19_ch 17 Mechanical Design budynas_SM_ch17

# 19_ch 17 Mechanical Design budynas_SM_ch17 - i 2 exp( f d )...

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438 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design For the ﬂat on ﬂywheel exp[0 . 13(3 . 791)] = 1 . 637 V = π dn 12 = π (11)(875) 12 = 2519 . 8 ft/min Table 17-13: Regression equation gives K 1 = 0 . 90 Table 17-14: K 2 = 1 . 15 Table 17-12: H tab = 7 . 83 hp/belt by interpolation Eq. (17-17): H a = K 1 K 2 H tab = 0 . 905(1 . 15)(7 . 83) = 8 . 15 hp Eq. (17-19): H d = H nom K s n d = 50(1 . 2)(1 . 1) = 66 hp Eq. (17-20): N b = H d H a = 66 8 . 15 = 8 . 1 belts Decision : Use 9 belts. On a per belt basis, ± F a = 63 025 H a n ( d / 2) = 63 025(8 . 15) 875(11 / 2) = 106 . 7 lbf/belt T a = ± F a d 2 = 106 . 7(11) 2 = 586 . 9 lbf per belt F c = 1 . 716 ± V 1000 ² 2 = 1 . 716 ± 2519 . 8 1000 ² 2 = 10 . 9 lbf/belt At fully developed friction, Eq. (17-9) gives F i = T d ³ exp( f θ d ) + 1 exp( f θ d ) 1 ´ = 586 . 9 11 ± 3 . 5846 + 1 3 . 5846 1 ² = 94 . 6 lbf/belt Eq. (17-10): F 1 = F c + F
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Unformatted text preview: i 2 exp( f d ) exp( f d ) + 1 = 10 . 9 + 94 . 6 2(3 . 5846) 3 . 5846 + 1 = 158 . 8 lbf/belt F 2 = F 1 F a = 158 . 8 106 . 7 = 52 . 1 lbf/belt n f s = N b H a H d = 9(8 . 15) 66 = 1 . 11 O.K. Ans . Durability: ( F b ) 1 = 145 . 45 lbf/belt, ( F b ) 2 = 76 . 7 lbf/belt T 1 = 304 . 4 lbf/belt, T 2 = 185 . 6 lbf/belt and t > 150 000 h Remember: ( F i ) drive = 9(94 . 6) = 851 . 4 lbf Table 17-9: C-section belts are 7 / 8 " wide. Check sheave groove spacing to see if 14 "-width is accommodating....
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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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