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18_ch 17 Mechanical Design budynas_SM_ch17

# 18_ch 17 Mechanical Design budynas_SM_ch17 -...

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Chapter 17 437 Eq. (17-21): F c = 3 . 498 V 1000 2 = 3 . 498 2722 . 7 1000 2 = 25 . 9 lbf/belt At fully developed friction, Eq. (17-9) gives F i = T d exp( f θ ) + 1 exp( f θ ) 1 = 2935 26 5 + 1 5 1 = 169 . 3 lbf/belt Eq. (17-10): F 1 = F c + F i 2 exp( f θ ) exp( f θ ) + 1 = 25 . 9 + 169 . 3 2(5) 5 + 1 = 308 . 1 lbf/belt F 2 = F 1 F a = 308 . 1 225 . 8 = 82 . 3 lbf/belt n f s = H a N b H d = (185 . 63) 84 = 1 . 109 Ans . Reminder: Initial tension is for the drive ( F i ) drive = N b F i = 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, H nom = 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 pre-
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