# 8 106 m bt 6 m bt 70 245 2 z tw 2 z 1 t

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Unformatted text preview: 50) 2 0.9 = 44 200 N 4 4 In order to find the thrust in the connecting rod (FQ), we should first find out the angle of inclination of the connecting rod with the line of storke (i.e. angle φ). We know that sin θ sin 35° = = 0.1275 l/r 4.5 φ = sin–1 (0.1275) = 7.32° sin φ = ∴ We know that thrust in the connecting rod, 44 200 44 200 FP = = = 44 565 N cos φ cos 7.32º 0.9918 Tangential force acting on the crankshaft, FQ = FT = FQ sin (θ + φ) = 44 565 sin (35º + 7.32º) = 30 × 103 N and radial force, FR = FQ cos (θ + φ) = 44 565 cos (35º + 7.32º) = 33 × 103 N Due to the tangential force (FT,) there will be two reactions at the bearings 1 and 2, such that HT1 = FT ( a + b ) 30 × 103 (239 + 470) = = 45 × 103 N 470 b FT × a 30 × 103 × 239 = = 15.3 × 103 N 470 b Due to the radial force (FR), there will be two reactions at the bearings 1 and 2, such that and HT2 = HR1 = FR ( a + b) 33 × 103 × (239 + 470) = = 49.8 × 103 N 470 b FR × a 33 × 103 × 239 = = 16.8 × 103 N 470 b Now t...
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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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