# Given 2 pbp bearing pressure 15 nmm given we know

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Unformatted text preview: end bearing to the maximum gas force, i.e. 13 (dc)2 = FL = 24 740 N ∴ (dc )2 = 24 740 / 13 = 1903 or dc = 43.6 say 44 mm Ans. and lc = 1.3 dc = 1.3 × 44 = 57.2 say 58 mm Ans. The big end has removable precision bearing shells of brass or bronze or steel with a thin lining (1mm or less) of bearing metal such as babbit. Again, let dp = Diameter of the piston pin or small end bearing, lp = Length of the piston pin or small end bearing = 2dp ...(Given) 2 pbp = Bearing pressure = 15 N/mm ..(Given) We know that the load on the piston pin or small end bearing = Project area × Bearing pressure = dp . lp . pbp = dp × 2 dp × 15 = 30 (dp)2 Since the piston pin or the small end bearing is designed for the maximum gas force (FL), therefore, equating the load on the piston pin or the small end bearing to the maximum gas force, i.e. 30 (dp)2 = 24 740 N ∴ (dp)2 = 24 740 / 30 = 825 or dp = 28.7 say 29 mm Ans. and lp = 2 dp = 2 × 29 = 58 mm Ans. Internal Combustion Engine Parts n 1159 The small end bearing is usually a phosphor bronze bush of about 3 mm thickness. 3. Size of bolts for secur...
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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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