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

2024
Views:

77

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 157. FBD Truss: ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0 L y = 0.75 kip FJK = 0 ! Inspection of joints K , J , and I , in order, shows that FIJ = 0 ! FHI = 0 ! Joint FBDs: FIK = FKL ; FHJ = FJL and FGI = FIK and that 0.75 F F = JL = KL 1 8 5 FJL = 2.1213 kips FKL = 1.6771 kips FJL = 2.12 kips C ! FKL = 1.677 kips T ! FHJ = 2.12 kips C ! and, from above: FGI = FIK = 1.677 kips T ! ΣFx = 0: 2 1 ( FFH + FGH ) − ( 2.1213 kips ) = 0 5 2 ΣFy = 0: 1 1 ( FGH + FFH ) + ( 2.1213 kips ) = 0 5 2 Solving: FFH = 2.516 kips FGH = − 0.8383 kips 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. FFH = 2.52 kips C ! FGH = 0.838 kips T ! COSMOS: Complete Online Solutions Manual Organization System ΣFx = 0: 2 ( FDF − 2.516 kips ) = 0 5 ΣFy = 0: FFG − 0.5 kip + FDF = 2.52 kips C 1 ( 2 )( 2.516 kips ) = 0 5 FFG = 1.750 kips 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.