HW08 Solution - CE 121 - Advanced Structural Analysis...

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CE 121 - Advanced Structural Analysis Homework 8 Filip C. Filippou a b c d e f g h 1 2 3 4 5 6 6 8 8 L a 8 = EA 30000 = L b 8 2 6 2 + = L b 10 = L c 6 = L d 8 2 6 2 + = L d 10 = L e 8 = L f 8 2 12 2 + = L f 14.422 = L g 6 = L h 8 2 6 2 + = L h 10 = Page 1
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CE 121 - Advanced Structural Analysis Homework 8 Filip C. Filippou There are 7 free global dof's (and thus equilibrium equations) and 8 basic forces. NOS = 1, therefore. 1 2 3 4 5 7 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 Q 8 Q 6 P 1 Q 1 0.8 Q 4 = We select Q 5 as redundant basic force P 2 Q 3 0.6 Q 4 + = P 3 0.8 Q 2 Q 5 = P 4 0.6 Q 2 Q 3 Q 7 + = P 5 0.8 Q 4 Q 5 + 0.8 Q 8 + = P 6 0.6 Q 4 0.6 Q 8 + = P 7 8 L f Q 6 0.8 Q 8 = With the redundant basic force equal to zero and the applied thermal loading all basic forces are zero. equilibrium state due to applied thermal loading is therefore: Q p 00000000 () T = Page 2
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CE 121 - Advanced Structural Analysis Homework 8 Filip C. Filippou Set redundant basic force Q 5 equal to unity with the applied loading equal to zero Q 5 1 = from 0Q 1 0.8 Q 4 = Q 1 0.5 = 3 0.6 Q 4 + = Q 3 0.36 0.96 = 0 0.8 Q 2 1 () = Q 2 1 0.8 = 0 0.6 Q 2 Q 3 Q 7 + = Q 7 0.6 0.8 0.36 0.96 + = 0 0.8 Q 4 1 + 0.8 Q 8 + = Q 8 0.6 0.96 = 0 0.6 Q 4 0.6 Q 8 + = Q 4 0.6 0.96 = Q 6 0.1 L f Q 8 = 0 8 L
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This note was uploaded on 02/27/2011 for the course CE 121 taught by Professor Filippou during the Fall '09 term at University of California, Berkeley.

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HW08 Solution - CE 121 - Advanced Structural Analysis...

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