Partial Exam Solutions for TAM 2101

# Partial Exam Solutions for TAM 2101 - Q2: BC, ED, FG, IH,...

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Partial Exam Solutions for TAM 210-211 SPRING 2009 exams: *Note solutions have not been doubled checked, this is just to help check final answers you have come up with. Spring 2009 Test 1A Q3: Ay = 150, Dx = 0, Dy = 450, -Cy = Cx = -Bx = 600, By = 0 Spring 2009 Test 1B Q1: |Fr|=12.4, angle 76 degrees below x axis Ma = -40, x = 3.96 m along BC Q2: 1 st section: -1000 2 nd section: -1500 Ma = -26250 By = 1750, Ay = 750 Total net force from distributed loads: -2500 at 10.5 TAM 211 Spring 2009 Q1: Ey = 25, Me = -40, DC = 47, By = -50, Bx = -40 Use equal mag and opp directions on each beam

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Unformatted text preview: Q2: BC, ED, FG, IH, JK DF = -0.667 EF = 0.373 FH = -0.333 FI = -0.373 AB FG is no longer a zero force member, by symmetry FD=FH and FE=FJ Q4: 167 Q5: 70.7 degrees 76 degrees T = 0 TAM 210 Spring 2009 Q1: By = 1200, Bx = -Ax = -Cx = 600, Cy = -800, Ay = -800 Equal magnitude and opposite directions on different members Q3: Ay = 600, Ma = 1400, V = -600 0&lt;x&lt;4, V=-200 at x=6, M = 600x-1400 0&lt;x&lt;4, M = -600 at x=6 Q4: BM, FH, FJ, EJ By symmetry AB = BC = -1.5P, BE = BG = -0.707P Q6: 0.018hb 3...
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## This note was uploaded on 04/27/2011 for the course TAM 211 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Partial Exam Solutions for TAM 2101 - Q2: BC, ED, FG, IH,...

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