Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 5M

Year: may 2016

$\theta(L) = 470^0 \hspace{3cm}\theta(L) =170^0c$

$\hspace{0.8cm}=(T_L-T_a)\hspace{3.cm} = (T_L-T_a)$

Governing d .e is,

$KA\frac{d^2\theta}{dx^2}-hp\theta = 0 \hspace{0.9cm} where , \theta = T_x-T_a$

$ \hspace{0.6cm}\frac{d^2\theta}{dx^2}-m^2\theta = 0 \hspace{0.9cm} where ,m^2 = \frac{hp}{KA}$

$Q_1^e\hspace{1.6cm}Q_2^e$

$\hspace{0.6cm}Q_1^e =-KA\frac{d\theta}{dx}$

$\hspace{0.6cm}Q_2^e =KA\frac{d\theta}{dx}$

Residue ,$ R = \frac{d^2\theta}{dx^2}-m^2\theta$

weighted integral form ,

$\int\limits_0^{he} w_i \hspace{0.1cm}R \hspace{0.1cm}dx = 0$ …