55_ch 03 Mechanical Design budynas_SM_ch03

# 55_ch 03 Mechanical Design budynas_SM_ch03 - π (2)(0 . 000...

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

68 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-83 ν 1 = 0 . 334, E 1 = 10 . 4 Mpsi, l = 2 in, d 1 = 1 in, ν 2 = 0 . 211, E 2 = 14 . 5 Mpsi, d 2 =− 8 in. With b = K c F 1 / 2 , from Eq. (3-73), K c = ± 2 π (2) (1 0 . 334 2 ) / [10 . 4(10 6 )] + (1 0 . 211 2 ) / [14 . 5(10 6 )] 1 0 . 125 ² 1 / 2 = 0 . 000 234 6 Be sure to check σ x for both ν 1 and ν 2 . Shear stress is maximum in the aluminum roller. So, τ max = 0 . 3 p max p max = 4000 0 . 3 = 13 300 psi Since p max = 2 F / ( π bl ) we have p max = 2 F π lK c F 1 / 2 = 2 F 1 / 2 π lK c So, F = ± π lK c p max 2 ² 2 = ±
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: π (2)(0 . 000 234 6)(13 300) 2 ² 2 = 96 . 1 lbf Ans . 3-84 Good class problem 3-85 From Table A-5, ν = . 211 σ x p max = (1 + ν ) − 1 2 = (1 + . 211) − 1 2 = . 711 σ y p max = . 711 σ z p max = 1 These are principal stresses τ max p max = 1 2 ( σ 1 − σ 3 ) = 1 2 (1 − . 711) = . 1445...
View Full 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