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Lab 09 - Solutions

# Lab 09 - Solutions - UNIVERSITY OF CALIFORNIA BERKELEY...

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UNIVERSITY OF CALIFORNIA, BERKELEY Engineering 7 – Spring 2009 Department of Civil and Environmental Engineering Instructor: Professor Rector Lab 9 (Solutions) Topics : Linear Equations Assigned : Monday, 4/13/2009, 8:00 am Due : Monday, 4/20/2009, 12:00 pm (noon) Type : Take-home 1. Solution of a Set of Linear Equations Exercise 1 – Truss Systems (5 pts) A truss is a structure whose components are connected by hinges. Hinges cannot support moments, and the loading can only be applied on the two ends of any truss component. The internal force in the truss component then is oriented only along the direction of the component, and is either tensile or compressive. F ( z ) S a S b H B V B V A H A A C a 5 m 1 3 . 5 m 2 4 m b B θ a c o s   θ a   =   0 . 6 c o s   θ b   =   0 . 8 θ b In order to determine the internal force of each member of the truss, we can form a system of linear equations and solve the system. The equations are: – 0.6 Sa – HA = 0 – VA – 0.8 Sa = 0 0.8 Sb – HB = 0 0 – VB – 0.6 Sb = 0 0.6 Sa – 0.8 Sb = 0 0.6 Sb + 0.8 Sa = – F(z) (a) Express the above equations in matrix form, Ax = b, and include them in your report. (1 point) A = -0.6000 0 -1.0000 0 0 0 -0.8000 0 0 0 -1.0000 0 1

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Lab 10 2 0 0.8000 0 -1.0000 0 0 0 -0.6000 0 0 0 -1.0000 0.6000 -0.8000 0 0 0 0 0.8000 0.6000 0 0 0 0 b = 0 0 0 0 0 -10 (b) Start a Matlab code for this problem by creating and populating matrix A and array b. Put your commands in your report. (1 point) fz= 10; % kN A = [ -.6 0 -1 0 0 0; -.8 0 0 0 -1 0; 0 .8 0 -1 0 0; 0 -.6 0 0 0 -1; .6 -.8 0 0 0 0; .8 .6 0 0 0 0] b= [0 0 0 0 0 -fz]' (c) Check the determinant of matrix A using the Matlab command det . Include the answer your report. (0.5 point) det(A) is -1 (d) Using the left division operator of Matlab to solve the problem, determine the reaction forces of HA, HB, VA, VB, and the internal force of members a and b for F(z) = 10 kN. Copy the answer to your lab report. (0.5 point) in the order: Sa, Sb, Ha, Hb, Va, Vb x = -8.0000 -6.0000 4.8000 -4.8000 6.4000 3.6000 They may use a different order (e) Plot Sa vs. F(z) and Sb vs. F(z) for F(z) between 30 and 150 kN at 1kN increments. (2 point) They may have one plot with two curves or two plots with one curve. If it is not written which one is Sa and which one is Sb (either in the title, or the legend, or anywhere), -1, and don’t check the curves.
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