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
= Ø n-1
Nmin
1800 = Ø 9-1
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
1320
1400
1500
1600
1700
1800
1900
150
355
850
160
170
180
190
200
212
224
375
400
425
450
475
500
530
900
950
1000
1060
1120
1180
1250
The selected speeds are;
180,236,315,425,560,750,1000,1320,1800
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
Selected speeds are;
1320
180,236,315,425,560,
750,1000,1320,1800
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
Stage - 2
1800
1320
1000
Lest find input speed for the
lowest output speed set.
For the first condition, possible
input speeds are 750 & 560.
For the second condition,
750
560
Nmin
425
Ni/p
315
Nmax
236
Ni/p
180
Stage 1
Stage 2
=
180
= 0.32
≥ 0.25
= 1.78
≤2
560
= 1000
560
The conditions are satisfied
Stage - 1
1800
1320
1000
Lest find input speed for the
lowest output speed set.
For the first condition, possible
input speeds are 1338 & 1790
For the second condition,
750
560
Nmin
425
Ni/p
315
Nmax
236
Ni/p
180
Stage 1
Stage 2
=
560
= 0.41
≥ 0.25
= 0.74
≤2
1320
= 1000
1320
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 ≅ 63
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 + 63 = 83
z12 = 63
z7 + z8 = 83
z7 = 0.76 z8
0.76 z8 + z8 = 83
z9 = 1.78 z10
z8 = 47.16 ≅ 48
z7 = 35
z9+ z10 = 83
1.78 z10+ z10 = 83
z10 = 29.79 ≅ 30
z9 = 53
Stage - 1 “First Pair - Maximum Speed Reduction”
Assume number of teeth in driver = 20
z5
z6
=
N6
N5
20
560
=
z6
1338
z6 = 47.14 ≅ 48
Stage - 1 “Second Pair – Intermediat Speed Reduction”
z1
z2
=
N2
z1
N1
z2
z1 = 0.57 z2
750
=
1338
Stage - 1 “Third Pair - Minimum Speed Increment”
z3
z4
=
N4
z3
N3
z4
z3 = 0.74 z4
1000
=
1338
Stage - 1
z1 + z2 = z3+ z4 = z5+ z6
z5 = 20
z1 + z2 = z3+ z4 = 20 + 42 = 68
z6 = 48
z3 + z4 = 68
z1 = 0.57 z2
0.76 z4 + z4 = 68
z3 = 0.74 z4
z4 = 38.64 ≅ 39
z3 = 29
z1 + z2 = 68
0.57 z2 + z2 = 68
z2 = 43.3 ≅ 44
z1 = 24
Solution
z1 = 24
z7 = 35
z2 = 44
z8 = 48
z3 = 29
z9 = 53
z4 = 39
z10 = 30
z5 = 20
z11 = 20
z6 = 48
z12 = 63