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7

/ Mechanical Properties

A

modern Rockwell hardness tester. (Photograph courtesy of

Wilson Instruments Division, Instron Corporation, originator of the Rockwell威 Hardness Tester.)

Why Study Mechanical Properties? It is incumbent on engineers to understand how the various mechanical properties are measured and what these properties represent; they may be called upon to design structures/components using prede-

termined materials such that unacceptable levels of deformation and/or failure will not occur. We demonstrate this procedure with respect to the design of a tensile-testing apparatus in Design Example 7.1.

147

Learning Objectives After studying this chapter you should be able to do the following: 1. Define engineering stress and engineering strain. 2. State Hooke’s law, and note the conditions under which it is valid. 3. Define Poisson’s ratio. 4. Given an engineering stress–strain diagram, determine (a) the modulus of elasticity, (b) the yield strength (0.002 strain offset), and (c) the tensile strength, and (d) estimate the percent elongation. 5. For the tensile deformation of a ductile cylindrical specimen, describe changes in specimen profile to the point of fracture. 6. Compute ductility in terms of both percent elongation and percent reduction of area for a material that is loaded in tension to fracture.

7. Compute the flexural strengths of ceramic rod specimens that have bent to fracture in threepoint loading. 8. Make schematic plots of the three characteristic stress–strain behaviors observed for polymeric materials. 9. Name the two most common hardness-testing techniques; note two differences between them. 10. (a) Name and briefly describe the two different microhardness testing techniques, and (b) cite situations for which these techniques are generally used. 11. Compute the working stress for a ductile material.

7.1 INTRODUCTION Many materials, when in service, are subjected to forces or loads; examples include the aluminum alloy from which an airplane wing is constructed and the steel in an automobile axle. In such situations it is necessary to know the characteristics of the material and to design the member from which it is made such that any resulting deformation will not be excessive and fracture will not occur. The mechanical behavior of a material reflects the relationship between its response or deformation to an applied load or force. Important mechanical properties are strength, hardness, ductility, and stiffness. The mechanical properties of materials are ascertained by performing carefully designed laboratory experiments that replicate as nearly as possible the service conditions. Factors to be considered include the nature of the applied load and its duration, as well as the environmental conditions. It is possible for the load to be tensile, compressive, or shear, and its magnitude may be constant with time, or it may fluctuate continuously. Application time may be only a fraction of a second, or it may extend over a period of many years. Service temperature may be an important factor. Mechanical properties are of concern to a variety of parties (e.g., producers and consumers of materials, research organizations, government agencies) that have differing interests. Consequently, it is imperative that there be some consistency in the manner in which tests are conducted, and in the interpretation of their results. This consistency is accomplished by using standardized testing techniques. Establishment and publication of these standards are often coordinated by professional societies. In the United States the most active organization is the American Society for Testing and Materials (ASTM). Its Annual Book of ASTM Standards comprises numerous volumes, which are issued and updated yearly; a large number of these standards relate to mechanical testing techniques. Several of these are referenced by footnote in this and subsequent chapters. The role of structural engineers is to determine stresses and stress distributions within members that are subjected to well-defined loads. This may be accomplished by experimental testing techniques and/or by theoretical and mathematical stress

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analyses. These topics are treated in traditional stress analysis and strength of materials texts. Materials and metallurgical engineers, on the other hand, are concerned with producing and fabricating materials to meet service requirements as predicted by these stress analyses. This necessarily involves an understanding of the relationships between the microstructure (i.e., internal features) of materials and their mechanical properties. Materials are frequently chosen for structural applications because they have desirable combinations of mechanical characteristics. This chapter discusses the stress–strain behaviors of metals, ceramics, and polymers and the related mechanical properties; it also examines their other important mechanical characteristics. Discussions of the microscopic aspects of deformation mechanisms and methods to strengthen and regulate the mechanical behaviors are deferred to Chapter 8.

7.2 CONCEPTS

OF

STRESS

AND

STRAIN

If a load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behavior may be ascertained by a simple stress–strain test; these are most commonly conducted for metals at room temperature. There are three principal ways in which a load may be applied: namely, tension, compression, and shear (Figures 7.1a, b, c). In engineering practice many loads are torsional rather than pure shear; this type of loading is illustrated in Figure 7.1d.

TENSION TESTS1 One of the most common mechanical stress–strain tests is performed in tension. As will be seen, the tension test can be used to ascertain several mechanical properties of materials that are important in design. A specimen is deformed, usually to fracture, with a gradually increasing tensile load that is applied uniaxially along the long axis of a specimen. A standard tensile specimen is shown in Figure 7.2. Normally, the cross section is circular, but rectangular specimens are also used. During testing, deformation is confined to the narrow center region, which has a uniform cross section along its length. The standard diameter is approximately 12.8 mm (0.5 in.), whereas the reduced section length should be at least four times this diameter; 60 mm (2 in.) is common. Gauge length is used in ductility computations, as discussed in Section 7.6; the standard value is 50 mm (2.0 in.). The specimen is mounted by its ends into the holding grips of the testing apparatus (Figure 7.3). The tensile testing machine is designed to elongate the specimen at a constant rate, and to continuously and simultaneously measure the instantaneous applied load (with a load cell) and the resulting elongations (using an extensometer). A stress–strain test typically takes several minutes to perform and is destructive; that is, the test specimen is permanently deformed and usually fractured. The output of such a tensile test is recorded on a strip chart (or by a computer) as load or force versus elongation. These load–deformation characteristics are dependent on the specimen size. For example, it will require twice the load to produce the same elongation if the cross-sectional area of the specimen is doubled. To minimize these geometrical factors, load and elongation are normalized to the respective parameters of engineering stress and engineering strain. Engineering 1

ASTM Standards E 8 and E 8M, ‘‘Standard Test Methods for Tension Testing of Metallic Materials.’’

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FIGURE 7.1 (a) Schematic illustration of how a tensile load produces an elongation and positive linear strain. Dashed lines represent the shape before deformation; solid lines, after deformation. (b) Schematic illustration of how a compressive load produces contraction and a negative linear strain. (c) Schematic representation of shear strain 웂, where 웂 ⫽ tan . (d) Schematic representation of torsional deformation (i.e., angle of twist ) produced by an applied torque T.

F F A0

l

l0

l0

l

A0

F F (a)

( b) T

A0

T

F

F θ

F (c)

(d)

stress is defined by the relationship

⫽

F A0

(7.1)

in which F is the instantaneous load applied perpendicular to the specimen cross section, in units of newtons (N) or pounds force (lbf ), and A0 is the original crosssectional area before any load is applied (m2 or in.2). The units of engineering

FIGURE 7.2 A standard tensile specimen with circular cross section.

Reduced section 1 2 " 4

3" Diameter 4

0.505" Diameter 2" Gauge length

3" 8

Radius

7.2 Concepts of Stress and Strain

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151

FIGURE 7.3 Schematic representation of the apparatus used to conduct tensile stress–strain tests. The specimen is elongated by the moving crosshead; load cell and extensometer measure, respectively, the magnitude of the applied load and the elongation. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2. Copyright 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

Load cell

Extensometer Specimen

Moving crosshead

stress (referred to subsequently as just stress) are megapascals, MPa (SI) (where 1 MPa ⫽ 106 N/m2), and pounds force per square inch, psi (Customary U.S.).2 Engineering strain ⑀ is defined according to ⑀⫽

li ⫺ l0 ⌬l ⫽ l0 l0

(7.2)

in which l0 is the original length before any load is applied, and li is the instantaneous length. Sometimes the quantity li ⫺ l0 is denoted as ⌬l, and is the deformation elongation or change in length at some instant, as referenced to the original length. Engineering strain (subsequently called just strain) is unitless, but meters per meter or inches per inch are often used; the value of strain is obviously independent of the unit system. Sometimes strain is also expressed as a percentage, in which the strain value is multiplied by 100.

COMPRESSION TESTS3 Compression stress–strain tests may be conducted if in-service forces are of this type. A compression test is conducted in a manner similar to the tensile test, except that the force is compressive and the specimen contracts along the direction of the stress. Equations 7.1 and 7.2 are utilized to compute compressive stress and strain, respectively. By convention, a compressive force is taken to be negative, which yields a negative stress. Furthermore, since l0 is greater than li , compressive strains computed from Equation 7.2 are necessarily also negative. Tensile tests are more common because they are easier to perform; also, for most materials used in structural applications, very little additional information is obtained from compressive tests. Compressive tests are used when a material’s behavior under large and perma2

Conversion from one system of stress units to the other is accomplished by the relationship 145 psi ⫽ 1 MPa. 3 ASTM Standard E 9, ‘‘Standard Test Methods of Compression Testing of Metallic Materials at Room Temperature.’’

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nent (i.e., plastic) strains is desired, as in manufacturing applications, or when the material is brittle in tension.

SHEAR AND TORSIONAL TESTS4 For tests performed using a pure shear force as shown in Figure 7.1c, the shear stress is computed according to

⫽

F A0

(7.3)

where F is the load or force imposed parallel to the upper and lower faces, each of which has an area of A0 . The shear strain 웂 is defined as the tangent of the strain angle , as indicated in the figure. The units for shear stress and strain are the same as for their tensile counterparts. Torsion is a variation of pure shear, wherein a structural member is twisted in the manner of Figure 7.1d; torsional forces produce a rotational motion about the longitudinal axis of one end of the member relative to the other end. Examples of torsion are found for machine axles and drive shafts, and also for twist drills. Torsional tests are normally performed on cylindrical solid shafts or tubes. A shear stress is a function of the applied torque T, whereas shear strain 웂 is related to the angle of twist, in Figure 7.1d.

GEOMETRIC CONSIDERATIONS OF THE STRESS STATE Stresses that are computed from the tensile, compressive, shear, and torsional force states represented in Figure 7.1 act either parallel or perpendicular to planar faces of the bodies represented in these illustrations. It should be noted that the stress state is a function of the orientations of the planes upon which the stresses are taken to act. For example, consider the cylindrical tensile specimen of Figure 7.4

FIGURE 7.4 Schematic representation showing normal ( ⬘) and shear ( ⬘) stresses that act on a plane oriented at an angle relative to the plane taken perpendicular to the direction along which a pure tensile stress () is applied.

p⬘ ⬘

⬘

p

4

ASTM Standard E 143, ‘‘Standard Test for Shear Modulus.’’

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153

that is subjected to a tensile stress applied parallel to its axis. Furthermore, consider also the plane p-p⬘ that is oriented at some arbitrary angle relative to the plane of the specimen end-face. Upon this plane p-p⬘, the applied stress is no longer a pure tensile one. Rather, a more complex stress state is present that consists of a tensile (or normal) stress ⬘ that acts normal to the p-p⬘ plane, and, in addition, a shear stress ⬘ that acts parallel to this plane; both of these stresses are represented in the figure. Using mechanics of materials principles,5 it is possible to develop equations for ⬘ and ⬘ in terms of and , as follows:

⬘ ⫽ cos2 ⫽

冉

冊 冉 冊

1 ⫹ cos 2 2

(7.4a)

sin 2 2

(7.4b)

⬘ ⫽ sin cos ⫽

These same mechanics principles allow the transformation of stress components from one coordinate system to another coordinate system that has a different orientation. Such treatments are beyond the scope of the present discussion.

ELASTIC DEFORMATION 7.3 STRESS –STRAIN BEHAVIOR The degree to which a structure deforms or strains depends on the magnitude of an imposed stress. For most metals that are stressed in tension and at relatively low levels, stress and strain are proportional to each other through the relationship

⫽ E⑀

(7.5)

This is known as Hooke’s law, and the constant of proportionality E (GPa or psi)6 is the modulus of elasticity, or Young’s modulus. For most typical metals the magnitude of this modulus ranges between 45 GPa (6.5 ⫻ 106 psi), for magnesium, and 407 GPa (59 ⫻ 106 psi), for tungsten. The moduli of elasticity are slightly higher for ceramic materials, which range between about 70 and 500 GPa (10 ⫻ 106 and 70 ⫻ 106 psi). Polymers have modulus values that are smaller than both metals and ceramics, and which lie in the range 0.007 and 4 GPa (103 and 0.6 ⫻ 106 psi). Room temperature modulus of elasticity values for a number of metals, ceramics, and polymers are presented in Table 7.1. A more comprehensive modulus list is provided in Table B.2, Appendix B. Deformation in which stress and strain are proportional is called elastic deformation; a plot of stress (ordinate) versus strain (abscissa) results in a linear relationship, as shown in Figure 7.5. The slope of this linear segment corresponds to the modulus of elasticity E. This modulus may be thought of as stiffness, or a material’s resistance to elastic deformation. The greater the modulus, the stiffer the material, or the smaller the elastic strain that results from the application of a given stress. The modulus is an important design parameter used for computing elastic deflections. Elastic deformation is nonpermanent, which means that when the applied load is released, the piece returns to its original shape. As shown in the stress–strain 5

See, for example, W. F. Riley, L. D. Sturges, and D. H. Morris, Mechanics of Materials, 5th edition, John Wiley & Sons, New York, 1999. 6 The SI unit for the modulus of elasticity is gigapascal, GPa, where 1 GPa ⫽ 109 N/m2 ⫽ 103 MPa.

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Table 7.1 Room-Temperature Elastic and Shear Moduli, and Poisson’s Ratio for Various Materials Modulus of Elasticity Material

GPa

6

10 psi

Metal Alloys 59 30 30 15.5 16 14 10 6.5

Shear Modulus GPa

106 psi

Poisson’s Ratio

160 83 76 45 46 37 25 17

23.2 12.0 11.0 6.5 6.7 5.4 3.6 2.5

0.28 0.30 0.31 0.34 0.34 0.34 0.33 0.35

Tungsten Steel Nickel Titanium Copper Brass Aluminum Magnesium

407 207 207 107 110 97 69 45

Aluminum oxide (Al2O3 ) Silicon carbide (SiC) Silicon nitride (Si3N4 ) Spinel (MgAl2O4 ) Magnesium oxide (MgO) Zirconia a Mullite (3Al2O3-2SiO2 ) Glass–ceramic (Pyroceram) Fused silica (SiO2 ) Soda–lime glass

Ceramic Materials 393 57 345 50 304 44 260 38 225 33 205 30 145 21 120 17 73 11 69 10

— — — — — — — — — —

— — — — — — — — — —

0.22 0.17 0.30 — 0.18 0.31 0.24 0.25 0.17 0.23

Polymersb 2.76–4.83 0.40–0.70 2.41–4.14 0.35–0.60 2.76–4.14 0.40–0.60 2.28–3.28 0.33–0.48 2.24–3.24 0.33–0.47

— — — — —

— — — — —

— 0.38 — 0.33 —

2.38 1.58–3.80 1.14–1.55 1.08

0.35 0.23–0.55 0.17–0.23 0.16

— — — —

— — — —

0.36 0.39 — —

0.40–0.55

0.058–0.080

—

—

0.46

0.17–0.28

0.025–0.041

—

—

—

Phenol-formaldehyde Polyvinyl chloride (PVC) Polyester (PET) Polystyrene (PS) Polymethyl methacrylate (PMMA) Polycarbonate (PC) Nylon 6,6 Polypropylene (PP) Polyethylene—high density (HDPE) Polytetrafluoroethylene (PTFE) Polyethylene—low density (LDPE) a

Partially stabilized with 3 mol% Y2O3 . Source: Modern Plastics Encyclopedia ’96. Copyright 1995, The McGraw-Hill Companies. Reprinted with permission.

b

plot (Figure 7.5), application of the load corresponds to moving from the origin up and along the straight line. Upon release of the load, the line is traversed in the opposite direction, back to the origin. There are some materials (e.g., gray cast iron, concrete, and many polymers) for which this initial elastic portion of the stress–strain curve is not linear (Figure

7.3 Stress–Strain Behavior

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155

FIGURE 7.5 Schematic stress–strain diagram showing linear elastic deformation for loading and unloading cycles.

Stress

Unload

Slope = modulus of elasticity

Load 0 0 Strain

7.6); hence, it is not possible to determine a modulus of elasticity as described above. For this nonlinear behavior, either tangent or secant modulus is normally used. Tangent modulus is taken as the slope of the stress–strain curve at some specified level of stress, while secant modulus represents the slope of a secant drawn from the origin to some given point of the –⑀ curve. The determination of these moduli is illustrated in Figure 7.6. On an atomic scale, macroscopic elastic strain is manifested as small changes in the interatomic spacing and the stretching of interatomic bonds. As a consequence, the magnitude of the modulus of elasticity is a measure of the resistance to separation of adjacent atoms/ions/molecules, that is, the interatomic bonding forces. Furthermore, this modulus is proportional to the slope of the interatomic force–separation curve (Figure 2.8a) at the equilibrium spacing: E앜

冉 冊 dF dr

(7.6)

r0

Figure 7.7 shows the force–separation curves for materials having both strong and weak interatomic bonds; the slope at r0 is indicated for each.

FIGURE 7.6 Schematic stress–strain diagram showing nonlinear elastic behavior, and how secant and tangent moduli are determined.

2

Stress

⌬ ⌬⑀

= Tangent modulus (at 2)

1

⌬ ⌬⑀

= Secant modulus (between origin and 1)

Strain ⑀

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Chapter 7 / Mechanical Properties

FIGURE 7.7 Force versus interatomic separation for weakly and strongly bonded atoms. The magnitude of the modulus of elasticity is proportional to the slope of each curve at the equilibrium interatomic separation r0 .

Strongly bonded

Force F

156

dF dr r 0 Separation r

0

Weakly bonded

Differences in modulus values between metals, ceramics, and polymers are a direct consequence of the different types of atomic bonding that exist for the three materials types. Furthermore, with increasing temperature, the modulus of elasticity diminishes for all but some of the rubber materials; this effect is shown for several metals in Figure 7.8. As would be expected, the imposition of compressive, shear, or torsional stresses also evokes elastic behavior. The stress–strain characteristics at low stress levels are virtually the same for both tensile and compressive situations, to include the magnitude of the modulus of elasticity. Shear stress and strain are proportional to each other through the expression

⫽ G웂

Temperature (⬚F) ⫺400

0

400

800

1200

1600

70

Tungsten 50 300 40

Steel

200

30

20 100

Aluminum 10

0

⫺200

0

200

400

Temperature (⬚C)

600

800

0

Modulus of elasticity (106 psi)

60

400 Modulus of elasticity (GPa)

FIGURE 7.8 Plot of modulus of elasticity versus temperature for tungsten, steel, and aluminum. (Adapted from K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering. Copyright 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

(7.7)

7.5 Elastic Properties of Materials

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157

where G is the shear modulus, the slope of the linear elastic region of the shear stress–strain curve. Table 7.1 also gives the shear moduli for a number of the common metals.

7.4 ANELASTICITY Up to this point, it has been assumed that elastic deformation is time independent, that is, that an applied stress produces an instantaneous elastic strain that remains constant over the period of time the stress is maintained. It has also been assumed that upon release of the load the strain is totally recovered, that is, that the strain immediately returns to zero. In most engineering materials, however, there will also exist a time-dependent elastic strain component. That is, elastic deformation will continue after the stress application, and upon load release some finite time is required for complete recovery. This time-dependent elastic behavior is known as anelasticity, and it is due to time-dependent microscopic and atomistic processes that are attendant to the deformation. For metals the anelastic component is normally small and is often neglected. However, for some polymeric materials its magnitude is significant; in this case it is termed viscoelastic behavior, 兵which is the discussion topic of Section 7.15.其

EXAMPLE PROBLEM 7.1 A piece of copper originally 305 mm (12 in.) long is pulled in tension with a stress of 276 MPa (40,000 psi). If the deformation is entirely elastic, what will be the resultant elongation?

S OLUTION Since the deformation is elastic, strain is dependent on stress according to Equation 7.5. Furthermore, the elongation ⌬l is related to the original length l0 through Equation 7.2. Combining these two expressions and solving for ⌬l yields

⫽ ⑀E ⫽ ⌬l ⫽

冉冊

⌬l E l0

l0 E

The values of and l0 are given as 276 MPa and 305 mm, respectively, and the magnitude of E for copper from Table 7.1 is 110 GPa (16 ⫻ 106 psi). Elongation is obtained by substitution into the expression above as ⌬l ⫽

7.5 ELASTIC PROPERTIES

OF

(276 MPa)(305 mm) ⫽ 0.77 mm (0.03 in.) 110 ⫻ 103 MPa

MATERIALS

When a tensile stress is imposed on virtually all materials, an elastic elongation and accompanying strain ⑀z result in the direction of the applied stress (arbitrarily taken to be the z direction), as indicated in Figure 7.9. As a result of this elongation,

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Chapter 7 / Mechanical Properties z

⌬lz 2

l0x

⌬lx 2

FIGURE 7.9 Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed tensile stress. Solid lines represent dimensions after stress application; dashed lines, before.

l0z

z

⑀z ⌬l /2 = z 2 l0z

⫺

z

y

⑀x ⌬l /2 = x 2 l0x

x

there will be constrictions in the lateral (x and y) directions perpendicular to the applied stress; from these contractions, the compressive strains ⑀x and ⑀y may be determined. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then ⑀x ⫽ ⑀y . A parameter termed Poisson’s ratio is defined as the ratio of the lateral and axial strains, or

⫽⫺

⑀x ⑀ ⫽⫺ y ⑀z ⑀z

(7.8)

The negative sign is included in the expression so that will always be positive, since ⑀x and ⑀z will always be of opposite sign. Theoretically, Poisson’s ratio for isotropic materials should be ; furthermore, the maximum value for (or that value for which there is no net volume change) is 0.50. For many metals and other alloys, values of Poisson’s ratio range between 0.25 and 0.35. Table 7.1 shows values for several common materials; a more comprehensive list is given in Table B.3, Appendix B. For isotropic materials, shear and elastic moduli are related to each other and to Poisson’s ratio according to E ⫽ 2G(1 ⫹ )

(7.9)

In most metals G is about 0.4E; thus, if the value of one modulus is known, the other may be approximated. Many materials are elastically anisotropic; that is, the elastic behavior (e.g., the magnitude of E) varies with crystallographic direction (see Table 3.7). For these materials the elastic properties are completely characterized only by the specification of several elastic constants, their number depending on characteristics of the crystal structure. Even for isotropic materials, for complete characterization of the elastic properties, at least two constants must be given. Since the grain orientation is random in most polycrystalline materials, these may be considered to be isotropic;

7.5 Elastic Properties of Materials

●

159

inorganic ceramic glasses are also isotropic. The remaining discussion of mechanical behavior assumes isotropy and polycrystallinity (for metals and crystalline ceramics) because such is the character of most engineering materials.

EXAMPLE PROBLEM 7.2 A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm (0.4 in.). Determine the magnitude of the load required to produce a 2.5 ⫻ 10⫺3 mm (10⫺4 in.) change in diameter if the deformation is entirely elastic.

S OLUTION This deformation situation is represented in the accompanying drawing. F

d0 di z

li

x

l0 ⌬l l0

⑀z =

=

⑀ x = ⌬d = d0

li ⫺ l0 l0 di ⫺ d 0 d0

F

When the force F is applied, the specimen will elongate in the z direction and at the same time experience a reduction in diameter, ⌬d, of 2.5 ⫻ 10⫺3 mm in the x direction. For the strain in the x direction, ⑀x ⫽

⌬d ⫺2.5 ⫻ 10⫺3 mm ⫽ ⫽ ⫺2.5 ⫻ 10⫺4 d0 10 mm

which is negative, since the diameter is reduced. It next becomes necessary to calculate the strain in the z direction using Equation 7.8. The value for Poisson’s ratio for brass is 0.34 (Table 7.1), and thus ⑀z ⫽ ⫺

⑀x (⫺2.5 ⫻ 10⫺4) ⫽⫺ ⫽ 7.35 ⫻ 10⫺4 0.34

The applied stress may now be computed using Equation 7.5 and the modulus of elasticity, given in Table 7.1 as 97 GPa (14 ⫻ 106 psi), as

⫽ ⑀z E ⫽ (7.35 ⫻ 10⫺4)(97 ⫻ 103 MPa) ⫽ 71.3 MPa

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Finally, from Equation 7.1, the applied force may be determined as F ⫽ A0 ⫽

冉冊 d0 2

2

앟

⫽ (71.3 ⫻ 106 N/m2)

冉

冊

10 ⫻ 10⫺3 m 2

2

앟 ⫽ 5600 N (1293 lbf )

MECHANICAL BEHAVIOR — METALS For most metallic materials, elastic deformation persists only to strains of about 0.005. As the material is deformed beyond this point, the stress is no longer proportional to strain (Hooke’s law, Equation 7.5, ceases to be valid), and permanent, nonrecoverable, or plastic deformation occurs. Figure 7.10a plots schematically the tensile stress–strain behavior into the plastic region for a typical metal. The transition from elastic to plastic is a gradual one for most metals; some curvature results at the onset of plastic deformation, which increases more rapidly with rising stress. From an atomic perspective, plastic deformation corresponds to the breaking of bonds with original atom neighbors and then reforming bonds with new neighbors as large numbers of atoms or molecules move relative to one another; upon removal of the stress they do not return to their original positions. This permanent deformation for metals is accomplished by means of a process called slip, which involves the motion of dislocations as discussed in Section 8.3.

7.6 TENSILE PROPERTIES YIELDING AND YIELD STRENGTH Most structures are designed to ensure that only elastic deformation will result when a stress is applied. It is therefore desirable to know the stress level at which Elastic Plastic

Upper yield point

Stress

y

Stress

FIGURE 7.10 (a) Typical stress–strain behavior for a metal showing elastic and plastic deformations, the proportional limit P, and the yield strength y , as determined using the 0.002 strain offset method. (b) Representative stress–strain behavior found for some steels demonstrating the yield point phenomenon.

P

y

Lower yield point

Strain

Strain

0.002 (a)

(b)

7.6 Tensile Properties

●

161

plastic deformation begins, or where the phenomenon of yielding occurs. For metals that experience this gradual elastic–plastic transition, the point of yielding may be determined as the initial departure from linearity of the stress–strain curve; this is sometimes called the proportional limit, as indicated by point P in Figure 7.10a. In such cases the position of this point may not be determined precisely. As a consequence, a convention has been established wherein a straight line is constructed parallel to the elastic portion of the stress–strain curve at some specified strain offset, usually 0.002. The stress corresponding to the intersection of this line and the stress–strain curve as it bends over in the plastic region is defined as the yield strength y .7 This is demonstrated in Figure 7.10a. Of course, the units of yield strength are MPa or psi.8 For those materials having a nonlinear elastic region (Figure 7.6), use of the strain offset method is not possible, and the usual practice is to define the yield strength as the stress required to produce some amount of strain (e.g., ⑀ ⫽ 0.005). Some steels and other materials exhibit the tensile stress–strain behavior as shown in Figure 7.10b. The elastic–plastic transition is very well defined and occurs abruptly in what is termed a yield point phenomenon. At the upper yield point, plastic deformation is initiated with an actual decrease in stress. Continued deformation fluctuates slightly about some constant stress value, termed the lower yield point; stress subsequently rises with increasing strain. For metals that display this effect, the yield strength is taken as the average stress that is associated with the lower yield point, since it is well defined and relatively insensitive to the testing procedure.9 Thus, it is not necessary to employ the strain offset method for these materials. The magnitude of the yield strength for a metal is a measure of its resistance to plastic deformation. Yield strengths may range from 35 MPa (5000 psi) for a low-strength aluminum to over 1400 MPa (200,000 psi) for high-strength steels.

TENSILE STRENGTH After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum, point M in Figure 7.11, and then decreases to the eventual fracture, point F. The tensile strength TS (MPa or psi) is the stress at the maximum on the engineering stress–strain curve (Figure 7.11). This corresponds to the maximum stress that can be sustained by a structure in tension; if this stress is applied and maintained, fracture will result. All deformation up to this point is uniform throughout the narrow region of the tensile specimen. However, at this maximum stress, a small constriction or neck begins to form at some point, and all subsequent deformation is confined at this neck, as indicated by the schematic specimen insets in Figure 7.11. This phenomenon is termed ‘‘necking,’’ and fracture ultimately occurs at the neck. The fracture strength corresponds to the stress at fracture. Tensile strengths may vary anywhere from 50 MPa (7000 psi) for an aluminum to as high as 3000 MPa (450,000 psi) for the high-strength steels. Ordinarily, when the strength of a metal is cited for design purposes, the yield strength is used. This is because by the time a stress corresponding to the tensile strength has been 7 ‘‘Strength’’ is used in lieu of ‘‘stress’’ because strength is a property of the metal, whereas stress is related to the magnitude of the applied load. 8 For Customary U.S. units, the unit of kilopounds per square inch (ksi) is sometimes used for the sake of convenience, where

1 ksi ⫽ 1000 psi It should be pointed out that to observe the yield point phenomenon, a ‘‘stiff ’’ tensiletesting apparatus must be used; by stiff is meant that there is very little elastic deformation of the machine during loading. 9

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Chapter 7 / Mechanical Properties

FIGURE 7.11 Typical engineering stress–strain behavior to fracture, point F. The tensile strength TS is indicated at point M. The circular insets represent the geometry of the deformed specimen at various points along the curve.

M TS

F

Stress

162

Strain

applied, often a structure has experienced so much plastic deformation that it is useless. Furthermore, fracture strengths are not normally specified for engineering design purposes.

EXAMPLE PROBLEM 7.3 From the tensile stress–strain behavior for the brass specimen shown in Figure 7.12, determine the following: (a) The modulus of elasticity. (b) The yield strength at a strain offset of 0.002. (c) The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm (0.505 in.). (d) The change in length of a specimen originally 250 mm (10 in.) long that is subjected to a tensile stress of 345 MPa (50,000 psi).

S OLUTION (a) The modulus of elasticity is the slope of the elastic or initial linear portion of the stress–strain curve. The strain axis has been expanded in the inset, Figure 7.12, to facilitate this computation. The slope of this linear region is the rise over the run, or the change in stress divided by the corresponding change in strain; in mathematical terms, E ⫽ slope ⫽

⌬ 2 ⫺ 1 ⫽ ⌬⑀ ⑀2 ⫺ ⑀ 1

(7.10)

Inasmuch as the line segment passes through the origin, it is convenient to take both 1 and ⑀1 as zero. If 2 is arbitrarily taken as 150 MPa, then ⑀2 will have

7.6 Tensile Properties

500

●

163

70 Tensile strength 450 MPa (65,000 psi) 60

400

300

40 200

30 Yield strength 250 MPa (36,000 psi)

200 20 100

30

Stress (103 psi)

Stress (MPa)

50

103 psi MPa 40

A

20 10

100

10 0 0

0

0

0

0.10

0.005 0.20

0 0.40

0.30

Strain

FIGURE 7.12 The stress–strain behavior for the brass specimen discussed in Example Problem 7.3.

a value of 0.0016. Therefore, E⫽

(150 ⫺ 0) MPa ⫽ 93.8 GPa (13.6 ⫻ 106 psi) 0.0016 ⫺ 0

which is very close to the value of 97 GPa (14 ⫻ 106 psi) given for brass in Table 7.1. (b) The 0.002 strain offset line is constructed as shown in the inset; its intersection with the stress–strain curve is at approximately 250 MPa (36,000 psi), which is the yield strength of the brass. (c) The maximum load that can be sustained by the specimen is calculated by using Equation 7.1, in which is taken to be the tensile strength, from Figure 7.12, 450 MPa (65,000 psi). Solving for F, the maximum load, yields F ⫽ A0 ⫽

冉冊 d0 2

2

앟

⫽ (450 ⫻ 106 N/m2)

冉

冊

12.8 ⫻ 10⫺3 m 2

2

앟 ⫽ 57,900 N (13,000 lbf )

(d) To compute the change in length, ⌬l, in Equation 7.2, it is first necessary to determine the strain that is produced by a stress of 345 MPa. This is accomplished by locating the stress point on the stress–strain curve, point A, and reading the corresponding strain from the strain axis, which is approximately 0.06. Inasmuch as l0 ⫽ 250 mm, we have ⌬l ⫽ ⑀l0 ⫽ (0.06)(250 mm) ⫽ 15 mm (0.6 in.)

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Chapter 7 / Mechanical Properties

DUCTILITY Ductility is another important mechanical property. It is a measure of the degree of plastic deformation that has been sustained at fracture. A material that experiences very little or no plastic deformation upon fracture is termed brittle. The tensile stress–strain behaviors for both ductile and brittle materials are schematically illustrated in Figure 7.13. Ductility may be expressed quantitatively as either percent elongation or percent reduction in area. The percent elongation %EL is the percentage of plastic strain at fracture, or %EL ⫽

冉 冊

lf ⫺ l 0 ⫻ 100 l0

(7.11)

where lf is the fracture length10 and l0 is the original gauge length as above. Inasmuch as a significant proportion of the plastic deformation at fracture is confined to the neck region, the magnitude of %EL will depend on specimen gauge length. The shorter l0 , the greater is the fraction of total elongation from the neck and, consequently, the higher the value of %EL. Therefore, l0 should be specified when percent elongation values are cited; it is commonly 50 mm (2 in.). Percent reduction in area %RA is defined as %RA ⫽

冉

冊

A0 ⫺ A f ⫻ 100 A0

(7.12)

where A0 is the original cross-sectional area and Af is the cross-sectional area at the point of fracture.10 Percent reduction in area values are independent of both l0 and A0 . Furthermore, for a given material the magnitudes of %EL and %RA will, in general, be different. Most metals possess at least a moderate degree of ductility at room temperature; however, some become brittle as the temperature is lowered (Section 9.8). A knowledge of the ductility of materials is important for at least two reasons. First, it indicates to a designer the degree to which a structure will deform plastically

Brittle

B

FIGURE 7.13 Schematic representations of tensile stress–strain behavior for brittle and ductile materials loaded to fracture.

Ductile

Stress

164

B'

A

C

C' Strain

10

Both lf and Af are measured subsequent to fracture, and after the two broken ends have been repositioned back together.

7.6 Tensile Properties

●

165

before fracture. Second, it specifies the degree of allowable deformation during fabrication operations. We sometimes refer to relatively ductile materials as being ‘‘forgiving,’’ in the sense that they may experience local deformation without fracture should there be an error in the magnitude of the design stress calculation. Brittle materials are approximately considered to be those having a fracture strain of less than about 5%. Thus, several important mechanical properties of metals may be determined from tensile stress–strain tests. Table 7.2 presents some typical room-temperature

Table 7.2 Room-Temperature Mechanical Properties (in Tension) for Various Materials Yield Strength Material

MPa

Molybdenum Titanium Steel (1020) Nickel Iron Brass (70 Cu–30 Zn) Copper Aluminum d

Zirconia (ZrO2 ) Silicon nitride (Si3N4 ) Aluminum oxide (Al2O3 ) Silicon carbide (SiC) Glass–ceramic (Pyroceram) Mullite (3Al2O3-2SiO2 ) Spinel (MgAl2O4 ) Fused silica (SiO2 ) Magnesium oxide (MgO) e Soda–lime glass Nylon 6,6 Polycarbonate (PC) Polyester (PET) Polymethyl methacrylate (PMMA) Polyvinyl chloride (PVC) Phenol-formaldehyde Polystyrene (PS) Polypropylene (PP) Polyethylene—high density (HDPE) Polytetrafluoroethylene (PTFE) Polyethylene—low density (LDPE) a

ksi Metal Alloys b 82 65 26 20 19 11 10 5

565 450 180 138 130 75 69 35 — — — — — — — — — —

Tensile Strength MPa 655 520 380 480 262 300 200 90

Ceramic Materials c — 800–1500 — 250–1000 — 275–700 — 100–820 — 247 — 185 — 110–245 — 110 — 105 — 69

44.8–82.8 62.1 59.3 53.8–73.1 40.7–44.8 — — 31.0–37.2 26.2–33.1 — 9.0–14.5

Polymers 6.5–12 9.0 8.6 7.8–10.6 5.9–6.5 — — 4.5–5.4 3.8–4.8 — 1.3–2.1

75.9–94.5 62.8–72.4 48.3–72.4 48.3–72.4 40.7–51.7 34.5–62.1 35.9–51.7 31.0–41.4 22.1–31.0 20.7–34.5 8.3–31.4

Ductility, %EL [in 50 mm (2 in.)] a

ksi 95 75 55 70 38 44 29 13

35 25 25 40 45 68 45 40

115–215 35–145 40–100 15–120 36 27 16–36 16 15 10

— — — — — — — — — —

11.0–13.7 9.1–10.5 7.0–10.5 7.0–10.5 5.9–7.5 5.0–9.0 5.2–7.5 4.5–6.0 3.2–4.5 3.0–5.0 1.2–4.55

15–300 110–150 30–300 2.0–5.5 40–80 1.5–2.0 1.2–2.5 100–600 10–1200 200–400 100–650

For polymers, percent elongation at break. Property values are for metal alloys in an annealed state. c The tensile strength of ceramic materials is taken as flexural strength (Section 7.10). d Partially stabilized with 3 mol% Y2O3 . e Sintered and containing approximately 5% porosity. b

●

Chapter 7 / Mechanical Properties

FIGURE 7.14 Engineering stress– strain behavior for iron at three temperatures.

120

800 ⫺200⬚C

100

Stress (MPa)

600 80 60

⫺100⬚C

400

40

Stress (103 psi)

166

25⬚C

200

20 0

0

0.1

0.2

0.3

0.4

0 0.5

Strain

values of yield strength, tensile strength, and ductility for several common metals (and also for a number of polymers and ceramics). These properties are sensitive to any prior deformation, the presence of impurities, and/or any heat treatment to which the metal has been subjected. The modulus of elasticity is one mechanical parameter that is insensitive to these treatments. As with modulus of elasticity, the magnitudes of both yield and tensile strengths decline with increasing temperature; just the reverse holds for ductility—it usually increases with temperature. Figure 7.14 shows how the stress–strain behavior of iron varies with temperature.

RESILIENCE Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. The associated property is the modulus of resilience, Ur , which is the strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding. Computationally, the modulus of resilience for a specimen subjected to a uniaxial tension test is just the area under the engineering stress–strain curve taken to yielding (Figure 7.15), or Ur ⫽

冕

⑀y

0

d⑀

(7.13a)

FIGURE 7.15 Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from the tensile stress–strain behavior of a material.

Stress

σy

0.002

y

Strain

7.7 True Stress and Strain

●

167

Assuming a linear elastic region, Ur ⫽ y⑀y

(7.13b)

in which ⑀y is the strain at yielding. The units of resilience are the product of the units from each of the two axes of the stress–strain plot. For SI units, this is joules per cubic meter (J/m3, equivalent to Pa), whereas with Customary U.S. units it is inch-pounds force per cubic inch (in.-lb f /in.3, equivalent to psi). Both joules and inch-pounds force are units of energy, and thus this area under the stress–strain curve represents energy absorption per unit volume (in cubic meters or cubic inches) of material. Incorporation of Equation 7.5 into Equation 7.13b yields Ur ⫽ y ⑀y ⫽ y

冉冊

y 2y ⫽ E 2E

(7.14)

Thus, resilient materials are those having high yield strengths and low moduli of elasticity; such alloys would be used in spring applications.

TOUGHNESS Toughness is a mechanical term that is used in several contexts; loosely speaking, it is a measure of the ability of a material to absorb energy up to fracture. Specimen geometry as well as the manner of load application are important in toughness determinations. For dynamic (high strain rate) loading conditions and when a notch (or point of stress concentration) is present, notch toughness is assessed by using an impact test, as discussed in Section 9.8. Furthermore, fracture toughness is a property indicative of a material’s resistance to fracture when a crack is present (Section 9.5). For the static (low strain rate) situation, toughness may be ascertained from the results of a tensile stress–strain test. It is the area under the –⑀ curve up to the point of fracture. The units for toughness are the same as for resilience (i.e., energy per unit volume of material). For a material to be tough, it must display both strength and ductility; and often, ductile materials are tougher than brittle ones. This is demonstrated in Figure 7.13, in which the stress–strain curves are plotted for both material types. Hence, even though the brittle material has higher yield and tensile strengths, it has a lower toughness than the ductile one, by virtue of lack of ductility; this is deduced by comparing the areas ABC and AB⬘C⬘ in Figure 7.13.

7.7 TRUE STRESS

AND

STRAIN

From Figure 7.11, the decline in the stress necessary to continue deformation past the maximum, point M, seems to indicate that the metal is becoming weaker. This is not at all the case; as a matter of fact, it is increasing in strength. However, the cross-sectional area is decreasing rapidly within the neck region, where deformation is occurring. This results in a reduction in the load-bearing capacity of the specimen. The stress, as computed from Equation 7.1, is on the basis of the original crosssectional area before any deformation, and does not take into account this diminution in area at the neck. Sometimes it is more meaningful to use a true stress–true strain scheme. True stress T is defined as the load F divided by the instantaneous cross-sectional area

●

Chapter 7 / Mechanical Properties

Ai over which deformation is occurring (i.e., the neck, past the tensile point), or

T ⫽

F Ai

(7.15)

Furthermore, it is occasionally more convenient to represent strain as true strain ⑀T , defined by ⑀T ⫽ ln

li l0

(7.16)

If no volume change occurs during deformation, that is, if A i li ⫽ A 0 l0

(7.17)

true and engineering stress and strain are related according to

T ⫽ (1 ⫹ ⑀)

(7.18a)

⑀T ⫽ ln(1 ⫹ ⑀)

(7.18b)

Equations 7.18a and 7.18b are valid only to the onset of necking; beyond this point true stress and strain should be computed from actual load, cross-sectional area, and gauge length measurements. A schematic comparison of engineering and true stress–strain behavior is made in Figure 7.16. It is worth noting that the true stress necessary to sustain increasing strain continues to rise past the tensile point M⬘. Coincident with the formation of a neck is the introduction of a complex stress state within the neck region (i.e., the existence of other stress components in addition to the axial stress). As a consequence, the correct stress (axial) within the neck is slightly lower than the stress computed from the applied load and neck crosssectional area. This leads to the ‘‘corrected’’ curve in Figure 7.16. For some metals and alloys the region of the true stress-strain curve from the onset of plastic deformation to the point at which necking begins may be approximated by

T ⫽ K⑀ Tn

True M⬘ Corrected Stress

168

M Engineering

Strain

(7.19)

FIGURE 7.16 A comparison of typical tensile engineering stress–strain and true stress–strain behaviors. Necking begins at point M on the engineering curve, which corresponds to M⬘ on the true curve. The ‘‘corrected’’ true stress–strain curve takes into account the complex stress state within the neck region.

7.7 True Stress and Strain

●

169

Table 7.3 Tabulation of n and K Values (Equation 7.19) for Several Alloys K Material Low-carbon steel (annealed) Alloy steel (Type 4340, annealed) Stainless steel (Type 304, annealed) Aluminum (annealed) Aluminum alloy (Type 2024, heat treated) Copper (annealed) Brass (70Cu–30Zn, annealed)

n

MPa

psi

0.26

530

77,000

0.15

640

93,000

0.45

1275

185,000

0.20 0.16

180 690

26,000 100,000

0.54 0.49

315 895

46,000 130,000

Source: From Manufacturing Processes for Engineering Materials by Seope Kalpakjian, 1997. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ.

In this expression, K and n are constants, which values will vary from alloy to alloy, and will also depend on the condition of the material (i.e., whether it has been plastically deformed, heat treated, etc.). The parameter n is often termed the strainhardening exponent and has a value less than unity. Values of n and K for several alloys are contained in Table 7.3.

EXAMPLE PROBLEM 7.4 A cylindrical specimen of steel having an original diameter of 12.8 mm (0.505 in.) is tensile tested to fracture and found to have an engineering fracture strength f of 460 MPa (67,000 psi). If its cross-sectional diameter at fracture is 10.7 mm (0.422 in.), determine: (a) The ductility in terms of percent reduction in area. (b) The true stress at fracture.

S OLUTION (a) Ductility is computed using Equation 7.12, as

%RA ⫽

⫽

冉

12.8 mm 2

冉

冊 冉 2

앟⫺

10.7 mm 2

12.8 mm 2

冊

2

冊

2

앟 ⫻ 100

앟

128.7 mm2 ⫺ 89.9 mm2 ⫻ 100 ⫽ 30% 128.7 mm2

(b) True stress is defined by Equation 7.15, where in this case the area is taken as the fracture area Af . However, the load at fracture must first be computed

170

●

Chapter 7 / Mechanical Properties

from the fracture strength as F ⫽ f A0 ⫽ (460 ⫻ 106 N/m2)(128.7 mm2)

冉

冊

1 m2 ⫽ 59,200 N 106 mm2

Thus, the true stress is calculated as

T ⫽

F ⫽ Af

59,200 N (89.9 mm2)

冉

冊

1 m2 106 mm2

⫽ 6.6 ⫻ 108 N/m2 ⫽ 660 MPa (95,700 psi)

EXAMPLE PROBLEM 7.5 Compute the strain-hardening exponent n in Equation 7.19 for an alloy in which a true stress of 415 MPa (60,000 psi) produces a true strain of 0.10; assume a value of 1035 MPa (150,000 psi) for K.

S OLUTION This requires some algebraic manipulation of Equation 7.19 so that n becomes the dependent parameter. This is accomplished by taking logarithms and rearranging. Solving for n yields n⫽ ⫽

log T ⫺ log K log ⑀T log(415 MPa) ⫺ log(1035 MPa) ⫽ 0.40 log(0.1)

7.8 ELASTIC RECOVERY DURING PLASTIC DEFORMATION Upon release of the load during the course of a stress–strain test, some fraction of the total deformation is recovered as elastic strain. This behavior is demonstrated in Figure 7.17, a schematic engineering stress–strain plot. During the unloading cycle, the curve traces a near straight-line path from the point of unloading (point D), and its slope is virtually identical to the modulus of elasticity, or parallel to the initial elastic portion of the curve. The magnitude of this elastic strain, which is regained during unloading, corresponds to the strain recovery, as shown in Figure 7.17. If the load is reapplied, the curve will traverse essentially the same linear portion in the direction opposite to unloading; yielding will again occur at the unloading stress level where the unloading began. There will also be an elastic strain recovery associated with fracture.

7.9 COMPRESSIVE, SHEAR, AND TORSIONAL DEFORMATION Of course, metals may experience plastic deformation under the influence of applied compressive, shear, and torsional loads. The resulting stress–strain behavior into

7.10 Flexural Strength

171

FIGURE 7.17 Schematic tensile stress–strain diagram showing the phenomena of elastic strain recovery and strain hardening. The initial yield strength is designated as y0 ; yi is the yield strength after releasing the load at point D, and then upon reloading.

D

yi

●

y0

Stress

Unload

Reapply load Strain Elastic strain recovery

the plastic region will be similar to the tensile counterpart (Figure 7.10a: yielding and the associated curvature). However, for compression, there will be no maximum, since necking does not occur; furthermore, the mode of fracture will be different from that for tension.

MECHANICAL BEHAVIOR — CERAMICS Ceramic materials are somewhat limited in applicability by their mechanical properties, which in many respects are inferior to those of metals. The principal drawback is a disposition to catastrophic fracture in a brittle manner with very little energy absorption. In this section we explore the salient mechanical characteristics of these materials and how these properties are measured.

7.10 FLEXURAL STRENGTH The stress–strain behavior of brittle ceramics is not usually ascertained by a tensile test as outlined in Section 7.2, for three reasons. First, it is difficult to prepare and test specimens having the required geometry. Second, it is difficult to grip brittle materials without fracturing them; and third, ceramics fail after only about 0.1% strain, which necessitates that tensile specimens be perfectly aligned in order to avoid the presence of bending stresses, which are not easily calculated. Therefore, a more suitable transverse bending test is most frequently employed, in which a rod specimen having either a circular or rectangular cross section is bent until fracture using a three- or four-point loading technique;11 the three-point loading scheme is illustrated in Figure 7.18. At the point of loading, the top surface of the specimen is placed in a state of compression, whereas the bottom surface is in tension. Stress is computed from the specimen thickness, the bending moment, and 11 ASTM Standard C 1161, ‘‘Standard Test Method for Flexural Strength of Advanced Ceramics at Ambient Temperature.’’

172

●

Chapter 7 / Mechanical Properties Possible cross sections F

b d

Rectangular Circular

Support L 2

= stress =

L 2

FIGURE 7.18 A three-point loading scheme for measuring the stress–strain behavior and flexural strength of brittle ceramics, including expressions for computing stress for rectangular and circular cross sections.

R

Mc I

where M = maximum bending moment c = distance from center of specimen to outer fibers I = moment of inertia of cross section F = applied load M

c

I

Rectangular

FL 4

d 2

bd3 12

3FL 2bd2

Circular

FL 4

R

R4 4

FL R3

the moment of inertia of the cross section; these parameters are noted in Figure 7.18 for rectangular and circular cross sections. The maximum tensile stress (as determined using these stress expressions) exists at the bottom specimen surface directly below the point of load application. Since the tensile strengths of ceramics are about one-tenth of their compressive strengths, and since fracture occurs on the tensile specimen face, the flexure test is a reasonable substitute for the tensile test. The stress at fracture using this flexure test is known as the flexural strength, modulus of rupture, fracture strength, or the bend strength, an important mechanical parameter for brittle ceramics. For a rectangular cross section, the flexural strength fs is equal to

fs ⫽

3Ff L 2bd 2

(7.20a)

where Ff is the load at fracture, L is the distance between support points, and the other parameters are as indicated in Figure 7.18. When the cross section is circular, then

fs ⫽

Ff L 앟R 3

(7.20b)

R being the specimen radius. Characteristic flexural strength values for several ceramic materials are given in Table 7.2. Since, during bending, a specimen is subjected to both compressive and tensile stresses, the magnitude of its flexural strength is greater than the tensile fracture strength. Furthermore, fs will depend on specimen size; as explained in Section 9.6, with increasing specimen volume (under stress) there is an increase in the probability of the existence of a crack-producing flaw and, consequently, a decrease in flexural strength.

7.13 Stress–Strain Behavior

●

173

FIGURE 7.19 Typical stress–strain behavior to fracture for aluminum oxide and glass.

40 250

30

200

150 20

100

Stress (103 psi)

Stress (MPa)

Aluminum oxide

10 50 Glass

0 0

0.0004

0.0008

0 0.0012

Strain

7.11 ELASTIC BEHAVIOR The elastic stress–strain behavior for ceramic materials using these flexure tests is similar to the tensile test results for metals: a linear relationship exists between stress and strain. Figure 7.19 compares the stress–strain behavior to fracture for aluminum oxide (alumina) and glass. Again, the slope in the elastic region is the modulus of elasticity; also, the moduli of elasticity for ceramic materials are slightly higher than for metals (Table 7.2 and Table B.2, Appendix B). From Figure 7.19 it may be noted that neither of the materials experiences plastic deformation prior to fracture.

7.12 INFLUENCE OF POROSITY ON THE MECHANICAL PROPERTIES OF CERAMICS (CD-ROM)

MECHANICAL BEHAVIOR — POLYMERS 7.13 STRESS –STRAIN BEHAVIOR The mechanical properties of polymers are specified with many of the same parameters that are used for metals, that is, modulus of elasticity, and yield and tensile strengths. For many polymeric materials, the simple stress–strain test is employed for the characterization of some of these mechanical parameters.12 The mechanical characteristics of polymers, for the most part, are highly sensitive to the rate of deformation (strain rate), the temperature, and the chemical nature of the environment (the presence of water, oxygen, organic solvents, etc.). Some modifications of the testing techniques and specimen configurations used for metals are necessary with polymers, especially for the highly elastic materials, such as rubbers. 12

ASTM Standard D 638, ‘‘Standard Test Method for Tensile Properties of Plastics.’’

Chapter 7 / Mechanical Properties 10 60

A 8

6

40 30 B

4

Stress (103 psi)

50

FIGURE 7.22 The stress–strain behavior for brittle (curve A), plastic (curve B), and highly elastic (elastomeric) (curve C ) polymers.

20 2 10 0

C 0

1

2

3

4

5

6

7

8

0

Strain

Three typically different types of stress–strain behavior are found for polymeric materials, as represented in Figure 7.22. Curve A illustrates the stress–strain character for a brittle polymer, inasmuch as it fractures while deforming elastically. The behavior for the plastic material, curve B, is similar to that found for many metallic materials; the initial deformation is elastic, which is followed by yielding and a region of plastic deformation. Finally, the deformation displayed by curve C is totally elastic; this rubberlike elasticity (large recoverable strains produced at low stress levels) is displayed by a class of polymers termed the elastomers. Modulus of elasticity (termed tensile modulus or sometimes just modulus for polymers) and ductility in percent elongation are determined for polymers in the same manner as for metals (Section 7.6). For plastic polymers (curve B, Figure 7.22), the yield point is taken as a maximum on the curve, which occurs just beyond the termination of the linear-elastic region (Figure 7.23); the stress at this maximum is the yield strength (y ). Furthermore, tensile strength (TS) corresponds to the stress at which fracture occurs (Figure 7.23); TS may be greater than or less than y . Strength, for these plastic polymers, is normally taken as tensile strength. Table 7.2 and Tables B.2, B.3, and B.4 in Appendix B give these mechanical properties for a number of polymeric materials.

FIGURE 7.23 Schematic stress–strain curve for a plastic polymer showing how yield and tensile strengths are determined.

TS y Stress

●

Stress (MPa)

174

Strain

7.14 Macroscopic Deformation

●

175

12

80 4°C (40°F) 70

10

8

20°C (68°F) 30°C (86°F)

50

6

40

40°C (104°F)

30

4 50°C (122°F)

20

To 1.30

2

60°C (140°F)

10 0

Stress (103 psi)

Stress (MPa)

60

0

0.1

0.2

0 0.3

Strain

FIGURE 7.24 The influence of temperature on the stress–strain characteristics of polymethyl methacrylate. (From T. S. Carswell and H. K. Nason, ‘‘Effect of Environmental Conditions on the Mechanical Properties of Organic Plastics,’’ Symposium on Plastics, American Society for Testing and Materials, Philadelphia, 1944. Copyright, ASTM. Reprinted with permission.)

Polymers are, in many respects, mechanically dissimilar to metals (and ceramic materials). For example, the modulus for highly elastic polymeric materials may be as low as 7 MPa (103 psi), but may run as high as 4 GPa (0.6 ⫻ 106 psi) for some of the very stiff polymers; modulus values for metals are much larger (Table 7.1). Maximum tensile strengths for polymers are on the order of 100 MPa (15,000 psi)—for some metal alloys 4100 MPa (600,000 psi). And, whereas metals rarely elongate plastically to more than 100%, some highly elastic polymers may experience elongations to as much as 1000%. In addition, the mechanical characteristics of polymers are much more sensitive to temperature changes within the vicinity of room temperature. Consider the stress–strain behavior for polymethyl methacrylate (Plexiglas) at several temperatures between 4 and 60⬚C (40 and 140⬚F) (Figure 7.24). Several features of this figure are worth noting, as follows: increasing the temperature produces (1) a decrease in elastic modulus, (2) a reduction in tensile strength, and (3) an enhancement of ductility—at 4⬚C (40⬚F) the material is totally brittle, whereas considerable plastic deformation is realized at both 50 and 60⬚C (122 and 140⬚F). The influence of strain rate on the mechanical behavior may also be important. In general, decreasing the rate of deformation has the same influence on the stress– strain characteristics as increasing the temperature; that is, the material becomes softer and more ductile.

7.14 MACROSCOPIC DEFORMATION Some aspects of the macroscopic deformation of semicrystalline polymers deserve our attention. The tensile stress–strain curve for a semicrystalline material, which was initially unoriented, is shown in Figure 7.25; also included in the figure are

●

Chapter 7 / Mechanical Properties FIGURE 7.25 Schematic tensile stress–strain curve for a semicrystalline polymer. Specimen contours at several stages of deformation are included. (From Jerold M. Schultz, Polymer Materials Science, copyright 1974, p. 488. Reprinted by permission of PrenticeHall, Inc., Englewood Cliffs, NJ.)

Stress

176

Strain

schematic representations of specimen profile at various stages of deformation. Both upper and lower yield points are evident on the curve, which are followed by a near horizontal region. At the upper yield point, a small neck forms within the gauge section of the specimen. Within this neck, the chains become oriented (i.e., chain axes become aligned parallel to the elongation direction, a condition that is represented schematically in Figure 8.27e), which leads to localized strengthening. Consequently, there is a resistance to continued deformation at this point, and specimen elongation proceeds by the propagation of this neck region along the gauge length; the chain orientation phenomenon (Figure 8.27e) accompanies this neck extension. This tensile behavior may be contrasted to that found for ductile metals (Section 7.6), wherein once a neck has formed, all subsequent deformation is confined to within the neck region.

7.15 VISCOELASTICITY (CD-ROM)

HARDNESS AND OTHER MECHANICAL PROPERTY CONSIDERATIONS 7.16 HARDNESS Another mechanical property that may be important to consider is hardness, which is a measure of a material’s resistance to localized plastic deformation (e.g., a small dent or a scratch). Early hardness tests were based on natural minerals with a scale constructed solely on the ability of one material to scratch another that was softer. A qualitative and somewhat arbitrary hardness indexing scheme was devised, termed the Mohs scale, which ranged from 1 on the soft end for talc to 10 for diamond. Quantitative hardness techniques have been developed over the years in which a small indenter is forced into the surface of a material to be tested, under controlled conditions of load and rate of application. The depth or size of the resulting indentation is measured, which in turn is related to a hardness number; the softer the material, the larger and deeper the indentation, and the lower the hardness index number. Measured hardnesses are only relative (rather than absolute), and care should be exercised when comparing values determined by different techniques.

7.16 Hardness

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177

Hardness tests are performed more frequently than any other mechanical test for several reasons: 1. They are simple and inexpensive—ordinarily no special specimen need be prepared, and the testing apparatus is relatively inexpensive. 2. The test is nondestructive—the specimen is neither fractured nor excessively deformed; a small indentation is the only deformation. 3. Other mechanical properties often may be estimated from hardness data, such as tensile strength (see Figure 7.31).

ROCKWELL HARDNESS TESTS13 The Rockwell tests constitute the most common method used to measure hardness because they are so simple to perform and require no special skills. Several different scales may be utilized from possible combinations of various indenters and different loads, which permit the testing of virtually all metal alloys (as well as some polymers). Indenters include spherical and hardened steel balls having diameters of , , , and in. (1.588, 3.175, 6.350, and 12.70 mm), and a conical diamond (Brale) indenter, which is used for the hardest materials. With this system, a hardness number is determined by the difference in depth of penetration resulting from the application of an initial minor load followed by a larger major load; utilization of a minor load enhances test accuracy. On the basis of the magnitude of both major and minor loads, there are two types of tests: Rockwell and superficial Rockwell. For Rockwell, the minor load is 10 kg, whereas major loads are 60, 100, and 150 kg. Each scale is represented by a letter of the alphabet; several are listed with the corresponding indenter and load in Tables 7.4 and 7.5a. For superficial tests, 3 kg is the minor load; 15, 30, and 45 kg are the possible major load values. These scales are identified by a 15, 30, or 45 (according to load), followed by N, T, W, X, or Y, depending on indenter. Superficial tests are frequently performed on thin specimens. Table 7.5b presents several superficial scales. When specifying Rockwell and superficial hardnesses, both hardness number and scale symbol must be indicated. The scale is designated by the symbol HR followed by the appropriate scale identification.14 For example, 80 HRB represents a Rockwell hardness of 80 on the B scale, and 60 HR30W indicates a superficial hardness of 60 on the 30W scale. For each scale, hardnesses may range up to 130; however, as hardness values rise above 100 or drop below 20 on any scale, they become inaccurate; and because the scales have some overlap, in such a situation it is best to utilize the next harder or softer scale. Inaccuracies also result if the test specimen is too thin, if an indentation is made too near a specimen edge, or if two indentations are made too close to one another. Specimen thickness should be at least ten times the indentation depth, whereas allowance should be made for at least three indentation diameters between the center of one indentation and the specimen edge, or to the center of a second indentation. Furthermore, testing of specimens stacked one on top of another is not recommended. Also, accuracy is dependent on the indentation being made into a smooth flat surface. 13

ASTM Standard E 18, ‘‘Standard Test Methods for Rockwell Hardness and Rockwell Superficial Hardness of Metallic Materials.’’ 14 Rockwell scales are also frequently designated by an R with the appropriate scale letter as a subscript, for example, RC denotes the Rockwell C scale.

178 ●

Hardness Testing Techniques Shape of Indentation Indenter

Side View

10-mm sphere of steel or tungsten carbide

D

Test Brinell

Vickers microhardness

Diamond pyramid

Knoop microhardness

Diamond pyramid

a

冦

Diamond cone , , , in. diameter steel spheres

Formula for Hardness Number a

Load P

HB ⫽

d1

P

HV ⫽ 1.854P/d 21

b

P

HK ⫽ 14.2P/l 2

d

2P 앟D[D ⫺ 兹D2 ⫺ d 2]

d 136⬚

d1

t

l/b = 7.11 b/t = 4.00

Rockwell and Superficial Rockwell

Top View

120⬚

l

冧 冧

60 kg 100 kg Rockwell 150 kg 15 kg 30 kg Superficial Rockwell 45 kg

For the hardness formulas given, P (the applied load) is in kg, while D, d, d1 , and l are all in mm. Source: Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior. Copyright 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.

Chapter 7 / Mechanical Properties

Table 7.4

7.16 Hardness

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179

Table 7.5a Rockwell Hardness Scales Scale Symbol

Indenter

Major Load (kg)

A B C D E F G H K

Diamond in. ball Diamond Diamond in. ball in. ball in. ball in. ball in. ball

60 100 150 100 100 60 150 60 150

Table 7.5b Superficial Rockwell Hardness Scales Scale Symbol

Indenter

Major Load (kg)

15N 30N 45N 15T 30T 45T 15W 30W 45W

Diamond Diamond Diamond in. ball in. ball in. ball in. ball in. ball in. ball

15 30 45 15 30 45 15 30 45

The modern apparatus for making Rockwell hardness measurements (see the chapter-opening photograph for this chapter) is automated and very simple to use; hardness is read directly, and each measurement requires only a few seconds. The modern testing apparatus also permits a variation in the time of load application. This variable must also be considered in interpreting hardness data.

BRINELL HARDNESS TESTS15 In Brinell tests, as in Rockwell measurements, a hard, spherical indenter is forced into the surface of the metal to be tested. The diameter of the hardened steel (or tungsten carbide) indenter is 10.00 mm (0.394 in.). Standard loads range between 500 and 3000 kg in 500-kg increments; during a test, the load is maintained constant for a specified time (between 10 and 30 s). Harder materials require greater applied loads. The Brinell hardness number, HB, is a function of both the magnitude of the load and the diameter of the resulting indentation (see Table 7.4).16 This diameter is measured with a special low-power microscope, utilizing a scale that is etched on the eyepiece. The measured diameter is then converted to the appropriate HB number using a chart; only one scale is employed with this technique. Maximum specimen thickness as well as indentation position (relative to specimen edges) and minimum indentation spacing requirements are the same as for Rockwell tests. In addition, a well-defined indentation is required; this necessitates a smooth flat surface in which the indentation is made. 15 16

ASTM Standard E 10, ‘‘Standard Test Method for Brinell Hardness of Metallic Materials.’’ The Brinell hardness number is also represented by BHN.

180

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Chapter 7 / Mechanical Properties

KNOOP AND VICKERS MICROHARDNESS TESTS17 Two other hardness testing techniques are Knoop (pronounced nup) and Vickers (sometimes also called diamond pyramid). For each test a very small diamond indenter having pyramidal geometry is forced into the surface of the specimen. Applied loads are much smaller than for Rockwell and Brinell, ranging between 1 and 1000 g. The resulting impression is observed under a microscope and measured; this measurement is then converted into a hardness number (Table 7.4). Careful specimen surface preparation (grinding and polishing) may be necessary to ensure a well-defined indentation that may be accurately measured. The Knoop and Vickers hardness numbers are designated by HK and HV, respectively,18 and hardness scales for both techniques are approximately equivalent. Knoop and Vickers are referred to as microhardness testing methods on the basis of load and indenter size. Both are well suited for measuring the hardness of small, selected specimen regions; furthermore, Knoop is used for testing brittle materials such as ceramics. There are other hardness-testing techniques that are frequently employed, but which will not be discussed here; these include ultrasonic microhardness, dynamic (Scleroscope), durometer (for plastic and elastomeric materials), and scratch hardness tests. These are described in references provided at the end of the chapter.

HARDNESS CONVERSION The facility to convert the hardness measured on one scale to that of another is most desirable. However, since hardness is not a well-defined material property, and because of the experimental dissimilarities among the various techniques, a comprehensive conversion scheme has not been devised. Hardness conversion data have been determined experimentally and found to be dependent on material type and characteristics. The most reliable conversion data exist for steels, some of which are presented in Figure 7.30 for Knoop, Brinell, and two Rockwell scales; the Mohs scale is also included. Detailed conversion tables for various other metals and alloys are contained in ASTM Standard E 140, ‘‘Standard Hardness Conversion Tables for Metals.’’ In light of the preceding discussion, care should be exercised in extrapolation of conversion data from one alloy system to another.

CORRELATION BETWEEN HARDNESS AND TENSILE STRENGTH Both tensile strength and hardness are indicators of a metal’s resistance to plastic deformation. Consequently, they are roughly proportional, as shown in Figure 7.31, on page 182, for tensile strength as a function of the HB for cast iron, steel, and brass. The same proportionality relationship does not hold for all metals, as Figure 7.31 indicates. As a rule of thumb for most steels, the HB and the tensile strength are related according to

17

TS(MPa) ⫽ 3.45 ⫻ HB

(7.25a)

TS(psi) ⫽ 500 ⫻ HB

(7.25b)

ASTM Standard E 92, ‘‘Standard Test Method for Vickers Hardness of Metallic Materials,’’ and ASTM Standard E 384, ‘‘Standard Test for Microhardness of Materials.’’ 18 Sometimes KHN and VHN are used to denote Knoop and Vickers hardness numbers, respectively.

7.18 Tear Strength and Hardness of Polymers FIGURE 7.30 Comparison of several hardness scales. (Adapted from G. F. Kinney, Engineering Properties and Applications of Plastics, p. 202. Copyright 1957 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

●

181

10,000 10

Diamond

5,000

2,000

Nitrided steels

1,000

80

1000 800

Cutting tools 60 40

110 400 300

100

20

200

200

100

100

80

0

60

Rockwell C

40 20 0

Knoop hardness

50

8

Topaz

7

Quartz

6

Orthoclase

5

Apatite

4 3

Fluorite Calcite

2

Gypsum

1

Talc

File hard

500

600

9

Corundum or sapphire

Rockwell B

20

Easily machined steels

Brasses and aluminum alloys

Most plastics

10

5 Brinell hardness

Mohs hardness

7.17 HARDNESS

OF

CERAMIC MATERIALS

One beneficial mechanical property of ceramics is their hardness, which is often utilized when an abrasive or grinding action is required; in fact, the hardest known materials are ceramics. A listing of a number of different ceramic materials according to Knoop hardness is contained in Table 7.6. Only ceramics having Knoop hardnesses of about 1000 or greater are utilized for their abrasive characteristics (Section 13.8).

7.18 TEAR STRENGTH

AND

HARDNESS

OF

POLYMERS

Mechanical properties that are sometimes influential in the suitability of a polymer for some particular application include tear resistance and hardness. The ability to resist tearing is an important property of some plastics, especially those used for thin films in packaging. Tear strength, the mechanical parameter that is measured,

Chapter 7 / Mechanical Properties Rockwell hardness

60 70 80 90

100 HRB 20

30

40

50 HRC

250

1500 200 Steels

150

1000

100

Tensile strength (103 psi)

●

Tensile strength (MPa)

182

FIGURE 7.31 Relationships between hardness and tensile strength for steel, brass, and cast iron. (Data taken from Metals Handbook: Properties and Selection: Irons and Steels, Vol. 1, 9th edition, B. Bardes, Editor, American Society for Metals, 1978, pp. 36 and 461; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th edition, H. Baker, Managing Editor, American Society for Metals, 1979, p. 327.)

500 Brass

Cast iron (nodular) 50

0

0

100

200

300

400

0 500

Brinell hardness number

is the energy required to tear apart a cut specimen that has a standard geometry. The magnitude of tensile and tear strengths are related. Polymers are softer than metals and ceramics, and most hardness tests are conducted by penetration techniques similar to those described for metals in the previous section. Rockwell tests are frequently used for polymers.19 Other indentation techniques employed are the Durometer and Barcol.20 Table 7.6 Approximate Knoop Hardness (100 g load) for Seven Ceramic Materials Material Diamond (carbon) Boron carbide (B4C) Silicon carbide (SiC) Tungsten carbide (WC) Aluminum oxide (Al2O3) Quartz (SiO2) Glass

Approximate Knoop Hardness 7000 2800 2500 2100 2100 800 550

19 ASTM Standard D 785, ‘‘Rockwell Hardness of Plastics and Electrical Insulating Materials.’’ 20 ASTM Standard D 2240, ‘‘Standard Test Method for Rubber Property—Durometer Hardness;’’ and ASTM Standard D 2583, ‘‘Standard Test Method for Indentation of Rigid Plastics by Means of a Barcol Impressor.’’

7.20 Design/Safety Factors

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183

PROPERTY VARIABILITY AND DESIGN / SAFETY FACTORS 7.19 VARIABILITY

OF

MATERIAL PROPERTIES

At this point it is worthwhile to discuss an issue that sometimes proves troublesome to many engineering students, namely, that measured material properties are not exact quantities. That is, even if we have a most precise measuring apparatus and a highly controlled test procedure, there will always be some scatter or variability in the data that are collected from specimens of the same material. For example, consider a number of identical tensile samples that are prepared from a single bar of some metal alloy, which samples are subsequently stress–strain tested in the same apparatus. We would most likely observe that each resulting stress–strain plot is slightly different from the others. This would lead to a variety of modulus of elasticity, yield strength, and tensile strength values. A number of factors lead to uncertainties in measured data. These include the test method, variations in specimen fabrication procedures, operator bias, and apparatus calibration. Furthermore, inhomogeneities may exist within the same lot of material, and/or slight compositional and other differences from lot to lot. Of course, appropriate measures should be taken to minimize the possibility of measurement error, and also to mitigate those factors that lead to data variability. It should also be mentioned that scatter exists for other measured material properties such as density, electrical conductivity, and coefficient of thermal expansion. It is important for the design engineer to realize that scatter and variability of materials properties are inevitable and must be dealt with appropriately. On occasion, data must be subjected to statistical treatments and probabilities determined. For example, instead of asking the question, ‘‘What is the fracture strength of this alloy?’’ the engineer should become accustomed to asking the question, ‘‘What is the probability of failure of this alloy under these given circumstances?’’ It is often desirable to specify a typical value and degree of dispersion (or scatter) for some measured property; such is commonly accomplished by taking the average and the standard deviation, respectively.

COMPUTATION OF AVERAGE AND STANDARD DEVIATION VALUES (CD-ROM)

7.20 DESIGN /SAFETY FACTORS There will always be uncertainties in characterizing the magnitude of applied loads and their associated stress levels for in-service applications; ordinarily load calculations are only approximate. Furthermore, as noted in the previous section, virtually all engineering materials exhibit a variability in their measured mechanical properties. Consequently, design allowances must be made to protect against unanticipated failure. One way this may be accomplished is by establishing, for the particular application, a design stress, denoted as d . For static situations and when ductile materials are used, d is taken as the calculated stress level c (on the basis of the estimated maximum load) multiplied by a design factor, N⬘, that is

d ⫽ N⬘c

(7.28)

where N⬘ is greater than unity. Thus, the material to be used for the particular application is chosen so as to have a yield strength at least as high as this value of d .

184

●

Chapter 7 / Mechanical Properties

Alternatively, a safe stress or working stress, w , is used instead of design stress. This safe stress is based on the yield strength of the material and is defined as the yield strength divided by a factor of safety, N, or

w ⫽

y N

(7.29)

Utilization of design stress (Equation 7.28) is usually preferred since it is based on the anticipated maximum applied stress instead of the yield strength of the material; normally there is a greater uncertainty in estimating this stress level than in the specification of the yield strength. However, in the discussion of this text, we are concerned with factors that influence yield strengths, and not in the determination of applied stresses; therefore, the succeeding discussion will deal with working stresses and factors of safety. The choice of an appropriate value of N is necessary. If N is too large, then component overdesign will result, that is, either too much material or a material having a higher-than-necessary strength will be used. Values normally range between 1.2 and 4.0. Selection of N will depend on a number of factors, including economics, previous experience, the accuracy with which mechanical forces and material properties may be determined, and, most important, the consequences of failure in terms of loss of life and/or property damage.

DESIGN EXAMPLE 7.1 A tensile-testing apparatus is to be constructed that must withstand a maximum load of 220,000 N (50,000 lbf). The design calls for two cylindrical support posts, each of which is to support half of the maximum load. Furthermore, plain-carbon (1045) steel ground and polished shafting rounds are to be used; the minimum yield and tensile strengths of this alloy are 310 MPa (45,000 psi) and 565 MPa (82,000 psi), respectively. Specify a suitable diameter for these support posts.

S OLUTION The first step in this design process is to decide on a factor safety, N, which then allows determination of a working stress according to Equation 7.29. In addition, to ensure that the apparatus will be safe to operate, we also want to minimize any elastic deflection of the rods during testing; therefore, a relatively conservative factor of safety is to be used, say N ⫽ 5. Thus, the working stress w is just

w ⫽ ⫽

y N 310 MPa ⫽ 62 MPa (9000 psi) 5

From the definition of stress, Equation 7.1, A0 ⫽

冉冊 d 2

2

앟⫽

F w

where d is the rod diameter and F is the applied force; furthermore, each of the two rods must support half of the total force or 110,000 N (25,000 psi). Solving for

Summary

●

185

d leads to d⫽2 ⫽2

冪앟F

w

N 冪앟 (62110,000 ⫻ 10 N/m ) 6

2

⫽ 4.75 ⫻ 10⫺2 m ⫽ 47.5 mm (1.87 in.) Therefore, the diameter of each of the two rods should be 47.5 mm or 1.87 in.

SUMMARY A number of the important mechanical properties of materials have been discussed in this chapter. Concepts of stress and strain were first introduced. Stress is a measure of an applied mechanical load or force, normalized to take into account cross-sectional area. Two different stress parameters were defined—engineering stress and true stress. Strain represents the amount of deformation induced by a stress; both engineering and true strains are used. Some of the mechanical characteristics of materials can be ascertained by simple stress–strain tests. There are four test types: tension, compression, torsion, and shear. Tensile are the most common. A material that is stressed first undergoes elastic, or nonpermanent, deformation, wherein stress and strain are proportional. The constant of proportionality is the modulus of elasticity for tension and compression, and is the shear modulus when the stress is shear. Poisson’s ratio represents the negative ratio of transverse and longitudinal strains. For metals, the phenomenon of yielding occurs at the onset of plastic or permanent deformation; yield strength is determined by a strain offset method from the stress–strain behavior, which is indicative of the stress at which plastic deformation begins. Tensile strength corresponds to the maximum tensile stress that may be sustained by a specimen, whereas percents elongation and reduction in area are measures of ductility—the amount of plastic deformation that has occurred at fracture. Resilience is the capacity of a material to absorb energy during elastic deformation; modulus of resilience is the area beneath the engineering stress–strain curve up to the yield point. Also, static toughness represents the energy absorbed during the fracture of a material, and is taken as the area under the entire engineering stress–strain curve. Ductile materials are normally tougher than brittle ones. For the brittle ceramic materials, flexural strengths are determined by performing transverse bending tests to fracture. 兵Many ceramic bodies contain residual porosity, which is deleterious to both their moduli of elasticity and flexural strengths.其 On the basis of stress–strain behavior, polymers fall within three general classifications: brittle, plastic, and highly elastic. These materials are neither as strong nor as stiff as metals, and their mechanical properties are sensitive to changes in temperature and strain rate. 兵Viscoelastic mechanical behavior, being intermediate between totally elastic and totally viscous, is displayed by a number of polymeric materials. It is characterized by the relaxation modulus, a time-dependent modulus of elasticity. The magnitude of the relaxation modulus is very sensitive to temperature; critical to the inservice temperature range for elastomers is this temperature dependence.其

186

●

Chapter 7 / Mechanical Properties

Hardness is a measure of the resistance to localized plastic deformation. In several popular hardness-testing techniques (Rockwell, Brinell, Knoop, and Vickers) a small indenter is forced into the surface of the material, and an index number is determined on the basis of the size or depth of the resulting indentation. For many metals, hardness and tensile strength are approximately proportional to each other. In addition to their inherent brittleness, ceramic materials are distinctively hard. And polymers are relatively soft in comparison to the other material types. Measured mechanical properties (as well as other material properties) are not exact and precise quantities, in that there will always be some scatter for the measured data. Typical material property values are commonly specified in terms of averages, whereas magnitudes of scatter may be expressed as standard deviations. As a result of uncertainties in both measured mechanical properties and inservice applied stresses, safe or working stresses are normally utilized for design purposes. For ductile materials, safe stress is the ratio of the yield strength and a factor of safety.

IMPORTANT TERMS AND CONCEPTS Anelasticity Design stress Ductility Elastic deformation Elastic recovery Elastomer Engineering strain Engineering stress Flexural strength

Hardness Modulus of elasticity Plastic deformation Poisson’s ratio Proportional limit 兵Relaxation modulus其 Resilience Safe stress Shear

Tensile strength Toughness True strain True stress 兵Viscoelasticity其 Yielding Yield strength

REFERENCES ASM Handbook, Vol. 8, Mechanical Testing, ASM International, Materials Park, OH, 1985. Billmeyer, F. W., Jr., Textbook of Polymer Science, 3rd edition, Wiley-Interscience, New York, 1984. Chapter 11. Boyer, H. E. (Editor), Atlas of Stress–Strain Curves, ASM International, Materials Park, OH, 1986. Boyer, H. E. (Editor), Hardness Testing, ASM International, Materials Park, OH, 1987. Davidge, R. W., Mechanical Behaviour of Ceramics, Cambridge University Press, Cambridge, 1979. Reprinted by TechBooks, Marietta, OH. Dieter, G. E., Mechanical Metallurgy, 3rd edition, McGraw-Hill Book Co., New York, 1986. Dowling, N. E., Mechanical Behavior of Materials, Prentice Hall, Inc., Englewood Cliffs, NJ, 1993.

Engineered Materials Handbook, Vol. 2, Engineering Plastics, ASM International, Materials Park, OH, 1988. Engineered Materials Handbook, Vol. 4, Ceramics and Glasses, ASM International, Materials Park, OH, 1991. Han, P. (Editor), Tensile Testing, ASM International, Materials Park, OH, 1992. Harper, C. A. (Editor), Handbook of Plastics, Elastomers and Composites, 3rd edition, McGrawHill Book Company, New York, 1996. Kingery, W. D., H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics, 2nd edition, John Wiley & Sons, New York, 1976. Chapters 14 and 15. McClintock, F. A. and A. S. Argon, Mechanical Behavior of Materials, Addison-Wesley Pub-

Questions and Problems

lishing Co., Reading, MA, 1966. Reprinted by TechBooks, Marietta, OH. Meyers, M. A. and K. K. Chawla, Mechanical Metallurgy, Principles and Applications, Prentice Hall, Inc., Englewood Cliffs, NJ, 1984. Modern Plastics Encyclopedia, McGraw-Hill Book Company, New York. Revised and published annually. Nielsen, L. E., Mechanical Properties of Polymers and Composites, 2nd edition, Marcel Dekker, New York, 1994. Richerson, D. W., Modern Ceramic Engineering, 2nd edition, Marcel Dekker, New York, 1992.

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187

Rosen, S. L., Fundamental Principles of Polymeric Materials, 2nd edition, John Wiley & Sons, New York, 1993. Tobolsky, A. V., Properties and Structures of Polymers, John Wiley & Sons, New York, 1960. Advanced treatment. Wachtman, J. B., Mechanical Properties of Ceramics, John Wiley & Sons, Inc., New York, 1996. Ward, I. M. and D. W. Hadley, An Introduction to the Mechanical Properties of Solid Polymers, John Wiley & Sons, Chichester, UK, 1993. Young, R. J. and P. Lovell, Introduction to Polymers, 2nd edition, Chapman and Hall, London, 1991.

QUESTIONS AND PROBLEMS Note: To solve those problems having an asterisk (*) by their numbers, consultation of supplementary topics [appearing only on the CD-ROM (and not in print)] will probably be necessary. 7.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram), derive Equations 7.4a and 7.4b. 7.2 (a) Equations 7.4a and 7.4b are expressions for normal ( ⬘) and shear ( ⬘) stresses, respectively, as a function of the applied tensile stress () and the inclination angle of the plane on which these stresses are taken ( of Figure 7.4). Make a plot on which is presented the orientation parameters of these expressions (i.e., cos2 and sin cos ) versus . (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) Also, at what inclination angle is the shear stress a maximum? 7.3 A specimen of aluminum having a rectangular cross section 10 mm ⫻ 12.7 mm (0.4 in. ⫻ 0.5 in.) is pulled in tension with 35,500 N (8000 lbf ) force, producing only elastic deformation. Calculate the resulting strain. 7.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 ⫻ 106 psi) and an original diameter of 3.8 mm (0.15 in.) will experience only elastic deformation when a tensile load of 2000 N (450 lbf ) is applied. Compute the maximum

7.5

7.6

7.7

7.8

length of the specimen before deformation if the maximum allowable elongation is 0.42 mm (0.0165 in.). A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lbf ), and experiences an elongation of 0.10 mm (4.0 ⫻ 10⫺3 in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel. Consider a cylindrical titanium wire 3.0 mm (0.12 in.) in diameter and 2.5 ⫻ 104 mm (1000 in.) long. Calculate its elongation when a load of 500 N (112 lbf ) is applied. Assume that the deformation is totally elastic. For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 ⫻ 106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm 2 (0.5 in. 2 ) without plastic deformation? (b) If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation? A cylindrical rod of copper (E ⫽ 110 GPa,

●

Chapter 7 / Mechanical Properties

FIGURE 7.33 Tensile stress–strain behavior for a plain carbon steel.

600 80

MPa

500

600

400 Stress ( MPa)

103 psi 60

80

400

60

300

40 40 200

200

20

Stress (103 psi)

188

20

100 0

0 0

0.005

0 0.15

0 0

0.05

0.10 Strain

16 ⫻ 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf ). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)? 7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10 mm (0.39 in.) in diameter and 75 mm (3.0 in.) long that is pulled in tension. Determine its elongation when a load of 23,500 N (5300 lbf ) is applied. 7.10 Figure 7.34 shows, for a gray cast iron, the tensile engineering stress–strain curve in the elastic region. Determine (a) the secant modulus taken to 35 MPa (5000 psi), and (b) the tangent modulus taken from the origin.

7.11 As was noted in Section 3.18, for single crystals of some substances, the physical properties are anisotropic, that is, they are dependent on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general [uvw] direction, E uvw , is described by the relationship

冉

(움 2 웁 2 ⫹ 웁 2 웂 2 ⫹ 웂 2 움 2 ) where E具100典 and E具111典 are the moduli of elasticity in [100] and [111] directions, respectively; 움, 웁, and 웂 are the cosines of the FIGURE 7.34 Tensile stress–strain behavior for a gray cast iron.

60 8

6

40 30

4

20 2 10

0

0.0002

0.0004 Strain

0.0006

0 0.0008

Stress (103 psi)

Stress (MPa)

50

0

冊

1 1 1 1 ⫽ ⫺3 ⫺ Euvw E具100典 E具100典 E具111典

Questions and Problems

angles between [uvw] and the respective [100], [010], and [001] directions. Verify that the E具110典 values for aluminum, copper, and iron in Table 3.7 are correct. 7.12 In Section 2.6 it was noted that the net bonding energy EN between two isolated positive and negative ions is a function of interionic distance r as follows: EN ⫽ ⫺

A B ⫹ r rn

(7.30)

where A, B, and n are constants for the particular ion pair. Equation 7.30 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, E앜

冉 冊 dF dr

r0

Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion system) using the following procedure: 1. Establish a relationship for the force F as a function of r, realizing that F⫽

dE N dr

2. Now take the derivative dF/dr. 3. Develop an expression for r0 , the equilibrium separation. Since r0 corresponds to the value of r at the minimum of the E N -versusr-curve (Figure 2.8b), take the derivative dEN /dr, set it equal to zero, and solve for r, which corresponds to r0 . 4. Finally, substitute this expression for r0 into the relationship obtained by taking dF/dr. 7.13 Using the solution to Problem 7.12, rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate A, B, and n parameters (Equation 7.30) for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers:

Material X Y Z

A 2.5 2.3 3.0

B

●

189 n

2 ⫻ 10 8 ⫻ 10⫺6 1.5 ⫻ 10⫺5 ⫺5

8 10.5 9

7.14 A cylindrical specimen of aluminum having a diameter of 19 mm (0.75 in.) and length of 200 mm (8.0 in.) is deformed elastically in tension with a force of 48,800 N (11,000 lbf ). Using the data contained in Table 7.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease? 7.15 A cylindrical bar of steel 10 mm (0.4 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Using the data in Table 7.1, determine the force that will produce an elastic reduction of 3 ⫻ 10⫺3 mm (1.2 ⫻ 10⫺4 in.) in the diameter. 7.16 A cylindrical specimen of some alloy 8 mm (0.31 in.) in diameter is stressed elastically in tension. A force of 15,700 N (3530 lbf ) produces a reduction in specimen diameter of 5 ⫻ 10⫺3 mm (2 ⫻ 10⫺4 in.). Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa (20.3 ⫻ 10 6 psi). 7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively. 7.18 Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm (0.31 in.). A tensile force of 1000 N (225 lbf ) produces an elastic reduction in diameter of 2.8 ⫻ 10⫺4 mm (1.10 ⫻ 10⫺5 in.). Compute the modulus of elasticity for this alloy, given that Poisson’s ratio is 0.30. 7.19 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an

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elastic modulus of 103 GPa (15.0 ⫻ 106 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why. 7.20 A cylindrical metal specimen 15.0 mm (0.59 in.) in diameter and 150 mm (5.9 in.) long is to be subjected to a tensile stress of 50 MPa (7250 psi); at this stress level the resulting deformation will be totally elastic. (a) If the elongation must be less than 0.072 mm (2.83 ⫻ 10⫺3 in.), which of the metals in Table 7.1 are suitable candidates? Why? (b) If, in addition, the maximum permissible diameter decrease is 2.3 ⫻ 10⫺3 mm (9.1 ⫻ 10⫺5 in.), which of the metals in Table 7.1 may be used? Why? 7.21 Consider the brass alloy with stress–strain behavior shown in Figure 7.12. A cylindrical specimen of this material 6 mm (0.24 in.) in diameter and 50 mm (2 in.) long is pulled in tension with a force of 5000 N (1125 lbf ). If it is known that this alloy has a Poisson’s ratio of 0.30, compute: (a) the specimen elongation, and (b) the reduction in specimen diameter. 7.22 Cite the primary differences between elastic, anelastic, and plastic deformation behaviors. 7.23 A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not experience either plastic deformation or a diameter reduction of more than 7.5 ⫻ 10⫺3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).

Material Aluminum alloy Brass alloy Steel alloy Titanium alloy

Modulus of Elasticity (GPa)

Yield Strength (MPa)

Poisson’s Ratio

70

200

0.33

101 207 107

300 400 650

0.35 0.27 0.36

7.24 A cylindrical rod 380 mm (15.0 in.) long, having a diameter of 10.0 mm (0.40 in.), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm (0.035 in.) when the applied load is 24,500 N (5500 lbf ), which of the four metals or alloys listed below are possible candidates? Justify your choice(s).

Material Aluminum alloy Brass alloy Copper Steel alloy

Modulus of Elasticity (GPa)

Yield Strength (MPa)

Tensile Strength (MPa)

70

255

420

100 110 207

345 250 450

420 290 550

7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the proportional limit? (c) What is the yield strength at a strain offset of 0.002? (d) What is the tensile strength? 7.26 A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensile load of 50,000 N (11,240 lbf ) is applied. Under these circumstances, what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior shown in Figure 7.12. 7.27 A load of 44,500 N (10,000 lbf ) is applied to a cylindrical specimen of steel (displaying the stress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm (0.40 in.). (a) Will the specimen experience elastic or plastic deformation? Why? (b) If the original specimen length is 500 mm (20 in.), how much will it increase in length when this load is applied? 7.28 A bar of a steel alloy that exhibits the stress– strain behavior shown in Figure 7.33 is subjected to a tensile load; the specimen is 300 mm (12 in.) long, and of square cross section 4.5 mm (0.175 in.) on a side.

Questions and Problems

(a) Compute the magnitude of the load necessary to produce an elongation of 0.46 mm (0.018 in.). (b) What will be the deformation after the load is released? 7.29 A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.000 in. (50.800 mm) is pulled in tension. Use the load–elongation characteristics tabulated below to complete problems a through f. Length

Load lbf 0 1,650 3,400 5,200 6,850 7,750 8,650 9,300 10,100 10,400 10,650 10,700 10,400 10,100 9,600 8,200

N

in.

0 2.000 7,330 2.002 15,100 2.004 23,100 2.006 30,400 2.008 34,400 2.010 38,400 2.020 41,300 2.040 44,800 2.080 46,200 2.120 47,300 2.160 47,500 2.200 46,100 2.240 44,800 2.270 42,600 2.300 36,400 2.330 Fracture

mm 50.800 50.851 50.902 50.952 51.003 51.054 51.308 51.816 52.832 53.848 54.864 55.880 56.896 57.658 58.420 59.182

(a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f ) Compute the modulus of resilience. 7.30 A specimen of ductile cast iron having a rectangular cross section of dimensions 4.8 mm ⫻ 15.9 mm ( in. ⫻ in.) is deformed in tension. Using the load-elongation data tabulated below, complete problems a through f.

0 4,740 9,140 12,920 16,540 18,300 20,170 22,900 25,070 26,800 28,640 30,240 31,100 31,280 30,820 29,180 27,190 24,140 18,970

191

Length

Load N

●

lbf 0 1065 2055 2900 3720 4110 4530 5145 5635 6025 6440 6800 7000 7030 6930 6560 6110 5430 4265 Fracture

mm

in.

75.000 75.025 75.050 75.075 75.113 75.150 75.225 75.375 75.525 75.750 76.500 78.000 79.500 81.000 82.500 84.000 85.500 87.000 88.725

2.953 2.954 2.955 2.956 2.957 2.959 2.962 2.968 2.973 2.982 3.012 3.071 3.130 3.189 3.248 3.307 3.366 3.425 3.493

(a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) Compute the modulus of resilience. (f) What is the ductility, in percent elongation? 7.31 A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm (0.260 in.), and the fractured gauge length is 72.14 mm (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation. 7.32 Calculate the moduli of resilience for the materials having the stress–strain behaviors shown in Figures 7.12 and 7.33. 7.33 Determine the modulus of resilience for each of the following alloys:

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Chapter 7 / Mechanical Properties

Yield Strength

Length

Load

Diameter

Material

MPa

psi

lbf

N

in.

mm

in.

mm

Steel alloy Brass alloy Aluminum alloy Titanium alloy

550 350 250 800

80,000 50,750 36,250 116,000

10,400 10,100 9,600 8,200

46,100 44,800 42,600 36,400

2.240 2.270 2.300 2.330

56.896 57.658 58.420 59.182

0.461 0.431 0.418 0.370

11.71 10.95 10.62 9.40

Use modulus of elasticity values in Table 7.1. 7.34 A brass alloy to be used for a spring application must have a modulus of resilience of at least 0.75 MPa (110 psi). What must be its minimum yield strength? 7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for a typical metal alloy. (b) Superimpose on this plot a schematic curve for the compressive true stress–strain behavior for the same alloy. Explain any difference between this curve and the one in part a. (c) Now superimpose a schematic curve for the compressive engineering stress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b. 7.36 Show that Equations 7.18a and 7.18b are valid when there is no volume change during deformation. 7.37 Demonstrate that Equation 7.16, the expression defining true strain, may also be represented by ⑀ T ⫽ ln

冉冊 A0 Ai

when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why? 7.38 Using the data in Problem 7.29 and Equations 7.15, 7.16, and 7.18a, generate a true stress–true strain plot for aluminum. Equation 7.18a becomes invalid past the point at which necking begins; therefore, measured diameters are given below for the last four data points, which should be used in true stress computations.

7.39 A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa (125,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi). 7.40 For some metal alloy, a true stress of 415 MPa (60,175 psi) produces a plastic true strain of 0.475. How much will a specimen of this material elongate when a true stress of 325 MPa (46,125 psi) is applied if the original length is 300 mm (11.8 in.)? Assume a value of 0.25 for the strain-hardening exponent n. 7.41 The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi)

True Strain

50,000 60,000

0.10 0.20

What true stress is necessary to produce a true plastic strain of 0.25? 7.42 For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking: Engineering Stress (MPa)

Engineering Strain

235 250

0.194 0.296

On the basis of this information, compute the engineering stress necessary to produce an engineering strain of 0.25. 7.43 Find the toughness (or energy to cause fracture) for a metal that experiences both elastic

Questions and Problems

and plastic deformation. Assume Equation 7.5 for elastic deformation, that the modulus of elasticity is 172 GPa (25 ⫻ 106 psi), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 MPa (1 ⫻ 106 psi) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs. 7.44 For a tensile test, it can be demonstrated that necking begins when dT ⫽ T d⑀ T

(7.31)

Using Equation 7.19, determine the value of the true strain at this onset of necking. 7.45 Taking the logarithm of both sides of Equation 7.19 yields log T ⫽ log K ⫹ n log ⑀ T

(7.32)

Thus, a plot of log T versus log ⑀ T in the plastic region to the point of necking should yield a straight line having a slope of n and an intercept (at log T ⫽ 0) of log K. Using the appropriate data tabulated in Problem 7.29, make a plot of log T versus log ⑀ T and determine the values of n and K. It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 7.18a and 7.18b. 7.46 A cylindrical specimen of a brass alloy 7.5 mm (0.30 in.) in diameter and 90.0 mm (3.54 in.) long is pulled in tension with a force of 6000 N (1350 lbf ); the force is subsequently released. (a) Compute the final length of the specimen at this time. The tensile stress–strain behavior for this alloy is shown in Figure 7.12. (b) Compute the final specimen length when the load is increased to 16,500 N (3700 lbf ) and then released. 7.47 A steel specimen having a rectangular cross section of dimensions 19 mm ⫻ 3.2 mm ( in. ⫻ in.) has the stress–strain behavior

●

193

shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500 lbf ), then (a) Determine the elastic and plastic strain values. (b) If its original length is 460 mm (18 in.), what will be its final length after the load in part a is applied and then released? 7.48 A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and width b 10 mm (0.4 in.); the distance between support points is 45 mm (1.75 in.). (a) Compute the flexural strength if the load at fracture is 290 N (65 lbf). (b) The point of maximum deflection ⌬y occurs at the center of the specimen and is described by ⌬y ⫽

FL3 48EI

where E is the modulus of elasticity and I the cross-sectional moment of inertia. Compute ⌬y at a load of 266 N (60 lbf). 7.49 A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425 N (95.5 lbf), the flexural strength is 105 MPa (15,000 psi), and the separation between load points is 50 mm (2.0 in.). 7.50 A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in.); the specimen fractured at a load of 950 N (215 lbf) when the distance between the support points was 50 mm (2.0 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm (0.47 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm (1.6 in.)? 7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the speci-

194

●

Chapter 7 / Mechanical Properties

men radius is 2.5 mm (0.10 in.) and the support point separation distance is 30 mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction. (b) Would you be 100% certain of the prediction in part a? Why or why not? 7.52* The modulus of elasticity for beryllium oxide (BeO) having 5 vol% porosity is 310 GPa (45 ⫻ 106 psi). (a) Compute the modulus of elasticity for the nonporous material. (b) Compute the modulus of elasticity for 10 vol% porosity. 7.53* The modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 ⫻ 106 psi). (a) Compute the modulus of elasticity for the nonporous material. (b) At what volume percent porosity will the modulus of elasticity be 235 GPa (34 ⫻ 106 psi)? 7.54* Using the data in Table 7.2, do the following: (a) Determine the flexural strength for nonporous MgO assuming a value of 3.75 for n in Equation 7.22. (b) Compute the volume fraction porosity at which the flexural strength for MgO is 62 MPa (9000 psi). 7.55* The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: fs (MPa) 100 50

P 0.05 0.20

(a) Compute the flexural strength for a completely nonporous specimen of this material. (b) Compute the flexural strength for a 0.1 volume fraction porosity. 7.56 From the stress–strain data for polymethyl methacrylate shown in Figure 7.24, determine the modulus of elasticity and tensile strength at room temperature [20⬚C (68⬚F)], and compare these values with those given in Tables 7.1 and 7.2.

7.57 When citing the ductility as percent elongation for semicrystalline polymers, it is not necessary to specify the specimen gauge length, as is the case with metals. Why is this so? 7.58* In your own words, briefly describe the phenomenon of viscoelasticity. 7.59* For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to

冉 冊

t (t) ⫽ (0) exp ⫺

(7.33)

where (t) and (0) represent the time-dependent and initial (i.e., time ⫽ 0) stresses, respectively, and t and denote elapsed time and the relaxation time; is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer the stress relaxation of which obeys Equation 7.33 was suddenly pulled in tension to a measured strain of 0.6; the stress necessary to maintain this constant strain was measured as a function of time. Determine Er (10) for this material if the initial stress level was 2.76 MPa (400 psi), which dropped to 1.72 MPa (250 psi) after 60 s. 7.60* In Figure 7.35, the logarithm of Er (t) versus the logarithm of time is plotted for polyisobutylene at a variety of temperatures. Make a plot of log Er (10) versus temperature and then estimate the Tg . 7.61* On the basis of the curves in Figure 7.26, sketch schematic strain-time plots for the following polystyrene materials at the specified temperatures: (a) Amorphous at 120⬚C. (b) Crosslinked at 150⬚C. (c) Crystalline at 230⬚C. (d) Crosslinked at 50⬚C. 7.62* (a) Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted. (b) For each of these tests, cite the experimental parameter of interest and how it is determined.

Questions and Problems

–80.8°C

–76.7°C –74.1°C

Relaxation modulus (MPa)

–70.6°C –49.6°C

–65.4°C

–40.1°C

195

FIGURE 7.35 Logarithm of relaxation modulus versus logarithm of time for polyisobutylene between ⫺80 and 50⬚C. (Adapted from E. Catsiff and A. V. Tobolsky, ‘‘Stress-Relaxation of Polyisobutylene in the Transition Region [1,2],’’ J. Colloid Sci., 10, 377 [1955]. Reprinted by permission of Academic Press, Inc.)

104

102

●

–58.8°C

1 0°C 25°C 10–2

50°C

10–4

1

10

102

103

104

105

106

Time (s)

7.63* Make two schematic plots of the logarithm of relaxation modulus versus temperature for an amorphous polymer (curve C in Figure 7.29). (a) On one of these plots demonstrate how the behavior changes with increasing molecular weight. (b) On the other plot, indicate the change in behavior with increasing crosslinking. 7.64 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 1.62 mm in diameter in a steel alloy when a load of 500 kg was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 450 HB when a 500 kg load is used? 7.65 Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stress– strain behavior is shown in Figure 7.12. (b) The steel for which the stress–strain behavior is shown in Figure 7.33.

7.66 Using the data represented in Figure 7.31, specify equations relating tensile strength and Brinell hardness for brass and nodular cast iron, similar to Equations 7.25a and 7.25b for steels. 7.67 Cite five factors that lead to scatter in measured material properties. 7.68* Below are tabulated a number of Rockwell B hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. 83.3 88.3 82.8 86.2 87.2

80.7 84.7 87.8 83.5 85.5

86.4 85.2 86.9 84.4 86.3

7.69 Upon what three criteria are factors of safety based? 7.70 Determine working stresses for the two alloys the stress–strain behaviors of which are shown in Figures 7.12 and 7.33.

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Chapter 7 / Mechanical Properties

Design Problems 7.D1 A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N (2500 lbf ). Determine the minimum required wire diameter assuming a factor of safety of 2 and a yield strength of 1030 MPa (150,000 psi). 7.D2 (a) Gaseous hydrogen at a constant pressure of 1.013 MPa (10 atm) is to flow within the inside of a thin-walled cylindrical tube of nickel that has a radius of 0.1 m. The temperature of the tube is to be 300⬚C and the pressure of hydrogen outside of the tube will be maintained at 0.01013 MPa (0.1 atm). Calculate the minimum wall thickness if the diffusion flux is to be no greater than 1 ⫻ 10⫺7 mol/m2-s. The concentration of hydrogen in the nickel, CH (in moles hydrogen per m3 of Ni) is a function of hydrogen pressure, pH2 (in MPa) and absolute temperature (T ) according to

冉

冊

12.3 kJ/mol RT (7.34)

CH ⫽ 30.8 兹pH2 exp ⫺

Furthermore, the diffusion coefficient for the diffusion of H in Ni depends on temperature as

冉

冊

39.56 kJ/mol RT (7.35)

DH(m2 /s) ⫽ 4.76 ⫻ 10⫺7 exp ⫺

(b) For thin-walled cylindrical tubes that are pressurized, the circumferential stress is a function of the pressure difference across the wall (⌬p), cylinder radius (r), and tube thickness (⌬x) as

⫽

r ⌬p 4 ⌬x

(7.36)

Compute the circumferential stress to which the walls of this pressurized cylinder are exposed.

(c) The room-temperature yield strength of Ni is 100 MPa (15,000 psi) and, furthermore, y diminishes about 5 MPa for every 50⬚C rise in temperature. Would you expect the wall thickness computed in part (b) to be suitable for this Ni cylinder at 300⬚C? Why or why not? (d) If this thickness is found to be suitable, compute the minimum thickness that could be used without any deformation of the tube walls. How much would the diffusion flux increase with this reduction in thickness? On the other hand, if the thickness determined in part (c) is found to be unsuitable, then specify a minimum thickness that you would use. In this case, how much of a diminishment in diffusion flux would result? 7.D3 Consider the steady-state diffusion of hydrogen through the walls of a cylindrical nickel tube as described in Problem 7.D2. One design calls for a diffusion flux of 5 ⫻ 10⫺8 mol/ m2-s, a tube radius of 0.125 m, and inside and outside pressures of 2.026 MPa (20 atm) and 0.0203 MPa (0.2 atm), respectively; the maximum allowable temperature is 450⬚C. Specify a suitable temperature and wall thickness to give this diffusion flux and yet ensure that the tube walls will not experience any permanent deformation. 7.D4 It is necessary to select a ceramic material to be stressed using a three-point loading scheme (Figure 7.18). The specimen must have a circular cross section and a radius of 2.5 mm (0.10 in.), and must not experience fracture or a deflection of more than 6.2 ⫻ 10⫺2 mm (2.4 ⫻ 10⫺3 in.) at its center when a load of 275 N (62 lbf) is applied. If the distance between support points is 45 mm (1.77 in.), which of the ceramic materials in Tables 7.1 and 7.2 are candidates? The magnitude of the centerpoint deflection may be computed using the equation supplied in Problem 7.48.

Chapter

8

/ Deformation and Strengthening Mechanisms

I

n this photomicro-

graph of a lithium fluoride (LiF) single crystal, the small pyramidal pits represent those positions at which dislocations intersect the surface. The surface was polished and then chemically treated; these ‘‘etch pits’’ result from localized chemical attack around the dislocations and indicate the distribution of the dislocations. 750ⴛ. (Photomicrograph courtesy of W. G. Johnston, General Electric Co.)

Why Study Deformation and Strengthening Mechanisms? With a knowledge of the nature of dislocations and the role they play in the plastic deformation process, we are able to understand the underlying mechanisms of the techniques that are used to strengthen and harden metals and their alloys; thus, it becomes possible to design and tailor the mechan-

ical properties of materials—for example, the strength or toughness of a metal-matrix composite. Also, understanding the mechanisms by which polymers elastically and plastically deform allows one to alter and control their moduli of elasticity and strengths (Sections 8.17 and 8.18).

197

Learning Objectives After studying this chapter you should be able to do the following: 1. Describe edge and screw dislocation motion from an atomic perspective. 2. Describe how plastic deformation occurs by the motion of edge and screw dislocations in response to applied shear stresses. 3. Define slip system and cite one example. 4. Describe how the grain structure of a polycrystalline metal is altered when it is plastically deformed. 5. Explain how grain boundaries impede dislocation motion and why a metal having small grains is stronger than one having large grains. 6. Describe and explain solid-solution strengthening for substitutional impurity atoms in terms of lattice strain interactions with dislocations. 7. Describe and explain the phenomenon of strain hardening (or cold working) in terms of dislocations and strain field interactions.

8. Describe recrystallization in terms of both the alteration of microstructure and mechanical characteristics of the material. 9. Describe the phenomenon of grain growth from both macroscopic and atomic perspectives. 10. On the basis of slip considerations, explain why crystalline ceramic materials are normally brittle. 11. Describe/sketch the various stages in the plastic deformation of a semicrystalline (spherulitic) polymer. 12. Discuss the influence of the following factors on polymer tensile modulus and/or strength: (a) molecular weight, (b) degree of crystallinity, (c) predeformation, and (d) heat treating of undeformed materials. 13. Describe the molecular mechanism by which elastomeric polymers deform elastically.

8.1 INTRODUCTION In this chapter we explore various deformation mechanisms that have been proposed to explain the deformation behaviors of metals, ceramics, and polymeric materials. Techniques that may be used to strengthen the various material types are described and explained in terms of these deformation mechanisms.

DEFORMATION MECHANISMS FOR METALS Chapter 7 explained that metallic materials may experience two kinds of deformation: elastic and plastic. Plastic deformation is permanent, and strength and hardness are measures of a material’s resistance to this deformation. On a microscopic scale, plastic deformation corresponds to the net movement of large numbers of atoms in response to an applied stress. During this process, interatomic bonds must be ruptured and then reformed. Furthermore, plastic deformation most often involves the motion of dislocations, linear crystalline defects that were introduced in Section 5.7. This section discusses the characteristics of dislocations and their involvement in plastic deformation. Sections 8.9, 8.10, and 8.11 present several techniques for strengthening single-phase metals, the mechanisms of which are described in terms of dislocations.

8.2 HISTORICAL Early materials studies led to the computation of the theoretical strengths of perfect crystals, which were many times greater than those actually measured. During the 1930s it was theorized that this discrepancy in mechanical strengths could be explained by a type of linear crystalline defect that has since come to be known as a dislocation. It was not until the 1950s, however, that the existence of such dislocation defects was established by direct observation with the electron microscope. Since then, a theory of dislocations has evolved that

198

8.3 Basic Concepts of Dislocations

●

199

explains many of the physical and mechanical phenomena in metals [as well as crystalline ceramics (Section 8.15)].

8.3 BASIC CONCEPTS

OF

DISLOCATIONS

Edge and screw are the two fundamental dislocation types. In an edge dislocation, localized lattice distortion exists along the end of an extra half-plane of atoms, which also defines the dislocation line (Figure 5.7). A screw dislocation may be thought of as resulting from shear distortion; its dislocation line passes through the center of a spiral, atomic plane ramp (Figure 5.8). Many dislocations in crystalline materials have both edge and screw components; these are mixed dislocations (Figure 5.9). Plastic deformation corresponds to the motion of large numbers of dislocations. An edge dislocation moves in response to a shear stress applied in a direction perpendicular to its line; the mechanics of dislocation motion are represented in Figure 8.1. Let the initial extra half-plane of atoms be plane A. When the shear stress is applied as indicated (Figure 8.1a), plane A is forced to the right; this in turn pushes the top halves of planes B, C, D, and so on, in the same direction. If the applied shear stress is of sufficient magnitude, the interatomic bonds of plane B are severed along the shear plane, and the upper half of plane B becomes the extra half-plane as plane A links up with the bottom half of plane B (Figure 8.1b). This process is subsequently repeated for the other planes, such that the extra half-plane, by discrete steps, moves from left to right by successive and repeated breaking of bonds and shifting by interatomic distances of upper half-planes. Before and after the movement of a dislocation through some particular region of the crystal, the atomic arrangement is ordered and perfect; it is only during the passage of the extra half-plane that the lattice structure is disrupted. Ultimately this extra half-plane may emerge

Shear stress

Shear stress A

B

C

D

Shear stress A

B

C

D

A

B

C

D

Slip plane

Unit step of slip

Edge dislocation line (a)

( b)

(c)

FIGURE 8.1 Atomic rearrangements that accompany the motion of an edge dislocation as it moves in response to an applied shear stress. (a) The extra halfplane of atoms is labeled A. (b) The dislocation moves one atomic distance to the right as A links up to the lower portion of plane B; in the process, the upper portion of B becomes the extra half-plane. (c) A step forms on the surface of the crystal as the extra half-plane exits. (Adapted from A. G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1976, p. 153.)

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Direction of motion

(a)

Direction of motion

(b)

FIGURE 8.2 The formation of a step on the surface of a crystal by the motion of (a) an edge dislocation and (b) a screw dislocation. Note that for an edge, the dislocation line moves in the direction of the applied shear stress ; for a screw, the dislocation line motion is perpendicular to the stress direction. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 70. Copyright 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

from the right surface of the crystal, forming an edge that is one atomic distance wide; this is shown in Figure 8.1c. The process by which plastic deformation is produced by dislocation motion is termed slip; the crystallographic plane along which the dislocation line traverses is the slip plane, as indicated in Figure 8.1. Macroscopic plastic deformation simply corresponds to permanent deformation that results from the movement of dislocations, or slip, in response to an applied shear stress, as represented in Figure 8.2a. Dislocation motion is analogous to the mode of locomotion employed by a caterpillar (Figure 8.3). The caterpillar forms a hump near its posterior end by pulling in its last pair of legs a unit leg distance. The hump is propelled forward by repeated lifting and shifting of leg pairs. When the hump reaches the anterior

FIGURE 8.3 Representation of the analogy between caterpillar and dislocation motion.

8.4 Characteristics of Dislocations

●

201

end, the entire caterpillar has moved forward by the leg separation distance. The caterpillar hump and its motion correspond to the extra half-plane of atoms in the dislocation model of plastic deformation. The motion of a screw dislocation in response to the applied shear stress is shown in Figure 8.2b; the direction of movement is perpendicular to the stress direction. For an edge, motion is parallel to the shear stress. However, the net plastic deformation for the motion of both dislocation types is the same (see Figure 8.2). The direction of motion of the mixed dislocation line is neither perpendicular nor parallel to the applied stress, but lies somewhere in between. All metals and alloys contain some dislocations that were introduced during solidification, during plastic deformation, and as a consequence of thermal stresses that result from rapid cooling. The number of dislocations, or dislocation density in a material, is expressed as the total dislocation length per unit volume, or, equivalently, the number of dislocations that intersect a unit area of a random section. The units of dislocation density are millimeters of dislocation per cubic millimeter or just per square millimeter. Dislocation densities as low as 103 mm⫺2 are typically found in carefully solidified metal crystals. For heavily deformed metals, the density may run as high as 109 to 1010 mm⫺2. Heat treating a deformed metal specimen can diminish the density to on the order of 105 to 106 mm⫺2. By way of contrast, a typical dislocation density for ceramic materials is between 102 and 104 mm⫺2; also, for silicon single crystals used in integrated circuits the value normally lies between 0.1 and 1 mm⫺2.

8.4 CHARACTERISTICS OF DISLOCATIONS Several characteristics of dislocations are important with regard to the mechanical properties of metals. These include strain fields that exist around dislocations, which are influential in determining the mobility of the dislocations, as well as their ability to multiply. When metals are plastically deformed, some fraction of the deformation energy (approximately 5%) is retained internally; the remainder is dissipated as heat. The major portion of this stored energy is as strain energy associated with dislocations. Consider the edge dislocation represented in Figure 8.4. As already mentioned, some atomic lattice distortion exists around the dislocation line because of the presence of the extra half-plane of atoms. As a consequence, there are regions in which compressive, tensile, and shear lattice strains are imposed on the neighboring atoms. For example, atoms immediately above and adjacent to the dislocation line

Compression Tension

FIGURE 8.4 Regions of compression (dark) and tension (colored) located around an edge dislocation. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 85. Copyright 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

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are squeezed together. As a result, these atoms may be thought of as experiencing a compressive strain relative to atoms positioned in the perfect crystal and far removed from the dislocation; this is illustrated in Figure 8.4. Directly below the half-plane, the effect is just the opposite; lattice atoms sustain an imposed tensile strain, which is as shown. Shear strains also exist in the vicinity of the edge dislocation. For a screw dislocation, lattice strains are pure shear only. These lattice distortions may be considered to be strain fields that radiate from the dislocation line. The strains extend into the surrounding atoms, and their magnitudes decrease with radial distance from the dislocation. The strain fields surrounding dislocations in close proximity to one another may interact such that forces are imposed on each dislocation by the combined interactions of all its neighboring dislocations. For example, consider two edge dislocations that have the same sign and the identical slip plane, as represented in Figure 8.5a. The compressive and tensile strain fields for both lie on the same side of the slip plane; the strain field interaction is such that there exists between these two isolated dislocations a mutual repulsive force that tends to move them apart. On the other hand, two dislocations of opposite sign and having the same slip plane will be attracted to one another, as indicated in Figure 8.5b, and dislocation annihilation will occur when they meet. That is, the two extra half-planes of atoms will align and become a complete plane. Dislocation interactions are possible between edge, screw, and/or mixed dislocations, and for a variety of orientations. These strain fields and associated forces are important in the strengthening mechanisms for metals.

C

C Repulsion

T

T (a)

C

T Dislocation annihilation

Attraction ;

+

=

(Perfect crystal) T

C (b)

FIGURE 8.5 (a) Two edge dislocations of the same sign and lying on the same slip plane exert a repulsive force on each other; C and T denote compression and tensile regions, respectively. (b) Edge dislocations of opposite sign and lying on the same slip plane exert an attractive force on each other. Upon meeting, they annihilate each other and leave a region of perfect crystal. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 75. Copyright 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons.)

8.5 Slip Systems

●

203

During plastic deformation, the number of dislocations increases dramatically. We know that the dislocation density in a metal that has been highly deformed may be as high as 1010 mm⫺2. One important source of these new dislocations is existing dislocations, which multiply; furthermore, grain boundaries, as well as internal defects and surface irregularities such as scratches and nicks, which act as stress concentrations, may serve as dislocation formation sites during deformation.

8.5 SLIP SYSTEMS Dislocations do not move with the same degree of ease on all crystallographic planes of atoms and in all crystallographic directions. Ordinarily there is a preferred plane, and in that plane there are specific directions along which dislocation motion occurs. This plane is called the slip plane; it follows that the direction of movement is called the slip direction. This combination of the slip plane and the slip direction is termed the slip system. The slip system depends on the crystal structure of the metal and is such that the atomic distortion that accompanies the motion of a dislocation is a minimum. For a particular crystal structure, the slip plane is that plane having the most dense atomic packing, that is, has the greatest planar density. The slip direction corresponds to the direction, in this plane, that is most closely packed with atoms, that is, has the highest linear density. 兵Planar and linear atomic densities were discussed in Section 3.14.其 Consider, for example, the FCC crystal structure, a unit cell of which is shown in Figure 8.6a. There is a set of planes, the 兵111其 family, all of which are closely packed. A (111)-type plane is indicated in the unit cell; in Figure 8.6b, this plane is positioned within the plane of the page, in which atoms are now represented as touching nearest neighbors. Slip occurs along 具110典-type directions within the 兵111其 planes, as indicated by arrows in Figure 8.6. Hence, 兵111其具110典 represents the slip plane and direction combination, or the slip system for FCC. Figure 8.6b demonstrates that a given slip plane may contain more than a single slip direction. Thus, several slip systems may exist for a particular crystal structure; the number of independent slip systems represents the different possible combinations of slip planes and directions. For example, for face-centered cubic, there are 12 slip systems: four unique 兵111其 planes and, within each plane, three independent 具110典 directions. The possible slip systems for BCC and HCP crystal structures are listed in Table 8.1. For each of these structures, slip is possible on more than one family of planes (e.g., 兵110其, 兵211其, and 兵321其 for BCC). For metals having these two crystal structures, some slip systems are often operable only at elevated temperatures.

A A C B

B

F

D

C

E

E D

(b) (a)

F

FIGURE 8.6 (a) A 兵111其具110典 slip system shown within an FCC unit cell. (b) The (111) plane from (a) and three 具110典 slip directions (as indicated by arrows) within that plane comprise possible slip systems.

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Table 8.1 Slip Systems for Face-Centered Cubic, Body-Centered Cubic, and Hexagonal Close-Packed Metals Metals

Slip Plane

Slip Direction

Number of Slip Systems

Cu, Al, Ni, Ag, Au

Face-Centered Cubic 兵111其 具110典

12

움-Fe, W, Mo 움-Fe, W 움-Fe, K

Body-Centered Cubic 兵110其 具111典 兵211其 具111典 兵321其 具111典

12 12 24

Hexagonal Close-Packed 兵0001其 具1120典 兵1010其 具1120典 兵1011其 具1120典

3 3 6

Cd, Zn, Mg, Ti, Be Ti, Mg, Zr Ti, Mg

Metals with FCC or BCC crystal structures have a relatively large number of slip systems (at least 12). These metals are quite ductile because extensive plastic deformation is normally possible along the various systems. Conversely, HCP metals, having few active slip systems, are normally quite brittle.

8.6 SLIP

IN

SINGLE CRYSTALS (CD-ROM)

8.7 PLASTIC DEFORMATION OF POLYCRYSTALLINE METALS For polycrystalline metals, because of the random crystallographic orientations of the numerous grains, the direction of slip varies from one grain to another. For each, dislocation motion occurs along the slip system that has the most favorable orientation (i.e., the highest shear stress). This is exemplified by a photomicrograph of a polycrystalline copper specimen that has been plastically deformed (Figure 8.10); before deformation the surface was polished. Slip lines1 are visible, and it appears that two slip systems operated for most of the grains, as evidenced by two sets of parallel yet intersecting sets of lines. Furthermore, variation in grain orientation is indicated by the difference in alignment of the slip lines for the several grains. Gross plastic deformation of a polycrystalline specimen corresponds to the comparable distortion of the individual grains by means of slip. During deformation, mechanical integrity and coherency are maintained along the grain boundaries; that is, the grain boundaries usually do not come apart or open up. As a consequence, each individual grain is constrained, to some degree, in the shape it may assume by its neighboring grains. The manner in which grains distort as a result of gross plastic deformation is indicated in Figure 8.11. Before deformation the grains are equiaxed, or have approximately the same dimension in all directions. For this 1 Surface steps or ledges produced by dislocations (Figure 8.1c) that have exited from a grain and that appear as lines when viewed with a microscope are called slip lines.

8.7 Plastic Deformation of Polycrystalline Metals

●

205

FIGURE 8.10 Slip lines on the surface of a polycrystalline specimen of copper that was polished and subsequently deformed. 173⫻. (Photomicrograph courtesy of C. Brady, National Bureau of Standards.)

FIGURE 8.11 Alteration of the grain structure of a polycrystalline metal as a result of plastic deformation. (a) Before deformation the grains are equiaxed. (b) The deformation has produced elongated grains. 170⫻. (From W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140. Copyright 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

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Chapter 8 / Deformation and Strengthening Mechanisms

particular deformation, the grains become elongated along the direction in which the specimen was extended. Polycrystalline metals are stronger than their single-crystal equivalents, which means that greater stresses are required to initiate slip and the attendant yielding. This is, to a large degree, also a result of geometrical constraints that are imposed on the grains during deformation. Even though a single grain may be favorably oriented with the applied stress for slip, it cannot deform until the adjacent and less favorably oriented grains are capable of slip also; this requires a higher applied stress level.

8.8 DEFORMATION

BY

TWINNING (CD-ROM)

MECHANISMS OF STRENGTHENING IN METALS Metallurgical and materials engineers are often called on to design alloys having high strengths yet some ductility and toughness; ordinarily, ductility is sacrificed when an alloy is strengthened. Several hardening techniques are at the disposal of an engineer, and frequently alloy selection depends on the capacity of a material to be tailored with the mechanical characteristics required for a particular application. Important to the understanding of strengthening mechanisms is the relation between dislocation motion and mechanical behavior of metals. Because macroscopic plastic deformation corresponds to the motion of large numbers of dislocations, the ability of a metal to plastically deform depends on the ability of dislocations to move. Since hardness and strength (both yield and tensile) are related to the ease with which plastic deformation can be made to occur, by reducing the mobility of dislocations, the mechanical strength may be enhanced; that is, greater mechanical forces will be required to initiate plastic deformation. In contrast, the more unconstrained the dislocation motion, the greater the facility with which a metal may deform, and the softer and weaker it becomes. Virtually all strengthening techniques rely on this simple principle: restricting or hindering dislocation motion renders a material harder and stronger. The present discussion is confined to strengthening mechanisms for singlephase metals, by grain size reduction, solid-solution alloying, and strain hardening. Deformation and strengthening of multiphase alloys are more complicated, involving concepts yet to be discussed.

8.9 STRENGTHENING

BY

GRAIN SIZE REDUCTION

The size of the grains, or average grain diameter, in a polycrystalline metal influences the mechanical properties. Adjacent grains normally have different crystallographic orientations and, of course, a common grain boundary, as indicated in Figure 8.14. During plastic deformation, slip or dislocation motion must take place across this common boundary, say, from grain A to grain B in Figure 8.14. The grain boundary acts as a barrier to dislocation motion for two reasons: 1. Since the two grains are of different orientations, a dislocation passing into grain B will have to change its direction of motion; this becomes more difficult as the crystallographic misorientation increases.

8.9 Strengthening by Grain Size Reduction

207

●

Grain boundary

Slip plane

Grain A

Grain B

FIGURE 8.14 The motion of a dislocation as it encounters a grain boundary, illustrating how the boundary acts as a barrier to continued slip. Slip planes are discontinuous and change directions across the boundary. (From A Textbook of Materials Technology by Van Vlack, 1973. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ.)

2. The atomic disorder within a grain boundary region will result in a discontinuity of slip planes from one grain into the other. It should be mentioned that, for high-angle grain boundaries, it may not be the case that dislocations traverse grain boundaries during deformation; rather, a stress concentration ahead of a slip plane in one grain may activate sources of new dislocations in an adjacent grain. A fine-grained material (one that has small grains) is harder and stronger than one that is coarse grained, since the former has a greater total grain boundary area to impede dislocation motion. For many materials, the yield strength y varies with grain size according to

y ⫽ 0 ⫹ ky d ⫺1/2

(8.5)

In this expression, termed the Hall-Petch equation, d is the average grain diameter, and 0 and ky are constants for a particular material. It should be noted that Equation 8.5 is not valid for both very large (i.e., coarse) grain and extremely fine grain polycrystalline materials. Figure 8.15 demonstrates the yield strength dependence on grain size for a brass alloy. Grain size may be regulated by the rate of solidification from the liquid phase, and also by plastic deformation followed by an appropriate heat treatment, as discussed in Section 8.14. It should also be mentioned that grain size reduction improves not only strength, but also the toughness of many alloys. Small-angle grain boundaries (Section 5.8) are not effective in interfering with the slip process because of the slight crystallographic misalignment across the boundary. On the other hand, twin boundaries (Section 5.8) will effectively block slip and increase the strength of the material. Boundaries between two different phases are also impediments to movements of dislocations; this is important in the strengthening of more complex alloys. The sizes and shapes of the constituent phases significantly affect the mechanical properties of multiphase alloys; these are the topics of discussion in Sections 11.7, 11.8, 兵and 15.1.其

208

●

Chapter 8 / Deformation and Strengthening Mechanisms Grain size, d (mm) 10

⫺1

10

⫺2

5 10

⫺3

30

150 20

100 10

Yield strength (ksi)

Yield strength (MPa)

200

FIGURE 8.15 The influence of grain size on the yield strength of a 70 Cu–30 Zn brass alloy. Note that the grain diameter increases from right to left and is not linear. (Adapted from H. Suzuki, ‘‘The Relation Between the Structure and Mechanical Properties of Metals,’’ Vol. II, National Physical Laboratory, Symposium No. 15, 1963, p. 524.)

50

0

0 4

8

12

16

d⫺1/2, (mm⫺1/2)

8.10 SOLID-SOLUTION STRENGTHENING Another technique to strengthen and harden metals is alloying with impurity atoms that go into either substitutional or interstitial solid solution. Accordingly, this is called solid-solution strengthening. High-purity metals are almost always softer and weaker than alloys composed of the same base metal. Increasing the concentration of the impurity results in an attendant increase in tensile and yield strengths, as indicated in Figures 8.16a and 8.16b for nickel in copper; the dependence of ductility on nickel concentration is presented in Figure 8.16c. Alloys are stronger than pure metals because impurity atoms that go into solid solution ordinarily impose lattice strains on the surrounding host atoms. Lattice strain field interactions between dislocations and these impurity atoms result, and, consequently, dislocation movement is restricted. For example, an impurity atom that is smaller than a host atom for which it substitutes exerts tensile strains on the surrounding crystal lattice, as illustrated in Figure 8.17a. Conversely, a larger substitutional atom imposes compressive strains in its vicinity (Figure 8.18a). These solute atoms tend to diffuse to and segregate around dislocations in a way so as to reduce the overall strain energy, that is, to cancel some of the strain in the lattice surrounding a dislocation. To accomplish this, a smaller impurity atom is located where its tensile strain will partially nullify some of the dislocation’s compressive strain. For the edge dislocation in Figure 8.17b, this would be adjacent to the dislocation line and above the slip plane. A larger impurity atom would be situated as in Figure 8.18b. The resistance to slip is greater when impurity atoms are present because the overall lattice strain must increase if a dislocation is torn away from them. Furthermore, the same lattice strain interactions (Figures 8.17b and 8.18b) will exist between impurity atoms and dislocations that are in motion during plastic deformation. Thus, a greater applied stress is necessary to first initiate and then continue plastic deformation for solid-solution alloys, as opposed to pure metals; this is evidenced by the enhancement of strength and hardness.

8.10 Solid-Solution Strengthening

209

●

180 25 60

40

Yield strength (MPa)

300

Tensile strength (ksi)

Tensile strength (MPa)

50

140

20

120 15

100

Yield strength (ksi)

160

400

80 200 0

10

20

30

40

30 50

10 60

0

10

20

30

Nickel content (wt%)

Nickel content (wt%)

(a)

( b)

40

50

Elongation (% in 2 in.)

60

50

40

30

20 0

10

20

30

40

Nickel content (wt%) (c)

50

FIGURE 8.16 Variation with nickel content of (a) tensile strength, (b) yield strength, and (c) ductility (%EL) for copper-nickel alloys, showing strengthening.

(a) (b)

FIGURE 8.17 (a) Representation of tensile lattice strains imposed on host atoms by a smaller substitutional impurity atom. (b) Possible locations of smaller impurity atoms relative to an edge dislocation such that there is partial cancellation of impurity–dislocation lattice strains.

210

●

Chapter 8 / Deformation and Strengthening Mechanisms FIGURE 8.18 (a) Representation of compressive strains imposed on host atoms by a larger substitutional impurity atom. (b) Possible locations of larger impurity atoms relative to an edge dislocation such that there is partial cancellation of impurity–dislocation lattice strains.

(a) (b)

8.11 STRAIN HARDENING Strain hardening is the phenomenon whereby a ductile metal becomes harder and stronger as it is plastically deformed. Sometimes it is also called work hardening, or, because the temperature at which deformation takes place is ‘‘cold’’ relative to the absolute melting temperature of the metal, cold working. Most metals strain harden at room temperature. It is sometimes convenient to express the degree of plastic deformation as percent cold work rather than as strain. Percent cold work (%CW) is defined as %CW ⫽

冉

冊

A0 ⫺ A d ⫻ 100 A0

(8.6)

where A0 is the original area of the cross section that experiences deformation, and Ad is the area after deformation. Figures 8.19a and 8.19b demonstrate how steel, brass, and copper increase in yield and tensile strength with increasing cold work. The price for this enhancement of hardness and strength is in the ductility of the metal. This is shown in Figure 8.19c, in which the ductility, in percent elongation, experiences a reduction with increasing percent cold work for the same three alloys. The influence of cold work on the stress–strain behavior of a steel is vividly portrayed in Figure 8.20. Strain hardening is demonstrated in a stress–strain diagram presented earlier (Figure 7.17). Initially, the metal with yield strength y0 is plastically deformed to point D. The stress is released, then reapplied with a resultant new yield strength, yi . The metal has thus become stronger during the process because yi is greater than y0 . The strain hardening phenomenon is explained on the basis of dislocation– dislocation strain field interactions similar to those discussed in Section 8.4. The dislocation density in a metal increases with deformation or cold work, due to dislocation multiplication or the formation of new dislocations, as noted previously. Consequently, the average distance of separation between dislocations decreases—the dislocations are positioned closer together. On the average, dislocation–dislocation strain interactions are repulsive. The net result is that the motion of a dislocation is hindered by the presence of other dislocations. As the dislocation density increases, this resistance to dislocation motion by other dislocations becomes more pronounced. Thus, the imposed stress necessary to deform a metal increases with increasing cold work.

8.11 Strain Hardening

●

211

140 120

900

800

1040 Steel 120

1040 Steel

800

80 500

Brass

400

60

700

100

600 Brass 80 500

Tensile strength (ksi)

600

Tensile strength (MPa)

100

Yield strength (ksi)

Yield strength (MPa)

700

Copper

300

40

60

400 Copper

200 300 40

20 100 0

10

20

30

40

50

60

70

200

0

10

20

30

40

Percent cold work

Percent cold work

(a)

( b)

50

60

70

70

60

Ductility (%EL)

50

40 Brass 30

20

1040 Steel

10

Copper 0

0

10

20

30

40

Percent cold work (c)

50

60

70

FIGURE 8.19 For 1040 steel, brass, and copper, (a) the increase in yield strength, (b) the increase in tensile strength, and (c) the decrease in ductility (%EL) with percent cold work. (Adapted from Metals Handbook: Properties and Selection: Irons and Steels, Vol. 1, 9th edition, B. Bardes, Editor, American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th edition, H. Baker, Managing Editor, American Society for Metals, 1979, pp. 276 and 327.)

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●

Chapter 8 / Deformation and Strengthening Mechanisms FIGURE 8.20 The influence of cold work on the stress–strain behavior for a lowcarbon steel. (From Metals Handbook: Properties and Selection: Irons and Steels, Vol. 1, 9th edition, B. Bardes, Editor, American Society for Metals, 1978, p. 221.)

Stress

P co erce ld nt wo rk

Strain

Strain hardening is often utilized commercially to enhance the mechanical properties of metals during fabrication procedures. The effects of strain hardening may be removed by an annealing heat treatment, 兵as discussed in Section 14.5.其 In passing, for the mathematical expression relating true stress and strain, Equation 7.19, the parameter n is called the strain hardening exponent, which is a measure of the ability of a metal to strain harden; the larger its magnitude, the greater the strain hardening for a given amount of plastic strain.

EXAMPLE PROBLEM 8.2 Compute the tensile strength and ductility (%EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm (0.60 in. to 0.48 in.).

S OLUTION It is first necessary to determine the percent cold work resulting from the deformation. This is possible using Equation 8.6:

%CW ⫽

冉

15.2 mm 2

冉

冊 冉 2

앟⫺

12.2 mm 2

15.2 mm 2

冊

2

冊

2

앟 ⫻ 100 ⫽ 35.6%

앟

The tensile strength is read directly from the curve for copper (Figure 8.19b) as 340 MPa (50,000 psi). From Figure 8.19c, the ductility at 35.6%CW is about 7%EL.

In summary, we have just discussed the three mechanisms that may be used to strengthen and harden single-phase metal alloys—strengthening by grain size reduction, solid solution strengthening, and strain hardening. Of course they may be used in conjunction with one another; for example, a solid-solution strengthened alloy may also be strain hardened.

8.13 Recrystallization

●

213

RECOVERY, RECRYSTALLIZATION, AND GRAIN GROWTH As outlined in the preceding paragraphs of this chapter, plastically deforming a polycrystalline metal specimen at temperatures that are low relative to its absolute melting temperature produces microstructural and property changes that include (1) a change in grain shape (Section 8.7), (2) strain hardening (Section 8.11), and (3) an increase in dislocation density (Section 8.4). Some fraction of the energy expended in deformation is stored in the metal as strain energy, which is associated with tensile, compressive, and shear zones around the newly created dislocations (Section 8.4). Furthermore, other properties such as electrical conductivity (Section 12.8) and corrosion resistance may be modified as a consequence of plastic deformation. These properties and structures may revert back to the precold-worked states by appropriate heat treatment (sometimes termed an annealing treatment). Such restoration results from two different processes that occur at elevated temperatures: recovery and recrystallization, which may be followed by grain growth.

8.12 RECOVERY During recovery, some of the stored internal strain energy is relieved by virtue of dislocation motion (in the absence of an externally applied stress), as a result of enhanced atomic diffusion at the elevated temperature. There is some reduction in the number of dislocations, and dislocation configurations (similar to that shown in Figure 5.12) are produced having low strain energies. In addition, physical properties such as electrical and thermal conductivities and the like are recovered to their precold-worked states.

8.13 RECRYSTALLIZATION Even after recovery is complete, the grains are still in a relatively high strain energy state. Recrystallization is the formation of a new set of strain-free and equiaxed grains (i.e., having approximately equal dimensions in all directions) that have low dislocation densities and are characteristic of the precold-worked condition. The driving force to produce this new grain structure is the difference in internal energy between the strained and unstrained material. The new grains form as very small nuclei and grow until they completely replace the parent material, processes that involve short-range diffusion. Several stages in the recrystallization process are represented in Figures 8.21a to 8.21d; in these photomicrographs, the small speckled grains are those that have recrystallized. Thus, recrystallization of cold-worked metals may be used to refine the grain structure. Also, during recrystallization, the mechanical properties that were changed as a result of cold working are restored to their precold-worked values; that is, the metal becomes softer, weaker, yet more ductile. Some heat treatments are designed to allow recrystallization to occur with these modifications in the mechanical characteristics 兵(Section 14.5).其 Recrystallization is a process the extent of which depends on both time and temperature. The degree (or fraction) of recrystallization increases with time, as may be noted in the photomicrographs shown in Figures 8.21a–d. The explicit time dependence of recrystallization is addressed in more detail in Section 11.3.

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(a)

(b)

(c)

(d)

FIGURE 8.21 Photomicrographs showing several stages of the recrystallization and grain growth of brass. (a) Cold-worked (33%CW) grain structure. (b) Initial stage of recrystallization after heating 3 s at 580⬚C (1075⬚F); the very small grains are those that have recrystallized. (c) Partial replacement of cold-worked grains by recrystallized ones (4 s at 580⬚C). (d) Complete recrystallization (8 s at 580⬚C). (e) Grain growth after 15 min at 580⬚C. ( f ) Grain growth after 10 min at 700⬚C (1290⬚F). All photomicrographs 75⫻. (Photomicrographs courtesy of J. E. Burke, General Electric Company.)

8.13 Recrystallization

(e)

●

215

(f)

FIGURE 8.21 (continued)

The influence of temperature is demonstrated in Figure 8.22, which plots tensile strength and ductility (at room temperature) of a brass alloy as a function of the temperature and for a constant heat treatment time of 1 h. The grain structures found at the various stages of the process are also presented schematically. The recrystallization behavior of a particular metal alloy is sometimes specified in terms of a recrystallization temperature, the temperature at which recrystallization just reaches completion in 1 h. Thus, the recrystallization temperature for the brass alloy of Figure 8.22 is about 450⬚C (850⬚F). Typically, it is between one third and one half of the absolute melting temperature of a metal or alloy and depends on several factors, including the amount of prior cold work and the purity of the alloy. Increasing the percentage of cold work enhances the rate of recrystallization, with the result that the recrystallization temperature is lowered, and approaches a constant or limiting value at high deformations; this effect is shown in Figure 8.23. Furthermore, it is this limiting or minimum recrystallization temperature that is normally specified in the literature. There exists some critical degree of cold work below which recrystallization cannot be made to occur, as shown in the figure; normally, this is between 2 and 20% cold work. Recrystallization proceeds more rapidly in pure metals than in alloys. Thus, alloying raises the recrystallization temperature, sometimes quite substantially. For pure metals, the recrystallization temperature is normally 0.3Tm , where Tm is the absolute melting temperature; for some commercial alloys it may run as high as 0.7Tm . Recrystallization and melting temperatures for a number of metals and alloys are listed in Table 8.2. Plastic deformation operations are often carried out at temperatures above the recrystallization temperature in a process termed hot working, 兵described in Section 14.2.其 The material remains relatively soft and ductile during deformation because it does not strain harden, and thus large deformations are possible.

Chapter 8 / Deformation and Strengthening Mechanisms Annealing temperature (°F) 400

600

600

800

1000

1200

60

Tensile strength 50 500 40

400

Ductility (%EL)

FIGURE 8.22 The influence of annealing temperature on the tensile strength and ductility of a brass alloy. Grain size as a function of annealing temperature is indicated. Grain structures during recovery, recrystallization, and grain growth stages are shown schematically. (Adapted from G. Sachs and K. R. Van Horn, Practical Metallurgy, Applied Metallurgy and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)

Tensile strength (MPa)

30 Ductility 20

300

Recovery

Recrystallization

Grain growth

Cold-worked and recovered grains New grains

0.040 0.030 0.020 0.010 100

200

300

400

500

600

700

Annealing temperature (°C)

900 1600

Recrystallization temperature (⬚C)

800 1400 700 1200 600 1000 500 800

400

300 0

10

20

Critical deformation

30 40 Percent cold work

50

60

600 70

Recrystallization temperature (⬚F)

●

Grain size (mm)

216

FIGURE 8.23 The variation of recrystallization temperature with percent cold work for iron. For deformations less than the critical (about 5%CW), recrystallization will not occur.

8.13 Recrystallization

●

217

Table 8.2 Recrystallization and Melting Temperatures for Various Metals and Alloys Recrystallization Temperature ⴗC ⴗF ⫺4 25 ⫺4 25 10 50 80 176 120 250 475 887 370 700 450 840 1200 2200

Metal Lead Tin Zinc Aluminum (99.999 wt%) Copper (99.999 wt%) Brass (60 Cu–40 Zn) Nickel (99.99 wt%) Iron Tungsten

Melting Temperature ⴗC ⴗF 327 620 232 450 420 788 660 1220 1085 1985 900 1652 1455 2651 1538 2800 3410 6170

DESIGN EXAMPLE 8.1 A cylindrical rod of noncold-worked brass having an initial diameter of 6.4 mm (0.25 in.) is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 MPa (50,000 psi) and a ductility in excess of 20%EL; in addition, a final diameter of 5.1 mm (0.20 in.) is necessary. Describe the manner in which this procedure may be carried out.

S OLUTION Let us first consider the consequences (in terms of yield strength and ductility) of cold working in which the brass specimen diameter is reduced from 6.4 mm (designated by d0) to 5.1 mm (di). The %CW may be computed from Equation 8.6 as

%CW ⫽

⫽

冉冊 冉冊 冉冊 冉 冊 冉 冊 冉 冊 d0 2

2

di 2

앟⫺ d0 2

2

6.4 mm 2

2

2

앟

⫻ 100

앟

앟⫺

6.4 mm 2

5.1 mm 2 2

2

앟 ⫻ 100 ⫽ 36.5%CW

앟

From Figures 8.19a and 8.19c, a yield strength of 410 MPa (60,000 psi) and a ductility of 8%EL are attained from this deformation. According to the stipulated criteria, the yield strength is satisfactory; however, the ductility is too low. Another processing alternative is a partial diameter reduction, followed by a recrystallization heat treatment in which the effects of the cold work are nullified. The required yield strength, ductility, and diameter are achieved through a second drawing step. Again, reference to Figure 8.19a indicates that 20%CW is required to give a yield strength of 345 MPa. On the other hand, from Figure 8.19c, ductilities greater

218

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Chapter 8 / Deformation and Strengthening Mechanisms

than 20%EL are possible only for deformations of 23%CW or less. Thus during the final drawing operation, deformation must be between 20%CW and 23%CW. Let’s take the average of these extremes, 21.5%CW, and then calculate the final diameter for the first drawing d⬘0 , which becomes the original diameter for the second drawing. Again, using Equation 8.6,

21.5%CW ⫽

冉冊 冉 冊 冉冊 d ⬘0 2

2

앟⫺

d ⬘0 2

5.1 mm 2 2

2

앟 ⫻ 100

앟

Now, solving from d ⬘0 from the expression above gives d ⬘0 ⫽ 5.8 mm (0.226 in.)

8.14 GRAIN GROWTH After recrystallization is complete, the strain-free grains will continue to grow if the metal specimen is left at the elevated temperature (Figures 8.21d–f ); this phenomenon is called grain growth. Grain growth does not need to be preceded by recovery and recrystallization; it may occur in all polycrystalline materials, metals and ceramics alike. An energy is associated with grain boundaries, as explained in Section 5.8. As grains increase in size, the total boundary area decreases, yielding an attendant reduction in the total energy; this is the driving force for grain growth. Grain growth occurs by the migration of grain boundaries. Obviously, not all grains can enlarge, but large ones grow at the expense of small ones that shrink. Thus, the average grain size increases with time, and at any particular instant there will exist a range of grain sizes. Boundary motion is just the short-range diffusion of atoms from one side of the boundary to the other. The directions of boundary movement and atomic motion are opposite to each other, as shown in Figure 8.24.

Atomic diffusion across boundary

Direction of grain boundary motion

FIGURE 8.24 Schematic representation of grain growth via atomic diffusion. (From Elements of Materials Science and Engineering by Van Vlack, 1989. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ.)

8.14 Grain Growth

●

219

800⬚C 850⬚C

Grain diameter (mm) (Logarithmic scale)

1.0

700⬚C

600⬚C 0.1 500⬚C

0.01 1

10

102 Time (min) (Logarithmic scale)

103

104

FIGURE 8.25 The logarithm of grain diameter versus the logarithm of time for grain growth in brass at several temperatures. (From J. E. Burke, ‘‘Some Factors Affecting the Rate of Grain Growth in Metals.’’ Reprinted with permission from Metallurgical Transactions, Vol. 180, 1949, a publication of The Metallurgical Society of AIME, Warrendale, Pennsylvania.)

For many polycrystalline materials, the grain diameter d varies with time t according to the relationship d n ⫺ d 0n ⫽ Kt

(8.7)

where d0 is the initial grain diameter at t ⫽ 0, and K and n are time-independent constants; the value of n is generally equal to or greater than 2. The dependence of grain size on time and temperature is demonstrated in Figure 8.25, a plot of the logarithm of grain size as a function of the logarithm of time for a brass alloy at several temperatures. At lower temperatures the curves are linear. Furthermore, grain growth proceeds more rapidly as temperature increases; that is, the curves are displaced upward to larger grain sizes. This is explained by the enhancement of diffusion rate with rising temperature. The mechanical properties at room temperature of a fine-grained metal are usually superior (i.e., higher strength and toughness) to those of coarse-grained ones. If the grain structure of a single-phase alloy is coarser than that desired, refinement may be accomplished by plastically deforming the material, then subjecting it to a recrystallization heat treatment, as described above.

DEFORMATION MECHANISMS FOR CERAMIC MATERIALS Although at room temperature most ceramic materials suffer fracture before the onset of plastic deformation, a brief exploration into the possible mechanisms is worthwhile. Plastic deformation is different for crystalline and noncrystalline ceramics; however, each is discussed.

220

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Chapter 8 / Deformation and Strengthening Mechanisms

8.15 CRYSTALLINE CERAMICS For crystalline ceramics, plastic deformation occurs, as with metals, by the motion of dislocations. One reason for the hardness and brittleness of these materials is the difficulty of slip (or dislocation motion). For crystalline ceramic materials for which the bonding is predominantly ionic, there are very few slip systems (crystallographic planes and directions within those planes) along which dislocations may move. This is a consequence of the electrically charged nature of the ions. For slip in some directions, ions of like charge are brought into close proximity to one another; because of electrostatic repulsion, this mode of slip is very restricted. This is not a problem in metals, since all atoms are electrically neutral. On the other hand, for ceramics in which the bonding is highly covalent, slip is also difficult and they are brittle for the following reasons: (1) the covalent bonds are relatively strong; (2) there are also limited numbers of slip systems; and (3) dislocation structures are complex.

8.16 NONCRYSTALLINE CERAMICS Plastic deformation does not occur by dislocation motion for noncrystalline ceramics because there is no regular atomic structure. Rather, these materials deform by viscous flow, the same manner in which liquids deform; the rate of deformation is proportional to the applied stress. In response to an applied shear stress, atoms or ions slide past one another by the breaking and reforming of interatomic bonds. However, there is no prescribed manner or direction in which this occurs, as with dislocations. Viscous flow on a macroscopic scale is demonstrated in Figure 8.26. The characteristic property for viscous flow, viscosity, is a measure of a noncrystalline material’s resistance to deformation. For viscous flow in a liquid that originates from shear stresses imposed by two flat and parallel plates, the viscosity is the ratio of the applied shear stress and the change in velocity dv with distance dy in a direction perpendicular to and away from the plates, or

⫽

F/A ⫽ dv/dy dv/dy

(8.8)

This scheme is represented in Figure 8.26. The units for viscosity are poises (P) and pascal-seconds (Pa-s); 1 P ⫽ 1 dynes/cm2, and 1 Pa-s ⫽ 1 N-s/m2. Conversion from one system of units to the other is

A

F

v y

FIGURE 8.26 Representation of the viscous flow of a liquid or fluid glass in response to an applied shear force.

8.17 Deformation of Semicrystalline Polymers

●

221

according to 10 P ⫽ 1 Pa-s Liquids have relatively low viscosities; for example, the viscosity of water at room temperature is about 10⫺3 Pa-s. On the other hand, glasses have extremely large viscosities at ambient temperatures, which is accounted for by strong interatomic bonding. As the temperature is raised, the magnitude of the bonding is diminished, the sliding motion or flow of the atoms or ions is facilitated, and subsequently there is an attendant decrease in viscosity. 兵A discussion of the temperature dependence of viscosity for glasses is deferred to Section 14.7.其

MECHANISMS OF DEFORMATION AND FOR STRENGTHENING OF POLYMERS An understanding of deformation mechanisms of polymers is important in order for us to be able to manage the mechanical characteristics of these materials. In this regard, deformation models for two different types of polymers— semicrystalline and elastomeric—deserve our attention. The stiffness and strength of semicrystalline materials are often important considerations; elastic and plastic deformation mechanisms are treated in the succeeding section, whereas methods used to stiffen and strengthen these materials are discussed in Section 8.18. On the other hand, elastomers are utilized on the basis of their unusual elastic properties; the deformation mechanism of elastomers is also treated.

8.17 DEFORMATION

OF

SEMICRYSTALLINE POLYMERS

Many semicrystalline polymers in bulk form will have the spherulitic structure described in Section 4.12. By way of review, let us repeat here that each spherulite consists of numerous chain-folded ribbons, or lamellae, that radiate outward from the center. Separating these lamellae are areas of amorphous material (Figure 4.14); adjacent lamellae are connected by tie chains that pass through these amorphous regions.

MECHANISM OF ELASTIC DEFORMATION The mechanism of elastic deformation in semicrystalline polymers in response to tensile stresses is the elongation of the chain molecules from their stable conformations, in the direction of the applied stress, by the bending and stretching of the strong chain covalent bonds. In addition, there may be some slight displacement of adjacent molecules, which is resisted by relatively weak secondary or van der Waals bonds. Furthermore, inasmuch as semicrystalline polymers are composed of both crystalline and amorphous regions, they may, in a sense, be considered composite materials. As such, the elastic modulus may be taken as some combination of the moduli of crystalline and amorphous phases.

MECHANISM OF PLASTIC DEFORMATION The mechanism of plastic deformation is best described by the interactions between lamellar and intervening amorphous regions in response to an applied tensile load. This process occurs in several stages, which are schematically diagrammed in Figure 8.27. Two adjacent chain-folded lamellae and the interlamel-

222 ●

Chapter 8 / Deformation and Strengthening Mechanisms

(a)

(b)

(c)

(d)

FIGURE 8.27 Stages in the deformation of a semicrystalline polymer. (a) Two adjacent chain-folded lamellae and interlamellar amorphous material before deformation. (b) Elongation of amorphous tie chains during the first stage of deformation. (c) Tilting of lamellar chain folds during the second stage. (d ) Separation of crystalline block segments during the third stage. (e) Orientation of block segments and tie chains with the tensile axis in the final deformation stage. (From Jerold M. Schultz, Polymer Materials Science, copyright 1974, pp. 500–501. Reprinted by permoission of Prentice-Hall, Inc., Englewood Cliffs, NJ.)

(e)

8.18b Factors That Influence the Mechanical Properties of Semicrystalline Polymers

●

223

lar amorphous material, prior to deformation, are shown in Figure 8.27a. During the initial stage of deformation (Figure 8.27b) the chains in the amorphous regions slip past each other and align in the loading direction. This causes the lamellar ribbons simply to slide past one another as the tie chains within the amorphous regions become extended. Continued deformation in the second stage occurs by the tilting of the lamellae so that the chain folds become aligned with the tensile axis (Figure 8.27c). Next, crystalline block segments separate from the lamellae, which segments remain attached to one another by tie chains (Figure 8.27d ). In the final stage (Figure 8.27e), the blocks and tie chains become oriented in the direction of the tensile axis. Thus appreciable tensile deformation of semicrystalline polymers produces a highly oriented structure. During deformation the spherulites experience shape changes for moderate levels of elongation. However, for large deformations, the spherulitic structure is virtually destroyed. Also, it is interesting to note that, to a large degree, the processes represented in Figure 8.27 are reversible. That is, if deformation is terminated at some arbitrary stage, and the specimen is heated to an elevated temperature near its melting point (i.e., annealed), the material will revert back to having the spherulitic structure that was characteristic of its undeformed state. Furthermore, the specimen will tend to shrink back to the shape it had prior to deformation; the extent of this shape and structural recovery will depend on the annealing temperature and also the degree of elongation.

8.18a FACTORS THAT INFLUENCE THE MECHANICAL PROPERTIES OF SEMICRYSTALLINE POLYMERS [DETAILED VERSION (CD-ROM)]

8.18b FACTORS THAT INFLUENCE THE MECHANICAL PROPERTIES OF SEMICRYSTALLINE POLYMERS (CONCISE VERSION) A number of factors influence the mechanical characteristics of polymeric materials. For example, we have already discussed the effect of temperature and strain rate on stress–strain behavior (Section 7.13, Figure 7.24). Again, increasing the temperature or diminishing the strain rate leads to a decrease in the tensile modulus, a reduction in tensile strength, and an enhancement of ductility. In addition, several structural/processing factors have decided influences on the mechanical behavior (i.e., strength and modulus) of polymeric materials. An increase in strength results whenever any restraint is imposed on the process illustrated in Figure 8.27; for example, a significant degree of intermolecular bonding or extensive chain entanglements inhibit relative chain motions. It should be noted that even though secondary intermolecular (e.g., van der Waals) bonds are much weaker than the primary covalent ones, significant intermolecular forces result from the formation of large numbers of van der Waals interchain bonds. Furthermore, the modulus rises as both the secondary bond strength and chain alignment increase. The mechanical behavior of polymers is affected by several structural/processing

224

●

Chapter 8 / Deformation and Strengthening Mechanisms 100 Hard plastics

Percent crystallinity

Brittle waxes 75 Tough waxes 50

Grease, liquids

25

0

Soft plastics

Soft waxes

0

500

2,000 5,000

20,000 Molecular weight (Nonlinear scale)

40,000

FIGURE 8.28 The influence of degree of crystallinity and molecular weight on the physical characteristics of polyethylene. (From R. B. Richards, ‘‘Polyethylene— Structure, Crystallinity and Properties,’’ J. Appl. Chem., 1, 370, 1951.)

factors that include molecular weight, degree of crystallinity, and predeformation (drawing). The magnitude of the tensile modulus does not seem to be influenced by molecular weight alterations. On the other hand, for many polymers, it has been observed that tensile strength increases with increasing molecular weight. For a specific polymer, the degree of crystallinity can have a rather significant influence on the mechanical properties, since it affects the extent of intermolecular secondary bonding. It has been observed that, for semicrystalline polymers, tensile modulus increases significantly with degree of crystallinity; in most cases, strength is also enhanced, and the material becomes more brittle. The influences of chain chemistry and structure (branching, stereoisomerism, etc.) on degree of crystallinity were discussed in Chapter 4. The effects of both percent crystallinity and molecular weight on the physical state of polyethylene are represented in Figure 8.28. On a commercial basis, one important technique that is used to improve mechanical strength and tensile modulus is permanently deforming the polymer in tension. This procedure, sometimes called drawing, produces the neck extension illustrated schematically in Figure 7.25. It is an important stiffening and strengthening technique that is employed in the production of fibers and films 兵(Section 14.15).其 During drawing, the molecular chains slip past one another and become highly oriented; for semicrystalline materials the chains assume conformations similar to those represented schematically in Figure 8.27e. It should also be noted that the mechanical properties of drawn polymers are normally anisotropic.

8.19 DEFORMATION

OF

ELASTOMERS

One of the fascinating properties of the elastomeric materials is their rubberlike elasticity. That is, they have the ability to be deformed to quite large deformations, and then elastically spring back to their original form. This behavior was probably first observed in natural rubber; however, the past few years have brought about the synthesis of a large number of elastomers with a wide variety of properties. Typical stress–strain characteristics of elastomeric materials are displayed in Figure 7.22, curve C. Their moduli of elasticity are quite small and, furthermore, vary with strain since the stress–strain curve is nonlinear.

8.19 Deformation of Elastomers

●

225

In an unstressed state, an elastomer will be amorphous and composed of molecular chains that are highly twisted, kinked, and coiled. Elastic deformation, upon application of a tensile load, is simply the partial uncoiling, untwisting, and straightening, and the resultant elongation of the chains in the stress direction, a phenomenon represented in Figure 8.29. Upon release of the stress, the chains spring back to their prestressed conformations, and the macroscopic piece returns to its original shape. The driving force for elastic deformation is a thermodynamic parameter called entropy, which is a measure of the degree of disorder within a system; entropy increases with increasing disorder. As an elastomer is stretched and the chains straighten and become more aligned, the system becomes more ordered. From this state, the entropy increases if the chains return to their original kinked and coiled contours. Two intriguing phenomena result from this entropic effect. First, when stretched, an elastomer experiences a rise in temperature; second, the modulus of elasticity increases with increasing temperature, which is opposite to the behavior found in other materials (see Figure 7.8). Several criteria must be met in order for a polymer to be elastomeric: (1) It must not easily crystallize; elastomeric materials are amorphous, having molecular chains that are naturally coiled and kinked in the unstressed state. (2) Chain bond rotations must be relatively free in order for the coiled chains to readily respond to an applied force. (3) For elastomers to experience relatively large elastic deformations, the onset of plastic deformation must be delayed. Restricting the motions of chains past one another by crosslinking accomplishes this objective. The crosslinks act as anchor points between the chains and prevent chain slippage from occurring; the role of crosslinks in the deformation process is illustrated in Figure 8.29. Crosslinking in many elastomers is carried out in a process called vulcanization, as discussed below. (4) Finally, the elastomer must be above its glass transition temperature (Section 11.16). The lowest temperature at which rubberlike behavior persists for many of the common elastomers is between ⫺50 and ⫺90⬚C (⫺60 and ⫺130⬚F). Below its glass transition temperature, an elastomer becomes brittle such that its stress–strain behavior resembles curve A in Figure 7.22.

Crosslinks

(b)

(a)

FIGURE 8.29 Schematic representation of crosslinked polymer chain molecules (a) in an unstressed state and (b) during elastic deformation in response to an applied tensile stress. (Adapted from Z. D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd edition. Copyright 1987 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

Chapter 8 / Deformation and Strengthening Mechanisms

VULCANIZATION The crosslinking process in elastomers is called vulcanization, which is achieved by a nonreversible chemical reaction, ordinarily carried out at an elevated temperature. In most vulcanizing reactions, sulfur compounds are added to the heated elastomer; chains of sulfur atoms bond with adjacent polymer backbone chains and crosslink them, which is accomplished according to the following reaction: H CH3 H H 兩 兩 兩 兩 UCUCuCUCU 兩 兩 H H

H CH3 H H 兩 兩 兩 兩 UCUCUCUCU 兩 兩 兩 兩 H 兩 兩 H 兩 兩 ⫹ (m ⫹ n) S 씮 (S)m (S)n 兩 兩 H H H 兩 兩 H 兩 兩 兩 兩 兩 兩 UCUCuCUCU UCUCUC CUCU 兩 兩 兩 兩 兩 兩 兩 兩 H CH3 H H H CH3 H H

(8.10)

in which the two crosslinks shown consist of m and n sulfur atoms. Crosslink main chain sites are carbon atoms that were doubly bonded before vulcanization, but, after vulcanization, have become singly bonded. Unvulcanized rubber is soft and tacky, and has poor resistance to abrasion. Modulus of elasticity, tensile strength, and resistance to degradation by oxidation are all enhanced by vulcanization. The magnitude of the modulus of elasticity is directly proportional to the density of the crosslinks. Stress–strain curves for vulcanized and unvulcanized natural rubber are presented in Figure 8.30. To produce a rubber that is capable of large extensions without rupture of the primary chain bonds, there must be relatively few crosslinks, and these must be widely separated. Useful rubbers result when about 1 to 5 parts (by weight) of sulfur is added to 100 parts of rubber. Increasing the sulfur content further hardens the rubber and also reduces its extensibility. Also, since they are crosslinked, elastomeric materials are thermosetting in nature.

10 60 8 50 6

40

30

Vulcanized

4

20 2 10

0

Unvulcanized

0

1

2

3 Strain

4

5

6

0

Stress (103 psi)

●

Stress (MPa)

226

FIGURE 8.30 Stress–strain curves to 600% elongation for unvulcanized and vulcanized natural rubber.

Summary

●

227

SUMMARY On a microscopic level, plastic deformation of metals corresponds to the motion of dislocations in response to an externally applied shear stress, a process termed ‘‘slip.’’ Slip occurs on specific crystallographic planes and within these planes only in certain directions. A slip system represents a slip plane–slip direction combination, and operable slip systems depend on the crystal structure of the material. 兵The critical resolved shear stress is the minimum shear stress required to initiate dislocation motion; the yield strength of a metal single crystal depends on both the magnitude of the critical resolved shear stress and the orientation of slip components relative to the direction of the applied stress.其 For polycrystalline metals, slip occurs within each grain along the slip systems that are most favorably oriented with the applied stress; furthermore, during deformation, grains change shape in such a manner that coherency at the grain boundaries is maintained. 兵Under some circumstances limited plastic deformation may occur in BCC and HCP metals by mechanical twinning. Normally, twinning is important to the degree that accompanying crystallographic reorientations make the slip process more favorable.其 Since the ease with which a metal is capable of plastic deformation is a function of dislocation mobility, restricting dislocation motion increases hardness and strength. On the basis of this principle, three different strengthening mechanisms were discussed. Grain boundaries serve as barriers to dislocation motion; thus refining the grain size of a polycrystalline metal renders it harder and stronger. Solid solution strengthening results from lattice strain interactions between impurity atoms and dislocations. And, finally, as a metal is plastically deformed, the dislocation density increases, as does also the extent of repulsive dislocation–dislocation strain field interactions; strain hardening is just the enhancement of strength with increased plastic deformation. The microstructural and mechanical characteristics of a plastically deformed metal specimen may be restored to their predeformed states by an appropriate heat treatment, during which recovery, recrystallization, and grain growth processes are allowed to occur. During recovery there is a reduction in dislocation density and alterations in dislocation configurations. Recrystallization is the formation of a new set of grains that are strain free; in addition, the material becomes softer and more ductile. Grain growth is the increase in average grain size of polycrystalline materials, which proceeds by grain boundary motion. Any plastic deformation of crystalline ceramics is a result of dislocation motion; the brittleness of these materials is, in part, explained by the limited number of operable slip systems. The mode of plastic deformation for noncrystalline materials is by viscous flow; a material’s resistance to deformation is expressed as viscosity. At room temperature, the viscosity of many noncrystalline ceramics is extremely high. During the elastic deformation of a semicrystalline polymer that is stressed in tension, the constituent molecules elongate in the stress direction by the bending and stretching of covalent chain bonds. Slight molecular displacements are resisted by weak secondary bonds. The mechanism of plastic deformation for semicrystalline polymers having the spherulitic structure was presented. Tensile deformation is thought to occur in several stages as both amorphous tie chains and chain-folded block segments (which separate from the ribbonlike lamellae) become oriented with the tensile axis. Also, during deformation the shapes of spherulites are altered (for moderate deforma-

228

●

Chapter 8 / Deformation and Strengthening Mechanisms

tions); relatively large degrees of deformation lead to a complete destruction of the spherulites. Furthermore, the predeformed spherulitic structure and macroscopic shape may be virtually restored by annealing at an elevated temperature below the polymer’s melting temperature. The mechanical behavior of a polymer will be influenced by both inservice and structural/processing factors. With regard to the former, increasing the temperature and/or diminishing the strain rate leads to reductions in tensile modulus and tensile strength, and an enhancement of ductility. In addition, other factors that affect the mechanical properties include molecular weight, degree of crystallinity, predeformation drawing, and heat treating. The influence of each of these factors was discussed. Large elastic extensions are possible for the elastomeric materials that are amorphous and lightly crosslinked. Deformation corresponds to the unkinking and uncoiling of chains in response to an applied tensile stress. Crosslinking is often achieved during a vulcanization process.

IMPORTANT TERMS AND CONCEPTS Cold working Critical resolved shear stress Dislocation density Grain growth Lattice strain

Recovery Recrystallization Recrystallization temperature Resolved shear stress Slip

Slip system Solid-solution strengthening Strain hardening Viscosity Vulcanization

REFERENCES Hirth, J. P. and J. Lothe, Theory of Dislocations, 2nd edition, Wiley-Interscience, New York, 1982. Reprinted by Krieger Publishing Company, Melbourne, FL, 1992. Hull, D., Introduction to Dislocations, 3rd edition, Pergamon Press, Inc., Elmsford, NY, 1984. Kingery, W. D., H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics, 2nd edition, John Wiley & Sons, New York, 1976. Chapter 14. Read, W. T., Jr., Dislocations in Crystals, McGrawHill Book Company, New York, 1953.

Richerson, D. W., Modern Ceramic Engineering, 2nd edition, Marcel Dekker, New York, 1992. Chapter 5. Schultz, J., Polymer Materials Science, PrenticeHall, Englewood Cliffs, NJ, 1974. Weertman, J. and J. R. Weertman, Elementary Dislocation Theory, The Macmillan Co., New York, 1964. Reprinted by Oxford University Press, Oxford, 1992.

QUESTIONS AND PROBLEMS Note: To solve those problems having an asterisk (*) by their numbers, consultation of supplementary topics [appearing only on the CD-ROM (and not in print)] will probably be necessary. 8.1 To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 104 mm⫺2. Suppose that all the dislocations in 1000 mm3 (1 cm3) were somehow removed and linked end to end. How far (in miles)

would this chain extend? Now suppose that the density is increased to 1010 mm⫺2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material? 8.2 Consider two edge dislocations of opposite sign and having slip planes that are separated

Questions and Problems

by several atomic distances as indicated in the diagram. Briefly describe the defect that results when these two dislocations become aligned with each other. —씮

씯— 8.3 Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer. 8.4 For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion. 8.5 (a) Define a slip system. (b) Do all metals have the same slip system? Why or why not? 8.6* (a) Compare planar densities (Section 3.14) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities for the (100), (110), and (111) planes for BCC. 8.7 One slip system for the BCC crystal structure is 兵110其具111典. In a manner similar to Figure 8.6b sketch a 兵110其-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different 具111典 slip directions within this plane. 8.8 One slip system for the HCP crystal structure is 兵0001其具1120典. In a manner similar to Figure 8.6b, sketch a 兵0001其-type plane for the HCP structure, and using arrows, indicate three different 具1120典 slip directions within this plane. You might find Figure 3.22 helpful. 8.9* Explain the difference between resolved shear stress and critical resolved shear stress. 8.10* Sometimes cos cos in Equation 8.1 is termed the Schmid factor. Determine the magnitude of the Schmid factor for an FCC single crystal oriented with its [100] direction parallel to the loading axis. 8.11* Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1⬚ and 47.9⬚, respectively, with the tensile axis. If the criti-

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cal resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa (6500 psi) cause the single crystal to yield? If not, what stress will be necessary? 8.12* A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1⬚ with the tensile axis. Three possible slip directions make angles of 62.4⬚, 72.0⬚, and 81.1⬚ with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi), determine the critical resolved shear stress for aluminum. 8.13* Consider a single crystal of silver oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101] direction, and is initiated at an applied tensile stress of 1.1 MPa (160 psi), compute the critical resolved shear stress. 8.14* The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension. 8.15* List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result. 8.16 Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries. 8.17 Briefly explain why HCP metals are typically more brittle than FCC and BCC metals. 8.18 Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid solution strengthening, and strain hardening). Be sure to explain how dislocations are involved in each of the strengthening techniques. 8.19 (a) From the plot of yield strength versus (grain diameter)⫺1/2 for a 70 Cu–30 Zn cartridge brass, Figure 8.15, determine values for the constants 0 and ky in Equation 8.5. (b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 ⫻ 10⫺3 mm.

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8.20 The lower yield point for an iron that has an average grain diameter of 5 ⫻ 10⫺2 mm is 135 MPa (19,500 psi). At a grain diameter of 8 ⫻ 10⫺3 mm, the yield point increases to 260 MPa (37,500 psi). At what grain diameter will the lower yield point be 205 MPa (30,000 psi)? 8.21 If it is assumed that the plot in Figure 8.15 is for noncold-worked brass, determine the grain size of the alloy in Figure 8.19; assume its composition is the same as the alloy in Figure 8.15. 8.22 In the manner of Figures 8.17b and 8.18b indicate the location in the vicinity of an edge dislocation at which an interstitial impurity atom would be expected to be situated. Now briefly explain in terms of lattice strains why it would be situated at this position. 8.23 When making hardness measurements, what will be the effect of making an indentation very close to a preexisting indentation? Why? 8.24 (a) Show, for a tensile test, that %CW ⫽

冉 冊

⑀ ⫻ 100 ⑀⫹1

if there is no change in specimen volume during the deformation process (i.e., A0 l0 ⫽ Ad ld). (b) Using the result of part a, compute the percent cold work experienced by naval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa (58,000 psi) is applied. 8.25 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and 11 mm, respectively. The second specimen, with an initial radius of 12 mm, must have the same deformed hardness as the first specimen; compute the second specimen’s radius after deformation. 8.26 Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is

rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows:

Original dimensions Deformed dimensions

Circular (diameter, mm) 15.2 11.4

Rectangular (mm) 125 ⫻ 175 75 ⫻ 200

Which of these specimens will be the hardest after plastic deformation, and why? 8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. If its coldworked radius is 10 mm (0.40 in.), what was its radius before deformation? 8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strength of 275 MPa (40,000 psi)? (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 690 MPa (100,000 psi)? 8.29 Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress crss is a function of the dislocation density D as

crss ⫽ 0 ⫹ A 兹D where 0 and A are constants. For copper, the critical resolved shear stress is 2.10 MPa (305 psi) at a dislocation density of 105 mm⫺2. If it is known that the value of A for copper is 6.35 ⫻ 10⫺3 MPa-mm (0.92 psi-mm), compute the crss at a dislocation density of 107 mm⫺2. 8.30 Briefly cite the differences between recovery and recrystallization processes. 8.31 Estimate the fraction of recrystallization from the photomicrograph in Figure 8.21c. 8.32 Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized. 8.33 Briefly explain why some metals (e.g., lead and tin) do not strain harden when deformed at room temperature. 8.34 (a) What is the driving force for recrystallization? (b) For grain growth?

Questions and Problems

8.35 (a) From Figure 8.25, compute the length of time required for the average grain diameter to increase from 0.01 to 0.1 mm at 500⬚C for this brass material. (b) Repeat the calculation at 600⬚C. 8.36 The average grain diameter for a brass material was measured as a function of time at 650⬚C, which is tabulated below at two different times: Time (min) 30 90

Grain Diameter (mm) 3.9 ⫻ 10⫺2 6.6 ⫻ 10⫺2

(a) What was the original grain diameter? (b) What grain diameter would you predict after 150 min at 650⬚C? 8.37 An undeformed specimen of some alloy has an average grain diameter of 0.040 mm. You are asked to reduce its average grain diameter to 0.010 mm. Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why. 8.38 Grain growth is strongly dependent on temperature (i.e., rate of grain growth increases with increasing temperature), yet temperature is not explicitly given as a part of Equation 8.7. (a) Into which of the parameters in this expression would you expect temperature to be included? (b) On the basis of your intuition, cite an explicit expression for this temperature dependence. 8.39 An uncold-worked brass specimen of average grain size 0.008 mm has a yield strength of 160 MPa (23,500 psi). Estimate the yield strength of this alloy after it has been heated to 600⬚C for 1000 s, if it is known that the value of ky is 12.0 MPa-mm1/2 (1740 psimm1/2). 8.40 Cite one reason why ceramic materials are, in general, harder yet more brittle than metals. 8.41 In your own words, describe the mechanisms by which semicrystalline polymers (a) elas-

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tically deform and (b) plastically deform, and (c) by which elastomers elastically deform. 8.42 Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why: (a) molecular weight; (b) degree of crystallinity; (c) deformation by drawing; (d) annealing of an undeformed material; (e) annealing of a drawn material. 8.43* Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline polymer and why: (a) molecular weight; (b) degree of crystallinity; (c) deformation by drawing; (d) annealing of an undeformed material. 8.44 Normal butane and isobutane have boiling temperatures of ⫺0.5 and ⫺12.3⬚C (31.1 and 9.9⬚F), respectively. Briefly explain this behavior on the basis of their molecular structures, as presented in Section 4.2. 8.45* The tensile strength and number-average molecular weight for two polymethyl methacrylate materials are as follows: Tensile Strength (MPa)

Number Average Molecular Weight (g/mol)

107 170

40,000 60,000

Estimate the tensile strength at a numberaverage molecular weight of 30,000 g/mol. 8.46* The tensile strength and number-average molecular weight for two polyethylene materials are as follows: Tensile Strength (MPa)

Number Average Molecular Weight (g/mol)

85 150

12,700 28,500

Estimate the number-average molecular weight that is required to give a tensile strength of 195 MPa. 8.47* For each of the following pairs of polymers, do the following: (1) state whether or not it

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is possible to decide if one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Syndiotactic polystyrene having a number-average molecular weight of 400,000 g/ mol; isotactic polystyrene having a numberaverage molecular weight of 650,000 g/mol. (b) Branched and atactic polyvinyl chloride with a weight-average molecular weight of 100,000 g/mol; linear and isotactic polyvinyl chloride having a weight-average molecular weight of 75,000 g/mol. (c) Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked. (d) Branched polyethylene with a numberaverage molecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150,000 g/mol. 8.48* For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to decide if one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strengh and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Syndiotactic polystyrene having a number-average molecular weight of 600,000 g/ mol; isotactic polystyrene having a numberaverage molecular weight of 500,000 g/mol. (b) Linear and isotactic polyvinyl chloride with a weight-average molecular weight of 100,000 g/mol; branched and atactic polyvinyl chloride having a weight-average molecular weight of 75,000 g/mol. (c) Graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked. (d) Network polyester; lightly branched polytetrafluoroethylene. 8.49 Would you expect the tensile strength of polychlorotrifluoroethylene to be greater than,

the same as, or less than that of a polytetrafluoroethylene specimen having the same molecular weight and degree of crystallinity? Why? 8.50* For each of the following pairs of polymers, plot and label schematic stress–strain curves on the same graph (i.e., make separate plots for parts a, b, c, and d). (a) Isotactic and linear polypropylene having a weight-average molecular weight of 120,000 g/mol; atactic and linear polypropylene having a weight-average molecular weight of 100,000 g/mol. (b) Branched polyvinyl chloride having a number-average degree of polymerization of 2000; heavily crosslinked polyvinyl chloride having a number-average degree of polymerization of 2000. (c) Poly (styrene - butadiene) random copolymer having a number-average molecular weight of 100,000 g/mol and 10% of the available sites crosslinked and tested at 20⬚C; poly(styrene-butadiene) random copolymer having a number-average molecular weight of 120,000 g/mol and 15% of the available sites crosslinked and tested at ⫺85⬚C. Hint: poly(styrene-butadiene) copolymers may exhibit elastomeric behavior. (d) Polyisoprene, molecular weight of 100,000 g/mol having 10% of available sites crosslinked; polyisoprene, molecular weight of 100,000 g/mol having 20% of available sites crosslinked. Hint: polyisoprene is a natural rubber that may display elastomeric behavior. 8.51 List the two molecular characteristics that are essential for elastomers. 8.52 Which of the following would you expect to be elastomers and which thermosetting polymers at room temperature? Justify each choice. (a) Epoxy having a network structure. (b) Lightly crosslinked poly(styrene-butadiene) random copolymer that has a glass-transition temperature of ⫺50⬚C. (c) Lightly branched and semicrystalline polytetrafluoroethylene that has a glasstransition temperature of ⫺100⬚C.

Questions and Problems

(d) Heavily crosslinked poly(ethylene-propylene) random copolymer that has a glasstransition temperature of 0⬚C. (e) Thermoplastic elastomer that has a glass-transition temperature of 75⬚C. 8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite) will not be an elastomer. 8.54 Ten kilograms of polybutadiene is vulcanized with 4.8 kg sulfur. What fraction of the possible crosslink sites is bonded to sulfur crosslinks, assuming that, on the average, 4.5 sulfur atoms participate in each crosslink? 8.55 Compute the weight percent sulfur that must be added to completely crosslink an alternating chloroprene-acrylonitrile copolymer, assuming that five sulfur atoms participate in each crosslink. 8.56 The vulcanization of polyisoprene is accomplished with sulfur atoms according to Equation 8.10. If 57 wt% sulfur is combined with polyisoprene, how many crosslinks will be associated with each isoprene mer if it is assumed that, on the average, six sulfur atoms participate in each crosslink? 8.57 For the vulcanization of polyisoprene, compute the weight percent of sulfur that must be added to ensure that 8% of possible sites will be crosslinked; assume that, on the average, three sulfur atoms are associated with each crosslink. 8.58 Demonstrate, in a manner similar to Equation 8.10, how vulcanization may occur in a chloroprene rubber.

Design Problems 8.D1 Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time have a ductility of at least 12%EL. Justify your decision. 8.D2 Determine whether or not it is possible to cold work brass so as to give a minimum Brinell hardness of 120 and at the same time

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have a ductility of at least 20%EL. Justify your decision. 8.D3 A cylindrical specimen of cold-worked steel has a Brinell hardness of 250. (a) Estimate its ductility in percent elongation. (b) If the specimen remained cylindrical during deformation and its uncold-worked radius was 5 mm (0.20 in.), determine its radius after deformation. 8.D4 It is necessary to select a metal alloy for an application that requires a yield strength of at least 345 MPa (50,000 psi) while maintaining a minimum ductility (%EL) of 20%. If the metal may be cold worked, decide which of the following are candidates: copper, brass, and a 1040 steel. Why? 8.D5 A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A coldworked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. 8.D6 A cylindrical rod of copper originally 16.0 mm (0.625 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked yield strength in excess of 250 MPa (36,250 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 11.3 mm (0.445 in.). Explain how this may be accomplished. 8.D7 A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a final diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20%, is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW.

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