14_ch 08 Mechanical Design budynas_SM_ch08

# 14_ch 08 Mechanical Design budynas_SM_ch08 - 2 60 ◦ 2 sin...

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Chapter 8 217 For the bolt, L T = 2(3 / 8) + (1 / 4) = 1in . So the bolt is threaded all the way. Since A t = 0 . 0775 in 2 k b = 0 . 0775(30) 0 . 5025 = 4 . 63 Mlbf/in Ans . 8-28 (a) F ± b = RF ± b ,max sin θ Half of the external moment is contributed by the line load in the interval 0 θ π. M 2 = ± π 0 F ± b R 2 sin θ d θ = ± π 0 F ± b , max R 2 sin 2 θ d θ M 2 = π 2 F ± b , max R 2 from which F ± b ,max = M π R 2 F max = ± φ 2 φ 1 F ± b R sin θ d θ = M π R 2 ± φ 2 φ 1 R sin θ d θ = M π R (cos φ 1 cos φ 2 ) Noting φ 1 = 75 , φ 2 = 105 F max = 12 000 π (8 / 2) (cos 75 cos 105 ) = 494 lbf Ans . (b) F max = F ± b , max R ±φ = M π R 2 ( R ) ² 2 π N ³ = 2 M RN F max = 2(12 000) (8 / 2)(12) = 500 lbf Ans . (c) F = F max sin θ M = 2 F max R [(1) sin 2 90 + 2 sin
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Unformatted text preview: 2 60 ◦ + 2 sin 2 30 ◦ + (1) sin 2 (0)] = 6 F max R from which F max = M 6 R = 12 000 6(8 / 2) = 500 lbf Ans . The simple general equation resulted from part ( b ) F max = 2 M RN 8-29 (a) Table 8-11: S p = 600 MPa Eq. (8-30): F i = . 9 A t S p = . 9(245)(600)(10 − 3 ) = 132 . 3 kN Table (8-15): K = . 18 Eq. (8-27) T = . 18(132 . 3)(20) = 476 N · m Ans ....
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