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

2021
Views:

456

Pages:

2
Author:

Carissa Thomas

PROBLEM 1.64 Two wooden members of uniform rectangular cross section of sides a = 100 mm and b = 60 mm are joined by a simple glued joint as shown. Knowing that the ultimate stresses for the joint are σ U = 1.26 MPa in tension and τ U = 1.50 MPa in shear, and that P = 6 kN, determine the factor of safety for the joint when (a) α = 20°, (b) α = 35°, (c) α = 45°. For each of these values of α, also determine whether the joint will fail in tension or in shear if P is increased until rupture occurs. SOLUTION Let θ = 90 ° − α as shown. From the text book: or σ = P cos2 θ A0 σ = P sin2 α A0 (1) τ = P sin α cosα A0 (2) τ = P sin θ cos θ A0 A0 = ab = (100 mm)(60 mm) = 6000 mm 2 = 6 × 10− 3m 2 τU = 1.50 × 10 6 Pa σU = 1.26× 106 Pa Ultimate load based on tension across the joint: (PU )σ = = (1.26 × 106 )(6 × 10−3 ) σ U A0 = 2 sin α sin 2 α 7560 7.56 = kN sin 2 α sin 2 α Ultimate load based on shear across the joint: (1.50 × 106 )(6 × 10 −3 ) τ U A0 = sin α cos α sin α cos α 9000 9.00 = = kN sin α cos α sin α cos α ( PU )τ = PROBLEM 1.64 (Continued) (a) 7.56 = 64.63 kN sin 2 20 ° 9.00 = ( PU ) τ = = 28.00 kN sin 20° cos 20° α = 20° : (PU )σ = The smaller value governs. The joint will fail in shear and PU = 28.00 kN. F .S . = (b) α = 35° : PU 28.00 = 6 P F .S . = 4.67 W 7.56 = 22.98 kN sin 2 35 ° 9.00 = 19.155 kN (PU )τ = sin 35° cos35° (PU )σ = The joint will fail in shear and PU = 19.155 kN. F .S . = (c) 19.155 PU = P 6 F .S . = 3.19 W 7.56 = 15.12 kN sin 2 45 ° 9.00 = 18.00 kN ( PU )τ = sin 45 ° cos 45 ° α = 45° : (PU )σ = The joint will fail in tension and PU = 15.12 kN. F .S . = 15.12 PU = P 6 F .S . = 2.52 W