7.5 极坐标中的平面问题

7.5 极坐标中的平面问题

平衡方程

$$\begin{align*} & \pdv{\sigma_r}{r} + \frac{1}{r} \pdv{\tau_{r \theta}}{\theta} + \frac{\sigma_r - \sigma_{\theta}}{r} + f_r = 0 \\ & \pdv{\tau_{r \theta}}{r} + \frac{1}{r} \pdv{\sigma_{\theta}}{\theta} + 2 \frac{\tau_{r \theta}}{r} + f_{\theta} = 0 \end{align*}$$

几何方程

$$\begin{align*} \varepsilon_r & = \pdv{u_r}{r} \\ \varepsilon_{\theta} & = \frac{1}{r} \pdv{v_{\theta}}{\theta} + \frac{u_r}{r} \\ \gamma_{r \theta} & = \frac{1}{r} \pdv{u_r}{\theta} + \pdv{v_{\theta}}{r} - \frac{v_{\theta}}{r} \end{align*}$$

$u_r$ — 径向位移; $v_{\theta}$ — 环向位移

本构方程

应变 - 应力

$$\begin{align*} \varepsilon_r & = \frac{1}{E} \pqty{\sigma_r - \nu \sigma_{\theta}} \\ \varepsilon_{\theta} & = \frac{1}{E} \pqty{\sigma_{\theta} - \nu \sigma_r} \\ \gamma_{r \theta} & = \frac{1}{G} \tau_{r \theta} \end{align*}$$

应力 - 应变

$$\begin{align*} \sigma_r & = \frac{2 G}{1 - \nu} \pqty{\varepsilon_r + \nu \varepsilon_{\theta}} \\ \sigma_{\theta} & = \frac{2 G}{1 - \nu} \pqty{\varepsilon_{\theta} + \nu \varepsilon_r} \\ \tau_{r \theta} & = G \gamma_{r \theta} \end{align*}$$

协调方程

$$\pdv[2]{\varepsilon_{\theta}}{r} + \frac{1}{r^2} \pdv[2]{\varepsilon_r}{\theta} - \frac{1}{r} \pdv{\gamma_{r \theta}}{r}{\theta} + \frac{2}{r} \pdv{\varepsilon_{\theta}}{r} - \frac{1}{r} \pdv{\varepsilon_r}{r} - \frac{1}{r^2} \pdv{\gamma_{r \theta}}{\theta} = 0$$

应力函数解法基本方程

$$\pqty{\pdv[2]{r} + \frac{1}{r} \pdv{r} + \frac{1}{r^2} \pdv[2]{\theta}} \pqty{\pdv[2]{r} + \frac{1}{r} \pdv{r} + \frac{1}{r^2} \pdv[2]{\theta}} \phi = 0$$

极坐标应力公式

$$\begin{align*} \sigma_r & = \frac{1}{r^2} \pdv[2]{\phi}{\theta} + \frac{1}{r} \pdv{\phi}{r} \\ \sigma_{\theta} & = \pdv[2]{\phi}{r} \\ \tau_{r \theta} & = \frac{1}{r^2} \pdv{\phi}{\theta} - \frac{1}{r} \pdv{\phi}{\theta}{r} = - \pdv{r}(\frac{1}{r} \pdv{\phi}{\theta}) \end{align*}$$

边界条件

位移边界 $\Gamma_u$

$$u_r = \overline{u}_r \qc v_{\theta} = \overline{v}_{\theta}$$

力边界 $\Gamma_{\sigma}$

$$\begin{gather*} \sigma_r \cos(v, r) + \tau_{r \theta} \cos(v, \theta) = \overline{R} \\ \tau_{r \theta} \cos(v, r) + \sigma_{\theta} \cos(v, \theta) = \overline{\Theta} \end{gather*}$$

在环向边界 ($r = const$) 上:

$$\sigma_r = \overline{N}_r(\theta) \qc \tau_{r \theta} = \overline{T}_r(\theta)$$

在径向边界 ($\theta = const$) 上:

$$\sigma_{\theta} = \overline{N}_{\theta}(r) \qc \tau_{\theta r} = \overline{T}_{\theta}(r)$$

7.6 轴对称问题

应力函数解法

应力公式

$$\sigma_r = \frac{1}{r} \pdv{\phi}{r} \qc \sigma_{\theta} = \pdv[2]{\phi}{r} \qc \tau_{r \theta} = 0$$

通解

$$\phi = A \ln r + B r^2 \ln r + C r^2 + D$$

应力分量

$$\begin{align*} \sigma_r & = \frac{A}{r^2} + B \pqty{1 + 2 \ln{r}} + 2 C \\ \sigma_{\theta} & = - \frac{A}{r^2} + B \pqty{3 + 2 \ln{r}} + 2 C \\ \tau_{r \theta} & = 0 \end{align*}$$

应变分量

$$\begin{align*} \varepsilon_r & = \frac{1}{E} \bqty{\pqty{1 + \nu} \frac{A}{r^2} + \pqty{1 - 3 \nu} B + 2 \pqty{1 - \nu} \times B \ln{r} + 2 \pqty{1 - \nu} C} \\ \varepsilon_{\theta} & = \frac{1}{E} \bqty{- \pqty{1 + \nu} \frac{A}{r^2} + \pqty{3 - \nu} B + 2 \pqty{1 - \nu} \times B \ln{r} + 2 \pqty{1 - \nu} C} \\ \tau_{r \theta} & = 0 \end{align*}$$

位移分量

$$\begin{align*} u_r & = \frac{1}{E} \bqty{- \pqty{1 + \nu} \frac{A}{r} + 2 \pqty{1 - \nu} B r \ln{r} - \pqty{1 + \nu} B r + 2 \pqty{1 - \nu} C r} + I \cos{\theta} + K \sin{\theta} \\ v_{\theta} & = \frac{4 B r \theta}{E} + H r + K \cos{\theta} - I \sin{\theta} \end{align*}$$

限制原点的刚体位移 — $I = K = 0$; 位移单值 — $B = 0$; 限制刚体转动 — $H = 0$

位移解法

限制原点的刚体位移和转动, 通解为

$$u_r = C_1 r + \frac{C_2}{r} \qc v_{\theta} = 0$$