16_ch 03 Mechanical Design budynas_SM_ch03

16_ch 03 Mechanical Design budynas_SM_ch03 -...

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Unformatted text preview: budynas_SM_ch03.qxd 11/28/2006 21:22 FIRST PAGES Page 29 29 Chapter 3 (d) cw 1 x 2 ( 12, 12cw) 12 + 22 = 17 2 R = 172 + 122 = 20.81 s 2 2 D CD = p C 1 R (22, 12ccw) y ccw −12 + 22 =5 2 C= σ1 = 5 + 20.81 = 25.81 σ2 = 5 − 20.81 = −15.81 2 17 1 90 + tan−1 = 72.39◦ cw 2 12 φp = 15.81 x 72.39 25.81 φs = 72.39◦ − 45◦ = 27.39◦ cw τ1 = R = 20.81, 5 x 27.39 5 20.81 3-11 (a) 14 2 1/3 1/2 2/3 5 2 y 4 3 7 y x 10 0 2 (b) x 0 + 10 =5 2 10 − 0 CD = =5 2 1/3 C= y 1/2 (0, 4cw) R 2/3 C 3 D 2 1 (10, 4ccw) x τ1/3 = R = 6.40, 1 τ1/2 = R= 52 + 42 = 6.40 σ1 = 5 + 6.40 = 11.40 σ2 = 0, σ3 = 5 − 6.40 = −1.40 11.40 = 5.70, 2 τ2/3 = 1.40 = 0.70 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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