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CLUTCHES AND BRAKES 8.17
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CLUTCHES AND BRAKES
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FIGURE 8.11 Equivalent inertia. I1, I2 = inertia of input and output shafts, respectively; I = equivalent inertia. (a) Original configuration; (b) equivalent system referred to input shaft; (c) equivalent system referred to output shaft. For a more extensive treatment of equivalent inertias, see Suggested Reading list, Mischke.
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Similarly, I = 1 In general, I l = i j
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(8.4)
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(8.5)
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where the equivalent inertia of the ith shaft is referred to the jth shaft. Equation (8.5) can be used to reduce a machine with several shafts connected by gears or flexible connectors to a single equivalent shaft. Example 2. For the two-shaft machine in Fig. 8.11a, the inertias are I1 = 2.88 pound-inch-square seconds (lb in s2) [0.3254 kilogram-square meters (kg m2)] and I2 = 0.884 lb in s2 (0.09988 kg m2). The pitch diameters of the gears are D1 = 4 in [0.102 meter (m)] and D2 = 7 in (0.178 m). What is the equivalent inertia of shaft 2 referred to shaft 1
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CLUTCHES AND BRAKES 8.18
MACHINE ELEMENTS THAT ABSORB AND STORE ENERGY
Solution. Equation (8.3) can be used once the speed ratio 2/ 1 is known. For spur or helical gears, 2 D1 4 = = = 0.5714 1 D2 7 Thus I2 = (0.5714)2(0.884) = 0.2887 lb in s2 (0.0326 kg m2)
8.2.3 Torque Requirement: Brakes Industrial Brakes. The torque to bring a rotating machine from an initial speed o to a lower one f (perhaps to rest) in a slowdown time of tS is T= I( o f) tS (8.6)
Vehicle Brakes. The braking torque to stop a vehicle of weight W at a deceleration rate a on a grade of b percent can be estimated as T= WR a b + g g 100 (8.7)
Here R is the tire-rolling radius. This is a conservative approach; both tire-rolling resistance and air resistance have been neglected. Of course, this torque capacity T must be allocated to the several brakes in a rational way (for example, in proportion to the weight of the vehicle supported by the corresponding wheel during a panic stop). For parking-brake capacity, simply set a = 0 in Eq. (8.7). The required acceleration rate a can be determined by setting either a total stopping time tS or a total stopping distance s: a= a= Vo tS td V2 o 2(s Votd) (8.8) (8.9)
In these equations, td is the combined delay time (about 1 s for a passenger car) for driver reaction and brake system reaction.
8.2.4 Energy Dissipation: Clutches A simple model of a clutch connecting a prime mover and a load is shown in Fig. 8.12. The clutch capacity is T, the driving torque provided by the prime mover is TP, and the load torque is TL. The inertias IP and IL include all rotating masses on their respective sides of the clutch. If the two sides of the clutch are initially rotating at P and L radians per second (rad/s) when the clutch is actuated, the duration of the slip period is
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CLUTCHES AND BRAKES 8.19
CLUTCHES AND BRAKES
FIGURE 8.12 Abstract model of a machine using a clutch.
tS =
IPIL( P L) T(IP + IL) (ILTP + IPTL)
(8.10)
The rate at which energy is dissipated during the slip period is, at t s from the beginning, q(t) = T ILTP + IPTL T(IP + IL)t IPIL + P L (8.11)
And the total energy dissipated in one actuation operation is E= T( P L)2 IP IL 2T(IP + IL) (ILTP + IPTL) (8.12)
8.2.5 Energy Dissipation: Brakes Vehicle Brakes. When a vehicle of weight W is slowed from an initial velocity Vo to a final velocity Vf, the heat energy that the brakes must dissipate is equal to the change of kinetic energy E: E= W 2 (V o V 2 ) f 2g (8.13)
In dealing with individual brakes, let W be that portion of the vehicle s weight for which the brake is responsible. Example 3. A sports car weighing 3185 lb [14.2 kilonewtons (kN)] has 62 percent of its weight on the front axle during an emergency stop. What energy must each of the front-wheel brakes dissipate in braking from 55 miles per hour (mph) [88 kilometers per hour (km/h)] to rest Local acceleration of gravity is 32.17 feet per second (ft/s) (9.81 m/s). Solution. Each front brake is responsible for a weight of W = 0.5(0.62)(3185) = 987 lb (4.39 kN) The initial velocity is Vo = 55 88 = 80.7 ft/s (24.6 m/s) 60
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