The pin is neither simply supported in fork nor

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Unformatted text preview: 743 / 6.25 = 279 or d1 = 16.7 say 17 mm l1 = 1.25 d1 = 1.25 × 17 = 21.25 say 22 mm Now let us check the average shear stress induced in the pin. Since the pin in double shear, therefore, load on the fulcrum pin (RF), π π 2 2 1743 = 2 × (d1 ) τ = 2 × (17) τ = 454 τ 4 4 ∴ τ = 1743 / 454 = 3.84 N/mm2 or MPa This induced shear stress is quite safe. Now external diameter of the boss, D1 = 2d1 = 2 × 17 = 34 mm Assuming a phosphor bronze bush of 3 mm thick, the internal diameter of the hole in the lever, dh = d1 + 2 × 3 = 17 + 6 = 23 mm and (d 1 )2 Internal Combustion Engine Parts n 1205 Now, let us check the induced bending stress for the section of the boss at the fulcrum which is shown in Fig. 32.30. Bending moment at this section, M = Fe × l = 885 × 150 N-mm = 132 750 N-mm Section modulus, ∴ 1 × 22 [(34) 3 − (23)3 ] 12 = 2927 mm3 Z= 34 / 2 Induced bending stress, M 132 750 = = 45.3 N/mm2 or MPa Z 2927 Fig. 32.30 The induced bending stress is quite safe. 2. Design for forked end Let d2= Diameter of the roller pin, and l2= Length of the roller pin =...
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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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