2. • Clutches, brakes, couplings, and flywheels are a group of elements usually
associated with rotation that have in common the function of storing
and/or transferring rotating energy
• In analyzing the performance of these devices we shall be interested in:
Background
• The actuating force
• The torque transmit
• The energy loss
• The temperature rise
2
3. Fig. 2 -A simplified dynamic representation of a friction clutch or brake
• Two inertias, I1 and I2, traveling at the respective angular velocities; ω1and ω2,
are to be brought to the same speed by engaging the clutch or brake.
• Slippage occurs because the two elements are running at different speeds and
energy is dissipated during actuation, resulting in a temperature rise.
3
Cont’d
4. • The varies types of devices to be studied may be classified as
following:
• Rim types with internal expanding shoes
• Rim types with external contracting shoes
• Band types
• Disk or axial types
• Cone types, and
• Miscellaneous types
4
Cont’d
5. Static Analysis of Clutches and Brakes
1. Estimate,or measure the pressure
distribution on the friction surfaces.
2. Find a relationship between the largest
pressure and the pressure at any point.
3. Use the conditions of static equilibrium
to find the braking force or torque and
the support reactions.
Analyzing general procedure:
• The net force in the y direction:
• Moment about C from the pressure:
= =
= =
5
6. • Sum of forces in the x-direction
• Sum of forces in the y-direction
•Sum of moment about the pin located at A:
= ∓ = 0
= ± = ± .
= − + + = 0
= − = − .
= − ( )(! + ) ∓ " = 0
= ( )(! + )) ± "
6
Cont’d
7. • F shall not be equal to or less than zero.
• Critical friction
! + − " ≤ 0
$% ≥
1
"
( ! +
(
=
1
"
! ( + (
(
$% ≥
! +
"
7
Cont’d
8. • Often used in industrial machinery, vehicles and machine tools where the clutch
may be located within the driving pulley.
• Depending upon the operating mechanism, such clutches are further classified
as expanding-ring, centrifugal, magnetic, hydraulic, and pneumatic.
• Expanding ring clutches benefit from centrifugal effects; transmit high torque,
even at low speeds; & require both positive engagement & ample release force.
• The internal-shoe rim clutch as shown in Fig. consists essentially of three
elements:
• The mating frictional surface.
• The means of transmitting the
torque to and from the surfaces.
• The actuating mechanism.
Internal Expanding Rim Clutches & Brakes
8
9. • To analyze an internal-shoe device, Fig. 7 shows a shoe pivoted at point A,
with the actuating force, F.
• Since the shoe is long, we cannot make the assumption that the
distribution of normal forces is uniform.
• The mechanical arrangement permits no pressure to be applied at the heel,
and we will therefore assume the pressure at this point to be zero.
Fig. 7: Internal friction shoe geometry 9
Cont’d
10. • It is the usual practice to omit the friction material for a short distance away
from point A.
• This eliminates interference, and the material would contribute little to
the performance anyway, as will be shown.
• Let us consider the pressure p acting upon an element of area of the frictional
material located at an angle θ from the hinge pin (Fig. 4).
• We designate the maximum pressure pa located at an angle θa from the hinge
pin.
10
Cont’d
11. • If, the shoe deforms by an infinitesimal rotation ∅ about the pivot point A,
To find the pressure distribution on the periphery of the internal shoe,
consider point B,
• Deformation perpendicular to AB = ℎ∆∅.
• From the isosceles ∆ AOB, ℎ = 2 . sin(2 2
⁄ )
• Deformation perpendicular to AB
ℎ∆∅ = 2 .∆∅ sin(2 2
⁄ )
• The deformation perpendicular to the rim is
ℎ ∆∅ !45(2 2
⁄ )
ℎ∆∅ !45(2 2
⁄ ) = 2 . ∆∅ 567 2 2
⁄ !45 2 2
⁄
= . ∆∅ 5672
• The deformation, and consequently the pressure, is proportional to sin θ.
11
Cont’d
12. • In terms of the pressure at B and where the pressure is a maximum, this
means
• The pressure distribution is sinusoidal with respect to the angle θ.
– If the shoe is short, as shown in Fig. 9a, the largest pressure on the
shoe is pa occurring at the end of the shoe, θ2.
– If the shoe is long, as shown in Fig. 9b, the largest pressure on the
shoe is pa occurring at θa = 900
Fig. 9: Defining the angle θa at which the maximum pressure pa occurs when
(a)shoe exists in zone θ1 ≤ θ2 ≤ π/2 (b) shoe exists in zone θ1 ≤ π/2 ≤ θ2
5672
=
5672
⇒ =
5672
5672
12
Cont’d
13. • When θ = 0, Eq. for pressure distribution shows that the pressure is zero.
• The frictional material located at the heel therefore contributes very little to the
braking action and might as well be omitted.
• A good design would concentrate as much frictional
material as possible in the neighborhood of the point
of maximum pressure.
• In Fig. 10 the frictional material begins at an angle θ1,
measured from the hinge pin A, & ends at an angle θ2.
• Any arrangement such as this will give a good
distribution of the frictional material.
• The actuating force F has components Fx and Fy and operates at distance c from
the hinge pin.
Fig. 10: Forces on the shoe
13
Cont’d
14. • Normal force
• The moment Mf of these frictional forces:
• The moment MN of normal force:
• The actuating force F
• When 9 = :, self locking condition (no actuation or self energize)
• Thus, the dimension ‘a’ in above figure must be such that 9 > :
= . 2 ⇒ = < . 5672 2 5672
⁄
: = . − "!452 =
.
5672
5672 . − "!452 2
=>
=
9 = "5672 =
."
5672
567 2 2
=>
=
=
9 − :
!
14
Cont’d
16. • The following assumptions are implied by the preceding analysis:
1. The pressure at any point on the shoe is assumed to be proportional to
the distance from the hinge pin to the heel.
• This should be considered from the standpoint that pressures
specified by manufacturers are averages rather than maxima.
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 this force (centrifugal) must be
considered in writing the equations of static equilibrium.
16
Cont’d
17. 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 coefficient of friction
that does not vary with pressure.
• Actually, the coefficient may vary with a number of
conditions, including temperature, wear, and environment.
17
Cont’d
18. External Contracting Rim Clutches & Brakes
• The moments of the frictional and normal forces about the hinge pin are the
same as for the internal expanding shoes. For CW rotation
Note:
• 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 F is
required.
: =
.
5672
5672 . − "!452 2
=>
=
9 =
."
5672
567 2 2
=>
=
=
9 + :
!
= !452 + 5672 −
= − 5672 + !452 +
18
19. • To get a pressure-distribution relation, we note that lining wear is such as to
retain the cylindrical shape.
• This means the abscissa component of wear is w0 for all positions θ.
• If wear in the radial direction is expressed as w(θ), …& rely on radial wear
equation with respect to pad parameters
• Normal force
K = a material constant
P = pressure
V = rim velocity, and
t = time
Symmetrically Located Pivot
2 = !452 = @ < A B
2 =
2
@AB
=
!452
@AB
C5, @AB
⁄ = constant
2 = (constant) !452 = !452
= . 2 = .!452 2 19
20. • 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.
: = 2 "!452 − . 2
=>
= 0
" =
4.5672
22 + 56722
0 = 2 . "!45 2 − .!452 2
=>
20
Cont’d
21. • Reaction, Rx
• Because of symmetry
• Reaction Ry
• where
• And we shall note too
• Torque
= 2 !452 =
.
2
22 + 56722
=>
5672 = 0
= 2 !452 =
.
2
22 + 56722
=>
5672 = 0
= − "7 = −
? = "
21
Cont’d
22. • Because of friction and the rotation of the drum, the actuating force P2 is less
than the pin reaction P1
• Summation of force in y-direction:
• Summation of force in x-direction:
• Further synthesis on above equations,
• Torque
• Normal Force
Band-Type Clutches and Brakes
< + < 567 2
2
F + <567 2
2
F − = 0
= < 2, … 567!H, 567 2
2
F = 2
2
F
< + < !45 2
2
F − <!45 2
2
F − = 0
< − = 0, 567!H, !45 2
2
F = 1
<
<
= 2
∅
I
I>
4. J7
<
<
= ∅;
<
<
= H:∅
? = (< − < )(L 2
⁄ )
= . 2 = < 2
=
<
.
=
2<
L
⇒ < =
2<
L 22
23. Frictional-Contact Axial Clutches
• An axial clutch is one in which the mating frictional members are moved in a
direction parallel to the shaft.
• Advantages of the disk clutch include
• the more effective heat-dissipation surfaces,
• the large frictional area that can be installed in a small space,
• the favorable pressure distribution.
• the freedom from centrifugal effects,
23
24. • Fig. 15 shows a friction disk having an outer (D) and inner (d) diameter.
• We are interested in obtaining the axial force F necessary to produce a certain
torque T and pressure p.
• Two methods of solving the problem, depending upon the construction of the
clutch, are in general use.
1. If the disks are rigid, then the greatest amount of wear will at first occur in the
outer areas, since the work of friction is greater in those areas.
– After a certain amount of wear has taken place,
the pressure distribution will change so as
to permit the wear to be uniform.
2. Another method of construction employs springs
to obtain a uniform pressure over the area.
Fig. 15 a friction disk 24
Cont’d
25. • After initial wear has taken place and the disks have worn down to a point
where uniform wear is established, the axial wear can be expressed by :
• By definition uniform wear is constant from place to place; therefore,
• The total normal force, F
• The torque is found by integrating the product of the frictional force and
the radius:
• Substituting the value of F
Case -1:Uniform Wear
= @<AB
<A = !475B"7B = M
.N = M
. = MO = P .Q = .Q =
2
= 2R . . = R .
S⁄
T⁄
S⁄
T⁄
=
R
2
(L − )
? = 2R . . = R . .
S⁄
T⁄
S⁄
T⁄
=
R
8
(L − )
? =
4
(L + ) 25
26. • When uniform pressure can be assumed over the area of the disk, the
actuating force F is simply the product of the pressure and the area. This
gives:
• The torque is found by integrating the product of the frictional force and
the radius:
• substituting the value of F and noting p = pa
Case 2: Uniform Pressure
=
R
2
(L − )
? = 2R . . =
S⁄
T⁄
R
12
(LO − O)
? =
3
LO − O
L −
26
27. Summary
27
• Machine systems can operate with intermittent motion. Starting and stopping operations
can cycle frequently.
• Motion control elements permit machine systems to achieve intermittent motion.
Motion Control elements:
• Clutches: Devices used to transmit power on an intermittent basis by connecting and/or
disconnecting a driven component to and/or from the prime mover.
– Motor operates efficiently at continuous speeds.
– Avoids accelerating and/or de-accelerating the rotor of the motor each time a driven
component of a machine needs to be cycled.
– Connect a rapidly turning shaft to one that is initially stationary.
– Cause two shafts to turn at the same speed and to do so in a manner that shock is not
produced.
– Limit torque that is transmitted or to prevent torque from being transmitted in a
reverse direction.
• Brakes: Device that absorbs the kinetic energy of a system and thus controls the motion
of the system by slowing down the system and/or bringing the system to rest.
28. Reference
• Open source
– Richard G. Budynas, J. Keith Nisbett, “Shigley’s
Mechanical Engineering Design,” 9th edition, McGraw-
Hill, New York, America
28
