MSE 3300-Lecture Note 05-Chapter 03 the Structure of Crystalline Solids 2
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Transcript of MSE 3300-Lecture Note 05-Chapter 03 the Structure of Crystalline Solids 2
MSE 3300 / 5300 UTA Fall 2014 Lecture 5 -
Lecture 5. The Structure of Crystalline Solids (2)
Learning Objectives After this lecture, you should be able to do the following:
1. Given point coordinates within a unit cell, locate the point; given the location of a point, specify its point coordinates. 2. Given index integers for a direction within a unit cell, draw the direction; given a direction, specify its direction indices. 3. Given the indices for a plane within a unit cell, draw the plane; given a plane, specify its indices.
Reading • Chapter 3: The Structure of Crystalline Solids (3.8–3.17)
Multimedia • Virtual Materials Science & Engineering (VMSE):
http://www.wiley.com/college/callister/CL_EWSTU01031_S/vmse/
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Crystallographic Points, Directions, and Planes
• How to specify a particular point within a unit cell, a crystallographic
direction, or some crystallographic plane of atoms? • Use indices (e.g., three number) • Right-handed coordinate system
1. Points 2. Directions 3. Planes
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1. Points: Point Coordinates
• Specify a point (lattice position) within a unit cell
• Use three point coordinate indices: q, r, and s
• These indices are fractional multiples of a, b, and c unit cell edge lengths
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Point Coordinates Point coordinates for unit cell
center are
a/2, b/2, c/2 ½ ½ ½
Point coordinates for unit cell
corner are 111 Translation: integer multiple of
lattice constants identical position in another unit cell
z
x
y a b
c
000
111
y
z 2c
b
b
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Crystallographic Directions
1. Determine coordinates of vector tail, pt. 1: x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2. 2. Tail point coordinates subtracted from head point coordinates. 3. Normalize coordinate differences in terms of lattice parameters a, b, and c:
4. Adjust to smallest integer values 5. Enclose in square brackets, no commas
[uvw] ex: pt. 1 x1 = 0, y1 = 0, z1 = 0
=> 1, 0, 1/2
=> [ 201 ]
z
x
Algorithm
y
=> 2, 0, 1
pt. 2 head pt. 1:
tail
pt. 2 x2 = a, y2 = 0, z2 = c/2
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Example Problem: Specification of Point Coordinate Indices
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Example Problem: Specification of Point Coordinate Indices
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2. Crystallographic Directions • Crystallographic direction is defined as a line directed between two points,
or a vector. 1. Construct a right-handed x-y-z coordinate system 2. Determine the coordinates of two points that lie on the direction vector (referenced to
the coordinate system)—Point 1 (vector tail): x1, y1, z1; Point 2 (vector head): x2, y2, z2 for the vector head
3. Subtract tail point coordinates from head point coordinates: x2-x1, y2-y1, and z2-z1. 4. Normalize (i.e., divide) the coordinate differences by using their respective lattice
parameters (a, b, and c): (x2-x1)/a, (y2-y1)/b, (z2-z1)/c. 5. If necessary, multiply or divide these three numbers by a common factor to reduce
them to the smallest integer values. 6. Enclose the three resulting indices in square brackets without comma: [uvw]. The u,
v, and w integers are the normalized coordinate differences referenced to the x, y, and z axes, respectively.
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Crystallographic Directions
-4, 1, 2
families of directions <uvw>
z
x
where the overbar represents a negative index
[ 412 ] =>
y
Example 2: pt. 1 x1 = a, y1 = b/2, z1 = 0 pt. 2 x2 = -a, y2 = b, z2 = c
=> -2, 1/2, 1
pt. 2 head
pt. 1: tail
Multiplying by 2 to eliminate the fraction
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Crystallographic Directions: [100], [110], and [111] directions
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Determine the Indices for the Direction
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Determine the Indices for the Direction
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Construct a Specified Crystallographic Direction
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]011[Draw a direction with its tail located at the origin of the coordinate system.
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Construct a Specified Crystallographic Direction
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]011[Draw a direction with its tail located at the origin of the coordinate system.
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VMSE Screenshot – [101] Direction
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Directions in Hexagonal Crystals
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Directions in Hexagonal Crystals
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)2(31 VUu −=
)2(31 UVv −=
)( vut +−=
Ww =
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Directions in Hexagonal Crystals
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Drawing HCP Crystallographic Directions (i)
1. Remove brackets 2. Divide by largest integer so all values are ≤ 1 3. Multiply terms by appropriate unit cell dimension a (for a1, a2, and a3 axes) or c (for z-axis) to produce projections 4. Construct vector by placing tail at origin and stepping off these projections to locate the head
Algorithm (Miller-Bravais coordinates)
Adapted from Figure 3.10, Callister & Rethwisch 9e.
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Drawing HCP Crystallographic Directions (ii) • Draw the direction in a hexagonal unit cell.
[1213]
4. Construct Vector
1. Remove brackets -1 -2 1 3
Algorithm a1 a2 a3 z
2. Divide by 3
3. Projections
proceed –a/3 units along a1 axis to point p –2a/3 units parallel to a2 axis to point q a/3 units parallel to a3 axis to point r c units parallel to z axis to point s
p
q r
s
start at point o
Adapted from p. 72, Callister & Rethwisch 9e.
[1213] direction represented by vector from point o to point s
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Determine Directions in Hexagonal Crystals
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)2(31 VUu −=
)2(31 UVv −=
)( vut +−=
Ww =
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1. Determine coordinates of vector tail, pt. 1: x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2. in terms of three axis (a1, a2, and z) 2. Tail point coordinates subtracted from head point coordinates and normalized by unit cell dimensions a and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas, for three-axis coordinates [UVW] 5. Convert to four-axis Miller-Bravais lattice coordinates using equations below: 6. Adjust to smallest integer values and enclose in brackets [uvtw]
Adapted from p. 72, Callister & Rethwisch 9e.
Algorithm
)2(31 VUu −= )2(
31 UVv −=
)( vut +−= Ww =
Determination of HCP Crystallographic Directions (ii)
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4. Brackets [110]
1. Tail location 0 0 0 Head location a a 0c
1 1 0 3. Reduction 1 1 0
Example a1 a2 z
5. Convert to 4-axis parameters
1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ]
6. Reduction & Brackets
Adapted from p. 72, Callister & Rethwisch 9e.
Determination of HCP Crystallographic Directions (ii)
Determine indices for green vector
2. Normalized
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Determine the indices (four-index system) for the direction
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3. Crystallographic Planes
Adapted from Fig. 3.11, Callister & Rethwisch 9e.
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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)
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Crystallographic Planes z
x
y a b
c
4. Miller Indices (110)
example a b c z
x
y a b
c
4. Miller Indices (100)
1. Intercepts 1 1 ∞ 2. Reciprocals 1/1 1/1 1/∞
1 1 0 3. Reduction 1 1 0
1. Intercepts 1/2 ∞ ∞ 2. Reciprocals 1/½ 1/∞ 1/∞
2 0 0 3. Reduction 2 0 0
example a b c
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Crystallographic Planes z
x
y a b
c
4. Miller Indices (634)
example 1. Intercepts 1/2 1 3/4
a b c
2. Reciprocals 1/½ 1/1 1/¾ 2 1 4/3
3. Reduction 6 3 4
“Family” of Planes: Crystallographically equivalent {hkl}
(001) (010), (100), (010), (001), Ex: {100} = (100),
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VMSE Screenshot – Crystallographic Planes
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Additional practice on indexing crystallographic planes
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Crystallographic Planes (HCP) • In hexagonal unit cells the same idea is used
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Crystallographic Planes (HCP) • In hexagonal unit cells the same idea is used
example a1 a2 a3 c
4. Miller-Bravais Indices (1011)
1. Intercepts 1 ∞ -1 1 2. Reciprocals 1 1/∞
1 0 -1 -1
1 1
3. Reduction 1 0 -1 1
a2
a3
a1
z
Adapted from Fig. 3.14, Callister & Rethwisch 9e.
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Determine the Miller-Bravais indices for the plane
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Summary of Equations
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Linear Density
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Linear Density (LD): Number of atoms per unit length whose centers lie on the direction vector for a specific crystallographic direction
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ex: linear density of Al in [110] direction a = 0.405 nm
Linear Density
• Linear Density of Atoms ≡ LD =
a
[110]
Adapted from Fig. 3.1(a), Callister & Rethwisch 9e.
Unit length of direction vector Number of atoms
# atoms
length
1 3.5 nm a 2
2 LD − = =
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Planar Density
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Planar Density (PD): Number of atoms per unit area that are centered on a particular crystallographic plane
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Planar Density
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Planar Density (PD): Number of atoms per unit area that are centered on a particular crystallographic plane
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Crystallographic Planes
• We want to examine the atomic packing of crystallographic planes
• Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important.
a) Draw (100) and (111) crystallographic planes for Fe. b) Calculate the planar density for each of these
planes.
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Planar Density of (100) Iron Solution: At T < 912°C iron has the BCC structure.
(100)
Radius of iron R = 0.1241 nm
R 3
3 4 a =
2D repeat unit
= Planar Density = a 2
1
atoms 2D repeat unit
= nm2
atoms 12.1 m2
atoms = 1.2 x 1019 1
2
R 3
3 4 area 2D repeat unit
Fig. 3.2(c), Callister & Rethwisch 9e [from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.]
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Planar Density of (111) Iron Solution (cont): (111) plane 1 atom in plane/ unit surface cell
atoms in plane
atoms above plane
atoms below plane
a h 2 3 =
a 2
1 = =
nm2 atoms 7.0
m2 atoms 0.70 x 1019
3 2 R 3
16 Planar Density =
atoms 2D repeat unit
area 2D repeat unit
3 3 3 2
2
R 3
16 R 3
4 2 a 3 ah 2 area = = = =
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VMSE Screenshot – Atomic Packing – (111) Plane for BCC
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• Single Crystals: The periodic arrangement of atoms extends throughout the entire specimen -Properties may vary with direction: anisotropic. -Example: the modulus of elasticity (E) in BCC iron:
Data from Table 3.4, Callister & Rethwisch 9e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)
• Polycrystals: Composed of a collection of many small crystals or grains; grain boundary -Properties may/may not vary with direction. -If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa) -If grains are textured (have a preferential crystallographic orientation), anisotropic.
200 μm Adapted from Fig. 4.15(b), Callister & Rethwisch 9e. [Fig. 4.15(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC (now the National Institute of Standards and Technology, Gaithersburg, MD).]
Single Crystals and Polycrystals E (diagonal) = 273 GPa
E (edge) = 125 GPa
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Summary 1. Crystallographic points, directions, and planes 2. Point coordinates 3. Crystallographic directions 4. Crystallographic planes
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Homework 2 • 3.3, 3.8, 3.12, 3.19 • 3.26, 3.35, 3.41, 3.42, 3.48, 3.53, 3.56, 3.60 * Problems from Callister, 9th Edition
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Densities of Material Classes ρ metals > ρ ceramics > ρ polymers
Why?
Data from Table B.1, Callister & Rethwisch, 9e.
ρ (g
/cm
)
3
Graphite/ Ceramics/ Semicond
Metals/ Alloys
Composites/ fibers Polymers
1
2
2 0
30 B ased on data in Table B1, Callister
*GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers
in an epoxy matrix). 10
3 4 5
0.3 0.4 0.5
Magnesium
Aluminum
Steels
Titanium
Cu,Ni Tin, Zinc
Silver, Mo
Tantalum Gold, W Platinum
G raphite Silicon Glass - soda Concrete
Si nitride Diamond Al oxide
Zirconia
H DPE, PS PP, LDPE PC
PTFE
PET PVC Silicone
Wood
AFRE * CFRE * GFRE* Glass fibers
Carbon fibers A ramid fibers
Metals have... • close-packing (metallic bonding) • often large atomic masses Ceramics have... • less dense packing • often lighter elements Polymers have... • low packing density (often amorphous) • lighter elements (C,H,O) Composites have... • intermediate values
In general
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• Some engineering applications require single crystals:
• Properties of crystalline materials often related to crystal structure.
(Courtesy P.M. Anderson)
-- Ex: Quartz fractures more easily along some crystal planes than others.
-- diamond single crystals for abrasives
-- turbine blades Fig. 8.34(c), Callister & Rethwisch 9e. (courtesy of Pratt and Whitney)
(Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.)
Crystals as Building Blocks
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• Most engineering materials are polycrystals.
• Nb-Hf-W plate with an electron beam weld. • Each "grain" is a single crystal. • If grains are randomly oriented, overall component properties are not directional. • Grain sizes typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers).
Fig. K, color inset pages of Callister 5e. (Courtesy of Paul E. Danielson, Teledyne Wah Chang Albany)
1 mm
Polycrystals
Isotropic
Anisotropic
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X-Ray Diffraction
• Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.
• Can’t resolve spacings < λ • Spacing is the distance between parallel planes of
atoms.
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X-Rays to Determine Crystal Structure
X-ray intensity (from detector)
θ θ c
d = n λ 2 sin θ c
Measurement of critical angle, c, allows computation of planar spacing, d.
• Incoming X-rays diffract from crystal planes.
Adapted from Fig. 3.22, Callister & Rethwisch 9e.
reflections must be in phase for a detectable signal
spacing between planes
d
θ λ
θ extra distance travelled by wave “2”
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X-Ray Diffraction Pattern
Adapted from Fig. 3.22, Callister 8e.
(110)
(200)
(211)
z
x
y a b
c
Diffraction angle 2θ
Diffraction pattern for polycrystalline α-iron (BCC)
Inte
nsity
(rel
ativ
e)
z
x
y a b
c z
x
y a b
c
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