5_ch 06 Mechanical Design budynas_SM_ch06

# 5_ch 06 Mechanical Design budynas_SM_ch06 -...

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

Chapter 6 151 σ max = S y n 6(800 × 10 3 N · mm) b 3 = 310 N/mm 2 1 . 5 b = 28 . 5mm Eq. (6-25): d e = 0 . 808 b Eq. (6-20): k b = ± 0 . 808 b 7 . 62 ² 0 . 107 = 1 . 2714 b 0 . 107 k b = 0 . 888 The remaining Marin factors are k a = 57 . 7(570) 0 . 718 = 0 . 606 k c = k d = k e = k f = 1 Eq. (6-18): S e = 0 . 606(0 . 888)(285) = 153 . 4MPa Eq. (6-14): a = [0 . 9(570)] 2 153 . 4 = 1715 . 6 Eq. (6-15): b =− 1 3 log 0 . 9(570) 153 . 4 =− 0 . 174 76 Eq. (6-13): S f = aN b = 1715 . 6[(10 4 ) 0 . 174 76 ] = 343 . 1MPa n = S f σ a or σ a = S f n 6(800 × 10 3 ) b 3 = 343 . 1 1 . 5 b = 27 . 6mm Check values for k b , S e , etc. k b = 1 . 2714(27 . 6) 0 . 107 = 0 . 891 S e = 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