Transformation Matrix 1-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: Transformation, Matrix, Local, Degree, Freedom, Global, Planar, Frame, Element

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Q # 6.8
The transformation matrix between local degrees of freedom and global degree of for the planer frame
element shown in the figure below is given by
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
ox ox
oz oz
ox ox
oz oz
lm
lm
lm
lm









Where
cos
ox
l
sin
ox
m
cos(90 ) sin
oz
l

sin(90 ) cos
oz
m

Using this generate the transformation matrix for the three elements shown in the figure below;
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Q # 6.

The transformation matrix between local degrees of freedom and global degree of for the planer frame element shown in the figure below is given by

ox ox oz oz

ox ox oz oz

l m l m

l m l m

Where

lox cos 

mox sin 

loz  cos(90  )  sin

moz  sin(90   ) cos

Using this generate the transformation matrix for the three elements shown in the figure below;

Solution ;

Putting values of

lox , mox , loz and moz we get the matrix in the form as given below;

cos sin 0 0 0 0 sin cos 0 0 0 0 0 0 1 0 0 0 0 0 0 cos sin 0 0 0 0 sin cos 0 0 0 0 0 0 1

Now applying this transformation matrix for each of the element one by one.

For Element # 1;

For Element # 3;

For the third element we have θ = 900 and the matrix becomes