2_ch 08 Mechanical Design budynas_SM_ch08

2_ch 08 Mechanical Design budynas_SM_ch08 - n = 1720 75 =...

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Chapter 8 205 8-4 Given F = 6kN, l = 5mm, and d m = 22 . 5mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) T R = 6(22 . 5) 2 ± 5 + π (0 . 08)(22 . 5) π (22 . 5) 0 . 08(5) ² + 6(0 . 05)(40) 2 = 10 . 23 + 6 = 16 . 23 N · m Ans . The torque required to lower the load, from Eqs. (8-2) and (8-6) is T L = 6(22 . 5) 2 ± π (0 . 08)22 . 5 5 π (22 . 5) + 0 . 08(5) ² + 6(0 . 05)(40) 2 = 0 . 622 + 6 = 6 . 622 N · m Ans . Since T L is positive, the thread is self-locking. The efficiency is Eq. (8-4): e = 6(5) 2 π (16 . 23) = 0 . 294 Ans . 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom seg- ment of the screws must be in compression. Where as tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. 8-6 Screws rotate at an angular rate of
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Unformatted text preview: n = 1720 75 = 22 . 9 rev/min (a) The lead is 0.5 in, so the linear speed of the press head is V = 22 . 9(0 . 5) = 11 . 5 in/min Ans . (b) F = 2500 lbf/screw d m = 3 . 25 = 2 . 75 in sec = 1 / cos(29 / 2) = 1 . 033 Eq. (8-5): T R = 2500(2 . 75) 2 . 5 + (0 . 05)(2 . 75)(1 . 033) (2 . 75) . 5(0 . 05)(1 . 033) = 377 . 6 lbf in Eq. (8-6): T c = 2500(0 . 06)(5 / 2) = 375 lbf in T total = 377 . 6 + 375 = 753 lbf in/screw T motor = 753(2) 75(0 . 95) = 21 . 1 lbf in H = Tn 63 025 = 21 . 1(1720) 63 025 = . 58 hp Ans ....
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