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

2021
Views:

391

Pages:

2
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 59. FBD Truss: Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60° outer members AC, CE, etc. are each 1.0 m radial members AB, CD, etc. are each 0.4 m By symmetry, Ay = By = 2.6 kN ΣFx = 0: A x = 0 FBD Section ABDFHGEC: ( resolve FGI at I ) ( 0.4 m ) FGI sin 80° + (1 m )( sin 80°)(1 kN ) ΣM J = 0: + (1 m ) sin 80° + (1 m ) sin 60°  ( 0.6 kN ) + (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° ( 0.4 kN ) − (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° + (1 m ) sin 20° ( 2.6 kN ) = 0 FGI = 10.8648 kN, 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. FGI = 10.86 kN C ! COSMOS: Complete Online Solutions Manual Organization System Note: so, σ = tan −1 ΣM H = 0: 0.4 m − (1 m ) cos80° = 12.9443° (1 m ) sin 80° ( 0.4 m ) cos 60° ( FGJ sin12.9443° ) − ( 0.4 m ) sin 60° ( FGJ cos12.9443° ) − ( 0.4 m ) sin 60° (10.8648 kN ) sin 80° + ( 0.4 m ) cos 60° (10.8648 kN ) cos80° + ( 0.4 m ) cos 60° (1 kN ) + (1 m ) sin 60° + ( 0.4 m ) cos 60° ( 0.6 kN ) + (1 m ) sin 60° + (1 m ) sin 40° + ( 0.4 m ) cos 60° ( 0.4 kN ) − (1 m )( sin 60° + sin 40° + sin 20° ) + ( 0.4 m ) cos 60° ( 2.6 kN ) = 0 FGJ = 0.93851 kN, FGJ = 939 N T ! FBD Joint I: By symmetry: ΣFy = 0: FIK = 10.8648 kN − FIJ − 1.2 kN + 2 (10.8648 kN ) cos80° = 0 FIJ = 2.5733 kN, 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. FIJ = 2.57 kN T !