Ch.1 Problem 1.69 Solution - Edubirdie

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

2021
Views:

103

Pages:

1
Author:

kundgetanufv

PROBLEM 1.69 The two portions of member AB are glued together along a plane forming an angle θ with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine the range of values of θ for which the factor of safety of the members is at least 3.0. SOLUTION A0 = (2.0)(1.25) = 2.50 in. 2 P = 2.4 kips PU = ( F. S.) P = 7.2 kips Based on tensile stress: σU = cos 2 θ = PU cos 2 θ A0 σU A0 PU cos θ = 0.93169 Based on shearing stress: τU = sin 2θ = 2θ = 64.52° Hence, = (2.5)(2.50) = 0.86806 7.2 θ = 21.3° θ > 21.3° PU P sinθ cosθ = U sin 2θ 2 A0 A0 2 A0τ U PU = (2)(2.50)(1.3) = 0.90278 7.2 θ = 32.3° θ < 32.3 ° 21.3° < θ < 32.3° W

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