A CST element has nodal co-ordinates (10, 10), (70, 35) and (75, 25) for node 1, 2 and 3 respectively. the element is 2mm thick and is of material with properties E= 70 GPa.Poission ratio is 0.3. After applying the load to the element the hodal defermation were found $u_1=0.01mm$ $u_2=0.03mm$ $u_3=-0.02mm$ $v_1=-0.04mm$ $v_2=0.02mm$ $v_3=-0.04mm$

Subject: Finite Element Analysis

Topic: Two Dimensional Vector Variable Problem

Difficulty: Medium

$(x_1,y_1)=(10,10)$
$(x_2,y_2)=(70,35)$
$(x_3,y_3)=(75,25)$

a) Jacobian matrix
$\begin{bmatrix}J\end{bmatrix}=\begin{bmatrix}(x_2-x_1)&(y_2-y_1) \\ (x_3-x_1)&(y_3-y_1) \end{bmatrix}$

$\hspace{1cm}=\begin{bmatrix} 60&25\\65&15 \end{bmatrix}$

b)Strain displacement relation matrix is given by,
$\begin{Bmatrix}e_x\\e_y\\\delta_{xy} \end{Bmatrix}=\frac{1}{2A}\begin{bmatrix}\beta_1&0&\beta_2&0&\beta_3&0 \\ 0&\gamma_1&0&\gamma_2&0&\gamma_3 \\ \gamma_1&\beta_1&\gamma_2&\beta_2&\gamma_3&\beta_3\end{bmatrix}\begin{Bmatrix}u_1\\v_1\\u_2\\v_2\\u_3\\v_3 \end{Bmatrix}$

$\beta_1=y_2-y_3=10$
$\beta_2=y_3-y_1=15$
$\beta_3=y_1-y_2=-25$

$\gamma_1=-(x_2-x_3)=5$
$\gamma_2=-(x_3-x_1)=-65$
$\gamma_3=-(x_1-x_2)=60$

$2A=\begin{vmatrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3 \end{vmatrix}$ $=\begin{vmatrix}1&10&10\\1&70&35\\1&75&25 \end{vmatrix}$
$\therefore 2A=-725$

$\begin{Bmatrix}e_x\\e_y\\\delta_{xy} \end{Bmatrix}=\frac{1}{-725}\begin{bmatrix}10&0&15&0&-25&0 \\ 0&5&0&-65&0&60 \\ 5&10&-65&15&60&-25 \end{bmatrix}\begin{Bmatrix}0.01 \\ -0.05 \\0.03 \\0.02 \\-0.02 \\ -0.04 \end{Bmatrix}$

c)
$\hspace{2cm} \therefore e_x=-0.1448*10^{-2}$
$\hspace{2cm} \therefore e_y=-0.5378*10^{-2}$
$\hspace{2cm} \therefore \delta_{xy}=0.3035*10^{-2}$

d) The elemental stress
$\begin{bmatrix}\delta \end{bmatrix}=\begin{bmatrix}D\end{bmatrix}\begin{Bmatrix}e\end{Bmatrix}$
$D=\frac{E}{1-v^2}\begin{bmatrix}1&v&0\\v&1&0\\0&0&\frac{1-v}{2}\end{bmatrix}$

$E=70\,GPa=0.7*10^5\,N/mm^2\,\,\,,\,\,v=0.09$

$\begin{bmatrix}D\end{bmatrix}=\frac{0.7*10^5}{1-0.3^2}\begin{bmatrix}1&0.3&0\\0.3&1&0\\0&0&\frac{1-0.3}{2}\end{bmatrix}$

$\begin{bmatrix}D\end{bmatrix}=10^3 \begin{bmatrix}76.92&23.077&0\\23.077&76.92&0\\0&0&26.92 \end{bmatrix}$

$\begin{Bmatrix} \sigma_x\\ \sigma_y \\ \tau_{xy} \end{Bmatrix}=10^3 \begin{bmatrix}76.92&23.077&0\\23.077&76.92&0\\0&0&26.92 \end{bmatrix}\begin{Bmatrix} -1.448\\-5.375\\ 3.035 \end{Bmatrix}$

$\hspace{1cm} \therefore \sigma_x=-235.544\,N/mm^2$
$\hspace{1cm} \therefore \sigma_y=-447.722\,N/mm^2$
$\hspace{1cm} \therefore \tau_{xy}=81.698\,N/mm^2$