6_ch 12 Mechanical Design budynas_SM_ch12

6_ch 12 Mechanical Design budynas_SM_ch12 - 0113 T = f Wr =...

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Chapter 12 309 12-8 c min = b min d max 2 = 75 . 10 75 2 = 0 . 05 mm l / d = 36 / 75 ˙= 0 . 5 (close enough) r = d / 2 = 75 / 2 = 37 . 5mm r / c = 37 . 5 / 0 . 05 = 750 N = 720 / 60 = 12 rev/s P = 2000 75(36) = 0 . 741 MPa Fig. 12-13: SAE 20, µ = 18 . 5mPa · s S = (750 2 ) ± 18 . 5(10 3 )(12) 0 . 741(10 6 ) ² = 0 . 169 From Figures 12-16, 12-18 and 12-21: h o / c = 0 . 29, fr / c = 5 . 1, P / p max = 0 . 315 h o = 0 . 29(0 . 05) = 0 . 0145 mm Ans . f = 5 . 1 / 750 = 0 . 0068 T = fWr = 0 . 0068(2)(37 . 5) = 0 . 51 N · m The heat loss rate equals the rate of work on the film H loss = 2 π TN = 2 π (0 . 51)(12) = 38 . 5W Ans . p max = 0 . 741 / 0 . 315 = 2 . 35 MPa Ans . Fig. 12-13: SAE 40, µ = 37 MPa · s S = 0 . 169(37) / 18 . 5 = 0 . 338 From Figures 12-16, 12-18 and 12-21: h o / c = 0 . 42, fr / c = 8 . 5, P / p max = 0 . 38 h o = 0 . 42(0 . 05) = 0 . 021 mm Ans . f = 8 . 5 / 750 =
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Unformatted text preview: . 0113 T = f Wr = . 0113(2)(37 . 5) = . 85 N · m H loss = 2 π T N = 2 π (0 . 85)(12) = 64 W Ans . p max = . 741 / . 38 = 1 . 95 MPa Ans . 12-9 c min = b min − d max 2 = 50 . 05 − 50 2 = . 025 mm r = d / 2 = 50 / 2 = 25 mm r / c = 25 / . 025 = 1000 l / d = 25 / 50 = . 5, N = 840 / 60 = 14 rev/s P = 2000 25(50) = 1 . 6 MPa...
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