I 6 here z 1 w t 2 ii 6 iii iv we

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Unformatted text preview: ect compressive stress due to the radial force FR ; and (iv) Shear stress due to the twisting moment of FT. We know that bending moment due to the tangential force, d MbT = FT r − 1 2 where d1 = Diameter of the bearing 1. Diesel, petrol and steam engines have crank shaft ∴ Bending stress due to the tangential force, MbT 6 MbT σbT = Z = t .w2 We know that bending moment due to the radial force, MbR = FR (0.75 lc + 0.5 t) ∴ Bending stress due to the radial force, MbR 6 MbR = σbR = Z w .t 2 We know that direct compressive stress, FR σd = w. t ∴ Total compressive stress, σc = σbT + σbR + σd ∴ Shear stress, where ZP = Polar section modulus = 1 × t · w2 ) ...(i) 6 ...(Here Z = 1 × w · t 2 ) ...(ii) 6 ...(iii) ...(iv) We know that twisting moment due to the tangential force, T = FT (0.75 lc + 0.5 t) 4.5 T T τ=Z = w .t 2 P ...(∵ Z = w .t 2 4.5 Internal Combustion Engine Parts n 1175 Now the total or maximum stress is given by σc 1 ( σ c ) 2 + 4τ 2 + 2 2 This total maximum stress should be less than the maximum allowable stress....
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