7. 平面问题

7.1 平面问题及其分类

$$\begin{array}{rlrl}& \text{平面应变}& & \text{平面应力}\\ & {\epsilon }_x={\epsilon }_x(x,y)& & {\sigma }_x={\sigma }_x(x,y)\\ & {\epsilon }_y={\epsilon }_y(x,y)& & {\sigma }_y={\sigma }_y(x,y)\\ & {\gamma }_{xy}={\gamma }_{xy}(x,y)& & {\tau }_{xy}={\tau }_{xy}(x,y)\\ & {\epsilon }_z={\gamma }_{zx}={\gamma }_{zy}=0& & {\sigma }_z={\tau }_{zx}={\tau }_{zy}=0\end{array}$$

1. 本构方程

应力 – 应变

$${\sigma }_{\alpha \beta }=2G{\epsilon }_{\alpha \beta }+\lambda {\epsilon }_{kk}{\delta }_{\alpha \beta }\left(\begin{array}{c}\alpha ,\beta =1,2\\ k=1,2,3\end{array}\right)$$$$\begin{array}{rlrl}& \text{平面应变}& & \text{平面应力}\\ & {\sigma }_x\frac{2G(1-\nu )}{1-2\nu }\text{pqty}{\epsilon }_x+\frac{\nu }{1-\nu }{\epsilon }_y& & {\sigma }_x=\frac{E}{1-{\nu }^2}({\epsilon }_x+\nu {\epsilon }_y)=\frac{2G}{1-\nu }({\epsilon }_x+\nu {\epsilon }_y)\\ & {\sigma }_y=\frac{2G(1-\nu )}{1-2\nu }\text{pqty}{\epsilon }_y+\frac{\nu }{1-\nu }{\epsilon }_x& & {\sigma }_y=\frac{E}{1-{\nu }^2}({\epsilon }_y+\nu {\epsilon }_x)=\frac{2G}{1-\nu }({\epsilon }_y+\nu {\epsilon }_x)\\ & {\tau }_{xy}=G{\gamma }_{xy}& & {\tau }_{xy}=G{\gamma }_{xy}\\ & {\sigma }_z=\nu ({\sigma }_x+{\sigma }_y)=\lambda ({\epsilon }_x+{\epsilon }_y)& & {\sigma }_z=0\\ & {\tau }_{zx}={\tau }_{zy}=0& & {\tau }_{zx}={\tau }_{zy}=0\end{array}$$

应变 – 应力

$${\epsilon }_{\alpha \beta }=\frac{1+\nu }{E}{\sigma }_{\alpha \beta }-\frac{\nu }{E}{\sigma }_{kk}{\delta }_{\alpha \beta }\left(\begin{array}{c}\alpha ,\beta =1,2\\ k=1,2,3\end{array}\right)$$$$\begin{array}{rlrl}& \text{平面应变}& & \text{平面应力}\\ & {\epsilon }_x=\frac{1-{\nu }^2}{E}\text{pqty}{\sigma }_x-\frac{\nu }{1-\nu }{\sigma }_y& & {\epsilon }_x=\frac{1}{E}\text{pqty}{\sigma }_x-\nu {\sigma }_y\\ & {\epsilon }_y=\frac{1-{\nu }^2}{E}\text{pqty}{\sigma }_y-\frac{\nu }{1-\nu }{\sigma }_x& & {\epsilon }_y=\frac{1}{E}\text{pqty}{\sigma }_y-\nu {\sigma }_x\\ & {\gamma }_{xy}=2{\epsilon }_{xy}=\frac{1}{G}{\tau }_{xy}& & {\gamma }_{xy}=2{\epsilon }_{xy}=\frac{1}{G}{\tau }_{xy}\\ & {\epsilon }_z=0& & {\epsilon }_z=-\frac{\nu }{E}({\sigma }_x+{\sigma }_y)=-\frac{\nu }{1-\nu }({\epsilon }_x+{\epsilon }_y)\\ & {\gamma }_{zx}={\gamma }_{zy}=0& & {\gamma }_{zx}={\gamma }_{zy}=0\end{array}$$

弹性常数替换

Tips

弹性常数替换考试给

平面应力 => 平面应变

$$\{\begin{array}{l}{E}^{\ast }=\frac{E}{1-{\nu }^2}\\ {\nu }^{\ast }=\frac{\nu }{1-\nu }\end{array}\text{qq}\mathrm{即}\{\begin{array}{l}{G}^{\ast }=G\\ {\nu }^{\ast }=\frac{\nu }{1-\nu }\end{array}$$

平面应变 => 平面应力

$$\{\begin{array}{l}{E}^{\prime }=\frac{E(1+2\nu )}{(1+\nu {)}^2}\\ {\nu }^{\prime }=\frac{\nu }{1+\nu }\end{array}\text{qq}\mathrm{即}\{\begin{array}{l}{G}^{\prime }=G\\ {\nu }^{\prime }=\frac{\nu }{1+\nu }\end{array}$$

2. 平衡方程

$$\begin{array}{rl}& \text{pdv}{\sigma }_{x}x+\text{pdv}{\tau }_{xy}y+f_x=0\\ & \text{pdv}{\tau }_{xy}x+\text{pdv}{\sigma }_{y}y+f_y=0\\ & f_z=0\end{array}$$

3. 协调方程

第一协调方程

$$\text{pdv}[2]{\epsilon }_{x}y+\text{pdv}[2]{\epsilon }_{y}x-\text{pdv}{\gamma }_{xy}xy=0$$

第二, 第三和第六协调方程

$$\text{pdv}[2]{\epsilon }_{z}y=0\text{qc}\text{pdv}[2]{\epsilon }_{z}x=0\text{qc}\text{pdv}{\epsilon }_{z}xy=0$$

4. 几何方程

$$\begin{array}{rl}& {\epsilon }_x=\text{pdv}ux\\ & {\epsilon }_y=\text{pdv}vy\\ & {\gamma }_{xy}=2{\epsilon }_{xy}=\text{pdv}uy+\text{pdv}vx\end{array}$$

5. 边界条件

$$\{\begin{array}{l}{\sigma }_x\cos (\nu ,x)+{\tau }_{xy}\cos (\nu ,y)=\bar{X}\\ {\tau }_{xy}\cos (\nu ,x)+{\sigma }_y\cos (\nu ,y)=\bar{Y}\\ 0=\bar{Z}\end{array}$$

7.2 平面问题的基本解法

平面应力问题的基本方程

平衡方程

$$\{\begin{array}{l}\text{pdv}{\sigma }_{x}x+\text{pdv}{\tau }_{xy}y+f_x=0\\ \text{pdv}{\tau }_{xy}x+\text{pdv}{\sigma }_{y}y+f_y=0\end{array}$$

几何方程

$$\{\begin{array}{l}{\epsilon }_x=\text{pdv}ux\\ {\epsilon }_y=\text{pdv}vy\\ {\gamma }_{xy}=\text{pdv}uy+\text{pdv}vx\end{array}$$

本构方程 (平面应力)

$$\{\begin{array}{l}{\epsilon }_x=\frac{1}{E}({\sigma }_x-\nu {\sigma }_y)\\ {\epsilon }_y=\frac{1}{E}({\sigma }_y-\nu {\sigma }_x)\\ {\gamma }_{xy}=\frac{1}{G}{\tau }_{xy}\end{array}$$$$\{\begin{array}{l}{\sigma }_x=\frac{E}{1-{\nu }^2}({\epsilon }_x+\nu {\epsilon }_y)\\ {\sigma }_y=\frac{E}{1-{\nu }^2}({\epsilon }_y+\nu {\epsilon }_x)\\ {\tau }_{xy}=G{\gamma }_{xy}\end{array}$$

协调方程

$$\text{pdv}[2]{\epsilon }_{x}y+\text{pdv}[2]{\epsilon }_{y}x+\text{pdv}{\gamma }_{xy}xy=0$$

边界条件

$$\{\begin{array}{l}l{\sigma }_x+m{\tau }_{xy}=\bar{X}\\ l{\tau }_{xy}+m{\sigma }_y=\bar{Y}\end{array}\text{qq}\mathrm{在}\mathrm{力}\mathrm{边}\mathrm{界}\${\mathrm{\Gamma }}_{\sigma }\$\mathrm{上}$$$$\{\begin{array}{l}u=\bar{u}\\ v=\bar{v}\end{array}\text{qq}\mathrm{在}\mathrm{位}\mathrm{移}\mathrm{边}\mathrm{界}\${\mathrm{\Gamma }}_u\$\mathrm{上}$$

1. 位移解法

平面应力位移法基本方程

$$\begin{array}{rl}& G\text{laplacian}u+G\frac{1+\nu }{1-\nu }\text{pdv}x(\text{pdv}ux+\text{pdv}vy)+f_x=0\\ & G\text{laplacian}v+G\frac{1+\nu }{1-\nu }\text{pdv}y(\text{pdv}ux+\text{pdv}vy)+f_y=0\end{array}$$

平面应变位移法基本方程

$$\begin{array}{rl}& G\text{laplacian}u+G\frac{1}{1-2\nu }\text{pdv}x(\text{pdv}ux+\text{pdv}vy)+f_x=0\\ & G\text{laplacian}v+G\frac{1}{1-2\nu }\text{pdv}y(\text{pdv}ux+\text{pdv}vy)+f_y=0\end{array}$$

2. 应力解法

协调方程 (B-M 方程)

$$\text{laplacian}({\sigma }_x+{\sigma }_y)=-(1+\nu )\text{pqty}\text{pdv}{f}_{x}x+\text{pdv}{f}_{y}y$$

平面应变的 B-M 方程

$$\text{laplacian}({\sigma }_x+{\sigma }_y)=-\frac{1}{1-\nu }\text{pqty}\text{pdv}{f}_{x}x+\text{pdv}{f}_{y}y$$

3. 应力函数解法

Warning

三次应力函数考试不给

设体力势为 $V$, 体力可表示为体力势的负梯度:

$$f_x=-\text{pdv}Vx\text{qc}{f}_y=-\text{pdv}Vy$$

平面问题的应力公式

$$\begin{array}{rl}& {\sigma }_x=\text{pdv}[2]\varphi y+V\\ & {\sigma }_y=\text{pdv}[2]\varphi x+V\\ & {\tau }_{xy}=-\text{pdv}\varphi xy\end{array}$$

应力函数解法的基本方程

Warning

重调和方程考试不给

$$\begin{array}{rlr}& \text{laplacian}\text{laplacian}\varphi =-(1-\nu )\text{laplacian}V& \text{(平面应力)}\\ & \text{laplacian}\text{laplacian}\varphi =-\frac{1-2\nu }{1-\nu }\text{laplacian}V& \text{(平面应变)}\end{array}$$