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

2022
Views:

429

Pages:

1
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 18. FBD Truss: A y = H y = 1600 lb By load symmetry, ΣFx = 0: Ax = 0 θ = tan −1 6.72 ft = 16.2602° 23.04 ft Joint FBDs: Joint H: ΣFy = 0: 1600 lb − 400 lb − FFH sin θ = 0 FFH = 4285.7 lb, ( 4285.7 lb ) cosθ ΣFx = 0: Joint F: − FGH = 0 FGH = 4114.3 lb ΣFy′ = 0: FGH = 4.11 kips T FFG − ( 800 lb ) cosθ = 0 FFG = 768.0 lb ΣFx′ = 0: FFH = 4.29 kips C FFG = 0.768 kips C FDF + ( 800 lb ) sin θ − 4285.7 lb = 0 FDF = 4061.7 lb FDF = 4.06 kips C Joint G: ΣFy = 0: FDG sin 2θ − (768 lb) cosθ = 0 FDG = 1371.4 lb ΣFx = 0: FDG = 1.371 kips T 4114.3 lb − (1371.4 lb ) cosθ − FEG − (768 lb) sin θ = 0 FEG = 2742.9 lb 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 = 2.74 kips T

COSMOS Chapter 6 Solution 18

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