POWER Transmission Questions- mechanical Principles

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : POWER TRANSMISSION

LESSON 3 : FRICTION CLUTCHES

MP - 3 - 3

© Teesside University 2011

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School of Science & Engineering

Teesside University

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

Clutches are one of the most important components in power transmission

systems. Clutches are used in devices with two rotating shafts: one shaft is

typically driven by a motor or pulley, and the other shaft is driving another

device, in order to transmit the power from the electrical device to the

mechanical device. FIGURE 1 shows clutch engagement and disengagement.

FIG. 1 Clutch engagement and disengagement

In a drill, one shaft is driven by a motor and the other is driving a drill chuck.

The clutch connects the two shafts so that they can either be locked together

and spin at the same speed, or be decoupled and spin at different speeds.

In a car, you need a clutch because the engine spins all the time and the car

wheels do not. In order for a car to stop without switching off the engine, the

wheels need to be disconnected from the engine somehow. The clutch allows

us to smoothly engage a spinning engine to a non-spinning transmission by

controlling the slippage between them. Also, the clutch helps to disengage the

engine from the manual gearbox (transmission) for changing of the gears.

Driven member

Flywheel driving member

Clutch engaged

Flywheel driving member

Clutch disengaged

Driven member

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The amount of force the clutch can hold depends on the friction between the

driving and driven members. In this lesson, therefore, to understand how a

clutch works, it helps to revise the basics of dry friction. Then, we will apply

the constant wear theory and the constant pressure theory to disc clutches and

cone clutches.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• understand how a flat plate and a conical clutch work

• explain the constant wear theory and the constant pressure theory

• determine the maximum power transmitted by a clutch.

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REVISION OF DRY FRICTION ________________________________________________________________________________________

Consider an object sliding on a surface, as shown in FIGURE 2(a).

FIG. 2

The force balance on the free body is shown in FIGURE 2(b). The pressing

force P, exerted normally, keeps the object surface and the support surface

together, so the support gives a reaction R on the object, i.e. R = P. When the

object is sliding, there will be a frictional force F acting on the object, which

is equal to the force required in the opposite direction to produce the

movement when the velocity is constant. The magnitude of this frictional

force can be determined by Coulomb's law of friction:

where F = the frictional force (N)

R = the reaction of the surface (N)

P = the pressing force (N)

µ = the coefficient of friction.

F R P= =µ µ

P

v

R

(a) (b)

F

P

W

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Thus, the frictional force F depends on the pressing force or pressure exerted

on the object, and the coefficient of friction, which is related to the materials of

the surface and the object.

There are two theories concerning the torque required to produce slip between

two contact surfaces:

(i) constant pressure theory – assume the pressure is uniformly

distributed over the surface;

(ii) constant wear theory – assume a uniform rate of wear even if the

pressure is not evenly distributed.

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CONE CLUTCHES ________________________________________________________________________________________

A cone clutch uses two conical (or cone-shaped) surfaces to transmit friction

and torque as shown in FIGURE 3.

FIG. 3 The clone clutch A mounted in an automobile overdrive for

(a) engagement and (b) disengagement

In the case shown in FIGURE 3(a), the hydraulically operated piston I acting

against bridge piece J moves forward, which causes the cone clutch A to engage

the brake ring B with sufficient load to hold the sun gear G at rest. Thus, the

planet carrier D can rotate with the input shaft H causing the planet gear F to

rotate, which then drives the annulus E rotating at a faster speed than the input

shaft, this being allowed by the free-wheeling action of the unidirectional clutch

C. The overdrive is engaged now as the power is transmitted from H to E.

(a)

(b)

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When the overdrive is required to be disengaged, the cone clutch holds the

annulus E at rest. To do that, the piston I is moved back to the position as

shown in FIGURE 3(b). The sun gear G is in the working condition. The

power is transmitted form the shaft H to G.

The cone clutch transfers torque with greater efficiency than the disc clutch

because it offers a greater amount of surface area.

CLUTCH EQUATIONS

A cone clutch has a half angle of α, called the coning angle. In the case where α = 90° the cone clutch becomes a disc clutch. Therefore, the cone clutch may be thought of as the general case whereas the disc clutch will be a special

case.

The equation for a cone clutch is derived in a similar way to that we have

already used when analysing V-belts. Consider the element ring with a width

dr in a simple cone clutch shown in FIGURE 4, the length of the ring along the

sloping surface is .

The area of this small ring dA is appr oximately the pr oduct of the

circumference 2πr and the width , which is

d d

sin A

r r =

2π α

............................................... 1( )

d sin

r

α

d sin

r

α

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FIG. 4 A simple cone clutch

If p is the normal pressure, then, the normal force dPn acting on this element is

and the corresponding force in the axial direction dP is

d sin d

then d d

P P

P rp r

n=

=

α

2π ............................................. 3( )

d d d

sin P p A p

r r n = =

2π α

................................. 2( )

P α

r r

o

dr

dr

r

α

dr sin α

r

α ri

dP n

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The total clamping force, P, acting on the area of the mating surface in the

axial direction is

Since the frictional force can be expressed by Coulomb’s law of friction, then

we have

The torque transmitted by the element, dT, is

Therefore, the total toque transmitted by the cone clutch is

Constant pressure theory

Now, applying the constant pressure theory, from equation (4), the total

clamping force along the axial direction P is

And, according to equation (7), the total torque that can be transmitted will be

T r p r p

r r r

r

o i i

o

= = ( )⌠ ⌡ ⎮ ⎮ µ α

µ α

2 2 3

2 3 3π πd

sin sin – ........ 9( )

P rp r p r r r

r

o i i

o

= = ( )⌠ ⌡ ⎮ ⎮ 2

2 2π πd – ................... 8( )

T r p r

r

r

i

o

= ⌠

⌡ ⎮ ⎮ µ α

2 2π d sin

..................................... 7( )

d d d

sin T r F

r p r = = µ

α 2 2π

................................ 6( )

d d d

sin F P

rp r n= =µ µ α

2π ................................ 5( )

P rp r r

r

i

o

= ⌠

⌡ ⎮ ⎮ 2π d ........................................... 4( )

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Substituting (8) into (9) gives:

Constant wear theory

If we apply the constant wear theory to the cone clutch, the wear rate is

considered as a constant. This rate is proportional to the pressure p exerted on

the contact surface, and to the rubbing velocity v. Since v depends on the

radius r, then we can say that the wear rate (constant) is proportional to the

product of the pressure p and the radius r. After introducing a constant of

proportionality c, we have

Substituting (11) into (4) and (7), respectively, gives

and T r p r cr r

r

r

r

r

i

o

i

o

= =

=

⌠

⌡ ⎮ ⎮

⌠

⌡ ⎮ ⎮µ α

µ α

2 22π πd sin

d sin

ππµ α c

r ro isin 2 2– ..........................( ) ........... 13( )

P rp r c r

c r r

r

r

r

r

o i

i

o

i

o

= =

= ( )

⌠

⌡ ⎮ ⎮

⌠

⌡ ⎮ ⎮2 2

2

π π

π

d d

– ......................................... 12( )

pr c= ........................................................... 11( )

T P r r

r r o i

o i

= ⎛ ⎝⎜

⎞ ⎠⎟

2 3

3 3

2 2

µ αsin

–

– ................................. 10( )

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Rearranging equation (12)

Substituting (14) into (13), the constant c can be replaced by the force acting

on the axial direction, P, thus

It is known that the greatest pressure on the friction surface of the cone clutch

occurs at the inside radius ri. Since pr = c and p = pmax at r = ri, then we

may write

p r cimax = .......................................................... 16( )

T c

r r

P

r r r r

o i

o i o i

= ( )

= ( ) (

π

π π

µ α

µ

α

sin

sin

2 2

2 2 2

–

– – ))

= ⎛ ⎝⎜

⎞ ⎠⎟

= ( )

µ α

µ α

P r r

r r

P r r

o i

o i

o i

2

2

2 2

sin

sin

–

–

– rr r

r r

P r r

o i

o i

o i

+( )⎛ ⎝ ⎜

⎞

⎠ ⎟

= +( )

–

......... µ

α2 sin ........................... 15( )

c P

r ro i = ( )2π – .......................................... 14( )

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Hence, equation (13) also can be expressed as

In the above equations

F = the frictional force (N)

P = the clamping force along the axial direction (N)

Pn = the normal force acting on the surface of the cone clutch due to

the normal pressure (N)

T = the total toque transmitted by the clutch (N m)

c = the constant of proportionality (N m–1)

p = the normal pressure on the friction surface of the clutch (Pa)

pmax = the maximum normal pressure (Pa)

r = the radius (m)

ri = the internal radius (m)

ro = the outer radius (m)

α = the coning angle µ = the coefficient of friction.

T c

r r

p r r r

o i

i o i

= ( )

= ( )

π

π

µ α

µ α

sin

sin max

2 2

2 2

–

– ................................ 17( )

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Example 1

A conical clutch has a coning angle of 60°. The outer and inner diameters are

100 mm and 20 mm respectively. Calculate the force required in the axial

direction in order to transmit 300 W at 600 rpm. The coefficient of friction is

0.3. Apply both the constant wear theory and the constrant pressure theory.

Solution

r

r

o

i

= ×

=

= ×

=

100 10 2

0 05

20 10 2

0 01

3

3

–

–

.

.

m

m

The powerr transmitted W

where

= =

= ×

T ω

ω

300

2 600 6

π 00

20=

∴

π rad s

the torque transmitted i

1–

ss

N m

T = =

=

300 300 20

4 775

ω π

.

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Applying constant wear theory

Applying constant pressure theory

T P r r

r r

P T

o i

o i

= ⎛ ⎝⎜

⎞ ⎠⎟

∴ =

2 3

3

3 3

2 2

µ αsin

–

–

ssin

sin

α

µ2

3 4 775 60

2

3 3

2 2

r r

r r o i

o i

–

–

.

⎛ ⎝⎜

⎞ ⎠⎟

= × × °

×× × ⎛ ⎝⎜

⎞ ⎠⎟

=

0 3 0 05 0 01 0 05 0 01

400 19

3 3

2 2 .

. – .

. – .

. N

T P

r r

P T

r r

o i

o i

= +( )

= +(

µ α

α µ

2

2

sin

then sin

))

= × × °

× +( )

=

2 4 775 60 0 3 0 05 0 01

459 47

. . . .

.

sin

N

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________________________________________________________________________________________

DISC CLUTCHES ________________________________________________________________________________________

FIGURE 5 shows an example of a disc clutch, in which we can see the two

states of the clutch: (a) disengaged and (b) engaged.

(a) Disengaged Clutch (b) Engaged Clutch

FIG. 5 Simple disc clutch

Similar to the cone clutch, the principle of operation is dependent on the

frictional force created between the flat discs in the clutch, which are pressed

together when an engagement is required. To apply the pressure on the discs,

we can use different methods, e.g. mechanical pressing or hydraulic pressing.

As already mentioned, the disc clutch is a special case of the cone clutch, with

the coning angle α = 90°. Therefore, the equations for the cone clutch can all be applied to the disc clutch with the condition that sin α = sin 90° = 1.

Pressure plate

Sprocket drum

Starter nut

Fixed plate

Springs

Levers

Drive hub Levers, turned out by centrifugal force, push the pressure plate forward to clamp the friction disc to the fixed plate

Friction disc

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Now, for p = constant (constant pressure theory) and sin α = 1, equation (10) becomes

For pr = c = constant (constant wear theory) and sin α = 1, equations (15) and (17) become the following expressions, respectively,

Clutch discs are designed with a special ratio of in order to maximise the

torque transmitted. It can be proved that the optimal ratio is

In engineering design, the ratios are approximately in the range of 0.5 ~ 0.8.

Note that equations (18) to (20) are applied to a single friction surface only.

In order to obtain more effective operation, multiple discs may be used to

compensate for the inevitable drop in the coefficient of friction over time,

when one set of flat plates is forced together with a mechanism.

r

r i

o

= 0 58.

r

r i

o

T P

r ro i= +( ) µ 2

..............................................

– .....

19

2 2

( )

= ( )T p r r ri o iπµ max ............................... 20( )

T P

r ro i= +( ) µ 2

.............................................. 18( )

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FIGURE 6 shows a hydraulically operated multi-disc clutch, in which the

driving discs rotate with the input shaft and the clutch is engaged using

hydraulic pressure. The driven discs are forced to rotate together with the

driving discs, so the power is transmitted from the driving shaft to the driven

shaft. In the state of disengagement, the discs separate themselves as the

hydraulic pressure is released.

FIG. 6 A hydraulically operated multi-disc clutch

From FIGURE 6, we can see that three driving discs (including the backing

plates) provide four friction surfaces, whereas two driven discs provide four

friction surfaces. Therefore, the maximum torque is increased 4 times. Note

that we need five discs to get four friction surfaces.

Drive shafts keyed to rotating assemblies

Three driving disks (four

friction surfaces)

Hydraulically actuated piston

Driving

Oil chamber

Piston seal assembly

Two driven disks (four friction surfaces)

Driven r

o r

i

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In the general case, if n is the number of friction surfaces, then, the disc

number will be n + 1. Thus, in a multi-disc clutch with n friction surfaces,

the maximum torque is increased n times. These are

for p = constant (and sin α = sin 90° = 1)

for pr = c = constant

Example 2

A multi-disc clutch is required to be able to transmit 200 N m torque. The

external and internal diameters of the friction disc are 120 mm and 50 mm

respectively. If the coefficient of friction in the rubbing surfaces is 0.3 and

pmax = 1 MPa, determine the total number of discs required and the axial

clamping force.

T nP

r ro imax = +( ) µ

2 ........................................... 22( )

=or max maxT np r riπµ oo ir2 2 23– ................................( ) (( )

T nP r r

r r o i

o i max =

⎛ ⎝⎜

⎞ ⎠⎟

2 3

3 3

2 2 µ

–

– ..................................... 21( )

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Solution

Apply the constant wear theory pmaxri = c

From the question, T = 200 N m, µ = 0.3, pmax = 1 MPa = 1 × 106 Pa,

then

Since n must be a whole number, then the nearest n above is n = 3, which

means that three friction surfaces are required. Thus, the disc number should

be 4.

Since

then

Therefore, we require 4 discs and a clamping force of 5228.8 N in the axial

direction.

T nP

r r

P T

n r r

o i

o i

= +( )

= +( )

= ×

× × ×

µ

µ

2

2

2 200

0 3 3 60 10. –– –

.

3 325 10

5228 8

+ ×( )

= N

n = × × × × × × ×( ) ×

200

0 3 1 10 25 10 60 10 25 106 3 3 2 3π . –– – –(( )( )

=

2

2 85.

r ri o= = × = = × 50 2

120 2

mm 25 10 m, mm 60 10 m3 3– –

T np r r r

n T

p r r r

i o i

i o

= ( )

∴ =

π

π

µ

µ

max

max

2 2

2

–

– ii 2( )

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NOTES ________________________________________________________________________________________

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SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. The following data is for a cone clutch:

inside diameter 60 mm

outside diameter 120 mm

coefficient of friction 0.25

clamping force in the axial direction 1 kN

coning angle 75°

rotation speed 1000 rpm

Calculate the torque and power that the clutch can transmit without

slipping using both p = constant and pr = constant conditions.

2. A multi-disc clutch consists of 5 discs, each with an internal and external

diameterof 120 mm and 200 mm respectively. The axial clamping force

exerted on the discs is 1.4 kN. If the coefficient of friction between the

rubbing surfaces is 0.25 and uniform wear conditions may be assumed,

determine the torque that can be transmitted by the clutch.

3. A multi-plate clutch must transmit 20 kW of power at 4000 rpm. The

coefficient of friction is 0.35. The inner and outer diameters are 80 mm

and 120 mm respectively. The axial clamping force applied is 500 N.

Determine the number of discs required using the constant pressure

theory.

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NOTES ________________________________________________________________________________________

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ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. From the question:

Using p = constant theory:

The torque transmitted

The power transmitted =

= ×

=

T ω

12 08 104 72

1264

. .

.883 W

T P r r

r r o i

o i

= ⎛ ⎝⎜

⎞ ⎠⎟

= × × ×

2 3

2 0 25 1 10

3 3

2 2

3

µ αsin

–

–

. 33 75

60 10 30 10

60 10 30

3 3 3 3

3 2sin ° ×

×( ) ×( ) ×( )

– –

–

–

– ××( )

=

10

12 08

3 2–

. N m

r ri o= = × = = ×

=

60 2

120 2

mm 30 10 m, mm 60 10 m,

0.25

3 3– –

µ ,, kN 1 10 N, 75

rad

3P = = × = °

= ×

=

1

2 1000 60

104 72

α

ω

,

. π

s 1–

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Using pr = constant theory:

The torque transmitted

The power transmitted =

= ×

=

T ω

11 65 104 72

1219

. .

.666 W

T P

r ro i= +( )

= × ×

× ° × ×

µ α2

0 25 1 10 2 75

60 10 3

3

sin

sin . – ++ ×( )

=

30 10

11 65

3–

. N m

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2. Since the uniform wear conditions are assumed, then the torque

transmitted is:

From the question:

3. Using the constant pressure theory the torque that can be transmitted is:

The number of friction surfaces is

n T

P r r

r r o i

o i

= ⎛ ⎝⎜

⎞ ⎠⎟

3

2 3 3

2 2 µ

–

–

T nP r r

r r o i

o i

= ⎛ ⎝⎜

⎞ ⎠⎟

2 3

3 3

2 2 µ

–

–

∴ = × × ×

× × + ×( ) T 0 25 4 1 4 10 2

100 10 60 10 3

3 3. . – –

== 112 N m

r r

P

o i= = × = = ×

=

200 2

120 2

1

mm 100 10 m, mm 60 10 m,3 3– –

.44 1 5 4kN 1.4 10 N, 0.25, then3= × = + = =µ n n

T nP

r ro i= +( ) µ

2

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From the question, the power transmitted is 20 kW, then

Since

then

Thus, the nearest whole number n above is n = 6. The disc number is

n + 1 = 7. We require 7 discs in the clutch to work together to get the

power of 20 kW transmitted.

n = ×

× × × ×( ) ×( )

3 47 75

2 0 35 500 60 10 40 10

60

3 3 3 3

.

. –– –

××( ) ×( )

=

10 40 10

5 38

3 2 3 2– ––

.

r ro i= = × = = ×

=

120 2

80 2

mm 60 10 m, mm 40 10 m,

0.35

3 3– –

µ aand NP = 500

20 10

2 4000 60

60 20 10 2 4

3

3

× =

= × ×

∴ = × ×

×

T

T

T

ω

π

π

0000

47 75= . N m

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson, we have applied two different theories to cone and disc friction

clutches. These are: (i) constant pressure theory; (ii) constant wear theory.

It can be seen that the study of this lesson is actually an extension of our

previous work on friction theory, angular motion, etc.

In the next lesson, we will move to our investigation on to another mechanical

component – the gear.

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