MIZAN TEPI UNIVERSITY

COLLEGE OF ENGINEERING &TECHNOLOGY

DEPARTMENT OF MECHANICAL ENGINEERING

Machine design project 2

Design of two stage speed reduction gearbox with bevel gear

setup

Students Name:
1.Gemechis Wendimu
2.Worku Berie
3.Solomon Ayu

Instructor: Mr. Nahom Lijalem

MTU, Tepi Campus

 MIZAN TEPI UNIVERSITY
COLLEGE OF ENGINEERING &TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING

Machine design project 2

Design of two stage speed reduction gearbox with bevel gear

setup

Students Name Id No.

1.Gemechis Wendimu NSR/0781/13
2.Worku Berie NSR/1923/12
3.Solomon Ayu NSR/1740/12

Instructor: Mr. Nahom Lijalem

 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Declaration
We, the undersigned declare that the project report entitled “Design of two speed
reduction gearbox with bevel gear setup” has been carried out and submitted in
partial fulfillment of the requirements for the Bachelor of Technology in Department
of Mechanical Engineering at Mizan Tepi University College of Engineering and
Technology.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Acknowledgment
First, we would like to thank to our almighty God for helping us in the successful
accomplishment of this project paper. We would like to express our heartfelt
appreciation and gratitude to our advisor and instructor, Mr. Nahom L. for his
invaluable advice, continuous support, encouragement, valuable guidance, ingenious
and constructive suggestion. The last but not the least, we would like to forward our
special gratitude to our friends for their grateful assistance and advice that brings the
project to success, and constructive ideas. throughout our work.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Abstract
This paper presents the design and analysis of a two-speed reduction gearbox utilizing
a bevel gear setup, aimed at enhancing torque output and operational efficiency in
various mechanical applications. The proposed gearbox configuration addresses the
increasing demand for versatile transmission systems that can adapt to varying load
conditions while maintaining compactness and reliability. The design process involves
selecting appropriate gear ratios to achieve two distinct speed ranges, enabling
optimal performance for both high-speed and low-speed operations.

The bevel gear arrangement allows for a compact layout, facilitating smooth power
transmission with minimal losses. Key design considerations include the selection of
materials, gear tooth profiles, and lubrication methods to ensure durability and
efficiency. Finite Element Analysis (FEA) is employed to evaluate stress distribution
and identify potential failure points within the gearbox under different loading
scenarios.

As per our design scope or span, the project contains introduction, main background
with brief description of its problem statement, general and specific objectives,
reduction gearbox, and scope of the project with its methodology in its instant chapter,
it further developed in chapter two within literature review and in chapter three
material selection method and design analysis. Again, on chapter four design of two
stage speed reduction and mathematical calculation each component. and chapter five
is result and discussion about the project.

Finally, on the fifth chapter it covers about its conclusion and recommendation.
Then, on last Utilizing CAD software, specifically Solid Works, the design process
includes detailed 2D drawings that illustrate the gearbox's components, assembly, and
dimensions. These technical drawings serve as essential references for manufacturing
and assembly, ensuring that all components meet specified tolerances and operational
requirements.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Table of Contents
CHAPTER ONE............................................................................................................1
1. INTRODUCTION......................................................................................................1
1.1 Definition of gear box.......................................................................................1
1.2 Gear Box Background.......................................................................................2
1.3 Reduction Gearbox...........................................................................................3
1.3.1 Types of Reduction Gearbox:.................................................................3
1.3.2 Components of A Gear Box:...................................................................3
1.3.3 Function of gear box...............................................................................9
1.3.4 How to Choose a Gear Reducer.............................................................9
1.4 Problem Statement..........................................................................................10
1.5 Objectives........................................................................................................11
1.5.1 Specific Objectives................................................................................11
1.6 Scope...............................................................................................................11
1.7 Methodology...................................................................................................12
CHAPTER TWO..........................................................................................................13
2.LITERATURE REVIEW..........................................................................................13
CHAPTER THREE......................................................................................................15
3. MATERIAL SELECTION METHOD AND DESIGN ANALYSIS......................15
3.1. Design Considerations:..................................................................................15
3.2 Required Specifications..................................................................................16
3.3 Material selection............................................................................................17
3.3.1 Material for Shafts:..............................................................................17
3.3.2 Material for Gears................................................................................18
3.3.3 For the Pinion.......................................................................................19
3.3.4 Material for Casing..............................................................................20
3.3.5 Aluminum Alloy-Grade 7071-T6:.........................................................20
3.4 Gear Oil Selection...........................................................................................22
3.5 Selection of Fits and Tolerances.....................................................................23
CHATER FOUR..........................................................................................................25
DESIGN OF TWO STAGE SPEED REDUCTION....................................................25
GEARBOX...................................................................................................................25
4.1 Designing of Gears:................................................................................................25

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

4.2 Speed Reduction Calculation..........................................................................25

4.2.1 Input Parameters:.................................................................................25
4.2.2 Calculate Output Speed and Gearbox Specifications:.........................25
4.2.3 Determine Overall Velocity Ratio (VR)................................................26
4.2.4 Split the Velocity Ratio Between Two Stages:......................................26
4.3 Design Analysis..............................................................................................27
4.3.1 Design of first stage (Bevel Gear)........................................................27
4.3.2 Design of a shaft for Bevel Gears.........................................................31
4.3.3 Design of intermediate shaft.................................................................36
4.3.4 Key Design Calculation........................................................................41
4.3.5 Design of a key for first stage (Bevel Gears)........................................42
4.3.6 Design of second stage (spur gear)......................................................43
4.3.7 Design of Shaft for Spur Gears.............................................................50
4.3.8 Design of a key for second stage (spur Gears).....................................54
4.3.9 Design of bearing.................................................................................56
CHAPTER FIVE..........................................................................................................58
5.1 RESULT AND DISCUSSION...............................................................................58
CHAPTER SIX............................................................................................................60
6.1 CONCLUSION AND RECOMMENDATION.....................................................60
Reference......................................................................................................................61

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MTU Design of Two Speed Reduction Gearbox 2017 E.C

List of Figures
Figure 1.1: A Gearbox contain various Gear Ratios...............................................2

Figure 1.2: Spur Gear..............................................................................................4

Figure 1.3: Helical Gear..........................................................................................4

Figure 1.4: Bevel Gear............................................................................................5

Figure 1.6: Spiroid Gears........................................................................................6

Figure 1.7: Epicyclical Gears..................................................................................6

Figure 1.8: Input Shaft............................................................................................7

Figure 1.9: Output Shaft.........................................................................................7

Figure 1.10: Intermediate Shafts.............................................................................7

Figure 1.11: Roller Bearings...................................................................................8

Figure 1.12: Ball bearings.......................................................................................8

Figure 1.13: Gearbox housing.................................................................................9

Figure 3.1: Forces acting on a shaft......................................................................18

Figure 4.1 Forces acting on a bevel gear..............................................................33

Figure 4.3 Load acting on the gear.......................................................................51

Figure 4.4 Standard designations of ball bearings................................................57

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

List of Table
Table 3.1: Minimum requirements a Gearbox should meet..................................16

Table 3.2: Mechanical properties of steels used for shafts Table 14.1[6].............18

Table 3.3: grey cast iron table 2.8(6)....................................................................19

Table 3.4: Choice of materials for gearbox casing................................................21

Table 3.5: Material Properties...............................................................................22

Table 3.6: Alloy Chemical Composition...............................................................22

Table 3.7: Properties of EP series Oils..................................................................23

Table 4.1: Proportions of standard parallel, tapered and gibe head keys, Table
13.1,(6)...........................................................................................................43

Table 4.2: Values of service factor Table 28.10, (6)..............................................45

Table 4.3: Values of deformation factor (C), Table 28.5, (6)................................48

Table 4.4: Values of tooth error in action (e) verses module, Table 28.7, (6).......48

Table 4.5: Values of surface endurance limit, Table 28.9, (6)...............................49

Table 4.6 Principal dimensions for radial ball bearings. Table 27.1,[6]...............58

Table 4.7: Principal dimensions for radial ball bearings.......................................63

List of Symbol
HP -Horse Power
NP = Speed of pinion
NG = Speed of gear

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TP = Teeth of pinion
TG = Teeth of gear
Y = Lewis form factor
WT = Tangential load
L = Length
Rm = Mean radius
WRH = Axial force
WRV = Radial force
M = Bending moment
DP = Diameter of pinion
DG = Diameter of gear
τ = Shear stress
α =¿ Pressure angle
δ = Crushing stress
∅ = Range of obliquity
V.R = Velocity ratio
G = Gear ratio
WD = Dynamic load
WN = Wear tooth gear
Q = Ratio factor
K = Load stress factor
WG = Weight of gear
w = width
t = Thickness
b = Face width

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MTU Design of Two Speed Reduction Gearbox 2017 E.C

CHAPTER ONE

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

1. INTRODUCTION

1.1 Definition of gear box

A gearbox is a mechanical device that is used to increase the output torque or change
the speed (RPM) of the motor.
The motor shaft is connected to one end of the gearbox and through the internal
configuration of the gears of the gearbox, it provides an output torque and a specific
speed determined by the gear ratio.
The most basic definition of a gearbox is that it is a mechanical unit or component
consisting of a gear train or set of gears embedded in a housing. In fact, the name
itself defines it as a box containing gears.
In the most basic sense, a gearbox works like a gear system. Change torque and speed
between a drive, such as a motor, and a load.
Gears in gearboxes can be one of several types, from bevel and helical gears to worm
gears and planetary gears.
The gear is mounted on a shaft supported by roller bearings and rotates around it.
A gearbox is a mechanical method of transferring energy from one device to another,
which is used to increase torque while decreasing speed.
Gearboxes are used in many applications, including machine tools, industrial
equipment, conveyor belts, and almost any power transmission application with rotary
motion that requires a change in torque and speed.

1.2 Gear Box Background

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

A Gearbox is a device that used for transmitting power from the Power source to the
output shaft. A gearbox has a set of gears that are enclosed in a casing. The gears are
mounted on shafts which rotate freely about their axis. The gears are fixed on the
shafts by Fits or by a key. These shafts are made to rotate freely on a support called
casing. Bearings are tightly fit between the shafts and the casing. Today’s cars have
various sets of gears which give different speeds and torque on different Gears. A
Gearbox is necessary because it is impractical to directly connect the input source to
the output shaft. The power source may not have enough torque to bear the whole
load at once. This will put a load on the power source which may cause overheating,
more fuel consumption or even failure of the components. Gearbox gives leverage to
the power source by enhancing the torque at initial gears and then delivering high
speeds at final stages. This reduces the capacity of the power source required and
hence less fuel consumption. Each Gearbox has its own set of Gear ratios that can be
selected by the driver or just one set of universal Gear Ratio that will work with the
help of a Torque converter or a Continuously Variable Transmission Major
components include gears, Casing, Shafts, and Bearing.

Figure 1.1: A Gearbox contain various Gear Ratios

1.3 Reduction Gearbox

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A reduction gearbox is a device by which an input speed can be lowered for a

requirement of slower output speed, with same or more output torque. Reduction gear
two assembly consists of a set of rotating gears connected to an output shaft. The
high-speed incoming motion from the wheel work is transmitted to the set of rotating
gears, wherein the motion or torque is changed. The number of gears used in the
reduction gear assembly depends on the output speed requirement of the application.
The reduction gear assembly is usually known as reduction gear box. Depending on
the Output speed required, the reduction may have single stage or two stage reduction.

1.3.1 Types of Reduction Gearbox:

There are mainly two types of reduction gears:

• Single reduction gear
• Double reduction gear
Single Reduction Gear: This arrangement consists of only one pair of gears. The
reduction gear box consists of ports through which the propeller shaft and engine
shaft enters the assembly. A small gear known as a pinion is driven by the incoming
engine shaft. The pinion directly drives a large gear mounted on the propeller shaft.
The speed is adjusted by making the ratio of the speed reduction to the diameter of
pinion and gear proportional. Generally, a single gear assembly has a gear double the
size of a pinion.
Double Reduction Gear: This is a type of mechanical gear system designed to
reduce the speed of an input shaft while simultaneously increasing the torque output.
It typically consists of two stages of gearing, each providing a separate reduction in
speed.

1.3.2 Components of A Gear Box:

A gearbox is a mechanical device used to transmit power and adjust the speed and
torque of an engine or motor. It consists of several key components, each crucial role
in its operation.

The main components of a typical gearbox are:

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A. Gears:
 Spur Gears:
Design: Spur gears consist of straight teeth that run parallel to the gear axis. The
simplicity of their design facilitates ease of manufacturing and reliable power
transmission.
Function: Primarily used for applications requiring continuous, uniform rotation.
Spur gears are found in various machinery, including simple gear trains and electric
screwdrivers.

Figure 1.2: Spur Gear

 Helical Gears:
Design: Helical gears feature angled teeth in a helix pattern, offering smoother
engagement and quieter operation compared to spur gears.
Function: Helical gears excel in applications demanding high precision and reduced
noise, such as automotive transmissions and industrial machinery.

Figure 1.3: Helical Gear

 Bevel Gears:
Design: A bevel gear is a type of gear designed to transmit power between shafts that
are oriented at an angle to each other, typically at 90 degrees. The teeth of bevel gears
are cut on a conical surface, which allows them to mesh smoothly with one another.
Bevel gears have conical teeth designed for transmitting motion between intersecting
shafts at right angles.

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Function: Commonly employed in differentials of automobiles, bevel gears are

crucial for changing the direction of rotation between shafts.

Figure 1.4: Bevel Gear

Types:
1. Straight Bevel Gears: These have straight teeth and are mounted on shafts that
are at right angles to each other. They are commonly used in applications where speed
and torque need to be changed.

2. Spiral Bevel Gears: These have teeth that are curved and angled, allowing for
smoother engagement and quieter operation compared to straight bevel gears. They
can handle higher loads and are often used in high-performance applications.

3. Hypoid Gears: A type of spiral bevel gear, hypoid gears can transmit power
between non-intersecting shafts and allow for offset shafts. They provide smooth
operation and high load capacity.

 Worm Gears:
Design: Worm gears consist of a screw-like gear (worm) and a mating gear (worm
wheel), providing a high reduction ratio.
Function: Widely used in applications requiring a significant speed reduction and
high torque output, such as conveyor systems and winches.

Figure 1.5: Worm Gear

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 Spiroid Gears:

Design: Spiroid gearboxes perform a similar function to worm boxes but the gears
have characteristics which combine those of the bevel and worm gears. Function:
High powers and speed ratios are possible and mechanical efficiencies higher than
worm boxes for equivalent ratios.

Figure 1.6: Spiroid Gears

 Epicyclic Gears:

Design: Epicyclic gearboxes are a versatile arrangement of spur or helical gears in

which the input and output shafts are concentric and either shaft or the casing may be
constrained to be the stationary element, the torque being transmitted between the
other two. The three main elements are thus a 'sun' gear, a 'ring' gear and a number of
'planet' gears meshing with both.

Figure 1.7: Epicyclical Gears

B. Shafts:

 Input Shaft:
Function: The input shaft receives rotational power from an external source, typically
an engine or motor. It serves as the entry point for power into the gearbox.

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Figure 1.8: Input Shaft

 Output Shaft:
Function: The output shaft transfers rotational power from the gearbox to the driven
components, such as wheels or other mechanical devices. It is the point where the
gearbox delivers its output.

Figure 1.9: Output Shaft

 Intermediate Shafts:
Function: Positioned within the gearbox, intermediate shafts facilitate the transfer of
rotational power between gears. They play a crucial role in distributing power
effectively and optimizing gear arrangement.

Figure 1.10: Intermediate Shafts

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C. Bearings:
 Roller Bearings:
Function: Roller bearings support rotating components within the gearbox, reducing
friction and ensuring smooth motion. They come in various types, including
cylindrical, tapered, and needle roller bearings.

Figure 1.11: Roller Bearings

 Ball Bearings:
Function: Ball bearings use spherical rolling elements to support rotating shafts and
gears, offering a balance between load capacity and rotational efficiency. They are
commonly employed in gearboxes to reduce friction.

Figure 1.12: Ball bearings

D. Gearbox Housing:

Material: Gearbox housing is typically constructed from durable materials such as

aluminum or cast iron.
Function: The housing encases all internal components, providing protection against
external elements and supporting the gearbox's structural integrity.

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Figure 1.13: Gearbox housing

1.3.3 Function of gear box

In the most basic sense, a gear box function like any system of gears.it alters torque
and speed between a driving device like a motor and a load. The gearbox is a
mechanical method of transferring energy from one device to another and is used to
increase torque while reducing speed. See in gear box design as we decrease speed we
increase the torque this mean speed and torque have inverse relationship in this case.

In general gear has three main functions in gear box or in any system

 To control speed at the same time to control torque

 To change the direction of motion
 To transfer power

1.3.4 How to Choose a Gear Reducer

There are certain factors that have to be evaluated before deciding to purchase a gear
reducer. The main purpose of a gear reducer is to adapt the characteristics of torque
and speed of the input and output axis of a mechanism. It is for this reason that it is
necessary to understand the torque and rotational speed of the application.

I. Torque of a Motor
A gear reducer increases the torque of a motor and creates a new torque for the
receiving application. As an assist to customers, manufacturers express maximum and
minimum torque in newton meters (Nm) for their products with the torque density
varying between different gear reducers.

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II. Speed of a Motor

The second purpose of a gear reducer is to reduce motor speed, which is expressed in
terms of the reduction ratio. The rotational speed of a motor is changed by the
rotational ratio to produce the output rotational speed, which is described in
revolutions per minute.

III. Gear Reducer Selection

At this stage of the selection process, an expert, engineer, or designer that is
knowledgeable in gear reducers is necessary since there are so many types of gear
reducers designed to meet the requirements of a wide assortment of functions and
parameters. The configuration of the input and output shafts are the first criteria
related to choosing the type of gear reducer.

IV. Gear Reducer Dimensions

To decide on the dimensions of a gear reducer, it is important to choose the right shaft,
which can be orthogonal, coaxial, and parallel. Each of these shaft types have a
different orientation in regard to the gear reducer, which are perpendicular, aligned
and parallel.

V. Motor Performance
In some applications, motors experience shock or cyclic loads. When choosing a gear
reducer, it is important to factor in these conditions in order to allow the gear reducer
to be able to deal with the increased torque.

1.4 Problem Statement

There are four principal shortcomings identified in the conventional gear box design.
1) Weight- Too high
2) Inadequate Compactness
3) Poor compatibility with the Briggs and Stratton, engine used at Baja event.
4) Low Torque output

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1.5 Objectives
The primary objective of this project is to design a two-stage speed reduction gearbox
that meets specific performance criteria while ensuring reliability, efficiency, and
manufacturability. The gearbox will be used in applications requiring a controlled
output speed from a higher input speed, particularly in industrial machinery.

1.5.1 Specific Objectives

 To determine the required reduction ratio based on the application

requirements.
 To select the appropriate gear types for each stage based on load capacity,
noise, and space constraints.
 To calculate gear ratio for each stage to achieve the desired overall reduction
ratio.
 To size gears based on torque and speed requirements, considering factors like
material strength, tooth, profile, and gear tooth contact ratio.

1.6 Scope
The Gearbox designed is lightweight, compact and gives more performance than the
present Reduction Gearboxes. However, there still scope for better and more precise
design in the following areas. For this Reduction Gearbox, the differential should be
connected externally through a chain drive. This will still make the power train
assembly little complicated. Assembling the Differential inside the Gearbox itself will
make the Power train assembly simpler. The Factor of Safety of the output Gear is too
high. This is done because the output shaft is connected to the wheel with the help of
a Knuckle Joint. Since the wheel is subjected to various loads from the road condition
it is believed that the loads may be transmitted to the out gear also. Considering the
loads from the drive shaft will result in better design. The Gear oil has been chosen
based upon the operating temperature of the gearbox. The flow simulation was done
to understand the flow. The behavior of oil changes with respect to the operating
temperature. Simulating Flow of Oil with varying temperatures with respect to time
will give more accurate results. The brackets for the Bolts to join the casing have been
protruded outwards to make sure that the oil flow inside the casing is efficient. The
brackets should be redesigned in such a way that they do not have to protrude outside

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completely. This reduces the height and makes the Gearbox look aesthetically
pleasing.

1.7 Methodology
This project has 4 chapters. And we can classify them in to two major parts. The first
part is an introductory part and the second part is literature review part and other part
will be analysis of the project by using the force analysis, geometrical analysis as well
as the detail drawing. Finally, this project design is focus on the gear box design
which is different parameters such as force: stresses etc. are calculated by numerical
method. Generally, describe by using block diagram of methodology as follow

NEED OR AIM

SYNTHESIZE

LITERATURE REVIEW

MATERIAL SELECTION

DESIGN ANALYSIS

RESULT AND DISCUSSION

CONCLUTION

RECOMMENDATION

DETAIL DRAWING

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

CHAPTER TWO

2.LITERATURE REVIEW

For design of a gear box, it is necessary to look into the design aspects and literature
available in order to better understand the designing aspects

Prof. K.Gopinath & Prof. M.M.Mayuram in which he conclude that History of

gears, Definition of gears, Types of gears and their applications. Spur Gear and
Helical Gear provide by using pdf.

Prof. A. V. Patil Faculty Department of Mechanical Engineering S.S.G.B.C.O.E

India work on "Stress and Design Analysis of Triple Reduction Gearbox Casing" This
work focuses on the stress analysis of triple reduction of gearbox casing designed and
manufactured by steward

Ramesh Banothu worked on 'Design and analysis of Gear Shaft in which he

concluded that Vibrations can be transmitted from the casing to the gears through the
gear shafts. He even concludes that the shafts are subjected to bending moments and
slight torsion when there are at least two gears mounted on one shaft.

Vilas Warudkar from his study 'Design and Optimization of 2-Stage Reduction
Gearbox has concluded the methodology to make a conventional Reduction gearbox
for Water pumps. This Gearbox did not use any alloys for materials and did not
consider vibrations or Thermal analysis inside the Gearbox.

Ralph E.Taggert made a detailed experimental study on "Forcex that affect the
operation and efficiency of reduction Gearbox that contributes the complete
information about a single stage reduction gearbox.
Machine Company These gearboxes are designed for high torque and low speed
applications for operating movable bridges, heavy hoisting machinery, or other lifting
mechanisms.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Chandresh Motka Assistant Professor Department of Mechanical Engineering Kalol

Institute of Technology and Research Center, Kalol "Analysis & Optimization of
Gearbox Efficiency A Review focus on the less efficiency of gear box of a machine
tool is a serious problem as it increases maintenance cost and also affects the
reputation of a firm. Hence its life has to be increased and should be made more
reliable.

The Volvo Group [1] introduced a concept of Automatic Transmissions. It is discussed

that Manual Gearboxes are complicated in design and is difficult for the users to drive
the vehicle. Usage of so many components in the manual gearbox also reduced the
efficiency and the clutch plates had to be replaced periodically. The company stated
that it is difficult to implement manual gearbox in smaller vehicles.

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CHAPTER THREE

3. MATERIAL SELECTION METHOD AND DESIGN

ANALYSIS

3.1. Design Considerations:

The gears and shafts present in the reduction gearbox undergo various forces acting
over them. The Designing a two-stage speed reduction gearbox using bevel gears
consists of three shafts and four gears. The input shaft is connected to a motor or
another power source. When the motor operates, it rotates the input shaft. First Stage
Operation Of The input bevel gear, which is fixed to the input shaft, rotates and
meshes with the intermediate bevel gear.
The meshing of these gears causes the intermediate bevel gear to rotate. The gear ratio
between these two gears determines the speed reduction and torque increase at this
stage.
Intermediate Shaft Rotation: As the intermediate bevel gear rotates, it drives the
intermediate shaft, which is connected to the second stage of gears.
The second stage Operation of intermediate bevel gear from the first stage now acts as
the input for the second stage, meshing with the output bevel gear mounted on the
output shaft. and the rotation of the intermediate bevel gear drives the output bevel
gear, further reducing speed and increasing torque.
The output shaft delivers this reduced speed and increased torque to whatever load is
connected (such as a conveyor, wheel, or other machinery).
All these components are positioned in a gearbox casing which is made of seven
series aluminum so that it could accommodate the load an also it comprises of less
weight. The shafts are supported with the grooves that are created on the casing of the
gearbox with the help of bearings. The bearings used could vary as per the
requirements, i.e., the load acting on them.
Now, considering all these parameters the calculations are done and the required
parameters are identified such as the load or the torque acting on the components and

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the material of the gear and shaft are selected. The selection of material includes
various parameters such as weight, cost, machinability, strength, etc. Then, based on
the heat dissipated and lubrication required the gear oil is selected.

3.2 Required Specifications

Many parameters have been gathered from the present ATVs. Table 3.1 mentions the
requirements that should be fulfilled by a Gearbox the final product should comply
with all these parameters. The components must structurally comply without any
failure.

Table 3.1: Minimum requirements a Gearbox should meet.

Design Specifications
Specification Value
3.3 1. Input power P 20 hp
2. N input 1500 rpm
3. Output Speed Required 25 m/s
4.safety factor More than 1.5
5.Design life 4,000 hours
Material selection
Material selection plays an important role. Material takes up most of the cost required
to make a gear box. So many alloys are available in the market at low prices. But
keeping the weight in mind as well as performance an optimal material that can
sustain the above requirements are selected for each component of the Gearbox. In 7
order to reduce the variety of materials that are being used, it was decided that only
one material should be used for all the shafts, another for Gear and another for the
casing.

3.3.1 Material for Shafts:

The material used for shafts should have the following properties:

1. It should have high strength.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

2. It should have good machinability.

3. It should have low notch sensitivity factor.

4. It should have good heat treatment properties.

5. It should have high wear resistant properties.

Figure 3.1: Forces acting on a shaft

[Adapted from Beswarick (1994a)]

The material used for ordinary shafts is carbon steel of grades 40 C 8, 45 C 8, 50 C 4

and 50 C12. The mechanical properties of these grades of carbon steel are given in the
following table.

Table 3.2: Mechanical properties of steels used for shafts Table 14.1[6]

Indian standard Ultimate tensile strength Yield strength, (MPa)

designation (MPa)

40 C 8 560 - 670 320

45 C 8 610 - 700 350

50 C 4 640 - 760 370

50 C 12 700 Min 390

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

When a shaft of high strength is required, then an alloy steel such as nickel, nickel-
chromium or chrome-vanadium steel is used 45C8 carbon steel is selected as shaft
material due to its better mechanical properties.

3.3.2 Material for Gears

For choosing the material and designing the gears, the following data should be given:
In the design of a gear drive, the following data is usually given:
1. The power to be transmitted.
2. The speed of the driving gear,
3. The speed of the driven gear or the velocity ratio, and
4. The center distance.

The following requirements must be met in the design of a gear drive:

(a) The gear teeth should have sufficient strength so that they will not fail under static

Loading or dynamic loading during normal running conditions.

(b) The gear teeth should have wear characteristics so that their life is satisfactory.

(c) The use of space and material should be economical.

(d) The lubrication of the gears must be satisfactory.

For the gear selected material for the same as shaft the material used for ordinary gear
is carbon steel of grades 40 C 8, 45 C 8, 50 C 4 and 50 C12. The mechanical
properties of these grades of carbon steel are given in the above table l.

When a shaft of high strength is required, then an alloy steel such as nickel, nickel-
chromium or chrome-vanadium steel is used 45C8 carbon steel is selected as shaft
material due to its better mechanical properties.

3.3.3 For the Pinion

For the pinion selected material, the cast iron is obtained by re-melting pig iron with
coke and limestone in a furnace known as cupola. It is primarily an alloy of iron and
carbon. The carbon contents in cast iron vary from 1.7 per cent to 4.5 per cent. It also
contains small amounts of silicon, manganese, phosphorous

19
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Since the cast iron is a brittle material, therefore, it cannot be used in those parts of
machines which are subjected to shocks. The properties of cast iron which make it a
valuable material for engineering purposes are its low cost, good casting
characteristics, high compressive strength, wear resistance and excellent
machinability. The compressive strength of cast iron is much greater than the tensile
strength. Following are the values of ultimate strength of cast iron.

 Tensile strength = (100 to 200 MPa)

 Compressive strength = (400 to 1000 MPa)

 Shear strength = (120 MPa)

Cast iron has a various types I selected grey cast iron it has a low tensile strength,
high compressive strength and no ductility. It can be easily machined. A very good
property of grey cast iron is that the free graphite in its structure acts as a lubricant.
Due to this reason, it is very suitable for those parts where sliding action is desired.

Table 3.3: grey cast iron table 2.8(6)

IS Designation Tensile strength (MPa or N/ Brinell hardness number
mm )
2
(B.H.N)

FG 150 150 130 to 180

FG 200 200 160 to 220

FG 220 220 180 to 220

FG 260 260 180 to 230

FG 300 300 180 to 230

FG 350 350 207 to 241

FG 400 400 207 to 270

3.3.4 Material for Casing

The casing should be light weigh as possible but even should comply with the
structure loads. Since Cast Irons are brittle and develop cracks due to constant

20
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

vibrations. Steel is a good replacement but the weight is still heavy. So, opting for
aluminum is a suitable option. Pure Aluminum has less Yield and has more chances
of failure so we opted for aluminum alloys that offer high strength to weight ratio.
Aluminum density is 2700 Kg/m3. The choice is between six series or 7 series
Aluminum Alloy. These alloys can be hardened after machining to get full strength.

Table 3.4: Choice of materials for gearbox casing.

MATERIAL YIELD(MPa)

8xxx Aluminum alloys 380

7xxx Aluminum alloys 520–620

6xxx Aluminum alloys 550

3.3.5 Aluminum Alloy-Grade 7071-T6:

Aluminum is used in aircraft construction. These super alloys are still quite
expensive for the aircraft. With its good strength to weight ratio and high cost,
aluminum is still used very widely in the industry. Current developments indicate that
more and more manufacturers (Boeing, Airbus) are using carbon fiber and other non-
metallic materials in aircraft construction. As time goes by these materials will have
to prove themselves to be as reliable as aluminum.

Aluminum alloys are identified by a four-digit number system. The first digit

Gives the alloy group and the others the alloys that are present. Below a list of the
most commonly used aircraft aluminum alloys and their respective properties.

The numbering is ended by a temper designation T3 is solution heat-treated and

cold-worked by the flattening process. T6 is solution heat-treated and artificially
aged.7071-T6:

This alloy has a very good corrosion resistance and finishing ability, welding goes
good too. Typical applications of this alloy is air crafts, Truck bodies and frames.

21
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Table 3.5: Material Properties

PROPERTY VALUE

Brinell Hardness Number 95

Density 2700kg/m

Elastic Modulus 89GPa

Poisson’s Ratio 033

Shear Modulus 30GPa

Strength to Weight Ratio 110kN-m/Kg

Tensile Strength (Ultimate) 750 MPa

Tensile Strength (Yield) 505 MPa

Table 3.6: Alloy Chemical Composition

Aluminum (Al) 90.7 to 94.7 %

Copper (Cu) 3.8 to 4.9 %`

Magnesium (Mg) 1.2 to 1.8 %

Manganese (MN) 0.3 to 0.9 %

Iron (Fe) 0 to 0.5%

Silicon (Si) 0 to 0.5%

Zinc (Zn) 0 to 0.25%

Zirconium (Zr) 0 to 0.2%

Residuals 0 to 0.15%

Titanium (Ti) 0 to 0.15%

Chromium (Cr) 0 to 1%

22
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

3.4 Gear Oil Selection

Gear oil plays an important role in cooling and lubricating the inside components of
the gearbox. From the previous Baja vehicle, it has been recorded that the Gearbox
reaches up to 180°c. So the major parameters that needs to be considered is Flash
point of the oil. Viscosity vary based on the temperatures. Figure 3.3 shows the rate of
change for different types of oils over a range of temperatures. The flash point should
be at least a minimum of 200°𝐶 on the safe side to remove the risk of fire accidents.
Through thorough research, we short-listed the oils as per the requirement.

They are EP 18, EP 90, and EP 140.

Table 3.7: Properties of EP series Oils

PROPERTIES EP 80 EP 90 EP 140

Kinematic
Viscosity at 100°? 10.5-12.5 16.5-18 28-33
VISCOSITY 90 90 90
INDEX, MIN

FLASH POINT, 165 180 190

(COC) C, MIN

POUR POINT, C, -27 -9 -3

MAX

CHANNEL POINT -35 -18 -7

MAX

23
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Figure 3.3: Viscosity for different type of oils

3.5 Selection of Fits and Tolerances

Since the bearings come in standard sizes, for the fit between inner race and shafts,
we choose the Hole basis system. This mean the shaft should be manufactured
accordingly to satisfy the fit. In between the outer race of the bearing and the Casing,
the outer race has a fixed dimension. Hence, we should follow Shaft basis and
machine our hole accordingly.

Regarding the type of fit that needed to be imposed, giving a clearance fit may
cause sliding between the mating parts. But if we give an Interference fit, the
interference may be so much that they may imply unnecessary forces on the mating
parts. Last year vehicle has a Tight Interference fit between the Inner race of bearing
and the Shaft. Due to this, a force was implied on the inner race in a radially outward
direction. This put radial pressure on the ball bearings because of which the Bearing
failed every 3 Kilometers of test run. Hence, it is decided to use a transition fit and
select a tolerance grade based on previous research. Figure 3.3 and 3.4 compares the
different types of bias systems used for mating components.

24
MTU Design of Two Speed Reduction Gearbox 2017 E.C

Figure 3.4: hole basis system is used between shaft and bearing.

25
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

CHATER FOUR

DESIGN OF TWO STAGE SPEED REDUCTION

GEARBOX

4.1 Designing of Gears:

The first criteria in designing the gears is to keep them simple, less weight and at
the same time to keep the cost as low as possible. So, the weight and cost have their
respective weight age during the design such that both the parameters could be worth
enough. The machinability is another important consideration.

4.2 Speed Reduction Calculation

Speed reduction calculation is a method used to determine the decrease in speed of an
object or vehicle due to various factors such as friction, air resistance, mechanical
limitations, or intentional deceleration.

4.2.1 Input Parameters:

1. Input power P = 20 hp (Convertible to Watts 20 × 746 = 14,920 W)

2. N input 1500 rpm

3. Output speed required = 25 m/s

4. Safety factor more than 1.5

5. Design Life of 4,000 hours

4.2.2 Calculate Output Speed and Gearbox Specifications:

The required output speed is given as 25 m/s, but we need to convert this to rpm to
calculate gear ratios or velocity ratio. Specified by the design requirements). For

26
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

simply, assume that the output speed is related to the rotational speed of the output
shaft using the circumference of the output gear, i.e.

Voutput = π DgearNoutput

Where:

 V out = 25 m/s (output linear velocity)

 D out = diameter of the output gear (assumed for the calculation)

Let’s assume an output gear diameter of 500 mm (0.5 m) and Rearrange to solve for N
output.

Voutput
N output =
π Dgear

25
N output = ≈ 15.93 rps (or 955.8 rpm)
π × 0.5

4.2.3 Determine Overall Velocity Ratio (VR)

Now, we need to calculate the gear ratios for each stage. For a two stage reduction,
the total reduction ratio (R) is the product of the individual gear ratios for each stage:

Ninput 1500 rpm

V Rtotal = = =1.57
Noutput 955.8 rpm

4.2.4 Split the Velocity Ratio Between Two Stages:

 V R1 = Velocity ratio for the first stage (bevel gear stage)

 V R2 = Velocity ratio for the second stage.

The product of these two must equal the total velocity ratio:

V Rtotal = V R1 × V R2

27
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

A typical assumption of is to divide the velocity ratio equally between stages for
balanced load distribution:

V R1 = V R2 = √ V Rtotal ≈√ 1.57 ≈ 1.42

So: V R1 = 1.25 ≈ 2
V R2 = 1.25 ≈ 2

4.3 Design Analysis

4.3.1 Design of first stage (Bevel Gear)

For satisfactory operation of the bevel gears, the face width should be from 6.3 m to
9.5 m, where m is the module. Also, the ratio L/ b should not exceed 3. For this, the
number of teeth in the pinion must not less than 48.

For the analysis we have to assume that teeth of pinion Tp = 48.

The velocity ratio rounds up to 2 because, The velocity ratio of the input speed to the
output speed. A VR of 1.25 means the output speed is 1.25 times slower than the input
speed, while a VR of 2 means the output speed is twice as slow as the input speed.

VR = 2: Lower output speed, higher output torque.

And we have the speed of pinion Ninput = 1500 rpm

TG =V. R × Tp

TG = 2 × 48 = 96

Now
Np = V. R × NG

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Np
NG =
V.R

1500
NG =
2
= 750
Pitch angle for the pinion

θp 1=tan−1 1 ( V 1. R ) = tan 1( 12 ) = 26.5°

−1

Pitch angle for the gear

θp 1=¿ 90° − 26.5° = 63.5°

We know that the equivalent (or formative) number of teeth

TEP = Tp sec θ p1 = 48 sec 26.5° = 53.8 = 54

TEG = TG sec θ p2 = 48 sec 63.5° = 214.9 = 215

Lewis factor Y = π × y where

Tooth form factor (or Lewis factor) for pinion

(
Yp = 0.124 −
0.686
T EP )
= 0.124 −
0.686
54
= 0.1112

Tooth form factor (or Lewis factor) for Gear

(
YG = 0.124 −
0.686
T EG )
= 0.124 −
0.628
215
= 0.1208

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Since the allowable static stress (σ o ) for both the pinion and gear is same (i.e. 140
MPa or N /mm 2 ) and YP is less than YG, therefore the pinion is weaker. Thus, the
design should be based upon the pinion.
Torque on the pinion

P× 60
T= NM
2× π × N P

Where,
P = power transmitted in watts
Np = Speed of the pinion in rpm

14,920 × 60
T= Nm = 95Nm = 95000Nmm
2× π × 1500

Tangential load on the pinion

2T 2× 95000 3958
WT = = = Nmm
M . TP M × 48 M

We know the pitch line

π × D × N π × D P× N P π × M × T P × N P π × M × 48× 1500
V=
60
=
60
=
60
=
60
= 3770M
M/min

Allowable working stress

8.5 8.5
σ W = 140 =140
8.5+V 8.5+3770 M

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

We know that length of the pitch cone element or slant height element or slant height

DP M ×T P M × 48
L=
2sin θ P 1
= 2sin θ P 1
= 2sin 26.5
= 54M mm

Since the face width (b) is 1/4th of the slant height of the pitch one should not exceed
1/3th, so let say 1/4th

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

L 54 M
B= 4 = 4 = 13.5M mm

We know the tangential load on the pinion

WT = σ o ( 3+V3 ) ×b × π × M ∗ YP ( L−L b )

3958
M
= 140 ( 8.5
8.5+ 4147 M )
×13.5 M × π × M ∗ 0.1112 (
54 M − 13.5 M
54 M )
2
3958 1485.6 M
=
M 3+ 4147 M

Solving this expression by hit and trial method, we find that

M≈2
Now
b = 13.5M = 27 = 27mm

L = 54M = 108= 108mm

V = 4.147M = 8.294 ≈ 9 m/s

3958
WT = = 1979
2

The proportions for the bevel gears may be taken as follows

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Addendum a = 1m = 2mm
Dedendum d = 1.2m = 2.4mm
Clearance = 0.2m = 0.4mm
Working depth = 2m = 4mm

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Thickness of tooth = 1.5708m = 3mm

Diameter of pinion = M ×Tp=2 ×48 = 96mm
Diameter of Gear = M ×T G=2 ×96 = 192mm

NG = 750 rpm
P = 14,920W

4.3.2 Design of a shaft for Bevel Gears

1
Let assume that pressure angle of teeth is 14 ° , the pair of bevel gears connect two
2
shafts at right angle and overhang = 150mm

Cone lengths L

L=
√ D G 2 D P2
2
2
+
2
2
+
√
1922 962
2
2
+ 2 = 107.4 ≈ 108 mm=0.108 m
2

In designing a pinion shaft, the following procedure may be adopted

1. First of all, find the torque acting on the pinion

P× 60
T=
2× π × N P

14,920 × 60
T= Nm = 95Nm
2× π × 1500

2. Find the tangential force (Wt) acting at the mean radius (Rm) of the pinion, We
know that
T
WT =
RM

34
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

But

RM = L ( b2 ) Dp
2L

(
RM = 108
27
)96
2 2×108
= 42

14920
WT = = 2262N
42

3. Now find the axial and radial forces (i.e. W RH and WRV) acting on the pinion shaft
as Discussed below.
Now the radial force (WR) acting at the mean radius may be further resolved into two
components, WRH and WRV, in the axial and radial directions as shown Fig.

Figure 4.1 Forces acting on a bevel gear

A little consideration will show that the axial force on the pinion shaft is equal to the
radial force on the gear shaft but their directions are opposite. Similarly, the radial
force on the pinion shaft is equal to the axial force on the gear shaft, but act in
opposite directions.

35
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

There are three loads acting on bevel gear which are tangential force, axial and radial
forces.
The force relationships are

WT
W=
cos ∅

WRH = WRsin θ P 1=¿ ¿WT tan∅ sin θ P 1

WRV = WRcos θ P 1=¿ ¿ WT tan∅ cos θ P 1

The commonly used pressure angle for bevel gears is 14.5 ° although pressure angles
of 22.5° and 25° are used for heavy-duty drives.
Therefore, the axial force acting on the pinion shaft,

WRH = WRsin θ P 1=¿ ¿WT tan∅ sin θ P 1

WRH = WRsin θ P 1=¿ ¿2262× tan 14.5 sin 26.5
WRH = 2262× tan 14.5 sin 26.5 = 261N

And the radial force acting on the pinion shaft

WRV = WRcos θ P 1=¿ ¿ WT tan∅ cos θ P 1

WRV = WRcos θ P 1=¿ ¿ 2262× tan 14.5 cos 26.5
WRV = 2262× tan 14.5 cos 26.5 = 524N

4. Find resultant bending moment on the pinion shaft as follows, The bending
moment due RA to and RB is given by

We know that the torque transmitted by the shaft

14,920 × 60
T= Nm = 95Nm
2× π × 1500

36
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

And we know that Diameter of pinion = M ×T P=2 ×48 = 96 = 0.096

Tangential force on the pinion

2T 2× 95
WT = = = 1979.1 = 1979N
D P 0.096

And the normal load acting on the tooth of the pinion

WT 1979 1979
W= = = = 2044N
cos ∅ cos 14.5 ° 0.968

Since the pinion is mounted at the end of the shaft, therefore maximum bending
moment at the end of the pinion
First of all, considering the vertical loading at W. Let R A and RB the reactions at the
bearings A and B respectively. We know that

RA + RB = W = 2044N

Taking moments about A,

RB ×0.075 = W × 0.15

RB × 0.075 = 2044× 0.15

RB = 4088N
Taking moments about B,

RA × 0.075 = - W × 0.075

RA × 0.075 = -2044 × 0.075

37
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

RA = -2044N

The summation of vertical loading is RA + RB = 4088 - 2044 = 2044N

Now considering horizontal loading at W. Let R A and RB be the reactions at the

bearings A and B respectively. We know that there is no applied force in horizontally
then the summation of horizontal force is zero

Therefore, Resultant bending moment

M = √ M vertical2 + M ℎorizontal 2

M = √ 20442 +0 2 = 2044Nm

5. Since the shaft is subjected to twisting moment (T) and resultant bending moment
(M), therefore equivalent twisting moment

TE = √ M 2+T 2

TE = √ 20442 +95 2 = 2046.21Nm = 2046210mm

6. Now the diameter 0f the pinion shaft may be obtained by using the torsion
equation. We know that
π 3
TE = 16 × τ × D P

DP =
√
3 16 ∗ TE
π ∗τ

38
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

DP =
√
3 16 ∗ 2046210
π ∗ 120
= 44mm

Where Dp = Diameter of the pinion shaft

τ = Shear stress for the material of the pinion shaft

7. Now the diameter of the gear shaft may be obtained by using the torsion equation.
We know that

π 3
TE = 16 × τ × DG

DG =
√
3 16 ∗ 2046210
π ∗ 42
= 63mm

Where DG = Diameter of the gear shaft,

4.3.3 Design of intermediate shaft

We know that the torque transmitted by the shaft

14,920× 60
T= = 189.967Nm
2× π × 750

And we know, Diameter of Gear = M ×T G=2 ×96 = 192mm = 0.192m

Diameter of pinion = M ×T P = 2.5 × 48=120 mm=0.12 m
∴ Tangential force on the gear

2T 2× 189.967
WT = = = 1979N
DG 0.192

39
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Assuming that the torque at C and D is same (i.e. 89967N-mm), therefore the
tangential force on the gear C, acting downward

∴ Tangential force on the pinion

2T 2× 189.967
WT = = = 3166N
DP 0.12

Since the pinion is mounted at the right side of the shaft and the gear is mounted at the
left side of the shaft therefore maximum bending moment at the both gear.

Now let us find the maximum bending moment for vertical and horizontal loading

WC = 1979 WD = 3166

RA
50mm 200mm 50mm

+2275.75
+296.75
−2869.2
+121.3 Nm

+84.1 Nm

First of all, considering the vertical loading at W. Let RA and RB be the reactions at
the bearings A and B respectively. We know that

40
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

RA + RB = 1979 + 3166 = 5145

Taking moments about A,

RB × 0.2=1979 ×0.05 +3166 ×0.15

RB × 0.2 = 573.85
RB = 2869.25

Taking moments about B,

RA × 0.2=1979 ×0.15 +3166 ×0.05

RA × 0.2 = 455.15

RA = 2275.75

The summation of vertical loading is RA + RB = 2275.75 + 2869.25 = 5145N

Now considering horizontal loading at W. Let RA and RB be the reactions at the

bearings A and B respectively. We know that there is no applied force in horizontally
then the summation of horizontal force is zero.

We know that shear force.

FA = +2275.75N

FCR = +2275.75N

FCL = +2275.75N - 1979N = +296.75N

FDR = +296.75N

FDL = +296.75 - 3166 = -2869.25N

FB = -2869.25N

41
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Now let us find the maximum bending moment for vertical and horizontal loading.
First of all, considering the vertical loading at C, Let R AV and RBV be the reaction at
the bearings A and B respectively, we know that

RAV +¿ RBV = 3166N

Taking moments about A, we get

RBV ×200=3166 ×150

RBV = 3166 ×150 /200=2375 N

RAV = 3166 - 2375 = 791N

We know that B.M, at A and B,

MAV = MBV = 0

B.M. at D, MDV = RAV × 50 = 791 ×50=39550 N − mm

B.M at C, MCV = RBV ×100=2375 ×50=118750 N −mm

Now considering horizontal loading at D. let R AH and RBH be the reactions at the
bearings A and B respectively. We know that

RAH + RBH = 1979N

Taking moments about A, we get

RBH ×200=50 ×1979 = 1979 ×50 /200=¿ 494.75N

RAH = 1979 - 494.75 = 1484.25N

We know that B.M. at A and B,

MAH = MBH = 0

B.M. at D, MDH = RAH × 50 = 1484.25 ×50=74212.5 N − mm

42
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

B.M at C, MCH = RBH ×50=494.75 × 50=24737.5 N −mm

We know that the resultant

MD = √ M DV 2 + M DH 2
MD = √ 395502 +74212.52 = 84093.386Nmm = 84.1Nm

And resultant B.M, at C,

MC = √ M CV 2 + M CH 2
MC = √ 1187502 +24737.52 = 121299.12Nmm

We see that the bending moment is maximum at C,

43
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

∴ Resultant bending moment

MC = 121299.12Nmm = +121.3Nm

Since the shaft is subjected to twisting moment (T) and resultant bending moment
(M), therefore equivalent twisting moment.

TE = √ M 2+T 2

TE = √ 121.32 +189.9672 = 225.391Nm = 225391Nmm

1. Now the diameter of the pinion shaft may be obtained by using the torsion
equation. We know that

π
TE = ∗ τ ∗D P3
16

DP =
√
3 16 ∗ 225391
π ∗120
= 21mm

Where DP = Diameter of the pinion shaft,

τ =¿Shear stress for the material of the pinion shaft,

2. Now the diameter of the gear shaft may be obtained by using the torsion equation.
We know that

π
TE = ∗ τ ∗DG3
16

DG =
√
3 16 ∗ 225391
π ∗ 42
= 30mm

Where DG = Diameter of the gear shaft,

4.3.4 Key Design Calculation

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

A key is a piece of mild steel inserted between the shaft and hub or boss of the pulley
to connect these together in order to prevent relative motion between them. It is
always inserted parallel to the axis of the shaft. Keys are used as temporary fastenings
and are subjected to considerable crushing and shearing stresses. A key way is a slot
or recess in a shaft and hub of the pulley to accommodate a key. There are different
types of keys but we well select key is rectangular sunk key.

Figure 4.2: Rectangular key

The sunken keys are provided half in the key way of the shaft and half in the key way
of the hub or boss of the pulley.

Width of key, w = d/4; and

thickness of key, t = 2w/3 = d/6

Where d = Diameter of the shaft or diameter of the hole in the hub

 The rectangular key is designed as discussed below:

Now we see proportions of standard parallel, tapered and gibe head keys table

45
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Table 4.1: Proportions of standard parallel, tapered and gibe head keys, Table 13.1,(6)

4.3.5 Design of a key for first stage (Bevel Gears)

where

DP = 21mm and DG = 30mm

Now from the table we gave the width and thickness of key (i.e., 6mm and 6mm
respectively)

We know that

Shear stress τ =42 Mpa

Crushing stress σc=70 Mpa

The length of key is obtained by considering the key in shearing and crushing. Let l =
1 Length of key. Considering shearing of the key. For gear

We know torsional shearing strength (or torque transmitted) of the shaft

π π
T= × τ × d 3= × 42 × 213 = 76372.51Nmm…………………………. (i)
16 16

And we know that shearing strength (or torque transmitted) of the key,

d 21
T = l ×w×τ× = l ×6 × 42 × = 2646lNmm………………………. (ii)
2 2

From equation (i) and (ii), we have

46
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

76372.51
l= = 28.9mm ≈ 29 mm
2646

Now considering crushing of the key. We know that shearing strength (or torque
transmitted) of the key,

t d 6 21
T=l× × σc × = l × ×70 × = 2205Nmm……………………………….
2 2 2 2
(iii)

From equation (ii) and (iii), we have

76372.51
l= = 34.6 ≈ 35mm
2205

Taking larger of the two values, we have length of key,

l = 35 mm

4.3.6 Design of second stage (spur gear)

From the first we have power transmitted of 14920 watts and the output speed of the
first stage is input speed of the second stage and the second stage of gear ratio is 3:1

And also, the pressure angel is 20°

The number of teeth on the pinion (T P) in order to avoid interference may be obtained
from the Following relation:

2 ∗ Aw
TP =
[√ 1+ ( )
1 1
G G
+2 sin∅ 2 −1
]

47
 MTU Design of Two Speed Reduction Gearbox 2017 E.C

where
Aw = I module
G = Gear ratio
∅ =¿ Pressure angle or angle of obliquity

2 ∗1
TP =
[√ 1+ ( )
1 1
2 2
+2 sin 202 − 1
]= 49.8 ≈ 50

TG = G ∗T P=2 ∗ 50 = 100

The following table shows the values of service factor for different types of loads

Table 4.2: Values of service factor Table 28.10, (6)

So, assume that the reduction transition gear works 10hour per day the value of
service factor became 1.54
In order to design spur gear the following producer may be followed

First of all, to design tangential tooth load is obtain from the power transmitted and
the pitch live velocity by using the following relation

P
WT = ∗C S
V

Where WT = Permissible tangential tooth load in newton,

P = power transmitted in watts
V = pitch line velocity in m/s
D = pitch circle diameter in meters

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

π ∗ M ∗ T ∗ N π ∗ M ∗ 50∗ 750
V= = = 1.963M/sec
60 60

14920
WT = ∗1.54 = 11704.94N/M
1.963

Apply the Lewis equation as follows

WT = σw × b× p × y

But
σw = σo ∗ σv
And we have σo = 140mpa

3
σv =
3+V
Now Apply the Lewis equation

WT = σo ∗ σv ×b × π m × y

3
WT = 140 ∗ ×b × π m × y
3+V

The face width (b) may be taken as 3pc to 4pc (9.5m to 12.5m) for cut teeth and 2pc
to 3pc (or 6.5m to 9.5m) for cast teeth for 3pc = 3 ∗ m∗ π = 9.425m
Tooth from factory y for 20°

0.684
Y = 0.124 −
T

0.684
Y = 0.124 − = 0.113
50
Then

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

3
WT = 140 ∗ ×9.425m × πm ×0.113
3+2.72 m

2
1405.27 m 11704.94 N
WT = =
3+2.72 m m

1405.27
3+¿ 2.72m =
11704.94

3 +2.72 m=0.12 m2

Solving this expression by hit and trial method, we find that

M ≈2
Now
WT = 11704.94N/M

11704.94 N
WT = = 5852.47N
2
Pc = 3 ∗ m∗ π =9.425 ∗2=23.56 mm

V = 2.72m=2.72 ∗ 2=6.8 m/sec

Calculate the dynamic load (WD) on the tooth by using Buckingham equation, i.e.

WD = WT +¿ WI

+ 21∗ v ∗(b . c +W T )
WD = WT
21∗ v+ √ b . c +W T

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

In calculating the dynamic load (W D), the value of tangential load (W T) may be
calculating by Neglecting the service factor (CS) i.e.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

P 14920
WT = = = 2194.12w
V 6.8

The value of deformation factor given below table

Table 4.3: Values of deformation factor (C), Table 28.5, (6)

Before that we need have you know the value of tooth error (e) in action also the
value taken from the table

Table 4.4: Values of tooth error in action (e) verses module, Table 28.7, (6)

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

From the above table we get the value of deformation factors C = 770Nmm and tooth
error in action e = 0.0700mm

Now

+ 21∗ V ∗(b . c+W T )

WD = WT
21∗ V + √ b . c +W T

+ 21∗ 6.8 ∗(23.56 ∗770+5852.47)

WD = 5852.47 = 6007N
21∗ 6.8+ √ 23.56 ∗ 770+5852.47

Find the static tooth load (i.e., beam strength or the endurance strength of the tooth)
by using
The relation
WS = σe ∗ b ∗ p ∗ y = σe ∗ b ∗ πm∗ y

For safety against breakage, WS should be greater than WD, WS ≥ 1.25W D

From table below we have the value of surface endurance.

Table 4.5: Values of surface endurance limit, Table 28.9, (6)

Late take the value 320mpa

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

WS = σe ∗ b ∗ p ∗ y = σe ∗ b ∗ πm∗ y

WS = σe ∗ b ∗ p ∗ y = 320 ∗23.56 ∗ π 2.5∗ 0.113 = 6691.1N

WS ≥ 1.25W D

Finally, find the wear tooth load by using the relation

By using this formula

WN = DP ∗ b ∗Q ∗ k

where

WN = Wear tooth gear

DP = Diameter of the pinion
b = Face width
G 3
Q = Ratio factor Q = 2 ∗
G+1
= 2∗ 3+1
= 1.5

K = Load stress factor (also known as material combination factor) in N/mm

( E1P + E1G )
2
(σe ) sin ∅
K=
1.4

( )
2
(320) sin20 ° 1 + 1 =0.543 N /m m2
K=
1.4 200 200

Then
WN = DP ∗ b ∗Q ∗ k

WN = 100 ∗ 23.56∗ 1.5 ∗ 0.543=1918.96

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

4.3.7 Design of Shaft for Spur Gears

In order to find the diameter of shaft for spur gears, the following procedure may be
followed
We know that the torque transmitted by the shaft

14,920× 60
T= = 149.1Nm
2× π × 955.8

And we know that Diameter of Gear = M ∗T G=2 ∗100 = 200mm = 0.2m

∴ Tangential force on the pinion

2T 2∗ 149.1
WT = = = 1491N
DG 0.2

First of all, find the normal load (WN), acting between the tooth surfaces. It is given
by
WT 1491 1491
W= = = = 1587N
cos ∅ cos 20 ° 0.9397

Where WT = Tangential load, and

∅ =Pressure angle=20°
A thrust parallel and equal to WN will act at the gear center as shown in fig 4.3

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MTU Design of Two Speed Reduction Gearbox 2017 E.C

Figure 4.3 Load acting on the gear

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

The weight of the gear is given by

WG = 0.00118TG ∗ b ∗m2

WG = 0.00118 ∗189 ∗ 23.56 ∗2.5 2 = 32.8N

Now the resultant load acting on the gear

WR = √ WN 2 +WG2 +2 WN ∗ WG cos ∅

WR = √ 1918.962 +32.82 +2 ∗1918.96 ∗ 32.8 cos 20 = 1950N

If the gear is overhung on the shaft, then bending moment on the shaft due to the
resultant,
Since the gear is mounted at the end of the shaft, therefore maximum bending
moment at the end of the gear

First of all, considering the vertical loading at W. Let R A and RB be the reactions at
the bearings A and B respectively, we know that

RA +¿ RB = 1950N

Taking moments about A,

RB∗ 0.2=1950 ∗0.05

RB = 487.5N
Taking moments about B,

RA ∗ 0.2=W ∗0.15

RA ∗ 0.2=1950 ∗0.15

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

RA = 1462.5N

The summation of vertical loading is RA+¿ RB = 487.5 +1462.5=1950 N

WT = 1950N

WT = 1950

RA RB

+1462.5 N
− 487.5 N

+97.5 Nm

Now considering horizontal loading at W. Let R A and RB be the reactions at the

bearings A and B respectively. We know that there is no applied force in horizontally
then the summation of horizontal force is zero

We know that Shear force.

FA = +1462.5 N

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

FCR = +1462.5 N

FCL = +¿ 1462.5N −1950N = −487.5N

FB = − 487.5N

We know that B.M. at A and B,

MAv = MBv = 0

B.M. at C, MCV = Wt ∗50 = 1950 ∗50 = 975000N-mm = 97.5Nm

∴Resultant bending moment

M = √ M vertical2 + M ℎorizontal 2

M = √ 97.52 +02 = 97.5Nm

1. Since the shaft is subjected to twisting moment (T) and resultant bending moment
(M), therefore equivalent twisting moment

TE = √ M 2+T 2

TE = √ 97.52 +149.12 = 178Nm = 178000Nmm

2. Now the diameter of the pinion shaft may be obtained by using the torsion
equation. We know that

π
TE = τ ∗ D P3
16

DP =
√
3 16 ∗ 178000
π ∗ 120
= 20mm

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Where DP = Diameter of the pinion shaft,

τ = stress for the material of the pinion shaft

3. Now the diameter of the gear shaft may be obtained by using the torsion equation.
We know that

π
TE = τ ∗ D g3
16

Dg =
√
3 16 ∗ 178000
π ∗ 42
= 29mm

Where Dg = Diameter of the gear shaft,

4.3.8 Design of a key for second stage (spur Gears)

The sunken keys are provided half in the key way of the shaft and half in the key way
of the hub or boss of the pulley.

Width of key, w = d/4; and thickness of key, t = 2w / 3 = d / 6

Where d = Diameter of the shaft or diameter of the hole in the hub

 The rectangular key is designed as discussed above:

where
D = DP = 20mm DG = 29mm

Now from the table we gave the width and thickens of key (i.e., 6mm and 6mm
respectively)

We know that

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

shear stress τ =42 Mpa

Crushing stress σc=70 Mpa

The length of key is obtained by considering the key in shearing and crushing.

Let l = Length of key. Considering shearing of the key. For gear

We know torsional shearing strength (or torque transmitted) of the shaft

π π 3
×τ×d = T =
3
T= × 42× 20 =65973 Nmm .................(i)
16 16

And we know that shearing strength (or torque transmitted) of the key,

d
T = l ×w × τ ×
2
= l ×6 × 42× 20
2
= 2520lNmm………….….…(ii)

From equations (i) and (ii), we have

65973
l= =26 mm
2520

Now considering crushing of the key, we know that shearing strength (or torque
transmitted) of the key,

t d 6 20
T=l × × σc × = l × ×70 × = 2100lNmm………………(iii)
2 2 2 2

From equations (i) and (ii), we have

65973
l= =26 mm
2520

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Now considering crushing of the key, we know that shearing strength (or torque
transmitted) of the key,

t d 6 20
T=l × × σc × = l × ×70 × = 2100lNmm………………(iii)
2 2 2 2

From equations (i) and (iii), we have

65973
l= = 31.5mm ≈ 32 mm
2100

Taking larger of the two values, we have length of key,

l = 31.5mm ≈ 32 mm

4.3.9 Design of bearing

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Standard Dimensions and Designations of Ball Bearings

The dimensions that have been standardized on an international basis are shown in
Fig. 4.4. These dimensions are a function of the bearing bore and the series of
bearing. The standard dimensions are given in millimeters. There is no standard for
the size and number of steel balls.

Figure 4.4 Standard designations of ball bearings

The bearings are designated by a number. In general, the number consists of at least
three digits. Additional digits or letters are used to indicate special features e.g. deep
groove, filling notch etc. The last three digits give the series and the bore of the
bearing. The last two digits from 04 onwards, when multiplied by 5, give the bore
diameter in millimeters. The third from the last digit designates the series of the
bearing. The most common ball bearings are available in four series as follows:

1. Extra light (100)

2. Light (200)

3. Medium (300), Select this

4. Heavy (400)

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

The following table shows the principal dimensions for radial ball bearings... from
table 27.1 [6]

Table 4.6 Principal dimensions for radial ball bearings. Table 27.1,[6]
Bearing no Bore mm Outside diameter Width mm
mm
200 10 30 9
304 20 52 14
404 20 72 19

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

CHAPTER FIVE

5.1 RESULT AND DISCUSSION

Design of first stage (bevel gear) Design of second stage (spur gear)
Velocity ratio V.R = 1.25≈ 2 Velocity ratio V.R = 1.25≈ 2
Teeth of pinion TP = 48mm Teeth of pinion TP = 50mm
Teeth of gear TG = 96mm Teeth of gear TG = 100mm
Speed of pinion NP = 1500rpm Speed of pinion NP = 750rpm
Speed of gear NG = 750rpm Speed of gear NG = 955.8rpm
Pitch angle for pinion θp 1=26.5 ° Pitch line velocity V = 26.8m/s
Pitch angle of gear θp 2=63.5 ° Lewis factor YP = 0.113
Formative no of pinion TEP = 54mm Tangential load WT = 5852.47N
Formative no of gear TEG = 215mm Face width b = 23.56mm
Lewis factor for pinion YP = 0.1112 Module m=2
Lewis factor for gear YG = 0.1208 Dynamic load Wd = 6007N
Torque of the pinion T = 95000Nmm Endurance strength WS = 6691N
Tangential of pinion WT = 1979N Design of shaft bevel gear
Pitch line velocity V = 9m/s Tangential force W = 2262N
Length of pitch cone L = 108mm Mean radius RM = 42mm
Face width b = 27mm Axial force on shaft WRH = 261N
Module m=2 Radial force on shaft WRV = 524N
Addendum a=¿2mm Total bending moment M = 2044Nm
Dedendum d = 2.4mm Twisting moment TE = 204610mm
Clearance c =0.4mm Dia of pinion shaft D p = 44mm
Working depth =4mm Dia of gear shaft D g=63 mm
Thickness of tooth t = 3mm Design of shaft spur gear
Diameter of pinion D p = 96mm Normal load WN = 1919N
Diameter of gear D g=192 mm Weight of gear WG = 32.8N
Total load on shaft WR = 1950N
Total bending moment M = 97.5Nmm

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Twisting moment TE = 178Nmm

Dia of pinion shaft D p 2 = 20mm
Dia of gear shaft D g 2=29 mm
Design of key bevel gear Design of key spur gear
Thickness of key t 1=¿6mm Thickness of key t 2=¿6mm
Width of key w 1=¿6mm Width of key w 2=6 mm
Length of key l 1=35mm Length of key l 2=¿32mm

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

CHAPTER SIX

6.1 CONCLUSION AND RECOMMENDATION

Conclusion

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

the conclusion from this project we design that the two-stage speed reduction gearbox
it to transmit 20hp, and has gear reduction of two, by using spur gear as output and
bevel gear as input also having intermediate shaft. It has bearing at input, intermediate
and output. Which is made from carbon steel can withstand any difficulty without
failure or it is safe at given power and rotational speed for a give life time.

Recommendation

When we design some part, we have to collect information from different source. This
means the design is performed by using many references and web sites. So, the
materials needed for design case must be fulfilled in order to perform applicable
design for all users.

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

Reference

1. Drago, R. J. (1988). Fundamentals of Gear Design. Boston, MA: Butterworths,

2. AGMA Design Manual for Bevel Gears, 2005

3. AGMA Rating Standard for Bevel Gears, 2003

4. http://www.skf.com/skf/productcatalogue/jsp/viewers/productTableViewer.jsp?&l

ang-en&newlink-1&tableName-1_1_1&presentationType-3&startnum-15

5. V. B. Bhandari "Design of Machine Elements", Tata MeGraw Hill Publication.

Third Edition 2010, p.p. 330-348,646-70

6. A Textbook of Machine Design by R.S.KHURMI AND J.K.GUPTA [tortuka]

7. PSG design data book

8. (https://www.apexdyna.nl/en/gear-history/)

9. Mechanics of materials, 9h edition by R.C.HIBBELER

10. Shigley's Mechanical Engineering Design - 8th edition

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 MTU Design of Two Speed Reduction Gearbox 2017 E.C

11. Design of Machine Elements By V B Bhandari 3Ed

12. Hid Shigley's Mechanical Engineering Design - 8th Edition with Solution of
Problems

13. Machine Design objective by VB Bhandari

14. Shigley's Mechanical Engineering Design 9th Edition

15. Machine_design_databook
Appendices

Table 4.7: Principal dimensions for radial ball bearings

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MTU Design of Two Speed Reduction Gearbox 2017 E.C

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MTU Design of Two Speed Reduction Gearbox 2017 E.C

72