# The pin is neither simply supported in fork nor

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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...
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## This note was uploaded on 09/30/2012 for the course MECHANICAL 403 taught by Professor A.thirumoorthy during the Spring '11 term at Anna University Chennai - Regional Office, Coimbatore.

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