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3_ch 16 Mechanical Design budynas_SM_ch16

3_ch 16 Mechanical Design budynas_SM_ch16 - a = 30(610(10...

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398 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 16-3 Given: θ 1 = 0°, θ 2 = 120°, θ a = 90°, sin θ a = 1, a = R = 90 mm, f = 0 . 30, F = 1000 N = 1 kN, r = 280 / 2 = 140 mm, counter-clockwise rotation. LH shoe : M f = f p a br sin θ a r (1 cos θ 2 ) a 2 sin 2 θ 2 = 0 . 30 p a (0 . 030)(0 . 140) 1 0 . 140(1 cos 120 ) 0 . 090 2 sin 2 120° = 0 . 000 222 p a N · m M N = p a bra sin θ a θ 2 2 1 4 sin 2 θ 2 = p a (0 . 030)(0 . 140)(0 . 090) 1 120° 2 π 180 1 4 sin 2(120°) = 4 . 777(10 4 ) p a N · m c = 2 r cos 180 θ 2 2 = 2(0 . 090) cos 30 = 0 . 155 88 m F = 1 = p a 4 . 777(10 4 ) 2 . 22(10 4 ) 0 . 155 88 = 1 . 64(10 3 ) p a p a = 1 / 1 . 64(10 3 ) = 610 kPa T L = f p a br 2 (cos θ 1 cos θ 2 ) sin θ a = 0 . 30(610)(10
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Unformatted text preview: a = . 30(610)(10 3 )(0 . 030)(0 . 140 2 ) 1 [1 − ( − . 5)] = 161 . 4 N · m Ans. RH shoe : M f = 2 . 22(10 − 4 ) p a N · m M N = 4 . 77(10 − 4 ) p a N · m c = . 155 88 m F = 1 = p a ± 4 . 77(10 − 4 ) + 2 . 22(10 − 4 ) . 155 88 ² = 4 . 49(10 − 3 ) p a p a = 1 4 . 49(10 − 3 ) = 222 . 8 kPa Ans. T R = (222 . 8 / 610)(161 . 4) = 59 . 0 N · m Ans ....
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