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7_ch 12 Mechanical Design budynas_SM_ch12

# 7_ch 12 Mechanical Design budynas_SM_ch12 - l d = 1 P =...

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310 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 12-13: SAE 30, µ = 34 mPa · s S = (1000 2 ) 34(10 3 )(14) 1 . 6(10 6 ) = 0 . 2975 From Figures 12-16, 12-18, 12-19 and 12-20: h o / c = 0 . 40, f r / c = 7 . 8, Q s / Q = 0 . 74, Q / ( rcNl ) = 4 . 9 h o = 0 . 40(0 . 025) = 0 . 010 mm Ans . f = 7 . 8 / 1000 = 0 . 0078 T = f Wr = 0 . 0078(2)(25) = 0 . 39 N · m H = 2 π T N = 2 π (0 . 39)(14) = 34 . 3 W Ans . Q = 4 . 9 rcNl = 4 . 9(25)(0 . 025)(14)(25) = 1072 mm 2 /s Q s = 0 . 74(1072) = 793 mm 3 /s Ans . 12-10 Consider the bearings as specified by minimum f : d + 0 t d , b + t b 0 maximum W : d + 0 t d , b + t b 0 and differing only in d and d . Preliminaries:
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Unformatted text preview: l / d = 1 P = 700 / (1 . 25 2 ) = 448 psi N = 3600 / 60 = 60 rev/s Fig. 12-16: minimum f : S ˙= . 08 maximum W : S ˙= . 20 Fig. 12-12: µ = 1 . 38(10 − 6 ) reyn µ N / P = 1 . 38(10 − 6 )(60 / 448) = . 185(10 − 6 ) Eq. (12-7): r c = ³ S µ N / P For minimum f : r c = ³ . 08 . 185(10 − 6 ) = 658 c = . 625 / 658 = . 000 950 . = . 001 in...
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