16_ch 17 Mechanical Design budynas_SM_ch17

# 16_ch 17 Mechanical Design budynas_SM_ch17 -...

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Unformatted text preview: budynas_SM_ch17.qxd 12/06/2006 17:29 Page 435 FIRST PAGES 435 Chapter 17 From Eq. (17-27), the number of belt passes is: 1193 150.1 NP = −10.929 −10.929 −1 1193 + 105.2 = 6.76(109 ) From Eq. (17-28) for N P > 109 , t= NP L p 109 (91.8) > 720V 720(5031.8) t > 25 340 h Ans. Suppose n f s was too small. Compare these results with a 2-belt solution. Htab = 4 hp/belt, Ta = 39.6 lbf · in/belt, Fa = 12.8 lbf/belt, nfs = Ha = 3.91 hp/belt Nb Ha Nb Ha 2(3.91) = = = 2.0 Hd Hnom K s 3(1.3) F1 = 40.8 lbf/belt, F2 = 28.0 lbf/belt, Fi = 9.99 lbf/belt, Also, Fc = 24.4 lbf/belt ( Fb ) 1 = 92.9 lbf/belt, ( Fb ) 2 = 48 lbf/belt T1 = 133.7 lbf/belt, T2 = 88.8 lbf/belt N P = 2.39(10 ) passes, t > 605 600 h 10 Initial tension of the drive: ( Fi ) drive = Nb Fi = 2(9.99) = 20 lbf 17-18 Given: two B85 V-belts with d = 5.4 in, D = 16 in, n = 1200 rev/min, and K s = 1.25 L p = 85 + 1.8 = 86.8 in Table 17-11: Eq. (17-17b): C = 0.25 86.8 − π (16 + 5.4) + 2 86.8 − π (16 + 5.4) 2 2 − 2(16 − 5.4) 2 = 26.05 in Ans. Eq. (17-1): θd = 180° − 2 sin−1 16 − 5.4 = 156.5° 2(26.05) From table 17-13 footnote: K 1 = 0.143 543 + 0.007 468(156.5°) − 0.000 015 052(156.5°) 2 = 0.944 Table 17-14: Belt speed: K2 = 1 π (5.4)(1200) = 1696 ft/min V= 12 ...
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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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