This document provides an overview of belt drives and chain drives. It discusses their applications and differences. Belt drives are used for high speed, low torque applications, while chain drives are used for low speed, high torque applications. The document then provides details on the design process for V-belt drives, including selecting belt type, calculating speed ratio, sizing sheaves, determining power rating, and selecting belt length. An example design problem is worked through step-by-step to illustrate the full V-belt drive design process. Finally, the document discusses chain drive design, including types of chains, construction, and provides an example design problem for sizing a chain drive.

1. UNIT: I
Belt Drives and Chain Drives
CVT

2. OVERVIEW – WHY USED?
1.) Transfer power (torque) from one location to
another. From driver: motor,peddles,
engine,windmill,turbine to driven: conveyor belt,
back wheels/bike,generator rock crusher,dryer.
2.) Used to span large distances or need flexible x-
mission elements. Gear drives have a higher
torque capability but not flexible or cheap.
3.) Often used as torque increaser (speed reducer),
max speed ratio: 3.5:1. Gear drives?? Virtually
unlimited!
Applications? Show rust abrader, glove factory, draw sample drive of rust abrader, show slides from mechanism book.

3. Sometimes
desirable to
have both chain
and belt drive
(Fig 7.1)
Belt: high
speed/low
torque
Chain: Low
speed/high
torque

4. BELTS VS. CHAINS
Belts
Chains
Use When:
Speed:
Dis:
Advs:
High Speed, Low T High T, Low Speed
2500 < Vt < 7000 ft./min. V < 1500 ft./min.
Must design with standard
lengths, wear, creep,
corrosive environment, slip,
temp., when must have
tension need idler
Must be lubricated,
wear, noise, weight,
vibration
Quiet, flexible, cost Strength, length
flexibility

5. TYPES OF BELTS:
a)V-belt most common for
machine design, several
types (Fig. 7.5 – 7.8)
•Timing belt (c & d) have
mating pulleys to
minimize slippage
•c) Pos retention due to
mating pulleys
•d) Pos retention due to
increased contact area
•Flat belt
(rubber/leather) not
shown, run on tapered
pulleys
Add notes

6. TYPES OF V-BELTS

7. V-BELT DRIVE DESIGN PROCESS
 Need rated power of the driving motor/prime mover. BASE sizing
on this.
 Service factor based on type of driver and driven load.
 Center distance (adjustment for center distance must be provided or
use idler pulley) nominal range D2 < C < 3(D2 + D1)
 Power rating for one belt as a function of size and speed of the
smaller sheave
 Belt length (then choose standard size)
 Sizing of sheaves (use standard size). Most commercially available
sheaves should be limited to 6500 ft/min belt speed.
 Belt length correction factor
 Angle of wrap correction factor. Angle of wrap on smaller sheave
should be greater than 120 deg.
 Number of belts
 Initial tension in belts

8. KEY EQUATIONS
1
2
2
1
2
2
1
1
2
2
1
1
2
2
D
D
D
D
R
R
b












Belt speed (no
slipping) =
Speed
ratio =
Pitch dia’s of
sheaves
12
1
1n
D
b

 
Belt speed
ft/min
Pitch dia
in inches rpm

9. KEY EQUATIONS
 Belt length:
 Center Distance:
 Where,
C
D
D
D
D
C
L
4
)
(
)
(
2
2
2
1
2
1
2






16
)
(
32 2
1
2
2
D
D
B
B
C




)
(
2
4 1
2 D
D
L
B 

 
Note: usually belt length standard
(use standard belt length table 7-2),
then calculate C based on fixed L
Recommended D2 < C < 3(D2+D1)

10. KEY EQUATIONS CONT…
 Angle of contact of belt on each sheave





 


 
C
D
D
2
sin
2
180 1
2
1
1






 


 
C
D
D
2
sin
2
180 1
2
1
2

Note: Select D’s and C’s so maximum contact (Ѳ1 + Ѳ2 = 180º). If
less then smaller sheave could slip and will need reduction factor
(Table 7-14).

11. V-BELT DESIGN EXAMPLE
 Given: 4 cylinder Diesel runs @ 80hp, 1800 rpm to
drive a water pump (1200 rpm) for less than 6
hr./day
 Find: Design V-belt drive

12. V-BELT DESIGN EXAMPLE CONT…
1.) Calculate design power:
Use table 7-1(<6h/day, pump, 4 cyl. Engine)
Design Power = input power x service factor
= 80 hp x 1.1
= 88 hp

13. V-BELT DESIGN EXAMPLE CONT…
2.) Select belt type, Use table 7-9
Design Power = 88 hp
Speed = 1800 rpm
Choose 5V

14. V-BELT DESIGN EXAMPLE CONT…
3.) Calculate speed ratio
SR = w1/w2
= 1800 rpm/1200 rpm
= 1.5

15. V-BELT DESIGN EXAMPLE CONT…
4.) Determine sheave sizes
Choose belt speed of 4000 ft/min
(Recall 2500ft./min. < vb < 7000 ft./min)
in
n
v
D
n
D
v b
b 488
.
8
1800
*
4000
*
12
12
12 1
1
1
1








So…
D1 = 8.488in
D2 = SR * D1 = 1.5 * 8.488
D2 = 12.732in

16. V-BELT DESIGN EXAMPLE CONT…
5.) Find sheave size (Figure 7-11)
Engine (D1)
8.4
8.4
8.9
X 1.5
12.6
12.6
13.35
Standard D2
12.4
13.1
13.1
Actual n2
1219
1154
1223
**All look OK, we will try the first one
Must find acceptable standard sheave 1, then corresponding
acceptable sheave 2
2
1
1
2
1
2
2
1
D
n
D
n
D
D
n
n




17. V-BELT DESIGN EXAMPLE CONT…
6.) Find rated power (use figure 7-11 again)
Rated Power = 21 hp

18. V-BELT DESIGN EXAMPLE CONT…
 Adjust for speed ratio to get total power/belt
Total power = 21hp +1.55hp = 21.55hp

19. V-BELT DESIGN EXAMPLE CONT…
7.) Find estimated center distance
D2 < C < 3(D2+D1)
12.4 < C < 3 (12.4 + 8.4)
Notice – using standard
sheave sizes found
earlier, not calculated
diameters
12.4 < C < 62.4
To provide service access will try towards long end,
try C = 40”

20. V-BELT DESIGN EXAMPLE CONT…
8.) Find belt length
C
D
D
D
D
C
L
4
)
(
)
(
2
2
2
1
2
1
2






in
in
in
in
in
L 765
.
112
)
40
(
4
)
4
(
)
8
.
20
(
57
.
1
)
40
(
2
2





21. V-BELT DESIGN EXAMPLE CONT…
9.) Select standard belt length
Lcalc = 112.765
Choose 112”

22. V-BELT DESIGN EXAMPLE CONT…
10.) Calculate actual center distance
)
(
2
4 1
2 D
D
L
B 

 
"
367
.
317
)
8
.
20
(
28
.
6
)
112
(
4



B
B
16
)
(
32 2
1
2
2
D
D
B
B
C




"
62
.
39
16
)
4
(
32
367
.
317
367
.
317 2
2




C
C

23. V-BELT DESIGN EXAMPLE CONT…
 11.) Find wrap angle, small sheave





 


 
C
D
D
2
sin
2
180 1
2
1
1











 
2
.
174
)
62
.
39
(
2
4
sin
2
180
1
1
1



24. V-BELT DESIGN EXAMPLE CONT…
12.) Determine correction factors
98
.


C 98
.

L
C

25. V-BELT DESIGN EXAMPLE CONT…
13.) Calculate corrected power
hp
hp
P
C
C
Power
Corrected L
61
.
21
)
5
.
22
)(
98
)(.
98
(.


 

26. V-BELT DESIGN EXAMPLE CONT…
14.) Belts needed
07
.
4
/
61
.
21
88
#
Power/belt
Corrected
(hp)
Power
Design
#



belt
hp
hp
belts
belts
Use 4 belts!

27. V-BELT DESIGN EXAMPLE CONT…
15.) Summary
D1=8.4”
D2=12.4”
Belt Length = 112”
Center Distance = 39.62”
4 Belts Needed

28. CHAIN DRIVES

29. CHAIN DRIVES
 Types of Chains

30. CHAIN DRIVES
 Roller Chain Construction (Most common Type)

31. CHAIN DESIGN PROCESS
 1.) # of sprocket teeth, N1 (smaller sprocket) > 17 (unless low
speed < 100 rpm.)
 2.) Speed ratio = n1/n2 7
 3.) 30 x Pitch Length < Center Distance < 50 x Pitch Length
 4.) Angle of contact of chain on smaller sprocket > 120°
 5.) # sprocket teeth, N2 (longer sprocket) < 120


32. CHAIN DRIVES

33. CHAIN DRIVES DESIGN EXAMPLE
 Given:
 Driver: Hydraulic Motor
 Driven: Rock Crusher
 ni = 625 rpm, 100 hp
 no = 225 rpm
 Find:
 Design belt drive

34. CHAIN DRIVES DESIGN EXAMPLE
 1.) Design Power
DP = SF x HP
DP = 1.4 ( Table 7-8) x 100 hp
DP = 140 hp

35. CHAIN DRIVES DESIGN EXAMPLE
 2.) Calculate Velocity Ratio
rpm
rpm
VR
N
N
n
n
VR
i
o
o
i
225
625



VR = 2.78
n = speed
N = teeth
Heavy Requirement!!

36. CHAIN DRIVES DESIGN EXAMPLE
 3.) Choose:
 Size - (40, 60, 80) 80 (1in)
 # Strands – use 4
Required HP/chain = 140hp/3.3
= 42.42 hp/chain
Number of Roller
Chain Strands
Multiple Strand
Factor
2 1.7
3 2.5
4 3.3
5 3.9
6 4.6
25
78
.
2 o
i
o N
N
N
VR 


No = 69.5  use 70 teeth

37. CHAIN DRIVES DESIGN EXAMPLE
 Conclusion:
 4 Strands
 No. 80 Chain
 Ni = 25 Teeth
 No= 70 Teeth

38. CHAIN DRIVE DESIGN EXAMPLE
Guess center distance: 40 Pitches
)
40
(
4
)
25
70
(
2
25
70
)
40
(
2
)
(
4
)
(
2
2
)
(
2
2
2
2
1
2
1
2












pitches
L
C
N
N
N
N
C
pitches
L
L = 128.8 pitches  use 130 pitches

39. CHAIN DRIVES DESIGN EXAMPLE
Actual Center Distance, C







 






 












 






 





2
2
2
2
2
1
2
2
1
2
1
2
4
)
25
70
(
8
2
25
75
130
2
25
70
130
4
1
)
(
4
)
(
8
2
2
4
1
)
(


pitches
C
N
N
N
N
L
N
N
L
pitches
C
C = 40.6  use 40 Pitches