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

2022
Views:

426

Pages:

1
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 56. FBD Truss: ΣFx = 0: ΣM N = 0: Ax = 0 ( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb ) ( ) + ( 33.6 ft ) 150 lb − Ay = 0 Ay = 2925 lb ΣFy = 0: 2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb − 900 lb − 1950 lb − 300 lb + N y = 0 FBD Section ABC: N y = 3075 lb ΣM B = 0: ( 3.6 ft ) FCE − ( 7.2 ft )( 2925 lb − 150 lb ) = 0 FCE = 5550 lb, ΣM A = 0: ( 7.2 ft ) FCE = 5.55 kips T ! 1 2 FBE + ( 3.6 ft ) FBE − ( 7.2 ft )(1050 lb ) = 0 5 5 FBE = 525 5 lb, ΣFx = 0: ( FBE = 1.174 kips C ! ) 2 FBD − 525 5 lb + 5550 lb = 0 5 FBD = −5031.2 lb By inspection of joint D, FDF = FBD , so 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. FDF = 5.03 kips C !

COSMOS Chapter 6 Solution 56

of 1

Description

Free up your schedule!

Take 5 seconds to unlock