We know that horizontal bending moment acting on the

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Unformatted text preview: (b) Design of shaft at the junction of crank Let ds1 = Diameter of the shaft at the junction of the crank. We know that bending moment at the junction of the crank, M = FQ (0.75lc + t) and twisting moment on the shaft T = FT × r ∴ Equivalent twisting moment, σmax = ...(v) M2 + T2 We also know that equivalent twisting moment, ...(i) Te = π (d )3 τ ...(ii) 16 s1 From equations (i) and (ii), the diameter of the shaft at the junction of the crank (ds1) may be determined. (c) Design of shaft under the flywheel Let ds = Diameter of shaft under the flywheel. The resultant bending moment (MR) acting on the shaft is obtained in the similar way as discussed for dead centre position. We know that horizontal bending moment acting on the shaft due to piston gas load, Te = 2 2 M1 = FP ( a + b) − ( H R1 ) + ( H T1 ) b2 and horizontal bending moment at the flywheel location due to belt pull, (T + T2 ) b1.b2 M2 = H1′ .b2 = H2 ′ .b1 = 1 b ∴ Total horizontal bending moment, MH = M1 + M2 Vertical bending moment due t...
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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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