hw2.sol.Fall06

hw2.sol.Fall06 - ME 452: Machine Design II Solution to...

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ME 452: Machine Design II Solution to Homework Set 2. Problem 1. (Problem 6-1, page 340) The minimum ultimate tensile strength can be found using Equation (2-17), page 37, as follows: 0.495 0.495 490 242.6 kpsi ut B SH == × = (Note that even though the book indicates this correlation is only valid for H B between 200 and 450, it still gives a reasonable answer. See, for example, page 46 and 47 in Norton, where the correlation is .5*H B , giving S ut = 245 kpsi plus or minus 14.7 kpsi, with no limitations on the values of H B .) Because 200 kpsi ut S > , then from Equation (6-8), page 274 we get: 100 kpsi e S = From Table (6-2), page 280, for ground surface finish we have: 1.34 a = and 0.085 b = − Therefore from Equation (6-19), page 279, the surface condition modification factor is given by: 0.085 1.34 (242.6) 0.840 b au t ka S ==× = For the given shaft diameter of d = 0.25 in, the size factor is given by Equation (6-20), page 280 as: 0.107 0.107 0.879 0.879 (0.25) 1.02 b kd −− × = For rotating bending case, the loading factor from Equation (6-26), page 282 is given by: 1 c k = Assuming that the operating temperature is 70 F ° the temperature factor from Table (6-4), page 283, is given by:
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue.

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hw2.sol.Fall06 - ME 452: Machine Design II Solution to...

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