eci130-w03-final_soln

# eci130-w03-final_soln - ECI 130 Winter 03 Holland...

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ECI 130 – Winter 03 Holland 1/10 Final (Open Book, Open Notes) Name: Problem 1 (5 points) Problem 2 (10 points) Problem 3 (15 points) Problem 4 (20 points) Problem 5 (30 points) Problem 6 (25 points) Problem 7 (25 points) TOTAL (125 points)

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ECI 130 – Winter 03 Holland 2/10 Final (Open Book, Open Notes) Name: 1. Using the principle of superposition for Moment Distribution allowing sidesway, write M as a function of M 1 , M 2 , M 3 , R 1 and R 2 . (Example: M = R 2 M 1 - M 2 ) (5 pts) (Dropped due to theoretical nature) M = M 1 + R 2 R 1 M 3 Note: M 1 is a result of the first moment distribution. M 2 is the FEM created by the sidesway (recall, we choose and arbitrary FEM and therefore get an arbitrary R 2 ). M 3 is a result of the second moment distribution and must be modified by R 2 /R 1 . 2. Given the following M/EI diagram, calculate the deflection at B with respect to the tangent of A using the Moment-Area method. (10 pts) Determine areas Areas = 1 2 1200 EI k - ft 3 ft () = 1800 EI k - ft 2 Deflection calculation t B / A = x M EI dx = 5 ft 1800 EI ⎟ − 1 ft 1800 EI ⎟ = 7200 EI k-ft 3 M M 1 M 2 M 3 R 1 R 2 1200 EI -1200 EI 6 ft A B (k-ft)
ECI 130 – Winter 03 Holland 3/10 Final (Open Book, Open Notes) Name: 3. Draw the conjugate beam and its loading for the following beam. Show supports and loading . (15 pts) Get M diagrams CD F y = 0: C y = D y M D = 6 + C y 3 m ( ) = 0 C y = 2 kN ABC M B = 0: 2 4m ( )− A y 5 m ( ) = 0 A y = 1.6 kN F y = 1.6 + B y + 2 = 0 B y = 3.6 kN

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## This note was uploaded on 03/24/2011 for the course ECI 130 taught by Professor Monti during the Spring '08 term at UC Davis.

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eci130-w03-final_soln - ECI 130 Winter 03 Holland...

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