## DEFLECTION CURVE EQUATION OF LINEAR STRENGTHENED ELASTIC-PLASTIC BENDING STRAIGHT BEAM1)

CHEN Yingjie,1), FENG Yongqiang, DONG Jingbo

College of Architectural Engineering and Mechanics, Yanshan University, Qinhuangdao 066004, Hebei, China

Hebei Province Civil Engineering Key Laboratory of Green Construction and Intelligent Operation and Maintenance, Qinhuangdao 066004, Hebei, China

Abstract

Aiming at the incomplete application of materials in elastic-plastic stage. In this paper, by using the principle of elastic-plastic partition minimum potential energy, thecriterion of potential energy and Euler equation of bending straight beam under linear strengthening model are derived. The deflection curve equation of cantilever beam and simply supported beam under concentrated load issolved. The deflection curve equation is put into Matlab software for numerical calculation, and the results are compared with ANSYS. The results show that both the numerical solution and the finite element value meet the allowable error range in practical engineering. The proposed method provides a new idea for solving practical engineering problems.

Keywords： principle of elastic-plastic zone variational minimum potential energy; elastic-plastic potential energy partition criterion; Euler's equation; deflection curve equation

CHEN Yingjie, FENG Yongqiang, DONG Jingbo. DEFLECTION CURVE EQUATION OF LINEAR STRENGTHENED ELASTIC-PLASTIC BENDING STRAIGHT BEAM1). Mechanics in Engineering, 2022, 44(2): 351-357 DOI:10.6052/1000-0879-21-378

## 1 线性强化弹塑性悬臂梁的欧拉方程

### 图1

Fig.1   Linearly strengthened cantilever beam under concentrated load

$\sigma_{{\rm s}} =E\varepsilon$

$\sigma =E_{1} \varepsilon +\sigma_{{\rm s}} \left( {1-\dfrac{E}{E_{1} }}\right)$

$M_{1} =2b\int_0^\eta E k_{1} z^{2}{\rm d}z+2b\int_\eta^{h/2}$

\begin{aligned}M_{1} &=2 b \int_{0}^{\eta} E k_{1} z^{2} \mathrm{~d} z+2 b \int_{\eta}^{h / 2} \sigma z \mathrm{~d} z=\\& \frac{3}{2} M_{\mathrm{e}}\left[1-\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}\right]+\\& \frac{3}{2} M_{\mathrm{e}} \frac{E_{1}}{E}\left[\frac{2}{3}\left(\frac{k_{1}}{k_{\mathrm{e}}}\right)+\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}-1\right]\end{aligned}

\begin{aligned}U_{\mathrm{b}}=& \int^{k_{1}}\left\{\frac{3}{2} M_{\mathrm{e}}\left[1-\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}\right]+\right.\\&\left.\frac{3}{2} M_{\mathrm{e}} \frac{E_{1}}{E}\left[\frac{2}{3}\left(\frac{k_{1}}{k_{\mathrm{e}}}\right)+\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}-1\right]\right\} \mathrm{d} k_{1}=\\& \frac{3}{2} M_{\mathrm{e}}\left(k_{1}+\frac{1}{3} \frac{k_{\mathrm{e}}^{2}}{k_{1}}\right)+\\& \frac{3}{2} M_{\mathrm{e}} \frac{E_{1}}{E}\left(\frac{1}{3} \frac{k_{1}^{2}}{k_{\mathrm{e}}}-\frac{1}{3} \frac{k_{\mathrm{e}}^{2}}{k_{1}}-k_{1}\right)+C\end{aligned}

$C=-\dfrac{3}{2}M_{{\rm e}} k_{{\rm e}} \left( {1-\dfrac{E_{1} }{E}} \right)$

\begin{aligned}\Pi_{p}=& \int_{0}^{\xi} \mathrm{d} x \int^{k_{1}}\left\{\frac{3}{2} M_{\mathrm{e}}\left[1-\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}\right]+\right.\\&\left.\frac{3}{2} M_{\mathrm{e}} \frac{E_{1}}{E}\left[\frac{2}{3}\left(\frac{k_{1}}{k_{\mathrm{e}}}\right)+\frac{1}{3}\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)^{2}-1\right]\right\} \mathrm{d} k_{1}+\\& \int_{\xi}^{1} \frac{1}{2} E J\left(\frac{\mathrm{d}^{2} \omega_{2}}{\mathrm{~d} x^{2}}\right) \mathrm{d} x-P \omega_{2(x=l)}\end{aligned}

$\delta \Pi=\delta_{\eta} \Pi_{\mathrm{p}}+\delta_{\xi} \Pi_{\mathrm{p}}+\delta_{\omega} \Pi_{\mathrm{p}}=0$

$\eta =\dfrac{\sigma_{{\rm s}} }{Ek_{1} }$

$\begin{array}{c}\delta_{\xi} \Pi_{p}=\left[\frac{3}{2} M_{\mathrm{e}}\left(k_{1}+\frac{1}{3} \frac{k_{\mathrm{e}}^{2}}{k_{1}}\right)+\right. \\\frac{3}{2} M_{\mathrm{e}} \frac{E_{1}}{E}\left(\frac{1}{3} \frac{k_{1}^{2}}{k_{\mathrm{e}}}-\frac{1}{3} \frac{k_{\mathrm{e}}^{2}}{k_{1}}-k_{1}\right)- \\\left.\frac{3}{2} M_{\mathrm{e}} k_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)\right]_{x=\xi}^{-}= \\\frac{1}{2} E J\left(\frac{\mathrm{d}^{2} \omega_{2}}{\mathrm{~d} x^{2}}\right)_{x=\xi}^{2}=0\end{array}$

$\dfrac{1}{2}M_{{\rm e}} k_{{\rm e}} -\dfrac{1}{2}EJ\left({\dfrac{{\rm d}^{2}\omega_{2} }{{\rm d}x^{2}}} \right)^{2}=0$

$\begin{array}{c}\frac{\mathrm{d}^{2}}{\mathrm{~d} x^{2}}\left[\frac{3}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)-\frac{1}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)+\right. \\\left.M_{\mathrm{e}} k_{\mathrm{e}} \frac{E_{1}}{E} \frac{k_{1}}{k_{\mathrm{e}}}\right]=0, \quad 0 \leqslant x \leqslant \xi\end{array}$
$\begin{array}{r}{\left[\frac{3}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)-\frac{1}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)+\right.} \\\left.M_{\mathrm{e}} k_{\mathrm{e}} \frac{E_{1}}{E} \frac{k_{1}}{k_{\mathrm{e}}}\right]_{x=\xi}+\left(E J \frac{\mathrm{d}^{2} \omega_{2}}{\mathrm{~d} x^{2}}\right)_{x=\xi}=0\end{array}$
$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{3}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)-\frac{1}{2} M_{\mathrm{e}}\left(1-\frac{E_{1}}{E}\right)\left(\frac{k_{\mathrm{e}}}{k_{1}}\right)+\right. \\\left.M_{\mathrm{e}} k_{\mathrm{e}} \frac{E_{1}}{E} \frac{k_{1}}{k_{\mathrm{e}}}\right]_{x=\xi}+\left(E J \frac{\mathrm{d}^{3} \omega_{2}}{\mathrm{~d} x^{3}}\right)_{x=\xi}=0\end{array}$
$E J \frac{\mathrm{d}^{4} \omega_{2}}{\mathrm{~d} x^{4}}=0, \quad \xi \leqslant x \leqslant l$
$\left(E J \frac{\mathrm{d}^{2} \omega_{2}}{\mathrm{~d} x^{2}} \delta \frac{\mathrm{d} \omega_{2}}{\mathrm{~d} x}\right)_{x=\xi}=0$
$E J\left(\frac{\mathrm{d}^{3} \omega^{2}}{\mathrm{~d} x^{3}}\right)_{x=l}+P=0$

$k_{1}=k_{\mathrm{e}} \sqrt{\frac{2}{3-\frac{2 M_{1}}{M_{\mathrm{e}}} \frac{E}{E-E_{1}}+\frac{2 E_{1}}{E-E_{1}} \frac{2 M_{\mathrm{e}}-M_{1}}{M_{\mathrm{e}}}}}$

## 2 集中载荷作用下线性强化弹塑性悬臂梁的计算

### 2.1 挠曲线方程推导

$M_{1} \left( x \right)=C_{1} x+C_{2}$

$M_{1} \left( x \right)=P\left( {l-x} \right)$

$M_{{\rm e}} =\dfrac{bh^{2}}{6}\sigma_{{\rm s}} =P\left( {l-\xi }\right)$

$\eta \left( x \right)=\dfrac{h}{2}\sqrt {3-\dfrac{2\left( {E+E_{1} }\right)\left( {l-x} \right)}{\left( {E-E_{1} } \right)\left( {l-\xi }\right)}+\dfrac{4E_{1} }{E-E_{1} }}$

\begin{aligned}\omega_{1}=& \frac{2 \sigma_{\mathrm{s}}}{E h}\left\{( E - E _ { 1 } ) ( l - \xi ) \left[\left(3 E+E_{1}\right)(l-\xi)-\right.\right.\\&\left.\left.2\left(E+E_{1}\right)(l-x)\right]^{3}\right\}^{1 / 2} /\left[3\left(E+E_{1}\right)^{2}\right]+\\& C_{3} x+C_{4}\end{aligned}

\begin{aligned}\omega_{1}=& \frac{2 \sigma_{\mathrm{s}}}{E h}\left\{( E - E _ { 1 } ) ( l - \xi ) \left[\left(3 E+E_{1}\right)(l-\xi)-\right.\right.\\&\left.\left.2\left(E+E_{1}\right)(l-x)\right]^{3}\right\}^{1 / 2} /\left[3\left(E+E_{1}\right)^{2}\right]-\\& \frac{2 \sigma_{\mathrm{s}}}{E h} x\left\{( E - E _ { 1 } ) ( l - \xi ) \left[\left(3 E+E_{1}\right)(l-\xi)-\right.\right.\\&\left.\left.2 l\left(E+E_{1}\right)\right]\right\}^{1 / 2} /\left(E+E_{1}\right)-\\& \frac{2 \sigma_{\mathrm{s}}}{E h}\left\{( E - E _ { 1 } ) ( l - \xi ) \left[\left(3 E+E_{1}\right)(l-\xi)-\right.\right.\\&\left.\left.2 l\left(E+E_{1}\right)\right]^{3}\right\}^{1 / 2} /\left[3\left(E+E_{1}\right)^{2}\right]\end{aligned}

$\dfrac{{\rm d}^{2}\omega_{2} }{{\rm d}x^{2}}=\dfrac{P}{EJ}\left( {l-x}\right)$

$P=bh^{2}\sigma_{s}/[6(l-\xi)]$,对式(22)进行积分得

$\omega_{2} =\dfrac{\sigma_{{\rm s}} }{Eh\left( {l-\xi } \right)}\left({lx^{2}-\dfrac{1}{3}x^{3}} \right)+C_{5} x+C_{6}$

\begin{aligned}\omega_{2}=& \frac{2 \sigma_{\mathrm{s}} l^{3}}{3 E h(l-\xi)}+\frac{2 \sigma_{\mathrm{s}} l}{E h} \frac{\left(E-E_{1}\right)(l-\xi)}{E+E_{1}}-\frac{2 \sigma_{\mathrm{s}} l^{2}}{E h(l-\xi)} \xi+\frac{\sigma_{\mathrm{s}} l}{E h(l-\xi)} \xi^{2}-\\& \frac{2 \sigma_{\mathrm{s}} l}{E h} \frac{\sqrt{\left(E-E_{1}\right)(l-\xi)\left[\left(3 E+E_{1}\right)(l-\xi)-2 l\left(E+E_{1}\right)\right]}}{E+E_{1}}+\frac{2 \sigma_{\mathrm{s}}}{E h} \frac{\left(E-E_{1}\right)^{2}(l-\xi)^{2}}{3\left(E+E_{1}\right)^{2}}-\\& \frac{2 \sigma_{\mathrm{s}}}{E h} \frac{\left(E-E_{1}\right)(l-\xi)}{E+E_{1}} \xi-\frac{2 \sigma_{\mathrm{s}}}{E h} \frac{\sqrt{\left(E-E_{1}\right)(l-\xi)\left[\left(3 E+E_{1}\right)(l-\xi)-2 l\left(E+E_{1}\right)\right]^{3}}}{3\left(E+E_{1}\right)^{2}}+\\& \frac{\sigma_{\mathrm{s}} l}{E h(l-\xi)} \xi^{2}-\frac{2 \sigma_{\mathrm{s}}}{3 E h(l-\xi)} \xi^{3}\end{aligned}

$\omega_{2(x=0)} =[d\omega_{2}/(dx)]_{x=0}=0$

$\omega_{2(x=l)} =\dfrac{2\sigma_{{\rm s}} l^{3}}{3Eh\left( {l-\xi }\right)}$

### 图2

Fig.2   Deflection distribution at different heights of elastic zone in beam section

## 3 集中载荷作用下线性强化弹塑性简支梁的计算

### 图3

Fig.3   Elastic-plastic simple-supported beam under concentrated load in mid-span

$M_{1} \left( x \right)=C_{1} x+C_{2}$

$M_{1} \left( x \right)=-\dfrac{P}{4}\left( {l-2x} \right)$

$M_{{\rm e}} =\dfrac{bh^{2}}{6}\sigma_{{\rm s}} =\dfrac{P}{4}\left( {l-2\xi }\right)$

$\eta \left( x \right)=\dfrac{h}{2}\sqrt{3-\dfrac{2\left( {E+E_{1} }\right)\left( {l-2x} \right)}{\left( {E-E_{1} } \right)\left( {l-2\xi }\right)}+\dfrac{4E_{1} }{E-E_{1}}}$

$\omega_{1}=\frac{2 \sigma_{\mathrm{s}}}{E h} \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2\left(E+E_{1}\right)(l-2 x)\right]^{3}}{12\left(E+E_{1}\right)^{2}}}+C_{3} x+C_{4}$

$\begin{array}{c}\omega_{1}=\frac{2 \sigma_{\mathrm{s}}}{E h} \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2\left(E+E_{1}\right)(l-2 x)\right]^{3}}{12\left(E+E_{1}\right)^{2}}}- \\\frac{2 \sigma_{\mathrm{s}}}{E h} x \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2 l\left(E+E_{1}\right)\right]}{2\left(E+E_{1}\right)}}+C_{4}\end{array}$

$\dfrac{{\rm d}^{2}\omega_{2} }{{\rm d}x^{2}}=\dfrac{P}{4EJ}\left( {l-2x}\right)$

$P=2 b h^{2} \sigma_{\mathrm{s}} /[3(l-2 \xi)]$,对式(32)进行积分计算,且$x={l}/{2}$处,有$\omega_{2(x={l}/{2})} =0$,得

$C_{6} =-\dfrac{\sigma_{{\rm s}} l^{3}}{6Eh\left( {l-2\xi }\right)}-\dfrac{1}{2}C_{5}$
$\omega_{2} =\dfrac{\sigma_{{\rm s}} }{Eh\left( {l-2\xi } \right)}\left({lx^{2}-\dfrac{2}{3}x^{3}} \right)+\\\qquad C_{5} \left( {x-\dfrac{l}{2}}\right)-\dfrac{\sigma_{{\rm s}} l^{3}}{6Eh\left( {l-2\xi } \right)}$

\begin{aligned}\omega_{1}=& \frac{2 \sigma_{\mathrm{s}}}{E h} \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2\left(E+E_{1}\right)(l-2 x)\right]^{3}}{12\left(E+E_{1}\right)^{2}}}+\\& \frac{\sigma_{\mathrm{s}}}{E h}(l-2 x) \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2 l\left(E+E_{1}\right)\right]}{2\left(E+E_{1}\right)}} \\& \frac{2 \sigma_{\mathrm{s}}}{E h} \frac{\left(E-E_{1}\right)^{2}(l-2 \xi)^{2}}{12\left(E+E_{1}\right)^{2}}-\frac{\sigma_{\mathrm{s}}}{E h} \frac{\left(E-E_{1}\right)(l-2 \xi)^{2}}{2\left(E+E_{1}\right)}-\frac{\sigma_{\mathrm{s}} l^{3}}{6 E h(l-2 \xi)}+\\& \frac{\sigma_{\mathrm{s}} l^{2}}{E h(l-2 \xi)} \xi-\frac{3 \sigma_{\mathrm{s}} l}{2 E h(l-2 \xi)} \xi^{2}+\frac{\sigma_{\mathrm{s}}}{3 E h(l-2 \xi)} \xi^{3}\end{aligned}
$\begin{array}{l}\omega_{2}=\frac{\sigma_{\mathrm{s}}}{E h(l-2 \xi)}\left(l x^{2}-\frac{2}{3} x^{3}\right)-\frac{\sigma_{\mathrm{s}} l^{3}}{6 E h(l-2 \xi)}+\\x\left[\frac{2 \sigma_{\mathrm{s}}}{E h} \frac{\left(E-E_{1}\right)(l-2 \xi)}{2\left(E+E_{1}\right)}-\frac{2 \sigma_{\mathrm{s}} l}{E h(l-2 \xi)} \xi+\frac{\sigma_{\mathrm{s}}}{E h(l-2 \xi)} \xi^{2}\right]+\\\frac{\sigma_{\mathrm{s}}}{E h}(l-2 x) \sqrt{\frac{\left(E-E_{1}\right)(l-2 \xi)\left[\left(3 E+E_{1}\right)(l-2 \xi)-2 l\left(E+E_{1}\right)\right]}{2\left(E+E_{1}\right)}}-\\\frac{\sigma_{\mathrm{s}} l}{E h} \frac{\left(E-E_{1}\right)(l-2 \xi)}{2\left(E+E_{1}\right)}+\frac{\sigma_{\mathrm{s}} l^{2}}{E h(l-2 \xi)} \xi-\frac{\sigma_{\mathrm{s}} l}{2 E h(l-2 \xi)} \xi^{2}\end{array}$

$C_{5} =0,\ \ C_{6} =-\dfrac{\sigma_{{\rm s}} l^{3}}{6Eh\left( {l-2\xi }\right)}$

$\omega_{2} =-\dfrac{\sigma_{{\rm s}} l^{3}}{6Eh\left( {l-2\xi }\right)}$

### 3.2 数值计算

Table 1  When $\eta_{x=0} \left( x \right)={h}/{4}$,theoretical and simulated deflection values

## 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

Liu Xiequan, Ni Xinhua, Liu Yunting.

Elastoplastic analysis of spatial bars of linear strengthening materials

Journal of Xi'an University of Science and Technology, 2002, 22(4):489-492 (in Chinese)

Zhang Xibin..

Deformation analysis of elastic plastic beam with linear strengthening material with yield platform

Journal of Changjiang University $($Natural Science Edition$)$, 2005, 2(10): 273-274, 364-368 (in Chinese)

Qian Weichang. Variational Method and Finite Element (Volume I). Beijing: Science Press, 1980 (in Chinese)

Qian Weichang.

Higher order Lagrange multiplier method and more general variational principle in elastic theory

Applied Mathematics and Mechanics, 1983, 4(2):137-150 (in Chinese)

Zhang Fufan. Elastic Thin Plate. Beijing: Science Press, 1984 (in Chinese)

Fu Baolian. Energy Principle and Its Application in Elasticity. Beijing: Science Press, 2004 (in Chinese)

Ding Guangtao.

Some advances in the study of inverse problems of variational method

Journal of Peking University $($Natural Science Edition$)$, 2016, 52(4):732-740 (in Chinese)

Hu Haichang.

On general variational principles in elastic mechanics and normed body mechanics

Journal of physics, 1954, 10(3):259-290 (in Chinese)

Tonti E.

Variational principles in elastostatics

Meccanica, 1967, 2(4):201-208

/

 〈 〉