We also know that bending moment due to the radial

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Unformatted text preview: earing 1, 27.14 × 106 = R1 = ( H T1 ) 2 + ( H R1 )2 = (42)2 + (47.3)2 = 63.3 kN ∴ Bending moment at the juncture of the right hand crank arm, l t t l MS1 = R1 b2 + c + − FQ c + 2 2 2 2 Internal Combustion Engine Parts n 1181 155 140 155 140 + − 126.5 + = 63.3 400 + 2 2 2 2 = 34.7 × 103 – 18.7 × 103 = 16 × 103 kN-mm = 16 × 106 N-mm and twisting moment at the juncture of the right hand crank arm, TS1 = FT × r = 84 × 300 = 25 200 kN-mm = 25.2 × 106 N-mm ∴ Equivalent twisting moment at the juncture of the right hand crank arm, Te = = ( M S1 ) 2 + (TS1 ) 2 (16 × 106 ) 2 + (25.2 × 106 ) 2 = 29.85 × 106 N-mm We know that equivalent twisting moment (Te), 29.85 × 106 = π π (d s1 )3 τ = (d s1 )3 42 = 8.25 (d s1 )3 16 16 ...(Taking τ = 42 MPa or N/mm2) ∴ (ds1)3 = 29.85 × 106 / 8.25 = 3.62 × 106 or ds1 = 153.5 say 155 mm Ans. (d) Design of right hand crank web Let σbR = Bending stress in the radial direction ; and σbT = Bending stress in the tangential direction. We also know that bendin...
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