Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 10M

Year: May 2015

For plane stress condition,

$[D] = \frac{E}{1 - v^2} \begin{bmatrix} \ 1 & v & 0 \\ \ v & 1 & 0 \\ \ 0 & 0 & \frac{1-0}{2} \\ \end{bmatrix} = \frac{200 \times 10^3}{1 - 0.3^2} \begin{bmatrix} \ 1 & 0.3 & 0 \\ \ 0.3 & 1 & 0 \\ \ 0 & 0 & 0.35 \\ \end{bmatrix}\\ [D] = 2.1978 \times 10^5\begin{bmatrix} \ 1 & 0.3 & 0 \\ \ 0.3 & 1 & 0 \\ \ 0 & 0 & 0.35 \\ \end{bmatrix}$

$\beta_1 = y_2 - y_3 = -2 \hspace{2cm} r_1 = x_3 - x_2 = -3\\ \beta_2 = y_3 - y_1 = 3 \hspace{2cm} r_2 = x_1 - x_3 = -2\\ \beta_3 = y_1 - y_2 = -1 \hspace{2cm} r_3 = x_2 - x_1 = 5$

(As no fig or coordinates of triangular element is not given. Assume above fig for this problem also.)

$2A = \begin{vmatrix} \ 1 & 2 & 3 \\ \ 1 & 7 & 4 \\ \ 1 & 4 & 6 \\ \end{vmatrix} = 13$

$[B] = \frac{1}{2A} \begin{bmatrix} \ \beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0 \\ \ 0 & r_1 & 0 & r_2 & 0 & r_3 \\ \ r_1 & \beta_1 & r_2 & \beta_2 & r_3 & \beta_3 \\ \end{bmatrix} \\ [B] = \frac{1}{13} \begin{bmatrix} \ 2 & 0 & 3 & 0 & -1 & 0 \\ \ 0 & -3 & 0 & -2 & 0 & 5 \\ \ -3 & -2 & -2 & 3 & 5 & -1 \\ \end{bmatrix}$

{$\sigma$} = [D] [B] [u]

$\begin{Bmatrix} \ \sigma_x \\ \ \sigma_y \\ \ Z_{xy} \\ \end{Bmatrix}= \frac{2.1978 \times 10^5}{13} \begin{bmatrix} \ 1 & 0.3 & 0 \\ \ 0.3 & 1 & 0 \\ \ 0 & 0 & 0.35 \\ \end{bmatrix} \begin{bmatrix} \ 2 & 0 & 3 & 0 & -1 & 0 \\ \ 0 & -3 & 0 & -2 & 0 & 5 \\ \ -3 & -2 & -2 & 3 & 5 & -1 \\ \end{bmatrix} \begin{Bmatrix} \ 0 \\ \ 0 \\ \ 0 \\ \ 0 \\ \ 2 \\ \ -0.1 \\ \end{Bmatrix}\\ \hspace{0.5cm} = \frac{2.1978 \times 10^5}{13} \begin{bmatrix} \ 1 & 0.3 & 0 \\ \ 0.3 & 1 & 0 \\ \ 0 & 0 & 0.35 \\ \end{bmatrix} \begin{bmatrix} \ -2 \\ \ -0.5 \\ \ 10.1 \\ \end{bmatrix}\\ \hspace{0.5cm} = \begin{bmatrix} \ -36348.23 \\ \ -18596.77 \\ \ 59763.25 \\ \end{bmatrix} \begin{matrix} \ N/mm^2 \\ \ N/mm^2 \\ \ N/mm^2 \\ \end{matrix}$