1_ch 16 Mechanical Design budynas_SM_ch16

# 1_ch 16 Mechanical Design budynas_SM_ch16 -...

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

Chapter 16 16-1 (a) θ 1 = 0°, θ 2 = 120°, θ a = 90°, sin θ a = 1, a = 5in Eq. (16-2): M f = 0 . 28 p a (1 . 5)(6) 1 ± 120° sin θ (6 5 cos θ ) d θ = 17 . 96 p a lbf · in Eq. (16-3): M N = p a (1 . 5)(6)(5) 1 ± 120° sin 2 θ d θ = 56 . 87 p a lbf · in c = 2(5 cos 30 ) = 8 . 66 in Eq. (16-4): F = 56 . 87 p a 17 . 96 p a 8 . 66 = 4 . 49 p a p a = F / 4 . 49 = 500 / 4 . 49 = 111 . 4 psi for cw rotation Eq. (16-7): 500 = 56 . 87 p a + 17 . 96 p a 8 . 66 p a = 57 . 9 psi for ccw rotation A maximum pressure of 111 . 4 psioccurs on the RH shoe for cw rotation. Ans. (b) RH shoe : Eq. (16-6): T R = 0 . 28(111 . 4)(1 . 5)(6) 2 (cos 0
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