Design of Gear Box

Using PSG Design Data Book

GOPINATH GASST.PROF - MECHANICAL

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

GOPINATH GASST.PROF - MECH

Step - 1 “Calculation of Step ratio”

Nmax

Nmin

= Ø n-1

1800

180 = 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 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

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 1700 1800 1900

No deed to check for deviation

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

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

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

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

Nmin

Ni/p

≥ 0.25

Nmax

Ni/p

≤ 2

180

560

1000

560

= 0.32

= 1.78

=

=

Stage - 2

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

Nmin

Ni/p

≥ 0.25

Nmax

Ni/p

≤ 2

560

1320

1000

1320

= 0.41

= 0.74

=

=

Stage - 1

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

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

20

z12

= 180

560

Assume number of teeth in driver = 20

z12 = 62.2 ≅ 63

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 + 63 = 83

z11 = 20

z12 = 63

z7 = 0.76 z8

z9 = 1.78 z10

z7 + z8 = 83

z9+ z10 = 83

0.76 z8 + z8 = 83

1.78 z10+ z10 = 83

z10 = 29.79 ≅ 30

z8 = 47.16 ≅ 48 z7 = 35

z9 = 53

Stage - 1 “First Pair - Maximum Speed Reduction”

z5

z6

= N6

N5

20

z6

= 560

1338

Assume number of teeth in driver = 20

z6 = 47.14 ≅ 48

Stage - 1 “Second Pair – Intermediat Speed Reduction”

z1

z2

= N2

N1

z1

z2

= 750

1338

z1 = 0.57 z2

Stage - 1 “Third Pair - Minimum 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 = 68

z5 = 20

z6 = 48

z1 = 0.57 z2

z3 = 0.74 z4

z3 + z4 = 68

z1 + z2 = 68

0.76 z4 + z4 = 68

0.57 z2 + z2 = 68

z2 = 43.3 ≅ 44

z4 = 38.64 ≅ 39 z3 = 29

z1 = 24

Solution

z1 = 24

z2 = 44

z3 = 29

z4 = 39

z5 = 20

z6 = 48

z8 = 48

z7 = 35

z10 = 30

z9 = 53

z11 = 20

z12 = 63

GOPINATH GASST.PROF - MECH