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hw3.sol.Spring07

hw3.sol.Spring07 - ME 452 Machine Design II Solution to...

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ME 452: Machine Design II Solution to Homework Set 3. Problem 1. (Problem 6-10, page 340) For AISI 1018 cold-drawn steel from Table (A-20), page 1020 we have 440 Mpa ut S = (64 kpsi) and 370 Mpa y S = . From Equation (6-8), page 274, for 1400 Mpa ut S < 0.5 e ut S S = Therefore 0.5 440 220.0 Mpa e S = × = From Table (6-2), page 280 for cold-drawn surface finish 4.51 a = and 0.265 b = - Then from Equation (6-19), page 279, the surface condition modification factor is given by: b a ut k aS = Therefore 0.265 4.51 (440) 0.899 a k - = × = From Equation (6-21), page 280, for axial loading there is no size adjustment factor, so: 1 b k = For axial loading, the loading factor is given by (see Equation (6-26), page 282) 0.85 c k = Assuming that the operating temperature is room temperature, the temperature factor from Table (6-4), page 283, is given by: 1 d k = Assuming 50% reliability, the reliability factor from Table (6-5), page 285 is given by: 1 e k = Assuming there are no miscellaneous effects (see Pages 285 and 286), the miscellaneous- effects factor is given by: 1 f k =
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The endurance strength is then given by the Marin equation (see Equation 6-18, page 279) : e a b c d e f e S k k k k k k S = Then 0.899 1 0.85 1 1 1 220.0 168.1 MPa e S = × × × × × × = For the given dimensions, from Figure (A-15-1), page 1006 for / 12/ 60 0.2 d w = = we have 2.5 t K = For the Neuber constant, note that we need to use S ut = 64 kpsi, in equation 6-35, page 288: 2 4 2 7 3 .245799 .307794(10 )64 .150874(10 )64 .266978(10 )64 0.1036 a in - - - = - + - = 0.1036 0.52133 a in mm = = and 6 2.449 mm r = =
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