written 6.1 years ago by
teamques10
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modified 6.0 years ago
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$(x_1,y_1)=(10,10)$
$(x_2,y_2)=(70,35)$
$(x_3,y_3)=(75,25)$
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 …
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