545 106 n mm piston and piston rod and shear stress

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: g moment due to the radial component of FQ, 155 140 l t MR = HR2 b1 − c − = 47.3 400 − kN-mm − 2 2 2 2 = 11.94 × 103 kN-mm = 11.94 × 106 N-mm ...(i) We also know that bending moment, MR = σbR × Z = σbR × 1 × t .w2 6 1 6 ... (∵ Z = × t .w2 ) 1 × 245 (140)2 = 800 × 103 σbR 6 ∴ σbR = 11.94 × 106 / 800 × 103 = 14.9 N/mm2 or MPa We know that bending moment due to the tangential component of FQ, 11.94 × 106 = σbR × 155 d MT = FT r − s1 = 84 300 − = 18 690 kN-mm 2 2 = 18.69 × 106 N-mm We also know that bending moment, 1 1 MT = σbT × Z = σbT × × t .w2 ... (∵ Z = × t.w2 ) 6 6 18.69 × 106 = σbT × 1 × 140(245)2 = 1.4 × 106 σbT 6 ∴ σbT = 18.69 × 106 / 1.4 × 106 = 13.35 N/mm2 or MPa Direct compressive stress, σb = FR 94.6 2 2 = = 1.38 × 10 −3 kN/mm = 1.38 N/mm 2w · t 2 × 245 × 140 1182 n A Textbook of Machine Design and total compressive stress, σc = σbR + σbT + σd = 14.9 + 13.35 + 1.38 = 29.63 N/mm2 or MPa We know that twisting moment on the arm, 155 l...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online