ON PRINCIPLE OF MINIMUM POTENTIAL ENERGY IN ELASTICITY FROM DEFORMATION OF BARS SUBJECTED TO TENSION

WU Zhigen,1), LI Xiaobao, MENG Zeng, ZHAN Chunxiao

School of Civil Engineering, Hefei University of Technology, Hefei 230009, China

Abstract

The principle of minimum potential energy in elasticity is comparatively difficult to understand. In this paper, by variational analyses on elastic potential energy, external force potential energy and total potential energy of elastic bars subjected to tension, it is obtained that the variation of total potential energy is the second order infinitesimal of the displacement variation or strain variation. Furthermore, when the displacement variation or strain variation equals to zero, the total potential energy reaches the minimum. The principle of minimum potential energy is thereby examined by the analysis of deformation for elastic bars under axial forces, which is helpful to profoundly understand the principle of minimum potential energy in general.

Keywords： elasticity; principle of minimum potential energy; elastic deformation; virtual displacement

WU Zhigen, LI Xiaobao, MENG Zeng, ZHAN Chunxiao. ON PRINCIPLE OF MINIMUM POTENTIAL ENERGY IN ELASTICITY FROM DEFORMATION OF BARS SUBJECTED TO TENSION. Mechanics in Engineering, 2021, 43(6): 964-966 DOI:10.6052/1000-0879-21-019

1 线弹性杆的变形分析

$\begin{eqnarray} V_{{\rm e}} =\frac{1}{2}ku^{2} \end{eqnarray}$

$\begin{eqnarray} V_{{\rm p}} =-Fu \end{eqnarray}$

图1

$\begin{eqnarray} \delta V_{{\rm e}} =\frac{1}{2}k(u+\delta u)^{2}-\frac{1}{2}ku^{2}=ku\delta u+\frac{1}{2}k(\delta u)^{2} \end{eqnarray}$

$\begin{eqnarray} \delta V_{{\rm p}} =-F(u+\delta u)+Fu=-F\delta u \end{eqnarray}$

$\begin{eqnarray} \delta (V_{{\rm e}} +V_{{\rm p}} )=ku\delta u+\frac{1}{2}k(\delta u)^{2}-F\delta u \end{eqnarray}$

$\begin{eqnarray} \delta (V_{{\rm e}} +V_{{\rm p}} )=\frac{1}{2}k(\delta u)^{2} \end{eqnarray}$

2 非线弹性杆的变形分析

$\begin{eqnarray} f=f(u_{f} ) \end{eqnarray}$

$\begin{eqnarray} V_{{\rm e}} =\int_0^u {f(u_{f} )} {\rm d}u_{f} \end{eqnarray}$

$\begin{eqnarray} &&\delta V_{{\rm e}} =\int_0^{u+\delta u} {f(u_{f} )}{\rm d}u_{f} -\int_0^u {f(u_{f} )} {\rm d}u_{f} =\\&&\qquad \int_u^{u+\delta u} {f(u_{f} )} {\rm d}u_{f} \end{eqnarray}$

$\begin{eqnarray} \int_u^{u+\delta u} {f(u_{f} )} {\rm d}u_{f} =f(\xi )\delta u \end{eqnarray}$

$u<\xi <u+\delta u\ (\delta u>0)$

$u+\delta u<\xi <u\ (\delta u<0)$

$\begin{eqnarray} &&\delta(V_{{\rm e}} +V_{{\rm p}} )=f(\xi )\delta u-f(u)\delta u=\\&&\qquad [f(\xi )-f(u)]\delta u \end{eqnarray}$

3 基于线弹性应变能的分析

$\begin{eqnarray} V_{\varepsilon } =\frac{1}{2}E\int_V {\varepsilon_{x}^{2} {\rm d}V} \end{eqnarray}$

$\begin{eqnarray} &&\delta V_{\varepsilon } =\frac{1}{2}E\int_V {[(\varepsilon_{x} +\delta \varepsilon_{x} )^{2}-\varepsilon_{x}^{2} ]{\rm d}V} =\\&&\qquad E\int_V {\varepsilon_{x} \delta \varepsilon_{x} {\rm d}V+} \frac{1}{2}E\int_V {(\delta \varepsilon_{x} )^{2}{\rm d}V} =\\&&\qquad\int_V {\sigma_{x} \delta \varepsilon_{x} {\rm d}V} +\frac{1}{2}E\int_V {(\delta \varepsilon_{x} )^{2}{\rm d}V} \end{eqnarray}$

$\begin{eqnarray} \delta V_{{\rm p}} =-F\delta u-\int_V {\rho g} \delta u_{x} {\rm d}V \end{eqnarray}$

$\begin{eqnarray} F\delta u+\int_V {\rho g} \delta u_{x} {\rm d}V=\int_V{\sigma_{x} } \delta \varepsilon_{x} {\rm d}V \end{eqnarray}$

$\begin{eqnarray} \delta(V_{\varepsilon } +V_{{\rm p}} )=\frac{1}{2}E\int_V{(\delta \varepsilon_{x} )^{2}{\rm d}V} \end{eqnarray}$

4 结束语

(1)通过给出弹性杆在轴向受拉情况下的弹性势能和外力势能以及它们的变分和总势能变分的准确形式,得出：无论是线性还是非线性,其总势能变分都是位移变分或应变变分的二阶无穷小量,只有在位移变分或应变变分等于零时总势能变分才取得极小值,均应验了最小势能原理。

(2)计算外力势能时外力是一个恒定的量,文中采用物体的重力作为外力,将重力势能作为外力势能,便于形成外力势能的概念,避免产生外力随物体变形而改变的思维习惯。

(3)虽然从弹性杆的轴向受力变形解释了最小势能原理,对于一般情况下的最小势能原理具有相同实质,有助于理解最小势能原理的真实含义。

参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Xu Zhilun. Elasticity (I), 5th edn. Beijing: Higher Education Press, 2016 (in Chinese)

Wu Jialong. Elasticity, 3rd edn. Beijing: Higher Education Press, 2016 (in Chinese)

Lu Mingwan, Luo Xuefu. Foundations of Elasticity (II), 2nd edn. Beijing: Tsinghua University Press, 2001 (in Chinese)

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