cee120-spring06-mt1-Mosalam-soln

cee120-spring06-mt1-Mosalam-soln - CE 120 Structural...

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Unformatted text preview: CE 120 Structural Engineering Examination #1 Solution 1 a) Cutting though C and between D and G: unknowns = U=6 (3 reactions, 2 internal forces at C and force in DG); Equations =E= 2x3 =6 for 2 FBDs n=U-E=0; Statically determinate and stable (reactions are not parallel or intersecting). If member DG is absent, it is unstable as n=-1. b) M A = 0 => B y = 39 K F y = 0 => A y = 51 K F x = 0 => A x = 12 K M C = 0 => 9 F DG + (12)(21) + (45)(7.5) (51)(15) = 0 => F DG = 19.5 K F y = 0 => C y = -6 K F x = 0 => C x = 31.5 K Bending Moment Diagram: D G E C F ) ) ) ( ) 144 K-ft 247.5 K-ft 427.5 K-ft 6 K-ft D G E C F ) ) ) ( ) 144 K-ft 247.5 K-ft 427.5 K-ft 6 K-ft D E C 12 K 51 K C x C y F DG 4 K 8 K 3 K/ft D G E C F A x A y B y Solution 2 a) Beam DH: Dead load from the slab = (150)(5/12)(5) = 312.5 lb/ft Area of cross section of the beam = [(2)(4)(1)+(8)(0.5)]/144 = 1/12 ft 2 Self weight of the beam = 490/12 = 40.83 lb/ft Beam CG: Dead load from the slab = (150)(5/12)(10) = 625 lb/ft...
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This note was uploaded on 07/11/2009 for the course CEE cee taught by Professor Monteiro during the Fall '05 term at University of California, Berkeley.

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cee120-spring06-mt1-Mosalam-soln - CE 120 Structural...

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