The document discusses different types of belting used to transmit power between rotating shafts in factories, including flat belts and V-belts. It provides objectives and formulas for calculating the length of open and closed belt drives, as well as the power transmitted by a belt based on the tension in the tight and slack sides and the belt velocity. Worked examples are included to demonstrate calculating belt length and transmitted power.

1. BELTING J 3010/6/1
UNIT 6
BELTING
OBJECTIVES
General Objective : To understand and apply the concept of belting
Specific Objectives : At the end of this unit you will be able to:
 state the difference between open and close belt.
 explain that power transmitted by the flat belt and V belt.
 explain that ratio tension for flat and V belt.
 calculate power transmitted by the belt with consider
centrifugal force.

2. BELTING J 3010/6/2
INPUT
6.0 INTRODUCTION
In factories, the power has to be
transmitted from one shaft to another,
then belt driving between pulleys on
the shafts may be used.
The pulley rotating shaft is called driver. The pulley intended to rotate is
known, as follower or driven. When the driver rotates, it carries the belt that
grip between its surface and the belt. The belt, in turn, carries the driven
pulley which starts rotating. The grip between the pulley and the belt is
obtained by friction, which arises from pressure between the belt and the
pulley.
(a) Types of belts
The flat belt is mostly used in the factories and work shops,
where a moderate amount of power is to be transmitted, from one
pulley to another, when the two pulleys are not more than 10 m a part.
The V-belt is mostly used in the factories and work shops
where a great amount of power is to be transmitted, from one pulley
to another, when the two pulleys are very near to each other.

3. BELTING J 3010/6/3
The types of belts:-
a. flat belt
b. V belt
6.1 LENGTH OF AN OPEN BELT DRIVE
A
B
K
α1 α2 C
O1 O2
F
r2
r1
D
E
d
Fig.6.1 Open belt drive
O1 and O2 = Centers of two pulleys
r1 and r2 = radius of the larger and smaller pulleys
d = Distance between O1 and O2
L = Total length of the belt.
Angle AO1O2 = α1
Angle BO2C = α2
Angle AFE = θ1 (radian)
Angle BCD = θ2 (radian)

4. BELTING J 3010/6/4
We know that the length of the belt,
L = Arc AFE + ED + Arc DCB + BA
r1  r2
Cos α1 =
d
r r
α1 = cos-1 1 2
d
θ1 = 2π - 2α1
= 2 (π - α1) (radian)
θ2 = α1 = 2 α2 (radian)
Arc AFE = r1θ1
Arc DCB = r2θ2
And ED = BA = KO2
KO2
Sin α1 =
d
KO2 = d Sin α1
Finally the total of length of belt,
L = Arc AFE + ED + Arc DCB + BA
= r1θ1 + d Sin α1 + r2θ2 + d Sin α1
= r1θ1 + r2θ2 + 2d Sin α1

5. BELTING J 3010/6/5
Example 6.1
Find the length of belt necessary to drive a pulley of 480 cm diameter running
parallel at a distance of 12 meter from the driving pulley of diameter 80 cm. This
system is an open belt drive.
Solution 6.1
A
B
K
F α1 O2 α2 C
O1
r2
r1 D
E
d
Fig.6.2 Open belt drive
O1 and O2 = Centers of two pulleys
r1 and r2 = radius of the larger and smaller pulleys
d = Distance between O1 and O2
L = Total length of the belt.
Angle AO1O2 = α1
Angle BO2C = α2
Angle AFE = θ1 (radian)
Angle BCD = θ2 (radian)
Radius of smaller pulley = 80/2 = 40 cm.
Radius of larger pulley = 480/2 = 240 cm.
Distance between the pulleys, d = 12m = 1 200 cm
We know that the length of belt is,
L = Arc AFE + ED + Arc DCB + BA
r1  r2 240  40
Cos α1 = =
d 1200

6. BELTING J 3010/6/6
240  40
α1 = cos-1
1200
= cos-1 0.16667
= 80º
α1 = 1.396 radian
θ1 = 2π - 2α1
= 2 (π - 1.396 ) (radian)
= 3.491 radian
θ2 = 2α1 = 2 α2 (radian)
= 2 (1.396 )
= 2.792 radian
Arc AFE = r1θ1
= 240 x 3.491
= 837.84 cm
Arc DCB = r2θ2
= 40 x 2.792
= 111.68 cm
And ED = BA = KO2
KO2
Sin 80º =
d
KO2 = d Sin α1 = 1 200 x 0.98481
= 1181.77 m
Finally the total of belt length is
L = Arc AFE + ED + Arc DCB + BA
= r1θ1 + d Sin α1 + r2θ2 + d Sin α1
= r1θ1 + r2θ2 + 2d Sin α1
= 240 x 3.491 + 40 x 2.792 +2 x 1181.77
= 837.84 + 111.68 + 2363.54
= 3313.06 cm
= 33.13 m

7. BELTING J 3010/6/7
6.2 LENGTH OF CLOSE BELT DRIVE
K
A
D
C
O1 α1 α2 O2
F
r2
r1
B
E
d
Fig.6.3 Cross belt drive
O1 and O2 = Centers of two pulleys
r1 and r2 = radius of the larger and smaller pulleys
d = Distance between O1 and O2
L = Total length of the belt.
Angle AO1O2 = α1
Angle DO2O1 = α2
Angle AFE = θ1 (radian)
Angle BCD = θ2 (radian)
We know that the length of the belt is
L = Arc AFE + ED + Arc DCB + BA
r1  r2
Cos α1 =
d
r r
α1 = cos-1 1 2
d
θ1 = 2π - 2α1
= 2 (π - α1) (radian)
θ1 = θ2

8. BELTING J 3010/6/8
Arc AFE = r1θ1
Arc DCB = r2θ2
And ED = BA = KO2
KO2
Sin α1 =
d
KO2 = d Sin α1
Finally the total of length belt,
L = Arc AFE + ED + Arc DCB + BA
= r1θ1 + d Sin α1 + r2θ2 + d Sin α1
= r1θ1 + r2θ2 + 2d Sin α1
Example 6.2
Find the length of belt for a cross belt drive system. The diameter of
the drive pulley is 480 cm which running parallel at a distance of 12
meter from the driving pulley which has a diameter of 80 cm.
Solution 6.2
K
A
D
α1 α2 C
F O1 O2
r2
r1
B
E
d
Fig.6.4 Cross belt drive

9. BELTING J 3010/6/9
O1 and O2 = Centers of two pulleys
r1 and r2 = radius of the larger and smaller pulleys
d = distance between O1 and O2
L = Total length of the belt.
Angle AO1O2 = α1
Angle DO2O1 = α2
Angle AFE = θ1 (radian)
Angle BCD = θ2 (radian)
Radius of smaller pulley = 80/2 = 40 cm.
Radius of larger pulley = 480/2 = 240 cm.
Distance between the pulleys, d = 12m = 1 200 cm
We know that the length of the belt,
L = Arc AFE + ED + Arc DCB + BA
r1  r2 240  40
Cos α1 = =
d 1200
240  40
α1 = cos-1
1200
-1
= cos 0.23333
= 76.5º
α1 = 1.335 radian
θ1 = 2π - 2α1
= 2 (π - α1) (radian)
= 2 (π - 1.335) (radian)
= 3.613 radian
θ1 = θ2
Arc AFE = r1θ1
= 240 x 3.613
= 867.12 cm
Arc DCB = r2θ2
= 40 x 3.613
= 144.52 cm
And
ED = BA = KO2
KO2
Sin α1 =
d
KO2 = d Sin 76.5º =1 200 x 0.9724
ED = BA = KO2 = 1166.88 cm

10. BELTING J 3010/6/10
Finally the total of length belt,
L= Arc AFE + ED + Arc DCB + BA
= r1θ1 + d Sinα1 + r2θ2 + d Sin α1
= r1θ1 + r2θ2 + 2d Sin α1
= 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.5º
= 867.12 + 144.52 + 2333.76
= 3345.4 cm
L = 33.45 m

11. BELTING J 3010/6/11
Activity 6A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mm
diameter are on parallel shaft of 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.
6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel
shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine
the size of pulley on the shaft B.
6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter running
parallel at a distance of 1.7 meter from the driving pulley of diameter 0.5 m.
It is connected by an open belt.
N1 d 2

N 2 d1
N1 = speed diver in rpm
N2 = speed follower in rpm
d1 ,d2 = diameter pulley driver and
follower.

12. BELTING J 3010/6/12
Feedback to Activity 6A
Have you tried the questions????? If “YES”, check your answers now
6.1 4.975 m; 199 or 3.473 radian.
6.2. 11.6 cm
6.3 6.51 m

13. BELTING J 3010/6/13
INPUT
6.3 POWER TRANSMITTED BY A BELT
Power = (T1 –T2) v
Where, T1 =Tension in tight side
in Newton.
T2 = Tension in slack side.
V = velocity of belt
Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B.
The driving pulley pulls the belt from one side, and delivers the same to the
other.The maximum tension in the tight side will be greater than that
slack side.
Fig. 6.5
Torque exerted on the driving pulley
= ( T1 - T2) r1
Torque exerted on the driven or follower
= ( T1 - T2) r2
Power transmitted = Force x distance
= ( T1 - T2) v

14. BELTING J 3010/6/14
Example 6.3
The tension in the two sides of the belt are 100 N and 80 N respectively. If
the speed of the belt is 75 meters per second. Find the power transmitted by
the belt.
Solution 6.3
Given,
Tension in tight side,
T1 = 100 N
Tension in slack side,
T2 = 80 N
Velocity of belt, v = 75 m/s
Let P = Power transmitted by the belt
Using the relation,
Power = (T1 –T2) v
= ( 100 – 80) 75
= 1500 watt
= 1.5 kw.

15. BELTING J 3010/6/15
Activity 6B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
6.4. Find the tension in the tight side, if the tension in slack side 150 N. The speed
of the belt is 58 meters per second and the power transmitted by this
belt is 2 kw.
6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotating
at 300 rev/ min. Calculate velocity of driver.
6.6 The tension in the two sides of the belt are 300 N and 180 N respectively. If
the speed of the belt is 50 meters per second, find the initial tension and
power transmitted by the belt.
6.7 A 100 mm diameter pulley is belt-driven from a 400 mm diameter pulley.
The 400 mm pulley rotates at 480 rev/min. The number of rev/sec of the
100 mm diameter pulley is
A. 2 B. 32 C. 120 D. 1920
T1  T2
T0 =
2
Where, T0 = Initial tension in the belt.

16. BELTING J 3010/6/16
Feedback to Activity 6B
Have you tried the questions????? If “YES”, check your answers now
6.4 115.5 N
6.5 0.785 m/s
6.6 6 000 watt or 6 kw; 240 N
6.7 B. 32

17. BELTING J 3010/6/17
INPUT
6.4 RATIO OF TENSIONS.
Fig. 6.6
Consider a driven pulley rotating in the clockwise direction as in fig 6.6.
Let T1 = Tension in the belt on the tight side.
T2 = Tension in the belt on the slack side.
ө = An angle of contact in radians (i.e , angle subtended by the arc
AB, along which the belt touches the pulley, at the centre)
Now consider a small position of the belt PQ, subtending an angle  at the
centre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the
following force:
i. Tension T in the belt at P.
ii. Tension T + T in the belt at Q.
iii. Normal reaction R, and
iv. Frictional force F =  x R
Where  is the coefficient of friction between the belt and pulley.
Resolving all the forces horizontally and equating the same,

18. BELTING J 3010/6/18
 
R = ( T + T) sin + T sin (6.1)
2 2
Since the  is very small, therefore, substituting,
 
sin = in equation 6.1
2 2
 
R = ( T + T) +T
2 2
T  T  T .
= + +
2 2 2
T 
= T. ( neglecting ) (6.2)
2
Now resolving the forces vertically,
 
 x R = (T + T) cos - T cos (6.3)
2 2

Since the angle . is very small, therefore substituting cos = 1 in
2
equation 6.3 or, xR = T + T – T = T
T
R = (6.4)

Equating the values of R from equation 6.2 and 6.4,
T
Or T. =

T
Or = .
T
Integrating both sides from A to B,

T
T1
T
T2
   
0
T1
or loge = 
T2
T1
= e  ration of tension for flat belt
T2
T1
= e /sin  ration of tension for V belt
T2
where  = half angle of groove

19. BELTING J 3010/6/19
Example 6.4
Find the power transmitted by a belt running over a pulley of 60 cm diameter
at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap
160 and maximum tension in the belt is 250 KN.
Solution 6.4
Given,
Diameter of pulley, d = 60 cm = 0.6 m
Speed of pulley N = 200 rpm
 dN  x 0.6 x 200
Speed of belt, v = = = 2 = 6.28 m/ sec
60 60
Coefficient of friction,
μ = 0.25
160 x 
Angle of contact,  = 160 = = 2.7926 radian.
180
Maximum tension in the belt,
T1 = 250 kN
Let P = power transmitted by the belt.
Using the relation,
T1
= e 
T2
= e 0.25 x 2.7926
T1
= 2.01
T2
T1 250
T2 = = = 124.38 kN
2.01 2.01
Now using the relation,
P = (T1 –T2 ) v
= ( 250 - 124.38 ) 6.28
= 788.89 watt
P = 0.79 kw

20. BELTING J 3010/6/20
INPUT
6.5 CENTRIFUGAL TENSION.
Fig.6.7
The tension caused by centrifugal force is called centrifugal tension. At lower
speeds the centrifugal tension is very small, but at higher speeds its effect is
considerable, and thus should be taken into account.
Consider a small portion PQ of the belt subtending an angle d at the centre
of the pulley as show in fig 6.7.
Let M = mass of the belt per unit length,
V = linear velocity of the belt,
r = radius of the pulley over which the belt runs,
Tc = Centrifugal tension acting tangentially at P and Q.
Length of the belt PQ,
= r d
Mass of the belt PQ,
M = M r d
We know that centrifugal force,
= M v2/r

21. BELTING J 3010/6/21
Centrifugal force of the belt PQ
M x r.d v 2
=
r
= M x d v 2
The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in
equilibrium.
Now resolving the force (i.e, centrifugal force and centrifugal tension)
horizontally and equating the same,
d
2 Tc sin = M x d v 2
2
since angle d is very small, therefore, substituting
d d
sin =
2 2
d
2 Tc = M x d v 2
2
Tc = M x d v 2 / d
Tc = M v2
T1  Tc
 e ratio tension for flat belt.
T2  Tc
T1  Tc
 e   / sin  ratio tension for V belt
T2  Tc
6.5.1 Condition for the transmission of maximum power
The maximum power,
When T c = ⅓T1
It shows that the power transmitted is maximum ⅓ of the maximum tension
is absorbed as centrifugal tension.
The velocity of belt for maximum transmission of power may be
obtained from equation T1 = 3Tc = M v2.

22. BELTING J 3010/6/22
3 T1
v2 =
M
T1
v =
3M
Example 6.5
An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on
parallel shafts 3.6 m apart. The belt has a mass of 0.9 kg/m length and the
maximum tension in it is not exceed 2 kN. The larger pulley runs at 200
rev/min. Calculate the torque on each of the two shafts and the power
transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller
pulley is 168° ( 2.947 radian ).
Solution 6.5
Given
Diameter of larger pulley,
d1 = 1.2 m
Radius of the larger pulley,
r1 = 0.6
Diameter of smaller pulley,
d2 =0.5 m
Radius of the smaller pulley,
r2 = 0.25 m
Distance between two shaft,
D = 3.6 m
Mass of the belt per meter length,
M = 0.9 kg/m
Maximum Tension, T1 = 2 kN = 2 000 N
Speed of the larger pulley
= 200 rpm
Velocity of the belt,
 x 1.2 x 200
V= = 12.57 m/s
60
T c = Mv2
= 0.9 x 12.572
= 142.2 N

23. BELTING J 3010/6/23
e µθ = e 0.3 x 2.947
= 2.421
Using the relation,
T1  Tc
 e
T2  Tc
2000  142.2
= 2.421
T2  142.2
(T2 – 142.2) 2.421 = 1857.8
1857.8
(T2 – 142.2) = = 767.37
2.421
T2 = 767.37 + 142.2
T2 = 909.57 N
We know that the torque on the larger pulley shaft (TL),
TL = ( T1 – T2) r1
= ( 2000 – 909.57) 0.6
= 654.26 Nm.
Torque on the smaller pulley shaft (Ts),
Ts = ( T1 – T2) r2
= (2000 – 909.57) 0.25
= 272.61 Nm.
Power transmitted, P = ( T1 – T2)v
= ( 2000 – 909.57) 12.57
= 13706.71 watt
= 13.71 kw

24. BELTING J 3010/6/24
Activity 6C
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT
INPUT…!
6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mm
diameter are on parallel of shafts 1.95 m apart. What power can be
transmitted by the belt when the larger pulley rotates at 200 rev/min, if the
maximum permissible tension in the belt is 1 kN and the coefficient of
friction between the belt and pulley is 0.25. Angle of contact between the belt
and larger pulley is 3.477 radian.
6.9 Find the power transmitted by a V drive from the following data:
Angle of contact = 84º
Pulley groove angle = 45º
Coefficient of friction = 0.25
Mass of belt per meter length = 0.472 kg/m
Permissible tension = 139 N
Velocity of V belt = 12.57 m/s.
6.10 An open belt drive connects two pulleys 1.2 m and 0.6 m, on parallel shafts 3
m apart. The belt has a mass 0f 0.56 kg/m and maximum tension is 1.5 kN. The
driver pulley runs at 300 rev/min. Calculate the inertial tension, power
transmitted and maximum power. The coefficient of friction between the belt
and the pulley surface is 0.3.

25. BELTING J 3010/6/25
Feedback to Activity 6C
Have you tried the questions????? If “YES”, check your answers now
6.8 2 740 watt or 2.74 kw
6.9 591 watt.
6.10 1074.5 N ; 8.0 kw ; 17.54 kw.

26. BELTING J 3010/6/26
SELF-ASSESSMENT 6
You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback on Self-Assessment 6
given on the next page. If you face any problems, discuss it with your lecturer.
Good luck.
1. A pulley is driven by a flat belt running at speed of 600 m/min. The
coefficient of friction between the pulley and the belt is 0.3 and the angle lap
is 160°. If the maximum tension in the belt is 700 N, find the power
transmitted by a belt.
2. A leather belt, 125 mm wide and 6 mm thick, transmits power from a pulley
of 750 mm diameter which run at 500 rpm. The angle of lap is 150° and
µ = 0.3. If the mass of 1 m3 of leather is 1 Mg and the stress in the belt does
not exceed 2.75 MN/m2, find the maximum power that can be transmitted.
3. A flat belt, 8 mm thick and 100 mm wide transmits power between two
pulleys, running at 1 600 m/min. The mass of the belt is 0.9 kg/m length. The
angle of lap in the smaller pulley is 165° and the coefficient of friction
between the belt and pulleys is 0.3. If the maximum permissible stress in the
belt is 2 MN/m2, find
(i) Maximum power transmitted; and
(ii) Initial tension in the belt.
4. The moment on a pulley, which produces the rotation of the pulley is called:
A. Momentum B. Torque C. Work D. Energy
5. If T1 and T2 are the tension in the tight and slack sides of a belt and θ is the
angle of contact between the belt and pulley. Coefficient of friction is μ, then
the ratio of driving tension will be:
T T T T
A. 2  e   B. 1  e   C. 1   D. log10 1  
T1 T2 T2 T2

27. BELTING J 3010/6/27
Feedback to Self-Assessment 6
Have you tried the questions????? If “YES”, check your answers now.
1. 3.974 kw ( see example 6.4)
2. 18.97 kw
3. (i) 14.281 kw
(ii) 1.3221 kN
4. B. Torque
T1
5. B.  e CONGRATULATIONS!!!!…..
T2 May success be with you
always….