Design of belt Rope and Chain drives

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 Chapter 11 : Belt, Rope and Chain Drives 325 11 Features (Main) 1. Introduction. Belt, Rope 4. Types of Belts. 5. Material used for Belts. 6. Types of Flat Belt Drives. 7. Velocity Ratio of Belt Drive. 11. Length of an Open Belt Drive. and Chain 13. Power Transmitted by a Belt. 14. Ratio of Driving Tensions for Flat Belt Drive. 16. Centrifugal Tension. Drives 17. Maximum Tension in the Belt. 19. Initial Tension in the Belt. 20. V-belt Drive. 11.1. Introduction 22. Ratio of Driving Tensions for V-belt. The belts or ropes are used to transmit power from 23. Rope Drive. one shaft to another by means of pulleys which rotate at the 24. Fibre Ropes. same speed or at different speeds. The amount of power trans- 26. Sheave for Fibre Ropes. mitted depends upon the following factors : 27. Wire Ropes. 28. Ratio of Driving Tensions for 1. The velocity of the belt. Rope Drive. 2. The tension under which the belt is placed on the 29. Chain Drives. pulleys. 30. Advantages and Disadvantages of Chain 3. The arc of contact between the belt and the smaller Drive Over Belt or Rope pulley. Drive. 4. The conditions under which the belt is used. 31. Terms Used in Chain Drive. 32. Relation Between Pitch and It may be noted that Pitch Circle Diameter. (a) The shafts should be properly in line to insure 33. Relation Between Chain uniform tension across the belt section. Speed and Angular Velocity of Sprocket. (b) The pulleys should not be too close together, in 34. Kinematic of Chain Drive. order that the arc of contact on the smaller pul- 35. Classification of Chains. ley may be as large as possible. 36. Hoisting and Hauling Chains. 37. Conveyor Chains. (c) The pulleys should not be so far apart as to cause 38. Power Transmitting Chains. the belt to weigh heavily on the shafts, thus in- 39. Length of Chains. creasing the friction load on the bearings. 325 326 Theory of Machines (d) A long belt tends to swing from side to side, causing the belt to run out of the pulleys, which in turn develops crooked spots in the belt. (e) The tight side of the belt should be at the bottom, so that whatever sag is present on the loose side will increase the arc of contact at the pulleys. ( f ) In order to obtain good results with flat belts, the maximum distance between the shafts should not exceed 10 metres and the minimum should not be less than 3.5 times the diameter of the larger pulley. 11.2. Selection of a Belt Drive Following are the various important factors upon which the selection of a belt drive depends: 1. Speed of the driving and driven shafts, 2. Speed reduction ratio, 3. Power to be transmitted, 4. Centre distance between the shafts, 5. Positive drive requirements, 6. Shafts layout, 7. Space available, and 8. Service conditions. 11.3. Types of Belt Drives The belt drives are usually classified into the following three groups : 1. Light drives. These are used to transmit small powers at belt speeds upto about 10 m/s, as in agricultural machines and small machine tools. 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. 3. Heavy drives. These are used to transmit large powers at belt speeds above 22 m/s, as in compressors and generators. 11.4. Types of Belts (a) Flat belt. (b) V-belt. (c) Circular belt. Fig. 11.1. Types of belts. Though there are many types of belts used these days, yet the following are important from the subject point of view : 1. Flat belt. The flat belt, as shown in Fig. 11.1 (a), is mostly used in the factories and workshops, where a moderate amount of power is to be transmitted, from one pulley to another when the two pulleys are not more than 8 metres apart. 2. V-belt. The V-belt, as shown in Fig. 11.1 (b), is mostly used in the factories and work- shops, where a moderate amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each other. 3. Circular belt or rope. The circular belt or rope, as shown in Fig. 11.1 (c), is mostly used in the factories and workshops, where a great amount of power is to be transmitted, from one pulley to another, when the two pulleys are more than 8 meters apart. Chapter 11 : Belt, Rope and Chain Drives 327 If a huge amount of power is to be transmitted, then a single belt may not be sufficient. In such a case, wide pulleys (for V-belts or circular belts) with a number of grooves are used. Then a belt in each groove is provided to transmit the required amount of power from one pulley to another. 11.5. Material used for Belts The material used for belts and ropes must be strong, flexible, and durable. It must have a high coefficient of friction. The belts, according to the material used, are classified as follows : 1. Leather belts. The most important material for the belt is leather. The best leather belts are made from 1.2 metres to 1.5 metres long strips cut from either side of the back bone of the top grade steer hides. The hair side of the leather is smoother and harder than the flesh side, but the flesh side is stronger. The fibres on the hair side are perpendicular to the surface, while those on the flesh side are interwoven and parallel to the surface. Therefore for these reasons, the hair side of a belt should be in contact with the pulley surface, as shown in Fig. 11.2. This gives a more intimate contact between the belt and the pulley and places the greatest tensile strength of the belt section on the outside, where the tension is maximum as the belt passes over the pulley. (a) Single layer belt. (b) Double layer belt. Fig. 11.2. Leather belts. The leather may be either oak-tanned or mineral salt tanned e.g. chrome tanned. In order to increase the thickness of belt, the strips are cemented together. The belts are specified according to the number of layers e.g. single, double or triple ply and according to the thickness of hides used e.g. light, medium or heavy. The leather belts must be periodically cleaned and dressed or treated with a compound or dressing containing neats foot or other suitable oils so that the belt will remain soft and flexible. 2. 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. These belts are woven also into a strip of the desired width and thickness. They are impregnated with some filler like linseed oil in order to make the belts water proof and to prevent injury to the fibres. The cotton belts are cheaper and suitable in warm climates, in damp atmospheres and in exposed positions. Since the cotton belts require little attention, therefore these belts are mostly used in farm machinery, belt conveyor etc. 3. Rubber belt. The rubber belts are made of layers of fabric impregnated with rubber com- position 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. One of the principal advantage of these belts is that they may be easily made endless. These belts are found suitable for saw mills, paper mills where they are exposed to moisture. 4. 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 effected by animal oils or alkalies. The balata belts should not be at temperatures above 40° C because at this temperature the balata begins to soften and becomes sticky. The strength of balata belts is 25 per cent higher than rubber belts. 328 Theory of Machines 11.6. Types of Flat Belt Drives The power from one pulley to another may be transmitted by any of the following types of belt drives: 1. Open belt drive. The open belt drive, as shown in Fig. 11.3, is used with shafts arranged parallel and rotating in the same direction. In this case, the driver A pulls the belt from one side (i.e. lower side RQ) and delivers it to the other side (i.e. upper side LM). Thus the tension in the lower side belt will be more than that in the upper side belt. The lower side belt (because of more tension) is known as tight side whereas the upper side belt (because of less tension) is known as slack side, as shown in Fig. 11.3. Fig. 11.3. Open belt drive. 2. Crossed or twist belt drive. The crossed or twist belt drive, as shown in Fig. 11.4, is used with shafts arranged parallel and rotating in the opposite directions. Fig. 11.4. Crossed or twist belt drive. In this case, the driver pulls the belt from one side (i.e. RQ) and delivers it to the other side (i.e. LM). Thus the tension in the belt RQ will be more than that in the belt LM. The belt RQ (because of more tension) is known as tight side, whereas the belt LM (because of less tension) is known as slack side, as shown in Fig. 11.4. Chapter 11 : Belt, Rope and Chain Drives 329 A little consideration will show that at a point where the belt crosses, it rubs against each other and there will be excessive wear and tear. In order to avoid this, the shafts should be placed at a maximum distance of 20 b, where b is the width of belt and the speed of the belt should be less than 15 m/s. 3. Quarter turn belt drive. The quarter turn belt drive also known as right angle belt drive, as shown in Fig. 11.5 (a), is used with shafts arranged at right angles and rotating in one definite direc- tion. In order to prevent the belt from leaving the pulley, the width of the 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 Fig. 11.5 (b), may be used. (a) Quarter turn belt drive. (b) Quarter turn belt drive with guide pulley. Fig. 11.5 4. Belt drive with idler pulleys. A belt drive with an idler pulley, as shown in Fig. 11.6 (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. (a) Belt drive with single idler pulley. (b) Belt drive with many idler pulleys. Fig. 11.6 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. 330 Theory of Machines 5. Compound belt drive. A compound belt drive, as shown in Fig. 11.7, is used when power is transmitted from one shaft to another through a number of pulleys. Fig. 11.7. Compound belt brive. 6. Stepped or cone pulley drive. A stepped or cone pulley drive, as shown in Fig. 11.8, is used for changing the speed of the driven shaft while the main or driving shaft runs at constant speed. This is accomplished by shifting the belt from one part of the steps to the other. 7. Fast and loose pulley drive. A fast and loose pulley drive, as shown in Fig. 11.9, 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. Fig. 11.9. Fast and loose pulley drive. Fig. 11.8. Stepped or cone pulley drive. 11.7. 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 d = Diameter of the driver, 1 d = Diameter of the follower, 2 Chapter 11 : Belt, Rope and Chain Drives 331 N = Speed of the driver in r.p.m., and 1 N = Speed of the follower in r.p.m. 2 ∴ Length of the belt that passes over the driver, in one minute = π d .N 1 1 Similarly, length of the belt that passes over the follower, in one minute = π d . N 2 2 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 d . N = π d . N 1 1 2 2 Nd 21 ∴ Velocity ratio, = Nd 12 When the thickness of the belt (t) is considered, then velocity ratio, Nd +t 21 = Nd +t 12 Note: The velocity ratio of a belt drive may also be obtained as discussed below : We know that peripheral velocity of the belt on the driving pulley, πdN . 11 v = m/s 1 60 and peripheral velocity of the belt on the driven or follower pulley, πdN . 22 v = m/s 2 60 When there is no slip, then v = v . 1 2 Nd ππ dN.. d N 21 11 2 2 = = ∴ or Nd 60 60 12 11.8. 11.8. 11.8. 11.8. 11.8. V V V V Velocity Ra elocity Ra elocity Ra elocity Ra elocity Ratio of a Compound Belt Dr tio of a Compound Belt Dr tio of a Compound Belt Dr tio of a Compound Belt Dr tio of a Compound Belt Driv iv iv iv ive e e e e Sometimes the power is transmitted from one shaft to another, through a number of pulleys as shown in Fig. 11.7. 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 d = Diameter of the pulley 1, 1 N = Speed of the pulley 1 in r.p.m., 1 d , d , d , and N , N , N = Corresponding values for pulleys 2, 3 and 4. 2 3 4 2 3 4 We know that velocity ratio of pulleys 1 and 2, Nd 21 = ...(i) Nd 12 Similarly, velocity ratio of pulleys 3 and 4, N d 4 3 = ...(ii) Nd 34 Multiplying equations (i) and (ii), d NN d 24 1 3 ×= × NN dd 13 2 4 332 Theory of Machines dd × N 4 13 = or ...(∵ N = N , being keyed to the same shaft) 2 3 Nd ×d 12 4 A little consideration will show, that if there are six pulleys, then Nd××d d 61 3 5 = Nd××d d 12 4 6 Speed of last driven Product of diameters of drivers or = Speed of first driver Product of diameters of drivens 11.9. Slip of Belt In the previous articles, we have discussed the motion of belts and shafts assuming a firm frictional grip between the belts and the shafts. But sometimes, the frictional grip becomes insufficient. This may cause some forward motion of the driver without carrying the belt with it. This may also cause some forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt and is generally expressed as a percentage. The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping of the belt is a common phenomenon, thus the belt should never be used where a definite velocity ratio is of importance (as in the case of hour, minute and second arms in a watch). Let s % = Slip between the 1 driver and the belt, and s % = Slip between the belt and the follower. 2 ∴ Velocity of the belt passing over the driver per second ππ dN.. d N sπd.N s 11 11 1 1 1 1 v=×–1=–  ...(i) 60 60 100 60 100  and velocity of the belt passing over the follower per second, πdN . s s  22 2 2 =× vv–1=v–  60 100 100  Substituting the value of v from equation (i), ππ dN d N s s  22 1 1 1 2 = 1– 1–  60 60 100 100  Nd s s   ss × 21 1 2 12 = 1– – ... Neglecting   Nd 100 100 100 × 100   12 ds +s d s 11 2 1 == 1– 1–   dd 100 100   22 ... (where s = s + s , i.e. total percentage of slip) 1 2 If thickness of the belt (t) is considered, then Nd +ts 21 = 1–  Nd +t 100  12 Chapter 11 : Belt, Rope and Chain Drives 333 Example 11.1. An engine, running at 150 r.p.m., drives a line shaft by means of a belt. The engine pulley is 750 mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameter pulley on the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Find the speed of the dynamo shaft, when 1. there is no slip, and 2. there is a slip of 2% at each drive. Solution. Given : N = 150 r.p.m. ; d = 750 mm ; d = 450 mm ; d = 900 mm ; d = 150 mm 1 1 2 3 4 The arrangement of belt drive is shown in Fig. 11.10. Let N = Speed of the dynamo shaft . 4 Fig. 11.10 1. When there is no slip N dd × N 750 × 900 4 13 4 We know that = or == 10 Nd ×d 150 450 × 150 12 4 ∴ N = 150 × 10 = 1500 r.p.m. Ans. 4 2. When there is a slip of 2% at each drive Ns dd × s  41 13 2 = 1– 1– We know that  Nd ×d 100 100  12 4 N 750 × 900 2 2   4 == 1– 1– 9.6   150 450 × 150 100 100   ∴ N = 150 × 9.6 = 1440 r.p.m. Ans. 4 11.10. Creep of Belt When the belt passes from the slack side to the tight side, a certain portion of the belt extends and it contracts again when the belt passes from the tight side to slack side. Due to these changes of length, there is a relative motion between the belt and the pulley surfaces. This relative motion is termed as creep. The total effect of creep is to reduce slightly the speed of the driven pulley or follower. Considering creep, the velocity ratio is given by E+σ Nd 21 2 =× Nd E+σ 12 1 where σ and σ = Stress in the belt on the tight and slack side respectively, and 1 2 E = Young’s modulus for the material of the belt. 334 Theory of Machines Example 11.2. The power is transmitted from a pulley 1 m diameter running at 200 r.p.m. to a pulley 2.25 m diameter by means of a belt. Find the speed lost by the driven pulley as a result of creep, if the stress on the tight and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. The Young’s modulus for the material of the belt is 100 MPa. 6 2 Solution. Given : d = 1 m ; N = 200 r.p.m. ; d = 2.25 m ; σ = 1.4 MPa = 1.4 × 10 N/m ; 1 1 2 1 6 2 6 2 σ = 0.5 MPa = 0.5 × 10 N/m ; E = 100 MPa = 100 × 10 N/m 2 Let N = Speed of the driven pulley. 2 Neglecting creep, we know that d 1 Nd 1 21 or NN=× = 200× = 88.9 r.p.m. = 21 d 2.25 Nd 12 2 Considering creep, we know that E+σ Nd 2 21 =× Nd E+σ 12 1 66 100×+ 10 0.5× 10 1 or N=× 200 × = 88.7 r.p.m. 2 66 2.25 100×+ 10 1.4× 10 ∴ Speed lost by driven pulley due to creep = 88.9 – 88.7 = 0.2 r.p.m. Ans. 11.11. Length of an Open Belt Drive Fig. 11.11. Length of an open belt drive. We have already discussed in Art. 11.6 that in an open belt drive, both the pulleys rotate in the same direction as shown in Fig. 11.11. Let r and r = Radii of the larger and smaller pulleys, 1 2 x = Distance between the centres of two pulleys (i.e. O O ), and 1 2 L = Total length of the belt. Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H as shown in Fig. 11.11. Through O , draw O M parallel to FE. 2 2 From the geometry of the figure, we find that O M will be perpendicular to O E. 2 1 Let the angle MO O = α radians. 2 1 Chapter 11 : Belt, Rope and Chain Drives 335 We know that the length of the belt, L =Arc GJE + EF + Arc FKH + HG = 2 (Arc JE + EF + Arc FK) ...(i) From the geometry of the figure, we find that OM O E–– EM r r 11 12 sinα= = = OO OO x 12 12 Since α is very small, therefore putting rr – 12 sin α = α (in radians) = ...(ii) x π  ∴ Arc JE=+ r α ...(iii) 1 2  π  Similarly Arc FK=α r – ...(iv) 2 2  22 2 2 and EF== MO () O O –(O M)= x–(r– r) 212 1 12 2 –  rr 12 = x 1–  x  Expanding this equation by binomial theorem, 2 2  1 rr–(rr–)  12 1 2 EF x 1– .... x – =+=  ...(v) 22 xx    Substituting the values of arc JE from equation (iii), arc FK from equation (iv) and EF from equation (v) in equation (i), we get 2  ππ (–rr)   12 Lr=+2–α+x +r–α  12   22 x 2     2  ππ (–rr) 12 =×2. rr+ α+x– +r×–r.α  11 2 2 22 x 2   2  π (–rr) 12 =+ 2(rr)+α(rr– )+x–  12 12 22 x   2 (–rr) 12 =π(rr + ) +2( αrr – ) +2–x 12 12 x rr – 12 Substituting the value of α= from equation (ii), x 2 (–rr) (rr – ) 12 1 2 Lr =π() +r +2× ×(r–r) +2x– 12 12 xx 22 2(rr – ) (rr – ) 12 12 =π() rr + + +2x– 12 xx 2 (–rr) 12 =π() rr + +2x + ...(In terms of pulley radii) 12 x 2 π (–dd) 12 =+() dd+2x+ ...(In terms of pulley diameters) 12 24x 336 Theory of Machines 11.12. Length of a Cross Belt Drive We have already discussed in Art. 11.6 that in a cross belt drive, both the pulleys rotate in opposite directions as shown in Fig. 11.12. Fig. 11.12. Length of a cross belt drive. Let r and r = Radii of the larger and smaller pulleys, 1 2 x = Distance between the centres of two pulleys (i.e. O O ), and 1 2 L = Total length of the belt. Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H, as shown in Fig. 11.12. Through O , draw O M parallel to FE. 2 2 From the geometry of the figure, we find that O M will be perpendicular to O E. 2 1 Let the angle MO O = α radians. 2 1 We know that the length of the belt, L =Arc GJE + EF + Arc FKH + HG = 2 (Arc JE + EF + Arc FK) ...(i) From the geometry of the figure, we find that OM OE++ EM r r 11 12 sinα= = = OO O O x 12 12 Since α is very small, therefore putting rr + 12 = sin α = α (in radians) ...(ii) x π  ∴ Arc =+α ...(iii) JE r 1 2   π Similarly Arc FK=+ r α ...(iv) 2 2  22 2 2 and EF== MO () O O –(O M)= x–(r+ r) 212 1 12 2  rr + 12 = x 1–  x  Chapter 11 : Belt, Rope and Chain Drives 337 Expanding this equation by binomial theorem, 2 2  rr++() rr 1 12 12 EF=+ x 1– ...= x –  ...(v) 22xx    Substituting the values of arc JE from equation (iii), arc FK from equation (iv) and EF from equation (v) in equation (i), we get 2  ππ () rr +   12 Lr=+2–α+x +r +α  12   22x 2     2  ππ () rr + 12 =×2. rr+ α+x– +r× +r.α  11 2 2 22x 2   2  π () rr + 12 =+ 2(rr)+α(r+r)+x–  12 1 2 22 x   2 () rr + 12 =π() rr + +2α(r +r) +2x– 12 12 x rr + 12 Substituting the value of α= from equation (ii), x 2 2(rr++ ) (rr ) 12 12 Lr =π() +r + ×() r +r +2x– 12 12 xx 22 2(rr++ ) (rr ) 12 12 =π() rr + + +2x– 12 xx 2 () rr + 12 =π() rr + +2x + ...(In terms of pulley radii) 12 x 2 π () dd + 12 =+() dd+2x+ ...(In terms of pulley diameters) 12 24x It may be noted that the above expression is a function of (r + r ). It is thus obvious that if sum of the radii of the two pulleys 1 2 be constant, then length of the belt required will also remain con- stant, provided the distance between centres of the pulleys remain unchanged. Example 11.3. A shaft which rotates at a constant speed of 160 r.p.m. is connected by belting to a parallel shaft 720 mm apart, which has to run at 60, 80 and 100 r.p.m. The smallest pulley on the driving shaft is 40 mm in radius. Determine the remaining radii of the two stepped pulleys for 1. a crossed belt, and 2. an open belt. Neglect belt thickness and slip. Solution. Given : N = N = N = 160 r.p.m. ; x = 720 mm ; 1 3 5 N = 60 r.p.m.; N = 80 r.p.m.; N = 100 r.p.m. ; r = 40 mm 2 4 6 1 Fig. 11.13. 338 Theory of Machines Let r , r , r , r and r be the radii of the pulleys 2, 3, 4, 5, and 6 respectively, as shown in Fig. 2 3 4 5 6 11.13. 1. For a crossed belt We know that for pulleys 1 and 2, Nr 21 = Nr 12 N 160 1 or rr=× = 40× = 106.7 mm Ans. 21 N 60 2 and for pulleys 3 and 4, N r N 160 4 3 3 = or rr=× =×r = 2r 43 3 3 Nr N 80 34 4 We know that for a crossed belt drive, r + r = r + r = r + r = 40 + 106.7 = 146.7 mm ...(i) 1 2 3 4 5 6 ∴ r + 2 r = 146.7 or r = 146.7/3 = 48.9 mm Ans. 3 3 3 and r = 2 r = 2 × 48.9 = 97.8 mm Ans. 4 3 Now for pulleys 5 and 6, N 160 Nr 5 65 or rr=× =r× = 1.6r = 65 5 5 N 100 Nr 6 56 From equation (i), r + 1.6 r = 146.7 or r = 146.7/2.6 = 56.4 mm Ans. 5 5 5 and r =1.6 r = 1.6 × 56.4 = 90.2 mm Ans. 6 5 2. For an open belt We know that for pulleys 1 and 2, Nr N 160 21 1 = or rr=× = 40× = 106.7 mm Ans. 21 Nr N 60 2 12 and for pulleys 3 and 4, N 160 N r 3 4 3 or rr=× =r× = 2r = 43 3 3 N 80 Nr 4 34 We know that length of belt for an open belt drive, 2 (–rr) 21 Lr =π() +r + +2x 12 x 2 (106.7 – 40) =π(40 + 106.7) + + 2× 720 = 1907 mm 720 Since the length of the belt in an open belt drive is constant, therefore for pulleys 3 and 4, length of the belt (L), 2 (–rr) 43 1907 =π(rr + ) + + 2x 34 x Chapter 11 : Belt, Rope and Chain Drives 339 2 (2rr – ) 33 =π(2 rr + ) + +2×720 33 720 2 = 9.426 r + 0.0014 (r ) + 1440 3 3 2 or 0.0014 (r ) + 9.426 r – 467 = 0 3 3 2 – 9.426±+ (9.426) 4× 0.0014× 467 r = ∴ 3 2 × 0.0014 – 9.426 ±9.564 = = 49.3 mm Ans. ...(Taking +ve sign) 0.0028 and r = 2 r = 2 × 49.3 = 98.6 mm Ans. 4 3 Now for pulleys 5 and 6, Nr 65 = or Nr 56 N 160 5 rr =× = ×r= 1.6r 65 5 5 100 N 6 and length of the belt (L), 2 (–rr) 65 1907 ( ) 2 =πrr + + + x 56 x 2 (1.6rr – ) 55 (1.6) 2 720 =πrr + + + × 55 720 2 = 8.17 r + 0.0005 (r ) + 1440 5 5 2 or 0.0005 (r ) + 8.17 r – 467 = 0 5 5 2 –8.17±+ (8.17) 4× 0.0005× 467 r = ∴ 5 2 × 0.0005 Milling machine is used for dressing surfaces by rotary cutters. –8.17 ±8.23 Note : This picture is given as additional = =60 mm Ans. information and is not a direct example of the 0.001 current chapter. ...(Taking +ve sign) and r =1.6 r = 1.6 × 60 = 96 mm Ans. 6 5 11.13. Power Transmitted by a Belt Fig. 11.14 shows the driving pulley (or driver) A and the driven pulley (or follower) B. We have already discussed that the driving pulley pulls the belt from one side and delivers the same to the other side. It is thus obvious that the tension on the former side (i.e. tight side) will be greater than the latter side (i.e. slack side) as shown in Fig. 11.14. Let T and T = Tensions in the tight and slack side of the belt respectively in 1 2 newtons, 340 Theory of Machines r and r = Radii of the driver and follower respectively, and 1 2 v = Velocity of the belt in m/s. Fig. 11.14. Power transmitted by a belt. The effective turning (driving) force at the circumference of the follower is the difference between the two tensions (i.e. T – T ). 1 2 ∴ Work done per second = (T – T ) v N-m/s 1 2 and power transmitted, P = (T – T ) v W ...(∵ 1 N-m/s = 1 W) 1 2 A little consideration will show that the torque exerted on the driving pulley is (T – T ) r . 1 2 1 Similarly, the torque exerted on the driven pulley i.e. follower is (T – T ) r . 1 2 2 11.14. Ratio of Driving Tensions For Flat Belt Drive Consider a driven pulley rotating in the clockwise direction as shown in Fig. 11.15. Fig. 11.15. Ratio of driving tensions for flat belt. Let T = Tension in the belt on the tight side, 1 T = Tension in the belt on the slack side, and 2 θ = Angle of contact in radians (i.e. angle subtended by the arc AB, along which the belt touches the pulley at the centre). Now consider a small portion of the belt PQ, subtending an angle δθ at the centre of the pulley as shown in Fig. 11.15. The belt PQ is in equilibrium under the following forces : 1. Tension T in the belt at P, 2. Tension (T + δ T) in the belt at Q, 3. Normal reaction R , and N 4. Frictional force, F = µ × R , where µ is the coefficient of friction between the belt and N pulley. Chapter 11 : Belt, Rope and Chain Drives 341 Resolving all the forces horizontally and equating the same, δθ δθ RT=+()δTsin +Tsin ...(i) N 22 Since the angle δθ is very small, therefore putting sin δ θ / 2 = δθ / 2 in equation (i), δθ δθTT.. δθ δ δθT.δθ RT=+()δT +T× = + + =T.δθ ...(ii) N 22 2 2 2 δδ T .θ  ... Neglecting  2  Now resolving the forces vertically, we have δθ δθ µ×RT = ( + δT) cos –T cos ...(iii) N 22 Since the angle δ θ is very small, therefore putting cos δ θ / 2 = 1 in equation (iii), δT µ × R = T + δT – T = δT or ...(iv) R = N N µ Equating the values of R from equations (ii) and (iv), N δT δT T.δθ = or =µ δ.θ µ T Integrating both sides between the limits T and T and from 0 to θ respectively, 2 1 T θ 1  T δ T 1 T 1 µθ . i.e. or log =µ θ. or ...(v) =µ δθ  e = e ∫∫ T T  2 T 0 T 2 2 Equation (v) can be expressed in terms of corresponding logarithm to the base 10, i.e.  T 1 2.3log =µ θ .  T  2 The above expression gives the relation between the tight side and slack side tensions, in terms of coefficient of friction and the angle of contact. 11.15. Determination of Angle of Contact When the two pulleys of different diameters are connected by means of an open belt as shown in Fig. 11.16 (a), then the angle of contact or lap (θ) at the smaller pulley must be taken into consideration. Let r = Radius of larger pulley, 1 r = Radius of smaller pulley, and 2 x = Distance between centres of two pulleys (i.e. O O ). 1 2 From Fig. 11.16 (a), OM O E–– ME r r 11 12 sinα= = = ...(∵ ME = O F = r ) 2 2 OO O O x 12 12 ∴ Angle of contact or lap, π θ= (180° – 2α) rad 180 342 Theory of Machines A little consideration will show that when the two pulleys are connected by means of a crossed belt as shown in Fig. 11.16 (b), then the angle of contact or lap (θ) on both the pulleys is same. From Fig. 11.16 (b), OM O E++ ME r r 11 12 sinα= = = OO O O x 12 12 π ∴ Angle of contact or lap, θ= (180°+ 2α) rad 180 (a) Open belt drive. (b) Crossed belt drive. Fig. 11.16 Example 11.4. 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 lap 160° 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 ; θ = 160° = 160 × π / 180 = 2.793 rad ; T = 2500 N 1 We know that velocity of the belt, ππ dN . × 0.6× 200 v== = 6.284 m/s 60 60 Let T = Tension in the slack side of the belt. 2  T 1 We know that 2.3log =µ θ . = 0.25× 2.793= 0.6982  T  2 Chapter 11 : Belt, Rope and Chain Drives 343  T 0.6982 1 log == 0.3036  T 2.3  2 T 1 = 2.01 ∴ ...(Taking antilog of 0.3036) T 2 T 2500 1 T== = 1244 N and 2 2.01 2.01 We know that power transmitted by the belt, P =(T – T ) v = (2500 – 1244) 6.284 = 7890 W 1 2 = 7.89 kW Ans. Another model of milling machine. Note : This picture is given as additional information and is not a direct example of the current chapter. Example 11.5. A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns round a drum of 300 mm diameter revolving at 20 r.p.m. The other end of the rope is pulled by a man. The coefficient of friction is 0.25. Determine 1. The force required by the man, and 2. The power to raise the casting. Solution. Given : W = T = 9 kN = 9000 N ; d = 300 mm = 0.3 m ; N = 20 r.p.m. ; µ = 0.25 1 1. Force required by the man Let T = Force required by the man. 2 Since the rope makes 2.5 turns round the drum, therefore angle of contact, θ =2.5 × 2 π = 5 π rad 344 Theory of Machines  T 1 We know that 2.3log . 0.25 5 3.9275 =µ θ= × π =  T  2  T 3.9275 T 1 1 log == 1.71 or = 51  2.3 T T  2 2 ...(Taking antilog of 1.71) T 9000 1 ∴ Ans. T== = 176.47 N 2 51 51 2. Power to raise the casting We know that velocity of the rope, ππ dN.0×.3×20 v== = 0.3142 m/s 60 60 ∴ Power to raise the casting, P =(T – T ) v = (9000 – 176.47) 0.3142 = 2772 W 1 2 = 2.772 kW Ans. Example 11.6. Two pulleys, one 450 mm diameter and the other 200 mm diameter are on parallel shafts 1.95 m apart and are connected by a crossed belt. Find the length of the belt required and the angle of contact between the belt and each pulley. What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if the maximum permissible tension in the belt is 1 kN, and the coefficient of friction between the belt and pulley is 0.25 ? Solution. Given : d = 450 mm = 0.45 m or r = 0.225 m ; d = 200 mm = 0.2 m or 1 1 2 r = 0.1 m ; x = 1.95 m ; N = 200 r.p.m. ; T = 1 kN = 1000 N ; µ = 0.25 2 1 1 We know that speed of the belt, πdN . π× 0.45 × 200 11 v== = 4.714 m/s 60 60 Length of the belt We know that length of the crossed belt, 2 () rr + 12 Lr =π() +r +2x + 12 x 2 (0.225 0.1) + Ans. =π(0.225 + 0.1) + 2 × 1.95 + = 4.975m 1.95 Angle of contact between the belt and each pulley Let θ = Angle of contact between the belt and each pulley. We know that for a crossed belt drive, rr + 0.225 + 0.1 12 or α = 9.6° sinα= = = 0.1667 x 1.95 ∴θ = 180° + 2 α = 180° + 2 × 9.6° = 199.2° π Ans. =× 199.2 = 3.477 rad 180

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