Chapter 7

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 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.

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 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

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

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

1

2

2

1

22112211 22

D

D

DDRRb

Belt speed (no slipping) =

Speed ratio =

Pitch dia’s of sheaves

1211nD

b

Belt speed ft/min

Pitch dia in inches

rpm

Key Equations Belt length: Center Distance:

Where,

C

DDDDCL

4

)()(

22

212

12

16

)(32 212

2 DDBBC

)(24 12 DDLB

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)

Key Equations cont… Angle of contact of belt on each

sheave

C

DD

2sin2180 121

1

C

DD

2sin2180 121

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).

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

Design Power = 88 hp

Speed = 1800 rpm

Choose 5V


3.) Calculate speed ratioSR = 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)

inn

vD

nDv b

b 488.81800*

4000*1212

12 11

11

So…D1 = 8.488in

D2 = SR * D1 = 1.5 * 8.488

D2 = 12.732in

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

112

1

2

2

1

D

nDn

D

D

n

n

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”


8.) Find belt length

C

DDDDCL

4

)()(

22

212

12

inin

inininL 765.112

)40(4

)4()8.20(57.1)40(2

2

V-belt Design Example Cont…9.) Select standard belt length

Lcalc = 112.765

Choose 112”


10.) Calculate actual center distance

)(24 12 DDLB

"367.317

)8.20(28.6)112(4

B

B

16

)(32 212

2 DDBBC

"62.3916

)4(32367.317367.317 22

C

C

V-belt Design Example Cont… 11.) Find wrap angle, small sheave

C

DD

2sin2180 121

1

2.174

)62.39(2

4sin2180

1

11


12.) Determine correction factors

98.C 98.LC

V-belt Design Example Cont…13.) Calculate corrected power

hp

hp

PCCPowerCorrected L

61.21

)5.22)(98)(.98(.


14.) Belts needed

07.4/61.21

88#

Power/belt Corrected

(hp)Power Design #

belthp

hpbelts

belts

Use 4 belts!


15.) Summary

D1=8.4”

D2=12.4”

Belt Length = 112”

Center Distance = 39.62”

4 Belts Needed

Chain Drives

Chain Drives Types of Chains

Chain Drives Roller Chain Construction (Most common Type)

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

Chain Drives

Chain Drives Design Example Given:

Driver: Hydraulic Motor Driven: Rock Crusher ni = 625 rpm, 100 hp no = 225 rpm

Find: Design belt drive

Chain Drives Design Example 1.) Design Power

DP = SF x HPDP = 1.4 ( Table 7-8) x 100 hpDP = 140 hp

Chain Drives Design Example 2.) Calculate Velocity Ratio

rpm

rpmVR

N

N

n

nVR

i

o

o

i

225

625

VR = 2.78

n = speedN = teeth

Heavy Requirement!!

Chain Drives Design Example 3.) Choose:

Size - (40, 60, 80) 80 (1in)

# Strands – use 4Required 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 4.625

78.2 o

i

o N

N

NVR

No = 69.5 use 70 teeth

Chain Drives Design Example Conclusion:

4 Strands No. 80 Chain Ni = 25 Teeth No= 70 Teeth

Chain Drive Design ExampleGuess center distance: 40 Pitches

)40(4

)2570(

2

2570)40(2)(

4

)(

22)(

2

2

2

21212

pitchesL

C

NNNNCpitchesL

L = 128.8 pitches use 130 pitches

Chain Drives Design Example

Actual Center Distance, C

2

22

2

212

2

1212

4

)2570(8

2

2575130

2

2570130

4

1)(

4

)(8

224

1)(

pitchesC

NNNNL

NNLpitchesC

C = 40.6 use 40 Pitches

Chapter 7

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