能量法(Energy Method 第十三章能量法(Energy Methods) 13-1概述(Introduction) 13-2杆件变形能的计算( Calculation of strain energy for various types of loading 13-3互等定理(Reciprocal theorems) 13-4单位荷载法·莫尔定理(Unit-load method&mohr' 's theorem) 13-5卡氏定理(Castigliano's' Theorem) 13-6计算莫尔积分的图乘法(The meth- od of moment areas for mohr's integration)
共1页 2 (Energy Method) 第十三章 能量法 (Energy Methods) §13-1 概述(Introduction) §13-2 杆件变形能的计算( Calculation of strain energy for various types of loading ) §13-3 互等定理(Reciprocal theorems) §13-4 单位荷载法 • 莫尔定理(Unit-load method & mohr’s theorem) §13-5 卡氏定理(Castigliano’s Theorem) §13-6 计算莫尔积分的图乘法 (The method of moment areas for mohr’s integration)
能量港( Energy Method) 513-1概述( Introduction) 、能量方法( Energy methods 利用功能原理V=W来求解可变形固体的位移变形和内力 等的方法 二、外力功( Work of the external force) 固体在外力作用下变形,引起力作用点沿力作用方向位移, 外力因此而做功,则成为外力功 、变形能( Strain energy) 在弹性范围内,弹性体在外力作用下发生变形而在体内积蓄 的能量称为弹性变形能简称变形能
共1页 3 (Energy Method) §13-1 概述(Introduction) 在弹性范围内,弹性体在外力作用下发生变形而在体内积蓄 的能量,称为弹性变形能,简称变形能. 一、能量方法 (Energy methods ) 三、变形能(Strain energy) 二、外力功(Work of the external force) 固体在外力作用下变形,引起力作用点沿力作用方向位移, 外力因此而做功,则成为外力功. 利用功能原理Vε = W 来求解可变形固体的位移,变形和内力 等的方法
能量滤 (Energy Method) 四、功能原理( Work-energy principle) 可变形固体在受外力作用而变形时,外力和内力均将作功对 于弹性体,不考虑其他能量的损失外力在相应位移上作的功在数 值上就等于积蓄在物体内的应变能 The formula V=N ( Work-Energy Principle We will not consider other forms of energy such as thermal energy, chemical energy, and electromagnetic energy. Therefore, if the stresses in a body do not exceed the elastic limit, all of work done on a body by external forces is stored in the body as elastic strain energy
共1页 4 (Energy Method) 可变形固体在受外力作用而变形时,外力和内力均将作功. 对 于弹性体,不考虑其他能量的损失,外力在相应位移上作的功,在数 值上就等于积蓄在物体内的应变能. Vε = W 四、功能原理(Work-energy principle) The formula: (Work-Energy Principle) We will not consider other forms of energy such as thermal energy, chemical energy, and electromagnetic energy. Therefore, if the stresses in a body do not exceed the elastic limit, all of work done on a body by external forces is stored in the body as elastic strain energy
能量滤( Energy Methoc 513-2杆件变形能的计算 Calculation of strain energy for various types of loading 、杆件变形能的计算( Calculation of strain energy for various types of loading) 1.轴向拉压的变形能( Strain energy for axial loads) 当拉力为F1时,杆件的伸长为81 当再增加一个dF1时,相应的变形增量为d(A1) 此外力功的增量为: dEl dW=F1d(Ml1)d(△1)=EA
共1页 5 (Energy Method) §13-2 杆件变形能的计算 ( Calculation of strain energy for various types of loading ) 一、杆件变形能的计算(Calculation of strain energy for various types of loading) 1.轴向拉压的变形能(Strain energy for axial loads) 此外力功的增量为: d d(Δ ) 1 1 W = F l EA F l l 1 1 d d(Δ ) = 当拉力为F1 时,杆件的伸长为l1 当再增加一个dF1时,相应的变形增量为d(Δl1 )
能量港 Energy Method) F dFI oL, dAf1 △l △l 积分得:W=[dW= F-d FF △Z 0E4 2E42 6
共1页 6 (Energy Method) F F l l F O l l l1 dl1 dF1 F1 积分得: l F EA F l F EA l W W F F Δ 2 2 d d 2 1 0 = = 1 = =
能量港( Energy Method) 根据功能原理 V=W,可得以下变形能表达式 VE=W=FA=-FNAZ 4l≈FFNl EA EA F Fml 2E42E4 当轴力或截面发生变化时:V ∑ 2E A
共1页 7 (Energy Method) 根据功能原理 当轴力或截面发生变化时: Vε= W , 可得以下变形能表达式 V W F l F Δl 2 1 Δ 2 1 ε = = = N EA F l EA Fl l N Δ = = EA F l EA F l V 2 2 2 N 2 ε = = = = n i i i i i E A F l V 1 2 N ε 2
能量滤( Energy Methoc d Fn(x)d 当轴力或截面连续变化时:V=J02EA(x) 比能( strain energy density): 单位体积的应变能记作U F =-o8 丿4l2 o= Ea 2E8 U522E2 (单位J/m3) 8
共1页 8 (Energy Method) (单位 J/m3 ) 比能 ( strain energy density): 单位体积的应变能. 记作u 当轴力或截面连续变化时: = l EA x F x x V 0 2 N ε 2 ( ) ( )d σε Al F l V U 2 1 Δ 2 1 uε = = = σ = Eε 2 2 2 1 2 2 ε Eε E σ u = σε = =
能量滤 (Energy Method) 2.扭转杆内的变形能( Strain energy for torsional loads) 1 MI M2I T2I E=W=M·Aq=M 2G 2G1 2GI 2 T(x) dx 2G,(x) 或V=∑ =12G: I
共1页 9 (Energy Method) 2.扭转杆内的变形能(Strain energy for torsional loads) 或 l p 2 p 2 e p e ε e e 2 2 2 1 Δ 2 1 GI T l GI M l GI M l V = W = M = M = = = l x GI x T x V d 2 ( ) ( ) p 2 ε = = n i i i i i G I T l V 1 p 2 ε 2 Me Me Me
能量港 Energy Method 3弯曲变形的变形能 Strain energy for flexural loads) M 纯弯曲( pure bending) 6 U=W=-M·0=-M MI MI 2E 2EI 横力弯曲( nonuniform bending) (x e 2E/(x) 10
共1页 10 (Energy Method) • 纯弯曲(pure bending ) • 横力弯曲(nonuniform bending ) 3.弯曲变形的变形能 (Strain energy for flexural loads) θ Me EI M l EI M l V W M θ M 2 2 1 2 1 2 = = = = e ε e e x EI x M x V l d 2 ( ) ( ) 2 e ε = Me Me Me