Subject: Finite Element Analysis

Topic: Two Dimensional Finite Element Formulations

Difficulty: Medium

$x,y=2.5,2.5 $
$x_1,y_1=2,2 $
$x_2,y_2=4,2 $
$x_3,y_4=4,3 $
$x_4,y_4=2,4 $

Converting into-local coordinate,

$l=2 \hspace{1cm}h=2$
$ \bar{x}=x-l\hspace{1cm}\bar{y}=y-h$
$ \bar{x}=2.5-2=0.5\hspace{1cm}\bar{y}=2.5-2=0.5$
$\bar{x_1}=0\hspace{1cm}\bar{y_1}=0$
$\bar{x_2}=2\hspace{1cm}\bar{y_2}=0$
$\bar{x_3}=2\hspace{1cm}\bar{y_3}=1$
$\bar{x_4}=0\hspace{1cm}\bar{y_4}=1$

Now, $\phi_1=\frac{1}{4}(1-\xi)(1-\eta)$
$\phi_2=\frac{1}{4}(1+\xi)(1-\eta)$
$\phi_3=\frac{1}{4}(1+\xi)(1+\eta)$
$\phi_4=\frac{1}{4}(1-\xi)(1+\eta)$

Now,$ \bar{x} =\sum \phi_j \bar{x_j} $
$\therefore \bar{x}=\phi_1\bar{x_1}+\phi_1\bar{x_2}+\phi_1\bar{x_3}+\phi_1\bar{x_4}$

$0.5=\frac{1}{4}(1-\xi)(1-\eta)\bar{x_1}+\frac{1}{4}(1+\xi)(1-\eta)\bar{x_2}+\frac{1}{4}(1+\xi)(1+\eta)\bar{x_3}+\frac{1}{4}(1-\xi)(1+\eta)\bar{x_4}$

i.e $\frac{1}{2}=\frac{1}{4}(1+\xi)(1-\eta)*2+\frac{1}{4}(1+\xi)(1+\eta)*2$
$\frac{1}{2}=\frac{1}{2}(1+\xi)(1-\eta)+\frac{1}{2}(1+\xi)(1+\eta)$
$1=1+\xi+\eta-\eta\xi+1+\xi+\eta+\eta\xi$
$1=2+2\xi$
$\therefore \xi=\frac{-1}{2}$

Now, $\bar{y}=\sum \phi_j\bar{y_j}$ $0.5=\frac{1}{4}(1-\xi)(1-\eta)\bar{y_1}+\frac{1}{4}(1+\xi)(1-\eta)\bar{x_2}+\frac{1}{4}(1+\xi)(1+\eta)\bar{x_3}+\frac{1}{4}(1-\xi)(1+\eta)\bar{x_4}$ $0.5=\frac{1}{4}(1+\xi)(1+\eta)+\frac{1}{4}(1-\xi)(1+\eta)$ $\therefore \frac{1}{2}=\frac{1}{4}[1+\xi+\eta+\eta\xi+1-\xi+\eta-\eta\xi$ $\therefore 2=2+2\eta\hspace{1cm}=\gt\hspace{1cm}2\eta=0\hspace{1cm}=\gt\hspace{1cm}\eta=0$

Therefore, point P is given by (-1/2,0) in $\xi-\eta$ coordinate,
To compute the result,
$\phi_1=\frac{1}{4}(1-\xi)(1-\eta)=\frac{1}{4}*(1+\frac{1}{2})*1=\frac{3}{8}$
$\phi_2=\frac{1}{4}(1+\xi)(1-\eta)=\frac{1}{4}*(1-\frac{1}{2})*1=\frac{1}{8}$
$\phi_3=\frac{1}{4}(1+\xi)(1+\eta)=\frac{1}{4}*(1-\frac{1}{2})*1=\frac{1}{8}$
$\phi_4=\frac{1}{4}(1-\xi)(1+\eta)=\frac{1}{4}*(1+\frac{1}{2})*1=\frac{3}{8}$

Now,
$\phi_1+\phi_2+\phi_3+\phi_4=\frac{3}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}=\frac{8}{8}$
$\phi_1+\phi_2+\phi_3+\phi_4=1$

Now, Temperature at point P,
$T_P=\phi_1T_1+\phi_2T_2+\phi_3T_3+\phi_4T_4$
$\therefore \,\,\, T_1=100^0C\,\,\,T_2=60^0C\,\,\,T_3=50^0C\,\,\,T_4=90^0C$
$\therefore\,\,T_P=\frac{3}{8}*100+\frac{1}{8}*60+\frac{1}{8}*50+\frac{3}{8}*90$
$\therefore\,\,T_P=85^0C$