12_ch 16 Mechanical Design budynas_SM_ch16

12_ch 16 Mechanical Design budynas_SM_ch16 - c 2 When the...

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

View Full Document Right Arrow Icon
Chapter 16 407 If friction is not fully developed P 1 / P 2 exp( f φ ) To help visualize what is going on let’s add a force W parallel to P 1 , at a lever arm of c 3 . Now sum moments about the rocker pivot. ± M = 0 = c 3 W + c 1 P 1 c 2 P 2 From which W = c 2 P 2 c 1 P 1 c 3 The device is self locking for ccw rotation if W is no longer needed, that is, W 0. It follows from the equation above P 1 P 2 c 2 c 1 When friction is fully developed 2 . 566 = 2 . 25 / c 1 c 1 = 2 . 25 2 . 566 = 0 . 877 in When P 1 / P 2 is less than 2.566, friction is not fully developed. Suppose P 1 / P 2 = 2 . 25, then c 1 = 2 . 25 2 . 25 = 1in We don’t want to be at the point of slip, and we need the band to tighten. c 2 P 1 / P 2 c 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: c 2 When the developed friction is very small, P 1 / P 2 → 1 and c 1 → c 2 Ans . (b) Rocker has c 1 = 1 in P 1 P 2 = c 2 c 1 = 2 . 25 1 = 2 . 25 f = ln( P 1 / P 2 ) φ = ln 2 . 25 3 π/ 2 = . 172 Friction is not fully developed, no slip. T = ( P 1 − P 2 ) D 2 = P 2 ² P 1 P 2 − 1 ³ D 2 Solve for P 2 P 2 = 2 T [( P 1 / P 2 ) − 1] D = 2(150)(12) (2 . 25 − 1)(8 . 25) = 349 lbf P 1 = 2 . 25 P 2 = 2 . 25(349) = 785 lbf p = 2 P 1 bD = 2(785) 2 . 125(8 . 25) = 89 . 6 psi Ans ....
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online