COSMOS Chapter 6 Solution 48 - Edubirdie

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

2021
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2
Author:

Lamghari6kx

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 48. FBD Truss: ΣFx = 0: A x = 0 By symmetry: A y = L y = 6 kN FBD Section: Notes: 15 m 6 2 5 5 yD = ⋅ = m 3 2 3 2 2 yE = ⋅ 1 = m 3 3 5 yF − yD = m 6 yG = 1 m yF = yD − yG = 2 m 3 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 ΣM D = 0: (1 m ) 6 FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0 37 FEG = 8 37 kN 3 FEG = 16.22 kN T !  1   3  ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m )  FDG  − (1 m )  FDG  = 0  10   10  FDG = ΣFx = 0: 4 10 kN 3 6 3 12 FEG − FDG − FDF = 0 13 37 10 FDG = 4.22 kN C ! 16 − 4 − FDF = 13 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. 12 FDF = 0 13 FDF = 13.00 kN C !

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