Aircraft Structures for Engineers

Official Lecture Notes for INME 4717, 5717

Aircraft Structures for Engineers Vijay K. Goyal, Ph.D. Associate Professor, Mechanical Engineering Department, University of Puerto Rico at Mayag¨ uez, Mayag¨ uez, Puerto Rico

Vinay K. Goyal, Ph.D. Sr. Member of the Technical Staff, Structural Mechanics Subdivision, The Aerospace Corporation, Los Angeles, California

M.S. / M.Eng. in Mechanical Engineering / Minor in Aerospace Engineering (for B.S.) This material is only for students enrolled at UPRM. All others must request permision from the author. ([email protected])

c 2008, V. K. Goyal and V. K. Goyal Copyright

Dedication

To the Almighty God: Father, Son Yeshua, and the Holy Spirit; Vijay: my wife Maricelis and my family Vinay: my wife Stacey and my family “I can do all things in Yeshua who strengthens me...” – BIBLE: Phillipians 4:13

Preface

Aerospace Structures is one of the most challenging courses to teach. It enclosed many advanced topics while introducing some fundamental thin-walled structural analysis. This book is written for students with a background in mechanical engineering, although the concepts are presented in a fundamental approach allowing students from all backgrounds to benefit from the material in this book. Intended Audience This book is intended to provide a foundation of the finite element and optimization techniques. The target audience are senior level undergraduate and first year graduate students who have had little, or no, exposure to thin-walled structural analysis. Practicing engineers will also benefit from the integration approach to obtain very impressive and useful results. Thus, we can assure that this book will fill up a void in the personal library of many engineers who are trying to, or planning, to design and analyze thin-walled structures. A background in solid mechanics, calculus, and basic programming knowledge is required. Motivation When writing this textbook, we have kept the reader in mind at all times. After years of using this manuscript, engineering graduates (from the University of Puerto Rico at Mayag¨ uez) have found the manuscript very useful in their respective jobs. In teaching and applying these subjects for years, we have come to the conclusion that students and engineers too often take a ”black-box” approach. The book also tries to bind traditional theoretical approaches with some modern numerical techniques. The original is proven to be an effective reference in the aerospace industry, such as Boeing and InfoTech Aerospace Services. The format of this book is student-friendly since each chapter begins with instructional objectives and ends with a chapter summary highlighting the most important aspects of the chapter with an outline of ongoing research within the topics presented in the chapter. The authors assume that the students have little experience with programming languages and numerical methods; thus, this is a reader-friendly book that enables the reader to self-learn the topics. It includes a variety of examples, specifically worked with a pedagogical approach, using a step-by-step procedure which is easy to apply to a wide range of engineering problems. At the end of each chapter one can find a variety of problems that are been carefully worked-out in an accompanying solution manual to the textbook, available online to the instructors. Importance was given to emphasis on application to keep the students interested in the subject. After the reader has completed this book, he/she will be able to: 1. Understand how and why the aircraft are designed the way they are. 2. Learn and apply the fundamentals of the linear elasticity to the analysis of thin-walled structures. 3. Use numerical techniques to approximate analytical solutions.

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Mathematical Level Readers need a strong background in linear algebra, calculus, differential equations, and programming. For those who have not been exposed to linear algebra, we have included an appendix that will enable the reader to self-study, or review, this topic. The authors assume that the readers have little experience with programming languages and numerical methods. Chapter Organization and Topical Coverage The format of this book is student-friendly since each chapter begins with instructional objectives and ends with a chapter summary highlighting the most important aspects of the chapter with an outline of ongoing research within the topics presented in the chapter. It includes a variety of examples, specifically worked with a pedagogical approach, using a step-by-step procedure which is easy to apply to a wide range of engineering problems. At the end of each chapter one can find a variety of problems that have been carefully worked-out in an accompanying solution manual to the textbook, available online to the instructors. Emphasis was placed on applications to keep the reader interested in the subject. The contents of this book are intended for a two semester term course. All examples have been solved using Mathematica and MATLABr (which are available to students and instructors through the book website). In short, this unique book will help the reader, whether a student or a practicing engineer, to independently learn the topics through carefully worked out examples and apply them to real aircraft engineering design problems. Any comments and suggestions can be sent to [email protected]. Vijay K. Goyal July 23, 2008

Course Syllabus

1. Instructor 1a. Dr. Vijay K. Goyal, Associate Professor of the Mechanical Engineering Department 1b. Office: L-207 1c. Office Hours: MW: 2:30p - 3:30p; Tu Th: 2:00p - 3:00p or by appointment 1d. Office Phone: (787) 832-4040 ext. 2111/3659 (Please do not call at home nor at my cell phone) 1e. E-mail: [email protected], [email protected] 2. General Information 2a. Course Number: INME 4717–5717 2b. Course Title: Aircraft Structural Analysis and Design Courses 2c. Credit-Hours: Three of lecture and lab included 2d. Classroom: L-236A 3. Course Description Aircraft Structural Analysis and Design (INME4717): Application of solid mechanics to analyze aerospace structures. Study of aircraft components and their design philosophy. Application of elasticity to describe the stress, strain, and displacement fields of one- and two-dimensional problems in aerospace structures. Exact and approximate solutions of two-dimensional structural problems. Analysis of bending, shear and torsional theories for arbitrary, multimaterial, and multicell wing cross-sections. Analysis of thin-walled single and multicell stiffened shell beams using analytical and numerical solutions. Advanced Aircraft Structural Design (INME5717): Application of work and energy principles, and numerical methods, to the design of flight vehicles. Study of deflection and load analysis using the Principle of Virtual Work, Principle of Complementary Virtual Work, analytical weak form solutions, and the finite element formulation. Wing design considering: fatigue, aeroelasticity, divergence, environmental loads, aerospace materials, dynamic stability of thin-walled compression members, and structural dynamics. 4. Pre/Co-requisites Aircraft Structural Analysis and Design (INME4717): 4a. Pre-requisites: Design of Machine Elements I (INME 4011) or DIR AUTHORIZATION Advanced Aircraft Structural Design (INME5717): 4a. Pre-requisites: Aircraft Structural Analysis and Design (INME 4717) or DIR AUTHORIZATION

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5. Textbook, Supplies and Other Resources 5a. Class notes are posted on the class website. The official course textbook is the course website: http://www.me.uprm.edu/vgoyal/inme4717.html http://www.me.uprm.edu/vgoyal/inme5717.html 5b. Other useful references: (a) Allen, D. H., Introduction to Aerospace Structural Analysis, 1985, John Wiley and Sons, New York, NY*. (b) Curtis, Howard D., Fundamentals Of Aircraft Structural Analysis, First Edition, 1997, Mc-Graw Hill, New York, NY* (c) Johnson, E. R., Thin-Walled Structures, 2006, Textbook at Virginia Polytechnic Institute and State University, Blacksburg, VA. (d) Keane, Andy and Nair, Prasanth, Computational Approaches for Aerospace Design: The Pursuit of Excellence, August 2005, John Wiley and Sons. (e) Newman, D., Interactive Aerospace Engineering And Design With CD-ROM, First Edition, Mass Institute Of Tech, 2004, Mcgraw-Hill. (f) Sun, C. T., Mechanics of Aircraft Structures, Second Edition 2006, John Wiley and Sons (g) Thomas, G. B., Finney R. L., Weir, M. D., and Giordano F. R., Thomas Calculus, Early Transcendentals Update, 2003, Tenth Edition, Addison-Wesley, Massachusetts. Entire book 6. Purpose Aircraft Structural Analysis and Design (INME4717): After completing this course students should be able to: (i) identify and understand the function of typical aircraft components, and discuss the behavior of monocoque and semi-monocoque structures; (ii) formulate multi-directional internal loads; (iii) formulate and analyze the state of point and state of stress; (iv) identify and evaluate various stress-strain formulations; (v) apply Hooke’s law including thermal effects; (vi) apply Euler-Bernoulli beam theory and Timoshenko beam theory; (vii) apply Airy Stress Function; (viii) apply the classical torsional theory for prismatic beams; (ix) analyze bending, shear, and torsion of arbitrary, multicell and multimaterial cross-sectional wings. Advanced Aircraft Structural Design (INME5717): After completing this course students should be able to: (i) analyze load and deflections of statically determinate and indeterminate structures using the Principle of Virtual Work and Principle of Complementary Virtual Work; (ii) analyze wings using the weak form and the finite element method; (iii) analyze and design wings based on fatigue, aeroelasticity, divergence, structural dynamics, and dynamic stability; (iv) learn and integrate environmental loads into aerospace design. 7. Course Goals The course will be divided into four modules. Each module has the purpose to help the student understand and grasp the basic concept in computer aided design for mechanical engineering problems. See website. 7a. ( %) Chapter 1. Learning About Aircraft Structures 7b. ( %) 7c. ( %)

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7d. ( %) In addition to the above topics, all students will demonstrate the ability analyze a recent journal paper provided by the instructor: the context of the report (introduction), describe clearly and precisely the procedures used (methodology), report verbally and visually the findings (results), interpret the findings (analysis of results), justify the solutions persuasively (conclusions), and provide final comments. The students will demonstrate the ability to make effective oral presentations and written reports using appropriate computer tools. 8. Requirements 8a. Requirements: In order to succeed in the course students are expected to: • should attend all class sessions and be punctual • on a daily basis check the class website • use a non-programmable calculator • do all homework

• practice all suggested problems • take all exams

• submit all work in English

• be ready to ask any questions at the beginning of every class session • and obtain a minimum of 69.50% in the course

8b. Grading Distribution: Total course points are 100% and are distributed as follows: Test I

25%

Test II

25%

Test III

25%

Test IV

25%

Students should take advantage of bonus homework and projects to improve their grade because there will be no “grade curving” at the end of the semester. Your grade will be determined by the following fixed grade scale: A B C D F

89.500 − 100+

78.500 − 89.499

69.500 − 78.499 59.500 − 69.499

0 − 59.499

Your final grade will be scaled based on the attendance. For an example, if you miss 3 classes and your final grade is 100% then your official final grade will be 100*(42/45)=93%. 8c. Students failing to provide a successful, high-standard, computer projects may not pass the course, as they are entitled to a grade of IF or D, regardless of their progress in the mid-term examinations, homework, small projects, among other evaluation criteria. By successful we mean obtaining a percentage higher than 80% in overall projects. Moreover, a successful projects do not entitle the student to pass the course either (see 8b).

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8d. Homework and Tests: Only your own handwritten solutions, written legibly on one side of an 8.500 × 1100 sheet of paper will be accepted for grading. In the case of computer assignment, a computer print out is acceptable whenever a copy of the code is included and well documented by hand. Students are encouraged to work together on the homework, but submissions must be the student’s own work. NO LATE HOMEWORK WILL BE ACCEPTED. 9. Laboratory/Field Work (If applicable) 9a. Cell phones/pagers: All students MUST turn off their cell phones and pagers at the beginning of each class session. By not doing so it is considered disrespectful and students will be asked to leave the class. Students who need to have their cell phones or pagers on at all time must inform the instructor at the beginning of the academic semester. 9b. Smoking: Smoking is not permitted in any area other than those areas designated for smoking. 9c. Electronic Devices: Radios, tape recorders, and other audio or video equipment are not permitted in the lab or classroom at any time. Students must consult with the professor at the beginning of the academic semester. 9d. Laptop Computers, Notebooks, PC-Tablets: Students can bring their personal computers to classroom. However this must not interfere with other student’s work nor with the class session. Students with their personal computers are responsible for any problems with software versions or differences with the one available in the classroom. 10. Department/Campus Policies 10a. Class attendance: Class attendance is compulsory. The University of Puerto Rico at Mayag¨ uez reserves the right to deal at any time with individual cases of non attendance. Professors are expected to record the absences of their students. Absences affect the final grade, and may even result in total loss of credits. Arranging to make up work missed because of legitimate class absence is the responsibility of the student. (Bulletin of Information Undergraduate Studies) Students with three unexcused absences or more may be subject to a one or two final grade letter drop. 10b. Absence from examinations: Students are required to attend all examinations. If a student is absent from an examination for a justifiable and acceptable reason to the professor, he or she will be given a special examination. Otherwise, he or she will receive a grade of zero of “F” in the examination missed. (Bulletin of Information Undergraduate Studies) In short, any student missing a test without prior notice or unexcused absence will be required to drop the course. There will be no reposition exam. At professor’s judgment, those students with a genuine excuse will be given an oral 15–20 minutes oral comprehensive final exam and it will substitute the missed examination(s). Under no circumstances should the students schedule interviews during previously set dates for examinations. 10c. Final examinations: Final written examinations must be given in all courses unless, in the judgment of the Dean, the nature of the subject makes it impracticable. Final examinations scheduled by arrangements must be given during the examination period prescribed in

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the Academic Calendar, including Saturdays. (see Bulletin of Information Undergraduate Studies). Because of the nature of this course, the comprehensive final exam is substituted with the course project. During their oral presentation, students should be ready to answer any question(s) from the course. 10d. Partial withdrawals: A student may withdraw from individual courses at any time during the term, but before the deadline established in the University Academic Calendar. (see Bulletin of Information Undergraduate Studies). 10e. Complete withdrawals: A student may completely withdraw from the University of Puerto Rico at Mayag¨ uez at any time up to the last day of classes. (see Bulletin of Information Undergraduate Studies). 10f. Disabilities: All the reasonable accommodations according to the Americans with Disability Act (ADA) Law will be coordinated with the Dean of Students and in accordance with the particular needs of the student. Those students with special needs must identify themselves at the beginning of the academic semester (with the professor) so that he/she can make the necessary arrangements according to the Office of Affairs for the Handicap. (Certification #44) 10g. Ethics: Any academic fraud is subject to the disciplinary sanctions described in article 14 and 16 of the revised General Student Bylaws of the University of Puerto Rico contained in Certification 018-1997-98 of the Board of Trustees. The professor will follow the norms established in articles 1-5 of the Bylaws. The honor code will be strictly enforced in this course. Students are encouraged to review the honor system policy which has been placed on the class website. All assignments submitted shall be considered graded work unless otherwise noted. Thus all aspects of the course work are covered by the honor system. Any suspected violations of the honor code will be promptly reported to the honor system. Honesty in your academic work will develop into professional integrity. The faculty and students of UPRM will not tolerate any form of academic dishonesty. MUST BE TAKEN SERIOUSLY. Any violation may result in an automatic “F” in the course and such behavior will be reported to the Dean’s office of the College of Engineering.

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11. General Topics 11a. Exam and Presentation Dates: (These dates may be subject to change) Test 1: Module Review session: Class Time Exam Date: Posted on class website Test 2: Module Review session: Class Time Exam Date: Posted on class website Test 3: Module Review session: Class Time Exam Date: Posted on class website Final Project: Presentations (Attendance compulsory) Date and Location to be posted on the website 11b. Course Outline:

Syllabus is subject to changes

Acknowledgments

There are many people who have made this work possible. First and foremost, I am mostly thankful to Yeshua, for giving me the opportunity to live in this time. All my success I give to him for He has been my strength and inspiration at all times. Secondly, I express my special appreciation to my wife Maricelis, my son Jeremiah, and my daughter Naarah for their support and inspiration behind this effort. I could not have completed this task without their prayers, love, understanding, encouragement, and support. Thirdly, I would like to thank all the graduate students who collaborated to complete this book: Juan Rein´es and Angel Quintero. In addition, many thanks to the invaluable inputs from the undergraduate students who used the manuscript form of this book during the 2004–2008 period at the University of Puerto Rico at Mayag¨ uez. Lastly, I cannot leave behind all the people who have given their suggestions to this work, such as Dr. Paul Sundaram, whom I consider my mentor. Special thanks to all the friends who encouraged and helped me achieve this goal. God bless and thank you all, Vijay K. Goyal

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Table of Contents

List of Figures

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List of Tables

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Chapter 1. Learning about Aircraft Structures 1.1 History of Aviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Pre-Wright Era: Early Aviation . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 The 19th Century . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.3 Era of Strut-and-Wire Biplanes: 1900 to World War I . . . . . . . . . . . . . 1.1.4 Before World War II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.5 Era of Propeller-Driven Airplane: During World War II . . . . . . . . . . . . 1.1.6 Era of Jet-Propelled Airplane: After World War II until end of 20th Century 1.2 What do we study in aircraft structure? . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Design Philosophy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Development of Aircraft Structures . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Loads Acting on an Aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Rotations Acting on an Airplane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Components of a typical Aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.1 Wings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Fuselage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.3 Horizontal stabilizer and Elevators . . . . . . . . . . . . . . . . . . . . . . . . 1.5.4 Stabilator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.5 Vertical Stabilizer and Rudder . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.6 Spoilers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.7 Ailerons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.8 Flaps and Slats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.9 Gas Turbine Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.10 Landing gear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Basic Structural Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.1 Wing Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 Fuselage Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.3 Semimonocoque Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Suggested Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 2. Principle of Aerodynamics 2.1 Aerodynamics . . . . . . . . . . . . . . 2.1.1 Continuity . . . . . . . . . . . 2.1.2 Newton’s Laws Of Motion . . . 2.1.3 Conservation laws . . . . . . .

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TABLE OF CONTENTS

2.2 2.3 2.4 2.5 2.6 2.7 2.8

Mach Number . . . . . . . . . . Dynamic pressure . . . . . . . . Aircraft Weight . . . . . . . . . Center of Gravity . . . . . . . . Center of Pressure . . . . . . . Aerodynamic Center . . . . . . Lift . . . . . . . . . . . . . . . . 2.8.1 How is lift generated . . 2.8.2 No Fluid, No Lift . . . . 2.8.3 No Motion, No Lift . . . 2.8.4 Factors That Affect Lift 2.8.5 Lift Equation . . . . . . 2.9 Drag . . . . . . . . . . . . . . . 2.10 Normal Load Factor . . . . . . 2.10.1 Equations of Motion . . 2.10.2 Steady, Level Flight . . 2.10.3 Level Turn: Pull-Up . . 2.10.4 Level Turn: Pull-Down 2.10.5 Banked Turns . . . . . 2.11 References . . . . . . . . . . . . 2.12 Suggested Problems . . . . . .

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Chapter 3. Load Analysis 3.1 Newton’s Laws . . . . . . . . . . . . . 3.2 Units . . . . . . . . . . . . . . . . . . . 3.2.1 Importance of Units . . . . . . 3.2.2 Systems of Units . . . . . . . . 3.3 Load Analysis . . . . . . . . . . . . . . 3.3.1 Internal Force Sign Convention 3.4 Load Diagrams . . . . . . . . . . . . 3.4.1 Sign Conventions . . . . . . . . 3.4.2 Linear Differential Equations of 3.5 Discrete Load Diagrams . . . . . . . . 3.6 References . . . . . . . . . . . . . . . . 3.7 Suggested Problems . . . . . . . . . .

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Chapter 4. Thin-Wall Cross-Sectional Properties 4.1 Geometric Properties of Plane Areas . . . . . . . 4.1.1 Area . . . . . . . . . . . . . . . . . . . . . 4.1.2 First Moments of Area . . . . . . . . . . 4.1.3 Centroid of an Area . . . . . . . . . . . . 4.1.4 Second Moments of Area . . . . . . . . . 4.1.5 Polar Moment of Inertia . . . . . . . . . . 4.1.6 Radius of Gyration . . . . . . . . . . . . . 4.2 Modulus-Weighted Properties of Plane Areas . . 4.2.1 Area . . . . . . . . . . . . . . . . . . . . . 4.2.2 First Moments of Area . . . . . . . . . .

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TABLE OF CONTENTS

4.3

4.4 4.5

4.2.3 Centroid of an Area . . . . . . . 4.2.4 Second Moments of Area . . . . 4.2.5 Polar Moment of Inertia . . . . . 4.2.6 Radius of Gyration . . . . . . . . Properties of Plane Areas of Thin-Walls 4.3.1 Area . . . . . . . . . . . . . . . . 4.3.2 First Moments of Area . . . . . 4.3.3 Centroid of an Area . . . . . . . 4.3.4 Second Moments of Area . . . . 4.3.5 Polar Moment of Inertia . . . . . 4.3.6 Radius of Gyration . . . . . . . . References . . . . . . . . . . . . . . . . . Suggested Problems . . . . . . . . . . .

xiv

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Chapter 5. Applied Linear Elasticity 5.1 Theory of Stresses . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 State of Stress at a Point . . . . . . . . . . . . . . . . . 5.1.2 Stress Convention and Signs . . . . . . . . . . . . . . . . 5.1.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Surface Equilibrium: Cauchy’s Stress Relation . . . . . 5.1.5 Principal Stresses and Principal Planes . . . . . . . . . 5.2 State of Plane Stress . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Principal stresses for Plane State of Stress . . . . . . . . 5.2.2 Principal stresses: Eigenvalue Approach . . . . . . . . . 5.2.3 Principal stresses: Transformation Equations Approach 5.2.4 Principal stresses: Mohr’s Circle Approach . . . . . . . 5.3 Important Stresses . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Octahedral Stresses . . . . . . . . . . . . . . . . . . . . 5.3.2 Von Mises Stress . . . . . . . . . . . . . . . . . . . . . . 5.4 Theory of Strains . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 State of Strain . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Strain compatibility equations . . . . . . . . . . . . . . 5.4.3 Cauchy’s relationship for Strains . . . . . . . . . . . . . 5.4.4 Principal Strains and Principal Planes . . . . . . . . . . 5.5 State of Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Principal strains for State of Plane Strain . . . . . . . . 5.5.2 Strain Measurements . . . . . . . . . . . . . . . . . . . . 5.6 Alternative Stress and Strain Quantities . . . . . . . . . . . . . 5.6.1 Green-Lagrange strains . . . . . . . . . . . . . . . . . . 5.6.2 Stress Measures . . . . . . . . . . . . . . . . . . . . . . . 5.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Suggested Problems . . . . . . . . . . . . . . . . . . . . . . . .

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126 127 128 128 138 138 139 139 140 141 141 182 183

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193 . 194 . 194 . 198 . 199 . 205 . 213 . 225 . 226 . 226 . 227 . 228 . 252 . 252 . 255 . 257 . 260 . 268 . 270 . 273 . 283 . 283 . 285 . 291 . 291 . 296 . 298 . 299

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Chapter 6. Mechanical Behavior of Aerospace Materials 6.1 Constitutive Equations for Elastic Materials . . . . . . . . . . . . . 6.1.1 Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Internal Strain Energy . . . . . . . . . . . . . . . . . . . . . 6.1.3 Anisotropic Materials . . . . . . . . . . . . . . . . . . . . . 6.1.4 Elastic Constitutive Relationship for Isotropic Materials . . 6.1.5 Elastic Stress-Strain Relationship for Orthotropic Materials 6.1.6 Temperature Strains in Isotropic Materials . . . . . . . . . 6.2 Plane Stress and Plane Strain . . . . . . . . . . . . . . . . . . . . . 6.2.1 Consequence of Plane Stress . . . . . . . . . . . . . . . . . . 6.2.2 Consequence of Plane strain . . . . . . . . . . . . . . . . . . 6.2.3 von Mises Stress in Plane Strain and Plane Stress . . . . . 6.3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Suggested Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 7. Advanced Beam Theories 7.1 Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Basic Considerations . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Principle of Saint-Venant . . . . . . . . . . . . . . . . . . . . 7.1.3 Internal Force Sign Convention . . . . . . . . . . . . . . . . . 7.1.4 Resultant Forces and Moments . . . . . . . . . . . . . . . . . 7.2 Euler-Bernoulli Beam Theory . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Curvatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3 Strains-displacement Equations . . . . . . . . . . . . . . . . . 7.2.4 Stress-Strain Equations . . . . . . . . . . . . . . . . . . . . . 7.2.5 Neutral Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.6 Axial Stresses for Linear Thermoelastic Heterogeneous Beams 7.2.7 Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . 7.2.8 Slope and deflection diagrams . . . . . . . . . . . . . . . . . . 7.3 Timoshenko Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Strains-displacement Equations . . . . . . . . . . . . . . . . . 7.3.3 Stress-Strain Equations . . . . . . . . . . . . . . . . . . . . . 7.3.4 Axial Stresses for Linear Thermoelastic Heterogeneous Beams 7.3.5 Slope and deflection diagrams . . . . . . . . . . . . . . . . . . 7.4 Plane Stress: Thick Beams . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Stress-Strain relationship . . . . . . . . . . . . . . . . . . . . 7.4.3 Compatibility Equations . . . . . . . . . . . . . . . . . . . . . 7.4.4 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . 7.4.5 Plane Stress Elasticity Problem . . . . . . . . . . . . . . . . . 7.4.6 Plane Stress Elasticity Solution via Airy Stress Function . . . 7.5 Classical (St. Venant0 s) Torsion Theory . . . . . . . . . . . . . . . . 7.5.1 Displacement field . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 Strains-displacement Equations . . . . . . . . . . . . . . . . . 7.5.3 Stress-strain Equations . . . . . . . . . . . . . . . . . . . . .

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382 383 383 383 384 385 386 388 389 391 394 395 395 400 401 414 414 415 416 417 422 435 435 436 436 437 437 438 472 473 475 475

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476 477 478 491 492

Chapter 8. Thin-Walled Beam Analysis 8.1 Thin-walled Beam Shear in Open Sections . . . . . . . . . . . . . . 8.1.1 No thermal loads . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 No axial and thermal loads . . . . . . . . . . . . . . . . . . 8.1.3 No axial and thermal loads and symmetric . . . . . . . . . 8.1.4 No axial, distributed, and thermal loads and symmetric . . 8.1.5 Procedure to Calculate the Shear Flow in Open Sections . . 8.1.6 Shear Flow in Multiweb Junctions . . . . . . . . . . . . . . 8.2 Shear Center in Thin-Walled Open Sections . . . . . . . . . . . . . 8.2.1 Definition of Shear Center . . . . . . . . . . . . . . . . . . . 8.2.2 Static Equivalence . . . . . . . . . . . . . . . . . . . . . . . 8.2.3 General Procedure . . . . . . . . . . . . . . . . . . . . . . . 8.3 Torsion in Open Thin-Walled Sections . . . . . . . . . . . . . . . . 8.3.1 Prandtl’s membrane analogy for torsion . . . . . . . . . . . 8.3.2 Torsion of a Narrow Rectangular Cross-Section . . . . . . . 8.3.3 Torsion of an Arbitrary Open Thin-walled Cross-Sections . 8.4 Cross-section Idealization . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Idealization of Semi-Monocoque construction . . . . . . . . 8.4.2 Typical method to Idealize of webs . . . . . . . . . . . . . . 8.4.3 Shear flow and shear center in Open Idealized Sections . . . 8.5 Closed Single-Cell Thin-Walled Sections . . . . . . . . . . . . . . . 8.5.1 Enclosed area . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Bredt’s formula . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Shear flow due to transverse shear . . . . . . . . . . . . . . 8.5.4 Solution procedure to obtain shear center in closed sections 8.6 Analysis of Thin-walled Multi-Cell Closed Sections . . . . . . . . . 8.6.1 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.2 Pure Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.3 Pure Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Analysis of Combined Open and Closed Thin-walled sections . . . 8.7.1 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.2 Pure Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.3 Pure Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Suggested Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

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494 495 502 502 502 502 503 512 521 521 522 523 531 531 534 536 544 544 545 548 561 561 566 584 585 601 601 601 609 610 610 610 611 616 617

Chapter 9. Virtual Work Principles 9.1 Differential Work and Virtual Work 9.1.1 Differential Work . . . . . . . 9.1.2 Virtual Work . . . . . . . . . 9.1.3 Complementary Virtual Work

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7.6 7.7

7.5.4 Equilibrium Equations 7.5.5 Boundary Conditions 7.5.6 Alternative Procedure References . . . . . . . . . . . Suggested Problems . . . . .

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9.2 9.3

9.4

9.5 9.6 9.7

9.8 9.9

xvii

Review of equations of linear elasticity . . . . . . . . . . . . PVW for a System of Particles . . . . . . . . . . . . . . . . 9.3.1 Virtual Displacements . . . . . . . . . . . . . . . . . 9.3.2 PVW of a particle . . . . . . . . . . . . . . . . . . . 9.3.3 PVW for rigid and deformable bodies . . . . . . . . 9.3.4 Procedure . . . . . . . . . . . . . . . . . . . . . . . . PVW for Deformable Continuous Structures . . . . . . . . . 9.4.1 PVW for an Elastic Bar . . . . . . . . . . . . . . . . 9.4.2 PVW for an Elastic Truss Bar . . . . . . . . . . . . 9.4.3 Stiffness Influence Coefficients . . . . . . . . . . . . . 9.4.4 Procedure . . . . . . . . . . . . . . . . . . . . . . . . 9.4.5 Strain Energy: Castigliano’s First Theorem . . . . . 9.4.6 PVW for Beams . . . . . . . . . . . . . . . . . . . . Principle of Complementary Virtual Work . . . . . . . . . PCVW for a System of Particles . . . . . . . . . . . . . . . PCVW for Continuous Deformable Bodies . . . . . . . . . . 9.7.1 PCVW for a Bar . . . . . . . . . . . . . . . . . . . . 9.7.2 PCVW for a Truss . . . . . . . . . . . . . . . . . . . 9.7.3 Procedure . . . . . . . . . . . . . . . . . . . . . . . . 9.7.4 Complementary Strain Energy: Castigliano’s Second 9.7.5 PCVW for a Beam . . . . . . . . . . . . . . . . . . . 9.7.6 PCVW for Frames . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Problems . . . . . . . . . . . . . . . . . . . . . .

Chapter 10. Failure Theories for Static Loading 10.1 Uncertainties in Design . . . . . . . . . . . . . . . . 10.1.1 Design Safety factors . . . . . . . . . . . . . 10.1.2 Margin of Safety . . . . . . . . . . . . . . . . 10.2 Ductile and Brittle Failure Theories . . . . . . . . . 10.3 3-D Stress State Failure Theories: Brittle Materials . 10.3.1 Maximum Normal Stress Criterion . . . . . . 10.3.2 Brittle Coulomb-Mohr Criterion . . . . . . . 10.3.3 Comparison of MNS and BCM Criterions . . 10.4 3-D Stress State Failure Theories: Ductile Materials 10.4.1 Aka Distortion Energy Criterion . . . . . . . 10.4.2 Maximum Shear Stress Criterion . . . . . . . 10.4.3 Comparison of DE and MSS Criterions . . . 10.4.4 Ductile Coulomb-Mohr Criterion . . . . . . . 10.5 Introduction to Fracture Mechanics . . . . . . . . . . 10.5.1 Fracture of Cracked Members . . . . . . . . . 10.5.2 Cracks as stress raisers . . . . . . . . . . . . . 10.5.3 Fracture toughness . . . . . . . . . . . . . . . 10.5.4 Fracture Mechanics: MODE I . . . . . . . . . 10.5.5 Fracture Mechanics: Tables and Plots . . . . 10.5.6 Fracture Mechanics: Mixed Modes . . . . . . 10.5.7 Plastic zone size in cracked metal plates . . .

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626 628 628 630 631 633 648 652 661 667 668 688 696 716 717 722 723 730 733 744 752 772 777 778

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xviii

10.5.8 Plastic zone . . . . . . . . . . . . . . . 10.5.9 Plane stress . . . . . . . . . . . . . . . 10.5.10 Plane strain . . . . . . . . . . . . . . . 10.5.11 Plasticity limitations on LEFM . . . . 10.5.12 Fracture toughness in plane strain and 10.5.13 Superposition of Combined Loading . 10.6 References . . . . . . . . . . . . . . . . . . . . 10.7 Suggested Problems . . . . . . . . . . . . . .

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835 836 836 836 838 846 856 857

Chapter 11. Failure Theories for Dynamic Loading 11.1 Vibration Analysis . . . . . . . . . . . . . . . . . . . 11.1.1 Fundamental Natural Frequency . . . . . . . 11.2 Impact . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Assumptions . . . . . . . . . . . . . . . . . . 11.2.2 Freely falling body . . . . . . . . . . . . . . . 11.2.3 Falling body with a velocity . . . . . . . . . . 11.2.4 Horizontally Moving Weight . . . . . . . . . . 11.2.5 Maximum Dynamic Load and Stress . . . . . 11.3 Fatigue . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 Cyclic Stresses . . . . . . . . . . . . . . . . . 11.3.2 Fluctuating . . . . . . . . . . . . . . . . . . . 11.3.3 Fully Reversed . . . . . . . . . . . . . . . . . 11.3.4 Repeated (Tension) . . . . . . . . . . . . . . 11.3.5 Repeated (Compression) . . . . . . . . . . . . 11.4 Alternate and mean stresses . . . . . . . . . . . . . . 11.4.1 Ductile materials . . . . . . . . . . . . . . . . 11.4.2 Brittle materials . . . . . . . . . . . . . . . . 11.5 S–N Diagrams . . . . . . . . . . . . . . . . . . . . . 11.5.1 Fatigue Regimens . . . . . . . . . . . . . . . . 11.5.2 Endurance Stress . . . . . . . . . . . . . . . . 11.5.3 Modified Endurance Stress . . . . . . . . . . 11.5.4 Stress concentration factor . . . . . . . . . . 11.5.5 Plotting S-N Diagrams . . . . . . . . . . . . 11.5.6 Fatigue Theories of Fatigue Failure . . . . . . 11.6 Cumulative fatigue damage . . . . . . . . . . . . . . 11.7 References . . . . . . . . . . . . . . . . . . . . . . . . 11.8 Suggested Problems . . . . . . . . . . . . . . . . . .

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863 864 864 870 870 870 872 873 873 880 880 881 882 883 883 884 884 885 885 886 886 887 889 891 893 917 933 934

Chapter 12. Structural Stability 12.1 Concept of Stability of Equilibrium . . . . . 12.1.1 Buckling . . . . . . . . . . . . . . . . 12.1.2 Stability of equilibrium . . . . . . . 12.1.3 Various Equilibrium Configurations 12.1.4 Methods of stability analysis . . . . 12.2 Stability of Rigid Bars . . . . . . . . . . . . 12.2.1 Analysis of a Perfect System . . . . 12.2.2 Analysis of an Imperfect System . .

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12.3 Stability of Beam-Columns . . . . . . . . . . . . . . . . . . . . . 12.3.1 Perfect Beam-Columns (Adjacent Equilibrium Method) . 12.3.2 Several Type of Column End Constraint . . . . . . . . . . 12.3.3 Imperfect Beam-Columns (Adjacent Equilibrium Method) 12.3.4 Perfect Beam-Column (Energy Approach) . . . . . . . . . 12.3.5 Inelastic Buckling . . . . . . . . . . . . . . . . . . . . . . 12.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Suggested Problems . . . . . . . . . . . . . . . . . . . . . . . . .

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961 961 975 986 992 1002 1003 1004

Chapter 13. Introduction to Aeroelasticity 13.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Static Aeroelasticity . . . . . . . . . . . . . . . . . . . . 13.2.1 Divergence Analysis of A Rigid Wing . . . . . . 13.2.2 Divergence Analysis of Flexible Straight Wings 13.2.3 Divergence Analysis of Flexible Swept Wings . . 13.2.4 Aileron Reversal speed . . . . . . . . . . . . . . . 13.3 The flight envelops . . . . . . . . . . . . . . . . . . . . . 13.3.1 Basic Maneuver V –n Diagram: No gust loads . . 13.3.2 Wing Design . . . . . . . . . . . . . . . . . . . . 13.3.3 Design Gust Load Factors . . . . . . . . . . . . . 13.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Suggested Problems . . . . . . . . . . . . . . . . . . . .

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1014 . 1015 . 1019 . 1019 . 1023 . 1028 . 1033 . 1034 . 1037 . 1045 . 1048 . 1055 . 1056

Appendix A. Math Review Using MatLab A.1 What is MATLABr . . . . . . . . . . . . . A.1.1 Getting Familiar with MATLABr . A.1.2 Basic commands and syntax . . . . . A.1.3 MATLABr Help command . . . . . A.1.4 M-Files . . . . . . . . . . . . . . . . A.1.5 Programming in MATLABr . . . . A.1.6 Diary on and diary off . . . . . . . . A.1.7 Graphical Display of Functions . . . A.1.8 Final Remarks on MATLABr . . . A.2 Linear Algebra . . . . . . . . . . . . . . . . A.2.1 Matrices . . . . . . . . . . . . . . . . A.2.2 Vectors . . . . . . . . . . . . . . . . A.2.3 Matrix and Vector Operations . . . A.2.4 General Rules for Matrix Operations A.2.5 Norm of a Vector . . . . . . . . . . . A.3 Solution to Linear System of Equations . . A.4 Polynomial Approximation . . . . . . . . . A.4.1 Lagrange Interpolation Functions . . A.4.2 Newton Interpolating Polynomial . . A.4.3 Hermite Interpolation Polynomial . A.5 Roots of polynomials . . . . . . . . . . . . . A.5.1 Linear Equations . . . . . . . . . . . A.5.2 Quadratic Equations . . . . . . . . .

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1060 . 1061 . 1061 . 1062 . 1063 . 1065 . 1067 . 1071 . 1072 . 1077 . 1078 . 1078 . 1086 . 1091 . 1117 . 1118 . 1121 . 1125 . 1125 . 1126 . 1127 . 1128 . 1128 . 1128

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A.5.3 Cubic Equations A.6 Eigenvalue Problem . . A.7 References . . . . . . . . A.8 Suggested Problems . .

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Appendix B. Overview Mohr’s Circle B.1 Mohr’s Circle in Stress Analysis . . . . . . . B.2 Procedure for the Mohr’s Circle . . . . . . B.3 Mohr’s Circle in Three-Dimensional Stresses B.4 Final Remarks . . . . . . . . . . . . . . . . B.5 References . . . . . . . . . . . . . . . . . . . B.6 Suggested Problems . . . . . . . . . . . . .

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1144 . 1144 . 1147 . 1151 . 1158 . 1158 . 1160

Appendix C.

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List of Figures

Figure 1.1 Typical forces acting on an airplane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.2 Airplane rotations and body axes.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.3 Airplane parts (in blue) and their functions (in red).. . . . . . . . . . . . . . . . . . . Figure 1.4 Geometry and nomenclature of a wing.. . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.5 The angle of attack is the angle between the chord of the airfoil and the relative wind.. Figure 1.6 Lift versus the angle of attack.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.7 Body of an airplane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.8 Horizontal stabilizer and elevator of an airplane.. . . . . . . . . . . . . . . . . . . . . Figure 1.9 Stabilator of a fighter aircraft.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.10 Vertical stabilizer and rudder of an airplane.. . . . . . . . . . . . . . . . . . . . . . . Figure 1.11 Spoilers of an airplane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.12 Ailerons of an airplane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.13 Flaps and slats of an airplane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.14 Flaps partially deployed (left), full flaps (middle), full flaps with spoilers deployed (right).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.15 The position of the leading edge slats on an airliner (Airbus A300).. . . . . . . . . . Figure 1.16 Gas turbine engines on various aircraft.. . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.17 The main undercarriage and nose undercarriage of a Qatar Airways A330-300 (A7ACA).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.18 The main undercarriage and nose undercarriage of a Qatar Airways A330-300 (A7ACA).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.19 Wing and fuselage undercarriages on a Boeing 747.. . . . . . . . . . . . . . . . . . . Figure 1.20 Landing gear parts of a Boeing 737-700.. . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.21 Different angles of a Boeing 757 landing gear (12 o’clock, 10 o’clock, 3 o’clock, 4 o’clock, 6 o’clock, 9 o’clock, 11 o’clock).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.22 Spar construction.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.23 Typical spar construction.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.24 (a) Spars only, (b) spars and stringers.. . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.25 Wing cross-sections with integrally stiffened skin.. . . . . . . . . . . . . . . . . . . . Figure 1.26 Fuselage structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.27 Monocoque and semi-monocoque structure.. . . . . . . . . . . . . . . . . . . . . . . . Figure 1.28 Typical semimonocoque aircraft structures.. . . . . . . . . . . . . . . . . . . . . . . . Figure 1.29 Idealization of semimonocoque structure: (a) actual structure, (b) idealized structure.. Figure 1.30 Idealization of monocoque shell: (a) actual structure, (b) idealized structure.. . . . . Figure 2.1 Newton’s third law applied to aerodynamics.. . . . . . . . . . . . . . . . . . . . . . . Figure 2.2 Aircraft weight.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 2.3 Center of pressure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 2.4 Forces acting on an aircraft. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 2.5 Forces acting on an aircraft during a vertical pull-up. . . . . . . . . . . . . . . . . . . Figure 2.6 Forces acting on an aircraft during a vertical pull-down. . . . . . . . . . . . . . . . . Figure 2.7 Forces acting on an aircraft during a banked turn. . . . . . . . . . . . . . . . . . . .

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LIST OF FIGURES

Figure Figure Figure Figure

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3.1 Positive sign convention.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 3.2 Three-dimensional bar-structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3.3 Free body diagrams for the three-dimensional bar-structure.. . . . . . . . . . . . . . 70 3.4 Equilibrium element supporting a general force system under the stress convention in the x-y plane.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Figure 3.5 Equilibrium element supporting a general force system under the structural convention in the x-y plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Figure 3.6 Equilibrium element supporting a general force system under the elasticity convention in the x-y plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Figure 3.7 Machine component for example below.. . . . . . . . . . . . . . . . . . . . . . . . . . 81 Figure 3.8 Dimensionless axial load distribution.. . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Figure 3.9 Dimensionless shear (in the y–axis) load distribution.. . . . . . . . . . . . . . . . . . 85 Figure 3.10 Dimensionless shear (in the z–axis) load distribution.. . . . . . . . . . . . . . . . . . 86 Figure 3.11 Dimensionless torsional load distribution.. . . . . . . . . . . . . . . . . . . . . . . . . 86 Figure 3.12 Dimensionless moment (about the y–axis) distribution.. . . . . . . . . . . . . . . . . 87 Figure 3.13 Dimensionless moment (about the z–axis) distribution.. . . . . . . . . . . . . . . . . 87 Figure 3.14 Cross-section of the helicopter blade.. . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Figure 3.15 Loading on the helicopter blade.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Figure 3.16 Replacing the pressure g(x, z) with a distributed load py (x).. . . . . . . . . . . . . . 89 Figure 3.17 Locating all loads at the weighted-modulus centroid.. . . . . . . . . . . . . . . . . . . 90 Figure 4.1 Arbitrary cross section of a structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Figure 4.2 Arbitrary cross section of a structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Figure 4.3 Typical cross section of a thin-walled structure.. . . . . . . . . . . . . . . . . . . . . 138 Figure 5.1 Solid body in equilibrium.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Figure 5.2 Solid body in equilibrium sliced with an arbitrary plane.. . . . . . . . . . . . . . . . 195 Figure 5.3 Complete definition of the state of stress at a point.. . . . . . . . . . . . . . . . . . . 199 Figure 5.4 Shear stresses on the faces of an element at a point in an elastic body about the z-axis.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Figure 5.5 Shear forces on the faces of an element at a point in an elastic body about the z-axis..203 Figure 5.6 This is an infinitesimal element representing the state of stress for the given problem (NOTE: Units are part of the answer). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 Figure 5.7 Principal state of stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Figure 5.8 Positive stresses on a two dimensional element.. . . . . . . . . . . . . . . . . . . . . . 225 Figure 5.9 a) Stresses acting on an element in plane stress. b) Stresses acting on an element oriented at an angle θ = α. c) Principal normal stresses. d) Maximum in-plane shear stresses..231 Figure 5.10 Mohr’s circle for plane stress in the y-z plane. . . . . . . . . . . . . . . . . . . . . . . 233 Figure 5.11 a) Stresses acting on an element in plane stress. b) Stresses acting on an element oriented at an angle θ = α. c) Principal normal stresses. d) Maximum in-plane shear stresses..236 Figure 5.12 Mohr’s circle for plane stress in the x-z plane. . . . . . . . . . . . . . . . . . . . . . . 238 Figure 5.13 a) Stresses acting on an element in plane stress. b) Stresses acting on an element oriented at an angle θ = α. c) Principal normal stresses. d) Maximum in-plane shear stresses..240 Figure 5.14 Mohr’s circle case for uniaxial state of stress. . . . . . . . . . . . . . . . . . . . . . . 242 Figure 5.15 Mohr’s circle case for triaxial state of stress. . . . . . . . . . . . . . . . . . . . . . . . 244 Figure 5.16 Mohr’s circle case for hydrostatic state of stress. . . . . . . . . . . . . . . . . . . . . 246 Figure 5.17 General state of stress for stresses acting on octahedral planes. . . . . . . . . . . . . 252 Figure 5.18 Tetrahedron element at O. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

LIST OF FIGURES

xxiii

Figure 5.19 Deformation of a solid body from the initial configuration, C 0 , to the current configuration, C 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Figure 5.20 The neighborhood of point P in the reference and deformed configurations... . . . . 258 Figure 5.21 Shear deformation in the reference and deformed configurations... . . . . . . . . . . . 262 Figure 5.22 Three strain gauges at the surface of a solid: 3-gage rosette.. . . . . . . . . . . . . . 285 Figure 5.23 Mohr’s circle for plane strain in the x-y plane. . . . . . . . . . . . . . . . . . . . . . . 288 Figure 6.1 Uniaxial loading-unloading stress-strain curves. . . . . . . . . . . . . . . . . . . . . . 308 Figure 6.2 Strain energy density.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 Figure 6.3 Mohr’s circle case for the principal state of stress. . . . . . . . . . . . . . . . . . . . . 329 Figure 7.1 Sign convention for stress resultants on a beam cross section.. . . . . . . . . . . . . . 385 Figure 7.2 Stresses acting on a beam’s cross-sectional differential volume.. . . . . . . . . . . . . 386 Figure 7.3 Decomposition of the axial displacement field.. . . . . . . . . . . . . . . . . . . . . . 388 Figure 7.4 Definition of curvature. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 Figure 7.5 Bending of a beam element of length dx in the x-y plane.. . . . . . . . . . . . . . . . 392 Figure 7.6 Cross-section of the helicopter blade.. . . . . . . . . . . . . . . . . . . . . . . . . . . 407 Figure 7.7 Shear deformation of a beam element about the z-axis.. . . . . . . . . . . . . . . . . 414 Figure 7.8 Cross-section of the helicopter blade.. . . . . . . . . . . . . . . . . . . . . . . . . . . 428 Figure 7.9 Cantilever rectangular beam carrying a point load at the tip.. . . . . . . . . . . . . . 441 Figure 7.10 Cantilever rectangular beam carrying a uniform shear load on upper surface.. . . . . 451 Figure 7.11 Simply-supported rectangular beam carrying a uniform normal load on upper surface..460 Figure 7.12 Cylindrical bar of arbitrary cross-section in pure torsion.. . . . . . . . . . . . . . . . 472 Figure 7.13 Cylindrical bar of arbitrary cross-section in pure torsion.. . . . . . . . . . . . . . . . 473 Figure 7.14 Representation of stress state along edge of solid cross-section under torsion.. . . . . 479 Figure 7.15 Representation of stress state at top cross-section of rod under torsion.. . . . . . . . 480 Figure 7.16 Linear elastic torsion of a shaft with a rectangular cross-section.. . . . . . . . . . . . 492 Figure 7.17 Linear elastic torsion of a shaft with a parabolic cross-section.. . . . . . . . . . . . . 493 Figure 8.1 Open thin-walled section.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 Figure 8.2 Shear stress in open thin-walled section.. . . . . . . . . . . . . . . . . . . . . . . . . . 496 Figure 8.3 Shear flow in arbitrary open thin-walled section.. . . . . . . . . . . . . . . . . . . . . 497 Figure 8.4 Differential element of a thin-walled beam showing shear flow and bending stress.. . 498 Figure 8.5 Shear flow convention in thin-walled open channel section.. . . . . . . . . . . . . . . 503 Figure 8.6 Unsymmetrically thin-walled channel section.. . . . . . . . . . . . . . . . . . . . . . . 505 Figure 8.7 Shear flow convention for the given thin-walled channel section.. . . . . . . . . . . . 506 Figure 8.8 Shear flow on a differential portion of a multiweb junction.. . . . . . . . . . . . . . . 512 Figure 8.9 Shear flow convention.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515 Figure 8.10 Tip-loaded cantilever beam: twisting and bending (first & two), and bending only (third).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 Figure 8.11 Vertical and horizontal applied shear forces.. . . . . . . . . . . . . . . . . . . . . . . 522 Figure 8.12 Location of the shear center of a thin-walled open section.. . . . . . . . . . . . . . . 523 Figure 8.13 Unsymmetrically thin-walled channel section.. . . . . . . . . . . . . . . . . . . . . . . 524 Figure 8.14 Suggested shear flow convention for statically equivalence by taking the torque at point 3.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 Figure 8.15 Suggested shear flow convention for statically equivalence by taking the torque at point 2.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 Figure 8.16 Shear flow convention.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 Figure 8.17 Membrane analogy: in-plane and transverse loading.. . . . . . . . . . . . . . . . . . . 531

LIST OF FIGURES

xxiv

Figure 8.18 Equilibrium of element of a membrane.. . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.19 Torsion of a narrow rectangular strip.. . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.20 Representation of cross-section for membrane analogy and the side-view of the membrane under pressure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.21 Arbitrary open thin-walled cross-Section.. . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.22 Actual thin-walled section and idealized section.. . . . . . . . . . . . . . . . . . . . . Figure 8.23 Actual thin-walled section and idealized section.. . . . . . . . . . . . . . . . . . . . . Figure 8.24 Idealization of a thin rectangular wall into two concentrated areas.. . . . . . . . . . . Figure 8.25 Idealization of a thin rectangular wall into two concentrated areas.. . . . . . . . . . . Figure 8.26 Actual thin-walled section and idealized section.. . . . . . . . . . . . . . . . . . . . . Figure 8.27 Symmetrically thin-walled channel section.. . . . . . . . . . . . . . . . . . . . . . . . Figure 8.28 Shear flow convention. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.29 Shear flow convention for statically equivalence when taking the torque at point 2.. . Figure 8.30 Shear flow convention for statically equivalence when taking the torque at point 3. . Figure 8.31 Area enclosed by the contour.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.32 Thin-walled, single cell beam with an arbitrary cross-sectional contour.. . . . . . . . Figure 8.33 Thin-walled element in its undeformed and deformed configurations.. . . . . . . . . . Figure 8.34 Geometry of the contour.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.35 Superposition of shear flows: problem consisting of an open section and a section with a constant shear flow.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.36 Symmetrical thin-walled monocoque closed section.. . . . . . . . . . . . . . . . . . . Figure 8.37 Shear flow convention. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.38 Shear flow convention. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.39 Shear flow convention for statically equivalence when taking the torque at point O.. Figure 8.40 Shear flow convention for statically equivalence when taking the torque at point O. . Figure 8.41 Shear flow convention for statically equivalence when taking the torque at point O. . Figure 8.42 Multicell thin-walled beam.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.43 Shear flow in a Multicell thin-walled beam cross-section.. . . . . . . . . . . . . . . . Figure 8.44 A two-cell thin-walled section under torsion.. . . . . . . . . . . . . . . . . . . . . . . Figure 8.45 A typical hybrid thin-walled wing section.. . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.46 A hybrid thin-walled section under torsion.. . . . . . . . . . . . . . . . . . . . . . . . Figure 8.47 Hybrid thin-walled wing section.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.1 Force vector and displacement vector at a location s.. . . . . . . . . . . . . . . . . . Figure 9.2 Differential work done.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.3 Virtual work done.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.4 Complementary Virtual work done.. . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.5 Virtual work done.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.6 Particle in equilibrium subject to n forces.. . . . . . . . . . . . . . . . . . . . . . . . Figure 9.7 System of particles showing both external and internal forces.. . . . . . . . . . . . . Figure 9.8 Rigid-bars configuration for Example 9.1.. . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.9 Rigid-bars configuration for Example 9.2.. . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.10 Rigid-bars configuration for Example 9.3.. . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.11 Elastic bar subject to a load P undergoing a virtual displacement.. . . . . . . . . . . Figure 9.12 Point force acting on a clamped bar. . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.13 Elastic truss bar.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.14 Undeformed and deformed states of the q th elastic truss bar.. . . . . . . . . . . . . .

532 534 534 536 544 545 545 547 548 551 552 555 556 561 566 567 569 584 586 590 591 593 594 599 601 602 604 610 611 613 624 625 625 626 629 630 632 636 640 644 652 655 661 662

LIST OF FIGURES

Figure Figure Figure Figure Figure Figure

9.15 Planar truss configuration.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.16 Idealized landing gear truss structure.. . . . . . . . . . . . . . . . . . . . . . . . . . 9.17 Four bar truss structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.18 Four bar truss structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.19 Representation of a single bay of a wing spar truss.. . . . . . . . . . . . . . . . . . 9.20 Virtual strain energy density per unit volume and virtual complementary strain energy density per unit volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.21 Four bar truss structure.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.22 Distributed load acting on a clamped beam. . . . . . . . . . . . . . . . . . . . . . . Figure 9.23 Distributed load acting on a simply-supported beam. . . . . . . . . . . . . . . . . . Figure 9.24 Nondimensional axial displacement with one and five term approximation. . . . . . Figure 9.25 Nondimensional transverse displacement with one and five term approximation. . . Figure 9.26 Nondimensional lateral displacement with one and five term approximation. . . . . Figure 9.27 System in its deformed and undeformed state. . . . . . . . . . . . . . . . . . . . . . Figure 9.28 Virtual loads acting on the system of particles. . . . . . . . . . . . . . . . . . . . . Figure 9.29 Two rigid pin-connected members joined by a spring. . . . . . . . . . . . . . . . . . Figure 9.30 Elastic bar subject to a virtual load δP .. . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.31 Uniform, homogeneous, elastic bar subject to an axially distributed load. . . . . . . Figure 9.32 Elastic truss bar.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.33 Planar truss configuration.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.34 Truss structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.35 Truss structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.36 Virtual strain energy density per unit volume and virtual complementary strain energy density per unit volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.37 Truss structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.38 Distributed load acting on a simply-supported beam. . . . . . . . . . . . . . . . . . Figure 9.39 Distributed load acting on a simply-supported beam. . . . . . . . . . . . . . . . . . Figure 9.40 Distributed load acting on a simply-supported beam. . . . . . . . . . . . . . . . . . Figure 9.41 Truss bar structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.42 Strut-braced wing subjected to a point load P . . . . . . . . . . . . . . . . . . . . . Figure 9.43 Distributed load acting on a simply-supported beam. . . . . . . . . . . . . . . . . . Figure 9.44 Elastic circular arch supporting a load P . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.45 Idealized truss-bar structure supporting a load P . . . . . . . . . . . . . . . . . . . . Figure 10.1 Three modes of fracture. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 11.1 Typical S-N diagram for ferrous materials.. . . . . . . . . . . . . . . . . . . . . . . Figure 12.1 Equilibrium states. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 12.2 One degree of freedom structural configuration, α = 0. . . . . . . . . . . . . . . . . Figure 12.3 Summary of the primary and secondary path stability. . . . . . . . . . . . . . . . . Figure 12.4 One degree of freedom structural configuration, α > 0. . . . . . . . . . . . . . . . . Figure 12.5 Linear response for various levels of imperfection. . . . . . . . . . . . . . . . . . . . Figure 12.6 Nonlinear response for various levels of imperfection. . . . . . . . . . . . . . . . . . Figure 12.7 One degree of freedom structural configuration. . . . . . . . . . . . . . . . . . . . . Figure 12.8 Summary of the primary and secondary path stability. . . . . . . . . . . . . . . . . Figure 12.9 Load and frequency plots against vertical deflection. . . . . . . . . . . . . . . . . . Figure 12.10 A simply-supported beam column subject to an axial load. . . . . . . . . . . . . . Figure 12.11 Cantilevered beam column subject to an axial load. . . . . . . . . . . . . . . . . . .

xxv

. . . . .

668 670 673 678 683

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

688 692 698 707 714 714 715 717 718 719 723 727 730 733 735 739

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

744 747 754 759 772 779 780 781 782 783 820 886 938 940 947 949 950 951 952 959 960 964 967

LIST OF FIGURES

Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure

xxvi

12.12 A clamped-spring supported beam column subject to an axial load. . . . . . . . . . . 970 12.13 Response for various levels of load imperfection. . . . . . . . . . . . . . . . . . . . . . 989 12.14 Response for various levels of geometric imperfection. . . . . . . . . . . . . . . . . . . 991 12.15 A simply-supported beam column subject to an axial load. . . . . . . . . . . . . . . 995 12.16 Beam column with unsymmetrical supports subject to an axial load. . . . . . . . . . 999 12.17 Configuration case A.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006 12.18 Configuration case B.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007 12.19 Configuration case C.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008 12.20 Rigid bar with a concentrated mass and spring system.. . . . . . . . . . . . . . . . . 1009 12.21 A spring-supported beam column subject to an axial load. . . . . . . . . . . . . . . . 1011 12.22 A simply-supported beam column subject to an axial load. . . . . . . . . . . . . . . 1012 13.1 Interdisciplinary nature of the field of aeroelasticity. . . . . . . . . . . . . . . . . . . 1015 13.2 The aeroelastic triangle of loads. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016 13.3 A two-dimensional rigid wing model to study divergence. . . . . . . . . . . . . . . . . 1019 13.4 The divergence dynamic pressure with respect the angle of attack. . . . . . . . . . . 1021 13.5 Slender straight wing subject to distributed torsional load. . . . . . . . . . . . . . . . 1023 13.6 Small element with the differential loads acting on the wing. . . . . . . . . . . . . . . 1024 13.7 Top view of an aircraft with swept wings. . . . . . . . . . . . . . . . . . . . . . . . . 1028 13.8 Reverse airflow: forward-swept wing vs. aft swept wing.. . . . . . . . . . . . . . . . . 1030 13.9 Aerodynamic axes on a two-dimensional model.. . . . . . . . . . . . . . . . . . . . . 1038 13.10 Lift coefficient vs. angle of attack.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1039 13.11 The V –n diagram for a typical aircraft. . . . . . . . . . . . . . . . . . . . . . . . . . 1040 13.12 Maneuver V –n diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043 13.13 Aircraft subject to gust loads. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048 13.14 Aircraft subject to gust loads. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1050 A.1 Basic MATLABr working environment.. . . . . . . . . . . . . . . . . . . . . . . . . . 1063 B.1 Typical Mohr’s circles for a given state of stress.. . . . . . . . . . . . . . . . . . . . . 1145 B.2 Sketch of the given information on the Mohr’s circle.. . . . . . . . . . . . . . . . . . 1148 B.3 a) Stresses acting on an element in plane stress. b) Stresses acting on an element oriented at an angle θ = α. c) Principal normal stresses. d) Maximum in-plane shear stresses..1151 Figure B.4 Mohr’s circle for plane stress in the x-y plane. . . . . . . . . . . . . . . . . . . . . . . 1154 Figure B.5 a) Stresses acting on an element in plane stress. b) Stresses acting on an element oriented at an angle θ = α. c) Principal normal stresses. d) Maximum in-plane shear stresses..1156

List of Tables

Table 7.1 Table 10.1 Table 10.2

Shear constant for various cross-sections (Shames and Dym).. . . . . . . . . . . . . . 420 Semiquantitative assessment of rating factors. . . . . . . . . . . . . . . . . . . . . . . 790

Table 12.1

Effective length coefficient CL for several type of column end constraints . . . . . . . . 975

Plane strain fracture toughness and corresponding tensile properties for representative metals at room temperature. “Mechanical Behavior of Materials”, by N.E. Dowling, Prentice-Hall Inc, NJ, 1999. Page 291. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 828

xxvii

Chapter 1 Learning about Aircraft Structures

Instructional Objectives of Chapter 1 After completing this chapter, the reader should be able to: 1. Understand how aerospace structures have evolved with history. 2. Identify the main aerodynamic loads acting on airplanes. 3. Explain the major structural components of an aircraft and understand their function(s). 4. Understand the use of semi-monocoque structures in modern aircraft.

Before we begin our journey to study aircraft structures let us understand what we mean by aircraft structures. We can define aircraft structures as the study of methods for designing and manufacturing aircraft, and ensuring they withstand any stress or strain. Although we could spend several books on the subject, for this introductory study we will limit our study to the few major structural components of an aircraft such as wing ribs, stringers or longerons, spars, heavy frames and bulkheads, skin, and truss components. These structural components play an important role most aircraft’s structural integrity. Their location, weight, design, material, etc. are crucial for an optimum design. In this chapter we will learn how structural components in modern aircraft have evolved. Followed by a brief explanation of the typical loads acting on an aircraft and what are roles of the various components on a typical aircraft. We conclude this chapter studying the difference of monocoque and semi-monocoque structures.

1

1.1. HISTORY OF AVIATION

1.1

2

History of Aviation

Before we begin the study of aircraft structures let us review the history of aviation at a glance. This will help us better understand the why-we-do-what-we-do. We define aviation as the design, manufacture, use, or operation of aircraft. By aircraft we mean any vehicle capable of flying. Mainly, we have two kinds of aircraft: (i) heavier-than-air (airplanes, autogiros, gliders, helicopters, and ornithopters), (ii) and lighter-than-air (balloons and airships). Man has always wanted to overcome the challenges to move through air, water and ground. When man succeeded to travel through water and on ground, he dreamed to soar with the birds. These dreams caused accidents due to structural failures. Icarus, from the Greek mythology, is famous for his death caused when the sun melt the wax holding his artificial wings together. Years later, inventors such as Leonardo Da Vinci, John Stringfellow, and Lawrence Hargrave designed intriguing flying machines long before the Wright brothers’ famous first flight at Kitty Hawk.

1.1.1

Pre-Wright Era: Early Aviation

Roughly speaking, the first type of aircraft was a kite. It was designed in China during the fifth century by Mozi and Lu Ban. These first kites were made with silk using bamboo as the framework. During the thirteenth century, Roger Bacon, famous Franciscan friars, concluded that air could support a craft just like water supports boats. Later during the sixteenth century, Leonardo da Vinci while studying the birds’ flight designed the airscrew, leading to the propeller later on and the parachute. Leonardo was a pioneer in the design of heavier-than-air crafts. Although his designs were not successful, we was the first to conclude that the human power was insufficient to generate flight. His three most important contributions are: (i) the helicopter powered by four men, (ii) the light hang glider, (iii) and the ornithopter (a machine with mechanical wings which flap to mimic a bird). In June of 1783, in Annonay (France), the Montgolfier brothers (Joseph and Jacques), were the first to succed in launching a human to air. Their design consisted in a hot air balloon made of silk and lined with paper to trap the gas, called the Montgolfiere. This first successful flight lifted 6,562 feet into the air, traveled more than a mile and stayed aloft for about ten minutes. The Montgolfiers believed they discovered a new gas (called Montgolfier gas) when they held a flame near the opening at the bottom, and the balloon expanded with hot air and floated upward. They thought that this gas was lighter than air and caused the inflated balloons to rise. The gas was merely air, which became more buoyant as it was heated.

1.1.2

The 19th Century

During the ninth century, new developments took place in the field of stability and trust generation. Lawrence Hargrave designed a box kite in 1893, followed by Alexander Graham Bell who experimented with box kites and wings built of multiple compound tetrahedral kites covered in silk (1907-1912). Bell named this tetrahedral kites Cygnet I, II and III. Jean Marie Le Bris designed a glider with movable

1.1. HISTORY OF AVIATION

3

wings. The greatest impact, during the ninth century, to the aviation industry was the integration of motors to aircraft. John Stringfellow designed a steam engine powered aircraft, Lawrence Hargrave designed a rigid-wing aircraft with flapping blades operated by compressed-air motor and the rotary engine, which powered many early aircraft up until about 1920. They realized that successful powered flight required light gasoline engines instead of the cumbersome steam previously used. Samuel Langley designed the first heavier-than air gasoline powered engine which actually flew, called the aerodrome. This aircraft was powered by a 53 horsepower 5-cylinder radial engine. Also, during this century, British Sir George Cayley designed a combined helicopter and horizontally propelled aircraft while Francis Herbert Wenham studied the behavior of multiple wings aircraft using wind tunnels.

1.1.3

Era of Strut-and-Wire Biplanes: 1900 to World War I

The first heavier-than-air machine powered flight took place on December 17, 1903 (10:35am). It was the famous Wright Bothers, Orville and Wilbur, first flight. This flight lasted 12 second and covered a distance of 120 feet. Their mayor breakthrough was the invention of the “three axis-control”, which enabled the pilot to steer the aircraft effectively and maintain its equilibrium. This has been become the standard on fixed wing aircraft of all kinds. Unfortunately, on September 17, their aircraft crashed injuring Orville and his passenger Lieutenant Thomas E. Selfridge. Selfridge later died due to complications, making him the first person to die in a powered airplane. In 1908, Wilbur completed a 2 hour and 20 minute flight, showing full control over his flyer. The flyer became the first successful military airplane and it remained in service for around two years. On July 4, 1908, Glenn H. Curtiss flew the “June Bug” 5090 ft in 1 minute and 42.5 seconds. Curtiss achieved the following: (i) the first American award the Scientific American Trophy, (ii) win the first international speed event, (ii) and the first American to develop and fly a seaplane. In 1913, A. V. Roe built the first tractor biplane. It consists of two-winged airplanes with engine and propeller in front of the wing. Years later, the military used the tractor biplanes with a closed fuselage as the first standard military aircraft. During the World War I, began the development of huge biplane bombers with two to four engines. Not only the military used aircraft, but the airmail also began using aircraft: on September 23, 1911 the pilot Earle Ovington completed the first airmail officially approved by the U.S. Post Office Department. Also in 1911, Calbraith P. Rogers completed the first transcontinental flight across the U.S.

1.1.4

Before World War II

During the period of 1910 to 1930, the aviation industry greatly grew. In 1919, Captain E. F. White made a nonstop flight from Chicago to New York; later in 1923, Lieutenants Oakley Kelly and John A. Macready made the first nonstop transcontinental flight from Roosevelt Field, Long Island to Rockwell Field. In 1924, Douglas World Cruiser was developed for the U.S. Army Air Service for an attempt to

1.2. WHAT DO WE STUDY IN AIRCRAFT STRUCTURE?

4

make the first flight around the world. One of the most successful designs during this period, was the Douglas DC-3 which became the first airliner to exclusively carry passengers, starting the modern era of passenger airline service. Also, mail delivery was impacted by the Kelly Air Mail act. This act authorized the postmaster general to contract for domestic airmail service with commercial air carriers. It also set airmail rates and the level of cash subsidies to be paid to companies that carried the mail. By transferring airmail operations to private companies, the government helped create the commercial aviation industry.

1.1.5

Era of Propeller-Driven Airplane: During World War II

By the beginning of World War II, many towns and cities had built airports, and there were numerous qualified pilots. The war brought many innovations to aviation such as the first jet aircraft and the first liquid-fueled rockets. The largest operator of all international airlines in operation was Pan American Airways, serving 46 countries and colonies. Before World War II, only about 193,000 people were employed in the aviation industry; and after 1941, the number increased to almost 450,000.

1.1.6

Era of Jet-Propelled Airplane: After World War II until end of 20th Century

In August of 1939, the Heinkel He 178 became the world’s first aircraft to fly under turbojet power, thus becoming the first practical jet plane. The first operational turbojet aircraft, the Messerschmitt Me 262 and the Gloster Meteor, entered service towards the end of World War II in 1944. Civilian aircraft orders drastically increased from 6,844 in 1941 to 40,000 by the end of 1945. One of the minor military contractors was the Boeing Company who later became the largest aircraft manufacturer in the world. New aerodynamic designs, metals, and power plants would result in high-speed turbojet airplanes. These planes would later be able to fly supersonically and make transoceanic flights regularly. During the 20th century, Burt Rutan designed hundreds of aircraft, including the now-famous Voyager, which was piloted by Dick, his brother, and Jeana Yeager in 1986 on a record-breaking nine-day non-stop flight around the world. The Voyager held 1,200 gallons of fuel in its 17 fuel tanks, and maintained an average speed of 115.8 mph. lasted 9 days, 3 minutes, 44 seconds and covered 25,012 miles.

1.2

What do we study in aircraft structure?

By aircraft structure we refer to study and analyze how the plane is built. An aircraft must be lightweight, but stress and strain resistant at the same time. The analysis of aircraft is quite complex, not only for its design but also due to the loads it is subject. The aircraft experiences many forces during flight and its structural components must be able to resist to these static and dynamics loads. What makes aircraft structural design unique t from other structural fields (such as buildings or ships) is that the aircraft

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must be both lightweight and strong.

1.2.1

Design Philosophy

The main forces acting on an aircraft during flight are lift, drag, thrust and weight. The structural components of an aircraft affect directly or indirectly all of these four forces. However, the structure determines the aircraft’s weight. The total weight of the plane consists in: the aircraft itself (empty weight) plus the passengers, crew, baggage and freight (payload), and the fuel. This known as the takeoff weight. There must be enough lift to get the total weight of the aircraft into the air. Engineers also consider cruising weight and landing weight. These weights are the totals of the empty weight, payload weight, and the weight of the fuel at the time. As an aircraft prepares to takeoff, we must keep the following in mind: (i) the aircraft must be able to lift the takeoff weight from the ground before the end of the runway; (ii) and the amount of fuel carried by the aircraft will depend on the traveling distance and the payload weight. Tradeoffs may have to be made. Lighter payloads for shorter runways; larger aircrafts with more fuel to carry heavy payloads long distances. Hence, weight becomes very important. The aircraft structural design teams are responsible to design aircraft to withstand all static and dynamic (transient and suddenly applied) loads. This team should keep the following goals in mind: 1. SAFE LIFE: Consists in designing each part for minimum weight and yet assuring they will last for a long time. 2. FAIL SAFE: Consists in designing the aircraft’s components in such a coordination that if one unit fails, the other units will take on the load. In other words, the overall airframe (structure) is designed so that failure in one component doesn’t cause the whole aircraft to fall apart.

1.2.2

Development of Aircraft Structures

Early aircraft were built from very lightweight materials such as bamboo, wood, and fabric. They design was similar to bridges, with beam and truss structures. As for an example, the wings on the Wright Flyer formed a truss: the two wings used wires and bars diagonally (at an angle) to strengthen the wing against aerodynamic forces. In general, inside the wings we had truss structures. The bars inside were called spars. The wires used on the diagonals strengthened the wing. The spars, plus the spar caps at each end, were shaped to give the wing aerodynamic features. This shape is often called the airfoil. As technology has grown, so have the manufacturing techniques. Hence, in the early twentieth century, metal rods and pieces began to replace the wooden components. Metal skins, rolled very thin were weather-resistant as opposed to the fabric skins. The ribs and spars of the plane were made by riveting many pieces together. When aluminum alloys became available at the end of the 1920’s, ribs and spars were often stamped (cut) out of whole aluminum sheets.

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At first, when the aircraft flew, the wood or metal frame took all of the stress. The fabric or thin metal skin could not withstand any of the load. Later, thicker metal skin was put on airplane frames. This thicker skin was able to share the stress. The metal frame could then be made of lighter metal. Hence, the aircraft’s final weight was lighter! Throughout the years, with the use of metals, the basic aircraft design changed. The original biplane design (two wings) with struts and bracing wires, was no longer efficient at the higher speeds. The spars and wires caused more drag at higher speeds. By using metal skins to carry some of the load (of the frame) made the biplane design no longer necessary. Monoplanes (single wing) create much less drag than a biplane. Also, the monoplane did not have the struts and wires sticking out, like the biplane. Nowadays, aircraft industry is moving towards developing aircraft with even materials that are even lighter such as composite materials. In fact, the design work continues in the field of aircraft structures for a better balance of weight and strength.

1.3

Loads Acting on an Aircraft

Figure 1.1: Typical forces acting on an airplane. A force is something that produces a change in a physical quantity. It is a vector quantity and thus has both a magnitude and a direction. Figure 1.1 shows the typical forces that act on an aircraft during flight. These four forces are: 1. WEIGHT (W ): Weight is a force directed toward the center of the earth (downwards force). Its

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magnitude will depend on the mass of all the aircraft’s components, the amount of fuel, and any payload on board (people, baggage, freight, etc.). Although the total weight may be distributed throughout the airplane, its resultant acts through the center of gravity. In fact, we want our aircraft to rotate about its center of gravity. 2. LIFT (L): For an aircraft to fly it must overcome the total weight of the aircraft. This force is called lift. Lift is generated by the aircraft’s motion through air and its an aerodynamic force. Aerodynamic is a combination of two words aero and dynamic: aero stands for the air, and dynamic denotes motion. Hence an aerodynamic load may be defined as a load produces as an aircraft moves though air. The aerodynamics load lift is perpendicular to the flight direction and its magnitude will depend on several factors such as the shape, size, and aircraft velocity. Each aircraft part will experience certain list and the sum of these “lift forces” gives the total lift acting on the aircraft. Most of the aircraft’s lift is generated by the wings. The aircraft lift acts through a single point called the center of pressure1 . 3. DRAG (D): During flight, there is another aerodynamic force which opposes the motion called drag. Drag acts along, but opposed, to the flight direction. As in the case of lift, many factors affect the magnitude of the drag force such as the shape of the aircraft, the “stickiness” of the air and the aircraft velocity. Just as the weight and lift, each of the of the individual components’ drags combine to produce the total aircraft drag. And like lift, drag acts through the aircraft center of pressure. 4. THRUST (T ): In order to overcome drag, an aircraft uses a propulsion system to generate thrust. Thrust is a propulsion force and its direction depends on how the engines are attached to the aircraft. On some aircraft, such as the Harrier, the thrust direction can change to help the aircraft take off in a very short distance. The magnitude of the thrust will depends on factors associated with the propulsion system such as the type of engine, the number of engines, and the throttle setting. For jet engines, it is often confusing to remember that aircraft thrust is a reaction to the hot gas rushing out of the nozzle. The hot gas goes out the back, but the thrust pushes towards the front. The aircraft motion will depend on the relative strength and direction of the forces shown in Fig. 1.1. If the forces are balanced, then it is said that the aircraft cruises at constant velocity. If the forces are unbalanced, the aircraft accelerates in the direction of the largest force.

1.4

Rotations Acting on an Airplane

The aircraft motion is a three-dimensional motion. Hence, we need to control the attitude or orientation of a flying aircraft in all directions. In flight, any aircraft will rotate about its center of gravity. We can define a three-dimensional coordinate system through the center of gravity with each axis of this coordinate system perpendicular to the other two axes. We can then define the orientation of the aircraft by the amount of rotation of the parts of the aircraft along these principal axes, as shown in Figure 1.2. 1 The center of pressure is defined just like the center of gravity, but using the pressure distribution around the body instead of the weight distribution.

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Figure 1.2: Airplane rotations and body axes. The yaw axis is perpendicular to the plane of the wings with its origin at the center of gravity and directed towards the bottom of the aircraft. A yaw motion is a movement of the nose of the aircraft from side to side. The pitch axis is perpendicular to the yaw axis and is parallel to the plane of the wings with its origin at the center of gravity and directed towards the right wing tip. A pitch motion is an up or down movement of the nose of the aircraft. The roll axis is perpendicular to the other two axes with its origin at the center of gravity, and is directed towards the nose of the aircraft. A rolling motion is an up and down movement of the wing tips of the aircraft.

1.5

Components of a typical Aircraft

Now that we have a fairly good idea about aircraft, let now us discuss its main parts and their functions. An aircraft is designed to move people and/or cargo from one place to another through air. Their shapes and sizes vary depending on the mission of the aircraft. The aircraft shown in Figure 1.3 is a turbine-powered aircraft used here to represent most civil transport aircraft. As previously mentioned, an aircraft flies by lifting its total weight. As the aircraft moves through air, the wings generate most of the lift to hold the plane in the air; the jet engines provide the necessary

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Figure 1.3: Airplane parts (in blue) and their functions (in red). thrust to push the airplane forward and overcome the aerodynamic drag. Some aircraft use propellers instead of jets as their propulsion system. The tail of the plane has built in smaller wings to help control and maneuver the aircraft. The tail usually has a fixed horizontal part known as the horizontal stabilizer, and a fixed vertical part known as the vertical stabilizer. The job of the stabilizers is to provide stability for the aircraft and to help maintain a straight flight. The vertical stabilizer keeps the nose of the plane from yawing, while the horizontal stabilizer prevents a pitching motion. The stabilizers are controlled through automatic controls. At the rear of the wings and stabilizers are small moving sections that are attached to the fixed sections by hinges. Changing the rear portion of a wing, will change the amount of force produced by the wing. This ability to change forces helps to control and maneuver the airplane. Let us understand the purpose of one of these four sections: 1. The hinged part of the vertical stabilizer is called the rudder and it is used to deflect the tail to the left and right as viewed from the front of the fuselage. 2. The hinged part of the horizontal stabilizer is called the elevator; it is used to deflect the tail up

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and down. 3. The outboard hinged part of the wing is called the aileron; it is used to roll the wings from side to side. 4. Most airplanes can also be rolled from side to side by using the spoilers. Spoilers are small plates that are used to disrupt the flow over the wing and to change the amount of force by decreasing the lift when the spoiler is deployed. The wings have additional hinged, rear sections near the body that are called flaps. Flaps are deployed downward on takeoff and landing to increase the amount of force produced by the wing. On some aircraft, the front part of the wing will also deflect and it is called slats. Slats are used at takeoff and landing to produce additional load. The spoilers are also used during landing to slow the plane down and to counteract the flaps when the aircraft is on the ground. The fuselage is the body of the aircraft and it holds all the components together. The pilots sit in the cockpit (front of the fuselage) and the passengers and cargo are carried in the rear of the fuselage. Some aircraft carry fuel in the fuselage, while others carry the fuel in the wings.

1.5.1

Wings

Figure 1.4: Geometry and nomenclature of a wing. Figure 1.4 helps us understand some fundamental definitions of a wing. The terms defined here are used throughout the aircraft industry. Although actual aircraft wings are complex three-dimensional

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structures, here we will start with some simple definitions for a two-dimensional wings. Figure 1.4 shows the three different views of the wing: the top view shows the view from the top looking down on the wing, the front view shows the view from the front looking at the wing leading edge, and the side view shows a view from the left looking in towards the centerline. The side view shows an airfoil shape with the leading edge to the left.

Top view The top view shows the wing geometry. The front of the wing is called the leading edge and the back of the wing is called the trailing edge. The leading edge is the front edge of the wing. When an aircraft is moving forward, the leading edge is that part of the wing that first contacts the air. The trailing edge of a wing is the rear edge of the wing, where the airflow separated by the leading edge rejoins after passing over and under the top and bottom surfaces of the wing. The distance from the leading edge to the trailing edge is called the chord and it will be denoted by the symbol c. The chord is measured in the direction of the normal airflow. Most wings change their chord over their width (or span): i.e., the wings are tapered. In general, we use the mean aerodynamics chord to give a characteristic figure which can be compared among various wing shapes. The ends of the wing are called the wing tips. The distance from one wing tip to the other is called the span and will be denoted by the symbol b. The wingspan, or simply span, of an aricraft is the distance from the left wing tip to the right wing tip and is always measured in a straight line, from wing tip to wing tip, independently of wing shape or sweep. As for an example:

In general, planes with a longer wingspan are more efficient because they suffer less induced drag and their wing tip vortices do not affect the wing as much. However, the long wings mean that the plane has a greater moment of inertia about its longitudinal axis and therefore cannot roll as quickly and is less maneuverable. Thus, combat aircraft and aerobatic planes usually opt for shorter wingspans to increase maneuverability. Since the amount of lift that a wing generates is proportional to the area of the wing, planes with short wings must correspondingly have a longer chord. An aircraft’s ratio of its wingspan to chord is

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therefore very important in determining its characteristics, and we call this value the aspect ratio of a wing (AR). The shape of the wing, when viewed from above looking down onto the wing, is called a planform. In Fig. 1.4, the planform is a rectangle. For a rectangular wing, the chord length at every location along the span is the same. For most other planforms, the chord length varies along the span. The wing area, A, is the projected area of the planform and is bounded by the leading and trailing edges and the wing tips. Note: The wing area is not the total surface area of the wing. The total surface area includes both upper and lower surfaces. The wing area is a projected area and is almost half of the total surface area. Different aircraft have different wing shapes and configurations. Configurations will depend on wether the wing is swept. A swept-wing is a wing planform used on high-speed aircraft that spend a considerable portion of their flight time in the transonic speed range. The swept-wing is a wing bent backwards or forward opposed to being at right angles to the fuselage. Aircraft wings can be placed into any of the following four type of swept-wing configurations: 1. STRAIGHT WING: Aircraft wings are straight with no swept angles 2. SWEPT-BACK: The swept-back wing extends backward from the fuselage at an angle.

3. FORWARD-SWEPT: The forward-swept wing gives an airplane the appearance of flying backward. The wing is angled toward the front of the aircraft and is usually attached to the airplane far back on the fuselage.

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4. VARIABLE-SWEPT: A variable-sweep wing can be moved during flight-usually between a sweptback position and a straight position.

Front view When we see the wing from the nose, front view, we can see that the left and right wing may or may not lie in the same plane as they might meet at an angle. The angle that the wing makes with the local horizontal is called the dihedral angle. Dihedral is added to the wings for roll stability. In fact, a wing with some dihedral will naturally return to its original position if it encounters a slight roll displacement. That is the reason for which most civil transport aircraft are designed with diherdral. The wing tips are farther off the ground than the wing root.

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On the other hand, highly maneuverable fighter planes do not have dihedral. In fact, some fighter aircraft have the wing tips lower than the roots giving the aircraft a high roll rate. A negative dihedral angle is called anhedral.

Side view A cut through the wing perpendicular to the leading and trailing edges will show the cross-section of the wing. This side view is called an airfoil. The straight line drawn from the leading to trailing edges of the airfoil is called the chord line. The chord line cuts the airfoil into an upper surface and a lower surface. If we plot the points that lie halfway between the upper and lower surfaces, we obtain a curve called the mean camber line. For a symmetric airfoil (upper surface the same shape as the lower surface) the mean camber line will fall on top of the chord line. But in most cases, these are two separate lines.

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The maximum distance between the two lines is called the camber, which is a measure of the curvature of the airfoil (high camber means high curvature). In other words, the difference between the upper and lower chamber is the chamber of the airfoil. The maximum distance between the upper and lower surfaces is called the thickness. Often you will see these values divided by the chord length to produce a nondimensional or “percent” type of number. Airfoils can come with all kinds of combinations of camber and thickness distributions. NACA has established a method of designating classes of airfoils and then wind tunnel tested the airfoils to provide lift coefficients and drag coefficients for designers.

Angle of Attack

Figure 1.5: The angle of attack is the angle between the chord of the airfoil and the relative wind. Angle of attack, α, is the term used in aerodynamics to describe the angle between the airfoil’s chord line and the direction of airflow wind, effectively the direction in which the aircraft is currently moving, as shown in Fig. 1.5. The angle of attack describes the angle between where the wing is pointing and where it is going. The amount of lift generated by a wing is directly related to the angle of attack, with greater angles generating more lift (and more drag). This remains true up to the stall point, where lift starts to decrease again because of airflow separation, as shown in Fig. 1.6. Planes flying at high angles of attack can suddenly enter a stall if, for example, a strong wind gust changes the direction of the relative wind. Also, to maintain a given amount of lift, the angle of attack must be increased as speed through the air decreases. This is why stalling is an effect that occurs more frequently at low speeds. Nonetheless, a wing (or any other airfoil) can stall at any speed.

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Figure 1.6: Lift versus the angle of attack.

1.5.2

Fuselage

Figure 1.7: Body of an airplane. The fuselage, or body of the airplane, is a long hollow tube which holds all aircraft components. Figure 1.7 shows the fuselage on a typical airplane. The fuselage is hollow to reduce weight. As for other components of airplanes, the aircraft’s mission is what will determine the shape of the fuselage. A supersonic fighter plane has a very slender, streamlined fuselage to reduce the drag associated with high speed flight. An airliner has a wider fuselage to carry the maximum number of passengers.

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Horizontal stabilizer and Elevators

Figure 1.8: Horizontal stabilizer and elevator of an airplane. The rear of the fuselage of most aircraft has a horizontal stabilizer and an elevator. The stabilizer is a fixed wing section and it is responsible for the stability of the aircraft, i.e., to keep it flying straight. The horizontal stabilizer prevents pitching motion of the aircraft nose. The elevator is the small moving section at the rear of the stabilizer that is attached to the fixed sections by hinges. As the elevator moves, it varies the amount of force generated by the tail surface and thus generates and controls the pitching aircraft’s motion. There is an elevator attached to each side of the fuselage. The elevators always work in pairs: when the right elevator goes up, the left elevator also goes up. Figure 1.8 shows the resulting motion when the pilot deflects the elevator. The main objective of the elevator is to control the position of the aircraft nose and the wing’s angle of attack. By changing the wing’s angle of attack, the amount of lift generated by the wing changes. Hence, causes the aircraft to climb (nose up) or dive (nose down). The elevators work by changing the effective shape of the horizontal stabilizer’s airfoil. With greater downward deflection of the trailing edge, lift increases. With greater upward deflection of the trailing edge, lift decreases and can even become negative. Suppose the lift force F is applied at center of pressure of the horizontal stabilizer which is some distance L from the aircraft center of gravity. This will produce a torque Tq = F × L on the aircraft and make the aircraft rotate about its center of gravity.

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Figure 1.9: Stabilator of a fighter aircraft.

1.5.4

Stabilator

On many fighter aircraft, to meet their high maneuvering requirements, the stabilizer and elevator are combined into one large moving surface called a stabilator. The change in force is then created by changing the inclination of the entire surface and not by changing its effective shape as is done with an elevator. As in the case of elevators, the stabilator moves to vary the amount of force generated by the tail surface and it generates and controls the pitching motion of the aircraft. There is usually a stabilator on each side of the fuselage and they also work in pairs. Figure 1.9 shows the resulting motion when the pilot deflects the stabilators. The stabilators follow the same mechanical behavior as elevators.

1.5.5

Vertical Stabilizer and Rudder

At the rear of the fuselage of most aircraft one finds a vertical stabilizer and a rudder. The stabilizer is a fixed wing section providing stability for the aircraft. The vertical stabilizer prevents yawing motion of the aircraft nose. The rudder is the small moving section at the rear of the stabilizer that is attached to the fixed sections by hinges. As the rudder moves, it varies the amount of force generated by the tail surface and it generates and controls the yawing motion of the aircraft. Figure 1.10 shows the resulting motion when the pilot deflects the rudder, a hinged section at the rear of the vertical stabilizer. The rudder is used to control the position of the nose of the aircraft but it is NOT used to turn the aircraft in flight. Aircraft turns are done by banking the aircraft to one side using either ailerons or spoilers. The banking creates an unbalanced side force component of the large wing lift force which

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Figure 1.10: Vertical stabilizer and rudder of an airplane. causes the aircraft’s flight path to curve. The rudder input insures that the aircraft is properly aligned to the curved flight path during the maneuver. Otherwise, the aircraft would encounter additional drag or even a possible adverse yaw condition in which, due to increased drag from the control surfaces, the nose would move farther off the flight path. The rudder works by changing the effective shape of the airfoil of the vertical stabilizer. With increased deflection, the lift will increase in the opposite direction. The rudder and vertical stabilizer are mounted so that they will produce forces from side to side, not up and down. The side force F is applied through the center of pressure of the vertical stabilizer which is some distance L from the aircraft center of gravity. This creates a torque: Tr = F × L on the aircraft and the aircraft rotates about its center of gravity. With greater rudder deflection to the left as viewed from the back of the aircraft, the force increases to the right. On all aircraft, the vertical stabilizer and rudder create a symmetric airfoil. This combination produces no side force when the rudder is aligned with the stabilizer and allows either left or right forces, depending on the deflection of the rudder. Some fighter planes have two vertical stabilizers and rudders because of the need to control the plane with multiple, very powerful engines.

1.5.6

Spoilers

Spoilers are small, hinged plates on the top portion of wings. Spoilers can be used to slow an aircraft, or to make an aircraft descend, if they are deployed on both wings. Spoilers can also be used to generate a rolling motion for an aircraft, if they are deployed on only one wing. Figure 1.11 shows what happens when the pilot only deflects the spoiler on the right wing.

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Figure 1.11: Spoilers of an airplane. Spoilers Deployed on Both Wings When we activate the spoilers, the plates flip up into the air stream. The flow over the wing is disturbed by the spoiler, the drag of the wing is increased, and thus the lift is decreased. Spoilers can be used to “dump” lift and make the airplane descend; or they can be used to slow the airplane down as it prepares to land. When the airplane lands on the runway, the pilot usually brings up the spoilers to kill the lift, keep the plane on the ground, and make the brakes work more efficiently. The friction force between the tires and the runway depends on the “normal” force, which is the weight minus the lift. The lower the lift, the better the brakes work. The additional drag of the spoilers also slows the plane down.

Spoilers Deployed on Only One Wing When we activate a single spoiler, the aircraft banks as one wing tip to move up and the other wing tip to move down. The banking creates an unbalanced side force component of the large wing lift force which causes the aircraft’s flight path to curve. On the figure, the airplane’s right wing spoiler is deployed, while the left wing spoiler is stored flat against the wing surface (as viewed from the rear of the airplane). The flow over the right wing will be disturbed by the spoiler, the drag of this wing will be increased, and the lift will decrease relative to the left wing. The lift force F is applied at the center of pressure of the segment of the wing containing the spoiler. This location is some distance L from the aircraft center of gravity which creates a torque T =F ×L about the center of gravity. The net torque causes the aircraft to roll (clockwise from the rear) about its center of gravity.

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Ailerons

Figure 1.12: Ailerons of an airplane. Ailerons can be used to generate a rolling motion for an aircraft. Ailerons are small hinged sections on the outboard portion of a wing. Ailerons usually work in opposition: as the right aileron is deflected upward, the left is deflected downward, and vice versa. Figure 1.12 shows the resulting motion when the pilot deflects the right aileron upwards and the left aileron downwards. The ailerons are also used to bank the aircraft by causing one wing tip to move up and the other wing tip to move down. The banking creates an unbalanced side force component of the large wing lift force which causes the aircraft’s flight path to curve. The ailerons work by changing the effective shape of the airfoil of the outer portion of the wing. Note that in figure the aileron on the left wing, as viewed from the rear of the aircraft, is deflected down and the aileron on the right wing is deflected up. Therefore, the lift on the left wing is increased, while the lift on the right wing is decreased. For both wings, the lift force (Fr or Fl ) of the wing section through the aileron is applied at the aerodynamic center of the section which is some distance L from the aircraft center of gravity. This creates a torque T =F ×L about the center of gravity. If the forces (and distances) are equal there is no net torque on the aircraft. But if the forces are unequal, there is a net torque and the aircraft rotates about its center of gravity. For the conditions shown in the figure, the resulting motion will roll the aircraft to the right (clockwise) as viewed from the rear.

1.5.8

Flaps and Slats

The amount of lift generated by a wing depends on the shape of the airfoil, the wing area, and the aircraft velocity. During takeoff and landing the airplane’s velocity is relatively low. To keep the lift high (to avoid objects on the ground!), airplane designers try to increase the wing area and change the airfoil shape by putting some moving parts on the wings’ leading and trailing edges. The part on the leading edge is called a slat, while the part on the trailing edge is called a flap. The flaps and slats move

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Figure 1.13: Flaps and slats of an airplane. along metal tracks built into the wings. Moving the flaps aft (toward the tail) and the slats forward increases the wing area. Pivoting the leading edge of the slat and the trailing edge of the flap downward increases the effective camber of the airfoil, which increases the lift. In addition, the large aft-projected area of the flap increases the drag of the aircraft. This helps the airplane slow down for landing. See Fig. 1.13. On takeoff, we want high lift and low drag, so the flaps will be set downward at a moderate setting. During landing we want high lift and high drag, so the flaps and slats will be fully deployed. See Fig. 1.13. When the wheels touch down, we want to decrease the lift (to keep the plane on the ground!), so often the spoilers are deployed on the top of the wing to kill the lift. Spoilers create additional drag to slow down the plane.

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Figure 1.14: Flaps partially deployed (left), full flaps (middle), full flaps with spoilers deployed (right).

Figure 1.15: The position of the leading edge slats on an airliner (Airbus A300).

1.5. COMPONENTS OF A TYPICAL AIRCRAFT

1.5.9

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Gas Turbine Engines

Figure 1.16: Gas turbine engines on various aircraft. Thrust is the force which moves any aircraft through the air. Thrust is generated by the propulsion system of the aircraft. Different propulsion systems develop thrust in different ways, but all thrust is generated through some application of Newton’s third law of motion. For every action there is an equal and opposite reaction. In any propulsion system, a working fluid is accelerated by the system and the reaction to this acceleration produces a force on the system. A general derivation of the thrust equation shows that the amount of thrust generated depends on the mass flow through the engine and the exit velocity of the gas. During World War II, a new type of airplane engine was developed independently in Germany and in England. This engine was called a gas turbine engine. We sometimes call this engine a jet engine. Early gas turbine engines worked much like a rocket engine creating a hot exhaust gas which was passed through a nozzle to produce thrust. But unlike the rocket engine which must carry its oxygen for combustion, the turbine engine gets its oxygen from the surrounding air. A turbine engine does not work in outer space because there is no surrounding air. For a gas turbine engine, the accelerated gas, or working fluid, is the jet exhaust. Most of the mass of the jet exhaust comes from the surrounding atmosphere. Most modern, high speed passenger and military aircraft are powered by gas turbine engines. Because gas turbine engines are so important for modern life, we will be providing a lot of information about turbine engines and their operation. Turbine engines come in a wide variety of shapes and sizes because of the many different aircraft missions. All gas turbine engines have some parts in common, however. Each aircraft has a unique mission and therefore a unique propulsion requirement.

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25

Landing gear

The undercarriage or landing gear is the structure (usually wheels) that supports an aircraft when it is taxiing or stationary. The assembly usually has wheels and some sort of shock absorber apparatus, but sometimes skis for snow or floats for water, and skids or pontoons (helicopters). To decrease drag in flight the undercarriages on many aircraft, particularly large modern ones, retract behind doors which close flush with the fuselage.

Figure 1.17: The main undercarriage and nose undercarriage of a Qatar Airways A330-300 (A7-ACA).

Figure 1.18: The main undercarriage and nose undercarriage of a Qatar Airways A330-300 (A7-ACA). Wheeled undercarriages come in two types: either taildragger (Fig. 1.18), where there are two main wheels towards the front of the aircraft and a single, much smaller, wheel or skid at the rear; or tricycle (Fig. 1.17) undercarriage where there are two main wheels (or wheel assemblies) under the wings and a third smaller wheel in the nose. Most modern aircraft have tricycle undercarriages or variants thereof. Taildraggers are considered harder to land and take off, and usually require special training. Sometimes

1.5. COMPONENTS OF A TYPICAL AIRCRAFT

26

a small tail wheel or skid is added to aircraft with tricycle undercarriage, in case the tail strikes the ground during take-off. See Fig. 1.17. As aircraft grow larger, they employ more wheels to cope with the increasing weights. The Airbus A340-500/-600 has an additional four-wheel undercarriage bogie on the fuselage centreline. The Boeing 747 has five sets of wheels, a nose-wheel assembly and four sets of four-wheel bogies. A set is located under each wing, and two inner sets are located in the fuselage, a little rearward of the outer bogies. Tricycle undercarriage aircraft are usually steered by the leading wheel(s) when taxiing. On the Boeing 747 the two inner bogies, and on the Boeing 777 the last two wheels on each leg, are also steerable with the nose wheels in order to reduce the lateral stresses on the undercarriage. See Fig. 1.19. The various parts of a typical landing gear are given in Fig. 1.20. Figure 1.21 shows different configurations of the Boeing 757 landing gear.

Figure 1.19: Wing and fuselage undercarriages on a Boeing 747.

1.5. COMPONENTS OF A TYPICAL AIRCRAFT

Figure 1.20: Landing gear parts of a Boeing 737-700.

27

1.5. COMPONENTS OF A TYPICAL AIRCRAFT

28

Figure 1.21: Different angles of a Boeing 757 landing gear (12 o’clock, 10 o’clock, 3 o’clock, 4 o’clock, 6 o’clock, 9 o’clock, 11 o’clock).

1.6. BASIC STRUCTURAL ELEMENTS

1.6

29

Basic Structural Elements

There are three basic types of materials or elements in the structure of an airplane: stiffened shells, stiffened plates, and I-beams. Stiffening means that the plate or shell has oddly shaped pieces of metal welded to the back side to strengthen it. This allows the plate or shell to carry more weight. On an airplane, the fuselage (body) and nacelles (outer covering of the engine) are covered with stiffened shells. The wing itself can be considered an I-beam. Spars are welded to the I-beam, at right angles, to form the wing. The top and bottom surfaces of the wings are covered with stiffened plates. Computing the loads on the different components (parts) of an airplane can be very difficult. Tension loads (pulling molecules apart) on simple parts can be fairly easy to compute. Compressive stresses (pushing molecules together) can be much more difficult to figure. Plates and shells tend to be thin. This means they can buckle or bend (deform) long before they reach the failure point. For this reason, engineers try to design stiffened shells and plates to delay permanent deformation (bending). This means much more stress can be applied before bending occurs. One way to understand buckling is to think of a thin rod standing on end on a solid surface. As more and more weight is placed on top of the rod, it will reach a critical point and bend or buckle. To compute where that critical point is the engineer must know the strength of the material (its elasticity), the length, shape and diameter of the rod.

1.6.1

Wing Structure

The main function of the wing is to pick up the air loads and transmit them to the fuselage. The wing cross-section takes the shape of an airfoil, which is designed based on aerodynamic considerations. The wing as a whole performs the combined function of a beam and a torsion member. It consists of axial members in stringers, bending members in spars and shear panels in the cover skin and webs of spars. The spar is a heavy beam running spanwise to take transverse shear loads and spanwise bending. It is usually composed of a thin shear panel (the web) with a heavy cap or flange at the top and bottom to take bending. A typical spar construction is depicted in Fig. 1.22. Wing ribs are planar structures capable of carrying in-plane loads. They are placed chordwise along the wing span. Besides serving as load redistributers, ribs also hold the skin stringer to the designed contour shape. Ribs reduce the effective buckling length of the stringers (or the stringer-skin system) and thus increase their compressive load capability. Figure 1.23 shows a typical rib construction. Note that the rib is supported by spanwise spars. The cover skin of the wing together with the spar webs form an efficient torsion member. For subsonic airplanes, the skin is relatively thin and may be designed to undergo postbuckling. Thus, the thin skin can be assumed to make no contribution to bending of the wing box, and the bending moment is taken by spars and stringers. Figure 1.24 presents two typical wing cross-sections for subsonic aircraft. One type consists only of spars (the concentrated flange type) to take bending. The other type (the distributed flange type) uses both spars and stringers to take bending. Supersonic airfoils are relatively thin compared with subsonic airfoils. To withstand high surface air

1.6. BASIC STRUCTURAL ELEMENTS

30

Figure 1.22: Spar construction. loads and to provide additional bending capability of the wing box structure, thicker skins are often necessary. In addition, to increase structural efficiency, stiffeners can be manufactured (either by forging or machining) as integral parts of the skin. Figure 1.25 shows an example of a wing cross-section with integrally stiffened skin for supersonic aircraft.

Ribs In an aircraft, ribs are forming elements of the structure of a wing, especially in traditional construction. By analogy with the anatomical definition of “rib”, the ribs attach to the main spar, and by being repeated at frequent intervals, form a skeletal shape for the wing. Usually ribs incorporate the airfoil shape of the wing, and the skin adopts this shape when stretched over the ribs.

Stringer In aircraft construction, a Longeron is a thin strip of wood or metal, to which the skin of the aircraft is fastened. Longerons are attached to formers, in the case of the fuselage, or ribs in the case of a wing, or empennage. In very early aircraft, a fabric covering was sewn to the longerons, and then stretched tight by painting it with ”dope”, which would make the fabric shrink, and become stiff. Sometimes the terms ”longeron” and ”stringer” are used interchangeably. Historically, though, there is a subtle difference between the two terms. If the longitudinal members in a fuselage are few in number (usually 4 to 8) then they are called ”longerons”. The longeron system also requires that the fuselage frames be closely spaced (about every 4 to 6 inches). If the longitudinal members are numerous (usually 50 to 100) then they are called ”stringers”. In the stringer system the longitudinal members are smaller and the frames are spaced farther apart (about 15 to 20 inches). On large modern aircraft the stringer system is more common because it is more weight efficient despite being more complex to construct and

1.6. BASIC STRUCTURAL ELEMENTS

31

Figure 1.23: Typical spar construction.

Figure 1.24: (a) Spars only, (b) spars and stringers. analyze. Some aircraft, however, use a combination of both stringers and longerons. The longitudinal members are known as longitudinals, stringers, or stiffeners. Longitudinals which have large cross-sectional areas are referred to as longerons. These members serve the following purposes: 1. They resist bending and axial loads along with the skin. 2. They divide the skin into small panels and thereby increase its buckling and failing stresses. 3. They act with the skin in resisting axial loads caused by pressurization.

1.6.2

Fuselage Structure

Unlike the wing, which is subjected to large distributed air loads, the fuselage is subjected to relatively small air loads. The primary loads on the fuselage include large concentrated forces from wing reactions,

1.6. BASIC STRUCTURAL ELEMENTS

32

Figure 1.25: Wing cross-sections with integrally stiffened skin.

Figure 1.26: Fuselage structure. landing gear reactions, and pay loads. For airplanes carrying passengers, the fuselage must also withstand internal pressures. Because of internal pressure, the fuselage often has an efficient circular cross-section. Fuselage frames often take the form of a ring. They are used to maintain the shape of the fuselage and to shorten the span of the stringers between supports in order to increase the buckling strength of the stringer, see Fig. 1.26. The loads on the frames are usually small and self-equilibrated. Consequently, their constructions are light. To distribute large concentrated forces such as those from the wing structure, heavy bulkheads are needed.

1.6. BASIC STRUCTURAL ELEMENTS

1.6.3

33

Semimonocoque Structures

Monocoque (French for ”single shell”) or unibody is a construction technique that uses the external skin of an object to support some or most of the load on the structure. This is as opposed to using an internal framework (or truss) that is then covered with a non-load-bearing skin. Monocoque construction was first widely used in aircraft, starting in the 1930s, and is the predominant automobile construction technology today. Prior to this time aircraft were built up from an internal frame, typically of wood or steel tubing, which was then covered (or skinned) with fabric to give it a smooth surface. The materials vary; some builders used sheet metal or plywood for the skin. In all of these designs the idea of load-bearing structure vs. skin remained. By the late 1920s the price of aluminum (specifically duralumin) started dropping considerably and many manufacturers started using it to replace the internal framing, and in some cases, the external skin. A classic example of such a design is the Ford Trimotor, which is an “old style” plane built of new materials. The structure of the plane consists of a latticework of U-shaped aluminum beams, with a thin skin of aluminum riveted on top. When these designs started appearing it was realized that the skin itself had significant structural properties of its own. With a sufficient thickness, one could do away with all of the internal structure. However this would be even heavier than the framing would have been. At thinner gauges the skin could easily provide the structure for tension and shear loads (metal resists being pulled apart quite well), and if it was bent into a curve or pipe, it became quite strong against bending loads as well. The only loading it could not handle on its own at least for thin “skins” was compression. Combining this sort of structural skin with a greatly reduced internal framing to provide strength against buckling in compression led to what is known as “semimonocoque”. The result was a structure that was just as strong as ones made with older methods, but weighed considerably less. For aircraft construction this is a very important consideration. As well, the monocoque structure has high torsional stiffness, important in reducing aeroelastic effects as aircraft speeds increased. At the beginning of WWII the technique was just starting to appear, and many aircraft still used mixed construction. By the end, all planes were monocoque. The fuselage and wing structures are semimonocoque constructions consisting of a thin shell stiffened by longitudinal axial elements (stringers and longerons) supported by many transverse frames or rings along its length, as seen Fig. 1.28. In semimonocoque structures the cover, or skin, has the following functions: 1. It transmits aerodynamic forces to the longitudinal and transverse supporting members by plate and membrane action. 2. It develops shearing stresses which react the applied torsional moments (Chap. 8) and shear forces 3. It acts with the longitudinal members in resisting the applied bending and axial loads

1.6. BASIC STRUCTURAL ELEMENTS

34

Figure 1.27: Monocoque and semi-monocoque structure. 4. It acts with the longitudinals in resisting the axial load and with the transverse members in reacting the hoop, or circumferential, load when the structure is pressurized. In addition to these structural functions, it provides an aerodynamic surface and cover for the content of the vehicle. Spar webs (Fig. 1.29b) play a role that is similar to function 2 of the skin. The spar caps in aerodynamic surface perform functions 1 and 2. The transverse members in body structures are called frames, rings, or if they cover all or most of the cross-sectional area, bulkheads. In aerodynamic surfaces they are referred to as ribs. These members are used to: 1. Maintain the cross-sectional shape. 2. Distribute concentrated loads into the structure and redistribute stresses around structural discontinuities 3. Establish the column length and provide end restraint for the longitudinals to increase their column buckling stress 4. Provide edge restraint for the skin panels and thereby increase the plate buckling stress of these elements

1.6. BASIC STRUCTURAL ELEMENTS

35

Figure 1.28: Typical semimonocoque aircraft structures. 5. Act with the skin in resisting the circumferential loads due to pressurization. The behavior of these structural elements is often idealized to simplify the analysis of the assembled component. The following assumptions are usually made: 1. The longitudinals carry only axial stress. 2. The webs (skin and spar webs) carry only shearing stresses. 3. The axial stress is constant over the cross section of each of the longitudinals, and the shearing stress is uniform through the thickness of the webs. 4. The transverse frames and ribs are rigid within their own planes, so that the cross section is maintained unchanged during loading. However, they are assumed to possess no rigidity normal to their plane, so that they offer no restraint to warping deformations out of their plane.

1.6. BASIC STRUCTURAL ELEMENTS

36

When the cross-sectional dimensions of the longitudinals are very small compare to the cross-sectional dimensions of the assembly, assumptions 1 and 3 result in little error. The webs in an actual structure carry significant axial stresses as well as shearing stresses, and it is therefore necessary to use an analytical model of the structure which includes this load-carrying ability. This is done by combining the effective areas of the webs adjacent to a longitudinal with the area of the longitudinal into a total effective area of material which is capable of resisting bending moments and axial forces. In the illustrative examples and problems on stiffened shells in this and succeeding chapters it may be assumed that his idealization has already been made and that areas given for the longitudinals are the total effective areas. The fact that the cross-sectional dimensions of most longitudinals are small when compared with those of the stiffened-shell cross section makes it possible to assume without serious error that the area of the effective longitudinal is concentrated at a point on the midline of the skin where it joins the longitudinal. The locations of these idealized longitudinals will be indicated by small circles, as shown in Fig. 1.29. In thin aerodynamic surfaces the depth of the longitudinals may not be small compared to the thickness of the cross section of the assembly, and more elaborate idealized model of the structure may be required.

Figure 1.29: Idealization of semimonocoque structure: (a) actual structure, (b) idealized structure.

Figure 1.30: Idealization of monocoque shell: (a) actual structure, (b) idealized structure. The fewer the number of longitudinals, the simpler the analysis, and in some cases several longitudinal may be lumped into a single effective longitudinal to shorten computations (Fig. 1.29). On the other hand, it is sometimes convenient to idealize a monocoque shell into an idealized stiffened shell by lumping the shell wall into idealized longitudinals, as shown in Fig. 1.30,and assuming that the skin between these

1.7. MATERIALS

37

longitudinals carries only shearing stresses.. The simplification of an actual structure into an analytical model represents a compromise, since elaborate models which nearly simulate the actual structure are usually difficult to analyze. Once the idealization is made, the stresses in the longitudinals due to bending moments, axial load, and thermal gradients can be computed from the equations of this chapter if the structure is long compared to its cross-sectional dimensions and if there are no significant structural or loading discontinuities in the region where the stresses are computed.

1.7

Materials

Most of the structural components of an airplane are made of metallic materials. An aluminum alloy is used on most metallic components, because it is relatively light weight. Remember, the lighter the plane, the farther it can fly, or the less fuel it will need. Yet, aluminum is strong enough to carry heavy loads. Steel is used for a smaller number of components that are exposed to heavy loads. Landing gears, engine fittings, and the tracks that the flaps move along are usually made of steel. Since aluminum and steel tend to lose their strength at high temperatures, titanium is used around engines, for firewalls and hot ducts. More and more, composite materials are being used for some components. Composite materials (two or more materials bonded together) are made of fibers of boron or graphite embedded in a layer of epoxy (glue). The strength along the fibers is very, very large, but is not very high across them. Most composite materials then, are created by layering the thin sheets with the fibers alternating directions. The resulting material is very strong in all directions. These materials are very light and stiff. They help reduce the weight of the airplane structure. The next generation of airplanes will be made significantly of composite materials.

1.8. REFERENCES

1.8

38

References

Allen, D. H., Introduction to Aerospace Structural Analysis , 1985, John Wiley and Sons, New York, NY. Curtis, H. D., Fundamentals of Aircraft Structural Analysis, 1997, Mc-Graw Hill, New York, NY. Dole, Charles E. and Lewis, James E., Flight Theory and Aerodynamics: A Practical Guide for Operational Safety, Second Edition, May 2000, John Wiley and Sons. Johnson, E. R., Thin-Walled Structures, 2006, Textbook at Virginia Polytechnic Institute and State University, Blacksburg, VA. Keane, Andy and Nair, Prasanth, Computational Approaches for Aerospace Design: The Pursuit of Excellence, August 2005, John Wiley and Sons. Kuethe, Arnold M., and Chow, Chuen-Yen, Foundations of Aerodynamics: Bases of Aerodynamic Design, Fifth Edition, November 1997, John Wiley and Sons. Newman, D., Interactive Aerospace Engineering And Design With CD-ROM, First Edition, Mass Institute Of Tech, 2004, Mcgraw-Hill.

http://en.wikipedia.org/wiki/Main Page http://www.grc.nasa.gov/WWW/K-12/airplane/guided.htm http://www.ndt.net/article/ecndt98/aero/001/001.htm

1.9. SUGGESTED PROBLEMS

1.9

39

Suggested Problems

Problem 1.1. Describe some of the most popular aircrafts configurations used during the history of aviation, its inventors and its function. 

Problem 1.2. What were the advantages of the utilization of metals in the aerospace’s structures? 

Problem 1.3. What is the principal engine purpose? Justify your answer with examples. 

Problem 1.4. Let us suppose that you was contracted by aircraft design company, who want a passenger airplane, fuel efficient, stable and need to resist transonic speed. What wing, fuselage and landing gear configuration you will be used? Explain your answer. 

Problem 1.5. What is a stringers and why are so important in the aerospace structure? 

Problem 1.6. What is the difference between stringers and longeron, and which help more to avoid failing stresses? 

Problem 1.7. Why semi-monocoque structures are widely used in current aircraft configurations? Give specific examples of actual aircraft configurations that are semi-monocoque structures. 

Chapter 2 Principle of Aerodynamics

Instructional Objectives of Chapter 2 After completing this chapter, the reader should be able to: 1. Understand and apply the fundamental aerodynamic equations. 2. Understand lift and drag. 3. Determine the load factor for various maneuvers.

Although the main focus of this book is on aircraft structures, these structures are subject to aerodynamics loads. Aerodynamics has its roots from two Greek words: aerios, concerning the air, and dynamis, which means force. Hence, aerodynamics can be defined as a branch of fluid dynamics concerned with the study of the motion of air and other gaseous fluids and other forces acting on objects in motion through the air (gases). In fact, aerodynamics is concerned with the object (aircraft), the movement (Relative Wind ), and the air (Atmosphere). In this chapter, we limit to the fundamental principles of aerodynamics as related to aerospace structures.

40

2.1. AERODYNAMICS

2.1

41

Aerodynamics

The solution of an aerodynamic problem normally involves calculating for various properties of the flow, such as velocity, pressure, density, and temperature, as a function of space and time. Understanding the flow pattern makes it possible to calculate or approximate the forces and moments acting on bodies in the flow. This mathematical analysis and empirical approximation form the scientific basis for heavierthan-air flight. Aerodynamic problems can be classified in a number of ways. The flow environment defines the first classification criterion. External aerodynamics is the study of flow around solid objects of various shapes. Evaluating the lift and drag on an airplane, the shock waves that form in front of the nose of a rocket or the flow of air over a hard drive head are examples of external aerodynamics. Internal aerodynamics is the study of flow through passages in solid objects. For instance, internal aerodynamics encompasses the study of the airflow through a jet engine or through an air conditioning pipe. Aerodynamics is important in a number of applications other than aerospace engineering. It is a significant factor in any type of vehicle design, including automobiles. It is important in the prediction of forces and moments in sailing. It is used in the design of small components such as hard drive heads. Structural engineers use aerodynamics, and particularly aeroelasticity, to calculate wind loads in the design of large buildings and bridges. Urban aerodynamics seeks to help town planners and designers improve comfort in outdoor spaces, create urban microclimates and reduce the effects of urban pollution. The field of environmental aerodynamics studies the ways atmospheric circulation and flight mechanics affects ecosystems. The aerodynamics of internal passages is important in heating/ventilation, gas piping, and in automotive engines where detailed flow patterns strongly affect the performance of the engine.

2.1.1

Continuity

Gases are composed of molecules which collide with one another and solid objects. In aerodynamics, however, gases are considered to have continuous quantities. That is, properties such as density, pressure, temperature, and velocity are taken to be well-defined at infinitely small points, and are assumed to vary continuously from one point to another. The discrete, molecular nature of a gas is ignored. The continuity assumption becomes less valid as a gas becomes more rarefied. In these cases, statistical mechanics is a more valid method of solving the problem than aerodynamics.

2.1.2

Newton’s Laws Of Motion

The motion of an aircraft through the air can be explained and described by physical principals discovered over 300 years ago by Sir Isaac Newton. The laws and their application to aerodynamics are given below. Newton’s three laws of motion are:

2.1. AERODYNAMICS

42

1. Inertia: The first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net force resulting from unbalanced forces acting on an object (if all the external forces cancel each other out), then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. And if an additional external force is applied, the velocity will change because of the force. The amount of the change in velocity is determined by Newton’s second law of motion. There are many excellent examples of Newton’s first law involving aerodynamics. The motion of an airplane when the pilot changes the throttle setting of the engine is described by the first law. The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton’s first law. The motion of a kite when the wind changes can also be described by the first law. We have created separate pages which describe each of these examples in more detail to help you understand this important physical principle. 2. Acceleration: The second law defines a force to be equal to the differential change in momentum per unit time as described by the calculus of mathematics, which Newton also developed. The momentum is defined to be the mass of an object m times its velocity v. So the differential equation for force F is: dm d (m v) dv F= =v +m = vm ˙ + ma dt dt dt If the mass is a constant and using the definition of acceleration a as the change in velocity with time, the second law reduces to the more familiar product of a mass and an acceleration: F=

d (m v) = ma dt

The important fact is that a force will cause a change in velocity; and likewise, a change in velocity will generate a force. The equation works both ways. The velocity, force, acceleration, and momentum have both a magnitude and a direction associated with them. Scientists and mathematicians call this a vector quantity. The equations shown here are actually vector equations and can be applied in each of the component directions. The motion of an aircraft resulting from aerodynamic forces and the aircraft weight and thrust can be computed by using the second law of motion. 3. Action/Reaction: - third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects. For aircraft, the principal of action and reaction is very important. It helps to explain the generation of lift from an airfoil. In this problem, the air is deflected downward by the action of the airfoil, and in reaction the wing is pushed upward. Similarly, for a spinning ball, the air is deflected to one side, and the ball reacts by moving in the opposite direction. A jet engine also produces thrust through action and reaction. The engine produces hot exhaust gases which flow out the back of the engine. In reaction, a thrusting force is produced in the opposite direction.

2.1. AERODYNAMICS

43

Figure 2.1: Newton’s third law applied to aerodynamics.

2.1.3

Conservation laws

Aerodynamic problems are solved using the conservation laws, or equations derived from the conservation laws. In aerodynamics, three conservation laws are used: 1. Conservation of mass: Matter is not created or destroyed. If a certain mass of fluid enters a volume, it must either exit the volume or increase the mass inside the volume. (a) Solid Mechanics: The mass of any object can be determined by multiplying the volume of the object by the density of the object. When we move a solid object, as shown at the top of the slide, the object retains its shape, density, and volume. The mass of the object, therefore, remains a constant between state “1” and state “2”. (b) Fluid Statics: Let us we consider an amount of a static fluid, liquid or gas. If we change the fluid from some state “1” to another state “2” and allow it to come to rest, we find that, unlike a solid, a fluid may change its shape. The amount of fluid, however, remains the same. We can calculate the amount of fluid by multiplying the density times the volume. Since the mass remains constant, the product of the density and volume also remains constant. (If the density remains constant, the volume also remains constant.) The shape can change, but the mass remains the same. (c) Fluid dynamics: Finally, let us consider the changes for a fluid that is moving through our domain. There is no accumulation or depletion of mass, so mass is conserved within the domain. Since the fluid is moving, defining the amount of mass gets a little tricky. Let’s consider an amount of fluid that passes through point “1” of our domain in some amount of time t. If the fluid passes through an area A at velocity v, we can define the volume V as: V = Avt Now the mass at time t = t1 is: m1 = (ρ A v t)1

2.1. AERODYNAMICS

44

where ρ is the fluid density, A the area normal to the velocity v and t the time. At a different time t = t2 the mass will be: m2 = (ρ A v t)2 From the conservation of mass, these two masses are the same and since the times are the same, we eliminate the time dependence: m1 = m2 = m (ρ A v)1 = (ρ A v)2 = ρ A v The conservation of mass gives us an easy way to determine the velocity of flow in a tube if the density is constant. If we can determine (or set) the velocity at some known area, the equation tells us the value of velocity for any other area. This information is used in the design of wind tunnels. The quantity density times area times velocity has the dimensions of mass/time and is called the mass flow rate. This quantity is an important parameter in determining the thrust produced by a propulsion system. As the speed of the flow approaches the speed of sound the density of the flow is no longer a constant and we must then use a compressible form of the mass flow rate equation. The conservation of mass equation also occurs in a differential form as part of the Navier-Stokes equations of fluid flow. For a fluid (a liquid or a gas) the density, volume, and shape of the object can all change within the domain with time. And mass can move through the domain. On the figure, we show a flow of gas through a constricted tube. There is no accumulation or destruction of mass through the tube; the same amount of mass leaves the tube as enters the tube. At any plane perpendicular to the center line of the tube, the same amount of mass passes through. We call the amount of mass passing through a plane the mass flow rate. The conservation of mass (continuity) tells us that the mass flow rate through a tube is a constant. We can determine the value of the mass flow rate from the flow conditions. We define mass flow as follows: m ˙ = ρvA How do engineers use this knowledge of the mass flow rate? From Newton’s Second Law of Motion, the aerodynamic forces on an aircraft (lift and drag) are directly related to the change in momentum of a gas with time. The momentum is defined to be the mass times the velocity, so we would expect the aerodynamic forces to depend on the mass flow rate past an object. The thrust produced by a propulsion system also depends on the change of momentum of a working gas. The thrust depends directly on the mass flow rate through the propulsion system. For flow in a tube, the mass flow rate is a constant. For a constant density flow, if we can determine (or set) the velocity at some known area, the equation tells us the value of velocity for any other area. If we desire a certain velocity, we know the area we have to provide to obtain that velocity. This information is used in the design of wind tunnels. Considering the mass flow rate equation, it would appear that for a given area, we could make the mass flow rate as large as we want by setting the velocity very high. However, in real fluids, compressibility effects limit the speed at which a flow can be forced through a given area. If there is a slight constriction in the tube, as shown in the nozzle graphics, the Mach number of the flow through the constriction cannot be greater than one. This is commonly referred to as flow choking and the details of the physics are given on a page considering compressible mass

2.1. AERODYNAMICS

45

flow rates. 2. Conservation of momentum: Also called Newton’s second law of motion. Momentum is defined to be the mass of an object multiplied by the velocity of the object. The conservation of momentum states that, within some problem domain, the amount of momentum remains constant; momentum is neither created nor destroyed, but only changed through the action of forces as described by Newton’s laws of motion. Dealing with momentum is more difficult than dealing with mass and energy because momentum is a vector quantity having both a magnitude and a direction. Momentum is conserved in all three physical directions at the same time. It is even more difficult when dealing with a gas because forces in one direction can affect the momentum in another direction because of the collisions of many molecules. Here we will present a very, very simplified flow problem where properties only change in one direction. The problem is further simplified by considering a steady flow which does not change with time and by limiting the forces to only those associated with the pressure. Be aware that real flow problems are much more complex than this simple example. Let us consider the flow of a gas through a domain in which flow properties only change in one direction, which we will call x. The gas enters the domain at station 1 with some velocity u and some pressure p and exits at station 2 with a different value of velocity and pressure. For simplicity, we will assume that the density ρ remains constant within the domain and that the area A through which the gas flows also remains constant. The location of stations 1 and 2 are separated by a distance ∆x1 . The velocity gradient is indicated by du/dx, change in velocity per change in distance. So at station 2, the velocity is given by the velocity at 1 plus the gradient times the distance: ∆u u2 = u1 + ∆x ∆x Similarly, the pressure at the exit: ∆p p2 = p1 + ∆x ∆x From Newton’s second law: du ∆u d(m u) =m =m F = dt dt ∆t The force comes from the pressure gradient. Since pressure is a force per unit area, the net force on our fluid domain is the pressure times the area at the exit minus the pressure times the area at the entrance: h i h i  F = − pA − pA 2

1

Thus,

h i h i u2 − u1 = − pA + pA ∆t 2 1 The minus sign at the beginning of this expression is used because gases move from a region of high pressure to a region of low pressure; if the pressure increases with x, the velocity will decrease. Substituting for our expressions for velocity and pressure:   ∆u   ∆x − u1 u1 + h i ∆p ∆x = − p1 + ∆x A2 + p A m ∆t ∆x 1 m

1A

change with distance is referred to as a gradient to avoid confusion with a change with time which is called a rate.

2.1. AERODYNAMICS

46

Assuming A1 = A2 = A and simplifying, we get:   ∆u   ∆x ∆p ∆x m =− ∆x A ∆t ∆x m



∆u ∆x



∆x ∆t



=−



∆p ∆x



∆x A

Recall, m = ρ V = ρ ∆x A Thus, −m



∆p ∆x



and

= ρu

u=

∆x ∆t

∆u ∆x

As we make ∆x → 0, we obtain the differential equation:   du dp = ρu −m dx dx This is a one dimensional, steady form of Euler’s Equation. It is interesting to note that the pressure drop of a fluid (the term on the left) is proportional to both the value of the velocity and the gradient of the velocity. A solution of this momentum equation gives us the form of the dynamic pressure that appears in Bernoulli’s Equation. 3. Conservation of energy: Although it can be converted from one form to another, the total energy in a given system remains constant. The most general form for the conservation of energy is given on the Navier-Stokes equation page. This formula includes the effects of unsteady flows and viscous interactions. Within some problem domain, the amount of energy remains constant and energy is neither created nor destroyed. Energy can be converted from one form to another (potential energy can be converted to kinetic energy) but the total energy within the domain remains fixed. Here we derive a useful form of the energy conservation equation for a gas beginning with the first law of thermodynamics. If we call the total energy of a gas E, the work done by the gas W , and the heat transferred into the gas Q, then the first law of thermodynamics indicates that between state “1” and state “2”: E2 − E1 = Q − W Aerospace engineers usually simplify a thermodynamic analysis by using intensive variables; variables that do not depend on the mass of the gas2 . We create a “specific” variable by taking a property whose value depends on the mass of the system and dividing it by the mass of the system. Many of the state properties listed on this slide, such as the work and total energy depend on the total mass of gas. We will use “specific” versions of these variables. Engineers usually use the lower case letter for the “specific” version of a variable. Our first law equation then becomes: e2 − e1 = q − w 2 Variables

that do not change with the mass of the gas are called specific variables.

2.2. MACH NUMBER

47

The total energy is composed of the internal and kinetic energy: e=u+k where the specific kinetic energy is defined as k = v 2 /2. The total specific work is defined as:     w = p vvol − p vvol + wshaft 2

1

Let us further define the total specific enthalpy as follows: ht = e + p vvol +

1 2 v 2

Now substituting and regrouping we get: ht2 − ht1 = q − wshaft For a compressor or power turbine, there is no external heat flow into the gas and the q term is set equal to zero. In the burner, no work is performed and the wshaft term is set to zero. All aerodynamic problems are therefore solved by the same set of equations. However, they differ by the assumptions made in each problem.

2.2

Mach Number

As an aircraft moves through the air, the air molecules near the aircraft are disturbed and move around the aircraft. If the aircraft passes at a low speed, typically less than 250 mph, the density of the air remains constant. But for higher speeds, some of the energy of the aircraft goes into compressing the air and locally changing the density of the air. This compressibility effect alters the amount of resulting force on the aircraft. The effect becomes more important as speed increases. Near and beyond the speed of sound, about 330 m/s or 760 mph at sea level, small disturbances in the flow are transmitted to other locations isentropically or with constant entropy. Sharp disturbances generate shock waves that affect both the lift and drag of the aircraft, and the flow conditions downstream of the shock wave. The ratio of the speed of the aircraft to the speed of sound in the gas determines the magnitude of many of the compressibility effects. This speed ratio is known as the Mach number and is defined as M=

V∞ a

where M is the Mach Number, V the speed of the aircraft and a the speed of sound. It allows us to define flight regimes in which compressibility effects vary. 1. Subsonic conditions occur for Mach numbers less than one, M < 1 . For the lowest subsonic conditions, compressibility can be ignored. 2. As the speed of the object approaches the speed of sound, the flight Mach number is nearly equal

2.3. DYNAMIC PRESSURE

48

to one, M ≈ 1 , and the flow is said to be transonic. At some places on the object, the local speed exceeds the speed of sound. Compressibility effects are most important in transonic flows and lead to the early belief in a sound barrier. Flight faster than sound was thought to be impossible. In fact, the sound barrier was only an increase in the drag near sonic conditions because of compressibility effects. Because of the high drag associated with compressibility effects, aircraft do not cruise near Mach 1. 3. Supersonic conditions occur for Mach numbers greater than one, 1 < M < 3. Compressibility effects are important for supersonic aircraft, and shock waves are generated by the surface of the object. For high supersonic speeds, 3 < M < 5, aerodynamic heating also becomes very important for aircraft design. 4. For speeds greater than five times the speed of sound, M > 5, the flow is said to be hypersonic. At these speeds, some of the energy of the object now goes into exciting the chemical bonds which hold together the nitrogen and oxygen molecules of the air. At hypersonic speeds, the chemistry of the air must be considered when determining forces on the object. The Space Shuttle re-enters the atmosphere at high hypersonic speeds, M ∼ 25. Under these conditions, the heated air becomes an ionized plasma of gas and the spacecraft must be insulated from the high temperatures. For supersonic and hypersonic flows, small disturbances are transmitted downstream within a cone. The trigonometric sine of the cone angle b is equal to the inverse of the Mach number M and the angle is therefore called the Mach angle. 1 sin(b) = M There is no upstream influence in a supersonic or hypersonic flow; disturbances are only transmitted downstream.

2.3

Dynamic pressure

2.4. AIRCRAFT WEIGHT

49

There are two ways to look at pressure: (1) the small scale action of individual air molecules or (2) the large scale action of a large number of molecules. On the the small scale, from the kinetic theory of gases, a gas is composed of a large number of molecules that are very small relative to the distance between molecules. The molecules of a gas are in constant, random motion and frequently collide with each other and with the walls of any container. During collisions with the walls, there is a change in velocity and therefore a change in momentum of the molecules. The change in momentum produces a force on the walls which is related to the gas pressure. The pressure of a gas is a measure of the average linear momentum of the moving molecules of a gas. On the large scale, the pressure of a gas is a state variable, like the temperature and the density. The change in pressure during any process is governed by the laws of thermodynamics. Although pressure itself is a scalar quantity, we can define a pressure force to be equal to the pressure (force/area) times the surface area in a direction perpendicular to the surface. If a gas is static and not flowing, the measured pressure is the same in all directions. But if the gas is moving, the measured pressure depends on the direction of motion. This leads to the definition of the dynamic pressure: 1 2 q∞ = ρ V∞ 2

2.4

Aircraft Weight

Figure 2.2: Aircraft weight. Weight is the force generated by the gravitational attraction of the earth on the airplane. Each part of the aircraft has a unique weight and mass, and for some problems it is important to know the distribution. But for total aircraft maneuvering, we only need to be concerned with the total weight and the location of the center of gravity. The center of gravity is the average location of the mass of any object. How do engineers determine the weight of an airplane which they are designing? An airplane is a combination of many parts; the wings, engines, fuselage, and tail, plus the payload and the fuel. Each part has a weight associated with it which the engineer can estimate, or calculate, using Newton’s weight equation: W = mg where W is the weight, m the mass, and g the gravitational constant (32.174 ft/s2 , English; 9.81 m/s2 , SI). The mass of an individual component can be calculated if we know the size of the component and

2.5. CENTER OF GRAVITY

50

its chemical composition. Every material (iron, plastic, aluminum, gasoline, etc.) has a unique density. If we can calculate the volume V , then the mass of each material is obtained by: m = ρV Then, the total weight W of the aircraft is simply the sum of the weight of all of the individual components: + ··· + w + w + w + w W = w fuselage

wing

engines

payload

fuel

In general. we may have a total of n discrete components and the weight of the aircraft is the sum of the individual ith component weights with the index i going from 1 to n: W =

i=n X

wi

i=1

This equation says that the weight of the airplane is equal to the sum of the weight of n discrete parts. What if the parts are not discrete? What if we had a continuous change of mass from front to back? The continuous change can be computed using integral calculus: Z W = w(x) dx The discrete weight is replaced with w(x) which indicates that the weight is some function of distance x. If we are given the form of the function, there are methods to solve the integration. If we don’t know the actual functional form, we can still numerically integrate the equation using a spread sheet by dividing the distance up into a number of small distance segments and determining the average value of the weight over that small segment, then summing up the value.

2.5

Center of Gravity

The center of gravity (CG) is a geometric property of the aircraft and is the average location of the weight of the aircraft. We can completely describe the motion of the aircraft through space in terms of the translation of the center of gravity of the aircraft from one place to another, and the rotation of the aircraft about its center of gravity if it is free to rotate. If the aircraft is confined to rotate about some other point, like a hinge, we can still describe its motion. In general, determining the CG) is a complicated procedure because the mass (and weight) may not be uniformly distributed throughout the aircraft. Thus, we can characterize the mass distribution by a function w(x) which indicates that the weight is some function of distance x from a reference line. Then the center of gravity can be determined from: Z x w(x) dx xCG = Z w(x) dx

2.6. CENTER OF PRESSURE

51

If we don’t know the actual functional form, we can numerically integrate the equation by dividing the distance into a number of small distance segments and determining the average value of the mass (or weight) over that small segment. Taking the sum of the average value times the distance times the distance segment divided by the sum of the average value times the distance segment will produce the center of gravity: n X xi wi xCG =

i

n X

wi

i

The weight of the airplane, pilot and passengers, fuel and baggage is distributed throughout the aircraft. However, the total weight can be considered as being concentrated at one given point. This point is the center of gravity. If the plane were suspended by a rope attached at the center of gravity it would be in balance. The center of gravity is affected by the way an aircraft is loaded. For example, if in a 4 place aircraft, there are 2 rather large individuals in the front seats, and no rear seat passengers or baggage, the CG will be somewhat toward the nose of the aircraft. If however, the 2 front seat passengers are smaller, with 2 large individuals in the rear seats, and a lot of baggage in the rear baggage compartment, the CG will be located more aft. Every aircraft has a maximum forward and rearward CG position at which the aircraft is designed to operate. Operating an aircraft with the CG outside these limits affects the handling characteristics of the aircraft. Serious “out of CG” conditions can be dangerous.

2.6

Center of Pressure

Figure 2.3: Center of pressure. As an object moves through a fluid, the velocity of the fluid varies around the surface of the object. The variation of velocity produces a variation of pressure on the surface of the object as shown by the the thin red lines on Figure 2.3. Integrating the pressure times the surface area around the body determines the aerodynamic force on the object: Z F =

p(x)dS

2.6. CENTER OF PRESSURE

52

where S is the surface area, p(x) the pressure variation and f the total force exerted by the pressure. We can consider this single force, F , to act through the average location of the pressure on the surface of the object. We call the average location of the pressure variation the center of pressure (CP) in the same way that we call the average location of the weight of an object the center of gravity. The aerodynamic force can then be resolved into two components, lift and drag, which act through the center of pressure in flight. Determining the center of pressure is very important for any flying object. To trim an airplane, or to provide stability for a model rocket or a kite, it is necessary to know the location of the center of pressure of the entire aircraft. How do engineers determine the location of the center of pressure for an aircraft which they are designing? In general, determining the center of pressure is a very complicated procedure because the pressure changes around the object. Determining the center of pressure requires a knowledge of the pressure distribution around the body. We can characterize the pressure variation around the surface as a function p(x) which indicates that the pressure depends on the distance x from a reference line usually taken as the leading edge of the object. If we can determine the form of the function, then the center of pressure can be determined from: Z x p(x) dx

xCP = Z

p(x) dx

If we don’t know the actual functional form, we can numerically integrate the equation using a spreadsheet by dividing the distance into a number of small distance segments and determining the average value of the pressure over that small segment. Taking the sum of the average value times the distance times the distance segment divided by the sum of the average value times the distance segment will produce the center of pressure: n X xi pi xCP =

i

n X

pi

i

There are several important problems to consider when determining the center of pressure for an airfoil. As we change angle of attack, the pressure at every point on the airfoil changes. And, therefore, the location of the center of pressure changes as well. The movement of the center of pressure caused a major problem for early airfoil designers because the amount (and sometimes the direction) of the movement was different for different designs. In general, the pressure variation around the airfoil also imparts a torque, or ”twisting force”, to the airfoil. If a flying airfoil is not restrained in some way it will flip as it moves through the air. (As a further complication, the center of pressure also moves because of viscosity and compressibility effects on the flow field. )

2.7. AERODYNAMIC CENTER

2.7

53

Aerodynamic Center

To resolve some of the above design problems, aeronautical engineers prefer to characterize the forces on an airfoil by the aerodynamic force, described above, coupled with an aerodynamic moment to account for the torque. It was found both experimentally and analytically that, if the aerodynamic force is applied at a location 1/4 chord back from the leading edge on most low speed airfoils, the magnitude of the aerodynamic moment remains nearly constant with angle of attack. Engineers call the location where the aerodynamic moment remains constant the aerodynamic center of the airfoil. Using the aerodynamic center as the location where the aerodynamic force is applied eliminates the problem of the movement of the center of pressure with angle of attack in aerodynamic analysis. (For supersonic airfoils, the aerodynamic center is nearer the 1/2 chord location.) When computing the trim of an aircraft, we usually apply the aerodynamic forces at the aerodynamic center of airfoils and compute the center of pressure of the vehicle as an area-weighted average of the centers of the components.

2.8

Lift

Lift is the force that directly opposes the weight of an airplane and holds the airplane in the air. Lift is generated by every part of the airplane, but most of the lift on a normal airliner is generated by the wings. Lift is a mechanical aerodynamic force produced by the motion of the airplane through the air. Because lift is a force, it is a vector quantity, having both a magnitude and a direction associated with it. Lift acts through the center of pressure of the object and is directed perpendicular to the flow direction. There are several factors which affect the magnitude of lift.

2.8.1

How is lift generated

Lift occurs when a moving flow of gas is turned by a solid object. The flow is turned in one direction, and the lift is generated in the opposite direction, according to Newton’s Third Law of action and reaction. Because air is a gas and the molecules are free to move about, any solid surface can deflect a flow. For an aircraft wing, both the upper and lower surfaces contribute to the flow turning. Neglecting the upper surface’s part in turning the flow leads to an incorrect theory of lift.

2.8.2

No Fluid, No Lift

Lift is a mechanical force. It is generated by the interaction and contact of a solid body with a fluid (liquid or gas). It is not generated by a force field, in the sense of a gravitational field,or an electromagnetic field, where one object can affect another object without being in physical contact. For lift to be generated, the solid body must be in contact with the fluid: no fluid, no lift. The Space Shuttle does not stay in space because of lift from its wings but because of orbital mechanics related to its speed. Space is nearly a vacuum. Without air, there is no lift generated by the wings.

2.8. LIFT

2.8.3

54

No Motion, No Lift

Lift is generated by the difference in velocity between the solid object and the fluid. There must be motion between the object and the fluid: no motion, no lift. It makes no difference whether the object moves through a static fluid, or the fluid moves past a static solid object. Lift acts perpendicular to the motion. Drag acts in the direction opposed to the motion. You can learn more about the factors that affect lift at this web site. There are many small interactive programs here to let you explore the generation of lift.

2.8.4

Factors That Affect Lift

All that is necessary to create lift is to turn a flow of air. An aerodynamic, curved airfoil will turn a flow. But so will a simple flat plate, if it is inclined to the flow. The fuselage of an airplane will also generate lift if it is inclined to the flow. For that matter, an automobile body also turns the flow through which it moves, generating a lift force. Any physical body moving through a fluid can create lift if it produces a net turning of the flow. There are many factors that affect the turning of the flow, which creates lift. We can group these factors into: (a) those associated with the object, (b) those associated with the motion of the object through the air, and (c) those associated with the air itself: 1. Object: At the top of the figure, aircraft wing geometry has a large effect on the amount of lift generated. The airfoil shape and wing size will both affect the amount of lift. The ratio of the wing span to the wing area also affects the amount of lift generated by a wing. 2. Motion: To generate lift, we have to move the object through the air. The lift then depends on the velocity of the air and how the object is inclined to the flow. 3. Air: Lift depends on the mass of the flow. The lift also depends in a complex way on two other properties of the air: its viscosity and its compressibility.

2.8.5

Lift Equation

We can gather all of this information on the factors that affect lift into a single mathematical equation called the Lift Equation. With the lift equation we can predict how much lift force will be generated by a given body moving at a given speed. Lift depends on the density of the air, the square of the velocity, the air’s viscosity and compressibility, the surface area over which the air flows, the shape of the body, and the body’s inclination to the flow. In general, the dependence on body shape, inclination, air viscosity, and compressibility is very complex. One way to deal with complex dependencies is to characterize the dependence by a single variable. For lift, this variable is called the lift coefficient, designated CL . This allows us to collect all the effects,

2.9. DRAG

55

simple and complex, into a single equation. The lift equation states that: L=

1 2 ρ V∞ S CL 2

where V∞ is the speed of the aircraft, S the wing area, and ρ the air density. We can further express the equation in terms of the dynamic pressure: L = q∞ S CL

2.9

Drag

In progress...

2.10. NORMAL LOAD FACTOR

2.10

56

Normal Load Factor

Load factor is the ratio of the total load supported by the airplane’s wing to the total weight of the airplane. In still air flight, the load on the wing equals the lift it generates. The load factor is expressed in “g” units and is defined as the ratio of lift and to weight: n=

L W

(2.1)

Hence,

1 2 ρ V∞ S CL.max 2 The load factor may be take positive or negative values: L = nW =

1. POSITIVE LOAD FACTOR: During normal flight, the load factor is 1 g or greater than 1 g. Whenever the load factor is one or greater the load factor is defined as positive. 2. NEGATIVE LOAD FACTOR: - Under certain conditions, an abrupt deviation from the airplane’s equilibrium can cause an inertial acceleration that in turn will cause the weight to become greater than the lift. For example, during a stall, the load factor may be reduced towards zero. This will cause the pilot to feel “weightless”. A sudden and forceful elevator control movement forward can cause the load factor to move into a negative region. Both excessive deviation from positive and negative load factor limits must be avoided because of the possibility of exceeding the structural load limits of the airplane. Let us consider four cases: 1. Level flight 2. Turning Flight 3. Vertical Pull-Ups 4. Vertical Pull-Downs

2.10.1

Equations of Motion

To derive the equations of motion, let us consider an aircraft in flight inclined at and angle with the horizon. Assume that the aircraft is a rigid body on which all four forces are acting at the center of mass. The forces acting on the aircraft, as shown in Fig. 2.4, are 1. Lift (L) which acts perpendicular to the flight path. 2. Drag (D) which acts parallel to the forward velocity vector.

2.10. NORMAL LOAD FACTOR

57

L T

D

W

Figure 2.4: Forces acting on an aircraft. 3. Weight (W ) which acts vertically downwards. 4. Thrust (T ) which is generally inclined at an angle T to the flight path (it is assumed zero here) Let the flight path angle be θ. Hence, the equation of motion are X  m v˙ = T − D − W sin θ Fk = m v˙  X m v2 = m v θ˙ m v θ˙ = L − W cos θ F⊥ = r

2.10.2

Steady, Level Flight

Since we need a specific amount of lift in order to maintain straight and level flight, turn, climb or descend, we must be able to control the amount of lift produced. Furthermore, by steady we mean unaccelerated level flight and thus the flight path angle is zero. Hence the equations reduce to 0=T −D

0=L−W Recall the lift equation:

1 2 ρ V∞ S CL 2 Generally, the wing span area, S is fixed, the air density, rho, is determined by the altitude we are flying at (assume it cannot be changed). This leaves with two options the coefficient of lift and the speed. L=

In level flight, the load acting on the wings is equal to the ratio of lift and to weight: n= Consequently, the load factor equals “1 g”.

L W

(2.2)

2.10. NORMAL LOAD FACTOR

2.10.3

58

Level Turn: Pull-Up

O

r

L θ

L

W v

W

Figure 2.5: Forces acting on an aircraft during a vertical pull-up. Consider the case of an aircraft initially in steady, straight, level flight and then subjected to a sudden increase in lift. This will result in a curved flight path in the vertical plane, as shown in Fig. 2.5. Due to the centripetal acceleration,  2  v L= + cos θ W gr

Note that the maximum lift (and therefore the load factor) requirement will be at the bottom of the curve, L v2 (2.3) n= = +1 W gr

2.10.4

Level Turn: Pull-Down

Consider the case of an aircraft initially in steady, straight, level flight and then subjected to a decrease in lift. This will result in a curved flight path in the vertical plane, as shown in Fig. 2.5. Typically, an aircraft is initially in level flight and rolled into an inverted position so both lift and weight are point downwards. It can be shown that the load factor is n=

2.10.5

L v2 =1− W gr

(2.4)

Banked Turns

Consider the case of an aircraft initially in steady, straight, level flight and then making a level turn. This will result in a level flight path as shown in Fig. 2.7. From force balance in the vertical and horizontal directions, m v2 W = L cos φ, Fr = = L sin φ r

2.10. NORMAL LOAD FACTOR

59

v L W r O Figure 2.6: Forces acting on an aircraft during a vertical pull-down. Hence, it can be shown that the load factor is L n= = W

s

v4 −1 g r 2 2

(2.5)

2.11. REFERENCES

60

Figure 2.7: Forces acting on an aircraft during a banked turn.

2.11

References

Dole, Charles E. and Lewis, James E., Flight Theory and Aerodynamics: A Practical Guide for Operational Safety, Second Edition, May 2000, John Wiley and Sons. Keane, Andy and Nair, Prasanth, Computational Approaches for Aerospace Design: The Pursuit of Excellence, August 2005, John Wiley and Sons. Kuethe, Arnold M., and Chow, Chuen-Yen, Foundations of Aerodynamics: Bases of Aerodynamic Design, Fifth Edition, November 1997, John Wiley and Sons. http://en.wikipedia.org/wiki/Main Page http://www.grc.nasa.gov/WWW/K-12/airplane/guided.htm http://www.ndt.net/article/ecndt98/aero/001/001.htm

2.12. SUGGESTED PROBLEMS

2.12

61

Suggested Problems

Problem 2.1. What happens when lift is higher than the weight? 

Problem 2.2. What happens when lift is equal to the weight? 

Problem 2.3. What happens when lift is less than the weight? 

Problem 2.4. A model wing of constant chord length is placed in a low-speed subsonic wind tunnel, spanning the test section, The wing has a NACA 2412 airfoil and a chord length of 1.3 m. The flow in the test section is at a velocity of 50 m/s at standard sea-level conditions. If the wing is at 4◦ angle of attack, calculate: 1. The lift coefficient, cl ; the drag coefficient, cd ; the moment coefficient, cm.c/4 . (Hint: use tables for the NACA airfoil given, use Reynolds number to determine cd ). 2. Determine the lift, drag, and moments about the quarter chord, per unit span. 3. Determine the lift to drag ration (L/D). 

Chapter 3 Load Analysis

Instructional Objectives of Chapter 3 After completing this chapter, the student should be able to: 1. Fully understand the importance of units. 2. Determine load analysis of airplane structural components under both static and dynamic loading. 3. Draw load diagrams using common analytical and discrete solutions.

This chapter presents a brief review of Newton’s laws and Euler’s equations as applied to dynamicallyloaded and steady-loaded systems in 3-D. The concepts and methods used in this chapter are usually presented in previous statics and dynamics courses. Students are encouraged to review their static and dynamic course contents.

3.1

Newton’s Laws

Most of the problems in structural analysis deal with static and dynamic analyses. In fact, static loading can be considered as a special case of the dynamic one. The most popular method for the dynamic analysis is the Newtonian approach based on Newton’s laws and is generally used to obtain information about internal forces. The three Newton’s Laws can be briefly summarized as follows: Newton’s First Law Newton’s First Law states that a body at rest tends to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force. Newton’s Second Law Newton’s Second Law states that the time rate of change of momentum of a body is equal to the magnitude of the applied force and acts in the direction of the force.

62

3.2. UNITS

63

Newton’s Third Law Newton’s Third Law states that when two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line, but have opposite sense.

3.2

Units

In all engineering problems, we must deal with units carefully. Each parameter in the problem may have a specific unit system. A unit may be defined as a specified amount of a physical quantity by which through comparison another quantity of the same kind is measured. It our job to ensure that we are working in the proper unit system and make the corresponding conversions, should it be necessary.

3.2.1

Importance of Units

Equations from physics and engineering that relate physical quantities are dimensionally homogeneous. Dimensionally homogeneous equations must have the same dimensions for each term. Newton’s second law relates the dimensions force, mass, length, and time: F

α

[F ] =

ma [M ] [L]

(3.1)

2

[T ]

If length and time are primary dimensions, Newton’s second law, being dimensionally homogeneous, requires that both force and mass cannot be primary dimensions without introducing a constant of proportionality that has dimension (and units). Because physical quantities are related by laws and definitions, a small number of physical quantities, called primary dimensions, are sufficient to conceive of and measure all others. Primary dimensions in all systems of dimensions in common use length and time. Force is selected as a primary dimension in some systems. Mass is taken as a primary dimension in others. For application in mechanics, we have four basic systems of dimensions: 1. force [F ], mass [M ], length [L], time [T ] 2. force [F ], length [L], time [T ] 3. mass [M ], length [L], time [T ] In system 1, length [L], time [T ], and both force [F ] and mass [M ] are selected as primary dimensions. In this system, in Newton’s second law ma F = gc

3.2. UNITS

64

where the constant of proportionality, gc , is not dimensionless. For Newton’s law to be dimensionally homogeneous, the dimensions of gc must be: gc =

[M ] [L] 2

[F ] [T ]

In system 2, mass [M ] is a secondary dimension, and in Newton’s second law the constant of proportionality is dimensionless. In system 3, force [F ] is a secondary dimension, and in Newton’s second law the constant of proportionality is again dimensionless. The measuring units selected for each primary physical quantities determine the numerical value of the constant of proportionality. Secondary dimensions are those quantities measured in terms of the primary dimensions. For example, if mass, length, and time are primary dimensions, area, density, and velocity would be secondary dimensions.

3.2.2

Systems of Units

Four different systems of units can be identified: 1. Syst` eme International d’Unit` es (SI) mass : kilogram (kg) length : meter (m) temperature : Celsius (◦ C) or Kelvin (◦ K) time : second (s) force : newton (N) 2. English Engineering

mass : pound mass (lbm) length : feet (ft) temperature : Rankine (◦ R) time : second (s) force : pound force (lb or lbf)

3. British Engineering: foot-pound-second (fps) mass : slug (slug) length : feet (ft) temperature : Fahrenheit (◦ F) time : second (s) force : pound force (lb)

3.2. UNITS

65

4. British Engineering: inch-pound-second (ips) mass : slug (slug) length : inch (in) temperature : Fahrenheit (◦ F) time : second (s) force : pound force (lb)

3.2. UNITS

66

Example 3.1. A special payload package is to be delivered to the surface of the moon. A prototype of the package, developed, constructed, and tested near Boston, has been determined to have a mass of 24.0 kg. Assume gBoston = 9.77 m/sec2 and gmoon = 1.7 m/sec2 . Show all your work. 3.1a) Estimate the weight of the package, using the international system, as measured near Boston. 3.1b) Estimate the weight of the package, using the international system, on the surface of the moon. 3.1c) Reexpress the weights using fps and ips systems.

Let’s use Eq. (3.1), F = ma

→

W = mg

(3.2)

Thus, 3.1a) Estimate the weight of the package, using the international system, as measured near Boston.     kg–m WBoston = m gBoston = 24.0 kg · 9.77 m/sec2 = 234.48 = 234.48 N sec2 3.1b) Estimate the weight of the package, using the international system, on the surface of the moon.     kg–m = 40.8 N Wmoon = m gmoon = 24.0 kg · 1.7 m/sec2 = 40.8 sec2 3.1c) Reexpress the weights using fps and ips systems. Using Tables,    1 lb  WBoston = 234.48 N · = 52.72 lb in both units 4.448 N    1 lb  Wmoon = 40.8 N · = 9.17 lb in both units 4.448 N Alternative approach:

3.2. UNITS

67

Note that the mass can be expressed as follows,    2.21 lbm   1.0 slug  lb–sec2 · = 1.65 slug = 1.65 mf ps = 24.0 kg · 1.0 kg 32.17 lbm ft mips =



lb–sec2 1.65 ft

  lb–sec2 1.0 ft  = 0.1375 · 12.0 in in

and the gravitational constant at Boston and at the moon are  m   1000.0 mm   1.0 in  in gBostonips = 9.77 · = 384.65 · 1.0 m 25.4 mm sec2 sec2

 ft in   1 ft  gBostonf ps = 384.65 2 · 12 in = 32.05 sec sec2

 in m   1000.0 mm   1.0 in  · = 66.93 gmoonips = 1.7 2 · 1.0 m 25.4 mm sec sec2

 ft in   1 ft  · = 5.58 gmoonf ps = 66.93 12 in sec2 sec2

Thus we could have also obtained the results by,

   ft  slug–ft Wmoon = 1.65 slugs · 5.58 = 9.20 lb 2 = 362.70 sec sec2  lb–sec2   in  Wmoon = 0.1375 = 9.20 lb · 66.93 in sec2    slug–ft ft  = 52.88 lb WBoston = 1.65 slugs · 32.05 2 = 52.88 sec sec2  lb–sec2   in  WBoston = 0.1375 · 384.65 = 52.88 lb in sec2 End Example 

3.3. LOAD ANALYSIS

3.3

68

Load Analysis

Many problems deal with constant velocity, or zero velocity (static), in such cases Newton’s Second Law reduces to: X X X Fx = 0 Fy = 0 Fz = 0 X

X

Mx = 0

X

My = 0

(3.3)

Mz = 0

Note that the above is just a special case of the dynamic loading situation but with zero accelerations.

3.3.1

Internal Force Sign Convention

Here we will always assumes all unknown forces and moments on the system to be positive in sign as shown in Figure 3.1, regardless of what one’s intuition or an inspection of the free-body diagram might indicate as to their probable directions. However, all known force components are given their proper signs to define their directions. The simultaneous solution of the set of equations that results will cause all the unknown components to have the proper signs when the solution is complete. If the loads act on the opposite direction it results in a sign reversal on that component in the solution.

Myy

Positively-oriented surface

y Vy Nxx x z

Mzz

Vz

Vz

Mxx Mxx

Mzz

Nxx Vy

Negatively-oriented surface

Myy

Figure 3.1: Positive sign convention.

We will need to apply the second law in order to solve for the forces on assemblies of elements that act upon one another. The six equations can be written in a 3-D system. In addition, as many (thirdlaw) reaction force equations as are necessary will be written and the resulting set of equations solved simultaneously for the forces and moments. The number of second-law equations will be up to six times the number of individual parts in a three-dimensional system (plus the reaction equations), meaning that even simple systems result in large sets of simultaneous equations. The reaction (third-law) equations are often substituted into the second-law equations to reduce the total number of equations to be solved simultaneously.

3.3. LOAD ANALYSIS

69

Example 3.2. For the given problem obtain all the reactions at point O (clamped-end). Use the shown sign convention (it is similar to the one used on class). The loads P and T act in the x-y plane. The length of bar CB is L, of bar BA is 2 L, and of bar OA is 3 L.

y

y3

z

z3 x3

O

3L

x

A

2L x2 y1 C

x1

y2

B

4

z2 z1

P 3

L

T

Figure 3.2: Three-dimensional bar-structure.

3.3. LOAD ANALYSIS

70

3.2a) Draw free-body diagrams of the each section OA, BA, CB. Mxx3

y3 z3

Mzz3 O

x3 Myy3

Myy3

Vy3

Mzz3

A Nxx3

Vz3

Mxx3

Mxx2

Vy2 A Nxx2 Myy2 Vz2 Mzz2

x2

Mzz2 y2

B Vz2

Vy2

Myy2

z2

Nxx2

Mxx2

Myy1 y1

Vy1

x1

C

Nxx1

4

z1 P

3

Mxx1

Vz1 T

B Mzz1

Figure 3.3: Free body diagrams for the three-dimensional bar-structure.

3.3. LOAD ANALYSIS

71

3.2b) Obtain the internal loads at B. First of all, the coordinate system can be changed from bar to bar whenever we are consistent are clear about what we are doing. In this context, let us use a local coordinate system such that x-axis always goes along the main axis of the bar. In order to do so, let us use subscript“1” to refer to the first bar, “2” for the second bar and “3” for the third bar. This will avoid any confusion as to what coordinate system we are working with. Next, we proceed to find the loads at B: +↑

X

Fy = 0

⇒

Vy1 (x1 ) +

4 P =0 5

4 Vy1 (x1 ) = − P 5 4 Vy1 (x1 = L) = − P 5

→ −X + Fx = 0

⇒

Nxx1 (x1 ) + 0 = 0

Nxx1 (x1 ) = 0 Nxx1 (x1 = L) = 0 ++

X

Fz = 0

⇒

Vz1 (x1 ) +

3 P =0 5

3 Vz1 (x1 ) = − P 5 3 Vz1 (x1 = L) = − P 5

3.3. LOAD ANALYSIS

72

X

+

MyB = 0

⇒

Myy1 (x1 ) +

3 P x1 = 0 5

3 Myy1 (x1 ) = − P x1 5 3 Myy1 (x1 = L) = − P L 5

X

+

MxB = 0

⇒

Mxx1 (x1 ) − T = 0

Mxx1 (x1 ) = T Mxx1 (x1 = L) = T +

X

MzB = 0

⇒

Mzz1 (x1 ) −

Mzz1 (x1 ) =

4 P x1 = 0 5

4 P x1 5

Mzz1 (x1 = L) =

4 P L 5

3.2c) Obtain the internal loads at A. From action reaction at B: 3 Mxx2 (x2 = 0) = Myy1 (x1 = L) = − P L 5 Myy2 (x2 = 0) = −Mxx1 (x1 = L) = −T Mzz2 (x2 = 0) = Mzz1 (x1 = L) =

4 PL 5

4 Nxx2 (x2 = 0) = Vy1 (x1 = L) = − P 5 Vy2 (x2 = 0) = −Nxx1 (x1 = L) = 0 3 Vz2 (x2 = 0) = Vz1 (x1 = L) = − P 5

3.3. LOAD ANALYSIS

73

Next, we proceed to find the loads at A: X +↑ Fy = 0 ⇒ −Vy2 (x2 = 0) + Vy2 (x2 ) = 0 Vy2 (x2 ) = Vy2 (x2 = 0) Vy2 (x2 = 2 L) = Vy2 (x2 = 0) = −Nxx1 (x1 = L) = 0 → −X + Fx = 0

⇒

−Nxx2 (x2 = 0) + Nxx2 (x2 ) = 0

Nxx2 (x2 ) = Nxx2 (x2 = 0) 4 Nxx2 (x2 = 2 L) = Nxx2 (x2 = 0) = Vy1 (x1 = L) = − P 5 ++

X

Fz = 0

⇒

−Vz2 (x2 = 0) + Vz2 (x2 ) = 0

Vz2 (x2 ) = Vz2 (x2 = 0) 3 Vz2 (x2 = 2 L) = Vz2 (x2 = 0) = Vz1 (x1 = L) = − P 5

3.3. LOAD ANALYSIS

+

74

X

MyA = 0

⇒

−Myy2 (x2 = 0) + Myy2 (x2 ) − Vz2 (x2 = 0) x2 = 0

Myy2 (x2 ) = Myy2 (x2 = 0) + Vz2 (x2 = 0) x2 Myy2 (x2 ) = −T −

3 P x2 5

Myy2 (x2 = 2 L) = −T − +

X

MxA = 0

⇒

6 PL 5

−Mxx2 (x2 = 0) + Mxx2 (x2 ) = 0

Mxx2 (x2 ) = Mxx2 (x2 = 0) = Myy1 (x1 = L) 3 Mxx2 (x2 = 2 L) = − P L 5 +

X

MzA = 0

⇒

−Mzz2 (x2 = 0) + Mzz2 (x2 ) + Vy2 (x2 = 0) x2 = 0

Mzz2 (x2 ) = Mzz2 (x2 = 0) + Vy2 (x2 = 0) x2 Mzz2 (x2 ) =

4 P L+0 5

Mzz2 (x2 = 2 L) =

4 PL 5

3.2d) Obtain the internal loads at O. From action reaction at A: 4 Mxx3 (x3 = 3 L) = −Mzz2 (x2 = 2 L) = − P L 5 Myy3 (x3 = 3 L) = −Mxx2 (x2 = 2 L) =

3 PL 5

Mzz3 (x3 = 3 L) = −Myy2 (x2 = 2 L) = T + Nxx3 (x3 = 3 L) = −Vz2 (x2 = 2 L) =

3 P 5

Vy3 (x3 = 3 L) = −Nxx2 (x2 = 2 L) = Vz3 (x3 = 3 L) = Vy2 (x2 = 2 L) = 0

6 PL 5

4 P 5

3.3. LOAD ANALYSIS

75

Next, we proceed to find the loads at O: X +↑ Fy = 0 ⇒ −Vy3 (x3 = 3 L) + Vy3 (x3 ) = 0 Vy3 (x3 = 3 L) = Vy3 (x3 ) Vy3 (x3 = 3 L) = Vy3 (x3 = 0) = −Nxx2 (x2 = 2 L) = → −X + Fx = 0

⇒

4 P 5

−Nxx3 (x3 = 3 L) + Nxx3 (x3 ) = 0

Nxx3 (x3 = 3 L) = Nxx3 (x3 ) Nxx3 (x3 = 3 L) = Nxx3 (x3 = 0) = −Vz2 (x2 = 2 L) = ++

X

Fz = 0

⇒

−Vz3 (x3 = 3 L) + Vz3 (x3 ) = 0

Vz3 (x3 = 3 L) = Vz3 (x3 ) Vz3 (x3 = 3 L) = Vz3 (x3 = 0) = Vy2 (x2 = 2 L) = 0

3 P 5

3.3. LOAD ANALYSIS

+

X

76

MyO = 0

⇒

−Myy3 (x3 = 3 L) + Myy3 (x3 ) − Vz3 (x3 = 3 L) x3 = 3 L

Myy3 (x3 ) = Myy3 (x3 = 3 L) + Vz3 (x3 = 3 L) x3 Myy3 (x3 ) =

3 P L+0 5

Myy3 (x3 = 0) = +

X

MxO = 0

⇒

3 PL 5

−Mxx3 (x3 = 3 L) + Mxx3 (x3 ) = 0

Mxx3 (x3 = 3 L) = Mxx3 (x3 ) = −Mzz2 (x2 = 2 L) 4 Mxx3 (x3 = 3 L) = Mxx3 (x3 = 0) = − P L 5 +

X

MzO = 0

⇒

−Mzz3 (x3 = 3 L) + Mzz3 (x3 ) + Vy3 (x3 = 3 L) x3 = 3 L

Mzz3 (x3 ) = Mzz3 (x3 = 3 L) + Vy3 (x3 = 3 L) x3 Mzz3 (x3 ) = T +

6 4 P L + P x3 5 5

Mzz3 (x3 = 0) = T +

18 PL 5

End Example 

3.4.

LOAD DIAGRAMS

3.4

77

Load Diagrams

In the most general case, a structural component may have all type of loadings: torsional, bending and axial. Before we proceed, let us discuss three different sign conventions, which are typically used.

3.4.1

Sign Conventions

Stress Convention In this course, problems will be solved using the following sign convention

y Vy

py(x)

Vy+dVy x O

Mzz

Mzz+dMzz dx

x

Figure 3.4: Equilibrium element supporting a general force system under the stress convention in the x-y plane.

Sum of forces in the y-direction, will give us an equation for the shear: X +↑ Fy = 0 ⇒ −Vy (x) + {Vy (x) + dVy (x)} + py (x) dx = 0

divide by dx and take limdx→0

dVy (x) dx

= −py (x)

(3.4)

Note that we can integrate the above equation over the domain where shear is interested: Z Vy (x) = − py (x) dx + Vy0 Sum of moment at O, will give us an equation for the moment: +

X

Mz = 0

⇒

Vy (x)dx + {Mzz (x) + dMzz (x)} − Mzz (x) − py (x) dx

dx =0 2

(3.5)

3.4.

LOAD DIAGRAMS

78

divide by dx and take limdx→0 d Mzz (x) dx

= −Vy (x)

(3.6)

Note that we can integrate the above equation over the domain where moment is interested: Z Mzz (x) = − Vy (x) dx + Mzz0

(3.7)

where Mzz0 is found from boundary conditions.

Structural Convention Problems can be solved using the following sign convention

y Vy

py(x)

Vy+dVy x O

Mzz

Mzz+dMzz dx

x

Figure 3.5: Equilibrium element supporting a general force system under the structural convention in the x-y plane

Sum of forces in the y-direction, will give us an equation for the shear: X +↑ Fy = 0 ⇒ Vy (x) − {Vy (x) + dVy (x)} + py (x) dx = 0

divide by dx and take limdx→0

dVy (x) dx

=

py (x)

(3.8)

Note that we can integrate the above equation over the domain where shear is interested: Z Vy (x) = py (x) dx + Vy0 Sum of moment at O, will give us an equation for the moment: +

X

Mz = 0

⇒

−Vy (x) dx + {Mzz (x) + dMzz (x)} − Mzz (x) − py (x) dx

dx =0 2

(3.9)

3.4.

LOAD DIAGRAMS

79

divide by dx and take limdx→0 d Mzz (x) dx

=

Vy (x)

(3.10)

Note that we can integrate the above equation over the domain where moment is interested: Z Mzz (x) = Vy (x) dx + Mzz0

(3.11)

where Mzz0 is found from boundary conditions.

Elasticity Convention Problems can be solved using the following sign convention

y Vy

py(x)

Vy+dVy x O

Mzz

Mzz+dMzz dx

x

Figure 3.6: Equilibrium element supporting a general force system under the elasticity convention in the x-y plane

Sum of forces in the y-direction, will give us an equation for the shear: X +↑ Fy = 0 ⇒ −Vy (x) + {Vy (x) + dVy (x)} + py (x) dx = 0

divide by dx and take limdx→0

dVy (x) dx

= −py (x)

(3.12)

Note that we can integrate the above equation over the domain where shear is interested: Z Vy (x) = − py (x) dx + Vy0 Sum of moment at O, will give us an equation for the moment: +

X

Mz = 0

⇒

Vy (x) dx − {Mzz (x) + dMzz (x)} + Mzz (x) − py (x) dx

dx =0 2

(3.13)

3.4.

LOAD DIAGRAMS

80

divide by dx and take limdx→0 d Mzz (x) dx

=

Vy (x)

(3.14)

Note that we can integrate the above equation over the domain where moment is interested: Z Mzz (x) = Vy (x) dx + Mzz0

(3.15)

where Mzz0 is found from boundary conditions.

3.4.2

Linear Differential Equations of Equilibrium

Consider a small differential element dx and construct a free body diagram with the actual stress distributions replaced by their statically equivalent internal resultants. Thus using the stress convention and applying Newton’s Second Law the differential equations for equilibrium are found as: dVy = −py (x) dx

dNxx = −px (x) dx dMxx = −mx (x) dx

dVz = −pz (x) dx

dMyy = −my (x) + Vz dx

(3.16)

dMzz = −mz (x) − Vy dx

where px (x) is the distributed load in the axial direction (x-axis), py (x) the distributed load in the transverse direction (y-axis), pz (x) the distributed load in the lateral direction (z-axis), mx (x) the distributed moments about the x-axis, my (x) the distributed moments about the y-axis, and mz (x) the distributed moments about the z-axis. These equations are the first order ordinary differential equations that may be solved by direct integration. The solution to these equations is: Z x Nxx (x) = Nxx (x1 ) − px (ζ) dζ (3.17) x1

Vy (x) = Vy (x1 ) − Vz (x) = Vz (x1 ) −

Z

Z

Mzz (x) = Mzz (x1 ) −

Z

Z

x

x1 x

x1

py (ζ) dζ

(3.18)

pz (ζ) dζ

(3.19)

x1 x

x1

Mxx (x) = Mxx (x1 ) − Myy (x) = Myy (x1 ) −

x

Z

x

mx (ζ) dζ

(3.20)

n o my (ζ) − Vz (ζ) dζ

(3.21)

x1

n o mz (ζ) + Vy (ζ) dζ

(3.22)

The first term on the right-hand side of the above equations are known as the boundary conditions; i.e.,

3.4.

LOAD DIAGRAMS

81

if the beam is statically determinate there will exist some point along the x-axis x = x1 at which the resultants are known. For the case of statically indeterminate, the boundary conditions may be found using compatibility equations.

Example 3.3. Aerospace engineers have idealized an aircraft structural component using the beam model as shown in Fig. 3.7. The cantilever beam’s squared cross section is uniform. These engineers need your help to analyze this component. Take a = 25 mm, b = 5 mm. Use the stress convention and show all your steps.

y

y

100 N/m 100 N/m

x z

z

a a

1000 N L

1000 N

b

b

Cross-sectional view Figure 3.7: Machine component for example below. 3.3a) Obtain axial load equation for Nxx (x) and shear equations for Vy (x) and Vz (x). First obtain the reactions at the fixed end: (used positive stress convention discussed in class) y VzR MxxR

MzzR 100 N/m

NxxR z

x VyR

MyyR

L

1000 N

3.4.

LOAD DIAGRAMS

82

The internal shear loads at the fixed end are (all in newtons, assuming L in meters) +↑ +→ +→

X

X

X

Fy = 0 →

−VyR − qo L = −VyR − 100 L = 0 →

Fx = 0 →

VyR = −100 L

−NxxR + 0 = 0 → NxxR = 0

Fz = 0 →

−VzR + P = −VzR + 1000 = 0 →

VzR = 1000

The internal moments at the fixed end are (all in N–m, assuming L in meters) + + +

X

X

X

My = 0 → Mx = 0 → Mz = 0 →

−MyyR − P L = −MyyR − 1000 L = 0 → MyyR = −1000 L −MxxR − P a = −MxxR − 1000 (0.025) = 0 → MxxR = −25 −MzzR + qo L

  L2 L =0 → = −MzzR − 100 2 2

MzzR = −50 L2

Now let us make an arbitrary cut at a distance x (used positive stress convention discussed in class) y VzR MxxR

Myy(x)

MzzR 100 N/m

Vy(x)

NxxR z

x VyR

MyyR

Nxx(x) Mxx(x)

Vz(x) x

Mzz(x)

The internal shear loads at the a distance x are (all in newtons, assuming L in meters) +↑

X

Fy = 0 →

−VyR + Vy (x) +

Z

x

py (ζ) dζ = 0

0

Vy (x) = −

Z

=−

Z

x

py (ζ) dζ + VyR

0

0

x

(−100) dζ − 100 L

= 100 x − 100 L n xo Vy (x) = 100 L −1 + L

3.4.

LOAD DIAGRAMS

83

+→

X

Fx = 0 →

−NxxR + Nxx (x) = 0 Nxx (x) = NxxR Nxx (x) = 0

X

+↑

Fz = 0 →

−VzR + Vz (x) = 0 Vz (x) = VzR Vz (x) = 1000

3.3b) Obtain moment equations for Mxx (x), Myy (x) and Mzz (x). The internal moments at the a distance x are (all in N–m, assuming L in meters) +

X

My = 0 →

Z

−MyyR + Myy (x) − −(−1000 L) + Myy (x) −

x

Vz (ζ) dζ = 0

0

Z

x

(1000) dζ = 0

0

1000 L + Myy (x) − 1000 x = 0

n xo Myy (x) = 1000 L −1 + L

+

X

Mx = 0 →

−MxxR + Mxx (x) = 0 25 + Mxx (x) = 0

Mxx (x) = −25 +

X

Mz = 0 →

−MzzR + Mzz (x) +
− −50 L2 + Myy (x) +

Z

x

Z

x

Vy (ζ) dζ = 0

0

(−100 L + 100 ζ) dζ = 0

0

50 L2 + Myy (x) − 100 L x + 50 x2 = 0   x   x 2  − Mzz (x) = 50 L2 −1 + 2 L L

3.3c) Plot all axial, shear, and moment equations.

In general, it is convenient to plot nondimensional quantities. Thus let the length be

3.4.

LOAD DIAGRAMS

84

normalize to one: η=

x L

06η61

and the nondimensional loads are: ¯ xx (η) = Mxx (x) = −25 M 1

¯xx (η) = 0 N

¯ yy (η) = Myy (x) = −1000 + 1000 η M L

Vy (x) V¯y (η) = = −100 + 100 η L

¯ zz (η) = Mzz (x) = −50 + 100 η − 50 η 2 M L2

Vz (x) = 1000 V¯z (η) = 1

3.4.

LOAD DIAGRAMS

85

Nxx(η) 1

0.5

η 0.2

0.4

0.6

0.8

1

-0.5

-1

Figure 3.8: Dimensionless axial load distribution.

Vy(η) 100

50

η 0.2

0.4

0.6

0.8

1

-50

-100

Figure 3.9: Dimensionless shear (in the y–axis) load distribution.

3.4.

LOAD DIAGRAMS

86

Vz(η) 1000

500

η 0.2

0.4

0.6

0.8

1

-500

-1000

Figure 3.10: Dimensionless shear (in the z–axis) load distribution.

Mxx(η) 20

10

η 0.2

0.4

0.6

0.8

1

-10

-20

Figure 3.11: Dimensionless torsional load distribution.

3.4.

LOAD DIAGRAMS

87

Myy(η) 1000

500

η 0.2

0.4

0.6

0.8

1

-500

-1000

Figure 3.12: Dimensionless moment (about the y–axis) distribution.

Mzz(η) 40

20

η 0.2

0.4

0.6

0.8

1

-20

-40

Figure 3.13: Dimensionless moment (about the z–axis) distribution. End Example 

3.4.

LOAD DIAGRAMS

88

Example 3.4.

Consider an idealization of the helicopter blade, show in Fig. 3.14, subject to the loading shown in Fig. 3.15. The following data is given:

y′

2 0.5 in 3 in 0.5 in

1

Hallow

4

z′

3 2 in

10 in

5 in

Cross-section Figure 3.14: Cross-section of the helicopter blade.

Figure 3.15: Loading on the helicopter blade.

3.4.

LOAD DIAGRAMS

89

L(x, z) = 4.0

 x 2 L

psi

D(x) = 4.0

 x 2

lb/in

L

P = 10000 lb

where L(x, z) is a pressure applied to the bottom surface. The total length of the beam is L = 200 in. Resolve all loads at the modulus-weighted centroid1 (as a function of x): (yc∗ = 0, zc∗ =5.767 in). First of all we concentrate the pressure load to a distributed load: x Fixed-end

pz(x)

200 in y′

z′

P

10 in py(x)

Figure 3.16: Replacing the pressure g(x, z) with a distributed load py (x).

py (x) =

Z

10

L(x, z) dz = 40.0

0

1 This

will be discussed in detail in the chapter 4.

 x 2 L

lb/in

3.4.

LOAD DIAGRAMS

90

Now move all loads to the modulus-weighted centroid:

x Fixed-end

pz(x)

mx(x) 200 in

yc

Mo

zc

P

10 in py(x)

Figure 3.17: Locating all loads at the weighted-modulus centroid. where, mx (x) = − (7 − zc∗ ) py (x) = −0.001233 x2 lb-in/in Thus the loads are:

Mo = 10000(5.767) = 57670.0 lb-in

px (x) = 0 py (x) = 40.0 pz (x) = 4.0

 x 2

 x 2 L

mx (x) = −49.32 my (x) = 0 mz (x) = 0

lb/in

L

lb/in

 x 2 L

lb-in/in

3.4.

LOAD DIAGRAMS

91

The loads at x = x1 = 0 come from equilibrium of a differential element just after x = 0. Using the stress convention, we get: = 10000 lb − 10000 = 0 → Nxx (x) Nxx (x) x1 =0

x1 =0

Vy (x) Vz (x)

Mxx (x)

Myy (x)

→

+0=0

→

+0=0

→

+ 57670 = 0

→

+0=0

→

x1 =0

x1 =0

x1 =0

x1 =0

Mzz (x)

+0=0

x1 =0

Vy (x)

x1 =0

Vz (x)

x1 =0

Mxx (x)

Myy (x)

Mzz (x)

x1 =0

x1 =0

x1 =0

=0

=0 =0 = −57670 lb-in =0

Now we proceed to obtain the internal loads. Integrating to obtain the axial internal load; Z

Nxx (x) = Nxx (x1 ) −

Vy (x) = Vy (x1 ) −

Vz (x) = Vz (x1 ) −

Z Z

x

px (ζ) dζ = Nxx (0) − 0

x

py (ζ) dζ = Vy (0) − 0

x

pz (ζ) dζ = Vz (0) − 0

Mxx (x) = Mxx (x1 ) −

Z

0

Z Z

Z x

0

x 0

x

(0) dζ = Nxx (0) = 10000 lb 0

(

 2 ) ζ 40.0 dζ = −0.000333333 x3 lb L

(

 2 ) ζ 4.0 dζ = −0.0000333333 x3 lb L

x

mx (ζ) dζ = Mxx (0) −

Z

x 0



−0.001233 ζ 2 dζ

= 0.000411 x3 lb-in Z xn Z x o Myy (x) = Myy (x1 ) − my (ζ) − Vz (ζ) dζ = Myy (0) − (0 − (−0.0000333333 ζ 3 )) dζ 0

0

= −57670 − 0.0000833333 x4 lb-in Z xn Z x o Mzz (x) = Mzz (x1 ) − mz (ζ) + Vy (ζ) dζ = Mzz (0) − (0 + (−0.000333333 ζ 3 )) dζ 0

0

= 0.0000833333 x4 lb-in

Note that in the above equations, x is measured in inches. End Example 

3.4.

LOAD DIAGRAMS

92

3.5. DISCRETE LOAD DIAGRAMS

3.5

93

Discrete Load Diagrams

For most aircraft, an analytical load expression may not available. The only information we may have is the experimental data obtained from sensors located throughout the aircraft. For such cases we can no longer obtain close-form load diagrams, but we have to use numerical techniques to obtain the load diagrams. The method explained here can also be used when the analytical expression is available but using a numerical method is desired. First, we need an array with all locations where the data is measured. For an example, suppose we want 10 intervals for the previous example, then we use the following locations x = {0, 20, 40, 60, . . . , 180, 200}

m

This is converted into small intervals: ∆x = {20, 20, 20, . . . , 20, 20}

m

In this example all intervals have the same interval, but they could be different. The smaller the ∆xi the better the approximation. At each location we calculate the distributed loads: px (xi )

py (xi )

pz (xi )

mx (xi )

my (xi )

mz (xi )

The next example will illustrate the approach.

Example 3.5.

Consider the idealized helicopter blade od Example 3.4. Use five interval elements approximation, to determine the load diagrams. From Example 3.4, we found that the loads acting on the helicopter blade are: px (x) = 0 mx (x) = −0.001233 x2 lb-in/in

py (x) = 0.001 x2 lb/in my (x) = 0

pz (x) = 0.0001 x2 lb/in mz (x) = 0

Since we want five element approximation, let us divide the interval of 0 < x < 200 into identical five elements: xroot − xtip 200 − 0 ∆x = = = 40 in 5 5

3.5. DISCRETE LOAD DIAGRAMS

94

Hence, ∆x1 = ∆x2 = ∆x3 = ∆x4 = ∆x5 = 40 The locations are

x1 = 0 x2 = 40 x3 = 80 x4 = 120 x5 = 160 x6 = 200

Let us proceed to obtain the discrete distributed loads at each location xi : x1 = 0

px1 = px (x1 ) = 0

py1 = py (x1 ) = 0.0

pz1 = pz (x1 ) = 0.00

x2 = 40

px2 = px (x2 ) = 0

py2 = py (x2 ) = 1.6

pz2 = pz (x2 ) = 0.16

x3 = 80

px3 = px (x3 ) = 0

py3 = py (x3 ) = 6.4

pz3 = pz (x3 ) = 0.64

x4 = 120

px4 = px (x4 ) = 0

py4 = py (x4 ) = 14.4

pz4 = pz (x4 ) = 1.44

x5 = 160

px5 = px (x5 ) = 0

py5 = py (x5 ) = 25.6

pz5 = pz (x5 ) = 2.56

x6 = 200

px6 = px (x6 ) = 0

py6 = py (x6 ) = 40.0

pz6 = pz (x6 ) = 4.00

Let us proceed to obtain the discrete distributed moments at each location xi : x1 = 0

mx1 = mx (x1 ) = 0.0000

my1 = my (x1 ) = 0

mz1 = mz (x1 ) = 0

x2 = 40

mx2 = mx (x2 ) = −1.9728

my2 = my (x2 ) = 0

mz2 = mz (x2 ) = 0

mx3 = mx (x3 ) = −7.8912

my3 = my (x3 ) = 0

mz3 = mz (x3 ) = 0

mx4 = mx (x4 ) = −17.7552

my4 = my (x4 ) = 0

mz4 = mz (x4 ) = 0

mx5 = mx (x5 ) = −31.5648

my5 = my (x5 ) = 0

mz5 = mz (x5 ) = 0

mx6 = mx (x6 ) = −49.3200

my6 = my (x6 ) = 0

mz6 = mz (x6 ) = 0

x3 = 80 x4 = 120 x5 = 160 x6 = 200

The tip load values are P = 10000 lb

Mo = 10000(5.767) = 57670 lb-in

Now in order to calculate the axial load along the wing’s major axis, let us use Simpson’s integration rule. We start we the solution of the axial differential equation: Nxx (x) = Nxx (0) −

Z

0

x

px (ζ) dζ

→

Nxx (xj ) ≈ Nxx0 −

j X i=1

px,avei ∆xi

3.5. DISCRETE LOAD DIAGRAMS

95

Hence, we need to obtain the average values for each distributed load: px,ave1 = px,ave2 = px,ave3 = px,ave4 = px,ave5 = Thus

px1 + px2 2 px2 + px3 2 px3 + px4 2 px4 + px5 2 px5 + px6 2

=0 =0 =0 =0 =0

∆Nxx1 = −px,ave1 ∆x1 = 0

∆Nxx2 = −px,ave2 ∆x2 = 0

∆Nxx3 = −px,ave3 ∆x3 = 0

∆Nxx4 = −px,ave4 ∆x4 = 0

∆Nxx5 = −px,ave5 ∆x5 = 0

Hence,

Nxx0 = Nxx (0) = P = 10000

In nondimensional form,

x1 = 0

Nxx1 = Nxx0 = 10000

x2 = 40

Nxx2 = ∆Nxx1 + Nxx1 = 10000

x3 = 80

Nxx3 = ∆Nxx2 + Nxx2 = 10000

x4 = 120

Nxx4 = ∆Nxx3 + Nxx3 = 10000

x5 = 160

Nxx5 = ∆Nxx4 + Nxx4 = 10000

x6 = 200

Nxx6 = ∆Nxx5 + Nxx5 = 10000 ¯xx = Nxxi N i P ¯xx = 1 η1 = 0.0 N 1 ¯xx = 1 η2 = 0.2 N 2 ¯xx = 1 η3 = 0.4 N

η=

x L

3

η4 = 0.6 η5 = 0.8 η6 = 1.0

¯xx = 1 N 4 ¯xx = 1 N 5 ¯xx = 1 N 6

Compare to the exact nondimensional equation ¯xx (η) = Nxx (x) = 1 N P The following plot shows the discrete method (points in red) and the exact solution (line in blue):

xx p1a = ListPlotATransposeA9 , Nxd=E, PlotStyle → [email protected], [email protected] 0 is applied through the shear center then no torsion exists and the shear flow distribution for this problem is given by

qi (s) = qoi +



A∗s

Q∗zs

Q∗ys

 1  A∗    0     0

0 −

1 ∗ Ryz

1 ∗ Rzz

0 −

1 ∗ Ryy

1 ∗ Ryz



 0       Vz       −Vy

          

where qoi is found by evaluating qi (s = 0). Substituting the cross-sectional properties in the above equation, we get     qi (si ) = qoi + 0.117658 Q∗ysi + 1.4687 Q∗zsi Vy + 92.5559 Q∗ysi + 0.117658 Q∗zsi Vz

(8.12) In order to calculate the shear flow in each flange, we need to first find the yi (si ) and zi (si ) about the centroid for each part.

8.1. THIN-WALLED BEAM SHEAR IN OPEN SECTIONS

516

Following the assumed flow convention in Fig. 8.9 and from the geometry of the crosssection we get: (about the centroid) For flange 32, s32 = 0 s32 = 0.25

⇒ z32 = 1.21651 − zc = 0.286553;

⇒ z32 = d − zc = 0.0700462 ;

y32 = 3.125 − yc = 1.54044

y32 = h − yc = 1.41544

For flange 12, s12 = 0 s12 = 0.5

⇒ z12 = 0.566987d − zc = −0.362967 ; ⇒ z12 = d − zc = 0.0700462 ;

y12 = 2.75 − yc = 1.16544 y12 = h − yc = 1.41544

For flange 52, s52 = 0 s52 = 3

⇒ z52 = d − zc = 0.0700462; ⇒ z52 = d − zc = 0.0700462;

y52 = 0 − yc = −1.58456 y52 = 3 − yc = 1.41544

For flange 45, s45 = 0 s45 = 0.5

⇒ z45 = −0.362967 − zc = −0.362967; ⇒ z45 = d − zc = 0.0700462;

y45 = 0.25 − yc = −1.33456 y45 = 0 − yc = −1.58456

For flange 65, s65 = 0 s65 = 0.25

⇒ z65 = 1.21651 − zc = 0.286553;

⇒ z65 = d − zc = 0.0700462;

y65 = −0.125 − yc = −1.70956 y65 = 0 − yc = −1.58456

Thus (d = 100 ), y12 = 1.16544 + 0.5 s12

z12 = −0.362967 + 0.866025 s12

y32 = 1.54044 − 0.5 s32

z32 = 0.286553 − 0.866025 s32

y52 = −1.58456 + s52

z52 = 0.0700462

y45 = −1.33456 − 0.5 s45

z45 = −0.362967 + 0.866025 s45

y65 = −1.70956 + 0.5 s65

z65 = 0.286553 − 0.866025 s65

(8.13)

8.1. THIN-WALLED BEAM SHEAR IN OPEN SECTIONS

517

The thicknesses are t12 (s12 ) = 0.1

t32 (s32 ) = 0.1

t52 (s52 ) = 0.05

t45 (s45 ) = 0.0666667

t65 (s65 ) = 0.0666667

The weighted modulus ratios are ξ12 = 2.0 For section 32, Z Q∗ys32 =

s32

0

Q∗zs32 =

Z

s32

0

ξ32 = 2.0

ξ52 = 1.0

ξ45 = 2.5

ξ65 = 2.5

ξ32 z32 (s32 ) t32 ds32 = 0.0573105 s32 − 0.0866025 s232 ξ32 y32 (s32 ) t32 ds32 = 0.308088 s32 − 0.05 s232

    ∆q32 (s32 ) = 0.117658 Q∗ys32 + 1.4687 Q∗zs32 Vy + 92.5559 Q∗ys32 + 0.117658 Q∗zs32 Vz

Hence,



= 0.459231 s32 − 0.0836243 s232 Vy + 5.34068 s32 − 8.02146 s232 Vz



q32 (s32 ) = qo32 + 0.459231 s32 − 0.0836243 s232 Vy + 5.34068 s32 − 8.02146 s232 Vz

Using the fact that the shear flow is zero at s32 = 0, qo32 = q32 (0) = 0 → qo32 = 0



q32 (s32 ) = 0.459231 s32 − 0.0836243 s232 Vy + 5.34068 s32 − 8.02146 s232 Vz

For section 12, Z Q∗ys12 =

s12

0

Q∗zs12 =

Z

0

s12

ξ12 z12 (s12 ) t12 ds12 = −0.0725933 s12 + 0.0866025 s212 ξ12 y12 (s12 ) t12 ds12 = 0.233088 s12 + 0.05 s212

    ∆q12 (s12 ) = 0.117658 Q∗ys12 + 1.4687 Q∗zs12 Vy + 92.5559 Q∗ys12 + 0.117658 Q∗zs12 Vz

Hence,



= 0.333794 s12 + 0.0836243 s212 Vy + −6.69152 s12 + 8.02146 s212 Vz



q12 (s12 ) = qo12 + 0.333794 s12 + 0.0836243 s212 Vy + −6.69152 s12 + 8.02146 s212 Vz

8.1. THIN-WALLED BEAM SHEAR IN OPEN SECTIONS

Using the fact that the shear flow is zero at s12 = 0, qo12 = q12 (0) = 0 → qo12 = 0

q12 (s12 ) = 0.333794 s12 + 0.0836243 s212 Vy + −6.69152 s12 + 8.02146 s212 Vz

For section 45, Z ∗ Qys45 =

s45

0

Qzs45 = ∗

Z

s45

0

ξ45 z45 (s45 ) t45 ds45 = −0.0604944 s45 + 0.0721688 s245 ξ45 y45 (s45 ) t45 ds45 = −0.222426 s45 − 0.0416667 s245

    ∆q45 (s45 ) = 0.117658 Q∗ys45 + 1.4687 Q∗zs45 Vy + 92.5559 Q∗ys45 + 0.117658 Q∗zs45 Vz

= −0.333794 s45 − 0.0527044 s245 Vy + −5.62529 s45 + 6.67475 s245 Vz

Hence,



q45 (s45 ) = qo45 + −0.333794 s45 − 0.0527044 s245 Vy + −5.62529 s45 + 6.67475 s245 Vz

Using the fact that the shear flow is zero at s45 = 0, qo45 = q45 (0) = 0 → qo45 = 0



q45 (s45 ) = −0.333794 s45 − 0.0527044 s245 Vy + −5.62529 s45 + 6.67475 s245 Vz

For section 65, Z ∗ Qys65 =

s65

0

Qzs65 = ∗

Z

0

s65

ξ65 z65 (s65 ) t65 ds65 = 0.0477588 s65 − 0.0721688 s265 ξ65 y65 (s65 ) t65 ds65 = −0.284926 s65 + 0.0416667 s265

    ∆q65 (s65 ) = 0.117658 Q∗ys65 + 1.4687 Q∗zs65 Vy + 92.5559 Q∗ys65 + 0.117658 Q∗zs65 Vz

Hence,



= −0.412851 s65 + 0.0527044 s265 Vy + 4.38683 s65 − 6.67475 s265 Vz



q65 (s65 ) = qo65 + −0.412851 s65 + 0.0527044 s265 Vy + 4.38683 s65 − 6.67475 s265 Vz

518

8.1. THIN-WALLED BEAM SHEAR IN OPEN SECTIONS

519

Using the fact that the shear flow is zero at s65 = 0, qo65 = q65 (0) = 0 → qo65 = 0

q65 (s65 ) = −0.412851 s65 + 0.0527044 s265 Vy + 4.38683 s65 − 6.67475 s265 Vz

For section 52, Z ∗ Qys52 =

s52

ξ52 z52 (s52 ) t52 ds52 = 0.00350231 s52

0

Qzs52 = ∗

Z

0

s52

ξ52 y52 (s52 ) t52 ds52 = −0.0792279 s52 + 0.025 s252

    ∆q52 (s52 ) = 0.117658 Q∗ys52 + 1.4687 Q∗zs52 Vy + 92.5559 Q∗ys52 + 0.117658 Q∗zs52 Vz

Hence,



= −0.11595 s52 + 0.0367174 s252 Vy + 0.314838 s52 + 0.00294146 s252 Vz



q52 (s52 ) = qo52 + −0.11595 s52 + 0.0367174 s252 Vy + 0.314838 s52 + 0.00294146 s252 Vz

In order to determine qo52 , we use the fact that the total shear flow going out minus the total shear flow going in at joint 2 must be zero: n X i=1

| {z }

n X

−

qi

Shear flow going out of a web junction



0 − q52

=0

qi

i=1

| {z }

Shear flow going into a web junction

s52 =3

+ q52

s52 =0.5

+ q32

s32 =0.25



=0

qo52 + 0.279992Vy + 0.464421Vz = 0

Hence qo52 = −0.279992Vy − 0.464421Vz and the shear flow in web 52 is
q52 (s52 ) = −0.279992 − 0.11595 s52 + 0.0367174 s252 Vy


+ −0.464421 + 0.314838 s52 + 0.00294146 s252 Vz

As a check, the total shear flow going out minus the total shear flow going in at joint 5

8.1. THIN-WALLED BEAM SHEAR IN OPEN SECTIONS

520

must be zero: n X i=1

| {z }

n X

−

qi

Shear flow going out of a web junction



q52

s52 =0



=0

qi

i=1

| {z }

Shear flow going into a web junction

 − q45

s45 =0.5

+ q65

s65 =0.25



=0

0=0

Thus the shear flow and stresses for each section are: For section 12,

q12 (x, s) = 0.333794 s12 + 0.0836243 s212 Vy + −6.69152 s12 + 8.02146 s212 Vz

q12 S12 (x, s) = = 3.33794 s12 + 0.836243 s212 Vy + −66.9152 s12 + 80.2146 s212 Vz t12

For section 32,



q32 (x, s) = 0.459231 s32 − 0.0836243 s232 Vy + 5.34068 s32 − 8.02146 s232 Vz

q32 = 4.59231 s32 − 0.836243 s232 Vy + 53.4068 s32 − 80.2146 s232 Vz S32 (x, s) = t32

For section 52,


q52 (x, s) = −0.279992 − 0.11595 s52 + 0.0367174 s252 Vy


+ −0.464421 + 0.314838 s52 + 0.00294146 s252 Vz
q52 S52 (x, s) = = −5.59984 − 2.31899 s52 + 0.734348 s252 Vy t52
+ −9.28842 + 6.29675 s52 + 0.0588292 s252 Vz

For section 45,



q45 (x, s) = −0.333794 s45 − 0.0527044 s245 Vy + −5.62529 s45 + 6.67475 s245 Vz

q45 = −5.00692 s45 − 0.790566 s245 Vy + −84.3793 s45 + 100.121 s245 Vz S45 (x, s) = t45

For section 65,



q65 (x, s) = −0.412851 s65 + 0.0527044 s265 Vy + 4.38683 s65 − 6.67475 s265 Vz

q65 S65 (x, s) = = −6.19277 s65 + 0.790566 s265 Vy + 65.8025 s65 − 100.121 s265 Vz t65 End Example 

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

8.2

521

Shear Center in Thin-Walled Open Sections

So far we have discussed the shear stress distribution along the thin-walled open cross-sections without considering any twisting. In other to ensure the above analysis is correct, we must find a point in the cross-section where there is no twisting due to the applied shear force(s). This point is known as the shear center.

8.2.1

Definition of Shear Center

Figure 8.10: Tip-loaded cantilever beam: twisting and bending (first & two), and bending only (third). Consider Fig. 8.10 showing a cantilever beam with a transverse force at the tip. Under the action of this load, the beam may twist as it bends. It is the line of action of the lateral force that is responsible for this bend-twist coupling. If the line of action of the force passes through the shear center of the beam section, then the beam will only bend without any twist. Otherwise, twist will accompany bending. The shear center is in fact the centroid of the internal shear force system. Depending on the beam’s cross-sectional shape along its length, the location of shear center may vary from section to section. A line connecting all the shear centers is called the elastic axis of the beam. When a beam is under the action of a more general lateral load system, then to prevent the beam from twisting, the load must be centered along the elastic axis of the beam. To understand this, the shear-stress distribution computed on the basis of bending without twisting must relate properly to the applied shear force from the standpoint of equilibrium. In fact, to eliminate all possibility of twisting at a section (Shames and Pitarresi): The applied shear force must have a line of action such that it develops an equal but opposite twisting moment about any axis parallel to the centerline of the beam as that stemming from the shear stress distribution at the exposed section computed on the basis of bending action only. Note that the shear forces may be applied or not to the structure. The calculation of the shear center is done by placing fictitious shear loads.

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

8.2.2

522

Static Equivalence y y

Vz Vy

s=0

s=0

ey ez z

z r

Os

r

Os

s = s1

s = s1

Vy > 0 and Vz = 0

Vz > 0 and Vy = 0

Figure 8.11: Vertical and horizontal applied shear forces. To determine the shear center, we use torque equivalence (not equilibrium) about a point, say Os , determines the location of the shear center, S = S(ey , ez ): Torque generated by the shear force = Torque generated by the shear flow Tapplied shear force =

X

Tinternal

To understand this concept, consider Fig. 8.11 and assume a counterclockwise positive torque. Thus the equivalent torque is defined as Tq =

n Z X i=1

0

ai

ri (si ) × qi (si ) dsi =

n Z X i=1

ai

ri (si ) qi (si ) dsi

(8.14)

0

Then we can determine the distance ey by taking Vy = 0 and Vz > 0, as follows: Vz ey = Tq

(8.15)

and the distance ez is obtained by taking Vz = 0 and Vy > 0, as follows −Vy ez = Tq

(8.16)

where i represents the ith branch, n the total number of branches, ai the length of the branch, and ri (si )

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

523

the distance perpendicular from Os to the contour si and it may be evaluated as follows: ri (si ) = −

∂zi ∂yi yi (si ) + zi (si ) ∂si ∂si

i = 1, 2, . . . , n

However, in some problems it may not be necessary to use this equation and the distance may be evaluated trivially. Note that the shear center depends only on the cross-section geometry and it is independent of the value of the applied loads. Once the line of action is found for no twisting this will be the same line of action for no twisting at that section no matter the value of the applied shear force. Thus, the intersection of the lines of action, where no twisting takes place, of the shear forces Vy and Vz determines the shear center, as shown in Fig. 8.12.

y Vz

S Shear center

ey

Vy ez z O

Figure 8.12: Location of the shear center of a thin-walled open section.

8.2.3

General Procedure

1. Calculate all second moments of area about the centroid. The location of the centroid will be needed. Determine the second moments of area ratios, as well. 2. Next, before we proceed let us decide where to take the torque equivalence. Recall that the torque equivalence about a point, say Os , determines the location of the shear center. Further, arbitrarily apply shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). 3. Torque equivalence can be taken about any point. However, for some points less calculation will be needed. For instance, if we take moment about the centroid then the shear flow acting at each branch will have to be calculated. If we choose an end-point, then only the shear flow acting in

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

524

branches not in the line of action will have to be calculated. Regardless about what point we take the torque equivalence, we should get the same answer.

Example 8.3. Shear Center in Unsymmetrical Open Sections Determine the shear center of the unsymmetrically thin-walled open section in Example 8.1.

a 4

3 2t y

2t 2a

zc z

C yc

t 2

1 2a

Figure 8.13: Unsymmetrically thin-walled channel section. One can often reduce the amount of computation by giving some thought to the problem. In the previous example we found all geometric properties about the centroid. Now we need to choose the point where to take torque equivalence, say Os , to determine the location of the shear center. Further, assume arbitrarily shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). Recall, torque equivalence can be taken about any point. However, for some points less calculation will be needed. For instance, if we take moment about the centroid then the shear flow acting at each branch will have to be calculated. If we choose point 1 or 4, then the shear flow acting in two branches will have to be calculated. For point 2 or 3, only the shear flow along one of the branches needs to be calculated. If torque equivalence is taken about a point 3, then only calculate q12 (si ). If torque equivalence is taken about a point 2, then only calculate q34 (si ).

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

525

If point Os = 3, then we should consider Fig. 8.14. Use the shear flow convention in such a way that only one branch needs to be evaluated, i.e., q12 . q34 4

3

4

3

Vy ez

=

Vz

q23

S ey 2

2

1

1 q12

Figure 8.14: Suggested shear flow convention for statically equivalence by taking the torque at point 3. If point Os = 2, then we should consider Fig. 8.15. Use the shear flow convention in such a way that only one branch needs to be evaluated, i.e., q43 . q43 4

3

4

3

Vy ez

=

Vz

q32

S ey 2

1

2

1 q21

Figure 8.15: Suggested shear flow convention for statically equivalence by taking the torque at point 2. Let us choose point 3 for our problem, Os = 3. BASIC EQUATIONS: Shear flow distribution Assume Vz and Vy is applied through the shear center, thus no torsion exists, and the shear

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

526

flow distribution is given by

qi (si ) = qoi +



A∗s

Q∗zs

Q∗ys

 1  A∗    0     0

0 −

0

1 ∗ Ryz

−

1 ∗ Ryy

1 ∗ Ryz

1 ∗ Rzz



 0       Vz       −Vy

          

(8.17)

where qoi is found by evaluating qi (si−1 = 0). When the thickness is constant along each branch: ti (si ) = ti . (The shear flow distributions where obtained in Example 8.1.) Torque equivalence Torque equivalence about a point, say Os = 3, determines the location of the shear center. For a positive torque counterclockwise (see Fig. 8.14): Tq =

n=3 X i=1

Z

ai

0

ri (si ) × qi (si ) dsi =

n=3 X i=1

Z

ai

ri (si ) qi (si ) dsi

0

For Vy = 0 and Vz > 0 −Vz (2a − ey ) = Tq

For Vz = 0 and Vy > 0

−Vy ez = Tq

(8.18)

Vy =0,Vz >0

(8.19)

Vy >0,Vz =0

where i represents the ith branch, ai the length of the branch, and ri (si ) the distance perpendicular from Os = 3 to the contour si . Shear flow distribution From Example 8.1: q12 (x, s) =



−

9 s212 21 s12 3 − 194a 388a2



Vy +



69 s12 24 s212 − 97a2 97a3



Vz

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

527

Torque due to shear flow Torque equivalence about a point 3, Os = 3, Tq =

n Z X

ai

0

i=1

ri (si ) × qi (si ) dsi =

Z

a12

r12 (s12 ) q12 (s12 ) ds12

0

For this problem, r12 (s12 ) = −2 a Hence, Tq = −

a12 = 2 a

148 Vz a 45 Vy a + 97 97

Vertical location of the shear center q34 4

3

4

3

Vy ez

=

Vz

q23

S ey 2

2

1

1 q12

Vertical location, ey , of the shear center is found by taking torque equivalence about a point 3, Os = 3, with Vy = 0 and Vz > 0, −Vz (2a − ey ) = Tq 148 Vz a −Vz (2a − ey ) = − 97 Hence, ey =

46 a 97

Vertical location of the shear center Horizontal location, ez , of the shear center is found by taking torque equivalence about a point 3, Os = 3, with Vy > 0 and Vz = 0, −Vy ez = Tq 45 Vy a −Vy ez = 97 Hence, ez = −

45 a 97 End Example 

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

528

Example 8.4. Determine the shear center of the cross-section of a composite beam of Example 8.2. y′ d b2 α

b1

E2 t2 E1 t1

h

E3 t3

c1

z′

α c2

Before we proceed let us decide where to take the torque equivalence. Recall that the torque equivalence about a point, say Os , determines the location of the shear center. Further, arbitrarily apply shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). Let us choose the centroid as our point, although point 2 would have been easier. Torque equivalence about a point, say Os (weighted centroid), determines the location of the shear center. For a positive torque counterclockwise (see Fig. 8.16), Tq =

n=5 X i=1

Z

0

ai

ri (si ) × qi (si ) dsi =

n=5 X i=1

Z

ai

ri (si ) qi (si ) dsi

0

for Vy = 0 and Vz > 0 Vz ey = Tq

(8.20)

Vy ez = Tq

(8.21)

for Vz = 0 and Vy > 0

where i represents the ith branch, ai the length of the branch, and ri (si ) the distance perpendicular from Os (weighted centroid) to the contour si and it is evaluated as follows: ri (si ) = −

∂zi ∂yi yi (si ) + zi (si ) ∂si ∂si

i = 1, 2, . . . , 5

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

529

yc Vz>0, Vy=0

q32

b2

3

b1

b2

ey

2

3

b1

q12

1

2 q52

ez

q12

1

q52

zc

Os

zc

Os

4 q45 Vz=0, Vy>0

q32

4 5

q45

c1

6 c2

5 c1

q65

6 c2

q65

Figure 8.16: Shear flow convention. For our problem, using Eq. (8.13) r12 = −1.1907800 r32 = 1.1907800 r52 = 0.070046200 r45 = 1.3372500 r65 = −1.3372500 The minus sign is due to the direction of the shear flow, in will create a negative torque. Thus, Tq =

n=5 X i=1

+

Z

ai

ri (si ) qi (si ) dsi =

0

Z

Z

a12

r12 (s12 ) q12 (s12 ) ds12 +

0

a52

r52 (s52 ) q52 (s52 ) ds52 +

0

= −0.151348Vy + 0.0452916Vz

Z

0

Z

a32

r32 (s32 ) q32 (s32 ) ds32

0

a65

r65 (s65 ) q65 (s65 ) ds65 +

Z

0

a45

r45 (s45 ) q45 (s45 ) ds45

8.2. SHEAR CENTER IN THIN-WALLED OPEN SECTIONS

530

Thus, for Vz = 0 and Vy > 0 Vy ez = Tq

→

ez = −0.15134800

Vz ey = Tq

→

ey = 0.045291600

for Vz > 0 and Vy = 0 The above is distances are given from the cross-section weighted centroid. End Example 

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

8.3

531

Torsion in Open Thin-Walled Sections

For a number of cross-sections, such as open thin-walled sections, it is not easy to obtain stress functions to study torsional effects. However, Prandtl (1903) suggested an extremely useful analogy relating torsion of an arbitrary cross-section to the deflected shape of a membrane. This is known as Prandtl’s membrane analogy for torsion.

8.3.1

Prandtl’s membrane analogy for torsion

To better understand this analogy, let us first derive the equation for the deflection of a membrane. A membrane may be defined as a structure whose thickness is small compared to surface dimensions and it has negligible bending rigidity. Thus this thin sheet of material relies for its resistance to transverse loads on internal in-plane or membrane forces. N y

N

N x

N

z z

N

Top view

N pr

Side view

Figure 8.17: Membrane analogy: in-plane and transverse loading. Suppose that a membrane has the same external shape as the cross-section of a torsional bar, as shown in Fig. 8.17. It supports a transverse uniform pressure pr and is restrained along its edges by a uniform tensile force unit length and it is denoted by N . Now consider the equilibrium of an element dy dz of the membrane and sum all the forces in the x direction (Fig. 8.18):

−N sin ϕz − N sin (ϕz + dϕz ) − N sin ϕy − N sin (ϕy + dϕy ) + pr dy dz = 0

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

532

x

z

φz+ dφz N

y

φy N N

pr

φz

φy+ dφy

N Figure 8.18: Equilibrium of element of a membrane. For small angles, pr dy dz − N

∂u dz − N ∂y



−

∂u ∂ 2 u − 2 dy ∂y ∂y

 N

dz − N

∂u dz − N ∂z



−

∂u ∂ 2 u − 2 dz ∂z ∂z



dy = 0

∂2u ∂2u dy dz + pr dy dz = 0 2 dy dz + N ∂y ∂z 2

Thus the governing partial differential equation for deflection, w, of a membrane is given by ∂2u ∂2u pr 2 2 + 2 = ∇ u = −N ∂y ∂z Note that since the membrane is attached at the boundary, u = 0 along the contour.

(8.22)

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

533

Note this is exactly the same as the torsional problem: Torsion

Membrane

∇2 Φp = 2 G φx

∇2 u = −

pr N

Φp = 0 on contour

u = 0 on contour

Φp

u

−φx

pr

1 2G

N

Sxy = − Sxz = T − 2

∂Φp ∂z

∂Φp ∂y

∂u ∂z ∂u ∂y Γ=

ZZ

u dA (Volume)

It should be noted that for orthotropic, would need a membrane to give different N 0 s in different directions in proportion to Gxz and Gyz . Thus membrane analogy only applied to isotropic materials.

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

8.3.2

534

Torsion of a Narrow Rectangular Cross-Section

y

x

z

N L b T h

Figure 8.19: Torsion of a narrow rectangular strip. Let us apply the membrane analogy to the torsion of narrow rectangular strip subjected to a torque, see Fig. 8.19. Assume that b  h. In order to better visualize the method, consider the narrow rectangular cross-sectional representation shown in Fig. 8.20. Consider a cross-section in the middle (away from the edges).

y x

z b

z pr -h/2

h/2

h

Cross-sectional view

Side view of membrane under uniform pressure

Figure 8.20: Representation of cross-section for membrane analogy and the side-view of the membrane under pressure. Using the Prandtl’s governing partial differential equation, Eq. (8.22): pr ∂2u ∂2u 2 + 2 = −N ∂y ∂z

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

535

Near the middle of the long strip (away from z = ±b/2), ∂2u 1 ∂y 2 Thus, integrating twice,

∂2u pr 2 = −N ∂z

→

∂2u pr =− N ∂z 2 ∂u pr = − z + C1 ∂z N pr 2 z + C1 z + C2 2N

u=− Applying boundary conditions: h @z = − , 2

u = 0,

h @z = + , 2

u = 0,

 2   h h − + C1 − + C2 2 2  2   pr h h u=− + C1 + C2 2N 2 2 u=−

pr 2N

Solving for the constants: C1 = 0 Thus

pr h2 u(z) ≈ 8N

The volume is given by Γ= Now use membrane analogy:

ZZ

u dA =

Z

(

h/2

−h/2

A

pr h2 8N

C2 =

1−

Z



b/2

2z h

u dy dz =

−b/2

(



T pr b h3 φx b h3 2 G = =− 2 12 N 12

→

pr h2 u = Φp = 8N

2 )

1−

2z h

pr b h3 12 N

2 )

pr = −φx N=

1 2G

Γ=−

T 2

Thus −

T =Γ 2

→

−

φx =

3T G b h3

→

φx =

T b h3 G 3

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

Recall φx =

536

dθx T = dx GJ

Thus, J=

b h3 3

and the stress function is: pr h2 Φp = 8N

(

1−



2z h

2 )

(

G φx h2 =− 4

1−



2z h

2 )

Although Φp does not disappear along the short-edges of the strip and therefore does not give an exact solution, the actual volume of the membrane differs only slight from the assumed volume so that the corresponding torque and shear stresses are reasonably accurate. Also, the maximum shear stress occurs along the long sides of the strip where the contours are closely spaced, indicating, in any case, that conditions in the end region of the strip are relatively unimportant. The stress distribution is then, Sxy = −

∂Φp dθx = −2 G z , ∂z dx

Sxz =

∂Φp =0 ∂y

Note that the maximum stress is twice that in a circular rod! The shear stress varying linearly across the thickness and attaining a maximum is Sxy

8.3.3

max

= −2 G z

T Th dθx dθx = ±G h =± = ±G h dx z=±h/2 dx GJ J

Torsion of an Arbitrary Open Thin-walled Cross-Sections

2

2

1

Sxs

y

s

q

1

t 2

n

1

s

Sxn z Vz > 0

Vy > 0

Figure 8.21: Arbitrary open thin-walled cross-Section. Using the membrane analogy, it can be shown that the previous theory for long, narrow rectangular

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

537

sections applies also to other shapes. In general, Sxs = 2 G n

dθx T = 2nG , dx (G J)eff

Sxn = 0

where n is the distance normal to the surface (in the thickness direction) and (G J)eff is the effective torsional rigidity and it is defined as (G J)eff = G0

n X Gi Ji , G 0 i=1

Ji =

ai t3i 3

If the thickness is not constant then the torsional constant is found through integration: Z 1 ai 3 ti dsi where ti = ti (si ) Ji = 3 0 Hence, the total shear stress with shear loads acting through the shear center will be Sshear (x, s) = SV (x, s) + ST (x) where SV (x, s) is the shear stress due to shear loads and ST (x) the shear stress due to torsion. Note that the torsional load is the same along the contour of the section.

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

538

Example 8.5. Assuming homogeneous cross-sections, determine the torsional constant. 8.5a)

h1, t1

h2, t2

J=

2 X ai t3

3

i=1

i

=

h1 t31 + h2 t32 3

8.5b)

h3, t3

h1, t1

h2, t2

J=

3 X ai t3 i=1

3

i

=

h1 t31 + h2 t32 + h3 t33 3

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

539

8.5c) h1 , t 1

h3 , t 3

h2, t2

J=

3 X ai t3 i=1

3

i

=

h1 t31 + h2 t32 + h3 t33 3

8.5d) t

r

J=

1 3

Z

0

a

t3 ds =

φ

1 3

Z

0

ϕ

t3 r dϕ =

ϕ r t3 3 End Example 

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

540

Example 8.6. Consider the beam of Example 8.4: y′ d b2 α

b1

E2 t2 E1 t1

h

E3 t3

c1

α

z′

c2

Assume ν12 = ν52 = ν23 = ν45 = ν56 = 0.3. If the only nonzero internal loads evaluated at the centroid and at the tip are: Vz = 0,

Vy = −1000 lb

Determine the value and the location of the maximum shear stress at the tip. This is an example where the shear loads do not pass through the shear center but through the centroid. If the shear force does not pass through the shear center, we will have torsional shear stresses, as well as those due to flexure alone. Basically, we can break a thin-walled open section problem of this kind into three parts: 1. First, the shear load is shifted to the shear center, and the flexure shear stress distribution is found in the manner illustrated in Example 8.4. 2. Then, the moment of the shear force (Tq ) in its actual location about the shear center is used as the applied torque to find the torsional shear stresses, using the procedure outlined in the previous section. 3. Superimpose the two solutions–pure flexure and pure torsion– to yield the net shear flow distribution throughout the cross-section. First, we move all the loads from the weighted centroid to the cross-sectional shear center:

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

541

2

2

2 1

3

3

3

1

1 1000 lb

1000 lb

Tq SC

SC

Oc

4

4

4

5

5

5 y

6

6

6

z

The the shear flow distribution due to shear load acting at the shear center were calculated in the Example 8.2: Max = 109.581 lb/in @ s23= 0.0

Max = 187.803 lb/in @ s21= 0.0

3 q21

Max = 297.385 lb/in @ s52= 3.0

q23

2

1 q52

Max = 371.531 lb/in @ s52= 1.57895

zc 4 q45 5

q65 6

Max = 279.992 lb/in @ s52= 0.0

Max = 99.918 lb/in @ s65= 0.25

Max = 180.073 lb/in @ s45= 0.5

From the figure we can see that the maximum shear stress occurs in the s52 = 1.57 (measured from point 5) in of the vertical web and has a value of Smax,shear =

qmax,52 = 7430.63 psi t52

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

542

Secondly, we determine the total torque at the shear center. Torque equivalence about a point, say Os (weighted centroid), will give us the torque due to shear loading acting at the location of the shear center: Tq =

n=5 X i=1

+

Z

ai

ri (si ) qi (si ) dsi =

a12

r12 (s12 ) q12 (s12 ) ds12 +

0

0

Z

Z

a52

r52 (s52 ) q52 (s52 ) ds52 +

Z

a32

Z

a65

r65 (s65 ) q65 (s65 ) ds65 +

= −0.151348Vy + 0.0452916Vz

Z

a45

r45 (s45 ) q45 (s45 ) ds45

0

0

0

r32 (s32 ) q32 (s32 ) ds32

0

= 151.348 lb–in

The total torque will be: T = Tq + Texternal However, for our problem there are no external loads applied: T = Tq + Texternal = 151.348 lb–in + 0 = 151.348 lb–in Hence, the moment about the shear center is 151.348 lb–in + , which is applied as a pure torque to the open section. Now to determine the shear stress we need to determine the torsional rigidity GJ of the section. For heterogeneous cross sections the shear stress is defined as Sxs = 2 n G

T , (G J)eff

Sxn = 0

where n is the distance normal to the surface (in the thickness direction) and (G J)eff is the effective torsional rigidity and it is defined as (G J)eff = G0

n X Gi Ji , G 0 i=1

Ji =

ai t3i 3

Hence, This is a problem is: (G J)eff = G0

n X Gi Ji G0 i=1

5 5 X Gi hi t3i G0 X Gi = hi t3i G 3 3 G 0 0 i=1 i=1   G0 G52 G G23 G45 G56 12 3 3 3 3 3 = h52 t52 + h12 t12 + h23 t23 + h45 t45 + h56 t56 3 G0 G0 G0 G0 G0

= G0

= 0.00243056 in4 where

Gi =

Ei , 2 (1 + νi )

i = 12, 52, 23, 45, 56

8.3. TORSION IN OPEN THIN-WALLED SECTIONS

543

Therefore, the maximum shear stress due to pure torsion is Smax,torsion = G0

Tt Tq t52 = G0 = 23949.60 psi (G J)eff (G J)eff

Note that the shear flow due to torque is uniformly distributed across the wall thickness. Hence, the shear flow distribution for the split problem looks like this

3

3 2

2

1

1 1000 lb

151.348 lb-in

SC

SC 4

4

5

5 6

6

Finally, superimposing the flexural and torsional shear stresses around the outer boundary of the section shows that the maximum shear stress is: Smax,shear = Smax,pure shear + Smax,pure torsion = 7430.63 + 23949.60 = 31380.3 psi ↑ at the s52 = 1.57 (measured from point 5) of the vertical web, on the right side, where the torsional and flexural stresses are in the same direction. End Example 

8.4. CROSS-SECTION IDEALIZATION

8.4

544

Cross-section Idealization

So far, we have concentrated on the analysis of beams subject to bending. Each point in the thin-walled section is subject to bending, stretching, and shearing. However, we often can simplify the analysis of semi-monocoque structures by idealizing the cross-section. This idealization of the thin-walled beam’s cross-section consists of two parts: 1. Parts that carry only extensional stress, Sxx (due to bending and axial loads). 2. Parts that carry only shear stress, Sxs (due to shear and torsional loads). The cross-section of an idealized beam consists of thin webs and concentrated flange areas. The flanges carry all of the bending load, while the webs transmit only shear between adjacent flanges.

8.4.1

Idealization of Semi-Monocoque construction

Recall that a monocoque construction is one in which the construction is done in one piece; i.e., the entire section carries all type of stresses. However, in semi-monocoque construction, only flanges and stringers, carry axial stresses, Sxx ; and the skins and webs carry only shear.

Figure 8.22: Actual thin-walled section and idealized section. The process of idealizing a beam cross-section involves lumping the area of the webs into discrete points, as suggested in Fig. 8.22. To better understand this idealization, consider the wing-box shown in Fig. 8.23. The concentrated areas include the areas of actual stiffeners (e.g., stringers, longerons, or spar caps) to which the skin is attached. Some concentrated areas may just be an outcome of the modeling and may not represent actual physical stiffeners. For example, the small brown dots in Fig. 8.23 are lumped skin areas, and not physical stiffeners. The larger brown dots on the other hand include the area of the spar caps and stiffeners. Since the webs of idealized beams are not active in bending, the shear flow in each wall of a section may be assumed constant (This greatly simplifies the shear flow analysis.) The idealization is done such that the area and the second moment of inertia remain identical.

8.4. CROSS-SECTION IDEALIZATION

545

MONOCOQUE CONSTRUCTION

SEMI-MONOCOQUE CONSTRUCTION

IDEALIZATION

Figure 8.23: Actual thin-walled section and idealized section.

8.4.2

Typical method to Idealize of webs

Thin rectangular wall Approximation Consider any thin rectangular wall of a beam cross-section. The total cross-sectional length of the wall is h and thickness t. The wall segment will be idealized as two stringer areas, each having having half the wall area and spaced to preserve the centroidal area moment of inertia, as shown in Fig. 8.24.

t Af1 h/2 d1 a

a d2 h/2 Af2

Actual wall

Idealized wall

Figure 8.24: Idealization of a thin rectangular wall into two concentrated areas.

8.4. CROSS-SECTION IDEALIZATION

546

The total area and the second moment of inertia about the aa axis are Aw = h t, The stringer areas are

Iaa =

1 t h3 12

As1 + As2 = Aw As + As = Aw

2 As = Aw = h t ht As = 2 Now we proceed to determine the distance d1 and d2 to preserve the second moment of inertia about the aa axis. Iaas1 + Iaas2 = Iaa 2

2

As1 (d1 ) + As2 (−d2 ) = Iaa 2

2

As (d1 ) + As (−d2 ) = Iaa But to keep the centroid unchanged d = d1 = d2 , thus, Iaas1 + Iaas2 = Iaa 2 As d2 = Iaa 1 ht 2 d = t h3 2 2 12 1 2 h d2 = 12 h d= √ 2 3 Stringer-skin Approximation Consider a beam’s cross-section composed with two flanges and a web, as shown in Fig. 8.24. The cross-section will be idealized as two stringer areas, each having having half the wall area and spaced to preserve the centroidal area moment of inertia. Take bf = tf = d, tw =  d, and h = 10 d. The total area and the second moment of inertia about the aa axis are Aactual = h tw + 2 bf tf = (2 + 10) d2 Iaa

1 = tw h3 + 12 =

1 bf t3f + bf tf 12



h + tf 2

2 !

+

250d4 182d4 + = d4 (60.6667 + 83.3333 ) 3 3

 2 ! 1 h + t f bf t3f + bf tf − 12 2

8.4. CROSS-SECTION IDEALIZATION

547

bf tf Af1

tw h/2 d1 a

a d2 h/2 Af2 tf Idealized cross-section

Actual cross-section

Figure 8.25: Idealization of a thin rectangular wall into two concentrated areas. The flange will be modeled as a stringer. thus stringer areas are Af 1 + Af 2 = Aactual As + As = Aactual 2 As = 2 d 2 As = d2 Note that here we do not choose the distance d1 and d2 to preserve the second moment of inertia about the aa axis, instead we choose the actual distance: d = d1 = d2 =

h tf + = 5.5 d 2 2

The moment of inertia for the idealized cross-section is:  2 1 h + tf 3 Iaa = t w h + 2 As = d4 (60.5 + 83.3333 ) 12 2

8.4. CROSS-SECTION IDEALIZATION

8.4.3

548

Shear flow and shear center in Open Idealized Sections

Representation of skins and stringer and associated loads and shear flows at a typical joint is shown in Fig. 8.26.

P+dP Stringer Skin qout

P

qin dx

x

Figure 8.26: Actual thin-walled section and idealized section. Recall that stringers only carry axial loads and skins carry shear flow. The shear flow at the end of the skin (where it is cut) must be the same as at the edge (the cross-section cut). This is due to equilibrium (Syz = Szy ). Adding forces in the x-direction: −P + (P + dP ) + qin dx − qout dx = 0 dP dx + qin dx − qout dx = 0 dx dP ∆qs = qout − qin = dx We know that P = σxx A then differentiating we get

dP σxx dA =A + σxx dx dx dx

8.4. CROSS-SECTION IDEALIZATION

549

For stringers with uniform cross-section:

dP σxx =A dx dx

Thus, ∆qs = qout − qin =

dP σxx =A dx dx

Now using the flexure equation for isothermal,

∆qs = qout − qin =

=





A

Asf

Ay

 1  A∗    Az  0    0

Qzsf

Qysf

 1  A∗    0     0

0 −



 0       Vz       −Vy

0

1 ∗ Ryz

−

1 ∗ Ryy

1 ∗ Ryz

1 ∗ Rzz 0 −

0

1 ∗ Ryz

−

1 ∗ Rzz

1 ∗ Ryy

1 ∗ Ryz



          

 0       Vz       −Vy

          

If we had more than one skin entering or leaving the joint then we may generalize the equation as follows

∆qs =

X

qout −

X

qin =



Asf

Qzsf

Shear across the skins are calculated as before.

Qysf

 1  A∗    0     0

0 −

1 ∗ Ryz

1 ∗ Rzz

0 −

1 ∗ Ryy

1 ∗ Ryz



 0       Vz       −Vy

          

Understanding the problem One can often reduce the amount of computation by giving some thought to the problem. 1. Obtain the cross-sectional properties. 2. Next, before we proceed let us decide where to take the torque equivalence. Recall that the torque equivalence about a point, say Os , determines the location of the shear center. Further, arbitrarily apply shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). 3. Torque equivalence can be taken about any point. However, for some points less calculation will be needed. For instance, if we take moment about the shear center then the shear flow acting at

8.4. CROSS-SECTION IDEALIZATION

550

each branch will have to be calculated. Determine the shear flows. 4. Remember that stiffeners only carry bending and not shear.

In general, consider a six-step solution procedure: 1. Calculate the centroid: yc , zc 2. Determine the second moments of area ratios about the centroid: Rzz , Ryy , Ryz 3. Determine the parametric equations about the centroid: yi (si ), zi (si ) 4. Find the shear flow distribution in each branch and stiffener, qi (si ): branch:  1 0 0  A∗    1 1  qi (si ) = qoi + A∗s Q∗zs Q∗ys  0 − ∗ − ∗ Ryz Ryy    1 1 0 ∗ ∗ Rzz Ryz stiffener:

∆qs = qout − qin =



Asf

Qzsf

Qysf

 1  A∗    0     0

0 −

1 ∗ Ryz

1 ∗ Rzz



 0       Vz       −Vy

0 −

1 ∗ Ryy

1 ∗ Ryz

Here Asf , Qzsf and Qysf are properties at the location of the stiffener. 5. Use torque equivalence to determine the shear center (same as before).



          

 0       Vz       −Vy

          

8.4. CROSS-SECTION IDEALIZATION

551

Example 8.7.

Shear Flow in Symmetrical Open Idealized Sections For the symmetrically open thin-walled section shown, determine the shear flow in all the panels and determine the location of the shear center.

h/2 t

Vy y

45° z

h S.C.

45° t h/2

Figure 8.27: Symmetrically thin-walled channel section.

Figure 3. Thin-walled cross section

All stringers have the same area Af = 0.01 t h and h = 100 ,

t = 0.00200

8.7a) Overview of the problem: i. Since this is a symmetrical thin-walled section, we do not need to calculate Iyz because it is zero. Therefore, only Izz and Iyy needs to be calculated. The location of the centroid will be needed. By inspection yc and zc needs to be calculated. ii. Next, before we proceed let us decide where to take the torque equivalence. Recall that the torque equivalence about a point, say Os , determines the location of the shear center. Further, arbitrarily apply shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). Note because of symmetry ey = 0.

8.4. CROSS-SECTION IDEALIZATION

552

iii. Torque equivalence can be taken about any point. However, for some points less calculation will be needed. For instance, if we take moment about the shear center then the shear flow acting at each branch will have to be calculated. If we choose point 1 or 4, then the shear flow acting in two branches will have to be calculated. For point 2 or 3, only the shear flow along one of the branches needs to be calculated. If torque equivalence is taken about a point 3, then only calculate q12 (s12 ). If torque equivalence is taken about a point 2, then only calculate q43 (s43 ). Regardless about what point we take the torque equivalence, we should get the same answer. Let us choose point 2 for these solutions. here we chose point Os = 2 and used the convention shown in Fig. 8.28

1

1 α

2

Vz

ez

2

α

3

α

=>

S.C.

α

3

4

4

Figure 8.28: Shear flow convention BASIC EQUATIONS: Shear flow distribution Since ey = 0, consider only the case when Vz = 0 and Vy > 0 is applied through the shear center then no torsion exists. Thus the shear flow distribution is given by

qi (si ) = qoi +

= qoi +





A∗s

A∗s

Q∗zs

Q∗zs

Q∗ys

Q∗ys

 1  A∗    0     0

        

0 −

0

Vy ∗ Ryy      Vy    − ∗ Ryz

0

1 ∗ Ryz

1 ∗ Rzz                 

−

1 ∗ Ryy

1 ∗ Ryz

= qoi +



            

0 0 −Vy

          

Vy ∗ Vy ∗ ∗ Qzs − ∗ Qys Ryy Ryz

8.4. CROSS-SECTION IDEALIZATION

553

where qoi is found by evaluating qi (si−1 = 0). When the thickness is constant along each branch: ti (si ) = ti . Further for a symmetric and homogenous cross-section, the shear flow expression reduced to: qi (si ) = qoi +

Vy Qzs Ryy

Shear flow across the stringer will be, 4qs =

Vy Qzsf Ryy

Torque equivalence Torque equivalence about a point, say Os = 2, determines the location of the shear center. For a positive torque counterclockwise: Tq =

n=3 X i=1

Z

0

ai

ri (si ) × qi (si ) dsi =

n=3 X i=1

Z

ai

ri (si ) qi (si ) dsi

0

For Vy = 0 and Vz > 0

For Vz = 0 and Vy > 0

Vz ey = Tq

Vy =0,Vz >0

Vy ez = Tq

Vy >0,Vz =0

(8.23)

(8.24)

where i represents the ith branch, ai the length of the branch, and ri (si ) the distance perpendicular from Os = 2 to the contour si and is defined as: ri (si )

i = 1, 2, . . . , n

8.7b) By placing the origin at point 2 and we can calculate zc and yc and it can be shown that the centroid is located at: 1 α

2

y zc

z C

3

α 4

Figureabout 3. Thin-walled cross section Since this is a symmetric cross section the z axis, the shear location in y is zero: ey = 0.

8.4. CROSS-SECTION IDEALIZATION

554

√ Qy 3 3 2 Qz zc = = h cos α = h yc = =0 A 16 32 A Second moments of area Izz , about the centroid, using thin-walled assumption is √ ! 8+3 2 Izz = Izz12 + Izz23 + Izz34 + Izzf 1 + Izzf 2 + Izzf 3 + Izzf 4 = h3 t 8 Second Moment of area Iyy , about the centroid, using thin-walled assumption is Iyy = Iyy12 + Iyy23 + Iyy34 + Iyyf 1 + Iyyf 2 + Iyyf 3 + Iyyf 4 =

37h3 t 384

Second Moment of area Iyz , about the centroid, using thin-walled assumption is Iyz = Iyz12 + Iyz23 + Iyz34 + Iyzf 1 + Iyzf 2 + Iyzf 3 + Iyzf 4 = 0 Second moments of area about the centroid are √ ! 8+3 2 37h3 t Izz = h3 t Iyy = 8 384

Iyz = 0

and the second moments of area ratios are  √  37 8 + 3 2 h6 t2 √  1 ∆I = Ryy = 8 + 3 2 h3 t 3072 8

Rzz =

37h3 t 384

Ryz = ∞

8.7c) Shear flow distributions When Vy > 0 and Vz = 0 is applied through the shear center then no torsion exists and the shear flow distribution for this problem is given by

qi (s) = qoi +



As

Qzs

Qys

 1  A    0     0

0 −

1 Ryz

1 Rzz

0 −

1 Ryy

1 Ryz



            

0 0 −Vy

          

where qoi is found by evaluating qi (s = 0). Substitute the moments of area in the above equation to get

qi (si ) = qoi +



As

Qzs

qi (s) = qoi + 

Qys

            

0

      

8Vy  √   8 + 3 2 h3 t      0

8Qzsi Vy √  8 + 3 2 h3 t

(8.25)

8.4. CROSS-SECTION IDEALIZATION

555

q10

q10a

q12 q20a

q20

q23

q30a q30 q34 q40a q40

Figure 8.29: Shear flow convention for statically equivalence when taking the torque at point 2. In order to calculate the shear flow in each skin, we need to first find the yi (si ) for each part. Following the assumed flow convention in Fig. 8.30 and from the geometry of the crosssection we get: (about the centroid)

For skin 12, 5h √ ; 16 2

s12 = 0

⇒

z12 =

h 2

⇒

z12 = −

s12 =

5h √ ; 16 2

y12 =

√  1 2+ 2 h 4

y12 =

h 2

8.4. CROSS-SECTION IDEALIZATION

556

x+∆x x q10

q10a

q12 q20a

q20

q23

q30a q30 q34 q40a q40

Figure 8.30: Shear flow convention for statically equivalence when taking the torque at point 3 For skin 23, s23 = 0

⇒

z23 = −

3h √ ; 16 2

y23 =

s23 = h

⇒

z23 = −

3h √ ; 16 2

y23 = −

h 2

s34 = 0

⇒

z34 = −

3h √ ; 16 2

y34 = −

h 2

h 2

⇒

z34 =

y34 = −

√  1 2+ 2 h 4

h 2

For skin 34,

s34 =

5h √ ; 16 2

Since these are straight lines, the equation of the line to can be used to calculate zi (si )

8.4. CROSS-SECTION IDEALIZATION

557

and yi (si ),

yi (si ) = zi (si ) =

yi zi

(si =ai )

− yi

(si =0)

ai − 0

(si =ai )

− zi

(si =0)

ai − 0

si + bi

where

si + ci

where

where ai is the length of each branch. For stringer 1, q10 = q10a + 

bi = y i

ci = zi

(si =0)

(si =0)

8Qzf1 Vy √  8 + 3 2 h3 t

√  1 2 + 2 h2 t 8 And because at free edge we have no shear Qzf1 =

q10a = 0 Thus q10 =



5+

√  2 Vy

23h

For skin 12, (t12 = t) y12 (s12 )

=

s h h √ + − √12 2 2 2 2

z12 (s12 )

=

5h s √ − √12 16 2 2

Qys12

=

Z

Qzs12

=

s12

0

Z

s12

0

q12 (s)

qo12

q12 (s)

z12 (s12 ) t12 ds12 =

=

=

=

qo12

t s2 ht s12 ht s12 y12 (s12 ) t12 ds12 = − √12 + √ + 2 2 2 2 2

 √ √ 2 2 s 2 2 s 4 s 2 12 12 12 + + + −  √  √  √   Vy 8 + 3 2 h3 8 + 3 2 h2 8 + 3 2 h2

q10 = 

5h s12 t s212 t √ − √ 16 2 2 2





5+

√  2 Vy

23h

 √ 2 √ √ 2 2 s 2 2 s 4 s 2 2 12 12 12 −  + + + √  √  √  √  + √   Vy 8 + 3 2 h3 8 + 3 2 h2 8 + 3 2 h2 8+3 2 h 8+3 2 h

8.4. CROSS-SECTION IDEALIZATION

558

For stringer 2, q20 = q20a + 

8Qzf2 Vy √  8 + 3 2 h3 t

Qzf2 =

h2 t 4

At the boundary q20a = q12 (s12 = h/2) = Thus q20 =



Vy 2h

√  39 − 6 2 Vy 46h

For skin 23, (t23 = t) h − s23 2

y23 (s23 )

=

z23 (s23 )

= −

Qys23

=

Qzs23

=

3h √ 16 2

Z

s23

0

Z

s23

z23 (s23 ) t23 ds23 = − y23 (s23 ) t23 ds23 =

0

q23 (s)

qo23

q23 (s)

=

qo23 +

s2 t h s23 t − 23 2 2

! √ √ 6 2 s223 16 s223 6 2 s23 16 s23 − − + Vy 23h3 23h3 23h2 23h2

= q23 (0) = q20 =

=

3h s23 t √ 16 2



√  39 − 6 2 Vy 46h

 2 4 s 4 s 3 6 23 23 −  + +√  √  √  √  + √   Vy 8 + 3 2 h3 8 + 3 2 h2 2 8+3 2 h 8+3 2 h 

For stringer 3,

q30 = q30a + 

8Qzf3 Vy √  8 + 3 2 h3 t

Qzf3 = − At the boundary q30a = q23 (s23 = h) =

h2 t 4

 √  3 13 − 2 2 Vy 46h

8.4. CROSS-SECTION IDEALIZATION

559

Thus q30 =

Vy 2h

For skin 34, (t34 = t) h s34 −√ 2 2

y34 (s34 )

= −

z34 (s34 )

=

s 3h √34 − √ 2 16 2

=

Z

Qys34

Qzs34

s34

s234 t 3hs34 t √ z34 (s34 ) t34 ds34 = √ − 2 2 16 2

s34

ts2 hts34 y34 (s34 ) t34 ds34 = − √34 − 2 2 2

0

=

Z

0



 √ 2 2 2s34 4s34 + −  √  3− √   Vy 8+3 2 h 8 + 3 2 h2

q34 (s)

=

qo34

qo34

=

q34 (0) = q30 = −

q34 (s34 )

30 Vz 42 Vy − 97a 97a



 √ 2 2 2s 4s 3 4 34 34 = −  √  + √  3− √  2+√  √   Vy 2 8+3 2 h 8+3 2 h 8+3 2 h 8+3 2 h

Check: Since point 4 is a free edge, we should get zero shear flow. For stringer 4, q40 = q40a +  Qzf4 = − At the boundary

8Qzf4 Vy √  8 + 3 2 h3 t

√  1 2 + 2 h2 t 8

q40a = q34 (s34 = h/2) =

√

2Vy 5Vy + 23h 23h

Thus q40 = 0 ...GOOD! 8.7d) Horizontal location, ez , of the shear center. As mentioned before, for torque equivalence

8.4. CROSS-SECTION IDEALIZATION

560

about a point 2, Os = 2, with Vy > 0 and Vz = 0, XZ

=

V y ez

Z

=

V y ez

ai

0

i

a34

ri (s) × qi (s) ds

r34 (s) q34 (s34 ) ds34

0

1 α

Vy 2

r34 = h cos α S.C.

ez α

3 q43

4

For this problem,

h r34 (s) = h cos(α) = √ 2

and a34 = h/2. Thus V y ez Thus ez =

=

√  1  5 + 24 2 h Vy 276

(8.26)

√  1  5 + 24 2 h = 0.141 h 276 End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

8.5

561

Closed Single-Cell Thin-Walled Sections

So far, we have discussed the analysis of open thin-walled cross-sections. The analysis of a closed single cell thin-walled section uses most of the equations derived so far. For the analysis of closed thin-walled sections, we operate at the shear center and split the problem in two parts: Sxs = Spure shear at the shear center + Spure twist at the shear center Let us begin by studying twisting and torsion in closed single cell thin-walled sections.

8.5.1

Enclosed area ds dAc r(s) O

Figure 8.31: Area enclosed by the contour. Let the geometry of the contour of an arbitrary thin-walled closed section be defined by Fig. 8.31. Let r(s) be the distance of the location of the infinitesimally small element ds relative to O. This distance is defined as ∂zi ∂yi ri (si ) = − yi (si ) + zi (si ) i = 1, 2, . . . , n ∂si ∂si where y(s) and z(s) are the contour coordinate functions and s the arc length of the contour. The infinitesimal enclosed area, denoted by dAc in Fig. 8.31 is given by: dAc =

1 r ds 2

For each differential arc length around the contour similar triangles can be formed. Then the total area is the sum of the differential areas of all these triangles: Ac =

n X 1 ri dsi 2 i=1

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

562

As we take the limit ds → 0 for each triangle, the enclosed area is given by 1 Ac = 2

I

c

n Z 1 X ai ri (si ) dsi r(s) ds = 2 i=1 0

(8.27)

Example 8.8. Determine the enclosed area of a thin-walled circular closed cross-section of radius R and constant thickness.

y t1, E1, ν1

T t2, E2, ν2

z

The contour is a circle of radius r(s) = R and the arc length ds = R dθ. Since, the enclosed area is measured about the centerline,

Using Eq. (8.27), we get 1 Ac = 2 which is expected.

R=

ro + ri 2

Z

(R) R dθ = π R2

2π

0

End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

563

Example 8.9. Determine the enclosed area of a thin-walled closed cross-section shown below.

t2

y t4

T t1 z

t5 t3 12"

29"

9"

The enclosed area is a composite of the triangle on the left, the rectangle in the middle, and the semicircle on the right. Thus, I Z Z Z 1 1 1 1 r(s) ds = r1 (s1 ) ds1 + r2 (s2 ) ds2 + r3 (s3 ) ds3 Ac = 2 c 2 1 2 2 2 3 1 1 = × 18 × 12 + 29 × 18 + π × 92 = 884.5 in2 | {z } 2 2 | {z } | {z } Area 1

Area 2

Area 3

Note that it is not always necessary to evaluate r(s), sometimes the enclosed area is trivial. However, the results are the same. End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

564

Example 8.10. Determine the enclosed area of a thin-walled closed cross-section of Example 4.8. Take h = 1 m.

y′

Af b2 3

Af E2 t2

α h/2

1

E4 t4

E1 t1 z′

h/2

2 α

Af

b3

E3 t3 4

E5 t5 Af

The enclosed area is a composite of the triangle on the left, the rectangle in the middle, and the semicircle on the right. Thus, I Z Z 1 1 1 r(s) ds = r12 (s12 ) ds12 + r24 (s24 ) ds24 Ac = 2 c 2 12 2 24 Z Z Z 1 1 1 + r45 (s45 ) ds45 + r53 (s53 ) ds53 + r31 (s31 ) ds31 2 45 2 53 2 31

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

565

The parametric equations about the centroid are found as: y12 = −0.45353 + s12 z12 = −1.81395

y24 = 0.54647 + 0.5 s24 z24 = −1.81395 + 0.866025 s24 y31 = −1.45353 + 0.5 s31

z31 = −0.0818989 − 0.866025 s31

y45 = 0.0464702 + 1.5 cos θ45 z45 = −0.0818989 + 1.5 sin θ45 y53 = 0.0464702 − 1.5 sin θ53

z53 = −0.0818989 + 1.5 cos θ53

The contour normal coordinate from the centroid to each section is: ∂y12 ∂z12 y12 (s12 ) + zi (s12 ) = −1.81395 ∂s12 ∂s12 ∂z24 ∂y24 r24 (s24 ) = − y24 (s24 ) + zi (s24 ) = −1.38023 ∂s24 ∂s24 ∂y31 ∂z31 y31 (s31 ) + zi (s31 ) = −1.29974 r31 (s31 ) = − ∂s31 ∂s31 1 ∂z45 1 ∂y45 r45 (θ45 ) = − y45 (θ45 ) + zi (θ45 ) = −1.5 − 0.0464702 cos θ45 + 0.0818989 sin θ45 R ∂θ45 R ∂θ45 1 ∂y53 1 ∂z53 y53 (θ53 ) + zi (θ53 ) = −1.5 + 0.0818989 cos θ53 + 0.0464702 sin θ53 r53 (θ53 ) = − R ∂θ53 R ∂θ53 r12 (s12 ) = −

where R=

h + b2 sin α + b3 sin α = 1.500 2

Use the absolute values in (recall that the negative values are due to the direction assumed) 1 Ac = 2

Z

0

h

Z 1 b2 r12 (s12 ) ds12 + r24 (s24 ) ds24 2 0 Z Z Z 1 π/2 1 π/2 1 b2 + r45 (θ45 ) R dθ45 + r53 (θ53 ) R dθ53 + r31 (s31 ) ds31 2 0 2 0 2 0

= 6.99839 m2

Note that the stringers do not play any role in the enclosed area. End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

8.5.2

566

Bredt’s formula

x y ds

q(s)

Vy dAc r(s) S.C.

Vz

z

T

Figure 8.32: Thin-walled, single cell beam with an arbitrary cross-sectional contour. Consider a single cell, thin-walled beam with a uniform cross-section along its length. This type of structure is known as a cylindrical shell. The contour of the cross-section has an arbitrary shape with the contour coordinate denoted by s and wall thickness by t(s), as shown in Fig. 8.32. The beam is made of an isotropic and homogeneous material that follows Hooke’s Law. The goal is to determine the twist per unit length, φx , in terms of the shear flow distribution around the contour. The total shear stress component tangent to the contour, Sxs =

q t

the total shear flow q(x, s) is the sum of the shear flow due to the torque T and the shear flow due to the transverse shear loads Vy and Vz acting through the shear center: q(x, s) = qT (x) + qV (x, s)

(8.28)

Note that the shear flow due to is not a function of the contour coordinate s. In order to determine the unit twist of the cross-section, we need to use the material law, the geometry of deformation, and displacements of the contour. We assume that the displacements are continuous and single-values functions of the coordinates x and s.

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

567

Hooke’s Law The shear stress and shear strain are directly related with the shear modulus G of the material (8.29)

Sxs = G γxs Multiplying the above equation by the wall thickness t to get the shear flow expression q = G t γxs

→

γxs =

q Gt

(8.30)

The strain-displacement relationship will express the shear strain γxs in terms of the deformation displacements.

Strain-Displacement Relationship

DEFORMED

UNDEFORMED

x, u y

C′

C

B′ A′

A

Vy

q(s)

S.C.

B

s, vt

Vz

T

Figure 8.33: Thin-walled element in its undeformed and deformed configurations. The shear strain γxs is the reduction in the right angle between line elements originally parallel to the s and x-directions in the undeformed shell wall, as shown in Fig. 8.33. The shear strain is related to the parallel derivatives of the displacement components. Let vt (x, s) be the displacement tangent to the contour (s-direction) of the particle at point (x, s) and u(x, s) be the displacement of this particle in the axial direction (x-direction).

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

568

DEFORMED

vt(x+dx,s) C′

E′ α

B′

u(x+dx,s) A′

u(x,s)

β

D′

C

u(x,s+ds) dx ds

A

B vt (x,s) vt(x,s+ds)

The three adjacent point labeled A, B, and C in the undeformed shell wall displace to position A0 , B0 , and C0 in the deformed shell wall. The reduction in the right angle between line elements originally parallel to s and x-directions is determined as follows π  π α+β+ − γxs = 2 2 α + β = γxs Using Taylor Series expansion for the infinitesimal deformations: tan β ≈ β +

β3 + ··· 3

(Ignore higher order terms)

u(x, s + ds) − u(x, s) tan β ≈ β = = vt (x, s + ds) − vt (x, s)

∂u ds ∂u ∂s = ∂vt ∂s ds + ds ∂s | {z } small

Similarly, tan α ≈ α +

α3 + ··· 3

(Ignore higher order terms)

vt (x + dx, s) − vt (x, s) tan α ≈ α = = u(x + dx, s) − u(x, s)

∂vt dx ∂vt ∂x = ∂u ∂x dx + dx ∂x | {z } small

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

Hence, the shear strain γxs becomes:

569

γxs = α + β ∂u ∂vt + = ∂s ∂x

(8.31)

From Eq. (8.30), q ∂u ∂vt = + Gt ∂s ∂x The tangential displacement vt can be expressed in terms of a rigid body translation vtT and rotation vtR : vt (x, s) = vtT (x, s) + vtR (x, s) Now take the origin of the contour coordinate (s = 0) at some point along the contour with s increasing clockwise. As s transverses the contour and returns to the origin its value is S, where S is the length of the perimeter of the contour, as shown in Fig. 8.34. y

z O

vo

dz

s

wo

θ dy

θ t

ds

A θ

n

A n

t

Figure 8.34: Geometry of the contour. A generic point on the contour is labeled A and has cartesian coordinates [y(s), z(s)]. The unit vectors tangent and normal to the contour at A are denoted by ˆ t and n ˆ , respectively, as is shown in Fig. 8.34, and the angle between the positive z-direction and the tangent to the contour is denoted by θx (s). For a differential length ds along the contour at A we have the trigonometric relations cos θx = The translation is given by

dz , ds

sin θx =

dy ds

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

570

y

z y

A' z

vtT vo

θ

wo

A

vtT (x, s) = wo cos θx + vo sin θx The rotation is given by

y

z θx

r(s)

r(s) vtR A

A'

vtR (x, s) = r(s) θx Hence the total tangential displacement of a typical point on the contour is vt (x, s) = vtT (x, s) + vtR (x, s) = wo cos θx + vo sin θx + r(s) θx

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

571

Now, the shear strain becomes q ∂u ∂vt = + Gt ∂s ∂x dvo dθx ∂u dwo + cos θx + sin θx + r(s) = ∂s dx dx dx

γxs =

(8.32)

Shear flow due to torque Integrating the Eq. (8.32) around the contour, gives I I I I I q dθx ∂u dwo dvo ds = ds + cos θx ds + sin θx ds + r(s) ds Gt ∂s dx dx dx I I I I I q ∂u dwo dvo dθx ds = ds + cos θx ds + sin θx ds + r(s) ds Gt ∂s dx dx dx

(8.33)

Let us work term by term, I I ∂u ds = du = u(x, s) − u(x, 0) = 0 (because the axial displacement of the material ∂s point at s = 0 and s = S is unique) I I I dz cos θx ds = ds = dz = z(s) − z(0) = 0 ds I I I dy sin θx ds = ds = dy = y(s) − y(0) = 0 ds Now recall that, Ac = Hence, Eq. (8.33) becomes,

1 2

I

r(s) ds I

→

I

r(s) ds = 2 Ac

q dθx ds = 2 Ac Gt dx

and the twist is related to the total shear flow as follows I dθx 1 q = ds dx 2 Ac Gt Using Eq. (8.28), we get

dθx 1 = dx 2 Ac

I

qT 1 ds + Gt 2 Ac

I

Since we choose Vy and Vz to act at the shear center, I 1 qV ds = 0 2 Ac Gt Hence

dθx 1 = dx 2 Ac

I

qT ds Gt

(8.34)

qV ds Gt

(8.35)

(8.36)

(8.37)

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

572

Equations (8.36) and (8.37) are valid for isotropic thin-walled cylindrical shell with an arbitrary shaped contour, constant cross-section, and subjected to only a torque T and transverse shear through the shear center. The material may change with the cross-section, thus G = G(s) Under uniform torsion assumption, the twist per unit length is constant φx =

dθx T = constant = dx GJ

(8.38)

where T is the constant torque, G the shear modulus, and J the torsional constant. The above is true whenever the torque, the cross-section and the material do not change with respect to x. Combining Eqs. (8.37) and (8.38), we get I 1 dθx qT = ds dx 2 Ac Gt I T 1 qT = ds GJ 2 Ac Gt From Eq. (8.28), we see that the qT is independent of the coordinate s; thus, I qT T ds = GJ 2 Ac Gt Rearranging the above equation,

qT (2 Ac ) T = GJ 4 A2c G0

I

ds G t G0 qT (2 Ac ) T 1 = GJ G0 4 A2c I ds G t G0 Hence, for a single cell section

4 A2 4 A2 J = I c = n Z ac i ds X dsi ∗ t c t∗ i=1

0

(8.39)

i

and the modulus-weighted thickness is given by

t∗ =

G(s) t G0

Shear modulus G0 is introduced in the definition of torsional stiffness as a reference shear modulus. It is selected for convenience in cross sections where the material can vary from branch to branch. If the cross section is composed of a single homogeneous material, then we take G0 = G, so that t∗ is the actual wall thickness t. It is important to remember that the formula for the torsion constant J above is only valid for single-cell, closed section.

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

573

Now Bredt’s formula, or Bredt-Batho formula, relates the torque and the shear flow via the area enclosed by the contour: T (8.40) T = qT (2 Ac ) → qT = 2 Ac

Example 8.11. The isotropic, uniform and homogeneous closed section in Example 8.9 is subject to a torque T . Determine total shear flow, maximum shear stress, the torsional constant, and the torsional stiffness.

t2

y t4

T t1 z

t5 t3 12"

29"

9"

The geometric and mechanical properties are: E = 3.5 × 106 psi

ν = 0.30

t = 0.02500

T = 50 kips-in

t1 = t2 = t3 = t4 = t5 = t 8.11a) The total shear flow is: qtotal = qpure twist + qpure shear = qpure twist

(there is no shear present in the problem)

The shear flow due to pure torsion is computed using Bredt’s formula, qT =

T = 28.26 lb/in 2 Ac

Note that one kip is equal to 1000 lbs. Since we have pure torsion, q = qT + qV = qT 8.11b) The shear stresses is the shear flow divided by the wall thickness. Since the walls of the

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

574

cross-section have uniform thickness: 28.26 q = = 1.13 ksi t 0.025

Sxs =

8.11c) In order to compute the torsional constant J, we use Eq. (8.39): 4 A2 J=I c ds ∗ c t where

I

c

ds = t∗

I

=

Z

c

ds = G(s) t G0 π/2

−π/2

I

9 dθ1 + t1

c

ds t

Z

0

29

ds2 + t2

Z

0

29

ds3 + t3

Z

0

12

ds4 + t4

1 = {9 π + 29 + 29 + 12 + 12} = 110.274 t Hence, the torsional constant is

Z

0

12

ds5 t5

4 A2 4 (884.5)2 J=I c = = 28378. in4 ds 110.274 ∗ c t 8.11d) The torsional stiffness is:
GJ = 2.5 × 106 (28378) = 7.094 × 109 lb-in2

8.11e) The twist per unit length is:

dθx T 50 × 103 = = dx GJ 7.094 × 109

rad/in End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

575

Example 8.12. Consider the 200 diameter circular section made of two different isotropic materials subject to a torque T . Determine the enclosed area, total shear flow, maximum shear stress, the torsional constant, and the torsional stiffness.

y t1, E1, ν1

T t2, E2, ν2

z

The geometric and mechanical properties are: E1 = 3.5 × 106 psi

E2 = 1.875 × 106 psi

ν1 = 0.30

t1 = 0.12500

ν2 = 0.25

t2 = 0.25000

T = 10 kips-in

8.12a) The enclosed area is: Ac =

1 2

Z

2π

(R) R dθ = π R2 = 3.14159 in2

0

8.12b) The total shear flow is: qtotal = qpure twist + qpure shear = qpure twist

(there is no shear present in the problem)

The shear flow due to pure twist is computed using Bredt’s formula, qT =

T = 1591.5 lb/in 2 Ac

Note that one kip is equal to 1000 lbs.

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

576

8.12c) The shear stresses in the walls number one and two are: q = 12.732 ksi t1 q = = 6.366 ksi t2

Sxs1 = Sxs2

8.12d) In order to compute the torsional constant J, we use Eq. (8.39): 4 A2 J=I c ds ∗ c t where

I

c

ds = t∗

I

c

ds = G(s) t G0

Z

π/2

−π/2

ds1 + G1 t1 G0

Z

−π/2

π/2

ds2 G2 t2 G0

The shear modulus for isotropic materials is defined as G1 =

E1 = 2.5 × 106 psi 2 (1 + ν1 )

G2

=

E2 = 5.0 × 106 psi 2 (1 + ν2 )

Take the reference shear modulus G0 as G0 = G1 = 2.5 × 106 psi Thus,

I

c

Z

Z −π/2 R1 dθ1 R2 dθ2 + G G2 1 −π/2 π/2 t1 t2 G0 G0 π (1) π (1) + = 8 π = 25.1327 G1 G2 t1 t2 G0 G0

ds = t∗

π/2

Hence, the torsional constant is 4 A2 4 (π (1)2 )2 π J=I c = = = 1.5708 in4 ds 8π 2 ∗ t c 8.12e) The torsional stiffness is: (GJ)eff = G0 J = 2.5 × 106 8.12f) The twist per unit length is:


π = 3.93 × 106 lb-in2 2

dθx T 10 × 103 = = dx (GJ)eff 2.5 × 106

rad/in

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

577

End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

578

Example 8.13. For the closed thin-walled section in Example 8.10, determine the torsional constant and the torsional stiffness. y′

Af b2 3

Af E2 t2

α h/2

1

E4 t4

E1 t1 z′

h/2

2 α

Af

b3

E3 t3 4

E5 t5 Af

Take: ν1 = ν2 = ν3 = ν4 = ν5 = 0.25 8.13a) In order to compute the torsional constant J, we use Eq. (8.39): 4 A2 J=I c ds ∗ c t where

I

c

ds = t∗

I

Z h Z b3 Z b2 ds ds12 ds24 ds31 = + + G G G G(s) 12 24 31 c 0 0 0 t12 t24 t31 t G G G G0 0 0 0 Z π/2 Z π/2 R dθ45 R dθ53 + + G45 G53 0 0 t45 t53 G0 G0

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

579

The shear modulus for isotropic materials is defined as G12 = G24 = G31 = G45 = G53 =

E1 2 (1 + ν1 ) E2 2 (1 + ν2 ) E3 2 (1 + ν3 ) E4 2 (1 + ν4 ) E5 2 (1 + ν5 )

= 53.3333 MPa = 80.0000 MPa = 80.0000 MPa = 133.333 MPa = 106.667 MPa

Take the reference shear modulus G0 as G0 = G12 = 53.3333 MPa The thickness are t12 = 0.05 m

t24 = 0.1 m

t31 = 0.1 MPa

t45 = 0.1 m

t53 = 0.0667 m

Thus, I

c

ds = t∗

Z

0

h

ds12 + G12 t12 G0

Z

b3

0

ds24 + G24 t24 G0

Z

0

b2

ds31 + G31 t31 G0

Z

0

π/2

R dθ45 + G45 t45 G0

Z

0

π/2

R dθ53 = 73.7629 G53 t53 G0

Hence, the torsional constant is 4 A2 4 (6.99839)2 = 2.65594 m4 J=I c = ds 73.7629 ∗ c t 8.13b) The torsional stiffness is : (GJ)eff = G0 J = (53.3333) (2.65594) = 141.65 MPa-m2 8.13c) The twist per unit length is: dθx T T = = dx (GJ)eff 141.65 × 106

rad/in End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

580

Example 8.14. From Example 4.7: The thin-walled cross-section shown has a circular contour with radius a. The wall thickness is t everywhere except the top-right quarter (section 1) where it is tapered with t at 2 and 2 t at 1. Assume, only a positive torque of 10000 lb-in is applied. Determine the enclosed area, total shear flow, maximum shear stress, the torsional constant, and the torsional stiffness.

2 Section 1 y z

Section 2

θ

1

C a t

t

The geometric and mechanical properties are: E = 3.5 × 106 psi

ν = 0.30

T = 10 kips-in

a = 100

t = 0.0200

The thickness for section 1-2 is t1 (θ1 ) = 2 t



1−

θ1 π



= 0.04 − 0.0127324θ1

and for section 2-1, t2 (θ2 ) = t = 0.02 8.14a) The enclosed area is: 1 Ac = 2

Z

2π

(a) a dθ = π a2 = 3.14159 in2

0

8.14b) The total shear flow is: qtotal = qpure twist + qpure shear = qpure twist

(there is no shear present in the problem)

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

581

The shear flow due to pure twist is computed using Bredt’s formula, qT =

T = 1591.5 lb/in 2 Ac

8.14c) The maximum shear stresses in the walls number one and two are: q q q = = = 79.578 ksi t (θ = π/2) t 1 1 t1 min q q = = = 79.578 ksi t2 t

Sxs1 = Sxs2

8.14d) In order to compute the torsional constant J, we use Eq. (8.39): 4 A2 J=I c ds ∗ c t where

I

c

ds = t∗

I

c

ds = G(s) t G0

I

c

ds = t

Z

0

π/2

a dθ1 + t1

Z

2π

π/2

a dθ2 = 290.059 t2

Thus, the torsional constant is 4 A2 4 (3.14159)2 J=I c = = 0.136105 in4 ds 290.059 ∗ c t 8.14e) The torsional stiffness is: T = G0 J dθx dx The shear modulus for isotropic materials is defined as (GJ)eff =

G0 = G1 = G2 = Thus

E = 5.0 × 106 psi 2 (1 + ν)


(GJ)eff = G0 J = 5.0 × 106 (0.136105) = 0.680 × 106 lb-in2

8.14f) The twist per unit length is:

dθx T 10 × 103 = = dx (GJ)eff 0.680 × 106

rad/in End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

582

Example 8.15. Compare the torsional shear stress and the torsional rigidity of two thin-walled tubes, on of them with a closed section and the other one with a small longitudinal gap in the wall, as shown in Figure.

R

R

t

t

8.15a) Torsional rigidity: For the closed section, Ac = π R 2

I

c

For the open section,

ds 2πR = t t

1 J= 3

Thus,

GJ

GJ

Z

2π

4 A2 J = I c = 2 π R3 t ds c t

t3 R dθ =

0

CLOSED

OPEN

2 π R t3 3

2 π R3 t = =3 2 π R t3 3



R t

2

Therefore, if R/t > 10, which must be true of a thin-walled tube,  2 GJ R CLOSED = 3 > 300 t GJ OPEN

We see that the closed torque tube has over 300 times the resistance to twist than the one with the lengthwise gap.

8.15b) Torsional shear stress: For closed section, SCLOSED =

q t

q=

T T = 2 Ac 2 π R2

SCLOSED =

T 2 π R2 t

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

583

For open section, SOPEN =

Tt J

Thus,

If R/t > 10, for the same torque,

J= S

S

2 π R t3 3

CLOSED

=

1 3



<

1 30

OPEN

S

S

CLOSED

SOPEN =

t R

3T 2 π R t2



OPEN

The shear stress in the open tube is over thirty times greater than it is in the closed tube. End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

8.5.3

584

Shear flow due to transverse shear

In the case of closed sections, the equations for the shear flow distribution for open sections, still applies, although no boundary condition is readily available to integrate this equation. Consider an arbitrary thin-walled closed section, as shown in Fig. 8.35.

y z

qv(s)

y s

q(s=0)=0

y

z

q(s)

z

qo

Figure 8.35: Superposition of shear flows: problem consisting of an open section and a section with a constant shear flow. First, we make a fictitious cut to the beam’s cross-section at an arbitrary point as shown in Fig. 8.35. Now, we can readily find the shear flow distribution, q(s), for this auxiliary problem using the procedures described for computing shear flow in open sections. In general, the shear forces Vy and Vz will not be applied at the shear center for this auxiliary problem, i.e., there are no a priori restrictions on where we make the cut and consequently, the beam will twist. Secondly, in order to find the unknown shear, q0 , in the closed section of Fig. 8.35, we impose the condition of zero unit twist in Eqs. (8.36). We can justify this assumption by the fact that the net torque

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

585

due to shear loads in closed section must be zero. Hence, we separate the problem into two parts: q = qV + qt = q(s) + qo In other words, I I I I 1 qV 1 qV 1 qV q(s) + q0 1 ds = ds ds = ds = 2 Ac c G t 2 Ac G0 c G 2 Ac G0 c t∗ 2 Ac G0 c t∗ t G0 I I I I 1 q(s) 1 q0 1 q(s) q0 ds = ds + ds = ds + =0 ∗ ∗ ∗ 2 Ac G0 c t 2 Ac G0 c t 2 Ac G0 c t 2 Ac G0 c t∗ Hence, the closing shear flow, q0 , is evaluated by I

q(s) ds t∗ q0 = − c I ds ∗ c t where

I

c

n

X q(s) ds = t∗ i=1

Z

0

ai

qi (si ) dsi , t∗ (si )

I

c

(8.41)

n

X dsi = t∗ i=1

Z

ai

0

dsi t (si ) ∗

Then the total shear flow acting on each cross-sectional branch will be given by qV (x, s) = q(x, s) + q0 If the cross-section is of homogeneous material, then t∗ = t.

8.5.4

Solution procedure to obtain shear center in closed sections

The evaluation of the shear center is identical as for the case of open-sections. In solving these problems, consider a eight-step solution procedure: 1. Calculate the centroid: yc , zc 2. Determine the second moments of area ratios about the centroid: Rzz , Ryy , Ryz 3. Determine the parametric equations about the centroid: yi (si ), zi (si ) 4. Find the shear flow distribution in each branch, qi (si ). 5. Set the unit twist to zero for the shear force acting through the S.C., to determine unknown q0 : use Eq. (8.41) 6. Determine the shear center. 7. Determine the shear due to torque alone. 8. Determine the total shear stress along the contour.

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

586

Example 8.16. Shear Flow in Symmetrical Closed Sections Consider the symmetrical thin-walled monocoque closed section shown. The beam’s total length is 100 inches. The following data is given:

y

500 lb 4 z

2

3

1 1

L

1 2

2

500 lb

Figure 8.36: Symmetrical thin-walled monocoque closed section. Portion (i) 12 23 34 41

Shear Modulus (Gi ) (psi) 12.0×106 3.75×106 12.0×106 3.75×106

Thickness (ti ) (in) 0.05 0.10 0.05 0.10

Length (ai ) (in) 2 1 2 1

8.16a) Determine the stress field for the beam. First of all the centroid is located at the center of the geometry, because of symmetry. The only loads applied are torsional loads at the centroid: Mxx = T = 1000 lb–in. All others are zero. Thus, all normal stresses are zero and the only remaining stresses in our problem are shear stresses due to torsion. These shear stresses are calculated as follows: qT qT Sxs1 = Sxs2 = t1 t2

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

587

The shear flow due to torsion is found using Bredt’s formula, Eq. (8.40): qT =

T 2 Ac

The total enclosed area is given by Eq. (8.27): Ac =

1 2

1 = 2

I

c

Z

4 Z 1 X ai ri (si ) dsi 2 i=1 0 Z Z Z 1 a23 1 a34 1 a41 r12 (s12 ) ds12 + r23 (s23 ) ds23 + r34 (s34 ) ds34 + r41 (s41 ) ds41 2 0 2 0 2 0

r(s) ds = a12

0

Measuring the ri about the centroid: r12 = 0.5

r23 = 1 r34 = 0.5

r41 = 1

Hence the enclosed area is Z Z Z Z 1 2 1 1 1 2 1 1 Ac = 0.5 ds12 + 1.0 ds23 + 0.5 ds34 + 1.0 ds41 2 0 2 0 2 0 2 0 1 = {1 + 1 + 1 + 1} = 2 in2 2 Thus, the shear flow due to torsion is qT =

T = 250 lb/in 2 Ac

Now, the shear stresses in each branch are found as follows: qT = 5000 psi, t12 qT = = 5000 psi, t34

qT = 2500 psi t23 qT = = 2500 psi t41

Sxs12 =

Sxs23 =

Sxs34

Sxs41

8.16b) Determine the strain field for the beam. The strains are found using Hooke’s Law. The nonzero strains are Sxs12 = 4.17 × 10−4 in/in, G12 Sxs34 = = 4.17 × 10−4 in/in, G34

Sxs23 = 6.67 × 10−4 in/in G23 Sxs41 = = 6.67 × 10−4 in/in G41

γxs12 =

γxs23 =

γxs34

γxs41

8.16c) Determine the torsional constant for the beam’s cross-section.

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

588

In order to compute the torsional constant J, we use Eq. (8.39): 4 A2 J=I c ds ∗ c t where I

c

ds = t∗

I

c

ds = G(s) t G0

Z

a12

0

ds12 + t∗12

Z

a23

0

ds23 + t∗23

Z

a34

ds34 + t∗34

0

Z

a41

0

ds41 t∗41

Take the reference shear modulus G0 as G0 = 3.75 Msi The modulus weighted thickness are G23 t23 = 0.1000 G0 G41 t41 ∗ = t41 = 0.1000 G0

G12 t12 = 0.1600 , G0 G34 t34 ∗ = t34 = 0.1600 , G0

t23 ∗ =

t12 ∗ =

Thus,

I

c

ds = t∗

Z

0

2

ds12 + 0.16

Hence, the torsional constant is

Z

0

1

ds23 + 0.10

Z

2

0

ds34 + 0.16

Z

0

1

ds41 = 45.0 0.10

4 A2 J = I c = 0.355556 in4 ds ∗ c t 8.16d) Determine the torsional rigidity of the beam’s cross-section. The torsional rigidity is defined as: (GJ)eff =

T lb-in2 = G0 J = (3.75 × 106 )(0.355556) = 1.3333 × 106 dθx rad dx

8.16e) Determine the angle of twist per unit length and the total angle of twist at the free end. The angle of twist per unit length is defined as: φx =

dθx T = = 7.5 × 10−4 rad/in dx (GJ)eff

The total angle of twist at the free end is θx

x=L

=


dθx x = 7.5 × 10−4 (100) = 7.5 × 10−2 rad dx x=L

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

589

End Example 

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

590

Example 8.17. Shear Flow in Symmetrical Closed Sections For the symmetrically thin-walled closed section shown, determine the shear flow in all the panels and determine the location of the shear center. The stringers carry only direct stresses but the skin is fully effective in carrying both shear and direct stresses. The area of each stringer is 100 mm2 . Take h = 80 mm and t = 0.32 mm.

3h 2

2t

1

y h

z

S.C.

t 3

2t

Vy 2t

4

Figure 8.37: Shear flow convention 8.17a) Understanding the problem. One can often reduce the amount of computation by giving some thought to the problem. i. Since this is a symmetrical thin-walled section, we do not need to calculate Iyz because it is zero. Therefore, only Izz and Iyy needs to be calculated. The location of the centroid will be needed. By inspection yc = 0 and zc needs to be calculated.

cross section ii. Next, before we Figure proceed 3. letThin-walled us decide where to take the torque equivalence. Recall that the torque equivalence about a point, say Os , determines the location of the shear center. Further, arbitrarily apply shear loads Vy and Vz , in the positive sense, at the shear center S = S(ey , ez ). Note because of symmetry ey = 0. Let us apply it at the middle of skin 23. If point Os middle of skin 23, then we should consider Fig. 8.38 BASIC EQUATIONS: Shear flow distribution

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

591

2

2t

y

1

Vy ez

z

S.C.

2t

t 3

4

2t

2

1

y O

z

t 4

3

Figure 8.38: Shear flow convention Since ey = 0, consider only the case when Vz = 0 and Vy > 0 is applied through the shear center then no torsion exists. Thus the shear flow distribution is given by

qi (si ) = qoi +

= qoi +





A∗s

A∗s

Q∗zs

Q∗zs

Q∗ys

Q∗ys

 1  A∗    0     0

        

0 −

0

Vy ∗ R yy      Vy    − ∗ Ryz

0

1 ∗ Ryz

−

1 ∗ Rzz                 

1 ∗ Ryy

1 ∗ Ryz

= qoi +



            

0 0 −Vy

          

Vy ∗ Vy ∗ Vy ∗ ∗ Qzs − ∗ Qys = qoi + ∗ Qzs Ryy Ryz Ryy

where qoi is found by evaluating qi (si−1 = 0). When the thickness is constant along each branch: ti (si ) = ti . Further for a symmetric cross-section: qi (si ) = qoi + Shear flow across the stiffener, 4qs =

Vy ∗ ∗ Qzs Ryy

Vy Qzsf Ryy

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

592

Twist per unit length of the beam Assuming the beam is made of the same homogeneous material, the condition of no unit twist of the section due to Vy is I

q(s) ds t q0 = − c I ds c t

(8.42)

Torque equivalence Torque equivalence about a point, say Os , determines the location ez of the shear center. For a positive torque counterclockwise (see Fig. 8.38), for Vz = 0 and Vy > 0 −Vy ez = Tq Tq =

n Z X

ai

0

i

ri (si ) × qi (si ) dsi =

XZ

ai

ri (si ) qi (si ) dsi

0

i

where i represents the ith branch, ai the length of the branch, and ri (si ) the distance perpendicular from Os to the contour si and is defined as:     dyi dzi yi + zi i = 1, 2, . . . , n ri (si ) = − dsi dsi 8.17b) Centroid 3h Vy

2

2t

1

y zc

h

z

2t

C

t 3

4

2t

By placing the origin at point 2 and we can calculate zc and yc and it can be shown that the centroid is located at: zc =

Qy = 0.123918 m A

yc

=

Qz =0 A

8.17c) Second Moment of area Izz , about the centroid, using thin-walled assumption is Izz = Izz12 + Izz23 + Izz34 + Izz41 + Izzf 1 + Izzf 2 + Izzf 3 + Izzf 4 = 1.17248 × 10−6 m4 Second Moment of area Iyy , about the centroid, using thin-walled assumption is Iyy = Iyy12 + Iyy23 + Iyy34 + Iyy41 + Iyyf 1 + Iyyf 2 + Iyyf 3 + Iyyf 4 = 8.32844 × 10−6 m4

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

593

Second Moment of area Iyz , about the centroid, using thin-walled assumption is Iyz = Iyz12 + Iyz23 + Iyz34 + Iyz41 + Iyzf 1 + Iyzf 2 + Iyzf 3 + Iyzf 4 = 0 The Second moments of area ratios are 2 ∆I = Izz Iyy − Iyz = 9.76493 × 10−12 m8

Ryy = 1.17248 × 10−6 m4

Rzz = 8.32844 × 10−6 m4

Ryz = ∞

8.17d) Shear flow distribution

q23a q12b q12a

q12 q23b

q41 q23

q41b

q34a

q34b

q41a q34

Figure 8.39: Shear flow convention for statically equivalence when taking the torque at point O. When Vy > 0 and Vz = 0 is applied through the shear center then no torsion exists and the shear flow distribution for this problem is given by

qi (s) = qoi +



As

Qzs

Qys

 1  A    0     0

0 −

1 Ryz

1 Rzz

0 −

1 Ryy

1 Ryz



            

0 0 −Vy

          

where qoi is found by evaluating qi (s = 0). Substitute the moments of area in the above

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

594

equation to get

qi (si ) = qoi +



As

Qzs

Qys

      

0

Vy  R  yy     0

qi (s) = qoi + 852893Qzsi Vy

            

(8.43)

In order to calculate the shear flow in each flange, we need to first find the yi (si ) for each part. Following the assumed flow convention in Fig. 8.40 and from the geometry of the crosssection we get:

q23a q12b q12a

q12 q23b

q41 q23

q41b

q34a

q34b

q41a q34

Figure 8.40: Shear flow convention for statically equivalence when taking the torque at point O

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

595

For skin 41, s41 = 0 s41 = 0.08

⇒

z41 = 0.24;

⇒

z41 = 0.24;

⇒

y41 = −0.04

(8.44)

y41 = 0.04

(8.45)

z12 = 0.24;

y12 = 0.04

(8.46)

⇒

z12 = 0.00;

y12 = 0.04

(8.47)

⇒

z23 = 0.00;

y23 = 0.04

(8.48)

⇒

z23 = 0.00;

y23 = −0.04

(8.49)

⇒

z34 = 0.00;

y34 = −0.04

(8.50)

For flange 12, s12 = 0 s12 = 0.24 For flange 23, s23 = 0 s23 = 0.08 For flange 34, s34 = 0 s34 = 0.24

z34 = 0.24;

⇒

y34 = −0.04

Let us make the cut at skin 41. Thus value the shear flow in skin 41 is q41 (s41 = 0) = q0 For skin 41, y41 (s41 )

=

−0.04 + s41

z41 (s41 )

=

0.24

Qys41

=

0.0001536 s41

=

Z

Qzs41

s41

0

q41 (s)

=

qo41

=

q41 (s)

=

y41 (s41 ) t41 ds41 = −0.0000256 s41 + 0.00032 s241


qo41 + −21.8341 s41 + 272.926 s241 Vy

q0


q0 + −21.8341 s41 + 272.926 s241 Vy

For stringer 1, (a41 = 0.08)

q12b = q12a + 852893 Qzf1 Vy

(8.51)

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

Qzf1 = 4 × 10−6 At the boundary q12a = q41 (s41 = a41 ) = q0 − 1.6 × 10−6 Vy Thus q12b = q0 + 3.41157 Vy For skin 12, (t12 = t) y12 (s12 )

=

0.04

z12 (s12 )

=

0.24 − s12

=

Z

Qys12

Qzs12

s12

0

=

Z

s12

z12 (s12 ) t12 ds12 = 0.0001536 s12 − 0.00032 s212 y12 (s12 ) t12 ds12 = 0.0000256 s12

0

q12 (s)

= qo12 + (21.8341 s12 ) Vy

qo12

= q12b = q0 + 3.41157 Vy

q12 (s)

= q0 + (3.41157 + 21.8341 s12 ) Vy

For stringer 2, (a12 = 0.24) q23b = q23a + 852893 Qzf2 Vy Qzf2 = 4 × 10−6 At the boundary q23a = q12 (s12 = a12 ) = q0 + 8.65175 Vy Thus q23b = qo + 12.0633 Vy

596

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

For skin 23, (t23 = t) y23 (s23 )

=

−0.04 + s23

z23 (s23 )

=

0.0

Qys23

=

Z

Qzs23

=

s23

z23 (s23 ) t23 ds23 = 0

0

Z

s23

0

q23 (s)

=

qo23

=

q23 (s)

=

y23 (s23 ) t23 ds23 = 0.0000128 s23 − 0.00016 s223


qo23 + 10.917 s23 − 136.463 s223 Vy

q23b = q0 + 12.0633Vy


q0 + 12.0633 + 10.917 s23 − 136.463 s223 Vy

For stringer 3, (a23 = 0.08)

q34b = q30a + 852893 Qzf3 Vy Qzf3 = −4 × 10−6 At the boundary q34a = q23 (s23 = h) = q0 + 12.0633 Vy Thus q34b = q0 + 8.65175 Vy For skin 34, (t34 = t) y34 (s34 )

=

−0.04

z34 (s34 )

=

s23

=

Z

Qys34

Qzs34

s34

0

=

Z

0

s34

z34 (s34 ) t34 ds34 = 0.00032 s234

y34 (s34 ) t34 ds34 = −0.0000256 s234

q34 (s)

=

qo34 + (−21.8341 s34 ) Vy

qo34

=

q34b = q0 + 8.65175 Vy

q34 (s)

=

q0 + (8.65175 − 21.8341 s34 ) Vy

597

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

598

Check: As we cross the stringer 4, we should get q0 , that is the shear flow we assumed acting on that point. Thus for stringer 4, (a34 = 0.24) q41b = q41a + 852893 Qzf4 Vy Qzf4 = −4 × 10−6 At the boundary q41a = q34 (s34 = a34 ) = q0 + 3.41157 Vy Thus q41b = q0 ...GOOD! Thus the shear flow acting on each skin is: q12 (s12 ) = q0 + (3.41157 + 21.8341 s12 ) Vy
q23 (s23 ) = q0 + 12.0633 + 10.917 s23 − 136.463 s223 Vy q34 (s34 ) = q0 + (8.65175 − 21.8341 s34 ) Vy


q41 (s41 ) = q0 + −21.8341 s41 + 272.926 s241 Vy

8.17e) Twist per unit length of the beam: Assuming the beam is made of the same homogeneous material, the condition of no unit twist of the section due to Vy is I I 1 qV qV 1 ds = ds ∗ 2 Ac G0 t 2 Ac G0 t n Z ai n Z ai X X qi (si ) qi (si ) 1 1 = dsi = dsi = 0 2 Ac G0 i 0 ti 2 Ac G0 i 0 ti 4 Z ai X qi (si ) = dsi = 0 ti i=1 0 Substitute the equations for the shear flow at the contour origin. Hence, the shear flows due to the transverse shear force are determined such that they are statically equivalent to and they result in no unit twist of the beam: Z a23 Z a34 Z a41 Z a12 q23 (s23 ) q34 (s34 ) q41 (s41 ) q12 (s12 ) ds12 + ds23 + ds34 + ds41 = 0 t t t t41 12 23 34 0 0 0 0 1125 q0 + 7539.57 Vy = 0 Thus, q0 = −6.70184 Vy

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

599

Thus the shear flow acting on each skin is:
q12 (s12 ) = −3.29027 + 21.8341 s212 Vy


q23 (s23 ) = 5.36148 + 10.917 s23 − 136.463 s223 Vy

q34 (s34 ) = (1.9499 − 21.8341 s34 ) Vy


q41 (s41 ) = −6.70184 − 21.8341 s12 + 272.926 s212 Vy

Note that same results would have been obtained if we ignored q0 all along (starting with q41 (s41 = 0) = 0) and obtained q0 using Eq. (8.42). 8.17f) Horizontal location, ez , of the shear center

q23a q12b q12a

q12 q23b r23

q41

r41 q23

q41b

q34a r34 q34b

q41a q34

Figure 8.41: Shear flow convention for statically equivalence when taking the torque at point O Torque equivalence about point Os determines the location of the shear center: −Vy ez = Tq

8.5. CLOSED SINGLE-CELL THIN-WALLED SECTIONS

600

where Tq is the torque about point Os due the shear flows (counterclockwise>0): Tq =

XZ

0

i

=

Z

0

ai

a12

ri (s) × qi (s) dsi

q12 (s12 )r12 ds12 +

Z

q23 (s23 )r23 ds23 +

r12 = −



r23 = −

Z

a34

q34 (s34 )r34 ds34 +

0

0

where

Thus,

a23

dz12 ds12





dz23 ds23

y12 +



r34 = −



dz34 ds34



r41 = −



dz41 ds41





y23 +

dy12 ds12





dy23 ds23

Z

a41

q41 (s41 )r41 ds41

0

z12 = 0.04



z23 = 0.0

y34 +



dy34 ds34



z34 = 0.04

y41 +



dy41 ds41



z41 = 0.24

Tq = −0.147132 Vy As mentioned before, for torque equivalence about Os , with Vy > 0 and Vz = 0, is −Vy ez −Vy ez

= Tq = −0.147132 Vy

Thus ez = 0.147 m End Example 

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

8.6

601

Analysis of Thin-walled Multi-Cell Closed Sections

In the previous sections, we analyzed closed, single-cell, thin-walled sections. The analysis can be extended to the case of cross-sections with closed, multi-cell, thin-walled sections.

8.6.1

Bending

Bending analysis does not depend on the shape of the beam’s cross-section, thus methods discussed in the previous chapter apply here.

8.6.2

Pure Torsion

First let us consider the analysis of closed, multi-cell, thin-walled sections subject to torsion. We showed that the shear stress due to torsion between the web and the skin junctions is constant.

Figure 8.42: Multicell thin-walled beam. Consider a thin-walled beam with n-cell closed sections subjected to a constant torque T , as shown in Fig. 8.42. Each cell will have constant shear flow.

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

602

skin

qi-1 q1

q2

qi+1 qn-1

qi

qn

web

Figure 8.43: Shear flow in a Multicell thin-walled beam cross-section. A slice of the beam of infinitesimal span, dx, is shown in Fig. 8.43. Let us define the unknown shear due to shear flow as shown in figure. Now we have n unknown shear flows to be determined in an n-cell thin-walled beam, and these shear flows correspond to the shear flows in the corresponding skins. As discussed in section 8.1.6, the shear flow continuity requirement must hold: the shear flow going into the connection must equal the sum of the outgoing flows. Thus, the shear flow in the each web is determined by using Eq. (8.11). skin

q2 q1 q1

q12 q1

q23 q21 q2

web

qskin1 = q1 , qskin2 = q2 ,

qweb12 = q1 − q2 qweb21 = q2 − q1

(q1 = q12 + q2 ) (q2 = q21 + q1 ),

qweb23 = q2 − q3

(q2 = q23 + q3 )

This result once again justifies the term shear flow to denoted the integration of the shear stresses through the wall thickness. If the walls of the cross-section are viewed as channels in which a fluid is circulating, the conservation of mass at connection points between channels requires the mass flow going into the connection to be equal to that flowing out of the connection. The very same continuity condition must hold for shear flows. In order to solve for the n unknowns, we need n independent equations. Clearly, the total sectional torque, T , carried by the section equals the sum of the torques carried by each individual cell, Ti , where

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

603

i is the cell number. Using the Bredt-Batho formula, Eq. (8.40), T =

n X i=1

Ti =

n X

qTi (2 Aci ) = 2

i=1

n X

(8.52)

qi Aci

i=1

where Ti is the torque in the ith cell, Aci the enclosed area of the ith cell, n the number of cells. Unfortunately, this provides a single equation only, but there are n unknown shear flows, one for each cell. Thus n − 1 set of equations is needed to complete the set. Since only one deformation of the cross-section is a rotational deformation, φx , each cross-section does not distort is its own place. Thus the n − 1 equations are obtained from the compatibility equation: the twist rates of the various cells must be identical. In other words, φx = φx 1 = · · · = φx i = · · · = φx n dθx dθx dθx dθx = = ··· = = ··· dx 1 dx i dx n I dx I I I 1 1 q1 qi qn q 1 1 ds = ds1 = · · · = dsi · · · = dsn 2 Ac G0 t∗ 2 Ac1 G01 1 t∗1 2 Aci G0i i t∗i 2 Acn G0n n t∗n Simplifying the above 1 Ac1 G01

I

1

q1 1 ds1 = · · · = t∗1 Aci G0i

I

i

qi 1 dsi · · · = t∗i Acn G0n

I

n

qn dsn t∗n

(8.53)

Equations (8.52) and (8.53) together provide the 1 + (n − 1) = n equations needed to solve for each of the n shear flows, qi , in the cells of a multi-cell section under torsion. For a uniform homogeneous cross-section, Eq. (8.53) becomes I I I q1 1 qi 1 qn 1 ds1 = · · · = dsi · · · = dsn Ac 1 1 t 1 Aci i ti Acn n tn

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

604

Example 8.18. Two-cell thin-walled section under torsion The thin-walled cross-section shown in Fig. 8.44 represents a highly idealized airfoil structure: the curved portion is the leading edge, the vertical, thicker web is the spar, and the trailing straight segments are the aft portion of the airfoil. The airfoil has a total length of L and consists of two closed cells. The x-axis is parallel to the length of the length of the airfoil, and the y- and z-axes are in a plane parallel to the cross-sections. The airfoil is subject to a case of pure torque T at the tip of the wing. (Thus, from equilibrium the airfoil is subject to equal and opposite directed torques T at each end.) Let us designate the left cell as cell 1 and the right cell as cell 2, as shown in figure.

L

3R 1 t2, G2 t1, G1

Cell 1 R

Cell 2 t12, G12

3 t2, G2

2 Figure 8.44: A two-cell thin-walled section under torsion. The geometric and mechanical properties are: Portion (i) 12 (skin) 12/21 (web) 23 (skin) 31 (skin)

Shear Modulus (Gi ) 2G 4G G G

Thickness (ti ) 2t 3t t t

Length (ai ) πR √2 R √10 R 10 R

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

605

Take G = 6.0 × 106 psi, t = 0.0500 , R = 4800 and T = 10000 lb–in. Take the reference coordinate the center of the semicircle. 8.18a) Determine the shear flows due to torsion.

1

1 q12

q1 q1

q2

3

q2 3

q21

q2

q1 2

2

The enclosed area of each cell is given by Ac 1 =

π R2 , 2

Ac2 =

(3 R)(2 R) = 3 R2 2

Hence the first equation is the total torque: T =2

2 X

→

qi Aci

i=1

T = π R2 q1 + 6 R2 q2

10000.0 = 7238.23 q1 + 13824.0 q2

→

Now, the unit twist compatibility equations is given by: dθx dθx dθx = = ··· dx dx 1 dx 2

The unit twist for cell 1 and for the common web shear flow is: Z a12s  I Z a21w dθx q1 q1 1 1 q21 ds1 = ds12s + ds21w = dx 1 2 Ac1 G01 1 t∗1 2 Ac1 G01 t∗12s t∗21w 0 0

where G01 = 2 G, the weighted modulus thicknesses are t∗12s =

2G G12s t12s = (2 t) = 2 t, G01 2G

t∗21w =

G21w 4G t21w = (3 t) = 6 t G01 2G

and the shear flow across the web is q1 = q21 + q2

→

q21 = q1 − q2

Thus, dθx = dx 1



1 2

(Z

πR

πR 0 (2 G) 2 (2 + 3 π) q1 q2 = − 12 G π R t 6GπRt 2

q1 ds12s + 2t

Z

0

2R

q1 − q2 ds12w 6t

)

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

606

The unit twist for cell 2 and for the common web shear flow is: Z a12w  I Z a23 Z a31 1 q12 dθx q2 1 q23 q31 ds2 = ds12w + ds23 + ds31 = dx 2 2 Ac2 G02 2 t∗2 2 Ac2 G02 t∗12w t∗23 t∗31 0 0 0 where G02 = G, the weighted modulus thicknesses are t∗12w =

4G G12w t12w = (3 t) = 12 t, G02 G

t∗23 =

G23 G t23 = (t) = t, G02 G

t∗31 =

G31 G t31 = (t) = t G02 G

and the shear flow across the web is q1 + q12 = q2

→

q12 = q2 − q1

Thus, ) (Z Z √10R Z √10R a2R dθx 1 q2 q2 q2 − q1
ds12w + ds23 + ds31 = dx 2 12 t t t 2 3 R2 (G) 0 0 0  √  1 + 12 10 q2 q1 = − 36 G R t 36 G R t Hence the second equation is

→

dθx dθx = ··· dx 1  dx 2√  1 + 12 10 q2 (2 + 3 π)q1 q2 q1 − = − 36 G R t 36 G R t  12 G πR t √6 G πR t −6 − π − 12 10 π (3 + 5 π) q1 + q2 = 0 18 G π R t 36 G π R t

or 2.10452 × 10−8 q1 − 3.684142 × 10−9 q2 = −1.92901 × 10−9 q1 + 7.512988 × 10−8 q2 →

2.29743 q1 − 7.8814 q2 = 0

Solving the system of equations: √     (3 + 5 π) (−6 − π − 12 10 π)  q1   0   18 G π R t  36 G π R t =       2 2 q T 2 πR 6R       T 0.887475 q1 0.204474 → = = q2 0.0596042 0.258699 R2 

Hence the shear flows q1 and q2 acting on the skins are: q1 = 0.204474

T = 0.887475 lb/in, R2

q2 = 0.0596042

T = 0.258699 lb/in R2

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

The shear flow acting on the web is q12 = −q21 = q2 − q1 = −0.14487

T = −0.628776 lb/in R2

8.18b) Determine the shear stresses due to torsion in each skin and web.

q1 T = 0.102237 2 = 8.87475 psi t12s R t T q2 = 0.0596042 2 = 5.17397 psi = t23 R t T q2 = 0.0596042 2 = 5.17397 psi = t31 R t q12 T = = −0.024145 2 = −2.09592 psi t12w R t

S12s = S23s S31s S12w

8.18c) Determine the shear strains due to torsion in each skin and web.

S12s T = 0.739563 µ in/in = 0.0511186 G12s G R2 t S23s T = = 0.862329 µ in/in = 0.0596042 G23 G R2 t S31s T = = 0.862329 µ in/in = 0.0596042 G31 G R2 t S12w T = = −0.00603625 = −0.0873301 µ in/in G12 G R2 t

γ12s = γ23s γ31s γ12w

8.18d) Determine the beam’s cross-sectional twist per unit length.

The twist per unit length is:

dθx dθx dθx = = dx dx 1 dx 2

(2 + 3 π) q1 q2 T dθx −8 − = 0.0588042 rad/in = 3 = 1.77241 × 10 dx 1 12 G π R t 6GπRt G R t   √ 1 + 12 10 q2 dθx q1 T − = 0.0588042 = 1.77241 × 10−8 rad/in = dx 2 36 G R t 36 G R t G R3 t

Thus,

dθx T = 0.0588042 = 1.77241 × 10−8 rad/in dx G R3 t

8.18e) Determine the beam’s cross-sectional torsional rigidity. The torsional rigidity is found using the fact that: dθx T = dx GJeff

607

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

Since

608

dθx dθx dθx = = dx dx 1 dx 2

and

T T T T = = 1.77241 × 10−8 rad/in = = 0.0588042 GJeff GJeff 1 GJeff 2 G R3 t

For cell 1

T T1 = 0.0588042 GJeff1 G R3 t

→

GJeff1 =

T1 0.0588042

T G R3 t

where T1 = 2 Ac1 q1 = 0.642375 T = 6423.75 lb–in Hence GJeff1 =

0.642375 T = 3.47721 π G R3 t = 362.431 Gsi T 0.0588042 G R3 t

For cell 2 T2 T = 0.0588042 GJeff2 G R3 t

→

GJeff2 =

T2 0.0588042

T G R3 t

where T2 = 2 Ac2 q2 = 0.357625 T = 3576.25 lb–in Hence GJeff1 =

0.357625 T = 1.93584 π G R3 t = 201.774 Gsi T 0.0588042 G R3 t

For the cross-section T T = 0.0588042 GJeff G R3 t

→

GJeff =

T T 0.0588042 G R3 t

= 5.41305 π G R3 t = 564.205 Gsi

Note that GJeff = GJeff1 + GJeff2 In fact, T =

n X i=1

Ti = T1 + T2 + · · · + Ti + Tn

φx GJeff = φx1 GJeff1 + φx2 GJeff2 + · · · + φxi GJeffi + · · · + φxn GJeffn GJeff = GJeff1 + GJeff2 + · · · + GJeffi + · · · + GJeffn n X GJeff = GJeffi i=1

8.6. ANALYSIS OF THIN-WALLED MULTI-CELL CLOSED SECTIONS

609

8.18f) Determine the beam’s cross-sectional torsional constant. The torsional constant is found by using the reference shear modulus per cell: J1 =

GJeff1 = 0.553415 π R3 t = 30202.6 in4 G01

J2 =

GJeff2 = 0.616198 π R3 t = 33629.0 in4 G02

The torsional constant for the cross-section is found by using a reference shear modulus of the cross-section (G0 = small[G01 , G02 ] = G): Jeff =

GJeff = 1.72303 π R3 t = 94034.2 in4 G0 End Example 

8.6.3

Pure Shear

For closed sections, the shear loads may be applied through any reference point in the cross-section, and it may or may not be the shear center. In progress...

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

8.7

610

Analysis of Combined Open and Closed Thin-walled sections

In many aerospace structures, the cross-sections of the beam is built by a combination of open and closed components. As an example of these hybrid sections is the wing section in the region of an undercarriage bay that can take the form shown in Fig. 8.45 in which the nose portion is a single cell section and the cut-out forms an open channel section. In general, such composite sections may be analyzed using a combination of the method previously described throughout this chapter.

Figure 8.45: A typical hybrid thin-walled wing section.

8.7.1

Bending

Bending analysis does not depend on the shape of the beam’s cross-section, thus methods discussed in the previous chapter apply here.

8.7.2

Pure Shear

Unlike the complete closed sections, shear loads must be applied through the shear center of the combined section, other wise shear stresses of due to torsion will arise. When the shear loads do not act through the shear center its location must be determined and the loading transferred to that location together with a torque. Now the analysis is done separately. We assume that the cross-section of the beam remains undistorted by the loading. In progress...

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

8.7.3

611

Pure Torsion

The behavior of open and closed sections are quite different. In practical situations, one is often confronted with cross-sections presenting a combination of open and closed sections and we are interested in determining the torsional rigidity, torsional constant, and the rate change of twist.

y i=3

i=2

t3, a3

t2, a2 i=1 t1, a1

Ac ti, ai i=i

z

T tc

th

tn, an i=n

Figure 8.46: A hybrid thin-walled section under torsion.

Torque Generally, in the torsion of composite sections, the closed portion is dominant since its torsional stiffness is far greater than that of the attached open section portion which may therefore be frequently ignored in the calculations of torsional stiffness; shear stresses, however, should be considered. To analyze these hybrid sections, let us consider a single closed cell and n open branches, as shown in Fig. 8.46. The total torque carried by the section is the sum of the torques carried by the closed cell and open branches: n X T = Tclosed + Topen = Tclosed + Ti 1=1

| {z } open

where Tclosed = GJclosed

dθx , dx

Ti = GJi

dθx dx

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

Torsional rigidity The torsional stiffness for the closed cell is GJclosed =

and for each open branch GJi = Thus, T = Tclosed +

n X

4 A2c I ds G0 t∗

1 Gi ai t3i 3 n

Ti = GJclosed

1=1

| {z }

dθx X dθx + GJi dx dx 1=1

open

n n o dθ X x GJi = GJclosed + dx 1=1

dθx dx In other words, the effective torsional stiffness for the cross-section is T = GJeff

GJeff = GJclosed + GJopen = GJclosed +

n X

GJi

1=1

Unit twist The cross-section unit twist is given by,

dθx T = dx GJeff

Torsional shear The shear stress for the closed cell is       Tclosed 1 dθx 1 T T GJclosed Sxs = = GJclosed = GJclosed = 2 Ac t 2 Ac t dx 2 Ac t GJeff 2 Ac t GJeff closed

and the shear flow for the closed cell is

qxs

= closed

T 2 Ac



GJclosed GJeff



The shear stress for the ith open branch      T Ti ti 3 Ti ti 3 dθx 3 1 T 3 Sxsi = = = G a t = Gi ti = GJ i i i i Ji dx GJeff GJeff ai t3i ai t2i ai t2i 3

612

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

and the shear flow for the ith open branch qxsi =

T Gi t2i GJeff

Example 8.19. Consider the thin-walled airfoil used in the region of an undercarriage bay, as shown in Fig. 8.47. The section is subject to pure torque T . Determine the torsional rigidity of the section, the unit twist, and total shear flow and shear stress in each section.

t23, a23

3

2

t34, a34 t12, a12

t43w, a43w

4

1

Figure 8.47: Hybrid thin-walled wing section. The geometric and mechanical properties are: Portion (i) 12 (skin) 23 (skin) 34 (skin) 43 (web)

Shear Modulus (Gi ) G G G G

Thickness (ti ) 2t 2t t t

Length (ai ) a 2a 3a a

Take G = 20 GPa, t = 2 mm, a = 100 mm, and T = 10 kN–m. The nose cell area is 10 a2 . 8.19a) The effective torsional torsional rigidity for the cross-section is GJeff = GJclosed + GJopen = GJclosed +

2 X 1=1

GJi

613

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

where

4 G A2c 4 G A2c 4 A2c I Z a43 = I = Z a34 ds ds ds34 ds43 G0 + t∗ t t t43 34 0 0
2 4 (20 × 109 ) 10 a2 = Z 0.3 = 1.600 × 107 N-m2 Z 0.1 ds34 ds43 + 0.002 0.002 0 0

GJclosed =

GJopen =

2 X

GJi =

1=1

= Thus

 G a12 t312 + a23 t323 3

 20 × 109  (0.100) (0.004)3 + (0.002) (0.004)3 = 128.0 N-m2 3 GJeff = GJclosed + GJopen = 1.60001 × 107 N-m2

8.19b) The effective torsional constant for the cross-section is Jo =

GJeff 1.6000128 × 107 4 = m = 8.00006 × 10−4 m4 G0 20 × 109

8.19c) In all unrestrained torsion problems the torque is related to the rate of twist by the expression dθx T = GJ dx Thus the angle of twist pe unit length is therefore given by φx =

dθx T 10 × 103 = = 0.000624995 rad/m = dx GJeff 1.60001 × 107

8.19d) The shear flow and shear stress in the closed section is   T GJclosed qxs = = 24999.8 N/m 2 Ac GJeff closed   T GJclosed Sxs = = 1.24999 × 107 Pa 2 Ac t34 GJeff 34   GJclosed T = 1.24999 × 107 Pa Sxs = 2 Ac t43w GJeff 43w The shear flow and shear stress for the open branches T GJeff T = GJeff T = GJeff T = GJeff

qxs12 =

G12 t212 = 199.998 N/m

Sxs12

G12 t12 = 49999.6 Pa

qxs23 Sxs23

G23 t223 = 199.998 N/m G23 t23 = 49999.6 × 107 Pa

614

8.7. ANALYSIS OF COMBINED OPEN AND CLOSED THIN-WALLED SECTIONS

We observer that the both the stresses and the stiffness of the closed section of the wing dominates. End Example 

615

8.8. REFERENCES

8.8

616

References

Allen, D. H., Introduction to Aerospace Structural Analysis , 1985, John Wiley and Sons, New York, NY. Curtis, H. D., Fundamentals of Aircraft Structural Analysis, 1997, Mc-Graw Hill, New York, NY. Johnson, E. R., Thin-Walled Structures, 2006, Textbook at Virginia Polytechnic Institute and State University, Blacksburg, VA. Keane, Andy and Nair, Prasanth, Computational Approaches for Aerospace Design: The Pursuit of Excellence, August 2005, John Wiley and Sons. Shames, I. H., and Dym, C. L., Energy and Finite Element Methods in Structural Mechanics, 1985, Taylor & Francis. Sun, C. T., Mechanics of Aircraft Structures, Second Edition 2006, John Wiley and Sons

8.9. SUGGESTED PROBLEMS

8.9

617

Suggested Problems

Problem 8.1. Consider the following idealization for an open-section. The flanges of the stringers carry only axial normal stress and their area is idealized with the flange area of Af . The following cross-section has six flanges, each one of the same area: π2 2 Af = R 600 Take the local coordinates as follows: 1→2

2→3

3→4

4→5

5→6

y′ E1 t1

E2 t2 2

R/2

θ

3

R

1

z′

R

E3 t3

6

4

R/2 5

E4 t4

E5 t5

The mechanical and geometrical properties are given as follows: E1 = 10 × 106 psi E2 = 20 × 106 psi

E3 = 5 × 106 psi E4 = 20 × 106 psi E5 = 10 × 106 psi R = 100

The circular sections are linearly tapered as follows: At 2: t = 0.02 π R At 3: t = 0.01 π R At 4: t = 0.04 π R At 5: t = 0.02 π R At the vertical skins: t1 = 0.02 π R

t5 = 0.02 π R

8.9. SUGGESTED PROBLEMS

618

Solve the problem by hand and write a MATLABr code to solve this problem. You may used any numerical integrating function such as QUAD. Verify your hand solutions with those obtained in MATLABr . Carefully document your MATLABr code. Explain any variable, function, and script file you will be using. Assume all materials are isotropic and take ν1 = ν2 = ν3 = ν4 = ν5 = 0.25 (a) Determine the shear flow due to shear for each section. (b) Determine the shear center. (c) Determine the shear flow due to torque for each section. (d) Determine the total shear stress for each section. Provide plots for Vy = 0, Vz = 1 and Vy = 1, Vz = 0. Plot both solutions on the same plot, but one plot for each section. (e) Determine the location of maximum shear stress. (f) Determine the torsional constant and shear rigidity of the cross-section. (g) Determine the total unit twist per length of the cross-section. 

8.9. SUGGESTED PROBLEMS

619

Problem 8.2. Consider the following idealization for a closed-section wing-box. The flanges of the stringers carry only axial normal stress and their area is idealized with the flange area of Af . The following cross-section has six flanges, each one of the same area: π2 2 Af = R 600 Take the local coordinates as follows: 1→2

2→3

3→4

4→5

5→1

y′ 3R

E2 t2 2

E1 t1 R/2

θ

3

R z′

R

E3 t3

1

4

E4 t4

R/2 E5 t5

5

The mechanical and geometrical properties are given as follows: E1 = 10 × 106 psi E2 = 20 × 106 psi

E3 = 5 × 106 psi E4 = 20 × 106 psi E5 = 10 × 106 psi R = 100

The circular sections are linearly tapered as follows: At 2: t = 0.02 π R At 3: t = 0.01 π R At 4: t = 0.04 π R At 5: t = 0.02 π R At the vertical skins: t1 = 0.02 π R

t5 = 0.02 π R

8.9. SUGGESTED PROBLEMS

620

The following empirical data is known for section 1 → 2: y=0

z = 3R

y = 0.5 R

z = 2R

y=R

z=0

The following empirical data is known for section 5 → 1: y=0

z = 3R

y = −0.5 R

z=R

y = −R

z=0

Solve the problem by hand and write a MATLABr code to solve this problem. You may used any numerical integrating function such as QUAD. Verify your hand solutions with those obtained in MATLABr . Carefully document your MATLABr code. Explain any variable, function, and script file you will be using. (Hint: you will need to use an interpolation function algorithm). Assume all materials are isotropic and take ν1 = ν2 = ν3 = ν4 = ν5 = 0.25 (a) Determine the shear flow due to shear for each section. (b) Determine the shear center. (c) Determine the shear flow due to torque for each section. (d) Determine the total shear stress for each section. Provide plots for Vy = 0, Vz = 1 and Vy = 1, Vz = 0. Plot both solutions on the same plot, but one plot for each section. (e) Determine the location of maximum shear stress. (f) Determine the torsional constant and shear rigidity of the cross-section. (g) Determine the total unit twist per length of the cross-section. 

8.9. SUGGESTED PROBLEMS

621

Problem 8.3. Consider a two rectangular squared cell with different isotropic materials subject to a torque T .

4 

3

2

Cell 1 

Cell 2 

5 

6

1

4 

3

2

The mechanical properties are:

E12 = E45 = E36 = 3.5 E0 psi

ν12 = ν45 = ν36 = 0.30

E34 = E Cell 1  23 = 2.5 E0 psi

ν34 = ν23 = 0.25 Cell 2 

E56 = E61 = 1.5 E0 psi

ν56 = ν61 = 0.20

The geometric properties are:

5 

6

1

a61 = a32 = a34 = a56 = 4 a

a45 = a36 = a12 = a

t61 = t32 = t34 = t56 = t

small slit  t45 = t36 = t12 = t

where a’s are the branch lengths and t’s the branch thicknesses. The torque is: T = 10 kips-in + (a) Determine the enclosed area. (b) Determine the total shear flow. (c) Determine the shear stresses in each wall. (d) Determine the torsional constant J. (e) Determine the torsional stiffness. (f) Determine the twist per unit length. 

4 

3

2

8.9. SUGGESTED PROBLEMS

622

Cell 1 

Cell 2 

Problem 8.4. Consider a two rectangular squared cell with different isotropic materials subject to a torque T . A small 5  branch in cell62. 1 slit is cut in the lower exterior

4 

3

Cell 1 

5 

2

Cell 2 

1

6 small slit 

The mechanical properties are: E12 = E45 = E36 = 3.5 E0 psi

ν12 = ν45 = ν36 = 0.30

E34 = E23 = 2.5 E0 psi

ν34 = ν23 = 0.25

E56 = E61 = 1.5 E0 psi

ν56 = ν61 = 0.20

The geometric properties are: a61 = a32 = a34 = a56 = 4 a

a45 = a36 = a12 = a

t61 = t32 = t34 = t56 = t

t45 = t36 = t12 = t

where a’s are the branch lengths and t’s the branch thicknesses. The torque is: T = 10 kips-in + (a) Determine the enclosed area. (b) Determine the total shear flow. (c) Determine the shear stresses in each wall. (d) Determine the torsional constant J. (e) Determine the torsional stiffness. (f) Determine the twist per unit length. 

Chapter 9 Virtual Work Principles

Instructional Objectives of Chapter 9 After completing this chapter, the student should be able to: 1. Understand the concept of Principle of Virtual Work (PVW) and Principle of Complementary Virtual Work (PCVW). 2. Apply PVW and PCVW to a system of particles. 3. Apply PVW and PCVW to a system of rigid and deformable bodies. 4. Apply PVW and PCVW to a deformable continuous structure. 5. Relate PVW to First Castigliano’s Theorem. 6. Relate PCVW to Second Castigliano’s Theorem.

Virtual work principles provide powerful methods to solve problems in elasticity. The virtual work principles are an alternative way—not approximate—of expressing the equations of motion and equilibrium. These principles convey the requirements of equilibrium and compatibility as integral equations, instead of the partial differential equations previously discussed. Virtual Work principles are powerful because: (i) are used to obtain rigorous bounds on the stiffness of an elastic structure or solid; (ii) provide new insight into general issues concerning uniqueness and existence of solutions to boundary value problems; (iii) and many numerical techniques for solving elastic boundary value problems are based on these principles. The concept is widely used to obtain loads and deflections in linear elastic problems. Throughout this chapter we shall discuss the two virtual work principles as applied to rigid and flexible bodies: the Principle of Virtual Work and the Principle of Complementary Virtual Work. The Principle of Virtual Work is the basis for the displacement method in structural analysis and the principle of complementary virtual work forms the basis of the classical force method of analysis.

623

9.1. DIFFERENTIAL WORK AND VIRTUAL WORK

9.1

624

Differential Work and Virtual Work

In mechanics a force F does work only when it undergoes a displacement in the direction of the force. For an example, consider the force F, which is located on the path s specified by the position vector r, as shown in Fig. 9.1. If the force moves along the path to a new position r0 , the displacement is dr and therefore the work dW is a scalar quantity defined by the dot product: dW = F · dr

F

θ ds dr

r

r′

Figure 9.1: Force vector and displacement vector at a location s.

9.1.1

Differential Work

Since dr is infinitesimal,

t dr ≈ ds ˆ

where ˆ t is the unit vector in the tangential direction at s. If the angle between the tails of dr and F is θ, ˆ + F cos θ ˆ t F = F sin θ n where n ˆ is the unit vector in the normal direction at s. Then the differential work done is dW = F cos θ ds From Fig. 9.2, we see that dW =

1 1 (Fs + (Fs + dFs )) ((s + ds) − s) = (2 Fs ds + dFs ds) = 2 2

Fs ds | {z }

first variation

+

1 dFs ds |2 {z }

higher order effect

9.1. DIFFERENTIAL WORK AND VIRTUAL WORK

625

F F′ θ ds

Fs Fs+dFs

dr

Fs

dW

r′

r

s

s+ds

s

Figure 9.2: Differential work done. Since we are working with infinitesimal displacements, the higher order effects are ignored, hence dW = Fs ds = F cos θ ds The definition of the differential work of a force have been presented in terms of actual movements expressed by differential displacement of magnitude ds.

9.1.2

Virtual Work F

F

Fs θ δs

θ

δr

r

Fs

δW

r′

s

s+δs

s

Figure 9.3: Virtual work done. Now, consider an imaginary or virtual displacement as shown in Fig. 9.3, which indicates a displacement that is assumed and does not actually exist. We use the symbol “δ” to denote the virtual displacements. These virtual displacements are done keeping the load constant. Hence the virtual work

9.2. REVIEW OF EQUATIONS OF LINEAR ELASTICITY

626

done by a force undergoing a virtual displacement δs is δW = Fs δs = F cos θ δs

9.1.3

Complementary Virtual Work

δF

Fs

F θ

δW*

Fs+δFs Fs

r

s

s

Figure 9.4: Complementary Virtual work done. Now, consider an imaginary or virtual load as shown in Fig. 9.4, which indicates a load that is assumed and does not actually exist. We use the symbol “δ” to denote the virtual loads. These virtual loads are done keeping the displacement constant. Hence the virtual work done by a virtual load δF undergoing a displacement s is δW ∗ = δFs s

9.2

Review of equations of linear elasticity

As we discussed in chapter 6, the elasticity field is completely describe by the following equations of elasticity: 1. The equations of equilibrium are the most fundamental equations. They were derived from Newtons law stating that the sum of all the forces acting on a differential element of the structure should

9.2. REVIEW OF EQUATIONS OF LINEAR ELASTICITY

vanish.

627

∂Sxx ∂Syx ∂Szx + + + bx = 0 ∂x ∂y ∂z ∂Syy ∂Szy ∂Sxy + + + by = 0 ∂x ∂y ∂z

(9.1)

∂Sxz ∂Syz ∂Szz + + + bz = 0 ∂x ∂y ∂z 2. The strain-displacement equations merely define the strain components that are used for the characterization of the deformation of the body at a point. The strain-displacement relationships were derived from purely geometric considerations. The strains associated with the displacement field are computed using the Green-Lagrange strains and were expressed in terms of the displacement gradients as follows

4


1 2 g1 + g22 + g32 2
1 2 = eyy = g5 + g4 + g52 + g62 2
1 2 g7 + g82 + g92 = ezz = g9 + 2 = 2 eyz = g6 + g8 + g4 g7 + g5 g8 + g6 g9

5

=

2 exz = g3 + g7 + g1 g7 + g2 g8 + g3 g9

(9.2e)

6

=

2 exy = g2 + g4 + g1 g4 + g2 g5 + g3 g6

(9.2f)

1 2 3

= exx = g1 +

3. The constitutive laws relate the stress    C11 S1            S2   C12       S3  C =  13  C14   S 4           C15 S 5       S6 C16

(9.2a) (9.2b) (9.2c) (9.2d)

and strain components. C12 C22 C23 C24 C25 C26

C13 C23 C33 C34 C35 C36

C14 C24 C34 C44 C45 C46

C15 C25 C35 C45 C55 C56

C16 C26 C36 C46 C56 C66

                  

1 2 3 4 5 6

                

(9.3)

A stress field S(x, y, z) is said to be statically admissible if it satisfies the equilibrium equations at all points in the domain, and the surface equilibrium equations at all points on the surface. A displacement field d(x, y, z) is said to be kinematically admissible if it is continuous and differentiable at all points in in the domain, and the geometric boundary conditions at all points on the surface. A strain field E(x, y, z) is said to be compatible if it is derived from a kinematically admissible displacement field through the strain-displacement relationships.

9.3. PVW FOR A SYSTEM OF PARTICLES

9.3

628

PVW for a System of Particles

In particle mechanics the incremental work of a force acting on a particle is the product of the force and the component of the incremental displacement of the particle in the direction of the force. If we have a system of particles, then the incremental work of the forces acting on their respective particles can be defined as the sum the incremental work of the forces acting on their respective particles.

9.3.1

Virtual Displacements

Before we proceed let us define various terms, 1. A displacement coordinate is a quantity used in specifying the change of configuration of a system. 2. A constraint is a kinematical restriction on the possible configurations a system may assume. 3. A virtual displacement is an infinitesimal imaginary change of configuration of a system consistent with its constraints. Virtual displacements are arbitrary displacements that satisfy the following conditions: (a) Infinitesimal (are very small) (b) kinematically admissible (compatible with kinematic boundary conditions) (c) time independent (d) forces remain constant during virtual displacements. The virtual displacements take place instantaneously. This is in direct contrast with actual displacements, which require a certain amount of time to evolve, during which time the forces and constraints may change. Lagrange introduced the special symbol δ to emphasize the virtual character of the instantaneous variations, as opposed to the symbol d designating actual differentials of positions taking place in the time interval dt. The virtual displacements, being infinitesimal, obey the rules of differential calculus. Suppose that the ith displacement vector can be written as follows qi = qi (x1 , x2 , . . . , xn ),

i = 1, 2, . . . , n

where xj ’s are the independent generalized coordinates. These generalized coordinates can be translations or rotations. From calculus, the differential displacement (using the chain rule is) dqi =

∂qi ∂x1

dx1 +

∂qi ∂x2

dx2 + · · · +

∂qi ∂xk

dxk + · · · +

∂qi ∂xn

dxn ,

i = 1, 2, . . . , n

Similarly, we can express the virtual displacement vectors in terms of the generalized virtual displacements as follows δqi =

∂qi ∂x1

δx1 +

∂qi ∂x2

δx2 + · · · +

∂qi ∂xk

δxk + · · · +

∂qi ∂xn

δxn

=

n X ∂qi

k=1

∂xk

δxk ,

i = 1, 2, . . . , n

9.3. PVW FOR A SYSTEM OF PARTICLES

629

To better understand the concept of virtual displacements, let us consider the displacement w and let it be a function of x, i.e., w = w(x). Then the virtual displacement of w is δw. The greek letter δ represents an infinitesimal change in the displacement for a fixed value of x, whereas the symbol d in calculus represents the change in a function with respect to a change in its independent variable x. The difference between the variation of a function and the differential of a function is shown in Fig. 9.5.

w(x) δw

dw dx

w(x)

x

x

x+dx Figure 9.5: Virtual work done.

In ordinary calculus the differential of the function w(x) is dw = w(x + dx) − w(x) Thus, the physical interpretation of dw is the difference in the displacements of two particles, one particle originally at x+dx and the second one originally at x in the undeformed body. The physical interpretation of the infinitesimal function δw(x) is change in the displacement of a single particle originally at x in the undeformed body. Thus the new function δ w(x) ¯ is δ w(x) ¯ = w(x) + δw(x)

→

δw(x) = δ w(x) ¯ − w(x)

The δ is considered an operator in a sense similar to the differential operator d in calculus. The δ operator means the variation in a function and not the differential of a function with respect to a change in its independent variable1 . 1 The distinction between the variation of a function and the differential of a function is essential in formulating the concept of work in a continuous system of particles, and the mathematics dealing with the variation of functions is called the calculus of variations. That is, the calculus of variations is a mathematical tool by which work and energy methods are applied to systems with infinitely many degrees of freedom.

9.3. PVW FOR A SYSTEM OF PARTICLES

630

The variational operator, δ, acts as a differential operator respect to dependent variables. The variational operator can commute with differential and integral operators (as long as the independent variables are fixed):   dy d δ = (δy) (9.4) dx dx δ

Z

x2

x1

9.3.2

 Z y(x) dx =

x2

(9.5)

δy(x) dx

x1

PVW of a particle

Consider system in equilibrium at some given position. Let us perturb the system by applying an arbitrary virtual displacement. Since the system is in equilibrium, the net force on the system is zero and hence the virtual work done during the virtual displacement must be zero. Since the virtual work done on the system is zero, we can determine the equilibrium paths or positions.

z F2 F3 F1 Fi Fn x y

Figure 9.6: Particle in equilibrium subject to n forces. To better understand this concept, let us begin by applying the Principle of Virtual Work (PVW) to a particle. Consider a single particle which is in equilibrium at some position subject to n applied forces as shown in Fig. 9.6; i.e, in cartesian coordinates: ˆ F1 = Fx1 ˆi + Fy1 ˆj + Fz1 k ˆ F2 = Fx2 ˆi + Fy2 ˆj + Fz2 k .. .. .=. ˆ Fi = Fxi ˆi + Fyi ˆj + Fzi k .. .. .=. ˆ Fn = Fxn ˆi + Fyn ˆj + Fzn k

9.3. PVW FOR A SYSTEM OF PARTICLES

Since the particle is in equilibrium,

631

n X

Fi = 0

i=1

In other words,

n X

Fi =

i=1

n X

Fxi

i=1

!

n X

Fxi ˆi +

i=1

ˆi +

n X

Fyi ˆj +

i=1

n X

Fyi

i=1

!

n X

ˆ=0 Fzi k

i=1

ˆj +

n X

!

Fzi

i=1

ˆ=0 k

Since each axis is independent from each other, the above requires that n X

i=1 n X

i=1 n X

Fxi = 0 Fyi = 0 Fzi = 0

i=1

Now let us make an arbitrary virtual displacement δq from the equilibrium state, where q = {u

v

T

w}

δq = {δu

→

δv

T

δw}

The total virtual work acting on the particle is then defined as δW = F1 · δq + F2 · δq + · · · + Fi · δq + · · · + Fn · δq = (F1 + F2 + · · · + Fi + · · · + Fn ) · δq n X = Fi · δq =

i=1 n X i=1

Fxi

!

δu +

n X

Fyi

i=1

!

δv +

n X i=1

Fzi

!

δw

Since the particle is in equilibrium, each of the sums is zero and thus for any virtual displacement δd δW = 0 Principle of Virtual Work states that a particle is in equilibrium if and only if the virtual work done on the particle is zero for any virtual displacement.

9.3.3

PVW for rigid and deformable bodies

We derived the above for only one particle; now let us expand the concept to a system of n particles, as shown in Fig. 9.7.

9.3. PVW FOR A SYSTEM OF PARTICLES

632

Q2

fn

Qn z, W

Q1 n

f2

2 1 f1

y, V x, U

Figure 9.7: System of particles showing both external and internal forces. Particle i is acted upon by a net external force Qi coming from outside the system (such as gravity or contact with an adjacent system) and an net internal force f i coming from other particles within the system. The resultant force acting on the particle QR , where If the particle undergoes a real, infinitesimal displacement dq, then the incremental work done by the forces in the particle is dW = QR · dq If instead we imagine that the particle is given a virtual displacement δq, while the forces are held constant, then the total work done on this particle is δW = QR · δq If the particle is in equilibrium, we showed that δW = 0 for any choice of the virtual displacement vector δq. Not, let the system of particles, shown in Fig. 9.7, undergo a small virtual distortion, so that that the virtual displacement is imparted to each particle. Since the virtual displacements are infinitesimal, the geometry of the system is essentially unchanged by the virtual deformation. During this virtual distortion, all forces—the true forces that exist in the system—are held fixed. In this case, the virtual work done by the external forces is n X δWext = Qi · δqi i=1

The virtual work done by the internal forces is δWint =

n X

n X

j=1 i=1;i6=j

Fij · δrij

where Fij = −Fji is the force on particle i due to particle j (j 6= i because a particle cannot exert an external force on itself) and rij is the position vector of point j relative to point i before deformation. Thus the virtual deformation is defined as δrij = δqj − δqi

9.3. PVW FOR A SYSTEM OF PARTICLES

633

Observe that the dot product Fij · δrij is the product of the true force mutually applied to points i and j times the virtual change in the distance between those points. Thus, δWint is the virtual work done on the material joining the points of the system. For example, if you imagine each particle to be joined to all of the others by elastic springs, then δWint is the virtual work done on all of those springs during the virtual distortion of the system. The term δWint is always zero for a rigid body2 since, by definition, the distance between points on a rigid body cannot change. Since the all the forces acting on a particle must vanish, the principle is stated as follows A system is in equilibrium if and only if δWext = δWint for any compatible virtual deformation. The Principle of Virtual Work is an exact, alternative statement of equilibrium and it does not depend on material properties. Note that the Principle of Virtual Work is based upon two major assumptions: 1. The system is equilibrium. 2. It is possible to make such virtual displacements.

9.3.4

Procedure

PVW is useful in relating forces and displacements. Four step when solving problems using PVW: Step 1. Give a virtual displacement from the equilibrium configuration Suppose our system depends on n generalized displacements: x1 , x2 , . . . , xn . Note that these generalized displacements can be either translations or rotations. Step 2. Virtual work due to external loads δWext =

n X i=1

Qi · δqi

2 Rigid body is a collection of particles that are constrained to move as one or more rigid bodies. For example, levers, pulleys, blocks, rigid bar linkages. The relative displacements between particles that make up the rigid body are zero.

9.3. PVW FOR A SYSTEM OF PARTICLES

634

(a) Position vector to the applied loads q1 = · · · q2 = · · · .. .. .=.

qn = · · · (b) Virtual displacement of the position vector to the applied load δq1 = δq2 =

∂ q1 ∂x1 ∂ q2

δx1 +

∂ q1 ∂x2 ∂ q2

δx2 + · · · +

∂ q1 ∂xn ∂ q2

δxn

δx1 + δx2 + · · · + δxn ∂x1 ∂x2 ∂xn .. .. .=. ∂ qn ∂ qn ∂ qn δx1 + δx2 + · · · + δxn δqn = ∂x1 ∂x2 ∂xn (c) External Loads Q1 = · · ·

Q2 = · · · .. .. .=.

QN = · · · (d) Virtual work due to external loads: δWext = Q1 · δq1 + Q2 · δq2 + · · · + Qn · δqn

= (. . .) δx1 + (. . .) δx2 + · · · + (. . .) δxn

Step 3. Virtual work due to internal loads

δWint =

n=1 X i=1

Fij · δrij

(a) Position vector rij = . . . (b) Virtual displacement of the position vector δrij =

∂ rij ∂ rij ∂ rij δx1 + δx2 + · · · + δxn ∂x1 ∂x2 ∂xn

(c) Internal Loads Fij = . . .

(9.6)

9.3. PVW FOR A SYSTEM OF PARTICLES

635

(d) Virtual work due to internal loads: δWint = Fij · δrij = (. . .) δx1 + (. . .) δx2 + · · · + (. . .) δxn Note that when the assemblage only involves rigid objects δWint = 0. Step 4. Apply the Principle of Virtual Work

δWext = δWint

→

δWext − δWint = 0

Collect all terms for each virtual displacement. δWext − δWint = (. . .)1 δx1 + (. . .)2 δx2 + · · · + (. . .)n δxn = 0 Since δxi 6= 0, (. . .)i = 0 for i = 1, 2, . . . , n and thus it gives n equations.

(9.7)

9.3. PVW FOR A SYSTEM OF PARTICLES

636

Example 9.1.

Consider two rigid links of length L and connected by a spring of stiffness k and to each other by a smooth pin. Determine angle θ for the equilibrium configuration using the principle of virtual work.

P

3

y 1

α

L

θ

L

k

2

x

Figure 9.8: Rigid-bars configuration for Example 9.1. (9.1-a) Give a virtual displacement from the equilibrium configuration

9.3. PVW FOR A SYSTEM OF PARTICLES

637

P P

v

δv α k

k

Just before the load is applied

u

δu

The problem can be easier to solve using α and knowing that α and θ are related as follows: π (9.8) 2 Note that we first find an equilibrium state and then apply the virtual displacement from that equilibrium state. Also, note that we place our reference coordinate at the pinned end. θ−α=

(9.1-b) Virtual work due to external loads δWext =

N X i=1

Qi · δqi

1) Position vector to the applied load q1 = L sin α ˆi + L cos α ˆj 2) Virtual displacement of the position vector to the applied load δq1 = 3) External Loads

∂ q1 ∂α

δα = L cos α δα ˆi − L sin α δα ˆj

Q1 = −P ˆj

9.3. PVW FOR A SYSTEM OF PARTICLES

638

4) Virtual work due to external loads: δWext = Q1 · δq1 = P L sin α δα (9.1-c) Virtual work due to internal loads δWint =

N X i=1

Fij · δrij

Note that since the bars are rigid they do no internal work because they do not stretch. The only component in the example that can deflect is the spring and thus is the only one contributing to the virtual work done due to internal loads. 1) Position vector

r12 = 2 L sin α ˆi

2) Virtual displacement of the position vector ∂ r12 δα = 2 L cos α δα ˆi ∂α

δr12 = 3) Internal Loads F12 = k s12 ˆi

s12 = |r12 | − sunstretched = 2 L sin α − 0 = 2 L sin α F12 = k 2 L sin α ˆi 4) Virtual work due to internal loads: δWint = F12 · δr12 = (2 k L sin α)(2 L cos α δα) = 4 k L2 sin α cos α δα (9.1-d) Apply the Principle of Virtual Work δWext − δWint = 0

P L sin α δα − 4 k L sin α cos α δα = 0 n o (P − 4 k L cos α) L sin α δα = 0 2

Since δα 6= 0:

L sin α

=

0

cos α

=

P 4kL

Using Eq. (9.8) we can rewrite the above as:  π  sin − + θ = 2  π  = cos − + θ 2

cos θ = 0 sin θ =

P 4kL

9.3. PVW FOR A SYSTEM OF PARTICLES

639

The solution to our problem is: θ=

π 2

θ = sin−1

(trivial solution) 

P 4kL

 End Example 

9.3. PVW FOR A SYSTEM OF PARTICLES

640

Example 9.2.

The four rigid bars are connected by smooth pins. Two rods 6-2, 1-4 are of length 2a, and rods 4-3 and 2-3 are of length a. Neglect the weight of the rods. Use PVW to determine the relationship between the load P and angle θ for equilibrium. Compare the results when the compressive spring is removed.

y 4

x 5 a

θ

a 1

Q1 =10 lbs

a

a

θ

3

k

a

a 2

Q2 = 5 lbs

Figure 9.9: Rigid-bars configuration for Example 9.2. (9.2-a) Give a virtual displacement from the equilibrium configuration

P

9.3. PVW FOR A SYSTEM OF PARTICLES

641

δu

u y x a

a

a

θ

θ

a

δv

k

P

a

a

v

Q2 = 5 lbs

Q1 =10 lbs

Note that we first find an equilibrium state and then apply the virtual displacement from that equilibrium state. Also, note that we place our reference coordinate at the pinned end. (9.2-b) Virtual work due to external loads δWext =

N X i=1

Qi · δqi

1) Position vector to the applied load q1 = −2 a sin θ ˆj (to the 10 lbs load)

q2 = 2 a cos θ ˆi − 2 a sin θ ˆj (to the 5 lbs load) q3 = 3 a cos θ ˆi − a sin θ ˆj (to the P load)

2) Virtual displacement of the position vector to the applied load δq1 = δq2 = δq3 =

∂ q1 ∂θ ∂ q2 ∂θ ∂ q3 ∂θ

δθ = (−2 a cos θ δθ) ˆj δθ = (−2 a sin θ δθ) ˆi + (−2 a cos θ δθ) ˆj δθ = (−3 a sin θ δθ) ˆi + (−a cos θ δθ) ˆj

9.3. PVW FOR A SYSTEM OF PARTICLES

3) External Loads

642

Q1 = −Q1 ˆj lbs Q2 = −Q2 ˆj lbs Q3 = P ˆi lbs

4) Virtual work due to external loads: δWext = Q1 · δq1 + Q2 · δq2 + Q3 · δq3 = (2Q1 a cos θ + 2Q2 a cos θ − 3P a sin θ)δθ = (2(Q1 + Q2 ) a cos θ − 3P a sin θ)δθ

(9.2-c) Virtual work due to internal loads δWint =

N X i=1

Fij · δrij

Note that since the bars are rigid they do no internal work because they do not stretch. The only component in the example that can deflect is the spring and thus is the only one contributing to the virtual work done due to internal loads. 1) Position vector r24 = r4 − r2 r4 = 2 a cos θ ˆi r2 = 2 a cos θ ˆi − 2 a sin θ ˆj r24 = 2 a sin θ ˆj 2) Virtual displacement of the position vector δr24 = 3) Internal Loads

∂ r24 δθ = 2 a cos θ δθ ˆj ∂θ

F24 = k s24 ˆj s24 = |r24 | − suncompressed = 2 a sin θ − 2 a F24 = 2 k a (sin θ − 1) ˆj

4) Virtual work due to internal loads:

δWint = F24 · δr24 = [4 a2 k cos θ(sin θ − 1)]δθ Note that when spring is not present δWint = 0, assemblage of rigid rods.

9.3. PVW FOR A SYSTEM OF PARTICLES

643

(9.2-d) Apply the Principle of Virtual Work δWext − δWint = 0 n o n o 2 (Q1 + Q2 ) a cos θ − 3 P a sin θ δθ − 4 a2 k cos θ(sin θ − 1) δθ = 0 n o 2 (Q1 + Q2 ) a cos θ − 3 P a sin θ + 4 a2 k cos θ(1 − sin θ) δθ = 0

Since δθ 6= 0:

    2 2ak P = (Q1 + Q2 ) cot θ 1 + (1 − sin θ) 3 Q1 + Q2

When the spring is not present: P =

2 (Q1 + Q2 ) cot θ 3

The load and the spring constant can be nondimensionalized as follows p=

s= U= Thus,

P 2 (Q1 + Q2 ) 3

dimensionless load

2ak Q1 + Q2

dimensionless spring constant

u = 3 cos θ a

dimensionless displacement

n o p = cot θ 1 + s (1 − sin θ) p = cot θ

with spring with no spring End Example 

9.3. PVW FOR A SYSTEM OF PARTICLES

644

Example 9.3.

Consider the two-bar system shown in Fig. 9.10. The two weightless rigid bars are connected by smooth pins. The rod 1-2 is of length a and the length of rod 2-3 is 2 a. All rotations are measured positive counter clockwise from the vertical. At node 2, the rods 1-2 and 2-3 are connect by a torsional spring of stiffness kt and at node 3, rod 2-3 is connected by a linear spring of stiffness ks . Use PVW to determine the relationship between the load P and the angles θ1 and θ2 for equilibrium. Assume small angle deformation.

x, u

1

L1

y, v

θ1

kt 2

θ2

4

ks

L2

3

P

Figure 9.10: Rigid-bars configuration for Example 9.3. (9.3-a) Give a virtual displacement from the equilibrium configuration

9.3. PVW FOR A SYSTEM OF PARTICLES

x, u

1

δθ1

y, v

θ1

645

Rod 2-3 remains unreformed (parallel to its original equilibrium position) because we keep θ2 fixed:

y, v

θ1

δθ2

2

θ2

θ2

4

Rod 1-2 remains unreformed because we keep θ1 fixed:

δθ1=0; δθ2≠0

δθ1≠0; δθ2=0

2

x, u

1

ks

P

3

4

ks

3

P

Note that we first find an equilibrium state and then apply the virtual displacement from that equilibrium state. Also, note that we place our reference coordinate at the pinned end. (9.3-b) Virtual work due to external loads δWext =

N X i=1

Qi · δqi

1) Position vector to the applied load at node 3 q1 = (L1 sin θ1 + L2 sin θ2 ) ˆi + (L1 cos θ1 + L2 cos θ2 ) ˆj 2) Virtual displacement of the position vector to the applied load δq1 =

∂ q1

δθ1 +

∂ q1

δθ2 ∂θ1 ∂θ2 = (L1 cos θ1 δθ1 + L2 cos θ2 δθ2 ) ˆi − (L1 sin θ1 δθ1 + L2 sin θ2 δθ1 ) ˆj

3) External Loads

Q1 = P ˆi

4) Virtual work due to external loads: δWext = Q1 · δq1 n o n o = P ˆi · (L1 cos θ1 δθ1 + L2 cos θ2 δθ2 ) ˆi − (L1 sin θ1 δθ1 + L2 sin θ2 δθ1 ) ˆj = P L1 cos θ1 δθ1 + P L2 cos θ2 δθ2

(9.3-c) Virtual work due to internal loads δWint =

N X i=1

Fij · δrij

Note that since the bars are rigid they do no internal work because they do not stretch.

9.3. PVW FOR A SYSTEM OF PARTICLES

646

The only component in the example that can deflect are the springs and thus are the only ones contributing to the virtual work done due to internal loads. 1) Position vector: due to linear spring

r43 = r3 − r4 r3 = (L1 sin θ1 + L2 sin θ2 ) ˆi r4 = 0 r43 = (L1 sin θ1 + L2 sin θ2 ) ˆi

due to torsional spring r12 = (θ2 − θ1 ) ˆ eθ 2) Virtual displacement of the position vector: due to linear spring ∂ r43 ∂ r43 δθ1 + δθ2 ∂θ1 ∂θ2 = (L1 cos θ1 δθ1 + L2 cos θ2 δθ2 ) ˆi

δr43 =

due to torsional spring

∂ r12 ∂ r12 δθ1 + δθ2 ∂θ1 ∂θ2 = −δθ1 ˆ eθ + δθ2 ˆ eθ

δr12 =

3) Internal Loads: due to linear spring F43 = ks s43 ˆi

s43 = |r43 | − sunstretched = (L1 sin θ1 + L2 sin θ2 ) − 0 F43 = (L1 ks sin θ1 + L2 ks sin θ2 ) ˆi due to torsional spring eθ M12 = kt s12 ˆ s12 = |r12 | − sunstretched = (θ2 − θ1 ) − 0

M12 = kt (θ2 − θ1 ) ˆ eθ 4) Virtual work due to internal loads: δWint = F43 · δr43 + M12 · δr12

= {(L1 ks sin θ1 + L2 ks sin θ2 ) (L1 cos θ1 δθ1 + L2 cos θ2 δθ2 )} + =

n

{kt (θ2 − θ1 ) (−δθ1 + δθ2 )}

o − kt (θ2 − θ1 ) + L1 cos θ1 (L1 ks sin θ1 + L2 ks sin θ2 ) δθ1 + n o kt (θ2 − θ1 ) + L2 cos θ2 (L1 ks sin θ1 + L2 ks sin θ2 ) δθ2

9.3. PVW FOR A SYSTEM OF PARTICLES

647

Note that when spring is not present δWint = 0, assemblage of rigid bars. (9.3-d) Apply the Principle of Virtual Work δWext − δWint = 0 n o P L1 cos θ1 + kt (θ2 − θ1 ) − L1 cos θ1 (L1 ks sin θ1 + L2 ks sin θ2 ) δθ1 + n o P L2 cos θ2 − kt (θ2 − θ1 ) − L2 cos θ2 (L1 ks sin θ1 + L2 ks sin θ2 ) δθ2 = 0

Since δθ1 and δθ2 are independent are arbitrary, take δθ1 6= 0, δθ2 = 0:

P L1 cos θ1 + kt (θ2 − θ1 ) − L1 cos θ1 (L1 ks sin θ1 + L2 ks sin θ2 ) = 0 and δθ1 = 0, δθ2 6= 0: P L2 cos θ2 − kt (θ2 − θ1 ) − L2 cos θ2 (L1 ks sin θ1 + L2 ks sin θ2 ) = 0 For small angle approximation, sin θ ≈ θ,

cos θ ≈ 1

Thus the equations are P + kt (θ2 − θ1 ) − (L1 ks θ1 + L2 ks θ2 ) = 0

P − kt (θ2 − θ1 ) − (L1 ks θ1 + L2 ks θ2 ) = 0

or

 kt P + (θ2 − θ1 ) − θ1 + L1 ks L1 ks  kt P − (θ2 − θ1 ) − θ1 + L1 ks L1 ks

Solving the above equation we get:

θ1 = θ2 =

 L2 θ2 = 0 L1  L2 θ2 = 0 L1

P ks (L1 + L2 )

End Example 

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

9.4

648

PVW for Deformable Continuous Structures

So far, we have worked with a system of particles. In this section, we extend the concept of Principle of Virtual Work for deformable continuous structures. A continuous structure can be thought as a system with infinite number of particles and the finite mass of the system becomes distributed continuously throughout the volume it occupies. For the problems of solid mechanics in which we assume the body is made of a continuum of particles, each infinitesimal particle, or material point, is identified by its coordinates in the undeformed or reference state of the body. The displacement components of the particle are continuous, single valued functions of its coordinates in the reference state so that the body in the deformed configuration remains a continuum; i.e., the deformation is such that no holes or overlapping of material points occur in the deformed configuration. For an arbitrary elastic body in equilibrium under applied body forces and surface tractions, we know the following is true ∂Sxx ∂Syx ∂Szx + + + bx = 0 ∂x ∂y ∂z ∂Sxy ∂Syy ∂Szy + + + by = 0 ∂x ∂y ∂z ∂Sxz ∂Syz ∂Szz + + + bz = 0 ∂z ∂y ∂z Tx = Tˆx Ty = Tˆy Tz = Tˆz

→ Tx − Tˆx = 0 → Ty − Tˆy = 0 → Tz − Tˆz = 0

Note that stress field is statically admissible, i.e. the equilibrium equations are satisfied at all points in Γ and the surface equilibrium equations at all points on Ω. Suppose it is possible to give an arbitrary virtual displacement δU , δV , δW , in the x- y- and z- directions, respectively. Now let us multiply each of the equilibrium equations by the arbitrary, virtual change in displacement, then integrated over the range of validity of the equation,  ZZZ  ∂Sxx ∂Syx ∂Szx + + + bx δU dΓ = 0 ∂x ∂y ∂z Γ

ZZZ 

∂Sxy ∂Syy ∂Szy + + + by ∂x ∂y ∂z

ZZZ 

∂Sxz ∂Syz ∂Szz + + + bz ∂z ∂y ∂z

Γ

Γ



δV dΓ = 0



δW dΓ = 0

Similarly, each of the three surface equilibrium equations are multiplied by an arbitrary, virtual change

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

649

in displacement, then integrated over the range of validity of the equation, ZZ  

Tx − Tˆx δU dΩ = 0 ZΩ Z  

Ty − Tˆy δV dΩ = 0 Ω

ZZ  

Tz − Tˆz δW dΩ = 0 Ω

Since the stress field is statically admissible, each term in parenthesis is zero, and multiplication by an arbitrary quantity results in a zero product. Each of the two integral then vanishes, as does their sum. Hence, the following is obtained    Z Z Z  ∂Syx ∂Szx ∂Syy ∂Szy ∂Sxy ∂Sxx + + + bx δU + + + + by δV ∂x ∂y ∂z ∂x ∂y ∂z Γ    ZZ n      o ∂Syz ∂Szz ∂Sxz + + + bz δW dΓ − Tx − Tˆx δU + Ty − Tˆy δV + Tz − Tˆz δW dΩ = 0 + ∂z ∂y ∂z Ω

Now let us integrate by parts the terms of the volume integral. Let us first integrate the first term: ZZ ZZZ ZZZ ∂Sxx ∂δU δU dΓ = nx Sxx δU dΩ − Sxx dΓ ∂x ∂x Ω

Γ

Γ

where nx is the component of the outward unit long the x-axis. Note that ∂U ∂δU =δ = δexx ∂x ∂x

Tx = nx Sxx , Thus ZZZ

∂Sxx δU dΓ = − ∂x

Γ

ZZZ

Sxx

Γ

ZZ ZZZ ZZ ∂δU dΓ + Tx δU dΩ = − Sxx δexx dΓ + Tx δU dΩ ∂x Ω

Γ

Ω

Generalizing and expressing in vectorial notation, we get: ZZZ ZZZ ZZ ZZ   T T ˆ δd dΩ = 0 − S δE dΓ + b δd dΓ + T δd dΩ − T − T Γ

Γ

−

ZZZ Γ

Ω

ST δE dΓ +

ZZZ Γ

Ω

ZZ ˆ δd dΩ = 0 bT δd dΓ + T Ω

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

650

where  Sxx      Syy    Szz S=  S yz     S xz    Sxy

                

,

 Exx      Eyy    Ezz E=  E yz     E xz    Exy

        

 exx      eyy    ezz =   2 eyz           2 exz     2 exy

                

,

   bx  , b= b  y  bz

   Tx  ˆ= T , T  y  Tz

   U  d= V   W

Since virtual displacements are now chosen to be kinematically admissible, the virtual compatible strain field is  δExx      δEyy    δEzz δE =  δE yz     δE xz    δExy

        

 δexx      δeyy    δezz =   2 δeyz           2 δexz     2 δexy

                

   δU  δd = δV   δW

,

Hence, for deformable bodies the Principle of Virtual Work is written as ZZZ ZZZ ZZ ˆ δd dΩ ST δE dΓ = bT δd dΓ + T Γ

Γ

(9.9)

Ω

The term on the left hand side of this expression can be interpreted as the virtual work done by the internal stresses whereas that on the right corresponds to the work done by the externally applied loads: ZZZ δWint = ST δE dΓ δWext =

Γ ZZ Z Γ

ZZ ˆ δd dΩ b δd dΓ + T T

Ω

where the internal virtual work is, ZZZ δWint = ST δE dΓ =

Γ ZZ Z

{Sxx δExx + Syy δEyy + Szz δEzz + Sxy δExy + Sxz δExz + Syz δEyz } dΓ

Γ

=

ZZZ

{Sxx δexx + Syy δeyy + Szz δezz + 2 Sxy δexy + 2 Sxz δexz + 2 Syz δeyz } dΓ

Γ

The Principle of Virtual Work for deformable bodies is defined as follows: A body is in equilibrium if the virtual work done by the internal stresses equals the virtual work done by the externally applied loads, for all arbitrary kinematically admissible virtual

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

651

displacements fields and corresponding compatible strain field. Equation (9.9) is equivalent to the set of equilibrium equations written for a differential element in the deformable body as well as of the stress boundary conditions on the part Ω of the surface. Since virtual displacements are automatically compatible when they are expressed in terms of continuous, singlevalued functions, we often mention only the need for consistency between strains and displacements. The principle of virtual work is also valid for large real displacements; however, we would write Eq. (9.9) using more complex measures of stresses and strains.

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

9.4.1

652

PVW for an Elastic Bar

P

L

C A

A′

δu(0) B ds

B′ P

δu(L)

Figure 9.11: Elastic bar subject to a load P undergoing a virtual displacement. To better understand this concept, let first derive the PVW for an elastic bar and truss structure. To do so, consider an axial bar of length L and give a virtual displacement from the equilibrium configuration, as shown in Fig. 9.11. The external virtual work acting on the elastic bar is given by: External Virtual Work at A : External Virtual Work at B :

−P δu(0)

+P δu(L)

Note that δu(0) 6= δu(L). Thus external virtual work due to a load P is A B δWext = δWext + δWext = P [δu(L) − δu(0)] 6= 0

Same results is produced when using the generalized expression for external virtual work acting on deformable bodies: ZZZ ZZ ˆ δd dΩ δWext = bT δd dΓ + T Γ

Ω

Since there are no body forces, the first integral vanishes. For an axial bar the displacement field is U (x, y, z) = u(x),

V (x, y, z) = 0,

W (x, y, z) = 0

Thus axial bars only stretch or shrink along its mayor axis,      δU   δU  δd = = δV 0     δW 0

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

653

Hence δWext

ZZ ZZ ZZ ZZ h i h i ˆ ˆ ˆ = T δd dΩ = Tx δu dΩ = Tx δu dΩ2 − Tˆx δu dΩ1 = Tˆx Ω δu − Tˆx Ω δu 2

Ω

Ω

Ω2

For our problem,

h

and

δWext

1

Ω1

Tˆx Ω

i

2

h i = Tˆx Ω = P 1

u1 = u(0), u2 = u(L) ZZ ˆ δd dΩ = P [δu(L) − δu(0)] = T Ω

To determine the internal virtual work acting in the elastic bar, let us consider a small element at C

s ds

σ σ

C

a

δu(s)

b

δu(s + ds)

Internal Virtual Work at a : Internal Virtual Work at b :

−σ dA δu(s)

+σ dA δu(s + ds)

Using Taylor series expansion and ignoring higher order terms: δu(s + ds) = δu(s) +

dδu(s) ds ds

The internal virtual work for a small element ds is dδWint =

a b dδWint + dδWint =σ

  dδu(s) d u(s) ds dA = σ δ ds dA = σ δε ds dA ds ds

Hence the internal virtual work for an elastic bar is: Z ZZ δWint = σ dA δε ds

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

654

Same results is produced when using the generalized expression for internal virtual work acting on deformable bodies ZZZ ZZZ δWint = ST δE dΓ = {Sxx δExx + Syy δEyy + Szz δEzz + Sxy δExy + Sxz δExz + Syz δEyz } dΓ Γ

Γ

For an axial bar the state of stress is



Sxx

  S=  0 

0 0

0

0

and the state of strain for isotropic materials is 

  e=  0 

exx =

Sxx , E

0

0

eyy = −ν



0

eyy

0

Note that (in absence of thermal loads)

  0   

0

exx



0

  0   

ezz

Sxx , E

ezz = −ν

Sxx E

When the rotations and deformations are small, cauchy’s stresses and Second Piola-Kirchhoff stresses coalesce: Sxx ≈ σxx The strains only have a stretching component, Exx = exx = εxx Hence δWint =

ZZZ

ST δE dΓ =

ZZZ

Sxx δexx dΓ =

Γ

Γ

ZZZ

Sxx δεxx dΓ

Γ

Using definitions for stress resultants, we see that δWint =

Z ZZ

σ dA δεxx ds =

Z

L

Nxx δεxx ds

0

where the material law for isotropic materials is defined as ZZ ZZ Nxx = σ dA = E εxx dA = EA εxx

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

655

Example 9.4.

x, u

P L/2

L/2

Figure 9.12: Point force acting on a clamped bar. Consider an elastic bar with fixed ends subject to an axial load P , as shown in Fig. 9.12. There no temperature change from the stress free state. The uniform and homogeneous axial bar has an axial stiffness of EA. (9.4-a) Determine the axial displacement and normal force of the bar by solving differential equations. This is a statically indeterminate problem, thus we will need to use all three set of equations: equilibrium, strain-displacement, and the constitutive law. The governing equation for the axial bar problem using concepts learned in chapter 6 is EA u00 = 0 and it is valid in the open intervals 0 < x < L/2 and L/2 < x < L, but no valid at the location of the point force since a point load is mathematically equivalent to an infinite value of the distributed load px (x) acting over a zero length. Hence, EA u001 = 0

0 < x < L/2

=0

L/2 < x < L

EA u002 Thus the solution is

EA u1 = c1 x + c2

0 < x < L/2

EA u2 = c3 x + c4

L/2 < x < L

We have for unknowns and thus we need four boundary conditions. We know that at the fixed end, u1 (x) = 0, u2 (x) =0 x=0

x=L

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

The displacement must be continuous at x = L/2: = u2 (x) u1 (x) x=L/2

656

x=L/2

The last boundary equation comes from them axial load equilibrium at the point of load application: x, u

dx

P N1

N2

L/2

−N1 + N2 +P =0 x=L/2 x=L/2 −EA u01 + EA u02 +P =0 x=L/2

x=L/2

Hence the displacement field is  P   x, 0 ≤ x ≤ L/2;  2 EA u(x) =    P (L − x), L/2 ≤ x ≤ L. 2 EA

The normal force in each segment is found through, N1 = EA u01 ,

N2 = EA u02

Hence

 P   , 0 ≤ x ≤ L/2;  2 N (x) =    − P , L/2 ≤ x ≤ L. 2 The maximum deflection occurs at the location of the applied load and the values is: umax =

PL 4 EA

The maximum value of the load is: Nmax =

P 2

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

(9.4-b) Determine the axial displacement and normal force of the bar using the principle of virtual work. 1. Kinematically admissible displacement function. In order to solve this using the Principle of Virtual Work, we need to assume a kinematically admissible displacement function u ¯(x). To be kinematically admissible the displacement function must be continuous in the interval 0 < x < L, and satisfy prescribed boundary conditions. The sine function is continuous in the interval, and its derivative with respect to x, or the strain, is also continuous. The prescribed boundary conditions on the axial displacement are u ¯(0) = 0 and u ¯(L) = 0. The function, π x sin L is zero at x = 0 and x = L, the prescribed displacement boundary conditions are satisfied. Thus, the assumed displacement takes the form π x u ¯(x) = a1 sin L in which the unknown parameter a1 represents the axial displacement at x = L/2. 2. Virtual displacement of the kinematically admissible function. The variation in the assumed displacement, or virtual displacement, is π x δu ¯(x) = δa1 sin L

where δa1 is the variation in the unknown parameter a1 . Note that we have taken the variation for a fixed value of x. 3. Virtual work due to external loads. The only external force acting on the bar is the point force P , so the external virtual work is π x δWext = P δ u ¯ = P δa1 sin = P δa1 L x=L/2 x=L/2 4. Virtual work due to internal loads. The internal virtual work is δWint =

Z

0

L

Nxx δexx dx

657

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

658

where the virtual strain is: π x π d¯ u = a1 cos dx L   L π x d¯ u dδ u ¯ π =δ = = δa1 cos dx dx L L

exx = δexx

Now, for linear elastic material with no temperature change, the material law is given by π x π Nxx = E A exx = EA u ¯0 = EA a1 cos L L Hence, the internal virtual work is δWint =

Z

L

Nxx δexx dx

0

= (a1 ) δa1

π L

Z

0

L



EA

 π x  π x π π2 cos δa1 cos dx = EA a1 L L L 2L

5. Application of the Principle of Virtual Work. Applying the PVW

δWext − δWint = 0

π2 P δa1 − EA a1 δa1 = 0 2L   π2 δa1 = 0 P − EA a1 2L

For all δa1 6= 0,

P − EA a1 Solving for a1

2 a1 = 2 π

Thus the displacement is 2 u ¯(x) = 2 π



π2 =0 2L



PL EA

PL EA 



sin

π x L

and the normal force is N = E A exx = EA u ¯0 =

π x 2P cos π L

(9.4-c) Now let us compare the results from both methods. The maximum displacement for the assumed function is 2P L u ¯max = 2 π EA

9.4. PVW FOR DEFORMABLE CONTINUOUS STRUCTURES

659

The percentage error in the approximate solution is given by umax − u ¯max × 100% = 18.9% umax So the approximate displacement at the center of the bar is 18.9% less than the exact value. The fact that the approximate displacement is less than the exact displacement means our assumed displacement function is too restrictive and results in a stiffer response than what the exact solution gives. The maximum value of the normal force in the approximate solution occurs at x = 0 and is ¯ = 2P N π The percentage error in the approximate solution for the normal force is ¯max Nmax − N × 100% = −27.3% Nmax The maximum normal force in the approximate solution exceeds the exact value by 27.3%. We can improve our approximate solution by adding more terms to the function:   π x 3πx u ¯(x) = a1 sin + a3 sin , 0 0

9.7. PCVW FOR CONTINUOUS DEFORMABLE BODIES

 P    2 sin θ − Q cos θ =   − P sin θ − Q cos θ 2

N (θ)

768

:

−α ≤ θ < 0

:

0 0

  

=

 

δP sin θ − δQ cos θ 2

:

−α ≤ θ < 0

− yδP 2 sin θ − δQ cos θ

:

01 σreq Thus the design safety factor is a ratio greater than one. That is, capacity must be greater than load and strength must be greater than stress. A large safety factor usually means a safer design, however, more material is used in the design with a corresponding increase in cost and weight. Therein lies one of the fundamental trade-offs in engineering design: cost vs. safety. Reducing cost is always a business goal, while the costumer demands increased safety. Thus it is highly important for the design engineer to choose an adequate safety factor to safeguard public safety at an affordable cost.

10.1. UNCERTAINTIES IN DESIGN

786

Factors in the Selection of a Safety Factor The selection of an appropriate safety factor is based primarily on the following five factors: 1. Degree of uncertainty about loading: In some situations loads can be determined with virtual certainty. The centrifugal forces in the rotor of an alternating-current motor cannot exceed those calculated for synchronous speed. The loads acting on an engine valve spring are definitely established by the valve open and valve closed positions (however, in a later chapter we will mention spring surge, which could introduce a degree of uncertainty). But what loads should be used for the design of automotive suspension components, whose loads can vary tremendously depending on the severity of use and abuse? And what about a comparable situation in a completely new kind of machine for which there is no previous experience to serve as a guide? The greater the uncertainty, the more conservative the engineer must be in selecting an appropriate design overload or safety factor. 2. Degree of uncertainty about material strength.: Ideally, the engineer would have available extensive data pertaining to the strength of the material as fabricated into the actual (or very similar) parts, and tested at temperatures and in environments similar to those actually encountered. But this is seldom the case. More often, the available material strength data pertain to samples smaller than the actual part, which have not experienced any cold working in part fabrication, and which have been tested at room temperature in ordinary air. Moreover, there is bound to be some variation is strength from one test specimen to another. Sometimes the engineer must work with material test data for which such information as specimen size and degree of data scatter (and the relationship between the reported single value and the total range of scatter) are unknown. Furthermore, the material properties may sometimes change significantly over the service life of the part, The greater the uncertainty about all these factors, the larger the safety factor that must be used. 3. Uncertainties in relating applied loads to material strength via stress analysis: At this point the reader is already familiar with a number of possible uncertainties, such as (a) validity of the assumptions involved in the standard equations for calculating nominal stresses, (b) accuracy I in determining the effective stress concentration factors, (c) accuracy in estimating residual stresses, if any, introduced in fabricating the I part, and (d) suitability of any failure theories and other relationships I used to estimate significant strength from available laboratory strength test data. 4. Need to conserve: The need to conserve material, weight, space, or dollars 5. Consequences of failurehuman safety and economics: If the consequences of failure are catastrophic, relatively large safety factors must, of course, be used. In addition, if the failure of some relatively inexpensive part could cause extensive shutdown of a major assembly line, simple economics dictates increasing the cost of this part severalfold (if necessary) in order to virtually eliminate the possibility of its failure An important item is the nature of a failure. If failure is caused I by ductile yielding, the consequences are likely to be less severe than I if caused by brittle fracture. Accordingly, safety factors recommended in handbooks are invariably larger for brittle materials. 6. Cost of providing a large safety factor: This cost involves a monetary I consideration and may also involve important consumption of re-1 sources. In some cases, a safety factor larger

10.1. UNCERTAINTIES IN DESIGN

787

than needed may have serious consequences. A dramatic example is a hypothetical aircraft with excessive safety factors making it too heavy to fly! With respect to the design of an automobile, it would be possible to increase safety factors on structural components to the point that a maniac driver could hardly cause a failure even when trying. But to do so would penalize sane drivers by requiring them to pay for stronger components than they can use. More likely, of course, it would motivate them to buy competitor’s cars! Consider this situation. Should an I automotive engineer increase the cost per car by $10 in order to avoid I 100 failures in a production run of a million cars, where the failures would not involve safety, but would entail a $100 repair? That is,

Selection of Design Safety Factor Selection of a design safety factor must be undertaken with care since there are unacceptable consequences associated with selected values that are either too low or too high. If the selected value is too small, the probability of failure will be too great. If the selected value is too large, the size, weight, or cost may be too high. Proper safety factor selection requires a good working knowledge of the limitations and assumptions in the calculation models or simulation programs used, the pertinent properties of the proposed materials, and operational details of the proposed application. Design experience is extremely valuable in selection of an appropriate design safety factor, but a rational selection may be made even with limited experience. The method suggested by Collins breaks the selection down into a series of semiquantitative smaller decisions that may be weighted and empirically recombined to calculate an acceptable value for the design safety factor, tailored to the specific application. To implement the selection of a design safety factor, consider separately each of the following eight rating factors: a) Accuracy of loads knowledge: The accuracy with which the loads, forces, deflections, or other failure-inducing agents can be determined b) Accuracy of stress calculation: The accuracy with which the stresses or other loading severity parameters can be determined from the forces or other failure-inducing agents c) Accuracy of strength knowledge: The accuracy with which the failure strengths or other measures of failure can be determined for the selected material in the appropriate failure mode d) Need to conserve: The need to conserve material, weight, space, or dollars e) Seriousness of failure consequences: The seriousness of the consequences of failure in terms of human life and/or property damage f) Quality of manufacture: The quality of workmanship in manufacture g) Conditions of operation: The conditions of operation h) Quality of inspection: The quality of inspection and maintenance available or possible during operation A semiquantitative assessment of these rating factors may be made by assigning a rating number, ranging in value from -4 to +4, to each one. These rating numbers (RNs) have the following meanings:

10.1. UNCERTAINTIES IN DESIGN

788

RN=1 mild need to modify nSF RN=2 moderate need to modify nSF RN=3 strong need to modify nSF RN=4 extreme need to modify nSF Further, if there is a need to increase the safety factor, the selected rating number is assigned a positive (+) sign; if it is to decrease the safety factor, the selected rating number is to assign a negative (−) sign. The next step is to calculate the algebraic sum of the eight rating numbers: t=

8 X

RNi

i=1

Now using the above result, the design safety factor may be empirically estimated from:  (10 + t)2    1+ , for t ≥ −6 100 nSF =    1.15, for t < −6

(10.2)

In general, the design safety factor will be bound by:

1.15 ≤ nSF ≤ 5

(10.3)

although for lightweight structures the safety factor is small as possible and for some machinery we might have safety factors higher than 5.

Recommended Values for a Safety Factor Having read through this much philosophy of safety factor selection, the reader is entitled to have, at least as a guide, some suggestions values of safety factor that have been found useful. For this purpose, the following recommendations of Joseph Vidosic are suggested. These safety factors are based on yield strength. 1. SF = 1.25 to 1.5 for exceptionally reliable materials used under controllable conditions and subjected to loads and stresses that can be determined with certaintyused almost invariably where low weight is a particularly important consideration. 2. SF = 1.5 to 2 for well-known materials, under reasonably constant environmental conditions, subjected to loads and stresses that can be determined readily. 3. SF = 2 to 2.5 for average materials operated in ordinary environments and subjected to loads and stresses that can be determined. 4. SF = 2.5 to 3 for less tried materials or for brittle materials under average conditions of environment, load, and stress.

10.1. UNCERTAINTIES IN DESIGN

789

5. SF = 3 to 4 for untried materials used under average conditions of environment, load, and stress. 6. SF = 3 to 4 should also be used with better-known materials that are to be used in uncertain environments or subjected to uncertain stresses. 7. Repeated loads: the factors established in items 1 to 6 are acceptable but must be applied to the endurance limit rather than to the yield strength of the material. 8. Impact forces: the factors given in items 3 to 6 are acceptable, but an impact factor should be included. 9. Brittle materials: where the ultimate strength is used as the theoretical maximum, the factors presented in items 1 to 6 should be approximately doubled. 10. Where higher factors might appear desirable, a more thorough analysis of the problem should be undertaken before deciding on their use.

Example 10.1. You have been asked to propose a value for the design safety factor to be used in determining the dimensions for the main landing gear support for a new executive jet aircraft. It has been determined that the application may be regarded as “average” in many respects, but the material properties are known a little better than for the average design case, the need to conserve weight and space is strong, there is a strong concern about threat to life and property in the event of a failure, and the quality of inspection and maintenance is regarded as excellent. What value would you propose for the design safety factor? Based on the information given, the rating numbers assigned to each of the eight rating factors might be Thus, t = 0 + 0 − 1 − 3 + 3 + 0 + 0 − 4 = −5 Since t ≥ −6,

(10 + t)2 (10 − 5)2 =1+ = 1.25 100 100 The recommended value for a design safety factor appropriate to this application would be: nSF = 1 +

nSF = 1.25 End Example 

10.1. UNCERTAINTIES IN DESIGN

790

Table 10.1: Semiquantitative assessment of rating factors Rating Factor

10.1.2

Selected Rating Number (RN )

1. Accuracy of loads knowledge

:

2. Accuracy of stress calculation

:

0 0

3. Accuracy of strength knowledge

:

-1

4. Need to conserve

:

-3

5. Seriousness of failure consequences

:

+3

6. Quality of manufacture

:

0

7. Conditions of operation

:

0

8. Quality of inspection

:

-4

Margin of Safety

Margins of Safety is an index indicating the amount beyond the minimum necessary; in other words, the margin of safety is the strength of the material minus the anticipated stress and is defined as: MS =

excess strength σall − σreq σall = = − 1 = nSF − 1 required strength σreq σreq

(10.4)

Lightweight structures have M S small as possible. A margin of safety of zero, implies no safety was considered in the design: nSF = 1

→

MS = 0

onset of failure

Negative margin of safety means failure has occurred: nSF < 1

→

MS < 0

failure occurred

Thus when designing for safety using deterministic approaches, we must always ensure nSF > 1

→

MS > 0

safe

Example 10.2. It is desired to design a shaft with a 20% margin of safety. What is the safety factor? The margin of safety is 20% or 0.20. Thus, nSF = M S + 1 = 0.20 + 1 = 1.20

10.1. UNCERTAINTIES IN DESIGN

791

The factor of safety is 1.20. End Example 

10.2. DUCTILE AND BRITTLE FAILURE THEORIES

10.2

792

Ductile and Brittle Failure Theories

Now let us see what governs a three-dimensional state of stress in yielding. Before we do so it is important to understand the behavior of structural metals. Structural metal behavior is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner.

σ

σ

Ductile materials:

Brittle materials:

εf ¥ 5%

εf < 5%

ε

ε

10.2. DUCTILE AND BRITTLE FAILURE THEORIES

793

Ductile materials

Brittle materials

The true strain at fracture is εf ≥ 5 % in 2 inches.

The true strain at fracture is εf < 5 % in 2 inches.

The yield strength is identifiable and is often the same in compression as in tension (Syt ≈ Syc = Syield ).

The yield strength is not identifiable. Typically classified by ultimate tensile strength Sut and ultimate compressive strength Suc , where Suc is a positive quantity.

A single tensile test is sufficient to characterize the material behavior of a ductile material, Sy and Sut .

Two material tests, a tensile test and a compressive test, are required to characterize the material behavior of a brittle material, Suc and Sut . The compressive strength is significantly higher than its tensile strength (Suc  Sut ).

Governed by Yielding Criteria Generally accepted theories for yielding criteria are:

Governed by Fracture Criteria

Generally accepted theories for fracture criteria are:

a) Distortion Energy Criterion

a) Maximum-Normal-Stress Criterion

. b) Maximum-Shear-Stress Criterion (Syt = Syc )

b) Brittle Coulomb-Mohr Criterion

c) Ductile Coulomb-Mohr Criterion (Syt < Syc )

10.3. 3-D STRESS STATE FAILURE THEORIES: BRITTLE MATERIALS

10.3

794

3-D Stress State Failure Theories: Brittle Materials

An important attribute in design with brittle materials is they do not possess defense against stress concentration opposed to ductile materials. Thus even small scratches and cracks as naturally occur in their fabrication can lead to brittle fracture. For this reason, brittle materials have to be used with extreme caution in tension structures. For such material one should evaluate the stress concentration factors.

10.3.1

Maximum Normal Stress Criterion

Postulate: Failure occurs when one of the three principal stresses equals the strength. The maximum stress criterion, also known as the normal stress, Coulomb, or Rankine criterion, is often used to predict the failure of brittle materials. The maximum stress criterion states that failure occurs when the maximum (normal) principal stress reaches either the uniaxial tension strength Sut , or the uniaxial compression strength Suc , −Suc < (σ1 , σ2 , σ3 ) < Sut where σ1 , σ2 , and σ3 are the principal stresses for 3-D stress. Recall: σ1 > σ2 > σ3 To better grasp this Criterion, consider the plane stress problem (σ3 = 0). Graphically, the maximum stress criterion requires that the two principal stresses lie within the green zone depicted below, σ2 Sut -Suc Sut

σ1

-Suc

The factor of safety can be obtain by: σ1 =

Sut nSF

σ1 ≥ σ2 ≥ 0 σ1 ≥ 0 ≥ σ2

σ2 Suc and ≤ σ1 Sut

(10.5)

10.3. 3-D STRESS STATE FAILURE THEORIES: BRITTLE MATERIALS

σ2 = −

Suc nSF

σ1 ≥ 0 ≥ σ2 0 ≥ σ1 ≥ σ2

795

σ2 Suc and > σ1 Sut

(10.6)

where Suc is the uniaxial ultimate compression strength and Sut the uniaxial ultimate tensile strength. In short, according to the Maximum-Normal-Stress Theory, as long as stress state falls within the box, the material will not fail. For a general three-dimensional state of stress, the design criteria is governed by   (a) σmax ≥ σmin ≥ 0   σmax 1 ≤ for σmin Suc  Sut nSF  ≤  (b) σmax ≥ 0 ≥ σmin and σmax Sut −

σmin 1 ≤ Suc nSF

for

   (c)  

(d)

σmax ≥ 0 ≥ σmin 0 ≥ σmax ≥ σmin

σmin Suc > and σmax Sut

where nSF = 1 at onset of failure (fracture begins). One should evaluate the corresponding case(s) and choose the design that falls inside the box.

10.3.2

Brittle Coulomb-Mohr Criterion

The Mohr Theory of Failure, also known as the Coulomb-Mohr criterion or internal-friction theory, is based on the famous Mohr’s Circle. Mohr’s theory is often used in predicting the failure of brittle materials, and is applied to cases of 2-D stress. Mohr’s theory suggests that failure occurs when Mohr’s Circle at a point in the body exceeds the envelope created by the two Mohr’s circles for uniaxial tensile strength and uniaxial compression strength. This envelope is shown in the figure below,

τ

-Sc

Uniaxial Tension

St

σ

Uniaxial Compression

The left circle is for uniaxial compression at the limiting compression stress Suc of the material. Likewise, the right circle is for uniaxial tension at the limiting tension stress Sut .

10.3. 3-D STRESS STATE FAILURE THEORIES: BRITTLE MATERIALS

796

The middle Mohr’s Circle on the figure (dash-dot-dash line) represents the maximum allowable stress for an intermediate stress state. Each case defines the maximum allowable values for the two principal stresses to avoid failure. For the plane stress problem:

σ2 Sut -Suc Sut

σ1

-Suc

Sut nSF

σ1 ≥ σ2 ≥ 0

σ2 1 σ1 − = Sut Suc nSF

σ1 ≥ 0 ≥ σ2

σ1 =

σ2 = −

Suc nSF

(10.7)

0 ≥ σ1 ≥ σ2

where Suc is the uniaxial ultimate compression strength and Sut the uniaxial ultimate tensile strength. For a general three-dimensional state of stress, the design criteria is governed by σmax σmin 1 Sut − Suc ≤ nSF

where nSF = 1 at onset of failure (fracture begins).

10.3.3

Comparison of MNS and BCM Criterions

Also shown on the figure is the maximum stress criterion (dashed line). The MNS theory is less conservative than Mohr’s theory since it lies outside Mohr’s boundary. The design safety factors are compared as nSF ≥ nSF MNS

Thus BCM Theory is the most conservative one.

BCM

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

797

Maximum Stress Mohr’s

σ2 Sut

-Suc Sut

σ1

-Suc

10.4

3-D Stress State Failure Theories: Ductile Materials

An important attribute in design with ductile materials is their capacity to accommodate stress concentrations through plastic deformation and hence to redistribute the stresses more evenly. Stress concentrations occur at stress raisers which are either geometric discontinuities (e.g., holes, sharp corners, cracks, fillets, etc.) and/or material discontinuities (notches). The capacity to redistribute stresses at stress riser make’s a ductile material “tough”, giving the material a defense mechanism against stress concentrations. Thus for static loads with high stress concentrations, we usually take the stress concentration factor as: Kt → 1 Ductile engineering materials are those for which static strength in engineering applications is limited by yielding and not fracture. The yield stress Syield is determined from the tensile test data, but the tensile test is designed to produce a uniaxial state of stress. However, we would like to know what governs yielding under combined states of stress that occur in structural components under service loads. Although no theoretical way to correlate yielding in a three-dimensional state of stress with yielding in the uniaxial tensile test exists, three empirical equations have been proposed: 1. Aka Distortion Energy Criterion. Also known as Mises Yield Criterion or Octahedral Shear-stress Criterion. Good for ductile materials but should not be used for brittle materials. 2. Maximum Shear Stress Criterion. Also known as Aka Tresca Criterion. Good for ductile materials with tensile yield strength approximately equal to compressive yield strength and should not be used for brittle materials. 3. Ductile Coulomb-Mohr. Good for ductile materials with tensile yield strength different to compressive yield strength and should not be used for brittle materials. Aka Distortion Energy Criterion and Maximum Shear Stress Criterion are based on:

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

798

1. State of stress can be completely described by the magnitude and direction of the principal stresses. For an isotropic material the principal stress directions are unimportant. 2. Experiments show that hydrostatic state of stress does NOT effect yielding.

10.4.1

Aka Distortion Energy Criterion

Postulate: Yielding will occur when the distortion-energy per unit volume equals the distortion-energy per unit volume in a uniaxial tension specimen stressed to its yield strength. τoct = τoct (10.8) 3-D

1-D

The basis for the maximum distortion energy theory of failure is that the overall strain energy is composed of two parts. The first part is the energy associated with merely changing the volume of the part while the second part is associated with the distortion of the part. Thus, the total strain energy per unit volume u can be written as (10.9) u = uv + ud

where uv is the energy of volume change per unit volume and ud is the energy of distortion per unit volume. It is this distortion part of the strain energy that is the basis for this failure theory. The hypothesis for this theory is that failure will occur in the complex part when the distortion energy per unit volume exceeds that for a simple uniaxial tensile test at failure. For purposes of describing this failure theory, the principal normal stresses can be thought of as being composed of two parts that are superimposed as follows

where σ1 , σ2 and, σ3 are the principal stresses. For this superposition, the relationships will be σ1 = σ10 + σav σ2 = σ20 + σav σ3 = σ30 + σav

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

and the state of stress can be written as:    σav σ1 0 0        0 σ2 0  =  0       0

0

0

σ3

0 σav 0

0





σ10

     0  + 0  

σav

0

0 σ20 0

799

0



  0   

σ30

Here σav represents the portion of the stress that causes volume change and σi0 represents the portion of the principal normal stresses that cause distortion: σ10 = σ1 − σav σ20 = σ2 − σav σ30 = σ3 − σav Now the hydrostatic state of stress can be shown to be: σav =

σ1 + σ2 + σ3 3

(10.10)

The distortional element is subject to pure angular distortion, that is no volume change. The total strain energy density for an element subject to the three principal stresses is: u=

o o 1n 1 n 2 σ1 ε1 + σ2 ε2 + σ3 ε3 = σ1 + σ22 + σ32 − 2 ν (σ1 σ2 + σ2 σ3 + σ1 σ3 ) 2 2E

The total strain energy density for an element subject to the hydrostatic stresses is: uv =

o 2 n
o 3σav 1 n 2 2 3σav − 2 ν 3σav 1 − 2ν = 2E 2E

Now substitute Eq. (10.10) into the above expression to obtain: uv =

o 1 − 2ν n 2 σ1 + σ22 + σ32 + 2 (σ1 σ2 + σ2 σ3 + σ1 σ3 ) 6E

Now the distortion energy is given by Eq. (10.9) ( ) 2 2 2 1 + ν (σ1 − σ2 ) + (σ2 − σ2 ) + (σ3 − σ1 ) ud = u − uv = 3E 2 Note that the distortion energy would be zero if σ1 = σ2 = σ3 . Now for a simple tensile test, at yield, ud =

1+ν 2 S 3 E yield

(σ1 = Syield , σ2 = σ3 = 0)

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

800

Further recall that von Mises stress was defined as: s 2 2 2 q (σ1 − σ2 ) + (σ2 − σ2 ) + (σ3 − σ1 ) Iσ21 − 3 Iσ2 = σeq = 2 Thus von Mises stress is a method to predict yielding in 3-D state of stress. ud

=

3-D

and it was defined as:

=

σeq

1+ν 2 σ 3 E eq

q

Iσ21 − 3 Iσ2

(10.11)

Now, the Aka Distortion Energy Criterion can be expressed in terms of the von Mises stress: σeq < Syield The safety factor can be derived as: σeq =

Syield nSF

nSF =

→

Syield σeq

To better explain the physical meaning of the DE Criterion let us consider a plane stress problem (σ3 = 0). Yielding will begin when σeq = Syield = Rewriting the above:

q

Iσ21 − 3 Iσ2 =

q

σ12 − σ1 σ2 + σ22

2 Syield = σ12 − σ1 σ2 + σ22

Now taking a transformation of θ = 45◦ : σ ¯1 =

σ1 + σ2 √ 2

σ ¯2 =

σ2 − σ1 √ 2

σ1 =

σ ¯1 − σ ¯ √ 2 2

σ2 =

σ ¯1 + σ ¯ √ 2 2

Solving for σ ¯1 and σ ¯2 ,

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

801

Then the criterion becomes 2 Syield =

1=

3σ ¯22 σ ¯12 + 2 2 σ ¯12 3σ ¯22 + 2 2 2 Syield 2 Syield

1=

1=

√ 

σ ¯1 2 Syield

!2

σ ¯1 1.4142 Syield



2

  σ ¯2  + r2  Syield 3 2

+



σ ¯2 0.8165 Syield

2

The above is an ellipse rotated at 45◦ :

σ2 σ2

σ1

Sy

1.41 Sy

0.816 Sy -Sy Sy -1.41 Sy

-Sy

σ1

-0.816 Sy

In short, according to the DE Theory, as long as stress state falls within the ellipse, the material will not fail. For a general three-dimensional state of stress, the design criteria is governed by σeq 1 ≤ Syield nSF where nSF = 1 at onset of failure (yielding begins).

10.4.2

Maximum Shear Stress Criterion

Postulate: Yielding begins whenever the maximum shear stress in a part becomes equal to the maximum shear stress in a tension test specimen that begins to yield.

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

802

τ

τ τ1/3=τmax

Sy/2

τ1/2 τ2/3 σ3

σ1=Sy σ1 σ

σ2

σ

σ2=σ3=0

Tensile Test Specimen

3-D Stress State in Part

Yielding begins in a 3-D stress state when the maximum shear stress τmax is equal to its value at the initiation of yielding in the tension test: τmax = τmax (10.12) 3-D

It is known that

τmax

τmax

= 1-D

3-D

1-D

Syield 2

σmax − σmin = 2

Note that YIELDING BEGINS when (10.12) is true. In other words,

τ τmax=Sy/2 σ1=Sy σ2=0 σ3=0

σ

τmax = 0.5 Syield However for design purposes this should not happen. It is wanted that τmax

< 3-D

Syield 2

(10.13)

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

803

Maximum Shear Stress is another method to predict yielding in 3-D state of stress and it is defined as: σmax − σmin = ±Syield

(10.14)

To better explain the physical meaning of the DE Criterion let us consider plane stress problem (σ3 = 0). Thus let us consider three different cases with σ3 = 0.

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

804

First Case: σmax − σmin = ±Syield

→

σ1 − σ3

= ±Syield

σ3 < σ2 < σ1

σ1 = ±Syield

→ σ2

σ2 σ3

σ1

-Sy

Sy

σ1

Second Case: σmax − σmin = ±Syield

→

σ1 − σ2 = ±Syield

→

σ2 < σ3 < σ1

σ1

= σ2 ± Syield

σ2 Sy -Sy

σ3 σ2

σ1

Sy

σ1

-Sy

Third Case σmax − σmin = ±Syield

→

σ3 − σ2

= ±Syield

σ2 < σ1 < σ3

→

σ2 = ∓Syield

σ2 Sy σ1

σ2

σ3

σ1 -Sy

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

805

Thus for a biaxial representation of the yield, the Maximum-Shear-Stress Theory can be represented as

σ2 Locus of failure states

Sy Case I

-Sy

σ1

Sy Case II Case III

-Sy

In short, according to the MSS Theory, as long as stress state falls within the green area, the material will not fail. For a general three-dimensional state of stress, the design criteria is governed by τmax

or

= 3-D

Syield 2 nSF

nSF =

→

Syield 2 τmax

σmax 1 2 τmax σmin ≤ = − Sy Syield Syield nSF

where nSF = 1 at onset of failure (yielding begins).

10.4.3

Comparison of DE and MSS Criterions

σ2 Maximum Shear Stress

Sy -Sy

B

A

C

D Sy

-Sy

Distorsion Energy

σ1

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

806

The blue line is the loading line. Let point B represent the actual loading condition. Then the design safety factor can be calculate as follows nSF =

AC AB

Maximum Shear Stress Criterion

nSF =

AD AB

Distortion Energy Criterion

From the above it is clear that, in general, nSF

Maximum Shear Stress Criterion

Thus MSS Theory is the most conservative one.

≤ nSF

Distortion Energy Criterion

The distortion-energy theory predicts no failure under hydrostatic stress and agrees well with all data for ductile behavior. Hence, it is the most widely used theory for ductile materials and is recommended for design purposes, although some engineers also apply the MSS Theory because of its simplicity and conservative nature.

10.4.4

Ductile Coulomb-Mohr Criterion

When the tensile yield strength and compressive yield strength are significantly different for ductile materials, a variation of the brittle Coulomb-Mohr Criterion is commonly used. The criterion for a general three-dimensional state of stress is governed by σmax 1 σmin Syt − Syc ≤ nSF where nSF = 1 at onset of failure (fracture begins), Syc the uniaxial compressive yield strength and Syt the uniaxial ultimate tensile strength.

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

Example 10.3.

A certain force F applied at D near the end of the 15-in lever shown in Figure, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel and it has a minimum (ASTM) yield strength of 81 ksi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find the force F . Solution: We will assume that lever DC is strong enough and hence not a part of the problem. Note that point O is the place of maximum bending moment but not necessarily maximum bending stress due to stress concentration at A. Points A and O have the same shear and torsional loads but not necessary the same shear stresses due to torque because of stress concentration at A. Thus a stress element at A will be most critical.

807

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

y

y A

A Vy

Mzz

808

F T x

z

x

z

14F

15F

Free Body Diagram at A

Sign Convention

For our sign convention: Mzz = −14 F

Vy = −F

T = 15 F

Let us consider the stress element on the top surface will be subjected to a tensile bending stress and a torsional stress (note that there is no shear stress due to shear load at A. This point is the weakest section, and governs the strength of the assembly. From Table 4.7, for a static load acting on a ductile material, the stress concentration factor is not important. The two stresses are σxx = −Ktb

Mzz y Izz

τxz = Kts

Tr Jxx

At the top: σxx = −Ktb

Mzz (c) Mzz = −Ktb Izz Z

τxz = Kts

T (c) T = Kts Jxx Q

Using Tables 4.2 and 4.4: σxx = −Ktb τxz = Kts

Mzz 32 Mzz = −Ktb Z π d3

T 16 T = Kts Q π d3

Since the stress concentration factor are not important here Ktb = 1.0 Thus (for d = 100 ) σxx = −(1.0) τxz = (1.0)

Kts = 1.0

32 (−14 F ) = 142.6 F π (1)3

16 (15 F ) = 76.4 F π (1)3

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

809

Now we proceed to find the principal stresses. The state of stress is 

σxx σ A =  τxy τxz

τxy σyy τyz

The stress invariants are

  τxz 142.6 F τyz  =  0 σzz 76.4 F

0 0 0

 76.4 F  psi 0 0

Iσ1

= σxx + σyy + σzz = 142.6 F

Iσ2

2 = −τxz = − (76.4 F ) = −5836.1 F 2

Iσ3

=

(10.15)

2

0

(Plane stress)

The characteristic equation is

λ3 − Iσ1 λ2 + Iσ2 λ = λ λ2 − Iσ1 λ + Iσ2 = λ λ2 − 142.6 F λ − 5836.1 F 2

=0

The principle stresses are obtained analytically as follows

1q 2 Iσ1 + Iσ1 − 4 Iσ2 = 76.4 F + 104.50 F = 175.8 F 2 2

λ1

=

λ2

=

1q 2 Iσ1 − Iσ1 − 4 Iσ2 = 76.4 F − 104.50 F = −33.20 F 2 2

λ3

=

0

σ1 = max[λ1 , λ2 , λ3 ] = 175.8 F

σ3 = min[λ1 , λ2 , λ3 ] = −33.20 F

σ2 = 0

The maximum stresses are σmax = max[σ1 , σ2 , σ3 ] = 175.8 F

σmin = min[σ2 , σ2 , σ3 ] = −33.20 F

σmax − σmin = 104.50 F τmax = 2

Distortion energy criterion: If we employ the DE criterion, we need to calculate the von Mises stress: q σeq = Iσ21 − 3 Iσ2 = 194.50 F The yielding criteria for DE criterion is σeq 1 ≤ Sy nSF

→

194.50 F 81000

≤1

(nSF = 1)

Thus a force of F = 416.38 lb will cause yielding. Maximum Shear Stress Criterion: If we employ the MSS criterion, we need to calculate the maximum overall shear stress: σmax − σmin = 104.50 F τmax = 2

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

810

The yielding criteria for MSS criterion is 1 2 τmax ≤ Sy nSF

→

104.50 F 40500

≤1

(nSF = 1)

Thus a force of F = 387.56 lb will cause yielding. We can see that the force F required found by MSS is about 6.9% less than the one found for the DE. As stated earlier, the MSS theory is more conservative than the DE theory. End Example 

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

Example 10.4.

A certain force F applied at D near the end of the 15-in lever shown in Figure, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel and it has a minimum (ASTM) yield strength of 81 ksi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. 1. If F = 400 lbs, will the component fail? 2. Certain safety factor was used to ensure that the structure would not fail. If F = 300 lbs was assumed to cause yielding, what is the realized margin of safety? Solution: We proceed as before. (1) If F = 400 lbs, will the component fail? The goal is to ensure that the margin of safety is positive and/or safety factor is greater than one. The two yielding theories will give slightly different solutions.

811

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

812

Distortion energy criterion: If we employ the DE criterion, we need to calculate the von Mises stress: q σeq = Iσ21 − 3 Iσ2 = 194.50 F The yielding criteria for DE criterion is σeq 1 ≤ Sy nSF

81000 nSF

194.50 F =

→

→

77814.0 =

81000 nSF

Thus a safety factor of nSF = 1.041 was used. Since the safety factor is greater than one (and M S = 0.041 > 0) the structure is likely to not fail. Although the design engineer should consider nSF = 1.15 for design, this design is acceptable since failure is not predicted. Maximum Shear Stress Criterion: If we employ the MSS criterion, we need to calculate the maximum overall shear stress: σmax − σmin = 104.5 F τmax = 2 The yielding criteria for MSS criterion is 2 τmax 1 ≤ Sy nSF

→

104.5 =

40500 nSF

→

41799.5 =

40500 nSF

Thus a safety factor of nSF = 0.97 was used. Since the safety factor is less than one (and M S = −0.03 < 0) the design will fail.

Note that the safety factor required found by MSS is smaller than the one found for the DE. As stated earlier, the MSS theory is more conservative than the DE theory. (2) Certain safety factor was used to ensure that the structure would not fail. If F = 300 lbs was assumed to cause yielding, what is the realized margin of safety? Distortion energy criterion: If we employ the DE criterion, we need to calculate the von Mises stress: q σeq = Iσ21 − 3 Iσ2 = 194.50 F The yielding criteria for DE criterion is σeq 1 ≤ Sy nSF

→

194.50 F =

81000 nSF

→

58360.50 =

81000 nSF

Thus a safety factor of nSF = 1.39 was used. Since the safety factor is greater than one the design will not fail. The margin of safety is M S = nSF − 1 = 0.39 = 39% Maximum Shear Stress Criterion: If we employ the MSS criterion, we need to calculate the maximum overall shear stress: σmax − σmin = 104.5 F τmax = 2

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

813

The yielding criteria for MSS criterion is 2 τmax 1 ≤ Sy nSF

→

104.5 =

40500 nSF

→

31349.6 =

40500 nSF

Thus a safety factor of nSF = 1.29 was used. Since the safety factor is greater than one the design will not fail. In addition it is greater than 1.15. The margin of safety is M S = nSF − 1 = 0.29 = 29% Note that the safety factor required found by MSS is smaller than the one found for the DE. As stated earlier, the MSS theory is more conservative than the DE theory. End Example 

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

Example 10.5.

A certain force F applied at D near the end of the 15-in lever shown in Figure, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is made of ASTM grade 30 cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. Find the force F . Solution: We assume that the lever DC is strong enough, and not part of the problem. The tensile ultimate strength is 31 ksi and the compressive ultimate strength is 109 ksi. The stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress (just as before). This location, on the l-in diameter section fillet, is the weakest location, and it governs the strength of the assembly. Since grade 30 cast iron is a brittle material and the load is static, we should find the stress concentration factors. Thus use Charts 4.17 of your textbook for the corresponding stress concentration factors. The two stresses are at the top are σxx = −Ktb

Mzz Z

τxz = Kts

T Q

814

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

815

Using Tables 4.2 and 4.4: Mzz 32 Mzz = −Ktb Z π d3

σxx = −Ktb

τxz = Kts

T 16 T = Kts Q π d3

From Charts 4.17: D = 1.5 d

r = 0.125 d

Thus, (for d = 100 )

⇒

Ktb ≈ 1.6

σxx = −(1.6)

32 (−14 F ) = 228.165 F π (1)3

τxz = (1.33)

16 (15 F ) = 101.605 F π (1)3

Kts ≈ 1.33

Now we proceed to find the principal stresses. The state of stress is 

σxx σ A =  τxy τxz

  τxz 228.165 F   = τyz 0 σzz 101.605 F

τxy σyy τyz

The stress invariants are

0 0 0

 101.605 F  psi 0 0

Iσ1

= σxx + σyy + σzz = 228.165 F

Iσ2

2 = −τxz = − (101.605 F ) = −10323.5 F 2

Iσ3

=

(10.16)

2

0

(Plane stress)

The characteristic equation is

λ3 − Iσ1 λ2 + Iσ2 λ = λ λ2 − Iσ1 λ + Iσ2 = λ λ2 − 228.165 F λ − 10323.5 F 2 = 0

The principle stresses are obtained analytically as follows

1q 2 Iσ1 + Iσ1 − 4 Iσ2 = 266.16 F 2 2

λ1

=

λ2

=

Iσ1 1q 2 − Iσ1 − 4 Iσ2 = −38.69 F 2 2

λ3

=

0

σ1 = max[λ1 , λ2 , λ3 ] = 266.16 F

σ3 = min[λ1 , λ2 , λ3 ] = −38.69 F

σ2 = 0

The maximum stresses are: σmax = max[σ1 , σ2 , σ3 ] = 266.16 F τmax

σmax − σmin = 152.77 F = 2

σmin = min[σ1 , σ2 , σ3 ] = −38.69 F

(Not needed for brittle failure criterions)

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

816

Maximum Normal Stress criterion: First let us assume that F > 0. If we employ the MNS criterion, we need to check what case we are working with: σmax ≥ 0 ≥ σmin Thus we need to evaluate: σmin −38.69 F = σmax 266.16 F = 0.145364

Since: σmin Suc σmax < Sut

→

σmax =

Sut nSF

Suc 109 = = 3.51613 Sut 31

266.85 F = 31000

→

(nSF = 1)

Thus a force of F = 116.17 lb will cause fracture.  σmax = 266.16 F = 31000 psi F = 116.17 ⇒ σmin = −38.69 F = −4494.18 psi For F both σmax and σmin fall on the box -limit because: −Suc < σmax , σmin = Sut

Thus the load for fracture is: F = 116.17

lb

Brittle Coulomb-Mohr Criterion: If we employ the BCM criterion, σmax 1 σmin − = → 0.008963 F = 1 (nSF = 1) Sut Suc nSF

Note we took nSF = 1 for failure. Thus a force of F = 111.57 lb will cause fracture.

We can see that the force F required found by BCM is about 4.0% less than the one found for the MNS. As stated earlier, the BCM theory is more conservative than the MNS theory. End Example 

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

Example 10.6.

A certain force F applied at D near the end of the 15-in lever shown in Figure, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is made of ASTM grade 30 cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. 1. If F = 115 lbs, will the component fail? 2. Certain safety factor was used to ensure that the structure would not fail. If F = 100 lbs was assumed to cause fracture, what is the realized margin of safety? Solution: We proceed as before. (1) If F = 115 lbs, will the component fail? The goal is to ensure that the margin of safety is positive and/or safety factor is greater than one. The two fracture theories will give slightly different solutions. The principal stresses are: σ1 = 266.16 F = 30687.8

σ2 = 0

σ3 = −38.69 F = −4448.93

817

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

818

Maximum Normal Stress criterion: If we employ the MNS criterion, If we employ the MNS criterion, we need to check what case we are working with: σmax ≥ 0 ≥ σmin Thus we need to evaluate and compare: σmin −38.69 F = σmax 266.16 F = 0.145364 σmin Suc σmax < Sut

→

σmax =

Suc 109 = = 3.51613 Sut 31

Sut nSF

→

30687.8 =

31000 nSF

Thus a safety factor of nSF = 1.012 was used. Since the safety factor is greater than one (and M S = 0.012 > 0) the structure is likely to not fail. Although the design engineer should consider nSF = 1.15 for design, this design is acceptable since failure is not predicted. Brittle Coulomb-Mohr Criterion: If we employ the BCM criterion, σmax 1 σmin 1 = − → 1.03 = Sut Suc nSF nSF

Thus a safety factor of nSF = 0.97 was used. Since the safety factor is less than one (and M S = −0.03 < 0) the design will fail.

We can see that the safety factor required found by BCM is smaller than the one found for the MNS. As stated earlier, the BCM theory is more conservative than the MNS theory. (2) Certain safety factor was used to ensure that the structure would not fail. If F = 100 lbs was assumed to cause yielding, what is the realized margin of safety? The goal is to ensure that the margin of safety is positive and/or safety factor is greater than one. The two fracture theories will give slightly different solutions. The principal stresses are: σ1 = 266.16 F = 26685.1

σ2 = 0

σ3 = −38.69 F = −3868.63

Maximum Normal Stress criterion: If we employ the MNS criterion, need to check what case we are working with: σmax ≥ 0 ≥ σmin Thus we need to evaluate and compare: σmin −38.69 F = σmax 266.16 F = 0.145364 σmin Suc Sut → σmax = σmax < Sut nSF

109 Suc = = 3.51613 Sut 31 →

26685.1 =

31000 nSF

Thus a safety factor of nSF = 1.16 was used. Since the safety factor is greater than

10.4. 3-D STRESS STATE FAILURE THEORIES: DUCTILE MATERIALS

819

one (and M S = 0.16 > 0) the structure will not fail and the design is acceptable. Brittle Coulomb-Mohr Criterion: If we employ the BCM criterion, σmax 1 σmin 1 = − → 0.8963 = Sut Suc nSF nSF

Thus a safety factor of nSF = 1.12 was used. Since the safety factor is greater than one (and M S = 0.12 > 0) the structure will not fail. However, the design engineer should consider nSF = 1.15 for design, thus this design is not acceptable since M S  0.15.

We can see that the safety factor required found by BCM is smaller than the one found for the MNS. As stated earlier, the BCM theory is more conservative than the MNS theory. End Example 

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5

820

Introduction to Fracture Mechanics

“every structure contains small flaws whose size and distribution are dependent upon the material and its processing. These may vary from nonmetallic inclusions and micro voids to weld defects, grinding cracks, quench cracks, surface laps, etc.”1 The objective of a Fracture Mechanics analysis is to determine if these small flaws will grow into large enough cracks to cause the component to fail catastrophically. Fracture Mechanics: 1. is the study of crack propagation in bodies. 2. is the methodology used to aid in selecting materials and designing components to minimize the possibility of fracture. 3. begins with the assumption that all real materials contain cracks of some size—even if only submicroscopically. 4. is based on three types of displacement modes. As shown in Fig. 10.1.

MODE I OPENING

MODE II SLIDING

MODE III TEARING

Tension

In-plane shear

Out-of-plane shear

Figure 10.1: Three modes of fracture

MODE I: Opening. The opening (or tensile) mode is the most often encountered mode of crack propagation. The crack faces separate symmetrically with respect to the crack plane. MODE II: Sliding. The sliding (or in-plane shearing) mode occurs when the crack faces slide relative to each other symmetrically with respect to the normal to the crack plane but asymmetrically with respect to the crack plane. MODE III: Tearing. The tearing (or antiplane) mode occurs when the crack faces slide asymmetrically with respect to both the crack plane and its normal. 1 T. J. Dolan, Preclude Failure: A Philosophy for Material Selection and Simulated Service Testing, SESA J. Exp. Mech., Jan. 1970.

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.1

821

Fracture of Cracked Members

1. The presence of a crack in a structure may weaken it so that it fails by fracturing in two or more pieces. 2. Fracture can occur at stresses below the material’s yield strength, where failure would not normally be expected.

10.5.2

Cracks as stress raisers

Consider the elliptic hole in an infinite plate loaded by an applied uniaxial stress σ in tension.

σyy τxy σxx

10.5. INTRODUCTION TO FRACTURE MECHANICS

822

The maximum stress occurs at (±a, 0) and has a value of  a σ = 1+2 σyy b max σyy

= Kt σ

max

where Kt is the dimensionless stress concentration factor. The radius at the tip of the ellipse can be defined as: b2 ρ= a

σyy τxy σxx

Ktσ σyy σ σxx

ρ 1

2

(x+a)/ρ

Thus the stress concentration factor becomes: r σyy a a max Kt = =1+2 =1+2 σ b ρ and can take values such as

a/b

1

2

3

Kt

3

5

7

Note that when a = b the ellipse becomes a circle can gives a stress concentration factor of 3. When b → 0 or ρ → 0: Kt → ∞ and this geometry is like a crack-like slot. Real materials cannot support infinite stresses.

10.5. INTRODUCTION TO FRACTURE MECHANICS

823

In ductile metals, large plastic deformation exists in the vicinity of the crack-tip. The stress is not ∞ and the sharp crack tip is blunted:

10.5.3

Fracture toughness

1. In fracture mechanics, one does not attempt to evaluate an effective stress concentration, rather a stress intensity factor K 2. After obtaining K, it is compared with a limiting value of K that is necessary for crack propagation in that material, called Kc 3. The limiting value Kc is characteristic of the material and is called fracture toughness 4. Toughness is defined as the capacity of a material to resist crack growth

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.4

824

Fracture Mechanics: MODE I

Stress Intensity Factor

S σy τxy

h y

σx

x a

b

y

a

r θ

σz

x

b

S

1. Observed that as a → b, the plate fractures into two pieces. 2. The stress intensity factor KI characterizes the magnitude of the stresses in the vicinity of an ideal sharp crack tip in a linear-elastic and isotropic material under mode I displacement. 3. Near the crack tip the dominant terms in the stress field are:   KI θ θ 3θ σxx = √ cos 1 − sin sin + ··· 2 2 2 2πr

(10.17)

  KI θ θ 3θ √ cos 1 + sin sin + ··· 2 2 2 2πr

(10.18)

=

K θ θ 3θ √ I cos sin cos + ··· 2 2 2 2πr

(10.19)

σzz

=

( 0

(10.20)

τyz

=

τxz = 0

σyy

=

τxy

ν(σxx + σyy )

(plane stress) (plane strain)

(10.21)

10.5. INTRODUCTION TO FRACTURE MECHANICS

825

4. KI measures the severity of the crack and it is generally expressed as: KI

=

CI σ

√

(10.22)

πa

CI is a dimensionless quantity accounting for the plate/specimen geometry and relative crack size for mode one, σ is the stress (σxx , σyy , ...) if no crack were present, a is half crack length √ √ √ 5. The dimensional units of KI are: [stress length], i.e., [MPa m] or [ksi in] Dowling2 gives various expressions for plate with any α = a/b. These are provided in the handout. As for an example, the dimensionless geometry constant for a crack-centered plate is CI =

1 − 0.5 α + 0.326 α2 √ 1− α

h ≥ 1.5 b

α = a/b

(10.23)

From the above expression it can be shown that CI = 1 for an infinite plate (b → ∞) and for 0  α  1. However, for a center-cracked plate with α ≤ 0.4, when taking CI = 1, the result is accurate within 10%. Critical Stress Intensity Factor 1. The calculated KI is compared to the critical stress intensity factor or fracture toughness KIc : KI < KIc , material will resist crack growth without brittle fracture (safe) KI = KIc , crack begins to propagate and brittle fracture occurs (fracture) √ 2. The critical value KIc is defined as KIc = Cσc πa Note that KIc is a material property but KI is not! The strength-to-ratio KIc /KI can be used to determine the safety factor: KIc nSF = KI Crack Length The following figure shows the relationship between the critical value of the remote stress and the crack length. Here at is the transition crack length, and it is defined as the approximate length above which strength is limited by brittle fracture; and σo = Syield . In other words, at is the crack length where σc = Syield : 1 at = 2 C π 2 “Mechanical



KIc σo

2

1 = π



KIc σo

2

(C = 1)

Behavior of Materials” (Prentice-Hall, 1999, pp. 301-304; or 1993, pp. 292)

10.5. INTRODUCTION TO FRACTURE MECHANICS

826

σc

σc =

σo

at

K Ic C πa

a

When a > at strength is limited by fracture, and when 0 < a < at yielding dominates strength.

10.5. INTRODUCTION TO FRACTURE MECHANICS

827

σc

σc =

σo

at

K Ic C πa

a

The above figure shows that materials with: 1. high KIc and low σo implies long at (red line); therefore small cracks are not a problem. In fact, the higher the fracture toughness, the lower the yield strength; and the material has a ductile-like behavior. 2. low KIc and high σo implies short at (blue line); therefore small cracks can be a problem. In fact, the lower the fracture toughness, the higher the yield strength; and the material has a brittle-like behavior. Thus when designing one should identify whether the yielding failure is more critical than fracture failure, or fracture failure is more critical than yielding. If yielding failure is more critical, one must ensure safety with the previously discussed three-dimensional theories of yielding failure.

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.5

828

Fracture Mechanics: Tables and Plots

Table 10.2: Plane strain fracture toughness and corresponding tensile properties for representative metals at room temperature. “Mechanical Behavior of Materials”, by N.E. Dowling, Prentice-Hall Inc, NJ, 1999. Page 291.

10.5. INTRODUCTION TO FRACTURE MECHANICS

829

Applications of KI to Design and Analysis: Stress intensity factors for three cases of cracked plates under tension. An additional expression for (a) can be found in fracture mechanic books and for (c) the load is centered on the uncracked width. (For our case C = F ) C

Values for small α = a/b and limits for 10 % accuracy: (a) KI = Sg

√

πa

(b) KI = 1.12 Sg

α ≤ 0.4 For KI = CI Sg

√

√

πa

α ≤ 0.6

(c) KI = 1.12 Sg

√

πa

α ≤ 0.13

π a, expressions for CI for any α: 2

CI =

1 − 0.5 α + 0.326 α √ 1− α

CI =

   s   2 πα πα 4 1 + 0.122 cos tan 2 πα 2 4

CI = 0.265 (1 − α) +

0.857 + 0.265 α (1 − α)

1.5

h ≥ 1.5 b

α = a/b

h ≥ 2.0 b

α = a/b

h ≥ 1.0 b

α = a/b

10.5. INTRODUCTION TO FRACTURE MECHANICS

830

Applications of KI to Design and Analysis: Stress intensity factors for three cases of concentrated load. Case (c) is the ASTM standard compact specimen.

Fp =

Cσt

√

πab

P

Values for small α = a/b and limits for 10 % accuracy: (a) KI =

t

P √ πa

P (b) 2.60 √ t πa

( α ≤ 0.3)

( α ≤ 0.08)

Expressions for Fp for any α:   πα 1.297 − 0.297 cos 2 Fp = q sin (π α) 5

Fp =

Fp =

2

2

0.92 + 6.12 α + 1.68 (1 − α) + 1.32 α (1 − α) √ 1.5 π α (1 − α) 2+ α 1.5

(1 − α)



2

3

0.886 + 4.64 α − 13.32 α + 14.72 α − 5.6 α

4



h ≥ 2.0 b

α = a/b

h → large b

α = a/b

α = a/b ≥ 0.2

10.5. INTRODUCTION TO FRACTURE MECHANICS

831

Stress intensities for a round shaft with a circumferential crack, including limits on the constant C for 10% accuracy and expressions for any α. For torsion (c), the stress intensity is for the shear Mode III.

C

C √ KI = CI σxx π a

KIII = CIII τxz

C

√

πa

α=

a b

β =1−α

Values for small α = a/b and limits for 10 % accuracy: (a) axial load P :

σxx =

C (b) bending moment M :

σxx =

(c) torsion T : C

τxz =

C

P 2

πb

4M 3

πb

C

2T 3

πb

C =1.12

(α ≤ 0.12)

C =1.12

(α ≤ 0.12)

C =1.00

(α ≤ 0.09)

Expressions for C for any α: (a) axial load P :

CI =

(b) bending moment M :

CI =

(c) torsion T :

CIII =

1 2β

1.5

3 8β

2.5

3 8β

2.5

C

n

1.0 + 0.5 β + 0.375 β − 0.363 β + 0.731 β

n

1.0 + 0.5 β + 0.375 β + 0.3125 β + 0.273438 β + 0.537 β

n

1.0 + 0.5 β + 0.375 β + 0.3125 β + 0.273438 β + 0.208 β

2

3

4

o

2

3

4

5

2

3

4

5

o o

or in term of α: (a) axial load P :

CI =

(b) bending moment M :

CI =

(c) torsion T :

CIII =

1.5

n o 2 3 4 1.1215 − 1.5425 α + 1.836 α − 1.2805 α + 0.3655 α

2.5

n o 2 3 4 5 1.12423 − 2.23734 α + 3.12117 α − 2.54109 α + 1.10941 α − 0.201375 α

2.5

n o 2 3 4 5 1.00085 − 1.62047 α + 1.88742 α − 1.30734 α + 0.492539 α − 0.078 α

1 1−α 1 1−α 1 1−α

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.6

832

Fracture Mechanics: Mixed Modes

Fracture under combined loading

In many cases, the structure is not only subjected to tensile stress σxx but also a contour shear stress τxy . Thus, the crack is exposed to tension and shear which leads to mixed mode cracking; i.e., a mixture of mode I and mode II. Whenever the crack length 2a is small with respect to the web length b, the geometric factor C is √ √ unity in formulas for the stress intensity factors from LEFM. That is, KI = σxx πa and KII = τxy πa, where σxx and τxy are the normal and shear stresses in the structure if there were no crack present. Mixed mode fracture is complicated by the fact that the crack extension takes place at an angle with respect to the original crack direction. If a crack propagates in the direction of the original crack, it is called self-similar crack growth. Under mixed mode fracture the crack growth is, in general, not self-similar. Various mixed mode criteria for crack growth have been proposed based on experiments. A common mixed mode criterion, at the initiation of the fracture, is 

KI KIc

2

+



KII KIIc

2

=1

(10.24)

where KIc is the fracture toughness for mode I loading only, and KIIc is the fracture toughness for mode II loading only. The plane strain fracture toughness for mode I loading is usually readily available in the literature, but the mode II fracture toughness is not usually available. Tests for mode II are more difficult to design than for mode I. To estimate KIIc knowing the value of KIc we use the maximum principal stress criterion3 . The maximum principal stress criterion postulates that crack growth will occur in the direction perpendicular to the maximum principal stress in the vicinity of the crack tip. Using this criterion it is possible to estimate KIIc as √ 3 KIc = 0.866 KIc (10.25) KIIc = 2 The mixed mode criterion given by Eq. (10.24) is plotted in the following figure: Under proportional loading, the stresses, and in turn the stress intensity factors, are proportional to the magnitude of the total lift acting on the wing. The stress intensity factors at the 80% limit load specified for the damage design condition determine the coordinates of the required strength in the plot, which is represented by the ray 0 − f . The quantity f denotes the dimensionless required strength. The excess strength with respect to fracture is represented by 1 − f , and if it is divided by the required strength we get the margin of safety 3 P´ erez,

N´ estor, Fracture Mechanics, Kluwer Academic Publishers, Boston, USA, 2004 (ISBN 1–4020–7745–9)

usually available. Tests for mode II are more difficult to design than for mode I. To of K Ic we use the maximum principal stress criterion1. The maximum principal

ck growth will occur in the direction perpendicular to the maximum principal stress ing this criterion it10.5. is possible to estimate K IIc (see Fig. 14.16 in Broek) as INTRODUCTION TO FRACTURE MECHANICS K IIc = 0.866K Ic

(3)

by eq. (2) is plotted in the figure at right. tresses, and in turn the stress intensity agnitude of the total lift acting on the at the 80% limit load specified for the ne the coordinates of the required esented by the ray 0 – f .The quantity f ed strength. The excess strength with re1 – f , and if it is divided by the reof safety as 1– f = -----------f

as

where

K II --------K IIc

2

KI  K II   ------- +  -------- K Ic  K IIc

1

2

= 1

f

0

K -------IK Ic

1

2

f =

833

KI  K II   ------- +  -------- K Ic  K IIc MS

2

=

σallowable 1−f = − 1 = nSF − 1 f σrequired

(4) (10.26)

agate if 0 ≤ f < 1 , and the initiation of fracture is predicted if f = 1 . The margin s 2  2 and the margin of safety is zero if f = 1 . K K f

=

I

KIc

+

II

KIIc

(10.27)

The crack is predicted not to propagate if 0 ≤ f < 1, and the initiation of fracture is predicted if f = 1. The margin of safety is positive if 0 < f < 1, and the margin of safety is zero if f = 1.

g fracture mechanics, Martinus Nijhoff Publishers, Dordecht, The Netherlands, 1987, pp. 374-379.

2

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.7

834

Plastic zone size in cracked metal plates S σy τxy

h y

σx

x a

y

a

σz

r θ

x

b

b

S The in-place stress are =

  KI θ θ 3θ √ cos 1 − sin sin + Higher Order Terms 2 2 2 2πr

(10.28)

σyy

=

  KI θ θ 3θ √ cos 1 + sin sin + H.O.T. 2 2 2 2πr

(10.29)

τxy

=

θ 3θ K θ √ I cos sin cos + H.O.T. 2 2 2 2πr

(10.30)

σxx

K √ I 2πr

= σ

r

a 2r



r √ Cσ πa a √ =Cσ 2 r 2πr

@ θ = 0 → x = r, y = 0

and

KI σyy = √ 2πr

 C=1

(10.31)

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.8

835

Plastic zone

If we neglecting higher order terms in the above expansion, we can show that the stress distribution near the crack tip is as follows

σyy

σyy x, r Plastic zone

2a

σ yy =

σo

rp

rp*

KI 2π r

x, r

t

where rp∗ is the estimate of plastic zone and is defined as rp∗ =

1 2π



KIc Syield

2

(10.32)

Experiments and analysis show the plastic zone size rp > rp∗ . Furthermore, we define plastic zone size as rp = c



KI Syield

2

c = constant of proportionality

(10.33)

The in-plane stresses near crack tip are very large, and εyy is large. The high stress region near the crack indicates the plastic core at the crack tip wants to contract in the thickness direction due to very large in-plane stresses (σxx , σyy , τxy ). The bulk elastic material surrounding plastic core does not contract in the thickness direction, or contracts a lesser amount, than the plastic core. The bulk elastic material constrains the contraction of the plastic core. Plastic zone is responsible to delay crack propagation, and as soon as the crack hits the elastic region the plane will fracture.

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.9

836

Plane stress

For this particular case σzz = 0 and the yielding occurs about Syield = σo rp∗ = roσ rp = 2 roσ =

1 π

coσ =

1 π

where the constant of proportionality is

10.5.10



KI σo

2

Plane strain

For this particular case σzz = ν(σxx + σyy ) and the yielding occurs about Syield =

√

3 σo :

rp∗ = roε rp = 2 roε where the constant of proportionality is

1 = π



K √ I 3 σo

coε =

Note that roε =

10.5.11

2

1 = 3π



KI σo

2

1 3π

roσ 3

Plasticity limitations on LEFM

Plastic zone size is characterized by KI only if first term in σyy dominates (recall H.O.T. were neglected in the in-plane stress). If plasticity spreads further then KI cannot be used to characterize the plastic zone size and the use of Linear Elastic Fracture Mechanics (LEFM) is invalid. In the following situations KI cannot be used to characterize the stress field because the plastic zone is too large: 1. relative to crack

2. relative to uncracked ligament

3. relative to specimen height

10.5. INTRODUCTION TO FRACTURE MECHANICS

837

The requirement for plane strain on the use of LEFM is: t, a, (b − a), h ≥ 2.5



KI Syield

2

Since 2 roσ > 2 roε , an overall limit for plane stress on the use of LEFM is: 4 a, (b − a), h ≥ π



KI σo

2

(10.34)

This must be satisfied for all three of a, (b − a), h. Note that KI → KIc only if it can be considered as plane strain.

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.12

838

Fracture toughness in plane strain and plane stress

Table 2.1 from the course textbook (as well of most books) provides plane strain fracture toughness KIc . However, for plane strain fracture toughness KIc to be a valid failure prediction criterion, plane strain conditions must exist at the crack tip. In other words, the material must be thick enough to ensure plane strain conditions. To better understand this concept, consider a plate. The plate thickness is the plastic core whose diameter is equal to rp :  2 KI rp = c σo

Thick Plate

Thin Plate

Plane Strain

Plane Stress

σzz = ν (σxx + σyy ) εzz = 0

σzz = 0 ν εzz = − (εxx + εyy ) 1−ν

t1 → Large rp The bulk material constrains

t2 → Small rp The bulk material provides

very thick core to large extent

little constraint to core



t KI σo

2 > Q



t KI σo

where Q ' 2.5, σo = Syield

2 < Q

Empirically it has been estimated that the minimum required material thickness for plane strain condition is given by (transition thickness between plane strain and plane stress) ts = 2.5



KIc Syield

2

For plane strain conditions the minimum material thickness t must be t ≥ ts If the material is not thick enough to meet the above criterion, plane stress better characterizes the state of stress at the crack tip, and Kc , the critical stress-intensity factor for failure prediction under plane stress conditions, may be estimated using a semiempirical relationship for Kc as a function of plane

10.5. INTRODUCTION TO FRACTURE MECHANICS

839

Plane stress dominates

Kc

Plane strain dominates

Kc= KIc

for plane strain

to

ts

strain fracture toughness KIc and thickness t. This relationship is s s  4  2 KIc 1.4 ts = KIc 1 + 0.224 Kc = KIc 1 + 2 S t t yield Note that when t = ts : Kc = KIc

√

1 + 0.224 = 1.106 KIc

t

(10.35)

(10.36)

which means that the minimum value of the thickness to assume plane strain problem is within 10 % accuracy for the plane strain fracture toughness, which seems to be commonly acceptable. As long as the crack-tip plastic zone is in the regime of small-scale yielding, this estimation procedure provides a good design approach. If the plastic zone size ahead of the crack tip becomes so large that the small-scale yielding condition is no longer satisfied, an appropriate elastic-plastic fracture mechanics procedure would give better results. The plane strain fracture toughness KIc use in design, of even thin plates, is conservative. In other words, KI is a minimum value for the material and the actual KIc may be higher, as a result an over-design of the structure is obtained. Thus for design purposes, the plane strain fracture toughness KIc is most commonly used. Different engineers and researchers decide on whether to use plane strain or plane stress fracture toughness. Here for design purposes we will always use plane strain assumption.

10.5. INTRODUCTION TO FRACTURE MECHANICS

840

Example 10.7. The panel of a structure is subject to tensile force P = 50 kN and is made of 2024-T351 P

h

2a h

t

2b P

Aluminum alloy and a crack is being propagated in the center. The length of the panel is 400 mm (h = 200 mm), the width 100 mm (b = 50 mm), and the thickness is 5 mm (t = 5 mm). The panel can be modeled as a plate. (1) Determine if plane strain is a good approximation for this problem. (2) Determine if the panel will break with a crack length of 20 mm. (3) Determine if the panel will break with a crack length of 60 mm. (4) Find the transition half crack length at . (5) Determine the load Pc for brittle fracture initiation if the critical crack length 2 ac = 50 mm. (6) The failure of the panel is governed by yielding or fracture? Solution: (1) Determine if plane strain is a good approximation for this problem. Let us determine if the plane strain fracture toughness is valid for our problem. From Table 2.1 √ KIc = 35 MPa m Syield = 375 MPa

10.5. INTRODUCTION TO FRACTURE MECHANICS

Then ts = 2.5



KIc Syield

2

= 2.5

The thickness of the panel is t = 5 mm and:



35 375

2

m = 21.7 mm

(t = 5) < (ts = 21.7)

⇒

t  ts

841

Since plane strain conditions are not satisfied region, the fracture toughness for our problem is: s  2 ts Kc = KIc 1 + 0.224 t = (35)

s

1 + 0.224



21.7 5

2

√ = 80 MPa m

(ts and t should have same units) (2) Determine if the panel will break with a crack length of 20 mm. The safety factor tells us if the panel will fail: nSF =

KIc KI

or

nSF =

σall σreq

Stress with no crack present: σyy =

P P 50000 = = = 100 MPa A 2bt 2 (0.050) (0.005)

Recall that a is half the crack length, thus a = 10 mm. Failure by both yielding and brittle fracture must be checked. For yielding: σyy

= true

P 50000 = = 125 MPa < Syield = 375 MPa 2 (b − a) t 2 (0.050 − 0.010) (0.005)

Thus failure by yielding is not predicted. Let us now check for brittle fracture. Recall that a is half the crack length, thus a = 10 mm. Then ratio α = a/b is (a = 10 mm): α=

a 10 = = 0.20 b 50

Now we use Chart 2.5 from the textbook: CI

√

1 − α = 0.93

→

CI = 1.04

(Since α < 0.4, we could have taken CI = 1.) Thus the apparent fracture for our problem is √ √ KI = CI σyy π a = 18.43 MPa m Recall that for design purposes we tend to be conservative, thus we use plane strain

10.5. INTRODUCTION TO FRACTURE MECHANICS

fracture toughness:

842

√ KIc = 35 MPa m

Thus the safety factor is nSF =

KIc 35 = = 1.90 KI 18.43

The hood has a 90 % of margin of safety. Since M S > 0, the panel will not break. If we were to use the plane stress fracture toughness √ Kc = 80 MPa m Thus the safety factor is nSF =

Kc 80 = = 4.34 KI 18.43

The hood has a 334 % of margin of safety. Since M S > 0, the panel will not break. It should be clear that the design is over-designed for this particular crack length. However, in practice it is common to stick with plane strain assumption, although each design engineer can make his/her own judgment. (3) Determine if the panel will break with a crack length of 60 mm. The safety factor tells us if the panel will fail: nSF =

KIc KI

or

nSF =

σall σreq

Stress with no crack present: σyy =

P 50000 P = = = 100 MPa A 2bt 2 (0.050) (0.005)

Recall that a is half the crack length, thus a = 30 mm. Failure by both yielding and brittle fracture must be checked. For yielding: σyy

= true

P 50000 = = 250 MPa < Syield = 375 MPa 2 (b − a) t 2 (0.050 − 0.030) (0.005)

Thus failure by yielding is not predicted. Let us now check for brittle fracture. Then ratio α = a/b is (a = 30 mm): α=

a 30 = = 0.60 b 50

Now we use Chart 2.5 from the textbook: CI

√

1 − α = 0.83

Thus KI = CI σyy

√

→

CI = 1.31

√ π a = 39.8 MPa m

Recall that for design purposes we tend to be conservative, thus we use plane strain

10.5. INTRODUCTION TO FRACTURE MECHANICS

fracture toughness:

843

√ KIc = 35 MPa m

Thus the safety factor is

KIc 35 = = 0.88 KI 39.8

nSF =

The hood has a negative 12 % of margin of safety, thus it has failed. If we were to use the plane stress fracture toughness √ Kc = 80 MPa m Thus the safety factor is nSF =

80 Kc = = 2.01 KI 39.8

The hood has a 101 % of margin of safety. Since M S > 0, the panel will not break. Interestingly, the design apparently has failed under the plain strain assumption. However, the real problem is a plane stress problem which indicated the contrary. Thus the panel in reality has not fractured. (4) Find the transition half crack length at . For transition length σc = Syield = 375 MPa Recall that for design purposes we tend to be conservative, thus we use plane strain fracture toughness: √ KIc = 35 MPa m The first approach is to use the transition crack length approximation: at =

1 π



KIc σo

2

=

1 π



35 375

2

= 0.00277 m = 2.77 mm

The second approach consists in not assuming C = 1. Thus, KIc = CI σc Let us rearrange:

Divide both sides by

√

√

π at

√ KIc √ = CI at σc π b: √ √ KIc at √ = CI √ = CI αt σc π b b

Note at = αt b. Thus the problem to solve is: √ KIc √ = 0.2355 = CI αt σc π b

10.5. INTRODUCTION TO FRACTURE MECHANICS

844

Since CI depends on αt from Eq. (10.23) CI =

1 − 0.5 α + 0.326 α2 √ 1− α

h ≥ 1.5 b

α = a/b

Thus, solve numerically: 1 − 0.5 αt + 0.326 αt2 √ √ αt 1 − αt

0.2355 =

and choose the positive real solution: αt = 0.0552984 Thus the transition half crack length is: at = 0.00276492 m = 2.76 mm. As we can see the results using both methods are very similar. (5) Determine the load Pc for brittle fracture initiation if the critical crack length 2 ac = 30 mm. The safety factor tells us if the panel will fail: nSF =

KIc KI

For fracture initiation nSF = 1 (at onset of failure or fracture initiation). Thus KI =

KIc = KIc nSF

and KI = CI σ

√

KIc = CI σc

√

Thus

πa π ac

Stress with no crack present: σyy =

Pc Pc Pc = = = 0.002 Pc MPa A 2bt 2 (0.050) (0.005)

Recall that for design purposes we tend to be conservative, thus we use plane strain fracture toughness: √ KIc = 35 MPa m Here σc = σyy = 0.002 Pc MPa. Recall that a is half the crack length, thus a = 15 mm. Let us rearrange: KIc √ Pc = CI 0.002 π ac Note αc = ac /b: αc = ac /b = 0.3

→

CI =

1 − 0.5 αc + 0.326 αc2 √ = 1.051 1 − αc

10.5. INTRODUCTION TO FRACTURE MECHANICS

845

Thus the critical load is Pc = 76.702 kN. Which means that a load of Pc = 76.702 kN will initiate brittle fracture. We should check for yielding failure: σyy

= true

P 76702 = = 219.15 MPa < Syield = 375 MPa 2 (b − a) t 2 (0.050 − 0.030) (0.005)

Thus failure by yielding is not predicted. (6) The failure of the panel is governed by yielding or fracture? Since ac > at it will fail due to fracture and not yielding. Indeed for the first two cases a > at , suggesting that failure will occur due to fracture. End Example 

10.5. INTRODUCTION TO FRACTURE MECHANICS

10.5.13

846

Superposition of Combined Loading

Stress intensity solutions for combined loading can be obtained by superposition, that is, by adding the contribution to K from the individual load components: K = K1 + K2 = C1 σ1

√

π a + C 2 σ2

√

π a = (C1 σ1 + C2 σ2 )

√

πa

Example 10.8. To better understand this concept consider the following problem: A 20 mm diameter shaft has a circumferential surface crack, as shown in Figure, of depth a = 1.5 mm. The shaft is made of Ti-6Al-4V Titanium. The shaft is loaded with an eccentric axial force of P , which produces a bending moment of P e, combined with a torque of T . Can we complete the operation without replacing the shaft? Note that KIIIc is unknown, and a reasonable and probably conservative assumption is to employ a relationship of the form: s 2  2 KI KIII (10.37) + =f KIc KIIIc where KIc is the fracture toughness for mode I loading only, and KIIIc is the fracture toughness for mode III loading only. The crack is predicted not to propagate if 0 ≤ f < 1, and the initiation of fracture is predicted if f = 1. Assume that KIIIc = 0.5 KIc . Take: P = 150 N

e = 5 mm

T = 300 N–m

Stress intensity solutions for combined loading can be obtained by superposition: KI = KI + KI axial

= CI σxx

√

bending

π a

+ CI σxx

axial

√

π a

bending

n o√ = CI σxx axial + CI σxx bending πa

Thus nominal stress for the combined loading is obtained by superposition of two states of stress for axial force P and moment Mzz , is expressed as CI σxx = CI axial σxx axial + CI bending σxx bending

(10.38)

We can continue the operation if the margin of safety bigger then zero. Thus our goal is to find the margin of safety; first for yielding then for fracture and ensure it is a positive

10.5. INTRODUCTION TO FRACTURE MECHANICS

847

quantity. From Table 2.1 √ KIc = 106 MPa m

Syield = 820 MPa

and from the problem statement: √ KIIIc = 0.5 KIc = 53 MPa m The margin of safety is defined as M S = nSF − 1 and the safety factor is defined as nSF =

1 f

or

nSF =

Sy σreq

Note that for the case of yielding we need to use a 3-dimensional theory of failure for yielding. Thus let us use the distortional energy. The loads at the cross-section are: Nxx = P = 150 N

Mzz = −P e = −0.75 N–m

Mxx = T = 300 N–m

and the radius is b = 0.010 m. The true stresses at the critical point in the cross-sectional element: Nxx P 4M Mzz + + = σxx = σxx bending + σxx axial = − 3 Z Ax π (b − a) π (b − a)2 = 1.55494 × 106 + 660851 = 2.21579 × 106 Pa

T 2T = 3.10989 × 108 Pa τxz = τ torsion = = Q π (b − a)3

The state of stress is  σxx τxy  σM = τxy σyy τxz τyz

  τxz 2.21579 × 106   τyz = 0 σzz 3.10989 × 108

The stress invariants are

0 0 0

 3.10989 × 108  Pa 0 0

Iσ1

= σxx + σyy + σzz = 2.21579 × 106 Pa

Iσ2

2 = −τxz = − − 9.6714 × 1016 Pa2

Iσ3

=

0

σeq =

q

The von Mises stress is Iσ21 − 3 Iσ2 = 5.38653 × 108 Pa

(10.39)

10.5. INTRODUCTION TO FRACTURE MECHANICS

848

The yielding criteria for DE criterion is σeq 1 = Sy nSF

→

nSF = 1.52232

→

M S = 0.52232

Since M S > 0, it will not fail due to yielding.

C

The ratio of the length to bracket width is C m a = 0.0015

α=

Using expressions for C:

C

a = 0.15 b C

C (a) axial load P :

(b) bending moment M : C (c) torsion T :

β = 0.85

CIa =1.18329 C CIb =1.27729

CIII =1.19511

Thus nominal stresses at the critical point in theC cross-sectional element are Mzz σxx bending = − = 954930 Pa Z

Nxx P σxx axial = = = 477465 Pa Ax π b2

T 2T τxz = τ torsion = = = 1.90986 × 108 Pa Q π b3

Thus nominal stress for the combined loading is obtained by superposition of two states of stress for axial force P and moment Mzz , is expressed as CI σxx = CIa σxx axial + CIb σxx bending = 1.78471 × 106 Pa CIII τxz = 2.2825 × 108 Pa

10.5. INTRODUCTION TO FRACTURE MECHANICS

KI = CI σxx

√

√ π a = 0.122514 MPa m

Thus f= and nSF =

s

KI KIc

2

1 = 3.38253 f

+



849

KIII = CIII τxz KIII KIIIc

2

→

√

√ π a = 15.6686 MPa m

= 0.295637

M S = 2.38253

Since M S > 0, it is save to continue the operation. End Example 

HOMEWORK 4

AOE 2001 3124 3124 – Spring 10.5. AOE INTRODUCTION TO FRACTURE MECHANICS DueAerospace Friday,Structures MarchAOE 2,3124 2001 by 12:30pm Aerospace Structures

Spring 8502001

Page 1

Aerospace Structures

AOE 3124 – SPRING 2001 – AEROSPA HOMEWORK 2B – DUE MONDAY, FEB

2 Homework 2 have been hired to select theHomework AOE3124 students best preliminary design of a prismatic, cant beam as an idealization of one wing on an given by

The prismatic, cantilever beam shown is an idealization subjected to an elliptic spanwise airload distribution given by airplane subjected to an elliptic spanwise s airload distrib  2 The prismatic, cantilever beam shown is an idealization 2xan airplane. 2Lof one wing on s 0 1− p(x) = spanwise airload distribution given by b  2 b π( 2 ) AOE 3124

Example 10.9. beam shown is an idealization 2L The prismatic, cantilever of one wing on 2x an airplane. It is subjected an elliptic b to 2L ------------------ 1 –per ( 2x ⁄ unit b ) 2 span, 0 ≤ xL≤ bis⁄ 2the p(x) = 1− ≤ p = intensity where p(x)0is≤ thex airload spanwise airload distribution given by π(b ⁄ 2) b b 2 π( 2 )

Homework wing span. The load factor at the bottom of a pull-up from maneuver L = 45 kN . Take b = 2 m. The cross section is the

where p is the airload intensity per unit span, L is the total lift (both wings), and b

2L the bottom a pull-uplift from (both a dive is specified as 2.5 and for this where p(x) is the airload L total wings), and bmaneuver is theL ------------------ 1 –per p = intensity ( 2x unit ⁄ b ) 2 span, 0 ≤tor x at≤is b ⁄ the 2 of (1) Consider the single cel π(b ⁄ 2) Consider design minimum weight of theofaluminum of the following beam. This span.the The loadforfactor at the bottom a pull-upalloy from a dive is specified as 2.5 and for this man Af4 p p( x) where the airload intensity per2unit L is the total of liftone (both wings), andairplane. b is the wing span. The load facprismatic, cantilever beam shown isspan, an idealization wing on yan Lp=is45 kN . (Take b= m.) y Af4

t4

r

tor at the bottom of a pull-up from a dive is specified as 2.5 and for this maneuver L = 45kN . Take b = 2m . The

S.C. z S.C.

Mdz

x

Af4

p

p( x)

30

b⁄2

y

y

S.C. z

C T

Vy

A fA1f 1

Af3

Negat Show all work.

cross section is theCsame as in homework one. In the cross section, the line of action r = 1001.mm mmdfrom the vertical line through one and four as shown. The distancethefrom Determine loca A fstringers 2 c the =beam 180is mm load to the shear center is denoted by d . The material of 2024-T4 alumin

x

2.

Determine the sec

= = A4.

Determine the she

of 325 MPa, an ultimate strength of 469 MPa, and mass of mm 2800Kg/m 3 . Tak h density = 50 Af1 3. factor Determine the max minimum yield divided by the of safety 30 of the c strength or the ultimate strength Af 1 = = Amine =maximum 325 mm com f 4 the fied as 1.5.

b⁄2

A

= 130 mm

2 the same asfin 3 homework o The dimensions of the cross section and the notationsfare Determineare the to she t 1 and t 2 are design variables. The margins of safety with respect to5. yielding

In the cross section, the line of action of the distributed a

webs at the wing root. The six points on the contour are, (1) s = hlow.(Partial ⁄ 2 , (2) sans. =q1 (h0 ),

2 one and four as shown. distance2 from cross section thecross same assection, in homework theaction cross section, the distributed line through of actionstringers ofairload the distributed airload is 30 The In is the theone. lineIn of of the is 30 mm from the vertica (6) scenter Use Mises criterion for the initiation ofthe yielding. s 4 =the 0 , and = πr ⁄is 2 .denoted shear by d. The material of beam is 4 It isfrom subjected to an airload distribution by mm the vertical lineelliptic throughspanwise stringers one and four as shown. given The distance from the line of action of the airS.C. t Oand strength of 325 MPa, an ultimate strength of 469 MPa, m through stringers one and four as shown. The distance from the line of action of the air-load load to the shear center is denoted by d . The s material of the beam is 2024-T4 aluminum with a yield strength 1. Determine thestress shear alloy force and bending moment at root. (Ans.strength V y = 22.5kN V allowable to be the minimum ofthe the yield or   2 shear center is denoted by 2d.L The dimensions of the cross section and theofnotations are the same b3 . of factor safety. The factor stress safety is specified as 1.5. of 325 MPa, an ultimate strength of 469MPa, and mass2 x density of 2800Kg/m Take the allowable to be the  p (x) = 1 − 0 ≤ x ≤ e 2. Take thicknesses and . = 0.75mm t = 1.0mm t y homework 2. The web 3124 thickness t1 andb t2 are the design variables. 1 2 The dimensions of the cross section and the notations areS Structures b strength minimum of the yield strengthAOE or the πultimate divided by the factorAerospace of2 safety. The factor of safety is specithat web thicknesses t and t are design variables. The mar 1 beam in2 Newtons. (Ans. 43.033 N) a. Compute the weight of the 6. Determine the loca fied as 1.5. to be computed at six pointalloy in the webs wing roo Material. The selected2material for the beam isare 2024-T351 aluminum withat athemass de y

b. Determine the margin of safety at the six points in the webs at the wing root

1.371, (1)1.701, s2 =1.639, h/2, (2) s21.470}) = h, (3)web s3 =thicknesses 0, (4) s3 = b3 , 7.(5)Determine s4 = the 0,shea 3 . (The The dimensions of the cross sectionyield and the notations are the same as in homework one, except of p(x) 2800is Kg/m strength, and mode Ithat fracture where the airload intensity perstrength, unit span,ultimate L is the total liftfor(both wings), b is the toughness can be the initiation ofand yielding. 1 variables. The margins of safety with respect to yielding 3. Determine thebe design variables such that the weight minimum and all six mar design are to computed at six point in isthe t 1 and t 2 are8.1 wingTable span. The.)load factor at the bottom of a pull-up from a non-negative. dive 1. is Determine specified asvariables 2.5 (L/W) For the design give the weight inmoment Newtons, at the the six m the shearselected, force and bending webs wing root. The six theThe contour are, cell (1) swing , (2) , (3) ⁄ 2 stresses =Mlocations h= 0margins. the ofs 3the= sixgeometric and at forthethis maneuver L points = 45 on kN. single hassat2 the 2 = h box 9.5493 kN −, (4) m) s 3 = b 3 , (5) z following

Homework 1

Strength constraints. As part of their design requirement the allowable stress must b

sproperties 4 = 0 , and (6) s 4 = πr ⁄ 2 . Use Mises criterion for the initiation of yielding. 2. divided Taking thicknesses t1factor = 0.75 mm t2 = 1.0 mm:fact minimum of the yield strength orcell thesection ultimate strength by the of and safety. Consider the single discussed in class with dimensions and section loads shown in The the sketch

(a)account Compute the of the beam in Newtons. safety has been specified as 1.5. The design should take into theweight margins of safety with[5rep (b) Determine the margin of safety at the six points f 4 to yielding at six points in the webs at the wing root: (1) s2 = h/2, (2) s2 = h, (3) s3 = 0, (4) s3 (Ans. margins = 2.148, 2.089, 1.701, 1.639, 1.371, Dimensions Loads t (5) s = 0, and (6) s 4= πr/2. They wantt 3to use Mises criterion for the initiation of yielding. 4 4 2. Take thicknesses t 1 = 0.75mm and t 2 = 1.0mm 1 of = 1 10 kN M y . r = 100 mm

1.

Determine the shear force and bending moment at the root. (Ans. V y = 22.5kN , M z = 9.5493kN-m .) A Af3

r

z

Damage tolerance constraint. Thez 43.033 designN)should also account V y = 10cond c = 180 mm for the design damage kN S.C. C (Ans. a. Compute the weight of the beam in Newtons. t2 h which is a crack centered at s = b /2 in web 1 that is 32 mm long with the crack faces perpendi 1 six points 1 = 50 mmmargins = {2.148, 2.089, T = 2.45kN m M z at the b. Determine the margin of safety in the webs at the wing hroot. (Ans. T A f 2 propagation t 1 = t 3 at = 0.75 to 1.701, the x-axis. It is 1.470}) specified that there be no crack 80%mm of the limit load. (The fra 1.639, 1.371, Vy

t 2 = t 4 = in 1.0the mm section titled Fracture u criterion and the margin of safety for thist 1constraint are discussed 3. Determine the design variables such that the weight is minimum and all six margins of safety at the2 root are A combined loading of the handout given in class.) f1 A f 1 = A f 4 = 325 mm

non-negative. For the design variables selected, give thec weight in Newtons, the six margins, and the Mises effective stresses at the locations of the six margins. A f 2margins = A f 3 = of 130safety mm 2 using t1 = 0.75mm Calibration. The students have calculated the seven Negative x-face

t2 = 1.0mm. These margins of safety at the six points on the contour and the margin of safet the damage condition are 2.154, 2.096, 1.707, 1.644, 1.376, 1.476, and 1.121, respectively. For Show all work. thicknesses, these margins of safety were found to be correct. r = 100 mm, 1.

c = 180 mm,

h = 50 mm

b=2m

Determine the location of the centroid C from the vertical line through stringers 1 and 4. (Ans. 39.65

Assignment. The AOE3124 students have the task to fully analyze six proposed designs. ever, they need some help in selecting the best design. As an expert in aircraft design 6 1 analysis, 4 1

2. Determine the second area moment I z about the z-axis through the centroid. (Ans. 9.5234 ×10 mm N.E. Dowling, Mechanical Behavior of Materials, Prentice-Hall Inc, NJ, 1999 3. Determine the maximum tensile bending normal stress and its location in the cross section. Likewise mine the maximum compressive bending normal stress and its location in the cross section.

10.5. INTRODUCTION TO FRACTURE MECHANICS

Af 1 = Af 4 = 325 mm2 ,

Af 2 = Af 3 = 130 mm2 ,

851

t1 = t3 ,

t2 = t4

In the cross section, the line of action of the distributed airload is 30 mm from the vertical line through stringers one and four as shown. The distance from the line of action of the air-load to the shear center is denoted by d. Design problem. The web thickness t1 and t2 are the design variables. Determine the design variables such that the weight is minimum and all margins of safety at the root (x = 0) are non-negative. Choose from the following proposed designs: I: t1 = 0.300 mm, t2 = 0.350 mm II: t1 = 0.350 mm, t2 = 0.370 mm III: t1 = 0.350 mm, t2 = 0.300 mm IV: t1 = 0.348 mm, t2 = 0.366 mm V: t1 = 0.345 mm, t2 = 0.365 mm Material. The beam is made of 2024-T351 aluminum alloy with a mass density of 2800 kg/m3 . (The yield strength, ultimate strength, and mode I fracture toughness can be obtained from Tables.) Strength constraints. As part of the design requirement the allowable stress must be the minimum of the yield strength or the ultimate strength divided by the factor of safety. The factor of safety has been specified as 1.5. The design should take into account the margins of safety with respect to yielding at six points in the webs at the wing root: (1) s2 = h/2, (2) s2 = h, (3) s3 = 0, (4) s3 = b3 , (5) s4 = 0, and (6) s4 = πr/2. Use Mises criterion for the initiation of yielding. Damage tolerance constraint. The design should also account for the design damage condition, which is a crack centered at s1 = b1 /2 in web 1 that is 32 mm long with the crack faces perpendicular to the x-axis. It is specified that there be no crack propagation at 80% of the limit load.

SOLUTION: The distance b1 and b3 are: s 2  h 2 b1 = b3 = c + r − = 195 mm 2 and b2 = h = 50 mm and b4 = π. The total wing span is b = 2000 mm. The total lift will be L = 45000 N. From Table 10.2, the yield strength is Sy = 325 MPa, the ultimate strength is √ Sut = 470 MPa, and mode I fracture toughness KIc = 34 MPa m. The loads at the root are (using techniques learned in chapter 3):

 

10.5. INTRODUCTION TO FRACTURE MECHANICS

852

 

Vy  20000

15000

10000

5000

200

400

600

1000

800

x 

Mzz  2. μ 107

1.5 μ 107

1. μ 107

5. μ 106

200

400

Vy0 = 22.5 × 103 N

600

800

1000

x 

Mzz0 = 22.5 × 106 N–mm

The parametric equations are (using techniques learned in chapters 4 and 8): y1 (s1 ) = −100 +

5 s1 13

y2 (s2 ) = −25 + s2 5 y3 (s3 ) = 25 + s3 13 y4 (θ4 ) = 100 cos (θ4 )

12 s1 13 z2 (s2 ) = 180 z1 (s1 ) =

12 s3 13 z4 (θ4 ) = −100 sin (θ4 ) z3 (s3 ) = 180 −

 

The normal coordinate (all is mm) from point O is (using techniques learned in chapter 8): ∂z1 ∂y1 1200 y1 (s1 ) + z1 (s1 ) = ∂s1 ∂s1 13 ∂z2 ∂y2 r2 (s2 ) = − y2 (s2 ) + z2 (s2 ) = 180 ∂s2 ∂s2 ∂z3 ∂y3 1200 r3 (s3 ) = − y3 (s3 ) + z3 (s3 ) = ∂s3 ∂s3 13 1 ∂z4 1 ∂y4 r4 (θ4 ) = − y4 (θ4 ) + z4 (θ4 ) = 100 R ∂θ4 R ∂θ4 r1 (s1 ) = −

Consider the single cell section discussed in class with dimensions and section loads shown in the sketch below. Af4 t4

Dimensions

t3

Loads M z = 10 kN m

y 10.5. INTRODUCTION TO FRACTURE MECHANICS r = 100 mm Af3 r

V y = 10 kN

c = 180 mm C t2 h h = 50 mm Mz A f 2 techniques TheTenclosed in mm chapter 8): t 1 =learned t 3 = 0.75 V area is (using z

S.C.

T = 2.45kN m

y

t1 Z t 2 = t 4 = 1.0 mm 4 1 X bi ri (si ) dsi A f 1 = A f 4 = 325 mm 2 Ac = c 2 i=1 0 Z b1 Z b2 A f 2 = A f 3 = 130 Z b3mm 2 Z Negative x-face = r1 (s1 ) ds1 + r2 (s2 ) ds2 + r3 (s3 ) ds3 + Af1

0

0

0

b4

r4 (θ4 ) R dθ4

0

= 38208 mm2

Show all work. 1.

Determine the location centroidas: C from the vertical line through stringers 1 and 4. (Ans. 39.65 mm) The total weightofisthe defined

2.

Determine the second area moment I z about the z-axis through the centroid. (Ans. 9.5234 ×10 mm 4 )b

6

W (t1 , t2 ) = ρ g (b1 t1 + b3 t3 + b2 t2 + π R t2 + Af 1 + Af 2 + Af 3 + Af 4 )

2

3. Determine the maximum tensile bending normal stress and its location in the cross section. Likewise, detert1 and t2 are the design variables, rhoitsthe material g the gravitational constant mine thewhere maximum compressive bending normal stress and location in thedensity, cross section.

(9.81×10−9 mm/s2 ). Hence,

4.

Determine the shear flow due to the torque only (Ans. 32.06 N/mm).

5.

Determine the shear flows q 1 ( s 1 ), q 2 ( s 2 ), q 3 ( s 3 ), q 4 ( s 4 ) due to V y at the S.C. as shown in the sketch be-

W (t1 , t2 ) = 24.9959 + 10.7125 t1 + 10.0027 t2

Theans. shear shear loads are (using12 techniques in mm chapter 8): low.(Partial ) q 1 ( 0flows ) = – due 10.37to – 0.07875s N/mm 0learned ≤ s 1 ≤ 195 1 + 0.00015145s

O S.C.

r

⇔

Vy

0 ≤ s1 ≤ b

q 3, s 3

θ

0 ≤ s2 ≤ h 0 ≤ s3 ≤ b

q 2, s 2

O

0 ≤ s 4 ≤ rπ

q 4, s 4

s 4 = rθ

q 1, s 1 b =

ez

c2 + ( r – h ⁄ 2 )2

6.

Determine the location of the center e z .t1(Ans. mm)t1 + 758.982 t2 + 3198.) ∆shear + 39-16.36 t2 ) (819 1 = (36.4159

7.

) = at75.6331 t1 s1τ = 28.93 N/mm 2 at s = b ). V (S1stress Determine theq1shear s 3 = b ,t1s 4s1= +0 81 andt1att2s 4s1=−r (39329.2 π ⁄ 2 ) . (Ans. xs 3

n

2 2

2

2

o − 42120 t1 t2 s1 + 658125. t21 − 1.08515 × 107 t1 + 931275 t1 t2 /∆1

n q2V (S2 ) = 210.6 t22 s22 + 196.646 t1 t2 s22 − 10530 t22 s2

− 9832.3 t1 t2 s2 − 4.13512 × 106 t21 − 1.21297 × 107 t1 − 4.2021 × 106 t1 t2 o 1 − 1.3689 × 106 t2 /∆1 n q3V (s3 ) = 75.6331 t21 s23 + 81 t1 t2 s23 + 9832.3 t21 s3

+ 10530 t1 t2 s3 − 4.13512 × 106 t21 − 1.08515 × 107 t1 o − 4.2021 × 106 t1 t2 /∆1 n q4V (θ4 ) = 658125. t21 + 3.93292 × 106 sin(θ) t2 t1

+ 931275 t2 t1 + 1.9305 × 106 t1 + 4.212 × 106 sin(θ) t22 o + 1.3689 × 107 t2 /∆1

853

10.5. INTRODUCTION TO FRACTURE MECHANICS

854

The equivalent torque acting through the shear center will be (using techniques learned in chapter 8): ∆2 = (5(1 + 2π) t1 + 39 t2 ) (819 t1 + 5(1 + 48π) t2 + 3198) Z b1 Z b2 Z b3 Z b4 Tq = r1 (s1 ) q1 (s1 ) ds1 + r2 (s2 ) q2 (s2 ) ds2 + r3 (s3 ) q3 (s3 ) ds3 + r4 (θ4 ) q4 (θ4 ) R ds4 0 0 0 n0 = − 963821 × 1010 t21 − 79960.5 × 109 t2 t1 − 4.39172 × 1011 t1 o + 8.34503 × 1010 t22 + 4.17733 × 1011 t2 /∆2 The shear center will be located at:

ez = −

Tq Vy0

Hence, the distance d is d = 30 + ez and the total torque experienced through the shear center then can be calculated as: T = −d Vy0 The shear flows due to pure torque is (using techniques learned in chapter 8): qT =

T 2 Ac

Hence, the total shear stresses across each branch are calculated as: q1 = q1V + qT q2 = q2V + qT q3 = q3V + qT q4 = q4V + qT and the total stress is through each branch at x = 0 (root) will be q1 t1 q2 = t2 q3 = t3 q4 = t4

S1xs = S2xs S3xs S4xs

The total axial stress at the root reduced to (using techniques learned in chapter 7): Sxx = −

Mzz0 y Izz

where Izz is calculated using techniques learned in chapter 4: Izz = 1.70625 × 106 t1 + 1.58121 × 106 t2 + 6.6625 × 106 mm4

10.5. INTRODUCTION TO FRACTURE MECHANICS

855

Hence the state of stress for the section becomes (using techniques learned in chapter 5): 

Sxx S =  Sixs 0

 0 0  0

Sixs 0 0

for i = 1, 2, 3, 4, 5, 6

The von Mises stress is found from Section 5.3.2 q 2 + 3 S2 Seq = Sxx ixs Hence the margin of safety is

M S = nsf − 1 =

Sall −1 Seq

where the allowable stress is defined as   Sut Sall = min Sy , = min [325, 313.333] = 313.333 MPa nSF Note that this is evaluated for each thickness and each location. Since we have two modes of fracture (mode I and mode II), we need to calculate fracture toughness for mode II: KIIc =

√

√ 3 KIc = 29.4449 MPa m 2

Hence, the margin of safety for fracture must include both modes and it is defined as: MS = where f=

s

KI KIc

1 −1 f

2

+



KII KIIc

2 End Example 

10.6. REFERENCES

10.6

856

References

Collins, J. A., Mechanical Design of Machine Elements and Machines, 2003, John Wiley and Sons, New York, NY. Hamrock, B. J., Schmid, S. R., and Jacobson, B., Fundamentals of Machine Elements, 2005, Second Edition, Mc-Graw Hill, New York, NY. Juvinall, R. C., and Marsheck, K. A., Fundamentals of Machine Component Design, 2000, John Wiley and Sons, New York, NY. Shigley, J. E., Mischke, C. R., and Budynas, R, G., Mechanical Engineering Design, 2004, Seventh Edition, Mc-Graw Hill, New York, NY. Thomas, G. B., Finney R. L., Weir, M. D., and Giordano F. R., Thomas Calculus, Early Transcendental Update, 2003, Tenth Edition, Addison-Wesley, Massachusetts. Entire book.

10.7. SUGGESTED PROBLEMS

10.7

857

Suggested Problems

Problem 10.1. The state of stress at a point is 

where p > 0 and τ > 0.

  σ=  

−p

τ

τ

−p

τ

τ

τ



  τ   

(10.40)

−p

a) If the true strain at fracture is 20 %, find the maximum allowable values of p and τ , according to each of the related failure criteria. Take Sy = 30 MPa. b) If the true strain at fracture is 2 %, find the maximum allowable values of p and τ , according to each of the related failure criteria. Take Sut = 200 MPa and Suc = 850 MPa. 

10.7. SUGGESTED PROBLEMS

858

Problem 10.2. A fracture toughness test was conducted on AISI 4340 steel having a yield strength of 1380 MPa. The standard compact specimen used had dimensions, as defined in the figure below, b = 50.0 mm, t = 15.0 mm, h/b = 0.6 (h = 30.0 mm), and a sharp precrack to a = 26.0 mm. Failure occurred suddenly at PQ = 15.0 kN. 1. Calculate KQ at fracture 2. Does this value qualify as a valid (plane strain) KIc value? 3. Estimate the plastic zone size at fracture



10.7. SUGGESTED PROBLEMS

859

Problem 10.3. The state of stress at the most critical point of a stucture is 

10000

  σ=  5000 

−6000

Calculate the margin of safety based on:

5000 15000 8000

−6000



  8000   

psi

4000

a) The true strain at fracture is 20 %. Take Sy = 25 ksi. b) The true strain at fracture is 2 %. Take Sut = 30 ksi and Suc = 120 ksi. 

10.7. SUGGESTED PROBLEMS

860

Problem 10.4. The state of stress at a point is 

100

  σ=  60  τ

60 −50 0

τ



  0   

MPa

75

a) If the true strain at fracture is 20 %, find the value of τ for a 12% margin of safety according to each of the related failure criteria. Take Sy = 30 MPa. b) If the true strain at fracture is 2 %, find the value of τ for a 12% margin of safety according to each of the related failure criteria. Take Sut = 200 MPa and Suc = 850 MPa. 

10.7. SUGGESTED PROBLEMS

861

Problem 10.5. Static/Quasi-Static Loading on a Shaft: The fundamental kinematic component of our mechanical universe is the wheel and axle. An essential part of this revolute joint is the shaft. It is a good example of a static, quasi-static, and dynamically loaded body. Application of the information developed to shafts is useful and necessary. It is left for the student to show that critical state of stress at an element located on the surface of a solid round shaft of diameter d subjected to bending, axial loading, and twisting is σxx =

32 M 4F + π d3 π d2

τxz = −

16 T π d3

1. Determine the principal stresses and von Mises stress. 2. Under many axial force F is either zero or so small that its effect may be neglected. Thus F = 0. a) If the true strain at fracture is 20 %, find the value of d for a 80% margin of safety according to each of the related failure criteria. Take Sy = 30 ksi. b) If the true strain at fracture is 2 %, find the value of d for a 80% margin of safety according to each of the related failure criteria. Take Sut = 30 MPa and Suc = 120 MPa. Take M = 1925 lb–in, T = 3300 lb–in. 

Taken from ”Mechanical Behavior of Materials”, by N.E. Dowling, Prentice-Hall Inc, NJ, 1993

You will need Figure 8.15 from class handout (page 4) to solve this homework. Turn in all your 10.7. SUGGESTED PROBLEMS 862 work stapled and place it in a box by Dr. Kapania’s office.

Problem 10.6.

PROBLEM # 1 (10 PTS) Problem 8.10 A standard compact fracture specimen, Fig 8.15(c) has the dimensions of b = 50 mm and = 25 mm, and itfracture is subjected to an load of of P b==22 A tstandard compact specimen hasapplied the dimensions 50KN. mm and t = 25 mm, and it is subjected to an applied load of P = 22 KN.

1. Plot the intensity factor K versus crack length a for an interval of crack lengths a = 15 to 35 mm. 1. Plot the intensity factor K versus crack length a for an interval of crack lengths a = 15 to 35 mm.

2. If the material is 2219-T851 aluminum, what is the longest crack that would permit the 22 kN 2. If to thebematerial 2219-T851 aluminum, what is the longest crack that would permit the 22 kN load appliediswithout brittle fracture occurring? load to be applied without brittle fracture occurring?



PROBLEM # 2. (20 PTS) Problem 10.7. Read section 8.6 pp. 313-319 from Dowling’s text. Problem areforgiven belowspecimens for compact specimens of 7075-T651 same Data are8.23 givenData below compact of 7075-T651 aluminum in the aluminum same sizes in as the those sizesphotographed as those photographed in Figure 8.44. All had dimensions, defined in figure of band = 50.8 in Figure 8.44. All had dimensions, as defined inasfigure 8.15(c), of b 8.15(c), = 50.8 mm mmhand h =mm, 30.5and mm, andsharp initialprecracks sharp precracks and as thickness asbelow. tabulated below. = 30.5 initial and thickness tabulated For each test:For each test: 1. Calculate KQ and determine where or not KQ qualifies as a valid (plane strain) KIc 1. Calculate KQ and determine where or not KQ qualifies as a valid (plane strain) KIc

2. Estimate the plastic zone size at KQ , using 2 roσ or 2 roε as applicable 2. Estimate the plastic zone size at KQ , using 2 roσ or 2 roε as applicable

3. Determine whether analysis by LEFM is applicable 3. Determine whether analysis by LEFM is applicable

4. Plot KQ versus thickness t and comment on the trend observed and its relationship to fracture 4. Plot K versus thickness t and comment on the trend observed and its relationship to fracture surfaces inQ Figure 8.44 surfaces in Figure 8.44



Chapter 11 Failure Theories for Dynamic Loading

Instructional Objectives of Chapter 11 After completing this chapter, the student should be able to: 1. Understand and solve structural problems under rapidly moving loads. 2. Understand and solve structural problems under time-dependent loads.

In the previous chapter we dealt exclusively with static loadings or, if time-varying, loads that are gradually and smoothly applied, with all parts continually in contact. The fact is that most of the mechanical engineering problems encounter dynamic loading. By dynamic loading we mean both impact and cyclic loading. What distinguishes static and dynamic loading is the time duration of the applied load: (i) if the load is applied slowly, it is considered static; (ii) if the load is applied rapidly, it is considered impact; (iii) if the load is is time-dependent, it is considered cyclic. Since for most problems, a fundamental knowledge in vibrations is crucial in the design of machine components, a brief discussion of fundamental natural frequency is included here. This chapter is then followed by impact dynamics and concludes with a throughout discussion of fatigue analysis in the design of machine components.

863

11.1. VIBRATION ANALYSIS

11.1

864

Vibration Analysis

Vibration may be defined as the oscillation or repetitive motion of a structure about an equilibrium position. The equilibrium position is the position the structure will attain when the force acting on it is zero. If the motion is the result of a disturbing force that is applied once and then removed, the motion is known as natural (or free) vibration. If a force of impulse is applied repeatedly to a system, the motion is known as forced vibration. Within both of the categories of natural and forced vibrations are the subcategories of damped and undamped vibrations. If there is no damping (i.e., no friction), a system will experience free vibrations indefinitely. This is known as free vibration and simple harmonic motion.

11.1.1

Fundamental Natural Frequency

Here we will focus on the most important information free (natural) vibrational analysis provide us. It is the information regarding the natural frequencies. Natural frequencies are frequencies at which the structure’s enters into resonance. We have experienced as the washing machine might suddenly and uncontrollably start shaking as a consequence of relocation of clothes within the machine, the automobile starts to shake and as you increase or decrease the speed the shaking disappears. All these are examples of resonances. As the system’s frequency enters in resonance with system’s natural frequency, it causes loss of structural stiffness. To better understand this topic, consider a simple spring-mass system:

k

k m

m

The frequency at which the system will became in resonance is defined as r r k g ω= = m δst where ω is the angular frequency of vibration and has units of radians per second, δst is the total static deflection with units of length, and g the gravitational constant with units of acceleration. The simple mass and ideal spring illustrated in the above Figure is an example of free vibration. After the mass is displaced and released, it will oscillate up and down. Since there is no friction (i.e., the vibration is undamped), the oscillations will continue forever. In design, we want to increase or decrease the natural frequencies to avoid the structures’ frequencies enter in resonance with the structure’s natural frequency. The problem reduces in trying to express all

11.1. VIBRATION ANALYSIS

865

structures in terms of the above spring-mass system. In free vibration analysis, besides the fundamental angular frequency of vibration (usually called as the fundamental natural frequency) we define the linear frequency of vibration as: ω = 2πf

→

f=

ω 2π

and the period of oscillation, the time to complete one cycle of oscillation, is defined as T =

1 f

The units for the linear frequency are Hertz (Hz), 1/sec, and the units for the period are seconds. An important concept used in calculating the behavior of a vibrating system is the static deflection, δst .

11.1. VIBRATION ANALYSIS

866

Example 11.1.

The steel right-angle support bracket with bar lengths L1 = 10 inches and L2 = 5 inches, as shown in Figure, is to be used to support the static load P = 1000 lb. The load is to be applied vertically at the free end of the cylindrical bar, as shown. Both bracket bar centerlines lie in the same horizontal plane. If the square bar has side s = 1.25 inches, and the cylindrical leg has diameter d = 1.25 inches. a) The total static deflection is defined as: δst =

P keff

The load P is known and the problem reduces to find the overall spring rate of the system. Note the square bar will be subject to both torsional and bending deflections, while the cylindrical bar is subject to bending only. This can be modeled as spring in series. Thus 1 1 keff = 3 = 1 1 1 X 1 + + k1 k2 k3 k i=1 i

where k1 is the spring rate caused by bending of the square bar, k2 the spring rate caused by torsion through of the square bar reflected to point O through rigid body rotation of cylinder bar length L2 , and k3 is the spring rate caused by bending of the cylindrical. For the bending of the squared cross-section, k1 =

P y1

11.1. VIBRATION ANALYSIS

867

Using tables, y1 =

P L31 P 3 EI → k1 = = 3 EI y1 L31

For a squared cross-section:

s4 12

I= Thus

E s4 4 L31

k1 =

Next, for the torsion of the square cross-section, k2 =

P y2

where y2 = L2 θ. The total rotation angle is calculated as θ=

P L2 L1 Kxx G

Using this information: k2 =

P P = = y2 L2 θ

L2



P K G  = xx 2 P L2 L1 L1 L2 Kxx G

Using tables for a squared cross-section: Kxx = 2.25 Thus k2 =

 s 4 2

= 0.14 s4

0.14 s4 G L1 L22

For the bending of the circular cross-section, k3 = Using tables, y3 =

P y3

P L32 P 3 EI → k3 = = 3 EI y3 L33

For a circular cross-section: I= Thus k3 =

πd4 64

3 π E d4 64 L32

11.1. VIBRATION ANALYSIS

868

Thus, the overall spring rate is keff =

=

Using tables,

1 1 1 1 + 4 + 4 Es 0.14 s G 3 π E d4 3 2 4 L1 L1 L2 64 L32

E s4 L31

    

4 + 0.14



E = 30 × 106



E G

psi

L2 L1

1 2

+

64 3π





  3    4 L2 s  L1 d

G = 11.5 × 106

keff = 7.70 × 103

psi

lb in

The total static deflection for the given structure is δst =

P = 0.13 keff

in = 0.010833

ft

b) Determine the fundamental natural frequency in rpm (revolutions per minute)

ω= where

r

g δst

ft sec2 and the total static deflection for the given structure, is g = 32.2

δst = Hence ω = 54.51887

P = 0.13 keff

in = 0.010833

rad  rev  sec 2 π rad



60 sec min



ft

= 520.62

rpm

c) Determine the period of oscillation when the structure enters in resonance with the fundamental frequency in seconds. The period of oscillation is defined as T =

1 f

where f is the fundamental linear frequency defined as; f=

ω = 8.676954 2π

1 = 8.676954 sec

Hz

11.1. VIBRATION ANALYSIS

869

Thus T = 0.12

sec End Example 

11.2. IMPACT

11.2

870

Impact

Impact loading is also known as shock, sudden, or impulsive loading. We constantly experience this types of loading: driving a nail with a hammer, hitting a baseball with a bat, automobile collisions, wheels dropping into potholes, jumping of a diving board, a bird striking an aircraft jet engine blade and the list goes on. Impact loads may be divided into three categories: 1. Rapidly moving loads of essentially constant magnitude, as produced by a vehicle crossing a bridge; 2. Suddenly applied loads, such as those in an explosion, or from combustion in an engine cycle. 3. direct-impact loads, as produced by a vehicle collision.

11.2.1

Assumptions

Although impact load causes elastic members to vibrate until equilibrium is reestablished, our concern here is with only the influence of impact or shock force on the maximum stress and deformation within the member. In engineering, the design of structures subject to impact loading may be far more complicated that the approach shown in textbook. However, few approximation greatly simplify the problem providing a qualitative guide in designing these structures. Here typical impact problems will use the energy approach of the mechanics of materials theory along with the following common assumptions: 1. The displacements is proportional to the loads. 2. The material behaves elastically, and a static stress-strain diagram is also valid under impact. 3. The inertia of the member resisting impact may be neglected. 4. No energy is dissipated because of local deformation at the point of impact or at the supports. Although there are many other types of impact loadings such as torsional loading, here we will limit to loadings that cause axial and bending stresses only.

11.2.2

Freely falling body

11.2. IMPACT

871

Consider the free-standing spring with a spring rate k, on which is dropped a body of mass m from a height h. The total energy in the system may be expressed as Ek + Ep = Es where Ek , Ep , and Es is the total change in kinetic, potential, and stored energy from its initial position to the instant of maximum deflection, respectively. For a freely falling body, the initial velocity is zero and again zero at the instant of maximum deflection of the spring (δmax ) and thus the change in kinetic energy of the system is zero. Therefore, the work done by gravity as its falls is equal to the resisting work done by the spring: m g ηm (h + δmax ) =

1 2 kδ 2 max

(11.1)

where ηm is a correction factor to account for the energy dissipation associated with the particular type of elastic member being struck and may defined for various cases. If the dissipation is negligible, ηm will be one. In general, 0 < ηm ≤ 1.0 and by taking ηm = 1.0 we are being conservative. Let the total weight of the mass be W = m g, and ignore the dissipation for derivation, and rearranging Eq. (11.1) we get: 1 2 W (h + δmax ) = k δmax 2 W 2 (h + δmax ) = δmax k From the previous section, the deflection corresponding to a static force is simply the total static deflection, δst . Thus the above may be expressed as 2

2 2 δst (h + δmax ) = δmax

Now let us solve for the maximum deflection of the spring 2 δmax − 2 δst δmax − 2 δst h = 0

Using the quadratic equation as the maximum dynamic deflection is defined as q 2 p − (−2 δst ) + (−2 δst ) − 4 (−2 δst h) (1) 2 + 2δ h = δst + δst δmax = st 2 (1) = δst

1+

r

2h 1+ δst

!

11.2. IMPACT

872

More generally let us define the impact factor as Km1 = 1 +

r

2h ηm δst

1+

(11.2)

Thus the maximum dynamic deflection is defined as δmax = Km1 δst

11.2.3

(11.3)

Falling body with a velocity

Consider the free-standing spring with a spring rate k, on which is a body of mass m is approach with a speed v from a height h. The total energy in the system may be expressed as Ek + Ep = Es where Ek , Ep , and Es is the total change in kinetic, potential, and stored energy from its initial position to the instant of maximum deflection, respectively. At impact the energy relationship is: Ek = Ep →

1 v2 m v2 = m g h → h = 2 2g

we can use the relationships for the free falling object and substitute h= Thus the impact factor will be Km2 = 1 +

v2 2g

s

1+

v2 ηm δst g

(11.4)

and the maximum dynamic deflection is defined as δmax = Km2 δst

(11.5)

11.2. IMPACT

11.2.4

873

Horizontally Moving Weight

Consider a mass (m) in horizontal motion with a velocity v, stopped by an elastic body. The total energy in the system may be expressed as Ek + Ep = Es where Ek , Ep , and Es is the total change in kinetic, potential, and stored energy from its initial position to the instant of maximum deflection, respectively. Since the mass is moving horizontally the potential energy is zero and the velocity is zero at the instant of maximum deflection of the spring (δmax ). Thus Ek = Es → Rearranging

1 2 1 m ηm v 2 = k δmax 2 2

1 W v2 1 2 2 2 2 m ηm g v 2 = k g δmax → ηm v 2 = g δmax → δst ηm = δmax 2 2 k g δst

Thus the maximum dynamic deflection can be written as s v2 = Km2 δst δmax = δst ηm g δst where the impact factor may be defined as Km3 =

11.2.5

s

ηm

v2 g δst

Maximum Dynamic Load and Stress

Since δmax =

Fmax k

we can show the following relationships are true Fmax = Km Fst

and

σmax = Km σst

11.2. IMPACT

874

Example 11.2. Freely falling object

W

x

h

k1

k2

L1

L2

An engineer has designed a machine component that can be modeled as a S3 × 5.7 beam on two identical springs, as shown in Figure. The 5 feet long beam is made of AISI 304 annealed steel. Just after the installation was completed, a 100 lb object 0.245 meters above the structural component suddenly falls at a distance 2 feet from the left spring. The spring rate are 100 lb/in. Using a dissipation correction factor of 0.95, determine if structural component needs to be replaced. First of all, AISI 304 annealed steel is a ductile material and thus we need to check for ductile failure. Let us consider the distortional energy theory. From tables: Syield = 35 × 103 psi For a S3 × 5.7 beam:

E = 28.0 × 106 psi

Izz = 2.5 in4

From the problem L1 = 2 ft = 24 in

L2 = 3 ft = 36 in

h = 0.245 m = 10 in

L = L1 + L2 = 60 in ηm = .95

Assume that the impact load is uniform at the location of impact at the beam. The maximum static deflection for the beam only is: p W L1 L2 (L1 + 2L2 ) 3 L1 (L1 + 2 L2 ) = 0.00608098 in δst,beam = 27 E Izz L and occurs at x=

r

L1 (L1 + 2 L2 ) = 27.71 in 3

11.2. IMPACT

875

The static deflection for the supporting springs only is: δst,springs =

W = 0.50 in k1 + k2

The total static deflection is: δst = δst,beam + δst,springs

p W L1 L2 (L1 + 2L2 ) 3 L1 (L1 + 2 L2 ) W = + = 0.506081 in 27 E Izz L 2k

The impact factor for freely falling object is r

1+

Km1 = 1 +

2h ηm = 7.20833 δst

Thus the maximum deflection is δmax = Km δst = 3.65 in and the maximum load is: Wmax = Km Wst = 720.833 lb As we can see from the figure the maximum bending moment will occur at the point of the load: Wmax L1 L2 = −10380 lb-in Mmax = − L and the maximum bending stress will occur at an element at the top y = c = 0.9+0.63 = 1.53 in (from tables for a S3 × 5.7 beam): σxx = −

Mmax c = 6352.56 psi Izz

Using the distortional energy, σeq = 6352.56 psi In order to determine if it is safe, we need to find the factor of safety: nSF =

Syield = 5.50959 σeq

and the margin of safety is 450%. Thus there is no need to change the structural component. End Example 

11.2. IMPACT

876

Example 11.3. Object falling with a speed

E, A

v

A 5-ton elevator is supported by a titanium cable with an effective modulus of elasticity of 18 × 106 psi and a cross-sectional area A. The titanium has a true strain at fracture higher than 5% in 2 inches. As the elevator is descending at a constant speed of 400 fpm, an accident causes the top of the cable, 70 ft above the elevator, to stop suddenly. What area A will ensure a 150% safety? For the design area what will be the maximum elongation the cable will experience. Be conservative. For a titanium cable: Syield = 128 × 103 psi For a 150% margin of safety, nSF = 2.5

σall =

→

Syield = 51200 psi nSF

From the information provided: W = 5 ton = 10000 lb

L = 70 ft = 840 in

v = 400 ft/min = 80 in/s

Assume: the mass of the cable is negligible (ηm = 1), neglect any stress concentrations, ignore damping due to internal friction within the cable, and the cable responds elastically to the impact. The static deflection is: δst =

7 WL = EA 15 A

The impact factor is: Km2 = 1 +

s

1+

√ v2 ηm = 1 + 1 + 426.254 A δst g

11.2. IMPACT

877

The maximum load is √ Wmax = Km Wst = 10000 + 10000 1 + 426.254 A lb The maximum axial stress is: σxx =

√ 10000 + 10000 1 + 426.254 A Wmax = psi A A

Since we want to be conservative and the material is ductile, let us use the maximum shear stress theory: √ 10000 + 10000 1 + 426.254 A σ1 = , σ2 = σ3 = 0 A √ 5000 + 5000 1 + 426.254 A τmax = A Thus, √ Sy 10000 + 10000 1 + 426.254 A 2 τmax = = 51200 = σall → nSF A Solving for area: A = 16.65 in2 A cross-sectional area of 16.65 in2 will ensure a 150% margin of safety. Thus the maximum elongation the cable will experience is: Km2 = 85.25

δst = 0.028 in

δmax = Km δst = 2.39 in End Example 

11.2. IMPACT

878

Example 11.4. Object moving horizontally v k, L

A car became stuck in sand at a waterfront. A pickup truck, of 1400-kg mass, has offered to help by attempting to jerk the stuck vehicle back onto the road using a 5-m steel tow cable of stiffness k = 5000 N/mm. The traction available to the pickup truck prevented it tight from exerting anystretched significant force on the cable. With the aid of a push from bystanders, the rescue car was able to back against the stuck car and then go forward and reach a speed of 4 km/h at the instant the cable became taut (stretched tight). If the cable is attached rigidly to the masses of the automobiles, estimate the maximum impact force that can be developed in the cable, and the resulting cable elongation. From the information provided: k = 5000 N/mm = 5 × 106 N/m 2

W = (1400 kg)(9.81 m/s ) = 13734 N

L=5m

v = 4 km/hr = 1.11 m/s

Assume: the mass of the rope is negligible (ηm = 1), neglect any stress concentrations, the rope is attached rigidly to the mass of the cars, ignore damping due to internal friction within the rope, and the rope responds to the impact elastically. The static deflection is: δst = The impact factor is: Km3 =

W = 0.00275 m k s

v2 ηm = 6.76 δst g

Thus the maximum impact for is: Fmax = Km W = 92.6 kN The maximum cable elongation is δmax = Km δst = 0.00186 m

11.2. IMPACT

879

End Example 

11.3. FATIGUE

11.3

880

Fatigue

Fatigue was first introduced in 1839 by Poncelet of France. Fatigue fractures begin with minute cracks at critical areas and propagate. Final fracture is largely “brittle” fracture and results from repeated plastic deformation. Fatigue failue may occur at stress levels far below the yield strength after thousands or millions of cycles. Strengthening vulnerable locations is often as effective as making the entire part from a stronger material. Most of the work depends on experimental data.

11.3.1

Cyclic Stresses

A cyclic stress is a time-dependent function where the variation is such that the stress sequence repeats itself. The cyclic stresses may be axial (compressive or tensile), flexural (bending), or torsional (twisting). There are several parameters used to characterize fluctuating cyclic stresses. First, let us define the life cycle with N . Note that one stress cycle (N = 1) constitutes a single application and removal of a load and then another application and removal of the load in the opposite direction.

σ σmax

σmin

t 0

1 cycle

Thus N = 1/2 means the load is applied once and then removed, which is the case with the simple tension test. The mean stress σm is the average of the maximum and minimum stresses in the cycle: σm =

σmax + σmin 2

The stress range σr is the difference range of the maximum and minimum stresses in the cycle: σr = ∆σ = |σmax − σmin | The stress amplitude σa is the one-half of the stress range in the cycle: σmax − σmin σr σa = = 2 2

11.3. FATIGUE

881

The stress ratio Rs is the ratio of minimum to maximum stress amplitudes: Rs =

σmin σmax

The amplitude ratio Aa is the ratio of the stress amplitude to the mean stress: Aa =

σa σmax − σmin 1 − Rs = = σm σmax + σmin 1 + Rs

The maximum and minimum stresses may also be calculated using: σmax = σm + σa

11.3.2

σmin = σm − σa

Fluctuating σ σo2

σmax σr

σa σa

σm σmin

σo1 t

0

σr = 2 σo

σa = σo

0

t

σmax σr

-σo1

σm

σa σa -σo2

σmin σ

σr = 2 σo

σa = σo

11.3. FATIGUE

11.3.3

882

Fully Reversed

It is also known as zero-mean or completely reversed. σ σo

σmax σr

σa σm

0

t

σa -σo

σmin

σmax = σo

σmin = −σo Aa = ∞

σm = 0

σr = 2 σo

Rs = −1

σa = σo

11.3. FATIGUE

11.3.4

883

Repeated (Tension) σ 2σo

σmax σa

σr σa

σm σmin

t

0

σmax = 2 σo

σmin = 0

σm = σo

Aa = 1

11.3.5

σr = 2 σo

σa = σo

Rs = 0

Repeated (Compression) σ 0

σmax

t

σm

σa

σr σa

-2σo

σmin

σmax = 0

σmin = −2 σo

σm = −σo

Aa = −1

Rs = ∞

σr = 2 σo

σa = σo

11.4. ALTERNATE AND MEAN STRESSES

11.4

884

Alternate and mean stresses

Alternate state of stress is found by determining the alternate loads and find the stresses for these loads: Ta = Va = Ma =

Pa =

Tmax − Tmin 2

→

Vmax − Vmin 2

→

Mmax − Mmin 2 Pmax − Pmin 2

→ →

τa

torsion

τxy,a

shear

σxx,a

σxx,a

bending

axial

Mean state of stress is found by determining the mean loads and find the stresses for these loads: Tm = Vm = Mm =

Pm =

Tmax + Tmin 2

→

Vmax + Vmin 2

→

Mmax + Mmin 2 Pmax + Pmin 2

→ →

τm

torsion

τxy,m

shear

σxx,m

σxx,m

bending

axial

Most real design situations involve fluctuating loads that produce multiaxial states of cyclic stress:     σxx,a τxy,a τxz,a σxx,m τxy,m τxz,m            σa =  σ = τ σ τ τ σ τ yy,a yz,a  yy,m yz,m  m  yx,a  yx,m     τzx,a τzy,a σzz,a τzx,m τzy,m σzz,m A consensus has not yet reached on the best approach to predict failure under multiaxial cyclic stress. However, the following techniques will be used in this book.

11.4.1

Ductile materials

Although there is little multiaxial fatigue data available, for ductile materials, the distortion energy multiaxial fatigue failure theory is the best theory to use. In it consists in determining the von Mises

11.5. S–N DIAGRAMS

885

stress for both mean state of stresses, σ m , and alternate state of stresses, σ m : σa = σeq,a =

= =

σm = σeq,m =

= =

s

(σ1,a − σ2,a ) + (σ2,a − σ3,a ) + (σ3,a − σ1,a ) 2

s

2 2 2 (σxx,a − σyy,a ) + (σyy,a − σzz,a ) + (σzz,a − σxx,a ) + 6 τxy,a + τyz,a + τxz,a 2

2

2

2

2

2

2




q Iσ21,a − 3 Iσ2,a

s

(σ1,m − σ2,m ) + (σ2,m − σ3,m ) + (σ3,m − σ1,m ) 2

s

2 2 2 (σxx,m − σyy,m ) + (σyy,m − σzz,m ) + (σzz,m − σxx,m ) + 6 τxy,m + τyz,m + τxz,m 2

q

2

2

2

2

2

2




Iσ21,m − 3 Iσ2,m

The equivalent maximum and minimum stresses may be found by: σmax = σm + σa

11.4.2

σmin = σm − σa

Brittle materials

Although there is little multiaxial fatigue data available, for brittle materials, the maximum normal multiaxial fatigue failure theory is the best theory to use. In it consists in determining the maximum stress for both mean and alternate state of stresses: σa = σ1,a

and

σm = σ1,m

The equivalent maximum and minimum stresses may be found by: σmax = σm + σa

11.5

σmin = σm − σa

S–N Diagrams

Fatigue test data are frequently presented in the form of a plot of fatigue strength S or completely reversed stress versus the number of cycles to failure or fatigue life N with a semi-logarithm scale; that is S–log N , as shown in Fig. 11.1.

11.5. S–N DIAGRAMS

886

SN, Sa

Low cycle

High cycle

Sut f Sut

Se Finite life 100

Infinite life

103

Ne=10g

N (log)

Figure 11.1: Typical S-N diagram for ferrous materials.

11.5.1

Fatigue Regimens

The S–N diagram has two basic regimens and these are: low-cycle fatigue and high-cycle fatigue. The low-cycle fatigue is any loading that causes failure below approximately 1000 cycles: 100 ≤ N ≤ 103 High-cycle fatigue is concerned with failure corresponding to stress cycle greater than 1000 cycles: N > 103

11.5.2

Endurance Stress

In the case of ferrous materials, a “knee” occurs in the S–N diagram, and beyond this knee failure will not occur, no matter how great the number of cycles. The strength corresponding to this knee is called the endurance limit Se0 , or the fatigue limit. The endurance limit is usually defined as the maximum stress a material can withstand “indefinitely” without fracture: Se0 :

N ≥ Ne

where

Ne → ∞

The endurance limit is therefore stated with no associated number of cycles to failure. Nonferrous materials, on the other hand, often exhibit no endurance limit. For some nonferrous materials, approximations for an endurance limit Se0 , using experimental data, has been suggested. Thus, for nonferrous materials we use the fatigue strength SN0 which is the fatigue limit at N cycles.

11.5. S–N DIAGRAMS

11.5.3

887

Modified Endurance Stress

Most of the data for Se0 is available for a single specimen test. If we want to use for other parts we need to use the correct value for Se . Thus in practice we do not use the endurance limit Se0 , but the modified endurance limit: Se = k∞ Se0 or the fatigue limit at N cycles and is also modified as follows SN = k∞ SN0 The factor k∞ accounts for the various influencing factors such as size, surface condition, reliability, loading, temperature, among others. This factor is expressed as the product of k∞ = kL kt ksr kr kg ke Loading factor To take into account the low-cycle effects, the loading factor is used:  1.00    0.85 kL =  0.59   1.00

bending axial torsion torsion combined with other stresses

Temperature factor When operating temperatures are below room temperature, brittle fracture is a strong possibility and should be investigated. When operating temperatures are higher than room temperature, yielding should be investigated first because yield strength drops off rapidly with temperature.

kt =

operating temperature room temperature

Surface finish factor Most parts of a machine do not usually have a high-quality surface finish (highly polished). Thus the surface finish factor incorporates the finish effects on the process used to generate the surface. The surface finish factor ksr can be obtained using charts or analytically by c ksr = e Sut

where Sut is the ultimate tensile strength of material and the coefficients e and c are defined as

11.5. S–N DIAGRAMS

888

Manufacturing Process

Factor e Sut [MPa] Sut [ksi]

Exponent c

Grinding

1.58

1.34

-0.085

Machining or cold drawing

4.51

2.70

-0.265

Hot rolling

57.7

14.4

-0.718

As forged

272.0

39.9

-0.995

For mirror-polished surfaces take ksr = 1.0.

Reliability factor Most of the data is empirical however we are often interested in the reliability of the probability of survival, that is the probability of surviving to the life indicated at a particular stress. Thus the reliability factor kr may be expressed as kr = 1 − 0.08 X where X is the transformation variate obtained from any table for a cumulative distribution function. However, the reliability factor for the most common probabilities of survival corresponding to 8% standard deviation of the endurance limit is Probability of Survival % 50

Transformation Variate X 0.000

Reliability factor kr 1.000

90

1.288

0.897

95

1.645

0.868

99

2.326

0.814

99.9

3.091

0.753

99.99

3.719

0.702

99.999

4.265

0.659

99.9999

4.753

0.620

11.5. S–N DIAGRAMS

889

Gradient size factor Choose kg as follows: Bending

Axial

Torsion

1

[0.7 0.9]

1

0.400 < d < 200 or 50 mm

0.9

[0.7 0.9]

0.9

d > 200 or 50 mm

[0.6 0.75]

1

[0.6 0.75]

d < 0.400 or 10 mm

If combined load usually choose 0.9, but in some cases we may choose 1.0.

Miscellaneous Factors 1. Residual stresses 2. Corrosion: There is no fatigue limit 3. Frettage corrosion: ke [0.24 0.90] 4. Operating speed: ke ∼ 0.9

11.5.4

Stress concentration factor

The stress concentration factor is a function of the type of discontinuity (hole, fillet, groove), the geometry of the discontinuity, and the type of loading being experienced. Some materials are not as sensitive to notches as implied by the theoretical stress concentration factor. For these materials a reduced value of Kt maybe used Kf . Not all ductile materials are ductile under all conditions, many become brittle under some circumstances. The most common cause of brittle behavior in materials normally considered to be ductile is being exposed to low temperatures. For ductile materials subjected to cyclic loading the stress concentration factor has to be included in the factors that reduce the fatigue strength of a component. Some materials are not as sensitive to notches as implied by the theoretical stress concentration factor. For these materials a reduced value of Kt is used: Kf . In these materials the maximum stress is: σt = Kf σN

(11.6)

Kf − 1 Kt − 1

(11.7)

The notch sensitivity, q, is defined as: q=

where q is the notch sensitivity factor and ranges between 0 (Kf = 1) and 1 (Kf = Kt ). This will be discussed later when working with fatigue analysis. Table 4.7 discusses the effects of stress concentration

11.5. S–N DIAGRAMS

890

of different type of loading, material, and failure. Depending the case we are working with is the stress concentration factor we choose. Different authors have chosen different approaches in solving these problems. In this book, we choose the methodology. Recall, for static loading the geometric stress concentration factor Kt is used for brittle materials but taken as one for ductile materials. For fatigue loading the fatigue stress concentration factor Kf may be used, depending whether it is ductile or brittle.

Ductile materials There is the method by Dowling for ductile materials, which, for materials with a pronounced yield point and approximated by an elastic-perfectly-plastic behavior model, quantitatively expresses the steady stress component stress-concentration factor Kfm as Kfm = Kf Kfm =

Kf |σmax,N | < Sy

Kf − Kf σa,N |σmax,N |

Kfm = 0

Kf |σmax,N | > Sy Kf |σmax,N − σmin,N | > 2 Sy

where σmax,N and σmin,N are the fluctuating maximum and minimum nominal stresses, and σa,N and σm,N are the alternate and mean nominal stresses. In this book the following approach will be used: a) When the plastic strain at the notch can be avoided, apply the fatigue stress concentration factor to the mean (Kfm = Kf ) and alternate stresses (Kf ). b) When the plastic strain at the notch cannot be avoided, apply fatigue stress concentration factor to the alternate stress (Kf ) and take it as one for the mean stresses (Kfm = 1, conservative approach).

Brittle materials For brittle materials a stress raiser increases the likelihood of failure under either steady or alternating stresses, and it is customary to apply a stress concentration factor to both. Thus, apply the fatigue stress concentration factor Kf to the alternating component of stress for ductile materials. In brittle materials, apply the geometric stress concentration factor Kt to the mean components of stress and fatigue stress concentration factor Kf to the alternating components of stress. Thus Kfm = Kt

11.5. S–N DIAGRAMS

891

Summary In general, σt,a axial = Kfa σN,a axial

σt,a bending = Kfb σN,a bending

σt,m axial = Kfma σN,m axial

σt,m bending = Kfmb σN,m bending

τt,a torque = Kfs τN,a torque

τt,m torque = Kfms τN,m torque

and only the mean fatigue stress-concentration factor Kfm changes for each material type.

11.5.5

Plotting S-N Diagrams SN, Sa

Low cycle

High cycle

Sut f Sut

Se Finite life 0

10

3

10

Infinite life Ne=10

g

N = 100 = 1,

S = Sut

N = 103 ,

Sf = f Sut

N = Ne = 10ge ,

S = Se

N (log)

For the case of pure bending use f = 0.9, for the case of pure axial loading use f = 0.75, and for the case of pure torsion use f = 0.72. For combined loading take f may be approximated as follows: f = 0.93 Sut = 60 ksi f = 0.86 Sut = 90 ksi f = 0.82 Sut = 120 ksi f = 0.77 Sut = 200 ksi

11.5. S–N DIAGRAMS

892

Stress-Cycle relationship The common empirical formula relating fatigue strength and number of cycles to failure is (11.8)

SN = a N b The constants a and b are derived from (11.8): log (SN ) = log a N b




log (SN ) = log (a) + log N b




log (SN ) = log (a) + b log (N ) Now for high-cycle fatigue use that fact that N = 103 ,

Sf = f Sut

N = Ne = 10ge ,

S = Se

to obtain two equations “linear” equations in a and b: log (Sf ) = log (a) + b log 103




log (Se ) = log (a) + b log (10ge ) Hence we have two equations and two unknown and the constants may be expressed as   ge 3 f Sut Sf Sf 1 log a = 3 b = 3 b = (f Sut ) ge −3 (Se ) 3−ge b= 3 − ge Se 10 10 Typically for ferrous materials ge = 6:   1 Sf b = − log 3 Se

a=

f Sut Sf 2 −1 = 3 b = (f Sut ) (Se ) 3b 10 10

In the S-N diagram, the fatigue stress is also the alternating stress. Thus for a given number of cycles 103 < N ≤ Ne , the fatigue stress may be evaluated as follows: S = a Nb Also, to obtain the number of cycles for a given alternating stress: S = aN

b

→

  1b S N= a

11.5. S–N DIAGRAMS

893

Finite and Infinite Life Infinite life begins for stresses bellow the endurance limit, that is Seq < Se When designing with materials that exhibit no endurance limit, the design will always be for finite life.

11.5.6

Fatigue Theories of Fatigue Failure Sa Syt Langer Criterion (first cycle yielding) Se

Goodman Criterion Soderberg Criterion -Syt

Syt

Sut

Sm

Goodman Criterion Among the fatigue theories, the Modified Goodman theory is the one widely used. The Modified Goodman criterion, which gives reasonably good results for brittle materials while conservative values for ductile materials is a realistic scheme for most materials. Goodman criterion is widely used because: 1. it is a straight line and the algebra is linear and easy. 2. it is easily graphed, every time for every problem. 3. It reveals subtleties of insight into fatigue problems. 4. Answers can be scaled from the diagrams as a check on the algebra.

11.5. S–N DIAGRAMS

894

For ductile materials, the fatigue equivalent stress is Seq =

Sa Sm 1− Sut

Seq = Sy

for

σm ≥ 0

and σmax ≤ Sy

for

σm ≥ 0

and σmax ≥ Sy

For brittle materials, the fatigue equivalent stress is Seq =

Sa Sm 1− Sut

In the above expressions: Sa = nSF σa

Sm = nSF σm

where nSF is the safety factor. The maximum and minimum stresses are Smax = Sm + Sa

Smin = Sm − Sa

Soderberg Criterion Among the fatigue theories, the Soderberg theory may the also be used for ductile materials. It gives conservative values for ductile materials. The criteria states that the fatigue equivalent stress is Seq =

Sa Sm 1− Syt

In the above expressions: Sa = nSF σa

Sm = nSF σm

where nSF is the safety factor. The maximum and minimum stresses are Smax = Sm + Sa

Smin = Sm − Sa

11.5. S–N DIAGRAMS

895

Example 11.5. Life Cycle Example: Brittle Material

The Class 60 gray cast-iron mounting arm is subjected to tension and torsion, as shown in Figure. It takes two minutes for completion of a full cycle. The Class 60 gray cast iron has an ultimate strength of 60 ksi in tension, and elongation in 2 inches of less than 0.5%. The design safety factor is 1.5. (a) The arm is subject to a static axial force of P = 50000 lb and a static torsional moment of T = 18000 lb–in. For the given dimensions, could the arm support the specified loading without failure? (b) During a different mode of operation, the axial force P cycles fluctuates from 50000 lb in tension to 10000 lb in compression, and the torsional moment remains zero at all times. What would be the estimated days of life for this mode of operation for a 99% reliability? Solution: (a) The arm is subject to a static axial force of P = 50000 lb and a static torsional moment of T = 18000 lb–in. For the given dimensions, could the arm support the specified loading without failure? From the problem it is known: Sut = 60 ksi Assuming the state of stress at the most critical location of the shaft’s cross-section occurs for an element located at the top:   σxx τxy τxz      σA =  τ σ τ yy yz   yx   τzx τzy σzz

Thus:

4P P = psi σxx = σxx axial = A π d2

11.5. S–N DIAGRAMS

896

and

T 16 T psi τxz = τ torsion = = Q π d3

Thus

 4P 16 T  0 2    πd π d3    σxx τxy τxz    0 0  σ =  τxy σyy τyz  =  0 psi    τxz τyz σzz   16 T 0 0 π d3 For the static conditions given, under combined loads of axial tension and torsional shear, the critical point will be at the root of the 0.125” radius fillet. Stress concentration factors must be separately determined for the tensile load and the torsional load. Thus using figures for stress concentration factor for a shaft with a fillet subject to axial (Figure 4.17b) and torsion (Figure 4.17c): d = 2 in d = 2 in

r = 0.125 in D = 2.25 in

→ →

r = 0.063 d D = 1.13 d

Thus Kta = 1.8 Thus the state of stress is  4P Kta  π d2   0 σ=    16 T Kts π d3

Kts = 1.15

modified as follows 16 T   0 Kts 28647.9 π d3      = 0 0 0      13178 0 0

0

13178

0

0

0

0

     

psi

Now we determine the principal stresses:

Iσ1 = σxx + σyy + σzz = 28647.9 psi 2 2 2 Iσ2 = σxx σyy + σzz σxx + σyy σzz − τxy − τyz − τzx = −1.7366 × 108 psi2 2 2 2 Iσ3 = σxx σyy σzz + 2 τxy τyz τzx − σxx τyz − σyy τzx − σzz τxy =0

λ3 − Iσ1 λ2 + Iσ2 λ = λ λ2 − Iσ1 λ + Iσ2 The principal stresses are:

σ1 = 33787.7 psi σ2 = 0 σ3 = −5139.76 psi




=0

11.5. S–N DIAGRAMS

897

σmax

=

max[σ1 , σ2 , σ3 ] = 33787.7 psi

σmin

=

min[σ1 , σ2 , σ3 ] = −5139.76 psi

τmax

σmax − σmin = 19463.7 psi = 2

Now, note that the strain at fracture is less than 0.5 % (εf < 0.005), thus the material is brittle, and we can use the Maximum Normal Stress Theory. Since no information is given for Suc , let us only use σmax 1 ≤ Sut nSF which leads to

σmax ? 1 ≤ Sut nSF

for σmax ≥ 0 0.563 < 0.667

→

Thus the arm can support the specified static load without failure for a safety factor of 1.5. (b) During a different mode of operation, the axial force P cycles fluctuates from 50000 lb in tension to 10000 lb in compression, and the torsional moment remains zero at all times. What would be the estimated days of life for this mode of operation for a 99% reliability? The state of stress is as follows 

4P  π d2   σ= 0   0

0 0 0

This is a fluctuating cyclic load problem. Thus, Pa =



   0    0

Pmax − Pmin (50000) − (−10000) = = 30000 lb 2 2

Pmax + Pmin 2  4 Pa 0  π d2   σa =  0 0   0 0

Pm =

0

(50000) + (−10000) = 20000 lb 2    4 Pm 0 0 0   π d2        σ =   0  0 0  m  0     0 0 0 0

=

Since it is brittle material, we need to apply the fatigue stress concentration factor to the alternate stresses and static stress concentration factor to the mean stresses. The

11.5. S–N DIAGRAMS

898

fatigue stress concentration factor Kfa is calculated as follows, Kfa = 1 + q (Kta − 1) For the mean stress Kfma = Kta From Table we find r = 0.12500

Sut = 60 ksi

q ≈ 0.75

→

Kfa = 1 + 0.75 (1.8 − 1) = 1.6 Thus



   σa =    

   σm =   

Kfa

Kta

4 Pa π d2

0

0

0

0

0

4 Pm π d2

0

0

0

0

0

The principal stresses are:

0





      = 0      0

15278.9 0

0

0

0



 11459.2       = 0 0      0 0

0

0

0 0 0

σ1,a = 15278.9

σ2,a = 0.0

σ3,a = 0.0

σ1,m = 11459.2

σ2,m = 0.0

σ3,m = 0.0

0



  0    0

0



  0    0

For brittle materials the mean and alternate stresses are σa = σ1,a = 15278.9 psi

σm = σ1,m = 11459.2 psi

The maximum and minimum stresses are σmax = σm + σa = 26738 psi σmin = σm − σa = −3819.72 psi The stress range σr is the difference range of the maximum and minimum stresses in the cycle: σr = ∆σ = σmax − σmin = 30557.7 psi The stress ratio Rs is the ratio of minimum to maximum stress amplitudes: Rs =

σmin = −0.142857 σmax

11.5. S–N DIAGRAMS

899

The amplitude or load ratio Aa is the ratio of the stress amplitude to the mean stress: σa Aa = = 1.333 σm Let us use the modified Goodman criterion: Seq =

Sa Sm 1− Sut

In the above expressions: Sa = nSF σa = 22918.3

Sm = nSF σm = 17188.7

where nSF is the safety factor. Thus Seq =

Sa = 32120 psi Sm 1− Sut

Now we need to find the modified Se . First of all at N = 1 life cycle plot S = Sut = 60 ksi. For N = 103 , Sf = 0.75 Sut = 45 ksi (for pure axial loading f = 0.75). Next for cast irons with Sut ≤ 88 ksi: Se0 = 0.4 Sut

at N = 106 cycles

So for the class 60 Gray cast iron alloy used for this mounting arm Se0 = 0.4 Sut = 24000 psi Since Ne = 106 , ge = 6. For axial loading kL = 0.85 and 99% reliability kr = 0.814. All others are 1.0. Thus k∞ = 0.6919 Thus the modified endurance limit is Se = k∞ Se0 = 16605.6 psi Since Seq > Se , the part has finite life. In order to find the life cycles, we use the equation  1   1b Seq b S → N= N= a a where

1 b= log 3 − ge



Sf Se



1 = log 3 − ge



f Sut Se



a=

ge 3 Sf f Sut = 3 b = (f Sut ) ge −3 (Se ) 3−ge 3b 10 10

Note that we took S = Seq because that is the fatigue stress at which we want to calculate the life cycles for failure.

11.5. S–N DIAGRAMS

900

For ge = 6: b=−

1 log 3



f Sut Se



= −0.144319

N=



Seq a

Thus

 1b

a=

f Sut = 121947. 103 b

= 10343.6 cycles

The component has 1.03 × 105 life cycles before failure. It is known:     2 minutes 1 hour 1 day 5 N = 1.03 × 10 cycles = 14.36 days 1 cycle 60 minutes 24 hours Thus the component has about 14 days of life. End Example 

11.5. S–N DIAGRAMS

901

Example 11.6. Life Cycle Example: Ductile Material

The wrought ferrous steel arm is subjected to tension and torsion, as shown in Figure. It takes two minutes for completion of a full cycle. It takes two minutes for completion of a full cycle. The steel has a yield strength of 70 ksi and ultimate strength of 90 ksi in tension, and elongation in 2 inches of greater than 0.5%. The design safety factor is 1.5. (a) The arm is subject to a static axial force of P = 50000 lb and a static torsional moment of T = 18000 lb–in. For the given dimensions, could the arm support the specified loading without failure? (b) During a different mode of operation, the axial force P cycles fluctuates from 50000 lb in tension to 10000 lb in compression, and the torsional moment remains zero at all times. What would be the estimated life for this cyclic mode of operation for a 99% reliability? Solution: (a) The arm is subject to a static axial force of P = 50000 lb and a static torsional moment of T = 18000 lb–in. For the given dimensions, could the arm support the specified loading without failure? From the problem it is known: Sut = 90 ksi

Sy

= 70 ksi

Assuming the state of stress at the most critical location of the shaft’s cross-section occurs for an element located at the top:   σxx τxy τxz      σ A =  τyx σyy τyz     τzx τzy σzz

11.5. S–N DIAGRAMS

902

Thus:

P 4P psi σxx = σxx axial = = A π d2

and

T 16 T τxz = τ torsion = = psi Q π d3

Thus

 4P 16 T  0    π d2 π d3    σxx τxy τxz      0 0  psi σ= = 0 τxy σyy τyz    τxz τyz σzz   16 T 0 0 π d3 For the static conditions given, under combined loads of axial tension and torsional shear, the critical point will be at the root of the 0.125” radius fillet. For ductile materials, stress concentration may be taken as one: Kta = 1.0

Kts = 1.0

Thus the state of stress is modified as follows  4P 16 T   0 15915.5 2  πd π d3        0 0 0 = σ= 0         16 T 11459.2 0 0 π d3

0

11459.2

0

0

0

0

     

psi

Now we determine the principal stresses:

σ1 = 21909. psi σ2 = 0 σ3 = −5993.52 psi σmax

=

max[σ1 , σ2 , σ3 ] = 21909. psi

σmin

=

min[σ1 , σ2 , σ3 ] = −5993.52 psi

τmax

σmax − σmin = 13951.3 psi = 2

Now, note that the strain at fracture is greater than 0.5 % (εf > 0.005), thus the material is ductile, and we may use the Distortional Energy Criterion: σeq 1 ≤ Sy nSF

11.5. S–N DIAGRAMS

903

which leads to (σeq = 25440.9 psi) σeq ? 1 ≤ Sy nSF

→

0.363442 < 0.667

Thus the arm can support the specified static load without failure for a safety factor of 1.5. (b) During a different mode of operation, the axial force P cycles fluctuates from 50000 lb in tension to 10000 lb in compression, and the torsional moment remains zero at all times. What would be the estimated life for this cyclic mode of operation for a 99% reliability? The state of stress is as follows 

4P  π d2   σ= 0   0

0 0 0

This is a fluctuating cyclic load problem. Thus, Pa =

0



   0    0

Pmax − Pmin (50000) − (−10000) = = 30000 lb 2 2

Pmax + Pmin 2  4 Pa 0  π d2   σa =  0 0   0 0

Pm =

(50000) + (−10000) = 20000 lb 2    4 Pm 0 0 0    π d2       σm =  0  0  0 0       0 0 0 0

=

Since it is ductile material, we apply the fatigue stress concentration factor to both alternate and mean stresses. The fatigue stress concentration factor Kfa is calculated as follows, Kfa = 1 + q (Kta − 1) From Tables and from the previous example: r = 0.12500

Sut = 60 ksi

Kfa = 1 + 0.75 (1.8 − 1) = 1.6

→

q ≈ 0.75 (Kta = 1.8)

11.5. S–N DIAGRAMS

Thus

904



   σa =    

   σm =   

Kfa

Kfa

4 Pa π d2

0

0

0

0

0

4 Pm π d2

0

0

0

0

0



 15278.9       = 0 0      0 0

0



 10185.9       = 0 0      0 0

0

0 0 0 0 0 0

0



  0    0

0



  0    0

For ductile materials the mean and alternate stresses are q q σa = σeq,a = Iσ21,a − 3 Iσ2,a = 15278.9 psi σm = σeq,m = Iσ21,m − 3 Iσ2,m = 10185.9 psi

The maximum and minimum stresses are

σmax = σm + σa = 25464.8 psi σmin = σm − σa = −5092.96 psi The stress range σr is the difference range of the maximum and minimum stresses in the cycle: σr = ∆σ = σmax − σmin = 30557.7 psi The stress ratio Rs is the ratio of minimum to maximum stress amplitudes: Rs =

σmin = −0.20 σmax

The amplitude or load ratio Aa is the ratio of the stress amplitude to the mean stress: σa Aa = = 1.5 σm Let us use the modified Goodman criterion: Sa = nSF σa = 22918.3

Sm = nSF σm = 15278.9

where nSF is the safety factor. Sm > 0

Smax = 38.2 ksi < Sy = 70 ksi

Thus Seq =

→

Seq =

Sa Sm 1− Sut

Sa = 27604.6 psi Sm 1− Sut

Now we need to find the modified Se . First of all at N = 1 life cycle plot S = Sut = 90

11.5. S–N DIAGRAMS

905

ksi. For N = 103 , Sf = 0.75 Sut = 67.5 ksi (for axial loading f = 0.75). Next for steels with Sut < 200 ksi: Se0 = 0.5 Sut

at N = 106 cycles

So for the steel used for this mounting arm Se0 = 0.5 Sut = 40500. psi For axial loading kL = 0.85 and 99% reliability kr = 0.814. All others are 1.0. Thus k∞ = 0.6919 Thus the modified endurance limit is Se = k∞ Se0 = 31135.5 psi Since Seq < Se , the part has infinite life. End Example 

11.5. S–N DIAGRAMS

906

Example 11.7. Infinite Life Cycle Design Example: Ductile Material

r

M

T

D

M d

T

A circular shaft is made of wrought-carbon steel with yield strength of 42 ksi and ultimate strength of 76 ksi in tension, and elongation in 2 inches of greater than 0.5%. The shaft’s current mode of operation is such that the loads are as follows: repeated bending moment of 50000 lb–in in tension and the repeated torsional moment of 10000 lb–in in compression. The fillet radius is 0.1 in and D = 1.5 d. Estimate the value of the diameter d, for infinite life. Consider the following operating conditions: 1. Operating speed is 3600 rev/min. 2. The design safety factor is 1.5. 3. A strength reliability level of 99.9%. 4. The part is to be lathe-turned from a bar of the wrought-steel alloy. Solution: First let us locate the most critical point in the shaft. It will occur at an element at the top (bending stress in compression) is or at the bottom (bending stress in tension). Let us consider the element at top. Thus,   M T  −Z 0 Q       0 0 0  σ=       T  0 0 Q where

π d3 π d3 Q= 32 16 This is a repeated load in tension for bending moment and repeated load in compression for Z=

11.5. S–N DIAGRAMS

907

the torsional load: Thus,

Thus



   σa =    

Ma =

(50000) − (0) Mmax − Mmin = = 25000 lb–in 2 2

Mm =

Mmax + Mmin (50000) + (0) = = 25000 lb–in 2 2

Ta =

Tmax − Tmin (0) − (−10000) = = 5000 lb–in 2 2

Tm =

Tmax + Tmin (0) + (−10000) = = −5000 lb–in 2 2

−

32 Ma π d3 0

0 0

16 Ta  π d3    0     0



   σm =    

−

32 Mm π d3 0

0 0

16 Tm  π d3     0    0

16 Ta 16 Tm 0 0 3 πd π d3 Since it is ductile material, we apply the fatigue stress concentration factor to both alternate and mean stresses. d = do in d = do in

r = 0.1 in D = 1.5 do in

r = 0.025(Assume do = 400 as an initial guess) d

→ →

D = 1.5 d

Thus Ktb = 2.6

Kts = 2.05

The fatigue stress concentration factor Kfa is calculated as follows, Kf = 1 + q (Kt − 1) From Tables and from the previous example: r = 0.100

Sut = 76 ksi

→

qb ≈ 0.775

Kfb = 1 + 0.775 (2.6 − 1) = 2.24 Kfs = 1 + 0.82 (2.05 − 1) = 1.861

qs ≈ 0.82

11.5. S–N DIAGRAMS

908

Thus, 

   σa =    

−Kfb

Kfs

32 Ma π d3

0

0

0

16 Ta π d3

0

Kfs

Substituting all values: 

   σa =    

−

570411 d3

0

0

0

47390 d3

0

The principal stresses are:

16 Ta  π d3     0    0



   σm =    

47390  d3     0    0



   σm =    

−Kfb

Kfs

−

32 Mm π d3

0

0

0

16 Tm π d3

0

570411 d3 0

−

47390 d3

0

−

0 0

σ1,a =

3910.37 psi d3

σ2,a = 0

σ3,a = −

574322 psi d3

σ1,m =

3910.37 psi d3

σ2,m = 0 σ3,m = −

574322 psi d3

Kfs

16 Tm  π d3     0    0

47390  d3     0    0

For ductile materials the mean and alternate stresses are q q 576287 576287 σa = σeq,a = Iσ21,a − 3 Iσ2,a = psi σ = σ = Iσ21,m − 3 Iσ2,m = psi m eq,m 3 d d3

The maximum and minimum stresses are σmax = σm + σa =

1.15257 × 106 psi d3

σmin = σm − σa = 0 The stress range σr is the difference range of the maximum and minimum stresses in the cycle: 1.15257 × 106 σr = ∆σ = σmax − σmin = psi d3 The stress ratio Rs is the ratio of minimum to maximum stress amplitudes: Rs =

σmin = 0.0 σmax

The amplitude or load ratio Aa is the ratio of the stress amplitude to the mean stress: Aa =

σa = 1.0 σm

11.5. S–N DIAGRAMS

909

Let us use the modified Goodman criterion: Sa = nSF σa =

864430 d3

Sm = nSF σm =

864430 d3

where nSF is the safety factor. Assume Sm > 0

Thus

→

Smax < Sy

Seq =

Sa Sm 1− Sut

864430 Sa 864430 d3 Seq = = 3 = psi 11.3741 Sm d − 5.10078 1− 1− Sut d3

Now we need to find the modified Se . First of all at N = 1 life cycle plot S = Sut = 76 ksi. For N = 103 and combined loading Sf = 0.9 Sut = 68.4 ksi (for combined loading f ≈ 0.90).

Next for steels with Sut < 200 ksi:

Se0 = 0.5 Sut

at N = 106 cycles

So for the steel used for this circular shaft Se0 = 0.5 Sut = 38000. psi The correction factors are: (a) Loading: kL = 1.0 (b) Temperature: kt = 1.0 (c) Surface finish factor: lathe-turned is a hot rolling treatment: c ksr = e Sut = 14.4 (76)−0.718 = 0.6426

(d) Reliability: 99.9%, kr = 0.753 (e) Gradient Size factor: Bending and torsion combined. Note assumed d = 400 thus kg = 0.7 (f) Miscellaneous, Operating speed: ke = 0.9 Thus k∞ = 0.304841 Thus the modified endurance limit is Se = k∞ Se0 = 11584 psi For infinite life take Seq < Se Solving for d:

→

864430 < 11584 d − 5.10078 3

d > 4.4139500

11.5. S–N DIAGRAMS

910

Choose d = 4.500 . End Example 

11.5. S–N DIAGRAMS

911

Example 11.8. Check for Safety

k

A spring board design is being evaluated by your company. You boss is requesting that you suggest a reasonable warrantee for the product. A typical diver jumps can jump 1 ft on the free end of a diving board before diving into the water. The maximum weight of a typical user is about 225 lbs. The dimensions of the diving board are L = 4 m (from the pin to the tip) and is 25.4 mm thick. Has a moment of inertia of Izz = 5 × 106 mm4 . The supported end of the diving board is fixed. Assume a 99.999% reliability and that material is a wrought steel with: E = 70 GPa Sy = 400 MPa Sut = 600 MPa The spring rate is: k = 16.4 kN/m 1. First determine, the margin of safety for infinite life using both yielding criterions: Modified Goodman and Soderberg. 2. If the board is to be used during day time (8hr/day) and it is expected that on the average a person uses it every 15 minutes. Suggest a warrantee period for a 400% margin of safety. Solution: First we need to calculate the maximum load due to impact and consider it as a repeated cyclic load. In order to do so we need to first obtain the static load which can be done using spring analysis. At the tip: δs =

Fs k

δb =

FB L 3 3 E Izz

Fs + Fb = F and δs = δb

→

Fs FB L3 = k 3 E Izz

→

FB =

3 E Izz Fs k L3

11.5. S–N DIAGRAMS

912

Thus

Fs + Fb = F Fs +

3 E Izz Fs =W k L3

Where W = 225 (4.45) = 1001.25 N. Thus Fs + 1.00038 Fs = 1001.25

→

Fs = 500.53 N

Thus the static deflection is: δst = δs =

Fs = 0.0305201 m k

For a freely falling object the impact factor is: r Km1 = 1 +

1+

2h ηm δst

For a conservative approximation, take ηm = 1.0 and h = 1(12)(25.4)/1000 = 0.3048 m. Thus impact factor is r 2h ηm = 5.57971 Km1 = 1 + 1 + δst and the impact load will be

Fe = W Km1 = 5586.68 N The most critical point is at the fixed end and at an element at the top. At the top the bending moment is: Mzz = −Fe L = −22346.7 N-m 1. First determine, the margin of safety for infinite life using both yielding criterions: Modified Goodman and Soderberg. Thus,



   σ=  

where

−

Mc I

0

0

0

0

0

0



   0    0

t 25.4/1000 = = 0.0127 m 2 2 This is a repeated load in compression for bending moment: Thus, c=

Ma =

Mmax − Mmin (0) − (−22346.7) = = 11173.4 N–m 2 2

Mm =

Mmax + Mmin (0) + (−22346.7) = = −11173.4 N–m 2 2

11.5. S–N DIAGRAMS

913

Thus 

  σa =   

−2.83803 × 107

0

0

0

0

0

0





  0   

  σm =   

0

No stress concentration factors are needed. The principal stresses are:

2.83803 × 107

0

0

0

0

0

σ1,a = 0

σ2,a = 0

σ1,m = 2.83803 × 107 Pa

σ2,m = 0 σ3,m = 0

0



  0    0

σ3,a = −2.83803 × 107 Pa

For ductile materials the mean and alternate stresses are q σa = σeq,a = Iσ21,a − 3 Iσ2,a = 2.83803 × 107 Pa σm = σeq,m =

q Iσ21,m − 3 Iσ2,m = 2.83803 × 107 Pa

The maximum and minimum stresses are

σmax = σm + σa = 5.67607 × 107 Pa σmin = σm − σa = 0 The stress range σr is the difference range of the maximum and minimum stresses in the cycle: σr = ∆σ = σmax − σmin = 5.67607 × 107 Pa The stress ratio Rs is the ratio of minimum to maximum stress amplitudes: σmin =0 σmax

Rs =

The amplitude or load ratio Aa is the ratio of the stress amplitude to the mean stress: σa = 1.0 Aa = σm Modified Goodman criterion: Sm > 0

Seq =

Smax < Sy

nSF σa nSF σm 1− Sut

→

→

Seq =

Sa Sm 1− Sut

σa σm 1 = + nSF Seq Sut

For infinite life take Seq = Se , the endurance limit.

11.5. S–N DIAGRAMS

914

Now we need to find the modified Se . First of all at N = 1 life cycle plot S = Sut = 600 MPa. For N = 103 and bending Sf = 0.9 Sut = 540 MPa (for bending f = 0.90). Next wrought steels under pure bending: Se0 = 0.5 Sut

at N = 106 cycles

So for the steel used for this spring board Se0 = 0.5 Sut = 300. MPa The correction factors are: (a) Loading: kL = 1.0 (b) Temperature: kt = 1.0 (c) Surface finish factor: ksr = 1.0 (d) Reliability: 99.999%, kr = 0.659 (e) Gradient Size factor: For rectangular cross-section: p √ d = 0.8 b h = 0.8 3.66(25.4/1000) = 0.243967 m kg = 0.7

(f) Miscellaneous:ke = 1.0 Thus k∞ = 0.4613 Thus the modified endurance limit is Se = k∞ Se0 = 138.39 MPa Thus

1 σa σm = + nSF Seq Sut

→

nSF = 3.96

and the margin of safety is 296%. Soderberg Criterion Seq =

nSF σa nSF σm 1− Sy

1 σa σm = + nSF Seq Sy

→

For infinite life take Seq = Se , the endurance limit. From above the modified endurance limit is Se = 138.39 MPa Thus

1 σa σm = + nSF Seq Sy

and the margin of safety is 262%.

→

nSF = 3.623

11.5. S–N DIAGRAMS

915

2. If the board is to be used during day time (8hr/day, 5 days) and it is expected that on the average a person uses it every 15 minutes. Suggest a warrantee period for a 400% margin of safety. The safety factor is nSF = 5.0: Sa = nSF σa = 141.902 MPa

Sm = nSF σm = 141.902 MPa

Using the modified Goodman criterion: σm > 0

Seq =

→

σmax < Sy

Sa = 185.858 MPa Sm 1− Sut

The modified endurance limit is Se = 138.39 MPa Since Seq > Se , the part has finite life. In order to find the life cycles, we use the equation N=

  1b S a

→

N=



Seq a

 1b

where 1 b= log 3 − ge



Sf Se



a=

ge 3 f Sut Sf = 3 b = (f Sut ) ge −3 (Se ) 3−ge 3b 10 10

Note that we took S = Seq because that is the fatigue stress at which we want to calculate the life cycles for failure. For ge = 6: b=−

1 log 3



f Sut Se



= −0.197096

Thus N=



Seq a

 1b

a=

f Sut = 2.10709 × 109 103 b

= 223969. cycles

The component has 2.24 × 105 life cycles before failure.      1 hour 1 day 1 week 1 year N = 2.24 × 105 cycles = 26.91 years 4 cycle 8 hours 5 days 52 weeks The part has 26.91 years of life. Thus warrantee could be for 25 years. Using the Soderberg criterion: Seq =

Sa = 220.00 MPa Sm 1− Sy

11.5. S–N DIAGRAMS

916

The modified endurance limit is Se = 138.39 MPa Since Seq > Se , the part has finite life. For ge = 6: 1 b = − log 3



f Sut Se



= −0.197096

Thus N=



Seq a

 1b

a=

f Sut = 2.10709 × 109 103 b

= 95365.9 cycles

The component has 9.54 × 105 life cycles before failure.      1 hour 1 day 1 week 1 year 5 N = 9.54 × 10 cycles = 11.46 years 4 cycle 8 hours 5 days 52 weeks Thus warrantee should be for ten years. End Example 

11.6. CUMULATIVE FATIGUE DAMAGE

11.6

917

Cumulative fatigue damage

Machine and structural members are not always subjected to the constant stress cycles. Many parts may be under different severe levels of reversed stress cycles or randomly varying stress levels. Examples include automotive suspension and aircraft structural components operating at stress levels between the fracture strength and endurance limit. When a machine part is to operate for a finite time at higher stress, the cumulative damage must be examined. It is important to note that predicting the cumulative damage of parts stressed above the endurance limit is at best a rough procedure. Clearly, for parts subjected to randomly varying loads, the damage prognosis is further complicated. The simplest and most widely accepted criterion used to explain cumulative damage is know as the Miner’s rule. The Palmgren-Miner cycle-ratio summation rule (known as Miner’s rule) for cumulative fatigue damage is defined as r X ni =c N i i=1 D = Bf

r X ni n1 n2 n3 nr = + + + ··· + Ni N1 N2 N3 Nr i=1

(11.9)

where Bf is the number of block or duty cycles, ni the number of cycles at stress levels σi , and Ni the number of cycles to fail at stress level σi . These life cycles are taken from the appropriate S-N diagram. The Miner’s equation assumes that the damage to the material is directly proportional to the number of cycles at a given stress. The rule also presupposes that the stress sequence does not matter and the rate of damage accumulation at a stress level is independent of the stress history. These have not been completely verified in tests. Sometimes specifications are used in which the right side of Eq. (11.9) is taken as 0.7 ≤ c ≤ 2.2 Failure is then determined by using D=c

onset of failure due to fatigue is predicted

D>c

failure due to fatigue has occurred

D Se

→ N1 = 1.082 × 103

S2eq = 76.0839 ksi > Se

→ N2 = 7.370 × 103

S3eq = 60.8672 ksi > Se

→ N3 = 2.631 × 104

S4eq = 30.4336 ksi < Se

→ N4 = ∞

Thus Bf 1 Bf = n1 n2 n3 n4 + + + N1 N2 N3 N4 =



a=



n2 n3 n4 n1 + + + N1 N2 N3 N4

1 = 742.057 duty cycles 0.00134761



30 seconds 1 duty

Thus there are approximately 6 hours of life.





=1

1 minute 60 seconds



1 hour 60 minutes



= 6.18

End Example 

11.7. REFERENCES

11.7

933

References

Collins, J. A., Mechanical Design of Machine Elements and Machines, 2003, John Wiley and Sons, New York, NY. Hamrock, B. J., Schmid, S. R., and Jacobson, B., Fundamentals of Machine Elements, 2005, Second Edition, Mc-Graw Hill, New York, NY. Juvinall, R. C., and Marsheck, K. A., Fundamentals of Machine Component Design, 2000, John Wiley and Sons, New York, NY. Shigley, J. E., Mischke, C. R., and Budynas, R, G., Mechanical Engineering Design, 2004, Seventh Edition, Mc-Graw Hill, New York, NY. Thomas, G. B., Finney R. L., Weir, M. D., and Giordano F. R., Thomas Calculus, Early Transcendental Update, 2003, Tenth Edition, Addison-Wesley, Massachusetts. Entire book.

11.8. SUGGESTED PROBLEMS

11.8

934

Suggested Problems

Problem 11.1. Consider a two rectangular squared cell with different isotropic materials subject to a torque T .

4 

3

2

Cell 1 

Cell 2 

5 

6

1

4 

3

2

The mechanical properties are:

E12 = E45 = E36 = 3.5 × 106 psi

ν12 = ν45 = ν36 = 0.30

E56 = E61 = 1.5 × 10 psi

ν56 = ν61 = 0.20

E34 = E23 = 2.5 × 10 psi Cell 1  6 6

Cell 2 ν34 = ν23 = 0.25

The geometric properties are:

5 

a12 = a45 = a36 = a t12 = t45 = t36 = t

6

a34 = a56 = a

1

a23 = a61 = 2 a

t34 = t23 =small slit  2t

where a’s are the branch lengths and t’s the branch thicknesses. The torque is: T = 10 kips-in + If the reference modulus E0 is made of Titanium Ti-6Al-4V and a = 10 t, determine the values of a and t for 30 years of life. Consider a 95% survivability. The wing box will be used during 10 hours a day and 5 days a week. During one year four week is used for maintenance; hence during that period no loads are applied. Take a margin of safety of 50%. A 30 second test shows that the wing box is subject to the following sequence of completely reversed torsional stress amplitudes: five 35 ksi torsional stress amplitude; four 25 ksi torsional stress amplitudes; one 20 ksi torsional stress amplitudes; ten 10 ksi torsional stress amplitudes. 

Chapter 12 Structural Stability

Instructional Objectives of Chapter 12 After completing this chapter, the student should be able to: 1. Understand the concept of stability. 2. Perform stability analysis on perfect and imperfect rigid bar structures. 3. Perform stability analysis on beam columns.

When a structure is subjected to loading it can fail because local stresses exceed the maximum allowable stress for the material. There exist, however, another type of failure mode where the entire structure suddenly collapses. The critical value of the applied load that triggers this failure mode primarily depends on the geometry of the structure and the stiffness of the material, not its strength. The study of this catastrophic failure mode is known as the theory of elastic stability. The main objective in stability analysis is to determine whether a system that is perturbed from an equilibrium state will return to that equilibrium steady state. If this is true for small perturbations from equilibrium, then we say that this equilibrium is stable. If a system always returns to that equilibrium, then we say it is globally stable.

935

12.1. CONCEPT OF STABILITY OF EQUILIBRIUM

12.1

936

Concept of Stability of Equilibrium

In order to understand why columns buckle it is necessary to understand the concept of stability. We are interested to study the stability of the equilibrium state.

12.1.1

Buckling

When a structure (subjected usually to compression) undergoes visibly large displacements transverse to the load then it is said to buckle. Buckling may be demonstrated by pressing the opposite edges of a flat sheet of cardboard towards one another. For small loads the process is elastic since buckling displacements disappear when the load is removed. Now consider increasing the load slowly. We are interested in the value of the load, called the critical load, at which buckling occurs. That is, we are interested in when a a sequence of equilibrium stable states as a function of the load, one state for each value of the load, ceases to be stable. Buckling of a structure means 1. Failure due to excessive displacements (loss of structural stiffness), and/or 2. Loss of stability of an equilibrium configuration of the structure.

Definition of Buckling Load The buckling load is the load at which the current equilibrium state of a structural element or structure suddenly changes from stable to unstable, and is, simultaneously, the load at which the equilibrium state suddenly changes from that previously stable configuration to another stable configuration with or without an accompanying large response. Thus, the buckling load is the largest load for which stability of equilibrium of a structural element or structure exists in its original (or previous) equilibrium configuration.

12.1.2

Stability of equilibrium

Stability of equilibrium means that the response of the structure due to a small disturbance from its equilibrium configuration remains small; the smaller the disturbance the smaller the resulting magnitude of the displacement in the response. If a small disturbance causes large displacement, perhaps even theoretically infinite, then the equilibrium state is unstable. Practical structures are stable at no load. Now consider increasing the load slowly. We are interested in the value of the load, called the critical load, at which buckling occurs. That is, we are interested in when a a sequence of equilibrium stable states as a function of the load, one state for each value of the load, ceases to be stable.

12.1. CONCEPT OF STABILITY OF EQUILIBRIUM

937

In general, modern aircraft structures consist of thin sheets attached to slender stiffeners. Thus, buckling of these lightweight members can occur at stresses well below the elastic limit. If buckling occurs before the elastic limit of the material, which is roughly the yield stress of the material, then it is called elastic buckling. If buckling occurs beyond the elastic limit, it is called inelastic buckling, or plastic buckling if the material exhibits plasticity during buckling (mainly metals). Most thin-walled structural components buckle in compression below the elastic limit. Therefore, buckling determines the limit state in compression rather than material yielding. In fact, about 50% of an airplane structure is designed based on buckling constraints. In short, prediction of resistance of compression and shear members to elastic buckling is very important. Note that buckling is a type of failure as is yielding and fracture.

12.1. CONCEPT OF STABILITY OF EQUILIBRIUM

12.1.3

938

Various Equilibrium Configurations

(a)

(b)

(c)

Figure 12.1: Equilibrium states. To illustrate the concept of stability of the equilibrium configuration, let us consider three cases as shown in Fig. 12.1. (a) Stable equilibrium: An equilibrium state or configuration of a structural element, structure, or mechanical system is stable if every small disturbance of the system results only in a small response after which the structure always returns to its original equilibrium state. The simplest example of a stable mechanical system is a rigid blue ball in a valley as in Figure. The ball can be disturbed slightly such as by tapping on it, but the ball always returns to the bottom of the valley. Thus, the ball is in a state of stable equilibrium at the bottom of the valley. The stable equilibrium state is stable with respect to both displacement and/or velocity disturbance. (b) Neutral equilibrium: After the rigid red ball has been slightly disturbed from the equilibrium position, it is still in equilibrium at the displaced position, and there is no tendency either to return to the previous position or to move to some other position. Equilibrium is always satisfied. The neutral equilibrium state is stable with respect to displacement but unstable with respect to velocity disturbance. (c) Unstable equilibrium: An equilibrium state or configuration of a structural element, structure, or mechanical system is unstable if any small disturbance of the system results in a sudden change in deformation mode after which the system does not return to its original equilibrium state. The simplest example of an unstable mechanical system is a rigid green ball precariously perched on the top of a hill as in Figure. If the ball is disturbed slightly (an infinitesimal disturbance suffices), the ball will immediately roll down the hill and will never return to the top of the hill. Thus, the ball is in a state of unstable equilibrium at the top of the hill. The unstable equilibrium state is unstable with respect to both displacement and/or velocity disturbance. All the above represent equilibrium paths (states). Two important concepts exist to study the stability of an equilibrium state:

12.1. CONCEPT OF STABILITY OF EQUILIBRIUM

939

1. System must be in equilibrium. 2. We study the stability of that equilibrium state by given the system a small disturbance.

12.1.4

Methods of stability analysis

Three methods of stability analysis for an equilibrium state exist: 1. Dynamic method. What is the value of the load for which the most general free motion of the perfect system in the vicinity of the equilibrium position ceases to be bounded? 2. Adjacent equilibrium method. What is the value of the load for which the perfect system admits nontrivial equilibrium configurations? 3. Energy method. What is the value of the load for which the potential energy ceases to be positive definite?

Only the dynamic method can be used to study the stability of the perfect system in the vicinity of the equilibrium position of both conservative and non-conservative systems.

12.2. STABILITY OF RIGID BARS

940

12.2

Stability of Rigid Bars

12.2.1

Analysis of a Perfect System

P

Io L q

a kt

q

kt

Initial configuration

Deflected configuration

kt

Infinitesimal de

Figure 12.2: One degree of freedom structural configuration, α = 0. To better understand the concept of buckling as a stability problem and failure by excessive displacements, let us consider the behavior of a simplified one-degree of freedom system. Consider a perfect weightless rigid bar of length L, as shown in Fig. 12.2. The bar is supported by a smooth fixed pin at its base and restrained by a linear elastic torsional spring of constant kt which produces a restoring moment kt θ. The bar-spring system is a perfect system. In a perfect system the rigid bar is perfectly straight, the applied load is perfectly aligned with the bar, and the unstretched position of the spring corresponds to α = 0 exactly. The model has one coordinate θ to describe the configuration of the model under the compressive deadweight1 load P . Since the rigid bar is weightless, the weight of the bar will be neglected. In order to study the stability of the equilibrium state, we need to the equation of motion. We can obtain the equations of motion using the extended principle of virtual work, which is stated as follows: δWext − δWint = −δWiner where Winer is the work done by the inertial forces. For the rigid bar: δWext = P L sin θ δθ

δWint = kt θ δθ

j

δWiner = −I0 θ¨ δθ

1 A deadweight load is a load that remains vertical at all times; i.e., it is specified load that is independent of time t and angle θ.

12.2. STABILITY OF RIGID BARS

941

Thus P L sin θ δθ − kt θ δθ = I0 θ¨ δθ For all θ 6= 0,

→

(

) ¨ P L sin θ − kt θ − I0 θ δθ = 0

P L sin θ − kt θ − I0 θ¨ = 0

Hence the equation of motion for rotation about the fixed hinge is ¨ P L sin θ − kt θ = I0 θ,

θ = θ(t),

t > 0,

−π < θ < π

where I0 is the moment of inertia of the bar about the fixed point and t represents the time.

Static Equilibrium The equilibrium configurations are obtained using static equilibrium. Hence the equilibrium states under the static, downward load P are characterized by θ being time-independent: θ(t) = θ0 Hence P L sin θ0 − kt θ0 = 0,

−π < θ0 < π

or in its dimensionless form pˆ sin θ0 − θ0 = 0, where pˆ =

−π < θ0 < π PL kt

There are two equilibrium paths: First path: Second path:

P1 : θ 0 = 0 P2 : pˆ =

Using L’Hopital’s rule: lim =

θ0 →0



for any P

θ0 sin θ0



θ0 =1 sin θ0

The two equilibrium paths are plotted in the load-deflection as follow

(12.1)

12.2. STABILITY OF RIGID BARS

942

p 2 1

2

2

1 1

θ -p

p

0

Note that the equilibrium path P1 coincides with the load axis in the plot and is called the primary equilibrium path, or the trivial equilibrium path. Equilibrium path P2 is called the secondary equilibrium path and we can see that it is symmetric about θ0 = 0. The two equilibrium paths intersect at (θ0 , pˆ) = (0, 1) This intersection is called a bifurcation point. When there is no load present, the bar is vertical and this corresponds to the origin in the loaddeflection diagram. As the load P slowly increases from zero the bar remains vertical (θ0 = 0) and at pˆ = 1 adjacent equilibrium configurations exist on the secondary path. The existence of adjacent equilibrium configurations in the vicinity of the primary equilibrium path is known as the onset of buckling. Hence, buckling is characterized by the bifurcation point on the load-deflection diagram. For this reason, the term bifurcation buckling is used to describe this condition. As we will show later, the bar will not remain vertical for loads pˆ > 1 if there are infinitesimal disturbances present (there always are), but will rotate either to the left or right depending on type of infinitesimal disturbance. We note that the magnitude of the angle θ0 becomes large as the load is increased from pˆ = 1 on the secondary path. The load at the bifurcation point is called the critical load and is denoted as Pcr . Hence,   kt θ0 pˆcr = 1 → Pcr = L sin θ0 Now, we may obtain the results for linearized buckling, where we assume small angles, sin θ0 ≈ θ0 , and the equilibrium equation becomes P L θ0 − kt θ0 = 0,

−π < θ0 < π

and the two equilibrium paths become First path: Second path:

P1 : θ 0 = 0

for any P

P2 : pˆ = 1

The two equilibrium paths are plotted in the load-deflection as follow

12.2. STABILITY OF RIGID BARS

943

p 2 1

2

2

1 1

θ -p

p

0

Strictly speaking the linearized solution, described above, is only valid for small θ0 , since sin θ0 ≈ θ0 was assumed. When θ0 becomes large, the post-buckling range starts, and the nonlinear approach must be used. However, the bifurcation point is the same for both approaches. From a design standpoint, it is often imperative to keep the applied load well below the critical buckling load as the collapse of the structure under the buckling load is a sudden and catastrophic phenomenon. The buckling load depends on the spring stiffness, kt , and the length of the bar, L. The strength of the system components are irrelevant in this analysis.

Stability Analysis Now we proceed to check the stability of the post-buckled configuration by holding θ0 fixed and giving the system a small perturbation:

Ds P d q(t)

qo

j(t)

kt

Infinitesimal deflected configuration

Hence, let the rotation angle be θ(t) = θ0 + ϕ(t)

(12.2)

where θ0 is time-independent and satisfies the equilibrium equation, i.e., P L sin θ0 − kt θ0 = 0

(12.3)

12.2. STABILITY OF RIGID BARS

944

The perturbation angle is a small angular orientation and a time-dependent quantity. In order words, we are considering small oscillations about an equilibrium state. Now, let us obtain the equations of motion for the perturbation ϕ(t). The equation of motion for rotation about the fixed hinge was derived as I0 θ¨ + kt θ − P L sin θ = 0,

θ = θ(t),

t > 0,

−π < θ < π

(12.4)

Substitute Eq. (12.2) into Eq. (12.4)   I0 θ¨0 + ϕ¨ + kt (θ0 + ϕ) − P L sin (θ0 + ϕ) = 0 Note that θ¨0 = 0. Using the trigonometric identity

sin (θ0 + ϕ) = sin θ0 cos ϕ + cos θ0 sin ϕ the equation of motion becomes n o I0 ϕ¨ + kt θ0 + kt ϕ − P L sin θ0 cos ϕ + cos θ0 sin ϕ = 0

(12.5)

Expand the trigonometric functions of angle ϕ in a Taylor Series about ϕ = 0: 1 3 ϕ + ··· 6 1 cos ϕ = 1 − ϕ2 + · · · 2 sin ϕ = ϕ −

Now arrange Eq. (12.5) in powers of ϕ to obtain n o n o nP L o nP L o I0 ϕ¨ + kt θ0 − P L sin θ0 + kt θ0 − P L cos θ0 ϕ + sin θ0 ϕ2 + cos θ0 ϕ3 + · · · = 0 2 6 (12.6) Note we keep only first order quantities because the perturbation is very small ϕ  1 and higher order terms vanish. Further using Eq. (12.3): n o n o I0 ϕ¨ + kt θ0 − P L sin θ0 + kt − P L cos θ0 ϕ = 0 | {z } =0

Hence, for very small perturbations about the equilibrium configuration, the equation of motion to study the stability of the equilibrium configurations is ϕ¨ +

n k − P L cos θ o t 0 ϕ = 0, I0

ϕ = ϕ(t),

t>0

(12.7)

The above equation is an second order ordinary differential equation and it may be written as follows ϕ¨ + ω 2 ϕ = 0, where ω2 =

ϕ = ϕ(t),

kt kt − P L cos θ0 = I0 I0

t>0

  PL 1− cos θ0 kt

12.2. STABILITY OF RIGID BARS

945

The solution will depend on the values of ω 2 :   > 0, ϕ¨ + ω 2 ϕ = 0     ω2 = = 0, ϕ¨ = 0      < 0, ϕ¨ − ω ¯2 ϕ = 0

ω 2 = −¯ ω2

The solution for each case is




a) When ω 2 > 0: The solution to ϕ¨ + ω 2 ϕ = 0,

ϕ = ϕ(t),

t>0

is ϕ(t) = A1 sin ωt + A2 cos ωt in which the constants A1 and A2 are determined by initial conditions for ϕ(0) and ϕ(0). ˙ The solution is a harmonic oscillation about the equilibrium configuration and ω is the natural frequency of the system. The initial conditions ϕ(0) and ϕ(0) ˙ are considered very small but arbitrary to simulate an arbitrary small initial disturbance.

j

t

The smaller the initial disturbance, the smaller the maximum amplitude of the oscillation in ϕ. Thus, ω 2 > 0 is a condition for stable equilibrium configuration with respect to infinitesimal disturbances. b) When ω 2 = 0: The solution to ϕ¨ = 0,

ϕ = ϕ(t),

t>0

is ϕ(t) = A1 + A2 t in which the constants A1 and A2 are determined by initial conditions for ϕ(0) and ϕ(0). ˙ The solution is a straight line about the equilibrium configuration.

12.2. STABILITY OF RIGID BARS

946

j

t

Thus, ω 2 = 0 is a condition for critical equilibrium configuration with respect to infinitesimal disturbances. c) When ω 2 < 0: The solution to ϕ¨ − ω ¯ 2 ϕ = 0,

ϕ = ϕ(t),

t>0

is ϕ(t) = A1 eω¯ t + A2 e−¯ωt in which the constants A1 and A2 are determined by initial conditions for ϕ(0) and ϕ(0). ˙ The term with positive exponent dominates the solution. This corresponds to large values of ϕ no matter how small the initial disturbance.

j

t

Thus, ω 2 < 0 is a condition for unstable equilibrium configuration with respect to infinitesimal disturbances. Hence an equilibrium state (P, θ0 ) is stable if critical if unstable if

ω2 > 0 ω2 = 0 ω2 < 0

Now, let us determine where does the rigid bar have stable, critical, and unstable configurations, if any. Before we proceed let us nondimensionalize the frequency squared:   I0 2 PL ω = 1− cos θ0 → ω ˆ 2 = (1 − pˆ cos θ0 ) kt kt

12.2. STABILITY OF RIGID BARS

947

The above is not necessary it just helps to understand the physical nature of the stability of the equilibrium path. Let us begin to analyze the stability of the primary path P1 , for which θ0 = 0 for any value of P : P1 : θ 0 = 0 ω ˆ 2 = (1 − pˆ) Hence, P1 :

P <

kt L

critical if ω ˆ 2 = 0 pˆ = 1 → P =

kt L

stable if

P1 : P1 :

unstable if

ω ˆ2 > 0

pˆ < 1

ω ˆ2 < 0

pˆ > 1

→

→ P >

kt L

Now, let us analyze the stability of the secondary path P2 :   θ0 θ0 P2 : pˆ = ω ˆ2 = 1 − sin θ0 tan θ0 Hence,

P2 : P2 : P2 :

stable if

ω ˆ2 > 0

0 < |θ0 | < π

critical if ω ˆ 2 = 0 θ0 = 0 unstable if

ω ˆ2 < 0

no θ0 satisfies this condition

p 2 1

(unstable) 2

2

(stable)

(stable) 1=

1

2

(critical) 1

(stable)

-p

0

p

Figure 12.3: Summary of the primary and secondary path stability.

θ

12.2. STABILITY OF RIGID BARS

948

Thus rigorous stability analysis confirms that the bifurcation point is the onset of buckling. Loss of stability for the system only occurs on path P1 for pˆ > 1. Hence the critical load is the load at the bifurcation point: kt pˆ = 1 → Pcr = L This illustrates why buckling is defined as loss of stability of an equilibrium configuration of the structure.

12.2. STABILITY OF RIGID BARS

12.2.2

949

Analysis of an Imperfect System

P

Io L q

a kt

q

kt

Initial configuration

Deflected configuration

j

kt

Infinitesimal de

Figure 12.4: One degree of freedom structural configuration, α > 0. Consider an imperfect weightless rigid bar of length L, as shown in Fig. 12.4. The bar is supported by a smooth fixed pin at its base and restrained by a linear elastic torsional spring of constant kt which produces a restoring moment kt θ. The model has one coordinate θ to describe the configuration of the model under the compressive deadweight2 load P . Since the rigid bar is weightless, the weight of the bar will be neglected. The system considered previously was a perfect system because the rigid bar was perfectly straight, the applied load is perfectly aligned with the bar, and the unstretched position of the spring corresponds to α = 0 exactly. However, every system contains some level of imperfections: (i) load application, (ii) and/or initially deformed bar (originally stretched). The actual system imperfections are often unknown because they are result of manufacturing inaccuracies or load misalignment. One way to treat this problem is to introduce the imperfection in the system by assuming that the unstretched position of the spring corresponds to θ = α. α is now a measure of the initial imperfection of the system. In this case the equation of motion becomes P L sin(θ) − kt (θ − α) − I0 θ¨ = 0 and thus the equilibrium equation is P L sin(θ0 ) − kt (θ0 − α) = 0 2 A deadweight load is a load that remains vertical at all times; i.e., it is specified load that is independent of time t and angle θ.

12.2. STABILITY OF RIGID BARS

950

The onset of buckling can be determined assuming θ0 to be a small quantity (P L − kt ) θ0 = kt α This equation possesses a non-zero right hand side in contrast with the homogeneous Eq. (12.1) found for the perfect system. Hence it contains a particular solution. The total solution to the ode is is θ0 =

α P 1− Pcr

=

α 1 − pˆ

→

pˆ =

P α =1− Pcr θ0

Load vs. Deflection (θ) 2

α=0: PERFECT SYSTEM

1

(unstable, perfect)

(stable, perfect)

1

α=0.0001

α=0.01 α=0.1

Load

0.8

0.6 α=0.001

αÆ0 0.4

1

(stable, perfect)

1

(the imperfect system has only one equilibrium path)

0.2

0

0

10

20

30

40 50 Deflection, θ [deg]

60

70

80

90

Load vs. Deflection (θ)

Figure 12.5: Linear response for various levels of imperfection. 2

1.6

When small angle approximation is not valid, the dimensionless (stable, response becomes, 1 perfect)

(unstable, perfect) 1.4

P θ0 − α pˆ = = Pcr sin θ0

1.2

Load

Though the critical 1buckling loads obtained for the perfect and imperfect systems are the same, their respective behaviors are quite different for applied loads below the critical buckling load. Indeed, 0.8 comparing the postbuckling plots Fig. 12.3 and 12.5 shows that the perfect system presents no lateral deflection for applied loads below pˆcr , whereas the imperfect system presents lateral deflections even for 0.6 the smallest applied load. In fact, failure can occur for applied loads far smaller that the buckling load. 1 0.4

αÆ0

1

(the imperfect system has only one equilibrium path)

(stable, perfect) 0.2 0 0

10

20

30

40 50 Deflection, θ [deg]

60

70

80

90

1

(stable, perfect)

1

(the imperfect system has only one equilibrium path)

0.2

0 12.2. STABILITY OF RIGID 0 BARS 10

20

30

40 50 Deflection, θ [deg]

60

70

80

90

951

Load vs. Deflection (θ) 2

1.6

(stable, perfect)

1

(unstable, perfect) 1.4 1.2

Load

1 0.8 0.6 1

0.4

αÆ0

1

(the imperfect system has only one equilibrium path)

(stable, perfect) 0.2 0 0

10

20

30

40 50 Deflection, θ [deg]

60

70

80

90

Figure 12.6: Nonlinear response for various levels of imperfection. First, the buckling load was defined as the critical load for which a perfect system admits a nontrivial solution. Second, the buckling load was defined as the load for which the response of an imperfect system grows without bounds. Though conceptually different, these two definitions give the same value of the critical buckling load for the simple rigid bar problem considered here. This illustrates why buckling is also defined as failure due to excessive displacements (loss of structural stiffness).

12.2. STABILITY OF RIGID BARS

952

Example 12.1.

P ks

ks

Mo

mb

Ds ks

Mo

mb

P

Mo

2L a

q

kt

kt

Infinitesimal deflected configu

Figure 12.7: One degree of freedom structural configuration. For the configuration shown in Fig. 12.7: obtain the equations of motion using the principle of virtual work; determine the equilibrium paths; (for each equilibrium path, plot P vs. δ, where P is the vertical load and δ is the vertical tip deflection (in the direction of the load); for each equilibrium path, plot ω 2 vs. P , where P is a deadweight load and ω 2 the square of the natural frequency.; for each equilibrium path, plot ω 2 vs. δ, where δ is the vertical tip deflection (in the direction of the load) and ω 2 the square of the natural frequency; for each equilibrium path, determine the stable and unstable ranges for P . Take: kt = ks L2 , mb = Mo /2, and α = 0◦ . The dimensionless quantities are defined as follows: pˆ =

P ks L

δ δˆ = L

ω ˆ2 =

(12.1-a) Deriving equation for the vertical displacement δ

mb

kt

Deflected configuration

Initial configuration

q

Mo 2 ω ks

12.2. STABILITY OF RIGID BARS

953

P ks

Ds ks

Mo

P d

mb

Mo

q

q

kt

mb

kt

Deflected configuration

Infinitesimal deflected configuration

δ = 2 L (cos α − cos θ) The total spring elongation is r10 = r1 − r0 where the spring deformation in the equilibrium position is r0 = 2 L sin α ˆi and at the deflected position

ri = 2 L sin θ ˆi

Thus the total spring deflection is r10 = r1 − r0 = 2 L (sin θ − sin α) ˆi and the magnitude is ∆s = |r10 | = 2 L (sin θ − sin α) Using the principle of virtual work: δWint − δWext = δWiner (12.1-b) (a) The virtual work done by internal forces is δWint = F1 · δr1 + F2 · δr2 a) The load related the linear spring is F1 = Fs ˆi = ks ∆s ˆi = 2 L ks (sin θ − sin α) ˆi

12.2. STABILITY OF RIGID BARS

954

and the virtual displacement is r1 = r10 = 2 L (sin θ − sin α) ˆi δr1 =

∂r1 δθ ∂θ

(r1 = r1 (θ))

= 2 L cos θ δθ ˆi Thus     F1 · δr1 = 2 L ks (sin θ − sin α) ˆi · 2 L cos θ δθ ˆi = 4 L2 ks cos θ (sin θ − sin α) δθ

b) The load related the torsional spring is

et = kt (θ − α) ˆ et F2 = M t ˆ and the virtual displacement is et r2 = (θ − α) ˆ δr2 =

∂r2 δθ ∂θ

(r2 = r2 (θ))

= δθ ˆ et Thus

    F2 · δr2 = kt (θ − α) ˆ et · δθ ˆ et = kt (θ − α) δθ

Then, the virtual work done by internal forces is δWint = F1 · δr1 + F2 · δr2

= 4 L2 ks cos θ (sin θ − sin α) δθ + kt (θ − α) δθ   = 4 L2 ks cos θ (sin θ − sin α) + kt (θ − α) δθ

(b) The virtual work done by external forces is

δWext = Q1 · δq1 The point load is

Q1 = −P ˆj

12.2. STABILITY OF RIGID BARS

955

and the virtual displacement is q1 = 2 L sin θ ˆi + 2 L cos θ ˆj δq1 =

∂q1 ∂θ

δθ



 q1 = q1 (θ)

= 2 L cos θ δθ ˆi − 2 L sin θ δθ ˆj Thus Q1 · δq1 =



   − P ˆj · 2 L cos θ δθ ˆi − 2 L sin θ δθ ˆj = 2 P L sin θ δθ

(c) The virtual work done by inertial forces is

δWiner = −I0 θ¨ δθ (d) Thus applying the principle of virtual work: δWint − δWext = δWiner 

where

 4 L2 ks cos θ (sin θ − sin α) + kt (θ − α) δθ − 2 P L sin θ δθ = −I0 θ¨ δθ

4 I0 = 4 Mo L + mb L2 = 4 Mo L2 3 2



mb 1+ 3 Mo



= 4 Mo L

2



1 1+ 6



=

14 Mo L 2 3

Rewriting the PVW   4 L2 ks cos θ (sin θ − sin α) + kt (θ − α) − 2 P L sin θ + I0 θ¨ δθ = 0 For all δθ 6= 0:

4 L ks cos θ (sin θ − sin α) + kt (θ − α) − 2 P L sin θ + 4 Mo L 2

2



mb 1+ 3 Mo



θ¨ = 0

Rearranging and substituting the values for mb and kt :   mb 2 θ¨ − 4 L2 ks cos θ (sin θ − sin α) − kt (θ − α) + 2 P L sin θ = 0 −4 Mo L 1+ 3 Mo −θ¨ −

4 L2 ks k   cos θ (sin θ − sin α) − t  (θ − α) m mb b 4 Mo L2 1 + 4 Mo L2 1 + 3 Mo 3 Mo 2P L   sin θ = 0 + mb 4 Mo L2 1 + 3 Mo

12.2. STABILITY OF RIGID BARS

−θ¨ −

956

  1 k P  s  cos θ (sin θ − sin α) + (θ − α) − sin θ = 0 mb 4 2 ks L Mo 1 + 3 Mo   3 3P ks 6 ¨ cos θ (sin θ − sin α) + (θ − α) − sin θ = 0 −θ − Mo 7 14 7 ks L

In terms of the dimensionless quantities: −θ¨ −

ω2 ω ˆ2



3 3 6 cos θ (sin θ − sin α) + (θ − α) − pˆ sin θ 7 14 7



=0

(12.1-c) To find the equilibrium path, we solve the static problem (θ¨ = 0): ω2 −θ¨ − 2 ω ˆ



6 3 3 cos θ (sin θ − sin α) + (θ − α) − pˆ sin θ 7 14 7



=0

6 3 3 cos θ (sin θ − sin α) + (θ − α) − pˆ sin θ = 0 7 14 7 

6 cos θ 7



sin θ − sin α θ−α



  3 3 sin θ + − pˆ (θ − α) = 0 14 7 θ−α

Thus there are two equilibrium paths: First path:

P1 :

(θ − α) = 0

Second path:

P2 :



6 cos θ 7



θo = α sin θ − sin α θ−α

2 cos θ



First path: Second path:

P1 :

+



P2 : pˆ = 2 cos θ0 +

θ → 1, sin θ Thus the secondary path, at bifurcation, has value of ⇒

P2 : pˆ =



  1 sin θ + − pˆ =0 2 θ−α

sin θ0 − sin α sin θ0

(θ − α) = 0

At bifurcation:

θ → 0◦

  3 3 sin θ − pˆ =0 14 7 θ−α

sin θ − sin α θ−α

pˆ = 2 cos θ0 For α = 0◦ :





1 + 2



θ0 − α sin θ0



θo = 0 1 2



θ0 sin θ0



cos θ → 1

5 2

(12.1-d) In order to study the stability using the dynamic criterion, lets perturb the system by

12.2. STABILITY OF RIGID BARS

957

an infinitesimal rotation: θ = θ0 + ϕ Thus ω ˆ2 ¨ 6 3 3 θ − cos θ (sin θ − sin α) − (θ − α) + pˆ sin θ = 0 7 14 7 ω2  6 3 3 ω ˆ2 ¨ (θ0 − α) + pˆ sin (θ0 + ϕ) = 0 − 2 θ0 + ϕ¨ − cos (θ0 + ϕ) (sin (θ0 + ϕ) − sin α) − 7 14 7 ω −

Rearranging the above: −

 7 ω 1 ˆ2 ¨ 1 θ + ϕ ¨ − cos (θ0 + ϕ) (sin (θ0 + ϕ) − sin α) − (θ0 − α) + pˆ sin (θ0 + ϕ) = 0 0 2 6 ω 4 2

Initially, the system does not vary with time:

θ¨0 = 0 Expanding about θ0 , the trigonometric quantities are written as follows: sin (θ0 + ϕ) = sin θ0 + cos θ0 ϕ + · · ·

cos (θ0 + ϕ) = cos θ0 − sin θ0 ϕ + · · · Note we keep only first order quantities because the perturbation is very small ϕ  1 and higher order terms vanish. Now substituting these equations into the equation of motion: o 1n (θ0 + ϕ) − α − cos (θ0 + ϕ) (sin (θ0 + ϕ) − sin α) − 4    ω ˆ2 7 ¨ pˆ θ0 + ϕ¨ = 0 + sin (θ0 + ϕ) − 2 2 6 ω n on o 1n o − cos θ0 − sin θ0 ϕ + · · · sin θ0 + cos θ0 ϕ + · · · − sin α − (θ0 + ϕ) − α 4   n o ω ˆ2 7 pˆ sin θ0 + cos θ0 ϕ + · · · − 2 ϕ¨ = 0 2 6 ω Expanding and keeping only linear terms in ϕ, i.e., ϕ(0) and ϕ(1) , we get n o − cos θ0 sin θ0 + cos θ0 cos θ0 ϕ − cos θ0 sin α − sin θ0 sin θ0 ϕ + sin θ0 sin α ϕ + · · ·   o pˆ n o ω 1n ˆ2 7 − (θ0 + ϕ) − α + sin θ0 + cos θ0 ϕ + · · · − 2 ϕ¨ = 0 4 2 6 ω n o n o − cos θ0 sin θ0 − cos θ0 sin α − cos θ0 cos θ0 − sin θ0 sin θ0 + sin θ0 sin α ϕ   o 1 o pˆ n o 1n pˆ n ω ˆ2 7 − θ0 − α − ϕ + sin θ0 + cos θ0 ϕ − 2 ϕ¨ = 0 4 4 2 2 6 ω

12.2. STABILITY OF RIGID BARS

958

Combining all terms: o pˆ n o 1n θ0 − α + sin θ0 − cos θ0 sin θ0 − cos θ0 sin α − 4 2   o n ω ˆ2 7 1 pˆ ϕ¨ = 0 − cos θ0 cos θ0 − sin θ0 sin θ0 + sin θ0 sin α + − cos θ0 ϕ − 2 4 2 6 ω Note that from the equilibrium path: n o 1n o pˆ − cos θ0 sin θ0 − cos θ0 sin α − θ0 − α + sin θ0 = 0 4 2

Thus, the linearized equation of motion to study the dynamic stability for the given system is n

o 1 pˆ ω ˆ2 − cos θ0 cos θ0 − sin θ0 sin θ0 + sin θ0 sin α + − cos θ0 ϕ − 2 4 2 ω

  7 ϕ¨ = 0 6

or substituting all values (including α = 0◦ ):   n o 1 pˆ ω ˆ2 7 ϕ¨ = 0 − cos θ0 cos θ0 − sin θ0 sin θ0 + − cos θ0 ϕ − 2 4 2 6 ω 6 ω 2 n pˆ 1o −ϕ¨ + cos θ0 − cos(2 θ0 ) − ϕ=0 2 2 4 7ω ˆ 1o 6 ω 2 n pˆ ϕ¨ + − cos θ0 + cos(2 θ0 ) + ϕ=0 2 2 4 7ω ˆ

The equation of motion to study the free vibrations of a system is given by ϕ¨ + ω 2 ϕ = 0

→

ϕ¨ +

ω2 2 ω ˆ ϕ=0 ω ˆ2

Thus, ω ˆ2 = −

o 1o 6 n pˆ 3 n cos θ0 − cos(2 θ0 ) − =− 2 pˆ cos θ0 − 4 cos(2 θ0 ) − 1 7 2 4 14

(12.1-e) To plot the deflection, note that from the system express P in terms of δ. From the secondary path:    θ0 1 4 cos θ0 + pˆ = 2 sin θ0 and

δ = 2 L (cos α − cos θ) δˆ = 2 − 2 cos θ0

→

δ = 2 cos α − 2 cos θ L

(α = 0◦ )

The above is a parametric plot. Now we plot δˆ and pˆ and not θ. Now we can substitute the above into the equation for frequency and do the same. In order to plot, ω 2 vs. δ, we substitute the expression for load into the expression for

12.2. STABILITY OF RIGID BARS

959

frequency squared to get:     3 2 θ0 3 {2 pˆ cos θ0 − 4 cos(2 θ0 ) − 1} = − 4 cos θ0 + cos θ0 − 4 cos(2 θ0 ) − 1 14 14 2 sin θ0     3 θ0 = −4 cos2 θ0 − cos θ0 + 4 cos(2 θ0 ) + 1 14 sin θ0

ω ˆ2 = −

Load vs. Angle 5 4

1

2

(stable)

3

1= 2 (critical)

2

(unstable)

(stable)

A

2 1 Load

2 2

(unstable)

0 -1

(unstable)

2

2

(critical)

(critical)

-2 -3

1

Point A is the bifurcation point

-4 -5

5 4 3

Frequency Squared

2 1 0 -1 -2

(stable)

-3

-2

-1

0 Angle

1

2

Frequency Squared Angle path stability. Figure 12.8: Summary of the primary and vs. secondary

3

12.2. STABILITY OF RIGID BARS

960

Frequency Squared vs. Deflection 5 4

CRITICAL ω2=0

3

Frequency Squared

2

UNSTABLE ω20

-1 -2 -3 -4 -5

0

0.5

1

1.5

2 Deflection

2.5

3

3.5

4

Load vs. Deflection 5 4 3

UNSTABLE ω20

-1 -2

CRITICAL ω2=0

-3 -4 -5

0

0.5

1

1.5

2 Deflection

2.5

3

3.5

4

Figure 12.9: Load and frequency plots against vertical deflection. As shown in Fig. 12.9, the system becomes unstable when we the natural frequency squared is negative; critical when it is zero and; stable otherwise. End Example 

12.3. STABILITY OF BEAM-COLUMNS

961

12.3

Stability of Beam-Columns

12.3.1

Perfect Beam-Columns (Adjacent Equilibrium Method)

Consider the equilibrium of the perfect column: The column is straight and subject to centric static compressive force P . The beam column obeys the Euler-Bernoulli Theory, is homogeneous, isotropic, and of uniform cross-section. Hence, at equilibrium: u0 (x) = −

P x, EA

v0 (x) = 0,

x ∈ (0, L)

Now let us study the stability of the equilibrium state by the method of adjacent equilibrium. Consider infinitesimal variation to the displacements represented by subscript “1” ( e.g., v1 (x) = δv(x)). If the column is fixed at x = 0 u(x)

=

u0 (x) + u1 (x)

u(x)

= u0 (x)

v(x)

= v0 (x) + v1 (x)

v(x)

= v1 (x)

⇒

u(0) = 0

∀ x ∈ (0, L)

u1 (x) = 0

⇒

v0 (x) = 0

∀ x ∈ (0, L)

The equilibrium of an element at a distance x is:

Vy+ dVy

Mzz+ dMzz

A

Mzz

Vy

P

dv

P

dx

Sum of force in the y direction gives X + ↑ Fy = 0, Now divide by dx and let dx → ∞,

→

−Vy + (Vy + dVy1 ) = 0

dVy1 =0 dx

Take moments at A +



X

MA = 0,

→

Vy1 dx − Mzz1 + Mzz1 + dMzz1 + P dv1 = 0

(12.8)

12.3. STABILITY OF BEAM-COLUMNS

962

Now divide by dx and let dx → ∞, dMzz1 dv1 + Vy1 + P =0 dx dx

(12.9)

Differentiate Eq. (12.9) with respect to x once d2 Mzz1 dVy1 d2 v1 + + 2 P =0 2 dx | dx {z } dx =0

Now use Eq. (12.8) to get

d2 Mzz1 d2 v1 + P =0 dx2 dx2 The material law is Mzz1 = −EIzz

(12.10)

d(−v10 ) dθz1 = −EIzz = EIzz v100 dx dx

(12.11)

Substituting Eq. (12.11) into Eq. (12.10) we get the governing ordinary differential equation for buckling: d2 dx2

  d 2 v1 d2 v1 =0 EIzz +P 2 dx dx2

v1 = v1 (x)

x ∈ (0, L)

(12.12)

For a column with EIzz = constant, we can re-write this differential equation as EIzz v10000 + P v100 =

0

⇒

v10000 +

P v 00 = 0 EIzz 1 | {z } λ2

Hence the homogeneous forth order ordinary differential equation to obtain the buckling load is v10000 + λ2 v100 =

0

v1 = v1 (x)

x ∈ (0, L),

λ2 =

P EIzz

(12.13)

General solution for λ2 > 0 v1 (x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4

(12.14)

and the buckling load is obtained from Eq. (12.13): λ2 =

P EIzz

→

P = λ2 EIzz

The constants are found by applying the boundary conditions. Before we proceed, the following will be needed v10 (x) = A1 λ cos(λ x) − A2 λ sin(λ x) + A3 v100 (x) = −A1 λ2 sin(λ x) − A2 λ2 cos(λ x)

v1000 (x) = −A1 λ3 cos(λ x) + A2 λ3 sin(λ x)

From the Hooke’s law Eq. (12.11)

Mzz1 = EIzz v100 (x) = −EIzz [A1 λ2 sin(λ x) + A2 λ2 cos(λ x)]

12.3. STABILITY OF BEAM-COLUMNS

963

From equilibrium Eq. (12.9) 0 Vy1 = −Mzz − P v10 = −EIzz [v1000 + λ2 v10 ] = −EIzz [A3 λ2 ] 1

Boundary conditions depend from problem to problem. However, there are three standard boundary condition evaluated at the boundary: PINNED: FREE : CLAMPED:

v1 = 0

and

Mzz1 = 0

Mzz1 = 0

and

Vy1 = 0

v1 = 0

and

v10 = 0

12.3. STABILITY OF BEAM-COLUMNS

964

Example 12.2.

y, v P x, u L Figure 12.10: A simply-supported beam column subject to an axial load. The uniform column with bending stiffness EIzz , shown in Fig. 12.10, is pinned at x = 0 and pinned at x = L. Determine the critical load Pcr and the associated buckling mode shape. (12.2-a) Perturb the system from its equilibrium state. This leads to the ordinary differential equation given by Eq. (12.13): v10000 + λ2 v100 =

0

v1 = v1 (x)

x ∈ (0, L),

λ2 =

The general solution for λ2 > 0 is v1 (x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4 (12.2-b) Now apply the boundary conditions: Pinned at x = 0: v1 (0) = 0 and Mzz1 (0) = EIzz v100 (0) = 0 Pinned at x = L: v1 (L) = 0 and Mzz1 (L) = EIzz v100 (L) = 0

P EIzz

12.3. STABILITY OF BEAM-COLUMNS

Thus

965

v1 (0) = A2 + A4 = 0 v100 (0) = −A2 λ2 = 0 v1 (L) = A1 sin(λ L) + A2 cos(λ L) + A3 L + A4 = 0 v100 (L) = −A1 λ2 sin(λ L) − A2 λ2 cos(λ L) = 0

Writing the boundary conditions in a matrix form in terms of the unknown coefficients A1 , A2 , A3 , A4         |

    A1   0                       2  0 −λ 0 0   A2   0   =       −λ2 sin(λ L) −λ2 cos(λ L) 0 0   0   A3                     sin(λ L) cos(λ L) L 1 A4 0 {z } C 0

1

0

1

(12.15)

(12.2-c) Obtain the characteristic equation.

Non-trivial solution (A 6= 0) requires det[C] =0. The determinant will lead to the characteristic equation: λ4 L sin(λ L) = 0 (12.16) (12.2-d) Obtain the buckling load. We proceed to solve the characteristic equation for λ. λ=0

⇒

P =0

sin(λ L) = 0

⇒

λL = mπ

⇒

Pm =

λm =

mπ L

The loads is obtained from:

leads to trivial solution

h m π i2 L

EIzz

m2 π 2 EIzz L2 Thus critical load is for a simply-supported beam is Pm =

Pcr = P1 =

π 2 EIzz L2

(12.2-e) Obtain the buckling mode shape. The buckling mode shapes associated with Pcr . Plug-in the value for λm in Eq. (12.15)

12.3. STABILITY OF BEAM-COLUMNS

966

to determine the coefficients. Recall that for our problem sin(λm L) = 0 

0

  0     0  0

Therefore the coefficient are row 2 : A2 = 0

1 −λ2m −λ2m 1

    A1   0                        0 0   A2   0   =   0   A3     0 0                      L 1 A4 0 0

1

row 1 : A4 = 0

(12.17)

row 3 : A3 = 0

Note that A1 remains indeterminate. Hence the buckling mode shape associated with Pcr is π x v1 (x) = sin A1 L End Example 

12.3. STABILITY OF BEAM-COLUMNS

967

Example 12.3.

y, v

P O x, u L Figure 12.11: Cantilevered beam column subject to an axial load. The uniform column with bending stiffness EIzz , shown in Fig. 12.11, is clamped at x = 0 and free at x = L. Determine the critical load Pcr and the associated buckling mode shape. (12.3-a) Perturb the system from its equilibrium state. This leads to the ordinary differential equation given by Eq. (12.13): v10000 + λ2 v100 =

0

v1 = v1 (x)

x ∈ (0, L),

λ2 =

The general solution for λ2 > 0 is v1 (x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4 (12.3-b) Now apply the boundary conditions: Clamped at x = 0: v1 (0) = 0 and v10 (0) = 0 Free at x = L: Mzz1 (L) = 0 ⇒ v100 (L) = 0 and Vy1 (L) = 0

P EIzz

12.3. STABILITY OF BEAM-COLUMNS

Thus

968

v1 (0) = A2 + A4 = 0 v10 (0) = A1 λ + A3 = 0 v100 (L) = −A1 λ2 sin(λ L) − A2 λ2 cos(λ L) = 0 Vy1 (L) = A3 λ2 = 0

Writing the boundary conditions in a matrix form in terms of the unknown coefficients A1 , A2 , A3 , A4 

0

1

  λ 0     −λ2 sin(λ L) −λ2 cos(λ L)  |

0

{z C

    A1   0                        0   A2   0   =      0  0   A3                     0 A4 0 }

0

0

1

1 0 λ2

(12.3-c) Obtain the characteristic equation.

(12.18)

Non-trivial solution (A 6= 0) requires det[C] =0. The determinant will lead to the characteristic equation: λ5 cos(λ L) = 0 (12.19) (12.3-d) Obtain the buckling load. We proceed to solve the characteristic equation for λ. λ=0

⇒

cos(λ L) = 0

⇒

λ L = (2 m − 1)

⇒

h π i2 Pm = (2 m − 1) EIzz 2L

λm = (2 m − 1)

π 2L

The loads is obtained from: Pm =

P =0

leads to trivial solution π 2

(2 m − 1)2 π 2 EIzz 4 L2

Thus critical load is for a cantilevered beam is Pcr = P1 =

π 2 EIzz 4 L2

(12.3-e) Obtain the buckling mode shape. The buckling mode shapes associated with Pcr . Plug-in the value for λm in Eq. (12.18)

12.3. STABILITY OF BEAM-COLUMNS

969

to determine the coefficients. Recall that for our problem cos(λm L) = 0 

0

1

  λm 0     −λ2m sin(λm L) 0  0

Therefore the coefficient are row 3 : A1 = 0

0 1 0

0 λ2m

    A1   0                        0   A2   0   =   0   A3     0                      0 A4 0 1

row 2 : A3 = 0

(12.20)

row 1 : A2 = −A4

Now the mode shapes are obtained by substituting these values into v1 (x) = A1 sin(λm x) + A2 cos(λm x) + A3 x + A4 v1 (x) = −A4 cos(λm x) + A4    (2m − 1) π x v1 (x) = A4 1 − cos 2L Note that A4 remains indeterminate. Hence the buckling mode shape associated with Pcr is π x v1 (x) = 1 − cos A4 2L End Example 

12.3. STABILITY OF BEAM-COLUMNS

970

Example 12.4.

y, v P O

kt

x, u

ks

L Figure 12.12: A clamped-spring supported beam column subject to an axial load. The uniform column with bending stiffness EIzz , shown in Fig. 12.12, is clamped at x = 0 and pinned to extensional and torsional springs at x = L. The linear extensional spring has a stiffness α and is unstretched when v1 (L) = 0. The linear torsional spring has a stiffness k and is unstretched when v10 (L) = 0. Determine the critical load Pcr and the associated buckling mode shape. Take: ks = α =

EIzz , L3

kt = k =

EIzz L

(12.4-a) Perturb the system from its equilibrium state. This leads to the ordinary differential equation given by Eq. (12.13): v10000 + λ2 v100 =

0

v1 = v1 (x)

x ∈ (0, L),

λ2 =

P EIzz

The general solution for λ2 > 0 is v1 (x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4 (12.4-b) Now apply the boundary conditions: Clamped at x = 0: v1 (0) = 0 and v10 (0) = 0 If Free at x = L: Mzz1 (L) = 0 ⇒ v100 (L) = 0 and Vy1 (L) = 0. However the springs change the boundary conditions. Replacing the springs by their loads and drawing the free body diagram:

12.3. STABILITY OF BEAM-COLUMNS

971

P Ms P Mzz

θz = v′ v

Fs

Vy

x x=0

x=L

Pinned with a linear linear spring at x = L (Vy1 + Fs )

=0

x=L

Pinned with a linear torsional spring at x = L =0 (Mzz1 − Ms ) x=L

Clamped at x = 0: v1 (0) = 0 and v10 (0) = 0

v1 (0) = A2 + A4 = 0 v10 (0) = A1 λ + A3 = 0 Extensional spring’s force is Fs = α v. Now, pinned with a linear extensional spring at x = L: (Vy1 + Fs ) =0 x=L


−EIzz A3 λ2 + α v

=0 x=L

−EIzz A3 λ2 + α A1 sin(λ L) + α A2 cos(λ L) + α A3 L + α A4 = 0   α A1 sin(λ L) + α A2 cos(λ L) + α L − EIzz λ2 A3 + α A4 = 0

Torsional spring’s force is Ms = k θ = k v 0 . Now, pinned with a linear torsional spring

12.3. STABILITY OF BEAM-COLUMNS

972

at x = L: (Mzz1

− Ms )

(EIzz v − k v ) 00

=0 x=L

=0

0

x=L

−EIzz A1 λ2 sin(λ L) − EIzz A2 λ2 cos(λ L) − k A1 λ cos(λ L) + k A2 λ sin(λ L) − k A3 = 0     −EIzz λ2 sin(λ L) − k λ cos(λ L) A1 + −EIzz λ2 cos(λ L) + k λ sin(λ L) A2 − k A3 = 0

Writing the boundary conditions in a matrix form in terms of the unknown coefficients A1 , A2 , A3 , A4 

0

  λ       −EIzz λ2 sin(λ L) − k λ cos(λ L)  |

α sin(λ L)

1

0

0

1

  −EIzz λ2 cos(λ L) + k λ sin(λ L)

−k

{z [C]

α cos(λ L)

  A1              A2 

(λ L)4 cos(λ L) − (λ L)3 sin(λ L) − (λ L)2 cos(λ L) + 2 (λ L) sin(λ L) + 2 cos(λ L) − 2 = 0 We proceed to solve the characteristic equation for λ. λL = 0 Using a numerical solver:

⇒

P =0

leads to trivial solution

λ1 L = 0.862312 λ1 =

0.862312 L

}

  0              0 

=  0       A 3                      0 A4 (12.21)

Non-trivial solution (A 6= 0) requires det[C] =0. The determinant will lead to the characteristic equation:

(12.4-d) Obtain the buckling load.



 0     0  

[α L − EIzz λ2 ] α

×

(12.4-c) Obtain the characteristic equation.

1

12.3. STABILITY OF BEAM-COLUMNS

Hence

973



0.862312 P1 = L

The loads is obtained from:

P1 = 0.743582

2

EIzz

4 π 2 EIzz EIzz 2 = 0.0188351 L L2

Thus critical load is for a cantilevered beam is Pcr = P1 = 0.743582

EIzz L2

(12.4-e) Obtain the buckling mode shape. The buckling mode shapes associated with Pcr . to determine the coefficients.  0 1 0   0.862312/L 0 1     0.759349 0.650684 0.256418 L  −1.12573

0.170959

−L

Therefore the coefficient are

A2 = 1.54083 A1 ,

Plug-in the value for λm in Eq. (12.21)     A1   0                        0   A2   0   =   0      A3  1                      A4 0 0

1

A4 = −1.54083 A1 ,

A3 = −

(12.22)

0.862312 A1 L

Now the mode shapes are obtained by substituting these values into v1 (x) = A1 sin(λ1 x) + A2 cos(λ1 x) + A3 x + A4 v1 (x) = A1 sin



   0.862312 0.862312 0.862312 x + 1.54083 A1 cos x − A1 x − 1.54083 A1 L L L

Note that A1 remains indeterminate. Hence the buckling mode shape associated with Pcr is     0.862312 0.862312 x v1 (x) = sin x + 1.54083 cos x − 0.862312 − 1.54083 A1 L L L (12.4-f) What does the buckling load represent? The buckling load is the largest load for which the stability of the equilibrium of a structure exists in its original equilibrium configuration. (12.4-g) What would be the buckling load of the column shown if α and k go to infinity? When α and k go to infinity, the boundaries at x = L become rigid. Meaning, v(L) = 0 and v 0 (L) = 0. These are same as for those for a clamped boundary condition.

12.3. STABILITY OF BEAM-COLUMNS

974

Therefore, the buckling load for a clamped-clamped column is: Pcr =

4 π 2 EIzz L2

End Example 

12.3. STABILITY OF BEAM-COLUMNS

12.3.2

975

Several Type of Column End Constraint

If the column ends are not pinned, the critical load and stress would be different from the pinned-pinned case. Often we use the critical load from the pinned-pinned case and substitute the actual length for an effective length: π 2 EIzz (12.23) Pcr = L2e The effective length Le of any column is defined as the length of a pinned-pinned column that would buckle at the same critical load as the actual column: that is the effective length is the overall column length minus the portion that takes into account the end conditions. Thus end conditions affect the effective length of the column. The critical load and critical stresses can be expressed in terms the effective length Le depending on the boundary conditions and is its values can be found in Table 12.1: (12.24)

Le = CL L

Table 12.1: Effective length coefficient CL for several type of column end constraints Theoretical

Recommended∗

1

1

pinned-fixed

0.7

0.8

fixed-fixed

0.5

0.65

fixed-free

2

2.1

Column End Conditions pinned-pinned

∗

Recommended effective column length by AISC (American Institute of Steel Construction, 1989)

Commonly two different equations are widely used, depending on the slender ratio of the beam. Let us define the critical slender ratio Rc as s s L 2 E π2 1 2 E π2 (12.25) Rc = = Le Sy CL Sy where E is the Young’s Modulus and Sy is the yield strength of the material. The actual effective slender ratio Ra of the column is: Le Ra = rg In the above equation rg is the radius of gyration and is defined as rg =

r

Izz A

where Izz is the area moment of inertia and A the cross-sectional area. The critical buckling stress will depend on the slender ratio of the column. Three cases need to be considered:

12.3. STABILITY OF BEAM-COLUMNS

976

Long Columns: Euler equation. If the slender ratio Ra ≥ Rc then the column will buckle elastically, and the Euler equation can be used to obtain the stress at buckling: σcr =

π2 E Pcr =  2 = E A Le rg



π Ra

2

(12.26)

Intermediate Columns: J. B. Johnson Equation. If the slender ratio 10 ≤ Ra ≤ Rc then the column will buckle inelastically, and the J. B. Johnson equation should be used to obtain the critical stress at buckling: σcr = Sy −

Sy2 4 π2 E



Le rg

2

= Sy −

For design we take σmax =

1 E



Sy Ra 2π

2

(12.27)

Sy 2

Short Columns: Yielding. If the slender ratio Ra < 10, in practice it is assumed that the column will fail due to yielding and the critical stress at buckling would be: σcr = Sy However, one may use the J. B. Johnson equation as well. For design we take σmax = Sy The critical load can be obtained by multiplying the stress by the cross-sectional area Pcr = σcr A

(12.28)

12.3. STABILITY OF BEAM-COLUMNS

977

Example 12.5. A column with one end fixed and the other end pinned is made of low-carbon steel. The column’s cross-section is rectangular with h = 0.5 in and b = 1.5 in. Determine the buckling load for the following three lengths: (a) L =0.5 ft (b) L =2.25 ft (c) L =4.0 ft Using the corresponding tables, for low-carbon steel, the material properties are: Sy = 43 × 103

psi E = 30 × 106

psi

The important parameters are: A = b h = 0.75

in2

Izz =

b h3 = 0.0156 12

rg =

r

CL =

L = 0.8 Le

L Rc = Le

Izz = 0.1442 A

s

Rectangular cross sectional area in4

Area moment of inertia

in

Radius of gyration

Effective length coefficient

2 E π2 = 146.75 Sy

Critical slender ratio

Now we proceed to solve the problem. (a) L = 0.5 ft First calculate the effective length: (we use the recommended effective length) Le = 0.8 L = 0.8 (6 in) = 4.8

in

Now evaluate the actual effective slender ratio Ra =

Le = 33.28 rg

Since 10 ≤ Ra < Rc , the J. B. Johnson equation should be used: (  2 ) 1 Sy Ra Pcr = A Sy − = 30950 lb E 2π

12.3. STABILITY OF BEAM-COLUMNS

978

(b) L = 2.25 ft First calculate the effective length: (we use the recommended effective length) Le = 0.8 L = 0.8 (27.0 in) = 21.60 in Now evaluate the actual effective slender ratio Ra =

Le = 149.792 rg

Since Ra > Rc , the Euler equation should be used: (   ) 2 π Pcr = A E = 9915.93 Ra

lb

(c) L = 4.0 ft First calculate the effective length: (we use the recommended effective length) Le = 0.8 L = 0.8 (48.0 in) = 38.4

in

Now evaluate the actual effective slender ratio Ra =

Le = 266.30 rg

Since Ra > Rc , the Euler equation should be used: (   ) 2 π Pcr = A E = 3132 Ra

lb

Observe that in part (a) the column was short, so that yield strength predominated and the J. B. Johnson equation was used; but in parts (b) and (c) the columns were long enough that the Euler equation was used. Also note as the column becomes larger, the critical load decreases significantly. End Example 

12.3. STABILITY OF BEAM-COLUMNS

979

Example 12.6.

An engineer is asked to design a safe round tubular column subject to static axial loads. The column has a length L an outside diameter do and inside diameter di . The diametral ratio is: di α= do Determine the outside diameter do such that failure of the tube is by buckling. First note that

Pcr P

→

Pcr = nSF P

σcr nSF

→

σcr = nSF σ

nSF = or in terms of stress

σ=

where the load and stresses are related as follows Pcr = σcr A Also, let

Le = CL → Le = C L L L where CL refers to the recommended value for the effective length to take into account several type of column end constraints. For our problem M S = 1.50

→

nSF = M S + 1 = 2.5

It is known that A= I=

rg2 = Thus


π
π 2 do − d2i = d2o 1 − α2 4 4





π 4 π 4 π 4 d − d4i = d 1 − α4 = d 1 − α2 1 + α2 64 o 64 o 64 o
I d2 = o 1 + α2 A 16 Ra2 =



Le rg

2

=

CL2 L2
d2o 1 + α2 16

=

16 CL2 L2
d2o 1 + α2

12.3. STABILITY OF BEAM-COLUMNS

980

The stress can be expressed as σcr = nSF σ = nSF

4 nSF P P P
= = nSF π
2 A π do 1 − α2 d2o 1 − α2 4

1. Assuming the actual slender ratio is smaller than the critical slender ratio [10pts] When Ra ≤ 10 the column is a short column, we can use: σcr = Sy Thus

σcr = Sy 4 nSF P
= Sy 1 − α2

π d2o Rearranging and solving for do d2o =

do =

4 nSF P
π 1 − α2 Sy

s

4 nSF P
π 1 − α2 Sy

When 10 ≤ Ra ≤ Rc the column is an intermediate column, we must use the J. B. Johnson Equation: Sy2 σcr = Sy − 4 π2 E



Le rg

2

Thus σcr = Sy −

1 = Sy − E Sy2 4 π2 E





Le rg

Sy Ra 2π

2

2

Sy2 16 CL2 L2 4 nSF P

2 = Sy − 2 2 1−α 4 π E do 1 + α2

π d2o Rearranging

Solving for do

4 Sy2 CL2 L2 4 nSF P 2

2 = do Sy − 2 π 1−α π E 1 + α2 d2o =

do =

4 nSF P 4 Sy CL2 L2

+ 2 2 π 1 − α Sy π E 1 + α2

s

4 nSF P 4 Sy CL2 L2

+ π 1 − α2 Sy π 2 E 1 + α2

2. Assuming the actual slender ratio is bigger than the critical slender ratio [10pts]

12.3. STABILITY OF BEAM-COLUMNS

981

When Ra ≥ Rc the column is a long column, we must use the Euler Equation: π2 E Pcr =  2 = E σcr = A Le rg Thus



π Ra

2

π2 E σcr =  2 Le rg π2 E 4 nSF P
=  2 π d2o 1 − α2 Le rg 4 nSF P
= 2 π do 1 − α2

Rearranging and solving for do

d4o =

64 CL L nSF P
π 3 E 1 − α4 2

2

→

π2 E 16 CL2 L2
d2o 1 + α2 #1 64 CL L nSF P 4
do = π 3 E 1 − α4 "

2

2

What happens when α = 0? Do you expect the column to be more stable when α > 0? [10pts] Note that 0 ≤ α < 1. It cannot take a value of one because there would be no cross-sectional area. When α = 0, di = 0 and the column becomes a solid column and A= I=

rg2 = Ra2 =

σcr =


π π 2 do 1 − α2 = d2o 4 4


π 4 π 4 d 1 − α4 = d 64 o 64 o
d2 I d2 = o 1 + α2 = o A 16 16 16 CL2 L2 16 CL2 L2
2 = 1+α d2o

d2o

4 nSF P 4 nSF P
2 = 1−α π d2o

π d2o

It is clear that the slender ratio is smaller and the critical buckling stress decreases in value. This implies the buckling stress would be smaller for a solid column than for a tubular column. Thus the column would be more stable when α > 0.

12.3. STABILITY OF BEAM-COLUMNS

When Ra ≤ 10 do =

s

When 10 ≤ Ra ≤ Rc do = When Ra ≥ Rc

s

982

4 nSF P
= π 1 − α2 Sy

s

4 nSF P 4 Sy CL2 L2

= + 2 2 π 1 − α Sy π E 1 + α2

4 nSF P π Sy

s

4 nSF P 4 Sy CL2 L2 + π Sy π2 E

1 #1   64 CL2 L2 nSF P 4 64 CL L nSF P 4
= do = π 3 E 1 − α4 π3 E "

2

2

End Example 

12.3. STABILITY OF BEAM-COLUMNS

983

Example 12.7.

It is desired to substitute the tubular column by a solid rectangular one. The column has rectangular cross-section with a height a and a width b (b = β a), determine the width b. It is known that

A = a b = β a2 I=

rg2 = Thus Ra2

=



Le rg

2

b a3 β a4 = 12 12 I a2 = A 12 =

12 CL2 L2 CL2 L2 = 2 a a2 12

The stress can be expressed as σcr = nSF σ = nSF

P nSF P = A β a2

1. Assuming the actual slender ratio is smaller than the critical slender ratio [10pts] When Ra ≤ 10 the column is a short column, we can use: σcr = Sy Thus

σcr = Sy nSF P = Sy β a2

Rearranging and solving for a a2 =

a= The width b is b = β a: b=

s

nSF P β Sy s

nSF P β Sy

β nSF P Sy

12.3. STABILITY OF BEAM-COLUMNS

984

When 10 ≤ Ra ≤ Rc the column is an intermediate column, we must use the J. B. Johnson Equation: σcr = Sy −

Sy2 4 π2 E

Thus



Le rg

2

= Sy −

1 E





Le rg

2

Sy2 σcr = Sy − 4 π2 E

Sy Ra 2π

2

Sy2 12 CL2 L2 nSF P = S − y β a2 4 π2 E a2 Rearranging nSF

3 Sy2 CL2 L2 P = a2 Sy − β π2 E

Solving for a a2 =

a=

3 Sy CL2 L2 nSF P + β π2 E s

3 Sy CL2 L2 nSF P + β π2 E

The width b is b = β a: b=

s

β nSF P 3 β 2 Sy CL2 L2 + Sy π2 E

2. Assuming the actual slender ratio is bigger than the critical slender ratio [10pts] When Ra ≥ Rc the column is a long column, we must use the Euler Equation: σcr =

Thus

Pcr π2 E =  2 = E A Le rg



π2 E σcr =  2 Le rg nSF P π2 E 2 = βa 12 CL2 L2 a2 nSF P π 2 E a2 = β a2 12 CL2 L2

π Ra

2

12.3. STABILITY OF BEAM-COLUMNS

985

Rearranging and solving for a

a4 =

12 CL L nSF P β π2 E

The width b is b = β a:

2

2

→

1 12 CL L nSF P 4 a= β π2 E 

2

2



1  12 β 3 CL2 L2 nSF P 4 b= π2 E 

End Example 

12.3. STABILITY OF BEAM-COLUMNS

12.3.3

986

Imperfect Beam-Columns (Adjacent Equilibrium Method)

In the previous section, we considered the perfect column, one that is initially straight and the axial load is perfectly aligned with the centroidal axis. Imperfect columns consist of geometric imperfection (initial deflection exists) and/or load misalignment (eccentric load). Let us begin by discussion imperfections due to load misalignment.

Eccentric load Columns used in applications rarely have the applied load aligned coincidentally with the centroidal axis of the cross section. The distance between the two axes is called eccentricity and is designated by e. Let us derive the equations for a simply-supported straight column subject to an eccentric static compressive force P . The beam column obeys the Euler-Bernoulli Theory, is homogeneous, isotropic, and of uniform cross-section.

y, v(x) x e

P

P

It is assumed that the load is always parallel with the centroid of the columns. Figure shows a pinned-end column subjected to forces acting at a distance e from the centerline of the undeformed column. It is assumed that the load is applied to the column at a short eccentric distance from the centroid of the cross section.

y, v(x)

P Me

x

P Me=Pe

This loading on the column is statically equivalent to the axial load and bending moment at the end points is: Me = −P e shown in above Figure. As when one is considering concentrically loaded columns, small deflections and linear elastic material behavior are assumed. The x–y plane is a plane of symmetry for the cross-sectional area.

12.3. STABILITY OF BEAM-COLUMNS

987

The eccentric axial load will simultaneously subject the column to compression and bending in the equilibrium state. The analysis for the equilibrium response of the column in compression (axial force P and axial displacement v(x)) is identical to the case of the perfect column, since the x-axis passes through the centroid of each cross section (decoupling the axial compression from bending in the material law). The equilibrium response of the column in bending, which includes the influence of axial compression on bending, will be determined by the same analysis that led to Eq. 12.13, except that we drop the subscript “1” on the transverse displacement, since in the eccentric load case the transverse displacement refers to an equilibrium state and not to a buckling mode. The ordinary differential equation for a column with EIzz = constant was derived as v 0000 +

P v 00 = 0 EIzz | {z } λ2

x ∈ (0, L)

(12.29)

v(x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4

(12.30)

General solution for λ2 > 0 is

v = v(x)

and the buckling load is obtained from Eq. (12.29): λ2 =

P EIzz

→

P = λ2 EIzz

The constants are found by applying the boundary conditions. The following will be needed v 0 (x)

=

A1 λ cos(λ x) − A2 λ sin(λ x) + A3

(12.31)

v 00 (x)

=

−A1 λ2 sin(λ x) − A2 λ2 cos(λ x)

(12.32)

v 000 (x)

=

−A1 λ3 cos(λ x) + A2 λ3 sin(λ x)

(12.33)

From the Hooke’s law: Mzz

= EIzz v 00 (x) = −EIzz [A1 λ2 sin(λ x) + A2 λ2 cos(λ x)]

(12.34)

0 = −Mzz − P v 0 = −EIzz [v 000 + λ2 v 0 ] = −EIzz [A3 λ2 ]

(12.35)

and from equilibrium: Vy

We want to determine the critical load Pcr and the associated buckling mode shape. We first start with the boundary conditions: Pinned at x = 0: v(0) = 0 and Mzz (0) = EIzz v 00 (0) = Me Pinned at x = L: v(L) = 0 and Mzz (L) = EIzz v 00 (L) = Me

12.3. STABILITY OF BEAM-COLUMNS

Thus

988

v(0) = A2 + A4 = 0 EIzz v 00 (0) = −EIzz A2 λ2 = −P e v 00 (0) = −A2 λ2 = −

Pe = −λ2 e EIzz

v(L) = A1 sin(λ L) + A2 cos(λ L) + A3 L + A4 = 0 EIzz v 00 (L) = −EIzz A1 λ2 sin(λ L) − EIzz A2 λ2 cos(λ L) = −P e v 00 (L) = −A1 λ2 sin(λ L) − A2 λ2 cos(λ L) = −

Pe = −λ2 e EIzz

Write the boundary conditions in a matrix form in terms of the unknown coefficients A, A2 , A3 , A4        

   A1   0                  2  −λ2 e   A2 0 −λ 0 0   =      A3  −λ2 e −λ2 sin(λ L) −λ2 cos(λ L) 0 0                  0 A4 sin(λ L) cos(λ L) L 1 0

1

0

1

       

(12.36)

      

Note that the boundary conditions are inhomogeneous. Thus these equation do not lead to trivial solution and we proceed to solve in term of the constants. The solution to the above equation is: A1 = −e cos (λ L) + e csc (λ L)

A2 = e

A3 = 0

A4 = −e

and replacing in Eq. (12.30) and simplifying we get: v(x) = e



tan



L λ 2



 sin (x λ) + cos (λ L) − 1

where

λ=

r

P EIzz

The maximum deflection occurs at: L x= : 2

vmax = e



sec



  L λ −1 2

Note that λL L = 2 2 rg

r

P = EA

(

L 2

r

P EIzz

)s

Thus, vmax = e

(

π 2 E Izz π = 2 Pcr L2e

sec

π 2

r

P Pcr

r

!

P Pcr

−1

)

where

Pcr = π 2

E Izz L2e

12.3. STABILITY OF BEAM-COLUMNS

989

Let us define the dimensionless quantities as pˆ = Hence,

P Pcr

n π p  o δ = vmax = e sec pˆ − 1 2

(12.37)

2

e=0: PERFECT SYSTEM

1

(unstable, perfect)

(stable, perfect)

1

e=0.01 e=0.1

Load

0.8

0.6 e=0.001

eÆ0 0.4

1

(stable, perfect)

1

(the imperfect system has only one equilibrium path)

0.2

0

0

0.5

1

1.5

2

2.5 3 Deflection, δ

3.5

4

4.5

5

Figure 12.13: Response for various levels of load imperfection. Figure 12.13 shows that as δ → ∞, P → Pcr for all e 6= 0. That is, no matter the magnitude of e, δ gets very large as P → Pcr of the perfect structure. Failure by excessive displacement or loss of structural stiffness.

12.3. STABILITY OF BEAM-COLUMNS

990

Geometric Imperfection Now consider the case of a uniform, pinned-pinned column that is slightly crooked under no load and it is subject to a centric, axial compressive load P . The initial shape under no load is described by the function vo (x). That is the transverse displacement of the column is such that v(x) = vo (x) when P = 0.

v(x) y, v(x) vo(x) P

x

P

Also, the bending moment in the column is zero under no load. Thus, the material law for bending is

n o Mzz = EIzz v 00

Hence, Eq. (12.10) becomes

 2  d v d2 vo d2 Mzz + + P =0 dx2 dx2 dx2     d2 vo d2 d2 v d2 v + EI + P =0 zz dx2 dx2 dx2 dx2

(12.38)

Hence, the ordinary differential equation for a column with EIzz = constant was derived as v 0000 +

P v 00 = −λ2 vo00 EIzz | {z } λ2

v = v(x)

x ∈ (0, L)

(12.39)

Note that we drop the subscript “1” on the transverse displacement, since the transverse displacement refers to an equilibrium state and not to a buckling mode. Let us assume that the initial shape of the bar is that of a sine function with amplitude a1 π x vo (x) = a1 sin L

where a1 denotes the amplitude at midspan of the slightly crooked column. hence the nonhomogeneous fourth order differential equation is v

0000

+ λ v = −λ 2

00

2

vo00

=−



λπ L

2

a1 sin

π x L

v = v(x)

x ∈ (0, L)

(12.40)

12.3. STABILITY OF BEAM-COLUMNS

991

General solution for λ2 > 0 is v(x) = A1 sin(λ x) + A2 cos(λ x) + A3 x + A4 + 

λπ L

1 2

a1 sin −1

π x L

(12.41)

The boundary conditions are same as those of a simply-supported beam: Pinned at x = 0: v(0) = 0 and Mzz (0) = EIzz v 00 (0) = 0 Pinned at x = L: v(L) = 0 and Mzz (L) = EIzz v 00 (L) = 0 The solution is v(x) =

where

π x π x a1 a1 = sin 2 sin  L L 1 − pˆ2 λπ 1− L pˆ =

P Pcr

Thus, δ = vmax =

a1 1 − pˆ2

(12.42)

2

a1=0: PERFECT SYSTEM

1

(unstable, perfect)

(stable, perfect)

1

a1=0.01 a1=0.1

Load

0.8

a1=0.001

0.6

a1Æ0 0.4

1

(stable, perfect)

1

(the imperfect system has only one equilibrium path)

0.2

0

0

0.5

1

1.5

2 2.5 3 Deflection, a1

3.5

4

4.5

5

Figure 12.14: Response for various levels of geometric imperfection. Figure 12.14 shows that as δ → ∞, P → Pcr for all a1 6= 0. That is, for a nonzero value of the imperfection amplitude, the displacement gets very large as the axial force approaches the buckling load

12.3. STABILITY OF BEAM-COLUMNS

992

of the perfect column. Also, the imperfect column deflects in the direction of imperfection; e.g., if a1 > 0, then δ > 0.

Summary of Beam-Column Imperfections In short, collectively the eccentric load and the geometric shape imperfection are called imperfections. All real columns are imperfect. Even for a well manufactured column whose geometric imperfections are small and with the load eccentricity small, the displacements become excessive as the axial compressive force P approaches the critical load Pcr of the perfect column. Hence, the critical load determined from the analysis of the perfect column is meaningful in practice.

12.3.4

Perfect Beam-Column (Energy Approach)

In engineering, we often want to find analytical or semi-analytical solutions for structural problems. However, exact solutions, if they exist, may be too expensive to find and derive from fundamental equations. For such problems we look for approximate analytical solutions for to obtain buckling loads. Here we employ the principle of virtual work to obtain the approximate buckling loads. It should be noted that is the basis for the finite element method. In chapter 9, we showed that for a straight cantilevered beam-column of length L the internal and external virtual work can be written as: Z L Z L δWint = Nxx δεxx dx + Mzz δκzz dx 0

0

δWext = −P δu

We are ignoring all lateral moments and loads. The beam is subject to an axial point load P in compression at x = L. Also, we know that ( 2  2  2 ) ∂U 1 ∂U ∂V ∂W εxx = + + + ∂x 2 ∂x ∂x ∂x Using Euler-Bernoulli Beam Theory, which assumes small rotations, the midplane strains are εxx ≈

du 1 + dx 2

and the midplane curvatures are κzz ≈ −



dv dx

2

d2 v dx2

Applying the principle of virtual work, we get δWext − δWint = −P δu −

Z

0

L

Nxx δεxx dx +

Z

0

L

Mzz δκzz dx = 0

12.3. STABILITY OF BEAM-COLUMNS

993

Since we are studying the stability of the equilibrium state u(x) = u0 (x) + u1 (x)

v(x) = v0 (x) + v1 (x)

Nxx (x) = Nxx0 (x) + Nxx1 (x)

Mzz (x) = Mzz0 (x) + Mzz1 (x)

From the equilibrium conditions we know: P P x → u00 = − EA EA v0 = 0 → v00 = v000 = 0

u0 = −

Nxx0 = −P

Mzz0 = 0

For a small perturbation about the equilibrium position, u1 = 0 v1 6= 0

Nxx1 = 0

u01 = 0

→

v10 = v100 6= 0

→

Mzz1 = −EIzz v100 The generalized virtual displacements are δu0 6= 0

δu1 = 0

Hence the virtual strains are   1 0 2 0 εxx0 = δ u0 + (v0 ) = δu00 2

δv0 = 0

εxx1 = δ



δv1 6= 0

u01

1 2 + (v10 ) 2



= v 0 δv 0

and the virtual curvatures are κzz0 = −δv000 = 0

κzz1 = −δv100

Hence the principle of virtual work becomes

n o −P δu0 + δu1

x=L n o −P δu0

−

Z

x=L

L

0

−

−P δu

x=L

Z n on o Nxx0 + Nxx1 δεxx0 + δεxx1 dx −

0

Z

0

L

Z n on o Nxx0 δεxx0 dx −

0

L

− L

Z

0

L

Nxx δεxx dx −

Z

L

Mzz δκzz dx = 0

0

n on o Mzz0 + Mzz1 δκzz0 + δκzz1 dx = 0

Z n on o Nxx0 δεxx1 dx −

0

L

n on o Mzz1 δκzz1 dx = 0

12.3. STABILITY OF BEAM-COLUMNS

994

Thus, n o −P δu0

x=L

−

Z

0

L

n

−P

Z on o δu00 dx −

L

0

x=L

Z

−P

Z on o v10 δv10 dx −

L

0

n

− EIzz v100

on

o − δv100 dx = 0

Z L L dδu00 EIzz v100 δv100 dx = 0 P v10 δv10 dx − dx + dx x=L 0 0 0 Z L x=L Z L + P v10 δv10 dx − EIzz v100 δv100 dx = 0 + P δu0 −P δu0 x=0 x=L 0 0 Z L Z L 0 0 + P v1 δv1 dx − − P δu0 EIzz v100 δv100 dx = 0 + P δu0

−P δu0

−P δu0

n

+

L

Z

P

x=0

x=L

0

0

Note that δu0 (x = 0) = 0 because it is a u0 is prescribed at x = 0. Thus Z

L

0

or

Z

0

L

P v10 δv10 dx −

Z

0

L

EIzz v100 δv100 dx = 0

EIzz v100 δv100 dx − Pcr

Z

0

L

v10 δv10 dx = 0

(12.43)

Although the above was derived for a cantilevered beam, the reader can derive and show this equation holds for all boundary conditions. Now the problem reduces in finding a kinematically admissible displacement function, usually in the form of a trigonometric or polynomial function. These functions obey the following statements: 1. must satisfy all displacement boundary conditions (known as essential or kinematical boundary conditions); i.e., must satisfy all values for: v1 (x)

and/or v10 (x)

2. may satisfy all non displacement boundary conditions, that is load condition (known as nonessential or natural boundary conditions); i.e., must satisfy all values for: v100 (x) =

Mzz EIzz

• When the assumed displacement functions satisfy all natural boundary conditions, we are using the Galerkin or Rayleigh-Ritz Approach. They both produce the same answer for conservative problems. • When the assumed displacement functions do not satisfy all natural boundary conditions, we are using the Rayleigh-Ritz Approach. 3. be capable of representing the buckling shape, expected from the physical grounds alone.

12.3. STABILITY OF BEAM-COLUMNS

995

Example 12.8.

y, v P x, u L Figure 12.15: A simply-supported beam column subject to an axial load. The uniform column with bending stiffness EIzz , shown in Fig. 12.15, is pinned at x = 0 and pinned at x = L. Using the principle of virtual work, determine the critical load Pcr . (12.8-a) Assuming a kinematically admissible trigonometric function. Let us choose the following function: v1 (x) = a1 sin

π x L

The form assumed must satisfy all displacement boundary conditions:   π (0) v1 (0) = 0 → v1 (x) = a1 sin =0 Good L x=0   π (L) v1 (L) = 0 → v1 (x) = a1 sin =0 Good L x=L It may satisfy the natural boundary condition, v100 (0)

=0

→

v100 (L) = 0

→



v100 (x)

v100 (x)

x=0

x=L

 π (0) = −a1 sin =0 L L    π 2 π (L) = −a1 sin =0 L L  π 2



X X

Since it satisfies both natural and essential boundary conditions, both Raleigh-Ritz and Galerkin give the same answer. Since the first buckling mode of a pinned-pinned column, computed by solving the actual differential equation of equilibrium, we should expect it to yield the same Euler buckling load.

12.3. STABILITY OF BEAM-COLUMNS

996

When using the principle of virtual work, we may use Eq. (12.43): Z

L

EIzz

0

v100

δv100

dx − Pcr

Z

L

0

v10 δv10 dx = 0

where π x cos L L π x  π 2 sin v100 (x) = −a1 L L v10 (x) = a1

Hence

π

Z

π x cos L L π x  π 2 δv100 (x) = −δa1 sin L L δv10 (x) = δa1

π

Z L EIzz v100 δv100 dx − Pcr v10 δv10 dx = 0 0 0  π 4  L   π 2  L  EIzz a1 δa1 − Pcr a1 δa1 = 0 L 2 L 2   π 2  L   π 4  L  − Pcr a1 δa1 = 0 EIzz L 2 L 2 L

Thus the buckling load is

Pcr =

π 2 EIzz L2

which is the Euler column formula! (12.8-b) Assuming a kinematically admissible polynomial function. First let us chose a linear function of the form v1 (x) = a0 + a1

x L

The form assumed must satisfy all displacement boundary conditions: v1 (0) = 0 → v1 (x) = a0 = 0 x=0 v1 (L) = 0 → v1 (x) = a1 = 0 x=L

and produce a zero deflection! Hence, we must choose a function with a higher order polynomial, say  x 2 x + a2 v1 (x) = a0 + a1 L L The form assumed must satisfy all displacement boundary conditions: v1 (0) = 0 → v1 (x) = a0 = 0 x=0 v1 (L) = 0 → v1 (x) = a1 + a2 = 0 x=L

Thus,

v1 (x) = −a2

x L

+ a2

 x 2 L

12.3. STABILITY OF BEAM-COLUMNS

997

When using the principle of virtual work, we may use Eq. (12.43): Z

L

EIzz

0

v100

δv100

dx − Pcr

Z

L

0

v10 δv10 dx = 0

where       1 1 x +2 L L L  2 1 00 v1 (x) = 2 a2 L v10 (x) = −a2

      1 1 x +2 L L L  2 1 00 δv1 (x) = 2 δa2 L δv10 (x) = −δa2

Note that although we know that the bending moment is zero at each end (natural boundary condition), we cannot satisfy this boundary condition because it would require a2 = 0 and it would produce a zero deflection. In other words, the assumed displacement field is too few degrees of freedom to satisfy both the essential and natural boundary conditions. Now applying the PVW: Z

L

0

EIzz 

Z

L

dx − Pcr v10 δv10 dx = 0 0   1 4 EIzz − Pcr a2 δa2 = 0 3L L3

v100

δv100

Thus the buckling load is Pcr =

12 EIzz π 2 EIzz = 1.216 L2 L2

which is 22% higher than the Euler buckling load and is considered a poor approximation. (12.8-c) Assuming a kinematically admissible polynomial function. Let us choose a polynomial that would satisfy both kinematic and natural boundary conditions. Hence, we must choose a function with a fourth order polynomial, say v1 (x) = a0 + a1

x L

+ a2

 x 2 L

+ a3

 x 3 L

+ a4

 x 4 L

The form assumed must satisfy all displacement boundary conditions: v1 (0) = 0 → v1 (x) = a0 = 0 x=0 v1 (L) = 0 → v1 (x) = a0 + a1 + a2 + a3 + a4 = 0 x=L 2 a2 v100 (0) = 0 → v100 (x) = 2 =0 x=0 L 2 a2 6 a3 12 a4 00 00 v1 (L) = 0 → v1 (x) = 2 + 2 + =0 x=L L L L2

12.3. STABILITY OF BEAM-COLUMNS

Thus,

998

   x 3  x 4  x −2 + v1 (x) = a4 L L L

When using the principle of virtual work, we may use Eq. (12.43): Z

L

EIzz

0

where

v100

δv100

dx − Pcr

Z

0

L

v10 δv10 dx = 0

         x 1 x 2 1 x 3 = a4 −6 +4 L L L L L          2 1 x x 1 x 3 −6 δv10 (x) = −δa4 +4 L L L L L         2 1 1 x x + v100 (x) = 12 a4 − 2 2 L L L L         1 x 1 x 2 δv100 (x) = 12 δa4 − + 2 2 L L L L v10 (x)

Hence,

Z

L

0

Thus the buckling load is

Z L EIzz v100 δv100 dx − Pcr v10 δv10 dx = 0  0   17 24 EIzz a4 δa4 = 0 3 − Pcr 35 L 5L

Pcr =

168 EIzz π 2 EIzz = 1.00129 2 17 L L2

which is 0.13% higher than the Euler buckling load and is considered an excellent approximation. End Example 

12.3. STABILITY OF BEAM-COLUMNS

999

Example 12.9.

y, v 0.6 L

P

x, u L Figure 12.16: Beam column with unsymmetrical supports subject to an axial load. The uniform column with bending stiffness EIzz , shown in Fig. 12.16, is pinned at x = 0 and pinned with a roller at x = 0.6L and x = L. Using the principle of virtual work, determine the buckling load and its associated mode shape. Note that the exact solution is Pcr = 3.729

π 2 EIzz L2

Let us choose a polynomial that would satisfy both kinematic and natural boundary conditions. Hence, we must choose a function with a sixth order polynomial, say v1 (x) = a0 + a1

x L

+ a2

 x 2 L

+ a3

 x 3 L

+ a4

 x 4 L

+ a5

 x 5 L

The form assumed must satisfy all displacement boundary conditions:

+ a6

 x 6 L

v1 (0) = a0 = 0 v1 (0.6 L) = a0 + a1 + a2 + a3 + a4 + a5 + a6 = 0 v1 (L) = a0 + 0.6 a1 + 0.36 a2 + 0.216 a3 + 0.1296 a4 + 0.07776 a5 + 0.046656 a6 = 0 2 a2 v100 (0) = 2 = 0 L 6 a3 12 a4 20 a5 30 a6 v100 (L) = 2 + =0 2 + 2 + L L L L2 Thus, v1 (x) = a5



1184 η 4 818 η 3 33 η η − + − 465 465 155 5



+ a6



3559 η 4 3243 η 3 459 η η − + − 775 775 775 6



12.3. STABILITY OF BEAM-COLUMNS

1000

where η=

x L

When using the principle of virtual work, we may use Eq. (12.43): Z

L

EIzz

0

v100

δv100

dx − Pcr

Z

L

0

v10 δv10 dx = 0

where 

 6 η5 14236 η 3 9729 η 2 459 − + − a6 L 775L 775L 775L   5   4 4736 η 3 818 η 2 33 6η 14236 η 3 9729 η 2 459 5η 0 − + − δa5 + − + − δa6 δv1 (x) = L 465L 155L 155L L 775L 775L 775L     20 η 3 4736 η 2 1636 η 30 η 4 42708 η 2 19458 η v100 (x) = − + a + − a6 5 2 2 2 2 2 + L 155L 155L L 775L 775L2     4736 η 2 1636 η 30 η 4 42708 η 2 19458 η 20 η 3 δv100 (x) = − + δa + − δa6 5 2 2 2 2 2 + L 155L 155L L 775L 775L2 v10 (x) =

5 η4 4736 η 3 818 η 2 33 − + − L 465L 155L 155L



a5 +



Hence, Z

L

Z

L

v10 δv10 dx = 0 0 0 n o (0.486452 − 0.0129431 λ) a5 + (1.45697 − 0.0386669 λ) a6 δa5 + n o (1.45697 − 0.0386669 λ) a5 + (4.41534 − 0.11609 λ) a6 δa6 = 0 EIzz v100 δv100 dx − Pcr

The above is mathematically simplified and uses λ=

L2 Pcr EIzz

Since δa5 and δa6 are arbitrary, (0.486452 − 0.0129431 λ) a5 + (1.45697 − 0.0386669 λ) a6 = 0 (1.45697 − 0.0386669 λ) a5 + (4.41534 − 0.11609 λ) a6 = 0

or in matrix form  0.486452 − 0.0129431 λ 1.45697 − 0.0386669 λ

1.45697 − 0.0386669 λ 4.41534 − 0.11609 λ



a5 a6



=



0 0



The trivial solution for this system of two linear and homogeneous equations is a5 = a6 = 0, but this would produce zero column deflection and hence no buckling. Since we are seeking the nontrivial solutions, we take the determinant of the 2 × 2 matrix equal to zero. Hence det



0.486452 − 0.0129431 λ 1.45697 − 0.0386669 λ

1.45697 − 0.0386669 λ 4.41534 − 0.11609 λ



= λ2 − 127.287 λ + 3369.52 = 0

12.3. STABILITY OF BEAM-COLUMNS

1001

The roots of this quadratic characteristic equation are L2 λ= Pcr = EIzz



37.5481 89.7388

Since we are interested in the most fundamental buckling load, we choose the smallest solution: π 2 EIzz EIzz L2 λ= Pcr = 37.5481 → Pcr = 37.5481 2 = 3.80441 EIzz L L2 Which is only two percent higher from the exact solution. To determine the buckling mode shape, we substitute the λ = 37.5481 into the system of two linear and homogeneous equations:      a5 0.486452 − 0.0129431 λ 1.45697 − 0.0386669 λ 0 = 1.45697 − 0.0386669 λ 4.41534 − 0.11609 λ a6 0      0.000462456 0.00510604 a5 0 = 0.00510604 0.0563764 a6 0 Since the determinant of this coefficient matrix is zero, the system of equations is not independent. Hence, let us choose the first row, 0.000462456 a5 + 0.00510604 a6

→

a5 = −11.0411 a6

and the approximate buckling mode shape for the fundamental buckling mode is  x  x 3  x 4  x 5  x 6  v(x) = a6 1.75843 − 15.2384 + 23.5211 − 11.0411 + L L L L L and a6 is the undetermined amplitude. The above is known as the Rayleigh-Ritz methods, in which the assumed displacement field is applied throughout the entire column. However, another method exists in which we assume displacements over subregions and then assemble them together to obtain the complete solution. This is known as the finite element approach, which is beyond the scope of this book. End Example 

12.3. STABILITY OF BEAM-COLUMNS

12.3.5

1002

Inelastic Buckling

The strength of a compression member (column) depends on its geometry (effective slenderness ratio Ra ) and its material properties (stiffness and strength). The Euler formula describes the critical load for elastic buckling and is valid only for long columns. The ultimate compression strength of the column material is not geometry-related and is valid only for short columns. In between, for a column with intermediate length, buckling occurs after the stress in the column exceeds the proportional limit of the column material and before the stress reaches the ultimate strength. This kind of situation is called inelastic buckling. Although we have previously discussed two widely accepted theories, in this section we will discuss inelastic buckling theories that fill the gap between short and long columns. Suppose that the critical stress st in an intermediate column exceeds the proportional limit of the material σp . Recall the proportional limit is defined as the stress where the compressive stress-strain curve of the material deviated from a straight line. For some materials the proportional limit is very difficult to obtain. Thus for intermediate column Young’s modulus at that particular stress-strain point is no longer E. Instead, the Young’s modulus decreases to the local tangent value, Et . Replacing the Young’s modulus E in the Euler’s formula with the tangent modulus E, the critical load becomes, Pcr =

π 2 Et Izz π 2 Et = L2e Ra2

where

Et =

dσ dε

Few comments on the Tangent-Modulus Theory: 1. The proportional limit σp , rather than the yield stress Sy , is used in the formula. Although these two are often arbitrarily interchangeable, the yield stress is about equal to or slightly larger than the proportional limit for common engineering materials. However, when the forming process is taken into account, the residual stresses caused by processing can not be neglected and the proportional limit may drop up to 50% with respect to the yield stress in some wide-flange sections. 2. The tangent-modulus theory tends to underestimate the strength of the column, since it uses the tangent modulus once the stress on the concave side exceeds the proportional limit while the convex side is still below the elastic limit. 3. The tangent-modulus theory oversimplifies the inelastic buckling by using only one tangent modulus. In reality, the tangent modulus depends on the stress, which is a function of the bending moment that varies with the displacement v.

12.4. REFERENCES

12.4

1003

References

Allen, D. H., Introduction to Aerospace Structural Analysis , 1985, John Wiley and Sons, New York, NY. Curtis, H. D., Fundamentals of Aircraft Structural Analysis, 1997, Mc-Graw Hill, New York, NY. Johnson, E. R., Thin-Walled Structures, 2006, Virginia Polytechnic Institute and State University, Blacksburg, VA.

12.5. SUGGESTED PROBLEMS Aerospace Structures

12.5

Spring 2001

1004

Suggested Problems

Homework 8

Problem 12.1.

Consider mechanical model in article 12.2clockwise of the text but with a of small rotation of θ model in article 12.2 of thethetext but with a small initial rotation theinitial bar clockwise

the bar θ¯ before the load P is applied; i.e., a geometrically imperfect structure. The spring is unstretched

¯ structure. The spring is unstretched when θ = θ . ., a geometricallywhen imperfect θ = θ.

aths for θ = 2° and k

ct model on the plane of

P ------P cr

≤ θ ≤ π ⁄ 2 . The equilibriθ

m θ = θ , and plot in the Note that P cr is the critical

P ------mP cr

L O

cture; i.e. θ = 0 .

ture in part (a), plot the

θ 0 θ

initial configuration

um load P m ⁄ P cr versus θ

π⁄2

odel of an arch shown below consists to two rigid bars of length L , a lumped mass 1. Plot the equilibrium paths for θ = 2◦ and θ = 5◦ of the imperfect model on the plane of P/Pcr versus θ for θ ≤ θto≤a π/2. The equilibrium paths emanate f stiffness K . The arch is subjected downward, deadweight load P from . θ = θ, and plot in the range 0 ≤ P/Pcr < 1. Note that Pcr is the critical load of the perfect structure; i.e., θ = 0◦ .

M structure in part (1), plot the scaled relative maximum load initial Pm /Pcr versus 2. For the imperfect θ for 0 ≤ θ ≤ 10◦ . 30°

nital configuration

cted configuration

L 30°

L



K

P L

L θ M

θ

K

ive when the bars are above the supports, the equation of motion is

K sin θ ( cos θ – cos α ) + PL cos θ = 0

θ = θ(t )

α = 30°

esponding to the load is ∆ = L ( sin α – sin θ ) . Define a dimensionless load mensionless displacement δ ≡ ∆ ⁄ L . Plot the static equilibrium states on the p – δ 0° .

( 1)

um paths emanate from θ = θ , and plot in the range 0 ≤ P ⁄ P cr < 1 . Note that P cr is the critical

------P cr

L O

load of the perfect structure; i.e. θ = 0 . b. For the imperfect structure in part (a), plot the 12.5. SUGGESTED PROBLEMS

θ 0 θ1005

initial configuration

scaled relative maximum load P m ⁄ P cr versus θ

π⁄2

Problem 12.2.

for 0 < θ ≤ 10° .

The one DOF structural model of an arch shown below consists to two rigid bars of length L, a lumped mass M , and a linear elastic spring of stiffness K. The perfect arch is subjected to a downward, deadDOF structural weight load P . model of an arch shown below consists to two rigid bars of length L , a lumped

2. The one M , and a linear elastic spring of stiffness K . The arch is subjected to a downward, deadweight load P .

mass

M L 30°

L 30°

inital configuration

K

P L

L θ M

deflected configuration

K

θ

For the angle θ measured positive when the bars are above the supports, the equation of motion is

For the angle θ measured positive when the bars are above the supports, the equation of motion is 2

d2 θ

d Mθ L2 dt22 − 4 L2 K sin θ (cos θ − cos α) + P L cos θ = 0 2 M L 2 – 4L K sin θ ( cos θ – cos α ) + PL cos θ = 0 dt

θ = θ(t)

θ = θ(t )

α = 30◦

(12.44)

α = 30°

(

1. The displacement corresponding to the load is ∆ = L (sin α − sin θ). Define a dimensionless load a. The displacement corresponding to the load is ∆ = L ( sin α – sin θ ) . Define a dimensionless load p ≡ P/(K L) and a dimensionless displacement δ ≡ ∆/L. Plot the static equilibrium states on the

p ≡ P ⁄ ( KL )p and a dimensionless δ ≡ ∆ ⁄ L . Plot the static equilibrium states on the p – δ - δ plane for −35◦ ≤ θ ≤displacement 30◦ .

plane for –2.35° θ ≤ motions 30° . about an equilibrium configuration let θ(t) = θo + ϕ(t), where θo For≤small b. For small

is the static equilibrium value related to P from part (1), and where ϕ(t) is an infinitesimal rotation about the motions about an equilibrium configuration equilibrium position. Equation (3.23) reduces to let θ ( t ) = θ 0 + ϕ ( t ) , where θ 0 is the static

equilibr

um value related to P from part (a), and whered2ϕϕ ( t ) is an infinitessimal rotation about the equilibrium pos 2

tion. Equation (1)that reduces to Show

dt2

+ ω2 ϕ = 0

dϕ + ω 2 ϕ = 0 . Show that
K dt2 cos α − cos3 θ sec θ ω2 = 4 M

3. Plot ω 2 (M/K) versus δ for2 −35◦ ≤ the stability of the equilibrium states. K θ ≤ 30◦ . Determine ---- ( cos α – cos3 θ ) sec θ ω = 4 What is the buckling load Pcr ? M 

(

c. Plot ω 2 ( M ⁄ K ) versus δ for – 35° ≤ θ ≤ 30° . Determine the stability of the equilibrium states. What is the

buckling load P cr ?

Page 1 of 1

12.5. SUGGESTED PROBLEMS

1006

Problem 12.3.

P Mo

Mo mb

mb ks a

L

ks q

L kt

kt

Initial configuration

Deflected configuration

Figure 12.17: Configuration case A.

For configuration case A: 1. obtain the equations of motion using the principle of virtual work. 2. determine the equilibrium paths. 3. for each equilibrium path, plot P vs. ∆, where P is the vertical load and ∆ is the vertical tip deflection (in the direction of the load). 4. for each equilibrium path, plot ω 2 vs. P , where P is the vertical load and ω 2 the square of the natural frequency. 5. for each equilibrium path, plot ω 2 vs. δ, where ∆ is the vertical tip deflection (in the direction of the load) and ω 2 the square of the natural frequency. 6. for each equilibrium path, determine the stable and unstable ranges for P . Take: P = 10 lb, L = 2 ft, ks = 5 lb/in, kt = 10 lb-in/in, Mo = 20 kg, mb = 10 kg, α = 0◦ , 30◦ . Plot dimensionless quantities only. Use MatLab for all plots and include a copy of the MatLab code. Show all your work by hand using a black/blue ink pen. 

12.5. SUGGESTED PROBLEMS

1007

Problem 12.4.

P Mo

Dh Mo

2L mb a

ks

mb

kt

q

ks

kt

Initial configuration

Deflected configuration

q

mb

ks

kt

Infinitesimal deflected config

Figure 12.18: Configuration case B.

For configuration case B: 1. obtain the equations of motion using the principle of virtual work. 2. determine the equilibrium paths. 3. for each equilibrium path, plot P vs. ∆, where P is the vertical load and ∆ is the vertical tip deflection (in the direction of the load). 4. for each equilibrium path, plot ω 2 vs. P , where P is the vertical load and ω 2 the square of the natural frequency. 5. for each equilibrium path, plot ω 2 vs. δ, where ∆ is the vertical tip deflection (in the direction of the load) and ω 2 the square of the natural frequency. 6. for each equilibrium path, determine the stable and unstable ranges for P . Take: P = 10 lb, L = 2 ft, ks = 5 lb/in, kt = 10 lb-in/in, Mo = 20 kg, mb = 10 kg, α = 0◦ , 30◦ . Plot dimensionless quantities only. Use MatLab for all plots and include a copy of the MatLab code. Show all your work by hand using a black/blue ink pen. 

12.5. SUGGESTED PROBLEMS

1008

Problem 12.5.

P L

mb

Mo

Dh Mo Mo mb

mb

L

D a

q

ks

kt

ks

kt

Initial configuration

Deflected configuration

q

ks

kt

Infinitesimal deflected confi

Figure 12.19: Configuration case C.

For configuration case C: 1. obtain the equations of motion using the principle of virtual work. 2. determine the equilibrium paths. 3. for each equilibrium path, plot P vs. ∆, where P is the vertical load and ∆ is the vertical tip deflection (in the direction of the load). 4. for each equilibrium path, plot ω 2 vs. P , where P is the vertical load and ω 2 the square of the natural frequency. 5. for each equilibrium path, plot ω 2 vs. δ, where ∆ is the vertical tip deflection (in the direction of the load) and ω 2 the square of the natural frequency. 6. for each equilibrium path, determine the stable and unstable ranges for P . Take: P = 10 lb, L = 2 ft, ks = 5 lb/in, kt = 10 lb-in/in, Mo = 20 kg, mb = 10 kg, α = 0◦ , 30◦ . Plot dimensionless quantities only. Use MatLab for all plots and include a copy of the MatLab code. Show all your work by hand using a black/blue ink pen. 

12.5. SUGGESTED PROBLEMS

1009

Problem 12.6.

B m g

k2=k

h k1=2k

2d

d . A

Figure 12.20: Rigid bar with a concentrated mass and spring system.

The structural model shown consists of a perfect rigid bar connected to a smooth pin at its base A and attached to two linear elastic restoring springs that remain horizontal as the rod rotates. The bar in its initial equilibrium configuration remains vertical supporting a box of mass m at the top. The distance from the center of gravity of the box to the pin is h. The weight of the bar is negligible with respect to the weight of the box. Spring with stiffness k1 = 2 k is located at a distance d from the pin, and on the opposite side spring with stiffness k2 = k is located at a distance 2 d from the pin. Each spring can act in either tension or compression. Let d = 150 mm, h = 500 mm, spring constant k = 2.5 kN/m, gravitational acceleration g = 9.81 m/sec2 , and P = m g. For this structural system: 1. obtain the equations of motion using the principle of virtual work. 2. determine the equilibrium paths.

12.5. SUGGESTED PROBLEMS

1010

3. for each equilibrium path, plot P vs. θ, where P is the weight of the mass and θ is the rotational angle measured positive clockwise. 4. for each equilibrium path, plot ω 2 vs. P , where P is the vertical load and ω 2 the square of the natural frequency. 5. for each equilibrium path, plot ω 2 vs. θ, where θ is the rotational angle measured positive clockwise and ω 2 the square of the natural frequency. 6. for each equilibrium path, determine the stable and unstable ranges for P . 7. What is the range of values of the mass m for which the equilibrium of the rod AB is stable, if any? Plot dimensionless quantities only. Use MatLab for all plots and include a copy of the MatLab code. Show all your work by hand using a black/blue ink pen. 

12.5. SUGGESTED PROBLEMS

1011

Problem 12.7.

y, v x, u

P kt

P kt

ks

ks

L

Figure 12.21: A spring-supported beam column subject to an axial load.

The uniform column with bending stiffness EIzz , shown in Fig. 12.21, is spring supported at x = 0 and x = L. Has both extensional linear springs and torsional springs. Take ks = α

EIzz L3

kt = β

EIzz L

1. Solve the differential equations to obtain the exact critical load Pcr and associated buckling mode. 2. Using the principle of virtual work, determine the approximate critical load Pcr and associated buckling mode. 3. What happens when α → ∞? What happens when β → ∞? What happens when both α, β → ∞? 

12.5. SUGGESTED PROBLEMS

1012

Problem 12.8.

y, v 0.6 L x, u

P ks

kt

ks

L Figure 12.22: A simply-supported beam column subject to an axial load.

The uniform column with bending stiffness EIzz , shown in Fig. 12.22, is pinned at x = 0. Has to extensional linear springs located at x = 0.6 L and at x = L. Take ks = α

EIzz L3

kt = β

EIzz L

1. Solve the differential equations to obtain the exact critical load Pcr and associated buckling mode. 2. Using the principle of virtual work, determine the approximate critical load Pcr and associated buckling mode. 3. What happens when α → ∞? What happens when β → ∞? What happens when both α, β → ∞? 

12.5. SUGGESTED PROBLEMS

1013

IT’S ALMOST OVER!!! One hwk left….

Chapter 13 Introduction to Aeroelasticity

Instructional Objectives of Chapter 13 After completing this chapter, the student should be able to: 1. Understand the concept of aeroelasticity. 2. Perform static aeroelasticity analysis. 3. Develop flight envelops for structural integrity.

We can devote a course or two on the subject of aeroelasticity but we will limit to introduce the most fundamental definitions and some applications to wing design. We will limit our discussion to static aeroelasticity and dynamic aeroelasticity. In aeroelasticity, loads depend on the deformation (aerodynamics), and the deformation depends on the loads (structural mechanics/dynamics); thus we has a coupled problem. Before we move on let us formally define some important definitions to better understand the field of aeroelasticity.

1014

13.1. DEFINITIONS

1015

Aerodynamics

Static Aeroelasticity

Flight Mechanics

Dynamic Aeroelasticity

Elasticity

Dynamics Structural Dynamics

Figure 13.1: Interdisciplinary nature of the field of aeroelasticity. Aeroelasticity is often defined as a science which studies the interaction between aerodynamics and elastic loads, and the influence of this interaction on structural design. The interdisciplinary nature of the field can be best illustrated by Fig. 13.1, which depicts the interaction of the three disciplines of aerodynamics, dynamics, and elasticity. Classical aerodynamic theories provide a prediction of the forces acting on a body of a given shape. Elasticity provides a prediction of the shape of an elastic body under a given load. Dynamics introduces the effects of inertial forces. With the knowledge of elementary aerodynamics, dynamics, and elasticity, we can look at problems in which two or more of these phenomena interact. Two of those areas of interaction are studied in flight mechanics and structural dynamics courses.

13.1

Definitions

Aeroelasticity can be defined as the study of static and dynamic behavior of structural elements in a flowing fluid. Aeroelasticity in aerospace engineering is chiefly concerned with the interaction between the deformation of an elastic structure in an airstream and the resulting aerodynamic force. Aeroelastic phenomena is not limited to applications to aerospace engineering but to a wider class of disciplines. As for an example, • Mechanical engineering: the flow of fluid in flexible piping systems and the pogo effect in rocket fuel lines in which structural vibrations contribute to inlet valve disturbances, which are in turn amplified by the fuel pump and produce fluctuations in the thrust.

13.1. DEFINITIONS

1016

• Civil engineering: wind-induced motion of suspension bridges and high-rise buildings. • Biomedical engineering: blood flow in flexible blood vessels In aerospace engineering, aeroelasticity is the study of mutual interaction between aerodynamic, inertial and elastic forces, their influence on aircraft design. Aeroelastic problems would not exist if airplane structures were perfectly rigid. Since modern aircraft are very flexible, aeroelasticity must be taken into account because aeroelastic phenomena arise when structural deformations induce additional aerodynamic loads.

A L DS D R F

B

Z

I

E V

Figure 13.2: The aeroelastic triangle of loads.

Static Aeroelasticity : Science which studies the mutual interaction between aerodynamic forces and elastic forces, and the influence of this interaction on airplane design. Dynamic Aeroelasticity : Phenomena involving interactions of inertial, aerodynamic, and elastic forces. Collar diagram : Describes the aeroelastic phenomena by means of a triangle of forces, shown in Fig. 13.2. in Fig. 13.2, Collar classified the aeroelastic problem by means of a triangle of loads. Three type of loads are considered: A

− Aerodynamic loads

E − Elastic loads I

− Inertial loads

The definitions for each aeroelastic phenomena can be defined concisely as follows:

13.1. DEFINITIONS

1017

DYNAMIC AEROELASTICITY - Phenomena involving all three type of forces: • F – Flutter: dynamic instability occurring for aircraft in flight at a speed called flutter speed Wing tip deflection

Oscillations of increasing amplitude

time

• B – Buffeting: transient vibrations of aircraft structural components due to aerodynamic impulses produced by wake behind wings, nacelles, fuselage pods, or other components of the airplane • Z – Dynamic response: transient response of aircraft structural components produced by rapidly applied loads due to gusts, landing, gun reactions, abrupt control motions, and moving shock waves STATIC AEROELASTICITY - Phenomena involving only elastic and aerodynamic forces: • L – Load distribution: influence of elastic deformations of the structure on the distribution of aerodynamic pressures over the structure • D – Divergence: a static instability of a lifting surface of an aircraft in flight, at a speed called the divergence speed, where elasticity of the lifting surface plays an essential role in the instability.

Wing angle of attack

time • R – Control system reversal: A condition occurring in flight, at a speed called the control reversal speed, at which the intended effect of displacing a given component of the control system are completely nullified by elastic deformations of the structure.

13.1. DEFINITIONS

1018

RELATED FIELDS : • V – Mechanical vibrations

• DS – Rigid-body aerodynamic stability

Before we move on let us review some basic airfoil definitions

EA CG AC

V

α AC CG EA d

e c

AC Aerodynamic center point about which the pitching moment MAC is independent of angle of attack α. Usually close to quarter chord, (0.25 c) CG Center of Gravity point of location of the net weight of the body. EA Elastic axis located by drawing a spanwise line through the shear centers of the cross sections of the beam V∞ Wind’s magnitude α Angle of attack For a uniform airfoil, the center of gravity, the elastic axis and the shear center are at the same location. In the next couple of sections we will study divergence and flutter of both rigid and flexible idealized wings.

13.2. STATIC AEROELASTICITY

13.2

1019

Static Aeroelasticity

Static aeroelasticity studies the interaction between aerodynamic and elastic forces on an elastic structure. Divergence characterizes the phenomenon where an initial deformation of the wing leads to aerodynamic loads that increase the deformation further, finally leading to failure of the structure. Even though the deformation increases with time, this phenomenon is commonly classified as a static problem. It falls under static aeroelasticity, mainly because there are no oscillations involved, and it is independent of the wing’s mass properties.

13.2.1

Divergence Analysis of A Rigid Wing

L

Mo

α V∞

Kt

W

Figure 13.3: A two-dimensional rigid wing model to study divergence. To better understand this concept let us consider a rigid wing segment, shown in Fig. 13.3. The airfoil is hinged about the elastic axis and has torsional spring of constant Kt . This rotational spring represents the actual torsional elasticity of the real elastic wing: Kt =

GJ L

Assume that the lift is far bigger that the drag, i.e., drag is negligible: L 1 D The angle of attack relative to zero lift angle is denoted by α. Assume small angles tan α ≈ α

sin α ≈ α

The total angle of attack is composed of a rigid and elastic components: α = α0 + θ where

cos α ≈ 1

13.2. STATIC AEROELASTICITY

1020

α0 – rigid angle of attack; initial wing incidence; angle if no aero- or gravity loads were present. θ – angle due to elastic deformation The lift force, primarily produced by pressure forces on vehicle surface, will be denoted by L and is defined as L = q∞ S CL where 2 q∞ = 0.5 ρ V∞ : is the dynamic pressure

ρ – fluid density V∞ – aircraft’s speed S = c b : is the wing planform area c – chord b – span CL = a0 α : is the dimensionless lift coefficient a0 is the lift curve slope (assumed constant between stall points) and is defined as a0 =

∂ CL ∂α

The wing’s pitching moment about the aerodynamic center is defined as M0 = q∞ S c CM0 where 2 q∞ = 0.5 ρ V∞ : is the dynamic pressure

S = c b : is the wing planform area CM0 : is the dimensionless pitching moment coefficient about the AC independent of α Now, our goal is obtain the divergence speed at which we get excessive rotations. Let us start by taking the moment about the elastic axis: X  MEA = 0 → e L cos | {zα} +M0 − d W cos | {zα} −Kt θ = 0 ≈1

≈1

13.2. STATIC AEROELASTICITY

1021

Assuming small angles of attack and using the definitions for the aerodynamic loads, we can write this equation as follows:   ∂ CL e q∞ S α + q∞ S c CM0 − d W − Kt (α − α0 ) = ∂α Rearranging,



 1 − 

q∞ S



∂ CL ∂α Kt



Let’s define the divergence dynamic pressure

 e  α = α0 − W d + q∞ S c CM0  Kt Kt

qD = S



Kt  ∂ CL e ∂α

(13.1)

(13.2)

Now Eq. (13.1) can be rewritten as α0 −

Wd + Kt



q∞ qD

α=



1− Wd α0 − + Kt

α=



q∞ qD

1−



q∞ qD

S



q∞ qD

Kt S c CM0  ∂ CL Kt e ∂α (13.3)

c CM0   e ∂ CL ∂α

q¶/qD

1

0

α0 – Wd/Kt

α

Figure 13.4: The divergence dynamic pressure with respect the angle of attack. Figure 13.4 shows that α → ∞ as q∞ → qD for 0 < q∞ < qD . In reality the wing will stall or twist off due to strength failure.

13.2. STATIC AEROELASTICITY

1022

The divergence speed is calculated from the definition of divergence dynamic pressure: r 2 qD 1 2 → VD = qD = ρ VD 2 ρ Divergence corresponds to a static instability and at V∞ = VD we get excessive rotation. The problem of torsional divergence is analogous to the problem of transverse deflection of a column with geometrical imperfection.

y, v vo(x)

x, u L vo (x) = δ sin

πx L

y, v P

v(x)

x, u L v(x) =

1 vo (x), 1 − pˆ

pˆ =

P Pcr

p

1

0

δ

v(L/2)

13.2. STATIC AEROELASTICITY

13.2.2

1023

Divergence Analysis of Flexible Straight Wings

Line of AC

V¶

q¶=0.5 ρ V¶2

dx T

tx dx

α

L

e T+ dT

EA x

differential element

Figure 13.5: Slender straight wing subject to distributed torsional load. Let us consider a straight, uniform, unswept, high aspect ratio, slender and cantilevered wing in steady incompressible flow subject to a constant distributed torsional load per unit span, tx , along its aerodynamic center axis, as shown in Fig. 13.5. Motivation to study this simplified flexible wing is that the simplest static aeroelastic problems of practical interest are posed by slender straight wings with straight elastic axes, essentially perpendicular to the fuselage center line. Here, we will assume that straight wings are characterized essentially by an elastic axis which is nearly perpendicular to the plane of symmetry of the airplane, and that chordwise segments of the wing remain rigid, that is, camber bending is not considered. Further we neglect airfoil weight, since we saw for the rigid wing segment that this factor played no role in the divergence dynamic pressure. The differential equation of torsional aeroelastic equilibrium of a straight wing about its elastic axis is obtained by relating rate of twist to applied torque by the St. Venant torsion theory as follows,   dT (−T ) + T + dx + tx (x) dx = 0 dx (13.4) dT + tx = 0 dx where tx denotes the external torque per unit span. In this case the external torque per unit span is due to the aerodynamic loads acting on the wing. St. Venant’s torsion theory relates torque to the unit twist as d θx (13.5) T = GJ0 dx where GJ0 is the torsional stiffness of the wing box: GJ0 = G0 Jeff

13.2. STATIC AEROELASTICITY

1024

Recall that the torsion constant for a single-cell, piecewise homogeneous cross section is given by 4 A2 Jeff = X I encl dsi i

i

t∗i

where the modulus-weighted thickness is defined as t∗i =

Gi ti G0

and G0 is the reference shear modulus. Substituting Eq. (13.5) into (13.4) and use the fact that the wing is uniform along the span we get GJ0

d2 θx + tx = 0 θx = θx (x) dx2

0 0) bends, its angle of attack in the streamwise direction is reduced. Bending causes nose-down twist in streamwise direction. ∆α = ∆α

torsion

+ ∆α

bending

= θx (x) + ϕz (x) tan Λ

where

n o n o = θx (x) + ϕz (x) tan Λ

ϕz (x) =

dv dx

13.2. STATIC AEROELASTICITY

1029

L>0 L B

A'

V¶

A

Reference axis Consider an upward force applied to reference axis above. Points A and B deflect upward about the same amount. Point A0 has less upward deflection. Streamwise segment A0 B will have a smaller angle of attack due to bending and a negative increment in left results. This negative left increment due to bending is a stabilizing influence, since it opposes the nose-up twist of segment A0 B. IV. For swept-forward wing (Λ < 0) bending causes an increase in the angle of attack for streamwise segment A0 B. This is a destabilizing influence. Bending causes nose-up twist increment for streamwise segment. ¯ then Let Λ = −Λ,

∆α = ∆α

torsion

+ ∆α

bending

¯ = θx (x) − ϕz (x) tan Λ

n o n o = θx (x) + ϕz (x) tan Λ

L

L