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## Chapter 7:

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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. 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. 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:**• 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…**3.) Calculate speed ratio SR = 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) 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…**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 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…**9.) Select standard belt length Lcalc = 112.765 Choose 112”**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…**14.) Belts needed Use 4 belts!**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**• 1.) Design Power DP = SF x HP DP = 1.4 ( Table 7-8) x 100 hp DP = 140 hp**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**Chain Drive Design Example**Guess center distance: 40 Pitches L = 128.8 pitches use 130 pitches**Chain Drives Design Example**Actual Center Distance, C C = 40.6 use 40 Pitches