7_ch 16 Mechanical Design budynas_SM_ch16

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

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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 p a 150 = 355 . 3 p a , M f = 12 800 p a 150 = 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 . 8 in 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°)
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Unformatted text preview: = 7940 lbf Â· in T total = 15 420 + 7940 = 23 400 lbf Â· in Ans. 16-8 M f = 2 Â± Î¸ 2 ( f dN )( a Â± cos Î¸ âˆ’ r ) where dN = pbr d Î¸ = 2 f pbr Â± Î¸ 2 ( a Â± cos Î¸ âˆ’ r ) d Î¸ = From which a Â± Â± Î¸ 2 cos Î¸ d Î¸ = r Â± Î¸ 2 d Î¸ a Â± = r Î¸ 2 sin Î¸ 2 = r (60Â°)( Ï€/ 180) sin 60Â° = 1 . 209 r Eq. (16-15) a = 4 r sin 60Â° 2(60)( Ï€/ 180) + sin[2(60)] = 1 . 170 r 16" 14 Â± F L Â² 1446 lbf F act Â² 361 lbf F R Â² 1491 lbf 4"...
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