9 speed gear box

B.Balavairavan
AP/Mech
KCET
ME 6601 -Design of Transmisson systems

1. DEPARTMENT OF MECHANICAL ENGINEERING By Mr. B.Balavairavan Assistant Professor Mechanical Engineering Kamaraj College of Engg and Tech Virudhunagar Design of NINE Speed Gear Box

2. Sample Problem Design a gearbox to give 9 speed output from a single input speed. The required speed range is 180 rpm to 1800 rpm. Given: n = 9 N min = 180 rpm N max = 1800 rpm

3. Step - 1 “Calculation of Step ratio” Refer PSG Data Book P. No : 7.20 to check whether, the calculated step ratio is a std. value N max N min = Ø n -1 1800 180 = Ø 9-1 Ø = 1.333

4. Since its not a std. value, Lets find a multiples of std. value come close to calculated step ratio 1.6 - 1.25 - 1.12 - 1.06 - Multiples of 1.06 gives nearest value of 1.333 As 1.06 is multiplied 4 times we skip 4 speed Hence std. Ø = 1.06 & R 40 series is selected Cannot be used Cannot be used 1.12 1.06 x 1.12 = 1.254 x 1.06 x 1.06 x 1.06 x 1.06 = 1.338

5. Step - 2 “Selection of Speeds” The selected speeds are; 180,236,315,425,560,750,1000,1320,1800 100 106 112 118 125 132 140 150 160 170 180 190 200 212 224 236 250 265 280 300 315 335 355 375 400 425 450 475 500 530 560 600 630 670 710 750 800 850 900 950 1000 1060 1120 1180 1250 1320 1400 1500 1600 170 1800 1900 No deed to check for deviation

6. Step - 3 “ Structural formula & Ray Diagram ” The structural formula for 9 speed gear box is 3 (1) 3 (3) Stage 1 - Single input is splitted into 3 speeds Stage 2 - 3 input is splitted into 9 speeds ie., each input is splitted into 3 speed

7. 1800 1320 1000 750 560 425 315 236 180 Selected speeds are; 180,236,315,425,560,750,1000,1320,1800 Lets group the final output speeds into 3, since the structural formula is 3 (1) 3 ( 3 ) Stage 1 Stage 2

8. Lets select the input speed of stage 2. For that the input speed should satisfy two following conditions. At Least one output speed should be greater than input speed. (1 for 3 o/p and 2 for 4 o/p) The input and output must satisfy the following ratios N max N i/p N min N i/p ≥ 0.25 ≤ 2

9. 1800 1320 1000 750 560 425 315 236 180 Stage 1 Stage 2 Lest find input speed for the lowest output speed set. For the first condition, possible input speeds are 750 & 560 . For the second condition, The conditions are satisfied Stage - 2 N min N i/p ≥ 0.25 N max N i/p ≤ 2 180 5 60 1000 560 = 0.32 = 1.78 = =

10. 1800 1320 1000 750 560 425 315 236 180 Stage 1 Stage 2 Lest find input speed for the lowest output speed set. For the first condition, possible input speeds are 1338 & 1790 For the second condition, The conditions are satisfied Stage - 1 N min N i/p ≥ 0.25 N max N i/p ≤ 2 560 1320 1000 1320 = 0.41 = 0.74 = =

11. Step - 4 “ Kinematic Arrangement ” Shaft - 1 / Input Shaft - 2 / Intermediate Shaft - 3 / Output 1 3 4 2 5 6 8 10 12 7 9 11

12. Step - 5 “ Calculation of number of number of teeth in gears ” Start from the final stage First find the number of teeth for maximum speed reduction pair. Assume the number of teeth in the driver gear (It should be above 17) The sum of number of teeth in meshing gears in a stage is always equal.

13. Stage - 2 “First Pair - Maximum Speed Reduction” Assume number of teeth in driver = 20 z 11 z 12 = N 12 N 11 20 z 12 = 180 560 z 12 = 62.2 ≅ 63

14. Stage - 2 “Second Pair - Minimum Speed Reduction” z 7 z 8 = N 8 N 7 z 7 z 8 = 425 560 z 7 = 0.76 z 8

15. Stage - 2 “Third Pair - Maximum Speed Increment” z 9 z 10 = N 10 N 9 z 9 z 10 = 1000 560 z 9 = 1.78 z 10

16. Stage - 2 z 7 + z 8 = z 9 + z 10 = z 11 + z 12 z 7 + z 8 = z 9 + z 10 = 20 + 63 = 83 z 11 = 20 z 12 = 63 z 7 = 0.76 z 8 z 9 = 1.78 z 10 z 7 + z 8 = 83 z 9 + z 10 = 83 0.76 z 8 + z 8 = 83 1.78 z 10 + z 10 = 83 z 10 = 29.79 ≅ 30 z 8 = 47.16 ≅ 48 z 7 = 35 z 9 = 53

17. Stage - 1 “First Pair - Maximum Speed Reduction” Assume number of teeth in driver = 20 z 5 z 6 = N 6 N 5 20 z 6 = 560 1338 z 6 = 47.14 ≅ 48

18. Stage - 1 “Second Pair – Intermediat Speed Reduction” z 1 z 2 = N 2 N 1 z 1 z 2 = 750 1338 z 1 = 0.57 z 2

19. Stage - 1 “Third Pair - Minimum Speed Increment” z 3 z 4 = N 4 N 3 z 3 z 4 = 1000 1338 z 3 = 0.74 z 4

20. Stage - 1 z 1 + z 2 = z 3 + z 4 = z 5 + z 6 z 1 + z 2 = z 3 + z 4 = 20 + 42 = 68 z 5 = 20 z 6 = 48 z 1 = 0.57 z 2 z 3 = 0.74 z 4 z 3 + z 4 = 68 z 1 + z 2 = 68 0.76 z 4 + z 4 = 68 0.57 z 2 + z 2 = 68 z 2 = 43.3 ≅ 44 z 4 = 38.64 ≅ 39 z 3 = 29 z 1 = 24

21. Solution z 1 = 24 z 2 = 44 z 3 = 29 z 4 = 39 z 5 = 20 z 6 = 48 z 8 = 48 z 7 = 35 z 10 = 30 z 9 = 53 z 11 = 20 z 12 = 63