Hwrk_11_Solution - CE121 - Advanced Structural Analysis...

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CE121 - Advanced Structural Analysis Homework 11 Filip C. Filippou Problem 1 Equilibrium equations: P 1 1 6 Q 2 1 6 Q 3 + 8 10 Q 5 = For this problem, we will choose Q 1 and Q 3 as the redundant forces P 2 Q 1 Q 2 + Q 4 + = B f 0 1 0 1 6 1 0 1 6 0 0 0 1 1 4 8 10 0 0 := P 3 1 4 Q 4 = Load Case 1: applied concentrated force Particular solution: For this we set Q 1 and Q 3 equal to zero, and then solve for the other values: Q 1 0 := Q 3 0 := applied force vector P f 0 0 25 := 0 1 6 Q 2 1 6 Q 3 + 8 10 Q 5 = from eq 3 Q 4 425 := 0Q 1 Q 2 + Q 4 + = from eq 2 Q 2 Q 4 := 25 1 4 Q 4 = from eq 1 Q 5 10 8 1 6 Q 2 := Page 11-1
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CE121 - Advanced Structural Analysis Homework 11 Filip C. Filippou summarize result in a vector of the basic forces for the particular solution Q p Q 1 Q 2 Q 3 Q 4 Q 5 := Q p 0 100 0 100 20.83 = confirm that this is correct (easy with matrix multiply!) B f Q p 0 0 25 = which matches P f First homogeneous solution: We set Q 1 equal to 1, and Q 3 equal to zero, with zero applied load, and solve for the other values: Q 1 1 := Q 3 0 := 0 1 6 Q 2 1 6 Q 3 + 8 10 Q 5 = from eq 3 Q 4 0 := 0Q 1 Q 2 + Q 4 + = from eq 2 Q 2 Q 1 := 0 1 4 Q 4 = from eq 1 Q 5 10 8 1 6 Q 2 := B x 1 ⟨⟩ Q 1 Q 2 Q 3 Q 4 Q 5 := B x 1 1 1 0 0 5 24 = Second homogeneous solution: We set Q 3 equal to 1, and Q 1 equal to zero, with zero applied load, and solve for the other values: Q 1 0 := Q 3 1 := 0 1 6 Q 2 1 6 Q 3 + 8 10 Q 5 = from eq 3 Q 4 0 := 1 Q 2 + Q 4 + = from eq 2 Q 2 Q 1 := 0 1 4 Q 4 = from eq 1 Q 5 10 8 1 6 Q 3 := B x 2 Q 1 Q 2 Q 3 Q 4 Q 5 := B x 2 0 0 1 0 5 24 = Page 11-2
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CE121 - Advanced Structural Analysis Homework 11 Filip C. Filippou Solve for the redundant force values: Solution A : by computer, i.e. use matrix operations define collection of element flexibility matrices stiffness properties EI 25000 := EA 5000 := element lengths L a 8 := L b 6 := L c 4 := L d 10 := F s L a 3EI 0 0 0 0 0 L b 3EI L b 6EI 0 0 0 L b 6EI L b 3EI 0 0 0 0 0 L c 0 0 0 0 0 L d EA := compatibility violation due to particular solution V gp B x T F s Q p := V gp 16.681 4.681 10 3
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This note was uploaded on 01/04/2010 for the course CE 121 taught by Professor Filippou during the Fall '09 term at University of California, Berkeley.

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Hwrk_11_Solution - CE121 - Advanced Structural Analysis...

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