应力平衡方程的推导

对于一点，已知其应力状态有：

$${\sigma }_x,{\tau }_{xy},{\tau }_{xz}$$

则其附近点的应力状态为：
$${\sigma }_x+\frac{\partial {\sigma }_x}{\partial x}dx,{\tau }_{xy}+\frac{\partial {\tau }_{xy}}{\partial x}dx,{\tau }_{xz}+\frac{\partial {\tau }_{xz}}{\partial x}dx$$

之所以如此的原因：

对于附近的一点：
$$\Delta y+\frac{dy}{dx}\Delta x$$
也可由泰勒公式的二阶导知道：

$$f(x)=f(x_0)+\frac{f^{\prime }(x_0)}{1!}(x-x_0)+o^n(x-x_0)$$

于是乎，x方向上的面积为$dydz$

根据外力平衡得：

$${\sigma }_{x}dydz+{\tau }_{xy}dxdz+{\tau }_{xz}dxdy=({\sigma }_x+\frac{\partial {\sigma }_x}{\partial x}dx)dydz+({\tau }_{xy}+\frac{\partial {\tau }_{xy}}{\partial x}dx)dxdz+({\tau }_{xz}+\frac{\partial {\tau }_{xz}}{\partial x}dx)dxdy$$

化简得：
$$\frac{\partial {\sigma }_x}{\partial x}dxdydz+\frac{\partial {\tau }_{xy}}{\partial y}dydxdz+\frac{\partial {\tau }_{xz}}{\partial z}dzdxdy=0$$

$$\frac{\partial {\sigma }_y}{\partial y}dxdydz+\frac{\partial {\tau }_{yx}}{\partial x}dydxdz+\frac{\partial {\tau }_{yz}}{\partial z}dzdxdy=0$$

$$\frac{\partial {\sigma }_z}{\partial z}dxdydz+\frac{\partial {\tau }_{xz}}{\partial x}dydxdz+\frac{\partial {\tau }_{yz}}{\partial x}dzdxdy=0$$

其中：$dxdydz$可以约掉。

当有外力作用$F$时，（这个外力是作用于微元体上的外力），其分量在三个方向上满足：
$$\frac{\partial {\sigma }_x}{\partial x}+\frac{\partial {\tau }_{xy}}{\partial y}+\frac{\partial {\tau }_{xz}}{\partial z}+F_x=0$$

$$\frac{\partial {\sigma }_y}{\partial y}+\frac{\partial {\tau }_{yx}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial z}+F_y=0$$

$$\frac{\partial {\sigma }_z}{\partial z}+\frac{\partial {\tau }_{xz}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial x}+F_z=0$$

当考虑时间变化时，在$x,y,z$三个方向上的位移$u,v,w$时：

$$\frac{\partial {\sigma }_x}{\partial x}+\frac{\partial {\tau }_{xy}}{\partial y}+\frac{\partial {\tau }_{xz}}{\partial z}=\rho \frac{{\partial }^{2}u}{\partial t^2}$$

$$\frac{\partial {\sigma }_y}{\partial y}+\frac{\partial {\tau }_{yx}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial z}=\rho \frac{{\partial }^{2}v}{\partial t^2}$$

$$\frac{\partial {\sigma }_z}{\partial z}+\frac{\partial {\tau }_{xz}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial x}=\rho \frac{{\partial }^{2}w}{\partial t^2}$$