Subject: Finite Element Analysis

Topic: One Dimensional Problems

Difficulty: Medium

a) Number of elements and nodes,

b) Element matrix Equation is given by,

$\begin{bmatrix} K \end{bmatrix}^e=\frac{AE}{h_e}\begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix}$

$For\,\, element \,\,1,\,\,A=250mm^2 \,\,,\,\, h_e=300mm$
$\frac{AE}{h_e}=\frac{250*2*10^5}{300}=167.67*10^3$
$\begin{bmatrix} K \end{bmatrix}^1=10^3\begin{bmatrix} 167.67&-167.67\\ -167.67&167.67 \end{bmatrix}$

$For\,\, element \,\,2,\,\,A=400mm^2 \,\,,\,\, h_e=200mm$
$\frac{AE}{h_e}=\frac{400*2*10^5}{200}=400*10^3$
$\begin{bmatrix} K \end{bmatrix}^2=10^3\begin{bmatrix} 400&-400\\ -400&400 \end{bmatrix}$

$For\,\, element \,\,3,\,\,A=400mm^2 \,\,,\,\, h_e=200mm$
$\frac{AE}{h_e}=\frac{400*2*10^5}{200}=400*10^3$
$\begin{bmatrix} K \end{bmatrix}^3=10^3\begin{bmatrix} 400&-400\\ -400&400 \end{bmatrix}$

c) Elemental matrix
$\begin{bmatrix} K \end{bmatrix}=10^3\begin{bmatrix} 167.67&-167.67&0&0\\-167.67&167.67+400&-400&0\\0&-400&400+400&-400 \\0&0&-400&400 \end{bmatrix}$

d) Global Matrix Equation,
$\begin{bmatrix} K \end{bmatrix}\begin{Bmatrix} u \end{Bmatrix}=\begin{Bmatrix} P \end{Bmatrix}$
$ 10^3 \begin{bmatrix} 167.67&-167.67&0&0\\-167.67&567.67&-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} P_1 \\ P_2\\P_3\\P_4 \end{Bmatrix}$

e) Imposing boundary condition,
$u_1=0\,\,,\,\,u_4=3.5mm\,\,,\,\,P_3=75*10^3N$
$P_2=0\hspace{3cm}$ -for balancing,

Frame the equation,
$10^3(-167.67*u_2)P_1\hspace{9cm}$ -(1)
$10^3(567.67u_2-400u_3)=0\hspace{8cm}$ -(2)
$10^3(-400u_2+800u_3-400*3.5)=75*10^3\hspace{5cm}$ -(3)
$10^3(-400u_3+400*3.5)=P_4\hspace{7.4cm}$ -(4)

Solving equation (2) and (3) we get,
$\therefore \,\,\,u_2=0.10389mm\hspace{2cm}u_3=0.1474mm$
Substituting in equation (1) and (4), we get,
$P_1=-17.41kN \,\,\,P_4=-57.56$
$\sum P_x=P_1+P_2+P_3+P+4=-17.41+0+75-57.56=0 \hspace{3cm} $ -verified