Design of Gear BoxUsing PSG Design Data Book

Sample ProblemDesign a gearbox to give 9 speed output from a single input speed. The required speed range is 180 rpm to 1800 rpm.Given:n = 9Nmin = 180 rpmNmax = 1800 rpm

Step - 1 “Calculation of Step ratio”Nmax

Nmin= Ø n-1

1800180

= Ø 9-1

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 ratio1.6 -1.25 -1.12 -1.06 -

Multiples of 1.06 gives nearest value of 1.333As 1.06 is multiplied 4 times we skip 4 speed

Hence std. Ø = 1.06 & R 40 series is selected

Cannot be usedCannot be used1.12 1.06

x 1.12 = 1.254x 1.06 x 1.06x 1.06 x 1.06 = 1.338

Step - 2 “Selection of Speeds”

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

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 1123 1191 1262

1338 1418 1503 1593 1689 1790 1898

Check for deviation

lets calculate the allowable deviation and actual deviation for the given range of speed.

Allowable deviation = ± 10 (Ø - 1) %

Nmin

Nmax

= (Nmax actual - Nmax) x

= ± 10 (1.333 - 1) % = ± 3.33 %

Actual deviation

1801800

= (1790 - 1800) x

= - 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 is3 (1) 3 (3)

Stage 1 - Single input is splitted into 3 speedsStage 2 - 3 input is splitted into 9 speeds

ie., each input is splitted into 3 speed

1790

1338

1000

750

560

425

315

236

180

Selected speeds are; 180,236,315,425,560,750,1000,1320,1790

Lets group the final output speeds into 3, since the structural

formula is 3 (1) 3 (3)

Stage 1 Stage 2

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

Nmax

Ni/p

Nmin

Ni/p

≥ 0.25 ≤ 2

1790

1338

1000

750

560

425

315

236

180Stage 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

Nmin

Ni/p

≥ 0.25

Nmax

Ni/p

≤ 2

180

660

1000560

= 0.32

= 1.78

=

=

Stage - 2

1790

1338

1000

750

560

425

315

236

180Stage 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

Nmin

Ni/p

≥ 0.25

Nmax

Ni/p

≤ 2

560

1338

10001338

= 0.41

= 0.74

=

=

Stage - 1

Step - 4 “ Kinematic Arrangement ”

Shaft - 1 / Input

Shaft - 2 / Intermediate

Shaft - 3 / Output

13

42

5

6

8 1012

79

11

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”

z11

z12=

N12

N11

20z12

= 180560

Assume number of teeth in driver = 20

z12 = 62.2 ≅ 62

Stage - 2 “Second Pair - Minimum Speed Reduction”

z7

z8=

N8

N7

z7

z8= 425

560

z7 = 0.76 z8

Stage - 2 “Third Pair - Maximum Speed Increment”

z9

z10=

N10

N9

z9

z10=1000

560

z9 = 1.78 z10

Stage - 2

z7 + z8 = z9+ z10 = z11+ z12

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

z11 = 20

z12 = 62

z7 = 0.76 z8

z9 = 1.78 z10

z7 + z8 = 82

z9+ z10 = 82

0.76 z8 + z8 = 82

1.78 z10+ z10 = 82

z10 = 29.49 ≅ 30

z8 = 46.5 ≅ 46 z7 = 36

z9 = 52

Stage - 1 “First Pair - Maximum Speed Reduction”

z5

z6=

N6

N5

20z6

= 5601338

Assume number of teeth in driver = 20

z6 = 41.8 ≅ 42

Stage - 1 “Second Pair - Minimum Speed Reduction”

z1

z2=

N2

N1

z1

z2= 750

1338

z1 = 0.56 z2

Stage - 1 “Third Pair - Maximum Speed Increment”

z3

z4=

N4

N3

z3

z4= 1000

1338

z3 = 0.74 z4

Stage - 1

z1 + z2 = z3+ z4 = z5+ z6

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

z5 = 20

z6 = 42

z1 = 0.56 z2

z3 = 0.74 z4

z3 + z4 = 62

z1 + z2 = 62

0.74 z4 + z4 = 62

0.56 z2 + z2 = 62

z2 = 39.74 ≅ 40

z4 = 35.63 ≅ 36 z3 = 26

z1 = 22

Solutionz1 = 22

z2 = 40

z3 = 26

z4 = 36

z5 = 20

z6 = 42

z8 = 46

z7 = 36

z10 = 30

z9 = 52

z11 = 20

z12 = 62