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# HW09sol - CE121 Advanced Structural Analysis Homework 9...

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CE121 - Advanced Structural Analysis Homework 9 Filip C. Filippou 8 8 6 6 1 2 3 4 5 a b c d e 25 w5 := L a 8 := L b 6 := L c 8 := L d 6 := L e 10 := With inextensible elements a through d, there are 4 free global dofs and 6 corresponding basic forces. The degree of static indeterminacy is therefore NOS=2. 8 8 6 6 dof 1 dof 2 dof 3 dof 4 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q Page 1

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CE121 - Advanced Structural Analysis Homework 9 Filip C. Filippou The equilibrium equations are: P 1 Q 1 Q 2 + Q 4 + Q 5 + = P 2 Q 2 Q 3 + 6 Q 5 6 0.8 Q 6 + = B f 1 0 0 0 1 1 6 0 0 0 1 6 0 0 1 0 1 8 0 1 1 6 0 1 6 0 0.8 0.6 0.8 := with P f 0 0 25 0 := P 3 Q 4 8 0.6 Q 6 = P 4 Q 5 6 0.8 Q 6 = we select Q 3 and Q 4 as redundant basic forces. with Q 3 0 := Q 4 0 := under the given loading we get from equation (3) we get: Q 6 1 0.6 25 := Q 6 41.67 = then from equation (4) Q 5 4.8 Q 6 := Q 5 200 = then from equation (2) Q 2 Q 5 4.8 Q 6 := Q 2 0 = then from equation (1) Q 1 Q 2 Q 5 := Q 1 200 = thus, the particular solution under the given loading is Q p Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 := Q p 200 0 0 0 200 41.67 = Now, we determine the homogeneous solution. For this case, there is no applied loading and each of the redundant basic forces is set equal to one, while the other(s) is equal to zero. We have Q 3 1 := Q 4 0 := from equation (3) we get: Q 6 0 := Q 6 0 = then from equation (4) Q 5 4.8 Q 6 := Q 5 0 = Bbar x 1 ⟨⟩ 1 1 1 0 0 0 := then from equation (2) Q 2 Q 3 := Q 2 1 = then from equation (1) Q 1 Q 2 := Q 1 1 = Page 2
CE121 - Advanced Structural Analysis Homework 9 Filip C. Filippou Q 3 0 := Q 4 1 := from equation (3) we get: Q 6 1 4.8 := Q 6 0.21 = then from equation (4) Q 5 4.8 Q 6 := Q 5 1 = Bbar x 2 ⟨⟩ 0 0 0 1 1 1 4.8 := then from equation (2) Q 2 Q 5 4.8 Q 6 := Q 2 0 = then from equation (1) Q 1 Q 2 Q 4 Q 5 := Q 1 0 = We check both particular and homogeneous solutions. B f Q p 0 0 25 0 = B f Bbar x 0 0 0 0 0 0 0 0 = it looks ok! Define element flexibility matrices EI 100000 := EA 30000 := set up collection of element flexibility matrices F s 1 EI L a 3 0 0 0 0 0 0 L b 3 L b 6 0 0 0 0 L b 6 L b 3 0 0 0 0 0 0 L c 3 0 0 0 0 0 0 L d 3 0 0 0 0 0 0 EI EA L e := Page 3

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CE121 - Advanced Structural Analysis
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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 Berkeley.

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HW09sol - CE121 Advanced Structural Analysis Homework 9...

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