# 25 400 1300 kn mm 13 106 n mm resultant bending moment

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Unformatted text preview: the balancing point of view, the dimensions of the right hand crank web (i.e. thickness and width) are made equal to the dimensions of the left hand crank web. (d) Design of shaft under the flywheel Let ds = Diameter of the shaft in mm. Since the lengths of the main bearings are equal, therefore 155 b l l1 = l2 = l3 = 2 − c − t = 2 400 − − 140 = 365 mm 22 2 Assuming width of the flywheel as 300 mm, we have c = 365 + 300 = 665 mm Hydrostatic transmission inside a tractor engine Internal Combustion Engine Parts n 1179 Allowing space for gearing and clearance, let us take c = 800 mm. c 800 ∴ c1 = c2 = = = 400 mm 2 2 We know that bending moment due to the weight of flywheel, MW = V3 · c1 = 25 × 400 = 10 000 kN-mm = 10 × 106 N-mm and bending moment due to the belt pull, MT = H3′ · c1 = 3.25 × 400 = 1300 kN-mm = 1.3 × 106 N-mm ∴ Resultant bending moment on the shaft, MS = ( MW )2 + ( MT )2 = (10 × 106 )2 + (1.3 × 106 )2 = 10.08 × 106 N-mm We also know that bending moment...
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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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