HW04sol - University of California, Berkeley Spring...

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University of California, Berkeley Dept. of Civil and Environmental Engineering Spring Semester 2009 1 CE 121 — Advanced Structural Analysis Homework Set #4 (due Feb. 20, 2009) 1. Problem (5 points) 1. Determine the collapse load factor ° for the truss in the figure under the given loading. The truss elements exhibit perfectly plastic behavior with axial force capacity in tension and compression equal to 50 units of force. 2. Determine all basic forces at collapse. 3. Determine the support reactions at collapse and check global equilibrium. 2. Problem (5 points) For the same structure in Problem 1: 1. Set up the compatibility relations between element deformation and free global DOF displacements in the form ? = ? ± ² ± . 2. Show that the compatibility matrix ? ± is equal to the transpose of the equilibrium matrix ³ ± . 3. Assume now that elements b and c are delivered longer than shown in the above figure: element b by 0.1 and element c by 0.2 units of length. Elements a and e are delivered with the required length. What should be the length of member d so as to avoid stresses due to restraint? 4. Suppose that you were given all element lengths and corresponding deformation values from the start. Show that they are compatible by using the relation ³ ´ µ ? = 0 (a good check for hand calculations).
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CE 121 - Advanced Structural Analysis Homework 3 Filip C. Filippou The following structural model and loading are given. Nodes and elements are numbered as shown. 4 4 3 3 a b c d e 1 2 3 4 λ 30 λ 50 For the lengths of elements a through e we have L a 5 := L b 5 := L c 4 2 6 2 + := L c 7.211 = L d 3 := L e L c := L e 7.211 = We number all unrestrained degrees of freedom (dofs) systematically, as shown in the following figure. For each node free body we need to write an equilibrium equation in the direction of the free dofs. The following figure shows the numbering of the complete set of unknown basic forces Q . 4 4 3 3 1 2 3 4 1 Q 2 Q 3 Q 4 Q 5 Q Page 1-1
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CE 121 - Advanced Structural Analysis Homework 3 Filip C. Filippou Equilibrium equations (node equilibrium) at free dof's 4 4 3 3 1 2 3 4 1 Q 2 Q 3 Q 5 Q 4 Q 5 Q 3 Q 1 Q 2 Q P 1 4 L a Q 1 4 L b Q 2 = P 2 3 L a Q 1 3 L b Q 2 + Q 4 = for the given applied loading P 3 4 L c Q 3 4 L e Q 5 = P 2 λ 50 = P 3 λ 30 = P 4 6 L c Q 3 Q 4 + 6 L e Q 5 + = From the sign of the applied forces and the signs of the basic forces in the corresponding equilibrium equations we "suspect" that Q 2 and Q 5 are in compression and may be the most highly stressed. Thus, we assume that Q 2 50 := Q 5 50 := The equilibrium equations to be solved are: 0 4 L a Q 1 4 L b Q 2 = λ 50 3 L a Q 1 3 L b Q 2 + Q 4 = λ 30 4 L c Q 3 4 L e Q 5 = 0 6 L c Q 3 Q 4 + 6 L e Q 5 + = Page 1-2
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CE 121 - Advanced Structural Analysis Homework 3 Filip C. Filippou from the first we conclude Q 1 Q 2 = Q 1 50 := there are now three equations in three unknowns; use simultaneuous linear equation solver Q 3 0 := Q 4 0 := λ 0 := Given Q 1 50 = Q 2 50 = Q 5 50 = λ 50 3 L a Q 1 3 L b Q 2 + Q 4 = λ
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HW04sol - University of California, Berkeley Spring...

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