Hwrk_9_Solution

Hwrk_9_Solution - UNIVERSITY OF CALIFORNIA BERKELEY Spring...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 a b c d e 25 20
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CE 121 - Advanced Structural Analysis Homework 9 Filip C. Filippou 8 8 6 1 2 3 4 a b 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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern