A constant strain triangle element has the nodal co-ordinates (1, 2), (4, 0.5) and (3,4) for i, j and k nodes respectively. The elements is 2mm thick and is of matrial with properties E = 70 GPa and Poisson's ratio 0.3 upon loading of model, the nodal deflections found to be:

$u_i = 100 \mu m \hspace{2cm} u_j = 75 \mu m \hspace{2cm} \mu k = 80 \mu m$

$v_i = -50 \mu m \hspace{2cm} v_j = -40 \mu m \hspace{2cm} v_k = -45 \mu m$

Determine

i) The Jacobian for $(x, y) - (\zeta - /eta)$ transformation

ii) The strain displacement relation

iii) The strams

iv) The element stresses

$(x_1,y_1)=(1,2)$
$(x_2,y_2)=(4,0.5)$
$(x_3,y_3)=(3,4)$

$u_1=100\mu m=0.1mm$
$u_2=75\mu m=0.075mm$
$u_3=80\mu m=0.8mm$

$v_1=-50\mu m=-0.05mm$
$v_2=-40\mu m=-0.04mm$
$v_3=-45\mu m=-0.045mm$

a) Jacobian matrix
$\begin{bmatrix}J\end{bmatrix}=\begin{bmatrix}(x_2-x_1)&(y_2-y_1) \\ (x_3-x_1)&(y_3-y_1) \end{bmatrix}$
$\hspace{1cm}=\begin{bmatrix} (4-1)&(0.5-2) \\ (3-1)&(4-2)\end{bmatrix}$
$\hspace{1cm}=\begin{bmatrix} 3&-1.5\\2&2 \end{bmatrix}$

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

$\beta_1=y_2-y_3=-3.5$
$\beta_2=y_3-y_1=2$
$\beta_3=y_1-y_2=1.5$

$\gamma_1=-(x_2-x_3)=-1$
$\gamma_2=-(x_3-x_1)=-2$
$\gamma_3=-(x_1-x_2)=3$

$2A=\begin{vmatrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3 \end{vmatrix}$ $=\begin{vmatrix}1&1&2\\1&4&0.5\\1&3&4 \end{vmatrix}$
$\hspace{1cm}=1(16-1.5)-1(4-0.5)+2(3-4)$
$\therefore 2A=9.5$

$\begin{Bmatrix}e_x\\e_y\\\delta_{xy} \end{Bmatrix}=\frac{1}{9.5}\begin{bmatrix}-3.5&0&2&0&1.5&0 \\ 0&-1&0&-2&0&3 \\ -1&-3.5&-2&2&3&1.5 \end{bmatrix}\begin{Bmatrix}0.1 \\ -0.05 \\0.075 \\-0.04 \\0.08 \\ -0.045 \end{Bmatrix}$

c) The Strains.
$\hspace{2cm} \therefore e_x=-8.421*10^{-3}$
$\hspace{2cm} \therefore e_y=-5.2631*10^{-4}$
$\hspace{2cm} \therefore \delta_{xy}=1.842*10^{-3}$

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

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

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

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

$\begin{Bmatrix} \sigma_x\\ \sigma_y \\ \tau_{xy} \end{Bmatrix}=10^3 \begin{bmatrix}76.92&23.077&0\\23.077&76.92&0\\0&0&26.92 \end{bmatrix}\begin{Bmatrix} -8.42*10^{-3}\\-5.26*10^{-4} \\ 1.842*10^{-3} \end{Bmatrix}$

$\hspace{1cm} \therefore \sigma_x=-659.80\,N/mm^2$
$\hspace{1cm} \therefore \sigma_y=-234.768\,N/mm^2$
$\hspace{1cm} \therefore \tau_{xy}=49.586\,N/mm^2$