# E dc 149 mm is less than the already calculated value

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online