13_ch 03 Mechanical Design budynas_SM_ch03

# 13_ch 03 Mechanical Design budynas_SM_ch03 -...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: budynas_SM_ch03.qxd 26 11/28/2006 21:22 FIRST PAGES Page 26 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design φs = 45◦ − 27.75◦ = 17.25◦ cw τ1 = R = 9.71, 0.5 9.71 x 17.25 0.5 (c) cw C= 1 x ( 8, 6cw) 2 2 D 2 CD = s p C −8 + 7 = −0.5 2 8+7 = 7.5 2 R= 1 7.52 + 62 = 9.60 σ1 = 9.60 − 0.5 = 9.10 R (7, 6ccw) σ2 = −0.5 − 9.6 = −10.1 y 2 ccw 7.5 1 90 + tan−1 = 70.67◦ cw 2 6 φp = 10.1 x 70.67 9.1 φs = 70.67◦ − 45◦ = 25.67◦ cw τ1 = R = 9.60, 0.5 x 25.67 0.5 9.60 (d) cw C= 2 C R CD = 9+6 = 7.5 2 x s (9, 3cw) 2 2 9−6 = 1.5 2 1 ( 6, 3ccw) p D 1 R= 7.52 + 32 = 8.078 σ1 = 1.5 + 8.078 = 9.58 y σ2 = 1.5 − 8.078 = −6.58 ccw 2 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online