Design and Analysis of an 8-Speed Gear box.

International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 10 Issue: 04 | Apr 2023 www.irjet.net p-ISSN: 2395-0072
© 2023, IRJET | Impact Factor value: 8.226 | ISO 9001:2008 Certified Journal | Page 678
Design and Analysis of an 8-Speed Gear box.
Abhishek Chavan1, Abhiraj Baji2, Yash Darakh3, Chinmay Inamdar4
145Department of Mechanical Engineering, Vishwakarma Institute of Technology, Pune-48, Maharashtra, India
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Abstract - Lathe Machine is used in all the engineering
applications and in the college Workshops. Lathe machine is
used to perform all the basic operations such as drilling,
cutting, tapping, turning, etc. with the help of different tools
placed in the work environment There are 2 main types of
Lathe machines: Capstan Lathe and Turret Lathe. Turret
Lathes use a turret head to perform operations. A turret head
is a vertical cylindrical revolving tool-holder for bringing
different tools into action successively in the lathe machine.
The turret head needs to rotate at different speeds to perform
different operations on the workpiece. This is controlled by a
gearbox. A gearbox which converts high speed input into
various output speeds is called a multi speed gearbox. A 6-
speed gearbox is usually used. But 8-speed and 9-speed
gearboxes give a wider range of speeds and efficiency. The
aim of this project is to design and analyze 8-speed
gearbox which meets the functional requirements of a
turret gearbox. Design and Analysis have been conducted
and a detailed study has been performed for various
components in the gearbox.
Key Words: Lathe, Gearbox, Turret, Gears, Shaft, Ray
Diagram, etc.
1. INTRODUCTION
Ever since the development of the first gearbox by
Sturtevant in 1904 for automobileapplication,theutilisation
of this technology has evolved marking its presence in
various industrial sectors like power generation, petroleum
refining, mining and material handling,metal processingand
so on. Considering the metal processing and machining
aspect, a wide range of industrial gearboxesareusedinlathe
machines. These gearboxes offer a vivid range of operating
speed, and the main aim of this project is to design an 8-
Speed gearbox for using in turret lathe machines.
1.1 Deciding Input Requirements
Speed Requirement: Aftera comprehensivestudyonspeed
ranges (Nmin and Nmax) for turret lathe, we arrived at a
generic average value of 100rpm to 1100rpm. These values
are optimal for further processing and give a better design
result as compared to other experimental values as well.
Torque Requirement: Turret lathe gearboxes usually
require a power of approximately 5-8kW [2].
Motor Selection: Motor selection has been done taking
power and input speed conditions into account. An
approximate 8.5HP(6.33kW) and 650rpm motor has been
selected. The values will be explained further on in the
project.
1.2 Deciding Material
After trial and error with various materials in the design
process and opinions of industry professionals,materialsfor
the following components were decided mainly based on
strength and durability.
Gear: All the gear are made from ‘Grey Cast Iron’.
Shaft: All the shafts have been made of the material ‘Steel
FeE 580’.
Keys and keyways: All the keys have been made out of
‘Steel 50C4.’
Casing: The casing has been made from ‘Cast Iron’.
2. DETAILED DESIGN PROCEDURE
2.1 Derivation of Structural Formula
The structural formula gives information about various
aspects of the gearbox like number of shafts, number of
gears and a basic idea of transmission pairs. It tells about
transmission range as well. The procedurefollowedtoreach
the structural formula is explained below.
Law of Discretizing Speeds: Discretizing speeds between
minimum speed and maximum speed is very important to
know our output speeds. These speeds are decided by three
basic laws as depicted in the table below.
Fig -1: Laws of discretizing speeds.

Justification of using Geometric Progression: The speeds
in gear boxes can be arranged in arithmetic progression
(A.P.), geometric progression (G.P.), harmonic progression
(H.P), and logarithmicprogression(L.P.).However,when the
speeds are arranged in G.P., it has the following advantages.
 The speed loss is minimum i.e., Speed loss=Desired
optimum speed – Available speed
 The number of gears to be employed is minimum.
 G.P. provides a more even rangeofspindlespeedsat
each step.
 The layout is comparatively very compact.
 Productivity of a machining operation, i.e., surface
area of the metal removed in unit time, is constant
in the whole speed range.
 G.P. machine tool spindle speeds can be selected
easily from preferred numbers. Because preferred
numbers are in geometric progression.
Deciding Speed Ratio from Standard Series: In gearbox
design a set of preferred step ratio or preferred numbers is
used to obtain the series of output speed of gearbox. The
preferred step ratio is mentioned as basic series named as
R5, R10, R20, R40 and R80. Each basic series has a specific
step ratio. In our case, we have Nmin as 100rpm and Nmax as
1100rpm.
Thus,
The Step Ratio 1.4 comes under R20 series.
Thus, the spindle speeds calculated are:
N1 = = 100rpm
N2 = = 140rpm
N3 = = 196rpm = 200rpm
N4 = = 274.4rpm = 280rpm
N5 = = 392rpm = 400rpm
N6 = = 560rpm
N7 = = 784rpm = 780rpm
N8 = = 1097rpm = 1100rpm
These are subject to change based on number of teeth
chosen.
Deriving Structural Formula:
Let,
n = Number of speeds available at the spindle.
p1, p2, p3….. = stage numbers in gearbox.
X1, X2, X3….. = Characteristics of the stages.
Then, structural formula is given as:
N = p1(X1).p2(X2).p3(X3).p4(X4)
Were,
X1 = 1; X2 = p1; X3 = p1 x p2; X4 = p1 x p2 x p3
Table -1:
SR NO: Number of speeds Preferred Structural Formula
1 6 Speeds 3 (1) . 2 (3)
2 (1) . 3 (2)
2 8 Speeds 2 (1) . 2 (2) . 2 (4)
4 (1) . 2 (4)
3 9 Speeds 3 (1) . 3 (3)
4 12 Speeds 3 (1) . 2 (3) . 2 (6)
2 (1) . 3 (2) . 2 (6)
2 (1) . 2 (2) . 3 (4)
For 8 – Speed Gearbox:
2 x 2 x 2 = 8
Therefore,
P1 = 2, X1 = 1
P2 = 2, X2 = 2
P3 = 2, X3 = 2 x 2 = 4
Then, structural formula = 2 (1) . 2 (2) . 2 (4)
Determining number of Shafts, Gears and
Transmissions: Calculation of No. of shafts to draw
kinematic arrangement:
No. of shafts = No. of stages + 1.
That is,
If 3 stages are there in gearbox design, then no. of shaft will
be = 3 + 1 = 4 shafts.
As per gear kinematics,
Shaft 1 = p1 gears = 2 gears.
Shaft 2 = (p1 + p2) gears = 4 gears.
Shaft 3 = (p2 + p3) gears = 4 gears.
Shaft 4 = p1 gears = 2 gears.
Limiting Value of (Ig)stage:
2.2 Structural Ray Diagram
Drawing Structural Ray Diagram: The ray diagram is
graphical representationofthedrivearrangementingeneral
from. In other words, the ray diagram is a graphical
representation of the structural formula, as shown in figure.
It provides the following data on the drive:
 The number of stages (a stage is a set of gear trains
arranged on two consecutive shafts).

 The number of speeds in each stage.
 The order of kinematic arrangement of the stages.
 The specific values of all the transmission ratios in
the drive.
 The total number of speeds available at the spindle.
Procedure: The following is the procedure to plot
structural ray diagram:
 In this diagram, shafts are shown by vertical
equidistant and parallel lines.
 The speeds are plotted vertical on a logarithmic
scale with log  as a unit.
 Transmission engaged at definite speeds of the
driving and driven shafts are shown on the diagram
by rays connecting the points on the shaft lines
representing these speeds.
 Figure shows the ray diagram for a 9-speed gear
box, having the structural formula, Z = 2 (1). 2 (2).
2(4).
Fig -2: Structural Ray Diagram
2.3 Speed Ray Diagram:
Significance: A speed ray diagram is a diagram depicting all
possible exact values of speeds at which the multi speed
gearbox can run. It gives us a clear picture of gear ratios as
well.
Plotting Speed Diagram: A basic procedure which is
followed is as follows-
 Draw u+1 vertical line atconvenientdistance where
u= Number of stages (for above formula u=4 three
transmission group) Note : first vertical line
represents the transmission from motor shaft and
the rest represent the transmission groupsofspeed
box. 2).
 Draw array of horizontal lines is equal to the
number of speed steps z of speed intersecting the
vertical lines distance of log Ø from each other.
Fig -3: Speed Ray Diagram
2.4 Preliminary Gearbox assembly:
Basic Assembly Diagram Optimisation: The following
diagram shows the basic gearbox assembly where Gn is the
number of the nth gear and Zn is number of teeth of Gn
th gear.
Fig -4: Preliminary Gearbox assembly
2.5 Determining Gear Ratios and Number of teeth:
Approximating Gear Ratio from Speed Ratio and
Constraints: As we have seen in the picture above,thereare
12 gears of which we need to find the numberofteeth.There

are a few constraints to take care of after which we can
easily select number of teeth from the design data book.
The summation of number of teeth of the meshing gears
between two shafts should be exactly equal.
m = constant for all
thus,
thus,
similarly,
and
Therefore, by selecting standard number of teeth from
design data book to satisfy the above given gear ration
equation, we get the following number of teeth per gear:
Table -2:
Gear Number Number of Teeth
01 Z01 = 20
02 Z02 = 25
03 Z03 = 45
04 Z04 = 40
05 Z05 = 24
06 Z06 = 34
07 Z07 = 34
08 Z08 = 24
09 Z09 = 20
10 Z10 = 40
11 Z11 = 40
12 Z12 = 20
Corrected final speeds:
Old Speed
(rpm)
New Speed
(rpm)
Difference
(rpm)
% Change
100 98 2 2%
140 137 3 2.1%
200 196 4 2%
280 276 4 1.4%
400 392 8 2%
560 551 9 1.6%
780 787 7 0.8%
1100 1106 6 0.5%
2.6 Determining final gear dimensions:
Determination of module:
Fig - 5: Determination of Module
By our calculations, we got a module of 2.8. Rounding this
off, we chose a module of 3mm.
Module m = 3mm.
Determination of Gear diameter:
Gear Number Diameter of Gear (mm)
1 D1 = m*Z1 = 60
2 D2 = m*Z2 = 75
3 D3 = m*Z3 = 135
4 D4 = m*Z4 = 120
5 D5 = m*Z5 = 72

6 D6 = m*Z6 = 102
7 D7 = m*Z7 = 102
8 D8 = m*Z8 = 72
9 D9 = m*Z9 = 60
10 D10 = m*Z10 = 120
11 D11 = m*Z11 = 120
12 D12 = m*Z12 = 60
Determination of Gear dimensions:
We have selected 20 full depth involute teeth system. As we
input our module, system of teeth and number of teeth into
Solid works, it gives us the remaining outputs.
Fig - 6: Types of teeth system
By considering 20-degree full depth system:
 Addendum (ha) = m = 3mm
 Dedendum (hf) = 1.25 m = 3.75mm
 Clearance (c) = 0.25 m = 0.75mm
 Working depth (hk) = 2 m = 6mm
 Whole depth (h) = 2.25 m = 6.75mm
 Tooth thickness (s) = 1.5708 m = 4.7mm
 Tooth space = 1.5708 m = 4.7mm
 Filet radius = 0.4 m = 1.2mm
Weights of individual gears:
Post obtaining the dimensions required for each gear, we
design them on modelling software and then obtain the
weights as follows:
Sr. No. Gear with teeth Weight(N)
1 20 4.28
2 24 6.76
3 25 7.595
4 34 14.98
5 40 21.98
6 45 28.616
Design of shaft:
Torque input to individual shafts: The selected motor
provides an input power of 6.33kW to the initial shaft. The
first shaft rotates at 625rpm. From the formula,
We get Torsional moment of first shaft T1 = 96.71 N-m.
Similarly, finding out all the possible torsional moments of
all shafts, we consider the maximum one for calculation
purposes. The results are tabulated as follows:
The torques marked in red are the highest torques acting on
the respective shafts. This be our torsional moments to be
considered during design.
Calculation and Direction of Gear Forces: There are 3
main types of forces acting when a gear pair meshes. They
are as follows:
 Tangential Force Ft: The force acting tangential to
the gear surface at the point of contact with
meshing gear.
 Radial force Fr: The force actingtowardsthecentre
of the gear at the point of contact with the meshing
gear.
 Axial Force Fa: The force that acts in thez-direction
along the shaft due to the angle of the teeth is the
axial force. As our teeth don’t have an angle with
respect to the gear base, this force for our
consideration, is 0.
As our gearbox is a spur gear system, we will consider the
tangential and radial forces of all gears for calculation
purposes.
For each gear

Gear Force Type Value
(N)
G1 Tangential force Ft 3223.6
Radial Force Fr 1173.3
Modified Weight of Gear: From Solid works, we have the
weight of the solid gear. As there will be a shaft of diameter
‘d’ through this, we must reduce that much weight.
Therefore, giving a new modified weight to work with.
Let,
Wnew = Modified Weight in Newton
Wold – Solid works output weight in Newton
Wr - Weight of Inner part to be removed in Newton.
d – diameter of shaft in m
f – face width of gear in m
ρ – density of gear in kg/m3
g – acceleration due to gravity in m/s2
Fig -1: Name of the figure
Wnew = Wold – Wr
Wr = Weight of cylindrical part of gear
Wr =
Calculating Maximum Bending Moment for each Shaft:
Our aim right now is to calculate maximum bendingmoment
acting on the shaft for the calculation of shaft diameter. This
can be done in terms of ‘d’ as all values in our SFD are in
terms of ‘d’.
The maximum bending moment isarea betweentheSFDand
the shaft axis above the shaft axis and the point of maximum
bending is the point where the SFD graph crosses the shaft
axis. Thus, calculating the areas, we get the followingresults.
Shaft Maximum Bending
Moment (Mb)
Maximum Torsional
Moment (Mt)
1 252261.7 346544.50
2 282504.30 264387.4
3 294134.50 357548.00
4 263150.23 346753.82
Determining Shaft Diameter: We will be designing the
shaft using ASME code of shaft design. According to this
code, the permissible shear stress τmax for the shaft without
keyways is taken as 30% of yield strength in tension or 18%
of the ultimate tensile strength of the material, whichever is
minimum. Therefore, τmax = 0.30 Syt or τmax = 0.18 Sut
(whichever is minimum).
If keyways are present, the above values are to be reduced
by 25 per cent. According to the ASME code, the bendingand
torsional moments are to be multiplied by factors kb and kt
respectively, to account for shock and fatigue in operating
condition. The ASME code is basedonmaximumshearstress
theory of failure.
were,
d

kb = combined shock and fatigue factor applied to bending
moment
kt = combined shock and fatigue factor applied to torsional
moment
Mb = Maximum Bending moment acting on shaft
Mt = Maximum torsional moment acting on shaft
Fig -7: Shock and fatigue factors
For the gearbox, it is a minor shock application. Thus, the
load is applied suddenly. We have taken as intermediate
value of kb = 1.5-2.0 and kt = 1.0-1.5.
Our shaft material is ‘Steel FeE 580 (Sut = 770 and Syt = 580
N/mm2).
Calculating permissible shear stress,
0.30 Syt = 0.30(580) = 174 N/mm2
0.18 Sut = 0.18(770) = 138.6 N/mm2
The lower of the two values is 138.6 N/mm2 and there are
keyways on the shaft.
τmax = 0.75(138.6) = 103.95 N/mm2
Combining values of τmax from both above equations,
Using max bendingmoment andtorsional momentvalues for
each shaft, we get the following shaft diameters which are
taken in accordance with the standard shaft table.
Fig -8: Shaft diameters
Shaft Number Shaft diameter (mm)
1 28
2 30
3 30
4 28
2.7 Design of keys and keyways:
Type of key: There are various types of keys. We have
selected Rectangular Saddle Key. Its design has been
conducted as follows.
Design on load basis: The material for key is Steel 50C4(Syt
= 460 N/mm2) and the factor of safety is taken as 3. For key
material, the yield strength in compression can be assumed
to be equal to the yield strength in tension. Keyways are on
the second and fourth shaft.
Syc = Syt = 460 N/mm2
N/mm2
According to maximum shear stress theory of failure,
Ssy = 0.5*Syt = 0.5(460) = 230 N/mm2
b – breadth of key
h – height of key
l – length of key
d – diameter of shaft
Industry standard is to select a basic dimension of
b = h = d/4 = 30/4 = 7.5 = 7mm
or
Taking maximum value using both equations, we get key
length = 28.7mm
Now selecting from standard table,

Fig -9: Determination of keys and keyways
Thus, we select the key with dimension (b,h) as (8,7) and a
length of 30mm.
Fig -10: Shafts
2.8 Selection of bearings:
Defining Bearing Life: To select bearings, it is necessary to
specify life of the bearings. This can be done from standard
bearing selection table. The recommended bearing life for
gear applications is given in the table below:
Fig -11: Bearing life.
The values given in the above table are only for general
guidance. For a particular application, the designer should
consider the experience, the difficulties faced by the
customer in replacing the bearing and the economics of
breakdown costs.
As we are relatively new to the designing aspect, we have
taken an estimated 15,000 hr( due to it being an
intermediate value for a service machine.
Selection from Manufacturer’s Catalogue: To select the
final bearings, we will need radial force acting. As thereis no
axial force, total force P = Fr. Diameter of all 4 shafts are
taken into consideration. Then using max speed of all shafts,
we calculate the rated bearing life L10 by the formula,
After we get this value for all shafts, we calculate the
dynamic load capacity C by,
Thus, we get the values of C as follows:
Shaft Number Dynamic Load Capacity
(N)
1 9682
2 12500
3 5411
4 3700
Based on these values of C, we must selecta bearingfromthe
manufacturer’s catalogue whose dynamic load capacity is
more than our requirement. The table for selection is given
below.

Fig -12: Bearing selection
The following table tells us which bearings are sufficient to
satisfy our requirement and are thus selected.
Shaft
Number
Bearing
Number
Principal
Dimensions(d,D,B) (all
mm)
1 6006 30,55,13
2 6006 30,55,13
3 6006 30,55,13
4 16006 30,55,9
Fig -13: Selected bearings
2.9 Casing Design:
Casing consideration: Followingstepshavebeentakeninto
consideration while designing the casing of the gearbox.
 Compact
 Weight optimisation
 Leak proof (bearing fitted in casing)
Visual Representation: The part of the casing which holds
the gearbox looks as follows.
Fig -14: Gearbox casing
2.10 Final Gearbox assembly:
Final Gearbox assembly: The final Gearbox assembly with
corrected values is as follows:
Fig -15: Schematic of final assembly
3. 3D Modelling and Analysis of gearbox.
Now that we had all the parameters required for designing
the gearbox on CAD tools, the CAD models for all the gears,
shafts, bearings, and the casing was created. The CAD was
modelled in Solid works software.

3.1 CAD Models – Gears and shafts
Fig -16: Shaft 1 Gear 1
Fig -18 Shaft 2 Gear 1

Fig -28: Shaft 1
Fig -29: Shaft 2
Fig -30: Shaft 3
Fig -31: Shaft 4

3.2 Final 3D Model:
Fig -32: Final Gear box assembly
3.3 Gearbox and shaft analysis:
Analysis of Gearbox
The gear pairs have been analyzed on ANSYS 18.1 software.
Analysis of components can be done in various ways like
static, dynamic, transient, etc. depending up on application.
Static analysis is concerned with determination of response
of a gear to steady loads whoseresponseremainsunchanged
with time. The response of the gear is expressed in terms of
stress, strain, displacement. Thetool usedin staticanalysisis
Static structural.
Procedure: The finite element analysis procedure of the
spur gear was given below:
1. A three-dimensional model of the spur gear was
created using the pro/engineer CAD software.
2. The material properties were defined for gears.
3. The model was meshed using finite element
software.
4. Boundary conditions for ANSYS Workbench as
mentioned below.
5. Fixed displacement constraint was applied on gear.
6. Moment was applied on gear.
7. To arrest the displacement on x, y, z directions and
rotations on x, y directions remote displacement
constraint is applied on pinion surface.
Contacts – Frictionless contacts between the gear pairs and
the gear and shaft pairs.
Boundary Conditions – Torques have been applied to the
input gear surfaces.
Material – Cast Iron has been considered for all the gear
pairs.
Deformation Results:
Fig -33: Total deformation for meshing pair 1

Equivalent Stress:
Fig -38: Equivalent stress for meshing pair 1
Factor of Safety:
Fig -43: Factor of safety analysis
The value obtained for the FOS for all the gear pairs is
mentioned below –
Gear Pair FOS Min FOS Max
1 1.3674 15
2 1.3896 15
3 1.2645 15
4 1.3564 15
5 1.2313 15

Hence, we can conclude that the design is within the
permissible limits for the given material and geometry.
Analysis of shafts:
The tool used in static analysis is Static structural.
Procedure: The finite element analysis procedure of the
shafts was given below:
1. A three-dimensional model of the shafts was
created using the pro/engineer CAD software.
2. The material properties were defined for shaft.
3. The model was meshed using finite element
software.
4. Boundary conditions for ANSYS Workbench as
mentioned below.
5. Fixed displacement constraint was applied on gear.
Contacts – Fixed contacts have been provided at the ends of
the shafts.
Boundary Conditions – Forces were applied on theshaft as
Point loads and UDL.
Total Deformation:
Fig -44: Shaft 1
Fig -45: Shaft 2
Fig -46: Shaft 3
Fig -47: Shaft 4
4.0 Analysis Results summary:
4.1 Gear Analysis:
Pair Min Def ( m) Max Def (m)
1 7.622e-6 5.202e-5
2 8.303e-6 4.303e-5
3 3.794e-6 1.446e-5
4 1.353e-6 7.533e-5
5 9.050e-6 4.197e-5
Pair Min stress Max stress
1 4536 1.45e8
2 2144.7 1.30e8
3 2101.6 6.01e8
4 3077.5 1.91e8
5 2153.4 1.01e8

4.2 Shaft Analysis
Shaft Min def (m) Max def (m)
1 0 5.51e-7
2 0 4.64e-7
3 0 4.59e-7
4 0 5.31e-7
Shaft Min stress (Pa) Max stress (Pa)
1 69.6 3.854e5
2 1590.6 2.763e5
3 189 3.017e5
4 666.45 2.814e5
Shaft Min strain Max strain
1 1.523e-9 1.938e-6
2 1.041e-9 1.394e-6
3 7.205e-9 1.571e-6
4 3.332e-9 1.414e-6
5. CONCLUSIONS
In conclusion, a standard design procedure for the designof
this 8-speed gearbox was followed considering various
research papers and textbooks. Normally, in lathe machine
applications, 6 speed gearboxes are used and using an 8-
speed gearbox will provide more operating speeds along
with offering more smoother and uniform machining
operation as the intervals between two successive speeds
will reduce. Post designing, thorough analysis considering
the working and static conditions of the gearbox was done
and all critical parameters were found to be well within the
permissible range.
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Design and Analysis of an 8-Speed Gear box.

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