Design of Flat belt, V belt and chain drives

Unit-5
Design of Belt and Chain Drives »
Belt Drives
Dr. L.K. Bhagi
Associate Professor
School of Mechanical Engineering
Lovely Professional University

Flexible Drives
There are two types of drives—
Rigid and Flexible.
Belt, chain and rope drives are called ‘flexible’
drives.
Gear drives are called rigid or non-flexible drives
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Flexible Drives vs Gear Drives
❖In gear drives, there is direct contact between
the driving and driven shafts through the gears.
❖In flexible drives, there is an intermediate link
such as belt, rope or chain between the driving
and driven shafts. Since this link is flexible, the
drives are called ‘flexible’ drives.
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Flexible Drives vs Gear Drives
❖In gear drives, rotary motion of the driving shaft
is directly converted into rotary motion of the
driven shaft by means of pinion and gear.
❖In flexible drives, the rotary motion of the driving
shaft is first converted into translatory motion of
the belt or chain and then again converted into
rotary motion of the driven shaft.
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Flexible Drives » Advantages
The advantages of flexible drives over rigid drives are
as follows:
(i) Flexible drives transmit power over a
comparatively long distance due to an intermediate
link between driving and driven shafts.
(ii) Since the intermediate link is long and flexible, it
absorbs shock loads and damps vibrations.
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Flexible Drives » Advantages
iii. Flexible drives provide considerable flexibility in
the location of the driving and driven shafts.
iv. Flexible drives are cheap compared to gear drives.
Their initial and maintenance costs are low.
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Flexible Drives » Disadvantages
The disadvantages of flexible drives are as follows:
(i) They occupy more space.
(ii) The velocity ratio is relatively small.
(iii)In general, the velocity ratio is not constant.
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Belt Drives
Belts are used to transmit power between two shafts by
means of friction.
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Elements of Belt drive
A belt drive consists of three elements—
Driving
Driven pulleys and
Endless belt.
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Belt Drives
It may be noted that
i. The shafts should be properly in line to insure
uniform tension across the belt section.
ii. The pulleys should not be too close together.
iii. The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus
increasing the friction load on the bearings.
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Belt Drives
iv. A long belt tends to swing from side to side,
causing the belt to run out of the pulleys.
v. 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.
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Belt Drives
vi. In order to obtain good results with flat belts, the
maximum distance between the shafts should not
exceed 10 meters and the minimum should not be
less than 3.5 times the diameter of the larger pulley.
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Velocity Ratio
For flat belt is up to 4:1.
For V-belts the velocity ratio is up to 7:1.
For chain drives it can be up to 15:1.
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Belt Construction
The desirable properties of belt materials are as follows:
(i) The belt material should have high coefficient of
friction with the pulleys.
(ii) The belt material should have high tensile strength
to withstand belt tensions.
(iii) The belt material should have high wear resistance.
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Belt Construction
iv. The belt material should have high flexibility and
low rigidity in bending in order to avoid bending
stresses while passing over the pulley.
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Belt Construction » Material Used
The belts, according to the material used, are classified
as follows:
✓ Leather belts
✓ Cotton or fabric belts
✓ Rubber belt
✓ Balata belts (These belts are similar to rubber belts
except that balata gum is used in place
of rubber.)
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Belt Construction » Belt Joints
When the endless belts are not available, then the belts
are cut from big rolls and the ends are joined together
by fasteners. The various types of joints are
1. Cemented joint,
2. Laced joint, and
3. Hinged joint.
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1. Cemented joint
Leather belts are cemented with a
tapered lap joint of length of 20 to 25
times the belt thickness.
Multiple ply belts are cemented in the
form of stepped joint.
Cementing is widely used for rubber and
leather belts.
The strength of cemented joint is 80 to
85% of the strength of the belt.
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2. Laced joint
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3. Hinged joint
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The strength of ……. joint is ………% of the strength of the
belt.

Types of Belt Drives
Open belt drive
used with shafts arranged parallel and rotating in the
same direction.
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Crossed belt drive
used with shafts arranged parallel and rotating in the
opposite direction.
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Open Belt Drive Crossed Belt Drive
Belt proceeds from the top of one
pulley to the top of another without
crossing.
Belt proceeds from the top of one
pulley to the bottom of another and
crosses over itself.
Both driving and driven pulleys rotate
in the same direction.
Driving and driven pulleys rotate in the
opposite direction
Angle of wrap is less Angle of wrap is more.
Power transmitting capacity ………. Power transmitting capacity of a
crossed belt drive is more than that of
an open belt drive
Wear is less wear is more and reduce the life of the
belt.
Belt Drives»
Open vs Crossed Belt Drives

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Open Belt Drive Crossed Belt Drive
when the centre distance is more, the
belt whips, i.e., vibrates in a direction
perpendicular to the direction of
motion.
When the centre distance is small, the
belt slip increases. Both these factors
limit the use of an open belt drive.
do not have these limitations
Open belt drives are more popular than crossed belt drives.
Belt Drives»
Open vs Crossed Belt Drives

Quarter Turn belt drive
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Belt Drive with Idler Pulleys
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Belt Drive with Idler Pulleys
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Compound Belt Drive
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is used when power is transmitted from one shaft to another
through a number of pulleys.

Compound Belt Drive
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is used when power is transmitted from one shaft to another
through a number of pulleys.

Stepped or cone pulley drive
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is used for changing the speed of the driven shaft while the
main or driving shaft runs at constant speed

Stepped or cone pulley drive
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is used for changing the speed of the driven shaft while the
main or driving shaft runs at constant speed

Fast and Loose pulley drive
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is used when the driven shaft is to be started or stopped
whenever desired without interfering with the driving shaft.

Velocity Ratio of a Belt Drive
It is the ratio between the velocities of the driver and the driven.
D = Diameter of the driver,
d = Diameter of the driven,
N1 = Speed of the driver in r.p.m.,
N2 = Speed of the follower in r.p.m.,
Peripheral velocity of the belt on the driving pulley,
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60
1
1
DN
v

=

Peripheral velocity of the belt on the driving pulley,
Peripheral velocity of the belt on the driven pulley,
When there is no slip, then
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60
1
1
DN
v

=
60
2
2
dN
v

=
21 vv =
6060
21 dNDN 
=
d
D
N
N
=
1
2

In case of a compound belt drive, the velocity ratio is
When thickness of the belt (t) is considered, then velocity ratio,
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42
31
1
4
dd
DD
N
N


=
td
tD
N
N
+
+
=
1
2

Slip of the Belt
When frictional grip between belt and pulley becomes
insufficient. This may cause some forward motion of the driver
without carrying the belt with it. This is called slip of the belt and
is generally expressed as a percentage.
s1 % = Slip between the driver and the belt, and (Slip over pulley
A)
s2 % = Slip between the belt and driven, (Slip over pulley B)
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Slip of the Belt
∴ Velocity of the belt passing over the driver per second,
and velocity of the belt passing over the driven per second
Substituting the value of ν
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





−=
100
1
60
11 sDN
v







−=





−=
100
1
10060
222 s
v
s
vv
dN






−





−=
100
1
100
1
6060
2112 ssDNdN 

Slip of the Belt
∴
s is total %age of slip
If thickness of the belt (t) is considered, then
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





−





−=
100
1
100
1
6060
2112 ssDNdN 






−−=
100100
1 21
1
2 ss
d
D
N
N






−=










 +
−=
100
1
100
1 21
1
2 s
d
Dss
d
D
N
N






−
+
+
=
100
1
1
2 s
td
tD
N
N

Creep of Belt
When the belt passes from the slack side to the tight side, a
certain portion of the belt extends and it contracts again when the
belt passes from the tight side to the slack side. Due to these
changes of length, there is a relative motion between the belt and
the pulley surfaces. This relative motion is termed as creep. The
total effect of creep is to reduce slightly the speed of the driven
pulley.
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Creep of Belt
The efficiency of the belt drive is reduced by 1 to 2 % as a result
of creep.
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Difference between Slip and
Creep of Belt
Both creep and slip lower the expected velocity of the driven
member.
However, there is a basic difference between the creep and slip.
Creep depends on the three factors.
1. Load
2. Time
3. Temperature
While slip is due to frictional grip between the pulley and belt.
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Power Losses in Belt Drive
The power losses in the belt drive are made up of the following
factors:
(i) Power loss due to belt slip and creep on the pulley.
(ii) Power loss due to internal friction between the particles of
the belt in alternate bending and unbending over the pulley.
(iii) Power loss due aerodynamic resistance to the motion of
pulleys and belt.
(iv) Power loss due to friction in bearings of pulleys.
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Length of an Open Belt Drive
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(degree)pulleysmallforangleWrap=s
(degree)pulleybigforangleWrap=b

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The area ogcb is a rectangle.
ob = gc
From triangle ogo1
1
1
1
1
1
1
sin
oo
obco
oo
gcco
oo
go −
=
−
==
C
dD
C
dD
2
22
sin
−
=
−
=
C
dD
2
sin
−
=

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Also,
( ) 2180 −=s
( ) 2180 +=b





 −
−= −
C
dD
s
2
sin2180 1






 −
+= −
C
dD
b
2
sin2180 1


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The length of the belt is
( ) ( ) efcdearcbcfabarcL +++=
( ) ( ) og
D
og
d
L bs +++= 
22
( ) ( )  cos2
2
cos2
2
C
D
C
d
L ++++−=
( ) ( ) 

cos2
2
CdD
dD
L +−+
+
=

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For small values of β
C
dD
2
sin
−
= 
2
1
2
sin21cos
2
2 
 −=





−=
( )
2
2
8
1cos
C
dD −
−=

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Substitute the value of β and cosβ
( ) ( ) 

cos2
2
CdD
dD
L +−+
+
=
( ) ( )( ) ( )







 −
−+−
−
+
+
= 2
2
8
12
22 C
dD
CdD
C
dDdD
L

( ) ( ) ( )
C
dD
C
C
dDdD
L
4
2
22
22
−
−+
−
+
+
=

( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=


Length of a Cross Belt Drive
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( ) ( )
C
dDdD
CL
42
2
2
+
+
+
+=

( ) 2180 +== bs 




 +
+== −
C
dD
bs
2
sin2180 1


Analysis of Belt Tensions »
Flat Belt Drive
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P1 = belt tension in the tight
side (N)
P2 = belt tension in the
loose side (N)
m = mass of the one meter
Length of belt (kg/m)
v = belt velocity (m/s)
f = coefficient of friction
α = angle of wrap for belt (radians)

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An element of the belt
subtending an angle (dϕ) is in
equilibrium under the action of
the following forces:
(i) tensions (P) and (P + dP) on
the loose and tight
(ii) the normal reaction (dN) and
the frictional force (f × dN) at
the interface and
(iii) centrifugal force in radially
outward direction.
Flat Belt Drive

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Centrifugal force = mass × acceleration
The length of element is (rdϕ) and
the mass per unit length is m.
Therefore,
Mass of element = mrdϕ
Centrifugal force =
( )  dmv
r
v
mrd 2
2
=







Flat Belt Drive

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Considering equilibrium of forces in
horizontal and vertical direction
( ) fdN
d
P
d
dPP =





−





+
2
cos
2
cos

( ) dNdmv
d
P
d
dPP +=





+





+ 
 2
2
sin
2
sin
1
2
cos 




 d
22
sin
 dd






Flat Belt Drive

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( ) fdN
d
P
d
dPP =





−





+
2
cos
2
cos

1
2
cos 




 d
( ) fdNPdPP =−+
f
dP
dNfdNdP ==
Flat Belt Drive

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( ) dNdmv
d
P
d
dPP +=





+





+ 
 2
2
sin
2
sin
22
sin
 dd






( ) dNdmv
d
P
d
dPP +=





+





+ 
 2
22
dNdmv
d
P
d
P +=





+ 
 2
22
dNdmvPd +=  2
Flat Belt Drive

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dNdmvPd +=  2
From
and
f
dP
dN =
f
dP
dmvPd +=  2
f
dP
dmvPd =−  2
( ) fd
mvP
dP
=
− 2
Integrating the above expression,
Flat Belt Drive

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( )  =
−


0
2
1
2
fd
mvP
dP
P
P
Integrating
( )   
 0
2 1
2
log fmvP
P
Pe =−
f
mvP
mvP
e =





−
−
2
2
2
1
log
f
e
mvP
mvP
=
−
−
2
2
2
1
Flat Belt Drive

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The force components
P
P + dP and
mv2dϕ
are the same
V-Belt Drive
The difference lies in the normal reaction dN. The normal reaction,
which acts on two sides of the V-belt, is assumed as on
each side.






dN
2
1

V-Belt Drive

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V-Belt Drive






=
−
− 2
sin
2
2
2
1

f
e
mvP
mvP

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Observation
The superiority of V-belt over flat belt can be explained by
comparing Eqs
and
The following observations are made:
f
e
mvP
mvP
=
−
−
2
2
2
1






=
−
− 2
sin
2
2
2
1

f
e
mvP
mvP

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Observation
The superiority of V-belt over flat belt can be explained by
comparing Eqs
and
The following observations are made:
(i) The equations of flat and V-belt are identical except that the
coefficient of friction f in flat belt drive is replaced by f
/sin(θ/2) in case of V-belt.
f
e
mvP
mvP
=
−
−
2
2
2
1






=
−
− 2
sin
2
2
2
1

f
e
mvP
mvP

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Observation
(ii) For a V-belt, θ = 40° or [f /sin (θ/2)] = 2.92 f
Therefore, for identical materials of belt and pulleys, the
coefficient of friction of V-belt is 2.92 times that of flat
belt.
Consequently, the power-transmitting capacity of V-belt is
much more than that of flat belt. Therefore, V-belts are
more powerful.

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Observation
(iii) Due to increased frictional force, the slip is less in V-belt
compared with flat belt.

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Condition for maximum power
Centrifugal force on the belt (CF)
= mass × acceleration
The length of belt element is (rdϕ) and
the mass per unit length is m.
Mass of element = mrdϕ
CF = ( )  dmv
r
v
mrd 2
2
=








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The centrifugal force CF induces belt
tension Pc. By symmetry, the
centrifugal force induces equal
tension on two sides of belt.
Resolving the forces acting on the
belt element in vertical direction,






=
2
sin2
d
PCF c

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So,
For small angle dϕ






=
2
sin22 

d
Pdmv c
22
sin
 dd
=





2
22 

d
Pdmv c= 2
mvPc =

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So, A little consideration will show
that the maximum tension in the belt
(Pmax) is equal to the total tension in
the tight side of the belt (P1)
Maximum tension in the belt,
Pmax = Maximum stress × Cross-
sectional area of belt = σ.b.t

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When centrifugal tension is neglected,
then
Pmax = P1
When centrifugal tension is
considered, then
Pmax = P1 + Pc

The power transmitted by belt is given by
Power =
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72
So,
P1 = Pmax − Pc
f
e
P
P
=
2
1
f
e
P
P 1
2 =
( ) 





−=





−=−
 ff
e
vPv
e
P
PvPP
1
11
1
121

The power transmitted by belt is given by
Power =
Power =
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(Dr. L K Bhagi)
74






−
f
e
vP
1
11
k
e f
=−

1
1
( ) ( )
( ) ( )kmvvPvkmvP
vkPPkvP c
3
max
2
max
max1
−=−=
−=

The power transmitted will be maximum when,
The optimum velocity of the belt for maximum power
transmission is given by,
or
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( ) ( ) 0
v
or0 3
max =−


=


mvvPpower
v
03 2
max =− mvP
m
P
v
3
max=
3
maxP
Pc =

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and,
P1 = Pmax − Pc
3
maxP
Pc =
cccc PPPPPP 23max1 =−=−=
cPP 21 =

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77
Condition for maximum power » Conclusion
The condition for maximum power transmission is,
(i) The maximum permissible tension in the belt should be
three times the tension due to centrifugal force (Pmax. = 3Pc).
(ii) Alternatively, the tension in the tight side of the belt should
be twice the tension due to centrifugal force (P1 = 2Pc).
(iii) It shows that when the power transmitted is maximum, 1/3rd
of the maximum tension is absorbed as centrifugal tension.
cPP 21 =
m
P
v
3
max=
3
maxP
Pc =

10-05-2019
(Dr. L K Bhagi)
78
Belt Drive»
Numerical 5.1
The layout of a leather belt drive
transmitting 15 kW of power. The
centre distance between the pulleys is
twice the diameter of the bigger
pulley. The belt should operate at a
velocity of 20 m/s approximately and
the stresses in the belt should not
exceed 2.25 N/mm2. The density of
leather is 0.95 g/cc and the coefficient
of friction is 0.35. The thickness of
the belt is 5 mm.

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79
Belt Drive»
Numerical 5.1
Calculate:
(i) the diameter of pulleys;
(ii) the length and width of the
belt; and
(iii) the belt tensions.

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80
Belt Drive»
Numerical 5.1 Solution
Power transmitting = 15 kW
Belt speed v = 20 m/s
C = 2D
t = 5 mm
ρ = 0.95 g/cc
σ = 2.25 N/mm2
f = 0.35

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81
Belt Drive»
(1) Calculate diameter of the pulleys
1440
6010002060
60 

=

==


n
v
d
dn
v
mmd 270=
d
D
N
N
=
1
2
270480
1440 D
= mmD 810270
480
1440
=





=

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82
Belt Drive»
(2) Calculate the length and width of the belt
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=

( ) ( ) ( )
( )16204
270810
2
270810
16202
2
−
+
+
+=

L
( ) mm162081022 === DC
mm46.4981=L

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83
Belt Drive»
(3) Calculate belt width and belt tensions
m is the mass of the belt per unit length (kg/m) and width b in mm
f
e
mvP
mvP
=
−
−
2
2
2
1
( )  == thicknesswidthlengthvolumem
( ) Kg/m1075.4
1000
1095.0
1000
5
1000
b
1 3
6
bm −
=







=

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(Dr. L K Bhagi)
84
Belt Drive»
α is the angle of wrap of the belt
f
e
mvP
mvP
=
−
−
2
2
2
1
=






−
−=




 −
−= −−
8.160
16202
270810
sin2180
2
sin2180 11
C
dD
s
rad18.2
180
8.160 =





=

s

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85
Belt Drive»
For corrected belt velocity v
f
e
mvP
mvP
=
−
−
2
2
2
1
m/s36.20
60
144010270
60
3
=

==
−
dn
v
( ) ( ) bbmv 97.136.201075.4 232
== − ( )( ) 67.281.235.0
== ee f

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86
Belt Drive»
Max stress in the belt =
f
e
mvP
mvP
=
−
−
2
2
2
1
67.2
97.1
97.1
2
1 =
−
−

bP
bP
029.367.2 21 =+− bPP
N25.11P
5
25.2 1
11 b
b
P
A
P
===

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87
Belt Drive»
Power transmitted by belt
So, now we have
( )vPPPower 21 −=
N74.736
36.20
1015 3
21 =

=− PP
029.367.2 21 =+− bPP
N25.11P1 b=
N74.73621 =− PP

10-05-2019
(Dr. L K Bhagi)
88
Belt Drive»
Solving for b, P1 and P2
Subtract eqs
029.367.2 21 =+− bPP N25.11P1 b= N74.73621 =− PP
029.367.225.11 2 =+− bPb ( ) 67.274.73621 =− PP
( ) 67.274.73625.11 2 =− Pb
0958.196767.20375.30 2 =− Pb
0958.19674975.15 =b
mm12792.126 =b N75.1428P1 = N01.692P2 =

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89
Belt Drive»
Procedure for selection of Flat Belt
For selection of belt for a given application, the following
information is required:-
1. Power to be transmitted
2. The input and output speeds
3. The centre distance depending upon availability of space
4. Type of load

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(Dr. L K Bhagi)
90
Belt Drive»
1. The optimum belt velocity of Dunlop belts is from 17.8 to 22.9
m/s. Assume some belt velocity such as 18 m/s in this range
and calculate diameter of smaller pulley by the following
relation
Diameter of bigger pulley
n
v
d

60
=






=





=
speedOutput
speedInput
pulleybiggerofSpeed
pulleysmallerofSpeed
ddD

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91
Belt Drive»
Modify the values of d and D to the nearest preferred diameters.
Determine the correct belt velocity for these preferred pulley
diameters and check whether the actual velocity is in the range of
optimum belt velocity.

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92
Belt Drive»
2. Determine the load correction factor Fa from Table 13.1. It
depends upon the type of load. Find out the maximum power
for the purpose of belt selection by following relationship,
( ) ( )kWkW max aF=

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93
Belt Drive»
3. Calculate angle of wrap for smaller pulley
Find the arc of contact factor Fd from table 13.2





 −
−= −
C
dD
s
2
sin2180 1


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94
Belt Drive»
4. Calculate corrected power by
(kW)corrected=(kW) max×Fd
5. Calculate Corrected Power Rating (CPR) for belt by (assume
belt speed 5.08 m/s)
For High speed belt, CPR
For Fort belt, CPR v is corrected belt speed m/s
08.5
0118.0 v
=
08.5
0147.0 v
=

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(Dr. L K Bhagi)
95
Belt Drive»
6. Calculate the product of (width × No. of plies) by dividing
the Corrected power by corrected power rating of belt.
( )
ratingbeltcorrected
powercorrected
pliesofNo.width =

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(Dr. L K Bhagi)
96
Belt Drive»
calculate diameter of smaller pulley
n
v
d

60
=






=





=
speedOutput
speedInput
pulleybiggerofSpeed
ddD
Determine the correct belt velocity
Find out the maximum power ( ) ( )kWkW max aF=
Calculate angle of wrap for smaller pulley 




 −
−= −
C
dD
s
2
sin2180 1

Find the arc of contact factor Fd
Calculate corrected power by
(kW)corrected=(kW) max×Fd

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(Dr. L K Bhagi)
97
Belt Drive»
Numerical 5.2
It is required to select a flat belt drive for a compressor
running at 720 rpm, which is driven by a 25 kW, 1440 rpm
motor. Space is available for a centre distance of 3 m. The
belt is open-type.

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(Dr. L K Bhagi)
98
Belt Drive»
Power transmitting = 25 kW
n1 = 1440 rpm n2 = 720 rpm
C = 3 meter

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(Dr. L K Bhagi)
99
Belt Drive»
Step-1:
Assume some belt velocity such as 18 m/s
Selecting the preferred pulley diameter d = 250 mm
( )1
60
n
v
d


=
( )
mm73.238
1440
601018 3
=

=

d

10-05-2019
(Dr. L K Bhagi)
100
Belt Drive»
Step-1:
Therefore, the diameter of bigger pulley (D)
The belt velocity for these dimensions is given by,






=
720
1440
dD mm5002250 ==D
( )( ) m/s85.18
100060
1440250
60
1 =

==
dn
v

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(Dr. L K Bhagi)
101
Belt Drive»
Step-2: Maximum power for selection
From Table 13.1, the load correction factor for the compressor is
1.3.
Therefore, Maximum power
=1.3×(25) = 32.5 kW
( ) ( )kWkW max aF=

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(Dr. L K Bhagi)
102
Belt Drive»
Step-3: Arc of contact factor (Fd)





 −
−= −
C
dD
s
2
sin2180 1
 =






−
−= −
23.175
30002
250500
sin2180 1
Assuming linear interpolation, the arc of contact factor is given
by, ( )( )
( )
019.1
170180
23.175180104.1
1 =
−
−−
+=dF

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103
Belt Drive»
Step-4: Corrected power policy
Calculate corrected power by
(kW)corrected=(kW) max×Fd= 1.019(32.5) = 33.12 kW

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(Dr. L K Bhagi)
104
Belt Drive»
Step-5: Corrected power rating of belt
Selecting HI-SPEED belt, the corrected rating at a belt speed
of 18.85 m/s is
kW0438.0
08.5
85.180118.0
=

=

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(Dr. L K Bhagi)
105
Belt Drive»
Step-6: Selection of belt
Belt widths
4 plies w =
5 Plies w =
6 Plies w =
( ) 17.756
0.0438
33.12
ratingbeltcorrected
powercorrected
pliesofNo.width ===
mm04.189
4
756.17
=
mm23.151
5
756.17
=
mm03.126
6
756.17
=
So, we select HI-
SPEED belt of 152
mm preferred width
with 5 plies.

10-05-2019
(Dr. L K Bhagi)
106
Belt Drive»
Step-7: Belt length
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=

( ) ( ) mm31.7183
30004
250500
2
250500
30002
2
=

−
+
+
+=

L

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(Dr. L K Bhagi)
107
V-Belt Drive»
Procedure for selection of V-Belts
In practice, the designer has to select a V-belt from the catalogue of
the manufacturer. The following information is required for the
selection:
i. Type of driving unit
ii. Type of driven machine
iii. Operational hours per day
iv. Power to be transmitted
v. Input and output speeds
vi. Approximate centre distance depending upon the availability
of space

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(Dr. L K Bhagi)
108
V-Belt Drive»
The following notations are used for the dimensions of the cross-
section:
Pitch Width (Wp )
It is the width of the belt at its pitch
zone.

10-05-2019
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109
V-Belt Drive»
section:
Nominal Top Width (W)
It is the top width of the trapezium
outlined on the cross-section of the belt.

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110
V-Belt Drive»
section:
Nominal Height (T)
It is the height of the trapezium outlined
on the cross-section of the belt.

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(Dr. L K Bhagi)
111
V-Belt Drive»
section:
Angle of Belt (A)
It is the included angle obtained by
extending the sides of the belt. The
included angle for the V-belt is usually
from 30° to 40°. The power is
transmitted by the wedging action
between the belt and the V-groove in the
pulley

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(Dr. L K Bhagi)
112
V-Belt Drive»
section:
Pitch Length (Lp)
This is the circumferential length of the
belt at the pitch width.

10-05-2019
(Dr. L K Bhagi)
113
V-Belt Drive»
❖ The manufacturers and the Bureau of Indian Standards have
standardized the dimensions of the cross-section. There are six
basic symbols— Z, A, B, C, D and E—for the cross-section of
V-belts.
Z-section belts are occasionally used for low power
transmission and small pulley diameters, while A, B, C, D and E
section belts are widely used as general purpose belts.

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114
V-Belt Drive»

10-05-2019
(Dr. L K Bhagi)
115
V-Belt Drive»
❖ The selection of the cross-section (Z, A, B, C, D and E) of V-
belts depends upon two factors, namely, the power to be
transmitted and speed of the faster shaft.

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116
V-Belt Drive»
The basic procedure for the selection of V-belts consists of the
following steps:
I. Determine the correction factor according to service (Fa)
from Table 13.15 (Design of Machine Elements by V B
Bhandari).
It depends upon the type of driving unit, the type of driven
machine and the operational hours per day.

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(Dr. L K Bhagi)
117
V-Belt Drive»

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(Dr. L K Bhagi)
118
V-Belt Drive»
following steps:
II. Calculate the design power by the following relationship:
Design power = Fa (Transmitted power)

10-05-2019
(Dr. L K Bhagi)
119
V-Belt Drive»
following steps:
III. Plot a point with design power as X co-ordinate and input
speed as Y co-ordinate in Fig. 13.24 (Design of Machine
Elements by V B Bhandari).
The location of this point decides the type of cross-section of
the belt.

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(Dr. L K Bhagi)
120
V-Belt Drive»

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(Dr. L K Bhagi)
121
V-Belt Drive»
following steps:
IV. Determine the recommended pitch diameter of the smaller
pulley (d) from Table 13.12 (Design of Machine Elements by V
B Bhandari).
It depends upon the cross-section of the belt.

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V-Belt Drive»

10-05-2019
(Dr. L K Bhagi)
123
V-Belt Drive»
Calculate the pitch diameter of the bigger pulley (D) by the
following relationship:
The above values of D and d are compared with the preferred
pitch diameters given in Table 13.13 (Design of Machine
Elements by V B Bhandari).
In case of non-standard value, the nearest values of d and D should
be taken from Table 13.13.






=





=
speedOutput
speedInput
pulleybiggerofSpeed
ddD

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(Dr. L K Bhagi)
124
V-Belt Drive»

10-05-2019
(Dr. L K Bhagi)
125
V-Belt Drive»
following steps:
V. Determine the pitch length of the belt (L)
VI. Compare the above value of (L) with preferred pitch length
(L) in Table 13.14 (Design of Machine Elements by V B
Bhandari).
In case of non-standard value, the nearest values of L should
be taken from Table 13.14.
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=


10-05-2019
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126
V-Belt Drive»
L = 3200 mm

10-05-2019
(Dr. L K Bhagi)
127
V-Belt Drive»
following steps:
VII. Find out the correct centre distance C by substituting the
Preferred pitch length (L) in the following equation:
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=


10-05-2019
(Dr. L K Bhagi)
128
V-Belt Drive»
following steps:
VIII. Determine the correction factor (Fc) for belt pitch length
from Table 13.21 (Design of Machine Elements by V B
Bhandari).
It depends upon the type of cross-section and the pitch length
of the belt.

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(Dr. L K Bhagi)
129
V-Belt Drive»

10-05-2019
(Dr. L K Bhagi)
130
V-Belt Drive»
following steps:
IX. Calculate the arc of contact for the smaller pulley (αs) by
the following relationship:
Determine the correction factor (Fd) for αs from Table
13.22. It is not advisable to use an arc of contact less than 120°
for V-belt drive. Therefore, the minimum arc of contact should
be 120°.





 −
−= −
C
dD
s
2
sin2180 1


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131
V-Belt Drive»

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(Dr. L K Bhagi)
132
V-Belt Drive»
following steps:
X. Depending upon the type of cross-section, refer to the
respective table from Table 13.16 to Table 13.20 and
determine the power rating (Pr) of single V-belt.
It depends upon three factors—speed of faster shaft, Pitch
diameter of smaller pulley and Speed ratio.

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133
V-Belt Drive»

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134
5.78 + 0.44
=Pr

10-05-2019
(Dr. L K Bhagi)
135

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(Dr. L K Bhagi)
136
V-Belt Drive»
following steps:
X. Find out the number of belts.
It depends upon the design power and the power transmitting
capacity of one belt. The number of belts is obtained by the
following relationship:
dcr
a
FFP
FP
beltsofNumber


=

10-05-2019
(Dr. L K Bhagi)
137
V-Belt Drive»
dcr
a
FFP
FP
beltsofNumber


= powerdTransmitteP =
ratio.Speed
andpulleysmallerofdiameterPitch
shaft,fasterofSpeed
beltsingleofratingPowerPr

=
dayperhoursloperationa
andmachinedrivenoftypeunit,drivingoftype
FactorCorrectionFa

=
lengthpitchPreferred
andsection-crossoftype
FactorCorrectionFc

=
shaftfasterofSpeedsversu
powerdTransmitteFpowerDesign
section-crossoftypedecideTo
a =
( )s
d
contactofarc
FactorCorrectionF

=

10-05-2019
(Dr. L K Bhagi)
138
V-Belt Drive»
Numerical 5.3
It is required to design a V-belt drive to connect a 7.5 kW,
1440 rpm induction motor to a fan, running at approximately
480 rpm, for a service of 24 h per day. Space is available for
a centre distance of about 1 m.
Design a V-belt drive
✓ Pitch diameter of smaller pulley (d)
✓ Pitch diameter of bigger pulley (D)
✓ Preferred pitch length (L)
✓ Correct centre distance (C)
✓ Number of belts

10-05-2019
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139
V-Belt Drive»
Power transmitting = 7.5 kW
n1 = 1440 rpm n2 = 480 rpm
C = 1 meter Service = 24 hr/day
Step 1: Correction factor (Fa)
Induction motor is driving and transmitting power at 7.5 kW
Fan is driven for 24 hr/day

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140
V-Belt Drive»

10-05-2019
(Dr. L K Bhagi)
141
V-Belt Drive»
Step 1: Correction factor (Fa)
Induction motor is driving and transmitting power at 7.5 kW
Fan is driven for 24 hr/day.
So, Fa = 1.3
Step 2: Design power
Design power = Fa (Transmitted power)
= 1.3 (7.5)
Design power = 9.75 kW

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142
V-Belt Drive»
Step 3: Type of cross-section for belt
Plot of design power versus speed of faster shaft
Point is located in the region of B-section
Cross-section of V-belt is B

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143
V-Belt Drive»
1440
9.75

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(Dr. L K Bhagi)
144
V-Belt Drive»
Step 3: Type of cross-section for belt
Plot of design power versus speed of faster shaft
Point is located in the region of B-section
Cross-section of V-belt is B

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(Dr. L K Bhagi)
145
V-Belt Drive»
Step 4: Pitch diameter of smaller (d) and bigger pulleys (D)
From Table 13.12: Dimensions of standard cross-sections
minimum pitch diameter for the smaller pulley = 200 mm

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146
V-Belt Drive»

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(Dr. L K Bhagi)
147
V-Belt Drive»
Step 4: Pitch diameter of smaller (d) and bigger pulleys (D)
From Table 13.12
minimum pitch diameter for the smaller pulley = 200 mm
Speed ratio = 1440/480 = 3
So, Pitch diameter of bigger pulley = 3×200 = 600 mm

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It is observed from Table
13.13 that both diameters
d and D have preferred
values.

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V-Belt Drive»
Step 5: Pitch length of belt (L)
Step 6: Preferred Pitch length of belt (L)
From Table 13.14, the preferred pitch length for B section
belt is:
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=

( ) ( ) ( )
( )
mm3296.64
10004
200600
2
200600
10002
2
=
−
+
+
+=

L

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L = 3296.64 mm
Preferred pitch length

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151
V-Belt Drive»
Step 6: Preferred Pitch length of belt (L)
So, from Table 13.14, the preferred pitch length for B section
belt is 3200 mm.
Step 7: Correct centre distance (C)
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=


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V-Belt Drive»
Step 7: Correct centre distance (C)
Correct centre distance (C) = 950.64 mm
( ) ( )
C
dDdD
CL
42
2
2
−
+
+
+=

( ) ( )
C
C
4
200600
2
200600
23200
2
−
+
+
+=

02000068.9712
=+− CC
( )
mmC 64.950
2
20000468.97168.971 2
=
−
=

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V-Belt Drive»
Step 8: Correction factor for belt Pitch length (Fc)
From Table 13.21, for B section and 3200 mm pitch length

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V-Belt Drive»
Correction factor (Fc)= 1.08

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155
V-Belt Drive»
Step 9: Correction factor for arc of contact (Fd)
First, calculate arc of contact for smaller pulley
For Fd refer Table 13.22





 −
−= −
C
dD
s
2
sin2180 1

=






−
−= −
15671.155
64.9502
200600
sin2180 1
s

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Fd is ≈ 0.94

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157
V-Belt Drive»
Step 10: Power rating of single V-belt
From Table 13.17, (1440 rpm, 200 mm pulley, B-
section) and (speed ratio = 3)

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158
5.9 + 0.46
Pr= 6.36 kW

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159
V-Belt Drive»
Step 10: Number of V-belts
dcr
a
FFP
FP
beltsofNumber


=
94.008.16.36
30.17.5
beltsofNumber


=
belts2or51.1beltsofNumber =

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160
V-Belt Drive»
V-belt drive
✓ Pitch diameter of smaller pulley (d) = 200 mm
✓ Pitch diameter of bigger pulley (D) = 600 mm
✓ Preferred pitch length (L) = 3200 mm
✓ Correct centre distance (C) = 950.64 mm
✓ Number of belts = 2

Unit-5
Design of Belt and Chain Drives »
Chain Drives
Dr. L.K. Bhagi
Associate Professor
School of Mechanical Engineering
Lovely Professional University

Chain Drive
Chain drives are a means of transmitting power like gears, shafts
and belt drives
❑ Characteristics
• High axial stiffness
• Low bending stiffness
• High efficiency
• Relatively cheap
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162

Chain Drive
In order to avoid slipping, steel chains are used.
The chains are made up of number of rigid links which are
hinged together by pin joints in order to provide the necessary
flexibility for wrapping round the driving and driven wheels.
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Chain Drive
These wheels have projecting teeth of special profile and fit into
the corresponding recesses in the links of the chain. The toothed
wheels are known as sprocket wheels or simply sprockets.
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164

Chain Drive
The sprockets and the chain are thus constrained to move
together without slipping and ensures perfect velocity ratio.
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165

Chain Drive » Advantages
The advantages of chain drives compared with belt and gear
drives are as follows:
I. Chain drives can be used for. They are particularly suitable
long as well as short centre distances.
II. A number of shafts can be driven in the same or opposite
direction by means of the chain from a single driving
sprocket.
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166

III. Chain drives have small overall dimensions than belt drives.
IV. The efficiency of chain drives is high. For properly
lubricated chain, the efficiency of chain drive is from 96% to
98%.
V. A chain does not slip and to that extent, chain drive is a
positive drive.
VI. Chain does not require initial tension.
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167

VII.Atmospheric conditions and temperatures do not affect the
performance of chain drives.
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Chain Drive » Disadvantages
The disadvantages of chain drives are as follows:
I. Chain drives operate without full lubricant film between the
joints unlike gears. This results in more wear at the joints.
The wear increases the pitch of the chain.
II. Chain drives are not suitable for non-parallel shafts. Bevel
and worm gears and quarter-turn belt drives can be used for
non-parallel shafts.
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169

Chain Drive » Disadvantages
III. Chain drive is unsuitable where precise motion is required
due to polygonal effect.
IV. Chain drives require housing.
V. Compared with belt drives, chain drives require precise
alignment of shafts.
VI. Chain drives require adjustment for slack. Compared with
the belt drive, chain drives require proper maintenance,
particularly lubrication.10-05-2019
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Chain Drive » Terms Used
VII.Chain drives generate noise.
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Chain Drive
Chain drives are used for velocity ratios less than 10 : 1 and
chain velocities of up to 25 m/s.
In general, they are recommended to transmit power up to 100
kW.
10-05-2019
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172

It is the distance between the
hinge centre of a link and the
corresponding hinge centre of
the adjacent link.
It is usually denoted by p.
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173
Roller chains are standardized and
manufactured on the basis of the pitch.
Pitch of Chain

It is the diameter of the circle
on which the hinge centers of
the chain lie, when the chain is
wrapped round a sprocket.
10-05-2019
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174
Pitch circle diameter of chain sprocket.

Chain Drive » Roller Chains
Roller chain consists of alternate links made of inner and outer link
plates. A roller chain consists of following five parts:
10-05-2019
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175
(i) Pin
(ii) Bushing
(iii) Roller
(iv) Inner link plate
(v) Outer link plate

The pin is press fitted to two outer link plates, while the bush is
press fitted to inner link plates.
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176

The bush and the pin form a swivel joint (called a hinge or pin
joint) and the outer link is free to swivel with respect to the inner
link.
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177

The rollers are freely fitted on bushes and, during engagement,
turn with the teeth of the sprocket wheels.
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178

The rolling friction reduces wear and frictional power loss and
improves the efficiency of the chain drive.
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179

The inner and outer link plates are made of medium carbon
steels having hardness up to 50 HRC.
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180

The pins, bushes and rollers are made of case carburizing alloy
steels and hardened to 50 HRC.
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181

The pins, bushes and rollers are made of case carburizing alloy
steels and hardened to 50 HRC.
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182

Roller chains are available in single strand or multi-strand
constructions such as simple, duplex or triplex chains.
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183

The roller chain number is designated as XXB or XXA, e.g., 08B or
16A.
The number in two digits expresses the ‘pitch’ in sixteenths of an
‘inch’ i.e. 8/16 inch or 16/16 inch.
A means American Standard ANSI series
B means British Standard series
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184

10-05-2019
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185
The pitch of chain (08/16) × (25.4) mm = 12.7 mm

Chain Drive » Roller Chains »
Geometric Relationships
D is the pitch circle diameter of the sprocket
α is called the pitch angle.
z is the no. of teeth on sprocket
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186
B
A
O
D/2
α/2






==
2
sin2

AOABp






=











=
2
sin
2
sin
2
2

D
D
p
z
360
=

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187
B
A
O
D/2
α/2






=
2
sin

Dp
z
360
=






=
z
p
D
180
sin

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Velocity Ratio of Chain Drives
The velocity ratio of the chain drives is given by,
n1, n2 = speeds of rotation of driving and driven shafts (rpm)
z1, z2 = number of teeth on driving and driven sprockets.
1
2
2
1
z
z
n
n
i ==
The average velocity of the chain is
33
10601060 
=

=
zpnDn
v


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Length of Chain and Centre Distance
The length of the chain (L) must be equal to the product of the
number of chain links (Ln) and the pitch of the chain (p).
pLL n =

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The length of the chain (L) must be equal to the product of the
number of chain links (Ln) and the pitch of the chain (p).
pLL n =

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191
( )
a
RR
RRaL
2
21
21 )(2
−
+++= 
and
2
11 z
p
R = 22
2
z
p
R = ( ) ( )
2
21
21
zzp
RR
+
=+
( ) 2
21
2
2
21
2
21
22
22






−=






−
=
−

 zz
a
p
a
pzpz
a
RR
2
21
2
21
22
)(
2 




 −
+
+
+=

zz
a
pzzp
aL

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192
2
21
2
21
22
)(
2 




 −
+
+
+=

zz
a
pzzp
aL
pLL n =
( )
2
2121
22
)(
2linkschainofNo. 




 −
+
+
+





=

zz
a
pzz
p
a
Ln
Chain must contain even integer number of links

Polygonal Effect
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193
Number of Teeth on the Smaller or Driving Sprocket
Suppose the sprocket has only 4 teeth and rotating at constant
speed of n rpm.
The chain link AB at distance D/2 from centre of sprocket.

Polygonal Effect
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194
Liner velocity of chain link is
sm
Dn
v /
1060 3max

=


Polygonal Effect
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195
As the sprocket rotates through an angle α/2. The link is at a
distance
sm
Dn
v /
1060
2
cos
3min







=









2
cos
2
D 

Polygonal Effect
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196
The linear speed of the chain is not uniform but varies from
vmax. to vmin. during every cycle of tooth engagement.
This results in a pulsating and jerky motion. The variation in
velocity is given by












−−
2
cos1minmax

vv












−−
z
vv
180
cos1minmax

Polygonal Effect
10-05-2019
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197
As the number of teeth (z) increases to ∞, (vmax. – vmin.) will
become zero. Therefore, the variation will be zero.
❖ In order to reduce the variation in chain speed, the
number of teeth on the sprocket should be increased.
❖ It has been observed that the speed variation is 4% for a
sprocket with 11 teeth, 1.6% for a sprocket with 17 teeth,
and less than 1% for a sprocket with 24 teeth.

Polygonal Effect
10-05-2019
(Dr. L K Bhagi)
198
❖ For smooth operation at moderate and high speeds, it is
considered a good practice to use a driving sprocket with at
least 17 teeth.
❖ From durability and noise considerations, the minimum
number of teeth on the driving sprocket should be 19 or
21.

10-05-2019
(Dr. L K Bhagi)
199
Chain Drive»
Numerical 5.4
A single-strand chain no. 12A is used in a mechanical drive.
The driving sprocket has 17 teeth and rotates at 1000 rpm.
What is the factor of safety used for standard power rating?
Neglect centrifugal force acting on the chain.
LoadBreakingMin.
)(PionChain tens 1=fos

10-05-2019
(Dr. L K Bhagi)
200
Chain Drive»
Z1 = 17
n1 = 1000 rpm
chain 12A
Step 1 chain tension (P1)
m/s4.5
1060
100005.1917
1060 33
11 =


=

=
pnz
v
1000
1 vP
kW

=

The power rating of chain 12A at 1000 rpm is 14.32.

10-05-2019
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204
Chain Drive»
Therefore, the chain tension P1 is
N
v
kW
P 85.2651
4.5
32.1410001000
1 =

=

=
73.11
85.2651
311000
==fos
Step 2 Factor of safety
Breaking load for 12A is 31100 N

10-05-2019
(Dr. L K Bhagi)
205
Chain Drive»
Numerical 5.5
It is required to design a chain drive to connect a 12 kW,
1400 rpm electric motor to a centrifugal pump running at
700 rpm. The service conditions involve moderate shocks.
(i) Select a proper roller chain and give a list of its
dimensions.
(ii) Determine the pitch circle diameters of driving and
driven sprockets.
(iii) Determine the number of chain links.
(iv) Specify the correct centre distance between the axes of
sprockets.

10-05-2019
(Dr. L K Bhagi)
206
Chain Drive»
kW = 12
n1 = 1400 rpm
n2 = 700 rpm
chain 12A
Step 1 kW rating of chain
21
edtransmittbeto
chainofrating
KK
KkW
kW s


=

10-05-2019
(Dr. L K Bhagi)
208
Chain Drive»
Step 1 kW rating of chain
21
edtransmittbeto
chainofrating
KK
KkW
kW s


=
kW6.15
11
3.112
chainofrating =


=kW

10-05-2019
(Dr. L K Bhagi)
209
Chain Drive»
Step 2 Selection of Chain
Power rating at 1400 rpm for 12B is 18.15 kW

Therefore chain no. 12B is selected

10-05-2019
(Dr. L K Bhagi)
211
Chain Drive»
Step 2 Selection of Chain
Power rating at 1400 rpm for 12B is 18.15 kW
So, the dimensions of 12B chain are from Table 14.1

p = 19.05 mm, d1 = 12.07, b1 = 11.68 mm

10-05-2019
(Dr. L K Bhagi)
213
Chain Drive»
Step 3 Pitch circle diameter of driving and driven pulleys
mm67.103
17
180
sin
05.19
180
sin
=






=






=
z
p
D
34
700
1400
17sprocket;drivenFor
2
1
12 =





=





=
n
n
zz
mm46.206
34
180
sin
05.19
2 =






=D

10-05-2019
(Dr. L K Bhagi)
214
Chain Drive»
Step 4 Number of chain links
The centre distance between sprocket wheels should be
between 30p to 50p. Take a mean value of 40p
a = 40 × p = 40 × 19.05 = 762 mm
( )
2
1221
22
)(
2linkschainofNo. 




 −
+
+
+





=

zz
a
pzz
p
a
Ln
links10668.105
2
1734
762
05.19
2
)3417(
05.19
762
2
2
=




 −
+
+
+





=

nL

10-05-2019
(Dr. L K Bhagi)
215
Chain Drive»
Step 5 Correct centre distance













 −
−













 +
−+










 +
−=
2
12
2
2121
2
8
224
a

zzzz
L
zz
L
p
nn













 −
−













 +
−+










 +
−=
22
2
1734
8
2
3417
106
2
3417
106
4
05.19
a

mm03.765a =

10-05-2019
(Dr. L K Bhagi)
216
Chain Drive»
Step 5 Correct centre distance
To provide small sag, the centre distance is reduced by
0.002a.
Therefore, the correct centre distance is given by,
a = 0.998 × 765.03 = 763.5 mm
mm03.765a =

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