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8.5.2 Centrifugal Clutches The simple centrifugal clutch shown in Fig. 8.18 has a number of shoes which can move radially and against the drum as the input shaft speed increases. A garter spring regulates the engagement speed. At engagement speed the weights contact the drum s inner surface and begin to drive it, and the attached pulley, by means of friction to bring it up to speed. Design Equations. The normal force between each shoe and the inner circumference of the drum is principally due to centrifugal force. However, the garter spring exerts some inward force. The net normal force [8.1] is Fn = mr 2 180 2S cos 90 gc Ns (8.39)
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Engagement occurs at the shaft speed e when F = 0. The proper initial tension S for the garter spring is found by setting = e and F = 0 in Eq. (8.39). The engagement speed is selected by the designer (for example, about 70 percent of running speed). The required value for S is S=
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2 mr e 2gc cos (90 180 /Ns)
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(8.40)
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FIGURE 8.18 Free-shoe centrifugal clutch with garter spring to regulate engagement speed. A, input shaft; B, output pulley; C, drum; D, weights; E, garter spring.
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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The torque capacity at any shaft speed is T = fFn RNs (8.41)
Example 7. A centrifugal clutch is being considered for an application where the running speed is 3000 rev/min and engagement is to begin at 1000 rev/min. It is planned to use four shoes, each with a mass of 140 grams (g). A coefficient of friction f = 0.3 can be achieved. The inner diameter of the drum is 75 mm, and the radius R to the center of gravity of each shoe is 25 mm. (1) What should be the initial tension of the garter spring (2) What is the normal force on each shoe at running speed (3) What is the torque capacity at running speed (4) What is the power capacity at running speed if a service factor of 2.25 is required Solution. 1. The initial tension of the garter spring is found from Eq. (8.40): S= 0.140(0.025)[1000(2 /60)]2 = 27.1 N 2(1) cos (90 180 /4)
2. The normal force on each shoe at 3000 rev/min is, by Eq. (8.39), Fn = (0.140)(0.025)[(2 /60)(3000)]2 2(27.1) cos (90 180 /4) 1
= 307.1 N 3. The torque capacity at running speed is T = fFn RNs = (0.3)(307.1) 0.075 (4) = 13.8 N m 2
4. The power capacity uncorrected for service factor is P= 13.8(3000) Tn = = 4.34 kW 9550 9550
Correcting for service factor, we see the power rating is Prating = P 4.34 = = 1.93 kW K S 2.25
8.6 BAND AND CONE BRAKES AND CLUTCHES
8.6.1 Band Brakes A typical design for a band brake is shown in Fig. 8.19. A flexible metal band lined with a friction material contacts the drum over an angle of wrap and exerts a braking torque T on the drum. This particular design is self-energizing, since the moment exerted on the lever by force P1 assists in actuating the brake.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
CLUTCHES AND BRAKES 8.35
CLUTCHES AND BRAKES
FIGURE 8.19 Forces on a band brake; b = bandwidth.
Four basic relationships are needed for analysis or design. For a band wrapped around a drum, the ratio of tensions is P1 = e f P2 (8.42)
where the notation is indicated in Fig. 8.19. The net torque exerted on the drum by the band is T = (P1 P2 D 2 (8.43)
The maximum contact pressure between the band and the drum occurs at the more taut P1 end of the band: pmax = 2 P1 bD (8.44)
Finally, it is necessary to sum moments on the lever about pivot A to get the relationship involving the actuating force Q:
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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