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

2024
Views:

9

Pages:

1
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 11. FBD Truss: ΣFx = 0: Ax = 0 ΣM G = 0: 3a Ay − 2a (3 kN) − a (6 kN) = 0 A y = 4 kN by inspection of joint C, FAC = FCE and FBC = 0 W by inspection of joint D, FBD = FDF and FDE = 6.00 kN C W Joint FBDs: Joint A: FAC FAB 4 kN = = 21 29 20 FAB = 5.80 kN C W FAC = 4.20 kN C W from above, Joint B: ΣFy = 0: 20 20 ( 5.80 kN ) − 3 kN − FBE = 0 29 29 FBE = ΣFx = 0: FCE = 4.20 kN C W 29 20 FBE = 1.450 kN T W 21  29  kN  − FBD = 0  5.80 kN + 29  20  FBD = 5.25 kN C W FDF = 5.25 kN C W from above, Joint F: ΣFx = 0: 5.25 kN − 21 FEF = 0 29 FEF = 7.25 kN T W ΣFy = 0: FFG − 20 (7.25 kN) − 1 kN = 0 29 FFG = 6.00 kN C W by inspection of joint G, 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 . FEG = 0 W

COSMOS Chapter 6 Solution 11

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