18_ch 17 Mechanical Design budynas_SM_ch17

18_ch 17 Mechanical Design budynas_SM_ch17 -...

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Unformatted text preview: budynas_SM_ch17.qxd 12/06/2006 17:29 Page 437 FIRST PAGES 437 Chapter 17 Eq. (17-21): Fc = 3.498 V 1000 2 = 3.498 2722.7 1000 2 = 25.9 lbf/belt At fully developed friction, Eq. (17-9) gives Fi = exp( f θ ) + 1 2935 = exp( f θ ) − 1 26 T d Eq. (17-10): F1 = Fc + Fi F2 = F1 − nfs = 5+1 5−1 = 169.3 lbf/belt 2(5) 2 exp( f θ ) = 25.9 + 169.3 = 308.1 lbf/belt exp( f θ ) + 1 5+1 Fa = 308.1 − 225.8 = 82.3 lbf/belt Ha Nb (185.63) = = 1.109 Ans. Hd 84 Reminder: Initial tension is for the drive ( Fi ) drive = Nb Fi = 5(169.3) = 846.5 lbf A 360 belt is at the right-hand edge of the range of center-to-center pulley distances. D ≤ C ≤ 3( D + d ) 26 ≤ C ≤ 3(26 + 26) 17-20 ˙ Preliminaries: D = 60 in, 14-in wide rim, Hnom = 50 hp, n = 875 rev/min, K s = 1.2, n d = 1.1, m G = 875/170 = 5.147, d = 60/5.147 = 11.65 in ˙ (a) From Table 17-9, an 11-in sheave exceeds C-section minimum diameter and precludes D- and E-section V-belts. Decision: Use d = 11 in, C270 belts L p = 270 + 2.9 = 272.9 in Table 17-11: π π 272.9 − (60 + 11) C = 0.25 272.9 − (60 + 11) + 2 2 = 76.78 in This fits in the range D < C < 3( D + d ) 60 < C < 3(60 + 11) 60 in < C < 213 in θd = π − 2 sin−1 60 − 11 = 2.492 rad 2(76.78) θ D = π + 2 sin−1 60 − 11 = 3.791 rad 2(76.78) exp[0.5123(2.492)] = 3.5846 2 − 2(60 − 11) 2 ...
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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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