# 5 14 2 3 24 mm 2 2 3 design for rocker arm cross

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Unformatted text preview: 1.25 d2 ...(Assume) Considering bearing of the roller pin. We know that load on the roller pin (Fc), 885= d2 × l2 × pb = d2 × 1.25 d2 × 7 = 8.75 (d2)2 ...(Taking pb = 7 N/mm2) ∴ (d2)2= 885 / 8.75 = 101.14 or d2 = 10.06 say 11 mm Ans. and l2= 1.25 d2 = 1.25 × 11 = 13.75 say 14 mm Ans. σb = Power transmission gears in an automobile engine 1206 n A Textbook of Machine Design Let us now check the roller pin for induced shearing stress. Since the pin is in double shear, therefore, load on the roller pin (Fc ), 885 = 2 × π π (d ) 2 τ = 2 × (11) 2 τ = 190 τ 42 4 ∴ τ = 885 / 190 = 4.66 N/mm2 or MPa This induced shear stress is quite safe. The roller pin is fixed in the eye and thickness of each eye is taken as one-half the length of the roller pin. ∴ Thickness of each eye, l2 14 = = 7 mm 2 2 Let us now check the induced bending stress in the roller pin. The pin is neither simply supported in fork nor rigidly fixed at the end. Therefore, the common practice is to assume the load distribution as sho...
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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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