1755年Lagrange的著作《Mecanique Analytique》(分析力学)问世^[1](1788年首次正式出版), 力学发展史上出现了与牛顿的矢量力学并驾齐驱的一个力学体系。Lagrange力学体系的特点是用对能量和功的分析代替对力和力矩的分析。1834年Hamilton^[2-3]建立了Hamilton原理和正则方程, 进一步发展了分析力学, 从而形成Lagrange体系和Hamilton体系。

如何将经典分析动力学应用于连续介质力学, 一直是各国学者关注的研究课题。

我国1958年出版的第一部分析力学专著《分析动力学》^[4], 注意到将分析力学从质点刚体力学扩展到连续介质力学、从离散系统扩展到连续系统的问题。2015年影印的《Classical Mechanics》^[5]仍然研究连续体分析动力学。一些学者成功地将Lagrange方程应用于机构动力学分析^[6-9]、振动系统^[10]、防护工程^[11]、电器系统和机电系统^[12]等领域, 也有学者研究如何将Lagrange方程应用于非惯性系统^[13-15], 以及弹性力学的Lagrange形式、弹性介质的Lagrange动力学和精确Cosserat弹性杆动力学的分析力学方法^[16-18]。另有一些学者研究完整系统三阶Lagrange方程、状态空间Lagrange函数和运动方程^[19-20]以及关于Birkhoff方程和Lagrange方程分析力学问题^[21]。

在一定的意义上, Lagrange方程和Hamilton原理都涉及变分学, 而Lagrange作为变分学的奠基人之一, 研究的就是基于变分学的基本理论, 所以借鉴变分学或许是一条可行的途径。Liang等^[22]提出变分的逆运算变积概念, 建立变积方法, 然后应用变积方法, 建立一般力学三类变量的广义变分原理^[23]。刘高联先生充分肯定了变积方法的首创性^[24]。以上研究使微积分学中的积分、微分和导数在变分学中有了对应的概念--变积、变分和变导, 初步地将变分学扩充为变积分学。

本文基于陈滨^[25]关于Lagrange力学完整的叙述, 应用变导的概念和运算法则, 研究Lagrange方程中求导的性质, 逐步将Lagrange方程应用于线性弹性动力学, 进而应用于非线性弹性动力学, 并且给出相关的算例, 对理论研究进行验证。

1 Lagrange方程中的导数的性质

首先, 明确变导的概念。设有定积分形式的泛函为

$ V = \int_a^b {F(x,y,y'){\rm{d}}x}, $

(1)

边界条件为

$ {y_{x = a}} = \alpha ,\quad {y_{x = b}} = \beta , $

(2)

其中, $y (x)$为自变函数, $x$为自变量。

对式(1)进行变分运算可得

$ {\rm{\delta }}V = \int_a^b {\left( {\frac{{\partial F}}{{\partial y}}{\rm{\delta }}y + \frac{{\partial F}}{{\partial y'}}{\rm{\delta }}y'} \right)} \;{\rm{d}}x。 $

(3)

应用分部积分:

$ \int_a^b {\frac{{\partial F}}{{\partial y'}}{\rm{\delta }}y'} {\rm{d}}x = \frac{{\partial F}}{{\partial y'}}\left. {{\rm{\delta }}y} \right|_a^b - \int_a^b {\frac{{\rm{d}}}{{{\rm{d}}x}}\frac{{\partial F}}{{\partial y'}}{\rm{\delta }}y} {\rm{d}}x。 $

(4)

将式(4)代入式(3), 考虑到边界条件(式(2)), 整理得

$ {\rm{\delta }}V = \int_a^b {\left( {\frac{{\partial F}}{{\partial y}} - \frac{{\rm{d}}}{{{\rm{d}}x}}\frac{{\partial F}}{{\partial y'}}} \right)\;} {\rm{\delta }}y{\rm{d}}x。 $

(5)

由于${\rm{\delta }}y$的任意性, 式(5)可以变换为

$ \frac{{{\rm{\delta }}V}}{{{\rm{\delta }}y}} = \int_a^b {\left( {\frac{{\partial F}}{{\partial y}} - \frac{{\rm{d}}}{{{\rm{d}}x}}\frac{{\partial F}}{{\partial y'}}} \right)} \;{\rm{d}}x。 $

(6)

在微分学中, 函数的微分表示为${\rm{d}}y$, 自变量的微分表示为${\rm{d}}x$, 微商表示为$\frac{{{\rm{d}}y}}{{{\rm{d}}x}}$, 又称导数。在变分学中, 泛函的变分表示为$ {\rm{\delta }}V$, 自变函数的变分表示为${\rm{\delta }}y$, 变商表示为$\frac{{{\rm{\delta }}V}}{{{\rm{\delta }}y}}$, 又称变导。

经典分析动力学中的Lagrange方程表示为

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}} - \frac{{\partial T}}{{\partial \mathit{\boldsymbol{q}}}} + \frac{{\partial U}}{{\partial \mathit{\boldsymbol{q}}}} = 0, $

(7)

其中, $\mathit{\boldsymbol{q}}=\mathit{\boldsymbol{q}}{\rm{(}}t{\rm{)}}$为广义坐标, 一般分析动力学中均将其处理为广义坐标列阵:

$ \begin{array}{l} {[{q_1}{\rm{(}}t{\rm{)}},\;\;{q_2}{\rm{(}}t{\rm{)}},\;\;{q_3}{\rm{(}}t{\rm{)}},\;\;...,\;\;{q_i}{\rm{(}}t{\rm{)}},\;\;...{q_n}{\rm{(}}t{\rm{)}}]^{\rm{T}}}\;\;\\ \;\;\;\;\;\;\;\;\;\;(i = 1,\;2,\;3,\;...i,\;...,\;n)。 \end{array} $

(8)

在变分学中, 基本上存在三级变量--自变量、可变函数和泛函。简单函数和泛函的区别在于, 简单函数是自变量的函数, 而泛函是可变函数的函数, 独立自主地变化的可变函数称为自变函数。从不独立的可变函数也是自变函数的函数的角度看问题, 不独立的可变函数也是泛函, 可称其为子泛函。明确变分学中的三级变量, 对区分微积分中的导数和变积分学中的变导很有帮助。对自变量求导为微积分中的导数, 对可变函数的求导则为变积分中的变导。

在Lagrange方程中, 有4个求导运算: $\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}, $$\frac{{\partial T}}{{\partial \mathit{\boldsymbol{q}}}}, $$\frac{{\partial U}}{{\partial \mathit{\boldsymbol{q}}}}$和$\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}$。一般情况, $\mathit{\boldsymbol{\dot q}}$和$\mathit{\boldsymbol{q}}$均为可变函数, 所以$\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}$, $\frac{{\partial T}}{{\partial \mathit{\boldsymbol{q}}}}$和$\frac{{\partial U}}{{\partial \mathit{\boldsymbol{q}}}}$均为变积分学中变导; 时间t为自变量, 所以$\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}$中对时间t求导为微积分学中的导数。

严格地说, 变导$\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}$, $\frac{{\partial T}}{{\partial \mathit{\boldsymbol{q}}}}, $$\frac{{\partial U}}{{\partial \mathit{\boldsymbol{q}}}}$和$\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial T}}{{\partial \mathit{\boldsymbol{\dot q}}}}$应写为$\frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{\dot q}}}}$, $\frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{q}}}}$, $\frac{{{\rm{ \mathsf{ δ} }}U}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{q}}}}$和$\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{\dot q}}}}$。但是, 考虑到目前分析力学界的习惯, 可仍然沿用原来的符号:变分符号用δ, 变导符号用$\frac{\partial }{\partial }$。

Lagrange已经注意到微分符号用d而变分符号用δ, 而且应用了符号$\frac{{\rm{\delta }}}{{\rm{\delta }}}$, 所以变导的概念已经隐含在其著作中。例如, 在文献[1]中, Lagrange方程表示为

$ {\rm{d}}\frac{{{\rm{\delta }}T}}{{{\rm{\delta }}\dot \xi }} - \frac{{{\rm{\delta }}T}}{{{\rm{\delta }}\xi }} + \frac{{{\rm{\delta }}U}}{{{\rm{\delta }}\xi }} = 0。 $

(9)

需要说明, Lagrange将$\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{\dot q}}}}$表示为${\rm{d}}\frac{{{\rm{\delta }}T}}{{{\rm{\delta }}\dot \xi }}$, 应用了微分符号d, 而没有应用求导符号$\frac{{\rm{d}}}{{{\rm{d}}t}}$。

如果将$\xi $表示为$\mathit{\boldsymbol{q}}$, 将d表示为$\frac{{\rm{d}}}{{{\rm{d}}t}}$, 则式(9)可以表示为

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{\dot q}}}} - \frac{{{\rm{ \mathsf{ δ} }}T}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{q}}}} + \frac{{{\rm{ \mathsf{ δ} }}U}}{{{\rm{ \mathsf{ δ} }}\mathit{\boldsymbol{q}}}} = 0。 $

(10)

按照目前的习惯, 将变导$\frac{{\rm{\delta }}}{{\rm{\delta }}}$表示为$\frac{\partial }{\partial }$, 则式(10)就变换为式(7)。

变积分学中变导与微积分学中导数的运算法则, 有时相同, 有时不同, 在后面研究具体问题时可以明显地表现出来。

2 Lagrange方程应用于线性弹性动力学 2.1 一类变量Lagrange方程应用于线性弹性动力学

线性弹性动力学的动能表示为

$ T=\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{\dot{u}}}\cdot \mathit{\boldsymbol{\dot{u}}}\rm{d}V}。 $

(11)

线性弹性动力学的势能表示为

$ \begin{align} & U={{U}_{1}}+{{U}_{2}} \\ & \ \ \ =\iiint\limits_{V}{\left[ \frac{1}{2}\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\ \mathit{\boldsymbol{a}}:\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right)- \right.\ } \\ & \ \ \ \left. \mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}} \right]\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\ \rm{d}S, \\ \end{align} $

(12)

其中, a为刚度系数张量, u为位移矢量, f为体力矢量, T为面力矢量, n为外法向矢量, $\nabla $为Hamilton算子, r为质量密度, S_s为力学边界面, S_u为位移边界面, V为空间体积域。${{U}_{1}}=\iiint\limits_{V}{\left[ \frac{1}{2}\left (\frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right): \right.}$ $\left. \mathit{\boldsymbol{a}}:\left (\frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right) \right]\rm{d}V$为弹性体的应变能; U₂=$\iiint\limits_{V}{-}\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}}\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\rm{d}S$为外力势能。

位移边界条件为

$ \mathit{\boldsymbol{u}}-\mathit{\boldsymbol{ }}\!\!\bar{\mathit{\boldsymbol{u}}}\!\!\rm{ }=\boldsymbol{0}。 $

(13)

Lagrange方程表示为

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=0。 $

(14)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial \mathit{\boldsymbol{u}}}=0, $

(15)

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{\dot{u}}}}=\frac{\rm{d}}{\rm{d}t}\frac{\partial }{\partial \mathit{\boldsymbol{\dot{u}}}}\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{\dot{u}}}\cdot \mathit{\boldsymbol{\dot{u}}}\rm{d}V}=\iiint\limits_{V}{\rho \mathit{\boldsymbol{\ddot{u}}}\rm{d}V}。 $

(16)

势能变导项的推导比较复杂:

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left\{ \iiint\limits_{V}{\left[ \frac{1}{2}\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\ \mathit{\boldsymbol{a}}: \right.} \right. \\ & \left. \left. \ \ \ \ \ \ \ \ \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right)-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}} \right]\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\ \rm{d}S \right\}, \\ \end{align} $

(17)

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{\left\{ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left[ \frac{1}{2}\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\ \mathit{\boldsymbol{a}}:\ \right. \right.}\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+ \right. \\ & \ \ \ \ \ \ \ \ \left. \left. \left. \frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right) \right]-\mathit{\boldsymbol{f}} \right\}\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S。 \\ \end{align} $

(18)

由于

$ \begin{align} & \iiint\limits_{V}{\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left[ \frac{1}{2}\left( \frac{1}{2}\nabla \mathit{u}+\frac{1}{2}\mathit{u}\nabla \right):\ \mathit{a}:\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right) \right]\ } \\ & =\iiint\limits_{V}{\mathit{\boldsymbol{a}}:\ \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right)}:\ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}\mathit{\boldsymbol{u}}\nabla \rm{d}\mathit{V}, \\ \end{align} $

(19)

应用Green定理:

$ \begin{align} & \ \ \ \ \iiint\limits_{V}{\left[ \left( \frac{1}{2}\nabla \mathit{\pmb{u}}+\frac{1}{2}\mathit{\pmb{u}}\nabla \right):\ \mathrm{a}:\frac{\partial }{\partial \mathit{\pmb{u}}}\mathit{\pmb{u}}\nabla \right]}\ \text{d}V \\ & =\iint\limits_{{{S}_{\sigma }}+{{S}_{u}}}{\left( \frac{1}{2}\nabla \mathit{\pmb{u}}+\frac{1}{2}\mathit{\pmb{u}}\nabla \right):\mathrm{a}\cdot \mathrm{n}\text{d}S}- \\ & \ \ \ \iiint\limits_{V}{\nabla \cdot \left[ \left( \frac{1}{2}\nabla \mathit{\pmb{u}}+\frac{1}{2}\mathit{\pmb{u}}\nabla \right):\ \mathrm{a} \right]\text{d}V,} \\ \end{align} $

(20)

将式(19)和(20)代入式(18), 则得

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=-\iiint\limits_{V}{\left\{ \nabla \cdot \left[ \frac{1}{2}(\nabla \mathit{\boldsymbol{u}}+\mathit{\boldsymbol{u}}\nabla ):\ \mathit{\boldsymbol{a}} \right]+\mathit{\boldsymbol{f}} \right\}}\ \rm{d}V+ \\ & \iint\limits_{{{S}_{\sigma }}}{\left[ \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\ \rm{d}S。 \\ \end{align} $

(21)

将相关各式代入Lagrange方程, 可得

$ \begin{align} & \ \ \ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{\dot{u}}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}} \\ & =\iiint\limits_{V}{\left\{ \rho \mathit{\boldsymbol{\ddot{u}}}-\nabla \cdot \left[ \frac{1}{2}(\nabla \mathit{\boldsymbol{u}}+\mathit{\boldsymbol{u}}\nabla ):\ \mathit{\boldsymbol{a}} \right]-\mathit{\boldsymbol{f}} \right\}\rm{d}V}+ \\ & \ \ \ \iint\limits_{{{S}_{\sigma }}}{\left[ \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\rm{d}S=0。 \\ \end{align} $

(22)

去掉积分号, 可得弹性动力学方程:

$ \rho \mathit{\boldsymbol{\ddot{u}}}-\nabla \cdot \left[ \frac{1}{2}(\nabla \mathit{\boldsymbol{u}}+\mathit{\boldsymbol{u}}\nabla ):\ \mathit{\boldsymbol{a}} \right]-\mathit{\boldsymbol{f}}=\bf{0}, $

(23)

$ \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right):\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}}=\bf{0}, $

(24)

先决条件为式(13)。

2.2 两类变量Lagrange方程应用于线性弹性动力学

两类变量Lagrange方程表示为

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=0, $

(25)

弹性动力学的动能表示为

$ T=\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{v}}\cdot \mathit{\boldsymbol{v}}\rm{d}V}, $

(26)

弹性动力学的势能表示为

$ \begin{align} & U={{U}_{1}}+{{U}_{2}} \\ & \ \ \ =\iiint\limits_{V}{(\frac{1}{2}\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }}-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}})\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\ \rm{d}S。 \\ \end{align} $

(27)

其中, ε为应变张量, v为速度适量, ${{U}_{1}}=$ $\iiint\limits_{V}{\frac{1}{2}\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }}\rm{d}V$为弹性体的应变能, ${{U}_{2}}=$ $-\iiint\limits_{V}{\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}}\rm{d}V}-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\rm{d}S$为外力势能。先决条件为

$ \mathit{\boldsymbol{v}}-\mathit{\boldsymbol{\dot{u}}}=\bf{0}, $

(28)

$ \mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }-\frac{1}{2}\nabla \mathit{\boldsymbol{u}}-\frac{1}{2}\mathit{\boldsymbol{u}}\nabla =\bf{0}, $

(29)

$ \mathit{\boldsymbol{u}}-\mathit{\boldsymbol{ }}\!\!\bar{\mathit{\boldsymbol{u}}}\!\!\rm{ }=\bf{0} 。 $

(30)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial \mathit{\boldsymbol{u}}}=0 , $

(31)

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}=\frac{\rm{d}}{\rm{d}t}\frac{\partial }{\partial \mathit{\boldsymbol{v}}}\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{v}}\cdot \mathit{\boldsymbol{v}}\rm{d}V}=\iiint\limits_{V}{\rho \mathit{\boldsymbol{\dot{v}}}\rm{d}V}。 $

(32)

势能变导项的推导比较复杂:

$ \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left[ \iiint\limits_{V}{\left( \frac{1}{2}\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}} \right)}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}\ }\rm{d}S \right], $

(33)

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{\left[ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left( \frac{1}{2}\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ } \right)-\mathit{\boldsymbol{f}} \right]}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S \\ & =\iiint\limits_{V}{\left( (\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\frac{\partial \mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }}{\partial \mathit{\boldsymbol{u}}}-\mathit{\boldsymbol{f}} \right)}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S, \\ \end{align} $

(34)

考虑到

$ \rm{ }\!\!\delta\!\!\rm{ }\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }-\rm{ }\!\!\delta\!\!\rm{ }\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla \right)=\bf{0}, $

(35)

$ \delta \mathit{\boldsymbol{u}}=\bf{0},\left( 在{\mathit{{S}_{u}}}上 \right), $

(36)

利用刚度系数张量的对称性, 则有

$ \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{(\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\frac{\partial \nabla \mathit{\boldsymbol{u}}}{\partial \mathit{\boldsymbol{u}}}}-\mathit{\boldsymbol{f}})\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S。 $

(37)

应用Green定理, 并考虑到式(36), 可得

$ \iiint\limits_{V}{\left( \mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}:\frac{\partial \nabla \mathit{\boldsymbol{u}}}{\partial \mathit{\boldsymbol{u}}} \right)}\rm{d}V=\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}\rm{d}S}-\iiint\limits_{V}{\nabla \cdot (\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}})\rm{d}V}。 $

(38)

将式(38)代入式(37), 则得

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=-\iiint\limits_{V}{[\nabla \cdot (\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}})+\mathit{\boldsymbol{f}}]}\ \rm{d}V+ \\ & \ \ \ \ \ \ \ \ \ \iint\limits_{{{S}_{\sigma }}}{(\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}})\ }\rm{d}S\ 。 \\ \end{align} $

(39)

将相关各式代入Lagrange方程, 可得

$ \begin{align} & \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{\left[ \rho \frac{\rm{d}\mathit{\boldsymbol{v}}}{\rm{d}t}-\nabla \cdot (\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\mathit{\boldsymbol{a}})-\mathit{\boldsymbol{f}} \right]\rm{d}V}+ \\ & \iint\limits_{{{S}_{\sigma }}}{(\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}})}\ \rm{d}S=\bf{0}, \\ \end{align} $

(40)

去掉积分号, 可得弹性动力学方程:

$ \rho \frac{\rm{d}\mathit{\boldsymbol{v}}}{\rm{d}t}-\nabla \cdot (\mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}})-\mathit{\boldsymbol{f}}=\bf{0}, $

(41)

$ \mathit{\boldsymbol{ }}\!\!\varepsilon\!\!\rm{ }:\ \mathit{\boldsymbol{a}}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}}=\bf{0}, $

(42)

先决条件为式(28), (29)和(30)。

2.3 算例

弹性平板是工程结构中的重要构件, 亦可以单独组成一个工程结构。将Lagrange方程应用于弹性平板的动力学比较有代表性。这里讨论线性弹性平板的动力学问题。

弹性平板的动能表示为

$ T=\iint{\frac{1}{2}}\rho h{{\left( \frac{\partial w}{\partial t} \right)}^{2}}\text{d}x\text{d}y。 $

(43)

弹性平板的势能包括两部分。第一部分为板的应变能:

$ \begin{align} & {{U}_{1}}=\iint{\frac{D}{2}\left\{ {{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)}^{2}}- \right.}\ \\ & \left. \ \ \ \ \ 2(1-\mu )\left[ \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}-{{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}} \right] \right\}\text{d}x\text{d}y; \\ \end{align} $

(44)

第二部分为横向分布载荷的势能:

$ {{U}_{2}}=\iint{-qw\text{d}x\text{d}y}。 $

(45)

总势能为

$ \begin{align} & U={{U}_{1}}+{{U}_{2}} \\ & \ \ \ =\iint{\frac{D}{2}\left\{ {{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)}^{2}}- \right.\ }2(1-\mu )\left[ \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}- \right. \\ & \ \ \ \ \left. \left. {{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}} \right] \right\}\ \text{d}x\text{d}y-\iint{qw\text{d}x\text{d}y}。 \\ \end{align} $

(46)

设为四边简支矩形板, 位移边界条件为

${{\left. w \right|}_{x=0}}=0,{{\left. w \right|}_{y=0}}=0,{{\left. w \right|}_{x=a}}=0,{{\left. w \right|}_{y=b}}=0,$

(47)

力学边界条件为

$ \left\{ \begin{align} & {{\left. \left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\mu \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right) \right|}_{x=0}}=0, \\ & {{\left. \left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\mu \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right) \right|}_{x=a}}=0, \\ & \left. \left( \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+\mu \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right) \right|_{y=0}^{{}}=0, \\ & {{\left. \left( \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+\mu \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right) \right|}_{y-b}}=0\ 。 \\ \end{align} \right. $

(48)

其中, w为板的挠度, D为板的抗弯刚度, h为板的厚度, m为泊松比, r为质量密度, q为分布载荷。

Lagrange方程表示为

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}-\frac{\partial T}{\partial w}+\frac{\partial U}{\partial w}=0。 $

(49)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial w}=0, $

(50)

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}=\frac{\text{d}}{\text{d}t}\frac{\partial }{\partial \dot{w}}\iint{\frac{1}{2}}\rho h{{(\dot{w})}^{2}}\text{d}x\text{d}y=\iint{\rho h\ddot{w}\text{d}x\text{d}y}。 $

(51)

势能的变导项的推导比较复杂:

$ \begin{align} & \frac{\partial U}{\partial w}=\frac{\partial }{\partial w}\iint{\frac{D}{2}\left\{ {{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)}^{2}}- \right.} \\ & \ \ \ \ \ \ \ \ \left. 2(1-\mu )\left[ \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}-{{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}} \right] \right\}\ \text{d}x\text{d}y- \\ & \ \ \ \ \ \ \ \ \frac{\partial }{\partial w}\iint{qw\text{d}x\text{d}y,} \\ \end{align} $

(52)

$ \begin{align} & \frac{\partial U}{\partial w}=\int_{0}^{a}{\int_{0}^{b}{\left\{ D\left[ \left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)- \right. \right.}} \\ & \ \ \ \ \ \ \ \ (1-\mu )\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+ \right.\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}- \\ & \ \ \ \ \ \ \ \ \left. \left. \left. 2\frac{{{\partial }^{2}}w}{\partial x\partial y}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial x\partial y} \right) \right]-q \right\}\text{d}x\text{d}y。 \\ \end{align} $

(53)

由于

$ \begin{align} & \int_{0}^{a}{\int_{0}^{b}{\left\{ D\left[ \left[ \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right]\frac{\partial }{\partial w}\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right)- \right. \right.}}(1-\mu )\cdot \\ & \left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+ \right.\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}-\ \left. \left. \left. 2\frac{{{\partial }^{2}}w}{\partial x\partial y}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial x\partial y} \right) \right] \right\}\ \text{d}x\text{d}y \\ & =\int_{0}^{a}{\int_{0}^{b}{\left\{ D\left[ \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+\mu \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}+ \right. \right.}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. \left. \mu \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}+2(1-\mu )\frac{{{\partial }^{2}}w}{\partial x\partial y}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial x\partial y} \right] \right\}\ \text{d}x\text{d}y, \\ \end{align} $

(54)

应用Green定理:

$ \int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}}}\text{d}x\text{d}y=\int_{0}^{b}{\left. \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{\partial w}{\partial x} \right|_{0}^{a}}\text{d}y-\int_{0}^{b}{\left. \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}\frac{\partial }{\partial w}w \right|_{0}^{a}\text{d}y+}\int_{0}^{b}{\int_{0}^{a}{\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y}}, $

(55)

$ \int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}}}\text{d}x\text{d}y=\int_{0}^{a}{\left. \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{\partial w}{\partial y} \right|_{0}^{b}}\text{d}x-\int_{0}^{a}{\left. \frac{{{\partial }^{3}}w}{\partial {{y}^{3}}}\frac{\partial }{\partial w}w \right|_{0}^{b}\text{d}x+}\int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y}}, $

(56)

$ \int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}}}\text{d}x\text{d}y=\int_{0}^{a}{\left. \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{\partial }{\partial w}\frac{\partial w}{\partial y} \right|_{0}^{b}}\text{d}x-\int_{0}^{a}{\left. \frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}\frac{\partial }{\partial w}w \right|_{0}^{b}\text{d}x}+\int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y}}, $

(57)

$ \int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}}}\text{d}x\text{d}y=\int_{0}^{b}{\left. \frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}\frac{\partial }{\partial w}\frac{\partial w}{\partial x} \right|_{0}^{a}}\text{d}y-\int_{0}^{b}{\left. \frac{{{\partial }^{3}}w}{\partial {{y}^{2}}\partial x}\frac{\partial }{\partial w}w \right|_{0}^{a}\text{d}y}+\int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y\ }}, $

(58)

$ \begin{align} & \int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{2}}w}{\partial x\partial y}\frac{\partial }{\partial w}\frac{{{\partial }^{2}}w}{\partial x\partial y}}}\text{d}x\text{d}y=\int_{0}^{a}{\left. \frac{{{\partial }^{2}}w}{\partial x\partial y}\frac{\partial }{\partial w}\frac{\partial w}{\partial x} \right|_{0}^{b}}\text{d}x- \\ & \int_{0}^{b}{\left. \frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}\frac{\partial }{\partial w}w \right|_{0}^{a}\text{d}y+}\int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y}} \\ & =\left. \frac{{{\partial }^{2}}w(x,b)}{\partial x\partial y}\frac{\partial }{\partial w}w(x,b) \right|_{0}^{a}-\left. \frac{{{\partial }^{2}}w(x,0)}{\partial x\partial y}\frac{\partial }{\partial w}w(x,0) \right|_{0}^{a}- \\ & \ \ \ \int_{0}^{a}{\left. \frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}\frac{\partial }{\partial w}w \right|_{0}^{b}\text{d}x}-\int_{0}^{b}{\left. \frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}\frac{\partial }{\partial w}w \right|_{0}^{a}\text{d}y+}\int_{0}^{a}{\int_{0}^{b}{\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}\frac{\partial }{\partial w}w\text{d}x\text{d}y,}} \\ \end{align} $

(59)

将式(54)~(59)代入式(53), 考虑到式(47)和(48), 则

$ \frac{\partial U}{\partial w}=\int_{0}^{a}{\int_{0}^{b}{\left[ D\left( \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}} \right)-q \right]\text{d}x\text{d}y}}。 $

(60)

将相关各式代入Lagrange方程, 可得

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}-\frac{\partial T}{\partial w}+\frac{\partial U}{\partial w}=\int_{0}^{a}{\int_{0}^{b}{\left[ \rho h\ddot{w}+D\left( \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}} \right)-q \right]}}\ \text{d}x\text{d}y=0。 $

(61)

去掉积分号, 可得弹性动力学方程：

$ \rho h\ddot{w}+D\left( \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}} \right)-q=0。 $

(62)

位移边界条件为式(47), 力学边界条件为式(48), 都是强制边界条件。当然, 也可以将力学边界条件处理为自然边界条件, 这需要通过变导运算获得。

3 Lagrange方程应用于非线性弹性动力学 3.1 一类变量Lagrange方程应用于非线性弹性动力学

非线性弹性动力学的动能表示为

$ T=\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{\dot{u}}}\cdot \mathit{\boldsymbol{\dot{u}}}\rm{d}V}, $

(63)

非线性弹性动力学的势能表示为

$ \begin{align} & U={{U}_{1}}+{{U}_{2}} \\ & \ \ \ =\iiint\limits_{V}{\left[ A\left( \frac{1}{2}\nabla \mathit{\pmb{u}}+\frac{1}{2}\mathit{\pmb{u}}\nabla +\frac{1}{2}\nabla \mathit{\pmb{u}}\cdot \mathit{\pmb{u}}\nabla \right)-\mathit{\pmb{f}}\cdot \mathit{\pmb{u}} \right]}\text{d}V- \\ & \ \ \ \ \ \iint\limits_{{{S}_{\sigma }}}{\mathit{\pmb{T}}\cdot \mathit{\pmb{u}}}\text{d}S, \\ \end{align} $

(64)

其中, ${{U}_{1}}=\iiint\limits_{V}{A\left (\frac{1}{2}\nabla \mathit{\pmb{u}}+\frac{1}{2}\mathit{\pmb{u}}\nabla +\frac{1}{2}\nabla \mathit{\pmb{u}}\cdot \mathit{\pmb{u}}\nabla \right)}\ \text{d}V$为弹性体的应变能; ${{U}_{2}}=-\iiint\limits_{V}{\mathit{\pmb{f}}\cdot \mathit{\pmb{u}}\text{d}V}-\iint\limits_{{{S}_{\sigma }}}{\mathit{\pmb{T}}\cdot \mathit{\pmb{u}}}\text{d}S$为外力势能。

位移边界条件为

$ \mathit{\boldsymbol{u}}-\mathit{\boldsymbol{ }}\!\!\bar{\mathit{\boldsymbol{u}}}\!\!\rm{ }=\bf{0} 。 $

(65)

Lagrange方程表示为

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{\dot{u}}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=0。 $

(66)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial \mathit{\boldsymbol{u}}}=0, $

(67)

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{\dot{u}}}}=\frac{\rm{d}}{\rm{d}t}\frac{\partial }{\partial \mathit{\boldsymbol{\dot{u}}}}\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{\dot{u}\dot{u}}}\rm{d}V}=\iiint\limits_{V}{\rho \mathit{\boldsymbol{\ddot{u}}}\rm{d}\mathit{V}}。 $

(68)

势能变导项的推导比较复杂:

$ \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left\{ \iiint\limits_{V}{\left[ A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}} \right]}\ \rm{d}V \right.\left. -\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\ \rm{d}S \right\}, $

(69)

$ \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{\left[ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)-\mathit{\boldsymbol{f}} \right]}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S。 $

(70)

由于

$ \begin{align} & \iiint\limits_{V}{\frac{\partial }{\partial \mathit{\boldsymbol{u}}}A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\ \rm{d}V= \\ & \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)} \right]}:\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\mathit{\boldsymbol{u}}\nabla \rm{d}V, \\ \end{align} $

(71)

应用Green定理, 并考虑到式(65), 可得

$ \begin{align} & \ \ \ \ \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)} \right]}:\ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}\mathit{\boldsymbol{u}}\nabla \rm{d}V \\ & =\iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \mathit{\boldsymbol{n}} \right]\rm{d}S}- \\ & \ \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \nabla \right]\rm{d}V}。 \\ \end{align} $

(72)

将式(71)和(72)代入式(70), 则得

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=-\iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \nabla +\mathit{\boldsymbol{f}} \right]\ }\rm{d}V+ \\ & \ \ \ \ \ \ \ \ \ \iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\ \rm{d}S。 \\ \end{align} $

(73)

将相关各式代入Lagrange方程, 可得

$ \begin{align} & \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{\dot{u}}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}= \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \iiint\limits_{V}{\left[ \rho \mathit{\boldsymbol{\ddot{u}}}-(\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \nabla -\mathit{\boldsymbol{f}} \right]\rm{d}V+} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\ \rm{d}S=0。 \\ \end{align} $

(74)

去掉积分号, 可得弹性动力学方程:

$ \rho \mathit{\boldsymbol{\ddot{u}}}-(\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \nabla -\mathit{\boldsymbol{f}}=\bf{0}, $

(75)

$ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A\left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}{\partial \left( \frac{1}{2}\nabla \mathit{\boldsymbol{u}}+\frac{1}{2}\mathit{\boldsymbol{u}}\nabla +\frac{1}{2}\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla \right)}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}}=\bf{0}, $

(76)

先决条件为式(65)。

3.2 两类变量Lagrange方程应用于非线性弹性动力学

非线性弹性动力学的动能表示为

$ T=\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{v}}\cdot \mathit{\boldsymbol{v}}\rm{d}\mathit{V}}。 $

(77)

非线性弹性动力学的势能表示为

$ U={{U}_{1}}+{{U}_{2}}=\iiint\limits_{V}{[A(\mathit{\boldsymbol{E}})}-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}}]\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\rm{d}\mathit{S}, $

(78)

其中, ${{U}_{1}}=\iiint\limits_{V}{A (\mathit{\boldsymbol{E}})}\ \rm{d}V$为弹性体的应变能, ${{U}_{2}}=$ $-\iiint\limits_{V}{\mathrm{f}\cdot \mathit{\pmb{u}}\text{d}V}-\iint\limits_{{{S}_{\sigma }}}{\mathit{\pmb{T}}\cdot \mathit{\pmb{u}}}\text{d}S$为外力势能。先决条件为

$ \mathit{\boldsymbol{v}}-\mathit{\boldsymbol{\dot{u}}}=\bf{0}, $

(79)

$ \mathit{\boldsymbol{E}}-\frac{1}{2}(\nabla \mathit{\boldsymbol{u}}+\mathit{\boldsymbol{u}}\nabla +\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla )=\bf{0}, $

(80)

$ \mathit{\boldsymbol{u}}-\mathit{\boldsymbol{ }}\!\!\bar{\mathit{\boldsymbol{u}}}\!\!\rm{ }=\bf{0}, $

(81)

其中, E为Green应变张量, u为位移矢量, f为体力矢量, T为面力矢量, v为速度矢量, ρ为质量密度, A(E)为应变能函数, n为法向矢量, ▽为Hamilton算子, S_σ为力的边界, S_u为位移边界, V为空间体积域。

Lagrange方程表示为

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=0。 $

(82)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial \mathit{\boldsymbol{u}}}=0, $

(83)

$ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}=\frac{\rm{d}}{\rm{d}t}\frac{\partial }{\partial \mathit{\boldsymbol{v}}}\iiint\limits_{V}{\frac{1}{2}\rho \mathit{\boldsymbol{v}}\cdot \mathit{\boldsymbol{v}}\rm{d}V}=\iiint\limits_{V}{\rho \mathit{\boldsymbol{\dot{v}}}\rm{d}\mathit{V}}。 $

(84)

势能变导项的推导较为复杂:

$ \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\left\{ \iiint\limits_{V}{[A(E)}-\mathit{\boldsymbol{f}}\cdot \mathit{\boldsymbol{u}}]\rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}\cdot \mathit{\boldsymbol{u}}}\rm{d}S \right\}, $

(85)

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=\iiint\limits_{V}{\left[ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}A(\mathit{\boldsymbol{E}})-\mathit{\boldsymbol{f}} \right]}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S \\ & \ \ \ \ \ =\iiint\limits_{V}{\left[ \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial \mathit{\boldsymbol{E}}}:\frac{\partial \mathit{\boldsymbol{E}}}{\partial \mathit{\boldsymbol{u}}}-\mathit{\boldsymbol{f}} \right]}\ \rm{d}V-\iint\limits_{{{S}_{\sigma }}}{\mathit{\boldsymbol{T}}}\rm{d}S。 \\ \end{align} $

(86)

考虑到

$ \rm{ }\!\!\delta\!\!\rm{ }\mathit{\boldsymbol{E}}-\rm{ }\!\!\delta\!\!\rm{ }\frac{1}{2}(\nabla \mathit{\boldsymbol{u}}+\mathit{\boldsymbol{u}}\nabla +\nabla \mathit{\boldsymbol{u}}\cdot \mathit{\boldsymbol{u}}\nabla )=0, $

(87)

则有

$ \begin{align} & \iiint\limits_{V}{\frac{\partial }{\partial \mathit{\boldsymbol{u}}}A(\mathit{\boldsymbol{E}})}\rm{d}V=\iiint\limits_{V}{\frac{\partial A(\mathit{\boldsymbol{E}})}{\partial \mathit{\boldsymbol{E}}}:\frac{\partial \mathit{\boldsymbol{E}}}{\partial \mathit{\boldsymbol{u}}}}\rm{d}V \\ & =\iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})} \right]}:\frac{\partial }{\partial \mathit{\boldsymbol{u}}}\mathit{\boldsymbol{u}}\nabla \rm{d}V。 \\ \end{align} $

(88)

应用Green定理, 并且考虑到

$ \delta \mathit{\boldsymbol{u}}=0\ \ \ \left( 在{{S}_{u}}上 \right), $

(89)

可得

$ \begin{align} & \ \ \ \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})} \right]}:\ \frac{\partial }{\partial \mathit{\boldsymbol{u}}}\mathit{\boldsymbol{u}}\nabla \rm{d}V \\ & =\iint\limits_{{{S}_{\sigma }}+{{S}_{u}}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \mathit{\boldsymbol{n}}\frac{\partial \mathit{\boldsymbol{u}}}{\partial \mathit{\boldsymbol{u}}} \right]\rm{d}S}- \\ & \ \ \ \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \nabla \right]\rm{d}V} \\ & =\iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \mathit{\boldsymbol{n}} \right]\rm{d}S}- \\ & \ \ \ \iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \nabla \right]\rm{d}\mathit{V}}。 \\ \end{align} $

(90)

将式(88)和(90)代入式(86), 则得

$ \begin{align} & \frac{\partial U}{\partial \mathit{\boldsymbol{u}}}=-\iiint\limits_{V}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \nabla +\mathit{\boldsymbol{f}} \right]\ }\rm{d}V+ \\ & \ \ \ \ \ \ \ \ \ \ \ \iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\ \rm{d}\mathit{S}。 \\ \end{align} $

(91)

将相关各式代入Lagrange方程, 可得

$ \begin{align} & \ \ \ \frac{\rm{d}}{\rm{d}t}\frac{\partial T}{\partial \mathit{\boldsymbol{v}}}-\frac{\partial T}{\partial \mathit{\boldsymbol{u}}}+\frac{\partial U}{\partial \mathit{\boldsymbol{u}}} \\ & =\iiint\limits_{V}{\left[ \rho \mathit{\boldsymbol{\dot{v}}}-(\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \nabla -\mathit{\boldsymbol{f}} \right]\rm{d}V}+ \\ & \ \ \ \iint\limits_{{{S}_{\sigma }}}{\left[ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}} \right]}\ \rm{d}\mathit{S}=0。 \\ \end{align} $

(92)

去掉积分号, 可得弹性动力学方程:

$ \rho \mathit{\boldsymbol{\dot{v}}}-(\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \nabla -\mathit{\boldsymbol{f}}=\bf{0}, $

(93)

$ (\mathit{\boldsymbol{I}}+\mathit{\boldsymbol{u}}\nabla )\cdot \frac{\partial A(\mathit{\boldsymbol{E}})}{\partial (\mathit{\boldsymbol{E}})}\cdot \mathit{\boldsymbol{n}}-\mathit{\boldsymbol{T}}=\bf{0}\ 。 $

(94)

3.3 算例

非线性弹性直梁也是工程结构中的重要构件, 亦可以单独组成一个工程结构。将Lagrange方程应用于非线性弹性直梁的动力学问题, 也具有一定的代表性。这里讨论非线性弹性Bernoulli直梁的动力学问题。

非线性弹性直梁的动能表示为

$ T=\int_{0}^{l}{\left[ \frac{1}{2}\rho A{{\left( \frac{\partial w}{\partial t} \right)}^{2}}+\frac{1}{2}\rho A{{\left( \frac{\partial u}{\partial t} \right)}^{2}} \right]}\ \text{d}x。 $

(95)

非线性弹性直梁的势能包括三部分。第一部分为梁的应变能:

${{U}_{1}}=\int_{0}^{l}{\frac{1}{2}EI{{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)}^{2}}\text{d}x}。$

(96)

第二部分是轴向拉力的势能, 包括拉伸应变能和拉弯耦合势能:

${{U}_{2}}=\int_{0}^{l}{\frac{1}{2}EA{{\left( \frac{\partial u}{\partial x} \right)}^{2}}\text{d}x},$

(97)

${{U}_{3}}=\int_{0}^{l}{\frac{1}{2}EA\left[ \frac{\partial u}{\partial x}{{\left( \frac{\partial w}{\partial x} \right)}^{2}}+\frac{1}{4}{{\left( \frac{\partial w}{\partial x} \right)}^{4}} \right]\text{d}x}。$

(98)

第三部分为横向分布载荷的势能:

${{U}_{4}}=-\int_{0}^{l}{qw\text{d}x}。$

(99)

总势能为

$ \begin{align} & U={{U}_{1}}+{{U}_{2}}+{{U}_{3}}+{{U}_{4}} \\ & \ \ \ =\int_{0}^{l}{\left[ \frac{1}{2}EI{{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)}^{2}}+\frac{1}{2}EA{{\left( \frac{\partial u}{\partial x} \right)}^{2}}+ \right.}\ \\ & \left. \ \ \ \ \ \frac{1}{2}EA\frac{\partial u}{\partial x}{{\left( \frac{\partial w}{\partial x} \right)}^{2}}+\frac{1}{2}EA\frac{1}{4}{{\left( \frac{\partial w}{\partial x} \right)}^{4}}-qw \right]\text{d}x\ 。 \\ \end{align} $

(100)

其中, w为梁的挠度, u为梁的轴向变形, E为梁的弹性模量, A为梁的横截面积, I为梁的抗弯截面系数, r为质量密度, q为分布载荷。

设为悬臂梁, 位移边界条件为

${{\left. w \right|}_{x=0}}=0,\ \ {{\left. \frac{\partial w}{\partial x} \right|}_{x=0}}=0,\ \ {{\left. u \right|}_{x=0}}=0,$

(101)

Lagrange方程表示为

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}-\frac{\partial T}{\partial w}+\frac{\partial U}{\partial w}=0, $

(102)

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{u}}-\frac{\partial T}{\partial u}+\frac{\partial U}{\partial u}=0, $

(103)

推导计算Lagrange方程中的各项:

$ \frac{\partial T}{\partial w}=0, $

(104)

$ \frac{\partial T}{\partial u}=0, $

(105)

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}=\frac{\text{d}}{\text{d}t}\frac{\partial }{\partial \dot{w}}\int_{0}^{l}{\frac{1}{2}\rho A{{(\dot{w})}^{2}}\text{d}x}=\int_{0}^{l}{\rho A\ddot{w}\text{d}x}, $

(106)

$ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{u}}=\frac{\text{d}}{\text{d}t}\frac{\partial }{\partial \dot{u}}\int_{0}^{l}{\frac{1}{2}\rho A{{(\dot{u})}^{2}}\text{d}x}=\int_{0}^{l}{\rho A\ddot{u}\text{d}x}。 $

(107)

势能的变导项的推导比较复杂:

$ \begin{align} & \frac{\partial U}{\partial w}=\frac{\partial }{\partial w}\int_{0}^{l}{\left[ \frac{1}{2}EI{{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)}^{2}}+\frac{1}{2}EA{{\left( \frac{\partial u}{\partial x} \right)}^{2}}+ \right.} \\ & \ \ \ \ \ \ \ \ \left. \frac{1}{2}EA\frac{\partial u}{\partial x}{{\left( \frac{\partial w}{\partial x} \right)}^{2}}+\frac{1}{2}EA\frac{1}{4}{{\left( \frac{\partial w}{\partial x} \right)}^{4}}-qw \right]\ \text{d}x, \\ \end{align} $

(108)

$ \begin{align} & \frac{\partial U}{\partial u}=\frac{\partial }{\partial u}\int_{0}^{l}{\left[ \frac{1}{2}EI{{\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)}^{2}}+\frac{1}{2}EA{{\left( \frac{\partial u}{\partial x} \right)}^{2}}+ \right.} \\ & \ \ \ \ \ \ \ \ \left. \frac{1}{2}EA\frac{\partial u}{\partial x}{{\left( \frac{\partial w}{\partial x} \right)}^{2}}-qw \right]\ \text{d}x。 \\ \end{align} $

(109)

式(108)和(109)可以进一步表示为

$ \begin{align} & \\ & \frac{\partial U}{\partial u}=\int_{0}^{l}{\left[ EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)+EA\frac{\partial u}{\partial x}\left( \frac{\partial u}{\partial x} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right) \right.} \\ & \left. \ \ \ \ \ \ \ \ \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}}\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right)-q \right]\text{d}x, \\ \end{align} $

(110)

$ \frac{\partial U}{\partial u}=\int_{0}^{l}{\left[ EA\left( \frac{\partial u}{\partial x} \right)\frac{\partial }{\partial u}\left( \frac{\partial u}{\partial x} \right)+\frac{1}{2}\frac{\partial }{\partial u}\left( EA\frac{\partial u}{\partial x} \right){{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right]\text{d}x}。 $

(111)

应用Green定理:

$ \begin{align} & \ \ \ \int_{0}^{l}{EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\text{d}x} \\ & =\left. EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right) \right|_{0}^{l}-\int_{0}^{l}{EI\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right)\text{d}x} \\ & =\left. EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right) \right|_{0}^{l}-EI\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right)\frac{\partial }{\partial w}w\left| _{0}^{l} \right.+ \\ & \int_{0}^{l}{EI\left( \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}} \right)\frac{\partial }{\partial w}w\text{d}x,} \\ \end{align} $

(112)

$ \begin{align} & \int_{0}^{l}{EA\frac{\partial u}{\partial x}\left( \frac{\partial w}{\partial x} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right)\text{d}x} \\ & =EA\frac{\partial u}{\partial x}\left( \frac{\partial w}{\partial x} \right)\frac{\partial }{\partial w}w\left| _{0}^{l} \right.-\int_{0}^{l}{\frac{\partial }{\partial x}\left( EA\frac{\partial u}{\partial x}\frac{\partial w}{\partial x} \right)\frac{\partial }{\partial w}w\text{d}x}, \\ \end{align} $

(113)

$ \begin{align} & \int_{0}^{l}{\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}}\frac{\partial }{\partial w}\left| \frac{\partial w}{\partial x} \right|\text{d}x} \\ & =\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}}\frac{\partial }{\partial w}w\left| _{0}^{l} \right.-\int_{0}^{l}{\frac{\partial }{\partial x}\left[ \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}} \right]\frac{\partial }{\partial w}w\text{d}x}, \\ \end{align} $

(114)

$ \begin{align} & \ \ \ \int_{0}^{l}{EA\left( \frac{\partial u}{\partial x} \right)\frac{\partial }{\partial u}\left( \frac{\partial u}{\partial x} \right)\text{d}x} \\ & =EA\left( \frac{\partial u}{\partial x} \right)\frac{\partial }{\partial u}u\left| _{0}^{l} \right.-\int_{0}^{l}{EA\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}\frac{\partial }{\partial u}u\text{d}x} \\ & =EA\frac{\partial u}{\partial x}\frac{\partial }{\partial u}u\left| _{0}^{l} \right.-\int_{0}^{l}{\frac{\partial }{\partial x}EA\frac{\partial u}{\partial x}\frac{\partial }{\partial u}u\text{d}x,} \\ \end{align} $

(115)

$ \begin{align} & \int_{0}^{l}{\frac{1}{2}\frac{\partial }{\partial u}\left( EA\frac{\partial u}{\partial x} \right){{\left( \frac{\partial w}{\partial x} \right)}^{2}}\text{d}x} \\ & =\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}}\frac{\partial }{\partial u}u\left| _{0}^{l} \right.-\int_{0}^{l}{\frac{\partial }{\partial x}\left[ \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right]\frac{\partial }{\partial u}u\text{d}x}\ 。 \\ \end{align} $

(116)

将式(112)~(116)代入式(110)和(111), 根据边界条件(式(101)), 并且考虑到在$l=x$处, $\frac{\partial }{\partial w}w, $$\frac{\partial }{\partial w}\left (\frac{\partial w}{\partial x} \right)$和$\frac{\partial }{\partial u}u$取定值, 可得

$ \begin{align} & \frac{\partial U}{\partial w}=\int_{0}^{l}{\left\{ EI\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}-\frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}\frac{\partial w}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}} \right]-q \right\}} \\ & \ \ \ \ \ \ \ \ \text{d}x+{{\left. EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left[ \frac{\partial w}{\partial x} \right] \right|}_{x=l}}-EI\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right)\frac{\partial }{\partial w}w\left| _{x=l} \right.+ \\ & {{\left. \ \ \ \ \ \ \ EA\frac{\partial u}{\partial x}\left( \frac{\partial w}{\partial x} \right)\frac{\partial }{\partial w}w \right|}_{x=l}}+\ {{\left. \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}}\frac{\partial }{\partial w}w \right|}_{x=l}}, \\ \end{align} $

(117)

$ \begin{align} & \frac{\partial U}{\partial u}=-\int_{0}^{l}{\frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right]\text{d}x+} \\ & {{\left. EA\frac{\partial u}{\partial x}\frac{\partial }{\partial u}u \right|}_{x=l}}+{{\left. \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}}\frac{\partial }{\partial u}u \right|}_{x=l}}。 \\ \end{align} $

(118)

将相关各式代入Lagrange方程, 可得

$ \begin{align} & \ \ \ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{w}}-\frac{\partial T}{\partial w}+\frac{\partial U}{\partial w} \\ & =\int_{0}^{l}{\left\{ \rho A\ddot{w}+EI\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}- \right.} \\ & \left. \ \ \ \frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}\frac{\partial w}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}} \right]-q \right\}\ \text{d}x+ \\ & {{\left. \ \ \ EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right)\frac{\partial }{\partial w}\left( \frac{\partial w}{\partial x} \right) \right|}_{x=l}}-{{\left. EI\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right)\frac{\partial }{\partial w}w \right|}_{x=l}}+ \\ & {{\left. \ \ \ EA\frac{\partial u}{\partial x}\left( \frac{\partial w}{\partial x} \right)\frac{\partial }{\partial w}w \right|}_{x=l}}+{{\left. \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{3}}\frac{\partial }{\partial w}w \right|}_{x=l}}=0, \\ \end{align} $

(119)

$ \begin{align} & \ \ \ \frac{\text{d}}{\text{d}t}\frac{\partial T}{\partial \dot{u}}-\frac{\partial T}{\partial u}+\frac{\partial U}{\partial u} \\ & =\int_{0}^{l}{\left\{ \rho A\ddot{u}-\frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right] \right\}\text{d}x}+ \\ & EA\frac{\partial u}{\partial x}\frac{\partial }{\partial u}u\left| _{x=l} \right.+{{\left. \frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}}\frac{\partial }{\partial u}u \right|}_{x=l}}=0\ 。 \\ \end{align} $

(120)

去掉积分号, 可得弹性直梁的动力学方程和自然边界条件:

$ \rho A\ddot{w}+EI\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}-\frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right]\frac{\partial w}{\partial x}-q=\text{0}, $

(121)

$ \rho A\ddot{u}-\frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}+\frac{1}{2}EA{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right]=0, $

(122)

$ {{\left. EI\left( \frac{{{\partial }^{2}}w}{\partial {{x}^{2}}} \right) \right|}_{x=l}}=0, $

(123)

$ -EI\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right)+{{\left. EA\left[ \frac{\partial u}{\partial x}\left( \frac{\partial w}{\partial x} \right)+\frac{1}{2}\left( \frac{{{\partial }^{3}}w}{\partial {{x}^{3}}} \right) \right] \right|}_{x=l}}=0, $

(124)

$ {{\left. EA\left[ \frac{\partial u}{\partial x}+\frac{1}{2}{{\left( \frac{\partial w}{\partial x} \right)}^{2}} \right] \right|}_{x=l}}=0。 $

(125)

式(121)和(122)是域中的控制方程, 式(123)~(125)是力的边界条件。