1) A belt drive uses a looped strip of flexible material to mechanically link rotating shafts. Belts are looped over pulleys to transmit power between shafts or track their relative movement. 2) Belt drives are simple, economical, tolerate some misalignment between shafts, dampen noise and vibration, and are highly efficient. However, they have limits on speed, power transmission, and operating temperatures. 3) Designing a v-belt drive involves determining factors like the application factor based on power and speed, selecting the proper belt type and pulley diameters and pitch length, calculating the center distance and contact angle, and determining the number of belts needed based on the rated power of each

1. TOPIC : Selection of belt drive
Made by:
Mr.Vaibhav Pardeshi

2. Introduction
1. A belt is a looped strip of flexible material, used to mechanically link
two or more rotating shafts.
2. They may be used as a source of motion, to efficiently transmit power,
or to track relative movement. Belts are looped over pulleys.
3. In a two pulley system, the belt can either drive the pulleys in the same
direction, or the belt may be crossed, so that the direction of the shafts
is opposite.

3. Advantages of belt drive
 They are simple. They are economical.
 Parallel shafts are not required.
 Noise and vibration are damped out. Machinery life is prolonged
because load fluctuations are cushioned (shock-absorbed).
 They are lubrication-free. They require only low maintenance.
 They are highly efficient (90–98%, usually 95%). Some
misalignment is tolerable.
 They are very economical when shafts are separated by large
distances.

4. Disadvantages of belt drive
 Heat buildup occurs. Speed is limited to usually 7000 feet per minute (35 meters per
second). Power transmission is limited to 370 kilowatts (500 horsepower).
 Operating temperatures are usually restricted to –31 to 185°F (–35 to 85°C).
 Some adjustment of center distance or use of an idler pulley is necessary for wear and
stretch compensation.
 A means of disassembly must be provided to install endless belts.

5. V-belt Drive Design Process
 Input power
Centre distance
Input speed
 Determine the application factor (Fa) from the table.
 Calculate the design power by using the following relation
 design power= Fa *transmitted power
 Plot a point with design power as x coordinate and input speed as y coordinate . the location
of this point decides the type of cross section of belt.
 Determine the recommended pitch dia. Of smaller pulley from the table. Calculate
the pitch dia. Of bigger pulley by following relationship.
D= i * d
where, i= speed ratio
take std. values of d & D from the table.
 determine the pitch length of belt
 L=2𝐶 +
𝜋(𝐷+𝑑)
2
+
(𝐷−𝑑)2
4𝐶

6.  Compare the above value of L with std. value from the given table
 Calculate correct centre distance C by substituting std. value of L .
L=2𝐶 +
𝜋(𝐷+𝑑)
2
+
(𝐷−𝑑)2
4𝐶
 Determine the correction factor(Fc) for belt pitch length from given table
 Calculate the arc of contact for smaller pulley by following relationship
𝛼 𝑠 = 180 − 2 sin−1 (𝐷−𝑑)
2𝐶
determine the correction factor Fd for arc of contact from the given table.
 depending upon the type of cross-section refer respective table to determine power rating
( Pr ) of single v belt.
 The last step is to find out no. of belts
𝑛𝑜. 𝑜𝑓 𝑏𝑒𝑙𝑡𝑠 =
𝑃∗𝐹𝑎
𝑃𝑟∗𝐹𝑐∗𝐹𝑑

7. V-Belt Design Example
 Given: Rated Power= 2 HP
N = 1400 rpm
n = 350 rpm
Center Distance = 500 mm
 Find: Design V-belt drive

8. V-belt Design Example Cont…
1.) Calculate design power:
Design Power = input power x service factor
= 2 HP x 1.2
= 2.4 x 746 W
= 1.7904 kW

9. V-belt Design Example Cont…
2.) Select belt type,
Design Power = 1.7904 kW
Speed = 1400 rpm
From the chart,
T=8mm
W=13mm
Wp=11mm
d=75mm

10. V-belt Design Example Cont…
3.) Calculate speed ratio
SR = i = w1/w2
= 1400/350
= 4
Therefore,
D = i * d
= 300mm

11. V-belt Design Example Cont…
4.) Pitch length of belt
L = 2𝐶 +
𝜋(𝐷+𝑑)
2
+
(𝐷−𝑑)2
4𝐶
Therefore 𝐿 = 1614.36𝑚𝑚
Selection of standard belt length from table,
𝐿 = 1640𝑚𝑚.

12. 5.) Find the correct center distance,
From the standard pitch length
and using same relation for pitch length we find the correct center distance
L = 2𝐶 +
𝜋(𝐷+𝑑)
2
+
(𝐷−𝑑)2
4𝐶
Therefore 𝐶 = 513.04𝑚𝑚
V-belt Design Example Cont…

13. V-belt Design Example Cont…
6.) Find correction factor from the standard table
For the obtained cross-section
and pitch length
the correction factor is
𝐹 𝑐 = 0.99

14. V-belt Design Example Cont…
7)Find Lap Angle
𝛼 𝑠 = 180 − 2 sin−1 𝐷−𝑑
2𝐶
Therefore,
𝛼 𝑠 = 154.67𝑚𝑚
8) From lap angle find correction factor for Arc of Contact (𝐹𝑑) from standard table
Therefore
𝐹𝑑= 0.73

15. V-belt Design Example Cont…
7.) Find Rated Power from table
𝑃𝑟 = 1.05 𝑘𝑊

16. V-belt Design Example Cont…
10) Number of belts,
𝑛𝑜. 𝑜𝑓 𝑏𝑒𝑙𝑡𝑠 =
𝑃∗𝐹𝑎
𝑃𝑟∗𝐹𝑐∗𝐹𝑑
= 2.

17. References
 Fig1 : http://nptel.ac.in/courses/112103024/module3/lec1/images/3.png
 Fig 2 :
https://ecoursesonline.icar.gov.in/pluginfile.php/1552/mod_page/content/3/283.
png
 V. B. Bhandari, Design of Machine Elements, McGraw Hill Publications, 4th
Edition, Pg.494-536.

18. THANK YOU