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
