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

Marks: 10M

Year: May 2015

$A_{st} = 200 mm^2 \hspace{1.5cm} E_{st} = 2 \times 10^5 N/mm^2\\ A_{At} = 370 mm^2 \hspace{1.5cm} E_{7 \times 10^4 N/mm^2}\\ A_{Br} = 370 mm^2 \hspace{1.5cm} E_{Br} = 8.8 \times 10^4 N/mm^2$

Elemental stiffness matrix,

$K_1 = \frac{AE}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{A_{AI} \times E_{AI}}{500} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = 10^3 \begin{bmatrix} \ 51.8 & -51.8 \\ \ -51.8 & 51.8 \\ \end{bmatrix}$

$K_2 = \frac{AE}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{A_{Br} \times E_{Br}}{500} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = 10^3 \begin{bmatrix} \ 65.12 & -65.12 \\ \ -65.12 & 65.12 \\ \end{bmatrix}$

$K_3 = \frac{AE}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{A_{st} \times E_{st}}{1000} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = 10^3 \begin{bmatrix} \ 40 & -40 \\ \ -40 & 40 \\ \end{bmatrix}$

$K_4 = \frac{AE}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{A_{st} \times E_{st}}{100} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = 10^3 \begin{bmatrix} \ 40 & -40 \\ \ -40 & 40 \\ \end{bmatrix}$

Global stiffness Matrix:

$k = 10^3 \begin{bmatrix} \ 131.8 & -51.8 & -80\\ \ -51.8 & 116.92 & -65.12 \\ \ -80 & -65.12 & 145.12 \\ \end{bmatrix}$

Global matrix Equation,

[k] {u} = {F}

$10^3 \begin{bmatrix} \ 131.8 & -51.8 & -80\\ \ -51.8 & 116.92 & -65.12 \\ \ -80 & -65.12 & 145.12 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ u_2 \\ u_3 \end{Bmatrix} = \begin{Bmatrix} \ F_1 \\ F_2 \\ F_3 \end{Bmatrix}$

Global Boundary condition $u_3 = 0, F_1 = 20 \times 10^3 N , F_2 = F_3 = 0$

$10^3 \begin{bmatrix} \ 131.8 & -51.8 & -80\\ \ -51.8 & 116.92 & -65.12 \\ \ -80 & -65.12 & 145.12 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ u_2 \\ 0 \end{Bmatrix} \begin{Bmatrix} \ 20 \\ 0 \\ 0 \end{Bmatrix} \times 10^3$

$u_1 = 0.1837 mm\\ u_2 = 0.0814 mm$

Stress:

$\sigma_1 = \sigma_{AI} = \frac{E_{AI}}{L_{AI}} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ u_1 \\ u_2 \\ \end{Bmatrix} = \frac{7 \times 10^4}{500} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ 0.1837 \\ \ 0.0814 \\ \end{Bmatrix} = -14.32 N/mm^2$

$\sigma_2 = \sigma_{Br} = \frac{E_{Br}}{L_{Br}} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ u_2 \\ u_3 \\ \end{Bmatrix} = \frac{8.8 \times 10^4}{500} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ 0.0814 \\ \ 0 \\ \end{Bmatrix} = -14.326 N/mm^2$

$\sigma_3 = \sigma_4 = \sigma_{st} = \frac{E_{st}}{L_{st}} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ u_1 \\ u_3 \\ \end{Bmatrix} = \frac{2 \times 10^5}{1000} [-1 \hspace{0.4cm} 1] \begin{Bmatrix} \ 0.1837 \\ \ 0 \\ \end{Bmatrix} = -36.74 N/mm^2$