Ch.1 Problem 1.39 Solution | Michigan State University - Edubirdie

University:
Michigan State University
Course:
ERG 100 | Introduction to Engineering Design
Academic year:

2021
Views:

294

Pages:

1
Author:

Shyanne Johnson

PROBLEM 1.39 A 43 -in.-diameter rod made of the same material as rods AC and AD in the truss shown was tested to failure and an ultimate load of 29 kips was recorded. Using a factor of safety of 3.0, determine the required diameter (a) of rod AC, (b) of rod AD. SOLUTION Forces in AC and AD. Joint C: ΣFy = 0: 1 FAC − 10 kips = 0 5 FAC = 22.36 kips T Joint D: ΣF y = 0: 1 17 F AD − 10 kips = 0 FAD = 41.23 kips T Ultimate stress. From test on 3 4 σU = -in. rod: σ all = Allowable stress: (a) Diameter of rod AC. (b) Diameter of rod AD. σ all = FAC 1π 2 4 d d2 = d2 = 29 kips PU = = 65.64 ksi 1 3 2 A 4 π (4 ) σU F. S. 4FAC πσall 4 FAD πσ all 65.64 ksi = 21.88 ksi 3.0 = = = 4(22.36) π (21.88) 4(41.23) π(21.88) = 1.301 = 2.399 d = 1.141 in. W d = 1.549 in. W

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