7_ch 16 Mechanical Design budynas_SM_ch16

# 7_ch 16 Mechanical Design budynas_SM_ch16 - = 7940 lbf ·...

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402 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (16-3): M N = 150(2)(10)(12 . 37) sin 90° ± 136° sin 2 θ d θ = 53 300 lbf · in LH shoe : c L = 12 + 12 + 4 = 28 in Now note that M f is cw and M N is ccw. Thus, F L = 53 300 12 800 28 = 1446 lbf Eq. (16-6): T L = 0 . 30(150)(2)(10) 2 (cos 6° cos 136°) sin 90° = 15 420 lbf · in RH shoe : M N = 53 300 ² p a 150 ³ = 355 . 3 p a , M f = 12 800 ² p a 150 ³ = 85 . 3 p a On this shoe, both M N and M f are ccw. Also c R = (24 2 tan 14°) cos 14° = 22 . 8in F act = F L sin 14° = 361 lbf Ans . F R = F L / cos 14° = 1491 lbf Thus 1491 = 355 . 3 + 85 . 3 22 . 8 p a p a = 77 . 2 psi Then T R = 0 . 30(77 . 2)(2)(10) 2 (cos 6° cos 136°) sin 90°
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Unformatted text preview: = 7940 lbf · in T total = 15 420 + 7940 = 23 400 lbf · in Ans. 16-8 M f = 2 ± θ 2 ( f dN )( a ± cos θ − r ) where dN = pbr d θ = 2 f pbr ± θ 2 ( a ± cos θ − r ) d θ = From which a ± ± θ 2 cos θ d θ = r ± θ 2 d θ a ± = r θ 2 sin θ 2 = r (60°)( π/ 180) sin 60° = 1 . 209 r Eq. (16-15) a = 4 r sin 60° 2(60)( π/ 180) + sin[2(60)] = 1 . 170 r 16" 14 ± F L ² 1446 lbf F act ² 361 lbf F R ² 1491 lbf 4"...
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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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