CHAPTER 6 - MECHANICAL PROPERTIES I PDF

Mechanical Properties: Chapter 6 Part One Have You Ever Wondered? Why can Silly Putty® be stretched a considerable amount when pulled slowly, but snaps when pulled fast?...

Mechanical Properties: Chapter 6 Part One Have You Ever Wondered? Why can Silly Putty® be stretched a considerable amount when pulled slowly, but snaps when pulled fast? Why can we load the weight of a fire truck on four ceramic coffee cups, yet ceramic cups tend to break easily when we drop them on the floor? What materials related factors played an important role in the sinking of the Titanic? What factors played a major role in the 1986 Challenger space shuttle accident? T he mechanical properties of materials depend on their composition and microstructure. In Chapters 2, 3, and 4, we learned that a material’s composi- tion, nature of bonding, crystal structure, and defects (e.g., dislocations, grain boundaries, etc.) have a profound influence on the strength and ductility of metallic materials. In this chapter, we will begin to evaluate other factors that affect the mechani- cal properties of materials, such as how lower temperatures can cause many metals and plastics to become brittle. Lower temperatures contributed to the brittleness of the plastic used for O-rings in the solid rocket boosters, causing the 1986 Challenger accident. In 2003, the space shuttle Columbia was lost because of the impact of debris on the ceramic tiles and failure of carbon–carbon composites. Similarly, the special chemistry of the steel used on the Titanic and the stresses associated with the fabrication and embrittlement of this steel when subjected to lower temperatures have been identified as factors contributing to the failure of the ship’s hull. Some researchers have shown that weak rivets and design flaws also contributed to the failure. The main goal of this chapter is to introduce the basic concepts associated with mechanical properties. We will learn terms such as hardness, stress, strain, elastic and plastic deformation, viscoelasticity, and strain rate. We will also review some of the testing procedures that engineers use to evaluate many of these properties. These concepts will be discussed using illustrations from real-world applications. 197 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 198 CHAPTER 6 Mechanical Properties: Part One 6-1 Technological Significance In many of today’s emerging technologies, the primary emphasis is on the mechanical properties of the materials used. For example, in aircraft manufacturing, aluminum alloys or carbon-reinforced composites used for aircraft components must be light weight, strong, and able to withstand cyclic mechanical loading for a long and predictable period of time. Steels used in the construction of structures such as buildings and bridges must have adequate strength so that these structures can be built without compromising safety. The plastics used for manufacturing pipes, valves, flooring, and the like also must have adequate mechanical strength. Materials such as pyrolytic graphite or cobalt chromium tungsten alloys, used for prosthetic heart valves, must not fail. Similarly, the performance of baseballs, cricket bats, tennis rackets, golf clubs, skis, and other sports equipment depends not only on the strength and weight of the materials used, but also on their abil- ity to perform under “impact” loading. The importance of mechanical properties is easy to appreciate in many of these “load-bearing” applications. In many applications, the mechanical properties of the material play an important role, even though the primary function is electrical, magnetic, optical, or biological. For example, an optical fiber must have a certain level of strength to withstand the stresses encountered in its application. A biocompatible titanium alloy used for a bone implant must have enough strength and toughness to survive in the human body for many years without failure. A scratch-resistant coating on optical lenses must resist mechanical abrasion. An aluminum alloy or a glass-ceramic substrate used as a base for building magnetic hard drives must have sufficient mechanical strength so that it will not break or crack during operation that requires rotation at high speeds. Similarly, electronic packages used to house semicon- ductor chips and the thin-film structures created on the semiconductor chip must be able to withstand stresses encountered in various applications, as well as those encountered during the heating and cooling of electronic devices. The mechanical robustness of small devices fabricated using nanotechnology is also important. Float glass used in automotive and build- ing applications must have sufficient strength and shatter resistance. Many components designed from plastics, metals, and ceramics must not only have adequate toughness and strength at room temperature but also at relatively high and low temperatures. For load-bearing applications, engineered materials are selected by matching their mechanical properties to the design specifications and service conditions required of the component. The first step in the selection process requires an analysis of the material’s application to determine its most important characteristics. Should it be strong, stiff, or ductile? Will it be subjected to an application involving high stress or sudden intense force, high stress at elevated temperature, cyclic stresses, and/or corrosive or abrasive conditions? Once we know the required properties, we can make a preliminary selection of the appro- priate material using various databases. We must, however, know how the properties listed in the handbook are obtained, know what the properties mean, and realize that the prop- erties listed are obtained from idealized tests that may not apply exactly to real-life engi- neering applications. Materials with the same nominal chemical composition and other properties can show significantly different mechanical properties as dictated by microstruc- ture. Furthermore, changes in temperature; the cyclical nature of stresses applied; the chemical changes due to oxidation, corrosion, or erosion; microstructural changes due to temperature; the effect of possible defects introduced during machining operations (e.g., grinding, welding, cutting, etc.); or other factors can also have a major effect on the mechanical behavior of materials. The mechanical properties of materials must also be understood so that we can process materials into useful shapes using materials processing techniques. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 2 Terminology for Mechanical Properties 199 Materials processing requires a detailed understanding of the mechanical properties of materials at different temperatures and conditions of loading. We must also under- stand how materials processing may change materials properties, e.g., by making a metal stronger or weaker than it was prior to processing. In the sections that follow, we define terms that are used to describe the mechan- ical properties of engineered materials. Different tests used to determine mechanical prop- erties of materials are discussed. 6-2 Terminology for Mechanical Properties There are different types of forces or “stresses” that are encountered in dealing with mechanical properties of materials. In general, we define stress as the force acting per unit area over which the force is applied. Tensile, compressive, and shear stresses are illustrated in Figure 6-1(a). Strain is defined as the change in dimension per unit length. Stress is typ- ically expressed in psi (pounds per square inch) or Pa (Pascals). Strain has no dimensions and is often expressed as in.> in. or cm> cm. Tensile and compressive stresses are normal stresses. A normal stress arises when the applied force acts perpendicular to the area of interest. Tension causes elongation in the direction of the applied force, whereas compression causes shortening. A shear stress arises when the applied force acts in a direction parallel to the area of interest. Many load- bearing applications involve tensile or compressive stresses. Shear stresses are often encoun- tered in the processing of materials using such techniques as polymer extrusion. Shear stresses Figure 6-1 (a) Tensile, compressive, and shear stresses. F is the applied force. (b) Illustration showing how Young’s modulus is defined for an elastic material. (c) For nonlinear materials, we use the slope of a tangent as a varying quantity that replaces the Young’s modulus. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 200 CHAPTER 6 Mechanical Properties: Part One are also found in structural applications. Note that even a simple tensile stress applied along one direction will cause a shear stress in other directions (e.g., see Schmid’s law, Chapter 4). Elastic strain is defined as fully recoverable strain resulting from an applied stress. The strain is “elastic” if it develops instantaneously (i.e., the strain occurs as soon as the force is applied), remains as long as the stress is applied, and is recovered when the force is withdrawn. A material subjected to an elastic strain does not show any permanent defor- mation (i.e., it returns to its original shape after the force or stress is removed). Consider stretching a stiff metal spring by a small amount and letting go. If the spring immediately returns to its original dimensions, the strain developed in the spring was elastic. In many materials, elastic stress and elastic strain are linearly related. The slope of a tensile stress-strain curve in the linear regime defines the Young’s modulus or modulus of elasticity (E) of a material [Figure 6-1(b)]. The units of E are measured in pounds per square inch (psi) or Pascals (Pa) (same as those of stress). Large elastic deformations are observed in elastomers (e.g., natural rubber, silicones), for which the relationship between elastic strain and stress is non-linear. In elastomers, the large elastic strain is related to the coiling and uncoiling of spring-like molecules (see Chapter 16). In dealing with such mate- rials, we use the slope of the tangent at any given value of stress or strain and consider that as a varying quantity that replaces the Young’s modulus [Figure 6-1(c)]. We define the shear modulus (G) as the slope of the linear part of the shear stress-shear strain curve. Permanent or plastic deformation in a material is known as the plastic strain. In this case, when the stress is removed, the material does not go back to its original shape. A dent in a car is plastic deformation! Note that the word “plastic” here does not refer to strain in a plastic (polymeric) material, but rather to permanent strain in any material. The rate at which strain develops in a material is defined as the strain rate. Units of strain rate are s-1. You will learn later in this chapter that the rate at which a material is deformed is important from a mechanical properties perspective. Many materials con- sidered to be ductile behave as brittle solids when the strain rates are high. Silly Putty® (a silicone polymer) is an example of such a material. When the strain rates are low, Silly Putty® can show significant ductility. When stretched rapidly (at high strain rates), we do not allow the untangling and extension of the large polymer molecules and, hence, the material snaps. When materials are subjected to high strain rates, we refer to this type of loading as impact loading. A viscous material is one in which the strain develops over a period of time and the material does not return to its original shape after the stress is removed. The development of strain takes time and is not in phase with the applied stress. Also, the material will remain deformed when the applied stress is removed (i.e., the strain will be plastic). A viscoelastic (or anelastic) material can be thought of as a material with a response between that of a vis- cous material and an elastic material. The term “anelastic” is typically used for metals, while the term “viscoelastic” is usually associated with polymeric materials. Many plastics (solids and molten) are viscoelastic. A common example of a viscoelastic material is Silly Putty®. In a viscoelastic material, the development of a permanent strain is similar to that in a viscous material. Unlike a viscous material, when the applied stress is removed, part of the strain in a viscoelastic material will recover over a period of time. Recovery of strain refers to a change in shape of a material after the stress causing deformation is removed. A qualitative description of development of strain as a function of time in relation to an applied force in elastic, viscous, and viscoelastic materials is shown in Figure 6-2. In viscoelastic materials held under constant strain, if we wait, the level of stress decreases over a period of time. This is known as stress relaxation. Recovery of strain and stress relaxation are different terms and should not be confused. A common example of stress relaxation is provided by the nylon strings in a tennis racket. We know that the level of stress, or the “tension,” as the tennis players call it, decreases with time. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 2 Terminology for Mechanical Properties 201 Figure 6-2 (a) Various types of strain response to an imposed stress where Tg = glass- transition temperature and Tm = melting point. (Reprinted from Materials Principles and Practice, by C. Newey and G. Weaver (Eds.), 1991 p. 300, Fig. 6-9. Copyright © 1991 Butterworth-Heinemann. Reprinted with permission from Elsevier Science.) (b) Stress relaxation in a viscoelastic material. Note the vertical axis is stress. Strain is constant. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 202 CHAPTER 6 Mechanical Properties: Part One While dealing with molten materials, liquids, and dispersions, such as paints or gels, a description of the resistance to flow under an applied stress is required. If the relationship # between the applied shear stress t and the shear strain rate (g) is linear, we refer to that material as Newtonian. The slope of the shear stress versus the steady-state shear strain rate curve is defined as the viscosity (␩) of the material. Water is an example of a Newtonian material. The following relationship defines viscosity: # t = hg (6-1) g The units of ␩ are Pa s (in the SI system) or Poise (P) or cm # s in the cgs system. # Sometimes the term centipoise (cP) is used, 1 cP = 10-2 P. Conversion between these units is given by 1 Pa # s = 10 P = 1000 cP. The kinematic viscosity (v) is defined as v = h>r (6-2) where viscosity (␩) has units of Poise and density (␳) has units of g> cm3. The kinematic viscosity unit is Stokes (St) or equivalently cm2> s. Sometimes the unit of centiStokes (cSt) is used; 1 cSt = 10-2 St. For many materials, the relationship between shear stress and shear strain rate is nonlinear. These materials are non-Newtonian. The stress versus steady state shear strain rate relationship in these materials can be described as # t = hgm (6-3) where the exponent m is not equal to 1. Non-Newtonian materials are classified as shear thinning (or pseudoplastic) or shear thickening (or dilatant). The relationships between the shear stress and shear strain rate for different types of materials are shown in Figure 6-3. If we take the slope of the line obtained by joining the origin to any point on the curve, we determine the apparent viscosity (␩app). Apparent viscosity as a function of steady-state shear strain rate is shown in Figure 6-4(a). The apparent viscosity of a Newtonian material will remain constant with changing shear strain rate. In shear thinning materials, the apparent viscosity decreases with increasing shear strain rate. In shear thickening materials, the apparent viscosity increases with increasing shear strain rate. If you have a can of paint sitting in Figure 6-3 Shear stress-shear strain rate relationships for Newtonian and non-Newtonian materials. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 2 Terminology for Mechanical Properties 203 # Figure 6-4 (a) Apparent viscosity as a function of shear strain rate (g). (b) and (c) Illustration of a Bingham plastic (Equations 6-4 and 6-5). Note the horizontal axis in (b) is shear strain. storage, for example, the shear strain rate that the paint is subjected to is very small, and the paint behaves as if it is very viscous. When you take a brush and paint, the paint is sub- jected to a high shear strain rate. The paint now behaves as if it is quite thin or less vis- cous (i.e., it exhibits a small apparent viscosity). This is the shear thinning behavior. Some materials have “ideal plastic” behavior. For an ideal plastic material, the shear stress does not change with shear strain rate. Many useful materials can be modeled as Bingham plastics and are defined by the following equations: t = G # g (when t is less than ty# s) (6-4) # t = ty # s + hg (when t Ú ty # s) (6-5) This is illustrated in Figure 6-4(b) and 6-4(c). In these equations, ty # s is the apparent yield strength obtained by interpolating the shear stress-shear strain rate data to zero shear strain rate. We define yield strength as the stress level that has to be exceeded so that the material deforms plastically. The existence of a true yield strength (sometimes also known as yield stress) has not been proven unambiguously for many plastics and dispersions such as paints. To prove the existence of a yield stress, separate measurements of stress versus strain are needed. For these materials, a critical yielding strain may be a better way to Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 204 CHAPTER 6 Mechanical Properties: Part One describe the mechanical behavior. Many ceramic slurries (dispersions such as those used in ceramic processing), polymer melts (used in polymer processing), paints and gels, and food products (yogurt, mayonnaise, ketchups, etc.) exhibit Bingham plastic-like behavior. Note that Bingham plastics exhibit shear thinning behavior (i.e., the apparent viscosity decreases with increasing shear rate). Shear thinning materials also exhibit a thixotropic behavior (e.g., paints, ceramic slurries, polymer melts, gels, etc.). Thixotropic materials usually contain some type of net- work of particles or molecules. When a sufficiently large shear strain (i.e., greater than the critical yield strain) is applied, the thixotropic network or structure breaks and the material begins to flow. As the shearing stops, the network begins to form again, and the resistance to the flow increases. The particle or molecular arrangements in the newly formed network are different from those in the original network. Thus, the behavior of thixotropic materials is said to be time and deformation history dependent. Some materials show an increase in the apparent viscosity as a function of time and at a constant shearing rate. These materials are known as rheopectic. The rheological properties of materials are determined using instruments known as a viscometer or a rheometer. In these instruments, a constant stress or constant strain rate is applied to the material being evaluated. Different geometric arrangements (e.g., cone and plate, parallel plate, Couette, etc.) are used. In the sections that follow, we will discuss different mechanical properties of solid materials and some of the testing methods to evaluate these properties. 6-3 The Tensile Test: Use of the Stress–Strain Diagram The tensile test is popular since the properties obtained can be applied to design different components. The tensile test measures the resistance of a material to a static or slowly applied force. The strain rates in a tensile test are typically small (10-4 to 10-2 s-1). A test setup is shown in Figure 6-5; a typical specimen has a diameter of 0.505 in. and a gage length of 2 in. The specimen is placed in the testing machine and a force F, called the load, is applied. A universal testing machine on which tensile and compressive tests can be Figure 6-5 A unidirectional force is applied to a specimen in the tensile test by means of the moveable crosshead. The crosshead movement can be performed using screws or a hydraulic mechanism. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 3 The Tensile Test: Use of the Stress–Strain Diagram 205 performed often is used. A strain gage or extensometer is used to measure the amount that the specimen stretches between the gage marks when the force is applied. Thus, the change in length of the specimen (⌬l ) is measured with respect to the original length (l0). Information concerning the strength, Young’s modulus, and ductility of a material can be obtained from such a tensile test. Typically, a tensile test is conducted on metals, alloys, and plastics. Tensile tests can be used for ceramics; however, these are not very popular because the sample may fracture while it is being aligned. The following discussion mainly applies to tensile testing of metals and alloys. We will briefly discuss the stress–strain behavior of polymers as well. Figure 6-6 shows qualitatively the stress–strain curves for a typical (a) metal, (b) thermoplastic material, (c) elastomer, and (d) ceramic (or glass) under relatively small strain rates. The scales in this figure are qualitative and different for each material. In practice, the actual magnitude of stresses and strains will be very different. The tempera- ture of the plastic material is assumed to be above its glass-transition temperature (Tg ). The temperature of the metal is assumed to be room temperature. Metallic and thermo- plastic materials show an initial elastic region followed by a non-linear plastic region. A separate curve for elastomers (e.g., rubber or silicones) is also included since the behavior of these materials is different from other polymeric materials. For elastomers, a large por- tion of the deformation is elastic and nonlinear. On the other hand, ceramics and glasses show only a linear elastic region and almost no plastic deformation at room temperature. When a tensile test is conducted, the data recorded includes load or force as a func- tion of change in length (⌬l ) as shown in Table 6-1 for an aluminum alloy test bar. These data are then subsequently converted into stress and strain. The stress-strain curve is ana- lyzed further to extract properties of materials (e.g., Young’s modulus, yield strength, etc.). Engineering Stress and Strain The results of a single test apply to all sizes and cross-sections of specimens for a given material if we convert the force to g Figure 6-6 Tensile stress–strain curves for different materials. Note that these are qualitative. The magnitudes of the stresses and strains should not be compared. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 206 CHAPTER 6 Mechanical Properties: Part One TABLE 6-1 The results of a tensile test of a 0.505-in. diameter aluminum alloy test bar, initial length (l0) = 2 in. Calculated Load (lb) Change in Length (in.) Stress (psi) Strain (in./in.) 0 0.000 0 0 1000 0.001 4,993 0.0005 3000 0.003 14,978 0.0015 5000 0.005 24,963 0.0025 7000 0.007 34,948 0.0035 7500 0.030 37,445 0.0150 7900 0.080 39,442 0.0400 8000 (maximum load) 0.120 39,941 0.0600 7950 0.160 39,691 0.0800 7600 (fracture) 0.205 37,944 0.1025 stress and the distance between gage marks to strain. Engineering stress and engineering strain are defined by the following equations: F Engineering stress = S = (6-6) A0 ¢l Engineering stress = e = (6-7) l0 where A0 is the original cross-sectional area of the specimen before the test begins, l0 is the original distance between the gage marks, and ⌬l is the change in length after force F is applied. The conversions from load and sample length to stress and strain are included in Table 6-1. The stress-strain curve (Figure 6-7) is used to record the results of a tensile test. Engineering stress S (psi) S e S e Engineering strain e (in./in.) Figure 6-7 The engineering stress–strain curve for an aluminum alloy from Table 6-1. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 3 The Tensile Test: Use of the Stress–Strain Diagram 207 Example 6-1 Tensile Testing of Aluminum Alloy Convert the change in length data in Table 6-1 to engineering stress and strain and plot a stress-strain curve. SOLUTION For the 1000-lb load: F 1000 lb > S = = = 4,993 psi A0 (p 4)(0.505 in)2 ¢l 0.001 in. e = = = 0.0005 in./in. l0 2.000 in. The results of similar calculations for each of the remaining loads are given in Table 6-1 and are plotted in Figure 6-7. Units Many different units are used to report the results of the tensile test. The most common units for stress are pounds per square inch (psi) and MegaPascals (MPa). The units for strain include inch> inch, centimeter> centimeter, and meter> meter, and thus, strain is often written as unitless. The conversion factors for stress are summarized in Table 6-2. Because strain is dimensionless, no conversion factors are required to change the system of units. TABLE 6-2 Units and conversion factors 1 pound (lb) = 4.448 Newtons (N) 1 psi = pounds per square inch 1 MPa = MegaPascal = MegaNewtons per square meter (MN> m2) = Newtons per square millimeter (N> mm2) = 106 Pa 1 GPa = 1000 MPa = GigaPascal 1 ksi = 1000 psi = 6.895 MPa 1 psi = 0.006895 MPa 1 MPa = 0.145 ksi = 145 psi Example 6-2 Design of a Suspension Rod An aluminum rod is to withstand an applied force of 45,000 pounds. The engineering stress–strain curve for the aluminum alloy to be used is shown in Figure 6-7. To ensure safety, the maximum allowable stress on the rod is limited to 25,000 psi, which is below the yield strength of the aluminum. The rod must be at least 150 in. long but must deform elastically no more than 0.25 in. when the force is applied. Design an appropriate rod. SOLUTION From the definition of engineering strain, ¢l e = l0 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 208 CHAPTER 6 Mechanical Properties: Part One For a rod that is 150 in. long, the strain that corresponds to an extension of 0.25 in. is 0.25 in. e = = 0.00167 150 in. According to Figure 6-7, this strain is purely elastic, and the corresponding stress value is approximately 17,000 psi, which is below the 25,000 psi limit. We use the def- inition of engineering stress to calculate the required cross-sectional area of the rod: F S = A0 Note that the stress must not exceed 17,000 psi, or consequently, the deflection will be greater than 0.25 in. Rearranging, F 45,000 lb A0 = = = 2.65 in.2 S 17,000 psi The rod can be produced in various shapes, provided that the cross-sectional area is 2.65 in.2 For a round cross section, the minimum diameter to ensure that the stress is not too high is pd2 A0 = = 2.65 in.2 or d = 1.84 in. 4 Thus, one possible design that meets all of the specified criteria is a suspension rod that is 150 in. long with a diameter of 1.84 in. 6-4 Properties Obtained from the Tensile Test Yield Strength As we apply stress to a material, the material initially exhibits elastic deformation. The strain that develops is completely recovered when the applied stress is removed. As we continue to increase the applied stress, the material eventually “yields” to the applied stress and exhibits both elastic and plastic deformation. The critical stress value needed to initiate plastic deformation is defined as the elastic limit of the material. In metal- lic materials, this is usually the stress required for dislocation motion, or slip, to be initiated. In polymeric materials, this stress will correspond to disentanglement of polymer molecule chains or sliding of chains past each other. The proportional limit is defined as the level of stress above which the relationship between stress and strain is not linear. In most materials, the elastic limit and proportional limit are quite close; however, neither the elastic limit nor the proportional limit values can be determined precisely. Measured values depend on the sensitivity of the equipment used. We, therefore, define them at an offset strain value (typically, but not always, 0.002 or 0.2%). We then draw a line parallel to the linear portion of the engineering stress-strain curve starting at this offset value of strain. The stress value corresponding to the intersection of this line and the engineering stress-strain curve is defined as the offset yield strength, also often stated as the yield strength. The 0.2% off- set yield strength for gray cast iron is 40,000 psi as shown in Figure 6-8(a). Engineers normally prefer to use the offset yield strength for design purposes because it can be reliably determined. For some materials, the transition from elastic deformation to plastic flow is rather abrupt. This transition is known as the yield point phenomenon. In these materials, Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 4 Properties Obtained from the Tensile Test 209 Engineering stress S Engineering stress S (psi) S S S e Engineering strain e Engineering strain e (in./in.) Figure 6-8 (a) Determining the 0.2% offset yield strength in gray cast iron, and (b) upper and lower yield point behavior in a low carbon steel. as plastic deformation begins, the stress value drops first from the upper yield point (S2) [Figure 6-8(b)]. The stress value then oscillates around an average value defined as the lower yield point (S1). For these materials, the yield strength is usually defined from the 0.2% strain offset. The stress-strain curve for certain low-carbon steels displays the yield point phe- nomenon [Figure 6-8(b)]. The material is expected to plastically deform at stress S1; how- ever, small interstitial atoms clustered around the dislocations interfere with slip and raise the yield point to S2. Only after we apply the higher stress S2 do the dislocations slip. After slip begins at S2, the dislocations move away from the clusters of small atoms and continue to move very rapidly at the lower stress S1. When we design parts for load-bearing applications, we prefer little or no plastic deformation. As a result, we must select a material such that the design stress is consider- ably lower than the yield strength at the temperature at which the material will be used. We can also make the component cross-section larger so that the applied force produces a stress that is well below the yield strength. On the other hand, when we want to shape materials into components (e.g., take a sheet of steel and form a car chassis), we need to apply stresses that are well above the yield strength. Tensile Strength The stress obtained at the highest applied force is the ten- sile strength (SUTS), which is the maximum stress on the engineering stress-strain curve. This value is also commonly known as the ultimate tensile strength. In many ductile mate- rials, deformation does not remain uniform. At some point, one region deforms more than others and a large local decrease in the cross-sectional area occurs (Figure 6-9). This locally deformed region is called a “neck.” This phenomenon is known as necking. Because the cross-sectional area becomes smaller at this point, a lower force is required to continue its deformation, and the engineering stress, calculated from the original area A0, decreases. The tensile strength is the stress at which necking begins in ductile metals. In compression testing, the materials will bulge; thus necking is seen only in a tensile test. Figure 6-10 shows typical yield strength values for various engineering materials. Ultra-pure metals have a yield strength of ⬃ 1 - 10 MPa. On the other hand, the yield strength of alloys is higher. Strengthening in alloys is achieved using different mechanisms Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 210 CHAPTER 6 Mechanical Properties: Part One Figure 6-9 Localized deformation of a ductile material during a tensile test produces a necked region. The micrograph shows a necked region in a fractured sample. (This article was published in Materials Principles and Practice, Charles Newey and Graham Weaver (Eds.), Figure 6.9, p. 300, Coyright Open University.) 14,500,000 1,450,000 145,000 MPa Yield strength (psi) 14,500 1,450 145 Figure 6-10 Typical yield strength values for different engineering materials. Note that values shown are in MPa and psi. (Reprinted from Engineering Materials I, Second Edition, M.F. Ashby and D.R.H. Jones, 1996, Fig. 8-12, p. 85. Copyright © 1996 Butterworth- Heinemann. Reprinted with permission from Elsevier Science.) Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 4 Properties Obtained from the Tensile Test 211 described before (e.g., grain size, solid solution formation, strain hardening, etc.). The yield strength of a particular metal or alloy is usually the same for tension and compres- sion. The yield strength of plastics and elastomers is generally lower than metals and alloys, ranging up to about 10 - 100 MPa. The values for ceramics (Figure 6-10) are for compressive strength (obtained using a hardness test). Tensile strength of most ceramics is much lower (⬃100–200 MPa). The tensile strength of glasses is about ⬃70 MPa and depends on surface flaws. Elastic Properties The modulus of elasticity, or Young’s modulus (E), is the slope of the stress-strain curve in the elastic region. This relationship between stress and strain in the elastic region is known as Hooke’s Law: S E = (6-8) e The modulus is closely related to the binding energies of the atoms. (Figure 2-18). A steep slope in the force-distance graph at the equilibrium spacing (Figure 2-19) indicates that high forces are required to separate the atoms and cause the material to stretch elastically. Thus, the material has a high modulus of elasticity. Binding forces, and thus the modulus of elasticity, are typically higher for high melting point materials (Table 6-3). In metallic materials, the modulus of elasticity is considered a microstructure insensitive property since the value is dominated by the stiffness of atomic bonds. Grain size or other microstructural features do not have a very large effect on the Young’s modulus. Note that Young’s modulus does depend on such factors as orientation of a single crystal material (i.e., it depends upon crystallographic direction). For ceramics, the Young’s modulus depends on the level of porosity. The Young’s modulus of a composite depends upon the stiffness and amounts of the individual components. The stiffness of a component is proportional to its Young’s modulus. (The stiffness also depends on the component dimensions.) A component with a high modulus of elasticity will show much smaller changes in dimensions if the applied stress causes only elastic defor- mation when compared to a component with a lower elastic modulus. Figure 6-11 compares the elastic behavior of steel and aluminum. If a stress of 30,000 psi is applied to each material, the steel deforms elastically 0.001 in./in.; at the same stress, aluminum deforms 0.003 in./in. The elastic modulus of steel is about three times higher than that of aluminum. Figure 6-12 shows the ranges of elastic moduli for various engineering materials. The modulus of elasticity of plastics is much smaller than that for metals or ceramics and glasses. For example, the modulus of elasticity of nylon is 2.7 GPa (⬃ 0.4 * 106 psi); the modulus of glass fibers is 72 GPa (⬃ 10.5 * 106 psi). The Young’s modulus of composites TABLE 6-3 Elastic properties and melting temperature (Tm) of selected materials Material Tm (°C) E (psi) Poisson’s ratio (␯) Pb 327 2.0 * 106 0.45 Mg 650 6.5 * 106 0.29 Al 660 10.0 * 106 0.33 Cu 1085 18.1 * 106 0.36 Fe 1538 30.0 * 106 0.27 W 3410 59.2 * 106 0.28 Al2O3 2020 55.0 * 106 0.26 Si3N4 44.0 * 106 0.24 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 212 CHAPTER 6 Mechanical Properties: Part One Figure 6-11 Comparison of the elastic behavior of steel and aluminum. For a given stress, aluminum deforms elastically three times as much as does steel Engineering stress S (psi) (i.e., the elastic modulus of aluminum is about three times lower than that of steel). Engineering strain e (in./in.) 145,000 14,500 1,450 E (GPa) E (ksi) 145 14.5 1.45 0.145 Figure 6-12 Range of elastic moduli for different engineering materials. Note: Values are shown in GPa and ksi. (Reprinted from Engineering Materials I, Second Edition, M.F. Ashby and D.R.H. Jones, 1996, Fig. 3-5, p. 35, Copyright © 1996 Butterworth- Heinemann. Reprinted with permission from Elsevier Science.) Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 4 Properties Obtained from the Tensile Test 213 such as glass fiber-reinforced composites (GFRC) or carbon fiber-reinforced composites (CFRC) lies between the values for the matrix polymer and the fiber phase (carbon or glass fibers) and depends upon their relative volume fractions. The Young’s modulus of many alloys and ceramics is higher, generally ranging up to 410 GPa (⬃60,000 ksi). Ceramics, because of the strength of ionic and covalent bonds, have the highest elastic moduli. Poisson’s ratio, ␯, relates the longitudinal elastic deformation produced by a sim- ple tensile or compressive stress to the lateral deformation that occurs simultaneously: -elateral n = e (6-9) longitudinal For many metals in the elastic region, the Poisson’s ratio is typically about 0.3 (Table 6-3). During a tensile test, the ratio increases beyond yielding to about 0.5, since during plastic deformation, volume remains constant. Some interesting structures, such as some honey- comb structures and foams, exhibit negative Poisson’s ratios. Note: Poisson’s ratio should not be confused with the kinematic viscosity, both of which are denoted by the Greek letter ␯. The modulus of resilience (Er), the area contained under the elastic portion of a stress-strain curve, is the elastic energy that a material absorbs during loading and subse- quently releases when the load is removed. For linear elastic behavior: Er = a b (yield strength)(strain at yielding) 1 (6-10) 2 The ability of a spring or a golf ball to perform satisfactorily depends on a high modulus of resilience. Tensile Toughness The energy absorbed by a material prior to fracture is known as tensile toughness and is sometimes measured as the area under the true stress–strain curve (also known as the work of fracture). We will define true stress and true strain in Section 6-5. Since it is easier to measure engineering stress–strain, engineers often equate tensile toughness to the area under the engineering stress–strain curve. Example 6-3 Young’s Modulus of an Aluminum Alloy Calculate the modulus of elasticity of the aluminum alloy for which the engineer- ing stress–strain curve is shown in Figure 6-7. Calculate the length of a bar of ini- tial length 50 in. when a tensile stress of 30,000 psi is applied. SOLUTION When a stress of 34,948 psi is applied, a strain of 0.0035 in.> in. is produced. Thus, S 34,948 psi Modulus of elasticity = E = = = 10 * 106 psi e 0.0035 Note that any combination of stress and strain in the elastic region will produce this result. From Hooke’s Law, = 0.003 in.> in. S 30,000 psi e = = E 10 * 106 psi Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 214 CHAPTER 6 Mechanical Properties: Part One From the definition of engineering strain, ¢l e = l0 Thus, ¢l = e(l0) = 0.003 in.> in.(50 in.) = 0.15 in. When the bar is subjected to a stress of 30,000 psi, the total length is given by l = ¢l + l0 = 0.15 in. + 50 in. = 50.15 in. Ductility Ductility is the ability of a material to be permanently deformed with- out breaking when a force is applied. There are two common measures of ductility. The percent elongation quantifies the permanent plastic deformation at failure (i.e., the elastic deformation recovered after fracture is not included) by measuring the distance between gage marks on the specimen before and after the test. Note that the strain after failure is smaller than the strain at the breaking point, because the elastic strain is recovered when the load is removed. The percent elongation can be written as lf - l0 % Elongation = * 100 (6-11) l0 where lf is the distance between gage marks after the specimen breaks. A second approach is to measure the percent change in the cross-sectional area at the point of fracture before and after the test. The percent reduction in area describes the amount of thinning undergone by the specimen during the test: A0 - Af % Reduction in area = * 100 (6-12) A0 where Af is the final cross-sectional area at the fracture surface. Ductility is important to both designers of load-bearing components and man- ufacturers of components (bars, rods, wires, plates, I-beams, fibers, etc.) utilizing materi- als processing. Example 6-4 Ductility of an Aluminum Alloy The aluminum alloy in Example 6-1 has a final length after failure of 2.195 in. and a final diameter of 0.398 in. at the fractured surface. Calculate the ductility of this alloy. SOLUTION lf - l0 2.195 - 2.000 % Elongation = * 100 = * 100 = 9.75% l0 2.000 A0 - Af % Reduction in area = * 100 A0 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 4 Properties Obtained from the Tensile Test 215 (p/4)(0.505)2 - (p/4)(0.398)2 = * 100 (p/4)(0.505)2 = 37.9% The final length is less than 2.205 in. (see Table 6-1) because, after fracture, the elastic strain is recovered. Effect of Temperature Mechanical properties of materials depend on temperature (Figure 6-13). Yield strength, tensile strength, and modulus of elasticity decrease at higher temperatures, whereas ductility commonly increases. A materials fab- ricator may wish to deform a material at a high temperature (known as hot working) to take advantage of the higher ductility and lower required stress. We use the term “high temperature” here with a note of caution. A high temper- ature is defined relative to the melting temperature. Thus, 500°C is a high temperature for aluminum alloys; however, it is a relatively low temperature for the processing of steels. In metals, the yield strength decreases rapidly at higher temperatures due to a decreased dis- location density and an increase in grain size via grain growth (Chapter 5) or a related process known as recrystallization (as described later in Chapter 8). Similarly, any strengthening that may have occurred due to the formation of ultrafine precipitates may also decrease as the precipitates begin to either grow in size or dissolve into the matrix. We will discuss these effects in greater detail in later chapters. When temperatures are reduced, many, but not all, metals and alloys become brittle. Increased temperatures also play an important role in forming polymeric mat- erials and inorganic glasses. In many polymer-processing operations, such as extrusion or the stretch-blow process (Chapter 16), the increased ductility of polymers at higher tem- peratures is advantageous. Again, a word of caution concerning the use of the term “high temperature.” For polymers, the term “high temperature” generally means a temperature Engineering stress S Engineering strain e Figure 6-13 The effect of temperature (a) on the stress–strain curve and (b) on the tensile properties of an aluminum alloy. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 216 CHAPTER 6 Mechanical Properties: Part One higher than the glass-transition temperature (Tg). For our purpose, the glass-transition temperature is a temperature below which materials behave as brittle materials. Above the glass-transition temperature, plastics become ductile. The glass-transition temperature is not a fixed temperature, but depends on the rate of cooling as well as the polymer molecular weight distribution. Many plastics are ductile at room temperature because their glass-transition temperatures are below room temperature. To summarize, many polymeric materials will become harder and more brittle as they are exposed to temper- atures that are below their glass-transition temperatures. The reasons for loss of ductil- ity at lower temperatures in polymers and metallic materials are different; however, this is a factor that played a role in the failures of the Titanic in 1912 and the Challenger in 1986. Ceramic and glassy materials are generally considered brittle at room tem- perature. As the temperature increases, glasses can become more ductile. As a result, glass processing (e.g., fiber drawing or bottle manufacturing) is performed at high temperatures. 6-5 True Stress and True Strain The decrease in engineering stress beyond the tensile strength on an engineering stress–strain curve is related to the definition of engineering stress. We used the original area A0 in our calculations, but this is not precise because the area continually changes. We define true stress and true strain by the following equations: F True stress = s = (6-13) A l = ln a b dl l True strain = e = (6-14) Ll0 l l 0 where A is the instantaneous area over which the force F is applied, l is the instantaneous sample length, and l0 is the initial length. In the case of metals, plastic deformation is essentially a constant-volume process (i.e., the creation and propagation of dislocations results in a negligible volume change in the material). When the constant volume assump- tion holds, we can write A0l0 A0l0 = Al or A = (6-15) l and using the definitions of engineering stress S and engineering strain e, Equation 6-13 can be written as a b = Sa b = S(1 + e) F F l l0 + ¢l s = = (6-16) A A0 l0 l0 It can also be shown that e = ln(1 + e) (6-17) Thus, it is a simple matter to convert between the engineering stress–strain and true stress–strain systems. Note that the expressions in Equations 6-16 and 6-17 are not valid Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 5 True Stress and True Strain 217 UTS UTS (a) (b) Figure 6-14 (a) The relation between the true stress–true strain diagram and engineering stress–engineering strain diagram. The curves are nominally identical to the yield point. The true stress corresponding to the ultimate tensile strength (UTS) is indicated. (b) Typically true stress–strain curves must be truncated at the true stress corresponding to the ultimate tensile strength, since the cross-sectional area at the neck is unknown. after the onset of necking, because after necking begins, the distribution of strain along the gage length is not uniform. After necking begins, Equation 6-13 must be used to cal- culate the true stress and the expression e = ln a b A0 (6-18) A must be used to calculate the true strain. Equation 6-18 follows from Equations 6-14 and 6-15. After necking the instantaneous area A is the cross-sectional area of the neck Aneck. The true stress–strain curve is compared to the engineering stress–strain curve in Figure 6-14(a). There is no maximum in the true stress–true strain curve. Note that it is difficult to measure the instantaneous cross-sectional area of the neck. Thus, true stress–strain curves are typically truncated at the true stress that corre- sponds to the ultimate tensile strength, as shown in Figure 6-14(b). Example 6-5 True Stress and True Strain Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 6-1 at (a) the maximum load and (b) fracture. The diameter at maximum load is 0.4905 in. and at fracture is 0.398 in. SOLUTION (a) At the maximum load, F 8000 lb (p> 4)(0.505 in)2 Engineering stress S = = = 39,941 psi A0 = 0.060 in.> in. ¢l 2.120 - 2.000 Engineering strain e = = l0 2.000 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 218 CHAPTER 6 Mechanical Properties: Part One True stress = s = S(1 + e) = 39,941(1 + 0.060) = 42,337 psi True strain = ln (1 + e) = ln (1 + 0.060) = 0.058 in.> in. The maximum load is the last point at which the expressions used here for true stress and true strain apply. Note that the same answers are obtained for true stress and strain if the instantaneous dimensions are used: F 8000 lb s = = = 42,337 psi A (p/4)(0.4905 in)2 (p> 4)(0.505 in2) e = ln Q R d = 0.058 in.> in. A0 (p> 4)(0.4905 in2) = Inc A Up until the point of necking in a tensile test, the engineering stress is less than the corresponding true stress, and the engineering strain is greater than the correspon- ding true strain. (b) At fracture, F 7600 lb (p> 4)(0.505 in)2 S = = = 37,944 psi A0 ¢l 2.205 - 2.000 e = = = 0.103 in./in. l0 2.000 F 7600 lb (p> 4)(0.398 in)2 s = = = 61,088 psi Af (p> 4)(0.505 in2) e = ln a b = ln c d = ln (1.601) = 0.476 in.> in. A0 Af (p> 4)(0.398 in2) It was necessary to use the instantaneous dimensions to calculate the true stress and strain, since failure occurs past the point of necking. After necking, the true strain is greater than the corresponding engineering strain. 6-6 The Bend Test for Brittle Materials In ductile metallic materials, the engineering stress–strain curve typically goes through a maximum; this maximum stress is the tensile strength of the material. Failure occurs at a lower engineering stress after necking has reduced the cross-sectional area supporting the load. In more brittle materials, failure occurs at the maximum load, where the tensile strength and breaking strength are the same (Figure 6-15). In many brittle materials, the normal tensile test cannot easily be performed because of the presence of flaws at the surface. Often, just placing a brittle material in the grips of the tensile testing machine causes cracking. These materials may be tested using the bend test [Figure 6-16(a)]. By applying the load at three points and causing bending, a tensile force acts on the material opposite the midpoint. Fracture begins at this loca- tion. The flexural strength, or modulus of rupture, describes the material’s strength: 3FL Flexural strength for three - point bend test sbend = (6-19) 2wh2 Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 6 The Bend Test for Brittle Materials 219 Figure 6-15 The engineering stress–strain behavior of brittle materials compared with that of more ductile Engineering stress S materials. Engineering strain e Figure 6-16 (a) The bend test often used for measuring the strength of brittle materials, and (b) the deflection ␦ obtained by bending. where F is the fracture load, L is the distance between the two outer points, w is the width of the specimen, and h is the height of the specimen. The flexural strength has units of stress. The results of the bend test are similar to the stress-strain curves; however, the stress is plotted versus deflection rather than versus strain (Figure 6-17). The corresponding bending moment diagram is shown in Figure 6-18(a). The modulus of elasticity in bending, or the flexural modulus (Ebend), is calcu- lated as L3F Flexural modulus Ebend = (6-20) 4wh3d where ␦ is the deflection of the beam when a force F is applied. This test can also be conducted using a setup known as the four-point bend test [Figure 6-18(b)]. The maximum stress or flexural stress for a four-point bend test is given by 3FL1 sbend = (6-21) 4wh2 for the specific case in which L1 = L> 4 in Figure 6-18(b). Note that the derivations of Equations 6-19 through 6-21 assume a linear stress–strain response (and thus cannot be correctly applied to many polymers). The four-point bend Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 220 CHAPTER 6 Mechanical Properties: Part One Figure 6-17 Stress-deflection curve for an MgO ceramic obtained from a bend test. Cross-section h w Figure 6-18 (a) Three-point and (b) four-point bend test setup. test is better suited for testing materials containing flaws. This is because the bending moment between the inner platens is constant [Figure 6-18(b)]; thus samples tend to break randomly unless there is a flaw that locally raises the stress. Since cracks and flaws tend to remain closed in compression, brittle materials such as concrete are often incorporated into designs so that only compressive stresses act on the part. Often, we find that brittle materials fail at much higher compressive stresses than tensile stresses (Table 6-4). This is why it is possible to support a fire truck on four coffee cups; however, ceramics have very limited mechanical toughness. Hence, when we drop a ceramic coffee cup, it can break easily. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 7 Hardness of Materials 221 TABLE 6-4 Comparison of the tensile, compressive, and flexural strengths of selected ceramic and composite materials Compressive Flexural Material Tensile Strength (psi) Strength (psi) Strength (psi) Polyester—50% glass fibers 23,000 32,000 45,000 Polyester—50% glass fiber fabric 37,000 27,000a 46,000 Al2O3 (99% pure) 30,000 375,000 50,000 SiC (pressureless-sintered) 25,000 560,000 80,000 aA number of composite materials are quite poor in compression. Example 6-6 Flexural Strength of Composite Materials The flexural strength of a composite material reinforced with glass fibers is 45,000 psi, and the flexural modulus is 18 * 106 psi. A sample, which is 0.5 in. wide, 0.375 in. high, and 8 in. long, is supported between two rods 5 in. apart. Determine the force required to fracture the material and the deflection of the sample at fracture, assum- ing that no plastic deformation occurs. SOLUTION Based on the description of the sample, w = 0.5 in., h = 0.375 in., and L = 5 in. From Equation 6-19: 3FL (3)(F)(5 in.) 45,000 psi = 2 = = 106.7F 2wh (2)(0.5 in.)(0.375 in.)2 45,000 F = = 422 lb 106.7 Therefore, the deflection, from Equation 6-20, is L3F (5 in.)3(422 lb) 18 * 106 psi = 3 = 4wh d (4)(0.5 in.)(0.375 in.)3d d = 0.0278 in. In this calculation, we assumed a linear relationship between stress and strain and also that there is no viscoelastic behavior. 6-7 Hardness of Materials The hardness test measures the resistance to penetration of the surface of a material by a hard object. Hardness as a term is not defined precisely. Hardness, depending upon the context, can represent resistance to scratching or indentation and a qualitative measure of the strength of the material. In general, in macrohardness measurements, the load applied is ⬃2 N. A variety of hardness tests have been devised, but the most commonly used are Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 222 CHAPTER 6 Mechanical Properties: Part One Figure 6-19 Indenters for the Brinell and Rockwell hardness tests. the Rockwell test and the Brinell test. Different indenters used in these tests are shown in Figure 6-19. In the Brinell hardness test, a hard steel sphere (usually 10 mm in diameter) is forced into the surface of the material. The diameter of the impression, typically 2 to 6 mm, is measured and the Brinell hardness number (abbreviated as HB or BHN) is calculated from the following equation: 2F HB = (6-22) pD cD - 3D2 - D2i d where F is the applied load in kilograms, D is the diameter of the indenter in millimeters, and Di is the diameter of the impression in millimeters. The Brinell hardness has units of kg> mm2. The Rockwell hardness test uses a small-diameter steel ball for soft materials and a diamond cone, or Brale, for harder materials. The depth of penetration of the indenter is automatically measured by the testing machine and converted to a Rockwell hardness number (HR). Since an optical measurement of the indentation dimensions is not needed, the Rockwell test tends to be more popular than the Brinell test. Several variations of the Rockwell test are used, including those described in Table 6-5. A Rockwell C (HRC) test is used for hard steels, whereas a Rockwell F (HRF) test might be selected for aluminum. Rockwell tests provide a hardness number that has no units. Hardness numbers are used primarily as a qualitative basis for comparison of materials, specifications for manufacturing and heat treatment, quality control, and TABLE 6-5 Comparison of typical hardness tests Test Indenter Load Application Brinell 10-mm ball 3000 kg Cast iron and steel Brinell 10-mm ball 500 kg Nonferrous alloys Rockwell A Brale 60 kg Very hard materials Rockwell B 1/16-in. ball 100 kg Brass, low-strength steel Rockwell C Brale 150 kg High-strength steel Rockwell D Brale 100 kg High-strength steel Rockwell E 1/8-in. ball 100 kg Very soft materials Rockwell F 1/16-in. ball 60 kg Aluminum, soft materials Vickers Diamond square pyramid 10 kg All materials Knoop Diamond elongated pyramid 500 g All materials Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.  6 - 8 Nanoindentation 223 correlation with other properties of materials. For example, Brinell hardness is related to the tensile strength of steel by the approximation: Tensile strength (psi) = 500HB (6-23) where HB has units of kg> mm2. Hardness correlates well with wear resistance. A separate test is available for meas- uring the wear resistance. A material used in crushing or grinding of ores should be very hard to ensure that the material is not eroded or abraded by the hard feed materials. Similarly, gear teeth in the transmission or the drive system of a vehicle should be hard enough that the teeth do not wear out. Typically we find that polymer materials are excep- tionally soft, metals and alloys have intermediate hardness, and ceramics are exceptionally hard. We use materials such as tungsten carbide-cobalt composite (WC-Co), known as “carbide,” for cutting tool applications. We also use microcrystalline diamond or diamond- like carbon (DLC) materials for cutting tools and other applications. The Knoop hardness (HK) test is a microhardness test, forming such small inden- tations that a microscope is required to obtain the measurement. In these tests, the load applied is less than 2 N. The Vickers test, which uses a diamond pyramid indenter, can be conducted either as a macro or microhardness test. Microhardness tests are suitable for materials that may have a surface that has a higher hardness than the bulk, materials in which different areas show different levels of hardness, or samples that are not macro- scopically flat. 6-8 Nanoindentation The hardness tests described in the previous section are known as macro or microhardness tests because the indentations have dimensions on the order of millimeters or microns. The advantages of such tests are that they are relatively quick, easy, and inexpensive. Some of the disadvantages are that they can only be used on macroscale samples and hardness is the only materials property that can be directly measured. Nanoindentation is hardness testing performed at the nanometer length scale. A small diamond tip is used to indent the material of interest. The imposed load and displacement are continuously measured with micro-Newton and sub-nanometer resolution, respectively. Both load and displacement are measured throughout the indentation process. Nanoindentation techniques are impor- tant for measuring the mechanical properties of thin films on substrates (such as for micro- electronics applications) and nanophase materials and for deforming free-standing micro and nanoscale structures. Nanoindentation can be performed with high positioning accuracy, permitting indentations within selected grains of a material. Nanoindenters incorporate optical microscopes and sometimes a scanning probe microscope capability. Both hardness and elastic modulus are measured using nanoindentation. Nanoindenter tips come in a variety of shapes. A common shape is known as the Berkovich indenter, which is a three-sided pyramid. An indentation made by a Berkovich indenter is shown in Figure 6-20. The indentation in the figure measures 12.5 ␮m on each side and about 1.6 ␮m deep. The first step of a nanoindentation test involves performing indentations on a calibration standard. Fused silica is a common calibration standard, because it has homogeneous and well-characterized mechanical properties (elastic modulus E = 72 GPa and Poisson’s ratio ␯ = 0.17). The purpose of performing indentations on the calibration Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 224 CHAPTER 6 Mechanical Properties: Part One Figure 6-20 An indentation in a Zr41.2Ti13.8Cu12.5Ni10.0Be22.5 bulk metallic glass made using a Berkovich tip in a nanoindenter. (Courtesy of Gang Feng, Villanova University.) standard is to determine the projected contact area of the indenter tip Ac as a function of indentation depth. For a perfect Berkovich tip, Ac = 24.5 h2c (6-24) This function relates the cross-sectional area of the indenter to the distance from the tip hc that is in contact with the material being indented. No tip is perfectly sharp, and the tip wears and changes shape with each use. Thus, a calibration must be performed each time the tip is used as will be discussed below. The total indentation depth h (as measured by the displacement of the tip) is the sum of the contact depth hc and the depth hs at the periphery of the indentation where the indenter does not make contact with the material surface, i.e., h = hc + hs (6-25) as shown in Figure 6-21. The surface displacement term hs is calculated according to Pmax hs = e (6-26)