Machine design, machine element, Belt drives and chain drives, selection of Belt - sheave and chain - sprocket, perancangan elemen mesin, transmisi sabuk dan rantai, pemilihan sabuk-puli dan rantai-sproket

1. BELT DRIVES and CHAIN DRIVES
Dodi Mulyadi, MT

2. Introduction
 Belts and chains are the major types of flexible power
transmission elements
 Belts operate on sheaves or pulleys
 Whereas chains operate on toothed wheels called sprockets
 Often used as torque increaser (speed reducer)
 (Gear drives have a higher torque capability but not flexible or
cheap)

3. Introduction
Belts
High Speed, Low Torque
2500 < v < 6500 ft/min
(4000 ft/min is generally ideal)
Standard lengths, wear, creep,
corrosive environment, slip,
temperature,
when must have tension need
idler
Quiet, flexible
Chains
Low Speed, High Torque
v < 1500 ft/min
Must be lubricated, wear, noise,
weight, vibration
Strength, length flexibility

4. TYPES OF BELT DRIVES
 Flat belt is the simplest type, often made
from leather or rubber-coated fabric
 The sheave surface is also flat and
smooth, and the driving force is therefore
limited by the pure friction between the
belt and the sheave
 Synchronous belts, sometimes called
timing belt
 Ride on sprockets having mating grooves
into which the teeth on the belt seat.
 This is a positive drive, limited only by the
tensile strength of the belt and the shear
strength of the teeth.

5. TYPES OF BELT DRIVES
 Cog belt, applied to standard V-grooved
sheaves
 The cogs give the belt greater flexibility and
higher efficiency compared with standard belts
 They can operate on smaller sheave diameters
Wrapped construction
 A widely used type of belt is the V-belt
drive
 The V-shape causes the belt to wedge
tightly into the groove, increasing friction
and allowing high torques to be
transmitted before slipping occurs.

6. TYPES OF BELT DRIVES
1. Protective fabric cover impregnated
with rubber
2. Tension Section - synthetic rubber
compounded to stretch as belt bends
3. Cords - synthetic fiber cords carry the
power and minimise stretch
4. Compression Section - synthetic rubber
compound to support cords evenly and
compress as belt bends

7. TYPES OF BELT DRIVES
Double angle V-beltPoly-rib belt
Other types
Vee-band

8. V - BELT DRIVES
 The typical arrangement
of the elements of a V-
belt drive is shown.
 The linear speed of the
pitch line of both sheaves
is the same as and equal to
the belt speed, then
vb = R1 ω1 = R2 ω2

9. V - BELT DRIVES

10. V - BELT DRIVES

11. V - BELT DRIVES
Standard Belt Cross Sections

12. V - BELT DRIVES
Standard Belt Cross Sections

13. V - BELT DRIVES
The basic data required for drive selection are the following:
1. The rated power of the driving motor or other prime mover
2. The service factor based on the type of driver and driven load
3. The center distance
4. The power rating for one belt as a function of the size and speed
of the smaller sheave
5. The belt length
6. The size of the driving and driven sheaves
7. The correction factor for belt length
8. The correction factor for the angle of wrap on the smaller
sheave
9. The number of belts
10. The initial tension on the belt

14. V - BELT DRIVES
Example
Design a V-belt drive that has the input sheave on the shaft of an
electric motor (normal torque) rated at 50.0 hp at 1160-rpm, full-
load speed. The drive is to a bucket elevator in a potash plant that
is to be used 12 hours (h) daily at approximately 675 rpm.
Solution
Given
Pin = 50 hp
N1 = 1160 rpm
N2 = 675 rpm

15. V - BELT DRIVES
Step 1. Calculate
design power
(PD)
PD = Pin x sf
= 50 x 1.3
= 65 hp

16. V - BELT DRIVES
Step 2. Select belt type.
for 65 hp at
1160-rpm
input speed.
5V belt is
recommended

17. V - BELT DRIVES
Step 3. Calculate the nominal speed ratio (SR)
SR = N1/N2
SR = 1160 / 675
= 1.72
Step 4. Determine sheave sizes
Choose belt speed of 4000 ft/min
vb = π D1 N1
D1 = 13.2 in  D2 = 22.7 in

18. V - BELT DRIVES
Step 5. Find “standard” sheave size
The two trials in boldface in Table 7-3 give only about 1 % variation
from the desired output speed of 675 rpm, and the speed of a bucket
elevator is not critical. Because no space limitations were given, let's
choose the larger size.

19. V - BELT DRIVES
Step 6a. Find
rated power (PR)
FIGURE - Power rating 5V belts
For small sheave Ø12.4 in
and 1160 rpm
PR= 26.4 hp

20. V - BELT DRIVES
Step 6b. Find “power added” to rated power
For SR = 1.72
and N1 = 1160 rpm
Power added
= 1.15 hp
∴ PR
= 26.4 hp + 1.15 hp
= 27.55 hp

21. V - BELT DRIVES
Step 7. Find estimated center distance
D2 < C < 3 (D2+D1)
21.2 < C < 3 (21.1+12.4)
21.2 < C < 100.5
In the interest of conserving space, let's try C = 24.0 in.
Step 8. Calculate belt length
L = 101.4 in

22. V - BELT DRIVES
Step 9. Select standard belt length
Lcalc = 101.4 in
The nearest
standard
length is
100 in

23. V - BELT DRIVES
Step 10. Calculate actual center distance
B = 4L – 6.28 (D2 + D1)
B = 4(100) – 6.28 (21.1+12.4) = 189.6
C = 23.3 in

24. V - BELT DRIVES
Step 11. Calculate wrap angle, small sheave
θ1 = 1580

25. V - BELT DRIVES
Step 12. Determine correction factors
Cθ = 0.94 CL = 0.96

26. V - BELT DRIVES
Step 13. Calculate corrected power
Corrected power = Cθ CL PR
Step 14. Calculate Belts needed
= (0.94)(0.96)(27.55 hp) = 24.86 hp
= 2.61 belts (Use 3 belts.)

27. V - BELT DRIVES
Step 15. Summary of Design
Input : Electric motor, 50.0 hp at 1160 rpm
Service factor : 1.3
Design power : 65.0 hp
Belt : 5V cross section, 100-in length, 3 belts
Sheaves : Driver, 12.4-in pitch diameter, 3 grooves, 5V
Driven, 21.1-in pitch diameter, 3 grooves, 5V
Actual output speed : 682 rpm
Center distance : 23.30 in

28. Chain DRIVES
 A chain is a power transmission element made as a series of pin-
connected links
 The design provides for flexibility while enabling the chain to
transmit large tensile forces
 When transmitting power between rotating shafts, the chain
engages mating toothed wheels, called sprockets

29. Chain DRIVES
Types of Chains

30. Chain DRIVES
Roller Chain Construction (Most common type)

31. Chain DRIVES
General recommendations for designing chain drives:
1. The minimum number of teeth in a sprocket should be 17
unless the drive is operating at a very low speed, under 100 rpm.
2. The maximum speed ratio should be 7.0, although higher ratios
are feasible. Two or more stages of reduction can be used to
achieve higher ratios.
3. The center distance between the sprocket axes should be
approximately 30 to 50 pitches (30 to 50 times the pitch of the
chain).
4. The larger sprocket should normally have no more than 120
teeth.

32. Chain DRIVES

33. Chain DRIVES

34. Chain DRIVES
General recommendations for designing chain drives:
9. The arc of contact, θ1, of the chain on the smaller sprocket
should be greater than 120°
θ1 = 180° - 2 sin-1 [(D2 – D1)/2C]
10. For reference, the arc of contact, θ2, on the larger sprocket is
θ2 = 180° + 2 sin-1 [(D2 – D1)/2C]

35. Chain DRIVES
Lubrication
 It is essential that adequate lubrication be provided for chain
drives.
 There are numerous moving parts within the chain, along with
the interaction between the chain and the sprocket teeth.
 Must be defined the lubricant properties and the method of
lubrication.
 Must be Recommended lubricant for chain drives

36. Chain DRIVES
Lubrication
Lubrication methods
(American Chain Association, Naples, FL)

37. Chain DRIVES
Lubrication
Lubrication methods
(American Chain Association, Naples, FL)

38. Chain DRIVES
Example
Design a chain drive for a heavily loaded coal conveyor to be
driven by a gasoline engine through a mechanical drive. The input
speed will be 900 rpm, and the desired output speed is 230 to 240
rpm. The conveyor requires 15.0 hp.
Solution
Given
Power transmitted, P = 15 hp
Speed of motor, N1 = 900 rpm
Output speed range, N2 = 230 to 240 rpm

39. Chain DRIVES
Type of driver
TABLE Service factors for chain drives
Step 1.
Calculate
design power
(PD)
PD = P x sf
= 15 x 1.4
= 21 hp

40. Chain DRIVES
Step 2. Calculate Speed Ratio (SR)
SR = N1 / N2
Using the middle of the required range of output speeds,
we have, SR = 900 / 235 = 3.83
Step 3.
 Refer to the tables for power capacity (Tables 7-5, 7-6, and 7-7),
and select the chain pitch.
 For a single strand, the no. 60 chain with p = 3/4 in seems best.
 A 17-tooth sprocket is rated at 21.96 hp at 900 rpm by
interpolation.
 At this speed, type B lubrication (oil bath) is required.

41. Chain DRIVES

42. Chain DRIVES

43. Chain DRIVES

44. Chain DRIVES
Step 4. Compute the required number of teeth on the large
sprocket:
Z2 = Z1 x SR
Z2 = 17 x 3.83 = 65.11
Let's use the integer: 65 teeth.
Step 5. Compute the actual expected output speed
N2 = N1 (Z1 / Z2)
= 900 (17/65)
= 235.3 rpm (Okay!)

45. Chain DRIVES
Step 6. Compute the pitch diameters of the sprockets
D1 = 4.082 in
D2 = 15.524 in

46. Chain DRIVES
Step 7. Specify the nominal center distance.
Let's use the middle of the recommended range,
C = 40 pitches.
Step 8. Compute the required chain length
L = 122.5 pitches, let’s use 122 pitches

47. Chain DRIVES
Step 9. Compute the actual theoretical center distance
C = 39.75 pitches
C = 39.75 x 0.75 in = 29.8 in

48. Chain DRIVES
Step 10. Compute the angle of wrap of the chain
θ1 = 180° - 2 sin-1 [(D2 – D1)/2C]
θ1 = 180° - 2 sin-1 [(15.524 – 4.082)/2(29.8)]
θ1 = 157.86°
Because this is greater than 120°, it is acceptable.
θ2 = 180° + 2 sin-1 [(D2 – D1)/2C]
θ2 = 180° + 2 sin-1 [(15.524 – 4.082)/2(29.8)]
θ2 = 202.14°

49. Chain DRIVES
Summary of Design
 Pitch: No. 60
chain. 3/4-in pitch
 Length: 122 pitches
= 122(0.75)
= 91.50 in
 Center distance:
C = 29.825 in (maximum)
 Sprockets: Single-strand, no. 60. 3/4-in pitch
Small: 17 teeth. D = 4.082 in
Large: 65 teeth, D = 15.524 in
 Type B lubrication is required. The large sprocket
can dip into an oil bath.