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5_ch 12 Mechanical Design budynas_SM_ch12

# 5_ch 12 Mechanical Design budynas_SM_ch12 - y 1 4 = 7 4 y 1...

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308 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The power loss due to friction is H = 2 π TN = 2 π (0 . 1125)(20) = 14 . 14 W Ans . Q s = 0 . 57 Q The side ﬂow is 57% of Q Ans . 12-7 c min = b min d max 2 = 30 . 05 30 . 00 2 = 0 . 025 mm r = d 2 = 30 2 = 15 mm r c = 15 0 . 025 = 600 N = 1120 60 = 18 . 67 rev/s P = 2750 30(50) = 1 . 833 MPa S = (600 2 ) ± 60(10 3 )(18 . 67) 1 . 833(10 6 ) ² = 0 . 22 l d = 50 30 = 1 . 67 This l / d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16). From Fig. 12-16, the h o / c values are: y 1 / 4 = 0 . 18, y 1 / 2 = 0 . 34, y 1 = 0 . 54, y = 0 . 89 Substituting into Eq. (12-16), h o c = 0 . 659 From Fig. 12-18, the fr / c values are:
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Unformatted text preview: y 1 / 4 = 7 . 4, y 1 / 2 = 6 . 0, y 1 = 5 . 0, y ∞ = 4 . Substituting into Eq. (12-16), f r c = 4 . 59 From Fig. 12-19, the Q / ( rcNl ) values are: y 1 / 4 = 5 . 65, y 1 / 2 = 5 . 05, y 1 = 4 . 05, y ∞ = 2 . 95 Substituting into Eq. (12-16), Q rcN l = 3 . 605 h o = . 659(0 . 025) = . 0165 mm Ans . f = 4 . 59 / 600 = . 007 65 Ans . Q = 3 . 605(15)(0 . 025)(18 . 67)(50) = 1263 mm 3 /s Ans ....
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