A 10 spar element having a linear shape function is as shown below. Find natural co-ordinate of point P. If temperature at node 1 is $50^0C$ and node 2 is $-20^C$, Find the temperature at point P.

$\therefore$ For linear element , $

$\phi_1=\frac{1}{2}=(1-\xi)\hspace{2cm}\phi_2=(1+\xi)$

$\hspace{1cm} x_1=25\hspace{2cm}x=30\hspace{2cm}x_2=50$

$\hspace{1cm} \therefore x=\sum \phi_i \,x_i$

$\hspace{1cm} \therefore x=\phi_1x_1+\phi_2x_2$

$\hspace{1cm} \therefore 30=\frac{1}{2}(1-\xi)*25+\frac{1}{2}(1+\xi)50$

$\hspace{1cm} \therefore 30=\frac{(25-25\xi+50+50\xi)}{2}$

$\hspace{1cm} \therefore 60=75+25\xi$

$\hspace{1cm} \therefore 25\xi=-15$

$\hspace{1cm} =\gt\xi=\frac{-15}{25}=-0.6$

Now,

$\hspace{1cm}T=\sum T_i\phi_i$
$\hspace{1cm}T=T_1\phi_1+T_2\phi_2$
$\hspace{1cm}\hspace{0.4cm}=50*0.5(1+0.6)+(-20)*0.5(1-0.6 )$
$\hspace{1cm}T=40-4$
$\hspace{1cm}T=36^0 C$