# 545 106 n mm piston and piston rod and shear stress

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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...
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