# E angle we know that sin sin 35 01275 lr 45 sin1

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Unformatted text preview: at the flywheel location due to piston gas load. M1 = FP (a + b2) – H1 · b2 V1 = b 470 470 3 = 103 × 103 239 + ... ∵ b2 = 2 − 155.4 × 10 × 2 2 = 48.8 × 106 – 36.5 × 106 = 12.3 × 106 N-mm Since there is no belt pull, therefore, there will be no horizontal bending moment due to the belt pull, i.e. M2 = 0. ∴ Total horizontal bending moment, MH = M1 + M2 = M1 = 12.3 × 106 N-mm We know that vertical bending moment due to the flywheel weight, W . b1 . b2 W × b × b W × b = = MV = 2×2×b 4 b 3 30 × 10 × 470 = = 3.525 × 106 N-mm 4 Inside view of a car engine 1186 n A Textbook of Machine Design ∴ Resultant bending moment, ( M H ) 2 + ( M V )2 = MR = (12.3 × 106 ) 2 + (3.525 × 106 ) 2 = 12.8 × 106 N-mm We know that bending moment (MR), π π (d )3 b = (d )3 60 = 5.9 (d s )3 32 s 32 s (ds)3 = 12.8 × 106 / 5.9 = 2.17 × 106 or ds = 129 mm 12.8 × 106 = ∴ Actually ds should be more than d1. Therefore let us take ds = 200 mm Ans. 2. Design of crankshaft when the crank is at an angtle of maximum twisting moment We know that piston gas load, FP = π π × D 2 × p ′ = (2...
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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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