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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...
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