23_ch 13 Mechanical Design budynas_SM_ch13

# 23_ch 13 Mechanical Design budynas_SM_ch13 -...

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

Chapter 13 355 For gears 2 and 3: φ t = tan 1 (tan φ n / cos ψ ) = tan 1 (tan 20° / cos 30 ) = 22 . 8°, For gears 4 and 5: φ t = tan 1 (tan 20° / cos 15°) = 20 . 6°, F t 23 = T 2 / r = 1200 / (2 . 021 / 2) = 1188 lbf F t 54 = 1188 5 . 196 3 . 106 = 1987 lbf F r 23 = F t 23 tan φ t = 1188 tan 22 . = 499 lbf F r 54 = 1986 tan 20 . = 746 lbf F a 23 = F t 23 tan ψ = 1188 tan 30° = 686 lbf F a 54 = 1986 tan 15° = 532 lbf Next, designate the points of action on gears 4 and 3, respectively, as points G and H , as shown. Position vectors are R CG = 1 . 553 j 3 k R CH =− 2 . 598 j 6 . 5 k R CD =− 8 . 5 k Force vectors are F 54 =− 1986 i 748 j + 532 k F 23 =− 1188 i + 500 j 686 k F C = F x C i + F y C j F D = F x D i + F y D j + F z D k Now, a summation of moments about bearing C gives ± M C = R CG × F 54 + R CH × F 23 + R CD × F D = 0 The terms for this equation are found to be R CG × F 54 =− 1412 i + 5961 j + 3086 k R CH × F
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