Evaluate the stresses $\delta x, \delta y$ and $tau xy$ for a rectangular element shown in fig at $\zeta=0$ and $\eta=0$. Assume plain stress conditions. Take $E = 2 \times 10^5 N/mm^2$ and $r^2 = 0.3$. Displacement at verticle are given as

$u_1 = 0 \hspace{2cm} u_2 = 0.05mm \hspace{2cm} u_3 = 0.15mm \hspace{2cm} u_4 = 0$

$v_1 = 0 \hspace{2cm} v_2 = 0.075mm \hspace{2cm} v_3 = 0.08mm \hspace{2cm} v_4 = 0$

We know that,
$\begin{bmatrix}\delta \end{bmatrix}=\begin{bmatrix}D\end{bmatrix}\begin{Bmatrix}e\end{Bmatrix}$
$D=\frac{E}{1-v^2}\begin{bmatrix}1&v&0\\v&1&0\\0&0&\frac{1-v}{2}\end{bmatrix}$

$\begin{bmatrix}D\end{bmatrix}=\frac{2*10^5}{1-0.3^2}\begin{bmatrix}1&0.3&0\\0.3&1&0\\0&0&\frac{1-0.3}{2}\end{bmatrix}$

$\begin{bmatrix}D\end{bmatrix}=10^5 \begin{bmatrix}2.2&0.66&0\\0.66&2.2&0\\0&0&0.77 \end{bmatrix}$

Now, $\begin{Bmatrix} e \end{Bmatrix}=\begin{bmatrix}B_1\end{bmatrix}\begin{bmatrix}B_2\end{bmatrix}\begin{bmatrix}B_3\end{bmatrix}\begin{bmatrix}U\end{bmatrix}$
Where,
$\begin{bmatrix}B_1\end{bmatrix}=\begin{bmatrix}1&0&0&0\\0&0&0&1\\0&1&1&0\end{bmatrix}$
$\begin{bmatrix}B_2\end{bmatrix}=\frac{1}{\begin{vmatrix}J\end{vmatrix}}\begin{bmatrix}J_{22}&-J_{12}&0&0\\-J_{21}&J_{11}&0&0\\0&0&J_{22}&-J_{12}\\0&0&-J_{21}&J_{11}\end{bmatrix}$

$\begin{bmatrix}J\end{bmatrix}=\begin{bmatrix} \sum_{1}^{4}\frac{\partial \phi_i}{\partial \xi}x_i & \sum_{1}^{4}\frac{\partial \phi_i}{\partial \xi}y_i \\ \sum_{1}^{4}\frac{\partial \phi_i}{\partial \eta}x_i & \sum_{1}^{4}\frac{\partial \phi_i}{\partial \eta}y_i \end{bmatrix}$

$J_{11}=\frac{\partial \phi_1}{\partial \xi}x_1+\frac{\partial \phi_2}{\partial \xi}x_2+\frac{\partial \phi_3}{\partial \xi}x_3+\frac{\partial \phi_4}{\partial \xi}x_4$
$=\frac{1}{4}[-(1-\eta)*0+(1-\eta)*50+(1+\eta)*50+(1+\eta)*0]$

$=\frac{1}{4}[50+50]$

$\therefore J_{11}|_{0,0}=25$

$J_{12}=\frac{\partial \phi_1}{\partial \xi}y_1+\frac{\partial …