# CH-32

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Unformatted text preview: on the shaft (MS), π π (d )3 42 = 4.12 (d s )3 10.08 × 106 = (d s )3 σb = 32 32 s ∴ (ds)3 = 10.08 × 106 / 4.12 = 2.45 × 106 or ds = 134.7 say 135 mm Ans. 2. Design of the crankshaft when the crank is at an angle of maximum twisting moment We also know that piston gas load, π π × D 2 × p ′ = (400)2 1 = 125 680 N = 125.68 kN FP = 4 4 In order to find the thrust in the connecting rod (FQ), we should first find out the angle of inclination of the connecting rod with the line of stroke (i.e. angle φ). We know that sin θ sin 35° = = 0.1147 sin φ = l/r 5 ∴ φ = sin–1 (0.1147) = 6.58° We know that thrust in the connecting rod, FP 125.68 125.68 FQ = cos φ = cos 6.58º = 0.9934 = 126.5 kN Tangential force acting on the crankshaft, FT = FQ sin (θ + φ) = 126.5 sin (35° + 6.58°) = 84 kN and radial force, FR= FQ cos (θ + φ) = 126.5 cos (35° + 6.58°) = 94.6 kN Due to the tangential force (FT), there will be two reactions at bearings 1 and 2, such that F × b1 84 × 400 = HT1 = T = 42 kN 800 b F × b2 84 × 400 = and HT2 = 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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