pinn求解固体力学问题（强形式）

弹性力学三类基本方程

• 平衡方程：该方程也称动量守恒方程或柯西第二运动定律，其表明物体内部应力的变化（散度）必须与作用在其上的体力相平衡
  $$\{\begin{array}{l}\frac{\partial {\sigma }_{xx}}{\partial x}+\frac{\partial {\tau }_{yx}}{\partial y}+\frac{\partial {\tau }_{zx}}{\partial z}+f_x=0\\ \frac{\partial {\tau }_{xy}}{\partial x}+\frac{\partial {\sigma }_{yy}}{\partial y}+\frac{\partial {\tau }_{zy}}{\partial z}+f_y=0\\ \frac{\partial {\tau }_{xz}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial y}+\frac{\partial {\sigma }_{zz}}{\partial z}+f_z=0\end{array}$$
  张量表示：${\sigma }_{ij,j}+f_i=0$
• 几何方程：描述材料形变与位移之间的关系
  $$\begin{array}{rlrlrl}{\epsilon }_{xx}& =\frac{\partial u}{\partial x},& {\epsilon }_{yy}& =\frac{\partial v}{\partial y},& {\epsilon }_{zz}& =\frac{\partial w}{\partial z},\\ {\gamma }_{xy}& =\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x},& {\gamma }_{yz}& =\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y},& {\gamma }_{zx}& =\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}\end{array}$$
  张量表示：${ϵ}_{ij}=\frac{1}{2}\left(u_{i,j}+u_{j,i}\right)$
• 本构方程：描述材料的应力-应变关系。对于线性弹性材料，这通常被表示为胡克定律
  $$\{\begin{array}{ll}{\epsilon }_{xx}=\frac{1}{E}\left[{\sigma }_{xx}-\nu \left({\sigma }_{yy}+{\sigma }_{zz}\right)\right],& {\gamma }_{xy}=\frac{1}{G}{\tau }_{xy}\\ {\epsilon }_{yy}=\frac{1}{E}\left[{\sigma }_{yy}-\nu \left({\sigma }_{xx}+{\sigma }_{zz}\right)\right],& {\gamma }_{yz}=\frac{1}{G}{\tau }_{yz}\\ {\epsilon }_{zz}=\frac{1}{E}\left[{\sigma }_{zz}-\nu \left({\sigma }_{xx}+{\sigma }_{yy}\right)\right],& {\gamma }_{zx}=\frac{1}{G}{\tau }_{zx}\end{array}$$
  张量表示：${\sigma }_{ij}=\lambda {\epsilon }_{kk}{\delta }_{ij}+2\mu {\epsilon }_{ij}$

算例

边界条件如下所示：

体力为
$$\begin{array}{rl}{f}_x& =\lambda [4{\pi }^2\cos (2\pi x)\sin (\pi y)-\pi \cos (\pi x)Q{y}^3\\ & +\mu \left(9{\pi }^2\cos (2\pi x)\sin (\pi y)-\pi \cos (\pi x)Q{y}^3\right)],\\ f_y& =\lambda [-3\sin (\pi x)Q{y}^2+2{\pi }^2\sin (2\pi x)\cos (\pi y)\\ & +\mu \left(-6\sin (\pi x)Q{y}^2+2{\pi }^2\sin (2\pi x)\cos (\pi y)+{\pi }^2\sin (\pi x)Q{y}^4/4\right)],\end{array}$$
式中：$\lambda =1，\mu =0.5，Q=4$

该方法采用多网络模型分别输出：$U,V,{\sigma }_{xx},{\sigma }_{yy},{\tau }_{xy}$

所有边界边界条件包含在损失函数中：
$$\begin{array}{rl}\mathcal{L}& =|u_x-u_x^{\ast }|+|u_y-u_y^{\ast }|+|{\sigma }_{xx}-{\sigma }_{xx}^{\ast }|+|{\sigma }_{yy}-{\sigma }_{yy}^{\ast }|+|{\sigma }_{xy}-{\sigma }_{xy}^{\ast }|\\ & +|{\sigma }_{xx,x}+{\sigma }_{xy,y}+f_x^{\ast }|+|{\sigma }_{xy,x}+{\sigma }_{yy,y}+f_y^{\ast }|\\ & +|(\lambda +2\mu ){\epsilon }_{xx}+\lambda {\epsilon }_{yy}-{\sigma }_{xx}|+|(\lambda +2\mu ){\epsilon }_{yy}+\lambda {\epsilon }_{xx}-{\sigma }_{yy}|+|2\mu {\epsilon }_{xy}-{\sigma }_{xy}|\end{array}$$
求解结果如下：

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