1. The document discusses different types of clutches including positive clutches and friction clutches. It describes the key components and operation of a single plate clutch commonly used in automotive applications. 2. Formulas are presented for calculating the torque capacity of clutches under uniform pressure and uniform wear conditions based on geometric parameters, pressure, and coefficient of friction. 3. The document provides an example problem demonstrating the use of the formulas to design a multi-plate clutch meeting specific torque and speed requirements.

1. 1
Design of Clutches
A Clutch is a machine member used to connect the driving shaft to a driven shaft, so that
the driven shaft may be started or stopped at will, without stopping the driving shaft. A clutch
thus provides an interruptible connection between two rotating shafts. Clutches allow a high
inertia load to be stated with a small power. A popularly known application of clutch is in
automotive vehicles where it is used to connect the engine and the gear box. Here the clutch
enables to crank and start the engine disengaging the transmission. Disengage the transmission
and change the gear to alter the torque on the wheels. Clutches are also used extensively in
production machinery of all types
Types of Clutches
Clutches are classified into
 Positive Clutches
 Friction clutches
Positive Clutches:
In this type of clutch, the engaging clutch surfaces interlock to produce rigid joint they
are suitable for situations requiring simple and rapid disconnection, although they must be
connected while shafts are stationery and unloaded, the engaging surfaces are usually of jaw
type. The jaws may be square jaw type or spiral jaw type. They are designed empirically by
considering compressive strength of the material used. The merits of the positive clutches are
 Simple
 No Slip
 No heat generated compact and low cost.
Friction Clutches:
Friction Clutches work on the basis of the frictional forces developed between the two or more
surfaces in contact. Friction clutches are usually – over the jaw clutches due to their better
performance. There is a slip in friction clutch. Their merits are
 They friction surfaces can slip during engagement which enables the driver to pickup
and accelerate the load with minimum shock.
 They can be used at high engagement speeds since they do not have jaw or teeth
 Smooth engagement due to the gradual increase in normal force.
The major types of friction clutches are
 Plate clutch (Single plate and multiple plate)
 Cone clutch
 Centrifugal clutch

2. 2
Frictional Contact Disc Clutches
An axial clutch is one in which the mating frictional members are moved in a direction
parallel to the shaft. A typical clutch is illustrated in the figure below. It consists of a driving disc
connected to the drive shaft and a driven disc connected to the driven shaft. A friction plate is
attached to one of the members. Actuating spring keeps both the members in contact and power
or motion is transmitted from one member to the other. When the power of motion is to be
interrupted the driven disc is moved axially creating a gap between the members as shown in the
figure.
Fig.1 A Single Plate Clutch
Method of Analysis
r2
W
Fig. 2 A single friction surface of a clutch disc
Area dA
Lining
dr
r1r
The torque that can be transmitted by a clutch is a function of its geometry and the
magnitude of the actuating force applied as well the condition of contact prevailing between the
members. The applied force can keep the members together with a uniform pressure all over its
contact area and the consequent analysis is based on uniform pressure condition. However as
the time progresses some wear takes place between the contacting members and this may alter or

3. 3
vary the contact pressure appropriately and uniform pressure condition may no longer prevail.
Hence the analysis here is based on uniform wear condition.
For the analysis following terms are defined. Let T is the torque transmitted in N-mm, ‘p’
is the pressure between the contact surfaces in N/mm2, r1 and r2 be the internal and external radii
of the friction plates,  is the coefficient of friction between the plates. Now consider a ring
element of thickness ‘𝑑𝑟’ and area ‘𝑑𝐴’ at a radius ‘r’ from the centre of the disc.
𝑑𝐴 = 2𝜋𝑟𝑑𝑟
The normal force acting on the ring is
𝑑𝑊 = 𝑝2𝜋𝑟𝑑𝑟
The frictional force on the ring is
𝑑𝐹 = 𝜇𝑝2𝜋𝑟𝑑𝑟
The friction torque of the ring element is
𝑑𝑇 = 𝜇𝑝2𝜋𝑟𝑑𝑟 × 𝑟 = 𝜇𝑝2𝜋𝑟2
𝑑𝑟
(a) Uniform pressure
In uniform pressure condition (for a new clutch) pressure is distributed uniformly over the entire
area of the friction plate, so that
𝑝 =
𝑊
𝜋(𝑟2
2−𝑟1
2)
, where W is the axial thrust with which the plates are held together.
The torque transmitted is
𝑇 = ∫ 2𝜋𝜇𝑝𝑟2
𝑑𝑟
𝑟2
𝑟1
(1)
𝑇 = 2𝜋𝜇𝑝 [
𝑟3
3
]
𝑟1
𝑟2
= 2𝜋𝜇𝑝(
𝑟2
3
−𝑟1
3
3
)
𝑇 = 2𝜋𝜇
𝑊
𝜋( 𝑟2
2
−𝑟1
2 )
(
𝑟2
3
−𝑟1
3
3
)
𝑇 = 𝜇𝑊
2
3
(
𝑟2
3
−𝑟1
3
( 𝑟2
2
−𝑟1
2)
) = 𝜇𝑊𝑅 (2)
Where 𝑅 =
2
3
(
𝑟2
3−𝑟1
3
𝑟2
2−𝑟1
2) is called the equivalent radius.
(b) Uniform wear
p p
r1
r2
Uniform pressure Uniform wear
Fig. 3

4. 4
For a used clutch, more wear on outside surface than the inside. The rate of wear depends on the
work done by the friction. If we assume that the coefficient of friction remains constant, wear is
proportional to the product of the normal pressure and velocity of rubbing; thus wear
proportional to the product of ‘𝑝𝑟’. Therefore in uniform wear condition
𝑝𝑟 = 𝐶 (𝑐𝑜𝑛𝑠𝑡𝑠𝑛𝑡)
The torque transmitted given in eqn. (1) is
𝑇 = 2𝜋𝜇 ∫ 𝑝𝑟2𝑟2
𝑟1
𝑑𝑟 = 2𝜋𝜇∫
𝐶
𝑟
𝑟2
𝑟1
𝑟2
𝑑𝑟
𝑇 = 2𝜋𝜇𝐶 ∫ 𝑟
𝑟2
𝑟1
𝑑𝑟 = 2𝜋𝜇𝐶 [
𝑟2
2
]
𝑟1
𝑟2
𝑇 = 2𝜋𝜇𝐶 (
𝑟2
2
−𝑟1
2
2
)
Now at the mean radius R, we have
𝑝𝑅 = 𝐶 𝑜𝑟
𝑊
𝜋( 𝑟2
2
−𝑟1
1)
𝑅 = 𝐶
𝐶 =
𝑊𝑅
2𝜋( 𝑟2− 𝑟1)
( 𝑟2+𝑟1)
2
=
𝑊
2𝜋 ( 𝑟2−𝑟1)
Substituting for ‘C’ in the torque equation gives
𝑇 = 2𝜋𝜇
𝑊
2𝜋( 𝑟2−𝑟1)
(
𝑟2
2
−𝑟1
2
2
)
𝑇 = 2𝜋𝜇
𝑊
2𝜋( 𝑟2−𝑟1)
( 𝑟2− 𝑟1)
( 𝑟2+𝑟1)
2
𝑇 = 𝜇 𝑊
( 𝑟2+𝑟1)
2
(3)
𝑇 = 𝜇 𝑊𝑅
In the above equation ‘𝑅’ is the mean radius or equivalent of the plate in uniform wear condition.
From the equation 𝐶 =
𝑊
2𝜋 ( 𝑟2−𝑟1)
= 𝑝𝑟
𝑝 =
𝑊
2𝜋𝑟( 𝑟2−𝑟1)
(4)
Pressure between the plates is maximum at the inner radius and minimum at the outer
radius and average pressure is at the mean radius.
Single plate dry Clutch – Automotive application
The clutch used in automotive applications (shown in Fig. 4) is generally a single plate
dry clutch. In this type the clutch plate is interposed between the flywheel surface of the engine
and pressure plate. Basically, the clutch needs three parts. These are the engine flywheel, a
friction disc called the clutch plate and a pressure plate (Fig.5). When the engine is running and
the flywheel is rotating, the pressure plate also rotates as the pressure plate is attached to the
flywheel. The friction disc is located between the two. When the driver has pushed down the
clutch pedal the clutch is released. This action forces the pressure plate to move away from the
friction disc. There are now air gaps between the flywheel and the friction disc, and between the
friction disc and the pressure plate. No power can be transmitted through the clutch.

5. 5
Fig. 4 A single Plate automobile clutch
When the driver releases the clutch pedal, power can flow through the clutch. Springs in
the clutch force the pressure plate against the friction disc. This action clamps the friction disk
tightly between the flywheel and the pressure plate. Now, the pressure plate and friction disc
rotate with the flywheel. As both side surfaces of the clutch plate is used for transmitting the
torque, a term ‘n’ is added to include the number of surfaces used for transmitting the torque. By
rearranging the terms the equations can be modified and a more general form of the equation can
be written as
𝑇 = 𝑛 𝜇 𝑊𝑅, for a single clutch n = 2. (5)
Springs
Outside discs
Driven
shaft
Driving
shaft
Feather key
Fig. 5 A single plate clutch
Pressure Plate
Friction disc
Fig. 6 A Multi plate clutch

6. 6
Friction discs
intermediate Plate
Clutch cover
Flywheel ring
Pressure
Plate
A Twin disc clutch Assembly
Fig. 7(a) Double Plate clutch
Fig. 7.(b) Details in a Clutch Assembly
A multi plate clutch shown in Fig. 6 may be used when large torque is to be transmitted. The
inside discs (n1) are of steel are connected to the driven shaft. The outside discs (n2) are of
bronze are held by bolts and are fastened to the housing which is keyed to the driving shafts. The
Total number of active friction surfaces is
𝑛 = 𝑛1 + 𝑛2
The inner details of a clutch assembly are shown in Fig. 7(a) and (b).
Clutch Materials:
1. Cotton and asbestos fiber-woven or molded together and impregnated with resins or other
binding agents.
2. Copper wires are woven or pressed into the facing to give added strength.
3. Ceramic or metallic facings.

7. 7
For the maximum torque transfer the ratio of the inner to outer radius of the clutch linings is
taken as 0.577. If not given pressure between the plates can be taken as 90 𝐾𝑃𝑎 to 150 𝐾𝑃𝑎.
Assume uniform pressure conditions for new clutch and uniform wear conditions for an old
clutch. Under uniform wear conditions pressure is maximum at the inner radius and minimum at
the outer radius.
Major Design steps:
1. Assuming power transmitted, speed, and space constraints are known.
2. Select suitable clutch liner and fix its friction factor (Table above).
3. Fix the allowable normal pressure between the friction surface (Table above)
4. Assume suitable d/D ratio, if not given select it as 0.577 (for maximum power
transmission.
5. If not specified assume uniform wear condition.
Example 1: A multi plate clutch is to be designed to transmit a power of 50 kW at 500 rpm.
Assuming suitable data, determine the main dimensions of the friction lining used in
the clutch.
Solution: Ferodo moulded lining material for which coefficient of friction 0.25 is recommended.
An average pressure of 0.1 N/mm2 (100 KPa) is assumed between the plates. For the
maximum torque transmission, the ratio of the outer radius to inner radius is taken as
0.577. Torque transmitted
𝑇 =
𝑃×1000×60
2×𝜋×𝑁
× 103
𝑁 − 𝑚𝑚
𝑇 =
50×1000 ×60
2×𝜋×500
× 103
= 954929.7 𝑁− 𝑚𝑚
Assuming 𝑟2 = 160 𝑚𝑚 outer radius for the friction plate, its inner radius must be 𝑟1 =
0.577 × 160 ≅ 92 𝑚𝑚, ∴ 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝑟𝑎𝑑𝑖𝑢𝑠 𝑅 = 126 𝑚𝑚. The axial force required can
be
𝑊 = 2𝜋𝑝𝑅( 𝑟2 − 𝑟1 ) = 2𝜋 × 0.10 × 126 × (160 − 92) = 5383.43 𝑁
Torque transmitted by the multi plate clutch is
𝑇 = 𝑛𝜇𝑊𝑅 = 𝑛 × 0.25 × 5383.43 × 126 = 954929.7

8. 8
𝑛 = 5.6
The next even number is
𝑛 = 6, ∴ 𝑛1 = 4, 𝑎𝑛𝑑 𝑛2 = 3
Where n1 is the no. plates on the driving member and n2 is the no. plated on the driven
member.
Example 2: A friction clutch is to transmit 15 hp at 3000 rpm. It is to be of a single plate with
both sides of the plate effective. The axial pressure p is limited to 0.9 kg/cm2. If the
external diameter of the friction lining is 1.4 at that of the internal diameter, find
the required dimensions of the frictional lining. Take  = 0.3.
Solution: (Uniform wear is assumed)
𝑇 =
𝑃×1000 ×60
2𝜋𝑁
=
15×0.736×1000 ×60
2𝜋×3000
= 35.141 𝑁 − 𝑚
Axial pressure is limited to 0.9 kg/cm2 indicates that
𝑝 𝑚𝑎𝑥 = 0.9
𝑘𝑔
𝑐𝑚2 = 0.9 × 9.81 × 10 −2
𝑁 𝑚𝑚2⁄ = 88.29× 10−3
𝑁 𝑚𝑚2⁄
The maximum pressure is at the inner radius of the plate, eqn. (4) is
𝑝 𝑚𝑎𝑥 𝑟1 =
𝑊
2𝜋( 𝑟2−𝑟1)
∴ 𝑊 = 2𝜋( 𝑟2 − 𝑟1 ) 𝑝 𝑚𝑎𝑥 𝑟1
Outer radius of the friction plate is 1.4 times the inner radius.
𝑟2 = 1.4 𝑟1`
Torque equation or the clutch is
𝑇 = 𝜇𝑊𝑅𝑛 = 𝜇𝑊 (
𝑟2+𝑟1
2
) 2
Substituting the numerical values
35.141 × 103
= 0.3 × 2𝜋 × ( 𝑟2 − 𝑟1 ) 𝑝𝑟1 × (
𝑟2+𝑟1
2
) 2
35.141 × 103
= 0.3 × 2𝜋 × 0.4𝑟1 × (88.29 × 10−3) 𝑟1 × 2.4 𝑟1
Simplifying
35.141 × 103
= 0.15977𝑟1
3
, ∴ 𝑟1 = 60.4 𝑚𝑚, 𝑟2 = 84.5 𝑚𝑚
Example 3: A single plate clutch is to be designed to transmit 10 Hp at 900 rpm. Find (1) the
diameter of the shaft, (2) mean diameter and width of the friction lining, assuming
ratio of mean radius to face width as 4, (3) dimensions of the spring if the number
of spring is 6, spring index C= 6 and the allowable shear stress is 4200 kg/cm2.
Solution: Assumed that the coefficient of friction  = 0.25, and pressure at the mean
𝑟𝑎𝑑𝑖𝑢𝑠 𝑝 = 0.7 𝑘𝑔/𝑐𝑚2. Torque transmitted through the shaft is
𝑇 =
𝑃×1000×60
2𝜋× 𝑁
=
10×0.736 ×1000×60
2𝜋×900
= 78.09 𝑁− 𝑚
𝑇 =
𝜋
16
𝜏 𝑎 𝑑3
Shaft Design
Given that 𝜏 𝑎 = 4200 𝑘𝑔 𝑐𝑚2⁄ = 412 𝑁 𝑚𝑚2⁄

9. 9
78.09 × 1000 =
𝜋
16
× 41.2 × 𝑑3
, ∴ 𝑑 = 9.99 𝑚𝑚 ≈ 10 𝑚𝑚
Clutch Plate Design
Assuming uniform wear, the mean radius is
𝑅 = ( 𝑟1 + 𝑟2 ) 2⁄
Face width of the plate b is
𝑏 = 𝑟1 − 𝑟2
Axial force is
𝑊 = 2𝜋𝑏𝑝𝑅
Torque equation
𝑇 = 𝜇𝑊𝑅𝑛
78.09 × 1000 = 𝜇(2𝜋𝑏𝑝𝑅) 𝑅𝑛 = 0.25 × 2𝜋 × 𝑏 × (0.7 × 9.81 × 10−2) 𝑅2
× 2
78.09 × 1000 = 0.2157× 𝑏 × 𝑅2
= 0.2157 ×
𝑅
4
× 𝑅2
𝑅 = 112.91 𝑚𝑚, 𝑏 = 28.23 𝑚𝑚
𝑟1 + 𝑟2 = 225.82
𝑟1 − 𝑟2 = 28.23 𝑆𝑜𝑙𝑣𝑖𝑛𝑔, 𝑟1 = 127.03 𝑚𝑚, 𝑟2 = 98.8 𝑚𝑚
Clutch Spring design (Finding D, d, free length, pitch)
Given that 𝑛 = 6, 𝐶 = 𝐷 𝑑⁄ = 6
𝜏 𝑎 = 𝐾𝑠
8𝑃𝐶
𝜋𝑑2
𝐾𝑠 =
4𝐶−1
$𝐶−4
+
0.615
𝐶
=
4×6−1
4×6−4
+
0.615
6
= 1.2525
Load on a spring 𝑃 = 𝑊 6⁄
𝑃 = (2 × 𝜋 × 28.23 × 0.7 × 9.81 × 10−2
× 112.91) 6⁄ = 229.21 𝑁
412 = 1.2525 ×
8×229.21×6
𝜋𝑑2 , ∴ 𝑑 = 3.26 ≈ 3.4 𝑚𝑚
𝐷 = 6 × 3.4 = 20.4 𝑚𝑚
Assuming effective number of turns as 5, the deflection under the given load is
𝑦 =
8𝑃𝐶3 𝑛
𝐺𝑑
Assuming modulus of rigidity 𝐺 = 8.5 × 104
𝑁 𝑚𝑚2⁄
𝑦 =
8×229.21 ×63×5
8.5×104×3.4
= 6.85 𝑚𝑚
The free length of the spring is
𝐿 = ( 𝑛 + 2) × 𝑑 + 𝑦 + (𝑛 + 1) × 0.1
𝐿 = (5 + 2) × 3.4 + 6.85 + (5 + 1) × 0.1 = 31.25 ≈ 32 𝑚𝑚
The pitch of the spring is
𝑝 = 32 7⁄ = 4.57 𝑚𝑚
Example 4: A multi disc clutch has 3 disc on driving shaft and 2 on the driven shaft. The outside
diameter of the contact surface is 240 mm and inside diameter is 120 mm. Assuming

10. 10
uniform wear and  = 0.3, find maximum intensity of pressure between the discs
for transmitting 25 kW at 1515 rpm.
Solution: Torque transmitted through the shaft is
𝑇 =
𝑃×1000×60
2𝜋× 𝑁
=
25×1000 ×60
2𝜋 ×1515
= 157.579 𝑁− 𝑚
Torque equation or the clutch is
𝑇 = 𝜇𝑊𝑅𝑛
For uniform wear
𝑟1 = 120 𝑚𝑚, 𝑟2 = 60 𝑚𝑚
𝑅 = (120+ 60) 2⁄ = 90 𝑚𝑚, 𝑛 = 3 + 2 − 1 = 4
157.579 × 10000 = 0.3 × 𝑊 × 90 × 4, ∴ 𝑊 = 1459.06 𝑁
Pressure is maximum at the inner radius of the disc
𝑝 𝑚𝑎𝑥 =
𝑊
2𝜋𝑟1( 𝑟2−𝑟1)
=
1459.06
2𝜋×60×(120−60)
= 0.065 𝑁 𝑚𝑚2⁄
𝑝 𝑚𝑎𝑥 = 65000 𝑁 𝑚2⁄
Example 5: Design a dry plate clutch having friction lining on both sides of the clutch plate to
transmit 15 kW at 1500 rpm. Six springs are provided around the clutch to maintain
the friction surfaces in contact. The toggle levers at equal angular distances are
provided to withdraw the two halves and part the friction surfaces. Ferodo molded
friction lining were used in the plate. The ratio of the mean radius of the friction
faces to the radial width of the friction face is 4.5. The average normal pressure
intensity between the faces is 60 𝐾𝑃𝑎. Design the clutch springs also.
Solution: Assume uniform wear condition, for the Ferodo liner  is assumed as 0.25 and for a
single plate clutch n = 2.
Torque transmitted
𝑇 =
𝑃×1000×60
2×𝜋×𝑁
=
15×1000×60
2×𝜋×1500
× 103
𝑁 − 𝑚𝑚 = 95493 𝑁 − 𝑚𝑚
The axial force required to engage the clutch based on the average pressure is
𝑊 = 2𝜋𝑝𝑅(𝑟2 − 𝑟1 )
Given that
𝑅
( 𝑟2−𝑟1)
= 4.5 𝑜𝑟 ( 𝑟2 − 𝑟1) =
𝑅
4.5
𝑊 = 2𝜋𝑝𝑅
𝑅
4.5
Torque transmitted by the clutch is
𝑇 = 𝜇𝑊𝑅𝑛
𝑇 = 𝜇2𝜋𝑝𝑅2 𝑅
4.5
𝑛
Substituting the known values
95493 × 4.5 = 0.25 × 2𝜋 × (60 × 103
) × 10−6
× 𝑅3
× 2
𝑅 = 131.6 = 135 𝑚𝑚

11. 11
𝑟2 + 𝑟1 = 270
𝑟2 − 𝑟1 = 30
𝑟2 = 150 𝑚𝑚, 𝑟1 = 120 𝑚𝑚
The outer diameter of the plate = 300 mm
The inner diameter of the plate = 240 mm
Springs
Fig. 8 The designed friction disc
300mm
240mm30
Spring design:
Load one spring is
𝑃 =
𝑊
6
=
2𝜋
6
× 𝑝 × 𝑅 × (𝑟2 − 𝑟1)
𝑃 =
2𝜋
6
× 0.06 × 135 × (150− 120) = 255 𝑁
Due to the space constraints, the coil diameter of the spring can be 24 mm. For clutch
springs spring index C = 6 (7.100).
𝐶 = 6 =
𝐷
𝑑
=
24
𝑑
; ∴ 𝑑 = 4 𝑚𝑚
𝐾𝑠 =
4𝐶−1
$𝐶−4
+
0.615
𝐶
=
4×6−1
4×6−4
+
0.615
6
= 1.2525
Allowable shear stress for C65 spring steel (1.9) is 𝜏 𝑎 = 365 𝑁 𝑚𝑚2⁄
𝜏 = 𝐾𝑠
8𝑃𝐶
𝜋𝑑2
𝜏 =
1.2525×8×255×6
𝜋42
= 305 𝑁 𝑚𝑚2⁄ < 𝜏 𝑎 Spring is safe.
Assuming effective number of turns as 5, the deflection under the given load is
𝑦 =
8𝑃𝐶3 𝑛
𝐺𝑑
For C65 steel modulus of rigidity 𝐺 = 8.9 × 104
𝑁 𝑚𝑚2⁄ (1.1)
𝑦 =
8×255×63×6
8.9×104×4
= 6.2 𝑚𝑚
The free length of the spring is
𝐿 = ( 𝑛 + 2) × 𝑑 + 𝑦 + (𝑛 + 1) × 0.1
𝐿 = (5 + 2) × 5 + 6.2 + (5 + 1) × 0.1 = 41.8 𝑚𝑚

12. 12
In order to accommodate wear, slipping, the compression in the spring may have to be
increased while the clutch is in service. Therefore the free length may be increased to 45
mm. The pitch of the spring is
p =
45
7
= 6.43 mm
Example 6: A multi plate clutch employs 3 steel and 2 bronze discs having outer diameter 300
mm and inner diameter 200 mm. For a coefficient of friction of 0.22, find the axial
pressure and power transmitted at 750 rpm, if the mean normal pressure is 0.13
N/mm2.
Solution: Uniform wear condition is assumed. The mean radius of the friction disc is
𝑟2 = 150 𝑚𝑚, 𝑟1 = 100 𝑚𝑚
𝑅 =
𝑟2+ 𝑟1
2
=
150+100
2
= 125 mm
No. friction discs are
𝑛 = 𝑛1 + 𝑛2 − 1 = 3 + 2 − 1 = 4
Axial force on the clutch must be
𝑊 = 2𝜋𝑝𝑅( 𝑟2 − 𝑟1 ) = 2𝜋 × 0.13 × 125 × (150 − 100) = 5105.1 𝑁
Torque transmitted by the multi plate clutch is
𝑇 = 𝑛𝜇𝑊𝑅 = 4 × 0.22 × 5105.1 × 125 = 561560 𝑁 − 𝑚𝑚
Power transmitted
𝑃 =
𝑇×(2𝜋𝑁)
1000×60
=
561.56×(2𝜋×750)
1000×60
= 44.1 𝑘𝑊
Example 7: A multi plate clutch having effective diameters 250 mm and 150 mm has to transmit
60 KW at 1200 rpm. The end thrust is 4.5 kN and coefficient of friction is 0.08.
Calculate the no. plates assuming (i) uniform wear condition and (ii) uniform
pressure condition.
Solution: Given 𝐷2 = 250 ∴ 𝑟2 = 125 𝑚𝑚; 𝐷1 = 150 ∴ 𝑟1 = 75 𝑚𝑚
𝑁 = 1200 𝑟𝑝𝑚, 𝜇 = 0.08, 𝑊 = 4.5 𝑘𝑁 = 4500 𝑁,
Torque transmitted
𝑇 =
𝑃×1000×60
2×𝜋×𝑁
× 103
𝑁 − 𝑚𝑚
𝑇 =
60×1000 ×60
2×𝜋×1200
× 103
= 477464.83 𝑁 − 𝑚𝑚
(ii) Assuming uniform wear
The mean radius
𝑅 =
125 +75
2
= 100 𝑚𝑚
Torque transmitted is

13. 13
𝑇 = 𝑛𝜇𝑊𝑅 = 𝑛 × 0.08 × 4500 × 100 = 477464.83
𝑛 = 13.3
The next even number is
𝑛 = 14, ∴ 𝑛1 = 8, 𝑎𝑛𝑑 𝑛2 = 7
Where n1 is the no. plates on the driving member and n2 is the no. plated on the driven
member.
(ii) Assuming uniform pressure
The mean radius
𝑅 =
2
3
×
𝑟2
3
−𝑟1
3
𝑟2
2
−𝑟1
2 =
2
3
×
1253
−753
1252 −752 = 102.1 𝑚𝑚
Torque transmitted is
𝑇 = 𝑛𝜇𝑊𝑅 = 𝑛 × 0.08 × 4500 × 102.1 = 477464.83
𝑛 = 12.99
The next even number is
𝑛 = 14, ∴ 𝑛1 = 8, 𝑎𝑛𝑑 𝑛2 = 7
Where n1 is the no. plates on the driving member and n2 is the no. plated on the driven
member.
Example 8: A multi plate clutch of alternate bronze and steel plates is to transmit 6 KW power at
800 rpm. The inner radius is 38 mm and outer radius is 70 mm. The coefficient of
friction is 0.1 and maximum allowable pressure is 350 kN/m2. Determine
(a) Axial force required to engage the clutch.
(b) Total no. plates required
(c) Average pressure
(d) Actual Max. pressure
Solution: Assuming uniform wear conditions, given 𝑟2 = 70 𝑚𝑚; 𝑟1 = 38 𝑚𝑚
𝑁 = 800 𝑟𝑝𝑚, 𝜇 = 0.10, 𝑝 𝑚𝑎𝑥 = 350 𝑘𝑁/𝑚2
= 0.35𝑁/𝑚𝑚2
,
The mean radius
𝑅 =
70+38
2
= 54 𝑚𝑚
Torque transmitted
𝑇 =
𝑃×1000×60
2×𝜋×𝑁
× 103
𝑁 − 𝑚𝑚
𝑇 =
6×1000×60
2×𝜋×800
× 103
= 71619.72 𝑁 − 𝑚𝑚
(a) The axial force required considering the maximum pressure is
𝑊 = 2𝜋𝑝 𝑚𝑎𝑥 𝑟1 ( 𝑟2 − 𝑟1 ) = 2𝜋 × 0.35 × 38 × (70− 38) = 2674.12 𝑁
(b) The total number of plates required:
𝑇 = 𝑛𝜇𝑊𝑅 = 𝑛 × 0.10 × 2674.12 × 54 = 71619.72
𝑛 = 4.96
The next even number is

14. 14
𝑛 = 6, ∴ 𝑛𝑜. 𝑠𝑡𝑒𝑒𝑙 𝑛1 = 4, 𝑎𝑛𝑑 𝑛𝑜. 𝑏𝑟𝑜𝑛𝑧𝑒 𝑛2 = 3
Where n1 is the no. plates on the driving member and n2 is the no. plated on the driven
member.
(c) The average pressure: The average pressure is at the mean radius
𝑊 = 2674.12 = 2𝜋𝑝 𝑎𝑣 𝑅( 𝑟2 − 𝑟1) = 2𝜋 × 𝑝 𝑎𝑣 × 54 × (70 − 38)
𝑝 𝑎𝑣 = 0.246 𝑁/𝑚𝑚2
(d) The actual maximum pressure is at the inner radius.
𝑇 = 𝜇 𝑊𝑎 𝑅𝑛 where 𝑊𝑎 is the actual axial force
71619.72 = 0.1 × 𝑊𝑎 × 54 × 6
𝑊𝑎 = 2210.5 𝑁
𝑊𝑎 = 2𝜋𝑝 𝑚𝑎𝑥 𝑟1(𝑟2 − 𝑟1)
2210.5 = 2𝜋 × 𝑝 𝑚𝑎𝑥 × 38(70 − 38)
𝑝 𝑚𝑎𝑥 = 0.29 𝑁/𝑚𝑚2
Example 9: An automobile plate clutch consists of two pair of contacting surfaces with an
asbestos lining. The torque transmitting capacity of the clutch is 550 N-m, 𝜇 = 0.25
and permissible intensity of pressure is 0.5 𝑁 𝑚𝑚2⁄ . Due to space limitations the
outer of the friction lining is fixed as 250 mm. Calculate (1) the inner diameter of
the friction disc, (2) spring force required to keep the clutch disc in engaged
position.
Example 10: A single plate clutch consists of one pair of contacting surfaces. Due to space
limitations the outer of the friction disc is fixed as D. Calculate (1) the inner
diameter of the friction disc, (2) spring force required to keep the clutch disc in
engaged position. The permissible intensity of pressure is 𝑝 and coefficient of
friction is .. Show that the torque transmission capacity of the clutch is maximum
when
𝑑
𝐷
= 0.577
Solution: Let the diameter ratio (Is the permissible pressure is the pressure at ri ?)
𝑑
𝐷
= 𝑥
The torque transmitted
𝑇 =
1
2
𝜇𝐹𝑎( 𝑟2 + 𝑟1)
Since permissible intensity of pressure is ‘p’
𝑇 =
1
2
𝜇2𝜋𝑝𝑟1( 𝑟2 − 𝑟1)( 𝑟2 + 𝑟1)
𝑇 =
1
2
𝜇2𝜋𝑝𝑟1( 𝑟2
2
− 𝑟1
2)
𝑇 =
1
2
𝜇2𝜋𝑝 𝑟2
3
(
𝑟1
𝑟2
)(1 − (
𝑟1
𝑟2
)
2
)
𝑇 =
1
2
𝜇2𝜋𝑝 𝑟2
3
𝑥(1 − 𝑥2)
Taking derivative with respect to x and equating to zero
𝑇 =
1
2
𝜇
𝜕
𝜕𝑥
( 𝑥(1 − 𝑥2
))

15. 15
1 − 3𝑥2
= 0 𝑜𝑟 𝑥 = √
1
3
= 0.577
∴ 𝑑 = 0.577 𝐷
Axial force on the clutch must be
𝑊 = 2𝜋𝑝𝑟1( 𝑟2 − 𝑟1) = 2𝜋𝑝𝑟2
2
(
𝑟1
𝑟2
)(1 −
𝑟1
𝑟2
)
𝑊 = 2𝜋𝑝𝑟2
2
0.577(1 − 0.577) = 1.813 𝑝𝑑2
Example 11: An oil-immersed multi-disk clutch with moulded asbestos on one side and steel
discs on other side is used in an application. The torque transmitted by the clutch is
75 N-m. The coefficient of friction between the asbestos lining and the steel plate is
0.1.The permissible intensity of pressure on asbestos lining is 500 𝑘𝑃𝑎. The outer
diameter of the friction lining is kept 100 mm due to space limitation. Calculate the
inside diameter of the discs, the required number of discs and clamping torque.
(Assume for Maximum torque transmission
𝒅
𝑫
= 𝟎. 𝟓𝟕𝟕)
Solution: (Assume uniform wear conditions) Given 𝑟2 = 50 𝑚𝑚;
For transmitting maximum torque
𝑟1 = 0.577 × 50 = 28.85 ≅ 29 𝑚𝑚
𝜇 = 0.10, 𝑝 𝑚𝑎𝑥 = 500 𝑘𝑁 𝑚2⁄ = 0.5 𝑁/𝑚𝑚2
,
The mean radius
𝑅 =
50+29
2
= 39.5 𝑚𝑚
Torque transmitted
𝑇 = 75 𝑁 − 𝑚 = 75000 𝑁− 𝑚𝑚
The axial force required considering the maximum pressure is
𝑊 = 2𝜋𝑝 𝑚𝑎𝑥 𝑟1 ( 𝑟2 − 𝑟1 ) = 2𝜋 × 0.5 × 29 × (50 − 29) = 1913.23 𝑁
The total number of plates required:
𝑇 = 𝑛𝜇𝑊𝑅 = 𝑛 × 0.10 × 1913.23 × 39.5 = 75000
𝑛 = 9.92 ≅ 10
The next even number is
𝑛 = 10, ∴ 𝑛𝑜. 𝑠𝑡𝑒𝑒𝑙 𝑝𝑙𝑎𝑡𝑒𝑠 𝑛1 = 6, 𝑎𝑛𝑑 𝑛𝑜. 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑝𝑙𝑎𝑡𝑒𝑠 𝑛2 = 5
Example 12: A single plate clutch consists of one pair of contacting surfaces. It is used for an
engine that develops a maximum torque of 120 N-m. Assume a factor of safety of
1.5 to account for slippage at full engine toque. The permissible intensity of
pressure is 350 𝑘𝑃𝑎 𝑎𝑛𝑑 𝜇 = 0.35, Calculate the dimensions of the clutch.

16. 16
Design of Cone Clutches
A simple form of a cone clutch is shown in fig 9. It consists of a driver or cup and the follower or
cone. The cup is keyed to the driving shaft by a sunk key and has an inside conical surface or
face which exactly fits the outside conical surface of the cone. The slope of the cone face is made
small enough to give a high normal force. The cone is fitted to the driven shaft by splines. The
follower may be shifted along the shaft by a forked shifting lever in order to engage the clutch by
bringing the two conical surfaces in contact. Advantages and disadvantages of cone clutch:
Fig. 9 A cone clutch
Driven shaft
Collar
Cone
Spline
Spring
Friction lining
Cup
Sunk key
Driving shaft
Advantages:
1. This clutch is simple in design.
2. Less axial force is required to engage the clutch.
Disadvantages:
1. There is a tendency to grab.
2. There is some reluctance in disengagement.
r1


T
r2
Fig. 10 Details of the cone
Fn
Fa
dr
r
b
sin
dr

17. 17
The slope of the cone clutch is small enough to give a high normal force between the cup and
cone. Let r1 be the inner radius of the cone, r2 be the outer radius of the cone, R is the mean
radius, Fn is the normal force and Fa is the axial force. From Fig.10, the axial force is
𝐹𝑎 = 𝐹𝑛 𝑠𝑖𝑛𝛼 (1)
𝐹𝑛 =
𝐹𝑎
𝑠𝑖𝑛𝛼
(2)
The area of the shaded elemental ring of the cone shown in Fig. 10 is
𝑑𝐴 = 2𝜋𝑟
𝑑𝑟
𝑠𝑖𝑛𝛼
The normal force on the area is
𝑑𝐹𝑛 = 𝑝2𝜋𝑟
𝑑𝑟
𝑠𝑖𝑛𝛼
where p is the normal pressure between the cup and cone. The axial component of the above
normal force can be obtained using (1) as
𝑑𝐹𝑎 = 𝑝2𝜋𝑟
𝑑𝑟
𝑠𝑖𝑛𝛼
𝑠𝑖𝑛𝛼 = 𝑝2𝜋𝑟𝑑𝑟
The total axial force of the entire cone is
𝐹𝑎 = ∫ 𝑝2𝜋𝑟𝑑𝑟
𝑟2
𝑟1
(3)
Friction force on the ring element is
𝑑𝐹 = 𝜇𝑑𝐹𝑛 = 𝜇𝑝2𝜋𝑟
𝑑𝑟
𝑠𝑖𝑛𝛼
Friction torque on the element is
𝑑𝑇 = 𝑟𝑑𝐹 = 𝜇𝑝2𝜋𝑟2 𝑑𝑟
𝑠𝑖𝑛𝛼
The total torque is
𝑇 = ∫ 𝜇𝑝2𝜋𝑟2 𝑑𝑟
𝑠𝑖𝑛𝛼
𝑟2
𝑟1
(4)
(a) Uniform pressure
Under uniform pressure condition pressure is a constant throughout the cone. Therefore eqn. (3)
can be integrated as
𝐹𝑎 = ∫ 𝑝2𝜋𝑟𝑑𝑟
𝑟2
𝑟1
= 2𝜋𝑝 (
𝑟2
2
−𝑟1
2
2
)
𝐹𝑎 = 𝜋𝑝( 𝑟2
2
−𝑟1
2) 𝑜𝑟 𝑝 =
𝐹𝑎
𝜋( 𝑟2
2
−𝑟1
2)
(5)
The torque transmitted is obtained by integrating the eqn.(4) as
𝑇 =
𝜇𝑝2𝜋
𝑠𝑖𝑛𝛼
∫ 𝑟2
𝑑𝑟
𝑟2
𝑟1
=
𝜇𝑝2𝜋
𝑠𝑖𝑛𝛼
[
𝑟2
3
− 𝑟1
3
3
]
Substituting for ‘p’
𝑇 =
𝜇2𝜋
𝑠𝑖𝑛𝛼
𝐹𝑎
𝜋( 𝑟2
2
−𝑟1
2)
[
𝑟2
3
− 𝑟1
3
3
] =
𝜇
𝑠𝑖𝑛𝛼
𝐹𝑎
2
3
[
𝑟2
3
− 𝑟1
3
𝑟2
2
−𝑟1
2 ]
𝑇 =
𝜇
𝑠𝑖𝑛𝛼
𝐹𝑎 𝑅 𝑒 = 𝜇𝐹𝑛 𝑅 𝑒 (6)
where the effective radius 𝑅 𝑒 =
2
3
[
𝑟2
3
− 𝑟1
3
𝑟2
2
−𝑟1
2 ]
Substituting (3) in (2) gives the normal force as

18. 18
𝐹𝑛 =
𝑝2𝜋
𝑠𝑖𝑛𝛼
∫ 𝑟
𝑟2
𝑟1
𝑑𝑟 =
𝑝2𝜋
𝑠𝑖𝑛𝛼
(
𝑟2
2
−𝑟1
2
2
)
𝐹𝑛 =
𝑝2𝜋
𝑠𝑖𝑛𝛼
( 𝑟2 − 𝑟1)(
𝑟2+𝑟1
2
) = 2𝜋𝑅𝑏𝑝 (7)
where R is the mean radius and ‘b’ is the face width of the cone.
Uniform wear:
Under uniform wear condition
𝑝𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛 = 𝐶
𝐹𝑎 = 2𝜋𝐶 ∫ 𝑑𝑟
𝑟2
𝑟1
= 2𝜋𝐶( 𝑟2 − 𝑟1) (7.1)
𝑜𝑟 𝐶 =
𝐹𝑎
2𝜋( 𝑟2−𝑟1)
Considering the pressure between the cone and the cup at the average radius
𝑝 𝑎𝑣 𝑅 =
𝐹𝑎
2𝜋( 𝑟2−𝑟1)
𝑜𝑟 𝐹𝑎 = 2𝜋𝑅(𝑟2 − 𝑟1 )𝑝 𝑎𝑣 (8)
𝑇 = ∫ 𝜇𝑝2𝜋𝑟2 𝑑𝑟
𝑠𝑖𝑛𝛼
𝑟2
𝑟1
Substituting 𝑝𝑟 = 𝐶 in the torque eqn. (4) gives
𝑇 =
𝜇2𝜋𝐶
𝑠𝑖𝑛𝛼
∫ 𝑟 𝑑𝑟
𝑟2
𝑟1
=
𝜇2𝜋𝐶
𝑠𝑖𝑛𝛼
(
𝑟2
2
−𝑟1
2
2
)
Substituting for ‘𝐶’ in terms of 𝐹𝑎 gives
𝑇 =
𝜇2𝜋
𝑠𝑖𝑛𝛼
𝐹𝑎
2𝜋( 𝑟2−𝑟1)
( 𝑟2 − 𝑟1 )(
𝑟2+𝑟1
2
) = 𝜇
𝐹𝑎
𝑠𝑖𝑛𝛼
𝑅
𝑇 = 𝜇𝐹𝑛 𝑅 (9)
The normal force obtained using eqn. (2) in (7.1)
𝐹𝑛 = 𝐹𝑎 𝑠𝑖𝑛𝛼 =⁄
2𝜋𝐶
𝑠𝑖𝑛𝛼
( 𝑟2 − 𝑟1)
Corresponding the average radius of the cone
𝐹𝑛 =
2𝜋
𝑠𝑖𝑛𝛼
𝑝 𝑎𝑣 𝑅 ( 𝑟2 − 𝑟1) = 2𝜋𝑅𝑏𝑝 𝑎𝑣 (10)
From Fig. (10), the face width 𝑏 = (𝑟2 − 𝑟1) 𝑠𝑖𝑛𝛼⁄
In uniform wear pressure is maximum at the inner radius and minimum at the outer
radius. In the design of cone clutch, the standard value of the cone angle is 12.5o, and the
ratio between mean radius to the face width is between 4.5 to 8. The average pressure
between the friction surfaces is 70 to 120 𝑲𝑷𝒂 and the coefficient of friction is in the range
of 0.09 to 0.25. The mean diameter of the cone is 6 to 10 times the diameter of the clutch
shaft.
Example 13: An engine developing 60 Hp at 1000 rpm is fitted with a cone clutch. The cone has
a face angle of 12.5o and mean diameter of 50 cm and  = 0.2. The normal pressure
on the clutch face is not to exceed 1 kgf/cm2. Determine (1) the face width and (2)
the dimensions of the spring if the allowable shear stress is 4000 kg/cm2.
Solution: The torque transmitted (assuming uniform wear) is

19. 19
𝑇 =
𝑃×106
×60
2𝜋𝑁
=
60×0.736×1000 ×60
2𝜋×1000
= 421.697 𝑁 − 𝑚
𝑝 𝑚𝑎𝑥 = 1 𝑘𝑔𝑓 𝑐𝑚2⁄ = 1× 9.81 × 10−2
= 98.1 × 10−3
𝑁 𝑚𝑚2⁄
Coefficient of friction is given as 𝜇 = 0.2
𝑇 = 𝜇𝐹𝑛 𝑅
Pressure is maximum at minimum radius, so that the normal force become
𝐹𝑛 = 2𝜋( 𝑝 𝑚𝑎𝑥 𝑟1) 𝑏
𝑟1 = ( 𝑅 − 0.5𝑏𝑠𝑖𝑛𝛼) = 250 − 0.5 × 𝑏 sin12.5
421.697 × 103
= 0.2 × 2𝜋 × 98.1 × 10−3(250 − 0.5 × 𝑏𝑠𝑖𝑛12.5)250 × 𝑏
421.697 × 103
= 30.819(250− 0.5 × 𝑏𝑠𝑖𝑛12.5) 𝑏
𝑏2
− 2310.1𝑏 + 126.437 × 103
= 0
On solving width of the cone is
𝑏 = 56.1 𝑚𝑚, 𝑟2 = 256.1 𝑚𝑚, 𝑟1 = 243.9 𝑚𝑚
b = 56.1 mm
R=250mm
256.1
243.9


Fig. 11
The dimensions of the cone
Spring Design
𝐹𝑛 = 2𝜋( 𝑝 𝑚𝑎𝑥 𝑟1) 𝑏
𝐹𝑛 = 2𝜋(98.1 × 10−1
× 243.9) × 56.1 = 8433.8 𝑁
𝐹𝑎 = 8433.8 × 𝑠𝑖𝑛12.5 = 1825.4 𝑁
For clutch springs spring index C = 6 (7.100).
𝐶 = 6 =
𝐷
𝑑
=
24
𝑑
; ∴ 𝑑 = 4 𝑚𝑚
𝐾𝑠 =
4𝐶−1
$𝐶−4
+
0.615
𝐶
=
4×6−1
4×6−4
+
0.615
6
= 1.2525
Allowable shear stress is 𝜏 𝑎 = 4000 𝑘𝑔𝑓 𝑐𝑚2⁄ ≅ 392 𝑁 𝑚𝑚2⁄
𝜏 = 𝐾𝑠
8𝑃𝐶
𝜋𝑑2
392 =
1.2525 ×8×1825 .4×6
𝜋𝑑2 ; 𝑑 = 9.4 𝑚𝑚 ≈ 10 𝑚𝑚.
𝐷 = 6 × 10 = 60 𝑚𝑚
Assuming effective number of the spring as 6 and modulus of rigidity as 8.7 ×
104
𝑁 𝑚𝑚2⁄ the deflection is

20. 20
𝑦 =
8𝑃𝐶3 𝑛
𝐺𝑑
𝑦 =
8×1825.4×63×6
8.7×104×10
= 21.75 𝑚𝑚
The free length of the spring is
𝐿 = ( 𝑛 + 2) × 𝑑 + 𝑦 + (𝑛 + 1) × 0.1
𝐿 = (6 + 2) × 10 + 21.75 + (5 + 1) × 0.1 = 102.35 ≈ 104 𝑚𝑚
p =
104
8
= 13 mm
Example 14: Design a cone clutch to transmit 10 KW at a maximum speed of 800 rpm. The outer
cone is of cast iron and forms the part of an I C engine flywheel. The overall
dimensions restrict the mean diameter of the cone to 300 mm. The semi-cone angle
is 15o. The inner cone is positioned by means of a centrally placed helical spring.
Solution: The torque transmitted is
𝑇 =
𝑃×106×60
2𝜋𝑁
=
10×106×60
2𝜋×800
= 119366.21 𝑁 − 𝑚𝑚
Assuming the coefficient of friction 𝜇 = 0.1 (𝑓𝑜𝑟 𝑐𝑎𝑠𝑡 𝑖𝑟𝑜𝑛 𝑜𝑛 𝑐𝑎𝑠𝑡 𝑠𝑡𝑒𝑒𝑙). Applying
𝑇 = 𝜇𝐹𝑛 𝑅
119366.21 = 0.1 × 𝐹𝑛 × 150
𝐹𝑛 = 7957.75 𝑁
𝐹𝑛 = 2𝜋𝑅𝑏𝑝 𝑎𝑣
Assuming average pressure between the cast iron cup and cast steel cone is 0.15 N/mm2
7957.75 = 2𝜋 × 150 × 𝑏 × 0.15
𝑏 = 56.39 𝑚𝑚 ≅ 58 𝑚𝑚
315
285


b = 58 mm

R=150mm
Fig. 12
Dimensions of the cone
Now
𝑟2 − 𝑟1 = 58 𝑠𝑖𝑛15 = 15 𝑚𝑚
𝑟2 + 𝑟1 = 300 𝑚𝑚

21. 21
On solving
𝑟2 = 157.5 𝑚𝑚
𝑟1 = 142.5 𝑚𝑚
Example 15: Determine the main dimensions of a cone clutch faced with leather to transmit 30
KW at 750 rpm from an electric motor to an air compressor. Assume an over load
factor of 1.75. Due to possibility of contamination of lining, a low value of 0.2 for
friction factor is recommended.
Solution: Considering the over load factor, the design power is
𝑃 𝑑 = 1.75 × 30 = 52.5 𝐾𝑊
𝑇 =
𝑃 𝑑 ×106
×60
2𝜋𝑁
=
52.5×106
×60
2𝜋 ×750
= 668450.8 𝑁 − 𝑚𝑚
Given that 𝜇 = 0.2 and assuming uniform wear condition, average pressure between the
cup and cone is 0.15 N/mm2 and 𝑅 𝑏⁄ ratio as 6.0.A semi cone angle of 12.5o is also
assumed.
𝑇 = 𝜇𝐹𝑛 𝑅
668450.8 = 0.2 × 𝐹𝑛 𝑅
𝐹𝑛 𝑅 = 3342254
b = 47 mm
574.2
553.8
R=282mm



Fig. 13
Dimensions of the cone
However the normal force
𝐹𝑛 = 2𝜋𝑅𝑏𝑝 𝑎𝑣 𝑜𝑟 𝐹𝑛 𝑅 = 2𝜋𝑅2
𝑏𝑝 𝑎𝑣 ,
For the selected 𝑅 𝑏⁄ ratio of 6.0
3342254 = 2𝜋 ×
𝑅3
6
× 0.15
𝑅 = 277.1 ≅ 282 𝑚𝑚 𝑎𝑛𝑑 ∴ 𝑏 = 47 𝑚𝑚
𝑟2 = 282 +
47
2
𝑠𝑖𝑛12.5 = 287.1 𝑚𝑚
𝑟1 = 282 −
47
2
𝑠𝑖𝑛12.5 = 276.9 𝑚𝑚
The main dimensions of the clutch cone is shown in Fig.13

22. 22
Example 16: A cone clutch is used to connect an electric motor running at 1440 rpm with a
machine that is stationary. The machine is equivalent to a rotor of mass 150 kg and
radius of gyration 250 mm. The machine has to bring to the full speed of 1440 rpm
from a stationary condition in 40 seconds. The semi cone angle is 12.5o. The mean
radius of the clutch is twice the face width;𝜇 = 0.25; pressure should not exceed
0.1.N/mm2. Assume uniform wear condition. Find (a) inner outer diameters of the
cone, (b) the face width, (c) forces required to engage the clutch, and (d) the heat
generated during each engagement.
Solution: The final and initial angular velocities are
ωf =
2π×1440
60
= 150.8 rad s⁄ , ωo = 0
The load torque during the starting condition can be calculated as
T = I ∝
The moment of inertia I of the machine is
I = 150 × (
250
1000
)
2
= 9.375 Kg− m2
α = (
ωf−ωo
t
) =
150.8−0
40
= 3.77 rad/s
𝑇 = 9.375 × 3.77 = 35.34375 𝑁 − 𝑚 = 35343.75 𝑁 − 𝑚𝑚
The normal force considering the maximum pressure is
𝐹𝑛 =
2𝜋𝐶
𝑠𝑖𝑛𝛼
( 𝑟2 − 𝑟1)
𝐹𝑛 = 2𝜋𝑝 𝑚𝑎𝑥 𝑟1 𝑏
𝑟1 = 𝑅 −
𝑏
2
𝑠𝑖𝑛𝛼
For an 𝑅 𝑏⁄ ratio of 2
𝑟1 = 2𝑏 −
𝑏
2
𝑠𝑖𝑛𝛼 =
𝑏
2
(4 − 𝑠𝑖𝑛𝛼)
𝐹𝑛 = 𝜋𝑝 𝑚𝑎𝑥 𝑏2
(4 − 𝑠𝑖𝑛𝛼)
𝑇 = 𝜇𝐹𝑛 𝑅 = 𝜇𝐹𝑛2𝑏
𝑇 = 2𝜇𝜋𝑝 𝑚𝑎𝑥(4 − 𝑠𝑖𝑛𝛼)𝑏3
35343.75 = 2𝜋 × 0.25 × 0.1 × (4 − 𝑠𝑖𝑛12.5)𝑏3
𝑏 = 39 .03 ≅ 40 𝑚𝑚, ∴ 𝑅 = 80 𝑚𝑚
𝑟2 = 80 + 20× 𝑠𝑖𝑛12.5 = 84.33 𝑚𝑚
𝑟1 = 80 − 20 × 𝑠𝑖𝑛12.5 = 75.67 𝑚𝑚
The forces required to engage the clutch
𝐹𝑎 = 2𝜋𝑟1(𝑟2 − 𝑟1)𝑝 𝑚𝑎𝑥
𝐹𝑎 = 2𝜋 × 75.67(8.66) × 0.1 = 412 𝑁
The heat generated during the engagement is equal to the work done during that period.
𝑊 = 𝑇𝜃 𝐽𝑢𝑙𝑒𝑠
𝜃 =
1
2
𝛼𝑡2
=
1
2
× 3.77 × 402
= 3016 𝑟𝑎𝑑
𝑊 = 35.34 × 3016 = 106.6 𝐾𝐽

23. 23


R=80mm
168.66
151.34

Fig. 14
Dimensions of the cone
b = 40 mm
E Example 17: A cone clutch is to transmit 7.5 kW at 900 rpm. The cone has a face angle of 12o.
The width is half of the mean radius and the normal pressure is 0.09 N/mm2.
Assume uniform wear and 𝜇 = 0.2, find the main dimensions of the clutch and the
axial force required
Solution:
The torque transmitted is
𝑇 =
𝑃×106
×60
2𝜋𝑁
=
7.5×106
×60
2𝜋×900
= 79577.47 𝑁 − 𝑚𝑚
𝜇 = 0.2
𝑇 = 𝜇𝐹𝑛 𝑅 = 𝜇(2𝜋 × 𝑅 × 𝑏 × 𝑝) × 𝑅
79577.47 = 0.2(2𝜋 × 0.5 × 0.09) × 𝑅3
𝑅 = 112.06 𝑚𝑚, 𝑏 = 56.03 𝑚𝑚
𝑟2 = 𝑅 +
𝑏
2
sin 12 = 112.06 +
56.03
2
sin 12 = 117.88 𝑚𝑚
𝑟2 = 𝑅 −
𝑏
2
sin 12 = 112.06 −
56.03
2
sin 12 = 106.24 𝑚𝑚
The normal force on the clutch is
𝐹𝑛 = 2𝜋( 𝑝𝑅) 𝑏
𝐹𝑛 = 2𝜋(0.09 × 112.06) × 56.03 = 3550.53 𝑁
The axial force required to engage the clutch is
𝐹𝑎 = 3550.53 × 𝑠𝑖𝑛12 = 738.2 𝑁
Example 18: Determine the principal dimensions of a cone clutch faced with leather to transmit
30 kW at 750 rpm, semi angle of cone is 12.5o,  = 0.2, mean diameter of the cone
to be 6 to 10d , where d is the diameter of the shaft. Allowable normal pressure is
0.1 N/mm2, take mean diameter to face width is 6.

24. 24
Design of Centrifugal Clutch
The inner details of a centrifugal clutch are shown in Fig. 15. It consists of a spider attached to a
driving shaft having radial slots in which heavy shoes are fitted. These shoes are connected to the
spider through springs. The opposite face of the shoes is covered with friction material. The
shoes moves radially in the guided slots are held against boss on driving shaft by means of
springs.
Fig. 15 Inner details of a centrifugal Clutch
The springs exert a radially inward force which is assumed to be constant. The shoes can move
radially outward due to centrifugal force. When the centrifugal force is less than the spring force
shoes remains in the same position. But when the centrifugal force is equal to the spring force,
the shoes are in a floating state. When the centrifugal force exceeds the spring force the shoes
move outward and come in contact with the driven member and presses against it. The increase
of speed causes the shoe to press harder and harder and enables more torque to be transmitted.
In Fig. 16 let
m – Mass of the shoe in kg.
n- Number of shoes
r- Distance from the centre of the spider to the center of gravity of the shoe
R- Inside radius of the cover plate or pulley
N-Running speed in rpm
𝜔 =
2𝜋𝑁
60
Angular velocity in rad/s
1-Speed at which engagement begins
- Coefficient of friction
The centrifugal force on the shoe at the running speed is
𝑃 𝐶 = 𝑚𝑟𝜔2

25. 25
Usually engagement begins at (
3
4
)
𝑡ℎ
of the running speed. Therefore force on the shoe by the
spring
𝑃𝑆 = 𝑚𝑟𝜔1
2
= 𝑚𝑟 (
3
4
𝜔)
2
=
9
16
𝑚𝑟𝜔2
The net outward radial force with which shoe presses against the rim at the running speed
𝑃 = 𝑃 𝐶 − 𝑃𝑆
𝑃 = 𝑚𝑟𝜔2
−
9
16
𝑚𝑟𝜔2
𝑃 =
7
16
𝑚𝑟𝜔2
L
PC
PS
F

r
RG
Fig. 16 Forces of the shoe
clearance
3-6 mm
The frictional force acting tangentially on each shoe is
𝐹 = 𝜇(𝑃 𝐶 − 𝑃𝑆)
Friction torque acting on each shoe is
𝑇 1 = 𝐹𝑅 = 𝜇(𝑃𝐶 − 𝑃𝑆)𝑅
The total torque transmitted
𝑇 = 𝑛 𝑇1 = 𝑛𝜇(𝑃𝐶 − 𝑃𝑆)𝑅
𝑇 =
7
16
𝜇𝑚𝑟𝜔2
𝑅𝑛
Solving the torque equation gives the mass of the shoe.
The size of the shoe
Let L be the contact length of the shoe, ‘b’ is the width of the shoe and  is the angle subtended
by the shoe at the centre of the spider. Usually 𝜽 𝒊𝒔 𝒇𝒊𝒙𝒆𝒅 𝒂𝒔 𝟔𝟎 𝒐
𝒐𝒓 𝝅 𝟑 𝒓𝒂𝒅⁄ . The contact
pressure ‘p’ between the shoe and the rim is 90 to 120 𝑲𝑷𝒂.
Contact length
𝐿 = 𝑅𝜃
Contact area of the shoe

26. 26
𝐴 = 𝐿𝑏
The force with which the shoe presses the rim
𝑃 =
7
16
𝑚𝑟𝜔2
= 𝐿𝑏𝑝
By solving the above equation width of the shoe can be found. The close coiled springs used in
the shoe can be designed by taking the load on the spring as
𝑃𝑆 =
9
16
𝑚𝑟𝜔2
Example 19: Design a centrifugal clutch having four shoes to transmit 15 KW at 1000 rpm. The
speed at which engagement begins is (3 4⁄ ) 𝑡ℎ
of the running speed. The inside
radius of the pulley is 140 mm. The shoes are lined with Ferodo lining. Assume 𝜇 =
0.25.
Solution: Running speed
𝜔 =
2𝜋×1000
60
= 104.7 𝑟𝑎𝑑/𝑠
Speed at which engagement begins
𝜔1 =
3
4
104.7 = 78.5 𝑟𝑎𝑑/𝑠
The toque transmitted
𝑇 =
15×103
×60
2𝜋 ×1000
= 143.24 𝑁 − 𝑚
Centrifugal force on each shoe
𝑃 𝐶 = 𝑚𝑟𝜔2
Inward force on the shoe from the spring
𝑃𝑆 =
9
16
𝑚𝑟𝜔2
Usually the CG of the shoe is taken 30 mm inside the rim radius.
∴ 𝑟 = 140 − 30 = 110 𝑚𝑚
The total driving torque
𝑇 =
7
16
𝜇𝑚𝑟𝜔2
𝑅𝑛
143.24 =
7
16
× 0.25 × 𝑚 ×
110
1000
× 104.72
×
140
1000
× 4
𝑚 = 1.94 𝑘𝑔
70
146.6
Fig. 17 Clutch shoe
Length of the shoe assuming  = 60o

27. 27
𝐿 = 140 ×
𝜋
3
= 146.6 𝑚𝑚
Assuming a normal pressure of 0.1 N/mm2 between the shoe and rim, the force with
which the shoe presses the rim is
𝑃 =
7
16
𝑚𝑟𝜔2
= 𝐿𝑏𝑝
7
16
× 1.94 × 110 × 104.7 2
= 146.6 × 𝑏 × 0.1
𝑏 = 69.81 ≅ 70 𝑚𝑚
Example 20: A centrifugal clutch is to be designed to transmit 15 KW at 900 rpm. The shoes are
four in numbers. The speed at which engagement begins is (3 4⁄ ) 𝑡ℎ
of the running
speed. The inner radius of the pulley rim is 150 mm. The shoes are lined with
friction material with 𝜇 = 0.25. Determine the weight of the shoe, size of the shoe
and spring dimensions.
Solution: Running speed
𝜔 =
2𝜋×900
60
= 94.25𝑟𝑎𝑑/𝑠
Speed at which engagement begins
𝜔1 =
3
4
94.25 = 70.69 𝑟𝑎𝑑/𝑠
The toque transmitted
𝑇 =
15×103
×60
2𝜋×900
= 159.16 𝑁 − 𝑚
Centrifugal force on each shoe
𝑃 𝐶 = 𝑚𝑟𝜔2
Inward force on the shoe from the spring
𝑃𝑆 =
9
16
𝑚𝑟𝜔2
Usually the CG of the shoe is taken 30 mm inside the rim radius.
∴ 𝑟 = 150 − 30 = 120 𝑚𝑚
The total driving torque
𝑇 =
7
16
𝜇𝑚𝑟𝜔2
𝑅𝑛
159.16 =
7
16
× 0.25 × 𝑚 ×
120
1000
× 94.252
×
150
1000
× 4
𝑚 = 2.28 𝑘𝑔
Length of the shoe
𝐿 = 150 ×
𝜋
3
= 157𝑚𝑚
Assuming a normal pressure of 98 𝐾𝑃𝑎, the force with which the shoe presses the rim

28. 28
𝑃 =
7
16
𝑚𝑟𝜔2
= 𝐿𝑏𝑝
7
16
× 2.28 × 120 × 94.25 2
= 157 × 𝑏 × 0.098
𝑏 = 69.1 ≅ 70 𝑚𝑚
Dimensions of the spring (close coiled)
Forces on the spring
𝑃𝑆 =
9
16
𝑚𝑟𝜔2
=
9
16
× 2.28 × 0.12 × 94.252
≅ 1368 𝑁
Allowable shear stress of the spring steel intended for dynamic loads (7.105) is
𝜎𝑢 = 1660 𝑁 𝑚𝑚2
,⁄ 𝜏 𝑎 = 415 𝑁 𝑚𝑚2⁄ , 𝑎𝑛𝑑 𝐺 = 8.9 × 104
𝑁 𝑚𝑚2⁄
For clutch springs
𝐶 = 6, 𝑎𝑛𝑑 𝑊ℎ𝑎𝑙 𝑠𝑡𝑟𝑎𝑠𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝐾𝑠 = 1.25 (1.100)
Assuming the total load on the extension spring as 1368 N, the design equation is
𝜏 𝑎 = 𝐾𝑠
8𝑃𝐶
𝜋𝑑2
415 = 1.25 ×
8×1368 ×6
𝜋𝑑2
𝑑 = 7.93 ≅ 8 𝑚𝑚 ( 𝑠𝑡𝑑. 𝑤𝑖𝑟𝑒 𝑑𝑖𝑎 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒), 𝑐𝑜𝑖𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝐷 = 48 𝑚𝑚,
Assuming an initial clearance between the shoe as 5 mm, and 𝐺 = 8.7 × 104
𝑁 𝑚𝑚2⁄ ,
the number of turns of the spring can be calculated.
𝑦 =
8𝑃𝐶3
𝑛
𝐺𝑑
5 =
8×1368 ×63
𝑛
8.7×104 ×8
, 𝑛 = 1.47 ≈ 2
1
1
2
turns are required for the end fixing of the extension spring, therefore the total
number of turns required is
𝑛 = 2 + 1.5 = 3.5
Example 21: A Centrifugal clutch having four shoes each having a mass of 1.5 kg. In the
engaged position, the radius to the CG of each shoe is 110 mm. While the inner
radius of the rim is 140mm, 𝜇 = 0.25, the preload on the spring is adjusted in
such a way that the spring force at the beginning of the engagement is 700 N. If the
running speed is 1440 rpm, calculate (i) The speed at which engagement begins,
(ii) The power transmitted by the clutch.

29. 29
Solution: Running speed
𝜔 =
2𝜋×1440
60
= 150.8 𝑟𝑎𝑑 𝑠⁄ , 𝑟 = 110 𝑚𝑚
Mass of the shoe 𝑚 = 1.5 𝑘𝑔, 𝑁𝑜. 𝑠ℎ𝑜𝑒𝑠 = 4, 𝑅 = 140 𝑚𝑚
Given that when engagement begins, the spring force is 700 N, so that
𝑚𝑟𝜔1
2
= 700, ∴ 𝜔1 = √700 (1.5 × 0.11)⁄ = 65.134 𝑟𝑎𝑑 𝑠⁄
∴ 𝑁1 = (60 × 65.134) (2𝜋)⁄ = 622 𝑟𝑝𝑚
The total driving torque
𝑇 = 𝜇𝑚𝑟( 𝜔2
− 𝜔1
2) 𝑅𝑛
𝑇 = 0.25 × 1.5 × 0.11(150.82
− 65.1342) × 0.14 × 4
𝑇 = 427.31 𝑁 − 𝑚
The power transmitted
𝑃 =
𝑇𝜔
1000
=
427 .31×150.8
1000
= 64.44 𝑘𝑊
Example 22: A Centrifugal clutch is transmitting 20 kW at 750 rpm consists of 4 shoes. The
clutch is to be engaged at 500 rpm. The inner radius of the drum is 165 mm. The
radius to the center of gravity of the shoe is 140 mm when the clutch is engaged.
Take 𝜇 = 0.3 𝑎𝑛𝑑 𝑝 = 0.1 𝑁 𝑚𝑚2⁄ . Design the clutch.