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