CH-32

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: say 245 mm Ans. 60 w (d) Design of shaft under the flywheel. Let ds = Diameter of shaft under the flywheel. First of all, let us find the horizontal and vertical reactions at bearings 1 and 2. Assume that the width of flywheel is 250 mm and l1 = l2 = 200 mm. Allowing for certain clearance, the distance ∴ 60 = l1 l2 + + clearance 2 2 200 200 + + 20 = 470 mm = 250 + 2 2 and a = 0.75 lc + t + 0.5 l1 = 0.75 × 92 + 70 + 0.5 × 200 = 239 mm We know that the horizontal reactions H1 and H2 at bearings 1 and 2, due to the piston gas load (FP) are b = 250 + Internal Combustion Engine Parts n 1185 FP ( a + b) 103 × 103 (239 + 470) = = 155.4 × 103 N 470 b F × a 103 × 10 3 × 239 and H2 = P = = 52.4 × 10 3 N 470 b Assuming b1 = b2 = b / 2, the vertical reactions V1 and V2 at bearings 1 and 2 due to the weight of the flywheel are H1 = W · b1 W × b / 2 W 30 × 103 = = = = 15 × 103 N 2 2 b b W · b2 W × b / 2 W 30 × 103 and V2 = = = = = 15 × 103 N 2 2 b b Since there is no belt tension, therefore the horizontal reactions due to the belt tension are neglected. We know that horizontal bending moment...
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