13_ch 17 Mechanical Design budynas_SM_ch17

# 13_ch 17 Mechanical Design budynas_SM_ch17 - = 301 3 lbf...

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432 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design For full friction development: exp( f θ d ) = exp[0 . 32(2 . 845)] = 2 . 485 V = π dn 12 = π (3)(1125) 12 = 883 . 6 ft/min S f = 56 670 psi From selection step 3 a = ± S f Et (1 ν 2 ) d ² t = ± 56 670 17(10 6 )(0 . 003) (1 0 . 22 2 )(3) ² (0 . 003) = 116 . 4 lbf/in ± F = 2 T d = 2(385 . 2) 3 = 256 . 8 lbf b min = ± F a ± exp( f θ d ) exp( f θ d ) 1 ² = 256 . 8 116 . 4 ³ 2 . 485 2 . 485 1 ´ = 3 . 69 in Decision #2 : b = 4in F 1 = ( F 1 ) a = ab = 116 . 4(4) = 465 . 6 lbf F 2 = F 1 ± F = 465 . 6 256 . 8 = 208 . 8 lbf F i = F 1 + F 2 2 = 465 . 6 + 208 . 8 2 = 337 . 3 lbf Existing friction f ± = 1 θ d ln ³ F 1 F 2 ´ = 1 2 . 845 ln ³ 465 . 6 208 . 8 ´ = 0 . 282 H = ( ± F ) V 33 000 = 256 . 8(883 . 6) 33 000 = 6 . 88 hp n fs = H 5(1 . 25) = 6 . 88 5(1 . 25) = 1 . 1 F i can be reduced only to the point at which f ± = f = 0 . 32. From Eq. (17-9) F i = T d ± exp( f θ d ) + 1 exp( f θ d ) 1 ² = 385 . 2 3 ³ 2 . 485 + 1 2 . 485 1 ´
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Unformatted text preview: = 301 . 3 lbf Eq. (17-10): F 1 = F i ± 2 exp( f θ d ) exp( f θ d ) + 1 ² = 301 . 3 ± 2(2 . 485) 2 . 485 + 1 ² = 429 . 7 lbf F 2 = F 1 − ± F = 429 . 7 − 256 . 8 = 172 . 9 lbf and f ± = f = . 32 17-16 This solution is the result of a series of ﬁve design tasks involving different belt thick-nesses. The results are to be compared as a matter of perspective. These design tasks are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions....
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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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