brakes - university of babylon · brakes caliper disk brakes are commonly used disk brakes. bicycle...

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38 Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction lining on the caliper contacts only a small portion of the disk surface, leaving the remainder exposed to dissipate heat. Figure (1) shows a hydraulically actuated caliper disk brake that uses a ventilated disk. Air circulation through the interior passages provides substantial additional cooling. Disk brakes can conveniently be examined on the front wheel of larger motorcycles. The cooling or heat dissipating characteristics of brakes are discussed her in. Fig.(1) Caliper disk brake, hydraulically operated Energy Absorption and Cooling The basic function of a brake is to absorb energy, that is, to convert kinetic and potential energy into friction heat, and to dissipate the resulting heat without developing destructively high temperatures . Where brakes are used more or less continuously for extended periods of time, provision must be made for rapid transfer of heat to the surrounding

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Page 1: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

38

Brakes

Caliper disk brakes are commonly used disk brakes. Bicycle brakes are

the best-known examples. The wheel rim constitutes the disk. Friction

lining on the caliper contacts only a small portion of the disk surface,

leaving the remainder exposed to dissipate heat. Figure (1) shows a

hydraulically actuated caliper disk brake that uses a ventilated disk. Air

circulation through the interior passages provides substantial additional

cooling. Disk brakes can conveniently be examined on the front wheel of

larger motorcycles. The cooling or heat dissipating characteristics of

brakes are discussed her in.

Fig.(1) Caliper disk brake, hydraulically operated

Energy Absorption and Cooling

The basic function of a brake is to absorb energy, that is, to convert

kinetic and potential energy into friction heat, and to dissipate the

resulting heat without developing destructively high temperatures.

Where brakes are used more or less continuously for extended periods of

time, provision must be made for rapid transfer of heat to the surrounding

Page 2: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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atmosphere. For intermittent operation, the thermal capacity of the parts

may permit much of the heat to be stored and then dissipated over a

longer period of time. Brake parts must be designed to avoid

objectionable thermal stresses and thermal distortion .The basic heat

transfer equation is:

(1)

H = time rate of heat dissipation (W or hp)

C = overall heat transfer coefficient (W per m2 per °C, or hp per in

2 per

°F)

A = exposed heat-dissipating surface area (m2 or in

2)

ts = average temperature of heat-dissipating surfaces (°C or °F)

ta = air temperature in the vicinity of the heat-dissipating surfaces (°C or

°F)

The ability of brakes to absorb large amounts of energy without reaching

destructive temperatures can be increased by

(1) Increasing exposed surface areas, as by fins and ribs.

(2) Increasing air flow past these surfaces by minimizing air flow

restrictions and maximizing the air pumping action of the rotating parts,

(3) increasing the mass and specific heat of parts in immediate contact

with the friction surfaces, thereby providing increased heat storage

capacity during short periods of peak braking load.

The sources of energy to be absorbed are primarily three.

1. Kinetic energy of translation:

(2)

2. Kinetic energy of rotation:

(3)

Page 3: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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3. Potential (gravitational) energy, as in an elevator being lowered or an

automobile descending a hill:

(Weight times vertical distance) (4)

The rate at which heat is generated on a unit area of friction interface is

equal to the product of the normal (clamping) pressure, coefficient of

friction, and rubbing velocity. Manufacturers of brakes and of brake

lining materials have conducted tests and accumulated experience

enabling them to arrive at empirical values of pV (normal pressure times

rubbing velocity) and of power per unit area of friction surface (as

horsepower per square inch or kilowatt per square millimeter) that are

appropriate for specific types of brake design, brake lining material, and

service conditions. Table (1) lists typical values of pV in industrial use.

Table(1) Typical Values of Pressure Times Rubbing Velocity Used

in Industrial Shoe Brakes

Short-Shoe Drum Brakes

Drum brakes are of two types:

(1) Those with external shoes that contract to bear against the outer

(cylindrical) drum surface

(2) Those with internal shoes that expand to contact the inner drum

surface.

Figure(2) shows a schematic representation of a simple external

drum brake with a “short shoe” that is, a shoe that contacts only a

small segment of the drum periphery. Force F at the end of the

lever applies the brake. Although the normal force (N) and the

Page 4: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

38

friction force (fN) acting between the drum and shoe are

distributed continuously over the contacting surfaces.

Fig.(2) Short shoe drum brake

The short-shoe analysis assumes these forces to be concentrated at

the center of contact. The complete brake assembly is shown in

Figure (2-a). Free-body diagrams of the basic components are

given in Figures (2-b and c). Drum rotation is clockwise.

Page 5: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Taking moments about pivot A for the shoe and lever assembly, we

have :

(5)

From summation of moments about O for the drum,

(6)

Solving equation (5) for N and substituting in equation (6) gives

(7)

(self energizing) (8)

Torque T is the inertial and load torque required for equilibrium, and it is

numerically equal to the friction torque developed by the brake.

Equation (8) is labeled “self-energizing” because the moment of the

friction force ( fNa) assists the applied force (F) in applying the brake.

For counter-clockwise drum rotation, the direction of the friction force

would be reversed. This would cause it to oppose the application of the

brake, making the brake self-de energizing. The derivation of the

equation for the de energizing brake is the same as for the self-energizing

brake, except that the signs of the friction force terms are reversed:

(self- de energizing) (9)

Returning now to self-energized braking (clockwise drum rotation), note

that the brake is self-locking if the denominator of Eq. (8) is zero or

negative. Thus, for self-locking,

(10)

For example, if f = 0.3, self-locking (for clockwise drum rotation) is

obtained if b 0.3a. This is illustrated in Figure (2-d). A self-locking

brake requires only that the shoe be brought in contact with the drum

(with F = 0) for the drum to be “locked” against rotation in one direction.

Because the brake in Figure (2)has only one shoe (block), the entire force

exerted on the drum by the shoe must be reacted by the shaft bearings.

Partly for this reason, two opposing shoes are almost always used.

Page 6: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Ex:

The two-shoe external drum brake shown in Figure(3) has shoes 80 mm

wide that contact 90° of drum surface. For a coefficient of friction of 0.20

and an allowable contact pressure of 400 kN per square meter of

projected area, estimate:

(a) The maximum lever force F that can be used, (b) the resulting braking

torque.

(c) The radial load imposed on the shaft bearings.

Assume short-shoe drum brake.

(a)

Page 7: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Fig.(3) Two shoe drum brake

The free body diagram for all the components of the two shoe drum brake

can be shown in figures (3-b) to (3-f).

1- The force analysis begins with floating lever 5 because it receives

the applied force F.

Taking moments about O25 establishes

2- For the link 4 ,

but in opposite direction.

Page 8: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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3- Link (3)

On the left shoe (link 3), the applied force is H43 (which is equal and

opposite to H34). The short-shoe analysis assumes that the normal and

friction forces applied by drum 6 act at the center of the shoe as shown.

The normal force is H63 and the friction force is H63 multiplied by the

given friction coefficient of 0.2.

Summation of moments about O13 gives:

0.08)=0

Or

Normal and friction forces H62 and V62 acting on shoe 2 are determined in

the same manner. The moment equation has an additional term because

both horizontal and vertical forces are applied by link 2:

Page 9: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Or

Horizontal and vertical forces applied to drum 6 are equal and opposite to

corresponding forces applied to the shoes. If as assumed the drum angular

acceleration is zero, the load torque T (which tends to continue the

clockwise direction of rotation) is equal to (2.11F + 1.41F) times the

drum radius, or 0.880F N.m.

Forces applied at fixed pivot O16 are H16 = 3.46F and V16 = 0.70F.

The allowable value of F is governed by the allowable pressure on the

self-energized shoe. The projected area of the shoe is the 80-mm width

multiplied by the chord length subtended by a 90° arc of the 250-mm-

radius drum:

A = 80[2(250 sin 45°)] = 28,284 mm2

The normal pressure on shoe 3 is

Equating this to the allowable value of pmax = 0.40 N/mm2 gives

The corresponding brake torque is

From figure (3-f)

The braking torque can be calculated as:

Or

The resultant radial load transmitted to the bearing is

√ F=3.53F=3791N

Page 10: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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External Long-Shoe Drum Brakes

If a brake shoe or block contacts the drum over an arc of about 45° or

more, errors introduced by the short-shoe equations are usually

significant. For this situation the following “long-shoe” analysis is

appropriate. Figure (4) shows the forces applied to a brake shoe

(including its associated lever).

1. Considering the shoe as a free body , we have

(*)

Fig.(4)

Where Mn is the moment of the normal forces,

(11)

And

Page 11: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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2. For long shoe brake the normal wear is proportional to friction

work, which for various locations on the shoe contact surface, is

proportional to the local pressure.

The maximum value of sinθ is, of course, unity, with = 90°. Hence, the

maximum pressure and the maximum wear occur at = 90°. So that

(12)

3. Combining equations (11) and (12) to get

(13)

4. Similarly, for the moment of friction forces, Mf ,we have

5. Substituting the relationship

gives

(14)

6. For equilibrium of moments acting on the drum

Page 12: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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(15)

7. Reaction forces at O2 and O3 can readily be obtained from

horizontal and vertical force equilibrium equations.

With reference to Eq.(*),

it is evident that a self-energizing brake is self- locking if Mf ≥ Mn. It is

often desired to make a brake shoe strongly self-energizing while staying

safely away from a self-locking condition. This can be done by designing

the brake so that the value of Mf , as calculated using a value of (f) that is

25 to 50 percent greater than the true value, is equal to the value of Mn.

Page 13: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Ex:

Figure (5) represents a double-shoe drum brake with spring force F

applied at distance c = 500 mm to both shoes. Design values of the

coefficient of friction and the allowable pressure are 0.3 and 600 kPa,

respectively. For occasional industrial use, determine an appropriate

value of spring force, and the resulting brake torque and power absorption

for 300-rpm drum rotation in either direction. The braking material has

pmaxV=2.1MPa.m/s

Fig.(5)

Solution

With reference to the right shoe shown in figure (6):

Fig.(6)

Page 14: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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(

)

Hence

(

) * (

) +

[

(

) ]

The absolute value of brake torque

Since the brake material has pmaxV=2.1MPa.m/s

(

)

Hence

Page 15: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Or

(Spring force)

The appropriate spring force was determined above on the basis of

allowable pressure on the right shoe, which proved to be mildly self-

energizing. This same spring force will produce a lower value of pmax on

the left shoe. Applying the following equation to the left shoe, the only

change is the sign of the friction moment term:

The total brake torque is the brake torque of the right plus the brake

torque of the left shoe

Since

Hence

The power can be calculated as

Band Brakes

Perhaps the simplest of the many braking devices is the band brake,

shown in Figure (7). The band itself is usually made of steel, lined with a

woven friction material for flexibility. For the clockwise drum rotation

shown, friction forces acting on the band serve to increase P1 and

Page 16: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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decrease P2. With the drum and band portion above the cutting plane

Figure (7) considered as a free body, brake torque T is equal to

Fig.(7) Band brake

(16)

With the lever and band portions below the cutting plane considered as a

free body, the applied lever force F is

(17)

Figure (8) shows the forces acting on an element of the band. For small

angle d .

Fig.(8) Element of band brake

(18)

(

) (19)

Page 17: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Also by definition

(20)

where p is the local contact pressure between drum and band. Substituting

Eq.(19) into Eq.(18) gives:

(21)

Or

(22)

Band force P varies from P1 to P2 over the band portion between = 0

and = . Hence, integrating Eq.(22) over the length of contact gives

(23)

(24)

Fig.(9)

(25)

Page 18: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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The maximum normal pressure pmax acting on the band occurs at

where P = P1.

and

Hence

(26)

The brake in Figure(7) is self-energized for clockwise drum rotation.

Greater self-energizing action can be obtained by attaching the previously

fixed end of the band to the lever at the opposite side of the pivot, as

shown in Figure (9). The tensile force of this attachment now serves to

assist in the application of the brake. Note also that distance s must be

less than distance a so that rotating the lever with force F tightens the end

of the band attached at distance a more than it loosens the end attached at

distance s. A study of the motion and forces involved at the two band

attachment points shows why the name differential band brake is

appropriate. For a differential band brake, Eq.(17) is replaced by

(27)

Ex:

A differential band brake shown in Figure (10) uses a woven lining

having a design value of f = 0.20. Dimensions are b = 80 mm, r = 250

mm, c = 700 mm, a = 150 mm, s = 35 mm, and = 240°. Find

(a) The brake torque if the maximum lining pressure is 0.5 MPa.

(b) The corresponding actuating force F.

(c) The values of dimension s that would cause the brake to be self-

locking.

Page 19: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Fig.(10) Differential band break

Solution

From equation (26)

From equation (25)

From equation (16)

From equation (27)

From equation (27)

For F=0,

The brake self-locked for (f=0.2)

Page 20: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Supplementary Problems(brakes)

(Q1) Consider the following dimensions for the short-shoe drum brake of

Figure (11) radius of drum = 5 in., shoe width = 2 in., shoe length = 4 in.,

c = 10 in., b = 6 in., a = 1.5 in., p = 100 psi, and f = 0.3. Determine the

value of the actuating force F.

Fig.(11)

Page 21: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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(Q2)Figure (12) shows a brake with only one shoe, being applied by a

1.5-kN force. (The complete brake would normally have a second shoe in

order to balance the forces, but only one shoe is considered here to keep

the problem short.) Four seconds after force F is applied, the drum comes

to a stop. During this time the drum makes 110 revolutions. Use the

short-shoe approximation and an estimated coefficient of friction of 0.35.

(a) Draw the brake shoe and arm assembly as a free body in equilibrium.

(b) Is the brake self-energizing or de-energizing for the direction of drum

rotation involved?

(c) What is the magnitude of the torque developed by the brake?

(d) What is the average braking power during the 4-second interval?

(e) How much work does the brake do in bringing the drum to a stop?

(f) How far below the drum center would the arm pivot need to be to

make the brake self-locking for f = 0.35?

Fig.(12)

Page 22: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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Page 23: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

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(Q3) The brake shown in Figure (13)is applied by the spring and released

by a hydraulic cylinder (not shown). Use the short-shoe equations and an

assumed coefficient of friction of 0.3.

(a) Draw as free bodies in equilibrium each of the brake shoe and arm

assemblies,

the spring, and the drum. Show forces in terms of spring force Fs .

(b) What spring force is required to produce a braking torque of 1200

N.m?

Fig.(13)

Page 24: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

998

(Q4) If the shoe in Figure (14) extends 45° on each side of the centerline,

(a) Estimate the brake torque using long-shoe equations.

(b) Determine the shoe width required for an allowable pressure equal to

the average value 620kPa and the coefficient of friction equal to 0.2.

Fig.(14)

Page 25: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

998

(Q5) The brake shown in Fig. (15) is 300 mm in diameter and is actuated

by a mechanism that exerts the same force F on each shoe. The shoes are

identical and have a face width of 32 mm. The lining is a molded asbestos

having a coefficient of friction of 0.32 and a pressure limitation of 1000

kPa. Estimate the maximum

(a) Actuating force F.

(b) Braking capacity.

.

Fig.(15)

Solution:

(a) The right-hand shoe is self-energizing, and so the force F is found on

the basis that the maximum pressure will occur on this shoe. Here θ1 = 0◦,

θ2 = 126◦, θa = 90◦, and

(sin θ)max = 1. Also,

Page 26: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

993

[

]

Or

The torque contributed by the left-hand shoe cannot be obtained until we

learn its maximum operating pressure. Equations (13) and (14) indicate

that the frictional and normal moments are proportional to this pressure.

Thus, for the left-hand shoe,

Page 27: Brakes - University of Babylon · Brakes Caliper disk brakes are commonly used disk brakes. Bicycle brakes are the best-known examples. The wheel rim constitutes the disk. Friction

998

The torque of the left shoe can be calculated as:

The total braking torque can be calculated as

T=T(right shoe)+ T(left shoe)

T= 366+162= 528 N.m