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

2022
Views:

316

Pages:

1
Author:

omiliar74be

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 50. For steel/steel contact, φ s1 = tan −1 µ s1 = tan −1 ( 0.3) = 16.6992° For steel/concrete interface, φ s2 = tan −1 µ s2 = tan −1 ( 0.6 ) = 30.964° FBD Plate CD: ΣFy = 0: N − 90 kN = 0, F = µ s1 N = 0.3 ( 90 kN ) = 27 kN Impending slip: ΣFx = 0: F = 90 kN F − Q = 0, Q = F = 27 kN FBD Top wedge assuming impending slip between wedges: ΣFy = 0: Rw = 100.74 kN Rw cos 26.699° − 90 kN = 0, ΣFx = 0: P − 27 kN − (100.74 kN ) sin 26.699° = 0 P = 72.265 kN, (a) P = 72.3 kN (b) Q = 27.0 kN To check above assumption; note that bottom wedge is a two-force member so the reaction of the floor on that wedge is Rw, at 26.699° from the vertical. This is less than φ s2 = 30.964°, so the bottom wedge doesn’t slip on the concrete. 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.