15 24 740 n 4 4 we know that the connecting rod is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (3t )3 = 12 t 12 Iyy = 2 × and ∴ 1 1 131 4 × t (4 t )3 + × 3t × t3 = t 12 12 12 I xx 419 12 × = 3.2 = 12 131 I yy I xx = 3.2, therefore the section chosen in quite satisfactory. I yy Now let us find the dimensions of this I-section. Since the connecting rod is designed by taking the force on the connecting rod (FC) equal to the maximum force on the piston (FL) due to gas pressure, therefore, Since π D2 π (100) 2 ×p= × 3.15 = 24 740 N 4 4 We know that the connecting rod is designed for buckling about X-axis (i.e. in the plane of motion of the connecting rod) assuming both ends hinged. Since a factor of safety is given as 6, therefore the buckling load, WB = F.C × F. S. = 24 740 × 6 = 148 440 N FC = FL = * Superfluous data Internal Combustion Engine Parts n 1157 We know that radius of gyration of the section about X-axis, kxx = I xx = A 41 9t 12 4 × 1 11t 2 = 1 .7 8 t Length of crank, Stroke of piston 190 = = 95 mm 2 2 Length of the connecting rod, l = 380 mm ∴ Equivalent len...
View Full Document

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.

Ask a homework question - tutors are online