20_ch 08 Mechanical Design budynas_SM_ch08

# 20_ch 08 Mechanical Design budynas_SM_ch08 - Ans. (b)...

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

Chapter 8 223 Since no stress concentration exists, use a load line slope of 1. From Table 7-10 for Gerber S a = 93 . 7 2 2(21 . 5) 1 + ± 1 + ² 2(21 . 5) 93 . 7 ³ 2 = 20 . 47 kpsi Note the mere 5 percent degrading of S e in S a n f = S a σ a = 20 . 47(10 3 ) 10 . 81 P = 1894 P Thread: Die cut. Table 8-17 gives 18.6 kpsi for rolled threads. Use Table 8-16 to ﬁnd S e for die cut threads S e = 18 . 6(3 . 0 / 3 . 8) = 14 . 7 kpsi Table 8-2: A t = 0 . 663 in 2 σ = P / A t = P / 0 . 663 = 1 . 51 P σ a = σ m = σ/ 2 = 1 . 51 P / 2 = 0 . 755 P From Table 7-10, Gerber S a = 120 2 2(14 . 7) 1 + ± 1 + ² 2(14 . 7) 120 ³ 2 = 14 . 5 kpsi n f = S a σ a = 14 500 0 . 755 P = 19 200 P Comparing 1894 / P with 19 200 / P , we conclude that the eye is weaker in fatigue.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ans. (b) Strengthening steps can include heat treatment, cold forming, cross section change (a round is a poor cross section for a curved bar in bending because the bulk of the mate-rial is located where the stress is small). Ans. (c) For n f = 2 P = 1894 2 = 947 lbf, max. load Ans . 8-34 (a) L ≥ 1 . 5 + 2(0 . 134) + 41 64 = 2 . 41 in. Use L = 2 1 2 in Ans . (b) Four frusta: Two washers and two members 1.125" D 1 0.134" 1.280" 0.75"...
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