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DEPARTMENT OF MECHANICAL ENGINEERING
By
Mr. B.Balavairavan
Assistant Professor
Mechanical Engineering
Kamaraj College of Engineering and
Technology
Virudhunagar
Problem on Design of 12 Speed Gear Box
Sample Problem
Design a gearbox with 12 speeds from a source of
motor with a speed of 1600. The required speed
range is from 160 rpm to 2000 rpm.
Given:
n = 12
Nmin = 160 rpm
Nmax = 2000 rpm
Step - 1 “Calculation of Step ratio”
Referring PSG Data Book P. No : 7.20 the
calculated step ratio is not a std. value
Nmax
Nmin
= Ø n-1
2000
160
= Ø 12-1
Ø = 1.258
Since its not a std. value, Using multiples of std.
value the required step ratio is calculated
1.6 -
1.25 -
1.12 -
1.06 -
Multiples of 1.12 gives nearest value of 1.258
As 1.06 is multiplied 1 times we skip 1 speed
Hence std. Ø = 1.12 & R 20 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
= 1.238
Step - 2 “Selection of Speeds”
The obtained speeds are;
160,200,250,315,400,500,630,800,
1000,1254,1573,1973
100 112 125 140 160 180 200 224 250 280 315 355 400 450 500
560 630 710 800 900
100
0
112
0
125
4
140
4
157
3
176
2
197
3
221
0
Check for deviation
The allowable deviation and actual deviation is
calculated,
Allowable deviation = ± 10 (Ø - 1) %
Nmin
Nmax
= (Nmax actual - Nmax) x
= ± 10 (1.258 - 1) %
= ± 2.58 %
Actual deviation
160
2000
= (1973 - 2000) x
= - 2.16 %
Since the deviation is within the allowable range
we can design for standard speeds.
The selected standard speeds are;
160, 200, 250, 315, 400, 500, 630, 800,
1000, 1254, 1573, 1973
Step - 3 “ Structural formula & Ray Diagram
”
The structural formula for 12 speed gear box is
3 (1) 2 (3) 2 (6)
Stage 1 - Single input is splitted into 3 speeds
Stage 2 - 3 input is splitted into 6 speeds
ie., each input is splitted into 2 speed
Stage 3 - 6 input is splitted into 12 speeds
ie., each input is splitted into 2 speed
1973
1573
1254
1000
800
630
500
400
315
250
200
160
3 (1) 2 (3) 2 (6)
Stage 1 Stage 2 Stage 3
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
160
400
630
400
= 0.4
= 1.57
=
=
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
400
630
800
630
= 0.63
= 1.27
=
=
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
630
1573
1000
1573
= 0.40
= 0.63
=
=
Stage - 3
Stage - 2
Stage - 1
Step - 4 “ Kinematic Arrangement ”
Shaft - 1 / Input
Shaft - 2 / Intermediate
Shaft - 3 / Intermediate
1
3
42
5
6
11
13
8
7
9
10
14
12 Shaft - 4 / Output
Step - 5 “ Calculation of number of
number of teeth in gears ”
Stage - 3 “First Pair - Lowest speed o/p
Assume number of teeth in driver, z13 = 20
z13
z14
=
N14
N13
20
z14
=
160
400
z14 = 50
Stage - 3 “Second Pair
z11
z12
=
N12
N11
z11
z12
=
630
400
z11 = 1.575 z12
z11 + z12 = z13+ z14
z11 + z12 = 20+ 50 = 70
1.575 z12 + z12 = 70
z12 = 27 .18 ≅ 27
z11 = 43
Stage - 2 “First Pair - Lowest speed o/p
Assume number of teeth in driver,z7 = 20
z7
z8
=
N8
N7
20
z8
=
400
630
z8 = 31.5≅ 32
Stage - 2 “Second Pair
z9
z10
=
N10
N9
z9
z10
=
800
630
z9 = 1.269 z10
z7 + z8 = z9+ z10
20+ 32 = z9 + z10 = 52
1.269 z10 + z10 = 52
z10 = 22.9 ≅ 23
z9 = 29
Stage - 1 “First Pair - Maximum Speed Reduction”
Assume number of teeth in driver, z5 = 20
z5
z6
=
N6
N5
20
z6
=
630
1573
z6 = 49.9 ≅ 50
Stage - 1 “Second Pair”
z1
z2
=
N2
N1
z1
z2
=
800
1573
z1 = 0.50 z2
Stage - 1 “Third Pair”
z3
z4
=
N4
N3
z3
z4
=
1000
1573
z3 = 0.63 z4
Stage - 1
z1 + z2 = z3+ z4 = z5+ z6
z1 + z2 = z3+ z4 = 20 + 50 = 70
z5 = 20
z6 = 50
z1 = 0.50 z2
z3 = 0.63 z4
z3 + z4 = 70
z1 + z2 = 70
0.63 z4 + z4 = 70
0.50 z2 + z2 = 70
z2 = 46.6 ≅ 47
z4 = 42.94 ≅ 43 z3 = 27
z1 = 23
Solution
z1 = 23
z2 = 47
z3 = 27
z4 = 43
z5 = 20
z6 = 50
z8 = 32
z7 = 20
z10 = 23
z9 = 29
z11 = 43
z12 = 27
z13 = 20
z14 = 50

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12 speed gear box

  • 1. DEPARTMENT OF MECHANICAL ENGINEERING By Mr. B.Balavairavan Assistant Professor Mechanical Engineering Kamaraj College of Engineering and Technology Virudhunagar Problem on Design of 12 Speed Gear Box
  • 2. Sample Problem Design a gearbox with 12 speeds from a source of motor with a speed of 1600. The required speed range is from 160 rpm to 2000 rpm. Given: n = 12 Nmin = 160 rpm Nmax = 2000 rpm
  • 3. Step - 1 “Calculation of Step ratio” Referring PSG Data Book P. No : 7.20 the calculated step ratio is not a std. value Nmax Nmin = Ø n-1 2000 160 = Ø 12-1 Ø = 1.258
  • 4. Since its not a std. value, Using multiples of std. value the required step ratio is calculated 1.6 - 1.25 - 1.12 - 1.06 - Multiples of 1.12 gives nearest value of 1.258 As 1.06 is multiplied 1 times we skip 1 speed Hence std. Ø = 1.12 & R 20 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 = 1.238
  • 5. Step - 2 “Selection of Speeds” The obtained speeds are; 160,200,250,315,400,500,630,800, 1000,1254,1573,1973 100 112 125 140 160 180 200 224 250 280 315 355 400 450 500 560 630 710 800 900 100 0 112 0 125 4 140 4 157 3 176 2 197 3 221 0 Check for deviation
  • 6. The allowable deviation and actual deviation is calculated, Allowable deviation = ± 10 (Ø - 1) % Nmin Nmax = (Nmax actual - Nmax) x = ± 10 (1.258 - 1) % = ± 2.58 % Actual deviation 160 2000 = (1973 - 2000) x = - 2.16 %
  • 7. Since the deviation is within the allowable range we can design for standard speeds. The selected standard speeds are; 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1254, 1573, 1973
  • 8. Step - 3 “ Structural formula & Ray Diagram ” The structural formula for 12 speed gear box is 3 (1) 2 (3) 2 (6) Stage 1 - Single input is splitted into 3 speeds Stage 2 - 3 input is splitted into 6 speeds ie., each input is splitted into 2 speed Stage 3 - 6 input is splitted into 12 speeds ie., each input is splitted into 2 speed
  • 9. 1973 1573 1254 1000 800 630 500 400 315 250 200 160 3 (1) 2 (3) 2 (6) Stage 1 Stage 2 Stage 3 Nmin Ni/p ≥ 0.25 Nmax Ni/p ≤ 2 160 400 630 400 = 0.4 = 1.57 = = Nmin Ni/p ≥ 0.25 Nmax Ni/p ≤ 2 400 630 800 630 = 0.63 = 1.27 = = Nmin Ni/p ≥ 0.25 Nmax Ni/p ≤ 2 630 1573 1000 1573 = 0.40 = 0.63 = = Stage - 3 Stage - 2 Stage - 1
  • 10. Step - 4 “ Kinematic Arrangement ” Shaft - 1 / Input Shaft - 2 / Intermediate Shaft - 3 / Intermediate 1 3 42 5 6 11 13 8 7 9 10 14 12 Shaft - 4 / Output
  • 11. Step - 5 “ Calculation of number of number of teeth in gears ” Stage - 3 “First Pair - Lowest speed o/p Assume number of teeth in driver, z13 = 20 z13 z14 = N14 N13 20 z14 = 160 400 z14 = 50
  • 12. Stage - 3 “Second Pair z11 z12 = N12 N11 z11 z12 = 630 400 z11 = 1.575 z12 z11 + z12 = z13+ z14 z11 + z12 = 20+ 50 = 70 1.575 z12 + z12 = 70 z12 = 27 .18 ≅ 27 z11 = 43
  • 13. Stage - 2 “First Pair - Lowest speed o/p Assume number of teeth in driver,z7 = 20 z7 z8 = N8 N7 20 z8 = 400 630 z8 = 31.5≅ 32
  • 14. Stage - 2 “Second Pair z9 z10 = N10 N9 z9 z10 = 800 630 z9 = 1.269 z10 z7 + z8 = z9+ z10 20+ 32 = z9 + z10 = 52 1.269 z10 + z10 = 52 z10 = 22.9 ≅ 23 z9 = 29
  • 15. Stage - 1 “First Pair - Maximum Speed Reduction” Assume number of teeth in driver, z5 = 20 z5 z6 = N6 N5 20 z6 = 630 1573 z6 = 49.9 ≅ 50
  • 16. Stage - 1 “Second Pair” z1 z2 = N2 N1 z1 z2 = 800 1573 z1 = 0.50 z2
  • 17. Stage - 1 “Third Pair” z3 z4 = N4 N3 z3 z4 = 1000 1573 z3 = 0.63 z4
  • 18. Stage - 1 z1 + z2 = z3+ z4 = z5+ z6 z1 + z2 = z3+ z4 = 20 + 50 = 70 z5 = 20 z6 = 50 z1 = 0.50 z2 z3 = 0.63 z4 z3 + z4 = 70 z1 + z2 = 70 0.63 z4 + z4 = 70 0.50 z2 + z2 = 70 z2 = 46.6 ≅ 47 z4 = 42.94 ≅ 43 z3 = 27 z1 = 23
  • 19. Solution z1 = 23 z2 = 47 z3 = 27 z4 = 43 z5 = 20 z6 = 50 z8 = 32 z7 = 20 z10 = 23 z9 = 29 z11 = 43 z12 = 27 z13 = 20 z14 = 50