modul 09 clutches, brakes, couplings and flywheel part i
TRANSCRIPT
1 MS3111 Elemen MesinMAK ยฉ 2021
Teknik Mesin - FTMD ITB
09.01. Introduction
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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Modul 09 Clutches, Brakes, Couplings and Flywheel Part I
IntroductionSegment
1
Static Analysis of Clutches
and Brakes
Segment
2
Internal Expanding Rim
Clutches and Brakes
Segment
3
External Contracting Rim
Clutches and Brakes
Segment
4
Band-Type Clutches and
Brakes
Segment
5
Frictional-Contact Axial
Clutches
Segment
6
Disk BrakesSegment
7
Cone Clutches and Brakes
Segment
8
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09.01. Introduction
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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Representation
๐2๐1
๐ผ1๐ผ2
Brakes or Clutch ๐ = angular velocity
๐ผ = Inertia
๐i, ๐i
Flywheel
๐ผ, ๐
๐o, ๐o
๐ = Torque๐ = angular displacement
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Performance Analysis
1. The actuating force (gaya aktivasi)
2. The torque transmitted (torsi diteruskan)
3. The energy loss (rugi-rugi)
4. The temperature rise (kenaikan temperatur)
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Various Types of Devices
1. Rim types with internal expanding shoes
2. Rim type with external contracting shoes
3. Band types
4. Disk or axial types
5. Cone types
6. Miscellaneous types
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09.02. Static Analysis of Clutches and Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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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 16โ2 A common doorstop.
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โข A normal pressure distribution ๐(๐) is shown under the friction pad as a function of position ๐ข, taken from the right edge of the pad.
โข Distribution of shearing frictional traction is on the surface, of intensity ๐๐(๐ข), in the direction of the motion of the floor relative to the pad, where ๐is the coefficient of friction.
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The net force in the y-direction:
The moment about C from the pressure:
๐ = ๐ค2เถฑ0
๐ค1
๐ ๐ข ๐๐ข = ๐๐๐ฃ๐ค1๐ค2
๐ค2เถฑ0
๐ค1
๐ ๐ข ๐ข๐๐ข = เดค๐ข๐ค2เถฑ0
๐ค1
๐ ๐ข ๐๐ข = ๐๐๐ฃ๐ค1๐ค2 เดค๐ข
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We sum the forces in the x-direction to obtain
๐น๐ฅ = 0 ๐ ๐ฅ โ๐ค2เถฑ0
๐ค1
๐๐ ๐ข ๐๐ข = 0
where โ or + is for rightward or leftward relative motion of the floor, respectively.
โข Assuming f constant, solving for Rx gives
๐ ๐ฅ = ยฑ๐ค2เถฑ0
๐ค1
๐๐ ๐ข ๐๐ข = ยฑ๐๐ค1๐ค2๐๐๐ฃ
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โข Summing the forces in the y-direction gives
from which
for either direction.
๐น๐ฆ = 0 โ๐น + ๐ ๐ฆ +๐ค2เถฑ0
๐ค1
๐ ๐ข ๐๐ข = 0
๐ ๐ฆ = ๐น โ ๐ค2เถฑ0
๐ค1
๐ ๐ข ๐๐ข = ๐น โ ๐๐๐ฃ๐ค1๐ค2
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Summing moments about the pin located at A we have
๐๐ด = 0
๐น๐ โ ๐ค2เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข โ ๐๐๐ค2เถฑ0
๐ค1
๐ ๐ข ๐๐ข = 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:
๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข ยฑ ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข
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๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข ยฑ ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข
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:
เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข โ ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข โค 0
from which:
๐๐๐ โฅ1
๐
0๐ค1 ๐ ๐ข ๐ + ๐ข ๐๐ข
0๐ค1 ๐ ๐ข ๐๐ข
=1
๐
๐ 0๐ค1 ๐ ๐ข ๐๐ข + 0
๐ค1 ๐ ๐ข ๐ข๐๐ข
0๐ค1 ๐ ๐ข ๐๐ข
๐๐๐ โฅ๐ + เดค๐ข
๐ where เดค๐ข is the distance of the center of pressure from the right edge of the pad.
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Some remarks:
โขThe conclusion that a self-acting or self-locking phenomenon is present and is independent of our knowledge of the normal pressure distribution ๐(๐ข).
โขOur ability to find the critical value of the coefficient of friction ๐๐๐ is dependent on our knowledge of ๐(๐ข), from which we derive เดค๐ข.
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EXAMPLE 16โ1
The doorstop depicted in Fig. 16โ2a 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 โcrushโ 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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๐ ๐ฅ = ๐๐ค1๐ค2๐๐๐ฃ = 0.4 1 0.75 ๐๐๐ฃ = 0.3๐๐๐ฃ
Solution
Leftward:
๐ ๐ฆ = ๐น โ ๐๐๐ฃ๐ค1๐ค2 = 10 โ 1 0.75 ๐๐๐ฃ = 10 โ 0.75๐๐๐ฃ
๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข ยฑ ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข
๐น =๐ค2
๐เถฑ0
1
๐๐๐ฃ ๐ + ๐ข ๐๐ข + ๐๐เถฑ0
1
๐๐๐ฃ๐๐ข
๐น =๐ค2
๐๐๐๐ฃ๐ เถฑ
0
1
๐๐ข + ๐๐๐ฃเถฑ0
1
๐ข๐๐ข + ๐๐๐๐๐ฃเถฑ0
1
๐๐ข
๐น =๐ค2๐๐๐ฃ๐
๐ + 0.5 + ๐๐
10 =(0.75)๐๐๐ฃ
21.6 + 0.5 + 4(0.4)
๐๐๐ฃ = 7.207 ๐๐ ๐
= 0.3 7.207 = 2.162 ๐๐๐
= 10 โ 0.75 7.207= 4.595 ๐๐๐
Answer
Answer
Answer
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b) Repeat part a for rightward relative movement of the floor.
๐ ๐ฅ = โ๐๐ค1๐ค2๐๐๐ฃ = โ0.4 1 0.75 ๐๐๐ฃ = โ0.3๐๐๐ฃ
๐ ๐ฆ = ๐น โ ๐๐๐ฃ๐ค1๐ค2 = 10 โ 1 0.75 ๐๐๐ฃ = 10 โ 0.75๐๐๐ฃ
๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข ยฑ ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข
๐น =๐ค2
๐เถฑ0
1
๐๐๐ฃ ๐ + ๐ข ๐๐ข โ ๐๐เถฑ0
1
๐๐๐ฃ๐๐ข
๐น =๐ค2
๐๐๐๐ฃ๐ เถฑ
0
1
๐๐ข + ๐๐๐ฃเถฑ0
1
๐ข๐๐ข โ ๐๐๐๐๐ฃเถฑ0
1
๐๐ข
๐น =๐ค2๐๐๐ฃ๐
๐ + 0.5 โ ๐๐
10 =(0.75)๐๐๐ฃ
21.6 + 0.5 โ 4(0.4)
๐๐๐ฃ = 53.33 ๐๐ ๐
= 0.3 53.33 = โ16 ๐๐๐
= 10 โ 0.75 5.33 = โ30 ๐๐๐
Answer
Answer
Answer
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c) Model the normal pressure to be the โcrushโ 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 ๐ฆ1๐1โ๐
=๐
๐1
๐ฆ2๐2โ๐
=๐ + ๐ค1๐2
๐ฆ1 = ๐โ๐ ๐ฆ2 = (๐ + ๐ค1)โ๐
This means that y is directly proportional to the horizontal distance from the pivot point A; that is, ๐ฆ = ๐ถ1๐, where ๐ถ1 is a constant
Assuming the pressure is directly proportional to deformation, then ๐ ๐ = ๐ถ2๐, where ๐ถ2 is a constant. In terms of ๐ข, the pressure is ๐ ๐ข = ๐ถ2 ๐ + ๐ข = ๐ถ2 1.6 + ๐ข
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๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐ + ๐ข ๐๐ข + ๐๐เถฑ0
๐ค1
๐ ๐ข ๐๐ข
๐น =๐ค2
๐เถฑ0
๐ค1
๐ ๐ข ๐๐๐ข + เถฑ0
๐ค1
๐ ๐ข ๐ข๐๐ข + ๐๐เถฑ0
๐ค1
๐(๐ข)๐๐ข
๐ ๐ข = ๐ถ2 ๐ + ๐ข = ๐ถ2 1.6 + ๐ข
๐น =0.75
2เถฑ0
1
๐ถ2 1.6 + ๐ข (1.6)๐๐ข + เถฑ0
1
๐ถ2 1.6 + ๐ข ๐ข๐๐ข + ๐๐เถฑ0
1
๐ถ2 1.6 + ๐ข ๐๐ข
10 = 0.375๐ถ2 1.6 + 0.5 1.6 + (0.8 + 0.3333) + 4(0.4)(1.6 + 0.5)
๐ถ2 = 3.396 ๐๐ ๐/๐๐ ๐ ๐ข = 3.396 1.6 + ๐ข
The average pressure is given by
๐๐๐ฃ =1
๐ค1เถฑ0
๐ค1
๐ ๐ข ๐๐ข =1
1เถฑ0
1
3.396(1.6 + ๐ข)๐๐ข = 3.396 1.6 + 0.5 = 7.132 ๐๐ ๐
The maximum pressure occurs at ๐ข = 1 ๐๐, and is ๐๐ = 3.396 1.6 + 1 = 8.83 ๐๐ ๐
๐ ๐ฅ = 0.3๐๐๐ฃ = 0.3 7.132 = 2.139 ๐๐๐ ๐ ๐ฆ = 10 โ 0.75๐๐๐ฃ = 10 โ 0.75 7.132 = 4.652 ๐๐๐
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 เดค๐ข we need to evaluate two integrations
เถฑ0
๐
๐ ๐ข ๐ข๐๐ข = เถฑ0
1
3.396(1.6 + ๐ข)๐ข๐๐ข = 3.396 0.8 + 0.3333 = 3.849 ๐๐๐
เถฑ0
๐
๐ ๐ข ๐๐ข = เถฑ0
1
3.396(1.6 + ๐ข)๐๐ข = 3.396 1.6 + 0.5 = 7.132 ๐๐๐/๐๐
Thus, เดค๐ข =3.849
7.132= 0.5397 ๐๐ ๐๐๐ โฅ
๐ + เดค๐ข
๐๐๐๐ โฅ
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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09.03. Internal Expanding Rim Clutches and Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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9.2. Internal Expanding Rim Clutches & Brakes
(a) Clutch (b) BrakeFigure 16โ3 (a) internal expanding centrifugal-acting rim clutch (b) internal expanding brake
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โ Depending upon the operating mechanism, such clutches are further classified as: expanding-ring, centrifugal, magnetic, hydraulic, and pneumatic.
โ Internal-shoe rim expanding type consist of 3 elements:
โข the mating frictional surface,
โข the means of transmitting the torque to and from the surfaces,
โข the actuating mechanism.
โ The expanding-ring clutch is often used: in textile machinery, excavators, and machine tools.
โ In braking systems, the internal-shoe or drum brake is used mostly for automotive applications.
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Force Analysisโช 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.
โช We designate the maximum pressure by pa
located at the angle a from the hinge pin A.
โช The mechanical arrangement permits no pressure to be applied at the heel (point A) โthe pressure at this point is assumed to be zero.
โช 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 16โ5 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 1Make the assumption that the pressure at any point is proportional to the vertical distance from the hinge pin.
This vertical distance is proportional to sinโ ๐ โ sin ๐ โ๐
sin ๐= ๐๐๐๐ ๐ก
Step 2
To find the pressure distribution on the periphery of the internal shoe, consider point B on the shoe.
As in Ex. 16โ1, if the shoe deforms by an infinitesimal rotation ฮ๐ about the pivot point A, deformation perpendicular to B is โฮ๐.
โช From triangle AOB, โ = 2๐ sin( ฮค๐ 2), so
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โช The deformation perpendicular to the rim is โฮ๐ cos ฮค๐ 2 , which is
Thus, the deformation, and consequently the pressure, is proportional to sin ๐.
โฮ๐ cos ฮค๐ 2 = 2๐ฮ๐ sin ฮค๐ 2 cos ฮค๐ 2 = ๐โ๐ sin ๐
๐
sin ๐=
๐๐sin ๐๐
๐ =๐๐
sin ๐๐sin ๐or
โช In terms of the pressure at B and where the pressure is a maximum, this means
(a) (1)
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โช The useful characteristics of the previous pressure distribution are:
โขThe pressure distribution is sinusoidal.
โข If the shoe is short, the largest pressure on the shoe is ๐๐ occurring at the end of the shoe, ๐๐ .
โข If the shoe is long, the largest pressure on the shoe is ๐๐ occurring at ๐๐ = 90ยฐ .
โช In choosing friction material, the designer should think in terms of ๐๐ and not about the amplitude of the sinusoidal distribution that addresses locations off the shoe.
Figure 16โ6
๐ =๐๐
sin ๐๐sin ๐
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Figure 16โ7Forces on the shoe
Drumโs rotation
Step 3
โช 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.
๐๐ = ๐๐๐๐๐ (b)
๐ =๐๐
sin ๐๐sin ๐
๐๐ =๐๐๐๐ sin ๐ ๐๐
sin ๐๐(c)
โช Substituting the value of the pressure, eq. (1), we find
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๐ ๐ต
๐
๐๐ cos ๐
๐๐ sin ๐๐๐๐ cos ๐
๐๐๐ sin ๐
๐ญ
๐ญ๐
๐ญ๐
Step 3 (continued)
โช The normal force, dN and other forces act on the friction material can be decomposed into horizontal and vertical component, as shown on Fig. 16โ7.
โช The actuating force F can be found by using the condition that the summation of the moments about the hinge pin is zero.
Figure 16โ7Forces on the shoe
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Figure 16โ7Forces on the shoe
(3)๐๐ = เถฑ๐๐ ๐ sin ๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin2 ๐ ๐๐
The moment arm of the normal force ๐๐ about the pin is ๐ sin ๐. Designating the moment of the normal forces by ๐๐ and summing these about the hinge pin give
๐น =๐๐ โ๐๐
๐(4)
The actuating force F must balance these two moments:
The frictional forces have a moment arm about the
pin of ๐ โ ๐ cos ๐ . The moment ๐๐ of these frictional forces is:
๐๐ = เถฑ๐๐๐ ๐ โ ๐ cos ๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin ๐ ๐ โ ๐ cos ๐ ๐๐
(2)
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Step 3 (continued)
โช If we make MN = Mf , self-locking is obtained, and no actuating force is required.
โช To obtain self-energizing condition, the dimension a in figure must be such that
(5)๐๐ > ๐๐
โช The torque ๐ applied to the drum by the brake shoe is the sum of the frictional forces ๐๐๐ times the radius of the drum:
๐ = เถฑ๐๐ ๐๐ =๐๐๐๐๐
2
sin ๐๐เถฑ
๐1
๐2
sin ๐ ๐๐ =๐๐๐๐๐
2 cos ๐1 โ cos ๐2sin ๐๐
(6)
Braking capacity
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Step 3 (continued)
โช The hinge pin reactions are found by taking a summation of the horizontal and vertical forces, thus we have
(d)๐ ๐ฅ = เถฑ๐๐ cos ๐ โ เถฑ๐๐๐ sin ๐ โ ๐น๐ฅ =๐๐๐๐
sin ๐๐๐ด โ ๐๐ต โ ๐น๐ฅ
(e)๐ ๐ฆ = เถฑ๐๐ sin ๐ + เถฑ๐๐๐ cos ๐ โ ๐น๐ฆ =๐๐๐๐
sin ๐๐๐ต + ๐๐ด โ ๐น๐ฆ
๐ด = เถฑ
๐1
๐2
sin ๐ cos ๐ ๐๐ = เธญ1
2sin2 ๐
๐1
๐2
๐ต = เถฑ
๐1
๐2
sin2 ๐ ๐๐ = เธญ๐
2โ1
4sin 2 ๐
๐1
๐2
where:
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Step 3 (continued)
โช The direction of the frictional forces is reversed if the rotation is reversed. Thus, for counterclockwise rotation of drum, the actuating force become
๐น =๐๐ +๐๐
๐(7)
โช Since both moments have the same sense, the self-energizing effect is lost and also self-locking.
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Step 3 (continued)
(g)
๐ ๐ฆ = เถฑ๐๐ sin ๐ โ เถฑ๐๐๐ cos ๐ โ ๐น๐ฆ
=๐๐๐๐
sin ๐๐๐ต โ ๐๐ด โ ๐น๐ฆ
(f)๐ ๐ฅ = เถฑ๐๐ cos ๐ + เถฑ๐๐๐ sin ๐ โ ๐น๐ฅ
=๐๐๐๐
sin ๐๐๐ด + ๐๐ต โ ๐น๐ฅ
For counterclockwise
rotation
โช 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 16โ8Brake with internal expanding shoes; dimensions in millimeters.
EXAMPLE 16โ2
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
๐ = 32 ๐๐๐ = 0.32๐๐ = 1000 ๐๐๐
๐1 = 0ยฐ๐2 = 126ยฐ๐๐ = 90ยฐ
๐ = 1122 + 502 = 122.7 ๐๐
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๐1 = 0ยฐ
๐2 = 126ยฐ
๐๐ = 90ยฐ
๐๐
๐๐๐
Solution
๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
0
๐2
sin ๐ ๐ โ ๐ cos ๐ ๐๐
The moment of the frictional force :
=๐๐๐๐๐
sin ๐๐โ๐ cos ๐ 0
๐2 โ ๐1
2sin2๐
0
๐2
=๐๐๐๐๐
sin ๐๐๐ โ๐ cos ๐2 โ
๐
2sin2๐2
=(0.32)(1000 ร 103) (0.032)(0.150)
sin 90ยฐ
ร 0.150โ0.150 cos 126ยฐ โ0.1227
2sin2126ยฐ
= 304 ๐๐
(a) Actuating force F.
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The moment of the normal forces
๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin2 ๐ ๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
0
๐2
sin2 ๐ ๐๐ =๐๐๐๐๐
sin ๐๐
๐
2โ1
4sin 2๐
0
๐2
=๐๐๐๐๐
sin ๐๐
๐22โ1
4sin 2๐2
=(1000 ร 103)(0.032)(0.150)(0.1227)
sin 90ยฐ
126ยฐ ร๐
180ยฐ2
โ1
4sin 2(126ยฐ)
= 788 ๐๐
The actuating force is
๐น =๐๐ โ๐๐
๐=788 โ 304
100 + 112= 2.28 ๐๐ Answer (a)
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(b) Braking capacity.
๐๐ =๐๐๐๐๐
2 cos ๐1 โ cos ๐2sin ๐๐
=(0.32)(1000 ร 103)(0.032) (0.150)2 cos 0 โ cos 126ยฐ
sin 90= 366 ๐๐
The torque contributed by the left-hand shoe cannot be obtained until we learn its maximum operating pressure. Equations (16โ2) and (16โ3) indicate that the frictional and normal moments are proportional to this pressure. Thus, for the left-hand shoe,
๐๐ =788๐๐1000
๐๐ =304๐๐1000
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2.28 =
788๐๐1000 +
304๐๐1000
100 + 112
๐น = 2.28 ๐๐
๐๐ = 443 ๐๐๐
๐๐ฟ =๐๐๐๐๐
2 cos ๐1 โ cos ๐2sin ๐๐
=(0.32)(443 ร 103)(0.032) (0.150)2 cos 0 โ cos 126ยฐ
sin 90= 162 ๐๐
The braking capacity is the total torque:
Answer (b)
๐ = ๐๐ + ๐๐ฟ = 366 + 162 = 528 ๐๐
๐น =๐๐ +๐๐
๐Left shoe:
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(c) Hinge-pin reactions.Right shoe:
๐ด = เธญ1
2sin2 ๐
0ยฐ
126ยฐ
= 0.3273
๐ต = เธญ๐
2โ1
4sin 2 ๐
0ยฐ
126ยฐ
= 1.3373
๐ท =๐๐๐๐
sin ๐๐=(1000 ร 103)(0.032)(0.15)
sin 90ยฐ= 4.8 ๐๐
๐น๐ฅ = 2.28 sin 24ยฐ = 0.93 ๐๐
๐น๐ฆ = 2.28 cos 24ยฐ = 2.08 ๐๐
๐ ๐ฅ = ๐ท ๐ด โ ๐๐ต โ ๐น๐ฅ = โ1.410 ๐๐
๐ ๐ฆ = ๐ท ๐ต + ๐๐ด โ ๐น๐ฆ = 4.839 ๐๐
๐ = ๐ ๐ฅ2 + ๐ ๐ฆ
2 = 5.04 ๐๐
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(c) Hinge-pin reactions.Left shoe:
๐ด = เธญ1
2sin2 ๐
0ยฐ
126ยฐ
= 0.3273
๐ต = เธญ๐
2โ1
4sin 2 ๐
0ยฐ
126ยฐ
= 1.3373
๐ท =๐๐๐๐
sin ๐๐=(443 ร 103)(0.032)(0.15)
sin 90ยฐ= 2.13 ๐๐
๐น๐ฅ = 2.28 sin 24ยฐ = 0.93 ๐๐
๐น๐ฆ = 2.28 cos 24ยฐ = 2.08 ๐๐
๐ ๐ฅ = ๐ท ๐ด + ๐๐ต โ ๐น๐ฅ = 0.678 ๐๐
๐ ๐ฆ = ๐ท ๐ต โ ๐๐ด โ ๐น๐ฆ = 0.538 ๐๐
๐ = ๐ ๐ฅ2 + ๐ ๐ฆ
2 = 0.866 ๐๐
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Teknik Mesin - FTMD ITB
09.04. External Contracting Rim Clutches
and Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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9.3. External Contracting Rim Clutches & Brakes
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
2. Levers, linkages, or toggle devices
3. Linkages with spring loading
4. Hydraulics and pneumatic devices
Figure 16โ10An 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
Figure 16โ11Notation of externalcontracting shoes.
(3)
๐๐ = เถฑ๐๐ ๐ sin ๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin2 ๐ ๐๐
The moment of the normal forces by ๐๐ and summing these about the hinge pin give
The moment ๐๐ of frictional forces is:
๐๐ = เถฑ๐๐๐ ๐ โ ๐ cos ๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin ๐ ๐ โ ๐ cos ๐ ๐๐
(2)
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(3)๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin2 ๐ ๐๐
The moment of the normal forces by ๐๐ and summing these about the hinge pin give
The moment ๐๐ of frictional forces is:
๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin ๐ ๐ โ ๐ cos ๐ ๐๐ (2)
Both these equations give positive values for clockwise moments (Fig. 16โ11) when used for external contracting shoes. The actuating force must be large enough to balance both moments:
๐น =๐๐ +๐๐
๐(11)
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The horizontal and vertical reactions at the hinge pin are:
(d)๐ ๐ฅ = เถฑ๐๐ cos ๐ + เถฑ๐๐๐ sin ๐ โ ๐น๐ฅ =๐๐๐๐
sin ๐๐๐ด + ๐๐ต โ ๐น๐ฅ
(e)๐ ๐ฆ = เถฑ๐๐๐ cos ๐ โ เถฑ๐๐ sin ๐ โ ๐น๐ฆ =๐๐๐๐
sin ๐๐๐๐ด โ ๐ต โ ๐น๐ฆ
๐ด = เถฑ
๐1
๐2
sin ๐ cos ๐ ๐๐ = เธญ1
2sin2 ๐
๐1
๐2
๐ต = เถฑ
๐1
๐2
sin2 ๐ ๐๐ = เธญ๐
2โ1
4sin 2 ๐
๐1
๐2
where:
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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
๐น =๐๐ โ๐๐
๐(13)
and self-energization exist for counterclockwise rotation.
(14b)๐ ๐ฆ =๐๐๐๐
sin ๐๐โ๐๐ด โ ๐ต + ๐น๐ฆ
(14a)๐ ๐ฅ =๐๐๐๐
sin ๐๐๐ด โ ๐๐ต โ ๐น๐ฅ
The horizontal and vertical reaction are found to be:
โขWhen 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 ๐น is required.
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dNf 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 16โ12(a) Brake with symmetricalpivoted shoe; (b) wear ofbrake lining.
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โข 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).
โข This means the abscissa component of wear is ๐ค0 for all positions ๐. If wear in the radial direction is expressed as ๐ค(๐), then.
๐ค(๐) = ๐ค0 cos ๐
โข The radial wear ๐ค(๐) can be expressed as
where K is a material constant, P is pressure, V is rim velocity, and t is time
๐ค(๐) = ๐พ๐๐๐ก
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โข Denoting ๐ as ๐(๐) above and solving for ๐(๐) gives
๐ ๐ =๐ค(๐)
๐พ๐๐ก=๐ค0 cos ๐
๐พ๐๐ก
โข Proceeding to the force analysis, we observe from Fig. 12 (a) that
โข Since all elemental surface areas of the friction material see the same rubbing speed for the same duration, ฮค๐ค0 (๐พ๐๐ก) is a constant and
where ๐๐ is the maximum value of ๐(๐).
๐ ๐ = ๐๐๐ ๐ก๐๐๐ก cos ๐ = ๐๐ cos ๐ (c)
๐๐ = ๐๐๐๐๐
๐๐ = ๐๐๐๐ cos ๐ ๐๐
or (d)
(e)
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โข The distance a to the pivot is chosen by finding where the moment of the frictional forces Mf is zero.
โข First, this ensures that reaction Ry is at the correct location to establish symmetrical wear.
โข Second, a cosinusoidal pressure distribution is sustained, preserving our predictive ability. Symmetry means ๐1 = ๐2, so
๐๐ = 2เถฑ
0
๐2
๐๐๐ ๐ cos ๐ โ ๐ = 0
Substituting Eq. (e) gives
2๐๐๐๐๐เถฑ
0
๐2
๐ cos2๐โ๐ cos ๐ ๐๐ = 0
๐ =4๐ sin ๐2
2๐2 + sin 2๐2from which : (15)
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โข The distance a depends on the pressure distribution. Mislocatingthe pivot makes Mf zero about a different location, so the brake lining adjusts its local contact pressure, through wear, to compensate.
โข With the pivot located according Eq.(15), the moment about the pin is zero.
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โข Note, too, that ๐ ๐ฅ = โ๐ and ๐ ๐ฆ = โ๐๐, as might be expected for the
particular choice of the dimension a.
โข Therefore the torque capacity is
๐ = ๐๐๐ (18)
เถฑ๐๐๐ sin ๐ = 0
where, because of symmetry :
เถฑ๐๐ sin ๐ = 0
๐ ๐ฆ = 2เถฑ0
๐2
๐๐๐ cos ๐ =๐๐๐๐๐
22๐2 + sin 2๐2 (17)
The horizontal and vertical reactions are
(16)๐ ๐ฅ = 2เถฑ0
๐2
๐๐ cos ๐ =๐๐๐๐
22๐2 + sin 2๐2
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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 16โ5
Known:
๐ = 1.25 ๐๐๐น = 90 ๐๐๐๐ = 0.25
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๐ = 62 + 82 = 10 ๐๐
๐1
๐ผ
๐1 = 45ยฐ โ ๐ผ
= 45ยฐ โ tanโ16
8= 8.13ยฐ
๐2 = 8.13ยฐ + 90ยฐ = 98.13ยฐ
๐๐ = 90ยฐ
Solution
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dN
fdN
๐
r
๐ cos ๐
๐ โ ๐ cos๐
๐ sin ๐
The moment of frictional force:
๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin ๐ ๐ โ ๐ cos ๐ ๐๐
=(0.25)๐๐(1.25)(6)
sin 90ยฐเถฑ
8.13ยฐ
98.13ยฐ
sin ๐ 6 โ 10 cos ๐ ๐๐
= 3.728๐๐ ๐๐๐. ๐๐
๐๐ =๐๐๐๐๐
sin ๐๐เถฑ
๐1
๐2
sin2 ๐ ๐๐
The moment of normal force:
=๐๐(1.25)(6)(10)
sin 90ยฐเถฑ
8.13ยฐ
98.13ยฐ
sin2 ๐ ๐๐ = 64.405๐๐ ๐๐๐. ๐๐
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๐น๐ = ๐๐ โ๐๐
(90)(2) = (69.405 โ 3.728)๐๐
๐๐ = 27.4 ๐๐ ๐ Answer
(b) Braking capacity.
๐๐ =๐๐๐๐๐
2 cos ๐1 โ cos ๐2sin ๐๐
=(0.25)(27.4)(1.25) (6)2 cos 8.13ยฐ โ cos 98.13ยฐ
sin 90ยฐ= 348.7 ๐๐๐. ๐๐ Answer
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Teknik Mesin - FTMD ITB
09.05. Band-Type Clutches and Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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Application, mostly:
โข Power excavators
โข Hoisting machinery
โข Others
9.4. Band-Type Clutches and Brakes
Broderson IC-80-1D Carry Deck
โข Engine: Continental TM27, 4 cyl., 165 C.I.D., 64 hp at governed speed.
โข Transmission: Borg Warner, 72T/T18, Reversing
Gearbox and Manual Shift Three Speed Gearbox.
โข Brakes: Service - 4-wheel drum type brakes, Parking - Band Type Mounted on Transmission
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(๐ + ๐๐) sin๐๐
2+ ๐ sin
๐๐
2โ ๐๐ = 0
or ๐๐ = ๐๐๐
(a)
(b)
for small angles,
โ) sin๐๐
2=๐๐
2โ)๐๐ ร ๐๐ โ 0
Note:Forces in vertical direction gives:
๐1 > ๐2
Why?
Figure 16โ13Forces on a brake band.
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Forces in horizontal direction gives
(๐ + ๐๐) cos๐๐
2โ ๐ cos
๐๐
2โ ๐๐๐ = 0
or๐๐ โ ๐๐๐ = 0
(c)
(d)
Substituting the value of dN from Eq. (b) in (d) and integrating gives
เถฑ
๐2
๐1๐๐
๐= ๐เถฑ
0
๐
๐๐ โ ln๐1๐2
= ๐๐
and๐1๐2
= ๐๐๐
(19)
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๐๐๐ = ๐๐๐๐๐
The torque may be obtained from the equation
๐ = ๐1 โ ๐2๐ท
2(20)
The normal force dN acting on element is
๐๐ = ๐๐๐๐๐ (e)
Substitution of dN from Eq. (b) gives
Therefore:
๐ =๐
๐๐=2๐
๐๐ท(21)
The maximum pressure will occur at the toe with the value ๐๐ =
2๐1๐๐ท
(22)
The pressure is therefore proportional to the tension in the band.
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Q 16โ11
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 270โฆ. Find the band tensions and the torque capacity.
Known:๐ท = 350 ๐๐๐ = 100 ๐๐๐๐ = 620 ๐๐๐๐ = 0.30๐ = 270ยฐ
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Solution
The band tensions
๐๐ =2๐1๐๐ท
Eq. (22)
๐1๐2
= ๐๐๐Eq. (19)
๐1 =๐๐๐๐ท
2=(620)(0.1)(0.35)
2= 10.85 ๐๐ Answer
๐2 =๐1๐๐๐
=10.85
๐(0.3)(270ยฐร๐
180ยฐ)= 2.64 ๐๐ Answer
The torque capacity.
๐ = ๐1 โ ๐2๐ท
2Eq. (20) = 10.85 โ 2.64
0.350
2= 1.437 ๐๐๐ Answer
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Teknik Mesin - FTMD ITB
09.06. Frictional-Contact Axial Clutches
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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โข Mating frictional members are moved in a direction parallel to shaft.
โข Most application: automotive
9.5. Frictional-Contact Axial Clutches
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Figure 16โ15An oil-actuated multiple-disk clutch-brake for operation in an oil bath or spray
Figure 16โ14Cross-sectional view of a single-plate clutch; A, driver; B, driven plate (keyed to driven shaft); C, actuator
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Advantages of the disk clutch:
Free from centrifugal effects
Large frictional area which can be installed in small space
More effective heat-dissipation surfaces
Favorable pressure distribution
Two methods for the analysis:
โข Uniform wear
โข Uniform pressure
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๐๐ = ๐๐๐
2or ๐ = ๐๐
๐
2๐
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 ๐ = ฮค๐ท 2 in order for wear to be uniform. Denoting the maximum pressure by ๐๐ , we can then write.(see the explanation in Shigley, page 847)
Figure 16โ16Disk friction member.(a)
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Figure 16โ16Disk friction member.
Referring to Fig. 16, we have an element of area of radius ๐ and thickness ๐๐ . The area of this element is 2๐๐๐๐ , so that the normal force acting upon this element is ๐๐น = 2๐๐๐๐๐. Thus, the total normal force become:
๐น = เถฑ
ฮค๐ 2
๐ท/2
2๐๐๐๐๐ = ๐๐๐๐ เถฑ
ฮค๐ 2
๐ท/2
๐๐ =๐๐๐๐
2๐ท โ ๐ (23)
The torque is found by integrating the product of the frictional force and the radius:
๐ = เถฑ
ฮค๐ 2
๐ท/2
2๐๐๐๐2๐๐ = ๐๐๐๐๐ เถฑ
ฮค๐ 2
๐ท/2
๐๐๐ =๐๐๐๐๐
8๐ท2 โ ๐2
๐ =๐น๐
4๐ท + ๐
(24)
(25)
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Uniform Pressure
When uniform pressure can be assumed over the area of the disk, the actuating force ๐น is simply the product of the pressure and the area
๐น =๐๐๐4
๐ท2 โ ๐2 (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:
๐ = 2๐๐๐๐ เถฑ
ฮค๐ 2
๐ท/2
๐2๐๐ =๐๐๐๐24
๐ท3 โ ๐3
๐ =๐น๐
3
๐ท3 โ ๐3
๐ท2 โ ๐2(28)
(27)
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Figure 16โ17Dimensionless plot of Eqs. (b) and (c).
Difference characteristic of uniform pressure and uniform wear
๐๐๐๐๐๐๐๐๐๐๐ (๐๐๐ ๐๐๐ข๐ก๐โ):๐
๐๐น๐ท=1
41 +
๐
๐ท
๐๐๐๐๐๐๐ ๐๐๐๐ ๐ ๐ข๐๐ (๐๐๐ค ๐๐๐ข๐ก๐โ):
๐
๐๐น๐ท=1
3
1 โ๐๐ท
3
1 โ๐๐ท
2
(b)
(c)
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Q 16โ16
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:
๐๐ท = 250 ๐๐๐ผ๐ท = 175 ๐๐๐ = 0.3๐น = 4 ๐๐
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Solution
a) Find the maximum pressure and the torque capacity using the uniform-wear model.
๐น =๐๐๐๐
2๐ท โ ๐Eq. (23)
๐ =๐น๐
4๐ท + ๐
Eq. (25)
๐๐ =2๐น
๐๐ ๐ท โ ๐=
2(4000)
๐(0.175) 0.250 โ 0.175= 194 ๐๐๐ Answer
=(4000)(0.3)
40.250 + 0.175 = 127.5 ๐๐ Answer
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b) Find the maximum pressure and the torque capacity using the uniform-pressure model.
๐น =๐๐๐4
๐ท2 โ ๐2
Eq. (26)
๐ =๐น๐
3
๐ท3 โ ๐3
๐ท2 โ ๐2
Eq. (28)
๐๐ =4๐น
๐ ๐ท2 โ ๐2=
4(4000)
๐ 0.2502 โ 0.1752= 159 ๐๐๐ Answer
=(4000)(0.3)
3
0.2503 โ 0.1753
0.2502 โ 0.1752= 128 ๐๐ Answer
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Teknik Mesin - FTMD ITB
09.07. Disk Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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โข There is no fundamental difference
between a disk clutch and a disk
brake.
โข The analysis of preceding section
applies to disk brakes too.
โข Disk brake has no self-energization,
and hence is not so susceptible to
changes in the coefficient of friction.
9.6. Disk Brakes
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โข Fig. 18 shows a floating caliper disk brake.
โข The caliper support a single floating piston
actuated by hydraulic pressure.
โข The action is like a screw clamp, with the
piston replacing the function of the screw.
โข The floating action compensates for wear
and ensures a fairly constant pressure over
the area of the friction pads.
Figure 16โ18An automotive disk brake.
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Figure 16โ19Geometry 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)
๐ค = ๐1๐2๐พ๐๐๐ก
Of interest also is the effective radius ๐๐ , which is the radius of an equivalent
shoe of infinitesimal radial thickness.
Annular-pad brake contact area.
The coordinate าง๐ locates the line of action of force F that intersects the y axis.
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If ๐ is the local contact pressure, the actuating force ๐น is
๐น = เถฑ๐1
๐2
เถฑ๐๐
๐๐
๐๐๐๐๐๐ = ๐2 โ ๐1 เถฑ๐๐
๐๐
๐๐๐๐
(29)
Figure 16โ19Geometry of contact area of an annular-pad segment of a caliper brake.
and the friction torque ๐ is
๐ = เถฑ๐1
๐2
เถฑ๐๐
๐๐
๐๐๐2๐๐๐๐ = ๐2 โ ๐1 ๐เถฑ๐๐
๐๐
๐๐2๐๐
(30)
The equivalent radius ๐๐ can be found from ๐๐น๐๐ = ๐, or
๐๐ =๐
๐๐น=๐๐๐๐ ๐๐2๐๐
๐๐๐๐ ๐๐๐๐
(31)
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The locating coordinate าง๐ of the activating force is found by taking moments about the x axis:
๐๐ฅ = ๐น าง๐ = เถฑ๐1
๐2
เถฑ๐๐
๐๐
๐๐ ๐ sin ๐ ๐๐๐๐ = cos ๐1 โ cos ๐2 เถฑ๐๐
๐๐
๐๐2๐๐
าง๐ =๐๐ฅ
๐น=
cos ๐1 โ cos ๐2๐2 โ ๐1
๐๐ (32)
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Uniform Wear
Using that pressure distribution ๐ = ฮค๐๐๐๐ ๐. Eqs. (29) to (32) become
๐น = ๐2 โ ๐1 เถฑ๐๐
๐๐ ๐๐๐๐๐
๐๐๐ = ๐2 โ ๐1 ๐๐๐๐ ๐๐ โ ๐๐ (33)
๐ = ๐2 โ ๐1 ๐เถฑ๐๐
๐๐ ๐๐๐๐๐
๐2๐๐ =1
2๐2 โ ๐1 ๐๐๐๐๐ ๐๐
2 โ ๐๐2 (34)
๐๐ =๐
๐๐น=๐๐๐๐ ๐๐๐๐
๐๐2๐๐
๐๐๐๐ ๐๐๐๐
๐๐๐๐
=๐๐ + ๐๐2
(35)
าง๐ =๐๐ฅ
๐น=
cos ๐1 โ cos ๐2๐2 โ ๐1
๐๐ =cos ๐1 โ cos ๐2
๐2 โ ๐1
๐๐ + ๐๐2
(36)
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Uniform Pressure
In this situation, approximated by a new brake, ๐ = ๐๐ , Eqs. (29) to (32) become
๐น = ๐2 โ ๐1 เถฑ๐๐
๐๐
๐๐๐๐๐ =1
2๐2 โ ๐1 ๐๐ ๐๐
2 โ ๐๐2 (37)
๐ = ๐2 โ ๐1 ๐เถฑ๐๐
๐๐
๐๐๐2๐๐ =
1
3๐2 โ ๐1 ๐๐๐ ๐๐
3 โ ๐๐3 (38)
๐๐ =๐
๐๐น=๐๐๐๐ ๐๐๐
2๐๐
๐๐๐๐ ๐๐๐๐๐
=2
3
๐๐3 โ ๐๐
3
๐๐2 โ ๐๐
2(39)
าง๐ =๐๐ฅ
๐น=
cos ๐1 โ cos ๐2๐2 โ ๐1
๐๐ =cos ๐1 โ cos ๐2
๐2 โ ๐1
2
3
๐๐3 โ ๐๐
3
๐๐2 โ ๐๐
2(40)
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EXAMPLE 16โ3
Two annular pads, ๐๐ = 3.875 in, ๐๐ = 5.50 in, subtend an angle of 108ยฐ, 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 ยท in. For uniform weara) Find the largest normal pressure ๐๐ .b) Estimate the actuating force ๐น.c) Find the equivalent radius ๐๐ and force location าง๐.d) Estimate the required hydraulic pressure.
Known ๐๐ = 3.875 ๐๐๐๐ = 5.500 ๐๐๐ = 0.37๐1 = 36ยฐ๐2 = 144ยฐ
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Solution Two annular pads, total torque = 13000 lbf.in.For each pad, ๐ = ฮค13000 2 = 6500 ๐๐๐. ๐๐
๐ =1
2๐2 โ ๐1 ๐๐๐๐๐ ๐๐
2 โ ๐๐2Eq. (34)
a) Find the largest normal pressure ๐๐ for uniform wear.
๐๐ =2๐
๐2 โ ๐1 ๐๐๐ ๐๐2 โ ๐๐
2
=2(6500)
(144ยฐ โ 36ยฐ) ร๐180
(0.37)(3.875)(5.52 โ 3.8752)= 315.8 ๐๐ ๐
Answer
b) Estimate the actuating force ๐น for uniform wear.
๐น = ๐2 โ ๐1 ๐๐๐๐ ๐๐ โ ๐๐Eq. (33)
= 144ยฐ โ 36ยฐ๐
180315.8 3.875 5.500 โ 3.875 = 3748 ๐๐๐
Answer
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c) Find the equivalent radius ๐๐ and force location าง๐ for uniform wear.
๐๐ =๐๐ + ๐๐2
=5.500 + 3.875
2= 4.688 ๐๐Eq. (35)
าง๐ =cos ๐1 โ cos ๐2
๐2 โ ๐1๐๐ =
cos 36ยฐ โ cos 144ยฐ
144ยฐ โ 36ยฐ ร๐180
4.688 = 4.024 ๐๐Eq. (36)
Answer
d) Estimate the required hydraulic pressure.
Each cylinder supplies the actuating force, 3748 lbf
๐โ๐ฆ๐๐๐๐ข๐๐๐ =๐น
๐ด๐=
3748๐4(1.5)2
= 2121 ๐๐ ๐ Answer
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Circular (Button) Pad Caliper Brake
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 16โ20Geometry of circular pad of acaliper brake.
Effective radius, ๐๐ = ๐ฟ๐
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Figure 16โ20Geometry of circular pad of acaliper brake.
Effective radius, ๐๐ = ๐ฟ๐
The effective radius is given by
๐๐ = ๐ฟ๐ (41)
The actuating force is given by
๐น = ๐๐ 2๐๐๐ฃ (42)
and the torque is given by๐ = ๐๐น๐๐ (43)
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EXAMPLE 16โ4
A button-pad disk brake uses dry sintered metal pads. The pad radius is ยฝ 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
๐ = 0.5 ๐๐๐ = 2 ๐๐๐๐๐๐ฅ = 350 ๐๐ ๐๐ = 0.31
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๐
๐=0.5
2= 0.25
By interpolation:
๐ฟ = 0.963๐๐๐๐ฅ
๐๐๐ฃ= 1.290
๐๐ = ๐ฟ๐ = 0.963 2 = 1.926 ๐๐Eq. (41)
๐๐๐ฃ =๐๐๐๐ฅ
1.290=
350
1.290= 135.7 ๐๐ ๐
๐น = ๐๐ 2๐๐๐ฃ = ๐ 0.5 2 135.7= 106.6 ๐๐๐
Eq. (42)
Answer
๐ = ๐๐น๐๐ = 0.31 106.6 1.926 = 63.65 ๐๐๐. ๐๐Eq. (43) Answer
Solution
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Teknik Mesin - FTMD ITB
09.08. Cone Clutches and Brakes
MS3111 - Elemen Mesin
Modul 09Clutches, Brakes, Couplings
and Flywheel Part I
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โข Consist of a cup and a cone.
โข Cone angle, the diameter and face width of the cone are the important geometric design parameters.
โข If the cone angle is too small, say, less than about 8o, then the force required to disengage the clutch may be quite large.
โข Wedging effect lessens rapidly when larger cone angles are used.
โข A good compromise can usually be found between 10o โ 15o.
8.7. Cone Clutch and Brakes
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Actuating force and torque transmittedcan be found using Fig. 22.
Uniform Wear
The pressure relation is the same as for the axial clutch
๐ = ๐๐๐
2๐(a)
The element area ๐๐ด of radius ๐ and width ฮค๐๐ sin๐ผ is
๐๐ด = ฮค(2๐๐๐๐) sin ๐ผ (b)
Figure 16โ22Contact area of a cone clutch.
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Figure 16โ22Contact 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 ๐๐๐ด. Thus
๐น = เถฑ๐๐๐ด sin ๐ผ = เถฑ
ฮค๐ 2
ฮค๐ท 2
๐๐๐
2๐
2๐๐ ๐๐
sin ๐ผ(sin ๐ผ)
๐น = ๐๐๐๐ เถฑ
ฮค๐ 2
ฮค๐ท 2
๐๐ =๐๐๐๐
2๐ท โ ๐ (44)
which is the same result as in Eq. (23)
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The differential friction force is ๐๐๐๐ด, and the torque is the integral of the product of this force with the radius. Thus:
๐ = เถฑ๐๐๐๐๐ด = เถฑ
ฮค๐ 2
ฮค๐ท 2
(๐๐) ๐๐๐
2๐
2๐๐๐๐
sin ๐ผ=๐๐๐๐sin ๐ผ
เถฑ
ฮค๐ 2
ฮค๐ท 2
๐๐๐ =๐๐๐๐๐
8 sin ๐ผ๐ท2 โ ๐2 (45)
Note that Eq. (24) is a special case of Eq. (45), with = 90o.
Using Eq. (44), the torque can also be written
๐ =๐น๐
4 sin ๐ผ๐ท + ๐ (46) Note that Eq. (25) is also a special case of Eq.
(46), with = 90o.
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Uniform Pressure
Using ๐ = ๐๐, the actuating force and torque are found to be
๐น = เถฑ๐๐๐๐ด sin๐ผ = เถฑ
ฮค๐ 2
ฮค๐ท 2
๐๐2๐๐๐๐
sin๐ผsin๐ผ =
๐๐๐4
๐ท2 โ ๐2 (47)
๐ = เถฑ๐๐๐๐๐๐ด = เถฑ
ฮค๐ 2
ฮค๐ท 2
๐๐๐๐2๐๐๐๐
sin๐ผ=
๐๐๐๐12 sin๐ผ
๐ท3 โ ๐3 (48)
(49)๐ =๐น๐
3 sin๐ผ
๐ท3 โ ๐3
๐ท2 โ ๐2
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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, ฮฑ, f, F, D, and d) parameters and four pi () terms:
As in Fig. 17, we plot T sin ฮฑ/( f FD) as ordinate and d/D as abscissa. The plots and conclusions are the same โ Students are recommended to do the plot.
๐ sin ๐ผ
๐๐น๐ท=1 +
๐๐ท
4
๐ sin ๐ผ
๐๐น๐ท=1
3
1 โ๐๐ท
3
1 โ๐๐ท
2
๐1 =๐
๐น๐ท๐2 = ๐ ๐3 = sin ๐ผ ๐4 =
๐
๐ท
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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 16โ19
Known
๐ท = 12 ๐๐๐ = 11 ๐๐๐ฟ = 2.25 ๐๐๐ = 0.28๐ = 1800 ๐๐๐. ๐๐
CLNot to scale
5.5
6.02.25
๐ผ๐ผ = tanโ1
0.5
2.25= 12.53ยฐ
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Solution
๐ =๐๐๐๐๐
8 sin ๐ผ๐ท2 โ ๐2Eq. (45)
Uniform Wear
(1800) =๐(0.28)๐๐(11)
8 sin 12.53ยฐ122 โ 112
๐น =๐๐๐๐
2๐ท โ ๐ =
๐(14.04)(11)
212 โ 1 = 243 ๐๐๐Eq. (44) Answer
๐๐ = 14.04 ๐๐ ๐ Answer
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Uniform Pressure
๐ =๐๐๐๐12 sin ๐ผ
๐ท3 โ ๐3Eq. (48)
1800 =๐(0.28)๐๐12 sin 12.53ยฐ
123 โ 113 ๐๐ = 13.42 ๐๐ ๐ Answer
๐น =๐๐๐4
๐ท2 โ ๐2 =๐(13.42)
4122 โ 112 = 242 ๐๐๐Eq. (47) Answer
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Thank YouLecturers
Faculty of Mechanical and Aerospace Engineering
Institut Teknologi Bandung
Modul 09 Clutches, Brakes, Couplings and Flywheel Part I