# We know that horizontal bending moment acting on the

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Unformatted text preview: ect compressive stress, σd = FR 33 × 103 = = 1.9 N/mm2 or MPa w · t 245 × 70 ∴ Total compressive stress, σc = σbT + σbR + σd = 2.6 + 17.1 + 1.9 = 21.6 MPa We know that twisting moment due to the tangential force, T = FT (0.75 lc + 0.5 t) = 30 × 103 (0.75 × 92 + 0.5 × 70) = 3.12 × 106 N-mm ∴ Shear stress, τ= 4.5 T T 4.5 × 3.12 × 106 = = ZP w · t2 245 (70) 2 2 ... ∵ Z P = w · t 4.5 = 11.7 N/mm2 or MPa We know that total or maximum stress, σc 1 21.6 1 (σ c ) 2 + 4 τ 2 = (21.6)2 + 4(11.7)2 + + 2 2 2 2 = 10.8 + 15.9 = 26.7 MPa Since this stress is less than the permissible value of 60 MPa, therefore, the design is safe. (b) Design of shaft at the junction of crank Let ds1 = Diameter of shaft at the junction of crank. We know that bending moment at the junction of crank, M = FQ (0.75lc + t) = 44 565 (0.75 × 92 + 70) = 6.2 × 106 N-mm and twisting moment, T = FT × r = 30 × 103 × 150 = 4.5 × 106 N-mm ∴ Equivalent twisting moment, σmax = 6 M 2 + T 2 = (6.2 × 10...
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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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