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# 75 92 70 05 200 246 106 n mm we also know that bending

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Unformatted text preview: gn of crankpin Let dc = Diameter of the crankpin in mm, and lc = Length of the crankpin = 0.8 dc ...(Assume) Considering the crankpin in bearing, we have FP = dc.lc.pb 103 × 103 = dc × 0.8 dc × 10 = 8 (dc)2 ...(Taking pb = 10 N/mm2) 2 = 103 × 103 / 8 = 12 875 or d = 113.4 say 115 mm ∴ ( d c) c and lc = 0.8 dc = 0.8 × 115 = 92 mm Let us now check the induced bending stress in the crankpin. We know that bending moment at the crankpin, FP = M= 3 3 F × l = × 103 × 103 × 92 = 7107 × 103 N-mm 4P c 4 and section modulus of the crankpin, π π ( d )3 = (115)3 = 149 × 103 mm3 32 c 32 ∴ Bending stress induced Z= M 7107 × 103 = = 47.7 N/mm 2 or MPa 3 Z 149 × 10 Since the induced bending stress is within the permissible limits of 60 MPa, therefore, design of crankpin is safe. = 1184 n A Textbook of Machine Design (b) Design of bearings Let d1 = Diameter of the bearing 1. Let us take thickness of the crank web, t = 0.6 dc= 0.6 × 115 = 69 or 70 mm and length of the bearing, l1 = 1.7 dc = 1.7 × 115 = 195.5 say 200 mm We...
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