5_ch 02 Mechanical Design budynas_SM_ch02

# 5_ch 02 Mechanical Design budynas_SM_ch02 - σ f = 106 kpsi...

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

10 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 2-12 Since | ε o |=| ε i | ± ± ± ± ln R + h R + N ± ± ± ± = ± ± ± ± ln R R + N ± ± ± ± = ± ± ± ± ln R + N R ± ± ± ± R + h R + N = R + N R ( R + N ) 2 = R ( R + h ) From which, N 2 + 2 RN Rh = 0 The roots are: N = R ² 1 ± ³ 1 + h R ´ 1 / 2 µ The + sign being signiﬁcant, N = R ² ³ 1 + h R ´ 1 / 2 1 µ Ans. Substitute for N in ε o = ln R + h R + N Gives ε 0 = ln R + h R + R ³ 1 + h R ´ 1 / 2 R = ln ³ 1 + h R ´ 1 / 2 Ans. These constitute a useful pair of equations in cold-forming situations, allowing the surface strains to be found so that cold-working strength enhancement can be estimated. 2-13 From Table A-22 AISI 1212 S y = 28 . 0 kpsi,
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: σ f = 106 kpsi, S ut = 61 . 5 kpsi σ = 110 kpsi, m = 0.24, ε f = . 85 From Eq. (2-12) ε u = m = . 24 Eq. (2-10) A A ² i = 1 1 − W = 1 1 − . 2 = 1 . 25 Eq. (2-13) ε i = ln 1 . 25 = . 2231 ⇒ ε i < ε u Eq. (2-14) S ² y = σ ε m i = 110(0 . 2231) . 24 = 76 . 7 kpsi Ans. Eq. (2-15) S ² u = S u 1 − W = 61 . 5 1 − . 2 = 76 . 9 kpsi Ans. 2-14 For H B = 250, Eq. (2-17) S u = 0.495 (250) = 124 kpsi = 3.41 (250) = 853 MPa 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