Consider the truss shown in figure.Given E = 210 GPa and cross section area A = 1 $cm^2$ for each element. Determine

1. Displacement at each node.

2. Stresses induced in each element.

3. Reaction at supports

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

Marks: 10M

Year: May 2015

$E = 210 GPa = 210 \times 10^3 N/mm^2\\ A = 1 cm^2 = 100 mm^2$

Sine rule,

$\frac{1000}{sin 105} = \frac{L_1}{sin 45} = \frac{L_2}{sin 30}\\ L_1 = 732.05 mm \\ L_2 = 517.638 mm$

Elemental stiffness Matrix:

$K = \begin{bmatrix} \ C^2 & cs & -c^2 & -cs \\ \ cs & s^2 & -cs & -s^2 \\ \ -c^2 & -cs & C^2 & cs \\ \ -cs & -s^2 & cs & s^2 \\ \end{bmatrix}$

$k_1 = 10^3 \begin{bmatrix} \ 20.01 & -11.35 & -20.01 & 11.55 \\ \ -11.55 & 6.67 & 11.55 & -6.67 \\ \ -20.01 & 11.55 & 20.01 & -11.55 \\ \ 11.55 & -6.67 & -11.55 & 6.67 \\ \end{bmatrix}$

$k_2 = 10^3 \begin{bmatrix} \ 20.285 & 20.285 & -20.285 & -20.285 \\ \ 20.285 & 20.285 & -20.285 & -20.285 \\ \ -20.285 & -20.285 & 20.285 & 20.285 \\ \ -20.285 & -20.285 & -20.285 & 20.285 \\ \end{bmatrix}$

$k_3 = 10^3 \begin{bmatrix} \ 21 & 0 & -21 & 0 \\ \ 0 & 0 & 0 & 0 \\ \ -21 & 0 & 21 & 0 \\ \ 0 & 0 & 0 & 0 \\ \end{bmatrix}$

Global stiffness Equation

[k] {u} = {F}

$10^3 \begin{bmatrix} \ 40.295 & 8.732 & -20.285 & -20.285 & -20.01 & 11.55 \\ \ 8.732 & 26.955 & -20.285 & -20.285 & 11.55 & -6.67 \\ \ -20.285 & -20.285 & 41.285 & 20.285 & -21 & 0 \\ \ -20.285 & -20.285 & 20.285 & 20.285 & 0 & 0 \\ \ -20.01 & 11.55 & -21 & 0 & 41.01 & -11.55 \\ \ 11.55 & -6.67 & 0 & 0 & -11.55 & 6.67 \\ \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}$ Boundary condition $x_2 = y_2 = x_3 = y_3 = 0, F_{x1} = 10KN, F_{y1} = -20KN$ $10^3 \begin{bmatrix} \ 40.295 & 8.732 \ \ 8.732 & 26.955 \ \end{bmatrix} \begin{bmatrix} \ x_1 \ \ y_1 \ \end{bmatrix} = \begin{bmatrix} \ 10 \ -20 \ \end{bmatrix} \times 10^3$ $x_1 = 0.44 mm\ y_1 = -0.884mm$ Support Reaction: $R_{x2} = -20.285(0.44) - 20.285(-0.884) = 9 KN\ R_{y2} = -20.285(0.44) - 20.285 (-0.884)\ R_{x3} = -20.01 (0.44) + 11.55(-0.884) = -19.01 KN\ R_{y3} = 11.55(0.44) - 6.67(-0.884) = 10.97 KN$ $\sum f_x = F_{x1} + R_{x2} + R_{x3} = 10 + 9 - 19.01 = -0.01 \approx 0\ \sum f_y = F_{y1} + R_{y2} + R_{y3} = -20 + 9 + 10.97 = -0.03 \approx 0$ Stresses: $\sigma = \frac{E}{L} \begin{bmatrix} \ -c & -s & c & s \ \end{bmatrix} \begin{Bmatrix} \ x_1 \ \ y_1 \ \ x_2 \ \ y_2 \ \end{Bmatrix}\\ \sigma_1 = \frac{210 \times 10^3}{732.05} \begin{bmatrix} \ 0.866 & -0.5 & -0.866 & 0.5 \ \end{bmatrix} \begin{Bmatrix} \ 0.44 \ \ -0.884 \ \ 0 \ \ 0 \ \end{Bmatrix} = 236.01 N/mm^2\ \sigma_2 = \frac{210 \times 10^3}{517.638} \begin{bmatrix} \ -0.707 & -0.707 & 0.707 & 0.707 \ \end{bmatrix} \begin{Bmatrix} \ 0.44 \ \ -0.884 \ \ 0 \ \ 0 \ \end{Bmatrix} = 127.34 N/mm^2\\ \sigma_3 = \frac{210 \times 10^3}{1000} \begin{bmatrix} \ 1 & 0 & -1 & 0 \ \end{bmatrix} \begin{Bmatrix} \ 0 \ \ 0 \ \ 0 \ \ 0 \ \end{Bmatrix} = 0 N/mm^2$