44_ch 03 Mechanical Design budynas_SM_ch03

# 44_ch 03 Mechanical Design budynas_SM_ch03 -...

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

Chapter 3 57 3-54 σ t l r τ max = ( σ t σ r ) / 2 at r = r i where σ l is intermediate in value. From Prob. 4-50 τ max = 1 2 ( σ t , max σ r , max ) τ max = p i 2 ± r 2 o + r 2 i r 2 o r 2 i + 1 ² Now solve for p i using r o = 75 mm, r i = 69 mm, and τ max = 25 MPa. This gives p i = 3 . 84 MPa Ans. 3-55 Given r o = 5in, r i = 4 . 625 in and referring to the solution of Prob. 3-54, τ max = 350 2 ³ (5) 2 + (4 . 625) 2 (5) 2 (4 . 625) 2 + 1 ´ = 2 424 psi Ans. 3-56 From Table A-20, S y = 57 kpsi; also, r o = 0 . 875 in and r i = 0 . 625 in From Prob. 3-52 σ t ,max =− 2 p o r 2 o r 2 o r 2 i Rearranging p o = ( r 2 o r 2 i ) (0 . 8 S y ) 2 r 2 o Solving, gives p o = 11 200 psi Ans. 3-57
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