The pin is neither simply supported in fork nor

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 τ = 2 × (30)2 τ = 1414 τ 41 4 ∴ τ = 4545 / 1414 = 3.2 N/mm2 or MPa This induced shear stress is quite safe. Now external diameter of the boss, D1 = 2d1 = 2 × 30 = 60 mm Assuming a phosphor bronze bush of 3 mm thick, the internal diameter of the hole in the lever, dh = d1 + 2 × 3 = 30 + 6 = 36 mm Let us now check the induced bending stress for the section of the boss at the fulcrum which is shown in Fig. 32.26. 4545 = 2 × 1198 n A Textbook of Machine Design Bending moment at this section, M = Fe × l = 2460 × 180 = 443 × 103 N-mm Section modulus, 1 × 37.5 [ (60)3 − (36)3 ] 12 = 17 640 mm3 Z= 60 / 2 ∴ Induced bending stress, 443 × 103 M = = 25.1 N/mm2 or MPa σb = 17 640 Z The induced bending stress is quite safe. 2. Design for forked end Let d2 = Diameter of the roller pin, Fig. 32.26 and l2 = Length of the roller pin = 1.25 d1 ...(Assume) Considering bearing of the roller pin. We know that load on the roller pin (Fc), 2460 = d2 × l2 × pb = d2 × 1.25 d2 × 7 = 8.75 (d2)2 ... (T...
View Full Document

Ask a homework question - tutors are online