4_ch 06 Mechanical Design budynas_SM_ch06

# 4_ch 06 Mechanical Design budynas_SM_ch06 - 88 7 = − 158...

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150 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The Marin factors are k a = 4 . 51(440) 0 . 265 = 0 . 899 k b = ± 20 7 . 62 ² 0 . 107 = 0 . 902 k c = 0 . 59, k d = 1, k e = 1 Eq. (6-18): S e = 0 . 899(0 . 902)(0 . 59)(220) = 105 . 3MPa Eq. (6-14): a = [0 . 9(295)] 2 105 . 3 = 669 . 4 Eq. (6-15): b =− 1 3 log 0 . 9(295) 105 . 3 =− 0 . 133 88 Eq. (6-16): N = ± 175 . 2 669 . 4 ² 1 / 0 . 133 88 N = 22 300 cycles Ans. (b) For an operating temperature of 450°C, the temperature modiﬁcation factor, from Table 6-4, is k d = 0 . 843 Thus S e = 0 . 899(0 . 902)(0 . 59)(0 . 843)(220) = 88 . 7MPa a = [0 . 9(295)] 2 88 . 7 = 794 . 7 b =− 1 3 log 0 . 9(295)
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Unformatted text preview: 88 . 7 = − . 158 71 N = ± 175 . 2 794 . 7 ² 1 / − . 15871 N = 13 700 cycles Ans. 6-9 f = . 9 n = 1 . 5 N = 10 4 cycles For AISI 1045 HR steel, S ut = 570 MPa and S y = 310 MPa S ± e = . 5(570 MPa) = 285 MPa Find an initial guess based on yielding: σ a = σ max = Mc I = M ( b / 2) b ( b 3 ) / 12 = 6 M b 3 M max = (1 kN)(800 mm) = 800 N · m F ± ² 1 kN b b 800 mm...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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