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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 .

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chapter 7

Chapter 7:

Belt Drives and Chain Drives

CVT

overview why used
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.

slide3

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

Belts vs. Chains

Use When:

High Speed, Low T

High T, Low Speed

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

Dis:

Advs:

Quiet, flexible, cost

Strength, length flexibility

types of belts
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

v belt drive design process
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
Key Equations

Belt speed (no slipping) =

Speed ratio =

Pitch dia’s of sheaves

Pitch dia in inches

rpm

Belt speed ft/min

key equations9
Key Equations
  • Belt length:
  • Center Distance:
    • Where,

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
Key Equations cont…
  • Angle of contact of belt on each sheave

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
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
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 cont13
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 cont14
V-belt Design Example Cont…

3.) Calculate speed ratio

SR = w1/w2

= 1800 rpm/1200 rpm

= 1.5

v belt design example cont15
V-belt Design Example Cont…

4.) Determine sheave sizes

Choose belt speed of 4000 ft/min

(Recall 2500ft./min. < vb < 7000 ft./min)

So…

D1 = 8.488in

D2 = SR * D1 = 1.5 * 8.488

D2 = 12.732in

v belt design example cont16
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

**All look OK, we will try the first one

v belt design example cont17
V-belt Design Example Cont…

6.) Find rated power (use figure 7-11 again)

Rated Power = 21 hp

v belt design example cont18
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 cont19
V-belt Design Example Cont…

7.) Find estimated center distance

Notice – using standard sheave sizes found earlier, not calculated diameters

D2 < C < 3(D2+D1)

12.4 < C < 3 (12.4 + 8.4)

12.4 < C < 62.4

To provide service access will try towards long end, try C = 40”

v belt design example cont20
V-belt Design Example Cont…

8.) Find belt length

v belt design example cont21
V-belt Design Example Cont…

9.) Select standard belt length

Lcalc = 112.765

Choose 112”

v belt design example cont22
V-belt Design Example Cont…

10.) Calculate actual center distance

v belt design example cont23
V-belt Design Example Cont…
  • 11.) Find wrap angle, small sheave
v belt design example cont24
V-belt Design Example Cont…

12.) Determine correction factors

v belt design example cont25
V-belt Design Example Cont…

13.) Calculate corrected power

v belt design example cont26
V-belt Design Example Cont…

14.) Belts needed

Use 4 belts!

v belt design example cont27
V-belt Design Example Cont…

15.) Summary

D1=8.4”

D2=12.4”

Belt Length = 112”

Center Distance = 39.62”

4 Belts Needed

chain drives29
Chain Drives
  • Types of Chains
chain drives30
Chain Drives
  • Roller Chain Construction (Most common Type)
chain design process
Chain Design Process
  • 1.) # of sprocket teeth, N1 (smaller sprocket) > 17 (unless low speed < 100 rpm.)
  • 2.) Speed ratio = n1/n27
  • 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 design example
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 example34
Chain Drives Design Example
  • 1.) Design Power

DP = SF x HP

DP = 1.4 ( Table 7-8) x 100 hp

DP = 140 hp

chain drives design example35
Chain Drives Design Example
  • 2.) Calculate Velocity Ratio

n = speed

N = teeth

VR = 2.78

Heavy Requirement!!

chain drives design example36
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

No = 69.5  use 70 teeth

chain drives design example37
Chain Drives Design Example
  • Conclusion:
    • 4 Strands
    • No. 80 Chain
    • Ni = 25 Teeth
    • No= 70 Teeth
chain drive design example
Chain Drive Design Example

Guess center distance: 40 Pitches

L = 128.8 pitches  use 130 pitches

chain drives design example39
Chain Drives Design Example

Actual Center Distance, C

C = 40.6  use 40 Pitches