COSMOS Chapter 6 Solution 14 | Michigan State University

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

2024
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2
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Apolenajz

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 14. FBD Truss: ΣFx = 0: A x = 0 ΣM A = 0: 4a H y − 3a (1.5 kN) − 2a (2 kN) − a (2 kN) = 0 ΣFy = 0: Ay − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN + 3.625 kN = 0 Joint FBDs: Joint A: A y = 3.875 kN FAB FAC 2.625 kN = = 1 29 26 FAB = 15.4823 kN, FAB = 15.48 kN C ! FAC = 14.6597 kN, FAC = 14.66 kN T ! By inspection of joint C: Joint B: H y = 3.625 kN ΣFy = 0: FCE = FAC = 14.66 kN T, 2 (15.4823 kN − FBD ) − 2 kN = 0 29 FBD = 10.0971 kN, ΣFx = 0: FBC = 0 ! FBD = 10.10 kN C ! 5 (15.4823 kN − 10.0971 kN ) − FBE = 0 29 FBE = 5.0000 kN, FBE = 5.00 kN 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: FDF = 10.0971 kN, By symmetry: ΣFy = 0: FDF = 10.10 kN C !  2  10.0971 kN  − 2 kN = 0 − FDE + 2   29  FDE = 5.50 kN T ! FGH FFH 2.625 kN = = 1 26 29 Joint H: By inspection of joint G: ΣFx = 0 : Joint F: FFH = 14.1361 kN FFH = 14.14 kN C ! FGH = 13.3849 kN FGH = 13.38 kN T ! FEG = FGH = 13.38 kN T FEF + and FFG = 0 ! 2 (10.0971 kN − 14.1361 kN ) = 0 29 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. FEF = 3.75 kN C !