B.Balavairavan AP/Mech KCET ME 6601 - DESIGN OF TRANSMISSION SYSTEMS

1. Design and Selection of V – Belt drives
using PSG design data book
DEPARTMENT OF MECHANICAL
ENGINEERING
By
Mr. B.Balavairavan
Assistant Professor
Mechanical Engineering
Kamaraj College of Engg and Tech
Virudhunagar

2. V – Belt Types
Based on the manufacture
specifications the v-belts are
classified into many types.
For example : Cog Belt, Wedge Belt,
Chipper Belt, Double angle belt, v-flat
belt etc..

3. Cross Section of V-Belt

4. Advantages of V-Belt drives
• Smooth starting and running.
• Permit a wide range of driven speeds, using
standard electric motors.
• They’re rugged and provide years of trouble-
free performance with minimal attention . . .
even under adverse conditions.
• Capable of transmitting power around
corners or out of plane drives.
• Clean—require no lubrication. • Highly
efficient.
• Extremely wide horsepower ranges.
• Dampen vibration between driver and driven
machines.
• Silent operation. • Long service life. • Easy
installation.
• Can be used as an effective means of
clutching.
• They act as a “safety fuse” refusing to
transmit severe power overload, except for a
very brief period.
• V-belts and sheaves wear gradually-making
preventive corrective maintenance simple and
easy.

5. V belt pulley
All pulley grooves are to be the same size.
Uneven wear of grooves causes belts
running on different diameter levels in the
pulley. That generates excessive slip of the
belts on one side.

6. Tight side and slack side

8. For the selection of a two-pulley power transmission
V - belt the following data are required:
Motor power or power to be transmitted (in kW)
Speed of drives
Number of revolutions of driving pulley (n)
Number of revolutions of driven pulley (N)
Diameter of pulley
Diameter of driving pulley (d)
Diameter of driven pulley (D)
Transmission ratio or velocity ratio (i)
Center distance (C)

9. Procedure Stages
1. Selection of belt section.
2. Selection of standard pulleys.
3. Approximate or Maximum centre distance.
4. Nominal pitch length selection.
5. Modification factors “length correction, Service factor, Arc of contact factor”
6. Calculating of maximum power capacity.
7. No of belts.
8. Actual centre distance
9. Pulley dimensions

10. Problem 1
Problem Statement:
A centrifugal pump running at 340
rpm is to be driven by a 100 kW
motor running at 1440 rpm. The drive
is to work for atleast 20 hours a day.
The centre distance between motor
and pump shaft is 1200 mm. Sugest
a suitable v-belt drive.
Given Data:
Power = 100 kW
Driver speed (n) = 1440 rpm
Driven speed (N) = 340 rpm
Centre Distance (c) = 1200 mm

11. Step - 1 “Selection of belt section”
From page no: 7.58
Based on the power of motor, the cross section of belt is selected, as D

12. Step - 2 “Selection of Std. Pulleys”
i =
340
1440
i = 4.235
4.235 =
355
D
D = 1503.33 mm
From page No: 7.58 for the selected cross section, d value is given
d = 355 mm

13. d = 355 mm
D = 1600 mm
from page no: 7.54
Standard pulley diameters are

14. Step - 3 “Centre distance”
If centre distance is not given, it can be approimated using the table in page no: 7.61
Given, Centre distance (C) = 1200 mm

15. Step - 4 “Nominal pitch length selection”
= 5793.83 mm
Length of belt =
4 x 1200
2 x 1200 +
2
𝜋 (1600+355)+ (1600 - 355)2

16. The standard nominal length is selected based on the cross section of belt from
page no:7.59 to 7.60
Standard Nominal Length = 6124 mm

17. Step - 5 “Modification factors”
Length Correction Factor (Fc)
from page no: 7.58 – 7.60
Based on cross section of belt
selected its choosen
Load Correcting factor = 1

18. Arc of Contact Factor (Fd)
from page no: 7.68
Arc of contact =
1200
1600 - 355
x 60 ൦180 ൦ -
= 117.75 ൦
Arc of contact factor = 0.81
for arc of contact 117.75 ൦

19. Service Factor (Fa)
from page no: 7.69
The working over 16 hrs
Service factor = 1.3

20. Step - 6 “Calculation of Maximum power capacity”
from page no: 7.62

21. Step - 6 “Calculation of Maximum power capacity”
from page no: 7.62
S =
= 26.76 m/s
x n
60
𝜋 x d
=
x 1440
60
𝜋 x 0.355
de= 355
= 404.7 mm
x 1.14
To find de

22. Power in kW=
= 25 kW
3.22 x 2.76-0.09 -
506.7
404.7
- 4.78 25 x 10 -4 x 26.762 26.76

23. Step – 7 “Determination of number of belts”
from page no. 7.70
= 7.486
=
21.44 x 1
100 x 1.3
x 0.81
= 8

24. Step - 8 “Actual Centre Distance”
from page no: 7.61

25. Step - 9 “Pulley Dimensions”
from page no: 7.70, for the selected “D” cros section v-belt, the pulley
dimensions follows
d = 355 mm
D = 1600 mm
Pitch Width (lp) = 27 mm
Angle (A) = 36 o
Centre to centre distance (e) = 37 mm
Edge of pulley to first grove (f) = 24 mm
Minimum distance down to pitch line = 8.1 mm

26. Result (dimensions in mm)
C =1386.83
D = 1600
d = 355
36 o
27
b = 8.1
A
37 24
lp
e f
t = 19
32
w
for “D” cross section V-Belt

27. Difference between Flat Belt and V- belt Drives
1. Frictional engagement between the
lateral wedge surfaces of the belt profile.
2. Large bending cross-section, large
mass.
3. Different effective diameters and thus
varying speeds of individual belts.
1. Frictional engagement on the outer
pulley diameter.
2. Small bending cross-section, small
mass.
3. Precisely defined effective diameters
across the belt width and therefore exact
speeds.

28. 1. Differences in the effective diameter
2. Uneven distribution of tension across
individual belts.
3. Unequal speed conditions in individual
belts cause tensions, vibrations and
fluttering.
4. Vibrations cause premature bearing
damages.
1. Precisely defined diameter across the
entire belt width.
2. Even distribution of tension across the
entire belt width.
3. Smooth running as a result of exactly
defined speed conditions across the entire
belt width, even at high speed.
4. Smooth running assures longer bearing
life.