1_ch 17 Mechanical Design budynas_SM_ch17

1_ch 17 Mechanical Design budynas_SM_ch17 - = 1 3 . 123 ln...

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Chapter 17 17-1 Given: F-1 Polyamide, b = 6in, d = 2in @ 1750 rev/min C = 9(12) = 108 in, vel. ratio 0.5, H nom = 2hp, K s = 1 . 25, n d = 1 Table 17-2: t = 0 . 05 in, d min = 1 . 0in, F a = 35 lbf / in, γ = 0 . 035 lbf / in 3 , f = 0 . 5 Table 17-4: C p = 0 . 70 w = 12 γ bt = 12(0 . 035)(6)(0 . 05) = 0 . 126 lbf/ft θ d = 3 . 123 rad, exp( f θ ) = 4 . 766 (perhaps) V = π dn 12 = π (2)(1750) 12 = 916 . 3 ft/min (a) Eq. ( e ), p. 865: F c = w 32 . 17 ± V 60 ² 2 = 0 . 126 32 . 17 ± 916 . 3 60 ² 2 = 0 . 913 lbf Ans . T = 63 025 H nom K s n d n = 63 025(2)(1 . 25)(1) 1750 = 90 . 0 lbf · in ± F = 2 T d = 2(90) 2 = 90 lbf Eq. (17-12): ( F 1 ) a = bF a C p C v = 6(35)(0 . 70)(1) = 147 lbf Ans . F 2 = F 1 a ± F = 147 90 = 57 lbf Ans . Do not use Eq. (17-9) because we do not yet know f ± . Eq. ( i ), p. 866: F i = F 1 a + F 2 2 F c = 147 + 57 2 0 . 913 = 101 . 1 lbf Ans . Eq. (17-7): f ± = 1 θ d ln ³ ( F 1 ) a F c F 2 F c ´
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Unformatted text preview: = 1 3 . 123 ln 147 . 913 57 . 913 = . 307 The friction is thus undeveloped. (b) The transmitted horsepower is, H = ( F ) V 33 000 = 90(916 . 3) 33 000 = 2 . 5 hp Ans . n f s = H H nom K s = 2 . 5 2(1 . 25) = 1 From Eq. (17-2), L = 225 . 3 in Ans . (c) From Eq. (17-13), dip = 3 C 2 w 2 F i where C is the center-to-center distance in feet. dip = 3(108 / 12) 2 (0 . 126) 2(101 . 1) = . 151 in Ans ....
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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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