平面四边形等参单元(Q4)的刚度矩阵

fem178

发布于 2020-05-08 09:14:10

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文章被收录于专栏：数值分析与有限元编程数值分析与有限元编程

$xOy$

坐标系(物理坐标系)下Q4单元的刚度矩阵为

$\mathbf k =t\iint_{A}\mathbf B^TDBdx dy$

此时应变矩阵

$B$

是

$x,y$

的函数

$\mathbf B= \begin{bmatrix} \mathbf B_1 & \mathbf B_2 & \mathbf B_3 & \mathbf B_4 \\\end{bmatrix}$

其中

$\mathbf B_i= \begin{bmatrix} \frac{\partial N_i}{\partial x} & 0 \\ 0 & \frac{\partial N_i}{\partial y} \\ \frac{\partial N_i}{\partial x} & \frac{\partial N_i}{\partial y}\\ \end{bmatrix} (i=1,2,3,4)$

单元刚度矩阵，等效节点荷载，单元应力，应变等物理量是通过

$xOy$

坐标系表达，而在计算时却是在

$\xi O \eta$

坐标系下。因此

$\mathbf k =t\int_{-1}^{1}\int_{-1}^{1}\mathbf B^T \mathbf D \mathbf B|\mathbf J|d\xi d\eta$

此时应变矩阵

$B$

是

$\xi,\eta$

的函数

两个坐标系下坐标转换的桥梁为

$x=\sum_{k=1}^4 N_k(\xi, \eta)x_k$

$y=\sum_{k=1}^4 N_k(\xi, \eta)y_k$

其中

$(x_k,y_k)$

是

$xOy$

坐标系中单元四个顶点坐标，

$N_k(k=1,2,3,4)$

是形函数

$N_1(\xi, \eta)=\frac{1}{4}(1-\xi)(1-\eta)$

$N_2(\xi, \eta)=\frac{1}{4}(1+\xi)(1-\eta)$

$N_3(\xi, \eta)=\frac{1}{4}(1+\xi)(1+\eta)$

$N_4(\xi, \eta)=\frac{1}{4}(1-\xi)(1+\eta)$

$xOy$

坐标系中任意函数

$f(x,y)$

在坐标系

$\xi O \eta$

的表达式为

$f(x(\xi, \eta),y(\xi, \eta))$

，根据链式求导法则

$\begin{cases} \frac{\partial f}{\partial \xi}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \xi}\\ \frac{\partial f}{\partial \eta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \eta} \end{cases}$

或者

$\begin{pmatrix} \frac{\partial f}{\partial \xi}\\ \frac{\partial f}{\partial \eta} \\ \end{pmatrix} =J \begin{pmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y} \\ \end{pmatrix}$

其中，

$J$

是雅可比矩阵

$J= \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\\frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta}\\\end{bmatrix}$

$= \begin{bmatrix} J_{11}& J_{12} \\J_{21}& J_{22}\\\end{bmatrix}$

$J^{-1}=\frac{1}{|J|} \begin{bmatrix} J_{22}& -J_{12} \\-J_{21}& J_{11}\\\end{bmatrix}$

其中

$J_{11}=\frac{1}{4}[-(1-\eta)x_1+(1-\eta)x_2+(1+\eta)x_3-(1+\eta)x_4]$

$J_{12}=\frac{1}{4}[-(1-\eta)y_1+(1-\eta)y_2+(1+\eta)y_3-(1+\eta)y_4]$

$J_{21}=\frac{1}{4}[-(1-\xi)x_1-(1+\xi)x_2+(1+\xi)x_3+(1+\xi)x_4]$

$J_{22}=\frac{1}{4}[-(1-\xi)y_1-(1+\xi)y_2+(1+\xi)y_3+(1+\xi)y_4]$

由此可得

$\begin{pmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y} \\ \end{pmatrix} =J^{-1} \begin{pmatrix} \frac{\partial f}{\partial \xi}\\ \frac{\partial f}{\partial \eta} \\ \end{pmatrix}$

将

$f$

换成形函数

$N_i$

$\begin{pmatrix} \frac{\partial N_i}{\partial x}\\ \frac{\partial N_i}{\partial y} \\ \end{pmatrix} =J^{-1} \begin{pmatrix} \frac{\partial N_i}{\partial \xi}\\ \frac{\partial N_i}{\partial \eta} \\ \end{pmatrix}$

例如

$\begin{pmatrix} \frac{\partial N_1}{\partial x}\\ \frac{\partial N_1}{\partial y} \\ \end{pmatrix} =J^{-1} \begin{pmatrix} -\frac{1}{4}(1-\eta)\\ -\frac{1}{4}(1-\xi)\\ \end{pmatrix}$

$=\frac{1}{4| \mathbf J|}\begin{pmatrix} -J_{22}(1-\eta)+J_{12}(1-\xi)\\ J_{21}(1-\eta)-J_{11}(1-\xi)\\ \end{pmatrix}$

在

$\xi O \eta$

坐标系下

$\mathbf B_1= \frac{1}{4| \mathbf J|} \begin{bmatrix} -J_{22}(1-\eta)+J_{12}(1-\xi)& 0 \\ 0 & J_{21}(1-\eta)-J_{11}(1-\xi) \\ -J_{22}(1-\eta)+J_{12}(1-\xi)& J_{21}(1-\eta)-J_{11}(1-\xi)\\ \end{bmatrix}$

则

$\mathbf k =t\int_{-1}^{1}\int_{-1}^{1} \begin{bmatrix} \mathbf B_1 \\ \mathbf B_2 \\ \mathbf B_3 \\ \mathbf B_4 \\ \end{bmatrix} \mathbf D \begin{bmatrix} \mathbf B_1& \mathbf B_2 & \mathbf B_3 & \mathbf B_4 \\\end{bmatrix} |\mathbf J|d\xi d\eta$

k是

$8\times8$

矩阵，若将

$\mathbf B^T \mathbf D \mathbf B|\mathbf J|$

看作函数

$g(\xi,\eta)$

，则

$g(\xi,\eta)$

也是

$8\times8$

列阵。

$k_{mn}=t\sum_{i=1}^2 \sum_{j=1}^2 w_i w_j g(\xi_i,\eta_j)_{mn}$