Coordinate Transformation Relation-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Cartesian, Cooridinates, Nodes, Quadratic, Quadrilateral, Isoparametric, Relation, Global, Transformational

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Assignment
Finite Element Methods
Problems 4.20 and 4.21
Problem 4.20
The Cartesian coordinates of the nodes of a quadratic quadrilateral isoparametric element are shown in
Figure 4.26. Determine the coordinate transformation relation between the local and global coordinates.
Using this relation, find the global coordinates corresponding to the point (r. s) = (0.0).
Solution:
As we know
1
2
8
1
2
8
[]
x
x
x
xN
yy
y
y





Where
1 2 8 0 0
[] 0 0 1 2 8
N N N
NN N N

According to page 106
1(1 )(1 )( 1)
4i i i i
Ni rr ss rr ss
for i= 1,2,3
And
2
1(1 )(1 )
2i
Ni r ss
for i = 5,7
2
1(1 )(1 )
2i
Ni rr s
for i=6,8
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Assignment

Finite Element Methods

Problems 4.20 and 4.

Problem 4.

The Cartesian coordinates of the nodes of a quadratic quadrilateral isoparametric element are shown in

Figure 4.26. Determine the coordinate transformation relation between the local and global coordinates.

Using this relation, find the global coordinates corresponding to the point (r. s) = (0.0).

Solution:

As we know

1 2 8 1 2 8

[ ]

x

x

x x

N

y y

y

y

Where

[ ]

N N N

N

N N N

According to page 106

i i i i

Ni   rrss rrss  for i= 1,2,

And

2

i

Ni   rss for i = 5,

2

i

Ni   rrs for i=6,

Now the given node points for the elements are

1 1 2 2 3 3 4 4

5 5 6 6 7 7 8 8

r s r s r s r s

r s r s r s r s

Therefore

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8

xN xn xN xN xN xN xN xN x

And

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8

yN yn yN yN yN yN yN yN y

Now using the values of xi and yi

1 2 3 4 5 6 7 8

x  2.2 N  3.3 N  3.8 N  1.2 N  0.8 N  3.9 N  1.4 N 0.2 N

1 2 3 4 5 6 7 8

y  1.7 N  1.9 N  2.2 N  3.2 N  2.4 N  0.2 N  2.1 NN

By substituting the values of Ns, we obtain an expression for the relation between the local and global

coordinates for the given quadrilateral element

Now for( , ) r s (0,0)

1 2 3 4

5 6 7 8

N N N N

N N N N

So we get,

x

y