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

2019
Views:

189

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 12. FBD Truss: ΣFx = 0: Ax = 0 By symmetry: A y = B y = 4.90 kN FAB = FEG , FAC = FFG , FBC = FEF and FBD = FDE , FCD = FDF 5 4 FAC − FAB = 0 5 29 2 3 FAC − FAB + 4.9 kN = 0 5 29 ΣFx = 0: Joint FBDs: Joint A: ΣFy = 0: FAC = 2.8 29 kN FAC = 15.08 kN T FAD = 17.50 kN C Joint B: ΣFx = 0: ΣFy = 0: 4 1 FBC = 0 (17.5 kN − FBD ) − 5 2 3 1 FBC − 2.8 kN = 0 (17.5 kN − FBD ) + 5 2 FBD = 15.50 kN C FBC = 1.6 2 kN; Joint C: ΣFy = 0: FBC = 2.26 kN C 4 1 FCD − 1.6 2 kN − 5 2 ΣFx = 0: FCF 2 (2.8 29 kN) = 0 29 FCD = 9.00 kN T 1 3 5 + 1.6 2 kN + (9 kN) − (2.8 29 kN) = 0 5 2 29 FCF = 7.00 kN T ( ( ) ) from symmetry, FEG = 17.50 kN C FFG = 15.08 kN T FEF = 2.26 kN C FDE = 15.50 kN C FDF = 9.00 kN T 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 .