internship report and design 5th year mechanical enginering

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ADIGRAT UNIVERSITY
COLLEGE OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
DESIGN OF COSTIC RECOVERY ROBOT MACHINE
A ProjectSubmitted to the college of technology, MechanicalEngineering in
partial Fulfillment of Requirement for the in internship Study in ofalmeda
textile factory p.l.c
Hosting company: Almeda Textile Factory PLC
PREPARED: ID NO
1 SHUSHAY HAILU 4142/07
2 ZEMEN ARAYA 4509/07
3G/WAHID G/HIWOT 3239/07
ADVISOR: REDAE (MSC)
SUBMITION DATE 28/02/2011E.C
DURATION FROMFBURARY29/6/10_JUN29/6/10

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The project Titled Design Of Costic Recovery Robot Machine by shushay hailu,zemen
araya,g/Wahid g/hiwot for the intern ship report of mechanical engineering.
Boardof Examiners
Name Signature date
Advisor……………. ………. ……….
Internal Examiner1 …………….. …………… …………
External Examiner1 ……………. …………… …………

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Declaration
Wedeclare that the work, which is being presented in this project entitled (Design of pneumatic
robot machine), is original work of our own and it is done by our effort. We want to declare that
throughout our stay from February29/06/ 2010--June 29/10/ 2010.we have done, understand,
get… different practical work atmosphere and those are written and included in this internship
report and all evidences, information or ideas, and pictures that used to bring into existence of
this internship report is done by ourselves exclusively and submitted to Adigrat University
Institute Of Technology Faculty of Mechanical engineering.
Ms.Redae
Name of the academic supervisor …………………………….
Signature
Mr .Gebrearegay
Name of company advisor ..............................................
Signature

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Acknowledgment
We would like to express our deepest gratitude to our advisor Ms. Redaefor his invaluable
advice and continuous guidance throughout the progress of this project. Also we would like to
express our heartfelt appreciation to Almeda Textile processing Department Heads, Mr.
Gaim,Mr temesgen,Mr.brkti wood work engineering,supervizers Mr.gebrearegay, Mr.shushay
and the whole operators for their keen cooperation in all part of the project.
we would like to use this opportunity to thank for Mr. hagos (Bsc in wood Eng.), Mr. Ftsum,
HR workers, Mr. Solomon and the management workers for their greatful contribution for the
completion of this project.

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Executive Summery
Costic Recovery Robot Machine is advice that lifts costic soda from one place to another plase
and it reduces wastage of time.Those chemical are hazards to human being. they can damage
seriously to different parts of body.The main objective of this project is designing and modeling
of pneumatic robot machine for industries.The focus of this project was to design a pneumatic
robot machine under certain constraints such as compact size, easy to operate, low cost.This
Costic recovery robotic machine for manual labors improves the speed of mixing of the
chemicals with steam.This report is written in almeda textile factory. It includes the brief history
and back ground of the company, the overall internship experience got, work flow of different
sections of the company, the major challenges and problems faced in almeda textile factory.

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Table of Contents
Board of Examiners............................................................................................................................ ii
Declaration...................................................................................................................................... iii
Signiture ................................................................................................Error! Bookmark not defined.
Acknowledgment...............................................................................................................................iv
Executive Summery ............................................................................................................................v
CHAPTER ONE....................................................................................................................................1
1. Back ground of Almeda Textile plc ...................................................................................................1
1.1 Brief history..............................................................................................................................1
1.2 Year of establishment ................................................................................................................2
1.3 Objective of the company ..........................................................................................................2
1.4 Mission of the company:............................................................................................................3
1.5 Vision.......................................................................................................................................3
1.5.1 Core values:........................................................................................................................3
1.6 Production program...................................................................................................................3
1.7 Pricing......................................................................................................................................4
1.8 Main products of Almeda textile plc ...........................................................................................4
1.9 Main customers or the end users of the company.........................................................................4
1.10 Local customers ......................................................................................................................5
1.11 RESPONSIBILITIES:.............................................................................................................7
1.12 Work Flow and Major Department of Alemda Textile Factory ...................................................7
1.12.1. Raw materials - cotton and polyester .................................................................................9
1.12.2 Spinning department..........................................................................................................9
1.12.3 Weaving department........................................................................................................13
1.12.4 Processing; .....................................................................................................................13
1.12.5 KNITTING DEPARTMENT...........................................................................................15
1.13.6 Knit dying ......................................................................................................................16
1.12.7 Garmenting.....................................................................................................................16
CHAPTER 2.......................................................................................................................................18
2. THE OVERALL BENEFITS THAT GAINED FROMTHE COMPANY..........................................................18
2.1. Improving practical skills........................................................................................................18
2.2 Upgrading of theoretical knowledge: ........................................................................................19

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2.3. Industrial problem solving capability .......................................................................................19
2.4. Interpersonal communication skills:.........................................................................................20
2.5. Team playing skills.................................................................................................................20
2.6. Improving leadership skills .....................................................................................................20
2.7. Understanding of work ethics ..................................................................................................21
2.8. Entrepreneurship Skills ...........................................................................................................21
CHAPTER 3.......................................................................................................................................23
CHALLENGES FACED IN THE FACTORY................................................................................................23
3.1 Challenges facing in work tasks................................................................................................23
3.2 Finding of problems ................................................................................................................24
CHAPTER FOUR.................................................................................................................................25
PROJECT (CASE STUDY) .....................................................................................................................25
4. COSTIC RECOVERY ROBOT MACHINE..............................................................................................25
4.1 INTRODUCTION...................................................................................................................25
4.1.1WORKINGPRINCIPLES OF COSTIC RECOVERY ROBOT MACHIN ............................26
4.2. Problem of the statement.........................................................................................................27
4.3. Objective of the Project...........................................................................................................28
4.3.1. General Objective ............................................................................................................28
4.3.2. Specific Objective ............................................................................................................28
4.4 Scope of the Study...................................................................................................................28
4.5. Limitation of study .................................................................................................................28
CHAPTER FIVE ..................................................................................................................................29
LETRECHER RIVEW............................................................................................................................29
5.1 COSTIC RECOVERY KNOW DAY IN ALMEDA TEXTAIL..................................................29
CHAPTRE SIX ....................................................................................................................................30
METHODOLOGY ...............................................................................................................................30
6.1 Primary Data: -........................................................................................................................30
6.2 Secondary Data: -....................................................................................................................30
CHAPTER SEVEN...............................................................................................................................31
DESIGNE CONSEPTS AND SELCTION...................................................................................................31
7.1. Concept Development Process ................................................................................................31
7.1.1. Identifying Opportunity ....................................................................................................31

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7.1.2. Identifying Company Need...............................................................................................31
7.1.3. Organizing Raw Data.......................................................................................................31
7.2. CONCEPT DESIGN ..............................................................................................................33
7.2.1. Model A ..........................................................................................................................33
CHAPTER EIGHT................................................................................................................................34
DESIGN ANALYSIS.............................................................................................................................34
8.1. Design of column support .......................................................................................................35
8.1.1 Design (trapezoidal support) ..............................................................................................35
8.2 DESIGN OF SHAFT SUPPORT..............................................................................................37
8.3. DESIGN OF SHAFT CONECTING BASE AND COLEMUN..................................................39
8.3.1 Shaft stress .......................................................................................................................39
8.3.2 Torque required rotating the shaft.......................................................................................40
8.3.3 Principal stresses...............................................................................................................40
8.4. DESIGN OF LEVER..............................................................................................................41
8.5 PART DESIGN OF PNUMATIC CYLINDERS .......................................................................42
8.6. Design of base........................................................................................................................45
8.6.1 Eccentric Loaded Riveted Joint..........................................................................................45
8.7 DESIGN OF HAND ARM SUPPORT......................................................................................52
8.8 Design of Welding ..................................................................................................................52
8.9 DESGHNE OF HAND LEVER.......................................................................................................56
8.9.1 Principal stresses...............................................................................................................57
8.9.2 Maximum shear stresses is as follows.................................................................................57
8.10. Result and Discussion ...........................................................................................................58
CHAPTER NIGN.................................................................................................................................59
CONCLUSION AND RECOMMENDATION.............................................................................................59
9.1. Conclusions ...........................................................................................................................59
9.2. Recommendation....................................................................................................................60
REFERENCES.....................................................................................................................................61
APPANDEXES....................................................................................................................................62

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List of figure
Figure Nodescription page No
Figure 1.1Almeda textile factory.................................................................................................... 1
Figure 1.2: Organizational structure of the alemda textile factory ................................................. 6
Figure1.3:work flow of alemda textile............................................................................................ 8
Figure1.4: raw cotton store ............................................................................................................ 9
Figure1. 5: unifloc machine .......................................................................................................... 10
Figure1. 6 Spinning yarn............................................................................................................... 10
Figure1. 7 roving........................................................................................................................... 12
Figure1. 8 weaving........................................................................................................................ 13
Figure 1.9: processing of different fabric...................................................................................... 15
Figure 1.10: processing of different fabrics.................................................................................. 15
Figure 1.11: Knitting machine ..................................................................................................... 15
Figure1.12: knit dyeing................................................................................................................ 16
Figure 1.13: garment of different cloths ...................................................................................... 16
Figure 2.1: industrial problem solving ......................................................................................... 19
Figure 4.1 pneumatic robot ........................................................................................................... 26
Figure 4.1 costic recovery............................................................................................................. 27
Figure 5.1costic recovery.............................................................................................................. 29
Figure 6.1general procedure in this design ................................................................................... 31
Figure8.2. Shaft............................................................................................................................. 37
Figure.8.5 Eccentric loaded riveted joint. ..................................................................................... 47
Figure. 8.7 pneumatic rivet system............................................................................................... 50
Figure 8.8.Welded joint of frame.................................................................................................. 53

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List of tables
Table No description page No
Table 1.1 global customers ............................................................................................................. 4
Table7.1. Preparing relative importance of needs......................................................................... 33
Table 8.1 the standard diameter of the rivet hole (d) and the rivet diameter ................................ 48
Table8.2.Stress concentration factor for welded joints................................................................. 54
Table8.3.Stress for weld joints...................................................................................................... 55
Table8.4. Result and discussion.................................................................................................... 58

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CHAPTER ONE
1.Back ground of Almeda Textile plc
1.1 Brief history
Almeda textile plc is suited 7 km from Adwa town on the main road to Axum and 1006 km from
Addis Ababa and 233km from Mekelle capital of Tigray region. Almeda textile plc is located at the
middle of the beautiful mountains of Adwa. There is an airport capable of handling small planes at
Axum near the mile site. Apart from other alternatives, the factory is directly linked with the main
port of Djibouti for import or export incidentals.
Almeda is the biggest textile manufacturing company in the country .it has 100 million birr sales
volume .that is 70 percent sales volume in local market and 30 percent sales volume in global
market. Almeda is enjoying from complete new production lines imported from Switzerland, Italy,
Germany and Japan that take up a total investment of 94 million USD. Employing expatriates to
ensure efficient, quality and low cost of production availability of local cotton of top quality and
prevalence of well trained and highly motivated employees. Favorite state policies and intensified
management support.
Figure 1.1Almeda textile factory
Almeda is one of the companies of the EFFORT (Endowment Fund For Re-habilitation of Tigray)
with an objective of enhance the investment opportunities .The group is ready to cooperate with
other companies and individual who have interests in investing in the region .Besides, the company
aims at getting acceptable rate of return on its investment for the provision of quality product and

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service through satisfying its customer.The company belongs to EFFORT Group (Endowment Fund
for Re-habilitation of Tigray), which is engaged in the following fields;
 Agriculture and agro processing
 Construction and Engineering
 Manufacturing (Textile, Leather, Ceramics, Pharmaceuticals, etc.)
 Mining, trade and Service
1.2 Year of establishment
The project team of Altex and Unionmatex in 1994 G.C has elaborated the technical and
technological concept of the project thoroughly. A detailed contract was negotiated in 1995 G.C
The civil construction works were substantially completed in 1997 G.C Installation of the
machinery and utilities was preceded in 1998 G.C. In middle of 1998 yarn and gray cloth were
produced in the factory. After a certain delay in the startup of the processing department and the
final optimization, the factor started complete work in October 1999 G.C Population, economy and
the natural condition of Adwa town
 Population - 50,000
 Economy: Commerce, Industry and Agriculture
 Natural Condition
 Altitude: 2200
 Annual mean temperature: Maximum 27oc and minimum 10 oc.
 Water resource: Gerea Dam
 Annual Rain fall: 800 mm
 Climate: Medium temperate
 Land escape: hilly and surrounded
1.3 Objective of the company
The company is established to accomplish the following business objectives:
 To produce threads yarns and all kinds of cloths from cotton, polyester and cotton and
polyester mix at any ratio.
 To produce clothing materials for personnel and house hold uses.

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 To engage in agricultural activities where becomes necessary for the realization of its
business.
 To engage generally in any other trade necessary or conducive to the achievement of its
business.
1.4 Missionof the company:
The mission statement originates from the total customer satisfaction for the value added products
based on R&D and creating a hub for excellence in the textile field in Africa in general and
Ethiopia in particular besides imparting training and skills for the human resource development of
the inhabitants so much required.
So generally the mission of the company includes:
 To make profit by competing in the global market
 To make profit by competing in the global market
 To generate foreign exchange by increasing export
 To create employment opportunities to the community
 To have sustainable production capacity and contribute in the development of the foreign
and the country.
1.5 Vision
Develop Almeda as a reference for Ethiopian Textile and Garmenting industry and create the
conditioned that Almeda can become leader in the Ethiopian market in terms of export and local
market.
1.5.1 Core values:
 Customer satisfaction
 Reliability
 Commitment
 Good governance
1.6 Production program
The spinning and weaving mills are designed to run 350days annually and the finishing mill run
300 days only. The spinning ,weaving ,&processing plants operate in three shifts daily ,working
8hrs per day ,but the garment department works in two shifts daily working 8hrs per day.

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1.7 Pricing
The selling prices of the product of the mill were determined on the basis of the following two
factors. The mills price should be competitive with those of other local mills producing similar
products. With respect to products not locally produced so far , the mills price is at least in parity
with the landed cost of imported similar products and/or ‘’second best’’ products available in the
local market.
1.8 Main products of Almeda textile plc
Product is the outcome of a production process supplied to the public at large. It should be delivered
to the society in the required amount quantity and time as well as at reasonable price. Almeda
textile plc is capable to produce any variety of fabric; however, the product mix ranges from Drill,
Twill, poplin and Sheeting Terry towels to heavy duty canvas.On the garment side, knit fabric of
various weighted as well as dyed fabric can be converted to value added articles for which all the
paraphernalia is present at the same location.
Typically Almeda textile produces the following products as listed bellow
 Trousers T-shirts
 Shirt
 Bed sheet
 Uniform
 Military/police
 Over all
 Yarn
 Gwen …etc.
1.9 Main customers or the end users of the company
Customers are the consumers of one company’s product and these also play an important role in
increasing the potential of the company’s production capacity by balancing the demand and supply
of the company. Almeda have an enough potential customers locally and globally these are:
Table 1.1 global customers
No of the customer Country
1 QC supply USA
2 Bern apparel USA
3 Edwards USA

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4 Tsi USA
5 Kopper man Germany
6 Vitconplc Sudan
7 Champro USA
8 Biggali Name Italy
9 Pinakle USA
1.10 Local customers
 Guna
 Tigray police
 Amhara police
 National defiance, Federal police
 Somali police

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Figure 1.2: Organizational structure of the alemda textile factory
Knit processing
Technical R& DEngineerin
g
department
Garment department
QC department
Knitting department
General Manager Legal advisor
Spinning department
Secretary
Weaving department
Production and technical
DGM
Woven processing
Planning and product
development
Internal
auditor
Finance and logical
Finance department
System development
department
Marketingdepartment
Supply department
Branch office
Almeda training
center department
Human resource
department
department
Administration DGM
General Service
department

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1.11 RESPONSIBILITIES:
Department: giving bus service, clean lines of the company, gardening and guarding.
ATC: General Manager: guides, controls and give commands for its branch and control
overall activity in the company.
Legal advisor’s for consulting/advising if there is a related issues around laws and rules.
Internal auditor: audit the financial and product of the company.
Production and technical DGM: works as behalf of manager for the technical and production
sector.
Commercial DGMcontrol all the activity of financial and supply of the company.
Planning and product development: plans the production capacity and assignment of each
department according production orders.
Engineering department: works on modification of materials, spare parts. Technique R&D dep’t
it is a research and development on techniques (ways of simple aspect) in the factory.
Finance department: control and study payment and custom clearance and cost analysis.
Marketing department: control sales, study cost or price of product and promotion of product.
Supplydepartment: purchasing of materials depend on department demand and distribute
accordingly.
Branch office:works on financial, supply and HR activities.
Humanresource development: promotions, annual leaves control discipline of community in the
company.
G/servicecoach and train the new operators and gives refreshment course for old operators, shift
leaders, supervisors and invite other company operators and train & entertain them.
Systemdevelopment:works on QMS (quality management Make ready for certification.
1.12 Work Flow and Major Department of Alemda Textile Factory
1. Raw material store
2. Spinning
3. Weaving
4. Processing
5. Knitting
6 Knit dyeing
7. Garment

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Processing
(finishing)
Knit dyeing
Knitting
Garment
Raw material store
Spinning
Weaving
Figure1.3: work flow of alemda textile

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1.12.1. Raw materials - cotton and polyester
Figure1.4: raw cotton store
Alemda textile are using the following raw materials
 Cotton
 Polyester
 Dyestuffs of different type
 LPG(LIQUID PETROLUEM GAS)
 Furnance oil
 Chemicals and auxilaries
 Electric energy
 Water
1.12.2 Spinning department

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Figure1. 5: unifloc machine
Figure1. 6 Spinning yarn
Spinning is the first step in the textile industry after ginning. It is the process of changing fiber
to the weave able or knittable yarn. And the raw material for spinning department is cotton.
Spinning department includes all the activities required to convert cotton in to yarn. And
there are different machines that are used in fabricating yarn from the receiving cotton
and capacity of the department is 20 tons per day.
A) Blow room
Blow room is the first of planning department where cotton bales or fibers are processed for
mixing, opening and cleaning, and to form a regular sheet of different votes according to the
requirement.
Objective of blow room
 To open the fiber
 Cleaning like(seed, leaves and dust)
 Blending/mixing two or more types of fibers which in a required ratio
B) CARDING
Carding is the second process of spinning department where small tufts of cotton are processed
for opening and cleaning and to form a regular strain of fibers called sliver carding. The
sliver. Carding plays the most significant role in determining yarn quality.
Objective of carding
 To open small tufts of cotton fiber in to fiber condition
 To remove the waste of fiber i.e. seed shells, motes, leaves and dust particles

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 To produce continuous strand of fiber in a very open form called sliver and
deposit in can form for subsequent process of draw frame
 To open disentangled naps.
 To form the parallel fiber and so on. and also We observed two carding lines in
Almeda, the first one is open end line which has line B, line C and line D. and the
second one is ring line i.e. line A.
C) DRAW FRAME
Draw frame is the 3rd process of the spinning department. The card sliver has hooked fiber and
Entangled fiber and those fibers are to be parallel to regularize the strand or sliver.
 The drafted sliver has more parallel fiber, weight per meter length is more regular.6 to 8
slivers are mixed together and drafted between the rollers to the required weight is the
main function of draw frame.
Objectives of draw frame
 improving material evenness
 parallelizing fibers
 blending
 dust removal
D) Lap forming (UNILAP)
 Lap former is the fourth process in the combed yarn production. it is a machine that
produces a lap of draw sliver in a suitable form for comber 24 numbers of first pass draw
sliver is reeled and Passed through drafting roller in sheet form and makes a lap under
calendar pressure. Tasks of the lap forming process
 To prepare a sheet of fiber called lap for onward in combed i.e. suitable package for
combing.
 To maintain the quality and regularity of product.
E) Comber
Comber is the 5th process in the combed yarn production. Is used for production super
quality yarn from cotton fibers. Naps and short fibers are removed and fibers are made parallel
to each other.
Objectives of comber
 To obtain uniformity in fiber length by removing short fibers from the carded material.

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 To remove naps and trash from the sliver.
 Parallelization of fibers.
F) ROVING (SPEED FRAME)
Roving is the 7th process in combed yarn production of the spinning department sliver is feed to
roving machine and passed through creel and drafting arrangement in speed frame sliver
received from draw frame is drafted, twisted and attenuated in the form of a thin rope
called roving.
Figure1. 7 roving
Objectives of roving
 To convert sliver in the form of thin rope. Its cross sectional fiber density will be much
less than feed sliver To wind it in on a package suitable for subsequent process.
G) RING FRAME
This is the 6th process in carded yarn production and 8th process in combed yarn production.
Roving bobbins are creeled over hungers and roving is passed through guide bars, drafting for
Required yarn count through ring and ring traveler for the formation of yarn which is wound on
Copes fitted over revolving spindle.
Functions of ring frame
• Drafting the roving until the required fineness is achieved.
•Imparting strength by inserting twist
•Winding the twisted yarn in a form suitable for storage, transportation and further processing
Generally, the total work flow is as the following order. Raw material receiving cotton.
 Blow room - for opening clearing and mixing of cotton.

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 Carding - for elimination of short fiber, carding formation
 1st drawing - for avoiding waste
 Combing - for combing lab 2nd drawing - for controlling silver thick
 Ring spinning - twisting the fiber strand to impart strength to yarn winding the
resulting yarn.
 The final yarn manufactured processed to weaving or knitting department.
1.12.3 Weaving department
The department is designed to produce different variety of fabric ranging from simple
plain to complex design of fabric. The designed production capacity of this department is 28,000
meters of fabric. The product types that can be produced in the department are; different type of
plain fabric (sheeting, shirting, rib stop and canvas) ,different type of twill (light to heavy twill
and French twill) and terry towel .
Figure1. 8 weaving
Alongside the above machines. There are also direct warping, sectional warping and sizing
machines which are capable of feeding the looms and employing strength for the yarns formed in
the spinning. These machines are computer aided and capable of providing detailed data of the
machine & operator efficiency.
1.12.4 Processing;
Processing department in almeda textile factory is to produce quality product which satisfies the
requirement of customer with a minimum cost of production. Processing is the process which
gives color to the textile material. The color given to the textile material was by dyeing,
bleaching and printing processes was responsible for the attractiveness of the textiles. Processing

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plays a pivotal role in value adding by products which contributes maximize profit in almeda
textile factory. The department can produce up to 800,000 meters per month.
Machine types
 Singeing desizing
 pad batch dyeing
 Scouring and bleaching
 dyeing
 Mercerizing
 washing
 Drying
 loop steamer jigger dyeing
 Printing
 yarn dyeing
 calendaring

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Figure 1.9: processing of different fabric
Figure 1.10: processing of different fabrics
1.12.5 KNITTING DEPARTMENT
This department converts the yarn produced in spinning in to smooth fabric that used to produce
t-shirts.it consists of the following steps
 The receiving yarn transfer to creel
 Quality check and transfer
 Knitting
 Inspection and numbering
Figure 1.11: Knitting machine

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1.13.6 Knit dying
It was the process of applying color to the knitted fabric or yarn. The machineries in this
department are described as follows
Dyeing and bleaching (Sclavos machine); chemicals are added to this machine with wax, salt etc.
and pH level adjustment in order to bring the required color and make it white.
Sample bleaching; the same as bleaching machine but used for sample production
Squeezer; removes water from knit fabric up to 60-70 %
Dryer; removes the remaining 30-40% of water from the knit fabric and make it dry
Figure1.12: knit dyeing
Compactor; this enables the knit fabric to sustain its strength after wash or avoids shrinkage and
adjusts width of the fabric.
1.12.7 Garmenting
Figure 1.13: garment of different cloths
 Garment in almeda textile factory were producing different types of wearing apparels for
the local and international market.
 The company has very flexible and brand-new state of the art technology in its design
and pattern making, make up, finishing and quality control units.
 Currently the factory has 1543 sewing machines.

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 The cutting unit with full size cutting table is strengthen by design and pattern making
unit equipped with computer aided design(CAD)
Product types which are produced in almeda garmenting factory are;
 Basic t-shirt
 bed sheet
 Other fashionable cloths
 jeans trouser
 Jeans shirt
 Classical T-shirt
 Pajama
 Casual trouser
 work wear
 Military uniform
 T-shirts

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CHAPTER 2
2. THE OVERALL BENEFITSTHAT GAINED FROM THE COMPANY
2.1. Improving practical skills
To improve practical skills first there must have theoretical, so our theoretical skills taken for
about three years and one semester were related with the practical. So, wehave got a chance to
improve our theoretical based practical skills. We have tried to put our practical skill that we
have got during internship in general: To list some of the practical skill that we had gained
during our internship period;
Material handling: there are material handling techniques
 All fabric should be covered by plastics because they can be contaminated by dust and
external color or chemicals.
 Chemicals should be put in the separate area because some are hazardous and toxic.
Chemicals and dye staff must be separated/segregated according their names to reduce
confusing what kind of chemicals to be used. Proper material handling gives the
following result: better quality of product, keep smooth running condition on of the
machine, reduce the waste, and minimize the cost of materials, equipment and spares like
cones, washers, springs, comb dents, reed dents.
On quality: quality of the end product is the customer satisfaction. The quality control must be a
culture and every employee participates for achieving quality objectives. In quality we have seen
how the raw material in the Wet processing department measures in the laboratory and how
correct it (by maintenance workers by taking the recorded data); to check the processed
parameters. There are two type of quality control of grey fabric:
Cost minimization and material utilization: cost can be increased by the following factors:
Shortage of awareness: some operators do not know what to do and what the cost of everything
in process. Also some operators do not feel sense of ownership and they are careless for every
cost losses.
Uncontrollable problems: in some case there will be failure of machines, steam leakage, water
leakage problem in processing department. So to minimize cost:
 An idle machine should not be opened light should be off to reduce electrical power
consumption.

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 In processing department water line should be closed during idle time of the
 Attention has to be payed when preparing of dye staffs, chemicals and sizing materials
preparation.
2.2 Upgrading of theoretical knowledge:
Our theoretical knowledge was upgraded by seeing practically and get more information through
observation. we have gained good experience on upgrading knowledge of organizational
structure, working process in industries and how to behave in industries by;
 Asking the operators and other responsible person to the different machines in different
department.
 Relating with theories, calculations, machine type differences, working principles,
efficiencies, and property of the machines.
 Using manuals and other materials from the company
2.3. Industrial problem solving capability
We have developed a skill to solve the industrial problem like this
Identify a solution
Examine result
Break problems
in to part
Select best solution Take an action
Gather informationEvaluate problem
Figure 2.1: industrial problem solving

20
2.4. Interpersonal communication skills:
In the internship program, we have got a chance to communicate with each department manager,
heads, operators, production heads shift leaders, training centers and people around marketing,
finance, supply that responsible for their own authority. As a result we have developed our;
 Verbal communication: during communicating, that the way of or the speaking must be
clear and light words to make easy for understand.
 Non-verbal communication:when communicate with someone, the act, sign and symbols
must be regular. Example: Eye contacts, gesture, space, touch.
 Listening skill, problem solving skill, decision making
2.5. Team playing skills
The amazing thing in our intern is our cooperation between the operators, confidence in the ,
respect seniority and before taking actions our individual opinions was come in to one great idea
(by sharing ideas), even to get permissions some facilities from the hosting company ALTEX, all
we have responsibility to ask.
 We have tried also to solve some problems with operators; we spent most of the times
together with pleasure.
In general we have developed our team work skills by
 Treating each other; even invite some lunch, tea… just to maintain working relationships.
 Encourage team members to share ideas.
 Sharing knowledge with other interns came from other universities and colleges
2.6. Improving leadership skills
We have tried to take some of the following good ideas about leadership skills:-
 A position to leader in any task must be given according to his/her profession.
 Some constructive ideas, too much advise gives interferes with two-way communication,
yet some advice can lead to improved performance.
 Good performance and behavior, may good performance come from the leader displays a
helpful constructive attitude and giving emotional support.
 The main thing that we have got is that ’’ a manager or a leader is a coach, not a player.’’
The leader can still performs some of the tasks but the emphasis is on showing other
peoples on how to accomplish work and then motivating them to sustain their
performance. Example:-for all in the side of production aspect, quality aspect, cost If a

21
team member makes a mistake, the manager should not take over, instead offers a
constructive suggestion.
 If there is conflict between members/co-workers the manager or leader must resolve the
conflict in a good manner with his problem solving method or negotiate them.
2.7. Understanding of work ethics:-In ALTEX there are job disciplines for operator to
increase or develop a good team sprite and friendly relationship among themselves and machine
safety. For example: eating, smoking inside the production area was prohibited. As we have
stayed, we have got practical work/job disciplines that used to us after graduation like:
 Punctuality: it has its own benefits.
 The company will be more productive
 Employee will be motivated morally and financially
 Bosses will have more confidence to the employee
 Employee will improve industrial culture
 Respecting others, social interaction, honesty, developing good behavior etc.
Accident prevention by using devices the accident or hazardous working condition can be
prevented. (E.g. safety devices on switch board, main switch, emergency push buttons) and keep
the chemicals in their proper area because they can make fire. And for operators they must use
safety materials like:
 Glove: to protect touching of chemicals and other hazardous material.
 Mask: used for covering of mouth and nose; and protect dust and tiny –particles
from entering.
 Safety rules during machine operation: for example before operate the machine
pay proper attention to the people around the machine, wear properly (don’t wear
scarf, high heal shoe, tied the long hair properly).
Generally, the work ethics helps minimization, work load, machine safety and good relation
between the communities in the company.
2.8. Entrepreneurship Skills
To be a competent entrepreneur, I must have fulfilled the following:
 Management skill (time, money, energy)
 Communication skill (e.g. the ability to sell ideas, search problems and know how to
solve

22
 The ability to work both as part of a team and independently.
 Able to research effectively (available markets, supplier, customers and the competition)
 Able to plan coordinate and organize effectively.
 Self-motivated and disciplined
 The ability to multi task and work under pressure
 The ability to network and make contacts So in our internship time, we have tried to
increase a good entrepreneurship skill.

23
CHAPTER 3
CHALLENGES FACED IN THE FACTORY
3.1 Challenges facing in work tasks
As we have expected we have faced so many problems in our stay in almeda Textile Company
some of them are as follows.
 Absence of regular supervisor that guide me the working condition and process of the
company; which means they roughly informed us about the whole organizational
structure but some supervisors were not willing to supervise us.
 Most operators had not have enough knowledge; when I ask them they do not replied
sufficient answer.
 The chemicals used in textiles had not scientific name so, it was difficult to know their
functions.
 Since the factory was very wide and vast the work flow was difficult to understand and
was being confused at the starting of my internship.
 Experience of the workers; when I ask them they lose a confidence.
 Rules and regulations of the company declared that moving from one section of
production to other was prohibited .unless changed by the human resource (HR)
 Absence of full library or enough manuals getting data was difficult
 Due to down time to know the whole process was hard.
Measures to solve the challenges
We have taken the following measures to solve the above problems
 To solve the challenge with regard to absence of supervisor that guide me in the factory
we have decided to make a good relation with both the operators, supervisor’s department
heads, shift leaders .then after a time everything was became very simple.
 We have solved our problem by searching internet, reading some of the manual of the
company. Generally most of the problems were solved by observation and
communicating with HR and training center of the respected authority.

24
3.2 Finding of problems
Throughout the internship period we found problems related with our knowledge of mechanical
engineering. The problems are on the following areas.
 Problem of the raw material store
the raw material cotton was naturally sensitive to moisture of the environment. The factory
stored the cotton in store at moisture of 3.8 %. But the standard of cotton fiber moisture content
is 7-8.5％.since the store was made up of metal and there is no air conditioning system on store
the cotton loss its moisture. This results in decreasing of the quality of cotton.
 The sequence of this kind of storing system will be lead to the following problems:-
 Decreasing the fiber strength, increasing of short fiber, increase the roller loping.
Problem in the weaving department
 In the sizing section the steam came from the boiler house and after the latent heat of the
steam exchanged to the yarn it forms a condensate .but disposed to the sewer
Problem in processing department
 In processing department we have observed different problems during production steps
were not followed to minimize effort and energy but; it has its own effect on the quality
of the fabric. For example the fabric came from weaving was sometimes directly enter
the scour-bleach process without getting pretreatment like singeing and desizing. For this
reason, customers might lose their trust on the company. The steam came from boiler and
exchanges its latent heat with process fluids. But there was no heat recovery for the
sensible heat.
Problem of heat losing in a boiler
 The steam available from the boiler is mostly at pressure of 9 bar and temperature of 190
0 C, but when the boiler works for long time without rest the level inside the boiler goes
down, the pressure of steam increase immediately and steam with pressure above 12 bar
is not needed in the factory. So the steam released to the atmosphere.
 In costic recovery mixing of the chemical with steam is performed by man power.

25
CHAPTER FOUR
PROJECT (CASE STUDY)
4. COSTIC RECOVERY ROBOT MACHINE
4.1 INTRODUCTION
Costic Recovery Robot Machine is advice that lifts costic soda from one place to another plase
and it reduces wastage of time. The textile industry is one of the most complicated industries
among the manufacturing industry. Costic recovery is one of the major problems faced in
Almeda textile factory. a study of processing department that there are many complicated
processes and chemicals used throughout the production. For the purposes of continuous
blanching in processing department they are used costic soda (NaoH).this chemical is strong and
hazard. In manufacturing industries robots have unlimited capacity to perform different tasks
with in limited time. Those robots have grate life time but should be use friction reducers like
griss.Robots can be classified based on their working fluid. Those are pneumatic robot and
hydraulic robot .the working fluid of pneumatic robot is called air. The working fluid of
hydraulic robot is called water.

26
4.1.1WORKING PRINCIPLES OF COSTIC RECOVERY ROBOT MACHIN
This machine installs in the house of chemical mixing room.This machine automatically carries
the chemical from the land to the reservoir using pneumatic system. The procedures are there is
utility this collects air in to one station. From this station the pneumatic robot sucks air through
hose and pick up the load in to the desired reservoir.
,
Figure 4.1 COSTIC RECOVERY ROBOTIC MACHIN

27
4.2. Problem of the statement
In every stages of alemda textile factory various types of chemicals are used. Those chemical
are hazards to human being. they can damage seriously to different parts of body such us
blindness of eye, skin,mantality retarded.eventhough this said effect those chemicals are being
performed by human labor in this factory in which is not suitable manner by caring through their
body. Thereforthis costic is very dangerous which causes dieses to human being do to their
strength.
Figure 4.1 costic recovery
The conventional treatment of this hazard representation of this machine is costic recovery robot.
This works instead of human energy. This type of machine is very easy to operate, time sever,
which working fluid is air. Costic recovery robot machines are used to left in to large height to
transport from one elevation to another. Those types of machines are used to transport goods
from one room to another room, but in some department there are not qualified machines to be
functional like in costic recovery. For those purposes I am starting to design simple pneumatic
robot which is operated with one person.

28
4.3. Objective of the Project
4.3.1. General Objective
The main objective of this project is designing and modeling of pneumatic robot machine for
industries.
4.3.2. Specific Objective
 To design mathematically for each components of pneumatic robot machine.
 To determine and calculate stress analysis, detail drawing of pneumatic robot machine
and assembly drawing of the machine by solid work software.
 To select proportional material
4.4 Scope of the Study
This project will be designed and modeled based on studies conducted by referring the existing
reference books and mechanical software. The main scope of this project is starting from
conducting literature review, design analysis of the components up to conducting modeling and
manufacturing drawing pneumatic robot machine.
4.5. Limitation of study
There is not internet assess
There is not enough cost to manufacture

29
CHAPTER FIVE
LETRECHER RIVEW
5.1 COSTIC RECOVERY KNOW DAY IN ALMEDA TEXTAIL
Textile industry is one of the most complicated industries among manufacturing industry (seicuk,
2005).costic recovery is the process of addition water to costic soda for the purpose of
continuous bleaching. In almeda textile factory the process of costic recovery is performed in
which is not suitable manner.
Figure 5.1 costic recovery

30
CHAPTRE SIX
METHODOLOGY
To conduct the design of the pneumatic robot machine the processes follows the product design
development (PDD) process.
Collecting data is done on existing parts of Tigray and Adwa areas; this information was
gathered from adigrat university mechanical department and almeda textile factory. The final
design is the output of data inputs, conceptual design, and analytic design. Data collection was
through informal interviews, different design books and from the internet. The final drawing is to
be prepared using solid work software.
Farther more, the necessary thing is examine the cost estimation of the pneumatic robot machine
by taking some information standard markets to know the cost of pneumatic robot component
and depending on the manufacturing process, assembly process, labor cost etc…After taking and
collecting some necessarily information for this project, process and analysis of that’s
information has done. Data collection was done through the two basic ways
6.1 Primary Data: -
 Direct observation,
 Formal & informal interview with different professionals and from written data.
6.2 Secondary Data: -
 Different research works and pneumatic robot manufacturers
 From internet.
Generally, after collecting the data by primary and secondary source, the process follows through
PDD sequence and procedure from identifying customer need, design analysis and cost
estimation. The methodology process follows PDD procedure as:

31
Figure 6.1 general procedure in this design
Need or Aim
Synthesis (mechanism)
Analysis of force
Material selection
Design of elements
Size and stress
Modification
Detailed drawing
Production

31
CHAPTER SEVEN
DESIGNE CONSEPTS AND SELCTION
7.1. Concept Development Process
7.1.1. Identifying Opportunity
The opportunity is the problem, which is lack of pneumatic robot machine, so the almeda textile
labors tend to work by their energy and hands by consuming a lot of time, capital and money.
7.1.2. Identifying Company Need
In Adwa tigray region most of the fabrics use traditional tools and ways to mix water and acid in
order to continuous blanching , so most of the Ethiopian fabrics need a machine that can
minimize the total cost, time and labor, which have an overall profit.
7.1.3. Organizing Raw Data
The needs that obtained by direct observation and interview. The needs are organized as follows:
 Less number of components
 Design simplicity
 Cost
 Strength
 Availability
 Manufacturability
 Durability
 Complexity
 Weight
 Operability
 Reliability
 Maintainability
Preparing relative importance of needs

33
Table7.1. Preparing relative importance of needs
7.2. CONCEPT DESIGN
Concept generation is when a product development team comes up with the ideas, is the most
critical step in the engineering design process, without it there is no design. It is the process of
creating, developing, communicating ideas, which are abstract, concrete or visual. The
processincludes constructing through the idea innovating the concept, developing the process and
bringing a concept to real worled.we willsee Different alternatives of robotic machine design.
Hence, the following are some of the competitive mechanisms anticipated to meet the quality
and feature needed.
7.2.1. Model A: Design of pneumatic machine with air as power source to the machine and that
uses collecting plates welded on it as collecting mechanis
Figure7.2 pneumatic robot machine
Product design needs Level of importance
Less number of components 3
Light weight 1
Low cost 10
Easy to operate and maintain 5
Safe and reliable 4
Strength 9
Design simplicity 6
Durability 8
Availability 5
Manufacturability 7

34
CHAPTER EIGHT
DESIGN ANALYSIS
Factors affecting performance of costic recovery pneumatic robotic machines:
1. Chemical factors
 Chemical change
 Ambient temperature
 Chemical moisture
 Chemical density
1. Machine factor
 Shape and size of pneumatic robot machine.
 Real position and speed
 Cutting blade shape and speed
 Arm speed
 Machine settings
2. Operational factor
 Height of cut
 Operational speed
In put parameters
 Weight to left =25 kg
 Minimum height to left =200mm
 Maximum height to left =950mm

35
8.1. Designof column support
8.1.1 Design(trapezoidal support)
Assumptions
 Height(h)=300mm
 Width(b2)=150mm
 Width(b1)=75mm b1
 Thickness of each top and bottom(t)=12mm
 Mass applied =250kg h
 Acceleration due to gravity=10 𝑚
𝑠2⁄
 Factor of safety= 5
 Length of trapezoidal beam support =1800mm b2
Elastic Modules (E) =210GpaFigure 8.1 trapezoidal support
Material
AISI steel 1030
 Ultimate tensile strength (𝜎ut) =848Mpa
 Yielding strength (𝜎y) =648Mpa
Analysis
First let’s determine the weight applied on it
W = mg = 25kg ∗ 10m
s2⁄ = 250N
Then let’s determine the allowable stress and allowable shear stress of the material
σall =
σy
F.S
σall =
648
5
σall = 129.6Mpa
τallow =
σy
2F. S
τallow =
648
2 ∗ 5

36
τallow = 64.8Mpa
Then let’s find the reaction force formed in the leg support
125N 125N
RA RB
Then summation of force in the vertical plane is zero. That is
∑ FY = 0
RA + RB = 250N
And summation of moment at point A is zero. That is
∑ MA = 0
1800 ∗ RB − 125 ∗ 300 − 125 ∗ 1500 = 0
𝑅 𝐵 = 125𝑁
Substituting
𝑅 𝐴 + 𝑅 𝐵 = 250𝑁
125+RA = 250N
RA=125N
Then let’s determine the maximum bending moment formed in the support beam.
σb =
MC
I
But moment is given by
M =
RA ∗ L
2
M =
250N ∗ 600mm
4
= 9.4 ∗ 103
Nmm
Also the moment of inertia is given by
Ixx =
ℎ2( 𝑎2
+4𝑎𝑏+𝑏2)
36( 𝑎+𝑏)

37
Ixx =
3002 (752
+4∗75∗150+1502 )
36(75+150)
Ixx = 9.03𝑚𝑚4
And the center of axis is given by
𝑐 𝑚𝑎𝑥=
𝑎+2𝑏
3( 𝑎+𝑏)
*h =
75+2∗150
3(75+150)
*300 = 166.7mm
Then the maximum bending moment formed in the support beam will be
σb=
M∗Cmax
I
𝜎𝑏 =
9.4∗103
𝑁𝑚𝑚 ∗166.7mm
9.03𝑚𝑚4 = 173534
𝑁
𝑚𝑚2
8.2 DESIGN OF SHAFT SUPPORT
A shaft is a rotating member, usually of circular cross section, used to transmit power or motion.
It provides the axis of rotation, or oscillation, of elements such as gears, pulleys, flywheels,
cranks, sprockets, and the like and controls the geometry of their motion. However, a shaft can
have a noncircular cross section and need not be rotating. An axle, a nonrotating member that
carries no torque, is used to support rotating members.
Figure8.2. Shaft
Material Selection
Mild steels general-purpose steel bars for machining, suitable for lightly stressed components
including studs, bolts, gears, handles and shafts.

38
It is less costly.
Mechanical Properties
Ultimate tensile strengthσut = 430Mpa
Yield strength σy = 220Mpa
Ultimate shear stress = 360 MPa
Factor of safety =5
Modulus of elasticity E =210Gpa
Density, ρ = 7860
kg
m3
Given Parameters and Assumptions
Design power, 𝐏 = 𝟎. 𝟖𝟓𝐤𝐰
Shaft speed, 𝐍 𝟐 = 𝟑𝟎𝟎𝐫𝐩𝐦
Shaft length =10mm
Required
 Diameter of shaft, ds
Analysis
The shaft is subjected to torque of, T =
60P
2πN
=
60×850
2π×300
= 27.056Nm
𝜏 =
𝜏𝑢
𝑓. 𝑠
𝜏 =
360
5
= 72
𝑁
𝑚𝑚2
Diameter of the solid shaft
Let d = Diameter of the solid shaft.
We know that torque transmitted by the shaft,
T =
𝑃∗60
2𝜋𝑁
,T=27.056Nmm

39
We also know that torque transmitted by the solid shaft (T), T=
𝜋
16
∗ 𝜏 ∗ 𝑑3
27.056*103
Nmm =
𝜋
16
∗ 72 ∗ 𝑑3
, 1913.82𝑚𝑚3
=𝑑3
,d=√1913.82
3
= 12.42𝑚𝑚
8.3. DESIGN OF SHAFT CONECTING BASE AND COLEMUN
Material specification selected for the shaft is plain carbon steel to British Standardspecification
BS 970 080M30, Hardened and Tempered, whose properties are as shown inAppendix Band the
material yield strength is 700MPa both in tension and pure compressionand 450MPa in shear.
Allowable compression=700Mpa
Allowable tensile stressσt = 700Mpa
Allowable shear stress (𝜏) = 450MPa
Factor of safety =5
Given Parameters
Weight (w) =mg=250N
Analysis
𝐰 = 𝛔𝐜 ∗ 𝐀𝐜,Where 𝛔𝐜=compressive stress
Ac=cross sectional area
𝐰
𝛔𝐜
=Ac, Ac=
𝛑
𝟒
𝐝 𝟐 ,
𝟐𝟓𝟎𝐍
𝟕𝟎𝟎
𝐍
𝐌𝐌 𝟐
=
𝛑
𝟒
𝐝 𝟐, 𝟎. 𝟑𝟓𝟕 =
𝛑
𝟒
𝐝 𝟐, 𝟒 × 𝟎. 𝟑𝟓𝟕 = 𝛑𝐝 𝟐
1.4285= 𝝅𝒅 𝟐,
𝟏.𝟒𝟐𝟖𝟓
𝝅
= 𝒅 𝟐, 𝟎. 𝟒𝟓𝟒𝟕 = 𝒅 𝟐, 𝒅 = √ 𝟎. 𝟒𝟓𝟒𝟕 = 𝟔𝟕𝒎𝒎
8.3.1 Shaft stress
Compressive stresses duo to axial loads using the new diameter is

40
𝜎𝑐 =
𝑤
𝜋
4
𝑑2
=
250𝑁
𝜋
4
67𝑚𝑚2
= 4.75
𝑁
𝑚𝑚2
8.3.2 Torque required rotating the shaft
𝑇1 = 𝑝 ×
𝑑𝑚
2
Where p is effort, dm is mean diameter
T1=200N×
67
2
𝑚𝑚 = 6700𝑁𝑚𝑚
The shear stress due to this torque using the new diameter is given
𝜏 =
𝑇1 × 𝑑
2𝑗
, 𝑗 =
𝜋
32
𝑑4
=
6700𝑁𝑚𝑚 × 67𝑚𝑚 × 32
𝜋 × 674 𝑚𝑚4
= 0.224
𝑁
𝑚𝑚2
8.3.3 Principal stresses
Maximum principal stress is as follows:
𝜎𝑐𝑚𝑎𝑥 =
1
2
⌈𝜎𝑐 + (√ 𝜎𝑐2 + √4𝜏2
⌉ =
1
2
⌈4.75 + (√4.752 + √4 × 0.2242
⌉ = 4.974
𝑁
𝑀𝑀2
Design value of 𝜎𝑐 =
𝜎𝑎𝑙𝑙
𝑓.𝑠
=
700
𝑁
𝑚𝑚2
5
= 140
𝑁
𝑚𝑚2
𝜎𝑐𝑚𝑎𝑥 ≤ 𝜎𝑐, 4.974
𝑁
𝑀𝑀2
≤ 140
𝑁
𝑚𝑚2
, 𝑤ℎ𝑖𝑐ℎ𝑖𝑠𝑠𝑎𝑓𝑒
Maximum shear stresses is as follows
𝜏𝑚𝑎𝑥 =
1
2
(√ 𝜎𝑐2 + 4𝜏2) =
1
2
(√4.752 + 4 × 0.2242) = 2.38
𝑁
𝑚𝑚2
Design value of 𝜏 =
𝜏𝑦
𝑓.𝑠
=
450
𝑁
𝑚𝑚2
5
= 90
𝑁
𝑀𝑀2
𝜏𝑚𝑎𝑥 ≤ 𝜏, 2.38
𝑁
𝑚𝑚2
≤ 90
𝑁
𝑀𝑀2
, 𝑤ℎ𝑖𝑐ℎ𝑖𝑠𝑠𝑎𝑓𝑒
Cheek; those maximum shear and compressive stresses are less than the permissible stresses
which is safe design.

41
8.4. DESIGN OF LEVER
A lever is a rigid rod or bar capable of turning about a fixed point called fulcrum. It is used as a
machine to lift a load by the application of a small effort. The ratio of load lifted to the effort
applied is called mechanical advantage. Sometimes, a lever is merely used to facilitate the
application of force in a desired direction. A lever may be straight or curved and the forces
applied on the lever (or by the lever) may be parallel or inclined to one another. The principle on
which the lever works is same as that of moments.
According to the principle of moments,
Material Selection
Mild steel is general-purpose steel bars for machining, suitable for lightly stressed components
including studs, bolts, gears, levers, handles and shafts.
Allowable tensile stressσt = 100Mpa
Allowable shear stress = 55 MPa
Factor of safety =5
Modulus of elasticity E =210Gpa
Given Parameters and Assumptions
Width of the lever =200mm
Effort (p) = 200 N
Height of the lever= 2800mm
Weight (w) =250N
Thickens=30mm

42
Length of the lever from the furculum to the weight is 200 mm, L1
Length of the lever from furculum to effort is 250 mm, L2
Required
 Furculum, (FR)
Analysis
The principle on which the lever works is same as that of moments.
According to the principle of moments,
W*L1=P*L2 or
𝑤
𝑝
=
𝑙2
𝑙1
Mechanical advantage,(M.A)=
𝑤
𝑝
=
𝑙2
𝑙1
P L2 FR L1 W
The forces acting on the lever areFigure. 8.3. Type of lever.
1. Load (W), L1
2. Effort (P), L2
3. Reaction at the fulcrum F (RF)
M.A=
𝑤
𝑝
=
𝑙2
𝑙1
, M.A=
𝑤
𝑝
=
250𝑁
200𝑁
=1.25
M.A=
𝑙2
𝑙1
= 1.25=
𝐿2
200𝑚𝑚
= 𝐿2 = 200𝑚𝑚 ∗ 1.25 = 250𝑚𝑚
Then summation of force in the vertical plane is zero. That is
∑ 𝐹𝑌 = 0
𝑃 + 𝑊 = 𝐹𝑅
200N+250N=FR
FR=450N
8.5 PART DESIGN OF PNUMATIC CYLINDERS
General specifications from geometrical analysis &assumptions;
A Design of piston rod

43
 It is the Cylinder retracted height =900 mm
 Cylinder extracted height=1500 mm
- column that pushes the boom in ordered to lift the load ;
- Subjected to higher compression stress
- Length of cylinder =600 mm<from geometrical analysis>
 Material selection; gray cast iron
-𝜎𝑢𝑡=400 𝑀𝑝𝑎, 𝜎𝑦=250 𝑀𝑝𝑎 , 𝜏 𝑦=145 𝑀𝑝𝑎 ,𝐸=200 𝐺𝑝𝑎, 𝐺=77 𝐺𝑝𝑎
F=design load =load * factor of safety
=250*2
=500 N,
𝑑 𝑝𝑟=𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑠𝑡𝑜𝑛 𝑟𝑜𝑎𝑑 ,
𝜎 𝑎𝑙𝑙 =
𝐹
𝐴
= 𝐹
3.14
4𝑑 𝑝𝑟∗2 ,
⁄⁄
𝜎𝑎𝑙𝑙 =
𝜎𝑦
4
=
250
4
= 62.5 𝑀𝑝𝑎
𝑑 𝑝𝑟 = √
4 × 500
3.14 × 62.5 × 106
= √4.01× 10−4
𝑑 𝑝𝑟 ≈ 20𝑚𝑚
- Cheeked for buckling of the piston road ;
 Apply Rankin’s formula for all types of columns;
𝑊𝑐𝑟 =
𝜎 𝑦×𝐴
1+⁄
𝑎( 𝐿
𝑘)⁄
2
=
250 × 106
× 3.14 × 0.032
1 + 1.266 × 10−4(
1.2
7.5×10−3)
𝑊𝑐𝑟 = 436.3 𝐾𝑁
Design factor /nd /=
𝑾 𝒄𝒓=𝟒𝟑𝟔.𝟑 𝑲𝑵
63080
= 𝟔. 𝟓so the design of piston road is safe.
𝐁) Design of cylinder
The cylinder is used to store the fluid or oil under pressure.
 Material selection ;
Steel standard < structural ASTM a 36>

44
𝜎𝑢𝑡=400 Mpa, 𝜎𝑦=250 𝑀𝑝𝑎 , 𝜏 𝑦=145 𝑀𝑝𝑎 , 𝐸=200 𝐺𝑝𝑎, 𝐺=77 𝐺𝑝𝑎
 To find the inner diameter of the cylinder/𝑑 𝑐𝑦,𝑜=𝑎𝑝𝑝𝑙𝑦𝑝𝑎𝑠𝑐𝑎𝑙𝑠𝑙𝑎𝑤 𝑃1 =𝑃2
 R/A1 = F2/A2
𝑑 𝑐𝑦,𝑖𝑛= 𝑑1√ 𝐹
𝑅⁄ And let Fh=240 N
𝑡𝑎𝑘𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑜; -Fh*(x+L) +R(Xcos 𝜃) = 0
R=
240 ∗0.5
0.04 cos30
= 3464N
SO, 𝑑 𝑐𝑦,𝑖𝑛=0.02√
19.6∗103
3464
= 0.3989
dcy,in=40 mm
 The outer diameter of the cylinder will be;
𝑑 𝑐𝑦,𝑜𝑢 = 𝑑 𝑐𝑦,𝑖𝑛 +2𝑡
= 40+2*3
𝑑 𝑐𝑦,𝑜𝑢= 46 mm
𝑓𝑜𝑟𝑠𝑎𝑓𝑒𝑡𝑦take dcy,ou=50mm
Cheek either the cylinder is thin or thick; Pin<<1/6(𝜎𝑎𝑙𝑙 )
Pin = 19600*4/3.14*0.052
= 9.98 Mpa 9.98 Mpa<<< 1/6(250 Mpa) =41.67 Mpa so it is thin cylinder.
This thin cylinder is subjected to tensile stresses;
1. Circumferential /hoop stress
𝜎
𝑡1 =
𝑝 𝑖∗𝑑 𝑐𝑦
2𝑡
⁄
So t =
𝑝 𝑖∗𝑑 𝑐𝑦
2𝜎𝑡1
=
9.98∗106
2∗125 ∗106 × 0.05
=2.5mm
Take t≈3mm
2. Longitudinal stress/
𝜎
𝑡2=
9.98∗0.05∗106
4∗3∗10−3
𝜎𝑡2=40.5*106
pa
 According to the maximum shear stress theory, the maximum shear stress is one half of the
algebraic different of the maximum & minimum principal stresses.
𝜏
𝑚𝑎𝑥 =
𝜎 𝑡2 −𝜎 𝑡1
2
=
125 − 40.5
2
𝜏 𝑚𝑎𝑥=42.25 Mpa

45
𝑠𝑖𝑛𝑒 𝛕 𝐦𝐚𝐱≪ 𝛕 𝐲where the design is safe.
C)Design of pipe
- Pre used to transport working fluid under pressure
- use flexible material for the pipe-rubber
 material selection; Rubber (grade Amg)
𝜎𝑢𝑡=200 Mpa, 𝜎𝑢𝑐=240 𝑀𝑝𝑎 , 𝜏 𝑢𝑡=350 𝑀𝑝𝑎 ,
Assumptions; internal diameter of the pipe, d=5mm, L=500 mm
Let us assume thin cylinder; this applied when
 The stress across the pipe section is uniform
 The internal diameter of the pipe is more than twenty times its wall thickness/t d/t > 20.
𝜎𝑡=
𝑝𝑑
2𝑡
, pressure is the same in the closed direction so p=9.8 Mpa
𝑡 =
𝑝𝑑
2𝜎𝑎𝑙𝑙
𝜎 𝑎𝑙𝑙=
𝜎 𝑡
𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑡𝑦
= 20/1.5=13.33 𝑁
𝑚𝑚2⁄
𝑡 =
𝑝𝑑
2𝜎 𝑎 𝑙𝑙
=
9.8∗5
2∗13.33
𝑡 = 2 𝑚𝑚
Cheek for thin;
 D/t>20
 5/2 >20 so it is not correct.
Therefore the pipe is thick and use lame’s equation;
𝑡 𝑝 = 𝑟
𝑝(√
𝜎 𝑎𝑙𝑙 +𝑝
𝜎 𝑎𝑙𝑙 −𝑝
−1)
= 2.5(√6.55-1)
𝑡 𝑝 ≈ 4 𝑚𝑚 Thickness of the pipe
8.6. Designof base
8.6.1 Eccentric Loaded RivetedJoint
When the line of action of the load does not pass through the centroid of the rivet system and
thus all rivets are not equally loaded, then the joint is said to be an eccentric loaded riveted joint.
The eccentric loading results in secondary shear caused by the tendency of force to twist the joint
about the center of gravity in addition to direct shear or primary shear.

46
Let P = Eccentric load on the joint, and e = Eccentricity of the load i.e. the distance between the
line of action of the load and the centroid of the rivet system i.e. G.
The following procedure is adopted for the design of an eccentrically loaded riveted joint
1. First of all, find the centre of gravity G of the rivet system.
Let A = Cross-sectional area of each rivet
X1, x2, x3 etc. = Distances of rivets from OY, and
y1, y2, y3 etc. = Distances of rivets from OX.
We know that 𝑋̅=
𝐴1 ×𝑋1+𝐴2×𝑋2+𝐴3𝑋3+....
𝐴1 +𝐴2+𝐴3+⋯
=
𝐴×𝑋1+𝐴×𝑋2+𝐴𝑋3+....
𝑛𝐴
=
𝑋1+𝑋2+𝑋3+....
𝑛
(where n = Number of rivets)
Similarly, 𝑦̅ =
𝑦1+𝑦2+𝑦3
𝑛

47
Figure.8.5 Eccentric loaded riveted joint.
2. Introduce two forces P1 and P2 at the centre of gravity ‘G’ of the rivet system. These
forcesare equal and opposite to P as shown in Fig. 8.5 (b).
3. Assuming that all the rivets are of the same size, the effect of P1 = P is to produce direct
shearload on each rivet of equal magnitude. Therefore, direct shear load on each rivet,
Ps =
𝑝
𝑛
, acting parallel to the load P.
4. The effect of P2 = P is to produce a turning moment of magnitude P × e which tends to
rotatethe joint about the centre of gravity ‘G’ of the rivet system in a clockwise direction. Due to
the turningmoment, secondary shear load on each rivet is produced. In order to find the
secondary shear load,
the following two assumptions are made :
(a) The secondary shear load is proportional to the radial distance of the rivet under
consideration from the centre of gravity of the rivet system.
(b) The direction of secondary shear load is perpendicular to the line joining the centre of the
rivet to the centre of gravity of the rivet system..
Let F1, F2, F3 ... = Secondary shear loads on the rivets 1, 2, 3...etc.
l1, l2, l3 ... = Radial distance of the rivets 1, 2, 3 ...etc. from the centre of
gravity ‘G’ of the rivet system.
From assumption (a),
F1 α l1 ; F2 α l2 and so on
𝐹1
𝐿1
=
𝐹2
𝐿2
=
𝐹3
𝐿3
= ⋯
F2=
𝐿2
𝐿1
× 𝐹1
F3=
𝑙3
𝑙1
× 𝐹1
We know that the sum of the external turning moment due to the eccentric load and of internal
resisting moment of the rivets must be equal to zero.
P.e=F1*L1+F2*L2+F3*L3
= F1*L1+𝐿2 ×
𝐿2
𝐿1
× 𝐹1 + 𝐿3 ×
𝑙3
𝑙1
× 𝐹1
=
𝐹1
𝑙1
[( 𝑙1)2
+ ( 𝑙2)2
+ ( 𝑙3)2
+ ⋯ ]
From the above expression, the value of F1 may be calculated and hence F2 and F3 etc.
areknown. The direction of these forces are at right angles to the lines joining the center of rivet
to theCentre of gravity of the rivet system, as shown in Fig. 5.2 (b), and should produce the
moment in thesame direction (i.e. clockwise or anticlockwise) about the center of gravity, as the
turning moment(P × e).
5. The primary (or direct) and secondary shear load may be added vectorially to determine the
Resultant shear load (R) on each rivet as shown in Fig. 5.2 (c). It may also be obtained by using
the relation
When the secondary shear load on each rivet is equal, then the heavily loaded rivet will be one in
which the included angle between the direct shear load and secondary shear load is minimum.
The maximum loaded rivet becomes the critical one for determining the strength of the riveted

48
joint. Knowing the permissible shear stress (𝜏), the diameter of the rivet hole may be obtained by
using the relation,
Maximum resultant shear load (R) =
𝜋
4
× 𝑑2
× 𝜏,
From Table 9.7, the standard diameter of the rivet hole (d) and the rivet diameter may be
Specified, according to IS : 1929 – 1982 (Reaffirmed 1996).(GUPTA)
Table 8.1 the standard diameter of the rivet hole (d) and the rivet diameter
Assumptions
Base (b) =400mm b=L squire cross section
Length (L) =400mm
Thickens (t) =50mm;
Load on the bracket (P) =250kN =25 ×
103
N;
Eccentricity (e) =400 mm;
Number of rivet (n)=7 ;
Rivet spacing (C) =100mm
Crushing stress (σc) =120MPa
Permissible shear stress (τ) = 65MPa = 65
N/mm2;
Required Figure. 8.6 rivet system
Size of the rivets

49
Analysis
First of all, let us find the centre of gravity (G) of the rivet system.
Let 𝑥̅ = Distance of centre of gravity from OY,
𝑦̅ = Distance of centre of gravity from OX,
𝑋̅=
𝑋1+𝑋2+𝑋3+𝑋4+𝑋5+𝑋6+𝑋7
𝑛
=
100+200+200+200
7
=100mm(x1=x6 = x7 = 0)
𝑦̅=
𝑦1+𝑦2+𝑦3+𝑦4+𝑦5+𝑦6+𝑦7
𝑛
=
200 +200+200+100+10
7
= 114.7𝑚𝑚(y5 = y6 = 0)
∴ the centre of gravity (G) of the rivet system lies at a distance of 100 mm from OY and 114.3
mm from OX, as shown in Fig. 8.6.
We know that direct shear load on each rivet,
Ps =
𝑝
𝑛
=
250×103
𝑁
7
= 35,714.28𝑁
The direct shear load acts parallel to the direction of load P i.e. vertically downward
as shown in Fig. 8.5. Turning moment produced by the load P due to eccentricity (e)
P × e=250× 103
𝑁 × 400𝑚𝑚 =1000000Nmm
This turning moment is resisted by seven rivets as shown in Fig.8.6

50
Figure. 8.7 pneumatic rivet system
Let F1, F2, F3, F4, F5, F6 and F7 be the secondary shear load on the rivets 1, 2, 3, 4, 5, 6 and 7
placed at distances l1, l2, l3, l4, l5, l6 and l7 respectively from the centre of gravity of the rivet
system as shown in Fig. 8.7
From the geometry of the figure, we find that
L1=L3=√(100)2 + (200 − 114.3)2 = 131.7𝑚𝑚
L2 = 200 – 114.3 = 85.7 mm

51
L4=L7 =√(100)2 + (114.3 − 100)2 = 101𝑚𝑚
L5=L6=√(100)2 + (114.3)2 = 152𝑚𝑚
Now equating the turning moment due to eccentricity of the load to the resisting moment of the
rivets, we have
250× 103
×400=
100× 106
× 131.7 = 𝐹1 × 108645 =
131.7×108
108645
= 121220.48𝑁 = 𝐹1
Since the secondary shear loads are proportional to their radial distances from the centre of
gravity, therefore F2 =F1×
𝐿2
𝐿1
=121220.48 ×
85.7
131.7
= 78880.75𝑁
F3=F1×
𝐿3
𝐿1
=121220.48𝑁, L3=L1
F4= F1×
𝐿4
𝐿1
=121220.48𝑁 ×
101
131 .7
= 92963.3𝑁
F5=F1×
𝐿5
𝐿1
=121220.48𝑁 ×
152
131 .7
= 139905.1857𝑁
F6= F1×
𝐿6
𝐿1
= 121220.48𝑁 ×
152
131 .7
= 139905.1857𝑁
F7= F1×
𝐿7
𝐿1
= 𝐹4 = 92963.3𝑁
By drawing the direct and secondary shear loads on each rivet, we see that the rivets 3, 4 and 5
are heavily loaded. Let us now find the angles between the direct and secondary shear load for
these three rivets. From the geometry of Fig. 8.6 we find that
cosθ3=
100
𝐿3
=
100
131.7
= 0.76
cosθ4=
100
𝐿4
=
100
101
= 0.99
cosθ5=
100
𝐿5
=
100
152
= 0.6578

52
Now resultant shear load on rivet 3,
R3=√(35,714.28)2 + (121220.48)2 + 2 × 35,714.28 × 12120.48 ∗ 0.76 = 150168.0853𝑁
R4=√( 𝑃𝑠)2 + ( 𝐹4)2 + 2 × 𝑃𝑠 × 𝐹4cosθ4
R4=√(35,714.28)2 + (92963.3)2 + 2 × 35,714.28 × 92963.3 × 0.99 =1284119.3025
R5=√( 𝑃𝑠)2 + ( 𝐹5)2 + 2 × 𝑃𝑠 × 𝐹5cosθ5
R5=√(35,714.28)2 + (139905.1857)2 + 2 × 35,714.28 × 139905.1857× 0.6578=165597.5
The resultant shear load may be determined graphically, as shown in Fig. 8.6. From above we
see that the maximum resultant shear load is on rivet 5. If d is the diameter of rivet hole, then
maximum resultant shear load (R5),
R5=
𝜋
4
𝑑2
× 𝜏 = R5=
𝜋
4
𝑑2
× 65
165597.5=
𝜋
4
𝑑2
× 65,165597.5× 4=𝜋𝑑2
662390
65𝜋
= 𝑑2
= 3243.77 = √3243.77=56.95mm=d
8.7 DESIGN OF HAND ARM SUPPORT
ASSUMPTION
Height (h) =200mm h
Width (b) =250mm b
8.8 Designof Welding
A welded joint is a permanent joint which is obtained by the fusion of the edges of the two parts
to be joined together, with or without the application of pressure and a filler material.
The welding processes may be broadly classified into the following two groups:

53
1. Welding processes that use heat alone e.g. fusion welding.2. Welding processes that use a
combination of heat and pressure e.g. forge welding.
The fusion welding, according to the method of heat generated, may be classified as:
1. Thermit welding,
2. Gas welding and
3. Electric arc welding.
In electric arc welding, the filler metal is supplied by metal welding electrode.
Therefore, this type of welding is going to be used in this design.
Given parameters and assumptions
Figure 8.8.Welded joint of frame
Eccentric load P =
wL
2
=
250N
m⁄ ∗4m
2
= 500N
 Length of eccentricity e = 600mm
 Electrode material is made of bare electrode
 (σy )weld=0.8σy = 0.8 × 220 = 176Mpa due to decrease in strength of the material
during welding and the type of electrode we use somewhat is multiplied by a factor
 Fillet type of weld is used
 b = 60mm = l

54
When the parts are subjected to fatigue loading, the stress concentration factor as given in the
following table should be taken into account.
Table8.2.Stress concentration factor for welded joints
Type of joint Stress concentration factor
1. Reinforced butt welds 1.2
2. Toe of transverse fillet weld 1.5
3. End of parallel fillet weld 2.7
4. T-butt joint with sharp corner 2.0
Stress concentration factor can be taken 2.7
Required
 Size of the weld, s
Analysis
The weld is supposed to join parts of the frame and other components in the machine. The part
subjected to higher load governs the design.
 The joint will be subjected to direct shear stress due to shear force (P)
 The joint will be subjected to bending stress due to a bending moment
The throat area for square fillet weld,
A = t(2b+ 2l) = 0.707s(2b+ 2b)
= 0.707s(2∗ 60 + 2 ∗ 60) = 170smm2
Then the direct shear stress will be,
τ =
P
A
=
500N
170s
=
2.94117
s
N
mm2⁄
Also the bending moment will be,

55
M = P ∗ e = 500N ∗ 600 = 0.3 ∗ 106
Nmm
The section of modulus for square section will be,
Z = t (b2
+
b2
3
) =
2.828sb2
3
Z =
2.828 ∗ s ∗ 602
3
= 3393.6smm3
Then bending moment will be,
σb =
M
Z
=
0.3 ∗ 106
Nmm
3393.6smm3
=
88.4
s
N
mm2⁄
Table8.3.Stress for weld joints
Type of weld Stress(MPa)
1 Fillet welds 80
2 Butts welds
Tension 90
Compression 100
Shear 55
Permissible shear stress =
stress
stress concentration factor
=
80
2.7
= 29.63Mpa
From maximum shear stress theory we have,
τmax =
1
2
√(σb)2 + 4(τ)2
29.63Mpa =
1
2
√(
88.4
s
N
mm2⁄ )
2
+ 4 (
2.94117
s
N
mm2⁄ )
2

56
s = 1.49mm
8.9 DESGHNE OF HAND LEVER
Material Selection
Mild steel is general-purpose steel bars for machining, suitable for lightly stressed components
including studs, bolts, gears, levers, handles and shafts.
Mechanical Properties g
Allowable tensile stressσt = 100Mpa a f
Yield strength σy = 220Mpa R=5
Allowable shear stress = 55 MPa b e
Factor of safety =5 d
Modulus of elasticity E =210Gpa c
Given Parameters and AssumptionsFigure 8.9 hand lever
Width of the lever (c) =400mm
Effort (p) = 200 N
Height of the lever (d) = 100mm
Height of the lever (b) =400mm
Height of the lever (e) =250mm
Weight (w) =250N
Thickens=100mm
Length of the lever is 500 mm, (a)
Analysis
First let’s determine the weight applied on it
𝑊 = 𝑚𝑔 = 25𝑘𝑔 ∗ 10 𝑚
𝑠2⁄ = 250𝑁
Then let’s determine the allowable stress and allowable shear stress of the material

57
𝜎𝑎𝑙𝑙 =
𝜎𝑦
𝐹. 𝑆
𝜎𝑎𝑙𝑙 =
220
5
𝜎𝑎𝑙𝑙 = 44𝑀𝑝𝑎
𝜏 𝑎𝑙𝑙𝑜𝑤 =
𝜎𝑦
2𝐹. 𝑆
𝜏 𝑎𝑙𝑙𝑜𝑤 =
220
2 ∗ 5
𝜏 𝑎𝑙𝑙𝑜𝑤 = 22𝑀𝑝𝑎
8.9.1 Principal stresses
Maximum principal stress is as follows:
𝜎𝑐𝑚𝑎𝑥 =
1
2
⌈𝜎𝑐 + (√ 𝜎𝑐2 + 4𝜏2⌉ =
1
2
⌈44 + (√442 + 4 ∗ 222⌉ = 53.11
𝑁
𝑀𝑀2
Design value of 𝜎𝑐 =
𝜎𝑎𝑙𝑙
𝑓.𝑠
=
700
𝑁
𝑚𝑚2
5
= 140
𝑁
𝑚𝑚2
𝜎𝑐𝑚𝑎𝑥 ≤ 𝜎𝑐, 53.11
𝑁
𝑀𝑀2
≤ 140
𝑁
𝑚𝑚2
, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑎𝑓𝑒
8.9.2 Maximum shear stresses is as follows
𝜏𝑚𝑎𝑥 =
1
2
(√ 𝜎𝑐2 + 4𝜏2) =
1
2
(√442 + 4 × 222) = 31.11
𝑁
𝑚𝑚2
Design value of 𝜏 =
𝜏𝑦
𝑓.𝑠
=
450
𝑁
𝑚𝑚2
5
= 90
𝑁
𝑀𝑀2
𝜏𝑐𝑚𝑎𝑥 ≤ 𝜏𝑐, 31.11
𝑁
𝑀𝑀2
≤ 90
𝑁
𝑚𝑚2
, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑎𝑓𝑒

58
8.10. Result and Discussion
This pneumatic robot machine for manual labors improves the speed of mixing of the chemicals
with steam. In addition, this system reduces the human effort. The workers are eliminated and
the ultimately the operation cost is reduced and profit is increased. This system is beneficial and
safety for the labors. By observing the cause of failure to the different parts of the hand arm like
shaft, column and other parts
Table8.4. Result and discussion
Capacity of the pneumatic robot 0.047kg/sec
Power supply 850W
Mechanical advantage of the machine(M.A) 1.25
diameter of rivet hole 56.96mm
Size of welding 1.49mm
velocity of the machine 1km/hr
Maximum lifting height 950mm

59
CHAPTER NIGN
CONCLUSION AND RECOMMENDATION
9.1. Conclusions
The focus of this project was to design a costing recovery rebooting machine under certain
constraints such as compact size, easy to operate, low cost etc. A model has been created
within prescribed boundaries to meet the demands of the factors and to make more
flexible operation. Based on the models and analyses presented in the previous sections,
various conclusions can be drawn. A model with design capability 0.047 kg/sec, use 850 watt
(0.85kw) power has been developed. Also a model, which is easy to operate and manipulate, has
been developed which can be used for efficient lifting of chemicals. Small-scale factors for
increasing their profits by decreasing labor and machinery cost can use the model so constructed.
The model can be carried out with affordable capacity of most of Ethiopians. And also
a better quality product can be produced. That in turn will save a huge amount of foreign
currency which was paid to import these huge machineries and able to provide the low factors to
use the machine in affordable cost. This model will offer a better and inexpensive method to
reduce man power and efforts involved in manufacturing industries .

60
9.2. Recommendation
The design capability of the costic recovery robot machine 0.047 kg/sec and the speed of the
machine must not exceed above measure therefore, we strongly recommend that the pneumatic
robot machine must not loaded above the capacity and the pneumatic robot machine is only left
one packed 25kg of chemical therefore design this to left 40 number of packs at a one time.
we also recommend that Different parts of the pneumatic robot like the connection between nut
and, between holder (bearing) and shaft, and other parts must be properly lubricated and handled
carefully.

61
REFERENCES
1 A text book of machine design 14th edition, R.S Khurmi and JK Gupta, Eurasia
publishing house (pvt. ltd) 2005.
2 Shigley's Mechanical Engineering Design (9th edition, Budynas, Nisbett).
3 Ashby, M. F., 2005. Material Selection in Mechanical Design. 3rd ed. New York:
Pergamon Press.
4 . Bhandari, V. B., 2010. Design of Machine Elements. Third Edition ed. New Delhi: Tata
McGraw-Hill Education.
5 . Gupta, R. K. &. J., 2005. Theory of Machines. Revised Edition ed. Punjab, India: S.
Chand and Company
6 Kern, process heat transfer. Mc Graw-Hill book company.

62
APPANDEXES
Materials
British
standards
Production
process
Maximum
section
size,
mm
Yield
Strength
MPa
Tensile
Strength,
MPa
Elongation
%
Hardness
number,
HB
0.20C 070M20 HR 152 215 430 22
126 –
179
254 200 400 20
116 –
170
CD 13 385 530 12 154
76 340 430 14 125
0.30C 080M30 HR 152 245 490 20
143 –
192
254 230 460 19
134 –
183
CD 13 470 600 10 174
63 385 530 12 154
H&T 63 385 550 - 700 13
152 –
207
0.40C 080M40 HR 150 280 550 16
152 –
207
CD 63 430 570 10 165
H&T 63 385 625 - 775 16
179 –
229
0.50C 080M50 HR 150 310 620 14
179 –
229
CD 63 510 650 10
202 –
255
H&T 150 430 625 – 775 11
248 –
302

63
1Cr 530M40 H&T 100 525
700 –
850
17
202 –
255
29 680 850 - 1000 13
248 –
302
1.5MnMo 605M36 H&T 150 525
700 –
850
17
202 –
255
29 755 925 - 1075 12
269 –
331
1.25NiCr 640M40 H&T 152 525
700 –
850
17
202 –
255
102 585 770 – 930 15
223 –
277
64 680 850 - 1000 13
248 –
302
29 755 930 - 1080 12
269 –
331
3NiCr 653M31 H&T 64 755
930 -
1080
12
269 –
331
- 680 850 – 000 12
248 –
302
8.3 Appendix B: Mechanical Properties of Steels (Nyangasi, 18 December, 2006)
Key: HR - Hot- Hot rolled and normalized
CD - Cold drawn
H&T - Hardened and tempered

64
Type of joint Stress concentration factor
1. Reinforced butt welds 1.2
2. Toe of transverse fillet weld 1.5
3. End of parallel fillet weld 2.7
4. T-butt joint with sharp corner 2.0
Table8.2.Stressconcentration factorfor welded joint
Type of weld Stress(MPa)
1 Fillet welds 80
2 Butts welds 80
3
4
5
Tension 90
Compression 100
Shear 55
Table8.3.Stressforweld joints

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