7. 平面问题

7.1 平面问题及其分类

$$\begin{align*} & \text{平面应变} & & \text{平面应力} \\ & \varepsilon_x = \varepsilon_x(x, y) & & \sigma_x = \sigma_x(x, y) \\ & \varepsilon_y = \varepsilon_y(x, y) & & \sigma_y = \sigma_y(x, y) \\ & \gamma_{xy} = \gamma_{xy}(x, y) & & \tau_{xy} = \tau_{xy}(x, y) \\ & \varepsilon_z = \gamma_{zx} = \gamma_{zy} = 0 & & \sigma_z = \tau_{zx} = \tau_{zy} = 0 \end{align*}$$

1. 本构方程

应力 – 应变

$$\sigma_{\alpha\beta} = 2 G \varepsilon_{\alpha\beta} + \lambda \varepsilon_{kk} \delta_{\alpha\beta} \quad \begin{pmatrix} \alpha, \beta = 1, 2 \\ k = 1, 2, 3 \end{pmatrix}$$

$$\begin{align*} & \text{平面应变} & & \text{平面应力} \\ & \sigma_x \frac{2 G (1 - \nu)}{1 - 2 \nu} \pqty{\varepsilon_x + \frac{\nu}{1 - \nu} \varepsilon_y} & & \sigma_x = \frac{E}{1 - \nu^2} (\varepsilon_x + \nu \varepsilon_y) = \frac{2 G}{1 - \nu} (\varepsilon_x + \nu \varepsilon_y) \\ & \sigma_y = \frac{2 G (1 - \nu)}{1 - 2 \nu} \pqty{\varepsilon_y + \frac{\nu}{1 - \nu} \varepsilon_x} & & \sigma_y = \frac{E}{1 - \nu^2} (\varepsilon_y + \nu \varepsilon_x) = \frac{2 G}{1 - \nu} (\varepsilon_y + \nu \varepsilon_x) \\ & \tau_{xy} = G \gamma_{xy} & & \tau_{xy} = G \gamma_{xy} \\ \hline & \sigma_z = \nu (\sigma_x + \sigma_y) = \lambda (\varepsilon_x + \varepsilon_y) & & \sigma_z = 0 \\ & \tau_{zx} = \tau_{zy} = 0 & & \tau_{zx} = \tau_{zy} = 0 \end{align*}$$

应变 – 应力

$$\varepsilon_{\alpha\beta} = \frac{1 + \nu}{E} \sigma_{\alpha\beta} - \frac{\nu}{E} \sigma_{kk} \delta_{\alpha\beta} \quad \begin{pmatrix} \alpha, \beta = 1, 2 \\ k = 1, 2, 3 \end{pmatrix}$$

$$\begin{align*} & \text{平面应变} & & \text{平面应力} \\ & \varepsilon_x = \frac{1 - \nu^2}{E} \pqty{\sigma_x - \frac{\nu}{1 - \nu} \sigma_y} & & \varepsilon_x = \frac{1}{E} \pqty{\sigma_x - \nu \sigma_y} \\ & \varepsilon_y = \frac{1 - \nu^2}{E} \pqty{\sigma_y - \frac{\nu}{1 - \nu} \sigma_x} & & \varepsilon_y = \frac{1}{E} \pqty{\sigma_y - \nu \sigma_x} \\ & \gamma_{xy} = 2 \varepsilon_{xy} = \frac{1}{G} \tau_{xy} & & \gamma_{xy} = 2 \varepsilon_{xy} = \frac{1}{G} \tau_{xy} \\ \hline & \varepsilon_z = 0 & & \varepsilon_z = - \frac{\nu}{E} (\sigma_x + \sigma_y) = - \frac{\nu}{1 - \nu} (\varepsilon_x + \varepsilon_y) \\ & \gamma_{zx} = \gamma_{zy} = 0 & & \gamma_{zx} = \gamma_{zy} = 0 \end{align*}$$

弹性常数替换

Tips

弹性常数替换考试给

平面应力 => 平面应变

$$\begin{dcases} E^* = \frac{E}{1 - \nu^2} \\ \nu^* = \frac{\nu}{1 - \nu} \\ \end{dcases} \qq{即} \begin{dcases} G^* = G \\ \nu^* = \frac{\nu}{1 - \nu} \\ \end{dcases}$$

平面应变 => 平面应力

$$\begin{dcases} E' = \frac{E (1 + 2 \nu)}{(1 + \nu)^2} \\ \nu' = \frac{\nu}{1 + \nu} \\ \end{dcases} \qq{即} \begin{dcases} G' = G \\ \nu' = \frac{\nu}{1 + \nu} \\ \end{dcases}$$

2. 平衡方程

$$\begin{align*} & \pdv{\sigma_x}{x} + \pdv{\tau_{xy}}{y} + f_x = 0 \\ & \pdv{\tau_{xy}}{x} + \pdv{\sigma_y}{y} + f_y = 0 \\ & f_z = 0 \end{align*}$$

3. 协调方程

第一协调方程

$$\pdv[2]{\varepsilon_x}{y} + \pdv[2]{\varepsilon_y}{x} - \pdv{\gamma_{xy}}{x}{y} = 0$$

第二, 第三和第六协调方程

$$\pdv[2]{\varepsilon_z}{y} = 0 \qc \pdv[2]{\varepsilon_z}{x} = 0 \qc \pdv{\varepsilon_z}{x}{y} = 0$$

4. 几何方程

$$\begin{align*} & \varepsilon_x = \pdv{u}{x} \\ & \varepsilon_y = \pdv{v}{y} \\ & \gamma_{xy} = 2 \varepsilon_{xy} = \pdv{u}{y} + \pdv{v}{x} \end{align*}$$

5. 边界条件

$$\begin{cases} \sigma_x \cos(\nu, x) + \tau_{xy} \cos(\nu, y) = \overline{X} \\ \tau_{xy} \cos(\nu, x) + \sigma_y \cos(\nu, y) = \overline{Y} \\ 0 = \overline{Z} \end{cases}$$

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

$$\begin{dcases} \pdv{\sigma_x}{x} + \pdv{\tau_{xy}}{y} + f_x = 0 \\ \pdv{\tau_{xy}}{x} + \pdv{\sigma_y}{y} + f_y = 0 \end{dcases}$$

几何方程

$$\begin{dcases} \varepsilon_x = \pdv{u}{x} \\ \varepsilon_y = \pdv{v}{y} \\ \gamma_{xy} = \pdv{u}{y} + \pdv{v}{x} \end{dcases}$$

本构方程 (平面应力)

$$\begin{dcases} \varepsilon_x = \frac{1}{E} (\sigma_x - \nu \sigma_y) \\ \varepsilon_y = \frac{1}{E} (\sigma_y - \nu \sigma_x) \\ \gamma_{xy} = \frac{1}{G} \tau_{xy} \end{dcases}$$

$$\begin{dcases} \sigma_x = \frac{E}{1 - \nu^2} (\varepsilon_x + \nu \varepsilon_y) \\ \sigma_y = \frac{E}{1 - \nu^2} (\varepsilon_y + \nu \varepsilon_x) \\ \tau_{xy} = G \gamma_{xy} \end{dcases}$$

协调方程

$$\pdv[2]{\varepsilon_x}{y} + \pdv[2]{\varepsilon_y}{x} + \pdv{\gamma_{xy}}{x}{y} = 0$$

边界条件

$$\begin{cases} l \sigma_x + m \tau_{xy} = \overline{X} \\ l \tau_{xy} + m \sigma_y = \overline{Y} \end{cases} \qq{在力边界 $\Gamma_{\sigma}$ 上}$$

$$\begin{cases} u = \overline{u} \\ v = \overline{v} \\ \end{cases} \qq{在位移边界 $\Gamma_u$ 上}$$

1. 位移解法

平面应力位移法基本方程

$$\begin{align*} & G \laplacian{u} + G \frac{1 + \nu}{1 - \nu} \pdv{x}(\pdv{u}{x} + \pdv{v}{y}) + f_x = 0 \\ & G \laplacian{v} + G \frac{1 + \nu}{1 - \nu} \pdv{y}(\pdv{u}{x} + \pdv{v}{y}) + f_y = 0 \end{align*}$$

平面应变位移法基本方程

$$\begin{align*} & G \laplacian{u} + G \frac{1}{1 - 2 \nu} \pdv{x}(\pdv{u}{x} + \pdv{v}{y}) + f_x = 0 \\ & G \laplacian{v} + G \frac{1}{1 - 2 \nu} \pdv{y}(\pdv{u}{x} + \pdv{v}{y}) + f_y = 0 \end{align*}$$

2. 应力解法

协调方程 (B-M 方程)

$$\laplacian(\sigma_x + \sigma_y) = - (1 + \nu) \pqty{\pdv{f_x}{x} + \pdv{f_y}{y}}$$

平面应变的 B-M 方程

$$\laplacian(\sigma_x + \sigma_y) = - \frac{1}{1 - \nu} \pqty{\pdv{f_x}{x} + \pdv{f_y}{y}}$$

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:

$$f_x = - \pdv{V}{x} \qc f_y = - \pdv{V}{y}$$

平面问题的应力公式

$$\begin{align*} & \sigma_x = \pdv[2]{\phi}{y} + V \\ & \sigma_y = \pdv[2]{\phi}{x} + V \\ & \tau_{xy} = - \pdv{\phi}{x}{y} \end{align*}$$

应力函数解法的基本方程

Warning

重调和方程考试不给

$$\begin{align*} & \laplacian{\laplacian{\phi}} = - (1 - \nu) \laplacian{V} & \text{(平面应力)} \\ & \laplacian{\laplacian{\phi}} = - \frac{1 - 2 \nu}{1 - \nu} \laplacian{V} & \text{(平面应变)} \end{align*}$$