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hw3csol.sp12

hw3csol.sp12 - Name of Student ME 452 Machine Design II...

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1 Name of Student ______________________ ME 452: Machine Design II Spring Semester 2012 Homework Set 3 Attempt by Wednesday, February 1st Solve the following problems from Chapter 6, Shigley’s Mechanical Engineering Design, Ninth Edition, R.G. Budynas and J.K. Nisbett. 1. Problem 6-1 page 348. 2. Problem 6-3 page 348. 3. Problem 6-6 page 348. 4. Problem 6-12 pages 348.

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2 Solution to Homework Set 3 Problem 6-1. Using Eq. (2-21), see page 41, the ultimate tensile strength of the drill rod is 3.4 3.4 300 1020 ut B SH == × = MPa (1) Using Eq. (6-8), see page 282, the uncorrected endurance strength of the drill rod is 0.5 0.5 1020 510 eu t SS ′= = × = MPa (2) From Table 6-2, see page 288, the constants are and 1.58 MPa 0.085 ab (3) From Eq. (6-19), see page 287, and Table 6.2, page 288, the surface factor for ground finish is 0.085 1.58 1020 0.877 b au t ka S ==× = (4) From Eq. (6-20), see page 288, the size factor is 0.107 0.107 1.24 1.24(10mm) 0.969 b kd −− = (5) Using Eq. (6-18), see page 287, the corrected endurance strength of the drill rod is 0.877 0.969 510 433 ea b e Sk k S =′ × = MPa Ans. (6) Problem 6-3. The ultimate tensile strength and the completely reversed stress, respectively, are 120 ut S = kpsi and 70 rev σ = kpsi (1) Using Eq. (6-8), see page 282, the corrected endurance limit of the steel specimen is 0.5 120 60 ee × = kpsi (2) From Fig. 6-18, see page 285, the fatigue strength fraction is 0.82 f = (3) Using Eq. (6-14), see page 285, the constant is 2 2 () [0.82 120] 161.4 60 ut e fS a S × = kpsi (4) Using Eq. (6-15), see page 285, the constant is 1 1 0.82 120 log log 0.0716 33 6 0
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hw3csol.sp12 - Name of Student ME 452 Machine Design II...

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