英汉双语材料力学课件.pptx
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1、,CHAPTER 11 ENERGY METHODS,材料力学,第十一章 能量方法,CHAPTER 11 ENERGY METHOD,111 GENERAL EXPRESSIONS OF THE STRAIN ENERGY112 MOHRS THEOREM(METHOD OF UNIT FORCE)113 CATIGLIANOS THEOREM,第十一章 能量方法,111 变形能的普遍表达式112 莫尔定理(单位力法)113 卡氏定理,111 GENERAL EXPRESSIONS OF THE STRAIN ENERGY,1、Principle of energy:,2、Calculatio
2、n of the strain energy of rods:,1). Calculation of the strain energy of rods in tension or compression:,Strain energy stored in the elastic body is equal to the work done by external forces,that is:,Method to analyze and calculate displacements 、deformations and internal forces of deformable bodies
3、by this kind of relation is called energy method.,ENERGY METHOD,or,Density of the strain energy:,111 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. Calculation of the strain energy of rods in torsion:,3. Calculation of strain ene
4、rgy of rods in bending:,ENERGY METHOD,or,Density of the strain energy:,or,Density of the strain energy:,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、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
5、 other.,For slender columns,the strain energy due to shearing forces may be neglected.,ENERGY METHOD,Deflection factor of shear,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:In energy method(work done by external forces is equal to the strain energy),Determine inter
6、nal forces,A,ENERGY METHOD,Bending 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.,MN,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),求内力,能量方法,
7、A,P,R,O,Work done by external forces is equal to the strain energy,Strain energy:,ENERGY METHOD,Let,then,外力功等于应变能,变形能:,能量方法,Example 2 Determine the deflection of point C by the energy method,where the beam is of equal section and straight.,Solution: Work done by external forces is equal to the strai
8、n energy,By using symmetry we get:,Thinking:For the distributed load ,can we determine the displacement of point C by this method?,C,a,a,A,P,B,f,ENERGY METHOD,Let,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,C,a,a,A,P,B,f,112 MOHRS THEOREM(METHOD OF UNIT FORCE),Determine the
9、displacement f A of an arbitrary point A.,1、Provement of the theorem:,a,A,Fig,fA,ENERGY METHOD,112 莫尔定理(单位力法),求任意点A的位移f A 。,一、定理的证明:,能量方法,a,A,图,fA,Mohrs theorem(method of unit force),2、General form of Mohrs theorem,ENERGY METHOD,莫尔定理(单位力法),二、普遍形式的莫尔定理,能量方法,3、What we must pay attention to as we apply
10、 Mohrs theorem:, Coordinate of M0(x) must be coincide with that of M(x). For each segment the coordinate may be set up freely.,Mohrs integrationmust be through the whole structure., M0: The internal force of the structure as we act a generalized unit force along the direction, of the generalized dis
11、placement that is to be determined, where the applied force is taken out., M(x):The internal force of the structure acted by original loads., The product of the applied generalized unit force and the generalized displacement to be determined determined must be of the dimension of work.,ENERGY METHOD
12、,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 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。, 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。,能量方法,Example 3 Determine the displacement and the angle of rotation of point C by the energy method .,Solution:Plot the diagram of the struc
13、ture acted by the unit load,Determine the internal force,ENERGY METHOD,例3 用能量法求C点的挠度和转角。梁为等截面直梁。,解:画单位载荷图,求内力,能量方法,B,q,x,Deformation,x,ENERGY METHOD,变形,能量方法,x,Determine the angle of rotation. Set up the coordinate again (as shown in the figure),ENERGY METHOD,=0,求转角,重建坐标系(如图),能量方法,=0,Solution: Plot t
14、he diagram of the structure acted by a unit load,Determine the internal force,5,20,A,300,P=60N,B,x,500,C,x1,ENERGY METHOD,Example 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,Dete
15、rmine the vertical displacement of point B.,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:画单位载荷图,求内力,能量方法,5,20,A,300,P=60N,B,x,500,C,x1,Determine the deformation,ENERGY METHOD,变形,能量方法,113 CATIGLIANOS THEOREM,Give Pn an increment dPn ,then:,1)First apply forces P1、 P2、 Pn on the b
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