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

Marks: 14M

Year: Dec 2015

$P_1 = 5KN\\ P_2 2KN\\ E = 180 GPa = 180 \times 1069 N/mm^2\\ A = 6 cm^2 = 6 \times 10^{-4}m^2\\ L_2 = L_3 = \sqrt{0.4^2 + 0.3^2} = 0.5 m$

$K = \frac{AE}{L} \begin{bmatrix} \ C^2 & CS & -C^2 & -cs \\ \ CS & S^2 & -CS & -S^2 \\ \ -C^2 & -CS & C^2 & CS \\ \ -CS & -S^2 & cs & S^2 \\ \end{bmatrix}$

$K_1 = 10^6 \begin{bmatrix} \ 180 & 0 & -180 & 0 \\ \ 0 & 0 & 0 & 0 \\ \ -180 & 0 & 180 & 0 \\ \ 0 & 0 & 0 & 0 \\ \end{bmatrix}$

$K_2 = 10^6 \begin{bmatrix} \ 77.76 & -103.68 & -77.76 & 103.68 \\ \ -103.68 & 138.24 & 103.68 & -138.24 \\ \ -77.76 & 103.68 & 77.76 & -103.68 \\ \ 103.68 & -138.24 & 103.68 & 138.24 \\ \end{bmatrix}$

$K_3 = 10^6 \begin{bmatrix} \ 77.76 & 103.68 & -77.76 & 103.68 \\ \ 103.68 & 138.24 & 103.68 & -138.24 \\ \ -77.76 & -103.68 & 77.76 & 103.68 \\ \ 103.68 & -138.24 & 103.68 & 138.24 \\ \end{bmatrix}$

$10^6 \begin{bmatrix} \ 257.76 & 103.68 & -180 & 0 & -77.76 & -103.68 \\ \ 103.68 & 138.24 & 0 & 0 & -103.68 & -138.24 \\ \ -180 & 0 & 257.76 & -103.68 & -77.76 & 103.68 \\ \ 0 & 0 & -103.68 & 138.24 & 103.68 & -138.24 \\ \ -77.76 & -103.68 & -77.76 & 103.68 & 155.52 & 0 \\ \ -103.68 & -138.24 & 103.68 & -138.24 & 0 & 276.48 \\ \end{bmatrix} \begin{bmatrix} \ x_1 \\ \ y_1 \\ \ x_2 \\ \ y_2 \\ \ x_3 \\ \ y_3 \\ \end{bmatrix} \begin{bmatrix} \ F_{x1} \\ \ F_{y1} \\ \ F_{x2} \\ \ F_{y2} \\ \ F_{x3} \\ \ F_{y3} \\ \end{bmatrix}$

B.C.: $X_1 = Y_1 = Y_2 = 0, F_{x3} = 2 \times 10^3 N, F_{y3} = -5 \times 10^3 N$

$10^6 \begin{bmatrix} \ 257.76 & -77.76 & 103.68 \\ \ -77.76 & 155.52 & 0 \\ \ 103.68 & 0 & 276.48 \\ \end{bmatrix} \begin{Bmatrix} \ X_2 \\ \ X_3 \\ \ Y_3 \\ \end{Bmatrix} = \begin{Bmatrix} \ 0 \\ \ -2 \\ \ -5 \\ \end{Bmatrix} \times 10^3$

$X_2 = 4.86 \times 10^{-6} m\\ X_3 = -10.4 \times 10^{-6} m\\ Y_3 = -19.9 \times 10^{-6} m$

Reaction:

$F_{x1} = (-180 X_2 - 77.76 X_3 - 103.68y_3) \times 10^6 = 1997.13 N\\ F_{y1} = (-103.68 X_3 - 138.24y_3) \times 10^6 = 3829.25 N\\ F_{y2} = (-103.68X_2 + 103.68 X_3 - 138.24 Y_3) \times 10^6 = 1168.82 N$

$\sum F_x = -2000 + 1997.13 = 2.87 \approx 0 \\ \sum F_x = -5000 + 3829.25 + 1168.82 = 1.932 approx 0$

Stress:

$\sigma = \frac{E}{L} \begin{bmatrix} \ -c & -s & C & s \\ \end{bmatrix} \begin{Bmatrix} \ X_1 \\ \ Y_1 \\ \ X_2 \\ \ Y_2 \\ \end{Bmatrix}\\ \sigma_1 = \frac{180 \times 10^9}{0.6} \begin{bmatrix} \ -1 & 0 & 1 & 0 \\ \end{bmatrix} \begin{Bmatrix} \ 0 \\ \ 0 \\ \ 4.86 \\ \ 0 \\ \end{Bmatrix} \times 10^6 = 1.458 \times 10^6 N/m^2 (Tensile)$

$\sigma_2 = \frac{180 \times 10^9}{0.5} \begin{bmatrix} \ 0.6 & -0.8 & -0.6 & -0.8 \\ \end{bmatrix} \begin{Bmatrix} \ 4.86 \\ \ 0 \\ \ -10.4 \\ \ -19.9 \\ \end{Bmatrix} \times 10^6 = 25.076 \times 10^6 N/mm^2 (Tensile)$

$\sigma_3 = \frac{180 \times 10^9}{0.5} \begin{bmatrix} \ -0.6 & -0.8 & 0.6 & 0.8 \\ \end{bmatrix} \begin{Bmatrix} \ 0 \\ \ 0 \\ \ -10.4 \\ \ -19.9 \\ \end{Bmatrix} \times 10^6 = - 22.16 \times 10^6 N/mm^2 = 22.16 \times 10^6 N/mm^2 (Compressive)$

Strain

$e_1 = \frac{\sigma_1}{E} = \frac{1.458 \times 10^6}{180 \times 10^9} = 8.16 \times 10^{-6}\\ e_2 = \frac{\sigma_2}{E} = \frac{25.076 \times 10^6}{180 \times 10^9} = 1.393 \times 10^{-4}\\ e_3 = \frac{\sigma_3}{E} = \frac{22.16 \times 10^6}{180 \times 10^9} = 1.23 \times 10^{-4}$