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

Marks: 10M

Year: Day 2016

$\hspace{1cm}E=200 GPa$

$\hspace{1.3cm}= 200\times 10^3 \frac{n}{mw^2}$

Elemental stiffness matrix.

$\hspace{1.6cm}k=\frac{AE}{L}$ $\begin{bmatrix} \ c^2 & cs & -C^2 & -cs \\ \ cs & s^2 & -cs & -s^2 \\ \ -c^2 & -cs & c^2 & cs \\ \ -cs & -s^2 & cs & s^2 \\ \end{bmatrix}$

$K_1=10^3$ $\begin{bmatrix} \ 70 & 0 & -70 & 0 \\ \ 0 & 0 & 0 & 0 \\ \ -70 & 0 & 70 & 0 \\ \ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ $k_2=10^3$ $\begin{bmatrix} \ 49.495 & -49.495 & -49.495 & +49.495 \\ \ -49.495 & 49.495 & 49.495 & -49.495 \\ \ -49.495 & 49.495 & 49.495 & -49.495 \\ \ 49.495 & -49.495 & -49.495 & 49.495 \\ \end{bmatrix}$

$K_3=10^3$ $\begin{bmatrix} \ 49.495 & 49.495 & -49.495 & -49.495 \\ \ 49.495 & 49.495 & -49.495 & -49.495 \\ \ -49.495 & -49.495 & +49.495 & 49.495 \\ \ -49.495 & -49.495 & +49.495 & 49.495 \\ \end{bmatrix}$

$[K]_3=10^3$ $\begin{bmatrix} \ 119.495 & 49.495 & -70 &0 &-49.495 & -49.495 \\ \ 49.495 & 49.495 & 0 & 0 & -49.495 & -49.495 \\ \ -70 & 0 & 119.495 & -49.495 & -49.495 & 49.495 \\ \ 0 & 0 & -49.495 & 49.495 & 49.495 & -49.495 \\ \ -49.495 & -49.495 & -49.495 & 49.495 & 98.99 & 0 \\ \ -49.495 & -49.495 & 49.495 & -49.495 & 0 & 98.99 \\ \end{bmatrix}$

$10^3$ $\begin{bmatrix} \ 119.495 & 49.495 & -70 &0 &-49.495 & -49.495 \\ \ 49.495 & 49.495 & 0 & 0 & -49.495 & -49.495 \\ \ -70 & 0 & 119.495 & -49.495 & -49.495 & 49.495 \\ \ 0 & 0 & -49.495 & 49.495 & 49.495 & -49.495 \\ \ -49.495 & 49.495 & -49.495 & 49.495 & 98.99 & 0 \\ \ -49.495 & 49.495 & 49.495 & -49.495 & 0 & 98.99 \\ \end{bmatrix}$ $\begin{bmatrix} \ x_1 \\ \ y_1 \\ \ x_2 \\ \ y_2 \\ \ x_3 \\ \ y_3 \\ \end{bmatrix}$= $\begin{bmatrix} \ x_1 \\ \ y_1 \\ \ 0 \\ \ y_2 \\ \ -15\times10^3 \\ \ -5\times10^3 \\ \end{bmatrix}$

$x_2$ = -0.0714mm

$x_3$ = -0.1872 mm

$y_3$ = -0.0148 mm

$\hspace{0.6cm}fx_1 = 15 \times 10^3 N.= 15KN$

$\hspace{0.6cm}fy_1 = 10 \times 10^3 N.=10KN $

$\hspace{0.6cm}fy_3 = 5 \times 10^3 N = .15KN$

$\hspace{1.7cm}\sum fx=15-15=0 KN$

$\hspace{1.7cm}\sum fy=10-5=0 KN$

Stress :

$6^1= \frac{F}{L} [-c \hspace{0.2cm}-s\hspace{0.2cm} c \hspace{0.2cm} s]$ $\begin{Bmatrix} \ y_1 \\ \ y_1 \\ \ y_2 \\ \ y_2 \\ \end{Bmatrix}$

$6_1 =\frac{200\times10^3}{1000}[-1\hspace{0.2cm}0\hspace{0.2cm}1\hspace{0.2cm}0]$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ -0.0714 \\ \ 0 \\ \end{Bmatrix} = 14.28 N/mm^2(compressive)$

$6_2 =\frac{200\times10^3}{707.1}[0.707\hspace{0.2cm}-0.707\hspace{0.2cm}-0.707\hspace{0.2cm}0.707]$ $\begin{Bmatrix} \ -0.0714 \\ \ 0 \\ \ -0.01872 \\ \ -0.0148 \\ \end{Bmatrix} = 20.19 N/mm^2(Tensile)$

$6_3 =\frac{200\times10^3}{-707.1}[0.707\hspace{0.2cm}-0.707\hspace{0.2cm}0.707\hspace{0.2cm}0.707]$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ -0.01872 \\ \ -0.0148 \\ \end{Bmatrix} = 40.4 N/mm^2(compressive)$

Strain

$e_1=\frac{\sigma_1}{E}= \frac{14.28}{200\times10^3}=7.14 \times10^{-5}$

$e_2=\frac{\sigma_1}{E}= \frac{2.-0.19}{200\times10^3}=1.0095 \times10^{-4}$

$e_3=\frac{\sigma_1}{E}= \frac{40.4}{200\times10^3}=2.02 \times10^{-4}$