Understanding Belt Drives in Mechanical Systems

First Semester
First Semester
chapter three
chapter three
 
Belt, Rope
Belt, Rope
 
Dr. Adil ABed Nayeeif
Dr. Adil ABed Nayeeif
2019-2020
2019-2020
 
Types of Belt Drives
Types of Belt Drives
 
The belt drives are usually classified into the following three groups :
1.
Light drives
. These are used to transmit small powers at belt speeds up to about 
10 m/s
, as in agricultural
machines and small machine tools .
2.
Medium drives
. These are used to transmit medium power at belt speeds over 
10 m/s 
but up to 
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.
Types of Belts
Types of Belts
 
Though there are many types of belts used these days, yet the following are important from the subject point of
view:
1.
Flat belt
. The flat belt, as shown in 
Figure (a)
, is mostly used in the factories and workshops, where a
moderate amount of power is to be transmitted, from one pulley to another when the two pulleys are not more
than 
8
 meters apart.
2.
V-belt.
 The V-belt, as shown in 
Figure (b)
, is mostly used in the factories and workshops where a moderate
amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each
other.
3.
Circular belt or rope
. The circular belt or rope, as shown in 
Figure (c)
, is mostly used in the factories and
workshops, where a great amount of power is to be transmitted, from one pulley to another, when the two
pulleys are more than 
8
 meters apart
 
Types of Flat Belt Drives
Types of Flat Belt Drives
 
Open belt drive
 
Crossed or twist belt drive
 
Open belt drive
Open belt drive
 
The open belt drive, as shown in 
Figure(1),
 is used with shafts arranged parallel and rotating in the 
same
direction
. In this case, 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). Thus 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 
whereas the upper side belt (because of less
tension) is known as 
slack side
, as shown in figure.
 
Figure(1)
 
Crossed or twist belt drive
Crossed or twist belt drive
 
The crossed or twist belt drive, as shown in 
Figure(2) 
, is used with shafts arranged parallel and rotating in the
opposite directions 
In this case, the driver pulls the belt from one side (i.e. RQ) and delivers it to the other side
(i.e. LM). Thus 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
, whereas the belt LM (because of less tension) is known as 
slack side
, as shown in
Figure (2).
 
Figure(2)
 
Velocity Ratio of Belt Drive
Velocity Ratio of Belt Drive
 
It is the 
ratio between the velocities of the driver and the follower or driven
. 
It may be expressed,
mathematically, as discussed below :
Let  
d
1
 = Diameter of the driver,                    
d
2
 = Diameter of the follower
        
N
1
 = Speed of the driver in r.p.m., and   
N
2
 
= Speed of the follower in r.p.m.
 Length of the belt that passes over the driver, in one minute 
= 
π 
d
1
.N
1
Similarly, length of the belt that passes over the follower, in one minute 
= π 
d2 . N2
Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the
follower in one minute, therefore, 
π d
1
 . N
1
 = π d
2
 . N
2
 Velocity ratio,
When the thickness of the belt (t) is considered
, 
then velocity ratio,
When there is no slip, then 
v
1
 = v
2
.
 
 
 
Velocity Ratio of a Compound Belt Drive
Velocity Ratio of a Compound Belt Drive
 
Sometimes the power is transmitted from one shaft to another, through a number of pulleys as shown in 
Figure (3)
.
Consider a pulley 
1 
driving the pulley 
2
. Since the pulleys 
2 
and 
3
 are keyed to the same shaft, therefore the pulley 
1
also drives the pulley 
3 
which, in turn, drives the pulley 
4
.
Let 
  d
1
 = Diameter of the pulley 
1
, N1 = Speed of the pulley 
1
 in r.p.m.,
           d
2
, d
3
, d
4
, N
2
 , N
3
 , N
4
 
= 
Corresponding values for pulleys 
2,
 
3
 and 
4
.
We know that velocity ratio of pulleys 
1 
and 
2
,                            …… (1)
Similarly, velocity ratio of pulleys 
3 
and 
4
,                            
……(2)
Multiplying equations (1) and (2),
 
 
                                       ...(
 N
2
 = N
3
, being keyed to the same shaft)
 
 
 
 
 
 
 
Figure (3)
 
Slip of Belt
Slip of Belt
 
In the previous articles, we have discussed the motion of belts and shafts assuming a firm frictional grip between the
belts and the shafts. But sometimes, the frictional grip becomes insufficient. This may cause some forward motion
of the driver without carrying the belt with it. This may also cause some  forward motion of the belt without
carrying the driven pulley with it. This is called 
slip
 of the belt and is generally expressed as a percentage.
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 (as in the case of
hour minute and second arms in a watch).
Let      s
1
 % = Slip between the driver and the belt, and
          s
2
 % = Slip between the belt and the follower.
 Velocity of the belt passing over the driver per second,                                                                              …..(i)
and velocity of the belt passing over the follower per second .
 
 
Substituting the value of v from equation (i),
 
 
 
 
 
 
 
 
 
... (where 
s 
= 
s
1
 + 
s
2
 
, 
i.e. 
total percentage of slip)
If thickness of the belt (t) is considered, then
 
 
 
 
 
 
 
 
Length of an Open Belt Drive
Length of an Open Belt Drive
 
An open belt drive, both the pulleys rotate in the 
same
 
direction as shown in 
Figure (4).
r
1
  and 
r
2
 = Radii of the larger and smaller pulleys
x
 = Distance between the centres of two pulleys (i.e
. O
1
 O
2
), and
L
 = Total length of the belt.
Let the belt leaves the larger pulley at 
E
 and 
G
 and the smaller pulley at 
F
 and 
H
 as shown in 
Figure (4).
 
Through 
O
2
,
draw 
O
2
 M
 parallel to 
FE
.
From the geometry of the figure, we find that 
O
2
 M 
will be perpendicular to 
O
1
 E
.
 Let the angle 
MO
2
 O
1
 
= 
α
 radians
We know that the length of the belt,
  
L = Arc GJE + EF + Arc FKH + HG
      = 2 (Arc JE + EF + Arc FK)             ......(1)
 
Figure (4)
 
 
Since α is very s mall, therefore putting, 
sin α = 
α
 (in radians)                       .... (2)
                                                  ….. (3)
Similarly
                                                ……(4)
 
 
Expanding this equation by binomial theorem,
 
                                                                                                 …… (5)
 
Substituting the values of arc JE from equation (3), arc FK from equation (4) and EF from equation (5) in equation
(1), we get
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Length of a Cross Belt Drive
Length of a Cross Belt Drive
 
In a cross belt drive, both the pulleys rotate in 
opposite 
directions as shown in 
Figure (5).
 
L = Arc GJE + EF + Arc FKH + HG
      = 2 (Arc JE + EF + Arc FK)             ......(1)
 
 
Figure (5)
 
 
 
 
 
 
 
Power Transmitted by a Belt
Power Transmitted by a Belt
 
Figure (6)
 shows the driving pulley (or driver) 
A 
and the driven pulley (or follower) 
B
. We have already discussed
that the driving pulley pulls the belt from one side and delivers the same to the other side. It is thus obvious that the
tension on the former side (i.e. tight side) will be greater than the latter side (i.e. slack side) as shown in 
Figure (6)
.
T
1
 
and 
T
2
 = Tensions in the tight and slack side of the belt respectively in 
newton's,
r
1
 and 
r
2
 = Radii of the driver and follower respectively, and
v 
= Velocity of the belt in 
m/s.
The effective turning (driving) force at the circumference of the follower is the difference between the two tensions
(i.e. 
T1
T2
).
 Work done per second = 
(T1 – T2) v     N-m/s
and power transmitted, 
P = (T1 – T2) v      W
 
Figure (6)
 
Ratio of Driving Tensions For Flat Belt Drive
Ratio of Driving Tensions For Flat Belt Drive
 
T
1
 
= Tension in the belt on the tight side,
T
2
 = Tension in the belt on the slack side, and
θ
 = Angle of contact in radians (
i.e. 
angle subtended by the arc 
AB
, along which the belt touches the pulley at the
centre).
 
 
The above expression gives the relation between the tight side and slack side tensions, in terms of coefficient of friction
and the angle of contact.
 
 
Angle of Contact
Angle of Contact
 
When the two pulleys of different diameters are connected by means of an open belt as shown in 
Figure(7)
,
then the angle of contact or lap 
(θ)
 at the smaller pulley must be taken into consideration..
r
1
 = Radius of larger pulley,
r
2
 = Radius of smaller pulley, and
x
 = Distance between centres of two pulleys (i.e. 
O
1
 O
2
).
 Angle of contact or lap
,
 
A little consideration will show that when the two pulleys are connected by means of a crossed belt as shown in
Figure (8)
,
 
then the angle of contact or lap (
θ
) on both the pulleys is same.
 
 Angle of contact or lap
,
 
Figure(7)
 
 
Figure (8)
 
 
Centrifugal Tension
Centrifugal Tension
 
Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to
increase the tension on both, 
tight
 as well as the 
slack
 sides. The tension caused by centrifugal force is called
centrifugal tension
.
 
At lower belt speeds (less than 
10 m/s
), the centrifugal tension is very small, but at higher
belt speeds (more than 
10 m/s
), its effect is considerable and thus should be taken into account.
m
 = Mass of the belt per unit length in kg,
v
 = Linear velocity of the belt in 
m/s
,
r
 = Radius of the pulley over which the belt runs in metres, and
T
C
 = Centrifugal tension acting tangentially at 
P
 and 
Q
 in newtons.
                                T
C
 = m . V
2
1.
When the centrifugal tension is taken into account, then total tension in the tight side
, T
t1
 = T
1
+T
C
2.
Power transmitted, 
P=(T
t1
 – T
t2
 ) v =[ (T
1
+T
C
) – (T
2
+T
C
)]v.
3.
The ratio of driving tensions may also be written as
 
 
 
Maximum Tension in the Belt
Maximum Tension in the Belt
 
A little consideration will show that the maximum tension in the belt (
T
) is equal to the total tension in the tight
side of the belt (
T
t1
).
σ 
= Maximum safe stress in 
N/mm
2
,
b 
= Width of the belt in mm, and
t
 = Thickness of the belt in mm.
We know that maximum tension in the belt,
T
 = 
Maximum stress × cross-sectional area of belt 
= 
σ. b. t
When centrifugal tension is neglected, then 
T (or T
t1
) = T
1
and when centrifugal tension is considered, then 
T (or T
t1
) = T
1
 + T
C
 
Condition For the Transmission of Maximum
Condition For the Transmission of Maximum
Power
Power
 
We know that power transmitted by a belt, 
P = (T
1
 – T
2
) v       ….. (1)
we have also seen that the ratio of driving tensions is                        or
Substituting the value of T
2
 in equation (
1
)
                                                                        ……(
2
)            Where
 
We know that  
T
1
= T - T
C
          
where  
T
 = Maximum tension to which the belt can be subjected in newton's,
Substituting the value of 
T
1
 in equation
 
 (
2
),
For maximum power, differentiate the above expression with respect to 
v
 and equate to 
zero
,
 
 
      
T – 3 m . v
2
 = 0             
T – 3 TC = 0 or
  
T = 3 TC
   ….(3)
1.
We know that 
T
1
 = T– TC  
and for maximum power,
2.
From equation (3), the velocity of the belt for the maximum power,
 
 
 
 
 
 
 
 
 
 
 
Initial Tension in the Belt
Initial Tension in the Belt
 
When a belt is wound round the two pulleys (
i.e
. driver and follower), its two ends are joined together ; so that the belt
may continuously move over the pulleys, since the motion of the belt from the driver and the follower is governed by a
firm grip, due to friction between the belt and the pulleys. In order to increase this grip, the belt is tightened up. At this
stage, even when the pulleys are stationary, the belt is subjected to some tension, called 
initial tension
.
T
0
 
= Initial tension in the belt,     
α 
= Coefficient of increase of the belt length per unit force
.
A little consideration will show that the increase of tension in the tight side, 
= T
1
 – T
0
and increase in the length of the belt on the tight side , 
= α (
T
1
 – T
0
)
Similarly, decrease in tension in the slack side, 
= T
0
 – T
2
and decrease in the length of the belt on the slack side, 
= α (
T
0
 – T
2
)
Assuming that the belt material is perfectly elastic such that the length of the belt remains constant, when it is at rest or
in motion, therefore increase in length on the tight side is equal to decrease in the length on the slack side
.
α (
T
1
 – T
0
) = 
α (
T
0
 – T
2
) or      T
1
 – T
0
 = T
0
 – T
2
                                 
                            ...(Neglecting centrifugal tension)
 
                                                            ...(Considering centrifugal tension)
 
 
 
 
 
V-belt drive
V-belt drive
 
We have already discussed that a 
V-belt
 is mostly used in factories and workshops 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.
Ratio of Driving Tensions for V-belt
2 β 
= Angle of the groove.
μ 
= Coefficient of friction between the belt and sides of the groove.
Thus the relation between 
T
1
 and 
T
2 
 
for the 
V-belt 
drive will be
 
 
Rope Drive
Rope Drive
The rope drives are widely used where a large amount of power is to be transmitted, from one pulley to another,
over a considerable distance.
Ratio of Driving Tensions for Rope Drive
The ratio of driving tensions for the rope drive may be obtained in the similar way as 
V-belts
.
 
 
Examples
Examples
 
EX(1):
EX(1):
 
An open belt running over two pulleys 
240 mm 
and 
600 mm 
diameter connects two parallel shafts
3 meters 
apart and transmits 
4 kW 
from the smaller pulley that rotates at 
300 r.p.m
. Coefficient of friction
between the belt and the pulley is 
0.3
 and the safe working tension is 
10N per mm width
. Determine :
1
. minimum width of the belt, 
2
. initial belt tension, and 
3
. length of the belt required.
Solution.
 Given :
d
2
 = 240 mm = 0.24 m ; 
d
1
 = 600 mm = 0.6 m ; 
x 
= 3 m ; 
P
 = 4 kW = 4000 W;
 
N
2
 = 300 r.p.m. ;
μ
 = 0.3 ; 
T
1
 = 10 N/mm width.
1.
Minimum width of belt
We know that velocity of the belt,
 Power transmitted (
P
), 
4000 = (T
1
 – T
2
) v = (T1 – T2) 3.77
T
1
 – T
2
 = 4000 / 3.77 = 1061 N         ….. (1)
 
and angle of lap on the smaller pulley, 
θ = 180° – 2α = 180° – 2 × 3.44° = 173.12°
 
    173.12 × π / 180 = 3.022 rad
We know that                                            ….(2)
From equations (1) and (2), 
T
1
 = 1779 N, and T
2
 = 718 N
Since the safe working tension is 
10 N per mm width
, therefore minimum width of the belt,
 
 
 
 
 
2.
Initial belt tension
We know that initial belt tension,                                                                      
Ans
 
3.
Length of the belt required
We know that length of the belt required,
 
 
 
 
 
= 1.32 + 6 + 0.01 = 7.33 m 
Ans
.
 
 
 
 
 
 
EX(2):  
EX(2):  
A pulley is driven by a flat belt, the angle of lap being 
120°
. The belt is 
100 mm 
wide by 
6 mm 
thick
and density
1000 kg/m
3
. If the coefficient of friction is 
0.3
 and the maximum stress in the belt is not to exceed
2 MPa
, find the greatest power which the belt can transmit and the corresponding speed of the belt.
Solution
. Given : 
θ 
= 120° = 120 × π / 180
 
= 2.1 rad ; 
b
 = 100 mm = 0.1 m ; 
t 
= 6 mm 
= 0.006 m ;
ρ
 = 1000 kg / m
3
 ; 
μ
 = 0.3 ; 
σ
 = 2 MPa = 2 × 106 N/m
2
 .
Speed of the belt for greatest power
We know that maximum tension in the belt, 
T = 
σ. 
b. t = 2 × 106 × 0.1 × 0.006 = 1200 N
and mass of the belt per metre length, 
m = Area × length × density = b. t. l. ρ
m = 0.1 × 0.006 × 1 × 1000 = 0.6 kg/m
 Speed of the belt for greatest power ,
Greatest power which the belt can transmit
We know that for maximum power to be transmitted, centrifugal tension, 
T
C
 = T/3 = 1200/3 = 400 N
and tension in the tight side of the belt, 
T
1
 = T – T
C
 = 1200 – 400 = 800 N
 
 
 Greatest power which the belt can transmit,
 
 
 
 
 
 
 
EX(2)
EX(2)
 
A compressor, requiring 
90 kW 
is to run at about 
250 r.p.m
. The drive is by 
V-belts
 from an electric motor
running at 
750 r.p.m
. The diameter of the pulley on the compressor shaft must not be greater than 
1 meter 
while the
center distance between the pulleys is limited to 
1.75 meter
. The belt speed should not exceed 
1600 m/min
.
Determine the number of 
V
-belts required to transmit the power if each belt has a cross-sectional area of 
375 mm
2
,
density 
1000 kg/m
3
 
and an allowable tensile stress of 
2.5 MPa
. The groove angle of the pulley is 
35°
. The coefficient
of friction between the belt and the pulley is 
0.25
. Calculate also the length required of each belt
Solution.
 Given :
P
 = 90 kW ; 
N
2
 
= 250 r.p.m. ;
 
N
1
 = 750 r.p.m. ; 
d
2
 = 1 m ; 
x
 = 1.75 m ; 
v
 = 1600 m/min = 26.67 m/s
; 
a
 = 375 mm
2
 = 375 × 10
-6
  m
2
 ;
ρ
 = 1000 kg/m
3
 ; 
σ
 = 2.5 M
p
a
 = 2.5 × 106 N/m
2
 ; 
2 
β
 = 35° 
or 
β
 = 17.5° ; 
μ 
= 0.25
First of all, let us find the diameter of pulley on the motor shaft (
d
1
). We know that.
 
 
We know that the mass of the belt per meter length, 
m = Area × length × density = 375 × 10–6 × 1 × 1000 = 0.375 kg
 Centrifugal tension, 
TC = m.v
2
 = 0.375 (26.67)2 = 267 N
and maximum tension in the belt, 
T = σ. a = 2.5 × 1
06
 × 375 × 10
-06
 = 937.5 N
 Tension in the tight side of the belt,
 
T1 = T – TC = 937.5 – 267 = 670.5 N
Let T2 = Tension in the slack side of the belt.
For an open belt drive, as shown in 
Figure (9)
,
 
 
 
 
and angle of lap on smaller pulley (
i.e. 
pulley on motor shaft),
 
θ = 180° – 2α = 180° – 2 × 11° = 158°
 
= 158 × π / 180 = 2.76 rad
 
 
Number of V-belts
We know that power transmitted per belt, 
= (T1 – T2) v = (670.5 – 67.36) 26.67 = 16 086 W 
= 
16.086 kW
 
 
Length of each belt
We know that length of belt for an open belt drive,
 
 
= 2.1 + 3.5 + 0.064 = 5.664 m    
Ans.
 
 
Figure (9)
,
 
 
 
 
 
 
Exercises
Exercises
 
Q1/
Q1/
 
 
A rope drive transmits 
600 kW 
from a pulley of effective diameter 
4 m
, which runs at a speed of 
90 r.p.m
.
The angle of lap is 
160°
 ; the angle of groove 
45°
 ; the coefficient of friction 
0.28
 ; the mass of rope 
1.5 kg / m
and the allowable tension in each rope 2400 N. Find the number of ropes required.
 
Q2/ 
Q2/ 
A pulley used to transmit power by means of ropes has a diameter of 
3.6 metres
 and has 
15 
grooves of
45°
 angle. The angle of contact is 
170°
 and the coefficient of friction between the ropes and the groove sides is
0.28
. The maximum possible tension in the ropes is 
960 N 
and the mass of the rope is 
1.5 kg 
per metre length.
What is the speed of pulley in r.p.m. and the power transmitted if the condition of maximum power prevail ?
 
Q3/ 
Q3/ 
A flat belt is required to transmit 
30 kW 
from a pulley of 
1.5 m
 effective diameter running at 
300 r.p.m
.
The angle of contact is spread over 
(11/24) 
of the circumference. The coefficient of friction between the belt and
pulley surface is 
0.3.
 Determine, taking centrifugal tension into account, width of the belt required. It is given
that the belt thickness is 
9.5 mm
, density of its material is 
1100 kg / m
3
 
and the related permissible working
stress is 
2.5 MPa
 
Q4/
Q4/
  
  
In a horizontal belt drive for a centrifugal blower, the blower is belt driven at 
600 r.p.m. 
by a 
15 kW
,
1750 r.p.m
. electric motor. The centre distance is twice the diameter of the larger pulley. The density of the belt
material = 
1500 kg/m
3
; maximum allowable stress = 
4 MPa
; 
μ
1
 = 
0.5
 (motor pulley); 
μ
2
 = 
0.4
 (blower pulley);
peripheral velocity of the belt = 
20 m/s
. Determine the following:
1.
 Pulley diameters diameters ; 
2.
 belt length; 3. cross-sectional area of the belt; 
4
. minimum initial tension for
operation without slip; and 
5. 
resultant force in the plane of the blower when operating with an initial tension 
50
per cent greater than the minimum value.
 
Q5/
Q5/
 
An open belt connects two flat pulleys. The pulley diameters are 
300 mm 
and 
450 mm 
and the corresponding
angles of lap are 
160°
 and 
210°
. The smaller pulley runs at 
200 r.p.m
. The coefficient of friction between the belt
and pulley is 
0.25
. It is found that the belt is on the point of slipping when 
3 kW 
is transmitted. To increase the
power transmitted two alternatives are suggested, namely (
i
) increasing the initial tension by 
10%
, and (
ii
)
increasing the coefficient of friction by 
10%
 by the application of a suitable dressing to the belt. Which of these two
methods would be more effective? Find the percentage increase in power possible in each case.
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Belt drives are essential components in mechanical systems for transmitting power between pulleys. They are classified based on power transmission capabilities and speed. Types of belts include flat belts, V-belts, and circular belts or ropes, each suitable for different applications. Belt drives can be open or crossed, impacting tension distribution. Velocity ratio in belt drives is crucial for determining the speed relationship between the driver and follower pulleys.


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  1. First Semester chapter three Belt, Rope Dr. Adil ABed Nayeeif 2019-2020

  2. Types of Belt Drives The belt drives are usually classified into the following three groups : 1. Light drives. These are used to transmit small powers at belt speeds up to about 10 m/s, as in agricultural machines and small machine tools . 2. Medium drives. These are used to transmit medium power at belt speeds over 10 m/s but up to 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. Types of Belts

  3. Though there are many types of belts used these days, yet the following are important from the subject point of view: 1. Flat belt. The flat belt, as shown in Figure (a), is mostly used in the factories and workshops, where a moderate amount of power is to be transmitted, from one pulley to another when the two pulleys are not more than 8 meters apart. 2. V-belt. The V-belt, as shown in Figure (b), is mostly used in the factories and workshops where a moderate amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each other. 3. Circular belt or rope. The circular belt or rope, as shown in Figure (c), is mostly used in the factories and workshops, where a great amount of power is to be transmitted, from one pulley to another, when the two pulleys are more than 8 meters apart Types of Flat Belt Drives Open belt drive Crossed or twist belt drive

  4. Open belt drive The open belt drive, as shown in Figure(1), is used with shafts arranged parallel and rotating in the same direction. In this case, 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). Thus 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 whereas the upper side belt (because of less tension) is known as slack side, as shown in figure. Figure(1)

  5. Crossed or twist belt drive The crossed or twist belt drive, as shown in Figure(2) , is used with shafts arranged parallel and rotating in the opposite directions In this case, the driver pulls the belt from one side (i.e. RQ) and delivers it to the other side (i.e. LM). Thus 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, whereas the belt LM (because of less tension) is known as slack side, as shown in Figure (2). Figure(2)

  6. Velocity Ratio of Belt Drive It is the ratio between the velocities of the driver and the follower or driven. It may be expressed, mathematically, as discussed below : Let d1 = Diameter of the driver, d2 = Diameter of the follower N1 = Speed of the driver in r.p.m., and N2= Speed of the follower in r.p.m. Length of the belt that passes over the driver, in one minute = d1.N1 Similarly, length of the belt that passes over the follower, in one minute = d2 . N2 Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the follower in one minute, therefore, d1. N1= d2. N2 d N N= Velocity ratio, 2 1 d + N d t 1 2 When the thickness of the belt (t) is considered, then velocity ratio, = 2 1 + N d t 1 2 When there is no slip, then v1 = v2.

  7. Velocity Ratio of a Compound Belt Drive Sometimes the power is transmitted from one shaft to another, through a number of pulleys as shown in Figure (3). Consider a pulley 1 driving the pulley 2. Since the pulleys 2 and 3 are keyed to the same shaft, therefore the pulley 1 also drives the pulley 3 which, in turn, drives the pulley 4. Let d1 = Diameter of the pulley 1, N1 = Speed of the pulley 1 in r.p.m., d2, d3, d4, N2 , N3 , N4= Corresponding values for pulleys 2,3 and 4. N= d We know that velocity ratio of pulleys 1 and 2, (1) 2 1 N d N= d 1 2 Similarly, velocity ratio of pulleys 3 and 4, (2) 4 3 N d 3 4 Multiplying equations (1) and (2), N N d d = 2 4 1 3 N N d d 1 3 2 4 d N d ...( N2 = N3, being keyed to the same shaft) d d N = 3 4 1 1 2 4 Figure (3)

  8. Slip of Belt In the previous articles, we have discussed the motion of belts and shafts assuming a firm frictional grip between the belts and the shafts. But sometimes, the frictional grip becomes insufficient. This may cause some forward motion of the driver without carrying the belt with it. This may also cause some forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt and is generally expressed as a percentage. 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 (as in the case of hour minute and second arms in a watch). Let s1 % = Slip between the driver and the belt, and s2 % = Slip between the belt and the follower. . . . 1 d N d N s d N s = = Velocity of the belt passing over the driver per second, ..(i) v 1 1 1 1 1 1 1 1 60 60 100 60 100 and velocity of the belt passing over the follower per second . . 1 v d N s s = = v v 2 2 2 2 60 100 100 Substituting the value of v from equation (i),

  9. . . 1 d N d N s s = 1 2 2 1 1 1 2 60 60 100 100 N d s s 1s s = 1 Neglecting 2 1 1 2 2 100 100 N d 100 1 2 + 1 N d s s d s = = 1 2 1 1 2 1 100 100 N d d 1 2 2 ... (where s = s1 + s2, i.e. total percentage of slip) If thickness of the belt (t) is considered, then + 100 2 1 t d N + 1 N d t s = 2 1

  10. Length of an Open Belt Drive An open belt drive, both the pulleys rotate in the same direction as shown in Figure (4). r1 and r2 = Radii of the larger and smaller pulleys x = Distance between the centres of two pulleys (i.e. O1 O2), and L = Total length of the belt. Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H as shown in Figure (4).Through O2, draw O2 M parallel to FE. From the geometry of the figure, we find that O2 M will be perpendicular to O1 E. Let the angle MO2 O1= radians Figure (4) We know that the length of the belt, L = Arc GJE + EF + Arc FKH + HG = 2 (Arc JE + EF + Arc FK) ......(1) O M O E EM r r = = = sin 1 1 1 2 O O 1 O O 1 x 2 2

  11. 1 r r = 2 Since is very s mall, therefore putting, sin = (in radians) .... (2) 2 x .. (3) + = JE Arc 1r Similarly (4) 2 = Arc FK 2r 2 x r r ( ) ( ) ( ) = = = = x 2 2 2 EF MO O O O M x r r 1 2 1 2 2 1 2 1 1 2 Expanding this equation by binomial theorem, ( ) 2 2 r r r r 1 = + = EF x x .......... .. 1 1 2 1 2 (5) x 2 x 2 Substituting the values of arc JE from equation (3), arc FK from equation (4) and EF from equation (5) in equation (1), we get ( ) 2 2 2 ( r 2 r r = + + + L r x r 2 ) 1 2 x 1 2 2 r ( ) ( ) = + + 2 + + L r r r r x 2 1 2 x 1 2 1 2

  12. Length of a Cross Belt Drive In a cross belt drive, both the pulleys rotate in opposite directions as shown in Figure (5). L = Arc GJE + EF + Arc FKH + HG = 2 (Arc JE + EF + Arc FK) ......(1) 2 2 ( ) 2 1 2 O O MO EF = = Figure (5) = + Arc JE 1r = + Arc FK 2r 2 + x r r ( ) ( ) = x 1 = + 2 2 2 O M x r r 1 2 2 1 1 2 ( ) 2 + + 2 r r r r 1 = + = EF x x .......... .. 1 1 2 1 2 x x 2 2 ( ) 2 + 2 2 2 r r = + + + + L r x r 2 1 2 x 1 2 ( ) + 2 r r ( ) ( ) = + + 2 + + L r r r r x 2 1 2 x 1 2 1 2

  13. Power Transmitted by a Belt Figure (6) shows the driving pulley (or driver) A and the driven pulley (or follower) B. We have already discussed that the driving pulley pulls the belt from one side and delivers the same to the other side. It is thus obvious that the tension on the former side (i.e. tight side) will be greater than the latter side (i.e. slack side) as shown in Figure (6). T1and T2 = Tensions in the tight and slack side of the belt respectively in newton's, r1 and r2 = Radii of the driver and follower respectively, and v = Velocity of the belt in m/s. The effective turning (driving) force at the circumference of the follower is the difference between the two tensions (i.e. T1 T2). Figure (6) Work done per second = (T1 T2) v N-m/s and power transmitted, P = (T1 T2) v W

  14. Ratio of Driving Tensions For Flat Belt Drive T1= Tension in the belt on the tight side, T2 = Tension in the belt on the slack side, and = Angle of contact in radians (i.e. angle subtended by the arc AB, along which the belt touches the pulley at the centre). e T 2 T= 1 The above expression gives the relation between the tight side and slack side tensions, in terms of coefficient of friction and the angle of contact.

  15. Angle of Contact When the two pulleys of different diameters are connected by means of an open belt as shown in Figure(7), then the angle of contact or lap ( ) at the smaller pulley must be taken into consideration.. r1 = Radius of larger pulley, r2 = Radius of smaller pulley, and Figure(7) x = Distance between centres of two pulleys (i.e. O1 O2). Angle of contact or lap, ( ) = 2 rad 180 o 180 A little consideration will show that when the two pulleys are connected by means of a crossed belt as shown in Figure (8),then the angle of contact or lap ( ) on both the pulleys is same. Figure (8) ( ) Angle of contact or lap, = + 2 rad 180 o 180

  16. Centrifugal Tension Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to increase the tension on both, tight as well as the slack sides. The tension caused by centrifugal force is called centrifugal tension.At lower belt speeds (less than 10 m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10 m/s), its effect is considerable and thus should be taken into account. m = Mass of the belt per unit length in kg, v = Linear velocity of the belt in m/s, r = Radius of the pulley over which the belt runs in metres, and TC = Centrifugal tension acting tangentially at P and Q in newtons. TC = m . V2 When the centrifugal tension is taken into account, then total tension in the tight side, Tt1 = T1+TC Power transmitted, P=(Tt1 Tt2 ) v =[ (T1+TC) (T2+TC)]v. 1. 2. T T 3. The ratio of driving tensions may also be written as C= e 1 T T C 2

  17. Maximum Tension in the Belt A little consideration will show that the maximum tension in the belt (T) is equal to the total tension in the tight side of the belt (Tt1). = Maximum safe stress in N/mm2, b = Width of the belt in mm, and t = Thickness of the belt in mm. We know that maximum tension in the belt, T = Maximum stress cross-sectional area of belt = . b. t When centrifugal tension is neglected, then T (or Tt1) = T1 and when centrifugal tension is considered, then T (or Tt1) = T1+ TC

  18. Condition For the Transmission of Maximum Power We know that power transmitted by a belt, P = (T1 T2) v .. (1) T T= 2= T 1 we have also seen that the ratio of driving tensions is or e 1 e T 2 Substituting the value of T2 in equation (1) T T P 1 = 1 1 1 (2) Where v e e = C = = v T T C . v . 1 1 e 1 1 We know that T1= T - TCwhere T = Maximum tension to which the belt can be subjected in newton's, Substituting the value of T1 in equation (2), For maximum power, differentiate the above expression with respect to v and equate to zero, ( ) ( )C . ( ) = = = P T T C . v . P T m v C . v T v m dp v . . . . 2 3 C = 0 dv d ( ) 3= T v m v . . 0 dv T 3 m . v2 = 0 T 3 TC = 0 or T = 3 TC .(3) T TC= We know that T1 = T TC and for maximum power, 1. 3 T 3 = v 2. From equation (3), the velocity of the belt for the maximum power, m

  19. Initial Tension in the Belt When a belt is wound round the two pulleys (i.e. driver and follower), its two ends are joined together ; so that the belt may continuously move over the pulleys, since the motion of the belt from the driver and the follower is governed by a firm grip, due to friction between the belt and the pulleys. In order to increase this grip, the belt is tightened up. At this stage, even when the pulleys are stationary, the belt is subjected to some tension, called initial tension. T0= Initial tension in the belt, = Coefficient of increase of the belt length per unit force. A little consideration will show that the increase of tension in the tight side, = T1 T0 and increase in the length of the belt on the tight side , = (T1 T0) Similarly, decrease in tension in the slack side, = T0 T2 and decrease in the length of the belt on the slack side, = (T0 T2) Assuming that the belt material is perfectly elastic such that the length of the belt remains constant, when it is at rest or in motion, therefore increase in length on the tight side is equal to decrease in the length on the slack side. (T1 T0) = (T0 T2) or T1 T0 = T0 T2 2 + T T ...(Neglecting centrifugal tension) = T 1 2 0 ...(Considering centrifugal tension) 2 + + T T C T 2 = T 1 2 0

  20. V-belt drive We have already discussed that a V-belt is mostly used in factories and workshops 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. Ratio of Driving Tensions for V-belt 2 = Angle of the groove. = Coefficient of friction between the belt and sides of the groove. Thus the relation between T1 and T2 for the V-belt drive will be T= e sec( Co ) 1 T 2 Rope Drive The rope drives are widely used where a large amount of power is to be transmitted, from one pulley to another, over a considerable distance. Ratio of Driving Tensions for Rope Drive The ratio of driving tensions for the rope drive may be obtained in the similar way as V-belts. T= e sec( Co ) 1 T 2

  21. Examples EX(1):An open belt running over two pulleys 240 mm and 600 mm diameter connects two parallel shafts 3 meters apart and transmits 4 kW from the smaller pulley that rotates at 300 r.p.m. Coefficient of friction between the belt and the pulley is 0.3 and the safe working tension is 10N per mm width. Determine : 1. minimum width of the belt, 2. initial belt tension, and 3. length of the belt required. Solution. Given :d2= 240 mm = 0.24 m ; d1= 600 mm = 0.6 m ; x = 3 m ; P = 4 kW = 4000 W; N2= 300 r.p.m. ; = 0.3 ; T1= 10 N/mm width. 1. Minimum width of belt N d v 60 60 . 0 24 300 = = = m s . / 3 77 2 2 We know that velocity of the belt, Power transmitted (P), 4000 = (T1 T2) v = (T1 T2) 3.77 T1 T2 = 4000 / 3.77 = 1061 N .. (1) d d x 2 2 r r . . 0 6 0 24 = = = = = rad or sin . . 0 06 3 44 o 1 2 1 2 x 3 and angle of lap on the smaller pulley, = 180 2 = 180 2 3.44 = 173.12 478 2 . = = e T 173.12 / 180 = 3.022 rad T 3 . 022 . 0 3 1 We know that .(2) 2 From equations (1) and (2), T1 = 1779 N, and T2 = 718 N Since the safe working tension is 10 N per mm width, therefore minimum width of the belt, T 10 1779 = = = b mm . 177 9 1 10

  22. 2. Initial belt tension + + T T 1779 718 We know that initial belt tension, Ans T 0 = = = N 8 . 124 1 2 2 2 3. Length of the belt required We know that length of the belt required, ( ) = 2 d d ( ) + + + L d d x 2 1 2 x 1 2 2 4 ( ) 2 6 . 24 . 0 0 ( ) = + + + L 6 . 24 . 0 0 2 3 2 4 3 = 1.32 + 6 + 0.01 = 7.33 mAns.

  23. EX(2): A pulley is driven by a flat belt, the angle of lap being 120. The belt is 100 mm wide by 6 mm thick and density1000 kg/m3. If the coefficient of friction is 0.3 and the maximum stress in the belt is not to exceed 2 MPa, find the greatest power which the belt can transmit and the corresponding speed of the belt. Solution. Given : = 120 = 120 / 180 = 2.1 rad ; b = 100 mm = 0.1 m ; t = 6 mm = 0.006 m ; = 1000 kg / m3; = 0.3 ; = 2 MPa = 2 106 N/m2 . Speed of the belt for greatest power We know that maximum tension in the belt, T = . b. t = 2 106 0.1 0.006 = 1200 N and mass of the belt per metre length, m = Area length density = b. t. l. m = 0.1 0.006 1 1000 = 0.6 kg/m T 3 1200 = = = v m s . / 25 82 Speed of the belt for greatest power , m . 3 0 6 Greatest power which the belt can transmit We know that for maximum power to be transmitted, centrifugal tension, TC = T/3 = 1200/3 = 400 N and tension in the tight side of the belt, T1= T TC= 1200 400 = 800 N T T 88 1 . = = e 88 . 1 = = 3 . 1 . 0 2 T N . 425 5 1 1 T 2 2 ( )v . = 1 Greatest power which the belt can transmit, P T T 2 ( ) = = P W 5 . 82 . 800 425 25 9670

  24. EX(2)A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by V-belts from an electric motor running at 750 r.p.m. The diameter of the pulley on the compressor shaft must not be greater than 1 meter while the center distance between the pulleys is limited to 1.75 meter. The belt speed should not exceed 1600 m/min. Determine the number of V-belts required to transmit the power if each belt has a cross-sectional area of 375 mm2, density 1000 kg/m3and an allowable tensile stress of 2.5 MPa. The groove angle of the pulley is 35 . The coefficient of friction between the belt and the pulley is 0.25. Calculate also the length required of each belt Solution. Given :P = 90 kW ; N2= 250 r.p.m. ; N1= 750 r.p.m. ; d2= 1 m ; x = 1.75 m ; v = 1600 m/min = 26.67 m/s ; a = 375 mm2= 375 10-6m2; = 1000 kg/m3; = 2.5 Mpa = 2.5 106 N/m2 ; 2 = 35 or = 17.5 ; = 0.25 First of all, let us find the diameter of pulley on the motor shaft (d1). We know that. d N d or d N 750 1 2 1 N d 250 1 = = = = m 33 . 0 2 1 2 21 N 1 We know that the mass of the belt per meter length, m = Area length density = 375 10 6 1 1000 = 0.375 kg Centrifugal tension, TC = m.v2= 0.375 (26.67)2 = 267 N and maximum tension in the belt, T = . a = 2.5 106 375 10-06= 937.5 N Tension in the tight side of the belt, T1 = T TC = 937.5 267 = 670.5 N Let T2 = Tension in the slack side of the belt. For an open belt drive, as shown in Figure (9),

  25. r r d d 33 . 1 0 = = = = = rad or sin . 0 1914 11 o 1 2 2 1 x x 75 . 2 2 1 and angle of lap on smaller pulley (i.e. pulley on motor shaft), = 180 2 = 180 2 11 = 158 = 158 / 180 = 2.76 rad T T 954 . Figure (9), = = = = = T N . e e 954 . 67 36 9 ec ec cos 25 . 75 . cos 5 . 0 2 17 1 1 T 2 9 2 Number of V-belts We know that power transmitted per belt, = (T1 T2) v = (670.5 67.36) 26.67 = 16 086 W = 16.086 kW Total power transmitte d 90 = = = Number of V - belts 6 . 5 6 Power transmitte per d belt 086 . 16 Length of each belt ( ) = 2 d d ( ) + + + L d d x 2 1 2 We know that length of belt for an open belt drive, x 1 2 2 4 ( ) = 2 33 . 1 0 ( ) + + + L 33 . 75 . 1 0 2 1 75 . 2 4 1 = 2.1 + 3.5 + 0.064 = 5.664 m Ans.

  26. Exercises Q1/ A rope drive transmits 600 kW from a pulley of effective diameter 4 m, which runs at a speed of 90 r.p.m. The angle of lap is 160 ; the angle of groove 45 ; the coefficient of friction 0.28 ; the mass of rope 1.5 kg / m and the allowable tension in each rope 2400 N. Find the number of ropes required. Q2/ A pulley used to transmit power by means of ropes has a diameter of 3.6 metres and has 15 grooves of 45 angle. The angle of contact is 170 and the coefficient of friction between the ropes and the groove sides is 0.28. The maximum possible tension in the ropes is 960 N and the mass of the rope is 1.5 kg per metre length. What is the speed of pulley in r.p.m. and the power transmitted if the condition of maximum power prevail ? Q3/ A flat belt is required to transmit 30 kW from a pulley of 1.5 m effective diameter running at 300 r.p.m. The angle of contact is spread over (11/24) of the circumference. The coefficient of friction between the belt and pulley surface is 0.3. Determine, taking centrifugal tension into account, width of the belt required. It is given that the belt thickness is 9.5 mm, density of its material is 1100 kg / m3and the related permissible working stress is 2.5 MPa

  27. Q4/ In a horizontal belt drive for a centrifugal blower, the blower is belt driven at 600 r.p.m. by a 15 kW, 1750 r.p.m. electric motor. The centre distance is twice the diameter of the larger pulley. The density of the belt material = 1500 kg/m3; maximum allowable stress = 4 MPa; 1 = 0.5 (motor pulley); 2 = 0.4 (blower pulley); peripheral velocity of the belt = 20 m/s. Determine the following: 1. Pulley diameters diameters ; 2. belt length; 3. cross-sectional area of the belt; 4. minimum initial tension for operation without slip; and 5. resultant force in the plane of the blower when operating with an initial tension 50 per cent greater than the minimum value. Q5/An open belt connects two flat pulleys. The pulley diameters are 300 mm and 450 mm and the corresponding angles of lap are 160 and 210 . The smaller pulley runs at 200 r.p.m. The coefficient of friction between the belt and pulley is 0.25. It is found that the belt is on the point of slipping when 3 kW is transmitted. To increase the power transmitted two alternatives are suggested, namely (i) increasing the initial tension by 10%, and (ii) increasing the coefficient of friction by 10% by the application of a suitable dressing to the belt. Which of these two methods would be more effective? Find the percentage increase in power possible in each case.

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