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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

Chains

Belts vs. ChainsUse 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:

- 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

- 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 =

Pitch dia’s of sheaves

Pitch dia in inches

rpm

Belt speed ft/min

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…

- 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

- 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…

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 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 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

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 Cont…

8.) Find belt length

V-belt Design Example Cont…

10.) Calculate actual center distance

V-belt Design Example Cont…

- 11.) Find wrap angle, small sheave

V-belt Design Example Cont…

12.) Determine correction factors

V-belt Design Example Cont…

13.) Calculate corrected power

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

- 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/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

- Given:
- Driver: Hydraulic Motor
- Driven: Rock Crusher
- ni = 625 rpm, 100 hp
- no = 225 rpm
- Find:
- Design belt drive

Chain Drives Design Example

- 2.) Calculate Velocity Ratio

n = speed

N = teeth

VR = 2.78

Heavy Requirement!!

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 Example

- Conclusion:
- 4 Strands
- No. 80 Chain
- Ni = 25 Teeth
- No= 70 Teeth

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