Fundamentals of Material Science
CHAPTER 3 Structure of Crystalline Materials
Fundamentals of Material Science Dr. Gamal Abdou Crystal Structure Atomic and Ionic Arrangement ISSUES TO ADDRESS... • How do atoms assemble into solid materials? (for now, focus on metals) • How does the density of a material depend on its structure? • When do material properties vary with the sample (i.e., part) orientation? • Arrangement of atoms and ions play an important role in determining the microstructure and properties of a material. • We can find many types of atomic or ionic arrangement • No Order, Short-Range Order and Long-Range Order. Atomic arrangement
Solid
Crystalline Amorphous
Crystalline – periodic arrangement of atoms: definite repetitive pattern Non-crystalline or Amorphous – random arrangement of atoms. The periodicity of atoms in crystalline solids can be described by a network of points in space called lattice. Lattice, basis and unit cell
• A lattice is a collection of points, called lattice points, which are arranged in periodic pattern so that the surroundings of each point in the lattice are identical. • A lattice may be one, two or three dimensional • The arrangement of ions and atoms is called lattice Space lattice
A space lattice can be defined as a three dimensional array of points, each of which has identical surroundings. If the periodicity along a line is a, then position of any point along the line can be obtained by a simple translation, ru = ua.
Similarly ruv = ua + vb will repeat the point along a 2D plane, where u and v are integers.
a
b Crystal Systems The space lattice points in a crystal are occupied by atoms. The position of any atom in the 3D lattice can be described by a vector ruvw = ua + vb + wc, where u, v and w are integers.
(a) (b)
Unit Cell The three unit vectors, a, b, c can define a cell as shown by the shaded region in Fig.(a) This cell is known as unit cell (Fig. b) which when repeated in the three dimensions generates the crystal structure. Crystal Systems Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal.
7 crystal systems
14 crystal lattices known as Bravais lattices a, b, and c are the lattice constants
Fig. 3.4, Callister
7e. Crystal Systems
Bravais Lattice
The unit vectors a, b and c are called lattice parameters. Based on their length equality or inequality and their orientation (the angles between them, , and ) a total of 7 crystal systems can be defined. With the centering (face, base and body centering) added to these, 14 kinds of 3D lattices, known as Bravais lattices, can be generated. Crystal Systems Cubic: a = b = c, = = = 90o
Simple Body-centered Face-centered cubic cubic (BCC) cubic (FCC) Tetragonal: a = b c, = = = 90o
Simple Body-centered Tetragonal Tetragonal (BCT) Crystal Systems
Orthorhombic: a b c, = = = 90o
Simple Body-centered Base-centered Face-centered
Monoclinic: a b c, = = 90o
Simple Base-centered monoclinic monoclinic Crystal Systems
Triclinic Rhombohedral Hexagonal a b c a = b = c a = b c 90o = = 90o = = 90o = 120o Crystal Systems
Crystal system Example
Triclinic K2S2O8,K2Cr2O7 Monoclinic As4S4, KNO2,CaSO4.2H2O, -S Rhombohedral Hg, Sb, As, Bi, CaCO3 Hexagonal Zn, Co, Cd, Mg, Zr, NiAs
Orthorhombic Ga, Fe3C, -S Tetragonal In, TiO2, -Sn Cubic Au, Si, Al, Cu, Ag, Fe, NaCl Cubic crystal system
• Face-centered cubic (FCC), and Body-centered cubic (BCC) • For illustration see Figure 3-7 page 57 • A specific number of lattice points define each of the unit cell. • Determining the number of lattice points in cubic crystal systems can be found • In BCC unit cell, lattice points are located at the corner and the center of the cube: (Lattice point/unit cell)= (8 corners)(1/8)+ (1center)(1)=2
Adapted from Fig. 3.2, Callister 7e. • In FCC unit cell, lattice points are located at the corner and the faces of the cube: (Lattice point/unit cell)= (8 corners)(1/8)+ (6 faces)(1/2)=4
Adapted from Fig. 3.1, Callister 7e. Coordination number
• Coordination number is the number of atoms touching a particular atom, or the number of nearest neighbors for that particular atom. Body Centered Cubic Structure (BCC)
• Atoms touch each other along cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. ex: Cr, W, Fe (), Tantalum, Molybdenum, Sodium, Niobium
• Coordination # = 8
Adapted from Fig. 3.2, Callister 7e.
(Courtesy P.M. Anderson) 2 atoms/unit cell: 1 center + 8 corners x 1/8 Atomic Packing Factor: BCC
3 a
a
2 a R Close-packed directions: a length = 4R = 3 a Adapted from Fig. 3.2(a), Callister 7e. atoms 4 volume unit cell 2 p ( 3a/4)3 3 atom APF = volume a3 unit cell Face Centered Cubic Structure (FCC)
Adapted from Fig. 3.1, Callister 7e.
--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.
ex: Al, Cu, Au, Pb, Ni, Pt, Ag, (g) Fe
• Coordination # = 12
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8 Atomic Packing Factor: FCC
maximum achievable APF Close-packed directions: 2 a length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 a = 4 atoms/unit cell
Adapted from atoms volume Fig. 3.1(a), 4 3 Callister 7e. unit cell 4 p ( 2a/4) 3 atom APF = volume a3 unit cell Hexagonal Close-Packed (HCP)
Ex: Zn, Ti, Mg, Co, Be, Zr • Coordination # = 12
Unit cell contains: 2 x 1/2 + 12 x 1/6 + 3 = 6 atoms/unit cell Atomic Packing Factor
• APF for a BCC cubic structure = 0.68
• APF for a FCC cubic structure = 0.74
• APF for a HCP Hexagonal str. = 0.74
Examples
Ex. 1: Theoretical density calculation from crystal structure. nA Theoretical density, V N C A n = number of atoms in the unit cell A = atomic weight
VC = volume of unit cell 23 NA = Avogadro’s number (6.023 x 10 atoms/mol) Calculate the theoretical density of Al. Al is FCC, lattice parameter, a = 4.05 Å, n = 4. Atomic weight of Al is 26.98 g/mol . g 4 26 98 . / cc . . 8 3 23 2 697 ︵ 4 0510 ︶ 6 02310 Atoms positions in atomic cells
• To locate atom position in cubic unit cells, we use rectangular x, y, and z axes. • In crystallography the positive y axis is usually the direction coming out of the paper, the positive y axis is the direction to the right of the paper, and the positive z axis is the direction to the top. • Negative directions are opposite to those just described. z 111
c
y 000 a b x
Crystallographic Directions z Algorithm 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a, b, and c y 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas x [uvw]
ex: 1, 0, ½ => 2, 0, 1 => [ 201 ] -1, 1, 1 => [ 111] where over-bar represents a negative index families of directions
1. Find the coordinate points of head and tail points. 2. Subtract the coordinate points of the tail from the coordinate points of the head. 3. Remove fractions. 4. Enclose in [ ] Indecies of Crystallographic Directions in Cubic System Direction A Head point – tail point (1, 1, 1/3) – (0,0,2/3) 1, 1, -1/3 Multiply by 3 to get smallest integers 3, 3, -1 A = [33Ī]
Direction B Head point – tail point (0, 1, 1/2) – (2/3,1,1) -2/3, 0, -1/2 Multiply by 6 to get smallest integers; _ _ C = [???] D = [???] B = [403] Miller indices of direction
• Because directions are vectors, a direction and its negative are not identical [100] is not equal to [-100] • A direction and its multiple are identical • The special brackets <> are used to indicate this collection of directions Crystallographic Planes
• Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.
• Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl) Crystallographic Planes z example a b c 1. Intercepts 1 1 c 2. Reciprocals 1/1 1/1 1/
3. Reduction 1 1 0 y a b 4. Miller Indices (110) x z example a b c 1. Intercepts 1/2 c 2. Reciprocals 1/½ 1/ 1/
3. Reduction 2 0 0 a y 4. Miller Indices (200) b x Crystallographic Planes z
example a b c c 1. Intercepts 1/2 1 3/4 2. Reciprocals 1/½ 1/1 1/¾ 2 1 4/3 y 3. Reduction 6 3 4 a b 4. Miller Indices (634) x
Family of Planes {hkl}
Ex: {100} = (100), (010), (001), (100), (010), (001) Miller indices for planes
• Plane and their negatives are identical (020)=(020) • Plane and their multiple are not identical • We represent the similar group of planes with the notation { } • In cubic system, a direction that has the same indices as the a plane is perpendicular to that plane Crystallographic Planes
Adapted from Fig. 3.9, Callister 7e.
It is very easy to visualize the location of a simple face given miller index, or to derive a miller index from simple faces
Fig. 2-23 Single vs Polycrystals • Single Crystals E (diagonal) = 273 GPaData from Table 3.3, -Properties vary with Callister 7e. (Source of data is R.W. direction: anisotropic. Hertzberg, Deformation and Fracture Mechanics -Example: the modulus of Engineering Materials, 3rd ed., John of elasticity (E) in BCC iron: Wiley and Sons, 1989.)
• Polycrystals E (edge) = 125 GPa -Properties may/may not 200 mm Adapted from Fig. 4.14(b), Callister 7e. vary with direction. (Fig. 4.14(b) is courtesy of L.C. Smith and C. -If grains are randomly Brady, the National Bureau of Standards, oriented: isotropic. Washington, DC [now the National Institute of (Epoly iron = 210 GPa) Standards and Technology, -If grains are textured, Gaithersburg, MD].) anisotropic. Crystals as Building Blocks • Some engineering applications require single crystals: --diamond single --turbine blades crystals for abrasives Fig. 8.33(c), Callister 7e. (Fig. 8.33(c) courtesy (Courtesy Martin Deakins, of Pratt and Whitney). GE Superabrasives, Worthington, OH. Used with permission.)
• Properties of crystalline materials often related to crystal structure. --Ex: Quartz fractures more easily along some crystal planes than others.
(Courtesy P.M. Anderson) Examples
Ex. 4: Draw the directions [236] and [203] and [21 1] in a cubic unit cell.
[211]
c/2
-b/2
Ex. 5: The atomic radius of Fe is 0.124 nm. Find the lattice parameter of Fe. Solution: Fe is BCC at roomR temperature.. Therefore, a R a 4 4 0 124 . 3 4 and 0 286 nm 3 3 Quiz 8. What is coordination number (CN)? Show that CN for FCC and HCP structure is 12 while it is 8 for BCC. 9. Show that packing efficiency of FCC is 74% and that of BCC is 68%. 10. Show that the ideal c/a ratio in a hexagonal unit cell is 1.633 and calculate the packing efficiency. 11. What are the coordinates of the center atom in the BCC unit cell. 12. What is miller index? How is it obtained? 13. Draw the planes (11 0), (12 1), (23 4), (1 12 ) and directions[1 11], [123], [120], [1 2 1] in a cubic unit cell. 14. Why it is necessary to include a fourth miller index in the hexagonal system? 15. Convert the directions [112], [12 3], [110], [111], [130] to four indices in a hexagonal lattice. Quiz 16. What is family of planes? Draw the {111} family of planes in cubic system? 17. What is linear density? What is planar density? 18. Find the planar of density {111} planes and linear density of <110> directions in FCC system. 19. What is the linear density of <111> directions in the BCC crystal. 20. What is interplanar spacing? Find the interplanar spacing of the vertical planes in the HCP system? 21. What is the stacking sequence of FCC and HCP crystals? 22. What is slip system? 23. Why FCC metals are ductile while BCC and HCP metals are not? 24. Calculate the theoretical density of Cu from its crystal structure. Quiz
25. Lattice constant of Al is 4.05 Å. What is the atomic radius of Al? 26. Calculate the theoretical density of Mg, Cu and Fe and compare them to the standard values. 27. A metal has a density of 10.22 g/cc, atomic weight of 95.94 and atomic radius of 0.136 nm. Is it BCC or FCC? 28. Calculate the volume of the unit cell of Zn crystal. a and c of Zn are 266.5 pm and 494.7 pm respectively. 29. Calculate the planar density of {110} planes in -Fe (BCC) crystal. a = 0.287 nm. 30. Calculate the linear density of [110] direction in a Cu crystal. a = 0.361 nm.