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

Marks: 10M

Year: May 2015

x coordinate of point $P = P_x = 4.5$

$N_1 = 0.3\\ N_2 = ?\\ N_3 = ?\\ P_y = ?$

Solution:-

$\alpha_1 = x_2y_3 - x_3y_2 = 7 \times 6 - 4 b\times 4 = 26\\ \alpha_2 = x_3y_1 - x_1y_3 = 4 \times 3 - 2 \times 6 = 0\\ \alpha_3 = x_1y_2 - x_2y_1 = 2 \times 4 - 7 \times 3 = -13\\ \beta_1 = y_2 - y_3 = 4 - 6 = -2\\ \beta_2 = y_3 - y_1 = 6 - 3 = 3\\ \beta_3 = y_1 - y-2 = 3 - 4 = -1\\ r_1 = - (x_2 - x_3) = - (7 - 4) = -3\\ r_2 = - (x_3 - x_1) = - (4 - 2) = -2\\ r_3 = -(x_1 - x_2) = -(2 - 7) = 5$

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

$N_1 = \frac{1}{2A} (\alpha_1 + \beta_1x + r_1 y)\\ 0.3 = \frac{1}{13}(26 + (-2) 4.5 + (-3)P_y)\\ \therefore P_y = 4.366$

$N_2 = \frac{1}{2A}(\alpha_2 + \beta_2x + r_2 y)\\ N_2 = \frac{1}{13}(0 + 3 \times 4.5 - 2 \times 4.366)\\ N_2 = 0.366$

$N_3 = \frac{1}{2A}(\alpha_3 + \beta_3x + r_3 y)\\ N_3 = \frac{1}{13}(-13 - 1 \times 4.5 + 5 \times 4.366)\\ N_2 = 0.333$

$N_1 + N_2 + N_3 = 0.3 + 0.366 + 0.333 = 0.999 \approx 1$