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1 MS3111 Elemen Mesin MAK © 2021 Teknik Mesin - FTMD ITB 09.01. Introduction MS3111 - Elemen Mesin Modul 09 Clutches, Brakes, Couplings and Flywheel Part I

# Modul 09 Clutches, Brakes, Couplings and Flywheel Part I

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and Flywheel Part I
Modul 09 Clutches, Brakes, Couplings and Flywheel Part I
Introduction Segment
Teknik Mesin - FTMD ITB
and Flywheel Part I
Representation
= Inertia
i, i
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Performance Analysis
3. The energy loss (rugi-rugi)
4. The temperature rise (kenaikan temperatur)
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Various Types of Devices
3. Band types
5. Cone types
6. Miscellaneous types
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General procedures of analysis steps.
1. Estimate or determine the distribution of pressure on the frictional surfaces.
2. Find a relation between the maximum pressure and the pressure at any point.
3. Apply the conditions of static equilibrium to find the actuating force, the torque, and the support reactions.
9.1. Static Analysis of Clutches and Brakes
Figure 16–2 A common doorstop.
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• A normal pressure distribution () is shown under the friction pad as a function of position , taken from the right edge of the pad.
• Distribution of shearing frictional traction is on the surface, of intensity (), in the direction of the motion of the floor relative to the pad, where is the coefficient of friction.
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The net force in the y-direction:
The moment about C from the pressure:
= 2 0
1
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We sum the forces in the x-direction to obtain
= 0 2 0
1
= 0
where − or + is for rightward or leftward relative motion of the floor, respectively.
• Assuming f constant, solving for Rx gives
= ±2 0
• Summing the forces in the y-direction gives
from which
1
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Summing moments about the pin located at A we have
= 0
1
= 0
A brake shoe is self-energizing if its moment sense helps set the brake, self-deenergizing if the moment resists setting the brake. Continuing:
= 2
1
= 2
1

Can F be equal to or less than zero? Only during rightward motion of the floor when the expression in brackets in Eq. (e) is equal to or less than zero. We set the brackets to zero or less:
0
1
0 1
= 1

1
≥ +
where is the distance of the center of pressure from the right edge of the pad.
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Some remarks:
The conclusion that a self-acting or self-locking phenomenon is present and is independent of our knowledge of the normal pressure distribution ().
Our ability to find the critical value of the coefficient of friction is dependent on our knowledge of (), from which we derive .
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EXAMPLE 16–1
The doorstop depicted in Fig. 16–2a has the following dimensions: a = 4 in, b = 2 in, c = 1.6 in, w1 = 1 in, w2 = 0.75 in, where w2 is the depth of the pad into the plane of the paper. a) For a leftward relative movement of the floor, an actuating force F of 10
lbf, a coefficient of friction of 0.4, use a uniform pressure distribution pav, find Rx , Ry , pav, and the largest pressure pa.
b) Repeat part a for rightward relative movement of the floor. c) Model the normal pressure to be the “crush” of the pad, much as if it
were composed of many small helical coil springs. Find Rx , Ry , pav, and pa for leftward relative movement of the floor and other conditions as in part a.
d) For rightward relative movement of the floor, is the doorstop a self- acting brake?
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= 12 = 0.4 1 0.75 = 0.3
Solution
Leftward:
= 2
1
1

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b) Repeat part a for rightward relative movement of the floor.
= −12 = −0.4 1 0.75 = −0.3
= − 12 = 10 − 1 0.75 = 10 − 0.75
= 2
1
1

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c) Model the normal pressure to be the “crush” of the pad, much as if it were composed of many small helical coil springs. Find Rx , Ry , pav, and pa for leftward relative movement of the floor and other conditions as in part a.
From similar triangles 1 1
=
1
1 = 2 = ( + 1)
This means that y is directly proportional to the horizontal distance from the pivot point A; that is, = 1, where 1 is a constant
Assuming the pressure is directly proportional to deformation, then = 2, where 2 is a constant. In terms of , the pressure is = 2 + = 2 1.6 +
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= 2
1
1
1
()
= 0.75
1
1
2 = 3.396 / = 3.396 1.6 +
The average pressure is given by
= 1
3.396(1.6 + ) = 3.396 1.6 + 0.5 = 7.132
The maximum pressure occurs at = 1 , and is = 3.396 1.6 + 1 = 8.83
= 0.3 = 0.3 7.132 = 2.139 = 10 − 0.75 = 10 − 0.75 7.132 = 4.652
The reaction at support:
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d) For rightward relative movement of the floor, is the doorstop a self- acting brake?
To evaluate we need to evaluate two integrations
0
1
0
1
Thus, = 3.849
+

1.6 + 0.5397
4 = 0.535
The doorstop friction pad does not have a high enough coefficient of friction to make the doorstop a self-acting brake. The configuration must change and/or the pad material specification must be changed to sustain the function of a doorstop.
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9.2. Internal Expanding Rim Clutches & Brakes
(a) Clutch (b) Brake Figure 16–3 (a) internal expanding centrifugal-acting rim clutch (b) internal expanding brake
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Depending upon the operating mechanism, such clutches are further classified as: expanding-ring, centrifugal, magnetic, hydraulic, and pneumatic.
Internal-shoe rim expanding type consist of 3 elements:
• the mating frictional surface,
• the means of transmitting the torque to and from the surfaces,
• the actuating mechanism.
The expanding-ring clutch is often used: in textile machinery, excavators, and machine tools.
In braking systems, the internal-shoe or drum brake is used mostly for automotive applications.
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Force Analysis Let us consider the unit pressure p acting upon
an element of area of the frictional material
located at an angle from the hinge A.
We designate the maximum pressure by pa
located at the angle a from the hinge pin A.
The mechanical arrangement permits no pressure to be applied at the heel (point A) the pressure at this point is assumed to be zero.
In some designs the hinge pin is made movable to provide additional heel pressure.
In this case, as long shoe, the uniform distribution of pressure is not valid anymore !
p
Figure 16–5 The geometry associated with an arbitrary point on the shoe.
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The following assumption are implied by the following analysis:
1. The pressure at any point on the shoe is assumed to be proportional to distance from the hinge pin, being zero at the heel.
2. The effect of centrifugal force has been neglected. In the case of brakes, the shoes are not rotating, and no centrifugal force exists. In clutch design, the effect of centrifugal force must be considered in writing the equations of static equilibrium.
3. The shoe is assumed to be rigid. Since this cannot be true, some deflection will occur, depending upon the load, pressure, and stiffness of the shoe. The resulting pressure distribution may be different from that which has been assumed.
4. The entire analysis has been based upon a constant coefficient of friction (does not vary with pressure). Actually, the coefficient may vary with a number of conditions including temperature, wear, and environment.
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Step 1 Make the assumption that the pressure at any point is proportional to the vertical distance from the hinge pin.
This vertical distance is proportional to sin ≈ sin ⇒
sin =
Step 2
To find the pressure distribution on the periphery of the internal shoe, consider point B on the shoe.
As in Ex. 16–1, if the shoe deforms by an infinitesimal rotation Δ about the pivot point A, deformation perpendicular to B is Δ.
From triangle AOB, = 2 sin( Τ 2), so
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The deformation perpendicular to the rim is Δ cos Τ 2 , which is
Thus, the deformation, and consequently the pressure, is proportional to sin .
Δ cos Τ 2 = 2Δ sin Τ 2 cos Τ 2 = sin

sin sin or
In terms of the pressure at B and where the pressure is a maximum, this means
(a) (1)
The useful characteristics of the previous pressure distribution are:
The pressure distribution is sinusoidal.
If the shoe is short, the largest pressure on the shoe is occurring at the end of the shoe, .
If the shoe is long, the largest pressure on the shoe is occurring at = 90° .
In choosing friction material, the designer should think in terms of and not about the amplitude of the sinusoidal distribution that addresses locations off the shoe.
Figure 16–6
Figure 16–7 Forces on the shoe
Drum’s rotation
Step 3
At the angle from the hinge pin, there acts a differential normal force dN whose magnitude is.
where b is the face width of the friction material.
= (b)
sin (c)
Substituting the value of the pressure, eq. (1), we find
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sin
Step 3 (continued)
The normal force, dN and other forces act on the friction material can be decomposed into horizontal and vertical component, as shown on Fig. 16–7.
The actuating force F can be found by using the condition that the summation of the moments about the hinge pin is zero.
Figure 16–7 Forces on the shoe
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Figure 16–7 Forces on the shoe
(3) = sin =
sin
sin2
The moment arm of the normal force about the pin is sin . Designating the moment of the normal forces by and summing these about the hinge pin give
= −
The actuating force F must balance these two moments:
The frictional forces have a moment arm about the
pin of − cos . The moment of these frictional forces is:
= − cos =
sin
(2)
Step 3 (continued)
If we make MN = Mf , self-locking is obtained, and no actuating force is required.
To obtain self-energizing condition, the dimension a in figure must be such that
(5) >
The torque applied to the drum by the brake shoe is the sum of the frictional forces times the radius of the drum:
= =
(6)
Step 3 (continued)
The hinge pin reactions are found by taking a summation of the horizontal and vertical forces, thus we have
(d) = cos − sin − =
sin − −
(e) = sin + cos − =
sin + −
=
2 sin2
2 − 1
Step 3 (continued)
The direction of the frictional forces is reversed if the rotation is reversed. Thus, for counterclockwise rotation of drum, the actuating force become
= +
(7)
Since both moments have the same sense, the self-energizing effect is lost and also self-locking.
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Step 3 (continued)
=
(f) = cos + sin −
=
For counterclockwise
rotation
Also, for counterclockwise rotation the signs of frictional terms in the equilibrium for the pin reactions change, and equation (d) and (e) become:
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Figure 16–8 Brake with internal expanding shoes; dimensions in millimeters.
EXAMPLE 16–2
The brake shown in Fig. 8 has 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. (c) Hinge-pin reactions.
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Known
1 = 0° 2 = 126° = 90°
= 1122 + 502 = 122.7
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The moment of the frictional force :
=
2 − 1

2 sin2126°
= 304
=
(b) Braking capacity.
sin 90 = 366
The torque contributed by the left-hand shoe cannot be obtained until we learn its maximum operating pressure. Equations (16–2) and (16–3) indicate that the frictional and normal moments are proportional to this pressure. Thus, for the left-hand shoe,
= 788 1000
= 304 1000
2.28 =
sin 90 = 162
= +
Left shoe:
(c) Hinge-pin reactions. Right shoe:
= 1
2 sin2
= − − = −1.410
= + − = 4.839
= 2 +
(c) Hinge-pin reactions. Left shoe:
= 1
2 sin2
= + − = 0.678
= − − = 0.538
= 2 +
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The patented clutch-brake of figure has external contracting friction elements, but the actuating mechanism is pneumatic.
The mechanism can be classified as:
1. Solenoids
4. Hydraulics and pneumatic devices
Figure 16–10 An external contracting clutch-brake
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The notation for external contracting shoes is shown in Figure 11.
The moments of friction and normal forces about the hinge pin are the same as for the internal expanding shoes.
Equations (2) and (3) apply and repeated here for convenience:
Force Analysis
(3)
sin
sin2
The moment of the normal forces by and summing these about the hinge pin give
The moment of frictional forces is:
= − cos =
sin
(2)
(3) =
sin2
The moment of the normal forces by and summing these about the hinge pin give
The moment of frictional forces is:
=
sin − cos (2)
Both these equations give positive values for clockwise moments (Fig. 16–11) when used for external contracting shoes. The actuating force must be large enough to balance both moments:
= +
The horizontal and vertical reactions at the hinge pin are:
(d) = cos + sin − =
sin + −
(e) = cos − sin − =
sin − −
=
2 sin2
2 − 1
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If the rotation is counterclockwise, the sign of the frictional term in each equation is reversed. Thus equation for actuating force becomes
= −
(14b) =
(14a) =
The horizontal and vertical reaction are found to be:
•When external contracting designs are used as clutches, the effect of centrifugal force is to decrease the normal force. Thus, as the speed increases, a larger value of the actuating force is required.
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dN f dN
A special case arise when the pivot is symmetrically located and also placed so that the moment of the friction forces about the pivot is zero. The geometry of such a brake will be similar to that of Fig. 12 (a).
Figure 16–12 (a) Brake with symmetrical pivoted shoe; (b) wear of brake lining.
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• To get a pressure-distribution relation, we assume that the lining wear is such to retain its cylindrical shape, much as a milling machine cutter feeding in the x direction would do to the shoe held in a vise, see Fig. 12 (b).
• This means the abscissa component of wear is 0 for all positions . If wear in the radial direction is expressed as (), then.
() = 0 cos
• The radial wear () can be expressed as
where K is a material constant, P is pressure, V is rim velocity, and t is time
() =
• Denoting as () above and solving for () gives
= ()

• Proceeding to the force analysis, we observe from Fig. 12 (a) that
• Since all elemental surface areas of the friction material see the same rubbing speed for the same duration, Τ0 () is a constant and
where is the maximum value of ().
= cos = cos (c)
=
or (d)
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• The distance a to the pivot is chosen by finding where the moment of the frictional forces Mf is zero.
• First, this ensures that reaction Ry is at the correct location to establish symmetrical wear.
• Second, a cosinusoidal pressure distribution is sustained, preserving our predictive ability. Symmetry means 1 = 2, so
= 2
Substituting Eq. (e) gives
= 4 sin 2
22 + sin 22 from which : (15)
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• The distance a depends on the pressure distribution. Mislocating the pivot makes Mf zero about a different location, so the brake lining adjusts its local contact pressure, through wear, to compensate.
• With the pivot located according Eq. (15), the moment about the pin is zero.
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• Note, too, that = − and = −, as might be expected for the
particular choice of the dimension a.
• Therefore the torque capacity is
= (18)
The horizontal and vertical reactions are
(16) = 2 0
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The block-type hand brake shown in the figure has a face width of 1.25 in and a mean coefficient of friction of 0.25. For an estimated actuating force of 90 lbf, find the maximum pressure on the shoe and find the braking torque.
Q 16–5
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= 62 + 82 = 10
1

dN
fdN

r
=
= (0.25)(1.25)(6)
= 3.728 .
= (1.25)(6)(10)
= −
(90)(2) = (69.405 − 3.728)
= (0.25)(27.4)(1.25) (6)2 cos 8.13° − cos 98.13°
sin 90° = 348.7 . Answer
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and Flywheel Part I
Application, mostly:
• Power excavators
• Hoisting machinery
Broderson IC-80-1D Carry Deck
• Engine: Continental TM27, 4 cyl., 165 C.I.D., 64 hp at governed speed.
• Transmission: Borg Warner, 72T/T18, Reversing
Gearbox and Manual Shift Three Speed Gearbox.
• Brakes: Service - 4-wheel drum type brakes, Parking - Band Type Mounted on Transmission
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( + ) sin
1 > 2
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Forces in horizontal direction gives
( + ) cos
(c)
(d)
Substituting the value of dN from Eq. (b) in (d) and integrating gives

2
=
= 1 − 2
= (e)
Therefore:
=
= 2
(21)
The maximum pressure will occur at the toe with the value =
21
(22)
The pressure is therefore proportional to the tension in the band.
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Q 16–11
The maximum band interface pressure on the brake shown in the figure is 620 kPa. Use a 350 mm diameter drum, a band width of 25 mm, a coefficient of friction of 0.30, and an angle-of-wrap of 270. Find the band tensions and the torque capacity.
Known: = 350 = 100 = 620 = 0.30 = 270°
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Solution
0.350
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09.06. Frictional-Contact Axial Clutches
and Flywheel Part I
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• Mating frictional members are moved in a direction parallel to shaft.
• Most application: automotive
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Figure 16–15 An oil-actuated multiple-disk clutch-brake for operation in an oil bath or spray
Figure 16–14 Cross-sectional view of a single-plate clutch; A, driver; B, driven plate (keyed to driven shaft); C, actuator
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Free from centrifugal effects
Large frictional area which can be installed in small space
More effective heat-dissipation surfaces
Uniform wear
Uniform pressure
=
Uniform wear
After initial wear has taken place and the disks have worn down to the point where uniform wear becomes possible, the greatest pressure must occur at = Τ 2 in order for wear to be uniform. Denoting the maximum pressure by , we can then write. (see the explanation in Shigley, page 847)
Figure 16–16 Disk friction member.(a)
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Figure 16–16 Disk friction member.
Referring to Fig. 16, we have an element of area of radius and thickness . The area of this element is 2 , so that the normal force acting upon this element is = 2. Thus, the total normal force become:
=
Τ 2
2 − (23)
The torque is found by integrating the product of the frictional force and the radius:
=
Τ 2
Uniform Pressure
When uniform pressure can be assumed over the area of the disk, the actuating force is simply the product of the pressure and the area
= 4
2 − 2 (26)
Note: Equations (26) and (28) are for single pair of mating surface. This value must be multiplied by the number of pairs of surfaces in contact.
As before, the torque is found by integrating the product of the frictional force and the radius:
= 2
Figure 16–17 Dimensionless plot of Eqs. (b) and (c).
Difference characteristic of uniform pressure and uniform wear
( ):
Q 16–16
A plate clutch has a single pair of mating friction surfaces 250-mm OD by 175-mm ID. The mean value of the coefficient of friction is 0.30, and the actuating force is 4 kN. a) Find the maximum pressure and the torque capacity using the
uniform-wear model. b) Find the maximum pressure and the torque capacity using the
uniform-pressure model.
Known:
= 250 = 175 = 0.3 = 4
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Solution
a) Find the maximum pressure and the torque capacity using the uniform-wear model.
=
= (4000)(0.3)
4 0.250 + 0.175 = 127.5 Answer
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b) Find the maximum pressure and the torque capacity using the uniform- pressure model.
= 4
= (4000)(0.3)
3
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and Flywheel Part I
• There is no fundamental difference
between a disk clutch and a disk
brake.
and hence is not so susceptible to
changes in the coefficient of friction.
9.6. Disk Brakes
• Fig. 18 shows a floating caliper disk brake.
• The caliper support a single floating piston
actuated by hydraulic pressure.
piston replacing the function of the screw.
• The floating action compensates for wear
and ensures a fairly constant pressure over
the area of the friction pads.
Figure 16–18 An automotive disk brake.
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Figure 16–19 Geometry of contact area of an annular- pad segment of a caliper brake.
Fig. 19 is the geometry of an annular-pad brake contact area. The governing axial wear equation is (see Eq. 12-27, p. 663 - Shigley)
= 12
Of interest also is the effective radius , which is the radius of an equivalent
The coordinate locates the line of action of force F that intersects the y axis.
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If is the local contact pressure, the actuating force is
= 1

(29)
Figure 16–19 Geometry of contact area of an annular- pad segment of a caliper brake.
and the friction torque is
= 1

2
(30)
The equivalent radius can be found from = , or
=
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The locating coordinate of the activating force is found by taking moments about the x axis:
= = 1
2

sin = cos 1 − cos 2

2
(32)
Uniform Wear
Using that pressure distribution = Τ . Eqs. (29) to (32) become
= 2 − 1

= 2 − 1

2 − 2 (34)
= cos 1 − cos 2
2 − 1
Uniform Pressure
In this situation, approximated by a new brake, = , Eqs. (29) to (32) become
= 2 − 1

2 − 2 (37)

3 − 3 (38)
= cos 1 − cos 2
2 − 1
EXAMPLE 16–3
Two annular pads, = 3.875 in, = 5.50 in, subtend an angle of 108°, have a coefficient of friction of 0.37, and are actuated by a pair of hydraulic cylinders 1.5 in in diameter. The torque requirement is 13000 lbf · in. For uniform wear a) Find the largest normal pressure . b) Estimate the actuating force . c) Find the equivalent radius and force location . d) Estimate the required hydraulic pressure.
Known = 3.875 = 5.500 = 0.37 1 = 36° 2 = 144°
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Solution Two annular pads, total torque = 13000 lbf.in. For each pad, = Τ13000 2 = 6500 .
= 1
2 − 2Eq. (34)
a) Find the largest normal pressure for uniform wear.
= 2
2
= 2(6500)
= 2 − 1 − Eq. (33)
= 144° − 36°
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c) Find the equivalent radius and force location for uniform wear.
= + 2
2 − 1 =
Each cylinder supplies the actuating force, 3748 lbf
=
Figure 20 displays the circular pad geometry. Numerical integration is necessary to analyze this brake since the boundaries are difficult to handle in closed form. Table 1 gives the parameters for this brake as determined by Fazekas.
Table 1 Parameters for a Circular-Pad Caliper Brake
Figure 16–20 Geometry of circular pad of a caliper brake.
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Figure 16–20 Geometry of circular pad of a caliper brake.
= (41)
= 2 (42)
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EXAMPLE 16–4
A button-pad disk brake uses dry sintered metal pads. The pad radius is ½ in, and its center is 2 in from the axis of rotation of the 312 -in-diameter disk. Using half of the largest allowable pressure, pmax = 350 psi, find the actuating force and the brake torque. The coefficient of friction is 0.31.
Known
= 0.5 = 2 = 350 = 0.31
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=
Eq. (42)
= = 0.31 106.6 1.926 = 63.65 . Eq. (43) Answer
Solution
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• Consist of a cup and a cone.
• Cone angle, the diameter and face width of the cone are the important geometric design parameters.
• If the cone angle is too small, say, less than about 8o, then the force required to disengage the clutch may be quite large.
• Wedging effect lessens rapidly when larger cone angles are used.
• A good compromise can usually be found between 10o – 15o.
8.7. Cone Clutch and Brakes
Actuating force and torque transmitted can be found using Fig. 22.
Uniform Wear
The pressure relation is the same as for the axial clutch
=
2 (a)
The element area of radius and width Τ sin is
= Τ(2) sin (b)
Figure 16–22 Contact area of a cone clutch.
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Figure 16–22 Contact area of a cone clutch.
As shown in Fig. 22, the operating force will be the integral of the axial component of the differential force . Thus
= sin =
Τ 2
Τ 2

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The differential friction force is , and the torque is the integral of the product of this force with the radius. Thus:
= =

8 sin 2 − 2 (45)
Note that Eq. (24) is a special case of Eq. (45), with = 90o.
Using Eq. (44), the torque can also be written
=
4 sin + (46) Note that Eq. (25) is also a special case of Eq.
(46), with = 90o.
Uniform Pressure
Using = , the actuating force and torque are found to be
= sin =
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As in the case of the axial clutch, we can write Eq. (43) dimensionlessly as
and write Eq. (49) as
This time there are six (T, α, f, F, D, and d) parameters and four pi () terms:
As in Fig. 17, we plot T sin α/( f FD) as ordinate and d/D as abscissa. The plots and conclusions are the same Students are recommended to do the plot.
sin

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A cone clutch has D = 12 in, d = 11 in, a cone length of 2.25 in, and a coefficient of friction of 0.28. A torque of 1800 lbf.in is to be transmitted. For this requirement, estimate the actuating force and pressure by both models.
Q 16–19
Known
= 12 = 11 = 2.25 = 0.28 = 1800 .
CL Not to scale
Solution
Uniform Wear
(1800) = (0.28)(11)
=
Uniform Pressure
3 − 3Eq. (48)
123 − 113 = 13.42 Answer
= 4
2 − 2 = (13.42)
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1 2 3 4 5 6 7 8 9 10 11 12 13 14
Thank You Lecturers
Institut Teknologi Bandung

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