31_ch 17 Mechanical Design budynas_SM_ch17

# 31_ch 17 Mechanical Design budynas_SM_ch17 - = 2 ft/s 2(a...

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450 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design This exceeds n d = 2 Decision #1 : d = 1 / 2in Decision #2 : m = 2 ropes supporting load. Rope should be inspected weekly for any signs of fatigue (broken outer wires). Comment :Table 17-25 gives n for freight elevators in terms of velocity. F u = ( S u ) nom A nom = 106 000 ± π d 2 4 ² = 83 252 d 2 lbf, each wire n = F u F t = 83 452(0 . 5) 2 (5620 / 2) + 162(0 . 5) 2 = 7 . 32 By comparison, interpolation for 120 ft/min gives 7.08-close. The category of construc- tion hoists is not addressed in Table 17-25. We should investigate this before proceeding further. 17-31 2000 ft lift, 72 in drum, 6 × 19 MS rope. Cage and load 8000 lbf, acceleration
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Unformatted text preview: = 2 ft/s 2 . (a) Table 17-24: ( S u ) nom = 106 kpsi; S u = 240 kpsi (p. 1093, metal area); Fig. 17-22: ( p / S u ) 10 6 = . 0014 F f = . 0014(240)(72) d 2 = 12 . 1 d kip Table 17-24: w l = 1 . 6 d 2 2000(10 − 3 ) = 3 . 2 d 2 kip Eq. (17-46): F t = ( W + w l ) ± 1 + a g ² = (8 + 3 . 2 d 2 ) ± 1 + 2 32 . 2 ² = 8 . 5 + 3 . 4 d 2 kip Note that bending is not included. n = F f F t = 12 . 1 d 8 . 5 + 3 . 4 d 2 ← maximum n Ans. d , in n 0.500 0.650 1.000 1.020 1.500 1.124 1.625 1.125 1.750 1.120 2.000 1.095...
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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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