6_ch 17 Mechanical Design budynas_SM_ch17

6_ch 17 Mechanical Design budynas_SM_ch17 -...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: budynas_SM_ch17.qxd 12/06/2006 17:29 Page 425 FIRST PAGES 425 Chapter 17 (b) See Prob. 17-4 statement. The final relation can be written bmin = = 1 Fa C p Cv − (12γ t /32.174)( V /60) 2 33 000 Ha exp( f θ ) V [exp( f θ ) − 1] 33 000(20.6)(11.17) 1 2 100(0.7)(1) − {[12(0.042)(0.13)]/32.174}(2749/60) 2749(11.17 − 1) = 4.13 in Ans. This is the minimum belt width since the belt is at the point of slip. The design must round up to an available width. Eq. (17-1): θd = π − 2 sin−1 D−d 2C = π − 2 sin−1 18 − 6 2(96) D−d 2C = π + 2 sin−1 18 − 6 2(96) = 3.016 511 rad θ D = π + 2 sin−1 = 3.266 674 Eq. (17-2): 1 L = [4(96) 2 − (18 − 6) 2 ]1/2 + [18(3.266 674) + 6(3.016 511)] 2 = 230.074 in Ans. (c) F= 2(742.8) 2T = = 247.6 lbf d 6 ( F1 ) a = bFa C p Cv = F1 = 4.13(100)(0.70)(1) = 289.1 lbf F2 = F1 − Fc = 25.6 Fi = F = 289.1 − 247.6 = 41.5 lbf 0.271 0.393 = 17.7 lbf F1 + F2 289.1 + 41.5 − Fc = − 17.7 = 147.6 lbf 2 2 Transmitted belt power H H= nfs = F (V ) 247.6(2749) = = 20.6 hp 33 000 33 000 H Hnom K s = 20.6 = 1.1 15(1.25) ...
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