Nine Speed Transmission Design Project Analyzed

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Nine Speed Transmission
Design
Final Project
POLITECHNIC UNIVERSITY OF PUERTO
RICO
DEPARTMENT OF MECHANICAL
ENGINEERING
HATO REY, PUERTO RICO
ME5240; MACHINE DESING ELEMENTS
II
SP-13
Submitted to:
Dr. Skrzypinski
May 28, 2013
Carlos J. Gutiérrez Román
#54543

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Abstract
A transmission as we know is the systemtransmits the power from one engine to the
point that is to be used. With the initial parameters set, we conduct a comprehensive
process of analysis, including the calculation of partial relations, speed, shaft distance,
acting forces and both shear and moment diagrams. For this case the transmission
design based on theories learned in class and also use knowledge learned in other
classes passed as solid mechanics II which applies in this design.

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Table of Contents
Abstract.................................................................................................................................2
Introduction...........................................................................................................................6
Theory...................................................................................................................................7
Calculations..........................................................................................................................12
Partial Relations................................................................................................................13
Total Relations..................................................................................................................14
Gear Analysis....................................................................................................................15
Torque.............................................................................................................................15
Real Forces.......................................................................................................................16
Lewis Forces.....................................................................................................................17
Results.................................................................................................................................18
Conclusions..........................................................................................................................48
References...........................................................................................................................49
Appendices ..........................................................................................................................50

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Tables and Figures
Table 1 Partial Relations........................................................................................................18
Table 2 Output gear relation .................................................................................................18
Table 3 Total Relations and Output speeds.............................................................................19
Table 4 Shaft Speeds, and distances between shafts...............................................................20
Table 5 Reducer Gear Analysis...............................................................................................21
Table 6 Real Forces and Lewis Forces comparison...................................................................22
Table 7 Real Forces and Lewis Forces Comparison..................................................................23
Table 8 Lewis Form Factor for N-Teeth...................................................................................24
Table 10 Gear Analysis..........................................................................................................25
Table 11 Real Forces.............................................................................................................25
Table 12 Failure Theories, deflection and shaft diammeter .....................................................26
Table 13 Bearing Design........................................................................................................46
Figure 1 Lewis Form Factor diagram.......................................................................................24
Figure 2 Shaft I FBD...............................................................................................................28
Figure 3 Shaft I Shear Diagram x-y.........................................................................................28
Figure 4 Shaft I Moment Diagram x-y.....................................................................................29
Figure 5 Shaft I FBD x-z..........................................................................................................29
Figure 6 Shaft I Shear diagramx-z..........................................................................................30
Figure 7 Shaft I Moment diagramx-z.....................................................................................30
Figure 8 Shaft II FBD x-y........................................................................................................31
Figure 9 Shaft II Shear Diagramx-y.........................................................................................31
Figure 10 Shaft II Moment diagramx-y...................................................................................32
Figure 11Shaft II FBD x-z........................................................................................................32
Figure 12 Shaft II Shear diagramx-z.......................................................................................33
Figure 13 Shaft II Moment diagramx-z...................................................................................33
Figure 14 Shaft III FBD x-y......................................................................................................34
Figure 15 Shaft III Shear diagram x-y......................................................................................34
Figure 16 Shaft III Moment diagram x-y..................................................................................35
Figure 17 Shaft III FBD x-z......................................................................................................35
Figure 18 Shaft III Shear diagram x-z......................................................................................36
Figure 19 Shaft III Moment diagram x-z..................................................................................36
Figure 20 Shaft IV FBD x-y.....................................................................................................37
Figure 21 Shaft IV Shear diagram x-y......................................................................................37
Figure 22 Shaft IV Moment diagram x-y.................................................................................38
Figure 23 Shaft IV FBD x-z......................................................................................................38
Figure 24 Shaft IV Shear diagram x-z......................................................................................39

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Figure 25 Shaft IV Moment diagram x-z..................................................................................39
Figure 26 Shaft V FBD x-y......................................................................................................40
Figure 27 Shaft V Shear diagramx-y.......................................................................................40
Figure 28 Shaft V Moment diagramx-y..................................................................................41
Figure 29 Shaft V FBD x-z.......................................................................................................41
Figure 30 Shaft V Shear diagramx-z.......................................................................................42
Figure 31 Shaft V Moment diagramx-z..................................................................................42
Figure 32 Shaft VI FBD x-y.....................................................................................................43
Figure 33 Shaft VI Shear diagram x-y......................................................................................43
Figure 34 Shaft VI Shear diagram x-y......................................................................................44
Figure 35 Shaft VI FBD x-z......................................................................................................44
Figure 36 Shaft VI Shear diagram x-z......................................................................................45
Figure 37 Shaft VI Moment diagram x-z..................................................................................45

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Introduction
A transmission is the system that transmits the power generated by the engine
or motor to the point where it is to be used. This document presents the design process
of a transmission meeting pre-established requirements, nine output velocities, a power
input of 30 kW, the motor’s angular velocity of 1600RPM, lowest angular velocity of 25
RPM and the geometric relation of 1.6. With the initial parameters already established,
a complete process of analysis was done, including but not limited to calculation of
partial relations, speed, distance between shafts, Lewis Forces and Shear and Moment
diagrams. The material selected to analyze the shaft was Steel A350. In this paper we
resume some theoretical background of a transmission, the mathematical procedure
made for the design, and a complete mechanical drawing, which specifies the
transmission design.

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Theory
In mechanics, a transmission or gearbox is the gear and/or hydraulic system that
transmits mechanical power from a prime mover (which can be an engine or electric
motor), to some form of useful output device. Early transmissions included right angle
drives and other gearing in windmills, horse-powered devices, and steam engines,
mainly in support of pumping, milling, and hoisting applications. Typically, the
rotational speed of an input shaft is changed, resulting in a different output speed.
However, some of the simplest gearboxes merely change the physical direction in which
power is transmitted. In daily life, individuals most often encounter the transmissions
used in automobiles, which cover an ever-expanding array of specific types. However,
gearboxes have found use in a wide variety of different-often stationary-applications.
Most mechanical transmissions function as rotary speed changers; the ratio of
the output speed to the input speed may be either constant (as in a gearbox) or
variable. On variable-speed transmissions, the speeds may be variable in individual
steps (as on an automobile or some machine-tool drives) or continuously variable
within a range. Step-variable transmissions, with some slip, usually use either gears or
chains and provide fixed speed ratios with no slip; step less transmissions use belts,
chains, or rolling-contact bodies.
Simple transmission
The simplest transmissions, often called gearboxes to reflect their simplicity
(although complex systems are also called gearboxes on occasion), provide gear
reduction (or, more rarely, an increase in speed), sometimes in conjunction with a right-
angle change in direction of the shaft. These are often used on PTO-powered
agricultural equipment, since the axial PTO shaft is at odds with the usual need for the
driven shaft, which is either vertical (as with rotary mowers), or horizontally extending
from one side of the implement to another (as with manure spreaders, flail mowers,

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and forage wagons). More complex equipment, such as silage choppers and snow
blowers, has drives with outputs in more than one direction. Regardless of where they
are used, these simple transmissions all share an important feature: the gear ratio
cannot be changed during use. It is fixed at the time the transmission is constructed.
Automotive basics
The need for a transmission in an automobile is a consequence of the
characteristics of the internal combustion engine. Engines typically operate over a range
of 600 to about 6000 revolutions per minute (though this varies from design to design
and is typically less for diesel engines), while the car's wheels rotate between 0 rpm and
around 2500 rpm. Furthermore, the engine provides its highest torque outputs
approximately in the middle of its range, while often the greatest torque is required
when the vehicle.
Multi-ratio systems
Many applications require the availability of multiple gear ratios. Often, this is to
ease the starting and stopping of a mechanical system, though another important need
is that of maintaining good fuel economy. e is moving from rest or traveling slowly.
Therefore, a system that transforms the engine's output so that it can supply high
torque at low speeds, but also operate at highway speeds with the motor still operating
within its limits, is required. Transmissions perform this transformation.
 Most transmissions and gears used in automotive and truck applications are
contained in a cast iron case, though sometimes aluminum is used for lower
weight. There are three shafts: a main shaft, a countershaft, and an idler
shaft.
 The main shaft extends outside the case in both directions: the input shaft
towards the engine, and the output shaft towards the rear axle (on rear
wheel drive cars). The shaft is suspended by the main bearings, and is split

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towards the input end. At the point of the split, a pilot bearing holds the
shafts together. The gears and clutches ride on the main shaft, the gears
being free to turn relative to the main shaft except when engaged by the
clutches.
 The countershaft is generally below the main shaft and turns in the opposite
direction, driven by a bevel gear on the input shaft.
Manual transmission
Manual transmissions come in two basic types: a simple unsynchronized system
where gears are spinning freely and must be synchronized by the operator to avoid
noisy and damaging "gear clash", and synchronized systems that will automatically
"mesh" while changing gears. It is hard and takes a long time to become familiar with
operating manual transmissions. A small mistake can lead to the car stalling or jumping
or to irreparable damage of the transmission. Manual transmissions tend to distract the
driver's attention from traffic and are associated with higher accident rates in cities and
stop and go traffic. For that reason insurance premiums are higher for cars with manual
transmissions in some countries. Manual transmissions have been popular in [Europe]
and less developed countries for the longest time.
Automatic transmission
Most modem cars have an automatic transmission that will select an appropriate
gear ratio without any operator intervention. They are primarily using hydraulics to
select gears, depending on pressure exerted by fluid within the transmission assembly.
Rather than using a clutch to engage the transmission, a torque converter is put in
between the engine and transmission. It is possible for the driver to control the number
of gears in use or select reverse, though precise control of which gear is in use is not

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possible. Automatic transmissions are easy to use. In the past, automatic transmissions
of this type have had a number of problems, they were complex and expensive, and
sometimes had reliability problems (which sometimes caused more expense in repair),
and often have been less fuel-efficient than their manual counterparts. With the
advancement of modem automatic transmissions this has changed. With computer
technology, considerable effort has been put into designing gearboxes based on the
simpler manual systems that use electronically-controlled actuators to shift gears and
manipulate the clutch, resolving many of the drawbacks of a hydraulic automatic
transmission.
Automatic transmissions have always been extremely popular in the United
States, where perhaps 19 of20 new cars are sold with them (most vehicles are not
available with manual gearboxes anymore). In Europe automatic transmissions are
finally gaining popularly as well. Attempts to improve the fuel efficiency of automatic
transmissions include the use of torque converters which lock-up beyond a certain
speed eliminating power loss, and overdrive gears which automatically actuate above
certain speeds; in older transmissions both technologies could sometimes became
intrusive, when conditions are such that they constantly cut in and out as speed and
such load factors as grade or wind vary slightly. Current computerized transmissions
possess very complex programming to both maximize fuel efficiency and eliminate any
intrusiveness. For certain applications, the slippage inherent in automatic transmissions
can be advantageous; for instance, in drag racing, the automatic transmission allows the
car to be stopped with the engine at a high rpm (the "stall speed") to allow for a very
quick launch when the brakes are released; in fact, a common modification is to
increase the stall speed of the transmission. This is even more advantageous for
turbocharged engines, where the turbocharger needs to be kept spinning at high rpm
by a large flow of exhaust in order to keep the boost pressure up and eliminate the
turbo lag that occurs when the engine is idling and the throttle is suddenly opened.

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Semi-automatic transmission
The creation of computer control also allowed for a sort of half-breed
transmission where the car handles manipulation of the clutch automatically, but the
driver can still select the gear manually if desired. This is sometimes called "clutch less
manual". Many of these transmissions allow the driver to give full control to the
computer. There are some specific types of this transmission, including Triptronic and
Direct Shift Gearbox. There are also sequential transmissions which use the rotation of a
drum to switch gears. A great example of this is the 7 -speed sequential transmission on
the Bugatti Veyron, a super car that puts out 1,001 horsepower (746 kW) and goes 254
miles per hour (409 km/h). You can see this at howstuffworks.com. Follow this link to
get to the howstuffworks.com article on sequential transmissions.

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Calculations
𝑛0 = 1600 [𝑅𝑃𝑀]
𝑖𝑝 = 1: 2
𝑛𝑝 = 800 [𝑅𝑃𝑀]
𝑖𝑔 = 1:2
(1, 𝜉, 𝜉2)(1,𝜉3𝜉6)
𝜉 = 1.6
𝑛1 = 25(𝜉)0 = 25(1.6)0 = 25[𝑅𝑃𝑀]
𝑛2 = 25(𝜉)1 = 25(1.6)1 = 40[𝑅𝑃𝑀]
𝑛3 = 25(𝜉)2 = 25(1.6)2 = 64[𝑅𝑃𝑀]
𝑛4 = 25(𝜉)3 = 25(1.6)3 = 102.4[𝑅𝑃𝑀]
𝑛5 = 25(𝜉)4 = 25(1.6)4 = 163.84[𝑅𝑃𝑀]
𝑛6 = 25(𝜉)5 = 25(1.6)5 = 262.14[𝑅𝑃𝑀]
𝑛7 = 25(𝜉)6 = 25(1.6)6 = 419.43[𝑅𝑃𝑀]
𝑛8 = 25(𝜉)7 = 25(1.6)7 = 671.08[𝑅𝑃𝑀]
𝑛9 = 25(𝜉)8 = 25(1.6)8 = 1073.7[𝑅𝑃𝑀]

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Partial Relations (Numberof Teethwas assumed)
𝒊𝒓 =
𝒁𝟏
𝒁𝟐
=
𝟐𝟏
𝟒𝟐
=
𝟏
𝟐
𝒊𝟏 =
𝒁𝟒
𝒁𝟓
=
𝟑𝟎
𝟏𝟐𝟎
=
𝟏
𝟒
𝒊𝟐 =
𝒁𝟐
𝒁𝟑
=
𝟒𝟐
𝟏𝟎𝟒
= 𝟎.𝟑𝟖𝟗
𝒊𝟑 =
𝒁𝟔
𝒁𝟕
=
𝟓𝟖
𝟗𝟐
= 𝟎.𝟔𝟑
𝒊𝟒 =
𝒁𝟏𝟑
𝒁𝟏𝟒
=
𝟑𝟎
𝟏𝟐𝟎
=
𝟏
𝟒
𝒊𝟓 =
𝒁𝟏𝟓
𝒁𝟏𝟔
=
𝟕𝟕
𝟕𝟑
= 𝟏. 𝟎𝟓𝟓
𝒊𝟔 =
𝒁𝟖
𝒁𝟗
×
𝒁𝟗
𝒁𝟏𝟎
=
𝟖𝟗
𝟏𝟗
= 𝟒.𝟔𝟖𝟒
𝒊𝟕 =
𝒁𝟏𝟏
𝒁𝟏𝟐
=
𝟑𝟖
𝟒𝟐
= 𝟎. 𝟗𝟎𝟓
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖2 ∗ 𝑖4 = 40[𝑅𝑃𝑀] → 𝑖2 =
1600∗0.5∗0.5∗0.25
40
= 0.4
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖3 ∗ 𝑖4 = 64[𝑅𝑃𝑀] → 𝑖3 =
1600 ∗ 0.5 ∗ 0.5 ∗ 0.25
64
= 0.64
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖1 ∗ 𝑖5 = 102.4[𝑅𝑃𝑀] → 𝑖5 =
1600 ∗ 0.5 ∗ 0.5 ∗ 0.25
102.4
= 1.024
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖2 ∗ 𝑖5 = 163.84[𝑅𝑃𝑀] = 1600 ∗ 0.5 ∗ 0.5 ∗ 0.4 ∗ 1.024 = 163.84 → 𝑜𝑘
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖3 ∗ 𝑖5 = 262.144[𝑅𝑃𝑀] = 1600 ∗ 0.5 ∗ 0.5 ∗ 0.64 ∗ 1.024 = 163.84 → 𝑜𝑘
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖1 ∗ 𝑖6 ∗ 𝑖7 = 419.43[𝑅𝑃𝑀] = 1600 ∗ 0.5 ∗ 0.5 ∗ 0.25 ∗ 4.684 ∗ 0.905 = 419.43
→ 𝑜𝑘
𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖2 ∗ 𝑖6 ∗ 𝑖7 = 671.08[𝑅𝑃𝑀] = 1600 ∗ 0.5 ∗ 0.5 ∗ 0.4 ∗ 4.684 ∗ 0.905 = 671.08
→ 𝑜𝑘

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𝑛0 ∗ 𝑖𝑝 ∗ 𝑖𝑟 ∗ 𝑖3 ∗ 𝑖6 ∗ 𝑖7 = 1073.74[𝑅𝑃𝑀] = 1600 ∗ 0.5 ∗ 0.5 ∗ 0.64 ∗ 4.684 ∗ 0.905
= 1073.74 → 𝑜𝑘
Total Relations
𝐽1 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖1 ∙ 𝑖4 = 0.5 ∗ 0.5 ∗ 0.25 ∗ 0.25 = 0.15625
𝐽2 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖2 ∙ 𝑖4 = 0.5 ∗ 0.5 ∗ 0.4 ∗ 0.25 = 0.025
𝐽3 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖3 ∙ 𝑖4 = 0.5 ∗ 0.5 ∗ 0.64 ∗ 0.25 = 0.04
𝐽4 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖1 ∙ 𝑖5 = 0.5 ∗ 0.5 ∗ 0.25 ∗ 1.024 = 0.064
𝐽5 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖2 ∙ 𝑖5 = 0.5 ∗ 0.5 ∗ 0.4 ∗ 1.024 = 0.1024
𝐽6 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖3 ∙ 𝑖5 = 0.5 ∗ 0.5 ∗ 0.64 ∗ 1.024 = 0.16384
𝐽7 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖1 ∙ 𝑖6 ∙ 𝑖7 = 0.5 ∗ 0.5 ∗ 0.25 ∗ 4.684 ∗ 0.905 = 0.26494
𝐽8 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖2 ∙ 𝑖6 ∙ 𝑖7 = 0.5 ∗ 0.5 ∗ 0.4 ∗ 4.684 ∗ 0.905 = 0.42390
𝐽9 = 𝑖𝑝 ∙ 𝑖𝑟 ∙ 𝑖3 ∙ 𝑖6 ∙ 𝑖7 = 0.5 ∗ 0.5 ∗ 0.64 ∗ 4.684 ∗ 0.905 = 0.67824

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Gear Analysis
 Gear I
 𝑍 = 21; 𝑚 = 5𝑚𝑚; 𝜃 = 20°
𝑑𝑝 = 𝑍 × 𝑚 = 21 × 5 = 105𝑚𝑚
𝑃𝑑 =
𝑍
𝑑𝑝
=
21
105
= 0.2𝑚𝑚
𝑑𝐵 = 𝑑𝑝𝐶𝑜𝑠𝜃 = 105𝐶𝑜𝑠20 = 98.67𝑚𝑚
𝑑𝐴 = (𝑍 + 2) × 𝑚 = (21 + 2) × 5 = 115𝑚𝑚
𝑑𝐷 = 𝑑𝑝 − 2.5(𝑚) = 105 − 2.5(5) = 92.50𝑚𝑚
𝑇𝑜𝑡𝑎𝑙 𝑇𝑒𝑒𝑡ℎ 𝐻𝑒𝑖𝑔ℎ𝑡 = 2.25 × 𝑚 = 2.25(5) = 11.25𝑚𝑚
𝑊𝑜𝑟𝑘 𝐻𝑒𝑖𝑔ℎ𝑡 = 2 × 𝑚 = 2 × 5 = 10𝑚𝑚
𝑃
𝑐 = 𝜋 × 𝑚 = 𝜋 × 5 = 15.71𝑚𝑚
𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 =
𝑃𝐶
2
=
15.71
2
= 7.86𝑚𝑚
𝐴𝑑𝑒𝑛𝑑𝑢𝑚 = 𝑚 = 5𝑚𝑚
𝐷𝑒𝑑𝑒𝑛𝑑𝑢𝑚 = 1.25 × 𝑚 = 1.25(5) = 6.25𝑚𝑚
𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 0.25 × 𝑚 = 0.25(5) = 1.25𝑚𝑚
Same procedure was done to obtain other gear dimensions
Torque → (𝑷 = 𝑻𝒏𝟎 → 𝑻 =
𝑷
𝒏𝟎
) 𝑷 = 𝑷𝒐𝒘𝒆𝒓; 𝒏𝟎 = 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔
 Shaft I
𝑇1 =
30000 × 60
800 × 2𝜋
= 358.1𝑁 ∙ 𝑚
Same procedure was done to obtain torques acting on other shafts

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Real Forces
𝐹𝐵1
=
2𝑇1
𝑑𝑝1
=
2(358.1𝑁∙𝑚)
0.105𝑚
= 6821𝑁 𝑊
𝑓 =
2𝑇
𝑑
𝐹𝑊1
=
𝐹𝐵1
𝐶𝑜𝑠𝜃
=
6821
𝐶𝑜𝑠20
= 7259𝑁
𝐹
𝑟1
= 𝐹𝑊1
𝑆𝑖𝑛𝜃 = 7259𝑆𝑖𝑛20 = 2483𝑁
𝐹1𝑡
= √(𝐹𝑅1
)
2
+ (𝐹𝐵1
)
2
= √(2483𝑁)2 + (6821𝑁)2 = 7259𝑁 → 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝐹𝑜𝑟𝑐𝑒
Same procedure was done to obtain real forces for the remaining gear pairs

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Lewis Forces
From table 2a and a BHN 302-351
𝑆𝑝𝑢𝑟 𝐻𝑒𝑙𝑖𝑐𝑎𝑙 (𝐵𝑒𝑛𝑑𝑖𝑛𝑔) = 151.7𝑀𝑃𝑎; 𝑘 = 4.49
𝑏 = 50𝑚𝑚 → 𝑎𝑠𝑠𝑢𝑚𝑒𝑑
 Bending
𝐹𝑏 = 𝜎𝑏𝑌𝑃
𝑃 = 𝜋 𝑥 𝑚 = 15.71𝑚𝑚 = 0.01571𝑚
𝐹𝑏1
; 𝑌 = 0.1130
𝐹𝑏1
= 151.7 × 106 𝑁
𝑚2
⁄ (0.050𝑚)(0.1130)(0.01571𝑚) = 13465𝑁
𝐹
𝑤𝐿1
= 𝑄𝑏𝑑𝑘
𝑄1 =
2𝑁2
𝑁1 + 𝑁2
=
2(120)
30 + 120
= 1.6
𝐹
𝑤𝐿1
= (1.6)(0.050𝑚)(0.005𝑚 ∗ 30) (4.49 × 106 𝑁
𝑚2
⁄ ) = 53880𝑁
Same procedure was done to obtain Lewis forces for other gears.

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Results
Table 1 Partial Relations
Input RPM Shaft 1 & 2 Shaft 2 & 3
400
1st
Gear
Pair M 5
4th
Gear
Pair M 5
Puley ratio Reducer gear ratio 30
i1 0.250
30
i4 0.250
0.5 0.5 120 120
Given: Z1=21, Z2=42,
2nd
Gear
Pair M 5
5th
Gear
Pair M 5
Engine RPM 42
i2 0.389
77
i5 1.055
1600 108 73
Power [kW]
3rd
Gear
Pair M 5
6th
Gear
Pair M 5
30 58
i3 0.630
89
i6 4.684
θ 92 61
1.6 19
Minoutput
[rpm]
25
Max output
[rpm]
1073.741824
Table 2 Output gear relation
output
gear M 5
38
i7 0.905
42

19 | P a g e
Table 3 Total Relations and Output speeds
1st
Gear Pair 2nd
Gear Pair Total ratioTotal ratio Actual Actual rpm
tooth tooth w/output rpm in order
30 30
120 120
30 77
120 73
30 89
120 61
42 30
108 120
42 77
108 73
42 89
108 61
58 30
92 120
58 77
92 73
58 89
92 19
0.097
0.410
0.264
1.060
J7
J8
J9
1.822
0.158 63.0
38.9
J6
0.665
1068.7
659.3
423.8
266.0
164.1
25.0
0.665
0.063
0.264
1.171
J1
J2
J3
J4
J5
2.953
25.0
105.5
423.8
38.9
164.1
659.3
63.0
266.0
1068.7
0.097
0.410
1.648
0.158
2.672
0.063
105.5

20 | P a g e
Table 4 Shaft Speeds, and distances between shafts
Speed per
shaft Rpm Torque Max
Shaft 1 n1 400.0 11464.97 [N.m]
Shaft 2 n1 100.0 Torque Min
n2 155.6 268.19 [N.m]
n3 252.2
Distance Between
Shafts
Shaft 3 n1 25.0 Shaft 1 & 2 [mm]
n2 105.5 1st Gear Pair 375
n3 423.8 2nd Gear Pair 375
n4 38.9 3rd Gear Pair 375
n5 164.1 Shaft 2 & 3 [mm]
n6 659.3 4th Gear Pair 375
n9 1068.7

21 | P a g e
Table 5 Reducer Gear Analysis
Reducer Gear
Cálculo de Fuerzas de Lewis
Potencia [W] 30000 Z1 21
N0 [RPM] 1600 Z2 42
N1 [RPM] 800 M 5
N2 [RPM] 400 dp1 0.105
degree] 20 dp2 0.21
Torque I [N.m] 358.1
Torque II [N.m} 716.2
Fuerzas Reales
Fuerza B 6820.9
Fuerza W 7258.7
Fuerza R 2482.6
Fuerzas de Lewis
[psi] 22000 55000
[MPa] 151.7 379.3
b = #*m ( 9 to 14) 9
P 15.7
Y 0.102
dp1 105
Q 1.333
K [MPa] 4.49
Fuerza BL 10939.2
Fuerza WL 28287.0

22 | P a g e
Table 6 Real Forces and Lewis Forces comparison
Shaft 1 & 2 [N]
1st
Gear Pair 9549.3 Fuerza BL PASS
30 10162.1 Fuerza WL PASS
120 3475.7 y Q 1.6
P 15.708 0.1130 b 50.0
2nd
Gear Pair 6820.9 Fuerza BL PASS
108 2482.6 y Q 1.4
P 15.708 0.1210 b 50.0
3rd
Gear Pair 4939.3 Fuerza BL PASS Torque at Shaft 2 for 3rd Gear Pair
92 1797.8 y Q 1.226667
P 15.708 0.1321 b 50.0
13465.5
53880.0
14413.8
67888.8
15741.4
79862.1
[N]
Fuerzas Reales [N] & Fuerzas de Lewis [N] Shaft 1 & 2 , 1st, 2nd, 3rd Gear Pairs
[N.m]
[N.m]
[N.m]
Fuerza W
Fuerza R
2864.8
1841.7
1136.0
Torque at Shaft 2 for 1st Gear Pair
Torque at Shaft 2 for 2nd Gear Pair
Fuerza R
Fuerza B
Fuerza W
Fuerza R
Fuerza B
Fuerza B
Fuerza W

23 | P a g e
Table 7 Real Forces and Lewis Forces Comparison
Shaft 2 & 3 [N]
4th
Gear PairFuerza B 38197.2 Fuerza BL 40396.5517 PASS
30 Fuerza W 40648.6 Fuerza WL 64656 PASS
120 Fuerza R 13902.6 b 60
15.70796 y 0.1130 Q 1.6
5th
77 Fuerza W 15837.1 Fuerza WL 84127.6333 PASS
73 Fuerza R 5416.6 b 50.0
15.70796 y 0.1369 Q 0.97333333
6th
Gear PairFuerza B 12875.5 Fuerza BL 16500 PASS
61 Fuerza R 4686.3 b 50.0
20 y 0.1385 Q 0.8133
P 15.70796
Shaft 2 & 3 [N]
4th
Gear PairFuerza B 24555.3 Fuerza BL 40396.5517 PASS Torque at Shaft 3 for 4th Gear Pair
120 Fuerza R 8937.4 b 60
15.70796 y 0.1130 Q 1.6
5th
73 Fuerza R 3482.1 b 50 [N.m]
15.70796 y 0.1369 Q 0.97333333
6th
Gear PairFuerza B 8277.1 Fuerza BL 16500 PASS Torque at Shaft 3 for 6th Gear Pair
61 Fuerza R 3012.6 b 50
20 y 0.1385 Q 0.8133
P 15.70796
Shaft 2 & 3 [N]
4th
120 Fuerza R 5513.1 b 60
15.70796 y 0.1130 Q 1.6
5th
73 Fuerza R 2148.0 b 50
15.70796 y 0.1369 Q 0.97333333
6th
Gear PairFuerza B 5105.8 Fuerza BL 16500 PASS
61 Fuerza R 1858.4 b 50
20 y 0.1385 Q 0.8133
P 15.70796
[N.m]
[N.m]
[N.m]
[N.m]
[N.m]
[N.m]
[N.m]
[N.m]
11459.2
2716.0
643.8
Torque at Shaft 3 for 4th Gear Pair
With 1st Gear Pair Torque
With 2nd Gear Pair Torque
With 2nd Gear Pair Torque
255.3
4544.1
1077.0
7366.6
1746.0
413.9

24 | P a g e
Table 8 Lewis Form Factor for N-Teeth
Lewis Form Factor for N-Teeth
N teeth Y y
30 0.355 0.11300001
43 0.38 0.12095776
58 0.415 0.1320986
77 0.43 0.13687325
89 0.435 0.1384648
Figure 1 Lewis Form Factor diagram

25 | P a g e
Table 9 Gear Analysis
Table 10 Real Forces
M= 5 θ= 20 All in mm
Z Number of teeth dp Pd db da dd width
1 21 105 0.2 98.66772518 115 92.5 45
2 42 210 0.2 197.3354504 220 197.5 45
3 108 540 0.2 507.4340152 550 527.5 45
4 30 150 0.2 140.9538931 160 137.5 45
5 120 600 0.2 563.8155725 610 587.5 45
6 58 290 0.2 272.51086 300 277.5 45
7 92 460 0.2 432.2586056 470 447.5 45
8 89 445 0.2 418.1632162 455 432.5 45
9 61 305 0.2 286.6062493 315 292.5 45
10 19 95 0.2 89.27079897 105 82.5 45
11 38 190 0.2 178.5415979 200 177.5 45
12 42 210 0.2 197.3354504 220 197.5 45
13 30 150 0.2 140.9538931 160 137.5 60
14 120 600 0.2 563.8155725 610 587.5 60
15 77 385 0.2 361.781659 395 372.5 45
16 73 365 0.2 342.9878066 375 352.5 45
Fuerzas Reales Engine Torque= 358.1
Shaft Torque Fb Fw Fr
I Par 1 (z1 y z2) 358.1 6820.926133 7258.677978 2482.614082
II Par 2 (z2 y z3) 716.1972439 6820.926133 7258.677978 2482.614082
II Par 3 (z4 y Z5) 716.1972439 9549.296586 10162.14917 3475.659715
II Par 4 (z6 y z7) 716.1972439 4939.291337 5256.284053 1797.755025
III Par 5 (z8 y z9) 2864.788976 12875.45607 13701.77416 4686.282762
V Par 6 (z9 y z10) 1963.507051 12875.45607 13701.77416 4686.282762
VI Par 7 (z11 y z12) 611.5841633 6437.728035 6850.88708 2343.141381
III Par 8 (z13 y z14) 2864.788976 38197.18634 40648.59668 13902.63886
III Par 9 (z15 y z16) 2864.788976 14882.02065 15837.11559 5416.612543
IV 11459.1559 38197.18634 40648.59668 13902.63886

26 | P a g e
Table 11 Failure Theories, deflection and shaft diammeter
Mt Mb
358.1 226.829405 I dia. (mm) 30.6 23.2 20.6 22.5
716.2 1521.20131 II dia. (mm) 36.3 39.2 35.0 35.5
2864.78 8737.51681 III dia. (mm) 51.4 69.6 62.0 62.6
11459.16 5681.47868 IV dia. (mm) 72.7 71.1 62.7 69.8
1963.5 565.183245 V dia. (mm) 46.8 37.8 32.6 37.9
611.58 983.929576 VI dia. (mm) 34.9 34.3 30.6 31.4
Minimal Allowed deflection: (0.010)module = 0.05mm
To comply with minimal deflections the following diameters were used:
SF 1.75 Shaft D (mm) Spline d (mm) a b Length (mm) Deflection (mm)
Sy 6.75E+08 I 28.00 14.0124 62.50 62.50 125.00 0.04775
tmax 192857143 II 84.00 82.1960 245.00 245.00 490.00 0.04972
S1 385714286 III 120.00 118.0647 253.00 1022.50 1275.50 0.34069
Splines 6 IV 132.00 115.2475 317.50 250.00 567.50 0.04924
E (Gpa) 205 V 40.00 #NUM! 82.50 82.50 165.00 0.04978
Pm (N/mm2
) 6.5 VI 58.00 54.7995 272.50 75.00 347.50 0.04821
4141 Properties
Shaft
Failure Theories for Shaft Design
ASME Code
for Shaft
Torsional
Rigidity
Max Principal
Stress Theory
Max Shear
Stress Theory

27 | P a g e
Shaft I
x-y
+↓ 𝛴𝑀𝑎 = 0
2,482.6(62.5) − 𝑅𝐵𝑦
(125) = 0
𝑅𝐵𝑦
= 1,241.3
+↑ 𝛴𝐹𝑦 = 0
𝑅𝐴𝑦
− 2,482.6 + 𝑅𝐵𝑦
= 0
𝑅𝐴𝑦
= 1,241.3𝑁
x-z
+↓ 𝛴𝑀𝑎 = 0
6,820(62.5)− 𝑅𝐵2
(125) = 0
𝑅𝐵𝑧
= 3,410 𝑁
+↑ 𝛴𝐹𝑧 = 0
𝑅𝐴𝑧
− 6,820 + 𝑅𝐵𝑍
= 0
𝑅𝐴𝑧
= 3,410 𝑁

28 | P a g e
Shaft I
 Plane X-Y
Figure 2 Shaft I FBD
Figure 3 Shaft I Shear Diagram x-y

29 | P a g e
Figure 4 Shaft I Moment Diagram x-y
Shaft I
 Plane X_Z
Figure 5 Shaft I FBD x-z

30 | P a g e
Figure 6 Shaft I Shear diagram x-z
Figure 7 Shaft I Moment diagram x-z

31 | P a g e
Shaft II
 Plane X-Y
Figure 8 Shaft II FBD x-y
Figure 9 Shaft II Shear Diagram x-y

32 | P a g e
Figure 10 Shaft II Moment diagram x-y
Shaft II
 Plane X-Z
Figure 11Shaft II FBD x-z

33 | P a g e
Figure 12 Shaft II Shear diagram x-z
Figure 13 Shaft II Moment diagram x-z

34 | P a g e
Shaft III
 Plane X-Y
Figure 14 Shaft III FBD x-y
Figure 15 Shaft III Shear diagram x-y

35 | P a g e
Figure 16 Shaft III Moment diagram x-y
Shaft III
 Plane X-Z
Figure 17 Shaft III FBD x-z

36 | P a g e
Figure 18 Shaft III Shear diagram x-z
Figure 19 Shaft III Moment diagram x-z

37 | P a g e
Shaft IV
 Plane X-Y
Figure 20 Shaft IV FBD x-y
Figure 21 Shaft IV Shear diagram x-y

38 | P a g e
Figure 22 Shaft IV Moment diagram x-y
Shaft IV
 Plane X-Z
Figure 23 Shaft IV FBD x-z

39 | P a g e
Figure 24 Shaft IV Shear diagram x-z
Figure 25 Shaft IV Moment diagram x-z

40 | P a g e
Shaft V
 Plane X-Y
Figure 26 Shaft V FBD x-y
Figure 27 Shaft V Shear diagram x-y

41 | P a g e
Figure 28 Shaft V Moment diagram x-y
Shaft V
 Plane X-Z
Figure 29 Shaft V FBD x-z

42 | P a g e
Figure 30 Shaft V Shear diagram x-z
Figure 31 Shaft V Moment diagram x-z

43 | P a g e
Shaft VI
 Plane X-Y
Figure 32 Shaft VI FBD x-y
Figure 33 Shaft VI Shear diagram x-y

44 | P a g e
Figure 34 Shaft VI Shear diagram x-y
Shaft VI
 Plane X-Z
Figure 35 Shaft VI FBD x-z

45 | P a g e
Figure 36 Shaft VI Shear diagram x-z
Figure 37 Shaft VI Moment diagram x-z
Bearings
Assumptions
𝐿𝐷 = 5,000 ℎ
𝐿𝑅 = 1 × 106
𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
𝑎 = 3
𝐶10 = 𝐹𝐷 (
𝐿𝐷 × 𝑁𝐷 × 60
𝐿𝑅
)
1
𝑎
𝐶10 = 7,258.67(
5,000 × 800 × 60
1 × 106
)
1
3
= 45,108.8 𝐾𝑁

46 | P a g e
Shaft C10 ND FD Bore Series OD Width
I 45108.8 800 7258.67 35 03-Series 80 17
II 50124.3 400 10162.2 60 02-Series 110 22
III 161512.2 252 38197.19 100 02-Series 180 34
IV 278148.3 1068 40648.6 130 02-Series 230 40
V 65710.9 367.67 13701.8 45 03-Series 100 25
VI 46882.4 1068.2 6850.9 55 03-Series 100 21
Bearing Design
The value of Bore, Outside Diameter (OD), Series and Width are obtain of the Table
11-3 of our textbook Shigley’s Mechanical Engineering Design 9th Edition.
Belt Drive
Design Power
= 𝑃𝑚𝑜𝑡𝑜𝑟 × 𝑆𝑒𝑟𝑣𝑖𝑐𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 = 30𝐾𝑊 × 1.34 × 1.4 = 56.32 ℎ𝑝
𝐵𝑒𝑙𝑡 = 3𝑉
𝑖 =
𝑛2
𝑛1
=
800
1600
=
1
2
𝐷1 =
12 × 𝑉𝑜𝑝𝑡𝑖𝑚𝑢𝑠
𝜋 × 𝑛1
=
12 × 4000
𝜋 × 1600
= 9.55"
𝑆𝑡𝑎𝑟𝑑𝑎𝑟𝑒𝑑 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟; 𝐷1 = 7.95"
𝐷2 = 𝐷1 × 2 = 7.95 × 2 = 15.9"
Table 12 Bearing Design

47 | P a g e
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑃𝑜𝑤𝑒𝑟 = 9.5 ℎ𝑝
𝐴𝑑𝑑𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 = 0.5 ℎ𝑝
15.9 < 𝐶 < 71.55
𝐶 = 44"
𝐿 = 2(44)+ 1.57(7.95 + 15.9) +
(15.9 − 7.95)2
4(44)
= 125.8"
𝐿𝑠𝑡𝑎𝑛𝑑𝑎𝑟 = 125"
𝐶𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑 =
𝐵 + √𝐵2 − 32(𝐷2 − 𝐷1)2
16
=
450.09 + √450.092 − 32(15.9 − 7.95)2
16
= 56.11"
𝐵 = 4 × 𝐿 − 6.28(𝐷2 − 𝐷1) = 4 × 125 − 6.28(15.9 − 7.95) = 450.09
𝜗1 = 171.9°, 𝜗2 = 188.1°, 𝐶𝜗 = 0.98, 𝐶𝐿 = 1.12
𝐵𝑒𝑙𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 =
𝐷𝑒𝑠𝑖𝑛𝑔 𝑃𝑜𝑤𝑒𝑟
𝑃 × 𝐶𝜗 × 𝐶𝐿
=
56.32
9.5 × 0.98 × 1.12
= 5.40 𝑏𝑒𝑙𝑡

48 | P a g e
Conclusions
After calculations and analysis we have designed a transmission that is fully
functional complying with the initial requirements. The 9 shift transmission design is
composed of 6 shafts and 16 gears. At the same time the transmission will have a
minimum output of 25RPM and a maximum of around 1,068 RPM. This explains the
proper function of the mechanism.
On the other hand, it was necessary to analyze the mechanism in order to draw
out the manner in which the transmission will look in a blue print. This drawing shows
the compilation of our analysis in one page were one can visualizes the behavior and
“connectivity” between the gears, shaft, pulleys, motor, etc. further explaining our
design.

49 | P a g e
References
 Shigley’sMechanical EngineeringDesign9th
Ed.(Budynas,Nisbett,2010)
 www.matweb.com
 www.wikipedia.com

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