28_ch 17 Mechanical Design budynas_SM_ch17

# 28_ch 17 Mechanical Design budynas_SM_ch17 - F t =(8 3 072...

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Chapter 17 447 17-29 The objective of the problem is to explore factors of safety in wire rope. We will express strengths as tensions. (a) Monitor steel 2-in 6 × 19 rope, 480 ft long Table 17-2: Minimum diameter of a sheave is 30 d = 30(2) = 60 in, preferably 45(2) = 90 in. The hoist abuses the wire when it is bent around a sheave. Table 17-24 gives the nominal tensile strength as 106 kpsi. The ultimate load is F u = ( S u ) nom A nom = 106 π (2) 2 4 = 333 kip Ans . The tensile loading of the wire is given by Eq. (17-46) F t = W m + w l 1 + a g W = 4(2) = 8 kip, m = 1 Table (17-24): w l = 1 . 60 d 2 l = 1 . 60(2 2 )(480) = 3072 lbf or 3 . 072 kip Therefore,
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Unformatted text preview: F t = (8 + 3 . 072) ³ 1 + 2 32 . 2 ´ = 11 . 76 kip Ans . Eq. (17-48): F b = E r d w A m D and for the 72-in drum F b = 12(10 6 )(2 / 13)(0 . 38)(2 2 )(10 − 3 ) 72 = 39 kip Ans . For use in Eq. (17-44), from Fig. 17-21 ( p / S u ) = . 0014 S u = 240 kpsi, p. 908 F f = . 0014(240)(2)(72) 2 = 24 . 2 kip Ans . (b) Factors of safety Static, no bending : n = F u F t = 333 11 . 76 = 28 . 3 Ans . Static, with bending : Eq. (17-49): n s = F u − F b F t = 333 − 39 11 . 76 = 25 . Ans ....
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