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

2021
Views:

127

Pages:

2
Author:

rehinhea0c

PROBLEM 1.24 Knowing that θ = 40° and P = 9 kN, determine (a) the smallest allowable diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the corresponding average bearing stress in each of the support brackets at B. SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here. Use joint A as a free body. Law of sines applied to force triangle P FAB FAC = = sin 20° sin110° sin 50° P sin110° FAB = sin 20° (9)sin110° = = 24.73 kN sin 20 ° PROBLEM 1.24 (Continued) (a) Allowable pin diameter. 2 τ = FAB = FπAB 2 = F AB2 where FAB = 24.73 × 103 N πd 2 AP 2 4 d d2 = 2 FAB πτ = (2)(24.73 × 103 ) = 131.18 × 10 −6 m 2 π (120 ×10 6) d = 11.45 × 10− 3m (b) 11.45 mm W Bearing stress in AB at A. Ab = td = (0.016)(11.45 × 10− 3 ) = 183.26 × 10− 6 m 2 σb = (c) F AB 24.73 × 10 3 = = 134.9 × 10 6 Ab 183.26 × 10 −6 134.9 MPa W Bearing stress in support brackets at B. A = td = (0.012)(11.45 × 10− 3 ) = 137.4 × 10− 6 m 2 σb = 1 2 FAB A = (0.5)(24.73 × 103 ) = 90.0 × 106 137.4 × 10−6 90.0 MPa W