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

Marks: 8M

Year: Dec 2015

Consider quadrilateral element with eight nodes,

$i) 2 - 3 : \xi \hspace{1.5cm} \therefore 1 - \xi = 0\\ ii) 3 - 4 : \eta = 1 \hspace{1.5cm} \therefore 1 - \eta = 0\\ iii) 8 - 5 : y = mx + c$

$\eta = \bigg( \frac{0 - (-1)}{-1-0} \bigg) \xi + (-1)\\ \eta = - \xi - 1\\ \therefore \eta + \xi + 1 = 0$

To find $\phi_1$, $\phi_1$ vanishes along lines,

Let, $\phi_1 = A(1 - \xi) (1 - \eta)(1 + \xi + \eta)$

At node 1: $\phi_1 = 1, \xi = -1$ and $\eta = -1$

$\therefore 1 = A(2)(2)(1-1-1)\\ A = -\frac{1}{4}\\ \therefore \phi_1 = - \frac{1}{4} (1 - \xi)(1 - \eta)(1 + \eta + \xi)$

Similarly,

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

To find $\phi_5, \phi_5$ vanishes along,

$i) 2 - 3 : \xi= 1 \hspace{1.5cm} \therefore 1 - \xi = 0\\ ii) 3 - 4 : \eta = 1 \hspace{1.5cm} \therefore 1 - \eta = 0\\ iii) 4 - 1 : \xi = -1 \hspace{1.5cm} \therefore 1 + \xi = 0$

$\therefore \phi_5 = A(1 - \xi)(1 - \eta)(1 + \xi)$

At node 5, $\phi_5 = 1, \xi = 0$ and $\eta = -1$

1 = A(1)(2)(1) $A = \frac{1}{2}\\ \phi_5 = \frac{1}{2}(1 - \xi^2)(1 - \eta)$

Similarly,

$\phi_6 = \frac{1}{2}(1 + \xi)(1 - \eta^2)\\ \phi_7 = \frac{1}{2}( 1 - \xi^2)(1 + \eta)\\ \phi_8 = \frac{1}{2}(1 - \xi)(1 - \eta^2)$