shi20396_ch17

# shi20396_ch17 - Chapter 17 17-1 Preliminaries to give some...

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Chapter 17 17-1 Preliminaries to give some checkpoints: VR = 1 Angular velocity ratio = 1 0 . 5 = 2 H nom = 2hp, 1750 rev/min, C = 9(12) = 108 in, K s = 1 . 25, n d = 1 d min = 1in, F a = 35 lbf/in, γ = 0 . 035 lbf/in 3 , f = 0 . 50, b = 6in, d = 2 in, from Table 17-2 for F-1 Polyamide: t = 0 . 05 in; from 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 ): F c = w 32 . 174 µ V 60 2 = 0 . 126 32 . 174 µ 916 . 3 60 2 = 0 . 913 lbf Ans . T = 63 025(2)(1 . 25)(1) 1750 = 90 . 0 lbf · in 1 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 1 F = 147 90 = 57 lbf Ans . Do not use Eq. (17-9) because we do not yet know f 0 . Eq. ( i ) F i = F 1 a + F 2 2 F c = 147 + 57 2 0 . 913 = 101 . 1 lbf Ans . f 0 = 1 θ d ln · ( F 1 ) a F c F 2 F c ¸ = 1 3 . 123 ln µ 147 0 . 913 57 0 . 913 = 0 . 307 The friction is thus undeveloped. (b) The transmitted horsepower is, H = ( 1 F ) V 33 000 = 90(916 . 3) 33 000 = 2 . 5hp Ans . n fs = H H nom K s = 2 . 5 2(1 . 25) = 1 From Eq. (17-2), L = 225 . 3in 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) = 0 . 151 in Ans .

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432 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Comment : The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f 0 . The limit of narrowing is b min = 4 . 680 in, whence w = 0 . 0983 lbf/ft ( F 1 ) a = 114 . 7 lbf F c = 0 . 712 lbf F 2 = 24 . 6 lbf T = 90 lbf · in (same) f 0 = f = 0 . 50 1 F = ( F 1 ) a F 2 = 90 lbf dip = 0 . 173 in F i = 68 . 9 lbf Longer life can be obtained with a 6-inch wide belt by reducing F i to attain f 0 = 0 . 50 . Prob. 17-8 develops an equation we can use here F i = ( 1 F + F c )exp( f θ ) F c exp( f θ ) 1 F 2 = F 1 1 F F i = F 1 + F 2 2 F c f 0 = 1 θ d ln µ F 1 F c F 2 F c dip = 3( CD / 12) 2 w 2 F i which in this case gives F 1 = 114 . 9 lbf F c = 0 . 913 lbf F 2 = 24 . 8 lbf f 0 = 0 . 50 F i = 68 . 9 lbf dip = 0 . 222 in So, reducing F i from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50, with a corresponding dip of 0.222 in. Having reduced F 1 and F 2 , the endurance of the belt is improved. Power, service factor and design factor have remained in tack. 17-2 There are practical limitations on doubling the iconic scale. We can double pulley diame- ters and the center-to-center distance. With the belt we could: • Use the same A-3 belt and double its width; • Change the belt to A-5 which has a thickness 0.25 in rather than 2(0 . 13) = 0 . 26 in, and an increased F a ; • Double the thickness and double tabulated F a which is based on table thickness. The object of the problem is to reveal where the non-proportionalities occur and the nature of scaling a Fat belt drive. We will utilize the third alternative, choosing anA-3 polyamide belt of double thickness, assuming it is available. We will also remember to double the tabulated F a from 100 lbf/in to 200 lbf/in.
Chapter 17 433 Ex. 17-2: b = 10 in, d = 16 in, D = 32 in, Polyamide A-3, t = 0 . 13 in, γ = 0 . 042, F a = 100 lbf/in, C p = 0 .

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## This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch17 - Chapter 17 17-1 Preliminaries to give some...

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