COSMOS Chapter 6 Solution 19

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 19. 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: Joint A: ΣFy = 0: 4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0 L = 10.3 kN 4.5 kN FAC F = = AB , 1 2 5 FAC = 9.00 kN T ! FAB = 4.5 5, Joint C: FAB = 10.06 kN C ! 9 kN FBC F = = CE , 16 5 281 FBC = 45 kN 16 FBC = 2.81 kN C ! FCE = Joint B: ΣFx = 0: ΣFy = 0: Solving: 9 281, 16 FCE = 9.43 kN T ! 2  16 FBD + (4.5 5) kN  + FBE = 0   5 265 1  FBD + (4.5 5) kN  −  5 3 45 kN = 0 FBE + 16 265 72 5 kN, 11 45 265 kN, FBE = 176 FBD = − 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. FBD = 14.64 kN C ! FBE = 4.16 kN T ! COSMOS: Complete Online Solutions Manual Organization System Joint E: ΣFx = 0: 16   9  FEG −  16 281   FEG = ΣFy = 0:  16  45   281  kN  − 265  kN = 0  265  176    9 281 kN, 11 FEG = 13.72 kN T ! 5 9 9  281 kN − 281 kN  +  16 281  11  3  45  265 kN   265  176  + FDE = 0, FDE = − Joint D: ΣFx = 0: ΣFy = 0: Solving: 45 kN, 22 FDE = 2.05 kN C ! 2  72 10  FDG = 0 5 kN  +  FDF + 11 5 101  1  72 1 45  FDG + 5 kN  − kN = 0  FDF + 11 22 5 101  7.5 101 kN, 22 FDG = 3.43 kN T ! FDF = − 8.25 5 kN, FDF = 18.45 kN C ! FDG = 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.