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

2024
Views:

93

Pages:

2
Author:

omiliar74be

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 39. (a) FBD Truss: Bx = 0 ΣM y = 0: Bx = 0 ΣM z = 0: (1.7 m ) Az + ( 0.6 m )(1700 N ) = 0, − (1.7 m ) Ax − (1.125 m )(1700 N ) = 0 ΣFx = 0: Cx − 1125 N = 0 Cx = 1125 N C y − 1700 N = 0 C y = 1700 N Bz − 600 N = 0 Bz = 600 N ΣM x = 0: ΣFy = 0: ΣFz = 0: Az = − 600 N Ax = −1125 N A = − (1125 Ν ) i − ( 600 N ) k so B = ( 600 Ν ) k C = (1125 Ν ) i + (1700 N ) j (b) Find zero force members: by inspection of joint B, FAB = 0 ! by inspection of joint D, FAD = 0 ! by inspection of joint C , FBC = 0 ! Joint FBDs: Joint E: FAE = FAE ΣFy = 0: −1.125 m i + 1.7 m j − 0.6 m k (1.125 m ) 2 2 + (1.7 m ) + ( 0.6 m ) 1.7 FAE − 1700 N = 0, 2.125 2 = FAE ( −1.125i + 1.7 j − 0.6k ) 2.125 FAE = 2125 N FAE = 2.13 kN T ! ΣFx = 0: − FBE − 1.125 ( 2125 N ) , FBE = −1125 N 2.125 FBE = 1.125 kN C ! ΣFz = 0: − FDE − 0.6 ( 2125 N ) = 0, 2.125 FDE = − 600 N FDE = 600 N C ! continued 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. COSMOS: Complete Online Solutions Manual Organization System Joint D: FBD = FBD ΣFz = 0: −1.125 m i + 0.6 m k (1.125 m )2 + ( 0.6 m )2 600 N − 0.6 FBD = 0, 1.275 = FBD ( −1.125i + 0.6k ) 1.275 FBD = 1275 N FBD = 1.275 kN T ! ΣFx = 0: − FCD − 1.125 (1275 N ) = 0, 1.275 FCD = −1125 N FCD = 1.125 kN C ! Joint C: By inspection: FAC = −1700 N 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. FAC = 1.700 kN C !