Engineering Mechanics.Statics

J.L. MERIAM • L.G. KRAIGE ENGINEERING MECHANICS STATICS Marwan and Waseem AI-Iraqi www.gigapedia.com ENCINEER...

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J.L.

MERIAM

•

L.G.

KRAIGE

ENGINEERING MECHANICS

STATICS

Marwan and Waseem AI-Iraqi

www.gigapedia.com

ENCINEERINC MECHANICS VO L U MEl

STATICS FIFTH EDITION

J. L. MERIAM L. G. KRAIGE Virginia Polytechnic Institute and State University

With Special Contributions by

WILLIAM J. PALM, III University of Rhode Island

GQ WILEY JOHN W ILEY & SONS, INC. Marwan and Waseem AI-Iraqi

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ACQUISITIONS EDITOR MARKETING MANAGER DEVELOPMENT EDITOR SEN IOR PRODUCTION EDITO R ART DIRECTION TEXT DESI GNER COVER DESIGNER ILLUSTRATION EDITOR ILLUSTRATION COORDINATOR ASSOCIATE PHOTO EDITOR FRON T COVER PHOTO BACK COVER PH OTO

Joseph Hayt on Katherine Hepburn Jo hnna Ba r to Christine Cervoni Da wn Stanley Nancy Fie ld J ean ette J acobs Design Sigm u nd Malinowski Gen e Aiello Lisa Gee Robert Fr erc kJSto ne P hot o by E rika Barahona Ede , cou rtesy of t he Guggenheim Museum , Bilbao, Spain. Rep roduced wit h per m ission.

Mechani cs in Action: An und erstanding of statics principles h elps engineers analyze and des ign both simple a nd complex structures . Innovative architectural structures like the Guggenheim Bilbao Muse um (shown on th e cover) r ely on statics analysis for t he beams, trusses, a nd cable s use d in constructing it. T his book was set in Cen tury Schoo lbook by UG / GGS Informat ion Services, Inc . a nd printed a nd bound by Von- Hoffman P re ss. T he cover was pri nted by Von-Hoffman press. T his book was pri nted on acid free paper. Copyr ight

@

CX)

2002 Jo hn Wiley & Son s , Inc. All rig hts reserved .

No part of th is publi cat ion may be re produced , stored in a re trieval system or t ransmitted in a ny form or by any mean s, electronic, mech an ical, photocOPyi.ilg, recording, scan ni ng or oth erwise , except as perm itted under Sections 107 or 108 of t he 1976 Un ited States Copyright Act , without either t he pr ior writ~n permission of t he Publisher, or au thorization through paym en t of t he appropriate per-copy fee to th e Copyright Clearance Center , 222 Rosew ood Drive, Danvers, MA 01923 , (5081750-8400, fax (508 )750-4470. Requests to t he Publisher fOr permiss ion sho uld be addressed to the Pe rmissions Departmen t , J ohn Wiley & !Jons, In c., 655 Th ir d Avenue , New York , NY 10158-0012 , (2121850·6011, fax (212/850·6008, E -Mail: [email protected]. To order books or for customer service pleas e call I-800-CALL-WILEY 1225-5945J.

ISBN 0-471-40646-5 Pr in ted in t he Un ite d States of America

10 9 8 7 6 5 4 3 2 1 Marwan and Waseem AI-Iraqi

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JAMES LATHROP MERIAM 1917-2000 Dr . J am es Lathrop Meriam , internati onally kn own autho r of engineering mechanics textbooks and distin guished professor of engi neer ing, died at his Sa nta Barbara home on July 18, 2000. Beca use of hi s numerous and sign ificant contributio ns to the engi neering profession, Dr. Meri am is regarded as one of the premi er engi neering educato rs of the twentieth century. Dr . Meri am (known as Lath to his fr iends ) received three degr ees from Yal e University , ending wit h the Ph.D. in 1942. He served in the U.S. Coast Guard du r ing World War II . His early industrial expe rience came at Pratt a nd Whitney Airc raft a nd t he General Electric Company. Dr . Meriam was a me mbe r of the faculty of the Un ivers ity of California-Berkeley for twe nty-one years . During this period he served as P r ofessor of Engin eering Mechanics, Ass istant Dean of Graduate St ud ies, and Chair ma n of the Division of Mechanics and Design . Fro m 1963 to 19 72, he was Dean of En gin eering at Duk e Univers ity. In 1972 he returned to full-t ime teaching at Californ ia Polytechnic State University-San Lui s Obispo , and retired in 1980. Subseque ntly , he was visiti ng professor at t he University of California-Santa Barbara and re t ire d for a second time in 1990 . Recognition of his su perb teaching ab ilities followed him whe rever he went . At Berkeley in 1963, he was the first r ecipient of t he Outstandin g Faculty Award of Tau Bet a Pi. In 1978 he received t he Distinguished Educator Awa rd from the Mechanics Division of the American Society for Engin eering Educati on (ASE E ). In 1992 he r eceived the Benjamin Garver Lamme Awa rd from ASE E. He was a fellow me mber of both ASEE and the American Society of Mech ani cal Enginee rs (ASME ). Dr. Meriam began his Engineering Mechanics textbook series in 1950 . The S tatics and Dynamics texts resh ap ed undergraduate mec hanics and becam e the defini ti ve t extb ooks in the field for the next five decad es. In addition to th e U.S. ver sion , the books have appeared in SI versions and have been translated into many foreign languages. His books have been characteri zed by clear and rigor ous presentatio n of the theory, inst ructive sample problems, and numerous and realistic homework exercises . From the outset, a high standard of illustration has distinguish ed the series . In th e early 1980s, Dr . Meri am design ed and hand-built, over a per iod of mor e than three years, a 23-foot wooden sa ilboat nam ed Mele Kai, which is Hawaii an for Song of the S ea. Ove r t he next severa l years, he and his fortunate sailing comp anions spe nt many happy hours sai ling off the coast of Santa Barbara. Dr . Meri am also designed and built four homes, including a vacation home on the island of Kauai, In addition to his many profession al accomplishme nts, Lath Meriam will be long remembered for his open friendliness, gentle manly demeanor , mature judgment and leadership, generosity , and absolute commitment to the hi ghest educational standards. Marwan and Waseem AI-Iraqi

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PREFACE

Engineering mechan ics is both a foundation and a framework for most of the branches of engineering. Many of the topics in such areas as civil, mechanical, ae rospace, and agricultural engineering, and of course engineeri ng mechan ics itsel f, are based upon the subjects of sta tics and dynamics. Even in a discipline such as electrical engineering, practitioners, in the course of considering the electrical component s of a robotic device or a manufacturing process, may find themselves first having to deal with the mechanics involved. Thus, the engineering mechanics se quence is critical to the engineering curriculum . Not only is this sequence needed in itself, but courses in engineering mechanics also se rve to solidify the student's understandin g of other important subjects, including applied mathem ati cs, physics, and gra phics. In addit ion, these courses se rve as excellent settings in which to stre ngthen problem-solving abilities . PHILOSOPHY

Th e primary purpose of th e study of enginee ring mechanics is to develop th e capacity to predict the effects of force and motion while carrying out the creative design functions of engine ering . This capacity requires more than a mere knowledge of the phy sical and mathematica l principles of mechani cs; also requ ired is th e ability to visualize physical configurations in terms of real materials, actual constraint s , a nd th e pr actical limitations which govern th e beh avior of machin es an d struct ures. One of the primary objectives in a mechanics course is to help the student develop this ability to visua lize, which is so vital to problem form u latio n. Indeed, t he construction of a meaningful mathematical model is often a more important experience th an its solut ion. Maximum pr ogress is mad e whe n th e principles and t heir limitations are learned together within the contex t of enginee ring application. There is a freque nt tendency in the presentation of mechanics to use problems main ly as a vehicle to illust rate th eory rather than to develop theo ry for the purp ose of solving problems. When the first view is allowed to predominate, problems tend to become over ly ideali zed and unrelat ed to enginee r ing with the result that the exercise becomes dull, academic, and unint erestin g. This approach deprives the

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Preface

student of valuable experience in formulating problems and thus of discovering the need for and meaning of theory. The second view provides by far the stronge r motive for learn ing t heory and leads to a better bal an ce between theory and application . T he crucial role played by interest and purpose in prov idin g the stronge st possible motive for learning cannot be overemphasized. Furthermore, as mechanics educators, we should stress the understanding that, at best, th eory can on ly ap proxi mate t he real world of mechanics rather than the view that th e real world approximates the theory. This difference in philosophy is indeed basic and distinguishes the engine ering of mechanics from the science of mechanics. Over the past several decades, several unfortunate tendencies have occurred in en gineeri ng educat ion. First, emphasis on the geometric and physical meanings of prerequisite mathematics appears to have diminished. Second, there has been a significant reduction and even elimination of instruction in graphics, which in the past enhanced the visualization and representation of mechanics problems. Third, in advancing the mathematical level of our treatment of mechanics, there has been a tendency to allow the notational manipulation of vector operations to mask or replace geometric visualization. Mechanics is inherently a subject which depends on geometric and physical perception , and we should increase our efforts to develop this ab ility . A special note on the use of computers is in order. The experience of formulating problems, where reason and judgment are developed, is vastly more important for the student than is the manipulative exercise in carrying out the solutio n. For this reason, computer usage must be carefully controlled. At present , constructing free- body diagrams and formulating governing equations are best done wit h penc il and pap er. On the ot her ha nd, there are in stances in wh ich the solution to the governing equations can best be carried out and displayed using the computer. Comp uter-orie nted pro blems sho uld be genuine in the sense that th ere is a condition of design or criticality to be found, rather than "makework" problems in which some parameter is varied for no apparent reason other than to force artificial use of the computer. T hese thought s have been kep t in mind duri ng the des ign of t he computer -or iented problems in the Fift h Edition. To conserve ad equate t ime for problem formulation , it is suggested t ha t t he student be ass igned on ly a limi ted number of the computer-oriented problems. As with previous editions, this Fifth Edition of Engin eering Mecha nics is written with the foregoin g philosop hy in mind. It is intend ed pr ima rily for t he first engineering course in mechanics, generally taught in the second year of study. Engi neering Mechan ics is written in a style which is both concise and friendly. The major emphasis is on basic principles and methods rather than on a multitude of special cases. Strong effort has been made to show both the cohesivenes s of the relatively few funda mental ideas and the great va riety of probl em s which th ese few ideas will solve. PEDAGOGICAL FEATURES

The basic structure of this textbook consists of an article which rigorously treats the particular su bject matter at hand , followed by one or more Sa mple P rob lems, followed by a gro up of P robl ems. T here is a Chapter Review at the end of each chapter which summarizes the main points in that chapter, followed by a Review Problem set. Marwan and Waseem AI-Iraqi

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Prefac e

Problems. The 80 Sa mple P roblems appear on specially colored pages by the mse lves . The solutions to typical statics problems are presented in detail. In addition, explana tory and cau tionary notes (He lpful Hin ts) in blue type are number -keyed to the main presentatio n. Ther e a re 963 homework exercises, of whi ch approximately 50 percent are new to the Fifth Edition. Th e prob lem sets are divided into Introductory Problems a nd Representative Problems. The first section consists of simple, uncomplicated problems designed to help st udents gain confidence with the new topic, whi le most of th e prob lems in the second sectio n are of ave rage difficulty a nd length. Th e problems are generally arranged in order of increasin g difficulty. More difficul t exercises ap pea r near t he end of the Representative Problems a nd a re ma rked wit h the symbol .... Computer-Oriented Problems, marked with an asterisk, appear in a special section at the conclus ion of th e Review Problems at t he end of each cha pter. Th e a nswers to all odd-numbered problems a nd to all difficult problems have been prov ided. In recogniti on of the need for emphasis on SI unit s, there are approximately two problems in SI uni ts for every one in U.S. custo mary units. This a pportionm ent between t he two sets of u nits permits anywh er e from a 50-50 emphasis to a 100per cent SI treat me nt. A notable fea ture of th e Fifth Edition, as with all previous editions, is the wealth of in teresti ng a nd important problems which apply to engineering design . Wh ether directly identified as such or not, virt ua lly all of the problems deal wit h principles and procedures inherent in the design and analysis of enginee ring structu res and mechanica l syste ms. Illustrations. In order to bring th e greatest possible degr ee of rea lism and clarity to th e illust rations, this textb ook ser ies continues to be produced in fu ll color. It is important to note that color is used consistently for the identi fication of certain quantities: · red for forces and moments, • green for velocity and acceleration arrows, · orange dashes for se lected trajectories of moving points. Subdued colors are used for those parts of an illustration which are not central to t he prob lem at hand. Whenever possibl e, mecha nism s or object s which commonly have a certain color will be por trayed in th at color. All of the fundame nta l eleme nts of techni cal illustration which have been an essential part of this Eng ineering Mechanics se ries of textbooks have been retai ned. The author wishes to resta te the conviction t hat a high standard of illust ration is cr itical to a ny written wor k in the field of mechanics.

Features New to this Edition. While retaining the hall mark features of all pr evious editions, we have incorporated these improvements: The theory portions were rewritten for clarity and readability, with a higher level of friendliness and a more active voice.

Sections have been sh orte ned and more subheads added to make informatio n eas ier to find. Marw an and W aseem AI-Iraq i

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Pr eface

Key Concepts areas within the theory pr esentation have been specially highlighted with blue screens . Th e Chapte r Reviews have been revised and highlighted, and feature itemized summaries. • Approximately 50 percent of th e hom ework probl ems are new to this Fifth Edition, and include new problems in th e area of biomechani cs. • New Sample Problems have been added . • All Sample Problems are printed on specially colored pages for quick identification. • The text has been printed in a new, larger format for a more open look. ORGANIZATION

In Chapter I, the fundamental concepts necessary for th e st udy of mechanics are established. In Chapte r 2, the prop erties of forces, mom ents, couple s, and resul ta nts are developed so that t he student may proce ed directly to the equ ilibrium of non concur re nt force sys te ms in Chapter 3 without unnecessarily belaboring the relatively trivial problem of the equilibrium of concurrent forces acting on a particle. In both Chapters 2 an d 3, analysis of two-dimensiona l pr oblems is presen ted in Section A before three-dimen sional problems are treated in Section B. With this a rrangeme nt, t he instructo r may cover all of Cha pter 2 before begin nin g Chapter 3 on equilibrium, or the instructor may cover the two chapte rs in the order 2A, 3A, 2B, 3B. The lat ter order treats force systems and equilibrium in two dimensions and then treats these topics in three dimensions. Application of equilibrium principles to sim ple trusses and to frames and machines is presented in Chapter 4 with primary attention given to two-dimen sion al syste ms. A sufficient number of three-dimensional examples are included, however, to enable students to exercise more general vector tools of analysis. Th e concepts and categories of distributed forces are introduced at the beginning of Cha pte r 5, with the balance of the chapter divided into two main sections . Section A treats centroids and mass center s; deta iled example s are presented to help students master ea rly applications of calculus to physical and geometrical problems. Section B includes the special topics of beam s, flexible cab les, and fluid forces, which may be omitted without loss of continuity of basic concept s. Cha pte r 6 on fr iction is divided into Section A on the phenomenon of dry fr iction a nd Section B on selecte d machin e a pplicat ions . Although Section B may be omitted if tim e is limited, thi s mat erial does pr ovide a valuable experi en ce for t he stude nt in dealing wit h both concentrated a nd distributed friction forces. Cha pte r 7 presents a consolidated introdu ction to virtual work with applications limit ed to single-degree-of-freedom syste ms. Special emphasis is placed on the advantage of the virtua l-work and energy meth od for interconnected sys tems and sta bility determinati on. Virtual work provides an excellent opportu nity to convince the student of the power of mathematical analysis in mechanics. Moments and products of inertia of areas are present ed in Appendix A. This topic helps to br idge the subjects of sta tics a nd solid mechanics. Appendix C con tains a summary review of selected topics of elementary mathematic s as well as se veral numerical techniques which the student should be prepared to use in corn puter-solved probl ems. Usefu l tables of physical const a nts, cent roids, and mom ents of inertia are contained in Appendix D. Marwan and W aseem Al-l raqt

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SUPPLEMENTS

Th e following items have been pre pared to complement this textbook: Instructor's Manual. Prepa red by the aut hors, fully wor ked solutions to all problems in th e text a re available to faculty by contacti ng th eir local Wiley representative.

Solving Mechanics Problems with. . .. A series of booklet s in troduces t he use of computationa l softwa re in t he solution of mechanics problems. Developed by Brian Harp er at Ohio State Un iver sity, the booklets are availab le for Matl ab, MathCAD, a nd Maple.

Wiley Website (www.wiley.com/college /meriam). Items on th is site in clude: Electronic figures for most of t he figures from the text are ava ilable electronically for use in creating lectures. Electronic tra nsparencies for over 100 solved problems, similar to those in the text , are ava ilable for use in lect ure or in self-study by students .

On-lin e problem solving , a progr am ca lled eGrade, pr ovides over 400 problems in mechanics for st udents to solve, featuring step-by-step procedures and imm ediate feedback. Th ese wer e developed by J oe Torok a t Rochester Institute of Techn ology. • Extension sample problems bu ild on sa mple problems from t he text a nd show how computat ional tools ca n be used to invest igat e a varie ty of " what if " scena rios. Availab le to both students a nd faculty, t hese were developed by Brian Ha rper at Ohio State Univer sity. ACKNOWLEDGMENTS

Th e contribution of Professor Willia m J . Palm, III , of the Univers ity of Rhode Island merits specia l acknowledgment for his excellent and carefu l review of the enti re text. Professor Palm has inspected the structure of every sen tence and, where necessary, has made modifications so that the presentation is clear, direct, concise, and friendly. He has carefully modified the heading structu re in order to make th e text more easily readable, and reorganized the Chapter Review sections so that the student can efficiently survey what has been presented. Professor Palm has worked under a number of constraints and has done so in a friendly and timely manner. Specia l recogn it ion is again due Dr. A. L. Hale, form erly of Bell Teleph one Laboratories , for his continuing contribution in the form of invaluable sugges tions and accurate checking of the manuscript. Dr. Hale has rendered s imilar se rvice for all previous versions of this entire ser ies of mechanics books, dating back to the I950s. He reviews all aspects of the books, including all old and new text and figures. Dr. Hale carries out an independent solution to each new homework exercise and provides the author wit h sugges tions and needed corrections to the solutions which appear in the Instructor's Manual . Dr. Hale is well known for being extre mely accurate in his wor k, a nd his fine know ledge of th e En glish la nguage is a great asset which aids every user of this text book. Professor J. Wa llace Gran t of VP I&SU has kindly provided severa l excellent equili brium problems in the area of biomechanics. These new problems se rve to st re ngt he n the textbook in t his impo rtant application field. Marwan and Waseem AI-Iraqi

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Prefa c e

The following in dividua ls (listed in alpha betical order) pr ovided feedback on th e Fourth Editio n, reviewed sa mples of the Fifth Edition, or othe rwise contributed to t he Fifth Edition : L, M, Brock , Univer sity of Kentucky Richard E. Dipp ery , Jr., Kettering University G, Stephe n Gipson, Oklahoma S tote Univer sity Dona ld A. Gran t, Univers ity of Maine A. Henry Ha gedoorn , University of Central Florida Brian Harper, Ohio State University William C. Hauser, Califo rnia State Polytechnic University, Pomona Pau l Heyliger, Colorado S tate University Michael Latcha , Oakland Uni versity Nels Madsen, Aubur n University Sally Steadman , Unioersity of Wyoming Daniel Suchora, Youngstown State University J osef S. Toro k, Rochester Institute ofTechnology David Whitman, University of Wyoming Gary E. Young, OIl/ah oma State University In addition to these individua ls, the autho r wish es to t ha nk t he 230 faculty members who responded to a market questionnaire concerning these textbooks and t he 30 faculty members who respond ed to a telephone survey concern ing the website materials. Th e contributions by the staff of J ohn Wiley & Sons, Inc., includ ing Ed itor J oe Hayton, Developmen t Editor J oh nna Barto, Senior P roduction Editor Chr isti ne Cervo ni, Copy Edit or Suzanne Ingrao, Senior Design er Dawn Stanley, and Illustration Editor Sigmund Malinowski, reflect a high degree of profession al competenc e and are du ly recognized. The talented illustrators of P recision Graph ics conti nue to maintain a high standard of illustration excellence. Fin ally, I wish to state t he ext re mely significa nt contribution of my family. In addition to providing patience and support for this project, my wife Dale has managed t he preparatio n of the ma nuscript for the Fifth Edition a nd has been a key individual in checking all stages of the proof. In additio n, my daughter Stepha nie and son David (both of whom are currently engineeri ng st udents) have contributed both problem ideas and have he lped wit h the illust rations. I a m extre mely pleased to particip at e in extending the time du ration of this text book series to th e fifty-year mark, In the inte rest of providing you wit h th e best possible educational materials over futu re yea rs, I encourage and welcome all comments and suggestions. Please address your comments to kraige@vt .edu.

t: GiRmm Krwge Blacksb ur g, Virginia


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BRIEF CONTENTS

CHAPTER 1

INTROOUCTION TO STATICS

CHAPTER 2

FORCE SYSTEMS

CHAPTER 3

EQUILIBRIUM

103

CHAPTER 4

STRUCTURES

165

CHAPTER 5

OISTRIBUTED FORCES

225

CHAPTER 6

FRICTION

327

CHAPTER 7

VIRTUAL WORK

385

3 23

APPENDICES A

AREA MOMENTS OF INERTIA

427

B

MASS MOMENTS OF INERTIA

463

C

SELECTED TOPICS OF MATHEMATICS

465

D

USEFUL TABLES

481

INDEX

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Chapter


1

INTRODUCTION TO STATICS

3

1/1 Mechanics 1/2 Bas ic Concepts 1/3 Scalars and Vectors Conventions for Equations and Diagr ams 5 Working wit h Vectors 6 1/4 Newton 's La ws 1/5 Units S! Un its 9 U.S. Customary Units 9 P rimary Standards 10 Unit Convers ions 11 1/6 Law of Gravitation Gravitatio na l Attraction of the Earth 12 1/7 Accuracy, Limits, and Approximations Differentials 14 Sm all-Angle Approx imations 14 1/8 Problem Solv ing in Statics Making the Appropriate Assumptions 15 Using Graphics 15 Formulating Pr oblems and Obtaining Solutions The Free-Body Diagram 16 Numerical Values ver sus Symbo ls 17 Solution Methods 17 Chapter Review

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12 13

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Chapter

2

FORCE SYSTEMS

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2 /1 Introduction 2 /2 Force

23 23

External and Internal Effects 24 Principle of Tra nsmissibility 24 Force Classificatio n 24 Action and Reaction 25 Concurrent Forces 25 Vector Compon en ts 26 A Special Case of Vector Addition 26 SECTION A . TWO·DIMENSIONAL FORCE SYSTEMS

27

2/1 Rectangular Components Conventions for Describing Vector Components

27

Determinin g t he Componen ts of a Force 2 /4 Moment Moment about a Point 37 Th e Cross Pr oduct 38 Varigno n's Th eorem 38 2/5 Couple Vector Algebra Meth od 48 Equivalent Couples 48 Force-Couple Systems 49 2 /6 Resultants Algebraic Method 56 Principle of Moments 57

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SECTION 8 . THREE-DIMENSIONAL FORCE SYSTEMS

64

2 /7 Rectangular Components

64

Dot Product 65 Angle bet ween Two Vectors 66 2 /8 Moment and Couple Moments in Three Dimensions 73 Evaluating the Cross Product 73 Moment about an Arbitrary Axis 74 Varignon's Th eorem in Three Dimensi ons Couples in Three Dimensions 75 2/9 Resultants Chapter Review

Chaeter

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3

EQUILIBRIUM

103

1 /1 Introduction

101

SECTION A . EQUILIBRIUM IN TWO DIMENSIONS

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3(2 System Isolation and the Free-Body Diagram Mode ling the Act ion of Forces 105 Con st ru ct ion of Free-Body Diagr am s 108 Examples of Free-Body Diagr am s 109 3(3 Equilibrium Conditions Cate gories of Equ ilibriu m 115 Two- a nd Three-Force Mem be rs 116 Alte rnative Equilibrium Equati ons 117 Constraints and Statical Determinacy 118 Adeq uacy of Constraints 119 Approach to Solving Problems 120

104

SECTION B. EQUILIBRIUM IN THREE DIMENSIONS

138

3(4 Eq uilibrium Conditions Free-Body Diagram s 139 Categories of Equ ilibriu m 139 Cons t rain ts a nd Statical Determin acy Chapter Review

138

ChaRter

115

14 1 156

4

5 T R U CTU RES

165

4(1 Introduction 4(2 Plane Trusses Simple Trusses 167 Truss Con nection s an d Su pports 168 4(3 Method of Joints Int ernal and External Redu ndancy 170 Sp ecia l Cond itio ns 170 4(4 Method of Sections Ill ust rat ion of the Method 179 Additional Considerations 180 4(S Spa ce Tr u s s e s Statically Deter minate Space Trusses 188 Me t hod of Joints for Space T ru sses 189 Method of Sections for Space Trusses 189 4(6 Frames and Machines In tercon nected Rigid Bodies with Mul tiforce Mem bers 195 Force Representation a nd Free-Body Diagr am s Ch a p t e r Review

165 167

ChaRter

168

179

188

195

196 215

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DI STR I B U T E D FORCES

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5(1 In t r o d u ct i o n

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SECTION A. CENTERS OF MASS AND CENTROiDS Marwan and Waseem AI-Iraqi

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5/2 Center of Mass Determinin g the Center of Gr avity 227 Cente r of Mass versus Center of Gravity 229 5/1 Centroids of Lines. Areas. and Volumes Choice of Elem en t for Integrat ion 231 5/4 Composite Bodies and Figures; Approximations An Appr oximation Method 246 Irregu lar Volu mes 247 5/5 Theorems of Pappus

227

SECTION B.

264

SPECIAL TOPICS

5/6 Beams-External Effects Typ es of Beam s 264 Distributed Loads 265 5/7 Beams- Internal Effects Shear, Bending, a nd Torsion 271 Shear-Force and Bending-Moment Diagrams 272 General Loading, Sh ear, and Moment Relationsh ips 5/B Flexible Cables General Relationsh ips 283 Parabolic Cable 285 Cate nary Cable 287 5/9 Fluid Statics Fluid Pressure 297 Hydro static Pressure on Submerged Rectangu lar Surfaces 298 Hydro static Pressure on Cylindrical Su rfaces 300 Hydrostatic Pressure on Flat Surfaces of Any Shape 301 Buoyan cy 302 Chapter Review

Cha~ter

Introduction

256 264

27 1

272 281

297

11 6

327 127

SECTION A . FRICTIONAL PHENOMENA

128

6/2 Type s of Friction 6 /1 Dry Friction Mechanism of Dry Friction 329 Sta tic Fri ction 330 Kinetic Fri ction 330 Friction Angles 33 1 Factors Affecting Friction 332 Types of Friction Problems 333

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FR IC TION 6 /1

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6/4 Wedges 6/5 Sc rews Force Analysis 350 Conditions for Unwinding 35 1 6/6 Journal Bearings 6/7 Thrust Bearings ; Disk Friction 6/8 Flexible Belts 6/9 Rolling Resistance Chapter Review

Chap.ter

359 360 368 369 376

7

VIRTUAL WORK

3B5

7/1 In t r o d u cti o n 7/2 Work Work of a Force 386 Work of a Couple 387 Dimensions of Work 388 Virtual Wor k 388 7/3 Equilibrium Equilibri um of a Particl e 389 Equilibrium of a Rigid Body 389 Equilibrium of Ideal Syst ems of Rigid Bodies Principle of Virtual Work 391 Degrees of Freedom 392 Sys tems with Fr ictio n 392 Mechanical Efficiency 393 7/4 Potential Energy and Stability Elasti c Potential Energy 405 Gravitational Potential Energy 406 Energy Equation 407 Activ e-For ce Diagrams 407 Principle of Virtual Work 408 Stability of Equilibrium 409 Chapter Review

385 385

389

390

405

420

Ap.p.endices APPENDIX A ARE A MOM ENT5 0 FIN ERTI A

427

A/I Int roduction A/2 Definitions Rectangular and Polar Moments of Inertia Radius of Gyration 429 Transfer of Axes 430 A/3 Compo si te Areas Marwan and Wasee m Al-lra qi

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Contents

A/4 Products of Inertia and Rotation of A xes Definition 449 Transfer of Axes 449 Rota tion of Axes 450 Mohr's Circle of In er tia 45 1 APPENDIX B MASS MOMENTS OF I N E RT I A APPENDIX C

44 9

463

ENGINEERING MECHANICS

STATICS


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Structures which support large forces must be designed with the principles of mechanics foremost in mind. In this view of Sydney Harbor, one can see several examples of such structures. Marwan and Waseem AI-Iraqi

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...C hapter

INTRODUCTIO'N TO STATICS

CHAPTER OUTLINE

1/1 1/2 1/ 3 1/4 I /S 1/6 1/7 1/8

Mechanics Basic Concepts Scalars and Vectors Newton's Laws Units Law of Gravitation Accuracy. Limits. and Approximations Problem Solving in Statics

Chapter Review

1 /1

MECHANICS

Mechanics is the phys ical science which dea ls with the effects of forces on objects. No other subject plays a greater role in engi nee ring analysis than mechani cs. Although the principles of mechan ics are few, they have wide application in enginee ring. The pr inc iples of mechanics are central to research a nd development in the fields of vibrations. stability a nd st re ngth of st ructures and machi nes, robot ics, rocket a nd spacecraft design, automatic control, engine performance, fluid flow, electrical machines and apparatus, and molecular, atomic, and subatomic behavior . A t horough under standing of this subject is an esse ntial prer equ isite for work in these and many ot her fields. Mechanics is the oldest of the physical sciences. The ea rly hist ory of this subject is synonymous wit h the very beginnings of enginee ring. Th e earliest recorded writings in mechan ics are those of Archimedes (287- 212 B.C.) on the pri nciple of the lever and the pr in ciple of bu oyancy. Substantial pr ogress came later with the formulation of the laws of vector combination of forces by Stevinus (1548- 1620), who also formulated most of the pr inciples of statics. The first investigatio n of a dynam ics probl em is cred ite d to Galileo (1564- 1642) for his exper ime nts wit h falling sto nes . Th e accurate formulation of the laws of motion, as Ma rwan and W aseem Al-lraqi

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3

4

Chapter 1

Introduction to Static s

well as t he law of gravitation, was mad e by Newton (1642- 1727), who also conceived the idea of the infinitesi mal in mathematical analysis. Substantial contribut ions to the development of mechanics were also made by da Vinci, Varign on , Eul er, D'Alembert, Lagrange, Lapl ace, and ot he rs. In this book we will be conce rned with both the development of th e principles of mechanics and their application. The principles of mech enics as a science are rigorously expresse d by mathemati cs, and thu s mathematics plays an important role in the application of these principles to th e solution of practical prob lem s. T he suhject of mecha nics is logically divided into two pa r ts: statics, which concerns the equilibrium of bodies under the action of forces, and dynamics, which concerns the motion of bodies. Engin eering Mechanics is divided into these tw o parts, Vol. 1 St ati cs a nd Vol. 2 Dyna m ics .

1 /2

BASIC CONC EPTS

The followin g conce pts a nd defin iti on s are basic to the study of mechanics, a nd th ey should be un der stood at the outset. Sp ace is the geometric region occupied by bodies whose positi ons are described by linear and angular measurement s relative to a coordinate sys te m. For three-dim en sion al problems, th ree ind ep end ent coor dinates are needed. For two-dimensional problems, only two coordinates are required. Ti m e is the measure of the succession of events and is a basic quantity in dynamics. Time is not directly involved in the analysis of statics prob lem s. Mas» is a measure of the inertia of a body, which is its resistance to a cha nge of velocity. Mass can also be t hought of as th e quantity of matter in a body. The ma ss of a body affects th e gravi tatio nal attraction force between it and othe r bodies. Thi s for ce appea rs in many applications in statics. Force is the action of one body on another. A force tends to move a body in the direction of its action. The actio n of a force is characterized by its magnitude, by the direction of its actio n, and by its point of application. Thus force is a vector quantity, and its properties are discussed in det ail in Chapte r 2. A particle is a body of negligible dim en sion s. In t he mathem atical sense, a particle is a body whose dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point. We often choose a particle as a differential element of a body. We may treat a body as a particle when its dimensions are irrelevant to the description of its posit ion or the action of forces applied to it. Rigid body. A body is cons ide re d rigid when th e change in dista nce betw een any two of its points is negligib le for the purpose at hand. For instance. the calculat ion of the tens ion in th e cable which su pports t he boom of a mobile crane under load is essenti al ly unaffected by the sma ll internal deformations in th e structural members of the boom. For t he purpose, the n, of det ermin ing the exte rnal for ces whi ch act on the boom, we may treat it as a rigid body. Stati cs deals pr imarily with the calculation of exte rnal forces which act on rigid bodies in equilibrium. DeterMarwan and Waseem AI-Iraqi

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Arti cle 1/3

Sc alars a nd Vec tors

mination of the internal deform ations belongs to t he study of th e mechanics of deformable bodies, which normally follows statics in the curriculum.

1 /3

SCALARS AND VECTORS

We use two kinds of quantities in mechanics- scalars and vectors. Scalar quantities are those with which only a magnitude is associated. Examples of scalar quantities are time, volume, density, speed, energy, and mass. Vector quantities, on the other hand, possess direction as well as magnitude, a nd must obey the parall elogram law of addition as described later in this article. Examples of vector quantities are displacement , velocity, acceleration, force, moment, and momentum. Speed is a scalar. It is the magnitude of velocity, which is a vector. Thus velocity is specified by a direct ion as well as a speed. Vectors representin g physical quantities can be classified as free, sliding, or fixed. A fr ee vec tor is one whose action is not confined to or associate d with a unique line in space. For exa m ple, if a body moves without rotation, then the movement or displacement of any point in the body may be tak en as a vector. Thi s vector describ es equa lly well the direction a nd magn itude of the displacement of every point in the body. Thus, we may represe nt t he displacement of such a body by a free vector. A sli d ing vec tor has a unique line of action in space but not a unique point of application. For example, when an external force acts on a rigid body, the force can be applied at a ny point along its line of actio n without changing its effect on the body as a whole: and thu s it is a sliding vector. A fixed vec tor is one for which a unique point of application is specified. The action of a force on a defor mab le or nonrigid body mu st be specified by a fixed vector at t he point of application of t he force. In thi s insta nce t he forces an d deform ations within the body depend on the point of application of t he force, as well as on its magnitude and line of action. Conventions for Equations and Diagrams

A vecto r quantity V is represented by a line segme nt, Fig. 1/1 , having the direction of the vector and having an arrowhead to indicate the sense. The lengt h of th e directed line segment represen ts to some conven ien t scale th e magn itude IVI of the vector and is printed with lightface italic type V. For example, we may choose a scale such that an arrow one inch long represen ts a force of twenty pounds. In scalar equations, a nd freq ue ntly on diagrams where only the mag nit ude of a vector is labeled, the symbol will appear in ligh tface ita lic type . Boldface type is used for vecto r quant ities whe never the directi ona l aspect of the vector is a part of its mathematical representation. When writing vector equations, al ways be certain to preserve the mathematical distinction between vectors and scalars. In handwritten work, use a dis*This is the prin ciple of trannmiesib ility , which is discussed in Art. 2/2. Marwan and Waseem AI-Iraqi www.gigapedia.com

Figure 1/ 1

5

6

Chap t er 1

Int roduc ti on t o

s te t tcs

Iv: ,t point O . For what value of (} wou ld t he result s of parts (a) a nd (b ) be ident ical? A n" , (a) F = 400 N, M A = 131.6 N vm CW lbl F = 400 N, M a = 214 N r m CW o ~ 0 ar 1800

Problem 2/67 2 /68 Ca lcu lat e the moment of the 1200-N force about pin A of the bra cket . Begin by re placing th e 1200-N force by a force- coup le system at point C.

600 mm 2

r-600 - o j , mm , I I

;;/

a

' --'

I

1200 N

500 mm

\

Problem 2/65

200 mm

2 /66 Replace the cou ple and force shown by a single force F a pplied at a point D. Locate D by determinin g the distance b.

Problem 2/68 1 /69 A force F of magnitude 50 N is exerted on th e automobil e parking-brake lever a t the pos it ion x = 250 mm . Replace the force by a n equivalent force-couple sys tem at the pivot point O. A ns. R ~ 50 N ,Mo = 17.29 Nvm CCW

60N'm F

Problem 2/66 2/67 T he wr ench is subjected to the 200 ·N force a nd t he force P as shown. If th e equivalent of the two forces is a force R at a and a couple expressed as th e vector l\l = 20k N . m. det erm ine th e vecto r expressions for

P and R.

An". P ~ 40j N R = - 160j N

\

;----

x ----- '-\

Problem 2/69


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Article 2 / 5

Prob lems

55

2/70 The figure represents two integr al gea rs subjecte d to the toot h-contact forces shown. Replace the two forces by an equivalent single force R at the rotat ion axis 0 a nd a corres ponding couple M. Specify the magn itud es of R and M. If the gears wer e to star t from rest un der th e action of th e tooth loads shown, in wha t directi on would rotation take place? 1.5 kN

Problem 2/71 2/72 The weld at a ca n su pport a max imum of 550 Ib of force along eac h of t he n- and z-directions and a max imum of 1000 lb-Ft of moment. Deter mine the allowable range for t he direction () of the 600-lb force ap plied at A. Th e angle e is rest rict ed to 0 s 0 s 90°.

)0

A

~ 2,~~ nl

I

i -,

y

Problem 2/70

600 1b

2/71 Th e combined drive whee ls of a fron t-wheel-dri ve automobile are acted on by a 7000-N normal react ion force and a friction force F , both of which are exerted by the road surface. If it is known that t he resu ltant of these two forces makes a 15° angle with t he vertical, determine the equivalen t force- cou ple system at th e car mass center G. Treat this as a twodimensional problem. Ans. R ~ 7250 N Ma


~

7940 Nv m CW

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Problem 2/72

56

Chap te r 2

For ce Sy st em s

2 /6

RESULTANTS

T he properties of force , moment, and cou ple were deve loped in the previous four articles. Now we are ready to describe the resultan t action ofa group or sys tem of for ces. Most problems in mechanics deal wit h a system offorces , and it is usually necessary to reduce the sys tem to it s simplest form to describe its action. The resulta nt of a system of forces is the s implest force combination which can replace the original forces without altering th e exte rnal etTect on t he rigi d body to which th e for ces are a pplied. Equilibrium of a body is t he cond ition in which t he resul tan t of all forces acting on the body is zero. This condition is studied in stat ics. When the resultan t of all for ces on a body is not zero, the accelerat ion of the body is obtained by equati ng t he force resulta nt to th e pro du ct of the mass and accelera tion of t he body. T his condition is st udied in dyn amics. Thus, the determination of resultant s is basic to both statics and dynam ics. The most common type of force system occurs when the forces all act in a single plane, say , the x-y plane, as illu st rated by th e system of th ree forces F " F 2, and F 3 in Fig. 2/ 130. We obta in t he magn it ud e a nd direction of the res ultant force R by for mi ng the force polygon shown in part b of the figure, where the forces are added head- to-tail in any sequence. Thus, for any sys tem of coplanar forces we may write

Rx

(2 / 9 )

la l

Graphica lly, t he correc t line of action of R may be obtained by prese rving th e corre ct lin es of act ion of t he forces and adding th em by the para llelogr am law. We see th is in pa rt 0 of the figure for t he case of th ree forces where the sum R , of F 2 a nd F 3 is added to F I to obtain R . The principle of transmi ssibility has been used in th is process.

y I I

Algebraic Method We ca n use a lgeb ra to obtain th e resultan t for ce a nd its line of action as follows:

In J

Figure 2/13

1. Choose a convenient reference point and move all forces to that poin t. This process is dep ict ed for a t hree-force sys te m in Figs. 2/ 140 a nd b, where M " M 2 , a nd M 3 are th e cou ples resul ti ng from t he transfer of forces F l ' F 2- and F 3 from their respective or iginal lines of action to lin es of act ion t hro ug h poin t O. 2. Add a ll for ces at 0 to for m the resultan t for ce R, and ad d all couples to form t he resu lta nt cou ple M o. We now have the single forcecou ple system, as shown in Fig. 2/140.

:J.

Ma rwan and W aseem AI- Iraqi

In Fig. 2/ 14d, find the lin e of action of R by requ iri ng R to have a moment of Ala abou t point O. Note that th e force sys tems of Figs . 2/ 140 a nd 2/1 4d are equivalent, a nd that ;:'(Fd ) in Fig. 2/1 40 is equa l to Rd in Fig. 2/ 14d. WNW.gigapedia.com

Ar t icl e 2 / 6

I I I I

F2

M~k:

,/

~' o 1

,/

,

,/

d Z / X, / ,/

, -,

Res ulta nts

-,

M Z = F2 d z

F3

-----

M 3 = F3 d 3 (bl

(a)

M ? d; :V

~R 0

0

o d: M R

(d l

(e)

Figure 2/14

Principle of Moments

This process is summarized in equation form by

R = 2:F M o = 2:M = 2:(Fd )

(2/10)

Rd = M O

The first two ofEqs. 2/10 red uce a given system of forces to a force-couple system at an arbitrarily chosen but convenient poin t O. The last equation specifies the distance d from point a to t he line of action of R , and states th at th e moment of th e resultant force about any point a equa ls t he sum of t he moments of the origina l forces of the system about t he same point . This extends Varignon's theorem to the case of nonconcurrent force sys tems; we call this extension the principle of moments . For a concurrent system of forces where the lines of action of all forces pass through a common point 0 , the momen t sum '2.M o about that point is zero . Thu s, the line of act ion of t he resu ltant R = 2:F , deter mined by t he first of Eqs. 2/10, passes through point O. For a parallel force syste m, select a coordinate axis in t he direction of th e forces. If the resultant force R for a given force system is zero, the resulta nt of the syste m need not be zero because the resultant may be a couple. The three forces in Fig. 2/ 15, for instance, have a zero resulta nt force but hav e a resultant clockwise couple M = F3d. Marwan and Waseem AI-Iraqi

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Figure 2/ 15

57

58

Chapter 2

Fo rce Systems

Sample Problem 2 /8

y

Determine the resultant of the four forces and one couple which act on the plate shown.

2m, 60N

I

5m - - i I I

50 N 45'

,

2m

Solution. Point 0 is selected as a convenient reference point for the forcecouple system that is to represent the given system.

= 40 + 80 cos 30' - 60 cos 45' = 66.9 N Ry = 50 + 80 sin 30' + 60 cos 45' = 132.4 N A n" . R = ) (66.9)2 + (132.4)2 = 148.3 N

[Rx

R,

[Ry [R =

o=

CD

[M o

=

~ (Fd)]

ta n

- 1

132.4 63 2' 66.9 = .

2m 40 N 1m y I

Ans.

Mo = 140 - 50(5) + 60 cos 45' (4)

60 sin 45' (7)

R = 148.3 N (0 )

0 =63.2'

[Mol =

= - 237 Nr m

237 N'm

- x

Th e for ce-couple system consisting of R an d 1.10 is shown in Fig. G. We now determine th e final line of action of R such t hat R alone represents th e original system. [Rd

= lMoll

148 .3d

= 237

d = 1.600 m

An s.

Henc e, the resul ta nt R may he applied at any point on t he line which mak es a 63.20 a ngle wit h the x-axis and is ta nge nt at point A to a circle of 1.6-m radius with center 0 , as shown in part b of t he figur e. We apply th e equa tion Rd = M o in an absolute-value sense (ign oring any sign of M o ) and let t he physics of th e situation, as depicted in Fig. G , dictate th e final placement of R. Had M o been cou nterclockwise, t he correct line of action of R wou ld have been th e tangent at point B . The resu ltant R may also be located by determini ng its intercept distance b to point C on the .r-axis, Fig. c . With R, and Ry act ing throug h point C, only Ry exerts a moment about 0 so that

RjJ = lMol

and

b

237 132.4

= - - = 1.792 m

(b)

(c)

132.4x - 66.9y = -237

I

y

I

I

- O+ - - -x '------++: - b-"-j+' R Hel p ful Hints

Alternat ively, the y -intercept could hav e been obtained by noting t hat th e me. ment about 0 would be du e to R, only. A more formal approach in determining the final line of action of R is to use the vector express ion r XR =l\I o where r = xi + yj is a position vector ru nni ng from point a to any poin t on the line of action of R . Substituting the vector expressions for r , H, and M o and carrying out th e cro ss product result in (xi + y j) x (66.9i + 132.4j ) = - 237k

(l32.4x - 66.!b')k = - 237k Thus, the desired line of act ion, Fig. c, is given by 132 .4x - 66.!b'

as a moment center eliminates any moment s due to the two forces whic h pass through O. Had the clockwise sign conventi on been adopted, M o wou ld have been + 237 N . m, wit h the plus sign indicating a sense which agrees with the sign convention. Either sign convention, of course, leads to the conclusion of a clockwise mementM o ·

Q) Note that the vector approach yields

= - 237

Q) By setting y

= 0, we obtai n x = -1.792 m, which agrees with our earlier calculat ion of t he distance b. Marwan and Was eem AI-Iraqi

CD We note that the choice of point 0

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sign information automatically, whe reas the scalar approach is more physically oriented. You should master both methods.

Article 2 /6

PROBLEMS Introductory Problems 2 /73 Deter mi ne the resul tant R cf'the t hr ee ten sion forces act ing on the eye bolt . Find the magnitud e of R and the a ngle Ox which R makes with the positive .r-axis. A ns. R ~ 17.43 kN, ex ~ 26 .1"

P ro bl em s

y

y

FI

FI

FI

d --..j

V

59

Y

(~)-j-x F-x F F 0

,

20 kN

(c!

(b l

(aJ

d

Problem 2/75

I

2/76 Determin e a nd locat e th e resultan t R of the two forces a nd one couple act ing on the I-beam . BkN

25 kjc-m

- - - -+-- x

•

8kN

Problem 2/76 4 kN

Problem 2/n 2/74 Determ ine the equivalen t force- cou ple system at th e center 0 for eac h of the th ree cases of forces being applied along the edges of a squa re plate of side d . y

y

I

F I

2/77 If t he resu lta nt of the t wo forces and cou ple M passes through poin t 0 , determine M . An s. M ~ 148.0 N· m CCW

Y

F I

FQj_x x T Q]-x if F 0!--d-j

!--d-!.j

!--d-j

(0 )

Ib l

Ie)

320N

--1

400 N

Problem 2/74 Problem 2/77

2/75 Determine the eq uivale nt force-coup le syste m at t he or igin 0 for each of t he t hree cases of forces bein g app lied along the edges of a regu lar hexagon of widt h d . If the resu ltant ca n be so expresse d, replace this force- coup le system wit h a single force.

A ns. (a) R (b)

(e )

R

~

~

2Fi along y

R = - 2Fi along y

F(- i + J3j l alongy

Marwan and Wasee m AI-Iraqi

~

~ - J3x

-

~

~ ~ +

~

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60

Chapter 2

Force Systems y

2 /78 A commercial a irliner with four jet engines, eac h producing 90 kN of forward thrust , is in a ste a dy, level cruise wh en engine number 3 sudde nly fails. Determine a nd locate the r esultant of th e three remaining engine thrust vectors. Treat this as a twodimen sional problem .

I IO m - - l

I I I ---- I ~ -

- x

90 kN _ --€lW/

Prob lem 2/80 2/81 T he flanged steel cantilever beam with riveted bracket is subjec te d to th e couple and two for ces shown, a nd their efTect on t he design of the attach me nt a t A mu st be det ermined. Replace th e tw o for ces a nd couple by a n equ ivalent couple .M a nd resultant force R at A . Ans. R ~ 1.644i + 1.159j kN Al A 2.22 kN'm CCW

" 9OkN _ -

-€l

9OkN _--G~

2 kN ' - -- -

Problem 2/78 2/79 Rep lace the three for ces acting on the ben t pipe by a single equ iva lent force R. Spec ify the distance x from point 0 to the point on th e x-ax is through whic h the line of action of H passes. An s. R ~ - 50i + 20j Ib, x ~ 65 in. (off pipe ) y

,,, ,,,

A

L _ _x

5"

J(Y'

L2kN

1 /82 Th e gear and attached V·belt pull ey are t u rning counterclockwise and a re subjected to the tooth load of 1600 N a nd th e 800-N a nd 450-N tens ions in th e V·belt. Rep resen t th e action of th ese three forces by a resul tant for ce R at 0 a nd a couple of magnitude M. Is th e unit slowing down or speeding up?

'0

-+-

500 N'm

Problem 2/81

40 lb

1--10"

{ I

50 lb

O.5m

1.5 m

60 lb

Problem 2/79

Representative Problems 2/80 Th e directions of th e two thru st vectors of a n expe rimental a ircraft can be inde pendently cha nged from th e conventio na l forw ard direction within lim its. For th e th rust con figuration shown, det ermine th e equiva lent force-coup le syste m a t point O. T he n replace th is force-eouple syste m by a single force a nd specify th e point on t he x-axi s through wh ich the line of action of this resu ltant pas ses. Th ese resu lts a re vital to assessing design performance. Marwa n and W aseem AI-Iraqi

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Problem 2/82

Art icle 2 /6 2 /83 Repl ace the t hree forces which act on t he ben t ba r by a force-couple sys tem at t he support point A . T hen det ermine the .r-intercept of t he line of action of t he sta nd-a lone resu lta nt force R . Ans. R ~ 400i - 30 10j Ib M A ~ 18,190 Ib-ft CW x ~ 6.05 ft

Pr ob lems

61

1 /85 While sliding a desk toward the doorway, th ree st u dents exert t he forces shown in the overhead view. Dete rm ine t he equiva lent force-couple sys tem at point A . T hen determ ine t he equation of the line of act ion of t he resultan t force. Ans. R ~ 45i - 15j Ib, MA ~ - 1650k Ib-in . 1 110 s ~ - 3x + :3

y y

I I I

2500 1b

I

--

I I

~~---

I

--

-

4' -

60" -

-

-

-

I

1200 lb

25 1b

151b

j 30"

---L

A

201b Problem 2/83 1 /84 Two integra l pulleys a re su bje cte d to t he belt ten sions shown . If th e resulta nt R of t hese forces passes t hrough the cente r 0 , det ermine T a nd the magn itude of R an d the coun te rclockwise an gle II it mak es with th e x-axis. 160 N 200 N

.------'-.",-- - -

Problem 2/85 2/86 T he asy mmet ric roof truss is of t he ty pe used when a nea r norma l angle of inciden ce of sun light onto t he sout h-faci ng su rface ABC is desirab le for sola r e nergy pu rposes. Th e five vert ica l load s repr esent the effect of th e weights of the t r uss an d supported roof ing materials. T he 400-N load repr esents the effect of'wind pressu re. Determine t he equivalent force-couple sys tem at A. Also, compute th e x-intercept of th e line of act ion of the system resultant t rea ted as a sing le force R.

-- x 500 N

150 N

250 N Problem 2/84

Prob lem 2/ 86


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62

Chapte r 2

For ce Systems

2/87 Determine th e .r- and y-axis intercepts of th e line of ac tion of th e resultant of th e three loads applied to th e gearset. A ns. x = 1.637 m , y = - 0.997 m y I

2/89 Th e rollin g rear wheel of a fro nt-wheel-dr ive automobile which is accelera ting to th e right is subjected to th e five forces and one moment shown . Th e for ces A x = 60 lb and A v = 500 lb are forces tran sm itt ed from th e axle to the wheel, F = 40 lb is th e frict ion force exerted by th e road su rface on the t ire, N = 600 Ib is th e normal react ion force exerted by th e road su rface, and W = 100 Ib is the weight of the wheel/ti re unit. Th e couple At = 2 lb-Ft is the bear ing frictio n moment. Determine and locate the resultant of the system. A ils. R = 20i Ib, cI = 2.40 ft above A

.,v, ,L __ -

x

Problem 2/87 2 /88 As part of a design test , the camshaft-drive sprocket is fixed and th en th e two forces shown are a pplied to a len gth of belt wrapped around th e sprocket. Find th e res ulta nt of th is syste m of two forces and det er min e whe re its line of action intersects bot h the x and y-axes. y

N

Problem 2/89 1 /90 A rea r-wheel-driv e car is stuck in th e snow bet ween other parked cars as shown. In an attempt to free the car , t hree students exert forces on the car at points A, B, and C while th e dri ver 's act ions re sul t in a forward thru st of 40 Ib acting parallel to the plan e of rotati on of each rear wheel. Treating the problem as two-dimensional , det ermine th e equ ivalent forcecouple syste m at th e car center of mass G and local e th e position x of th e point on the car cente rline through which th e resu ltant passes. Neglect all forces not shown.

I I

y I

T, = 500 N

[

Problem 2/88

Problem 2/90


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A rti cl e 2/ 6

2 /91 An exhaust system for a pickup tru ck is shown in the figure. The weights W,!' Will' and W, of t he headpipe,

o

l

I

IVII I

t he figure. T he left foot of the rider exerts t he 40-lb force, while the use of toe clips allows the right foot to exert th e nearly upward 20-lb force. Deter mine the equ ivalent force- couple system at point O. Also, determ ine the equation of the line of action of th e system resultan t treated as a single force R . Treat th e problem as two-di men sional.

FIJ

B

A

I

IVIII

IVI

I.,

J~ r

I

0.3.5

- 0 .5 ---.02 1 0.65 - -,- 0.65 -,..,.-- 0.5 --'I 0.4 · Dimens ions in me ters

Problem 2/91

Problem 2/92


63

2 /92 Th e pedal-chai nwheel unit of a bicycle is shown in

muffler , and tailpipe are 10, 100, and 50 N, respectively, and act at the indicated point s. If the exhaus tpipe ha nger at point A is adju ste d so that its ten sion FA is 50 N, determine th e requ ired forces in the ha ngers at point s B , C, and D so that the force- couple syste m at point a is zero . Why is a zero forcecouple system at a desi rable? Ail s. F n = 98.9 N, Fe ~ FIJ ~ 6.42 N Fe

Probl e m s

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64

Cha p t e r 2

Fo rc e Syst e m s

SECTION B. THREE-DIMENSIONAL FORCE SYS T E MS

2 /7

RECTANGULAR COMPONENTS

Man y pro blem s in mecha nics requ ire a nalysis in t hree dimensions, a nd for such problems it is often necessary to resolve a force in to it s three mu tu ally perpe ndicular compone nts. Th e force F acting at point o in Fig. 2/ 16 has th e rectangular components Fx' F'..,., Fz' where z I I

Fx

I

=

F cos Ox

Fy = F cos By

F z = F cos

e,

F

+ Fy 2 + Fz2

= ' IFx 2

F = F) + Fyi + Fzk

(2/ 11)

F(i cos Bx + j cos By + k cos Bz )

F

Th e unit vectors i ,j, an d k a re in t he x-,y-, and z-direc tions, respect ively. Usin g the direction cosin es of F , which are l = cos Ox. m = cos fly, and n = cos Hz, wher e [ '2 + m 2 + n '2 = 1, we may write the force as'

F(/i

+ mj + nk ) )

(2/ 12)

Figure 2/16

We may rega rd th e rig ht-side expressio n of Eq. 2/ 12 as the force magn itude F ti mes a unit vect or " T-' which characte rize s th e di rect ion of F , ol' (2 / 12a)

It is clear from Eqs. 2/12 and 2/120 t hat n F = I i + mj + Il k, wh ich shows that the sca lar components of th e unit vector n F are th e direct ion cosin es of t he line of action of F . In solving th ree-dime nsional prob lem s, one must usually find th e .r, y, an d z sca la r compone nt s of a force. In most cases , the direct ion of a force is describ ed (a) by tw o points on the line of actio n of the force or (b) by two angles which or ient th e line of actio n. (a) Specification by two points on the line of action of the force. If th e coordinates of points A and B of Fig. 2/ 17 are kno wn, th e force F may be written as

F

Figure 2/17 Marwan and Waseem AI-Iraqi

F nF

=F

-

AB AB

F

(X2 / (X2 -

xl )i + XI )2

+

(Y2 -

(Y2 -

YIlj + y l l2

(z 2 -

+

(Z2 -

z l lk Z I )2

T hu s the x , y, and z sca lar components of F are the scala r coefficients of th e un it vectors i , j , and k , respect ively. www .gigapedia.com

Articl e 2/7

(b) Specification by two angles which orient the line of action of the force. Consider the geometry of Fig. 2/ 18. We ass ume that the angles e and '" are kn own . First reso lve F into hori zontal and vert ical components .

Rectan gular Compo n ents

z I I F,

Fxy ~ F cos '" F z = F sin '" Then resolve the horizontal component FXY into x- and y-components.

Fx Fy

= =

FXY cos Fxy sin

e~ e=

F cos '" cos F cos '" sin

e e

The quantiti es Fx ' Fy , and F; are the desired scalar components of F. The choice of orientation of the coordinate system is arbitrary, wit h convenience being the primary consideration. However, we must use a right-handed se t of axes in our th ree-dimensional work to be consistent with the r ight-hand-ru le definition of the cross pr oduct . When we rot at e from the x - t o the y-axis t hr ough t he 90" angle, the positive direction for the z-axis in a right-han ded system is that of the advancement of a r ighthanded scre w rota te d in t he same sense. This is equivalent to the r ightha nd rule. Dot Product

We can express the rectangular components of a force F (or any other vect or ) with the aid of the vecto r opera tion known as the dot or scal ar prod uct (see item 6 in Art. C/7 of Appendix C). Th e dot pr oduct of two vectors P and Q, Fig. 2/ 190, is defined as the pr oduct of their magnitudes times the cosine of the angle ex between them. It is writt en as

p.Q = PQ cos ex We can view t his pr odu ct either as t he orthogona l proj ection P cos ex of P in the direction of Q multiplied by Q, or as the orthogonal pr ojection Q cos ex of Q in t he direction of P multiplied by P. In eit her case the dot product of the two vectors is a scalar quant ity. Thus, for instance, we can express the scalar component Fx = F cos Ox of the force F in Fig. 2/16 as Fx = F · i, whe re i is t he uni t vecto r in the x-direction.

F/ \

- --,.. Q

~-1" 1\1\ \

F~

--\.=

- \ - ---

1.6 01

1.2m

__________

/"

0.8 m

7"-

/ 0.8 m

-x

Problem 2/117

x-

"' y

Problem 2/11 5 1 /116 The tu rn bu ckle is tightened unti l t he te nsio n in ca ble AB is 1.2 kN. Ca lcu late the magn itu de of the moment about point 0 of the force act ing un point A .

1 /118 In an a tt empt to pull down a nearly sa wn- through branch. th e tree su rgeo n exerts a 400 -N pull on the line which is looped around th e branch a t A. Det ermine the momen t about point C of th e force exerted on th e branch a nd sta te the magnit ude of th is mo me nt . z J J

12 m 2m--:-_ 4 m "

~ I

- l~

11 m

-,...------2m

l

o

...... -_L... .../

r" "

.

.. .,.

»

I

1.5 m

J--.

I

8m ---::'iI1,-"..'"

>

:

-

-~~'( IJ 611~ ~

2m B


Marwan and W aseem AI-Iraqi

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Article 2 / 8 2 /119 The figure of Prob. 2/101 is shown again her e. If th e magnitude of the moment of F a bout line CD is 50 N· m , determine th e magnitude of F. Ans. F = 228 N B I I

r

I

Prob le m s

81

2/121 A 50-lb force is a pplied to th e cont rol ped al as shown. The force lies in a plane parallel to th e x -z plan e a nd is perpendicular to BC. Det ermine th e moments of this force about point 0 and about th e shaft OA. Ans. M o = - 90.6i - 690j - 338k Ib-in . M O, \ ~ - 690 Ib-in.

/

O.2m

,Ie------- _

~

0.2 m

-

D

A

O.2m

Problem 2/119

Representative Problems 2 /120 Two 1.2-lb thrusters on th e non rotating sa tellite are simu ltaneo us ly fired as sho.....n. Compute th e mome nt associated with this couple an d state about wh ich satellite a xes rotations will begin to occur.

Problem 2/121 2/122 A mom ent (torque ) M applied to t he shaft a nd at tach ed a rm causes a te nsio n T of 120 Ib a pplied to A by t he restraini ng cab le AB. Det ermi ne the moment 1\1 0 of the ten sion about point O.

I- - 20·:L;'-I

.--y 25" I

/ X

-I..

/

1.21b

Problem 2 / 120

Problem 2/122

Marw an and W aseem Al-lraqi

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82

Ch ap te r 2

Forc e Sy st ems

2 /125 A 50- N hor izontal force is a pplied to th e handle of th e industrial water valve as shown. The force is perpendicular to the vertical pla ne containing lin e OA of th e ha nd le. Deter mine th e equivalent forcecoup le sys tem at point O. Ail s. R = - 38.3i - 32.1j N 1\I" = 643 i - 766j + 6250 k N . mm

2/123 A 300-N force is applied to the handle of the winch as shown. The for ce lies in a plane which is parallel to th e y -s: plane and is perpendicu lar to line AB of th e handle. Determine th e momen ts of this force about point 0 a nd about the x-axis. Ails. M() = - 98.7i + 17.25j + 29.9k Nr m M x ~ - 98.7 N r m

~

1$i'R?i~i::i

-

~ 40 mm

" "75 mm

~75 mm

B

/=--=:~

p =' Problem 2/125 2/126 A space shuttle orbit er is subj ected to th ru st s from five of t he e ngi nes of it s reaction cont rol system. Four of th e thru st s a re show n in th e figu re; th e fifth is an 850-N upw ard thru st a t the righ t rear, symmetric to the 850-N t hru st shown on the left rear. Com pute t he mome nt of these force s a bout point G a nd show that t he forces have th e sa me moment ab out a ll points.

Problem 2/123

2/124 Compute the mom en t l\1() of t he 250-lb force about the axi s 0 -0 . I

o

1 2" ~ r-------------I

------.""-

250 lb

I

~m

)'

I 8"

j

o

I

1700 N

1700 N 1700 N

where r l is a vector from 0 to any point on the line of action of F l' When all forces are shifted to 0 in this manner , we hav e a syste m of concurrent forces at 0 and a syste m of couple vectors, as represent ed in part b of th e figur e. Th e concur rent forces may then be added vectorially to produ ce a resultant force R, and the couples may also be add ed to produce a resulta nt coupl e M , Fig. 2/2&. T he gen era l force system , then, is reduced to R

= F1 +

M = M)

=

F2 + F3 +

+ M2 + M3 +

~F

=

(2 /20) ~( r

x F)

T he couple vectors are shown through point 0 , but becau se they are free vectors, they may be represen ted in any parallel positions. The magnitudes of the resu ltant s and their component s are

n, = ss,

Ry

= ~Fy

R,

= ~F,

R = J( ~Fx )2 + (~Fy)2 + (~F,l2 Mx

= ~ (r

M" = ~ ( r x

x F ),

M

F )y

M,

= ~ (r

(2/ 2 1) x F),

= "1Mx 2 + My 2 + M z2

/. ' a)

' bJ

(e )

Figure 2/28 Marwan and Wa see m AI- Iraqi

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Result ant s

85

86

Chapter 2

For ce Systems

Th e point 0 selected as th e point of concur re ncy for th e forces is arbitrary , and th e magnitude and directi on of M depend on the particular point 0 selected, Th e magnitude and direct ion of R, however , are the same no matter which point is selected. In general, any system of forces may be replaced by its resu lt ant force R and th e resultant couple M. In dyn ami cs we usually select the mass cente r as the reference point. The change in the linear motion of the body is determin ed by th e resultant force, and th e change in th e a ngular motion of th e body is det ermined by th e resul tan t couple. In statics, the body is in complete equilibrium when the res ultant force R is zero and the resultant couple M is also zero. Thus, the determination of resultants is ess ential in both statics and dyn ami cs. \Ve now examine the result ants for several special force systems. Concurrent Forces. \Vhen forces are concurrent at a point, only the first of Eqs. 2/20 needs to be used becau se there are no moment s about the point of concurrency.

Parallel Farces. For a syste m of par allel forces not a ll in t he same plan e, th e magn itu de of the para llel resu ltant force R is simply the magnitude of t he algebra ic sum of the given for ces. Th e position of its line of action is obtai ned from the prin ciple of moments by requiring that r x R = Mo . Here r is a posit ion vector extending from the forcecouple refe re nce point 0 to the fina l lin e of actio n of R, a nd M o is the sum of the moments of th e ind ividual forces about O. See Sample P roblem 2/ 14 for a n exa mple of parall el-force systems. Coplanar Forces. Article 2/6 was devot ed to this force syste m. Wrench Resultant. When the resul tan t couple vecto r M is parallel to th e res ultant force R, as shown in Fig. 2/29, the resultant is called a wrench. By definition a wrench is posit ive if t he couple and force vectors point in the same direction and negative if they point in opposite directions. A common example of a positive wrench is found wit h the application of a screwdriver, to drive a right-handed screw. Any gen eral force system may be represen ted by a wrenc h applied alon g a un ique line of act ion. Thi s reductio n is illustrat ed in Fig. 2/ 30, where part a of t he figure represent s, for the general force system, the resultant force R act ing at some poin t 0 a nd the correspond ing resu lta nt coup le M. Although 1\1 is a free vector, for convenie nce we represent it as acting th rough O. In part b of the figure , M is resolved into components M, along t he direct ion of R and M 2 norma l to H. In par t c of t he figu re, the coup le

Pos iti ve wrench

Negative wrench Figure 2/29

M arwan and W aseem AI-Iraqi

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Articl e 2/9 M

I

;

t 7"1.---'

II

----!C, '-iIiIII (a)

fbi

fe)

fd)

- II

Figure 2/30

M 2 is repl aced by its equivalent of two forces R and - R se parated by a distan ce d = M 2 /R with - R appli ed at a to cancel th e or iginal R. This ste p leaves th e resul tant R, which act s a long a new and unique line of actio n, and the parallel couple M i , which is a free vector, as shown in part d of the figure. Th us, the resultants of the origi na l general force sys tem have been transformed into a wrench (positive in this illustration ) with its un ique axis defined by th e new posit ion of R. We see from Fig. 2/ 30 that the axis of the wrench resu ltant lies in a plan e th roug h a nor mal to the plan e defined by R and M. Th e wrench is th e simplest form in which th e resultant of a gen er al force system may be expres sed. Thi s form of the resultant, however, has limited a pplication , because it is us ually more conve nient to use as the reference point some point 0 such as the mass center of the body or another convenient origin of coordinates not on the wrench axis.

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Resultants

87

88

Ch ap t e r 2

Fo rc e Syst ems

Sample Problem 2 /13

z I I I

Determine th e resul tan t of the force and couple syste m which acts on th e rectangular solid.

50lb

1000I b·in .

cb Solution. We choose point 0 as a convenient reference point for t he initi al step

'--,.

of reducing th e given forces to a force-c ouple syste m. Th e resultant force is

Q)

I

R ~ ~F ~ (80 - 80li + (100 - 100)j + (50 - 50)k ~ 0 Ib

12"

The sum of the momen ts about 0 is l'tI o = [50(16 ) - 700J; + [80(12) - 960ij + [100(10) - 1000]k lb-in . = 1001 Ib-in .

/

/'

Hence, t he resu lta nt consist s of a couple, which of course may be applied at any point on the body or the body extended.

10" 100lb

300 111m

/

- x Probl em 2/155 2 {156 Rep resen t t he resultant of t he t hree forces a nd cou ple by a force-cou ple system loca ted a t point A . 3 kN

- - - - - - -A

-

- - 2 111

5 kN 4kN

Pro blem 2/156 2{157 Reduce t he given loadi ng syst em to a force-couple sys tem at point A . Then det ermine the dist an ce .r to the righ t of point A at which the resul tant of t he t hree forc es acts. A il s. R ~ SO Ib down, M A ~ 1240 Ib-in . CW .r = 15 .5 in .

Prob lem 2/ 158 2{159 The direct ion s of rotat ion of t he inpu t sha ft A a nd out put sha ft B of t he wor m-gea r reduce r are ind icated by the curved da shed arrows . An inpu t torque (cou ple) of 80 N · m is applied to sha ft A in the di rection of ro tation . The output shaft B su pplies a to rque of 320 N . m to t he mac hine whic h it drives (not shown). The shaft of t he d riven machine exerts a n equal a nd opposite reacting torque on t he output sha ft of t he red ucer . Determ ine the resu lt a nt M of the t wo couples which act on t he reducer u nit a nd calcul at e t he directi on cosine of M with respec t t o t he x-axis. Ans. M ~ - 320i - SOj N · m cos fI% = - 0. 970 r

IS0lb

200 lb

20"

S"

B)(

-:

A

IS"

150 111111......,"'-

.i-:

x

3001b

Pro ble m 2/157 Problem 2/159


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Rev iew Prob l e m s

2 /160 Replac e th e force P a pplied at point A by an equ iv-

alent force- coupl e system at point B.

,--

z

I I AI

I I

P

I

~ ' II _ 3

99

2/162 When the pole OA is in the position shown , the te nsion in cab leAB is 600 lb. (0 ) Wr ite th e tens ion force exe rte d on point A as a vector usin g th e coordi nat es shown. (b) Dete rmi ne the momen t of this force about point 0 a nd state t he moments abo ut the X - , y- , and z-axes. (c ) Det ermin e the projection of thi s te ns ion force onto line AD .

--J

OA = 50' OB = 40' OC = 30'

b b

b X /

/

- y

B

40'

Problem 2/ 160

- .r

2(161 Th ree couples a re form ed by t he three pairs of equal a nd opposit e forces. Det ermine the resu lta nt M of the th ree couples. Ails. M = - 201 - 6.77j - 37.2k N· m

2/163 The combined action of the three forces on the base at 0 may be obtained by esta blishing their resultant through O. Determine the magn it ude s of R and the accompa nying coup le M. Ails. R ~ 1093 lb, M = 9730 lb-tt

~~{I -, 7;~=_-:'" 120

N

Pro blem 2/161

Problem 2/163


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100

Ch ap t er 2

Forc e Systems

1 /164 During a drilling operation, the small robotic device is subjected to a n 800·N force a t point C as shown. Replace this force by an equivalent force-cou ple systern at poin t O.

-1 / 166 The trap door OA is raised by th e cabl e AB wh ich passes over the sma ll guid e pull eys at R. The ten sia n everywhere in t he cable is T , and this ten sion applied at A cau ses a momen t Al 0 a bout t he hi nge . . . Al() at O. P lot th e qu an tity T ' wh ich I S t he momen t a rm rela tive to 0 of the ten sio n a pplied a t A , as a fun ction of t he door a ngle (J for the range 0 es II :=:; 90~ . Determin e th e maximum a nd mi nim um valu es of th e moment arm over thi s ra nge of O.

-0

•' C

I 250mm

T ~-i'r.i"::"-'-

\ •

L

I

0.4

;;00\ F = 800 N

III

1

0.5

III

Problem 2/ 166 Problem 2/ 164

'!

' Computer-Oriented Problems

' 2/ 165 Four force s are exerted on t he eyebolt as shown. If the net effect on the bolt is a direct pull of 600 Ib in the y -di rect ion, det e rmine the necessa ry values of T a nd fl. All", T ~ 204 lb, /I = 21.70 240lb

T

I

\

' 2 / 167 Wit h th e 300-lb cylindrical part P in it s gr ip, th e rob otic a r m pivot s ab out 0 through th e range - 4 5~ :=:; (I :=:; 45° with th e an gle at A locked at 120°. Determine a nd plot (as a fun ction of m th e mom ent at 0 due to the combined effect s of t he 300-lb part P, t he 120·lb weight of member OA (mass cente r a t G , l, a nd the 50·lb weight of memb er AB (mass cent e r a t G 2 l. Th e en d gri p is include d as a pa rt of member AB. T he lengt hs 1.. 1 a nd 1.. 2 a re 3 It and 2 Ft, respectively. Wha t is th e maxim um value of Al() and at what value of Ii does this maximum occur? Ail S. M a : : : 1230 cos lJ + 650 cosl60° - OJ lb-It (M U )m ll x ::::: 1654 lb-Ft at tJ ::::: 19.90

/f'

y

I I

F,

L _ _x

X

F. ~ Fa

y

3. Parallel

_ F, _ a F _-

4. General

y

I

Ft

LFx = 0 'L Fy = 0

LFx = 0

LMz

=0

LFx = 0

LAll

=0

I

L _ _x F,

/

F, Fa Y I I

M)

t

'LFy = 0

L __ x

F,

Figure 1/1

Ca tegory' 4, equilibrium of a general system of forces in a plane (r -y ), requi res th e two force equations in the plane and on e momen t equa t ion about an axis (z-axis) normal to the plan e. Two- and Three-Force Members p

-p

-p Two-force mem bers Figure 1/4

Marw an and W aseem AI-I raqi

You shou ld be alert to two frequ en t ly occur ring equilibr iu m situa tions. The first situation is the equilibrium of a body under the act ion of two forces on ly. Two examples are shown in Fig. 3/ 4, and we see that for such a two-force member to be in equilibrium, the forces must be equal, opposite, a nd collinear . The s hape of t he memb er does not affect this simple requiremen t. In the illustrations cited, we cons ider the weights of the members to be negligible compared wit h th e a pplied forces. The second situation is a three-force member , which is a body under the action of three forces, Fig. 3/00. We see that equil ibrium requires the lines of action of the t hree forces to be concurrent. If t hey were not concurrent, then one of the forces would exert a resultant moment about th e point of intersection of the ot her two, which would violate the requirement of zero moment about every point. The only exception occurs when the three forces are para llel. In th is case we may consider the point of concurrency to be at infinity . www .gigapedia.c om

Article 3 / 3

Equ i librium Cond itio n s

Th e pri nciple of t he concurrency of three forces in equilibrium is of cons iderable use in carrying out a graphical solution of the force equation s. In this case th e polygon of forces is drawn and made to close, as shown in Fig. 3/ 5b. Frequently, a body in equilibrium under the action of more than three forces may be reduced to a three-force mem ber by a combination of two or more of the know n forces. Alternative Equilibrium Equations

F3

In addition to Eqs. 3/2, t here are two ot her ways to express t he general condi tions for th e equilibrium of forces in two dimen sions. Th e first way is illustrated in Fig. 3/ 6, parts (a) a nd (b ). For the body shown in Fig. 3/6a, if"i.MA = 0, the n t he resul tan t, if it still exists , cannot be a couple, but must be a force R passing through A . If now th e equation ";;;F x = 0 holds, where the x-direction is arbitrary, it follows from Fig. 3/ 6b that th e resultant force R, if it st ill exists, not only must pass through A , but also must be perpendicular to the x-direction as shown . Now, if ";;;AlB = 0, where B is any point such th at th e line AB is not perpendicular to th e x-direction, we see that R must be zero, and thus th e body is in equilibrium. Therefore, an alternative set of equilibrium equations is "i.MA

=

0

where the two poin ts A a nd B must not lie on a line perp endi cular to t he x-direction. A third formu lat ion of the equilibrium conditions may be made for a coplanar force syste m. Thi s is illustrated in Fig. 3/ 6, parts (c ) a nd (d ). Again, if ";;;M A = 0 for any body such as th at shown in Fig. 3/6c, th e resulta nt , if a ny, must be a force R through A . In additio n, if "i.MB = 0, th e resultant, if one still exists, must pass through B as shown in Fig. 3/6d . Such a force cannot exist, however, if "i.Mc = 0, where C is not

I .\fA = 0 sa tis fied

--- ~ •

__ x

B

A

(a)

(b)

l:MA=O} . fi d LAIn = 0 satls Ie

LAtA = 0 s atis fied

1 (e)

(d)

Figure3/6 Marwan and Wa see m AI- Iraqi

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(u )

(b )

Three-force member

Clas ed polygaa satis fies I F = 0

Figure3/5

117

118

Chapt er:}

Equ ilibrium

collinea r with A and B. Thus , we may write the equations of equilibrium as ":::M c = 0 where A , B , and C ar e an y three points not on the same stra ight lin e. When equilibrium equations are written which are not independent, redundant information is obtained, and a correct solution of the equat ions will yield 0 = O. For example, for a genera l problem in two d imens ions with three unknowns, three moment equations written about three points which lie on the same straight line are not indepe ndent. Such equations will conta in duplicated information, and solution of tw o of th em can at best det erm ine two of t he u nknowns , with the third equat ion merely ver ifying the identity 0 = o.

Constraints and Statical Determinacy The equilibr ium equations developed in this arti cle a re both necessa ry a nd su fficient condit ions to establish the equilibr ium of a body. However, they do not necessarily provide all the informat ion required to calculate all the unknown forces whi ch may act on a body in equ ilibrium. Whether the equatio ns are adequate to det ermine a ll th e unknowns depends on the characteristics of the constra ints against possible movement of the body provided by its su pports. By constraint we mean the restriction of movement. In Example 4 of Fig. 3/ 1 the roller , ba ll, and rocker provide constraint normal to the surface of contact, but none tangent to the surface. Thus, a tangential for ce cannot be su pported. For the collar a nd slide r of Example 5, cons traint exis ts only norm al to the guide. In Example 6 the fixed-pin connection provides constraint in both directions, but offers no resistance to rotation about the pin unless the pin is not free to turn. T he fixed su pport of Example 7, however, offers constraint agai nst rotation as well as lateral movement. If th e rocker which supports t he truss of Example 1 in Fig. 3/2 were replaced by a pin joint, as at A , th ere would be on e additiona l constraint beyond those required to support an equilibrium configuration with no freedom of movement. The three scalar conditions of equilibrium, Eq s. 3/ 2, would not pro vide sufficient information to det ermine all four unkn owns, since Ax and Bx could not be solved for se parate ly; only their sum could be determined. These two components of force would be de pen den t on th e defor mation of th e membe rs of t he t russ as influenced by their corresponding stiffness properties. The horizonta l reactions Ax and Bx would also depend on any initial deformation required to fit th e dimensions of th e stru cture to those of t he foundation between A and B. Thus, we ca nnot det ermine Ax and Bx by a rigid -body analysis. Again referring to Fig. 3/2, we see that if the pin B in Examp le 3 were not free to turn , the support could tra nsmit a couple to the beam t h ro ugh th e pin . Therefor e, th ere would be four unknown sup port ing reactions acting on the beam, namely, the force at A, the two componen ts of force at B, an d the cou ple at B. Conseq ue ntly th e t hree indeMa rwan and W aseem AI-I raqi

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Articl e 3 / 3

pendent scalar equations of equilibrium would not provide enough information to compute all four unknowns. A rigid body, or rigid combi nat ion of eleme nts treated as a single body, which possesses more external supports 0 1' const raints than are necessary to maintain an equilibrium position is called statically indeterminate. Supports which can be removed without destroying the equilibrium condition of the body are said to be redunda nt . The number of redundant supporting elements present corresponds to the degree of statical indeterminacy and equals the total number of unk nown external forces, minus the number of available independent equat ions of equilibd um. On th e oth er hand, bodies which are su pported by the mi nim um number of constraints necessary to ensure an equilibrium configuration are called stati cally determinate , and for such bodies the equilibrium equat ions are sufficient to determine the unknown external forces. T he prob lem s on equilibr ium in this article and throughout Vol. 1 S tatics are gene ra lly rest r ict ed to statically deter minate bodies where the const raints are just sufficient to ensure a stable equilibrium configuration and where the unknown supporting forces can be completely determ ined by the avai lable inde pen den t equations of equilibrium . We must be aware of the nature of the constraints before we atte mpt to solve an equilibr ium problem . A body can be recogni zed as stat ically indeterminate when there are more unknown exte rnal reactions than there are available independent equilibrium equat ions for the force sys tem involved. It is always well to coun t the nu mber of u nk now n va riables on a given body an d to be cert ain that an equal numb er of independent equations can be written; otherwise, effort might be wasted in attempt ing a n impo ssible solution with the aid of the equilibriu m equations only. The unknown variables may be forces, couples, distan ces, or angles.

Equil ibr ium Conditio n s

(u )

Complet e fixity Adequa te constraints

Adequacy of Constraints In discussing the relationship between const raints and equilibrium, we should look further at the question of t he ade quacy of con stra ints. The existence of three constraints for a two-dimensional problem does not always guarantee an equilibrium configuration. Figure 3/ 7 shows fou r differe nt types of cons tr aints. In part a of t he figu re, point A of th e r igid body is fixed by the two links and can not move, a nd the thi rd link prevents any rotat ion about A . Thu s, this body is completely fixed with three adequ ate (prope r ) const rai nts. In part b of the figu re, the third link is positioned so t ha t th e force tra nsmitted by it passes t h rough point A where t he other tw o cons t raint forces act. Thus, this configuration of constraints can offer no initia l resistance to rotatio n about A, which would occur when exte rnal loads were applied to th e body. We conclude, the refore , tha t this body is incompletely fixed under pa rtial const raints. The configuration in part c of the figure gives us a similar condition of incomplete fixity because the three para llel lin ks could offer no initial resistance to a small vertical movement of the body as a result of external loads app lied to it in t his dire ct ion . T he constra ints in thes e tw o examples are often termed imp roper . Marwan and Waseem AI-Iraqi

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(bl Incomplete fixity Partial constraints

(c)

Incomplete fixity Partial cons traints

(dJ Excessive fixity

Redundant cons traint

Figure 3/7

11 9

120

Chapter 3

Equ il ibrium

In part d of Fig. 3/ 7 we have a condition of complete fixity, with link 4 acting as a fourth constraint which is unnecessary to maintain a fixed positi on. Link 4, the n, is a redu ndant constraint , and th e body is statically indeter minate. Ai; in the four examples of Fig. 3/7, it is genera lly possible by dir ect observation to conclude whethe r the constraints on a body in twodimensional equilibrium are adequate (proper), partial (improper), or redun dan t . Ai; indicated pr eviously, the vast majority of problems in this book are statically det ermi nate with adequa te (proper) cons t raints.

Approach to Solving Problems The sam ple pr oblems at the end of this article illus trate th e application of free-body diagrams and the equations of equilibrium to typical statics pr oblems. Th ese solutions should be studied th oroughly. In th e problem work of this cha pter an d throu ghout mechanics, it is important to develop a logical a nd syste mat ic approach which includes the following steps: 1. Ident ify clearly th e quan tities which are known a nd u nkn own.

2. Mak e an un ambiguous choice of the body (or syste m of connected bodies treat ed as a single body) to be isolated and dr aw its complete free -body diagram , labelin g all exte rnal known and unknown but identifi able forces and couples which act on it. 3. Choose a convenient set of reference axes , always usin g righthanded axes when vector cross products are employed. Choose momen t centers with a view t o simplifying the calculatio ns. Gen erally the best choice is one through which as many u nknown forces pass as possible. Simultaneous solut ions of equilibriu m equations are frequen tly necessary, bu t can be minimized or avoid ed by a careful choice of reference axes and moment centers. 4. Identify and state the appli cable force and moment principles or equa tions which govern th e equilibrium conditio ns of th e problem . In th e following sample pr oblems these relations are shown in brackets an d pr ecede each major calcu lati on. 5. Mat ch the number of independent equations with th e number of unknowns in each problem. 6. Carry out th e solution and check the results. In many problems engineering judgment can be developed by first mak ing a reasonable guess or estimate of the result pri or to th e calculat ion and th en comparing th e estimate with the calculated value.

Ma rwan and W aseem AI-Iraqi

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Article 3/3

Equil ib rium Conditions

Sample Problem 3 /1

y I

Determine the magnitudes of the forces C and T, which, along with the other three forces shown, act on the brid ge.tru ss joint.

CD

I

c

Solution. Th e given sketch constitutes the free-body diagram of the isolated

y~

section of th e join t in que stion and shows th e five forces which are in equi librium.

I

,

T -,

16kN

Solution I (scalar algebra). For the x-y axes as shown we have [LFx

= 0]

8 + T cos 40· + C sin 20· - 16

(a )

T sin 40" - C cos 20" - 3 = 0 0.643T - 0.940C

Simultaneous solut ion of Eqs. (a) and

T

~

9.09 kN

(b )

(b )

produces

Ans.

C = 3.03 kN

axes x' ~y ' with the firs t su mma tion in th e y ' -direction to elimi nate refe rence to

@ T. Thus,

0]

II

I

\\11 1

/

/

\ 1,1"/

/

- ~~--

40°

8 kN

--x

- C cos 20" - 3 cos 40" - 8 sin 400 + 16 sin 40" = 0 C

~

T

+ 8 cos 40" - 16 cos 40" - 3 sin 40" - 3.03 sin 20°

Helpful Hints

CD Since this is a prob lem of concu rrent

=3

Solution II (scalar algebra). To avoid a simultan eous solu tion, we may use

[LFy ' = 0]

\

=0

0.766T + 0.342C = 8

OJ

121

3.03 kN

forces, no necessary.

momen t

equation

is

@ The select ion of reference axes to Iacilitate computati on is always an important consideration. Alternatively in this exa mple we could tak e a set of axes along and normal to the direction of C and emp loy a force summati on normal to C to eliminat e it.

An s. = 0

Ans.

T = 9.09 kN

Solution "' (vedor algebra). With unit vectors i and j in t he x - an d y -dir ectio ns , the zero summation of forces for equilibrium yields th e vector equ ation [LF = OJ

8i + (T cos 40·)i + (T sin 40·)j - 3j + (C sin 20·)i - (C cos 20· )j - 16i

o

Equ ating th e coefficients of the i· and j -ter ms to zero gives 8 + T eas 40· + C sin 20· - 16

0

T sin40· - 3 - Ccos20·

0

which are the sa me, of course , as Eqs. (a) and (b) , wh ich we solved above.

@

Solution IV(geometric). Th e polygon representing the zero vector sum of the five forces is shown. Equations (a) and (b) are seen imm ediately to give the projection s of th e vecto rs onto the ,X- and y-directions. Simil arly, projections onto the x '- and y ' -directions give the alte rn at ive equations in Solution II. A graphical solution is easily obtained. The known vectors are laid off headto-tail to some conven ient scale, and the directions of T and C are then drawn to close t he polygon . The r esult ing inte rsec tion at point P completes the solut ion, thu s enabling us to measure th e magnitu des of T and C directly from the drawing to whatever degree of accuracy we incorporate in th e construction. Marwan and Waseem AI-Iraqi

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8 kN 3 kN

@ Th e known vectors may be add ed in any order desired , but they mu st be added before the u nknown vectors .

122

Chapte r 3

Equ ilibrium

Sample Problem 3 /2

T

~ = 30'

Calculate the ten sion T in th e cable which supports t he lOOO·lb load with the pulley arra ngement shown. Each pulley is free to rotate about its beari ng, and t he weights of all parts ar e small compared with the load. Find th e magn itu de of the total force on t he bearing of pulley C.

B

Solution. The free -body diagram of ea ch pull ey is drawn in it s re lative position to t he othe rs. We begin with pulley A, which includes th e only known force. With the unspecified pulley rad ius designated by r, th e equilibr iu m of momen ts abou t its cen ter 0 and the equilibriu m of forces in the vertical direction requ ire

= 01

cW A Fx

T 1r - T 2r = 0

[LMo = OJ [ ~Fy

1000 1b

T, + T 2

1000

-

~

1 and th e t hr ee forces exerte d on the cut members by t he right-hand sect ion which has been removed. We can usually draw the forces with their proper senses by a visual approximation of the equilibrium requirements. Thus , in balancing the momen ts about point B for th e left-hand section, t he force EF is clearly to the left, which makes it compressive, because it acts toward the cut section of member EF. The load L is greater than the reaction R 11 so that the for ce BE must be up a nd to t he right to supply the needed upward component for vertical equilibrium. Force BE is therefore tensile, since it acts away from the cut section. With the approximate magni tudes of R , and L in mind we see th at th e balan ce of moments about point E requires that BC be to the right. A casua l glance at the truss should lead to t he sa me conclusion whe n it is reali zed that t he lower horizontal memb er will st retch under the tension cau sed by bending. Th e equat ion of momen ts ab out joint B eliminates three forces from the relation, and EF can be determined directly. The force BE is calculated fr om the equilibrium equa t ion for the y-direction. Fin ally, we deter mine BC by bala ncing moments about point Marwan and W aseem AI-Iraqi

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F

A

\

E

\

R,

C

r

R,

L ( a)

F

E / EF

E

I I / I RE 1/

C

R,

L

8,

y I I

L __ x (b )

Figure 4/12

18 0

Chapter 4

Structure s

E. In this way each of th e three u nknown s has been determined independen tly of the ot he r two. The rig ht-han d section of t he tru ss, Fig. 4/1 26, is in equilibrium und er the action of R 2 and the same three forces in the cut memb ers applied in the direction s opposite to t hose for the left sect ion . The pr oper sense for the horizontal forces can easily be seen from the bala nce of moments about points B and E .

Additional Considerations It is essen tial to understand t hat in t he method of sections an entire portion of t he tru ss is considered a single body in equilibr ium. Thus, the forces in members internal to the section are not invo lved in the ana lysis of the sect ion as a whole. To clarify the free body and the forces acting exte rnally on it, th e cutt ing sect ion is prefera bly passed th rough the members and not the join ts. \Ve may use either portion of a truss for the calcula tions, but the one involvin g the smaller number of forces will usu ally yield th e simpler solution. In some cases the methods of sections and joints can be combined for an efficient solution. For example, suppose we wish to find the force in a central member of a large truss. Furthermore, suppose that it is not possible to pass a sectio n through thi s mem ber without passin g through at least four unknown mem bers. It may be possible to determine t he forces in nearby memb ers by the method of sections a nd th en progress to th e unknown memb er by the method of joints. Such a combination of the two met hods may be more expedient than exclus ive use of either meth od. The moment equations are used to great advantage in the meth od of sections . One should choose a moment center, either on or ofT the section, through which as many unknown forces as possible pass. It is not always poss ible to assign the proper sense of an unknown force when the free-body diagram of a section is ini tially drawn . Once an arbitrary assignment is made, a positive answer will verify the assumed sense and a negative result will indicate that the force is in the se nse opposite to that assumed. An alternative notation preferred by some is to ass ign all unknown forces arbitrarily as positi ve in the te nsion directi on (away from the sectio n) and let the algebraic sign of th e an swer disti nguish betw een tension and compression. Thus, a plus sign would sign ify te nsion and a minu s sign compr ession. On th e othe r hand , the advantage of assigning forces in their correct sen se on the free-body diagram of a section wherever possible is that doing so emphasizes the physica l act ion of the forces more directly, and th is pra ctice is t he one which is pr eferred here.


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Arl icle 4/ 4

Mel h o d of Sec tio n s

181

Sample Problem 4 /2 Calculate the forces induced in memb er s KL, CL , and CR by the 20-too load on the cantilever tru ss.

CD

Solution. Although the vertical components of th e re actions at A a nd M are stat ically ind et ermi nate wit h the tw o fixed supports, all memb ers oth er than AM are statically determinate. We may pass a section dir ectly th rough member s KL, CL , and CR and ana lyze the porti on of th e truss to the left of this section as a statically determinate rigid body. The free -body diagram of the portion of the truss to the left of the section is shown. A momen t su m about L qu ickly ver ifies the assignment of CR as compr ession, and a moment sum about C quickly discloses that KL is in ten sion . The direction of CL is no t quite so obviou s un til we observe that KL and CB inter sect at a point P to the right of G. A moment sum about P elimina tes re fere nce to KL an d CB a nd shows that CL mu st be compressive to balance the moment of the 20-ton force about P . With these consi dera tions in mind th e solution becomes straight forwa rd, as we now see how to solve for each of the three unknown s inde pend ently of the other two. Su mming moments abou t L require s finding the mom ent arm BL = 16 + (26 - 16)/ 2 21 ft. Thus, 20(5)(12) - CB (21)

~

0

CB

~

57.1 Ions C

Ans.

Next we tak e moments about C, wh ich requires a calculation of cos 9. From the given dimensions we see 9 = tan - 1{5/12) so that cos 8 = 12/13. Th er efore, [LAIc =

OJ

20(4)(12) - ¥SKL(16) = 0

KL = 65.0 Ions T

Ans.

H

11 16' i-G f---.'F:---:E~-D!':--C~,--+-+---,.·t:J -

-

- 6 pa nels at 12'-

-

-

20 tons 8

KL L ~-~""7 ~ ""'ZfCL

y I

I G fl

Ill' I -....:------'--"~_ p

C

- .r

CB

20 to ns

Helpful Hinls

CD We note tha t a na lysis by the method of j oints wou ld necessit ate working wit h eight j oints in order to calcu lat e t he t hree forces in question . Th us, the me thod of sections offers a consider able advantage in t his case.

@ We could have sta rted with mo ments Final ly, we may find CL by a moment sum about P, whose distance from C is give n by PC/16 ~ 24/ (26 - 16) or PC ~ 38.4 ft. We also need fl, which is given by fl ~ ta n - 1(CB/ BL) ~ lan - 1(12/ 21) ~ 29.7' and cos fl ~ 0.868 . We now ha ve

@

[LAIp =

OJ

20(48 - 38.4 ) - CL,"",r T

D

Dimensions in millimeters

Solution. Because of symmetry we a na lyze only one of t he two hin ged memo

CD

bers. Th e upper part is chosen, an d its free-body diagram along with th at for th e connecti on at D is dra wn . Becau se of symmetry th e forces at S a nd A have no zcompone nts . Th e two-force members BD an d CD exert forces of eq ual magn itude B = C on the conn ect ion a t D. Equilibrium of the connec tion gives

B cos O + C cos 8 -T =0 B

~

)" ,

2B cos O ~T

I

From the free-body diagram of t he up per pa rt we express th e eq uilibriu m of moments about point A . Substituting S = 800 N and th e express ion for B gives

(cos 0)(50) + -2 T (sin 0)(36) - 36 (800) cos 0 cos 0

-2 T

f

0

T(25 + 5(36) _ 13) = 28 800 2(12)

T = 1.477 kN

Ans.

F inally, eq uilibriu m in th e y-direc ticn gives us

Ir Fy

~

OJ 800

S - BsinO - A

o

14 77 5 _ A 2(12/ 13) 13

o

Marwan and Wase em AI-Iraqi

A

492 N

•

A

T

2

S A

Helpful Hints (26)

Subst itu ti ng sin Oleos fJ = ta n () = 5/ 12 a nd solving for T give

or

~

B B .....

~c I

T / (2 cos 0)

T = 1477 N

Released positi on

An s.

CD It is alwa ys useful to recognize symmetry. Here it tells u s that t he forces acting on the two parts behave as mirror images of each other with r e· spect to the .r-axis . Thus, we cannot ha ve a n action on one member in t he plus z-direction and its reacti on on th e ot her member in the negative .rdirecti on. Consequently the forces a t S and A have no x-compo nents.

@ Be care ful not to forget th e moment of th e y -cornponent of B. Note tha t ou r uni ts here are newt onmillim et ers .

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200

Chapter 4

Structur e s

PROBLEMS (Unless otherwi se inst ructed, neglect th e ma ss of the various members a nd all fri ctio n in th e pr oblems which follow.I

Introductory Problems 4/65 Det ermine th e magnitudes of all pin reacti ons for th e fram e loaded as shown . A ns. A = 5 12 N , B = D = 10 13 N

301b

Problem 4/67

4/68 Deter mine the com ponents of a ll forc es act ing on ea ch membe r of th e load ed fram e.

SOON

p

p

c

R/

kA_ _

----'4:50e. .~~,.4;"-50---

kA

y

I I I

Problem 4/65

x

L

4/66 For a n BD-N squeeze on the handle s of th e pliers, determi ne th e force F applied to the round rod by each ja w. In addit ion , ca lculate th e force supported by t he pin at A. BON

-1- - - - - 95 mm - - -- -1

Problem 4/68

4/69 Det ermine t he compone nts of a ll force s act ing on each member of the loaded truss. Wha t is th e prima ry difference between th is pr oblem a nd P rob. 4/ 6B? Ans. Ax = e x = B, = 0 A y = 0.707P, B, - 0.707P, C, = 0.293P p

p

C A

BON

1.J,-----4"':. ~/ y I I

Problem 4/66

IL

4/67 Compu te th e force support ed by t he pin at A for the slip-join t plier s un der a gr ip of 3D lb. An s. A = 157.6 Ib


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x

Problem 4/69

Article 4 /6 4/70 A force P is a pplied to t he midpoint D of link BC. State t he value of th e couple AI which would render (a) t he hori zon tal force tran sm itted by pin B zero a nd tb) the vertical force transmitted by pin B zero.

Problem.

201

4/72 Determine th e react ion at t he ro ller F for t he fra me loaded as show n . !

a .2m

~======~===jJ C A B

I

~-.- - O,5 m - -r- O,3 m-.-j

250 N

Problem 4/ 12

c

4/73 Th e device shown in th e figur e is designed to drive brad s int o picture-framing mat erial. For a gr ipping force of 10 Ib on th e ha ndle s, determine th e force F exerted on th e brad . Ans. F = 25 1b

p

Problem 4/ 70 4 /71 The automobile bu mper jack is designed to su pport a 4000-N down ward load . Begin with a free -body diagram of BCD a nd determine t he force su pporte d by roller C. Note th at roller B does not contact t he ver tical colu mn. Ans. C = 6470 N 150mm

150 mm

~"GfO

o

400

"H+-

A

0.75"

, , "" 1 1 2 ,0 -,,

-

mml

C 4000 N

- t-C=-:'::"':'-rI 1' 00 mm I .

,

~-- 24 0 m m

10 Ib

Problem 4/73

D

' F

~ lOoo mm--1 Probl em 4/7 1


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202

Chapter 4

Struc tur es

4/74 Th e device shown is used to st ra ighten bowed decking boards j ust prior to fina l nailing to th e joists. Th er e is a lower bracket (not shown ) at 0 which fixes the part OA to a joist, so th at th e pivot A may be considered fixed. For a given for ce P exerted perpe ndicular to th e handle ABC as shown, determine th e corre sponding norm al force N applied to th e bent board nea r point B . Neglect friction.

4/76 T he wingnut B of th e collapsible buck saw is tig ht ened until th e ten sion in rod AB is 200 N. Determ ine th e force in th e sa w blad e EF and th e magnitude F of the force supported by pin C.

25°:\ I \ , \,

..,

..,-_-'.::'I=====\~. B

E

p

Problem 4/76

4 /77 Determin e the magn itude of t he pin reaction at A and t he magnitude and directi on of th e for ce reaction at th e roller s. Th e pull eys at C and D are small. AilS . A = 999 N. F ~ 314 N up

Problem 4/74

D

Representative Problems 4 /75 The "jaws-of-life" device is utilized by rescuers to pry

apart wrecka ge, th us helping to free accident vict ims. If a pressure of 500 Ib/ in.2 is developed behind th e pist on P of area 20 in.2 , det ermine th e vertical force R which is exerted by th e jaw tip s on th e wreckage for th e positi on shown. Note tha t link AB and its cou nterpart a re bot h horizont al in th e figu re for thi s posit ion. Ans. R ~ lllllb

T 1

0.5m

·c

60 kg

-r-~ " '~

0.4 m - --, --- 0.4 m

- +- 0.4 m

Problem 4/77

~.J+~~~ 18"~-Problem 4/75

Marwan and Wa seem AI-Iraqi

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Art icl e 4/6 4 /78 The figure illustrates a common problem ass ociated with simple stru ctures. Un der the loadings L, the rafters can rotate, t he ridge bea m at A ca n lower, and the walls Be and DE can rota te ou tward, as shown in part b of th e figure. Th is phenomenon is sometimes clearl y observed in old wooden farm struc tu res as a centra l sa.gging of th e rid ge beam when viewed from th e side. A simple remedy is shown in part a of th e figur e. A cha in or cable is st retched between fast eners at Band E , and t he tu rn buckle F is tightened u ntil a proper tension is achieved, thereby prev enti ng the outward tilti ng of th e walls. For given values of the dimension d and t he point loads L (wh ich resu lt fro m th e distributed loads of th e rafter and roofing weights and any additio na l loads such as snow), ca lculate th e tension T required so th at there are no out wa rd for ces on the walls at B and E. Assu me th at the support of th e rafters at the ridge beam is pu rely ho rizontal and that all joints are free to rotate. d

d

6

31 2l

A

t~7

4/79 Compou nd-lever snips, shown in th e figure, are designed to rep lace regu lar ti nne rs' snips when large cutting forces are requ ired. For the gri pping force of 150 N, wha t is the cutti ng force P at a dist an ce of 30 mm along th e blade fro m the pin at A ? Ans. P = 1467 N

Problem 4/79

L

t

20 lb

12

B

E

F (a)

O.6

Z I

.~

-

\ ,.---

,, '.....---' ,

'x (b l

'a l

Figure 5/4

tant gravitational force lVabout any axis equals the sum of t he moments about th e sa me axis of th e gravita tiona l forces d lV acti ng on all particles treat ed as in finitesim al elements of th e body. Th e res ultant of the gra vitational forces acting on a ll eleme nts is th e weigh t of th e body a nd is given by the sum IV = f dW. If we apply th e moment principle about th e y-ax is. for example, t he moment abo ut this axis of the eleme nta l weight is x dW, and the sum of th ese mom ents for all eleme nts of th e body is f x dW This sum of moments mu st equa l IVx, th e mom ent of the sum. Thu s, XW = f x d W. With similar expression s for the other two components, we may express t he coordinates of the center of gravity G as

JXdW x

=

\V

_ JydW y

=

IV

J zdlV z =

lV

(5/ 1" )

To visualize the physical moments of the gravity forces appearing in the t hird equa tion, we may reori ent the body a nd attached axes so th at t he z-axis is horizonta l. It is essential to recognize that the numerator of each of these expressions represent s the Slim of the m om ent s , whereas the product of W a nd th e corres ponding coordina te of G re presents t he moment of the Slim . This moment principle finds repeated use throu ghout mechanics. Wit h the substitution of W = IIIg an d d W = g dill , t he express ions for the coordinates of the center of gravity become

y 111

(5/ 1bl

Equ at ions 5/ 1b may be expressed in vector form with t he aid of Fig. 5/ 4b, in which the eleme ntal mass and th e mass center G are located Ma rwan and W aseem AI- Iraqi

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Ar t icl e 5 / 2

Ce nte r of M a ss

by their respective position vectors r = xi + yj + zk and r = xi + yj + Ek. Thus, Eqs. 5/ 1b are th e components of t he single vector equation

(5/2 )

The density p of a body is its mass per unit volume . Thus, the mass of a differ ential element of volume dV becomes dm = p dV. If p is not const ant throughout th e body but can be expressed as a function of t he coordinates of the body, we must account for thi s variation when calculating th e num era tors and denominators of Eqs . 5/1b . We may then write these expressions as

x

JxpdV J pdV

Y

JYfl dV JpdV

Jzp dV JpdV

(5/ 3)

Center of Mass versus Center of Cravity

Equ ati ons 5/ 1b, 5/2, and 5/ 3 are independent of gravitatio nal effects since g no longer appears. They the refore define a unique point in the body which is a fun ction solely of the distribution of mass. Thi s point is called the center of mass , and clearly it coincides with the center of gravity as long as t he gravity field is treated as uniform and parallel. It is meaningless to speak of t he cen te r of gravity of a body which is removed from the gravitational field of the earth , since no gravitational forces would act on it . Th e body would , however , still have its un ique center of mass. We will usually refer henceforth to the center of mass rather than to the center of gravity. Also, the center of mass has a special sign ifica nce in calculat ing the dynamic respon se of a body to unbalanced forces. This class of problems is discussed at lengt h in Vol. 2 Dy namics . In most pr oblems th e calculation of the position of the cente r of mass may be simplified by an intelligent choice of refer ence axes. In gene ra l th e axes shou ld be placed so as to simplify the equations of t he boundaries as much as possible. Thus, polar coordinates will be useful for bodies with circular boundaries. Another important clue may be taken from considerations of symmetry. Whenever there exists a line or plane of symmetry in a homogeneous body, a coordinate axis or plane should be chosen to coincide with this line or plane. The center of mass will always lie on such a line or plane, since the moments due to symmetrically located elements will always cancel, and the body may be considered composed of pairs of t hese eleme nts. Thus, th e center of mass G of the homogeneous righ tcircular cone of Fig. 5 /5a will lie somewhere on its central axis, which is a line of symmetry. The center of mass of the half right -circular cone lies on its plan e of symmet ry, Fig. 5/5 b. Th e cente r of mass of the half rin g in Fig. 5/5 c lies in both of it s plan es of symmetry and th erefore is Marwan and Waseem AI-Iraqi

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(b l

(e l Figure 5 /5

22 9

230

Chapter 5

Distributed Forces

situate d on line AB. It is eas iest to find the locat ion of G by using symmet ry when it exists.

5 /3 CENTROIDS OF LINES , AREAS, AND VOLUMES When t he density p of a body is uniform throughout , it will be a cons tan t factor in both the numera tors and den ominators of Eqs . 5/3 and will therefore cancel. The remaining expressions define a purely geometrical pr oper ty of t he body, since any reference to its mass prop erties has disappeared. The term centroid is used whe n the calculation concerns a geomet rica l sha pe only. When speaking of an actual physica l body, we use the te rm center of mas s . If the den sity is uniform throughout th e body, the positions of the centroid a nd center of ma ss are identical, whereas if the density varies, these two point s will, in general, not coincide. Th e calculation of centroids falls with in three dist inct cat egories, depending on wheth er we can model th e shape of th e body involved as a line, an area , or a volume. (I) Lines. For a slender rod or wire of lengt h L , cross-sectional area A , and density p, Fig. 5/6, t he body approximates a line segme nt , and dm = pA dL. If P a nd A are constant over t he lengt h of the rod, t he

coordina te s of th e center of mass also become th e coordinat es of the cent roid C of the line segme nt , which, fro m Eqs. 5/ 1b, may be written

_= -f - -

xdL

x

L

y =

f

ydL

L

z =

f

zdL L

(5/4)

Figure 5/6

Note that , in general, the cen troid C will not lie on th e line. If the rod lies on a single plane, such as the x-y plane, only two coordinates need to be calculated.

A

(2) Areas. When a body of den sity p has a small but constant t hickness t, we can model it as a su rface a rea A, Fig. 5/ 7. Th e mass of a n element becomes dm = pi dA . Again, if p and t are consta nt over the ent ire area, t he coordinates of the center of mass of the body a lso become the coordinates of th e centro id C of th e surface a rea , an d from Eqs. 5/ 1b the coordinates may be written

_f

xdA x = - - A x

Figure 5 /7

y

f

ydA A

z =

f

zdA A

(5/ 5)

The nume ra tors in Eqs. 5/5 are called t he first moments of c rec ." If t he surface is curved, as illust ra ted in Fig. 5/7 wit h the she ll segme nt, all "Second momen ts of areas (moments of first moment s ) appear later in our discussion of second mom enta of urea, also called area moment s of inertia, in Appendi x A.

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Art icl e 5 /3

Centro ids of Lines , Areas , an d Vo l u m e s

three coordinates will be involved . Th e centroid C for the curved surface will in general not lie on the surface. If the area is a flat surface in, say , the x-y plane, only the coordina tes of C in th at plan e need to be calculated.

(3) Volumes. FOI' a general body of volume V and density 1', th e element has a mass dm = I' dV. Th e den sity I' cancels if it is cons ta nt over the entire volume, and the coordinates of the center of mass also become the coordinates of the centroid C of t he body. From Eqs. 5/3 or 5/ 1b th ey become

_f

xdV x = - - V

y

f

y dV V

f

zdV V

(5/ 6 )

Choice of Element for Integration Th e principal difficulty with a theory often lies not in its concepts but in t he procedures for applying it . With mass centers and cent roids th e concept of th e mom ent principle is simple enough; the difficult steps a re th e choice of the differentia l eleme nt and setti ng up th e integrals. Th e following five guid elin es will be useful.

(I) Order of Element Whenever possible, a first -order differen tial element should be selected in prefer ence to a higher-or der eleme nt so that only one int egration will be required to cover the entire figure. Thus , in Fig. 5/ Sa a first-order horizont al st rip of area dA = I dy will require only one integration with respect to y to cover the ent ire figure. The second-order element dx dy will require two integrations, first with respect to x and second with respect to y, to cover the figure. As a further example, for th e solid cone in Fig. 5/ Sb we choose a first -ord er eleme nt in the form of a circular slice of volume dV = r.r 2 dy . Th is choice requires only one integration. and thus is preferable to choosing a thirdorder eleme nt d V = dx dy dz, which would require three awkward integrations.

(2) Continuity. Whenever possibl e, we choose a n eleme nt which can be integrated in one continuous operation to cover the figure. Thu s, the horizontal st rip in Fig. 5/Sa would be pr eferable to the vertica l strip in Fig. 5/9, which, if used, would require two sepa rate integr als because of the discontinu ity in t he expression for the height of the strip at x = X l'

(a)

Iy I

IY 1

(3) Discarding Higher-Order Terms. Higher-order terms may always be dropped compared with lower-order term s (see Art. 1/7 ). Th us, the vertical st r ip of a rea under the curve in Fig. 5/10 is given by the first -ord er ter m dA = y dx , and the second-order t ria ngu lar area ~dx dy is discarded. In the limit , of course, there is no error. (4) Choice of Coordinates. As a general rul e, we choose th e coordin at e system whi ch best matches the bou ndar ies of the figur e. Thus, th e boundaries of th e a rea in Fig. 5/ 110 a re most easily describ ed in Marwan and Waseem AI-Iraqi

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(b)

Figure 5/8

231

232

Chap t e r 5

Distribu ted Force s

dy

.r T L _LLJ

x

L_.LL-'--_ ----' __ X

-1 f- dx

xI

Figure 5/9

Figure 5/10

rectangular coordinates, whereas the boundaries of the circular sector of Fig. 5/ 11b are best suited to polar coordinates.

(5) Centroidol Coordinate of Element When a first - or secondorder differential element is chosen, it is essential to use the coordinate of the centroid of the element for the moment arm in expressing the moment of the differential element. Thus, for the horizontal strip of area in Fig. 5/ 12a, the moment of ciA about the y-axis is Xc dA, where Xc is th e x-coordinate of the centroid C of the element . Note that Xc is not th e X which describes either boundary of the ar ea . In the y -direction for this element th e moment ar m Yc of the cent roid of t he element is t he same, in t he limit , as th e y-coordinates of the two boun daries. As a second exam ple, cons ider th e solid half-cone of Fig. 5/ 12b with th e semicircular slice of differential thickn ess as th e element of volume. The moment arm for the element in the x-direction is the distance X c to th e cent roid of th e face of th e element and not the x-distance to the bound a ry of the element. On th e other hand , in t he z-direction th e momen t arm Zc of th e cent roid of th e element is th e sa me as t he z-coordinat e of th e element . With these exampl es in mind , we rewrite Eqs. 5/5 and 5/6 in th e form

x =

y

f

yc dA

y I ~ I I .,:1 ~ 1 1I I

y

l

lLJ I

L _ _ -.J __ x

(b)

Figure 5/11 WNW.gigapedia.com

r

: 0

(a)


(5/5 a)

z =

A

--x

Article 5 /3

Centroids of Lines, Areas , and Vo lumes

/ /

Y

Figure 5/12

and

Y =

f

Ye d V

V

Z =

(5/ Gal

It is essential to recognize that the subscript c serves as a reminder that the moment arms appearing in the numerators of the integral expressions for moments are always the coordinates of the centroids of the particular elements chosen. At this point you shou ld be certain to under st and clearly the principle of moments, which was intr oduced in Art. 2/4. You should recognize t he physical mea nin g of t his pri nciple as it is applied to the syste m of parallel weight forces depicted in Fig. 5/40. Keep in mind th e equivalence between t he moment of the resu ltant weight W and t he sum (integrall of th e moments of the eleme nta l weights dW, to avoid mistakes in setting up the necessary mathem at ics. Recognition of the pr inc iple of moments will help in obtaining the correct expression for the moment arm XC' Ye' or ze of the cen troid of the chosen different ial element. Keepi ng in mind the physica l pictu re of the pr inciple of mome nts, we will recognize that Eqs. 5/4, 5/5, a nd 5/6, which a re geometric relationships, are descriptive also of homogeneous physical bodies, because the density p cancels. If the density of the body in qu estion is not constant but varies througho ut t he body as some func tion of t he coordinates , then it will not cancel from the numerator and denominator of the mass-center expressions . In this event , we must use Eqs. 5/3 as explained earlier.

Sa mple Problems 5/ 1 through 5/ 5 which follow have been carefully chose n to illustra te the applicati on of Eqs. 5/ 4, 5/5, an d 5/ 6 for ca lcula tin g the loca tion of t he centro id for line segme nts (slender rods), areas (t hin flat plates), and volumes (homogeneous solids). Th e five integration considerations listed above are illustrated in detail in these sample problems. Secti on C/ lD of Append ix C conta ins a tab le of integr als which in cludes those needed for the problems in thi s a nd subseque nt cha pte rs. A su mmary of the centroida l coord ina tes for some of the comm only used shapes is given in Tables D/ 3 a nd D/ 4, Appendix D. Marwa n and Waseem AI-Iraqi

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233

234

Chapter 5

Distr ibu t ed Fo rc es

Sample Problem 5 /1 Centroid of a circular arc. Locate th e centro id of a circu lar arc as shown in the figure.

CD

Solution. Choosing th e axis of sym metry as th e .r-a:xis makes y = O. A differ ential element of arc has the length dL = r d o expressed in polar coordina tes, and th e x-coordina te of t he element is r cos H. Applying th e first of Eqs. 5/4 and substituting L = 2ar give [LX =

J

.r dL I

~

(2nr li

f..

(r cos 0) r

de

2urx = 2r 2 sin a

x=

r sin

(~

- - a

Ans.

--

For a se micircular arc 2a = 1T, which gives x = 2r/ 1T. By symmetry we see immed iate ly tha t this res u lt also applies to the quarter-circular arc when the measu rement is made as shown.

! r

--

x

a

Helpful Hin t

CD It should be perfectl y evident that polar coordinates are preferable to rectangu lar coordinates to express th e lengt h of a circular arc.

Sample Problem 5/2 Centroid of a triangular area. Det ermine t he distance Ii from the base of a tria ngle of alt itude h to the cent roid of its area.

CD

Solution. The x-axis is taken to coincide with the base. A differential strip of area dA = x dy is chosen. By similar tri an gles x / (h - y ) = b/h . Applyin g th e second of Eqs. 5/00 gives lAy

and

~

J

Yc dAI

bh _ 2"Y = y =

r

2 b(h - Y) d = bh oY h Y 6

h

A ns.

3

H elpful Hint

This sa me result holds wit h respect to either of th e ot her two sides of th e trian gle consider ed a new base wit h cor res ponding new altitude. Thus, th e cent roid lies at t he intersect ion of th e medians, since the distance of this point from an y side is one-third th e altitude of th e t rian gle wit h that side considered th e bas e.

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CD We save one integr ation here by using th e first-order element of ar ea. Recogni ze that dA mu st be expressed in term s of th e integration varia ble y; hence. x = f ty ) is required.

Article 5/1

Centro ids of Lin es. Area s. and Volum es

235

Sample Problem 5/3 r

Centroid of the area of a circular sedor. Locate th e centroid of the area of a circular sector wit h respect to its vertex.

CD CV

Solution I. Th e .r-axis is chose n as th e axis of symmetry. an d y is th erefore automat ically zero. We may cover th e area by moving an elemen t in t he form of a' partial circular rin g, as shown in t he figure, from the center to the outer periph ery. Th e ra dius of the rin g is "o and its thi ckn ess is dro, so that its area is dA ~ 21'0" dl'o· Th e .r-cocrdina te to th e centroid of the element from Sample Probl em 5/ 1 is Xc = "0 sin nl u, whe re "o replaces r in th e formul a. Thus, th e first of Eqs. 5/00 gives

2" ( 1l'T2 )x-

-

21T

=

i' (1'0

-sin --,,) (2roa

0

_ X

a

dro )

2 r sin a

Ans.

= ----

3

"

Solution I

Solution II. Th e area may also be covered by swinging a t rian gle of differe ntial area about th e vertex and through th e total angle of t he sector. Thi s tr iangle, shown in th e illustrati on , has an area dA = (r I 2)( r d m, where higher-order term s are neglected. From Sample Problem 5/2 the centroid of the triangu lar element of area is two-thirds of its alt itude from its vertex, so th at the .r-coordina te to th e centro id of th e element is Xc = ~ r cos it, Applying th e first of Eq s. 5/00 gives

IAi =

Jx, dA ]

Helpful Hints

CD Note

carefu lly that we mu st distin gu ish betwe en t he vari able "o and t he consta nt r .

CV Be carefu l not

to use "o as th e centroi dal coordi na te for the eleme nt. .I'

I

2

I-

3

1 .t"(' =- r cos 8 and as before

_ x

2r sina

An s.

= ----

3

"

I I I

a

For a semicircular area 20 = 1T, which gtves F 4r13rr. By symmet ry we see immediately that thi s resu lt also app lies to the quart er-circular area where the measur ement is made as shown. It should be noted t hat, if we had chose n a second-or der element "o d r o dO, one int egr ation with res pect to (J wou ld yield the ring with which Solution I began . On th e other ha nd, int egration with respect to "o initially would give th e t riangu lar element with which Solution II began .

- x

Soluti on II

41'13H) ~ c

J \-\vt': ~ I--- r


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I

'_

_

1'---1

236

Chapt er 5

Dis tr ibut ed For ces

Sample Problem 5 /4 Locate th e centroid of t he area u nder the curve x x = a.

CD

hi' from

x

010

Y I I

x = ky 3

I I I -X

Solution I. A vertical element of area ciA = y dx is chosen as shown in the

I I I I

figur e. Th e x-cocr dinate of th e cent roid is foun d from th e first ofEqs. 5/00 . Thus,

'---

[AX =

J

Xc

; ; - _.L----'

X

a

dA ]

Y I

bL - - - - - - - I

Subst itut ing y = (x / k ) IJ3 and k = a /b 3 and integrating give 3ab _ 4

3a'b 7

-

-x = - -

4

Ans.

= 'i a

X

I x = ky 3 I I I I I I

In the solution for y from the second of Eqs. 5/00 , the coordina te to the cen troi d of th e rectangular element is Yr: = y /2, where y is the height of the strip govern ed by th e equa tion of th e curve x = kY. Thus, th e moment principle becomes lAy =

b

I-- x----1 r- dx

J dAl Yc

Substitutingy = b(x /a )1/3 and integr ating give 3ab _

3ab'

Ans.

""4 Y = W

Yh- x = '!..!2:

--1

b ~ _~ __2__

Solution II. The hori zont al element of area shown in the lower figure may be employed in place of t he vertical element . Th e x-eoordina te to th e centroid of th e rectan gu lar element is seen to be Xc = x + ~ (a - x ) = (a + x )/2, wh ich is simply the average of t he coordina tes a and x of th e ends of the strip. Hence, [AX

=

J

Xc

dAl

x

r

(a - x) dy

=

r

(a ;

x) (a

- .r) dy

Th e valu e of y is fou nd from [Ay

=

J

Yc dA ]

y

t o

(a - x )

dy =

Jr"o y (a

-

.r }

dy

where Yc = Y for th e hori zontal str ip. The evalua tion of t hese integrals will check t he previous results for i and y.


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t --

I x = ky 3

:I I

I

dy

---.L

x+--a-x--ll

L---~aJ-x Helpful Hint

CD Note

that element.

Xc

x for the vertical

Article 5/3

Cen troids of Lines. Are as. a n d Volumes

237

Sample Problem 5/5 Hemispherical volume. Locate the centroid of the volu me of a hem isph er e of radius r with respect to its base .

z = 0 by symmetry. Th e most convenient element is a circu lar slice of thickness dy parall el to the x -e plan e. Since th e hemisph ere int ersectj t he y- z plan e in the circle y2 + z2 = r 2 , th e radius of th e circula r slice is z = + r 2 - l . Th e volume of t he elemental slice becomes

Solution I. With the axes chosen as shown in th e figure , x =

CD

dV = mr' - y') dy The second of Eq s. 5/ 00 requ ires

[Vy

=

I

y, dV]

whe re Yc = y . Inte grating gives

~m-3y =

±m-

4

Solut ion I

y = ~r

Ans. z

Solution II. Altern ati vely we may use for our differential element a cylind rical shell of length y , radius z, and thickness dz, as shown in th e lower figure. By

I~ YJ'

y,=."/2 ,-

.

expanding th e ra dius of th e shell from zero to r, we cover the entire volume. By symmet ry th e centroid of the elemental shell lies at its center, so t hat y, = y / 2. Th e volume of the element is d V = dzHy ), Exp ressin g y in term s of z from th e equation of the circle gives y = + r 2 - z2. Uelng tb e velu e of'{er- compute d in Solution I for th e volum e of the hemisph er e and subst itut ing in th e second of Eqs. 5/00 give us

, wher e x is in feet. Determine th e location of the center of mass of the rod .

5/17 Determ ine the .r- and y-coordinates of t he centroid of th e shaded area. An s. :< = 0.762, Y = 0.533

{

[L

_

Iy = kx II.3

I

I I I

I

oo -- - ---- --- - !.--x 2 Problem 5/14


Problem 5/17

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Art icl e 5/ 3 5 /18 Deter mine t he y-coordinate of the centroid of t he

shade d area.

Pro b lems

2 41

5/10 Determine the y -coordinate of the centroid of t he shaded area. y I I

y I I I I

b

a

I I

I I

b

x= ky 2 l b I

a

2"

I _ __ _ _ _ _ _ _ _ _ LI

/ 45" -

Problem 5/18

-

-

-

-

-

-

- -- r

Problem 5/20

5 /19 Deter mine t he x- and y-coordi nates of t he centroid of

the shaded area.

/

/

r

a

/ /

_ 7a _ Ans . x = 6 (77" _ 1) ,y =

5 /11 Determine the x- and y-coordinat es of the centroid of

th e sha ded area.

a 1T -

A ns. i = _ a_ y ,, - 1 '

1

y

=

7b 6(,, - 0

y

I

1

1

1

1

1

T b

2"

t: QI

21

'----

r- ~--1

-

-

--' - - - -

1:

x

_

- - -x

I'

Problem 5/19 Problem 5/2 t

5/11 Determine th e coordina tes of t he centroid of th e shaded area. y

y = bsi n ~ -...... 2a

1

I

---

" Problem 5/28 _~:::""_~_ _...J

-

-

-

x

Problem 5/25

Marwa n and W aseem AI-Iraqi

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_x

Articl e 5 /3 5/19 The th ickness of the t riangular plate varies linearly with y from a valu e to along its base y = 0 to 2/ 0 at y = II . Determi ne the y-coordinate of the ce nter of mass of the plate. All s. 5' ~ 3,./8

P roblems

243

5/31 Deter mine th e a-coordina te of the centroid of th e volume obtained by revo lving th e shaded a rea within the parabola abou t the z-axis through 180°.

z I

s

I

I I I I I I I I

b-

-

I I

.u b

Problem 5/29 5/30 Ca lcu la te th e dist ance h measured from th e base to th e centro id of the volume of th e fru stum of th e right-circular cone.

Problem 5/32 5/33 Th e homogeneous slender rod has a uniform cross sect ion an d is bent into the sha pe shown. Calculate the y -coordinate of the mas s center of th e rod. (Reminder: A differential arc lengt h is d L = ,f(dx )2 + (dy )2 ~ h + (dx / dy )2 dy.) AilS. Y = 57.4 mm y

I I I

I I I 100 mm :

I

Problem 5/30

I

5 /3 1 Deter mine the a-coordinate of the cent roid of the solid genera ted by revolvin g the qu arter-circu lar area

through 9W about the z-axis.

,_ /Vt S .

z

_ Z

=

lOOmm

1Iu 2(4 + 3 ,, )

Problem 5/ll 5/34 Determine the z-coord inate of th e centroid of th e volum e obta ined by revolving th e shaded t ria ngu lar area about th e z-axis through 360".

I I

z

y -, 1_ -

_

.... 5 /38 Determine t he coordinates of th e cen troid of th e volume obtai ned by revolving th e shaded area a bout th e z-ax is t hrough th e 90 0 angle.

r

Ans. x ::: y = (.; -

~)a, Z = a/4

Problem 5/35

z 5 /36 Locate th e ma ss center of t he hom ogeneou s solid body wh ose volume is determined by revolving the shaded a rea through 360 0 about th e a-ax is.

1

---1- - -

---3 / 1

r

I

r--

I I I I 1 1 I

a

30 0 m m

I

y-

I

1 1 1

X

/ /

200mm

r = kz 3

Problem 5/38

z

.... 5 /39 T he cylindr ica l shell of u niform small th ickness has a rad ius r a nd height z which varies fro m zero at o = 0 to h a t lJ = 11' according to z ::: klJ wh ere k is a constant. Dete rmine the X - , y o, a nd a-coordinates of t he mas s center of th e she ll. 2r _ 4r _ Ii A ns. x ::: - ,y 11'2 ' Z 1T 3

Problem 5/36 .... 5 /37 Deter mine t he y -coordina te of t he centroid of t he pla ne a rea shown. Set h = 0 in your result a nd com -

pare with t he resu lt

y :::

4a for a full semicircula r 3 1T

a rea (see Sa mple Problem 5/3 a nd Tab le D/ 3). Also eva lua te you r result for the conditions h ~

/

I

00

Ii

/

a ;;;.

~ a nd

4

Problem 5/39

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Art icle 5/3 .,. 5/40 Locat e th e center of mas s of the homogen eous bellshaped she ll of uniform but negligible thickne ss.

Ans. z = __ o_ n - 2

Problem 5/42

.... 5/41 Locate t he center of mass G of the steel half r ing. (H i nt: Choose an element of volume in th e form of a cylindrical sh ell whose intersection with th e pla ne of th e ends is shown.) a2 + 4R 2 Ans. = -:;--;;-2rrR

~ , ----- It.( ----- - i

_R--l

R~

Problem 5/41


245

... 5/41 Determi ne the x-coordina te of t he mass center ufthe homogeneo us hem isph ere with th e sma ller hemispherical portion remov ed. 45 Ans.i = R 112

Problem 5/40

r

Pr oblems

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246

Chapter 5

Distr ibuted Forces

5 /4 COMPOSITE BODIES AND FIGURES; ApPROXIMATIONS \Vhen a body or figure can be conveniently divided into several parts whose mass centers are easily determined, we use the principle of moments a nd treat each part as a finite element of the whole. Such a body is illustrat ed schematica lly in Fig. 5/ 13. Its par ts have masses "' 1' m2. ma with the respective mass-center coordinates XI. X2. X3 in the x-direction. The moment principle gives

where X is the x-coordinate of the center of mass of the whole. Sim ilar relations hold for the other two coordinate directions. We generalize, th en, for a body of any number of parts and express the sums in condensed form to obtain the mass-cent er coordinates

(x -

-

=-};mi }; ",

"_ my-

y =-

(5/ 7)

}; m

Analogous relation s hold for composite lines, areas, and volumes, where the nr's are replaced by L's, A's, an d V's, respecti vely. Note that if a hole or cavity is considered one of the component parts of a composite body or figure, the corresponding mass represent ed by the cavity or hole is treat ed as a negat ive quantity.

An Approximation Method In pract ice the boundaries of an area 0 1' volu me might not be express ible in terms of simple geometrical shapes or as shapes which can be represen ted mathematically. For such cases we must resort to a method of approx ima t ion. As an exa mple, consider th e pro blem of 10-

f---- x, --~

- x,---.j -;:j ' _.~-+---;-\J I

I \

• G2

•

I I GI

\

\

G3 l1/ :l

~--- X ---l

Figure 5/13


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Articl e 5 {4

Composite Bodies and Figures; Approximations

eati ng the centroid C of th e ir regu lar area shown in Fig. 5/ 14. Th e a rea is divided into stri ps of width .lx an d variable height h . The a rea A of eac h strip, such as the one show n in red, is h ..h and is mu lt iplied by the coordinates Xc and Y c of its centroid to obtain the momen ts of the eleme nt of area. The sum of the momen ts for all strips divided by th e tota l area of the strips will give the corresponding centroidal coordinate. A sys tematic tabulation of the result s will permit an orderly evaluation of the total area ~A , the sums ~A\"(" and ~Ayc' and the centroidal coordinates y

Figure 5/14

We can increase the accuracy of the approximation by decreasing th e widths of the st rips. In all cases th e average height of th e st rip should be estimated in approximating th e a reas . Alth ough it is usu ally advantageous to use elements of constant width, it is not necessary. In fact, we may use ele ments of any size and shape which approximate the given area to sat isfactory accuracy. Irregular Volumes

To locate the centroid of an irregular volume, we may reduce the problem to one of locating the centroid of an area. Consider the volume shown in Fig. 5/15, wher e the magn itudes A of th e cross-sectio na l area s normal to the x-direction are plotted against x as shown. A vertica l strip of area under the curve is A ~x, which equals the corresponding ele ment of volume .l V. Thus, the area under the plotted curve represents the volume of th e body, and th e x-coordinate of the centroid of the a rea under the curve is given by

whic h equa ls for the centroid of t he act ua l volume.

xA

I

.

C

Figure 5/15 Marwan and Wa see m AI- Iraqi

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247

248

Chapter 5

Distributed Forces

Sample Problem S/6

y I ~- 1 2"'------j

Locate the centroid of the shaded area.

, 4-

Solution . The composite area is divided into the four elementary shapes shown

41 2"

in the lower figure. The centroid locations of all these shapes may be obtained from Table D/ 3. Note that th e areas of the " holes" (parts 3 and 4) are taken as negative in the following table:

A

i

2

PART

in.

1 2 3 4

120 30 - 14. 14 -8

Y

];A

in.

in.

in.

6 14 6 12

5 10/ 3 1.273 4

3

720 420 - 84 .8 -96

c-t-t - ---': 1 -

I

3"

yA 3

in.

600 100 - 18 -32

2

L-_--"

L-_---,_-----'

TOTAl.";

127.9

959

- I

650

The area counte rparts to Eqs. 5/ 7 are now applied and yield

[-X _- -LAX] U\

X

7.50 in.

Am.

[y ~ ~]

650 Y ~ 127.9 ~ 5.08 in.

Am.

~

959 -127.9

~

Sample Problem S/7

6

Approximate the x-coordinate of the volume centroid of a body whose length is 1 m and whose cross-sec tional area varies with x as shown in the figure.

~s

.,; Solution. The body is divided into five sections . For each section, the average area, volume , and centroid location are determi ned and entered in the following ta ble: A. v m2

Volume V m'

i

Vi

INT ERVAL

m

m'

0-0.2 0.2-0.4 0.4-0.6 0.6-0.8 0.6-1.0

3 4.5 5.2 5.2 4.5

0.6 0.90 1.04 1.04 0.90

0.1 0.3 0.5 0.7 0.9

0.060 0.270 0.520 0.728 0.810

TOTAl.";

4.48

I

5

V

4

3

2

a

~-- - - - - -I ,

:45 0 '

1

1 1 1

-,

-- x

Problem 5/74

-, -,

" , '

Problem 5/72

5/75 Th e body shown in cross section is a complete circu la r ri ng formed by revolving the octagonal area about the a-axis. The entire su rface is to be cover ed with a specia l coating. Determine th is su rface a rea . An s. A = 177 100 mm 2

Dimensions in millimeters

Problem 5/75 Marwan and Waseem AI-Iraqi

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260

Chapter 5

Distributed Forc es

5/76 Det erm ine th e volu me V ge ne rated by revolving t he qua rt er-cir cular a rea a bout the z-ax is t hrough a n a ngle of 90°.

5/79 Det ermine t he total surface a rea of th e body ge ne rated by revolving t he shaded a rea t hro ugh 360 0 about the x-a xis.

z I

AilS . A

= m ·2 ( 1T

-

1)

y I I

="-' __ x

L __

Problem 5/79

Problem 5 /76

Representative Problems 5 /77 Ca lculate t he volu me V of the space r in the form of t he comp lete rin g of the semicircula r sect ion shown . Also compute th e total su rface a rea A of the space r. Ans. V = 6.56 in .a, A = 29.3 in.:!

z

"""""""

5/80 T he water storage tank is a shell of revolution a nd is to be spra yed with two coats of paint wh ich ha s a coverage of 500 ft2 per gal lon. The engi neer (who remem be rs mechanics ) cons ults a sca le drawing of t he ta nk a nd determ ines t hat the curved line AB C has a lengt h of 34 ft a nd t hat its cent ro id is 8.2 ft from th e ce nter line of th e tank. How man y ga llons of paint will be used for t he ta nk includi ng t he vert ica l cylindrical column ?

I

:.",..,~~ ::?'C"'l

A

I I I I I

B

I I

J

I - - 12"."- 1

I S'

Problem 5/77

S'

5/78 Compute th e volume V a nd tot a l surfa ce a rea A of the complete circu lar ri ng whose cross sect ion is shown.

Problem 5 /80

z I

I

I--J

so '

mm

Problem 5 /78

Marwan and W aseem AI-Iraq i

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Articl e 5 /5 5/81 The t wo circu lar a rcs AB a nd Be a re revolved about the vertical ax is to obtain the surface of re volution shown. Compute th e area A of t he outs ide of this sur face. Ans. A = 15 7.9 in. 2

,

261

5/84 A steel die, shown in secti on, has th e form of a solid ge ne ra te d by revolving the sha ded a rea a rou nd t he zaxis. Calcu late t he mass m of t he die.

z

I

f-- 2"-1 ..-_ ;.,'. .. IA - -2,71

1

I

I I,

Problems

IB

200 mm

c Problem 5/81

5/82 Ca lculate t he weig ht \V of th e a lu minu m cast ing shown. Th e solid is generat ed by revo lving the trapezoida l area shown about t he z-axis through 180".

z, I ~--I---

I I

=t

1 "

I I I

'": I

----- r I

Problem 5/ 84

5/85 Determi ne t he volume V and total sur face a rea A of th e solid gene rated by revolving t he a rea shown thro ugh 1800 about th e a-axis. An s. V ~ 1.775(0 6 ) m m'' , A = 105 800 mm 2

2"

»>:":""''---'~-J

r--

I

2" ~

3"

~

J-1" J30 mm I

I

I

- ,,-

Problem 5/82

»

5 /S3 Determine t he sur face a rea of on e side of th e bellshaped shell of P rob. 5/40, shown again here, using t he theorem of Pappu s. A ns. A = 1TQ 2( 'IT - 2) x

:\ I

I I I

a '\

Prob lem 5/81


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>:

_-- I

--1

~ 75 m m

Problem 5/ 8 5

262

Chapte r 5

Distributed Forces

5 /86 A hand-operated control wheel mad e of a lu minu m has th e proportions shown in th e cross-sectional view. T he area of the total sect ion shown is 15 200 mm 2 , a nd th e wheel has a mass of 10.0 kg. Calcu la te th e dist an ce to t he centroid of th e half-section. T he a luminu m has a den sity of 2.69 Mg/m 3 .

r

5/89 A sur face is gene ra te d by revolving th e circular a rc of 0.8-m rad ius a nd subte nded a ng le of 120" com pletely about t he a-ax is. T he diameter of th e neck is 0.6 m. Dete rmine t he outside ar ea A ge ne ra ted. A ns. A = 4.62 m 2

z

I , O.8 m \

/\

120" ) ~I

O.6m

I Problem 5/86 Problem 5/89 5/87 Det ermine th e volume genera ted by rotating th e se micircu la r area t hro ug h 180°. AlIS. V = 361 000 m m3

z I

5 /90 The shaded a rea is bounded by one hal f-cycle of a sine wav e a nd th e axis of th e sine wave. Det erm ine th e volume gene rated by com plete ly revolvin g t he area about t he x-ax is. y

I

r:

b I

II

I ' r- 75 mm ~ I

+, a

---;:---'1 c

( I -

-'-- 1-- -

Problem 5/87 5 /88 Calcu lat e the volume V of the large neo prene washer in th e form of th e complete r ing of sect ion shown . Also compute th e overa ll surface a rea A.

Problem 5/90

-;"--

2" - . j

Problem 5/88


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---1- -

X

Arti cl e 5/ 5 5/91 Calcu late th e mass m of concrete requ ired to const ruct the arched dam shown. Concrete has a density

"

Ans. m = L 126(lOB) Mg

of 2.40 Mg/ rn .

- -I I-- 10

III

: 70 01 I

/

/ /

10 m

/

Sect ion A -A

/' /' 'i\ 60 0

" J -,

.J...!!

300 mm

IA

12 ' -

--+,- 6,J

Probl em 5/96

--1

5/91 Det ermine th e reaction s a t A for th e ca ntileve r beam subjected to th e distributed a nd conce ntrated load s. An s. Ax = 0, A.v 8 kN , M A = 21 kN · m

Probl em 5/93

5/94 Det ermine th e reactions a t th e support s A a nd B for th e beam loaded as shown.

.~

(b)

(c )

Figure 5/35

su rface, as illustrated by the irregular dashed boundary in Fig. 5/300. If the body of the fluid could be sucked out fro m with in the closed cavity and replaced simulta neously by the forces which it exerted on the boundary of the cavity , Fig. 5/35b , th e equilibrium of the sur rounding fluid wou ld not be distu rbed. Furthe rmo re, a free-body diagram of th e fluid portion before remova l, Fig. 5/350, shows that th e res ultant of th e pr essure forces distributed over its surface must be equal and opposite to its weight mg and must pass through the cente r of mass of th e fluid element. If we replace th e fluid element by a body of th e same dimen sions, th e surface forces act ing on the body held in th is positi on will be ident ical to th ose acting on the fluid element. Th us, the resu ltan t force exerted on t he su rface of an object immersed in a fluid is equal and opposite to th e weight of fluid displaced a nd passes th rough th e center of mass of the displaced fluid. Thi s result ant force is called the force of buoyancy (5 / 2 7)

where p is th e density of the fluid, g is th e accelera tion due to gravity , and V is the volum e of th e fluid displaced. In th e case of a liquid whose density is constant, the center of mass of the displaced liqu id coincides with th e centroid of the displaced volume . Thus when the density of an object is less th an th e density of the fluid in which it is fully immersed, there is a n imbalance of force in th e vertical direction, a nd th e object rises. When th e immersing fluid is a liquid , th e object continues to rise u nt il it comes to the surface of the liquid and then comes to rest in an equilibrium position, ass uming that th e density of th e new fluid above th e su rface is less tha n the density of th e object . In th e case of th e su rface boundary between a liquid and a gas, such as wat er and air, th e effect of the gas pressu re on that portion of th e floating object above the liquid is ba lan ced by the added pressu re in the liquid due to th e action of th e gas on its surface. An important problem involving buoyan cy is the det ermination of th e stability of a floating object, such as a ship hull shown in cross section in a n upri ght position in Fig. 5/360 . Poin t B is t he centroid of the displaced volum e and is called th e center of buoyancy . The resultant of th e forces exerted on the hull by th e water pressu re is the bu oyancy force F which passes through B a nd is equa l and opposite to the weight Waf the ship. If the ship is caused to list through an a ngle c, Fig. 5/3 6b, MarNan and W aseem AI-Iraqi

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Fluid Statics

:50 :5

304

Chapter 5

Dis tributed Forc es

I

I I

I

I I (a ) F

(b)

F

(e ) F

Figure 5/16

th e shape of the displaced volume changes, a nd the center of buoyancy shifts to B'. Th e point of int ersection of t he vertical line t hrough B ' with the centerline of the ship is called t he metacenter M, a nd th e dist an ce h of M from the cente r of mass G is called th e me/acentric height. For most hull sha pes h remains practically constant for an gles of list up to about 20°. When M is above G, as in Fig. 5/ 36b, t here is a righting moment which tends to bring th e ship back to its upright position. If M is below G, as for th e hull of Fig. 5/360 , the moment accompa nying the list is in th e direction to increase the list . This is clea rly a condition of instability a nd must be avoided in th e design of any sh ip.

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Art icl e 5/9

Sample Problem 5/18

A

,- "

1m

A rectangular plate, shown in vertical section AB , is 4 m high and 6 m wide (normal to the plane of t he pap er ) and blocks the end of a fres h -water channel 3 m deep . Th e plat e is hinged about a horizontal axis along its upper edge through A and is re strained from openi ng by th e fixed ridge B which bears hor izonta lly agai ns t th e lower edge of t he plat e. Find th e force B exerted on the plate by the r idge.

I

3m

B

Solution. Th e free-body diagr am of the plate is shown in section and include s the vertical and horizo ntal componen ts of t he force at A, th e u nspecified weight W = mg of th e plate, th e unkn own horizon tal force B, an d the resultant R of the triangular distribution of pressure against th e vertical face. Th e den sity of fres h water is p = 1.000 Mgjm:J so th at the average pressure is

l

2

Pay ~ 1.000(9. 8 1 ) (~ ) = 14.72 kP a

R

(14.72)(3)(6)

~

i--. ~ B

= 265 kN

H elpful Hint

Ans.

B = 198.7 kN

CD Note that the units of pressure pgh are ( 10'

~;) (~) (m) = ( 10' k~~m)C~2) ~ kN/m 2 ~ kP a.

Sample Problem 5 /19

Al

The ai r space in the closed fresh-wat er tank is maintained at a pressure of 0.80 Ib/ in.2 (above atmosp her ic). Determine th e res u ltant force R exerted by th e air and water on the end of the tank.

Solution . Th e pressu re distribu tion on th e end surface is shown, where Po = 0.80 Th e specific weight of fresh wat er is I' ~ pg ~ 62.4/1728 = 0.0361 Ib/ in.3 so tha t th e incre ment of pressure up due to the wat er is

Ib/in .2

t:.p ~ I' ::'11 = 0.0361(30) = 1.083 Ib/in

CD

~

PoA,

~

R 2 = ::'P.,A2

Th e res ultant is t hen R

~

0.80(38)(25) ~

R 1 + R2

~

~

760 + 406

~

~

406 Ib 1166 lb.

Ans.

We locat e R by applying th e momen t pr inciple about A noti ng th at R 1 acts th rough the center of th e 38-in. depth an d t hat R 2 acts through t he cent roid of th e trian gu lar pressu re dist rib ution 20 in . below t he sur face of th e wat er and 20 + 8 =: 28 in . below A. Thu s, 116611 = 760(19) + 406(28) Marwan and Waseem AI-Iraqi

8"

Water

30"

h ::: 22.1 in. www.gigapedia.com

25"---c1

T

1

B

Side view Po

--

End view

A

R

R-

I~I-

~m I. 28" II

~ --L I[

R,

760 Ib

") -1.083 2- (30 )(20

Air

2

Th e resul tan t forces R 1 and R 2 due to the rectangul ar and trian gu lar dist ributions of pressure, respectively, are

R1

1

4m

-I-I R-

This force acts through the cen troid of t he triangular distribution of pressu re , which is 1 m above th e bottom of t he plate. A zero mom ent summation about A establishes the unknown force B. Thus, 3(265) - 4B = 0

1

mu

III

Im

The resultant R of the pressure forces aga inst the plate becomes [R ~ Pa,A1

305

Fluid Statics

An s.

Po B

Helpful H int

CD Dividing

the pressure distri bu tion into these two parts is decided ly th e simplest way in which to make the calculati on.

306

Chap ter 5

Distr ibu ted Fo rc e s

Sample Problem 5/10

a

Determine completely the resultant force R exerted on the cylindrical dam surface by the wate r. The density of fresh water is 1.000 Mg/ m3 , and the dam has a length b, normal to the paper, of 30 m.

Solution. The circular block of water BDO is isolated and its free-body diagram

D

~

I r =4m

Bi

is drawn. The force Px is ., pgr b (1.000)(9.81)(4) 30 () 23 kN Px -_ I'g -,~ = ""2 r = 2 ( )4 = 50

(j)

ar---,=-

J

The weight W of the water passes through the mass center G of the quartercircular sect ion and is

D - .,- - .r

I

L-_~~ ""'

Equilibrium of the section of water requires [ ~ F,

01

n;

P, = 2350 kN

[ ~F,

01

R,

mg = 3700 kN

c

- -- --\

mg = pgV = (1.000)(9.81) m: )2 (30) = 3700 kN

B"_ ' -4 t

A

/I

y

The resultant force R exerted by the fluid on the dam is equal and opposite to that shown acti ng on the fluid and is

[R

=

J R x 2 + Ry 2)

R

=

/( 2350)2 + (3700)2 - 4380 kN

Ans.

The x-coordinate of the point A through which R passes may be found from the principle of moment s. Using B as a moment ce nter gives

r

2350

4r

Px 3 + mg 3" - R,x = 0, x =

W

+ 3700 3700

(~) = 2.55 m Ans.

Helpful Hints

@ Altemative Solution. The force acting on the dam surface may be obtained

by a direct

integration of the component s dR;r

wher e p

=p

ciA cos H

= I'gh = I'gr sin

e,

=

Ry

=

R

Thus , = values gives

f'2

0 and dA

dRy

and

"2

0

pgrb sin 2 0 dO = pgr2b

JR/

if there is any question about the units for- pgli.

(1

= bir dOl. Thus,

pgr'b sin 0 cos 0 d o = _ pgr2b

f

P dA sin

=

[0

2-

Q) This approach by integra tion is fea-

[co~ 20]:,2 = ~pgr2b . 20] , ,2 0 4

S10

sible here mainly becau se or the simple geometry of the circular arc.

= ~ 1Tpgr2b

+ R/ ' = ~pgr2b,h + r?/4. Substituting the numerical

R = 4(1.000)(9.81)(42J(30)J! + . '/4 = 4380 kN

Ans.

Since dR always passes through point 0 , we see that R also passes through about 0 mus t cancel. So we write R;rYl = R y X l , which gives us

o and, therefore. the moments of R;r and Ry x, /y,

=

Rr/ R y = (~I'gr'b)/(~ "I'gr'b )

=

2/"

By similar triangles we see that

x/ r = x, /y, = 2/"

and

x = 2r/" = 2(4)/" = 2.55 m

Marwan and Was eem AI-Iraqi

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(j) See note (j) in Sample P roblem 5/ 18

Ans.

Article 5/9

Flu id Stati cs

307

Sample Problem 5/21 Determine t he resu ltant force R exer ted on th e semicircular end of the wat er tank shown in the figur e if the tank is filled to cap acity. Express the result in terms of th e radiu s r an d the water den sity p.

Solution I. We will obtain R first by a direct integr ation. Wit h a horizon tal stri p of a rea dA = 2x dy acte d on by the pressure p = pgy, th e increm ent of th e resu lt an t for ce is dR = p dA so that R

~

f

P dA

~

f

h--:

r

pgyl2x dy)

=:-1

An s.

In tegr ating gives

Th e locat ion of R is determ ined by usin g th e pri nciple of mom ents. Taking mome nts about the x-ax is gives

fRY

~

f

y dR]

In tegr atin g gives

-y

a nd

~

3nT

A ns.

16

SolutIon II. We may use Eq. 5/ 25 directly to find R, where th e average pressu re is pgh a nd h is the coordinate to t he centroid of th e area over which t he pre ssu re acts. For a se micirc ular a rea Ii = 4r / (31T). 4r 1Tr2 2 ~ -pgr' 31T 2 3

[R ~ pghA ]

CD

R ~ pg -

Helpful Hint

CD Be very careful not to

make the mis take of assuming that R passes through the centroid of the area over which the pressure acts .

A ns.

which is the volume of the pre ssure-ar ea figure. The resu ltan t R acts t hrough t he cent roi d C of the volu me defined by th e pressure-area figure. Ca lculation of the cent roida l distance Y invo lves th e sa me integra l obtained in Solution I.

Sample Problem 5/22 A buoy in th e form of a uniform 8-rn pole 0.2 m in diameter ha s a mass of 200 kg and is secured at its lower end to th e bot to m of a fres h-water lak e wit h 5 m of cable. If the depth of t he water is 10 m, ca lculate t he angle 0 mad e by th e pole wit h the hori zontal.

Solution. T he free-bo dy diagram of the bu oy shows it s weight acti ng th roug h G, th e vertical ten sion T in t he an chor cable, a nd the buoya ncy force B which passes through centroid C of the submerged porti on of the buoy . Let x be t he distan ce from G to th e waterline. Th e density of fresh wate r is p = 10 3 kg/m 3 , so th at the bu oyancy force is [B

~

pgV]

Moment equilibri um ,

B

+

~ 103(9.8 1),,(0 .1)2(4

~ MA

x) N

= 0, about A gives 4 +x

200(9.8 1)(4 cos OJ - [103 (9.8 1),,(0.1)2(4 + xJ] -2T hus ,

x = 3.14 m

an d

I L+5fi .L

tJ = sin - 1 (4


5

+ 3.14

cos 8 = 0

) = 44.50

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Ans.

T

308

Chapter 5

Distributed Forces

PROBLEMS Introductory Problems 5/ 163 The submersible diving cha mber has a total mass of 6.7 Mg inclu ding per sonnel, equipment , and ballast. When th e chamber is lowered to a depth of 1.2 km in the ocean, the cabl e ten sion is 8 kN . Compute the tot al volu me V displaced by the cha mber. A n s. V = 5.71 m :J

5/1 65 A recta ngu lar block of den sity 1'1 floats in a liquid of den sity P.2' Deter mine the ra t io r = h i e. wher e h is th e submerged depth of block. Evalu at e r for an oak block floating in fresh wat er and for steel floatin g in mercu ry. An s. r = !2 r = 0.8, 0.577 /}.2·

P2

Problem 5/165

5/ 166 The form s for a small concre te ret ainin g wall are shown in section. Th er e is a brace Be for every 1.5 m of wall lengt h. Assu ming that the joints at A . B, and C act as hinged conn ect ions, comp ut e th e compression in each bra ce BC. Wet concrete may be treat ed us a liquid with a densit y of 2400 kg/ m:J• Pro blem 5/163

r-r.,..,---~:--------,-

5/164 Specify th e magnitu de and locati on of the resulta nt force which acts on each side and t he bottom of the aqua riu m due to the fres h water inside it. 0.301r---

/ -----

3m

0.7 m

~..:..........

A

0.4 01

·- -

Problem 5/166

Problem 5/164


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Article 5 /9 5/167 A deep-submersible diving cham ber designed in the for m of a spherical she ll 1500 mm in diameter is ballasted wit h lead so that its weight slightly exceeds its buoyancy . Atmospheric pressur e is maintai ned within the sphere during an ocean dive to a depth of 3 km . The t hick ness of t he shell is 25 mm. For this depth calculate the compressive stress a whic h acts on a diametral section of the shell, as indicated in the rig ht- hand view. Ans. (J' = 463 MP a

(J

Proble ms

~09

5 /169 When t he sea-water level ins ide the hemisp herical

cha mbe r reac he s t he C.B-m level shown in the figure, the plunger is lifted , allowing a su rge of sea water to enter the vertical pipe. For this flu id level (a) det erm ine the average pressur e (T su pported by the seal area of the valve befor e for ce is app lied to lift th e plu nger and (b) determine the force P (in addition to th e force needed to support its weight) required to lift the plunger. Assume atmospheric pressu re in all airspaces and in the sea l area when cont act ceases under t he act ion of P. Ans. a ~ 10.74 kPa, P ~ 1.687 kN

Seawater supply -

-=::\

.

I

I ~

Problem 5/167

a .6m

5 /168 Fresh water in a channel is contained by the uniform 2.5-m plat e freely hi nged at A. If the gate is

designed to open when the dept h of the water reach es 0.8 m as shown in the figur e, what must be the weight w (in newtons per meter of horizontal lengt h into t he paper) of the gate?

-r =-0---==-==;:--;0;:;:::--

Problem 5/169

5/170 One end of a u niform pole of length L and density p' is secured at C to the bottom of a tank of liquid of density p and dept h h . For t he conditions p' < p and h < L , find the angle () assumed by the pole.

- -\'"

a .8m

-L

B 30'

Problem 5/16B

Problem 5/170


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310

Chapter 5

Distributed Forces

5/111 The cross section of a fresh-wate r tank with a slanted bottom is shown . A rectangular door 1.6 m by 0.8 m (normal to th e pla ne of th e figure ) in th e bottom of t he tank is hi nged at A a nd is ope ned against th e pr essu re of the water by the cabl e u nder a ten sion P as show n. Ca lcu late P. Ans. P = 12.57 kN

5/113 The solid concrete cylinde r 6 ft long a nd 4 ft in diameter is supported in a half- submerged position in fre sh water by a cab le which passes over a fixed pul ley at A. Compute th e te nsion T in the cab le. The cylinder is waterproofed by a plastic coating. (Consu lt T ab le D/ 1, Appe ndix D, as needed .) An s. T = 8960 lb

P

T

1.2 m

Problem 5 /171

Problem 51tH

Representative Problems 5/112 A block of wood in t he form of a waterproofed 16· in . cube is floa tin g in a tank of salt water with a a-In. layer of oil floating on the water. Assu me t hat the cube floats in t he a ttit ude shown, and calculate the height h of th e block above th e surface of th e oil. The specific weights of oil, salt water. a nd wood are 56 , 64, a nd 50 Ib/n 3 , respecti vely.

5/114 A ma rk er bu oy cons isting of a cylinder a nd cone has t he dim en sions shown an d weighs 625 Ib whe n out of th e water. Det er mine th e protrusion h whe n t he buoy is floating in sa lt wa te r. T he buoy is weighted so that a low center of mas s ensures sta bility.

r 'l 2

I

6'

--r

Oil

Salt wate r

16"

~~3 _-----,l Problem 5/172 Problem 5/174

Marwan and W aseem Al-l raqi

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Article 5 /9

5/175 A channel-marker buoy consists of an 8-ft hollow stee l cylinder 12 in. in diamete r weighing 180 lb and anchored to th e bott om with a cable as shown. If h = 2 ft at high t ide, calculate the ten sion T in th e cable. Also find t he value of h when the cable goes slack as th e t ide drops. Th e specific weight of sea water is 64 Ib/ ft 3 . Assume the buoy is weighted at its base so t hat it remains vert ical. Ans. T ~ 121.6 Ib, h ~ 4.4 2 ft

Probl ems

311

5/177 The hinged gate ABC closes an opening of widt h b (perpendicular to t he paper ) in a wate r cha n nel. The wat er has free access to the u nder side as well as the right side of th e gate. When t he wat er level rises above a certain value of h, th e gate will open. Determine t he critical value of h . Neglect the ma ss

Ans. h ~ a ./3

of t he gate. C

12"

-I I-

'~l 1~

h

8'

J

ru--l A

B

~ a-I~ r Prob lem 5/177

Problem 5/ 175

5/176 A fre sh-water channel 10 ft wide (normal to the plan e of the paper ) is blocked at its end by a rectangular barr ier, shown in section ABD. Su ppor ting struts Be are spaced every 2 ft along the lO-ft width. Det er mine th e compre ssion C in each st rut. Neglect the weight s of t he members.

5/178 Th e rectangular gate shown in section is 10 ft long (perpendicular to the paper ) and is hin ged about its upp er edge B. The gat e divides a chann el leading to a fresh-water lak e on th e left and a salt -water tidal basin .on the righ t , Calculate the torque M on the shaft of the gate at B requi red to pre vent the gate from opening when th e sa lt-water level drops to h ~3 ft.

Problem 5/178

Problem 5/176


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312

Cha pte r 5

Dis tr ibu ted Force s

5 /179 Th e hydraulic cylinder operates the toggle which

closes t he vertical gate agai nst t he pressure of fresh water on the opposite side . Th e gat e is rectangular with a horizon ta l widt h of 2 m perpendicular to t he pa per. For a depth h = 3 m of water, calculate t he requ ired oil pressure p which acts on the 150·mm· diam eter pist on of the hydraulic cylinder. Ans. p ~ 7.49 MPa

5 / 18 1 The barge crane of rectangular proportions has a 12·ft by 30·ft cross section over its ent ire lengt h of 80 ft. If the maximum permissible submerge nce and list in sea water ar e represented by th e position shown, determine the corresponding maximum sa fe load w whic h the bar ge can handle at the 20-ft exte nded position of th e boom . Also find t he tot al displacement W in lon g tons of th e un loaded barge (I long ton equ als 2240 lb). Th e distribution of ma ch inery and ballast places t he cente r of gravity G of th e barge, minu s the load w. at th e center of t he

hul l.

A ns. w = 100,800 lb, IV = 366 long tons

Probl em 5/179 5 /180 The triangu lar and rectangu lar sections are being

conside red for the design of a sma ll fresh-water concrete dam . Fr om the sta ndpoint of resistan ce to overtu rning abou t C, which section will require less concrete, and how much less per foot of dam length? Concrete weighs 150 Ib/ ft 3 .

Problem 5/181 5 / 182 Th e cast-iron plug seals th e drainpipe of an open fresh-w ater tank which is tilled to a dept h of 20 ft. Deter mine the tension T required to remove t he

plug from its ta pered hole. Atmospheric pressure exists in th e drainp ipe an d in t he seal area as t he plug is being removed. Neglect mechan ical frict ion between th e plug and its su pport ing sur face.

c

20'

f--a Problem 5/180

Prab lem 5/182


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A rticl e 5 / 9

5/183 The Quonset hut subjected to a hor izon tal wind a nd the pr essu re P agains t the circu la r roof is approximated by Po cos O. Th e pr essu re is positive on the wind ward side of the hut and is negat ive on the leeward side. Det ermine the tota l hori zon tal shea r Q on the foundat ion per un it length of roof measured 1 normal to the paper . All s. Q

=

Prob lem s

313

5/185 Th e up stream side of an a rch ed da m has the form of a verti ca l cylindrical surface of 500-ft radius an d subte nds an a ngle of 60°. If th e fres h wate r is 100 ft deep , det ermine th e total force R exer ted by th e w atel' on the dam face . An". R ~ 156.01l0 6 11b

FT1'PO

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,

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,/

Problem 5/183 5 /184 T he semicylindrica l stee l shell with closed ends has a mass of 26.6 kg. Determine th e mass m of th e lead ba llast which mu st be placed in the shell so that it floa ts in fres h water a t its hal f-rad ius depth of 150 mm.

Problem 5/185 5/186 Th e fresh-wat er side of a concrete da m has the sha pe of a vertical parabola with vertex at A. Determine t he positi on b of th e base point B throug h wh ich acts the resultant force of t he wate r against th e dam face C. 27 m

l

Problem 5/184

Problem 5/1 86


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314

Chapter 5

Distributed Forces

5/187 A st ru ct u re design ed for observat ion of sea life benea th th e ice in pola r wate rs consists of the cylindrical viewing cham ber connected to th e sur face by th e cylindrical s haft open a t the top for ingr ess a nd egress. Ballast is ca rr ied in t he rack below the cha mber. To e ns ure a sta ble condition for th e st ructure, it is necessa ry th at its legs bear on th e ice with a force that is at least 15 percent of th e tot al bu oya ncy force of t he subme rged st ructure . If th e st ructu re less ballast has a mass of 5.7 Mg, calcu lat e th e requi red mass m of lead ballast. Th e de nsity of lead is 11.37 Mg/m". Ans. m = 4 .24 Mg

, ((I )

r

( ,

x

:'iI O.375 I m j,.~,~.,.."."",,_~L ' O.55m

12 m

tb)

Problem 5/188

5/189 Th e 3-m plank shown in section has a den sit y of800 kg/ m 3 an d is h inged abou t a hor izont al axis through its upper edge O. Calcu late the a ngle H as su med by th e pla nk with t he hori zontal for t he level of fre sh wate r shown. A ns. /I = 48.2°

/~""t-T-'

.> / ·0-'1-

3m

1m I

!

Problem 5/187

Problem 5 /189

5/188 The small access hole A allows maintenance workers to enter the storage ta nk at gro un d level whe n it is e mpty . Two designs, (0) and (b), are shown for th e hole cover . If th e tank is full of fresh wa te r, est ima te t he ave rage pressure (7 in t he seal a rea of design (a) a nd th e ave rage increase u 'I' in th e initial ten sion in eac h of t he 16 bolts of design (b). You may take th e pre ssure over the hole a rea to be consta nt, an d the pressure in the sea l a rea of design (b) may be assu med to be at mospheric.

5/190 Th e deep-submer sible research vesse l ha s a passen ger compart ment in t he for m of a sp herical steel shell wit h a mea n ra dius of 1.000 m a nd a thickness of 35 mm . Calcul ate the mass of lead ballast wh ich th e vessel mu st ca rry so tha t th e combined weight of th e steel shell a nd lead ball ast exact ly ca ncels t he combin ed buoya ncy of th ese two part s alo ne. (Consu lt Tabl e D/l, Append ix D, as needed.I

Problem 5 /190 Marwan and W aseem AI-Iraqi

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Article 5 /9 5/191 The elem ents of a new method for constructing concr et e foundation walls for new hou ses ar e shown in th e figure. Onc e the footin g F is in place, polyst yren e forms A are erected and a thin con cr et e mixture B is poured between the form s. Ties T pr event th e form s from se pa ra ting. After the concr et e cures, th e forms a re left in place for insul ati on . As a design exercise, make a con servat ive es t ima te for the unifor m tie spacing d if th e ten sion in each ti e is not to exceed 6.5 kN. The hori zontal tie spac ing is th e sa rne as the vertica l spaci ng. St at e a ny assu mptions. Th e den sity of wet concre te is 2400 kg jm3 • A ns. d ~ 0.300 m

Water level --------------------------------,-----

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Ramp

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Problem 5/192

d

I

-J,1/2 r

T A

B

A

F

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Problem 5/191


315

5 /192 T he trapezoidal viewi ng wind ow in a sea-life aquarium has the dimensions shown. Wit h th e aid of a ppropriate diagram s a nd coordin at es, describe two methods by which the resultant forc e R on th e glass due to water pr essure, a nd th e vertica l locat ion of R , could be found if numeri cal values were supplied.

I

:301

Problem s

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316

Chapt er 5

Distr i bu t ed Forces

CHAPTER REVIEW In Cha pter 5 we have studied variou s common examples of forces distributed throughout volumes, over a reas , an d along lines. In all these problems we ofte n need to determine the resu ltant of th e distributed forces and the locati on of th e resu ltant.

Finding Resultants of Distributed Forces To find the resultant and line of action of a dist ributed force : 1. Begin by multiplying the intensity of the force by the appropriate

element of volume, area, or length in terms of which the intensity is expressed. Then sum (integrate ) the incremental forces over the region involved to obtain their resultant.

2. To locat e t he line of action of th e resultant, use the principle of moments. Evaluate the sum of the moments, about a conve nient axis, of all of th e increments of force. Equate th is sum to th e moment of th e resultan t about t he sa me axis. Then solve for th e un known moment arm of the resultant.

Gravitational Forces When force is distribu ted th roughout a mass, as in the case of gravitational attraction, the intensity is the force of att raction pg per unit of volume, where p is th e den sity a nd g is th e gravitat ional accelera tion. For bodies whose density is constant, we saw in Section A that PC cancels when t he momen t pri nciple is applied. Thi s leaves us wit h a strict ly geometric prob lem of findin g th e centroid of the figure, which coincides with th e mass cente r of th e physical body whose boundary defines th e figure. 1. For fla t plates a nd shells which are homogeneous a nd have constant

thickness, th e prob lem becomes one of findin g th e properties of an area. 2. For slender rods and wires of uniform density and constant cross secti on, th e problem becomes one of findin g th e properties of a lin e segment.

Integration of Differential Relationships For problems which require th e integration of dilTerential relationships, keep in mind the following considerations. 1. Select a coordinate system which provides the simplest description

of th e boundaries of t he region of integration. 2. Elim inate higher-o rd er dilTerential quantities whenever lower-order different ial quantities will remain. 3. Choose a first-order differential eleme nt in preference to a secondorder element and a second-order element in preference to a thirdorder element. Marwan and W aseem AI-Iraqi

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Chapter Rev iew

4. Wherever possible, choose a differential element which avoids discontinuities within the region of integration.

Distributed Forces in Beams, Cables, and Fluids In Sectio n B we used th ese guidelines along with the principles of equ ilibrium to solve for the effects of distributed forces in beams , cables, a nd fluids. Rememb er th at: 1. For beam s and cables th e force intensity is expressed as force per unit len gth.

2. For fluids the force intensity is expressed as force per unit area, or pressure.

Alth ough beams, cabl es, and fluids are physically quite different appli cations, their problem formulations share the common elements cited above.


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31 7

318

Chapter 5

Distr ibu ted Forces

REVIEW PROBLEMS 5/193 Determine th e x-coordinate of the centroid of th e shaded area. Ans. X = 3.66 in .

{ o

To

7.5"

4.5"

+

Dimensions in millim et ers

4 .5"

~

Problem 5/1 95 5/196 Det ermine t he volu me V and total su rface a rea A of the complete rin g wh ich is shown in sect ion. All four corne r radii of t he cross section are 10 mm . z

es. 1

Problem 5/ 191

1

5/1 94 Determ ine th e x- a nd y-coc rdine tes of the cent roid of th e sha ded area. y I

IOmm , """"""",",,:-+,,,,-

L2 0 I

o o

mm

10

y = kx l/3

~Jmm

40--1

mm

o

Problem 5/ 196

I o o o o o

I

5/197 Th e assembly consists of four rods cut from th e same bar stock . The curved me mbe r is a circu lar a rc of radius b. Det ermine t he y- a nd a-coordina tes of t he mass center of th e assemb ly.

o o

Ans. Y ~ 0.461b, Z = 0.876b

I I

I

°0L------OJ.5-- - --~----x Problem 5/194 5{195 Calculate t he y -coordinate of the centroid of the sha ded area. Ans. Y ~ 99.7 mm

b .J--

6/4 The coefficien ts of static and kinetic friction between the lOO.kg block and the inclined plane are 0.30 and 0.20, respectively. Determine (0) the friction force F acting on th e block whe n P is applied with a magni tude of 200 N to th e block at rest, (b) the force P requ ired to init iat e motion up the incline from rest . and (c) the friction force F act ing on th e block if P 600 N.

Introductory Pr oblems P

6/1 Th e 85-lb force P is a pplied to th e 200-lb crate, which is stationary before the for ce is applied. Determine th e magnitude and dir ect ion of th e fricti on force F exert ed by th e hori zontal surface on th e crate . An s. F = 85 lb to th e left P ., = 0.30 P. = 0.20

200 lb P = 85 1b

Ilil ;I

is-

Jp = 0.50

Problem 6/4

/ )p; - 0.40 r

Problem 6 /1 6 /2 Th e 700-N force is a pplied to th e 100-kg block, which

is stationary befor e th e force is applied. Determ ine th e magnitude a nd directi on of th e fricti on force F exerted by the hor izontal su rface on th e block.

6/5 The 1.2. kg wooden block is used for level support of the 9-kg ca n of paint. Determine the magnitude and direction of (0 ) th e fr iction force exert ed by th e roof surface on t he wooden block and (b) th e tot al force exert ed by th e roof sur face on th e wooden block. Ans. (0) F = 3 1.6 N , Ib) P = 100.1 N up

P = 700 N

100 kg

l

.,/ 30'

I' = 0.80 = 0.60 ' \

II;

4

r--=----... 12 Problem 6/5

Problem 6/2

6/3 The design er of a ski re sort wishes to have a portion of a beginn er's slope on which th e skier 's speed will remain fairly consta nt. Tests indicate the average coefficients of friction between skis and snow to be 1', = 0.10 a nd 1'. = 0.08. What should be th e slope angle tJ of th e const an t-speed section? Ans. (I = 4.57 0

6/6 Th e magnitude of force P is slowly increased. Does th e homogeneou s box of mass III slip or t ip first? St at e th e value of P wh ich would cau se each occurrence. Neglect any effect of the size of the sma ll feet. P ----,

A ~~

9

c Problem 6 /3

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2d - -

-

-

Problem 6/6 MalWan and Waseem AI-Iraqi

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- -I

Ar ticle 6 /3

Pr oblems

339 F

6/7 T he light bar is used to suppor t t he 50-kg block in its ver tical gu ides . If the coefficient of stat ic frict ion is 0.30 at the upp er end of t he bar a nd 0.40 at t he lower end of the bar, find th e frict ion force act ing a t each end for x = 75 mm. Also find th e maximu m value of x for wh ich the ba r will not slip. Ans. FA = F n = 126.6 N, Xm a .x = 86.2 m m

50 kg F B

Problem 6 /9

6/10 T he 300·lb cra te wit h mass center a t G is supported on the horizon ta l su rfaces by a skid at A a nd a roller a t B . If a force P 0[60 Ib is requ ired to in it ia te motion of the crate , determ ine the coefficient of sta t ic frictio n at A.

A

-t----- --r i -+ \/ i I 12" AI-- 24"--+- 24T

Problem 6/7

P

G,

6 /8 Th e 3D-kg homogeneous cylinder of 400·mm diam eter re st s aga inst the vertica l an d inclin ed sur faces as shown. If t he coefficient of sta t ic frict ion between the cylind er a nd the su rfaces is 0.30, calculate the a pplied clockwise couple M whi ch wou ld ca use the cylind er to slip.

12"

B

Problem 6/10

6/11 T he illust ration shows the design of a ca m-typ e locking device, which , in th e presence of su fficient friction , limit s the movem ent of body B to be to th e left only; rightwa rd movem e nt is pr evented. Th e su rface of the cam near point D is circula r with center a t C. Given the distan ce L , specify th e ca m offset d so t ha t th e device will work if t he coefficient of st atic fr ict ion fJ...~ is 0.20 or gr ea ter. Ans. d es 0.2L

30°

d

Problem 6/8 6/9 The to ngs are design ed to ha ndle hot ste el tu bes wh ich a re being heat-treated in a n oil bath . For a 20 0 jaw open ing, wha t is the min imum coefficient of sta tic friction between the jaws a nd the tube which will ena ble th e to ngs to gri p t he t ube wit hout slipping? A ns. fJ..:; m in ::::. 0.176

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0

I I A

/ Y I Allowable No / - ,

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7>-

,

- - - 8>-

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D

E

Problem 7 /29 7 /10 Th e antitorque wre nch is design ed for use by a crew

member of a spacecraft where no sta ble platform exists agai nst which to push as a bolt is turn ed. The pin A fits into an adjacent hole in th e space st ruc tu re which conta ins th e bolt to be turned. Successive oscillations of th e gear and handle unit turn the socket in one direct ion through th e action of a ra tchet mechani sm. Th e reac tion against th e pin A provides the " a nt itorque " characteris t ic of the tool. For a gr ipping for ce P = 150 N, det erm ine t he torq ue M transmitted to th e bolt. mg/2b if h < mg/ 2b

A ns.

Thus, if the spr ing is sufficiently stiff, the bar will re tur n to the vertical positi on even though there is no force in t he spri ng at that position.

@ We might not have anticipated thi s re sult without the mathematical analysis of th e sta bility.

Solution II. cos 0 = ~~, 0 ~ cos - 1 ~~

~:~ = hb' [2(;;,~ )' - 1]- ~mgb (;~~) ~ lib' [(;~~)' @

1] Ans.

Since t he cosine mu st be less than u nity , we see th at th is solution is limited to th e case where k > mg /2b, which mak es the second deriv at ive of V negative. Thus, equ ilibrium for Solution II is never stable. If Ii < mg / 2b, we no longer hav e Solution II since the spring will be too wea k to maint ain equ ilibrium at a value of (1 between 0 and 90 Q

•


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@ Again , without the benefit of the mathematical analysis of th e stability we might have supposed er roneously that the bar could come to rest in a st able equ ilibrium positio n for some value of (J between 0 and 90°.

414

Chapter 7

Vir tual Work

PROBLEMS (Assume th at the nega ti ve work of fri ction is negligib le in the follow ing prob lems. I

7/36 The sma ll cylinder of ma ss In a nd radius r is confined to roll on the circ u lar surface of radius R. By t he met hod s of th is a rt icle, prov e tha t t he cylinder is unsta ble in case (a ) a nd sta ble in case (bl.

Introdudory Problems

f

7/33 The potential e nergy of a mech anical sys te m is given by V = 6x" - 3x 2 + 5, wh ere .r is the position coordi nate whic h defines t he configuration of t he single-degree-of-freedom system. Determine the equ ilibrium value s of x and th e sta bility condition of each . A il s. x = 0, un stable; x = sta ble; x = stable

4,

rei R

If I

1

Ie

/..-

I

I1

'-

'I

(a l

(bl

Problem 7/36 7/37 For the mechanism shown the spring is uncompressed whe n (J = O. Determ ine th e a ngle 0 for th e equilibrium position and specify th e min imu m spri ng stiffness k whic h will limit (} to 30°. T he rod DE passes freely through the pivoted colla r C, a nd th e cylinde r of ma ss m slides freely on th e fixed vertica l shaft. - 1 mg Ans. fJ = cos 2kb' k mm

L

Probl em 7/34 7/35 Th e bar of mass m with cente r of ma ss at G is pivoted about a horizontal axi s through O. Prove the stability conditions for the tw o positions of equilibr ium. Ail S. II = 0, u nstable; (J = 1800, stable

A

Probl em 7/37

Problem 7/35 Marwan and W aseem AI-Iraqi

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7/34 The uniform bar of mass m a nd length L is supported in the vert ical plane by two identical springs each of . stiffness h a nd compressed a distance f, in the vertical position /} = O. Determine the minimum stiffness k which will ens ure a sta ble equi libr ium position with tJ = O. T he spr ings may be assumed to act in t he horizontal direction during sma ll a ngu lar motion of th e bar.

I

I

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Article 7/ 4 7/38 The figu re shows th e cross sect ion of a un iform 60-kg venti lat or door hinged along its upp er hor izontal edge at O. Th e door is contro lled by the spring-loaded cable which passes over th e small pulley at A. The spring has a st iffness of 160 N per met er of stre tch an d is undeform ed when (J = O. Deter mine the an gle fJ for equilibri u m.

I II

Pr o b le m s

415

7/40 Deter min e t he equ ilibriu m value of x for the springsupported bar. Th e spring has a st ifTness h and is unst re tched when x = O. Th e force F acts in the directi on of t he bar , and the mass of t he ba r is negligible.

k

11I1 1

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Problem 7/40 7 /41 One of the critical requirements in th e design of an

i-f Problem 7/38 7/39 For the device shown the spr ing would be u nstretc hed in the positio n (J = O. Specify the st iffness h of the spring which w ill establish an equilibrium posit ion II in the vertical plane . The mass of the links is negligible compa red with m. mg cot fJ Ans. h = -26 1 - cos (J

artificial leg for an amput ee is to prevent th e knee joint fro m buckling u nder load when t he leg is st ra ight. As a first app roximation, simulate the artiflcial leg by t he two light links wit h a torsion spr ing at their comm on joint. The spr ing develops a torqu e M = K{3 , which is proport ional to th e angle of bend {3 at the joinl. Deter mine the min imum value of K wh ich will ens ure stability of the kn ee joint for

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A n s. K min = ~mgl

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m

II

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Problem 7/39

.JfJ. Prob lem 7/41


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416

Chapter 7

Virtual Work

Representative Problems 7/42 Th e cylinder of mass M and radius R rolls with out slipping on th e circu lar surface of radius 3R. Attac hed to th e cylinder is a small body of mass 11l . Determine th e required relationship betwee n M and m if th e body is to be stable in th e equili brium position s hown.

7/44 Determine the max imum height Ii of th e mass I1l for which the inve rted pendulum will be stable in th e vertical position shown. Each of th e springs has a stiffness k, and th ey have equal precompressions in this positi on. Neglect the mas s of th e remaind er of the mechani sm .

/~I I

R /4 3R /4 ---- /~,--- -

Problem 7/42

Problem 7/44

7/43 Each of the two gears ca rries an eccentric mass m and is free to rot at e in the vertical plan e about its bearing. Determine th e values of fJ for equi librium and identify th e type of equilibrium for eac h positi on. Ans. (J = 0, un stable (} = 120°, stable o = 180°, unsta ble (J = 240°, stable

7/45 One en d of th e torsion spr ing is secured to t he gro u nd at A , and th e othe r end is faste ned to th e shaft at B. Th e torsional st iffness K of th e elasti c spr ing is the torq ue required to twi st the spr ing through an angle of one radian . The spring resist s th e moment about the shaft axis caused by th e tension mg in the cable wrapped aroun d the dru m of ra dius r, Determ ine the equ ilibriu m value of h measured from the dashed positio n, where the spring is untwist ed . mg ,:J. An s. li = K

Problem 7/43

Problem 7/45


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Articl e 7/4 7/46 For th e mechani sm shown, th e sprin g of stiffness k ha s an un stretch ed length of esse ntially zero, and th e larger link has a mass m with mass center at B. Th e mas s of t he smaller lin k is negligible. Determ ine th e equ ilibriu m angle (} for a given downward force P.

Problem s

417

7/48 Th e 3-tb pendulum swings about axis 0-0 and has a mass center at G. When (} = 0, each sprin g ha s an initial stretch of 4 in. Calculate th e maximum stifTness k of eac h of the parall el spr ings which will allow t he pend ulum to be in sta ble equilibrium at the bottom positio n 8 = O.

p

k

·~--\ \ \ \ \ \ \ \ \ \ \ \ \ -----~1 b

b

•

8

B

b • A

Problem 7/46

7/47 The cross sect ion of a tra p door hinged at A and having a mass m and a center of mass at G is shown in th e figur e. Th e spring is compressed by th e rod wh ich is pinn ed to th e lower end of t he door and which passes through th e swivel block at B. When II = 0, th e spring is undeformed. Show that with th e proper stiffness k of the spring, the door will be in equilibrium for any angle 8. mgb A lI S.ll = - 2a

Problem 7/48 7/49 Th e solid he misph er e of diameter 2b and concen tric cylindrical knob of diam eter b are resting on a hori zontal su rface. Determine the maximum heigh t h whic h the kn ob may have wit hout caus ing th e un it to be u nstable in the upright positio n shown. Both parts are made fro m the same material. An s. h < b ,,''2

Problem 7J47

Problem 7/49


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418

Chapt er 7

Virtual Work

mine th e angle of tilt (I for equ ilibrium. (Hi nt: Th e defor mati on of th e spring may be visualized by a llow. ing th e base to tilt th rou gh th e required angle fl about a while th e seat is held in a fixed posi tio n. I

7 /50 Pr edict thro ugh calculatio n wheth er t he homogeneo us semicylinder an d the half-cylindrical shell will remain in th e positions shown or wheth er t hey will roll ofTt he lower cylinders .

~8 ----j \...--:-- -,.,:3 00

r

III III

FO'~ ~8 ~~

\

Problem 7/50 7 / 5 1 Th e uni form lin k AB has a mass m , and its left end A travels freely in t he fixed hor izont al slot. End B is

attached to the vertical plunger , wh ich compresses the spring as B falls. The spring would be uncompressed at th e position (I = O. Determine th e angle 11 for equilibriu m (other th an the impossible position corr espo nding to /J = 90°) and designat e the condition wh ich will ensu re sta bility.

Problem 7/52 7/53 A proposed parallelogr am link age for an adjus ta ble-

position lamp is shown. If th e un stretched length of the spring is b/2 , determine t he necessa ry sp ring stifTness k for equilibrium at a given an gle fl with th e vert ica l. Th e mass of th e lamp and triangu la r fixture is m . Check the stability withi n th e working range from fI = 2 sin " ! ~ es 29° to (J = 180°. mgl 1

Aus. sin II = mg /". > mg 2i1l' 21

A TIS.

= l7" -1-...,...'----,-

~ esc 0/ 2 sta ble wit hin specified range

1

b

I

8

V,

Probl em 7/51

7/51 The figur e shows a tiltin g desk chair toget her with t he design detail of the spr ing-loaded til ti ng mecha nism . Th e fra me of the seat is pivoted about th e fixed point 0 on the base. The incre ase in distance between A and B as th e chair tilt s back about 0 is th e increase in compress ion of th e spri ng. Th e spring, which has a stifTness of 96 kN/ m, is un compressed when H = O. For sma ll angles of ti lt it may be assu med with negligible erro r that the axis of the spri ng remains parallel to the seat. Th e center of mass of an 80. kg perso n who sits in the chair is at G on a line through 0 perpe ndicu lar to t he seat. DeterWNW.gigapedia.com

~\

I111 k 11 I11

~-Hl~ ./


I;

b/

Problem 7/53

Art icle 7 / 4

.. 7/54 Th e fro nt -end sus pe ns ion of P rob . 4/108 is repeated here. In a test of th e designed action, th e fram e F mu st be jacked up so th at h = 350 mm in orde r to relieve the compress ion in th e coil springs. Determine t he va lue of Jz whe n th e jack is re moved. Ea ch spr ing has a stiffness of 120 kN /m . Th e load L is 12 kN , and t he central frame F has a mass of 40 kg. Each wheel and a ttached link has a ma ss of 35 kg with a center of mass 680 mm from th e vert ical cen te rline . A ns. h = 265 mm

Prob lem s

4 19

.. 7/56 The un iform garage door AB shown in section has a mass m and is equipped with two of th e spring-loade d mech anism s shown, on e on each side of t he door . The a rm DB has negligible ma ss, and t he upper corner A of the door is free to move horizontally on a roller. Th e u nstretched len gt h of th e spri ng is r - a , so that in the to p posit ion wit h fJ = 1T th e spr ing force is zero. To en sure smoot h action of the door as it reach es the vertica l closed position 0 = 0, it is desirabl e that the door be insen sitive to movement in thi s position. Deter mine th e spri ng stiffness k req uired for t his design . mg tr + a J An s. k = &z2

A

" -, .... \

\ \ \

, I

Probl em 7 /54

I I

.. 7/55 The portable roller stand for supporting board s ejected from a wood plan er is designed wit h a microfine height adjustment produced by turning th e kn url ed knob of the adjus ting scr ew wit h a torque M . T he single-th rea d screw wit h square threads has a pit ch p (adva nce ment per revolution ) a nd is threaded into th e collar a t B to control th e distance between A a nd B (a nd hen ce C a nd D ). The roller E a nd supporting box ha ve a mass m i. a nd th e four uniform lin ks (two on eac h side) have a combined mass "' 2 a nd a le ngth 2h for eac h. Neglect a ll frict ion a nd find the to rque AI necessa ry to raise t he roller for a given va lue of fJ. 2")p,,,g -(:.:2::.m::..!.1_+;-:n:':'''cot fJ Ail S. AI

4"

E

EEi 0

b A

•

C Partial end view

8 b

-«.

B

• Probl em 7 /55

Marwan and Waseem Al-l raqi

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I /' /' /'

Problem 7 /56

B

420

Chapte r 7

Vir t ual Wo rk

C H APT E R REVIEW In this chapte r we have developed the principle of virtual work. This principle is use ful for determining th e possible equilibriu m configurations of a body or a syst em of inte rconnected bodies where the external forces are known. To app ly the met hod successfully, you must understa nd th e concepts of virtual displacement, degr ees of freedom, and pote ntial energy .

Method of Virtuol Work When various configuration s are possible for a body or a syste m of interconnect ed bodies und er th e act ion of applied forces, we ca n find t he equilibrium position by applying the principle of virtua l work. When usin g this method, keep the following in mind . 1. Th e only forces which need to be cons idered when determining the equilibrium position a re those which do work (active forces) during th e assumed differential movement of t he body or system away from its equilibriu m posit ion.

2. Those exte rnal forces which do no work (reactive forces ) need not be involved. 3. For thi s reason th e acti ve-force diagram of the body or system (rather than th e free-body diagram) is useful to focus attention on only th ose external forces which do work during th e virtual displacemen ts.

Virtual Displacements A virtual displacement is a first-order differential change in a linear or angular position. This change is fictitious in that it is an assumed movement which need not take place in reality. Mat hematically, a virtual displacement is treated th e same as a differential cha nge in an actual movemen t . We use th e symbol S for t he differen tial virtual change and t he usual symbol d for the different ial change in a rea l moveme nt . Relating th e linear an d angular virtual displacements of the parts of a mechanical sys tem during a virtual movement cons istent with the constraints is often t he most difficult part of th e analysis. To do th is, 1. Write th e geometric relationships which describe th e configuration of the system.

2. Esta blish th e differential cha nges in the positions of parts of the syste m by differentiating th e geometric relationship to obtain expressions for the differen tial virtual movement s.

Degrees of Freedom In Chapter 7 we have restricted our atte ntion to mechanical sys tems for which th e positions of the memb ers can be specified by a single vari ab le (single-degree-of-freedom syste ms). For two or more degr ees of freeMa rwa n and W as eem AI- Iraqi

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Chapter Revie w

dom, we wou ld apply the virtual-wo rk equation as many times as th ere ar e degr ees of freedom, allowing one variable to change at a time while holding the remaini ng ones constant.

Potential Energy Method The concept of potential energy, both gravitational (Vg ) and elastic (Ve), is useful in solving equilibrium problems where virtual displacements caus e cha nges in the vertical positions of th e mass cente rs of th e bodies and changes in the len gths of elastic members (springs). To apply this met hod, 1. Obtain an expressio n for the total potential energy V of the system

in te rms of the variable which specifies the possible position of the system. 2. Exa mine the first and second derivat ives of V to establish, respect ively, th e positi on of equilibriu m an d th e corresponding stability condition.

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421

422

Chapter 7

Virtual Work

REVIEW PROBLEMS

Plane of each figure is vertical. Size and mass of each member and applied force are known.

7/57 A control mechanism consist s of a n input shaft at A which is turned by a pplying a couple M and an out put slide r B which moves in the x-direction again st th e action of force P. Th e mechanism is designed so that th e linear movement of B is proportional to the angu lar movement of A, with .r increasing 60 mm for every complete turn of A. If M = 10 N · m, det ermine P for equ ilibrium. Neglect int ernal fricti on and assu me all mechan ical componen ts are ideally connect ed rigid bodies. Ail s. P

~

p

Find B for equilibrium

p

Find reactions at A and B

fal

fb i

1047 N

f - - x--'1 Find r for equilibrium

Find forces at A. B. and C fdl

fe)

Problem 7/57

(n

7/58 Iden ti fy which of the problems (0 ) t hrough are best solved (A ) by the force and momen t equilibr iu m equations an d (!3 J by virt ua l work. Outline br iefly the procedu re for each solution.

f---- x----I Find x for equilibrium. fel

Determine maximum k for stable equilibrium at B = 0

ff!

Problem 7 j 58

7/59 The semicylindr ical shell of radiu s r is pivoted about a shaft through points 0 as shown. Th e mass of th e two support tab s is small compared with the mass of th e shell. Determine the maximum value of Ii for which equilibriu m in th e position shown is sta ble. Ans. h mllJf. = O.363r

Problem 7/59 MarNan and Waseem AI-Iraqi

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Re view P robl em s

4 23

7 /60 Neglect the mass of th e crossed links an d det ermi ne

the a ng le II for th e equilibrium position of th e symmetrical mechanism in the vertical plan e. Ea ch of th e identical rectangu lar blocks of mass m is homogeneous with mass cente r a t G. Evaluate 1/ for equilibr iu m whe n b = a.

Problem 7/62 7/63 Determine th e eq uilibri um values of

(J and th e stability of equ ilibrium a t eac h posit ion for t he un bala nced wheel on th e 10" incline. Stat ic friction is sufficient to preven t slipping. Th e mas s center is at G. Ans. II = - 6.82", sta ble; fI = 207", un stable

Problem 7/60

7/61 Th e sketch shows th e approxima te design configura t ion of one of th e four toggle-act ion hold-down as· semblies whi ch clamp the base flange of th e Saturn V rocket vehicl e to th e pedestal of it s platform prior to lau nching. Calcu late the pre set clamp ing force F at A if th e link CE is under ten sion produ ced by a flu id pressure of 2000 Ib/in .2 act ing on t he left side of t he pist on in the hydra ulic cylinder. T he piston has a net area of 16 in. 2 The weight of th e assembly is considerable, but it is small compa red wit h th e clampi ng force produced a nd is therefore neglected here. A ns. F = 960 ,000 Ib

='7'--

, Saturn V ~_ L __ ..:\_ -----~_..!.

I

Prob le m 7/63 7/64 Two semicylindr ical shells with equ a l pr ojecting rec-

I I

\:;:.4-- - ""'t'!f' B -

--

- T

= 60 mm A

60" - ~

6"

basc tlan ge

,. = 100 mm

r

I

tangles are form ed from sheet metal, one with configurat ion (a) an d th e oth e r wit h configurat ion (bl. Both shells rest on a horizontal sur face. For case (0) det ermine the maximum valu e of h for wh ich th e shell will remain sta ble in th e position shown. For case (b> prove th at sta bility in th e positi on shown is not alTecte d by the dimen sion II.

40" E

c• 2"

IL

40" .L

Problem 7/61 7/61 Th e figure shows the cross section of a contai ner

composed of a hem ispher ical she ll of radiu s r a nd a cylindrical shell of heigh t h, both made from th e sa me ma te rial. Speci fy th e lim itat ion of h for stab ility in t he upri ght posit ion when th e containe r is placed on th e horizontal sur face.


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(al

(bl

Problem 7/64

424

Chapter 7

Vir tual Work

7/65 An exploration device, which unfo lds from the body A of an unmanned space vehicle resting on th e moon 's surface, consist s of a spring-loaded pantograph with detector head B . It is desired to select a spri ng that will limit th e vertical contact force P to 100 N in the position for which 0 = 120°. If th e mass of the arms and head is negligible, specify the neeessary spring st ifTness k. Th e spring is uncompressed when 0 = 30°. A ns. k = 1.664 kN/m

I.. 1/61

The platform of mas s m is support ed by equa l legs and braced by the two springs as shown. If th e masses of the legs and springs are negligible , design th e springs by det ermining th e minimum stiffness k of each spring which will en sure stability of th e plat . form in the position shown. Each spri ng ha s a ten sile preset deflection equal to .l. Ans. k min = mg 2b ( 1

2

b ) + (i

A

;--

- -Problem 7/67

B

b = 300 mm

p

Problem 7/65

7/66 Th e uniform alu min um disk of ra dius R an d mass m rolls without slipping on the fixed circu lar surface of radi us ZR. Fastened to the disk is a lead cylinder also of mass m with its center located a distance b from th e center 0 of th e disk . Determine the minimum value of b for which th e disk will remain in sta ble equilibrium on th e cylindrica l sur face in the top posit ion shown.

... 1/68 In t he mechanism shown the spring of stiffness k is un compressed when 0 = 60°. Also the masses of th e parts are small compared wit h t he su m m of th e masses of th e two cylinders. The mechan ism is constr ucted so that the arms may swing past the vertical, as seen in the right-ha nd side view. Determine th e values of 0 for equi libri um a nd invest igate the stability of the mechanism in eac h position. Neglect friction . Ans. tI = 0, stable if k < mgf a un stable if k > mg/a

l(

g O = cos - I - 1 +m- ) 2 ka only if k > mg /a, sta ble

a

/

a

-u

/

III

~ A

/ ~ Ik

B k

Prablem 7/66

m

m

2

2

Prablem 7/68

Marwa n and W aseem AI-Iraqi

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Re view Probl ems

e

=,

' Computer-Oriented Problems

-7/ 69 Th e bar OA , wh ich weighs 50 Ib with center of gravity at G, is pivoted about its end 0 and swings in th e vertical pla ne under th e constra int of th e 20-lb cou n terweight. Write th e express ion for th e total potential energy of the system, taking Vg = 0 when II = 0, and compute Vg as a func tion of U from II = 0 to (} = 360 0 • Fr om your plot of the resul ts, determine the positio n or positions of equilibrium and the stability of equilibri u m at each position. A ns. 0 = 78.0 0 , stable; 0 = 2600 , u nstable

425

"7/71 Determine th e equi libri um value of th e coordina te x for the mechani sm und er the action of th e 60-N force applied. normal to the ligh t bar. Th e spring has a st ifTness of 1.6 kN /m and is un st retched when x = O. (Hint: Replace t he applied force by a forcecouple system at poin t B.> Ans. X = 130.3 mm

k = 1.6 Iu'l/m

B

i j "-! 3'

.~

Problem 7/71

20 lb Problem 7 / 6 9

- 7/7 0 The toggle mechani sm is use d to lift the 80· kg mass to a locked position whe n OB moves to OB ' in the 30 posit ion. To evaluate the design act ion of the toggle, plot the valu e of P required to oper ate t he toggle as a functi on of H from (J = 20 to (J = _30.

-7 / 71 The u niform link OA ha s a mass of 20 kg and is support ed. in th e vertical pla ne by the spring AB whose unst retched lengt h is 400 mm . Plot the tot al pote ntial ener gy V and its derivative dVl d l1as fun ctions of () from 0 = 0 to (} = 1200. Fr om t he plot s identify t he equilibrium values of (J a nd the corr esponding stability of equilibrium. Tak e Vg = 0 on a level throug h O.

0

k = 550 N /m

A

Problem 7 /72

Problem 7/70


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426

Chapt e r 7

Vir t u a l Wo r k

·7 /73 Deter mine the equ ilibr iu m angle

f} for the mechanism shown. Th e spr ing of stiffness I~ = 12 Ib/ in. has an unst retched length of 8 in. Each of the u nifor m links AB and CD ha s a weight of 10 lb, and member BD wit h its load weighs 100 lb. Motion is in t he vertical plane. A il s. f) = 71.7°

·7/74 Th e unifor m 25-kg trap door is freely h inged along

D

•

T 1

its bottom edge 0 -0 and is att ached to the two spr ings eac h of stiffness k = 800 N/ m. Th e spr ings are u nst retched when (1 = 90°. Take Vg = 0 on th e hor izontal plane through 0 -0 and plot th e pot ential energy V = Vg + V(:' as a fun ction of 0 from (I = 0 to fJ = 90°. Also deter mine the an gle fJ for equ ilibriu m and determine the stability of this position . D

l 00 lb

16"

B

.

-

Iii



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Appendix

AREA MOMENTS OF INERTIA

ApPENDIX OUTLINE

All A/2 A/3 A/4

Introduction Definitions Composite Areas Products of Inertia and Rotation of Axes

All

INTRODUCTION

When forces are distributed continuously over an area on which they act, it is often necessa ry to calculate t he moment of these forces about some axis either in or perpendicular to t he plane of th e area . Frequ ently t he inten sit y of th e force (pressure or stress ) is proportional to the d istance of the line of action of t he force from the moment axis. The element al for ce act ing on an eleme nt of area, then, is proport ional to distan ce tim es differential area, a nd t he eleme ntal momen t is proport iona l to distan ce squa red times differential area. \Ve see , t he refo re, t hat t he to ta l moment involves a n int egr al of form J (dista nce)2 d (area ). This int egral is called th e moment of inertia or th e second moment of t he a rea . Th e integral is a fun cti on of the geo metry of th e a rea a nd occurs frequ ent ly in the applicati ons of mechan ics. Thu s it is usefu l to develop its proper ties in some detail and to have t hese properties ava ilabl e for ready use when t he integral ar ises. Figure Al l illustrates the physical origin of these int egrals. In part a of th e figur e, t he surface area ABe D is subjec ted to a distributed pr essure p whose inten sity is prop ortional to th e dist an ce y from th e axis AB. Thi s situation was treated in Art. 5/9 of Chapter 5, whe re we descri bed t he actio n of liqu id pressure on a plan e surface . T he moment abou t AB du e to the pressure on the element of area dA is py dA = ky 2 dA. Th us, the integral in question appears when the total moment M = k J y 2 dA is evaluated. Marwan and W aseem AI-Iraqi

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427

428

Appendi x A

Are a Moments of Inertia

A

In Fig. A/ lb we show the distribution of stress acting on a t ransverse section of a simple elastic beam bent by equal and opposite couples applied to its ends. At any section of the beam, a linear distr ibution of force intensity or stress CT, give n by a = ky , is present. The stress is positive (tensile) below the axis 0 - 0 and negative (compressive) above th e axis. We see that the eleme ntal moment about t he axis 0-0 is dM = y (tr dA ) = hy 2 dA. Thus, the same integral appears when the tot a l momen t M = II Jy2 dA is evaluated. A third exa mple is given in Fig. A/Ie, which s hows a cir cular shaft subjected to a twist or torsional moment. With in the elast ic limit of the material, thi s moment is resisted at each cross section of the shaft by a distribution of tangential or shear st ress T, which is proportional to the radial distance r from the center. Thus , T = kr, and the total moment abou t the central axis is M = J r ( T dA ) = h J r 2 dA . Here t he integral differs from tha t in the preceding two examples in that the area is normal instead of parallel to t he momen t axis and in t hat r is a radial coordinate instead of a rectangular one. Although the integral illustrated in the pr ecedi ng examples is generally called the moment of inertia of t he area about the axis in question, a more fitt ing term is the second moment ofarea , since the first moment y dA is mult iplied by t he moment ar m y to obtain t he second moment for th e element dA . Th e word inertia appears in the ter minology by reason of the similar ity bet ween the mathema tical form of th e integra ls for second moments of areas and those for the resultant moments of the so-called inertia forces in the case of rotating bodies. The moment of inertia of an area is a purely mathematical property of the area and in itself has no physica l sign ificance.

A /2

DEFINITIONS

Th e following definit ions form t he basis for the a nalys is of area moments of inertia.

Rectangular and Polar Moments of Inertia ,

(c)

Consider th e area A in the x -y plane, Fig. A/2. The moments of inert ia of th e element dA about t he x- and y-axes a re, by definition, -u, = y" dA and dl y = x 2 dA , respectively. Th e momen ts of inertia of A about the same axes are therefore

Figure All

y I I 1

A

~f-- x

- ' f . ] dA

I 1

o

1/

r

(A/l)

/1

y

l

x

Figure A/2 Marwa n and Waseem AI- Iraqi

where we carry out the integration over the entire area. WNW.gigapedia.com

Article A/2

Th e moment of inertia of dA about the pole a (a-axis) is, by similar definition, dl z = 1''2 dA . Th e moment of inertia of th e entire area about a is (A/ 2)

Th e expressions defined by Eqs . A/I are called rectangular moments of inertia, whereas the expression of Eq . A/2 is called th e polar moment of inert ia: Becau se x 2 + y2 = ,-2, it is clear tha t (A/ 3l For an a rea whose boundaries are more simply described in rectangular coordina tes th an in polar coordinates, its polar moment of inertia is eas ily calculat ed with th e a id of Eq . A/3. Th e moment of inertia of an elem ent involves t he squa re of the dist anc e from the in ertia axis to th e eleme nt. Thus an eleme nt whose coordinat e is negative contributes as muc h to the momen t of inertia as does a n equa l eleme nt with a positive coordinate of the same magn itu de. Conseq uently th e area moment of inertia about any axis is always a positive qua ntity. In con trast , the first moment of t he area, whi ch was involved in t he comp ut atio ns of centroids, could be either positive, negative, or zero. Th e dimensions of momen ts of inertia of a reas a re clea rly L4 , whe re L sta nds for the dimension of lengt h . Thus, th e SI uni ts for a rea moments of inertia are exp res sed as qu artic met er s (m'') or qu artic rnillimeters (m rn"). The U.S. customary units for area mome nts of iner tia a re quartic feet (ft4 ) or quartic inches (in." ). T he choice of th e coord inates to use for th e calcu lation of moments of inertia is important. Rectangular coordinates should be used for shapes whose bounda ries are most easily expressed in these coordinates. Polar coordina tes will usually simplify problems involving boundaries which are eas ily described in I' and H. Th e choice of an eleme nt of area which simplifies th e int egration as much as possibl e is also impo rt ant. Th ese consideratio ns are quite a nalogous to tho se we discussed and illustrat ed in Cha pte r 5 for the calculat ion of cent roids.

Radius of Gyration Consider a n area A , Fig. A/ 3a , which has rectangular moments of inertia Ix and I and a polar moment of inertia Iz about O. \Ve now " as concentrated into a long narrow st rip of area A a visua lize thi s area dist an ce !Ix from th e .r-axis, Fig. A/ 3b. By definition th e moment of iner tia of th e st rip about th e r -axis will be th e sa me as that of the original area if !I/ A = Ix. The distance !Ix is called th e radiu s af gyration of the "T he pola r mome nt of inerti a of an area is so met imes den oted in mecha nics literatu re by the sy mbol.' .


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Definitions

429

430

App endi x A

Ar e a M om en t s of I n er ti a

y

.Y

1

1

l-r1

y I

Il l '

A

---"-j

.

1

A

A

1

- - ;I:

1

o

-- x

L __

(a )

--x

(ell

(c)

Figure A/3

a rea about the a-axis. A simila r relation for th e y-ax is is wr itten by consider ing the area as concen trated int o a narrow st rip parall el to the y-axis as shown in Fig. A/3c. Also, if we visua lize t he area as concentrat ed into a narrow r ing of ra dius I~z as shown in Fig. A / 3d , we may express the polar moment of iner tia as h z2A = lz. In sum ma ry we wr ite

t,

=

k/A

Iy

=

ky 2A

k x = J Ix/A

or

k y = ) Iy/A

I , = k;A

k, =

(A /4 )

jIJA

Th e radius of gyrat ion, then, is a measure of t he distribution of t he area from the axis in questi on. A rectangul ar or polar momen t of inertia may be expressed by specifying t he ra dius of gyra t ion and the a rea. When we substitute Eq s, A/ 4 into Eq . A/3, we have

(k;

y I f-o-

)'u

- d,. -~If.- Xo

1

1 1

A

1

I dA OI ~~ I Yo

Cl

1

_

- - xu

=

kx 2

+

k/ )

(A/5)

Thus, th e square of t he radius of gyration about a polar axis equals the sum of the squ ares of the radii of gyration abo ut t he two correspond ing rectan gu lar axes. Do not con fuse the coordinate to t he centroid C of a n area with the radius of gyration. In Fig. A / 3a the square of th e centro ida l dist ance from t he .r-axis, for exa mple, is y 2 , which is the squa re of the mean value of th e dist an ces from t he element s of the a rea to th e x-axis. T he qu an ti ty hx 2 , on t he other hand, is the mean of t he squa res of t hese distan ces. Th e moment of iner tia is not equal to Ay2 , since t he square of t he mean is less than th e mean of the squa res.

1

:

1/

(1~

d/

ok - - - - - - - - - - - - -

Transfer of Axes - - ;I:

Figure A/4

Marwan and Waseem Al-lraqi

The mome nt of ine rtia of a n area about a no ncentroidal axis may be eas ily express ed in te rms of th e moment of inertia about a parallel cent roida l axis. In Fig. A/4 the Xo-Yo axes pass t hrough the centroid C of t he a rea. Let us now determin e t he mom ents of iner tia of the area www.gigapedia.com

Article A /2

about th e parallel x-y axes. By definiti on, th e moment of in er tia of th e element dA about t he x-axis is

Expanding and integrating give us

t,

=

f

y02

dA + 2d x

f

Yo dA + d/

f

dA

We see that th e first integral is by definition the mom ent of inertia Ix about the centroidal xo-axis. The second int egral is zero, since J Yo dA ~ Ayo and Yo is au tom atically zero with the cen troid on the xoaxis. Th e third term is simply Ad/ . Thus, t he expression for Ix a nd the similar express ion for I y become I; = I, Iy

_ - Iy

+ Adx 2 + Ady

(A / 6 )

2

By Eq . A / 3 th e sum of these two equations gives (A/ 6a)

Eq uations A/ B and A/Ba are the so-called parallel-axis theorems . Two points in particular should be noted. Fir st , the axes betw een which th e transfer is made must be pa rall el , and second, one of the axes must pass through the centroid of th e area . If a tra nsfer is desired between two parallel axes neither of which passes through the centroid, it is first necessary to transfer from one axis to the parallel cent roida l axis and the n to transfer from t he centroidal axis to the second axis. The parallel-axis theo re ms also hold for rad ii of gyrat ion . With substitution of the definition of " in to Eqs. A/B , th e transfer relation becomes (A / 6b)

whe re Ii is the rad ius of gyrat ion about a cent roida l axis parallel to th e axis about which" applies and d is th e dist anc e betw een the two axes. Th e axe s may be eit her in th e plan e or norm al to t he plan e of the area. A summary of the moment -of-inertia relations for some common plan e figures is given in Table D/ 3, Appendix D.


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Defi ni t i on s

431

432

Appendi x A

Area Moments of Inertia

Sample Problem All

,,

y

Determine the moments of inertia of the rectangu lar area about the centroidal xo- and Yo-axes, th e centroidal polar axis Zo th rough C, th e x-axis, and t he polar axis z t hrough O.

Th

Yo ,

, ,

,,

2

Solution . For t he calcu lation of the moment of inertia Ix about the xo-axis, a

Q) hor izontal stri p of area b dy is chosen so that all elements of th e strip have th e sa me y-coordina te . Thus,

+

Ans. Th e centroidal polar moment of ine rtia is

An s. By the parallel-axis theorem the moment of inertia about the z -axis is I, = f-"bh 3 + bh

1, + Ad/l

J

c,

_L

- - xo

2

J.- !",o_-,--_~ __ x

By int erc hange of symbo ls, t he mome nt of inertia abou t th e centroida l Yo-axis is

~

I

h

r - b ------1

An s.

II,

dy

(~)

2

= l bh 3 = ¥h2

Ans.

Helpful Hin t

Q) If we had started with the second or der element dA = dx dy, integration with respect to x holding y constant amounts simply to multi plication by b and gives us the expression y 'J.b dy, wh ich we chose at t he outset.

We also obtain the polar moment of inertia about 0 by t he parallel-axis theorem, which gives us [I, =

I , = f-,A(b 2

1, + Ad 2 ]

+ h 2) + A [ ( b2 ) 2 + ( _h 2] 2)

I, = lA(b 2 + h 2 )

Ans.

Sample Problem A ll Determine the moments of inertia of the triangular area about its ba se and abou t parallel axes through its centroid and vertex.

CD

Solution. A strip of area parallel to the base is selected as shown in th e figure,

(1) and it has the area dA

II,

~

fy 2dAJ

I = ,

~

x dy

~

fh0 y 2 h

[(h - y )b/ hJ dy . By defin ition - Y b dy = b h

[i _y4]h 4h

3

0

= bh

3

12

Ans. Helpful Hin ts

By th e parallel-axis theor em th e momen t of iner tia 1 about an axis through t he

[1

~I

- A d 2j

1=

b~3

An s.

An s.

@ Expressing x in terms of y should

2 _ (bnm

= b;:

A tran sfer from the centroidal axis to the x ' -axis thro ugh the vertex gives [I ~

1 + Ad2 J

3 3 I . = bh + ( bh ) ( 2h)2 = bh x 36 2 3 4

Marwa n and Waseem AI-Iraqi

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CD Here

agai n we choose the simp lest possib le clement. If we had chose n dA = dx dy, we would have to integrate J ''J. dx ely with respect to .r first. This gives usy~x dy . which is the expression we chose at the outset.

centroid, a distance hl3 above the x-ax ie, is

cause no difficulty if we observe the proportional relationship between the similar triangles.

Articl e A/ 2

Sample Problem A/3

Defin itio ns

433

y I

Calcu late th e moments of inertia of the area of a circle abou t a diametral axis and about the polar axis through the center. Specify the radii of gyr ation.

CD

Solution. A differential element of area in th e form of a circular rin g may be used for t he calculation of the moment of inertia about th e polar a-axis thro ugh since al l elements of the rin g are equ idistant from O. T he elementa l a rea is dA = 21Tro dro, and thu s,

o

[I ,

~

Jr

2

dA )

I,

~

r r0

J

2(

o

2rrro dro )

~

4

";

~

0 ,-2

Ans.

Th e polar radiu s of gyration is

= .!:-

k By sym met ry i , [lz

= t, +

Iyl

~

Ans.

,/2

Z

I y • so that from Eq . A/3 rrr 4 2 I x ~ 2'"z lr = _ 4 = l4 A r

Ans.

Th e radiu s of gyrat ion about th e diam etral axis is

k

x

= !:.2

Ans.

T he foregoin g det ermination of Ix is the simplest possible . T he resu lt may also be obta ined by direct integr ation, usin g the eleme nt of are a dA = "o dro d fJ shown in th e lower figur e. By defin ition

Helpfu l H ints

CD Polar coordinates are certainly indi cated her e. Also, as be fore, we choose th e simplest and lowest-order element possible, which is th e differ ential rin g. It should be evident immediately from the definition t ha t the polar moment of inertia of th e r ing is its area 27Tro dro times r02 .

@ Thi s int egration is st ra ightfor wa rd, Ans.


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but th e use of Eq . A/3 along with th e resul t for I z is certainly simpler.

434

App endix A

Ar ea Moments of Inerti a

Sample Problem A /4

,

)'

3 ,L - - - - - - -

Det ermine th e moment of inertia of the area und er th e parabola about th e x· axis. Solve b.... using (a ) a horizontal strip of area and (b ) a verti cal strip of area.

Solution. Th e constant k = ~ is obtained firs t by substituting x )' = 3 into th e equation for the parabola.

4

~ Jy2 dAl

t,

=

f

I

an d

O ~--------' -

o

(a) Horizontal strip. Since all parts of the ho rizontal stri p are the same distance from the .r-axis, th e moment of ine rtia of th e st rip about t he .r-axis is)'2 dA where dA ~ (4 - x ) dy ~ 4(1 - y2/9 ) dy . Integrating with respect to y gives us

IIx

, , ,I

4

,

)'

f- x

-)1':::::= :::::;:=1

52

Solution (0)

y L-

4y 2 ( 1 - y:) dy = 7 = 14.40 (u nits)'

-- x

-'-___'

X

A ns .

,

)'

(b) Vertical strip. Her e all parts of th e element are at different distances from th e .r-axis, so we mu st use th e correct expression for th e moment of ine rti a of th e elementa l rectangle about its base, which, from Sample Problem AyLvis bIl 3 / 3. For the width dx and th e height y the express ion becomes

To int ewate with respect to .r, we must express y in terms of x, which gives y = 3Jx / 2, and the integral becomes

Ix ~ 1 i'

CD

3

0

2

~

72 -5

~

Solution (bl

, I

y

I

1-- x- -++,---'- - - x .... six ~

dl , ~ ~(dr)y3

(3 Jx)3 dx

I I

. , 14 .40 (u nits)

Ans.

Sample Problem A I5

Helpful Hint

CD Ther e is litt le prefer ence bet ween Soluticns (a ) and (h) . Solut ion (b ) requi res knowing the moment of inertia for a rectangular area about its base.

6~-l\---x:,

Find t he moment of inertia about the x-axis of th e semicircular area.

- -- x

I

Solution. Th e moment of inertia of the semicircu lar area about th e x' -axis is one- half of t hat for a complete circle abou t th e sa me axis. Thus, from the results of Sa mple P roblem A/3 2 4

CD

15 0101

l

----- ---- -~----- - x

8

We obtai n t he moment of ine rtia I about th e parall el cen troidal axis Xo next . Transfer is made th rough the distance r ~ 4r/ (3,,) ~ (4)(20)/( 3 ,,) ~ 80/ (3 ,,) mm by th e parallel-axis theorem. Hen ce,

Helpful Hint

[1 = I - Ad2)

CD This problem

Finally, we transfer from th e centroi dal xo-axis to the x-axis. Thus, [l

~ 1 + Ad 2

J

I,

~ 1.75500' ) + ( 20;,,)( 15 + :~)2 = 1.75500' ) + 34.7(10' ) ~ 36.4(10') mOl'

Ma rwan and W aseem AI-Iraqi

W'HW .gigap edia.com

An s.

illustrates the caution we should observe in using a double transfer of axes since neither the x ' nor the .r-axis passes through the centroid C of the area. If the circle were complete with the centroid on the x ' -axis. only one transfer would be needed.

Article A / 2

Defini tion s

4 35

Sample Problem A/6 Calculate the mome nt of inertia about t he x-axis of the area enclosed between t he y-ax is and t he circu lar arc s of ra dius a whose centers are at 0 and A .

Solution. The choice of a vertica l differential str ip of area permits one integr a-

a

t ion to cover th e entire area . A hori zontal str ip would require two integrat ions wit h respect to y by virtue of the discontinuity. The moment of inertia of the strip about th e x-axis is t hat of a strip of height Y2 min us that of a str ip of heig ht )' 1' Thu s, from t he results of Sample P roblem A/ I we write d lx

CD

= ~( )'2 dX)Yl - ! (Yt dX)Yt 2 = ~(y~?

o

- Y 13) dx

Th e values of Y2 and Y l are obtained from the equ ations of the two curves, which are x 2 + yl = a 2 and (x - a )2 + YI 2 = a 2 , and which give Y 2 J a 2 - x 2 and Yl = J a 2 - (x - a )2. Thus, Ix

l

=:\ o

{ (a 2 - x 2) Ja 2 - x 2 - [a 2 - (x - aj2 ] J a 2 - (x - a )2}

dx

ra dicals here since both Yl and)'2 lie above the x-axis.

Simultan eou s solution of th e two equ ations which define the two circles gives the .r-coordinate of th e intersection of the two curves, which, by inspect ion, is a/ 2. Evaluation of the integrals gives

_( t2a2/ a2 [ 12

(r

_

a )2J a 2

Collect ion of t he integrals with the factor of ~ gives a4 Ix ~ 96 (9}3 - 2,,) ~ 0.09690'

An s.

If we had sta rted from a second-or der element dA = dx dy , we wou ld write y2 dx dy for the mom en t of iner tia of th e element about th e .r-axis. Integr atin g from Y I to)'2 holding x constant produces for the ver tical strip

whic h is th e expression we st arted with by having the moment-of-inertia result for a rectan gle in mind.


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Helpful Hint

CD We choose the posit ive signs for t he

ar2

436

Appendi x A

Area Moment s of In ertia

PROBLEMS

A/4 Th e t hin qu arter-c ircular ring has a n area of 1600 mm 2 • Determine th e moment of ine rt ia of the ri ng abou t the .r-axis to a close a pproximation.

Introductory Problems

j'

All If the mome nt of inertia of th e t hin strip of area about th e x-axis is 2.56(06 ) mm''. determine th e a rea A of th e stri p to wit hin a close a pprox ima tio n. Ails. A = 1600 mm 2

I I

- - x

o Problem A /4 Problem A/ I

All Determine by direct integratio n th e momen t of inerti a of the triangu lar area about t he y-axis.

A/5 Det erm ine the polar moments of inert ia of the sem icircula r area about points A and B. 3 , • III An s. IA = 4"rrr

Problem A /2

>_-

-

- ;::"--

-

r' (34" - :!3 )

- -'- - - x

A

All The moments of inerti a of the area A about the parallel p- a nd p '·axes differ by 50 in." Compute th e a rea A, which has its centroid a t C. Ans. A = 10 in.2

Problem A j5

A/6 Calculate th e mom ent of inert ia of th e sha ded area about th e y -axis.

p p'

c.

-: A

Problem A/6

Prob lem A/3

Marwa n and Waseem AI-Iraqi

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Art icl e A / 2

A/7 Show that the moment of inertia of the rectan gular area a bout t he x-ax is t hrough one end may be used for its polar momen t of inertia about point 0 when b is small compared wit h a . What is the percen ta ge er ror n wh en b [a ~ 1/1O? Ans. n = - 0.249% 1

o!- - ----''------- - Ib a

1

Problems

4~ 7

Representative Problems A/lO T he area of th e narrow st rip of u niform widt h is 750 mm 2 . Using t he fact th at th e widt h is small compared wit h t he len gth of the st rip, approximate its moment of inertia about t he x-axi s. Compare your answer wit h t he erroneous result uf multiplying the ar ea by t he square of its distance from its centroid to th e x-axis.

-1-

1

1 .r

\00 ",,,,

- T-

Problem A/7

l OO m m

~r:

A/8 Deter mine the moments of iner tia Ix and l y of th e ar ea of the t hin semicircular ring about t he x- and y-axes. Also find t he polar moment of inertia I e of th e r ing about its centroid C.

l_x Problem All 0

y

Alll Determine the polar radii of gyration of th e t riangu lar area about points 0 a nd A. o

1 1

A ns. k o

1

~ t«

~ 0,

kA

~

jij

1

Ct

""'r r""" I:

A

____ ___ .'::>L

_

°

0

I I I I I I

---x o

Problem AlB

I 1

A/9 Ca lculate th e moment of inertia of th e rectan gu lar area about t he x-axis and find t he polar momen t of iner tia abo ut point O. A ns. Ix = 7.2(10 6 ) mm", 10 = 15.95(0 6) mm" y 1

o

1

25mm

60mm

1

--- - ----- . J O Problem AliI

Alll Determine the radius of gyration about a polar axis thro ugh the midpoin t A of the hypotenuse of the righ t-t rian gular area . (H int: Simplify your calculation by observi ng the results for a 30 x 40- mm rectangu lar area.)

lOOm m

A

Problem A/9

40 mm

30 mm Problem A/12


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438

App en d i x A

Ar ea Mo ments of Inert ia

A/1J Determine th e rectangu la r and polar ra dii of gyra-

A/16 Th e moment of inertia about the .r-axis of th e rec-

tion of the shaded area abo ut the axes shown. Ans. hx ~ 0.754, h.y ~ 1.673, h, ~ 1.835

ta ngle of area A is approxi mate ly equal to Ad'l. if II is small compa red with d. Deter min e an d plot th e percentage error n of th e ap prox ima te value for hId ratios fro m 0.1 to 1. What is the percent age er ror for h ~ d /4?

.v I I

p~l- -

b

- xu

d

1

;-

o --~'::- L-_---'--_ x o 2

--1-_

Problem A/16

Problem A/ll A/14 In two different ways show that t he mome nts of iner tia of th e square area about the x- and x ' -axes are

A/17 Calcu late the moment of inertia of th e shaded area about the .r-axis .

.v

the same.

;-

;-

-::...---

a I I

afl

I I

I 30 nun

I I

«n

I I

---- -40mm ---- -- ------ x Proble m A/17

Problem A/14 A/15 Determine the polar rad ius of gyration of the area of th e equilateral triangle about the midpoi nt M of its

base. An s. k,\1

=

A/18 Determin e t he momen t of inert ia of the quartercircular area about th e tan gen t .r ' -ax is.

a

./6

.v I

I

- - - - - -x·

a ""-- - - - - ' - - - - x

Problem All 8 afl

a fl

Problem A/15

MarNan and W aseem AI-Iraqi

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Ar ti cle A /2 A/19 Determine the momen t of inertia of t he shaded area ab out t he x-axis usin g (a ) a hor izontal strip of area an d (b ) a vert ical strip of area . 16ab3 An s. Ix = 105

Pr o bl ems

439

A/l1 Calculate by direct integration th e momen t of iner t ia of the sha ded area about t he x-axis . Solve, first , by usin g a hor izontal str ip having differen tia l area and, second, by usin g a vert ical st rip of differential area. y

.v

I I

4"

x2

y = b (1 - a 2

)

4"

b

--------- x l-

-x

-----'

Problem A/22

a Problem A/19

A/20 Determine th e moment of iner tia abou t th e .r-axis

and the polar rad ius of gyration about point 0 for the semicircular area shown. y I I I I

All1 The plane figu re is symmet rical with respect to the 45° line a nd ha s an ar ea of 1600 mm 2. It s polar moment of inertia about its cent roid C is 40( 04 ) rnm''. Compute (a ) the polar rad ius of gyr at ion ab out 0 and (b ) the radiu s of gyration ab out the xo·axis. Ans. (a) I~ o = 45 .3 m m, (b ) I~ xo = 11.18 mm y

r

I

I

Yo

I

I I

I

1---- 30 mm -r--l I I I

I I

I I

AI- - - - -

i

B

I I I

L

_

o

r

I I

1, I

A/11 Determine the mom ents of inertia of the shaded area about the x- and y-axes. Use the same differe nt ial eleme nt for both calculati ons. An s. t, = a 4 /28 , ly = a4 / 20

-- - x

I I

/

I

,,

I

I 0'

x2

a

8 -8 --------

R .r

a

Problem A/24

Problem A/2! Marwan and Waseem AI-Iraqi

1

y

1/

, : (1 ,, ,, _________ L

/

A/14 Deter mine the momen ts of iner tia of th e circu lar sector about the x- an d y-axes .

,

;-.v = ,

30mm

Problem A/l3

I /

,I

'

o~~~~ - - - - - - -

Problem A/20

.v

/

,?'-----r 'o

i

- - - - --x

./

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- -x

440

App endi x A

Ar e a Momen ts o f Inertia

AilS Determ ine t he rad ius of gyra tion about the y-axis of the shaded area shown. Ans . k)' = 53.1 mm

A/lS Calculate t he momen ts of ine rtia of th e shaded area about th e x- and y-axe s, and find the polar moment of ine rtia about point O.

y

y I

I I I

1

- _ _

'!::

I I I

I I I

_

,

I I I I I I I I I

,,, I I

' 4"

I I I I

y =k,r' , I

0' ------ -- - - Problem A/2S

y

I I I

r

I I I I I I II

A/l9 Determine th e rectan gular moments of ine rtia of the sha ded area about th e .r- and y-axes a nd the pola r radius of gyration abou t point O. b b3 Ans. Ix = ~h 4 - a4 ) , Iy ~ - -3(h 4 - a 4 ) 4h 4810 ko

-' -0

r- ~

Problem A/26

All1 Determine t he moment of inertia of t he shaded area about the y -ax is. y. mm I I

40 r - - - - - - - - - - - - -· I ' I 'f.~ : I 1I I '$: I ' I : I :

o o

JH

1 +

1~2)(h2 + a

: I

2

)

~ --j

iI

1 h

------~~l~~~ l__, o

Problem A /29

I

- - - - .J.. - · x, mm 20 40 Problem A /27


~

y

- - - x

0

x

Problem A/2S

A/l6 From considerations of symmetry show tha t Ix' l.v· == Ix = 1)' for the semicircular area regardless of the angle 0' .

y;

I _---l _ _

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Art icle A /2 A/10 By the methods of this art icle, det ermine the rectan-

gula r and polar radii of gyra tion of t he shaded area abou t the axes shown. .l' 1 1

a

L -_ _

~

..

:

~ ~I+

L -_ _

~

__ x

--- x

Problem A/30

.... A/ll Calculate the moment of inertia of the shaded area of th e two over lapping circles about the .r-axis. Ans. Ix = 0.19881'4 .l' 1

,,_--- __ 1,-----_' /

/""

r

:

I

---:---- --+-\

1

\

I I

"

,

,

/"\

\

- - --:--- x I

/ /

"

',,-----~I-- -----_/ 1

Problem Al31


441

.... Ajll A narrow st rip of area of constant width b ha s the form ora spiral r = k O. Afte r one comp lete tu rn fro m o = 0 to 0 = 27T, t he end radius to the spira l is R . Determine t he polar moment of ine rt ia and the ra dius of gyration of th e area about O. A ns. / 0 ~ 1.609R 3b, li o ~ O.690R

o

/

Problem,

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\ Problem A/ 32

442

Append i x A

Are a Moment s of I ne rt ia

A/3

COMPOSITE AREAS

It is frequently necessary to calculat e the moment of inertia of an a rea composed of a number of distinct parts of simple and calculable geometric shape. Because a moment of inertia is the integral or sum of th e products of dist an ce squared times eleme nt of a rea, it follows t ha t the moment of inertia of a positi ve area is always a positi ve quantity. The moment of inertia of a composite area about a particular axis is therefore simply the sum of the moment s of inertia of its component parts about the same axis . It is often convenient to regard a composite area as being composed of positive and negative parts. \Ve may then treat the moment of inertia of a negative area as a negative quantity. \Vhen a composite area is composed of a large number of parts, it is conveni ent to tabulate the results for each of the parts in terms of its area A, its centroidal moment of inertia 1, the distance d from its cen troidal axis to the axis about which the moment of inerti a of the entire section is being computed, a nd the product Ad 2 . For an y one of th e parts t he moment of iner tia about the desired axis by th e tran sfer -of-axis theorem is 1 + Ad 2 . Thus, for the entire section the desired moment of inertia becomes I = ~1 + ~Ad2. For such an area in the x-y plane, for example. and wit h the notation of Fig. A/4 , wher e Ix is the same as Ixo an d I,. is the sa me as I,. 0 the ta bu lat ion would include

Part

Sum s

Area, A

dx

dy

~A

Ad}

Ad, 2

Ix

I,

~Adx 2

~Ady 2

~I

~I .

From the sums of the four columns, t hen , the moments of inert ia for th e composit e area about the x - a nd y-axes become

t,

si,

I.v

~1J' + ~AdJ' 2

+ 2.Adx 2

Although we may add th e mom ents of in ertia of th e individua l parts of a composite area about a given axis, we may not add their radii of gyration. Th e radius of gyration for t he composit e area about th e axis in questi on is given by k = , II/ A , where I is the tot al moment of iner tia a nd A is th e tot al area of the composite figur e. Similarly, th e radiu s of gyration k about a polar axis through some point equals Hz/A, wher e I , = Ix + I , for x -y axes through that point.


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Article A I]

Composite Ar eas

443

Sample Problem A /7 Calcu late th e moment of ine rtia and radius of gyration about the s -axis for th e sha ded area shown . 30mm

~- 40 mm

Solution. Th e composite area is composed of th e positive a rea of the rectangle and the negative areas of th e qu arter circle (2) and triangle (3). For the rectangle the momen t of inertia about the .r-axis, from Sample P roblem All (or Table D/ 3), is

(1 )

1% = 0h 2 = ! (80)(60)(60J" = 5.76(106 ) mm'

I

X'~0f'J.[J LV

Xo

From Sa mple P roblem A/3 (or Table D/ 3), th e moment of inertia of t he negative quarter-circu lar area about its base axis x ' is

III ---

I . = - -I (""') = - -16 u-) " mm' ". (30), = - 0 1590Il " % 4 4 . We now tra nsfe r thi s result through the distance F = 4r / (37T) = 4(30)/
1

r- - - -

I ,. -

-

-

---1

1

~ lmin

1 1 1 1 1

>--- -

I

1 1 1 1

r-

I,.

D

I.' I 1- - 1.',. - -, ll max - - - - - -.J Figure A/7

and Ixv the corresponding values of Ix·, I,.., and Ix'v· may be determin ed from th e diagram for a ny desired a ngle rioA hori zontal axis for th e measurement of moments of inertia and a vertical axis for the measurement of prod ucts of inertia are first selected, Fig. A/7. Next, poin t A , whic h has th e coordina tes (1" Ixv )' a nd point H, which has the coordinates (1, ,, - Ix,Y )' are located. . We now draw a circle with these two points as the extremities of a diameter. The angle from th e radius OA to t he horizontal axis is 2a or twice the angle from the x-axis of the area in quest ion to the axis of maximum moment of inertia. Th e angle on th e diagram and th e a ngle on the area are both measured in the same sense as shown. The coordinates of a ny point Care (1x" Ix'y ' )' and those of the corresponding point D are (1,Y., - Ixy l. Also the a ngle between OA a nd OC is 28 or twice t he angle from the .r-axis to the x' -ax is. Again we measure both angles in the same sense as shown. \Ve may verify from the trigonometry of the circle that Eqs. A/9, A/90, and A/l0 agree with th e statements made.

Ma rwan and W aseem AI- Iraqi

WNW.gigapedia.com

Article A/4

Products of Inertia and Rotation of Axes

453

Sample Problem A/B Det ermine th e product of inerti a of the rectangu lar area with centr oid at C with respect to th e x -y axes parallel to its sides.

Solution. Since the product of inertia IX)' about th e axes :Co-Yo is zero by symmetry, th e t rans fer -of-axis theor em gives us II", = / ", + d.,d,AJ

An s.

In this exam ple both d x and dy ar e shown positive. We mu st be carefu l to be consiste nt with th e positive direct ions of d, and dy as defined . so that th eir proper signs are observed .

Sample Problem A/9

Y I

Determine the product of inertia about the x-y axes for the area under t he parabola.

br - - - - - - - - - I

I I I

dx x --!jJdy

r--o

Solution. With t he substitu tion of x = a wheny = b, the equat ion of t he curve becomes x = ay2/ b2.

=

dx dy , we have

xy dx dy . Th e int egral over the enti re are a is

Ix, =

fbfa, o

xy dx dy =

ay 'ljb 2

(a 2_ab2~4) y dy Jo

02

0

20

:c

6a

b

I

I

:cal

0

Helpful Hin t

CD If

Yo

- - x

I I I

r" -21

sa ve one int egration by using t he resul ts of Sam ple Problem A/ S. Ta king a vertical st rip dA = y dx gives dIn = 0 + (h )(x )(y dxs, where the distan ces to the centroidal axes of the elemental rectangle ar e d, = y / 2 and dy = .r. Now we have o 2 2 Ixy = dx = [:Cb dx = b = ~a2b2 Ans .

[i:c

y I I

-

I I

Solution II. Alternatively we can st art wit h a first-order elementa l strip and

CD

y

I

'---------'--~ a-

Solution I. If we sta rt wit h the second-order element dA d Ixy

I

,'---- - - ----'--'---'--!.a- - - - x dx Y

a +x

Yo

1-- - - - - . ; I 2

r ---j/:::::~i:=::~~ x

I I I

we had chosen a horizontal strip. our expression would have become + x J[(a - xl dyJ. which when integrated. of course, gives us the same result as before .

dIn =

.dfa

Sample ProblemA /l 0

Y I

Determine t he product of inertia of th e semicircular area with respect to th e .r-y axes.

CD

Solution. We use the tran sfer-of-axis theor em, Eq . A/ S, to write

r =- -2 ( 311'4r") (m-') 2 3

:

Yo

I

I

4r

0-9:::· ( -

4

I

xy

= 0

+ --

(r)

-

Ans.

whe re the .r- and y-coordinates of the cent roid Care d y = + r and d x = - 4r/

2. Logarith ms

bX = y , x = log, Y

cos u = q / (p / p ), 0

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