Chapter 7:

Report
Chapter 7:
Belt Drives and Chain Drives
CVT
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 xmission 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.
Sometimes
desirable to
have both chain
and belt drive
(Fig 7.1)
Belt: high
speed/low
torque
Chain: Low
speed/high
torque
Belts vs. Chains
Belts
Use When:
Speed:
Dis:
Advs:
High Speed, Low T
2500 < Vt < 7000 ft./min.
Chains
High T, Low Speed
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
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
Types of V-Belts
V-belt Drive Design Process
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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
Key Equations
Belt speed (no
slipping) = 
Speed
ratio =
D11 D22

b  R11  R2 2 
2
2
1 D2
Pitch dia’s of

sheaves
Pitch dia
2 D1
in inches
Belt speed
ft/min
b 
rpm
D1n1
12
Key Equations


Belt length:
( D2  D1 ) 2
L  2C  ( D2  D1 ) 

4C
Center Distance:

2
B  B 2  32( D2  D1 ) 2
C
16
Where, B  4L  2 ( D2  D1 )
Recommended D2 < C < 3(D2+D1)
Note: usually belt length standard
(use standard belt length table 7-2),
then calculate C based on fixed L
Key Equations cont…

Angle of contact of belt on each sheave
 D2  D1 
1  180  2 sin 
 2C 
1
 D2  D1 
 2  180  2 sin 
 2C 
1
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).
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
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
V-belt Design Example Cont…
2.) Select belt type, Use table 7-9
Choose 5V
Speed = 1800 rpm
Design Power = 88 hp
V-belt Design Example Cont…
3.) Calculate speed ratio
SR = w1/w2
= 1800 rpm/1200 rpm
= 1.5
V-belt Design Example Cont…
4.) Determine sheave sizes
Choose belt speed of 4000 ft/min
(Recall 2500ft./min. < vb < 7000 ft./min)
D1n1
12vb 12* 4000
vb 
 D1 

 8.488in
12
n1  *1800
So…
D1 = 8.488in
D2 = SR * D1 = 1.5 * 8.488
D2 = 12.732in
V-belt Design Example Cont…
5.) Find sheave size (Figure 7-11)
Must find acceptable standard sheave 1, then corresponding
acceptable sheave 2
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
n1 D2
Dn

 n2  1 1
n2 D1
D2
**All look OK, we will try the first one
V-belt Design Example Cont…
6.) Find rated power (use figure 7-11 again)
Rated Power = 21 hp
V-belt Design Example Cont…

Adjust for speed ratio to get total power/belt
Total power = 21hp +1.55hp = 21.55hp
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”
V-belt Design Example Cont…
8.) Find belt length
( D2  D1 ) 2
L  2C  ( D2  D1 ) 

4C
(4in) 2
L  2(40in)  1.57(20.8in) 
 112.765in
4(40in)
2
V-belt Design Example Cont…
9.) Select standard belt length
Lcalc = 112.765
Choose 112”
V-belt Design Example Cont…
10.) Calculate actual center distance
B  4L  2 ( D2  D1 )
B  4(112)  6.28(20.8)
B  317.367"
B  B 2  32( D2  D1 ) 2
C
16
317.367  317.3672  32(4) 2
C
16
C  39.62"
V-belt Design Example Cont…

11.) Find wrap angle, small sheave
 D2  D1 
1  180  2 sin 
 2C 

4
1 
1  180  2 sin 

2
(
39
.
62
)


1
1  174.2
V-belt Design Example Cont…
12.) Determine correction factors
C  .98
CL  .98
V-belt Design Example Cont…
13.) Calculate corrected power
Corrected Power  C CL P
 (.98)(.98)(22.5hp)
 21.61hp
V-belt Design Example Cont…
14.) Belts needed
Design P ower (hp)
# belts 
CorrectedP ower/belt
88hp
# belts 
 4.07
21.61hp / belt
Use 4 belts!
V-belt Design Example Cont…
15.) Summary
D1=8.4”
D2=12.4”
Belt Length = 112”
Center Distance = 39.62”
4 Belts Needed
Chain Drives
Chain Drives
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Types of Chains
Chain Drives
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Roller Chain Construction (Most common Type)
Chain Design Process
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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
Chain Drives
Chain Drives Design Example
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Given:
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Driver: Hydraulic Motor
Driven: Rock Crusher
ni = 625 rpm, 100 hp
no = 225 rpm
Find:
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Design belt drive
Chain Drives Design Example
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1.) Design Power
DP = SF x HP
DP = 1.4 ( Table 7-8) x 100 hp
DP = 140 hp
Chain Drives Design Example
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2.) Calculate Velocity Ratio
ni N o
VR 

no N i
n = speed
N = teeth
625rpm
VR 
225rpm
VR = 2.78
Heavy Requirement!!
Chain Drives Design Example
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3.) Choose:
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Size - (40, 60, 80)
80 (1in)
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# Strands – use 4
Required HP/chain = 140hp/3.3
= 42.42 hp/chain
VR  2.78 
No No

Ni 25
No = 69.5  use 70 teeth
Number of Roller
Chain Strands
Multiple Strand
Factor
2
1.7
3
2.5
4
3.3
5
3.9
6
4.6
Chain Drives Design Example
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Conclusion:
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4 Strands
No. 80 Chain
Ni = 25 Teeth
No= 70 Teeth
Chain Drive Design Example
Guess center distance: 40 Pitches
N 2  N1 ( N 2  N1 ) 2
L( pitches)  2C 

2
4 2C
70  25 (70  25) 2
L( pitches)  2(40) 

2
4 2 (40)
L = 128.8 pitches  use 130 pitches
Chain Drives Design Example
Actual Center Distance, C
1
N  N1
C ( pitches)   L  2

4
2

1
70  25
C ( pitches)  130

4
2

N 2  N1  8( N 2  N1 ) 2 



 L 
2 
4 2


2
75  25 8(70  25)

130



2 
4 2
2
C = 40.6  use 40 Pitches
2





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