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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,

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Presentation on theme: "Chapter 7: Belt Drives and Chain Drives CVT. Overview – why used? 1.) Transfer power (torque) from one location to another. From driver: motor,peddles,"— Presentation transcript:

1 Chapter 7: 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 BeltsChains Use When: Speed: Dis: Advs: High Speed, Low T High T, Low Speed 2500 < V t < 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 Belt speed (no slipping) = Speed ratio = Pitch dia’s of sheaves Belt speed ft/min Pitch dia in inches rpm

9 Key Equations Belt length: Center Distance: Where, Note: usually belt length standard (use standard belt length table 7-2), then calculate C based on fixed L Recommended D 2 < C < 3(D 2 +D 1 )

10 Key Equations cont … Angle of contact of belt on each sheave Note: Select D ’ s and C ’ s so maximum contact ( Ѳ 1 + Ѳ 2 = 180 º ). If less then smaller sheave could slip and will need reduction factor (Table 7-14).

11 V-Belt Design Example Given: 4 cylinder Diesel 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 = w 1 /w 2 = 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. < v b < 7000 ft./min) So … D 1 = 8.488in D 2 = SR * D 1 = 1.5 * D 2 = in

16 5.) Find sheave size (Figure 7-11) Engine (D 1 ) X Standard D Actual n V-belt Design Example Cont … **All look OK, we will try the first one Must find acceptable standard sheave 1, then corresponding acceptable sheave 2

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 D 2 < C < 3(D 2 +D 1 ) 12.4 < C < 3 ( ) 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

21 V-belt Design Example Cont … 9.) Select standard belt length L calc = Choose 112 ”

22 V-belt Design Example Cont … 10.) Calculate actual center distance

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

24 V-belt Design Example Cont … 12.) Determine correction factors

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

26 V-belt Design Example Cont … 14.) Belts needed Use 4 belts!

27 V-belt Design Example Cont … 15.) Summary D 1 =8.4 ” D 2 =12.4 ” Belt Length = 112 ” Center Distance = ” 4 Belts Needed

28 Chain Drives

29 Types of Chains

30 Chain Drives Roller Chain Construction (Most common Type)

31 Chain Design Process 1.) # of sprocket teeth, N 1 (smaller sprocket) > 17 (unless low speed < 100 rpm.) 2.) Speed ratio = n 1 /n ) 30 x Pitch Length < Center Distance < 50 x Pitch Length 4.) Angle of contact of chain on smaller sprocket > 120° 5.) # sprocket teeth, N 2 (longer sprocket) < 120

32 Chain Drives

33 Chain Drives Design Example Given: Driver: Hydraulic Motor Driven: Rock Crusher n i = 625 rpm, 100 hp n o = 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 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 = hp/chain Number of Roller Chain Strands Multiple Strand Factor No = 69.5  use 70 teeth

37 Chain Drives Design Example Conclusion: 4 Strands No. 80 Chain N i = 25 Teeth N o = 70 Teeth

38 Chain Drive Design Example Guess center distance: 40 Pitches L = pitches  use 130 pitches

39 Chain Drives Design Example Actual Center Distance, C C = 40.6  use 40 Pitches


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