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Find nodal displacement and element stress for the bar as shown in figure using FEM. Take $E=200GP_a$

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$k_1 =\frac{AE}{L}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}= \frac{200\times200\times10^3}{300}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}=10^3$ $\begin{bmatrix} \ 166.67 & -166.67 \\ \ -166.67 & 166.67 \\ \end{bmatrix}_2^1$

$k_2 =\frac{AE}{L}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}= \frac{400\times200\times10^3}{300}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}=10^3$ $\begin{bmatrix} \ 400 & -400 \\ \ -400 & 400 \\ \end{bmatrix}_3^2$

$k_1 =\frac{AE}{L}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}= \frac{400\times200\times10^3}{300}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}=10^3$ $\begin{bmatrix} \ 400 & -400 \\ \ -400 & 400 \\ \end{bmatrix}_4^3$

$[K]=10^3$ $\begin{bmatrix} \ 166.67 & -166.67 & 0 & 0 \\ \ -166.67 & 566.67 & -400 & 0 \\ \ 0 & -400 & 800 & -400 \\ \ 0 & 0 & -400 & 400 \\ \end{bmatrix}$

without considering Gap :

$\hspace{1.3cm}$[k][u]=[f]

$\begin{bmatrix} \ 166.67 & -166.67 & 0 & 0 \\ \ -166.67 & 566.67 & -400 & 0 \\ \ 0 & -400 & 800 & -400 \\ \ 0 & 0 & -400 & 400 \\ \end{bmatrix}$ $\begin{bmatrix} \ u_1 \\ \ u_2 \\ \ u_3 \\ \ u_4 \\ \end{bmatrix}$= $\begin{bmatrix} \ 0 \\ \ 0 \\ \ 75 \times10^3 \\ \ 0 \\ \end{bmatrix}$

$\hspace{3cm}u_2= 0.449 mm$

$\hspace{3cm}u_3= 0.637 mm$

$\hspace{3cm}u_4= 0.637 mm$

As space provided for expansion is 3.5 mm that is more than actual deformation (u.0.637 mm)

stress

$\sigma_1 =\frac{ E(u_2-u_1)}{L_1}=29.93 N/mm^2$

$\sigma_2 =\frac{ E(u_3-u_2)}{L_2}=18.8 N/mm^2$

$\sigma _3=\frac{ E(u_4-u_3)}{L_3}=0 N/mm^2$

strain

$e_1=\frac{ u_4-u_3}{L_3}= 1.49 \times10^3$

$\begin{vmatrix} \ e_2\frac{u_3-u_3}{L_2} \\ \ =9.4\times 10^{-4} \\ \end{vmatrix}$

$e_3 = \frac{u_4-u_3}{L_2}=0$

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