Global Coordinate System-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, Global, Coordinates, Corner, Nodes, Quadrilateral, Element, Transformation, Point, Local

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Problem 4.3
The Cartesian global coordinates of the corner nodes of a quadrilateral element
are given by (0,-1), (-2, 3),(2,4) and (5,3). Find the coordinate transformation
between the global and local (natural) coordinates. Using this determine the
Cartesian coordinates of the point defined by (r, s) = (0.5, 0.5) in the global
coordinate system.
SOLUTION
}= }
Where (xi, yi ) are the (x ,y) coordinates of node i(1,2,3,4)
Ni =1/4 (1+rri) (1+ssi) ;I =1,2,3,4
And (r1, s1) = (-1,-1)
(r2, s2) = (1,-1)
(r 3,s3)= (1, 1)
(r4, s4)= (-1, 1)
By substituting values of (xi, yi); i= 1,2,3,4 in equation 4.32, we obtain
X=N1x1+N2x2+N3x3+N4x4= 5N2+2N3-2N4
Y=N1y2+N2y2+N3y3+N4y4=N1+3N2+4N3+3N4
And N1=1/4(1-r) (1-s)
N2=1/4(1+r) (1-s)
N3=1/4(1+r) (1+s)
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Problem 4.

The Cartesian global coordinates of the corner nodes of a quadrilateral element are given by (0,-1), (-2, 3),(2,4) and (5,3). Find the coordinate transformation between the global and local (natural) coordinates. Using this determine the Cartesian coordinates of the point defined by (r, s) = (0.5, 0.5) in the global coordinate system.

SOLUTION

Where (xi, yi ) are the (x ,y) coordinates of node i(1,2,3,4)

Ni =1/4 (1+rri) (1+ssi) ;I =1,2,3,

And (r 1 , s 1 ) = (-1,-1)

(r 2 , s 2 ) = (1,-1) (r 3 ,s 3 )= (1, 1) (r 4 , s 4 )= (-1, 1)

By substituting values of (xi, yi); i= 1,2,3,4 in equation 4.32, we obtain

X=N 1 x 1 +N 2 x 2 +N 3 x 3 +N 4 x 4 = 5N 2 +2N 3 -2N 4

Y=N 1 y 2 +N 2 y 2 +N 3 y 3 +N 4 y 4 =N 1 + 3 N 2 +4N 3 +3N 4

And N 1 =1/4(1-r) (1-s)

N 2 =1/4(1+r) (1-s) N 3 =1/4(1+r) (1+s)

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N 4 =1/4(1-r) (1+s)

By substituting values of N 1 , N 2 , N 3 , N 4 in above relations, we obtained

X= 1/4 [5(1+r) (1-s) +2(1+r) (1+s)-2(1-r) (1+s)]

Y=1/4 [-1(1-r) (1-s) +3(1+r) (1-s) +4(1+r)(1+s)+3(1-r)(1+s)]

This is the coordination transformation between global and natural coordinates.

For global coordinates of point (r, s) = (0.5, 0.5)

Substitute the values of r and s in above relations

X=1/4[5(1.5)(0.5)+2(1.5)(1.5)-2(0.5)(1.5)]

X=1.

Y=1/4[-(0.5)(0.5)+3(1.5)(0.5)+4(1.5)(1.5)+3(0.5)(1.5)]

Y=3.

So global coordinates of point (x, y) = (1.6875,3.3125)

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