hw4csol.sp12

hw4csol.sp12 - Name of Student ME 452 Machine Design II...

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2 Solution to Homework Set 4. Problem 6-15. From Table A-20 for AISI 1095 HR steel, see page 1040, the ultimate tensile strength and the yield strength of the solid steel round bar, respectively, are 120 ut S = kpsi and 66 y S = kpsi (1) Using Eq. (6-8), see page 282, the uncorrected endurance strength of the solid steel round bar is 0.5 0.5 120 60 eu t SS == × = kpsi (2) From Table 6-2, see page 288, the constants are an 2.70 kpsi 0 2 d. 6 5 ab (3) From Eq. (6-19), see page 287, the surface factor is 0.265 2.70 120 0.76 b au t ka S ==× = (4) From Eq. (6-24), see page 289, the equivalent diameter is 0.370 0.370 1.8 0.666 e dd = = in (5) Therefore, from Eq. (6-20), see page 288, the size factor is 0.107 0.107 0.879 0.879 0.666 0.92 be kd −− × = (6) From Eq. (6-26), see page 290, the loading factor is 1 c k = (7) Using Eq. (6-18), see page 287, the corrected endurance strength of the solid steel round bar is 0.76 0.92 1 60 42.0 ea c Sk k k S =′ = × × × = kpsi (8)
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hw4csol.sp12 - Name of Student ME 452 Machine Design II...

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