11_ch 16 Mechanical Design budynas_SM_ch16

# 11_ch 16 Mechanical Design budynas_SM_ch16 - c 1 < c 2 ....

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406 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (16-14): P 2 = P 1 exp( f φ ) = 1680 exp( 0 . 942) = 655 lbf T = ( P 1 P 2 ) D 2 = (1680 655) 16 2 = 8200 lbf · in Ans. H = Tn 63 025 = 8200(200) 63 025 = 26 . 0hp Ans. P = 3 P 1 10 = 3(1680) 10 = 504 lbf Ans. (b) The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with the drum at center span, the bearing radial load is 1803 / 2 = 901 lbf . (c) Eq. (16-22): p = 2 P bD p | θ = = 2 P 1 3(16) = 2(1680) 3(16) = 70 psi Ans . As it should be p | θ = 270° = 2 P 2 3(16) = 2(655) 3(16) = 27 . 3 psi Ans . 16-15 Given: φ = 270°, b = 2 . 125 in, f = 0 . 20, T = 150 lbf · ft, D = 8 . 25 in, c 2 = 2 . 25 in Notice that the pivoting rocker is not located on the vertical centerline of the drum. (a) To have the band tighten for ccw rotation, it is necessary to have
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Unformatted text preview: c 1 < c 2 . When fric-tion is fully developed, P 1 P 2 = exp( f φ ) = exp[0 . 2(3 π/ 2)] = 2 . 566 Net torque on drum due to brake band: T = T P 1 − T P 2 = 13 440 − 5240 = 8200 lbf · in 1803 lbf 8200 lbf•in 1680 lbf 655 lbf Force of shaft on the drum: 1680 and 655 lbf T P 1 = 1680(8) = 13 440 lbf · in T P 2 = 655(8) = 5240 lbf · in 1680 lbf 1803 lbf 655 lbf 13,440 lbf•in 5240 lbf•in Force of belt on the drum: R = (1680 2 + 655 2 ) 1 / 2 = 1803 lbf 1680 lbf 655 lbf...
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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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