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

2021
Views:

452

Pages:

2
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 17. FBD Truss: By load symmetry, ΣFx = 0: θ = tan −1 Joint FBDs: A y = H y = 1600 lb Ax = 0 6.72 ft = 16.2602° 23.04 ft ΣFy′ = 0: (1600 lb − 400 lb) cosθ − FAC sin θ = 0 FAC = 4114.3 lb Joint A: ΣFx = 0: FAC cos 2θ − FAB cosθ = 0 FAB = 3613.5 lb ΣFx′ = 0: Joint B: Joint C: FBC = 0.768 kips C ! FCD sin 2θ − (768 lb) cosθ = 0 FCD = 1371.4 lb ΣFx′′ = 0: FBD = 3.84 kips C ! FBC − (800 lb) cosθ = 0 FBC = 768.00 lb ΣFy′′ = 0: FAB = 3.61 kips C ! 3613.5 lb − FBD + (800 lb)sin θ = 0 FBD = 3837.5 lb ΣFy′ = 0: FAC = 4.11 kips T ! FCD = 1.371 kips T ! FCE + (1371.4 lb) cos2θ − (768 lb)sin θ − 4114.3 lb = 0 FCE = 2742.9 lb FCE = 2.74 kips T ! continued 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 © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Joint E: ΣFy = 0: (2742.9 lb)sin 2θ − FDE cosθ = 0 FDE = 1536.01 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 © 2007 The McGraw-Hill Companies. FDE = 1.536 kips C !