LSECTION1.1Introduction to Mechanics of Materials1.1INTRODUCTIONTOMECHANICSOFMATERIALSMechanicsof materials is abranch of applied mechanics thatdealswith the behavior of solid bodies subjected to various types of loading.Other names for this field of study are strength of materials andmechanics of deformable bodies.The solid bodies considered in thisbook include bars with axial loads,shafts in torsion,beams in bending,andcolumns in compression.The principal objective of mechanics of materials is to determinethe stresses,strains,and displacements in structures and their compo-nents dueto the loads acting on them.If wecanfind thesequantities forall values of the loads up to the loads that cause failure, we will have acompletepictureof themechanical behavior of these structures.An understanding of mechanical behavior is essential for the safedesign of all types of structures, whether airplanes and antennas, buildingsand bridges,machines and motors, or ships and spacecraft.That is whymechanics of materials is a basic subject in so manyengineering fields.Stat-ics and dynamics are also essential, but those subjects deal primarily withthe forces and motions associated with particles and rigid bodies. Inmechanicsofmaterialswegoonestepfurtherbyexaminingthestressesandstrains inside real bodies, that is, bodies of finite dimensions that deformunderloads.Todetermine the stresses and strains,weuse the physical prop-erties of the materials as well as numerous theoretical laws and concepts.Theoretical analyses andexperimental results haveequallyimportantroles in mechanics of materials. We use theories to derive formulasand equations for predicting mechanical behavior, but these expressionscannot be used in practical design unless thephysical properties of thematerials areknown.Such properties are available only after carefulexperiments havebeen carried out inthelaboratory.Furthermore,not allpracticalproblems are amenabletotheoretical analysis alone,and insuch cases physical testing is a necessity.The historical development of mechanics of materials is a fascinatingblend ofboth theory and experiment-theoryhas pointed the waytouseful results in some instances, and experiment has done so in others.Such famous persons as Leonardo da Vinci (1452-1519)and GalileoGalilei (1564-1642)performed experimentsto determinethe strength ofwires,bars,andbeams,although theydid not develop adequatetheories(by today's standards) to explain their test results.By contrast, thefamousmathematicianLeonhardEuler(1707-1783)developedthemath-ematicaltheoryofcolumnsandcalculatedthecriticalloadofacolumnin1744,longbeforeanyexperimental evidenceexisted to showthesignifi-cance of his results. Without appropriate tests to back up his theories,Euler's results remained unused for over a hundred years, although todaythey arethebasis for the design and analysis ofmost columns.The history of mechanics of materials, beginning with Leonardo and Galileo, is given inRefs. 1-1, 1-2, and 1-3
SECTION 1.1 Introduction to Mechanics of Materials 5 1.1 INTRODUCTION TO MECHANICS OF MATERIALS Mechanics of materials is a branch of applied mechanics that deals with the behavior of solid bodies subjected to various types of loading. Other names for this field of study are strength of materials and mechanics of deformable bodies. The solid bodies considered in this book include bars with axial loads, shafts in torsion, beams in bending, and columns in compression. The principal objective of mechanics of materials is to determine the stresses, strains, and displacements in structures and their components due to the loads acting on them. If we can find these quantities for all values of the loads up to the loads that cause failure, we will have a complete picture of the mechanical behavior of these structures. An understanding of mechanical behavior is essential for the safe design of all types of structures, whether airplanes and antennas, buildings and bridges, machines and motors, or ships and spacecraft. That is why mechanics of materials is a basic subject in so many engineering fields. Statics and dynamics are also essential, but those subjects deal primarily with the forces and motions associated with particles and rigid bodies. In mechanics of materials we go one step further by examining the stresses and strains inside real bodies, that is, bodies of finite dimensions that deform under loads. To determine the stresses and strains, we use the physical properties of the materials as well as numerous theoretical laws and concepts. Theoretical analyses and experimental results have equally important roles in mechanics of materials. We use theories to derive formulas and equations for predicting mechanical behavior, but these expressions cannot be used in practical design unless the physical properties of the materials are known. Such properties are available only after careful experiments have been carried out in the laboratory. Furthermore, not all practical problems are amenable to theoretical analysis alone, and in such cases physical testing is a necessity. The historical development of mechanics of materials is a fascinating blend of both theory and experiment—theory has pointed the way to useful results in some instances, and experiment has done so in others. Such famous persons as Leonardo da Vinci (1452–1519) and Galileo Galilei (1564–1642) performed experiments to determine the strength of wires, bars, and beams, although they did not develop adequate theories (by today’s standards) to explain their test results. By contrast, the famous mathematician Leonhard Euler (1707–1783) developed the mathematical theory of columns and calculated the critical load of a column in 1744, long before any experimental evidence existed to show the significance of his results. Without appropriate tests to back up his theories, Euler’s results remained unused for over a hundred years, although today they are the basis for the design and analysis of most columns.* * The history of mechanics of materials, beginning with Leonardo and Galileo, is given in Refs. 1-1, 1-2, and 1-3
CHAPTER1Tension, Compression,andShearProblemsWhen studying mechanics of materials,you will find that your effortsare divided naturally into two parts:first,understanding the logicaldevelopment of theconcepts,and second,applyingthoseconceptstopractical situations. The former is accomplished by studying the deriva-tions, discussions, and examples that appear in each chapter, andthelatterisaccomplishedby solvingtheproblemsatthe ends of thechapters.Some of the problems are numerical in character,and othersare symbolic (or algebraic).An advantage of numerical problems is that themagnitudes of allquantities are evident at every stage of the calculations,thus providingan opportunityto judge whether the values are reasonable or not.Theprincipal advantage of symbolic problems is that theylead togeneral-purposeformulas.A formula displaysthe variables that affectthe final results; for instance, a quantity may actually cancel out of thesolution, a fact that would not be evident from a numerical solution.Also,an algebraic solution shows themanner inwhich eachvariableaffects the results,as when one variable appears in the numerator andanotherappearsin thedenominator.Furthermore,a symbolic solutionprovides the opportunity to check the dimensions at every stage of thework.Finally, the most important reason for solving algebraically is toobtainageneral formulathat canbeusedformanydifferentproblems.Incontrast,a numerical solution appliesto onlyone set ofcircumstances.Because engineers must be adept at bothkinds of solutions,you will findamixture of numeric and symbolic problemsthroughout this book.Numerical problemsrequirethat you workwith specific units ofmeasurement.In keeping with current engineering practice, this bookutilizes both the Intermational System of Units (SI) and the U.S.CustomarySystem(USCS).Adiscussion of both systems appears inAppendixAwhere you will also find many useful tables,including a table ofconversionfactors.Allproblemsappearattheendsofthechapters,withtheproblemnumbers and subheadings identifying the sections to which theybelongIn the case of problems requiring numerical solutions, odd-numberedproblems are in USCS units and even-numbered problems are in SI units.The techniques for solving problems are discussed in detail inAppendix B.In addition to a list of sound engineering procedures,Appendix B includes sections on dimensional homogeneity and signifi-cantdigits.Thesetopics are especiallyimportant,becauseeveryequationmustbedimensionallyhomogeneous and everynumerical resultmustbeexpressed with the proper number of significant digits.In this book, finalnumerical results are usuallypresented with three significant digits whena number begins with the digits 2 through 9, and with four significantdigits when a number begins with the digit 1.Intermediate values areoften recorded with additional digits to avoid losing numerical accuracydue to rounding of numbers
6 CHAPTER 1 Tension, Compression, and Shear Problems When studying mechanics of materials, you will find that your efforts are divided naturally into two parts: first, understanding the logical development of the concepts, and second, applying those concepts to practical situations. The former is accomplished by studying the derivations, discussions, and examples that appear in each chapter, and the latter is accomplished by solving the problems at the ends of the chapters. Some of the problems are numerical in character, and others are symbolic (or algebraic). An advantage of numerical problems is that the magnitudes of all quantities are evident at every stage of the calculations, thus providing an opportunity to judge whether the values are reasonable or not. The principal advantage of symbolic problems is that they lead to general-purpose formulas. A formula displays the variables that affect the final results; for instance, a quantity may actually cancel out of the solution, a fact that would not be evident from a numerical solution. Also, an algebraic solution shows the manner in which each variable affects the results, as when one variable appears in the numerator and another appears in the denominator. Furthermore, a symbolic solution provides the opportunity to check the dimensions at every stage of the work. Finally, the most important reason for solving algebraically is to obtain a general formula that can be used for many different problems. In contrast, a numerical solution applies to only one set of circumstances. Because engineers must be adept at both kinds of solutions, you will find a mixture of numeric and symbolic problems throughout this book. Numerical problems require that you work with specific units of measurement. In keeping with current engineering practice, this book utilizes both the International System of Units (SI) and the U.S. Customary System (USCS). A discussion of both systems appears in Appendix A, where you will also find many useful tables, including a table of conversion factors. All problems appear at the ends of the chapters, with the problem numbers and subheadings identifying the sections to which they belong. In the case of problems requiring numerical solutions, odd-numbered problems are in USCS units and even-numbered problems are in SI units. The techniques for solving problems are discussed in detail in Appendix B. In addition to a list of sound engineering procedures, Appendix B includes sections on dimensional homogeneity and significant digits. These topics are especially important, because every equation must be dimensionally homogeneous and every numerical result must be expressed with the proper number of significant digits. In this book, final numerical results are usually presented with three significant digits when a number begins with the digits 2 through 9, and with four significant digits when a number begins with the digit 1. Intermediate values are often recorded with additional digits to avoid losing numerical accuracy due to rounding of numbers
7SECTION1.2 Normal StressandStrain1.2NORMALSTRESSANDSTRAINThe mostfundamental concepts in mechanics of materials are stress andstrain.These concepts can be illustrated in their most elementary formby considering a prismatic bar subjected to axial forces. A prismaticbar is a straight structural member having the same cross sectionthroughout its length, and an axial force is a load directed along the axisof themember, resulting ineithertension or compression inthe bar.Examples are shown in Fig.1-1,where thetow bar isa prismaticmember intension andthe landinggear strut is amember in compres-sion.Otherexamplesarethemembersof abridgetruss,connecting rodsin automobileengines,spokes of bicycle wheels,columns inbuildings,and wing struts in small airplanesFor discussion purposes,wewill consider thetowbar of Fig.1-1and isolate a segment of it as a free body (Fig.1-2a).When drawing thisfree-body diagram,we disregard the weight of the bar itself and assumethat the only active forces are the axial forces P at the ends. Next weconsidertwoviewsof thebar,thefirst showingthesamebarbeforetheloads are applied (Fig.1-2b) and the second showing it after the loadsareapplied (Fig.1-2c).Notethattheoriginal lengthof thebar is denotedby the letter L, and the increase in length due to the loads is denoted bythe Greek letter (delta).The internal actions in the bar are exposed if we make an imaginarycut through the bar at section mn (Fig.1-2c).Because this section is takenperpendicular to the longitudinal axis of thebar, it is called a cross section.Wenowisolatethepartof thebartotheleftof cross sectionmn asafree body (Fig.1-2d).At the right-hand end of this free body (section mn)we show the action of the removed part of the bar (that is, the part to theright of section mn)upon the part that remains.This action consists ofcontinuously distributed stressesacting over the entire cross section,andthe axial force P acting at the cross section is the resultant of thosestresses.(Theresultant force is shownwitha dashed line inFig. 1-2d.)Stress has units of force per unit area and is denoted by the Greekletter(sigma).Ingeneral,thestresses acting onaplane surfacemaybeuniformthroughout thearea or mayvary in intensity from onepointto another.Letus assume thatthe stresses acting on cross section mnFIG.1-1 Structural members subjected toaxial loads. (The tow bar is in tensionand the landing gear strut is incompression.)Landing gear strutTow bar
SECTION 1.2 Normal Stress and Strain 7 1.2 NORMAL STRESS AND STRAIN The most fundamental concepts in mechanics of materials are stress and strain. These concepts can be illustrated in their most elementary form by considering a prismatic bar subjected to axial forces. A prismatic bar is a straight structural member having the same cross section throughout its length, and an axial force is a load directed along the axis of the member, resulting in either tension or compression in the bar. Examples are shown in Fig. 1-1, where the tow bar is a prismatic member in tension and the landing gear strut is a member in compression. Other examples are the members of a bridge truss, connecting rods in automobile engines, spokes of bicycle wheels, columns in buildings, and wing struts in small airplanes. For discussion purposes, we will consider the tow bar of Fig. 1-1 and isolate a segment of it as a free body (Fig. 1-2a). When drawing this free-body diagram, we disregard the weight of the bar itself and assume that the only active forces are the axial forces P at the ends. Next we consider two views of the bar, the first showing the same bar before the loads are applied (Fig. 1-2b) and the second showing it after the loads are applied (Fig. 1-2c). Note that the original length of the bar is denoted by the letter L, and the increase in length due to the loads is denoted by the Greek letter d (delta). The internal actions in the bar are exposed if we make an imaginary cut through the bar at section mn (Fig. 1-2c). Because this section is taken perpendicular to the longitudinal axis of the bar, it is called a cross section. We now isolate the part of the bar to the left of cross section mn as a free body (Fig. 1-2d). At the right-hand end of this free body (section mn) we show the action of the removed part of the bar (that is, the part to the right of section mn) upon the part that remains. This action consists of continuously distributed stresses acting over the entire cross section, and the axial force P acting at the cross section is the resultant of those stresses. (The resultant force is shown with a dashed line in Fig. 1-2d.) Stress has units of force per unit area and is denoted by the Greek letter s (sigma). In general, the stresses s acting on a plane surface may be uniform throughout the area or may vary in intensity from one point to another. Let us assume that the stresses acting on cross section mn FIG. 1-1 Structural members subjected to axial loads. (The tow bar is in tension and the landing gear strut is in compression.) Tow bar Landing gear strut
8CHAPTER1Tension, Compression,andShear(a)O-L1(b)mnL+8(c)FIG.1-2 Prismatic bar in tension:m(a) free-body diagram of a segment ofthe bar, (b) segment of the bar beforenPloading,(c)segment of the bar after[d:Aloading, and (d) normal stresses in the(d)bar(Fig.1-2d) are uniformlydistributed over the area.Then the resultant ofthose stresses must be equal to the magnitude of the stress times thecross-sectional area A of thebar,that is,P=A.Therefore,we obtainthefollowing expressionforthemagnitudeof the stresses:P(1-1)AThis equation gives theintensity of uniform stress in an axially loaded.prismaticbarofarbitrarycross-sectional shape.Whenthebaris stretched bytheforcesP,thestresses aretensilestresses; if the forces are reversed in direction, causing the bar to becompressed,weobtaincompressivestresses.Inasmuchasthestressesact in a direction perpendicular to the cut surface,they are called normalstresses. Thus, normal stresses may be either tensile or compressive.Later, in Section 1.6, wewill encounter another type of stress, calledshear stress, that acts parallel tothe surface.Whena sign convention for normal stresses is required,it iscustomary todefine tensile stresses as positive and compressive stressesas negative.Because the normal stress is obtained by dividing the axial forceby the cross-sectional area, it has units of force per unit of area. WhenUsCs units areused, stress is customarily expressed inpoundspersquare inch (psi) or kips per square inch (ksi).For instance, suppose“One kip, or kilopound, equals 1000 1b
(Fig. 1-2d) are uniformly distributed over the area. Then the resultant of those stresses must be equal to the magnitude of the stress times the cross-sectional area A of the bar, that is, P sA. Therefore, we obtain the following expression for the magnitude of the stresses: (1-1) This equation gives the intensity of uniform stress in an axially loaded, prismatic bar of arbitrary cross-sectional shape. When the bar is stretched by the forces P, the stresses are tensile stresses; if the forces are reversed in direction, causing the bar to be compressed, we obtain compressive stresses. Inasmuch as the stresses act in a direction perpendicular to the cut surface, they are called normal stresses. Thus, normal stresses may be either tensile or compressive. Later, in Section 1.6, we will encounter another type of stress, called shear stress, that acts parallel to the surface. When a sign convention for normal stresses is required, it is customary to define tensile stresses as positive and compressive stresses as negative. Because the normal stress s is obtained by dividing the axial force by the cross-sectional area, it has units of force per unit of area. When USCS units are used, stress is customarily expressed in pounds per square inch (psi) or kips per square inch (ksi).* For instance, suppose s P A 8 CHAPTER 1 Tension, Compression, and Shear d P (a) P P (c) (d) P P P —A P = (b) m n m n L L + d s FIG. 1-2 Prismatic bar in tension: (a) free-body diagram of a segment of the bar, (b) segment of the bar before loading, (c) segment of the bar after loading, and (d) normal stresses in the bar * One kip, or kilopound, equals 1000 lb.
9SECTION1.2Normal StreSSandStrainthat thebarof Fig.1-2hasadiameterdof 2.0 inchesandtheloadPhasamagnitudeof6kips.Thenthestressinthebaris6kpP=1.91 ksi (or1910psi)Amd=/4(2.0 in.)/4In this examplethe stress is tensile, orpositive.When SI units are used, force is expressed in newtons (N) and areainsquaremeters(m).Consequently,stresshasunitsof newtonspersquare meter (N/m), that is, pascals (Pa).However, the pascal is such asmall unit of stress that it is necessary to work with largemultiples,usually the megapascal (MPa).Todemonstrate that a pascal is indeed small,we have only to notethatittakesalmost7000pascalstomakeIpsi.Asanillustration,thestress inthebar described inthepreceding example(1.91ksi)convertsto13.2MPa,which is13.2×10°pascals.Although it isnot recommendedin SI,you will sometimes find stress given in newtonsper squaremillimeter(N/mm),which isaunitequal tothemegapascal (MPa).LimitationsThe equation = P/A is valid only if the stress is uniformly distributed overthecross sectionof thebar.Thisconditionisrealizediftheaxial forcePacts through the centroidofthecross-sectional area,asdemonstratedlaterinthis section.WhentheloadPdoesnotactatthecentroid,bendingofthebarwill result,and a more complicated analysis is necessary (see Sections 5.12and 11.5).However,in this book (as in common practice)it is understoodthat axial forces areapplied at the centroids of the cross sections unlessspecifically stated otherwise.The uniform stress condition pictured inFig.1-2d exists throughoutthelengthof the barexcept near the ends.The stress distribution at theendofabardependsuponhowtheloadPistransmittedtothebar.Iftheload happens to be distributed uniformly over the end,then the stresspattern at the end will be the same as everywhere else.However, it ismore likely that the load is transmitted through a pin or a bolt,producinghighlocalized stressescalled stress concentrations.One possibility is illustrated by theeyebar shown in Fig.1-3.In thisUinstancetheloadsParetransmittedtothebarbypinsthatpassthroughthe holes (or eyes) at the ends of the bar.Thus, the forces shown in thefigure areactuallytheresultants of bearingpressures betweenthepinsand theeyebar,and thestressdistributionaroundtheholesis quitecomplex.However,aswemoveawayfromtheendsandtowardtheFIG.1-3Steeleyebarsubjectedtotensilemiddle of the bar,the stress distribution gradually approaches theloads Puniformdistributionpictured inFig.1-2d.As a practical rule,the formula o=P/Amaybe used withgoodaccuracy at any point within a prismatic bar that is at least as far away'Conversion factors between USCS units and SI units are listed in Table A-5, Appendix A
SECTION 1.2 Normal Stress and Strain 9 that the bar of Fig. 1-2 has a diameter d of 2.0 inches and the load P has a magnitude of 6 kips. Then the stress in the bar is s P A pd P 2 /4 p(2. 6 0 k in.)2 /4 1.91 ksi (or 1910 psi) In this example the stress is tensile, or positive. When SI units are used, force is expressed in newtons (N) and area in square meters (m2 ). Consequently, stress has units of newtons per square meter (N/m2 ), that is, pascals (Pa). However, the pascal is such a small unit of stress that it is necessary to work with large multiples, usually the megapascal (MPa). To demonstrate that a pascal is indeed small, we have only to note that it takes almost 7000 pascals to make 1 psi.* As an illustration, the stress in the bar described in the preceding example (1.91 ksi) converts to 13.2 MPa, which is 13.2 106 pascals. Although it is not recommended in SI, you will sometimes find stress given in newtons per square millimeter (N/mm2 ), which is a unit equal to the megapascal (MPa). Limitations The equation s P/A is valid only if the stress is uniformly distributed over the cross section of the bar. This condition is realized if the axial force P acts through the centroid of the cross-sectional area, as demonstrated later in this section. When the load P does not act at the centroid, bending of the bar will result, and a more complicated analysis is necessary (see Sections 5.12 and 11.5). However, in this book (as in common practice) it is understood that axial forces are applied at the centroids of the cross sections unless specifically stated otherwise. The uniform stress condition pictured in Fig. 1-2d exists throughout the length of the bar except near the ends. The stress distribution at the end of a bar depends upon how the load P is transmitted to the bar. If the load happens to be distributed uniformly over the end, then the stress pattern at the end will be the same as everywhere else. However, it is more likely that the load is transmitted through a pin or a bolt, producing high localized stresses called stress concentrations. One possibility is illustrated by the eyebar shown in Fig. 1-3. In this instance the loads P are transmitted to the bar by pins that pass through the holes (or eyes) at the ends of the bar. Thus, the forces shown in the figure are actually the resultants of bearing pressures between the pins and the eyebar, and the stress distribution around the holes is quite complex. However, as we move away from the ends and toward the middle of the bar, the stress distribution gradually approaches the uniform distribution pictured in Fig. 1-2d. As a practical rule, the formula s 5 P/A may be used with good accuracy at any point within a prismatic bar that is at least as far away b P P FIG. 1-3 Steel eyebar subjected to tensile loads P * Conversion factors between USCS units and SI units are listed in Table A-5, Appendix A.