COLLEGE OF ENIGINEERING AND TECHNOLOGY

DEPARTMENT OF MECHANICAL ENGINEERING

STREAM OF MOTOR VEHICLE
INTERNSHIP REPORT AND PROJECT
ON POWER SCISSOR JACK
HOSTING COMMPANY ETHIOPIA ROAD
CONSTRUCTION CORPORATION ADIGRAT DISTRICT
(ERCC)
Prepared by
Group members ID NO
1. Adhanom W/gebriel 2667/07
2 .Letezgiher Hailu 3635/07
3. Libanos Tsegay 4544/07
4. Mebrahtu Mehari 3714/07
5. Mihreteab K/mariam 3818/07

Duration of Internship; February 29/2010E.C – July 06/2010E.C

Company Advisor: Girmay Hagos (MSc)
University Advisor: Melaku (MSc)
Submission Date: 27/02/2011 E.C
Tel: 034 445 2226/04 Post No: 19
Fax: 034 445 0740 Web Site: www.ecwc.gov.et
Place: Ethiopia, Tigray, Adigrat
Internship Report and Project on Power Scissor Jack Stream of Motor Vehicle

DECLARATION
We declare this written sub mission represents our ideas in our own words and others’ ideas or
words have been included. We also declare that we have adhered to all principles of academic
honesty and integrity and have not misrepresented and fabricated or falsified any idea, data or
source in our submission. We understand that any violation of the above will be cause for
disciplinary action by the institute and can also evoke penal action from the sources which have
thus not been properly cited or from whom proper permission has not been taken when needed.

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ACKNOWLEDGMENT
First of all we would like to take this opportunity to give our special thanks to our dedicated
supervisor, Melaku (M.Sc.) for guiding this project at every stage with clarity, spending much
time to discuss and help with this project, and that priceless gift of getting things done by sharing
his valuable ideas as well as share his knowledge. We express our sincere thanks to the Head of
Mechanical and Automation Engineering Department Manager Girmay Hagos, for his most
valuable guidance, advice and encouragement in all stages of our internship work. We extend our
thanks to all mechanical department staff members and all mechanics all who directly and
indirectly helped us to bring this internship report and project successfully completed in time.
Finally, we would like to express our deepest and special long lasting thanks to our friends,
everyone who helped us for everything that they have done and sacrificed a lot for this internship
report and project in the given months.

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EXECUTIVE SUMMARY
This internship program introduces out to the real world and test our performance. This report
and project is detailed covered all the activities in the internship program from the beginning
until the end. Those are brief history of the company (E.C.W.C) Company, overall internship
experience and the overall benefits of the internship. In addition our problem defines are
included in the ERCC Company Automobile Gear Box Holder. This report and project is listed
the company problem, cause of the problem and solutions of the problems.
Generally, this internship experience we have got good knowledge by observing the outside
environment and the overall benefits that we gained from this internship experience including:
Improving our practical skill, Upgrading our theoretical knowledge, Improving our interpersonal
communication skills, Improving our team playing skill, Improving our leadership skill we are
understanding about work ethics and related issues (problems) and Upgrading entrepreneurship
skill. Finally the internship program is tested our performance in the real work.

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TABLE OF CONTENTS
DECLARATION ........................................................................................................................................... i
ACKNOWLEDGMENT............................................................................................................................... ii
EXECUTIVE SUMMARY ......................................................................................................................... iii
LIST OF TABLES ...................................................................................................................................... vii
LIST OF FIGURES .................................................................................................................................... vii
NOMENCLATURE AND ABBREVIATIONS ........................................................................................ viii
PART-ONE ................................................................................................................................................... 1
1. Back Ground and History of the Company ............................................................................................... 1
2. Vision ........................................................................................................................................................ 2
3. Mission...................................................................................................................................................... 2
4. Company Goal .......................................................................................................................................... 2
5. Company Customers ................................................................................................................................. 2
6. The Main Purpose and Future Plan of the Company ................................................................................ 3
7. The Main Product and Services of the Company...................................................................................... 3
7.1 Overall Organization and Work Flow of the Company ...................................................................... 3
8. Overall Internship Experience .................................................................................................................. 5
8.1. What Is An Internship ........................................................................................................................ 5
8.2. Upgrading practical skills .................................................................................................................. 5
8.3. Upgrading theoretical knowledge ...................................................................................................... 6
8.4. Industrial problem solving capability ................................................................................................ 6
8.5. Improving interpersonal skills ........................................................................................................... 6
8.6. Improving interpersonal skills ........................................................................................................... 6
8.7. Leadership skills ................................................................................................................................ 7
8.8. Work Ethics ....................................................................................................................................... 7
8.9. Entrepreneurship ................................................................................................................................ 7
PART TWO .................................................................................................................................................. 9
CHAPTER ONE ......................................................................................................................................... 10
1.1 Introduction ....................................................................................................................................... 10
1.2 Problem Statement ............................................................................................................................ 10
1.3 Objective of the Project .................................................................................................................... 11
1.3.1 General Objectives ..................................................................................................................... 11

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1.3.2 Specific objects .......................................................................................................................... 11

1.4 Scope of the Project .......................................................................................................................... 12
1.5. Limitation......................................................................................................................................... 12
1.6. Significant of the project .................................................................................................................. 12
CHAPTER TWO ........................................................................................................................................ 13
2. LITERATURE REVIEW ....................................................................................................................... 13
CHAPTER THREE .................................................................................................................................... 16
3. METHODOLOGY AND MATERIAL SELECTION............................................................................ 16
3.1 Methodology ..................................................................................................................................... 16
3.2. Material Selection ............................................................................................................................ 18
3.3. Manufacturing Process..................................................................................................................... 19
3.3.1 Manufacture Bracket .................................................................................................................. 19
3.3.2 Manufacture of Power Screw ..................................................................................................... 20
3.3.3 Manufacturing Process of Rivet ................................................................................................. 21
3.3.4 Mass of Rivet ............................................................................................................................. 21
3.3.5 Manufacturing Process of Base Plate ......................................................................................... 21
3.3.6 Mass of Base Plate ..................................................................................................................... 21
3.3.7 Manufacture Process of the Handle ........................................................................................... 22
3.3.8 Mass of Hand Power .................................................................................................................. 22
3.3.9 Manufacture Process of Link’s .................................................................................................. 22
3.4 Cost Analysis .................................................................................................................................... 24
3.4.1 Material Cost .............................................................................................................................. 24
3.4.2 Labor Cost .................................................................................................................................. 24
3.4.3 Standard Cost ............................................................................................................................. 24
CHAPTER FOUR....................................................................................................................................... 26
4. DESIGN ANALYSIS ............................................................................................................................. 26
4.1. Design Concept ................................................................................................................................ 26
4.1.1. Concept Generation................................................................................................................... 26
4.2 Given Specification ........................................................................................................................... 28
4.3 The Force Analysis on the Screw ..................................................................................................... 32
4.5 Design of Power Screw Housing ...................................................................................................... 39
4.7 Analysis for the Four Rivets (Nuts) .................................................................................................. 40

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4.8 Analysis for the Base Plate ............................................................................................................... 41

4.9 Force Distribution for the Fastener (Connecting Member) at the Top ............................................. 42
4.10 Design for Cup at the Top (Break).................................................................................................. 43
4.11 Analysis of Arm’s ........................................................................................................................... 45
4.11.1. Analysis of Upper Arm’s ........................................................................................................ 45
4.11.2. Analysis for the Lower Arms .................................................................................................. 47
4.12 Design of the Driving Handle ......................................................................................................... 48
CHAPTER FIVE ........................................................................................................................................ 50
5. CONCLUTION AND RECOMENDATION ......................................................................................... 50
5.1 Conclusion ........................................................................................................................................ 50
5.2 Recommendation .............................................................................................................................. 51
APPENDIX ................................................................................................................................................. 53
DETAIL DRAWING .................................................................................................................................. 53
REFERENCE.............................................................................................................................................. 67

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LIST OF TABLES
Table 1. Length and angle on different given angle.................................................................................... 17
Table 2. Cost analysis ................................................................................................................................. 25
Table 3. Concept Generation ...................................................................................................................... 28
Table 4. Length and angle on different given angle.................................................................................... 30
Table 5. Change of height jack lift at d/t angle ........................................................................................... 30
Table 6. Force on links at different angle ................................................................................................... 36

LIST OF FIGURES
Figure 1. Forced connecting members lifting members, pins and power screw ......................................... 15
Figure 2. Psm Flow chart ............................................................................................................................ 16
Figure 3. Concept I ..................................................................................................................................... 26
Figure 4. Concept II .................................................................................................................................... 27
Figure 5. Concept III ................................................................................................................................... 27
Figure 6. Concept IV................................................................................................................................... 28
Figure 7. Dimension analysis ...................................................................................................................... 29
Figure 8. Angle between links .................................................................................................................... 31
Figure 9. Force analysis FBD ..................................................................................................................... 31
Figure 10. Power screw FBD ...................................................................................................................... 32
Figure 11. Force distribution on base plate ................................................................................................. 32
Figure 12. Power screw............................................................................................................................... 33
Figure 13. Power screw FBD ...................................................................................................................... 33
Figure 14. Power screw housing ................................................................................................................. 39
Figure 15. Rivet force analysis ................................................................................................................... 40
Figure 16. Base plate................................................................................................................................... 41
Figure 17. Force distribution on fastener .................................................................................................... 42
Figure 18. Top fastener strength analysis ................................................................................................... 43
Figure 19. Design of cup at top ................................................................................................................... 44
Figure 20. Upper arm strength analysis ...................................................................................................... 45
Figure 21. Bearing stress by upper arm ...................................................................................................... 46
Figure 22. Lower arm strength analysis ...................................................................................................... 47
Figure 23. Bending stress analysis .............................................................................................................. 47
Figure 24. Driving handle analysis ............................................................................................................. 48

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NOMENCLATURE AND ABBREVIATIONS

Hmax Maximum height of the jack, mm
Hmin Minimum height of the jack, mm

σult Ultimate stress, N/mm2

tc Cutting time, sec
tm Drilling time, sec
V Speed of cutting, m/sec2
I Moment inertia, mm4
Ia Moment inertia at a distance, mm4
Ib Moment inertia at base, mm4
F Applied force or load, N
W Width of base plate, mm
b Base of the base plate, mm.
h Height of base plate, mm

σb Bearing stress, N/mm2

σn Normal stress, N/mm2

σy Yield strength, N/mm2

F.S (n) Factor of safety
τ Allowable shear stress, N/mm2
τmax Maximum shear stress, N/mm2
m
g= Gravitational acceleration, 9.81 s2

τu Ultimate shear stress, N/mm2

M Bending moment, N-m.
Mmax Maximum bending moment at the saddle, N-m.
W Weight, N
g Acceleration due to gravity, m/sec2

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P Intensity of fluid pressure, N/m2.

ρ Density, kg/m3

Pb Bearing pressure, N/mm2.

V Volume, m3
A Area, m2
P Pressure, Pa
P Applied load, N
Fp Force on different links, N
L Length, mm
ri Internal radius, mm.
R External radius, mm.
t Thickness, mm.
VT Total volume occupied on the jack, m3.

β, θ Angles on the different links of the jack, degree.

σall Allowable stress, Mpa

σb , σall Allowable stress on the base plate, Mpa

σt Tangential (tensile stress), Mpa.

µ Poisson Ratio.
di Inner diameter, mm
do Outer diameter, mm
W Weight of the lifting links, N
τall Allowable shear stress, N/mm2

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PART-ONE
1. Back Ground and History of the Company
The history of ECWC in Ethiopia can be traced back into the 17th and 18th century. During that
time there were a number of small roads, trails and foot paths, in addition to the traditional
shoulder porter age, different animals were used as a means of transportation in Ethiopia. In the
18th century, especially during the reign of Emperor Tewodros, although the technology was not
modern, it was believed that planned road construction efforts were made. It is also believed that
Emperor Yohannes IV, who succeeded Tewodros, was engaged in road building. However due
to the danger of invasion by foreign invaders, the Emperor was not able to achieve his desires.
However construction of modern roads was started during the era of emperor Menelik. Specially,
in 1907, following the advent of cars in Addis Ababa, expansion of road networks had become
vital. In 1903 the road from Eritrea to Addis Ababa and the road from Addis Ababa to Addis
Alem were built. In addition it was during this time that the fist Asphalt roads constructed in
Addis. Though it was not effective, in 1930, road construction contracts were made with the
American-based ‘White Engineering Corporation’. Road construction works were, however,
highly expanded during the five years occupation of fascist Italy to meet the requirement of the
military control rather than to promote the overall development of the country’s economy.
Nevertheless, after the liberation of the country, many of the roads were highly deteriorated and
went out of service. Therefore, after eviction of the Italian occupiers, there was a huge task of
maintaining the already damaged roads and also building new roads. Thus, on January 26, 1951
by Proc. No. 115/1951, the ‘Imperial Highway Authority, IHA’ was established as a semi-
autonomous agency. Then after, series of road construction programs were formulated. The IHA
has been renamed three times and substantially restructured seven times by law in its history.
Finally, in 1978, IHA was dissolved and re-established as ‘Ethiopian construction works
corporation, (E.C.W.C) according to Proc. No. 133/1978.
In 1997, the country embarked on a new era of road sector development. The road sector
development program (RSDP) was officially launched, under which the road network increased
from 26,550 Km to 53,997 Km within 13 years.

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2. Vision
 To becoming world renowned and competent construction company by 2026.
3. Mission
By using competent professionals and modern technologies, the corporation engages on:
 construction and maintenance of roads and bridges,
 Production and sales of construction materials,
 Rental and maintenance of construction equipment in local and foreign countries to
support the economic and social development of the nation.
4. Company Goal
Consistent with its mission and responsibilities, the E.W.C.C’s efforts are guided by the
following core values:
1. Cost Effectiveness 5. Quality fist
2. Transparency and accountability 6. Safety
3. Team work- 7. Environmental friendly
4. Ethics
These core values are elaborated briefly as follows:
 Cost Effectiveness: E.C.W.C strives to create a culture of effectiveness that appreciates
value for money.
 Quality first: To maintain the quality of its product, E.C.W.C understands precisely what
Customers will aim at and consistently delivers accurate solutions within budget and on
time.
 Safety: In a high hazard industry like construction, safety is an instrument that provides
real benefits.
5. Company Customers
The main customers or the end users of E.CWC are;

 Internal customers :
 Road construction equipment and machinery
 Ethiopian road authority
 External customers :

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 Federal police commotion service

 Ethiopian military defense force service
6. The Main Purpose and Future Plan of the Company
 To be a leading and competent construction company in construction sector in the world.
 Avoiding old machines and vehicles that have been given a service above twenty years
 Introducing and replacing modernized machines
 Become compatible and known in the real world
7. The Main Product and Services of the Company
Currently, the major services provided by the corporation are:
 Designing and building roads and bridges,
 Maintaining roads and bridges,
 Maintenance and rental of construction equipment’s,
 Production and Sales of construction materials,
 Production and installment of sign posts, and road marking,
 Other civil works like construction of airfields, dams, irrigation canals, access roads, etc.
 Maintenance of construction equipment and machinery
 Maintenance of federal police services
 Road maintenance (externally)

7.1 Overall Organization and Work Flow of the Company

ECWC is newly established public enterprises with the aim of becoming a leading, preferred and
competitive construction contractor. ECWC is governed by the public enterprises proclamation
No 25/1992.its supervising authority is the ministry of public enter praises and its policy making
body is the board of the corporation whose members are appointed by the government selected
from different organizations. The executive manager of ECWC is accountable to board of
director’s .as per the new organization structural, Deputy executive managers lead. Water
infrastructure construction sector, transport infrastructure construction sector, building
technology & construction, dams and irrigation development administration sector, construction
machinery &equipment sector. The overall organization structure is the following diagram

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Work flow of the department:-The work flow of the section in which we were participating have
various work units. In the project engineer work section there are many working sub sections and
other assistance working sections that play a great role in the success of the project.
Due to this each working groups will perform their job on time that the next group don’t have to
stop working until they finish.
The steps are goes as follows:
1. Request for service (RFS): the driver or operator asks to maintain the problem created on the
machine or vehicle after that the inspector identify the problem of the vehicle or machine
weather by direct observation or by asking the driver or operator.
2. Shop repair order (SRO): in this part the vehicle or construction equipment machine repair
properly depending up on the given problem. After disassembling and maintaining the vehicle or
construction equipment machine the check again weather the problem created on the machine if
normal shop order is closed there is no additional process.
3. Internal repair order (IRO): in this part the labor and time cost required maintaining the
problem is considered by the mechanic.
4. Parts requisition and issue ticket (PRI): in this part if the spare part of the machine or vehicle
is not normal parts requisition and issue ticket prepared to buy the required spare part.
8. Overall Internship Experience
8.1. What Is An Internship?
An internship is an opportunity to apply the knowledge you have gained from your academic
studies in a practical, workplace setting. Internships may be part of a formal internship program,
but many students create their own internships. All internship experiences should provide
exposure to an occupation, industry or career field, have a clear purpose/focus, and a specific
project for you to complete.

8.2. Upgrading practical skills

During the intern the most beneficiary thing is that upgrading of practical knowledge that we
earned their theoretical in the class. Most of the courses that we had earned theoretically are
highly applicable in what we have been working in, and so we have got a chance to relate the
theory with practical knowledge. We almost saw the practical aspects of the courses and their

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application in real work. The practical skills that we gained during the intern are maintaining
and operate the machines, assembly by welding etc…
8.3. Upgrading theoretical knowledge
The contribution of the internship program is not only for practical also it upgrades our
theoretical knowledge. And it increases the ability to apply systematic design procedure to open
ended problem. And how to analyses something technically. So it upgrades our theoretical
knowledge the same as the practical one. Also Internship memorizes the important conceptual
theoretical parts that have learned. Among the important ones that the internship upgrades our
theoretical parts includes
1) Understanding the wiring diagram of machines.
2) Assembling and disassembling engines.
3) The power distribution of the company.
4) The operation of different machine devices.
8.4. Industrial problem solving capability
This capability is very important for us to have good knowledge about the organization situation,
used to know where and when the problem is happened and how it can be solved with in short
time. So we develop such skills in a good manner.
8.5. Improving interpersonal skills
This skill is very important for every person to smooth the relation between the co-workers. Due
to this reason we develop such skills in a good manner like:
1) The ability to speak clearly and confidentially.
2) The ability to listen and understand others (co-workers).
3) The ability giving a solution for a problem
So as exact implementation of this skill it will provide the worker to have good interpersonal
skill and resulting good output in the experience of technical job.

8.6. Improving interpersonal skills

Among the most important skills that we developed during the intern are our communication and
working together /team sprit/ skill. In our company all of the workers’ are sociable. Most of the
time we discussed about the works and other related issues wisely.

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During the internship there were many job responsibilities that we have been done with the
worker’s. So we have learned the following points from them:
1) Listening the idea of others whatever it is.
2) Be patient and wise and respect fullness of the voice of others.
3) Open minded to other point of view.
4) Able to share information freely.
5) Able to be an optimistic and attractive person.
6) Able to celebrate the successes and learn the setbacks.
7) Focus on the positive and take the negative in stride not let them down.
8.7. Leadership skills
Generally, we have noticed the following important points from the leaders:
1) Ability to encourage and motivate workers to do their best
2) The ability to communicate with all workers without any force
3) The ability to listen problems and focus on its solution.
4) As leader to have clear and defined goals missions and vision.
5) Be willing to admit and learn from failures and weaknesses.
8.8. Work Ethics
The benefit of the internship program was interesting in improving ethical work habit. Although
it is difficult to memorize all benefits that are gained about work ethics, it is important to
mention the following
1) Learning and respecting of any work tasks that were important in order to do those tasks
effectively & efficiently.
2) Be on time at work and leave work by the right time (punctuality).
3) Be responsible in analyzing and interpreting gathering of data.
4) Being loyal and honest when doing judgments on the issues of: - Cost vs. benefit
analyzing, Safety and also Quality and Quantity when doing any project.

8.9. Entrepreneurship
Internship plays a great role for improving our entrepreneurship skills in multidirectional ways.
Among these ways some of that we have gained benefits includes:
1) Increases our attitude to be creative and innovative through practical knowledge
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2) How organize and control resources to ensure the profit for our business
3) The way how to manage and own our business
4) How to identify new products (services) opportunities
5) Allows too willing to take calculated risks is risk eliminates
6) Desire for immediate feed backs
 Major Problems of the Company
 Lack of simple machine that are used to left and lower materials from their positions
 Lack of specified areas to store the idle vehicles
 Complex maintenance areas even it gives a full and quality fixings
 During maintenance all most all it uses a new replacement rather than renew the first
one this leads to avoid such parts without giving any service.

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PART TWO

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Internship Report and Project on Power Scissor Jack Stream of Motor Vehicle

CHAPTER ONE
1.1 Introduction
A jack is a mechanical device used as lifting device to lift heavy loads or apply great forces.
Jacks employ a screw thread or hydraulic cylinder to apply linear forces. Car jacks use
mechanical advantage to allow us to lift a vehicle by manual force alone. More powerful jack use
hydraulic power to provide more lift over greater distance. A scissor jack is a device constructed
with a cross-hatch mechanism, much like a scissor, to lift up a vehicle for repair or storage. It
typically works in just a vertical manner. The jack opens and folds closed, applying pressure to
the bottom supports along the crossed pattern to move the lift. When closed, they have a
diamond shape. Scissor jacks are simple mechanisms used to drive large loads short distances.
The power screw design of a common scissor jack reduces the amount of force required by the
user to drive the mechanism. Most scissor jacks are similar in design, consisting of four main
members driven by a power screw. A scissor jack is operated simply by turning a small crank
that is inserted into one end of the scissor jack. This crank is usually "Z" shaped. The end fits
into a ring hole mounted on the end of the screw, the scissor jack can lift a vehicle that is several
thousand pounds. The early man apply a crude way of lifting objects to great Heights
through the use of ropes and rollers, which was mostly applied in the construction industry,
where, it was used to raise mortar (cement, sand & water).After the industrial revolution, with
the advent of automobile, the automotive industry was also faced with the challenge of load
lifting, because of the bulkiness of some automotive parts. This report presents the study of a
scissors lift for the automotive industry.

1.2 Problem Statement

Available jacks are typically large, heavy and also difficult to store, transport, carry or move
into the proper position under an automobile and also jacks present difficulties for the elderly
people and women and are especially disadvantageous under adverse weather conditions.
Presently available jacks further require the operator to remain in prolonged bent or squatting
position to operate the jack which is not ergonomic to human body. It will give physical
problems in course of time. Moreover, the safety features are also not enough for operator to
operate the present jack. Furthermore, the purpose of this project is to overcome these problems.

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Since the aim of the project is to design mechanical scissor jack, the main statement of the
project is to modify the handle and the head of the power screw of the jack to reduce power loss
and to increase efficiency. Before we design the machine, the head of the power screw is circular
plate with a hole which used to connect the handle and the screw. But in our design, we are
modifying the head of the power screw and handle. The outside of the head of the screw is
hexagonal and the handle is solid circular shaft. The shaft has an internal hexagonal hole which
is connected to the head of the screw. An electric car jack which has a frame type of design by
using electricity from the car will be developed. Operator only needs to press the button from the
controller without working in a bent or squatting position for a long period of time to change the
tire.

1.3 Objective of the Project

1.3.1 General Objectives
To design a power scissor jack which is safe and reliable to raise and lower the load easily and
also to design an efficient scissor jack that can be used in the automobile sector.

1.3.2 Specific objects

 To show detail design analysis of each components of scissor jack:
 To design cup at the top brake
 To design fastner
 To design link
 To design power screw
 To design Power screw housing
 To design nut
 To design rivet
 To design handle
 To design base
 To design top view
 To design front view at minimum height
 To design assembled view

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1.4 Scope of the Project

Scope of this Jack is to assess Jack for vehicles. Here the necessity lies in reducing the human
effort applied during operation of the jacks and hence the need of the invention. In day to day life
it is very tedious job to operate the jack and it is also a very time consuming work as well. So
that to make it easier for everyone especially for aged person and for lady drivers. But the
general scope of the project is to minimize the human effort while operating the jack.

1.5. Limitation
During designing scissor jack, we have got different limitation.
Some of these are;
 Lack of sample worked scissor jack in the company
 Limitation in finding enough literatures concerning with the scissor jack.
 Limitation of internet accesses.
 Limitation of references about scissor jack.
1.6. Significant of the project
 To reduce the weight of the jack by changing the manufacturability.
 To reduce the no. of parts for simplifying the assembly process
 To select proper material selection for each part of Scissor jack,
 To show detail design analysis of each components of scissor jack,
 To show assembly drawing of the scissor jack,
 To show the flow of manufacturing process of scissor jack.

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CHAPTER TWO
2. LITERATURE REVIEW
A lifting device is a system that allows small force (effort) to overcome a large force or
load. There are practically hundreds of uses for lift tables in manufacturing, warehousing and
distribution facilities. The addition of this device (lift table) makes job faster, safer and easier.
Some typical applications include; machine feeding and off-loading, product assembly,
inspection quality control repair, feeding and offloading conveyor levels. The
commonest method for operating a scissors jack is the use of a power screw. Power screws are
devices that provide means for obtaining large mechanical advantage. Other researchers define
the power screw as a device used in converting rotary motion into uniform longitudinal motion
the manually operated scissor jack is a device that makes use of a horizontally placed power
screw to overcome large load through less effort applied on the power screw, by turning the
power screw with the aid of a ratchet handle on one side of the device. The device is capable of
lifting an average load with little effort applied. One of the most important factors of lift platform
are its stability. Knowing that stability is a source of concern for a lift platform, its positioning
should be on a flat surface and the load should be place or concentrated at the center of gravity of
the table. Other constraint to be considered is the deflection of the unit. Deflection in scissor jack
can be defined as the resulting change in elevation of all parts of a scissor jack assembly,
typically measured from the floor to the top of the platform deck, whenever load is applied to or
removed from the lift. Safety requirement for industrial scissor jack states that “All industrial
scissor lift will deflect under load” [1]. A scissor jack is a mechanical device used as a lifting
device to lift heavy loads or apply great forces. Car jacks use mechanical advantage to allow us
to lift a vehicle by manual force alone. More powerful jacks use hydraulic power to provide
more lift over greater distance. A scissor jack is a device constructed with a cross-hatch
mechanism, much like a scissor. A scissor jack is operated by turning a lead screw. It is
commonly used as car-jacks. In the case of a scissor jack, a small force applied in the horizontal
plane is used to raise or lower large load. A scissor jack’s compressive force is obtained through
the tension force applied by its lead screw [2].

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A scissor lift contain multiple stages of cross bars which can convert a linear displacement
between any two points on the series of cross bars (provided they are not on the same bar) in to a
vertical displacement multiplied by a mechanical advantage factor the main disadvantage of the
scissor lift is that when the initial actuator force required to begin rising the load is much greater
to the other lift mechanisms. The actuator force is applied to the base of the scissor lift as shown
in the figure below. When the fully loaded position the angle θ approaches zero. And tan θ value
of a small number becomes even smaller as a result the initial force required to begin rising the
load become very large and therefore the stress on the linkage and joints becomes very high. In
order to withstand the large stress the component of the lift would have to be tough and heavy to
withstand the given stress [3]:
Stress = actuator force applied to the middle linkage,
W= weight = F*a
F*a = [W sin θ], the compression force applied to the link this axially which determines
the design of the link.
Fx =F*a*cos θ, =the force applied to the screw
The upper brake cup and the base carries full of the load, since the scissor jack is symmetric the
first two half links carry half of the weight equally. So the design of the screw will depend on the
resolved force which is applied tensile to the screw.

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Figure 1. Forced connecting members lifting members, pins and power screw
The life span of the jack will depend greatly on the type of materials used for each component to
avoid failure. The contact members, connecting members lifting members, pins and power screw
will all use the high strength low-alloy steel with due to the following reasons:
Good machinability
Good ductility
High strength
Wear resistance
Ease of producing component parts
Economical

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CHAPTER THREE
3. METHODOLOGY AND MATERIAL SELECTION
3.1 Methodology
Methodology is one of the most important things to be considered to ensure that the project will
run and achieve the objective. Project methodology will describe the flow of the project progress.
This shows as how the project started, how data was collected and how the next steep done. The
following methodology will be used to accomplish the goals/objectives of this project by
referring different literatures concerning with the scissor jack to collect the most important and
required data’s/information’s. Preparing different possible mechanisms and comparing them by
different criteria through design matrix and selecting the most efficient one.
Select the title

Determine the Objectives and Scopes

Problem analysis

Literature review
Study on previous research Study on previous research

Create new design

Analysis of result

Problem analysis and solutions

Discussion

Conclusion
Figure 2. Psm Flow chart
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This procedure is done by the method of the following system

 Primary data collection
 Interviews to operators
 Interviews to technical assistance
 Interviews to guidance and advisors
 Secondary data collection
 Internet access & different text books

 Given Specification
 Max height = 380mm
 Min height = 160mm
 Length of the link does not vary (constant)
 Capacity (load to be lift) = 1tone =1000kg=9,810N
 Factor of safety =2.5

Maximum height jack stand is:-

H max = Δh + 2a
= Δh + 60; where H max = 380mm
Δhmax= 320mm hmax
For maximum and minimum height length of the like should be calculated

Δh = Lsinθ1 + Lsinθ2; when θ1 = θ2 = θ

Δh =2Lsinθ
L = Δhmax/2sinθ hmax=320mm
L = 320/2sinθ
Table 1. Length and angle on different given angle

θ1=θ2= 17.7600 200 300 400 450 500 550 600 700
θ
L(mm) 170.26 170. 170.26 170.268 170.268 170.268 170.268 170.268 170.268
8 268 8
β 72.9240 700 600 500 450 400 350 300 200



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Depend on the above value of L, we have to get the value that approximate or equal to the
change of weight by making change of height unknown and using the above length at each
angle of inclination.

Δh = 2Lsinθ
Let at L =170.268mm and θ =700
Δh = 2(170.268) sin700 = 320mm

3.2. Material Selection

 Material selection for rivets
The material for rivet should be tough and ductile
 ductile cast iron
 oil quenched (temped)
σy
=621Mpa, σnt = 827Mpa and F.S= 2.5

 Material selection for base plate fastner

 steel alloy AISI-1030
 normalized(8400)
 σy = 655Mpa, σnt = 1025Mpa and F.s= 2.5
Now, σall = σy/F.s= 655 / 2.5 =262Mpa
 Material selection for cup
 steel alloy 4340
 normalized (@8700)
σult =862Mpa, σy = 1200Mpa
σall = σy / F.S= 862 / 2.5 =344.8Mpa
max = σsy /F.S = 0.577σy /F.s
max = 0.577 * 862 / 2.5 =497.374Mpa
Now axial compressive stress
σall= F / A; where A = b * w =50mm * 10mm = 500mm2=0.0005m2
σall = 9381N / 0.0005m2 =18.76Mpa
Since σ<σall it safe
When σall = 344.8Mpa and F = 9810N area should be
A = F / σall = 9810/ 344.8Mpa = 24.45mm2
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 AISI-1030
 Normalized steel
σy =345Mpa, σall = 521Mpa and F.s = 2

t = 3mm @d = 5mm hole

 Material selection for arms
 Steel alloy 4340
 Normalized
 σy = 862Mpa and σult = 1280Mpa and
 F.s= 2.5
 Material selection for driving handle
 steel 1030
 normalized
 σult =520Mpa, σy = 345Mpa and F.s= 1.5
 The required power and efficiency of the scissor jack would be:
 P=10.4W
 η = 0.9851 = 98.51%

3.3. Manufacturing Process

3.3.1 Manufacture Bracket
To manufacture this plate what we should have to be known is this plate must be strong because
it holds the heavy weight. This plate can be manufactured as one plate as shown on part drawing
by using mining and make drill on place must be drilled.
The volume, mass the bracket plat should be required is:-
Volume of bracket
V = A * h = w * b *h
V = 3000m3; where b =50mm, w = 2mm, and h = 30mm, A=100mm
Mass of bracket
M=ρ*v
M =0.5kg; where ρ =1.66*10-4kg/mm3 (from property of material)
Drilling time
tm = l /Fr,
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Where feeding rate, fr = N * f =140.1mm/min

Rotation speed, N = 1000*V / πd =700rpm; d=10mm
Speed of cutting V =22m/min
Feeding = 0.2mm/rev both velocity and feeding are selected
Then tm = 100 / 140.1 = 0.713min

3.3.2 Manufacture of Power Screw

To manufacture power screw we should have to be known the method must be used. Then to
manufacture this part first we select material have round shape and cut by using cutting machine
on appropriate place and use milling machine to prepare the place where the hand driving can be
Inters (head of the power screw) and drill the head of this power screw by using drilling machine
and make the tread by using milling machine.
During manufacturing this material volume, weight and time takes to manufacture should be
considered.
Volume of power screw
V = A*h = π/4*d2 * h
V = 35325.9mm3; where d = 10mm, A =78.5mm2 and length h =450mm
Mass of power screw
M = ρ * V =0.353259mm3*7.85
M = 2kg; where ρ = 5.65g/cm3 (from property of material)
Cutting time
tc = A / as, where specific cross section As = 5000mm2/min area of surface cutting machine cute
within time.
tc= 78.5/ 5000 = 0.0157min
Drilling time
tm= l /fr,
Where feeding rate, fr = N * f =93.62mm/min
Rotation speed, N = 1000*V / π*d = 374.5rpm;
Speed of cutting (v) = 20m/min
Feeding = 0.25mm/rev both velocity and feeding are selected
Then tm = 23.5/ 93.62 = 0.25min

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3.3.3 Manufacturing Process of Rivet

Manufacture of rivet can be the same as the other means it not required new procedure to
manufacture it use milling machine to manufacture after selecting appropriate material used to
rivet.
The volume, mass the rivet should be required is:-
Volume of rivet
V = A * h =π /4 * d2 *h
V =785mm3; where d = 5mm, h =40mm and A =19.63mm2
3.3.4 Mass of Rivet
M=ρ*v
M =178.632g; where ρ =0.0127g/cm3 (from property of material)
Cutting time
tc = A / as, where specific cross section as = 5000mm2/min area of surface cutting machine cute
within time.
tc = 785mm3/ 5000 = 0.57min

3.3.5 Manufacturing Process of Base Plate

Material selected for the bottom plate is steel by having AISI-1030.
Manufacture of this base plate can be may have different type but I use to manufacture this plate
flat material have some tackiness. To manufacture this plate first cut this metal by using power
hack saw in different wanted part as we seen on figure scissor jack part drawing and use milling
for some surface finishing.
The volume, mass the base plat should be required is:-
Volume of base plate
V = A * h = w * b *h
V = 8100mm3; where b=134mm, w=2mm, h=30mm and A = 270mm2

3.3.6 Mass of Base Plate

M=ρ*v
M =0.5g; where ρ =6.172g/cm3 (from property of material)
Cutting time
tc = A / As, where specific cross section As = 5000mm2/min area of surface cutting machine
cute within time.
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tc = 8100/ 5000 = 1.62min

Drilling time
tm = l /fr,
Where feeding rate, fr = N * f = 127.33mm/min
Rotation speed, N = 1000*V / π*d =636.62rpm; d=10mm and
Speed of cutting V =20m/min

Feeding =0.2mm/rev both velocity and feeding are selected

Then tm = 65 / 127.33 = 0.51min

3.3.7 Manufacture Process of the Handle

This hand power can be manufactured from round shape metal means that we select the
appropriate metal which resist the load applied on it considering the scissor jack size. It
manufactured by making it easy to care and to use.
The volume, mass the hand power should be required is:-
Volume of hand power
V = A * h =π /4 * d2 *h; where d =15mm, h =80mm A = 176.714mm2
V =14137.16mm3

3.3.8 Mass of Hand Power

M=ρ*v
M =0.2kg; where ρ =1.4147kg/mm3 (from property of material)
Cutting time
tc = A / As , where specific cross section As = 5000mm2/min area of surface cutting machine
cute within time.
tc = 176.714mm2 / 5000 =0.035min

3.3.9 Manufacture Process of Link’s

The upper link and lower links are equal of the scissor jack can be manufactured from flat plate,
by cutting on its appropriate dimension by using power hack saw and then to get wanted shape
using driller machine.
To manufacture both upper and lower link we should know the volume, mass and time it take to
manufacture each link.

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3.3.9.1 Upper Link

Volume of upper link
V = A * h = b * w * h; where b=26mm, w=22mm, h=170.268mm and A= 572mm2
V = 97393.296mm3
Mass of upper link
M=ρ*V
M =0.5kg; where ρ =5.133g/mm3 (from property of material)
Cutting time
tc = A / As , where specific cross section As = 5000mm2/min area of surface cutting machine
cute within time.
tc = 572/ 5000 = 0.1144min
Drilling time
tm = l /fr,
Where feeding rate, fr = N * f =227,591.6mm/min
Rotation speed, N = 1000*V / πd = 910,366.3rpm; d=13mm and Speed of cutting V=22m/min
Feeding = 25mm/rev both velocity and feeding are selected
Then tm = 320/227,591.6 =0.00141min

3.3.9.2 Lower Link

Volume of lower link
V = A * h = b*w* h; where b=26mm, w=22mm, h=170.268mm and A= 572mm2
V = 97393.296mm3
Mass of lower link
M=ρ*V
M =0.5kg; where ρ =5.133g/mm3 (from property of material)
Cutting time
tc = A / As, where specific cross section As = 5000mm2/min area of surface cutting machine
cute within time.
tc = 572/ 5000 = 0.1144min
Drilling time
tm = l /fr,

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Where feeding rate, fr = N * f = 227,591.6mm/min

Rotation speed, N = 1000*V / π*d =910,366.3rpm; d= 13mm and
Speed of cutting V = 22m/min, Feeding = both velocity and feeding are selected. Then tm =
320/227,591.6 =0.00141min
Generally, to fulfil this manufacturing process we can use different technique but from them the
above is the main use full manufacturing process for our scissor jack project. This manufacturing
process effectiveness and manufacturing cost can vary from one industry area to another area due
to machines we used.

3.4 Cost Analysis

In this section we see the all cost the machine take to manufacture means cost for labor, cost for
each material, and cost of standard components of scissor jack.
Total cost of scissor jack:
Total Cost = Manufactured Cost + Labor Cost + Standard Cost

3.4.1 Material Cost

Material cost means the cost of each material in project. Means the cost each material have
regarding its material property, the mass of each material have and cost it be sold per kilogram.

3.4.2 Labor Cost

Labor cost means the cost of all components should have regarding its process of manufacturing
and time it take to manufactured. Not only this one labor cost means expenditure made of on the
salaries, wages, overtime, bonuses, etc. of the employer of the inter price.

3.4.3 Standard Cost

From the let mechanical scissor jacks 300 birr were sold, now we have minimized it in to 271
birr.
Standard cost means the cost of standard material gained from shop.
As shown on table bellow of cost analysis below; the labor cost, material cost and standard
components cost can be listed.
Generally, the total cost of scissor jack must be as our project design is 271.0979birr.

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Table 2. Cost analysis

Part Qty Mass materia Cos Cost Machin process Time Cos Total Total
name (kg) l t/kg of e (sec) (birr) manuf cost(bir
raw acturi r)
mtl ng
cost
rivet 5 0.1786 Cast 14 0.25 cutting cutting 0,57 15 2 2.025
32 iron
bracket 1 0.5 Steel 14 6.02 drilling drilling 0.713 45 9 15.021
alloy 175 75
AISI43
40
handle 1 0.2 AISI10 14 2.70 Cutting Cutting 18 3 5.7068
30 683 37
7
Power 1 2 Carbon 14 17.0 cutting cutting 0.0157 20 5 30.025
screw steelAI 2526
drilling drilling 0.25 40 8
SI1030
Upper 2 0.5 Steel 14 19.8 cutting cutting 0.1144 25 6 28.859
link alloy 5984 84
drilling drilling 0.0011 10 3
4340
41
Lower 2 0.5 Steel 14 29.0 Drillin drilling 0.0014 10 3 40.025
link alloy 2592 g 1 9
4340 cutting cutting 0.1144 30 8

base 1 0.5 Carbon 14 35.6 drill drilling 0.51 50 10 65.607

plate steel 076 6
AISI cutter cutting 1.62 60 20
1030

Standard cost

bolt 2 Cast 0.0 18.0 Pc Pc Pc Pc Pc 18.067

steel 019 67
Fixed 1 Cast 0.9 20.6 Pc Pc Pc Pc Pc 20.67
nut steel 25 7
Main 1 Cast 1.0 35.8 Pc Pc Pc Pc Pc 35.89
nut steel 92 9
Total Cost 271.097
9 ETB

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CHAPTER FOUR
4. DESIGN ANALYSIS
4.1. Design Concept
4.1.1. Concept Generation
It is the recommended that many options should be to come up with different options before
moving in to other step. It should be generated to give value addition or new ideas. We have
prepared a table below for sub function of jack and four concepts and also its evaluation to
choose that which is the best.

CONCEPT –I
The jack should have a rectangular metal and rough surface(1), its single lifting mechanism, and
(1)power screw, (2)pin connector and (2)portable also medium length, (1)rectangular metal foot
as how n figure(2) below

Figure 3. Concept I

CONCEPT –II
A circular metal, smooth surface(1), its single lifting mechanism, (2)power screw with (2)pin
jack connected by belt and (1)fixed crank with power screw and short length(1) also a
rectangular metal foot as shown on figure(3) below.

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Figure 4. Concept II
CONCEPT –III
A rectangular wood and smooth surface(1), it has multiple lifting mechanism and (1)a power
screw also have (2)pin connected jack, (1)fixed crank with power screw, short length(1) and
rectangular metal foot as shown on figure(4) below.

Figure 5. Concept III

CONCEPT –IV
A square shape plastic and smooth surface (1) with a single lifting mechanism and (1) power
screw. It also have (1) jack connector connected by bolt, (1) fixed crank with power screw. It
have (1) short length and rectangular shape foot as shown on figure(5) below remember the
figure is the same as on concept-I but the top plate is flat(squared).

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Figure 6. Concept IV
Table 3. Concept Generation
Criteria Concept-I Concept-II Concept-III Concept-IV
Manufacturing cost 5 5 8 9
Material cost 6 6 9 8
Simplification 9 7 3 5
Durability 9 7 5 3
Reliability 10 8 5 2
Time to produce 8 7 2 5
Handling 9 6 6 7
Weight 5 5 2 8
Stability 8 8 10 8
Strength 10 8 6 4
Functionality 9 9 9 8
TOTAL 88 76 64 67
Generally, as we have seen above, from the four (4) concepts the best concept according to the
data which meets the criteria is concept one (1) and we have choose it. And we choose it to

4.2 Given Specification

Max height = 380mm
Min height = 160mm
Length of the link does not vary (constant)
Capacity (load to be lift) = 1tone =1000kg=9,810N
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Factor of safety =2.5

The whole dimension of the machine the minimum and maximum height should be calculated
first before we began force and strength analysis of the machine. When the jack is on its max
height

Figure 7. Dimension analysis

Assume: a = 30mm is common for the standard jacks for bottom and upper plates. Maximum
height jack stand is:-
H max = Δh + 2a
= Δh + 60; where H max = 380mm
Δhmax= 320mm hmax
For maximum and minimum height length of the like should be calculated
Δh = Lsinθ1 + Lsinθ2; when θ1 = θ2 = θ
Δh =2Lsinθ
L = Δhmax/2sinθ hmax=320mm
L = 320/2sinθ

We have seen above that how to calculate the values of links at maximum height and minimum
height; but in order to get exact values of the links, we should have to iterate the values of Δh of
links when angle can be varied.
Let’s θ1 = θ2 = θ = 700 Will be the maximum angle we specified
L = 320/2sin700 =170.268mm

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When β + θ + 900 = 1800

β = 900–θ
To find the minimum angle we use: hmin=2Lsinθ = (160-60) mm since L=170.268mm constant
then θ=sin1 (hmin/2L) =17.076
The iteration at different angle’s like; θ=17.0760, 200, 300, 400, 500, 600, 700
Table 4. Length and angle on different given angle
Θ1= 17.7600 200 300 400 450 500 550 600 700
θ2=θ
L(m 170.26 170. 170.26 170.268 170.268 170.268 170.268 170.268 170.268
m) 8 268 8
β 72.9240 700 600 500 450 400 350 300 200

Depend on the above value of L, we have to get the value that approximate or equal to the
change of weight by making change of height unknown and using the above length at each
angle of inclination.

Δh = 2Lsinθ
Let at L =170.268mm and θ =700
Δh = 2(170.268) sin700 = 320mm
Then iterate change of height for different angle’s like θ = 17.0760,200,300,400,500,600,700
Table 5. Change of height jack lift at d/t angle

Θ1=θ2=θ 17.0760 200 300 400 500 600 700

β1=β2=β 72.9240 700 600 500 400 300 200
L(mm) 170.268 170.268 170.268 170.268 170.268 170.26 8 170.2 68
Δh(mm) 100 116.46 170.26 218.89 260.88 294.89 8 319.9 99
Then from the above table we have seen that different change of height ,β and θ for constant L
which means that for the maximum height jack stand the length of the link should be L
=170.268mm and the maximum angle the jack should be stand is θ = 700.
Now for the jack when it is on its minimum height we have only on inclination because the link
length would not change.
H min= 2a + Δh
=2(30) + 2LSin θ; when H min =160mm and L=170.268mm

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Figure 8. Angle between links

Θ =Sin-l (100/2(170.268))
θ = 17.07680 = θ1= θ2 and also the angle of β =180-90-θ; when θ=17.07680, β=72.9230
Therefore, we have found the values of links and angles of inclination
L = 170.268mm, θ =17.07680 and β=72.9230
The force analysis consideration is based on the assumption that, the scissor jack holds vertically
symmetrical

Figure 9. Force analysis FBD

F = FA + FB
AO = OB= CP = PD
FA + FB =9810N

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4.3 The Force Analysis on the Screw

Figure 10. Power screw FBD

FE = FAEx + FCEx, FF = FBFX + FDFX
|FAE| = |FCE| = |FBF| = |FDF|
FAE x = FCE x = FAE cos θ = FCE cos θ
FBF x = FDF x = FBF cos θ = FDF cos θ
FAE y = FCE y = FBE sin θ = FDE sin θ
FBFy = FDFy = FBF sin θ = FDF sin θ
All the force which we have taken about are going to be transferred in to the base plate by using
rivets and should with stand the whole forces.

Figure 11. Force distribution on base plate

4.4 Power Screw Design

A power screw is advice used in machinery to change angular motion in to linear and usually, to
transmit power. This must have adequate strength to with stand axial load and the applied torque.
Again this screw is used to synchronize by its oppositely traded portion.
For power screw four types of stresses is going to be calculated.

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Figure 12. Power screw

Those are:-
 Direct (compressive and tensile) stress due to axial load.
 Torsional shear stress.
 Shear stress due to axial loading.
 Bearing stress (pressure).

Figure 13. Power screw FBD

FE = FAEx + FCEx, FF = FBFX + FDFX
|FAE| = |FCE| = |FBF| = |FDF|
FAE x = FCE x = FAE cos θ = FCE cos θ
FBF x = FDF x = FBF cos θ = FDF cos θ
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FAE y = FCE y = FBE sin θ = FDE sin θ

FBFy = FDFy = FBF sin θ = FDF sin θ
For screw, when β = 200(at max lifting height)
FAE = 4905cos β =|FAE| = 4905Cos200 = 4609.192N
Then, |FAE| = |FCE| = |FBF| = |FDF|=4609.192N
|FAE| = |FDE| = |FBF| = |FDF| =4609.192N
FAEx =|FCE| = |4609.192Sin200| = -1576.43N
`FBFx = FDFx = 14609.192sin200 = 1576.43N
FAEy =| FBF | = |4609.192Cos200| = -4331.2N
FCE = FDF = 4609.192N
FE = FAE + FCE = 4609.192N+4609.192N=9218.38N
FF = FBF + FDF = 4609.192N+4609.192N=9218.38N
|FAE|= |FCE| = |FBF| = |FDF|: are the compression load applied to the member.

|FAE| = |FCE| = |FBF| = |FDF|: are tensile load (tensile axial loads transferred to power screw).
AT β = 300
FAE = 4905Cos30 FAE = 4247.850N
|FAE| = |FCE| = |FBF| = |FDF|= 4247.850N
FAEx = FCEx = 4247.850Ncos600 = 2123.92N
FBFx= FDFx = 4247.850Ncos600 =2123.92N
FAEy =| FBF |=4247.850Nsin600=3678.7
FBF =FDF =4905Cos600 = 4247.85
FCE = FDF = 4905Cos600 = 4247.85
FE = FAE + FCE = 4247.85+4247.85=8495.7N
FF = FBF + FDF =8495.7N
AT β = 400
|FAE| = 4905Cos400 =3757.44N
|FAE| = |FCE| = |FBF| = |FDF|= 3757.44N
FAEx = FCEx = 3757.44cos500 =2415.2N
FBFx = FDFx=3757.44cos500 =2415.2N
FAEy =| FBF |=3757.44sin500 =-2878.366N
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FAE = FBF =3757.44

FCE = FDF = 3757.44
FE = FAE + FCE = -7514.89N
FF = FBF + FDF =7514.89N
AT β = 500
|FAE| = 4905Cos500 =3152.87N
|FAE| = |FCE| = |FBF| = |FDF|= 3152.87N

FAEx= FCEx = 3152.87Sin500 = 2415.24N

FBFx = FDFx =3152.87Sin500 = 2415.24N
FAEy =| FBF |=3252.8sin400 =-2026.66N
FAE = FBF =- 3152.87N
FCE = FDF = 3152.87N
FE = FAE + FCE = 3152.87N+3152.87N=-6305.74N
FF = FBF + FDF = 6305.74N
AT β =600
|FAE| = 4905Cos600 =2452.5KN
|FAE| = |FCE| = |FBF| = |FDF|=2452.5N
FAEx= FCEx = |2452.5Ncos300| = -2123.9N
FBFx = FDFx = |2452.5Ncos300| = -2123.9N
FAEy =| FBF |=2452.5cos600 =-1226.25N
FAE = FBF = 2452.5KN
FCE = FDF = 2452.5KN
FE = FAE + FCE = -4905N
FF = FBF + FDF =4905N
AT β =730(β when the jack is on its own minimum height)
|FAE| = 4905Cos170 =4690.67N
|FAE| = |FCE| = |FBF| = |FDF|=4690.67N
FAEx = FCEx = 4690.67Ncos170 = -4485.7N
FBFx = FDFx = 4690.67Ncos170| = -4485.7N
FAEy =| FBF |=4690.6cos730 =-1371.39N
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FAE = FBF = -4690.67N

FCE = FDF = 4690.67N
FE = FAE + FCE = -4690.67N+4690.67N=9381.34N
FF = FBF + FDF =9381.34N
Now let’s collect all the calculated loads in the table form and the design analysis is analyzed at
those points of action.
Table 6. Force on links at different angle

Θ (0 ) Β (0 ) FAEX (KN) FAEY (KN) FE(KN)

17 73 4485.7 1371.39 9381.34
30 60 2123.9 1226.25 4905
40 50 2415.24 2026.66 6305.74
50 40 2415.24 2878.366 7514.89
50 30 2123.92 3678.7 8495.7
70 20 1576.43 4331.2 4609.192

Then we select FAE = 4690.67N and FE = 9381.34N.

I:-DIRECT TENSILE STRESS (σt):

σt = axial load / Ac (but Ac = πdc2 / 4)

Design stress (σd) = yield stress / F.S = σy / F.S

σd = σall = 345Mpa / 2.5

σd = σall = 138Mpa
Now σd > σt = |Ft| / Ac =4FE / πdc2

Dc > √ {(4*9381.34)/ (π*138*106)}

dc> 0.0093035m ≈10mm
Therefore take pitch (p) = 3mm
Major diameter (do) = dc + p = 10+ 3 =13mm
Mean diameter (dm) = dc – p/2 = 10 – 3/2 = 8.5mm
Ac= πdc2 / 4 =78.539mm2 or 7.85 * 10-5m2

Tensile stress (σt) = |FE| / AC

= 9381.34/7.85 * 10-5m2
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σt = 119.44Mpa
Which implies σd > σt; which means it safe the power screw.
II – TORSIONAL SHEAR STRESS:
=16T / πdc2; where T is the maximum one of Tright and Tleft
Tright = |FE|dm /2{(L + πμdm) / (πdm - μL)}
Where μ = coefficient of friction
ɸ =helical angle
λ= load angle
L =np; n=1(single thread) and P =3 then L = P, d=8.5mm
tanλ = L / πdm and
tanɸ = μ > tanλ
μ > L / πdm =0.112344
Since μ > L / πdm then μ = 0.112
Now Tright= {((9381.34)*(8.5*10-3)/2) ((3+π0.174*8.5)/ (π8.5-0.174*3))}
=8916.2Nmm
Tleft=|FE|dm /2{(πdmμ-L) / (πdm+μL)}
= (4.6446*103 * 5.5*10-3)/2{(π*5.5*0.174 - 3)/ (π*5.5 + 0.174*3)}
= 0.2575Nmm
=16TR / πdc3
= (16*8916.2Nmm/ (π*(10*10-3)3)
 = 45.409Mpa
Maximum principal stress (tensile):-
σt (max) = (σt/2) +0.5√(σt2 +4(2 ))
= (119.44/2) +0.5√ (119.442 + 4(45.4092))
σt (max) = 134.74Mpa
Maximum shear stress (max):-
max = σt /2 +1/2√ (σt2 + 4)
= (119.44/2) + 0.5√ (119.44)2 + (45.409)2
max = 63.8Mpa
Since the stresses are with in safe mode or limit, therefore the design of square threaded screw is
safe.
III- SHEAR STRESS DUE TO AXIAL LOADING:
Shear stress (= |FE|/πndct; but t = pitch / 2 = 3 / 2 = 1.5
Where n = length of nut / pitch = 20 / 3 = 6.66
= (9381.34) / (π* 6.66*10*1.5)
= 29.89Mpa
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Assume the load is uniformly distributed over threaded contact.

IV – BEARING STRESS (PRESSURE-Pb):
In the design of power screw, the bearing pressure depends on:
 Material of the screw and nut.
 Relative velocity between the nut and screw.
 Nature of lubrication.
In order to reduce wear of the screw and nut, the bearing pressure on the thread surfaces must be
within the limit.
 Assume: the load is distributed uniformly over thread contact.
Pb = (FE / (π/4) (do2 – dc2))/F.s
=|FE|/πdptn; where t = p/2 =1.5
NB: - From the table 17.7(machine design) limiting value of bearing pressure between thread
and nut is from 17.5 – 24.5N/mm2
Let’s take Pb =22Mpa
n = FE /πdptpb = (9381.34)/ (π*1.5*22*106*8.5*10-3)
=0.01064 ≈ 10.6.……..safe
The length of power screw is must
L= [L* cos θ+2D (for nut length) +2D (for screw housing length)]*2+D (added for handle
connection)
Where L=length of link
D=diameter of screw
L= [170.268cos17+2*10+2*10]*2+10=430.9mm≈450mm
The screw should be lubricated using grease to achieve smooth operation, so the friction between
the driving of the screw is too small, the jack is efficient.
As the screw is rotated once the height of the jack lifts 3mm.
For this type of scissor jack, the arms should be synchronized with each other by the application
of oppositely traded (clockwise and counter clockwise) power screw. The main reason to select
the screw is there is no simple mechanisms which will with stand the applied load. They are
needed to actuate the jack and to hold at specified height. They are good for smooth operation

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and they are easy to manufacture save manufacturing cost, they can easily assembled and
disassembled and to maintain also.

4.5 Design of Power Screw Housing

The connecting members take the applied load and transfer to the power screw and leave the
power screw as tensile load. The connecting members at point E will act on a not having internal
square turned and also transferred to the lower arms.
Screw housing have two important function:-
 It save as nut for the power screw.
 As a rivet for the links (arms).

Figure 14. Power screw housing

4.6 Nut (For the Power Screw) Design

Material selection:-
Cast iron
Safe bearing pressure 12.6 – 17.6Mpa

σy =276Mpa and σnt = 414Mpa

Pb = |FE| / (π/4) (do2– dc2) n = |FE|/ (πdptn); where width of thread t = p/2 =3/2 = 1.5 and
Pb=16Mpa
n = |FE| / dp * t *pb
F.s = (4905) / (π * 8.5*10-3 * 1.5*10-3 * 16*106)
= 7.653; take F.s ≈8
To prevent rocking and to have good stability of screw in the next (F.s = 14) taken.
 SHEAR STESS ON THE NUT THREADS
max = |FE| / (π do tn)

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=9381.34N / (π * 14 * 10 * 1.5) standard outer diameter D=10mm

= 14.21Mpa
σsy = 0.577σy
= 0.577 * 276Mpa
= 159.252Mpa
max = σsy /F.S;
F.s = 159.3 / 5.469 =11.21 ≈12 –it is satisfactory
Length of nut thread (H) =2*D inner np = 10*2 = 20mm?

 shear stress on the part of supporting arms

= |FE| / A = 4905N /(πd2/4)
But σall = σy / F.S =276 / 2.5 =110.4Mpa
σsy =0.577 * σy = 0.577 * 276 = 159.25Mpa
max = σsy /F.s=159.25 / 2.5 =63.7Mpa
= 4*4905N /π (63.7*106)2 <max=63.7Mpa
Then d > √ (4 * 4905 / π (63.7*106)2) > 9.9mm
d is approximating or taking any value above 9.9mm make it sensible let D=10mm.and the outer
2D=20mm hexagonal nut.

4.7 Analysis for the Four Rivets (Nuts)

In most case the threat is zero over proof loading of fasteners and this is best addressed by
statically methods. The threat from fatigue is lower and deterministic method can be adequate.
Material selection
The material for rivet should be tough and ductile
 ductile cast iron
 oil quenched (temped)
σy =621Mpa, σnt = 827Mpa and F.S= 2.5

Figure 15. Rivet force analysis

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σall = σy /F.s=621/2.5 =248.4Mpa

σsy = 0.577σy = 0.577*621 = 358.317Mpa

max = σsy /F.s = 358.317 / 2.5 = 143.32Mpa
SHEAR STRESS:

<max
|F/2| / (πd2/4) <max
d > √ (2F/π max); where F = 9381.34N
d > 4.667mm
Which is going to take d = 5mm

BEARING STRESS:
σb= F/2 / (td) <σall
t > F / 2d σall
t> 9381.34 / 2 * 248.4 * 5
t> 3.776mm or t =4mm
σ = MmaxC / I; but C = 5 / 2 = 2.5 and I = bh3/ 12 = 5 * 53/ 12 = 52.083mm4 and
Mmax @center of cup
σ = 279.8 * 2.5 / 52.083
= 13.43Mpa it’s safe
4.8 Analysis for the Base Plate
Here the base plate should with stand the applied load plus the whole component or linkage load
by itself, the force, we have to select the material during the design process in order to resist the
above all loads. And the base must be grooved to protect sliding of the jack during application.

Figure 16. Base plate

Material
 steel alloy AISI-1030
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 normalized(8400)
 σy = 655Mpa, σnt = 1025Mpa and F.s= 2.5
Now, σall = σy/F.s= 655 / 2.5 =262Mpa
BEARING STRESS:
σb = (F/4) /td = F / 4td;
Where d = diameter of the rivet through the fastener hole
t = thickness of the fastener hole
σb = F / 4td <σall
t > F / 4t σall
t> 9381.34 / 4 * 5 * 262
t> 1.7903mm
Then take any value above 1.7903mm, let t = 2mm it will be safe.

4.9 Force Distribution for the Fastener (Connecting Member) at the Top
These fasteners as used to fully transfer the applied load from the break, to the rivet, which
connects the upper arm with the fastener. The applied force at two points will have same
magnitude (FA = FB =) and this loads FA and FB on the two rivets will be transferred to the
arms.

Figure 17. Force distribution on fastener

The load (F) is applied on break direction so the total load applied on a break is 0.2KN and this
force is totally transferred to the fasteners which are connected to the cup by pin.

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Figure 18. Top fastener strength analysis

Material selection
 steel alloy 4340
 normalized(8400)
σy = 655Mpa, σnt = 1025Mpa and n = 2.5
Now, σall = σy / n = 655 / 2.5 =262Mpa
BEARING STRESS:
σb = (F/4) /td = F / 4td;
Where d = diameter of the rivet through the fastener hole
t = thickness of the fastener hole
σb = F / 4td < σall
t > F / 4t σall
t > 9381.34 / 4 * 5 * 262
t > 1.7903mm
Then take any value above 1.7903mm, let t = 2mm it will be safe.

4.10 Design for Cup at the Top (Break)

This cup is subjected to compressive stress and bending stress which is placed at the top of the
frame and should with stand the applied load without failure.

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Figure 19. Design of cup at top

Material selection
steel alloy 4340
normalized (@8700)
σult =862Mpa, σy = 1200Mpa
σall= σy / F.S= 862 / 2.5 =344.8Mpa
max = σsy /F.S = 0.577σy /F.s
max = 0.577 * 862 / 2.5 =497.374Mpa
Now axial compressive stress
σall= F / A; where A = b * w =50mm * 10mm = 500mm2=0.0005m2
σall= 9381N / 0.0005m2 =18.76Mpa
Since σ<σall it safe
When σall = 344.8Mpa and F = 9810N area should be
A = F / σall = 9810/ 344.8Mpa = 24.45mm2
BENDING STRESS:
The maximum bending stress is taken place at Mmax = F (10-5) =9381.44 * 5 = 49156.7Nmm
Material selection
AISI-1030
Normalized steel
Σy =345Mpa, σall = 521Mpa and F.s = 2
σsy = 0.577* σy = 0.577 * 345 = 199.065Mpa
max = σsy /F.s=199.065 / 2 = 99.53Mpa
σb = F/4 / (td) = F / 4td;
Where d = diameter of the rivet through the base plate hole
t = thickness of the base plate hole
n =factor of safety fracture
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To be safe σb < σall

Therefore, F/ 4td < σall
t > F / 4d σall
t > 9381.34 / 4 * 5 * 521
t > 0.90032mm
Any value for thickness greater than 9.0032*10-4m is safe. Therefore let t = 1mm so, using
t=5mm is safe.
t = 3mm @d = 5mm hole

4.11 Analysis of Arm’s

4.11.1. Analysis of Upper Arm’s
Used to transfer the load applied to the screw driver housing (connected mechanism).The links
(arms) are subjected to bending stress due to the applied load on the counter of gravity of those
links at a point. And we should assume that the load is distributed over the surface of the links.

Material selection
Steel alloy 4340
Normalized
σy = 862Mpa and σult = 1280Mpa and
F.s= 2.5

Figure 20. Upper arm strength analysis

BENDING STRESS:
When θ = 700and L = 170.628mm at maximum height
X = L Cos θ = 170.268 * Cos 700= 58.23mm
Y = L Sin θ = 170.26 * Sin 700= 159.998mm
+↺ME = M + FAEyX – FAExY = 0
M = 4905cos 20*58.23– 4905cos700*159.998
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=4407.78-1677.6
=2730.17N-mm
M =2.73N-m

Figure 21. Bearing stress by upper arm

Aa= 50 * 30 = 1500mm2
Ab = 40 * 25 = 1000mm2
A = Aa + Ab = 1500 – 1000 =500mm2
Ȳa = 15mm and Ȳb = 12.5
Now, Ȳ = (Aa Ȳa - Ab Ȳb) / A = {(1500*15)-(1000*12.5)} / 500 = 20mm =c2
C1 = 30mm – c2 =10mm
Ia = bh / 12 = 50 * 253 / 12 = 112,500mm4
3

Ib = bh3 / 12 = 35 * 253 / 12 = 45,572.9mm4

Total moment of inertia
I = {(Ia + Aa (Ȳ - Ȳa) 2) – (Ib + Ab (Ȳ - Ȳb) 2}
= {(112,500 + 1500(20 - 15)2) - (45,572.9 + 1000(20 - 12.5)2)}
=48,177.1mm4
σc = MC2 / I = (2730.17 * 20) / 48,177.1
= 1.1340 N/mm2
σb = M(C1 – 5) / I = (2730.17*(10 - 5)) / 48,177.1
= 0.2833N/mm2
σa = MC1 / I = (2730.17 / 48,177.1)
=0.0566N/mm2
NB: - Generally, in order to be the design is going to be safe σ < σall and as we have seen above
in the analysis it is safe.
BEARING STRESS:
σb = |FAE|/2 / (td) =|FAE| / (2td); where d = 5mm and t = 3mm
σb = 4690 /(2 * 3 * 5) = 23.02Mpa
Where σb <σall it is safe.

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4.11.2. Analysis for the Lower Arms

Figure 22. Lower arm strength analysis

BENDING STRESS:
When θ = 700 and L = 170.268mm at maximum height
X = L Cos θ = 170.268 * Cos 700 = 58.37mm
Y = L Sin θ = 170.268 * Sin 700 = 159.998mm
+↺ME = M + FECX Y – FECYX = 0
M = 4690N (58.37) – 4690(159.998)
= -2.73N-m; change assumption +↻M = 0
Then M = 2.73N-m

Figure 23. Bending stress analysis

σb = MC / I =where M=2.73N-m, I= pi*d4/64,c=d/2 , σb =70Mpa

D=7.35mm
BEARING STRESS:
σb = |FAE|/2 / (td) =|FAE| / (2td); where d = 5mm and t = 3mm
σb = 9831.34 /(2 * 3 * 5 ) = 3.2770N/mm2
For σy = 1620Mpa and F.S = 2.5
σall =σy / F.S = 1620 / 2.5 =648Mpa
Since σb <σall; the design and material selection is good for the design to with stand the load
applied.

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4.12 Design of the Driving Handle

The handle is used for lifting mechanism to those applied loads by engaging the handle and
power screw driver in order to lifting and lowering the desired (certain load).
The normal person can apply a force from 70N-80N. For our case we have choose a force of (F =
75N)

Figure 24. Driving handle analysis

Material selection
steel 1030
normalized
σult =520Mpa, σy = 345Mpa and F.s= 1.5
σsy = 0.577* σy = 0.577 * 345 = 199.065
σall = σy / F.s =345 / 1.5 =132.66Mpa
Considering bending moment about A
ΣMA = 0
MA = F * r = 75N * 50mm
=3750Nmm
=3.75Nm
NB: - It is also taken as torque since it’s due to rotational effect about A.
The combined stress due to the bending moment and torque developed can be determined as
follows
σb = MC / I; where I = πd2 /64 , C= d /2 , let d = 15mm
σb = 32M /πd3
= 32(6Nm) / (π* 0.00153)
= 18.108Mpa

Which implies 346.6Mpa> 18.108Mpa σb <σall; therefore it safe.

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The efficiency of the mechanical scissor jack will be:

As the screw is rotated once the hydraulic jack lifts 6mm, at 53.3 rotations the jack lifts 320mm
total length.
And the input power to drive the jack is: `
2πnTright 8.912
p= = 2π ∗ 11 ∗ = 10.4W
60 60
Again the power by the jack to the load: `
(F.s)
p= = (9.81*1000kg/m3)/t =10.4W then t=300sec
t
η= 100% theoretically efficient.
idealeffort tanα
Actually, η = or η =
actualeffort tan(α+Φ)

w(sinα+µcosα)
p0 = w tan(α − Φ) = p0 = ,
cosα−µsinα

α=600, µ=0.00476 coefficient of friction nut and screw for lubricated screw

The effort applied to the circumference of the jack:

sinα + µcosα 9810(sin60 + 0.00476cos60)
p0 = w =
cosα − µsinα cos60 − 0.00476sin60
p0 = 16738.4532W
p0 = w tan(α − Φ)
p0
tanα − tanΦ − tanα
p0 = w ( ) tanΦ = w
1 + tanαtanΦ −1 − tanα
Φ = 0.3773530
tanα tan60
η= = = 0.9851 = 98.51% or
tan(α − Φ) tan(60 − 0.377353)
We can check the theoretical efficiency @which the μ=0, α= 00

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CHAPTER FIVE
5. CONCLUTION AND RECOMENDATION
5.1 Conclusion
Scissor jack is device which is used to lift a load and designed depending on the desired load to
lift and cost. This design is also done depending on the given specification to with stand the load
applied. It also needs lubricating of the power screw. The cost of this scissor jack as shown on
cost analysis is 271.0979birr.
In this work, a scissor jack is design analytically and then standardize. In this jack, Nut and
Screw are the one of the most significant components. A screw is designed based on a maximum
tensile stress and maximum shear stress.
For maximum load it is very necessary to keep both the values within limit for safe design. Nut
is a stationary part in which a screw rotates. Therefore a bearing pressure is also needs to be
considered. For both the components, if we take combination of different material for each pair
of screw and nut so we can find best suitable material for designate maximum load. For
simplicity of current design, parameters of screw like, core diameter, helix angle and mean
diameter are taken same throughout design for all loads. Also for link design standard
dimensions of link cross-section are considered. For material of link, same kind of material i.e.
Mild steel is taken.
From this project we have concluded that Scissor jack are devices which are used to lift a load
and designed depending on the desired load to lift and this design is also done depending on the
given specification to with stand it. It also needs oiling of the screw.

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5.2 Recommendation
Recommendation to the University:
 The university should arrange some practical visit in some courses before joining the
schooling. Because it helps the students to understand easily what they were learnt in
laboratory and lecture class. In this internship we saw how it is easy to understand on
practical work. For example, courses like Automotive Maintenance, IC engine and Fluid
Power.
 The university must be advising to the students at the inspire time they spent in visiting
different maintenance works in nearby place or in their hood because this may give the
students a good experience. Generally, the school must full fill the following important
things: Evaluating and supervising student at least weekly in the company. Attendance
for the student and the supervisor at company should be mandatory. In short, the
university should have closely relationship with the manager of the company because the
manager able to manage and control all workers in the company and those workers have
willing for intern students.

Recommendation to the Hosting Company:

The company we have been working as an intern student was very good in creating an
opportunity of work (sharing experience) for the students who want to work as an intern student.
The company also provides some of our basic needs. So, we would like to recommend the
company to continue its assistance for the internship students to create learned and experienced
future generations. All the workers in this company were positive to help us during execution of
our tasks. They also taught us each and every step of maintenance. So, we would like thank and
recommend all the workers to continue their positive behavior for all the students they would
get as an intern student in their future working life. We will recommend that the company
should once in a while organize more team building exercises to its staff in order to bring them
closer. This would give staff time to re-examine what they have achieved so far, share personal
experiences and ideas, learn to together and appreciate the importance of team work at work.
Beside these, the company has to work more to satisfy its own customers and employers,
purchasing units, small firm supplying materials for the company especially in facilitating
means of payment for the finished woks.
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The quality of material is the major one to fulfill the design strength so, the material needed by
the company should store well and place in the appropriate position. If this is done our work
becomes smart enough so that we can have better quality.
Most of company workers, particularly site engineers, Formal and daily labors do not have
safety wears like safety shoes, helmets, etc. to protect them from sudden injuries. Due to this
problem, they are exposed to different injuries while working. Therefore, we recommend the
company to arrange safety tools for site workers.

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APPENDIX
DETAIL DRAWING

Fig 6.1 cup at the top brake

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Fig 6.2 fastner

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6.3 link

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Fig 6.4 power screw

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6.5 Power screw housing

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6.6 nut

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6.7 rivet

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6.8 handle

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6.9 base

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6.10 top view

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6.12 front view at minimum height

6.13 assembled view

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REFERENCE
[1]. Shigly’s (mechanical engineering design book). 9th- edition.
[2]. Hand book of mechanical engineering (design).
[3]. R.S. Kurmi and J.K.Gupta (design book)
[4]. ERCC Company (www.ecwc.gov.et)
[5]. Internet (www.wikpedia.com)
[6]. Atlas engineering bar hand book

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