# 2 to 04 mm let us adopt g1 128 mm and g2 03 mm ans 4

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Unformatted text preview: he gap between the free ends of the ring, G1 = 3.5 t1 to 4 t1 = 3.5 × 3.4 to 4 × 3.4 mm = 11.9 to 13.6 mm and the gap when the ring is in the cylinder, G2 = 0.002 D to 0.004 D = 0.002 × 100 to 0.004 × 100 mm = 0.2 to 0.4 mm Let us adopt G1 = 12.8 mm ; and G2 = 0.3 mm Ans. 4. Piston barrel Since the radial depth of the piston ring grooves (b) is about 0.4 mm more than the radial thickness of the piston rings (t1), therefore, b = t1 + 0.4 = 3.4 + 0.4 = 3.8 mm We know that the maximum thickness of barrel, t3 = 0.03 D + b + 4.5 mm = 0.03 × 100 + 3.8 + 4.5 = 11.3 mm and piston wall thickness towards the open end, t4 = 0.25 t3 to 0.35 t3 = 0.25 × 11.3 to 0.35 × 11.3 = 2.8 to 3.9 mm Let us adopt t4 = 3.4 mm 5. Piston skirt Let l = Length of the skirt in mm. We know that the maximum side thrust on the cylinder due to gas pressure ( p ), t2 = R = µ× π D2 π (100) 2 × p = 0.1 × × 5 = 3928 N 4 4 ...(Taking µ = 0.1) We also know that the side thrust due to bearing pressure on the piston barrel ( pb ), R = pb × D × l = 0.45 × 100 × l = 45 l N ...(Taking pb = 0.45 N/mm2) F...
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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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