13_ch 03 Mechanical Design budynas_SM_ch03

13_ch 03 Mechanical Design budynas_SM_ch03 -...

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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 ...
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