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FACULTY OF MECHANICAL AND MANUFACTURING ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA

BDA 31203 - Mechanical Component Design (Lecture Slides)

Semester II 2017 / 2018 Course Coordinator : MR. MOHD AZWIR BIN AZLAN Lecturers :

1) Assoc. Prof. Dr. Sia Chee Kiong – (S1 & S2) 2) Mr. Mohd Azwir Bin Azlan (S3 & S4) 3) Dr. Koh Ching Theng - (S5 & S6) 4) Mr. Mohd Nizam bin Katimon- (S7 & S8) 5) Dr. Ahmad Mubarak Bin Tajul Ariffin – (S9 & S10)

FACULTY : SEMESTER : SESION : COURSE :

FACULTY OF MECHANICAL AND MANUFACTURING ENGINEERING (FKMP) 2 2017 / 2018 BDA 31203 - MECHANICAL COMPONENT DESIGN (MCD)

TIME SCHEDULE DAY / TIME 8.00 - 8.50 9.00 - 9.50 10.00 - 10.50 11.00 - 11.50 12.00 - 12.50 1.00 - 2.00 2.00 - 2.50 3.00 - 3.50 4.00 - 4.50 5.00 - 5.50 6.00 - 6.50 7.00 - 8.00 8.00 - 8.50 9.00 - 10.00 10.00 - 11.00

SUNDAY ITEMS BDA 31203 (S9 & S10) (Dr. Ahmad Mubarak)

LOCATION

MONDAY ITEMS

LOCATION

TUESDAY ITEMS

LOCATION

BDA 31203 BDA 31203 (S5 & S6) (S1 & S2) PERP - BT10 PERP - BT10 PERP - BT10 (Dr. Koh (PM Dr. Sia [max - 60] [max - 60] [max - 60] Ching Theng) Chee Kiong)

WEDNESDAY ITEMS

LOCATION

THURSDAY ITEMS

LOCATION

BDA 31203 (S3 & S4) (En. Azwir) / (S7 & S8) (En. Nizam)

PERP - BT10 / G3-BKB2 [max - 60]

FRIDAY ITEMS

LOCATION

Sem 2 Sesi 2017/2018

Borang Akuan Mengikuti Kuliah Ganti Kursus BDA 31203 - Rekabentuk Komponen Mekanikal

Saya pelajar yang bernama dari seksyen

no matrik

,

mengakui bahawa saya telah mengikuti kuliah ganti yang telah diajar oleh

(PM Dr. Sia / Dr. Koh / Dr. Mubarak / En. Azwir / En. Nizam ) pada

bagi

(Sila potong yang tidak berkenaan)

menggantikan kuliah yang tidak dapat saya hadiri pada

.

Topik yang dipelajari adalah:Sila tandakan () bagi topik / kuliah ganti yang telah diikuti.

Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8 Week 9 Week 10 Week 11 Week 12 Week 13 Week 14

Lesson Plan, Introduction To Design Process, Analysis And Synthesis Static Design Failure Of Theories Fatigue Design Failure Of Theories (Topic 4.1 ~ 4.7) Fatigue Design Failure Of Theories (Topic 4.8 ~ 4.11) Gear – Part 1 (Topic 5.1 ~ 5.13) Design Project (Project Briefing, Review on Best Example Project Report) Gear – Part 2 (Topic 5.14 ~ 5.19) Gear – Part 2 (Example & Discussion) Shaft Design Bearing Non-Permanent Joints (Topic 8.1 ~ 8.5) Non-Permanent Joints (Topic 8.6 ~ 8.10) Non-Permanent Joints (Topic 8.11 ~ 8.13) Project Supervision

Tandatangan Pelajar,

Disahkan oleh,

...........................................

...........................................

Tarikh :

Nama Pensyarah : Tarikh :

Lampiran A

UNIVERSITI TUN HUSSEIN ONN MALAYSIA FAKULTI KEJURUTERAAN MEKANIKAL DAN PEMBUATAN PERANCANGAN KULIAH LECTURE PLAN MAKLUMAT KURSUS (COURSE INFORMATION) SEMESTER / SESI (SEMESTER / SESSION) : II / 2017 - 2018 KOD KURSUS (COURSE CODE)

: BDA 31203

NAMA KURSUS (COURSE TITLE)

: REKABENTUK KOMPONEN MEKANIKAL (MECHANICAL COMPONENT DESIGN)

BEBAN AKADEMIK PELAJAR (STUDENT ACADEMIC LOAD) : Kategori Aktiviti (Category of Activities) Pembelajaran bersemuka (Face-to-face learning)

Pembelajaran kendiri (Independent study)

Pentaksiran rasmi (Formal assessment)

Aktiviti Pembelajaran Jumlah Jam/ Semester (Learning Activities) (Total Hours/ Semester) Kuliah (Lecture) 36 Tutorial / Amali (Tutorial / Practical) Aktiviti pembelajaran berpusatkan pelajar lain 6 (Other student centered learning activities) Penyediaan tugasan, projek dan lain-lain 24 (Preparing assignment, project and others) Ulangkaji 36 (Revision) Persediaan bagi pentaksiran 12 (Preparation for assessment) Pentaksiran berterusan 6 (Continuous assessment) Peperiksaan akhir (Final examination) 120 JUMLAH JAM BELAJAR PELAJAR (JBP) (TOTAL STUDENT LEARNING TIME (SLT))

Kursus Pra-syarat (Prerequisite course)

: BDA 10703 – Lukisan Teknikal & CAD / Technical Drawing & CAD BDA 20903 – Mekanik Pepejal II / Solid Mechanics II

Nama Staf Akademik (Name of Academic Staff) :  Assoc. Prof. Dr. Sia Chee Kiong (S1 & S2)  Mr. Mohd Azwir bin Azlan (S3 & S4) – coordinator  Dr. Koh Ching Theng (S5 & S6)  Mr. Mohd Nizam bin Katimon (S7 & S8)  Dr. Ahmad Mubarak bin Tajul Arifin (S9 & S10)

Disediakan oleh (Prepared by):

Disahkan oleh (Approved by):

Tandatangan (Signature):

Tandatangan (Signature):

Nama (Name): MOHD AZWIR BIN AZLAN

Nama (Name): Assoc. Prof. Dr. SUFIZAR AHMAD

Tarikh (Date): 05th February 2018

Tarikh (Date): 05th February 2018

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

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) Menilai beberapa komponen mesin (iaitu gear, aci dan galas) yang berfungsi dalam satu sistem mekanikal atau mesin. (C5, LO10) Mencari sumber maklumat yang paling sesuai bagi pemilihan komponen dalam projek reka bentuk. (A3, LO6) Menghasilkan model dan simulasi dengan menggunakan perisian kejuruteraan dalam mengesahkan rekabentuk projek. (C4, LO2)

After completing this course, the students are able to:    

Calculate factor of safety by using static and fatigue failure of theories. (C3, LO1) Evaluate several machine components (i.e. gears, shafts and bearing) that function in one mechanical system or machine. (C5, 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. (C4, LO2)

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

Lampiran A

ISI KANDUNGAN (CONTENT): MINGGU

KANDUNGAN

PENTAKSIRAN

(WEEK)

(CONTENT)

(ASSESSMENT)

W1 (18th ~ 24th Feb 2018)

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)

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 2.2 2.3 2.4 2.5

Ujian 1 (1st Test)

Kekuatan dan Kekerasan Bahan (Material Strength and Stiffness) Keseimbangan dan GBB (Equilibrium and FBD) Jenis-Jenis Daya (Types of Load) Tegasan (Stress) 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)

W2

3.0 TEORI-TEORI KEGAGALAN REKABENTUK STATIK (STATIC DESIGN FAILURE OF THEORIES) ---- (3 hours)

th

(25 Feb ~ 03rd Mac 2018)

3.1 3.2 3.3 3.4 3.5 3.6 3.7

Pengenalan (Introduction) Kenapa Perlu Teori Kegagalan (Why Need Failure Theories) Teori Kegagalan Statik (Static Failure Theories) Teori Tegasan Ricih Maksimum (Maximum Shear Stress Theory) Teori Tenaga Herotan (Distortion Energy Theory) Teori Column-Mohr (Column-Mohr Theory) Teori Tegasan Normal Maksimum (Maximum Normal Stress Theory)

3.8 Teori Pengubahsuaian Column-Mohr (Modification of Mohr Theory)

Tugasan 1 – Hantar Sebelum 08/03/2018 ; 3:30 pm (1st Assignment – Submit before 08/03/2018 ; 3:30 pm) RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

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

Lampiran A

W3 & W4 4.0 TEORI-TEORI KEGAGALAN BAGI REKABENTUK LESU (FATIGUE DESIGN FAILURE OF THEORIES) ---- (6 hours) th

th

(04 ~ 17 Mac 2018)

4.1 4.2 4.3 4.4 4.5 4.6

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

Pengenalan kepada Lesu (Introduction to Fatigue) Kegagalan dan Beban Lesu (Fatigue Load and Failure) Hayat dan Kekuatan Lesu (Life and Fatigue Strength) Rajah S-N (S-N Diagram) Had Ketahanan (Endurance Limits) Faktor Berubah Had Ketahanan (Endurance Limit Modifying Factors)

4.7

Penumpuan Tegasan dan Kepekaan Takuk (Stress

Concentration and Notch Sensitivity) Kekuatan Lesu (Fatigue Strength) Ciri-ciri Tegasan Berulang (Characterizing Fluctuating stressess) 4.10 Kombinasi Mod Beban (Combination of Loading Modes) 4.11 Faktor Keselamatan (Safety Factor)

4.8 4.9

Tugasan 2 – Hantar Sebelum 22/03/2018 ; 3:30 pm (2nd Assignment – Submit before 22/03/2018 ; 3.30 pm)

W5 (18th ~ 24th Mac 2018)

W6 (25th ~ 31st Mac 2018)

5.0 GEAR – BAHAGIAN 1 (GEAR – PART 1) ---- (3 hours) 5.1

Pengenalan: tatanama, jenis-jenis giar dan kegunaannya

5.2 5.3 5.4 5.5 5.6

(Introduction: terminology, types of gears and its application) Pembinaan Giar (Construction of gears) Sistem Gigi (Tooth systems) Nisbah Giar (Gear ratio) Barisan gear (Gear train) Analisis Daya pada Gigi (Gear Tooth Analysis)

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

PROJEK REKABENTUK (DESIGN PROJECT)  Penerangan Projek – Tajuk Projek, Tema, Jadual Perancangan, Format Laporan, Penilaian Laporan dan Kemahiran Insaniah. (Project Briefing – Project Title & Theme, Planning Schedule, Report Format, Report & Soft Skill Assessment)

 Perbincangan dan Semakan Contoh Laporan Projek Terbaik. (Discussion and Review on Best Example Project Report)  Pembahagian Kumpulan (Group Distribution)  Pendaftaran Kumpulan (Group Registration)

W6 - UJIAN 1 (1st TEST) --- (1.5 hours) (27/03/2018 ; 8:00 – 9:30 pm) W7 – W8 (01st ~ 21st April 2018)

5.0 GEAR – BAHAGIAN 2 (GEAR – PART 2) ---- (6 hours) 5.7 5.8 5.9

Analisis Lenturan Gigi Giar (Gear Tooth Bending Analysis) Analisis Kehausan Gigi Giar (Gear Tooth Wear Analysis) Faktor Keselamatan (Factor of Safety)

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

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

Lampiran A

W9 (22nd ~ 28th April 2018)

W10 (29th April ~ 05th May 2018)

6.0 REKABENTUK ACI (SHAFT DESIGN) ---- (3 hours) 6.1 6.2 6.3 6.4 6.5

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

7.0 GALAS (BEARING) ---- (3 hours) 7.1 7.2 7.3

Pengenalan (Introduction) Jenis-Jenis Galas (Bearing Types) Perletakan dan Pemasangan galas (Bearing Mounting and

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

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

Enclosures)

7.4 7.5

Hayat Galas (Bearing Life) 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

7.8

Thrust Loading) Pelinciran (Lubrication)

W10 - UJIAN 2 (2nd Test) --- (2.5 hours) (02/05/2018 ; 8:00 – 10:30 pm) W11 – W13 (06th ~ 26th May 2018)

8.0 PENYAMBUNGAN SEMENTARA (NON-PERMANENT JOINTS) ---- (9 hours) 8.1 8.2

Pengenalan (Introduction) Definasi dan piawaian bebenang (Thread standard and definition)

8.3 8.4 8.5

Mekanik skru kuasa (The mechanic of power screw) Bebenang pengikat (Threaded fasteners) Penyambung: Kekukuhan pengikat (Joints: Fastener stiffness)

8.6

Penyambung: Kekukuhan anggota (Joints: Member stiffness)

8.7 8.8

Kekuatan bolt (Bolt strength) Ketegangan sambungan : Beban luaran (Tension joints : The external load)

8.9

Perkaitan daya kilas bolt dengan ketegangan bolt (Relating bolt torque to bolt tension)

8.10 8.11 8.12

Ketegangan sambungan beban statik berserta pra beban (Statically loaded tension joint with preload) Sambungan gasket (Gasketed joints)

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)

W14 (27th May ~ 02nd June 2018)

PROJEK REKABENTUK (DESIGN PROJECT)  Penyeliaan Projek – (Project Supervision)

W14 - UJIAN 3 (3rd Test) --- (2.0 hours) (30/05/2018 ; 8:00 - 10:00pm) RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

Ujian 3 (3rd Test)

Lampiran A

W15

PENGHANTARAN LAPORAN PROJEK REKABENTUK (SUBMISSION OF DESIGN PROJECT REPORT – 05th June 2018)

(03rd June 2018

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.

PENTAKSIRAN (ASSESSMENT): 1. Kuiz (Quiz)

:

0 %

2. Tugasan (Assignment)

:

10 %

3. Ujian (Test)

:

50 %

4. Project (Project)

:

40 %

5. Lain-lain (Others)

:

0 %

6. Peperiksaan Akhir (Final Examination) :

0 %

Jumlah (Total) :

100 %

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

Lampiran A

RUJUKAN (REFERENCES): 1. BDA 31203 Lecturer Notes 2. Richard. G. Budynas, J. Keith Nisbett, (2015), “Shigley’s Mechanical Engineering Design”, Tenth Edition in SI Units, McGraw Hill Education. (ISBN: 978-981-4595-28-5) 3. Robert L. Mott, (2014), “Machine Elements in Mechanical Design”, New Jersey: Pearson. (ISBN: 978-013-3349-07-8) 4. Robert L. Norton, (2011), “Machine Design – An Integrated Approach (Fourth Edition)”, New Jersey: Pearson. (ISBN: 978-013-6123-70-5) ; (Call number: TJ230 .N67 2011)

KEHADIRAN / PERATURAN SEMASA KULIAH (ATTENDANCE / REGULATIONS DURING LECTURE)

1. Pelajar mesti hadir tidak kurang dari 80% masa pertemuan yang ditentukan bagi sesuatu kursus termasuk kursus Hadir Wajib (HW) dan kursus Hadir Sahaja (HS). Students must attend not less than 80% of the contact hours specified for a course including Compulsory Attendance (HW) course and Attendance Only (HS) course. 2. Pelajar yang tidak memenuhi perkara 1 di atas tidak dibenarkan menghadiri kuliah dan menduduki sebarang bentuk pentaksiran selanjutnya. Markah sifar ‘0’ akan diberikan kepada pelajar yang gagal memenuhi perkara 1. Manakala untuk kursus HW, pelajar yang gagal memenuhi perkara 1 akan diberi gred Hadir Gagal (HG). Students who do not fulfill item 1 of the above are not allowed to attend further lecture and sit for any form of assessment. Zero ‘0’ mark will be given to students who fail to comply with item 1. As for HW course, students who fail to comply with item 1 will be given Failure Attendance (HG) grade. 3. Pelajar perlu patuh kepada peraturan berpakaian yang berkuatkuasa dan menjaga disiplin diri masingmasing untuk mengelakkan dari tindakan tatatertib diambil terhadap pelajar. Students should adhere to the dress regulations in effect and must discipline themselves to avoid from any disciplinary actions. 4. Pelajar perlu mematuhi peraturan keselamatan semasa aktiviti pembelajaran dan pengajaran. Students should adhere to the safety regulations during the learning and teaching activities. MATRIK HASIL PEMBELAJARAN KURSUS DAN HASIL PEMBELAJARAN PROGRAM (MATRIX OF COURSE LEARNING OUTCOMES AND PROGRAMME LEARNING OUTCOMES) Dilampirkan. (Attached).

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: 4 / No. Semakan: 0

Lampiran A

MATRIK HASIL PEMBELAJARAN KURSUS DAN HASIL PEMBELAJARAN PROGRAM MATRIX OF COURSE LEARNING OUTCOMES AND PROGRAMME LEARNING OUTCOMES Fakulti (Faculty)

: FACULTY OF MECHANICAL AND MANUFACTURING ENGINEERING

Program (Programme)

: BACHELOR OF MECHANICAL ENGINEERING WITH HONOURS

Kod Kursus (Course Code)

: BDA 31203

Nama Kursus (Course Title)

: MECHANICAL COMPONENT DESIGN

Matrik ini perlu digunakan bersama: (This matrix is to be used together with) 1. Objektif Pendidikan Program (PEO) (Programme Educational Objectives (PEO)) 2. Hasil Pembelajaran Program (PLO) (Programme Learning Outcomes (PLO))

2

3

4

C1 C2 C3 C4 C5 C6

PLO13

PLO12

PLO11

PLO10

PLO9

PLO8

PLO7

PLO6

PLO5

C5

PO-PBL

Project - Soft Skill.

PO-PBL

Project - Drawing.

A3

C4 1

Kognitif (Cognitive) Pengetahuan (Knowledge) Pemahaman (Comprehension) Aplikasi (Application) Analisis (Analysis) Sintesis (Synthesis) Penilaian (Evaluation)

RPP-04 / Prosedur Pelaksanaan Kuliah Edisi: / No. Semakan:

Lecture, Case Study, PO-PBL

Kaedah Pentaksiran (Method of Assessment) Assignment 1, Assignment 2, Project – Report, Test 1. Test 2, Test 3, Project - Report.

Kaedah Penyampaian (Method of Delivery) Lecture, Case Study,

C3

Evaluate several machine components (i.e. gears, shafts and bearing) that function in one mechanical system or machine. (C5, 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. (C4, LO2) Jumlah (Total)

PLO4

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

PLO3

1

PLO2

Hasil Pembelajaran Kursus (Course Learning Outcomes)

PLO1

Pemetaan kepada PLO (Mapping to PLO) Bil. (No.)

1

P1 P2 P3 P4 P5 P6 P7

1

1

Taksonomi Pembelajaran (Learning Taxonomy) Psikomotor (Psychomotor) Persepsi (Perception) A1 Set (Set) A2 Respons Berpandu (Guided Response) A3 Mekanisme (Mechanism) A4 Respons Ketara Kompleks (Complex Overt Response) A5 Adaptasi (Adaptation) Lakuan Tulen (Origination)

KPI* 100% students pass 100% students pass 100% students pass 100% students pass

*KPI – Petunjuk prestasi utama (Key performance indicator)

Afektif (Affective) Menerima (Receiving) Memberikan Maklum Balas (Responding) Menilai (Valuing) Mengorganisasi (Organising) Menghayati Nilai (Internalising)

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

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

Week 1

• recognize the approach and the process of engineering design.

Chapter 1

Introduction to Engineering Design

• practice standards and codes, ethics and professionalism of mechanical engineer. Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

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

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

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.

3

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

4

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.2 - Mechanical Engineering Design

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

• 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.

5

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

Recognition of need

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

Recognition of Need

Definition of Problem

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

Definition of Problem

7

• 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.

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

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

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

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

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BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

BDA 31203 – Mechanical Component Design

1.3 - Design Process

Synthesis

Analysis and Optimization

• The combination of ideas into a complex whole. Synthesis

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

• Is sometimes called the invention of the concept or concept design. • Generate concept ⇒ variant concept ⇒ concept selection ⇒ concept improvement ⇒ detailing concept.

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

Analysis and Optimization

BDA 31203 – Mechanical Component Design

10

CHAPTER 1 – Introduction to Engineering Design

1.3 - Design Process

Presentation • Presentation is a selling job.

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

• 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.

• 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? Presentation

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

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

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

Evaluation

Evaluation

CHAPTER 1 – Introduction to Engineering Design

11

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

• Undoubtedly, many great designs, inventions and creative works, have been simple lost because the originators were unable to explain their accomplishment to others. 12

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.4 - Design Tools and Resources

1.5 - Design Engineer’s Professional Responsibilities

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

•

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:

¾ Computational Tools ƒ CAD (Computer Aided Design) – AutoCAD, I-Deas, SolidWorks, ProEngineer ƒ CAE (Computer Aided Engineering) – Cosmos, Algor, Fluent, ADAMS ƒ CAM (Computer Aided Manufacturing) – MasterCam, UniGraphic, SolidCAM

– Sharpen your communication skills either oral or writing – Keep a neat and clear journal / logbook of your activities, entering dated entries frequently

¾ Acquiring Technical Information ƒ ƒ ƒ ƒ ƒ

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

– 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.

categories listed above.

– Conduct activities in an ethical manner. 13

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

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

BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

1.6 - Standards and Codes

1.6 - Standards and Codes – cont…

•

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

•

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.

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

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ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ

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

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BDA 31203 – Mechanical Component Design

CHAPTER 1 – Introduction to Engineering Design

BDA 31203 – Mechanical Component Design

1.7 - Economics

1.7 - Economics – cont…

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

Breakeven Points

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

Large Tolerances ƒ Tolerances, manufacturing processes and surface finish are interrelated and influence the producibility of the end product in many way. ƒ Large tolerances can often be produced by machines with higher production rates; costs will be significantly smaller.

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

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CHAPTER 1 – Introduction to Engineering Design

ƒ Use when two or more design approaches are compared to cost. ƒ 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. ƒ The point corresponding to equal cost known as the breakeven point.

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

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BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

Learning Learning Outcomes Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

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

WEEK 1

Chapter 2

Analysis and Synthesis Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

What you will be learn here? • • • • • • • • • •

What you will be learn here?

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

CHAPTER 2 – Analysis and Synthesis

• • • • • • •

3

2.11 - Torsion 2.12 - Stress Concentration 2.13 - Part Failure 2.14 - Design Factor 2.15 - Factors Effecting Design Factor 2.16 - Predictions of Failure for Static Load 2.17 - Predictions of Failure for Fluctuating Load

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

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.1 – Material Strength and Stiffness

2.1 – Material Strength and Stiffness – cont… • pl ⇒ Proportional limits

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

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

– 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

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.

• 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

5

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

2.1 – Material Strength and Stiffness – cont… Stress-strain diagram obtained from the standard tensile test for Brittle material

CHAPTER 2 – Analysis and Synthesis

2.2 – Equilibrium and Free Body Diagram

• 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

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.

• 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

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

∑F = 0 ∑M = 0 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

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BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.2 – Equilibrium and Free Body Diagram – cont…

Info given:

Use to simplify the analysis of a very complex structure or machine

Input torque, Ti = 240 Ibf

by isolating or freeing a portion of the total system in order to study

Pitch radii of gear ; Gearbox

G1 ⇒ r1= 0.75 in

the behaviour of one of its segments. •

2.2 – Equilibrium and Free Body Diagram – cont… FBD Example – Gear reducer

Free Body Diagram (FBD) •

CHAPTER 2 – Analysis and Synthesis

G2 ⇒ r2= 1.50 in

Thus FBD is essentially a means of breaking a complicated

Gear pressure angle, ∅ = 20o

Input Shaft

problem into manageable segments, analyzing these simple problems, and then usually putting information together again.

Output Shaft

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

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.2 – Equilibrium and Free Body Diagram – cont…

10

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

CHAPTER 2 – Analysis and Synthesis

2.3 – Types of Load

Bending load Tension load Answers: To = 480 Ibf RAY = 192 Ibf RAZ = 69.9 Ibf

Torsion load

RBY = 128 Ibf RBZ = 46.6 Ibf RCY = 192 Ibf RCZ = 69.9 Ibf RDY = 128 Ibf

Compression load

RDZ = 46.6 Ibf

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

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

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

2.4 – Stress

CHAPTER 2 – Analysis and Synthesis

2.4 – Stress – cont…

Stress components on surface normal to x direction

• The force distribution will not be uniform across the surface.

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.

• 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). 13

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

2.5 – Principle Stress for Plane Stress

14

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

CHAPTER 2 – Analysis and Synthesis

2.6 – Mohr’s Circle for Plane Stress σave

σy τxy

τ1,2

σave φ

σx θo τ1,2

τxy Stress components on surface normal to ‘x’ and ‘y’ direction

σ1,σ 2 =

σ x +σ y 2

Maximum and minimum normal stresses are called principle stresses which have zero shear stresses 2

2

⎛ σ x −σ y ⎞ ⎟⎟ + τ xy2 ± ⎜⎜ ⎝ 2 ⎠

⎛ σ x −σ y ⎞ ⎟⎟ + τ xy2 ⎝ 2 ⎠

τ 1 ,τ 2 = ± ⎜⎜

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

Maximum shear stresses have average normal stresses

σ ave =

σ x +σ y 2 15

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.7 – General Three Dimensional (3D) Stress

2.8 – Uniformly Distributed Stresses

• In design, 3D transformations are rarely

Simple tension, compression and shear loads that always perform this uniform

performed since most maximum stress

distribution of stress which results

states occur under plane stress conditions. • But if there is need to be countable, make A

sure the principle normal stress are always ordered so that σ1 > σ2 > σ3

F σ= A

: tensile and compression stress

F A

: shear stress e.g. a bolt in shear

F

F

• Therefore τmax = τ1/3 where

τ 1/ 3 =

σ1 − σ 3 2

; τ 1/ 2 =

σ1 − σ 2 2

; τ 2/3 =

σ 2 −σ3

τ=

2

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

A

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CHAPTER 2 – Analysis and Synthesis

2.9 – Normal Stresses for Beams in Bending – cont…

Bending stress, σ is directly proportional to the distance, c

Bending Moment Diagram

from the neutral axis and bending moment, M.

Is sometimes needed to determine:-

Mc σ= I

F

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

BDA 31203 – Mechanical Component Design

2.9 – Normal Stresses for Beams in Bending

F

where; M – moment c – distance from neutral axis I – second moment of area

• location where moment is maximum

Loading diagram

• moment in specified location. E.g.

A

Shear-force diagram

B

MB MA

Mmax =

A

B

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

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

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.10 – Shear Stresses for Beams in Bending

2.10 – Shear Stresses for Beams in Bending – cont… Formula for Maximum Shear Stress Due to Bending Beam Shape

Formula

τ max =

Beam Shape

3ν 2A

τ max =

Rectangular

τ max =

• As it move away from the neutral axis, the shear stress decrease parabolically until it zero at the outer surfaces

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

Structural I beam (thin-walled)

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

CHAPTER 2 – Analysis and Synthesis

• When USC units is used, the equation is :

H=

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

• When SI units is used, the equation is :

H = Tω

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

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

πTn FV 2πTn Tn = = = 33,000 33,000(12) 198,000 63,025

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

where ; T = torque r = bar radius J = polar second moment of area

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

τ max

A web

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

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

T ⎛ 1.8 ⎞ = 2 ⎜3 + ⎟ bc ⎝ b / c ⎠

ν

2.11 – Torsion – cont…

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

τ max

τ max =

Web

BDA 31203 – Mechanical Component Design

2.11 – Torsion

Tr = J

4ν 3A

Circular

where y = ± c

2ν A

Hollow, thin-walled round

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

Formula

where ; H = power (W) T = torque (Nm) ω = angular velocity (rad/s) 23

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

• The torque T corresponding to the power in watts is given approximately by

T = 9.55

H n 24

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.12 – Stresses Concentration

2.12 – Stresses Concentration – cont…

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

σ

Stress distribution • Such discontinuities are called stress raisers,

and the regions in which they occur are called areas of stress concentration.

σ0= F/A0 σmax A

A0 =d (w-d )t

A

Stress trajectories

B

B

A = wt w

σ= F/A

σ

σmax >σ0 > σ

Stress Stressdistribution distributionnear nearaahole holeininaa plate plateloaded loadedinintension. tension.The Thetensile tensilestress stress on onaasection sectionB-B, B-B,remote remotefrom fromthe thehole holeisis σ= σ=F/A F/Awhere whereAA==wt wt and andt tisisthe theplate platethickness. thickness.On Onaasection section atatA-A, A-A,through throughthe thehole, hole,the thearea area AA0 ==(w-d)t (w-d)t 0 and andnominal nominalstress, stress, σσo ==F/A F/Ao.o. o Note Notethat thatthe thestress stressare areincreases increaseswhen when move movetowards towardsto tothe thehole holeand and maxsimum stress occur at the edge maxsimum stress occur at the edgeofof the thehole holewhere wherethe theload loadlines linesbecome become very verycompact compactthere. there.

• The factors are define by the equations:

Kt =

σ max σo

K ts =

τ max τo

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

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

CHAPTER 2 – Analysis and Synthesis

26

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.13 – Part Failure

P

• 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.

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

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

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.14 – Design Factor

Parts Fail When?

• Analysis

Example :

Crack initiation site

P

N=

Failure Strength Applied Stress Sy σ

• Design

This crack in the part is very small. If the level of stress in the part is SMALL, the crack will remain stable and not expand. If the level of stress in the part is HIGH enough, the crack will get bigger (propagate) and the part will eventually fail.

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

Factor of Safety =

Failure Strength Design Factor Sy σ ALLOW = N

Allowable Stress = Example : 27

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BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.14 – Design Factor – cont…

2.14 – Design Factor – cont… Factor of Safety

Application

1.25 - 1.5

Material properties known in detail. Operating conditions known in detail Loads and resultant stresses and strains known with high degree of certainty. Material test certificates, proof loading, regular inspection and maintenance. Low weight is important to design.

1.5 - 2

Known materials with certification under reasonably constant environmental conditions, subjected to loads and stresses that can be determined using qualified design procedures. Proof tests, regular inspection and maintenance required

2 - 2.5

Materials obtained for reputable suppliers to relevant standards operated in normal environments and subjected to loads and stresses that can be determined using checked calculations.

2.5 - 3

For less tried materials or for brittle materials under average conditions of environment, load and stress.

3-4

For untried materials used under average conditions of environment, load and stress.

3-4

Should also be used with better-known materials that are to be used in uncertain environments or subject to uncertain stresses.

Adapted from R. B. Englund

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

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

30

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.15 – Factors Effecting Design Factor

2.15 – Factors Effecting Design Factor

• • • • • •

• Application

• How many will be produced?

• • • • •

• What manufacturing methods will be used?

Application Environment Loads Types of Stresses Material Confidence

Environment Loads Types of Stresses Material Confidence

• What are the consequences of failure? ‰ Danger to people ‰ Cost

• Size and weight important? • What is the life of the component? • Justify design expense? Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

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

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.15 – Factors Effecting Design Factor

2.15 – Factors Effecting Design Factor

• Application

• Application • Environment

• Environment • • • •

Loads Types of Stresses Material Confidence

• Temperature range. • Exposure to electrical voltage or current.

• Loads • Types of Stresses • Material • Confidence

• Susceptible to corrosion • Is noise control important? • Is vibration control important?

• Load characteristic ‰ Static, repeated & reversed, fluctuating, shock or impact

• Magnitudes

‰ Guard ‰ Housing

‰

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

‰ Startup, shutdown, normal operation, any foreseeable overloads

• Variations of loads over time.

• Will the component be protected?

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

• Nature of the load considering all modes of operation:

Maximum, minimum, mean

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

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

2.15 – Factors Effecting Design Factor

2.15 – Factors Effecting Design Factor

• Application • Environment • Loads

• • • •

• What kind of stress? ‰ ‰ ‰ ‰

• Types of Stresses • Material • Confidence

Direct tension or compression Direct shear Bending Torsional shear

Application Environment Loads Types of Stresses

• Application

• Material

‰ Uniaxial ‰ Biaxial ‰ Triaxial

• Confidence

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

• Material properties • Ultimate strength, yield strength, endurance strength, • Ductility

‰ Ductile: (εf ≥ 5%) ‰ Brittle: (εf < 5%)

• Ductile materials are preferred for fatigue, shock or impact loads.

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

36

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

2.15 – Factors Effecting Design Factor

2.16 – Predictions of Failure for Static Loads

• • • • •

• Brittle Materials:

Application Environment Loads Types of Stresses Material

• Confidence

• Reliability of data for ‰ Loads ‰ Material properties ‰ Stress calculations

– Maximum Normal Stress – Modified Mohr

• How good is manufacturing quality control

• Ductile Materials:

• Will subsequent handling, use and environmental conditions affect the safety or life of the component?

BDA 31203 – Mechanical Component Design

– Yield Strength – Maximum Shear Strength – Distortion Energy 37

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

CHAPTER 2 – Analysis and Synthesis

- Uniaxial - Biaxial - Biaxial or Triaxial 38

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

BDA 31203 – Mechanical Component Design

2.17 – Predictions of Failure for Fluctuating Loads

CHAPTER 2 – Analysis and Synthesis

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.

• Brittle Materials: – Not recommended

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.

• Ductile Materials: – Goodman – Gerber – Soderberg

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

- Uniaxial - Biaxial

39

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

40

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

Solution :

Solution :

(a) The results are:-

(b) Maximum torsional and bending stress at arm BC

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

•

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

Mc M (h / 2) 6 M = = 2 bh 3 I bh 12 6(1200) = = 18,400 psi 0.25(1.252 )

σ=

(300 Ibf)

•

For rectangular section having torsional stress

(450 Ibf.in)

τ= (1200 Ibf.in) (450 Ibf.in)

41

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CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

42

CHAPTER 2 – Analysis and Synthesis

Solution :

Stress element on the top surface of the shaft at A

τxz A

σx

•

x

τxz

(d) Maximum normal and shear stresses at A. The bending is tensile and is

σx

σ=

z

=

The maximum normal stress is given by

Mc M (d / 2) 32 M = = πd 4 I πd 3 64

32(1950) = 47,100 π (0.753 )

σ1 =

psi

•

The torsional stress is

(1950 Ibf.in)

τ=

(1200 Ibf.in) (300 Ibf) (450 Ibf.in)

= (1200 Ibf.in) Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

2

2

⎛σ −σ z ⎞ 2 + ⎜ x ⎟ + τ xz ⎝ 2 ⎠

47.1 + 0 ⎛ 47.1 − 0 ⎞ 2 + ⎜ ⎟ + 14.5 2 ⎝ 2 ⎠

= 51.2 kpsi

Tr T (d / 2) 16T = = J πd 4 / 32 πd 3

16(1200) = 14,500 π (0.75)3

σ x +σ z

2

=

(300 Ibf)

1.8 450 ⎛ ⎞ ⎜3 + ⎟ = 19,400 psi 1.25(0.252 ) ⎝ 1.25 / 0.25 ⎠

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

BDA 31203 – Mechanical Component Design

Solution : (c)

=

(300 Ibf)

T ⎛ 1.8 ⎞ ⎜3 + ⎟ bc 2 ⎝ b / c ⎠

The maximum shear stress is 2

⎛ σ x −σ z ⎞ 2 ⎟ + τ xz ⎝ 2 ⎠

τ1 = ⎜

2

⎛ 47.1 − 0 ⎞ 2 = ⎜ ⎟ + 14.5 ⎝ 2 ⎠

= 27.7

kpsi

psi

43

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

44

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

Example 2:

Solution :

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.

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

•

CHAPTER 2 – Analysis and Synthesis

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

•

•

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.

•

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

2

2

Tr T (d / 2) 16T 16(1600) = = = = 2414 J πd 4 / 32 πd 3 π (1.53 ) σ x +σ z 2

2

2

24,890 ⎛ σ −σ z ⎞ ⎛ 24,890 ⎞ 2 2 + ⎜ x + ⎜ ⎟ + τ xz = ⎟ + 2414 = 25,120 psi 2 ⎝ 2 ⎠ ⎝ 2 ⎠

The extreme shear stress τ1 is given by:

τ1 = ⎜ Ibf.in

M C = 4000 + 4000 = 5657 Ibf.in

psi

2

2

⎛σ x −σ z ⎞ ⎛ 24,890 ⎞ 2 2 ⎟ + 2414 = 12,680 psi ⎟ + τ xz = ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠

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

M B = 2000 2 + 8000 2 = 8246

psi

Maximum tensile stress σ1 is given by:

σ1 = •

Mc M (d / 2) 32 M 32(8246) = = = = 24,890 I πd 4 / 64 πd 3 π (1.53 )

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

τ=

M = M y2 + M z2

CHAPTER 2 – Analysis and Synthesis

Solution :

σ=

•

46

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

BDA 31203 – Mechanical Component Design

Solution : •

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

45

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

BDA 31203 – Mechanical Component Design

CHAPTER 2 – Analysis and Synthesis

BDA 31203 – Mechanical Component Design

47

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

48

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

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

Week 2

Chapter 3

Static Design Failure of Theories Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

What you will be learn here? •

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

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

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

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

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

4

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.1 – Introduction – cont…

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.2 – Why needs failure theories?

Safety and Failure:¾ Failure can mean a part has separated into two or more pieces and become permanently deformed

¾

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

¾

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

¾ Why parts fail ⇒ stresses exceed its strength. ¾ Can be categories :

⇒ under static loading ⇒ under dynamic loading

Static Load:¾ is a stationary force or couple applied to a member ¾ the force or couple must be unchanging in magnitude, point or points of application, and direction. ¾ can produce axial tension or compression, a shear load, a bending load, a torsional load or any combination of these. 5

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

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.3 – Static Failure Theories

6

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.3 – Static Failure Theories – cont… Static Failure Theories

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

Ductile Materials

™ Over the years, several hypotheses have been formulated and -

tested, leading today’s engineering practice.

-

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

ε f ≥ 0.05 S yt = S yc = S y Yield Criteria

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

Brittle Materials -

ε f < 0.05 Sut & Suc Fracture Criteria

-Maximum Normal Stress -Brittle-Coulomb-Mohr

( example 1 , example 2 , example 3 ) Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

7

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

8

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.4 – Maximum Shear Stress (MSS) Theory ™

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

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

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

τ max =

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

Sy 2

from the tensile surface with a magnitude of:

τ max = or at yield,

σ 2

σ −σ

∴ 1

τmax = S y 2

9

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

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

S sy = 0.50S y Therefore, taking

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

BDA 31203 – Mechanical Component Design

S 3= y 2 2 10

CHAPTER 3 – Static Design Failure of Theories

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

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

N as safety factor; Maximum Allowable Load or Stress

Usually safety factor, N is defined by;

Case 1 : σA ≥ σB ≥ 0.

Applied Load or Stress

For this case, σ1 = σA , σ3 = 0.

N=

S sy

τ max

=

0.50S y

τ max

=

Sy / 2 (σ 1 − σ 3 ) / 2

=

Case 2 : σA ≥ 0 ≥ σB.

Sy

Here, σ1 = σA , σ3 = σB.

(σ 1 − σ 3 )

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

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

11

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

12

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.5 – Distortion Energy (DE) Theory

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.5 – Distortion Energy (DE) Theory – cont…

¾ 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. ¾ Ud at element in specimen ≥ Ud for yield in simple tension. ¾ Also known as von Mises or von Mises-Hencky Theory ¾ Developed by studying a unit volume in a three-dimensional

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

stress state

(b) Element under hydrostatic tension undergoes only volume change. (c) Element has angular distortion without volume change. 13

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

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.5 – Distortion Energy (DE) Theory – cont… strain energy = U =

1 σε 2

U=

14

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

σ av =

1 (σ 1ε1 + σ 2ε 2 + σ 3ε 3 ) 2

where

U = Uv + Ud

3.5 – Distortion Energy (DE) Theory – cont…

for 3-D analysis;

U=

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

σ1 + σ 2 + σ 3 3

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:-

1 (σ 1 − vσ 2 − vσ 3 ) E 1 ε 2 = (σ 2 − vσ 1 − vσ 3 ) E 1 ε 3 = (σ 3 − vσ 1 − vσ 2 ) E

ε1 =

Uv =

3σ 2 av (1 − 2v ) 2E

------- (2)

1 ⎡σ 12 + σ 22 + σ 32 − 2v (σ 1σ 2 + σ 2σ 3 + σ 1σ 3 ) ⎤⎦ ------- (1) 2E ⎣ Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

15

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

16

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.5 – Distortion Energy (DE) Theory – cont…

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.5 – Distortion Energy (DE) Theory – cont… Von Mises Effective Stress

Ud = U −Uv

σ ' = σ 12 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 1σ 3

2

⎛σ +σ2 +σ3 ⎞ ⎟ (1 − 2v ) ⎧ 1 ⎫ 3⎜ 1 ⎡⎣σ 12 + σ 22 + σ 32 − 2v (σ 1σ 2 + σ 2σ 3 + σ 1σ 3 ) ⎤⎦ ⎬ =⎨ 3 ⎝ ⎠ ⎩ 2E ⎭ − 2E

1+ v 2 ⎡⎣σ 1 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 1σ 3 ⎤⎦ Ud = 3E Ud =

(1 + v ) 3E

S y2

σ = '

--- for element in speciment

(σ

− σ y ) + (σ y − σ z ) + (σ z − σ x ) + 6 (τ xy2 + τ yz2 + τ zx2 ) 2

x

2

2

2

σ ' = σ 12 − σ 1σ 3 + σ 32

(for 2D principal stress)

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

σ ' = σ x2 + σ y2 − σ xσ y + 3τ xy2 17

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.5 – Distortion Energy (DE) Theory – cont…

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

For Pure Shear:

1+υ 2 1+υ 2 σ 1 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 1σ 3 ≥ Sy 3E 3E

]

2 S y2 = σ 12 + σ 1σ 1 + σ 12 = 3σ 12 = 3τ max

σ1 =

σ 12 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 1σ 3 ≥ S y

Sy 3

= 0.577 S y = τ max

σ average

σ ' ≥ S y2 Therefore, safety factor N is:

18

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

3.5 – Distortion Energy (DE) Theory – cont…

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

[

(for 2D plane stress)

N=

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

Sy

σ'

σ average 19

τ max

n=

S sy

τ max

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

S sy = 0.577 S y 20

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.5 – Distortion Energy (DE) Theory – cont…

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.6 – Coulomb-Mohr Theory - used when Syt ≠ Syc - based on Mohr’s theory, whereby the failure line is assumed to be straight

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

Three Mohr circle, one for the

σ’ = Sy.

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

S sy = 0.577 S y

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

21

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.6 – Coulomb-Mohr Theory – cont…

22

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.7 – Maximum Normal Stress (MNS) Theory

From the diagram, equation developed can be simplified to:

σ1 σ 3 St

−

Sc

=1

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

Incorporating the safety factor;

σ1 > Sut

σ1 σ 3

1 − = St Sc n

S sy

τ max

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

σ3 < - Suc

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

For pure torsional shear strength;

n=

or

S sy =

S yt S yc S yt + S yc 23

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

24

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.7 – Maximum Normal Stress (MNS) Theory – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.8 – Modification of Mohr Theory - 3 modifications of the existing Mohr theory are applicable in analyzing brittle materials.

Graph of maximum-normal

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

stress (MNS) theory of failure for plane stress states. Stress states that plot inside the failure locus

a) Brittle Coulomb Mohr Theory

are safe.

b) Modified I-Mohr c) Modified II-Mohr

25

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

CHAPTER 3 – Static Design Failure of Theories

BDA 31203 – Mechanical Component Design

3.8 – Modification of Mohr Theory – cont…

26

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

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

3.8 – Modification of Mohr Theory – cont…

Equations Theories

σA ≥σB ≥ 0

σA ≥ 0 ≥σB σA

S ut n

σB

0 ≥σA ≥σB

1 n

Brittle CoulombMohr

σA ≥

Modified I-Mohr

σA ≥

S ut n

( Suc − Sut )σ A σ B 1 − = Suc Sut Suc n

Modified II-Mohr

σA ≥

S ut n

nσ A ⎛ nσ B + Sut ⎞ ⎟ =1 +⎜ Sut ⎜⎝ Sut − Suc ⎟⎠

Sut

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

−

Suc

=

σB ≥ −

S uc n

σB ≥ −

S uc n

σB ≥ −

S uc n

2

27

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

28

BDA 31203 – Mechanical Component Design

CHAPTER 3 – Static Design Failure of Theories

Conclusion

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

29

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 [MPa]

500

200 - 500

•A

- 50

500

[MPa]

B

••C

X Y - 500

FOS

DE ⇒ n =

y 200

x MSS ⇒ n = 200

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.

σx =

MC ( Fl ) c 1 000 ( 6)( 0.75) = = = 18 108 psi 0.249 I I

τ xz =

Tr ( Fa ) r 1 000 ( 8)( 0.75) = = = 12 072 psi J J 0.497

τ max

⎛σ −σ z ⎞ ⎛ 18108 − 0 ⎞ 2 = ⎜ x ⎟ + τ 2 xy = ⎜ ⎟ + 12072 = 15090 psi 2 ⎠ 2 ⎝ ⎠ ⎝

2

2.

σ1 =

σ x +σ z 2

2

+ τ max =

18108 + 15 090 = 24144 psi 2

− τ max =

18108 − 15090 = −6036 psi 2

σ2 = 0 σ3 =

3.

σx +σz 2

σ ' = σ 1 2 − σ 1σ 3 + σ 3 2

σ ' = 24144 2 − 24144(−6036) + (−6036) 2 = 27 661 psi

4.

N=

5.

N=

Sy

σ'

=

47000 = 1.7 ------------- DE theory 27661

0.50 S y

τ max

=

0.50(47000) = 1.6 --------------- MSS theory 15090

Element at point B

6.

τ bending =

4V 4(1000) = = 755 psi 3 A 3(1.767)

τ max = τ torsion + τ bending = 12072 + 755 = 12827 psi

7.

N=

N=

0.577 S y

τ max 0.50 S y

τ max

=

=

0.577(47000) = 2.1 --------- DE theory 12827

0.50(47000) = 1.8 -------- MSS theory 12827

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

τ=

Tr 16T 16(230) = = = 75MPa J πd 3 π 2.53

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

∴n =

σ1 S yt

1 −

σ3

=

1 = 1.10 75 / 160 − (−75) / 170

S yc

Alternatively;

S sy =

S yt S yc S yt + S yc

∴n =

S sy

τ max

=

82.4 = 1.10 75

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

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

Week 3 & 4

Chapter 4

Fatigue Design Failure of Theories

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

Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

What you will be learn here? •

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.1 – Introduction

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

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

3

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

4

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.2 – Fatigue Load and Failure

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.2 – Fatigue Load and Failure – cont… Example of repeated load

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

1st Case – Bending a steel wire repeatedly

2nd Case – Impact and Vibration on vehicle axle NG

repeated or fluctuating stresses; yet the most careful analysis reveals that the actual maximum stresses were well below the ultimate

Too much repeatedly Impact and Vibration will break the axle

NG

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

Too much repeatedly Bending that beyond the limit will break the steel wire

strength. The most distinguishing characteristic of these failures is

3rd Case – Steel Bridge

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

4th Case - Vehicle Suspension

Hence the failure is called a fatigue failure.

5

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.2 – Fatigue Load and Failure – cont…

4.2 – Fatigue Load and Failure – cont… 2nd Case – Vehicle Axle

Wire Wireisisbend bendfrom fromthe thetop top ↓↓ Top of the wire is suffered Top of the wire is sufferedtoto “Tension” “Tension”(+σ) (+σ) ↓↓ Meanwhile Meanwhilethe thebottom bottomisissuffered sufferedtoto “Compression” “Compression”(-σ) (-σ)

Stress : Bending Stress σ and Shear Stress τ

τ

+

σa σm = 0

Tension

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.

Graph Graphplotting plottingfor foroverall overallwire wirebending bending process: process: σ

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

1st Case – Continuously bending the steel wire Stress : Bending stress; σ

6

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

σa

t

σr

τm

−

Compression

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

σ

This type of stress is known as Completely This type of stress is known as Completely Reverse Stress Reverse Stress

Shear Stress cause by torsion

7

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

8

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.2 – Fatigue Load and Failure – cont… Fatigue Fatiguefailure failureofofaabolt boltdue duetotorepeated repeated unidirectional unidirectionalbending. bending.Fatigue Fatiguefailure failurestart startwith with small crack that unseen with naked small crack that unseen with nakedeyes eyesand and also difficult to detect with X-ray at the thread also difficult to detect with X-ray at the thread root rootatatA, A,propagated propagatedacross acrossmost mostofofthe thecross cross section sectionshown shownby bythe thebeach beachmarks marksatatB, B,before before final fast fracture at C. final fast fracture at C.

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.3 – Life and Fatigue Strength

Crack Crackoften oftenstart startatat weak weakpart partgeometries geometries which whichhave havediscontinuity discontinuityininmaterial materialsuch suchas as atatholes, holes,keyways, keyways,notch, notch,fillet filletand andothers others(at (at this thislocation, location,the thestress stressisishigh highbecause becauseofof high highstress stressconcentration) concentration)

•

Three major fatigue life models

•

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

i.

Stress-life method 9 Least accurate, particularly for low cycle applications 9 Most traditional, easiest to implement

ii. Strain-life method 9 Detailed analysis of plastic deformation at localized regions 9 Several idealizations are compounded, leading to uncertainties in results iii. Linear-elastic fracture mechanics method 9 Assumes crack exists 9 Predicts crack growth with respect to stress intensity

9

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.3 – Life and Fatigue Strength – cont… Fatigue Fatiguestrength strengthalso alsohave haveits itsmaximum maximumlimits. limits. compression

10

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.4 – Stress Life Method An AnS-N S-Ndiagram diagramplotted plottedfrom fromthe theresults resultsofofcompletely completelyreversed reversedaxial axialfatigue fatiguetest. test.Material Material UNS strength,NN: :cycle cycle UNSG4100 G4100steel, steel,normalized. normalized.––SS: :strength,

R. R Moore Test Sample ujian

tension motor

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

specimen motor Endurance limit , Se or known as fatigue limit

F

Low-cycle fatigue High-cycle fatigue

Rotating Beam Test

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

S-N Diagram

11

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12

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.5 – Endurance Limit, Se

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.5 – Endurance Limit, Se – cont…

Basically, Basically,the thefatigue fatigueendurance endurancelimits limitsSSe eare aredetermine determinethrough throughthe thetest. test.Yet Yetthe thedata dataare arealso alsoavailable available on onthese thesestandard: standard: i.i.American AmericanSociety SocietyofofTesting Testingand andMaterials Materials(ASTM) (ASTM) ii.ii.American AmericanIron Ironand andSteel SteelInstitute Institute(AISI) (AISI) iii. Society of Automotive Engineer (SAE) iii. Society of Automotive Engineer (SAE) ItItisisunrealistic unrealistictotoexpect expectthe theendurance endurancelimit limitofofaamechanical mechanicalororstructural structuralmember membertotomatch matchthe thevalues values obtained obtainedininthe thelaboratory laboratory

So, So,the thevalues valuesobtain obtainfrom fromlab labtest testare areknown knownas asRotary Rotarybeam beamtest testspecimen specimenendurance endurancelimit, limit,SSe’e.’ However . Howeverthere there isisaarelation relationexist existbetween betweenSSe’e’ and andSSutut. . Value Valueofofaamechanical mechanicalororstructural structuralmember membertotomatch matchthe thevalues valuesobtained obtainedininthe thelaboratory laboratoryafter afterconsidering considering other otherfactors factorsthat thatinfluence influencethe thefatigue fatiguelife lifeisisknown knownas asendurance endurancelimits, limits,SSe e. . Se’ – refer to the endurance limit of the controlled laboratory specimen Se – refer to the endurance limit of an actual machine element subjected to any kind of loading

13

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

BDA 31203 – Mechanical Component Design

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.

CHAPTER 4 – Fatigue Design Failure of Theories

4.5 – Endurance Limit, Se – cont…

14

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.6 – Endurance Limit Modifying Factors Factors Factorsthat thatinfluence influencethe thefatigue fatiguelife lifeand andendurance endurancelimits. limits.

Assumption AssumptionofofSSe’e’ value valuefor forsteel steel For Forstudent studentapplication: application:

SSe’ ’ ==0.5 0.5SSutut →→ SSutut≤≤1400 1400MPa MPa[[200 200kpsi kpsi]] e SSe’ ’ ==700 700MPa MPa[[100 100kpsi kpsi]] →→ SSutut>>1400 1400MPa MPa e For Forreal realengineering engineeringpractice: practice:

SSe’ ’ ==0.4 0.4SSutut →→ SSutut≤≤1400 1400MPa MPa e SSe’ ’ ==550 550MPa MPa[[84.1 84.1kpsi kpsi]→ ]→ SSutut>>1400 1400MPa MPa e Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

15

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

16

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.6 – Endurance Limit Modifying Factors – cont…

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.6 – Endurance Limit Modifying Factors – cont… Surface SurfaceCondition ConditionModification ModificationFactor, Factor,kkaa

AAMarin MarinEquation Equationisistherefore thereforewritten writtenthe theendurance endurancelimit limitSSeeas: as:

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:

Se = kakbkckdkekfSe’

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

Where, Where,

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

SSe’ =’ =rotary beam test endurance limit e rotary beam test endurance limit kka ==surface condition surface conditionmodification modificationfactor factor a kkb ==size modification factor b size modification factor kkc ==load modification factor c load modification factor kkd ==temperature modification temperature modificationfactor factor d kke ==reliability factor e reliability factor kkf ==miscellaneous effect modification factor f miscellaneous effect modification factor

17

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

BDA 31203 – Mechanical Component Design

Surface Finish

CHAPTER 4 – Fatigue Design Failure of Theories

Factor a Sut, kpsi

Sut, MPa

Exponent b

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 18

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.6 – Endurance Limit Modifying Factors – cont…

4.6 – Endurance Limit Modifying Factors – cont…

Surface SurfaceCondition ConditionModification ModificationFactor, Factor,kkaa

Size SizeFactor, Factor,kkbb

EXAMPLE 1

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

⎧ (d / 0.3)−0.107 = 0.879d −0.107 ⎪ − 0.157 ⎪ 0.91d kb = ⎨ − 0.107 = 1.24d −0.107 ⎪( d / 7.62) ⎪ 1.51d −0.157 ⎩

A steel has a minimum ultimate strength of 520 MPa and a machined surface. Estimate ka. Solution From Table 4–1, a = 4.51 and b =−0.265. Then, Answer

- depends by the types of the load

0.11 ≤ d ≤ 2 in 2 < d ≤ 10 in 2.79 ≤ d ≤ 51 mm 51 < d ≤ 254 mm

where d – effective dimension

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

ka = 4.51(520)−0.265 = 0.860

D

de= 0.370d

h

de = 0.808 (hb)1/2 b

~~there thereisisno nosize sizeeffect, effect,so sosize sizefactor factorkkbb==1.0 1.0 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

19

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20

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.6 – Endurance Limit Modifying Factors – cont…

4.6 – Endurance Limit Modifying Factors – cont…

Size SizeFactor, Factor,kkbb

Loading LoadingFactor, Factor,kkc c

EXAMPLE 2 A 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.

When Whenfatigue fatiguetests testsare arecarried carriedout outwith withrotating rotatingbending, bending,axial axial (push-pull), and torsional loading, the endurance (push-pull), and torsional loading, the endurancelimits limitsdiffer differ with withSSutut..Here Hereare arethe thevalues valuesof ofthe theload loadfactor factoras as

Solution (a)

(b)

⎛ d ⎞ kb = ⎜ ⎟ ⎝ 7.62 ⎠

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

−0.107

⎛ 32 ⎞ =⎜ ⎟ ⎝ 7.62 ⎠

⎧ 1 ⎪ kc = ⎨0.85 ⎪0.59 ⎩

−0.107

= 0.858

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

⎛ 11.84 ⎞ kb = ⎜ ⎟ ⎝ 7.62 ⎠

−0.107

= 0.954

Axial Torsion

** If there is a combination of loads, use kc = 1 21

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

BDA 31203 – Mechanical Component Design

Bending

CHAPTER 4 – Fatigue Design Failure of Theories

22

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.6 – Endurance Limit Modifying Factors – cont…

4.6 – Endurance Limit Modifying Factors – cont…

Temperature TemperatureFactor, Factor,kkdd

Temperature TemperatureFactor, Factor,kkdd

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)

EXAMPLE 3 A 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 :

⎛ 450 − 400 ⎞ k d = (ST / S RT )450o = 1.018 + (0.995 − 1.018)⎜ ⎟ = 1.007 ⎝ 500 − 400 ⎠

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

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

23

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

24

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.6 – Endurance Limit Modifying Factors – cont…

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.6 – Endurance Limit Modifying Factors – cont…

Reliability ReliabilityFactor, Factor,kkee

Miscellaneous-Effects Miscellaneous-EffectsFactor, Factor,kkf f

ke = 1 − 0.08 zα

the thereliability reliabilitymodification, modification,kkeefactor factorisiswritten writtenas as

kkf factor is intended to account for the reduction in endurance limit due to all other f factor is intended to account for the reduction in endurance limit due to all other effects effectssuch suchas, as,Corrosion, Corrosion,Electrolytic ElectrolyticPlating, Plating,Metal MetalSpraying, Spraying,Cyclic CyclicFrequency, Frequency,and and other othermore. more.

where zα values can be determined from Table A–10 in Appendix A. Table below gives reliability factors for some standard specified reliabilities.

However Howeverthe theactual actualvalues valuesofofkkf fare arenot notalways alwaysavailable. available.IfIfthis thisfactor factorisisnot notimportant, important, assume assume; ;

kkf ==1.0 1.0 f ItItisisreally reallyintended intendedas asaareminder reminderthat thatthis thisfactor factormust mustbe beaccounted accountedininreal realengineering engineering practice. practice. Reliability Factors ke Corresponding to 8% Standard Deviation of the Endurance Limit 25

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.7 – Stress Concentration and Notch Sensitivity σ Stress distribution

σ0= F/A0 σmax A

A

A0 = d (w-d )t

Stress trajectories

B

A = wt w

B

• 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

Regular feature

σ

AAtheoritical theoriticalor orgeometric geometricstress stressconcentration concentrationfactor factorKKt t––isis used usedto torelate relatethe theactual actualmaximum maximumstress stressat atthe thediscontinity discontinityto to the thenominal nominalstress. stress.

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

F

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.7 – Stress Concentration and Notch Sensitivity – cont…

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

F

26

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

Changes in cross section

Kt =

σ max σo

or

K ts =

τ max τo

where Kt : is used for normal stresses Kts : is used for shear stresses • Kt or Kts depends on geometry of the part

σ= F/A σmax > σ0 > σ

F

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

F notch

F

F

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

hole

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

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

27

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28

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.7 – Stress Concentration and Notch Sensitivity – cont…

K f −1

q=

t

or

4.7 – Stress Concentration and Notch Sensitivity – cont… Notch sensitivity q is defined by the equation :

For Forvariable variableloading loadingcases casesthat thatcause causefatigue, fatigue,some somematerials materialsare arenot notfully fully sensitive to the presence of notches and hence, for these, a reduced sensitive to the presence of notches and hence, for these, a reducedvalue value of ofKKt can canbe beused. used.For Forthese thesematerials, materials,the themaximum maximumstress stressis, is,ininfact, fact,

σ max = K f σ o

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

τ max = K fsτ o

Kt −1

qshear =

or

K fs − 1 K ts − 1

where q is usually between 0 ~ 1. IfIfqq==00, ,then thenKKf f==11→→material materialhas hasno nosensitivity sensitivityto tonotches notchesat atall. all. IfIfqq==11, ,then thenKKf =K =Kt →→material materialhas hasfull fullnotch notchsensitivity. sensitivity. f

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

t

In analysis or design work : Find Kt from geometry of the part

σ0 is the nominal stress.

Specify the material

K f = 1 + q(K t − 1) Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

BDA 31203 – Mechanical Component Design

29

CHAPTER 4 – Fatigue Design Failure of Theories

4.7 – Stress Concentration and Notch Sensitivity – cont… Notch sensitivity q

Solve for Kf

K fs = 1 + qshear (K ts − 1) 30

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.7 – Stress Concentration and Notch Sensitivity – cont… Notch sensitivity q for torsion load

for bending and axial load

Figure 4–1 Notch-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. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

or

Find q

31

Figure 4–2 Notch-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). Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

32

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.7 – Stress Concentration and Notch Sensitivity – cont… • •

•

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

K f = 1+ • •

•

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.7 – Stress Concentration and Notch Sensitivity – cont… EXAMPLE 4 A 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).

Kt − 1 1+ a / r

Solution From 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.

where a is defined as the Neuber constant and is a material constant. Then notch sensitivity equation become q = 1 a 1+ r For steel, with Sut in kpsi, the Neuber constant can be approximated by a third-order polynomial fit of data as -3

-5

2 ut

-8

(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,

a = 0.246 - 3.08(10-3 )100 + 1.51(10-5 )100 2 - 2.67(10-8 )1003

3 ut

9Bending or Axial:

a = 0.246 - 3.08(10 )Sut + 1.51(10 )S - 2.67(10 )S

9Torsion:

a = 0.190 - 2.51(10-3 )Sut + 1.35(10-5 )S2ut - 2.67(10-8 )S3ut

= 0.0622 in = 0.313 mm Substituting this into next Eq. with r = 3 mm gives 33

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.8 – Fatigue Strength

K f = 1+

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

Kt −1 1.65 − 1 = 1+ = 1.55 0.313 1+ a / r 1+ 3

34

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.8 – Fatigue Strength – cont… Sut, MPa

Fatigue stress Sf

S f = S ut N

S-N Diagram

490

560

630

700

770

840

910

980

1050

1120

1190

1260

1330 1400

Sut

(log f ) / 3

fSut

Figure 4–3 Fatigue strength fraction, f,

S f = aN

of Sut at 103 cycles for

b

Se = Se’ = 0.5Sut .

Low cycle, Finite life

a=

( f Sut ) 2

High cycle,

Se

1 ⎛ f Sut b = − log⎜⎜ 3 ⎝ Se

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

Finite life

Se

High Cycle, Infinite Life

⎞ ⎟⎟ ⎠ 100

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

103

106 Number of stress cycle (N)

35

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36

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.8 – Fatigue Strength – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.8 – Fatigue Strength – cont…

EXAMPLE 5

Solution

Given a 1050 HR steel, estimate

(a) From Table A–20, Sut = 620 MPa.

Fatigue stress Sf

(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.

(c) With S’f = 385 MPa,

∴ Se’ = 0.5(630) = 310 MPa

⎛ S' f N = ⎜⎜ ⎝ a

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

Sut fSut

a=

Sf

[0.86(620)]

2

310

1/ b

⎞ ⎟⎟ ⎠

1 / − 0.0785

⎛ 385 ⎞ =⎜ ⎟ ⎝ 917 ⎠

= 63.3(103 ) cycles

= 917

[0.86(620 )] = −0.0785 1 b = − log 3 310

Se

100

103

104

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

BDA 31203 – Mechanical Component Design

S ' f = 917 N −0.0785 = 917(10 4 ) −0.0785 = 445 MPa

106

37

CHAPTER 4 – Fatigue Design Failure of Theories

38

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.9 – Characterizing Fluctuating Stresses

4.9 – Characterizing Fluctuating Stresses

™ Fluctuating stresses in machinery often take the form of sinusoidal pattern

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

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

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

™ Other patterns some quite irregular do occur.

™ 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:

Fm =

Fmax + Fmin 2

Fa =

Fmax − Fmin 2

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

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

39

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

40

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

Some Somestress-time stress-timerelations: relations:

Stress

4.9 – Characterizing Fluctuating Stresses – cont… σa σa

σmax

Stress

σr σm

σmin

(R>0 ) Time

(d) sinusoidal fluctuating stress

Stress

Stress

(a) fluctuating stress with highfrequency ripple

σa σa σmin = 0

Time

(c) nonsinusoidal fluctuating stress

σm

σr

9

Mean (Midrange Stress)

9

Stress Amplitude (Alternating Stress)

9

Stress Ratio

9

Amplitude Ratio

Time

(e) repeated stress Stress

Stress

(b) nonsinusoidal fluctuating stress

Stress Range

4.9 – Characterizing Fluctuating Stresses – cont…

(R =0)

σmax

Time

Time

9

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

σmax

(R =-1)

σa

σm = 0 σmin

Time

σa

σr

(f) Completely reversed sinusoidal stress

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

BDA 31203 – Mechanical Component Design

41

CHAPTER 4 – Fatigue Design Failure of Theories

4.9 – Characterizing Fluctuating Stresses – cont…

42

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.9 – Characterizing Fluctuating Stresses – cont…

For Foraacase casethat thatinvolved involvedfluctuating fluctuatingbending bendingstress stressor orfluctuating fluctuatingshear shear stress stressin inthe thepresent presentof ofaanotch, notch,the theequation equationfor foramplitude amplitudeand andmean meanstress stress have havebecome: become:

σ a = K f σ ao

τ a = K fsτ ao

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43

and

σ m = K f σ mo

and

τ m = K fsτ mo

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44

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.10 – Combination of Loading Modes

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

4.11 – Factor of Safety 5 theories of fatigue failure

For Foraacase casethat thatinvolved involvedcombinations combinationsof ofdifferent differenttypes typesof ofloading, loading,such suchas as combined combinedbending, bending,torsion, torsion,and andaxial. axial.

2

σa= '

σ 'm =

Soderberg

(σ a ) axial ⎤ ⎡ ⎢( K f ) bending (σ a ) bending + ( K f ) axial 0.85 ⎥ + 3 ( K fs ) torsion (τ a ) torsion ⎣ ⎦

[( K

f

) bending (σ m ) bending + ( K f ) axial (σ m ) axial

]

2

[

[

+ 3 ( K f ) torsion (τ m ) torsion

]

σa Se

+

Sy

=

1 n

σa Se

+

σm Sut

=

1 n

Larger static yield

2

nσ a ⎛ nσ m ⎞ ⎟ =1 +⎜ S e ⎜⎝ S ut ⎟⎠

Sy n

For Fortorsion torsion(shear) (shear)load loadonly. only. Use Usethe thesame sameequations equationsas asapply applyfor forσσmm≥≥ 0,0, except replace σ and σ with except replace σmm and σa a withττmmand andττa a, ,use use kkc ==0.59 0.59for forSSe e, ,replace replaceSSututwith withSSsusu==0.67 0.67SSutut c and replace S wih S = 0.577 S . and replace Sy wih Ssy = 0.577Sy . y

BDA 31203 – Mechanical Component Design

σa +σm =

2

Gerber

45

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

2

2

⎛ nσ a ⎞ ⎛ nσ m ⎞ ⎟ =1 ⎜⎜ ⎟⎟ + ⎜ ⎜ ⎟ ⎝ Se ⎠ ⎝ S y ⎠

Asme-elliptic

2

Modified-Goodman

]

σm

CHAPTER 4 – Fatigue Design Failure of Theories

4.11 – Factor of Safety – cont…

sy

y

46

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

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

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

Figure 4–4

Load line which intersection with fatigue criterion Load line which intersection with static Larger criterion

Intersection of static and fatigue criterion

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47

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

48

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

CHAPTER 4 – Fatigue Design Failure of Theories

4.11 – Factor of Safety – cont…

4.11 – Factor of Safety – cont…

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

Load line which intersection with fatigue criterion

Load line which intersection with fatigue criterion

Load line which intersection with static Larger criterion

Load line which intersection with static Larger criterion

Intersection of static and fatigue criterion

Intersection of static and fatigue criterion

49

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

CHAPTER 4 – Fatigue Design Failure of Theories

BDA 31203 – Mechanical Component Design

Other Examples of Relevant Fatigue Exercise

Example 6-8 – page 298 Example 6-9 – page 299 Example 6-10 – page 308

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

Example 6-14 – page 318

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51

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50

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

9 recognize types of gear that available and know the function Week 5

9 identify nomenclature of spur gear

Chapter 5

9 construct a gear

Gears – Part I

9 perform load and power calculations analytically as applied to a gears components. Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

What you will be learn here?

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.1 – Introduction

In this chapter (part 1), generally we will learn:learn:-

¾ To transmit power, torque and speed ⇒ ideally, power is constant

•

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

– Exact velocity ratio

•

5.10 - Interference

– May transfer large power

•

5.11 - Force Analysis (Spur Gear)

– High efficiency

•

5.12 - Force Analysis (Bevel Gear)

•

5.13 - Force Analysis (Helical Gear)

¾ To reduce and increase speed and torque ¾ To increase efficiency and reliability of a system (no slip) compare using other mechanism such as belt and pulley or chain and sprocket. Advantages:

– Reliable – Compact

Disadvantages: – High cost in manufacturing (requires special tools) – Vibration & noise – Lubricant required

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

3

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

4

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

5.2 – Type of Gears

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

5.2 – Type of Gears – cont… SPUR GEAR

• • • •

Spur Gear Helical Gear Bevel Gear Worm Gear

• have teeth parallel to the axis of rotation. • used to transmit rotary motion between parallel shafts. • the simplest gear.

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

BDA 31203 – Mechanical Component Design

5

CHAPTER 5 – Gears

5.2 – Type of Gears – cont…

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BDA 31203 – Mechanical Component Design

6

CHAPTER 5 – Gears

5.2 – Type of Gears – cont…

HELICAL GEAR

BEVEL GEAR

• have teeth inclined to the axis of rotation.

ƒ

are essentially conically shaped.

ƒ

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

ƒ

Bevel gear may be classified as follow:-

• Not make much noise during meshing. • Inclined tooth also develops thrust loads and bending couples.

9 Straight bevel gears

• Sometimes used to transmit motion between nonparallel shafts.

9 Spiral bevel gears 9 Zerol bevel gears 9 Hypoid gears 9 Spiroid gears

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

7

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

8

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.2 – Type of Gears – cont…

5.2 – Type of Gears – cont…

DIFFERENTIAL

STRAIGHT BEVEL GEAR

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

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

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

BDA 31203 – Mechanical Component Design

10

CHAPTER 5 – Gears

ZEROL BEVEL GEAR

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.

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

•

5.2 – Type of Gears – cont…

SPIRAL BEVEL GEAR

•

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

9

5.2 – Type of Gears – cont…

•

•

CHAPTER 5 – Gears

11

•

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

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12

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.2 – Type of Gears – cont…

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

5.2 – Type of Gears – cont…

HYPOID AND SPIROID GEARS

•

BDA 31203 – Mechanical Component Design

WORM GEARS

ƒ Used to transmit rotary motion between nonparallel and not intersecting shafts. Spiroid Ring gear

Hypoid

Spiral gear

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BDA 31203 – Mechanical Component Design

13

CHAPTER 5 – Gears

5.2 – Type of Gears – cont…

ƒ 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. ƒ The worm can always drive the gear. However, if the gear attempts to drive the worm, it may or may not succeed. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

BDA 31203 – Mechanical Component Design

14

CHAPTER 5 – Gears

5.3 – Nomenclature of Spur Gear

RACK & PINION

ƒ A rack is a toothed bar or rod that can be thought of as a sector gear with an infinitely large radius of curvature. ƒ Torque can be converted to linear force by meshing a rack with a pinion: the pinion turns; the rack moves in a straight line. ƒ 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). Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

15

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

16

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.3 – Nomenclature of Spur Gear – cont…

5.3 – Nomenclature of Spur Gear – cont…

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

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

m=

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

πD p N

;

π Pd

1 Pd

(mm or inch)

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

=

N Dp

b = 1.25m ;

(Teeth per inch)

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

BDA 31203 – Mechanical Component Design

17

CHAPTER 5 – Gears

5.3 – Nomenclature of Spur Gear – cont…

1.25 Pd

(mm or inch)

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

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

(mm or inch)

i)

Calculate each pitch circle

d1 =

N1 = mN1 Pd

d2 =

N2 = mN 2 Pd

Where center distance of the gear is, CD =

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

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 pressure 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

vi)

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

rb = r cos φ

Then, draw the addendum and dedendum circle for addendum diameter,

19

d1 + d 2 2

ii)

or radius of base gear is,

• Backlash – is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circles. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

18

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

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

c =b−a

This is a commonly used sizes for spur gear tooth. Refer tooth system for more info.

5.4 – Construction of Gear

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

ht = a + b

(mm)

N

a=m ;

(mm or inch)

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

Dp

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

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

pc =

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

d a = d + 2a

d b = d − 2b 20

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.4 – Construction of Gear – cont…

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.4 – Construction of Gear – cont…

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.

vii)

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

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

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

ix)

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

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

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.

t=

t

αo

pc πm π = = 2 2 2 Pd

(mm or inch)

v) From tooth thickness we can determine the angle needed, α to make a mirror line between it. 180 ⋅ t o ( ) α=

π d

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

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

21

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.5 – Forming of Gear Teeth

22

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

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.6 – Tooth Systems

Milling • Gear teeth may be cut with a milling cutter shaped to conform to the tooth space.

•

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

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.

•

Planned to attain interchangeability.

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

Shaping – Pinion cutter

Tooth Sysem Full Depth

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

Pressure Angle φ, deg

Addendum a

Dedendum b

20

1/Pd or 1m

1.25/Pd or 1.25m

22½

1/Pd or 1m

1.25/Pd or 1.25m

25

1/Pd or 1m

1.25/Pd or 1.25m

20

0.8/Pd or 0.8m

1.35/Pd or 1.35m 1.35/Pd or 1.35m

Shaping – Rack cutter

1.35/Pd or 1.35m Stub

23 Hobbing

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

1/Pd or 1m

24

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.6 – Tooth Systems

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.7 – Gear Ratio •

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:

Table 5-2 Tooth Sizes in General Uses

Diametral Pitch Coarse

2, 2¼, 2½, 3, 4, 6, 8, 10, 12, 16

Fine

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

Modules Preffered

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

ni N o d o To = = = no N i d i Ti Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

25

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.8 – Gear Train •

where

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

•

To calculate overall gear or speed ratio from multiple gearing system

Overall Gear Ratio; mG -O = Final gear

•

N 2 N3 N5 n2 N3 N 4 N6

Train value,

n e= L nF

26

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

5.8 – Gear Train – cont…

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

n6 =

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

where

where

To ni 1 = = Ti no e TF = Torque at the first gear TL = Torque at the final gear nF = speed of the first gear (rev/min) nL = speed of the last gear (rev/min) e = train value

N = number of teeth n = gear speed (rev/min) First gear

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

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

27

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

28

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.8 – Gear Train – cont…

5.8 – Gear Train – cont… Example 5-1

Motor 5 Hp 2000 rpm

Component

Example 5-1

Motor 5 Hp 2000 rpm

Fill the table below:-

Pd = 4 teeth/in

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

Pitch Diameter (in)

Speed (rpm)

Fill the table below:-

Torque (ib-in)

Torque (ib-in)

Pd = 4 teeth/in

motor

-

2000

157.5

-

2000

157.5

-

2 (Pulley)

-

2 (Pulley)

3 (Pulley)

-

3 (Pulley)

Pd = 4 teeth/in

5 (Bevel Gear) 6 (Spur Gear) 7 (Spur Gear)

Pd = 6 teeth/in

Speed (rpm)

motor

4 (Bevel Gear) Pd = 4 teeth/in

Pitch Diameter (in)

Component

8 (Worm Gear)

Pd = 6 teeth/in

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

29

Train value, e = ?

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

5.9 – Contact Ratio

-

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

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

Train value,

e=

nL 19.7 = = 0.00985 nF 2000

30

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.9 – Contact Ratio – cont… rb (gear)

ra (gear)

Base circle (gear)

Dedendum circle (gear) Clearance circle (gear) s Pre

ra (pinion)

• If you were to follow the contact point, it would describe a straight line that starts near one

rb (pinion)

Pitch circle (gear) Addendum circle (gear)

Addendum circle (pinion)

Pitch circle (pinion) Clearance circle (pinion)

Dedendum circle (pinion)

gear and ends up near the other.

Base circle (pinion)

• This means that the radius of the contact point gets larger as the teeth engage. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

ine

L ab act ont of C e L in

• 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.

eL su r

motion

31

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

32

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.9 – Contact Ratio – cont…

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

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

• 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).

• 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.

CR =

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

Lab pc cos φ

Base circle

where Lab = length of the line of action pc = circular pitch φ = pressure angle 2

CR =

2

2

2

ra1 − rb1 + ra 2 − rb 2 − CD sin φ pc cos φ Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

where ra = addendum circle radius rb = base circle radius CD = center distance of the gear

33

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.10 – Interference – cont…

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

• Causing the involute tip or face of the driven gear tends to dig out the noninvolute flank of the driver. 34

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears

5.11 – Force Analysis (Spur Gear)

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

NP =

2k

⎛ m + m 2 + (1 + 2m ) sin 2 φ ⎞ ⎜ G ⎟ G G ⎠ (1 + 2mG ) sin φ ⎝ 2

For spur pinion Where:

NP = NP =

2k 2

sin φ

For spur pinion that operate with a rack

2k cosψ

⎛ m + m 2 + (1 + 2m ) sin 2 φ ⎞ ⎜ G G G t ⎟ ⎠ (1 + 2mG ) sin 2 φt ⎝

k = 1 (for full depth teeth) k = 0.8 (for stub teeth) mG = mating gear ratio φ = pressure angle

For helical pinion Where:

NP =

2k cosψ sin 2 φt

For helical pinion that operate with a rack

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

ψ = helix angle φt = tangential pressure angle (refer slide 44)

35

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

36

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.11 – Force Analysis (Spur Gear) – cont… Gear 3

Fb3

Tb3

3

b

2

English unit

t F 32

Ta2

b t F a2

V=

a r F a2

φ

n2

F32

n2

φ

φ

5.11 – Force Analysis (Spur Gear) – cont…

r F 32

n3

F32

φ

a

SI unit

πdn

V=

12

2

Fa2

F23

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

Wt = 33000

d2 (c)

H V

Wt =

πdn 60

60000 H πdn

Ta2 Pinion

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

Fa2

V = pitch-line velocity, ft/min Wt = transmitted load, Ibf d = gear diameter, in n = gear speed, rev/min H = power, hp

Gear force have been resolved into tangential and radial components.

a

(a)

2

Wt = F32t =

(b) Free body diagrams of the forces and moments acting upon two gears of a simple gear train.

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

T d /2

Wr = F32r = Wt tan φ

r W W = F32 = t kosφ

37

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.11 – Force Analysis (Spur Gear) – cont…

V = pitch-line velocity, mm/s Wt = transmitted load, kN d = gear diameter, mm n = gear speed, rev/min H = power, kW 38

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

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.11 – Force Analysis (Spur Gear) – cont…

Example 5-2

Solution 5-2 The pitch diameters for gears 2 and 3 are :

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.

d2 = mN2 = 2.5(20) = 50 mm d3 = mN3 = 2.5(50) = 125 mm The transmitted load to be :

Wt =

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

60000 H 60000 ( 2 .5) = = 0 .546 kN πd 2 n π (50 )(1750 )

t Thus, the tangential force F23 = 0.546 kN, as shown in free body diagram figure, Therefore

n = 1750 rpm

F23r = F23r tan 20 o = 0.546 tan 20 o = 0.199 kN

H = 2.5 kW

φ = 20o

FREE BODY DIAGRAM

and so

m = 2.5 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

39

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

F23 =

F23t 0.546 = = 0.581 kN cos 20o cos 20o 40

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.11 – Force Analysis (Spur Gear) – cont…

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.12 – Force Analysis (Bevel Gear)

Solution 5-2 – cont… 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

F43t = 0.546 kN

F43r = 0.199 kN

Pinion pitch angle :

F43 = 0.581 kN

NP NG

tan γ =

The shaft reactions in the x and y directions are

Fbx3 = −( F23t + F43r ) = −(−0.546 + 0.199) = 0.347 kN Fby3 = −( F23r + F43t ) = −(0.199 − 0.546) = 0.347 kN

Gear pitch angle :

The resultant shaft reaction is

NG NP

tan Γ =

Fb 3 = 0.347 + 0.347 = 0.491 kN

Terminology of bevel gears

41

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

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.12 – Force Analysis (Bevel Gear) – cont…

42

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

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.13 – Force Analysis (Helical Gear)

Transmitted load : y

Normal circular pitch

T Wt = rav

x

Wt

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

W φ

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). z

Wa

γ

Wr

Axial Force :

Wa = Wt tan φ sin γ Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

Axial pitch

px =

pn = pt cosψ

pt tanψ

rav

Radial Force :

Wr = Wt tan φ cos γ

Represent a portion of top view of a helical rack

Since pnPn = π, the normal diametral pitch is

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

Pt cosψ

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:

cosψ = 43

Pn =

tan φn tan φt 44

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.13 – Force Analysis (Helical Gear) – cont…

5.13 – Force Analysis (Helical Gear) – cont…

Example 5-3

Solution 5-3 – cont…

pn = pt cosψ = 15.71cos 25o = 14.24 mm

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

px =

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

pt 15.71 = = 33.69 mm tanψ tan 25o

(d) The transverse pressure angle

d = Nmt = 18(5) = 90 mm

(b)

pt = πmt = π (5) = 15.71 mm 45

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

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

5.13 – Force Analysis (Helical Gear) – cont…

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

Wr = W sin φn Wt = W cos φn cosψ

Wa = W cos φn sinψ 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 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

• ψ = helix angle

47

pn

(c)

mn =

(d)

φt = tan −1 ⎜⎜

Solution 5-3 (a)

CHAPTER 5 – Gears

BDA 31203 – Mechanical Component Design

π

=

14.24

π

= 4.53 mm

or

mn = mt cosψ = 5 cos 25o = 4.53 mm

⎛ tan 20o ⎞ ⎛ tan φn ⎞ ⎟ = 21.88o ⎟⎟ = tan −1 ⎜⎜ o ⎟ ⎝ cosψ ⎠ ⎝ cos 25 ⎠

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

46

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Learning Outcomes

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

At the end of this topic, the students would be able to apply BDA 31203 Notes – Mechanical Component Design

and appreciate the knowledge to:

Week 7 & 8

9 analyses and design of spur and helical gears to resist

Chapter 5

bending failure of the teeth as well as pitting failure of

Gears – Part II

tooth surfaces.

Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

What you will be learn here?

5.14 – Introduction GEAR FAILURE

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

5.14 - Introduction

•

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Teeth Bending Failure Occur when significant tooth stress equals or exceeds either the yield strength or bending endurance strength.

Tooth Surface Pitting Failure Occur when significant contact stress equals or exceeds the surface 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)

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

3

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

4

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.14 – Introduction – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.15 – Lewis Bending Equation

Example of Gear Catalog Source: 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?

•

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):

σ=

σ= 5

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.15 – Lewis Bending Equation – cont…

W t Pd FY

(U.S. customary units) Where;

t

W FmY

(SI units)

Wt Pd m F Y

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

= transmitted force (Ibf or N) = diametral pitch (teeth per inch) = module (mm) = width of gear tooth (inch or mm) = Lewis form factor (dimensionless) (refer Table 5-3)

6

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.15 – Lewis Bending Equation – cont…

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

Dynamic Effects, Kv

Number of Teeth

Y

Number of Teeth

Y

12

0.245

28

0.353

(These Values are for a Normal Pressure Angle

13

0.261

30

0.359

of 20o, Full Depth Teeth, and a Diametral Pitch

14

0.277

34

0.371

of Unity in the Plane of Rotation)

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

Where;

21

0.328

150

0.460

22

0.331

300

0.472

x P

24

0.337

400

0.480

26

0.346

Rack

0.485

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

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

Y=

2xP 3

Interpolation for Y value is needed if number of teeth is not in the table. 7

SI units (Velocity in meters per second)

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

Cast iron, cast profile

Kv =

600 + V 600

Kv =

3.05 + V 3.05

• Gear under dynamic loading;

Cut or milled profile

Kv =

1200 + V 1200

Kv =

6.1 + V 6.1

Hobbed or shaped profile

Kv =

50 + V 50

Kv =

3.56 + V 3.56

Shaved or ground profile

Kv =

78 + V 78

Kv =

5.56 + V 5.56

σ=

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

US customary units (Velocity in feet per minute)

K vW t Pd FY

K Wt σ= v FmY

(U.S. customary units)

(SI units)

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

** Where V is the pitch line velocity 8

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.15 – Lewis Bending Equation – cont…

5.15 – Lewis Bending Equation – cont…

Example Example5-4 5-4

Solution Solution5-4 5-4- -cont… cont…

D p = mN = 3(16) = 48 mm

The pitch diameter is,

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.

So the pitch velocity is,

V = πdn = π (0.048)(20) = 3.02 m/s 6.1 + V 6.1 + 3.02 = = 1.5 6.1 6.1 mFYσ allow where Y = 0.296 for 16 teeth Wt = Kv

Velocity factor is found to be, Solution Solution5-4 5-4

So the transmitted force is,

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

σ allowable =

Sy nd

=

210 = 70.0 MPa 3

Wt =

0.003(0.038)0.296(68.7)106 = 1545.5 N 1.5

The power that can be transmitted is, H = W tV = 1545.5(3.02) = 4667 W This is a rough estimation, and that this approach must not be used for important application

9

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Kv =

10

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.16 – Surface Durability

5.16 – Surface Durability – cont…

™ Because a gear tooth experiences into and out engagement ⇒ causes contact stress ⇒ wear and pitting 9 Wear depends on surface hardness

™ Surface Compressive stress (Hertzian stress), σC

⎡ K W t ⎛ 1 1 ⎞⎤ ⎜⎜ + ⎟⎟⎥ σ c = −C p ⎢ ν ⎣ F cos φ ⎝ r1 r2 ⎠⎦

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

1/ 2

Where; Prolong operation after pitting Cp Kν Wt F

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

11

φ

r1 r2

= elastic coefficient = dynamic or velocity factor = transmitted force (Ibf or N) = width of gear tooth (inch or mm) = Pressure angle d P sin φ = radii of curvature on the pinion profile ⇒ r1 = 2 d G sin φ = radii of curvature on the gear tooth profile ⇒ r2 = 2

™ Sign is negative because σC is a compressive stress. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

12

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.16 – Surface Durability – cont…

5.16 – Surface Durability – cont… Table 5-4: Elastic Coefficient Cp,

™ AGMA defines an elastic coefficient ,Cp by the equation

⎡ ⎤ ⎢ ⎥ 1 ⎥ Cp = ⎢ ⎢ ⎛ 1 −ν P2 1 −ν G2 ⎞ ⎥ ⎟⎥ + ⎢ π ⎜⎜ EG ⎟⎠ ⎥⎦ ⎢⎣ ⎝ E P

psi

(

MPa

)

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

1/ 2

psi

(

MPa

Pinion Material

)

Steel

Where;

νP &νG

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

= Poisson’s ratio for pinion and gear

EP & EG = Modulus of Elasticity for pinion and gear

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

Pinion Modulus of Elasticity EP psi (MPa)*

Malleable Iron 25 x 106 (1.7 x 105)

Nodular Iron 24 x 106 (1.7 x 105)

22 x 106 (1.5 x 105)

Aluminum Bronze 17.5 x 106 (1.2 x 105)

Tin Bronze

30 x 106 (2 x 105)

Steel

Cast Iron

30 x 106 (2 x 105)

2300 (191)

2180 (181)

2160 (179)

2100 (174)

1950 (162)

1900 (158)

6

16 x 106 (1.1 x 105)

Malleable iron

25 x 10 (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

6

17.5 x 10 (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)

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

13

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.16 – Surface Durability – cont…

5.16 – Surface Durability – cont…

Example Example5-5 5-5

Solution Solution5-5 5-5- -cont… cont…

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.

14

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

Calculated the elastic coefficient as

⎤ ⎡ ⎥ ⎢ 1 ⎥ Cp = ⎢ ⎢ ⎛ 1 − (0.292) 2 1 − (0.211) 2 ⎞ ⎥ ⎟ + ⎥ ⎢ π ⎜⎜ 9 100(109 ) ⎟⎠ ⎦ ⎣ ⎝ 207(10 )

1/ 2

= 150927.3

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

Solution Solution5-5 5-5

The gear pitch diameter is, d G = mN G = 3(50) = 150 mm

From Table A-5 EP = 207 GPa

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

EG = 100 GPa

νP = 0.292 νG = 0.211

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

rP = 15

48 sin 20 = 8.2 mm 2

rG =

150 sin 20 = 25.7 mm 2

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

16

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.16 – Surface Durability – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation 2 fundamental stress equations

Solution Solution5-5 5-5- -cont… cont…

• Bending stress

⎡ 1.5(1700) ⎛ 1 1 ⎞⎤ + σ c = −150927.3 ⎢ ⎟⎥ o ⎜ ⎣ 0.038 cos 20 ⎝ 0.0082 0.0257 ⎠⎦

Where ;

⎧ t ⎪⎪W K o K v K s σ =⎨ ⎪W t K o K v K s ⎩⎪

The face width is given as F = 38 mm. Use Kν = 1.5 from example 5-4. Substituting all these values to calculate contact stress as: 1/ 2

= −511.5 MPa

Pd K m K B F J 1 Km KB Fmt J

Wt = tangential transmitted load (ibf or N) (U.S. customary units)

2 C 2 C

Kv = dynamic factor Ks = size factor

(SI units)

Pd = transverse diameteral pitch F

KB = rim thickness factor J

σ c = C p W t Ko Kv K s

2

loss _ of _ function _ load S ⎛ 578 ⎞ = =⎜ n= ⎟ = 1.28 imposed _ load σ ⎝ 511.5 ⎠

Km C f d pF I

= geometry factor for bending strength

mt = transverse metric module

• Pitting resistance (contact stress)

∴ S c = 2.206(262) = 578 MPa

Where ; (U.S. customary or SI units)

Cp = elastic coefficient (√ ibf/in2 or √ N/mm2) Cf = surface condition factor (still not been establish) used Cf = 1 dp = pitch diameter of the pinion (in or mm) I

17

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

= face width (in or mm)

Km = load distribution factor

The surface endurance strength of cast iron can be estimated from S c = 2.206 H B MPa From table A24, ASTM No. 50 cast iron ⇒ HB = 262

Ko = overload factor

5.17 – AGMA Stress Equation – cont…

= geometry factor for pitting resistance

18

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

Overload Factor, Ko

Dynamic Factor, Kv ƒ Used to account for inaccuracies in the manufacture and meshing of gear teeth in action.

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

ƒ AGMA has define a set of quality numbers (Qv) to specify the tolerances for gears of various sizes manufactured to a specified accuracy.

• Similar factors such as application factor or service factor

¾ Qv = 3 to 7 ⇒ for most commercial-quality gears

Table 5-5: Overload Factors, Ko

Power Source

Uniform Generator, Centrifugal compressor, pure liquid mixer

Driven Machine Moderate shock Heavy shock Machine tool main drive, multi-cylinder compressor or pump, liquid + solid mixer

Ore crusher, rolling mill, power 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

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

¾ Qv = 8 to 12 ⇒ for precision quality ƒ Dynamic factor equation :

⎧⎛ A + V ⎞ B ⎟ ⎪⎜ ⎪⎪⎜⎝ A ⎟⎠ Kv ⎨ ⎪⎛ A + 200V ⎪⎜⎜ A ⎪⎩⎝

where ; V in ft/min

⎞ ⎟ ⎟ ⎠

B

V in m/s

A = 50 + 56(1 − B) B = 0.25(12 − Qv ) 2 / 3

ƒ Maximum velocity, (at the end point of the Qv curve) ; 19

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

(Vt )max

⎧[A + (Q − 3] 2 ft/min v ⎪ =⎨ [A + (Qv − 3] 2 m/s ⎪ 200 ⎩ 20

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont… Size Factor, Ks ƒ The size factor reflects nonuniformity of material properties due to size. It depends upon ¾ ¾ ¾ ¾ ¾ ¾ ¾

Tooth size Diameter of part Ratio of tooth size to diameter of part Face width Area of stress pattern Ratio of case depth to tooth size Hardenability and heat treatment

ƒ AGMA size factor equation :

⎛F Y K s = 1.192⎜⎜ ⎝ Pd

⎞ ⎟ ⎟ ⎠

0.0535

(U.S. customary units)

(

K s = 0.904 Fm Y

)

0.0535

(SI units)

ƒ If Ks is less than 1, use Ks = 1 Figure 5-1: Dynamic factor Kv

21

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

22

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

Load Distribution Factor, Km

Load Distribution Factor, Km (cont…)

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

where for uncrowned teeth ⎧1 Cmc ⎨ ⎩0.8 for crowned teeth

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

Examples of crowned teeth

ƒ However this is not always possible. The following procedure is applicable to: ¾ Net face width to pinion pitch diameter ratio F / d ≤ 2 ¾ Gear elements mounted between the bearings

C pf

¾ Face widths up to 40 in ¾ Contact, when loaded, across the full width of the narrowest member

ƒ Face load distribution factor :

K m = Cmf = 1 + Cmc (C pf C pm + Cma Ce ) Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

⎧ F ⎪10d − 0.025 ⎪ ⎪ F =⎨ − 0.0375 + 0.0125 F ⎪10d ⎪ F 2 ⎪10d − 0.1109 + 0.0207 F − 0.000228 F ⎩ for values of

23

F ≤ 1 in 1< F ≤ 17 in 17 < F ≤ 40 in

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

F F = 0.05 is used. < 0.05 , 10d 10d

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

24

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

Load Distribution Factor, Km (cont…)

Load Distribution Factor, Km (cont…)

for straddle-mounted pinion with S1/S < 0.175 ⎧1 C pm ⎨ 1 . 1 for straddle-mounted pinion with S1/S ≥ 0.175 ⎩

Cma = A + BF + CF 2 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)

Figure 5-2: Definition of distances S and S1 used in evaluating Cpm Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

25

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

26

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

Load Distribution Factor, Km (cont…)

Load Distribution Factor, Km (cont…)

⎧ Ce = ⎨ ⎩

0.8 1

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

Figure 5-3: Mesh alignment factor Cma Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

27

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

28

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

Rim-Thickness Factor, KB

Rim-Thickness Factor, KB (cont…)

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

2.242 ⎧ ⎪1.6 ln mB KB = ⎨ ⎪1 ⎩

mB < 1.2 mB ≥ 1.2

where mB is a function of the backup ratio

mB =

tR ht

Figure 5-4: Rim thickness factor KB

29

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

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

BDA 31203 – Mechanical Component Design

30

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

Bending-Strength Geometry Factor, J

Bending-Strength Geometry Factor, J – cont… Used Fig. 5-5 to obtain the geometry factor J for 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.

Figure 5-5: Spur-gear geometry factors J. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

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

31

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

32

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.17 – AGMA Stress Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.17 – AGMA Stress Equation – cont…

Bending-Strength Geometry Factor, J – cont…

Surface-Strength Geometry Factor, I ƒ Also known as pitting-resistance geometry factor by AGMA ƒ For both spur and helical gears:-

⎧ cos φt sin φt mG ⎪⎪ 2m mG + 1 N I =⎨ cos φt sin φt mG ⎪ ⎪⎩ 2mN mG − 1

For helical gear

mN =

external gears

PN = Normal base pitch Pn = Normal circular pitch φn = Normal pressure angle

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

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

mG = Gear ratio (never less than 1)

[

φt = Transverse pressure angle

33

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation

p N = pn cos φn

and

where

internal gears

where:

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

pN 0.95Z

Z = (rP + a ) − rb2 P 2

]

1/ 2

[

+ (rG + a ) − rb2 G 2

]

1/ 2

− (rP + rG )sin φt

r P,G = pitch radius (pinion or gear) a = addendum rb P,G = base circle radius r = r cos φ b t (pinion or gear)

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

34

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

AGMA introduced 2 types of strength:

Equation for allowable bending stress:

⎧ St YN ⎪⎪ S K K σ all ⎨ F T R S Y ⎪ t N ⎪⎩ S F Yθ YZ

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)

(U.S. customary units)

(SI units)

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

Both allowable stress numbers (strength) are for:

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

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

9 Unidirectional loading

YN = Stress cycle factor for bending stress

9 10 million stress cycles

Kt (Yθ) = Temperature factor

9 99 % reliability

KR (YZ) = Reliability factor

(refer Figs. 5-12)

(refer Tables 5-12)

SF = AGMA factor of safety, stress ratio Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

35

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

36

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

Equation for contact bending stress:

⎧ Sc ⎪⎪ S σ c ,all ⎨ H S ⎪ c ⎪⎩ S H

Z N CH KT K R Z N ZW Yθ YZ

(U.S. customary units)

(SI units)

where

Sc = Contact fatigue strength, Ibf/in2 (MPa) ZN = Stress cycle life factor

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

(refer Figs. 5-13)

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

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

Kt (Yθ) = Temperature factor KR (YZ) = Reliability factor

Figure 5-8: Allowable bending stress number for through-hardened steels. The SI equations are St = 0.533HB + 88.3 MPa, grade 1, and St = 0.703 HB + 113 MPa, grade 2.

(refer Tables 5-12)

SH = AGMA factor of safety, stress ratio

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

37

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

38

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

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.)

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.)

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

39

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

40

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

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

Material Designation

Heat Treatment

Steel

Through-hardened

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

Minimum Surface Hardness See Fig. 5-8

Allowable Bending Stress Number, St2 psi (MPa) Grade 1

Grade 2

Grade 3

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)

Carburized and hardened

55 000 (380)

Nitrided (throughhardened steels)

83.5 HR15N

Nitrided

87.5 HR15N

See Fig. 5-9

22 000 (151)

65 000 or (448 or 70 0006 482) See Fig. 5-9

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

Material ASTM A48 gray cast iron

-

75 000 (517)

ASTM A536 ductile (nodular) Iron

-

Material Designation

See Fig. 5-10

As cast

-

5000 (35)

As cast

174 HB

8500 (58)

Class 40

As cast

201 HB

13 000 (90)

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)

Grade 100-70-03

Quenched and tempered

229 HB

27 000 - 40 000 (186 - 275)

Grade 120-90-02

Quenched and tempered

269 HB

31 000 - 44 000 (213 - 275)

Sand cast

Minimum tensile strength 40 000 psi

5700 (39)

Heat treated

Minimum tensile strength 90 000 psi

23 600 (163)

BDA 31203 – Mechanical Component Design

See Fig. 5-10

41

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

Allowable Bending Stress Number, St psi (MPa)

Class 30

ASTM B-148 Alloy 954

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

Typical Minimum Surface Hardness

Class 20

Bronze See Fig. 5-10

Heat Treatment

42

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

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

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

Temperature Before Nitriding, oF

Nitriding, oF

Case

Core

Nitralloy 135

1150

975

62 - 65

30 - 35

Sc = 2.41HB + 237 MPa, grade 2.

Nitralloy 135M

1150

975

62 - 65

32 - 36

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

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

Sc = 2.22HB + 200 MPa, grade 1,

Steel

and

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

43

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

Hardness, Rockwell C Scale

44

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

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

Allowable Contact Stress Number, Sc2 psi (MPa)

Minimum Surface Hardness

Material Designation

Heat Treatment

Steel

Through-hardened

See Fig. 5-11

Flame or induction hardened

50 HRC 54 HRC

180 000 (1240)

Carburized and hardened

Grade 1

Grade 2

Grade 3

See Fig. 5-11

See Fig. 5-11

─

170 000 (1172) 175 000 (1206)

190 000 (1310) 195 000 (1344)

─ ─

225 000 (1551)

275 000 (1896)

Nitrided (throughhardened steels)

83.5 HR15N 84.5 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)

45

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

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

Material ASTM A48 gray cast iron ASTM A536 ductile (nodular) Iron

Material Designation

Heat Treatment

Typical Minimum Surface Hardness

Allowable Contact Stress Number, Sc psi (MPa)

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)

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)

Grade 100-70-03

Quenched and tempered

229 HB

92 000 – 112 000 (634 – 772)

Grade 120-90-02

Quenched and tempered

Bronze ASTM B-148 Alloy 954

269 HB

103 000 – 126 000 (710 – 868)

Sand cast

Minimum tensile strength 40 000 psi

30 000 (206)

Heat treated

Minimum tensile strength 90 000 psi

65 000 (448)

46

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

Stress Cycle Factor for Bending Stress, YN

Stress Cycle Life Factor, ZN

Figure 5-13: Pitting resistance stress-cycle factor ZN.

Figure 5-12: Repeatedly applied bending strength stress-cycle factor YN.

(Source: ANSI/AGMA 2001-D04)

(Source: ANSI/AGMA 2001-D04)

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

47

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

48

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.18 – AGMA Strength Equation – cont…

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;

Both pinion and gear are through hardened

C H = 1 + A' (mG − 1) See figure 10-14 (slide 50)

Hardness ⎛ H B (P) ⎞ ⎜ ⎟ Ratio ⎜H ⎟

⎝

H B (P) H B (G) 1.2 ≤

Surface hardened pinion is mated with through hardened gear

H B (P)

H B (P)

C H = 1 + B ' (450 − H BG ) See figure 10-15 (slide 51)

B (G)

< 1.2

H B (G)

H B (G) Surface hardened pinions with hardnesses of 48 Rockwell C scale or harder mated with through hardened gears (180400 Brinell)

≤ 1.7

⎠ A’ = 0

⎛ H B (P) A' = 8.98(10 −3 )⎜ ⎜ H B (G) ⎝

⎞ ⎟ − 8.29(10 −3 ) ⎟ ⎠

A’ = 0.00698

> 1.7

Figure 5-14: Hardness ratio factor CH (through-hardened steel).

B’ = 0.00075 exp [ 0.0112 fP ]

(Source: ANSI/AGMA 2001-D04)

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

49

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

Hardness Ratio Factor, CH

Hardness Ratio Factor, CH

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

50

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

5.18 – AGMA Strength Equation – cont…

Hardness Ratio Factor, CH

Reliability Factor, KR (YZ)

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

Reliability 0.9999

KR (YZ) 1.50

0.999

1.25

0.99

1.00

0.90

0.85

0.50

0.70

Temperature Factor, KT (Yθ)

temperatur e ≤ 250 o F (120 o C)

K T = Yθ = 1 For higher temperature, heat exchangers may be used to ensure that operating temperatures are considerably below this value.

Figure 5-15: Hardness ratio factor CH (surface-hardened steel pinion). (Source: ANSI/AGMA 2001-D04)

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

51

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

52

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

5.19 – Safety Factors SF and SH

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

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

Safety factor SF guarding against bending failure:

•

Roadmap of spur gear bending equations based on AGMA standards in U.S. customary units. (ANSI/AGMA 2001D04) – 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

* Important:

•

When deciding whether bending or wear is the threat to function, compare SF with S2H. For crowned gears compare SF with S3H.

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

SF =

fully corrected bending strength bending stress

=

S t YN /( K T K R )

σ

Safety factor SH guarding against pitting failure:

SH * =

fully corrected contact strength contact stress

=

S c Z N C H /( K T K R )

σc

53

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Roadmap of Gear Analysis (Spur Gear Bending)

54

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Roadmap of Gear Analysis (Spur Gear Wear)

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

AGMA standards in U.S. customary units. (ANSI/AGMA 2001-D04) Slide 47

N dP = P Pd Wt =

33000 H V

V=

Gear bending endurance strength equation

12

Slide 29, 30

σ all

σ = W t Ko Kv Ks

Slide 22

(slide 36)

Pd K m K B F J

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

SF =

S t YN /( K T K R )

σ

Gear contact stress equation (slide 53)

Gear contact endurance strength equation

12 1

σ c, all =

Slide 19

(slide 37)

Slide 52

⎛ t K C ⎞ ⎜⎜W K o K v K s m f ⎟⎟ dP F I ⎠ ⎝

σ c = Cp

Sc Z N CH S H KT K R

1 if T < 250o F

Slide 23

(slide 18)

Slide 34

Slide 20, 21

Gear only

Wear factor of safety

SH =

S c Z N C H /( K T K R )

σc

(slide 53)

Slide 22

Remember when deciding whether bending or wear is the threat to function, compare SF with S2H. For crowned gears compare SF with S3H. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

33000 H W = V Slide 13, 14

(slide 18) Bending factor of safety

V=

Slide 43 ~ 46

πdn

t

Slide 52

Slide 19 Slide 20, 21

N dP = P Pd

1 if T < 250o F

Slide 23

Gear bending stress equation

S YN = t S F KT K R

Slide 49 (Gear only)

Slide 48

Slide 38 ~ 42

πdn

Remember when deciding whether bending or wear is the threat to function, compare SF with S2H. For crowned gears compare SF with S3H. 55

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

56

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples

Examples – cont…

Example Example5-6 5-6

Solution Solution5-6 5-6

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.

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

d P = N P / Pd = 17 / 10 = 1.7 in

⎛ 59.77 + 801.1 ⎞ ⎟ Kv = ⎜ ⎜ ⎟ 59.77 ⎝ ⎠

d G = 52 / 10 = 5.2 in

V=

πd P n P 12

Wt = a) Find the factor of safety of the gears in bending.

=

π (1.7)1800 12

= 801.1 ft/min

33000 H 33000(4) = = 164.8 ibf V 801.1

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

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.

B = 0.25(12 − 6) 2 / 3 = 0.8255 57

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples – cont…

(C ) (C )

pf P

= 1.5 /[10(1.7)] − 0.0375 + 0.0125(1.5) = 0.0695

pf G

= 0.05 − 0.0375 + 0.0125(1.5) = 0.0313

C pm = 1 (Bearing immediately adjacent) C ma = 0.15 (Commercia l enclosed gear unit) Ce = 1

(K m )P

= 1 + C mc ( C pf C pm + C ma C e ) = 1 + 1[ 0 . 0695 (1) + 0 . 15 (1)]

(K m )G

= 1 . 22 = 1 . 18 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

= 1.377

⎛

(K s )P = 1.192⎜⎜ 1.5 + ⎝

(K s )G = 1.192⎜⎜ 1.5 + ⎝

0.303 ⎞⎟ ⎟ 10 ⎠

0.0535

0.412 ⎞⎟ ⎟ 10 ⎠

= 1.043 0.0535

= 1.052 58

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

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

Examples – cont…

Solution Solution5-6 5-6––cont… cont…

C mc = 1 (uncrowned)

0.8255

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 ,

⎛

A = 50 + 56(1 − 0.8255) = 59.77

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Solution Solution5-6 5-6––cont… cont… 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: 8 −0.0178

(YN ) P = 1.3558 (10 )

= 0.977

(YN ) G = 1.3558 (10 8 / 3.059 ) − 0.0178 = 0.996 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;

I=

cos 20 sin 20 3.059 = 0.121 2 3.059 + 1 o

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

psi.

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) ;

(S t )P (S t )G

= 77 .3( 240 ) + 12 ,800 = 31,350 psi = 77 .3( 200 ) + 12 ,800 = 28 , 260 psi

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

(S c )P (S c )G

= 322 ( 240 ) + 29 ,100 = 106 , 400 psi

Form figure 5-13 (slide 48) ;

(Z N )P = 1.4488 (10 8 ) −0.023 = 0.948 (Z N )G = 1.4488 (10 8 / 3.059 ) − 0.023 = 0.973 For the hardness ratio factor CH, the hardness ratio is HB(P) / HB(G) = 240/200 = 1.2. Then, from slide 49

A' = 8 .98 (10 −3 )(1 .2 ) − 8 .29 (10 −3 ) = 0 .00249 ∴ C H = 1 + 0 .00249 (3 .059 − 1) = 1 .005

= 322 ( 200 ) + 29 ,100 = 93 ,500 psi

o

59

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

60

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples – cont…

Examples – cont… Gear tooth bending

Solution Solution5-6 5-6––cont… cont…

⎛ 10 ⎞ ⎛ 1 . 18 (1) ⎞ (σ ) G = 164 . 8 (1)(1 . 377 )(1 . 052 ) ⎜ ⎟⎜ ⎟ ⎝ 1 . 5 ⎠ ⎝ 0 . 40 ⎠ = 4695 psi

(a) Pinion tooth bending

P K K ⎞ ⎛ (σ ) P = ⎜ W t K o K v K s d m B ⎟ F J ⎝ ⎠P ⎛ 10 ⎞ ⎛ 1 .22 (1) ⎞ = 164 .8 (1)(1 .377 )(1 .043 ) ⎜ ⎟⎜ ⎟ ⎝ 1 .5 ⎠ ⎝ 0 .30 ⎠ = 6417 psi

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

⎛ 28 , 260 ( 0 .996 ) / 1( 0 .85 ) ⎞ (S F )G = ⎜ ⎟ 4695 ⎝ ⎠ = 7.05

⎛ S Z C / KT K R ⎞ ⎟⎟ ( S H ) G = ⎜⎜ c N H σc ⎝ ⎠G ⎛ 93,500 ( 0 .973 )(1 .005 ) / 1( 0 .85 ) ⎞ =⎜ ⎟ 69496 ⎝ ⎠ = 1.55

Solution Solution5-6 5-6––cont… cont… ⎛ S Z / KT K R ( S H ) P = ⎜⎜ c N σc ⎝

⎞ ⎟ ⎟ ⎠P

⎛ 106 , 400 ( 0 .948 ) / 1( 0 .85 ) ⎞ = ⎜⎜ ⎟⎟ 70 ,360 ⎝ ⎠ = 1.69

(c) Threat Comparison (tooth bending or wear) Bending FOS ( SF )

Wear FOS ( SH )2

Threat

Pinion

5.62

1.692 = 2.86

wear

Gear

7.05

1.552 = 2.39

wear

(b) Pinion tooth wear

⎛ S Y / KT K R ⎞ (S F ) P = ⎜ t N ⎟ σ ⎝ ⎠P

Gear tooth wear

1/ 2

⎛ K Cf ⎞ ⎟ (σ c ) P = ⎜⎜ W t K o K v K s m d P F I ⎟⎠ P ⎝

⎛ 31,350 ( 0 .977 ) / 1( 0 .85 ) ⎞ =⎜ ⎟ 6417 ⎝ ⎠ = 5.62

⎡ 1 .22 1 ⎤ = 2300 ⎢164 .8(1)1 .377 (1 .043 ) ⎥ 1 .7 (1 .5) 0 .121 ⎦ ⎣ = 70 ,360 psi

61

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

BDA 31203 – Mechanical Component Design

1/ 2

⎡ 1 .18 1 ⎤ (σ c ) G = 2300 ⎢164 .8(1)1 .377 (1 .052 ) 1 .7 (1 .5) 0 .121 ⎥⎦ ⎣ = 69 , 496 psi

CHAPTER 5 – Gears (Part II)

Examples – cont…

62

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

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples – cont…

Example Example5-7 5-7

Solution Solution5-7 5-7

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.

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 For helical gears, the transverse diametral pitch;

Pt = Pn cosψ = 10(cos 30 o ) = 8.660 teeth/in 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:

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.

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

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63

πd P n P 12

=

π (1.963)1800 12

= 925 ft/min

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

Wt =

33000 H 33000(4) = = 142.7 ibf V 925

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

⎛ 59.77 + 925 ⎞ ⎟ Kv = ⎜ ⎜ ⎟ 59.77 ⎝ ⎠

0.8255

= 1.404 64

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples – cont…

Examples – cont…

Solution Solution5-7 5-7––cont… cont…

Solution Solution5-7 5-7––cont… cont…

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

⎛ tan φ n φt = tan ⎜ ⎝ cosψ −1 ⎜

o ⎞ ⎛ ⎟ = tan −1 ⎜ tan 20 ⎜ cos 30 o ⎟ ⎝ ⎠

⎛ p ⎞ ⎛ 0.2952 ⎞ ⎟⎟ = 0.6895 m N = ⎜ N ⎟ = ⎜⎜ ⎝ 0.95Z ⎠ ⎝ 0.95(0.4507 ) ⎠

Z = (0.9815 + 0.1) 2 − 0.9048 2 +

⎞ ⎟ = 22.80 o ⎟ ⎠

(3.004 + 0.1) 2 − 2.769 2

The surface strength geometry factor :

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

( p N ) = p n cos φ n =

(rb ) G = rG cos φt = 3.002(cos 22.80 o ) = 2.767 in

=

π 10

π PN

cos φ n

(a) Pinion tooth bending

P K K ⎞ ⎛ (σ ) P = ⎜ W t K o K v K s d m B ⎟ F J ⎝ ⎠P ⎛ 8 .66 ⎞ ⎛ 1 .208 (1) ⎞ = 142 .7 (1)(1 .404 )(1 .043 ) ⎜ ⎟⎜ ⎟ ⎝ 1 .5 ⎠ ⎝ 0 .423 ⎠ = 3445 psi

⎛ 31,350 ( 0 .977 ) / 1( 0 .85 ) ⎞ =⎜ ⎟ 3445 ⎝ ⎠ = 1 0 .5

=

(C )

= 0.05 − 0.0375 + 0.0125(1.5) = 0.0313

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:

pf G

C pm = 1 (Bearing immediately adjacent)

C ma = 0.15 (Commercia l enclosed gear unit) Ce = 1

(K m )P

= 1 + C mc ( C pf C pm + C ma C e ) = 1 + 1[ 0 . 0577 (1) + 0 . 15 (1)]

(K m )G

= 1 . 208 = 1 . 181

66

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

CHAPTER 5 – Gears (Part II)

Examples – cont… Solution Solution5-7 5-7––cont… cont…

⎞⎛ 3.06 ⎞ ⎟⎜ = 0.195 ⎟⎝ 3.06 + 1 ⎟⎠ ⎠

J G = 0.54(0.98) = 0.529 65

BDA 31203 – Mechanical Component Design

⎛ sin 22.80 o cos 22.80 o I =⎜ ⎜ 2(0.6895) ⎝

1. 5 − 0.0375 + 0.0125(1.5) = 0.0577 10(1.963)

(C )

J P = 0.45(0.94) = 0.423

cos 20 o = 0.2952 in

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

C mc = 1 (uncrowned) pf P

− (0.9815 + 3.004) sin 22.80 o = 0.5924 + 1.4027 - 1.544.4 = 0.4507 in

(rb ) P = rP cos φt = 0.9815(cos 22.80 o ) = 0.9048 in

The load-distribution factor Km is estimated :

The load sharing ratio :

Then the surface strength geometry factor

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:

⎛ S Y / KT K R ⎞ (S F ) P = ⎜ t N ⎟ σ ⎝ ⎠P

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

CHAPTER 5 – Gears (Part II)

BDA 31203 – Mechanical Component Design

Examples – cont… Gear tooth bending

Solution Solution5-7– 5-7–cont… cont…

⎛ 8 . 66 ⎞ ⎛ 1 . 181 (1) ⎞ (σ ) G = 142 . 7 (1)(1 . 404 )(1 . 052 ) ⎜ ⎟⎜ ⎟ ⎝ 1 . 5 ⎠ ⎝ 0 . 529 ⎠ = 2717 psi

⎛ S Z / KT K R ( S H ) P = ⎜⎜ c N σc ⎝

⎛ 28 , 260 ( 0 .996 ) / 1( 0 .85 ) ⎞ (S F )G = ⎜ ⎟ 2717 ⎠ ⎝ = 1 2 .2

⎛ S Z C / KT K R ⎞ ( S H ) G = ⎜⎜ c N H ⎟⎟ σc ⎝ ⎠G ⎛ 93,500 ( 0 .973 )(1 .005 ) / 1( 0 .85 ) ⎞ =⎜ ⎟ 47 ,888 ⎝ ⎠ = 2 .2 5

⎞ ⎟ ⎟ ⎠P

⎛ 106 , 400 ( 0 .948 ) / 1( 0 .85 ) ⎞ = ⎜⎜ ⎟⎟ 48 , 230 ⎝ ⎠ = 2.46

(c) Threat Comparison (tooth bending or wear)

(b) Pinion tooth wear Gear tooth wear

1/ 2

⎛ K Cf ⎞ ⎟ (σ c ) P = ⎜⎜ W t K o K v K s m d P F I ⎟⎠ P ⎝

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

⎡ 1 .208 1 ⎤ = 2300 ⎢142 .7 (1)1 .404 (1 .043 ) ⎥ 1 .963 (1 .5) 0 .195 ⎦ ⎣ = 48 , 230 psi

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1/ 2

⎡ 1.181 1 ⎤ (σ c ) G = 2300 ⎢142 .7 (1)1.404 (1.052 ) 1.963(1.5) 0.195 ⎥⎦ ⎣ = 47,888 psi

1/ 2

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

Pinion Gear

Bending FOS ( SF )

Wear FOS ( SH )2

Threat

10.5

2.462 = 6.05

wear

12.2

2.252 = 5.05

wear

68

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

Learning Outcomes

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

At the end of this topic, the students would be able to apply and appreciate the knowledge to: BDA 31203 Notes – Mechanical Component Design

Week 9

Chapter 6

Shaft Design

•

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 shaft design applications

Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

What you will be learn here?

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.1 – Introduction What is shaft?!

• • • • • •

6.1 6.2 6.3 6.4 6.5 6.6

- Introduction - Shaft Materials - Shaft Layout - Shaft Design for Stress - Deflection Considerations - Limits and Fits

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

~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, cranks and the like, and controls the geometry of their motion. 3

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

4

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.1 – Introduction – cont…

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.1 – Introduction – cont…

What is axle?! What is spindle?! An axle is a nonrotating member that carries no torque and

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

What it is used for?! is used to support rotating wheels, 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.

Train wheels are affixed to a straight axle, such that both wheels rotate in unison. 5

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.1 – Introduction – cont…

Material Selection

•

Geometric Layout

•

Stress and strength

• 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.

– Fatigue strength

• Strength controlled by material property.

Deflection and rigidity

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

– Bending deflection – Torsional deflection

•

CHAPTER 6 – Shaft Design

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

– Static strength

•

BDA 31203 – Mechanical Component Design

6.2 – Shaft Materials

Considerations for Shaft Design

•

6

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

– Slope at bearings and shaft-supported elements

• Cold drawn steel typical for d < 3 in.

– Shear deflection due to transverse loading of short shafts

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

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

7

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

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BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.2 – Shaft Materials – cont…

6.3 – Shaft Layout

• 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.

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

9

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

•

Issues to consider for shaft layout – Axial layout of components – Supporting axial loads – Providing for torque transmission – Assembly and Disassembly

10

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.3 – Shaft Layout – cont… Various Variousmethod methodtotoattach attachelement elementon onshaft. shaft.

Axial AxialLayout LayoutofofComponents Components

snap ring

clamp collar

y Axial loads must be supported through a bearing to the frame. y Generally best for only one bearing to carry axial load to shoulder y Allows greater tolerances and prevents binding

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

11

key hub bearing

taper pin

hub shaft

step

bearing

step

press fit

step

step press fit

axial clearance

sheave frame

frame sprocket

gear

Assembly/Disassembly → progressively smaller diameter toward the ends Axial clearance → to allow machinery vibration Keys/pins/rings → to secure rotating elements ( gear, pulley, etc) Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

12

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.3 – Shaft Layout – cont…

6.3 – Shaft Layout – cont… • 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.

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

14

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.3 – Shaft Layout – cont…

Common Commontorque torquetransfer transferelements: elements:

Pins: Pins:

Keys Splines Setscrews Pins Press or shrink fits Tapered fits

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

Common Commonshaft shaft loading loadingmechanism: mechanism:

13

6.3 – Shaft Layout – cont…

• • • • • •

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

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 15

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

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BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

Keys Keysand andkeyseats: keyseats:

Spline Splineshaft shaftand andHub: Hub:

- Used when large amounts of torque are to be transferred Keys are used to transmit torque from a component to the shaft. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

BDA 31203 – Mechanical Component Design

- Stress concentration is generally quite moderate 17

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

BDA 31203 – Mechanical Component Design

18

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

Locational Locationaldevice: device:

• • • • • • • •

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

Nut Nutand andWasher: Washer:

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

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

19

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

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BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.3 – Shaft Layout – cont…

Sleeve: Sleeve:

Shaft Shaftshoulder shoulder: :

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.

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

Example:

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

BDA 31203 – Mechanical Component Design

21

CHAPTER 6 – Shaft Design

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

BDA 31203 – Mechanical Component Design

6.3 – Shaft Layout – cont…

6.3 – Shaft Layout – cont…

Spring Springloaded loadedRetaining RetainingRing Ring: :

Set SetScrew Screw: :

• • •

CHAPTER 6 – Shaft Design

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.

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.

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

22

23

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

24

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.3 – Shaft Layout – cont…

6.3 – Shaft Layout – cont…

Split SplitHub Hub: :

Collar Collarand andScrew Screw: :

CHAPTER 6 – Shaft Design

• 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. 25

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

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

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress

6.4 – Shaft Design for Stress – cont…

Critical CriticalLocation Location: :

Critical CriticalLocation Location: :

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

27

26

CHAPTER 6 – Shaft Design

• 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.

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

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CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

6.4 – Shaft Design for Stress – cont…

Shaft ShaftStresses Stresses: :

Shaft ShaftStresses Stresses: :

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

σa = K f

M ac I

; σm = K f

M mc I

τ a = K fs

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

T c Ta c ; τ m = K fs m J J

1/ 2

σ 'a = (σ + 3τ )

⎡⎛ 32 K f M a ⎞ 2 ⎛ 16 K fsTa ⎞ 2 ⎤ ⎟⎟ + 3⎜⎜ ⎟⎟ ⎥ = ⎢⎜⎜ 3 3 ⎢⎣⎝ πd ⎠ ⎝ πd ⎠ ⎥⎦

1/ 2

σ 'm = (σ + 3τ )

⎡⎛ 32 K f M m ⎞ 2 ⎛ 16 K fsTm ⎞ 2 ⎤ ⎟⎟ ⎥ ⎟⎟ + 3⎜⎜ = ⎢⎜⎜ 3 3 ⎢⎣⎝ πd ⎠ ⎥⎦ ⎝ πd ⎠

2 a

2 1/ 2 a

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

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

σa = K f

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

32 M a 32 M m ; σm = K f 3 πd πd 3

τ a = K fs

16Ta 16T ; τ m = K fs 3m 3 πd πd

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

BDA 31203 – Mechanical Component Design

29

CHAPTER 6 – Shaft Design

6.4 – Shaft Design for Stress – cont…

2 m

2 1/ 2 m

30

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

Shaft ShaftStresses Stresses: :

DE-Goodman DE-Goodman: : Fatigue failure curve on the modified Goodman diagram

1 16 ⎧ 1 = 3 ⎨ 4( K f M a ) 2 + 3( K fsTa ) 2 n πd ⎩ S e

[

]

1/ 2

+

[

1 4( K f M m ) 2 + 3( K fsTm ) 2 Sut

]

1/ 2

⎫ ⎬ ⎭

Equation for the minimum diameter

⎛ 16n ⎧ 1 2 2 d = ⎜⎜ ⎨ 4( K f M a ) + 3( K fsTa ) π S ⎩ e ⎝

[

]

1/ 2

[

1 4( K f M m ) 2 + 3( K fsTm ) 2 + Sut

]

1/ 2

⎫⎞ ⎬ ⎟⎟ ⎭⎠

1/ 3

This criteria does not guard against yielding, so required separate check for possibility of static failure (yield occur) in the first load cycle. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

31

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

32

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

6.4 – Shaft Design for Stress – cont…

DE-Gerber DE-Gerber: :

DE-ASME DE-ASMEElliptic Elliptic: :

Fatigue failure curve on the Gerber diagram

Fatigue failure curve on the ASME Elliptic diagram

where

2 1/ 2 ⎫ ⎧ 1 8 A ⎪ ⎡ ⎛ 2 BS e ⎞ ⎤ ⎪ ⎟ ⎥ ⎬ = ⎨1 + ⎢1 + ⎜ n πd 3 S e ⎪ ⎢ ⎜⎝ ASut ⎟⎠ ⎥ ⎪ ⎦ ⎭ ⎩ ⎣

2 2 2 2 ⎛ K fsTm ⎞ ⎤ ⎛ K f Mm ⎞ ⎛ K fsTa ⎞ 1 16 ⎡⎢ ⎛ K f M a ⎞ ⎟ ⎥ ⎟ + 3⎜ ⎟ + 4⎜ ⎟ + 3⎜⎜ 4⎜ = ⎜ S ⎟ ⎥ ⎟ ⎜ S n πd 3 ⎢ ⎜⎝ S e ⎟⎠ S e ⎟⎠ y y ⎝ ⎠ ⎦ ⎝ ⎠ ⎝ ⎣

A = 4( K f M a ) 2 + 3( K fsTa ) 2 B = 4( K f M m ) 2 + 3( K fsTm ) 2

1/ 2

Equation for the minimum diameter

1/ 3

Equation for the minimum diameter

1/ 3

2 1/ 2 ⎫ ⎞ ⎛ ⎧ ⎜ 8nA ⎪ ⎡ ⎛ 2 BS e ⎞ ⎤ ⎪ ⎟ ⎟⎟ ⎥ ⎬ ⎟ d =⎜ ⎨1 + ⎢1 + ⎜⎜ ⎜ πS e ⎪ ⎢⎣ ⎝ ASut ⎠ ⎥⎦ ⎪ ⎟ ⎩ ⎭⎠ ⎝

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

33

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

1/ 2 ⎧ ⎡ ⎛ K M ⎞2 ⎛ K T ⎞2 ⎛ K M ⎞2 ⎛ K T ⎞2 ⎤ ⎫ n 16 ⎪ ⎢4⎜ f a ⎟ + 3⎜ fs a ⎟ + 4⎜ f m ⎟ + 3⎜ fs m ⎟ ⎥ ⎪⎬ d =⎨ ⎜ S ⎟ ⎜ S ⎟ ⎜ S ⎟ ⎥ π ⎢ ⎜⎝ S e ⎟⎠ y ⎝ e ⎠ ⎪ ⎝ ⎠ ⎝ y ⎠ ⎦ ⎪ ⎣ ⎭ ⎩

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. 34

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

DE-Soderberg DE-Soderberg: :

Check Checkfor foryielding yielding: :

Fatigue failure curve on the Soderberg diagram

1 16 = n πd 3

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

⎧⎪ 1 2 2 ⎨ 4( K f M a ) + 3( K fsTa ) S ⎪⎩ e

[

]

1/ 2

+

[

1 4( K f M m ) 2 + 3( K fsTm ) 2 S yt

]

1/ 2

⎫⎪ ⎬ ⎪⎭

Factor of safety

Equation for the minimum diameter

⎛ 16n ⎧⎪ 1 d =⎜ 4( K f M a ) 2 + 3( K fsTa ) 2 ⎜ π ⎨⎪ S e ⎩ ⎝

[

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

]

1/ 2

+

[

1 4( K f M m ) 2 + 3( K fsTm ) 2 S yt

]

1/ 2

⎫⎪ ⎞ ⎬ ⎟⎟ ⎪⎭ ⎠

1/ 3

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

35

where

ny =

Sy

σ 'max

σ 'max = (σ 'a2 +σ 'm2 )

1/ 2

⎡⎛ 32 K f ( M a + M m ) ⎞ 2 ⎛ 16 K fs (Ta + Tm ) ⎞ 2 ⎤ ⎟⎟ + 3⎜⎜ ⎟⎟ ⎥ = ⎢⎜⎜ πd 3 πd 3 ⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦ Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

1/ 2

36

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont… Example Example6-1 6-1: :

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

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 Nm and 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.

τa = 0

σm = 0

(b) Determine the yielding factor of safety.

These will simply drops out some of previously terms.

37

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

Ta = 0 Nm

M m = 0 Nm

Tm = 124.3 Nm

r 2.8 = = 0.10 d 28

Kt = 1.68 (figure A-15-9) Kts = 1.42 (figure A-15-8)

k a = 4.51(735) −0.265 = 0.787

S e = (0.787)(0.87)(0.814)(367.5) = 205 MPa

⎛ 28 ⎞ kb = ⎜ ⎟ ⎝ 7.62 ⎠

1 16 ⎧ 1 2 = ⎨ 4( K f M a ) n πd 3 ⎩ S e

[

r = 2.8

q = 0.85 (figure 4-1)

Sut = 0.735 GPa

qs = 0.92 (figure 4-2)

K f = 1 + 0.85(1.68 − 1) = 1.58

=

16 π (0.028)3

[

]

1/ 2

+

[

]

= 0.87

k c = k d = k f = 1.0

1 3( K fsTm ) 2 Sut

⎧ 4(1.58(142.4)) 2 ⎪ ⎨ 205 x106 ⎪⎩

−0.107

1/ 2

+

]

1/ 2

⎫ ⎬ ⎭

ke = 0.814

[3(1.39(124.3)) ] 2

735 x10

6

1/ 2

⎫ ⎪ ⎬ ⎪⎭

= 232004(2.195 x10 −6 + 0.407 x10 −6 ) = 0.604

∴n = 1.65

K fs = 1 + 0.92(1.42 − 1) = 1.39 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

S e ' = 0.5(735) = 367.5 MPa

Applying Eq. DE-Goodman criteria gives

a) Determine the fatigue factor of safety of the design: D 42 = = 1.50 d 28

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

Solution Solution6-1 6-1: :

M a = 142.4 Nm

38

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

39

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

40

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.4 – Shaft Design for Stress – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.4 – Shaft Design for Stress – cont… Estimating EstimatingStress StressConcentrations Concentrations

Similarly, apply same technique for other failure criteria,

∴ n = 1.87

DE-Gerber

• Stress analysis for shafts is highly dependent on stress concentrations.

∴n = 1.88

DE-ASME Elliptic

• Stress concentrations depend on size specifications, which are not known the first time through a design process.

∴n = 1.56

DE-Soderberg

• Standard shaft elements such as shoulders and keys have standard proportions, making it possible to estimate stress concentrations factors before determining actual sizes.

b) Determine the Yield factor of safety : 2

2

⎛ 32(1.58)(142.4) ⎞ ⎛ 16(1.40)(124.3) ⎞ σ 'max = ⎜⎜ 3 ⎟⎟ + ⎜⎜ π (0.028) 3 ⎟⎟ = 125.4 ⎝ π (0.028) ⎠ ⎝ ⎠

∴ ny =

Sy

σ 'max

=

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

574 = 4.58 125.4

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

BDA 31203 – Mechanical Component Design

41

CHAPTER 6 – Shaft Design

6.4 – Shaft Design for Stress – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.5 – Deflection Considerations

Reducing ReducingStress StressConcentration ConcentrationatatShoulder ShoulderFillet 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) (b) (c)

42

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

Large radius undercut into shoulder Large radius relief groove into back of shoulder Large radius relief groove into small diameter of shaft

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

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

43

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

44

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.5 – Deflection Considerations –cont…

6.6 – Limits and Fits

•

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

•

Each point of interest requires entirely new deflection analysis.

3 types of fitting

Clearance Fits. No interference occur.

•

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:9 Superposition 9 The moment-area method 9 Singularity functions 9 Numerical integration

45

CHAPTER 6 – Shaft Design

6.6 – Limits and Fits – cont… Capital letters always refer to the hole; lowercase letters are used for the shaft.

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.

Refer in text book chapter 4 page 147

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

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

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

BDA 31203 – Mechanical Component Design

46

CHAPTER 6 – Shaft Design

6.6 – Limits and Fits – cont… Definitions applied to a cylindrical fit.

Table 6–3 Descriptions of Preferred Fits Using the Basic Hole System Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BS 4500.

D = basic size of hole d = basic size of shaft δu = upper deviation δl = lower deviation δF = fundamental deviation D = tolerance grade for hole d = tolerance grade for shaft Note that these quantities are all deterministic. Thus, for the hole, Dmin = D Dmax = D + ∆D For shafts with clearance fits c, d, f, g, and h, dmin = d + δF − ∆d dmax = d + δF For shafts with interference fits k, n, p, s, and u, dmax = d + δF + ∆d dmin = d + δF Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

47

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

48

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.6 – Limits and Fits – cont…

BDA 31203 – Mechanical Component Design

CHAPTER 6 – Shaft Design

6.6 – Limits and Fits – cont… Table A–12

Table A–11 A 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)

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.

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

49

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

CHAPTER 6 – Shaft Design

BDA 31203 – Mechanical Component Design

6.6 – Limits and Fits – cont… Example Example6-2 6-2: : Find the shaft and hole dimensions for a loose running fit with a 34-mm basic size. Solution Solution6-2 6-2: : From Table 6–3, the ISO symbol is 34H11/c11. From Table A–11, we find that tolerance grade 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 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

51

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

50

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

9 recognize types of available bearings and know the function Week 10

9 identify nomenclature of ball bearing

Chapter 7

9 select suitable bearing and know how to apply it

Bearings

9 perform load and bearing life calculations analytically

Prepared by: Mohd Azwir Bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

What you will be learn here? •

7.1 - Introduction

•

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

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.1 – Introduction •

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

•

2 types of bearing: sliding & rolling

sliding

3

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)

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

4

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.1 – Introduction – cont…

7.2 – Bearing Types

•

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:

Nomenclature NomenclatureofofBall BallBearing Bearing

• outer ring

– Timken (USA)

• inner ring

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

• balls or rolling element

ISO (International Standard Organization)

–

ABMA (American Bearing Manufacturers Association)

Outside diameter (D)

Bore (d)

• seperator

Bearing standard :–

Inner ring

4 essential parts of a bearing:

– NTN (Jepun)

•

Width (B) Outer ring

– SKF (Sweden)

•

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

Separator

Face 5

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

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.2 – Bearing Types – cont…

7.2 – Bearing Types – cont…

Various Varioustypes typesofofball ballbearings bearings

Various Varioustypes typesofofroller rollerbearings bearings

Deep Groove ball bearing

6

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

CHAPTER 7 – Bearings

Straight roller bearing Shielded ball bearing

Double row

Needle roller bearing

Angular Contact ball bearing

Tapered roller bearing

Sealed ball bearing

Spherical roller thrust bearing Self-Aligning ball bearing

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

Trust ball bearing

Tapered roller thrust bearing

7

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

8

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.2 – Bearing Types – cont…

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.3 – Bearing Mounting & Enclosure

Bearing BearingCharacteristics Characteristics

A common bearing mounting.

Mounting for a washing machine spindle.

9

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

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.4 – Bearing Life

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

10

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.5 – Bearing Load Life at Rated Reliability

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

Application factor (assume af = 1 if not given)

L10 life (number of revolution)

C10 ( LR nR 60)1/ a = a f FD ( LD nD 60)1/ a Catalog load rating, ibf or kN

Desired speed, rpm Desired life, hours

Rating life, hours

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

Desired radial load, ibf or kN

Rating speed, rpm

where :-

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

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.

See

a = 3 for ball bearing a = 10/3 for roller bearing

Solving for C10 gives

11

⎛ L n 60 ⎞ C10 = a f FD ⎜⎜ D D ⎟⎟ ⎝ LR nR 60 ⎠

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

Table 7.1 Data for Bearing Manufacturer. 1/ a

Table 7.2 Equivalent Radial Load Factor. Table 7.3 Deep-Groove and Angular-Contact Ball Bearings Catalog. Table 7.4 Cylindrical Roller Bearings Catalog.

12

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.5 – Bearing Load Life at Rated Reliability – cont… Example Example7-1 7-1 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?

Solution Solution7-1 7-1

⎛ L n 60 ⎞ C10 = a f FD ⎜⎜ D D ⎟⎟ ⎝ LR nR 60 ⎠

1/ a

⎡ 5000(1725)60 ⎤ = 2⎢ ⎥⎦ 106 ⎣

1

3

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.6 – Relating Load, Life and Reliability • It is always the desired parameters (load, speed and reliability) is not the manufacturer’s test parameter or catalog entry. • To used the catalog data to comply with the desired parameters, one needs to determine equivalent catalog load rating by using next formula.

= 16.05 kN

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.

13

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

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.6 – Relating Load, Life and Reliability – cont… ⎡ ⎤ xD C10 = a f FD ⎢ 1/ b ⎥ ⎣ x0 + (θ − x0 )(ln(1 / RD )) ⎦

Reliability Reliabilityfor forbearing bearingisisgiven givenby bythe theequation: equation:

1/ a

b a ⎛ ⎧ ⎫ ⎞⎟ ⎛ a f FD ⎞ ⎜ ⎜ ⎪ x D ⎜⎜ C ⎟⎟ − x 0 ⎪ ⎟ ⎪ ⎪ ⎟ 10 ⎠ R = exp⎜ − ⎨ ⎝ ⎬ ⎜ ⎪ θ − x0 ⎪ ⎟ ⎜ ⎪ ⎪ ⎟⎟ ⎜ ⎩ ⎭ ⎠ ⎝

where:-

RD = Desired reliability a

= 3 (for ball bearing) = 10/3 (for roller bearing)

x0 ⎫ ⎪ θ ⎬ = Weibull parameters (refer table 6-1). b ⎪⎭ Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.6 – Relating Load, Life and Reliability – cont…

where:af = Application factor (assume af = 1 if not given) FD = Desired radial load, ibf / kN 60LD n D L = xD = Dimensionless desired life = L10 60LR n R a

14

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

See Table 7.1 Data for Bearing Manufacturer. Table 7.2 Equivalent Radial Load Factor. Table 7.3 Deep-Groove and Angular-Contact Ball Bearings Catalog. Table 7.4 Cylindrical Roller Bearings Catalog.

af FD

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

xD

= Dimensionless desired life =

60LD n D L = L10 60LR n R

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

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

R = RA RB x0 ⎫ ⎪ θ⎬ b ⎪⎭

= Weibull parameters (refer table 6-1).

See Table 7.1 Data for Bearing Manufacturer. Table 7.2 Equivalent Radial Load Factor. Table 7.3 Deep-Groove and Angular-Contact Ball Bearings Catalog.

15

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

16 Table 7.4 Cylindrical Roller Bearings Catalog.

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.6 – Relating Load, Life and Reliability – cont…

7.6 – Relating Load, Life and Reliability – cont…

Example Example7-2 7-2

Example Example7-3 7-3

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?

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.

Solution Solution7-3 7-3

Solution Solution7-2 7-2

xD =

60 LD n D 60(30000)(300) L = = = 540 L10 60 LR n R 10 6

•

Calculate result, C10 = 29.7 kN

•

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

•

Therefore, the selected bearing size is:-

hold 29.77 kN load is 30.7 kN.

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

⎡ ⎤ 540 C10 = (1.2)(1.84) ⎢ 1 / 1.483 ⎥ ⎣ 0.02 + (4.439)(ln 1 / 0.99) ⎦

1/ 3

= 29.7 kN

•

CHAPTER 7 – Bearings

7.7 – Combined Radial and Thrust Loading

,

Bore = 40 mm

&

Width = 18 mm

1.483 3 ⎞ ⎛ ⎧ ⎟ ⎜ 540⎛ (1.2)(1.84) ⎞ − 0.02 ⎫ ⎟ ⎜ ⎪ ⎪ ⎟ ⎜ ⎪ 30.7 ⎠ ⎪ ⎝ And the new reliability is R = exp⎜ − ⎨ ⎟ = 0.9914 ⎬ 4.439 ⎟ ⎜ ⎪ ⎪ ⎟ ⎜ ⎪ ⎪⎭ ⎠ ⎝ ⎩

17

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

BDA 31203 – Mechanical Component Design

OD = 80 mm

18

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

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.7 – Combined Radial and Thrust Loading – cont… Equivalent Equivalentradial radialload load

•

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:

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

Fe = X iVFr + Yi Fa where Fa Fr Xi Yi V

-

axial thrust radial load radial load factor (table 7-2) thrust load factor (table 7-2) rotation factor ⎧= 1.0 - Inner ring rotates

•

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

•

Table 7-2 list X1, Y1 and X2, Y2 as a e function based on the ratio of the thrust component to the bearing static load catalog rating Fa / C0.

⎪ V = ⎨= 1.2 - Outer ring rotates ⎪= 1.0 - Self aligning bearing ⎩

19

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

20

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.7 – Combined Radial and Thrust Loading – cont…

7.7 – Combined Radial and Thrust Loading – cont…

Example Example7-4 7-4

Solution Solution7-4 7-4––cont… cont…

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.

Fe = X 2VFr + Y2 Fa = 0.56(1)2225 + 1.527(1780) = 3964N With LD = L10 and FD = Fe, solving for L10 gives:

Solution Solution7-4 7-4 V = 1 and

Fa 1780 = = 0.090 . C0 19800

Interpolate for e in Table 7-2

Fa 1780 = = 0.8 > 0.285 (VFr ) 1(2225)

e

Fa / C0

Y2

0.084

0.28

0.084

1.55

e

0.110

0.30

from which e = 0.285

60 LR nR 60nD

3

a

⎛ C10 ⎞ 106 ⎛ 35150 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ = 16139.5 h 60(720) ⎝ 3964 ⎠ ⎝ Fr ⎠

Thus, interpolate for Y2 :

Fa / C0 0.090

L10 =

.

from which Y2 = 1.527

0.090

Y2

0.110

1.45

** See more example on bearing analysis – example 7-5 & 7-6 21

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

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.8 – Lubrication

22

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

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.9 – Appendix

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

Table 7-1 : Data for Bearing Manufacturer

Manufacturer

Rules of selecting grease and oil as a lubricant Use Grease When

1. Timken

Use Oil When

1. 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.

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

2. SKF and most bearing manufacturer

23

Rating life, revolution

Rating life, Weibull Parameter

X0

θ

b

90(10)6

0

4.48

1.5

(10)6

0.02

4.459

1.483

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

24

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.9 – Appendix – cont…

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

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

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

dS

dH

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

* use 0.014 if Fa / C0 < 0.014 25

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

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.9 – Appendix – cont…

26

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

BDA 31203 – Mechanical Component Design

CHAPTER 7 – Bearings

7.9 – Appendix – cont…

Table 7-4 : Dimensions and Basic Load Ratings for Cylindrical Roller Bearings. Table 7-5 : Bearing-Life Recommendations for Various Classes of Machinery

Types of Application Instruments and apparatus for infrequent use Aircraft engines

27

Up to 0.5 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

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

Life, k hour

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

100 - 200

28

CHAPTER 7 – Bearings

BDA 31203 – Mechanical Component Design

7.9 – Appendix – cont… 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

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

29

CHAPTER 7 – Bearings

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

P = Tω = T

∴n =

2π n 60

60 P 9.55 P 9.55(18.7) = = = 655 rpm 2πT 0.2727 T

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

1

CHAPTER 7 – Bearings

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 4 0.96 = 0.98985 = 0.99 . The dimensionless design life for both bearings is :

xD =

60 LD n D 60(10000)655 L = = = 393 L10 60 LR n R 10 6

(a) Selection of roller bearing at D. (Info obtained af = 1.2 and a = 10/3) 1

3 ⎤a ⎡ 10 ⎡ ⎤ ⎥ ⎢ xD 393 ⎥ = 16.0 kN ⎥ = 1.2(1406) ⎢ C10 = a f FD ⎢ 1/ b 1 / 1.483 ⎥ ⎢ ⎢ ⎛ 1 ⎞ ⎥ 0.02 + (4.439) ln 1 0.99 ⎣⎢ ⎦⎥ ⎢⎣ x0 + (θ − x 0 )⎜⎝ ln R D ⎟⎠ ⎥⎦

(

)

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) ii) iii) iv) v) vi) vii)

Choose Y2 from table 7-2 Find C10 Tentatively identify a suitable bearing from Table 7-3, note C0 Using Fa / C0 enter Table 7-2 to obtain a new value of Y2. Find C10 If the same bearing obtained, stop. 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 Fe = X 2VFr + Y2 Fa = 0.56(1.0)(2067) + 1.63(1531) = 3.65 kN 1

⎡ ⎤3 393 ⎥ = 53.1 kN C10 = 1.2(3650 ) ⎢ 1 / 1.483 ⎢ 0.02 + ( 4.439) ln 1 ⎥ 0.99 ⎣⎢ ⎦⎥

(

)

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,

Fa 1.531 = = 0.0431 C 0 35.5

2

CHAPTER 7 – Bearings

F

1531

a = = 0.74 , which is greater Which makes e from Table 7-2 approximately 0.24. Now 1.0(2067) VF r than 0.24, so we find Y2 by interpolation:

Fa / C0

e

Fa / C0

Y2

0.042

0.24

0.042

1.85

0.043

e

0.043

Y2

0.056

0.26

0.056

1.71

Therefore

from which e ≈ 0.24

from which Y2 = 1.84

Fe = X 2VFr + Y2 Fa = 0.56(1.0)(2067) + 1.84(1531) = 3.97 kN

The prior calculation for C10 changes only in Fe, so

C10 =

3.97 (53.1) = 57.8 kN 3.65

From Table 7-3, angular contact bearing 02-65 mm has C10 = 63.7 kN and C0 is 41.5 kN. Again,

Fa 1.531 = = 0.0369 C 0 41.5

F

1531

a Making e approximately 0.23. Now from before = = 0.74 , which is greater than 0.23, VFr 1.0(2067) so we find Y2 again by interpolation:

Fa / C0

e

Fa / C0

Y2

0.028

0.22

0.028

1.99

0.0369

e

0.0369

Y2

0.042

0.24

0.042

1.85

Therefore

from which e ≈ 0.23

from which Y2 = 1.90

Fe = X 2VFr + Y2 Fa = 0.56(1.0)(2067) + 1.90(1531) = 4.065 kN

The prior calculation for C10 changes only in Fe, so

C10 =

4.07 (53.1) = 59.2 kN 3.65

From Table 7-3, angular contact bearing 02-65 mm is still selected, so the iteration is complete.

3

CHAPTER 7 – Bearings

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. y 500 400

O

FC

250

20o

z B

Gear 3 600 D

2 Gear 4 250 D

FA

x

20o

Solution FBD Oy Oz

TA = (FA cos 20o)(0.3) FC sin 20o

FA sin 20o FA cos 20o

Bz

FC cos 20o By

TC = (FC cos 20o) (0.125)

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. ƒ ƒ ƒ ƒ ƒ

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

4

CHAPTER 7 – Bearings

For a combined reliability goal of 0.90, use

0.90 = 0.95 for the individual bearings.

At O (angular contact ball bearing)

xD =

60 LD n D 60(50000)(480) = = 1440 60 LR n R 10 6

⎡ ⎤ 1440 ∴C10 = (1.4)(2727.2) ⎢ 1 / 1.483 ⎥ ⎣ 0.02 + (4.439)(ln 1 / 0.95) ⎦ ∴ Suitable size;

1/ 3

= 50.6 kN

bore = 60 mm, OD = 110 mm, width = 22 mm

At B (cylindrical roller bearing)

xD =

60 LD n D 60(50000)(480) = = 1440 60 LR n R 10 6

⎡ ⎤ 1440 ∴C10 = (1.4)(7407.4) ⎢ 1 / 1.483 ⎥ ⎣ 0.02 + (4.439)(ln 1 / 0.95) ⎦ ∴ Suitable size;

3 / 10

= 106.1 kN

bore = 85 mm, OD = 150 mm, width = 28 mm

5

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

Learning Outcomes

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

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

BDA 31203 Notes – Mechanical Component Design

9 recognize types of available bolt and screw Week 11, 12 & 13

9 know bolt / screw thread standards and definition

Chapter 8

9 perform load and stress calculations that acting on bolt / screw

Nonpermanent Joints

9 select suitable application of bolt / screw

Prepared by: Mohd Azwir bin Azlan 2

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

What you will be learn here? •

8.1 - Introduction

•

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.1 – Introduction • 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. 3

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

4

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.1 – Introduction – cont…

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.1 – Introduction – cont…

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

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

• 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.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

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

• 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.

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

Figure 8-1: Several types of fastener

5

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.1 – Introduction – cont…

6

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.2 – Thread Standards and Definitions • Pitch – distance between adjacent thread forms measured parallel to the thread axis. • Major diameter (d) – largest diameter of screw thread.

(a)

(b)

(c)

(d)

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

(e)

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

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

(a)

(b)

(c)

(d)

(e)

• Lead – distance parallel to the screw axis when the nut moves one turn. A doublethreaded 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.

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

7

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

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.2 – Thread Standards and Definitions – cont…

8.2 – Thread Standards and Definitions – cont…

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

1)

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

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 1.75 M = Basic Metric, J = greater root radius for fatigue applications; 12 = nominal major diameter (mm); 1.75 = pitch (mm)

• 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

2)

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

The American National (Unified) thread (Table 8-2) Thread standards is used mainly in the US and GB. Series designation used UN or UNR

Figure 8-9: Regular or Flat Thread

UN (regular thread), UNR (greater root radius for fatigue applications)

Specifications: e.g.: 5/8”-18 UN, UNC, UNF, UNR, UNRC, UNRF 5/8” = nominal major diameter (inch) ; 18 = Number of threads per inch (N) UN = Unified, F = fine, C = Coarse, R = Round Root

• Metric bolt designation determined by: M12 x 1.75

pitch

3)

Nominal diameter (mm) Figure 8-8: Basic profile for metric M and MJ threads.

9

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.2 – Thread Standards and Definitions – cont…

Figure 8-10: Rounded Thread Square and The ACME Threads Square 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.

10

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.2 – Thread Standards and Definitions – cont…

™Coarse series UNC • • • •

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

Thread series

Threads per inch

General assembly Frequent disassembly Not good for vibrations The “normal” thread to specify

Material grade

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

™Fine series UNF

Nominal diameter

• Good for vibrations • Good for adjustments • Automotive and aircraft

length

Metric

Class fit

Pitch

Head type

Head type

™Extra Fine series UNEF • • • •

Good for shock and large vibrations High grade alloy Instrumentation Aircraft

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

M12 x 1.75 ISO 4.8 Hex head bolt Nominal diameter 11

Material class

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

12

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

79 5 16

8.2 – Thread Standards and Definitions – cont… Nominal Major Diameter d (mm)

Table 8-1 Diameter and Areas of CoarsePitch and FinePitch Metric Threads.

1.6 2 2.5 3 3.5 4 5 6 8 10 12 14 16 20 24 30 36 42 48 56 64 72 80 90 100 110

Coarse – Pitch Series Pitch p (mm)

Minor Diameter Area Ar (mm2)

0.35 0.40 0.45 0.5 0.6 0.7 0.8 1 1.25 1.5 1.75 2 2 2.5 3 3.5 4 4.5 5 5.5 6 6 6 6 6

1.27 2.07 3.39 5.03 6.78 8.78 14.2 20.1 36.6 58.0 84.3 115 157 245 353 561 817 1120 1470 2030 2680 3460 4340 5590 6990

1.07 1.79 2.98 4.47 6.00 7.75 12.7 17.9 32.8 52.3 76.3 104 144 225 324 519 759 1050 1380 1910 2520 3280 4140 5360 6740

BDA 31203 – Mechanical Component Design

8.2 – Thread Standards and Definitions – cont…

Fine – Pitch Series

Tensile Stress Area At (mm2)

Pitch p (mm)

Tensile Stress Area At (mm2)

Coarse Series – UNC Size Designation

Minor Diameter Area Ar (mm2)

0 1 2 3 4 5 6 8 10 12 ¼

Table 8-2 1 1.25 1.25 1.5 1.5 1.5 2 2 2 2 2 2 2 2 1.5 2 2 2

39.2 61.2 92.1 125 167 272 384 621 915 1260 1670 2300 3030 3860 4850 6100 7560 9180

36.0 56.3 86.0 116 157 259 365 596 884 1230 1630 2250 2980 3800 4800 6020 7470 9080

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

⅜ ½ ⅝ ¾ ⅞ 1 1¼ 1½

13

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

79 5 16

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BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.2 – Thread Standards and Definitions – cont…

Fine Series - UNF

Nominal Major Diameter in

Threads per Inch N

Tensile Stress Area At in2

Minor Diamater Area Ar in2

0.0600 0.0730 0.0860 0.0990 0.1120 0.1250 0.1380 0.1640 0.1900 0.2160 0.2500 0.3125 0.3750 0.4375 0.5000 0.5625 0.6250 0.7500 0.8750 1.0000 1.2500 1.5000

64 56 48 40 40 32 32 24 24 20 18 16 14 13 12 11 10 9 8 7 6

0.002 63 0.003 70 0.004 87 0.006 04 0.007 96 0.009 09 0.014 0 0.017 5 0.024 2 0.031 8 0.052 4 0.077 5 0.106 3 0.141 9 0.182 0.226 0.334 0.462 0.606 0.969 1.405

0.002 18 0.003 10 0.004 06 0.004 96 0.006 72 0.007 45 0.011 96 0.014 50 0.020 6 0.026 9 0.045 4 0.067 8 0.093 3 0.125 7 0.162 0.202 0.302 0.419 0.551 0.890 1.294

Threads per Inch N

Tensile Stress Area At in2

Minor Diameter Area Ar in2

80 72 64 56 48 44 40 36 32 28 28 24 24 20 20 18 18 16 14 12 12 12

0.001 80 0.002 78 0.003 94 0.005 23 0.006 61 0.008 80 0.010 15 0.014 74 0.020 0 0.025 8 0.036 4 0.058 0 0.087 8 0.118 7 0.159 9 0.203 0.256 0.373 0.509 0.663 1.073 1.581

0.001 51 0.002 37 0.003 39 0.004 51 0.005 66 0.007 16 0.008 74 0.012 85 0.017 5 0.022 6 0.032 6 0.052 4 0.080 9 0.109 0 0.148 6 0.189 0.240 0.351 0.480 0.625 1.024 1.521

14

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

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8.2 – Thread Standards and Definitions – cont…

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

Figure 8-11 (a) Square thread (b) Acme thread

• 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.

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

Table 8-3: Preferred Pitches for Acme Threads

15

d, in

1\4

5\16

3\8

1\2

5\8

3\4

7\8

1

p, in

1\16

1\14

1\12

1\10

1\8

1\6

1\6

1\5

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

11\4 13\4 1\5

1\4

2

21\2

3

1\4

1\3

1\2

16

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

• 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: 9 to lift weight – jack for cars 9 to exert large forces – home compactor or a press

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

Figure 8-12a: The Joyce wormgear screw jack.

• 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. Figure 8-13: Portion of Power Screw

Figure 8-12b: Some applications of power screw

17

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BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.3 – The Mechanics of Power Screw – cont…

• 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:

Fx = PR − N sin λ − fN cos λ = 0 y

= − F − fN sin λ + N cos λ = 0

• Then, divide the numerator and the denominator of these equations by cos λ and use the relation tan λ = l / πdm. • Finally, noting that the torque is the product of the force P and the mean radius dm/2,

Torque required for raising the load to overcome thread friction and to raise the load

Torque required for lowering the load

TL

Fx = − PL − N sin λ + fN cos λ = 0 y

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

TR

• In a similar manner, for lowering the load;

∑ ∑F

to overcome part of the thread friction in lowering the load

= − F + fN sin λ + N cos λ = 0

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CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

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

∑ ∑F

18

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19

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TR =

Fd m 2

⎛ l + π fd m ⎞ ⎟ ⎜ ⎜ π d − fl ⎟ m ⎠ ⎝

TL =

Fd m 2

⎛ π fd m − l ⎜ ⎜ π d + fl m ⎝

where, F = force dm = mean screw diameter l = lead distance f = coefficient of friction

⎞ ⎟ ⎟ ⎠

20

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

8.3 – The Mechanics of Power Screw – cont…

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

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

spin without any external effort. In such cases the torque TL

will be negative or zero.

be unscrewed by a definite negative torque; Clearly self-locking behavior is essential for threaded fasteners.

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

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

locking Condition for Self Locking:

handle was released.

π fd m > l

‰ 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.

‰ Dividing both sides of the above inequality by π d m and recognizing that l

π dm = tan λ ,

f > tan λ

we get

‰ If we let f = 0 , we obtain T0 =

which is the torque,

required to raise the load.

‰ The critical coefficient of friction for the lead concerned, ‰ If f = fcr

Fl 2π

the nut is on the point of moving down the thread without any torque applied. 21

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

‰ The efficiency is therefore

e=

T0 Fl = T R 2π T R 22

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CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

• 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.

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

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

1. Shearing stress in screw body.

τ =

2. Axial stress in screw body σ =

F 4F = A πd r2

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

16T πd r3

• For raising the load:

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

Fd m = 2

⎛ l + π fd m sec α ⎞ ⎜⎜ ⎟⎟ ⎝ π d m − fl sec α ⎠

3. Thread bearing stress

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

σB = −

• If fc is the coefficient of collar friction, assuming the load is concentrated at the mean collar diameter dc Ff c d c TC = 2 Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

F 2F =− πd m nt p / 2 πd m nt p

where nt is the number of engaged threads. Figure 8-16: Thrust collar has frictional diameter, dc

23

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

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

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BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

8.3 – The Mechanics of Power Screw – cont…

Stresses in Threads of Power Screws

4. Bending stress at root of thread.

σ

b

=

Mc ( p / 4) ⎛ Fp ⎞ = ⎜ ⎟ I ⎝ 4 ⎠ ( π d r n t )( p / 2 ) 3

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

= 12

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

6F πd r nt p

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

τ =

F 3V 3 3F = = πd r nt p 2A 2 πd r nt ( p / 2) Transform to von Mises stress

25

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.3 – The Mechanics of Power Screw – cont…

8.3 – The Mechanics of Power Screw – cont… Table 8-4: Screw bearing Pressure pb.

• Power screw thread is in compression, causing elastic shortening of screw

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

Bronze

5.5 – 9.7

100-200 mm/s

Cast iron

4.1 – 6.9

100-200 mm/s

Bronze

1.0 – 1.7

≥ 250 mm/s

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

Steel

experiment shows that:

Steel

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

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

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

Thread Deformation in Screw-Nut Combination

9 9 9 9

26

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

27

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

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

28

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont… Example Example8-1 8-1

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

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;

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

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

Hard steel on bronze

0.06

0.08

b) the torque required to raise and lower the load. c) the efficiency during lifting the load.

‰

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

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.

29

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

TL =

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

Fd m 2

⎛ πfd m − l ⎞ Ff c d c 6.4(30) ⎛ π (0.08)30 − 8 ⎞ 6.4(0.08)( 40) ⎜ ⎟ ⎜ ⎟+ ⎜ πd + fl ⎟ + 2 = 2 ⎜⎝ π (30) + 0.08(8) ⎟⎠ 2 ⎝ m ⎠

∴ T L = 9 . 77 Nm

thread depth = thread width = p/ 2 = 4 / 2 = 2 mm pitch diameter; d m = d − p / 2 = 32 − 4 / 2 = 30 mm

c) the efficiency during lifting the load.

minor diameter; d r = d − p = 32 − 4 = 28 mm

e=

lead for double thread, l = np = 2(4) = 8 mm b) the torque required to raise and lower the load.

Fd m 2

6.4(8) Fl = = 0.311 2πTR 2π (26.18)

d) the body stresses, torsional and compressive.

⎛ l + πfd m ⎞ Ff c d c 6.4(30 ) ⎛ 8 + π (0.08)(30 ) ⎞ (6.4)(0.08) 40 ⎜ ⎟ ⎜ ⎟+ ⎜ πd − fl ⎟ + 2 = 2 ⎜⎝ π (30 ) − (0.08)(8) ⎟⎠ 2 ⎝ m ⎠

The body shear stress τ due to torsional moment TR

∴ TR = 26.18 N .m Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

Solution Solution8-1 8-1

TR =

30

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

The axial nominal stress σ is 31

Td Tr 16T 16(26.18) = 42 = 3 = = 6.07 MPa J πd r πd r π (0.028)3 32 F −6400 −6400 = −10.39 MPa σ= = 2 = A πd r / 4 π (0.028)2 / 4

τ=

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

32

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BDA 31203 – Mechanical Component Design

8.3 – The Mechanics of Power Screw – cont…

8.3 – The Mechanics of Power Screw – cont…

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

For the von Misses stress,

2(0.38 F ) 2(0.38)(6400) σB = − =− = −12.9 MPa πd m (1) p π (30 )(1)4

σ '=

σ y = −10.39 MPa

τ yz = 6.07 MPa

σz = 0

τ zx = 0

⎛σ y +σ z ⎝ 2

σ 2,3 = ⎜⎜

]

1/ 2

⎞ ⎟⎟ ± ⎠

⎛σ y −σ z ⎜⎜ ⎝ 2

2

⎞ ⎟⎟ + τ yz2 ⎠

2

=

y

z

− 10.39 ⎛ − 10.39 ⎞ 2 ± ⎜ ⎟ + 6.07 = 2.79 MPa & - 13.18 MPa 2 ⎝ 2 ⎠

⎛ σ − σ 3 ⎞ 41.5 − (−13.18) ∴τ max = ⎜ 1 = 27.3 MPa ⎟= 2 ⎝ 2 ⎠

x

33

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

BDA 31203 – Mechanical Component Design

+ (0 − (−10.39)) 2 + (−10.39 − 41.5) 2 + 6(6.07) 2

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.

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:

τ xy = 0

2

h) the maximum shear stress

6(0.38 F ) 6(0.38)(6400) = = 41.5 MPa πd r (1) p π (28)(1)4

σ x = 41.5 MPa

[(41.5 − 0) 2

1

= 48.7 MPa

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

σb =

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.4 – Threaded Fastener

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BDA 31203 – Mechanical Component Design

34

CHAPTER 8 – Nonpermanent Joints

8.4 – Threaded Fastener – cont…

A- BOLTS: BOLTS Threaded Lengths

Purpose:

to clamp two or more members together. Metric

Parts:

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

Figure 8-18: Hexagon-head bolt

Hexagon-Head Bolt • • • •

English

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

Where d is the nominal diameter 35

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

36

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.4 – Threaded Fastener – cont…

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.4 – Threaded Fastener – cont…

B- NUTS: NUTS Same material as that of a screw

• Hexagon head bolt

Table AA-31 gives dimensions of Hexagonal nuts

Head Type of Bolts

– Usually uses nut – Heavy duty

• Hexagon head cap screw – Thinner head – Often used as screw (in threaded hole, without nut)

• Socket head cap screw End view

Washer-faced, regular

Chamfered both Washer-faced, sides, regular jam nut

– Usually more precision applications – Access from the top

Chamfered both sides, jam nut

Figure 8-19: Types of hexagon-head nut

Good Practice:

• Machine screws

1.First three threads of nut carry majority of load 2.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. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

BDA 31203 – Mechanical Component Design

– Usually smaller sizes – Slot or philips head common – Threaded all the way 37

CHAPTER 8 – Nonpermanent Joints

8.4 – Threaded Fastener – cont…

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

BDA 31203 – Mechanical Component Design

38

CHAPTER 8 – Nonpermanent Joints

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.

Machine Screws

‰ 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.

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

39

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

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.

40

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.5 – Joints : Fastener Stiffness

8.5 – Joints : Fastener Stiffness – cont…

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

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.

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

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.

‰ 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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

‰

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.

‰ Studs are regarded as permanent, and so the joint can be disassembled merely by removing the nut and washer. 41

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

42

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.5 – Joints : Fastener Stiffness – cont…

8.5 – Joints : Fastener Stiffness – cont…

Procedure to find bolt stiffness

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

8-1

(a)

(b)

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

(a)

43

8-2

(b)

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

44

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.5 – Joints : Fastener Stiffness – cont…

8.6 – Joints : Member Stiffness

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

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

1 1 1 = + kb kd kt

Joint pressure distribution theoretical models Model compressed members as if they are frusta spreading from the bolt head and nut to the midpoint of the grip

kd kt kb = kd + kt kd

km

A E AE ; kt = t = d ld lt

A d At E kb = A d l t + At l d

For short bolts kb= kt 45

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

‰

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. 46

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.6 – Joints : Member Stiffness – cont…

8.6 – Joints : Member Stiffness ‰

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

The area of element is

We choose a simpler approach using a fixed cone angle.

Substituting this Eq. and integrating gives a total contraction of 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.

‰

Each frustum has a half-apex angle of α

‰

The contraction of an element of the cone of thickness dx is subjected to a compressive force P is, Pdx dδ = EA Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

47

Integrating from 0 to t ; δ =

P

π Ed tan α

Thus the stiffness of this frustum is

ln

k =

( 2 t tan α + D − d )( D + d ) ( 2 t tan α + D + d )( D − d )

P

δ

=

π Ed tan α ( 2 t tan α + D − d )( D + d ) ln ( 2 t tan α + D + d )( D − d )

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

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CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.6 – Joints : Member Stiffness – cont…

8.6 – Joints : Member Stiffness – cont… For Members made of Aluminum, hardened steel and cast iron 25 < α < 33° For α = 30° 0 .5774 πEd k = (1 .155 t + D − d )( D + d ) ln (1 .155 t + D + d )( D − d )

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

km = A exp (Bd/l) Ed

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 diameter Using standard washer face diameter of 1.5d, and with α = 30º,

km =

2 ln

(

Combine all frusta as springs in series , km

0 . 5774 π Ed 5

0 . 5774 l + 0 . 5 d 0 . 5774 l + 2 . 5 d

1 1 1 1 1 = + + + .... k m k1 k 2 k 3 ki

) 49

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.6 – Joints : Member Stiffness – cont…

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.

50

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BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.6 – Joints : Member Stiffness – cont… Example Example8-2 8-2

Table 8-7:

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. 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.

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CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.6 – Joints : Member Stiffness – cont…

8.6 – Joints : Member Stiffness – cont… Solution Solution8-2 8-2

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

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

0 .75 + 2( 0 .595 ) tan 30 o = 1 . 437 in

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

1 ( 0 . 5 + 0 . 75 + 0 . 095 ) = 0 . 6725 in 2

k1 k2

The outer diameter at the midpoint of the entire joint is

k3

0 .75 + 2( 0 .6725 ) tan 30 = 1 . 527 in o

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

The spring rate of the steel is

0.6725 - 0.5 - 0.095 = 0.0775 in

k1 =

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

0.5774π (30)(10 6 )0.5 = 30.80(10 6 ) Ibf/in [ 1.155(0.595) + 0.75 − 0.5](0.75 + 0.5) ln [1.155(0.595) + 0.75 + 0.5](0.75 − 0.5)

steel with E = 30(10)6 psi, they can be considered a single frustum of 0.595 in thick. 53

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

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

For the upper cast-iron frustum

0.5774π (14.5)(10 6 )0.5

⎧ [1.155(0.0775) + 1.437 − 0.5](1.437 + 0.5) ⎫ ln ⎨ ⎬ ⎩ [1.155(0.0775) + 1.437 + 0.5](1.437 − 0.5) ⎭

= 285.5(10 6 ) Ibf/in

km =

For the lower cast-iron frustum

k3 =

0.5774π (14.5)(10 6 )0.5

⎧ [1.155(0.6725) + 0.75 − 0.5](0.75 + 0.5) ⎫ ln ⎨ ⎬ ⎩ [1.155(0.6725) + 0.75 + 0.5](0.75 − 0.5) ⎭

The three frusta are in series, so

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.6 – Joints : Member Stiffness – cont…

8.6 – Joints : Member Stiffness – cont…

k2 =

54

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

(

5

0 . 5774 (1 . 345 ) + 0 . 5 ( 0 . 5 ) 0 . 5774 (1 . 345 ) + 2 . 5 ( 0 . 5 )

)

= 14 . 64 (10 6 ) ibf/in

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

= 14.15(10 6 ) Ibf/in

⎡ 0 . 62873 ( 0 . 5 ) ⎤ 6 k m = 30 (10 6 )( 0 . 5 )( 0 . 78715 ) exp ⎢ ⎥ = 14 . 92 (10 ) ibf/in 1 . 345 ⎣ ⎦ In this case, the different between results is less than 2%.

1 1 1 1 = + + k m 30 . 80 (10 6 ) 285 . 5 (10 6 ) 14 . 15 (10 6 )

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

L T = 2 ( 0 . 5 ) + 0 . 25 = 1 . 25 in

This results in km = 9.378 (106) ibf/in Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

2 ln

0 . 5774 π ( 30 )( 10 6 ) 0 . 5

55

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8.6 – Joints : Member Stiffness – cont…

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

l d = 1 . 5 − 1 . 25 = 0 . 25 in

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

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

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

l t = 1 . 345 − 0 . 25 = 1 . 095 in The major diameter area is

Ad

⎞ ⎟ = 0 . 1963 in ⎟ ⎠

– Corresponds to proportional limit 2

– Slightly lower than yield strength – Typically used for static strength of bolt

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

kb =

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

6

A d At E 0 . 1963 ( 0 . 1599 ) 30 (10 ) = 0 . 1963 (1 . 095 ) + 0 . 1599 ( 0 . 25 ) A d l t + At l d

If Sp not available use Sp =0.85 Sy Fp = At Sp

6

= 3 . 69 (10 ) Ibf/in 57

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.7 – Bolts Strength

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

⎛ 0 .5 2 =π⎜ ⎜ 4 ⎝

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.7 – Bolts Strength – cont…

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BDA 31203 – Mechanical Component Design

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

58

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8.7 – Bolts Strength – cont… Table 8-8: SAE Specifications for Steel Bolts

• 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

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8.7 – Bolts Strength – cont…

BDA 31203 – Mechanical Component Design

Table 8-9: ASTM Specifications for Steel Bolts.

61

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8.7 – Bolts Strength – cont…

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BDA 31203 – Mechanical Component Design

62

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8.7 – Bolts Strength – cont…

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

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

CHAPTER 8 – Nonpermanent Joints

8.7 – Bolts Strength – cont…

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

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

BDA 31203 – Mechanical Component Design

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

63

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8.7 – Bolts Strength – cont…

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

65

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

CHAPTER 8 – Nonpermanent Joints

8.8 – Tension Joints : The External Load Let us consider what happens when an external tensile load to a bolt connection.

66

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BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

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)

P, is applied

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

– bolt is stretched – members in grip are compressed

Fi = preload P = external tensile load Pb = portion of P taken by bolt Pm = portion of P taken by members Fb = Pb + Fi = resultant bolt load Fm = Pm – Fi = resultant load on the members C = fraction of external load P carried by bolt 1-C = fraction of external load P carried by members

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

CHAPTER 8 – Nonpermanent Joints

8.7 – Bolts Strength – cont…

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

BDA 31203 – Mechanical Component Design

BDA 31203 – Mechanical Component Design

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

67

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8.8 – Tension Joints : The External Load – cont…

BDA 31203 – Mechanical Component Design

8.8 – Tension Joints : The External Load – cont…

• Since External Load P is shared by bolt and members, therefore

Resultant Bolt and Member Load : Fb & Fm

P = Pb + Pm

Fb = Pb + Fi = C P + Fi Fm = Pm − Fi = (1 − C ) P − Fi

• 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.

Fm < 0

‰ 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-use Fi = 0.90 Fp for permanent joint

Table 8-11: Computation of Bolt and Member Stiffnesses. Steel members clamped using a ½ in – 13 NC steel bolt.

69

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70

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.9 – Relating Bolt Torque to Bolt Tension

8.9 – Relating Bolt Torque to Bolt Tension – cont…

• 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

• Assuming a washer face diameter of 1.5d, the collar diameter is dc = (d + 1.5d)/2 = 1.25d, giving

Torque required to give preload Fi

• Define term in brackets as torque coefficient; K

• From the power screw equations, Table 8-12: Torque factor K

• Applying tan λ = l / π dm

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71

•

Some recommended values for K for various bolt finishes is given in Table 8–12

•

Use K = 0.2 for other cases Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

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BDA 31203 – Mechanical Component Design

8.9 – Relating Bolt Torque to Bolt Tension – cont…

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.9 – Relating Bolt Torque to Bolt Tension – cont… Solution Solution8-3 8-3

Example Example8-3 8-3

From Table 8-2, At = 0.373 in2. 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) (b) (c)

a) The preload is σ i =

Fi 25 = = 67 . 02 kpsi At 0 . 373

The stiffness constant is C =

Determine the preload and service load stresses in the bolt. Compare these to the minimum proof strength of the bolt. Specify the torque necessary to develop the preload using equation T = Kfid Specify the torque necessary to develop the preload if given f = fc = 0.15

kb 6 .5 = = 0 . 320 kb + km 6 . 5 + 13 . 8

The stress under the service load is σ b =

=

Fb CP + Fi CP = = +σi At At At 0 . 320 ( 6 ) + 67 . 02 = 72 . 17 kpsi 0 . 373

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. 73

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.9 – Relating Bolt Torque to Bolt Tension – cont…

T = KF i d = 0 . 2 ( 25 x10 3 )( 0 . 75 ) = 3750 Ibf.in c) The minor diameter can be determined from the minor area in Table 8-2. Thus

4 Ar

π

=

4 ( 0 . 351 )

π

BDA 31203 – Mechanical Component Design

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8.9 – Relating Bolt Torque to Bolt Tension – cont… For α = 30o

b) The torque necessary to achieve the preload is

dr =

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

⎧⎪ ⎡ 0 . 7093 ⎤ ⎡ tan 1 . 6066 o + 0 . 15 (sec 30 o ) ⎤ ⎫⎪ T = ⎨⎢ + 0 . 625 ( 0 . 15 ) ⎬ 25 (10 3 )( 0 . 75 ) ⎥⎢ o o ⎥ 2 ( 0 . 75 ) 1 − 0 . 15 (tan 1 . 6066 )(sec 30 ) ⎪⎩ ⎣ ⎪⎭ ⎦ ⎣⎢ ⎦⎥ = 3551 Ibf.in

= 0 . 6685 in Which is 5.3% less than the value found in part (b)

Than the mean diameter can be calculated as follow

dm =

0 . 75 + 0 . 6685 = 0 . 7093 in 2

The lead angle is

λ = tan −1

l

πd m

= tan −1

1

πd m N

= tan −1

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

1

π ( 0 . 7093 )(16 )

= 1 . 6066

o

75

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

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8.10 – Statically Loaded Tension Joint with Preload Failure of joints occurs when i) Bolt Yields

σb =

Failure starts

σb = Sp

a) Yielding Factor of safety:

b) Load Factor:

8.10 – Statically Loaded Tension Joint with Preload ii) Joint separates Let

At: Tensile stress area

Fb CP Fi = + At At At

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

Fm = (1 − C ) P − Fi Fm = 0

Proof strength

np =

Sp

σb

CP + Fi = S p At

=

Sp (CP + Fi ) / At

∴ nL =

=

n0 =

S p At CP + Fi

For n bolts n =

CP

(1 − C ) P0 − Fi = 0

P0 Fi = P P (1 − C )

S p At − Fi

Fi P / N (1 − C )

no : factor of safety against joint separation 77

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BDA 31203 – Mechanical Component Design

P0 be external load causing separation Fm= 0

CHAPTER 8 – Nonpermanent Joints

8.10 – Statically Loaded Tension Joint with Preload

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8.10 – Statically Loaded Tension Joint with Preload Solution Solution8-4 8-4

Example Example8-4 8-4 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. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

Figure 8-25:

79

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8.10 – Statically Loaded Tension Joint with Preload

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BDA 31203 – Mechanical Component Design

8.10 – Statically Loaded Tension Joint with Preload

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8.10 – Statically Loaded Tension Joint with Preload

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BDA 31203 – Mechanical Component Design

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

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8.11 – Gasketed Joints If a full gasket is present in joint, The gasket pressure p is:

p=−

Fm Ag / N

No. of bolts

With load factor n Fm = (1 − C ) nP − Fi p = [ Fi − nP (1 − C )]

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

83

N Ag

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

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IMPORTANT:

• Fatigue methods of Ch. 4 are directly applicable

1. To maintain uniformity of pressure adjacent bolts should not be placed more than 6 nominal diameters apart on bolt circle.

• Distribution of typical bolt failures is – 15% under the head – 20% at the end of the thread

2. To maintain wrench clearance bolts should be placed at least 3 d apart.

– 65% in the thread at the nut face

• Fatigue stress-concentration factors for threads and fillet are given in Table 8–13

3. A rough rule for bolt spacing around a bolt circle is

3≤

Nd

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints

8.11 – Gasketed Joints – cont…

π Db

BDA 31203 – Mechanical Component Design

Table 8-13: Fatigue Stress-Concentration Factors Kf for Threaded Elements

≤6

where Db is the diameter of the bolt circle and N is the number of bolts.

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BDA 31203 – Mechanical Component Design

85

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8.12 – Fatigue Loading of Tension Joints – cont…

BDA 31203 – Mechanical Component Design

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8.12 – Fatigue Loading of Tension Joints – cont… Fatigue Stresses

Endurance Strength for Bolts • Bolts are standardized, so endurance strengths are known by experimentation, including all modifiers. • 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 in Table 8-14.

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

Table 8-14: Full Corrected Endurance Strengths for Bolts and Screws with Rolled Threads*

• 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 stress-concentration factor Kf is also included as a reducer of the endurance strength, so it should not be applied to the bolt stresses. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

87

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

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8.12 – Fatigue Loading of Tension Joints – cont…

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.12 – Fatigue Loading of Tension Joints – cont…

Typical Fatigue Load Line for Bolts

Typical Fatigue Load Line for Bolts

‰ Typical load line starts from constant preload, then increases with a constant slope

Next, find the strength components Sa of the fatigue failure locus. These depend on the failure Criteria.

¾ Goodman Sa Sm + =1 Se Sut

‰ On the designer’s fatigue diagram, shown in Figure 8-26, the load line is .

Sa = Se −

Sa =

¾ Gerber 2

Sa ⎛ Sm ⎞ +⎜ ⎟ =1 Se ⎝ Sut ⎠

‰ High Preload is especially important in fatigue. σi is a constant the load line at Fi/At has a unit slope, r = 1.0

¾ ASMEASME-Elliptic 2

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.

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

BDA 31203 – Mechanical Component Design

2

⎛ Sa ⎞ ⎛ Sm ⎞ ⎜ ⎟ + ⎜⎜ ⎟⎟ = 1 ⎝ Se ⎠ ⎝ Sp ⎠ 89

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont… Typical Fatigue Load Line for Bolts

Equation of a typical fatigue load line:

Se Sm S ut

⎛S S a = S e − ⎜⎜ m ⎝ S ut

2

⎞ ⎟ Se ⎟ ⎠

⎛S Sa = Se 1 − ⎜ m ⎜ Sp ⎝

⎞ ⎟ ⎟ ⎠

σa (S m − σ i ) σ m −σi

∴ Sm =

S a (σ m − σ i )

σa

+σi

2

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BDA 31203 – Mechanical Component Design

90

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8.12 – Fatigue Loading of Tension Joints – cont… Repeated Load Special Case

• Solving (a) and (b) for Goodman line intersection point,

• Bolted joints often experience repeated load, where external load fluctuates between 0 and Pmax • Setting Pmin = 0; equation in slide 88

• Fatigue factor of safety based on Goodman line and constant preload load line,

nf =

Sa

• With constant preload load line,

σa

• Other failure curves can be used, following the same approach.

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

• Load line has slope of unity for repeated load case 91

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

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8.12 – Fatigue Loading of Tension Joints – cont… Repeated Load Special Case Fatigue factor of safety equations for repeated loading, constant preload load line, with various failure curves:

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont… Yield Check with Fatigue Stresses

Substitute σa and σi into any of the fatigue factor of safety equations

• 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

¾ Goodman

¾ Gerber

• This is equivalent to the yielding factor of safety from slide 77.

¾ ASME – Elliptic 93

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont…

94

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BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont… Solution Solution8-5 8-5

Example Example8-5 8-5

(a) For the symbols of Figs. 8-22 and 8-27,

h = t1 + t w = 0.6875 in l = h + d / 2 = 1 in D2 = 1.5d = 0.9375 in The joint is composed of three frusta; the upper two frusta are steel and the lower one is cast iron. Figure 8-27 Pressure-cone frustum member model for a cap screw. For this model the significant sizes are

For the upper frustrum; using Eq. in slide 49:

t = l / 2 = 0.5 in D = 0.9375 in E = 30 Mpsi

⎧h + t2 2 t2 < d l=⎨ (Refer to slide 43) ⎩h + D 2 t2 ≥ d D1 = dw + l tan α = 1.5d + 0.577l D2 = dw = 1.5d

⎫ ⎪ ⎬ k1 = 46.46 MIbf/in ⎪ ⎭

Where l = effective grip. The solutions are for α=30o and dw=1.5d. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

95 Figure 8-27:

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

96

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.12 – Fatigue Loading of Tension Joints – cont… For the middle frustrum;

For the lower frustrum;

t = h − l / 2 = 0.1875 in

⎫ ⎪ D = 0.9375 + 2(l − h) tan 30 = 1.298 in ⎬ k 2 = 197.43 MIbf/in ⎪ E = 30 Mpsi ⎭ o

t = l − h = 0.3125 in D = 0.9375 in Eci = 30 Mpsi

⎫ ⎪ ⎬ k 3 = 32.39 MIbf/in ⎪ ⎭

BDA 31203 – Mechanical Component Design

(b) Equation in slide 70 gives the preload as,

Fi = 0.75 F p = 0.75 At S p = 0.75(0.226)(85) = 14.4 kip 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

S p At CP + Fi

=

85(0.226) = 1.22 0.280(5) + 14.4

This is the traditional factor of safety, which compares the maximum bolt stress to the proof strength. Using Eq. in slide 77

nL =

kb 6.78 = = 0.280 k b + k m 6.78 + 17.40

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

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont…

np =

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

C=

BDA 31203 – Mechanical Component Design

S p At − Fi CP

=

85(0.226) − 14.4 = 3.44 0.280(5)

This factor is an indication of the overload on P that can be applied without exceeding the proof strength. 97

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont…

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

BDA 31203 – Mechanical Component Design

98

CHAPTER 8 – Nonpermanent Joints

8.12 – Fatigue Loading of Tension Joints – cont…

Next, using Eq. in slide 78, we have

n0 =

Fi 14.4 = = 4.00 P (1 − C ) 5(1 − 0.280)

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 defines a 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.

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

99

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

100

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.12 – Fatigue Loading of Tension Joints – cont… 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.

σi =

Fi 14.4 = = 63.72 kpsi At 0.226

Point B

σa =

CP 0.280(5) = = 3.10 kpsi 2 At 2(0.226)

The factor of safety resulting from this is n p =

S e ( Sut − σ i ) 18.6(120 − 63.72) = = 2.44 kpsi σ a ( Sut + S e ) 3.10(120 + 18.6)

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.12 – Fatigue Loading of Tension Joints – cont… 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 overproof loading. These two factors should always be compared to determine where is the greatest danger lies. Point E For the Gerber criterion, from Eq. in slide 93, the safety factor is

2σ a S e

[S

ut

Sut2 + 4 S e ( S e + σ i ) − Sut2 − 2σ i S e

[

]

1 = 120 120 2 + 4(18.6)(18.6 + 63.72) − 120 2 − 2(63.73)(18.6) 2(3.10)(18.6) = 3.65

]

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

2

=

=

85 − 63.72 = 10.64 kpsi 2

10.64 = 3.43 3.10

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.

101

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

nf =

Sa

σa

Sp −σi

which, of course, is identical to the result previously obtained by using Eq. in slide 77.

Point C This 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

1

Point D This is on the proof-strength line where S m + S a = S p

Solving Eqs. above simultaneously results in S a =

σ m = σ a + σ i = 3.10 + 63.72 = 66.82 kpsi

nf =

8.12 – Fatigue Loading of Tension Joints – cont…

In addition, the horizontal projection of the load line AD is S m = σ i + S a

The quantities shown in the caption of Fig. 8-28 are obtained as follows: Point A

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

103

102

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear y Shear loaded joints are handled the same for rivets, bolts, and pins y 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

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

Fig. 8–29: Modes of failure in shear loading of a bolted or riveted connection.

104

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CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear Failure by Bending

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear Failure by Shear of Bolt

•

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).

•

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.

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

BDA 31203 – Mechanical Component Design

105

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BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear Failure by Tensile Rupture of Member

•

•

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear Failure by Bearing Stress

Simple tensile failure

Use the smallest net area of the member, with holes removed

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

106

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

107

•

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

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

108

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BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear Failure by Shear-out or Tear-out •

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear Example Example8-6 8-6

Edge shear-out or tear-out is avoided by spacing bolts at least 1.5 diameters away from the edge

The bolted connection shown in Figure 8-30 uses SAE grade 5 bolts. The members are hotrolled 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. F = 4000 Ibf

Fig. 8–30

F = 4000 Ibf 109

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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear Solution Solution8-6 8-6

8.13 – Bolted and Riveted Joints Loaded in Shear

n=

Bolts: Sy = 92 kpsi (from Table 8-8) ; Ssy = 0.577Sy = 0.577(92) = 53.08 kpsi

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

Bearing of members

Members: Sy = 32 kpsi (from Table A-20)

S sc

σb

=

32 = 1.50 − 21.3

Tension on members

At = (2.375 − 0.75)(1 / 4) = 0.406 in 2

n=

S sy

τ

=

53.08 = 2.93 18.1

4 = 9.85 kpsi 0.406 Sy 32 n= = = 3.25 At 9.85

σt =

Shear of bolts

⎡ π (0.375) 2 ⎤ 2 As = 2 ⎢ ⎥ = 0.221 in 4 ⎣ ⎦ Fs 4 τ= = = 18.1 kpsi As 0.221

110

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

Bearing on bolts

Ab = 2(0.25)(0.375) = 0.188 in 2

Ab

−4 σb = = −21.3 kpsi 0.188 Sy 92 n= = = 4.32 σ b − 21.3

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

At

111

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

112

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BDA 31203 – Mechanical Component Design

8.13 – Bolted and Riveted Joints Loaded in Shear 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

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

BDA 31203 – Mechanical Component Design

•

Secondary Shear, due to moment load around centroid

(a) Example of eccentric loading (b) Free body diagram

Fig. 8–31: Centroid of pins, rivets or bolts.

113

CHAPTER 8 – Nonpermanent Joints

Shear Joints with Eccentric Loading Primary Shear

8.13 – Bolted and Riveted Joints Loaded in Shear

(c) Close up of bolt pattern

8.13 – Bolted and Riveted Joints Loaded in Shear

•

CHAPTER 8 – Nonpermanent Joints

Shear Joints with Eccentric Loading

•

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 ith area center.

BDA 31203 – Mechanical Component Design

114

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

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear Example Example8-7 8-7 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. Department of Material and Engineering Design, Faculty of Mechanical and Manufacturing Engineering, University of Tun Hussein Onn Malaysia (UTHM) Johor.

115

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

116

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CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear

8.13 – Bolted and Riveted Joints Loaded in Shear The primary shear load per bolt is F ' =

Solution Solution8-7 8-7

;

F"=

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

75 mm

75 = 51.3o 60

α = 90o − θ = 90o − 51.3o = 38.7 o

FC = FD = (17.7 cos 38.7 o − 4) 2 + (17.7 sin 38.7 o ) 2 = 14.8 kN αo

117

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear (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

BDA 31203 – Mechanical Component Design

F 21.0(10) =− = −131 MPa Ab 160

119

CHAPTER 8 – Nonpermanent Joints

8.13 – Bolted and Riveted Joints Loaded in Shear (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

M = 16(300 + 50) = 5600 Nm The second moment of area through this section is obtained by the use of the transfer formula, as follows:

I = I bar − 2( I holes + d 2 A) =

(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 3

118

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

F 21.0(10)3 τ= = = 146 MPa 144 As

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

θ = tan −1

FA = FB = (4 + 17.7 cos 38.7 o ) 2 + (17.7 sin 38.7 o ) 2 = 21.0 kN

Fig. 8–33

σ =−

60 mm

θo

r = (60) 2 + (75) 2 = 96.0 mm

BDA 31203 – Mechanical Component Design

Mr M 6800 = = = 17.7 kN 4r 2 4r 4(96.0)

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

M = 16(425) = 6800 Nm

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

V 16 = = 4 kN n 4

Since the secondary shear forces are equal, Eq. in slide 115 becomes

(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

CHAPTER 8 – Nonpermanent Joints

BDA 31203 – Mechanical Component Design

⎤ ⎡15(16) 3 15(200) 3 − 2⎢ + 60 2 (15)(16)⎥ = 8.26(10) 6 mm 4 12 ⎥⎦ ⎢⎣ 12

Then

σ=

Mc 5600(100) = = 67.8 MPa I 8.26(10) 6

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

120

BDA 31203 – Mechanical Component Design

CHAPTER 8 – Nonpermanent Joints

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.

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

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A

Appendix

Useful Tables Appendix Outline A–1

Standard SI Prefixes

A–2

Conversion Factors

A–3

Optional SI Units for Bending, Torsion, Axial, and Direct Shear Stresses

A–4

Optional SI Units for Bending and Torsional Deflections

A–5

Physical Constants of Materials

A–6

Properties of Structural-Steel Angles

A–7

Properties of Structural-Steel Channels

A–8

Properties of Round Tubing

A–9

Shear, Moment, and Deflection of Beams

961 962 963

963

963 964 966

968 969

A–10

Cumulative Distribution Function of Normal (Gaussian) Distribution

A–11

A Selection of International Tolerance Grades—Metric Series

A–12

Fundamental Deviations for Shafts—Metric Series

A–13

A Selection of International Tolerance Grades—Inch Series

A–14

Fundamental Deviations for Shafts—Inch Series

A–15

Charts of Theoretical Stress-Concentration Factors K t

A–16

Approximate Stress-Concentration Factors K t and K ts for Bending a Round Bar or Tube with a Transverse Round Hole 987

A–17

Preferred Sizes and Renard (R-series) Numbers

A–18

Geometric Properties

A–19

American Standard Pipe

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

A–22

Results of Tensile Tests of Some Metals

A–23

Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels

A–24

Mechanical Properties of Three Non-Steel Metals

A–25

Stochastic Yield and Ultimate Strengths for Selected Materials

A–26

Stochastic Parameters from Finite Life Fatigue Tests in Selected Metals

977

978

979 980

981 982

989

990 993

995

997 998

1000 1002 1003 959

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A–27

Finite Life Fatigue Strengths of Selected Plain Carbon Steels

A–28

Decimal Equivalents of Wire and Sheet-Metal Gauges

A–29

Dimensions of Square and Hexagonal Bolts

A–30

Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws

A–31

Dimensions of Hexagonal Nuts

A–32

Basic Dimensions of American Standard Plain Washers

A–33

Dimensions of Metric Plain Washers

A–34

Gamma Function

1012

1004

1005

1007

1009

1011

1010

1008

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Useful Tables

Table A–1

Name ∗†

Standard SI Prefixes

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 countries

of using commas for decimal points. ‡ Not recommended but sometimes encountered.

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Table A–2 Conversion Factors A to Convert Input X to Output Y Using the Formula Y = AX ∗ Multiply Input X

By Factor A

To Get Output Y

Multiply Input X

By Factor A

To Get Output Y

British thermal unit, Btu

1055

joule, J

mile/hour, mi/h

1.61

kilometer/hour, km/h

mile/hour, mi/h

0.447

meter/second, m/s

moment of inertia, lbm ·ft2

0.0421

kilogram-meter2, kg · m2

Btu/second, Btu/s

1.05

kilowatt, kW

calorie

4.19

joule, J

centimeter of mercury (0◦ C)

1.333

kilopascal, kPa

moment of inertia, lbm · in2

centipoise, cP

0.001

pascal-second, Pa · s

degree (angle)

0.0174

radian, rad

moment of section (second moment of area), in4

foot, ft

0.305

meter, m

ounce-force, oz

0.278

newton, N

foot2, ft2

0.0929

meter2, m2

ounce-mass

0.0311

kilogram, kg

4.45

newton, N

1.36

newton-meter, N·m

foot/minute, ft/min

0.0051

meter/second, m/s

pound, lbf †

foot-pound, ft · lbf

1.35

joule, J

pound-foot, lbf · ft

foot-pound/ second, ft · lbf/s

1.35

watt, W

foot/second, ft/s

0.305

meter/second, m/s

gallon (U.S.), gal

3.785

liter, L

horsepower, hp

0.746

kilowatt, kW

pound/inch, lbf/in

inch, in

0.0254

meter, m

pound/inch , psi (lbf/in2)

inch, in

25.4

inch2, in2

645

millimeter, mm millimeter2, mm2

inch of mercury (32◦ F)

3.386

kilopascal, kPa

kilopound, kip

4.45

kilonewton, kN

kilopound/inch2, kpsi (ksi)

6.89

megapascal, MPa (N/mm2)

2

mass, lbf · s /in mile, mi

175 1.610

kilogram, kg kilometer, km

pound/foot2, lbf/ft2

293 41.6

47.9

kilogram-millimeter2, kg · mm2 centimeter4, cm4

pascal, Pa

pound-inch, lbf · in

0.113

joule, J

pound-inch, lbf · in

0.113

newton-meter, N·m

175

2

6.89

newton/meter, N/m kilopascal, kPa

pound-mass, lbm

0.454

kilogram, kg

pound-mass/ second, lbm/s

0.454

kilogram/second, kg/s

quart (U.S. liquid), qt 946

milliliter, mL

section modulus, in3

16.4

centimeter3, cm3

slug

14.6

ton (short 2000 lbm) 907 yard, yd 0.914

∗ 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.

kilogram, kg kilogram, kg meter, m

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Useful Tables

Table A–3

Axial and Direct Shear

Bending and Torsion

Optional SI Units for Bending Stress σ = Mc/l, Torsion Stress τ = Tr/J, Axial Stress σ = F/A, and Direct Shear Stress τ = F/A

M,T

I,J

c, r

σ, τ

F

A

σ, τ

N · m∗

m4

m

Pa

N∗

m2

Pa

N·m

cm4

cm

MPa (N/mm2)

N†

mm2

MPa (N/mm2)

2

kPa GPa

4

†

N·m

mm

mm

GPa

kN

m

kN · m

cm4

cm

GPa

kN†

mm2

4

N · mm

mm

†

2

mm

MPa (N/mm )

∗ Basic relation. † Often preferred.

Table A–4

Bending Deflection

Optional SI Units for Bending Deflection y = f (Fl 3/El ) or y = f (wl 4/El ) and Torsional Deflection θ = Tl/GJ

F, wl

l

I

N∗

m

m4

Torsional Deflection

E

y

T

l

J

G

Pa

m

N · m∗

m

m4

θ

Pa

rad

kN

mm

mm

GPa

mm

N·m

mm

mm4

GPa

rad

kN

m

m4

GPa

µm

N · mm

mm

mm4

MPa (N/mm2)

rad

MPa (N/mm2)

rad

†

N

mm

4

4

mm

kPa

m

†

N·m

cm

cm

4

∗ Basic relation. † Often preferred.

Table A–5

Physical Constants of Materials Modulus of Elasticity E

Modulus of Rigidity G

Unit Weight w

Material

Mpsi

GPa

Mpsi

GPa

Poisson’s Ratio v

Aluminum (all alloys)

10.4

71.7

3.9

26.9

0.333

0.098

169

26.6

Beryllium copper

18.0

124.0

7.0

48.3

0.285

0.297

513

80.6

Brass

15.4

106.0

5.82

40.1

0.324

0.309

534

83.8

lbf/in3

lbf/ft 3

kN/m3

Carbon steel

30.0

207.0

11.5

79.3

0.292

0.282

487

76.5

Cast iron (gray)

14.5

100.0

6.0

41.4

0.211

0.260

450

70.6

Copper

17.2

119.0

6.49

44.7

0.326

0.322

556

87.3

1.6

11.0

0.6

4.1

0.33

0.016

28

4.3

Douglas fir Glass Inconel

6.7

46.2

2.7

18.6

0.245

0.094

162

25.4

31.0

214.0

11.0

75.8

0.290

0.307

530

83.3

Lead

5.3

36.5

1.9

13.1

0.425

0.411

710

111.5

Magnesium

6.5

44.8

2.4

16.5

0.350

0.065

112

17.6

Molybdenum

48.0

331.0

17.0

117.0

0.307

0.368

636

100.0

Monel metal

26.0

179.0

9.5

65.5

0.320

0.319

551

86.6

Nickel silver

18.5

127.0

7.0

48.3

0.322

0.316

546

85.8

Nickel steel

30.0

207.0

11.5

79.3

0.291

0.280

484

76.0

Phosphor bronze

16.1

111.0

6.0

41.4

0.349

0.295

510

80.1

Stainless steel (18-8)

27.6

190.0

10.6

73.1

0.305

0.280

484

76.0

Titanium alloys

16.5

114.0

6.2

42.4

0.340

0.160

276

43.4

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Mechanical Engineering Design

Table A–6 Properties of StructuralSteel Angles∗†

w = weight per foot, lbf/ft m = mass per meter, kg/m A = 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

1 × 1 × 18 × 14 1 12 × 1 12 × 18 × 14 2 × 2 × 18 × 14 × 38 1 1 2 2 × 2 2 × 14 × 38 3 × 3 × 14 × 38 × 12 3 12 × 3 12 × 14 × 38 × 12 4 × 4 × 14 × 38 × 12 × 58 6 × 6 × 38 × 12 × 58 × 34

0.80

3 1

1 y 3

l1−1

k1−1

Z1−1

y

k3−3

0.234

0.021

0.298

0.029

0.290

0.191

1.49

0.437

0.036

0.287

0.054

0.336

0.193

1.23

0.36

0.074

0.45

0.068

0.41

0.29

2.34

0.69

0.135

0.44

0.130

0.46

0.29

1.65

0.484

0.190

0.626

0.131

0.546

0.398

3.19

0.938

0.348

0.609

0.247

0.592

0.391

4.7

1.36

0.479

0.594

0.351

0.636

0.389

4.1

1.19

0.703

0.769

0.394

0.717

0.491

5.9

1.73

0.984

0.753

0.566

0.762

0.487

4.9

1.44

1.24

0.930

0.577

0.842

0.592

7.2

2.11

1.76

0.913

0.833

0.888

0.587

9.4

2.75

2.22

0.898

1.07

0.932

0.584

5.8

1.69

2.01

1.09

0.794

0.968

0.694

8.5

2.48

2.87

1.07

1.15

1.01

0.687

11.1

3.25

3.64

1.06

1.49

1.06

0.683

6.6

1.94

3.04

1.25

1.05

1.09

0.795

9.8

2.86

4.36

1.23

1.52

1.14

0.788

12.8

3.75

5.56

1.22

1.97

1.18

0.782

15.7

4.61

6.66

1.20

2.40

1.23

0.779

14.9

4.36

15.4

1.88

3.53

1.64

1.19

19.6

5.75

19.9

1.86

4.61

1.68

1.18

24.2

7.11

24.2

1.84

5.66

1.73

1.18

28.7

8.44

28.2

1.83

6.66

1.78

1.17

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Useful Tables

Table A–6 Properties of StructuralSteel Angles∗† (Continued)

Size, mm

m

A

25 × 25 × 3

1.11

1.42

×4

1.45

1.85

l1−1

965

k1−1

Z1−1

y

k3−3

0.80

0.75

0.45

0.72

0.48

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

34.9

1.78

8.41

1.85

1.16

55.8

2.44

9.57

2.17

1.57

80 × 80 × 6 ×8 × 10 100 ×100 × 8

7.34 9.63 11.9

11.1 9.35 12.3

72.2

2.43

12.6

2.26

1.56

15.1

87.5

2.41

15.4

2.34

1.55

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.

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Mechanical Engineering Design

Table A–7 Properties of Structural-Steel Channels∗ 2

a, b = size, in (mm) w = weight per foot, lbf/ft m = mass per meter, kg/m t = 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)

t 1

1 a

x 2 b

a, in

b, in

t

A

w

3

1.410

0.170

1.21

4.1

3

1.498

0.258

1.47

5.0

3

1.596

0.356

1.76

4

1.580

0.180

1.57

4

1.720

0.321

5

1.750

0.190

5

1.885

6

1.920

6 6

l1−1

k1−1

Z1−1

l2−2

k2−2

Z2−2

x

1.66

1.17

1.10

0.197

0.404

0.202

0.436

1.85

1.12

1.24

0.247

0.410

0.233

0.438

6.0

2.07

1.08

1.38

0.305

0.416

0.268

0.455

5.4

3.85

1.56

1.93

0.319

0.449

0.283

0.457

2.13

7.25

4.59

1.47

2.29

0.433

0.450

0.343

0.459

1.97

6.7

7.49

1.95

3.00

0.479

0.493

0.378

0.484

0.325

2.64

9.0

8.90

1.83

3.56

0.632

0.489

0.450

0.478

0.200

2.40

8.2

13.1

2.34

4.38

0.693

0.537

0.492

0.511

2.034

0.314

3.09

10.5

15.2

2.22

5.06

0.866

0.529

0.564

0.499

2.157

0.437

3.83

13.0

17.4

2.13

5.80

1.05

0.525

0.642

0.514

7

2.090

0.210

2.87

9.8

21.3

2.72

6.08

0.968

0.581

0.625

0.540

7

2.194

0.314

3.60

12.25

24.2

2.60

6.93

1.17

0.571

0.703

0.525

7

2.299

0.419

4.33

14.75

27.2

2.51

7.78

1.38

0.564

0.779

0.532

8

2.260

0.220

3.36

11.5

32.3

3.10

8.10

1.30

0.625

0.781

0.571

8

2.343

0.303

4.04

13.75

36.2

2.99

9.03

1.53

0.615

0.854

0.553

8

2.527

0.487

5.51

18.75

44.0

2.82

11.0

1.98

0.599

1.01

0.565

9

2.430

0.230

3.91

13.4

47.7

3.49

10.6

1.75

0.669

0.962

0.601

9

2.485

0.285

4.41

15.0

51.0

3.40

11.3

1.93

0.661

1.01

0.586

9

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

0.634

10

2.739

0.379

5.88

20.0

78.9

3.66

15.8

2.81

0.693

1.32

0.606

10

2.886

0.526

7.35

25.0

91.2

3.52

18.2

3.36

0.676

1.48

0.617

10

3.033

0.673

8.82

30.0

103

3.43

20.7

3.95

0.669

1.66

0.649

12

3.047

0.387

7.35

25.0

144

4.43

24.1

4.47

0.780

1.89

0.674

12

3.170

0.510

8.82

30.0

162

4.29

27.0

5.14

0.763

2.06

0.674

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Page 967

Useful Tables

967

Table A–7 Properties of Structural-Steel Channels (Continued) 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

2.95

19.46

10.66

1.12

4.07

1.19

102 × 51

10.42

6.1

13.28

207.7

3.95

40.89

29.10

1.48

8.16

1.51

127 × 64

14.90

6.4

18.98

482.5

5.04

67.23

1.88

15.25

1.94

152 × 76

17.88

6.4

22.77

851.5

6.12

111.8

113.8

2.24

21.05

2.21

152 × 89

23.84

7.1

30.36

1166

6.20

153.0

215.1

2.66

35.70

2.86

178 × 76

20.84

6.6

26.54

1337

7.10

150.4

134.0

2.25

24.72

2.20

178 × 89

26.81

7.6

34.15

1753

7.16

197.2

241.0

2.66

39.29

2.76

203 × 76

23.82

7.1

30.34

1950

8.02

192.0

151.3

2.23

27.59

2.13

203 × 89

29.78

8.1

37.94

2491

8.10

245.2

264.4

2.64

42.34

2.65

229 × 76

26.06

7.6

33.20

2610

8.87

228.3

158.7

2.19

28.22

2.00

229 × 89

32.76

8.6

41.73

3387

9.01

296.4

285.0

2.61

44.82

2.53

254 × 76

28.29

8.1

36.03

3367

9.67

265.1

162.6

2.12

28.21

1.86

254 × 89

35.74

9.1

45.42

4448

350.2

302.4

2.58

46.70

2.42

305 × 89

41.69

10.2

53.11

7061

11.5

463.3

325.4

2.48

48.49

2.18

305 × 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.

74.14

9.88

75.99

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Mechanical Engineering Design

Table A–8 Properties of Round Tubing

wa = unit weight of aluminum tubing, lbf/ft ws = unit weight of steel tubing, lbf/ft m = unit mass, kg/m A = 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 1× 1× 1 12 × 1 12

×

2× 2× 2 12 × 2 12

×

3× 3× 4× 4×

1 8 1 4 1 8 1 4 1 8 1 4 1 8 1 4 1 4 3 8 3 16 3 8

wa

ws

A

l

k

Z

J

0.416

1.128

0.344

0.034

0.313

0.067

0.067

0.713

2.003

0.589

0.046

0.280

0.092

0.092

0.653

1.769

0.540

0.129

0.488

0.172

0.257

1.188

3.338

0.982

0.199

0.451

0.266

0.399

0.891

2.670

0.736

0.325

0.664

0.325

0.650

1.663

4.673

1.374

0.537

0.625

0.537

1.074

1.129

3.050

0.933

0.660

0.841

0.528

1.319

2.138

6.008

1.767

1.132

0.800

0.906

2.276

2.614

7.343

2.160

2.059

0.976

1.373

4.117

3.742 2.717 5.167

10.51 7.654 14.52

3.093

2.718

0.938

1.812

5.436

2.246

4.090

1.350

2.045

8.180

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

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Useful Tables

Table A–9 Shear, Moment, and Deflection of Beams (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

969

1 Cantilever—end load y

R1 = V = F l

M = F(x − l) F x

M1

M1 = Fl

R1

y=

Fx2 (x − 3l) 6E I

ymax = −

Fl 3 3E I

V

+ x M x –

2 Cantilever—intermediate load y

R1 = V = F l a

M1 = Fa

M A B = F(x − a)

b

MBC = 0

F A

B

C x

yA B =

F x2 (x − 3a) 6E I

yB C =

Fa 2 (a − 3x) 6E I

ymax =

Fa 2 (a − 3l) 6E I

M1 R1 V

+ x M –

x

(continued)

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Mechanical Engineering Design

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

3 Cantilever—uniform load y

R1 = wl

l

M1 =

w

V = w(l − x)

x M1

y=

R1 V

wl 2 2 M =−

w (l − x)2 2

wx 2 (4lx − x 2 − 6l 2 ) 24E I

ymax = −

wl 4 8E I

+ x M x

–

4 Cantilever—moment load y

R1 = 0

l M1

MB

y=

A x

B R1 V x M

x

MB x 2 2E I

M1 = M B ymax =

M = MB MB l2 2E I

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Page 971

Useful Tables

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

971

5 Simple supports—center load y

R1 = R2 =

l l/2

F

A

B

V A B = R1

C x

V A B = R1

VB C = −R2

Fx F M B C = (l − x) 2 2 Fx = (4x 2 − 3l 2 ) 48E I

MA B =

R2

R1

F 2

yA B

V

ymax = −

+

Fl 3 48E I

x –

M

+ x

6 Simple supports—intermediate load y

R1 =

l a

b F

A

B

V A B = R1

C x

R1

R2 =

Fa l

VB C = −R2

Fbx Fa MB C = (l − x) l l Fbx 2 = (x + b2 − l 2 ) 6E I l Fa(l − x) 2 = (x + a 2 − 2lx) 6E I l

MA B =

R2

yA B

V

yB C

+ –

Fb l

x

M

+ x

(continued)

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Mechanical Engineering Design

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

7 Simple supports—uniform load y

R1 = R2 =

l w

V =

wl − wx 2

wx (l − x) 2 wx y= (2lx 2 − x 3 − l 3 ) 24E I

M=

x R1

wl 2

R2

V

ymax = −

5wl 4 384E I

+ x

–

M

+ x

8 Simple supports—moment load y

R1 = R2 =

l b

a

A

C x

B

yA B

R1

yB C

V

+ x M

+ –

x

MB l MB = (x − l) l

V =

MB x MB C l MB x 2 = (x + 3a 2 − 6al + 2l 2 ) 6E I l MB 3 = [x − 3lx 2 + x(2l 2 + 3a 2 ) − 3a 2 l] 6E I l

MA B =

R2

MB

MB l

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Useful Tables

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

973

9 Simple supports—twin loads y

R1 = R2 = F

l a

F

A

F

VB C = 0

VC D = −F

a

B

VA B = F

C

MA B = F x

D x R2

R1

M B C = Fa

MC D = F(l − x)

Fx 2 (x + 3a 2 − 3la) 6E I Fa = (3x 2 + a 2 − 3lx) 6E I Fa = (4a 2 − 3l 2 ) 24E I

yA B =

V

yB C +

ymax x –

M

+ x

10 Simple supports—overhanging load y a

l

F

R1 B

A

C x

R2

VA B MA B

V

yA B +

yB C

x

–

Fa F R2 = (l + a) l l Fa =− VB C = F l Fax =− M B C = F(x − l − a) l Fax 2 = (l − x 2 ) 6E I l F(x − l) = [(x − l)2 − a(3x − l)] 6E I

R1 =

yc = −

M

Fa 2 (l + a) 3E I

x –

(continued)

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Mechanical Engineering Design

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

11 One fixed and one simple support—center load y l

R1 =

F

l/2 A

V A B = R1

C

B

11F 16

R2 =

5F 16

M1 =

3Fl 16

VB C = −R2

x

MA B =

R2

M1 R1

F (11x − 3l) 16

MBC =

5F (l − x) 16

F x2 (11x − 9l) 96E I F(l − x) = (5x 2 + 2l 2 − 10lx) 96E I

yA B =

V

yB C + x –

M

+ x

–

12 One fixed and one simple support—intermediate load y l

F

a A

C

B

x R2

M1 R1 V

+ –

Fb 2 (3l − b2 ) 2l 3 Fb M1 = 2 (l 2 − b2 ) 2l R1 =

b

x

V A B = R1

+ –

x

Fa 2 (3l − a) 2l 3

VB C = −R2

MA B =

Fb 2 [b l − l 3 + x(3l 2 − b2 )] 2l 3

MBC =

Fa 2 2 (3l − 3lx − al + ax) 2l 3

yA B =

Fbx 2 [3l(b2 − l 2 ) + x(3l 2 − b2 )] 12E I l 3

yB C = y A B −

M

R2 =

F(x − a)3 6E I

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Useful Tables

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

975

13 One fixed and one simple support—uniform load y

5wl 3wl R2 = 8 8 5wl − wx V = 8 w M = − (4x 2 − 5lx + l 2 ) 8

R1 =

l

x M1

0.4215l R1

R2 ymax

y=

V 5l / 8

M1 =

wl 2 8

wx 2 (l − x)(2x − 3l) 48E I

ymax = − +

wl 4 185E I

x

–

M l /4 + x

–

14 Fixed supports—center load y

R1 = R2 =

l l/2

F

A

B

C x

M1

M2 R1

R2

V A B = −VB C = MA B = yA B =

V

x

M1 = M2 =

Fl 8

F 2

F (4x − l) 8

MBC =

F (3l − 4x) 8

F x2 (4x − 3l) 48E I

ymax = −

+

F 2

Fl 3 192E I

–

M

+ –

–

x

(continued)

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Mechanical Engineering Design

Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

15 Fixed supports—intermediate load y l a

Fb2 (3a + b) l3

R2 =

M1 =

Fab2 l2

Fa 2 b l2

b

F

A

R1 =

B

C x

M1

V A B = R1

M2 R1

R2

MA B =

V

M2 =

Fa 2 (3b + a) l3

VB C = −R2

Fb2 [x(3a + b) − al] l3

M B C = M A B − F(x − a) + x

–

M

yA B =

Fb2 x 2 [x(3a + b) − 3al] 6E I l 3

yB C =

Fa 2 (l − x)2 [(l − x)(3b + a) − 3bl] 6E I l 3

+ –

–

x

16 Fixed supports—uniform load y

R1 = R2 =

l

x M1

M2 R1

R2

V

x

–

M 0.2113l + –

–

x

M1 = M2 =

w (l − 2x) 2 w M= (6lx − 6x 2 − l 2 ) 12 V =

y=− +

wl 2

ymax = −

wx 2 (l − x)2 24E I wl 4 384E I

wl 2 12

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Useful Tables

977

Table A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution
(z α ) =

−∞

 =

zα

 2 u 1 du exp − √ 2 2π

α 1−α

f(z) ⌽(z␣)

zα ≤ 0 zα > 0

␣

0 z␣

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

0.4641

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.4247

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3897

0.3859

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.3557

0.3520

0.3483

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3238

0.3192

0.3156

0.3121

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.2877

0.2843

0.2810

0.2776

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.2546

0.2514

0.2483

0.2451

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

0.0985

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.0838

0.0823

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

0.0694

0.0681

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

0.0582

0.0571

0.0559

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0.0475

0.0465

0.0455

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

2.0

0.0228

0.0222

0.0217

0.0212

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

2.1

0.0179

0.0174

0.0170

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

2.2

0.0139

0.0136

0.0132

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

2.3

0.0107

0.0104

0.0102

0.00990

0.00964

0.00939

0.00914

0.00889

0.00866

0.00842

2.4

0.00820

0.00798

0.00776

0.00755

0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

(continued)

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Mechanical Engineering Design

Table A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution (Continued) Zα 3

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.00135

0.03968

0.03687

0.03483

0.03337

0.03233

0.03159

0.03108

0.04723

0.04481

4

4

0.0 317

4

0.0 207

4

0.0 133

5

0.0 854

5

0.0 541

5

0.0 340

5

0.0 211

5

0.0 130

6

0.0 793

0.06479

5

0.06287

0.06170

0.07996

0.07579

0.07333

0.07190

0.07107

0.08599

0.08332

0.08182

6

9

0.0 987

9

0.0 530

9

0.0 282

9

0.0 149

10

0.0 777

10

0.0 402

10

0.0 206

10

0.0 104

11

0.011260

zα

−1.282

−1.643

−1.960

−2.326

−2.576

−3.090

−3.291

−3.891

F(zα)

0.10

0.05

0.025

0.010

0.005

0.001

0.005

0.00005

0.000005

R(zα)

0.90

0.95

0.975

0.999

0.995

0.999

0.9995

0.9999

0.999995

Table A–11 A 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.

0.0 523 −4.417

Tolerance Grades

Basic Sizes

IT6

IT7

IT8

IT9

IT10

IT11

0–3

0.006

0.010

0.014

0.025

0.040

0.060

3–6

0.008

0.012

0.018

0.030

0.048

0.075

6–10

0.009

0.015

0.022

0.036

0.058

0.090

10–18

0.011

0.018

0.027

0.043

0.070

0.110

18–30

0.013

0.021

0.033

0.052

0.084

0.130

30–50

0.016

0.025

0.039

0.062

0.100

0.160

50–80

0.019

0.030

0.046

0.074

0.120

0.190

80–120

0.022

0.035

0.054

0.087

0.140

0.220

120–180

0.025

0.040

0.063

0.100

0.160

0.250

180–250

0.029

0.046

0.072

0.115

0.185

0.290

250–315

0.032

0.052

0.081

0.130

0.210

0.320

315–400

0.036

0.057

0.089

0.140

0.230

0.360

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Useful Tables

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.

Upper-Deviation Letter

Lower-Deviation Letter

Basic Sizes

c

d

f

g

h

0–3

−0.060

−0.020

−0.006

−0.002

0

k 0

n

p

s

u

+0.004

+0.006

+0.014

+0.018

3–6

−0.070

−0.030

−0.010

−0.004

0

+0.001

+0.008

+0.012

+0.019

+0.023

6–10

−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

+0.033

14–18

−0.095

−0.050

−0.016

−0.006

0

+0.001

+0.012

+0.018

+0.028

+0.033

18–24

−0.110

−0.065

−0.020

−0.007

0

+0.002

+0.015

+0.022

+0.035

+0.041

24–30

−0.110

−0.065

−0.020

−0.007

0

+0.002

+0.015

+0.022

+0.035

+0.048

30–40

−0.120

−0.080

−0.025

−0.009

0

+0.002

+0.017

+0.026

+0.043

+0.060

40–50

−0.130

−0.080

−0.025

−0.009

0

+0.002

+0.017

+0.026

+0.043

+0.070

50–65

−0.140

−0.100

−0.030

−0.010

0

+0.002

+0.020

+0.032

+0.053

+0.087

65–80

−0.150

−0.100

−0.030

−0.010

0

+0.002

+0.020

+0.032

+0.059

+0.102

80–100

−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

+0.144

120–140

−0.200

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.092

+0.170

140–160

−0.210

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.100

+0.190

160–180

−0.230

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.108

+0.210

180–200

−0.240

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.122

+0.236

200–225

−0.260

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.130

+0.258

225–250

−0.280

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.140

+0.284

250–280

−0.300

−0.190

−0.056

−0.017

0

+0.004

+0.034

+0.056

+0.158

+0.315

280–315

−0.330

−0.190

−0.056

−0.017

0

+0.004

+0.034

+0.056

+0.170

+0.350

315–355

−0.360

−0.210

−0.062

−0.018

0

+0.004

+0.037

+0.062

+0.190

+0.390

355–400

−0.400

−0.210

−0.062

−0.018

0

+0.004

+0.037

+0.062

+0.208

+0.435

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Mechanical Engineering Design

Table A–13 A Selection of International Tolerance Grades—Inch Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–11)

Basic Sizes

Tolerance Grades IT6

IT7

IT8

IT9

IT10

IT11

0–0.12

0.0002

0.0004

0.0006

0.0010

0.0016

0.0024

0.12–0.24

0.0003

0.0005

0.0007

0.0012

0.0019

0.0030

0.24–0.40

0.0004

0.0006

0.0009

0.0014

0.0023

0.0035

0.40–0.72

0.0004

0.0007

0.0011

0.0017

0.0028

0.0043

0.72–1.20

0.0005

0.0008

0.0013

0.0020

0.0033

0.0051

1.20–2.00

0.0006

0.0010

0.0015

0.0024

0.0039

0.0063

2.00–3.20

0.0007

0.0012

0.0018

0.0029

0.0047

0.0075

3.20–4.80

0.0009

0.0014

0.0021

0.0034

0.0055

0.0087

4.80–7.20

0.0010

0.0016

0.0025

0.0039

0.0063

0.0098

7.20–10.00

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

0.0126

12.60–16.00

0.0014

0.0022

0.0035

0.0055

0.0091

0.0142

−0.0016 −0.0020 −0.0026 −0.0026 −0.0031 −0.0031 −0.0039 −0.0039 −0.0047 −0.0047 −0.0057 −0.0057 −0.0057 −0.0067 −0.0067 −0.0067 −0.0075 −0.0075 −0.0083 −0.0083

−0.0031 −0.0037 −0.0043 −0.0043 −0.0047 −0.0051 −0.0055 −0.0059 −0.0067 −0.0071 −0.0079 −0.0083 −0.0091 −0.0094 −0.0102 −0.0110 −0.0118 −0.0130 −0.0142 −0.0157

0.72–0.96

0.96–1.20

1.20–1.60

1.60–2.00

2.00–2.60

2.60–3.20

3.20–4.00

4.00–4.80

4.80–5.60

5.60–6.40

6.40–7.20

7.20–8.00

8.00–9.00

9.00–10.00

10.00–11.20

11.20–12.60

12.60–14.20

14.20–16.00

−0.0012

−0.0028

0.24–0.40

−0.0008

−0.0024

0–0.12

0.12–0.24

0.40–0.72

d

−0.0024

−0.0024

−0.0022

−0.0022

−0.0020

−0.0020

−0.0020

−0.0017

−0.0017

−0.0017

−0.0014

−0.0014

−0.0012

−0.0012

−0.0010

−0.0010

−0.0008

−0.0008

−0.0006

−0.0005

−0.0004

−0.0002

f

g

−0.0007

−0.0007

−0.0007

−0.0007

−0.0006

−0.0006

−0.0006

−0.0006

−0.0006

−0.0006

−0.0005

−0.0005

−0.0004

−0.0004

−0.0004

−0.0004

−0.0003

−0.0003

−0.0002

−0.0002

−0.0002

−0.0001

Upper-Deviation Letter

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

h

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

0

0

0

0

k

+0.0015

+0.0015

+0.0013

+0.0013

+0.0012

+0.0012

+0.0012

+0.0011

+0.0011

+0.0011

+0.0009

+0.0009

+0.0008

+0.0008

+0.0007

+0.0007

+0.0006

+0.0006

+0.0005

+0.0004

+0.0003

+0.0024

+0.0024

+0.0022

+0.0022

+0.0020

+0.0020

+0.0020

+0.0017

+0.0017

+0.0017

+0.0015

+0.0015

+0.0013

+0.0013

+0.0010

+0.0010

+0.0009

+0.0009

+0.0007

+0.0006

+0.0005

+0.0002

p

s

+0.0082

+0.0075

+0.0067

+0.0062

+0.0055

+0.0051

+0.0048

+0.0043

+0.0039

+0.0036

+0.0031

+0.0028

+0.0023

+0.0021

+0.0017

+0.0017

+0.0014

+0.0014

+0.0011

+0.0009

+0.0007

+0.0006

Lower-Deviation Letter +0.0002

n

+0.0171

+0.0154

+0.0130

+0.0124

+0.0112

+0.0102

+0.0093

+0.0083

+0.0075

+0.0067

+0.0057

+0.0049

+0.0040

+0.0034

+0.0028

+0.0024

+0.0019

+0.0016

+0.0013

+0.0011

+0.0009

+0.0007

u

3:43 PM

c

6/3/03

Basic Sizes

Fundamental Deviations for Shafts—Inch Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–12)

Table A–14

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Mechanical Engineering Design

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t Figure A–15–1

3.0 d

Bar in tension or simple compression with a transverse hole. σ0 = F/A, where A = (w − d )t and t is the thickness.

2.8

w

2.6 Kt 2.4

2.2

2.0

Figure A–15–2

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

3.0

Rectangular bar with a transverse hole in bending. σ0 = Mc/I, where 3 I = (w − d )h /12.

0.7

0.8

d d/h = 0

w

2.6 0.25

M

M

0.5

2.2

h

1.0

Kt

2.0

1.8

⬁ 1.4

1.0

Figure A–15–3

0

0.1

0.2

0.3

0.4 d/w

0.5

3.0

0.6

0.7

0.8

r w /d = 3

Notched rectangular bar in tension or simple compression. σ0 = F/A, where A = dt and t is the thickness.

w

2.6

d

1.5 2.2 1.2 Kt

1.1 1.8 1.05 1.4

1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

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Useful Tables

983

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued) Figure A–15–4

3.0 w/d = ⬁

1.10

Notched rectangular bar in bending. σ0 = Mc/I, where c = d/2, I = td 3 /12, and t is the thickness.

2.6

r M

1.5

1.05

w

M

d

2.2 1.02

Kt 1.8

1.4

1.0

Figure A–15–5

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

0.25

0.30

3.0

Rectangular filleted bar in tension or simple compression. σ0 = F/A, where A = dt and t is the thickness.

r

D/d = 1.50 2.6

d

D 1.10 2.2 Kt

1.05 1.8 1.02 1.4

1.0

Figure A–15–6

0

0.05

0.10

0.15 r/d

0.20

3.0 r

Rectangular filleted bar in bending. σ0 = Mc/I, where 3 c = d/2, I = td /12, t is the thickness.

2.6

M

1.05

d

D

M

3 2.2

1.1 1.3

Kt 1.8

D/d = 1.02

1.4

1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

(continued)

*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.

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Mechanical Engineering Design

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued) Figure A–15–7

2.6

Round shaft with shoulder fillet in tension. σ0 = F/A, where A = πd 2 /4.

r 2.2

d

D

Kt 1.8

D/d 1.0

= 1.

50

1.10

5

1.4

1.02 1.0

0

Figure A–15–8

3.0

Round shaft with shoulder fillet in torsion. τ0 = Tc/J, where 4 c = d/2 and J = πd /32.

2.6

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r d

D

T

T

2.2 Kts 1.8

D/d =

1.4

1.0

Figure A–15–9

1.20 1.33

2

1.09

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

3.0 r

Round shaft with shoulder fillet in bending. σ0 = Mc/I, where c = d/2 and I = πd 4 /64.

2.6 M

d

D

M

2.2 Kt 1.8

D/d

=3

1.5 1.4

1.10

1.02

1.05 1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

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Useful Tables

985

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued)

Figure A–15–10

4.0 d

Round shaft in torsion with transverse hole.

3.6

Kts, B

2.8

Figure A–15–11

B A

J ␲D3 dD2 c = 16 – 6 (approx)

Kts, A

Kts 3.2

2.4

D

T

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

3.0 d

Round shaft in bending with a transverse hole. σ0 = 2 M/[(πD3 /32) − (dD /6)], approximately.

D

2.6 M

M

2.2 Kt 1.8

1.4

1.0

Figure A–15–12 Plate loaded in tension by a pin through a hole. σ0 = F/A, where A = (w − d)t . When clearance exists, increase Kt 35 to 50 percent. (M. M. Frocht and H. N. Hill, “Stress Concentration Factors around a Central Circular Hole in a Plate Loaded through a Pin in Hole,” J. Appl. Mechanics, vol. 7, no. 1, March 1940, p. A-5.)

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

11

h

9

d

h/w = 0.35 w

7

t

Kt 5 h/w = 0.50 3

1

h/w ⱖ 1.0

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

0.7

0.8

(continued)

*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.

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Mechanical Engineering Design

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued)

Figure A–15–13

3.0 r

1.15

Grooved round bar in tension. σ0 = F/A, where A = πd 2 /4.

2.6 D

1.05

d

2.2 Kt 1.02

D/d = 1.50

1.8

1.4

1.0

0

Figure A–15–14

3.0

Grooved round bar in bending. σ0 = Mc/l, where 4 c = d/2 and I = πd /64.

2.6

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r

M

D

M

d

1.05 2.2 Kt D/d = 1.50

1.02 1.8

1.4

1.0

0

Figure A–15–15

2.6

Grooved round bar in torsion. τ0 = Tc/J, where c = d/2 4 and J = πd /32.

2.2

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r T

T D

1.8

d

1.05

Kts

D/d = 1.30 1.4 1.02 1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

*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.

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Useful Tables

a

Table A–16 Approximate StressConcentration Factor Kt for Bending of a Round Bar or Tube with a Transverse Round Hole

987

D

d

M

M

The nominal bending stress is σ0 = M/Z net where Z net is a reduced value of the section modulus and is defined by

Source: R. E. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. 146, 235.

Z net =

πA (D 4 − d 4 ) 32D

Values of A are listed in the table. Use d = 0 for a solid bar

d/D 0.9

0.6

0

a/D

A

Kt

A

Kt

A

Kt

0.050

0.92

2.63

0.91

2.55

0.88

2.42

0.075

0.89

2.55

0.88

2.43

0.86

2.35

0.10

0.86

2.49

0.85

2.36

0.83

2.27

0.125

0.82

2.41

0.82

2.32

0.80

2.20

0.15

0.79

2.39

0.79

2.29

0.76

2.15

0.175

0.76

2.38

0.75

2.26

0.72

2.10

0.20

0.73

2.39

0.72

2.23

0.68

2.07

0.225

0.69

2.40

0.68

2.21

0.65

2.04

0.25

0.67

2.42

0.64

2.18

0.61

2.00

0.275

0.66

2.48

0.61

2.16

0.58

1.97

0.30

0.64

2.52

0.58

2.14

0.54

1.94 (continued)

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Mechanical Engineering Design

Table A–16 (Continued) Approximate Stress-Concentration Factors Kts for a Round Bar or Tube Having a Transverse Round Hole and Loaded in Torsion Source: R. E. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. 148, 244.

D

a

d

T

T

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(D 4 − d 4 ) 32

Values of A are listed in the table. Use d = 0 for a solid bar. d/D 0.9

0.8 A

0.6 Kts

A

0.4 Kts

A

0

a/D

A

Kts

Kts

A

Kts

0.05

0.96

1.78

0.95

1.77

0.075

0.95

1.82

0.10

0.94

1.76

0.93

1.74

0.92

1.72

0.92

1.70

0.93

1.71

0.92

1.68

0.125

0.91

1.76

0.91

1.74

0.90

1.70

0.90

0.15

0.90

1.77

0.89

1.75

0.87

1.69

0.87

1.67

0.89

1.64

1.65

0.87

1.62

0.175

0.89

1.81

0.88

1.76

0.87

1.69

0.20

0.88

1.96

0.86

1.79

0.85

1.70

0.86

1.64

0.85

1.60

0.84

1.63

0.83

1.58

0.25

0.87

2.00

0.82

1.86

0.81

0.30

0.80

2.18

0.78

1.97

0.77

1.72

0.80

1.63

0.79

1.54

1.76

0.75

1.63

0.74

1.51

0.35

0.77

2.41

0.75

2.09

0.40

0.72

2.67

0.71

2.25

0.72

1.81

0.69

1.63

0.68

1.47

0.68

1.89

0.64

1.63

0.63

1.44

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Useful Tables

Table A–17 Preferred Sizes and Renard (R-Series) Numbers (When a choice can be made, use one of these sizes; however, not all parts or items are available in all the sizes shown in the table.)

989

Fraction of Inches 1 , 1 , 1 , 3 , 1 , 5 , 3 , 1 , 5 , 3 , 7 , 1 , 9 , 5 , 11 , 3 , 7 , 1, 1 14 , 1 12 , 1 34 , 2, 2 14 , 64 32 16 32 8 32 16 4 16 8 16 2 16 8 16 4 8 2 12 , 2 34 , 3, 3 14 , 3 12 , 3 34 , 4, 4 14 , 4 12 , 4 34 , 5, 5 14 , 5 12 , 5 34 , 6, 6 12 , 7, 7 12 , 8, 8 12 , 9, 9 12 , 10, 10 12 , 11, 11 12 , 12, 12 12 , 13, 13 12 , 14, 14 12 , 15, 15 12 , 16, 16 12 , 17, 17 12 , 18,

18 12 , 19, 19 12 , 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, 10 2d choice, R10: 1.25, 2, 3.15, 5, 8 3d choice, R20: 1.12, 1.4, 1.8, 2.24, 2.8, 3.55, 4.5, 5.6, 7.1, 9 4th 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.

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Page 990

Mechanical Engineering Design

Table A–18 Geometric Properties

Part 1 Properties of Sections A = area G = location of centroid
Ix = x 2 d A = second moment of area about x axis
Ix y = x y d A = mixed moment of area about x and y axes

JG = r 2 d A = (x 2 + y 2 ) d A = Ix + I y = second polar moment of area about axis through G k x2

= Ix /A = squared radius of gyration about x axis

Rectangle

y b 2

h

h 2

G

x

b

A = bh

bh 3 Ix = 12

b3 h Iy = 12

Circle

Ix y = 0

y

D x

G

A=

π D2 4

Ix = I y =

π D4 64

Ix y = 0

Hollow circle

y

d

D

G

A=

π 2 (D − d 2 ) 4

Ix = I y =

π (D 4 − d 4 ) 64

x

Ix y = 0

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Useful Tables

991

Table A–18 Geometric Properties (Continued)

y

Right triangles

y

b 3

b h 3 G

h h 3

G

x b 3

b

A=

bh 2

Ix =

bh 3 36

Iy =

x

h

b3 h 36

−b2 h 2 72

Ix y =

y

y

Right triangles

b 3

b h 3 h

h 3

h x

G

b 3

b

A=

bh 2

Ix =

bh 3 36

Iy =

x

G

b3 h 36

Ix y =

b2 h 2 72 y

y

Quarter-circles 4r 3␲

r 4r 3␲ G

4r 3␲

r

A=

πr 2 4

 Ix = I y = r 4

π 4 − 16 9π



 Ix y = r 4

1 4 − 8 9π



y

y

Quarter-circles

x

G

x

4r 3␲

4r 3␲

r 4r 3␲ G

x 4r 3␲

4r 3␲

πr 2 A= 4

 Ix = I y = r

4

π 4 − 16 9π

G

x

r



 Ix y = r

4

4 1 − 9π 8



(continued)

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Mechanical Engineering Design

Table A–18

Part 2 Properties of Solids (ρ = Density, Weight per Unit Volume)

Geometric Properties (Continued)

Rods

y

d z

l x

m=

πd 2 lρ 4g

I y = Iz =

ml 2 12 y

Round disks t

d

x

z

m=

πd 2 tρ 4g

Ix =

md 2 8

md 2 16

I y = Iz =

y

Rectangular prisms

b

c

z

m=

abcρ g

Ix =

m 2 (a + b2 ) 12

a

Iy =

x

m 2 m 2 (a + c2 ) Iz = (b + c2 ) 12 12

y

Cylinders

d z

m=

πd 2 lρ 4g

Ix =

md 2 8

l

x

m (3d 2 + 4l 2 ) 48

I y = Iz = y

Hollow cylinders

di do z

m=

  π do2 − di2 lρ 4g

Ix =

l

 m 2 do + di2 8

x

I y = Iz =

 m  2 3do + 3di2 + 4l 2 48

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Useful Tables

Table A–19 American Standard Pipe

993

Wall Thickness, in Nominal Size, in 1 8 1 4 3 8 1 2 3 4

Outside Diameter, in

Threads per inch

Standard No. 40

Extra Strong No. 80

Double Extra Strong

0.405

27

0.070

0.098

0.540

18

0.090

0.122

0.675

18

0.093

0.129

0.840

14

0.111

0.151

0.307

1.050

14

0.115

0.157

0.318

1.315

11 12

0.136

0.183

0.369

1 14 1 12

1.660

0.143

0.195

0.393

0.148

0.204

0.411

1

2

2.375

11 12 11 12 11 12

0.158

0.223

0.447

2 12

2.875

8

0.208

0.282

0.565 0.615

1.900

3

3.500

8

0.221

0.306

3 12

4.000

8

0.231

0.325

4

4.500

8

0.242

0.344

0.690

5

5.563

8

0.263

0.383

0.768

6

6.625

8

0.286

0.441

0.884

8

8.625

8

0.329

0.510

0.895

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Mechanical Engineering Design

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 ( 34 to 1 14 in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification. See Table 1–1 for certain ASTM steels.] Source: 1986 SAE Handbook, p. 2.15. 1

2

3

UNS No.

SAE and/or AISI No.

Processing

G10060

1006

HR

G10100

1010

G10150

1015

G10180

1018

G10200

1020

G10300

1030

G10350

1035

G10400

1040

G10450

1045

G10500

1050

G10600

1060

4 5 Tensile Yield Strength, Strength, MPa (kpsi) MPa (kpsi) 300 (43)

170 (24)

6

7

8

Elongation in 2 in, %

Reduction in Area, %

Brinell Hardness

30

55

86

CD

330 (48)

280 (41)

20

45

95

HR

320 (47)

180 (26)

28

50

95

CD

370 (53)

300 (44)

20

40

105

HR

340 (50)

190 (27.5)

28

50

101

CD

390 (56)

320 (47)

18

40

111

HR

400 (58)

220 (32)

25

50

116

CD

440 (64)

370 (54)

15

40

126

HR

380 (55)

210 (30)

25

50

111

CD

470 (68)

390 (57)

15

40

131

HR

470 (68)

260 (37.5)

20

42

137

CD

520 (76)

440 (64)

12

35

149

HR

500 (72)

270 (39.5)

18

40

143

CD

550 (80)

460 (67)

12

35

163

HR

520 (76)

290 (42)

18

40

149

CD

590 (85)

490 (71)

12

35

170

HR

570 (82)

310 (45)

16

40

163

CD

630 (91)

530 (77)

12

35

179

HR

620 (90)

340 (49.5)

15

35

179

CD

690 (100)

580 (84)

10

30

197

HR

680 (98)

370 (54)

12

30

201

G10800

1080

HR

770 (112)

420 (61.5)

10

25

229

G10950

1095

HR

830 (120)

460 (66)

10

25

248

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995

Useful Tables

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. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American Society for Metals, Metals Park, Ohio, 1983.

1

2

3

AISI No.

Treatment

Temperature °C (°F)

1030

1040

1050

1060

1095

1141

4 5 Tensile Yield Strength Strength, MPa (kpsi) MPa (kpsi)

6

7

8

Elongation, %

Reduction in Area, %

Brinell Hardness

Q&T*

205 (400)

848 (123)

648 (94)

17

47

495

Q&T*

315 (600)

800 (116)

621 (90)

19

53

401

Q&T*

425 (800)

731 (106)

579 (84)

23

60

302

Q&T*

540 (1000)

669 (97)

517 (75)

28

65

255

Q&T*

650 (1200)

586 (85)

441 (64)

32

70

207

Normalized

925 (1700)

521 (75)

345 (50)

32

61

149

Annealed

870 (1600)

430 (62)

317 (46)

35

64

137

Q&T

205 (400)

779 (113)

593 (86)

19

48

262

Q&T

425 (800)

758 (110)

552 (80)

21

54

241

Q&T

650 (1200)

634 (92)

434 (63)

29

65

192

Normalized

900 (1650)

590 (86)

374 (54)

28

55

170

Annealed

790 (1450)

519 (75)

353 (51)

30

57

149

Q&T*

205 (400)

1120 (163)

807 (117)

9

27

514

Q&T*

425 (800)

1090 (158)

793 (115)

13

36

444

Q&T*

650 (1200)

717 (104)

538 (78)

28

65

235

Normalized

900 (1650)

748 (108)

427 (62)

20

39

217

Annealed

790 (1450)

636 (92)

365 (53)

24

40

187

Q&T

425 (800)

1080 (156)

765 (111)

14

41

311

Q&T

540 (1000)

965 (140)

669 (97)

17

45

277

Q&T

650 (1200)

800 (116)

524 (76)

23

54

229

Normalized

900 (1650)

776 (112)

421 (61)

18

37

229

Annealed

790 (1450)

626 (91)

372 (54)

22

38

179

Q&T

315 (600)

1260 (183)

813 (118)

10

30

375

Q&T

425 (800)

1210 (176)

772 (112)

12

32

363

Q&T

540 (1000)

1090 (158)

676 (98)

15

37

321

Q&T

650 (1200)

896 (130)

552 (80)

21

47

269

Normalized

900 (1650)

1010 (147)

500 (72)

9

13

293

Annealed

790 (1450)

658 (95)

380 (55)

13

21

192

Q&T

315 (600)

1460 (212)

1280 (186)

9

32

415

Q&T

540 (1000)

896 (130)

765 (111)

18

57

262 (continued)

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Mechanical Engineering Design

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. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. Unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American Society for Metals, Metals Park, Ohio, 1983.

1

2

3

AISI No.

Treatment

Temperature °C (°F)

4130

4140

4340

*Water-quenched

4 5 Tensile Yield Strength Strength, MPa (kpsi) MPa (kpsi)

6

7

8

Elongation, %

Reduction in Area, %

Brinell Hardness

Q&T*

205 (400)

1630 (236)

1460 (212)

10

41

467

Q&T*

315 (600)

1500 (217)

1380 (200)

11

43

435

Q&T*

425 (800)

1280 (186)

1190 (173)

13

49

380

Q&T*

540 (1000)

1030 (150)

910 (132)

17

57

315

Q&T*

650 (1200)

814 (118)

703 (102)

22

64

245

Normalized

870 (1600)

670 (97)

436 (63)

25

59

197

Annealed

865 (1585)

560 (81)

361 (52)

28

56

156

Q&T

205 (400)

1770 (257)

1640 (238)

8

38

510

Q&T

315 (600)

1550 (225)

1430 (208)

9

43

445

Q&T

425 (800)

1250 (181)

1140 (165)

13

49

370

Q&T

540 (1000)

951 (138)

834 (121)

18

58

285

Q&T

650 (1200)

758 (110)

655 (95)

22

63

230

Normalized

870 (1600)

1020 (148)

655 (95)

18

47

302

Annealed

815 (1500)

655 (95)

417 (61)

26

57

197

Q&T

315 (600)

1720 (250)

1590 (230)

10

40

486

Q&T

425 (800)

1470 (213)

1360 (198)

10

44

430

Q&T

540 (1000)

1170 (170)

1080 (156)

13

51

360

Q&T

650 (1200)

965 (140)

855 (124)

19

60

280

Aluminum alloy Aluminum alloy

2024

7075

T6

T4

T6

Annealed

Annealed

Q&T 600°F

Q&T 600°F

HR

Annealed

Annealed

Condition

542 (78.6)

296 (43.0)

169 (24.5)

276 (40.0)

241 (35.0)

1720 (250)

1520 (220)

193 (28.0)

358 (52.0)

220 (32.0)

593 (86.0)

446 (64.8)

324 (47.0)

568 (82.4)

601 (87.3)

1930 (210)

1580 (230)

424 (61.5)

646 (93.7)

341 (49.5)

1270 (185) 620 (90) 689 (100) 882 (128)

1600 (233)† 325 (47.2)† 533 (77.3)† 706 (102)†

1520 (221)

0.13

0.15

0.28

0.45

0.51

0.048

1760 (255)† 1410 (205)

0.24

†

2340 (340)

2380 (345)

0.041

†

0.14

0.25 758 (110)

992 (144)

898 (130)†

Strain Strength, Exponent m

1880 (273)†

620 (90.0)

628 (91.1)† 729 (106)

Coefficient σ0, MPa (kpsi)

Fracture, σf, MPa (kpsi)

*Values from one or two heats and believed to be attainable using proper purchase specifications. The fracture strain may vary as much as 100 percent. † Derived value.

Aluminum alloy

2011

Steel

4142

Stainless steel

Steel

1045

304

Steel

1212

Stainless steel

Steel

1144

303

Steel

Material

1018

Number

Ultimate Su, MPa (kpsi)

0.18

0.18

0.10

1.67

1.16

0.43

0.81

0.85

0.49

1.05

Fracture Strain εf

3:43 PM

Yield Sy, MPa (kpsi)

Strength (Tensile)

Source: J. Datsko, “Solid Materials,” chap. 7 in Joseph E. Shigley and Charles R. Mischke (eds.-in-chief), Standard

Handbook of Machine Design, 2nd ed., McGraw-Hill, New York, 1996, pp. 7.47–7.50.

Results of Tensile Tests of Some Metals*

6/3/03

Table A–22

shi20361_app_A.qxd Page 997

997

998

L LT LT L

10B62

1005-1009

1005-1009

1005-1009

L L L

1045

1045

1144

L L

1045

1045

L L

1045

1045

L L

1020

1040

L

L

L

CD sheet

LT

RQC-100 (c)

RQC-100 (c)

1005-1009

Q&T HR sheet

L

1015

HR plate

L

Gainex (c)

H-11

CD

CDSR

Q&T

Q&T

Q&T

Q&T

Q&T

Q&T

As forged

HR plate

Normalized

HR sheet

CD sheet

HR plate

Ausformed

HR sheet

HR sheet

L LT

AM-350 (c)

Gainex (c)

HR, A

STA

L L

A538C (b)

405

265

595

500

450

390

410

225

225

108

80

90

125

125

90

430

290

290

660

496

480

460

1515 220

74

77

90

64

60

50

60

68

52

930 135

2240 325

1825 265

1585 230

1345 195

1450 210

725 105

620

440

415

345

415

470

360

1640 238

930 135

940 136

2585 375

510

530

1905 276

1315 191

2000 290

1860 270

67

33

41

51

55

59

51

65

60

62

68

80

64

66

73

38

67

43

33

64

58

20

52

55

56

1.10

0.51

0.52

0.71

0.81

0.89

0.72

1.04

0.93

0.96

1.14

1.6

1.02

1.09

1.3

0.89

1.02

0.56

0.40

1.02

0.86

0.23

0.74

0.81

0.82

185

195

205

205

205

205

200

200

200

205

205

200

200

205

205

195

205

205

205

200

200

180

195

180

185

27

28.5

30

30

30

30

29

29

29

29.5

30

29

29

30

30

28

30

30

30

29.2

29.2

26

28

26

27

1655 240

93

78

75

84

1000 145

2725 395

2275 330

1795 260

1585 230

1860 270

1225 178

1540 223

895 130

825 120

640

540

515

580

1780 258

1240 180

1240 180

3170 460

805 117

805 117

2690 390

2800 406

2240 325

2135 310

0.25 0.07 0.32

−0.08

0.35 −0.08

0.45

−0.074 −0.07 −0.081

1.00 0.60

−0.095 −0.073

0.41 0.61

0.95 −0.12

0.10

−0.109 −0.11 −0.14

0.30 0.11

0.15 −0.059

0.32

−0.067 −0.09 −0.073

0.66 0.66

−0.07 −0.07

0.86 0.08

0.86 −0.071

0.10

−0.102 −0.07 −0.077

0.60 0.33

−0.07 −0.14

0.30 0.80

−0.065 −0.071

−0.62

−0.58

−0.60

−0.68

−0.69

−0.68

−0.70

−0.66

−0.57

−0.51

−0.64

−0.39

−0.41

−0.51

−0.43

−0.56

−0.69

−0.69

−0.74

−0.68

−0.65

−0.42

−0.84

−0.75

−0.71

3:43 PM

AM-350 (c)

STA STA

L L

A538A (b)

A538B (b)

6/3/03

Grade (a)

Tensile Strength Sut

Source: ASM Metals Reference Book, 2nd ed., American Society for Metals, Metals Park,

Fatigue True Strength Strain Fatigue Fatigue Fatigue Modulus of Coefficient Strength HardReduction at Ductility Ductility σ f
Orienta- Description ness in Area Fracture Elasticity E Exponent Coefficient Exponent 4 tion (e) (f) HB MPa ksi % εf GPa 10 psi MPa ksi b εF
c

Ohio, 1983, p. 217.

Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels

Table A–23

shi20361_app_A.qxd Page 998

L

950X (d)

DAT

Plate channel

HR plate

Plate channel

HR bar

HR plate

Q&T

Q&T

A

SH, Q&T

Q&T

Q&T

Q&T

HR, A

Q&T

Q&T

Q&T and deformed

Q&T and deformed

Q&T

Q&T and deformed

Q&T

225

156

150

150

159

410

280

260

518

430

350

409

243

560

475

450

475

450

400

380

335

310

310

77

64

82

82

695 101

530

440

565

565

565 227

1000 145

925 134

2015 292

1670 242

1240 180

1470 213

825 120

2240 325

1930 280

1930 280

2035 295

1760 255

1550 225

1415 205

1250 181

1060 154

1075 156

1425 207

895 130

890 129

950 138

1035 150

68

72

65

69

64

32

33

14

11

42

57

38

43

27

35

37

20

42

47

48

28

29

60

55

67

60

49

25

1.15

1.24

1.06

1.19

1.03

0.38

0.41

0.16

0.12

0.87

0.84

0.48

0.57

0.31

0.43

0.46

0.22

0.54

0.63

0.66

0.34

0.35

0.69

0.79

1.12

0.93

0.68

0.29

195

205

205

205

205

200

195

205

205

195

195

200

195

205

205

200

200

205

200

205

200

200

200

200

220

205

205

200

28.2

29.5

30

30

29.6

29

28

30

30

28

28

29

28

30

30

29

29

30

29

30

28.9

29

29.2

29

32

29.9

29.9

28.8

91 1055 153

1005 146

625

970 141

1170 170

1855 269

1220 177

1040 151

2585 375

1930 280

1655 240

2000 290

1200 174

2655 385

2170 315

2105 305

2070 300

2000 290

1895 275

1825 265

1250 181

1450 210

1825 265

1695 246

1275 185

1275 185

1275 185

1585 230

0.95 0.85 0.35 0.85 0.21

−0.12 −0.11 −0.075 −0.10 −0.08

0.41 0.38

−0.073 −0.057

0.18 0.16

−0.09 −0.071

0.73 0.40

−0.076 −0.071

0.45

0.07

−0.089

0.48

0.09

−0.081 −0.095

0.60

−0.09

−0.091

0.40 0.20

−0.082

0.50

−0.09 −0.08

0.06 0.45

−0.08 −0.08

1.2 0.22

−0.08 −0.10

0.92 0.89

−0.083 −0.081

0.68 0.93

−0.076 −0.071

0.27

−0.09

−0.53

−0.61

−0.54

−0.59

−0.61

−0.65

−0.60

−0.47

−0.56

−0.57

−0.62

−0.60

−0.54

−0.76

−0.61

−0.76

−0.77

−0.73

−0.75

−0.75

−0.62

−0.51

−0.59

−0.69

−0.63

−0.65

−0.65

−0.53

Notes: (a) AISI/SAE grade, unless otherwise indicated. (b) ASTM designation. (c) Proprietary designation. (d) SAE HSLA grade. (e) Orientation of axis of specimen, relative to rolling direction; L is longitudinal (parallel to rolling direction); LT is long transverse (perpendicular to rolling direction). (f) STA, solution treated and aged; HR, hot rolled; CD, cold drawn; Q&T, quenched and tempered; CDSR, cold drawn strain relieved; DAT, drawn at temperature; A, annealed. From ASM Metals Reference Book, 2nd edition, 1983; ASM International, Materials Park, OH 44073-0002; table 217. Reprinted by permission of ASM International ®, www.asminternational.org.

L L

950X (d)

L

950X (d)

LT

950C (d)

950C (d)

L L

L

52100

9262 L

L

9262

L

4340

5160

9262

L L

4340

4340

L L

4142

L

4142

4142

L

L

4142

L

L

4142

L

4142

4142

Q&T, DAT DAT

365

258

260

290

305

3:43 PM

4142

L L

4140

Q&T

Q&T

Q&T forging

Q&T forging

DAT

6/3/03

4142

L L

4130

4130

L L

1541F

1541F

L

1144

shi20361_app_A.qxd Page 999

999

1000

22 26 31 36.5 42.5 52.5 62.5

20

25

30

35

40

50

60

187.5

164

140

124

109

97

83

Compressive Strength Suc, kpsi

88.5

73

57

48.5

40

32

26

20.4–23.5

18.8–22.8

16–20

14.5–17.2

13–16.4

11.5–14.8

9.6–14

Tension†

7.8–8.5

7.2–8.0

6.4–7.8

5.8–6.9

5.2–6.6

4.6–6.0

3.9–5.6

Torsion

Modulus of Elasticity, Mpsi

24.5

21.5

18.5

16

14

11.5

10

Endurance Limit* Se, kpsi

*Polished or machined specimens. † The modulus of elasticity of cast iron in compression corresponds closely to the upper value in the range given for tension and is a more constant value than that for tension.

Tensile Strength Sut, kpsi

ASTM Number

302

262

235

212

201

174

156

Brinell Hardness HB

1.50

1.35

1.25

1.15

1.10

1.05

1.00

Fatigue StressConcentration Factor Kf

3:43 PM

Shear Modulus of Rupture Ssu, kpsi

6/3/03

Mechanical Properties of Three Non-Steel Metals (a) Typical Properties of Gray Cast Iron [The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.]

Table A–24

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Page 1001

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 12 in; similar properties can be obtained by using proper purchase specifications. The values given for fatigue strength correspond to 50(107) cycles of completely reversed stress. Alluminum alloys do not have an endurance limit. Yield strengths were obtained by the 0.2 percent offset method.] Aluminum Association Number

Strength Temper

Yield, Sy, MPa (kpsi)

Tensile, Su, MPa (kpsi)

Fatigue, Sf, MPa (kpsi)

Elongation in 2 in, %

Brinell Hardness HB

O

70 (10)

179 (26)

90 (13)

22

45

O

76 (11)

186 (27)

90 (13)

22

47

T3

345 (50)

482 (70)

138 (20)

16

120

Wrought: 2017 2024 3003 3004 5052

H12

117 (17)

131 (19)

55 (8)

20

35

H16

165 (24)

179 (26)

65 (9.5)

14

47

H34

186 (27)

234 (34)

103 (15)

12

63

H38

234 (34)

276 (40)

110 (16)

6

77

H32

186 (27)

234 (34)

117 (17)

18

62

H36

234 (34)

269 (39)

124 (18)

10

74

319.0*

T6

165 (24)

248 (36)

69 (10)

2.0

80

†

T5

172 (25)

234 (34)

83 (12)

1.0

100

T6

207 (30)

289 (42)

103 (15)

1.5

105

T6

172 (25)

241 (35)

62 (9)

3.0

80

T7

248 (36)

262 (38)

62 (9)

0.5

85

Cast: 333.0

335.0*

*Sand casting. † Permanent-mold casting.

(c) Mechanical Properties of Some Titanium Alloys Yield, Sy (0.2% offset) MPa (kpsi)

Strength Tensile, Sut MPa (kpsi)

Elongation in 2 in, %

Hardness (Brinell or Rockwell)

Titanium Alloy

Condition

Ti-35A†

Annealed

210 (30)

275 (40)

30

135 HB

Ti-50A†

Annealed

310 (45)

380 (55)

25

215 HB

Ti-0.2 Pd

Annealed

280 (40)

340 (50)

28

200 HB

Ti-5 Al-2.5 Sn

Annealed

760 (110)

790 (115)

16

36 HRC

Ti-8 Al-1 Mo-1 V

Annealed

900 (130)

965 (140)

15

39 HRC

Ti-6 Al-6 V-2 Sn

Annealed

970 (140)

1030 (150)

14

38 HRC

900 (130)

830 (120)

14

36 HRC

1207 (175)

1276 (185)

8

40 HRC

Ti-6Al-4V

Annealed

Ti-13 V-11 Cr-3 Al

Sol. + aging

†

Commercially pure alpha titanium 1001

1002

87.6

CD HR CD CD CD CD HT bolts

1018

1035

1045

1117

1137

12L14

1038

53.4

7075

T6 .025”

75.5

64.9 67.5

T4 T6

2024

28.1

175.4 0

2024

Ti-6AL-4V

198.8 149.1

AM350SS

A

17-7PSS

84.8 105.3

A

85.0

195.9

403SS

310SS

105.0

A A

CD

301SS

304SS

191.2

CD

201SS

64.8 122.2

Nodular Nodular

100-70-04

93.9

604515

Malleable Pearlitic

Malleable

53.3

35018

32510

44.5 Malleable

ASTM40

133.4

79.6

106.5

3.92

2.10

1.50

1.64

1.73

7.91

8.29

9.51

3.09

4.23

4.14

5.68

5.82

7.76

7.65

3.77

3.83

2.68

1.59

4.34

3.38

6.92

6.15

5.25

7.13

68.8

55.9

60.2

24.2

141.8

101.8

163.3

95.7

71.6

66.6

92.3

151.9

180.7

47.6

53.7

80.1

44.7

48.7

27.7

122.3

70.3

96.2

73.0

90.2

72.6

30.8

x0

76.2

68.1

65.5

28.7

178.5

152.4

202.3

106.4

86.3

86.6

106.6

193.6

197.9

125.6

66.1

95.3

54.3

53.8

46.2

134.6

80.4

107.7

84.4

120.5

87.5

90.1

θ

3.53

9.26

3.16

2.43

4.85

6.68

4.21

3.44

3.45

5.11

2.38

8.00

2.06

11.84

3.23

4.04

3.61

3.18

4.38

3.64

1.36

1.72

2.01

4.38

3.86

12

b

63.7

53.4

40.8

163.7

63.0

189.4

78.5

37.9

46.8

166.8

79.3

49.0

60.2

34.9

38.5

78.1

98.1

81.4

95.5

49.6

78.4

µSy

1.98

1.17

1.83

9.03

5.05

11.49

3.91

3.76

4.70

9.37

4.51

4.20

2.78

1.47

1.42

8.27

4.24

4.71

6.59

3.81

5.90

σSy

58.9

51.2

38.4

101.5

38.0

144.0

64.8

30.2

26.3

139.7

64.1

33.8

50.2

30.1

34.7

64.3

92.2

72.4

82.1

39.5

56

x0

64.3

53.6

41.0

167.4

65.0

193.8

79.9

38.9

48.7

170.0

81.0

50.5

61.2

35.5

39.0

78.8

98.7

82.6

97.2

50.8

80.6

2.63

1.91

1.32

8.18

5.73

4.48

3.93

2.17

4.99

3.17

3.77

4.06

4.02

3.67

2.93

1.72

1.41

2.00

2.14

2.88

4.29

b

0.0278

0.0222

0.0253

0.0616

0.0451

0.0556

0.0478

0.0293

0.0499

0.0487

0.0541

0.0304

0.0396

0.0626

0.0582

0.0408

0.0502

0.0298

0.0975

0.0253

0.0869

0.0577

0.0632

0.0606

0.0455

0.0655

CSut

0.0311

0.0219

0.0449

0.0552

0.0802

0.0607

0.0498

0.0992

0.1004

0.0562

0.0569

0.0857

0.0462

0.0421

0.0369

0.1059

0.0432

0.0579

0.0690

0.0768

0.0753

CSy

3:43 PM

83.1

117.7

5.74

σSut

6/3/03

86.2

µSut

Material

(March 1992), pp. 29–34.

θ

Source: Data compiled from “Some Property Data and

Corresponding Weibull Parameters for Stochastic Mechanical Design,” Trans. ASME Journal of Mechanical Design, vol. 114

Stochastic Yield and Ultimate Strengths for Selected Materials

Table A–25

shi20361_app_A.qxd Page 1002

5.2 26.3 (3.82) 143 (20.7) 39.6 (5.75)

b σ µ N σ

Aluminum

Ti-6A1-4V

992 (144)

365 (53)

599 (87)

N

W

µ

684 (99.3)

38.1 (5.53)

116 (16.9)

21.4 (3.11)

5.0

528 (76.7)

657 (95.4)

36.6 (5.31)

95 (13.8)

17.4 (2.53)

5.5

463 (67.2)

393 (57)

4.1

496 (72.0)

420 (61)

2.85

425 (61.7)

391 (56.7)

106

493 (71.6)

35.1 (5.10)

77 (11.2)

14.0 (2.03)

107

9

Statistical parameters from a large number of fatigue tests are listed. Weibull distribution is denoted W and the parameters are x0, “guaranteed” fatigue strength; θ, characteristic fatigue strength; and b, shape factor. Normal distribution is denoted N and the parameters are µ, mean fatigue strength; and σ, standard deviation of the fatigue strength. The life is in stress-cycles-to-failure. TS = tensile strength, YS = yield strength. All testing by rotating-beam specimen.

1040 (151)

489 (71)

744 (108)

712 (108)

604 (87.7)

θ

HT-46

455 (66)

510 (74)

x0

T-4

3.4

4.3

2024

588 (85.4)

699 (101.5)

θ b OQ&T, 1300°F

510 (74)

3140

2.75

W x0

579 (84)

661 (96)

2.60

799 (116)

b OQ&T 1200°F

2340

503 (73.0)

462 (67)

105

7 8 Stress Cycles to Failure

594 (86.2)

104

6

544 (79)

W

Distribution

5

θ

565 (82)

YS MPa (kpsi)

TS MPa (kpsi) 723 (105)

4

3

x0

WQ&T, 1210°F

Condition

Number

1046

2

1

Source: E. B. Haugen, Probabilistic Mechanical Design, Wiley, New York, 1980,

3:43 PM

Appendix 10–B.

Stochastic Parameters for Finite Life Fatigue Tests in Selected Metals

6/3/03

Table A–26

shi20361_app_A.qxd Page 1003

1003

1004 Source: Compiled from Table 4 in H. J. Grover, S. A. Gordon,

860

OQT

1200

OQT

As Rec.

1200

369

224

227

162

67 Rb

193 277

WQT

196

N

1200

164

*BHN = Brinell hardness number; RA = fractional reduction in area.

10120

1095

1060

.56 MN

N, AC

1050 WQT

HR, N

1045

209 195

WQT Forged

1040

132

Normal

1035

135

Air-cooled

1030

BHN*

Furnace cooled

Condition

180

117

115

84

134

111

98

97

92

107

92

103

72

80

58

130

59

65

33

65

84

47

70

47

63

53

87

35

45

30

0.15

0.12

0.40

0.37

0.20

0.57

0.42

0.58

0.40

0.49

0.23

0.65

0.54

0.62

0.63

RA*

77

50

65

94

61

50

80

10

4

102

60

68

43

60

81

55

60

48

70

80

51

4(10 )

4

95

56

64

40

55

73

51

57

46

56

72

44

47

37

105

91

51

57

34

50

62

47

52

40

47

40

65

40

42

34

4(105)

91

50

56

31

48

57

43

50

38

47

47

60

37

38

30

106

91

50

56

30

48

55

41

50

34

47

33

57

34

38

28

4(106)

Stress Cycles to Failure

91

50

56

30

48

55

41

50

34

47

33

57

33

38

25

107

56

30

55

41

50

57

33

108

3:43 PM

1020

Material

Yield Strength kpsi

6/3/03

Tensile Strength kpsi

and L. R. Jackson, Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960.

Finite Life Fatigue Strengths of Selected Plain Carbon Steels

Table A–27

shi20361_app_A.qxd Page 1004

0.324 9 0.289 3 0.257 6 0.229 4 0.204 3 0.181 9 0.162 0 0.144 3 0.128 5 0.114 4 0.101 9 0.090 74 0.080 81 0.071 96 0.064 08 0.057 07 0.050 82 0.045 26

0 1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17

0.065 0.058

0.120 0.109 0.095 0.083 0.072

0.203 0.180 0.165 0.148 0.134

0.340 0.300 0.284 0.259 0.238 0.220

0.454 0.425 0.380

Principal Use:

0.062 5 0.056 25

0.125 0.109 357 0.093 75 0.078 125 0.070 312 5

0.203 125 0.187 5 0.171 875 0.156 25 0.140 625

0.312 5 0.281 25 0.265 625 0.25 0.234 375 0.218 75

0.500 0.468 75 0.437 5 0.406 25 0.375 0.343 75

Nonferrous Sheet, Wire, and Rod

0.059 8 0.053 8

0.119 6 0.104 6 0.089 7 0.074 7 0.067 3

0.194 3 0.179 3 0.164 4 0.149 5 0.134 5

0.239 1 0.224 2 0.209 2

Ferrous Sheet

0.062 5 0.054 0

0.120 5 0.105 5 0.091 5 0.080 0 0.072 0

0.192 0 0.177 0 0.162 0 0.148 3 0.135 0

0.306 5 0.283 0 0.262 5 0.243 7 0.225 3 0.207 0

0.490 0.461 5 0.430 5 0.393 8 0.362 5 0.331 0

Ferrous Wire Except Music Wire

Steel Wire or Washburn & Moen

0.037 0.039

0.026 0.029 0.031 0.033 0.035

0.016 0.018 0.020 0.022 0.024

0.009 0.010 0.011 0.012 0.013 0.014

0.004 0.005 0.006 0.007 0.008

Music Wire

Music Wire

0.175 0.172

0.188 0.185 0.182 0.180 0.178

0.201 0.199 0.197 0.194 0.191

0.227 0.219 0.212 0.207 0.204

Steel Drill Rod

Stubs Steel Wire

(continued)

0.177 0 0.173 0

0.191 0 0.189 0 0.185 0 0.182 0 0.180 0

0.204 0 0.201 0 0.199 0 0.196 0 0.193 5

0.228 0 0.221 0 0.213 0 0.209 0 0.205 5

Twist Drills and Drill Steel

Twist Drill

3:43 PM

0.580 0 0.516 5 0.460 0 0.409 6 0.364 8

Ferrous Sheet and Plate, 480 lbf/ft3

Tubing, Ferrous Strip, Flat Wire, and Spring Steel

Manufacturers Standard

6/3/03

7/0 6/0 5/0 4/0 3/0 2/0

United States Standard †

Birmingham or Stubs Iron Wire

American or Brown & Sharpe

Name of Gauge:

Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches)

Table A–28

shi20361_app_A.qxd Page 1005

1005

1006

Nonferrous Sheet, Wire, and Rod 0.040 30 0.035 89 0.031 96 0.028 46 0.025 35 0.022 57 0.020 10 0.017 90 0.015 94 0.014 20 0.012 64 0.011 26 0.010 03 0.008 928 0.007 950 0.007 080 0.006 305 0.005 615 0.005 000 0.004 453 0.003 965 0.003 531 0.003 145

Principal Use:

18 19 20

21 22 23 24 25

26 27 28 29 30

31 32 33 34 35

36 37 38 39 40

0.007 031 25 0.006 640 625 0.006 25

0.010 937 5 0.010 156 25 0.009 375 0.008 593 75 0.007 812 5

0.018 75 0.017 187 5 0.015 625 0.014 062 5 0.012 5

0.034 375 0.031 25 0.028 125 0.025 0.021 875

*Specify sheet, wire, and plate by stating the gauge number, the gauge name, and the decimal equivalent in parentheses. † Reflects present average and weights of sheet steel.

0.004

0.010 0.009 0.008 0.007 0.005

0.018 0.016 0.014 0.013 0.012

0.032 0.028 0.025 0.022 0.020

0.006 7 0.006 4 0.006 0

0.010 5 0.009 7 0.009 0 0.008 2 0.007 5

0.017 9 0.016 4 0.014 9 0.013 5 0.012 0

0.032 9 0.029 9 0.026 9 0.023 9 0.020 9

0.047 8 0.041 8 0.035 9

Ferrous Sheet

0.009 0 0.008 5 0.008 0 0.007 5 0.007 0

0.013 2 0.012 8 0.011 8 0.010 4 0.009 5

0.018 1 0.017 3 0.016 2 0.015 0 0.014 0

0.031 7 0.028 6 0.025 8 0.023 0 0.020 4

0.047 5 0.041 0 0.034 8

Ferrous Wire Except Music Wire

Steel Wire or Washburn & Moen

0.085 0.090 0.095

0.063 0.067 0.071 0.075 0.080

0.047 0.049 0.051 0.055 0.059

0.041 0.043 0.045

Music Wire

Music Wire

0.106 0.103 0.101 0.099 0.097

0.120 0.115 0.112 0.110 0.108

0.146 0.143 0.139 0.134 0.127

0.157 0.155 0.153 0.151 0.148

0.168 0.164 0.161

Steel Drill Rod

Stubs Steel Wire

0.106 5 0.104 0 0.101 5 0.099 5 0.098 0

0.120 0 0.116 0 0.113 0 0.111 0 0.110 0

0.147 0 0.144 0 0.140 5 0.136 0 0.128 5

0.159 0 0.157 0 0.154 0 0.152 0 0.149 5

0.169 5 0.166 0 0.161 0

Twist Drills and Drill Steel

Twist Drill

3:43 PM

0.05 0.043 75 0.037 5

Ferrous Sheet and Plate, 480 lbf/ft3

Tubing, Ferrous Strip, Flat Wire, and Spring Steel

Manufacturers Standard

6/3/03

0.049 0.042 0.035

United States Standard †

Birmingham or Stubs Iron Wire

American or Brown & Sharpe

Name of Gauge:

Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches) (Continued)

Table A–28

shi20361_app_A.qxd Page 1006

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6/3/03

3:43 PM

Page 1007

Useful Tables

1007

Table A–29 Dimensions of Square and Hexagonal Bolts H

W R

Head Type Square

Regular Hexagonal

Heavy Hexagonal

Structural Hexagonal

W

H

Rmin

W

H

Rmin

0.01

7 8

5 16

0.009

25 64

0.021

Nominal Size, in

W

H

W

H

Rmin

1 4

3 8

11 64

7 16

11 64

0.01

5 16

1 2

13 64

1 2

7 32

0.01

3 8

9 16

1 4

9 16

1 4

0.01

7 16

5 8

19 64

5 8

19 64

0.01

1 2

3 4

21 64

3 4

11 32

0.01

7 8

11 32

5 8

15 16

27 64

15 16

27 64

0.02

1 1 16

27 64

0.02

1 1 16

1 18

1 2

1 18

1 2

0.02

1 14

1 2

0.02

1 14

15 32

0.021

1

1 12

21 32

1 12

43 64

0.03

1 58

43 64

0.03

1 58

39 64

0.062

1 18

1 11 16

3 4

0.03

1 13 16

3 4

0.03

1 13 16

11 16

0.062

1 14

1 78

27 32

1 78

27 32

0.03

2

27 32

0.03

2

25 32

0.062

1 38

1 2 16

29 32

1 2 16

29 32

0.03

3 2 16

29 32

0.03

3 2 16

27 32

0.062

1 12

2 14

1

2 14

1

0.03

2 38

1

0.03

2 38

15 16

0.062

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

3 4

3 4

1 11 16

Nominal Size, mm M5

shi20361_app_A.qxd

1008

6/3/03

3:43 PM

Page 1008

Mechanical Engineering Design

Table A–30 Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws (W = Width across Flats; H = Height of Head; See Figure in Table A–29)

Nominal Size, in

Minimum Fillet Radius

Type of Screw Cap W

Heavy W

Height H

1 4 5 16

0.015

7 16

5 32

0.015

1 2

3 8

0.015

9 16

7 16

0.015

5 8

13 64 15 64 9 32

1 2

0.015

5 8

0.020

3 4 15 16

3 4 7 8

0.020

1 18

1 14

0.040

5 1 16

7 1 16

1

0.060

1 12

1 18

1 14

0.060

1 78

2

35 64 39 64 25 32

1 38

0.060

1 2 16

3 2 16

27 32

1 12

0.060

2 14

2 38

15 16

7 8

5 16

1 1 16

25 64 15 32

Nominal Size, mm M5

0.2

8

3.65

M6

0.3

10

4.15

M8

0.4

13

5.50

M10

0.4

16

6.63

M12

0.6

18

21

7.76

M14

0.6

21

24

9.09

M16

0.6

24

27

10.32

M20

0.8

30

34

12.88

M24

0.8

36

41

15.44

M30

1.0

46

50

19.48

M36

1.0

55

60

23.38

shi20361_app_A.qxd

6/3/03

3:43 PM

Page 1009

Useful Tables

Table A–31 Dimensions of Hexagonal Nuts

Height H Nominal Size, in

Width W

Regular Hexagonal

Thick or Slotted

JAM

1 4 5 16

7 16

7 32

9 32

5 32

1 2 9 16

21 64 13 32

3 16

3 8 7 16

11 16

17 64 21 64 3 8

1 2

1 4 5 16

9 16

3 4 7 8

29 64 9 16

5 16

5 8

15 16

39 64 23 32

3 4 7 8

1 18

13 16

7 16 31 64 35 64 41 64 3 4 55 64 31 32

5 1 32

7 32

3 8

1

1 12

1 18

1 11 16

1 14

1 78

1 1 16

1 14

27 64 31 64 35 64 39 64 23 32

1 38

1 2 16

1 11 64

1 38

25 32

1 12

2 14

9 1 32

1 12

27 32

5 1 16

29 32

1

Nominal Size, mm M5

8

4.7

5.1

2.7

M6

10

5.2

5.7

3.2

M8

13

6.8

7.5

4.0

M10

16

8.4

9.3

5.0

M12

18

10.8

12.0

6.0

M14

21

12.8

14.1

7.0

M16

24

14.8

16.4

8.0

M20

30

18.0

20.3

10.0

M24

36

21.5

23.9

12.0

M30

46

25.6

28.6

15.0

M36

55

31.0

34.7

18.0

1009

shi20361_app_A.qxd

6/3/03

Table A–32 Basic Dimensions of American Standard Plain Washers (All Dimensions in Inches)

3:43 PM

Page 1010

Fastener 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

0.049

3 16

0.188

0.250

0.562

0.049

#12

0.216

0.250

0.562

0.065

1 N 4 1 W 4 5 N 16 5 W 16 3 N 8 3 W 8 7 N 16 7 W 16 1 N 2 1 W 2 9 N 16 9 W 16 5 N 8 5 W 8 3 N 4 3 W 4 7 N 8 7 W 8

0.250

0.281

0.625

0.065

1N

0.250

0.312

0.734

0.065

0.312

0.344

0.688

0.065

0.312

0.375

0.875

0.083

0.375

0.406

0.812

0.065

0.375

0.438

1.000

0.083

0.438

0.469

0.922

0.065

0.438

0.500

1.250

0.083

0.500

0.531

1.062

0.095

0.500

0.562

1.375

0.109

0.562

0.594

1.156

0.095

0.562

0.625

1.469

0.109

0.625

0.656

1.312

0.095

0.625

0.688

1.750

0.134

0.750

0.812

1.469

0.134

0.750

0.812

2.000

0.148

0.875

0.938

1.750

0.134

0.875

0.938

2.250

0.165

1.000

1.062

2.000

0.134

1W

1.000

1.062

2.500

0.165

1 18 N

1.125

1.250

2.250

0.134

1 18 W

1.125

1.250

2.750

0.165

1 14 N

1.250

1.375

2.500

0.165

1 14 1 38 1 38 1 12 1 12 1 58 1 34 1 78

W

1.250

1.375

3.000

0.165

N

1.375

1.500

2.750

0.165

W

1.375

1.500

3.250

0.180

N

1.500

1.625

3.000

0.165

W

1.500

1.625

3.500

0.180

1.625

1.750

3.750

0.180

1.750

1.875

4.000

0.180

1.875

2.000

4.250

0.180

2

2.000

2.125

4.500

0.180

2 14

2.250

2.375

4.750

0.220

2 12

2.500

2.625

5.000

0.238

2 34

2.750

2.875

5.250

0.259

3

3.000

3.125

5.500

0.284

N = narrow; W = wide; use W when not specified. 1010

Diameter

Washer Size

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6/3/03

3:43 PM

Page 1011

Useful Tables

1011

Table A–33 Dimensions of Metric Plain Washers (All Dimensions in Millimeters) Washer Size*

Minimum ID

Maximum OD

Maximum Thickness

Washer Size*

Minimum ID

Maximum OD

Maximum Thickness

1.6 N

1.95

4.00

0.70

10 N

10.85

20.00

2.30

1.6 R

1.95

5.00

0.70

10 R

10.85

28.00

2.80

1.6 W

1.95

6.00

0.90

10 W

10.85

39.00

3.50

2N

2.50

5.00

0.90

12 N

13.30

25.40

2.80

2R

2.50

6.00

0.90

12 R

13.30

34.00

3.50

2W

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

2.80

2.5 R

3.00

8.00

0.90

14 R

15.25

39.00

3.50

2.5 W

3.00

10.00

1.20

14 W

15.25

50.00

4.00

3N

3.50

7.00

0.90

16 N

17.25

32.00

3.50

3R

3.50

10.00

1.20

16 R

17.25

44.00

4.00

3W

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

4.00

3.5 R

4.00

10.00

1.40

20 R

21.80

50.00

4.60

3.5 W

4.00

15.00

1.75

20 W

21.80

66.00

5.10

4N

4.70

10.00

1.20

24 N

25.60

44.00

4.60

4R

4.70

12.00

1.40

24 R

25.60

56.00

5.10

4W

4.70

16.00

2.30

24 W

25.60

72.00

5.60

5N

5.50

11.00

1.40

30 N

32.40

56.00

5.10

5R

5.50

15.00

1.75

30 R

32.40

72.00

5.60

5W

5.50

20.00

2.30

30 W

32.40

90.00

6.40

6N

6.65

13.00

1.75

36 N

38.30

66.00

5.60

6R

6.65

18.80

1.75

36 R

38.30

90.00

6.40

6W

6.65

25.40

2.30

36 W

38.30

110.00

8.50

8N

8.90

18.80

2.30

8R

8.90

25.40

2.30

8W

8.90

32.00

2.80

N = narrow; R = regular; W = wide. *Same as screw or bolt size.

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Page 1012

Mechanical Engineering Design

Table A–34




Values of (n) =

e−x x n−1 dx; (n + 1) = n(n)

0

Gamma Function* Source: Reprinted with permission from William H. Beyer (ed.), Handbook of Tables for Probability and Statistics, 2nd ed., 1966. Copyright CRC Press, Boca Raton, Florida.

∞

n

(n)

n

(n)

n

(n)

n

(n)

1.00

1.000 00

1.25

.906 40

1.50

.886 23

1.75

.919 06

1.01

.994 33

1.26

.904 40

1.51

.886 59

1.76

.921 37

1.02

.988 84

1.27

.902 50

1.52

.887 04

1.77

.923 76

1.03

.983 55

1.28

.900 72

1.53

.887 57

1.78

.926 23

1.04

.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 38

1.06

.968 74

1.31

.896 00

1.56

.889 64

1.81

.934 08

1.07

.964 15

1.32

.894 64

1.57

.890 49

1.82

.936 85

1.08

.959 73

1.33

.893 38

1.58

.891 42

1.83

.939 69

1.09

.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 61

1.11

.947 39

1.36

.890 18

1.61

.894 68

1.86

.948 69

1.12

.943 59

1.37

.889 31

1.62

.895 92

1.87

.951 84

1.13

.939 93

1.38

.888 54

1.63

.897 24

1.88

.955 07

1.14

.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 77

1.16

.929 80

1.41

.886 76

1.66

.901 67

1.91

.965 23

1.17

.936 70

1.42

.886 36

1.67

.903 30

1.92

.968 78

1.18

.923 73

1.43

.886 04

1.68

.905 00

1.93

.972 40

1.19

.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 88

1.21

.915 58

1.46

.885 60

1.71

.910 57

1.96

.983 74

1.22

.913 11

1.47

.885 63

1.72

.912 58

1.97

.987 68

1.23

.910 75

1.48

.885 75

1.73

.914 66

1.98

.991 71

1.24

.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   1 2x 139 571 1 1+ − − + ··· + x x e−x 2 3 4 x 12x 288x 51 840x 2 488 320x