应力系统基础

$ 给定二维应力系统 \begin{bmatrix}\sigma_x & \tau_y\\\tau_x & \sigma_y\end{bmatrix}，求正、切应力最值. $

 

$ 设 \begin{bmatrix}\sigma_\theta & \tau_{\theta + 90^o} \\\tau_{\theta + 90^o} & \sigma_\theta\end{bmatrix} 为应力单元旋转\theta后得到的等价的应力系统，显然有：$

$\begin{bmatrix}\sigma_\theta & \tau_{\theta + 90^o} \\\tau_{\theta + 90^o} & \sigma_\theta\end{bmatrix} = \begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix} \cdot \begin{bmatrix}\sigma_x & \tau_y\\\tau_x & \sigma_y\end{bmatrix} \cdot \begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}$

 

$1.\quad 求正应力最值\sigma_{max}, \sigma_{min}$

$\because \tau_x = \tau_y$

$\therefore \begin{bmatrix}\sigma_x & \tau_y\\\tau_x & \sigma_y\end{bmatrix}为对称矩阵，可正交对角化为: \begin{bmatrix}\sigma_{max} & 0\\0 & \sigma_{min}\end{bmatrix}$

$\quad 令\begin{vmatrix}\sigma_x - \lambda & \tau\\\tau & \sigma_y - \lambda\end{vmatrix} = 0，得：$

$\quad \lambda = {\frac{\sigma_x + \sigma_y}{2}} \pm \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + {\tau_x}^2}$

$\therefore \sigma_{max} = {\frac{\sigma_x + \sigma_y}{2}} + \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + {\tau_x}^2}$

$\quad \sigma_{min} = {\frac{\sigma_x + \sigma_y}{2}} - \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + {\tau_x}^2}$

 

$2.\quad 求切应力最值\tau_{max}$

$\because \begin{bmatrix}\sigma_\theta & \tau_{\theta + 90^o}\\\tau_\theta & \sigma_{\theta + 90^o}\end{bmatrix} \sim \begin{bmatrix}\sigma_x & \tau_y\\\tau_x & \sigma_y\end{bmatrix} \sim \begin{bmatrix}\sigma_{max} & 0\\0 & \sigma_{min}\end{bmatrix}$

$\therefore 有:$

$\ 1)\quad 迹tr(A) 不变，即\ \sigma_\theta + \sigma_{\theta+90^o} = \sigma_x + \sigma_y = \sigma_{max} + \sigma_{min}$

$\ 2)\quad 模|A| 不变， 即\ \sigma_\theta \cdot \sigma_{\theta + 90^o} - \tau_\theta \cdot \tau_{\theta + 90^o} = \sigma_x \cdot \sigma_y - {\tau_x}^2 = \sigma_{max} \cdot \sigma_{min}$

$\ \Rightarrow {\tau_\theta}^2 = \tau_\theta \cdot \tau_{\theta + 90^o} = \sigma_\theta \cdot \sigma_{\theta + 90^o} - \sigma_{max} \cdot \sigma_{min}$

$\quad \leqslant (\frac{\sigma_\theta + \sigma_{\theta + 90^o}}{2})^2 - \sigma_{max} \cdot \sigma_{min}$

$\quad = (\frac{\sigma_{max} + \sigma_{min}}{2})^2 - \sigma_{max} \cdot \sigma_{min}$

$\quad = (\frac{\sigma_{max} - \sigma_{min}}{2})^2$

$\ \Rightarrow \tau_\theta \leqslant {\frac{1}{2}}(\sigma_{max} - \sigma_{min})$

$\quad 等号当且仅当 \sigma_\theta = \sigma_{\theta + 90^o} = {\frac{1}{2}}(\sigma_x + \sigma_y) 成立$

$\therefore \tau_{max} = {\frac{1}{2}}(\sigma_{max} - \sigma_{min}) = \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + {\tau_x}^2}$