COSMOS Chapter 6 Solution 46 | Michigan State University - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2019
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COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 46. \ FBD Truss: ΣFx = 0: By symmetry, Ax = 0 A y = M y = 8.5 kN FBD Section ABDC: ΣM D = 0: (1 m )(1.5 kN − 8.5 kN ) + ( 0.5 m ) 40 FCE = 0 41 FCE = 14.35 kN T W ΣM E = 0: (1 m )(1.5 kN ) + ( 2 m )(1.5 kN − 8.5 kN ) + ( 0.5 m ) 40 FDF = 0, 41 FDF = 25.625 kN FDF = 25.6 kN C W ΣFx = 0: 40 40 FDE = 0 (14.35 kN − 25.625 kN ) + 41 1721 FDE = 11.4084 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. FDE = 11.41 kN T W

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