1、Mechanics of Materials 111 GENERAL EXPRESSIONS OF THE STRAIN ENERGY112 MOHRS THEOREM(METHOD OF UNIT FORCE)113 CATIGLIANOS THEOREM 111 变形能的普遍表达式变形能的普遍表达式112 莫尔定理莫尔定理(单位力法单位力法)113 卡氏定理卡氏定理111 GENERAL EXPRESSIONS OF THE STRAIN ENERGY1、Principle of energy：2、Calculation of the strain energy of rods：1). 1
2、). Calculation of the strain energy of rods in tension or compression：LxEAxNUd2)( 2niiiiiAELNU12221u Strain energy stored in the elastic body is equal to the work done by external forces，that is:WU Method to analyze and calculate displacements 、deformations and internal forces of deformable bodies b
3、y this kind of relation is called energy method.orDensity of the strain energy:111 变形能的普遍表达式变形能的普遍表达式一、能量原理：一、能量原理：二、杆件变形能的计算：二、杆件变形能的计算：1.1.轴向拉压杆的变形能计算：轴向拉压杆的变形能计算：LxEAxNUd2)( 2niiiiiAELNU122 或21:u比能 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WU 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. 2. Calculation of the strain e
4、nergy of rods in torsion：LPnxGIxMUd2)( 2niPiiiniIGLMU122 21u3. 3. Calculation of strain energy of rods in bending：LxEIxMUd2)( 2niiiiiIELMU122 21uor Density of the strain energy:orDensity of the strain energy:2.2.扭转杆的变形能计算：扭转杆的变形能计算：LPnxGIxMUd2)( 2niPiiiniIGLMU122 或21:u比能3.3.弯曲杆的变形能计算：弯曲杆的变形能计算：LxEIx
5、MUd2)( 2niiiiiIELMU122 或21:u比能3、General expressions of the strain energy： Strain energy is independent of the order of loading. Deformations due to mutually independent load may be summed up each other.For slender columns，the strain energy due to shearing forces may be neglected.xEIxMxGIxMxEAxNULLPn
6、Ld2)(d2)(d2)(222LxEAxQd2)( 2SSxEIxMxGIxMxEAxNULLPnLd2)(d2)(d2)(222Deflection factor of shear三、变形能的普遍表达式：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。LxEAxQd2)( 2S剪切挠度因子SxEIxMxGIxMxEAxNULLPnLd2)(d2)(d2)(222xEIxMxGIxMxEAxNULLPnLd2)(d2)(d2)(222Solution：In energy method（work done by e
7、xternal forces is equal to the strain energy）sin)(PRMT)cos1 ()( PRMNABending moment:Torque:Example 1 A semicircle rod as shown in the figure is lie in horizontal plane. A vertical force P act at its point A. Determine the displacement of point A in vertical direction.PROQMNMTAAPNB TOMN 例例1 1 图示半圆形等截
8、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）求内力sin)(:PRMT弯矩)cos1 ()(: PRMN扭矩APROQMTAAPNB TO：LLPLxEIxMxGIxMxEAxNUd2)( d2)( d2)( 22n202220222d2)(sind2)cos1(REIRPRGIRPPEIRPGIRPP4433232,2UfPWALetEIPRGIPRfPA22333then外力功等于应变能变形能：LLPLxEIxMxGIxMxEAxNUd2)( d2)( d2)( 22n202220222d2)(sind2)cos1(REIRPRGIRPPE
9、IRPGIRPP4433232UfPWA2EIPRGIPRfPA22333Example Example 2 Determine the deflection of point C by the energy method,where the beam is of equal section and straight. CPfW21Solution： LxEIxMUd2)( 2)0( ; 2)(axxPxMBy using symmetry we get：EIaPxxPEIUa12d)2(2123202EIPathenfUWC6,3Thinking：For the distributed lo
10、ad ,can we determine the displacement of point C by this method？qCaaAPBfLet 例例2 用能量法求C点的挠度。梁为等截面直梁。CPfW21解：外力功等于应变能LxEIxMUd2)( 2)0( ; 2)(axxPxM应用对称性，得：EIaPxxPEIUa12d)2(2123202EIPafUWC63思考：分布荷载时，可否用此法求C点位移？qCaaAPBf112 MOHRS THEOREM(METHOD OF UNIT FORCE)ACfUUU10LxEIxMUd2)( 2LxEIxMUd2)( 200LCxEIxMxMUd2
11、)()( 20LAxEIxMxMfd)()( 0Determine the displacement f A of an arbitrary point A.1、Provement of the theorem：aAFigfAq(x)Figc A0P =1q(x)fAFigb A=1P0112 莫尔定理莫尔定理(单位力法单位力法)ACfUUU10LxEIxMUd2)( 2LxEIxMUd2)( 200LCxEIxMxMUd2)()( 20LAxEIxMxMfd)()( 0求任意点A的位移f A 。一、定理的证明：一、定理的证明：aA图fAq(x)图c A0P =1q(x)fA图b A=1P0
12、Mohrs theorem(method of unit force)2、General form of Mohrs theoremxEIxMxMfLAd)()(0LPnnLAxGIxMxMxEAxNxNd)()(d)()(00 xEIxMxMLd)()(0 莫尔定理莫尔定理( (单位力法单位力法) )二、普遍形式的莫尔定理二、普遍形式的莫尔定理xEIxMxMfLAd)()(0LPnnLAxGIxMxMxEAxNxNd)()(d)()(00 xEIxMxMLd)()(03、What we must pay attention to as we apply Mohrs theorem：M0(x)
13、 must be coincide with that of M(x). For each segment the coordinate may be set up freely.M0: The internal force of the structure as we act a generalized unit force along the direction, of the generalized displacement that is to be determined, where the applied force is taken out.M(x)：The internal f
14、orce of the structure acted by original loads. 三、使用莫尔定理的注意事项：三、使用莫尔定理的注意事项：M0(x)与M(x)的坐标系必须一致，每段杆的坐标系可 自由建立。莫尔积分必须遍及整个结构。M0去掉主动力，在所求 点，沿所求的方向加时，结构产生的内力。M(x)：结构在原载荷下的内力。所加广义单位力与所求广义位移之积，必须为功的量纲。Example 3 3 Determine the displacement and the angle of rotation of point C by the energy method .2)(2qxaqx
15、xM)2( ; )2(2)0( ; 2)(0axaxaxaxxxMSolution： Determine the internal forceBAaaCqBAaaC0P =1x 例例3 3 用能量法求C点的挠度和转角。梁为等截面直梁。2)(2qxaqxxM)2( ; )2(2)0( ; 2)(0axaxaxaxxxM解：画单位载荷图求内力BAaaCqBAaaC0P =1x d)()(d)()(2000aaaCxEIxMxMxEIxMxMfaxEIxMxM00d)()(2SymmetrySymmetryEIqaxxqxqaxEIa245d2)2(2402DeformationBAaaC0P =1
16、BAaaCqx( ) d)()(d)()(2000aaaCxEIxMxMxEIxMxMfaxEIxMxM00d)()(2对称性对称性EIqaxxqxqaxEIa245d2)2(2402变形BAaaC0P =1BAaaCqx( )）aaxaxqxqaxEIxaxqxqaxEI022222011211d2)2(1d2)2(12)( :211qxqaxxMAC axxM2)( 10 2)( :222qxqaxxMBCaxxM2)(20qBAaaCx2x1BAaaCMC0=1 d)()( )()()(00)(00aBCaABxEIxMxMdxEIxMxMc=0求转角，重建坐标系（如图）aaxaxqxq
17、axEIxaxqxqaxEI022222011211d2)2(1d2)2(12)( :211qxqaxxMAC axxM2)( 10 2)( :222qxqaxxMBCaxxM2)(20qBAaaCx2x1BAaaCMC0=1 d)()( )()()(00)(00aBCaABxEIxMxMdxEIxMxMc=0PxxMAB)(xxMAB)(0PxMnCA3 . 0)(13 . 0)(10 xMCAnSolution： Determine the internal force510 20A300P=60NBx500Cx1510 20A300Bx500C=1P0Example 4 Example
18、4 A folding rod is shown in the figure. A bearing is at position A and the rod may rotate freely in the bearing but can not move up and down. Knowing：E=210Gpa，G=0.4E，Determine the vertical displacement of point B. 例例4 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。PxxMAB)(xxMAB)(0PxMn
19、CA3 . 0)(13 . 0)(10 xMCAn解：画单位载荷图求内力510 20A300P=60NBx500Cx1510 20A300Bx500C=1P0PxxMAB)(xxMAB)(0PxMnCA3 . 0)(13 . 0)(10 xMCAnLLPnnBxEIxMxMxGIxMxMd)()( d)()( 011013 . 0025 . 001dd3 . 03 . 0 xEIPxxGIPPPACABABABGILPLLEIPL33333101052103123 . 0603410202104 . 0325 . 03 . 0603 . 0mm22. 8( )PxxMAB)(xxMAB)(0P
20、xMnCA3 . 0)(13 . 0)(10 xMCAnLLPnnBxEIxMxMxGIxMxMd)()( d)()( 011013 . 0025 . 001dd3 . 03 . 0 xEIPxxGIPPPACABABABGILPLLEIPL33333101052103123 . 0603410202104 . 0325 . 03 . 0603 . 0mm22. 8变形( )113 CATIGLIANOS THEOREMGive Pn an increment dPn ，then：),.,(21nPPPUU nnPPUUUd11)First apply forces P1、 P2、 Pn on
21、 the body ，then：2). First apply the force dPn on the body，then：)d()d(212nnPU1、Provement of the theorem 1P2P nnP113 卡氏定理卡氏定理给Pn 以增量 dPn ，则：),.,(21nPPPUU nnPPUUUd11. 先给物体加P1、 P2、 Pn 个力，则：2.先给物体加力 dPn ，则：)d()d(212nnPU一、定理证明一、定理证明 1P2P nnPAgain apply forces P1、 P2、Pn ，then：)d(21nnPUUUnnPU 1P2P nnPn nPU
22、Second Castiglianos theorem Italian engineer Alberto Castigliano， 18471884 再给物体加P1、 P2、Pn 个力，则：)d(21nnPUUUnnPU 1P2P nnPn nPU 第二卡氏定理第二卡氏定理 意大利工程师阿尔伯托卡斯提安诺(Alberto Castigliano， 18471884)2、what we must pay attention to as we apply Catiglianos theorem：ULinear elastic strain energy of the whole structure
23、 acted by external loads Pn is considered as a variable. The reactions and the strain energy of the structure and so on must be all expressed as the function of Pn.is the deformation of the point acted by Pn and it isalong the direction of Pn .Pn corresponding to we may first act a Pn along and dete
24、rmine the partial derivative and then let Pn be zero.1P2P nnP二、使用卡氏定理的注意事项：二、使用卡氏定理的注意事项：U整体结构在外载作用下的线 弹性变形能 Pn 视为变量，结构反力和变形能 等都必须表示为 Pn的函数为 Pn 作用点的沿 Pn 方向的变形。当无与 对应的 Pn 时，先加一沿 方向的 Pn ，求偏导后， 再令其为零。1P2P nnP3、Castiglianos theorem for special structures(rods)：LLPLxEIxMxGIxMxEAxNUd2)( d2)( d2)( 22n2LnLn
25、nPLnnnxPxMEIxMxPxMGIxMxPxNEAxNPUd)()( d)()( d)()( n三、特殊结构（杆）的卡氏定理：三、特殊结构（杆）的卡氏定理：LLPLxEIxMxGIxMxEAxNUd2)( d2)( d2)( 22n2LnLnnPLnnnxPxMEIxMxPxMGIxMxPxNEAxNPUd)()( d)()( d)()( nExample 5 Example 5 The structure is shown in the figure. Determine the deflection and the angle of rotation of the section A
26、 by Catiglianos theorem.Determine the deformationDetermine the internal forceSolution：Determine the deflection. Set up the coordinatexPxPxMA)(EIPL33PAxPxMA)(LAAAxPxMEIxMPUfd)()( LxEIPx02dALPEIxO ( ) 例例5 5 结构如图，用卡氏定理求A 面的挠度和转角。变形求内力解：求挠度，建坐标系xPxPxMA)(EIPL33将内力对PA求偏导xPxMA)(LAAAxPxMEIxMPUfd)()( LxEIPx0
27、2dALPEIxO ( )Determine the angle A of rotationAMxPxM)(There is no the generalized force corresponding to A. we may act one.EIPL22“Negative sign”expresses that A is contrary to the direction of the acted generalized force MA( )MA and let M A =0.1)(0AMAMxMLAAxMxMEIxMd)()( LxEIPx0dDetermine the deforma
28、tion（ Note：M A=0）LxO APMAEIPLA22求转角 A求内力AMxPxM)(没有与A向相对应的力（广义力），加之。EIPL22 “负号”说明 A与所加广义力MA反向。( )EIPLA22 将内力对MA求偏导后，令M A=01)(0AMAMxMLAAxMxMEIxMd)()( LxEIPx0d求变形（ 注意：M A=0）LxO APMAExample 6 Determine the deflection curve of the beam shown in the figure by Castiglianos theorem.Solution：Determine the de
29、flection curvethe deflection of an arbitrary point on the beam f（x）.Determine the internal forcesPx and let Px =0. There is no the generalized force corresponding to f（x）.we may act one. )()()(111xxPxLPxMxAB)()(11xLPxMBCxxPxMPxAB10)(x0)(0 xxBCPPxMPALxBPx CfxOx1 例例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）
30、求内力将内力对Px 求偏导后，令Px=0没有与f（x）相对应的力，加之。)()()(111xxPxLPxMxAB)()(11xLPxMBCxxPxMPxAB10)(x0)(0 xxBCPPxMPALxBPx CfxOx1（ Note：Px=0）LxxxPxMEIxMPUxfd)()( )(xxxxxLPEI0111d)(1)2)(3(223LxxxLxEIP变形（ 注意：Px=0）LxxxPxMEIxMPUxfd)()( )(xxxxxLPEI0111d)(1)2)(3(223LxxxLxEIPExample 7 A beam with equal section is shown in th
31、e figure. Determine the deflection f(x) of point B by Catiglianos theorem.determine internal forcesSolution：1.Determine redundant reactions according to0CfRC. )5 . 0()()(xLPxLRxMCAB)()(xLRxMCBCxLRxMCAB)(xLRxMCBC)(PCAL0.5 LBfxOPCAL0.5 LBRC 例例7 等截面梁如图，用卡氏定理求B 点的挠度。求内力解：1.依 求多余反力，0 Cf将内力对RC求偏导)5 . 0()(
32、)(xLPxLRxMCAB)()(xLRxMCBCxLRxMCAB)(取静定基如图xLRxMCBC)(PCAL0.5 LBfxOPCAL0.5 LBRCDeformationLCCCxRxMEIxMRUfd)()( LCLxxLRxxLxLPEI025 .00d)(d)()5 .0(10)3485(133LRPLEIC165PRCSo变形LCCCxRxMEIxMRUfd)()( LCLxxLRxxLxLPEI025 .00d)(d)()5 .0(10)3485(133LRPLEIC165PRC2.DetermineBf)5 .0()(165)(xLPxLPxMAB)(165)(xLPxMBC1
33、6311)(LxPxMAB16)(5)(xLPxMBCP.2.求Bf将内力对P求偏导)5 .0()(165)(xLPxLPxMAB)(165)(xLPxMBC16311)(LxPxMAB16)(5)(xLPxMBC求内力DeformationLBxPxMEIxMPUfd)()( LLLxxLPxLxPEI5 .0225 .002d)()165(d)16311(1EIPL76873( )变形LBxPxMEIxMPUfd)()( LLLxxLPxLxPEI5 .0225 .002d)()165(d)16311(1EIPL76873( )Solution：Plot the diagram of th
34、e structure acted by unit loadExample 8 A frame is shown in the figure. Determine the distance between section A and section B after the deformation.PPAB11变形解：画单位载荷图求内力 例例8 结构如图，求A、B两面的拉开距离。PPAB1159 Chapter 11 Exercises1. A straight rod with the tension (compression) rigidity EI is subjected forces
35、shown in the figure. May the strain energy be expressed as 2. Try to explain how to determine the deflection of the free end of the beam shown in the figure by Castiglianos theorem.3. As shown in the figure, a rigid frame is subjected to forces. Knowing EI is a constant. Try to determine the relativ
36、e displacement between point A and point B by Mohrs theorem (neglecting the tensile deformation of Section CD). ?22222121EALPEALPU60 第十一章第十一章 练习题练习题 一、抗拉（压）刚度为一、抗拉（压）刚度为EIEI的等直杆，受力如图，的等直杆，受力如图，其变形能是否为：其变形能是否为： 二、试述如何用卡氏定理求图示梁自由端的挠度。二、试述如何用卡氏定理求图示梁自由端的挠度。 三、刚架受力如图，已知三、刚架受力如图，已知EIEI为常数，试用莫尔为常数，试用莫尔定理求
37、定理求A A、B B两点间的相对位移（忽略两点间的相对位移（忽略CDCD段的拉伸变段的拉伸变形）。形）。 ?22222121EALPEALPU61 Solution: aaEIxMxMEIxMxMABdxdx02/0212021012EIPadxdxaaEIaPaEIxPx302/0213521162解：解： aaEIxMxMEIxMxMABdxdx02/0212021012EIPadxdxaaEIaPaEIxPx302/0213521163 4. A beam with the bending rigidity EI is shown in the figure. The rigidity
38、of the spring at the end B is k. Try to determine the deflection of the point where the force P is applied by Castiglianos theorem. Solution: The strain energy of the system is The deflection of Section C is kPEILPkPPLLPPEIdxxdxU18243233213/03/202323221232 kPEIPLPUcf92434364 四、抗弯刚度为四、抗弯刚度为EIEI的梁如图，的梁如图，B B端弹簧刚度为端弹簧刚度为k k，试用卡氏定理求力试用卡氏定理求力P P作用点的挠度。作用点的挠度。 解：解： 系统的变形能系统的变形能 C C截面的挠度截面的挠度 kPEILPkPPLLPPEIdxxdxU18243233213/03/202323221232 kPEIPLPUcf9243436566
