17_ch 03 Mechanical Design budynas_SM_ch03

# 17_ch 03 Mechanical Design budynas_SM_ch03 -...

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Unformatted text preview: budynas_SM_ch03.qxd 30 11/28/2006 21:22 FIRST PAGES Page 30 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) ( 2, 4cw) x 2 circles −2 − 8 = −5 2 C= Point is a circle 3 D 8−2 =3 2 CD = C 12 32 + 42 = 5 R= ( 8, 4ccw) y σ1 = −5 + 5 = 0, σ2 = 0 σ3 = −5 − 5 = −10 10 = 5, 2 τ1/3 = τ1/2 = 0, (d) τ2/3 = 5 10 − 30 = −10 2 10 + 30 CD = = 20 2 1/3 C= 2/3 y ( 30, 10cw) 1/2 R C D 2 3 1 202 + 102 = 22.36 R= (10, 10ccw) x σ1 = −10 + 22.36 = 12.36 σ2 = 0 σ3 = −10 − 22.36 = −32.36 τ1/3 = 22.36, τ1/2 = 12.36 = 6.18, 2 τ2/3 = 32.36 = 16.18 2 3-12 (a) 1/3 C= 2/3 x ( 80, 20cw) 1/2 CD = C 3 D 2 R 1 −80 − 30 = −55 2 80 − 30 = 25 2 252 + 202 = 32.02 R= ( 30, 20ccw) y σ1 = 0 σ2 = −55 + 32.02 = −22.98 = −23.0 σ3 = −55 − 32.0 = −87.0 τ1/2 = 23 = 11.5, 2 τ2/3 = 32.0, τ1/3 = 87 = 43.5 2 ...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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