23_ch 17 Mechanical Design budynas_SM_ch17

# 23_ch 17 Mechanical Design budynas_SM_ch17 -...

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

442 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Equating and solving for n 1 gives n 1 = ± 0 . 25(10 6 ) K r N 0 . 42 1 p (2 . 2 0 . 07 p ) ² 1 / 2 . 4 Ans . (b) For a No. 60 chain, p = 6 / 8 = 0 . 75 in, N 1 = 17, K r = 17 n 1 = ³ 0 . 25(10 6 )(17)(17) 0 . 42 0 . 75 [2 . 2 0 . 07(0 . 75)] ´ 1 / 2 . 4 = 1227 rev/min Ans . Table 17-20 conﬁrms that this point occurs at 1200 ± 200 rev/min. (c) Life predictions using Eq. (17-40) are possible at speeds greater than 1227 rev/min. Ans. 17-25 Given: a double strand No. 60 roller chain with p = 0 . 75 in, N 1 = 13 teeth at 300 rev/min, N 2 = 52 teeth . (a) Table 17-20: H tab = 6 . 20 hp Table 17-22: K 1 = 0 . 75 Table 17-23: K 2 = 1 . 7 Use K s = 1 Eq. (17-37): H a = K 1 K 2 H tab = 0 .
This is the end of the preview. Sign up to access the rest of the 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