9_ch 06 Mechanical Design budynas_SM_ch06

# 9_ch 06 Mechanical Design budynas_SM_ch06 -...

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

Chapter 6 155 6-15 σ ± max = σ ± a = 3(207) = 358 . 5MPa, σ ± m = 0 Yield: 358 . 5 = 413 n y n y = 1 . 15 Ans . (a) Modiﬁed Goodman, Table 6-6 n f = 1 (358 . 5 / 276) = 0 . 77 Ans . (b) Gerber criterion of Table 6-7 does not work; therefore use Eq. (6-47). n f σ a S e = 1 n f = S e σ a = 276 358 . 5 = 0 . 77 Ans . (c) ASME-Elliptic, Table 6-8 n f = ± ² 1 358 . 5 / 276 ³ 2 = 0 . 77 Ans . Let f = 0 . 9 to assess the cycles to failure by fatigue Eq. (6-14): a = [0 . 9(551)] 2 276 = 891 . 0MPa Eq. (6-15): b =− 1 3 log 0 . 9(551) 276 =− 0 . 084 828 Eq. (6-16): N = ² 358 . 5 891 . 0 ³ 1 / 0 . 084 828 = 45 800 cycles Ans . 6-16 σ ± max = [103 2 + 3(103) 2 ] 1 / 2 = 206 MPa n y = S y σ ± max = 413 206 = 2 . 00 Ans . σ ± a = 3(103)
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