hw2.sol.Fall.2010x

# Hw2.sol.Fall.2010x - ME 452 Machine Design II Solution to Homework Set 2 Problem 1(Problem 6-1 page 348 A 10-mm drill rod was heat-treated and

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ME 452: Machine Design II Solution to Homework Set 2. Problem 1. (Problem 6-1, page 348) A 10-mm drill rod was heat-treated and ground. The measured hardness was found to be 300 Brinell. Estimate the endurance strength in MPa if the rod is used in rotating bending. SOLUTION: The minimum ultimate tensile strength of steels can be estimated using Equation (2-21), page 41, as follows: 3.4 3.4 300 1020 MPa ut B S H = = × = Because 1400 MPa ut S < , then from Equation (6-8), page 282 we get: 0.5 = 0.5(1020) = 510 MPa e ut S S = This is the endurance limit for a test specimen made from the given material. This is not the endurance limit for the given drill rod, however, because it needs to be adjusted for size, surface finish and the other Marin factors. From Table (6-2), page 288, for ground surface finish we have: 1.58 a = and 0.085 b = - Therefore from Equation (6-19), page 287, the surface condition modification factor is given by: 0.085 1.58 (1020) 0.8768 b a ut k aS - = = × = For the given shaft diameter of d = 10 mm, the size factor is given by Equation (6-20), page 288 as: 0.107 0.107 1.24 1.24 (10) 0.9692 b k d - - = = × = For rotating bending case, the loading factor from Equation (6-26), page 290 is given by: 1 c k = Assuming that the operating temperature is 70 F ° the temperature factor from Table (6-4), page 291, is given by: 1 d k = Assuming 50% reliability, the reliability factor from Table (6-5), page 293 is given by:

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Hw2.sol.Fall.2010x - ME 452 Machine Design II Solution to Homework Set 2 Problem 1(Problem 6-1 page 348 A 10-mm drill rod was heat-treated and

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