12_ch 17 Mechanical Design budynas_SM_ch17

# 12_ch 17 Mechanical Design budynas_SM_ch17 - d = 1 . 1 Belt...

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Chapter 17 431 Decision #1 : b = 4 . 5in F 1 = ( F 1 ) a = ab = 16 . 5(4 . 5) = 74 . 25 lbf F 2 = F 1 ± F = 74 . 25 45 . 38 = 28 . 87 lbf F i = F 1 + F 2 2 = 74 . 25 + 28 . 87 2 = 51 . 56 lbf Existing friction f ± = 1 θ d ln ± F 1 F 2 ² = 1 3 . 010 ln ± 74 . 25 28 . 87 ² = 0 . 314 H t = ( ± F ) V 33 000 = 45 . 38(916 . 3) 33 000 = 1 . 26 hp n fs = H t H nom K s = 1 . 26 1(1 . 2) = 1 . 05 This is a non-trivial point. The methodology preserved the factor of safety corresponding to n d = 1 . 1 even as we rounded b min up to b . Decision #2 was taken care of with the adjustment of the center-to-center distance to accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this. Remem- ber to subsequently recalculate θ d and θ D . 17-15 Decision set: A priori decisions • Function: H nom = 5hp, N = 1125 rev/min, VR = 3, C ˙= 20 in, K s = 1 . 25, N p = 10 6 belt passes • Design factor: n
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Unformatted text preview: d = 1 . 1 Belt material: BeCu, S y = 170 000 psi, E = 17(10 6 ) psi, = . 220 Belt geometry: d = 3 in, D = 9 in Belt thickness: t = . 003 in Design decisions Belt loop periphery Belt width b Preliminaries : H d = H nom K s n d = 5(1 . 25)(1 . 1) = 6 . 875 hp T = 63 025(6 . 875) 1125 = 385 . 2 lbf in Decision #1 : Choose a 60-in belt loop with a center-to-center distance of 20.3 in. d = 2 sin 1 9 3 2(20 . 3) = 2 . 845 rad D = + 2 sin 1 9 3 2(20 . 3) = 3 . 438 rad...
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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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