5_ch 16 Mechanical Design budynas_SM_ch16

5_ch 16 Mechanical Design budynas_SM_ch16 - R H = . 658 ....

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400 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) Primary shoes : R x = p a br sin θ a ( C fB ) F x = (10 6 )(0 . 075)(0 . 200) sin 75° [0 . 4514 0 . 24(0 . 528)](10) 3 5 . 70 =− 0 . 658 kN R y = p a br sin θ a ( B + fC ) F y = (10 6 )(0 . 075)(0 . 200) sin 75° [0 . 528 + 0 . 24(0 . 4514)](10) 3 0 = 9 . 88 kN Secondary shoes : R x = p a br sin θ a ( C + fB ) F x = [0 . 619(10) 6 ](0 . 075)(0 . 200) sin 75° [0 . 4514 + 0 . 24(0 . 528)](10) 3 5 . 70 =− 0 . 143 kN R y = p a br sin θ a ( B fC ) F y = [0 . 619(10) 6 ](0 . 075)(0 . 200) sin 75° [0 . 528 0 . 24(0 . 4514)](10) 3 0 = 4 . 03 kN Note from figure that + y for secondary shoe is opposite to + y for primary shoe. Combining horizontal and vertical components,
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Unformatted text preview: R H = . 658 . 143 = . 801 kN R V = 9 . 88 4 . 03 = 5 . 85 kN R = (0 . 801) 2 + (5 . 85) 2 = 5 . 90 kN Ans. 16-5 Preliminaries: 1 = 45 tan 1 (150 / 200) = 8 . 13, 2 = 98 . 13 a = 90, a = [(150) 2 + (200) 2 ] 1 / 2 = 250 mm Eq. (16-8): A = 1 2 ( sin 2 ) 98 . 13 8 . 13 = . 480 Let C = 2 1 sin d = ( cos ) 98 . 13 8 . 13 = 1 . 1314 y y x x R R V R H...
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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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