J3010 Unit 6

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J3010 - Mechanics of Machines 1

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J3010 Unit 6

  1. 1. BELTING J 3010/6/1 UNIT 6 BELTING OBJECTIVES General Objective : To understand and apply the concept of belting Specific Objectives : At the end of this unit you will be able to:  state the difference between open and close belt.  explain that power transmitted by the flat belt and V belt.  explain that ratio tension for flat and V belt.  calculate power transmitted by the belt with consider centrifugal force.
  2. 2. BELTING J 3010/6/2 INPUT 6.0 INTRODUCTION In factories, the power has to be transmitted from one shaft to another, then belt driving between pulleys on the shafts may be used. The pulley rotating shaft is called driver. The pulley intended to rotate is known, as follower or driven. When the driver rotates, it carries the belt that grip between its surface and the belt. The belt, in turn, carries the driven pulley which starts rotating. The grip between the pulley and the belt is obtained by friction, which arises from pressure between the belt and the pulley. (a) Types of belts The flat belt 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 not more than 10 m a part. The V-belt is mostly used in the factories and work shops where a great amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each other.
  3. 3. BELTING J 3010/6/3 The types of belts:- a. flat belt b. V belt 6.1 LENGTH OF AN OPEN BELT DRIVE A B K α1 α2 C O1 O2 F r2 r1 D E d Fig.6.1 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian)
  4. 4. BELTING J 3010/6/4 We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ2 = α1 = 2 α2 (radian) Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length of belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1
  5. 5. BELTING J 3010/6/5 Example 6.1 Find the length of belt necessary to drive a pulley of 480 cm diameter running parallel at a distance of 12 meter from the driving pulley of diameter 80 cm. This system is an open belt drive. Solution 6.1 A B K F α1 O2 α2 C O1 r2 r1 D E d Fig.6.2 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of belt is, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200
  6. 6. BELTING J 3010/6/6 240  40 α1 = cos-1 1200 = cos-1 0.16667 = 80º α1 = 1.396 radian θ1 = 2π - 2α1 = 2 (π - 1.396 ) (radian) = 3.491 radian θ2 = 2α1 = 2 α2 (radian) = 2 (1.396 ) = 2.792 radian Arc AFE = r1θ1 = 240 x 3.491 = 837.84 cm Arc DCB = r2θ2 = 40 x 2.792 = 111.68 cm And ED = BA = KO2 KO2 Sin 80º = d KO2 = d Sin α1 = 1 200 x 0.98481 = 1181.77 m Finally the total of belt length is L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.491 + 40 x 2.792 +2 x 1181.77 = 837.84 + 111.68 + 2363.54 = 3313.06 cm = 33.13 m
  7. 7. BELTING J 3010/6/7 6.2 LENGTH OF CLOSE BELT DRIVE K A D C O1 α1 α2 O2 F r2 r1 B E d Fig.6.3 Cross belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) We know that the length of the belt is L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ1 = θ2
  8. 8. BELTING J 3010/6/8 Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 Example 6.2 Find the length of belt for a cross belt drive system. The diameter of the drive pulley is 480 cm which running parallel at a distance of 12 meter from the driving pulley which has a diameter of 80 cm. Solution 6.2 K A D α1 α2 C F O1 O2 r2 r1 B E d Fig.6.4 Cross belt drive
  9. 9. BELTING J 3010/6/9 O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200 240  40 α1 = cos-1 1200 -1 = cos 0.23333 = 76.5º α1 = 1.335 radian θ1 = 2π - 2α1 = 2 (π - α1) (radian) = 2 (π - 1.335) (radian) = 3.613 radian θ1 = θ2 Arc AFE = r1θ1 = 240 x 3.613 = 867.12 cm Arc DCB = r2θ2 = 40 x 3.613 = 144.52 cm And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin 76.5º =1 200 x 0.9724 ED = BA = KO2 = 1166.88 cm
  10. 10. BELTING J 3010/6/10 Finally the total of length belt, L= Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sinα1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.5º = 867.12 + 144.52 + 2333.76 = 3345.4 cm L = 33.45 m
  11. 11. BELTING J 3010/6/11 Activity 6A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel shaft of 1.95 m apart and are connected by a cross belt. Find the length of the belt required and the angle of contact between the belt and each pulley. 6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine the size of pulley on the shaft B. 6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter running parallel at a distance of 1.7 meter from the driving pulley of diameter 0.5 m. It is connected by an open belt. N1 d 2  N 2 d1 N1 = speed diver in rpm N2 = speed follower in rpm d1 ,d2 = diameter pulley driver and follower.
  12. 12. BELTING J 3010/6/12 Feedback to Activity 6A Have you tried the questions????? If “YES”, check your answers now 6.1 4.975 m; 199 or 3.473 radian. 6.2. 11.6 cm 6.3 6.51 m
  13. 13. BELTING J 3010/6/13 INPUT 6.3 POWER TRANSMITTED BY A BELT Power = (T1 –T2) v Where, T1 =Tension in tight side in Newton. T2 = Tension in slack side. V = velocity of belt Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B. The driving pulley pulls the belt from one side, and delivers the same to the other.The maximum tension in the tight side will be greater than that slack side. Fig. 6.5 Torque exerted on the driving pulley = ( T1 - T2) r1 Torque exerted on the driven or follower = ( T1 - T2) r2 Power transmitted = Force x distance = ( T1 - T2) v
  14. 14. BELTING J 3010/6/14 Example 6.3 The tension in the two sides of the belt are 100 N and 80 N respectively. If the speed of the belt is 75 meters per second. Find the power transmitted by the belt. Solution 6.3 Given, Tension in tight side, T1 = 100 N Tension in slack side, T2 = 80 N Velocity of belt, v = 75 m/s Let P = Power transmitted by the belt Using the relation, Power = (T1 –T2) v = ( 100 – 80) 75 = 1500 watt = 1.5 kw.
  15. 15. BELTING J 3010/6/15 Activity 6B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.4. Find the tension in the tight side, if the tension in slack side 150 N. The speed of the belt is 58 meters per second and the power transmitted by this belt is 2 kw. 6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotating at 300 rev/ min. Calculate velocity of driver. 6.6 The tension in the two sides of the belt are 300 N and 180 N respectively. If the speed of the belt is 50 meters per second, find the initial tension and power transmitted by the belt. 6.7 A 100 mm diameter pulley is belt-driven from a 400 mm diameter pulley. The 400 mm pulley rotates at 480 rev/min. The number of rev/sec of the 100 mm diameter pulley is A. 2 B. 32 C. 120 D. 1920 T1  T2 T0 = 2 Where, T0 = Initial tension in the belt.
  16. 16. BELTING J 3010/6/16 Feedback to Activity 6B Have you tried the questions????? If “YES”, check your answers now 6.4 115.5 N 6.5 0.785 m/s 6.6 6 000 watt or 6 kw; 240 N 6.7 B. 32
  17. 17. BELTING J 3010/6/17 INPUT 6.4 RATIO OF TENSIONS. Fig. 6.6 Consider a driven pulley rotating in the clockwise direction as in fig 6.6. Let T1 = Tension in the belt on the tight side. T2 = Tension in the belt on the slack side. ө = An 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 position of the belt PQ, subtending an angle  at the centre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the following force: i. Tension T in the belt at P. ii. Tension T + T in the belt at Q. iii. Normal reaction R, and iv. Frictional force F =  x R Where  is the coefficient of friction between the belt and pulley. Resolving all the forces horizontally and equating the same,
  18. 18. BELTING J 3010/6/18   R = ( T + T) sin + T sin (6.1) 2 2 Since the  is very small, therefore, substituting,   sin = in equation 6.1 2 2   R = ( T + T) +T 2 2 T  T  T . = + + 2 2 2 T  = T. ( neglecting ) (6.2) 2 Now resolving the forces vertically,    x R = (T + T) cos - T cos (6.3) 2 2  Since the angle . is very small, therefore substituting cos = 1 in 2 equation 6.3 or, xR = T + T – T = T T R = (6.4)  Equating the values of R from equation 6.2 and 6.4, T Or T. =  T Or = . T Integrating both sides from A to B,  T T1 T T2     0 T1 or loge =  T2 T1 = e  ration of tension for flat belt T2 T1 = e /sin  ration of tension for V belt T2 where  = half angle of groove
  19. 19. BELTING J 3010/6/19 Example 6.4 Find the power transmitted by a belt running over a pulley of 60 cm diameter at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap 160 and maximum tension in the belt is 250 KN. Solution 6.4 Given, Diameter of pulley, d = 60 cm = 0.6 m Speed of pulley N = 200 rpm  dN  x 0.6 x 200 Speed of belt, v = = = 2 = 6.28 m/ sec 60 60 Coefficient of friction, μ = 0.25 160 x  Angle of contact,  = 160 = = 2.7926 radian. 180 Maximum tension in the belt, T1 = 250 kN Let P = power transmitted by the belt. Using the relation, T1 = e  T2 = e 0.25 x 2.7926 T1 = 2.01 T2 T1 250 T2 = = = 124.38 kN 2.01 2.01 Now using the relation, P = (T1 –T2 ) v = ( 250 - 124.38 ) 6.28 = 788.89 watt P = 0.79 kw
  20. 20. BELTING J 3010/6/20 INPUT 6.5 CENTRIFUGAL TENSION. Fig.6.7 The tension caused by centrifugal force is called centrifugal tension. At lower speeds the centrifugal tension is very small, but at higher speeds its effect is considerable, and thus should be taken into account. Consider a small portion PQ of the belt subtending an angle d at the centre of the pulley as show in fig 6.7. Let M = mass of the belt per unit length, V = linear velocity of the belt, r = radius of the pulley over which the belt runs, Tc = Centrifugal tension acting tangentially at P and Q. Length of the belt PQ, = r d Mass of the belt PQ, M = M r d We know that centrifugal force, = M v2/r
  21. 21. BELTING J 3010/6/21 Centrifugal force of the belt PQ M x r.d v 2 = r = M x d v 2 The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in equilibrium. Now resolving the force (i.e, centrifugal force and centrifugal tension) horizontally and equating the same, d 2 Tc sin = M x d v 2 2 since angle d is very small, therefore, substituting d d sin = 2 2 d 2 Tc = M x d v 2 2 Tc = M x d v 2 / d Tc = M v2 T1  Tc  e ratio tension for flat belt. T2  Tc T1  Tc  e   / sin  ratio tension for V belt T2  Tc 6.5.1 Condition for the transmission of maximum power The maximum power, When T c = ⅓T1 It shows that the power transmitted is maximum ⅓ of the maximum tension is absorbed as centrifugal tension. The velocity of belt for maximum transmission of power may be obtained from equation T1 = 3Tc = M v2.
  22. 22. BELTING J 3010/6/22 3 T1 v2 = M T1 v = 3M Example 6.5 An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on parallel shafts 3.6 m apart. The belt has a mass of 0.9 kg/m length and the maximum tension in it is not exceed 2 kN. The larger pulley runs at 200 rev/min. Calculate the torque on each of the two shafts and the power transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller pulley is 168° ( 2.947 radian ). Solution 6.5 Given Diameter of larger pulley, d1 = 1.2 m Radius of the larger pulley, r1 = 0.6 Diameter of smaller pulley, d2 =0.5 m Radius of the smaller pulley, r2 = 0.25 m Distance between two shaft, D = 3.6 m Mass of the belt per meter length, M = 0.9 kg/m Maximum Tension, T1 = 2 kN = 2 000 N Speed of the larger pulley = 200 rpm Velocity of the belt,  x 1.2 x 200 V= = 12.57 m/s 60 T c = Mv2 = 0.9 x 12.572 = 142.2 N
  23. 23. BELTING J 3010/6/23 e µθ = e 0.3 x 2.947 = 2.421 Using the relation, T1  Tc  e T2  Tc 2000  142.2 = 2.421 T2  142.2 (T2 – 142.2) 2.421 = 1857.8 1857.8 (T2 – 142.2) = = 767.37 2.421 T2 = 767.37 + 142.2 T2 = 909.57 N We know that the torque on the larger pulley shaft (TL), TL = ( T1 – T2) r1 = ( 2000 – 909.57) 0.6 = 654.26 Nm. Torque on the smaller pulley shaft (Ts), Ts = ( T1 – T2) r2 = (2000 – 909.57) 0.25 = 272.61 Nm. Power transmitted, P = ( T1 – T2)v = ( 2000 – 909.57) 12.57 = 13706.71 watt = 13.71 kw
  24. 24. BELTING J 3010/6/24 Activity 6C TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT…! 6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel of shafts 1.95 m apart. 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. Angle of contact between the belt and larger pulley is 3.477 radian. 6.9 Find the power transmitted by a V drive from the following data: Angle of contact = 84º Pulley groove angle = 45º Coefficient of friction = 0.25 Mass of belt per meter length = 0.472 kg/m Permissible tension = 139 N Velocity of V belt = 12.57 m/s. 6.10 An open belt drive connects two pulleys 1.2 m and 0.6 m, on parallel shafts 3 m apart. The belt has a mass 0f 0.56 kg/m and maximum tension is 1.5 kN. The driver pulley runs at 300 rev/min. Calculate the inertial tension, power transmitted and maximum power. The coefficient of friction between the belt and the pulley surface is 0.3.
  25. 25. BELTING J 3010/6/25 Feedback to Activity 6C Have you tried the questions????? If “YES”, check your answers now 6.8 2 740 watt or 2.74 kw 6.9 591 watt. 6.10 1074.5 N ; 8.0 kw ; 17.54 kw.
  26. 26. BELTING J 3010/6/26 SELF-ASSESSMENT 6 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 6 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. 1. A pulley is driven by a flat belt running at speed of 600 m/min. The coefficient of friction between the pulley and the belt is 0.3 and the angle lap is 160°. If the maximum tension in the belt is 700 N, find the power transmitted by a belt. 2. A leather belt, 125 mm wide and 6 mm thick, transmits power from a pulley of 750 mm diameter which run at 500 rpm. The angle of lap is 150° and µ = 0.3. If the mass of 1 m3 of leather is 1 Mg and the stress in the belt does not exceed 2.75 MN/m2, find the maximum power that can be transmitted. 3. A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at 1 600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulley is 165° and the coefficient of friction between the belt and pulleys is 0.3. If the maximum permissible stress in the belt is 2 MN/m2, find (i) Maximum power transmitted; and (ii) Initial tension in the belt. 4. The moment on a pulley, which produces the rotation of the pulley is called: A. Momentum B. Torque C. Work D. Energy 5. If T1 and T2 are the tension in the tight and slack sides of a belt and θ is the angle of contact between the belt and pulley. Coefficient of friction is μ, then the ratio of driving tension will be: T T T T A. 2  e   B. 1  e   C. 1   D. log10 1   T1 T2 T2 T2
  27. 27. BELTING J 3010/6/27 Feedback to Self-Assessment 6 Have you tried the questions????? If “YES”, check your answers now. 1. 3.974 kw ( see example 6.4) 2. 18.97 kw 3. (i) 14.281 kw (ii) 1.3221 kN 4. B. Torque T1 5. B.  e CONGRATULATIONS!!!!….. T2 May success be with you always….

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