Chapter_8_Solutions

Chapter_8_Solutions - Chapter 8 Note to the Instructor for...

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Chapter 8 Note to the Instructor for Probs. 8-41 to 8-44 . These problems, as well as many others in this chapter are best implemented using a spreadsheet. 8-1 (a) Thread depth= 2 . 5 mm Ans. Width = 2 . 5 mm Ans. d m = 25 - 1 . 25 - 1 . 25 = 22 . 5 mm d r = 25 - 5 = 20 mm l = p = 5 mm Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. d m = 22 . 5 mm d r = 20 mm l = p = 5 mm Ans. ______________________________________________________________________________ 8-2 From Table 8-1, 1.226 869 0.649 519 1.226 869 0.649 519 0.938 194 2 r m dd p p dp    p 2 2 ( 0.938 194 ) . 44 t d Ad p   A n s ______________________________________________________________________________ 8-3 From Eq. ( c ) of Sec. 8-2, tan 1t a n tan 22 1 t a n R Rm m R f PF f Pd Fd f T f 0 / (2 ) 1 tan 1 tan tan . /2 tan tan TF l f f eA d f f n s   Chap. 8 Solutions - Rev. A, Page 1/69
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Using f = 0 . 08, form a table and plot the efficiency curve. , deg. e 0 0 0 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 ______________________________________________________________________________ 8-4 Given F = 5 kN, l = 5 mm, and d m = d p /2 = 25 5/2 = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6)    5 22.5 5 0.09 22.5 5 0.06 45 15.85 N m . 2 22.5 0.09 5 2 R TA     n s The torque required to lower the load, from Eqs. (8-2) and (8-6) is   5 22.5 0.09 22.5 5 5 0.06 45 7.83 N m . 2 22.5 0.09 5 2 L n s Since T L is positive, the thread is self-locking. From Eq.(8-4) the efficiency is 55 0.251 . 2 15.85 e Ans  ______________________________________________________________________________ 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screws must be in compression. Whereas, 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 1720 28.67 rev/min 60 n Chap. 8 Solutions - Rev. A, Page 2/69
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(a) The lead is 0.25 in, so the linear speed of the press head is V = 28.67(0 . 25) = 7.17 in/min Ans. (b) F = 2500 lbf/screw o 2 0.25 / 2 1.875 in sec 1 / cos(29 / 2) 1.033 m d   Eq. (8-5): 2500(1.875) 0.25 (0.05)(1.875)(1.033) 221.0 lbf · in 2 (1.875) 0.05(0.25)(1.033) R T    Eq. (8-6): 2500(0.08)(3.5 / 2) 350 lbf · in 350 221.0 571 lbf · in/screw 571(2) 20.04 lbf · in 60(0.95) 20.04(1720) 0.547 hp . 63 025 63 025 c total motor T T T Tn HA  n s ______________________________________________________________________________ 8-7 Note to the Instructor: The statement for this problem in the first printing of this edition was vague regarding the effective handle length. For the printings to follow the statement “The overall length is 4.25 in.” will be replaced by “ A force will be applied to the handle at a radius of 1 2 3 in from the screw centerline.” We apologize if this has caused any inconvenience. 33 3.5 in 3.5 3.5 3.125 88 41 kpsi 32 32(3.125) 41 000 (0.1875) 8.49 lbf y y L TF M LF F S MF S d F   F n s 3.5(8.49) 29.7 lbf · in . TA ( b ) Eq. (8-5), 2 = 60 , l = 1 / 10 = 0 . 1 in, f = 0 . 15, sec = 1 . 155, p = 0.1 in Chap. 8 Solutions - Rev. A, Page 3/69
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 clamp clamp clamp 3 0.649 519 0.1 0.6850 in 4 (0.6850) 0.1 (0.15)(0.6850)(1.155) 2 (0.6850) 0.15(0.1)(1.155) 0.075 86 29.7 392 lbf . 0.075 86 0.075 86 m R R R d F T TF T FA      n s ( c ) The column has one end fixed and the other end pivoted. Base the decision on the mean diameter column. Input: C = 1.2, D = 0.685 in, A = (0.685 2 )/4 = 0.369 in 2 , S y = 41 kpsi, E = 30(10 6 ) psi, L = 6 in, k = D/ 4 =0 . 171 25 in, L/k = 35.04. From Eq. (4-45),  1/2 1/2 26 2 1 2 1.2 30 10 2 131.7 41 000 y lC E kS  
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_8_Solutions - Chapter 8 Note to the Instructor for...

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