SECTION 24 MECHANICAL AND ELECTRICAL BRAKES Brake Selection for a Known Load 24.1 Mechanical Brake Surface Area and Cooling Time 24.3 Band Brake Heat Generation, Temperature Rise, and Re
Trang 1SECTION 24 MECHANICAL AND ELECTRICAL
BRAKES
Brake Selection for a Known Load 24.1
Mechanical Brake Surface Area and
Cooling Time 24.3
Band Brake Heat Generation,
Temperature Rise, and Required
Area 24.6
Designing a Brake and Its Associated Mechanisms 24.8
Internal Shoe Brake Forces and Torque Capacity 24.15
Analyzing Failsafe Brakes for Machinery 24.17
BRAKE SELECTION FOR A KNOWN LOAD
Choose a suitable brake to stop a 50-hp (37.3-kW) motor automatically when power
is cut off The motor must be brought to rest within 40 s after power is shut off The load inertia, including the brake rotating member, will be about 200 lb䡠ft2 (82.7 N䡠m2); the shaft being braked turns at 1800 r / min How many revolutions will the shaft turn before stopping? How much heat must the brake dissipate? The brake operates once per minute
Calculation Procedure:
1. Choose the type of brake to use
Table 1 shows that a shoe-type electric brake is probably the best choice for stop-ping a load when the braking force must be applied automatically The only other possible choice—the eddy-current brake—is generally used for larger loads than this brake will handle
2. Compute the average brake torque required to stop the load
Use the relation Ta⫽Wk2n / (308t), where T a⫽average torque required to stop the load, lb䡠ft; Wk2⫽load inertia, including brake rotating member, lb䡠ft2, n⫽shaft
speed prior to braking, r / min; t ⫽ required or desired stopping time, s For this
brake, Ta⫽(200)(1800) / [308(40)]⫽29.2 lb䡠ft, or 351 lb䡠in (39.7 N䡠m)
3. Apply a service factor to the average torque
A service factor varying from 1.0 to 4.0 is usually applied to the average torque
to ensure that the brake is of sufficient size for the load Applying a service factor
of 1.5 for this brake yields the required capacity ⫽ 1.5(351) ⫽ 526 in䡠lb (59.4
N䡠m)
Trang 2TABLE 1 Mechanical and Electrical Brake Characteristics
4. Choose the brake size
Use an engineering data sheet from the selected manufacturer to choose the brake size Thus, one manufacturer’s data show that a 16-in (40.6-cm) diameter brake will adequately handle the load
5. Compute the revolutions prior to stopping
Use the relation R s⫽tn / 120, where R⫽number of revolutions prior to stopping;
other symbols as before Thus, R s⫽ (40)(1800) / 120⫽600 r
6. Compute the heat the brake must dissipate
Use the relation H⫽ 1.7FWk2(n / 100)2, where H⫽heat generated at friction sur-faces, ft䡠lb / min; F⫽number of duty cycles per minute; other symbols as before
Thus, H⫽1.7(1)(200)(1800 / 100)2⫽110,200 ft䡠lb / min (2490.2 N䡠m / s)
7. Determine whether the brake temperature will rise
From the manufacturer’s data sheet, find the heat dissipation capacity of the brake while operating and while at rest For a 16-in (40.6-cm) shoe-type brake, one
man-ufacturer gives an operating heat dissipation Ho⫽150,000 ft䡠lb / min (3389 5 N䡠
m / s) and an at-rest heat dissipation of H v⫽35,000 ft䡠lb / min (790.9 N䡠m / s) Apply the cycle time for the event; i.e., the brake operates for 400 s, or 40 / 60
of the time, and is at rest for 20 s, or 20 / 60 of the time Hence, the heat dissipation
of the brake is (150,000)(40 / 60)⫹(35,000)(20 / 60)⫽111,680 ft䡠lb / min (2523.6
N䡠m / s) Since the heat dissipation, 111,680 ft䡠lb / min (2523.6 N䡠m / s), exceeds the heat generated 110,200 ft䡠lb / min (2490.2 N䡠m / s), the temperature of the
Trang 3MECHANICAL AND ELECTRICAL BRAKES 24.3
brake will remain constant If the heat generated exceeded the heat dissipated, the brake temperature would rise constantly during the operation
Brake temperatures high than 250⬚F (121.1⬚C) can reduce brake life In the 250
to 300⬚F (121.1 to 148.9⬚C) range, periodic replacement of the brake friction sur-faces may be necessary Above 300⬚F (148.9⬚C), forced-air cooling of the brake is usually necessary
Related Calculations. Because electric brakes are finding wider industrial use, Tables 2 and 3, summarizing their performance characteristics and ratings, are pre-sented here for easy reference
The coefficient of friction for brakes must be carefully chosen; otherwise, the brake may ‘‘grab,’’ i.e., attempt to stop the load instantly instead of slowly Usual values for the coefficient of friction range between 0.08 and 0.50
The methods given above can be used to analyze brakes applied to hoists,
ele-vators, vehicles, etc Where Wk2is not given, estimate it, using the moving parts
of the brake and load as a guide to the relative magnitude of load inertia The
method presented is the work of Joseph F Peck, reported in Product Engineering.
MECHANICAL BRAKE SURFACE AREA AND
COOLING TIME
How much radiating surface must a brake drum have if it absorbs 20 hp (14.9 kW), operates for half the use cycle, and cannot have a temperature rise greater than
300⬚F (166.7⬚C)? How long will it take this brake to cool to a room temperature
of 75⬚F (23.9⬚C) if the brake drum is made of cast iron and weighs 100 lb (45.4 kg)?
Calculation Procedure:
1. Compute the required radiating area of the brake
Use the relation A ⫽ 42.4hpF / K, where A ⫽ required brake radiating area, in2;
hp⫽power absorbed by the brakes; F⫽brake load factor⫽operating portion of
use cycle; K⫽constant⫽Ct r, where C⫽radiating factor from Table 4, tr⫽brake temperature rise, ⬚F For this brake, assuming a full 300⬚F (166.7⬚C) temperature
rise and using data from Table 4, we get A ⫽ 42.4(20)(0.5) / [(0.00083)(300)] ⫽
1702 in2(10,980.6 cm2)
2. Compute the brake cooling time
Use the relation t ⫽ (cW ln t r ) / (K c A), where t ⫽ brake cooling time, min; c ⫽ specific heat of brake-drum material, Btu / (lb䡠 ⬚F); W⫽weight of brake drum, lb;
t r ⫽ drum temperature rise,⬚F; ln ⫽ log to base e ⫽ 2.71828; Kc ⫽ a constant
varying from 0.4 to 0.8; other symbols as before Using Kc⫽ 0.4, c ⫽0.13, t ⫽ (0.13⫻100 ln 300) / [(0.4)(1702)]⫽ 0.1088 min
Related Calculations. Use this procedure for friction brakes used to stop loads that are lifted or lowered, as in cranes, moving vehicles, rotating cylinders, and similar loads
Trang 4T
Trang 524.5
Trang 6TABLE 4 Brake Radiating Factors Temperature
rise of brake
⬚F ⬚C Radiating factor C
100 200 300 400
55.6 111.1 166.7 222.2
0.00060 0.00075 0.00083 0.00090
BAND BRAKE HEAT GENERATION,
TEMPERATURE RISE, AND REQUIRED AREA
A construction-industry hoisting engine is to be designed to lower a maximum load
of 6000 lb (2724 kg) using a band brake, Fig 1, having a 48-in (121.9-cm) drum diameter and a drum width of 8-in (20.3-cm) The brake band width is 6-in (15.2-cm) with an arc of contact of 300⬚ Maximum distance for lowering a load is 200
ft (61 m) The cycle of the engine is 1.5 min hoisting and 0.75 min lowering, with 0.3 min for loading and unloading If a temperature rise of 300⬚F (166.5⬚C) is permitted, determine the heat generated, radiating surface required, and the actual temperature rise of the drum if the brake is made of cast iron and weighs 600 lb (272.4 kg)
Calculation Procedure:
1. Determine the heat generated, equivalent to the power developed
Use the relation, H g⫽ (wd / T L )(33,000), where H g⫽ heat generated during
low-ering of the load of w, lb (kg), hp (kW); d⫽distance the load is lowered, ft (m);
T L ⫽ time for lowering, minutes Substituting for this hoisting engine, we have,
H g⫽(6000)(200) / (0.75)(33,000)⫽ 48.48 hp (36.2 kW)
2. Find the load factor for the brake
The load factor, q, for a brake⫽T L / (TH⫹T L⫹ T LU), where TH⫽hoisting time,
minutes; TLU ⫽ time to load and unload, minutes; other symbols as before
Sub-stituting, q⫽(0.75) / (1.5⫹ 0.75⫹0.3)⫽0.294
3. Compute the required radiating area for the brake
Use the relation, AR⫽ 42.4(q)(Hg) / Ctr ), where AR ⫽ required radiating area, in2 (cm2); Ctr⫽brake radiating factor from Table 4 Assuming a temperature rise of
300⬚F (166.7⬚C), and substituting, AR ⫽ (42.4)(0.294)(48.5) / 0.25 ⫽ 2418.3 in2 (15,601.9 cm2)
It is necessary to assume a temperature rise when analyzing a brake because the rise is a factor in the computation Without such an assumption the required radi-ating area cannot be determined
Using the given dimensions for this brake, we can find the area of the drum
from Ad⫽2()(48⫻ 8)⫺()(48⫻6)(300⬚/ 360⬚)⫽1657.9 in2(10,696.1 cm2) Since the required radiating area is 2418 in2., as computed above, the excess area required will be 2418⫺1658⫽760 in2(4903.2 cm2) This area can be provided
Trang 7MECHANICAL AND ELECTRICAL BRAKES 24.7
FIGURE 1 (a) Self-locking band brake (b) Pressure
var-iation along the surface of a band brake.
by the brake flanges and web To be certain that sufficient area is available for heat radiation, check the physical dimensions of the brake flanges and webs to see if the needed surface is present
4. Find the temperature rise during brake operation
The actual temperature of the brake drum will vary slightly above and below the assumed 300⬚F (166.7⬚C) temperature rise during operation The reason for this is because heat is radiated during the whole cycle of operation but is generated only during the lowering cycle, which is 29.4 percent of the total cycle
The temperature change of the drum during the braking operation will be Tr⫽
[1 / (778)(Wr)(c)][Wh ⫺ Ct r A r m(778)], where c ⫽ specific heat of the drum material ⫽ 0.13 for cast iron; m ⫽ lowering time in minutes; other symbols
as given earlier Substituting, Tr ⫽ [1 / (778)(600)(0.13)][(6000 ⫻ 200) ⫺ (0.25)(2418)(0.75)(778)] ⫽14⬚F (7.78⬚C) Thus, the drum temperature can range
Trang 8about 14⬚, or about 7⬚ above and 7⬚ below the average operating temperature of
300⬚F (148.9⬚C)
Related Calculations. The actual temperature attained by a brake drum, and the time required for it to cool, cannot be accurately calculated But the method given here is suitable for preliminary calculations In the final design of a new brake, heating should be checked by a proportional comparison with a brake already known to give good performance in actual service
An approximation of the time required for a brake to cool can be made using
the relation given in step 2 of the previous procedure Note that the value of K
selected in that relation will directly influence brake cooling time Thus, a lower
value chosen for K will increase the estimated cooling time while a higher value will decrease the time For safety reasons, engineers will often select lower K values
so the brake will be given more time to cool, or will be provided with a larger capacity cooling mechanism
This procedure is the work of Alex Vallance, Chief Designer, Reed Roller Bit
Co and Venton L Doughtie, Professor of Mechanical Engineering, University of Texas
DESIGNING A BRAKE AND ITS ASSOCIATED
MECHANISMS
Design a hoist for the building industry to lift a cubic yard (0.76 m3) of concrete
at the rate of 200 ft/min (61 m / min) A cubic yard of concrete weighs approxi-mately 400 lb (1816 kg) and the bucket weighs 1250 lb (568 kg) Since the hoist may be used for other construction purposes, the design capacity will be 6000 lb (2725 kg) There will be no counterweights in this hoist and the cable drum will
be connected to a 1750-r/min electric motor through a reduction gear train Low-ering will be controlled by manual operation of a brake This brake must automat-ically hold the load at any position when the motor is not driving the hoist Figure
2 shows the proposed arrangement of parts and the limiting dimensions of this hoist control It has been proposed that a spring-loaded band brake be used that will be manually released by the operator during lowering of the load An overrunning clutch is to be provided to disengage the brake automatically when the torque from
the motor through the gear train is sufficient to raise the load Determine (a) the dimensions q and n, Fig 2, for maximum self-energization without self-locking if
the coefficient of friction, , varies between 0.20 and 0.50—the wide range is selected to cover possible changes in service due to unavoidable entrance of small
amounts of water, oil, or dirt into the brake; (b) the spring force required to ensure
that the brake will not slip whenis between 0.20 and 0.50; (c) the force exerted
by the operator when the rated load first starts to lower under normal conditions; i.e., when ⫽0.35; (d) the minimum width of brake lining; and (e) the maximum
lowering speed for a reasonable wear life
Calculation Procedure:
1. Determine the actuating arm dimensions
The force F1and F2must act as shown, Fig 2, for the braking torque to oppose
the load torque This brake will be self-energizing when the F1moment tends to
Trang 9MECHANICAL AND ELECTRICAL BRAKES 24.9
SI Values
12 in 30.5 cm
48 in 121.9 cm
24 in 60.9 cm
6000 lb 2724 kg
FIGURE 2 Band brake used in hoisting application.
apply the brake, that is q is less than n, and becomes self-locking when F1qⱖF2n.
The appropriate equation is
⫽e
F2 and the critical design condition will be when F1/ F2 is a maximum, ⫽ 0.50 Thus,
F1 n 0.50⫻(3 / 2)
This unit will be most compact when q is as small as possible The strengths of
the pin and the lever will be the major factors in determining the minimum
dimen-sion for q However, if q is estimated to be 1.5 in (3.8 cm), it can be seen that there is not enough space in which to place the lever, as shown, when n⫽10.55q.
Therefore, under the specified conditions, it is not practical to use self-energization,
Trang 10SI Values
48 in 121.9 cm
12 in 30.5 cm
2 in 5.1 cm
FIGURE 3 Lever for band brake actuation.
and the proposed design will be modified to make q ⫽ 0, as shown in Fig 3
Choose the dimension n as 2.0 in (5.08 cm).
2. Find the required spring force
Maximum spring force will be required when the coefficient of friction is a mini-mum, that is when ⫽0.20 For this case,
F1 0.20⫻(3 / 2)
⫽e ⫽e ⫽2.57
F2
and
F1⫽2.57F2
D
T⫽(F1⫺F )2
2 and
Trang 11MECHANICAL AND ELECTRICAL BRAKES 24.11
Dcable drum
T⫽Fcable ⫽6000⫻24 / 2⫽72,000 lb䡠in (8136 N䡠m)
2
D
⫽30 / 2⫽15 in (38.1 cm)
2
Solving for F1, we find
F1⫽F2⫹4800 lb Solving the two equations above simultaneously, we find
F2⫽3060 lb (13,611 N) and
F1⫽7860 lb (34,961 N)
兺M O⫽0
F s⫻12⫺F2⫻2⫽F s⫻12⫺3060⫻2⫽0 Solving for the spring force we find it to be 510 lb (2268 N)
3. Compute the operating force for rated load under normal conditions
The operator must push the lever to the right to lower the load When ⫽0.35,
F1 0.35⫻(3 / 2)
⫽e ⫽e ⫽5.20
F2
and
F1⫽5.20F2
Solving the equations for force simultaneously, we find
F1⫽5940 lb (26421 N) F2⫽1140 lb (5071 N)
Then,
兺M O⫽0
⫺P⫻48⫹F s⫻12⫺F2⫻2⫽0
⫺P⫻48⫹510⫻12⫺1140⫻2⫽0
Solving for P, we find P⫽80 lb (356 N) This force is too large for an operator
to exert and the situation becomes even worse when the hoist is lightly loaded For example, if the load is considered to be negligible, the force required to release the brake will be 510 / 4⫽127.5 lb (567 N) A reasonable design solution would be to use a compound lever, Fig 4, in place of the single lever we have been analyzing A force of 30 lb (133 N) will be satisfactory for releasing the brake when the load is negligible
Based on the analysis above, we shall consider 2 in (5.08 cm) as the value for the distance from the point of spring attachment to the pin B on lever 2, Fig 4 It