The Structure of Materials
1.1 STRUCTURE OF METALS AND ALLOYS
1.1.1 Crystal Structures
Our description of atomic packing leads naturally into crystal structures. While some of the simpler structures are used by metals, these structures can be employed by heteronuclear structures, as well. We have already discussed FCC and HCP, but there are 12 other types of crystal structures, for a total of 14 space lattices or Bravais lattices. These 14 space lattices belong to more general classifications called crystal systems, of which there are seven.
STRUCTURE OF METALS AND ALLOYS 31
β γ
α
a
b c
y z
x
Figure 1.19 Definition of a coordinate system for crystal structures.
Before describing each of the space lattices, we need to define a coordinate system.
The easiest coordinate system to use depends upon which crystal system we are looking at. In other words, the coordinate axes are not necessarily orthogonal and are defined by the unit cell. This may seem a bit confusing, but it simplifies the description of cell parameters for those systems that do not have crystal faces at right angles to one another. Refer to Figure 1.19. For each crystal system, we will define the space lattice in terms of three axes,x, y, andz, with interaxial anglesα, β, γ. Note that the interaxial angleαis defined by the angle formed between axeszandy, and also note that angles β and γ are defined similarly. Only in special circumstances are α, β, γ equal to 90◦. The distance along they axis from the origin to the edge of the unit cell is called thelattice translation vector,b. Lattice translation vectorsaandcare defined similarly along the axesx andz, respectively. The magnitudes (lengths) of the lattice translation vectors are called thelattice parameters,a,b, andc. We will now examine each of the seven crystal systems in detail.
1.1.1.1 Crystal Systems. The cubic crystal system is composed of three space lattices, or unit cells, one of which we have already studied:simple cubic(SC),body- centered cubic (BCC), andface-centered cubic (FCC). The conditions for a crystal to be considered part of the cubic system are that the lattice parameters be the same (so there is really only one lattice parameter,a) and that the interaxial angles all be 90◦.
The simple cubic structure, sometimes called the rock salt structure because it is the structure of rock salt (NaCl), is not a close-packed structure (see Figure 1.20). In fact, it contains about 48% void space; and as a result, it is not a very dense structure.
The large space in the center of the SC structure is called an interstitial site, which is a vacant position between atoms that can be occupied by a small impurity atom or alloying element. In this case, the interstitial site is surrounded by eight atoms. All eight atoms in SC are equivalent and are located at the intersection of eight adjacent unit cells, so that there are 8×(1/8)=1 total atoms in the SC unit cell. Notice that
Crystal Structure
Lattice Parameters
Interaxial Angles
Simple Body-centered Face-centered
Simple Base-centered Body-centered Face-centered
Simple Body-centered
Simple Base-centered
Cubic a = b = g =
a = b = c 90°
Orthorhombic a = b = g =
90° a ≠ b ≠ c
Rhombohedral a = b = c a = b = g ≠ 90°, < 120 °
Tetragonal a = b ≠ c a = b = g = 90°
Monoclinic a ≠ b ≠ c a = g = 90°, b ≠ 90°
a = b = g ≠ 90° Triclinic a ≠ b ≠ c
a = b = 90°, g = 120° Hexagonal a = b, a ≠ c
Figure 1.20 Summary of the 14 Bravais space lattices.
STRUCTURE OF METALS AND ALLOYS 33
Figure 1.20 shows the atoms as points, but the atoms actually occupy a larger space than that. In fact, for SC, the atoms touch along the edge of the crystal.
Body-centered cubic (BCC) is the unit cell of many metals and, like SC, is not a close-packed structure. The number of atoms in the BCC unit cell are calculated as follows:
1×1 =1 center atom 8×(1/8)=1 corner atom 2 total atoms
Finally, face-centered cubic (FCC) has already been described (Figure 1.18). Even though FCC is a close-packed structure, there are interstitial sites, just as in SC. There are actually two different types of interstitial sites in FCC, depending on how many atoms surround the interstitial site. A group of four atoms forms atetrahedral interstice, as shown in Figure 1.21. A group of six atoms arranged in an octahedron (an eight-sided geometric figure), creates anoctahedral interstice(Figure 1.22). Figure 1.23 shows the locations of these interstitial sites within the FCC lattice. Note that there are eighttotal tetrahedral interstitial sites in FCC and there are four total octahedral interstitial sites in FCC (prove it!), which are counted in much the same way as we previously counted the total number of atoms in a unit cell. We will see later on that these interstitial sites play an important role in determining solubility of impurities and phase stability of alloys.
Interstitial sites are the result of packing of the spheres. Recall from Figure 1.18 that the spheres touch along the face diagonal in FCC. Similarly, the spheres touch along the body diagonal in BCC and along an edge in SC. We should, then, be able to calculate the lattice parameter,a, or the length of a face edge, from a knowledge of the sphere radius. In SC, it should be evident that the side of a unit cell is simply 2r. Application of a little geometry should prove to you that in FCC,a=4r/√
2. The relationship between a and r for BCC is derived in Example Problem 1.4; other geometric relationships, including cell volume for cubic structures, are listed in Table 1.8. Finally, atomic radii for the elements can be found in Table 1.9. The radius of an atom is not an exactly defined quantity, and it can vary depending upon the bonding environment in which
Figure 1.21 A tetrahedral interstice. From W. D. Kingery, H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics. Copyright1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.
Figure 1.22 An octahedral interstice. From W. D. Kingery, H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics. Copyright1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.
Figure 1.23 Location of interstitial sites in FCC. From W. D. Kingery, H. K. Bowen, and D. R. Uhlmann,Introduction to Ceramics. Copyright1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.
STRUCTURE OF METALS AND ALLOYS 35
Example Problem 1.4
Molybdenum has a BCC structure with an atomic radius of 1.36 ˚A. Calculate the lattice parameter for BCC Mo.
H
G
a
a a b
C A
E
B
E
A
G
C 2r a
r
r
a √2 Reprinted, by permission, from Z. Jastrzebski, The Nature and Properties of Engineering Materials, p. 47, 2nd ed. Copyright 1976, John Wiley & Sons, Inc.
Answer: We know that the molybdenum atoms touch along the body diagonal in BCC, as shown in the projec- tion at right. The length of the body diagonal, then, is 4r, and is related to the lattice parameter, a (which is the length of the cube edge, not the length of the face diagonal, which is a√
2) by
application of the Pythagorean theorem:
(4r)2=a2+(a√
2)2=3a2 a=4r/√
3=4(1.36)/√ 3=3.14
The lattice parameter for BCC Mo is 3.14 ˚A, which is consistent with the value in Table 1.11.
Table 1.8 Summary of Important Parameters in the Cubic Space Lattices
Simple Cubic Face-Centered Cubic Body-Centered Cubic
Unit cell side,a 2r 4r/√
2 4r/√
3
Face diagonal √
2(2r) 4r √
2/3(4r)
Body diagonal √
3(2r) √
3/2(4r) 4r
Number of atoms 1 4 2
Cell volume 8r3 32r3
√2
64r3 3√
3 r=atomic radius.
it finds itself. As a result, three types of radii are listed for each element in Table 1.9:
an atomic radius of an isolated atom, an ionic radius, and ametallic radius. Just as in Figure 1.2 for electronic structure, there are some important trends in the atomic radii. The atomic radius tends to increase as one goes down the column in a series.
This is due to the addition of energy levels and more electron density. Radii tend to decrease as we move across a row, because there is less shielding from inner electrons
Table1.9Atomic,Ionic,andMetallicRadiioftheElements 1 Pb4−2.15Mn2+0.91Mn3+0.70Fe2+0.87 Pb2+1.32Pt4+0.55Co2+0.82W6+0.65 S6+0.34Rh4+0.65Mo4+0.68Cr3+0.64 Te4+0.89V3+0.65V4+0.61Sn4−2.15 Si4−1.98Nb4+0.74Tl+1.49 Ti3+0.69Ti2+0.76Se6+0.30–0.04
2 HManyelementshavemultiplevalencestates.Additionalionicradiiarelisted below.He Atomic0.462+3+4+5+2−1−— Ionic1.54ionsionsionsionsionsions— Metallic—— 345678910 LiBeBCNOFNe Atomic1.521.140.970.770.710.6—1.60 Ionic0.780.540.2<0.20.11.321.33— Metallic1.230.890.81————— 1112131415161718 NaMgAlSiPSClAr Atomic1.861.603+4+5+6+4+4+3+2+1+2+1.431.171.091.061.071.92 Ionic0.980.78ionsionsionsionsionsionsionsionsionsions0.570.390.31.741.81— Metallic1.571.361.251.17———— 192021222324252627282930313233343536 KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr Atomic2.311.971.601.471.321.251.121.241.251.251.281.331.351.221.251.161.191.97 Ionic1.331.060.82+0.640.45+0.30.520.673+0.650.780.960.830.620.440.691.911.96— Metallic2.031.741.441.321.221.181.171.171.161.151.171.251.251.22———— 373839404142434445464748495051525354 RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe Atomic2.512.151.811.581.431.36—1.341.341.371.441.501.571.581.611.431.362.18 Ionic1.491.271.060.890.740.65—0.650.680.501.131.030.910.740.93+0.892.20— Metallic2.161.911.621.451.341.301.271.251.251.281.341.411.501.40———— 555657727374757677787980818283848586 CsBaLaHfTaWReOsIrPtAuHgTIPbBiPoAtRn Atomic2.652.171.871.591.471.371.381.351.351.381.441.501.711.751.821.40—— Ionic1.651.31.220.840.680.680.720.670.664+0.521.371.121.490.841.23+0.76+0.67+— Metallic2.351.981.691.441.341.301.281.261.271.301.341.441.551.54———— Source:MaterialsScienceandEngineeringHandbook;Pauling,NatureoftheChemicalBond.Allvaluesinangstroms,˚ A(1˚ A=0.1nm).
STRUCTURE OF METALS AND ALLOYS 37
and the outer-core electrons are drawn more tightly toward the nucleus. There are some notable exceptions, or “jumps,” in the row trend (why?). In general, the ionic radius is much smaller for positive ions and much larger for negative ions than the corresponding isolated atom (why?), and it follows the same general trend as for the isolated atoms. For the discussion of elemental, crystalline solids, the metallic radius is most appropriate. We will find later that the ionic values will be equally important for heteronuclear structures. There are other types of radii, such as covalent radii and van der Waals radii. The former is highly dependent upon the type of covalent bond.
For example, a carbon atom in a carbon–carbon single bond has a covalent radius of 1.54 ˚A, whereas the same atom in a carbon–carbon triple bond is only 1.35 ˚A.
Continuing with our survey of the seven crystal systems, we see that thetetragonal crystal system is similar to the cubic system in that all the interaxial angles are 90◦. However, the cell height, characterized by the lattice parameter, c, is not equal to the base, which is square (a=b). There are two types of tetragonal space lattices:
simple tetragonal, with atoms only at the corners of the unit cell, and body-centered tetragonal, with an additional atom at the center of the unit cell.
Orthorhombic crystals are similar to both tetragonal and cubic crystals because their coordinate axes are still orthogonal, but now all the lattice parameters are unequal.
There are four types of orthorhombic space lattices:simple orthorhombic, face-centered orthorhombic, body-centered orthorhombic, and a type we have not yet encountered, base-centered orthorhombic. The first three types are similar to those we have seen for the cubic and tetragonal systems. The base-centered orthorhombic space lattice has a lattice point (atom) at each corner, as well as a lattice point only on the top and bottom faces (called basal faces). All four orthorhombic space lattices are shown in Figure 1.20.
There is only one space lattice in the rhombohedral crystal system. This crystal is sometimes called hexagonal R or trigonal R, so don’t confuse it with the other two similarly-named crystal systems. The rhombohedral crystal has uniform lattice parameters in all directions and has equivalent interaxial angles, but the angles are nonorthogonal and are less than 120◦.
The crystal descriptions become increasingly more complex as we move to themon- oclinicsystem. Here all lattice parameters are different, and only two of the interaxial angles are orthogonal. The third angle is not 90◦. There are two types of monoclinic space lattices: simple monoclinic and base-centered monoclinic. The triclinic crystal, of which there is only one type, has three different lattice parameters, and none of its interaxial angles are orthogonal, though they are all equal.
Finally, we revisit the hexagonal system in order to provide some additional details.
The lattice parameter and interaxial angle conditions shown in Figure 1.20 for the hexagonal cell refer to what is called the primitive cell for the hexagonal crystal, which can be seen in the front quadrant of the extended cell in Figure 1.17. The primitive hexagonal cell has lattice points only at its corners and has one atom in the center of the primitive cell, for a basis of two atoms. A basis is a unit assembly of atoms identical in composition, arrangement, and orientation that is placed in a regular manner on the lattice to form a space lattice. You should be able to recognize that there are three equivalent primitive cells in the extended HCP structure. The HCPextended cell, which is more often used to represent the hexagonal structure, contains a total of six atoms, as we calculated earlier. In the extended structure, the ratio of the height of
Table 1.10 Axial Ratios for Some HCP Metals
Metal c/a
Be, Y 1.57
Hf, Os, Ru, Ti 1.58
Sc, Zr 1.59
Tc, Tl 1.60
La 1.61
Co, Re 1.62
Mg 1.63
Zn 1.85
Cd 1.89
Ideal (sphere packing) 1.633
the cell to its base,c/a, is called the axial ratio. Table 1.10 lists typical values of the axial ratio for some common HCP crystals.
A table of crystal structures for the elements can be found in Table 1.11 (excluding the Lanthanide and Actinide series). Some elements can have multiple crystal struc- tures, depending on temperature and pressure. This phenomenon is called allotropy and is very common in elemental metals (see Table 1.12). It is not unusual for close- packed crystals to transform from one stacking sequence to the other, simply through a shift in one of the layers of atoms. Other common allotropes include carbon (graphite at ambient conditions, diamond at high pressures and temperature), pure iron (BCC at room temperature, FCC at 912◦C and back to BCC at 1394◦C), and titanium (HCP to BCC at 882◦C).
1.1.1.2 Crystal Locations, Planes, and Directions. In order to calculate such important quantities as cell volumes and densities, we need to be able to specify loca- tions and directions within the crystal.Cell coordinates specify a position in the lattice and are indicated by the variablesu,v,w, separated by commas with no brackets:
u distance along the lattice translation vectora v distance along the lattice translation vectorb w distance along the lattice translation vectorc
HISTORICAL HIGHLIGHT On warming, gray (orα) tin, with a cubic
structure changes at 13.2◦C into white (or β) tin, the ordinary form of the metal, which has a tetragonal structure. When tin is cooled below 13.2◦C, it changes slowly from white to gray. This change is affected by impuri- ties such as aluminum and zinc and can be prevented by small additions of antimony or
bismuth. The conversion was first noted as growths on organ pipes in European cathe- drals, where it was thought to be the devils work. This conversion was also speculated to be caused by microorganisms and was called
“tin plague” or “tin disease.”
Source: www.webelements.com/
webelements/elements/text/key/Sn.html
Table1.11CommonCrystalStructures,Densities,andLatticeParametersoftheElements 12 HHe Struct.hcpfcc ρ,g/ccn/an/a a,˚ A
4.704.24 345678910 LiBeBCNOFNe Struct.bcchcprhomdiamhcpmonmonfcc ρ,g/cc0.5331.852.473.51n/an/an/an/a a,˚ A
3.512.295.063.563.865.405.504.42 1112131415161718 NaMgAlSiPSClAr Struct.bcchcpfccdiamtriclorthorthfcc ρ,g/cc0.9661.742.702.331.822.09n/an/a a,˚ A
4.293.214.055.4311.510.46.225.26 192021222324252627282930313233343536 KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr Struct.bccfcchcphcpbccbcccubicbcchcpfccfcchcporthdiamrhommonorthfcc ρ,g/cc0.8621.532.994.516.097.197.477.878.88.918.937.135.915.325.784.81n/an/a a,˚ A
5.335.563.312.953.032.888.912.862.513.523.612.664.525.653.769.056.725.71 373839404142434445464748495051525354 RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe Struct.bccfcchcphcpbccbcchcphcpfccfccfcchcptetragtetragtrigtrigorthfcc ρ,g/cc1.532.584.486.518.5810.2211.512.3612.4212.010.58.657.297.296.696.254.95n/a a,˚ A
5.596.083.653.613.303.142.742.713.803.884.082.983.255.834.314.467.186.20 555657727374757677787980818283848586 CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn Struct.bccbcchcphcpbccbcchcphcpfccfccfccrhomhcpfccrhomcubn/an/a ρ,g/cc1.913.596.1713.316.719.321.022.5822.5521.419.28n/a11.8711.349.809.2n/an/a a,˚ A
6.145.013.773.203.303.172.762.733.843.924.073.013.464.954.743.36n/an/a Notethatmanyelementshaveallotrones.Citedstructuresaregenerallythemoststable. Source:MaterialsScienceandEngineeringHandbook;Kittel,SolidStatePhysics;http://www.webelements.com/
39
Table 1.12 Some Metal Allotropes
Metal
R.T. Crystal Structure
Structure at Other Temperatures
Ca FCC BCC(>447◦C)
Co HCP FCC(>427◦C)
Hf HCP BCC(>1742◦C)
Fe BCC FCC(>912◦C)
BCC(>1394◦C)
Li BCC BCC(<−193◦C)
Na BCC BCC(<−233◦C)
Sn BCT Cubic(<13◦C)
Tl HCP BCC(>234◦C)
Ti HCP BCC(>883◦C)
Y HCP BCC(>1481◦C)
Zr HCP BCC(>872◦C)
For example, the center atom in the BCC space lattice (see Figure 1.20) has cell coordinates of 1/2, 1/2, 1/2. Any two points are equivalent if the fractional portions of their coordinates are equal:
1/2,1/2,1/2≡ −1/2,−1/2,−1/2(center) 0,0,0≡1,0,1(corner)
A cell direction is designated by the vector r, which is a combination of the lattice translation vectorsa,b, andc:
r=ua+vb+wc (1.27)
A direction can also be specified with the cell coordinates in square brackets, with commas and fractions removed:
[1 1 1]≡[1/2 1/2 1/2]≡1/2a+1/2b+1/2c
Negative directions are indicated by an overbar [1-1] and are called the “one negative one one” direction. All directions are relative to the origin where the three lattice translation vectors originate (see Figure 1.19).
Thecell volume,V, can be calculated using the lattice translation vectors:
V = |a×b·c| (1.28)
Mathematically, this is a triple scalar product and can be used to calculate the volume of any cell, with only a knowledge of the lattice translation vectors. If the lattice parameters and interaxial angles are known, the following expression for V can be derived from the vector expression:
V =abc(1−cos2α−cos2β−cos2γ+2 cosαcosβcosγ )1/2 (1.29)
STRUCTURE OF METALS AND ALLOYS 41
This looks complicated, but for orthogonal systems (α, β, γ =90◦) it reduces quite nicely to the expected expression:
V =abc (1.29a)
Now that we know how to find the cell volume, we can use some previous infor- mation to calculate an important property of a material, namely, itsdensity, which we represent with the lowercase Greek letter rho,ρ. For example, aluminum has an FCC space lattice. Recall that there are four atoms in the FCC unit cell. We know that each aluminum atom has an atomic weight of 27 g/mol. From Table 1.11, the cubic lattice parameter for aluminum is 4.05 ˚A, or 0.405 nm (4.05×10−8 cm). This gives us a volume ofa3=6.64×10−23cm3. You should confirm that the theoretical density for aluminum is then:
ρ = (4 atoms/unit cell)(27 g/mol)
(6.02×1023atoms/mol)(6.64×10−23 cm3) =2.70 g/cm3 per unit cell
Cooperative Learning Exercise 1.2
The actinide-series element protactinium(Pa, AW=231.04)has a body-centered tetrago- nal structure with cell dimensionsa=0.3925 nm, c=0.3238 nm.
Person 1: Determine the weight, in grams, of a single unit cell of Pa.
Person 2: Calculate the volume of single Pa unit cell.
Combine your answers appropriately to arrive at the density.
Answer :15.38
3 g/cm
In addition to cell coordinates and directions,crystal planes are very important for the determination and analysis of structure. We begin with the cell’s coordinate system, with axes x, y, and z. Recall that the axes are not necessarily orthogonal and that a, b, and care the lattice parameters. Look at Figure 1.24. The equation of an arbitrary plane with intercepts A, B, and C, relative to the lattice parameters is given by
1 A
x a + 1
B y b + 1
C z
c =1 (1.30)
We designate a plane by Miller indices, h, k, andl, which are simply the reciprocals of the intercepts,A,B, andC:
1 A
1 B
1 C
=(hkl) (1.31)
Note that the Miller indices are enclosed in parentheses andnotseparated by commas.
Miller indices are determined as follows:
ž Remove all indeterminacy; that is, the planes should have nonzero intercepts.
ž Find intercepts along three axes of the crystal system.
ž Take the reciprocals of the intercepts in terms ofa,b, andc.
O g b
a C
A
x
y z
D B
Figure 1.24 Definition of Miller indices for an arbitrary plane (shaded area). From Z. Jastrzebski,The Nature and Properties of Engineering Materials, 2nd ed. Copyright1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.
ž Multiply reciprocals by the smallest integer necessary to convert them into a set of integers.
ž Enclose resulting integers in parentheses, (hkl), without commas.
Any planes that have common factors are parallel. For example, a (222) and a (111) plane are parallel, as are (442) and (221) planes. As with cell directions, a minus sign (in this case, indicating a negative intercept) is designated by an overbar. The (221) plane has intercepts at 1/2,−1/2, and 1 along thex,y, andzaxes, respectively. Some important planes in the cubic crystal system are shown in Figure 1.25.
In a manner similar to that used to calculate the density of a unit cell, we can calcu- late the density of atoms on a plane, orplanar density. The perpendicular intersection of a plane and sphere is a circle, so the radius of the atoms will be helpful in calculat- ing the area they occupy on the plane. Refer back to Example Problem 1.4 when we calculated the lattice parameter for a BCC metal. The section shown along the body diagonal is actually the (110) plane. The body-centered atom is entirely enclosed by this plane, and the corner atoms are located at the confluence of four adjacent planes, so each contributes 1/4 of an atom to the (110) plane. So, there are a total of two atoms on the (110) plane. If we know the lattice parameter or atomic radius, we can calculate thearea of the plane,Ap, thearea occupied by the atoms, Ac, and the corresponding