1. Mechanical Engineering Design I Mechanical Engineering Design I MEEN :3391 Gears Gears

2. Gears In precision machines, where a definite velocity ratio is of importance (as in watch mechanism), the only positive drive is by means of gears or toothed wheels.

3. Gears Wheel B will be rotated (by the wheel A) so long as the tangential force exerted by the wheel A does not exceed the maximum frictional resistance between the two wheels. But when the tangential force (P) exceeds the frictional resistance (F), slipping will take place between the two wheels. Thus the friction drive is not a positive drive…….. So Gears are usually used for +ve drive

4. Advantages & Disadvantages of Gear Drives The following are the advantages an Disadvantages of Gear drives as compared to belt, rope and chain drives :

5. Advantages & Disadvantages of Gear Drives The following are the advantages an Disadvantages of Gear drives as compared to belt, rope and chain drives : Advantages 1. It transmits exact velocity ratio. 2. It may be used to transmit large power. 3. It has high efficiency. 4. It has reliable service. 5. It has compact layout. Disadvantages 1. The manufacture of gears require special tools and equipment. 2. The error in cutting teeth may cause vibrations and noise during operation.

6. Spur Types of Gear Helical Double Helical

7. Bevel & Worm Types of Gear http://www.learnengineering.org/2014/ 05/working-of-differential.html

8. Types of Gear

9. Terminology of Gears Pitch circle. It is an imaginary circle which by pure rolling action, would give the same motion as the actual gear. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter.

10. Terminology of Gears Pitch point. It is a common point of contact between two pitch circles. Addendum. It is radial distance of a tooth from pitch circle to top of tooth. Dedendum. It is radial distance of a tooth from pitch circle to bottom of tooth.

11. Terminology of Gears Circular pitch. It is distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth denoted by p c two gears will mesh together if two wheels have same P c

12. Terminology of Gears Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter denoted by p d Module. It is the ratio of the pitch circle diameter to the number of teeth denoted by m.

13. Terminology of Gears Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in a meshing gear. Total depth. It is the radial distance between the addendum and the dedendum circles of a gear. It is equal to the sum of the addendum and dedendum. Tooth thickness. It is width of tooth measured along pitch circle.

14. Terminology of Gears Working depth. It is the radial distance from the addendum circle to the clearance circle. It is equal to the sum of the addendum of the two meshing gears. Tooth space. It is the width of space between the two adjacent teeth measured along the pitch circle. Backlash. It is the difference between the tooth space and the tooth thickness, as measured along the pitch circle. Theoretically, the backlash should be zero, but in actual practice some backlash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion.

15. Terminology of Gears Face of tooth. It is the surface of the gear tooth above the pitch surface. Flank of tooth. It is the surface of the gear tooth below the pitch surface. Top land. It is the surface of the top of the tooth. Face width. It is the width of the gear tooth measured parallel to its axis.

16. 16 Fundamental Law of Gearing The Angular Velocity (ω) Ratio is Constant at All Positions on mating gears V 2 = V 1 R 2 ω 2 = R 1 ω 1 (D 2 /2) ω 2 = (D 1 /2) ω 1 ω 1 / ω 2 = D 2 / D 1 2πN 1 / 2πN 2 =D 2 / D 1 N 1 / N 2 =D 2 / D 1

17. Gear Trains When two or more gears are made to mesh with each other to transmit power from one shaft to another. Such a combination is called gear train Types 1.Simple gear train: one gear on each shaft

18. Gear Trains 1.Simple gear train: Motion of the driven gear is opposite to the motion of driving gear

19. Gear Trains 1.Simple gear train: When distance between the two gears is large. Either use large sized gear OR Use intermediate gears

20. Gear Trains 1.Simple gear train: Speed ratio and train value is independent of size and number of intermediate gears. These intermediate gears are called idle gears. 1. Idle gears are used where a large centre distance is required 2. To obtain the desired direction of motion of the driven gear

21. Example A spur gear having 30 teeth and a diametral pitch of 6 in-1, is rotating at 200 rpm. Determine the circular pitch and the magnitude of the pitch line velocity Solution: T= 30, N = 200 = 6 Pitch Line Velocity = D/2 * ω ω = 2π N

22. Example Two spur gears are in mesh. Driven gear has a magnitude of rotational speed that is one half that of driver gear. The driver gear rotates at 500 rpm, has a module of 3 mm and has 48 teeth. Determine (a) The number of teeth of the driven gear (b) The magnitude of the pitch line velocity N2 = N1 /2 N1 = 500 rpm therefore N2 = 250 rpm Solution: m 1 = D 1 / T 1 = 3 T 1 = 48 therefore D 1 = 144 mm N 1 / N 2 = T 2 / T 1 …………………….. T 2 = 96 Pitch Line Velocity = D 1 /2 * ω 1 ω 1 = 2π N 1

23. Gear Trains 2.Compound gear train: There are more than one gear on a shaft. Distance between the driver and the driven or follower has to be bridged over by intermediate gears and at the same time a great ( or much less ) speed ratio is required

24. Gear Trains 2.Compound gear train:

25. Gear Trains 2.Compound gear train: Gears 2 and 3 are mounted on one shaft B, therefore N 2 = N 3 Gears 4 and 5 are mounted on shaft C, therefore N 4 = N 5

26. Gear Trains 2.Compound gear train: Advantage of a compound train over a simple gear train:  Much larger speed reduction can be obtained with small gears.  If a simple gear train is used : the last gear has to be very large.

27. Problem The gearing of a machine tool is shown in Fig. The motor shaft is connected to gear A and rotates at 975 r.p.m. The gear wheels B, C, D and E are fixed to parallel shafts rotating together. The final gear F is fixed on the output shaft. What is the speed of gear F ? The number of teeth on each gear are as given below :

28. Problem

29. Design of Spur Gears Spur gears (both driver and driven) are usually designed for given velocity ratio and distance between the centres of their shafts. Two gears will mesh together if they have same Circular pitch P c

30. Problem Two parallel shafts, about 600 mm apart are to be connected by spur gears. One shaft is to run at 360 r.p.m. and the other at120 r.p.m. Design the gears, if the circular pitch is to be 25 mm.

31. Problem 1 Number of teeth on both the gears are to be in complete numbers, therefore let us make the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teeth on the second gear should be 38 × 3 = 114.

32. Problem Exact pitch circle diameter of the first gear Exact distance between the two shafts

33. Gear Trains 3.Reverted gear train: Shafts of the first gear and the last gear are co-axial. Direction of motion of the first gear and the last gear is same. These have applications in automotive transmissions, lathe back gears and in clocks (where the minute and hour hand shafts are co-axial)

34. Gear Trains 3.Reverted gear train: Shafts of the first gear and the last gear are co-axial. Direction of motion of the first gear and the last gear is same. These have applications in automotive transmissions, lathe back gears and in clocks (where the minute and hour hand shafts are co-axial)

35. Gear Trains 3.Reverted gear train:

36. Gear Trains 3.Reverted gear train: If module is same for all gears

37. Problem The speed ratio of the reverted gear train, as shown in Fig. is to be 12. The module pitch of gears A and B is 3.125 mm and of gears C and D is 2.5 mm. Calculate the suitable numbers of teeth for the gears. No gear is to have less than 24 teeth.

38. Problem

40. Gear Trains 4.Epicyclic Gear Train: Axes of the shafts, over which the gears are mounted, may move relative to a fixed axis

41. Gear Trains 4.Epicyclic Gear Train: Axes of the shafts, over which the gears are mounted, may move relative to a fixed axis

42. Gear Trains 4.Epicyclic Gear Train: useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space These have applications in differential gears, lathe back gears, wrist watches, Wankle engines.

43. Gear Trains 4.Epicyclic Gear Train: useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space These have applications in differential gears, lathe back gears, wrist watches, Wankle engines.

44. Gear Trains 4.Epicyclic Gear Train: Conventional 1-DOF Gearset Epicyclic 2-DOF Gearset

45. Gear Trains 4.Epicyclic Gear Train:

46. Gear Trains 4.Epicyclic Gear Train: If arm is fixed, the gear train is simple and gear A can drive gear B or vice- versa, but if gear A is fixed and the arm is rotated about the axis of gear A (i.e. O 1 ), then the gear B is forced to rotate upon and around gear A. Such a motion is called epicyclic Velocity Ratio 1.Tabular method 2. Algebraic/Analytical method

47. Gear Trains Velocity Ratio using Tabular method Suppose arm is fixed When the gear A makes one revolution anticlockwise, the gear B will make Assuming the anticlockwise rotation as positive and clockwise as negative, when gear A makes + 1 revolution, then the gear B will make (– TA/ TB ) revolutions

48. Gear Trains Velocity Ratio using Tabular method A makes + x revolutions, then the gear B will make – x × TA / TB Revolutions (Row 2) Each element of an epicyclic train is given + y revolutions (Row 3)

49. Gear Trains Velocity Ratio using Tabular method Motion of each element of the gear train is added up from Row 2 to row 3 (Row 4)

50. Gear Trains Velocity Ratio using Tabular method

51. Problems N A = ? X+Y = -100, Y = -200 N A = -200 -100 x 40 / 80 = -250 rpm

52. Problems

53. Tabular method

54. Problems Tabular method

55. In a reverted epicyclic gear train, the arm A carries two gears B and C and a compound gear D - E. The gear B meshes with gear E and the gear C meshes with gear D. The number of teeth on gears B, C and D are 75, 30 and 90 respectively. Find the speed and direction of gear C when gear B is fixed and the arm A makes 100 r.p.m. clockwise. Problems Solution. Given : T B = 75 ; T C = 30 ; T D = 90 ; N A = 100 r.p.m. (clockwise)

56. Problems

57. x

59. In an epicyclic gear of the ‘sun and planet’ type shown in Fig., the pitch circle diameter of the internally toothed ring is to be 224 mm and the module 4 mm. When the ring D is stationary, the spider A, which carries three planet wheels C of equal size, is to make one revolution in the same sense as the sun wheel B for every five revolutions of the driving spindle carrying the sun wheel B. Determine suitable numbers of teeth for all the wheels. Solution. Given : d D = 224 mm ; m = 4 mm ; N A = N B / 5

60. Problems Solution. Given : d D = 224 mm ; m = 4 mm ; N A = N B / 5

61. Problems Solution. Given : d D = 224 mm ; m = 4 mm ; N A = N B / 5

62. Torque transmitted by gears P = T ω If no frictional losses P1 = P2 T1 ω1 = T2 ω2 ω1/ ω2 = T2/ T1 2π N1 / 2π N2 = T2/ T1 N1 /N2 = T2/ T1 ( T is torqur) N1 /N2 = D2 / D1 N1 /N2 = T2/ T1 ( T = no of teeth)