shi20396_ch08 - Chapter 8 8-1 (a) 2.5 mm 25 mm 5 mm 2.5...

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Chapter 8 8-1 (a) Thread depth = 2 . 5mm Ans. Width = 2 . Ans. d m = 25 1 . 25 1 . 25 = 22 . d r = 25 5 = 20 mm l = p = Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. d m = 22 . d r = 20 mm l = p = Ans. 8-2 From Table 8-1, d r = d 1 . 226 869 p d m = d 0 . 649 519 p ¯ d = d 1 . 226 869 p + d 0 . 649 519 p 2 = d 0 . 938 194 p A t = π ¯ d 2 4 = π 4 ( d 0 . 938 194 p ) 2 Ans . 8-3 From Eq. ( c ) of Sec. 8-2, P = F tan λ + f 1 f tan λ T = Pd m 2 = Fd m 2 tan λ + f 1 f tan λ e = T 0 T = Fl / (2 π ) m / 2 1 f tan λ tan λ + f = tan λ 1 f tan λ tan λ + f Ans . Using f = 0 . 08, form a table and plot the ef±ciency curve. λ , deg. e 00 10 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 1 05 0 l , deg. e 5 mm 5 mm 2.5 2.5 2.5 mm 25 mm 5 mm
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212 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 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) ¸ + 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 . 08)22 . 5 5 π . 5) + 0 . 08(5) ¸ + 6(0 . 05)(40) 2 = 0 . 622 + 6 = 6 . 622 N · m Ans . Since T L is positive, the thread is self-locking. The efFciency 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 n = 1720 75 = 22 . 9rev/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 0 . 25 = 2 . 75 in sec α = 1 / cos(29 / 2) = 1 . 033 Eq. (8-5): T R = 2500(2 . 75) 2 µ 0 . 5 + π . 05)(2 . 75)(1 . 033) π (2 . 75) 0 . 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 = 0 . 58 hp Ans .
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Chapter 8 213 8-7 The force F is perpendicular to the paper. L = 3 1 8 1 4 7 32 = 2 . 406 in T = 2 . 406 F M = µ L 7 32 F = µ 2 . 406 7 32 F = 2 . 188 F S y = 41 kpsi σ = S y = 32 M π d 3 = 32(2 . 188) F π (0 . 1875) 3 = 41 000 F = 12 . 13 lbf T = 2 . 406(12 . 13) = 29 . 2 lbf · in Ans . (b) Eq. (8-5), 2 α = 60 , l = 1 / 14 = 0 . 0714 in, f = 0 . 075, sec α = 1 . 155, p = 1 / 14 in d m = 7 16 0 . 649 519 µ 1 14 = 0 . 3911 in T R = F clamp . 3911) 2 µ Num Den Num = 0 . 0714 + π . 075)(0 . 3911)(1 . 155) Den = π . 3911) 0 . 075(0 . 0714)(1 . 155) T = 0 . 028 45 F clamp F clamp = T 0 . 028 45 = 29 . 2 0 . 028 45 = 1030 lbf Ans . (c) The column has one end Fxed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 in, S y = 41 kpsi, E = 30(10 6 ) psi, L = 4 . 1875 in, k = D / 4 = 0 . 097 75 in, L / k = 42 . 8. ±or this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, F clamp = P cr = 4663 lbf . (d) This is a subject for class discussion. 8-8 T = 6(2 . 75) = 16 . 5 lbf · in d m = 5 8 1 12 = 0 . 5417 in l = 1 6 = 0 . 1667 in, α = 29 2 = 14 . 5 , sec 14 . 5 = 1 . 033 1 4 " 3 16 D . " 7 16 " 2.406" 3"
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214 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (8-5): T = 0 . 5417( F / 2) · 0 . 1667 + π (0 . 15)(0 . 5417)(1 . 033) π . 5417) 0 . 15(0 . 1667)(1 . 033) ¸ = 0 . 0696 F Eq. (8-6): T c = 0 . 15(7 / 16)( F / 2) = 0 . 032 81 F T total = . 0696 + 0 . 0328) F = 0 . 1024 F F = 16 . 5 0 . 1024 = 161 lbf Ans .
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This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch08 - Chapter 8 8-1 (a) 2.5 mm 25 mm 5 mm 2.5...

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