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

2024
Views:

71

Pages:

1
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 58. 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 ABDC: ΣM F = 0: ( resolve FCE at E )  1   1   2 ( 0.4 m )  FCE sin 40° +  2 ( 0.4 m )  ( FCE cos 40° )     1   + (1 m ) sin 40° + ( 0.4 m ) ( 0.4 kN ) 2   1   − (1 m ) sin 20° + (1 m ) sin 40° + ( 0.4 m ) ( 2.6 kN ) = 0, 2   FCE = 7.3420 kN, FCE = 7.34 kN C ! (1 m ) cos 40° − ( 0.4 m ) sin 45° + ( 0.4 m ) sin 22.5° (1 m ) sin 40° + ( 0.4 m ) cos 45° − ( 0.4 m ) cos 22.5° (1 m ) cos 40° − ( 0.4 m ) sin 45° = 27.566° = tan −1 (1 m ) sin 40° + ( 0.4 m ) cos 45° note σ = tan −1 and ε Then, ΣM C = 0: = 48.848° ( 0.4 m )  FDF cos ( 90° − 22.5° − 48.848°) − (1 m ) sin 20° ( 2.6 kN ) = 0, FDF = 2.3464 kN, and, ΣFx = 0: FDF = 2.35 kN T ! FCF ( cos 27.566° ) + ( 2.3464 kN )( cos 48.848° ) − ( 7.3420 kN ) sin 40° = 0 FCF = 3.5819 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. FCF = 3.58 kN T !