COSMOS Chapter 6 Solution 15

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 15. FBD Truss: ΣFx = 0: A x = 0 ΣM A = 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: − 1 kN − 1 kN + 6.7 kN = 0 Joint FBDs: Joint A: Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN ΣFy = 0: 5 kN − A y = 6.0 kN 3 5 13 FAB = 0, FAB = kN 3 13 FAB = 6.01 kN C ! ΣFx = 0: Joint B: ΣFx = 0: FAC − 2 13  5 13  kN  = 0,   3  FAC = 3.33 kN T !  2  5 13 kN − FBC − FBD  = 0,  13  3  5 13 kN 3  3  5 13 kN + FBC − FBD  − 1 kN = 0,  13  3  4 13 FBD − FBC = kN 3 3 13 kN FBD = FBD = 5.41 kN C ! 2 1 13 kN FBC = FBC = 0.601 kN C ! 6 FBC + FBD = ΣFy = 0: 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: Joint E: FDE = FEG By symmetry: ΣFy = 0: 2 1 FDE − 4.5 kN = 0, 17 FDE = 9 17 kN 4 FDE = 9.28 kN C ! ΣFx = 0: 2 3 4 9   13 kN  − 17 kN  = 0   13  2 17  4   FDF + Joint D: FDF = 6.00 kN T ! ΣFy = 0: 3 3 1 9   13 kN  − 1.4 kN − FCD − 17 kN  = 0   13  2 17  4   FCD = 0.850 kN T ! Joint C: ΣFy = 0: 0.850 kN − FCG = ΣFx = 0: FCI − 3 13  13  3 kN  − FCG = 0   5  6  1.75 kN 3 FCG = 0.583 kN C !  10 4  1.75 2  13  kN  − kN  − kN = 0   5 3 13  6   3 FCI = 3.47 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.