Belts under Engineering Mechanics, Exercises of Mechanics

Download Belts under Engineering Mechanics and more Exercises Mechanics in PDF only on Docsity! 6 Belts, ropes and chain drives. Pulley drive systems depend on friction to enable the belt to grip the wheel and pull it around with it. To enable this, the belt must be tensioned, even when the wheels are stationary. This is unlike positive chain drive systems where teeth mesh with the chain and slip is not possible so no initial tension is required. Pulley drives are most often used to produce speed reduction between a motor and the machine being driven (e.g. a motor driving an air compressor). Many other applications exist from small rubber band drives in video recorders to large multi belt systems on heavy industrial equipment. On many modem systems, toothed belts are used (e.g. timing belt on a car engine) to prevent the belt slipping. This tutorial is only concerned with smooth belts. 6.1 Material used for Belts a) Leather belts. The most important material for the belt is leather. The best leather belts are made from 1.2 meters to 1.5 meters. b) Cotton or fabric belts. Most of the fabric belts are made by folding canvass or cotton duck to three or more layers (depending upon the thickness desired) and stitching together. c)Rubber belt. The rubber belts are made of layers of fabric impregnated with rubber composition and have a thin layer of rubber on the faces. These belts are very flexible but are quickly destroyed if allowed to come into contact with heat, oil or grease. d)Balata belts. These belts are similar to rubber belts except that balata gum is used in place of rubber. These belts are acid proof and water proof and it is not affected by animal oils or alkalizes. The strength of balata belts is 25 per cent higher than rubber belts. 6.2 Types of Belt Drives • The belt drives are usually classified into the following three groups depending on the amount of load transferred: 1. Light drives. These are used to transmit small powers at belt speeds up to about 10 m/s, as in agricultural machines and small machine tools. Agricultural machines 2. Medium drives. These are used to transmit medium power at belt speeds over 10 m/s but up to 22 m/s, as in machine tools. Machine tools 3. Heavy drives. These are used to transmit large powers at belt speeds above 22 m/s, as in compressors and generators. face of the pulley should be greater or equal to 1.4 b, where b is the width of belt. In case the pulleys cannot be arranged, as shown in Fig. 11.5 (a), or when the reversible motion is desired, then a quarter turn belt drive with guide pulley, as shown in Figure (b), may be used. Crossed or twist belt drive. d)Belt drive with idler pulleys. A belt drive with an idler pulley, as shown in Figure (a), is used with shafts arranged parallel and when an open belt drive cannot be used due to small angle of contact on the smaller pulley. This type of drive is provided to obtain high velocity ratio and when the required belt tension cannot be obtained by other means. When it is desired to transmit motion from one shaft to several shafts, all arranged in parallel, a belt drive with many idler pulleys, as shown in Fig. 11.6 (b), may be employed. e) Compound belt drive. A compound belt drive, as shown in Figure below, is used when power is transmitted from one shaft to another through a number of pulleys. f) Fast and loose pulley drive. A fast and loose pulley drive, as shown in Figure below, is used when the driven or machine shaft is to be started or stopped when ever desired without interfering with the driving shaft. A pulley which is keyed to the machine shaft is called fast pulley and runs at the same speed as that of machine shaft. A loose pulley runs freely over the machine shaft and is incapable of transmitting any power. When the driven shaft is required to be stopped, the belt is pushed on to the loose pulley by means of sliding bar having belt forks. 6.4. Velocity Ratio of Belt Drive. It is the ratio between the velocities of the driver and the follower or driven. It may be expressed, mathematically, as discussed below: Let: d1 = Diameter of the driver, d2 = Diameter of the follower, N1 = Speed of the driver in r.p.m., and N2 = Speed of the follower in r.p.m. ∴ Length of the belt that passes over the driver, in one minute: 𝑣1 = 𝜋 ∙ 𝑛1 ∙ 𝑑1 Similarly, length of the belt that passes over the follower, in one minute: 𝑣2 = 𝜋 ∙ 𝑛2 ∙ 𝑑2 Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the follower in one minute, therefore: 𝜋 ∙ 𝑛1 ∙ 𝑑1 = 𝜇 ∙ 𝑛2 ∙ 𝑑2 ∴ Velocity ratio 𝑛1 𝑛2 = 𝑑2 𝑑1 When the thickness of the belt (t) is considered, then velocity ratio: 𝑛1 𝑛2 = 𝑑2 + 𝑡 𝑑1 + 𝑡 6.5. Velocity Ratio of a Compound Belt Drive Sometimes the power is transmitted from one shaft to another, through a number of pulleys as shown in Figure below. Consider a pulley 1 driving the pulley 2. Since the pulleys 2 and 3 are keyed to the same shaft, therefore the pulley 1 also drives the pulley 3 which, in turn, drives the pulley 4. Let: d1 = Diameter of the pulley 1, n1 = Speed of the pulley 1 in r.p.m., d2, d3, d4, and n2, n3, n4= Corresponding values for pulleys 2, 3 and 4. We know that velocity ratio of pulleys 1 and 2: 𝑛2 𝑛1 = 𝑑1 𝑑2 (1) Similarly, velocity ratio of pulleys 3 and 4, 𝑛4 𝑛3 = 𝑑3 𝑑4 (2) Multiplying equations (1) and (2), 𝑛2 𝑛1 ∙ 𝑛4 𝑛3 = 𝑑1 𝑑2 ∙ 𝑑3 𝑑4 ; (… (∵ 𝑛2 = 𝑛3, being keyed to the same shaft) 𝑛4 𝑛1 = 𝑑1∙𝑑3 𝑑2 ∙ 𝑑4 A little consideration will show, that if there are six pulleys, then 𝑛6 𝑛1 = 𝑑1∙𝑑3∙𝑑5 𝑑2 ∙ 𝑑4 ∙ 𝑑6 sin 𝛼 = 𝑂1𝑀 𝑂1𝑂2 = 𝑂1𝐸 −𝑀𝐸 𝑂1𝑂2 = 𝑟1 − 𝑟2 𝑥 ∴ 𝐴𝑛𝑔𝑙𝑒 𝑜𝑝𝑒𝑛 𝑏𝑒𝑙𝑡 𝑑𝑟𝑖𝑣𝑒 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑜𝑟 𝑙𝑎𝑝 𝜃 = (180 − 2 ∙ 𝛼) ∙ 𝜋 180 𝑟𝑎𝑑 A little consideration will show that when the two pulleys are connected by means of a crossed belt as shown in Figure then the angle of contact or lap (θ) on both the pulleys is same. From Figure: sin 𝛼 = 𝑟1 + 𝑟2 𝑥 ∴ 𝐴𝑛𝑔𝑙𝑒 𝑐𝑟𝑜𝑠𝑠 𝑏𝑒𝑙𝑡 𝑑𝑟𝑖𝑣𝑒 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑜𝑟 𝑙𝑎𝑝 𝜃 = (180 + 2 ∙ 𝛼) ∙ 𝜋 180 𝑟𝑎𝑑 6.8. Belt tension: The tension in the belt will be the same along its whole length, and equal to T Or 𝑇 = 𝑇1 + 𝑇2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑡 And the initial tension is: 𝑇0 = 𝑇1 + 𝑇2 2 6.9. Power transmitted 𝑃𝑊 = 𝑣 ∙ (𝑇1 − 𝑇2); 𝑓𝑜𝑟 𝑣 = 𝜋 ∙ 𝑛 ∙ 𝐷 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑃𝑊 = 𝜋 ∙ 𝑛 ∙ 𝐷 ∙ (𝑇1 − 𝑇2) Where T1 is tension for driving pulley, T2 is tension for driven pulley 6.10 Relation of Driving Tension 6.10.1. For flat Belt. Consider a pulley wheel with a belt passing around it as shown. In order for the belt to produce a torque on the wheel (whether or not it is rotating), there must be tension in both ends. If this was not so, the belt would not be pressed against the wheel and it would slip on the wheel. The belt depends upon friction between it and the wheel in order to grip and produce torque. For a belt to produce torque on the wheel, the force in one end must be greater than the force in the other end. otherwise the net torque is zero. Let T1 be the larger force and T2 the smaller force. θ is the angle of lap. Now consider an elementary length of the belt on the wheel. The force on one end is T and on the other end is slightly larger and is T + dT. The angle made by the small length is dθ. R is the reaction between the belt and the pulley. The forces in the Radial direction (vertically) are: 𝑇 ∙ 𝑠𝑖𝑛 𝑑𝜃 2 + (𝑇 + 𝑑𝑇) ∙ 𝑠𝑖𝑛 𝑑𝜃 2 = 𝑅; 𝑆𝑖𝑛𝑐𝑒 𝑑𝜃 2 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 𝑠𝑜 𝑠𝑖𝑛 𝑑𝜃 2 = 𝑑𝜃 2 Hence 𝑇 ∙ 𝑑𝜃 2 + 𝑇 𝑑𝜃 2 + 𝑑𝑇 ∙ 𝑑𝜃 2 = 𝑅; since 𝑑𝑇 ∙ 𝑑𝜃 2 is very small so it can be neglecting So 𝑇 ∙ 𝑑𝜃 = 𝑅 (1) The forces in the Tangential direction (Horizontally) are: 𝑇 + 𝑑𝑇 ∙ 𝑐𝑜𝑠 𝜃 2 − 𝑇 ∙ 𝑐𝑜𝑠 𝜃 2 = 𝜇 ∙ 𝑅 , 𝑠𝑖𝑛𝑐𝑒 sin 𝜃 2 is very small 𝑠𝑜 𝑐𝑜𝑠 𝜃 2 = 1 Then: 𝑑𝑇 = 𝜇 ∙ 𝑅 (2) Subst. eqn. (1) in eqn. (2) gives: 𝑑𝑇 = 𝜇 ∙ 𝑇 ∙ 𝑑𝜃 𝑜𝑟 𝑑𝑇 𝑇 = 𝜇 ∙ 𝑑𝜃 ∫ 𝑑𝑇 𝑇 𝑇1 𝑇2 = ∫𝜇 ∙ 𝑑𝜃 𝜃 0 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑙𝑛 𝑇1 𝑇2 = 𝜇 ∙ 𝜃 𝑇1 𝑇2 = 𝑒𝜇∙𝜃 (3) Where T1 is tension for driving pulley, T2 is tension for driven pulley, μ is coefficient friction between contact surface belt and pulley, θ is the angle of lap. Example2 A rope is hung over a stationary drum as shown with weights on each. The rope is just on the point of slipping. What is coefficient of frication? Solution Angle of lap θ=1800=π; T1=400 N; T1=100 N. 𝑇1 𝑇2 = 𝑒𝜇∙𝜃 𝑦𝑖𝑒𝑙𝑑𝑠 → 400 100 = 𝑒𝜇∙𝜋 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑙𝑛4 = 𝜋 ∙ 𝜇 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝜇 = 𝑙𝑛4 𝜋 μ=0.441 Example3 Find the power transmitted by a belt running over a pulley of 600 mm diameter at 200 r.p.m. The coefficient of friction between the belt and the pulley is 0.25, angle of lap160° and maximum tension in the belt is 2500 N. Solution. Given: d = 600 mm = 0.6 m; n = 200 r.p.m.; μ = 0.25; the smaller wheel. The wheel is 240 mm diameter and the coefficient of friction is 0.32. The angle of lap is 1650. The wheel speed is 1500 rev/min. Solution Belt velocity: 𝑣 = 𝜋 ∙ 𝑛 ∙ 𝐷 = 𝜋 ∙ 1500 60 ∙ 0.24 = 18.85 𝑚 𝑠⁄ Lap Angle: 𝜃 = 165 180 ∙ 𝜇 = 2.87 𝑟𝑎𝑑 Initial tensions: 𝑇1 = 𝑇2 = 110 𝑁 𝑤ℎ𝑒𝑛 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑠𝑜 𝑇1 + 𝑇2 = 220𝑁 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇2 = 220𝑁 − 𝑇1 Tension in belts: 𝑇1 𝑇2 = 𝑒𝜇∙𝜃 = 𝑒0.32∙2.87 = 2.513 𝑇1 = 2.513 ∙ 𝑇2 = 2.513 ∙ (220𝑁 − 𝑇1) = 552.9 − 2.513 ∙ 𝑇1 𝑇1 + 2.513 ∙ 𝑇1 = 552.9 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇1 = 157.4𝑁; 𝑇2 = 157.4 2.513 = 62.6 𝑁 Power: 𝑃 = 𝑣 ∙ (𝑇1 − 𝑇2) = 18.85 ∙ ( 157.4 − 62.6 ) = 1786.𝑊 Check: 𝑃𝑊 = 𝑣 ∙ 𝑇1 ∙ (1 − 𝑒 −𝜇∙𝜃) = 18.85 ∙ 157.4 ∙ (1 − 0.398) = 1786.13 𝑊 6.12. The effect of centrifugal tension(force): Since the belt continuously runs over the pulleys, therefore, some centrifugal force is caused, whose effect is to increase the tension on both, tight as well as the slack sides. The tension caused by centrifugal force is called centrifugal tension. At lower belt speeds (less than 10 m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10 m/s), its effect is considerable and thus should be taken into account. Consider a belt of mass m per unit length, and an element of this belt can be shown in the figure bellow: Consider a small portion PQ of the belt subtending an angle dθ the centre of the pulley as shown in Figure below: Let: m = Mass of the belt per unit length in kg, v = Linear velocity of the belt in m/s, r = Radius of the pulley over which the belt runs in meters, and TC = Centrifugal tension acting tangentially at P and Q in newtons. We know that length of the belt PQ.: 𝑃𝑄 = 𝑟 ∙ 𝑑𝜃 ∴ Centrifugal force acting on the belt: 𝑚𝑃𝑄 = 𝑚 ∙ 𝑃𝑄 = 𝑚 ∙ 𝑟 ∙ 𝑑𝜃 ∴ Centrifugal force acting on the belt PQ: 𝐹𝑐 = 𝑚 ∙ 𝑟 ∙ 𝑑𝜃 ∙ 𝑣2 𝑟 = 𝑚 ∙ 𝑣2 ∙ 𝑑𝜃 The centrifugal tension TC acting tangentially at P and Q keeps the belt in equilibrium. Now resolving the forces (i.e. centrifugal force and centrifugal tension) horizontally and equating the same, we have: 𝑇𝑐 ∙ sin 𝑑𝜃 2 + 𝑇𝑐 ∙ sin 𝑑𝜃 2 = 𝐹𝑐 = 𝑚 ∙ 𝑣 2 ∙ 𝑑𝜃 Since the angle dθ is very small, therefore, putting: 𝑠𝑖𝑛 𝑑𝜃 2 = 𝑑𝜃 2 in the above expression 2 ∙ 𝑇𝑐 ∙ 𝑑𝜃 2 = 𝑚 ∙ 𝑣2 ∙ 𝑑𝜃 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇𝑐 = 𝑚 ∙ 𝑣 2 Notes: 1. When the centrifugal tension is taken into account, then total tension in the tight side: 𝑇𝑡1 = 𝑇1 + 𝑇𝑐 and total tension in the slack side, 𝑇𝑡2 = 𝑇2 + 𝑇𝑐 2. Power transmitted: 𝑃𝑊 = 𝑣 ∙ (𝑇𝑡1 − 𝑇𝑡2) = 𝑣 ∙ [(𝑇1 + 𝑇𝑐) − (𝑇2 + 𝑇𝑐)] = 𝑣 ∙ (𝑇1 − 𝑇2) Thus, we see that centrifugal tension has no effect on the power transmitted. 3. The ratio of driving tensions may also be written as: 𝑇𝑡1 − 𝑇𝑐 𝑇𝑡2 − 𝑇𝑐 = 𝑒𝜇∙𝜃 (for flat belts. ) 𝑇𝑡1 − 𝑇𝑐 𝑇𝑡2 − 𝑇𝑐 = 𝑒 𝜇∙𝜃 sin𝛽 (for VEE − Belts. ) 6.13 Condition for the Transmission of Maximum Power We know that power transmitted by a belt: 𝑃 = 𝑣 ∙ (𝑇1 − 𝑇2) (I) Where T1 = Tension in the tight side of the belt in newtons, T2 = Tension in the slack side of the belt in newtons, and v = Velocity of the belt in m/s. we have also seen that the ratio of driving tensions is: 𝑇1 𝑇2 = 𝑒𝜇∙𝜃 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇2 = 𝑇1 𝑒𝜇∙𝜃 (II) Substituting the value of T2 in equation (&) 𝑃 = 𝑣 ∙ (𝑇1 − 𝑇1 𝑒𝜇∙𝜃 ) = 𝑣 ∙ 𝑇1 ∙ (1 − 1 𝑒𝜇∙𝜃 ) = 𝑣 ∙ 𝑇1 ∙ 𝐶 (III) Where: 𝐶 =∙ (1 − 1 𝑒𝜇∙𝜃 ) We know that: 𝑇1 = 𝑇 − 𝑇𝑐 Where: T = Maximum tension to which the belt can be subjected in newtons, and TC = Centrifugal tension in newtons. Substituting the value of T1 in equation (III), 𝑃 = 𝑣 ∙ 𝐶 ∙ (𝑇 − 𝑇𝑐) = 𝑣 ∙ 𝐶 ∙ (𝑇 − 𝑚 ∙ 𝑣 2), (𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑇𝑐 = 𝑚 ∙ 𝑣 2) 𝑃 = 𝐶 ∙ (𝑇 ∙ 𝑣 − 𝑚 ∙ 𝑣3) For maximum power, differentiate the above expression with respect to v and equate to zero, 𝑑𝑃 𝑑𝑣 = 0 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑑 𝑑𝑣 𝐶 ∙ (𝑇 ∙ 𝑣 − 𝑚 ∙ 𝑣3) = 0 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇 − 3 ∙ 𝑚 ∙ 𝑣2 = 0 𝑇 = 3 ∙ 𝑇𝑐 (IV) It shows that when the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension. Notes: 1. We know that: 𝑇1 = 𝑇 − 𝑇𝑐 , 𝑎𝑛𝑑 𝑓𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑤𝑒𝑟𝑇𝑐 = 𝑇 3 ∴ 𝑇1 = 𝑇 − 𝑇 3 = 2 ∙ 𝑇 3 2. From equation (IV), the velocity of the belt for the maximum power, 𝑣 = √ 𝑇 3 ∙ 𝑚 Example5 A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits 6 kW through a belt. The belt is 100 mm wide and 10 mm thick. The distance between the shafts is 4m. The smaller pulley is 0.5 m in diameter. Calculate the 𝑇𝑐 = 𝑚 ∙ 𝑣 2 = 10 ∙ 𝑏 ∙ 15.712 = 157.1 ∙ 𝑏 = 2468 ∙ 𝑏 and maximum tension in the belt 𝑇 = 𝜎. 𝑏. 𝑡 = 1.5 × 106 ∙ 𝑏 ∙ 0.01 = 15 000 ∙ 𝑏 𝑁 We know that: 𝑇 = 𝑇 1 + 𝑇𝐶 𝑜𝑟 15000 ∙ 𝑏 = 824.6 + 2468 𝑏 𝑦𝑖𝑒𝑙𝑑𝑠 → 15 000 ∙ 𝑏 – 2468 ∙ 𝑏 = 824.6 𝑦𝑖𝑒𝑙𝑑𝑠 → 12 532 ∙ 𝑏 = 824.6 𝑏 = 824.6 12 532 = 0.0658 𝑚 = 65.8 𝑚𝑚 Example 7 A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by V-belts from an electric motor running at 750 r.p.m. The diameter of the pulley on the compressor shaft must not be greater than 1 metre while the centre distance between the pulleys is limited to 1.75 metre. The belt speed should not exceed 1600 m/min. Determine the number of V-belts required to transmit the power if each belt has a crosssectional area of 375 mm2, density 1000 kg/m3 and an allowable tensile stress of 2.5 MPa. The groove angle of the pulley is 35°. The coefficient of friction between the belt and the pulley is 0.25. Calculate also the length required of each belt. Solution. Given : P = 90 kW ; n2 = 250 r.p.m. ; n1 = 750 r.p.m. ; d2 = 1 m ; x = 1.75 m ; v = 1600 m/min = 26.67 m/s ; a = 375 mm2 = 375 × 10–6 m2 ; ρ = 1000 kg/m3 ; σ = 2.5 MPa = 2.5 × 106 N/m2 ; 2 β = 35° or β = 17.5° ; μ = 0.25 First of all, let us find the diameter of pulley on the motor shaft (d1). We know that: 𝑛1 𝑛2 = 𝑑2 𝑑1 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑑1 = 𝑛2 ∙ 𝑑2 𝑛1 = 250 ∙ 1 750 = 0.33𝑚 We know that the mass of the belt per metre length: 𝑚 = 𝐴𝑟𝑒𝑎 ∙ 𝑙𝑒𝑛𝑔𝑡ℎ ∙ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 375 × 10–6 ∙ 1 ∙ 1000 = 0.375 𝑘𝑔 ∴ Centrifugal tension: 𝑇𝑐 = 𝑚 ∙ 𝑣 2 = 0.375 ∙ 26.67 2 = 267 𝑁 and maximum tension in the belt: 𝑇 = 𝜎 ∙ 𝐴 = 2.5 ∙ 106 = 937.5 𝑁 ∴ Tension in the tight side of the belt: 𝑇1 = 𝑇 − 𝑇𝐶 = 937.5 − 267 = 670.5 𝑁 For an open belt drive, as shown in Figure: sin 𝛼 = 𝑟1 − 𝑟2 𝑥 = 𝑑1 − 𝑑2 2 ∙ 𝑥 = 1 − 0.33 2 ∙ 1.75 = 0.1914 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝛼 = 11° and angle of lap on smaller pulley (i.e. pulley on motor shaft): ∴ 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑜𝑟 𝑙𝑎𝑝 𝜃 = (180 − 2 ∙ 𝛼) ∙= (180 − 2 ∙ 11) ∙= 158° 𝜃 = 158 ∙ 𝜋 180 = 2.76𝑟𝑎𝑑 We know that: 𝑇1 𝑇2 = 𝑒𝜇∙𝜃 = 𝑒0.25∙2.7 = 9.954 670.5 𝑇2 = 9.954 𝑦𝑖𝑒𝑙𝑑𝑠 → 𝑇2 = 670.5 9.954 = 67.36𝑁 Number of V-belts We know that power transmitted per belt: 𝑃 = 𝑣 ∙ (𝑇1 − 𝑇2) = 26.67 ∙ (670.5 − 67.36) = 16 086 𝑊 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑉 − 𝑏𝑒𝑙𝑡𝑠 = 𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑝𝑒𝑟 𝑏𝑒𝑙𝑡 = 90000 16 086 = 5.6 ≈ 6 Length of each belt We know that length of belt for an open belt drive: 𝐿 = 𝜋 ∙ (𝑟1 + 𝑟2) + 2 ∙ 𝑥 + (𝑟1 − 𝑟2) 2 𝑥 = 𝜋 2 ∙ (𝑑1 + 𝑑2) + 2 ∙ 𝑥 + (𝑑1 − 𝑑2) 2 4 ∙ 𝑥 𝐿 = 𝜋 2 ∙ (1 + 0.33) + 2 ∙ 1.75 + (1 − 0.33)2 4 ∙ 1.75 = 𝐿 = 2.1 + 3.5 + 0.064 = 5.664 𝑚