B.Balavairavan AP/Mech KCET ME 6601 -Design of Transmisson systems

1. DEPARTMENT OF MECHANICAL ENGINEERING
By
Mr. B.Balavairavan
Assistant Professor
Mechanical Engineering
Kamaraj College of Engg and Tech
Virudhunagar
Design of NINE Speed Gear Box

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

3. Step - 1 “Calculation of Step ratio”
Refer PSG Data Book P. No : 7.20 to check
whether, the calculated step ratio is a std.
Nmax
Nmin
= Ø n-1
1800
180
= Ø 9-1
Ø = 1.333

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 1.06x 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
100
0
106
0
112
0
118
0
125
0
132
0
140
0
150
0
160
0
170
0
180
0
190
0
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;
180,236,315,425,560,75
0,1000,1320,1800
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
Stage - 2
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
180
560
1000
560
= 0.32
= 1.78
=
=

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

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”
Assume number of teeth in driver = 20
z11
z12
=
N12
N11
20
z12
=
180
560
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”
Assume number of teeth in driver = 20
z5
z6
=
N6
N5
20
z6
=
560
1338
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