11_ch 03 Mechanical Design budynas_SM_ch03

11_ch 03 Mechanical - budynas_SM_ch03/28/2006 21:21 FIRST PAGES Page 24 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical

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Unformatted text preview: budynas_SM_ch03.qxd 24 11/28/2006 21:21 FIRST PAGES Page 24 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design φp = 26.22 69.7 7 1 90 + tan−1 = 69.7◦ ccw 2 6 7.78 x φs = 69.7◦ − 45◦ = 24.7◦ ccw τ1 = R = 9.22, 17 9.22 17 24.7 x (d) x 1 cw C= (9, 8cw) 2 s 2 D 2 1 R (19, 8ccw) 2 19 − 9 =5 2 R = 52 + 82 = 9.434 CD = p C ccw σ1 = 14 + 9.43 = 23.43 σ2 = 14 − 9.43 = 4.57 y φp = 4.57 5 1 90 + tan−1 = 61.0◦ cw 2 8 x 61 23.43 τ1 = R = 9.434, φs = 61◦ − 45◦ = 16◦ cw 14 x 16 14 9.434 9 + 19 = 14 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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