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,
$\begin{bmatrix}B_1\end{bmatrix}=\begin{bmatrix}1&0&0&0\\0&0&0&1\\0&1&1&0\end{bmatrix}$
$\begin{bmatrix}B_2\end{bmatrix}=\frac{1}{\begin{vmatrix}J\end{vmatrix}}\begin{bmatrix}J_{22}&-J_{12}&0&0\\-J_{21}&J_{11}&0&0\\0&0&J_{22}&-J_{12}\\0&0&-J_{21}&J_{11}\end{bmatrix}$

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

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

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

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

$J_{12}=\frac{\partial \phi_1}{\partial \xi}y_1+\frac{\partial \phi_2}{\partial \xi}y_2+\frac{\partial \phi_3}{\partial \xi}y_3+\frac{\partial \phi_4}{\partial \xi}y_4$
$=\frac{1}{4}[-(1-\eta)*0+(1-\eta)*0+(1+\eta)*25-(1+\eta)*25]$

$=\frac{1}{4}[25-25]$

$\therefore J_{12}|_{0,0}=0$

$J_{21}=\frac{\partial \phi_1}{\partial \eta}x_1+\frac{\partial \phi_2}{\partial \eta}-x_2+\frac{\partial \phi_3}{\partial \eta}x_3+\frac{\partial \phi_4}{\partial \eta}x_4$
$=\frac{1}{4}[-(1-\xi)*0-(1+\xi)*50+(1+\xi)*50+(1-\xi)*0]$

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

$\therefore J_{21}|_{0,0}=0$

$J_{22}=\frac{\partial \phi_1}{\partial \eta}y_1+\frac{\partial \phi_2}{\partial \eta}-y_2+\frac{\partial \phi_3}{\partial \eta}y_3+\frac{\partial \phi_4}{\partial \eta}y_4$
$=\frac{1}{4}[-(1-\xi)*0-(1+\xi)*0+(1+\xi)*25+(1-\xi)*25]$

$=\frac{1}{4}[25+25]$

$\therefore J_{22}|_{0,0}=12.5$

$ \therefore \begin{bmatrix}J\end{bmatrix}=\begin{bmatrix} J_{11}&J_{12}\\ J_{21}&J_{22} \end{bmatrix}=\begin{bmatrix} 25&0\\ 0&12.5 \end{bmatrix}$

$ \therefore \begin{vmatrix}J\end{vmatrix}=\begin{vmatrix} 25&0\\ 0&12.5 \end{vmatrix}=25*12.5=312.5$

$\begin{bmatrix}B_2\end{bmatrix}=\frac{1}{312.5}\begin{bmatrix}12.5&0&0&0\\0&25&0&0\\0&0&12.5&0\\0&0&0&25\end{bmatrix}$

$\begin{bmatrix}B_2\end{bmatrix}=\frac{1}{25}\begin{bmatrix}1&0&0&0\\0&2&0&0\\0&0&1&0\\0&0&0&2\end{bmatrix}$

$\therefore \begin{bmatrix}B_1\end{bmatrix}\begin{bmatrix}B_2\end{bmatrix}=\frac{1}{25}\begin{bmatrix}1&0&0&0\\0&0&0&2\\0&2&1&0 \end{bmatrix}$

$\begin{bmatrix}B_3\end{bmatrix}=\begin{bmatrix} \frac{\partial \phi_1}{\partial \xi}&0&\frac{\partial \phi_2}{\partial \xi}&0&\frac{\partial \phi_3}{\partial \xi}&0&\frac{\partial \phi_4}{\partial \xi}&0 \\ \frac{\partial \phi_1}{\partial \eta}&0&\frac{\partial \phi_2}{\partial \eta}&0&\frac{\partial \phi_3}{\partial \eta}&0&\frac{\partial \phi_4}{\partial \eta}&0 \\ 0&\frac{\partial \phi_1}{\partial \xi}&0&\frac{\partial \phi_2}{\partial \xi}&0&\frac{\partial \phi_3}{\partial \xi}&0&\frac{\partial \phi_4}{\partial \xi} \\ 0&\frac{\partial \phi_1}{\partial \eta}&0&\frac{\partial \phi_2}{\partial \eta}&0&\frac{\partial \phi_3}{\partial \eta}&0&\frac{\partial \phi_4}{\partial \eta}\end{bmatrix}$

$\begin{bmatrix}B_3\end{bmatrix}=\frac{1}{3} \begin{bmatrix} -1&0&1&0&1&0&-1&0 \\-1&0&-1&0&1&0&1&0 \\0&-1&0&1&0&1&0&-1 \\0&-1&0&-1&0&1&0&1 \end{bmatrix}$

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

$\begin{bmatrix}B\end{bmatrix}=\frac{1}{100} \begin{bmatrix} -1&0&1&0&1&0&-1&0 \\0&-2&0&-2&0&2&0&2\\ -2&-1&-2&1&2&1&2&-1 \end{bmatrix}$

$\begin{bmatrix}D\end{bmatrix}\begin{bmatrix}B\end{bmatrix}=\frac{105}{100} \begin{bmatrix}2.2&0.66&0\\0.66&2.2&0\\0&0&0.77\end{bmatrix}\begin{bmatrix} -1&0&1&0&1&0&-1&0 \\0&-2&0&-2&0&2&0&2\\ -2&-1&-2&1&2&1&2&-1 \end{bmatrix}$

$\therefore \,\,\begin{bmatrix}D\end{bmatrix}\begin{bmatrix}B\end{bmatrix}=10^3\begin{bmatrix} -2.2&-1.32&2.2&-1.32&2.2&1.34&-2.2&1.32 \\ -1.32&-4.4&0.66&-4.4&0.66&4.4&-0.66&4.4 \\ -1.54&-0.77&-1.54&0.77&1.54&0.77&1.54&-0.77 \end{bmatrix}$

$\therefore \,\,\begin{Bmatrix}\sigma\end{Bmatrix}=\begin{bmatrix}D\end{bmatrix}\begin{bmatrix}B\end{bmatrix}\begin{Bmatrix}U\end{Bmatrix}=10^3\begin{bmatrix} -2.2&-1.32&2.2&-1.32&2.2&1.34&-2.2&1.32 \\ -1.32&-4.4&0.66&-4.4&0.66&4.4&-0.66&4.4 \\ -1.54&-0.77&-1.54&0.77&1.54&0.77&1.54&-0.77 \end{bmatrix}\begin{Bmatrix}0\\0\\0.05\\0.075\\0.15\\0.08\\0\\0 \end{Bmatrix}$

$\therefore \,\,\begin{Bmatrix}\sigma\end{Bmatrix}=\begin{Bmatrix}\sigma_x \\ \sigma_x \\ \tau_{xy}\end{Bmatrix}=\begin{Bmatrix} 446.8\\154\\273.5 \end{Bmatrix}$

$\therefore \,\,\sigma_x=446.8\,N/mm^2$
$\therefore \,\,\sigma_y=154\,N/mm^2$
$\therefore \,\,\tau_{xy}=273.5\,N/mm^2$