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Chapter 7:

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  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 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

  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 Pitch dia in inches rpm Belt speed ft/min

  9. 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

  10. 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).

  11. 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

  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 Choose 5V Speed = 1800 rpm Design Power = 88 hp

  14. V-belt Design Example Cont… 3.) Calculate speed ratio SR = w1/w2 = 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. < vb < 7000 ft./min) So… D1 = 8.488in D2 = SR * D1 = 1.5 * 8.488 D2 = 12.732in

  16. 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

  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 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”

  20. V-belt Design Example Cont… 8.) Find belt length

  21. V-belt Design Example Cont… 9.) Select standard belt length Lcalc = 112.765 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 D1=8.4” D2=12.4” Belt Length = 112” Center Distance = 39.62” 4 Belts Needed

  28. Chain Drives

  29. Chain Drives • Types of Chains

  30. Chain Drives • Roller Chain Construction (Most common Type)

  31. 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

  32. Chain Drives

  33. Chain Drives Design Example • Given: • Driver: Hydraulic Motor • Driven: Rock Crusher • ni = 625 rpm, 100 hp • no = 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 n = speed N = teeth VR = 2.78 Heavy Requirement!!

  36. 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

  37. Chain Drives Design Example • Conclusion: • 4 Strands • No. 80 Chain • Ni = 25 Teeth • No= 70 Teeth

  38. Chain Drive Design Example Guess center distance: 40 Pitches L = 128.8 pitches  use 130 pitches

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