Design of gear box for Machine Tool Application (3 stage & 12 speed ) by Sagar Dhotare

Design of Gear Box
for
M/c Tool Application
(3-Stage & 12 Speeds)
Presented by
Prof. Sagar A. Dhotare,
Assistant Professor,
ViMEET, Khalapur

Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Introduction of Gear Box
for Machine Tool Application
• Gear box is a mechanism used for transmitting a power from prime mover to the machine with change in torque
and speed.
• Machine tool expect to perform various operations such as turning, facing, milling, boring etc. For best
performance of each operation is depend on given cutting speed, feed and depth of cut.
• Based on the diameter to be machined of the job/workpiece, different spindle speeds are necessary to determine
optimum cutting speed and feed rates.
• Different possible spindle speeds for different diameter of the job need to be generated by multi-speed gear box.
This multi-speed gear box is used for stepped regulation of speeds.
Types of Gear Boxes
1. Sliding Mesh Gear Box.
2. Constant Mesh Gear Box.
3. Synchromesh Gear Box.

1. The speed range between to increments should be 10 to 15%. A large range, would further reduce the
optimum use of the tool.
2. A no. of speed ratio should be available to the operator without stopping the machine.
3. The operator should able to change speeds without going through any intermediate sequence.
4. Due to compact system, axial length of the gear train is the least, if the gear cluster are made sliding.
But, the sliding gear cluster should not have more than three gears.
5. Only the gears which are used to achieve the gear ratio should be rotating, whereas other gears of the
gear box should not be engaged.
6. The no. of shaft, gears and operating levers should be minimum. Lesser the no. of components,
machine reliability is higher.
7. Control unit should be designed ergonomically. It should be centralised and should have all possible
controls at one place.

The ideal speed range in a machine tool would be the one were there would be infinity no. of speed ratios.
This would allow an optimum operation of a machine for any job diameter. Such a speed regulation would be called as
a Stepless Regulation.
Higher the speed ratio available, higher would be the cost of the gear box. Hence, from the cost and utility
point of view, the optimum gear ratios should be available to get the different speed ranges between the maximum and
minimum speed. These intermediate discrete values can be obtained by different laws.
Let Nmin. and Nmax. be the minimum and maximum spindle speeds to be achieved in ‘z’ no. of intermediate
speed steps.
These steps can be arranged by the following laws :
1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)

In Arithmetic Progression (AP) the difference between two adjacent speed steps is constant.
Various intermediate speeds can be found using following sequence :
𝑵𝟏 = 𝑵𝒎𝒊𝒏
𝑵𝟐 = 𝑵𝟏 + 𝒂 = 𝑵𝒎𝒊𝒏 + 𝒂
Where, a - Difference between two adjacent speed steps,
which is constant.
𝑵𝟑 = 𝑵𝟐 + 𝒂 = 𝑵𝒎𝒊𝒏 + 𝟐𝒂
𝑵𝒌 = 𝑵𝒌−𝟏 + 𝒂 = 𝑵𝒎𝒊𝒏 + (𝒌 − 𝟏)𝒂
𝑵𝒌+𝟏 = 𝑵𝒌 + 𝒂 = 𝑵𝒎𝒊𝒏 + 𝒌𝒂
𝑵𝒛 = 𝑵𝒛−𝟏 + 𝒂 = 𝑵𝒎𝒊𝒏 + 𝒛 − 𝟏 𝒂 = 𝑵𝒎𝒂𝒙
𝒂 =
𝑵𝒎𝒂𝒙 − 𝑵𝒎𝒊𝒏
𝒛 − 𝟏
Where, z – no. of steps.

𝑫𝒌 =
𝟏𝟎𝟎𝟎 𝑽
𝝅 𝑵𝒌
Considering any two intermediate speed ranges Nk and Nk+1. Corresponding to these, the diameters
that can be machined would be as follows,
Upper Limit in mm
𝑫𝒌+𝟏 =
𝟏𝟎𝟎𝟎 𝑽
𝝅 𝑵𝒌+𝟏
Lower Limit in mm
The diameter range that can be accommodated at a particular speed would be as follows,
∆𝑫𝒌 = 𝑫𝒌 − 𝑫𝒌−𝟏 =
𝟏𝟎𝟎𝟎 𝑽
𝝅
𝟏
𝑵𝒌
−
𝟏
𝑵𝒌+𝟏

Example of AP :-
Considering a gear box to be designed for a minimum and maximum speed of 40rpm to 400rpm in
6 steps. Consider the optimum velocity for the tool be 20mmin.
𝑵𝟏 = 𝑵𝒎𝒊𝒏 = 40 rpm
𝑵𝒛 = 𝑵𝟔 = 𝑵𝒎𝒂𝒙 = 400 rpm
z = 6
V = 20 mmin.
𝒂 =
𝒛−𝟏
=
400 −40
6−1
= 72rpm
Let the gear ratio between adjacent speeds be ϕ.
Hence,
ϕ𝑘
=
𝑁𝑘+1
𝑁𝑘
Sr.
No.
N (rpm) ϕ𝑘 =
𝑁𝑘+1
𝑁𝑘
𝐷𝑘 ∆𝐷𝑘
1 40 (112/40) = 2.8 159.15 102.3
2 40 + 72 = 112 (184/112) = 1.64 56.84 22.2
3 112 + 72 = 184 (256/184) = 1.4 34.6 9.73
4 184 + 72 = 256 (328/256) = 1.28 24.87 5.46
5 256 + 72 = 328 (400/328) = 1.22 19.4 3.5
6 328 + 72 = 400 - 15.9
The diameter and ranges with their gear ratio are
calculated as follows(table),
Note : In the Arithmetic Progression, it is seen that at low speed steps, more speed ranges are desired than
what is available. Where as at high speed steps, some of the speed ranges available are not necessary.

In Geometric Progression (GP) the difference between two consecutive speed ranges is same.
Considering the similar example as discussed for AP
𝑵𝟐 = 𝑵𝒎𝒂𝒙
As per definition of GP
𝑁2
𝑁1
=
𝑁3
𝑁2
=
𝑁4
𝑁3
= … =
𝑁𝑘+1
𝑁𝑘
=
𝑁𝑧−1
𝑁𝑧
= ϕ
𝑁𝑧
𝑁1
=
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
= ϕ
ϕ =
𝑵𝒎𝒂𝒙
𝑵𝒎𝒊𝒏
𝟏
(𝒛−𝟏)

Example of GP :-
z = 6
V = 20 mmin.
Sr.
No. N (rpm) ϕ𝑘 =
𝑁𝑘+1
𝑁𝑘
1 40 1.585 159.15 58.73
2 40 X 1.585 = 63.4 1.585 100.4 37.05
3 63.4 X 1.585 = 100.5 1.585 63.33 23.4
4 100.5 X 1.585 = 159.6 1.585 40 14.78
5 159.6 X 1.585 = 252.5 1.585 25.22 9.3
6 252.5 X 1.585 = 400 - 15.9
The diameter and ranges with their gear ratio are
calculated as follows(table),
Note : In the Geometric Progression, it is seen that at low speed steps, still more speed ranges are desired
than what is available. Where as at high speed steps, some of the speed ranges available are not necessary.
ϕ =
𝑵𝒎𝒂𝒙
𝑵𝒎𝒊𝒏
𝟏
(𝒛−𝟏)
ϕ =
400
40
1
(6−1)
ϕ = 1.585

In Harmonic Progression (HP) the diameter changes between two consecutive speed ranges remains same.
Considering the similar example as discussed for AP and GP
∆𝑫𝒌 =
𝟏𝟎𝟎𝟎 𝑽
𝝅
𝟏
𝑵𝒌
−
𝟏
𝑵𝒌+𝟏
= Constant
𝟏
𝑵𝒌
−
𝟏
𝑵𝒌+𝟏
= C
𝟏
𝑵𝒌+𝟏
=
𝟏
𝑵𝒌
- C =
𝟏 −𝑪𝑵𝒌
𝑵𝒌
𝑵𝒌+𝟏 =
𝑵𝒌
𝟏−𝑪𝑵𝒌
𝑵𝟐 =
𝑵𝟏
𝟏 − 𝑪𝑵𝟏
𝑵𝟑 =
𝑵𝟐
𝟏 − 𝟐𝑪𝑵𝟏
𝑵𝒌 =
𝑵𝟏
𝟏 − 𝒌 − 𝟏 𝑪𝑵𝟏
𝑵𝒌 =
𝑵𝒎𝒊𝒏
𝟏 − 𝒌 − 𝟏 𝑪𝑵𝒎𝒊𝒏
𝑵𝒛 =
𝑵𝒛−𝟏
𝟏 − 𝒛 − 𝟏 𝑪𝑵𝟏
𝑵𝒛 =
𝑵𝒎𝒊𝒏
𝟏 − 𝒛 − 𝟏 𝑪𝑵𝒎𝒊𝒏
= 𝑵𝒎𝒂𝒙
𝑪 =
𝒛 − 𝟏 𝑵𝒎𝒊𝒏 𝑵𝒎𝒂𝒙
ϕ𝒌
=
𝑵𝒌+𝟏
𝑵𝒌
=
𝟏
𝟏−𝑪𝑵𝒌

Example of HP :-
z = 6
V = 20 mmin.
Sr.
No.
ϕ𝑘
=
𝑁𝑘+1
𝑁𝑘
1 40 (48.78 / 40) = 1.22 159.15 28.65
2 40
1 − 4.5 × 10−3 × 40
= 48.78 (62.5 / 48.78) = 1.28 130.5 28.65
3 48.78
1 − 4.5 × 10−3 × 48.78
= 62.5 (86.96 / 62.5) = 1.39 101.86 28.65
4 62.5
1 − 4.5 × 10−3 × 62.5
= 86.96 (142.86 / 86.96) = 1.643 73.21 28.65
5 86.96
1 − 4.5 × 10−3 × 86.96
= 142.86 (400 / 142.86) = 2.8 44.56 28.65
6 142.86
1 − 4.5 × 10−3 × 142.86
= 400 - 15.9 -
The diameter and ranges with their gear ratio are calculated as follows(table),
Note : In the comparison with AP & GP series, the HP gives a better distribution of speeds in the lower speed
range. At higher speeds, the values are spaced too wide apart which results in non-optimum use of tool.
𝑪 =
=
400 − 40
6 − 1 𝑋 400 𝑋 40
𝑪 = 4.5 X 10−3
𝑁𝑘 =
𝑁𝑚𝑖𝑛
1 − 𝑘 − 1 𝐶𝑁𝑚𝑖𝑛

Sr.
No. Arithmetic Progression (AP) Geometric Progression (GP) Harmonic Progression (HP)
1
Constant Change in speed at every stage
is constant
Constant gear ratio at every stage
Constant diameter change at every
stage
2
𝒂 =
𝒛 − 𝟏
ϕ =
𝑵𝒎𝒂𝒙
𝑵𝒎𝒊𝒏
𝟏
(𝒛−𝟏) 𝑪 =
3
𝑵𝒌 = 𝑵𝒎𝒊𝒏 + (𝒌 − 𝟏)𝒂 𝑵𝒌 = 𝑵𝒎𝒊𝒏 ϕ(𝒌−𝟏)
𝑵𝒌 =
𝑵𝒎𝒊𝒏
𝟏 − 𝒌 − 𝟏 𝑪𝑵𝒎𝒊𝒏
4 Low speed zone requires much more steps
for optimised machining.
Low speed zone, better than AP, but still
needs few more steps for optimised
machining.
High speed zone requires much more
steps for optimised machining.
5 Not based on a preferred no. of series. Based on preferred no. of series Not based on a preferred no. of series.
A. Constant loss of Economic cutting speed in the whole rpm range
B. Better gear box design feature
Advantages of GP

𝑅𝑁 =
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
𝑁𝑚𝑎𝑥 =
1000 × 𝑉
𝑚𝑎𝑥
𝜋 𝐷𝑚𝑖𝑛
𝑁𝑚𝑖𝑛 =
1000 × 𝑉𝑚𝑖𝑛
𝜋 𝐷𝑚𝑎𝑥
𝑅𝑁 =
𝑉
𝑚𝑎𝑥
𝑉𝑚𝑖𝑛
×
𝐷𝑚𝑎𝑥
𝐷𝑚𝑖𝑛
= 𝑅𝑣 × 𝑅𝐷
The ratio 𝑅𝑁 is defind as ratio of maximum spindle speed to minimum spindle speed
• Fixing the limit on higher cutting speeds results in loss of productivity.
• Machining operation can be done at higher speeds, but they are not available with machine.
• At the same time very high limits of upper cutting speed is not practical, nor economical from machine
building point of view.
• The lower cutting speed is determined by the considering that below a particular cutting speed, the tool life
reduces.
• In general purpose machine such as lathes , boring and milling machines, the magnitude of 𝑅𝑁 is large to
account of different of jobs that can be machined on them.

To find the geometric progression ratio ∅ we refer Preferred Number Series. This is because following resons;
1. Each size is larger than the preceding size by a fixed percentage.
2. Small size will differ from each other by small amounts and large sizes will differ from each other by larger
amount
The series of numbers have been standardised in order that the various
sizes of a series can be determined in an orderly fashion and these are
called as Preferred Number which shows in tabular form
Generally ∅ 𝑖𝑠 𝑖𝑛 𝑟𝑎𝑛𝑔𝑒 𝑓𝑟𝑜𝑚 1 𝑡𝑜 2
ϕ =
𝑵𝒎𝒂𝒙
𝑵𝒎𝒊𝒏
𝟏
(𝒛−𝟏)
= 𝑹𝑵
𝟏
(𝒛−𝟏)

The values obtained from above equation may not be a whole no. and hence it should be rounded off to the
nearest whole no. with a preferred given to the no.
𝑧 =
log(𝑅𝑁 ∅)
log ∅
Range of z z Rounded Value
5 to 6.49 6 (2 × 3)
7 to 8.49 8 (2 × 2 × 2) = 23
8.5 to 10 9 (3 × 3) = 32
10 to 13 12 (2 × 2 × 3) = 22
× 3
13 to 16 16 (2 × 2 × 2 × 2) = 24
Generally the value of z are selected on the basis of following equation,
𝑧 = 2𝑛1 × 3𝑛2
𝑛1- No. stages having two speed steps
𝑛2- No. stages having three speed steps
Total no. of steps (n) = 𝑛1+ 𝑛2
Standard Values of z and n
z 𝟐𝒏𝟏 × 𝟑𝒏𝟐 𝒏𝟏 𝒏𝟐 n
2 21 × 30 1 0 1
4 22 × 30 2 0 2
6 21
× 31
1 1 2
8 23
× 30
3 0 3
9 20 × 33 0 2 2
12 21
× 32
2 1 3

It can be defind as an expression which gives the distribution of the no. of transmission stages and the
difference between the adjacent speed step in each of these stages.
Following is the procedure to calculate Structural Formula,
Consider “z” be the total no. speed steps
Let 𝑝1, 𝑝2,𝑝𝑛 are the no. of speed steps in first, second and nth stage of transmission.
z = 𝑝1 × 𝑝2 × 𝑝3 × ………. × 𝑝𝑛 where n – no. of stages of the gear box
Consider a case where we analyse the difference between two adjacent speeds ‘x’ in a given stage, x is referred
to as the characteristics of transmission stage.
Consider any two adjacent speeds available in this stage. If the difference between these two speeds is 1, then
x = 1, where as if the value is 2, then x = 2. Thus for each stage there will be particular value of ‘x’.
As the speed of the o/p shaft is GP, there must be one transmission stage which would give x = 1. This group is called as
Main Transmission group and has a progression ratio of ∅𝒙 = ∅𝟏 = ∅ .
The next group would have 𝒙𝟐 = 𝒑𝟏 and the progression ratio of ∅𝒙𝟐 = ∅𝒑𝟏
Next transmission group, 𝒙𝟑 = 𝒑𝟏𝒑𝟐 and the progression ratio of ∅𝒙𝟑 = ∅𝒑𝟏𝒑𝟐

The values of 𝑥1, 𝑥2, 𝑥3, … . . , 𝑥𝑛 can be calculated as,
𝑥1 = 1
𝑥2 = 𝑝1
𝑥3 = 𝑝1 𝑝2
.
.
𝑥𝑛 = 𝑝1 × 𝑝2 × 𝑝3 × ⋯ × 𝑝𝑛−1
So final structural formula give as,
𝒛 = 𝒑𝟏(𝒙𝟏) 𝒑𝟐(𝒙𝟐) 𝒑𝟑(𝒙𝟑) ………….. 𝒑𝒏(𝒙𝒏)

The representation of structural formula in the form of special graph is called as Structure Diagram.
Which describes the kinematic structure of the gear box.
z = 2(1) 2(2) z = 2(2) 2(1)
Rules for drawing structure diagram:-
1. If the no. of stages in a gear box are ‘n’, draw
‘n+1’ vertical lines spaced at a convenient
distance.
2. 1st vertical line indicate I/P shaft and last line
indicates O/P shaft. Other vertical lines
indicates intermediate shafts which forms
transmission groups or stages of the gear
box.
3. ‘m’ horizontal lines are drawn to represent
total no. of speeds steps of the gear box.
Spacing between the horizontal lines is taken
as log∅.
4. Using structural formula
𝒛 = 𝒑𝟏(𝒙𝟏) 𝒑𝟐(𝒙𝟐).. 𝒑𝒏(𝒙𝒏)
Plot 𝑝1(𝑥1) 𝑝2(𝑥2) etc
𝑝1 = 2 𝑥1 = 1 𝑥2 = 𝑝1 = 2
Z=4 n=2

Flow of transmission to get the 4 speeds
Table shows Preferred structural formula
Few Structure diagram combinations

To Designer what information get from Structural Diagram :-
1. The no. of shafts in the gear box.
2. The no. of gears on each shaft. The approximate no. of gears would be found by following equation:
𝐺 = 2 × 𝑝1 + 𝑝2+. . . . +𝑝𝑛
3. The order of changing transmissions at each stage to get the required spindle speed.
4. The transmission range and the characteristics of each group
Characteristics of a structure diagram are as follows:
1. Each line must connect another follow-up line till the final O/P shaft is reached.
2. At any stage the arrow must terminate at one and only point. However at any input there can be more than one
arrow. Generally this no. is limited to a maximum of three.
3. On the final shaft, all the points must be connected by arrows.
4. No arrow should go beyond the bounds of the speed required speeds at the final O/P shaft.

Structural diagram gives the information about the transmission flow only not about the speed. To
determine the speeds of various gears, the ray diagram and speed diagram are drawn. This diagram allows
the designer to read the speeds of all shafts gears and finally spindle speeds.
Following are the steps to draw the ray and speed diagram:-
1. If the no. of steps in a gear box is ‘z’, and the no. of stages of the gear box is ‘n’, then draw ‘n+2’ vertical lines at
a convenient distance. The one extra line as compared to the structural diagram would represent the motor shaft.
2. If the I/P speed from motor is less then the maximum speed, draw ‘z’ horizontal lines, where as if the I/P speed
from motor is more then the maximum speed, draw as many horizontal lines as necessary to locate the motor
speed on the diagram . The spacing between the horizontal lines take as log ∅ , like structure diagram. Thus,
If the 𝑁𝑚 < 𝑁𝑚𝑎𝑥 , z no. of Horizontal lines and
If the 𝑁𝑚 > 𝑁𝑚𝑎𝑥 , z no. of Horizontal lines + additional lines to locate 𝑁𝑚 on the chart.
3. Using the value GP ratio '∅′
, mark the spindle speeds on the vertical shaft. Mark the various spindle spedds
starting with minimum speed at the lowermost line.
4. Draw the ray depicting transmission between the last shft and the shaft preceding it. The ray are drawn for the
lowest speed of the last shaft.

After completion of structure and speed diagram, a rough layout of the gears should be drawn. This layout is
used to determine the no. of teeth on each gear based on speed calculation referring speed diagram. The gears,
shafts and other component are designed on the basis of strength consideration.
While preparing the gear layout, the following rules should be consider :
1. Minimum Number of Teeth on Gears.
2. Sum of Number of Teeth on Gear Pairs for Parallel Shafts Should be The same.
3. Check on Percent Deviation of Actual and Theoretical Speeds.
4. The spacing between to adjacent Gears should be proper.

Thank You

Design of gear box for Machine Tool Application (3 stage & 12 speed ) by Sagar Dhotare

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Design of gear box for Machine Tool Application (3 stage & 12 speed ) by Sagar Dhotare