Global Stiffness Matrices 1-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

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Finite Element Methods

Assignment No. 6

Problem No. 6.

Problem No.6.

Solution: As, we know that

 K ( ) e^^   ^ ( ) e^^ ^ T^  K ( ) e ^ ( ) e    (^)     ^ ………………(1)

Now putting values in equation (1), we get,

( ) ( )^ ( ) ( )

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

e^ e^ e e

K A^ E

L

   ^    

( ) ( )^ ( ) ( )

cos 0 cos 0 sin 0 sin 0 cos sin 0 0 0 0 cos 0 0 cos sin 0 sin 0 sin

e^ e^ e e

K A^ E

L cos

 ^ ^  

   ^  

 

K G AE

 ^ ^ ^  

As A (1)^ E (1)= A (2)^ E (2)= AE

 

K G AE

 ^  

 ^  

And,

1 1 1 1 2 2 3 3

u Px v Py u P AE v u v

 ^   

 ^ ^   

 ^ ^  ^ 

2 2 3 3

x Py Px Py

Where, Px 1 = Py 1 = Px 2 = Px 3 = Py 3 =0 , Py 2 = -

2 2

u AE v

  ^  

  ^ ^ ^  

and A=1 in, E=30x10^6 psi, S

2 2

u v

  ^ ^ ^  

or

966000 u 2 (^)  288000 v 2  0 and

288000 u 2 (^)  384000 v 2   100

By solving, we get,

u 2 =6.34* -5 (^) in

v 2 = -2.13* -4 (^) in