48_ch 03 Mechanical Design budynas_SM_ch03

48_ch 03 Mechanical Design budynas_SM_ch03 -...

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

View Full Document Right Arrow Icon
Chapter 3 61 Inner member, from Prob. 3-52 ( σ t ) i =− p o r 2 o r 2 o r 2 i ± 1 + r 2 i r 2 o ² =− 2 . 25(10 4 )(0 . 5 2 ) 0 . 5 2 0 ³ 1 + 0 0 . 5 2 ´ =− 22 500 psi Ans. 3-69 ν i = 0 . 292, E i = 30(10 6 ) psi, ν o = 0 . 211, E o = 14 . 5(10 6 ) psi δ = 1 2 (1 . 002 1 . 000) = 0 . 001 in, r i = 0, R = 0 . 5, r o = 1 Eq. (3-56) 0 . 001 = µ 0 . 5 14 . 5 ( 10 6 ) ³ 1 2 + 0 . 5 2 1 2 0 . 5 2 + 0 . 211 ´ + 0 . 5 30 ( 10 6 ) ³ 0 . 5 2 + 0 0 . 5 2 0 0 . 292 ´¶ p p = 13 064 psi Ans. Eq. (3-50) for outer member at r i = 0 . 5in ( σ t ) o = 0 . 5 2 (13 064) 1 2 0 . 5 2 ³ 1 + 1 2 0 . 5 2 ´ = 21 770 psi Ans. Inner member, from Prob. 3-52 t ) i =− 13 064 ( 0 . 5 2 ) 0 . 5 2 0 ³ 1 + 0 0 . 5 2 ´ =− 13 064 psi Ans. 3-70 δ max = 1 2 (1 . 003 1 . 000) = 0 . 0015 in r i = 0, R = 0 . 5 in, r o = 1in δ min = 1 2 ( 1 . 002 1 . 001 ) = 0 . 0005 in Eq. (3-57) p max = 30(10 6 )(0 . 0015) 0 . 5 3 µ (1 2 0 . 5 2 )(0 . 5 2 0) 2(1 2 0) = 33 750 psi
Background image of page 1
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