9_ch 16 Mechanical Design budynas_SM_ch16

# 9_ch 16 Mechanical Design budynas_SM_ch16 - sin 2(45° = 2...

This preview shows page 1. Sign up to view the full content.

404 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) The direction of brake pulley rotation affects the sense of S y , which has no effect on the brake shoe lever moment and hence, no effect on S x or the brake torque. The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression (column loading). The right lever is de- signed and used as a left lever, producing interchangeable levers (identical levers). But do not infer from these identical loadings. 16-10 r = 13 . 5 / 2 = 6 . 75 in, b = 7 . 5in , θ 2 = 45° From Table 16-3 for a rigid, molded nonasbestos use a conservative estimate of p a = 100 psi, f = 0 . 31 . In Eq. (16-16): 2 θ 2 + sin 2 θ 2 = 2( π/
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + sin 2(45°) = 2 . 571 From Prob. 16-9 solution, N = S x = 4 . 174 P = p a br 2 (2 . 571) = 1 . 285 p a br P = 1 . 285 4 . 174 (100)(7 . 5)(6 . 75) = 1560 lbf Ans. Applying Eq. (16-18) for two shoes, T = 2 af N = 2(7 . 426)(0 . 31)(4 . 174)(1560) = 29 980 lbf · in Ans. 16-11 From Eq. (16-22), P 1 = p a bD 2 = 90(4)(14) 2 = 2520 lbf Ans . f φ = . 25( π )(270° / 180°) = 1 . 178 Eq. (16-19): P 2 = P 1 exp( − f φ ) = 2520 exp( − 1 . 178) = 776 lbf Ans. T = ( P 1 − P 2 ) D 2 = (2520 − 776)14 2 = 12 200 lbf · in Ans . Ans. 1.252 P 2.049 P 4.174 P 2.68 P Right shoe lever 2.125 P 1.428 P 2.049 P 4.174 P 1.252 P 1.68 P Left shoe lever 2.125 P 0.428 P...
View Full Document

Ask a homework question - tutors are online