Catrainas Y Cadenas

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

Thread depth = 2.5 mm

Ans.

Width = 2.5 mm Ans. dm = 25 − 1.25 − 1.25 = 22.5 mm dr = 25 − 5 = 20 mm l = p = 5 mm Ans. Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans.

(b)

5 mm 2.5 5 mm

8-2 From Table 8-1, dr = d − 1.226 869 p dm = d − 0.649 519 p ¯ d − 1.226 869 p + d − 0.649 519 p = d− 0.938 194 p d= 2 2 ¯ πd π At = = (d − 0.938 194 p) 2 Ans. 4 4 8-3 From Eq. (c) of Sec. 8-2, P=F T = e= tan λ + f 1 − f tan λ

Pdm Fdm tan λ + f = 2 2 1 − f tan λ Ans.

Fl/(2π) 1 − f tan λ 1 − f tan λ T0 = = tan λ T Fdm /2 tan λ + f tan λ + f

Using f = 0.08, form a table and plot the efficiency curve. λ, deg. 0 10 20 30 40 45
1

e 0 0.678 0.796 0.838 0.8517 0.8519
e

0 , deg.

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205

8-4 Given F = 6 kN, l = 5 mm, and dm = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) TR = 6(0.05)(40) 6(22.5) 5 + π(0.08)(22.5) + 2 π(22.5) − 0.08(5) 2

= 10.23 + 6 = 16.23 N · m Ans. The torque required to lower the load, from Eqs. (8-2) and (8-6) is TL = 6(0.05)(40) 6(22.5) π(0.08)22.5 − 5 + 2 π(22.5) + 0.08(5) 2

= 0.622 + 6 =6.622 N · m Ans. Since TL is positive, the thread is self-locking. The efficiency is Eq. (8-4): e= 6(5) = 0.294 Ans. 2π(16.23)

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. 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 anangular rate of n= 1720 = 22.9 rev/min 75

(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 dm = 3 − 0.25 = 2.75 in sec α = 1/cos(29/2) = 1.033 Eq. (8-5): TR = Eq. (8-6): Tc = 2500(0.06)(5/2) = 375 lbf · in Ttotal = 377.6 + 375 = 753 lbf · in/screw Tmotor = H= 753(2) = 21.1 lbf · in 75(0.95) Tn 21.1(1720) = = 0.58 hp Ans. 63025 63 025 2500(2.75) 2 0.5 + π(0.05)(2.75)(1.033) π(2.75) − 0.5(0.05)(1.033) = 377.6 lbf · in

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

8-7 The force F is perpendicular to the paper.
1" 4 7" 16 3" D. 16 2.406"

3"

1 1 7 − − = 2.406 in 8 4 32 T = 2.406F L =3− M= L− 7 32 F = 2.406 − 7 32 F = 2.188F

S y = 41 kpsi σ = Sy = 32M32(2.188) F = = 41 000 3 πd π(0.1875) 3

F = 12.13 lbf T = 2.406(12.13) = 29.2 lbf · in Ans. (b) Eq. (8-5), 2α = 60◦ , l = 1/14 = 0.0714 in, f = 0.075, sec α = 1.155, p = 1/14 in dm = 7 1 = 0.3911 in − 0.649 519 16 14 Fclamp (0.3911) Num TR = 2 Den

Num = 0.0714 + π(0.075)(0.3911)(1.155) Den = π(0.3911) − 0.075(0.0714)(1.155) T = 0.028 45Fclamp T 29.2 Fclamp = = = 1030 lbf Ans. 0.028 45 0.02845 (c) The column has one end fixed 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(106 ) psi, L = 4.1875 in, k = D/4 = 0.097 75 in, L/k = 42.8. For this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, Fclamp = Pcr = 4663 lbf. (d) This is a subject for class discussion. 8-8 T =6(2.75) = 16.5 lbf · in dm = 5 1 − = 0.5417 in 8 12 1 29◦ l = = 0.1667 in, α = = 14.5◦ , 6 2

sec 14.5◦ = 1.033

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207

Eq. (8-5): Eq. (8-6):

T = 0.5417( F/2)

0.1667 + π(0.15)(0.5417)(1.033) = 0.0696F π(0.5417) − 0.15(0.1667)(1.033)

Tc = 0.15(7/16)( F/2) = 0.032 81F Ttotal = (0.0696 + 0.0328) F = 0.1024F F= 16.5 = 161 lbf Ans. 0.1024

8-9 dm = 40 − 3 = 37 mm, l = 2(6)= 12 mm From Eq. (8-1) and Eq. (8-6) TR = 10(0.15)(60) 10(37) 12 + π(0.10)(37) + 2 π(37) − 0.10(12) 2

= 38.0 + 45 = 83.0 N · m Since n = V /l = 48/12 = 4 rev/s ω = 2πn = 2π(4) = 8π rad/s so the power is H = T ω = 83.0(8π) = 2086 W 8-10 (a) dm = 36 − 3 = 33 mm, l = p = 6 mm From Eqs. (8-1) and (8-6) 0.09(90) F 33F 6 + π(0.14)(33) + T = 2 π(33) − 0.14(6) 2 = (3.292 + 4.050) F = 7.34F N · m ω =...