsection)
1)Purebendingexperiments(1)Thetransverselineremainsstraight,butisturnedbyasmallangle(2)Thelongitudinallinebecomescurved,butremainsperpendiculartothetransverseline(3)Thelongitudinallinelocatedattheconcaveedgeshortensandthelongitudinallineattheconvexedgeelongates(4)Inthebeamwidthdirection,theupperpartelongatesandthelowerpartshortensmnabxDabnmnmbbaamnMMqD(a)(b)Assumptions
(1)Planesectionhypothesis:eachcrosssectionofthebeamremainsasaplaneperpendiculartotheaxisofthedeformedbeamafterdeformation.Allcrosssectionsonlyturnedbyananglearoundoneoftheaxesonthecrosssection.
(2)Uniaxialforceassumption:allthefibersparalleltotheaxisofthebeamdonotsqueezeeachother.Allthelongitudinalfibersareaxiallystretchedorcompressed.Deformationmodelofabeamneutralsurface:Alayeroflongitudinalfibersinthebeamwhichisneitherelongatednorshortened.neutralaxis:TheintersectionlinebetweentheneutralsurfaceandthecrosssectionWhenthebeamisdeformed,thecrosssectionisrotatedarounditsneutralaxis.neutralaxisneutralsurfaceTransverseaxisofsymmetryLongitudinalaxisofsymmetry2)Deformationgeometryequationdx12bby121O2O(a)(b)dqr2¢b¢2¢b¢1O2Odx1¢1¢yThelinestrainoflongitudinalfiberb-bis:(a)Note:Thelongitudinalnormalstrainεatanypointonthecrosssectionisproportionaltothedistanceyfromthatpointtotheneutralaxis.3)Physicalequations
Hooke'slaw
(b)Thenormalstressinbendingislinearlydistributedalongtheheightofthecrosssection.Mzmaxs+(1)Thenormalstressateachpointatthesamedistancefromtheneutralaxisisequal(2)Thenormalstressontheneutralaxisisequaltozeroneutralaxis4)StaticrelationsThecombinedforceofnormalstressisequaltotheaxialforceyzNote:theneutralaxismustpassthroughthecentroidofthecrosssection.(c)MzsdAyzoyneutralaxisneutralaxisisthemomentofinertiaofthecrosssectionareawithrespecttotheneutralzaxis.Thecombinedforceofthenormalstressonthez-axis(neutralaxis)isequaltothebendingmomentinthecrosssectionyz(c)MzsdAyzoyneutralaxisneutralaxisImportantformula:yzmaxs-(c)Mzmaxs+EIziscalledtheflexuralrigidityofthebeam,whichindicatestheabilityofthebeamtoresistbendingdeformationThebasicformulafordeterminingthecurvature1/ρoftheneutralsurfaceofthebeam.neutralaxisneutralaxis5)Sectionmodulusofbendingthemaximumnormalstressoccursatthefarthestdistancefromtheneutralaxis,i.e.mergethetwogeometricquantitiesIzandymaxofthecrosssection,weobtainthen
whereWziscalledthesectionmodulusofbendingandisageometricquantityofthebeamstrengthwiththedimensionof[length]3.Forarectangularsection(Fig.a)Foracircularsection(Fig.b)2bydyzyczydc··(a)(b)2b6)NormalstressinthecrosssectionduringshearbendingCrosssectionnolongerremainsflatwhenshearbending.Experimentsandelastictheoryshowthatthecontributionofshearforcetothenormalstressisnegligibleforlongandslenderbeams(span/height>5).ExampleAsimplysupportedbeamisshowninpicture.Wehaveknownl=3m,q=40kN/m.Find:(1)Normalstressatpointsaandbonthedangeroussectionwhenthebeamisplacedvertically.
(2)Maximumnormalstressinthedangeroussectionwhenthebeamisplacedhorizontally(b)(d)·ya·b120z18050yzABlq(a)solution:(1)Makebendingmomentdiagram.Thedangeroussectionisthespansection.ABlq(a)(c)+28ql(2)CalculationofmomentofinertiaIz.·ya·b120z18050Pointa:compressivestress;pointb:tensilestress.(3)Findthenormalstressatpointsa,b.·ya·b120z18050z(b)maxs-maxs+sWhenthebeamisplacedhorizontally,they-axisistheneutralaxisThemaximumnormalstressislocatedatyz(d)maxs-maxs+·ya·b120z18050z10.6Shearstressinbending1.ShearstressesinrectangularbeamsInderivingtheshearstressformula,thefollowingassumptionsaremadeaboutitsdistributionlaw:(1)ThedirectionofshearstressatanypointonthecrosssectionisparalleltothedirectionofshearforceFQ.(2)Theshearstressisuniformlydistributedalongthesectionwidth.Themagnitudeoftheshearstressisonlyrelatedtothecoordinatey.Theshearstressisequalateachpointofequaldistancetotheneutralaxis.dxxm1m1nn(a)ozbxm1mn1ndxyxytFQ(b)nMM+dMm1mn1ndxxt¢rpxzybm1mn1ntpydxFN1FN21yqdAs1At¢r(d)(c)dxxm1m1nn(a)ozbxm1mn1ndxyxytFQ(b)ncombinedforceoftheinternalforcesystem(rightsection):(a)(b)MM+dMm1mn1ndxxt¢rpxzybm1mn1ntpydxFN1FN21ydAs1At¢r(d)(c)isstaticmomentofthepartialcross-sectionalareaA1totheneutralaxis.Thisvaluevarieswiththepositionofthelongitudinalsectionpr.MM+dMAccordingtothetheoremofcomplementaryshearingstressesandtheaboveassumptions,itisknownthatτ'isnumericallyequaltoτy,andisalsouniformlydistributedalongthecross-sectionalwidth.Similarly,thecombinedforceoftheinternalforcesystemontheleftsiderncanbefoundas(c)Duetoand
arenotequal,theremustbeashearforceonthetopsurfacerpofthehexahedron.shearstresscomposedofτ'is(d)tomaintaintheequilibriumofthexdirectionofthelowermicro-segment,i.e.(e)Substituting(a),(b)and(c)intoequation(e),wegetisnumericallyequaltoτ(y).TheshearstressinthecrosssectionatyfromtheneutralaxisisTheaboveequationistheformulafortheshearstressinarectangularbeam.Sz*isthestaticmomentoftheareaA1abouttheneutralaxis.Aftersimplification,weget10.7Calculationofstrength1.StrengthconditionofbendingstressinpurebendingThenormalstressstrengthconditionofthebeam:
beamofequalsection:
Forthematerialswithequaltensileandcompressivestrength(e.g.carbonsteel),themaximumabsolutenormalstressinthebeamshouldnotexceedtheallowablestress.Formaterialswithunequaltensileandcompressivestrengths(e.g.castiron),thestrengthshouldbecalculatedseparately,i.e.whereσmax+andσmax-arethemaximumtensilestressandmaximumcompressivestress,respectively.[σ+]and[σ-]arethecorrespondingallowablestresses.
Thestrengthconditionsofthebendingstresscangenerallybeusedforstrengthverification,sectiondesignanddeterminationoftheallowableload.Thegeneralprocessofstrengthcalculationofnormalstressis:(1)Findthesupportreactionforceofthebeamanddrawthebendingmomentdiagram.(2)Findoutthedangeroussectionanddeterminethedangerouspointaccordingtothebendingmomentdiagram.(3)Accordingtothenormalstressstrengthcondition,performthecorrespondingstrengthcalculation.2.Strengthconditionofshearstressinbendingtheshearstressstrengthcondition:forstraightbeamsofequalsection:Forslendersolidsectionbeams,theanalysiscanbedoneaccordingtothenormalstressstrengthconditionwithoutcheckingtheshearstress.Forbeamswithshortspanlengthsandlargeloadsnearthebearingsorrivetedandweldedbeamswiththinwebs,theshearstressmustbecheckedExample
Thereisacastironbeamwithslottedsectionandneutralz-axis.Itisknownthatq=10N/mm,F=20KNTrytocheckthenormalstressstrengthofthebeam.FABCDqAFBF2m2m2m(a)yz2y1ysolution
FindthesupportreactionforceThemaximumbendingmomentisshowninthefigureAFBFqFABCD2m2m2m+-10kNm20kNmInsectionB,themaximumtensilestressisateachpointontheupperedgeAFBFqFABCD2m2m2m+-10kNm20kNmyz2y1yAFBFqFABCD2m+-10kNm20kNmyz2y1yThemaximumtensilestressisateachpointontheloweredge2m2mAFBFqFABCD+-10kNm20kNmyz2y1yMaximumtensilestressinsectionC2m2m2mInsummary,thestrengthofthebeamissufficientdiscussionIfthecrosssectionofthebeaminthisquestionisinverted,isitreasonableintermsofthestrengthofthebeam?yz10.8Displacementandrigidityconditionofbeams1.DisplacementofbeamTakethesimplysupportedbeamasanexample.Theaxisofthebeambeforedeformationissetasxaxisandtheleftendissetastheoriginofthecoordinateyaxis.yxqqxFvrOAfterthebeamisdeformed,itsaxiswillbebentintoacurveinthex-yplane,whichiscalledthedeflectioncurveofthebeam.Twobasicquantitiesusedtomeasurethedisplacementofthebeam:1)deflectionvofpoint:Thelineardisplacementofthepointonthebeamaxisinthedirectionperpendiculartothexaxis.Deflectionispositivewhendirectedupwardsandnegativewhendirecteddownwards;2)angulardisplacementθofsection:theangulardisplacementofthecrosssectionarounditsneutralaxis.Counterclockwise----positive,clockwise----negative.yxqqxFvrOThedeflectionofeachpointonthebeamvarieswithx.Thisvariationruleisexpressedinthedeflectioncurveequationasυ=f(x)
Angleθisalsoequaltotheanglebetweenthetangentlineoftheflexurecurveatthatsectionandtheaxis.θisaverysmallangle.WegetyxqqxFvrO2.rigidityconditionofbeamsInordertoensurethenormaloperationofthebeam,wenotonlyrequirethebeamtohavesufficientstrength,butalsoneedtocontrolthedeformationofthebeam:
wherevmaxandθmaxarethemaximumdeflectionandmaximumangleofrotationofthebeam.The[v]and[θ]aretheallowabledeflectionandallowableangleofrotation.Theirspecificvaluesaredeterminedbytheworkingconditionsandcanbefoundinthecode.
10.9DeflectionofbeamsCurvatureequationforbeaminpurebendingSincethebendingmomentMisfunctionofx,thecurvatureatdifferentlocationsalongthebarlengthisdifferent.Sowehave(b)Fromadvancedmathematics
(c)ApproximatedifferentialequationsofdeflectioncurveSubstitutingequation(c)intoequation(b),weget
(d)Thisisthedifferentialequationforthedeflectioncurveofthebeam.Undersmalldeformationv'≪1,theaboveequationcanbeapproximatedas(e)Thisequationhasapositiveornegativesign.Butafterspecifyingthesignofthebendingmomentandselectingthex-ycoordinatesystem,thesignisdetermined.RegulationsWhenthebendingmomentispositive,thebeamaxisisbentintoadownwardconvexcurve;whenthebendingmomentisnegative,thebeamaxisisbentintoanupwardconvexcurve.Itisalsoknownthatwhenthepositivedirectionofvisspecifiedupward,thesecond-orderderivativev''ispositiveforevertpointonadownwardconvexcurve;whilethesecond-orderderivativev''isnegativeforeverypointonanupwardconvexcurve.Sincev''andM(x)
havethesamesign,theequation(e)shouldtakeapositivesign.
Forthevariablesectionbeam,theIshouldbeunderstoodasI(x).Forequal-sectionbeams,Iisaconstant.equation(a)isalsooftenwrittenas
Theaboveequationisusuallyreferredtoastheapproximatedifferentialequationofthedeflectioncurveofbeams,fromwhichtheequationofθandtheequationoffcanbederived.2.Integrationmethod(e)Thenmultiplybothendsofequationbydxandintegratetoobtaintheequationofthedeflectioncurve(f)WhereCandDareintegrationconstants.Integratingoverequation,wegettheangleequationForaspecificbeam,thedeflectionorangleofrotationofsomesectionsissometimesknown.
Forexample,atthefixedend,boththedeflectionandtheangleofrotationareequaltozero(Fig.a).Thedeflectioni
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