MATERIALS OF IMPORTANCE Catalysts (and Surface Defects)
4.11 Grain Size Determination • 113
The grain sizeis often determined when the properties of a polycrystalline mate- rial are under consideration. In this regard, there exist a number of techniques by which size is specified in terms of average grain volume, diameter, or area. Grain size may be estimated by using an intercept method, described as follows. Straight lines all the same length are drawn through several photomicrographs that show the grain structure. The grains intersected by each line segment are counted; the line length is then divided by an average of the number of grains intersected, taken over all the line segments. The average grain diameter is found by dividing this re- sult by the linear magnification of the photomicrographs.
Probably the most common method, however, is that devised by the Amer- ican Society for Testing and Materials (ASTM).7The ASTM has prepared sev- eral standard comparison charts, all having different average grain sizes. To each is assigned a number ranging from 1 to 10, which is termed the grain size num- ber. A specimen must be properly prepared to reveal the grain structure, which is photographed at a magnification of 100. Grain size is expressed as the grain size number of the chart that most nearly matches the grains in the micrograph.
Thus, a relatively simple and convenient visual determination of grain size num- ber is possible. Grain size number is used extensively in the specification of steels.
The rationale behind the assignment of the grain size number to these various charts is as follows. Let n represent the grain size number, and N the average num- ber of grains per square inch at a magnification of 100. These two parameters are related to each other through the expression
(4.16)
Concept Check 4.2
Does the grain size number (n of Equation 4.16) increase or decrease with decreasing grain size? Why?
[The answer may be found at www.wiley.com/college/callister(Student Companion Site).]
EXAMPLE PROBLEM 4.4
Computations of ASTM Grain Size Number and Number of Grains per Unit Area
(a) Determine the ASTM grain size number of a metal specimen if 45 grains per square inch are measured at a magnification of 100.
(b)For this same specimen, how many grains per square inch will there be at a magnification of 85?
Solution
(a) In order to determine the ASTM grain size number (n) it is necessary to employ Equation 4.16. Taking logarithms of both sides of this expression leads to
N2n1
4.11 Grain Size Determination • 113
grain size
7 ASTM Standard E 112, “Standard Test Methods for Determining Average Grain Size.”
Relationship between ASTM grain size number and number of grains per square inch (at 100)
Solving for n yields
From the problem statement, N 45, and therefore
(b)At magnifications other than 100, use of the following modified form of Equation 4.16 is necessary:
(4.17) In this expression NM the number of grains per square inch at magnifica- tion M. In addition, the inclusion of the (M/100)2term makes use of the fact that, while magnification is a length parameter, area is expressed in terms of units of length squared. As a consequence, the number of grains per unit area increases with the square of the increase in magnification.
Solving Equation 4.17 for NM, realizing that M 85 and n 6.5, leads to
216.512a100
85 b262.6 grains/in.2 NM2n1a100
M b2 NMa M
100b22n1 n log45
log2 16.5 n logN
log2 1 logN1n12log2
S U M M A R Y
Vacancies and Self-Interstitials
• Point defects are those associated with one or two atomic positions; these include vacancies (or vacant lattice sites) and self-interstitials (host atoms that occupy interstitial sites).
• The equilibrium number of vacancies depends on temperature according to Equation 4.1.
Impurities in Solids
• An alloy is a metallic substance that is composed of two or more elements.
• A solid solution may form when impurity atoms are added to a solid, in which case the original crystal structure is retained and no new phases are formed.
• For substitutional solid solutions, impurity atoms substitute for host atoms.
• Interstitial solid solutions form for relatively small impurity atoms that occupy interstitial sites among the host atoms.
• For substitutional solid solutions, appreciable solubility is possible only when atomic diameters and electronegativities for both atom types are similar, when both elements have the same crystal structure, and when the impurity atoms have a valence that is the same as or less than the host material.
Specification of Composition
• Composition of an alloy may be specified in weight percent (on the basis of mass fraction, Equation 4.3) or atom percent (on the basis of mole or atom fraction, Equation 4.5).
• Expressions were provided that allow conversion of weight percent to atom per- cent (Equation 4.6a) and vice versa (Equation 4.7a).
• Computation of average density and average atomic weight for a two-phase alloy are possible using other equations cited in this chapter (Equations 4.10a, 4.10b, 4.11a, and 4.11b).
Dislocations—Linear Defects
• Dislocations are one-dimensional crystalline defects of which there are two pure types: edge and screw.
An edge may be thought of in terms of the lattice distortion along the end of an extra half-plane of atoms.
A screw is as a helical planar ramp.
For mixed dislocations, components of both pure edge and screw are found.
• The magnitude and direction of lattice distortion associated with a dislocation are specified by its Burgers vector.
• The relative orientations of Burgers vector and dislocation line are (1) perpen- dicular for edge, (2) parallel for screw, and (3) neither perpendicular nor parallel for mixed.
Interfacial Defects
• Within the vicinity of a grain boundary (which is several atomic distances wide), there is some atomic mismatch between two adjacent grains that have different crystallographic orientations.
• For a high-angle grain boundary, the angle of misalignment between grains is rel- atively large; this angle is relatively small for small-angle grain boundaries.
• Across a twin boundary, atoms on one side reside in mirror-image positions of atoms on the other side.
Microscopic Techniques
• The microstructure of a material consists of defects and structural elements that are of microscopic dimensions. Microscopy is the observation of microstructure using some type of microscope.
• Both optical and electron microscopes are employed, usually in conjunction with photographic equipment.
• Transmissive and reflective modes are possible for each microscope type; pref- erence is dictated by the nature of the specimen as well as the structural element or defect to be examined.
• In order to observe the grain structure of a polycrystalline material using an op- tical microscope, the specimen surface must be ground and polished in order to produce a very smooth and mirrorlike finish. Some type of chemical reagent (or etchant) must then be applied in order to either reveal the grain boundaries or produce a variety of light reflectance characteristics for the constituent grains.
• The two types of electron microscopes are transmission (TEM) and scanning (SEM).
For TEM an image is formed from an electron beam that, while passing through the specimen, is scattered and/or diffracted.
SEM employs an electron beam that raster-scans the specimen surface; an image is produced from back-scattered or reflected electrons.
Summary • 115
• A scanning probe microscope employs a small and sharp-tipped probe that raster- scans the specimen surface. Out-of-plane deflections of the probe result from interactions with surface atoms. A computer-generated and three-dimensional image of the surface results having nanometer resolution.
Grain Size Determination
• With the intercept method, used to measure grain size, a series of straight-line segments (all having the same length) are drawn on a photomicrograph. Line length is divided by the average number of grain intersections on a per-line basis.
Average grain diameter is taken as this result divided by the magnification of the photomicrograph.
• Comparison of a photomicrograph (taken at a magnification of 100) with ASTM standard comparison charts may be used to specify grain size in terms of a grain size number.
• The average number of grains per square inch at a magnification of 100is related to grain size number according to Equation 4.16.
Equation Summar y
Equation Page
Number Equation Solving For Number
4.1 Number of vacancies per unit volume 92
4.2 Number of atomic sites per unit volume 93
4.3 Composition in weight percent 95
4.5 Composition in atom percent 96
4.6a Conversion from weight percent to atom percent 96
4.7a Conversion from atom percent to weight percent 96
4.9a Conversion from weight percent to mass per unit volume 97
4.10a Average density of a two-component alloy 97
4.11a Average atomic weight of a two-component alloy 97
4.16 N2n1 Number of grains per in.2 at 100magnification 113
Aave 100 C1
A1
C2
A2
rave 100 C1
r1 C2
r2 C–
1
° C1
C1 r1 C2
r2
¢ 103 C1 C¿1A1
C1¿A1 C2¿A2
100 C¿
1 C1A2
C1A2 C2A1
100 C¿
1 nm1
nm1nm2100 C1 m1
m1m2
100 N NAr
A NyNexpaQy
kTb
List of Symbols
Symbol Meaning
A Atomic weight
k Boltzmann’s constant (1.38 ⫻10⫺23 J/atom K, 8.62 ⫻10⫺5 eV/atom K)
m1, m2 Masses of elements 1 and 2 in an alloy
n ASTM grain size number
NA Avogadro’s number (6.022 ⫻1023atoms/mol) nm1, nm2 Number of moles of elements 1 and 2 in an alloy
Q Energy required for the formation of a vacancy Density
Processing/Structure/Properties/
Performance Summar y
In this chapter we discussed several schemes used to specify concentration of one element in another; equations were also provided to convert from one scheme to an- other. During the processing of silicon to form integrated circuit components (Chap- ters 5 and 18), it is imperative that specification and control of impurity concentration be extremely precise. These relationships are represented in the following diagram:
r
#
#
Summary • 117
Specification of composition (Chapter 4)
Diffusion in semiconductors (Chapter 5)
Fabrication of integrated circuits (Chapter 18) Silicon
Semiconductors (Processing)
Concept of solid solution (Chapter 4)
Solid solution strengthening (Chapter 10) Iron–Carbon Alloys
(Processing)
Concept of dislocation defects
(Chapter 4)
Strengthening mechanisms (Chapter 10) Iron–Carbon Alloys
(Properties)
The concept of a solid solution was also discussed. One form of solid solution in an iron–carbon alloy, or steel (martensite), derives its high strength and hardness from the formation of an interstitial solid solution (carbon dissolved in iron). The following diagram represents this relationship:
With a knowledge of the characteristics of dislocation defects, we are able to understand the mechanisms by which metals [i.e., iron–carbon alloys (steels)] per- manently deform (Chapter 7), and, in addition, techniques that are used to improve the mechanical properties of these materials. The following diagram notes this relationship.
Important Terms and Concepts alloy
atomic vibration atom percent
Boltzmann’s constant Burgers vector composition dislocation line edge dislocation grain size imperfection
interstitial solid solution microscopy
microstructure mixed dislocation photomicrograph point defect
scanning electron microscope (SEM)
scanning probe microscope (SPM)
screw dislocation self-interstitial solid solution solute solvent
substitutional solid solution transmission electron
microscope (TEM) vacancy
weight percent
R E F E R E N C E S
ASM Handbook, Vol. 9, Metallography and Mi- crostructures, ASM International, Materials Park, OH, 2004.
Brandon, D., and W. D. Kaplan, Microstructural Characterization of Materials, 2nd edition, Wiley, Hoboken, NJ, 2008.
Clarke, A. R., and C. N. Eberhardt, Microscopy Techniques for Materials Science, CRC Press, Boca Raton, FL, 2002.
Kelly, A., G. W. Groves, and P. Kidd, Crystallography and Crystal Defects, Wiley, Hoboken, NJ, 2000.
Tilley, R. J. D., Defects in Solids, Wiley-Interscience, Hoboken, NJ, 2009.
Van Bueren, H. G., Imperfections in Crystals, North- Holland, Amsterdam (Wiley-Interscience, New York), 1960.
Vander Voort, G. F., Metallography, Principles and Practice, ASM International, Materials Park, OH, 1984.
Q U E S T I O N S A N D P R O B L E M S
Vacancies and Self-Interstitials
4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom.
4.2 Calculate the number of vacancies per cubic meter in iron at 850C.The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm3(at 850ºC) and 55.85 g/mol, respectively.
4.3 Calculate the activation energy for vacancy formation in aluminum, given that the equi- librium number of vacancies at 500C (773 K) is 7.57 1023m3.The atomic weight and den- sity (at 500C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3.
Impurities in Solids
4.4 Atomic radius, crystal structure, electronega- tivity, and the most common valence are
tabulated in the following table for several el- ements; for those that are nonmetals, only atomic radii are indicated.
Atomic
Radius Crystal Electro-
Element (nm) Structure negativity Valence
Cu 0.1278 FCC 1.9 2
C 0.071
H 0.046
O 0.060
Ag 0.1445 FCC 1.9 1
Al 0.1431 FCC 1.5 3
Co 0.1253 HCP 1.8 2
Cr 0.1249 BCC 1.6 3
Fe 0.1241 BCC 1.8 2
Ni 0.1246 FCC 1.8 2
Pd 0.1376 FCC 2.2 2
Pt 0.1387 FCC 2.2 2
Zn 0.1332 HCP 1.6 2
Which of these elements would you expect to form the following with copper:
Questions and Problems • 119
(a) A substitutional solid solution having complete solubility
(b) A substitutional solid solution of incom- plete solubility
(c) An interstitial solid solution
4.5 For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0 positions—that is, lying on {100}
faces and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site.
For both FCC and BCC crystal structures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.
Specification of Composition
4.6 Derive the following equations:
(a) Equation 4.7a (b) Equation 4.9a (c) Equation 4.10a (d) Equation 4.11b
4.7 What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu?
4.8 What is the composition, in weight percent, of an alloy that consists of 6 at% Pb and 94 at%
Sn?
4.9 Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 14.6 kg aluminum, and 9.7 kg vanadium.
4.10 What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g lead?
4.11 What is the composition, in atom percent, of an alloy that contains 99.7 lbmcopper, 102 lbm
zinc, and 2.1 lbmlead?
4.12 What is the composition, in atom percent, of an alloy that consists of 97 wt% Fe and 3 wt% Si?
4.13 Convert the atom percent composition in Problem 4.11 to weight percent.
1 4 1 2
4.14 Calculate the number of atoms per cubic meter in aluminum.
4.15 The concentration of carbon in an iron–
carbon alloy is 0.15 wt%. What is the con- centration in kilograms of carbon per cubic meter of alloy?
4.16 Determine the approximate density of a high- leaded brass that has a composition of 64.5 wt% Cu, 33.5 wt% Zn, and 2 wt% Pb.
4.17 Calculate the unit cell edge length for an 85 wt% Fe–15 wt% V alloy. All of the vanadium is in solid solution, and at room temperature the crystal structure for this alloy is BCC.
4.18 Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27 and 6.35 g/cm3, respectively, whereas their respec- tive atomic weights are 61.4 and 125.7 g/mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of 0.395 nm.
4.19 For a solid solution consisting of two elements (designated as 1 and 2), sometimes it is desir- able to determine the number of atoms per cubic centimeter of one element in a solid solution, N1, given the concentration of that element specified in weight percent, C1. This computation is possible using the following expression:
(4.18)
where
NA Avogadro’s number
and densities of the two elements A1 the atomic weight of element 1 Derive Equation 4.18 using Equation 4.2 and expressions contained in Section 4.4.
4.20 Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver–gold alloy that contains 10 wt% Au and 90 wt%
Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively.
4.21 Germanium forms a substitutional solid solu- tion with silicon. Compute the number of germanium atoms per cubic centimeter for a
r2 r1
N1 NAC1
C1A1
r1 A1
r21100C12
germanium–silicon alloy that contains 15 wt%
Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm3, respectively.
4.22 Sometimes it is desirable to determine the weight percent of one element, C1, that will produce a specified concentration in terms of the number of atoms per cubic centimeter, N1, for an alloy composed of two types of atoms.
This computation is possible using the fol- lowing expression:
(4.19)
where
NA⫽Avogadro’s number
and ⫽densities of the two elements A1and A2⫽the atomic weights of the two
elements
Derive Equation 4.19 using Equation 4.2 and expressions contained in Section 4.4.
4.23 Molybdenum forms a substitutional solid so- lution with tungsten. Compute the weight per- cent of molybdenum that must be added to tungsten to yield an alloy that contains 1.0 ⫻ 1022Mo atoms per cubic centimeter. The den- sities of pure Mo and W are 10.22 and 19.30 g/cm3, respectively.
4.24 Niobium forms a substitutional solid solution with vanadium. Compute the weight percent of niobium that must be added to vanadium to yield an alloy that contains 1.55 ⫻1022Nb atoms per cubic centimeter. The densities of pure Nb and V are 8.57 and 6.10 g/cm3, respectively.
4.25 Silver and palladium both have the FCC crys- tal structure, and Pd forms a substitutional solid solution for all concentrations at room temperature. Compute the unit cell edge length for a 75 wt% Ag–25 wt% Pd alloy. The room-temperature density of Pd is 12.02 g/cm3, and its atomic weight and atomic radius are 106.4 g/mol and 0.138 nm, respectively.
Dislocations—Linear Defects
4.26 Cite the relative Burgers vector–dislocation line orientations for edge, screw, and mixed dislocations.
r2 r1
C1⫽ 100
1⫹ NAr2 N1A1⫺r2
r1
Interfacial Defects
4.27 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? (Note:
You may want to consult the solution to Prob- lem 3.54 at the end of Chapter 3.)
4.28 For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why?
(Note: You may want to consult the solution to Problem 3.55 at the end of Chapter 3.) 4.29 (a) For a given material, would you expect the
surface energy to be greater than, the same as, or less than the grain boundary energy? Why?
(b) The grain boundary energy of a small- angle grain boundary is less than for a high- angle one. Why is this so?
4.30 (a) Briefly describe a twin and a twin boundary.
(b) Cite the difference between mechanical and annealing twins.
4.31 For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:
(a) . . . A B C A B C B A C B A . . . (b) . . . A B C A B C B C A B C . . .
Now, copy the stacking sequences and indi- cate the position(s) of planar defect(s) with a vertical dashed line.
Grain Size Determination
4.32 (a) Using the intercept method, determine the average grain size, in millimeters, of the spec- imen whose microstructure is shown in Fig- ure 4.14(b); use at least seven straight-line segments.
(b) Estimate the ASTM grain size number for this material.
4.33 (a) Employing the intercept technique, deter- mine the average grain size for the steel spec- imen whose microstructure is shown in Fig- ure 9.25(a); use at least seven straight-line segments.
(b) Estimate the ASTM grain size number for this material.
4.34 For an ASTM grain size of 8,approximately how many grains would there be per square inch
Design Problems • 121 (a) at a magnification of 100, and
(b) without any magnification?
4.35 Determine the ASTM grain size number if 25 grains per square inch are measured at a mag- nification of 600.
4.36 Determine the ASTM grain size number if 20 grains per square inch are measured at a mag- nification of 50.
Spreadsheet Problems
4.1SS Generate a spreadsheet that allows the user to convert the concentration of one element of a two-element metal alloy from weight percent to atom percent.
4.2SS Generate a spreadsheet that allows the user to convert the concentration of one element of a two-element metal alloy from atom per- cent to weight percent.
4.3SS Generate a spreadsheet that allows the user to convert the concentration of one element of a two-element metal alloy from weight percent to number of atoms per cubic centimeter.
4.4SS Generate a spreadsheet that allows the user to convert the concentration of one element of a two-element metal alloy from number of atoms per cubic centimeter to weight percent.
D E S I G N P R O B L E M S
Specification of Composition
4.D1 Aluminum–lithium alloys have been devel- oped by the aircraft industry to reduce the weight and improve the performance of its aircraft. A commercial aircraft skin material having a density of 2.55 g/cm3 is desired.
Compute the concentration of Li (in wt%) that is required.
4.D2 Iron and vanadium both have the BCC crys- tal structure, and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt% V at room tempera- ture. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm.