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

2024
Views:

419

Pages:

2
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 20. FBD Truss: ΣFx = 0: ΣM L = 0: Ax = 0 (1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN) − (11 m) Ay = 0 A y = 4.5 kN Joint FBDs: ΣFy = 0: 4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0 Joint L: L = 10.3 kN 10.3 kN FKL FJL = = , 1 2 5 FKL = 20.6 kN T ! FJL = 10.3 5 kN, FJL = 23.0 kN C ! 20.6 kN FJK F = = IK , 16 5 281 Joint K: FJK = 51.5 kN 8 FJK = 6.44 kN C ! FIK = Joint J: ΣFx = 0: ΣFy = 0: Solving: − 10.3 281 kN, 8 ( FIK = 21.6 kN T! ) 2 16 FHJ + 10.3 5 kN − FIJ = 0 5 265 ( ) 1 FHJ + 10.3 5 kN − 5 FHJ = − FIJ = − 3 51.5 kN = 0 FIJ − 6.6 kN + 8 265 112 5 kN, 11 1.3 265 kN, 88 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. FHJ = 22.8 kN C ! FIJ = 0.240 kN C ! COSMOS: Complete Online Solutions Manual Organization System Joint I: 16  10.3 16  1.3   281 kN  − 265 kN  = 0  − FGI +  8 88 281  265    ΣFx = 0: FGI = ΣFy = 0: ΣFx = 0: ΣFy = 0: − 10.4 kN, 88 1  112 1 10.4  FGH − 2.2 kN − 5 kN  − kN = 0  FFH + 11 88 5 101  FFH = − 8.25 5 kN, FGH = − 34 101 kN, 88 FFH = 18.45 kN C ! FGH = 3.88 kN C ! FDF = 8.25 5 kN By symmetry: ΣFy = 0: FHI = 0.1182 kN T ! 2  112 10  FGH = 0 5 kN  −  FFH + 11 5 101  Solving: Joint F: FGI = 21.3 kN T ! 5  112 10.3  281 kN − 281 kN  + FHI  8 281  88  3  1.3  265 kN  = 0 −  265  88  FHI = Joint H: 112 281 kN, 88 ( ) 2 8.25 5 kN − 6 kN − FFG = 0 5 FFG = 10.50 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 © 2007 The McGraw-Hill Companies.