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

2021
Views:

454

Pages:

1
Author:

5prinzg1s

PROBLEM 1.55 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. SOLUTION Statics: Use ABC as free body. ΣM B = 0 : 0.20 FA − 0.18 P = 0 P= 10 FA 9 ΣM A = 0 : 0.20 FBD − 0.38P = 0 P = 10 FBD 19 Based on double shear in pin A: A = π 4 d2 = π 4 (0.008) 2 = 50.266 × 10− 6m 2 6 6 (2)(100 × 10 )(50.266 × 10− ) = = 3.351 × 103 N 3.0 F .S . 10 P = FA = 3.72 × 103 N 9 FA = 2τU A Based on double shear in pins at B and D: A = π 4 d2 = π 4 (0.012)2 = 113.10 × 10− 6 m 2 (2)(100 × 106 )(113.10 × 10−6 ) = = 7.54 × 103 N 3.0 F .S . 10 P = F = 3.97 × 10 3 N 19 BD F BD = 2τU A Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10 −6m 2 2σU A (2)(250 × 106 )(160 × 10 −6 ) = = 26.7 × 103 N 3.0 F .S. 10 P = FBD = 14.04 × 103 N 19 FBD = Allowable value of P is smallest, ∴ P = 3.72 × 10 3 N P = 3.72 kN W