E dc 149 mm is less than the already calculated value

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Unformatted text preview: 42 kN 800 b Due to the radial force (FR), there will be two reactions at bearings 1 and 2, such that F × b1 94.6 × 400 = HR1 = R = 47.3 kN 800 b F × b2 94.6 × 400 = HR2 = R = 47.3 kN 800 b Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin Let dc = Diameter of crankpin in mm. 1180 n A Textbook of Machine Design We know that the bending moment at the centre of the crankpin, MC = HR1 × b2 = 47.3 × 400 = 18 920 kN-mm and twisting moment on the crankpin, TC = HT1 × r = 42 × 300 = 12 600 kN-mm ∴ Equivalent twisting moment on the crankpin, Te = ( M C ) 2 + (TC ) = (18 920)2 + (12 600) 2 = 22 740 kN-mm = 22.74 × 106 N-mm We know that equivalent twisting moment (Te), 22.74 × 106 = π π (dc )3 τ = (dc )3 35 = 6.873 (dc )3 16 16 ...(Taking τ = 35 MPa or N/mm2) ∴ (dc)3 = 22.74 × 106 / 6.873 = 3.3 × 106 or dc = 149 mm Since this value of crankpin diameter (i.e. dc = 149 mm) is less than the already calculated value of dc = 205 mm, therefore, we shall take dc = 205 mm. Ans. (b) Design of shaft under the flywheel Let ds = Diameter of the shaft in mm. The resulting b...
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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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