BDA 30803 Notes (Student Version - Printable) Sem 2 2012_2013.pdf

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 (Lecturer’s 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 Engineer’s 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 (Mohr’s 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 PLO1

PLO2

PLO3

PLO4

PLO5

PLO6

PLO7

PLO8

PLO9

PLO10

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) PBL


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 Engineer’s 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.”

• Webster’s 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


1.3 - Design Process





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 ToolsCAD (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 InformationLibraries – 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 thecategories listed above.



1.5 - Design Engineer’s 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 SizesUse 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 TolerancesTolerances, 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 PointsUse 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 - Mohr’s 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 Young’s 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 (ε = 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 (ε = 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 in

G2 ⇒ 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 (σ) and tangential shear stress (τ).



2.4 – Stress – cont…


Stress components on surface normal to x direction

General three-dimensional stress

Plane stress with “cross-shears”equal

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


σx

σy

τxy

τxy

σave

σave

τ1,2

τ1,2

φ

θ 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 τσσσσ

σσ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −±

+= 2

2

21 2, xy

yx τσσ

ττ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −±=


2yx

ave

σσσ

+=



2.6 – Mohr’s 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 σ1 > σ2 > σ3

• Therefore τmax = τ1/3 where


2.7 – General Three Dimensional (3D) Stress

221

2/1σστ −

= 232

3/2σστ −

=


231

3/1σστ −

= ; ;


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


2.8 – Uniformly Distributed Stresses

AF

=σ


AF

=τ

: tensile and compression stress

: shear stress e.g. a bolt in shear

F F

A

FF

A


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


2.9 – Normal Stresses for Beams in Bending


IMc

=σwhere;

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

MA

MBMmax =


• 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

maxντ =

Aντ 2

max =

A34

maxντ =

Beam Shape Formula Beam Shape Formula

Rectangular

Circular

Hollow, thin-walled round

Structural I beam (thin-walled)

Web

webAντ =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

=maxτ


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:

⎟⎠⎞

⎜⎝⎛ +=

cbbcT

/8.132maxτ

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 ====

ππ

ωTH =


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)ω = 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

σmax

σ

σ

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

σ= 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,σo = 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

σ= 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,σo = 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.

σ= F/A

σ0= F/A0

A = wt

A0 = (w-d )t

σmax >σ0 > σ

• 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

σσmax=


• 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 (εf ≥ 0.05) – not usually applied since has a strengthening effect in plastic region

√ Brittle material (εf < 0.05) – applied to the nominal stress before comparing it with strength

otsK

ττmax=



Example 1:


Figure below shows a crank loaded by a force F = 300 Ibf that causes twisting and bending of a ¾ 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===σ

⎟⎠⎞

⎜⎝⎛ +=

cbbcT

/8.132τ


(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 ==

π

3432

64

)2/(dM

ddM

IMc

ππσ ===

34

1632/

)2/(dT

ddT

JTr

ππτ ===


(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 ==π

(300 Ibf)(450 Ibf.in)

(300 Ibf)

(1200 Ibf.in)

(1950 Ibf.in)

(1200 Ibf.in)

A

z

x

σxσx τxz

τxz



Solution :

22

5.142

01.47+⎟

⎠⎞

⎜⎝⎛ −

=

22

1 22 xzzxzx τσσσσσ +⎟⎠⎞

⎜⎝⎛ −

++

= 22

1 2 xzzx τσστ +⎟⎠⎞

⎜⎝⎛ −

=


(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.in

Ibf.in34Department of Material and Engineering Design,

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


Solution :

890,24)5.1()8246(3232

64/)2/(

334 =====πππ

σdM

ddM

IMc

2414)5.1()1600(1616

32/)2/(

334 =====πππ

τdT

ddT

JTr

120,2524142890,24

2890,24

222

22

2

1 =+⎟⎠⎞

⎜⎝⎛+=+⎟

⎠⎞

⎜⎝⎛ −

++

= xzzxzx τσσσσσ


• 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 σ1 is given by:

680,1224142890,24

22

22

2

1 =+⎟⎠⎞

⎜⎝⎛=+⎟

⎠⎞

⎜⎝⎛ −

= xzzx τσστ

psi

• The extreme shear stress τ1 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 today’s 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

-

-

05.0≥fεyycyt SSS ==

Brittle Materials

-

- ucut SS &05.0<fε

Yield Criteria-Maximum Shear Stress-Distortion Energy-Ductile Coulomb-Mohr

Fracture Criteria-Maximum Normal Stress-Brittle-Coulomb-Mohr

( example 1 , example 2 , example 3 )



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 σ, max. shear stress occurs on a surface 450

from the tensile surface with a magnitude of:

or at yield,2max

yS=τ

2maxστ =



3.4 – Maximum Shear Stress (MSS) Theory – cont…


2231 yS=

−∴

σσ

2maxyS

=τ





Therefore, taking N as safety factor;

ysy SS 50.0=

)(2/)(2/50.0

3131maxmax σσσσττ −=

−=== yyysy SSSS

N

StressorLoadAppliedStressorLoadAllowableMaximumUsually safety factor, N is defined by;





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

Case 1 : σA ≥ σB ≥ 0. For this case, σ1 = σA , σ3 = 0.

Case 2 : σA ≥ 0 ≥ σB. Here, σ1 = σA , σ3 = σB.

Case 3 : 0 ≥ σA ≥ σB. For this case, σ1 = 0 , σ3 = σ 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





σε1U =2

( )1 1 2 2 3 3σ ε σ ε σ ε+ +1U =2

( )1 2 31 v vE

ε σ σ σ= − −1

( )2 1 31 v vE

ε σ σ σ= − −2

( )3 1 21 v vE

ε σ σ σ= − −3

( )2 2 21 2 3 1 2 2 3 1 32vσ σ σ σ σ σ σ σ σ⎡ ⎤+ + − + +⎣ ⎦

1U =2E

strain energy =

for 3-D analysis;

where

------- (1)





3321 σσσσ ++

=av

( )vE

U avv 21

23 2

−=σ

The strain energy for producing only volume change Uv can be obtained by substituting σav for σ1, σ2, and σ3 in Eq. (1). The result is:-

------- (2)





( )2 2 21 2 3 1 2 2 3 1 32vσ σ σ σ σ σ σ σ σ⎧ ⎫⎡ ⎤= + + − + +⎨ ⎬⎣ ⎦⎩ ⎭

12E

2 2 21 2 3 1 2 2 3 1 3

13d

vUE

σ σ σ σ σ σ σ σ σ+ ⎡ ⎤= + + − − −⎣ ⎦

( ) 213d y

vU S

E+

=

vd UUU −=

( )

E

v

2

213

32

321 −⎟⎠⎞

⎜⎝⎛ ++

−

σσσ

--- for element in speciment

--- for yield in simple tension where σ1 = Sy , σ2 = σ3 = 0





Von Mises Effective Stress

' 2 2 21 2 3 1 2 2 3 1 3σ σ σ σ σ σ σ σ σ σ= + + − − −

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

6

2x y y z z x xy yz zxσ σ σ σ σ σ τ τ τ

σ− + − + − + + +

=

' 2 21 1 3 3σ σ σ σ σ= − +

' 2 2 23x y x y xyσ σ σ σ σ τ= + − +

(for 2D principal stress)

(for 2D plane stress)




2' yS≥σ

[ ] 2313221

23

22

21 3

13

1yS

EEυσσσσσσσσσυ +

≥−−−+++


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

yS≥−−−++ 31322123

22

21 σσσσσσσσσ

Therefore, safety factor N is:'yS

Nσ

=





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

1 max0.577 3y

y

SSσ τ= = =

For Pure Shear:

τ max

σ average

σ average

ysy SS 577.0=maxτsyS

n =





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

σ’ = Sy.

ysy SS 577.0=



3.6 – Coulomb-Mohr Theory


- used when Syt ≠ Syc

- based on Mohr’s 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

σσ

nSS ct

131 =−σσ

ycyt

ycytsy SS

SSS

+=

From the diagram, equation developed can be simplified to:

Incorporating the safety factor;

For pure torsional shear strength;

maxτsyS

n =



3.7 – Maximum Normal Stress (MNS) Theory


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

σ1 > Sut or σ3 < - 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 σσ BA σσ ≥≥ 0BA σσ ≥≥0

nSut

A ≥σnSS uc

B

ut

A 1=−

σσn

SucB −≥σ

nSut

A ≥σ

nSSSSS

uc

B

utuc

Autuc 1)(=−

− σσ

nSuc

B −≥σ

nSut

A ≥σ

12

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+

+ucut

utB

ut

A

SSSn

Sn σσ

nSuc

B −≥σ



3.8 – Modification of Mohr Theory – cont…




Conclusion


Exercise 1: Problem Certain stresses are applied at one object which σ1 = 200 MPa and σ2 = -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 0.7518 108

0.249x

Fl cMC psiI I

σ = = = =

( ) ( )( )1 000 8 0.75

12 0720.497xz

Fa rTr psiJ J

τ = = = =

2. psixyzx 1509012072

2018108

22

22

2

max =+⎟⎠⎞

⎜⎝⎛ −

=+⎟⎠⎞

⎜⎝⎛ −

= τσσ

τ

psizx 24144090152

181082 max1 =+=++

= τσσ

σ

02 =σ

psizx 6036150902

181082 max3 −=−=−+

= τσσ

σ

3. 2331

21' σσσσσ +−=

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

4. 7.12766147000

'===

σyS

N ------------- DE theory

5. 6.115090

)47000(50.050.0

max

===τ

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

Element at point B

6. psiAV

bending 755)767.1(3)1000(4

34

===τ

psibendingtorsion 1282775512072max =+=+= τττ

7. 1.212827

)47000(577.0577.0

max

===τ

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

8.112827

)47000(50.050.0

max

===τ

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

MPadT

JTr 75

5.2)230(1616

33 ====ππ

τ The two nonzero principal stresses are 75 and -75 MPa, making the ordered principal stresses σ1 = 75, σ2 = 0, and σ3 = -75 MPa.

10.1170/)75(160/75

1131

=−−

=−

=∴

ycyt SS

n σσ Alternatively;

ycyt

ycytsy SS

SSS

+= 10.1

754.82

max

===∴τ

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






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;σ

Tension

Compression

Wire is bend from the top↓

Top of the wire is suffered to“Tension” (+σ)

↓Meanwhile the bottom is suffered to

“Compression” (-σ)

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 to“Tension” (+σ)

↓Meanwhile the bottom is suffered to

“Compression” (-σ)

Graph plotting for overall wire bending process:

This type of stress is known as Completely Reverse Stress

σ

+

−

σm = 0

σa

σa

σrt





2nd Case – Vehicle Axle

Stress : Bending Stressσ and Shear Stress τ

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.

τ

τm

σShear 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 methodLeast accurate, particularly for low cycle applicationsMost traditional, easiest to implement

ii. Strain-life methodDetailed analysis of plastic deformation at localized regionsSeveral idealizations are compounded, leading to uncertainties in results

iii. Linear-elastic fracture mechanics methodAssumes crack existsPredicts 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 MPa

Se’ = 550 MPa [ 84.1 kpsi ]→ Sut > 1400 MPa

For real engineering practice:

Se’ = 0.4 Sut → Sut≤ 1400 MPa

Se’ = 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 Sbut

where 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 4–1, 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

( )

⎪⎪⎩

⎪⎪⎨

⎧

≤<≤≤=

≤<≤≤=

=

−

−−

−

−−

mmddmmddd

inddinddd

kb

2545151.15179.224.1)62.7/(

10291.0211.0879.03.0/

157.0

107.0107.0

157.0

107.0107.0

For NONCIRCULAR CROSS SECTION , or NONROTATING ROUND BAR, the effective dimension de is: D

h

b

de= 0.370d de = 0.808 (hb)1/2

- depends by the types of the load


EXAMPLE 2A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm in diameter. The shaft material has a mean ultimate tensile strength of 690 MPa.Estimate the Marin size factor kb if the shaft is used in

(a) A rotating mode.(b) A nonrotating mode.

Solution(a)

(b) de = 0.37d = 0.37(32) = 11.84 mm



858.062.7

3262.7

107.0107.0

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

−−dkb


Size Factor, kbSize Factor, kb

954.062.784.11 107.0

=⎟⎠⎞

⎜⎝⎛=

−

bk





Loading Factor, kcLoading Factor, kc

When fatigue tests are carried out with rotating bending, axial (push-pull), and torsional loading, the endurance limits differwith Sut. Here are the values of the load factor as

When fatigue tests are carried out with rotating bending, axial (push-pull), and torsional loading, the endurance limits differwith Sut. Here are the values of the load factor as

⎪⎩

⎪⎨

⎧=

59.085.01

ckBending

Axial

Torsion

** If there is a combination of loads, use the higher value of loading factor





Temperature Factor, kdTemperature Factor, kd

If Se’ is known at room temperature, then use kd = ST / SRT .If not, compute the ultimate strength at the elevated temperature obtained by using the factor from below table, then use kd = 1

Effect of Operating Temperature on the Tensile Strength of Steel.*

(ST = tensile strength at operating temperature;

SRT = tensile strength at room temperature; 0.099 ≤ˆσ ≤ 0.110)





Temperature Factor, kdTemperature Factor, kd

EXAMPLE 3A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service. Estimate the Marin temperature modification factor and (Se)450◦ if

(a) The room-temperature endurance limit by test is (Se )70◦ = 39.0 kpsi(b) Only the tensile strength at room temperature is known.

Solution(a) Interpolating from previous Table gives :

( ) 007.1400500400450)018.1995.0(018.1/ 450 =⎟

⎠⎞

⎜⎝⎛

−−

−+== oRTTd SSk

Thus, (Se)450◦ = kd (Se )70◦ =1.007(39.0) = 39.3 kpsi

(b) Since the rotating-beam specimen endurance limit is not known at room temperature, we determine the ultimate strength at the elevated temperature first, which the ultimate strength at 450° is

(Sut )450◦ = (ST /SRT )450◦(Sut )70◦= 1.007(70) = 70.5 kpsi

Then, (Se)450◦ = 0.5 (Sut )450◦ = 0.5(70.5) = 35.2 kpsi





Reliability Factor, keReliability Factor, ke

the reliability modification, ke factor is written asthe reliability modification, ke factor is written as

Reliability Factors ke Corresponding to 8% Standard Deviation of the Endurance Limit

ke = 1 − 0.08 zαwhere zα values can be determined from Table A–10 in Appendix A. Table below gives reliability factors for some standard specified reliabilities.





Miscellaneous-Effects Factor, kfMiscellaneous-Effects Factor, kf

kf factor is intended to account for the reduction in endurance limit due to all other effects such as, Corrosion, Electrolytic Plating, Metal Spraying, Cyclic Frequency, and other more.

kf factor is intended to account for the reduction in endurance limit due to all other effects such as, Corrosion, Electrolytic Plating, Metal Spraying, Cyclic Frequency, and other more.

However the actual values of kf are not always available. If this factor is not important, assume ;

kf = 1.0

It is really intended as a reminder that this factor must be accounted in real engineering practice.

However the actual values of kf are not always available. If this factor is not important, assume ;

kf = 1.0

It is really intended as a reminder that this factor must be accounted in real engineering practice.



4.7 – Stress Concentration and Notch Sensitivity


w

B B

A A

σmax

σ

σ

d

Stress trajectories

Stress distribution

σ= F/A

σ0= F/A0

A = wt

A0 = (w-d )t

σmax > σ0 > σ

• 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 discontinity to the nominal stress.

whereKt : is used for normal stresses Kts : is used for shear stresses

otK

σσmax=

otsK

ττmax=or

• Kt or Kts depends on geometry of the part

• The analysis of geometric shapes is a difficult problem and not many solutions can be found

• Kt or Kts for a variety of geometries may be found in Appendix A Tables A-15 and A-16





where Kf is a reduced value of Kt and also known asfatigue stress-concentration factor.

σ0 is the nominal stress.

For variable loading cases that cause fatigue, some materials are not fully sensitive to the presence of notches and hence, for these, a reduced value of Kt can be used. For these materials, the maximum stress is, in fact,

For variable loading cases that cause fatigue, some materials are not fully sensitive to the presence of notches and hence, for these, a reduced value of Kt can be used. For these materials, the maximum stress is, in fact,

ofK σσ =max ofsK ττ =maxor





Notch sensitivity q is defined by the equation :

where q is usually between 0 ~ 1.

If q = 0, then Kf = 1 → material has no sensitivity to notches at all.If q = 1, then Kf =Kt→ material has full notch sensitivity.

If q = 0, then Kf = 1 → material has no sensitivity to notches at all.If q = 1, then Kf =Kt→ material has full notch sensitivity.

In analysis or design work :

Find Kt from geometry of the part Specify the material Find q Solve for Kf

or11

−

−=

t

f

KK

q11

−

−=

ts

fsshear K

Kq

or( )11 −+= tf KqK ( )11 −+= tsshearfs KqK





Figure 4–1Notch-sensitivity charts for steels and UNS A92024-T wrought aluminum alloys subjected to reversed bending or reversed axial loads. For larger notch radii, use the values of q corresponding to the r = 0.16-in (4-mm) ordinate.

Notch sensitivity qfor bending and axial load





Figure 4–2Notch-sensitivity curves for materials in reversed torsion. For larger notch radii, use the values of q shear corresponding to r = 0.16 in (4 mm).

Notch sensitivity q for torsion load


• If there is any doubt about the true value of q, it is always safe to use Kf = Kt .• The notch sensitivity of the cast irons is very low, varying from 0 to about 0.20,

depending upon the tensile strength. To be on the conservative side, it is recommended that the value q = 0.20 be used for all grades of cast iron.

• Figure 4–1 has as its basis the Neuber equation, which is given by

• where is defined as the Neuber constant and is a material constant.• Then notch sensitivity equation become

• For steel, with Sut in kpsi, the Neuber constant can be approximated by a third-orderpolynomial fit of data as

Bending or Axial:

Torsion:



raKK t

f /111

+−

+=

ra

q+

=1

1

3ut

-82ut

-5ut

-3 )S2.67(10 -)S1.51(10 )S3.08(10 - 0.246 +=a


a

3ut

-82ut

-5ut

-3 )S2.67(10 -)S1.35(10 )S2.51(10 - 0.190 +=a





EXAMPLE 4A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using:(a) Figure 4-1.(b) Equations (6–33) and (6–35).

SolutionFrom Fig. A–15–9, using D/d = 38/32 = 1.1875, r/d = 3/32 = 0.09375, we read the graph to find Kt = 1.65.

(a) From Fig. 4–1, for Sut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Eq.

Kf = 1 + q (Kt − 1)= 1 + 0.84(1.65 − 1)= 1.55

(b) From below Eq. with Sut = 690 MPa = 100 kpsi,

= 0.0622 = 0.313

Substituting this into next Eq. with r = 3 mm gives

in

3-82-5-3 )1002.67(10 -)1001.51(10 )1003.08(10 - 0.246 +=amm

55.1

3313.01

165.11/111 =

+

−+=

+−

+=ra

KK tf



4.8 – Fatigue Strength

3/)(log futf NSS =


fSut

Se

103 106

Number of stress cycle (N)

Fatigue stress Sf

High cycle,

Finite life

S-N Diagram

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

e

ut

SSf

b log31

e

ut

SSf

a2)(

=

bf aNS =

Sut

High Cycle,Infinite Life

Low cycle,

Finite life

100

36

490 560 630 700 770 840 910 980 1050 1120 1190 1260 1330 1400

Sut, MPa



4.8 – Fatigue Strength – cont…


Figure 4–3

Fatigue strength fraction, f,

of Sut at 103 cycles for

Se = Se’ = 0.5Sut .

If Sut < 70 kpsi (490 MPa),let f = 0.9


EXAMPLE 5

Given a 1050 HR steel, estimate

(a) the rotating-beam endurance limit at 106 cycles.(b) the endurance strength of a polished rotating-beam specimen corresponding to 104 cycles to failure(c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 385 MPa.




fSut

Se

103 106

Number of stress cycle (N)

Fatig

ue s

tres

s S f

Sut

100 104

Sf


Solution

(a) From Table A–20, Sut = 620 MPa.

∴ Se’ = 0.5(630) = 310 MPa



( )[ ] 917310

62086.0 2

==a

( )[ ] 0785.0310

62086.0log31

−=−=b

445)10(917917' 0785.040785.0 === −−NS f


(b) From Fig. 4–3, for Sut = 620 MPa, f .= 0.86.

0785.0/1/1

917385' −

⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

bf

aS

N

(c) With S’f = 385 MPa,

)10(3.63 3=

MPa

cycles



4.9 – Characterizing Fluctuating Stresses


Fluctuating stresses in machinery often take the form of sinusoidal pattern

because of the nature of the nature of some rotating machinery.

Other patterns some quite irregular do occur.


where Fm is the midrange steady component of force, and Fa is the amplitude of thealternating component of force.


4.9 – Characterizing Fluctuating Stresses

2minmax FFFm

+=


2minmax FFFa

−=

In periodic patterns exhibiting a single maximum and single minimum of force, the shape of the wave is not important.

The peaks on both sides (maximum, minimum) are important.

Fmax and Fmin in a cycle can be used to characterize the force pattern.

A steady component and an alternating component can be constructed as follows:



4.9 – Characterizing Fluctuating Stresses – cont…


Some stress-time relations:Some stress-time relations:

(a) fluctuating stress with highfrequency ripple

Stre

ss

Time

(b) nonsinusoidal fluctuating stress

Stre

ss

Time

(c) nonsinusoidal fluctuating stress

Stre

ss

Time

(d) sinusoidal fluctuating stress

Stre

ss

Time

σa

σa

σr

σmσmin

σmax

(e) repeated stress

Stre

ssTime

σa

σrσm

σmin = 0

σmax

σa

(f) Completely reversed sinusoidal stress

Stre

ss

Time

σa

σrσm = 0σmin

σmaxσa

Stress RangeStress Range

Mean (Midrange Stress)Mean (Midrange Stress)

Stress Amplitude Stress Amplitude (Alternating Stress)(Alternating Stress)

Stress RatioStress Ratio

Amplitude RatioAmplitude Ratio

(R =-1)

(R =0)

(R>0 )













mofsm K ττ =

For a case that involved fluctuating bending stress or fluctuating shear stress in the present of a notch, the equation for amplitude and mean stresshave become:

For a case that involved fluctuating bending stress or fluctuating shear stress in the present of a notch, the equation for amplitude and mean stresshave become:

mofm K σσ =aofa K σσ = and

andaofsa K ττ =



4.10 – Combination of Loading Modes

[ ] [ ] 22' )()(3)()()()( torsionmtorsionfaxialmaxialfbendingmbendingfm KKK τσσσ ++=


For a case that involved combinations of different types of loading, such as combined bending, torsion, and axial.

For a case that involved combinations of different types of loading, such as combined bending, torsion, and axial.

[ ] 22

' )()(385.0)()()()( torsionatorsionfs

axialaaxialfbendingabendingfa KKK τσσσ +⎥⎦

⎤⎢⎣⎡ +=



4.11 – Factor of Safety

nSS y

m

e

a 1=+

σσ

nSS ut

m

e

a 1=+

σσ

12

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

ut

m

e

a

Sn

Sn σσ


122

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

y

m

e

a

Sn

Sn σσ

nSy

ma =+σσ

Soderberg

Modified-Goodman

Gerber

Asme-elliptic

Larger static yield

5 theories of fatigue failure

For torsion (shear) load only.Use the same equations as apply for σm ≥ 0, except replace σm and σa with τm and τm , use kc = 0.59 for Se , replace Sut with Ssu = 0.67Sutand replace Sy wih Ssy = 0.577Sy .

For torsion (shear) load only.Use the same equations as apply for σm ≥ 0, except replace σm and σa with τm and τm , use kc = 0.59 for Se , replace Sut with Ssu = 0.67Sutand replace Sy wih Ssy = 0.577Sy .



4.11 – Factor of Safety – cont…


Figure 4–4





Load line which intersection with fatigue criterion

Load line which intersection with static Larger criterion

Intersection of static and fatigue criterion

Table 4–1 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Modified Goodman and Langer Failure Criteria








Table 4–2 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria








Table 4–3 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for ASME-Elliptic and Langer Failure Criteria



Other Examples of Relevant Fatigue Exercise


Example 6-8 – page 298




“Fatigue additional notes”From Engineering Design text book (Shigley-Ninth Edition)

Chapter 5Gears – Part I




Week 5

2


Learning Outcomes


recognize types of gear that available and know the function

identify nomenclature of spur gear

construct a gear

perform load and power calculations analytically as applied to a gears components.


CHAPTER 5 – Gears

3



In this chapter (part 1), generally we will learn:In this chapter (part 1), generally we will learn:--• 5.1 - Introduction• 5.2 - Types of Gears• 5.3 - Nomenclature of Spur Gear• 5.4 - Construction of Gear• 5.5 - Forming of Gear Teeth• 5.6 - Tooth Systems• 5.7 - Gear Ratio• 5.8 - Gear Train• 5.9 - Contact Ratio• 5.10 - Interference• 5.11 - Force Analysis (Spur Gear)• 5.12 - Force Analysis (Bevel Gear)• 5.13 - Force Analysis (Helical Gear)


CHAPTER 5 – Gears

Gear manufacturing video

4




CHAPTER 5 – Gears

To transmit power, torque and speed ⇒ ideally, power is constant

To reduce and increase speed and torque

To increase efficiency and reliability of a system (no slip) compare using other mechanism such as belt and pulley or chain and sprocket.

Advantages:– Exact velocity ratio– May transfer large power– High efficiency– Reliable– Compact

Disadvantages:

– High cost in manufacturing

(requires special tools)

– Vibration & noise

– Lubricant required

5


5.2 – Type of Gears


CHAPTER 5 – Gears

• Spur Gear• Helical Gear• Bevel Gear• Worm Gear

6


5.2 – Type of Gears – cont…


CHAPTER 5 – Gears

SPUR GEAR

• have teeth parallel to the axis of rotation.

• used to transmit rotary motion between parallel shafts.

• the simplest gear.

7




CHAPTER 5 – Gears

HELICAL GEAR

• have teeth inclined to the axis of rotation.

• Not make much noise during meshing.

• Inclined tooth also develops thrust loads and bending couples.

• Sometimes used to transmit motion between nonparallel shafts.

8




CHAPTER 5 – Gears

BEVEL GEAR

are essentially conically shaped.

Used to transmit rotary motion between intersecting shafts. The angle between the shafts can be anything except zero or 180 degrees ⇒ eg. Differential

Bevel gear may be classified as follow:-Straight bevel gearsSpiral bevel gearsZerol bevel gearsHypoid gearsSpiroid gears

9




CHAPTER 5 – Gears

DIFFERENTIAL

10

• Usually used for pitch line velocities up to 1000 ft/min (5 m/s) when noise level is not an important consideration.

• Available in many stock sizes and less expensive to produce than other bevel gears.




CHAPTER 5 – Gears

STRAIGHT BEVEL GEAR

11

• have teeth that are both curved along their (the tooth's) length; and set at an angle, analogously to the way helical gear teeth are set at an angle compared to spur gear teeth.

• Recommended for higher speeds and where the noise level is an important consideration.




CHAPTER 5 – Gears

SPIRAL BEVEL GEAR

12

• Have teeth which are curved along their length, but not angled.

• Permissible axial thrust load are not large as those for the spiral bevel gear




CHAPTER 5 – Gears

ZEROL BEVEL GEAR

13

• Hypoid gear - Similar to bevel gears but with the relatively small shafts offset.

• Spiroid gear – larger shaft offset, the pinion begins to resemble a tapered worm




CHAPTER 5 – Gears

HYPOID AND SPIROID GEARS

Spiral gear

Hypoid

Spiroid

Ring gear

14




CHAPTER 5 – Gears

WORM GEARS

Used to transmit rotary motion between nonparallel and not intersecting shafts.

Worm is a gear that resembles a screw. It is a species of helical gear, but its helix angle is usually somewhat large and its body is usually fairly long in the axial direction.

The worm can always drive the gear. However, if the gear attempts to drive the worm, it may or may not succeed.

15




CHAPTER 5 – Gears

RACK & PINION

A rack is a toothed bar or rod that can be thought of as a sector gear with an infinitely large radius of curvature.

Torque can be converted to linear force by meshing a rack with a pinion: the pinion turns; the rack moves in a straight line.

Such a mechanism is used in automobiles to convert the rotation of the steering wheel into the left-to-right motion of the tie rod(s).

16


5.3 – Nomenclature of Spur Gear


CHAPTER 5 – Gears


5 3 – Nomenclature of Spur Gear – cont

CHAPTER 5 – Gears

5.3 Nomenclature of Spur Gear cont…

• Pitch circle – is a theoretical circle upon which all calculations are usually based.

• Pitch diameter (Dp) – is a diameterPitch diameter (Dp) is a diameter of pitch circle

• Circular pitch (pc)– is the distance, from a point on one tooth to afrom a point on one tooth to a corresponding point on an adjacent tooth (measured on the pitch circle).

D ππ

d

pc PN

Dp ππ

; = (mm or inch)

• Diametral pitch (Pd) – is the ratio of the number of teeth on the gear to the pitch diameter. d D

NP = (Teeth per inch)


pD



CHAPTER 5 – Gears


• Module (m) – is the ratio of the pitch diameter to the number of teeth. D

m p= (mm)

• Addendum (a) – is a radial distance

N

Addendum (a) is a radial distance between the top land and the pitch circle.

Pma 1 ; = (mm or inch)

• Dedendum (b) – is the radial distance from the bottom land to the

dP

distance from the bottom land to the pitch circle.

dPmb 25.1 ; 25.1= (mm or

inch) This is a commonly used sizes for spur gear tooth Refer tooth s stem for more info


d tooth. Refer tooth system for more info.



CHAPTER 5 – Gears


• Whole Depth (ht) – is the radial distance from the bottom land to the pitch circle.

baht += (mm or inch)

• Clearance circle – is a circle that is tangent to the addendum circle of the mating gear

t

the mating gear.

• Clearance (c) – is the amount by which the dedendum in a given gear exceeds the dum of its mating gear.

abc −= (mm or inch)

• Backlash – is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circles

19

measured on the pitch circles.



5 4 – Construction of Gear

CHAPTER 5 – Gears

5.4 Construction of Gear

It is necessary that you actually be able to draw the teeth on a pair of meshing gear to obtain an understanding of the problem involved in the meshing of the mating teeth

First – Draw the circle of gear layout

i) Calculate each pitch circle

problem involved in the meshing of the mating teeth.

11

1 mNPNd

d

== 22

2 mNPNd

d

==

221 ddCD +

=Where center distance of the gear is,

ii) Draw the pitch circle, d1 & d2

iii) Draw line ab

iv) Draw line cd through point P at an angle φ to the common tangent ab. This line is also known as co o ta ge t ab s e s a so o aspressure line, generating line or line of action.

v) Next, on each gear, draw a circle tangent to the pressure line which known as base circle

φcosrr =or radius of base gear is

vi) Then, draw the addendum and dedendum circle

φcosrrb =or radius of base gear is,

adda 2+=bdd 2

for addendum diameter,

f d d d di t


bddb 2−=for dedendum diameter,

21

t


5.4 – Construction of Gear – cont…


CHAPTER 5 – Gears

Second - Construct an involute curve

i) Divide the base circle into a number of equal parts and construct radial lines OA0, OA1, OA2, OA3, OA4, etc.

ii) Beginning at A1, construct perpendicular line A1B1, A2B2, A3B3 etc.

where distance A1B1 = A1A0, A2B2 = 2A1B1, A3B3 = 3A1B1 etc.

iii) Then draw a curve at each end point of the perpendicular line to construct the involute curve. Make sure the involute curve has exceed the addendum circle.

iv) To draw a tooth, we must know the thickness. Therefore the tooth thickness is half the distance of the circular pitch which measured on pitch circle.

v) From tooth thickness we can determine the angle needed, α to make a mirror line between it.

vi) Then the involute is mirror to another halve at previous mirror line.

d

c

Pmpt

222ππ

=== (mm or inch)

αo

dt

πα ⋅

=180 (o)

22


5.4 – Construction of Gear – cont…


CHAPTER 5 – Gears

vii) Trim the involute line at the needed range between addendum and dedendum.

viii) Finally, array the complete tooth according to the required number.

ix) Repeat the step (i) to the step (viii) to draw the other tooth at the other mating gear.

23


5.5 – Forming of Gear Teeth


CHAPTER 5 – Gears

Milling• Gear teeth may be cut with a milling cutter shaped to conform to

the tooth space.

Shaping• May be generated with either a pinion cutter or rack cutter.• Pinion cutter reciprocates along the vertical axis and slowly fed

into the gear blank to the required depth.• Rack cutter reciprocates into the gear blank and roll slightly on

their pitch circle.

Hobbing• Is simply a cutting tool that is shaped like a worm.• Both the hob and the blank must be rotated when the hob fed

slowly across the face of the blank.

Finishing• Shaving or burnishing is needed to diminish error after cutting

until the surface become smooth.• Grinding and lapping are used for hardened gear teeth after heat

treatment.

Shaping – Pinion cutter

Shaping – Rack cutter

Hobbing

24


5.6 – Tooth Systems


CHAPTER 5 – Gears

• Is a standard that specifies the relationships involving addendum, dedendum, working depth, tooth thickness and pressure angle.

• Planned to attain interchangeability.

Tooth Sysem Pressure Angle φ, deg

Addendum a

Dedendum b

Full Depth 20 1/Pd or 1m 1.25/Pd or 1.25m

1.35/Pd or 1.35m

22½ 1/Pd or 1m 1.25/Pd or 1.25m

1.35/Pd or 1.35m

25 1/Pd or 1m 1.25/Pd or 1.25m

1.35/Pd or 1.35m

Stub 20 0.8/Pd or 0.8m 1/Pd or 1m

Table 5-1 Standard and Commonly Used Tooth Systems for Spur Gears

25


5.6 – Tooth Systems


CHAPTER 5 – Gears

Diametral PitchCoarse 2, 2¼, 2½, 3, 4, 6, 8, 10, 12, 16

Fine 20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200

Table 5-2 Tooth Sizes in General Uses

ModulesPreffered 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50

Next Choice 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18,22, 28, 36, 45

26


5.7 – Gear Ratio


CHAPTER 5 – Gears

• The gear ratio is a number, usually expressed as a decimal, representing how many turns of the input gear cause one revolution of the output gear. Gear ratio is suggested max. 1:10.

• Also known as a Speed ratio.

• It can be a relationship between the numbers of teeth on two gears that are meshed or two sprockets connected with a common roller chain, or the circumferences of two pulleys connected with a drive belt.

• Consider a pinion (input) driving gear (output), speed ratio is given by:

i

o

i

o

i

o

o

i

TT

dd

NN

nn

===where N = number of teeth

n = gear speed (rev/min)d = pitch diameter (mm or in)T = torque (Nm or Ib-in)

27


5.8 – Gear Train


CHAPTER 5 – Gears

• Consist of multiple gears in the train, as shown in the two figures below :

26

5

4

3

3

26 n

NN

NN

NNn =

numberstoothdrivenofproductnumberstoothdrivingofproducte =

F

L

nne =

• Train value,

where• nL = speed of the last gear• nF = speed of the first gear

where N = number of teethn = gear speed (rev/min)

28


5.8 – Gear Train – cont…


CHAPTER 5 – Gears

• To calculate overall gear or speed ratio from multiple gearing system

enn

TTm

o

i

i

o 1 Ratio;Gear Overall O-G ===

where TF = Torque at the first gearTL = Torque at the final gearnF = speed of the first gear (rev/min)nL = speed of the last gear (rev/min)e = train value

First gear

Finalgear

29




CHAPTER 5 – Gears

Motor5 Hp2000 rpm Fill the table below:-

Component Pitch Diameter (in)

Speed (rpm)

Torque (ib-in)

motor -

2 (Pulley) -

3 (Pulley) -

4 (Bevel Gear)

5 (Bevel Gear)

6 (Spur Gear)

7 (Spur Gear)

8 (Worm Gear)

9 (Spur Gear)

Pd = 4 teeth/in

Pd = 4 teeth/in

Pd = 6 teeth/in

Example 5-1

Train value, e = ? 30




CHAPTER 5 – Gears

Motor5 Hp2000 rpm Fill the table below:-

Component Pitch Diameter (in)

Speed (rpm)

Torque (ib-in)

motor - 2000 157.5

2 (Pulley) - 2000 157.5

3 (Pulley) - 1200 262.5

4 (Bevel Gear) 4.5 1200 262.5

5 (Bevel Gear) 9.5 568.4 554.2

6 (Spur Gear) 5 568.4 554.2

7 (Spur Gear) 12 236.8 1330.1

8 (Worm Gear) 0.5 236.8 1330.1

9 (Spur Gear) 6 19.7 15961.2

Pd = 4 teeth/in

Pd = 4 teeth/in

Pd = 6 teeth/in

Example 5-1

Train value, 00985.02000

7.19===

F

L

nne

31


5.9 – Contact Ratio


CHAPTER 5 – Gears

• On an involute profile gear tooth, the contact point starts closer to one gear, and as the gear spins, the contact point moves away from that gear and toward the other.

• If you were to follow the contact point, it would describe a straight line that starts near one gear and ends up near the other.

• This means that the radius of the contact point gets larger as the teeth engage. 32


5.9 – Contact Ratio – cont…


CHAPTER 5 – Gears

Lab

Line of ContactAddendum circle (pinion)

Pitch circle (pinion)

Clearance circle (pinion)

Dedendum circle (pinion)Base circle (pinion)

Addendum circle (gear)

Pitch circle (gear)

Clearance circle (gear)

Dedendum circle (gear)Base circle (gear)

Pressure Line

ra (pinion) rb (pinion)

ra (gear)rb (gear)

motion

33


5.9 – Contact Ratio – cont…

φcosc

ab

pLCR =


CHAPTER 5 – Gears

• contact ratio (CR) is a number that indicates the average number of pairs of teeth in contact.

• In general gear arrangement, the number of teeth in contact (CR) determine the smooth transmission → CR must be 1.2 or more to reduce teeth impact as well as noise level.

whereLab = length of the line of actionpc = circular pitchφ = pressure angle

φφ

cossin2

22

22

12

1

c

baba

pCDrrrr

CR−−+−

=where

ra = addendum circle radiusrb = base circle radiusCD = center distance of the gear

34


5.10 – Interference


CHAPTER 5 – Gears

• Interference – the contact of portions of tooth profiles that are not conjugate

• Occur when first gear contact at the tip of the driven tooth touch the flank of the driving tooth before the involute portion of the driving tooth comes within the range (below the base circle of gear 2).

• Causing the involute tip or face of the driven gear tends to dig out the noninvolute flank of the driver.

Base circle

35


5.10 – Interference – cont…


CHAPTER 5 – Gears

• To avoid interference – ones has to calculate smallest number of pinion teeth can exist without interference is NP where NP is:-

⎟⎠⎞⎜

⎝⎛ +++

+= φ

φ22

2 sin)21(sin)21(

2GGG

GP mmm

mkN

φ2sin2kNP =

⎟⎠⎞⎜

⎝⎛ +++

+= tGGG

tGP mmm

mkN φ

φψ 22

2 sin)21(sin)21(

cos2

tP

kNφψ

2sincos2

=

For spur pinion

For spur pinion that operate with a rack

For helical pinion

For helical pinion that operate with a rack

Where:k = 1 (for full depth teeth)k = 0.8 (for stub teeth)mG = mating gear ratioφ = pressure angle

Where:ψ = helix angleφt = tangential pressure angle

(refer slide 44)

36


5.11 – Force Analysis (Spur Gear)


CHAPTER 5 – Gears

37


5.11 – Force Analysis (Spur Gear) – cont…


CHAPTER 5 – Gears

φkosWFW t== 32

r

2/32 dTFW t

t ==

φtan32 tr

r WFW ==Free body diagrams of the forces and moments acting upon two gears of a simple gear train.

F32F23

Fa2

Fb3Tb3

Ta2

3

2

a

b

φ

φ

φ

(b)

Shows a pinion mounted on a shaft a rotating clockwise at n2 rev/min and driving a gear on a shaft b at n3 rev/min

n3

n2

3

2

a

b

φ

(a)

Gear

Pinion Gear force have been resolved into tangential and radial components.

F32

Fa2

a

(c)

n2

Ta2

2

d2

F 32t

F a2t

F a2r

F 32r

38



English unit SI unit

V = pitch-line velocity, ft/minWt = transmitted load, Ibfd = gear diameter, inn = gear speed, rev/minH = power, hp

V = pitch-line velocity, mm/sWt = transmitted load, kNd = gear diameter, mmn = gear speed, rev/minH = power, kW

12dnV π

=60dnV π

=


CHAPTER 5 – Gears

VHWt 33000=

dnHWt π

60000=

39




CHAPTER 5 – Gears

Example 5-2

Pinion 2 in this figure runs at 1750 rev/min and transmits 2.5 kW to idle gear 3. The teeth are cut on the 20o full-depth system and have a module of m = 2.5 mm. Draw a free body diagram of gear 3 and show all the forces that act upon it.

From above statement, it can be summarizes that below are the info given :

n = 1750 rpm

H = 2.5 kW

φ = 20o

m = 2.5

40

The pitch diameters for gears 2 and 3 are :

d2 = mN2 = 2.5(20) = 50 mm

d3 = mN3 = 2.5(50) = 125 mm

The transmitted load to be :




CHAPTER 5 – Gears

Solution 5-2

FREE BODY DIAGRAM

546.0)1750)(50(

)5.2(6000060000

2

===ππ nd

HWt kN

546.023 =tF kN

199.020tan546.020tan2323 === oorr FF kN

581.020cos

546.020cos23

23 === oo

tFF kN

Thus, the tangential force , as shown in free body diagram figure, Therefore

and so

41

Since gear 3 is idler, it transmits no power (torque) to its shaft, and so the tangential reaction of gear 4 on gear 3 is also equal to Wt. Therefore




CHAPTER 5 – Gears

Solution 5-2 – cont…

347.0)199.0546.0()( 43233 =+−−=+−= rtxb FFF kN

546.043 =tF kN

491.0347.0347.03 =+=bF kN

The shaft reactions in the x and y directions are

The resultant shaft reaction is

199.043 =rF kN 581.043 =F kN

347.0)546.0199.0()( 43233 =−−=+−= tryb FFF kN

42


5.12 – Force Analysis (Bevel Gear)

P

G

NN

=Γtan


CHAPTER 5 – Gears

Terminology of bevel gears

G

P

NN

=γtan

Pinion pitch angle :

Gear pitch angle :

43


5.12 – Force Analysis (Bevel Gear) – cont…


CHAPTER 5 – Gears

rav

x

y

z

Wt

W

Wa Wrγ

φ

avt r

TW =

Transmitted load :

where• T is the torque• rav is the pitch radius at the midpoint• Wt is also known as tangential force

Radial Force :

γφ costantr WW =

Axial Force :

γφ sintanta WW =

44


5.13 – Force Analysis (Helical Gear)


CHAPTER 5 – Gears

Normal circular pitch ψcostn pp =

ψtant

xpp =

ψcost

nPP =

t

n

φφψ

tantancos =

Axial pitch

Since pnPn = π, the normaldiametral pitch is

Since φn is a pressure angle in the normal direction, φt pressure angle in the direction of rotation (transverse pressure angle) and ψ is the helix angle, these angle are related by the equation:

Represent a portion of top view of a helical rack

Lines ab and cd are the centerlines of two adjacent helical teeth taken on the same pitch plane. The angle ψ is the helix angle. The distance ac is the transverse circular pitch pt in the plane of rotation (usually called the circular pitch).

45


5.13 – Force Analysis (Helical Gear) – cont…


CHAPTER 5 – Gears

Example 5-3

A stock helical gear has a normal pressure angle of 20o, a helix angle of 25o and a transverse module of 5.0 mm and has 18 teeth. Find:

(a) The pitch diameter

(b) The transverse, the normal and the axial pitch

(c) The normal module

(d) The transverse pressure angle

Solution 5-3

(a) 90)5(18 === tNmd mm

(b) 71.15)5( === ππ tt mp

mm

mm

46




CHAPTER 5 – Gears

Solution 5-3 – cont…

or(c)

oo

on

t 88.2125cos20tantan

costantan 11 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

ψφφ

24.1425cos71.15cos === otn pp ψ mm

69.3325tan71.15

tan=== o

tx

ppψ

mm

53.424.14===

ππn

npm mm 53.425cos5cos === o

tn mm ψ mm

(d)

47




CHAPTER 5 – Gears

3 components of total (normal) tooth force W are :

nr WW φsin=

ψφ coscos nt WW =

ψφ sincos na WW =

where

• W = total force

• Wv = radial component

• Wt = tangential component, also called transmitted load

• Wa = axial component, also called thrust load

• φn = normal pressure angle

• ψ = helix angle

Chapter 5Gears – Part II




Week 6 & 7

2


Learning Outcomes

At the end of this topic, the students would be able to apply

and appreciate the knowledge to:

analyses and design of spur and helical gears to resist bending failure of the teeth as well as pitting failure of tooth surfaces.


CHAPTER 5 – Gears (Part II)

3



In this chapter, generally we will learn:In this chapter, generally we will learn:--


• 5.15 - Lewis Bending Equation

• 5.16 - Surface Durability

• 5.17 - AGMA Stress Equations

• 5.18 - AGMA Strength Equations

• 5.19 - Safety Factors SF and SH



4




Occur when significant tooth stressequals or exceeds either the yield strength or bending endurance strength.


GEAR FAILURE

Teeth Bending Failure

Tooth Surface Pitting Failure

Occur when significant contact stress equals or exceeds thesurface endurance strength.

Gear Standard and Quality• AGMA (American Gear Manufacturers Association)

• BGA (British Gear Association)

• JGMA (Japanese Gear Manufacturers Association)

• EUROTRANS (European Committee of Associations of Manufacturers of Gears and Transmission Parts)

5





Example of Gear CatalogSource: KHK Stock Gears Website: http://www.qtcgears.com/RFQ/SpurGears.htm

In gear selection, Allowable torque is important criteria you need to know. Why? How to

calculate it?

6


5.15 – Lewis Bending Equation

FYPW d

t

=σ



• If power is transmitted between mating gears, there are transmitted force W t and radial force W r.

• Maximum stress in a gear tooth is assume occurs at point a as shown in the figure.

• Bending stress in gear tooth according to Lewis (1892):

Where;

Wt = transmitted force (Ibf or N)Pd = diametral pitch (teeth per inch)m = module (mm)F = width of gear tooth (inch or mm)Y = Lewis form factor (dimensionless)

(refer Table 5-3)

FmYW t

=σ

(U.S. customary units)

(SI units)

7


5.15 – Lewis Bending Equation – cont…

Number of Teeth Y Number

of Teeth Y

12 0.245 28 0.353

13 0.261 30 0.359

14 0.277 34 0.371

15 0.290 38 0.384

16 0.296 43 0.397

17 0.303 50 0.409

18 0.309 60 0.422

19 0.314 75 0.435

20 0.322 100 0.447

21 0.328 150 0.460

22 0.331 300 0.472

24 0.337 400 0.480

26 0.346 Rack 0.485



Table 5-3: Values of the Lewis Form Factor Y

(These Values are for a Normal Pressure Angle of 20o, Full Depth Teeth, and a Diametral Pitch of Unity in the Plane of Rotation)

Lewis Form Factor, Y also can be calculate by using this formula:-

32xPY =

Where;

x = distance refer figure slide 6P = diametral pitch (teeth per inch)

Interpolation for Y value is needed if number of teeth is not in the table.

8



US customary units(Velocity in feet per

minute)

SI units(Velocity in meters

per second)

Cast iron, cast profile

Cut or milled profile

Hobbed or shaped profile

Shaved or ground profile



Dynamic Effects, Kv

• Present when a pair of gears is driven at moderate or high speed and noise is generated.

• Gear under dynamic loading;

FYPWK d

tv=σ

FmYWK t

v=σ 7878 VKv

+=

600600 VKv

+=

1.61.6 VKv+

=

05.305.3 VKv

+=

56.356.3 VKv

+=

56.556.5 VKv

+=

12001200 VKv

+=

5050 VKv

+=

** Where V is the pitch line velocity


(SI units)

9





A stock spur gear is available having a module of 3mm, a 38-mm face, 16 teeth, and a pressure angle of 20o with full-depth teeth. The material is AISI 1020 steel in as-rolled condition. Use a design factor of nd = 3 to rate the power output of the gear corresponding to a speed of 20 rev/s and moderate applications.

Example 5-4Example 5-4

Solution 5-4Solution 5-4

The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure. From table A-20 in appendix, we find Sut = 380 MPa and Sy = 210 MPa. A design factor of 3 means that the allowable bending stress is

MPa0.703

210 ===d

yallowable n

Sσ

10



48)16(3 === mNDp

02.3)20)(048.0( === ππdnV

5.11.6

02.31.61.6

1.6=

+=

+=

VKv



Solution 5-4 - cont…Solution 5-4 - cont…

The pitch diameter is, mm

So the pitch velocity is, m/s

Velocity factor is found to be,

So the transmitted force is, v

allowt

KmFYW σ

=

5.15455.1

10)7.68(296.0)038.0(003.0 6

==tW

The power that can be transmitted is, 4667)02.3(5.1545 === VWH t W

where Y = 0.296 for 16 teeth

N

This is a rough estimation, and that this approach must not be used for important application

11

Because a gear tooth experiences into and out engagement ⇒ causes contact stress ⇒ wear and pitting

Wear depends on surface hardness

Pitting ⇒ small particles are removed due to high contact stress in gear

Prolong operation after pitting

⇒ roughen (deteriorate) the teeth surface and causing failure

To prevent pitting ⇒ computed contact stress in the gear must not exceed the allowable contact stress given by manufacturer.

Other factors can also be included in the contact stress calculation, namely: reliability factor, velocity factor, size factor, etc.


5.16 – Surface Durability



12

Surface Compressive stress (Hertzian stress), σC


5.16 – Surface Durability – cont…



2/1

21

11cos ⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

rrFWKC

t

pc φσ ν

Where;

Cp = elastic coefficientKν = dynamic or velocity factorWt = transmitted force (Ibf or N)F = width of gear tooth (inch or mm)φ = Pressure angler1 = radii of curvature on the pinion profile ⇒

r2 = radii of curvature on the gear tooth profile ⇒

2sin

1φPdr =

2sin

2φGdr =

Sign is negative because σC is a compressive stress.

13

AGMA defines an elastic coefficient ,Cp by the equation

Value of Cp may be computed directly from above equation or obtained from Table 5-4





2/1

22 111

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

−=

G

G

P

Pp

EE

Cννπ

Where;

νP &νG = Poisson’s ratio for pinion and gear

EP & EG = Modulus of Elasticity for pinion and gear

( )MPapsi

14





Gear Material and Modulus of Elasticity EG,Ibf/in2 (MPa)*

Pinion Material

Pinion Modulus of Elasticity EPpsi (MPa)*

Steel

30 x 106

(2 x 105)

Malleable Iron

25 x 106

(1.7 x 105)

Nodular Iron

24 x 106

(1.7 x 105)

Cast Iron

22 x 106

(1.5 x 105)

Aluminum Bronze

17.5 x 106

(1.2 x 105)

Tin Bronze

16 x 106

(1.1 x 105)

Steel 30 x 106

(2 x 105)2300(191)

2180(181)

2160(179)

2100(174)

1950(162)

1900(158)

Malleable iron

25 x 106

(1.7 x 105)2180(181)

2090(174)

2070(172)

2020(168)

1900(158)

1850(154)

Nodular iron

24 x 106

(1.7 x 105)2160(179)

2070(172)

2050(170)

2000(166)

1880(156)

1830(152)

Cast iron 22 x 106

(1.5 x 105)2100(174)

2020(168)

2000(166)

1960(163)

1850(154)

1800(149)

Aluminum bronze

17.5 x 106

(1.2 x 105)1950(162)

1900(158)

1880(156)

1850(154)

1750(145)

1700(141)

Tin bronze 16 x 106

(1.1 x 105)1900(158)

1850(154)

1830(152)

1800(149)

1700(141)

1650(137)

Table 5-4: Elastic Coefficient Cp, ( )MPapsi

Poisson’s ratio = 0.30* When more exact values for modulus of elasticity are obtained from roller contact test, they may be used.

15





The pinion of Examples 5-4 is to be mated with a 50-tooth gear manufactured of ASTM No. 50 cast iron. Using the tangential load of 1700N, estimate the factor of safety of the drive based on the possibility of a surface fatigue failure.



From Table A-5

EP = 207 GPa

EG = 100 GPa

νP = 0.292

νG = 0.211

16





Solution 5-5 - cont…Solution 5-5 - cont…

From Example 5-4, the pinion pitch diameter is dP = 48 mm.

Calculated the elastic coefficient as 3.150927

)10(100)211.0(1

)10(207)292.0(1

1

2/1

9

2

9

2=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

−=

πpC

The gear pitch diameter is, mm150)50(3 === GG mNd

Then the radii curvature of the tooth profiles at the pitch point.

2.82

20sin48==Pr 7.25

220sin150

==Grmm mm


5 16 – Surface Durability – cont


5.16 Surface Durability cont…

Solution 5-5 - cont…

The face width is given as F = 38 mm. Use Kν = 1.5 from example 5-4. Substituting all th l t l l t t t tthese values to calculate contact stress as:

5.51111)1700(5.13.1509272/1

−=⎥⎤

⎢⎡ ⎞

⎜⎛ +−=σ MPa5.511

0257.00082.020cos038.03.150927 ⎥

⎦⎢⎣ ⎠

⎜⎝

+ocσ

The surface endurance strength of cast iron can be estimated from Bc HS 206.2= MPag Bc

From table A24, ASTM No. 50 cast iron ⇒ HB = 262 578)262(206.2 ==∴ cS MPa

28.15.511

578_

___ 2

2

2

=⎟⎠⎞

⎜⎝⎛===

C

CSloadimposed

loadfunctionoflossnσ



5 17 – AGMA Stress Equation


5.17 AGMA Stress Equation

2 fundamental stress equationsq

• Bending stress

⎪⎧ KKPKKKW Bmdt

Where ;Wt = tangential transmitted load (ibf or N)Ko = overload factor(U.S.

t

⎪⎪

⎪⎪⎨= KKKKKW

JFKKKW

Bmsvo

t

svo

1σ

o

Kv = dynamic factorKs = size factorPd = transverse diameteral pitchF = face width (in or mm)

customary units)

(SI units)⎪⎩ JFmt

svo F = face width (in or mm)Km = load distribution factorKB = rim thickness factorJ = geometry factor for bending strength

( )

• Pitting resistance (contact stress) mt = transverse metric module

CKKKKWC fmtσ(U.S.

Where ;Cp = elastic coefficient (√ ibf/in2 or √ N/mm2)

IFdKKKWC f

p

msvopc =σ customary or

SI units)

p ( )Cf = surface condition factor (still not been

establish) used Cf = 1dp = pitch diameter of the pinion (in or mm)I = geometry factor for pitting resistance


g y p g


5 17 – AGMA Stress Equation – cont


5.17 AGMA Stress Equation cont…

Overload Factor, Ko, o

• Is intended to make allowance for all externally applied loads in excess of the nominal tangential load, Wt in a particular application.

Table 5-5: Overload Factors, Ko

• Similar factors such as application factor or service factor

Driven MachinePower Source Uniform Moderate shock Heavy shock

Generator, C t if l

Machine tool main drive, lti li d

Ore crusher, rolling mill, power h l i l li dCentrifugal

compressor, pure liquid mixer

multi-cylinder compressor or pump, liquid + solid mixer

shovel, single cylinder compressor or pump, punch press

Uniform Electric motor, steam turbine, gas turbine 1.00 1.25 1.75

Light shock Multi cylinder internal combustion engine with many cylinder 1.25 1.50 2.00

Medium shock Single cylinder internal combustion engine 1.50 1.75 2.25



5 17 – AGMA Stress Equation – cont


5.17 AGMA Stress Equation cont…

Dynamic Factor, Kv

Used to account for inaccuracies in the manufacture and meshing of gear teeth in action.

AGMA has define a set of quality numbers (Qv) to specify the tolerances for gears of i i f t d t ifi d

y , v

various sizes manufactured to a specified accuracy.

Qv = 3 to 7 ⇒ for most commercial-quality gears

Qv = 8 to 12 ⇒ for precision qualityQv 8 to 12 ⇒ for precision quality

Dynamic factor equation :

⎪⎧ ⎞⎜⎛ +

BVA

where ;

⎪

⎪⎪⎪

⎨⎞

⎜⎛ +

⎟⎠

⎞⎜⎜⎝

⎛ +

Bv

VA

AVA

K200

V in ft/min

V i /

)1(5650 BA −+=3/2)12(25.0 vQB −=

Maximum velocity, (at the end point of the Qv curve) ;

⎪⎪⎪

⎩⎟⎠

⎞⎜⎜⎝

⎛ +A

VA 200

( )[ ][ ]⎪⎨

⎧ −+= 3(

3(2

2v

t QA

QAV

V in m/s

ft/min

20

y, ( p Qv ) ;


( ) [ ]⎪⎩

⎨ −+200

3( 2maxv

t QAVm/s

21


5.17 – AGMA Stress Equation – cont…



Figure 5-1: Dynamic factor Kv

22

The size factor reflects nonuniformity of material properties due to size. It depends uponTooth sizeDiameter of partRatio of tooth size to diameter of partFace widthArea of stress patternRatio of case depth to tooth sizeHardenability and heat treatment

AGMA size factor equation :

If Ks is less than 1, use Ks = 1





Size Factor, Ks

0535.0

192.1 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ds P

YFK (U.S. customary units)

( ) 0535.0904.0 YFmKs =

(SI units)

23

The load modification factor modified the stress equations to reflect nonuniform distribution of load across the line of contact.

The ideal is to locate the gear “midspan” between two bearings at the zero slope place when the load is applied.

However this is not always possible. The following procedure is applicable to:

Net face width to pinion pitch diameter ratio F / d ≤ 2

Gear elements mounted between the bearings

Face widths up to 40 in

Contact, when loaded, across the full width of the narrowest member

Face load distribution factor :





Load Distribution Factor, Km

)(1 emapmpfmcmfm CCCCCCK ++==

24

where



⎩⎨⎧

8.01

mcC

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−+−

+−

−

=

2000228.00207.01109.010

0125.00375.010

025.010

FFd

F

Fd

Fd

F

Cpf



Load Distribution Factor, Km (cont…)

for uncrowned teeth

for crowned teeth

F ≤ 1 in

1< F ≤ 17 in

17 < F ≤ 40 in

Examples of crowned teeth

for values of , is used.05.010

<d

F 05.010

=d

F

d and F must be in US customary units (in)

25






for straddle-mounted pinion with S1/S < 0.175

⎩⎨⎧

1.11

pmCfor straddle-mounted pinion with S1/S ≥ 0.175

Figure 5-2: Definition of distances S and S1 used in evaluating Cpm

26



2CFBFACma ++=




Table 5-6: Empirical Constants A, B, and C for Face Width F in Inches

Condition A B C

Open Gearing 0.247 0.0167 -0.765(10-4)

Commercial, enclosed units 0.127 0.0158 -0.930(10-4)

Precision, enclosed units 0.0675 0.0128 -0.926(10-4)

Extra precision enclosed gear units 0.00360 0.0102 -0.822(10-4)

27






Figure 5-3: Mesh alignment factor Cma

28






for gearing adjusted at assembly, or compatibility is improved by lapping, or both

for all other conditions⎩⎨⎧=

8.0

1eC

29

Rim thickness factor adjusts the estimate bending stress for the thin-rimmed gear.

If the rim thickness is not sufficient to provide full support for the tooth root, the location of bending fatigue failure may be through the gear rim rather than at the tooth fillet.

Rim thickness factor :





Rim-Thickness Factor, KB

⎪⎩

⎪⎨

⎧=

1

242.2ln6.1BB mK

mB < 1.2

mB ≥ 1.2

where mB is a function of the backup ratio

t

RB h

tm =

30





Rim-Thickness Factor, KB (cont…)

Figure 5-4: Rim thickness factor KB

31





Bending-Strength Geometry Factor, J

Figure 5-5: Spur-gear geometry factors J.

Used Fig. 5-5 to obtain the geometry factor Jfor spur gears having a 20o pressure angle and full-depth teeth. Use Fig. 5-6 and 5-7 for helical gears having a 20o normal pressure angle and face contact ratios of mF = 2 or greater. For other gears, consult the AGMA standard.

32





Bending-Strength Geometry Factor, J – cont…

Figure 5-6: Helical-gear geometry factors J.

33





Bending-Strength Geometry Factor, J – cont…

Figure 5-7: J factor multipliers for use withFig. 10-6 to find J.

34





Surface-Strength Geometry Factor, I

Also known as pitting-resistance geometry factor by AGMA

For both spur and helical gears:-

external gears

⎪⎪⎩

⎪⎪⎨

⎧

−

+=

12sincos

12sincos

G

G

N

tt

G

G

N

tt

mm

m

mm

mI φφ

φφ

internal gears

mN = Load sharing ratio ( = 1 for spur gear)

mG = Gear ratio (never less than 1)

φt = Transverse pressure angle

where:

Zpm N

N 95.0= nnN pp φcos=

PN = Normal base pitch

Pn = Normal circular pitch

φn = Normal pressure angle

Z = Length line of action in transverse plane. (distance Lab in slide chapter 9 page 35)

and

where

For helical gear

( )[ ] ( )[ ] ( ) tGPGbGPbP rrrarrarZ φsin2/1

222/1

22 +−−++−+=

r P,G = pitch radius (pinion or gear)a = addendum

rb P,G = base circle radius (pinion or gear)

tb rr φcos=

35


5.18 – AGMA Strength Equation



AGMA introduced 2 types of strength:

1) St - Gear bending strength (Allowable bending stress number)(refer Figs. 5-8, 5-9, 5-10 and Tables 5-7, 5,8)

2) Sc - Contact fatigue strength (Allowable contact stress number)(refer Fig. 5-11 and Tables 5-9, 5,10, 5-11)

Both allowable stress numbers (strength) are for:Unidirectional loading10 million stress cycles99 % reliability

36


5.18 – AGMA Strength Equation – cont…



Equation for allowable bending stress:

⎪⎪⎩

⎪⎪⎨

⎧

Z

N

F

t

RT

N

F

t

all

YYY

SS

KKY

SS

θ

σ(U.S. customary units)

(SI units)

St = Gear bending strength, Ibf/in2 (MPa)

YN = Stress cycle factor for bending stress

Kt (Yθ) = Temperature factor

KR (YZ) = Reliability factor

SF = AGMA factor of safety, stress ratio

where(refer Figs. 5-8, 5-9, 5-10 and Tables 5-7, 5,8)

(refer Tables 5-12)

(refer Figs. 5-12)

37





Equation for contact bending stress:

⎪⎪⎩

⎪⎪⎨

⎧

Z

WN

H

c

RT

HN

H

c

allc

YYZZ

SS

KKCZ

SS

θ

σ ,


(SI units)

Sc = Contact fatigue strength, Ibf/in2 (MPa)

ZN = Stress cycle life factor

CH (ZW) = Hardness ratio factors for pitting resistance (used for gear only)

Kt (Yθ) = Temperature factor

KR (YZ) = Reliability factor

SH = AGMA factor of safety, stress ratio

where(refer Fig. 5-11 and Tables 5-9, 5,10, 5-11)

(refer Figs. 5-14 & 5-15)

(refer Tables 5-12)

(refer Figs. 5-13)

38





Figure 5-8: Allowable bending stress number for through-hardened steels. The SI equations areSt = 0.533HB + 88.3 MPa, grade 1, and St = 0.703 HB + 113 MPa, grade 2.(Source: ANSI/AGMA 2001-D04 and 2101-D04.)

39





Figure 5-9: Allowable bending stress number for nitrided through-hardened steels gears (i.e., AISI 4140, 4340) St. The SI equations are St = 0.568HB + 83.8 MPa, grade 1, and St = 0.749 HB + 110 MPa, grade 2.(Source: ANSI/AGMA 2001-D04 and 2101-D04.)

40





Figure 5-10: Allowable bending stress number for nitriding steels gears St. The SI equations are :

St = 0.594HB + 87.76 MPa, Nitralloy grade 1, St = 0.784HB + 114.81 MPa, Nitralloy grade 2, St = 0.7255HB + 63.89 MPa, 2.5% chrome grade 1, St = 0.7255HB + 153.63 MPa, 2.5% chrome grade 2 St = 0.7255HB + 201.91 MPa, 2.5% chrome grade 3.

(Source: ANSI/AGMA 2001-D04 and 2101-D04.)

41





MaterialDesignation Heat Treatment

Minimum Surface

Hardness

Allowable Bending Stress Number, St2

psi (MPa) Grade 1 Grade 2 Grade 3

Steel Through-hardened See Fig. 5-8 See Fig. 5-8 See Fig. 5-8 -

Flame or induction hardened with type A pattern

45 000 (310) 55 000 (380) -

Flame or induction hardened with type B pattern

22 000 (151) 22 000 (151) -

Carburized and hardened

55 000 (380) 65 000 or (448 or 70 0006 482)

75 000 (517)

Nitrided (through-hardened steels)

83.5 HR15N See Fig. 5-9 See Fig. 5-9 -

Nitralloy 135M,Nitralloy N, and 2.5% chrome (no aluminum)

Nitrided 87.5 HR15N See Fig. 5-10 See Fig. 5-10 See Fig. 5-10

Table 5-7: Repeatedly Applied Bending Strength St at 107 Cycles and 0.99 Reliability for Steel GearsSource: ANSI/AGMA 2001-D04

42





Material MaterialDesignation

Heat Treatment Typical Minimum Surface Hardness

Allowable Bending Stress Number, St

psi (MPa)

ASTM A48 gray cast iron

Class 20 As cast - 5000 (35)

Class 30 As cast 174 HB 8500 (58)

Class 40 As cast 201 HB 13 000 (90)

ASTM A536 ductile (nodular) Iron

Grade 60-40-18 Annealed 140 HB 22 000 - 33 000 (151 - 227)

Grade 80-55-06 Quenched and tempered 179 HB 22 000 - 33 000 (151 - 227)



Bronze Sand cast Minimum tensile strength 40 000 psi

5700 (39)

ASTM B-148 Alloy 954

Heat treated Minimum tensile strength 90 000 psi

23 600 (163)

Table 5-8: Repeatedly Applied Bending Strength St for Iron and Bronze Gears at 107 Cycles and 0.99 ReliabilitySource: ANSI/AGMA 2001-D04

43





Figure 5-11: Contact-fatigue strength Sc at 107 cycles and 0.99 reliability for through-hardened steel gears. The SI equations are :

Sc = 2.22HB + 200 MPa, grade 1,

and

Sc = 2.41HB + 237 MPa, grade 2.

(Source: ANSI/AGMA 2001-D04 and 2101-D04.)

44





SteelTemperature

Before Nitriding, oF

Nitriding, oFHardness, Rockwell

C ScaleCase Core

Nitralloy 135 1150 975 62 - 65 30 - 35

Nitralloy 135M 1150 975 62 - 65 32 - 36

Nitralloy N 1000 975 62 - 65 40 - 44

AISI 4340 1100 975 48 - 53 27 - 35

AISI 4140 1100 975 49 - 54 27 - 35

31 Cr Mo V 9 1100 975 58 - 62 27 - 33

Table 5-9: Nominal Temperature Used in Nitriding and Hardnesses ObtainedSource: Darle W. Dudley, Handbook of Practical Gear Design, rev. ed., McGraw-Hill. New York, 1984

45





MaterialDesignation Heat Treatment

Minimum Surface

Hardness

Allowable Contact Stress Number, Sc2

psi (MPa) Grade 1 Grade 2 Grade 3

Steel Through-hardened See Fig. 5-11 See Fig. 5-11 See Fig. 5-11 ─

Flame or induction hardened

50 HRC 54 HRC

170 000 (1172)175 000 (1206)

190 000 (1310)195 000 (1344)

──

Carburized and hardened

180 000 (1240) 225 000 (1551) 275 000 (1896)

Nitrided (through-hardened steels)

83.5 HR15N84.5 HR15N

150 000 (1035)155 000 (1068)

163 000 (1123)168 000 (1158)

175 000 (1206)180 000 (1240)

2.5% chrome (no aluminum)

Nitrided 87.5 HR15N 155 000 (1068) 172 000 (1186) 189 000 (1303)

Nitralloy 135M Nitrided 90.0 HR15N 170 000 (1172) 183 000 (1261) 195 000 (1344)

Nitralloy N Nitrided 90.0 HR15N 172 000 (1186) 188 000 (1296) 205 000 (1413)

2.5% chrome (no aluminum)

Nitrided 90.0 HR15N 176 000 (1213) 196 000 (1351) 216 000 (1490)

Table 5-10: Repeatedly Applied Contact Strength Sc at 107 Cycles and 0.99 Reliability for Steel GearsSource: ANSI/AGMA 2001-D04

46





Material MaterialDesignation

Heat Treatment Typical Minimum Surface Hardness

Allowable Contact Stress Number, Sc

psi (MPa)

ASTM A48 gray cast iron

Class 20 As cast - 50 000 – 60 000 (344 – 415)

Class 30 As cast 174 HB 65 000 – 75 000 (448 – 517)

Class 40 As cast 201 HB 75 000 – 85 000 (517 – 586)

ASTM A536 ductile (nodular) Iron

Grade 60-40-18 Annealed 140 HB 77 000 – 92 000 (530 – 634)

Grade 80-55-06 Quenched and tempered 179 HB 77 000 – 92 000 (530 – 634)



Bronze Sand cast Minimum tensile strength 40 000 psi

30 000 (206)

ASTM B-148 Alloy 954

Heat treated Minimum tensile strength 90 000 psi

65 000 (448)

Table 5-11: Repeatedly Applied Contact Strength Sc for Iron and Bronze Gears at 107 Cycles and 0.99 ReliabilitySource: ANSI/AGMA 2001-D04

47





Figure 5-12: Repeatedly applied bending strength stress-cycle factor YN. (Source: ANSI/AGMA 2001-D04)

Stress Cycle Factor for Bending Stress, YN

48





Figure 5-13: Pitting resistance stress-cycle factor ZN. (Source: ANSI/AGMA 2001-D04)

Stress Cycle Life Factor, ZN

49





Hardness Ratio Factor, CH

• Pinion is subjected to more cycles of contact stress (due to smaller size)

• The hardness-ratio factor is used only for gear to adjust the gear surface strength due to different condition of pinion and gear hardness and size;

)1('1 −+= GH mAC

7.12.1(G)

(P) ≤≤B

B

HH


Hardness Ratio

Both pinion and gear are through hardened

Surface hardened pinion is mated with through hardened gear

See figure 10-14 (slide 50)

A’ = 0

A’ = 0.00698

Surface hardened pinions with hardnesses of 48 Rockwell C scale or harder mated with through hardened gears (180-400 Brinell)

See figure 10-15 (slide 51)

B’ = 0.00075 exp [ 0.0112 fP ]

where fP is the surface finish of the pinion expressed as root-mean-square roughness Ra in µ in.

2.1(G)

(P) <B

B

HH

7.1(G)

(P) >B

B

HH

⎟⎟⎠

⎞⎜⎜⎝

⎛

(G)

(P)

B

B

HH

)10(29.8)10(98.8' 3

(G)

(P)3 −− −⎟⎟⎠

⎞⎜⎜⎝

⎛=

B

B

HH

A

)450('1 BGH HBC −+=

50





Figure 5-14: Hardness ratio factor CH

(through-hardened steel). (Source: ANSI/AGMA 2001-D04)


51





Figure 5-15: Hardness ratio factor CH

(surface-hardened steel pinion). (Source: ANSI/AGMA 2001-D04)


52





Reliability Factor, KR (YZ)

Reliability KR (YZ)

0.9999 1.50

0.999 1.25

0.99 1.00

0.90 0.85

0.50 0.70

Table 5-12: Realibility Factor KR (YZ)Source: ANSI/AGMA 2001-D04

Temperature Factor, KT (Yθ)

)120(250 C Fetemperatur oo≤

1== θYKT

For higher temperature, heat exchangers may be used to ensure that operating temperatures are considerably below this value.

53


5.19 – Safety Factors SF and SH



Safety factor SF guarding against bending failure:

σ)/( RTNt

FKKYS

S ==stress bending

strength bending correctedfully

Safety factor SH guarding against pitting failure:

c

RTHNcH

KKCZSS

σ)/(* ==

stress contact

strength contact correctedfully

* Important:

When deciding whether bending or wear is the threat to function, compare SF with S2H. For

crowned gears compare SF with S3H.

54


Roadmap of Gear Analysis



Refer also roadmap given in the Shigley’s Mechanical Engineering Design book (9th edition):

• Roadmap of spur gear bending equations based on AGMA standards in U.S. customary units. (ANSI/AGMA 2001-D04) – Figure 14-17 page 766

• Roadmap of spur gear wear equations based on AGMA standards in U.S. customary units. (ANSI/AGMA 2001-D04) – Figure 14-17 page 767

• Roadmap of spur gear bending equations based on AGMA standards in SI units. (ANSI/AGMA 2001-D04) – Figure B-1 page 1061

• Roadmap of spur gear wear equations based on AGMA standards in SI units. (ANSI/AGMA 2001-D04) – Figure B-2 page 1062

• Roadmap summary of principal straight-bevel gear wear equations and their parameters in U.S. customary units. (ANSI/AGMA 2003-B97). – Figure 15-14 page 801

• Roadmap summary of principal straight-bevel gear bending equations and their parameters in U.S. customary units. (ANSI/AGMA 2003-B97). . – Figure 15-15 page 802

• Roadmap summary of principal straight-bevel gear wear equations and their parameters based on AGMA standards in SI units. – Figure B-3 page 1063

• Roadmap summary of principal straight-bevel gear bending equations and their parameters based on AGMA standards in SI units. – Figure B-4 page 1064

55


Roadmap of Gear Analysis (Spur Gear Bending)



Gear bending stress equation

AGMA standards in U.S. customary units. (ANSI/AGMA 2001-D04)

d

PP P

Nd =12dnV π

=

VHW t 33000

=

JKK

FP

KKKW Bmdsvo

t=σ

σ)/( RTNt

FKKYS

S =

RT

N

F

tall KK

YSS

=σ

Gear bending endurance strength equation

Bending factor of safety

Remember when deciding whether bending or wear is the threat to function, compare SF with S2H.

For crowned gears compare SF with S3H.

Slide 19

Slide 20, 21

Slide 22

Slide 23

Slide 29, 30

Slide 31 (Spur Gear)Slide 32, 33 (Helix gear)

Slide 38 ~ 42

Slide 47

1 if T < 250o F

Slide 52

(slide 18)

(slide 36)

(slide 53)

56


Roadmap of Gear Analysis (Spur Gear Wear)



Gear contact stress equation

AGMA standards in U.S. customary units. (ANSI/AGMA 2001-D04)

d

PP P

Nd =12dnV π

=

VHW t 33000

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

IC

FdKKKKWC f

P

msvo

tpcσ

c

RTHNcH

KKCZSS

σ)/(

=

RT

HN

H

callc KK

CZSS

=,σ

Gear contact endurance strength equation

Wear factor of safety

Remember when deciding whether bending or wear is the threat to function, compare SF with S2H.

For crowned gears compare SF with S3H.

Slide 19

Slide 20, 21

Slide 22

Slide 23

1

Slide 34

Slide 43 ~ 46

Slide 48

1 if T < 250o F

Slide 52

(slide 18)

(slide 37)

(slide 53)

Slide 13, 14

Slide 49 (Gear only)

Gear only

57


Examples



A 17-tooth 20o pressure angle spur pinion rotates at 1800 rev/min and transmits 4 hp to a 52-tooth disk gear. The diametral pitch is 10 teeth/in, the face width 1.5 in, and the quality standard is No. 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion is grade 1 steel with a hardness of 240 Brinell tooth surface through-hardened core. The gear is steel, through-hardened also, grade 1 material, with a Brinell hardness of 200, tooth surface and core. Poisson’s ratio is 0.30, JP=0.30, JG=0.40, and Young’s modulus is 30(106) psi. The loading is smooth because of motor and load. Assume a pinion life of 108 cycles and a reliability of 0.90, and use YN=1.3558N -0.0178, ZN=1.4488N -0.023. The tooth profile is uncrowned. This is a commercial enclosed gear unit.

a) Find the factor of safety of the gears in bending.b) Find the factor of safety of the gears in wear.c) By examining the factors of safety, identify the threat to each gear and to the mesh.


58


Examples – cont…




in 7.110/17/ === dPP PNd

in 2.510/52 ==Gd

ft/min 801.112

1800)7.1(12

===ππ PPndV

ibf 8.6411.801

)4(3300033000===

VHW t

Assuming uniform loading, Ko = 1. To evaluate Kv , with a quality number Qv = 6,

There will be many term to obtain, so use roadmap in slide 54 & 55 as guides to what is needed:

8255.0)612(25.0 3/2 =−=B

77.59)8255.01(5650 =−+=A

377.177.59

1.80177.598255.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +=vK

To determine the size factor, Ks , the Lewis form factor is needed. From table 5-3, with NP = 17 teeth, YP = 0.303. Interpolation for the gear with NG = 52 teeth yields YG = 0.412. Thus, with F = 1.5 in ,

( ) 043.110

303.05.1192.10535.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +=PsK

( ) 052.110

412.05.1192.10535.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +=GsK

59





Solution 5-6 – cont…Solution 5-6 – cont…

( ) 0695.0)5.1(0125.00375.0)]7.1(10/[5.1 =+−=PpfC

The load distribution factor Km is determined where 5 terms are needed: They are

)(uncrowned 1=mcC

adjacent)y immediatel (Bearing 1=pmC

unit)gear enclosed l(Commercia 15.0=maC1=eC

( )

( ) 18.122.1

)]1(15.0)1(0695.0[11

)(1

==

++=

++=

Gm

emapmpfmcPm

K

CCCCCK

Assuming constant thickness gears, the rim thickness factor KB = 1. The speed ratio is mG = NG / NP = 52 / 17 = 3.059.

The load cycle factors given in the problem statement, with N (pinion) = 108 cycles and N (gear) = 108 / mG = 108 / 3.059 cycles are:

996.0)059.3/10(3558.1)(

977.0)10(3558.1)(0178.08

0178.08

==

==−

−

GN

PN

Y

Y

From table 5-12 (slide 52), with a reliability of 0.9, KR = 0.85, From slide 56, the temperature and surface condition factor; KT = 1 & Cf = 1. From slide 34, with mN = 1 for spur gears;

121.01059.3

059.32

20sin20cos=

+=

ooI

( ) 0313.0)5.1(0125.00375.005.0 =+−=GpfC

60






From table 5-4 (slide 14), CP = 2300 .

Next, the terms of gear endurance strength need to be calculate. From table 5-7 (slide 41) for grade 1 through-hardened steel with HB(P) = 240 and HB(G) = 200 select figure 5-8 (slide 38) ;

( )( ) psi 260,28800,12)200(3.77

psi 350,31800,12)240(3.77=+=

=+=

Gt

Pt

SS

psi

Similarly, from table 5-10 (slide 45), use figure 10-11 (slide 43)

( )( ) psi 500,93100,29)200(322

psi 400,106100,29)240(322=+=

=+=

Gc

Pc

SS

Form figure 5-13 (slide 48) ;

( )( ) 973.0)059.3/10(4488.1

948.0)10(4488.1023.08

023.08

==

==−

−

GN

PN

Z

Z

For the hardness ratio factor CH, the hardness ratio is HB(P) / HB(G) = 240/200 = 1.2. Then, from slide 49

00249.0)10(29.8)2.1)(10(98.8' 33 =−= −−A

005.1)1059.3(00249.01 =−+=∴ HC

61






(a) Pinion tooth bending

psi 6417 30.0

)1(22.15.1

10)043.1)(377.1)(1(8.164

)(

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

⎟⎠

⎞⎜⎝

⎛=P

Bmdsvo

tP J

KKFPKKKWσ

5.62 6417

)85.0(1/)977.0(350,31

/)(

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠

⎞⎜⎝

⎛=P

RTNtPF

KKYSS

σ

Gear tooth bending

psi 4695 40.0

)1(18.15.1

10)052.1)(377.1)(1(8.164)(

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=Gσ

7.05 4695

)85.0(1/)996.0(260,28)(

=

⎟⎠⎞

⎜⎝⎛=GFS

(b) Pinion tooth wear

psi 360,70 121.01

)5.1(7.122.1)043.1(377.1)1(8.1642300

)(

2/1

2/1

=

⎥⎦

⎤⎢⎣

⎡=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

P

f

P

msvo

tPc I

CFd

KKKKWσ

62






1.69 360,70

)85.0(1/)948.0(400,106

/)(

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Pc

RTNcPH

KKZSS

σ

Gear tooth wear

psi 496,69 121.01

)5.1(7.118.1)052.1(377.1)1(8.1642300)(

2/1

=

⎥⎦

⎤⎢⎣

⎡=Gcσ

1.55 69496

)85.0(1/)005.1)(973.0(500,93

/)(

=

⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Gc

RTHNcGH

KKCZSSσ

(c) Threat Comparison (tooth bending or wear)

Bending FOS ( SF )

Wear FOS

( SH )2Threat

Pinion 5.62 1.692 = 2.86 wear

Gear 7.05 1.552 = 2.39 wear

63





A 17-tooth 20o normal pitch-angle helical pinion with a right hand helix angle of 30o rotates at 1800 rev/min and transmits 4 hp to a 52-tooth helical gear. The normal diametral pitch is 10 teeth/in, the face width 1.5 in, and the set has a quality number of 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion and gear are made from a through-hardened steel with surface and core hardnesses of 240 Brinell on the pinion and surface and core hardnessess of 200 Brinell on the gear. The transmission is smooth, connecting an electric motor and a centrifugal pump. Assume a pinion life of 108 cycles and a reliability of 0.90, and use the upper curve in Figs. 10-12 and 10-13.

a) Find the factor of safety of the gears in bending.b) Find the factor of safety of the gears in wear.c) By examining the factors of safety, identify the threat to each gear and to the mesh.


64






All of the parameters in this example are the same as in example 5-6 with the exception that we are using helical gears. Thus, several terms will be the same as example 5-6. The reader should verify that the following terms remain unchanged: Ko = 1, YP = 0.303, YG = 0.412, mG = 3.059, (Ks)P = 1.043, (Ks)G = 1.052, (YN)P = 0.977, (YN)G = 0.996, KR = 0.85, KT = 1, Cf = 1, CP = 2300 √psi, (St)P = 31,350 psi, (St)G = 28,260 psi, (Sc)P = 106,380 psi, (Sc)G = 93,500 psi, (ZN)P = 0.948, (ZN)G = 0.973 and CH = 1.005

ft/min 25912

1800)963.1(12

===ππ PPndV

ibf 7.241925

)4(3300033000===

VHW t

404.177.59

92577.598255.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +=vK

For helical gears, the transverse diametral pitch;

teeth/in660.8)30(cos10cos === ont PP ψ

Thus the pitch diameters are dP = NP / Pt = 17 / 8.660 =1.963 in and dG = 52 / 0.8660 = 6.005 in. The pitch line velocity and transmitted force are:

As in previous example; for the dynamic factor, B = 0.8255 and A = 59.77. Thus

65






The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by :

oo

on

t 80.2230cos20tantan

costan

tan 11 =⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛= −−

ψφ

φ

The radii of the pinion and gear are rP = 1.963 / 2 = 0.9815 in and rG = 6.004 / 2 = 3.002 in. The addendum is a = 1 / Pn = 1 / 10 = 0.1, and the base circle radii of the pinion and gear are given by:

in 767.2)80.22(cos002.3cos)(

in 9048.0)80.22(cos9815.0cos)(

===

===o

tGGb

otPPb

rr

rr

φ

φ

Then the surface strength geometry factor

in 0.45071.544.4-1.40270.5924 80.22sin)004.39815.0(

769.2)1.0004.3(

9048.0)1.09815.0(22

22

=+=+−

−+

+−+=

o

Z

Since the first two terms are less than 1.5444, the equation for Z stands. The normal base pitch pN is

in 2952.020cos10

coscos)(

==

==

o

nN

nnN Ppp

π

φπφ

66






The load sharing ratio :

6895.0)4507.0(95.0

2952.095.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛=Z

pm N

N

The surface strength geometry factor :

195.0106.3

06.3)6895.0(2

80.22cos80.22sin=⎟

⎠⎞

⎜⎝⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

ooI

From slide 32, The geometry factors JP’ = 0.45 and JG’ = 0.54. Also from slide 33, the J-factor multipliers are 0.94 and 0.98, correcting JP’ and JG’ to:

529.0)98.0(54.0423.0)94.0(45.0

====

G

P

JJ

The load-distribution factor Km is estimated :

( ) 0577.0)5.1(0125.00375.0)963.1(10

5.1=+−=

PpfC

)(uncrowned 1=mcC

adjacent)y immediatel (Bearing 1=pmC

unit)gear enclosed l(Commercia 15.0=maC

1=eC

( )

( ) 181.1208.1

)]1(15.0)1(0577.0[11

)(1

==

++=

++=

Gm

emapmpfmcPm

K

CCCCCK

( ) 0313.0)5.1(0125.00375.005.0 =+−=GpfC

67






(a) Pinion tooth bending

psi 3445 423.0

)1(208.15.1

66.8)043.1)(404.1)(1(7.142

)(

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

⎟⎠

⎞⎜⎝

⎛=P

Bmdsvo

tP J

KKFPKKKWσ

5.01 3445

)85.0(1/)977.0(350,31

/)(

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠

⎞⎜⎝

⎛=P

RTNtPF

KKYSS

σ

Gear tooth bending

psi 2717 529.0

)1(181.15.1

66.8)052.1)(404.1)(1(7.142)(

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=Gσ

2.21 2717

)85.0(1/)996.0(260,28)(

=

⎟⎠⎞

⎜⎝⎛=GFS

(b) Pinion tooth wear

psi 230,48 195.01

)5.1(963.1208.1)043.1(404.1)1(7.1422300

)(

2/1

2/1

=

⎥⎦

⎤⎢⎣

⎡=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

P

f

P

msvo

tPc I

CFd

KKKKWσ

68





Solution 5-7– cont…Solution 5-7– cont…

2.46 230,48

)85.0(1/)948.0(400,106

/)(

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Pc

RTNcPH

KKZSS

σ

Gear tooth wear

psi 888,47 195.01

)5.1(963.1181.1)052.1(404.1)1(7.1422300)(

2/1

=

⎥⎦

⎤⎢⎣

⎡=Gcσ

52.2 888,47

)85.0(1/)005.1)(973.0(500,93

/)(

=

⎟⎠

⎞⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Gc

RTHNcGH

KKCZSSσ

(c) Threat Comparison (tooth bending or wear)

Bending FOS ( SF )

Wear FOS

( SH )2Threat

Pinion 10.5 2.462 = 6.05 wear

Gear 12.2 2.252 = 5.05 wear

Chapter 6Shaft Design




Week 8


2

Learning Outcomes


• select suitable material for shaft design

• perform load, stress, and power calculations analytically as applied to a shaft components.

• design a shaft with some consideration on static and fatigue failure.

• do tolerance analysis and specify appropriate tolerances for shaftdesign applications


CHAPTER 6 – Shaft Design




• 6.1 - Introduction• 6.2 - Shaft Materials• 6.3 - Shaft Layout• 6.4 - Shaft Design for Stress• 6.5 - Deflection Considerations• 6.6 - Limits and Fits

CHAPTER 6 – Shaft Design BDA 30803 – Mechanical Engineering Design




What is shaft?!

~a rotating member, usually of circular cross section

What it is used for?!

~to transmit power or motion~It provides the axis of rotation, or oscillation, of elements such as gears, pulleys, flywheels, cranksand the like, and controlsthe geometry of their motion.





Train wheels are affixed to a straight axle, such that both wheels rotate in unison.

An axle is a nonrotating memberthat carries no torque and

What is axle?!

What it is used for?!

is used to support rotatingwheels, pulleys and etc.





Tapered roller bearings used in a mowing-machine spindle. This design represents good practice for situations where one or more torque-transfer elements must be mounted outboard.

A spindle is a short shaft. Terms such as lineshaft, headshaft, stub shaft, transmission shaft, countershaft, and flexible shaftare names associated with special usage.

What is spindle?!





Considerations for Shaft DesignConsiderations for Shaft Design

• Material Selection

• Geometric Layout

• Stress and strength– Static strength– Fatigue strength

• Deflection and rigidity– Bending deflection– Torsional deflection– Slope at bearings and shaft-supported elements– Shear deflection due to transverse loading of short shafts

• Vibration due to natural frequency



6.2 – Shaft Materials


• Many shafts are made from low carbon, cold-drawn or hot-rolled steel, such as ANSI 1020-1050 steels.

• A good practice is to start with an inexpensive, low or medium carbon steel for the first time through the design calculations.

• Deflection primarily controlled by geometry, not material.

• Stress controlled by geometry, not material.

• Strength controlled by material property.

• Shafts usually don’t need to be surface hardened unless they serve as the actual journal of a bearing surface.

• Cold drawn steel typical for d < 3 in.

• Hot rolled steel common for larger sizes. Should be machined all over.



6.2 – Shaft Materials – cont…


• For low production - turning is the suitable process (minimum material removal may be design goal).

• For High production - Forming or casting is common where minimum material may be design goal. Cast iron may be specified if the production quantity is high, and the gears are to be integrally cast with the shaft.

• Stainless steel may be appropriate for some environments – e.g. Involved in food processing.



6.3 – Shaft Layout


• Issues to consider for shaft layout– Axial layout of

components– Supporting axial

loads– Providing for torque

transmission– Assembly and

Disassembly



6.3 – Shaft Layout – cont…


Axial Layout of ComponentsAxial Layout of ComponentsAxial Layout of Components

Axial loads must be supported through a bearing to the frame.

Generally best for only one bearing to carry axial load to shoulder

Allows greater tolerances and prevents binding





Various method to attach element on shaft.Various method to attach element on shaft.

Assembly/Disassembly → progressively smaller diameter toward the endsAxial clearance → to allow machinery vibrationKeys/pins/rings → to secure rotating elements ( gear, pulley, etc)

clamp collar

snap ring

taper pinkeyhubhub

step

step stepstep

shaft

gearsprocket

axial clearance

bearing bearing

press fit

press fit

frame framesheave





• Significant detail is required to completely specify the geometry needed to fabricate a shaft.

• The geometry of a shaft is generally that of a stepped cylinder.

• The use of shaft shoulders is an excellent means of axially locating the shaft elements and to carry any thrust loads.





Common shaft loading mechanism:

Common shaft loading mechanism:





Common torque transfer elements:Common torque transfer elements:

• Keys• Splines• Setscrews• Pins• Press or shrink fits• Tapered fits





Pins:Pins:

Round pins Taper pins Split tubular spring pins

- Pins are used for axial positioning and for the transfer of torque or thrust or both.

- Some pins should not be used to transmit very much torque

- Weakness – will generate stress concentration to the shaft





Keys and keyseats:Keys and keyseats:

Keys are used to transmit torque from a component to the shaft.





- Used when large amounts of torque are to be transferred

- Stress concentration is generally quite moderate

Spline shaft and Hub:Spline shaft and Hub:





Locational device:Locational device:

• Nut and washer• Sleeve• Shaft shoulder• Ring and groove• Setscrew• Split hub or tapered two-pieces hub• Collar and screw• Pins





Nut and Washer:Nut and Washer:





Sleeve:Sleeve:

is a tube or enclosure used to couple two mechanical components together, or to retain two components together; this permits two equally-sized appendages to be connected together via insertion and fixing within the construction.





Shaft shoulder :Shaft shoulder :

The use of shaft shoulders is an excellent means of axially locating the shaft elements and to carry any thrust loads.

(a) Choose a shaft configuration to support and locate the two gears and two bearings. (b) Solution uses an integral pinion, three shaft shoulders, key and keyway, and sleeve.

Example:





Spring loaded Retaining Ring :Spring loaded Retaining Ring :

• Most popular used because give an economical solution to some problem.

• “Bowed” retaining rings provide restoring forces to the components being held.

• Flat retaining rings allow small amounts of axial motion of the held component.





Set Screw :Set Screw :

is a type of screw generally used to secure an object within another object. The set screw passes through a threaded hole in the outer object and is tightened against the inner object to prevent it from moving relative to the outer object.





Split Hub :Split Hub :





Collar and Screw :Collar and Screw :

• is a simple, short ring fastened over a rod or shaft

• found in many power transmission applications -most notably motors and gearboxes.

• used as mechanical stops, locating components, and bearing faces. The simple design lends itself to easy installation - no shaft damage.

• Since the screws compress the collar, a uniform distribution of force is imposed on the shaft, leading to a holding power that is nearly twice that of set screw collars.



6.4 – Shaft Design for Stress


• It is not necessary to evaluate the stresses in a shaft at every point; a few potentially critical locations will be adequate.

• Critical locations will usually be on the outer surface, at axial locations where the bending moment is large, where the torque is present, and where stress concentrations exist.

• Most shafts will transmit torque through a portion of the shaft. Typically the torque comes into the shaft at one gear and leaves the shaft at another gear. The torqueis often relatively constant at steady state operation.

• The bending moments on a shaft can be determined by shear and bending moment diagrams. Since most shaft problems incorporate gears or pulleys that introduce forces in two planes, the shear and bending moment diagrams will generally be needed in two planes.

Critical Location :Critical Location :



6.4 – Shaft Design for Stress – cont…


• Resultant moments are obtained by summing moments as vectors at points of interest along the shaft. In situations where a bearing is located at the end of the shaft, stresses near the bearing are often not critical since the bending moment is small.

• Axial stresses on shafts due to the axial components transmitted through helical gears or tapered roller bearings will almost always be negligibly small compared to the bending moment stress. They are often also constant, so they contribute little to fatigue.

• Consequently, it is usually acceptable to neglect the axial stresses induced by the gears and bearings when bending is present in a shaft. If an axial load is applied to the shaft in some other way, it is not safe to assume it is negligible without checking magnitudes.

Critical Location :Critical Location :



;

;;


IcMK a

fa =σI

cMK mfm =σ

JcTK a

fsa =τ


• The fluctuating stresses due to bending and torsion are given by: -

Shaft Stresses :Shaft Stresses :

Under many conditions, the axial components F is either zero or so small that it can be neglected.

JcTK m

fsm =τ

• Assuming a solid shaft with round cross section, appropriate geometry terms can be introduced for c, I, and J resulting in

3

32dMK a

fa πσ = 3

32dMK m

fm πσ = ;3

16dTK a

fsa πτ = 3

16dTK m

fsm πτ =




2/12

3

2

32/122 16

332

)3('⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=+=

dTK

dMK afsaf

aaa ππτσσ


• Combining bending and shear stresses accordance to the von Misses stress at two stress element are given by: -


2/12

3

2

32/122 16

332

)3('⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=+=

dTK

dMK mfsmf

mmm ππτσσ









[ ] [ ]⎭⎬⎫

⎩⎨⎧

+++= 2/1222/1223 )(3)(41)(3)(41161

mfsmfut

afsafe

TKMKS

TKMKSdn π


DE-Goodman :DE-Goodman :

[ ] [ ]3/1

2/1222/122 )(3)(41)(3)(4116⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

+++= mfsmfut

afsafe

TKMKS

TKMKS

ndπ

Fatigue failure curve on the modified Goodman diagram

Equation for the minimum diameter

This criteria does not guard against yielding, so required separate check for possibility of static failure (yield occur) in the first load cycle.




⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++=

2/12

3

21181

ut

e

e ASBS

SdA

n π

3/12/122118

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++=

ut

e

e ASBS

SnAdπ

22 )(3)(4 mfsmf TKMKB +=


DE-Gerber :DE-Gerber :

Fatigue failure curve on the Gerber diagram


This criteria does not guard against yielding, so required separate check for possibility of static failure (yield occur) in the first load cycle.

22 )(3)(4 afsaf TKMKA +=

where




2/12222

3 3434161⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

y

mfs

y

mf

e

afs

e

af

STK

SMK

STK

SMK

dn π


DE-ASME Elliptic :DE-ASME Elliptic :

Fatigue failure curve on the ASME Elliptic diagram


This criteria takes yielding into account, but is not entirely conservative, so also required separate check for possibility of static failure (yield occur) in the first load cycle.

3/12/12222

343416

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

y

mfs

y

mf

e

afs

e

af

STK

SMK

STK

SMKnd

π





DE-Soderberg :DE-Soderberg :

Fatigue failure curve on the Soderberg diagram

This criteria inherently guards against yielding, so it is not required to check for possibility of static failure (yield occur) in the first load cycle.

[ ] [ ]⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++= 2/1222/1223 )(3)(41)(3)(41161

mfsmfyt

afsafe

TKMKS

TKMKSdn π

[ ] [ ]3/1

2/1222/122 )(3)(41)(3)(4116⎟⎟⎠

⎞⎜⎜⎝

⎛

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++= mfsmfyt

afsafe

TKMKS

TKMKS

ndπ





max'σy

y

Sn =

( ) 2/122max ''' ma σσσ +=


Check for yielding :Check for yielding :

where

2/12

3

2

3

)(163

)(32

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ ++⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

dTTK

dMMK mafsmaf

ππ

Factor of safety

Always necessary to consider static failure, even in fatigue situation•Soderberg criteria inherently guards against yielding•ASME-Elliptic criteria takes yielding into account, but is not entirely conservative•Gerber and modified Goodman criteria require specific check for yielding




0=mσ


For a rotating shaft with constant bending and torsion, the bending stress is completely reversed and the torsion is steady. Therefore

0=aτ

These will simply drops out some of previously terms.





Example 6-1 :Example 6-1 :

At a machined shaft shoulder the small diameter d is 28 mm, the large diameter D is 42 mm, and the fillet radius is 2.8 mm. The bending moment is 142.4 Nmand the steady torsion moment is 124.3 Nm. The heat-treated steel shaft has an ultimate strength of Sut = 735 MPa and a yield strength of Sy = 574 MPa. The reliability goal is 0.99.

(a) Determine the fatigue factor of safety of the design using each of the fatigue failure criteria described in this section.

(b) Determine the yielding factor of safety.





Solution 6-1 :Solution 6-1 :

a) Determine the fatigue factor of safety of the design:

4.142=aM

0=aT

0=mM

3.124=mT

Nm

NmNm

Nm

10.028

8.2==

dr

50.12842

==dD

Kt = 1.68 (figure A-15-9)

Kts = 1.42 (figure A-15-8)

r = 2.8 q = 0.85 (figure 4-1)

Sut = 0.735 GPa qs = 0.92 (figure 4-2)

58.1)168.1(85.01 =−+=fK

39.1)142.1(92.01 =−+=fsK




5.367)735(5.0' ==eS


205)5.367)(814.0)(87.0)(787.0( ==eS

Applying Eq. DE-Goodman criteria gives

MPa

[ ] [ ]⎭⎬⎫

⎩⎨⎧

+= 2/122/123 )(31)(41161

mfsut

afe

TKS

MKSdn π

[ ] [ ]⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+=2/1

6

22/1

6

2

3 10735))3.124(39.1(3

10205))4.142(58.1(4

)028.0(16

xxπ

604.0)10407.010195.2(232004 66 =+= −− xx

65.1=∴n

787.0)735(51.4 265.0 == −ak

87.062.7

28 107.0

=⎟⎠⎞

⎜⎝⎛=

−

bk

0.1=== fdc kkk

814.0=ek

MPa




4.125)028.0(

)3.124)(40.1(16)028.0(

)4.142)(58.1(32'2

3

2

3max =⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ππσ


87.1=∴n DE-Gerber

Similarly, apply same technique for other failure criteria,

88.1=∴n DE-ASME Elliptic

56.1=∴n DE-Soderberg

b) Determine the Yield factor of safety :

58.44.125

574'max

===∴σ

yy

Sn





Estimating Stress ConcentrationsEstimating Stress Concentrations

• Stress analysis for shafts is highly dependent on stress concentrations.• Stress concentrations depend on size specifications, which are not known

the first time through a design process.• Standard shaft elements such as shoulders and keys have standard

proportions, making it possible to estimate stress concentrations factors before determining actual sizes.

Table 6–1First Iteration Estimates for Stress-Concentration Factor Kt and Kts





Reducing Stress Concentration at Shoulder FilletReducing Stress Concentration at Shoulder Fillet

• Bearings often require relatively sharp fillet radius at shoulder

• If such a shoulder is the location of the critical stress, some manufacturing techniques are available to reduce the stress concentration

(a) Large radius undercut into shoulder(b) Large radius relief groove into back of shoulder(c) Large radius relief groove into small diameter of shaft



6.5 – Deflection Considerations


• Shaft subject to bending produces deflection (δ or y)• Deflection analysis at even a single point of interest requires complete

geometry information for the entire shaft.• Deflection of the shaft, both linear and angular should be checked at

gears and bearings.

Table 6–2Typical Maximum Ranges for Slopes and Transverse Deflections



6.5 – Deflection Considerations –cont…


SuperpositionThe moment-area methodSingularity functionsNumerical integration

Refer in text bookchapter 4 page 147

• Deflection analysis is straightforward, but lengthy and tedious to carry out manually.

• Each point of interest requires entirely new deflection analysis.

• Consequently, shaft deflection analysis is almost always done with the assistance of software.

• Options include specialized shaft software, general beam deflection software, and finite element analysis software.

• Some popular methods to solve the integration problem for beam deflection:-



6.6 – Limits and Fits


Clearance Fits.No interference occur.

3 types of fitting

Interference Fits.An interference fit is the condition that exist when, due to the limits of the dimensions, mating parts must be pressed together.

Transition Fits.The fit can have either clearance or interference.



6.6 – Limits and Fits – cont…


Definitions applied to a cylindrical fit.Capital letters always refer to the hole; lowercase letters are used for the shaft.

D = basic size of holed = basic size of shaftδu = upper deviationδl = lower deviationδF = fundamental deviationD = tolerance grade for holed = tolerance grade for shaft

Note that these quantities are all deterministic. Thus, for the hole,Dmax = D + ∆D Dmin = D

For shafts with clearance fits c, d, f, g, and h,dmax = d + δF dmin = d + δF − ∆d

For shafts with interference fits k, n, p, s, and u,dmin = d + δF dmax = d + δF + ∆d





Table 6–3Descriptions of PreferredFits Using the BasicHole SystemSource: Preferred Metric Limitsand Fits, ANSI B4.2-1978.See also BS 4500.





Table A–11A Selection of International Tolerance Grades—Metric Series(Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters)Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BSI 4500.





Table A–12

Fundamental Deviations for Shafts—Metric Series(Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters)

Source: Preferred Metric Limits and Fits , ANSI B4.2-1978. See also BSI 4500.





Example 6-2 :Example 6-2 :

Find the shaft and hole dimensions for a loose running fit with a 34-mm basic size.

Solution 6-2 :Solution 6-2 :

From Table 6–3, the ISO symbol is 34H11/c11. From Table A–11, we find that tolerancegrade IT11 is 0.160 mm. The symbol 34H11/c11 therefore says that ∆D = ∆d = 0.160 mm.

Using Eq. (Dmax = D + ∆D) for the hole, we get

Dmax = 34 + 0.160 = 34.160 mm Dmin = D = 34.000 mm

The shaft is designated as a 34c11 shaft. From Table A–12, the fundamental deviation is δF = −0.120 mm. Using Eq. “for shaft with clearance fits”, we get the shaft dimensions

dmax = d + δF = 34 + (−0.120) = 33.880 mm

dmin = d + δF − ∆d = 34 + (−0.120) − 0.160 = 33.720 mm

Chapter 7Bearings




Week 9

2


recognize types of available bearings and know the function

identify nomenclature of ball bearing

select suitable bearing and know how to apply it

perform load and bearing life calculations analytically


Learning Outcomes


CHAPTER 7 – Bearings

3




• 7.2 - Bearing Types

• 7.3 - Bearing Mounting and Enclosure

• 7.4 - Bearing Life

• 7.5 - Bearing Load Life at Rated Reliability

• 7.6 - Relating Load, Life, and Reliability

• 7.7 - Combined Radial and Thrust Loading

• 7.8 - Lubrication

• 7.9 - Appendix



4





• Bearing acting as a support and allow rotational and sliding motion in mechanism.

• 2 types of bearing: sliding & rolling

• Sliding friction is the resistance that takes place when one object slides against another. Sliding friction can be reduced by using smooth machined surface and lubrication.

• Another way to eliminate sliding friction is the introduction of rolling elements (ball, roller), because rolling elements have smallest contact surface that produces low friction value (0.001 - 0.005)

sliding rolling

5





• It as been designed to take pure radial load, pure axial load or combination of these load. Today, all machines use this bearing such as in vehicle engine, shaft, fan, bicycle and many more.

• Among the famous bearing manufacturer are:

– SKF (Sweden)

– NTN (Jepun)

– Timken (USA)

• For more information, please refer to book and catalogue from the bearing manufacturer company.

• Bearing standard :-– ISO (International Standard Organization)

– ABMA (American Bearing Manufacturers Association)

6


7.2 – Bearing Types



Nomenclature of Ball BearingNomenclature of Ball Bearing

Outer ring

Inner ring

Width (B)

Outside diameter

(D)Bore (d)

Separator

Face

4 essential parts of a bearing:

• outer ring

• inner ring

• balls or rolling element

• seperator

7


7.2 – Bearing Types – cont…



Various types of ball bearingsVarious types of ball bearings

Deep Groove ball bearing

Angular Contact ball bearing

Self-Aligning ball bearing

Shielded ball bearing

Sealed ball bearing

Trust ball bearing

Double row

8





Various types of roller bearingsVarious types of roller bearings

Straight roller bearing

Needle roller bearing

Tapered roller bearing

Tapered roller thrust bearing

Spherical roller thrust bearing

9





Bearing Characteristics Bearing Characteristics

10


7.3 – Bearing Mounting & Enclosure



A common bearing mounting.

An alternative method of bearing mounting in which the both inner races are backed up against the shaft shoulder. Disadvantage –may destroy the bearing if shaft expand when temperature rise during operation.

Mounting for a washing machine spindle.

11


7.4 – Bearing Life



Commonly life used term is bearing life and measures are• Number of revolutions of the inner ring (outer ring stationary) until the first tangible

evidence of fatigue• Number of hours of use at a standard angular speed until the first tangible

evidence of fatigue

Fatigue criterion used by the Timken Company Laboratories• Spalling or pitting of an area of 6.45 mm2 (first evidence of fatigue failure)• useful life may extend considerably beyond this point

Rating life, (LR)• Is a term certified by the ABMA and used by most manufacturers• Is defined as the number of revolutions (or hours at constant speed) that 90

percent of a group of bearings will achieve or exceed before the failure criterion develops

• The term minimum life, L10 life and B10 life are also used as synonym

12


7.5 – Bearing Load Life at Rated Reliability



Catalog load rating, ibf or kN

Rating life, hours

Rating speed, rpm Desired radial load, ibf or kN

Desired life, hours

Desired speed, rpm

aDDDf

aRR nLFanLC /1/1

10 )60()60( =

where :-a = 3 for ball bearinga = 10/3 for roller bearing

a

RR

DDDf nL

nLFaC/1

6060

10 ⎟⎟⎠

⎞⎜⎜⎝

⎛=Solving for C10 gives

Application factor (assume af = 1 if not given)L10 life (number of revolution)

Table 7.3 Deep-Groove and Angular-Contact Ball Bearings Catalog.

Table 7.4 Cylindrical Roller Bearings Catalog.

See

Table 7.2 Equivalent Radial Load Factor.

Table 7.1 Data for Bearing Manufacturer.

13


7.5 – Bearing Load Life at Rated Reliability – cont…




Consider SKF, which rates its bearings for 1 million revolutions. Timken for example, uses 90(106) revolutions. If you desire a life of 5000 hour at 1725 rev/min with a load of 2 kN with a reliability of 90 percent, for which catalog rating would you search in a SKF catalog?


05.1610

60)1725(500026060 3

1

610

/1

=⎥⎦⎤

⎢⎣⎡=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

a

RR

DDDf nL

nLFaC kN

14


7.6 – Relating Load, Life and Reliability



Constant reliability contours. Point A represents the catalog rating C10

at X = L / L10 = 1. Point B is on the target reliability design line RD, with load of C10. Point D is a point on the desired reliability contour exhibiting the design life XD = LD / L10 at the design load FD.

• It is always the desired parameters (load, speed and reliability) is not the manufacturer’s test parameter or catalog entry.

• To used the catalog datato comply with the desired parameters, one needs to determine equivalent catalog load rating by using next formula.

15


7.6 – Relating Load, Life and Reliability – cont…



a

bD

DDf Rxx

xFaC/1

/100

10 ))/1)(ln(( ⎥⎦

⎤⎢⎣

⎡−+

=θ

RR

DD

nLnL

LL

6060

10

=

⎪⎭

⎪⎬

⎫

b

xθ

0

where:-af = Application factor (assume af = 1 if not given)FD = Desired radial load, ibf / kN

xD = Dimensionless desired life =

RD = Desired reliabilitya = 3 (for ball bearing)a = 10/3 (for roller bearing)

= Weibull parameters (refer table 6-1).Table 7.3 Deep-Groove and Angular-Contact Ball

Bearings Catalog.


See



16

where:-af = Application factor (assume af = 1 if not given)

FD = Desired radial load, ibf / kN

xD = Dimensionless desired life =

RD = Desired reliabilitya = 3 (for ball bearing)a = 10/3 (for roller bearing)





⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−

−⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

baDf

D

x

xC

Fax

R0

010exp

θ

Reliability for bearing is given by the equation:Reliability for bearing is given by the equation:

RR

DD

nLnL

LL

6060

10

=⎪⎭

⎪⎬

⎫

b

xθ

0

= Weibull parameters (refer table 6-1).

Table 7.3 Deep-Groove and Angular-Contact Ball Bearings Catalog.


See



Overall Reliability• Shaft generally have 2 bearings. Often

these bearings are different.• If the individual reliability for each

individual bearing is RA and RB, therefore overall bearing reliability, R is

BARRR =

17



54010

)300)(30000(606060

610

====RR

DDD nL

nLLLx




The design load on a ball bearing is 1840N and an application factor of 1.2 is appropriate. The speed of the shaft s to be 300 rev/min, the life to be 30 kh with a reliability of 0.99. What is the C10 catalog entry to be sought (or exceeded) when searching for a deep-groove bearing in a SKF catalog?


Thus, the design life is 540 times the L10 life. For a ball bearing, a = 3, thus from previous equation.

7.29)99.0/1)(ln439.4(02.0

540)84.1)(2.1(3/1

483.1/110 =⎥⎦

⎤⎢⎣

⎡+

=C kN

18






Based on example 7-2, choose the suitable dimension for Single-Row 02-Series Deep-Groove Ball Bearings from table 7-3 and calculate the new reliability.


• Calculate result, C10 = 29.7 kN

• The closest, C10 on the Single-Row 02-Series Deep-Groove Ball Bearings catalog that suitable to

hold 29.77 kN load is 30.7 kN.

• Therefore, the selected bearing size is:-

OD = 80 mm , Bore = 40 mm & Width = 18 mm

9914.0439.4

02.07.30

)84.1)(2.1(540exp

483.13

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧−⎟

⎠⎞

⎜⎝⎛

−=R• And the new reliability is

19


7.7 – Combined Radial and Thrust Loading



• Ball bearing is capable to resist radial load, thrust load or combine of these loading.

• However, the capability of ball bearings to withstand the thrust load had certain limits and not good enough such as thrust bearing or tapered roller bearing.

• While the straight roller bearing is just capable to withstand large radial load compare than ball bearing.

• ABMA has come out guidelines to determine equivalent radial load for ball bearing that acts with thrust load as follow:

20


7.7 – Combined Radial and Thrust Loading – cont…



• X and Y value need to be select from table 7-2

• Table 7-2 list X1, Y1 and X2, Y2as a e function based on the ratio of the thrust component to the bearing static load catalog rating Fa / C0.

airie FYVFXF +=

whereFa - axial thrustFr - radial loadXi - radial load factor (table 7-2)Yi - thrust load factor (table 7-2) V - rotation factor

- Inner ring rotates

- Outer ring rotates

- Self aligning bearing⎪⎩

⎪⎨

⎧

===

=0.12.10.1

V

Equivalent radial loadEquivalent radial load

21






An SKF 6210 angular-contact ball bearing has an axial load Fa of 1780 N and a radial load Fr of 2225 N applied with the outer ring stationary. The basic static load rating C0 is 19,800 N and the basic load rating C10 is 35,150 N. Estimate the L10 life at a speed of 720 rev/min.


090.0198001780

0

==CFaV = 1 and .

Interpolate for e in Table 7-2

Fa / C0 e

0.084 0.28

0.090 e from which e = 0.285

0.110 0.30

285.08.0)2225(1

1780)(

>==r

a

VFF.

Thus, interpolate for Y2 :

Fa / C0 Y2

0.084 1.55

0.090 Y2 from which Y2 = 1.527

0.110 1.45

22






N

With LD = L10 and FD = Fe, solving for L10 gives:

3964)1780(527.12225)1(56.022 =+=+= are FYVFXF

h5.16139396435150

)720(6010

6060 36

1010 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

a

rD

RR

FC

nnLL

** See more example on bearing analysis – example 7-5 & 7-6

23


7.8 – Lubrication

Purpose of Lubrication• To provide a film of lubricant between the sliding and rolling surfaces• To help distribute and dissipate heat• To prevent corrosion of the bearing surfaces• To protect the parts from the entrance of foreign matter



Rules of selecting grease and oil as a lubricant

Use Grease When Use Oil When1. The temperature is not over 200OF. 1. Speed are high.

2. The speed is low. 2. Temperatures are high.

3. Unusual protection is required from the entrance of foreign matter.

3. Oil tight seals are readily employed.

4. Simple bearing enclosures are desired. 4. Bearing type is not suitable for grease lubrication.

5. Operation for long periods without attention is desired

5. The bearing is lubricated from a central supply which is also used for other machine parts.

24


7.9 – Appendix



Manufacturer Rating life, revolution

Rating life, Weibull Parameter

X0 θ b1. Timken 90(10)6 0 4.48 1.5

2. SKF and most bearing manufacturer

(10)6 0.02 4.459 1.483

Table 7-1 : Data for Bearing Manufacturer

25


7.9 – Appendix – cont…



* use 0.014 if Fa / C0 < 0.014

Table 7-2 : Equivalent Radial Load Factors for Ball Bearings

26





Shaft and housing shoulder diameter, dS and dH, should be adequate to ensure good bearing support.

Table 7-3 : Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearings.

dSdH

27





Table 7-4 : Dimensions and Basic Load Ratings for Cylindrical Roller Bearings.

28





Table 7-5 : Bearing-Life Recommendations for Various Classes of Machinery

Types of Application Life, k hour

Instruments and apparatus for infrequent use Up to 0.5

Aircraft engines 0.5 – 2

Machines for short or intermittent operation where service interruption is of minor importance

4 – 8

Machine for intermittent service where reliable operation is of great importance 8 – 14

Machines for 8-h service that are not always fully utilized 14 – 20

Machines for 8-h service that are fully utilized 20 – 30

Machines for continuous 24-h service 50 – 60

Machines for continuous 24-h service where reliability is of extreme importance 100 - 200

29





Table 7-6 : Load Application Factor

Types of Application Load Factor

Precision gearing 1.0 – 1.1

Commercial gearing 1.1 – 1.3

Application with poor bearing seals 1.2

Machinery with no impact 1.0 – 1.2

Machinery with light impact 1.2 – 1.5

Machinery with moderate impact 1.5 – 3.0


1

More Examples on Bearing Analysis

Example 7-5 The second shaft on a parallel-shaft 18.7 kW foundry crane speed reducer contains a helical gear with a pitch diameter of 206 mm. Helical gear transmit components of force in the tangential, radial and axial direction. The components of the gear force transmitted to the second shaft are shown in figure below, at point A. The bearing reactions at C and D, assuming simple-support, are also shown. A ball bearing is to be selected for location C to accept the thrust, and a cylindrical roller bearing is to be utilized at location D. The life goal of the speed reducer is 10 kh, with a reliability factor for the ensemble of all four bearings (both shaft) to equal or exceed 0.96. The application factor is to be 1.2. i) Select the roller bearing for location D. ii) Select the ball bearing (angular contact) for location C, assuming the inner ring rotates. Solution The torque transmitted is T =2.648 (0.103) = 0.2727 kNm. The speed at the rated power is

rpm The radial load at D is (1324 2 + 474 2 )0.5 = 1406 N,

602 nTTP πω ==

6552727.0

)7.18(55.955.9260

====∴T

PTPn

π


2

( ) 1.5399.0

1ln)439.4(02.0

393)3650(2.131

483.1/110 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+=C

and the radial load at C is (1587 2 + 1324 2 )0.5 = 2067 N. The individual bearing reliabilities, if equal, must be at least 99.098985.096.04 == . The dimensionless design life for both bearings is :

39310

655)10000(606060

610

====RR

DDD nL

nLLLx

(a) Selection of roller bearing at D. (Info obtained af = 1.2 and a = 10/3)

( ) 0.16

99.01ln)439.4(02.0

393)1406(2.11ln)(

103

483.1/1

1

/1

00

10 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞⎜

⎝⎛−+

=

a

b

D

DDf

Rxx

xFaC

θkN

The absence of a thrust component makes the selection procedure simple. Choose a 02-25 mm series, or a

03-25 mm series cylindrical roller bearing from Table 7-4.

(b) Selection of angular contact ball bearing at C. (Info obtained af = 1.2 and a = 3). The ball bearing at C

involves a thrust component. This selection procedure requires an iterative procedure. Assuming Fa / (V Fr) > e.

i) Choose Y2 from table 7-2 ii) Find C10 iii) Tentatively identify a suitable bearing from Table 7-3, note C0 iv) Using Fa / C0 enter Table 7-2 to obtain a new value of Y2. v) Find C10 vi) If the same bearing obtained, stop. vii) If not, take next bearing and go to step iv. As a first approximation, take the middle entry from Table 7-2: X2 = 0.56 Y2 = 1.63 With V = 1 and kN

kN From Table 7-3, angular contact bearing 02-60 mm has C10 = 55.9 kN. C0 is 35.5 kN. Step 4 becomes, with Fa in kN,

0431.05.35

531.1

0

==CFa

65.3)1531(63.1)2067)(0.1(56.022 =+=+= are FYVFXF


3

Which makes e from Table 7-2 approximately 0.24. Now , which is greater than 0.24, so we find Y2 by interpolation: Therefore kN The prior calculation for C10 changes only in Fe, so kN From Table 7-3, angular contact bearing 02-65 mm has C10 = 63.7 kN and C0 is 41.5 kN. Again, Making e approximately 0.23. Now from before , which is greater than 0.23, so we find Y2 again by interpolation: Therefore kN The prior calculation for C10 changes only in Fe, so kN From Table 7-3, angular contact bearing 02-65 mm is still selected, so the iteration is complete.

0.26

e

0.24

e

0.056

from which e ≈ 0.24 0.043

0.042

Fa / C0

74.0)2067(0.1

1531==

r

a

VFF

97.3)1531(84.1)2067)(0.1(56.022 =+=+= are FYVFXF

8.57)1.53(65.397.3

10 ==C

0369.05.41

531.1

0

==CFa

1.85

Y2

1.99

Y2

0.042

from which Y2 = 1.90 0.0369

0.028

Fa / C0

74.0)2067(0.1

1531==

r

a

VFF

1.71

Y2

1.85

Y2

0.056

from which Y2 = 1.84 0.043

0.042

Fa / C0

0.24

e

0.22

e

0.042

from which e ≈ 0.23 0.0369

0.028

Fa / C0

2.59)1.53(65.307.4

10 ==C

065.4)1531(90.1)2067)(0.1(56.022 =+=+= are FYVFXF


4

Example 7-6 The figure shown is a geared countershaft with an overhanging pinion at C. Select an angular contact ball bearing from table 7-3 for mounting at O and a 02-series cylindrical roller bearing for mounting at B. The force on gear A is FA = 2700 N, and the shaft is to run at a speed of 480 rev/min. Specify the bearings required, using an application factor of 1.4, a desired life of 50,000 hour, and a combined reliability goal of 0.90. All dimension stated are in millimeter. Solution FBD Solution of the static problem gives force of bearings against the shaft at O as RO = -1740j + 2100k N, and at B as RB = 1420j – 7270k. N.

60LR nR = 106 LD = 50 khours nD = 480 rpm FD at O = [(-1740)2 + (2100)2]1/2 = 2727.2 N FD at B = [(1420)2 + (-7270)2]1/2 = 7407.4 N

500

400

250

20o

20o

2

O

BGear 3 600 D

Gear 4 250 D

z

x

y

FA

FC

Oz

Bz

Oy

By

FC sin 20o

FC cos 20o

FA sin 20o

FA cos 20o

TA = (FA cos 20o)(0.3)

TC = (FC cos 20o) (0.125)


5

For a combined reliability goal of 0.90, use 95.090.0 = for the individual bearings. At O (angular contact ball bearing)

144010

)480)(50000(606060

6 ===RR

DDD nL

nLx

6.50)95.0/1)(ln439.4(02.0

1440)2.2727)(4.1(3/1

483.1/110 =⎥⎦

⎤⎢⎣

⎡+

=∴C kN

∴ Suitable size; bore = 60 mm, OD = 110 mm, width = 22 mm At B (cylindrical roller bearing)

144010

)480)(50000(606060

6 ===RR

DDD nL

nLx

1.106)95.0/1)(ln439.4(02.0

1440)4.7407)(4.1(10/3

483.1/110 =⎥⎦

⎤⎢⎣

⎡+

=∴C kN

∴ Suitable size; bore = 85 mm, OD = 150 mm, width = 28 mm

Chapter 8Nonpermanent Joints

Prepared by: Mohd Azwir bin Azlan

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


Week 10, 11 & 12

2


recognize types of available bolt and screw

know bolt / screw thread standards and definition

perform load and stress calculations that acting on bolt / screw

select suitable application of bolt / screw


Learning Outcomes


CHAPTER 8 – Nonpermanent Joints

3




• 8.2 - Thread Standards and Definitions

• 8.3 - The Mechanics of Power Screws

• 8.4 - Threaded Fasteners

• 8.5 - Joints : Fastener Stiffness

• 8.6 - Joints : Member Stiffness

• 8.7 - Bolt Strength

• 8.8 - Tension Joints : The External Load

• 8.9 - Relating Bolt Torque to Bolt Tension

• 8.10 - Statically Loaded Tension Joint with Preload

• 8.11 - Gasketed Joints

• 8.12 - Fatigue Loading of Tension Joints

• 8.13 - Bolted and Riveted Joints Loaded in Shear



4





• The fundamental operation in manufacture is the creation of shape - this includes

assembly, where a number of components are fastened or joined together either

permanently by welding (Chapter 9) or detachably (nonpermanent) by screws, nuts and

bolts and so on.

• Since there is such a variety of shapes in engineering to be assembled, it is hardly

surprising that there is more variety in demountable fasteners than in any other machine

element.

• Fasteners based upon screw threads are the most common, so it is important that their

performance is understood, and the limitations of the fastened assemblies are

appreciated.

• Bolts, screws and nuts are common fastener used to join between one part to another.

This type of joining is a temporary connection in which it is easy to open again.

5





• Although the bolts and screws have very similar functions, but there are differences in the

application.

• Bolt used where it thread is designed to get through and past the hole in the part to be

connected and then tied with a nut at the end of the bolt.

• While the screws are used where it thread is designed to bind the connection with the

internal threaded screw. Figure 8-1 shows the difference in application of bolts and screws.

Figure 8-1: Several types of fastener

6





• There are many types of bolts are available in the market. Figure 8-2 below shows the

types of bolts that are commonly used.

• Then, screws also can be further categorized into several types according to it use i.e.,

machine screws, sheet metal & lag screw and set screw. Figure 8-3, 8-4 and 8-5 shows the

types of screws have been categorized according to it application.

Figure 8-2: Several types of bolt (a) Carriage, (b) Elevator, (c) Countersunk, (d) Plow, (e) Track, (f) Stud, (g) Stove, (h) Stove.

(a) (b) (c) (d) (e) (f) (g) (h)

Figure 8-3: Type of machine screw: (a) Flat, (b) Button, (c) Fillister, (d) Flat Fillister, (e) Round, (f) Socket.(a) (b) (c) (d) (e) (f)

7





Figure 8-4: Several types of metal & lag screw; (a) Round head, (b) Elliptical head, (c) Countersunk head, (d) Phillips head, (e) Lag screw.

(a) (b) (c) (d) (e)

(a) (b) (c) (d) (e) (f)

Figure 8-5: Several types of set screw; (a) Headless flat point, (b) Square head cup point, (c) Hex socket head, cone point, (d) Fluted socket head, dual point, (e) Full-dog point, (f) Half-dog point.

8


8.2 – Thread Standards and Definitions


• Pitch – distance between adjacent threadforms measured parallel to the thread axis.

• Major diameter (d) – largest diameter of screw thread.

• Minor or root diameter – smallest diameter of screw thread.

• Lead – distance parallel to the screw axis when the nut moves one turn. A double-threaded screw has a lead equal a twice the pitch (figure 8-7b), a triple-threaded screw has a lead equal to 3 times pitch (figure 8-7c), and so on.


Figure 8-6: Terminology of screw threads. Sharp veethreads shown for clarity; the crests and roots are actually flattened or rounded during the forming operation.

Figure 8-7: Single, double & triple threaded screw

9


8.2 – Thread Standards and Definitions – cont…


• For threaded that have been produced according to metric standard is known as “M series”.

• Metric bolt designation determined by:M12 x 1.75

pitch

Nominal diameter (mm)


• All threads are made according to the right-hand rule unless otherwise noted.

• All threads size on bolt and screw were followed according to the inch series or metric series standard.

• Inch series referred to the American standard where it has been approved by Great Britain, Canada and United States. Threaded that have been produced according to this standard is also known as “Unified threads”. Two major series are: UN and UNR

Figure 8-8: Basic profile for metric M and MJ threads.

10





1)1) Metric thread Metric thread ((Table 8-1) Major diameter (mm), 2α = 60° Standard thread is RH (Right Handed)Specifications: e.g.: M12 x 1.75 or MJ12 x 1.75M = Basic Metric, J = greater root radius for fatigue applications; 12 = nominal major diameter (mm); 1.75 = pitch (mm)

2)2) The American National (Unified) thread The American National (Unified) thread (Table 8-2) Thread standards is used mainly in the US and GB. Series designation used UN or UNRUN (regular thread), UNR (greater root radius for fatigue applications)Specifications: e.g.: 5/8”-18 UN, UNC, UNF, UNR, UNRC, UNRF5/8” = nominal major diameter (inch) ; 18 = Number of threads per inch (N)UN = Unified, F = fine, C = Coarse, R = Round Root

3)3) Square Square and The ACME ThreadsThe ACME ThreadsSquare and Acme threads are used when the threads are intended to transmit power.Used mainly in power screws Table 8-3 gives preferred pitches for ACME threads.

Figure 8-9: Regular or Flat Thread

Figure 8-10: Rounded Thread

11





Coarse series UNC• General assembly• Frequent disassembly• Not good for vibrations• The “normal” thread to specify

Fine series UNF• Good for vibrations• Good for adjustments• Automotive and aircraft

Extra Fine series UNEF• Good for shock and large vibrations• High grade alloy• Instrumentation• Aircraft

12





Nominal diameter

¼-20 x ¾ in UNC-2 Grade 5 Hex head bolt

Threads per inch

length

Thread series

Class fit

Material grade

Head type

M12 x 1.75 ISO 4.8 Hex head bolt

Metric

Nominal diameter

Pitch

Material class

Head type

13





Nominal Major

Diameter d (mm)

Coarse – Pitch Series Fine – Pitch Series

Pitch p (mm) Tensile Stress Area At (mm2)

Minor Diameter Area Ar (mm2) Pitch p (mm) Tensile Stress

Area At (mm2)Minor Diameter Area Ar (mm2)

1.6 0.35 1.27 1.072 0.40 2.07 1.79

2.5 0.45 3.39 2.983 0.5 5.03 4.47

3.5 0.6 6.78 6.004 0.7 8.78 7.755 0.8 14.2 12.76 1 20.1 17.98 1.25 36.6 32.8 1 39.2 36.0

10 1.5 58.0 52.3 1.25 61.2 56.312 1.75 84.3 76.3 1.25 92.1 86.014 2 115 104 1.5 125 11616 2 157 144 1.5 167 15720 2.5 245 225 1.5 272 25924 3 353 324 2 384 36530 3.5 561 519 2 621 59636 4 817 759 2 915 88442 4.5 1120 1050 2 1260 123048 5 1470 1380 2 1670 163056 5.5 2030 1910 2 2300 225064 6 2680 2520 2 3030 298072 6 3460 3280 2 3860 380080 6 4340 4140 1.5 4850 480090 6 5590 5360 2 6100 6020100 6 6990 6740 2 7560 7470110 2 9180 9080

Table 8-1

Diameter and Areas of Coarse-Pitch and Fine-Pitch Metric Threads.

14





Size Designation

Nominal Major

Diameter in

Coarse Series – UNC Fine Series - UNF

Threads per Inch N

Tensile Stress Area At in2

Minor Diamater

Area Ar in2

Threads per Inch N

Tensile Stress Area At in2

Minor Diameter

Area Ar in2

0 0.0600 80 0.001 80 0.001 511 0.0730 64 0.002 63 0.002 18 72 0.002 78 0.002 372 0.0860 56 0.003 70 0.003 10 64 0.003 94 0.003 393 0.0990 48 0.004 87 0.004 06 56 0.005 23 0.004 514 0.1120 40 0.006 04 0.004 96 48 0.006 61 0.005 665 0.1250 40 0.007 96 0.006 72 44 0.008 80 0.007 166 0.1380 32 0.009 09 0.007 45 40 0.010 15 0.008 748 0.1640 32 0.014 0 0.011 96 36 0.014 74 0.012 85

10 0.1900 24 0.017 5 0.014 50 32 0.020 0 0.017 512 0.2160 24 0.024 2 0.020 6 28 0.025 8 0.022 6¼ 0.2500 20 0.031 8 0.026 9 28 0.036 4 0.032 6

0.3125 18 0.052 4 0.045 4 24 0.058 0 0.052 4⅜ 0.3750 16 0.077 5 0.067 8 24 0.087 8 0.080 9

0.4375 14 0.106 3 0.093 3 20 0.118 7 0.109 0½ 0.5000 13 0.141 9 0.125 7 20 0.159 9 0.148 6

0.5625 12 0.182 0.162 18 0.203 0.189⅝ 0.6250 11 0.226 0.202 18 0.256 0.240¾ 0.7500 10 0.334 0.302 16 0.373 0.351⅞ 0.8750 9 0.462 0.419 14 0.509 0.4801 1.0000 8 0.606 0.551 12 0.663 0.625

1¼ 1.2500 7 0.969 0.890 12 1.073 1.0241½ 1.5000 6 1.405 1.294 12 1.581 1.521

16516

716

9

Table 8-2

Diameter and Areas of Unified Screw Threads UNC and UNF*.

15




CHAPTER 8 – Nonpermanent Joints16516

716

9

• The tensile stress area, At , is the area of an unthreaded rod with the same tensile strength as a threaded rod.

• It is the effective area of a threaded rod to be used for stress calculations.

• The diameter of this unthreaded rod is the average of the pitch diameter and the minor diameter of the threaded rod.

Tensile Stress Area

16




CHAPTER 8 – Nonpermanent Joints16516

716

9

Figure 8-11

(a) Square thread

(b) Acme thread

d, in 1\4 5\16 3\8 1\2 5\8 3\4 7\8 1 11\4 13\4 2 21\2 3p, in 1\16 1\14 1\12 1\10 1\8 1\6 1\6 1\5 1\5 1\4 1\4 1\3 1\2

Table 8-3: Preferred Pitches for Acme Threads

17


8.3 – The Mechanics of Power Screw



• A power screw is a device in machinery to change angular motion into linear motion and usually to transmit power.

• Thread usually of square or ACME profile

• More specifically, power screw are used:

to lift weight – jack for cars

to exert large forces – home compactor or a press

Figure 8-12a: The Joyce worm-gear screw jack.

Figure 8-12b: Some applications of power screw

18


8.3 – The Mechanics of Power Screw – cont…



Figure 8-13: Portion of Power Screw

Figure 8-14: Force diagrams; (a) lifting the load; (b) lowering the load

• Imagine a single thread of the screw is unrolled

• Then one edge of the thread will form of a right triangle with the base is the circumference of the mean thread diameter circle with a high is a lead.

• Angle λ is the lead angle of the thread.

19





• To raise a load, a force PR acts to the right, and to lower the load, PL acts to the left.

• The friction force is the product of the coefficient of friction f with the normal force N, and act oppose the motion.

• The system is equilibrium under the action of these forces and hence, for raising the load:

∑∑

=+−−=

=−−=

0cossin

0cossin

λλ

λλ

NfNFF

fNNPF

y

Rx

• In a similar manner, for lowering the load;

∑∑

=++−=

=+−−=

0cossin

0cossin

λλ

λλ

NfNFF

fNNPF

y

Lx

20





⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=fldlfdFd

Tm

mmL π

π2

• Since the normal force N, is not interested, eliminate it from the equation.

• Then, divide the numerator and the denominator of these equations by cos λ and usethe relation tan λ = l / πdm.

• Finally, noting that the torque is the product of the force P and the mean radius dm/2,

where, F = forcedm = mean screw diameterl = lead distancef = coefficient of friction

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=

fldfdlFd

Tm

mmR π

π2

Torque required for raising the loadto overcome thread friction and to raise the loadRT

Torque required for lowering the loadto overcome part of the thread friction in lowering the loadLT

21





If the lead is large or the friction is low, the load will lower itself by causing the screw to

spin without any external effort. In such cases the torque will be negative or zero.

When a positive torque is obtained from this equation, the screw is said to be self

locking

LT

Condition for Self Locking: mfd lπ >

Dividing both sides of the above inequality by and recognizing that ,

we get

The critical coefficient of friction for the lead concerned,

If f = fcr the nut is on the point of moving down the thread without any torque applied.

mdπ tanml dπ λ=

tanf λ>

22





If f > fcr then the thread is self-locking in that the nut cannot undo by itself, it needs to

be unscrewed by a definite negative torque; Clearly self-locking behavior is essential

for threaded fasteners.

Car lifting jacks would not be of much use if the load fell as soon as the operating

handle was released.

If f < fcr then the thread is overhauling in that the nut will unscrew by itself under the

action of the load unless prevented by a positive tightening torque.

If we let , we obtain which is the torque,

required to raise the load.

0 2FlTπ

=

The efficiency is thereforeRR T

FlTTe

π20 ==

0f =

23





Figure 8-15: Normal thread force is increased because of angle α

• For ACME or other threads, the normal thread load is inclined to the axis because of the thread angle 2α and the lead angle λ.

• Since lead angle is small, this inclination can be neglected.

• Just consider angle α, which increase the frictional force.

• For raising the load:

• An additional component of torque is often needed to account for the friction between a collar and the load.

• If fc is the coefficient of collar friction, assuming the load is concentrated at the mean collar diameter dc

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=

απαπ

secsec

2 fldfdlFdT

m

mmR

Figure 8-16: Thrust collar has frictional diameter, dc

2cc

CdFfT =

3. Thread bearing stress

Figure 8-17: Geometry of square thread useful in finding bending and transverse shear stresses at the thread root

2/ 2B

m t m t

F Fπd n p πd n p

σ = − = −

where nt is the number of engaged threads.

24





The following stresses should be checked on both the nut and the screw:

1. Shearing stress in screw body.

2. Axial stress in screw body 2

4

r

F FA πd

σ = =

3

16

r

Tπd

τ =

25





4. Bending stress at root of thread.

5. Transverse shear stress at center of root of thread (due to bending)

pndF

pndpFp

IMc

trtrb ππ

σ 6

12)2/)((

)4/(4 3 =⎟

⎠⎞

⎜⎝⎛==

pndF

pndF

AV

trtr ππτ 3

)2/(23

23

===

26





Stresses in Threads of Power Screws

Transform to von Mises stress

Consider stress element at the top of the root “plane”

27





Thread Deformation in Screw-Nut Combination• Power screw thread is in compression, causing elastic shortening of screw

thread pitch.

• Engaging nut is in tension, causing elastic lengthening of the nut thread pitch.

• Consequently, the engaged threads cannot share the load equally. Some

experiment shows that:

• To find the largest stress in the first thread of a screw-nut combination, use

0.38F in place of F, and set nt = 1.

the first engaged thread carries 0.38 of the loadthe second engaged thread carries 0.25 of the loadthe third engaged thread carries 0.18 of the loadthe seventh engaged thread is free of load

28





Screw Material

Nut Material Safe pb, MPa Notes

Steel Bronze 17.2 – 24.1 Low speed

Steel Bronze 11.0 – 17.2 ≤ 50 mm/s

Cast iron 6.9 – 17.2 ≤ 40 mm/s

Steel Bronze 5.5 – 9.7 100-200 mm/s

Cast iron 4.1 – 6.9 100-200 mm/s

Steel Bronze 1.0 – 1.7 ≥ 250 mm/s

Table 8-4: Screw bearing Pressure pb.

Screw Material

Nut Material

Steel Bronze Brass Cast Iron

Steel, dry 0.15 - 0.25 0.15 - 0.23 0.15 - 0.19 0.15 - 0.25

Steel, machine oil 0.11 - 0.17 0.10 - 0.16 0.10 - 0.15 0.11 - 0.17

Bronze 0.08 - 0.12 0.04 - 0.06 - 0.06 - 0.09

Table 8-5: Coefficients of friction f for Threaded Pairs.

29





Coefficients of friction around 0.1 to 0.2 may be expected for common materials under conditions of ordinary service and lubrication.

Combination Running Starting

Soft steel on cast iron 0.12 0.17

Hard steel on cast iron 0.09 0.15

Soft steel on bronze 0.08 0.10

Hard steel on bronze 0.06 0.08

Table 8-6: Thrust Collar friction coefficient, fc.

30






A square thread power screw has a major diameter of 32 mm and a pitch of 4 mm with double threads. The load of 6.4 kN used in this application. If f = fc =0.08, dc=40mm, determine;

a) thread depth, thread width, pitch diameter, minor diameter and lead.

b) the torque required to raise and lower the load.

c) the efficiency during lifting the load.

d) the body stresses, torsional and compressive.

e) the bearing stress.

f) the thread bending stress at the root of the thread.

g) the Von Misses stress at the root of the thread.

h) the maximum shear stress at the root of the thread.

31






a) thread depth, thread width, pitch diameter, minor diameter and lead.

mm 302/4322/ diameter;pitch =−=−= pddm

mm 28432 diameter;minor =−=−= pddr

mm 2242 width threaddepth thread / p/ ====

mm 8)4(2 thread,doublefor lead === npl

b) the torque required to raise and lower the load.

240)08.0)(4.6(

)8)(08.0()30()30)(08.0(8

2)30(4.6

22+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+=+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+=

ππ

ππ cc

m

mmR

dFffld

fdlFdT

mNTR .18.26=∴

32





2)40)(08.0(4.6

)8(08.0)30(830)08.0(

2)30(4.6

22+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=ππ

ππ cc

m

mmL

dFffldlfdFdT

Nm 77.9=∴ LT

c) the efficiency during lifting the load.

311.0)18.26(2

)8(4.62

===ππ RT

Fle

d) the body stresses, torsional and compressive.

( )( )

MPa 07.6028.0

18.261616

32

2334 =====

πππτ

rr dT

d

dT

JTrThe body shear stress τ due to

torsional moment TR

The axial nominal stress σ is( )

MPa 39.104/028.0

64004/

640022 −=

−=

−==

ππσ

rdAF

33





e) the bearing stress, σB is, with one thread carrying 0.38F.

( ) MPa 9.124)1(30

)6400)(38.0(2)1(

)38.0(2−=−=−=

ππσ

pdF

mB

f) the thread-root bending stress, σb with one thread carrying 0.38F is

( ) MPa 5.414)1(28

)6400)(38.0(6)1(

)38.0(6===

ππσ

pdF

rb

g) the transverse shear at the extreme of the root cross section due to bending is zero. However, there is a circumferential shear stress at the extreme of the root cross section of the thread as shown in part (d) of 6.07 MPa. The three-dimensional stresses are:

00

MPa 07.6MPa 39.10

0MPa 5.41

z ==

=−=

==

zx

yzy

xyx

τσ

τσ

τσ

x

y

z

34





For the von Misses stress,

[ ]MPa 7.48

)07.6(6)5.4139.10())39.10(0()05.41(2

1'2/12222

=

+−−+−−+−=σ

h) the maximum shear stress

MPa 3.272

)18.13(5.412

31max =

−−=⎟

⎠⎞

⎜⎝⎛ −

=∴σστ

Note that, there are no shear stresses on the x face. This means that σx is a principal stress. The remaining stress can be transformed by using the plane stress equation.

MPa 13.18- & MPa 79.207.62

39.102

39.10

22

22

22

3,2

=+⎟⎠⎞

⎜⎝⎛ −

±−

=

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±⎟⎟

⎠

⎞⎜⎜⎝

⎛ += yz

zyzy τσσσσ

σ

35


8.4 – Threaded Fastener



• Hexagon-head bolts are one of the most common for engineering applications• Standard dimensions are included in Table A–29• W is usually about 1.5 times nominal diameter• Bolt length L is measured from below the head

AA-- BOLTSBOLTS:

Purpose:Purpose:to clamp two or more members together.

Parts:Parts:(1) Head (Square or Hexagonal)(2) Washer (dw=1.5d)(3) Threaded part (4) Unthreaded part

Hexagon-Head Bolt

Figure 8-18: Hexagon-head bolt

36


8.4 – Threaded Fastener – cont…



Metric

English

Threaded Lengths

Where d is the nominal diameter

BB-- NUTSNUTS: Same material as that of a screwTable ATable A--31 gives dimensions of Hexagonal nuts31 gives dimensions of Hexagonal nuts

Good Practice:Good Practice:1.First three threads of nut carry majority of load2.Localized plastic strain in the first thread is likely, so nuts should not be re-used in critical

applications.3.Tightening should be done such that 1 or 2 threads come out of the nut;4.Washers should always be used under bolt head to prevent burr stress concentration.

37





End view Washer-faced, regular

Chamfered both sides, regular

Washer-faced, jam nut

Chamfered both sides, jam nut

Figure 8-19: Types of hexagon-head nut

38





Head Type of Bolts• Hexagon head bolt– Usually uses nut– Heavy duty

• Hexagon head cap screw– Thinner head– Often used as screw (in threaded

hole, without nut)

• Socket head cap screw– Usually more precision

applications– Access from the top

• Machine screws– Usually smaller sizes– Slot or philips head common– Threaded all the way

39





Machine Screws

40


8.5 – Joints : Fastener Stiffness



When a connection is desired that can be disassembled without destructive methods and that is strong enough to resist external tensile loads, moment loads, and shear loads, or a combination of these, then the simple bolted joint using hardened steel washers is a good solution.

Twisting the nut stretches the bolt to produce the clamping force. This clamping force is called the pretention or bolt preload.

This force exists in the connection after the nut has been properly tightened.

Grip length l, includes everything being compressed by bolt preload, including washers.

Washer under head prevents burrs at the hole from gouging into the fillet under the bolt head. Figure 8-20:

A bolted connection loaded in tension by the force P. Note the use of two washers. Note how the threads extend into the body of the connection. This is usual and is desired. l is the grip of the connection.

41


8.5 – Joints : Fastener Stiffness



• Hex-head cap screw in tapped hole used to fasten cylinder head to cylinder body.

• Only part of the threaded length of the bolt contributes to the effective grip l.

Figure 8-21: Section of cylindrical pressure vessel. Hexagon-head caps crews are used to fasten the cylinder head to the body. Note the use of an O-ring seal. L’G is the effective length of the connection.

Figure 8-21 shows another tension-loaded connection. This joint uses cap screws threaded into one of the members.

An alternative approach to this problem (of not using a nut) would be to use studs.

A stud is a rod threaded on both ends. The stud is screwed into the lower member first; then the top member is positioned and fastened down with hardened washers and nuts

Studs are regarded as permanent, and so the joint can be disassembled merely by removing the nut and washer.

42


8.5 – Joints : Fastener Stiffness – cont…



Spring Rate : The ratio between the force applied to the member and the deflection produced by that force.

The grip l of a connection is the total thickness of the clamped material.

Total distance between the underside of the nut to the bearing face of the bolt head; includes washer, gasket thickness etc. The grip l here is the sum of

the thicknesses of both members and both washers.

43





Procedure to find bolt stiffness

• Given fastener diameter d and pitch p in mm or number of threads per inch• Washer thickness: t from Table A-32 or A-33• Nut thickness [Fig. (a) only]: H from table A-31

(a) (b)

44





Procedure to find bolt stiffness

(a) (b)

8-1 8-2

45





t

tt

d

dd

td

tdb

tdb

lEAk

lEAk

kkkkk

kkk

==

+=

+=

;

111

In joint under tension the members are under compressionand the bolt under tension: kb = equivalent spring constant of bolt composed of threaded (kt) and unthreaded (kd) parts acting as springs in series

For short bolts kb= kt

km

dttd

tdb lAlA

EAAk+

=

46


8.6 – Joints : Member Stiffness



• Stress distribution spreads from face of bolt head and nut• Model as a cone with top cut off• Called a frustum

Model compressed members as if they are frusta spreading from the bolt head and nut to the midpoint of the grip

Joint pressure distribution theoretical models

Ito used ultrasonic techniques to determine pressure distribution at the member interface. Results show that pressure stays high out to about 1.5 bolt radii.

Ito suggested the use of Rotscher’s pressure cone method for stiffness calculations with a variable cone angle. This method is quite complicated.

We choose a simpler approach using a fixed cone angle.

Each frustum has a half-apex angle of αThe contraction of an element of the cone of thickness dx is subjected to a compressive force P is,

47


8.6 – Joints : Member Stiffness



Figure 7-22: Compression of a member with the equivalent elastic properties represented by frustum of a hollow cone. Here l represents the grip length.

Pdxd

EAδ =

48


8.6 – Joints : Member Stiffness – cont…



Integrating from 0 to t

))(tan2())(tan2(ln

tan dDdDtdDdDt

EdP

−+++−+

=αα

απδ

))(tan2())(tan2(ln

tan

dDdDtdDdDt

EdPk

−+++−+

==

αα

απδ

The area of element is

Substituting this Eq. and integrating gives a total contraction of

Thus the stiffness of this frustum is

49





For Members made of Aluminum, hardened steel and cast iron 25 < α < 33°For α = 30°

))(155.1())(155.1(ln

5774.0

dDdDtdDdDt

Edk

−+++−+

=π

If the grip consists of any number of members all of the same material, two identical frusta can be added in series. The entire joint can be handled with one equation,

dw is the washer face diameterUsing standard washer face diameter of 1.5d, and with α = 30º, )(

5.25774.05.05774.05ln

5774.0

2dldl

Edk m

++

=π

50





From Finite element analysis results, A and B from table 8.7 for standard washer Faces and membersof same material

(Bd/l)AEdk m exp=

Figure 8-23: The dimensionless plot of stiffness versus aspect ratio of the members of a bolted joint, showing the relative accuracy of methods of Rotscher, Mischke, and Motosh, compared to a finite-element analysis (FEA) conducted by Wileman, Choudury and Green.

im kkkkk1....1111

321

+++=

Combine all frusta as springs in series , km

51





Table 8-7:

52






a) Determine the member spring rate km if the top plate is steel and the bottom plate is gray cast iron.

b) Using the method of conical frusta, determine the member spring rate km if both plates are steel.

c) Using Finite Element Approach to Member Stiffness, determine the member spring rate km if both plates are steel. Compare the results with part (b)

d) Determine the bolt spring rate kb.

As shown in figure below, two plates are clamped by washer-faced ½ in-20 UNF x 1½ in SAE grade 5 bolts each with a standard ½ N steel plain washer.

53






From Table A-32, the thickness of a standard ½ N washer is 0.095 in.

a) As shown in figure below, the frusta extend halfway into the joint the distance

in 6725.0)095.075.05.0(21

=++

The distance between the joint line and the dotted frusta line is

in 0.07750.095-0.5-0.6725 =

Thus, the top frusta consist of the steel washer, steel plate and 0.0775 in of the cast iron. Since the washer and top plate are both

steel with E = 30(10)6 psi, they can be considered a single frustum of 0.595 in thick.

54





The outer diameter of the frustum of the steel member at the joint interface is

in 437.130tan59502750 =+ o).(.

The outer diameter at the midpoint of the entire joint is

in 527.130tan672502750 =+ o).(.

The spring rate of the steel is

[ ][ ]

Ibf/in )10(80.30

)5.075.0(5.075.0)595.0(155.1)5.075.0(5.075.0)595.0(155.1ln

5.0)10)(30(5774.0 66

1 =

−+++−+

=πk

k1

k2

k3

55





For the upper cast-iron frustum

[ ][ ]

Ibf/in )10(5.285

)5.0437.1(5.0437.1)0775.0(155.1)5.0437.1(5.0437.1)0775.0(155.1ln

5.0)10)(5.14(5774.0 66

2 =

⎭⎬⎫

⎩⎨⎧

−+++−+

=πk

For the lower cast-iron frustum

[ ][ ]

Ibf/in )10(15.14

)5.075.0(5.075.0)6725.0(155.1)5.075.0(5.075.0)6725.0(155.1ln

5.0)10)(5.14(5774.0 66

3 =

⎭⎬⎫

⎩⎨⎧

−+++−+

=πk

The three frusta are in series, so )10(15.14

1)10(5.285

1)10(80.30

11666 ++=

mk

This results in km = 9.378 (106) ibf/in

56





b) If the entire joint is steel, therefore l = 2(0.6725) = 1.345 in gives,

ibf/in )10(64.14

)5.0(5.2)345.1(5774.0)5.0(5.0)345.1(5774.05ln

5.0)10)(30(5774.0 66

)(2=

++

=π

mk

c) From table 8.7, A = 0.78715, B = 0.62873.

ibf/in )10(92.14345.1

)5.0(62873.0exp)78715.0)(5.0)(10(30 66 =⎥⎦⎤

⎢⎣⎡=mk

In this case, the different between results is less than 2%.

d) Following the procedure of slide 36, the threaded length of a 0.5-in bolt is

in 25.125.0)5.0(2 =+=TL

57





The length of the unthreaded portion is (refer slide 44)

in 095.125.0345.1 =−=tl

in 25.025.15.1 =−=dl

The length of the unthreaded portion in grip is (refer slide 44)

The major diameter area is

22

in 1963.045.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛= πdA

From table 8-2 (slide 14) the tensile stress area is At = 0.1599 in.

Therefore

Ibf/in )10(69.3

)25.0(1599.0)095.1(1963.0)10(30)1599.0(1963.0

6

6

=

+=

+=

dttd

tdb lAlA

EAAk

58


8.7 – Bolts Strength



Bolt strength is specified by minimum proof strength Sp or minimum proof load, Fpand minimum tensile strength, Sut

• Proof load is the maximum load that a bolt can withstand without acquiring a permanent set

• Proof strength is the quotient of proof load and tensile-stress area

– Corresponds to proportional limit

– Slightly lower than yield strength

– Typically used for static strength of bolt

• Good bolt materials have stress-strain curve that continues to rise to fracture

If Sp not available use Sp =0.85 SyFp = At Sp

Figure 8-24: Typical stress-strain diagram for bolt materials

59


8.7 – Bolts Strength – cont…



• Grades specify material, heat treatment, strengths– Table 8–8 for SAE grades (The SAE specifications are numbered according to

minimum tensile strength)

– Table 8–9 for ASTM designations (ASTM are mostly deals with structural)

– Table 8–10 for metric property class

• Grades should be marked on head of bolt

60





Table 8-8: SAE Specifications for Steel Bolts

61





Table 8-8: SAE Specifications for Steel Bolts (cont…)

62





Table 8-9: ASTM Specifications for Steel Bolts.

63





Table 8-9: ASTM Specifications for Steel Bolts (cont…)

64





Table 8-9: ASTM Specifications for Steel Bolts (cont…)

65





Table 8-10: Metric Mechanical-Property Classes for Steel Bolts

66





Table 8-10: Metric Mechanical-Property Classes for Steel Bolts (cont…)

67


8.8 – Tension Joints : The External Load



Let us consider what happens when an external tensile load P, is applied to a bolt connection.

Assuming a clamping force, (preload Fi )is applied by tightening the nut before external force, P is applied.

Fi = preloadP = external tensile loadPb = portion of P taken by boltPm = portion of P taken by membersFb = Pb + Fi = resultant bolt loadFm = Pm – Fi = resultant load on the membersC = fraction of external load P carried by bolt1-C = fraction of external load P carried by members

68


8.8 – Tension Joints : The External Load – cont…



• During bolt preload; Fi (clamping force produced by tightening the nut before external load; P is applied)– bolt is stretched– members in grip are compressed

• When external load P is applied– Bolt stretches an additional amount δ– Members in grip uncompress same amount δ

69





• Since External Load P is shared by bolt and members, thereforeP = Pb + Pm

• C is defined as the stiffness constant of the joint;

• C indicates the proportion of external load P that the bolt will carry. A good design target is around 0.2.

Table 8-11:

Computation of Bolt and Member Stiffnesses. Steel members clamped using a ½ in – 13 NC steel bolt.

70





Resultant Bolt and Member Load : Fb & Fm

(1 )b b i i

m m i i

F P F CP FF P F C P F

= + = += − = − −

0mF <

These results are only valid if the load on the members remains negative, indicating the members stay in compression.

Fi is preload; high preload is desirable in tension connections.

Fi = 0.75 Fp for re-useFi = 0.90 Fp for permanent joint

71


8.9 – Relating Bolt Torque to Bolt Tension



• Best way to measure bolt preload is by relating measured bolt elongation and calculated stiffness

• Usually, measuring bolt elongation is not practical• Measuring applied torque is common, using a torque wrench• Need to find relation between applied torque and bolt preload

Torque required to give preload Fi

• From the power screw equations,

• Applying tan λ = l / π dm

72


8.9 – Relating Bolt Torque to Bolt Tension – cont…



• Assuming a washer face diameter of 1.5d, the collar diameter isdc = (d + 1.5d)/2 = 1.25d, giving

• Define term in brackets as torque coefficient; K

Table 8-12: Torque factor K

• Some recommended values for K for various bolt finishes is given in Table 8–12

• Use K = 0.2 for other cases

73





A ¾ in-16 UNF x 2 ½ in SAE grade 5 bolt is subjected to a load P of 6 kip in a tension joint. The initial bolt tension is Fi = 25 kip. The bolt and joint stiffness are 6.50 and 13.8 Mlbf/in, respectively.

(a) Determine the preload and service load stresses in the bolt. Compare these to the minimum proof strength of the bolt.

(b) Specify the torque necessary to develop the preload using equation T = Kfid(c) Specify the torque necessary to develop the preload if given f = fc = 0.15


74






From Table 8-2, At = 0.373 in2.

a) The preload is kpsi 02.67373.025

===t

ii A

Fσ

The stiffness constant is 320.08.135.6

5.6=

+=

+=

mb

b

kkkC

The stress under the service load is

kpsi 17.7202.67373.0

)6(320.0 =+=

+=+

== itt

i

t

bb A

CPA

FCPAF σσ

From Table 8-8, the SAE minimum proof strength of the bolt is Sp = 85 kpsi. The preload and service load stresses are respectively 21 and 15 percent less than the proof strength.

75





b) The torque necessary to achieve the preload is

Ibf.in 3750)75.0)(1025(2.0 3 === xdKFT i

c) The minor diameter can be determined from the minor area in Table 8-2. Thus

in 6685.0)351.0(44===

ππr

rAd

Than the mean diameter can be calculated as follow

in 7093.02

6685.075.0=

+=md

The lead angle is

o

mm Nddl 6066.1

)16)(7093.0(1tan1tantan 111 ==== −−−

πππλ

76





Ibf.in 3551

)75.0)(10(25)15.0(625.0)30)(sec6066.1(tan15.01

)30(sec15.06066.1tan)75.0(2

7093.0 3

=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+

⎥⎦

⎤⎢⎣

⎡= oo

ooT

For α = 30o

Which is 5.3% less than the value found in part (b)

77


8.10 – Statically Loaded Tension Joint with Preload



Failure of joints occurs wheni) Bolt Yields

Proof strength

At: Tensile stress area

t

i

tt

bb A

FA

CPAF

+==σ

Failure starts pb S=σ

a) Yielding Factor of safety:i

tp

ti

p

b

pp FCP

ASAFCP

SSn

+=

+==

/)(σ

b) Load Factor:CP

FASnASFCP itp

Ltpi−

=∴=+

78





ii) Joint separates

Let P0 be external load causing separation Fm= 0

( )

( )

00

(1 )0

1

/ 1

m i

m

i

i

F C P FF

P FnP P C

FFor n bolts nP N C

= − −

=

= =−

=−

0(1 ) 0iC P F− − =

no : factor of safety against joint separation

79






Figure 8-25 is a cross section of a grade 25 cast-iron pressure vessel. A total of N bolts are to be used to resist a separating force of 36 kip.

(a) Determine kb, km, and C(b) Find the number of bolts required for a

load factor of 2 where the bolts may be reused when the joint is taken apart.

(c) With the number of bolts obtained in part (b), determine the realized load factor for overloading, the yielding factor of safety, and the load factor for joint separation. Figure 8-25:

80






81





82





83





84


8.11 – Gasketed Joints



If a full gasket is present in joint, The gasket pressure p is:

gi

im

g

m

ANCnPFp

FnPCFnfactorloadWith

NAFp

)]1([

)1(

/

−−=

−−=

−= No. of bolts

85


8.11 – Gasketed Joints – cont…



IMPORTANT:1. To maintain uniformity of pressure

adjacent bolts should not be placed more than 6 nominal diameters apart on bolt circle.

2. To maintain wrench clearance bolts should be placed at least 3 d apart.

3. A rough rule for bolt spacing around a bolt circle is

where Db is the diameter of the bolt circle and N is the number of bolts.

3 6bDNd

π≤ ≤

86


8.12 – Fatigue Loading of Tension Joints



• Fatigue methods of Ch. 4 are directly applicable

• Distribution of typical bolt failures is– 15% under the head

– 20% at the end of the thread

– 65% in the thread at the nut face

• Fatigue stress-concentration factors for threads and fillet are given in Table 8–13

Table 8-13: Fatigue Stress-Concentration Factors Kf for Threaded Elements

87


8.12 – Fatigue Loading of Tension Joints – cont…



• Bolts are standardized, so endurance strengths are known by experimentation, including all modifiers.

Table 8-14: Full Corrected Endurance Strengths for Bolts and Screws with Rolled Threads*

• Fatigue stress-concentration factor Kf is also included as a reducer of the endurance strength, so it should not be applied to the bolt stresses.

• In thread-rolling the amount of cold-work and strain strengthening is unknown to the designer; therefore, fully corrected (including Kf) axial endurance strength is reported inTable 8-14.

Endurance Strength for Bolts

88





• With an external load on a per bolt basis fluctuating between Pmin and Pmax,

• The alternating stress experienced by a bolt is

• The midrange stress experienced by a bolt is

Fatigue Stresses

89





Typical load line starts from constant preload, then increases with a constant slope

On the designer’s fatigue diagram, shown in Figure 8-26, the load line is .

High Preload is especially important

in fatigue. σi is a constant the load line at Fi/At has a unit slope, r = 1.0 Figure 8-26: Designer’s fatigue diagram showing a Goodman failure

locus and how a load line is used to define failure and safety in preloaded bolted joints in fatigue.

Typical Fatigue Load Line for Bolts

90






Next, find the strength components Sa of the fatigue failure locus. These depend on the failure Criteria.

GoodmanGoodman

1a m

e ut

S SS S

+ =

GerberGerber2

1a m

e ut

S SS S

⎛ ⎞+ =⎜ ⎟

⎝ ⎠

ASMEASME--EllipticElliptic22

1a m

e p

S SS S

⎛ ⎞⎛ ⎞+ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

mut

eea S

SS

SS −=

eut

mea S

SS

SS2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

p

mea S

SSS

)( imim

aa SS σ

σσσ

−−

=

ia

imam

SS σ

σσσ

+−

=∴)(

Equation of a typical fatigue load line:

91






• Solving (a) and (b) for Goodman line intersection point,

• Fatigue factor of safety based on Goodman line and constant preload load line,

• Other failure curves can be used, following the same approach.

af

a

Sn

σ=

92





Repeated Load Special Case

• Bolted joints often experience repeated load, where external load fluctuates between 0 and Pmax

• Setting Pmin = 0; equation in slide 88

• With constant preload load line,

• Load line has slope of unity for repeated load case

Fatigue factor of safety equations for repeated loading, constant preload load line, with various failure curves:

93





Goodman

Gerber

Repeated Load Special Case

Substitute σa and σi into any of the fatigue factor of safety equations

ASME –Elliptic

94





Yield Check with Fatigue Stresses

• As always, static yielding must be checked.

• In fatigue loading situations, since σa and σm are already calculated, it may be convenient to check yielding with

• This is equivalent to the yielding factor of safety from slide 77.

95






Figure 8-27:

Figure 8-27Pressure-cone frustum member model for a cap screw. For this model the significant sizes are

Where l = effective grip. The solutions are for α=30o and dw=1.5d.

2 2

2

1

2

2

2

tan 1.5 0.577

1.5w

w

h t t dl

h D t d

D d l d l

D d d

α

+ <⎧= ⎨ + ≥⎩

= + = += =

(Refer to slide 43)

96






(a) For the symbols of Figs. 8-22 and 8-27,

The joint is composed of three frusta; the upper two frusta are steel and the lower one is cast iron.

For the upper frustrum; using Eq. in slide 49:

in 9375.05.1in 12/

in 6875.0

2

1

===+=

=+=

dDdhltth w

MIbf/in 46.46 Mpsi 30

in 9375.0in 5.02/

1 =⎪⎭

⎪⎬

⎫

==

==k

ED

lt

97





For the middle frustrum;

For the lower frustrum;

Substituting these three stiffnesses gives km = 17.40 MIbf/in. The cap screw is short and threaded all the way. Using l = 1 in for the grip and At = 0.226 in2 from Table 8-2, we find the stiffness to be kb = AtE / l = 6.78 MIbf/in. Thus the joint constant is


in 1.29830tan)(29375.0

in 1875.02/

2 =⎪⎭

⎪⎬

⎫

==−+=

=−=

kE

hlD

lhto


in 9375.0in 3125.0

3 =⎪⎭

⎪⎬

⎫

==

=−=k

ED

hlt

ci

280.040.1778.6

78.6=

+=

+=

mb

b

kkk

C 98





(b) Equation in slide 70 gives the preload as,

where from Table 8-8, Sp = 85 kpsi for an SAE grade 5 cap screw. Using Eq. in slide 77, we obtain the load factor as the yielding factor of safety is

This is the traditional factor of safety, which compares the maximum bolt stress to the proof strength. Using Eq. in slide 77

This factor is an indication of the overload on P that can be applied without exceeding the proof strength.

kip 4.14)85)(226.0(75.075.075.0 ==== ptpi SAFF

22.14.14)5(280.0

)226.0(85=

+=

+=

i

tpp FCP

ASn

44.3)5(280.0

4.14)226.0(85=

−=

−=

CPFAS

n itpL

99





Next, using Eq. in slide 78, we have

If the force P gets too large, the joint will separate and the bolt will take the entire load. This factor guards against that event.

For the remaining factors, refer to Fig. 8-28 at next slide. This diagram contains the modified Goodman line, the Gerber line, the proof-strength line, and the load line. The intersection of the load line L with the respective failure lines at points C, D, and E definesa set of strengths Sa and Sm at each intersection. Point B represents the stress state σa , σm . Point A is the preload stress σi . Therefore the load line begins at A and makes an angle having a unit slope. The angle is 45o only when both stress axes have the same scale.

00.4)280.01(5

4.14)1(0 =

−=

−=

CPF

n i

100





Fig. 8–28: Designer’s fatigue diagram for preload bolts, drawn to scale, showing the modified Goodman line, the Gerber line, and the Larger proof-strength line, with an exploded view of the area of interest. The strengths used are Sp = 85 kpsi, Se = 18.6 kpsi, and Sut = 120 kpsi. The coordinates are;

A, σi = 63.72 kpsi ; B, σa = 3.10 kpsi, σm = 66.82 kpsi ; C, Sa = 7.55 kpsi, Sm = 71.29 kpsi ; D, Sa = 10.64 kpsi, Sm = 74.36 kpsi ; E, Sa = 11.32 kpsi , Sm = 75.04 kpsi.

101





The factors of safety are found by dividing the distances AC, AD and AE by the distance AB. Note that this is the same as dividing Sa for each theory by σa.

The quantities shown in the caption of Fig. 8-28 are obtained as follows:

Point A

Point B

Point CThis is the modified Goodman criteria. From Table 8-14, we find Se = 18.6 kpsi. Then, using Eq. in Slide 93, the factor of safety is found to be

kpsi 72.63226.0

4.14===

t

ii A

Fσ

kpsi 82.6672.6310.3

kpsi 10.3)226.0(2)5(280.0

2

m =+=+=

===

ia

ta A

CP

σσσ

σ

kpsi 44.2)6.18120(10.3)72.63120(6.18

)()(

=+−

=+−

=euta

iutef SS

SSnσ

σ

102





Point DThis is on the proof-strength line where

In addition, the horizontal projection of the load line AD is

Solving Eqs. above simultaneously results in

The factor of safety resulting from this is

which, of course, is identical to the result previously obtained by using Eq. in slide 77.

A similar analysis of a fatigue diagram could have been done using yield strength instead of proof strength. Though the two strengths are somewhat related, proof strength is a much better and more positive indicator of a fully loaded bolt than is the yield strength. It is also worth remembering that proof-strength values are specified in the design codes; yields strengths are not.

pam SSS =+

aim SS +=σ

kpsi 64.102

72.63852

=−

=−

= ipa

SS

σ

43.310.364.10

===a

ap

Snσ

103





We found nf = 2.44 on the basis of fatigue and the modified Goodman line, and np = 3.43 on the basis of proof strength. Thus the danger of failure is by fatigue, not by overproofloading. These two factors should always be compared to determine where is the greatest danger lies.

Point EFor the Gerber criterion, from Eq. in slide 93, the safety factor is

which is greater than np = 3.43 and contradicts the conclusion earlier that the danger of failure is fatigue. Figure 8-28 clearly shows the conflict where point D lies between points C and E. Again, the conservative nature of Goodman criterion explains the discrepancy and the designer must form his or her own conclusion.

[ ][ ]

65.3

)6.18)(73.63(2120)72.636.18)(6.18(4120120)6.18)(10.3(2

1

2)(42

1

22

22

=

−−++=

−−++= eiutieeututea

f SSSSSSS

n σσσ

104


8.13 – Bolted and Riveted Joints Loaded in Shear



Shear loaded joints are handled the same for rivets, bolts, and pins

Several failure modes are possible(a) Joint loaded in shear

(b) Bending of bolt or members

(c) Shear of bolt

(d) Tensile failure of members

(e) Bearing stress on bolt or members

(f) Shear tear-out

(g) Tensile tear-out

Fig. 8–29: Modes of failure in shear loading of a bolted or riveted connection.

105





Failure by Bending

• Bending moment is approximately M = Ft / 2, where t is the grip length, i.e. the total thickness of the connected parts.

• Bending stress is determined by regular mechanics of materials approach, where I/c is for the weakest member or for the bolt(s).

106





Failure by Shear of Bolt

• Simple direct shear

• Use the total cross sectional area of bolts that are carrying the load.

• For bolts, determine whether the shear is across the nominal area or across threaded area. Use area based on nominal diameter or minor diameter, as appropriate.

107





Failure by Tensile Rupture of Member

• Simple tensile failure

• Use the smallest net area of the member, with holes removed

108





Failure by Bearing Stress

• Failure by crushing known as bearing stress

• Bolt or member with lowest strength will crush first

• Load distribution on cylindrical surface is non-trivial

• Customary to assume uniform distribution over projected contact area, A = td

• t is the thickness of the thinnest plate and d is the bolt diameter

109





Failure by Shear-out or Tear-out

• Edge shear-out or tear-out is avoided by spacing bolts at least 1.5 diameters away from the edge

110






The bolted connection shown in Figure 8-30 uses SAE grade 5 bolts. The members are hot-rolled AISI 1018 steel. A tensile shear load F = 4000 Ibf is applied to the connection. Find the factor of safety for all possible modes of failure.

Fig. 8–30

F = 4000 Ibf

F = 4000 Ibf

111






Members: Sy = 32 kpsi (from Table A-20)

Bolts: Sy = 92 kpsi (from Table 8-8) ; Ssy = 0.577Sy = 0.577(92) = 53.08 kpsi

Shear of bolts

93.21.1808.53

kpsi 1.18221.04

in 221.04

)375.0(2 22

===

===

=⎥⎦

⎤⎢⎣

⎡=

τ

τ

π

sy

s

s

s

Sn

AF

A Bearing on bolts

32.43.21

92

kpsi 3.21188.04

in 188.0)375.0)(25.0(2 2

=−

==

−=−

=

==

b

y

b

b

Sn

A

σ

σ

Ab

112





Bearing of members

50.13.21

32=

−==

b

scSnσ

Tension on members

25.385.9

32

kpsi 85.9406.04

in 406.0)4/1)(75.0375.2( 2

===

==

=−=

t

y

t

t

AS

n

A

σ

At

113





Shear Joints with Eccentric Loading

• Eccentric loading is when the load does not pass along a line of symmetry of the fasteners.

• Requires finding moment about centroid of bolt pattern

• Centroid location

Fig. 8–31: Centroid of pins, rivets or bolts.

where A1 to A5 is the group of pins, rivets or bolts respective cross sectional area and xi and yi are the distances to the itharea center.

114






(a) Example of eccentric loading

(b) Free body diagram

(c) Close up of bolt pattern

115






• Primary Shear

• Secondary Shear, due to moment load around centroid

116






Shown in Fig. 8-32 is a 15 – by 200-mm rectangular steel bar cantilevered to a 250-mm steel channel using four tightly fitted bolts located at A, B, C, and D.

For a F = 16 kN load find

(a) The resultant load on each bolt

(b) The maximum shear stress in each bolt

(c) The maximum bearing stress

(d) The critical bending stress in the bar

Fig. 8–32: Dimensions in millimeters.

117






(a) Point O, the centroid of the bolt group in Fig. 8-32, is found by symmetry. If a free-body diagram of the beam were constructed, the shear reaction V would pass through O and the moment reactions M would be about O. These reaction are

V = 16 kN ; M = 16(425) = 6800 Nm

In Fig. 8-33, the bolt group has been drawn to a larger scale and the reactions are shown. The distance from the centroid to the center of each bolt is

mm 0.96)75()60( 22 =+=rFig. 8–33

118





The primary shear load per bolt is

Since the secondary shear forces are equal, Eq. in slide 115 becomes

The primary and secondary shear forces are plotted to scale in Fig. 8-33 and the resultants obtained by using parallelogram rule. The magnitudes are found by measurement (or analysis) to be

kN 44

16' ===nVF

kN 17.7)0.96(4

680044

" 2 ====r

Mr

MrF

kN .012)7.38sin7.17()7.38cos7.174( 22 =++== ooBA FF

75 mm

θ o

α o

60 mmo3.51

6075tan 1 == −θ oooo 7.383.519090 =−=−= θα

kN .814)7.38sin7.17()47.38cos7.17( 22 =+−== ooDC FF

119





(b) Bolts A and B are critical because they carry the largest shear load. Does this shear act on the threaded portion of the bolt, or on the unthreaded portion? The bolt length will be 25 mm plus the height of the nut plus about 2 mm for a washer. Table A-31 gives the nut height as 14.8 mm. Including two threads beyond the nut, this adds up to a length of 43.8 mm, and so a bolt 45 mm long will be needed. From Eq. in slide 36, we compute the thread length as LT = 38mm. Thus the unthreaded portion of the bolt is 45 – 38 = 7 mm long. This is less than the 15 mm for the plate in Fig. 8-33, and so the bolt will tend to shear across its minor diameter. Therefore the shear-stress area is As = 144 mm2, and so the shear stress is

(c) The channel is thinner than the bar, and so the largest bearing stress is due to the pressing of the bolt against the channel web. The bearing area is Ab = td = 10(16) = 160 mm2. Thus the bearing stress is

MPa 146144

)10(0.21 3

===sA

Fτ

MPa 131160

)10(0.21 3−=−=−=

bAFσ

120





(d) The critical bending stress in the bar is assumed to occur in a section parallel to the y axis and through bolts A and B. At this section the bending moment is

The second moment of area through this section is obtained by the use of the transfer formula, as follows:

Then

Nm 5600)50300(16 =+=M

46233

2

mm )10(26.8)16)(15(6012

)16(15212

)200(15

)(2

=⎥⎥⎦

⎤

⎢⎢⎣

⎡+−=

+−= AdIII holesbar

MPa 8.67)10(26.8

)100(56006 ===

IMcσ

121


References



1. Richard G. Budynas, J. Keith Nisbett. Shigley's Mechanical Engineering Design

(Ninth Edition). Singapore : McGraw‐Hill Companies, Inc., 2011. ISBN 978‐007‐

131113‐7.

2. Bazoune, Dr. A. Aziz. KFUPM Open Courseware. [Online] King Fahd University

of Petroleum & Minerals. [Cited: February 28, 2012.]

http://opencourseware.kfupm.edu.sa/colleges/ces/me/me307/lectures.asp.

Chapter 9Permanent Joints

Prepared by: Mohd Azwir bin Azlan



Week 13 & 14

2


acquaintance with the terminology, and types of permanent joints.

design, analysis, and sizing of welded joints


Learning Outcomes


CHAPTER 9 – Permanent Joints

3




• 9.2 - Welding Symbols

• 9.3 - Butt and Fillet Welds

• 9.4 - Stresses in Welded Joints in Torsion

• 9.5 - Stresses in Welded Joints in Bending

• 9.6 - The Strength of Welded Joints

• 9.7 - Static Loading

• 9.8 - Fatigue Loading

• 9.9 - Resistance Welding

• 9.10 - Adhesive Bonding



4




• Form can more readily pursue function with the help of joining processes such as welding, brazing, soldering, cementing, and gluing - processes that are used extensively in manufacturing today.

• Particularly when sections to be joined are thin, the elimination of individual fasteners, with their holes and assembly costs lead to significant savings.

• Welding is the process of joining two pieces of metal together by hammering, pressure or fusion. Filler metal may or may not be used.

• Welding is the strongest and most common method of permanently joining steel components together.

• Arc welding is the most important since it is adaptable to various manufacturing environments and is relatively cheap.


5


9.2 – Welding Symbols


• A weldment is fabricated by welding together a collection of metal shapes, cut to particular configurations.

• The welding symbol is standardized by the American Welding Society (AWS).

• The weld must be precisely specified on working drawing and this is done by welding symbol, Fig. 9-1.

• The arrow of this symbol points to the joint to be welded.

• The body of the symbol contains as many of the following elements as are deemed necessary:


Reference lineArrowBasic weld symbols in Fig. 9-2Dimensions and other data

Supplementary symbolsFinish symbolsTailSpecification or process.

6


9.2 – Welding Symbols – cont…



Figure 9-1 : The AWS standard welding symbol showing the location of the symbol elements

7





Figure 9-2 : Arc and gas-weld symbols

There 2 general types of welds:

1.1. Fillet weldsFillet welds for general machine elements.

2. Butt or groove welds for pressure vessels, piping systems,...

There are also others such as: ,

Fillet weldsFillet welds

groove welds

BeadBead Plug or slotPlug or slot

8





• Parts to be joined must be arranged so that there is sufficient clearance for welding operation.

• Due to heat, there are metallurgical changes in the parent metal in the vicinity of the weld.

• Residual stresses may be introduced because of clamping or holding.

• These residual stresses are not severe enough to cause concern.

• A light heat treatment after welding is done to relive these stresses.

• When the parts to be welded are thick, a preheating will also be of benefit.

9





Welding Symbol Examples

Weld leg size of 5 mmFillet weldBoth sides

Intermittent and staggered 60 mm along on 200 mm centers

Leg size of 5 mmOn one side only (outside)Circle indicates all the way around

10






11






12






13


9.3 – Butt and Fillet Welds



• Where h is the weld throat and l is the length of the weld. Notice that the value of h does not include the reinforcement.

• Reinforcement adds some strength for static loaded joints.• The reinforcement can be desirable, but it varies somewhat and does produce stress

concentration at point A in the figure. If fatigue loads exist, it is good practice to grind ormachine off the reinforcement.

hlF

=σ stress; NormalhlF

=τ ; stressShear

14


9.3 – Butt and Fillet Welds – cont…



• At angle θ the forces on each weldment consists of a normal force Fn and a shear force Fs

sin ,cosn

sF FF F

θθ

==

• Fig. 9-3 illustrates a typical transverse fillet weld.

• In Fig. 9-4 a portion of the welded joint has been isolated from Fig. 9-3

Figure 9-3 : A transverse fillet weld

Figure 9-4 : Free Body Diagram from Fig. 9-4

15





sin (cos sin )

cos (cos sin )n

sF FA hlF FA hl

θ θ θτ

θ θ θσ

+= =

+= =

• The nominal stresses at the angle θ in the weldment, τ and σ, are

• The von Mises stress σ’at angle θ is

• σ’max occurs at θ = 62.5o with a value of

• The corresponding values of τ and σ, are and

• τ max can be found by solving the equation [d(τ)/dθ] = 0.

• The stationary point occurs at θ = 67.5o with a corresponding &

hlF16.2'max =σ

hlF196.1=τ

hlF623.0=σ

hlF5.0=σ

hlF207.1max =τ

( ) ( ) ( )[ ] 2/122222/122 cossinsin3cossincos3' θθθθθθτσσ +++=+=

hlF

16





• No analytical approach accurately predicts the experimentally measured stresses.• Standard practice is to use a simple and conservative model• Assume the external load is carried entirely by shear forces on the minimum throat

area. Equation for simpler case of simple shear loading in fillet weld

• By ignoring normal stress on throat, the shearing stresses are inflated sufficiently to render the model conservative.

• By comparison with previous maximum shear stress model, this inflates estimated shear stress by factor of 1.414/1.207 = 1.17.

Transverse Fillet Weld Simplified Model

hlF

hlF 414.1

707.0==τ

Figure 9-3 : Parallel fillet welds

17


9.4 – Stresses in Welded Joints in Torsion



Figure 9-4 : This is a moment connection; such a connection produces torsion in the weld

• Figure 9-4 illustrates a cantilever of length lwelded to a column by 2 fillet welds.

• The reaction at the support of a cantilever always consists of shear force V and a moment reaction M.

• The shear force produces a primary shear in the welds of magnitude

where A is the throat area of the welds.

• The moment at the support produces secondary shear or torsion of the welds, and this stress is given by

' VA

τ =

" MrJ

τ = wherer: distance from the centroid of the weld

group to the point in the weld of interest.

J: second polar moment of area of the group about the centroid of the group.

Fillet Welds Loaded in Torsion

18


9.4 – Stresses in Welded Joints in Torsion – cont…



• Figure 9-5 shows 2 welds in a group. The rectangles represent the throat areas of the welds.

• Weld 1 has a throat width t1 = 0.707 h1

• Weld 2 has a throat width t2 = 0.707 h2

• Throat area of both welds together is A = A1 + A2 = t1d + t2b

• The x-axis passes through the Centroid G1 of the weld 1.

• The second moment of area about this axis is

• Similarly, the second moment of area about an axis passing through G1 parallel to the y-axis is

Figure 9-5 : Second polar moment of area on two welds

12

31dtI x =

12

31dtI y =

Example of Finding A and J

19





Figure 9-5 : Second polar moment of area on two welds

• Thus the second polar moment of areas of weld 1 and weld 2 about their Centroid are

• The Centroid G of the weld group is located at

• The distances r1 and r2 from G1 and G2 are respectively given by

• Using the parallel axis theorem, the second polar moment of area of the weld group is

1212

12123

232

2

31

31

1

btbtIIJ

dtdtIIJ

yxG

yxG

+=+=

+=+=

AyAyAy

AxAxAx 22112211 ; +

=+

=

( )[ ] ( ) ( )[ ] 1/22

22

22

1/222

11 ; yyxxryxxr −+−=+−=

)()( 222111 rAJrAJJ GG +++=

20





• Note that t3 terms will be very small compared to b3 and d3 ⇒ Usually neglected

• Leaves JG1 and JG2 linear in weld width

• Can normalize by treating each weld as a line with unit thickness t

• Results in unit second polar moment of area, Ju

• Since t = 0.707h,

J = 0.707hJu

• in which Ju is found by conventional methods for an area having unit width.

12/ ; 12/ 322

311 btJdtJ GG ==

( )u

GG

JtJ

brbdrdtJ

tbrtbtdrtdJ

rAJrAJJ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

+++=

+++=

2

3

1

3

2

3

1

3222111

1212

1212

)()(

Ju : is found from table 9.1 on next slide

21





Table 9-1 : Common Torsional Properties of Fillet Welds*

22





Table 9-1 : Common Torsional Properties of Fillet Welds* - cont…

23





Example 9-1

A 50-kN load is transferred from a welded fitting into a 200-mm steel channel as illustrated in Figure below. Estimate the maximum stress in the weld.

Figure 9-6 : Dimensions in millimeters

24





Solution 9-1

a) Label the ends and corners of each weld by letter.

b) Finding the primary shear stress for 1 side:

Figure 9-7 : Diagram showing the weld geometry on a single plate. Note that V and M represent the reaction loads applied by the welds to the plate.

A = 0.707(6)[2(56) + 190] = 1280 mm2

Then the primary shear stress is

MPa 5.191280

)10(25'3

===AVτ

• Weld joint is under torsion → primary shear and torsion shear

• Weld joint is in form of weld group → centroidO position and r

• 2 side joints (left and right) → for one side, F= 25 kN

25





Solution 9-1

c) Draw the τ’ stress, to scale, at each lettered corner or end. See Fig. 9-8

d) Locate the centroid of the weld pattern. Using case 4 of Table 9-1, we find

e) Find the distance ri (see Fig. 9-8):

These distance can also be scale from the drawing.

f) Find J. Using case 4 of Table 9-1 again, we get

mm 4.10190)56(2

)56( 2

=+

==AVx

mm 6.95])4.10()2/190[(

mm 105])4.1056()2/190[(2/122

2/122

=+==

=−+==

DC

BA

rr

rr

464323

mm )10(07.7190)56(2

)56(12

)190()190)(56(6)56(8)6(707.0 =⎥⎦

⎤⎢⎣

⎡+

−++

=J

26





Solution 9-1

g) Find M:

h) Estimate the secondary shear stress τ” at each lettered end or corner;

i) Draw the τ” stress at each corner end. See Fig 9-8. Note that this is a free body diagram of one of the side plates, and therefore the τ’ and τ” stresses represent what the channel is doing to the plate (through the welds) to hold the plate in equilibrium.

j) At each point labeled, combine the two stress components as vectors (since they apply to the same area). At point A, the angle that τA” makes with the vertical, α, is also the angle rA makes with the horizontal, which is α = tan-1 (45.6/95) = 25.64o and θ = 90o –25.64o = 64.36o. This angle also applies to point B.

Nm 2760)4.10100(25 =+== FlM

MPa 3.37)10(07.7

)6.95()10(2760""

MPa 0.41)10(07.7

)105()10(2760""

6

3

6

3

===

====

DC

BA JMr

ττ

ττ

α

27





Solution 9-1

Figure 9-8 : Free-body diagram of one of the side plates.

τA”

τA’

θ

τA

αA

τB’

τB”

γ

τB

β D

β

28





Solution 9-1

Thus:

MPa 0.37)36.64cos()0.41)(5.19(20.415.19 22 =−+== oBA ττ

Similarly, for C and D, β = tan-1 (10.4/95) = 6.25o and γ = 90o + 6.25o = 96.25o.

Thus :

MPa 9.43)25.96cos()3.37)(5.19(23.375.19 22 =−+== oDC ττ

k) Identify the most highly stressed point:

MPa 9.43max === DC τττ

29


9.5 – Stresses in Welded Joints in Bending



• A FBD diagram of the beam would show a shear force V and a moment diagram M.

• Bending cause 2 shear stresses :

i. primary shear ⇒

ii. Secondary shear ⇒

• Bending moment cause normal bending in base metal, BUT it cause SHEAR stress in weld THROAT.

• The second moment of area I, based on weld throat area, is

• in which Iu is found from table 9.2

• The 2 shear stress are then combine as vectors. Because the 2 shear vector are PERPENDICULAR ⇒ use Pythagoras rules.

AV

='τ

uhIMc

IMc

707.0" ==τ

uhII 707.0=

22 )"()'( τττ +=30


9.5 – Stresses in Welded Joints in Bending – cont…



• Fig. 9-9 shows a cantilever welded to a support by fillet welds at top and bottom.

Figure 9-9 : A rectangular cross-section cantilever welded to a support at the top and bottom edges.

Example of Finding τ”

• Treating the two welds of Fig. 9-9 as lines we find the second moment of area to be (from table 9-2)

• The nominal throat shear stress is now found to be

12

2bdIu =

bdhM

bdhdM

hIMc

u

414.1)2/(707.0

)2/(707.0

" 2 ===τ

31





Table 9-2 : Bending Properties of Fillet Welds

Pattern

32





Pattern

33


9.6 – The Strength of Welded Joints



• Must check for failure in parent material and in weld

• Safety factor in welding is based on YIELD strength of parent material NOT ultimate strength.

• Weld strength is dependent on choice of electrode material

• Weld material is often stronger than parent material

• Parent material experiences heat treatment near weld

• Cold drawn parent material may become more like hot rolled in vicinity of weld

• Often welded joints are designed by following codes rather than designing by the conventional factor of safety method

34


9.6 – The Strength of Welded Joints – cont…



• The properties of electrodes vary considerably, but Table 9-3 lists the minimum properties for some electrode classes.

Table 9-3 : Minimum Weld-Metal Properties

35





E 7 0 X X – H 4 R

Electrode

Tensile in Kpsi

Welding Position1=All Position, 2=Flat & Horizontal Position, 3=Flat Welding, 4=Vertical

Type of Current and Coating

*HydrogenH4=Less than 4ml/100g weld metal, H8=Less than 8ml/100g weld metalor H16=Less than 16ml/100 weld metal

*Meets Requirement of Absorbed Moisture Test

*Optional Designator

Electrode Welding Code

36





Table 9-4 : Stresses Permitted by the AISC Code for Weld Metal

• The table below show the recommended safety factor for welded structure.

• Safety factor n* is estimated by using Distortion Energy Theory (DET) or Von Mises.

• According DET, permissible effective stress

• Note: Look how to calculate Von Mises effective stress.

nyield

effective

σσ ≤

37





Table 9-5 : Fatigue Stress-Concentration Factors, Kfs

• The fatigue stress concentration factors listed in Table 9-5 are suggested for use. These factors should be used for the parent metal as well as for the weld metal.

• Kfs appropriate for application to shear stresses

• Use for parent metal and for weld metal

38





Table 9-6 : Allowable Steady Loads and Minimum Fillet Weld Sizes

39


9.7 – Static Loading



A 12 -mm by 50mm rectangular-cross-section 1015 bar carries a static load of 73kN. It is welded to a gusset plate with a 10-mm fillet weld 50 mm long on both sides with an E70XX electrode as depicted in Fig. 9-10. Use the welding code method.

(a) Is the weld metal strength satisfactory?

(b) Is the attachment strength satisfactory?

Example 9-2

Figure 9-10

40


9.7 – Static Loading – cont…



Solution 9-2

(a) From Table 9-6, allowable force per unit length for a 10-mm E70 electrode metal is 1025 N/mm of weldment; thus

Fallow = 1025l = 1025(100) = 102.5 kN

Since 102.5 kN > 73 kN, weld metal is satisfactory.

(b) Check shear in attachment adjacent to the welds. From Table 9-4 and Table A-20, from which Sy = 190 MPa, the allowable attachment shear stress is

τall = 0.4 Sy = 0.4(190) = 76 MPa

The shear stress τ on the base metal adjacent to the weld is

Since τall ≥ τ , the attachment is satisfactory near the weld beads.

MPa 73)05.0)(01.0(2

730002

===hlFτ

41





Solution 9-2

The tensile stress in the shank of the attachment σ is

The allowable tensile stress σall , from Table 9-4, is 0.60 Sy and, with welding code safety level preserved,

σ all = 0.6 Sy = 0.60(190) = 114 MPa

Since σall ≥ σ , the shank tensile is satisfactory

MPa 122)05.0)(012.0(

73000===

tlFσ

42





Example 9-3

The attachment shown in Figure 9-11 is made of 1018 HR steel 12 mm thick. The static force is 100kN. The member is 75 mm wide. Specify the weldment ( give the pattern, electrode number, type of weld, length of weld, and leg size ).

Figure 9-11 : Dimension in millimeters.

Materials:

Attachment (1018 HR) Sy = 220 MPa ; Sut = 400 MPa (refer Table A-20)

Members (A36) Sy = 250 MPa ; Sut = 400 ~ 550 MPa, use 400 MPa(refer to www.matweb.com for ASTM A36 Steel, plate)

The member and attachment are weak compared to the E60XX Electrode. (refer Table 9-3)

Decision : Specify E6010 electrode (refer slide 35 and electrode code)

Controlling property : τall = min [0.3(400),0.4(220)] = min (120,88) = 88 MPa --- (refer Table 9-4)

43





Solution 9-3

(weld metal) , (base metal)Stress permitted Critical location – (base metal)

44





Solution 9-3

For a static load the parallel and transverse fillets are the same. If n is the number of beads,

Make a table

mm 43.21)88)(75)(707.0(

100000)707.0(

)707.0(

===⇒

==

all

all

lFnh

hlnF

τ

ττ

Number of beads Leg size

n h

1 21.43 → 22 mm

2 10.72 → 11 mm

3 7.14 → 8 mm

4 5.36 → 6 mm

Decision: Specify 6 mm leg sizeDecision: Well all-roundWeldment Specifications:Pattern: All-round squareElectrode: E6010Type: Two parallel filletsTwo transverse filletsLength of bead: 300 mmLeg size: 6 mm

Answer

45





Example 9-4

Perform an adequacy assessment of the statically loaded welded cantilever carrying 2.2 kN depicted in Fig. 9–12. The cantilever is made of AISI 1018 HR steel and welded with a 10-mm fillet weld as shown in the figure. An E6010 electrode was used, and the design factor was 3.0.

(a) Use the conventional method for the weld metal.

(b) Use the conventional method for the attachment (cantilever) metal.

(c) Use a welding code for the weld metal.

Figure 9-12 46





Solution 9-4

(a) From Table 9-3, Sy = 345 MPa, Sut = 427 MPa. From Table 9-2, second pattern, b = 10mm, d = 50 mm, so

Primary shear:

Secondary shear:

The shear magnitude τ is the Pythagorean combination

The factor of safety based on minimum strength and the distortion-energy criterion is

4

333

2

mm 289,147)883,20)(10(707.0707.0

mm 883,206/506/ mm 707)50)(10(414.1414.1

===

===

===

u

u

hII

dIhdA

MPa 1.37072200' ===

AFτ

MPa 56147289

)25)(150(2200" ===I

Mrτ

( ) ( ) MPa 1.56561.3"' 2/1222/122 =+=+= τττ

55.31.56

)345(577.0===

τsyS

n Since n > nd, that is, 3.39 > 3.0, the weld metal has satisfactory strength.Since n > nd, that is, 3.39 > 3.0, the weld metal has satisfactory strength.

47





Solution 9-4

(b) From Table A-20, minimum strengths are Sut =400 MPa and Sy = 220 MPa. Then

(c) From part (a), τ = 56.1 MPa. For an E6010 electrode Table 9-6 gives the allowable shear stress τall as 124 MPa, Since τ < τall , the weld is satisfactory. Since the code already has a design factor of 0.577(345)/124 = 1.6 included at the equality, the corresponding factor of safety to part (a) is

which is consistent.

78.22.79

220

MPa 2.796/)50(10)150(2200

6// 22

===

====

σ

σ

ySn

bdM

cIM

Since n < nd, that is, 2.78 < 3.0, the joint is unsatisfactory as to the attachment strength.Since n < nd, that is, 2.78 < 3.0, the joint is unsatisfactory as to the attachment strength.

54.31.56

1246.1 =⎟⎠⎞

⎜⎝⎛=n

48


9.8 – Fatigue Loading



• The conventional methods will be provided here.

• In fatigue, the Gerber criterion is best; however, you will find that the Goodman criterion is in common use.

• Recall, that the fatigue stress concentration factors are given in Table 9-5.

• For Welding codes, see the fatigue stress allowable in the AISI manual.

49


9.8 – Fatigue Loading – cont…



Example 9-5

The 1018 steel strap of Fig. 9-13 has a 4500-N, completely reversed load applied. Determine the factor of safety of the weldment for infinite life.

Figure 9-1350





Solution 9-5

From Table A-20 for the 1018 attachment metal the strengths are Sut = 400 MPa and Sy = 220 MPa. For the E6010 electrode, Sut = 430 MPa and Sy = 340 MPa. The fatigue stress-concentration factor, from Table 9-5, is Kfs = 2.7. From Table 6-2, ka = 272(400)-0.995 = 0.7. The shear area is:

For a uniform shear stress on the throat, kb = 1

For torsion (shear), kc = 0.59 kd = ke = kf = 1

Sse = 0.70(1)(0.59)(1)(1)(1) [(0.5)(400)] = 82.8 MPa

Kfs = 2.7 Fa = 4500 N Fm = 0

Only primary shear is present:

In the absence of a midrange component, the fatigue factor of safety nf is given by

2mm 707)50)(10)(707.0(2 ==A

MPa 0' ; MPa 2.17707

)4500(7.2' ==== mafs

a AFK

ττ

81.42.178.82

'===

a

sef

Snτ

51





Example 9-6

The 1018 steel strap of Fig. 9-14 has a repeatedly applied load of 9000 N (Fa = Fm = 4500 N). Determine the fatigue factor of safety fatigue strength of the weldment.

Figure 9-14

52





Solution 9-6

From Table 6-2, ka = 272(400)-0.995 = 0.7. The shear area is:

For uniform shear stress on the throat, kb = 1

For torsion (shear), kc = 0.59 kd = ke = kf = 1

Sse = 0.70(1)(0.59)(1)(1)(1) [(0.5)(400)] = 82.8 MPa

From Table 9-5, Kfs = 2. Only primary shear is present:

From Eq. (6-53), p. 317 (in text book). Ssu = 0.67 Sut. This together with the Gerber Fatigue Failure criterion for shear stresses from Table 6-7, p. 307 (in text book) gives,

2mm 707)50)(10)(707.0(2 ==A

MPa 7.12707

)4500(2'' ====AFK afs

ma ττ

99.57.12)400(67.0

8.82)7.12(2118.827.12

7.12)400(67.0

21

67.021167.0

21

2222

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++−⎟

⎠⎞

⎜⎝⎛=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛++−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

aut

sem

se

a

m

utf S

SS

Snτ

τττ

53


9.9 – Resistance Welding



• The heating and consequent welding that occur when an electric current is passed through several parts that are pressed together is called resistance welding. Spot welding and seam welding are forms of resistance welding most often used.

• Failure of a resistance weld occurs either by shearing of the weld or by tearing of the metal around the weld.

• Somewhat larger factors of safety should be used when parts are fastened by spot welding rather than by bolts or rivets, to account for the metallurgical changes in the materials due to the welding.

Figure 9-15 : (a) Spot welding ; (b) seam welding

54


9.10 – Adhesive Bonding



• Adhesive bonding has unique advantages

• Reduced weight, sealing capabilities, reduced part count, reduced assembly time, improved fatigue and corrosion resistance, reduced stress concentration associated with bolt holes

Figure 9-16 : Diagram of an automobile body showing at least 15 locations at which adhesives and sealants could be used or are being used.

55


9.10 – Adhesive Bonding – cont…



Types of Adhesive

Structural adhesives are relatively strong adhesives that are normally used well below their glass transition temperature (e.g. epoxies and certain acrylic).

Contact adhesives, provides a solution or emulsion containing an elastomericadhesive is coated onto both adherends where failure would be less critical such for aesthetic purposes (e.g. rubber cement).

Pressure-sensitive adhesives are very low modulus elastomers that deform easily under small pressures (e.g. tapes and labels for nonstructural application).

Anaerobic adhesives cure within narrow spaces poor of oxygen (e.g. thread locking, pipe sealing and gasketing).

56


9.10 – Adhesive Bonding – cont…



Table 9-7 : Mechanical Performance of Various Types of Adhesives

Useful Tables Appendix AAppendix Outline

A–1 Standard SI Prefixes 961

A–2 Conversion Factors 962

A–3 Optional SI Units for Bending, Torsion, Axial, and Direct Shear Stresses 963

A–4 Optional SI Units for Bending and Torsional Deflections 963

A–5 Physical Constants of Materials 963

A–6 Properties of Structural-Steel Angles 964

A–7 Properties of Structural-Steel Channels 966

A–8 Properties of Round Tubing 968

A–9 Shear, Moment, and Deflection of Beams 969

A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution 977

A–11 A Selection of International Tolerance Grades—Metric Series 978

A–12 Fundamental Deviations for Shafts—Metric Series 979

A–13 A Selection of International Tolerance Grades—Inch Series 980

A–14 Fundamental Deviations for Shafts—Inch Series 981

A–15 Charts of Theoretical Stress-Concentration Factors Kt 982

A–16 Approximate Stress-Concentration Factors Kt and Kts for Bending a Round Baror Tube with a Transverse Round Hole 987

A–17 Preferred Sizes and Renard (R-series) Numbers 989

A–18 Geometric Properties 990

A–19 American Standard Pipe 993

A–20 Deterministic ASTM Minimum Tensile and Yield Strengths for HR and CD Steels 994

A–21 Mean Mechanical Properties of Some Heat-Treated Steels 995

A–22 Results of Tensile Tests of Some Metals 997

A–23 Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels 998

A–24 Mechanical Properties of Three Non-Steel Metals 1000

A–25 Stochastic Yield and Ultimate Strengths for Selected Materials 1002

A–26 Stochastic Parameters from Finite Life Fatigue Tests in Selected Metals 1003

959

shi20361_app_A.qxd 6/3/03 3:42 PM Page 959

A–27 Finite Life Fatigue Strengths of Selected Plain Carbon Steels 1004

A–28 Decimal Equivalents of Wire and Sheet-Metal Gauges 1005

A–29 Dimensions of Square and Hexagonal Bolts 1007

A–30 Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws 1008

A–31 Dimensions of Hexagonal Nuts 1009

A–32 Basic Dimensions of American Standard Plain Washers 1010

A–33 Dimensions of Metric Plain Washers 1011

A–34 Gamma Function 1012

960 Mechanical Engineering Design


Useful Tables 961

Name Symbol Factor

exa E 1 000 000 000 000 000 000 =1018

peta P 1 000 000 000 000 000 =1015

tera T 1 000 000 000 000 =1012

giga G 1 000 000 000 =109

mega M 1 000 000 =106

kilo k 1 000 =103

hecto‡ h 100 =102

deka‡ da 10 =101

deci‡ d 0.1 =10−1

centi‡ c 0.01 =10−2

milli m 0.001 =10−3

micro µ 0.000 001 =10−6

nano n 0.000 000 001 =10−9

pico p 0.000 000 000 001 =10−12

femto f 0.000 000 000 000 001 =10−15

atto a 0.000 000 000 000 000 001 =10−18

∗ If possible use multiple and submultiple prefixes in steps of 1000.†Spaces are used in SI instead of commas to group numbers to avoid confusion with the practice in some European countriesof using commas for decimal points.‡Not recommended but sometimes encountered.

Table A–1

Standard SI Prefixes∗†



Multiply Input By Factor To Get Output Multiply Input By Factor To Get OutputX A Y X A Y

British thermal 1055 joule, Junit, BtuBtu/second, Btu/s 1.05 kilowatt, kWcalorie 4.19 joule, Jcentimeter of 1.333 kilopascal, kPamercury (0◦C)centipoise, cP 0.001 pascal-second,

Pa · sdegree (angle) 0.0174 radian, radfoot, ft 0.305 meter, mfoot2, ft2 0.0929 meter2, m2

foot/minute, 0.0051 meter/second, m/sft/minfoot-pound, ft · lbf 1.35 joule, Jfoot-pound/ 1.35 watt, Wsecond, ft · lbf/s foot/second, ft/s 0.305 meter/second, m/sgallon (U.S.), gal 3.785 liter, Lhorsepower, hp 0.746 kilowatt, kW inch, in 0.0254 meter, m inch, in 25.4 millimeter, mm inch2, in2 645 millimeter2, mm2

inch of mercury 3.386 kilopascal, kPa(32◦F)kilopound, kip 4.45 kilonewton, kNkilopound/inch2, 6.89 megapascal, MPakpsi (ksi) (N/mm2)mass, lbf · s2/in 175 kilogram, kgmile, mi 1.610 kilometer, km

∗Approximate.†The U.S. Customary system unit of the pound-force is often abbreviated as lbf to distinguish it from the pound-mass, which is abbreviated as lbm.

Table A–2

Conversion Factors A to Convert Input X to Output Y Using the Formula Y = AX∗

mile/hour, mi/h 1.61 kilometer/hour, km/hmile/hour, mi/h 0.447 meter/second, m/smoment of inertia, 0.0421 kilogram-meter2,lbm ·ft2 kg · m2

moment of inertia, 293 kilogram-millimeter2,lbm · in2 kg · mm2

moment of section 41.6 centimeter4, cm4

(second moment of area), in4

ounce-force, oz 0.278 newton, Nounce-mass 0.0311 kilogram, kgpound, lbf† 4.45 newton, Npound-foot, 1.36 newton-meter,lbf · ft N · mpound/foot2, lbf/ft2 47.9 pascal, Papound-inch, lbf · in 0.113 joule, Jpound-inch, lbf · in 0.113 newton-meter,

N · mpound/inch, lbf/in 175 newton/meter, N/mpound/inch2, psi 6.89 kilopascal, kPa(lbf/in2)pound-mass, lbm 0.454 kilogram, kgpound-mass/ 0.454 kilogram/second,second, lbm/s kg/squart (U.S. liquid), qt 946 milliliter, mLsection modulus, in3 16.4 centimeter3, cm3

slug 14.6 kilogram, kgton (short 2000 lbm) 907 kilogram, kgyard, yd 0.914 meter, m


Axial andBending and Torsion Direct Shear

M , T I , J c, r σ, τ F A σ, τ

N · m∗ m4 m Pa N∗ m2 PaN · m cm4 cm MPa (N/mm2) N† mm2 MPa (N/mm2)N · m† mm4 mm GPa kN m2 kPakN · m cm4 cm GPa kN† mm2 GPaN · mm† mm4 mm MPa (N/mm2)

∗Basic relation.†Often preferred.

Bending Deflection Torsional DeflectionF, w l l I E y T l J G θ

N∗ m m4 Pa m N · m∗ m m4 Pa radkN† mm mm4 GPa mm N · m† mm mm4 GPa radkN m m4 GPa µm N · mm mm mm4 MPa (N/mm2) radN mm mm4 kPa m N · m cm cm4 MPa (N/mm2) rad

∗Basic relation.†Often preferred.

Table A–4

Optional SI Units forBending Deflectiony = f (Fl3/El ) or y = f (wl4/El ) andTorsional Deflectionθ = Tl/GJ

Table A–3

Optional SI Units forBending Stressσ = Mc/l, Torsion Stressτ = Tr/J, Axial Stress σ= F/A, and DirectShear Stressτ = F/A

Table A–5

Physical Constants of Materials

Modulus of Modulus ofElasticity E Rigidity G Poisson’s Unit Weight w

Material Mpsi GPa Mpsi GPa Ratio v lbf/in3 lbf/ft3 kN/m3

Aluminum (all alloys) 10.4 71.7 3.9 26.9 0.333 0.098 169 26.6Beryllium copper 18.0 124.0 7.0 48.3 0.285 0.297 513 80.6Brass 15.4 106.0 5.82 40.1 0.324 0.309 534 83.8Carbon steel 30.0 207.0 11.5 79.3 0.292 0.282 487 76.5Cast iron (gray) 14.5 100.0 6.0 41.4 0.211 0.260 450 70.6Copper 17.2 119.0 6.49 44.7 0.326 0.322 556 87.3Douglas fir 1.6 11.0 0.6 4.1 0.33 0.016 28 4.3Glass 6.7 46.2 2.7 18.6 0.245 0.094 162 25.4Inconel 31.0 214.0 11.0 75.8 0.290 0.307 530 83.3Lead 5.3 36.5 1.9 13.1 0.425 0.411 710 111.5Magnesium 6.5 44.8 2.4 16.5 0.350 0.065 112 17.6Molybdenum 48.0 331.0 17.0 117.0 0.307 0.368 636 100.0Monel metal 26.0 179.0 9.5 65.5 0.320 0.319 551 86.6Nickel silver 18.5 127.0 7.0 48.3 0.322 0.316 546 85.8Nickel steel 30.0 207.0 11.5 79.3 0.291 0.280 484 76.0Phosphor bronze 16.1 111.0 6.0 41.4 0.349 0.295 510 80.1Stainless steel (18-8) 27.6 190.0 10.6 73.1 0.305 0.280 484 76.0Titanium alloys 16.5 114.0 6.2 42.4 0.340 0.160 276 43.4

Useful Tables 963



w = weight per foot, lbf/ftm = mass per meter, kg/mA = area, in2 (cm2)I = second moment of area, in4 (cm4)k = radius of gyration, in (cm)y = centroidal distance, in (cm)Z = section modulus, in3, (cm3)

Size, in w A l1−1 k1−1 Z1−1 y k3−3

1 × 1 × 18 0.80 0.234 0.021 0.298 0.029 0.290 0.191

× 14 1.49 0.437 0.036 0.287 0.054 0.336 0.193

1 12 × 1 1

2 × 18 1.23 0.36 0.074 0.45 0.068 0.41 0.29

× 14 2.34 0.69 0.135 0.44 0.130 0.46 0.29

2 × 2 × 18 1.65 0.484 0.190 0.626 0.131 0.546 0.398

× 14 3.19 0.938 0.348 0.609 0.247 0.592 0.391

× 38 4.7 1.36 0.479 0.594 0.351 0.636 0.389

2 12 × 2 1

2 × 14 4.1 1.19 0.703 0.769 0.394 0.717 0.491

× 38 5.9 1.73 0.984 0.753 0.566 0.762 0.487

3 × 3 × 14 4.9 1.44 1.24 0.930 0.577 0.842 0.592

× 38 7.2 2.11 1.76 0.913 0.833 0.888 0.587

× 12 9.4 2.75 2.22 0.898 1.07 0.932 0.584

3 12 × 3 1

2 × 14 5.8 1.69 2.01 1.09 0.794 0.968 0.694

× 38 8.5 2.48 2.87 1.07 1.15 1.01 0.687

× 12 11.1 3.25 3.64 1.06 1.49 1.06 0.683

4 × 4 × 14 6.6 1.94 3.04 1.25 1.05 1.09 0.795

× 38 9.8 2.86 4.36 1.23 1.52 1.14 0.788

× 12 12.8 3.75 5.56 1.22 1.97 1.18 0.782

× 58 15.7 4.61 6.66 1.20 2.40 1.23 0.779

6 × 6 × 38 14.9 4.36 15.4 1.88 3.53 1.64 1.19

× 12 19.6 5.75 19.9 1.86 4.61 1.68 1.18

× 58 24.2 7.11 24.2 1.84 5.66 1.73 1.18

× 34 28.7 8.44 28.2 1.83 6.66 1.78 1.17

Table A–6

Properties of Structural-Steel Angles∗†

1 1

3

3

y


Useful Tables 965

Size, mm m A l1−1 k1−1 Z1−1 y k3−3

25 × 25 × 3 1.11 1.42 0.80 0.75 0.45 0.72 0.48× 4 1.45 1.85 1.01 0.74 0.58 0.76 0.48× 5 1.77 2.26 1.20 0.73 0.71 0.80 0.48

40 × 40 × 4 2.42 3.08 4.47 1.21 1.55 1.12 0.78× 5 2.97 3.79 5.43 1.20 1.91 1.16 0.77× 6 3.52 4.48 6.31 1.19 2.26 1.20 0.77

50 × 50 × 5 3.77 4.80 11.0 1.51 3.05 1.40 0.97× 6 4.47 5.59 12.8 1.50 3.61 1.45 0.97× 8 5.82 7.41 16.3 1.48 4.68 1.52 0.96

60 × 60 × 5 4.57 5.82 19.4 1.82 4.45 1.64 1.17× 6 5.42 6.91 22.8 1.82 5.29 1.69 1.17× 8 7.09 9.03 29.2 1.80 6.89 1.77 1.16× 10 8.69 11.1 34.9 1.78 8.41 1.85 1.16

80 × 80 × 6 7.34 9.35 55.8 2.44 9.57 2.17 1.57× 8 9.63 12.3 72.2 2.43 12.6 2.26 1.56× 10 11.9 15.1 87.5 2.41 15.4 2.34 1.55

100 ×100 × 8 12.2 15.5 145 3.06 19.9 2.74 1.96× 12 17.8 22.7 207 3.02 29.1 2.90 1.94× 15 21.9 27.9 249 2.98 35.6 3.02 1.93

150 × 150 × 10 23.0 29.3 624 4.62 56.9 4.03 2.97× 12 27.3 34.8 737 4.60 67.7 4.12 2.95× 15 33.8 43.0 898 4.57 83.5 4.25 2.93× 18 40.1 51.0 1050 4.54 98.7 4.37 2.92

∗Metric sizes also available in sizes of 45, 70, 90, 120, and 200 mm.†These sizes are also available in aluminum alloy.

Table A–6

Properties of Structural-Steel Angles∗†

(Continued)



a, b = size, in (mm)w = weight per foot, lbf/ftm = mass per meter, kg/mt = web thickness, in (mm)

A = area, in2 (cm2)I = second moment of area, in4 (cm4)k = radius of gyration, in (cm)x = centroidal distance, in (cm)Z = section modulus, in3 (cm3)

a, in b, in t A w l1−1 k1−1 Z1−1 l2−2 k2−2 Z2−2 x

3 1.410 0.170 1.21 4.1 1.66 1.17 1.10 0.197 0.404 0.202 0.4363 1.498 0.258 1.47 5.0 1.85 1.12 1.24 0.247 0.410 0.233 0.4383 1.596 0.356 1.76 6.0 2.07 1.08 1.38 0.305 0.416 0.268 0.4554 1.580 0.180 1.57 5.4 3.85 1.56 1.93 0.319 0.449 0.283 0.4574 1.720 0.321 2.13 7.25 4.59 1.47 2.29 0.433 0.450 0.343 0.4595 1.750 0.190 1.97 6.7 7.49 1.95 3.00 0.479 0.493 0.378 0.4845 1.885 0.325 2.64 9.0 8.90 1.83 3.56 0.632 0.489 0.450 0.4786 1.920 0.200 2.40 8.2 13.1 2.34 4.38 0.693 0.537 0.492 0.5116 2.034 0.314 3.09 10.5 15.2 2.22 5.06 0.866 0.529 0.564 0.4996 2.157 0.437 3.83 13.0 17.4 2.13 5.80 1.05 0.525 0.642 0.5147 2.090 0.210 2.87 9.8 21.3 2.72 6.08 0.968 0.581 0.625 0.5407 2.194 0.314 3.60 12.25 24.2 2.60 6.93 1.17 0.571 0.703 0.5257 2.299 0.419 4.33 14.75 27.2 2.51 7.78 1.38 0.564 0.779 0.5328 2.260 0.220 3.36 11.5 32.3 3.10 8.10 1.30 0.625 0.781 0.5718 2.343 0.303 4.04 13.75 36.2 2.99 9.03 1.53 0.615 0.854 0.5538 2.527 0.487 5.51 18.75 44.0 2.82 11.0 1.98 0.599 1.01 0.5659 2.430 0.230 3.91 13.4 47.7 3.49 10.6 1.75 0.669 0.962 0.6019 2.485 0.285 4.41 15.0 51.0 3.40 11.3 1.93 0.661 1.01 0.5869 2.648 0.448 5.88 20.0 60.9 3.22 13.5 2.42 0.647 1.17 0.583

10 2.600 0.240 4.49 15.3 67.4 3.87 13.5 2.28 0.713 1.16 0.63410 2.739 0.379 5.88 20.0 78.9 3.66 15.8 2.81 0.693 1.32 0.60610 2.886 0.526 7.35 25.0 91.2 3.52 18.2 3.36 0.676 1.48 0.61710 3.033 0.673 8.82 30.0 103 3.43 20.7 3.95 0.669 1.66 0.64912 3.047 0.387 7.35 25.0 144 4.43 24.1 4.47 0.780 1.89 0.67412 3.170 0.510 8.82 30.0 162 4.29 27.0 5.14 0.763 2.06 0.674

Table A–7

Properties of Structural-Steel Channels∗

b

x

a

t

1

2

2

1


Useful Tables 967

a × b, mm m t A I1−1 k1−1 Z1−1 I2−2 k2−2 Z2−2 x

76 × 38 6.70 5.1 8.53 74.14 2.95 19.46 10.66 1.12 4.07 1.19102 × 51 10.42 6.1 13.28 207.7 3.95 40.89 29.10 1.48 8.16 1.51127 × 64 14.90 6.4 18.98 482.5 5.04 75.99 67.23 1.88 15.25 1.94152 × 76 17.88 6.4 22.77 851.5 6.12 111.8 113.8 2.24 21.05 2.21152 × 89 23.84 7.1 30.36 1166 6.20 153.0 215.1 2.66 35.70 2.86178 × 76 20.84 6.6 26.54 1337 7.10 150.4 134.0 2.25 24.72 2.20178 × 89 26.81 7.6 34.15 1753 7.16 197.2 241.0 2.66 39.29 2.76203 × 76 23.82 7.1 30.34 1950 8.02 192.0 151.3 2.23 27.59 2.13203 × 89 29.78 8.1 37.94 2491 8.10 245.2 264.4 2.64 42.34 2.65229 × 76 26.06 7.6 33.20 2610 8.87 228.3 158.7 2.19 28.22 2.00229 × 89 32.76 8.6 41.73 3387 9.01 296.4 285.0 2.61 44.82 2.53254 × 76 28.29 8.1 36.03 3367 9.67 265.1 162.6 2.12 28.21 1.86254 × 89 35.74 9.1 45.42 4448 9.88 350.2 302.4 2.58 46.70 2.42305 × 89 41.69 10.2 53.11 7061 11.5 463.3 325.4 2.48 48.49 2.18305 × 102 46.18 10.2 58.83 8214 11.8 539.0 499.5 2.91 66.59 2.66

∗These sizes are also available in aluminum alloy.

Table A–7

Properties of Structural-Steel Channels (Continued)



wa = unit weight of aluminum tubing, lbf/ftws = unit weight of steel tubing, lbf/ftm = unit mass, kg/mA = area, in2 (cm2)I = second moment of area, in4 (cm4)J = second polar moment of area, in4 (cm4)k = radius of gyration, in (cm)Z = section modulus, in3 (cm3)

d, t = size (OD) and thickness, in (mm)

Size, in wa ws A l k Z J

1 × 18 0.416 1.128 0.344 0.034 0.313 0.067 0.067

1 × 14 0.713 2.003 0.589 0.046 0.280 0.092 0.092

1 12 × 1

8 0.653 1.769 0.540 0.129 0.488 0.172 0.257

1 12 × 1

4 1.188 3.338 0.982 0.199 0.451 0.266 0.399

2 × 18 0.891 2.670 0.736 0.325 0.664 0.325 0.650

2 × 14 1.663 4.673 1.374 0.537 0.625 0.537 1.074

2 12 × 1

8 1.129 3.050 0.933 0.660 0.841 0.528 1.319

2 12 × 1

4 2.138 6.008 1.767 1.132 0.800 0.906 2.276

3 × 14 2.614 7.343 2.160 2.059 0.976 1.373 4.117

3 × 38 3.742 10.51 3.093 2.718 0.938 1.812 5.436

4 × 316 2.717 7.654 2.246 4.090 1.350 2.045 8.180

4 × 38 5.167 14.52 4.271 7.090 1.289 3.544 14.180

Size, mm m A l k Z J

12 × 2 0.490 0.628 0.082 0.361 0.136 0.163

16 × 2 0.687 0.879 0.220 0.500 0.275 0.440

16 × 3 0.956 1.225 0.273 0.472 0.341 0.545

20 × 4 1.569 2.010 0.684 0.583 0.684 1.367

25 × 4 2.060 2.638 1.508 0.756 1.206 3.015

25 × 5 2.452 3.140 1.669 0.729 1.336 3.338

30 × 4 2.550 3.266 2.827 0.930 1.885 5.652

30 × 5 3.065 3.925 3.192 0.901 2.128 6.381

42 × 4 3.727 4.773 8.717 1.351 4.151 17.430

42 × 5 4.536 5.809 10.130 1.320 4.825 20.255

50 × 4 4.512 5.778 15.409 1.632 6.164 30.810

50 × 5 5.517 7.065 18.118 1.601 7.247 36.226

Table A–8

Properties of RoundTubing


Useful Tables 969

Table A–9

Shear, Moment, andDeflection of Beams(Note: Force andmoment reactions arepositive in the directionsshown; equations forshear force V andbending moment Mfollow the signconventions given inSec. 4–2.)

1 Cantilever—end load

R1 = V = F M1 = Fl

M = F(x − l)

y = Fx2

6E I(x − 3l)

ymax = − Fl3

3E I

2 Cantilever—intermediate load

R1 = V = F M1 = Fa

MA B = F(x − a) MBC = 0

yA B = Fx2

6E I(x − 3a)

yBC = Fa2

6E I(a − 3x)

ymax = Fa2

6E I(a − 3l)

x

F

l

y

R1

M1

x

V

+

x

M

–

x

F

CBA

l

y

R1

M1

a b

x

V

+

x

M

–

(continued)



Table A–9

Shear, Moment, andDeflection of Beams(Continued)(Note: Force andmoment reactions arepositive in the directionsshown; equations forshear force V andbending moment Mfollow the signconventions given inSec. 4–2.)

3 Cantilever—uniform load

R1 = wl M1 = wl2

2

V = w(l − x) M = −w

2(l − x)2

y = wx2

24E I(4lx − x2 − 6l2)

ymax = − wl4

8E I

4 Cantilever—moment load

R1 = 0 M1 = MB M = MB

y = MB x2

2E Iymax = MB l2

2E I

x

l

w

y

R1

M1

x

V

+

x

M

–

MB

xB

A

l

y

R1

M1

x

V

x

M


Useful Tables 971

5 Simple supports—center load

R1 = R2 = F

2VA B = R1

VA B = R1 VBC = −R2

MA B = Fx

2MBC = F

2(l − x)

yA B = Fx

48E I(4x2 − 3l2)

ymax = − Fl3

48E I

6 Simple supports—intermediate load

R1 = Fb

lR2 = Fa

l


MA B = Fbx

lMBC = Fa

l(l − x)

yA B = Fbx

6E Il(x2 + b2 − l2)

yBC = Fa(l − x)

6E Il(x2 + a2 − 2lx)

Table A–9


x

F

CBA

l

y

R1 R2

l / 2

x

V

+

–

x

M

+

x

F

CB

a

A

l

y

R1 R2

b

x

V

+

–

x

M

+

(continued)



7 Simple supports—uniform load

R1 = R2 = wl

2V = wl

2− wx

M = wx

2(l − x)

y = wx

24E I(2lx2 − x3 − l3)

ymax = − 5wl4

384E I

8 Simple supports—moment load

R1 = R2 = MB

lV = MB

l

MA B = MB x

lMBC = MB

l(x − l)

yA B = MB x

6E Il(x2 + 3a2 − 6al + 2l2)

yBC = MB

6E Il[x3 − 3lx2 + x(2l2 + 3a2) − 3a2l]

Table A–9


x

l

w

y

R1 R2

x

V

+

–

x

M

+

xC

BA

a

l

y

R1

R2

b

MB

x

V

+

x

M

+

–


Useful Tables 973

Table A–9


9 Simple supports—twin loads

R1 = R2 = F VA B = F VBC = 0

VC D = −F

MA B = Fx MBC = Fa MC D = F(l − x)

yA B = Fx

6E I(x2 + 3a2 − 3la)

yBC = Fa

6E I(3x2 + a2 − 3lx)

ymax = Fa

24E I(4a2 − 3l2)

10 Simple supports—overhanging load

R1 = Fa

lR2 = F

l(l + a)

VA B = − Fa

lVBC = F

MA B = − Fax

lMBC = F(x − l − a)

yA B = Fax

6E Il(l2 − x2)

yBC = F(x − l)

6E I[(x − l)2 − a(3x − l)]

yc = − Fa2

3E I(l + a)

x

F F

DB C

a

A

l

y

R1 R2

a

x

V

+

–

x

M

+

x

F

CBA

y

R2

R1

al

x

V

+

–

x

M

–

(continued)



11 One fixed and one simple support—center load

R1 = 11F

16R2 = 5F

16M1 = 3Fl

16


MA B = F

16(11x − 3l) MBC = 5F

16(l − x)

yA B = Fx2

96E I(11x − 9l)

yBC = F(l − x)

96E I(5x2 + 2l2 − 10lx)

12 One fixed and one simple support—intermediate load

R1 = Fb

2l3(3l2 − b2) R2 = Fa2

2l3(3l − a)

M1 = Fb

2l2(l2 − b2)


MA B = Fb

2l3[b2l − l3 + x(3l2 − b2)]

MBC = Fa2

2l3(3l2 − 3lx − al + ax)

yA B = Fbx2

12E Il3[3l(b2 − l2) + x(3l2 − b2)]

yBC = yA B − F(x − a)3

6E I

Table A–9


xCA

ly

R2

B

F

R1

M1

l / 2

x

V

+

–

x

M

+

–

xCA

ly

R2

B

Fa b

R1

M1

x

V

+

–

x

M

+

–


Useful Tables 975

13 One fixed and one simple support—uniform load

R1 = 5wl

8R2 = 3wl

8M1 = wl2

8

V = 5wl

8− wx

M = −w

8(4x2 − 5lx + l2)

y = wx2

48E I(l − x)(2x − 3l)

ymax = − wl4

185E I

14 Fixed supports—center load

R1 = R2 = F

2M1 = M2 = Fl

8

VA B = −VBC = F

2

MA B = F

8(4x − l) MBC = F

8(3l − 4x)

yA B = Fx2

48E I(4x − 3l)

ymax = − Fl3

192E I

Table A–9


x

l

y

R1

R2M1ymax

0.4215l

x

V

+

–

5l / 8

x

M

+

–

l /4

x

l

y

A B

F

C

R1 R2

M1 M2

l / 2

x

V

+

–

x

M

+

– –

(continued)



15 Fixed supports—intermediate load

R1 = Fb2

l3(3a + b) R2 = Fa2

l3(3b + a)

M1 = Fab2

l2M2 = Fa2b

l2


MA B = Fb2

l3[x(3a + b) − al]

MBC = MA B − F(x − a)

yA B = Fb2 x2

6E Il3[x(3a + b) − 3al]

yBC = Fa2(l − x)2

6E Il3[(l − x)(3b + a) − 3bl]

16 Fixed supports—uniform load

R1 = R2 = wl

2M1 = M2 = wl2

12

V = w

2(l − 2x)

M = w

12(6lx − 6x2 − l2)

y = − wx2

24E I(l − x)2

ymax = − wl4

384E I

Table A–9


l

a

y

A B

F

xC

R1 R2

M1 M2

b

x

V

+

–

x

M

+

– –

x

l

y

R1 R2

M1 M2

x

V

+

–

M

x+

– –

0.2113l


Useful Tables 977

Zα 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.46410.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.42470.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.38590.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.34830.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3238 0.3192 0.3156 0.31210.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.27760.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.24510.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.21480.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.18670.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.16111.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.13791.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.11701.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.09851.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.08231.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.06811.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.05591.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.04551.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.03671.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.02941.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.02332.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.01832.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.01432.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.01102.3 0.0107 0.0104 0.0102 0.00990 0.00964 0.00939 0.00914 0.00889 0.00866 0.008422.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.006392.5 0.00621 0.00604 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.004802.6 0.00466 0.00453 0.00440 0.00427 0.00415 0.00402 0.00391 0.00379 0.00368 0.003572.7 0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.002642.8 0.00256 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.00199 0.001932.9 0.00187 0.00181 0.00175 0.00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139

Table A–10

Cumulative Distribution Function of Normal (Gaussian) Distribution

�(zα ) =∫ zα

−∞

1√2π

exp

(−u2

2

)du

={

α zα ≤ 01 − α zα > 0

�(z�)

f (z)

�

0 z�

(continued)



Table A–10

Cumulative Distribution Function of Normal (Gaussian) Distribution (Continued)

Zα 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

3 0.00135 0.03968 0.03687 0.03483 0.03337 0.03233 0.03159 0.03108 0.04723 0.044814 0.04317 0.04207 0.04133 0.05854 0.05541 0.05340 0.05211 0.05130 0.06793 0.064795 0.06287 0.06170 0.07996 0.07579 0.07333 0.07190 0.07107 0.08599 0.08332 0.081826 0.09987 0.09530 0.09282 0.09149 0.010777 0.010402 0.010206 0.010104 0.011523 0.011260

zα −1.282 −1.643 −1.960 −2.326 −2.576 −3.090 −3.291 −3.891 −4.417F(zα) 0.10 0.05 0.025 0.010 0.005 0.001 0.005 0.00005 0.000005R(zα) 0.90 0.95 0.975 0.999 0.995 0.999 0.9995 0.9999 0.999995

Basic Tolerance GradesSizes IT6 IT7 IT8 IT9 IT10 IT11

0–3 0.006 0.010 0.014 0.025 0.040 0.0603–6 0.008 0.012 0.018 0.030 0.048 0.0756–10 0.009 0.015 0.022 0.036 0.058 0.090

10–18 0.011 0.018 0.027 0.043 0.070 0.11018–30 0.013 0.021 0.033 0.052 0.084 0.13030–50 0.016 0.025 0.039 0.062 0.100 0.16050–80 0.019 0.030 0.046 0.074 0.120 0.19080–120 0.022 0.035 0.054 0.087 0.140 0.220

120–180 0.025 0.040 0.063 0.100 0.160 0.250180–250 0.029 0.046 0.072 0.115 0.185 0.290250–315 0.032 0.052 0.081 0.130 0.210 0.320315–400 0.036 0.057 0.089 0.140 0.230 0.360

Table A–11

A Selection ofInternational ToleranceGrades—Metric Series(Size Ranges Are forOver the Lower Limitand Including the UpperLimit. All Values Arein Millimeters)Source: Preferred Metric Limitsand Fits, ANSI B4.2-1978.See also BSI 4500.


Useful Tables 979

Table A–12

Fundamental Deviations for Shafts—Metric Series(Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters)Source: Preferred Metric Limits and Fits , ANSI B4.2-1978. See also BSI 4500.

Basic Upper-Deviation Letter Lower-Deviation LetterSizes c d f g h k n p s u

0–3 −0.060 −0.020 −0.006 −0.002 0 0 +0.004 +0.006 +0.014 +0.0183–6 −0.070 −0.030 −0.010 −0.004 0 +0.001 +0.008 +0.012 +0.019 +0.0236–10 −0.080 −0.040 −0.013 −0.005 0 +0.001 +0.010 +0.015 +0.023 +0.028

10–14 −0.095 −0.050 −0.016 −0.006 0 +0.001 +0.012 +0.018 +0.028 +0.03314–18 −0.095 −0.050 −0.016 −0.006 0 +0.001 +0.012 +0.018 +0.028 +0.03318–24 −0.110 −0.065 −0.020 −0.007 0 +0.002 +0.015 +0.022 +0.035 +0.04124–30 −0.110 −0.065 −0.020 −0.007 0 +0.002 +0.015 +0.022 +0.035 +0.04830–40 −0.120 −0.080 −0.025 −0.009 0 +0.002 +0.017 +0.026 +0.043 +0.06040–50 −0.130 −0.080 −0.025 −0.009 0 +0.002 +0.017 +0.026 +0.043 +0.07050–65 −0.140 −0.100 −0.030 −0.010 0 +0.002 +0.020 +0.032 +0.053 +0.08765–80 −0.150 −0.100 −0.030 −0.010 0 +0.002 +0.020 +0.032 +0.059 +0.10280–100 −0.170 −0.120 −0.036 −0.012 0 +0.003 +0.023 +0.037 +0.071 +0.124

100–120 −0.180 −0.120 −0.036 −0.012 0 +0.003 +0.023 +0.037 +0.079 +0.144120–140 −0.200 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.092 +0.170140–160 −0.210 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.100 +0.190160–180 −0.230 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.108 +0.210180–200 −0.240 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.122 +0.236200–225 −0.260 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.130 +0.258225–250 −0.280 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.140 +0.284250–280 −0.300 −0.190 −0.056 −0.017 0 +0.004 +0.034 +0.056 +0.158 +0.315280–315 −0.330 −0.190 −0.056 −0.017 0 +0.004 +0.034 +0.056 +0.170 +0.350315–355 −0.360 −0.210 −0.062 −0.018 0 +0.004 +0.037 +0.062 +0.190 +0.390355–400 −0.400 −0.210 −0.062 −0.018 0 +0.004 +0.037 +0.062 +0.208 +0.435



Basic Tolerance GradesSizes IT6 IT7 IT8 IT9 IT10 IT11

0–0.12 0.0002 0.0004 0.0006 0.0010 0.0016 0.00240.12–0.24 0.0003 0.0005 0.0007 0.0012 0.0019 0.00300.24–0.40 0.0004 0.0006 0.0009 0.0014 0.0023 0.00350.40–0.72 0.0004 0.0007 0.0011 0.0017 0.0028 0.00430.72–1.20 0.0005 0.0008 0.0013 0.0020 0.0033 0.00511.20–2.00 0.0006 0.0010 0.0015 0.0024 0.0039 0.00632.00–3.20 0.0007 0.0012 0.0018 0.0029 0.0047 0.00753.20–4.80 0.0009 0.0014 0.0021 0.0034 0.0055 0.00874.80–7.20 0.0010 0.0016 0.0025 0.0039 0.0063 0.00987.20–10.00 0.0011 0.0018 0.0028 0.0045 0.0073 0.0114

10.00–12.60 0.0013 0.0020 0.0032 0.0051 0.0083 0.012612.60–16.00 0.0014 0.0022 0.0035 0.0055 0.0091 0.0142

Table A–13

A Selection ofInternational ToleranceGrades—Inch Series(Size Ranges Are forOver the Lower Limit and Including the UpperLimit. All Values Are inInches, Converted fromTable A–11)


981

Basi

cU

pper

-Dev

iation L

ette

rLo

wer

-Dev

iation L

ette

rSi

zes

cd

fg

hk

np

su

0–0.

12−0

.002

4−0

.000

8−0

.000

2−0

.000

10

0+0

.000

2+0

.000

2+0

.000

6+0

.000

70.

12–0

.24

−0.0

028

−0.0

012

−0.0

004

−0.0

002

00

+0.0

003

+0.0

005

+0.0

007

+0.0

009

0.24

–0.4

0−0

.003

1−0

.001

6−0

.000

5−0

.000

20

0+0

.000

4+0

.000

6+0

.000

9+0

.001

10.

40–0

.72

−0.0

037

−0.0

020

−0.0

006

−0.0

002

00

+0.0

005

+0.0

007

+0.0

011

+0.0

013

0.72

–0.9

6−0

.004

3−0

.002

6−0

.000

8−0

.000

30

+0.0

001

+0.0

006

+0.0

009

+0.0

014

+0.0

016

0.96

–1.2

0−0

.004

3−0

.002

6−0

.000

8−0

.000

30

+0.0

001

+0.0

006

+0.0

009

+0.0

014

+0.0

019

1.20

–1.6

0−0

.004

7−0

.003

1−0

.001

0−0

.000

40

+0.0

001

+0.0

007

+0.0

010

+0.0

017

+0.0

024

1.60

–2.0

0−0

.005

1−0

.003

1−0

.001

0−0

.000

40

+0.0

001

+0.0

007

+0.0

010

+0.0

017

+0.0

028

2.00

–2.6

0−0

.005

5−0

.003

9−0

.001

2−0

.000

40

+0.0

001

+0.0

008

+0.0

013

+0.0

021

+0.0

034

2.60

–3.2

0−0

.005

9−0

.003

9−0

.001

2−0

.000

40

+0.0

001

+0.0

008

+0.0

013

+0.0

023

+0.0

040

3.20

–4.0

0−0

.006

7−0

.004

7−0

.001

4−0

.000

50

+0.0

001

+0.0

009

+0.0

015

+0.0

028

+0.0

049

4.00

–4.8

0−0

.007

1−0

.004

7−0

.001

4−0

.000

50

+0.0

001

+0.0

009

+0.0

015

+0.0

031

+0.0

057

4.80

–5.6

0−0

.007

9−0

.005

7−0

.001

7−0

.000

60

+0.0

001

+0.0

011

+0.0

017

+0.0

036

+0.0

067

5.60

–6.4

0−0

.008

3−0

.005

7−0

.001

7−0

.000

60

+0.0

001

+0.0

011

+0.0

017

+0.0

039

+0.0

075

6.40

–7.2

0−0

.009

1−0

.005

7−0

.001

7−0

.000

60

+0.0

001

+0.0

011

+0.0

017

+0.0

043

+0.0

083

7.20

–8.0

0−0

.009

4−0

.006

7−0

.002

0−0

.000

60

+0.0

002

+0.0

012

+0.0

020

+0.0

048

+0.0

093

8.00

–9.0

0−0

.010

2−0

.006

7−0

.002

0−0

.000

60

+0.0

002

+0.0

012

+0.0

020

+0.0

051

+0.0

102

9.00

–10.

00−0

.011

0−0

.006

7−0

.002

0−0

.000

60

+0.0

002

+0.0

012

+0.0

020

+0.0

055

+0.0

112

10.0

0–11

.20

−0.0

118

−0.0

075

−0.0

022

−0.0

007

0+0

.000

2+0

.001

3+0

.002

2+0

.006

2+0

.012

411

.20–

12.6

0−0

.013

0−0

.007

5−0

.002

2−0

.000

70

+0.0

002

+0.0

013

+0.0

022

+0.0

067

+0.0

130

12.6

0–14

.20

−0.0

142

−0.0

083

−0.0

024

−0.0

007

0+0

.000

2+0

.001

5+0

.002

4+0

.007

5+0

.015

414

.20–

16.0

0−0

.015

7−0

.008

3−0

.002

4−0

.000

70

+0.0

002

+0.0

015

+0.0

024

+0.0

082

+0.0

171

Table

A–1

4

Fund

amen

tal D

evia

tions

for S

hafts

—In

ch S

erie

s (S

ize

Rang

es A

re fo

r Ove

rthe

Low

er L

imit

and

Incl

udin

gth

e U

pper

Lim

it. A

ll Va

lues

Are

inIn

ches

, Con

verte

d fro

m T

able

A–1

2)



Table A–15

Charts of Theoretical Stress-Concentration Factors K*t

Figure A–15–1

Bar in tension or simplecompression with a transversehole. σ0 = F/A, whereA = (w − d )t and t is thethickness.

Kt

d

d/w0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

2.0

2.2

2.4

2.6

2.8

3.0

w

Figure A–15–2

Rectangular bar with atransverse hole in bending.σ0 = Mc/I, whereI = (w − d )h3

/12.

Kt

d

d/w0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

1.0

1.4

1.8

2.2

2.6

3.0

w

MM0.25

1.0

2.0

�

d /h = 0

0.5h

Kt

r

r /d0

1.5

1.2

1.1

1.05

1.0

1.4

1.8

2.2

2.6

3.0

dww /d = 3

0.05 0.10 0.15 0.20 0.25 0.30

Figure A–15–3

Notched rectangular bar intension or simple compression.σ0 = F/A, where A = dt and tis the thickness.


Useful Tables 983

Table A–15

Charts of Theoretical Stress-Concentration Factors K*t (Continued)

1.5

1.10

1.05

1.02

w/d = �

Kt

r

r /d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0

dw MM

1.02

Kt

r/d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0

r

dD

D/d = 1.50

1.05

1.10

Kt

r/d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0

r

dD

D/d = 1.02

3

1.31.1

1.05 MM

Figure A–15–4

Notched rectangular bar inbending. σ0 = Mc/I, wherec = d/2, I = td 3/12, and t isthe thickness.

Figure A–15–5

Rectangular filleted bar intension or simple compression.σ0 = F/A, where A = dt and tis the thickness.

Figure A–15–6

Rectangular filleted bar inbending. σ0 = Mc/I, wherec = d/2, I = td3

/12, t is thethickness.

*Factors from R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951,p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

(continued)



Table A–15


Figure A–15–7

Round shaft with shoulder filletin tension. σ0 = F/A, whereA = πd 2/4.

Figure A–15–8

Round shaft with shoulder filletin torsion. τ0 = Tc/J, wherec = d/2 and J = πd4

/32.

Figure A–15–9

Round shaft with shoulder filletin bending. σ0 = Mc/I, wherec = d/2 and I = πd4/64.

Kt

r/d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

r

1.05

1.02

1.10

D/d = 1.50

dD

Kts

r/d

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.4

1.8

2.2

2.6

3.0

D/d = 21.09

1.20 1.33

r

TTD d

Kt

r/d

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.4

1.8

2.2

2.6

3.0

D/d = 3

1.02

1.5

1.10

1.05

r

MD dM


Useful Tables 985

Table A–15


Figure A–15–10

Round shaft in torsion withtransverse hole.

Figure A–15–11

Round shaft in bending witha transverse hole. σ0 =M/[(πD3/32) − (dD2

/6)],approximately.

Kts

d /D0 0.05 0.10 0.15 0.20 0.25 0.30

2.4

2.8

3.2

3.6

4.0

Jc

TB

d

�D3

16dD2

6= – (approx)

AD

Kts, A

Kts, B

Kt

d /D0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0d

D

MM

Figure A–15–12

Plate loaded in tension by apin through a hole. σ0 = F/A,where A = (w − d)t . Whenclearance exists, increase Kt

35 to 50 percent. (M. M. Frochtand H. N. Hill, “Stress Concentration Factorsaround a Central Circular Hole in a PlateLoaded through a Pin in Hole,” J. Appl.Mechanics, vol. 7, no. 1, March 1940, p. A-5.)

dh

t

Kt

d /w0 0.1 0.2 0.3 0.4 0.60.5 0.80.7

1

3

5

7

9

11

w

h/w = 0.35

h/w � 1.0

h/w = 0.50

(continued)



Table A–15




Figure A–15–13

Grooved round bar in tension.σ0 = F/A, whereA = πd 2/4.

Figure A–15–14

Grooved round bar inbending. σ0 = Mc/l, wherec = d/2 and I = πd4

/64.

Figure A–15–15

Grooved round bar in torsion.τ0 = Tc/J, where c = d/2and J = πd4

/32.

Kt

r /d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0

D/d = 1.50

1.05

1.02

1.15

d

r

D

Kt

r /d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

3.0

D/d = 1.501.02

1.05

d

r

D MM

Kts

r /d0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

1.4

1.8

2.2

2.6

D/d = 1.30

1.02

1.05

d

r

D

TT


Useful Tables 987

Table A–16

Approximate Stress-Concentration Factor Kt

for Bending of a RoundBar or Tube with aTransverse Round HoleSource: R. E. Peterson, StressConcentration Factors, Wiley,New York, 1974, pp. 146,235.

The nominal bending stress is σ0 = M/Znet where Znet is a reduced valueof the section modulus and is defined by

Znet = π A

32D(D4 − d4)

Values of A are listed in the table. Use d = 0 for a solid bar

d/D0.9 0.6 0

a/D A Kt A Kt A Kt

0.050 0.92 2.63 0.91 2.55 0.88 2.420.075 0.89 2.55 0.88 2.43 0.86 2.350.10 0.86 2.49 0.85 2.36 0.83 2.270.125 0.82 2.41 0.82 2.32 0.80 2.200.15 0.79 2.39 0.79 2.29 0.76 2.150.175 0.76 2.38 0.75 2.26 0.72 2.100.20 0.73 2.39 0.72 2.23 0.68 2.070.225 0.69 2.40 0.68 2.21 0.65 2.040.25 0.67 2.42 0.64 2.18 0.61 2.000.275 0.66 2.48 0.61 2.16 0.58 1.970.30 0.64 2.52 0.58 2.14 0.54 1.94

M M

D d

a

(continued)



Table A–16 (Continued)

Approximate Stress-Concentration Factors Kts for a Round Bar or Tube Having a Transverse Round Hole andLoaded in Torsion Source: R. E. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. 148, 244.

TTD a d

The maximum stress occurs on the inside of the hole, slightly below the shaft surface. The nominal shear stress is τ0 = T D/2Jnet ,where Jnet is a reduced value of the second polar moment of area and is defined by

Jnet = π A(D4 − d4)

32

Values of A are listed in the table. Use d = 0 for a solid bar.

d/D0.9 0.8 0.6 0.4 0

a/D A Kts A Kts A Kts A Kts A Kts

0.05 0.96 1.78 0.95 1.770.075 0.95 1.82 0.93 1.710.10 0.94 1.76 0.93 1.74 0.92 1.72 0.92 1.70 0.92 1.680.125 0.91 1.76 0.91 1.74 0.90 1.70 0.90 1.67 0.89 1.640.15 0.90 1.77 0.89 1.75 0.87 1.69 0.87 1.65 0.87 1.620.175 0.89 1.81 0.88 1.76 0.87 1.69 0.86 1.64 0.85 1.600.20 0.88 1.96 0.86 1.79 0.85 1.70 0.84 1.63 0.83 1.580.25 0.87 2.00 0.82 1.86 0.81 1.72 0.80 1.63 0.79 1.540.30 0.80 2.18 0.78 1.97 0.77 1.76 0.75 1.63 0.74 1.510.35 0.77 2.41 0.75 2.09 0.72 1.81 0.69 1.63 0.68 1.470.40 0.72 2.67 0.71 2.25 0.68 1.89 0.64 1.63 0.63 1.44


Useful Tables 989

Table A–17

Preferred Sizes andRenard (R-Series)Numbers (When a choice can bemade, use one of thesesizes; however, not allparts or items areavailable in all the sizesshown in the table.)

Fraction of Inches1

64 , 132 , 1

16 , 332 , 1

8 , 532 , 3

16 , 14 , 5

16 , 38 , 7

16 , 12 , 9

16 , 58 , 11

16 , 34 , 7

8 , 1, 1 14 , 1 1

2 , 1 34 , 2, 2 1

4 ,

2 12 , 2 3

4 , 3, 3 14 , 3 1

2 , 3 34 , 4, 4 1

4 , 4 12 , 4 3

4 , 5, 5 14 , 5 1

2 , 5 34 , 6, 6 1

2 , 7, 7 12 , 8, 8 1

2 , 9, 9 12 ,

10, 1012 , 11, 111

2 , 12, 1212 , 13, 131

2 , 14, 1412 , 15, 151

2 , 16, 1612 , 17, 171

2 , 18,

1812 , 19, 191

2 , 20

Decimal Inches

0.010, 0.012, 0.016, 0.020, 0.025, 0.032, 0.040, 0.05, 0.06, 0.08, 0.10, 0.12, 0.16,0.20, 0.24, 0.30, 0.40, 0.50, 0.60, 0.80, 1.00, 1.20, 1.40, 1.60, 1.80, 2.0, 2.4, 2.6, 2.8, 3.0, 3.2, 3.4, 3.6, 3.8, 4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 7.0, 7.5,8.5, 9.0, 9.5, 10.0, 10.5, 11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5,16.0, 16.5, 17.0, 17.5, 18.0, 18.5, 19.0, 19.5, 20

Millimeters

0.05, 0.06, 0.08, 0.10, 0.12, 0.16, 0.20, 0.25, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80,0.90, 1.0, 1.1, 1.2, 1.4, 1.5, 1.6, 1.8, 2.0, 2.2, 2.5, 2.8, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5,6.0, 6.5, 7.0, 8.0, 9.0, 10, 11, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 35, 40, 45, 50,60, 80, 100, 120, 140, 160, 180, 200, 250, 300

Renard Numbers*

1st choice, R5: 1, 1.6, 2.5, 4, 6.3, 102d choice, R10: 1.25, 2, 3.15, 5, 83d choice, R20: 1.12, 1.4, 1.8, 2.24, 2.8, 3.55, 4.5, 5.6, 7.1, 94th choice, R40: 1.06, 1.18, 1.32, 1.5, 1.7, 1.9, 2.12, 2.36, 2.65, 3, 3.35, 3.75,4.25, 4.75, 5.3, 6, 6.7, 7.5, 8.5, 9.5

*May be multiplied or divided by powers of 10.



Part 1 Properties of Sections

A = area

G = location of centroid

Ix =∫

x2 d A = second moment of area about x axis

Ix y =∫

xy d A = mixed moment of area about x and y axes

JG =∫

r2 d A =∫

(x2 + y2) d A = Ix + Iy

= second polar moment of area about axis through G

k2x = Ix /A = squared radius of gyration about x axis

Rectangle

A = bh Ix = bh3

12Iy = b3h

12Ix y = 0

Circle

A = π D2

4Ix = Iy = π D4

64Ix y = 0

Hollow circle

A = π

4(D2 − d2) Ix = Iy = π

64(D4 − d4) Ix y = 0

Table A–18

Geometric Properties

b

h x

yb2

h2

G

x

y

G

D

x

y

G

Dd


Useful Tables 991

Table A–18

Geometric Properties(Continued)

Right triangles

A = bh

2Ix = bh3

36Iy = b3h

36Ix y = −b2h2

72

Right triangles

A = bh

2Ix = bh3

36Iy = b3h

36Ix y = b2h2

72

Quarter-circles

A = πr2

4Ix = Iy = r4

(π

16− 4

9π

)Ix y = r4

(1

8− 4

9π

)

Quarter-circles

A = πr2

4Ix = Iy = r4

(π

16− 4

9π

)Ix y = r4

(4

9π− 1

8

)

x

x

G

Gh h

b

b

y y

h3

b3

b3

h3

h

x

xh

b

b

y y

h3

b3

b3

h3

G

G

r

y

x

4r3�

4r3�

r

y

x

4r3�

4r3�

GG

r

y

x

4r3�

4r3�

r

y

x

4r3�

4r3�

G

G

(continued)



Table A–18

Geometric Properties(Continued)

Part 2 Properties of Solids (ρ = Density, Weight per Unit Volume)

Rods

m = πd2lρ

4gIy = Iz = ml2

12

Round disks

m = πd2tρ

4gIx = md2

8Iy = Iz = md2

16

Rectangular prisms

m = abcρ

gIx = m

12(a2 + b2) Iy = m

12(a2 + c2) Iz = m

12(b2 + c2)

Cylinders

m = πd2lρ

4gIx = md2

8Iy = Iz = m

48(3d2 + 4l2)

Hollow cylinders

m = π(d2

o − d2i

)lρ

4gIx = m

8

(d2

o + d2i

)Iy = Iz = m

48

(3d2

o + 3d2i + 4l2

)

y

z

x

dl

y

td

zx

ca

b

xz

y

y

zx

d

l

y

zx

do

di

l


Useful Tables 993

Table A–19

American Standard Pipe

Wall Thickness, inNominal Outside Extra Double

Size, Diameter, Threads Standard Strong Extrain in per inch No. 40 No. 80 Strong

18 0.405 27 0.070 0.09814 0.540 18 0.090 0.12238 0.675 18 0.093 0.12912 0.840 14 0.111 0.151 0.30734 1.050 14 0.115 0.157 0.318

1 1.315 1112 0.136 0.183 0.369

1 14 1.660 111

2 0.143 0.195 0.3931 1

2 1.900 1112 0.148 0.204 0.411

2 2.375 1112 0.158 0.223 0.447

2 12 2.875 8 0.208 0.282 0.565

3 3.500 8 0.221 0.306 0.6153 1

2 4.000 8 0.231 0.3254 4.500 8 0.242 0.344 0.6905 5.563 8 0.263 0.383 0.7686 6.625 8 0.286 0.441 0.8848 8.625 8 0.329 0.510 0.895



Table A–20

Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm (3

4 to 114 in). These

strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform toASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numberingsystem is not a specification. See Table 1–1 for certain ASTM steels.] Source: 1986 SAE Handbook, p. 2.15.

1 2 3 4 5 6 7 8Tensile Yield

SAE and/or Proces- Strength, Strength, Elongation in Reduction in BrinellUNS No. AISI No. sing MPa (kpsi) MPa (kpsi) 2 in, % Area, % Hardness

G10060 1006 HR 300 (43) 170 (24) 30 55 86CD 330 (48) 280 (41) 20 45 95

G10100 1010 HR 320 (47) 180 (26) 28 50 95CD 370 (53) 300 (44) 20 40 105

G10150 1015 HR 340 (50) 190 (27.5) 28 50 101CD 390 (56) 320 (47) 18 40 111

G10180 1018 HR 400 (58) 220 (32) 25 50 116CD 440 (64) 370 (54) 15 40 126

G10200 1020 HR 380 (55) 210 (30) 25 50 111CD 470 (68) 390 (57) 15 40 131

G10300 1030 HR 470 (68) 260 (37.5) 20 42 137CD 520 (76) 440 (64) 12 35 149

G10350 1035 HR 500 (72) 270 (39.5) 18 40 143CD 550 (80) 460 (67) 12 35 163

G10400 1040 HR 520 (76) 290 (42) 18 40 149CD 590 (85) 490 (71) 12 35 170

G10450 1045 HR 570 (82) 310 (45) 16 40 163CD 630 (91) 530 (77) 12 35 179

G10500 1050 HR 620 (90) 340 (49.5) 15 35 179CD 690 (100) 580 (84) 10 30 197

G10600 1060 HR 680 (98) 370 (54) 12 30 201G10800 1080 HR 770 (112) 420 (61.5) 10 25 229G10950 1095 HR 830 (120) 460 (66) 10 25 248


Useful Tables 995

Table A–21

Mean Mechanical Properties of Some Heat-Treated Steels[These are typical properties for materials normalized and annealed. The properties for quenched and tempered(Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. Inall cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gaugelength 2 in. unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American

Society for Metals, Metals Park, Ohio, 1983.


Temperature Strength Strength, Elongation, Reduction BrinellAISI No. Treatment °C (°F) MPa (kpsi) MPa (kpsi) % in Area, % Hardness

1030 Q&T* 205 (400) 848 (123) 648 (94) 17 47 495Q&T* 315 (600) 800 (116) 621 (90) 19 53 401Q&T* 425 (800) 731 (106) 579 (84) 23 60 302Q&T* 540 (1000) 669 (97) 517 (75) 28 65 255Q&T* 650 (1200) 586 (85) 441 (64) 32 70 207Normalized 925 (1700) 521 (75) 345 (50) 32 61 149Annealed 870 (1600) 430 (62) 317 (46) 35 64 137

1040 Q&T 205 (400) 779 (113) 593 (86) 19 48 262Q&T 425 (800) 758 (110) 552 (80) 21 54 241Q&T 650 (1200) 634 (92) 434 (63) 29 65 192Normalized 900 (1650) 590 (86) 374 (54) 28 55 170Annealed 790 (1450) 519 (75) 353 (51) 30 57 149

1050 Q&T* 205 (400) 1120 (163) 807 (117) 9 27 514Q&T* 425 (800) 1090 (158) 793 (115) 13 36 444Q&T* 650 (1200) 717 (104) 538 (78) 28 65 235Normalized 900 (1650) 748 (108) 427 (62) 20 39 217Annealed 790 (1450) 636 (92) 365 (53) 24 40 187

1060 Q&T 425 (800) 1080 (156) 765 (111) 14 41 311Q&T 540 (1000) 965 (140) 669 (97) 17 45 277Q&T 650 (1200) 800 (116) 524 (76) 23 54 229Normalized 900 (1650) 776 (112) 421 (61) 18 37 229Annealed 790 (1450) 626 (91) 372 (54) 22 38 179

1095 Q&T 315 (600) 1260 (183) 813 (118) 10 30 375Q&T 425 (800) 1210 (176) 772 (112) 12 32 363Q&T 540 (1000) 1090 (158) 676 (98) 15 37 321Q&T 650 (1200) 896 (130) 552 (80) 21 47 269Normalized 900 (1650) 1010 (147) 500 (72) 9 13 293Annealed 790 (1450) 658 (95) 380 (55) 13 21 192

1141 Q&T 315 (600) 1460 (212) 1280 (186) 9 32 415Q&T 540 (1000) 896 (130) 765 (111) 18 57 262

(continued)




Temperature Strength Strength, Elongation, Reduction BrinellAISI No. Treatment °C (°F) MPa (kpsi) MPa (kpsi) % in Area, % Hardness

4130 Q&T* 205 (400) 1630 (236) 1460 (212) 10 41 467Q&T* 315 (600) 1500 (217) 1380 (200) 11 43 435Q&T* 425 (800) 1280 (186) 1190 (173) 13 49 380Q&T* 540 (1000) 1030 (150) 910 (132) 17 57 315Q&T* 650 (1200) 814 (118) 703 (102) 22 64 245Normalized 870 (1600) 670 (97) 436 (63) 25 59 197Annealed 865 (1585) 560 (81) 361 (52) 28 56 156

4140 Q&T 205 (400) 1770 (257) 1640 (238) 8 38 510Q&T 315 (600) 1550 (225) 1430 (208) 9 43 445Q&T 425 (800) 1250 (181) 1140 (165) 13 49 370Q&T 540 (1000) 951 (138) 834 (121) 18 58 285Q&T 650 (1200) 758 (110) 655 (95) 22 63 230Normalized 870 (1600) 1020 (148) 655 (95) 18 47 302Annealed 815 (1500) 655 (95) 417 (61) 26 57 197

4340 Q&T 315 (600) 1720 (250) 1590 (230) 10 40 486Q&T 425 (800) 1470 (213) 1360 (198) 10 44 430Q&T 540 (1000) 1170 (170) 1080 (156) 13 51 360Q&T 650 (1200) 965 (140) 855 (124) 19 60 280

*Water-quenched

Table A–21 (Continued)

Mean Mechanical Properties of Some Heat-Treated Steels[These are typical properties for materials normalized and annealed. The properties for quenched and tempered(Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. Inall cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gaugelength 2 in. Unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American

Society for Metals, Metals Park, Ohio, 1983.


997

Table

A–2

2

Resu

lts o

f Ten

sile

Tests

of S

ome

Met

als*

Sour

ce:J

. Dat

sko,

“So

lid M

ater

ials,

” ch

ap. 7

in Jo

seph

E. S

higl

ey a

nd C

harle

s R.

Misc

hke

(eds

.-in-

chie

f), S

tand

ard

Han

dboo

k of

Mac

hine

Des

ign,

2nd

ed.,

McG

raw

-Hill,

New

Yor

k, 1

996,

pp.

7.4

7–7.

50.

Stre

ngth

(Te

nsi

le)

Yie

ldU

ltim

ate

Fract

ure

,Coef

fici

ent

Stra

inS y

,S u

,σ

f,σ

0,

Stre

ngth

,Fr

act

ure

Num

ber

Mate

rial

Conditio

nM

Pa

(kpsi

)M

Pa

(kpsi

)M

Pa

(kpsi

)M

Pa

(kpsi

)Ex

ponen

t m

Stra

in ε

f

1018

Stee

lA

nnea

led

220

(32.

0)34

1(4

9.5)

628

(91.

1)†

620

(90.

0)0.

251.

0511

44St

eel

Ann

eale

d35

8(5

2.0)

646

(93.

7)89

8(1

30)†

992

(144

)0.

140.

4912

12St

eel

HR

193

(28.

0)42

4(6

1.5)

729

(106

)†75

8(1

10)

0.24

0.85

1045

Stee

lQ

&T

600°

F15

20(2

20)

1580

(230

)23

80(3

45)

1880

(273

)†0.

041

0.81

4142

Stee

lQ

&T

600°

F17

20(2

50)

1930

(210

)23

40(3

40)

1760

(255

)†0.

048

0.43

303

Stai

nles

sA

nnea

led

241

(35.

0)60

1(8

7.3)

1520

(221

)†14

10(2

05)

0.51

1.16

steel

304

Stai

nles

sA

nnea

led

276

(40.

0)56

8(8

2.4)

1600

(233

)†12

70(1

85)

0.45

1.67

steel

2011

Alu

min

umT6

169

(24.

5)32

4(4

7.0)

325

(47.

2)†

620

(90)

0.28

0.10

allo

y20

24A

lum

inum

T429

6(4

3.0)

446

(64.

8)53

3(7

7.3)

†68

9(1

00)

0.15

0.18

allo

y70

75A

lum

inum

T654

2(7

8.6)

593

(86.

0)70

6(1

02)†

882

(128

)0.

130.

18al

loy

*Valu

es fr

om on

e or t

wo he

ats an

d beli

eved

to be

attai

nable

using

prop

er pu

rchas

e spe

cifica

tions

. The

frac

ture s

train

may v

ary as

muc

h as 1

00 pe

rcent.

† Deriv

ed va

lue.


Table

A–2

3

Mea

n M

onot

onic

and

Cyc

lic S

tress

-Stra

in P

rope

rties

of S

elec

ted

Stee

lsSo

urce

: ASM

Met

als

Refe

renc

e Bo

ok,2

nd e

d., A

mer

ican

Soc

iety

for M

etal

s, M

etal

s Pa

rk,

Ohi

o, 1

983,

p. 2

17.

True

Fatigue

Tensi

leSt

rain

Stre

ngth

Fatigue

Fatigue

Fatigue

Hard

-St

rength

Red

uct

ion

at

Modulu

s of

Coef

fici

ent

Stre

ngth

Duct

ility

Duct

ility

Ori

enta

-D

escr

iption

nes

sS u

tin

Are

aFr

act

ure

Elast

icity E

σ′ f

Exponen

tCoef

fici

ent

Exponen

tG

rade

(a)

tion (

e)(f

)H

BM

Pa

ksi

%ε

fG

Pa

10

4psi

MPa

ksi

bε

′ Fc

A53

8A (b

)L

STA

405

1515

220

671.

1018

527

1655

240

−0.0

650.

30−0

.62

A53

8B (b

)L

STA

460

1860

270

560.

8218

527

2135

310

−0.0

710.

80−0

.71

A53

8C (b

)L

STA

480

2000

290

550.

8118

026

2240

325

−0.0

70.

60−0

.75

AM

-350

(c)

LH

R, A

1315

191

520.

7419

528

2800

406

−0.1

40.

33−0

.84

AM

-350

(c)

LC

D49

619

0527

620

0.23

180

2626

9039

0−0

.102

0.10

−0.4

2G

aine

x (c

)LT

HR

shee

t53

077

580.

8620

029

.280

511

7−0

.07

0.86

−0.6

5G

aine

x (c

)L

HR

shee

t51

074

641.

0220

029

.280

511

7−0

.071

0.86

−0.6

8H

-11

LA

usfo

rmed

660

2585

375

330.

4020

530

3170

460

−0.0

770.

08−0

.74

RQC

-100

(c)

LTH

R pl

ate

290

940

136

430.

5620

530

1240

180

−0.0

70.

66−0

.69

RQC

-100

(c)

LH

R pl

ate

290

930

135

671.

0220

530

1240

180

−0.0

70.

66−0

.69

10B6

2L

Q&

T43

016

4023

838

0.89

195

2817

8025

8−0

.067

0.32

−0.5

610

05-1

009

LTH

R sh

eet

9036

052

731.

320

530

580

84−0

.09

0.15

−0.4

310

05-1

009

LTC

D s

heet

125

470

6866

1.09

205

3051

575

−0.0

590.

30−0

.51

1005

-100

9L

CD

she

et12

541

560

641.

0220

029

540

78−0

.073

0.11

−0.4

110

05-1

009

LH

R sh

eet

9034

550

801.

620

029

640

93−0

.109

0.10

−0.3

910

15L

Nor

mal

ized

8041

560

681.

1420

530

825

120

−0.1

10.

95−0

.64

1020

LH

R pl

ate

108

440

6462

0.96

205

29.5

895

130

−0.1

20.

41−0

.51

1040

LA

s fo

rged

225

620

9060

0.93

200

2915

4022

3−0

.14

0.61

−0.5

710

45L

Q&

T22

572

510

565

1.04

200

2912

2517

8−0

.095

1.00

−0.6

610

45L

Q&

T41

014

5021

051

0.72

200

2918

6027

0−0

.073

0.60

−0.7

010

45L

Q&

T39

013

4519

559

0.89

205

3015

8523

0−0

.074

0.45

−0.6

810

45L

Q&

T45

015

8523

055

0.81

205

3017

9526

0−0

.07

0.35

−0.6

910

45L

Q&

T50

018

2526

551

0.71

205

3022

7533

0−0

.08

0.25

−0.6

810

45L

Q&

T59

522

4032

541

0.52

205

3027

2539

5−0

.081

0.07

−0.6

011

44L

CD

SR26

593

013

533

0.51

195

28.5

1000

145

−0.0

80.

32−0

.58

998


1144

LD

AT30

510

3515

025

0.29

200

28.8

1585

230

−0.0

90.

27−0

.53

1541

FL

Q&

T fo

rgin

g29

095

013

849

0.68

205

29.9

1275

185

−0.0

760.

68−0

.65

1541

FL

Q&

T fo

rgin

g26

089

012

960

0.93

205

29.9

1275

185

−0.0

710.

93−0

.65

4130

LQ

&T

258

895

130

671.

1222

032

1275

185

−0.0

830.

92−0

.63

4130

LQ

&T

365

1425

207

550.

7920

029

1695

246

−0.0

810.

89−0

.69

4140

LQ

&T,

DAT

310

1075

156

600.

6920

029

.218

2526

5−0

.08

1.2

−0.5

941

42L

DAT

310

1060

154

290.

3520

029

1450

210

−0.1

00.

22−0

.51

4142

LD

AT33

512

5018

128

0.34

200

28.9

1250

181

−0.0

80.

06−0

.62

4142

LQ

&T

380

1415

205

480.

6620

530

1825

265

−0.0

80.

45−0

.75

4142

LQ

&T

and

400

1550

225

470.

6320

029

1895

275

−0.0

90.

50−0

.75

defo

rmed

4142

LQ

&T

450

1760

255

420.

5420

530

2000

290

−0.0

80.

40−0

.73

4142

LQ

&T

and

475

2035

295

200.

2220

029

2070

300

−0.0

820.

20−0

.77

defo

rmed

4142

LQ

&T

and

450

1930

280

370.

4620

029

2105

305

−0.0

90.

60−0

.76

defo

rmed

4142

LQ

&T

475

1930

280

350.

4320

530

2170

315

−0.0

810.

09−0

.61

4142

LQ

&T

560

2240

325

270.

3120

530

2655

385

−0.0

890.

07−0

.76

4340

LH

R, A

243

825

120

430.

5719

528

1200

174

−0.0

950.

45−0

.54

4340

LQ

&T

409

1470

213

380.

4820

029

2000

290

−0.0

910.

48−0

.60

4340

LQ

&T

350

1240

180

570.

8419

528

1655

240

−0.0

760.

73−0

.62

5160

LQ

&T

430

1670

242

420.

8719

528

1930

280

−0.0

710.

40−0

.57

5210

0L

SH, Q

&T

518

2015

292

110.

1220

530

2585

375

−0.0

90.

18−0

.56

9262

LA

260

925

134

140.

1620

530

1040

151

−0.0

710.

16−0

.47

9262

LQ

&T

280

1000

145

330.

4119

528

1220

177

−0.0

730.

41−0

.60

9262

LQ

&T

410

565

227

320.

3820

029

1855

269

−0.0

570.

38−0

.65

950C

(d)

LTH

R pl

ate

159

565

8264

1.03

205

29.6

1170

170

−0.1

20.

95−0

.61

950C

(d)

LH

R ba

r15

056

582

691.

1920

530

970

141

−0.1

10.

85−0

.59

950X

(d)

LPl

ate

chan

nel

150

440

6465

1.06

205

3062

591

−0.0

750.

35−0

.54

950X

(d)

LH

R pl

ate

156

530

7772

1.24

205

29.5

1005

146

−0.1

00.

85−0

.61

950X

(d)

LPl

ate

chan

nel

225

695

101

681.

1519

528

.210

5515

3−0

.08

0.21

−0.5

3

Notes

:(a)

AISI/

SAE g

rade,

unles

s othe

rwise

indic

ated.

(b) A

STM

desig

natio

n. (c)

Prop

rietar

y des

ignati

on. (

d) S

AE H

SLA g

rade.

(e) O

rienta

tion o

f axis

of sp

ecim

en, r

elativ

e to r

olling

direc

tion;

L is l

ongit

udina

l (pa

rallel

to ro

lling d

irecti

on);

LT is

long t

ransv

erse (

perpe

ndicu

larto

rollin

g dire

ction

). (f)

STA

, solu

tion t

reated

and a

ged;

HR, h

ot rol

led; C

D, co

ld dra

wn; Q

&T, q

uenc

hed a

nd te

mpere

d; CD

SR, c

old dr

awn s

train

reliev

ed; D

AT, dr

awn a

t tem

perat

ure; A

, ann

ealed

.

From

ASM

Metal

s Refe

rence

Boo

k, 2n

d edit

ion, 1

983;

ASM

Intern

ation

al, M

ateria

ls Pa

rk, O

H 44

073-0

002;

table

217

. Rep

rinted

by pe

rmiss

ion of

ASM

Intern

ation

al ®

, www

.asmi

nterna

tiona

l.org.

999


Fatigue

Shea

rSt

ress

-Te

nsi

leCom

pre

ssiv

eM

odulu

sM

odulu

s of

Endura

nce

Bri

nel

lConce

ntr

ation

AST

MSt

rength

Stre

ngth

of

Ruptu

reEl

ast

icity,

Mpsi

Lim

it*

Hard

nes

sFa

ctor

Num

ber

S ut,

kpsi

S uc, k

psi

S su, k

psi

Tensi

on

†To

rsio

nS e

, k

psi

HB

Kf

2022

8326

9.6–

143.

9–5.

610

156

1.00

2526

9732

11.5

–14.

84.

6–6.

011

.517

41.

0530

3110

940

13–1

6.4

5.2–

6.6

1420

11.

1035

36.5

124

48.5

14.5

–17.

25.

8–6.

916

212

1.15

4042

.514

057

16–2

06.

4–7.

818

.523

51.

2550

52.5

164

7318

.8–2

2.8

7.2–

8.0

21.5

262

1.35

6062

.518

7.5

88.5

20.4

–23.

57.

8–8.

524

.530

21.

50

*Poli

shed

or m

achin

ed sp

ecim

ens.

† The m

odulu

s of e

lastic

ity of

cast

iron i

n com

pressi

on co

rresp

onds

clos

ely to

the u

pper

value

in th

e ran

ge gi

ven f

or ten

sion a

nd is

a mo

re co

nstan

t valu

e tha

n tha

t for

tensio

n.

Table

A–2

4

Mec

hani

cal P

rope

rties

of T

hree

Non

-Ste

el M

etal

s(a

) Typ

ical

Pro

perti

es o

f Gra

y C

ast I

ron

[The

Am

eric

an S

ocie

ty fo

r Tes

ting

and

Mat

eria

ls (A

STM

) num

berin

g sy

stem

for g

ray

cast

iron

is su

ch th

at th

e nu

mbe

rs c

orre

spon

d to

the

min

imum

tens

ile s

treng

thin

kps

i. Th

us a

n A

STM

No.

20

cast

iron

has

a m

inim

um te

nsile

stre

ngth

of 2

0 kp

si. N

ote

parti

cula

rly th

at th

eta

bula

tions

are

typi

calo

f sev

eral

hea

ts.]

1000


Table A–24

Mechanical Properties of Three Non-Steel Metals (Continued)(b) Mechanical Properties of Some Aluminum Alloys [These are typical properties for sizes of about 1

2 in; similar properties can be obtained by using proper purchase specifications. The values given for fatigue strength correspond to 50(107) cycles of completelyreversed stress. Alluminum alloys do not have an endurance limit. Yield strengths were obtained by the0.2 percent offset method.]

Aluminum Strength Elongation BrinellAssociation Yield, Sy, Tensile, Su, Fatigue, Sf, in 2 in, Hardness

Number Temper MPa (kpsi) MPa (kpsi) MPa (kpsi) % HB

Wrought:2017 O 70 (10) 179 (26) 90 (13) 22 452024 O 76 (11) 186 (27) 90 (13) 22 47

T3 345 (50) 482 (70) 138 (20) 16 1203003 H12 117 (17) 131 (19) 55 (8) 20 35

H16 165 (24) 179 (26) 65 (9.5) 14 473004 H34 186 (27) 234 (34) 103 (15) 12 63

H38 234 (34) 276 (40) 110 (16) 6 775052 H32 186 (27) 234 (34) 117 (17) 18 62

H36 234 (34) 269 (39) 124 (18) 10 74Cast:319.0* T6 165 (24) 248 (36) 69 (10) 2.0 80333.0† T5 172 (25) 234 (34) 83 (12) 1.0 100

T6 207 (30) 289 (42) 103 (15) 1.5 105335.0* T6 172 (25) 241 (35) 62 (9) 3.0 80

T7 248 (36) 262 (38) 62 (9) 0.5 85

*Sand casting.†Permanent-mold casting.

1001

(c) Mechanical Properties of Some Titanium Alloys

Yield, Sy Strength Elongation Hardness(0.2% offset) Tensile, Sut in 2 in, (Brinell or

Titanium Alloy Condition MPa (kpsi) MPa (kpsi) % Rockwell)

Ti-35A† Annealed 210 (30) 275 (40) 30 135 HBTi-50A† Annealed 310 (45) 380 (55) 25 215 HBTi-0.2 Pd Annealed 280 (40) 340 (50) 28 200 HBTi-5 Al-2.5 Sn Annealed 760 (110) 790 (115) 16 36 HRCTi-8 Al-1 Mo-1 V Annealed 900 (130) 965 (140) 15 39 HRCTi-6 Al-6 V-2 Sn Annealed 970 (140) 1030 (150) 14 38 HRCTi-6Al-4V Annealed 900 (130) 830 (120) 14 36 HRCTi-13 V-11 Cr-3 Al Sol. + aging 1207 (175) 1276 (185) 8 40 HRC

†Commercially pure alpha titanium


Table

A–2

5

Stoc

hasti

c Yi

eld

and

Ulti

mat

e St

reng

ths

for S

elec

ted

Mat

eria

lsSo

urce

:Dat

a co

mpi

led

from

“So

me

Prop

erty

Dat

a an

d

Cor

resp

ondi

ng W

eibu

ll Pa

ram

eter

s fo

r Sto

chas

tic M

echa

nica

l Des

ign,

” Tr

ans.

ASM

E Jo

urna

l of M

echa

nica

l Des

ign,

vol.

114

(Mar

ch19

92),

pp. 2

9–34

.

Mate

rial

µSu

tσ

Sut

x0

θb

µSy

σSy

x0

θb

CSu

tC

Sy

1018

CD

87.6

5.74

30.8

90.1

1278

.45.

9056

80.6

4.29

0.06

550.

0753

1035

HR

86.2

3.92

72.6

87.5

3.86

49.6

3.81

39.5

50.8

2.88

0.04

550.

0768

1045

CD

117.

77.

1390

.212

0.5

4.38

95.5

6.59

82.1

97.2

2.14

0.06

060.

0690

1117

CD

83.1

5.25

73.0

84.4

2.01

81.4

4.71

72.4

82.6

2.00

0.06

320.

0579

1137

CD

106.

56.

1596

.210

7.7

1.72

98.1

4.24

92.2

98.7

1.41

0.05

770.

0432

12L1

4C

D79

.66.

9270

.380

.41.

3678

.18.

2764

.378

.81.

720.

0869

0.10

5910

38H

T bo

lts13

3.4

3.38

122.

313

4.6

3.64

0.02

53A

STM

4044

.54.

3427

.746

.24.

380.

0975

3501

8M

alle

able

53.3

1.59

48.7

53.8

3.18

38.5

1.42

34.7

39.0

2.93

0.02

980.

0369

3251

0M

alle

able

53.4

2.68

44.7

54.3

3.61

34.9

1.47

30.1

35.5

3.67

0.05

020.

0421

Mal

leab

lePe

arlit

ic93

.93.

8380

.195

.34.

0460

.22.

7850

.261

.24.

020.

0408

0.04

6260

4515

Nod

ular

64.8

3.77

53.7

66.1

3.23

49.0

4.20

33.8

50.5

4.06

0.05

820.

0857

100-

70-0

4N

odul

ar12

2.2

7.65

47.6

125.

611

.84

79.3

4.51

64.1

81.0

3.77

0.06

260.

0569

201S

SC

D19

5.9

7.76

180.

719

7.9

2.06

0.03

9630

1SS

CD

191.

25.

8215

1.9

193.

68.

0016

6.8

9.37

139.

717

0.0

3.17

0.03

040.

0562

A10

5.0

5.68

92.3

106.

62.

3846

.84.

7026

.348

.74.

990.

0541

0.10

0430

4SS

A85

.04.

1466

.686

.65.

1137

.93.

7630

.238

.92.

170.

0487

0.09

9231

0SS

A84

.84.

2371

.686

.33.

450.

0499

403S

S10

5.3

3.09

95.7

106.

43.

4478

.53.

9164

.879

.93.

930.

0293

0.04

9817

-7PS

S19

8.8

9.51

163.

320

2.3

4.21

189.

411

.49

144.

019

3.8

4.48

0.04

780.

0607

AM

350S

SA

149.

18.

2910

1.8

152.

46.

6863

.05.

0538

.065

.05.

730.

0556

0.08

02Ti

-6A

L-4V

175.

47.

9114

1.8

178.

54.

8516

3.7

9.03

101.

516

7.4

8.18

0.04

510.

0552

2024

028

.11.

7324

.228

.72.

430.

0616

2024

T464

.91.

6460

.265

.53.

1640

.81.

8338

.441

.01.

320.

0253

0.04

49T6

67.5

1.50

55.9

68.1

9.26

53.4

1.17

51.2

53.6

1.91

0.02

220.

0219

7075

T6 .0

25”

75.5

2.10

68.8

76.2

3.53

63.7

1.98

58.9

64.3

2.63

0.02

780.

0311

1002


Table

A–2

6

Stoc

hasti

c Pa

ram

eter

s fo

r Fin

ite L

ife F

atig

ue T

ests

in S

elec

ted

Met

als

Sour

ce:E

. B. H

auge

n,Pr

obab

ilisti

c M

echa

nica

l Des

ign,

Wile

y, N

ew Y

ork,

198

0,

App

endi

x 10

–B.

12

34

56

78

9TS

YS

Dis

tri-

Stre

ss C

ycl

es t

o F

ailu

reN

um

ber

Conditio

nM

Pa

(kpsi

)M

Pa

(kpsi

)bution

10

410

510

610

7

1046

WQ

&T,

121

0°F

723

(105

)56

5(8

2)W

x 054

4(7

9)46

2(6

7)39

1(5

6.7)

θ59

4(8

6.2)

503

(73.

0)42

5(6

1.7)

b2.

602.

752.

8523

40O

Q&

T 12

00°F

799

(116

)66

1(9

6)W

x 057

9(8

4)51

0(7

4)42

0(6

1)θ

699

(101

.5)

588

(85.

4)49

6(7

2.0)

b4.

33.

44.

131

40O

Q&

T, 1

300°

F74

4(1

08)

599

(87)

Wx 0

510

(74)

455

(66)

393

(57)

θ60

4(8

7.7)

528

(76.

7)46

3(6

7.2)

b5.

25.

05.

520

24T-

448

9(7

1)36

5(5

3)N

σ26

.3(3

.82)

21.4

(3.1

1)17

.4(2

.53)

14.0

(2.0

3)A

lum

inum

µ14

3(2

0.7)

116

(16.

9)95

(13.

8)77

(11.

2)Ti

-6A

1-4V

HT-

4610

40(1

51)

992

(144

)N

σ39

.6(5

.75)

38.1

(5.5

3)36

.6(5

.31)

35.1

(5.1

0)µ

712

(108

)68

4(9

9.3)

657

(95.

4)49

3(7

1.6)

Stati

stica

l para

meter

s from

a lar

ge nu

mber

of fat

igue t

ests

are lis

ted. W

eibull

distr

ibutio

n is d

enote

d Wan

d the

param

eters

are x 0

, “gu

arante

ed” f

atigu

e stre

ngth;

θ, c

harac

terist

ic fat

igue s

treng

th; an

d b, s

hape

facto

r. Norm

al dis

tributi

on is

deno

ted N

and t

hepa

ramete

rs are

µ, m

ean f

atigu

e stre

ngth;

and σ

, stan

dard

devia

tion o

f the

fatig

ue st

rength

. The

life i

s in s

tress-

cycle

s-to-f

ailure

. TS =

tensile

stren

gth, Y

S =

yield

stren

gth. A

ll tes

ting b

y rota

ting-b

eam

spec

imen

.

1003


Table

A–2

7

Fini

te L

ife F

atig

ue S

treng

ths

of S

elec

ted

Plai

n C

arbo

n St

eels

Sour

ce: C

ompi

led

from

Tab

le 4

in H

. J. G

rove

r, S.

A. G

ordo

n,

and

L. R

. Jac

kson

,Fat

igue

of M

etal

s an

d St

ruct

ures

,Bur

eau

of N

aval

Wea

pons

Doc

umen

t NAV

WEP

S 00

-25-

534,

196

0.

Tensi

leY

ield

Stre

ngth

Stre

ngth

Stre

ss C

ycl

es t

o F

ailu

reM

ate

rial

Conditio

nBH

N*

kpsi

kpsi

RA

*10

44(1

04)

10

54(1

05)

10

64(1

06)

10

710

8

1020

Furn

ace

5830

0.63

3734

3028

25co

oled

1030

Air-

cool

ed13

580

450.

6251

4742

3838

3810

35N

orm

al13

272

350.

5444

4037

3433

33W

QT

209

103

870.

6580

7265

6057

5757

1040

Forg

ed19

592

530.

2340

4733

3310

45H

R, N

107

630.

4980

7056

4747

4747

1050

N, A

C16

492

470.

4050

4846

4038

3434

WQ

T12

0019

697

700.

5860

5752

5050

5050

.56

MN

N19

398

470.

4261

5551

4743

4141

41W

QT

277

111

840.

5794

8173

6257

5555

5512

0010

60A

s Re

c.67

Rb

134

650.

2065

6055

5048

4848

1095

162

8433

0.37

5043

4034

3130

3030

OQ

T22

711

565

0.40

7768

6457

5656

5656

1200

1012

022

411

759

0.12

6056

5150

5050

OQ

T36

918

013

00.

1510

295

9191

9191

860

*BHN

=Br

inell h

ardne

ss nu

mber;

RA =

fracti

onal

reduc

tion i

n area

.

1004


Table

A–2

8

Dec

imal

Equ

ival

ents

of W

ire a

nd S

heet

-Met

al G

auge

s* (A

ll Si

zes

Are

Giv

en in

Inch

es)

Stee

l Wir

eN

am

eA

mer

ican

Bir

min

gham

United

Manu-

or

Stubs

of

or

Bro

wn

or

Stubs

State

sfa

cture

rsW

ash

burn

Musi

cSt

eel

Twis

tG

auge:

& S

harp

eIr

on W

ire

Standard

†St

andard

& M

oen

Wir

eW

ire

Dri

llTu

bin

g,

Ferr

ous

Ferr

ous

Ferr

ous

Nonfe

rrous

Stri

p, Fl

at

Shee

t and

Wir

eSt

eel

Twis

tPri

nci

pal

Shee

t, W

ire,

Wir

e, a

nd

Pla

te,

Ferr

ous

Exce

pt

Musi

cD

rill

Dri

lls a

nd

Use

:and R

od

Spri

ng S

teel

480 lb

f/ft

3Sh

eet

Musi

c W

ire

Wir

eRod

Dri

ll St

eel

7/0

0.50

00.

490

6/0

0.58

0 0

0.46

8 75

0.46

1 5

0.00

45/

00.

516

50.

437

50.

430

50.

005

4/0

0.46

0 0

0.45

40.

406

250.

393

80.

006

3/0

0.40

9 6

0.42

50.

375

0.36

2 5

0.00

72/

00.

364

80.

380

0.34

3 75

0.33

1 0

0.00

8

00.

324

90.

340

0.31

2 5

0.30

6 5

0.00

91

0.28

9 3

0.30

00.

281

250.

283

00.

010

0.22

70.

228

02

0.25

7 6

0.28

40.

265

625

0.26

2 5

0.01

10.

219

0.22

1 0

30.

229

40.

259

0.25

0.23

9 1

0.24

3 7

0.01

20.

212

0.21

3 0

40.

204

30.

238

0.23

4 37

50.

224

20.

225

30.

013

0.20

70.

209

05

0.18

1 9

0.22

00.

218

750.

209

20.

207

00.

014

0.20

40.

205

5

60.

162

00.

203

0.20

3 12

50.

194

30.

192

00.

016

0.20

10.

204

07

0.14

4 3

0.18

00.

187

50.

179

30.

177

00.

018

0.19

90.

201

08

0.12

8 5

0.16

50.

171

875

0.16

4 4

0.16

2 0

0.02

00.

197

0.19

9 0

90.

114

40.

148

0.15

6 25

0.14

9 5

0.14

8 3

0.02

20.

194

0.19

6 0

100.

101

90.

134

0.14

0 62

50.

134

50.

135

00.

024

0.19

10.

193

5

110.

090

740.

120

0.12

50.

119

60.

120

50.

026

0.18

80.

191

012

0.08

0 81

0.10

90.

109

357

0.10

4 6

0.10

5 5

0.02

90.

185

0.18

9 0

130.

071

960.

095

0.09

3 75

0.08

9 7

0.09

1 5

0.03

10.

182

0.18

5 0

140.

064

080.

083

0.07

8 12

50.

074

70.

080

00.

033

0.18

00.

182

015

0.05

7 07

0.07

20.

070

312

50.

067

30.

072

00.

035

0.17

80.

180

0

160.

050

820.

065

0.06

2 5

0.05

9 8

0.06

2 5

0.03

70.

175

0.17

7 0

170.

045

260.

058

0.05

6 25

0.05

3 8

0.05

4 0

0.03

90.

172

0.17

3 0

(con

tinue

d)

1005


1006

Table

A–2

8

Dec

imal

Equ

ival

ents

of W

ire a

nd S

heet

-Met

al G

auge

s* (A

ll Si

zes

Are

Giv

en in

Inch

es) (

Con

tinue

d)

Stee

l Wir

eN

am

eA

mer

ican

Bir

min

gham

United

Manu-

or

Stubs

of

or

Bro

wn

or

Stubs

State

sfa

cture

rsW

ash

burn

Musi

cSt

eel

Twis

tG

auge:

& S

harp

eIr

on W

ire

Standard

†St

andard

& M

oen

Wir

eW

ire

Dri

llTu

bin

g,

Ferr

ous

Ferr

ous

Ferr

ous

Nonfe

rrous

Stri

p, Fl

at

Shee

t and

Wir

eSt

eel

Twis

tPri

nci

pal

Shee

t, W

ire,

Wir

e, a

nd

Pla

te,

Ferr

ous

Exce

pt

Musi

cD

rill

Dri

lls a

nd

Use

:and R

od

Spri

ng S

teel

480 lb

f/ft

3Sh

eet

Musi

c W

ire

Wir

eRod

Dri

ll St

eel

180.

040

300.

049

0.05

0.04

7 8

0.04

7 5

0.04

10.

168

0.16

9 5

190.

035

890.

042

0.04

3 75

0.04

1 8

0.04

1 0

0.04

30.

164

0.16

6 0

200.

031

960.

035

0.03

7 5

0.03

5 9

0.03

4 8

0.04

50.

161

0.16

1 0

210.

028

460.

032

0.03

4 37

50.

032

90.

031

70.

047

0.15

70.

159

022

0.02

5 35

0.02

80.

031

250.

029

90.

028

60.

049

0.15

50.

157

023

0.02

2 57

0.02

50.

028

125

0.02

6 9

0.02

5 8

0.05

10.

153

0.15

4 0

240.

020

100.

022

0.02

50.

023

90.

023

00.

055

0.15

10.

152

025

0.01

7 90

0.02

00.

021

875

0.02

0 9

0.02

0 4

0.05

90.

148

0.14

9 5

260.

015

940.

018

0.01

8 75

0.01

7 9

0.01

8 1

0.06

30.

146

0.14

7 0

270.

014

200.

016

0.01

7 18

7 5

0.01

6 4

0.01

7 3

0.06

70.

143

0.14

4 0

280.

012

640.

014

0.01

5 62

50.

014

90.

016

20.

071

0.13

90.

140

529

0.01

1 26

0.01

30.

014

062

50.

013

50.

015

00.

075

0.13

40.

136

030

0.01

0 03

0.01

20.

012

50.

012

00.

014

00.

080

0.12

70.

128

5

310.

008

928

0.01

00.

010

937

50.

010

50.

013

20.

085

0.12

00.

120

032

0.00

7 95

00.

009

0.01

0 15

6 25

0.00

9 7

0.01

2 8

0.09

00.

115

0.11

6 0

330.

007

080

0.00

80.

009

375

0.00

9 0

0.01

1 8

0.09

50.

112

0.11

3 0

340.

006

305

0.00

70.

008

593

750.

008

20.

010

40.

110

0.11

1 0

350.

005

615

0.00

50.

007

812

50.

007

50.

009

50.

108

0.11

0 0

360.

005

000

0.00

40.

007

031

250.

006

70.

009

00.

106

0.10

6 5

370.

004

453

0.00

6 64

0 62

50.

006

40.

008

50.

103

0.10

4 0

380.

003

965

0.00

6 25

0.00

6 0

0.00

8 0

0.10

10.

101

539

0.00

3 53

10.

007

50.

099

0.09

9 5

400.

003

145

0.00

7 0

0.09

70.

098

0

*Spe

cify s

heet,

wire

, and

plate

by st

ating

the g

auge

numb

er, th

e gau

ge na

me, a

nd th

e dec

imal

equiv

alent

in pa

renthe

ses.

† Refle

cts pr

esen

t ave

rage a

nd w

eights

of sh

eet s

teel.


Table A–29

Dimensions of Square and Hexagonal Bolts

Head Type

Nominal Square Regular Hexagonal Heavy Hexagonal Structural HexagonalSize, in W H W H Rmin W H Rmin W H Rmin

14

38

1164

716

1164 0.01

516

12

1364

12

732 0.01

38

916

14

916

14 0.01

716

58

1964

58

1964 0.01

12

34

2164

34

1132 0.01 7

81132 0.01 7

85

16 0.009

58

1516

2764

1516

2764 0.02 1 1

162764 0.02 1 1

162564 0.021

34 1 1

812 1 1

812 0.02 1 1

412 0.02 1 1

41532 0.021

1 1 12

2132 1 1

24364 0.03 1 5

84364 0.03 1 5

83964 0.062

1 18 1 11

1634 1 11

1634 0.03 1 13

1634 0.03 1 13

161116 0.062

1 14 1 7

82732 1 7

82732 0.03 2 27

32 0.03 2 2532 0.062

1 38 2 1

162932 2 1

162932 0.03 2 3

162932 0.03 2 3

162732 0.062

1 12 2 1

4 1 2 14 1 0.03 2 3

8 1 0.03 2 38

1516 0.062

NominalSize, mm

M5 8 3.58 8 3.58 0.2

M6 10 4.38 0.3

M8 13 5.68 0.4

M10 16 6.85 0.4

M12 18 7.95 0.6 21 7.95 0.6

M14 21 9.25 0.6 24 9.25 0.6

M16 24 10.75 0.6 27 10.75 0.6 27 10.75 0.6

M20 30 13.40 0.8 34 13.40 0.8 34 13.40 0.8

M24 36 15.90 0.8 41 15.90 0.8 41 15.90 1.0

M30 46 19.75 1.0 50 19.75 1.0 50 19.75 1.2

M36 55 23.55 1.0 60 23.55 1.0 60 23.55 1.5

H

R

W

Useful Tables 1007



Table A–30

Dimensions ofHexagonal Cap Screwsand Heavy HexagonalScrews (W = Widthacross Flats; H = Heightof Head; See Figurein Table A–29)

Minimum Type of ScrewNominal Fillet Cap Heavy HeightSize, in Radius W W H

14 0.015 7

165

325

16 0.015 12

1364

38 0.015 9

161564

716 0.015 5

89

3212 0.015 3

478

516

58 0.020 15

16 1 116

2564

34 0.020 1 1

8 1 14

1532

78 0.040 1 5

16 1 716

3564

1 0.060 1 12 1 1

83964

1 14 0.060 1 7

8 2 2532

1 38 0.060 2 1

16 2 316

2732

1 12 0.060 2 1

4 2 38

1516

NominalSize, mm

M5 0.2 8 3.65M6 0.3 10 4.15M8 0.4 13 5.50M10 0.4 16 6.63M12 0.6 18 21 7.76M14 0.6 21 24 9.09M16 0.6 24 27 10.32M20 0.8 30 34 12.88M24 0.8 36 41 15.44M30 1.0 46 50 19.48M36 1.0 55 60 23.38


Useful Tables 1009

Table A–31

Dimensions ofHexagonal Nuts

Height HNominal Width Regular Thick orSize, in W Hexagonal Slotted JAM

14

716

732

932

532

516

12

1764

2164

316

38

916

2164

1332

732

716

1116

38

2964

14

12

34

716

916

516

916

78

3164

3964

516

58

1516

3564

2332

38

34 1 1

84164

1316

2764

78 1 5

1634

2932

3164

1 1 12

5564 1 35

64

1 18 1 11

163132 1 5

323964

1 14 1 7

8 1 116 1 1

42332

1 38 2 1

16 1 1164 1 3

82532

1 12 2 1

4 1 932 1 1

22732

NominalSize, mm

M5 8 4.7 5.1 2.7M6 10 5.2 5.7 3.2M8 13 6.8 7.5 4.0M10 16 8.4 9.3 5.0M12 18 10.8 12.0 6.0M14 21 12.8 14.1 7.0M16 24 14.8 16.4 8.0M20 30 18.0 20.3 10.0M24 36 21.5 23.9 12.0M30 46 25.6 28.6 15.0M36 55 31.0 34.7 18.0


Table A–32

Basic Dimensions ofAmerican StandardPlain Washers (AllDimensions in Inches)

Fastener Washer DiameterSize Size ID OD Thickness

#6 0.138 0.156 0.375 0.049#8 0.164 0.188 0.438 0.049#10 0.190 0.219 0.500 0.0493

16 0.188 0.250 0.562 0.049#12 0.216 0.250 0.562 0.06514 N 0.250 0.281 0.625 0.06514 W 0.250 0.312 0.734 0.0655

16 N 0.312 0.344 0.688 0.0655

16 W 0.312 0.375 0.875 0.08338 N 0.375 0.406 0.812 0.06538 W 0.375 0.438 1.000 0.0837

16 N 0.438 0.469 0.922 0.0657

16 W 0.438 0.500 1.250 0.08312 N 0.500 0.531 1.062 0.09512 W 0.500 0.562 1.375 0.1099

16 N 0.562 0.594 1.156 0.0959

16 W 0.562 0.625 1.469 0.10958 N 0.625 0.656 1.312 0.09558 W 0.625 0.688 1.750 0.13434 N 0.750 0.812 1.469 0.13434 W 0.750 0.812 2.000 0.14878 N 0.875 0.938 1.750 0.13478 W 0.875 0.938 2.250 0.1651 N 1.000 1.062 2.000 0.1341 W 1.000 1.062 2.500 0.1651 1

8 N 1.125 1.250 2.250 0.1341 1

8 W 1.125 1.250 2.750 0.1651 1

4 N 1.250 1.375 2.500 0.1651 1

4 W 1.250 1.375 3.000 0.1651 3

8 N 1.375 1.500 2.750 0.1651 3

8 W 1.375 1.500 3.250 0.1801 1

2 N 1.500 1.625 3.000 0.1651 1

2 W 1.500 1.625 3.500 0.1801 5

8 1.625 1.750 3.750 0.1801 3

4 1.750 1.875 4.000 0.1801 7

8 1.875 2.000 4.250 0.1802 2.000 2.125 4.500 0.1802 1

4 2.250 2.375 4.750 0.2202 1

2 2.500 2.625 5.000 0.2382 3

4 2.750 2.875 5.250 0.2593 3.000 3.125 5.500 0.284

N = narrow; W = wide; use W when not specified.

1010


Table A–33

Dimensions of Metric Plain Washers (All Dimensions in Millimeters)

Washer Minimum Maximum Maximum Washer Minimum Maximum MaximumSize* ID OD Thickness Size* ID OD Thickness

1.6 N 1.95 4.00 0.70 10 N 10.85 20.00 2.301.6 R 1.95 5.00 0.70 10 R 10.85 28.00 2.801.6 W 1.95 6.00 0.90 10 W 10.85 39.00 3.50

2 N 2.50 5.00 0.90 12 N 13.30 25.40 2.802 R 2.50 6.00 0.90 12 R 13.30 34.00 3.502 W 2.50 8.00 0.90 12 W 13.30 44.00 3.50

2.5 N 3.00 6.00 0.90 14 N 15.25 28.00 2.802.5 R 3.00 8.00 0.90 14 R 15.25 39.00 3.502.5 W 3.00 10.00 1.20 14 W 15.25 50.00 4.00

3 N 3.50 7.00 0.90 16 N 17.25 32.00 3.503 R 3.50 10.00 1.20 16 R 17.25 44.00 4.003 W 3.50 12.00 1.40 16 W 17.25 56.00 4.60

3.5 N 4.00 9.00 1.20 20 N 21.80 39.00 4.003.5 R 4.00 10.00 1.40 20 R 21.80 50.00 4.603.5 W 4.00 15.00 1.75 20 W 21.80 66.00 5.10

4 N 4.70 10.00 1.20 24 N 25.60 44.00 4.604 R 4.70 12.00 1.40 24 R 25.60 56.00 5.104 W 4.70 16.00 2.30 24 W 25.60 72.00 5.60

5 N 5.50 11.00 1.40 30 N 32.40 56.00 5.105 R 5.50 15.00 1.75 30 R 32.40 72.00 5.605 W 5.50 20.00 2.30 30 W 32.40 90.00 6.40

6 N 6.65 13.00 1.75 36 N 38.30 66.00 5.606 R 6.65 18.80 1.75 36 R 38.30 90.00 6.406 W 6.65 25.40 2.30 36 W 38.30 110.00 8.50

8 N 8.90 18.80 2.308 R 8.90 25.40 2.308 W 8.90 32.00 2.80

N = narrow; R = regular; W = wide.*Same as screw or bolt size.

Useful Tables 1011


Table A–34

Gamma Function*Source: Reprinted withpermission from William H.Beyer (ed.), Handbook ofTables for Probability andStatistics, 2nd ed., 1966.Copyright CRC Press, BocaRaton, Florida.

Values of �(n) =∫ ∞

0e−x xn−1dx;�(n + 1) = n�(n)

n �(n) n �(n) n �(n) n �(n)

1.00 1.000 00 1.25 .906 40 1.50 .886 23 1.75 .919 061.01 .994 33 1.26 .904 40 1.51 .886 59 1.76 .921 371.02 .988 84 1.27 .902 50 1.52 .887 04 1.77 .923 761.03 .983 55 1.28 .900 72 1.53 .887 57 1.78 .926 231.04 .978 44 1.29 .899 04 1.54 .888 18 1.79 .928 77

1.05 .973 50 1.30 .897 47 1.55 .888 87 1.80 .931 381.06 .968 74 1.31 .896 00 1.56 .889 64 1.81 .934 081.07 .964 15 1.32 .894 64 1.57 .890 49 1.82 .936 851.08 .959 73 1.33 .893 38 1.58 .891 42 1.83 .939 691.09 .955 46 1.34 .892 22 1.59 .892 43 1.84 .942 61

1.10 .951 35 1.35 .891 15 1.60 .893 52 1.85 .945 611.11 .947 39 1.36 .890 18 1.61 .894 68 1.86 .948 691.12 .943 59 1.37 .889 31 1.62 .895 92 1.87 .951 841.13 .939 93 1.38 .888 54 1.63 .897 24 1.88 .955 071.14 .936 42 1.39 .887 85 1.64 .898 64 1.89 .958 38

1.15 .933 04 1.40 .887 26 1.65 .900 12 1.90 .961 771.16 .929 80 1.41 .886 76 1.66 .901 67 1.91 .965 231.17 .936 70 1.42 .886 36 1.67 .903 30 1.92 .968 781.18 .923 73 1.43 .886 04 1.68 .905 00 1.93 .972 401.19 .920 88 1.44 .885 80 1.69 .906 78 1.94 .976 10

1.20 .918 17 1.45 .885 65 1.70 .908 64 1.95 .979 881.21 .915 58 1.46 .885 60 1.71 .910 57 1.96 .983 741.22 .913 11 1.47 .885 63 1.72 .912 58 1.97 .987 681.23 .910 75 1.48 .885 75 1.73 .914 66 1.98 .991 711.24 .908 52 1.49 .885 95 1.74 .916 83 1.99 .995 81

2.00 1.000 00

*For large positive values of x, �(x) approximates the asymptotic series

xx e−x

√2x

x

[1 + 1

12x+ 1

288x2− 139

51 840x3− 571

2 488 320x4+ · · ·

]



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BDA 30803 Notes (Student Version - Printable) Sem 2 2012_2013.pdf