ME 6601 -DESIGN OF TRANSMISSION SYSTEMS DESIGN OF 14 SPEED GEAR BOX AND PROBLEMS

1. By
Mr. B.Balavairavan
Assistant Professor
Mechanical Engineering
Kamaraj College of Engg and Tech
Virudhunagar
PSG Design Data Book
DEPARTMENT OF MECHANICAL ENGINEERING
Problem on Design of 14 Speed Gear Box

2. Sample Problem
A 14 speed gear boxis required to furnish output
speeds in the range of 125 to 2500 rpm. Draw the
speed diagram and kinematic arrangement.
Given:
n = 14
Nmin = 125 rpm
Nmax = 2500 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
2500
125
= Ø 14 -1
Ø = 1.259

4. • Since Ø is 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 1.12 is multiplied 1 time, so 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,
1000,1250,1600,2000,2500
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
0
140
0
160
0
180
0
200
0
224
0
250
0
The speeds are withing the range hence no need to
check for deviation

6. Step - 3
“ Structural formula & Ray Diagram ”
The structural formula for 14 speed gear box is
3 (1) 3 (3) 2 (5)

7. 3 (1) 3 (3) 2 (5)
Stage 1 Stage 2 Stage 3 Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
125
315
400
315
= 0.4
= 1.27
=
=
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
315
800
1250
800
= 0.39
= 1.56
=
=
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
800
1600
1250
1600
= 0.5
= 0.78
=
=
Stage - 3
Stage - 2
Stage - 1
2500
2000
1600
1250
1000
800
630
500
400
315
250
200
160
125

8. Step - 4 “ Kinematic Arrangement ”
Shaft - 1 / Input
Shaft - 2 / Intermediate
Shaft - 3 / Intermediate
1
3
42
5
6
13
15
8
7
9
10
16
14 Shaft - 4 / Output
11
12