Engineering fundamentals : an introduction to engineering

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Third Edition

Engineering Fundarnerntals An Introduction to Engineering Saeed Moaveni M i n nesota State U n iversity,

Mankato THOTVtSON

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Tt{rotvtsoNEngineering Fundamentals: An Introdudion to Engineering, Third Edition by Saeed Moaveni

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Preface

CHANGES IN THE THIRD EDITION The third edition, consisting of 20 chapters, indudes a number of new additions that were incorporated in response to sugestions and requests made by professors and students using the second edition of the book The major changes include:

. A new section on Engineering Technology . Additional Ethics Case Studies . Additiond secdons on M,ffif-AB " " " " "

Additional Professional Profiles Additiond Impromptu Designs Additional Engineering Marvels Case Studies Additional problems A new website to offer additional information for instructors and students including PowerPoint slides for each chapter

ORGANIZATIOl'I This book is organized into six parts andZ} chaprcrs; Each chapter begrns by stating its objectives and concludes by summarizing what the reader should have gained from studying that chapter. I have included enough m*erial for two semeter-long courses. The reason for this approach is to give tfre instructor sufficient materials and the fexibiliry to choose specific topics to meet his or her nee&. Relevant, weryday examples with which students can associate easily are provided in each chapter. Many of the problems at the conclusion of each chapter are hands-on, requiring the student to gather and analyze information. Moreover, information collection and proper utilization of that information are encour€ed in this book by asking studene to do a number of assignments that require information gathering by using the Internet as well as employing uaditional methods. Many of the problems at the end of each chapter require studena to make briefrepoms so that they learn that successfirl engineers need to have good written and oral communication skills. To emphasize t}re importance of teamwork in engineering and to encourage group panicipation, many of the assignment problems require group work; some require the panicipation of the entire class. The main parts. of the book are:

Pr.rrlcs

Part 0ne: Engineering-An Exciting Profession In Pan One, consisting of Chapters I through

5, we

inuoduce the students to the engineering

profession, how to prepare for an exciting engineering career, the design process, engineering communication, and ethics. Chapter 1 provides a comprehensive inuoduction to the engineering profession and its branches. It introduces the students to what the engineering profession is and explains some of the common uaia of good engineers. Various engineering disciplines and engineering organizations are discussed. In Chapter 1, we also emphasize the fact that engineers are problem solvers. They have a good grasp of fundamental physical and chemical lan's and mathematics, and apply these fundamenel lars and principles to design, develop, test, and supervise the manufacnrre of millions of products and services. Through the use of examples, we also show that there are many satisfying and challenging jobs for engineers. W'e pointed out that atthough the activities of engineers can be quite varied, rhere are some personaliry uaits and work habia that

typifr most of today's successfi.rl engineers:

. Engineers are problem solvers. " Good engineers hane a firm grasp of the fundamental principles that can be used to solve many different problems. . Good engineers are analytical, detailed oriented, and creative. . Good engineers have a desire to be lifeJong learners. For example, they uke continuing edu. . " . . . "

cation classes, seminars, and workshops to stay abreast of new innovations and technologres. Good engineers have written and oral communication skills that equip them to work well with their colleagues and to convey their expenise to a wide range of clients. Good engineers have time management skills that enable tlem to work productively and efficiendy. Good engineers hane good "people skills" that allow them to interact and communicate effectively with various people in their organization. Engineers are required to write reports. These reports might be l""gthy, detailed, and technical, containing graphs, charts, and engineering dt*i"gs. Or the may take the form of a brief memorandum or an executive sutnmary. Engineers are adept at using computers in many &fferent ways to model and analyze various practical problems. Good engineers aaively panicipate in local and national discipline-specific organizations by attending seminars, worlchops, and meetings. Many even make presentations at professional meetings. Engineers generally work in a team environment where they consult each other to solve complo< problems. Good inteqpersonal and communication skills have become increasingly im-

portant now because of the global market. 1, we also enplain the difference berween an Engineer and an engineering technologist, and the difference in their career options. In Chaprcr 2, the transition from high school to college is explained in terms of the need to form good study habits and suggestions are provided on how to budget time effectively. In Chapter 3, an introduction to engineering desbn, 'We show that engineers, regardless of their teamwork, and standards and codes is provided. background, follow cerain steps when designing the products and services we use in our everyday lives. In Chapter 4, we explain that presentations are an integral pan of any engineering pro1ecr. Depending on the size of the project, presentations might be brief, lengthy, freguent,

In Chapter

Pnsrecr and may follow a certain format requiring calculations, graphs, charts, and engineering draltrings. In Chapter 4, various forms of engineering communication, including homework presentation, brief technical memos, progless reports, detailed technicd reports, and research papers are explained. A brief introduction to PowerPoint is also provided. In Chapter 5, engineering ethics is emphasized by noting that engineers design many products and provide many services that affect our quality of life and safety. Therefore, engineers must perform under a standard ofprofessional behavior that requires adherence to the highest principles ofethical conduct. A large number ofengineering ethics related case studies are also presented in this chapter.

Part Two: Engineering FundamentalsConcepts Every Engineer Should Know In Part Two, consisting of Chapters 6 duough 13, we focus on engineering fundamenals and introduce students to the basic principles and ph1'sical larvs that th.y *ill see over and over in some form or other during the next four years. Successfirl engneers hane a good grasp ofFundamentals, which they can use to understand and solve many different problems. These are concepts that wery engineer, regardless ofhis or her area ofspecialization, should know In these chapteis, we emphasize that, from 6ur obseryation of our surioundings, we hare learned that we need only a few physical quandties to flrlly describe events and our surroundings. These are length, time, mass, force, temperature, mole, and electric current. We also sr-

plain that we need not only physical dimensions to describe our surroundings, but also some way to scale or divide these physical dimensions. For example, time is considered a physical dimension, but it can be divided into both small and large portions, such as seconds, minutes, hours, days, years, decades, centuries, and millennia'W'e discuss comrnon s;rctems of unin and emphasize that engineers must know how to conveft from one qystem of units to another and alwErs show the appropriare units that go with their calculations. \V'e also explain that the phlnical lars and formulas that engineers use are based on observations of our surroundings. W'e show that we use mathematics and basic physical quantities to express our observations, In these chapters, we also explain that ttrere are many engineering design variables that arg related to the fundamental dimensions (quantities). To become a successfi.rl engineer a student must first firlly understand these fundamental and related variables and the penaining governing larvs and formulas. Then it is imporant for the student to know how these variables are measured, approximated, calculated, or used in pracdce. Chapter 6 explains the role and imponance of fundamental dimension and units in analof engineering problems. Basic steps in the analysis of any engineering problem are disysis cussed in detail. Chapter 7 inroduces length and length-related variables and explains their imponance in engineering work For example, the role of area in heat transfer, aerodynamics, load distribution, and stress analysis is discussed. Measurement of length, area, and volume, dong with numerical estimation (such as trapezoidal nrle) of these values, are presented. Chapter 8 conslders time arid time-related engineering paiameters. Periods, ftequencies, linear and angular velocities and acceleradons, volumetric flow rates and fow of traffic are also discussed in Chapter 8.

Pnuecs Mass and mass-related parameters such density, specific weight, mass florv rate, and mass moment of inenia and their role in engineering analpis, are presented in Chapter 9. Chapter 10 covers the imporance of force and force-related parameters in engineering. 'W'hat is meant by force, pressure, modulus of elasticity, impulsive force (force acting over time), work (force acting over a distance) and moment (force acting at a distance) are discussed

in detail. Temperarure and temperature-related parameters are presented in Chapter 11. Concepts such as temperafure difference and heat transfer, specific heat, and thermal conductivity also are covered in Chapter 11. Chapter 12 considers topics such as direct and alternating current, elecuicity, basic circuits components, pov/er sources, and the tremendous role of electric motors in our wery day life. Chapter 13 presents energy and power and explains the distinction berween these two topics. The importance ofunderstandingwhat is meantbywork, energy, power'watm, horsepo$/er,

and efficiency is emphasized in Chapter 13.

Part Three: Computational Engineering ToolsUsing Available Software to Solve Engineering Problems 15, we inuoduce Microsoft Excel and MAI1AB computational tools that are used commonly by engineers to solve engineering prob-two lems. These compuationd tools are used to record, organize, analyze data using formule", and present the resulm ofan analysis in chart forms. MAILAB is also versatile enough that students cln use it to write their own programs to solve complor problems.

In Part Three, consisting of Chapters 14 and

Part Four: Engineering Graphical eommunication: Conveying Information to 0ther Engineers, Machinists, Technicians, and Managers In Part Four, consisting of Chapter 16, we inuoduce students to the principles and nrles of engineering graphical communication and engineering sfmbols. A good grasp of these principles '!fe explain that engiwill enable students ro convey and undentand information effectively. neers use technical drawings to convey useful informadon to others in a standard manner. An engineering drawing provides information, such as the shape of a product, its dimensions, materials from which ro fabricate the product, and the assembly steps. Some engineering drawings are specific to a particular discipline. For enample, civil engineers deal with land or boundary' topographic, consrrucdon, and route survey drarings. Electrical and electronic engineers, on the other hand, could deal with printed circuit board assembly drawings, printed circuit board drill plans, and wiring diagrams.'W'e also show that engineers use special symbols and signs to convey their ideas, analyses, and solutions to problems.

Part Five: Engineering Material SelectionAn lmportant Design Decisioll fu

engineers, whether you are designing a machine pa$, a toy, a frame of a car, or a structruer the selection of materids is an important design decision. In Pan Five, Chapter 17, we look

Pnrrecn

vtl

more closely at materials such as metals and their allop, plastics, glass, wood, composites, and concrete rhat commonly are used in various engineering applications.'We also drscuss some of the basic characteristics ofthe materials that are considered in design.

fart

Six: Mathematics, Statistics, and Engineering EconomicsWhy Are They lmportant? In Part Six, consisting of Chapters 18 through 20,we introduce students to important mathematical, statistical, and economical concepts. We orplain that engineering problems are mathematical models of physical situations. Some engineering problems lead to linear models, whereas

others result in nonlinear models. Some engineering problems are formulated in the form of differential equations and some in the form of integrals. Therefore, a good undersanding of mathematical concepts is essential in dre formulation and solution of many engineering problems. Moreover, statistical models are becoming common tools in the hands of practicing engineers to solve quahty control and reliability issues, and to perform failure andyses. Civil engineers use statistical models to study the reliabfity of construction materials and stnrctures, and to design for flood control, for example. Elecuical engineers use statistical models for signal processing and for developing voice-recognition software. Manufacturing engineers use statistics for qualiry control assurance of the products they produce. Mechanical engineers use statistics to study the failure of materials and machine parts. Economic factors also play important roles in engineering desigrr decision making. Ifyou design a product that is too expensive to manufacnue, then it qrn not be sold at a price that consumers can afford and still be profitable to your company.

CASE STUDIES-ENGINEERING MARVELS To emphasize that engineers are problem solvers and that engineers apply physical and chemical laws and principles, along with mathematics, to design products and services that we use in our everyday lives, seven case studies are placed throughout the book These projects are ruly engineering marvels. Following Chapter 7, the design of New York City W'ater Tunnel No. 3 is discussed. The design of the Cateqpillar 797 Mining Tiuck, the largest mining mrck in the world, follows Chaprcr 10. FollowingChapter 13, relwantinformationaboutthe design ofthe Hoover Dam is discussed. The design of the Boeing 777 :spresented.in a case sudyfollowing Chapter 16. Finally, the Pratt and Whimey Jet Engine is discussed at the end of Chapter 17. There are assigned problems at the end of those case studies. The solutions to these problems incorporate the engineering concepts and laws that are discussed in the preceding chapters. There are also a number of engineering ethics case studies, from the National Society of ProFessional Engineers, in Chapter 5, to promote the discussion on engineering ethics.

IMPROMPTU DESIGNS I have included seven inexpensive impromptu designs that could be done during class times. The basic idea behind some of the Imprompru Designs have come from theASME.

viii

Pnsrecn

REFERENCES In writing this book, sweral engineering books, Veb pages, and other marerials were consulted. Rather than giving you a list that contains hundreds of resources, I will cite some of the sources that I believe to be usefirl to you. I think all freshman engineering students should ov"n a handbook in their chosen field. Currendy, there are many engineering handbooks anailable in print or electronic format, including chemical engineering handbools, civil engineering handbools, electrical and electronic engineering handbooks, and mechanical engineering handboolis. I also beliwe all engineering students should own chemistry phpics, and mathematics handbooks. These texts c:rn serve as supplementary resources in all your classes. Many engineers may also

findusefirl theASHRAEhandbook, the,FzndammalVolume,byrheAmerican SocieryofHeating, Refrigeratin& and Ah Conditioning Engineers. In rhis book, some data and diagrams were adapted with permission from the following sources:

Baumeisrer, T., et al., Marh's Handbooh,8th ed., McGravr Hill, 1978. Electrical Viing, 2nd ed., AA VIM, 1981. Electric Motors,5th ed., AAVIM, 1982. Gere, J. M, Mechanics ofMateriak,6th ed., Thomson, 2004. Hibbler, C., Mechanics ofMateriak,6th ed., Pearson Prentice Hall'

k

SnndardAtrnoEhne, Vashington D.C., U.S. Government Printing Office, 1962, 'Weston, K C,, Energt Conuersion, Vest Publishing, 1992. U.S.

AEKNOWLEDGMENTS my sincere gratitude to the editing and production tenm at Thomson Engineering, especially Hilda Gown. I am also gratefirl to Rose Kernan of RPK Editorial Services, Inc., and Newgen-Austin I would also like to thank Dr. Karen Chou of Minnesota State University, who reviewed the second edition carefirlly and made valuable suggestions. I am also thankful to the follouring reviewers who offered general and specific comments: N.lly M.Abboud, Universiry of Connecticut; Barbara Engerer, Valparaiso Universiqt Alex J. Fowler, Universiry of Massachusetrs, Darrmouth; Peter Golding, University of Texas at El Paso; Fadh Oncul, Fairleigh Dickinson University. I would also like to thank the following individuals for graciously providing their insighu for our Student and Professional Profiles sections: Katie McCullough, Celeste Baine, NahidAfsari, Dominique L. Green, Susan Thomas, and Ming Dong. Thank you for considering this book and I hope you enjoy the third edition.

I would like to

enpress

Saeed

Moaueni

Gontents

Preface nl

PART 01{E:

| 1.1 1.2 1.3 1.4 t5

ENGTNEERTNG-AN ExcrTrNG

pRoFEssroN z

flntroduction to the Engineering Profession 4 Englneerlnq Work lg All Around

You

5

Englneerlng as a Professlon and Conmon Tnlts of Good Enqlneers 8 Conmon Traltg of Good Englneers 9 Englneerlng Dlsclpllnes 12

Accredltotlon Board for Englneerlng ond Technolory OBET) Professlonal

ltl

Proflle 23

Summary 23

Prcblem

23

lmBromptu Deslgn

2 2J 2.2 23 2,4 2,5 2.6

I

26

Freparing for an Engineering

eareer

Maklng the Transltlon from Hlgh School to

Colleqe 28

BudgetlngYourllne 28 Dally Studylng and Prepantlon

31

Get lnvolyed wlth an Englneerlng

0rganlzetlon 36

Your Graduatlon

Plan

?7

37

Gher Conslderatlons 3? Student

Proflle 38

ProfesslonalProflle 39

Sumnary 40

Problens 40

3 3.1 3.2 3.3 3.4 3.5 3.6

fintroduction to Engineering Design 4l EnglnecrlngDeslgnPfocess 42 Englneerlng Economlss 4i8

Materlal Selectlon 49

Teamwork 5l Common Tralts of Good

Teams 52

Confllct Resolutlon 53

lx

X

Coxrrrrs

3.7 3.8 3.9 3.10 3.ll 3.12 3.13 3.14 3.15

Prolect Schedullng and Task

Chart

53

Alternetlves 55 Patent, Trademark, and Copyrlght 56 Englneerlng Standards and Godes 5l Evaluatlng

Eramples of Standards snd Codes 0rganlzatlons ln the Unfted Exomples

ol lnternatlonal Standards

and

Drlnklng lTater Standards In the Unlted

States 60

Codes 62

States 67 States 68

0utdoor Alr Quallty Standards In the Unlted lndoor Alr Quallty Standards In the Unlted

states

70

Sunmary 72 Problems ?3 lmpromptu Deslgn

4

75

Engineering Communication 81

41 4,2 4.3 4.4 4.5 4.6 41 4.8

5

ll

Communlcatlon Skllls end Presentetlon of Englneerlng Baslc Steps lnvolved In the Solutlon of Englneerlng

Work 82

Problems 82

Homework Presentatlon 85 Progress Report, Erecutlve Sunmary end Short DetalledTeshnlcal

Report

Menos

0ral Gommunlcatlon and Presentatlon 90 PowerPolnt Presentatlon 91 Englneerlng Graphlcal Communlcatlon 100

Summary

101

Problens

102

Engineering

5.1 52 5.3 5.4

Ethics

Englneerlng

Ethlcs

105

106

The Gode of Ethlcs of the l{otlonal Socle$ of Prolerslonal .|07 Gode ol Ethlcs for Englneers

Greed

Englneer's

111

Summary 1l?

Problem

117

Englneerlng Ethlcs: A Case Study from NSPE& 120

PART TWO:

ENGINEERING FUNDAMENTALS_

CONCEPTS EVERY ENGINEER SHOULD

5

87

87

Fundamental Dimensions and

6.1 6.2 6.3 6.4 65

KNOW 124

Units

126

Englneerlng Problemsand Fundamental Dlmenslons 127

Systens of

Unlts

128

Unlt Converslon 134 Dlmenslonal Homogenelty 136 Numerlcal versus Symbollc

Solutlons

138

Englneers l0?

CoNrprrs 6.6 6:l 6.8

Slgnlflcant Dlglts(Flgures) 139

Systems l4l

Englneerlng Components 8nd

Phyrlcal Laws of 0bservatlons In Englneerlng

ltlil

Summary t46 Problems

7

147

Length and Length-Related Parametcrs tr52

7.1 7.2 73 7.4 7.5 7.6 7:I 7.8

Length as a Fundamental Dlmenslon 15il

length

Measurement of

157

llomlnal Slzes versus Actual

Slzes

160

lko lengths l6il StralnasRatloofTlrolengths 163 Radlans as Ratlo of

Area

16:!

Volume

172

Second Moment of

Summary

Areas 17

182

Problems 18:l lmpromptu Design

lll

188

An Englneerlng Marvel: The Nerv York Clty Water Tunnel No.

8 Time and Time-Related Parameters 195 8.1 8.2 8.3 8.4 8.5 8.6

Tlme os e Fundanentol Dlmenslon 196 .|98 Measurement ot nme

PerlodsandFrequencles Flow of

Trofflc

Parameten Inyolvlng length and

AngularMotlon Summary

215

Problems

215

Mass as a Fundamental

Measurement of

Parameters 218

Dlmenslon

Mass

219

222

Denslty, Specltlc Volume, snd Speclfls Mass

FlowRate 224

lnertla 2A

Mass Moment of

Momentum

Conservatlon of

205

214

Mass and Mass'Related

9.1 9,2 93 9.4 95 9.6 9.1

llme

212

ProfesslonalProflle

9

201

203

224

Mass 229

Sumnary 232 Problems 232 IMPROMPTU DESIGN

IV

235

onvlty

222

3*

189

xi

xii

Covrenrs

l0

Parameters 236

Force and Force-Related

10.1 102 10.3 10.4

What We Mean by

Force

237

Newtonl Laws In Mechanlcs

241

Pressure snd Stress-Force Actlng Over an

Area

244

Modulus of Elastlclty, Modulus ot Rlgldlty, and Bulk Modulug of Gompresslblllty 256

105 10.6 l0:l

Dlstence 264

Moment, Torque-Forces Actlng at a

Work-Force Actlnq 0ver a Dlstance 268 Llnear lmpulse-Forces Actlng 0ver

Summary

271

Problemr

271

IMPROMPTU OESIGII

V

Tlme

269

2r/

An Englneerlng Marvel: Caterplllar 797 Mlning

lf

278

Temperature and Temperature-Related Parameters 282

It.l 1L2 ll.3 11.4 tls 11.6

Tempenture os a Fundrmentol Dlmemlon 283 Measurement of Temperature and lts

Unlts

286

Tenperature Dlfference and HeotTransfer 293 lhermal Comfort, Metobollc Ratg and Clothlng Insuletlon 306 Some Tenpemture'Related Materlals

Heatlng Values of

Fuels

ProlesslonolProflle

n2

Truck*

Summary

315

Problems

315

Propertles 309

312

314

Electric Currcnt and Related Parameters 319

121 12.2 123 l2A 12.5

Electrlc Gunent as a Fundenental Dlmenslon 320

Voltage

321

Dlrect Cunent and Alternatln0

Cunent 323

Electrlc Clrcults and Components 326 Electdc

Motors 334

ProfesslonalProflle 3il7 Summary 337 Problems 338

13 Energy and Power 341 l3.l 13.2 13.3 13.4 135

Work, Mechanlcal Energy,Thermal

Energy 342

Gonservatlon of Energy-Flrst Lrw of Thermodynonlcs 348 Understandlng What tfe Mean by

lfottsand Horsepower

Efflclency 355

351

Power 350

CoNrsNrs Student

Proflle 362

ProfesslonalProf,le 363 Summary 364 Problems 364 IMPROMPTU DESIGN

V[

366

An Englneerlng Marvel: Hoover

PART THREE:

Dam

367

coMpuTATroNAL ENGTNEERTNG

T00Ls-

USING AVAILABLE SOFTWARE TO SOLVE ENGINEERING

14 Efectronic 14,1 14.2 l/t3 14.4 t4J 14.6 14:l 14.8

PROBLEMS 370

Spreadsheets 372

Mlcrosoft Ercel-Baslc Cells and

ldeas 37il

lhelr Addresses 374

Crestlng Formulas In

Ercel

375

Functlons 383 Uslng Ercel Loglcal Functlons 387 Uslng Excel

Ploftlng wlth

Ercel

389

Matrlx Computatlon wlth

Ercel

400

CurveFlttlngwlthErcel 407 Sumnary 4ll Problems

15

412

MATLAE 419

15.1 152 15.3 15.4 155 15.6 15.7

MAT[AB-Baslc

ldeas 420

Uslng MATLAB Bullt-ln

Functlons 429

Plottlng wlth MATUIB tl38 lmportlng Excel and Other Data Flles Into MATLAB 445

MatrhGomputatlongwlthMATLAB 447 Curve Flttlng wlth MATLAB 450 Symbollc Mathematlcs wlth MATLAB 451

ProfesslonalProflle 454 Summary 455 Problems 455

PART F0UR:

ENcTNEERTNG

cRApHrcAL coMMUNrcATroN-

CONVEYING INFORMATION TO OTHER ENGINEERS, MACHINISTS,

TECHNICANS,ANDMANAGERS 460

16

Engineering Drawings and

16.1 16.2 16.3 16.4

lmportance ot Englneerlng

Symbols 462

Dnwlng 46il

0rthognphlc Vlews 4&l Dlmenslonlng and Toleranclng 467

lsometrlsVlew 469

xiii

xiv

CoxrrNrs

16.5 16.6 16:I 16.8 16,9

Sectlonal

Vlews

472

Clvll, Electrlcal, and Electronlc Solld

Dnwlngs

476

trlodellng 476

Why Do We l'leed Englneerlng

Symbols? tl83

Examples of Common Symbols used In Clvll, Electrlcal,

and Mechanlcal Englneerlng tl85

summary rts? Problems 487 An Engineerlng Marvel: Boelng 777* Commerclal

PART FIVE:

ENGTNEERTNG MATERIAL

Alrplane 493

sELEcTtoN-

AN IMPORTANT DESIGN DECISION 498

l7

Engineering

17.1 172 t73 17.4

Materials 500

MaterlalSelectlon

501

Electrlcal, Mechanlcal, and lhermophyslcal Propertles of Materlals 503

Materlals 509

Some Common Solld Englneerlng

ilaterlals

Some Common Fluld

ProfesslonalProflle

519

521

Summary 522 Problems 522

VII

IMPROMPTU DESIGI'I

525

An Englneerlnq Marvel: The Jet

PART SIX: WHY ARE THEY

fB

Englne* 526

NnrnrMATrcs, sTATlsTtcs, AND IMPORTANT? 530

ENGTNEERING

Mathematics in Engineerinq 532

18.1 182 18.3 18.4 18.5 t8.5 18.7

trlathematlcal Symbols and Greek

LlnearModels

Alphabet 533

5:15

Nonllnearilodels

541

Erponentlal and logarlthmlc Models 546

MatrlxAlgebn

Glculus

551

562

Dlfferentlal Eguatlons 570

Summery 572 Problems 57!l

EcoNoMtcs-

Corrslrrs

19

Probability and Statistics in Engineering 577

19.1 19.2 193 19.4 t95

Probablllty-Baslc ldeas 510

Statlstlcs-B$lc ldeos 5|9 FrequencyDlstrlbutlons 580 Heasuras ol Gntral Tendency Yarlotlon-lllean, Medlon, and Standard

t{ormilDbtrllutlon

587

Sumn8rt 594 hoblems 594

20 Engineering 20.1 mA 20.3 2OA 20J 20.6 ZOJ 20.8 20.9

Carh Flow

Economics 597 Dlagnm 598

SlmpleandCompoundlntenit 599 FutureUlorthofaPnsentAmount 600 Efrectlve lnteregt

nde

60ll

PresentWorthofFutureAmount Present Worth of Serles

605

hynent or

hnutty

605

FuturEl{orthofSerlshynnnt 606 SummaryofEnglnesrlngEconomlcaAndysb 609 Chooslng the Best Alternatlves-Declslon

Summory

617

PruUem

617

AppEndlr 620

Credlts 625 Index 626

l{aklng

613

Devlatlon 582

XV

Part

ENCTNEERING Al. Excmttc PnoFEssron

1

j@@ltryq'@qt

i

In Part One of this book, we will introduce you to the engineering profession. Engineers

.---aregroblem solvers. They have a good grasp of fundamental physical and chemical laws and mathematics and apply these laws and principles to design, develop, test, and

supervise the manufacture of millions of products and services. Engineers, regardless of

their background, follow certain steps when designing the products and services we use in our everyday lives. Successful engineers possess good communication skills and are

team players. Ethics plays a very important role in engineering. As eloguently stated by

the National Society of Professional Engineers (NSPE) code of ethics, "Engineering is an important and learned profession. As members of this profession, engineers are

expected to exhibit the highest standards of honesty and integrity. Engineering has a

direct and vital impact on the quality of life for all people. Accordingly, the services provided by engineers require honesty, impartiality, fairness and equity, and must be dedicated to the protection of the public health, safety and welfare. Engineers must

perform under a standard of professional behavior which requires adherence to the highest principles of ethical conduct." In the next five chapters, we will introduce you to the engineering profession, how to prepare for an exciting engineering career, the design process, engineering communication, and ethics.

Chapter I

Introductlon to the Englneering Professlon

Chapter 2

Preparlng for an Engineerlng Career

Chapter 3

Introductlon to Englneering Design

Chapter 4

Englneering Communlcatlon

Chapter 5

Englneering Ethlcs

CHAPTE,R.

1

INrnoDUcrroN To THE ENcTNEERTNG PnoFEssroN [f

ngineers are problem solvers.

|1|

Successful engineers possess good

L

communication skills and are team

players. They have a good grasp of

fundamental physical laws and mathematics. Engineers apply physical and chemical laws and mathematics to design, develop, test, and supervise the

manufacture of millions of products and services. They consider important

factors such as efficiency, cost, reliability, and safety when designing products. Engineers are dedicated to lifelong

learning and service to others.

1.1

ENcTNEERTNc

Wom Is Ar.r. Anomlo You

ofloa are not yet ceTtain loa uant to study mginening during the nextfour years in college and may haue questions simikr to thefollowing

Possibfu sorne

I

really uant to study mgineering? is mgineeringand what do mgineers do? 'Vhat are some ofthe areas ofspecialization in mgineering? How many differmt mgineering disciplines are tbere? Do I want to become a rnechanical enginen, or shoull.I pursue ciuil mgineering? Or would I be happier becoming an electrical mginen? How will I hnow rhat I haue piched the bestfeldfor rue? Wll the dernand far rn! area of speciakm.tian be hi.gh whm I gradaate, and'

Do

Vhat

bryond that? The main objectiues of this chapter Are to prouidz sorne answers to these and other questi.ons loa may haue, and to i.nnodace you to tlte engineeri.ng profasion

and its aaious brancltes.

1.1 Engineering Work ls All Around

You

Engineers make products and provide services that make our lives beter (see Figue 1.1). To see how engineers contribute to the comfon and the betterment of our weryday lives, tomor-

row morning when you get up, just look around you more carefirlly. During the night, your bedroom was kept at the right temperature thanks to some mechanical engineers who designed the heating air-conditioning, and ventilating q/stems in your home. When you get up in the morning and turn on the lights, be assured that thousands of mechanical and elecuical engineers and technicians at po$'er plants and power stations around the country are making certain the fow of electricity remains uninterrupted so that you have enough pov/er to turn the lights on or flun on your TV to watch the morning news and weather repon for the dty. Th" TVyou are using to get your morning news was designed by electrical and elecronic engrneers. There are, ofcourse, engineers from other disciplines involved in creating the final product; for example, manu6curing and industrial engineers. When you are gening ready to take your morning shower, the clean water you are about to use is coming to yor[ home thanks rc civil

f,

Figurel.l

Eramples of products and serviees designed by engineers.

CnerrsR

I

IxrnopucrroNTo rnrExcwnsRrNc Pnorsssrox and mechanical engineers. Even if you live out in the country on a farm, the pump you use to bring water from the well to your home was designed by mechanical and civil engineers. The water could be heated by natural gas that is brought to your home thanls to the work and effon of chemical, mechanical, civil, and peuoleum engineers. After your moming shower, when you ger ready to dry yourself with a towel, think about what types of engineer worked behind the scenes to produce the towels. Yes, the cotton towel was made with the help of agriculturd, industrial, manufacnuing chemical, petroleum, civil, and mechanical engineers. Think about the machines that were used to pick the cotton, transport the cotton m a factory clean it, and dye it to a pretty color that is pleasing to your eyes. Then other machines were used to weave the fabric and send it to sewing machines that were designed by mechanical engineers. The same is uue of the clothing you are about to wear. Your clothing may contain some polyester, which was made possible with the aid of petroleum and chemical engineers. "V'ell," you may say, "I can at least sit down and eat my breakfast and not wonder whether some engineers made this possible as well." But the food you are about to eat rras made with the help and collaboration ofvarious engineering disciplines, from agricultural to mechanicd. Lett sayyou are about to have some cereal. The milk was kept fresh in your refrigerator thanls to the efforts and work of mechanical engineers who designed the refrigerator components and chemical engineers who investigated alternative refrigerant fluids with appropriate thermal properties and other environmentally friendly propenies that can be used in your refrigerator. Furthermore, electrical engineers designed the control and the electrical power units. Now you are ready to get into your car or take the bus to go to school. The car you are about to drive was made possible with the help and collaboration of automotive, mechanical, electricd, electronic, material, chemical, and peuoleum engineers. So, you see there is not much that you do in your daily life that has not involved the work of engineers. Be proud of the decision you have made to become an engineer. Soon you will become one ofthose whose behindthe-scenes efforts will be taken for granted by billions of people around the world. But you will accept that fact gladly, knowing that what you do will make peoplet lives bener.

Engineers Deal with an Increasing World Population 'V'e

as people, regardless ofwhere we live, need the following things: food, dothing, shelter, and warer for drinking or cleaning purposes. In addition, we need various modes of transportation to get ro different places, because we may live and work in different cities or wish to visit friends and relatives who may live elsewhere. Ve also like to have some sense of security, to be able to 'We need to be liked and apprecixedby our friends and family, as well. relax and be entertained. At the turn of the 20th century, there were approximately six billion of us inhabiting the eafth. As a means of comparison, it is imporent to note that dre world population 100 years ago, at the turn of the 19th century, was one billion. Think about it. It took us since the beginning of human enistence to reach a populadon of one billion. It only took 100 years to increase the population by fivefold. Some of us have a good standard of living, but some of us living in developing countries do not. You will probably agree that our world would be a bemer place if wery one of us had enough to eat, a comfortable and safe place to live, meaningfirl work to do, and some time for relaxation. According to'the Iatest estimates and projections of the U.S. Census Bureau, the world

population will reach 9.3 billion people by the year 2050. Not only will the number of people inhabiting the eanh continue to rise but the age strucrure of the wodd population will also

1.1

EncnrBBnnlc W'onr Is Ar.r. A,nouuu You

1

0 1950

1960

lwo

1980 1990 2W

20LO

Year (a)

Over- 1 00 population (thousands)

* estimate (b)

E figure t.2 (d ilre

latest pmjection of world population growth. (b) Ihe latest estimate of U.S. elderly population growth.

Smrce: Dacr hom U.S. Census Bureau.

people at least 65 yars of age-will more than 1.2). Figure double in the next 25 yers How is this information relevant? Well, now that you harre decided to study to become an engineer, you need to realize that what you do in a few years after your graduation is very importanr ro all ofus. You will design producs and provide services especially suited to the needs and demands of an increasing elderly population as well as increased numbers of people of all ages. So prepare well to become a good engrneer and be proud that you have chosen the engineering profession in order to conuibute to raising the living standard for weryone. Todry't world economy is very dynamic. Coqporations continually employ new technologies to maximize efficiency and piofits. Because ofthis ongoing change and emerging technologies, new jobs change. The world's eldedy

population-the

(see

are created and orhers are eliminated. Computers and smart

elecronic devices are continuously

reshaping our way of life. Such devices infuence the way we do things and help us provide the

Cger"rsn

I

lNrnopucrroN To nrs

ENcrNnnRrNG PRoFEssroN

necessities of our lives-clean water, food, and shelrcr. You need to become a lifelong leamer so that you can make informed decisions and anticipate as well as react to the global changes

caused by technological innovations as well as population and environmental changes. According to the Bureau of labor Statistics, U.S. Department of Labor, among the fastestgrowing occupations are engineers, computer specialists, and qystems analysts.

1.2

Engineering as a Profession and Common Traits of Good Engineers In this section, we will first discuss engineering in a broad sense, and then we will focus on se'W'e lected aspecs of engineering. will also look at the traits and characteristics common to many engineers. Next we will discuss some specific engineering disciplines. As we said earlier in this chapter, perhaps some ofyou have not yet decided what you want to study during yolu college years and consequendy may hane many questions, including: Wh.at is engineering and what do engineers do? !7hat are some of the areas of specialization in engineering? Do I really want to study engineering? How will I know that I have picked the best field for me? \Xrril the demand for my area of specialization be high when I graduate, and beyond that? The following sections are intended to help you make a decision that you will be happy with; and dorit worry about finding answers to all these questions right now You have some time to ponder them becluse most of the coursework during the fust year of engineering is similar for all engineering students, regardless oftheir specific discipline. So you have at least a year to consider various possibilities. This is true at most educational institutions. Even so, you should mlk to your advisor early to determine how soon you must choose an area of speciallzation. And dont be concerned about your chosen profession changrng in a way that makes your education obsolete. Most companies assist their engineers in acquiring further training and education to keep up with technologies. A good engineering education will en"h*grg able you to become a good problem solver throughout your life, regardless ofthe particular problem or situation. You maywonder during the ner>.Amount = 1000:250:3000; >>

IntereEt_Rate = 0.06;0. 01:0.08; Interest_Earaed. = (.Anount' ) * (IatereEt_Rate), fprl-nrf (' \n\n\r\L\r\r\r\r\r rnrereEr Ratet ) ;fprLntf (' \a\t .Anount\t\t' ) I . .. fprLatf ( ' \t\t e"g[' , Interest-Rate) ; fprl-ntf ( ' \n' ; ; dtsp ( lihount ! ,IDterest Earned] ) >> >> >>

Interest Rate

Dollar Amount

0.06

o.a7

1000

50

t250

/)

70 87.5

80 100

1500

90

105

t20

t750

105

t22.5

140

2000 ?250 2500

120 135

t40

160

r57.5

180

150

175

?:750

r65

r92.5

3000

180

2t0

204 220 240

15.2

Usu.rc MATLAB Bur.r-nr Frnrcrrorvs

429

,i,

[l

Figurc

15.5

The commands and result

for tmmple

14.2 (Revisited).

t:t

On the last command line, note that the three periods . . . (an ellipsis) represents a continuation marker in MAILAB. The ellipsis means there is more to follow on this command line. Note the use of fprinLf and disp commands. The final result for Example 14.2 (Revisited) is shown in Figure 15.5. ,l

15.2 [fsing MATLAB Built'in MATLAB offers

a large selection

Functions of built-in functions that you qln use to analyze data. fu we

discussed in the prwious &apter, by built-in funcdons we mean standard functions such as the sine or cosine of an angle, as well as formulas that calculate the totd value, the average value,

or the sandard deviation of a set of data points. The MlfILAB functions are available in various categories, including mathematical, trigonometric, statistical, and logical functions. In this chapter, we will discuss some of the common firnctions. MA|I,AB offers a Help menu that you can use to obain information on various commands and functions. The Help button is marked by a question mark ? located to the left of the current directory. You can also rype help followed by a command name to leam how to use the command. Some examples of commonly used MATLAB functions, alongwith their proper use and descriptions, are shown in Thble 15.7. Refs to Example 15.2 when studying Tirble 15.7.

The following set of values will be used to introduce some of MAlf-ABt built-in functions. Mass: U02 lts 99 rc6 rc3 95 97 102 98 96l.WrenstudyingTableL5.T,theresults of the executed functions are shown under the "Result of the Example" column. More examples of MlflLAB's Functions are shown in Thble 15.8.

430

Cnu'rrn 15 MAILAB

'#;;

i meqn "'q i j i

;**;oor

;t

*i;i:;

It

sums the values

It

calculates the average value

dre data in

It daermines

max

*n"

a

in

a

,;

*orr1* ogthu n="-pl" ,,,i til;lilrdfn

gilen array.

1013:

of

101.3

given array.

the largest value

in

the given array; ,It determines ttre smalle* value irt the given array, It calculates the standard deviadon fot the values in the giv-eqrena)r It sorts the values in the give$' ,,, , array in ascending order. It returns the value of zr,

i

r min ,l

sEq

:

joir; I,:

3.14151.926535897.. i

'

,.,'i;

.

It retums tangent value of the

9@l

argument. The argument must be

in radians,

cos

It

renrrns cosinevalue ofthe argument. The argument must be

in

It

returns sine value of the argument.

in radians, s

,',,,,,

frg*su**l**rh,h:ifT:,

;l

sqrt (x) f actorial (x)

Retums the square root ofvalue.t

ky"^ *:.:*e of factorial of*. For example, factorial(5) wlll return: (5)(4)(3)(2)(r) : 120.

Trigonornzt ix Functions (x) This is the inverse cosine function ofx. It is used to determine the value of an angle when its cosine value is kno\trn. asin (x ) This is the inverse sine function of*. It is used to determine the value of an angle when its sine value is known. atan (x) This is the inverse tangent function of r. It is used to determine the value :. of an angle rylieg its glgent value is knor,rn. '',1 ,,

acos

,

',

'

kponcntial and Ingarithmic Functiot s exp

(x)

los (x) 1og10 (x) 1og2 (x)

Retums the value of a". Retums rhe value of the natural logarithm of *. Note that

* must

greater than 0.

Returns the value ofthe common (base 10) logarithm of.r. Returns the value of the base 2 loSanthm of x.-

be

15,2 Usnc MAIIAB Bur.r-nv FuNcrroxs

431

Usiag MAil.AB, compure the aver4ge (ari hmecic mean) and the sandard deviation of the density of water data given in Thble 15.9. Refer to Chapter 19, Section t9.4, to learn about what the vdue of the standard deviation for a set of data points represents.

1,,

Group-AFindingq

GroupB Flndings

1.,

ffi* The final resuks for Example 14.4 (Revisite$ are shown in Figure 15.6. The commands leading to these results are:

MAILAB

>>DensJ.ty-A = [1020 1015 990 1060 1030 950 975 1020 980 950] t >>DeasLty-B = [950 9AO 890 1080 1120 900 ].040 1150 910 10201; >> DensLty_A_Avela![e = nean (Denstty_A) Density-A_Av€rdg€ = 1000.00 >> DenEl.ty_B_Aver?![€ = mean (DensLty_B) Densi-Ly-B-Av€rdg€ = 1000.00 > > Standard._Xlevlat lon_For_croup_l\ = std ( Dengity_A) Standard-Deviat i on_For_Group_A = 34.56 > > Standard_DevLat Lon_For_Group_B = std ( Deagity_B ) SLandard_Deviat. ion_For_Group_B = 93 .22

432

CHer"rsR

15

MAILAB

e

Edt uew w& wln*w

He

rdi&'ce@-*l&jt

I

figumls.O

Command l{indow

i|ATIAB'S

for trample

14.4

(Revhited).

Li-.!l:l:r-.':.t:

I--::

:l-:t:--,---:r

tlt: ;l:a:a:-lt-: _t:

:-:ri.,::::::-

The Loop eontrol

.

- forand while commands

When writing a computer program, often it becomes necessarl to execute a line or a block ofyour computer code many times. MAIf,AS providesprand ubilocnmmands for such situations.

forloop

Using theprloop, you can execute number of times. The sfnax of aforloop is

The

a

line or a block of code a specified (defined)

for index = sLart-value : incremenL : end-va1ue a line or a block of your computer code end For example, suppose you want to evaluate the function ! : x2 * 10 for r values of 22.00 22.50,23.00,23.50, and24.00. This operation will result in correspondingTvalues of 494.00, 516.25,539.00,562.25,and 586.00. The MlftI-AB code forthis enample then couldhave the following form:

x = 22.0; Fnr

i

-

1.1.R

Y=xn2+IO i

disp(lx',y'l) x = x + 0.5;

end

I5.2

UsrNc MAII-A,B Bunr-rN FuNc"rrors

433

Note that in the preceding example, the index is the integer i and its start-value is 1, it is incremented by a value of 1, and ia end-value is 5.

rfile Loop Using the whilzloop,you c:ur execute condition is met. The qyntax of a while loop is

The

whi

a

line or a block of code until

a

specified

Le control- I ing - expres s ion

a 1i-ne or a block of vour comouLer

code

end 'With the while command,

as

long

as

the controlling-expression is true, the line or a block of code using rhe whib com-

will be enecuted. For the preceding example, the MIIILAB

code

mand becomes:

x = 22.0; while x >pressur6=[20 tS 22 26 L9 1,9 2L L27i >>fprlotf('\t Line Pressure (pgt) \t VaLve FosLtlon\n\n');

for L=1:8 if pressure(t) >=20 fPrLntf (' \t e"9 \t\t\ts\t\t el-se

fpriatf ( '\t

end end

I

rigurel5.8

The solution

ol Ennple

14.5 (Revhited).

?"9

\t\t\t\t\t

OFEN\n'rBreEsure(L) CITOSED\n'

)

,pregsure (L) )

436

Cneprsn

15

MAILAB

The M-file As explained previously, for simple operations you c:m use M,{'l,AB's Command lTindow to enter variables and issue commands. However, when you write a program that is more than a few lines long, you use an M-frle. It is called an M-file because of its .zz extension. You can cre-

M-file using any text editor or using MAIT,ABI Editor/Debugger. To crqrte an M-fiIe, open the M-file Editor and MlfIl,AB opens a new window in which you can type your program. fu you type your program, you will notice that MAILAB assigns line numbers in the left column of the window. The line numbers are quite useftrl for debugging your program. To save the file, simply click File -* Save and type in the file-name. The name of your file must begin with a letter and may include other characters such as underscore and digits. Be carefirl not to name your file the same as a MAILAB command. To see if a file-name is used by a MATI-AB command, type exist ('file-namd) in the M,$LAB's Command'S7indow. To run your program, click on Debug + Run (or use the function key F5). Don't be discouraged to find mistakes in your program the first time you attempt to run it. This is quite normal! You can use the Debugger to find your mistakes. To learn more about debugging opuons, type help ate an

dtbug

It

n

the

MATU.B Command !7indow.

has been said that when Pascal was 7 years old, he qrme up with the formula

rdn -f l\ t-tt

determine the sum of 1,2,3, . . . , through n. The story suggests that one day he was asked by his teacher to add up numbers I through 100, and Pascal came up with the answer in few minutes. It is believed that Pascal solved the problem in the following manner: First, on one line he wrote the numbers I *uough 100, similar to

1 2 3 4.. ,........99

100

Then on the second line he wrote the numbers backward

100 99 98

97

...........2

1

Then he added up the numbers in the nvo lines, resulting in one hundred identical values

of

101

101 101 101

101

...........

101

101

Pascal also realized that the result should be divided by

1 through 100 twice-leading to the answer:

.

approach and came up with the formula

100(10 I )

--;:

n(n+ r)

-- ,

2-since

he wrote down the numbers

5050. l,ater, he generalized his

15.2

Flle Edt

Tqt

Bdffitr ,

Cell T@h Dehrg De$ktop Wrdow

&EB-

UsrNc MAILAB Burr,t-rN Funcrrous

437

Ftalp

"leitmTl aelc*e-areeli:

F{iF

rtil 4.. ":,i :

g Ask the usar lo input the uFper vaLue ,, upper_val-ue=input('Please input th€ upper vlaue of the nunLters.'), . 4;'t -,:

,1i

:

:l

6:, i,

q iFt

1: for

'8,.:

a! 1,:

1.0

i i

;:1 Figure

fhF

crrn onil:l

r^

zero

surn=o,

k=1:1:upper_value sum=sum+kt

end

g Print the results

fprintf('\n

?he eum of numbors from 1 to $g is eqaul to: tg\nr,upper_value'sum)

15.9 lhe lr|-file for Enmple

15.3.

Dcktop Wftew Heb

rFilei

rgjeFi3e d.H|? shortcuts E Howto Add m

Figure

15.10

uJffis tgfd

The results 0f Example 15.3.

Nexr, we will wdte a computer program using an M-file that asls a user to input a value for a and computes the sum of 1 through n. To make the program interesting, we will not make use of Pascal's formula; instead, we will use a for loop to solve the problem. \(e hane used MATI"A.B'S Editor to create the program and have named it For-Ioop-F,xample"m, as shown in Figure 15.9. In r:he shown program, the 7o symbol denotes comments, and any text following tle %o symbol will be ffeated as comments by MAII,AB. Also, note that you can find the Line (Ln) and Column (Col) numbers corresponding to a specific location in your program by moving the cursor. The line and the column numbers are shown in the right side, bocom corner of the Editor window. As you will see, the knowledge of line and column numbers are use'We frrl for debugging your program. run the program by clicking on Debug -* Run, and the result is shown in Figure 15.10. , ..,:ri:i:1rtrj.I i:

. -.i

438

Cneprun

15

MAIL.AB

tr5.3 Plotting with MATLAB M.{flAB

offers many choices when it comes to creating chats. For example, you c:m crezrte columl charts (or histograms), contour, or surface plorc. As we mentioned in Chapter 14, as an engineering student, and later as a practicing engineer, most of the charts that you will create will be tclt We charts. Therefore, we will explain in deeil how to ueztte an

x-y

charts,

x-!

chart.

Starting with a 10 cm X 10 cm sheet of paper, what is the largest volume you c:rn create by cuning out r cm X r cm from each corner ofthe sheet and then folding up the sides? See Figure 15.11. Use Ml[f,AB to obtain the solution. The volume created by cutting out r cm X r cm from each corner of the 10 cm X 10 cm sheet of paper is given by V: (10 - 2r)(10 - 2.a)*. Moreover, we know that, for r : 0 and x : 5, the volume will be zero. Therefore, we need to create a range of x values from 0 to 5 using some small increment, such as 0.1. We then plot the volume versus r and look for the maximum value of volume. The MAII-AB commands that lead to the solution are:

I The

>>x = 0:0.1:51 >>volune = (10-2*x) .*(10-2*x) .*xi >> Blot (x,vohme) >>tLtle ('VoJ.ume aE a functlon of x') >> xLabel- ( 'x (cn) ' )

rigurct5.ll l0 cm

x

l0 cn

sheet in

Erample 15.4.

>> >>

I

y1abeL ( 'VoLu.me

grid. mLnor

figure

15.12

(cm^3 ) ' )

The II|AIIAB Command l|indow

for txample 15.4.

The MAILAB CommandWindow for ExampleL5.4 is shown in Figure 15.12. The plot of volume versus r is shown in Figure 15.13.

ld

Figurc

fi1ii.lftrxT,ff{idi.iRBiEt:

15.13

nIF,Il

The plot ot volune versus

r

for Erample 15.4.

?j

Let us now discuss the MAIT.AB commands that commonly are used when plotting data. command plots yvalues versus xvalues. You can use various line types, plot symbols, or colors with the command pLoE(x,x,E), where s is a characer suing that defines a particular line type, plot symbol, or line color. The E can take on one of the properties shown in Thble 15.12.

The pLot (x, y)

DataSvmbol

lb

lo ,o

ic im iv ik

Blue Green

o

Point Circle

Red

x

r-mark

+

Plus Star

cy* Magenta

ltllow Black

Solid

Doned Dashdot Dashed

Square

d

Diamond Triangle (down)

Triangle (up) Tiiangle (left) Triangle (right)

439

440

I

Cnaprsn

15 MAILAB

Figurcl5.l4

l'lATLABl Line Property Editor.

For example, ifyou issue the cornmand pL

ot, ( x, y t t k* - | ),

MAILAB will plot the curve using a black solid line with an * marker shown at each dara point. If you do not spec' ify a line color, MAlf.AB automatically asslgns a color to the plot. Using the tttle ( 'text' ) commandr /ou crrn add text on top of the plot. The

xlabel

(

t

text' )

command creates the tide for the x-aris. The text that you enclose will be shown below the x-axis. Similady, the ylabeL ( 'text' ) command creates the tide for the y-axis. To turn on the grid lines, type the command grLd on (or just grtd). The command grLd off removes the grid lines. To flrn on the minor grid lines, as shown in Figure 15.13, type the command grLd between single quotation marks

mlnor.

L5.3 Plorrrrcvrrn MAILAB

I

441

tigurelS.tS

The plot of

tranple

15.3 with

nodified properties.

Generally, it is easier to use the Graph Property Editor. For enample, to make the curve line thicker, change the line color, and to add markers to the data points (with the rnouse

pointer on the curve) double-click the left mouse button. Make sure you are in the picking mode first. You may need to click on the arrow next to the print icon to activate the picking mode. After double-clicking on the line, you should see the line and the Marker Editor window. As shovrn in Figure t5.T4,weincreased dre line thickness from 0.5 to 2, c,hanged the Iine color to black, and set the daa-point marker sryle to Diamond. These new seaings are reflected in Figure I 5.15. Next, we will add an arrow pointing to the maximum value of the volume by selecting the TixtArrow under *re Insert option (see Figure l5.LG), and add the text'Maximnm volume occurs at x: 1.7 cm." These additions are reflected in Figure 15.17. V'e can also change the font size and style and make the title or the axes labels boldface. To do so, we pick the object that we want to modify and then, from the Menu bar, select Edit and then Current Object Properties . . . . Then, using the Propercy Editor shown in Figure 15.18, we c:rn modi& the propenies of the selected object.'We have changed the font size and the font weight of dre tide and the labels for F,xample 15.3 and shown the changes in Figure 15.19.

\firh MAILAB, you c:ln generate other types of plots, including contour and surface plos. You can also control the r- and y-axis scales. For enample, the MAIT,AB's

Loglog(x,y)

usesthebase-l0logarithmicscalesforr-andy-axes.Note.randyarethevariplot. The command LogLog (x, y) is identical to the plot (x, y) ,

ables that you want to

K

figurets.t6

Using the lnsert Tert Arow

options, you can add anows or text to the plot.

I

Figurel5.lT

The solution of

M2

tnmple

15.4.

15.3 Pr.orrrNcwrnr MAILAB

t

443

Figurel5.l0

MAIIAB5

Ie*

Property Editor.

it

uses logarithmic axes. The command semlLogx(x,y) or semilogy(xry) plot with base-l0 logarithmic scales for either only dre r-axis ory-axis. Finally, it is wonh noting that you can use the hold command to plot more than one set of daa on the

except

creates a

same chan.

A reminder, when creating an engineering chart, whether you are using MAIT,AB, Excel, other drawing software, or a free-hand dranring; an engineering chan must contain proper labels with proper units for each axis. The chart must also contain a figure number with a tide exptaining what the chart rcpresents. If more than one set of data is plomed on the same chart, the chan must also contain a legend or list showing symbols used for different data sets,

43

Cueprnn

I

15 MAIAB

FigurelS.lg

The result of Erample 15.4.

Using the resula of F-:rample 14.1, create a graph showing the value of the air density as a funcdon of temperature. The Command'l7indow and the plot of the density of air as a function of temperature are shown in Figures 15.20 and 15.21, respectivelp

I

Figure

15.20

Command $lindow

for Erample t4.6 (Revhited).

15.4 Iuponrrxc E:rcrr,aNo Orrrrn Dere

L

Figure

15.?l

Frr.ss rNro

MAil,AB

445

Plot of density of air for Erample 14.6 (Revhited).

i] t1

IttA,.i,

r:ff1&1i.lilr,,t,t -

.i,-r"lllta?nji:11-lftlfiil!:

15.4

--ii,!liijitii!.'i iiirxlitiijliirrrtii

--

---ii-,rXitilrl!ii1-11ta-- lr,iilril'liTrlTi1.ir'rrira

l:

Jli:1ii-,1ir,irlrirl..- -.'lTlLr.li4lltl

Nmporting Exeel and Other Data Filcs into MATLAB At times, it might

be

convenient to impon data files tlat were generated by other programs, such

MAil-AB for additional analpis. To demonsrate how we go about imponing a daa file into MATLAB, consider the Fxcel file shown in Figure I5,22.The Excel file was created for Example 15.4 with two columru: the r values and the corresponding volume. To impon this file into MAII-AB, from the Menu bar, we select File *1sn Irnport Data . . . , t}ren go to the appropriate directory, and open the file we want. The Impon'Wizard window, as

Excel, into

in Figure L5.23,r ,il|appear next. Select the "Create vectors from each column using column nameso and the \Pizard will i-port the data and will sare them as x and voXu:ne

as shown

variables.

Now lett the

MIIIIAB

Figure 15.25.

'We

that we want to plot the volume as a function of r. then simply type commands that are shoiyn in.Figtire 15,24.Tlte resulting plot is shown in

say

-*r; ; .';I-"-**-*-"

t11

:A

,

,:.9.i,

:'l

x

:.t:,

2 0.1

0.0 9.6

A:

na

18.4

sl

0.3

20.5

6

9.4

l,1.9 40.5

8l

0 0.

116.€

0.€

58.4

r:,$

10

I

[:]

D:I

rrolumo it:,;

iril!,

51.8

lr;

m,{ 6{.(

iii

fa\=rX.i;, r"#-?fizr;; ri

I

Figurel5.Z

The Ercel data

L

;rr:

file used in tranple 15.4.

Figurul5.Z3

i|AIIABI lmport ltizad.

@:-..:

n aiI*, a

fo';

-4'g.i'1i;;**l.rryffii!

Ar Sat rtst d, el,el ..qdf|l!8 Ealtn trs l&r E ls &m $Frg 61ffid o!e*d @l,ablo, ln rhe ol@9 @tt!D€R. > al@ ta.wlwel > td.tla lr9ol@ e a 6Mt@ ot r'l I ilib€t (,a a@l'l > Itd!€r, (190r@ le.3lrl

il

t *ld Figurel5.Z4

The commands leading to the plot

shown in Figure 15.25.

46

>l

"

"

.

t.

::

""ll.f

:';: "'

15.5 Mernor CouputetroNs wrrg MAILAB

L

Figure

15.25

Plot of uolume venus

r

using data imported frorn an Ercel file.

15.5 lVatrix Computations with MATLAB fu

explained earlier, MATI/,B offers many tools for matrix operations and manipuladons. Thble 15.4 shows er) A=[0 5 0;8 3 7;9 -2

91

A-

0 50 8 37 9-29 >> B=tA 6 -2;7 2 3;1 3 El-

4 0 -z

72 3 1 3 -4 >> C=[-1; 2; 5l -12 5

>> A+B

ans

=

4It-2 1-5 5 10 1

10 5

>> A-B alfD

_A

-

_1

')

r14 tt

-f

-LJ

>>3*A

0 15 0 24 9 21zt

-o

zt

>>A*B

35 10 15 60 75 -35 31 77 -50 >>A*C AIfD

-

10 33

>>det (.4)

ans

I

_4tr' Figurel5.26

Ihe solution to Example 15.5.

448

=

-41

15.5 Mernnr CoupurerroNs wrrH MAILAB The formulation of many engineering problems leads to

a

449

qrstem of dgebraic equations. As you

will learn later in your math and engineering classes, there are a number ofwaln to solve a set of linear equations. Solve the following set of equations using the Gauss elimination, by invening dte [r4] matrix (the coefficients of unknowns), and multiplying it by i6] matrix (the values on the right hand side of equations. The Gauss elimination method is discussed in detail in Section 18.5. Here, our intent is to show how to use MAIT-AB to solve a set of linear equations.

2qI x2* 4:13 3x1*2x2*4xu=i2 5q-x2t34:17 For this problem, the coefficient matrix [r4]

"nd

the right-hand side matrix

Ibl

are

",:l', ii].,,u,,, ={1i\

I(e will first use the Mltil,AB matrix left division operator \ to

solve this problem. The

\

operator solves the problem using the Gauss elimination. W'e then solve the problem using the inv command.

To get started, select "MATLAB Help" from the Help menu.

>>A

=

12 7.

Li3 2 4;5 -1 3l

11 IJ

24 -l_ 3

J

>>b = lL3t32iL71

1-\ -

l_

-t

JZ L7 >>

X=

x = .4,\b

2.0000 5.0000 4.0000 >>x = invla'1 *5 X= 2 .0000 5.0000 4.0000

450

CHaprsn

15

MAILAB

:

:

Note that if you substitute the solution xr: 2, x2 5, and, x3 4 into each equation, you find that they satisfy them. That 2(2) + 5 + 4= 13,3Q) + 2(5) + 4(4) 32, and

k

:

5Q)-5+3(4)=17. l----lr.-:--------- -- ---- ---- :-----il ------r.r:i-

15.6

-----"----..:--.------l

Curve Fitting with MATLAB In Section 14.8, we discussed the concept of curve fitti"g. MAII-A.B offers a variety of curve' fitting options. V'e will use Example L4.ll to show how you can also use MAlf-AB to obtain an equation that closely fits a set of data points. For Example 14.11 (Revisited), we will use the command POLYFIT (x, y, n) , which determines the coefEcients (ca, c1, c21 . . . t e) of a polynomial of order a that best fits the data accordrng to:

!= cox'*

c1x"-l

*

c2xo-2

*

car"-3

* "' *

co

Find the equation that best fits the following set of data poins in Thble 15.13. In Section 14.8, ploa of data points revealed that the relationship betrveeny and x is quadradc (second order polynomial). To obtain the coefficients of the second order polynomial that best fie the given daa, we will rype the following sequence of commands. The MAII-A,B Command Window for Example 14.11 (Revisited) is shown inFigte 15.27:

>>format compact

))x=0:0.5:3 ))! = t2 O.75 0 -0.25 0 0.75 2J >> COefficients = poLtrfit (x,y,2)

0.00

2.00

0.50

4.75

1.00 1.50

0.00

2.00 2.50 3.00

-0.25 0.00 Q,75

2.04

I

Figuru

15.27

T[e Comnand Window for Example

l4.ll (Rwhited).

15.7 Upon execution of the d0 :

iji 1i

,l

1, c1 : -3t

polyfit

and c2

=

Snvmor.rc M.mrBuerrcs wrrH

command,

MAILAB

Mdfl-{B will return the following

2, which leads to the equationy

:

x2

-

451

coefficients,

3x -t 2.

.''..-.'.-''

fl5.7 Symbolic Mathematies with

MATLAB

use MA|I.AB to solve engineering problems with numerical values. In this section, we briefy explain the syrnbolic capabilities of MATLAB. In rymbolic mathematics, as the name implies, the problem and the solution are presented using symbols, such as x instead of numerical values. \Wb will demonstrate MA|LAB's symbolic capabilities using F.xamples 15.7 and 15.8.

In the previous sections, we discussed how to

\7e will use the fo[owing functions to perform MAn-A.Bt qymbolic operations,

as

Thble 15.14.

f(*):*-5x+6 rtQ): *- t

: (x+ 5)2 fsk):5*-t-r2x-Y

fi(r)

I

Function

Description of the Function

ExamFle

It

Flx = qfm('x^2-5\+6') F2x = sfm('x-3')

creates a

sfmbolic

function.

'When

F3x F4x possible,

it facorizes

the function into simpler terms. simplifies the funcdon.

simplify

It

expand co11ect,

It expands the function. It simplifies a s)'mbolic expression

solve

It

aryn]

It

vavlve

nt- \| f!,

min,

max)

by collecting like coefficients. solves tfie orpression for its roots. plots the function f in the range of min and max.

= sfm('(x+5)^2') = sym('51-y+2\-y')

Flx = x^2*5*x*6

F2x: x-3 = (x*5)^2

p31

F4x=

5\-y+2\-y

faao(Fxl)

(x-2).(x-3)

simplifr(Flx/F2x) expand(F3x)

x-2 x^2*10\*25

collea(F4x)

7*x-2*y

solve(Flx)

x=2andx=3

ezplot(F1x,0,2)

See

Figure 15.27

shown in

452

Cru.prnn

15

MAil-AB

t

Figuru

15.28 lhe ezplot

for Erample l5J;

see lhe last row in

labh

1514.

-::-:-:l--.-'I

Solutions of Simuftaneous Linear Equations In this secdon, we will show how you to a set oflinear equations.

c:rn use

MAII-ABI symbolic

Consider the following three linear equations with three unknowns:

2x*y*z=t3 3xt 2y * 4z:32 5x- !-t3z=t7 In MAILAB, the solve equations. The basic form

solvers to obtain solutions

I

y, arrdz

to obein solutions to symbolic algebraic of the solve command is solve('eqal r, teqn2t,

command is used

. , 'eqnr ) . fu shown below, we define each equation first and then use the goLve command to obtain the solution.

>>equatioa 1 = '2*x+y+z=13'i >>equatJ.on 2 = I 3*2i1!*y+4*z=32' i >>equat,Lon-3 = '5*x-y+3*z=L7' i >>[x,Yrz] = golve(equatl-on l,equatlon 2,equation 3)

15.7 The soludon is given

by, :

2,

!:

Svmnor,rc

5, and a

:

Menrpuencs wrrn MAILAB

453

4. The MAIL"EB Command Vindow for

Example 15.8 is shown in Figure 15.29.

H '::

Figure

15.29 I[e

:--:1r: -:

solution of the set of linear equations discussed in Eranple 15.8.

ll

As we said at the beginning of this chapter, there are manygood textbools that discuss the capabilities of MAIf-AB to solve a firll range of problems. Here, our intent was to introduce only some basic id6as so that you can perform some essential operadons,'fu you continue your engineering education in other classes, you will lsarn more about how to use MAILAB effectively to solve a wide range of engineering problems.

454

CrrnPrsR

15 MAILAB

Pnonr-eMs

455

SUMMARY Now that you hane reached this point in the text you should:

.

MAILAB is a tool that can be used to solve engineering problems. Moreover, you MAILAB to present the results of an analysis in chan form. You can input your own formulas or use the buih-in funcdons provided by MAII,AB. know that can use

o know how to edit the content of a MIILAB file.

.

be familiarwith MAILABI built-in functions. o know how to import data files into MAILAB.

"

. . .

r5J.

Using the

know how to creirte a proper engineering chart using MAILAB. know how to perform mauix computations with MAIIAS. be

familiarwith MAII-A.8I curve-fiting capabilities.

be familiar with

MAILABI symbolic mathematics capabilities.

MAfLAB Help menu, discuss howthe fo[-

lowing functions are used. Create a simple example, and demonsuate the proper use of the function. a. ABS (X)

b. TIC, TOC

c. SIZE (x) d. FD( (x) e. FLOOR

(x)

f.

CEIL (x) g. CALENDAR r5.2.

Create a table that shovss the water pressure in lb/in'z in a pipe located at the base of the water tower as you

varythe heightofthewater in incremen$ of 10 ft.Also, plot the water pressrue 0b/in1 versus the height of water in feet. V'hat should the water level in the water tower be to create 80 psi of water pressrue in a pipe at the base of the water tower? 15.3. As we explained in Chapter 10, viscosit,' is a measure of how easily a fuid fows. The viscosity of water can be determined from the following correlation.

In

Chapcer 10, we discussed fluid pressure and the role of water towers in small towns. IJse MAILAB to create a able that shows the relationship berween the height of water above ground in the water tower and the water pressure in a pipeline located at the base of the water tower. The relationship is given by

p: :

p

:

g: h:

/-:-\ : ,r1g\T-"/

where ;r,

=

viscosiry (Nls. rn'z)

T = temperature (IQ ct=2.414x 10-5NA.m2

pgh

where

P

11

c2= 247.8K water pressure at the base of the water tower in pounds per square foot (lbiff)

density ofwater in slugs per cubic foot (p 1.94 slugs/ff) acceleration due to gravity (S

= 32.2 ftl*)

height ofwater above ground in feet (ft)

:

cz= l40K Using MAIT-A3, create a able that shows the viscosas a function of temperature in the range of 0'CQ73.15 K) to 100"C(373.15 K) inincrements of 5'C. Also, create a graph showing the value of viscosity as a function of temperature.

ity of water

456 15.4.

CHePrsR

15 MAILAB

Using MAII-AB, create a table that shows the relationship between the unia of temperature in degrees Celsius and Fahrenheit in the range of -50oC to

The frontal area./4 represents the frontal projecdon area and could be approximated simply by multiplying 0.85 times the width and the height of a rectangle that oudines the ftont of the car. This is the area that you see when you view the car from a direction normal to the front grill. The 0.85 factor is used to adjust for rounded corners, open space below the bumper, and so on. To give you some idea, rypi"d drag coefficient values for sports c:us are between 0.27 to 0.38 and for sedans are between 0.34 to 0.5. The power requirement to overcome air resistance is computed by

of the cart

150"C. Use increments of 10"C. 15.5.

t5.6.

15.7.

t5.8.

t5.9.

1510.

Using M,{ILAB, create a table that shows the relationship among units of the height of people in centimeters, inches, and feet in the range of 150 cm to 2 m. Use increments of 5 cm. Using MAILAB, create a table that shows the relationship among the unim of mass to describe people's mass in kilograms, slugs, and pound mass in the range of 20I

iE

Integral Facorialr forexample,

!

Not equal to

i


, ,,

,

*f: !?q

it' i*-,"

Frequency Midpoint Rarye

fn

50-59 6A-69

3

7A-79

80-89 9A-99

i

tc-

)4.>

74.9

64.5

74.9

9

//+.>

6

84.5

74.9 74,9

3

94.5

74.9

tc

(*

- 7)'f

*20.4 1248.5 540.8 -10.A *0.4 1.44 552.96 9.6 19.6 1152.5 i ' E(r-V)2f=34g6

Using Equation (I9.7), the mean of the scores is

>("/)= re47 = 74.9 -i: -T 26 Similarly, using Equation (19.8), we cdculate the sandard deviation, as shown in Table 19.8.

n: 2f: n-t:25

26

It4%

\25

------. l-,,11'rarit..I,-

Normal disuibution is discussed next.

11.8

-.---l .-: .:f.rli::,:lli.',i.I

- --

rtii

:,

-

--

_:l

19.5 NonrvrarDrsrnrsurroN

587

19.5 NormalDistribution In Section 19.1, we explained what we mean by a statistical e"Feriment and outcome. Recall that the result ofan experiment is called an outcome. In an engineering situation, we often perform enperiments that could have many outcomes. To organize the outcomes of an experiment, it is customary to make use of probability distributions. A probability distribution shows the probability values for the occurrence of the outcomes of an ."T'eriment. To better undersand the concept of probability disuibution, let's turn our affenuon to Example 19.2. Ifwe were to consider the chemistryrcst as an experimentwith outcomes represented bystudent scores, then we can calculate a probability value for each range of scores by divifing each frequency by 26 (the total number of scores). The probability disuibution for F-:rample L9.2 is given in Tirble 19.9. From examining Tirble I9.9, you should note ttrat the sum of probabilities is 1, which is rue for any probability disuibution. The plot of t}re probability distribution for Example 19.2 is shown in Figure 19.4. Moreover, if this was a grpical chemistry test with rypical

Range

50-59

60-69 70-79 80-89

90-99

Probabilitv

Frequency 3

0.1 15

26

0.192

26

1

9

0.346

26 5

0.231

26 3

0.115

26

}p=l

h

0.250 0.200

P

0.150 0.100 0.050

D

Figurelg.4

Plot of probability distribution

for Elample

19.2.

0.0m

i

:

1

I

588

CH^IITSR

19

PnosABrlrryeNo Sransrrcs rN Er.IcrNEsRrNc students, ttren we might be able to use the probability distribudon for this class to predict how srudenrs might do on a similar test next year. Often, it is difficult to define what we mean by a typical class or a tfpical test. However, ifwe had a lot more students take this test and incorporate their scores into our analysis, we might be able to use the results of this eirperiment to

predict rhe ourcomes of a similar tesr to be given later. As the number of students taking the tesr increases (leading to more scores), the line connecting the midpoint of scores shown in Figure 19.4 becomes smoother and approaches a bell-shaped curve.'We use the next example to funher explain this concept.

In order to improve the production time, the supervisor of assembly lines in

a manufacturing sening of computers has studied rhe time that it takes to assemble ceftain parts of a computer at various stations. She measures the time that it takes to assemble a specific part by 100 people at different shifts and on different dqrs. The record of her study is organized and shown in Thble 19.10.

TABLE

I

1

19.19

Dgta Pgrtaininq to Example 19.4

TimeThatlt t"ker aPenson

itoAssemblethe

.

Part (minutes) Frequency

Probability 0.05

5

8

0.08

I

7 8

II

0.I

t5

0.15

9

17

0.17

10

t4

o.r4

1l

t1

0.13

l2

8

0.08

t3

6

0.06

r4

3

0.03

Based on data provided, we have calculated the probabilities corresponding to the time intervals that people took to assemble the parts. The probability distribution for F-xample 19.4 is shown in Thble 19.10 and Figure 19.5. Again, note that the sum of probabilities is equal to 1. Also note that if rve were to connect dre midpoints of time results (as shown in Figure 19.5), we would have a curve that approximates a bell shape. As the number of data points increases and the intervals decrease, the probability-distribution curve becomes smoother. A probability distribution that has a bell-shaped cun'e is called a nornaal distribution. The probability distribution for many engineering experiments is approximated by a normal

distribution.

19.5 NonruarDrsrnrsurroN

589

0.18 0.16 0.1"4

o.L2 0.10 0.08 0.06 0.04

I

0.02

tigurelg.S

0.00

8910rL Time (minutes) ---

.--.--.----tdrnn

The detailed shape of a normal-distribution curve is determined by its mean and standard deviation values. For example, as shown in Figwe 19.6, an enperiment with a small standard deviation will produce a tall, narrow curve; whereas a large sandard deviarion will result in a shon, wide curve. However, it is important to note that since the normal probabiliry distribution represents all possible outcomes of an experiment (with the total of probabilities equal to 1), the area under any given normal distribution should always be equal to 1. Also, note normal disaibution is qfmmeuical about the mean. In statistics, it is custornary and easier to normalize the mean and the sundard deviation values of an enperiment and work with what is called the standard norrnal dis*ibutioa which has a mean value of zero (x : 0) and a standard deviation value of 1 (s = l). To do this, we define what commonly is referred to as a z score acrr,rding to

z:- x-V

fi9.e)

J

In Equation (19.9), arepresenc the number of sandard deviations from the mean. The mathematical function that describes a normal-distribution curve or a standard normal curve is ratfier complicated and may be beyond the level of your qurent understanding. Most of you will learn about it later in your statistics or engineering classes. For now, using Excel, we have generated a table that shows the areas under ponions of the standard normal-disaibudon curve, shown in Thble 19.1 1 . At this stage ofyour education, it is imponant for you to know how to use the table and solve some problems. A more detailed explanation will be provided in your future classes.

f

Ihe shape of a nonnal dishibntion curve

[y its

will next demonsffaJe how to

use

Thble 19. 1 I , using

A

Figurelg.6

determined

'We

as

mean and

standard deviation.

/\

-/*:""\ Small standard

large standard

deviation

deviation

a

number of example problems.

iiii

ffi

Note that the standard normd curve is qmrmetrical about the mean.

z=1.@

Mean:0

z =3.N

z=2.@

t:i o

e t

0.0000 i O,AL 0.0040 i, a.oz 0.00s0 1 ,0.03 0.0120 i ,'A.M 0.0160 1 i0.05 0.0199 10.06 a.0239 iit':a.97 0.0279 I ]O.OS 0.0319 1

i

Z

A

0.51 0,1950 r.o2 a.3461

O.52 0.1985 r.03 0.3485 0.53 0.2019 La4 a350s 0.54 0.2054 1.05 0.3531

Z 1.53

r.54 t.55

t,56

0.2088 r.06 0.3554 1.57 0.2123 r.07 a3577 1.58 0.2157 r.08 0.3599 t.59 0.2190 1.09 0.3621 1.6 0.2224 1.1 0.3643 1.6r i ioos 0.0is9 0.6 0.2252 r.rl 0.i665 r.62 i lO.r 0.0398 0.61 0.2291 r.r2 0.3686 1.63 i)t ,0.11 0.04i8 a.62 0.2324 t,r3 0.3708 l.g* ,o.tz 0.a478 0.63 0.2357 L.r4 0.3729 r.65 r 0.13 0.0517 O.g+ 0,2i89 r.15 0.3749 r.65 t o.r4 0.0557 a,65 0.2422 1.16 0.3770 r.67 i o.t5 0.0596 0.66 0.2454 r.r7 0.3790 1.6s i .,o.ta '0.06i6 0.67 0.2486 t.tg 0.3810 1.69 ',,,,a.L7 '0.0675 0.68 0.2517 1.r9 0.3t30 t,7 i.r0.r8 ,0.A714 0.69 0.2549 rA 0.3849 r.7r i ',0.19 0.0753 oJ 0.258a L.zr 0.3869 lJz ', 0a' :,0,029s o,7l 0.26il 1.22 0.3s88 L:73

0.55 0.56 o,57 o.58 0.59

A

0.4370 0.4382 0.4394 0.4406 0.4415 0.4429 0.4441 0.4452 0,4463 0.4474 0.44s4 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 0.4554 0.4564 0.4573 0.4582

z

Ai

A

z

2.O4 0.4793 '::2,55 0.4946 ,.a6 2.O5 0.4798 2.56 0.4948 ?,O7 2.06 0.48A3 '2;57 0.4949 3,A8 2,O7 0.4805 ,2,58 0.4951 3.09 2.A8 0:4812 2.59 0.4952 3.L 2-09 0.4817 2.6 0.4953 3.r1 2.t 0.4821 2.61 0.4955 3.r2 2.rr 0..4826, 2.62 0.4956 3.r3 2,t2 0;483A " 2161' 0.4957 !.L4 2.13 0;4814' \& 0.4959 3.15 2,14 0.4s35 2,65 0.4960 3.16 2.L5 0:4842 ,',2.66 0.4961 5.L7 2.16 0,4s46 2,67 ' 0.4962, 3.18 2,17 0,;4850' '2.68, 0.4963,' 3.r9 2.18 0,4854 2.69 0.4964 3.2 2.t9 0,4857 :2:7 0.4965 3.21 22 0,4861 2.7t ' 0.4966 3.22 2,2t 0,4864 2.72 0.4967 3.23 2;22 0.4s65' 2.73 0.4968 324 223 0,4871 ',2,74 0.4969 3,25 2.24 0.4875 2J:5 0.4970, ,.26

0.4989 0.4959 0.4990 0.4990 0.4990 A.4991 0,4991

0.4991 0.4992 0.4992 0.4992 0:4992 0.4993 014993

0.4993 0.4993 0.4994 0.4994 0.4994 0,4994 0.4994 Continucd

590

19.5 Nonruar DrsrnrsurroaT

o.2t 0.0832 0.72 022 0.0871 0,73 o.23 0.091A O,74 0.24 0.0948 0.75 0.25 0.0987 0.76 0.26 0.1026 0,77 a.z7 0.1064 0.7& 0.28 0.1103 0.79 a.29 0.1141 0.8 0.3 0.1179 0.81 0.31 0.1217 0.82 o.32 0.1255 0.83 0.33 0.1293 0.84 a.3.4 0,1i31

0.85

o.35 0.1368 0.86 0.36 0J4A6 0.87 a$7 0.1443 0.88 0.38 0.1480 0.E9 439 4.15t7 0.9 o.4 0.1554 0.91 o.At 0.1591 0.92 0.42 0.1628 0.93 o.43 0.1664 0.94 o.u 0,1700 a.95 a,45 0.1736 0.96 0.46 0.1772 0.97 o,47 0.1808 0.98 0.48 0.1844 0.99 o.49 0.1879 I 0.5 0.1915 1.0r

a.2642 0.2673 0.27A4 0.2734 0.2764 0.2794 0.2823 0.2852 0.2881 0.2910

L,23

t.24

L25 L.26

r.27 1.28

L29 L.3

t,'r 1,32

0,29i9

L.33

0.2967 0.2995 a.3023 0.3051 0.3078

t.Yt

0.ir06

r,35 t.36 r.37 r.38 1.39

0.3133 L,4 0.3159 r.4l 0.3186 1.42

0.j212

r.43

0.3238 t.M 0.3264 r.45 0.3259 t,46 0.3315 L.47 0,3340 1.48 0.3365 t.49 0.3389 t.5 0,3413 1.51 0.3438 t.52

0.3907 t,74 0.3925 r.75 0.3944 rJ6 0.3962 r,77 0.3980 1.78 0.3997 L:79 0.4015 1.8 0.4032 1.8r 0.4049 t.82 0.4066 1.83 0.4082 t.84 0.4099 1.85 0.4115 1.86 0.4131 r.87 0.4147 1.88 0.4162 1.89 0.4177 r.9 0.4192 1.91 0.42A7 L92 0.4222 19t 0,4236 t.94 0.4251 t.95 0.4265 r.96 a.4279 r.97 0.4292 1.98 0.4306 r.99 0.4319 2 0.4332 2.Or 0.4345 2.02

a.457

2.O5

0.4591 0.4599 0,4608 0.4616 0.4625

2.29

0.463i

2.3

0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 0.4713 0.4719 0.4726 0.4732

0.47i8

2.25 2,26

2,27 2.28

2.3t 2.32 2.33

2.34 2.35

2.36 2.37

238 2.39 2.4

2.4t 2.42 2.43

2.4* 2.45

0.4744 2.46 0.4750 2.47 0.4756 2.4 0.4761 2.49 0.4767 2.5 0.4772 2.5r 0.4778 2.52 0.4783 2.53 0.4788 2.54

591

2J6 0.4971 ?.27 0.4995 2,77 0.4972 3,28 0.4995 2.78 0,4973 3.29 0.4995 2.79 0.4974 33 0.4995 2,8 0.4974 3.?t 0.4995 2.8r 0.4975 3.32 A.4995 2.82 0.4976 3.33 0.4996 2.83 0.4977 3.34 0.4996 2.84 0.4977 3.35 0.4996 2.85 0.4978 3.36 0.4996 2.86 0.4979 3.37 0.4996 2.87 0.4979 3.38 0.4996 0.49r t 2.88 0.4980 3.39 0.4997 0.4913 2.89 0.4981 3.4 0.4997 0.4916 2.9 0.4981 3.4r 0.4997 0.4918 2.9r 0.4952 3,42 0.4997 0.4920 2.92 0.4982 3.43 0.4997 0.4922 2.93 0.4983 3.4* 0.4997 0.4925 2.94 0.4984 0.4997 '.45 0.4997 0.4927 2.95 0.4984 1.46 0.4929 2.96 0.4955 3,47 0.4997 0.4878 0.4881 0.4884 0.4887 0.4890 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909

0.4931 2.97 0.4985 5,48 0.49i2 2.98 0.4956 3.49 0.4934 2.99 0.4986 3.5 0.4936 0.4987 3,5r ' 0.4938 3.0r 0.4987 3.52 0.4940 0.4987 3.53 '.O2 0.4988 0.4941 3.O3 0.4943 3.04 0.4958 0.4%5 3.05 0.4989 3.9

0.4997 0.4995 0.4998 0.4998 0.4998 0.4998

0.5000

Using Table 19. 1 1, show tiat for a standard normal distribution of a data set, approximately 68%o of the data will fall in the interval of -s to s, about 95olo of the data falls between -2s to 2s, and approximately all of the data points lie benreen -3s to 3r. In Thble t9.ll, z: 1 represents one standard deviation above the mean and 34.13o/o of the total area under a sandard normal curve. On the other hand, s, : -I represen$ one sran; dard deviation below the mean and 34.73o/o of the total area, as shown in Figure 19.7. There, fore, for a standard normal distribution, 6870 of the data fall in the interval of a : - I to g = 1 (-r to s). Similarly, z: -2 and z = 2 (nvo standard dwiations below and above the

s92

Cnepren

19

PnosABrr-rrv.oxo Srarrsrrcs nr ENcnvssRrNc mean) each represent 0.4772o/o of the toal area under the normal curve. Then, as shown in Figure 19.2 95o/o of the data fall in the interval of -2s to 2s. In the same way, we can show

rhat99,7o/o(forz= -3then.r4 =0.4987and.2:3thenA:0.4987) ordmostallofthe

data points lie between

-3s to 3s. A -- 0.4772 + 0.4772*0.95

A=03413+0.3413=0.68

z:-1.00 e:-1.00 e=1.00 e=1.00

;

tigure

19.7

The area under a norlnal curve

for

z=-2.00

z

=2.ffi

Example 19.5.

For Example 19.4, calculate the mean and standard deviation, and determine the probability that it will take a person between 7 and, ll minutes to assemble the computer parts. Refer to Table 19.12 when following solution steps.

r Time r (minuts) ;x

Frequency

*f

tc- fr

5

25 48

-4.22

89.04 82.95

77 r20

-2,.22

-t.22

54.2t' 22.33

x

6 7

- v)'f

f

1l

8

(*

-4 )1

9

1'7

r53

-Q.22

a.82

10

r4

140

1l t2

L3

r43

0.78 1.78

4r.r9

B

96

14

6

t4

3

78 42

2.78 47R 4.78

2*f=gzz

-x: 2*f n -:

K:9.22

minutes

': ,ffi:'{#:2'28minutes

8.52

6r.83

.

s5.73 168.55

2(x-x)2f=515.16

19.5 NonuelDrstnnrrrron

593

The value 7 is below the mean valae (9.22) and the z value corresponding to 7 is deter-

minedfrom

x-V 7-9.22 ,: --;-: --rE-:

-o.97

From Table l9.ll, A = A334O. Similarly, tlre vdue 1 1 is above the mean value and the sscore corresponding to 11 is computed from

It - 9.22 :0,78 z: x-V t 2.28 -: From Thble l9.ll, A= 0.2823, Therefore,

the probability that it will take a penon be+ 0.2823 0.6163 as shoqrn

tween 7 and I L minutes to assemble the compffer paft is 0.3340

:

in Figure 19.8.

A

z--0.97

I

:

03340 + 0.2823 = 0.6t63

z=0.78

figuretg.g

Area under tfie probability dishibuthn curtte for Emtnple 19.6.

For Example 19.4, deterrnine the probability that it will take a person longsr than 10 minutes to assemble the computer parts. For this problem, the .6score is

x-V ro-9.22 -Zff:0.34 =;=: ": From Thble 19.11, A = 0.1331. Since we wish to determine the probabiliry that it tak"s longer than 10 minutes to assemble the part, we need to calculate the area 0.5 - 0.1331 : 0.3669, as shov"n in Figure 19.9. The probability t6o il lyill take a person longer than 10 minutes to assemble tfre compurer pan is approximately 0.37.

594

Cneprsn

19

Pnonasrr,rrverp Sterrsrrcs

rN ENcrrvnsR$Ic

A:0.L331"

z = 0.34

I

Figurelg.g

Areas under the probability distribution curve for Erample 19.7. ril.

In closing, keep in mind that the pqpose of this chapter was to make you arvare of the importance ofprobability and statistics in engineering, not to provide a detailed coverage ofstatistics. As you ake setistics classes and advanced classes in engineering, you will learn much more about stadsticd concepts and models.

SUMMARY Now that you have reached this point in the text n You should undersand the imporant role of statistics in various engineering disciplines. . You should be familiar with basic probability and statistics terminologies. . You should have a good underssnding of frequency distribution and cumulative frequenry disnibution and what kind of information they provide. . You should have a good grasp of statistical measures of cenual tendenry and variation. . You should know how to compute basic statistical information such as mean, variance, and sandard deviation for a set of daa points. . You should know what is meant by normal disuibution and standard normal distribution. . You should know how to use Thble 19.1 1.

of a test for an

engineering class of 30 students is shown here. Organize the data in a manner similar to Thble 19.1 and use Excel to creare a

l9J. The

scores

19.2.

t9.3.

histogram.

For Problem 19.1, calculate the cumulative frequency and plot a cumulative-frequency polygon. ForProblem 19.1, usingEquations (19.1) and (L9.6), calculate the mean and sandard deviation of the class scores.

Scores: 57, 94, 8I, 77, 66, 97, 62, 86, 7 5, 87, 9I, 7 8, 61, 82, 74,72,70, 88, 66,75, 55, 66, 58, 73, 79,

51,63,77,52,94

t9.4.

For Problem 19.1, using Equations (19.7) and (19.8), calculate the mean and standard deviation of the class scores.

PnosLEnIs 595 r9.5.

For Problem 19.1, calculate the probability distribu-

19.9.

tion and plot the probability-distribution curve. r9.6.

In order to improve the production time, the supervisor of assembly lines in a manufacturing setting of cellular phones has studied the time that it akes to assemble cerain parts of a phone at various stations. She

| i

2 x 4 Lumber Steel Spherical Balle (cm) Width (ir.)

The record of her study is organized and shown

3.50

3.40

1.00 0.95 1.05 1.10 1.00 0.90 0.85

20

3.65 3.35

0.95

28

3.60

0.90

in the accompanying able.

3.55

Time that it takes a person to assemble the part (minutes)

Frequency

4

r5

5

6

are

]':-"""1

measures the time that it akes to assemble a specific pan by 165 people at different shifts and on different days.

Determine the average, variance, and standard devia-

tion for the following parts. The measured values given in the accompanying table.

3.45 3.60 3.55 3.40

',

I

1.05

34

191.

t9.8.

8

28

9

24

10

L6

Plot the data and calculate the mean and sandard deviation. For Problem 19.6, calculate the probabiliry disuibution and plot the probability-disuibution curve. Determine the average, variance, and standard dwi-

ation for the following parr. The measured values are given in the accompanying able.

Screw l-ength

(cm)

19.X0.

ance, and standard deviation for the cereal boxes. Does

the manufacturer's information noted on the box fall

within your meruurement? l9.ll. 19J2.

Pipe Diameter (in )

t.25

2.55 2.45 2,55 2.35

1.18 1.22 1.15

2.60

r.l7

2.40 2.30

L.r9

2.40 2.50

1.18 1 la

2.50

t.25

The next time you make a trip to a supermarket ask the manager if you can measure the mass of at least 10 cereal boxes ofyour choice. Choose the same brand and the same size boxes. Tell the manager this is an assignment for a class. Repon the anerage mass, vari-

Repeat Problem 19.10 using tluee other products, such as cans ofsoup, tuna, or peanu6. Obtain the height, age, and mass ofplayers for your favorite professional basketball team. Determine the average, variance, and sandard deviation for the height, age, and mass. Discuss your findings. Ifyou do not like basketball, perform the experiment using daa from a soccer team, football team, or a sports team of your choice.

r9J3.

For Example 19.4, determine the probability that it will take a person benrreen 5 and 10 minutes to assemble the computer parts.

1914.

For Example 19.4, determine the probabiliry that

will take a person longer than 7 minutes to

1.22

it

assemble

the computer parts. 1915.

For Problem 19.6 (assuming normal distribution), determine the probability that it will take a person between 5 to 8 minutes to assemble the phone.

596 Class

Csaprcn

PnosABrlrrr.al.tn Sranrsrrcs rx ErvcrxnrnrNc

are

pafonned in

class.

Your instructor will pass along an unopened bag of Hershey's kisses. You are to estimate the number of kisses in the bag and write it down on a piece of paper. Your instructor will then collect the daa and share the results with the class. Your assignment is to organize the data per your insructor's suggestion and calculate the mean and standard dwiation. Compute the prob-

ability distribution. Does your data disuibution ap-

19.17.

proximate a normal distribution? Answer any additional questions that your instructor might ask. Your instructor will ask for a volunteer in class. You are to estimate his or her height in inches (or in cm) and write it down on a piece of paper. Your instructor will then collect the daa and share the results with the class. Your assignment is to organize the data per your

instrucort suggestion and calculate the mean and sandard deviation of the data. Compute the probability

1918.

a normal disuibution? Answer any additional ques-

Experi.ments-Problems I 9. I 6 througb I 9.20 are acpn-

immx that 19.16.

19

distribution. Does your data disuibution approximate a normal disuibution? Answer any additional quesrions that your instructor might ask Your instrucor will ask for a volunteer from class. You are to estimate his or her mass in lb. (or in kg) and write it down on a piece of paper. Your instructor will then collect the data and share the results with the class. Your assignment is to organize the data per your insmrctor's suggestion and calculate the mean and sandard dwiation of the data. Compute the probability disuibution. Does your data disuibution approximate

19.t9.

$"20,.

t9.2r.

tions that your instructor might aslc You are to write down on a piece of paper the number ofcredits you are taking this semester. Your irutructor will then collect the data and share the resuhs with the class. Calculate the mean and standard deviation of the data. Assuming a normal distribution, determine the probability that a student is taking benreen 12 to 15 credits this semester. !?hat is the probability that a student is taking less than 12 credits? You are to write down on apiece of paper how much (to the nearest penny) money you have on you. Your instnrctor will then collect the data and share the results with the class. Your assignment is to organize the daa per your instructort suggestion and cdculate the mean and scandard deviation of the data. Assuming a normal distribution, determine the probability that a student has between $5 to $tO.'What is the probabiliry that a student has less than $10? You are to ffiite down your waist size on a piece of pa-

per.

If you don't know your waist size, ask your in-

structor for

a

measuring tape. Your instructor will then

collect the dam and share the results with the class. Your assignment is to organize the data per your instruccort suggestion and calculate the mean and standard deviation of the data. Assuming a normal disuibution, determine the probability that a student will '\V'hat have a waist size that is less than 34 inches. is the probability that a student will have a waist size that is between 30 in. to 36'n.?

CHAPTER

ENcTNEERTNG

|r lr r

conomic considerations play

20

EcoNoMrcs

a

vitat role in product and service development and in engineering

design decision making pfocess.

597

598

Cnepren20 ExcrxnrnrNcEconourcs

in Chapter 3, economicfactors alwals play important roles i.n mgineering daign decision making. Ifyou drsigrc a product that is too erpmsiae to rtantr.faaure, then it cAn not be sold at a price that consurners can afford and still be

As we explained

prof.table to your colnpanJ/. Thefact is that companies design producx andprouide smtices not only to rnahe our liaes bener but ako to make rnone/ In this section, we will discuss the basics of engineering econornics. The information proaidcd here not only applies to mgineering projecx bwt can ako be applied to financing A car or A house or borcowingfrorn or inaesting mons! i.n bank* Some of you rnd! want to apply the knowledge gained here to dzterrnine lour srudent ban payrnmts or lour credit card paltments. Therefof€, t!)e aduise you to deuelop a good und.erstanding of mgineering economics; the information presmted here could help you manage yoar rnone! more wisely.

20.1 Gash Flow Diagrams Cash flow diagrams are visual aids drat show the flow of costs and revenues over a period of time. Cash flow diagrams show whm the cashfuw occurs, tlte cashfuw magnitudc, and' whaher the cash fnw is out ofyur pocha (cost) or into your pocha (reumae). It is an imponant visual tool that shows the timing, the magnitude, and the direction of cash flow. To shed more light on the concept of the cash fow diagram, imagine rhatyouare interested in purchasing a new car. Being a first-year engineering student, you mqf not hane too much money in your sarings account at this time; for the sake of this example, let us say that you have $1200 to yotu name in a sarings account. The car that you are interested in buying costs $ 15,500; let us further assume that including the sales tax and other fees, the total cost of the car would be $16,880. Assumingyou can afford to put down $1000 as a down payment for your new shiny car, you askyour bank for a loan. The bank decides to lend you the remainder, which is $15,880 at 87o interest. You will sign a contract that requires you to pay $315.91 every month for the next five years. You will soon learn how to calculate these monthly payments, but for now let us focus on how to dravr the cash fow di"Sr-. The cash fow diagram for this activity is shown in Figure 20.1 . Note in Figure 20.1 the direction of the arrows representing the money given to you by the bank and rhe payments that you must make to the bank over the next five years (60 months).

$1s,880

I

Figure2o.t

A cash flow diagram

for honowed

money and the monthly payments.

$315.91

20.2

Srtvrpr,s AND

CoupouNp [NrsREsr

599

cash fow diagram for an investment that includes purchasing a machine that costs $50,000 with a maintenance and operating cost of $1000 per year. It is expected that the machine will generate revenues of$15,000 per year for five years. The expected salvage value of the machine at the end of five years is $8000. The cash flow diagram for the investment is shown in Figure 20.2. Agaun, note the directions of anows in the cash flow diagram. Ve hare represented the initial cost of $50,000 and tfi.e maintenance cost by arrows pointing down, while the revenue and the sdvage value of the machine are shown by arrows poindng up.

Drav the

$15,000

$8000

I

0

2

3

4 $1000

t

Figure?0.Z

T[e cash flow diagram

lor Erample 20J.

$50,000

i,

it--- - ---ii.1rlfi:lii:ilrllltirriinllilf,illi'riitl

ili.i.liri1ii,r,.l:ir',r.::i

---.,:iirlrr,i,rililllii'i.li

20.2 Simple and Compound Interest Interest is the exua money in addirion to the borrowed amount thar one must pa)r for the purofhadng access to the borrowed money. Simph interastis rhe interest that would be paid only on the initial borrowed or deposited amount. For simple interest, the interest accumulated on the principle each year will not collect interest itself. Only the initial principal will collect interest. For example, if you deposit $100.00 in a bank at 6% simple interest, after six years you will have $136 in your account. In general, ifyou deposit the amount P ar a rate of io/o for a period of n years, then the total future valu e F of the P at the end of the ath year is grven by pose

F:P+(PXrXtx):P(r+ni\

(20r)

Compute the future value of a $ 1500 deposit, after eight years, in an account that pqrs a simple interest rate of 7o/o. How much interest will be paid to this account? You can determine the future value ofthe deposited amount using Equation (20.1), which results

in

F:

P(r

*

ni) = r5oo[1 + 8(0.07)]

:

$2340

And the total interest to be paid to this account is

intere*: (p)(n)(i) l:.- ::::- .

:

(1500x8x0.07) = 634s

600

Cno,prsn20 Excnwrnruc Ecoxourcs

Interestfor BalanceattheEnd Bqfnring of the Year (dollars and cents)

t

tfieYear 60lo

at

(dollars

and

oftheYear, Includirg the Interest (dollare

cents)

and cents)

106.00

,

t12.36

'

3

r12.36

6.00 6.36 6.74

4

I 19.10

7.r4

rv,6.24

,

5

126.24

r33.8r

'

6

133.81

7.57 8.02

100.00 106.00

I 19.10

t4 t.65

Simple interests are very rare these days! Almost all interest charged to borrow accounm or interest earned on money deposited in a bank is computed using conqpound interest The concept of compound interest is discussed next.

Compound Interest Under the compounding interest scheme, t-he interest paid on the initial principal will also collect interest. To better understand how the compound interest earned or paid on a principal works, consider the following example. Imagine that you put $100.00 in a bank that pq/s you 67o interest compounding annually. At the end of the first year (or the beginning of the second year) you will have $106.00 in your bank account. You have earned inrerest in the amount of $6.00 during the first year. Howwer, the interest earned during the second year is determined by ($106.00X0.06) : $6.36. That is because the $6.00 interest of the first year also collecs 6%o interest, which is 36 cena itself. Thus, the total interest earned during the second year is $6.36, and the total amount available in your account at the end ofthe second year is $1 12.36. Computing the interest and the total amount for the thfud, fouffh, fifth and the sixth year in a similar fashion will lead ro $141.83 in your account at the end of the sixth year. Refer to Thble 20.1 for detailed calculations. Note the difference between $100.00 invested at 6%o simple interest and 60/o interest compounding annually for a duration of six years. For the simple interest case, the total interest eamed, after six years, is $36.00, whereas the total interest accumulated under the annual compounding case is $41.83 for the same duration.

20.3

Future Worth of a Present Amount Now we will dwelop a general formula that you can use to compute the future value Fof any present amount (principd) 4 after nyears collecting zolo interest compounding annually. The cash

fow

di"S"-

for this situation is shorrn in Figure 20.3.In order to demonsffate, step-by-

step, the compounding effect of the interest each year, Table 20.2 has been developed. As shown

in Thble 20.2, sardng with the principal P, at the end of the first year we will hare P * Pi or P(l + i). During the second year, dre P(l * i) colleca interest in an amount ofP(l + i)i' au;.d. by adding the interest to the P(l * i) amount that we stamed with in the second year, we will have a toal amount of P(L + i) + P(l * i)i. Fae'oling out the P(1 * /) term, we will hare

2O.3 Ftmrnr Wonmr or e PnEseNT Arvrorlxr

I

601

Figure20.3

The cash flow diagram for future

wort[ of a deposit made in the bank today.

Balance at the

Year

Begimiryof

Interest for

theYear

the Year

i1

P

t3 :4 i-

P(l + r) P(t + ilz PU + i)3

ll ,A

5

Balance at the Eqd

P+(P)(i)=P(r+i) P0+i) +P(l + DQ)=P(L+i)z

(P)(t P(t + t)(i) P(r

+

P0 + il2 + P(l + i)2(t) : P(r + il3 P0 + il3 + I{1 + i)3(i): P(L + il4 P$+ il4 +P(l + i)4{i\: ?(1 +rt

ilz?)

P(l+

t)4

p0 + i)'Q) PIL + il4(il

P(r +

i),-l

P(r +

il-t(i)

ofthe Year,

Inclu,f ing the Interest

P(l + i)-L + P(l + i)"-11;1: P(l + i)

P0 + il2 doilars at the end of the second year. Now by following Tirbl e 2A.2 youcan see how the interest earned and the total amounr are computed for the third, fourth, fifth, . . . , and the nth yan. Consequently, you c:rn see that the relationship beween the present worth P and the future value.Fof an amount collecting iolo interest compounding annually after a years is given by

F: P(r + i)"

(20.21

Compute the future value of a $1500 deposit made today, after eight years, in an account that of 7o/o that compounds annually. How much interest will be paid to this

pays an interest rate

account?

P,

i,

The future value of the $1500 deposit is computed by subsdruting in Equation Q0.2) for and zl, which results in the amount that follows:

F= p(L + i)n:1500(t + .07)8 = $2577.27 The total interest earned during the eight-year life of dris account is determined by calculating the difference beween the future value and the presenr deposit value.

L ..,._

inwest:

$2577.27

-

$1500

:

$1O77.27 :--:.-:,--:

,:--

:

:-'::---:-

--r---::-I

602

CHrprsR20 ENcrNsrRrNcEcoxourcs Many financial instirutions pay interest that compounds more than once a year. For example, a bank may pay you an interest rate that compounds semiannudly (mice a year), or quanerly (four times ayqi, or monthly (12 periods ayar).If the principal Pis deposited for a duration of n yars and the interest given is compounded n periods (or zz dmes) Per year, then the future value.Fof the prin"tpal P is determined from

r=n(t.*)

(20.3)

Compute the future value of a $1500 deposit, after eight years, in an account that paln an interesr rate of 7o/o that compounds monthly. How much interest will be paid to this account?

To determine t'he future value of the $1500 deposit, we substitute in Equation (20.3) for P, i, rn and z. The substitution results in the future value shown next.

/ q{Z) = / s.s7\(8)(tz) I':1500[1+ =1500t1+ J 12/ r2l \ -

\

s262r.73

And the total interest is n

interest: $262I.73

iL,,r*,,r.lr.,l,-..,,.

-

$1500

=

$11.21.73 _i ..

:r

_-_.i_1

The results of ExamplCI 20,2, 20,3, and 20.4 are compared and summarized in Table 20.3. Note the effecs of simple interest, interest compounding annually, and interest compounding monthly on the total future value of the $1500 deposit.

Exanple Numbet

D qration

Principal (dollars)

Example 20.2 Example 20.3

1500 1500

Example 20.4

1500

lnterest Rate

(y€ars)

Futrrre Yalue (dollarc and centr)

Interest Earned (dollars and cents)

7olo simple

8

2340,00

840.00

compounding annually 7%o compounding

8

2577.27

1077.27

2621.73

112r.73

7olo

monthly

20.4 Errucrrw

20.4

603

INrsREsr RArs

Effeetive Interest Rate If you deposit $1OO.OO in a savings account, at 60/o compounding monthly, then, using Equation (20.3), at the end ofone year you will have $106.16 in your account. The $6.16 earned during the first year is higher than the stated 60lo interest, which could be understood as $6.00 for a $100.00 deposit over a period of one year. In order to aroid confusion, the stared or the quoted interest rate is called the noninal interest rate, and" the actual earned interest rate is called the effeaioe i,ntercet rate. The relationship benveen the nominal nte, i, and the effective rate, ia,, is given by

t&:

('*t)--'

(20.4)

where rn represents the number of compounding periods per year. To bener understand the compounding effect of interest, let us see what happens if we deposit $100.00 in an account for a year based on one of the following quoted interests: 6%o compounding annually, 67o semiannually, 60/o qwterly,6olo monthly, and 6o/o daily. Thble 20.4 shows the difference among these compounding periods, the total amount of money at the end of one year, the interest sarned, and the effective interest rates for each case. \[hen comparing t]re five different interest compounding frequencies, the difference in the interests earned on a $100.00 investment, over a period of ayear, may not seem much to you, but as the principal and the time of deposit are increased this value becomes significant. To better demonstrate the effect of principal and time of deposit, consider the following orample.

Totel Number of Conpounding Compounding

Period

Periods

TotalAmount after I Year (dollars and cents)

6.00

60/o

10011+ - l:106.09 2/ \

\'

6.09

6.090/o

/ I+--=-l:106.13 o.otr\4

6.13

6.130/o

6.16

6.160/o

6.18

6.180/o

Annually

I

100(1+0.0O:106.00

Semiannually

2

Monthly

D*ily

/

o.o6

1001

Quanerly

4/

\

12

365

lnterest Effective (dollars Interest andcents) Rate

1oo( r

*

Y)": 106.16 roo(r *#)"' = 106.18

604

CneprsR

20

ENcrNErRrNcEcoNourcs Determine the interest earned on $5000 deposited in a savings account, for 10 years, based on one ofdre following quoted interest rates: 670 compounding annually, semiannually, quarterly, mondrly, and daily. The solution to this problem is presented in Thble 20.5.

Period

Periods

Annually

5000(1

10

8954.23

/ *i), o.oe \m =9030.55 / o.oe \& : eo7o.oe 5o0o(1 .; ) 1 0.06\'20 5ooo(1 *i),,:eoe6e8

Qu".t"rly

40

Motrthlv

120

/ *d) o.o6 \ trto , = e11o'14

i

I

+ 0.00,lo:

and cents)

5000(1

Semiannually

ii'

(dollan

UsrngEq. (16,8) (dollars and cents)

Compounding Compounding

Dnily

5ooo[1

3650

':_'

,-.

: :

::.:.::L:.-,-.:-.:,

3954.23

'4030.55

4o7o.oe

4ae6.e8

4710'14

t-'

:,.--

Determine the effective interest rates corresponding to the nominal rates: (a) 7olo compounding monthly, (b) 16.50/o compounding monthly, (c) 6Vo compounding semiannually, (d) 9o/o

compounding quanerly. We can compute the (a)

i.r=

(b)i.n

i"6.

for each

case by

substituting for

/ * ; \- 1 - / *t/o.oz \'2 - I (t \, ;) -

i and m n

E4uation

(20 ,4) .

0.0722or7.22o/o

( o.t6j\'2 | 0.1780 orr7.80o/o = . *. i \, ) - = (t - ,:o.o6oe or6.oeo/o

" ry)'

(d) z"s'

: (t .

?)n - ,:o.oe3o ore.3oo/o iirr-lfifilii.lii.:liirrif.EJ

i

20.6 PnrsmmVoRrrr

20.5 Prescnt

oF Senrss PeuvrsNT

onArvNurrr

605

Worth of a Future Amount

Let us now consider the following situation. You would like to hare $2000 available to you for down payment on a car when you graduate from college in, say, five years. How much money do you need to put in a cenificate of deposit (CD) with an interesr rate o f 6.50/o (compounding annually) today? The relationship between the future and present value was dweloped earlier and is given by Equation (20.2). Rearanging Equation Q\.Z),wehxe a

' --!(r+;)" p=

(20.5)

and substituting in Equation (20,5) for the future value we hane

2N9p ' = -(1 + 0.065)t

:

4

the interest rate

i,

and the period

re,

$1459.76

This may be

a relatively large sum to put aside all at once, especially for a fust-year engineering student. A more realistic option would be to put aside some money each year. Then rhe question becomes, how much money do you need to put aside every year for the next 6ve years ar the given interest rate to hare that $2000 available ro you at the end of the fifth year? To answer this question, we need to develop the formula that deals with a series of payments or series of deposits. This situation is discussed next.

20.6

Present Worth of Series Payment or Annuity In this section, we will firsr formulate the relationship between a present lump sum, 4 and future uniform series payments,r4, and then from ttrat relationship we will develop the formula that relates the uniform series payments,4 to a future lump sum F. This approach is much easier to follow as you will see. To derive these reladonships, let us first consider a situation where we hane borrowed some money, denoted by P, at an annual interest rate i from a bank, and we are planning to pay the loan yearly, in equal amouns l, in n years, as shown in Figure 20.4.

I

Figure20.n

The casi-flory diagram for a

borrowed sum of money and its equivalent seiles payntents.

605

Cnqprsn20

ElvcrrvsERnrc EcoNourcs To obtain the relationship between P andA, we will treat each future payment separately and relate each payment to its present equivalent value using Equation Q0.5); we then add all the resulting terms together. This approach leads to the following relationship:

AAAAA -P--+_ (r+4 (1 +;y*G;t+"'+OiAa*0*ny

Q0''61

As you can see, Equation (20.O is not very user-friendly, so we need to simplify it somehow. 'What ifwe were to multiply borh sides ofEquation Q0.6) by the term (1 + z)? This operation results in the following relationship:

P(t +

A,=+, A,r+"'+ . n.-.+;4* + i)=A+, "' (r !, (t + i)'-z (t + i)"^' + i)' (r + ;)z' (t + 4r

QL.tt

Now ifwe subtract Equation (20.6) from Equation (20.7), we have

P(l

+i)-P=A+-L-+r,A-.+-A * + i) rY1t*

\1

_

1r* o,

+"'+

A L(l + 4' (1 + i)',', (r + ;72'

I a

A

l_l-!-r---I

A

A

O;;l;;* 1t* o'''A A I + (r + i)',-t (l 4"1 I-l

(20.8)

Simplifying the right-hand side of Equadon (20.8) leads to the following relationship:

P(r+i)-P-A-

i

€0.8b)

(r + i)"

And after simplifying the left-hand side of Equation (20.8), we have

P(i) =

A((L+i)"-r)

(20.8c)

Q+i)"

Now if we divide both sides of Equation (20.8c) by i, wehave

tl p=nl\!il'. ' '^L i(r + i)'

ao.,

I

Equation (20.9) esablishes the relationship bewveen the present value of a lump sum P and its equivalent uniform series payments ,4. We can also rearrange Equadon (20.9) , to represent -r4 in terms of Pdirecdy, as given by the following formula:

.l r)' ! 0 + i)' - t: 'L0 + i)" - r)

. P(i)(r + i)" :,l A:

.(t)(r

o.r,

?0.7 Future Worth of Series Payment To dwelop a formula for computing dre future worth of a series of uniform payments, we begin with the relationship betrveen the present worth and the furure worth, Equation (20.2), and then we substitute for P in Equation Q0.2) in terms of

.r4,

using Equation (20.9). This

20.7

Ftrnrns'WonrH or Srnrss PeuvrrNr

607

procedure is demonsuated step-by+tep, next. The relation berween a presenr value and a future value is given by Equation (20.2):

F

= P(r

* i)"

(20.2)

And the relationship between the present wonh and a uniform series

is

given by Equation (20.9):

[(t+z')'-tl P=Al --n * I| L i(l+i)"

CO.tt

Substituting into Equation (20.2) for P in terms ofr4 using Equation (20.9),we have

P F

= P(r

*

i)" =

lfl + t)'-

^L- 4r;

1.l

n

l(1

+

,)'

(20.10

Simplifying ftuation (20.11) rCIults in the direct relationship between the future wonh Fand the uniform payments or depositsr4, which follows:

,=^lsTl

G0.ta

And by rearranging Equation Q0.12), we c:rn obain a formula for A in terms of future worth F:

.s=Fl I L(l + 4'- ll

aomt

Now that we have all the necessary tools, we fllrn our attention to the question we asked eadier about how much money you need to put aside wery year for the next five years to have $2000 for the down payment of your car when you graduate. Recall that the interest rate is 6,50/o ampounding annually. The annual deposits are calculated from Equation (20.13), which leads to the following amounr:

--

s'oe\

n = ,oool L(l + 0.065),

-

I

:

$351.25

Puttingaside $351.26 in abankweryyear for the nortfiveyears maybe more manageable than depositing a lump sum of $1459.76 today, especially if you don't currendy have access ro that large a sum!

It is importanr to note thar Equations (20.9), (20.10), (20.12), and (20.13) apply to a situation wherein the uniform series of paymen$ or revenues occur annua$t, Vell, the next question is, how do we handle situations where the payments are made mondrly? For example, qu or a house loan paymens occur monthly. Let us now modi& our findings by considering the relationship benareen present value ? and uniform series paymenrs or revenuer4 that occur more than once a year at the same frequency as the frequency of compounding interest per year. a

608

CllePrsn

20

ENcrNsERrNc EcoNourcs For this situation, Equation Q0.9) is modified to incorporate the frequency of compounding interest per year, m, inthe following maruler:

l(r*.)-_]'l tv *)" *l'* | )

n: nl \tT!/

(20.14)

Note that in order to obtain Equation Q\.I ),we simply substituted in Equation (20.9) for i, ilm. andfor n" nm.Equation (20.14) can be rearranged to solve for.r4 in terms ofPaccording to

(*)('.

,="[

*)-

(' * t)* -,

Similarly, Equations Q0,12) and (20.13) can be modified for situations wherer4 occurs more than once a y@r_at the same frequency as the compounding interest-leading to the following relationship:

a= rL

('**)''""'-'

I (20.16)

t 7n

ltl

ml .-l n:,17---

t.7m| -'_l

(2017)

L\'*;)

Finally, when the frequency of uniform series is different from the frequency of compounding interest, i"6 must first be calculated to match the frequency of the uniform series. Ler us retum to the question we asked earlier about how much money you need to put aside for the next five ycrfs to have $2000 for the down payment on your car when you graduate. Now consider the situation where you make your deposis every month, and the interest rate is 6.50/o compounfing monthly. The deposits are calculated from Equation (20.17), which leads to the following:

* | lt :^"1 ^:'15;7ll

0.065 _-_..v-

('*

T2

o.o6t

\('

n)

=

2)(5)

$28.29

-1

Putting aside $28.29 in the bank every month for the next five years is even more manageable than depositing $l5t.Ze in a bank every year for the next five years, and it is cenainly more manageable than depositing a lump sum of $1459.76inthe bank todal L,,,u,,r,,,,.,,.,.,.,,,,

-::-,'',-,-- .-l- --

-l -------

:---t,r:al:l

2A.8 Suvuany or

ENcrNpsRrNc EcoNourcs ANer,ysrs

609

Determine the mon$Iy paymeng for a five-year, $10,000 loan ar an interest rare of 87o com-

pounding monthly. To calculate the monthlypayments, we use Equation

^:,lV)l:?)-.1 L\'.;) -rJ 20.8 Summary

:,o,ooo[(;,

LT

(2O.Ir.

x-* )-1 ):$20276

of Engineerinq Economies Analysis

The engineering economics formulas that we have dweloped so far are summarized in Thbles 20.6 and20.7. The definitions of the terms in the formulas are given here:

P: F:

cost-lump sum ($) future wofth, or fuguss ssst-lump sum ($) I : uniform series payment, or uniform series revenue ($) i : nominal interest rare z"n'= effecrive interest rate a : number of years rn = number of interest compounding periods pet yeet present worth, or present

The interest-time factors shown in the fourth column of Thble 2A.6 arc used as shoncuts to artoid writing long formulas when waluating equivalent values ofvarious cash fow occurrences.

F

P

P

F

F= P(l * i)

(FIP, i, n)

o- F '-(1+i), f(t+i)"-ll P:AI' i(r+i)' , , L l

I

(rxr + i)" I

F

:6+S

ffl+;\t-1'1 (PtA,i.n):Ld;1

| A=Pl:--l

(Atp,,,o)=l#++]

': ^L---i-)

(FtA,i,n):lga#]

LQ+ir-rJ f(l + r)'- 1l

A

(PIF,i,n)

= (l + i)

.-f

(4

I

^- "l1r * ;1, - t1

(AtTi,n)=

ltr#_,]

Cnerrsn20

ExcrNsERrNcEcoNomrcs

ur"

ffi, ro"-J"

r*,= (,r + *))-

i

-t

r= i"(r * *)*

,:G# I

(,. *)* ,:^l -17*t\ry

,1

;\1+;)

r /;\/ A= ?l

P

L

,=

^f L

I ,="L

(;/(1+;)

|

J ;\ou-1

-7--76-

\'*;)

|

-'J

|

1 ;\@)t") I -' \'*;)

-=-l

I

ln

l:

nl

t:

rl,

('**)'-""-'I1

For example, when evaluating the series payment equivalence of apresent p.itt"ip"l, instead

of

writing,

O0 + i)" A- pl 'L(l +i)"-r)f wewtiteA: P(A/4 (Atp, L

t, n),where, of course,

J,il" .f o' :1,!'),0 L(r +i)"-rJ

In this example,the (AIP, a z) term is cdledt\einterest-timefaaor,andit reads.d gSven P at io/o inreresr rate, for a duradon of n years,It is used to findA, when the present principal value P is given, by multiplying P by the value of the interest-time fa*or (AIP, A n). As an otample' the numerical values of interest-time factors for i = 8o/o are calculated and shown in Table 20.8.

I

1.08000000 1.16640000 7,219V1200 4 ,. 1.3664s396 r.4.6932808 t 6 r.58687432 r.71382427 7 1.85093021 2 5

t,

I

lo

,.

Ti; t2

0.92592593 0.85733882

4,92592593

p.79r53224

2.5770if)699

g:/3t02985

3.31212684 3,99271004 4,62287966

0.65058320 0.63016963 0.58349040 0.54026888 r.99900463 0.50024897 2.t5592500 0.46319349 2.33163900 0.42888286

251817012 0.397rr376 2.71962373 036769792

T3

2.93719362 0.34046104 3.L72169rr A3r524r7o 3.42594264 0.29189047

L4 15

l6

t7',, 3]0001805 0.27026895

l8

3.99601950 0.25024903

' 43\57On6,:

A.23171206

4.66095714 zt.: 5.03383372 ',?2 ;, 5.43654041 2t 5.87146365 6.34118074 2/t 25 6.84847520 26 7.39635321 7.98806147 2l 8.62710639 29 9.31727490 30 10.06265689 31 , 10.86766944

0.2145482r

19

I

2A:,,

',

n

:

1,7832&7'

5,VA697A06

5:/4663894 6.2468:8791

6.71008140 7,13896426 7.536fr7802 7.94377594

8.2U23698 8sr:94786e 8.85136916-

9-1216381t

937188714 e.6fi35?92a

0.r0r8522r

0.19865575

9.818r4V41 r0,01680316

0.1539405r

rc20A7$66

o.og8o32w

0.09642217 ra.12875525 0.09497796 0J4601790 'r0.674n619 0.09367878

a.17045&5 a.j.36nr559 o.t:tz,arz^o a.094a1476

a,a80wg7r 0$6902949 0.060a7&4 ,0n52695.02

0.04652t8r ,0.04129685 0,a3682954 ,a,03297697

0,a2962943

0.a267\2n '0.a24r276' a,a2$522t:' O.OtgglZZS"

0'9199??07

10.37105895

60.89329557 66.76475922

0,ar642217 a:frt497796

73.10593995

0,01367878

0.13520176 0.12518682 0.11591372 0.10732752 0.09937733 0.09201605

10.80997795

0.0925W13 79.95U15r5 a.ar2507l'3 0.09144810 87.35076836 0.01144810 0.09048891 95.33882983 b.olor)-)rO logarithmic models, 549-55 I

outd@r air quality slandaids, 68-70 Safe Drinking VaterAct, 65 Sufae Varer Tiernent Rule

(svTR),68 Enos, typs of in *pedmtal obsemtion,582 E*iq, L05-I22. See afu Code ofethio

dsignprm,45 -

32, 54L -546,

normal distdbution, 587 -594

372

role of, 52 418, 445 - 447.

prcbab{rty,5V-579

mlc K% oL J/J-Jla

@J1s.374-376,380-382 we fiaing 407-411 FilIomlffi4 380-382

sK6, )//-)/y stopping sight

disuna,

532, 541-543

temperatue distribution aaos

a

plain

wall,536-538 time, rcle of in, 197-198 Engineing profesi on, 2 -l 22 Accrediation Bord for Enginedng and Technologr

(ABE$,

A-n

262

@.S.),

47-48 Fibs (qptic) gl,s' 517-518 Fill ommd,380-382 Filet, dwings,468-469

Fihn rcistane (o#cimt), 302-303 Finishs, tm rcle o(,52 Fim lm of themodynmic,348-350

emn, 582 nte,210-211,224 volme,210-211

Fixed

Fls

M,224

Fluid friction, re Vrmsity Fluid matqials, 5 1 9-521 Foot (ft), U, S. Cutomaryuit oflength,

l)o

MAILAB,432

pmetm, 236-

z8l

ra,

mingover

a,244-2i6

bu& modulu of omprcibility, 263264 Caterpillar 797 miaiag t ,r"ls *Ctneerng |Wg, z/d-l6r omprshe (ultimate) strength, 261

d6^d,237 disvne, aarng at a, 2& -267 distane, acting ovo a 268-269 facmr of ufety @,S,), 262

fnaton,240-241 gaitanoml 237,242-244 See

ako

imFul*, 269-271 modulu of elasticity, 256 -259, 260 modulu of rigidity, 259-263 momertof a,?14-267 Nswton! lM, 241-244

fomulo,376-380 tunaons376-3W,387-389 imponane in ngindhg, 372-373 ioponing fila into MAfl-{B, 445447

insningells, olmm, ad rom,376

eeprepmtion for,27-40

logiel functiom, 387-389

ommon uaits of 8-12

mari: compuradom,

400 -

Hnke\lm,238,258-259

lia6

Cells: Ercel fuoaiom uithmetic operatioro, 376

slution o[,82-84

of*fety

&fuenheit fF), U.S. Cretomarymitof tmpemtue, 291-293 Fmd (F), boic uit for qpacitors, 334 Faible rclution region, dcign prre,

xtage racrion,2.69-271.

112

dsign altematim, 55-56

ErqL

5

-122

Evalvdot,45,55-56

nonlina

models,

LzO

112

tm

Factor

Fora ad forerelated

enginert med, 111-l12 enginsin& 106-109 Ndonal Sciery of Profesional Enginen (NSPE), 107, 120 -122 plagiaisn, 112 prcfmional mpomibility, I 12

Evalutoc

385

MATLAB,43O Extmion linc,468

for omd,

for euginm aademic dishoncty, 112 @e studis fo!, ll2-1I7, ode of, 107-111

mthematia h,532- 576

)4tJ-ttu

Exc€I, 1

lryel onaminmt goal

@ntram

Pste,383

Exponential functiorc, 385, 430

lwel ontamimt (MCL)

onfliaofinterct,

logiel,387-389

nigonomeaic 385

EPA, ree Environmqtal Protection Agqcy (EPA)

546-549,549-

togmmc, 16)

Xxmtiresrmic,87

(Ms-G),67

-57 6,

-.h ffowdiagm,598-599

onomio m4 597-619

r95-217 See

Enginering organiatiom, involvement with,36

ooling

time-rdated

mim

r97 -198, 530 -53r, 577-596,597-619

pamet6,

t52-

tmpemtm od

lroimm

Enginering papu, 85-86 Enginering prcblem, 82-84, 127 -128,

236-281

innoduction to, 12J length md length-relatd

st€I,511-512 water,520-527

pmetes,

specialiatioro of, ll, I3-L4 technologr progm, 2l -22 U.S. Buru of Iabor Satistia for I world PoPrl'iion' altr€ts of 6-8

-387,387-389 rc of, 3M

engin@irg arab{is qponential,3S5

relational opemton, 388 todsy ( ), 384

5-6

Air Act, 65, 68-69 drinking water mdads, 57-68 function of,65 indor air qualiry GACD, 70 hdmr air quality sandarls, /0-l2

rcod,514-515

-l5l

o6,

Clm

503-508

tianim,509-510

128,3t9-340 enerry md power, 341-36! fore md forcrelared llmetes,

497.

o4,

wie

Enviromental enginering prcfesion o4, 18-19 Enviromental Protation Agacy (EPA), 65,67-72

soli4 509-519

of labor Statisia for,

dimmiom md unis,

ad

l0

ofmatter, 502-503

prcpenic

\Feb sitc for, 12-13 Enginering findmenels, 124 -369

time

wel,526-

silion,5l

spaialiaion md, 13-14

128,

mgineing

plasic,5l5-516

nucla, 19-20 petloleu, 19

Buru

(water vapor), 520

lightweight neals, 509-5

phm

of Enginming

15

l5

mgn$im,510

19

materials,20-21 minir& f,$

U.S.

F.sm,

ircn,511 Jet enghe,

14

ud Pmaie

Principla

)r/->rd

hmidity

wg6,5/)-5/O Ercel functiom, 376

Fmdanentals of Enginering Eem (FE), 15 inuoduction m, 2-3, 4-26 Ntional Sociery of Profesioual Engins (NSPE),3

@nqetg'5L2-514 opper md its allqre, 5 10-5 1 I dcign pr>2, >>5-))+ mr:ftipliccion of, 553-554 time, a a phpiel prcperty of, 206 Srundmommtof m,177-182

bem, mmpls

of in, 180-182

akoBn-

(EPA)

Amaim

National Standtrds Insdtute

terials

(ASIM,6l-62 FmFis & nomalistim

(ATNOR),63

599

British Stadrds Instituts (BSt), 63 C-e standarcls, 63 Chim Sate Bua of cllatry *4 Tirhniel Supwisioa (CSBTS), 63 @nmirms, methods of nmaging

nrle, 159, 621 Singu&r marix, 559 Sir, nomiasl w actnal, 160-162

Deutschc Initut fiu Nomung @IN), 63

139

l4I

Silion,51 Simple

intact,

Sine

Slidc, PwuPoint pwntarioro, 93-97, 97-98, IOO mimation of, 97-98 imting ftom otha PmrPoint 6lc,

mcta,94 of, 131

Solid linc, onhogra.phic dwings, 465 Solid marerials, 509-5 19 Solid modeling, 476 -482,493-497 Bcing 777 omerciat airylane, engi-

nuingmd493-497 Bmlru

openriom for, 479, 481 bonom-up, 478-479

onputer nmaielly onrolled (CNC) mchines 47 imporane of in enginedng

42-

84. 138-138, 197-198. See aho Engineing Problem; Linear

muatiom malFis of,83 defining the prcblm, 82-83

mple ofpmtation

fur,84

ronoiel,

138-139 pametric fom' 83

simpli$ing the prcblm, 83 197-198

of, 82-83 symbolic, 138-139 time rcle of in, 197-198 steps

(EPL),65,6712 of U.S., 65-67

mmpls

(NFPA),62 D@Afor,58-60 orymiations for, 60 -62, 0-64 outdmr air quality, 68-70 Star