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 deriva- tions,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 signifi- cant 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
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
SECTION 1.2 Normal Stress and Strain 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 compres- sion.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 6(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 o(sigma).In general,the stresses o 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.) Landing gear strut- 四 Tow 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
8 CHAPTER 1 Tension,Compression,and Shear (a) (b) L+6 (c) FIG.1-2 Prismatic bar in tension: (a)free-body diagram of a segment of m the bar,(b)segment of the bar before loading,(c)segment of the bar after P O= loading,and (d)normal stresses in the A bar (d) (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 oA.Therefore,we obtain the following expression for the magnitude of the stresses: U= (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 o 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 '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.
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 PP 6k == Amd24π2.0in.)24 =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/m )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 I 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 X 10 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 o=PA 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 FIG.1-3 Steel eyebar subjected to tensile complex.However,as we move away from the ends and toward the loads P middle of the bar,the stress distribution gradually approaches the uniform distribution pictured in Fig.1-2d. As a practical rule,the formula o P/A may be used with good accuracy 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.
10 CHAPTER 1 Tension,Compression,and Shear from the stress concentration as the largest lateral dimension of the bar.In other words,the stress distribution in the steel eyebar of Fig.1-3 is uniform at distances b or greater from the enlarged ends,where b is the width of the bar,and the stress distribution in the prismatic bar of Fig.1-2 is uniform at distances d or greater from the ends,where d is the diameter of the bar (Fig.1-2d).More detailed discussions of stress concentrations produced by axial loads are given in Section 2.10. Of course,even when the stress is not uniformly distributed,the equation o P/A may still be useful because it gives the average normal stress on the cross section. Normal Strain As already observed,a straight bar will change in length when loaded axially,becoming longer when in tension and shorter when in compression. For instance,consider again the prismatic bar of Fig.1-2.The elongation 6 of this bar (Fig.1-2c)is the cumulative result of the stretching of all elements of the material throughout the volume of the bar.Let us assume that the material is the same everywhere in the bar.Then,if we consider half of the bar (length L2),it will have an elongation equal to 62,and if we consider one-fourth of the bar,it will have an elongation equal to 8/4. In general,the elongation of a segment is equal to its length divided by the total length L and multiplied by the total elongation 6.Therefore,a unit length of the bar will have an elongation equal to 1/L times 8.This quantity is called the elongation per unit length,or strain,and is denoted by the Greek letter e(epsilon).We see that strain is given by the equation (1-2) If the bar is in tension,the strain is called a tensile strain,representing an elongation or stretching of the material.If the bar is in compression,the strain is a compressive strain and the bar shortens.Tensile strain is usually taken as positive and compressive strain as negative.The strain e is called a normal strain because it is associated with normal stresses. Because normal strain is the ratio of two lengths,it is a dimension- less quantity,that is,it has no units.Therefore,strain is expressed simply as a number,independent of any system of units.Numerical values of strain are usually very small,because bars made of structural materials undergo only small changes in length when loaded. As an example,consider a steel bar having length L equal to 2.0 m When heavily loaded in tension,this bar might elongate by 1.4 mm, which means that the strain is 1.4mm E= L2.0m =0.0007=700×10-6
from the stress concentration as the largest lateral dimension of the bar. In other words, the stress distribution in the steel eyebar of Fig. 1-3 is uniform at distances b or greater from the enlarged ends, where b is the width of the bar, and the stress distribution in the prismatic bar of Fig. 1-2 is uniform at distances d or greater from the ends, where d is the diameter of the bar (Fig. 1-2d). More detailed discussions of stress concentrations produced by axial loads are given in Section 2.10. Of course, even when the stress is not uniformly distributed, the equation s P/A may still be useful because it gives the average normal stress on the cross section. Normal Strain As already observed, a straight bar will change in length when loaded axially, becoming longer when in tension and shorter when in compression. For instance, consider again the prismatic bar of Fig. 1-2. The elongation d of this bar (Fig. 1-2c) is the cumulative result of the stretching of all elements of the material throughout the volume of the bar. Let us assume that the material is the same everywhere in the bar. Then, if we consider half of the bar (length L/2), it will have an elongation equal to d/2, and if we consider one-fourth of the bar, it will have an elongation equal to d/4. In general, the elongation of a segment is equal to its length divided by the total length L and multiplied by the total elongation d. Therefore, a unit length of the bar will have an elongation equal to 1/L times d. This quantity is called the elongation per unit length, or strain, and is denoted by the Greek letter e (epsilon). We see that strain is given by the equation (1-2) If the bar is in tension, the strain is called a tensile strain, representing an elongation or stretching of the material. If the bar is in compression, the strain is a compressive strain and the bar shortens. Tensile strain is usually taken as positive and compressive strain as negative. The strain e is called a normal strain because it is associated with normal stresses. Because normal strain is the ratio of two lengths, it is a dimensionless quantity, that is, it has no units. Therefore, strain is expressed simply as a number, independent of any system of units. Numerical values of strain are usually very small, because bars made of structural materials undergo only small changes in length when loaded. As an example, consider a steel bar having length L equal to 2.0 m. When heavily loaded in tension, this bar might elongate by 1.4 mm, which means that the strain is e 5 L d 1 2 .4 .0 m m m 0.0007 700 106 e 5 L d 10 CHAPTER 1 Tension, Compression, and Shear