32_ch 05 Mechanical Design budynas_SM_ch05

32_ch 05 Mechanical Design budynas_SM_ch05 - 9) 2 + [9 (...

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

View Full Document Right Arrow Icon
146 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-46 σ t = p d 2 t = 6000 N (1, 0 . 083 33)(0 . 75) 2(0 . 125) = 18 N (1, 0 . 083 33) kpsi σ l = p d 4 t = 6000 N (1, 0 . 083 33)(0 . 75) 4(0 . 125) = 9 N (1, 0 . 083 33) kpsi σ r =− p =− 6000 N (1, 0 . 083 33) kpsi These three stresses are principal stresses whose variability is due to the loading. From Eq. (5-12), we find the von Mises stress to be σ ± = ± (18
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 9) 2 + [9 ( 6)] 2 + ( 6 18) 2 2 1 / 2 = 21 . 0 kpsi = C p = . 083 33(21 . 0) = 1 . 75 kpsi z = S ( 2 S + 2 ) 1 / 2 = 50 21 . (4 . 1 2 + 1 . 75 2 ) 1 / 2 = 6 . 5 The reliability is very high R = 1 (6 . 5) = 1 4 . 02(10 11 ) . = 1 Ans....
View Full Document

Ask a homework question - tutors are online