2_ch 17 Mechanical Design budynas_SM_ch17

2_ch 17 Mechanical Design budynas_SM_ch17 - ters and the...

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Chapter 17 421 Comment : The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f ± . 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 ± = f = 0 . 50 ± 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 . 50 . Prob. 17-8 develops an equation we can use here F i = ( ± F + F c )exp( f θ ) F c exp( f θ ) 1 F 2 = F 1 ± F F i = F 1 + F 2 2 F c f ± = 1 θ d ln ± F 1 F c F 2 F c ² 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 . 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-
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Unformatted text preview: 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) = . 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 at 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....
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