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June 14 Gondar, Ethiopia

EXAMINER’S APRROVAL PAGE UNIVERSITY OF GONDAR INSTITUTE OF TECHNOLOGY SCHOOL OF MECHANICAL ENGINEERING We certify that the project entitled “Design and Modeling of Manually Hydraulic Hand Lift Jib Crane Machine” is written by ______________________ and ____________________________. We have examined the final copy of this project and in our opinion; it is fully adequate in terms of scope and design for the award of the degree of Bachelor of Engineering. We here with recommend that it be accepted in partial fulfillment of the requirements for the degree of Bachelor of Mechanical Engineering. APRROVED BY: 1. EXTERNAL EXAMINER

SIGNATURE

INTERNAL EXAMINER

SIGNATURE

DATE

2. DATE

3. ADVISOR

SIGNATURE

DATE

4. CHAIRPERSON

SIGNATURE

i

DATE

ACKNOWLEDGEMENT First of all, we would like to thank the almighty God for blessing us with strength, aptitude and patience for successfully completing our project. Our special appreciation goes to our advisor Mr. Ayele Hailu (Msc.) for giving us the opportunity to work with him during this project. We have been able to compile and complete this project report on comprehensive manner due to guidance, support and counseling that he has provided to us. Finally yet importantly, our sincere thanks go to each and every one who has helped and supported us significantly indifferent stages during this project time.

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ABSTRACT Jib crane is a device used for loading and unloading heavy material from trucks in the open yard and carry them inside the shop. We have seen great challenge at the time of transportation of heavy machine parts and equipment within and outside the workshop has been a source of concern and needs urgent attention because of the hazard it exhibits. This negative effect on the health of the workers which drives us to design manually hydraulic hand lift jib crane machine. This project is mainly focused on design and modeling of manually hydraulic hand lift jib crane machine in garages, micro and small enterprises in Ethiopia. This crane is suitable for loading and unloading heavy material workshops with a maximum load of 3 ton (3000 kg), maximum lifting height of 2500mm and minimum lifting height of 600mm. Beside these advantages in most garages, micro and small enterprises in Ethiopia, this machine is not present. Instead of this they use fork lift, overhead crane and others. So main objects of the project is to make them to use this machine for accelerate and make easily their work activity. To achieve these purposes, we distributed questionnaires, interviews, observation and also design selection has been taken. Then after the responses are tabulated and analyzed using qualitative and quantitative method of data analysis. The main findings of the project are all most all the respondents were interested to have this machine in their area and most of the respondent approved that this machine could help for the employee’s safety and to accelerate and make easily the work activity in the area. This device is very efficient with its high safety, reduces labor force, and reduces working time and workers fatigue.

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TABLE OF CONTENT Content

page

ACKNOWLEDGEMENT .............................................................................................................. ii ABSTRACT ................................................................................................................................... iii TABLE OF CONTENT ................................................................................................................. iv LIST OF FIGURES ..................................................................................................................... viii LIST OF TABLES .......................................................................................................................... x NOMENCLATURE AND ABBREVIATION .............................................................................. xi CHAPTER ONE ............................................................................................................................. 1 1.

INTRODUCTION ................................................................................................................... 1 1.1.

Background of the Study ................................................................................................ 1

1.2.

Statement of the Problem ................................................................................................ 2

1.3.

Objective of the Project .................................................................................................. 3

1.3.1.

General Objective: .................................................................................................... 3

1.3.2.

Specific Objective ..................................................................................................... 3

1.4.

Significance of the Project .............................................................................................. 3

1.5.

Scope of the Project ........................................................................................................ 3

1.6.

Limitation of the Project ................................................................................................. 4

1.7.

Assumptions.................................................................................................................... 4

CHAPTER TWO ............................................................................................................................ 5 2.

LITERATURE REVIEW ........................................................................................................ 5 2.1.

Introduction to Design and Development of Product ..................................................... 5

2.1.1.

Application of Cranes ............................................................................................... 7

2.1.2.

Fluid Power Systems................................................................................................. 8

2.2.

Current Theories and Design Done On Jib Crane .......................................................... 9

iv

2.3.

Importance of Our Design in Engineering Applications .............................................. 14

2.4.

Design Approach With Regard to Our Project ............................................................. 15

2.5.

Available Standards Pertinent to Our Design ............................................................... 15

2.6.

REASON FOR SELECTING OUR TOPIC ................................................................. 15

CHAPTER THREE ...................................................................................................................... 16 3.

METHODOLOGY OF THE PROJECT ............................................................................... 16 3.1.

Data Collection Method ................................................................................................ 17

3.2.

Comparison of Previous Work Related to Our Project List in the Literature Review . 18

3.2.1.

Design Matrix ......................................................................................................... 29

3.2.2.

Summary of Literature Review ............................................................................... 31

3.3.

Selection of Materials for Engineering Purposes .......................................................... 31

3.4. 3.5.

Design Part and Stress Analysis ................................................................................. 32 Draw Part and Assembly Drawing by using Solid Work Software and Check the

Deformation by using ANSYS Software. ................................................................................. 32 3.6.

Result and Discussion ................................................................................................... 32

3.7.

Cost Analysis of our Design ......................................................................................... 32

3.8.

Recommendation and Conclusion ................................................................................ 32

CHAPTER FOUR ......................................................................................................................... 33 4.

DESIGN ANALYSIS ............................................................................................................ 33

4.1.

Data Gathering .................................................................................................................. 33 4.1.1.

Interview ................................................................................................................. 33

4.1.2.

Questionnaires......................................................................................................... 34

4.1.3.

Observation ............................................................................................................. 34

4.2.

Selection of Hydraulic Bottle Jack ............................................................................... 36

4.2.1.

Oil Type .................................................................................................................. 37

v

4.3.

Design Specification ..................................................................................................... 37

4.3.1.

Geometrical Analysis of the Machine..................................................................... 37

4.3.2.

Parameters of the Cylinder from Standard Table ................................................... 38

4.4.

Design of Components of Machines ............................................................................. 43

4.4.1.

Design of Horizontal Boom .................................................................................... 43

4.4.2.

Design of Back Hoist Body .................................................................................... 49

4.4.3.

Design of the Horizontal Frame and Base Body .................................................... 60

4.4.4.

Design of Base Body .............................................................................................. 62

4.4.4.6.

Deflection of Base Body ..................................................................................... 69

4.4.5.

Design of Back Hoist Support ................................................................................ 70

4.4.6.

Design of Horizontal Frame.................................................................................... 76

4.4.7.

Design of Connecting Frame .................................................................................. 82

4.4.8.

Design of Hook ....................................................................................................... 88

4.4.9.

Design of Bolt and Nut ........................................................................................... 93

4.4.10.

Wheel Selection................................................................................................. 104

4.4.11.

Welding ............................................................................................................. 106

4.4.12.

Manufacturing Process ...................................................................................... 109

CHATER FIVE ........................................................................................................................... 112 5.

RESULT AND DISCUSSION ............................................................................................ 112 5.1.

Results ......................................................................................................................... 112

5.2.

ANSYS results for horizontal boom [input from appendix table 0:8.A] .................... 114

5.3.

ANSYS results for back hoist [input data from appendix table 0:7.A] ...................... 115

5.4.

ANSYS results for hook [input data from appendix table 0:8.A]............................... 116

5.5.

Comparisons of ANSYS and manual result................................................................ 117

5.6.

Discussion ................................................................................................................... 120

vi

CHAPTER SIX ........................................................................................................................... 121 6.

COST ANALYSIS .............................................................................................................. 121 6.1.

Material Cost for Manual Hydraulic Hand Lift Jib Crane .......................................... 121

6.2.

Manufacturing Cost and Labor Cost ........................................................................... 123

6.3.

Purchase Cost; ............................................................................................................. 123

6.4.

Total Cost for Manually Hydraulic Hand Lift Jib Crane Machine ............................. 124

CHAPTER SEVEN .................................................................................................................... 125 7.

CONCLUSION AND RECOMMENDATION .................................................................. 125 7.1.

Conclusion .................................................................................................................. 125

7.2.

Recommendation ........................................................................................................ 125

REFERENCE .............................................................................................................................. 126 DECLARATION ........................................................................................................................ 128 APPENDICES ............................................................................................................................ 129 PART AND ASSEMBLY DRAWING ...................................................................................... 143

vii

LIST OF FIGURES FIGURE 2:1.HEAVY DUTY FOLDING WORKSHOP CRANE .................................................................. 10 FIGURE 2:2.HEAVY DUTY FOLDING WORKSHOP CRANE .................................................................. 10 FIGURE 2:3.MANUAL MOBILE MATERIAL HANDLING MACHINE ...................................................... 11 FIGURE 2:4.MOBILE JIB CRANE ...................................................................................................... 12 FIGURE 2:5.ROTATING FLOOR CRANE MACHINE ............................................................................. 12 FIGURE 2:6.CANTILEVER BEAM OF JIB CRANE. ............................................................................... 13 FIGURE 2:7.SIDE VIEW OF MOBILE FLOOR CRANE ........................................................................... 14 FIGURE 3:1.HEAVY DUTY FOLDING WORKSHOP CRANE .................................................................. 18 FIGURE 3:2.HYDRAULIC BOTTLE JACK ........................................................................................... 19 FIGURE 3:3.HEAVY DUTY FOLDING WORKSHOP CRANE .................................................................. 20 FIGURE 3:4.MANUAL MOBILE MATERIAL HANDLING MACHINE ...................................................... 21 FIGURE 3:5.MOBILE JIB CRANE ...................................................................................................... 22 FIGURE 3:6.ROTATING FLOOR CRANE MACHINE ............................................................................. 24 FIGURE 3:7.WORKING PRINCIPLE OF HYDRAULIC JACK .................................................................. 25 FIGURE 3:8.CANTILEVER BEAM OF JIB CRANE. ............................................................................... 26 FIGURE 3:9.SIDE VIEW OF GENERAL ASSEMBLY ............................................................................. 27 FIGURE 4.1:1.INTERVIEWING FROM DIFFERENT COMPANY ............................................................. 33 FIGURE 4.1:2.OBSERVATION OF MANUALLY HYDRAULIC HAND LIFT FIXED JIB CRANE FROM JOVANI GARAGE ................................................................................................................................. 34

FIGURE 4.1:3.OBSERVATION OF BARANCO FROM JOVANI GARAGES ............................................... 35 FIGURE 4.1:4.3D VIEW OF THE MACHINE ........................................................................................ 37 FIGURE 4.1:5.FRONT VIEW OF OUR MACHINE ................................................................................. 38 FIGURE 4.1:6.WHEN THE CYLINDER IS RETRACTED AND EXTRACTED TRIANGLE............................. 39 FIGURE 4.1:7.WHEN THE CYLINDER IS RETRACTED ........................................................................ 39 FIGURE 4.1:8.WHEN THE CYLINDER IS AT NORMAL POSITION ........................................................ 40 FIGURE 4.1:9.WHEN THE CYLINDER IS EXTRACTED ....................................................................... 41 FIGURE 4.1:10.TRIANGLE OF BACK HOIST BODY WITH BASE BODY (∆ACO) .................................. 41 FIGURE 4.1:11.TRIANGLE OF BACK HOIST BODY WITH HORIZONTAL FRAME AND BACK HOIST SUPPORT (∆AHO) ................................................................................................................... 42

viii

FIGURE 4.1:12.FBD OF BOOM AT NORMAL POSITION ..................................................................... 42 FIGURE 4.1:13. 3D VIEWS OF HORIZONTAL BOOM........................................................................... 44 FIGURE 4.1:14.FBD OF BOOM WHEN THE CYLINDER AT EXTRACTED POSITION .............................. 44 FIGURE 4.1:15.3D VIEWS OF BACK HOIST BODY ............................................................................. 50 FIGURE 4.1:16.FBD OF THE HOIST BODY ....................................................................................... 51 FIGURE 4.1:17.FRONT VIEW OF MACHINE ....................................................................................... 60 FIGURE 4.1:18.TOP VIEW OF MACHINE ........................................................................................... 61 FIGURE 4.1:19.3D VIEW OF BASE BODY .......................................................................................... 62 FIGURE 4.1:20.FBD OF BASE BODY ............................................................................................... 63 FIGURE 4.1:21.3D VIEW OF BACK HOIST SUPPORT .......................................................................... 70 FIGURE 4.1:22.FBD OF BACK HOIST SUPPORT ................................................................................ 71 FIGURE 4.1:23.SIDE VIEW OF BACK HOIST SUPPORT ....................................................................... 71 FIGURE 4.1:24.3D VIEW OF HORIZONTAL FRAME............................................................................ 77 FIGURE 4.1:25.FBD OF HORIZONTAL FRAME ................................................................................. 77 FIGURE 4.1:26.3D VIEW OF CONNECTING FRAME ........................................................................... 83 FIGURE 4.1:27.FBD OF BASE BODY ............................................................................................... 83 FIGURE 4.1:28.3D VIEW OF HOOK................................................................................................... 88 FIGURE 4.1:29.FBD HOOK ............................................................................................................. 90 FIGURE 4.1:30.3D VIEW OF BOLT 2 ................................................................................................. 93 FIGURE 4.1:31.3D VIEW OF NUT 1 .................................................................................................. 93 FIGURE 4.1:32.HEXAGONAL HEADED BOLT WITH A NUT AND WASHER IN POSITION [14] ............... 98 FIGURE 4.1:33.3D VIEW OF FRONT WHEEL ................................................................................... 104 FIGURE 4.1:34.3D VIEW OF REAR CASTER WHEEL ........................................................................ 105 FIGURE 4.1:35.THE CROSS-SECTION OF A WELDED BUTT JOINT .................................................... 109

ix

LIST OF TABLES TABLE 3:1.COMPARISON CRITERIA FOR EACH MODEL .................................................................... 29 TABLE 4.1:1.INTERPRETATION OF INTERVIEW AND DISTRIBUTED QUESTIONNAIRES ...................... 35 TABLE 4.1:2.STANDARD TABLE OF THE RELATION BETWEEN EQUIVALENT LENGTH (L) AND ACTUAL LENGTH (L) ............................................................................................................................ 57

TABLE 4.1:3.STANDARD VALUE OF “C” ......................................................................................... 58 TABLE 4.1:4.VALUE OF CRUSHING STRESS (ΣC) AND RANKINE’S CONSTANT (A)............................ 59 TABLE 4.1:5: STANDARD DIMENSIONS OF CRANE HOOK [13] ....................................................... 89 TABLE 4.1:6.STANDARD TABLE OF FRONT WHEEL ........................................................................ 105 TABLE 4.1:7.STANDARD TABLE OF REAR WHEEL .......................................................................... 106 TABLE 5:1.RESULT OF ALL PARTS ................................................................................................ 113 TABLE 6:1.MATERIAL COST OF EACH COMPONENT ....................................................................... 121 TABLE 6:2.MANUFACTURING & LABOR COST OF EACH MATERIAL ............................................... 123 TABLE 6:3.PURCHASE COST ......................................................................................................... 124 TABLE 0:1.A. TYPICAL MECHANICAL AND PHYSICAL PROPERTIES FOR ENGINEERING METALS ..... 129 TABLE 0:2.A. STANDARD TABLE OF HYDRAULIC CYLINDER ......................................................... 130 TABLE 0:3.A. DIMENSIONS AND CROSS-SECTIONAL PROPERTIES SQUARE HOLLOW SECTIONS ..... 131 TABLE 0:4.A. SPECIFICATION FOR STEELED IN MILLIMETER SERIES SCREWS AND BOLTS .............. 133 TABLE 0:5.A. BASIC DIMENSIONS FOR SQUARE THREADS IN MM (NORMAL SERIES) ACCORDING TO IS: 4694 – 1968 (REAFFIRMED 1996)................................................................................... 133 TABLE 0:6.A. INPUT DATA FOR ANSYS RESULTS OF HORIZONTAL BOOM .................................... 135 TABLE 0:7.A. INPUT DATA FOR ANSYS RESULTS OF BACK HOIST BODY ....................................... 137 TABLE 0:8.A. INPUT DATA FOR ANSYS RESULTS OF BACK HOIST BODY ....................................... 140

x

NOMENCLATURE AND ABBREVIATION Symbol

Full name

Unit

Le

Cylinder extracted length

mm

Lr

Cylinder retracted length

mm

θ1

Cylinder extracted angle

°

θ2

Cylinder retracted angle

°

kg

Kilogram

kg

UK

United kingdom

-

Max

Maximum

-

Min

Minimum

-

TÜV

Technical Control Board

-

mm

Millimeter

mm

g

Gravity

m/s 2

kN

Kilo newton

kN

Lb

boom length

mm

FBD

Free body diagram

-

@

At

-

W

Weight

kN

Ap

Area of the piston

mm2

D

Bore diameter

mm

xi

Fp

Force acting on the piston

kN

P

Working pressure of the piston.

MPa

E

Young’s modulus of the elasticity

GN/m2

G

Shear modules

MN/m2

σt

Tensile strength

MN/m2

σy

Yield strength

MN/m2

ρ

Density

kg/m3

3d

Three dimensional view

-

I

Moment of inertia

kNmm

M

bending Moment

kNmm

m

mass

Kg

σb

bending stress,

MN/m2

F. s

Factor of safety

-

σd

Direct stress

N/m2

τ

Shear stress

Mpa

σp

principal stress

Mpa

ymax

Maximum deflection

Mm

Vmax

Maximum shear stress

kN

l

effective length of the back hoist body

mm

xii

K

The least radius of gyration of the section

mm

Wcr

Crippling load by Rankine’s formula

kN

Wc

Ultimate crushing load for column

kN

WE

Crippling load obtain by Ewer’s formula

kN

σc

Crushing yield stress

Mpa

a

Ranking constant

-

A

cross section area of the column

mm2

Ltb

total length of the base body

mm

LT

The total length of the horizontal frame

mm

b

Base of the section

mm

h

Height of the section

mm

t

Thickness of the section

mm

xiii

DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

CHAPTER ONE 1. INTRODUCTION 1.1.

Background of the Study

Jib crane is suitable for loading and unloading heavy material from trucks in the open yard and carry them inside the shop. Established in 1869, The David Round Company is one of the oldest and most experienced hoist manufacturers in the world. The David Round Company specializes in the production of standard handling products such as chain hoists, jib cranes, winches, and tractor drives. We offer cuttingedge efficiency solutions like engineered wire rope hoists, stainless steel chain hoists, and jib crane motorization kits [1]. Manual hydraulic hand lift mobile floor crane with extendable boom in 1, 2 and 3ton capacities and vertical height reaching up to 4 meters. These are also provided base in the rear for putting counter weight, in case the boom is extended beyond the base wheels. These are extensively used in engineering workshops for loading and unloading jobs on the machine tools where there is no provision of an overhead crane or it is busy else-where in repair garages these are required for handling engines and its parts. These are equipped with high precision hydraulic cylinder and hard chrome plated Ram which provides upward thrust to the boom while lifting. The lowering is effected by feather touch foot pedal or wheel valve. Lowering is hydraulically cushioned to avoid jerks. The base frame can be wide enough to take the load between the two outstanding legs. All the 4 number wheels are two ball bearings each. A single handle will pull, push and steer the wheels as well as operate the pump for lifting which is a great advantage in our design. The jib crane is also available in electro hydraulic lifting and battery hydraulic lifting version in case of heave loads or more frequency of operations lifting done by a hand lever mounted on power pack place on the front of jib crane [2]. The crane for lifting heavy loads was invented by the Ancient Greeks in the late 6th century BC. The archaeological record shows that no later than c.515 BC distinctive cuttings for both lifting tongs and Lewis irons begin to appear on stone blocks of Greek temples. Since these holes point at the use of a lifting device, and since they are to be found either above the center of gravity of UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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THESIS PROJECT

DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

the block, or in pairs equidistant from a point over the center of gravity, they are regarded by archaeologists as the positive evidence required for the existence of the crane [3]. The introduction of winch and pulley hoist then leads to a widespread replacement of ramps as the main means of vertical motion. For the next two hundred years, Greek building sites witnessed a sharp drop in the weights handled, as the new lifting technique made the use of several smaller stones more practical than of fewer larger ones. In contrast to the archaic period with its tendency to ever increasing block sizes, Greek temples of the classical age like the Parthenon invariably featured stone blocks weighing less than 15-20 metric tons. Also, the practice of erecting large monolithic columns was practically abandoned in favor of using several column drums [4]. Although the exact circumstances of the shift from the ramp to the crane technology remain unclear, it has been argued that the volatile social and political conditions of Greece were more suitable to the employment of small, professional construction teams than of large bodies of unskilled labor, making the crane more preferable to the Greek polis than the more labor-intensive ramp which had been the norm in the autocratic societies of Egypt or Assyria [5]. The first unequivocal literary evidence for the existence of the compound pulley system appears in the mechanical problems attributed to Aristotle (384–322 BC), but perhaps composed at a slightly later date. Around the same time, block sizes at Greek temples began to match their archaic predecessors again, indicating that the more sophisticated compound pulley must have found its way to Greek construction sites by then [5].

1.2.

Statement of the Problem

Since manufacturing company are needed to create methods that can be accelerating their work activity which reflect to get better production and life efficiency. Sound professional best practice aimed at workshops injury and accident prevention and reduction ensures the utilization of appropriate tools for any job. As we observed in UOG work shop, the garages, and micro and small enterprises in Ethiopia, Motor vehicle repair and maintenance often require the lifting of the entire vehicle or sub-assembly part of it or the lifting up and down of its heavy components. Also in other industries and welding shops, there are items and repairs which need the employment of shop cranes. The lack of shop crane utilization in our industries and UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

automobile repair shops not only leads to injury and accident but also to poor repair and maintenance occurrence. They are practically use machines like overhead crane, forklift, and other kind of cranes. Even though, those machines are working on rigid condition (i.e. fixed or not having a roller) and the structure we have seen in a company’s are exposed to stress and fracture. So this was the reason, which drives as to design of manually hydraulic hand lift jib crane machine.

1.3.

Objective of the Project

1.3.1. General Objective: The general objective of this project is to design and modelling of manually hydraulic hand lift jib crane machine with a maximum lifting capacity up to 3 tons, minimum and maximum height of 600mm and 2500mm respectively. 1.3.2. Specific Objective:  Design the components of manual hydraulic hand lift jib crane machine.  Selection of proper material for our manufacturing process.  To save time and reduce cost of material handling equipment.  To minimize human load carrying risk.  Select standard parts such as hook, wheel and hydraulic bottle jack.  Check the deformation by using ANSYS software.

1.4.

Significance of the Project

Manually hydraulic hand lift Jib crane machine is a manual mobile material handling machine suitable for indoor and outdoor services and it can be handle by the single person. Jib crane, a free standing or portable jib crane is an economical solution for moving materials on working and/or for use an auxiliary lifting device under an overhead crane. In addition to this, this machine is mechanical in nature and does not pollute the environment. Finally, it is a good opportunity for small and micro enterprises to take advantage by utilizing this machine because one of the current developmental strategies in Ethiopia is to support these enterprises.

1.5.

Scope of the Project

The project is conducted on the title “design and modelling of manual hydraulic hand lift jib crane machine” and mainly focusing on small and micro enterprise and garages in Ethiopia.

UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

This project is started with literature review on the product specification in order to satisfy the project objectives. After obtaining the product specification, this project is done based on the scope listed below;  Design of jib crane includes structural design, mechanical design and electrical equipment design part. But in our case, we focused on mechanical design and structural design of jib crane, such as material selection, horizontal boom design, back hoist body design, bolt and nut design, etc.  Completing the project with specified time.  The machine uses in labeled land and friction less surface.

1.6.

Limitation of the Project 

Shortage of time to limit the work.



The data’s and observations are taken from few companies.



We observe problem of the machine only in UOG work shop and Gondar town not in other town of Ethiopia.



1.7.

Internet connection problem to limit the work.

Assumptions

We are gather different information from different company through interviewing to be maximum and minimum weight of the material is 3000kg and 200kg respectively for moving, loading and unloading purpose. So our design analysis depends on by taking maximum weight of material to be lifted. By reading different journals and researchers of literature review (maximum height = 2445 mm minimum height = Floor level)

we take maximum lifted height to be 2500mm in our design

analysis and from standard table of cylinder for three ton read from the appendix table Maximum height extracted position (Le ) =1176.274 mm, (𝜃1 = 65°), the maximum lifting height is 2500mm and Minimum height retract position (Lr ) =668.274 mm (𝜃2 = 50°) and the minimum lowering height is 600mm. Because of the extracted position is greater than retracted position ( 𝜃1 < 𝜃2 ). And also the length of the boom is 1250mm.

UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

CHAPTER TWO 2. LITERATURE REVIEW Crane in ancient times applied during the Roman Empire, when construction activity soared and buildings reached enormous dimensions. The Romans adopted the Greek crane and developed it further. There are also two surviving reliefs of Roman tread wheel cranes, with the Haterii tombstone from the late first century AD being particularly detailed. The simplest Roman crane, the trespasses, consisted of a single beam jib, a winch, a rope, and a block containing three pulleys.

2.1.

Introduction to Design and Development of Product

Product design and development is the set of activity beginning with the perception of market opportunity and ending in the product. It is an interdisciplinary activity requiring contribution from nearly all the function of a firm however three function are almost always central to product and design development project those are, marketing, design and manufacturing (Hayes, Robert, Steven C., New York, 1988). A generic development process is a sequence of steps that transforms asset of inputs into asset of out puts and it is the sequence of steps or activity which an enterprise employs to conceive, design, and commercialize product. Therefore it is useful because of the following reason these are quality assurance, coordination, planning, management and improvement. In the product design and development there are six phases. These phases are as follows (Cooper, Robert’s, MA, 2001): Phase 0.plannig: this phase is begins with corporate strategy and includes assessment of technology developments and market objectives. Phase1.concept development: it adscription of the form, function, and feature of product and is usually accompanied by asset of specification, an analysis of competitive product, and an economic justification of the product. Phase 2.system level design: the output of this phase is geometry layout of the product, functional specification of each of the product subsystem and preliminary process flow diagram for the final assembly process.

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DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

Phase 3.detail design: it includes the complete specification of the geometry, material and tolerance of all of the unique part in the product and the identification of all of the standard parts to be purchased from supplier. Phase 4.testing and refinement: this phase is involves the construction and evaluation of multiple preproduction versions of the product. Phase 5: production ramp-up: at this phase the product is made using the intended production system. The purpose of the ramp up is to train the work force and to work out any remaining problems in the production processes. Product planning process takes place before product development project is formally approved, before substantial resources are applied and before the larger development team is formed. Product planning is an activity that considers the portfolio of projects that an organization might pursue and determines what subset of these projects will be pursued over what time period (Christensen, 1997).in general there are four type of product development projects, those are, new product platforms, derivatives of existing product platforms, incremental improvements to existing product and fundamental new products. There are many methods to help managers balance an organization portfolio of development project. so that managers

may consider the strategic implications of

their planning decision(Cooper et

al.(2001))one particularly useful mapping, suggested by Wheelwright and Clark(1992),plots the portfolio of projects along two specific dimension. Machine Design is the creation of new and better machines and improving the existing ones. A new or better machine is one which is more economical in the overall cost of production and operation. The process of design is a long and time consuming one. From the study of existing ideas, a new idea has to be conceived. The idea is then studied keeping in mind its commercial success and given shape and form in the form of drawings. In the preparation of these drawings, care must be taken of the availability of resources in money, in men and in materials required for the successful completion of the new idea into an actual reality. In designing a machine component, it is necessary to have a good knowledge of many subjects such As Mathematics, Engineering Mechanics, Strength of Materials, Theory of Machines, Workshop Processes and Engineering Drawing. Historically, a variety of factors, both internal and external to a company have influenced its product design goals. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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DESIGN AND MODELING OF HYDRAULIC HAND LIFT JIB CRANE

| 2010 E.C

Material Handling is the movement, storage, control and protection of materials, goods and products throughout the process of manufacturing, distribution, consumption and disposal. The focus is on the methods, mechanical equipment, systems and related controls used to achieve these functions. Hydraulic cranes are an important part of the material handling equipment’s. The hydraulic cranes that are being used work on electrical supply or manual power. A crane is a type of machine, generally equipped with a hoist, wire ropes or chains, and sheaves, that can be used both to lift and lower materials and to move them horizontally. It is mainly used for lifting heavy things and transporting them to other places. It uses one or more simple machines to create mechanical advantage and thus move loads beyond the normal capability of a man. Cranes are commonly employed in the transport industry for the loading and unloading of freight, in the construction industry for the movement of materials and in the manufacturing industry for the assembling of heavy equipment. 2.1.1. Application of Cranes Cranes exist in an enormous variety of forms – each tailored to a specific use. Sometimes sizes range from the smallest jib cranes, used inside workshops, to the tallest tower cranes, used for constructing high buildings. For a while, mini - cranes are also used for constructing high buildings, in order to facilitate constructions by reaching tight spaces. Finally, we can find larger floating cranes, generally used to build oil rigs and salvage sunken ships. These days hydraulics principle is being used extensively in material handling processes through cranes. Depending on the loads to be handled and the operations to be performed there are different types of cranes like Crawler Cranes, Truck Cranes, and Floor Cranes. Hydraulic Crawler cranes are used for picking and moving huge amount of loads. Generally loads are kept in containers for Bulk loading. Hydraulic truck cranes have good flexibility with high load carrying capacities. Hydraulic workshop foldable crane are used in industries for moving small to medium sized materials from one place to other. The load carrying capacity can vary from half ton to 2 ton or more.

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2.1.2. Fluid Power Systems The transmission of power by fluid power system is most convenient and highly efficient. Due to this, the present conventional power transmission system are being replaced and changed over to fluid transmission system. In this prime mover supplies mechanical energy to a pump which is used to pressurize fluid. Then the pressurized fluid is transmitted to different parts of the system through special piping’s or tubing’s. At desired places pressure energy is converted back to mechanical energy by the devices called actuators consisting of hydraulic cylinders, hydraulic motors etc. Since the power is transferred through the fluid as a medium, such a system is called as fluid power system. 2.1.2.1.

Components of Fluid Power System

1. Hydraulic Power Pack: A hydraulic power pack is the device used to supply pressurized fluid to the piston cylinder so as to extend or contract it. It is required to move the arm in the intended direction. A hydraulic power maybe electrically powered or manually powered. 2. Base: Base is the bottom most part of the hydraulic floor crane which supports all the other parts. It should be designed so as to sustain the weight of the crane as well as the weight to be lifted. 3. Support Column: Support column is fixed to the base and it supports the arm of the crane, also the cylinder is hinged to it. In the rotary design of crane the support column is connected to the output gear through which rotation force is transmitted, it also contains the bearings. Support column can be put vertically on the base or can be inclined at an angle with the base for better support. 4. Arm: It is fixed to the upper end of the support column and a hook is fixed to the other end of the arm to lift the weight. The arm is crucial in design as it takes the bending loads and may bend due to weight. For variable load distribution, the arm can be split up into two parts that can slide in each other.

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5. Piston Cylinder: Hydraulic floor crane has a piston cylinder arrangement to provide the movement of the arm. The maximum and minimum height of the arm is achieved by the piston cylinder. Thus it acts as an actuator to do the intended work. 6. Steer wheel: It is rear wheel of the hydraulic floor crane. It is situated near the steering so the name steer wheels. These are responsible for the turning of the crane. 7. Load wheels: It is the front wheel of the hydraulic floor crane. The weight to be lifted acts downward where the load wheels are mounted. Thus they are called as load wheels and are to be designed so as to sustain the load lifted. 8. Hook: It is attached to the arm of the hydraulic floor crane. It is used to lift the weight. The design of hook is also crucial. Standard hooks are available in the market for different loads.  Identifying customer need is an integral part of the larger product development process and is most closely related to concept generation, concept selection, competitive benchmarking and the establishment of product specification.

2.2.

Current Theories and Design Done On Jib Crane

These days hydraulics principle is being used extensively in material handling processes through cranes. Depending on the loads to be handled and the operations to be performed there are different types of cranes like Crawler Cranes, Truck Cranes, and Floor Cranes. Hydraulic Crawler cranes are used for picking and moving huge amount of loads. Generally loads are kept in containers for Bulk loading. Hydraulic truck cranes have good flexibility with high load carrying capacities. Hydraulic workshop foldable crane used in industries for moving small to medium sized materials from one place to other. The load carrying capacity can vary from half ton to 3 ton or more. There are so many theories and designs are done by many researchers and companies on design and developing of jib crane. We have listed them with their pictures below; 1. The UK business market LTD [6], “design and manufacturing of heavy duty Folding Workshop Crane machine (2018)”. This paper presents for self-standing and completely mobile when in the folded position, saving valuable workshop space. Heavy duty 3 position UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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telescopic jib double acting pump unit Swivel hook with safety catch Pressure relief valve to prevent overloading Self standing on 4 wheels. And it has capacity of 1-2 tons only and Max Height = 2445 mm Min Height = Floor level. But this crane is expensive due to the two extra wheel and have minimum load capacity relatively to our design.

Figure 2:1.Heavy duty folding workshop crane 2. Nationwide industrial supply [7], “design and manufacturing of beech counterweighted crane machine (2018)”. This paper stands without front legs, thus eliminating interference with load lifting or equipment. Maximum height of 2100mm and minimum height of 20mm at full capacity of load lifting of 2000lbs at a counter weight of 1000lbs with a price of $7059.6 And Safety swivel hook is attached directly to extension boom and Safety bypass relief valve is standard to prevent dangerous overloading. For ease of movement, crane has phenolic wheels with roller bearings. But this type of crane is expensive, needs extra load for safety, minimum height and have minimum load capacity relatively to our design.

Figure 2:2.Heavy duty folding workshop crane

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3. Rockwell Hoisto Cranes Private Limited [8], “design and manufacturing of manual mobile material handling machine (2018)”. This paper stands in suitable for Indoor and Outdoor services. It is the basic requirement in all the motor garages and machine shop for loading and unloading of the engines and materials from the vehicles. But it is tedious during operation and it is costly due to rope, pulley gear mechanism and teeth wear.

Figure 2:3.Manual mobile material handling machine 4. Selby Engineering and Lifting Safety LTD [9], “design and manufacturing of mobile jib crane machine (2018)”. This paper was developed in collaboration with users to ensure that it is practical manual handling aid, approved by the Technical Control Board (TÜV) in Germany this is a top quality piece of lifting equipment as totally unique in its design. The crane features easy handling and increased work place safety through high quality products standard. It is a collapsible unit which enables easier transportation through low headroom areas such as doorways and it rotate 360 degree to lift load. This design uses generator or motor for its loading system due to this it may have unwanted noise and high output cost. Also it uses technical control leads to increase its manufacturer cost.

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Figure 2:4.Mobile jib crane 5. Asmita Jadhav, B.E. Mechanical (Pune University) [10], “Design and development of Rotating Floor Crane (2018)”. This paper stands for working with Applied Hydrotech, PuneMember of Entrepreneurs Club Pune. City-Pune, India. This crane needs gear box to transmit motion and power from hand to the rotating column and the structure is exposed to stress and fracture.

Figure 2:5.Rotating floor crane machine 6. Chaitra C. Danavatimath, Prof. H. D. Sarode (P.G.student.Dpartment of Mechanical Engineering (2017)) [11], “Finite Element Analysis and Optimization of Jib Crane Boom”. This paper stands for structural analysis of cantilever beam of jib crane includes an

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investigation of the stresses, deflections, shear capacity and lateral-torsional buckling behavior of regular I Section Cantilever Beam of jib crane subjected to UDL (self-weight) and concentrated load at free end. The beam fails due to Lateral Torsional buckling. Different shapes of cantilever are proposed in this study with different cross-sections, web shapes and materials. FEA and experimental study are carried out for regular and proposed beam to calculate and validate results. Thickness of web and flange is constant for all specimens with length 2.54 and tested for 500kg load lifting capacity. Structural analysis is done to examine the influence of the section dimension due to point load at the free end and uniformly distributed load on cantilever. Using the study it is observed that not only the web thickness, but also the shape of web and sectional cross section of cantilever beam influences the resistance to lateral torsional buckling and bending.

a

Figure 2:6.Cantilever beam of jib crane. 7. Okolie Paul Chukwulozie et al [12], “Design and Analysis of a mobile floor crane (2015)”. Design and fabrication of a mobile floor crane equipped with a facility to lock the load at any level as a special feature, to tackle the issue of failure due to static load. The mobile crane is

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designed to bear a maximum load of about 1000 kg and a maximum height of two meter, with a counter weight of 2.6 KN which gave the crane a 3.034 factor of safety [11]. But the counter weights at the rear base for cases where the boom is extended beyond the front base wheel to balance the weight of the material to be lifted and the load locking device highly stressed.

Figure 2:7.side view of mobile floor crane

2.3.

Importance of Our Design in Engineering Applications

 Important of manual hydraulic hand lift jib crane in engineering application  Our design is used for picking and moving huge amount of loads from one place to another place.  Our design have good flexibility with high load carrying capacities.  Our design is quite easy to operate.  We can use the hydraulic bottle jack for multiple purpose in this design. For example it can be used for picking, carrying and moving any loaded object up to this design maximum loading capacity (3 ton).  It can reduce the risks of human beings from carrying high load object such as disk prolapse.

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 It can’t use motor or generator power for moving purpose so that it helps to reduce the output cost.

2.4.

Design Approach With Regard to Our Project  We collect the necessary Data through interview, questionnaires and observation.  We select proper material for each parts of the machine.  We design parts and stress analysis.  We estimate cost analysis.  We draw parts and assembly drawing by using solidwork software.  We check the deformation of each parts by using ANSYS software.

2.5.

Available Standards Pertinent to Our Design

 Standards pertinent to our design are:  Hydraulic bottle jack  Wheel  Crane hook

2.6.

Reason for Selecting Our Topic

 The reason we select this topic is that:  The UOG workshop have already finished the building and it’s waiting for the availability of necessary machines for the workshop so that we selected this design to solve the problems of picking, carrying and moving any loaded objects from one place to other place in a safe way, since this service are not given adequately by the product supply company that means it may give the service of putting loaded objects to one place but not moving to the place we needed.  In addition to this, our design can also reduce the problems of picking, carrying and moving loaded objects on garages and other small enterprise in our country.  We can use the machine in the workshop without disturbing those taking services in the workshop (students) and workshop lab assistance.  It can also reduce risk of polluting the environment from smoke, since it operates hydraulically, it can’t release any smoke.

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CHAPTER THREE 3. METHODOLOGY OF THE PROJECT Methodology is the basic part of any scientific research because it gives detail about the data. Thus from this process on wards we will be able to get our design requirements and start designing accordingly.

Need or Aim

(

)

Fig. design flow chart To come up with adequate solution for the problem what we proposed; as possible as we could we were following an exact technical procedures throughout this paper according to what we gain from mechanical engineering knowledge back ground.

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In order to conduct this study, methods and procedures have great contribution for reaching the final result of the paper. The methods used are discussed below in detail.

3.1.

Data Collection Method  Interview: in order to strengthen the design, we will use this method. So some employees that are concern to our project will involve in responding the interview.  Observation: in order to check and to be sure our project this method is very crucial. During this observation, we will observe in person in the place what kinds of machines are present and how they have used the machines.  Questionnaire: in order to achieve the purpose of this project we will distribute questioners for concerned persons.

Data collection method

Secondary data

Primary data

Quantitative data

Example: read from previous journals, researches, internet, etc.

Qualitative data

v

Survey

Experiments

Focus group

v

 Personal interview  Companies  Email

Mechanical observation

Individual depth interview Human observation

Simulation Case studies Figure: data collection method flow chart

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3.2.

| 2010 E.C

Comparison of Previous Work Related to Our Project List in the Literature Review 1. MODEL -1

Figure 3:1.Heavy duty folding workshop crane  Component of model 1  Bolt and nut  Boom  Hydraulic jack  Frame  Double acting pump

 unit Swivel hook with safety catch  Pressure relief valve to prevent overloading  Self-standing on 4 wheels

 Operating instructions of model 1 1. Insert the jack handle into the hole of the handle socket. 2.

Make sure the saddle is correctly positioned. To prevent damage to the jack, do not move the jack while the handle is intact in the socket.

3.

To raise load, use one hand to hold the front part of the handle, and use the other hand to turn at the rear end of the handle clockwise.

4. To lower load, use one hand to hold the front part of the handle and use the other to turn at the rear end of the handle counterclockwise slowly  Working principle of model 1 Hydraulic jack works on the principle of ―Pascal‘s law. When the handle is operated, the plunger reciprocates then the oil from the reservoir is sucked UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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into the plunger cylinder during upward stroke of the plunger through the suction valve. The oil in the plunger cylinder is delivered into the ram cylinder during the downward stroke of the plunger through the delivery valve. This pressurized oil lifts the load up, which is placed on top plate of the ram. After the work is completed the pressure in the ram cylinder is released by unscrewing the lowering screw thus the pressure releases and the ram is lowered, then the oil is rushed into the reservoir. It consists of plunger cylinder on one side and ram cylinder on the other side. These two cylinders are mounted on base which is made of mild steel. Plunger cylinder consists of plunger which is used to build up the pressure by operating the handle. Plunger cylinder consists of two non-return valves i.e. one for suction and other for delivery. Ram cylinder consists of ram which lifts the load. The ram cylinder connected to delivery valve of plunger cylinder. It is also consists of lowering screw this is nothing but a hand operated valve used for releasing the pressure in the ram cylinder for get down the load.

Figure 3:2.hydraulic bottle jack  Advantage of model 1  Heavy duty Folding Workshop Crane  Use in business market

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 It is self-standing  completely mobile when in the folded position,  Saving valuable workshop space.  Heavy duty 3 position telescopic jib  Double acting pump unit Swivel hook with safety catch  Pressure relief valve to prevent overloading Self standing on 6 wheels.  Disadvantage of model 1 

It has capacity of 1-2 tons only

 Crane is expensive  Have minimum load capacity relatively to our design.  It has 6 wheels. 2. MODEL -2

Figure 3:3.Heavy duty folding workshop crane  Components of model 2  Extension boom

 Roller bearings

 Safety bypass relief valve

 Balancing load at the back.

 Crane has phenolic wheels

 Hook

 Horizontal frame

 Bolt and nut

 Working principle of model 2 The working principle of model -2 is the same as that of model -1 except balancing load at the back. The lifting load in the crane is must be balance at the back load. If the lifting load is greater than the balancing load at the back, the crane is not safe.

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 Advantage of model 2  Beech Counterweighted Crane is designed without front legs.  Thus eliminating interference with load lifting or equipment.  Safety swivel hook is attached directly to extension boom and  Safety bypass relief valve is standard to prevent dangerous overloading.  For ease of movement, crane has phenolic wheels with roller bearings.  Disadvantage of model 2  But this type of crane is expensive,  Needs extra load for safety,  Minimum height  Limited capacity  That is it can’t be load greater than the balancing load at the back. 3. MODEL-3

Figure 3:4.Manual mobile material handling machine  Component of model 3  Boom

 Handle

 Hook

 Rope

 Wheel

 Pulley

 Base frame

 Coiler

 Bolt and nut

 Gear box

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 Working principle of model 3 The working principle is when the operator turns clock wise the Handle at back the rope coiled at the coiler by the help of gear box. So the Rope reduce its length. The Rope lift the load at the front by using hook in the help of pulley to multiply force to lift the load. To move the crane from place to place apply pushing force at back. When the crane reaches at desired place and to lower the load, rotate the handle anti clockwise.  Advantage of model 3  It is suitable for Indoor and Outdoor services.  It is the basic requirement in all the motor garages.  Machine shop for loading and unloading of the engines Materials from the vehicles.  Disadvantage of model 3  But it is tedious during operation.  It requires strong man.  It is costly due to rope, pulley gear mechanism.  The teeth wear. 4. MODEL-4

Figure 3:5.Mobile jib crane  Component of model 4  Extension boom

 Cable or remove control

 Wheel

 Bolt and nut

 Motor UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 Hydraulic creation, manual or powered

| 2010 E.C

 Powered treble trolley  Optional powered rotation

 Two speed electric hoist  Working principle of model 4 When the Motor turn on, hydraulic creation powered at the back to retract in order to lower Powered treble trolley to pick the load at lower position. Powered treble trolley to pick the load. Then the hydraulic creation powered to extract in order to lift the load and rotate to place the load at desired position and the load place when the hydraulic creation powered to extract again.  Advantage of model 4  It is practical manual handling aid,  Approved by the Technical Control Board (TÜV) in Germany,  A top quality piece of lifting equipment as totally unique in its design.  The crane features easy handling and increased work place safety through high quality products standard.  It is a collapsible unit which enables easier transportation through low headroom areas such as doorways and it rotate 360 degree to lift load.  Disadvantage of model 4  This design uses generator or motor for its loading system due to this it may have unwanted noise and high output cost.  Also it uses for huge loads which makes it to increase its manufacturer cost.

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5. MODEL-5

Figure 3:6.Rotating floor crane machine  Component of model 5  Boom

 Hydraulic jack

 Base frame

 Bolt and nut

 Hook

 Wheel

 Operating instructions of model 5 1. Insert the jack lever into the hole of the lever socket. 2. Make sure the saddle is correctly positioned. To prevent damage to the jack, do not move the jack while the lever is intact in the socket. 3. To raise load, use hand to hold the part of the lever, and use the hand to move up and down at the rear end of the lever. 4. To lower load, use one hand to hold the front part of the handle and use the other to turn at the rear end of the lever slowly.  Working principle of model 5 Hydraulic jack works on the principle of ―Pascal‘s law‖. When the lever is operated, the plunger reciprocates then the oil from the reservoir is sucked into the plunger cylinder during upward stroke of the plunger through the suction valve. The oil in the

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plunger cylinder is delivered into the ram cylinder during the downward stroke of the plunger through the delivery valve. This pressurized oil lifts the load up, which is placed on top plate of the ram. After the work is completed the pressure in the ram cylinder is released by unscrewing the lowering screw thus the pressure releases and the ram is lowered, then the oil is rushed into the reservoir. It consists of plunger cylinder on one side and ram cylinder on the other side. These two cylinders are mounted on base which is made of mild steel. Plunger cylinder consists of plunger which is used to build up the pressure by operating the lever. Plunger cylinder consists of two non-return valves i.e. one for suction and other for delivery. Ram cylinder consists of ram which lifts the load. The ram cylinder connected to delivery valve of plunger cylinder. It is also consists of lowering screw this is nothing but a hand operated valve used for releasing the pressure in the ram cylinder for get down the load.

Figure 3:7.Working principle of hydraulic jack  Advantage of model 5  On Design of Rotating Floor Crane, Working with Applied Hydrotech.  This crane needs gear box to transmit motion and power from hand to the rotating column.  Disadvantage of model 5  The structure is exposed to stress and fracture. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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6. MODEL-6

Figure 3:8.Cantilever beam of jib crane.  Component of model 6  Boom  Trolley  Mast  Gusset  Trapezoidal web  Working principle of model 6 The applied forces diagram details the relative position and direction of the forces that this jib crane applies to the supporting structure when a load is picked up. 

Free Standing

When a load is applied to the crane, the front of the head assembly, the front of the base plate, and the front gussets are in compression (exerting thrust); the back boom plate, the back of the head and the back of the gussets are placed in tension (pulling). These forces put a moment on the foundation and exert significant thrust & pull on the crane which must be of sufficient size to resist the forces.

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

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Wall/Column Mounted

When a load is applied, the top wall bracket applies a downward and outward force on its support. This places the support in tension (pulling). The bottom wall bracket applies a downward and inward force on its support, placing it in compression (exerting thrust). These Thrust & Pull forces are significantly higher than the capacity of the crane! Be sure to have a qualified structural engineer verify the adequacy of the supporting structure.  Advantage of model 6  FEA and experimental study are carried out for regular and proposed beam to calculate and validate results.  Uniformly distributed load on cantilever.  Thickness of web and flange is constant for all specimens with length 2.4m and tested for 500kg load lifting capacity.  Provides a versatile and economical solution where 360º rotation is desired.  Often used when the thrust and pull exerted by other crane types is too great.  The key to Gorbel’s superior Mast Type jib lies in the quality bearing design.  Crane is easy to level during installation.  Disadvantage of model 6  The beam fails due to Lateral Torsional buckling.  Sectional cross section of cantilever beam influences the resistance to lateral torsional buckling and bending. 7. MODEL 7

Figure 3:9.Side view of general assembly UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 Component of model 7  Base

 Pump Lever

 Mast

 Lift cylinder

 Metal Roller

 Hook

 Pump

 Boom

 Handle

 Load locking device

 Working principle of model 7 A mobile floor crane is equipment with portable features which makes it admirable and recommended for both indoor (workshop/ warehouse) and outdoor purposes, for the sole aim of lifting and moving heavy materials from one place to another. Some of these features found in them include; adjustable boom, hydraulic height and balance due to rest base design. These adjustable features are to accommodate various heights and sizes of materials to be lifted. This project gains its uniqueness with the design of a dead stop incorporated in its mast, to tackle the issue of failure due to static load. There are also provisions for counter weights at the rear base for cases where the boom is extended beyond the front base wheel to balance the weight of the material to be lifted.  Advantage of model 7  Transportation of heavy machine parts within and outside the workshop.  It can also be used to load and unload machine parts on trucks.  For the sole aim of lifting and moving heavy materials from one place to another.  Dead stop incorporated in its mast to tackle the issue of failure due to static load.  These adjustable features are to accommodate various heights and sizes of materials to be lifted.  Disadvantage of model 7  For counter weights at the rear base for cases where the boom is extended beyond the front base wheel to balance the weight of the material to be lifted.  For permanent joints, the arc welding process was employed.  High fracture of teeth and lock device

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3.2.1. Design Matrix Design matrix a method used to compare the different models according their material property, function, operation, availability, cost required etc. and is used to choose the most acceptable, excellent, preferable, model for design purpose. Table 3:1.Comparison criteria for each model Function No

Model designation

requirement Model

Model

Model

Model

Model

Model

Model

1

2

3

4

5

6

7

Complexity

4

4

3

5

5

3

4

Weight

5

4

2

5

5

3

3

Operability

5

5

3

2

4

2

3

4

3

3

2

4

2

2

failure reduction

4

3

2

3

4

3

3

Deflection and jerk

3

3

2

2

4

2

2

4

4

3

3

4

3

2

Reliability

4

4

2

2

4

3

3

Durability

4

4

2

2

5

3

3

Stability

3

5

3

3

4

4

3

Physical functionality 1

Mechanical functionality Wear failure reduction Fatigue and creep 2

reduction Corrosion failure reduction

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Materials functionality Strength

4

3

3

3

4

4

4

Availability

4

4

4

3

4

4

3

Affordability

4

3

3

2

4

4

3

Manufacturability

4

4

4

3

4

4

3

Maintenance

4

3

4

2

4

4

3

Safety

5

4

3

5

5

4

4

Cost required

5

4

4

2

4

3

3

Total +ve weight

70

62

50

49

72

55

51

82.35

72.94

58.82

57.64

84.70

64.70

60

3

for each of Model % weight= Total +ve weight for each of Model / Total weight

i.

Safety: Considerations must be carried out through the entire life cycle of the machine includes Hazards which occurs during 

The process of making the product,



The expected use of the product,



Foreseeable misuse & abuse of the product

 Since each design is different, the designer needs to give full consideration to the safety aspect of the product even if it is the modification of an existing product. ii.

Complicity of mechanisms: the design should not be complex in parts & operation mechanism; also it can be operated by unskilled person

iii.

Manufacturability: the condition that any damaged part of the machine can be replaced easily with the appropriate spare part when failure is occurred.

iv.

Maintainability: the condition that any damaged part of the machine can be replaced easily with the appropriate spare part when failure is occurred.

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v.

| 2010 E.C

Cost: Are the basic criteria in designing a machine, low total cost must be the first target for manufacturing process

 Description: 2. Poor, small, few, low difficult 3. Acceptable, average 4. Great, high, large, possible 5. Excellent, very high 3.2.2. Summary of Literature Review Our project is mainly focused on modelling and design analysis of manual hydraulic hand lift jib crane in garages and micro and small enterprises in Ethiopia. This crane is suitable for loading and unloading heavy material in workshops. From the above required criteria table 3-1 the model are listed in its % weight for each of Model are 1st (84.70 %) MODEL 5, 2nd (82.35% total) Model 1, 3rd (72.94% total) Model 2. The structure of our design is effective in distribution of load and minimizes stress comparing with the above table 3-1 (model 5 is the best one). So by taking model 5 we modify to decrease stress of the machine by adding some components, increase balancing during lifting or carrying loaded object by tilting horizontal frame 45° to the connecting frame and inclined the back hoist body. This jib crane has four rolling wheels so it can move simply comparing with beech Counterweighted Crane and it has the capacity to load up to 3 tons, maximum height 2500mm and minimum height up to 600mm.our design have good capacity to load and lift materials rather than the above designs. And our design can be manufactured easily in our country by our product.

3.3.

Selection of Materials for Engineering Purposes

 The selection of a proper material, for engineering purposes, is one of the most difficult problem for the designer. The best material is one which serve the desired objective at the minimum cost.  The following factors should be considered while selecting the material:  Availability of the materials,  Suitability of the materials for the working conditions in service, and

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 The cost of the materials

3.4.

Design Part and Stress Analysis

In order to achieve the project, material handling mechanisms were considered. In this condition, needed to consider:  Selection of hydraulic bottle jack

 Design of back hoist support

 Design of horizontal boom

 Design of connecting frame

 Design of back hoist body

 Selection of hook

 Design of base body

 Design of bolt and nut

 Design of horizontal frame

 Selection of wheel

3.5.

Draw Part and Assembly Drawing by using Solid Work Software and Check the Deformation by using ANSYS Software. 

3.6.

The parts are draw according to the correct dimension get from design analysis of part.

Result and Discussion

When we compare our design to the existing manually hydraulic hand lift jib crane the stress and fracture of component is reduced and the stability of the machine is increase by aligned of horizontal frame by 45 degree. The carrying capacity and lifting height of the machine is increasing by adding some component.

3.7.

Cost Analysis of our Design

A system which systematically recodes all the expenditure to determine the cost of manufactured products. Generally good, easily, available material selection of horizontal boom, back hoist support, connecting frame, horizontal fame, base body, nuts and bolts are compatible each other to resist the direct stress, bending moment and shear stress that occurred during assemble and disassemble of manually hydraulic hand lift jib crane and to reduce the total cost of the design.

3.8.

Recommendation and Conclusion

Because of this machine becomes manual, simple to lift up loads by pumping and it doesn’t ask qualification of employees, those garages and small and micro enterprises should be use it to accelerate and to make easily the work activity. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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CHAPTER FOUR 4. DESIGN ANALYSIS 4.1.

Data Gathering

4.1.1. Interview In order to strengthen the design, we have added this method. So some employees that are concern to our project are involved in responding the interview. Totally we interview about some garages and small and micro enterprise employees. After taking the interviews the customer statement were understand and analyzed.

Figure 4.1:1.interviewing from different company

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4.1.2. Questionnaires To achieve the purpose of the project about 7 questionnaires are distributed to garages and small and micro enterprise. Then after the response were tabulated and analyzed. The distributed questionnaires shows that all most all have lifting machine. 4.1.3. Observation In order to check and to be sure our project this method is very important. During this observation, we have observed in person in the place what kinds of machines are present and how they have used the machines.

Figure 4.1:2.observation of manually hydraulic hand lift fixed jib crane from Jovani garage

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Figure 4.1:3.observation of baranco from Jovani garages Table 4.1:1.Interpretation of interview and distributed questionnaires Question

Customer statement  We used some times fixed cranes

 Typical uses of lifting machine

 We used cricks  We used forklifts  We used overhead crane to lift heavy loads  We lift a load from 2000-3000kg

 Load lifting

 We lift a load of about 1000-2000kg

limitation  Height lifting

 We lift a height from 600- 2500mm

limitation  Accidence on employees when they carry  Effects through no use of lift machine

 It takes huge manpower to carry when it is heavy load  Lose off safety of the material when the employees carry the object

 Kind of lift machine needed to have.

 We need to have Mobil hydraulic lift, best than others such as mechanical lifts, fixed cranes...etc.

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 Need to have mobile

| 2010 E.C

 We need to have this lift in our work shop

hydraulic jib crane.  We prevent accidence of the employees (keep safety  Solving problems

of the employees)  We decrease the number of manpower during carrying

through using hydraulic jib crane

 High safety of the material.

lift machine

4.2.

heavy loads

Selection of Hydraulic Bottle Jack

The cylinder capacity for three ton, the stroke rang can be from 3 up to 20 Inch and the type is aluminum and steel jacks with a model of series JHA/JH EBJL-3GC. Because of the following advantage are included in the industrial bottle jacks we select it.  Lower handle effort, reduces operator fatigue  Fully service cable  Cast beam and cast pump linkage  Safety relief valve to prevent over load  Pumping handle included on all models  Automatic by pass port to prevent over extension

Figure 4.1.3d view of hydraulic bottle jack

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4.2.1. Oil Type The hydraulic crane is based on a simple concept-the transmission of forces from point to point through a fluid. Most hydraulic machines use some sort of incompressible fluid, a fluid that is at its maximum density. Oil is the most commonly fluid for hydraulic machines, including hydraulic cranes. In a simple hydraulic system, when a piston pushes down on the oil, the oil transmits all of the original force to another piston, which is driven up. Therefore the oil used for this machine is hydraulic oil with its viscosity of 22 up to 32.

4.3.

Design Specification

 Maximum height=2500mm  Minimum height =600mm  Maximum mass=3 ton=3000Kg  Gravity= g = 9.81m⁄s 2  Load = mass ∗ g = 3000kg ∗ 9.81 m⁄s 2 = 29430N = 𝟐𝟗. 𝟒𝟑𝟎𝐤𝐍 4.3.1. Geometrical Analysis of the Machine

Figure 4.1:4.3d view of the machine

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Figure 4.1:5.Front view of our machine  The variables are: 

Lb=L1+L2, boom length



L=L3+L4, back (hoist) body length



Lm=length of the cylinder at horizontal (normal) position



Le=extracted cylinder length



Lr=retracted cylinder length



θ1=the angle of the boom when extracted from the normal position



θ2=the angle of the boom when retracted from the normal position



θ3=the angle between the cylinder and the back hoist body



θ4=the angle between the back hoist body and horizontal frame

 Therefore from the above geometrical we can assume that Lm is perpendicular to Lb when it is at normal position. From the above we can draw the following geometrics. 4.3.2. Parameters of the Cylinder from Standard Table  For three ton read from the appendix table 2.A.  Maximum height (Le ) =1176.274 mm  Minimum height (Lr ) =668.274 mm  Bore diameter =65mm 

Assume

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

θ1 = 65°



θ2 = 50°



Lb = 1250 mm

| 2010 E.C

Figure 4.1:6.when the cylinder is retracted and extracted triangle  From the above figure 4.1:3 we can calculate L2 by using cosine law L2 2 L2 2 L2 L2 (Le − Lr ) = + −2 cos(θ1 + θ2 ) 2 2 cos θ1 cos θ2 cos θ1 ∗ cos θ2 2

L2 2 L2 2 L2 L2 (1176.274 − 668.274) = + −2 cos(65 + 50) 2 2 cos 65 cos 50 cos 65 ∗ cos 50 2

(508)2 = 5.6L2 2 + 2.42L2 2 + 3.11L2 2 = 11.13L2 2  L2 = 𝟏𝟓𝟐. 𝟐𝟕 𝐦𝐦

Figure 4.1:7.When the cylinder is retracted UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 From the above triangle figure 4.1:4 we will get θ4 − θ2 + γ + θ3 = 180°, but θ3 = 90° − θ4 θ4 − θ2 + γ + (90° − θ4) = 180° γ − θ2 + 90 − 180° = 0, γ = 90° + θ2, but θ2 = 50°  γ = 90° + θ2 = 90° + 50° = 𝟏𝟒𝟎°  Also from figure 4.1:4 using cosine law we can calculate L3 L3 2 = Lr 2 + L2 2 − 2 ∗ Lr ∗ L2 ∗ cos γ L3 2 = (668.274)2 + (152.27)2 − 2(668.274)(152.27)cos(140°)  L3 = 𝟕𝟗𝟏𝐦𝐦

Figure 4.1:8.When the cylinder is at normal position  From the above triangle figure 4.1:5 we will get θ4 + 90° + θ3 = 180° θ3 = 180° − 90° − θ4 = 90° − θ4 L

 cosθ4 = L2 3

, 𝛉𝟒 = Cos −1 (

152.27 791

) = 𝟕𝟖. 𝟗°

 𝛉𝟑 = 90 − θ4 = 90 − 78.9° = 𝟏𝟏. 𝟏°

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Figure 4.1:9.When the cylinder is extracted  From the above triangle figure 4.1:6 we will get X by using sine law X

sin θ1 = L

b

, X=Lb ∗ sin θ1 = 1250 ∗ sin 65°

 X=1132.884 mm

Figure 4.1:10.Triangle of back hoist body with base body (∆ACO)  To find the value of h,L1,L4 we use the above figure 4.1:7  To get, L1 

L1 = Lb − L2 = 1250mm − 152.27mm = 1097.73mm

 To get the value of the height of the column, h 

h = Lb − X = 2500mm − 1132.884mm = 1367.1mm h

 Then, sin θ4 = L 

h

L = sin θ = 4

1367.1mm sin 78.9°

= 1393.16mm

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 Now, L = L3 + L4 

L4 = L − L3 = 1393.16mm − 791mm = 602.16mm

From the above figure 4.1:2 we can calculate Lh and Xh

Figure 4.1:11.triangle of back hoist body with horizontal frame and back hoist support (∆Aho)  From the above figure 4.1:8 use sine and cosine law  Then,

sin θ4 =

Lh 2 3

cos θ4 =

,

( ∗L3 +L4 )

2

Xh 2 3

( ∗L3 +L4 )

2

 Lh = (3 ∗ L3 + L4 ) ∗ sin θ4 = (3 ∗ 791mm + 602.16mm) ∗ sin 78.9° 

Lh = 1108.364mm 2

2

 Xh = (3 ∗ L3 + L4 ) ∗ cos θ4 = (3 ∗ 791mm + 602.16mm) ∗ cos 78.9° 

Xh = 217.452mm

 To find the pressure of the piston we use moment at point c

Figure 4.1:12.FBD of boom at normal position

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 Take moment at point c and counter clockwise positive from the above figure 4.1:9  ∑M@C =0  FP*L2-W*(L2+L1) =0  FP =

W∗(L2+L1) L2

Where, W=mass*acceleration due to gravity  W = 3000Kg ∗ 9.8m/s 2 = 𝟐𝟗. 𝟒𝟑𝟎𝐤𝐍

 Fp =

29.430KN∗1250mm 152.27mm

 Fp = 𝟐𝟒𝟏. 𝟓𝟗𝟒𝐤𝐍  Therefore the pressure of the piston will become,  Fp = P ∗ Ap , Ap = 

πD2 4

Where Ap area of the piston, D=bore diameter =65mm, Fp is force acting on the piston and P is working pressure of the piston. Fp

 P=A = p

241.594kN π(65mm)2 4

 P = 72.806MPa

4.4.

Design of Components of Machines

4.4.1. Design of Horizontal Boom 4.4.1.1.

Material Selection for Horizontal Boom The material selection for horizontal boom is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80 m2



Tensile strength (σt) = 480 m2



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN MN

MN

Kg

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Figure 4.1:13. 3d views of horizontal boom 4.4.1.2.

Horizontal Boom Subjected to Bending Moment

 Parameters that solved from the previous are as shown below:

Figure 4.1:14.FBD of boom when the cylinder at extracted position  From the above we get 

L1 = 1097.73mm



L2 = 152.27mm



Lb = 1250mm



W=29.430KN



Fp =241.594kN

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First find reaction force  Force along y-axis from the above figure 4.1:11 

+↑ Σfy = 0



−fyc − W + Fp ∗ sinβ2 = 0



fyc = Fp ∗ sinβ2 − W = 241.594kN ∗ sin25° − 29.430kN

but ,

β2 = 90 − θ1 = 90 − 65 = 𝟐𝟓°

 fyc = 𝟕𝟐. 𝟔𝟕𝟐𝐤𝐍  Force along x-axis from the above figure 4.1-11 

→ +Σfx = 0,



−fxc + Fp ∗ cosβ2 = 0



fxc = Fp ∗ cosβ2 = 241.594 ∗ cos25°  fxc = 𝟐𝟏𝟖. 𝟗𝟓𝟗𝐤𝐍

 Now find Mc, take anti clockwise bending moment at point c are positive 

∑ M@c = 0, Mc + Fp ∗ sin β2 ∗ L2 − W ∗ Lb = 0

Mc = W ∗ Lb − Fp ∗ sin β2 ∗ L2 = 29.43kN ∗ 1250mm − 241.594kN ∗ sin 25° ∗ 152.27𝑚𝑚  Mc = 𝟐𝟏𝟐𝟒𝟎. 𝟒𝟐𝟑𝐤𝐍 Bending moment and shear force  Bending moment and shear force at section x1-x1

Where 0 ≤ x1 ≤ L2 = 152.27mm To find bending moment at each point take bending moment at point x1 anti clockwise is zero 

∑ M@x = 0, Mx + Mc + fy ∗ x1 = 0 Mx = −fy ∗ x1 − 𝑀𝑐

 𝐌𝐜@(𝐱𝟏 = 𝟎) = −72.672kN ∗ 0mm − 21240.423kN = −𝟐𝟏𝟐𝟒𝟎. 𝟒𝟐𝟑𝐤𝐍𝐦𝐦 = 𝟐𝟏𝟐𝟒𝟎. 𝟒𝟐𝟑𝐤𝐍𝐦𝐦(𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐞)

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 𝐌𝐝@(𝐱𝟏 = 𝟏𝟓𝟐. 𝟐𝟕𝐦𝐦) = −72.672kN ∗ 152.27mm − 21240.423kNmm = −𝟑𝟐, 𝟑𝟎𝟔. 𝟏𝟖𝟖𝐤𝐍𝐦𝐦 = 𝟑𝟐, 𝟑𝟎𝟔. 𝟏𝟖𝟖𝐤𝐍𝐦𝐦(𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞) To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, Vx − fy = 0 Vx = fy = 72.672kN  Vc@(x1 = 0) = Vd@(x1 = 0) = fy = 𝟕𝟐. 𝟔𝟕𝟐𝐤𝐍

 Bending moment and shear force at section x2-x2

Where L2 ≤ x2 ≤ Lb To find bending moment at each point take bending moment at point x2 anti clockwise is zero 

∑ M@x2 = 0, Mx + fy ∗ x2 − Fp ∗ sin β2 ∗ (x2 − L2) + Mc = 0



Mx = Fp ∗ sin β2 ∗ (x2 − L2) − fy ∗ x2 − Mc



𝐌𝐝@(𝐱𝟐 = 𝐋𝟐) = 241.594kN ∗ sin 25° ∗ (L2 − L2) − fy ∗ L2 − Mc

 𝐌𝐝@(𝐱𝟐 = 𝟏𝟓𝟐. 𝟐𝟕𝐦𝐦) = −72.672kN ∗ 152.27mm − Mc = −11065.765kNmm − 21240.423kNmm = −32,306.188kNmm = 𝟑𝟐, 𝟑𝟎𝟔. 𝟏𝟖𝟖𝐤𝐍𝐦𝐦(𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞) 

𝐌𝐞@(𝐱𝟐 = 𝐋𝐛) = Fp ∗ sin β2 ∗ (Lb − L2) − fy ∗ Lb − Mc

 𝐌𝐞@(𝐱𝟐 = 𝐋𝐛) = 241.594kN ∗ sin 25° ∗ (1250mm − 152.27mm) − 72.672kN ∗ 1250mm − 21240.423 = 𝟎. 𝟎𝟒𝟓𝟑𝐤𝐍𝐦𝐦 To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, Vx − fy + Fp ∗ sin β2 = 0 Vx = fy − Fp ∗ sin β2 = 72.672kN − 241.594 ∗ sin 25° = −29.43kN  Vd@(x2 = L2) = Ve@(x2 = Lb) = −𝟐𝟗. 𝟒𝟑𝐤𝐍

 Now maximum bending moment and shear force get at point

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

𝐌𝐦𝐚𝐱@𝐝 = 𝟑𝟐𝟑𝟎𝟔. 𝟏𝟖𝟖𝐤𝐍𝐦𝐦



𝐕𝐦𝐚𝐱@𝐜 = 𝟕𝟐. 𝟔𝟕𝟐𝐤𝐍

 Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I = σy

 Therefore, σb = 

b4 −h4 12 MN

 σb = σall = S.F  σb =

| 2010 E.C

, let S.F=1.5 and σy for mild steel is 280 m2

280MN⁄ 2 m 1.5

= 186.667 m2 = 186.667 N⁄mm2

b 2 b4 −h4 12

but, M = 32306.188kNmm and b=h+2*t, h=b-2*t

M∗

MN

Let, b=20*t, t=

b 20

and h=0.9*b b 2

 186.667 N⁄mm2 =

32306.188kNmm∗

 186.667 N⁄mm2 =

193837.128kNmm∗b

b4 −(0.9∗b)4 12

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 563,643,873.2Nmm  b3 = 3,019,515.357mm3  b = 𝟏𝟒𝟒. 𝟓𝟑𝟕𝐦𝐦 b

 t = 20 =

144.537mm 20

= 𝟕. 𝟐𝟐𝟕𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 144.537mm and t ≥ 7.227mm. 

b=203mm,



t=9.5mm,



h=b-2t=203mm-2*9.5mm=184mm

 Therefore, checking for the safety:  σb =

b 2 b4 −h4 12

M∗

=

203mm 2 (203mm)4 −(184mm)4 12

32306.188kNmm∗

= 𝟕𝟏. 𝟐𝟗 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 71.29Mpa = 𝟑. 𝟗𝟐𝟖 > F. s old = 𝟏. 𝟓, so the design of horizontal boom b

is safe due to bending load when the hydraulic cylinder at extracted position.

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4.4.1.3.

| 2010 E.C

Horizontal Boom Subjected to Direct Stress

 Direct stress is given by σd =

fxc=Fp∗cos β2 A

Where 

fxc=218.959kN



A= area of the cross section of the material=b2 − h2 218.959kN

 𝛔𝐝 = (203mm)2 −(184mm)2 = 29.778 𝐍⁄ 𝐦𝐦𝟐  The horizontal boom subjected to direct and bending stress  4.4.1.4. 

𝛔𝒕 = σd + σb = 29.778Mpa + 71.29Mpa = 𝟏𝟎𝟏. 𝟎𝟔𝟖𝐌𝐩𝐚

Horizontal Boom Subjected to Shear Stress

The shear stress on the square hollow section on the boom when cylinder is extracted will become:

 Shear stress on the square hollow section on the boom is  τ=

Vmax A

,

Where

F

 τ=A=

Fyc A

=



Vmax = 72.672kN



A=area of the cross section of the material=b2 − h2

72.672kN b2 −h2

72.672kN

= 2032 −1842 = 𝟗. 𝟖𝟖𝟑𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σ𝑡 ± √σ𝑡 2 + 4 ∗ τ2 ) = 2 (101.068Mpa ± √(101.068Mpa)2 + 4 ∗ (9.883Mpa)2 )  𝛔𝟏 = 𝟏𝟎𝟐. 𝟎𝟐𝟔𝐌𝐩𝐚 And 𝛔𝟐 = −𝟎. 𝟗𝟓𝟖𝐌𝐩𝐚, 𝛔𝟑 = 𝟎  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (102.026Mpa − (−0.958Mpa))  τmax = 𝟓𝟏. 𝟒𝟗𝟐𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

max

280Mpa

= 2∗51.492Mpa = 2.719 > 1.5, it is 𝐬𝐚𝐟𝐞!

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 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (102.026)2 + (−0.958)2 − 2 ∗ 102.026 ∗ −0.958 = ( f.s )2 σy 2

 ( f.s ) = 10,605.704Mpa2 

σy F.s

= 102.984Mpa σy

280Mpa

 F. s new = 102.984Mpa = 102.984Mpa = 2.719 > 1.5, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐡𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 𝐛𝐨𝐨𝐦 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.2. Design of Back Hoist Body 4.4.2.1.

Material Selection for Back Hoist Body The material selection for back hoist body is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80 m2



Tensile strength (σt) = 480 m2



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN MN

MN

Kg

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Figure 4.1:15.3d views of back hoist body 4.4.2.2.

Back Hoist Body Subjected to Bending Moment

 Parameters that solved from the previous are as shown below:

=

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Figure 4.1:16.FBD of the hoist body  From horizontal boom we get the following values 

Fxc=218.959kN



Fyc=72.672kN



Fp=241.594kN



Mc=21240.423kNmm

 From the above figure 4.1:13  Fy ′ c = Fxc ∗ cos 53.9° + Fyc ∗ sin 53.9° = 218.959kN ∗ cos 53.9° + 72.672kN ∗ sin 53.9°  Fy ′ c = 187.728kN  Fx ′ c = Fxc ∗ sin 53.9° − Fyc ∗ cos 53.9° = 218.959kN ∗ sin 53.9° − 72.672kN ∗ cos 53.9°  Fx ′ c = 134.099kN  Fy ′ b = Fp ∗ sin θ3 = 241.594kN ∗ sin 11.1° = 46.512kN  Fx ′ b = Fp ∗ cos θ3 = 241.594kN ∗ cos 11.1° = 237.074kN

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First find reaction force and moment  Force along y-axis from the above free body diagram





+↑ ΣFy′ = 0



Fy ′ c − Fy ′ h − Fy ′ b + Fy′a = 0

Fy ′ h − Fy ′ a = Fy ′ c − Fy ′ b = 187.728kN − 46.512kN  Fy ′ h − Fy ′ a = 141.216kN  Fy ′ h = Fy ′ a + 141.216kN………………….[1]

 Force along x-axis from the above figure 4.1-13 

→ +ΣFx′ = 0,



−Fx ′ c + Fx ′ b − Fx′a = 0



Fx ′ a = Fx ′ b − Fx ′ c = 237.074kN − 134.099kN

 Fx ′ a = 102.975kN  To find Ma, Fy’a and Fy’h , we use Macaulay's method for statically indeterminate beams 

Bending moment at section x-x is



Mxx = Mc + Fy ′ c ∗ x + Mh − Fy ′ h(x − 263.667) − Fy′b(x − 791)



Mxx = Mc + Fy ′ c ∗ x + Mh − Fy ′ h ∗ x + 263.667 ∗ Fy ′ h − Fy ′ b ∗ x + 791 ∗ Fy ′ b



Mxx = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b) + (Fy ′ c − Fy ′ h − Fy ′ b) ∗ x d2 y

 EI dx2 = Mxx = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b) + (Fy ′ c − Fy ′ h − Fy ′ b) ∗ x  Integrate both sides and then as follow dy

 EI dx = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b)x + (Fy ′ c − Fy ′ h − Fy ′ b)  EIy = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b)

x2 2

+ (Fy ′ c − Fy ′ h − Fy ′ b)

x2 2 x3 6

+ C1 +

C1x + C2

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 Now consider the boundary condition 

x=0, y=0 then

 EI ∗ 0 = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b)

02 2

+ (Fy ′ c − Fy ′ h − Fy ′ b)

03 6

+

C1 ∗ 0 + C2  C2 = 0 

dy

x=0, dx = 0 then

 EI ∗ 0 = (Mc + Mh + 263.667 ∗ Fy ′ h + 791 ∗ Fy ′ b) ∗ 0 + (Fy ′ c − Fy ′ h − Fy ′ b)

02 2

+

C1  C1 = 0 

x=1393.16, y=0 then

 EI ∗ 0 = (21240.423 + Mh + 263.667 ∗ Fy ′ h + 791 ∗ 46.512) Fy ′ h − 46.512)

1393.162 2

+ (187.728 −

1393.163 6

 970,447.393Mh − 194787877.2Fy ′ h + 1,199,572,376,00 = 0 … … … [2] 

dy

x=1393.16, dx = 0, then

 EI ∗ 0 = (21240.423 + Mh + 263.667 ∗ Fy ′ h + 791 ∗ 46.512) ∗ 1393.16 + (187.728 − Fy ′ h − 46.512)

1393.162 2

 1393.16Mh − 603117.075Fy ′ h + 217889745.1 = 0 … … … [3]  Now find Mh and Fy’h by using simultaneous equation, 

2

Multiply equation 2 by 1393.16 2

 (970,447.393Mh − 194787877.2Fy ′ h + 119,957,237,600 = 0) ∗ 1393.16  1393.16Mh − 279,634.61Fy ′ h + 172,208,845.5 = 0 … … … … … … . [4]  Now subtract equation 4 and equation 2 

1393.16Mh − 279,634.61Fy ′ h + 172,208,845.5 = 0



1393.16Mh − 603,117.075Fy ′ h + 217889745.1 = 0

 Then 

323,482.465Fy ′ h − 45,680,899.6 = 0



323,482.465Fy ′ h = 45,680,899.6

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 Fy ′ h = 141.216kN  From equation 1 

Fy ′ h = Fy ′ a + 141.216kN



141.216kN = Fy ′ a + 141.216kN  Fy ′ a = 0kN

 From equation 3 

1393.16Mh − 603,117.075Fy ′ h + 217,889,745.1 = 0



1393.16Mh = 603,117.075 ∗ 141.216 − 217,889,745.1



1393.16Mh = −132,719,964.2kNmm  Mh = −95,265.414kNmm = 95,265.414kNmm(anticlockwise)

 To find Ma Taking clockwise bending moment at point c are positive 

∑ M@c = 0, Mc + Fy ′ h ∗ 263.667 + Mh + Fy ′ b ∗ 791 − Fy ′ a ∗ 1393.16 − Ma = 0



Ma = 21240.423 + 141.216 ∗ 263.667 − 95,265.414 + 46.512 ∗ 791 − 0 ∗ 1393.16  Ma = 0.000072kNmm



Maximum bending moment is occur at point h  𝐌𝐦𝐚𝐱@𝐡 = 𝟗𝟓𝟐𝟔𝟓. 𝟒𝟏𝟒𝐤𝐍𝐦𝐦

 Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I = σy

 Therefore, σb =

12 MN

 σb = σall = S.F  σb =

b4 −h4

, let S.F=1.5 and σy for mild steel is 280 m2

280MN⁄ 2 m 1.5

= 186.667 m2 = 186.667 N⁄mm2

b 2 b4 −h4

but, M = 95,265.414kNmm and b=h+2*t, h=b-2*t

M∗

MN

12



b

Let, b=20*t, t=20 and h=0.9*b b 2

 186.667 N⁄mm2 =

95,265.414kNmm∗

 186.667 N⁄mm2 =

571,592.484kNmm∗b

b4 −(0.9∗b)4 12

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 1,662,089,224Nmm  b3 = 8,904033.512mm3 UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 b = 𝟐𝟎𝟕. 𝟐𝟔𝟔𝐦𝐦 b

 t = 20 =

207.266mm 20

= 𝟏𝟎. 𝟑𝟔𝟑𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 207.266mm and t ≥ 10.363mm. 

b=250mm,



t=12mm,



h=b-2t=250mm-2*12mm=226mm

 Therefore, checking for the safety:  σb =

b 2 4 b −h4

M∗

=

250mm 2 4 (250mm) −(226mm)4 12

95265.414kNmm∗

12

= 𝟏𝟏𝟎. 𝟏𝟑𝟒 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 110.134Mpa = 𝟐. 𝟓𝟒𝟐 > F. s old = 𝟏. 𝟓, so the design of back hoist body b

is safe due to bending load. 4.4.2.3.

The Back Hoist Body Subjected to Direct Stress

 Direct stress is given by 

σd =

Fx′a A

Where 

Fx ′ a = 102.975𝑘𝑁



A= area of the cross section of the material=b2 − h2 102.975kN

 𝛔𝐝 = (250mm)2 −(226mm)2 = 𝟗. 𝟎𝟏𝟒 𝐍⁄ 𝐦𝐦𝟐  The back hoist body subjected to direct and bending stress o 𝛔𝒄 = σd + σb = 9.014Mpa + 110.134Mpa = 𝟏𝟏𝟗. 𝟏𝟒𝟖𝐌𝐩𝐚 4.4.2.4. 

Back Hoist Body Subjected to Shear Stress

The shear stress on the square hollow section on back hoist body when the boom is extracted will become:

 Shear stress on the square hollow section on back hoist body is  τ=

Vmax A

,

Where

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F

 τ=A=

Fy′c A

| 2010 E.C



Vmax = Fy ′ c = 187.728kN



A=area of the cross section of the material=b2 − h2

=

187.728kN b2 −h2

187.728kN

= 2502 −2262 = 𝟏𝟔. 𝟒𝟑𝟑𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σ𝑐 ± √σ𝑐 2 + 4 ∗ τ2 ) = 2 (119.148Mpa ± √(119.148Mpa)2 + 4 ∗ (9.014Mpa)2 )  𝛔𝟏 = 𝟏𝟏𝟗. 𝟖𝟐𝟔𝐌𝐩𝐚 And 𝛔𝟐 = −𝟎. 𝟔𝟕𝟖𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (119.826Mpa + 0.678Mpa)  τmax = 𝟔𝟎. 𝟐𝟓𝟐𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

max

280Mpa

= 2∗60.252Mpa = 2.324 > 1.5, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (119.826)2 + (−0.678)2 − 2 ∗ 119.826 ∗ −0.678 = ( f.s )2 σy 2

 ( f.s ) = 14,521.214Mpa2 

σy F.s

= 120.504Mpa σy

280Mpa

 F. s new = 120.504Mpa = 120.504Mpa = 2.324 > 1.5, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐡𝐨𝐢𝐬𝐭 𝐛𝐨𝐝𝐲 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.2.5.

Back Hoist Body Subjected to Buckling and Crushing Load

 Therefore we can find the force acting on the back hoist body axially.  axial load acting on the back hoist body 

F𝑎𝑥𝑖𝑎𝑙 = Fx ′ b + Fx ′ c = 237.074kN + 134.099kN 

F𝑎𝑥𝑖𝑎𝑙 = 𝟑𝟕𝟏. 𝟏𝟕𝟑𝐤𝐍

 Hence if the slenderness ratio is less than 80, Euler’s formula for a mild steel column is not valid.

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 Sometimes, the columns whose slenderness ratio is more than 80, are known as long columns, and those whose slenderness ratio is less than 80 are known as short columns. It is thus obvious that the Euler’s formula holds good only for long columns.  Now find slenderness ratio in order to the back hoist body is long or short column. effective length of the back hoist body

𝑙

 slenderness ratio = the least radius of gyration of the section = 𝑘 Table 4.1:2.Standard table of the relation between equivalent length (l) and actual length (L) No

End condition

Relationship

1

Both ends hinged

l=L

2

Both ends fixed

l=2

3

One end fixed and other hinged

l=

L

L

√2

One end fixed and other end free

4

l=2L

 Then from the above table 4.1:2 we select that one end fixed and other end hinged 𝑙=

L √2

and the value of L is 1393.16mm then 𝑙 =

1393.16mm √2

= 𝟗𝟖𝟓. 𝟏𝟏𝟑𝐦𝐦

 The least radius of gyration of the square hollow section (k) is given by 

k = 0.289√h2 + b 2 = 0.289√2502 + 2262 = 𝟗𝟕. 𝟑𝟗𝟔𝐦𝐦

 Then 

l

slenderness ratio = k =

985.113mm 97.396mm

= 𝟏𝟎. 𝟏𝟏𝟒 < 𝟖𝟎, so the back hoist

body is a short column  In column ((hoist) design we concede about the Ewers column theory and Rankine’s columns theory formula. Note: Euler’s formula gives correct results only for very long columns. Though Rankine’s formula is applicable for columns, ranging from very long to short ones, yet it does not give reliable results. So, we use Rankine’s formula because the column is short. 

The column will fall by crushing and the load will be known as crushing load and the load at which the column tends to have lateral displacement or tender to buckler

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called buckling load, critical load or crippling load (Wcr) and the column is said to be have developed on elastic instability. Rankine’s column theory  According to Rankine’s column theory, crippling load (Wcr) as follows: 

1 Wcr

1

1

= WC + WE Wc∗WE

 Wcr = WE+Wc =  Wcr =

Wc 1+

σc∗A

Wc WE

=

σC A∗L2 1+ 2 ∗ cπ ∗E A∗K2

σc∗A σC L2 1+ 2 ∗ 2 cπ ∗E K

=

σc∗A a 𝑐

L K

1+ ∗( )2

Where 

Wcr=crippling load by Rankine’s formula



Wc=ultimate crushing load for column= σc ∗ A



WE=crippling load obtain by Ewer’s formula=

cπ2 ∗E∗A L ( )2 k



I=A*K 2 moment inertia



σc =Crushing yield stress



A=cross section area of the column= 𝐡𝟐 − 𝐛𝟐



a=Ranking constant= π2 ∗E



L=equivalent length of the column



K=least radius of gyration, 𝐊 = 𝟎. 𝟐𝟖𝟗√𝐡𝟐 + 𝐛 𝟐 (for the square hollow section)



C=constant, representing the end condition of the column or end fixity coefficient.

σC

 The following value of C as showing in the table below: Table 4.1:3.Standard value of “c” NO

End condition

coefficient

1

Both end hinged

1

2

Both end fixed

4

3

One end fixed and other hinged

2

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One end fixed and other end free

4 

| 2010 E.C

0.25

From the above table 4.1:3 the value of C is select 2 because of back hoist body is one end fixed and other is hinged.

Table 4.1:4.Value of crushing stress (σc) and Rankine’s constant (a) 𝛔𝐜

NO

Materials

𝛔𝐜 in M pa

a=𝛑𝟐 ∗𝐄

1

Wrought iron

250

1⁄ 9600

2

Cast iron

550

1⁄ 1600

3

Mild steel

320

1⁄ 7500

4

Timber

50

1⁄ 750

 using mild steel read from above table 4.1:4, 

a=1/7500



σC =320Mpa

 crushing load is given by Wc = σc ∗ A =

320N ∗ ((250mm)2 − (226mm)2 ) = 3,655,680𝑁 = 3,655.680𝐤𝐍 mm2 Wc

 F. s new = Fx′a =

3,655.680𝐤𝐍 371.173kN

= 9.849 > 1.5, it is 𝐬𝐚𝐟𝐞!

 crippling load is given by Wcr =

320N/mm2 ∗ ((250mm)2 − (226mm)2 ) 3,655.680𝑘𝑁 = 1 1.00682 2 1+ ∗ (10.114mm) 2 ∗ 7500 Wcr = 𝟑, 𝟔𝟑𝟎. 𝟗𝟏𝟕𝟏𝟒𝟓𝐤𝐍 Wcr

 F. s new = Fx′a =

3630.917145 371.173kN

= 9.782 > 1.5, it is 𝐬𝐚𝐟𝐞!

∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐡𝐨𝐢𝐬𝐭 𝐛𝐨𝐝𝐲 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐫𝐢𝐩𝐩𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐨𝐫 𝐛𝐮𝐜𝐤𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐜𝐫𝐮𝐬𝐡𝐢𝐧𝐠 𝐥𝐨𝐚𝐝.

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4.4.3. Design of the Horizontal Frame and Base Body  Using the geometric method from the free body diagram below when the cylinder is at normal position, the total length of the horizontal frame will become:

Figure 4.1:17.front view of machine  To find m use cosine law  Therefore,

cos θ4=

m L

, m = L ∗ cos θ4 , but L = 1393.16mm and θ4 = 78.9°



m = L ∗ cos θ4 = 1393.16 ∗ cos 78.9° = 𝟐𝟔𝟖. 𝟐𝟏𝟒𝐦𝐦



n = Lb − m = 1250 − 268.214 = 𝟗𝟖𝟏. 𝟕𝟖𝟔𝐦𝐦, which is value from center to center

 But, to find the lengths of both sides of the horizontal frame from the FBD below is:

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| 2010 E.C

Figure 4.1:18.Top view of machine  Therefore from the above FBD the values of “x” and “y” are y

y



tan 45° =



y = 268.214mm ∗ tan 45° = 𝟐𝟔𝟖. 𝟐𝟏𝟒𝐦𝐦



sin 45° = x , x = sin 45° =

m

=

y

268.214mm

y

268.214 sin 45°

= 𝟑𝟕𝟗. 𝟑𝟏𝟐𝐦𝐦

 Hence, the total length of the base body (Ltb ) is equal to the total length of the connecting frame(Lc ) will become: 

Ltb = Lc = 400mm + 2 ∗ y = 400 + 2 ∗ 268.214mm = 𝟗𝟑𝟔. 𝟒𝟐𝟖𝐦𝐦

 And also the total length of the horizontal frame(LT ) using equivalent triangle is:   

m n+m

x

=L

LT = LT =

T

(n+m)∗x m (981.786mm+268.214)∗379.312mm 268.214mm

= 𝟏𝟕𝟔𝟕. 𝟕𝟔𝟖𝐦𝐦

 Therefore the total length of the horizontal frame is 1767.768mm.

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4.4.4. Design of Base Body 4.4.4.1.

Material Selection for Base Body The material selection for base body is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80 m2



Tensile strength (σt) = 480



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN MN m2

MN

Kg

Figure 4.1:19.3d view of base body 4.4.4.2.

Base Body Subjected to Bending Moment

 Parameters that solved from the previous are as shown below:

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Figure 4.1:20.FBD of base body  From horizontal boom we get the following values 

Fx ′ a = 102.975kN



θ4 = 78.9°



Mi =

Fx′ a ∗sin θ4∗Ltb 8

= 11,828.094kNmm(anticlockwise)[see from beam design

formula page 15] 

Mj = −Mi = −11,828.094kNmm = 11,828.094kNmm(clockwise)

First find reaction force and moment  Force along y-axis from the above figure 4.1:17 

+↑ ΣFy = 0



Fyi + Fyj − Fx ′ a ∗ sin θ4 = 0



Fyi + Fyj = Fx ′ a ∗ sin θ4 = 102.975kN ∗ sin 78.9°



Fyi + Fyj = 101.049kN, because the force is at the center, so Fyi = Fyj,



Fyi + Fyi = 2Fyi = 101.049kN  Fyi = Fyj = 𝟓𝟎. 𝟓𝟐𝟓𝐤𝐍

 Force along x-axis from the above figure 4.1:17 

← +ΣFx = 0,



Fxi + Fxj − Fx ′ a ∗ cos θ4 = 0

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

| 2010 E.C

Fxi + Fxj = Fx ′ a ∗ cos θ4 = 102.975kN ∗ cos 78.9° = 𝟏𝟗. 𝟖𝟐𝟓𝐤𝐍 because the force is at the center, so Fxi = Fxj  Fxi = Fxj =

19.825kN 2

= 𝟗. 𝟗𝟏𝟐𝐤𝐍

 Bending moment and shear force  Bending moment and shear force at section x1-x1

Where 0 ≤ x1 ≤

Ltb 2

= 468.214mm

To find bending moment at each point take bending moment at point x1 anti clockwise is zero 

∑ M@x = 0, Mx + Mi − Fyi ∗ x1 = 0 Mx = Fyi ∗ x1 − Mi

 𝐌𝐢@(𝐱𝟏 = 𝟎) = 50.525kN ∗ 0mm − 11,828.094kNmm = −𝟏𝟏, 𝟖𝟐𝟖. 𝟎𝟗𝟒𝐤𝐍𝐦𝐦  𝐌𝐀@(𝐱𝟏 = 𝟒𝟔𝟖. 𝟐𝟏𝟒𝐦𝐦) = 50.525kN ∗ 468.214mm − 11,828.094kNmm = −𝟏𝟏, 𝟖𝟐𝟖. 𝟎𝟗𝟒𝐤𝐍𝐦𝐦 To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, Vx + Fyi = 0 Vx = −Fyi = −50.525kN  Vi@(x1 = 0mm) = VA@(x1 = 468.214mm) = −Fyi = −𝟓𝟎. 𝟓𝟐𝟓𝐤𝐍

 Bending moment and shear force at section x2-x2

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Where 0 ≤ x2 ≤

Ltb 2

| 2010 E.C

= 468.214mm

To find bending moment at each point take bending moment at point x2 clockwise is zero 

∑ M@x2 = 0, Mx + Mj − Fyj ∗ x2 = 0



Mx = Fyj ∗ x2 − Mj



𝐌𝐣@(𝐱𝟐 = 𝟎) = 50.525kN ∗ 0mm − 11,828.094kNmm = −𝟏𝟏, 𝟖𝟐𝟖. 𝟎𝟗𝟒𝐤𝐍𝐦𝐦



𝐌𝐀@(𝐱𝟐 = 𝟒𝟔𝟖. 𝟐𝟏𝟒𝐦𝐦) = 50.525kN ∗ 468.214mm − 11,828.094kNmm = −𝟏𝟏, 𝟖𝟐𝟖. 𝟎𝟗𝟒𝐤𝐍𝐦𝐦

To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, −Vx + Fyj = 0 Vx = Fyj = 50.525kN  Vj@(x2 = 0mm) = VA@(x2 = 468.214mm) = 𝟓𝟎. 𝟓𝟐𝟓𝐤𝐍

 Now maximum bending moment and shear force 

𝐌𝐦𝐚𝐱 = 𝟏𝟏, 𝟖𝟐𝟖. 𝟎𝟗𝟒𝐤𝐍𝐦𝐦 

𝐕𝐦𝐚𝐱 = 𝟓𝟎. 𝟓𝟐𝟓𝐤𝐍

 Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I = σy

 σb = σall = S.F  σb =  Therefore, σb =

b4 −h4 12 MN

, let S.F=1.5 and σy for mild steel is 280 m2

280MN⁄ 2 m 1.5

= 186.667 m2 = 186.667 N⁄mm2

b 2 b4 −h4

but, M = 11,828.094kNmm and b=h+2*t, h=b-2*t

M∗

MN

12



b

Let, b=20*t, t=20 and h=0.9*b

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b 2

 186.667 N⁄mm2 =

11,828.094kNmm∗

 186.667 N⁄mm2 =

70968.564kNmm∗b

b4 −(0.9∗b)4 12

| 2010 E.C

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 206,363,954.6Nmm  b3 = 1,105,519.211mm3  b = 𝟏𝟎𝟑. 𝟒𝐦𝐦 b

 t = 20 =

103.4mm 20

= 𝟓. 𝟏𝟕𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 103.4mm and t ≥ 5.17mm. 

b=120mm,



t=6mm,



h=b-2t=120mm-2*6mm=108mm

 Therefore, checking for the safety:  σb =

b 2 b4 −h4

M∗

=

120mm 2 (120mm)4 −(108mm)4 12

11,828.094kNmm∗

12

= 𝟏𝟏𝟗. 𝟒𝟐𝟒 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 119.424Mpa = 𝟐. 𝟑𝟒𝟓 > F. s old = 𝟏. 𝟓, so the design of base body is safe b

due to bending load. 4.4.4.3.

The Base Body Subjected to Direct Stress

 Direct stress is given by 

σd =

Faxial A

Where Fxi+Fxj

Fx′ a∗cos θ4

19.824kN



Faxial =



A= area of the cross section of the material=b2 − h2

2

=

2

=

2

= 9.912kN

9.912kN

 𝛔𝐝 = (120mm)2 −(108mm)2 = 𝟑. 𝟔𝟐𝟑 𝐍⁄ 𝐦𝐦𝟐  The base body subjected to direct and bending stress o 𝛔𝐜 = σd + σb = 3.623Mpa + 119.424Mpa = 𝟏𝟐𝟑. 𝟎𝟒𝟕𝐌𝐩𝐚

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4.4.4.4. 

| 2010 E.C

Base Body Subjected to Shear Stress

The shear stress on the square hollow section on base body.

 Shear stress on the square hollow section on the base body is  τ=

Vmax A

,

Where

F

 τ=A=

Fy′A A



Vmax = Fyj = 50.525kN



A=area of the cross section of the material=b2 − h2

=

50.525kN b2 −h2

50.525kN

= 1202 −1082 = 𝟏𝟖. 𝟒𝟔𝟕𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σ𝑐 ± √σ𝑐 2 + 4 ∗ τ2 ) = 2 (123.047Mpa ± √(123.047Mpa)2 + 4 ∗ (18.467Mpa)2 )  𝛔𝟏 = 𝟏𝟐𝟓. 𝟕𝟓𝟗𝐌𝐩𝐚 And 𝛔𝟐 = −𝟐. 𝟕𝟏𝟐𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (125.759Mpa + 2.712Mpa)  τmax = 𝟔𝟒. 𝟐𝟑𝟓𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

max

280Mpa

= 2∗64.235Mpa = 2.18 > 1.5, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (125.759)2 + (−2.712)2 − 2 ∗ 125.759 ∗ −2.712 = ( f.s )2 σy 2

 ( f.s ) = 16,504.798Mpa2 

σy F.s

= 128.471Mpa σy

280Mpa

 F. s new = 128.471Mpa = 128.471Mpa = 2.18 > 1.5, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐬𝐞 𝐛𝐨𝐝𝐲 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.4.5.

Base Body Subjected to Buckling and Crushing Load

 Therefore we can find the force acting on the base body axially. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 axial load acting on the base body 

F𝑎𝑥𝑖𝑎𝑙 = Fxi + Fxj = Fx ′ a ∗ sin 𝜃4 = 9.912kN + 9.912kN 

F𝑎𝑥𝑖𝑎𝑙 = 𝟏𝟗. 𝟖𝟐𝟒𝐤𝐍

 Hence if the slenderness ratio is less than 80, Euler’s formula for a mild steel column is not valid.  Now find slenderness ratio in order to the back hoist body is long or short column. effective length of the back hoist body

𝑙

 slenderness ratio = the least radius of gyration of the section = 𝑘 Then from the above table 4.1:2 we select that both end fixed 936.428mm then l =

936.428mm 2

L

l = 2 and the value of L is

= 𝟒𝟔𝟖. 𝟐𝟏𝟒𝐦𝐦

 The least radius of gyration of the square hollow section (k) is given by 

k = 0.289√h2 + b 2 = 0.289√1202 + 1082 = 𝟒𝟔. 𝟔𝟓𝟕𝐦𝐦

 Then l

 slenderness ratio = k =

468.214mm 46.657mm

= 𝟏𝟎. 𝟎𝟑𝟓 < 𝟖𝟎, so the back hoist body is a

short column Note: Euler’s formula gives correct results only for very long columns. Though Rankine’s formula is applicable for columns, ranging from very long to short ones, yet it does not give reliable results. So, we use Rankine’s formula because the column is short. Rankine’s column theory  According to Rankine’s column theory, crippling load (Wcr) as follows: 

1 Wcr

1

1

= WC + WE Wc∗WE

 Wcr = WE+Wc =  Wcr = 

Wc 1+

σc∗A σC A∗L2 1+ 2 ∗ cπ ∗E A∗K2

Wc WE

=

σc∗A σC L2 1+ 2 ∗ 2 cπ ∗E K

=

σc∗A a 𝑐

L K

1+ ∗( )2

From the above table 4.1:3 the value of C is select 4 because of base body is both end fixed.  using mild steel read from above table 4.1:4, 

a=1/7500



σc = 320Mpa

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 crushing load is given by Wc = σc ∗ A =

320N ∗ ((120mm)2 − (108mm)2 ) = 875,520N = 𝟖𝟕𝟓. 𝟓𝟐𝐤𝐍 mm2

 Crushing load (Wc = 𝟖𝟕𝟓. 𝟓𝟐𝐤𝐍) >>> axial load (Faxial = 𝟏𝟗. 𝟖𝟐𝟒𝐤𝐍), it is safe!  crippling load is given by 320N/mm2 ∗ ((120mm)2 − (108mm)2 ) 875,520N Wcr = = 1 1.00336 1+ ∗ (10.035)2 4 ∗ 7500 Wcr = 872.591𝐤𝐍  Crippling load (Wcr = 𝟖𝟕𝟐. 𝟓𝟗𝟏𝐤𝐍) >>> axial load (Faxial = 𝟏𝟗. 𝟖𝟐𝟒𝐤𝐍), it is safe! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐬𝐞 𝐛𝐨𝐝𝐲 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐫𝐢𝐩𝐩𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐨𝐫 𝐛𝐮𝐜𝐤𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐜𝐫𝐮𝐬𝐡𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 4.4.4.6.

Deflection of Base Body

 Maximum deflection is given by when beam fixed at both ends-concentrated load at the center,  ∆max (at the center) =

Fx′ a∗sin θ4∗Ltb 3 192∗EI

 Where b4 −h4

1204 −1084



I=



Ltb = 936.428mm



Fx ′ a ∗ sin θ4 = 102.975kN ∗ sin 78.9° = 101.049kN = 101049N



E = 207 ∗ 103 N/mm2



EI = E ∗ I = 207 ∗ mm2 ∗ 5,942,592mm4 = 1230116544000Nmm2

12

=

12

= 5,942,592mm4

103 N

 Now ∆max (at the center) =

101049N ∗ 936.428mm3 192 ∗ 1230116544000Nmm2

 ∆𝐦𝐚𝐱 = 𝟎. 𝟑𝟓𝟏𝐦𝐦 ∴ As compared to the length 936.428mm, the deflection of 0.351mm is very negligible. Therefore, the design is safe!

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4.4.5. Design of Back Hoist Support 4.4.5.1.

Material Selection for Back Hoist Support The material selection for back hoist support is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80 m2



Tensile strength (σt) = 480



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN MN m2

MN

Kg

Figure 4.1:21.3d view of back hoist support 4.4.5.2.

Back Hoist Support Subjected to Bending Moment

 Parameters that solved from the previous are as shown below:

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Figure 4.1:22.FBD of back hoist support

Figure 4.1:23.Side view of back hoist support Now find , Leh, θh , Lkk  From horizontal frame 

m = 268.214mm



Xh = 217.452mm



Lh = 1108.364mm

 Lkk = 400 + 2 ∗ (m − Xh) ∗ tan 45° = 400 + 2 ∗ (268.214 − 217.452) ∗ tan 45°  Lkk = 501.524mm  Leh = √(

Lkk 2 ) 2

+ Lh2 = √(

501.524 2 ) 2

+ (1108.364)2

 Leh = 1136.377mm UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 tan θh =

Lh Lkk ( ) 2

Lh

, θh = tan−1[

| 2010 E.C

1108.364

Lkk ( ) 2

] = tan−1 [

(

501.524 ) 2

]

 θh = tan−1 [4.41998] = 77.25°  From back hoist body we get the following values 

Fy′h = 141.216kN



Mh = 95,265.414kNmm

First find reaction force and moment  Force along y-axis from the above figure 4.1:19 

+↑ ΣFy = 0



−Fyk +



Fyk =

Fy′ h 2

Fy′ h 2

∗ cos θh = 0 141.216kN

∗ cos θh =

2

∗ cos 77.25° = 15.583kN

 Force along x-axis from the above figure 4.1:19 

+→ ΣFx = 0,



−Fxk +



Fxk =

Fy′ h 2

Fy′ h 2

∗ sin θh = 0

∗ sin θh =

141.216kN 2

∗ sin 77.25° = 68.867kN

 To find Mk use clockwise moment at point k equal to zero 

∑ M@k = 0,



Mk =

Mh 2

−

Mh 2

Fy′ h 2

− Mk −

Fy′ h 2

∗ cos θh ∗ Leh = 0

∗ cos θh ∗ Leh =

95,265.414 2

−

141.216 2

∗ 1136.377 ∗

cos 77.25° Mk = 29,924.544kNmm  Now maximum bending moment and shear force get at point ‘h’ 

𝐌𝐦𝐚𝐱 == 

Mh 2

=

95,265.414kNmm 2

= 𝟒𝟕, 𝟔𝟑𝟐. 𝟕𝟎𝟕𝐤𝐍𝐦𝐦

𝐕𝐦𝐚𝐱 = 𝟏𝟓. 𝟓𝟖𝟑𝐤𝐍

 Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I = σy

 σb = σall = S.F  σb =

280MN⁄ 2 m 1.5

b4 −h4 12 MN

, let S.F=1.5 and σy for mild steel is 280 m2 MN

= 186.667 m2 = 186.667 N⁄mm2

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 Therefore, σb = 

b 2 b4 −h4 12

M∗

| 2010 E.C

but, M = 47,632.707kNmm and b=h+2*t, h=b-2*t b

Let, b=20*t, t=20 and h=0.9*b b 2

 186.667 N⁄mm2 =

47,632.707kNmm ∗

 186.667 N⁄mm2 =

285,796.242kNmm∗b

b4 −(0.9∗b)4 12

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 831,044,611.8Nmm  b3 = 4,452,016.756mm3  b = 𝟏𝟔𝟒. 𝟓𝟎𝟕𝐦𝐦 b

 t = 20 =

164.507mm 20

= 𝟖. 𝟐𝟐𝟓𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 164.507mm and t ≥ 8.225mm. 

b=175mm,



t=9mm,



h=b-2t=175mm-2*9mm=157mm

 Therefore, checking for the safety:  σb =

b 2 b4 −h4

M∗

=

12

175mm 2 (175mm)4 −(157mm)4 12

47,632.707kNmm ∗

= 𝟏𝟓𝟏. 𝟒𝟏𝟑 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 151.413Mpa = 𝟏. 𝟖𝟒𝟗 > F. s old = 𝟏. 𝟓, so the design of back hoist b

support is safe due to bending load. 4.4.5.3.

The Back Hoist Support Subjected to Direct Stress

 Direct stress is given by 

σd =

Faxial A

Where Fy′ h



Faxial =



A= area of the cross section of the material=b2 − h2

2

∗ sin θh = 68.867kN

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| 2010 E.C

68.867kN

 𝛔𝐝 = (175mm)2 −(157mm)2 = 𝟏𝟏. 𝟓𝟐𝟒 𝐍⁄ 𝐦𝐦𝟐  The back hoist support subjected to direct and bending stress o 𝛔𝒄 = σd + σb = 11.524Mpa + 151.413Mpa = 𝟏𝟔𝟐. 𝟗𝟑𝟕𝐌𝐩𝐚 4.4.5.4. 

The Back Hoist Support Subjected to Shear Stress

The shear stress on the square hollow section on back hoist support.

 Shear stress on the square hollow section on the back hoist support is  τ=

Vmax A

,

Where

 τ=

Vmax A

=

Fy′ h



Vmax =



A=area of the cross section of the material=b2 − h2

15.583kN b2 −h2

2

∗ cos θh = 15.583kN

15.583kN

= 1752 −1572 = 𝟐. 𝟔𝟎𝟖𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σ𝑐 ± √σ𝑐 2 + 4 ∗ τ2 ) = 2 (162.937Mpa ± √(162.937Mpa)2 + 4 ∗ (2.608Mpa)2 )  𝛔𝟏 = 𝟏𝟔𝟐. 𝟗𝟕𝟗𝐌𝐩𝐚 And 𝛔𝟐 = −𝟎. 𝟎𝟒𝟐𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (162.979Mpa + 0.042Mpa)  τmax = 𝟖𝟏. 𝟓𝟏𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

max

280Mpa

= 2∗81.51Mpa = 1.718 > 1.5, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (162.979)2 + (−0.042)2 − 2 ∗ 162.979 ∗ −0.042 = ( f.s )2 σy 2

 ( f.s ) = 26,575.846Mpa 

σy F.s

= 163.021Mpa σy

280Mpa

 F. s new = 163.021Mpa = 163.021Mpa = 1.718 > 1.5, it is 𝐬𝐚𝐟𝐞! UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐡𝐨𝐢𝐬𝐭 𝐬𝐮𝐩𝐩𝐨𝐫𝐭 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.5.5.

Back Hoist Support Subjected To Buckling and Crushing Load

 Therefore we can find the force acting on the back hoist support axially.  axial load acting on the back hoist support 

F𝑎𝑥𝑖𝑎𝑙 =

Fy′ h 2

∗ sin θh = 𝟔𝟖. 𝟖𝟔𝟕𝐤𝐍

 Hence if the slenderness ratio is less than 80, Euler’s formula for a mild steel column is not valid.  Now find slenderness ratio in order to the back hoist body is long or short column. effective length of the back hoist body

l

 slenderness ratio = the least radius of gyration of the section = k  Then from the above table 4.1:2 we select that one end fixed and the other free 𝑙 = 2L and the value of L is 1136.377mm then 𝑙 = 2 ∗ 1136.377mm = 𝟐, 𝟐𝟕𝟐. 𝟕𝟓𝟒𝐦𝐦  The least radius of gyration of the square hollow section (k) is given by 

k = 0.289√h2 + b 2 = 0.289√1752 + 1572 = 𝟔𝟕. 𝟗𝟒𝟓𝐦𝐦

Then l

slenderness ratio = k =

2,272.754mm 67.945mm

= 𝟑𝟑. 𝟒𝟓 < 𝟖𝟎, so the back hoist support is a short

column Note: Euler’s formula gives correct results only for very long columns. Though Rankine’s formula is applicable for columns, ranging from very long to short ones, yet it does not give reliable results. So, we use Rankine’s formula because the column is short. Rankine’s column theory  According to Rankine’s column theory, crippling load (Wcr) as follows: 

1 Wcr

1

1

= WC + WE Wc∗WE

 Wcr = WE+Wc =  Wcr =

Wc 1+

σc∗A σC A∗L2 1+ 2 ∗ cπ ∗E A∗K2

Wc WE

=

σc∗A σC L2 1+ 2 ∗ 2 cπ ∗E K

=

σc∗A a 𝑐

L K

1+ ∗( )2

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

| 2010 E.C

From the above table 4.1:3 the value of C is select 0.25 because of back hoist support is one end fixed and other is free.  using mild steel read from above table 4.1:4, 

a=1/7500



σC =320Mpa

 crushing load is given by Wc = σc ∗ A =

320N ∗ ((175mm)2 − (157mm)2 ) = 𝟏, 𝟗𝟏𝟐. 𝟑𝟐𝐤𝐍 mm2

 Crushing load (Wc = 𝟏, 𝟗𝟏𝟐. 𝟑𝟐𝐤𝐍) >>> axial load (Faxial = 𝟔𝟖. 𝟖𝟔𝟕𝐤𝐍), it is safe!  crippling load is given by Wcr =

320N/mm2 ∗ ((175mm)2 − (157mm)2 ) 1,912.32kN = 1 1.597 2 1+ ∗ (33.45) 0.25 ∗ 7500 Wcr = 𝟏, 𝟏𝟗𝟕. 𝟔𝟑𝟒𝐤𝐍

 Buckling load (Wcr = 𝟏, 𝟏𝟗𝟕. 𝟔𝟑𝟒𝐤𝐍) >>> axial load (Faxial = 𝟔𝟖. 𝟖𝟔𝟕𝐤𝐍), it is safe! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐡𝐨𝐢𝐬𝐭 𝐬𝐮𝐩𝐩𝐨𝐫𝐭 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐫𝐢𝐩𝐩𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐨𝐫 𝐛𝐮𝐜𝐤𝐥𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐜𝐫𝐮𝐬𝐡𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 4.4.6. Design of Horizontal Frame 4.4.6.1.

Material Selection for Horizontal Frame The material selection for horizontal frame is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80



Tensile strength (σt) = 480 m2



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN m2 MN

MN

Kg

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| 2010 E.C

Figure 4.1:24.3d view of horizontal frame 4.4.6.2.

Horizontal Frame Subjected to Bending Moment

Bending moment Along x-y plane  And then the free body diagram of the horizontal frame with its forces:

Figure 4.1:25.FBD of horizontal frame  From the above we gate 

Mk = 29,924.544kNmm



Mi = 11,828.094kNmm

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

Fyi = 50.525kN



Fyk = 15.583kN



Fxk = 68.867kN



θh = 77.25°

| 2010 E.C

 Now find Fy ′ k = Fyk ∗ cos θh + Fxk ∗ sin θh = 15.583 ∗ cos 77.25° + 68.867 ∗ sin 77.25°  Fy ′ k = 70.608kN  First find reaction force and moment  Force along y-axis from the above figure 4.1-22 

+↑ ΣFy = 0



Fy ′ k + Fyl − Fyi − Fyo = 0



Fyo = Fyl + Fy ′ k − Fyi = Fyl + 70.608kN − 50.525kN

 Fyo = Fyl + 20.083kN………………….[1]  Sum of bending moment at point ‘0’ clockwise is equal to zero  ∑ M@o = 0,−Fy ′ k ∗ 71.788 + Mk + Mi + Fyi ∗ 379.312 − Fyl ∗ 1767.768 = 0  −70.608 ∗ 71.788 + 29,924.544 + 11,828.094 + 50.525 ∗ 379.312 − Fyl ∗ 1767.768 = 0 

Fyl ∗ 1767.768mm = 55,649.431kNmm  Fyl = 31.48kN

 Then Fyo = Fyl + 20.083kN = 31.48kN + 20.083kN = 51.563kN  Bending moment at section x1-x1  ∑ M@x1 = 0, , Fyo ∗ x1 + Mx1 = 0 

Mx1 = −Fyo ∗ x1  M@x1 = 0 = M@o = −51.563 ∗ 0 = 0kNmm  M@x1=71.788 = M@k = −51.563 ∗ 71.788 = −3,701.605kNmm

 Bending moment at section x2-x2  ∑ M@x2 = 0, , Fyo ∗ x2 − Fy ′ k ∗ (x2 − 71.788) − Mk + Mx2 = 0 

Mx2 = −Fyo ∗ x2 + Fy ′ k ∗ (x2 − 71.788) + Mk

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

| 2010 E.C

M@x2=71.788 = M@k = −51.563 ∗ 71.788 + 70.608 ∗ (71.788 − 71.788) + 29,924.544

 M@k = 26,222.939kNmm 

M@x2=379.312 = M@i = −51.563 ∗ 379.312 + 70.608 ∗ (379.312 − 71.788) + 29,924.544

 M@i = 32,079.734kNmm  Bending moment at section x3-x3  ∑ M@x3 = 0, , Fyl ∗ x3 − Mx3 = 0 

Mx3 = Fyl ∗ x3

 M@x3 = 0 = M@l = 31.48 ∗ 0 = 0kNmm  M@x1=1388.456 = M@i = 31.48 ∗ 1388.456 = 43,708.595kNmm There fore 

Mxymax@k = 𝟐𝟗, 𝟗𝟐𝟒. 𝟓𝟒𝟒𝐤𝐍𝐦𝐦



Mxymax@i = 𝟒𝟑, 𝟕𝟎𝟖. 𝟓𝟗𝟓𝐤𝐍𝐦𝐦

Bending moment Along x-z plane

 From the above we gate 

Fxi = 9.912kN



Fyk = 15.583kN



Fxk = 68.867kN



θh = 77.25°

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| 2010 E.C

 Now find Fx ′ k = Fyk ∗ sin θh − Fxk ∗ cos θh = 15.583 ∗ sin 77.25° − 68.867 ∗ cos 77.25°  Fx ′ k = −0.0000109 ≈ 0 First find reaction force and moment  Force along y-axis from the above figure 4.1:22 

+↑ ΣFz = 0



Fzo − Fxi = 0



Fzo = Fxi

 Fzo = 9.912kN  Bending moment at section x1-x1  ∑ M@x1 = 0, , − Fz0 ∗ x1 + Mx1 = 0 

Mx1 = Fzo ∗ x1



M@x1=0 = M@o = 9.912kN ∗ 0mm = 0kNmm

 M@x1=71.788 = M@k = 9.912kN ∗ 71.788mm = 711.563kNmm  M@x1=379.312 = M@i = 9.912kN ∗ 379.312mm = 3,759.741kNmm There fore 

Mxzmax@k = 𝟕𝟏𝟏. 𝟓𝟔𝟑𝐤𝐍𝐦𝐦



Mxzmax@i = 𝟑, 𝟕𝟓𝟗. 𝟕𝟒𝟏𝐤𝐍𝐦𝐦

Resultant moment at each point  M@k = √Mxy@k 2 + Mxz@k 2 = √29,924.5442 + 711.5632 = 29,933.003kNmm  M@i = √Mxy@i 2 + Mxz@i 2 = √43,708.5952 + 3,759.7412 = 43,870kNmm  Vmax@o = √Vy@0 2 + Vz@0 2 = √51.5632 + 9.9122 = 𝟓𝟐. 𝟓𝟎𝟕𝐤𝐍  Now maximum bending moment is take place at point ‘k’  Mmax = 𝟒𝟑, 𝟖𝟕𝟎𝐤𝐍𝐦𝐦  Vmax = 𝟓𝟐. 𝟓𝟎𝟕𝐤𝐍  Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I =

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σy

MN

 σb = σall = S.F  σb =  Therefore, σb = 

| 2010 E.C

, let S.F=1.5 and σy for mild steel is 280 m2

280MN⁄ 2 m 1.5

= 186.667 m2 = 186.667 N⁄mm2

b 2 b4 −h4 12

but, M = 43,870kNmm and b=h+2*t, h=b-2*t

M∗

MN

b

Let, b=20*t, t=20 and h=0.9*b b 2

 186.667 N⁄mm2 =

43,870kNmm∗

 186.667 N⁄mm2 =

263,220kNmm∗b

b4 −(0.9∗b)4 12

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 765,396,917.7Nmm  b3 = 4,100,333.309mm3  b = 𝟏𝟔𝟎𝐦𝐦 b

 t = 20 =

160mm 20

= 𝟖. 𝟎𝟎𝟑𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 160mm and t ≥ 8.003mm. 

b=175mm,



t=9mm,



h=b-2t=175mm-2*9mm=157mm

 Therefore, checking for the safety:  σb =

b 2 b4 −h4

M∗

=

175mm 2 (175mm)4 −(157mm)4 12

43,870kNmm∗

12

= 𝟏𝟑𝟗. 𝟒𝟓𝟐 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 139.452Mpa = 𝟐. 𝟎𝟎𝟖 > F. s old = 𝟏. 𝟓, so the design of horizontal b

frame is safe due to bending load. 4.4.6.3. 

Horizontal Frame Subjected to Shear Stress

The shear stress on the square hollow section on horizontal frame.

 Shear stress on the square hollow section on the horizontal frame is  τ=

Vmax A

,

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Where

 τ=

Vmax A

=



Vmax = 52.507kN



A=area of the cross section of the material=b2 − h2

52.507kN b2 −h2

52.507kN

= 1752 −1572 = 𝟖. 𝟕𝟖𝟔𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σb ± √σb 2 + 4 ∗ τ2 ) = 2 (139.452Mpa ± √(139.452Mpa)2 + 4 ∗ (8.786Mpa)2 )  𝛔𝟏 = 𝟏𝟒𝟎. 𝟎𝟎𝟒𝐌𝐩𝐚 And 𝛔𝟐 = −𝟎. 𝟐𝟕𝟔𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (140.004Mpa + 0.276Mpa)  τmax = 𝟕𝟎. 𝟏𝟒𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

280Mpa

max

= 2∗70.14Mpa = 1.996 > 1.5, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (140.004pa)2 + (−0.276Mpa)2 − 2 ∗ 140.004Mpa ∗ −0.276Mpa = ( f.s )2 σy 2

 ( f.s ) = 19,678.478Mpa2 

σy F.s

= 140.28Mpa σy

280Mpa

 F. s new = 140.28Mpa = 140.28Mpa = 1.996 > 1.5, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐡𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 𝐟𝐫𝐚𝐦𝐞 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.7. Design of Connecting Frame 4.4.7.1.

Material Selection for Connecting Frame The material selection for connecting frame is low carbon (mild steel) with mechanical and physical property as follows: GN



Young’s modulus of the elasticity (E) =207m2



Shear modules (G) =80 m2

MN

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MN



Tensile strength (σt) = 480 m2



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3



The selected hollow section is square steel.

MN

Kg

Figure 4.1:26.3d view of connecting frame 4.4.7.2.

Connecting Frame Subjected to Bending Moment

 Parameters that solved from the previous are as shown below:

Figure 4.1:27.FBD of base body

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 From horizontal frame we get the following values 

Fyo = 51.563kN



Fzo = 7.785kN

First find reaction force and moment  Force along y-axis from the above figure 4.1:24 

+↑ ΣFy = 0



Fym − Fyo − Fyo + Fym = 0



2Fym = 2Fy0  Fym = Fyo = 𝟓𝟏. 𝟓𝟔𝟑𝐤𝐍

 Bending moment and shear force  Bending moment and shear force at section x1-x1

Where 0 ≤ x1 ≤ 268.214mm To find bending moment at each point take bending moment at point x1 anti clockwise is zero 

∑ M@x1 = 0, Mx − Fym ∗ x1 = 0 Mx = Fym ∗ x1

 𝐌𝐦@(𝐱𝟏=𝟎) = 51.563kN ∗ 0mm = 𝟎𝐤𝐍𝐦𝐦  𝐌𝐨@(𝐱𝟏=𝟐𝟔𝟖.𝟐𝟏𝟒𝐦𝐦) = 51.563kN ∗ 268.214mm = 𝟏𝟑, 𝟖𝟐𝟗. 𝟗𝟏𝟖𝐤𝐍𝐦𝐦 To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, Vx + Fym = 0 Vx = −Fym = −51.563kN  Vm@(x1=0mm) = Vo@(x1=468.214mm) = −Fym = −𝟓𝟏. 𝟓𝟔𝟑𝐤𝐍

 Bending moment and shear force at section x2-x2

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Where 268.214mm ≤ x2 ≤ 668.214mm To find bending moment at each point take bending moment at point x2 anticlockwise is zero 

∑ M@x2 = 0, Mx − Fym ∗ x2 + 51.563kN ∗ (x2 − 268.214) = 0



Mx = Fym ∗ x2 − 51.563kN ∗ (x2 − 268.214mm)



𝐌𝐨@(x2=268.214mm) = 51.563kN ∗ 268.214mm − 51.563kN ∗ (268.214mm − 268.214mm) = 𝟏𝟑, 𝟖𝟐𝟗. 𝟗𝟏𝟖𝐤𝐍𝐦𝐦



𝐌𝐨@(𝐱𝟐=𝟔𝟔𝟖.𝟐𝟏𝟒𝐦𝐦) = 51.563kN ∗ 668.214mm − 51.563kN ∗ (668.214mm − 268.214mm) = 𝟏𝟑, 𝟖𝟐𝟗. 𝟗𝟏𝟖𝐤𝐍𝐦𝐦

To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, Vx + Fym − 51.563kN = 0 Vx = Fym − 51.563kN = 51.563kN − 51.563kN = 0kN  Vo@(x2=268.214mm) = Vo@(x2=668.214mm) = 𝟎𝐤𝐍  Bending moment and shear force at section x3-x3

Where 0mm ≤ x3 ≤ 268.214mm To find bending moment at each point take bending moment at point x2 anticlockwise is zero 

∑ M@x3 = 0, Mx − Fym ∗ x3 = 0

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

Mx = Fym ∗ x2



𝐌𝐦@(x2=0mm) = 51.563kN ∗ 0mm = 𝟎𝐤𝐍𝐦𝐦



𝐌𝐨@(x2=268.214mm) = 51.563kN ∗ 268.214mm = 𝟏𝟑, 𝟖𝟐𝟗. 𝟗𝟏𝟖𝐤𝐍𝐦𝐦

To find shear force at each point take sum of force along y-axis is zero 

+↑ Σfy = 0, −Vx + Fym = 0 Vx = Fym = 51.563kN Vm@(x2=0mm) = Vo@(x2=268.214mm) = 𝟓𝟏. 𝟓𝟔𝟑𝐤𝐍

 Now maximum bending moment and shear force 

𝐌𝐦𝐚𝐱 = 𝟏𝟑, 𝟖𝟐𝟗. 𝟗𝟏𝟖𝐤𝐍𝐦𝐦 

𝐕𝐦𝐚𝐱 = 𝟓𝟏. 𝟓𝟔𝟑𝐤𝐍

 Hence, 𝛔𝐛 =

𝐌∗𝐲 𝐈

b

where, σb =bending stress, y = 2 and I = σy

 Therefore, σb =

12 MN

 σb = σall = S.F  σb =

b4 −h4

, let S.F=1.5 and σy for mild steel is 280 m2

280MN⁄ 2 m 1.5

= 186.667 m2 = 186.667 N⁄mm2

b 2 b4 −h4

but, M = 13,829.918kNmm and b=h+2*t, h=b-2*t

M∗

MN

12



b

Let, b=20*t, t=20 and h=0.9*b b 2

 186.667 N⁄mm2 =

13,829.918kNmm∗

 186.667 N⁄mm2 =

82,979.508kNmm∗b

b4 −(0.9∗b)4 12

0.3439∗b4

 186.667 N⁄mm2 ∗ b3 = 241,289,642.3Nmm  b3 = 1,292,620.776mm3  b = 𝟏𝟎𝟖. 𝟗𝟑𝟐𝐦𝐦 b

 t = 20 =

108.932mm 20

= 𝟓. 𝟒𝟒𝟕𝐦𝐦

 But this value of “b” and “t” is not required in standard table, so we take b and t from standard table by using the value of b ≥ 108.932mm and t ≥ 5.447mm. 

b=175mm,



t=6mm,

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

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h=b-2t=175mm-2*6mm=163mm

 Therefore, checking for the safety:  σb =

b 2 b4 −h4 12

M∗

=

175mm 2 (175mm)4 −(163mm)4 12

13,829.918kNmm∗

= 𝟔𝟐. 𝟓𝟗𝟖 𝐍⁄ 𝐦𝐦𝟐

 Now find the new factor of safety (F.s new) σy

280Mpa

 F. s new = σ = 62.598Mpa = 𝟒. 𝟒𝟕𝟑 > F. s old = 𝟏. 𝟓, so the design of connecting frame b

is safe due to bending load. 4.4.7.3. 

Connecting Frame Subjected to Shear Stress

The shear stress on the square hollow section on connecting frame.

 Shear stress on the square hollow section on the connecting frame is  τ=

Vmax A

,

Where

F

 τ=A=

Fy′A A



Vmax = Fym = 51.563kN



A=area of the cross section of the material=b2 − h2

=

50.525kN b2 −h2

51.563kN

= 1752 −1632 = 𝟏𝟐. 𝟕𝟏𝟑𝐌𝐩𝐚

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σ𝑏 ± √σ𝑏 2 + 4 ∗ τ2 ) = 2 (62.598Mpa ± √(62.598Mpa)2 + 4 ∗ (12.713Mpa)2 )  𝛔𝟏 = 𝟔𝟓. 𝟎𝟖𝟐𝐌𝐩𝐚 And 𝛔𝟐 = −𝟐. 𝟒𝟖𝟒𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (65.082Mpa + 2.484Mpa)  τmax = 𝟑𝟑. 𝟕𝟖𝟑𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

max

280Mpa

= 2∗33.783Mpa = 4.144 > 1.5, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( f.s )2 σy

 (65.082)2 + (−2.484)2 − 2 ∗ 65.082 ∗ −2.484 = ( f.s )2 σy 2

 ( f.s ) = 4,565.164Mpa2 UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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

σy F.s

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= 67.566Mpa σy

280Mpa

 F. s new = 67.566Mpa = 67.566Mpa = 4.144 > 1.5, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐜𝐨𝐧𝐧𝐞𝐜𝐭𝐢𝐧𝐠 𝐟𝐫𝐚𝐦𝐞 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. 4.4.8. Design of Hook 4.4.8.1.

Material Selection for Hook The material selection for connecting frame is low carbon (mild steel) with mechanical and physical property as follows: 

Young’s modulus of the elasticity (E) =207



Shear modules (G) =80 m2



Tensile strength (σt) = 480 m2



Yield strength (σy) = 280 m2



Density (ρ) = 7800 m3

GN m2

MN MN

MN

Kg

Figure 4.1:28.3d view of hook

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4.4.8.2.

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Bending Moment Subjected to the Hook  Selection of hook

 All Crosby 320 Eye Hoist Hooks incorporate the following features: • Designed with a 5:1 Design Factor. • Every Crosby Eye Hook has a pre-drilled cam which can be equipped with a latch. Even years after purchase of the original hook, latch assemblies can be added. • Eye hooks are load rated. • Available in carbon steel and alloy steel.

Table 4.1:5: Standard Dimensions of Crane Hook [13]

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 By using work load limit (3ton) from the above standard table we have A=200.66mm

J=37.338mm

B=74.676mm

K=33.274mm

C=146.558mm

M=28.702mm

D=122.174mm

O=34.544mm

F=50.8mm

P=89.154mm

G=36.576mm

Q=39.624mm

H=41.402mm

T=38.862mm

 Load= total mass*gravity*service factor  W = 3000kg ∗ 9.81m/s2 ∗ 1.25 = 36.787kN

Figure 4.1:29.FBD hook  From the above figure 4.1:26 we have F

 Ri = 2 =

50.8mm 2

= 25.4mm

F

 R o = 2 + H = R i + H = 25.4mm + 41.402 = 66.802mm,  h = H = 41.402mm  b = T = 38.862mm  Then the area of the cross section X-X is  A = b ∗ h = 38.862 ∗ 41.402mm = 1,608.965mm2 UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 Radius of curvature of the neutral axis,  Rn =

h R log𝑒 ( o )

, for rectangular cross section

h R log𝑒 ( o )

=

Ri

 Rn =

41.402mm 66.802mm ) 25.4mm

log𝑒(

Ri

=

41.402𝑚𝑚 0.967

= 42.815𝑚𝑚

 Radius of curvature of the centroid axis, h

 R = R i + 2 , for rectangular cross section h

41.402mm

2

2

 R = R i + = 25.4mm +

= 46.101𝑚𝑚

 The distance between the centroid axis and neutral axis,  e = R − R n = 46.101mm − 42.815mm = 3.286mm  Distance between the load and the centroid axis,  x = R = 46.101mm  Bending moment about the centroid axis,  M = W ∗ x = 36.787kN ∗ 46.101mm = 1,695.917MNmm  Distance from the neutral axis to the inside fiber,  yi = R n − R i = 42.815mm − 25.4mm = 17.415mm  Distance from the neutral axis to outside fiber,  yo = R o − R n = 66.802mm − 42.815mm = 23.987mm  We know that direct tensile stress at section X-X,  σdt =

W A

36.787kN

= 1,608.965mm2 = 22.864Mpa

 We know that maximum bending stress at the inside fiber, M∗y

1,695.917kNmm∗17.415mm

 σbi = A∗e∗Ri = 1,608.965mm2 ∗3.286mm∗25.4mm i

 σbi = 219.928Mpa  We know that maximum bending stress at the outside fiber, M∗y

1,695.917kNmm∗23.987mm

 σbo = A∗e∗Ro = 1,608.965mm2 ∗3.286mm∗66.802mm o

 σbo = 115.18Mpa  Resultant stress at the inside fibre  σRi = σdt + σbi = 22.864Mpa + 219.928Mpa  σRi = 𝟐𝟒𝟐. 𝟕𝟗𝟐𝐌𝐩𝐚(𝐭𝐞𝐧𝐬𝐢𝐥𝐞) UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 Resultant stress at the outside fiber  σRo = σdt − σbo = 22.864Mpa − 115.18Mpa  σRo = −92.316Mpa = 𝟗𝟐. 𝟏𝟑𝟔𝐌𝐩𝐚(𝐜𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐯𝐞)  We know that Shear stress at section X-X W

W

 τ = TANGENTIAL CROSS SECTIONAL AREA = A

𝑡

Where 

W=36.787kN



A𝑡 = T ∗ K = b ∗ K = 38.862mm ∗ 33.274mm = 1,293.094mm2

Then W

36.787kN

 τ = A = 1,293.094mm2 = 𝟐𝟖. 𝟒𝟒𝟗𝐌𝐩𝐚 𝑡

According the principal stress the following equation can be solving the principal stress. 1

1

 σp = 2 (σRi ± √σRi 2 + 4 ∗ τ2 ) = 2 (242.792Mpa ± √(242.792Mpa)2 + 4 ∗ (28.449Mpa)2 )  𝛔𝟏 = 𝟐𝟒𝟔. 𝟎𝟖𝟏𝐌𝐩𝐚 And 𝛔𝟐 = −𝟑. 𝟐𝟖𝟗𝐌𝐩𝐚, 𝛔𝟑 = 𝟎𝐌𝐩𝐚  Therefore the maximum shear stress theory is: 1

 τmax = 2 (σ1 − σ2) = 1⁄2 (246.081Mpa + 3.289Mpa)  τmax = 𝟏𝟐𝟒. 𝟔𝟖𝟓𝐌𝐩𝐚 σy

 𝐅. 𝐬 𝐧𝐞𝐰 = 2∗τ

280Mpa

max

= 2∗124.685Mpa = 1.123 > 1, it is 𝐬𝐚𝐟𝐞!

 And to find the safety factor use of the maximum distortion energy theory σy

 σ12 + σ22 − 2 ∗ σ1 ∗ σ2 = ( )2 f.s

σy

 (246.081)2 + (−3.289)2 − 2 ∗ 246.081 ∗ −3.289 = ( f.s )2 σy 2

 ( f.s ) = 62,185.397Mpa2 

σy F.s

= 249.37Mpa σy

280Mpa

 F. s new = 249.37Mpa = 249.37Mpa = 1.123 > 1, it is 𝐬𝐚𝐟𝐞! ∴ 𝐝𝐞𝐬𝐢𝐠𝐧 𝐨𝐟 𝐡𝐨𝐨𝐤 𝐢𝐬 𝐬𝐚𝐟𝐞 𝐝𝐮𝐞 𝐭𝐨 𝐛𝐞𝐧𝐝𝐢𝐧𝐠 𝐥𝐨𝐚𝐝, 𝐚𝐱𝐢𝐚𝐥 𝐥𝐨𝐚𝐝 𝐚𝐧𝐝 𝐬𝐡𝐞𝐚𝐫 𝐥𝐨𝐚𝐝. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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4.4.9. Design of Bolt and Nut 4.4.9.1.

Material Selection

 From the standard table SAE class 10.9 steel is the selected material for bolt and nut with the following parameters  Proof load=830Mpa  Yield strength=940Mpa  Tensile strength=1040Mp

Figure 4.1:30.3d view of bolt 2

Figure 4.1:31.3d view of nut 1

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4.4.9.1.1.

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Design of Bolt and Nut for Connecting Back Hoist Body and Horizontal Boom

4.4.9.1.1.1.

Bolt subjected to Bending moment

Along x-z plane



From the horizontal boom we gate 

Fxc = 218.959kN

First find reaction force and moment  Force along z-axis from the above figure 

+↑ ΣFz = 0



R1z + R2z − Fxc = 0



R1z + R2z = Fxc = 218.959kN

 Sum of bending moment at point ‘1’ clockwise is equal to zero 

∑ M@1 = 0,125mm ∗ 218.959kN − R2z ∗ 250mm = 0



125mm ∗ 218.959kN = R2z ∗ 250mm  R2z = 109.48kN  R1z = Fxc − R2z = 218.959kN − 109.48 = 109.48kN

Therefore  Mxzmax@c =

Fxc∗L 4

=

218.959kN∗250mm 4

= 13,684.938kNmm

 Vzmax = R1z = R2z = 109.48kN

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Along x-y plane



From the above we gate 

Fyc = 72.672kN



Mc = 21240.423kNmm

First find reaction force and moment  Force along y-axis from the above figure 

+↑ ΣFy = 0



−R1y + R2y − Fyc = 0



R2y − R1y = Fyc = 72.672kN

 Sum of bending moment at point ‘1’ clockwise is equal to zero 

∑ M@1 = 0,125mm ∗ 72.672kN + Mc − R2y ∗ 250mm = 0



125mm ∗ 72.672kN + 21240.423kNmm = R2y ∗ 250mm



R2y ∗ 250mm = 30,324.423kNmm  R2y = 121.298kN  R1y = R2y − Fyc = 121.298kN − 72.672kN = 48.626kN

Bending moment at section x1-x1 

∑ M@x1 = 0, , − R2y ∗ x1 + Mx1 = 0



Mx1 = R1y ∗ x1



M@x1 = 0 = M@1 = 48.626kN ∗ 0mm = 0kNmm



M@x1=125mm = M@c = 48.626kN ∗ 125mm = 6,078.25kNmm

Bending moment at section x2-x2 

∑ M@x2 = 0, , − R2y ∗ x2 + Mx2 = 0

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

Mx2 = R2y ∗ x2



M@x2 = 0 = M@2 = 121.298kN ∗ 0mm = 0kNmm



M@x1=125mm = M@c = 121.298kN ∗ 125mm = 15,162.25kNmm



Vmax = R2y = 121.298kN

Therefore  Mxymax@c = 21240.423kNmm  Vymax = R2y = 121.298kN  Resultant moment at each point  Mmax@c = √Mxz@c 2 + Mxy@c 2 = √15,162.252 + 21,240.4232 = 26,096.923kNmm  Vmax@2 = √Vy@2 2 + Vz@2 2 = √121.2982 + 109.482 = 163.399kN  Now maximum bending moment is take place at point ‘c’  Mmax = 𝟐𝟔, 𝟎𝟗𝟔. 𝟗𝟐𝟑𝐤𝐍𝐦𝐦  Vmax = 𝟏𝟔𝟑. 𝟑𝟗𝟗𝐤𝐍  Hence, 𝛔𝐛 =

𝐌∗𝐲

d

where, σb =bending stress, y = 2 and I =

𝐈

σy

 σb = σall = S.F  σb =  σb =

940MN⁄ 2 m 1.5 M∗y I

=

64

, let S.F=1.5 MN

= 626.667 m2 = 626.667 N⁄mm2

d 2 πd4

M∗

πd4

=

32∗M πd3

64



32∗26,096.923kNmm πd3

= 626.667 N⁄mm2

 d3 = 424,182.341mm3  𝐝 = 𝟕𝟓. 𝟏𝟑𝟔𝐦𝐦 4.4.9.1.1.2.  τall =

Shear load subjected to bolt Vmax πd2 4

τy

σy

, but τall = F.s = 2∗F.s =

 313.334Mpa =

940Mpa 2∗1.5

= 313.334Mpa

163.399kN πd2 4

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 d2 = 663.975mm2  𝐝 = 𝟐𝟓. 𝟕𝟔𝟖𝐦𝐦  For safe design take the larger one so 𝐝 = 𝟕𝟓. 𝟏𝟑𝟔𝐦𝐦  But this value of “d” is not required in standard table, so we take “d” from standard table by using the value of d ≥ 75.136mm  𝐝 = 𝟖𝟎𝐦𝐦

 𝐡 = 𝟓𝐦𝐦

 𝐃 = 𝟖𝟎. 𝟓𝐦𝐦

 𝐇 = 𝟓. 𝟐𝟓𝐦𝐦

 𝐝𝐜 = 𝟕𝟎𝐦𝐦

 𝐏 = 𝟏𝟎𝐦𝐦

Where 

d=major diameter of bolt



h=depth of bolt thread



D=major diameter of nut



H=depth of nut thread



dc =minor or core diameter of bolt



p=pitch

Now find bending moment  σb =

32∗M πd3

=

32∗26,096.923kNmm π(80mm)3

= 519.182Mpa

Then find the new factor of safety F. S𝑛𝑒𝑤 =

σy 940Mpa = = 1.81 > 1.5 it is safe! σb 519.182Mpa

 Therefore the design of bolt and nut is safe due to bending load.

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Figure 4.1:32.Hexagonal headed bolt with a nut and washer in position [14] Now the outer diameter of the bolt and nut equal is given by from the above figure 4-23 as follow  dout = 2 ∗ d = 2 ∗ 80mm = 160mm  Then thickness of the bolt and nut is given by from the above figure 4-23 as follow  wbolt = 0.75 ∗ d = 0.75 ∗ 80 = 60mm  wnut = d = 80mm  Number of thread in the nut is given by  nnut =

wnut p

80mm

= 10mm = 𝟖

 Length of thread bolt portion is given by  L𝑡 = 2 ∗ d = 2 ∗ 80mm = 160mm  Number of thread in the bolt is given by  nbolt =

L𝑡 p

=

160mm 10mm

= 𝟏𝟔

 Total Length of bolt is given by  Ltotal = wbolt + 250mm + wnut + 34.5mm = 60mm + 250mm + 80mm + 34.5mm  Ltotal = 424.5mm  The inner, outer diameter and width of washer is given by from the above figure 4-23 as follow  d𝑖𝑛 = d = 80mm  d𝑜𝑢𝑡 = 2 ∗ d + 4 = 2 ∗ 80mm + 4mm = 164mm  wwasher = 0.15 ∗ d = 0.15 ∗ 80mm = 12mm  Shear Stress Across the Threads  The average thread shearing stress for the bolt (τs ) is obtained by using the relation : τs =

FRc πd ∗ b ∗ nbolt

Where 

b = width of thread section at the root

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

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FRc = √Fxc 2 + Fyc 2 = √218.9592 + 72.6722 = 230.704kN σy

F

b = πd∗τ Rc ∗n s

, but τs = τall = 2∗F.S =

bolt

940Mpa 2∗1.5

= 313.334Mpa

Now b=

230704N = 0.183mm π ∗ 80mm ∗ 313.334N/mm2 ∗ 16

 The average thread shearing stress for the nut (τs ) is obtained by using the relation : τs = σy

F

b = πD∗τ Rc∗n s

FRc πD ∗ b ∗ nnut

, but τs = τall = 2∗F.S =

nut

940Mpa 2∗1.5

= 313.334Mpa

Now b=

230704N = 0.364mm π ∗ 80.5mm ∗ 313.334N/mm2 ∗ 8

So take the larger one for safe design b =0.364mm  Compression or crushing stress on threads  The compression or crushing stress between the threads for bolt (σc ) may be obtained by using the relation : σc =

FRc 2

π(d2 − dc ) ∗ nbolt 4

=

230704N = 12.239Mpa π(80mm2 − 70mm2 ) ∗ 16 4

σc ≪≪ σy = 12.239Mpa ≪≪ 940Mpa it is safe due to crushing in the bolt  The compression or crushing stress between the threads for nut(σc ) may be obtained by using the relation : σc =

FRc 2

π(D2 − dc ) ∗ nnut 4

=

230704N = 23.235Mpa π(80.5mm2 − 70mm2 ) ∗8 4

σc ≪≪ σy = 23.235Mpa ≪≪ 940Mpa it is safe due to crushing in the nut

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4.4.9.1.2.

Design of Bolt and Nut for Connecting Hook and Horizontal Boom

4.4.9.1.2.1.

Bolt subject to bending moment

First find reaction force and moment  Force along y-axis from the above figure 

+↑ ΣFy = 0



R1y + R2y − W = 0



R1y + R2y = W = 29.43kN

 Sum of bending moment at point ‘1’ clockwise is equal to zero 

∑ M@1 = 0,40mm ∗ 29.43kN − R2y ∗ 80mm = 0



40mm ∗ 29.43kN = R2y ∗ 80mm  R2y = 14.715kN  R1y = W − R2y = 29.43kN − 14.715 = 14.715kN

Therefore  Mmax@c =

W∗L 4

=

29.43kN∗80mm 4

= 588.6kNmm

 Vmax = R1y = R2y = 14.715kN  Hence, 𝛔𝐛 =

𝐌∗𝐲

d

where, σb =bending stress, y = 2 and I =

𝐈

σy

 σb = σall = S.F  σb =  σb =

940MN⁄ 2 m 1.5 M∗y I

=

64

, let S.F=1.5 MN

= 626.667 m2 = 626.667 N⁄mm2

d 2 πd4 64

M∗

πd4

=

32∗M πd3

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

32∗588.6kNmm

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= 626.667 N⁄mm2

πd3

 d3 = 9,567.171mm3  𝐝 = 𝟐𝟏. 𝟐𝟐𝟗𝐦𝐦 4.4.9.1.2.2.  τall =

Bolt Subject to Shear Load Vmax πd2 4

τy

σy

, but τall = F.s = 2∗F.s =

 313.334Mpa =

940Mpa 2∗1.5

= 313.334Mpa

14.715kN πd2 4

 d2 = 59.795mm2  𝐝 = 𝟕. 𝟕𝟑𝟑𝐦𝐦  For safe design take the larger one so 𝐝 = 𝟐𝟏. 𝟐𝟐𝟗𝐦𝐦  But this value of “d” is not required in standard table, so we take “d” from standard table by using the value of d ≥ 21.229mm and in order to fit the bolt with eye hook (Q=39.624mm), so take the following values  𝐝 = 𝟑𝟔𝐦𝐦

 𝐡 = 𝟑𝐦𝐦

 𝐃 = 𝟑𝟔. 𝟓𝐦𝐦

 𝐇 = 𝟑. 𝟐𝟓𝐦𝐦

 𝐝𝐜 = 𝟑𝟎𝐦𝐦

 𝐏 = 𝟔𝐦𝐦

Where 

d=major diameter of bolt



h=depth of bolt thread



D=major diameter of nut



H=depth of nut thread



dc =minor or core diameter of bolt



p=pitch

Now find bending moment  σb =

32∗M πd3

=

32∗588.6kNmm π(36mm)3

= 128.503Mpa

Then find the new factor of safety F. Snew =

σy 940Mpa = = 7.315 > 1.5 σb 128.503Mpa

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 Therefore the design of bolt and nut is safe due to bending load.

Figure 4- 1. Hexagonal headed bolt with a nut and washer in position [14]  Now the outer diameter of the bolt and nut equal is given by from the above figure 4-24 as follow  dout = 2 ∗ d = 2 ∗ 36mm = 72mm  Then thickness of the bolt and nut is given by from the above figure 4-24 as follow  wbolt = 0.75 ∗ d = 0.75 ∗ 36 = 27mm  wnut = d = 36mm  Number of thread in the nut is given by  nnut =

wnut p

=

36mm 6mm

=𝟔

 Length of thread bolt portion is given by  L𝑡 = 2 ∗ d = 2 ∗ 36mm = 72mm  Number of thread in the bolt is given by  nbolt =

L𝑡 p

=

72mm 6mm

= 𝟏𝟐

 Total Length of bolt is given by  Ltotal = wbolt + 80mm + wnut + 30.6mm = 27mm + 80mm + 36mm + 30.6mm  Ltotal = 173.6mm

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 The inner, outer diameter and width of washer is given by from the above figure 4-24 as follow  din = d = 36mm  dout = 2 ∗ d + 4 = 2 ∗ 36mm + 4mm = 76mm  wwasher = 0.15 ∗ d = 0.15 ∗ 36mm = 5.4mm  Shear stress across the threads  The average thread shearing stress for the bolt (τs ) is obtained by using the relation : τs =

W πd ∗ b ∗ nbolt

Where 

b = width of thread section at the root

b = πd∗τ

σy

W

, but τs = τall = 2∗F.S =

s ∗nbolt

940Mpa 2∗1.5

= 313.334Mpa

Now b=

29430N = 0.07mm π ∗ 36mm ∗ 313.334N/mm2 ∗ 12

 The average thread shearing stress for the nut (τs ) is obtained by using the relation : τs =

W πD ∗ b ∗ nnut

σy

W

b = πD∗τ

, but τs = τall = 2∗F.S =

s ∗nnut

940Mpa 2∗1.5

= 313.334Mpa

Now b=

29430N = 0.14mm π ∗ 36.5mm ∗ 313.334N/mm2 ∗ 6

So take the larger one for safe design b =0.14mm  Compression or Crushing Stress on Threads

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 The compression or crushing stress between the threads for bolt (σc ) may be obtained by using the relation: σc =

W π(d2 − dc 2 ) ∗ nbolt 4

=

29430N = 7.885Mpa π(36mm2 − 30mm2 ) ∗ 12 4

σc ≪≪ σy = 7.885Mpa ≪≪ 940Mpa it is 𝐬𝐚𝐟𝐞 due to crushing in the bolt  The compression or crushing stress between the threads for nut (σc ) may be obtained by using the relation: σc =

W 2

π(D2 − dc ) ∗ nnut 4

=

29430N = 14.448Mpa π(36.5mm2 − 30mm2 ) ∗6 4

σc ≪≪ σy = 14.448Mpa ≪≪ 940Mpa it is 𝐬𝐚𝐟𝐞 due to crushing in the nut 4.4.10. Wheel Selection Four wheels are need for the machine two wheels are at the back base of the machine and two wheels are at the front base of the machine and the following parameters are as follows:  Front wheel: Material selection for front wheel is Steel-ductile

Figure 4.1:33.3d view of front wheel  From the above total load of the machine the load carrying capacity of connecting frame, UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 Fyl = 31.48 𝑘𝑁 Applied at each end of the connecting frame.  The mass to carry the wheel =  m=

31480 N 9.81

applied force gravity

= 3208.97 kg

In order to carry the load from the connecting frame we are selecting the wheel from the standard. Table 4.1:6.standard table of front wheel Wheel Bore(mm)

30

Tread(mm)

100

Wheel Hub(mm)

108

Bearing type

Tapper

Wheel material

Steel Ductile

Wheel Dia (mm)

200

Load capacity( kg)

3600

Weight (kg)

11.35

 Rare Wheels: Material selection for rare wheel is steel –Ductile plate swivel

Figure 4.1:34.3d view of rear caster wheel UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 From the above total load of the machine the load carrying capacity of base body.  Fym = 5156.3kN Applied at each end of the base body.  The mass to carry the wheel =  m=

51563 N 9.81

applied force gravity

= 5256.17 kg

In order to carry the load from the horizontal frame we are selecting the wheel from the standard. Table 4.1:7.standard table of rear wheel Castor swivel radius (mm)

235

Castor fixing hole(mm)

16

Castor plate size(mm)

225*160

Castor hole center (mm)

210*160

Castor height (mm)

395

Tread (mm)

100

Wheel Dim (mm)

300

Load capacity(kg)

5500

Castor fixing

Plate

Castor option

Swivel

Wheel material

Steel –Ductile

Bearing type

Tapper

4.4.11. Welding Welding is a fabrication process that joins materials, usually metals or thermoplastics, by causing coalescence. This is often done by melting the work pieces and adding a filler material to form a pool of molten material (the weld puddle) that cools to become a strong joint, but sometimes

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pressure is used in conjunction with heat, or by itself, to produce the weld. This is in contrast with soldering and brazing, which involve melting a lower-melting-point material between the work pieces to form a bond between them, without melting the work pieces. Mild steel is routinely welded, but it must be done under an inert gas atmosphere. The most reliable method for Mild steel welding is the Tungsten Inert Gas (TIG) process. Gas tungsten arc welding (GTAW), or tungsten inert gas (TIG) welding, is a manual welding process that uses a nonconsumable electrode made of tungsten, an inert or semi-inert gas mixture, and a separate filler material. Especially useful for welding thin materials, this method is characterized by a stable arc and high quality welds, but it requires significant operator skill and can only be accomplished at relatively low speeds. TIG welding has the advantage of a small weld head, lower heat input is required and filler metal is optional. The welder must be careful not to apply too much heat for too long during welding. Welding of thin gauge Mild steel requires a definite skill. Producing defect-free welds without overheating the steel takes years of practice, no matter which welding process is used. Bad welds are difficult to correct in Mild steel. It is more economical to get things done right the first time. In addition, an experienced welder will know how to produce a good weld without overheating it. Overheating causes precipitation of the chromium atoms away from the grain boundaries to form chromium carbides, depleting the steel of its corrosion resistance. Welding Filler Metals The selection of the filler metal alloy for welding the Mild steels is based on the composition of the mild steel. The various mild steel filler metal alloys are normally available as covered electrodes and as bare solid wires. Filler metal alloy for welding the various mild steel base metals are: Cr-Ni-Mn (AISI No. 308); Cr-Ni-Austenitic (AISI No. 309, 310, 316, 317, and 347); Cr-Martensitic (AISI No. 410, 430); Cr-Ferritic (AISI No. 410, 430, 309, 502). It is possible to weld several different mild steel base metals with the same filler metal alloy. Welding Procedures UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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For shielded metal arc welding; Covered electrodes for shielded metal arc welding must be stored at normal room temperatures in dry area. These electrode coatings, of low hydrogen type, are susceptible to moisture pickup. Once the electrode box has been opened, the electrodes should be kept in a dry box until used. The gas tungsten arc welding; process is widely used for thinner sections of mild steel. The 2% tungsten is recommended and the electrode should be ground to a taper. Argon is normally used for gas shielding; however, argon-helium mixtures are sometimes used for automatic applications. The gas metal arc welding process is widely used for thicker materials since it is a faster welding process. The spray transfer mode is used for flat position welding and this requires the use of argon for shielding with 2% or 5% oxygen or special mixtures. The oxygen helps producing better wetting action on the edges of the weld. For all welding operations, the weld area should be cleaned and free from all foreign material, oil, paint, dirt, etc. The welding arc should be as short as possible when using any of the arc processes. Common welding joint types

(1) Square butt joint,

(2) Single-V preparation joint,

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(3) Lap joint, (4) T-joint.

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Figure 4.1:35.The cross-section of a welded butt joint  To calculate the heat input for arc welding procedures, the following formula can be used:

Where Q = heat input (kJ/mm), V = voltage (V), I = current (A), and S = welding speed (mm/min). The efficiency is dependent on the welding process used, with shielded metal arc welding having a value of 0.75, gas metal arc welding and submerged arc welding, 0.9, and gas tungsten arc welding, 0.8. 4.4.12. Manufacturing Process The knowledge of manufacturing processes is of great for a design engineer. The following are the various manufacturing processes used in Mechanical Engineering. 1. Primary shaping processes: The processes used for the preliminary shaping of the machine component are known as primary shaping processes. The common operations used for this process are casting, forging, extruding, rolling, drawing, bending, shearing, spinning, powder metal forming, squeezing, etc. 2. Machining processes: The processes used for giving final shape to the machine component, according to planned dimensions are known as machining processes. The common operations used for this process are turning, planning, shaping, drilling, boring, reaming, sawing, broaching, milling, grinding, hobbling, etc. UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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3. Surface finishing processes: The processes used to provide a good surface finish for the machine component are known as surface finishing processes. The common operations used for this process are polishing, buffing, honing, lapping, abrasive belt grinding, barrel tumbling, electroplating, super finishing, sherardizing, etc. 4. Joining processes: The processes used for joining machine components are known as joining processes. The common operations used for this process are welding, riveting, soldering, brazing, screw fastening, pressing, sintering, etc.  Casting: It is one of the most important manufacturing process used in Mechanical Engineering. The castings are obtained by remitting of ingots* in a cupola or some other foundry furnace and then pouring this molten metal into metal or sand molds. Table: manufacturing process for each components of the machine Part name

Horizontal Boom

Back hoist body

Material

Mild steel

Mild steel

Required operation

Required machine

Casting

Sand casting

Facing

Lath machine

Drilling

Drill machine

Welding

Welding machine

Finishing

Sand paper

Painting

Painting machine

Casting

Sand casting

Facing

Lath machine

Drilling

Mild steel

Drill machine

Tapering

Lath

Welding

Welding machine

Painting

Painting machine

Casting

Sand casting

Facing

Lath machine

Tapering

Lath machine

Welding

Welding machine

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Horizontal frame

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Painting

Painting machine

Casting

Sand casting

Facing

Lath machine

Tapering

Lath machine

Welding

Welding machine

Casting

Sand casting

Drilling

Drill machine

Facing

Lath

Welding

Electric arc

Painting

Painting machine

Casting

Sand mold

Tapering

Lath machine

Shaping

Lath machine

Finishing

Milling machine

Painting

Painting machine

Facing

Lath machine

Turning

Lath machine

Tapering

Lath machine

Drilling

Dill machine

Threading

Milling machine

Painting

Painting machine

and base body

Back hoist support

Connecting frame

Hook

Bolt and Nut

Mild steel

Mild steel

Mild steel

SAE class 10.9 steel

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CHATER FIVE 5. RESULT AND DISCUSSION 5.1.

Results

In this chapter, design results of parts and discussing the way how to install manual hydraulic hand lift jib crane system.  Design summary  Design specification 

Maximum lifting height=2500mm



Minimum lifting height =600mm



θ1 = 65° and θ2 = 50°



Boom length, Lb = 1250 mm



Maximum mass to be lifted=3 ton=3000Kg Gravity= g = 9.81m⁄s2  Load=mass*g=3000kg*9.81m⁄s 2 =29430N=29.430kN



 Parameters of the cylinder from standard table 

for three ton read from the appendix table 

Maximum height (Le ) =1176.274 mm



Minimum height (Lr ) =668.274 mm



Bore diameter =65mm

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Table 5:1.result of all parts Part name

Material

Horizontal

mild

boom

steel

Back hoist

mild

body

steel

Horizontal

mild

frame

steel

Base body

mild

Mmax

Vmax

(kNmm)

(KN)

32306.188

Dimension (mm) B

H

t

L

72.672

203

184

9.5

1250

95265.414

187.728

250

226

12

1393.16

43870

52.507

175

157

9

1767.768

11828.o94

50.525

120

108

6

936.428

47623.707

15.583

175

157

9

1136.377

13829.918

51.563

175

163

6

936.428

Ri

Ro

25.4

66.802

d

D

dc

80

80.5

36

36.5

steel Back hoist

mild

support

steel

Connecting

mild

frame

steel mild

Hook

steel

SAE Bolt and nut1

class

1695.917

124.685

26,096.923 163.399

Rn

R

Yi

Yo

17.415

23.685

h

H

P

70

5

5.25

10

30

3

3.25

6

42.815 46.101

10.9 steel Bolt and nut

SAE

2

class

588.6

14.715

10.9 steel

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5.2.

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ANSYS results for horizontal boom [input from appendix table 0:8.A]

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5.3.

| 2010 E.C

ANSYS results for back hoist [input data from appendix table 0:7.A]

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5.4.

| 2010 E.C

ANSYS results for hook [input data from appendix table 0:8.A]

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5.5.

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Comparisons of ANSYS and manual result 1. Discussion on equivalent stress on the horizontal boom  From Ansys equivalent stress result as follow  σmax = 802.19 Mpa  σmin = 3.914Mpa  From manual maximum stress result as follow  σmax = σ1 = 102.026 Mpa

 The maximum value of manual result is found between the maximum and minimum value of ANSYS result. So the design is safe!  From Ansys equivalent strain result as follow  ∈max = 0.00388 m/m  ∈min = 3.215 ∗ 10−5 m/m  From manual maximum strain result as follow  ∈max =

𝜎𝑚𝑎𝑥 E

=

102.026Mpa 207Gpa

 ∈𝑚𝑎𝑥 = 0.000493  The maximum value of manual strain result is found between the maximum and minimum value of ANSYS strain result. So the design is safe!  From Ansys the deflection result as follow  ∆max = 3.64mm  ∆min = 0mm  From manual maximum deflection result as follow is  ∆max =∈max ∗ L  ∆max = 0.000493 ∗ 1250mm  ∆max = 0.616mm  The maximum value of manual deflection result is found between the maximum and minimum value of ANSYS deflection result. So the design is safe! 2. Discussion on equivalent stress on the back hoist body  From Ansys equivalent stress result as follow  σmax = 6167Mpa  σmin = 0.7459Mpa UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

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 From manual calculation maximum stress result is;  σmax = σ1 = 119.826Mpa  The maximum value of manual stress result is found between the maximum and minimum value of ANSYS result. So the design is safe!  From ansys equivalent strain result as follow m

 ∈max = 0.03658 m  ∈min = 3.828 ∗ 10−6 m/m  From manual maximum strain result as follow  ∈max =

σmax E

=

119.826Mpa 207Gpa

 ∈max = 0.000579  The maximum value of manual strain result is found between the maximum and minimum value of ANSYS strain result. So the design is safe!  From ANSYS the deflection result as follow  ∆max = 16.78mm  ∆min = 0mm  From manual calculation the maximum deflection result of back hoist body is;  ∆max =∈max ∗ L  ∆max = 0.000579 ∗ 1393.16mm  ∆max = 0.806mm  The maximum value of manual deflection result is found between the maximum and minimum value of ANSYS deflection result. So the design is safe! 3. Discussion on equivalent stress on the hook  from Ansys equivalent stress result is;  σmax = 167.19Mpa  σmin = 0.000677Mpa  From manual calculation maximum stress result is;  σmax = 242.792Mpa

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 The maximum value of manual stress result is not found between the maximum and minimum value of ANSYS result. So the design of hook is highly stress cross section. So the designer recommended that avoid over load more than 3 ton.  From Ansys equivalent strain result for hook is; m

 ∈max = 8.186 ∗ 10−4 m  ∈min = 3.276 ∗ 10−9 m/m  From manual maximum strain result for hook is; as follow  ∈max =

σmax E

=

242.792Mpa 207Gpa

 ∈max = 0.00117  The maximum value of manual strain result is not found between the maximum and minimum value of ANSYS strain result. So the design of hook is highly strain cross section. So the design is safe!  From ANSYS the deflection result as follow  ∆max = 0.0003226mm  ∆min = 0mm  From manual calculation the maximum deflection result of back hoist body is;  ∆max =∈max ∗

D−25 2

 ∆max = 0.0017 ∗

80−25 2

 ∆max = 0.04675mm  The maximum value of manual deflection result is not found between the maximum and minimum value of ANSYS deflection result. So the design of hook is highly strain. We recommend that another designer select above our factor of safety in order maintain this problems. And also our machine is flexible or movable we recommend to use this machine garages specially Jovani garages it is located on autoparko in Gondar when we at observation time in Jovani garage, we saw our machine but it is fixed and out of use instead the company uses baranco.

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5.6.

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Discussion

Generally the result from the above table tells us, good, easily, available material selection of horizontal boom, back hoist support, connecting frame, horizontal fame, base body, nuts and bolts are compatible each other to resist the direct stress, bending moment and shear stress that occurred during assemble and disassemble of manually hydraulic hand lift jib crane. It also tells us to reduce the deformation of components compatibility of each components must be considered (i.e horizontal boom, back hoist support, bolt with nut). So the component of the manually hydraulic hand lift jib crane assembly can be aligned properly. When we compare our design to the existing manually hydraulic hand lift jib crane the stress and fracture of component is reduced and the stability of the machine is increase by aligned of horizontal frame by 45 degree. The carrying capacity and lifting height of the machine is increasing by adding some component. Bolt is long and strong to resist the axial load, the selection of caster wheel is best and standard to resist the maximum applied load caused by the manually hydraulic hand lift jib crane machine. In order to move the loading machine at most two people are required.

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CHAPTER SIX 6. COST ANALYSIS A system which systematically recodes all the expenditure to determine the cost of manufactured products is called costing. The total cost on manual hydraulic hand lift jib crane made up of four main elements; 1. material cost 2. expense cost (manufacturing cost) 3. labor cost 4. purchase cost

6.1.

Material Cost for Manual Hydraulic Hand Lift Jib Crane Table 6:1.material cost of each component Material cost

Part name

Material

Qty

type

Cost

ρ(kg

b or

h or

L

V

M

Cost

(U)

/m3 )

Do

Di

(m)

(m3)

(kg)

($)

(m)

(m)

0.203

0.184

1.25

0.009191

71.6917

186.399

($/kg) Horizontal

Mild

Boom

steel

Back hoist

Mild

1

2.6

7800

5 1

2.6

7800

0.25

0.226

1.3932

0.015915

steel Base body

Mild

124.140

322.766

5 1

2.6

7800

0.12

0.108

0.9364

0.002562

19.984

51.958

2

2.6

7800

0.175

0.157

1.1364

0.01358

105.939

275.441

2

2.6

7800

0.175

0.157

1.7678

0.02113

164.794

428.464

1

2.6

7800

0.175

0.163

0.9364

0.003798

29.6255

77.025

steel Back hoist

Mild

support

steel

Horizontal

Mild

frame

steel

Connectin

Mild

g frame

steel

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Bolt 1

SAE

| 2010 E.C

1

4

7920

0.16

-

0.4245

0.008531

67.5635

270.254

1

4

7920

0.072

-

0.1736

0.000706

5.595

22.380

1

4

7920

0.16

0.08

0.08

0.1206

9.549

38.196

1

4

7920

0.072

0.036

0.036

0.00011

0.87

3.48

1

2.6

7800

-

-

-

2

5.2

class 10.9 steel Bolt 2

SAE class 10.9 steel

Nut 1

SAE class 10.9 steel

Nut 2

SAE class 10.9 steel

Hook

Mild steel

$ 1681.563

Total cost  Where  b= base of each material  h= height of each material  L=length of each material  D= diameter of each material  Qty= quantity of each material  ρ =density of each material  U=($/kg)

 V = (b2 − h2 ) ∗ L for square hollow cross-section  V=

π 4

∗ (Do2 − Di2 ) ∗ L for circular hollow cross-section

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 V=

π 4

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∗ Do2 ∗ L for circular solid cross-section

 M= ρ∗V  Cost = M ∗ U

6.2.

Manufacturing Cost and Labor Cost

 Manufacturing cost are cost required when different operation are performed on the raw marital.  Labor cost are cost to pay the labor worker. Table 6:2.manufacturing & labor cost of each material manufacturing

6.3.

Part name

cost($ )

Labor cost($)

Horizontal Boom

8

2

Back hoist

8

2

Base body

4

1

Back hoist support

4

1

Horizontal frame

6

1.75

Connecting frame

4

1

Bolt 1

6

2

Bolt 2

3

1

Nut 1

2

0.5

Nut 2

1

0.25

Hook

5

2

Total cost

$ 47.00

$14.50

Purchase Cost;  The wheel and hydraulic cylinder of manually hydraulic hand lift jib crane are Purchase.

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Table 6:3.purchase cost Part name

Load range(kg)

Qty

Purchase cost

Hydraulic

3000

1

$500

Rare Wheel

5,256.2

2

$100

Front Wheel

3,209

2

$70

cylinder

Total

6.4.

Total Cost for Manually Hydraulic Hand Lift Jib Crane Machine  Total cost is the sum of all cost required.  𝐓𝐨𝐭𝐚𝐥 𝐜𝐨𝐬𝐭 = material cost + manufacturing cost + labor cost + purchase cost  𝐓𝐨𝐭𝐚𝐥 𝐜𝐨𝐬𝐭 = $ 1681.563 + $47 + $14.5 + $500 + $100 + $70  𝐓𝐨𝐭𝐚𝐥 𝐜𝐨𝐬𝐭 = 𝟐𝟒𝟏𝟑. 𝟎𝟔𝟑$

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CHAPTER SEVEN 7. CONCLUSION AND RECOMMENDATION 7.1.

Conclusion

The design of manual hydraulic hand lift jib crane machine with a maximum lift mass of 3 ton (3000kg), maximum lift height of 2500mm & minimum lift height of 600mm is important for the application of garages and small and micro enterprises in Ethiopia. This crane is suitable for loading and unloading heavy material workshops and it is busy else-where in rapid garages these are required for handling engines and its parts. These cranes can also be used to carry liquids in containers and moving parts from place to place by pumping the hydraulic handle part and pushing the machine manually.

7.2.

Recommendation

Because of this machine becomes manual, simple to lift up loads by pumping and it doesn’t ask qualification of employees, those garages and small and micro enterprises should be use it to accelerate and to make easily the work activity. Some components of the machine are made up of from different steel class therefore the user should be prevent from rest in order to keep their life time. Because if they have got rest, they could be malfunction. Capacity of the machine is three tons therefore the users should not use above its capacity. If the load is above three ton there may be bending of the component such as boom and buckling of hoist body. Finally, we recommend that IOT director should see this project and install this project for the benefit of material handling. Its benefit can be seen in terms of high safety of material during lifting and moving. This project paper can also be used as guideline for design of manually hydraulic hand lift jib crane and we do all the Ansys results in static load condition so we recommend another designer uses dynamic load for real results.

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REFERENCE [1] www.davidround.com/company-history/. [2] https://www.indiamart.com/proddetail/mobile-floor-jib-crane-3684882555.html. [3] Coulton, J. J. (1974), "Lifting in Early Greek Architecture", The Journal of Hellenic Studies, 94: 1–19, doi:10.2307/630416,JSTOR 630416, p.7. [4] Coulton, J. J. (1974), "Lifting in Early Greek Architecture", The Journal of Hellenic Studies, 94: 1–19, doi:10.2307/630416,JSTOR 630416, pp.14f. [5] Coulton, J. J. (1974), "Lifting in Early Greek Architecture", The Journal of Hellenic Studies, 94: 1–19, doi:10.2307/630416,JSTOR 630416, p.16. [6] The UK business market LTD, “design and manufacturing of heavy duty Folding Workshop Crane machine”, 2018. Or https://www.UKL TD- cranes.com/English/workshop- cranes. [7] Nationwide industrial supply, “design and manufacturing of beech counterweighted crane machine”,

2018.

Or

https://www.nationwideindustrial

supply

cranes.com/English/beech

counterweighted - cranes. [8] Rockwell Hoisto Cranes Private Limited, “design and manufacturing of manual mobile material

handling

machine”.

Or

https://www.

RockwellHoistoCranes(Pv.t)Ltd

supply

cranes.com/mobile material handling- cranes. [9] Selby Engineering and Lifting Safety LTD, “design and manufacturing of mobile jib crane machine” Or https://www.liftingsafety.co.uk/product/transportable-mini-cranes-3141.html. [10] Asmita Jadhav, B.E. Mechanical (Pune University), “Design and development of Rotating Floor Crane”. [11] Chaitra C. Danavatimath, Prof. H. D. Sarode (P.G.student.Dpartment of Mechanical Engineering), “Finite Element Analysis and Optimization of Jib Crane Boom” international Journal of Innovative Research in Science, Engineering and Technology (IJIRSET), Volume: 6, Issue: 07, Page 1-8 | July -2017.

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[12] Okolie Paul Chukwulozie et al (2015), “Design and Analysis of a mobile floor crane” British Journal of Applied science and Technology (BJAST), Volume: 13, Issue: 05, Page 1-9 | Nov -2015. [13] https://www.grosbygroupink.co.uk. [14] Department of Mechanical Engineering Indian Institute of Technology, Machine drawing, third edition. K.L.Narayana-et al, 2006, chapter 5, page 88, fig 5.15.

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DECLARATION We undersigned, declare that the thesis comprises our own work. In compliance with internationally accepted practice, we have acknowledge and referred all materials used in this work. We understand that non-adherence to the principle of academic honesty and integrity, misrepresentation/fabrication of any idea/ data/ fact/source will constitute sufficient ground for disciplinary action by the university and can also evoke penal action from the source which have not been properly cited or acknowledged. NAME

ID NO

1. YINAGER YALEW

0858/06

2. ZEKIROSE MIHERETAB

0891/06

SIGNATURE

DATE

This thesis has been submitted for examination with my approval as a university adviser. ADVISOR NAME: ADVISOR’S SIGNATURE:

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APPENDICES Table 0:1.A. typical mechanical and physical properties for engineering metals material

Young’s

Shear

Elastic

Shear

Tensile

Shear

modulus

modulus

limit 𝛔𝐲

Yield

strength

Ultimate

of

G

strength

𝛔𝐮𝐭

strength

elasticity E(𝐆𝐍⁄ 𝟐 𝐦

(𝐆𝐍⁄ 𝟐 ) 𝐦

(𝐌𝐍⁄ 𝟐 𝐦

𝛕𝐲

)

(𝐌𝐍⁄ 𝟐 ) 𝐦

𝛕𝐮𝐭 (𝐌𝐍⁄ 𝟐 ) 𝐦

Density 𝛒(𝐤𝐠 /𝐦𝟑 )

(𝐌𝐍⁄ 𝟐 𝐦

)

) 69

26

230

-

390

240

2770

brass

102

38

-

-

350

-

8350

bronze

115

45

210

-

310

-

7650

Cast iron

90

41

-

-

210

-

7640

170

83 248

166

370

330

7640

280

175

480

350

7800

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Aluminum alloy

Grey

Cast iron Malleable

Low

207

80

carbon (mild steel)

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208

82

1200

650

1700

950

7800

titanium

107

40

480

-

551

-

4507

magnesiu

45

17

262

-

379

165

1791

Nickelchrome steel

m [Source: University of Warwick and United Kingdom, introduction to the mechanics of elastic and plastic deformation of solid and structural material, third edition. Ox28DPE.JHearn, 1997, Apendix1, page 534] Table 0:2.A. standard table of hydraulic cylinder Jack Capa city

Strok

Minimu

Maxi

Plun

Base

e

m

mum

ger

dimension

(In)

Height

Heigh

Dia

(In)L*W

t

mete

(ton)

(In)

(In)

Saddle Diame ter

ght

Model Number

(Ibs )

(in)

r

Wei

bore diamete r (In)

(In) 1.5

18

21.72

39.72

.88

3.63*5

.75

12.8

*EBJL-

1.275

15GC 2

3.74

6.89

13.39

.87

4.02*3.78

.83

6.6

EBJL-

1.7

2GC 3

20

26.31

46.31

1.12

4.25*5.5

1.12

22

*EBJL-

2.55

3GC 4

4.72

7.68

15.16

1.11

4.41*4.13

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1.03

9.3

130

EBJ-4GC

2.55

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6

5.12

8.27

16.54

1.34

4.72*4.49

1.19

12.1

EBJ-6GC

2.57

8

5.51

8.66

17.32

1.50

4.92*4.69

1.34

13.7

EBJ-8GC

2.57

12

6.10

9.45

18.70

1.7

5.31*5.12

1.58

17.6

EBJ12G

2.9

C 12

3.03

6.10

10.83

1.7

5.31*5.12

1.58

14.6

*EBJS-

2.9

12GC 15

5.91

9,45

18.50

1.89

5.71*5.43

1.70

20.7

EBJ-

3

15GC 20

6.10

9.84

19.09

2.09

6.10*5.71

1.82

25.1

EBJ-

3.149

20GC 20

3.11

6.50

11.22

2.09

6.10*5.71

1.82

19.8

*EBJS-

3.149

20GC 20

6.89

11.22

18.11

2.80

7.48*5.91

2.72

56.9

*EBJ-

3.149

30GC 50

4.13

9.25

13.78

3.35

10.04*7.48

3.15

92.6

*EBJ-

3.228

50GC 100

5.91

12.28

18.38

4.89

11.81*9.45

3.94

198.

*EBJ-

9

100GC

3.349

[Source: www.enerpac.com] Table 0:3.A. Dimensions and cross-sectional properties Square hollow sections H= height, B=base T=thickness

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H*B

T

H*B

T

H*B

T

(mm*mm)

(mm)

(mm*mm)

(mm)

(mm*mm)

(mm)

25*25

2

60*60

5

120*120

3

25*25

2.5

70*70

2.5

120*120

4

25*25

3

70*70

3

120*120

5

30*30

2

70*70

4

120*120

5.6

30*30

2.5

70*70

5

140*140

4

30*30

3

80*80

2.5

140*140

5

40*40

2

80*80

3

140*140

5.6

40*40

2.5

80*80

4

140*140

6

40*40

3

80*80

5

150*150

4

40*40

4

80*80

6

150*150

5

50*50

2

90*90

2.5

150*150

6

50*50

2.5

90*90

3

150*150

7.1

50*50

3

90*90

4

160*160

4

50*50

4

90*90

5

160*160

5

50*50

5

90*90

6

160*160

6

60*60

2

100*100

2.5

160*160

12.5

60*60

2.5

100*100

3

175*175

6

60*60

3

100*100

4

175*175

9

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60*60

4

100*100

5

180*180

5

70*70

3

120*120

5

180*180

6

70*70

3.6

120*120

6

180*180

10

75*75

3.2

120*120

6.3

203*203

9.5

75*75

4

120*120

8

250*250

12

[Source: www.consteel.com.sg] Table 0:4.A. specification for steeled in millimeter series screws and bolts SAE

Proof load

Yielded

Tensile

Elongatio

Reductio

Core

class

Strength(

strength(Ma

strength(Ma

n

n of area

hardness

Map)

p)

p)

Minimum

minimu

Rockwell

(%)

m (%)

min max

4.6

225

240

400

22

35

B67 B87

4.8

310

-

420

-

-

B71 B87

5.8

380

-

520

-

-

B82 B95

8.8

600

660

830

12

35

C23 C34

9.8

650

900

-

-

C27 C36

10.9

830

940

1040

9

35

C33 C39

12.9

970

1100

1220

8

35

C38 C44

[Source: society of automotive engineers standardJ1199 (1979)] Table 0:5.A. Basic dimensions for square threads in mm (Normal series) according to IS: 4694 – 1968 (Reaffirmed 1996)

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Normal series Major diameter

Minor diameter (dc)

Bolt (d)

Nut (D)

22

22.5

17

24

24.5

19

26

26.5

21

28

28.5

23

30

30.5

24

32

32.5

26

34

34.5

28

36

36.5

30

38

38.5

31

40

40.5

33

42

42.5

35

44

44.5

37

46

46.5

38

48

48.5

40

50

50.5

42

52

52.5

44

55

55.5

46

Pitch (p)

UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

Depth of thread Bolt (h)

Nut (H)

5

2.5

2.75

6

3

3.25

7

3.5

3.75

8

4

4.25

9

4.5

5.25

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58

58.5

49

60

60.5

51

62

62.5

53

65

65.5

55

68

68.5

58

70

70.5

60

72

72.5

62

75

75.5

65

78

78.5

68

80

80.5

70

82

82.5

72

| 2010 E.C

10

5.25 5

[Source: a text book of machine design by R.S.KHURMI and J.K.GUPTA, 2005; chapter 17 table 17.2, page 628] Table 0:6.A. Input Data for Ansys results of Horizontal Boom Model (A4) > Geometry > Parts Object Name Solid Solid State

Meshed

Graphics Properties Visible

Yes

Transparency

1 Definition

Suppressed

No

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Stiffness Behavior

Flexible

Coordinate System

Default Coordinate System

Reference Temperature

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By Environment Material

Assignment

Mild steel

Nonlinear Effects

Yes

Thermal Strain Effects

Yes

Bounding Box Length X

0.203 m

Length Y

0.203 m

Length Z

1.0977 m

0.15227 m

Properties Volume 8.0716e-003 m³ Mass

1.1196e-003 m³

62.959 kg

8.7332 kg

Centroid X

-2.2476e-002 m

Centroid Y

7.6148e-002 m

Centroid Z

-0.47323 m

0.15177 m

Moment of Inertia Ip1

6.716 kg·m²

7.1504e-002 kg·m²

Moment of Inertia Ip2

6.716 kg·m²

7.1504e-002 kg·m²

Moment of Inertia Ip3 0.78766 kg·m²

0.10926 kg·m²

Statistics Nodes

4416

UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

2080

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Elements

608

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260

Mesh Metric

None

Model (A4) > Static Structural (A5) > Loads Object Name Fixed Support Force State

Force 2

Fully Defined Scope

Scoping Method Geometry

Geometry Selection 1 Face

1 Edge Definition

Type Fixed Support

Force

Suppressed

No

Define By

Components

Coordinate System

Global Coordinate System

X Component

0. N (ramped)

Y Component

1.021e+005 N (ramped) -29430 N (ramped)

Z Component

-2.1896e+005 N (ramped)

0. N (ramped)

Mild steel > Constants Density 7800 kg m^-3 Mild steel > Isotropic Elasticity Temperature C Young's Modulus Pa Poisson's Ratio Bulk Modulus Pa Shear Modulus Pa 2.07e+011

0.3

1.725e+011

7.9615e+010

Table 0:7.A. Input Data for Ansys results of back hoist body Model (A4) > Geometry > Parts Solid Solid Object Name

UOG/IOT/SCHOOL OF MECHANICAL ENGINEERING

Solid

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State

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Meshed Graphics Properties

Visible

Yes

Transparency

1 Definition

Suppressed

No

Stiffness Behavior

Flexible

Coordinate System

Default Coordinate System

Reference Temperature

By Environment Material

Assignment

Mild steel

Nonlinear Effects

Yes

Thermal Strain Effects

Yes Bounding Box

Length X Length Y

0.25 m 0.60216 m

0.52733 m

Length Z

0.26367 m

0.25 m Properties

Volume 6.8791e-003 m³ 6.0243e-003 m³ 3.0121e-003 m³ Mass

53.657 kg

Centroid X Centroid Y

46.989 kg

23.495 kg

-2.2263e-002 m -0.83784 m

Centroid Z

-0.2731 m

0.1224 m

0.27639 m

Moment of Inertia Ip1

2.1292 kg·m²

1.5336 kg·m²

0.35848 kg·m²

Moment of Inertia Ip2

1.0157 kg·m²

0.88947 kg·m²

0.44474 kg·m²

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Moment of Inertia Ip3

2.1292 kg·m²

| 2010 E.C

1.5336 kg·m²

0.35848 kg·m²

Statistics Nodes

4224

3916

2808

Elements

572

528

364

Mesh Metric

Object Name

Fixed Support

None

Model (A4) > Static Structural (A5) > Loads Fixed Force Force 2 Support 2

State

Moment

Fully Defined Scope

Scoping Method

Geometry Selection

Geometry

1 Edge Definition

Type Suppressed

Fixed Support

Force

Moment

No

Define By

Components

Coordinate System

Global Coordinate System

X Component

Vector

0. N (ramped)

Y Component

1.341e+005 N (ramped)

-2.3707e+005 N (ramped)

Z Component

1.8773e+005 N (ramped)

-46512 N (ramped) -32306 N·m (ramped)

Magnitude Direction

Defined

Behavior

Deformable

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Advanced Pinball Region

All

Mild steel > Constants Density 7800 kg m^-3 Mild steel > Isotropic Elasticity Temperature C Young's Modulus Pa Poisson's Ratio Bulk Modulus Pa Shear Modulus Pa 2.07e+011

0.3

1.725e+011

7.9615e+010

Table 0:8.A. Input Data for Ansys results of back hoist body Model (A4) > Geometry > Parts Solid Object Name State

Meshed

Graphics Properties Visible

Yes

Transparency

1

Definition Suppressed

No

Stiffness Behavior

Flexible

Coordinate System Default Coordinate System Reference Temperature

By Environment

Material Assignment

Mild steel

Nonlinear Effects

Yes

Thermal Strain Effects

Yes

Bounding Box Length X

0.14448 m

Length Y

0.20066 m

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Length Z

| 2010 E.C

3.8862e-002 m

Properties Volume

4.6659e-004 m³

Mass

3.6394 kg

Centroid X

-1.6981e-003 m

Centroid Y

2.5643e-002 m

Centroid Z

-1.9431e-002 m

Moment of Inertia Ip1

1.1718e-002 kg·m²

Moment of Inertia Ip2

4.1471e-003 kg·m²

Moment of Inertia Ip3

1.4953e-002 kg·m²

Statistics Nodes

2041

Elements

999

Mesh Metric

None

Model (A4) > Static Structural (A5) > Loads Force Object Name Fixed Support State

Fully Defined Scope

Scoping Method Geometry

Geometry Selection 2 Faces

1 Face

Definition Type Fixed Support Suppressed Define By Coordinate System

Force No Components Global Coordinate System

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X Component

0. N (ramped)

Y Component

-29430 N (ramped)

Z Component

0. N (ramped)

| 2010 E.C

Mild steel > Constants Density 7800 kg m^-3

Mild steel > Isotropic Elasticity Temperature C Young's Modulus Pa Poisson's Ratio Bulk Modulus Pa Shear Modulus Pa 2.07e+011

0.3

1.725e+011

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142

7.9615e+010

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PART AND ASSEMBLY DRAWING

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