Subject: Finite Element Analysis

Topic: One Dimensional Problems

Difficulty: Medium

a) Number of elements and nodes.

b) Elemental matrix,
$\begin{bmatrix}K\end{bmatrix}^e=\frac{GJ}{h_e}\begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix}$

$For\,\, element \,\,1,\,\,d=100mm \,\,,\,\, h_e=450mm$
$J=\frac{\pi}{32}d^4=\frac{\pi}{32}*100^4=9.817*10^6$
$\frac{GJ}{h_e}=\frac{1*10^5*9.817*10^6}{450}=2181.7*10^6$
$\begin{bmatrix} K \end{bmatrix}^1=10^6\begin{bmatrix} 2181.7&-2181.7\\ -2181.7&2181.7 \end{bmatrix}$

$For\,\, element \,\,2,\,\,d=80mm \,\,,\,\, h_e=400mm$
$J=\frac{\pi}{32}d^4=\frac{\pi}{32}*80^4=4.02*10^6$
$\frac{GJ}{h_e}=\frac{1*10^5*4.02*10^6}{400}=1005.3*10^6$
$\begin{bmatrix} K \end{bmatrix}^2=10^6\begin{bmatrix} 1005.3&-1005.3\\ -1005.3&1005.3 \end{bmatrix}$

$For\,\, element \,\,3,\,\,d=50mm \,\,,\,\, h_e=500mm$
$J=\frac{\pi}{32}d^4=\frac{\pi}{32}*50^4=0.613*10^6$
$\frac{GJ}{h_e}=\frac{1*10^5*0.613*10^6}{500}=122.7*10^6$
$\begin{bmatrix} K \end{bmatrix}^3=10^6\begin{bmatrix} 122.7&-122.7\\ -122.7&122.7 \end{bmatrix}$

c) Elemental matrix
$ \hspace{2cm}=10^6\begin{bmatrix} 2181.7&-2181.7&0&0\\-2181.7&3187&-1005.3&0\\0&-1005.3&1128&-122.7\\0&0&-122.7&122.7 \end{bmatrix}$

d) Global Matrix Equation,
$\begin{bmatrix} K \end{bmatrix}\begin{Bmatrix} \theta \end{Bmatrix}=\begin{Bmatrix} T \end{Bmatrix}$
$ 10^6\begin{bmatrix} 2181.7&-2181.7&0&0\\-2181.7&3187&-1005.3&0\\0&-1005.3&1128&-122.7\\0&0&-122.7&122.7 \end{bmatrix}\begin{Bmatrix} \theta_1\\ \theta_2 \\ \theta_3 \\ \theta_4 \end{Bmatrix}=\begin{Bmatrix} T_1 \\ T_2\\T_3\\ T_4 \end{Bmatrix}$

e) Imposing boundary condition,
$\theta_1=0\,\,,\, \theta_4=0 \,\,,\,\,T_3=-2*10^6\,N/mm$
$T_2=3*10^6\,N/mm$

f) Frame the equation,
$10^6(-2181.7\,\theta_2)=T_1\hspace{7cm}$ -(1)
$10^6(3187\,\theta_2-1005.3\,\theta_3)=3*10^6\hspace{5cm}$ -(2)
$10^6(-1005.3\,\theta_2+1128\,\theta_3)=-2*10^6\hspace{4.3cm}$ -(3)
$10^6(-122.7\,\theta_3)=T_4\hspace{7cm}$ -(4)

Solving equation (2) and (3) we get,
$\therefore \,\,\,\theta_2=5.314*10^{-4}\,rad\hspace{2cm}\theta_3=-1.3*10^{-3}\,rad$
Substituting in equation (1) and (4), we get,
$T_1=-1.159 KNm\hspace{2cm}T_4=0.159\,KNm$
$\sum T_x=T_1+T_2+T_3+T_4=-1.159+3-2+0.159=0 \hspace{3cm} $ -verified