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

2020
Views:

260

Pages:

2
Author:

Sean Horton

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 23. FBD Truss: ΣFx = 0: ΣM A = 0: Ax = 0 (12 m ) (M y − 1 kN) − (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0 M y = 5.05 kN ΣFy = 0: Ay − 2(1 kN) − 5(1.5 kN) + M y = 0 A y = 4.45 kN Joint FBDs: ΣFy = 0: 4.45 kN − 1 kN − Joint A: ΣFx = 0: ΣFx = 0: Joint C: ΣFy = 0: ΣFy = 0: Joint B: FAC − 5 FAB = 0, 13 12 (8.97 kN ) = 0, 13 FAC = 8.28 kN T ! 12 FCE − 8.28 kN = 0, 13 FCE = 8.97 kN T ! 5 (8.97 kN) − FBC = 0, 13 FBC = 3.45 kN C ! 5 5 (8.97 kN ) − 1.5 kN + 3.45 kN − FBD = 0 13 13 FBD = 14.04 kN ΣFx = 0: FAB = 8.97 kN C ! FBD = 14.04 kN C ! 12 12 (8.97 kN) − (14.04 kN) + FBE = 0 13 13 FBE = 4.68 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. FBE = 4.68 kN T ! COSMOS: Complete Online Solutions Manual Organization System Joint E: ΣFx = 0: 6 12 FEH − 4.68 kN − (8.97 kN) = 0 13 37 FEH = 13.1388 kN ΣFy = 0: FDE − 5 (8.97 kN ) − 13 or 1 (13.1388) = 0 37 FDE = 5.6100 kN Joint D: ΣFx = 0: ΣFy = 0: FEH = 13.14 kN T ! or FDE = 5.61 kN T ! 12 12 1 (14.04 kN ) − FDG − FDH = 0 13 13 2 5 5 (14.04 kN ) − FDG − 1.5 kN − 5.61 kN 13 13 1 + FDH = 0 2 Solving: FDG = 8.60 kN C ! FDH = 7.10 kN C ! 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.