written 7.7 years ago by | • modified 2.8 years ago |

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

**Marks:** 5M

**Year:** May 2015

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Derive shape function for id quadratic element in natural coordinates.

written 7.7 years ago by | • modified 2.8 years ago |

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

**Marks:** 5M

**Year:** May 2015

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written 7.7 years ago by |

Shape function for D quadratic element in natural cordinates.

Let $\phi = A \xi (\xi - 1)(\xi + 1)$

At node 1, $(\xi + 1)$ vanishes.

$\phi_1 = A \xi (\xi - 1)$

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

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

Similarly at node 2, turn $(\xi - 1)$ vanishes.

$\therefore \phi_2 = A \xi (\xi + 1)$

At node 2, $\phi_2 = 1$ and $\xi = 1$

$\therefore 1 = A.1(1+1) = 2A\\ A = \frac{1}{2}\\ \therefore \phi_2 = \frac{1}{2} \xi(\xi + 1)$

At node 3, $\phi_3 = 1$ and $\xi = 0$

$\therefore 1 = A(-1)(1)\\ \therefore A = -1\\ \ therefore \phi_3 = (1 - \xi)(1 + \xi)$

Shape function:

$\phi_1 = \frac{1}{2} \xi (\xi - 1) \hspace{1cm} \phi_2 = \frac{1}{2} \xi (\xi + 1) \hspace{1cm} \phi_3 = (1 - \xi)(1 + \xi)$

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