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

2021
Views:

429

Pages:

2
Author:

sonalhava9qv

PROBLEM 1.28 The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 58 in. thick and is connected to the vertical rod by a 38 -in.-diameter bolt. Determine (a) the average shearing stress in the bolt, (b) the bearing stress at C in member BD. SOLUTION Use member BCD as a free body, and note that AB is a two force member. lAB = 82 + 1.82 = 8.2 in. § 8 · § 1.8 · ΣM C = 0: (4 cos 20 °) ¨ FAB ¸ − (4sin 20 °) ¨ FAB ¸ © 8.2 ¹ © 8.2 ¹ −(7 cos 20 °)(400sin 75 °) − (7sin 20 °)(400 cos 75 °) = 0 3.36678FAB − 2789.35 = 0 ∴ FAB = 828.49 lb 1.8 FAB + Cx + 400cos 75 ° = 0 8.2 (1.8)(828.49) − 400cos75° = 78.34 lb Cx = 8.2 ΣFx = 0: − 8 F AB + C y − 400sin 75 ° = 0 8.2 (8)(828.49) + 400sin 75° = 1194.65 lb Cy = 8.2 ΣF y = 0: − C = Cx2 + Cy2 = 1197.2 lb PROBLEM 1.28 (Continued) 2 (a) (b) Shearing stress in the bolt: P = 1197.2 lb A= π 2 π §3· d = ¨ ¸ = 0.11045 in 2 4 4 ©8¹ τ= P 1197.2 = = 10.84 × 103 psi = A 0.11045 10.84 ksi W § 3 ·§ 5 · Bearing stress at C in member BCD: P = 1197.2 lb Ab = dt = ¨ ¸¨ ¸ = 0.234375 in 2 © 8 ¹© 8 ¹ σb = P 1197.2 = = 5.11 × 103 psi = 5.11 ksi W Ab 0.234375