written 6.6 years ago by
teamques10
★ 67k

•
modified 6.5 years ago

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(\xi1)4+\frac{1}{2}\xi(\xi+1)8+(1\xi^2)6 $
$ =2\xi^22\xi+4\xi^2+4\xi+66\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(\xi1)4+\frac{1}{2}\xi(\xi+1)8+(1\xi^2)5 $
$ =2\xi^22\xi+4\xi^2+4\xi+55\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 nonlinear.
Such a transformation is useful to formulate finite elements having curved edges so that curved structural geometry can be modelled.