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

2021
Views:

458

Pages:

1
Author:

esfredint7qo

PROBLEM 1.56 In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter is to be used at A. Assuming that all other specifications remain unchanged, determine the allowable load P if an overall factor of safety of 3.0 is desired. PROBLEM 1.55 In the structure shown, an 8-mmdiameter 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. Σ M B = 0: 0.20 FA − 0.18P = 0 ΣM A = 0: 0.20 FBD − 0.38 P = 0 Based on double shear in pin A: A = 10 F 9 A P= π 4 P = d2 = π 4 10 F 19 BD (0.010)2 = 78.54 × 10−6 m 2 − 2τ UA (2)(100 × 106 )(78.54 × 10 6 ) = = 5.236 × 103 N F .S . 3.0 10 P = F A = 5.82 × 10 3 N 9 FA = 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 − 2τ UA (2)(100 × 106 )(113.10 × 10 6 ) = = 7.54 × 103 N F .S . 3.0 10 P = F BD = 3.97 × 10 3 N 19 FBD = Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10 −6m 2 − 2σ U 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 = 3.97 × 10 3 N . A student using thismanual isusing itwithout permission. P = 3.97 kN W