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Derive the shape function for a linear quadrilateral element and show its variation over the element.

shape function

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Shape function for Quadratic Element

quadratic element

The element has 3 nodes.

Pilot function will be $\phi = A \bar x \left[ \bar x - \frac{h_e}{2} \right] [\bar x - h_e] \space \space \space.... (1)$

For node 1, $\bar x$ vanishes.

$ \therefore \phi_1 = A \left[ \bar x - \frac{h_e}{2} \right] [\bar x - h_e] $

At node 1, $\phi_1 = 1$ & $\bar x = 0$, above equation becomes

$1 = A \left[ \frac{-h_e}{2} \right] (-h_e)$

$\therefore A = \frac{2}{h_e^2}$

$\therefore \phi_1 = \frac{2}{h_e^2} \left( \bar x - \frac{h_e}{2} \right) (\bar x - h_e)$

Rearranging the term

$\phi_1 = \left( 1 - \frac{2\bar x}{h_e} \right) \left( 1 - \frac{\bar x}{h_e} \right)$

diagram

For node 2, term $\left( \bar x - \frac{h_e}{2} \right) $ vanishes.

Equation (1) becomes $\phi_2 = A \bar x (\bar x - h_e)$

At node 2, $\bar x = \frac{h_e}{2} \space \space \space \space \phi_2 = 1$

$\therefore 1 = A. \frac{h_e}{2} \left( \frac{h_e}{2} - h_e \right) $

$\space \space \space = A \left( \frac{h_e}{2} \right) \left( \frac{-h_e}{2} \right) $

$A = \frac{-4}{h_e^2}$

$\therefore \phi_2 = \frac{-4}{h_e^2} \bar x (\bar x - h_e)$

Rearranging the term we get

$\phi_2 = \frac{4 \bar x}{h_e} \left( 1 - \frac{\bar x}{h_e} \right) $

diagram

For node 3, $(\bar x - h_e)$ vanishes.

Equation (1) becomes $\phi_3 = A \bar x \left( \bar x - \frac{h_e}{2} \right)$

At node 3, $\phi_3 = 1 \space \space \space \bar x = h_e$

$\therefore 1 = A. h_e \left(h_e - \frac{h_e}{2} \right) $

$\space \space \space = A (h_e) \left( \frac{h_e}{2} \right) $

$\therefore A = \frac{2}{h_e^2} $

$\phi_3 = \frac{2}{h_e^2}. \bar x (\bar x - h_e) $

Rearranging the term we get

$ \phi_3 = \frac{-\bar x}{h_e} \left( 1 - \frac{2\bar x}{h_e} \right) $

diagram

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