Structure and Manufacturing Properties of Metals

Transcription

1 Chapter 3 Structure and Manufacturing Properties of Metals Questions 3.1 What is the difference between a unit cell and a single crystal? A unit cell is the smallest group of atoms showing the characteristic lattice structure of a particular metal. A single crystal consists of a number of unit cells; some examples are whiskers, chips for semiconductor devices, and turbine blades. 3.2 Explain why we should study the crystal structure of metals. By studying the crystal structure of metals, information about various properties can be inferred. By relating structure to properties, one can predict processing behavior or select appropriate applications for a metal. Metals with face-centered cubic structure, for example, tend to be ductile whereas hexagonal close-packed metals tend to be brittle. 3.3 What effects does recrystallization have on the properties of metals? As shown in Figs on p. 96 and 3.18 on p. 97, strength and hardness are reduced, ductility is increased, and residual stresses are relieved. 3.4 What is the significance of a slip system? The greater the number of slip systems, the higher the ductility of the metal. Also, the slip system and the number of active slip systems give direct understanding of the material s plastic behavior. For example, an hcp material has few slip systems. Thus, in a bulk material, few grains will be preferentially oriented with respect to a slip system and high stresses will be required to initiate plastic deformation. On the other hand, fcc materials, have many slip systems and thus a lower stress will be required for plastic deformation. See also Section starting on p Explain what is meant by structure-sensitive and structure-insensitive properties of metals. As described in Section starting on p. 89, those properties that depend on the structure of a metal are known as structure-sensitive properties (yield and fracture strength, electrical conductivity). Those that are not (other physical properties and elastic constants) are called structure-insensitive properties. 3.6 What is the relationship between nucleation rate and the number of grains per unit volume of a metal? This relationship is described at the beginning of Section 3.4 starting on p. 91. Generally, rapid cooling produces smaller grains, whereas slow cooling produces larger grains. 29

2 3.7 Explain the difference between recovery and recrystallization. These phenomena are described in Section 3.6 on p. 96. Recovery involves relief of residual stresses, reduction in the number of dislocations, and increase in ductility. In recrystalization, new equiaxed and stress-free grains are formed, replacing the older grains. 3.8 (a) Is it possible for two pieces of the same metal to have different recrystallization temperatures? Explain. (b) Is it possible for recrystallization to take place in some regions of a workpiece before other regions do in the same workpiece? Explain. (a) Two pieces of the same metal can have different recrystallization temperatures if the pieces have been cold worked to different amounts. The piece that was cold worked to a greater extent will have more stored energy to drive the recrystallization process, and hence its recrystallization temperature will be lower. See also Fig on p. 97. (b) Recrystallization may occur in some regions before others if i. the workpiece was unevenly worked, as is generally the case in deformation processing of materials, since varying amounts of cold work have different recrystallization temperatures, or ii. the part has varying thicknesses; the thinner sections will heat up to the recrystallization temperature faster. 3.9 Describe why different crystal structures exhibit different strengths and ductilities. Different crystal structures have different slip systems, which consist of a slip plane (the closest packed plane) and a slip direction (the closepacked direction). The fcc structure has 12 slip systems, bcc has 48, and hcp has 3. The ductility of a metal depends on how many of the slip systems can be operative. In general, fcc and bcc structures possess higher ductility than hcp structures, because they have more slip systems. The shear strength of a metal decreases for decreasing b/a ratio (b is inversely proportional to atomic density in the slip plane and a is the plane spacing), and the b/a ratio depends on the slip system of the chemical structure. (See also Section starting on p. 87.) 3.10 Explain the difference between preferred orientation and mechanical fibering. Preferred orientation is anisotropic behavior in a polycrystalline workpiece that has crystals aligned in nonrandom orientations. Crystals become oriented nonrandomly in a workpiece when it is deformed, because the slip direction of a crystal tends to align along the general deformation direction. Mechanical fibering is caused by the alignment of impurities, inclusions, or voids during plastic working of a metal; hence, the properties vary with the relative orientation of the stress applied to the orientation of the defect. (See also preferred orientation in Section 3.5 on p. 95.) 3.11 Give some analogies to mechanical fibering (such as layers of thin dough sprinkled with flour). This is an open-ended problem with many acceptable answers. Some examples are plywood, laminated products (such as countertops), winter clothing, pastry with layers of cream or jam, and pasta dishes with layers of pasta and cheese A cold-worked piece of metal has been recrystallized. When tested, it is found to be anisotropic. Explain the probable reason for this behavior. The anisotropy of the workpiece is likely due to preferred orientation resulting from the recrystallization process. Copper is an example of a metal that has a very strong preferred orientation after annealing. As shown in Fig on p. 97, no recrystallization occurs below a critical deformation, being typically five percent Does recrystallization completely eliminate mechanical fibering in a workpiece? Explain. Mechanical fibering involves the alignment of impurities, inclusions, and voids in the workpiece during deformation. Recrystallization generally modifies the grain structure, but will not eliminate mechanical fibering. 30

3 3.14 Explain why we may have to be concerned with the orange-peel effect on metal surfaces. Orange peel not only influences surface appearance of parts, which may or may not be desirable, but also affects their surface characteristics such as friction, wear, lubrication, and corrosion and electrical properties, as well as subsequent finishing, coating, and painting operations. (See also surface roughness in practice in Section 4.3 on p. 137.) 3.15 How can you tell the difference between two parts made of the same metal, one shaped by cold working and the other by hot working? Explain the differences you might observe. Note that there are several methods that can be used to determine the differences between the two parts. Some of the methods of distinguishing hot vs. cold worked parts are: (a) The surface finish of the cold-worked part would be smoother than the hot-worked part, and possibly shinier. (b) If hardness values could be taken on the parts, the cold-worked part would be harder. (c) The cold-worked part would likely contain residual stresses and exhibit anisotropic behavior. (d) Metallographic examination of the parts can be made: the hot-worked part would generally have equiaxed grains due to recrystallization, while the cold-worked part would have grains elongated in the general direction of deformation. (e) The two parts can be subjected to mechanical testing and their properties compared Explain why the strength of a polycrystalline metal at room temperature decreases as its grain size increases. Strength increases as more entanglements of dislocations take place with grain boundaries and with each other. Metals with larger grains have less grain-boundary area per unit volume, and hence they are not be able to generate as many entanglements at grain boundaries, thus the strength will be lower. (See also Eq. (3.8) on p. 92.) 3.17 What is the significance of some metals, such as lead and tin, having recrystallization temperatures at about room temperature? For these metals, room temperature is sufficiently high for recrystallization to occur without heating. These metals can be cold worked to large extent without requiring a recrystallization cycle to restore their ductility, hence formability. However, as the strain rate increases, their strength at room temperature increases because the metal has less time to recrystallize, thus exhibiting a strain hardening behavior You are given a deck of playing cards held together with a rubber band. Which of the material-behavior phenomena described in this chapter could you demonstrate with this setup? What would be the effects of increasing the number of rubber bands holding the cards together? Explain. (Hint: Inspect Figs. 3.5 and 3.7.) The following demonstrations can be made with a deck of cards sliding against each other: (a) Slip planes; permanent slip of cards with no rubber band, similar to that shown in Fig. 3.5a on p. 86. (b) Surface roughness that develops along the edges of the deck of cards, similar to the lower part of Fig. 3.7 on p. 88. (c) Friction between the cards, simulating the shear stress required to cause slip, similar to Fig. 3.5 on p. 86. Friction between the cards can be decreased using talcum powder, or increased by moisture or soft glue (that has not set yet). (d) Failure by slip, similar to Fig. p b on (e) Presence of a rubber band indicates elastic behavior and recovery when unloaded. (f) The greater the number of rubber bands, the higher the shear modulus, G, which is related to the elastic modulus, E. (g) The deck of cards is highly anisotropic. 31

4 3.19 Using the information given in Chapters 2 and 3, list and describe the conditions that induce brittle fracture in an otherwise ductile piece of metal. Brittle fracture can be induced typically by: (a) high deformation rates, (b) the presence of stress concentrations, such as notches and cracks, (c) state of stress, especially high hydrostatic tension components, (d) radiation damage, and (e) lower temperatures, particularly for metals with bcc structure. In each case, the stress required to cause yielding is raised above the stress needed to cause failure, or the stress needed for crack propagation is below the yield stress of the metal (as with stress concentrations) Make a list of metals that would be suitable for a (1) paper clip, (2) bicycle frame, (3) razor blade, (4) battery cable, and (5) gas-turbine blade. Explain your reasoning. In the selection of materials for these applications, the particular requirements that are significant to these components are briefly outlined as follows: (a) Yield stress, elastic modulus, corrosion resistance. (b) Strength, toughness, wear resistance, density. (c) Strength, resistance to corrosion and wear. (d) Yield stress, toughness, elastic modulus, corrosion resistance, and electrical conductivity. (e) Strength, creep resistance, resistance to various types of wear, and corrosion resistance at high temperature. Students are encouraged to suggest a variety of metals and discuss the relative advantages and limitations with regard to particular applications Explain the advantages and limitations of cold, warm, and hot working of metals, respectively. These are explained briefly in Section 3.7 on p. 98. Basically, cold working has the advantages of refining the materials grain structure while increasing mechanical properties such as strength, but it does result in anisotropy and reduced ductility. Hot working does not result in strengthening of the workpiece, but the ductility of the workpiece is preserved, and there is little or no anisotropy. Warm working is a compromise Explain why parts may crack when suddenly subjected to extremes of temperature. Thermal stresses result from temperature gradients in a material; the temperature will vary significantly throughout the part when subjected to extremes of temperature. The higher the temperature gradient, the more severe thermal stresses to which the part will be subjected, and the higher stresses will increase the probability of cracking. This is particularly important in brittle and notch-sensitive materials. (See also Section starting on p. 107 regarding the role of coefficient of thermal expansion and thermal conductivity in development of thermal stresses.) 3.23 From your own experience and observations, list three applications each for the following metals and their alloys: (1) steel, (2) aluminum, (3) copper, (4) magnesium, and (5) gold. There are numerous acceptable answers, including: (a) steel: automobile bodies, structural members (buildings, boilers, machinery), fasteners, springs, bearings, knives. (b) aluminum: aircraft bodies, baseball bats, cookware, beverage containers, automotive pistons. (c) copper: electrical wire, cookware, battery cable terminals, printed circuit boards. (d) lead: batteries, toy soldiers, solders, glass crystal. (e) gold: jewelry, electrical connections, tooth fillings, coins, medals List three applications that are not suitable for each of the following metals and their alloys: 32

5 (1) steel, (2) aluminum, (3) copper, (4) magnesium, and (5) gold. There are several acceptable answers, including: (a) steel: electrical contacts, aircraft fuselage, car tire, portable computer case. (b) aluminum: cutting tools, shafts, gears, flywheels. (c) copper: aircraft fuselage, bridges, submarine, toys. (d) lead: toys, cookware, aircraft structural components, automobile body panels. (e) gold: any part or component with a large mass and that requires strength and stiffness Name products that would not have been developed to their advanced stages, as we find them today, if alloys with high strength and corrosion and creep resistance at elevated temperatures had not been developed. Some simple examples are jet engines and furnaces. The student is encouraged to cite numerous other examples Inspect several metal products and components and make an educated guess as to what materials they are made from. Give reasons for your guess. If you list two or more possibilities, explain your reasoning. This is an open-ended problem and is a good topic for group discussion in class. Some examples, such as an aluminum baseball bat or beverage can, can be cited and students can be asked why they believe the material is aluminum List three engineering applications each for which the following physical properties would be desirable: (1) high density, (2) low melting point, and (3) high thermal conductivity. Some examples are given below. (a) High density: adding weight to a part (such as an anchor for a boat), flywheels, counterweights. (b) Low melting point: Soldering wire, fuse elements (such as in sprinklers to sense fires). (c) High thermal conductivity: cookware, car radiators, precision instruments that resist thermal warping. The student is encouraged to site other examples Two physical properties that have a major influence on the cracking of workpieces, tools, or dies during thermal cycling are thermal conductivity and thermal expansion. Explain why. Cracking results from thermal stresses that develop in the part during thermal cycling. Thermal stresses may be caused both by temperature gradients and by anisotropy of thermal expansion. High thermal conductivity allows the heat to be dissipated faster and more evenly throughout the part, thus reducing the temperature gradient. If the thermal expansion is low, the stresses will be lower for a given temperature gradient. When thermal stresses reach a certain level in the part, cracking will occur. If a material has higher ductility, it will be able to undergo more by plastic deformation before possible fracture, and the tendency for cracking will thus decrease Describe the advantages of nanomaterials over traditional materials. Since nanomaterials have fine structure, they have very high strength, hardness, and strength-to-weight ratios compared to traditional materials. The student is encouraged to review relevant sections in the book; see, for example, pages , as well as nanoceramics and nanopowders Aluminum has been cited as a possible substitute material for steel in automobiles. What concerns, if any, would you have prior to purchasing an aluminum automobile? By the student. Some of the main concerns associated with aluminum alloys are that, generally, their toughness is lower than steel alloys; thus, unless the automobile is properly designed and tested, its crashworthiness could suffer. A perceived advantage is that weight savings with aluminum result in higher fuel efficiencies, but steel requires much less energy to produce from ore, so these savings are not as high as initially believed. 33

6 3.31 Lead shot is popular among sportsmen for hunting, but birds commonly ingest the pellets (along with gravel) to help digest food. What substitute materials would you recommend for lead, and why? Obviously, the humanitarian concern is associated with the waterfowl ingesting lead and, therefore, perishing from lead poisoning; the ideal material would thus be one that is not poisonous. On the other hand, it is important for the shot material to be effective for its purpose, as otherwise a bird is only wounded. Effective shot has a high density, thus a material with a very high density is desired. Referring to Table 3.2 on p. 98, materials with a very high density but greater environmental friendliness are gold and tungsten, but obviously tungsten would be the more logical choice What are metallic glasses? Why is the word glass used for these materials? These materials are described in Section starting on p They are produced through such processes as rapid solidification (described in Section starting on p. 235) so that the material has no grain structure or orientation. Thus, none of the traditional metallic characteristics are present, such as deformation by slip, anisotropy, or grain effects. Because this is very similar to the microstructure and behavior of glass, hence the term Which of the materials described in this chapter has the highest (a) density, (b) electrical conductivity, (c) thermal conductivity, (d) strength, and (e) cost? As can be seen from Table 3.3 on p. 106, the highest density is for tungsten, and the highest electrical conductivity and thermal conductivity in silver. The highest ultimate strength mentioned in the chapter is for Monel K-500 at 1050 MPa, and the highest cost (which varies from time to time) is usually is associated with superalloys What is twinning? How does it differ from slip? This is illustrated in Fig. 3.5 on p. 86. In twinning, a grain deforms to produce a mirror-image about a plane of twinning. Slip involves sliding along a plane. An appropriate analogy to differentiate these mechanisms is to suggest that twinning is similar to bending about a plane, and slip is similar to shearing. Problems 3.35 Calculate the theoretical (a) shear strength and (b) tensile strength for aluminum, plain-carbon steel, and tungsten. Estimate the ratios of their theoretical strength to actual strength. Equation (3.3) and Eq. (3.5) give the shear and tensile strengths, respectively, as τ G 2π σ E 10 The values of E and ν are obtained from Table 2.1 on p. 32, and G is calculated using Eq. (2.24) on p. 49, E G 2(1 + ν) Thus, the following table can be generated: Mat- E G τ σ erial (GPa) ν (GPa) (GPa) (GPa) Al Steel W A technician determines that the grain size of a certain etched specimen is 6. Upon further checking, it is found that the magnification used was 150, instead of 100 as required by ASTM standards. What is the correct grain size? To answer this question, one can either interpolate from Table 3.1 on p. 93 or obtain the data for a larger number of grain sizes, as well as the grain diameter as a function of the ASTM No. 34

7 The following data is from the Metals Handbook, ASM International: ASTM Grains per Grains per Avg. grain No mm 2 mm 3 dia., mm , , Since the magnification ratio is 150/1001.5, the diameter was magnified 1.5 times more than it should have been. Thus, the grains appeared larger than they actually are. Because the grain size of 6 has an average diameter of mm, the actual diameter is thus d mm mm As can be seen from the table, this corresponds to a grain size of about Estimate the number of grains in a regular paper clip if its ASTM grain size is 9. As can be seen in Table 3.1 on p. 93, an ASTM grain size of 9 has 185,000 grains/mm 3. An ordinary paper clip (although they vary depending on the size of paper clip considered) hass a wire diameter of 0.80 mm and a length of 100 mm. Therefore, the paper clip volume is V πd2 l 4 π(0.80)2 (100) mm 3 The number of grains can thus be calculated as (50.5)(185,000)9.34 million The natural frequency f of a cantilever beam is given by the expression EIg f 0.56 wl 4, where E is the modulus of elasticity, I is the moment of inertia, g is the gravitational constant, w is the weight of the beam per unit length, and L is the length of the beam. How does the natural frequency of the beam change, if any, as its temperature is increased? Let s assume that the beam has a square cross section with a side of length h. Note, however, that any cross section will result in the same trends, so students shouldn t be discouraged from considering, for example, circular cross sections. The moment of inertia for a square cross section is I h4 12 The moment of inertia will increase as temperature increases, because the cross section will become larger due to thermal expansion. The weight per length, w, is given by w W L where W is the weight of the beam. Since L increases with increasing temperature, the weight per length will decrease with increasing temperature. Also note that the modulus of elasticity will decrease with increasing temperature (see Fig. 2.9 on p. 41). Consider the ratio of initial frequency (subscript 1) to frequency at elevated temperature (subscript 2): f 1 f E1I 1g w 1L 4 1 E2I 2g w 2L 4 2 Simplifying further, f 1 f 2 E 1 I 1 L 3 2 E 2 I 2 L 3 1 E1I 1 (W/L 1)L 4 1 E2I 2 (W/L 2)L 4 2 E 1 h 4 1 L3 2 E 2 h 4 2 L3 a E1I 1 L 3 1 E2I 2 Letting α be the coefficient of thermal expansion, we can write h 2 h 1 (1 + α T ) L 2 L 1 (1 + α T ) Therefore, the frequency ratio is f 1 E 1 h 4 1 L3 2 f 2 E 2 h 4 2 L3 1 E 1 h 4 1 L3 1 (1 + α T )3 E 2 h 4 1 (1 + α T )4 L 3 1 E 1 E 2 (1 + α T ) L

8 To compare these effects, consider the case of carbon steel. Figure 2.9 on p. 41 shows a drop in elastic modulus from 190 to 130 GPa over a temperature increase of 1000 C. From Table 3.3 on p. 106, the coefficient of thermal expansion for steel is 14.5 µm/m C (average of the extreme values given in the table), so that the change in frequency is: f 1 f 2 E 1 E 2 (1 + α T ) [1 + ( ) (1000)] or f 1 /f Thus, the natural frequency of the beam decreases when heated. This is a general trend (and not just for carbon steel), namely that the thermal changes in elastic modulus plays a larger role than the thermal expansion of the beam A strip of metal is reduced in thickness by cold working from 25 mm to 15 mm. A similar strip is reduced from 25 mm to 10 mm. Which one of these strips will recrystallize at a lower temperature? Why? In the first case, reducing the strip from 25 to 15 mm involves a true strain (absolute value) of ( ) 25 ɛ ln and for the second case, ( ) 25 ɛ ln A review of Fig will indicate that, because of the higher degree of cold work and hence higher stored energy, the second case will involve recrystallization at a lower temperature than the first case A 1-m long, simply-supported beam with a round cross section is subjected to a load of 50 kg at its center. (a) If the shaft is made from AISI 303 steel and has a diameter of 20 mm, what is the deflection under the load? (b) For shafts made from 2024-T4 aluminum, architectural bronze, and 99.5% titanium, respectively, what must the diameter of the shaft be for the shaft to have the same deflection as in part (a)? (a) For a simply-supported beam, the deflection can be obtained from any solid mechanics book as δ P L3 48EI For a round cross section with diameter of 20 mm, the moment of inertia is I πd4 64 π(0.020) m 4 64 From Table 2.1, E for steel is around 200 GPa. The load is 50 kg or 490 N; therefore, the deflection is δ P L3 48EI (490 N)(1 m) 3 48(200 GPa)( m 4 ) or δ m 6.5 mm. (b) It is useful to express the diameter as a function of deflection: δ P L3 48EI Solving for d, we have ( 4P L 3 d 3πEδ 64P L3 48πEd 4 ) 1/4 Thus, the following table can be constructed, with the elastic moduli taken from Table 2.1 on p. 32. Material E (GPa) d (mm) 2024-T4 Al Arch. bronze % Ti If the diameter of the aluminum atom is 0.5 nm, estimate the number of atoms in a grain with an ASTM size of 5. If the grain size is 5, there are 2900 grains per mm 3 of aluminum, and each grain has a volume of 1/ mm 3. Recall that for an fcc material there are four atoms per unit cell, with a total volume of 16πR 3 /3, and that the diagonal, a, of the unit cell is given by ( a 2 ) 2 R 36

9 400 APFfcc Note that as long as all the atoms in the unit cell are of the same size, the atomic packing factors do not depend on the atomic radius. Therefore, the volume of the grain taken up by atoms is ( )(0.74) mm3. (Recall that 1 mm106 nm.) The diameter of an aluminum atom is 0.5 nm, thus its radius is 0.25 nm or mm. The volume of an aluminum atom is 4π( )3 4πR3 V or mm. Dividing the volume of aluminum in the grain by the volume of an aluminum atom gives the total number of atoms in the grain as ( )/( ) Yield stress (MPa) 1000 Steel Stainless steel Copper Titanium Aluminum 50 0 Lead Magnesium ,000 15,000 20,000 Density (kg/m3) Molybdenum Steel Nickel 150 Copper 100 Aluminum 50 Titanium Magnesium The following data is obtained in tension tests of brass: Grain Size (µm) Molybdenum Yield stress (MPa) Nickel Does this material follow the Hall-Petch effect? If so, what is the value of k? Tungsten Copper 400 First, it is obvious from this table that the material becomes stronger as the grain size decreases, which is the expected result. However, it is not clear whether Eq. (3.8) on p. 92 is applicable. It is possible to plot the yield stress as a function of grain diameter, but it is better to plot it as a function of d 1/2, as follows: Aluminum Nickel 150 Titanium Steel Relative Cost The plots are shown below, based on the data given in Tables 2.1 on p. 32, 3.3 on p. 106, and 16.4 on p Average values have been used to obtain these plots Molybdenum Plot the following for the materials described in this chapter: (a) yield stress versus density, (b) modulus of elasticity versus strength, and (c) modulus of elasticity versus relative cost. Hint: See Table Tungsten Elastic modulus (GPa) 16πR / R 2 3 Elastic modulus (GPa) Hence, Magnesium Lead ,000 15,000 Density (kg/m3) 20,000 37

10 Yield strength (MPa) d -1/2 The least-squares curve fit for a straight line is Y d 1/2 Material k α k/α Plastics Wood Glasses Lead Graphite Ti alloys Pb alloys Ti Ceramics Steels Ni alloys Mg alloys Mg Iron Nickel Columbium Tantalum Aluminum Al Alloys Cu alloys Gold Berylium Si Silver Copper Molybdenum Tungsten This data is shown graphically as follows: with an R factor of This suggests that a linear curve fit is proper, and it can be concluded that the material does follow the Hall- Petch effect, with a value of k 458 MPa- µm It can be shown that thermal distortion in precision devices is low for high values of thermal conductivity divided by the thermal expansion coefficient. Rank the materials in Table 3.3 according to their suitability to resist thermal distortion. The following table can be compiled, using maximum values of thermal conductivity and minimum values of thermal expansion coefficient (to show optimum behavior for low thermal distortion): Tungsten Molybdenum Copper Silver Silver alloys Berylium Cu-alloys Al-alloys Aluminum Tantalum Columbium Nickel Magnesium Mg-alloys Ni-alloys Steel Ceramics Titanium Lead alloys Ti-alloys Graphite Lead Glasses Wood Plastics k/ (10 6 N/s) Increasing performance 3.45 Assume that you are asked to give a quiz to students on the contents of this chapter. Prepare three quantitative problems and three qualitative questions, and supply the answers. By the student. This is a challenging, openended question that requires considerable focus and understanding on the part of the students, and has been found to be a very valuable homework problem

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