Co-ordinates of the nodes of finite element are given by P(4, 0) and Q(8, 0). Find the expression of x in terms of $\zeta$ when:

i) Third node R is taken at (6, 0)

ii) Third node R is taken at (5,0)

Comment on the result.

Subject: Finite Element Analysis

Topic: Two Dimensional Finite Element Formulations

Difficulty: Medium

i) Third node R is taken at (6,0)

$ x= x_j \phi_j = \phi_1 x_1 + \phi_2 x_2+\phi_3 x3 $
$ =\frac{1}{2}\xi(\xi-1)4+\frac{1}{2}\xi(\xi+1)8+(1-\xi^2)6 $
$ =2\xi^2-2\xi+4\xi^2+4\xi+6-6\xi^2 $
$ \,\,\,x=2\xi+6$

ii) Third node R is taken at (5,0)

$ x=\sum x_j\phi_j=\phi_1x_1+\phi_2 x_2+\phi_3 x3 $
$ =\frac{1}{2}\xi(\xi-1)4+\frac{1}{2}\xi(\xi+1)8+(1-\xi^2)5 $
$ =2\xi^2-2\xi+4\xi^2+4\xi+5-5\xi^2 $
$ \,\,\,x=\xi_2+2\xi+5$

Comment:- When C is taken at midpoint of the element the transformation $x$ and $\xi$ is linear, but when C is taken away from the midpoint the transformation becomes non-linear.
Such a transformation is useful to formulate finite elements having curved edges so that curved structural geometry can be modelled.