38_ch 03 Mechanical Design budynas_SM_ch03

38_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 51 ± ² M D ³ z = 7 C x 4 . 3(92 . 8) 3 . 9(362 . 8) = 0 C x = 259 . 1 lbf ± ² M C ³ z =− 7 D x 2 . 7(92 . 8) + 3 . 9(362 . 8) = 0 D x = 166 . 3 lbf ± ² M D ³ x C z = 4 . 3 7 808 = 496 . 3 lbf ± ² M C ³ x D z = 2 . 7 7 808 = 311 . 7 lbf Torque : T = 808(3 . 9) = 3151 lbf · in Bending Q : M = ´ 699 . 6 2 + 1340 2 = 1512 lbf · in Torque: τ = 16 T π d 3 = 16(3151) π (1 . 25 3 ) = 8217 psi Bending: σ b 32(1512) π . 25 3 ) 7885 psi Axial: σ a F A 362 . 8 ( π/ 4)(1 . 25 2 ) 296 psi | σ max | = 7885 + 296 = 8181 psi τ max = µ 8181 2 · 2 + 8217 2 = 9179 psi Ans. M y 496.3 lbf
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