20_ch 13 Mechanical Design budynas_SM_ch13

20_ch 13 Mechanical Design budynas_SM_ch13 - t = 22 . 8 d 2...

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

View Full Document Right Arrow Icon
352 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-37 Gear 3 : P t = P n cos ψ = 7 cos 30° = 6 . 062 teeth/in tan φ t = tan 20° cos 30° = 0 . 4203, φ t = 22 . d 3 = 54 6 . 062 = 8 . 908 in W t = 500 lbf W a = 500 tan 30° = 288 . 7 lbf W r = 500 tan 22 . = 210 . 2 lbf W 3 = 210 . 2 i + 288 . 7 j 500 k lbf Ans . Gear 4 : d 4 = 14 6 . 062 = 2 . 309 in W t = 500 8 . 908 2 . 309 = 1929 lbf W a = 1929 tan 30° = 1114 lbf W r = 1929 tan 22 . = 811 lbf W 4 =− 811 i + 1114 j 1929 k lbf Ans . 13-38 P t = 6 cos 30° = 5 . 196 teeth/in d 3 = 42 5 . 196 = 8 . 083 in
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t = 22 . 8 d 2 = 16 5 . 196 = 3 . 079 in T 2 = 63 025(25) 1720 = 916 lbf in W t = T r = 916 3 . 079 / 2 = 595 lbf W a = 595 tan 30 = 344 lbf W r = 595 tan 22 . 8 = 250 lbf W = 344 i + 250 j + 595 k lbf R DC = 6 i , R DG = 3 i 4 . 04 j T 3 C A B D T 2 y 3 2 x z y x W t W r W a W t W r W a r 4 r 3...
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