University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
Views:

496

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 10. FBD Truss: ΣFx = 0: H x = 0 By symmetry: A y = H y = 4 kips FAC = FCE and FBC = 0 W by inspection of joints C and G : FEG = FGH and FFG = 0 W also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF Joint FBDs: Joint A: FAB FAC 3 kips = = 5 4 3 FAB = 5.00 kips C W so FAC = 4.00 kips T W FFH = 5.00 kips C W and, from above, and Joint B: FCE = FEG = FGH = 4.00 kips T W 4 4 10 (5 kips) − FBE − FBD = 0 5 5 109 ΣFx = 0: ΣFy = 0: 3 3 3 FBD + FBE = 0 ( 5 kips ) − 2 − 5 5 109 so FBD = 3.9772 kips, FBE = 0.23810 kips FBD = 3.98 kips C W or Joint E: FBE = 0.238 kips C W FDF = 3.98 kips C W and, from above, FEF = 0.238 kips C W ΣFy = 0 : FDE − 2 3 (0.23810 kips) = 0 5 FDE = 0.286 kips T W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell .

COSMOS Chapter 6 Solution 10

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