19_ch 13 Mechanical Design budynas_SM_ch13

19_ch 13 Mechanical Design budynas_SM_ch13 -...

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

View Full Document Right Arrow Icon
Chapter 13 351 13-35 Sketch gear 2 pictorially. P t = P n cos ψ = 4 cos 30° = 3 . 464 teeth/in φ t = tan 1 tan φ n cos ψ = tan 1 tan 20° cos 30° = 22 . 80° Sketch gear 3 pictorially, d P = 18 3 . 464 = 5 . 196 in Pinion (Gear 2) W r = W t tan φ t = 800 tan 22 . 80° = 336 lbf W a = W t tan ψ = 800 tan 30° = 462 lbf W =− 336 i 462 j + 800 k lbf Ans . W = [( 336) 2 + ( 462) 2 + 800 2 ] 1 / 2 = 983 lbf Ans . Gear 3 W = 336 i + 462 j 800 k lbf Ans . W = 983 lbf Ans . d G = 32 3 . 464 = 9 . 238 in T G = W t r = 800(9 . 238) = 7390 lbf · in 13-36 From Prob. 13-35 solution, Notice that the idler shaft reaction contains a couple tending to turn the shaft end-over-
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online