Hwrk_9_Solution

Hwrk_9_Solution - UNIVERSITY OF CALIFORNIA, BERKELEY Spring...

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1 UNIVERSITY OF CALIFORNIA, BERKELEY Dept. of Civil and Environmental Engineering Spring Semester 2008 Instructor: Filip C. Filippou CE 121 – Advanced Structural Analysis Homework Set #9 (due April 4, 2008) 1. Problem (5 points) The truss model in the following figure is the same as the model used in Examples 1 and 8 of the CE121 class notes. Thus, you are welcome to use the equilibrium and compatibility equations directly without re-deriving them. All truss elements have the same axial stiffness of EA=10,000 units. You are asked to 1. Determine the basic element forces for the two applied forces separately. What do you observe about the force values for each load case? (note the symmetry in geometry of the truss model). 2. Determine the element deformations of the truss elements for each load case separately. 3. Determine the horizontal translation of node 4 and the vertical translation of node 2 for each load case separately. What do you observe? 4. Draw the deformed shape of the truss model under the combined loading. 5. Determine the flexibility coefficients in the direction of the applied forces (note that a flexibility coefficient has two subscripts i and j: the first subscript refers to the dof for the displacement value and the second subscript refers to the dof for the force value; a flexibility coefficient ij F represents the displacement at dof i under a unit force at dof j). 8 8 6 1 2 3 4 ab c d e 25 20
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CE 121 - Advanced Structural Analysis Homework 9 Filip C. Filippou 8 8 6 1 2 3 4 ab c d e 25 20 L a 8 := L a 8 = L b 8 := L b 8 = L c 6 2 8 2 + := L c 10 = L d 6 := L d 6 = L e 6 2 8 2 + := L e 10 = EA 10000 := Equilibrium equations (dof numbering) and basic forces P 1 Q 1 Q 2 = P 2 Q 4 = P 3 Q 2 0.8 Q 5 + = B f 1 0 0 0 0 1 0 1 0 0 0 0 0 0.8 0.6 0 1 0 0 1 0 0 0.8 0.8 0.6 := with P fa 0 0 0 20 0 := P fb 0 25 0 0 0 := P 4 0.8 Q 3 0.8 Q 5 = P 5 0.6 Q 3 Q 4 + 0.6 Q 5 + = We know, of course, already that this is a statically determinate structure with NOS=0. There are as many equations available as unknown basic element forces. The two applied forces are dealt with in two
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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 Berkeley.

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Hwrk_9_Solution - UNIVERSITY OF CALIFORNIA, BERKELEY Spring...

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