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4_ch 17 Mechanical Design budynas_SM_ch17

# 4_ch 17 Mechanical Design budynas_SM_ch17 - = 456 1(4775 33...

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Chapter 17 423 17-3 As a design task, the decision set on p. 873 is useful. A priori decisions: • Function: H nom = 60 hp, n = 380 rev/min, C = 192 in, K s = 1 . 1 • Design factor: n d = 1 • Initial tension: Catenary • Belt material: Polyamide A-3, F a = 100 lbf / in, γ = 0 . 042 lbf / in 3 , f = 0 . 8 • Drive geometry: d = D = 48 in • Belt thickness: t = 0 . 13 in Design variable: Belt width of 6 in Use a method of trials. Initially choose b = 6in V = π dn 12 = π (48)(380) 12 = 4775 ft/min w = 12 γ bt = 12(0 . 042)(6)(0 . 13) = 0 . 393 lbf/ft F c = w V 2 g = 0 . 393(4775 / 60) 2 32 . 174 = 77 . 4 lbf T = 63 025 H nom K s n d n = 63 025(60)(1 . 1)(1) 380 = 10 946 lbf · in ± F = 2 T d = 2(10 946) 48 = 456 . 1 lbf F 1 = ( F 1 ) a = bF a C p C v = 6(100)(1)(1) = 600 lbf F 2 = F 1 ± F = 600 456 . 1 = 143 . 9 lbf Transmitted power H H = ± F ( V ) 33 000
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Unformatted text preview: = 456 . 1(4775) 33 000 = 66 hp F i = F 1 + F 2 2 − F c = 600 + 143 . 9 2 − 77 . 4 = 294 . 6 lbf f ± = 1 θ d ln ± F 1 − F c F 2 − F c ² = 1 π ln ± 600 − 77 . 4 143 . 9 − 77 . 4 ² = . 656 Eq. (17-2): L = [4(192) 2 − (48 − 48) 2 ] 1 / 2 + . 5[48( π ) + 48( π )] = 534 . 8 in Friction is not fully developed, so b min is just a little smaller than 6 in (5.7 in). Not having a ﬁgure of merit, we choose the most narrow belt available (6 in). We can improve the 48" 192"...
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