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section_5_beta0

Course Number: E 344, Fall 2009

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Chapter 5. Mechanical Properties After having studied this chapter you will be able to: Describe the importance of mechanical properties of materials in the utilization and in the fabrication of machine elements. Draw a stress-strain curve and identify elastic and <a href="/keyword/plastic-deformation/" >plastic deformation</a> , yield stress, ultimate tensile stress, ductility,...

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Chapter 5. Mechanical Properties After having studied this chapter you will be able to: Describe the importance of mechanical properties of materials in the utilization and in the fabrication of machine elements. Draw a stress-strain curve and identify elastic and <a href="/keyword/plastic-deformation/" >plastic deformation</a> , yield stress, ultimate tensile stress, ductility, work-hardening. Describe the effect of residual stresses on the strength of structural materials. Perform a hardness test; describe the different hardness tests and their significance Perform work hardening and annealing and describe what effect they have on the strength of the metal and explain in terms of the creation and disappearance of defects. Explain why the defects make a material strong. Name and explain the main means of strengthening of a metal: strain hardening (i.e. work hardening), solution strengthening, precipitation-strengthening, fine grain structure. Describe annealing how it changes mechanical properties and why. 5.1. Structural Materials Let us take a straight wire and bend it around a mandrel into a helical spring. We have performed a <a href="/keyword/plastic-deformation/" >plastic deformation</a> of the material in order to give it a new permanent shape. If we fix one end of the spring and now hang a weight on from the other end of the spring, the spring stretches. If the weight is not too large, this stretch is reversible, and the spring assumes its original shape when the weight is removed. This corresponds to an elastic deformation that is proportional to the weight we applied. If the weight is large and exceeds some threshold, the spring elongates permanently, and the spring does not return to its original shape when the weight is removed. The non-recoverable part of the deformation corresponds to <a href="/keyword/plastic-deformation/" >plastic deformation</a> . <a href="/keyword/plastic-deformation/" >plastic deformation</a> defeats the purpose of making a spring. Even before bending the wire into a helical spring, the straight wire displays these same elastic and <a href="/keyword/plastic-deformation/" >plastic deformation</a> behaviors. If a weight is hung from one end of the straight wire, it stretches elastically but can support the weight. If the latter is too heavy, the wire will not hold its shape. It will yield, become longer, and finally break. If we can perform these experiments with different metals, such as a hard or a soft steel wire or a copper wire, we will observe that the threshold weight which each can support before deforming plastically as well as the total amount of <a href="/keyword/plastic-deformation/" >plastic deformation</a> each can withstand before breaking is different for each metal. We know that we could not shape a glass or ceramic rod into a spring, it would break and we know that a polymer bar, would not keep the shape as a metal does. What we have just done with this simple wire and spring example is repeated in countless ways throughout industry. The use of strength of materials and of elastic deformation is obvious. Most metallic objects are 1 shaped through <a href="/keyword/plastic-deformation/" >plastic deformation</a> : metallic sheets are made by rolling a bloc of material, wires by drawing through a dye, and are bodies are manufactured by pressing a metallic sheets. In this chapter, we shall first describe how one measures the mechanical properties of a metal and learn how to read the specifications given by manufacturers. We shall then examine what happens in a metal during elastic and <a href="/keyword/plastic-deformation/" >plastic deformation</a> ; this will suggest several methods by which we can strengthen the metal for its use or soften it for easier shaping. 5.2. The Tensile Test The tensile strength is the most widely quoted measure of a material s ability to support load. In the tensile test, a bar or a plate of material is prepared in accordance with American Society for Testing and Materials (ASTM) standards. The specimen is more massive at the ends and has a reduced section known as the gauge length (See figure 5.1.) Figure 5.1. Tensile test bar according to ASTM Standards E 8 and E 8M Standard Test Methods for Tension Testing of Metallic Materials 2 Figure 5.2. Tensile Testing Machine It is then gripped in a tensile testing machine (Figure 5.2) and loaded in uniaxial tension until it fails. What is recorded is the instantaneous specimen load F and extension Dl as the crosshead moves at a constant rate. The length l and the cross section A are those of the thinner gauge. In modern machines, one enters the cross sectional area A and the gauge length l into the computer; the data obtained have the form of a graph reporting the strain Dl/l on the horizontal axis and the stress F/A on the vertical axis as in figure 5.6. 5.3.Stresses and Strains When we pull on a bar of material with a force F, it will be stretched by the amount Dl. This is shown in Figure 5.3. The force required for a given deformation is proportional to the cross section A of the bar; the elongation Dl of the bar is proportional to its original length l. In order to express the deformation in terms of the material properties, independent of size, we define the tensile stress s as the force per unit area 3 s = F/A and the strain as the relative elongation e = Dl/l (5.1) (5.2) These are normal stresses and strains because they are perpendicular to the area to which they are applied. The stresses can be applied in three directions, s x, sy and s z; similarly, there are three strain directions, ex, ey and ez. The international units for the stress are Pascal or N/m2 and the American units are pounds per square inch. Practical stresses are too large for these units to be convenient and one expresses the stress in Megapascals (1 MPa = 106 N/m2 = 1 N/mm2) or in thousands of pounds per square inch (ksi). The strain is dimensionless; it is often expressed in percent %. Instead of pulling on the bar, we can compress it. Compressive stresses have the same units. By convention, one uses positive stresses and strains for tensile deformation and negative values for compression. Figure 5.4 illustrates the application of a shear stress and the resulting shear deformation. A shear stress is caused by a couple of forces F1 and F2 acting in opposite directions. A single force couple (F1, F2) disrupts the equilibrium; therefore a second couple (F3, F4) must be applied to achieve a balance of moments. To obtain the resultant shear stresses, we divide the forces by the areas to which they are applied. The relevant area for the force F1 is shaded in the figure. The shear stress is tyz = F1/ A (5.3) The second couple of forces generates the shear stress tzy, which has the same magnitude tyz = tzy (5.4) 4 Figure 5.3. Normal stresses and strains applied to a solid. (From Ohring). Note that the strains ex, ey and ez are called ex, ey and ez in the text. The figure on the right shows that the surface to which the stresses are applied can be changed and that sz applied to the surface in the left figure is now decomposed into a normal stress (perpendicular to the new, inclined, surface) and a shear stress t that is parallel to the new surface. This new surface need not be a real cut but an imaginary surface. The decomposition of the stress sz into a normal stress sn and a shear stress t on the inclined surface at right is wrong in Ohring. (The two must add to reconstitute the original sz. It has been crudely corrected here. . 5 Figure 5.4. Shear stresses and shear strains. Note that the deformation Dl is not an elongation, but a twisting of the solid. The shear strain is defined as g = Dl/l = tan a (5.5) where Dl is now perpendicular the l; a is the angle by which the solid is deformed. The stresses are defined by the surfaces to which they apply. It is possible to consider other surfaces, oriented in different directions, to which the stresses are now different. The right side of figure 5.3 shows a surface on which the original tensile stress is now decomposed into a stress sn normal to the new surface and a shear stress t. The shear stress in Figure 5.4 can also be decomposed into a tensile stress along one diagonal of the cube and a compressive normal stress along the other. The analysis of the stresses in solids is an important aspect of mechanical design and is treated in specialized texts. 5.4 Elastic deformation: Hooke s law, Young s modulus, Poisson s ratio. When applied the stresses are not too large, elastic deformation takes place. The strains are proportional to the stresses; this is Hooke s law sz = ez E (5.6) E is the elastic modulus also called Young s modulus. Elastic deformation is reversible: when the stress is removed, the strain disappears and the solid regains its original shape. Hooke s law applies to compression as well as tension. A solid that is elongated by the strain ez becomes smaller in the two other directions (the bar in figure 5.3 becomes thinner as it elongates). 6 TABLE 5.1 ELASTIC PROPERTIES FOR SELECTED ENGINEERING MATERIALS AT ROOM TEMPERATURE Elastic modulus, E (106 psi, GPa) Shear modulus, G (106 psi, GPa) Poisson's ratio n Density g/cm3 Material Metals Aluminum alloys Copper alloys Nickel Steels (low alloy) Stainless steel (8-18) Titanium Tungsten Ceramics Diamond Alumina (Al2O3) Zirconia (ZrO2) Silicon carbide Titanium carbide Tungsten carbide Quartz (SiO2) Pyrex glass Fireclay brick Plastics Polyethylene PMMA Polystyrene Nylon 10.5 17 30 30.0 28.0 16.0 56.0 145 53 29 65 55 80 13.6 10 14 0.058-0.19 0.35-0.49 0.39-0.61 0.17 72.4 117 207 207 193 110 386 1000 390 200 450 379 550 94 69.0 96.6 0.4-1.3 2.4-3.4 2.7-4.2 1.2 4.0 6.4 11.3 11.3 9.5 6.5 22.8 3.51 31.8 4.5 27.6 44 77.7 77.7 65.6 44.8 157.3 0.31 0.33 0.30 0.33 0.28 0.31 0.27 2.7 8.9 8.9 7.8 7.9 4.5 19.3 0.22 0.17 0.4 0.4 0.4 0.49 0.4-0.8 0.4-0.8 3.9 5.8 2.9 7.2 15.5 2.6 0.91-0.97 1.2 1.1 1.2 2.5 Other materials Concrete-cement Common bricks Rubbers Common wood (R grain) Common wood ( grain) 6.9 45-50 1.5-2.5 10.4-17.2 0.01 -0.1 9-16 0.6-1.0 Property values for ceramics and polymers vary widely depending on structure and processing. Ranges and average values are given. (Data taken from many sources). 7 ex = ey = - n ez (5.7) n is called Poisson s ratio. Note the minus sign: the strains due to Poisson s ratio are opposite to the strain that is parallel to the stress. Negative strains denote shrinkage. Elastic shear strains are also proportional to shear stresses t=gG (5.8) G is called the shear modulus. It is related to Young s modulus; E = 2(1+n) G. There is no Poisson s ratio for shear deformation: application of a shear stress t causes the shear deformation g shown in figure 5.4 and no other. Table 5.1 gives the elastic modulus, the shear modulus and Poisson s ratio for a number of important structural materials. Note that the elastic and shear modulus are given in gigapascals (Gpa). 1 GPa = 109 Pau = 1000 MPa. Elastic deformation is caused by the elasticity of the chemical bond. We have seen in chapter 2 that the chemical bond, whatever its nature, establishes an equilibrium distance between the atoms. If a tensile force is applied to the atoms, their distance increases; if a compressive force is applied, the distance between the atoms decreases. The bonds between the atoms act like springs. When we apply a tensile stress to a solid, its bonds are stretched, the distance between its atoms increases. If we apply a compressive stress, the distance between the atoms decreases. In single crystals, the elastic modulus changes with crystallographic direction. In copper, for instance, it varies by as much as a factor 2. In poycrystalline materials, which form the vast majority of the structural materials, the random orientation of the grains renders the elastic modulus isotropic. The values of Table 5.1 apply to polycrystalline materials. Some industrial applications of elasticity are springs, bells, springboards and diving boards, and the quartz crystals in watches. 5.5. <a href="/keyword/plastic-deformation/" >plastic deformation</a> s When the stress applied increases beyond a certain value, called the yield strength sy, it causes <a href="/keyword/plastic-deformation/" >plastic deformation</a> , of the material. <a href="/keyword/plastic-deformation/" >plastic deformation</a> can be much larger than elastic deformation and it is permanent: when the stress is removed, the material does not return to its original shape. 5.5.1. The Stress-Strain Curve Figure 5.6 shows a stress-stain curve obtained by the instrument shown in figure 5.2. The steep straight line at left represents elastic deformation. Above the yield stress 8 sy, <a href="/keyword/plastic-deformation/" >plastic deformation</a> occurs. In most metals, a further increase in strain requires an increase in stress, in other words, the metal becomes stronger as it is deformed; this is called strain hardening. Figure 5.6. Schematic engineering stress-strain curve of a metal. In the elastic range, the slope is Young s modulus and the strain is reversible: removal of the stress removes the strain. In the plastic region, deformation is permanent. If at a certain point, we reduce the stress to zero, the sample shortens a little because of elastic deformation but the strain marked eP is permanent. The inserts show the deformation of the sample; necking starts before point M. The stress eventually reaches a maximum, called the ultimate tensile stress, or tensile strength sUTS. From this point, the force required for further deformation decreases. If we observe the sample, we find that the decrease in force is not due to a weakening of the material, but to necking: a localized deformation decreases the cross section at some random point. Even if the stress s necessary for deformation keeps increasing, the force F = sA decreases and all further <a href="/keyword/plastic-deformation/" >plastic deformation</a> takes place in this neck. Ultimately, the sample breaks. The strain at which the sample breaks is called elongation at fracture eF or simply elongation. The mechanical properties are thus defined by the elastic modulus E, the yield strength sy, the tensile strength sUTS and the elongation eF. 5.5.2. Ductility In the manufacture of metallic objects, one needs to ask the question: How much can I deform or shape my metal before it breaks? This is the ductility of the material. The measure of ductility given by the tensile test is the elongation at fracture 9 l f - lo (5.9). lo Since most practical deformation does not involve tensile deformation, a more common definition of ductility is the percent reduction in area ef = .100 (5.10) Ao The two measures are related because, in <a href="/keyword/plastic-deformation/" >plastic deformation</a> , the volume of the material remains constant: Aflf = Aolo. %RA = A material with ductility less than 5 % is considered brittle. 5.5.3. Resilience and Toughness Resilience is the ability of a material to absorb and release elastic strain energy. The energy absorbed in deformation is calculated as E = XFdl. If we replace the force F by s = F/A and the deformation Dl by the strain e = Dl/l, we obtain the energy per unit volume, namely the energy density. (V = A.l). Ao - A f 1 2 E R = sde = sds = 1 s 0 E 2 E 0 0 sy sy (5.11) It is represented by the area of the triangle below the elastic part of the stress-strain curve. Toughness is the total energy per unit volume that the material can absorb before it fractures. It is measured as the total area under the stress-strain curve. It is related to the ductility. It is an important property of material as it is associated with the safety of its use. In the construction of bridges, buildings, airplanes, for instance, one utilizes 10 Figure 5.7. The toughness of a brittle and a soft metal. While the brittle material is stronger (it resists <a href="/keyword/plastic-deformation/" >plastic deformation</a> ), it absorbs less energy than the weaker metal. Note that strength of a metal is its resistance to deformation, not to fracture. ductile and tough metals. In cutting tools, where keeping the edge , i.e. avoiding <a href="/keyword/plastic-deformation/" >plastic deformation</a> is more important than avoiding fracture, one utilizes the hardest possible materials, which tend to be brittle. 5.5.4. True Stress-True Strain and Engineering Stress and Strain It is obvious that the true stresses and strains in the deformation of the necking region are not the ones given by the machine and indicated in figure 5.6. True stress is defined by equation (5.1) where the force is divided by the actual area, which evolves during the test, either by conservation of volume or in necking. The true strain is defined by equation (5.2) where l in the denominator is the actual, evolving length. The engineering stress and strain are defined by equations (5.1) and (5.2) by dividing the force by the initial cross section Ao and the deformation by the initial length lo . True stress and strain are of importance in scientific investigations: it presents an accurate representation of the material s behavior. It is also of importance in manufacturing, where large deformations, for instance by rolling or forging, do not cause necking. In these cases, reduction in area (eq. 5.10) is a more useful measure of ductility than elongation at fracture. The engineering stress and strain are simpler and more convenient and are generally used to characterize materials. 5.5.5 Residual Stresses Residual stresses are internal stresses that are not caused by the application of an external force. They result from the processing of the material; uneven rapid cooling or the extensive <a href="/keyword/plastic-deformation/" >plastic deformation</a> of pieces with complex shape leave internal stresses in which part of the material is compressed and a neighboring area is in tension. These stresses add themselves to the stresses caused by external forces. They can cause failure of the piece at unexpected low load, or they can increase the resistance of the material to failure depending whether they are compressive or tensile. Later in this text, we shall encounter examples where the deliberate introduction of compressive residual stresses produces safe glass for car windows (tempered glass) or increases the strength of concrete (pre-stressed concrete). 5.5.6 Hardness The hardness of a material is its resistance to scratching or to indentation caused by the penetration of a hard object. Scratching and indentation are <a href="/keyword/plastic-deformation/" >plastic deformation</a> s: the measurement of hardness is therefore related to the yield and tensile strength. Hardness is measured by the indentation left after a hard sphere, cone or pyramid is 11 pressed into the surface with a known force. This is a simple, rapid, non-destructive measurement and is by far the most widely performed of the mechanical tests. Several hardness test methods have been devised, some for the rapidity of the measurement and others for the quality of the information they provide. These are illustrated in figure 5. 8. The Rockwell hardness tests are the most rapid and most popular. They measures the depth of penetration of hard indenter under a given, known, load. Several Rockwell hardness tests exist, with different indenter geometries and applied loads so that accurate hardness measurements can be obtained for materials that vary from the softest aluminum to the hardest steel. Indenters are a diamond cone or steel spheres of 1/16, 1/8, _ and _ inch diameter. Applied loads are 60, 100 and 150 kg. (1 kg = 9.81 N). When measuring Rockwell hardness, one first presses the indenter into the surface with a load of 10 kg then one increases the load to 60, 100 or 150 kg and measures the increment in depth of penetration. Each combination of indenter and load is designated by a letter as shown in Table 5.2. The hardness is reported by an R followed by the letter corresponding to the test performed. For instance, a bearing ball has hardness RC 60. Table 5.2. Rockwell Hardness Scales Symbol Indenter Applied Load (kg) ____________________________________________________ A Diamond cone 60 B 1/16-in ball 100 C Diamond cone 150 D Diamond cone 100 E 1/8-in ball 100 F 1/16-in ball 60 G 1/8-in ball 150 H 1/8-in ball 60 K 1/8-in ball 150 A different set of tests, marked as N, T or W serve to measure the hardness of thin sheets of material. These Superficial Rockwell Hardness Tests use smaller loads and larger spherical indenters. Modern hardness instruments apply the load automatically for the correct length of time and compute the hardness value. Each measure requires only a few seconds. Rockwell hardness values have been established arbitrarily and are not simply related to the properties measured in the tensile test. They are specified in ASTM Standard Test E 18 Standard Test Methods for Rockwell Hardness and Rockwell Superficial Hardness of Metallic Materials . 12 When performing the Brinell hardness test, one presses a sphere into the surface and measures the diameter of the indent left after the load has been removed. Figure 5.8 shows the calculation of the hardness value. 13 14 The Vickers Hardness Test is the preferred method for the scientific characterization of materials. In this test one presses a diamond pyramid into the surface with a load that is chosen for the material and the type of information sought. Loads from 1 g to 1000 g are generally used, although loads as high as 20 kg are applied for special measurements. After removal of the load, one measures the diagonal of the square indentation left by the diamond. The hardness is the ratio of the load to the surface of the indentation HV = 1.854P d2 Vickers hardness is usually reported in kg/mm2 although it is often given without dimensions. In the scientific literature, it is sometimes reported in the international units of MPa or GPa. Note that 1 kg/mm2 = 9.81 MPa. It is usual to specify the load P when reporting Vickers hardness, such as HV1kg 500. (This is 500 kg/mm2 measured at a load of 1kg). Since hardness is a measure of a material s resistance to <a href="/keyword/plastic-deformation/" >plastic deformation</a> , it is roughly proportional to the yield strength. As a rule of thumb, the Vickers hardness value is about 3 times the yield strength. This correspondence, however, is not exact because the deformation in hardness measurements has a complex geometry and strain hardening takes place. A modern Vickers hardness tester applies the selected load automatically. It is equipped with an optical microscope for the measurement of the indentation diagonal and computes the hardness. The Knoop Hardness test is a variant of the Vickers designed to measure the hardness of narrow samples (such as cross sections of thin sheets or coatings). For this purpose the pyramid is replace by a diamond knife as shown in figure 5.8. The measurement is performed on the same instrument, with the same method, as the Vickers test. 5.6. Strengthening Metals. In this section we will learn that we can modify the mechanical properties of metals to suit the applications. A steel, for instance, can be made quite soft, with a yield strength of 210 MPa, or hard and brittle, with a yield strength as high as 1620 MPa; the variation of strength in aluminum is even larger, the softest aluminum has a yield strength of 34 MPa, but a yield strength or 500 MPa (15 times higher!) can be achieved by suitable modifications and treatments. First, we need to analyze how metals deform plastically; this knowledge will suggest several methods for increasing the resistance of the metal to such deformation. It will also show the way to softening a metal, lowering 15 its resistance to deformation when necessary. The remarkable thing is that metals are made stronger by the introduction of defects. 5.6.1. The <a href="/keyword/plastic-deformation/" >plastic deformation</a> of metals. We have seen that elastic deformation occurs by the stretching of bonds; when the stress is removed, the atoms return to their equilibrium positions. Large permanent deformations require a different mechanism. In order to understand <a href="/keyword/plastic-deformation/" >plastic deformation</a> , let us first examine the deformation of a single crystal. If we observe the <a href="/keyword/plastic-deformation/" >plastic deformation</a> of a crystal, the first thing we find is that, after the deformation, the material is still a single crystal with the same structure. Entire planes of atoms slide over each other until all atoms have resumed crystal positions. Figure 5.9 shows such a deformation. All <a href="/keyword/plastic-deformation/" >plastic deformation</a> s are shear deformations in response to a corresponding shear stress. Figure 5.9. Model of <a href="/keyword/plastic-deformation/" >plastic deformation</a> in a metal. The upper plane slides from position A to position C where atoms occupy crystal sites as before. 5.6.1.1 Slip Systems Crystallographic planes that are smooth and dense require a smaller shear stress to slide than planes that contain steps or in which the atoms are further apart. <a href="/keyword/plastic-deformation/" >plastic deformation</a> takes place only by the sliding of these dense planes, which are known as slip planes. The sliding of these slip planes occurs only along the directions that offer the smallest obstacles, these are the slip directions. The slip directions and the slip planes form the slip systems of the crystal structure. These are illustrated in Table 5.3. In face-centered crystals (FCC), the {111} planes are the densest and the slip directions are along the dense lines of atoms, which are in the &lt;110&gt; family of directions. There are four planes in the {111} family and each has three directions in the &lt;110&gt; 16 family. The FCC crystal can slide in 12 different directions. Since the {111} planes in the FCC structure are very dense, the minimum shear stress necessary for slip in these materials is quite low. FCC metals are ductile and are preferred where extensive <a href="/keyword/plastic-deformation/" >plastic deformation</a> is required in manufacture. The densest planes of the body centered crystals (BCC) are the {110} family of planes and the slip directions are the &lt;111&gt; family. There are six {110}-type planes and in each one, there are two &lt;111&gt; directions. Thus there are 12 slip systems in the BCC crystal. The {110} planes in the BCC structure are not as dense and smooth as the {111} planes of the FCC structure; the minimum shear stress for <a href="/keyword/plastic-deformation/" >plastic deformation</a> in BCC metals is therefore higher than in FCC materials. Hexagonal Close Packed crystals (HCP) have fewer slip systems than the cubic metals. In Zn, for instance, the slip plane is the basal plane (0001), which has three slip directions belonging to the &lt;1120&gt; family. In titanium, the slip systems consist of the three prismatic planes of the {1010} family, which have only on slip direction each. Thus HCP crystals have only 3 slip systems. Table 5.3. The slip systems of crystals. Note that the minimum shear stress for FCC metals is much smaller than for BCC metals. The HCP metals have fewer slip systems. 17 5.6.1.2. Deformation of a single crystal in a tensile test. If <a href="/keyword/plastic-deformation/" >plastic deformation</a> is always a shear deformation responding to a shear stress, how does a bar deform in a tensile test? As illustrated in figure 5.10, deformation takes place along a slip direction that lies in a slip plane at an angle with the tensile force. a is the angle between the normal to the slip plane and the tensile force direction, and b the angle between the slip direction and the tensile force. The tensile force F can be decomposed into a slip force Fs = F cosb and a force normal to the slip plane Fn = F sinb. Figure 5.10. The slip plane and slip direction in a tensile test. The area of the slip plane is As = A/cosa so that the tensile force F generates the resolved shear stress tRSS = (F/A) cosa cosb = scosa cosb (5.12) <a href="/keyword/plastic-deformation/" >plastic deformation</a> will occur on the plane and along the slip directions that provide the largest value of cosa cosb. Table 5.3 indicates the critical resolved shear stress tCRSS for the <a href="/keyword/plastic-deformation/" >plastic deformation</a> of several metals. Obviously the minimum tensile stress is then s = tCRSS/(cosa cosb) (5.13) This is illustrated in the deformation of a zinc crystal, figure 5.11. Table 5.3 shows that the slip plane illustrated is the only possible for zinc. 18 Figure 5.11. <a href="/keyword/plastic-deformation/" >plastic deformation</a> of a zinc crystal. 5.6.1.2. <a href="/keyword/plastic-deformation/" >plastic deformation</a> of polycrystalline materials. Most metals in practical use are polycrystalline: they consist of a large number of small grains, which are crystals oriented in random directions. When a polycrystalline material is subjected to a tensile stress, each grain responds in the same manner as the single crystal described above: it slips along the slip plane and the slip direction that results in the largest possible value of (cosa cosb). Figure 5.13 shows the slip lines in such a polycrystalline material. During the process, the individual grains are deformed as shown in Figure 5.14 (marked as 7.11). In the deformation of these grains, mechanical integrity is maintained: the grains deform in such a way that they remain in contact. The requirement that the grains do not separate restricts the possible slip directions to smaller values of (cosa cosb) and therefore larger tensile stresses s than in a single crystal. For this reason, polycrystalline materials are stronger than single crystals. 5.5.2. Dislocations and strengthening of metals. In this section we explore why the critical resolved shear stresses for <a href="/keyword/plastic-deformation/" >plastic deformation</a> are as low as shown in Table 5.3 and how we can obtain metals with tensile strengths as large as 1600 MPa. During <a href="/keyword/plastic-deformation/" >plastic deformation</a> , the crystallographic planes do not slide rigidly, all at once, over each other; they slide by the movement of dislocations. This is shown in figure 5.15 for the movement of an edge dislocation. 19 Figures 5.13 and 5.14 taken from Callister. 20 Under the effect of the applied shear stress t, atom A moves slightly to the right and atom 2 moves to the left. The result is that the bond A-2 predominates over the bond B-2 and the dislocation moves to the right. The dislocation then moves to atom C and further until it emerges on the surface of the crystal and forms a step. The result is that the plane containing the atoms A,B,C has slipped over the plane containing atoms 1,2,3. This movement of dislocations is easy, it requires the very low stresses indicated in Table 5.3 because the positions at atoms A and B are equivalent. Figure 5.15. Atom movements associated with edge dislocation movement across a stressed crystal. When the dislocation emerges at the surface, a step is produced. After A.G. Guy, Essentials of Materials Science,McGraw-Hill, New York (1976) Figure 5.16. <a href="/keyword/plastic-deformation/" >plastic deformation</a> by the movement of an edge dislocation (a) and of a screw dislocation (b). Adapted from H.W. Hayden, W. G&gt; Moffat, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, J. Wiley &amp;Sons, New York, 1965 21 In a similar vein, the movement of a screw dislocation through the crystal results in the sliding of one crystallographic plane over the other. Figure 5.16 illustrates <a href="/keyword/plastic-deformation/" >plastic deformation</a> by the movement of edge and screw dislocations. It is important to note that this easy movement of dislocations is possible only in a perfect crystal. The easy movement of dislocations also requires that the chemical bonds do not have preferred directions; this is only the case in metals. In amorphous solids, dislocations cannot be defined, and this mechanism for easy <a href="/keyword/plastic-deformation/" >plastic deformation</a> does not exist: Metallic glasses do not deform plastically. The following analogy helps to understand how dislocation movement is possible with much lower stresses than the rigid sliding of a whole plane: consider a large rug that must be moved on the ground by a small distance. Pulling the whole rug all at once would require a large force. If, however, we form a dislocation in the shape of a bulge as shown in Figure 5.17, we can move the rug with a small force simply by moving the dislocation to the right. Figure 5.17. A rug is displaced easily to the right when a bulge is made and pushed. How do we strengthen a metal? A metal is stronger when it requires a higher stress to deform it. Any method that makes it more difficult for a dislocation to move strengthens the metal. We have seen that the easy movement of dislocations is a consequence of the perfect periodicity of the crystal in which positions A, B and C are equivalent. Therefore, crystal defects impede the movement of dislocations and strengthen the metal. Practically any kind of defect will be effective; the most useful types of defects are: Impurities; o these can be atoms of a different element that are dissolved at random in the solid, they result in solution strengthening or solution hardening. o they can be entire aggregates of matter that is not soluble in the host crystal and result in precipitation hardening Grain boundaries, which lead to strengthening by grain refinement Dislocations that are introduced by <a href="/keyword/plastic-deformation/" >plastic deformation</a> ; they lead to strain hardening also called work hardening or cold working. 22 A locally distorted lattice due to the presence of atoms not soluble in the metal but frozen in by rapid cooling. This is the case of quenched and tempered steels. We shall examine these strengthening methods in more detail below. It is worth noting that the production of dislocation-free metals is not a strengthening option: not only would it be difficult and costly, but dislocations are created easily in metals by the slightest deformation. 5.6.2.1. Solution strengthening Almost all metals in modern use are alloys. Adding impurity atoms, either in interstitial or substitutional positions increases the stress needed for dislocation motion. A look at figure 5.15 will show that the dislocation introduces a large elastic compression on the top half and an equally large tensile field below the dislocation. Any foreign atom that is larger than the host atoms positioned in the region of compressive strain will exacerbate the compression and increases the elastic stress energy: it will repel the dislocation. A large atom in the tensile strain region decreases the elastic stress energy and attracts the dislocation. The atom will impede dislocation movement either by impeding its approach or its escape. A smaller atom will have the same effect, increasing the tensile strain or decreasing the compressive strain. Figure 5.18. Stress-strain curves of copper alloyed with 0, 10%, 20% and 50% nickel. 23 5.6.2.2. Precipitation Strengthening Precipitates, which are small particles with different composition, structure and often bonding type from the host metal are much more powerful obstacles to the propagation of dislocations than dissolved alloy atoms. An example is afforded by aluminum alloyed with as little as a heat copper after 4.4% treatment that will be discusses in Chapter. This heat treatment produces microscopic precipitates of the compound CuAl2. As figure 5.19 shows, these precipitates increase the yield strength of aluminum from 35 MPa to 345 MPa, which is 10 times higher! Figure 5.19. Precipitation hardening of aluminum. 5.6.2.3. Strengthening by grain refinement. The grain of metals, namely the fact that they consist of many small crystals (grains) with random crystallographic orientation, increases their strength in two different ways. First, as we have already discussed in section 5.6.1.2, the deformation of the grains occurs in such a way that the grains remain attached to each other. This restricts the choice of slip system such that the term cosacosb is not maximized; this increases the tensile stress necessary for deformation according to equation 5.13. Secondly, grain boundaries are areas of quite severe lattice disorder; they present a strong obstacle to the movement of the dislocations. A metal with a fine grain structure contains a larger total grain boundary area and is stronger than a coarse grained metal. The Hall-Petch relation 24 sy = so + k d (5.14) describes how the yield strength sy of metals increases as the average size d of the grains decreases. Here s o is the yield strength of a metal with very large grains and k is a constant. Figure 5.21 is an example of Hall-Petch behavior. Figure 5.20. Dependence of the yield strength of brass on the average grain size d. The upper scale denotes the grain diameter. 5.6.2.4. Strain hardening In figures 5.l8 and 5.19, we can observe that the stress needed for <a href="/keyword/plastic-deformation/" >plastic deformation</a> increases with the strain; the ultimate tensile strength is larger than the yield strength. This phenomenon is called strain hardening, strain strengthening or work hardening, or also cold working because the deformation is performed at room temperature. Strain hardening is the result of forces between dislocations. <a href="/keyword/plastic-deformation/" >plastic deformation</a> creates large numbers of dislocations by mechanisms that are beyond the scope of this text. These dislocations inhibit each other s movement for the following reasons. Every edge dislocation is surrounded with a region of elastic compression and tension. Screw dislocations are surrounded by regions of large shear strain. When two dislocations approach each other, the local strain is the sum of the strains surrounding the two dislocations. When the dislocations have the same sign (i.e. the extra plane of atoms is on the same side), the resultant strain increases and so does the elastic stress energy: the two dislocations repel each other. When the dislocations have opposite signs, the resultant strain and the elastic stress energy decreases and the dislocations attract each other. In either case, they impede each other s motion and the result is strain hardening. 25 As figure 5.22 shows, the strength increase is quite large. Increases in strength by a factor 3 are common. Figure 5.22 Strain hardening of brass (a Cu-Zn alloy). Stress-strain curve A: without cold working. B): 30% cold working, C): after 70% cold working. Strain hardening is a widespread method for the strengthening of metals. It is commonly accomplished by rolling, where a piece of metal is deformed between two rolls; or by forging, where the metal is deformed by hammering. These methods allow larger strains than tensile deformation because they avoid necking. Percent cold working is defined as the reduction in area, equation 5.10. Strain hardening presents problems in manufacturing. Most metallic objects are shaped by <a href="/keyword/plastic-deformation/" >plastic deformation</a> during which the metal becomes hard and brittle. It is desirable to undo the effects of strain hardening and recover the soft and ductile material. This is achieved by annealing. 5.6.2.5. Annealing. Annealing is a processing method in which a material is kept at elevated temperatures for a certain time and cooled slowly with the purpose of removing residual stresses and of undoing the effects of strain hardening. The strong and brittle brass that presented the stress-strain curve of figure 5.22 C, for instance, would recover the lower strength and high ductility of curve A after annealing. The thermal energy at elevated temperatures facilitates the movement of atoms and dislocations. Annealing proceeds in three successive stages: recovery, recrystallization and grain growth. During recovery, the dislocations move under the action of the internal stresses. Slight <a href="/keyword/plastic-deformation/" >plastic deformation</a> removes residual stresses that had been introduced by processing. 26 Dislocations also move and rearrange themselves in such a way as to reduce the total elastic strain energy that surrounds them. By aligning themselves as shown in figure Figure 5.23. Grain boundary formed by the alignment of edge dislocations. The tilt between the two grains accommodates the half planes of atoms without compressive and tensile stresses. 5.23, they create boundaries between new grains in which the elastic strain is minimized. When strain hardening has been extensive, the new crystal structure is much finer than it had been before annealing. This phase of annealing is recrystallization. The final phase of annealing is grain growth. At elevated temperatures, on observes that the size of the larger grains increases and that the small grains disappear. This is driven by the decreases in total grain boundary energy. 27 Figure 5.24. Model of grain growth: atoms jump to the larger grain on the right. Figure 5.25. Cold working and annealing of 99.97% pure iron. A) Before deformation. B) after 60% cold working. C) after annealing at 5100C for 1 hour; D) after annealing at 510oC for 10 hours; E) after annealing at 510oC for 100 hours; F) after annealing at 675oC for 1 hour; G) after annealing at 9000 C for 1 hour. 28 Figure 5.25 continued. Annealing of 99.97% pure Fe. F) after annealing at 675oC for 1 hour; G) after annealing at 9000 C for 1 hour. Consider the curved grain boundary, figure 5.24, between a small grain on the left and a larger grain on the right. The surface energy of the concave (large) grain is lower than that of the convex, smaller grain. The atoms move to the low-energy positions and the grain boundary moves to the left, increasing the size of the larger grain and leading to the eventual disappearance of the smaller grain. In short, annealing removes residual stresses, decreases the density of dislocations and a lowers the grain boundary area. The result is a lowering of the yield strength and an increase in ductility. 29
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