Design of 9 Speed Gear Box

Design of Gear Box Using PSG Design Data Book

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 Nmin = 180 rpm Nmax = 1800 rpm

Step - 1 “Calculation of Step ratio” Nmax Nmin

= Ø

n-1

1800 9-1 = Ø 180 Ø = 1.333

Refer PSG Data Book P. No : 7.20 to check whether, the calculated step ratio is a std. value

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

-

Cannot be used Cannot be used 1.12 x 1.12 = 1.254 1.06 x 1.06 x 1.06 x 1.06 x 1.06 = 1.338

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

Step - 2 “Selection of Speeds” 100

106

112

118

125

132

140

236

250

265

280

300

315

335

560

600

630

670

710

750

800

1338

1418

1503

1593

1689

1790

1898

150 355 850

160

170

180

190

200

212

224

375

400

425

450

475

500

530

900

950

1000

1060

1123

1191

1262

The speeds are; 180,236,315,425,560,750,1000,1320,1790 Check for deviation

lets calculate the allowable deviation and actual deviation for the given range of speed. Allowable deviation = ± 10 (Ø - 1) % = ± 10 (1.333 - 1) % = ± 3.33 % Actual deviation

= (Nmax actual - Nmax) x

Nmin Nmax

= (1790 - 1800) x 180 1800 = - 1.0 %

Since the deviation is within the allowable range we can design for standard speeds.

The selected standard speeds are; 180,236,315,425,560,750,1000,1320,1790

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

1790

Selected speeds are;

1338

180,236,315,425,560, 750,1000,1320,1790

1000 750 560 425 315 236 180 Stage 1

Stage 2

Lets group the final output speeds into 3, since the structural formula is 3 (1) 3 (3)

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 Nmin Ni/p

≥ 0.25

Nmax Ni/p

≤2

1790 1338 1000 750

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,

560

Nmin

425

Ni/p

315

Nmax

236

Ni/p

180 Stage 1

Stage - 2

=

180

= 0.32

≥ 0.25

= 1.78

≤2

660 = 1000 560

The conditions are satisfied

1790 1338 1000 750

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,

560

Nmin

425

Ni/p

315

Nmax

236

Ni/p

180 Stage 1

Stage - 1

=

560

= 0.41

≥ 0.25

= 0.74

≤2

1338 = 1000 1338

The conditions are satisfied

Step - 4 “ Kinematic Arrangement ” 1

3

5 Shaft - 1 / Input 7

2

9 11

Shaft - 2 / Intermediate

4 6

8

Shaft - 3 / Output

10 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.

Stage - 2 “First Pair - Maximum Speed Reduction” Assume number of teeth in driver = 20 z11 z12

=

N12 N11

20 180 = z12 560 z12 = 62.2 ≅ 62

Stage - 2 “Second Pair - Minimum Speed Reduction” z7 z8

=

N8

z7

N7

z8 z7 = 0.76 z8

=

425 560

Stage - 2 “Third Pair - Maximum Speed Increment” z9 z10

=

N10

z9

N9

z10 z9 = 1.78 z10

=

1000 560

Stage - 2 z7 + z8 = z9+ z10 = z11+ z12

z11 = 20

z7 + z8 = z9+ z10 = 20 + 62 = 82

z12 = 62

z7 + z8 = 82

z7 = 0.76 z8

0.76 z8 + z8 = 82

z9 = 1.78 z10

z8 = 46.5 ≅ 46

z7 = 36

z9+ z10 = 82 1.78 z10+ z10 = 82 z10 = 29.49 ≅ 30

z9 = 52

Stage - 1 “First Pair - Maximum Speed Reduction” Assume number of teeth in driver = 20 z5 z6

=

N6 N5

20 560 = z6 1338 z6 = 41.8 ≅ 42

Stage - 1 “Second Pair - Minimum Speed Reduction” z1 z2

=

N2

z1

N1

z2 z1 = 0.56 z2

=

750 1338

Stage - 1 “Third Pair - Maximum Speed Increment” z3 z4

=

N4

z3

N3

z4 z3 = 0.74 z4

=

1000 1338

Stage - 1 z1 + z 2 = z 3 + z 4 = z 5 + z 6

z5 = 20

z1 + z2 = z3+ z4 = 20 + 42 = 62

z6 = 42

z3 + z4 = 62

z1 = 0.56 z2

0.74 z4 + z4 = 62

z3 = 0.74 z4

z4 = 35.63 ≅ 36

z3 = 26

z1 + z2 = 62 0.56 z2 + z2 = 62 z2 = 39.74 ≅ 40

z1 = 22

Solution z1 = 22

z7 = 36

z2 = 40

z8 = 46

z3 = 26

z9 = 52

z4 = 36

z10 = 30

z5 = 20

z11 = 20

z6 = 42

z12 = 62