Friction Clutches:

The friction clutch is another most important type of clutch. It is used to transmit the
rotary motion of one shaft to another when desired. The axes of the two shafts are
coincident.

The surfaces can be pressed firmly against one against when engaged and the
clutch tends to rotate as a single unit.

Single Plate clutch:

A disc clutch consists of a Clutch plate attached to a hub that has splines cut on it
and which is free to slide a on splines cut on the driven shaft in the axial direction
means parallel to the shaft.

The clutch plate is made of metal generally of steel and has a ring of friction lining
on each side which has large coefficient friction. The engine shaft supports a
Flywheel.

A spring-loaded pressure plate presses the clutch plate firmly against the Flywheel
when the clutch is Engaged. the disengaged position, the springs press the cover
attached to the Flywheel.

Thus both the Flywheel and the pressure plate rotate with the driving shaft. The
movement of the clutch pedal is the movement of the pressure plate through a
thrust bearing.

The pressure plate pull by the release levers and the friction linings on the clutch
plate is in no contact with the pressure plate or the Flywheel. The Flywheel rotates
without driving the clutch plate and thus, the driven shaft.

When we pressed off the foot from the pedal, the pressure on the thrust bearing is
released. As a result, the springs become free to move the pressure plate to bring in
contact with the clutch plate.

The clutch plate slides on the splined hub and is gripped between the pressure plate
and the Flywheel. The friction between the linings on the clutch plate and the
Flywheel on one side and the pressure plate on the other cause the clutch and
hence the driven shaft to rotate.

In case if the resisting torque on the driver shaft more than the torque at the clutch,
a clutch slip will occur.
Multi-Plate clutch:
In the multi-plate clutch, the number of frictional linings and the metal plates is
increased which increases the capacity of the clutch to transmit torque.

Friction rings have splined on outer boundary and Engage with corresponding
splines on the Flywheel. They are free to slide axially. The friction material thus
rotates with the Flywheel and the engine shaft. The number of friction rings depends
upon the torque to be transmitted.

The driven shaft also supports disc on the splines and them which rotate with the
driven shaft and can slide in the axial direction If the adding force on the pedal is
removed. If n is the total number of plates both on the driving and the driven
members, the number of active surfaces will be n-1 because 1 surface will be
common.

Cone Clutch:
In a cone clutch the contact surfaces in the form of cones. In the Engaged position,
the friction surfaces of the two cones are in complete contact due to spring pressure
which will make in touch all the time. When the clutch is engaged.

If F is the axial Force, Fn the normal force and £ the semi–cone angle of the clutch
then for a conical collar with uniform wear theory.
The main disadvantage of the Cone clutch is If the angle of the cone is made
smaller than 200 the male cone tends to adhere in the female cone and it becomes
difficult to disengage the clutch.

Cone clutch is used in low-speed applications. Cone clutch is commonly used in ix

engines and automobiles.

It is also used in very specialist transmissions in racing, rallying, or in extreme off-

road vehicles. Cone clutches are used in powerboats.

Centrifugal Clutch:
Centrifugal Clutches are being increasingly used in automobiles and machines. A
Centrifugal clutch has a driving member consisting of four slipping blocks and blocks
kept by position means of flat springs for this purpose.
As the speed of the shaft increases, the Centrifugal force on the shoe increases.
The condition when centrifugal force more than the resisting force of springs, then
the shoes will move forward and press against the inside of the rim and thus the
torque is transmitted to the rim.
The only clutch is engaged when the motor gets sufficient speed to take up the load
inefficient manner. The outer surfaces of the shoes are lined with some friction
material.
On the input shaft, there are large extension springs, which connect to a clutch
shoe. When driving spins fast enough, the springs extend causing the clutch shoes
to engage the friction face.
When the engine shaft reaches a certain RPM (Revolutions Per Minute), the clutch
activates, working almost as it will gradually increase As the load increases the
R.P.M. drops thereby disengaging the clutch and letting the RPM rise again and
reengaging the clutch.
These results in a fair bit of waste heat, but over a broad range of speeds, it is much
more useful than a direct drive in many applications like mopeds and go-karts, etc.
Weaker spring or heavier shoes will cause the clutch to engage at a lower R.P.M.
while a stronger spring or lighter shoes will cause the clutch to engage at a higher
R.P.M.
Advantages:

 Centrifugal Clutch has less maintenance.

 It is not expensive.
 It does not need clutch pedal because it’s automatic.
 It helps to prevent the engine from stop running.
Disadvantages:

 Due to slipping and friction, there is a loss of power.

 It will transfer a high amount of power.
 It causes overheating problems.
 Its engagement and disengagements depend upon the speed of the driving shaft.

Torque transmitted by plate or disc clutch

The following notations are used in the derivation

T= Torque transmitted by the clutch
P= intensity of axial pressure
r1&r2=external and internal radii of friction faces
µ= co-efficient of friction
Consider an elemental ring of radius r and thickness dr
Friction surface = 2πrdr
Axial force on the dw= pressure *area
= P*2πrdr
Frictional force acting on the ring tangentially at radius r
Fr= µdw=µ*p*2πrdr
Frictional torque acting on the ring T r=Fr*r=µp*2πr*dr*r=2πµpr2dr
Considering uniform pressure

When the pressure is uniformly distributed over the entire area of the friction face, then
the intensity of pressure,

P=W/π[(r1)2-(r2)2]-------------(i)

Where W = Axial thrust with which the contact or friction surfaces are held together.
We have discussed above that the frictional torque on the elementary ring of radius r and
thickness dr is
Tr = 2 πμ.p.r2 dr

Integrating this equation within the limits from r 2 to r1 for the total frictional
torque.
Substituting the value of p from equation (i),

2. Considering uniform wear

Let p be the normal intensity of pressure at a distance r from the axis of the
Clutch. Since the intensity of pressure varies inversely with the distance, therefore

p.r. = C (a constant) or p = C/r

and the normal force on the ring,

We know that the frictional torque acting on the ring,

Total frictional torque on the friction surface,

R = Mean radius of the friction surface = (r1+ r2)/2

CONE CLUTCH
A cone clutch, as shown in Fig. 10.24, was extensively used in automobiles but
now-a-days it has been replaced completely by the disc clutch

It consists of one pair of friction surface only. In a cone clutch, the driver is keyed to the
driving shaft by a sunk key and has an inside conical surface or face which exactly fits
into the outside conical surface of the driven.
Consider a pair of friction surface as shown in Fig. Since the area of contact of a pair
of friction surface is a frustum of a cone, therefore the torque transmitted by the cone clutch
maybe determined in the similar manner as discussed.
Let p n = Intensity of pressure with which the conical friction surfaces are held
together (i.e. normal pressure between contact surfaces),
r 1 and r2 = Outer and inner radius of friction surfaces respectively
R = Mean radius of the friction surface=(r1+ r2)/2
α= Semi angle of the cone (also called face angle of the cone) or the angle of the friction
surface with the axis of the clutch,
μ = Coefficient of friction between contact surfaces, and
b = Width of the contact surfaces (also known as face width or clutch face).
Consider a small ring of radius r and thickness dr, as shown in Fig. 10.25 (b). Let
dl is length of ring of the friction surface, such that
dl = dr.cose α
Area of the ring= A = 2π r.dl = 2πr.dr cosec α

We shall consider the following two cases :

1. When there is a uniform pressure, and
2. When there is a uniform wear.
Considering uniform pressure

We know that normal load acting on the ring,

δW n = Normal pressure × Area of ring = p n × 2 πr.dr.cosec α
The axial load acting on the ring,
δW = Horizontal component of δW n (i.e. in the direction of W)
= δW n × sin α= p n × 2π r.dr. cosec α× sin α = 2π × p n .r.dr
Total axial load transmitted to the clutch or the axial spring force required,

We know that frictional force on the ring acting tangentially at radius r,

Fr = μ.δW n = μ.pn × 2 π r.dr.cosec α
Frictional torque acting on the ring,

T r = Fr× r = μ.p n × 2 πr.dr. cosec α.r = 2 πμ.p n.cosec α.r2 dr

Integrating this expression within the limits from r2 to r1 for the total frictional torque on the
clutch.
Considering uniform wear

In Fig. 10.25, let pr be the normal intensity of pressure at a distance r from the axis
of the clutch. We know that, in case of uniform wear, the intensity of pressure varies inversely
with the distance.
P r .r = C (a constant) or pr = C / r
We know that the normal load acting on the ring,
δW n = Normal pressure × Area of ring = pr × 2πr.dr cosecα
The axial load acting on the ring ,
δW = δW n × sin α = pr.2 π r.dr.cosec α.sin α= pr × 2π r.dr

We know that frictional force acting on the ring,

Fr = μ.δW n = μ.pr × 2πr × dr cosecα

Frictional torque acting on the ring,

T r = Fr × r = μ.pr × 2π r.dr.cosecα × r
SOLVED PROBLEMs

1. A single plate clutch with both sides effective; is required to transmit 25

kw at 900 rpm. t hw outer diameter of the plate is 350 mm. the maximum
intensity of pressure over the friction surface is not to exceed 0.1 N/mm2 .
considering uniform wear criteria and assuming coefficient of friction as
0.25; determine (i) the inner diameter of the plate (ii) axial force required
to engage the clutch.
Given;
P= 25 kw ; N =900 rpm ;2r =350 mm ; P max = 0.1 N/mm2 ; µ = 0.25
Sol:-
I.Inner diameter of the plate

Subtracting the relation (ii) in (i) and using the values,

r2 [(0.175)2 – r2] = 0.001689
0.031r2 - r2 3 = 0.001689 By trial and error method,
r2 = 0.136m
II.Axial force required to engage the clutch (F)
= 2 π* 0.1 *106 * 0.136* (0.175 – 0.136 ) * 2
= 6665.2 N
2. A multiple – disc clutch transmits 50 kw of power at 1400 rom. Axial
intensity of pressure not to exceed 0.12 1 N/mm2, and the coefficient of
friction of the friction surfaces is 0.12. the inner radius of the discs is 80
mm, and is .7 times the outer radius. Determine number of disc required to
transmit the given power. Assume uniform wear condition.
Given;
P = 60 kw; N = 1400rpm; P max = 0.15 N/mm2; µ = 0.12; r2 = 80mm; r1 = r2/
0.7

3. A cone clutch with asbestos friction lining, transmits 25 kw at 600 rpm.

The coefficient is 0.25 and the maximum intensity of pressure is 0.25 N/
mm2. The semi cone angle is 12.50. the outer diameter of friction lining is
250 mm . Considering uniform wear theory. Determine
i. the inner diameter of the friction lining,
ii. the face width of friction lining and
iii.the force required to engage to the clutch.
Given;
P = 25kw; N = 600rpm; P max = 0.25 N/mm2; µ = 0.25; α = 12.50; 2r1 = 250mm
Sol;
