Find the expression of x in terms of when: i) Third node R is taken at (6, 0)

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

iii)Comment on the result.

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

$x = xjφj = φ1x1 + φ2x2 + φ3x3$

=$ \frac{1}{2}ξ(ξ − 1)4 +\frac{1}{2}ξ (ξ + 1)8 + (1 −\zeta^{2} )6$

=2$\xi^{2}-2\xi +4\xi^{2}+4\xi+6+6\xi^{2}$

x = $2ξ + 6$

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

x = $∑x_{j}φ_{j} = φ_{1}x_{1} + φ_{2}x_{2} + φ_{3}x3$

= $\frac{1}{2}4+ \frac{1}{2}ξ(ξ − 1) 8 + (1 −\xi^{2} )5$

=2$\xi^{2}-2\xi+4\xi^{2}+4\xi+5-5\xi^{2}$

x=ξ2 + 2ξ + 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.