英汉双语材料力学11

CHAPTER11

ENERGYMETHODS

材料力学第十一章能量方法CHAPTER11ENERGYMETHOD

§11–1GENERALEXPRESSIONSOFTHESTRAINENERGY§11–2MOHR’STHEOREM(METHODOFUNITFORCE)§11–3CATIGLIANO’STHEOREM第十一章能量方法

§11–1

变形能的普遍表达式§11–2莫尔定理(单位力法)§11–3卡氏定理§11–1

GENERALEXPRESSIONSOFTHESTRAINENERGY1、Principleofenergy：2、Calculationofthestrainenergy

ofrods：1).Calculationofthestrainenergyofrodsintensionorcompression：

Strainenergystoredintheelasticbodyisequaltotheworkdonebyexternalforces，thatis:

Methodtoanalyzeandcalculatedisplacements、deformationsandinternalforcesofdeformablebodiesbythiskindofrelationiscalledenergymethod.ENERGYMETHODorDensityofthestrainenergy:§11–1

变形能的普遍表达式一、能量原理：二、杆件变形能的计算：1.轴向拉压杆的变形能计算：能量方法

弹性体内部所贮存的变形能，在数值上等于外力所作的功，即

利用这种功能关系分析计算可变形固体的位移、变形和内力的方法称为能量方法。2.Calculationofthestrainenergyofrodsintorsion：3.Calculationofstrainenergyofrodsinbending：ENERGYMETHODorDensityofthestrainenergy:orDensityofthestrainenergy:2.扭转杆的变形能计算：3.弯曲杆的变形能计算：能量方法3、Generalexpressionsofthestrainenergy：

Strainenergyisindependentoftheorderofloading.Deformationsduetomutuallyindependentloadmaybesummedupeachother.Forslendercolumns，thestrainenergyduetoshearingforcesmaybeneglected.ENERGYMETHODDeflectionfactorofshear三、变形能的普遍表达式：

变形能与加载次序无关；相互独立的力（矢）引起的变形能可以相互叠加。细长杆，剪力引起的变形能可忽略不计。能量方法Solution：Inenergymethod（workdonebyexternalforcesisequaltothestrainenergy）①DetermineinternalforcesAENERGYMETHODBendingmoment:Torque:Example1

Asemicirclerodasshowninthefigureislieinhorizontalplane.AverticalforcePactatitspointA.

DeterminethedisplacementofpointAinverticaldirection.PROQMNMTAAPNBjTOMN[例1]图示半圆形等截面曲杆位于水平面内，在A点受铅垂力P的作用，求A点的垂直位移。解：用能量法（外力功等于应变能）①求内力能量方法APROQMTAAPNBjTO③Workdonebyexternalforcesisequaltothestrainenergy②Strainenergy：ENERGYMETHODLetthen③外力功等于应变能②变形能：能量方法Example2DeterminethedeflectionofpointCbytheenergymethod,wherethebeamisofequalsectionandstraight.

Solution：WorkdonebyexternalforcesisequaltothestrainenergyByusingsymmetryweget：Thinking：Forthedistributedload,canwedeterminethedisplacementofpointCbythismethod？qCaaAPBfENERGYMETHODLet[例2]用能量法求C点的挠度。梁为等截面直梁。解：外力功等于应变能应用对称性，得：思考：分布荷载时，可否用此法求C点位移？能量方法qCaaAPBf

§11–2

MOHR’STHEOREM(METHODOFUNITFORCE)

DeterminethedisplacementfAofanarbitrarypointA.1、Provementofthetheorem：aAFigfAq(x)Figc

A0P=1q(x)fAFigb

A=1P0ENERGYMETHOD§11–2

莫尔定理(单位力法)求任意点A的位移fA。一、定理的证明：能量方法aA图fAq(x)图c

A0P=1q(x)fA图b

A=1P0

Mohr’stheorem(methodofunitforce)2、GeneralformofMohr’stheoremENERGYMETHOD

莫尔定理(单位力法)二、普遍形式的莫尔定理能量方法3、WhatwemustpayattentiontoasweapplyMohr’stheorem：④CoordinateofM0(x)mustbecoincidewiththatofM(x).Foreachsegmentthecoordinatemaybesetupfreely.⑤Mohr’sintegrationmustbethroughthewholestructure.②M0:Theinternalforceofthestructureasweactageneralizedunitforcealongthedirection,ofthegeneralizeddisplacementthatistobedetermined,wheretheappliedforceistakenout.①M(x)：Theinternalforceofthestructureactedbyoriginalloads.③Theproductoftheappliedgeneralizedunitforceandthegeneralizeddisplacementtobedetermineddeterminedmustbeofthedimensionofwork.

ENERGYMETHOD三、使用莫尔定理的注意事项：④M0(x)与M(x)的坐标系必须一致，每段杆的坐标系可自由建立。⑤莫尔积分必须遍及整个结构。②M0——去掉主动力，在所求广义位移

点，沿所求

广义位移

的方向加广义单位力

时，结构产生的内力。①M(x)：结构在原载荷下的内力。③所加广义单位力与所求广义位移之积，必须为功的量纲。能量方法Example3

DeterminethedisplacementandtheangleofrotationofpointCbytheenergymethod.Solution：①Plotthediagramofthestructureactedbytheunitload

②DeterminetheinternalforceBAaaCqBAaaC0P=1xENERGYMETHOD[例3]用能量法求C点的挠度和转角。梁为等截面直梁。解：①画单位载荷图②求内力能量方法BAaaCqBAaaC0P=1xSymmetry③DeformationBAaaC0P=1BAaaCqxENERGYMETHOD()③变形能量方法BAaaC0P=1BAaaCqx()④Determinetheangleofrotation.Setupthecoordinateagain(asshowninthefigure）

qBAaaCx2x1BAaaCMC0=1

d)()(

)()()(00)(00òò+=aBCaABxEIxMxMdxEIxMxMENERGYMETHOD=0④求转角，重建坐标系（如图）

能量方法qBAaaCx2x1BAaaCMC0=1

d)()(

)()()(00)(00òò+=aBCaABxEIxMxMdxEIxMxM=0Solution：①Plotthediagramofthestructureactedbyaunitload

②Determinetheinternalforce510

20A300P=60NBx500Cx1510

20A300Bx500C=1P0ENERGYMETHODExample4Afoldingrodisshowninthefigure.AbearingisatpositionAandtherodmayrotatefreelyinthebearingbutcannotmoveupanddown.Knowing：E=210Gpa，G=0.4E，DeterminetheverticaldisplacementofpointB.[例4]拐杆如图，A处为一轴承，允许杆在轴承内自由转动，但不能上下移动，已知：E=210Gpa，G=0.4E，求B点的垂直位移。解：①画单位载荷图②求内力能量方法510

20A300P=60NBx500Cx1510

20A300Bx500C=1P0③DeterminethedeformationENERGYMETHOD()③变形能量方法()§11–3CATIGLIANO’STHEOREMGive

Pn

anincrementdPn

，then：1)FirstapplyforcesP1、

P2、•••、Pn

onthebody，then：2).FirstapplytheforcedPn

onthebody，then：1、Provementofthetheorem

dnENERGYMETHOD§11–3

卡氏定理给Pn

以增量dPn

，则：1.先给物体加P1、

P2、•••、

Pn

个力，则：2.先给物体加力dPn

，则：一、定理证明

能量方法dnAgainapplyforces

P1、

P2、•••、Pn，then：

dnn=nPU¶¶dSecondCastigliano’stheoremItalianengineer—AlbertoCastigliano，1847～1884ENERGYMETHOD再给物体加P1、

P2、•••、Pn个力，则：

能量方法dnn=nPU¶¶d第二卡氏定理

意大利工程师—阿尔伯托·卡斯提安诺(AlbertoCastigliano，1847～1884)2、whatwemustpayattentiontoasweapplyCatigliano’stheorem：①U—Linearelasticstrainenergyofthewholestructureactedbyexternalloads②

Pnisconsideredasavariable.ThereactionsandthestrainenergyofthestructureandsoonmustbeallexpressedasthefunctionofPn.③

nisthedeformationofthepointactedbyPn

anditisalongthedirectionofPn.④Ifthereisno

Pn

correspondingto

nwemay firstactaPn

along

nanddeterminethepartialderivativeandthenletPn

bezero.dnENERGYMETHOD二、使用卡氏定理的注意事项：①U——整体结构在外载作用下的线弹性变形能②

Pn视为变量，结构反力和变形能等都必须表示为Pn的函数③

n为Pn

作用点的沿Pn

方向的变形。④当无与

n对应的Pn

时，先加一沿

n

方向的Pn

，求偏导后，再令其为零。能量方法dn3、Castigliano’stheoremforspecialstructures(rods)：ENERGYMETHOD三、特殊结构（杆）的卡氏定理：能量方法Example5Thestructureisshowninthefigure.DeterminethedeflectionandtheangleofrotationofthesectionA

byCatigliano’stheorem.③Determinethedeformation①DeterminetheinternalforceSolution：Determinethedeflection.Setupthecoordinate②Determinethepartialderivativeoftheinternalforcewithrespectto

PAALPEIxO

ENERGYMETHOD()[例5]结构如图，用卡氏定理求A

面的挠度和转角。③变形①求内力解：求挠度，建坐标系②将内力对PA求偏导能量方法ALPEIxO

()Determinetheangle

Aofrotation①DeterminetheinternalforceThereisnothegeneralizedforcecorrespondingto

A.wemayactone.“Negativesign”expressesthat

AiscontrarytothedirectionoftheactedgeneralizedforceMA()②DeterminethepartialderivativeoftheinternalforceM(x)withrespecttoMAandletMA=0.③Determinethedeformation（Note：MA=0）LxO

APMAENERGYMETHOD求转角

A①求内力没有与

A向相对应的力（广义力），加之。“负号”说明

A与所加广义力MA反向。()②将内力对MA求偏导后，令M

A=0③求变形（注意：MA=0）能量方法LxO

APMAExample

6

DeterminethedeflectioncurveofthebeamshowninthefigurebyCastigliano’stheorem.Solution：Determinethedeflectioncurve—thedeflectionofanarbitrarypointonthebeamf（x）.①Determinetheinternalforces②DeterminethepartialderivativeoftheinternalforceM(x)withrespecttoPxandletPx=0.

Thereisnothegeneralizedforcecorrespondingtof（x）.wemayactone.

PALxBPx

CfxOx1ENERGYMETHOD[例6

]结构如图，用卡氏定理求梁的挠曲线。解：求挠曲线——任意点的挠度f（x）①求内力②将内力对Px求偏导后，令Px=0没有与f（x）相对应的力，加之。能量方法PALxBPx

CfxOx1③Determinethedeformation（Note：Px=0）ENERGYMETHOD③变形（注意：Px=0）能量方法Example7

Abeamwithequalsectionisshowninthefigure.Determinethedeflectionf(x)ofpointBbyCatigliano’stheorem.②determineinternalforcesSolution：1.Determineredundantreactionsaccordingto③DeterminethepartialderivativeoftheinternalforcewithrespecttoRC.

①TakeaprimarybeamasshowninthePCAL0.5LBfxOPCAL0.5LBRCENERGYMETHODfigure.[例7]等截面梁如图，用卡氏定理求B

点的挠度。②求内力解：1.依求多余反力，③将内力对RC求偏导①取静定基如图能量方法PCAL0.5LBfxOPCAL0.5LBRC④DeformationENERGYMETHODSo④变形能量方法2.Determine②Determinethepartialderivativeoftheinternalforcewithrespect①DeterminetheinternalforcesENERGYMETHODtoP.2.求②将内力对P求偏导①求内力能量方法③DeformationENERGYMETHOD()③变形能量方法()③DeterminethedeformationSolution：①Plotthediagramofthestructureactedbyunitload②DeterminetheinternalforceExample

8

Aframeisshowninthefigure.DeterminethedistancebetweensectionAandsectionB

afterthedeformation.PPAB11ENERGYMETHOD③变形解：①画单位载荷图②求内力[例8

]结构如图，求A、B两面的拉开距离。PPAB能量方法1159

Chapter11Exercises1.Astraightrodwiththetension(compression)rigidityEIissubjectedforcesshowninthefigure.Maythestrainenergybeexpressedas