COSMOS Chapter 6 Solution 24

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 24. FBD Truss: ΣFx = 0: Ax = 0 ΣM A = 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 Joint FBDs: ΣFy = 0: Ay − 2(1 kN) − 5(1.5 kN) + M y = 0 A y = 4.45 kN Joint M: ΣFy = 0: 5.05 kN − 1 kN − 5 FKM = 0, 13 ΣFx = 0: 12 (10.53 kN) − FLM = 0, 13 ΣFx = 0: 9.72 kN − FKM = 10.53 kN C ! FLM = 9.72 kN T ! 6 FJL = 0, 37 FJL = 1.62 37 Joint L: FJL = 9.85 kN T ! ΣFy = 0: ΣFx = 0: Joint K: ΣFy = 0: 1 (1.62 37 kN) − FKL = 0, 37 12 FIK − 13 FKE = 1.620 kN C ! 24 12 FJK − (10.53 kN) = 0 13 577 5 5 (10.53 kN) − FIK − 13 13 1 FJK 577 − 1.5 kN + 1.62 kN = 0 Solving: FIK = 10.8136 kN, FIK = 10.81 kN C ! FJK = 0.26205 kN, FJK = 0.262 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. COSMOS: Complete Online Solutions Manual Organization System Joint J: ΣFx = 0: 24 (0.26205 kN) − 577 6 ( FHJ − 9.85 kN) = 0 37 FHJ = 10.1154 kN, ΣFy = 0: 1 (10.1154 kN ⋅ 9.85 kN) − 37 FHJ = 10.12 kN T ! 1 (0.26205 kN) − FIJ = 0 577 FIJ = 0.054541 kN, Joint I: ΣFx = 0: ΣFy = 0: Solving: Joint G: 12 24 ( FGI − 10.8136 kN ) + FHI = 0 13 25 5 7 (10.8136 kN − FGI ) + FHI − 1.5 kN + 0.05454 kN = 0 13 25 FGI = 8.6029 kN, FGI = 8.60 kN C ! FHI = 2.1257 kN, FHI = 2.13 kN C ! By symmetry: ΣFy = 0: FIJ = 54.5 N C ! FOG = 8.60 kN C !  5 2   (8.6029 kN) − 1.5 kN − FGH = 0  13  FGH = 5.12 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.