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

2021
Views:

368

Pages:

2
Author:

kundgetanufv

PROBLEM 1.25 Determine the largest load P which may be applied at A when θ = 60°, knowing that the average shearing stress in the 10-mm-diameter pin at B must not exceed 120 MPa and that the average bearing stress in member AB and in the bracket at B must not exceed 90 MPa. 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 F = = AC sin 30 ° sin 120 ° sin 30 ° P = FAB sin 30 ° = 0.57735 FAB sin 120° P = FAC sin 30 ° = FAC sin 30 ° PROBLEM 1.25 (Continued) If shearing stress in pin at B is critical, A= π 4 d2 = π 4 (0.010)2 = 78.54 × 10 −6 m 2 F AB = 2Aτ = (2)(78.54 × 10−6 )(120 × 106 ) = 18.850 × 103 N If bearing stress in member AB at bracket at A is critical, Ab = td = (0.016)(0.010) = 160 × 10 −6 m 2 FAB = Abσ b = (160 ×10 − 6)(90 × 10 6) = 14.40 × 10 3N If bearing stress in the bracket at B is critical, Ab = 2td = (2)(0.012)(0.010) = 240 × 10− 6 m 2 FAB = Abσ b = (240 × 10− 6)(90 × 10 6) = 21.6 × 10 3N Allowable FAB is the smallest, i.e., 14.40 × 103N Then from Statics Pallow = (0.57735)(14.40 × 10 3) = 8.31 × 10 3N 8.31 kN W