GEAR BOX DESIGN - 9SPEED

1. Design of Gear Box
Using PSG Design Data Book
GOPINATH G
ASST.PROF - MECHANICAL

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
Nmin = 180 rpm
Nmax = 1800 rpm
GOPINATH G
ASST.PROF - MECH

3. 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

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 1.06x 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 1700 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
Nmax
Ni/p
Nmin
Ni/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
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
180
560
1000
560
= 0.32
= 1.78
=
=
Stage - 2

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
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
560
1320
1000
1320
= 0.41
= 0.74
=
=
Stage - 1

11. Step - 4 “ Kinematic Arrangement ”
Shaft - 1 / Input
Shaft - 2 / Intermediate
Shaft - 3 / Output
1
3
42
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”
z11
z12
=
N12
N11
20
z12
=
180
560
Assume number of teeth in driver = 20
z12 = 62.2 ≅ 63

14. Stage - 2 “Second Pair - Minimum Speed Reduction”
z7
z8
=
N8
N7
z7
z8
=
425
560
z7 = 0.76 z8

15. Stage - 2 “Third Pair - Maximum Speed Increment”
z9
z10
=
N10
N9
z9
z10
=
1000
560
z9 = 1.78 z10

16. 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

17. 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

18. Stage - 1 “Second Pair – Intermediat Speed Reduction”
z1
z2
=
N2
N1
z1
z2
=
750
1338
z1 = 0.57 z2

19. Stage - 1 “Third Pair - Minimum Speed Increment”
z3
z4
=
N4
N3
z3
z4
=
1000
1338
z3 = 0.74 z4

20. 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

21. 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 G
ASST.PROF - MECH