BELT and ROPE DRIVE

Belt Drive
Overview
Problem: To Transfer power from one location to another
From driver: motor, peddles, engine, windmill, turbine
To driven: conveyor belt, back wheels/bike, generator, rock crusher,
dryer

Belt Drive
Belt and rope drive
The belts and ropes are used to
transmit power from one shaft to
another by means of pulleys which
rotate at same speed or at different
speed

Amount of power transmitted

depends upon the following factors:

1. The velocity of the belt

2. The tension under which the belt
is placed on the pulleys
3. The arc of contact between the
belt and the pulleys
4. The conditions under which the
belt is used

Belt Drive
Belt Drive
• When a belt is used for power transmission it is called a belt drive
• A belt drives are used to transmit power from one shaft to another
shaft
• To transmit power from one shaft to another, pulleys are mounted
on the two shafts
• The connecting belt is kept in tension so that motion of one pulley
is transferred to the other without slip
• The speed of the driven shaft can be varied by varying the
diameters of the two pulleys
• Belts are the cheapest utility for power transmission between
shafts that may not be parallel.
• Power transmission is achieved by specially designed belts and
pulleys.

Belt Drive
Important consideration
• The shafts should be properly in line to ensure uniform tension
across the belt section.
• The pulleys should not be too close together, in order that the arc
of contact on the smaller pulley may be as large as possible
• The pulleys should not be so far apart as to cause the belt to weigh
heavily on the shafts, thus in-creasing the friction load on the
bearings
• A long belt tends to swing from side to side, causing the belt to run
out of the pulleys.
• The tight side of the belt should be at the bottom, so that
whatever sag is present on the loose side will increase the arc of
contact at the pulleys
• In order to obtain good results with flat belts, the maximum
distance between the shafts should not exceed 10 m and the
minimum should not be less than 3.5 times the diameter of the
larger pulley Belt Drive
Advantages
• Cheap
• Allows misalignment
(parallel shafts)
• Protects from overload
• Absorbs noise and
vibrations
• Cushion load fluctuations
• Needs little maintenance

Belt Drive
Disadvantages
• Speed ratio is not constant (slip & stretch)
• Heat accumulation
• Speed limited – 2000 m/min,
• Power limited - 700 kW
• Endless belts needs special attention to install

Belt Drive
Selection of Belt drives
1. Speed of driving and driven shafts

2. Speed reduction ratio

3. Power to be transmitted

4. Centre distance between the shafts.

5. Shafts layout

6. Space available

7. Service conditions

Belt Drive
Belt
• A Belt is a looped strip of flexible
material, used to mechanically
link two or more rotating shafts.
• They may be used as a source of
motion, to efficiently transmit
POWER, or to track relative
movement.
• Belts are looped over pulleys

Belt Drive
Types of Belt drives
1. Light Drives
These are used to transmit small powers at
belt speeds upto about 10 m/s, as in
agricultural machines and small machine
tools.
2. Medium Drives
These are used to transmit medium powers
at belt speeds over 10 m/s but upto 22
m/s, as in machine tools.

3. Heavy Drives
These are used to transmit large powers at
belt speeds above 22 m/s, as in compressors
and generators.
Belt Drive
Types of Belt
1. Flat belt
• mostly used in the factories
and workshops

• For a moderate amount of

power transmission

• Normally used when pulleys

are not more than 8 meters
apart.

• Pulleys for flat belt

applications may be either
flanged or crowned.
Belt Drive
Types of Belt
2. V belt
• mostly used in the factories
and workshops

• For a moderate amount of

power transmission

• Used when two pulleys are

very near to each other.

Belt Drive
Types of Belt
3. Round or Circular belt
• mostly used in the factories and
workshops

• For a great amount of power

transmission

• When two pulleys are more than 8

meters apart.

• Pulleys for round belt applications

have a concaved face to fit the
belt cross-section.

Belt Drive
Materials used for belts
1. Leather belts

2. Cotton or fabric belts

3. Rubber belt

4. Balata belts ( Acid proof

and water proof)

Belt Drive
Types of flat belt drive
1. Open belt drive
• used with shaft arranged parallel and rotating
in the same direction

• For a large pulley distance, tight side is kept on

lower side to increase the contact of the belt

• The driver A pulls the belt from one side (i.e.

lower side RQ) and delivers it to the other side
(i.e. upper side LM).

• The tension in the lower side belt will be more

than that in the upper side belt.

• The lower side belt (because of more tension)

is known as tight side, the upper side belt
(because of less tension) is known as slack side
Belt Drive
Types of flat belt drive
2. Cross belt drive
• used with shaft arranged parallel and rotating
in the opposite direction

• The driver pulls the belt from one side (i.e. RQ)
and delivers it to the other side (i.e.LM).

• The tension in the belt RQ will be more than

that in the belt LM.

• The belt RQ(because of more tension) is

known as tight side, the belt LM(because of
less tension) is known as slack side

A point where the belt crosses, it rubs against each other and there will be excessive
wear and tear. In order to avoid this, the shafts should be placed at a maximum
distance of 20 b, where b is the width of belt and the speed of the belt should be less
than 15 m/s. Belt Drive
Types of flat belt drive
3. Belt drive with idler pulley
• Idler pulley is used to increase
tension in the belt when it is
loose.

• Idler pulley reduces the chance of

belt coming over the pulleys

Belt Drive
Types of flat belt drive
4. Compound Belt drive
• Compound belt drive is used
when it is required to have large
velocity ratio.

• The pulley used in between drive

and driven pulley is called
Intermediate pulley

Belt Drive
Types of flat belt drive
5. Guide pulley drive
• A guide pulley is used to connect
two non parallel shaft arranged at
right angles

• Rotation of pulley is in one definite

direction

• In order to prevent the belt from

leaving the pulley, the width of the
face of the pulley should be
greater or equal to 1.4 b, where b
is the width of belt.

Belt Drive
Types of flat belt drive
6. Cone pulley drive

• It is used when it is required to run

the driven shaft at different speeds.

• A cone pulley has different sets

pulley radii to give varying speeds
of the driven shaft.

• The radii of different steps are so

chosen that the same belt can be
used at different sets of the cone
pulley

Belt Drive
Velocity Ratio of Belt Drive
The ratio between the velocities of driver and the follower or driven is known as
Velocity Ratio.
Let, d1 = diameter of driver pulley Also we know that,
d 2 = diameter of driven pulley
N1 = speed of driver pulley in rpm
N2 = speed of driven pulley in rpm
So, velocity ratio
Then, velocity of belt passing over driver

and velocity of belt passing over driven pulley

When the thickness of belt (t)
is considered, then velocity
Assuming no slip between the belt and pulley, ratio

Belt Drive
Slip of Belt
• Frictional grip between belt and pulley may insufficient in
some cases
• This may cause some forward motion of the driver
without carrying the belt with it
• Also, This may cause some forward motion of the belt
without carrying the driven pulley with it
• This is called the slip of belt, generally expressed as a
percentage
• Slip may cause difference between the linear speed of
the pulley and the belt on it

Belt Drive
Effect of Slip on Velocity Ratio
• The result of the belt slipping is to reduce the velocity ratio of the system.
• As the slipping of the belt is a common phenomenon, thus the belt should never
be used where a definite velocity ratio is of importance
Let, s1 = percentage slip between driver pulley and the belt.
s2 = percentage slip between belt and driven pulley
We know, velocity of the belt passing over the driving pulley,

velocity of the belt passing over the driven pulley,

Substituting value of v,

If thickness of belt
is considered, then
If s is the total percentage slip, then

Belt Drive
 Example 1:
An engine running at 150 rpm drives a line shaft by means of belt. The engine pulley is 150
mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameter pulley on
the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Find the speed of the
dynamo shaft when
a.There is no slip (1500 rpm)
b.There is a slip of 2% at each drive (1440 rpm).
Example 2:
A shaft runs at 80 rpm and drives another shaft at 150 rpm through belt drive. The diameter
of the driving pulley is 600 mm. Determine the diameter of the driven pulley in the following
cases:
a.Neglecting belt thickness (320 mm)
b.Taking belt thickness as 5 mm (317.7 mm)
c.Assuming for case (b) a total slip of 4% (304.8 mm)
d.Assuming for case (b) a slip of 2% on each pulley (304.9 mm)

Example 3:
With the help of a belt, an engine running at 200 rpm, drives a line shaft. The diameter of the
pulley on the engine is 80 cm and the diameter of the pulley on the line shaft is 40 cm. A 100
cm diameter pulley on the line shaft drives a 20 cm diameter pulley keyed to a dynamo shaft.
Find the speed of the dynamo shaft when:
a.There is no slip (2000 rpm)
b.There is a slip of 2.5 % at each drive. (1901.25 rpm)
Belt Drive
 Creep of Belt

Self Study

Belt Drive
 Length of Belt- Open Belt
Driver
F Driven
Pulley G Pulley

r1 r2
E A α B H
N
C
D
x
Let, r1 and r2 = Radii of the bigger and smaller pulleys
x = centre distance of two pulleys
L = total length of the belt
From B, draw a line BN parallel to DC meet at N on AD. From the geometry of the figure BN is
perpendicular to AD.
Let, angle ABN = α
Then, total length of the belt, L = Arc ED + Arc EF + FG + DC+ Arc GH + Arc CH.
But, Arc ED = Arc EF; FG = DC; Arc GH = Arc CH.
So total length L can be written as, L = 2(Arc EF + Arc GH + FG)…………….(1)
Belt Drive
From triangle ABN,

For small angle,

So,

From figure, we know that,

Expanding the above equation by binomial theorem

Neglecting higher terms

Belt Drive
Now we have,

Substituting the above values in equation 1, we get

Substituting the value of

OR

Belt Drive
 Angle of contact for open belt drive
• The angle of lap is smaller on the smaller pulley than on larger
pulley
• So, the belt will begin to slip on the smaller pulley
• The angle of contact should be taken as minimum angle of
contact
• The angle of contact of lap at the smaller pulley must be taken
into consideration
F Driven
Angle GHC is Angle of G Pulley
contact
r1 r2
E A α B H
N
But, the value of α is C
D
x

Belt Drive
Length of Cross Belt Drive
Driver
F Driven
Pulley G Pulley

r1 r2
E A B H
N
C
D

Belt Drive
 Angle of contact for crossed belt drive
• For the crossed belt drive, the angle of lap on both the pulleys
is same
Driver
F Driven
Pulley G Pulley

r1 r2
E A B H
N
C
D

Angle GHC or angle FED is Angle of contact

But, the value of α is

Belt Drive
 Example 4 :
Two parallel shafts 6 m apart are to be connected by a belt running over pulleys of diameters
60 cm and 40 cm respectively. Determine the exact and approximate lengths of the belt
required
i)If the belt is open
i)If the belt is crossed

Belt Drive
Power Transmitted by Belt
Also, we know that,
Let, T1 = Tension in the tight side of the belt
T2 = Tension in the slack side of the belt
r1 = Radius of the driver pulley in m
r2 = Radius of the driven pulley in m
v = the velocity of the belt in m/sec

The effective tension or force acting at the circumference of the driven pulley is the
difference between the two tensions

Effective driving force, Also,

Power = Torque X angular velocity
So, Power
Torque at driver pulley,
P = Net Force X Distance moved/second

And Angular Velocity,

Belt Drive
Ratio of Driving Tensions R
Consider driven pulley rotating in clockwise direction
Let, T1 = tension in tight side F
T2 = tension in slack side T T+δT
P Q
θ = angle of contact of belt with the pulley
δθ
μ = coefficient of friction between the belt and
pulley θ
Now, considering small portion PQ subtending an
angle δθ at the centre of pulley and is in equilibrium
under the forces
a. Tension T in the belt at P
b. Tension T+ δT in the belt at Q
c. R normal reaction T2 T1
d. Frictional force F

Now, solving forces in radial direction

Since, δθ is very small, putting

Neglecting term δT δθ/2

Belt Drive
Now, solving forces in
tangential direction
Putting cosδθ/2 =1

Now, from the above calculation

or

Now, Integrating both sides between the limits T2 and T1 and from 0 to θ respectively

Belt Drive
Example 5. A belt is running over a pulley of diameter 120 cm at 200
rpm. The angle of contact is 165֯ and coefficient of friction between the
belt and pulley is 0.3. if the maximum tension in the belt is 3000 N, find
the power transmitted by the belt.

Example 6. An open belt drive connects two pulleys 120 cm and 50 cm

diameters, on parallel shafts 4 m apart. The maximum tension in the
belt is 1855.3 N. The coefficient of friction is 0.3. The driver pulley of
diameter 120 cm runs at 200 rpm. Calculate (a) the power transmitted,
and (b) torque on each of the two shafts.

Example 7. A flat belt runs on a pulley 1 m in diameter and transmits

7.5 kW at a speed of 200 rpm. Taking angle of lap as 170 and
coefficient of friction as 0.2. Find the necessary width of the belt, if pull
is not to exceed 196 N/cm width of belt.
Belt Drive
Example 7 Solution.

Solution
Given Now,
R1 = 100 cm
P = 7.5 kW
N = 200 rpm
θ = 170° = 170 X π/180 radian
T1/b = 196 N; b=width of belt

Now, Then,

Also,

Belt Drive
Example 8. Two pulleys, one 450 mm diameter and the other 200 mm diameter are on parallel
1.95 m apart and are connected by a cross belt. Find the length of the belt required and the
angle of contact between the belt and each pulley. What power can be transmitted by the belt
when the larger pulley rotated at 200 rpm, if the maximum permissible tension in the belt is
1X103 N and the coefficient of friction between the belt and pulley is 0.25

Solution
Given, d1 = 0.45 m; r1 = 0.225 m
d2 = 0.2 m; r2 = 0.1m
x = 1.95 m, N1 = 200 rpm, T1 = 1 kN, μ= 0.25
We know that for cross belt system, the length is given by

For angle of contact

between belt and pulley, θ
α = 9.594°
θ = 180°+2α = 199.188°
For tension on slack side, T2
Form this relation, T2 = 419.36 N

Power Transmitted, P = (T1-T2)v =2734.76 W

Belt Drive
Centrifugal Tension
Belt is subjected to centrifugal force as it runs over the pulley with some speed having certain
mass
The tension in the belt caused by that centrifugal force is known as Centrifugal Tension
The force tends to lift the belt from the pulley thereby reducing the normal reaction and
hence the frictional resistance. Fc
Its effect is to reduce the driving power capacity.
C
Consider an elementary length of belt A-B δθ/2 A B
subtending an angle δθ at the centre Tc
δθ
Let, m= mass of the belt per unit length, kg/m Tc
v = linear velocity of belt, m/s
r = radius of pulley, m O
Tc= centrifugal tension acting at A and B, N
Fc = Centrifugal force, N
Now, mass of the elementary length AB = m(r δθ )
Centrifugal force is given by,

Belt Drive
The elementary portion of the belt is in equilibrium under the forces
Fc, Tc and Tc
Resolving the forces,

When δθ is very small, sin(δθ/2) = δθ/2

Hence,

Substituting the value of Fc

Belt Drive
Centrifugal Tension
• Centrifugal tension is Independent of tight side and slack side
tension
• Depends only on the velocity of the belt
• When centrifugal tension is to be taken into consideration
then total tensions on tight side and the slack side of the belt
is Tension on Tight side = T1+Tc
Tension on slack side = T2+Tc
• Hence belt has to design to carry a maximum permissible
tension Where,
f = maximum safe stress in the belt
b = width of the belt
t = thickness of the belt
• Within permissible limit of tension in the belt, the centrifugal
tension has no effect on power transmitted.
• Then, Tm = T1+TC…if centrifugal tension is to be considered
Tm = T1……..if centrifugal tension is to be neglected.
Belt Drive
Maximum Power transmitted by a Belt
We know the power transmitted by belt drive,

Also We have,
Where,
From the above equations

Maximum tension in the belt, Tm = T1+Tc

So,

Differentiating the above equation with respect to v, we get

For maximum power to be transmitted

From Calculation

Maximum Velocity
and maximum Power

Belt Drive
Example 9. The following data relate to a flat belt drive:
Power transmitted 18 kW, pulley diameter 180 cm, angle of contact 175°, speed
of pulley, 300 rpm, coefficient of friction between belt and pulley surface 0.3,
permissible stress for belt 300 N/cm2, thickness of belt 8 mm, density of belt
material 0.95X10-3 gm/cm3.
Determine the width of the belt required taking centrifugal tension into account.

From power equation find b

Belt Drive
Initial tension in the belt
A belt fitted in the pulleys is having some tension because of which it can run continuously
over the pulleys.
This tension is known as Initial Tension To

Let, T1 = Tension in the belt on tight side

T2 = Tension in the belt on slack side
Then,

Considering centrifugal tension

Design of Belt

Let, σt = Tensile stress in Belt, N/m

t = thickness of belt, m
b = width of belt, m
T = maximum tension to which the belt can be subjected

If centrifugal tension is considered

This expression is useful in the design of belt.

Belt Drive
Cone Pulleys
i. Cone pulleys are used when the driven
shaft is required to run at different speeds
ii. A stepped or cone pulley consists of a
number of sets of pulleys of different
diameters cast together as one piece.
iii. The driving shaft runs at constant speed.
Let N = speed of driving shaft and N1, N2,
N3, N4 are speed of driven shaft for pairs 1,
2, 3 and 4 respectively.
If R1, R2, R3 and R4 and r1, r2, r3 and r4
are the radii of the driving and driven
pulleys respectively, then,

For Open Belt For crossed Belt

Use relation L1 = L2=L3=L4 and so on The relation L1 = L2=L3=L4 and so on is valid if

Belt Drive
Example 10: Design a set of stepped pulleys to drive a machine from a countershaft that runs
at 220 rpm. The distance between centers of two sets of pulleys is 2 m. the diameter of the
smallest step on the countershaft is 160 mm. the machine is to run at 80, 100 and 130 rpm
and should be able to rotate in either direction.
Solution:
As the driven shaft is to rotate in either direction, both the cases of a crossed
belt and open belt are to be considered
i. For Crossed belt System:
The smallest step on the countershaft will correspond to the biggest step on the machine
shaft (i.e. minimum speed of the machine shaft.
We have, n1=n2=n3=220 rpm, r1 = 80mm
N1, N2, N3 = 80, 100, 130 rpm respectively
a. For first step

a. For Second
step

Also we have,

Belt Drive
a. For third step

Also we have,

For Open Belt

a. For first step, r1 = 80 mm R1 = 220 mm as before
b. For second step,

From calculation, r2 = 94 mm
R2 = 206.8 mm
Similarly for third step, r3= 112 mm
R3 = 189.3 mm

Belt Drive