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

2021
Views:

30

Pages:

1
Author:

Prince Sampson

PROBLEM 1.42 A steel loop ABCD of length 1.2 m and of 10-mm diameter is placed as shown around a 24-mm-diameter aluminum rod AC. Cables BE and DF, each of 12-mm diameter, are used to apply the load Q. Knowing that the ultimate strength of the steel used for the loop and the cables is 480 MPa and that the ultimate strength of the aluminum used for the rod is 260 MPa, determine the largest load Q that can be applied if an overall factor of safety of 3 is desired. SOLUTION Using joint B as a free body and considering symmetry, 2⋅ 3 6 FAB − Q = 0 Q = F AB 5 5 Using joint A as a free body and considering symmetry, 4 FAB − FAC = 0 5 8 5 3 ⋅ Q − F AC = 0 ∴ Q = F AC 5 6 4 2⋅ Based on strength of cable BF: QU = σ UA = σ U π 4 d 2 = (480 × 106 ) π 4 (0.012)2 = 54.29 × 103 N Based on strength of steel loop: 6 6 6 π FAB, U = σU A = σU d 2 5 5 5 4 6 6 π 2 = (480 × 10 ) (0.010) = 45.24 × 103 N 5 4 QU = Based on strength of rod AC: π π 3 3 3 3 QU = FAC ,U = σU A = σU d 2 = (260 × 10 6 ) (0.024) 2 = 88.22 × 10 3 N 4 4 4 4 4 4 Actual ultimate load QU is the smallest, ∴ QU = 45.24 × 103 N Allowable load: Q= QU 45.24 ×10 3 = = 15.08 ×10 3N F. S. 3 Q = 15.08 kN W