Module
13
Belt drives
Version 2 ME , IIT Kharagpur
Lesson
1
Introduction to Belt
drives
Version 2 ME , IIT Kharagpur
Instructional Objectives:
At the end of this lesson, the students should be able to understand:
•
Uses and advantages of belt drives
•
Types of belt drives and their nomenclature
•
Relationship between belt tensions
•
Some commonly used design parameters
13.1.1 Flexible Machine Elements
Belt drives are called flexible machine elements. Flexible machine elements are
used for a large number of industrial applications, some of them are as follows.
1. Used in conveying systems
Transportation of coal, mineral ores etc. over a long distance
2. Used for transmission of power.
Mainly used for running of various industrial appliances using prime movers
like electric motors, I.C. Engine etc.
3. Replacement of rigid type power transmission system.
A gear drive may be replaced by a belt transmission system
Flexible machine elements has got an inherent advantage that, it can absorb a
good amount of shock and vibration. It can take care of some degree of
misalignment between the driven and the driver machines and long distance
power transmission, in comparison to other transmission systems, is possible.
For all the above reasons flexible machine elements are widely used in industrial
application.
Although we have some other flexible drives like rope drive, roller chain drives
etc. we will only discuss about belt drives.
13.1.2 Typical belt drives
Two types of belt drives, an open belt drive, (Fig. 13.1.1) and a crossed belt drive
(Fig. 13.1.2) are shown. In both the drives, a belt is wrapped around the pulleys.
Let us consider the smaller pulley to be the driving pulley. This pulley will transmit
motion to the belt and the motion of the belt in turn will give a rotation to the
larger driven pulley. In open belt drive system the rotation of both the pulleys is in
the same direction, whereas, for crossed belt drive system, opposite direction of
rotation is observed.
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13.1.3 Nomenclature of Open Belt Drive
d
L
- Diameter of the larger pulley
d
S
– Diameter of the smaller pulley
α
L
- Angle of wrap of the larger pulley
α
S
– Angle of wrap of the smaller pulley
L
C
- Center distance between the two pulleys
Basic Formulae
α
L
=
180
ο
+ 2
β
α
S
=
180
ο
- 2
β
Where angle
β
is,
L
0
= Length of open belt
This formulae may be verified by simple geometry.
13.1.4 Nomenclature of Cross Belt Drive
d
L
- Diameter of the larger pulley
d
S
– Diameter of the smaller pulley
α
L
- Angle of wrap of the larger pulley
α
S
– Angle of wrap of the smaller pulley
C
- Center distance between the two pulleys
α
S
α
C
β
d
s
d
L
Fig.13.1.1 Open belt drive
1
L
S
d
d
sin
2C
−
−
⎛
⎞
β =
⎜
⎟
âŽ
âŽ
(
)
(
)
2
o
L
S
L
1
L
d
d
2C
d
d
2
4C
Ï€
=
+
+
+
−
S
Fig. 13.1.2 Cross belt drive
L
α
S
α
β
C
d
s
d
L
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Basic Formulae
α
L =
α
S
= 180
ο
+ 2
β
Where angle
β
is,
1
L
S
d
d
sin
2C
−
−
⎛
⎞
β =
⎜
⎟
âŽ
âŽ
Length of cross belt
(
)
(
)
2
c
L
S
L
1
L
d
d
2C
d
d
2
4C
Ï€
=
+
+
+
+
S
13.1.5 Belt tensions
The belt drives primarily operate on the friction principle. i.e. the friction between
the belt and the pulley is responsible for transmitting power from one pulley to the
other. In other words the driving pulley will give a motion to the belt and the
motion of the belt will be transmitted to the driven pulley. Due to the presence of
friction between the pulley and the belt surfaces, tensions on both the sides of
the belt are not equal. So it is important that one has to identify the higher tension
side and the lower tension side, which is shown in Fig. 13.1.3.
Belt motion
Fig.13.1.3 Belt tensions
Driving pulley
Driven pulley
Friction
on pulley
Friction
on belt
T
2
T
2
T
1
T
1
T
1
>T
2
When the driving pulley rotates (in this case, anti-clock wise), from the
fundamental concept of friction, we know that the belt will oppose the motion of
the pulley. Thereby, the friction,
f
on the belt will be opposite to the motion of the
pulley. Friction in the belt acts in the direction, as shown in Fig. 13.1.3, and will
impart a motion on the belt in the same direction. The friction
f
acts in the same
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direction as T
2
. Equilibrium of the belt segment suggests that T
1
is higher than
T
2
. Here, we will refer T
1
as the tight side and
T
2
as the slack side, ie, T
1
is higher
tension side and
T
2
is lower tension side.
Continuing the discussion on belt tension, the figures though they are
continuous, are represented as two figures for the purpose of explanation. The
driven pulley in the initial stages is not rotating. The basic nature of friction again
suggests that the driven pulley opposes the motion of the belt. The directions of
friction on the belt and the driven pulley are shown the figure. The frictional force
on the driven pulley will create a motion in the direction shown in the figure.
Equilibrium of the belt segment for driven pulley again suggests that T
1
is higher
than T
2
.
It is observed that the slack side of the belt is in the upper side and the tight side
of the belt is in the lower side. The slack side of the belt, due to self weight, will
not be in a straight line but will sag and the angle of contact will increase.
However, the tight side will not sag to that extent. Hence, the net effect will be an
increase of the angle of contact or angle of wrap. It will be shown later that due to
the increase in angle of contact, the power transmission capacity of the drive
system will increase. On the other hand, if it is other way round, that is, if the
slack side is on the lower side and the tight side is on the upper side, for the
same reason as above, the angle of wrap will decrease and the power
transmission capacity will also decrease. Hence, in case of horizontal drive
system the tight side is on the lower side and the slack side is always on the
upper side.
13.1.6 Derivation of relationship between belt tensions
The Fig.13.1.4 shows the free body diagram of a belt segment.
d
φ
2
d
φ
d
2
φ
r
dN
dN
μ
CF
α
2
2
v
centrifugal force(CF) m(rd
φ
)
r
mv d
φ
where, m
bt
Ï
,
=
=
=
T
1
T
2
T
T+dT
Fig.13.1.4
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The belt segment subtends an angle
d
φ
at the center. Hence, the length of the
belt segment,
dl = r d
φ
(13.1.1)
At the impending condition, ie., when the belt is in just in motion with respect to
the pulley, the forces acting on the belt segment are shown in Fig.13.1.4. This
belt segment is subjected to a normal force acting from the pulley on the belt
segment and due to the impending motion the frictional force will be acting in the
direction as shown in the figure.
f =
μ
dl
(13.1.2)
where
μ
is the coefficient of friction between the belt and the pulley.
The centrifugal force due to the motion of the belt acting on the belt segment is
denoted as CF and its magnitude is,
CF = [m(rd
φ
)x v
2
]/r = mv
2
d
φ
(13.1.3)
Where,
v
is the peripheral velocity of the pulley
m
is the mass of the belt of unit
length,
m = bt
Ï
(13.1.4)
where,
b
is the width,
t
is the thickness and
Ï
is the density of the belt material.
From the equation of equilibrium in the tangential and normal direction,
(
)
d
d
Tcos
T dT cos
dN 0
2
2
φ
φ
−
+
+ μ
=
(13.1.5)
t
F
0
∑ =
(
)
2
mv d
dN T sin
d
d
T dT sin
0
2
2
φ
φ
⎛
⎞
φ +
+
−
+
=
⎜
⎟
âŽ
âŽ
(13.1.6)
n
F
0
∑
=
For small angle, d
φ
,
d
cos
and
1
2
φ
≈
d
d
sin
φ
2
2
φ
≈
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Therefore, simplified form of (13.1.5) is,
dT
dN
=
μ
(13.1.7)
From (13.1.6) and using (13.1.7),
2
dT
mv d
Td
0
φ +
−
φ =
μ
or,
2
dT
d
T mv
= μ φ
−
(13.1.8)
Considering entire angle of wrap,
2
1
T
2
T
dT
d
T mv
α
=
μ φ
−
∫
∫
0
(13.1.9)
The final equation for determination of relationship between belt tensions is,
2
1
2
2
T
mv
e
T
mv
μα
−
=
−
(13.1.10)
It is important to realize that the pulley, driven or driver, for which the product,
μα
of (13.1.10) is the least, should be considered to determine the tension ratio.
Here,
α
should be expressed in radians.
13.1.7 Elastic Creep and Initial Tension
Presence of friction between pulley and belt causes differential tension in the
belt. This differential tension causes the belt to elongate or contract and create a
relative motion between the belt and the pulley surface. This relative motion
between the belt and the pulley surface is created due to the phenomena known
as elastic creep.
The belt always has an initial tension when installed over the pulleys. This initial
tension is same throughout the belt length when there is no motion. During
rotation of the drive, tight side tension is higher than the initial tension and slack
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side tension is lower than the initial tension. When the belt enters the driving
pulley it is elongated and while it leaves the pulley it contracts. Hence, the driving
pulley receives a larger length of belt than it delivers. The average belt velocity
on the driving pulley is slightly lower than the speed of the pulley surface. On the
other hand, driven pulley receives a shorter belt length than it delivers. The
average belt velocity on the driven pulley is slightly higher than the speed of the
pulley surface.
Let us determine the magnitude of the initial tension in the belt.
Tight side elongation
âˆ
(T
1
– T
i
)
Slack side contraction
âˆ
(T
i –
T
2
)
Where,
T
i
is the initial belt tension .
Since, belt length remains the same, ie, the elongation is same as the
contraction,
1
2
i
T
T
T
2
+
=
(13.1.11)
It is to be noted that with the increase in initial tension power transmission can be
increased. If initial tension is gradually increased then
T
1
will also increase and at
the same time
T
2
will decrease. Thus, if it happens that
T
2
is equal to zero, then
T
1
= 2T
i
and one can achieve maximum power transmission.
13.1.8 Velocity ratio of belt drive
Velocity ratio of belt drive is defined as,
(
S
L
S
L
d
t
N
1 s
N
d
t
)
+
=
+
−
(13.1.12)
where,
N
L
and
N
S
are the rotational speeds of the large and
the small pulley
respectively,
s
is the belt slip and
t
is the belt thickness.
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13.1.9 Power transmission of belt drive
Power transmission of a belt drive is expressed as,
P = ( T
1
– T
2
)v
(13.1.13)
where,
P
is the power transmission in Watt and
v
is the belt velocity in m/s.
Sample problem
A pump is driven by an electric motor through a open type flat belt drive.
Determine the belt specifications for the following data.
Motor pulley diameter(d
S
) = 300 mm, Pump pulley diameter(d
L
) = 600 mm
Coefficient of friction (
μ
S
) for motor pulley = 0.25
Coefficient of friction (
μ
L
) for pump pulley = 0.20
Center distance between the pulleys=1000 mm; Rotational speed of the
motor=1440 rpm;
Power transmission = 20kW; density of belt material (
Ï
)= 1000 kg/m
3
; allowable
stress for the belt material (
σ
) = 2 MPa; thickness of the belt = 5mm.
Solution
(
)
(
)
(
)
(
)
1
0
L
s
0
L
0
S
2
o
L
S
L
S
2
D e t e r m i n a t i o n o f a n g l e o f w r a p
d
d
s i n
(
)
8 . 6 3
2 C
1 8 0
2
1 9 7 . 2 5
3 . 4 4 r a d
1 8 0
2
1 6 2 . 7 5
2 . 8 4 r a d
L e n g t h o f o p e n b e l t
1
L
d
d
2 C
d
d
2
4 C
1
=
6 0 0
3 0 0
2 0 0 0
6 0 0
3 0 0
= 3 4 3 6 m m
2
4 0 0 0
β
α
β
α
β
Ï€
Ï€
−
−
=
=
=
+
=
=
=
−
=
=
=
+
+
+
−
+
+
+
−
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3
3
3
2
s
s
L
L
0.688
1
2
300 1440
v
22.62 m / s
60 1000
b
5
m
bt
10
0.005kg / m
10
10
mv
2.56 b N
Now,
0.25 2.84
0.71
0.20 3.44
0.688
l arg er pulley governs the design
T
2.56b
e
1.99.........................
T
2.56b
Ï€
Ï
μ α
μ α
×
×
=
=
×
=
=
×
×
=
=
×
=
×
=
=
×
=
∴
−
=
=
−
1
2
1
2
1
( 1 )
power equation
P
( T
T ) v
putting data,
( T
T )
884.17 N
................................( 2 )
again,T
2 b 5N
10bN ( from permissible stress )................( 3 )
From
=
−
×
∴
−
=
= × ×
=
( 1 ),( 2 ) and ( 3 ), solving for b,
b
240 mm
Hence,the required belt dim ensions are,
Length
3436 mm; breadth
240mm and thickness
5mm
≈
=
=
=
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Questions and answers
Q1. What are the advantages of a belt drive?
A1. The advantages of a belt drive are that, it can absorb a good amount of
shock and vibration. It can take care of some degree of misalignment
between the driven and the driver machines and long distance power
transmission, in comparison to other transmission systems, is possible.
Q2. Why the slack side of the belt of a horizontal belt drive is preferable to place
on the top side?
A2. The slack side of the belt is preferably placed on the top side because, the
slack side of the belt, due to its self weight, will sag. For this reason the
angle of contact between the belt and the pulleys will increase. However,
the tight side will not sag to that extent. Hence, the net effect will be an
increase in the angle of contact or angle of wrap. Thus, due to the increase
in angle of contact, the power transmission capacity of the drive system will
increase.
Q3. Which one should be the governing pulley to calculate tension ratio?
A3. The pulley, driven or driver, for which the product,
μα
of equation for belt
tension is the least, should be considered to determine the tension ratio.
References
1. V.Maleev and James B. Hartman , Machine Design, CBS Publishers And
Distributors.3
rd
Edition. 1983.
2. J.E Shigley and C.R Mischke , Mechanical Engineering Design , McGraw
Hill Publication, 5
th
Edition. 1989.
3. M.F Spotts, Design of Machine Elements, Prentice Hall India Pvt. Limited,
6
th
Edition, 1991.
4. Khurmi, R.S. and Gupta J.K., Text book on Machine Design, Eurasia
Publishing House, New Delhi.
5. Sharma, C.S. and Purohit Kamalesh, Design of Machine Elements,
Prentice Hall of India, New Delhi, 2003.
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