bda 30803 notes (student version - printable) sem 2 2012_2013

BDA 30803 - Mechanical Engineering Design (Lecture Slides)

Semester I 2013 / 2014

FACULTY OF MECHANICAL AND MANUFACTURING ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA

Course Coordinator : MR. MOHD AZWIR BIN AZLAN

Lecturer :

1) Mr. Mohd Azwir bin Azlan (S1, S2 & S3 BDA 30803) ; (S1 BDA 3083)

2) Dr. Sia Chee Kiong (S4, S5 & S6 BDA 30803)

Lampiran A

RPP-04 / Prosedur Perlaksanaan Kuliah Edisi: 3 / No. Semakan: 0

PERANCANGAN KULIAH LECTURE PLAN

MAKLUMAT MATA PELAJARAN (COURSE INFORMATION)

SEMESTER / SESI (SEMESTER / SESSION)

: I / 2013 - 2014

KOD MATA PELAJARAN (COURSE CODE)

: BDA 30803 / BDA 3083

NAMA MATA PELAJARAN (COURSE NAME)

: REKABENTUK KEJURUTERAAN MEKANIKAL / (MECHANICAL ENGINEERING DESIGN)

BEBAN AKADEMIK PELAJAR (COURSE ACADEMIC LOAD)

:

Aktiviti Pembelajaran (Learning Activity)

Minggu (Week)

Jam / Minggu (Hours / Week)

Bilangan Jam / Semester

(Hours / Semester) Kuliah (Lecture) 14 3 42 Tutorial (Tutorial) 0 0 0 Amali (Practical) 0 0 0 Pembelajaran Kendiri (Independent Study) 14 3 42 Lain-lain (Others) 1. Projek (Project)

2. Tugasan (Assignment)

32 4

JUMLAH JAM BELAJAR (JJB) TOTAL STUDENT LEARNING TIME (SLT) 120

Matapelajaran Pra-syarat (Pre requisite courses) : BDA 10203 Statik / Statics

BDA 20103 Dinamik / Dynamics BDA 30303 Mekanik Pepejal II / Solid Mechanics II

BDA 20402 Pemilihan Bahan / Material Selection Nama Pensyarah (Lecturers name)

: Mr. Mohd Azwir Azlan - coordinator (S1, S2 & S3 BDA 30803) (S1 BDA 3083) Dr. Sia Chee Kiong (S4, S5, & S6 BDA 30803)

Disediakan oleh (Prepared by) : Tandatangan (Signature) : Nama (Name) : MOHD AZWIR BIN AZLAN Tarikh (Date) : 19th August 2013

Disahkan oleh (Approved by) : Tandatangan (Signature) : Nama (Name) : Dr. NUR AZAM

BADARULZAMAN Tarikh (Date) : 19th August 2013

UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FAKULTI KEJURUTERAAN MEKANIKAL DAN PEMBUATAN

Lampiran A


MATLAMAT (GOALS) :

Matlamat kursus ini adalah untuk menyediakan para pelajar dengan keupayaan untuk mengaplikasi, menganalisis dan merekabentuk komponen mesin yang lazim seperti, aci, galas, giar dan skru yang menekankan kepada kekuatan, ketegaran, kegagalan statik dan lesu. The goal of this course is to provide the student with the capability to apply, analyze and design of standard machine components such as shaft, bearing, gears and screws etc. which are emphasized on strength, rigidity, static and fatigue failure.

SINOPSIS (SYNOPSIS) :

Kursus ini terdiri daripada analisis, sintesis dan reka bentuk bagi komponen mekanikal asas dan kompleks iaitu galas, aci, giar, sambungan kekal dan tidak kekal, spring, skru dan pengikat dengan mengambil kira faktor kekuatan, ketegaran, keboleharapan serta kegagalan statik dan lesu.

This course consists of analysis, synthesis and design basic and complex mechanical component i.e. bearings, shafts, gears, permanent and non permanent joining, springs, screw and fastener with consideration of strength, rigidity, reliability, static and fatigue failure.

HASIL PEMBELAJARAN (LEARNING OUTCOMES) :

Di akhir kursus ini, pelajar dapat :

Mengira faktor keselamatan dengan menggunakan teori-teori kegagalan statik dan lesu. (C3, LO1) Menganalisa beberapa komponen mesin (iaitu gear, aci dan galas) yang berfungsi dalam satu sistem

mekanikal atau mesin. (C4, LO10) Mencari sumber maklumat yang paling sesuai bagi pemilihan komponen dalam projek reka bentuk. (A3,

LO6) Menghasilkan model dan simulasi dengan menggunakan perisian kejuruteraan untuk pengesahan reka

bentuk projek. (P4, LO2)

After completing this course, the students are able to:

Calculate factor of safety by using static and fatigue failure of theories. (C3, LO1) Analyze several machine component (i.e. gears, shafts and bearing) that function in one mechanical system

or machine. (C4, LO10) Seek for the most appropriate information source for component selection in the design project. (A3, LO6) Produce model and simulate by using engineering software for project design validation. (P4, LO2)

ISI KANDUNGAN (CONTENT) :

MINGGU (WEEK)

KANDUNGAN (CONTENT)

PENTAKSIRAN(ASSESSMENT)

W1

(9th ~ 13th Sept 2013)

1.0 PENGENALAN KEPADA PROSES REKABENTUK

(INTRODUCTION TO DESIGN PROCESS) ---- (1 hours)

1.1 Definisi Rekabentuk (Design Definition) 1.2 Rekabentuk Kejuruteraan Mekanikal (Mechanical Engineering Design) 1.3 Proses Rekabentuk (Design Process) 1.4 Sumber Rujukan dan Peralatan Rekabentuk (Design Tools and Resources)

Lampiran A


1.5 Tanggungjawab Professional Jurutera Rekabentuk (Design Engineers Professional Responsibilities) 1.6 Kod dan Piawaian (Standards and Codes) 1.7 Ekonomik (Economics)

2.0 ANALISIS DAN SINTESIS (ANALYSIS AND SYNTHESIS) ---- (2 hours)

2.1 Kekuatan dan Kekerasan Bahan (Material Strength and Stiffness) 2.2 Keseimbangan dan GBB (Equilibrium and FBD) 2.3 Jenis-Jenis Daya (Types of Load) 2.4 Tegasan (Stress) 2.5 Prinsip Tegasan untuk Tegasan Satah (Principle Stress for Plane Stress) 2.6 Bulatan Mohr bagi Tegasan Satah (Mohrs Circle for aPlane Stress) 2.7 Asas Tegasan 3 Dimensi (General 3 Dimensional Stress) 2.8 Tegasan Tertabur Seragam (Uniformly Distributed Stresses) 2.9 Tegasan Normal pada Rasuk akibat Lenturan (Normal Stress for Beam in Bending) 2.10 Tegasan Ricih pada Rasuk akibat Lenturan (Shear Stress for Beam in Bending) 2.11 Kilasan (Torsion) 2.12 Penumpuan Tegasan (Stress Concentration)

Ujian 1 (1st Test)

W2

(16th ~ 20th Sept 2013)

3.0 TEORI-TEORI KEGAGALAN REKABENTUK STATIK (STATIC

DESIGN FAILURE OF THEORIES) ---- (3 hours)

3.1 Pengenalan (Introduction) 3.2 Kenapa Perlu Teori Kegagalan (Why Need Failure Theories) 3.3 Teori Kegagalan Statik (Static Failure Theories) 3.4 Teori Tegasan Ricih Maksimum (Maximum Shear Stress Theory) 3.5 Teori Tenaga Herotan (Distortion Energy Theory) 3.6 Teori Column-Mohr (Column-Mohr Theory) 3.7 Teori Tegasan Normal Maksimum (Maximum Normal Stress Theory) 3.8 Teori Pengubahsuaian Column-Mohr (Modification of Mohr Theory)

Tugasan 1 (1st Assignment), Ujian 1 (1st Test)

W3 & W4

(23rd Sept ~

4th Oct 2013)

4.0 TEORI-TEORI KEGAGALAN BAGI REKABENTUK LESU

(FATIGUE DESIGN FAILURE OF THEORIES) ---- (6 hours)

4.1 Pengenalan kepada Lesu (Introduction to Fatigue) 4.2 Kegagalan dan Beban Lesu (Fatigue Load and Failure) 4.3 Hayat dan Kekuatan Lesu (Life and Fatigue Strength) 4.4 Rajah S-N (S-N Diagram) 4.5 Had Ketahanan (Endurance Limits) 4.6 Faktor Berubah Had Ketahanan (Endurance Limit Modifying Factors) 4.7 Penumpuan Tegasan dan Kepekaan Takuk (Stress Concentration and Notch Sensitivity) 4.8 Kekuatan Lesu (Fatigue Strength) 4.9 Ciri-ciri Tegasan Berulang (Characterizing Fluctuating stressess) 4.10 Kombinasi Mod Beban (Combination of Loading Modes) 4.11 Faktor Keselamatan (Safety Factor)

Tugasan 2 (2nd Assignment), Ujian 1 (1st Test)

W5 W7

(7th ~ 25th Oct 2013)

5.0 GEAR (GEAR) ---- (9 hours)

5.1 Pengenalan: tatanama, jenis-jenis giar dan kengunaannya (Introduction: terminology, types of gears and its application)

5.2 Pembinaan Giar (Construction of gears)

Projek berkumpulan (Group Project), Ujian 2 (2nd Test)

Lampiran A


5.3 Sistem Gigi (Tooth systems) 5.4 Nisbah Giar (Gear ratio) 5.5 Barisan gear (Gear train) 5.6 Analisis Daya pada Gigi (Gear Tooth Analysis) 5.7 Analisis Lenturan Gigi Giar (Gear Tooth Bending Analysis) 5.8 Analisis Kehausan Gigi Giar (Gear Tooth Wear Analysis) 5.9 Faktor Keselamatan (Factor of Safety)

PROJEK REKABENTUK (DESIGN PROJECT)

Pembahagian Kumpulan, Penerangan Ringkas Projek Rekabentuk. (Group distribution, Short Briefing of Design Project)

UJIAN 1 (1st TEST) ---- (1.5 hours) (08/10/2013 ; 8:00 9:30 pm)

W8

(28th Oct ~

1st Nov 2013)

6.0 REKABENTUK ACI (SHAFT DESIGN) ---- (3 hours)

6.1 Pengenalan (Introduction) 6.2 Bahan-Bahan Aci (Shaft Materials) 6.3 Aturan pada Aci (Shaft Layout) 6.4 Rekabentuk Aci untuk Tegasan (Shaft Design for Stress) 6.5 Had dan Padanan (Limits and Fits)

Projek berkumpulan (Group Project), Ujian 2 (2nd Test)

W9

(11th ~ 15th Nov 2013)

7.0 GALAS (BEARING) ---- (3 hours)

7.1 Pengenalan (Introduction) 7.2 Jenis-Jenis Galas (Bearing Types) 7.3 Perletakan dan Pemasangan galas (Bearing Mounting and Enclosures) 7.4 Hayat Galas (Bearing Life) 7.5 Hayat Galas Berbeban Pada Kadar Keboleharapan (Bearing Load, Life at Rated Reliability) 7.6 Perhubungan Hayat, Beban dan Keboleharapan (Relating Load, Life and Reliability) 7.7 Kombinasi Beban Jejarian dan Paksi (Combined Radial and Thrust Loading) 7.8 Pelinciran (Lubrication)

UJIAN 2 (2nd Test) ---- (2.5 hours) (12/11/2013 ; 8:00 10:30 pm)

Projek berkumpulan (Group Project), Ujian 3 (3rd Test)

W10 W12

(18th Nov ~

6th Dec 2013)

8.0 PENYAMBUNGAN SEMENTARA (NON-PERMANENT JOINTS) ---- (9 hours)

8.1 Pengenalan (Introduction) 8.2 Definasi dan piawaian bebenang (Thread standard and definition) 8.3 Mekanik skru kuasa (The mechanic of power screw) 8.4 Bebenang pengikat (Threaded fasteners) 8.5 Penyambung: Kekukuhan pengikat (Joints: Fastener stiffness) 8.6 Penyambung: Kekukuhan anggota (Joints: Member stiffness) 8.7 Kekuatan bolt (Bolt strength) 8.8 Ketegangan sambungan : Beban luaran (Tension joints : The external load)

Ujian 3 (3rd Test)

Lampiran A


8.9 Perkaitan daya kilas bolt dengan ketegangan bolt (Relating bolt torque to bolt tension) 8.10 Ketegangan sambungan beban statik berserta pra beban (Statically loaded tension joint with preload) 8.11 Sambungan gasket (Gasketed joints) 8.12 Beban lesu pada ketegangan sambungan (Fatigue loading of tension joints) 8.13 Bolt dan penyambungan rivet dibebankan dalam ricihan (Bolted and riveted joints loaded in shear)

W13 W14

(9th ~ 20th Dec 2013)

9.0 PENYAMBUNGAN KEKAL (PERMANENT JOINTS) ---- (6 hours)

9.1 Simbol Kimpalan (Welding symbol) 9.2 Tegasan pada Penyambungan Kimpalan di dalam Kilasan dan Lenturan (Stresses in Welded Joint in Torsion and Bending) 9.3 Kekuatan Penyambungan Kimpalan (The Strength of Welded Joints) 9.4 Beban Statik dan Lesu (Satic and Fatigue loading)

UJIAN 3 (3rd Test) ---- (2.0 hours) (12/12/2013 ; 8:00 10:00 pm) PENGHANTARAN LAPORAN PROJEK REKABENTUK

(SUBMISSION OF DESIGN PROJECT REPORT 20th Dec 2013)

Laporan Akhir Projek (Final Project report)

TUGASAN / PROJEK (ASSIGNMENT / PROJECT) :

Projek rekabentuk merupakan antara aspek penting di dalam kursus ini. Ia membawa pemberat bernilai 40% di mana ianya bertujuan untuk memenuhi kehendak Universiti yang menawarkan pengajaran dan pembelajaran berkualiti berpusatkan pelajar dengan melaksanakan aktiviti PBL (Problem Based Learning).

Projek ini akan dijalankan di dalam kumpulan di mana setiap kumpulan mempunyai ahli antara 3 hingga 5 orang pelajar. Projek ini berhubungkait dengan merekabentuk sebuah kotak transmisi yang bersesuaian yang akan digunakan pada sebuah mesin. Pelajar perlu menganalisis semua komponen-komponen mekanikal di dalam sistem gearbox/transmisi seperti aci, galas dan giar dari segi kekuatan, keselamatan statik dan lesu, keboleharapan, pergerakan dinamik, jangka hayat dan lain-lain seperti apa yang telah dipelajari dalam teori bagi meramalkan sistem fizikal dan tingkah laku sebenar produk. Kemudian, pelajar perlu membuat pemodelan 3D rekabentuk tersebut beserta dengan lukisan kejuruteraannya dengan menggunakan perisian CAD yang bersesuaian.

Design project is one of the important aspects in this course where it brings 40% of marks. This design project is target to

meet the requirements of the University which offer high quality learning through student-centered learning by implementation

of a PBL (Problem Based Learning) activity.

The project will be carried out in groups where each group has 3 to 5 members. In this project, students have to design an

appropriate gearbox that will apply to a machine. Students must analyze all mechanical components inside the gearbox /

transmission system such as shaft, bearing and gears in term of their strength, static and fatigue safety, reliability, dynamic

motion, life estimation and others like what have been learned in the theory to predict the real physical system and product

behaviour. Then students also need to make a 3D model of their design include with the engineering drawing by using suitable

CAD software.

Lampiran A


PENILAIAN (ASSESSMENT) :

1. Kuiz (Quiz) : 0 %

2. Tugasan (Assignment) : 10 %

3. Ujian 1 (Test 1) : 10 %

4. Ujian 2 (Test 2) : 20 %

5. Ujian 3 (Test 3) : 20 %

5. Projek (Project) Laporan Akhir (Final Report) 35 %

Kemahiran Insaniah (Soft Skill) 5 %

: 40 %

6. Peperiksaan Akhir (Final Examination)

: 0 %

Jumlah (Total) : 100 % RUJUKAN (REFERENCES) :

1. BDA 30803 Lecturer notes

2. Shigley, J. E., Mischke, C. R. & Budynas, R. G., (2010), Mechanical Engineering Design, Ninth Edition, McGraw Hill.

3. Hamrock, Bernard J., Steven R. Schmid, Bo O. Jacobson, (2005), Fundamentals of Machine Elements, 2nd Edition, Boston: McGraw-Hill,

Lampiran A


KEHADIRAN / PERATURAN SEMASA KULIAH (LECTURE ATTENDANCE / REGULATION)

(1) Pelajar mesti hadir tidak kurang dari 80% masa pertemuan yang ditentukan bagi sesuatu mata pelajaran termasuk mata pelajaran Hadir Wajib (HW) dan mata pelajaran Hadir Sahaja (HS). Students must attend lectures not less than 80% of the contact hours for every subject including Compulsory Attendance Subjects (Hadir Wajib HW) and Attendance Only Subjects (Hadir Sahaja HS).

(2) Pelajar yang tidak memenuhi perkara (1) di atas tidak dibenarkan menghadiri kuliah dan menduduki

sebarang bentuk penilaian selanjutnya. Markah sifar (0) akan diberikan kepada pelajar yang gagal memenuhi perkara (1). Manakala untuk mata pelajaran Hadir Wajib (HW), pelajar yang gagal memenuhi perkara (1) akan diberi Hadir Gagal (HG). Students who do not fulfill (1) will not be allowed to attend further lectures and sit for any further examination. Zero mark (0) will be given to students who fail to comply with (1). While for Compulsory Attendance Subjects (Hadir Wajib HW), those who fail to comply with (1) will be given Failure Attendance (Hadir Gagal HG).

(3) Pelajar perlu mengikut dan patuh kepada peraturan berpakaian yang berkuatkuasa dan menjaga disiplin diri masing-masing untuk mengelakkan dari tindakan tatatertib diambil terhadap pelajar. Students must obey all rules and regulations of the university and must discipline themselves in order to avoid any disciplinary actions against them.

(4) Pelajar perlu mematuhi peraturan keselamatan semasa pengajaran dan pembelajaran.

Student must obey safety regulations during learning and teaching process. MATRIK HASIL PEMBELAJARAN KURSUS DAN HASIL PEMBELAJARAN PROGRAM (COURSE

LEARNING OUTCOMES AND PROGRAMME LEARNING OUTCOMES MATRIX)

Dilampirkan (Attached)

Faculty: Faculty of Mechanical and Manufacturing Engineering

Programme:

Course: Mechanical Engineering DesignCode: BDA 30803

No PLO1PLO

2PLO

3PLO

4PLO

5PLO

6PLO

7PLO

8PLO

9PLO10

PLO11

PLO12

PLO13 Delivery Assessment KPI

C3

C4

A3

P4

x x - - - x - - - x - - -

P1 Perception C1 Remembering A1 ReceivingP2 Set C2 Understanding A2 RespondingP3 Guided Response C3 Applying A3 ValuingP4 Mechanism C4 Analyzing A4 OrganisingP5 Complex Overt Response C5 Evaluating A5 Internalising

Revised date: 21/03/2013 P6 Adaptation C6 CreatingPrepared by: MOHD AZWIR BIN AZLAN P7 Origination

Course Learning Outcome, Delivery and Assessment Template

4

Assignment,Test, Project

Compliance to PLO

Bachelor of Mechanical Engineering with Honours

6

Course Learning Outcomes

Seek for the most appropriate information source for component selection in the design project. (A3, LO6)

Total

Psychomotor AffectiveLevel of Learning Taxonomy

Cognitive

5

Analyze several machine components (i.e. gears, shafts and bearing) that function in one mechanical system or machine. (C4, LO10)

Calculate factor of safety by using static and fatigue failure of theories. (C3, LO1)

Assignment,Test, Project

PBL

Lecture, Case Study, PBL

1

2

3

Lecture, Case Study, PBL

100% of students must get 40% of marks and

above


above

ProjectProduce model and simulate by using engineering software for project design validation. (P4, LO2) PBL100% of students must get 40% of marks and

above


aboveProject

C3

A3

P4

C4

Chapter 1Introduction to Engineering Design

Prepared by: Mohd Azwir Bin Azlan

Department of Material and Engineering Design,Faculty of Mechanical and Manufacturing Engineering,University of Tun Hussein Onn Malaysia (UTHM) Johor.

BDA 30803 Notes Mechanical Engineering Design

Week 1

2

BDA 30803 Mechanical Engineering Design

Learning Outcomes

At the end of this topic, the students would be able to apply and appreciate the knowledge to:

understand the basic principles of mechanical engineering and its applications in engineering design.

recognize the approach and the process of engineering design.

practice standards and codes, ethics and professionalism of mechanical engineer.


CHAPTER 1 Introduction to Engineering Design


3Department of Material and Engineering Design,Faculty of Mechanical and Manufacturing Engineering,University of Tun Hussein Onn Malaysia (UTHM) Johor.

What you will be learn here?


1.1 - Design Definition

1.2 - Mechanical Engineering Design

1.3 - Design Process

1.4 - Design Tools and Resources

1.5 - Design Engineers Professional Responsibilities

1.6 - Standards and Codes

1.7 - Economics



1.1 Design Definition

Word design is derived from the Latin designare, which means to designate, or mark out.

Websters gives several definitions, to outline, plot, or plan, as action or work to conceive, invent contrive.

To design is either to formulate a plan for satisfaction of a specified need or to solve a problem.

Design is an innovative and highly iterative process. It is also a decision making process.




1.2 - Mechanical Engineering Design

Mechanical engineering design involves all disciplines of mechanical engineering.

Examples:

A simple journal bearing involves fluid flow, heat transfer, friction, energy transport, material selection, thermomechanical treatments, and so on.

CHAPTER 1 Introduction to Engineering Design BDA 30803 Mechanical Engineering Design







Recognition of Need


Often consist highly creative act, because the need may be only a vague discontent, a feeling of uneasiness, or a something is not right.

Usually triggered by a particular adverse circumstance or a set of random circumstances that arises almost simultaneously.

Recognition of need




Definition of Problem


Is more specific and must include all the specifications for the object that is to be designed.

The specifications are the input and output quantities, the characteristics and dimensions of the space the object must occupy, and all the limitations on these quantities.

Definition of Problem




Synthesis


The combination of ideas into a complex whole.

Is sometimes called the invention of the concept or concept design.

Generate concept variant concept concept selection concept improvement detailing concept.

Synthesis




Analysis and Optimization


Construct or devise abstract models of the system that will admit some form of mathematical analysis.

Carry out to simulate or predict real physical system very well.

Analysis and Optimization




Evaluation


Is the final proof of a successful design and usually involves the testing of a prototype in the laboratory.

Intent to discover if the design really satisfies the needs.

Is it reliable? Will it compete successfully with similar products? Is it economical to manufacture and to use? Is it environmental friendly?

Evaluation




Presentation


Presentation is a selling job.

When designers sell a new idea, they also sell themselves. If they are repeatedly successful in selling ideas, designs and new solutions to management, they begin to receive salary increases and promotions; in fact, this is how anyone succeeds in their profession.

Undoubtedly, many great designs, inventions and creative works, have been simple lost because the originators were unable to explain their accomplishment to others.

Presentation



1.4 - Design Tools and Resources

Today engineer has a great variety of tools and resources available to assist in the solution of design problems:-

Computational Tools CAD (Computer Aided Design) AutoCAD, I-Deas, SolidWorks, ProEngineer CAE (Computer Aided Engineering) Cosmos, Algor, Fluent, ADAMS CAM (Computer Aided Manufacturing) MasterCam, UniGraphic, SolidCAM

Acquiring Technical Information Libraries Encyclopaedia, Monographs, Handbooks, Journals Government sources U.S. Patent and Trademarks Office, SIRIM Professional societies ASME, SAE, SME, ASTM, AWS Commercial vendors Catalogs, Test data, Samples, Cost information Internet the computer network gateway to website associated with most of the

categories listed above.



1.5 - Design Engineers Professional Responsibilities

Required to satisfy the needs of customers and is expected to do so in a competent, responsible, ethical and professional manner.

The way to develop professional work ethic and skills: Sharpen your communication skills either oral or writing

Keep a neat and clear journal / logbook of your activities, entering dated entries frequently

Develop a systematic approach when working on a design problem

Must keep current in the field of expertise by being an active member of a professional society, attending meetings, conferences and seminar of societies, manufacturers, universities, etc.

Conduct activities in an ethical manner.




1.6 - Standards and Codes

Standard is a set of specifications for parts, materials, or processes intended to achieve uniformity, efficiency and a specific quality.

aim to place a limit on the number of items in the specification so as to provide a reasonable inventory of tooling, sizes, shapes and varieties.

Code is a set of specifications for the analysis, design, manufacture and construction of something.

aim to achieve a specific degree of safety, efficiency and performance or quality.



1.6 - Standards and Codes cont

Some organizations or societies that interest to mechanical engineers are:-

Aluminum Association (AA) American Gear Manufacturers Association (AGMA) American Institute of Steel Construction (AISC) American Iron and Steel Institute (AISI) American Society of Mechanical Engineers (ASME) American Society of Testing Materials (ASTM) American Welding Society (AWS) American Bearing Manufacturers Association (ABMA) British Standards Institute (BSI) Industrial Fasteners Institute (IFI) Institution of Mechanical Engineers (I. Mech. E.) International Bureau of Weights and Measure (BIMP) International Standard Organization (ISO) Society of Automotive Engineer (SAE)




1.7 - Economics

The consideration of cost plays such an important role in the design decision process. Some general concepts and simple rules of cost factor study involves:

Standard Sizes Use of standard or stock sizes is a first principle of cost reduction. Specify a parts that are readily available. Select a part that are made and sold in large quantities because of

usually the cost is somewhat less.

Large Tolerances Tolerances, manufacturing processes and surface finish are

interrelated and influence the producibility of the end product in many way.

Large tolerances can often be produced by machines with higher production rates; costs will be significantly smaller.



1.7 - Economics cont

Breakeven Points Use when two or more design

approaches are compared to cost.

The choice between the two depends on a set of conditions such as the quantity of production, the speed of the assembly lines, or some other condition.

The point corresponding to equal cost known as the breakeven point.


Chapter 2Analysis and Synthesis




Week 1

2


Learning Outcomes


perform and analyse load, stress and strain, which applied in standard machine components.


CHAPTER 2 Analysis and Synthesis





2.1 - Material Strength and Stiffness 2.2 - Equilibrium and Free Body Diagram 2.3 - Types of Load 2.4 - Stress 2.5 - Principle Stress for Plane Stress 2.6 - Mohrs Circle for Plane Stress 2.7 - General Three Dimensional (3D) Stress 2.8 - Uniformly Distributed Stresses 2.9 - Normal Stresses for Beams in Bending 2.10 - Shear Stress for Beam in Bending 2.11 - Torsion 2.12 - Stress Concentration



2.1 Material Strength and Stiffness


A typical tension-test specimen. Some of the standard dimensions used for do are 2.5, 6.25 and 12.5 mm and 0.505 in, but other sections and sizes are in use. Common gauge length lo used are 10, 25 and 50 mm and 1 and 2 in.

The standard tensile test is used to obtain a variety of material characteristics and strengths that are used in design.



2.1 Material Strength and Stiffness cont


Stress-strain diagram obtained from the standard tensile test for Ductile material

pl Proportional limits Curve begins to deviate from a straight line No permanent set observable The slope of the linear known as Youngs modulus or the

Modulus of elasticity; E.

el Elastic limit Beyond this limit, plastic deformation will occur and material

will take on permanent set when load is removed

y Yield point Strain begins to increase very rapidly without a

corresponding increase in stress Point a is define by offset method usually about 0.2% from

original gauge length ( = 0.002) Stress at this point known as yield strength, Sy

u maximum stress Stress at this point known as Ultimate or tensile strength, Su

f fracture point Stress at this point known as fracture strength, Sf



2.1 Material Strength and Stiffness cont


Stress-strain diagram obtained from the standard tensile test for Brittle material

y Yield point Strain begins to increase very rapidly without a

corresponding increase in stress Point a is define by offset method usually about 0.2% from

original gauge length ( = 0.002) Stress at this point known as yield strength, Sy

u maximum stress Stress at this point known as Ultimate or tensile strength, Su

f fracture point Stress at this point known as fracture strength, Sf

There is little deformation occurs for brittle material before it fail.

For brittle material ultimate strength is sometimes called as fracture strength



2.2 Equilibrium and Free Body Diagram

= 0F


= 0M

Equilibrium

Assume that the system to be studied is motionless or at most have constant velocity then the system has zero acceleration.

Under this condition, the system is said to be in equilibrium.

For equilibrium, the forces and moments acting on the system balance such that:



2.2 Equilibrium and Free Body Diagram cont


Free Body Diagram (FBD)

Use to simplify the analysis of a very complex structure or machine by isolating or freeing a portion of the total system in order to study the behaviour of one of its segments.

Thus FBD is essentially a means of breaking a complicated problem into manageable segments, analyzing these simple problems, and then usually putting information together again.





FBD Example Gear reducer

Gearbox

Input Shaft

Output Shaft

Info given:

Input torque, Ti = 240 Ibf

Pitch radii of gear ;

G1 r1= 0.75 inG2 r2= 1.50 in

Gear pressure angle, = 20o





Answers:

To = 480 Ibf

RAY = 192 Ibf

RAZ = 69.9 Ibf

RBY = 128 Ibf

RBZ = 46.6 Ibf

RCY = 192 Ibf

RCZ = 69.9 Ibf

RDY = 128 Ibf

RDZ = 46.6 Ibf



2.3 Types of Load


Tension load

Compression load

Bending load

Torsion load

Shear load



2.4 Stress


The force distribution will not be uniform across the surface.

The force distribution at a a point will have components in the normal an tangential direction giving rise to a normal stress () and tangential shear stress ().



2.4 Stress cont


Stress components on surface normal to x direction

General three-dimensional stress

Plane stress with cross-shearsequal

If the stresses in one face is zero, the state of stress is called plane stress.


x

yxy

xy

ave

ave

1,2

1,2

o

Stress components on surface normal to x and y direction

Maximum and minimum normal stresses are called principle stresses which have zero shear stresses

Maximum shear stresses have average normal stresses


2.5 Principle Stress for Plane Stress

22

21 22, xy

yxyx +

+= 2

2

21 2, xy

yx +

=


2yx

ave

+=



2.6 Mohrs Circle for Plane Stress



In design, 3D transformations are rarely performed since most maximum stress states occur under plane stress conditions.

But if there is need to be countable, make sure the principle normal stress are always ordered so that 1 > 2 > 3

Therefore max = 1/3 where


2.7 General Three Dimensional (3D) Stress

221

2/1 = 2

323/2

=


231

3/1 = ; ;


Simple tension, compression and shear loads that always perform this uniform distribution of stress which results


2.8 Uniformly Distributed Stresses

AF=


AF=

: tensile and compression stress

: shear stress e.g. a bolt in shear

F F

A

FF

A


Bending stress, is directly proportional to the distance, cfrom the neutral axis and bending moment, M.


2.9 Normal Stresses for Beams in Bending


IMc=

where;M momentc distance from neutral axisI second moment of area


Bending Moment DiagramIs sometimes needed to determine:-

location where moment is maximum

moment in specified location.

E.g.


2.9 Normal Stresses for Beams in Bending cont


A B

Loading diagram

Shear-force diagram

Bending-moment diagram

A B

MAMB

Mmax =


Maximum shear stress exists when y1= 0, which is at bending neutral axis

As it move away from the neutral axis, the shear stress decrease parabolically until it zero at the outer surfaces where y = c


2.10 Shear Stresses for Beams in Bending



Formula for Maximum Shear Stress Due to Bending


2.10 Shear Stresses for Beams in Bending cont


A23

max =

A 2max =

A34

max =

Beam Shape Formula Beam Shape Formula

Rectangular

Circular

Hollow, thin-walled round

Structural I beam (thin-walled)

Web

webA =max


Any moment vector that is collinear with an axis of a mechanical element is called a torque vector or torsion.

For solid round bar, the shear stress is zero at the center at maximum at the surface.


2.11 Torsion

JTr=max


where ;T = torquer = bar radiusJ = polar second moment of area

For noncircular cross-section members especially rectangular b x csection bar which use to transmit torque, maximum shearing stress is:

+=cbbc

T/8.132max

where ;b = is the longer sidec = is the shorter side


Obtain the torque T from a consideration of the power and speed of a rotating shaft


2.11 Torsion cont

025,63000,198)12(000,332

000,33TnTnTnFVH ====

TH =


nHT 55.9=

where ;H = power (hp)

(1 hp = 33,000 ft.Ib/s)T = torque (Ibf.in)

where ;H = power (W)T = torque (Nm) = angular velocity (rad/s)

When USC units is used, the equation is :

When SI units is used, the equation is : The torque T corresponding to the power in watts is given approximately by

n = shaft speed (rev/min)F = force (Ibf)V = velocity (ft/min)



2.12 Stresses Concentration


w

B B

A A

max

d

Stress trajectories

Stress distribution

Stress distribution near a hole in a plate loaded in tension. The tensile stress on a section B-B, remote from the hole is

= F/A where A = wt and t is the plate thickness. On a section

at A-A, through the hole, the area A0 = (w-d)t

and nominal stress,o = F/Ao.

Note that the stress are increases when move towards to the hole and maxsimum stress occur at the edge of the hole where the load lines become very compact there.

Stress distribution near a hole in a plate loaded in tension. The tensile stress on a section B-B, remote from the hole is

= F/A where A = wt and t is the plate thickness. On a section

at A-A, through the hole, the area A0 = (w-d)t

and nominal stress,o = F/Ao.

Note that the stress are increases when move towards to the hole and maxsimum stress occur at the edge of the hole where the load lines become very compact there.

= F/A

0= F/A0

A = wt

A0 = (w-d )t

max >0 >

Any discontinuity in a machine part alter a stress distribution in the neighbourhood of the discontinuity

Such discontinuities are called stress raisers, and the regions in which they occur are called areas of stress concentration.



2.12 Stresses Concentration cont

otK

max=


A theoretical or geometric, stress concentration factor Kt or Kts is used to relate the actual maximum stress at the discontinuity to the nominal stress.

The factors are define by the equations:

where Kt is used for normal stress and Kts for shear stress.

The stress concentration factor depends on the geometry of the part which cause difficult problem since not many analysis of geometric shapes solutions can be found.

However stress concentration factors for a variety of standard geometries may be found in Tables A-15 and A-16.

In static loading, stress concentration factors are applied as follow to predict critical stress;

x Ductile material (f 0.05) not usually applied since has a strengthening effect in plastic region Brittle material (f < 0.05) applied to the nominal stress before comparing it with strength

otsK

max=



Example 1:


Figure below shows a crank loaded by a force F = 300 Ibf that causes twisting and bending of a in diameter round bar fixed to a support at the origin of the reference system. In actuality, the support may be an inertia that we wish to rotate, but for the purpose of a stress analysis we can consider this is a static problem.

a) Draw separate FBD of the shaft AB and the arm BC, and compute the values of all forces, moment, and torques that act. Label the directions of the coordinate axes on these diagram.

b) Compute the maxima of the torsional stress and the bending stress in the arm BC.

c) Locate a stress element on the top surface of the shaft at A, and calculate all the stress components that act upon this element.

d) Determine the maximum normal and shear stresses at A.



Solution :


(a) The results are:-

At C ; F = -300j Ibf, T = -450k Ibf.in

At end B of arm BC ; F = 300j Ibf, M = 1200i Ibf.in, T = 450k Ibf.in

At end B of shaft AB ; F = -300j Ibf, T = -1200i Ibf.in, M = -450k Ibf.in

At A ; F = 300j Ibf, T = 1200i Ibf.in, M = 1950k Ibf.in


(300 Ibf)

(450 Ibf.in)

(300 Ibf)

(1200 Ibf.in)

(450 Ibf.in)


Solution :

400,1925.0/25.1

8.13)25.0(25.1

4502 =

+=

236

12

)2/(bh

Mbh

hMI

Mc ===

+=cbbc

T/8.132


(b) Maximum torsional and bending stress at arm BC

The bending moment will reach a maximum near the shaft at B which is 1200 Ibf.in

psi

For rectangular section having torsional stress

psi

400,18)25.1(25.0

)1200(62 ==



Solution :

500,14)75.0()1200(163 ==

3432

64

)2/(dM

ddM

IMc

===

34

1632/)2/(

dT

ddT

JTr

===


(c) Stress element on the top surface of the shaft at A

The bending is tensile and is

psi

The torsional stress is

psi

100,47)75.0()1950(32

3 ==

(300 Ibf)(450 Ibf.in)

(300 Ibf)

(1200 Ibf.in)

(1950 Ibf.in)

(1200 Ibf.in)

A

z

x

xx xz

xz



Solution :

22

5.142

01.47 +

=

22

1 22 xzzxzx +

++= 22

1 2 xzzx +

=


(d) Maximum normal and shear stresses at A.

The maximum normal stress is given by

kpsi

The maximum shear stress is

kpsi

22

5.142

01.472

01.47 +

++=

2.51= 7.27=



Example 2:


The 1.5-in diameter solid steel shaft shown in figure below is simply supported at the ends. Two pulleys are keyed to the shaft where pulley B is of diameter 4.0 in and pulley C is of diameter 8.0 in. Considering bending and torsional stress only, determine the locations and magnitudes of the greatest tensile, compressive, and shear stresses in the shaft.



Solution :


(a) Figure below shows the FBD of the net forces, reactions and torsional moments on the shaft.-

Although this is a 3D problem, the components of the moment vector is perform in a two plane analysis.



Solution :

22zy MMM +=

565740004000 22 =+=CM


Thus the moment are label as My versus x for xy plane and Mz versus x for xz plane:-

The net moment on a section is the vector sum of the components. That is

824680002000 22 =+=BM Ibf.inIbf.in



Solution :

890,24)5.1()8246(3232

64/)2/(

334 ===== dM

ddM

IMc

2414)5.1()1600(1616

32/)2/(

334 ===== dT

ddT

JTr

120,2524142890,24

2890,24

222

22

2

1 =+

+=+

++= xzzxzx


In this case where the shaft diameter is same along the axis, maximum bending stress occurs at location where the bending moment is maximum which is at point B.

psi

The maximum torsional shear stress occurs between B and C and is:

psi

Maximum tensile stress 1 is given by:

680,1224142890,24

22

22

2

1 =+

=+

= xzzx

psi

The extreme shear stress 1 is given by:

psi

Chapter 3Static Design Failure of Theories




Week 2

2


Learning Outcomes


explain and apply the various static failure theories, including the use of safety factors and reliability in mechanical engineering design.


CHAPTER 3 Static Design Failure of Theories





3.1 - Introduction

3.2 - Why needs Failure Theories?

3.3 - Static Failure Theories

3.4 - Maximum Shear Stress (MSS) Theory

3.5 - Distortion Energy (DE) Theory

3.6 - Colomb-Mohr Theory

3.7 - Maximum Normal Stress (MSN) Theory

3.8 - Modification of Mohr Theory



3.1 Introduction


Tacoma Bridge Failure 1940

It was built with shallow plate girders for the aesthetics purposes.This vibration motion lasted 3 hours and the bridge collapsed. The failure caused millions fund loss.

In 1950, the bridge was rebuilt and truss-girders were used to increase the stiffness of the bridge.



3.1 Introduction cont


Static Load:-

is a stationary force or couple applied to a member the force or couple must be unchanging in magnitude, point or points of

application, and direction.

can produce axial tension or compression, a shear load, a bending load, a torsional load or any combination of these.

Safety and Failure:-

Failure can mean a part has separated into two or more pieces and become permanently deformed

Why parts fail stresses exceed its strength. Can be categories : under static loading

under dynamic loading



3.2 Why needs failure theories?


To design parts or components that meet it requirements and functions as it suppose to be.

It suppose to test the real components exactly the same loading conditions to obtain precise information => increase cost



3.3 Static Failure Theories


Actually, there is no universal theory of failure for the general case of material properties and stress state.

Over the years, several hypotheses have been formulated and tested, leading todays engineering practice.

Being accepted worldwide, these practices are used as theoriesas most designers do.



3.3 Static Failure Theories cont


Static Failure Theories

Ductile Materials

-

-

05.0fyycyt SSS ==

Brittle Materials

-

- ucut SS &05.0



3.4 Maximum Shear Stress (MSS) Theory


predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test of the same material.

also known as Tresca or Guest Theory. for a simple tensile stress , max. shear stress occurs on a surface 450

from the tensile surface with a magnitude of:

or at yield,2maxyS=2max =



3.4 Maximum Shear Stress (MSS) Theory cont


2231 y

S=

2maxyS=





Therefore, taking N as safety factor;

ysy SS 50.0=

)(2/)(2/50.0

3131maxmax ====yyysy SSSSN

StressorLoadAppliedStressorLoadAllowableMaximumUsually safety factor, N is defined by;





The maximum-shear stress (MSS) theory for plane stress, where A and B are the two nonzero principal stresses

Case 1 : A B 0. For this case, 1 = A , 3 = 0.Case 2 : A 0 B. Here, 1 = A , 3 = B.Case 3 : 0 A B. For this case, 1 = 0 , 3 = B.



3.5 Distortion Energy (DE) Theory


Predicts that yielding begins when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension for the same material.

Ud at element in specimen Ud for yield in simple tension. Also known as von Mises or von Mises-Hencky Theory

Developed by studying a unit volume in a three-dimensional stress state



3.5 Distortion Energy (DE) Theory cont


(a) Element with triaxial stresses; this element undergoes both volume change and angular distortion.

(b) Element under hydrostatic tension undergoes only volume change.

(c) Element has angular distortion without volume change.

U = Uv + Ud





1U =2

( )1 1 2 2 3 3 + +1U = 2( )1 2 31 v vE = 1( )2 1 31 v vE = 2( )3 1 21 v vE = 3

( )2 2 21 2 3 1 2 2 3 1 32v + + + + 1U = 2E

strain energy =

for 3-D analysis;

where

------- (1)





3321 ++=av

( )vE

U avv 2123 2 =

The strain energy for producing only volume change Uv can be obtained by substituting av for 1, 2, and 3 in Eq. (1). The result is:-

------- (2)





( )2 2 21 2 3 1 2 2 3 1 32v = + + + + 1

2E

2 2 21 2 3 1 2 2 3 1 3

13dvUE

+ = + + ( ) 21

3d yv

U SE+=

vd UUU =( )

E

v

2

213

32

321

++

--- for element in speciment

--- for yield in simple tension where 1 = Sy , 2 = 3 = 0





Von Mises Effective Stress

' 2 2 21 2 3 1 2 2 3 1 3 = + +

( ) ( ) ( ) ( )2 2 2 2 2 2' 62

x y y z z x xy yz zx + + + + +=

' 2 21 1 3 3 = +

' 2 2 23x y x y xy = + +

(for 2D principal stress)

(for 2D plane stress)




2' yS

[ ] 2313221232221 3131 ySEE ++++


Ud at element in specimen Ud for yield in simple tension.

yS++ 313221232221

Therefore, safety factor N is:'ySN =





2 2 2 2 21 1 1 1 1 max3 3yS = + + = =

1 max0.577 3y

y

SS = = =

For Pure Shear:

max

average

averageysy SS 577.0=

maxsySn =





The distortion-energy (DE) theory for plane stress states. This is plot of points with

= Sy.

ysy SS 577.0=



3.6 Coulomb-Mohr Theory


- used when Syt Syc- based on Mohrs theory, whereby the failure line is assumed to be

straight

Three Mohr circle, one for the

unaxial compression test, one

for the test in pure shear, and

one for the unaxial tension test,

are used to define failure by the

Mohr hypothesis. The strengths

Sc and St are the compressive and tensile strengths,

respectively, they can be used for yield or ultimate strength.



3.6 Coulomb-Mohr Theory cont


131 =ct SS

nSS ct131 =

ycyt

ycytsy SS

SSS +=

From the diagram, equation developed can be simplified to:

Incorporating the safety factor;

For pure torsional shear strength;

maxsySn =



3.7 Maximum Normal Stress (MNS) Theory


- states that failure occurs whenever one of the three principal stresses equals or exceeds the strength

1 > Sut or 3 < - Suc

- where Sut and Suc are the ultimate tensile and compressive strength respectively



3.7 Maximum Normal Stress (MNS) Theory cont


Graph of maximum-normal

stress (MNS) theory of

failure for plane stress

states. Stress states that

plot inside the failure locus are safe.



3.8 Modification of Mohr Theory


a) Brittle Coulomb Mohr Theory

b) Modified I-Mohr

c) Modified II-Mohr

- 3 modifications of the existing Mohr theory are applicable in analyzing brittle materials.

- by limiting the discussion to plane stresses, those theories are as follows:



3.8 Modification of Mohr Theory cont


Theories

Equations

Brittle Coulomb-Mohr

Modified I-Mohr

Modified II-Mohr

0 BA BA 0 BA 0

nSut

A nSS ucB

ut

A 1= nSuc

B

nSut

A

nSSSSS

uc

B

utuc

Autuc 1)( =

nSuc

B

nSut

A

12

=

++

ucut

utB

ut

A

SSSn

Sn

nSuc

B



3.8 Modification of Mohr Theory cont




Conclusion


Exercise 1: Problem Certain stresses are applied at one object which 1 = 200 MPa and 2 = -50 MPa. This object is made by steel that it has a yield strength of 500 MPa. Find the factor of safety of this object by using DE and MSS theory. Solve FOS by using graph method. Answer

500

500

- 500

- 500

[MPa]

[MPa]

200

- 50 A

B C

X

Y

FOS

200xnMSS =

200ynDE =

Exercise 2: Problem Determine the safety factors for the bracket rod shown in figure above based on both the distortion-energy theory and the maximum shear theory and compare them. Given: The material is 2024-T4 aluminum with a yield strength of 47 000 psi. The rod length l = 6 in and arm a = 8 in. The rod outside diameter d = 1.5 in. Load F=1 000 lb. Assumptions: The load is static and the assembly is at room temperature. Consider shear due to transverse loading as well as other stresses. Answer

Element at point A 1. ( ) ( )( )1 000 6 0.75 18 108

0.249xFl cMC psi

I I = = = =

( ) ( )( )1 000 8 0.75 12 072

0.497xzFa rTr psi

J J = = = =

2. psixyzx 15090120722

0181082

22

22

max =+

=+

=

psizx 24144090152

181082 max1

=+=++=

02 =

psizx 6036150902

181082 max3

==+=

3. 23312

1' +=

psi66127)6036()6036(2414424144' 22 =+=

4. 7.12766147000

'===

ySN ------------- DE theory

5. 6.115090

)47000(50.050.0

max

=== ySN --------------- MSS theory

Element at point B

6. psiAV

bending 755)767.1(3)1000(4

34 ===

psibendingtorsion 1282775512072max =+=+=

7. 1.212827

)47000(577.0577.0

max

=== ySN --------- DE theory

8.112827

)47000(50.050.0

max

=== ySN -------- MSS theory

Exercise 3: Problem A 25-mm diameter shaft is statically torqued to 230 Nm. It is made of cast 195-T6 aluminium, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft. Answer MPa

dT

JTr 75

5.2)230(1616

33 ==== The two nonzero principal stresses are 75 and -75 MPa, making the ordered principal stresses 1 = 75, 2 = 0, and 3 = -75 MPa.

10.1170/)75(160/75

1131

===

ycyt SS

n Alternatively;

ycyt

ycytsy SS

SSS += 10.175

4.82

max

=== sySn

Chapter 4Fatigue Design Failure of Theories




Week 3 & 4

2


Learning Outcomes


explain and apply the fatigue failure of theories, including the use of safety factors and reliability in mechanical engineering design.

confidently apply this technique in the selection and analysis of machine components, and make decision on material selection.


CHAPTER 4 Fatigue Design Failure of Theories




4.1 - Introduction to Fatigue 4.2 - Fatigue load and failure 4.3 - Life and fatigue strength 4.4 - Stress Life Method 4.5 - Endurance limits, Se 4.6 - Endurance Limit Modifying Factors (Marin Factor) 4.7 - Stress Concentration and Notch Sensitivity 4.8 - Fatigue Strength 4.9 - Characterizing Fluctuating Stresses 4.10 - Combination of Loading Modes 4.11 - Safety Factor




4.1 Introduction


Cause by the action of static load or load that acts only once until a component destruct such as in tensile test. However this phenomena is rarely occur.

Cause by the action of variable, repeated, alternating or fluctuating load and this load are often found in many failure cases that occurs.



4.2 Fatigue Load and Failure


Often, machine members are found to have failed under the action of

repeated or fluctuating stresses; yet the most careful analysis reveals

that the actual maximum stresses were well below the ultimate

strength of the material, and quite frequently even below the yield

strength. The most distinguishing characteristic of these failures is

that the stresses have been repeated a very large number of times.

Hence the failure is called a fatigue failure.



4.2 Fatigue Load and Failure cont


1st Case Bending a steel wire repeatedly 2nd Case Impact and Vibration on vehicle axle

3rd Case Steel Bridge 4th Case - Vehicle Suspension

NG

NG

Too much repeatedlyBending that beyond the limit will break the steel wire

Too much repeatedly Impact and Vibrationwill break the axle

Example of repeated load





1st Case Continuously bending the steel wire

Stress : Bending stress;

Tension

Compression

Wire is bend from the top

Top of the wire is suffered toTension (+)

Meanwhile the bottom is suffered to

Compression (-)

Graph plotting for overall wire bending process:

This type of stress is known as Completely Reverse Stress

Wire is bend from the top

Top of the wire is suffered toTension (+)

Meanwhile the bottom is suffered to

Compression (-)

Graph plotting for overall wire bending process:

This type of stress is known as Completely Reverse Stress

+

m = 0

a

art





2nd Case Vehicle Axle

Stress : Bending Stress and Shear Stress Torsion shear stress: is cause when power

from the engine is transmit to the tire. Torque are required to overcome tire

friction and vehicle weight and this stress is always assume as constant.

Shaft axle is suffered to bending and shear stress while running.

m

Shear Stress cause by torsion





Fatigue failure of a bolt due to repeated unidirectional bending. Fatigue failure start with small crack that unseen with naked eyes and also difficult to detect with X-ray at the thread root at A, propagated across most of the cross section shown by the beach marks at B, before

final fast fracture at C.

Fatigue failure of a bolt due to repeated unidirectional bending. Fatigue failure start with small crack that unseen with naked eyes and also difficult to detect with X-ray at the thread root at A, propagated across most of the cross section shown by the beach marks at B, before

final fast fracture at C.

Crack often start at weak part geometrieswhich have discontinuity in material such as at holes, keyways, notch, fillet and others (at

this location, the stress is high because of high stress concentration)

Crack often start at weak part geometrieswhich have discontinuity in material such as at holes, keyways, notch, fillet and others (at

this location, the stress is high because of high stress concentration)



4.3 Life and Fatigue Strength


Three major fatigue life models

Methods predict life in number of cycles to failure, N, for a specific level of loading

i. Stress-life method9 Least accurate, particularly for low cycle applications9 Most traditional, easiest to implement

ii. Strain-life method9 Detailed analysis of plastic deformation at localized regions9 Several idealizations are compounded, leading to

uncertainties in results

iii. Linear-elastic fracture mechanics method9 Assumes crack exists9 Predicts crack growth with respect to stress intensity



4.3 Life and Fatigue Strength cont


Fatigue strength also have its maximum limits.Fatigue strength also have its maximum limits. R. R Moore Test

Procedure of rotating beam testProcedure of rotating beam test

Constant bending load is applied on test sample and rotate it at high rpm.

Stress that have been applied on first test is an ultimate strength value Sut

compression

tension

motor

Sample ujian

Test samples are rotate until failure and the failure number of cycle is then be record.

The tests are repeat with new stress that lower than before.

Then S-N diagram graph which indicate number of cycle (N) and Fatigue Strength (Sf) is plotted.

motor

specimen

F

Rotating Beam Test



4.4 Stress Life Method


An S-N diagram plotted from the results of completely reversed axial fatigue test. Material UNS G4100 steel, normalized. S : strength, N : cycle

An S-N diagram plotted from the results of completely reversed axial fatigue test. Material UNS G4100 steel, normalized. S : strength, N : cycle

S-N Diagram

Low-cycle fatigueHigh-cycle fatigue

Endurance limit , Se or known as fatigue limit



4.5 Endurance Limit, Se


Basically, the fatigue endurance limits Se are determine through the test. Yet the data are also available on these standard:

i. American Society of Testing and Materials (ASTM)ii. American Iron and Steel Institute (AISI)

iii. Society of Automotive Engineer (SAE)

Basically, the fatigue endurance limits Se are determine through the test. Yet the data are also available on these standard:

i. American Society of Testing and Materials (ASTM)ii. American Iron and Steel Institute (AISI)

iii. Society of Automotive Engineer (SAE)

It is unrealistic to expect the endurance limit of a mechanical or structural member to match the values obtained in the laboratory

It is unrealistic to expect the endurance limit of a mechanical or structural member to match the values obtained in the laboratory

So, the values obtain from lab test are known as Rotary beam test specimen endurance limit, Se . However there is a relation exist between Se and Sut .

So, the values obtain from lab test are known as Rotary beam test specimen endurance limit, Se . However there is a relation exist between Se and Sut .

Value of a mechanical or structural member to match the values obtained in the laboratory after considering other factors that influence the fatigue life is known as endurance limits, Se .

Value of a mechanical or structural member to match the values obtained in the laboratory after considering other factors that influence the fatigue life is known as endurance limits, Se .

Se refer to the endurance limit of the controlled laboratory specimenSe refer to the endurance limit of an actual machine element subjected to any kind of loading



4.5 Endurance Limit, Se cont


Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought irons and steels. Ratios of Se /Sut of 0.60, 0.50, and 0.40 are shown by the solid and dashed lines.



4.5 Endurance Limit, Se cont


Assumption of Se value for steelAssumption of Se value for steel

For student application:

Se = 0.5 Sut Sut 1400 MPa [ 200 kpsi ]Se = 700 MPa [ 100 kpsi ] Sut > 1400 MPa

For student application:

Se = 0.5 Sut Sut 1400 MPa [ 200 kpsi ]Se = 700 MPa [ 100 kpsi ] Sut > 1400 MPa

For real engineering practice:

Se = 0.4 Sut Sut 1400 MPaSe = 550 MPa [ 84.1 kpsi ] Sut > 1400 MPa

For real engineering practice:

Se = 0.4 Sut Sut 1400 MPaSe = 550 MPa [ 84.1 kpsi ] Sut > 1400 MPa



4.6 Endurance Limit Modifying Factors


Factors that influence the fatigue life and endurance limits.Factors that influence the fatigue life and endurance limits.



4.6 Endurance Limit Modifying Factors cont


A Marin Equation is therefore written the endurance limit Se as:A Marin Equation is therefore written the endurance limit Se as:

Se = kakbkckdkekfSe

Where,Se = rotary beam test endurance limitka = surface condition modification factorkb = size modification factorkc = load modification factorkd = temperature modification factorke = reliability factorkf = miscellaneous effect modification factor

Where,Se = rotary beam test endurance limitka = surface condition modification factorkb = size modification factorkc = load modification factorkd = temperature modification factorke = reliability factorkf = miscellaneous effect modification factor





Surface Condition Modification Factor, kaSurface Condition Modification Factor, ka

The surface modification factor depends on the quality of the finish of the actual part surface and on the tensile strength of the part material. The data can be represented by:

ka = a Sbutwhere Sut is the ultimate strength and a and b value are to be found using below table

TABLE 4-1: Parameters for Marin surface modification factor.

Surface FinishFactor a Exponent

bSut, kpsi Sut, MPa

Ground 1.34 1.58 -0.085

Machine or cold drawn 2.70 4.51 -0.265

Hot-rolled 14.4 57.7 -0.718

As-forged 39.9 272 -0.995





Surface Condition Modification Factor, kaSurface Condition Modification Factor, ka

EXAMPLE 1

A steel has a minimum ultimate strength of 520 MPa and a machinedsurface. Estimate ka.

SolutionFrom Table 41, a = 4.51 and b =0.265. Then,

Answer ka = 4.51(520)0.265 = 0.860





Size Factor, kbSize Factor, kb

~ there is no size effect, so size factor kb = 1.0 ~ there is no size effect, so size factor kb = 1.0

Size factor for ROTATING ROUND bar is given by below equation :

whered effective dimension

( )

Any discontinuity in a machine part alter a stress distribution in the neighbourhood of the discontinuity

Such discontinuities are called stress raisers, and the regions in which they occur are called areas of stress concentration.

Existence of irregularities or discontinuities, such as holes, grooves, or notches, in a part increases the theoretical stresses significantly in the immediate at nearby region of the discontinuity.

F F F F

F FF F

Regular feature Changes in cross section

notch hole

Load lines at several types of bar that has been suffered by axial force



4.7 Stress Concentration and Notch Sensitivity cont


A theoritical or geometric stress concentration factor Kt is used to relate the actual maximum stress at the discontinity to the nominal stress.

A theoritical or geometric stress concentration factor Kt is used to relate the actual maximum stress at the disc

bda 30803 notes (student version - printable) sem 2 2012_2013

Documents