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

2021
Views:

152

Pages:

1
Author:

Alma Stephenson

PROBLEM 1.70 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 (a) the value of θ for which the factor of safety of the member is maximum, (b) the corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors of safety with respect to normal stress and shear.) SOLUTION A0 = (2.0)(1.25) = 2.50 in 2 At the optimum angle, (F .S .)σ = (F.S .)τ σ = σU A0 P cos2 θ ∴ PU , σ = A0 cos2 θ Normal stress: ( F. S.) σ = Shearing stress: τ = σU A0 P cos2 θ Solving: (b) PU = P = σU A 0 P cos2 θ τU A 0 P sin θ cos θ ∴ PU ,τ = A0 sin θ cos θ ( F. S.)τ = Equating: PU ,σ = PU ,τ P = τ U A0 Psinθ cosθ τU A0 P sin θ cos θ τ sinθ 1.3 = tanθ = U = = 0.520 cosθ 2.5 σU (a ) θ opt = 27.5° W (12.5)(2.50) σ U A0 = = 7.94 kips cos 2 θ cos 2 27.5 ° F. S. = PU 7.94 = P 2.4 F. S. = 3.31 W