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

2020
Views:

462

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 16. \ FBD Truss: ΣFx = 0: ΣM A = 0: Ax = 0 8a( J y − 1 kN) − 7a(1 kN) − 6a(2.8 kN) − 4a(4.5 kN) − 2a(4 kN) − a(1 kN) = 0 J y = 6.7 kN ΣFy = 0: Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN −1 kN − 1 kN + 6.7 kN = 0 Joint FBDs: Joint J: ΣFy = 0: (6.7 − 1) kN − 3 FHJ = 0, 13 A = 6.0 kN FHJ = 1.9 13 kN FHJ = 6.85 kN C ! ΣFx = 0: Joint H: ΣFx = 0: ΣFy = 0: 2 (1.9 13 kN) − FIJ = 0, 13 2 ( FGH + FHI − 1.9 13 kN) = 0 13 3 ( FHI − FGH + 1.9 13 kN) − 1 kN = 0 13 26 13 kN, FGH = FGH = 6.25 kN C ! 15 FHI = Joint I: ΣFx = 0: 3.80 kN − FGI − 13 kN, 6 FHI = 0.601 kN C !  2  13 kN  − FCI = 0  13  6  FCI = ΣFy = 0: FIJ = 3.80 kN T ! 10.4 kN, 3  3  13 kN  = 0,   13  6  FCI = 3.47 kN T ! FGI = 0.500 kN T ! 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 FEF = 0 ! By inspection of joint F, By symmetry FDE = FEG Joint E: ΣFy = 0:  1  FEG  − 4.5 kN = 0, 2 17   FEG = 9 17 kN 4 FEG = 9.28 kN C ! Joint G: ΣFy = 0: 3  26 3 1 9  1  13 kN  − kN − FCG − 17 kN    5 13  15 17  4  2  − 2.8 kN = 0 FCG = − ΣFx = 0: 1.75 kN 3 FCG = 0.583 kN C ! 4 9 4  1.75   17 kN  − FFG −  −   5 3  17  4  − 2  26  13 kN  = 0  13  15  FFG = 6.00 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.