1 400 62 840 kn mm i we also know that h2 d

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Unformatted text preview: W × c1 W × c / 2 W 50 = = = = 25 kN 2 2 c c W × c2 W × c / 2 W 50 = = = = 25 kN 2 2 c c Due to the resultant belt tension (T1 + T2) acting horizontally, there will be two horizontal reactions H2′ and H3′ respectively, such that and V3 = (T1 + T2 ) c1 (T1 + T2 ) c / 2 T1 + T2 6.5 = = = = 3.25 kN 2 2 c c (T + T2 ) c2 (T1 + T2 ) c / 2 T1 + T2 6.5 = = = = 3.25kN and H3′ = 1 2 2 c c Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin Let dc = Diameter of the crankpin in mm ; lc = Length of the crankpin in mm ; and σb = Allowable bending stress for the crankpin. It may be assumed as 75 MPa or N/mm2. We know that the bending moment at the centre of the crankpin, MC = H1 · b2 = 157.1 × 400 = 62 840 kN-mm ...(i) We also know that H2′ = π π ( d ) 3 σb = (d )3 75 = 7.364(dc )3 N-mm 32 c 32 c = 7.364 × 10–3 (dc)3 kN-mm Equating equations (i) and (ii), we have (dc)3 = 62 840 / 7.364 × 10–3 = 8.53 × 106 dc = 204.35 say 205 mm Ans. We kn...
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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.

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