1 Body Centered Cubic (BCC) CHAPTER 3: THE STRUCTURE OF CRYSTALLINE SOLIDS Face Centered Cubic (FCC) Hexagonal Close-Packing (HCP) CHBE213 – Dan Samborsky Introduction to Crystals Crystalline Materials •Atoms are arranged in a repeating array over large atomic distances When a liquid solidifies, the atoms move to a specific 3-dimensional structure •What materials can assume a crystalline structure? All metals, most ceramics and some polymers •Materials without crystalline structure don’t have a large scale atomic order •Many different crystal structures Metals are relatively simple Ceramics and polymers can be quite complex Atomic Hard Sphere Models The description of crystalline structures is often done through “solid spheres”, this model is also known as the Atomic Hard Sphere Model Lattice Center of atom Lattice The term lattice is being used to describe a three-dimensional array of points coinciding with atomic positions each sphere represents an ion core UNIT CELL - Basic structural unit/building block of crystal structure Most are parallelpipeds or prisms with three sets of parallel faces 14 General Crystal Lattices Simple cubic 1811-1863 Auguste Bravais
Transcript
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Body Centered Cubic (BCC)
CHAPTER 3: THE STRUCTURE OF CRYSTALLINE SOLIDS
Face Centered Cubic (FCC)
Hexagonal Close-Packing (HCP)
CHBE213 – Dan Samborsky
Introduction to Crystals
Crystalline Materials
•Atoms are arranged in a repeating array over large atomic distancesWhen a liquid solidifies, the atoms move to a specific 3-dimensional structure
•What materials can assume a crystalline structure?All metals, most ceramics and some polymers
•Materials without crystalline structure don’t have a large scale atomic order
•Many different crystal structuresMetals are relatively simpleCeramics and polymers can be quite complex
Atomic Hard Sphere ModelsThe description of crystalline structures is often done through
“solid spheres”, this model is also known as the Atomic Hard Sphere Model
Lattice
Center of atom
LatticeThe term lattice is being used to describea three-dimensional array of points coinciding with atomic positions
each sphere represents an ion core
UNIT CELL - Basic structural unit/building block of crystal structureMost are parallelpipeds or prisms with three sets of parallel faces
14 General Crystal LatticesSimplecubic
1811-1863
Auguste Bravais
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Metallic Crystal Structures
Remember:Atomic bonding in metals is “metallic” and is non-directional in nature (equal bonding in all directions)
No restriction as to the numbers and position of nearest-neighbor atomsg
In the Hard Sphere Model, each sphere represents an ion core
Nanometer (nm) = 1 x 10-9 m = 1 x 10-7 cmso the atomic radius of Al = 1.43 x 10-8 cm
Watch radius versus diameter mistakes
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Li Li+
Na+
Mg2+
Be2+
Na
K Ca
Mg
Be
K+ Ca2+
Pauling-Ahrens Radii
Ti V Cr Mn Fe
Cs
Rb
Ba
Sr
Cs2+ Ba2+
Rb+ Sr2+
BB3+
Al
CC4+
SiSi4+
NN3+
P P5+
OO2
SS6+
FF
Cl Cl
Ti4+ V3+ Cr3+ Mn2+ Fe3+Fe2+
Ions can be smaller or larger than the neutral atom
Atomic radius, and more generally the size of an atom, is not a precisely defined physical quantity, nor is it constant in all circumstances. The value assigned to the radius of a particular atom will always depend on the definition chosen for "atomic radius", and different definitions are more appropriate for different situations.
Best Discussion of "atomic radius" found on Wikipedia
The atomic radius is determined entirely by the electrons: The size of the atomic nucleus is measured in femtometres, 100,000 times smaller than the cloud of electrons. However the electrons do not have definite positions—although they are more likely to be in certain regions than others—and the electron cloud does not have a sharp edge.
Face Centered Cubic (FCC)
FCC is found in many metals (Al Cu Au Pt Pb Ni )
The FCC unit cell has atoms locatedon each of the corners and the centersof all faces
(Al, Cu, Au, Pt, Pb, Ni,...)
FCC
The ion spheres touch one another across the face diagonal
The cube edge length “a” and the atomic radius “R” are related through:
22Ra
aR 24
Equation 3.1, p.48
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•Each corner atom is shared among 8 unit cells
•Face-centered atoms only belong to two unit cells
•Therefore only 1/8 of each of the 8 corner atomsmay be assigned to a FCC unit cell
FCC
The total number of atoms per FCC unit cell is:The total number of atoms per FCC unit cell is:
462
18
8
1
xx
FCC
The coordination number (C.N.) is the number of nearest neighbor or touching atoms
The C.N. for FCC is 12
Atomic Packing Factor (APF)
APF is the fraction of solid sphere volume in a unit cell
llll
cellunitperatomsofvolumeAPF
Equation 3.2
cellunitinvolumetotal
The APF for FCC is 0.74.(spheres are all the same diameter)
74% of the unit cell volume are atoms
Metals typically have a relatively large APF to maximizethe shielding provided by the free electron cloud
Body Centered Cubic (BCC)
BCC is found in some metals, (eg. Cr, W, Ta...)
The BCC unit cell has atoms located on each of the corners and a single atom at the cube center.
The APF is 0 68 for BCC and the C N is 8The APF is 0.68 for BCC and the C.N. is 8
atoms/unit cell = (8 x 1/8) + (1 x 1) = 2
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BCC
Center and corner atoms touch one another along cube diagonal
The cube edge length “a” and the atomicradius “R” are related through: aR 34
3
4 Ra
Equation 3.3, p. 48
Hexagonal Close-Packed (HCP)
Not all metals have a cubic symmetry
HCP is found in Mg, Be, Cd, Co, Ti, Zn...
APF = 0.74 and C.N. = 12
atoms/unit cell = (12 x 1/6) + (2 x ½) + (3 x 1) = 6
Hexagonal Close-Packed (HCP)
Basal Planes (red)
The top and bottom faces of the unit cell consists of 6 atomsthat form a regular hexagon (basal plane).These 6 atoms surround one atom in the center of the facesAnother plane of 3 atoms is between these faces.
Atom c/aO ratioCd 1.886Zn 1.856ideal 1.633Mg 1.624C 1 621
Meltingpoint oC
321420
6491495
HCP - “ideal” in this case is defined as the closest packing possible with hard spheres of the same size
Co 1.621Zr 1.593Ti 1.587Be 1.568
1495185216681278
c/a > 1.633 : the atoms are farther apart in the c-direction.c/a < 1.633 : the atoms are closer together in the c-direction.
Closer together = stronger bonds = higher melting temperatures
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Atoms in
each cell
Atomic Packing Factor
CN Unit cell dimension,
“a”
Unit Cell Volume (a3),
VC
BCC 2 0.68 8 4R
Structure Summary(This is a great slide for homework and exams)
64 3R
FCC 4 0.74 12
HCP 6 0.74(ideal)
12 2R = aOc/aO = 1.63
(ideal)
3a
22Ra 216 3R
33
2
33 2ca
Theoretical Atomic Density
g/cm3
(6 022 1023 t / l)
Equation 3.5, page 51
(perfect atomic packing, no voids or impurities)
How to solve homework and exam problems:1) from front cover: BCC, FCC, HCP; and Atomic weight2) find “n” and “VC” from Ch.2 or “structure summary slide”3) NA = 6.022 x 1023 atoms/mol
+27% volume change rips the solid material apart on cooling
• Most engineering materials are polycrystals (more than one).
1 mm
3.14 - Polycrystals
polished and etchedwith acid to show grains(different unit cell orientations show as different colors)
• Nb-Hf-W plate with an electron beam weld (center of photo).• Each "grain" is a single crystal. Where they meet is called a
grain boundary.• If crystals are randomly oriented, overall component properties
are not directional, but “averaged”.• Crystal sizes typically range from 1 nm to 2 cm
(i.e., from a few, to millions of atomic layers).
• Single Crystals
-Properties vary with direction: anisotropic.
-Example: the modulus of elasticity (E) in BCC iron:
• Polycrystals
E (diagonal) = 273 GPa
E (edge) = 125 GPa
Single versus Polycrystals
y y-Properties may or may not vary with direction.-If grains are randomly oriented: isotropic.
(Epoly iron = 210 GPa)-If grains are textured, anisotropic.
200 mm
REMEMBER THE GLOSSARY IN THE BACK OF THE BOOKanisotropic – properties vary with direction
isotropic – properties are the same in each direction
Processing hasan effect on properties
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Modulus of Elasticity for Several Metals at VariousCrystallographic Orientations at “room temperature”
Metal Modulus of Elasticity (GPa)
[100] Direction
[110] Direction
[111] Direction
Aluminum (FCC) 64 73 76
E (GPa)
(listed)
69
Copper (FCC) 68 130 195
Iron (BCC) 134 211 283Tungsten (BCC) 385 385 385
Remember atomic spacing – closer the atoms are, the stronger the bonding, the stiffer the structure will be in that direction
115
207
385
3-D “smeared” properties
Crystallographic Indices:Point Coordinates
Figure 3.5. The manner in which the q, r and s coordinates at point P within the unit cell are determined. The q coordinate (which is a fraction) corresponds to the distance qa along the x axis, where a is the unit cell edge length. The respective r and s coordinates for the yand z axes are determined similarly.
Y
Z
Example problem 3.4, page 56
X
Crystallographic Directions → [ h j k]
Steps to follow1 - Position the vector so that it passes through the origin
(vectors can be translated parallel)OR Move the origin to the tail of the direction vector
calculate [ ΔX ΔY ΔZ ]
2 - Determine the length of the vector projections on each axis,a,b,c, and reduce to smallest integer values
3 - replace negative integers with a bar over the number
4 - write the indices in square brackets without commas : [h j k]
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calculate [ ΔX ΔY ΔZ ]
The origin (O) has been arbitrarily selected as the bottom left back corner of the unit cell
OD: [ ΔX=1 ΔY=0 ΔZ=1/2 ] , multiply by 2 to get rid of fractions, OD = [2 0 1]OE: [ ΔX=0 ΔY=-1 ΔZ=0 ] , OE = [0 ¯1 0] (Over line is supposed to be over the “1”)OF: [ ΔX=1/2 ΔY=1/2 ΔZ=1 ] , multiply by 2 to get rid of fractions, OF = [1 1 2]
CB: [ ΔX=0 ΔY=0 ΔZ=-1 ] ,OD = [0 0 ¯1 ] (Over line is supposed to be over the “1”)
Example Problem 3.6, page 52
Show all these steps for full credit on homework and exams
The textbook uses “a”, “b” and “c”....... just use “x”, “y” and “z”Capital X, Y and Z = axis. Lower case x, y and z = intersects
Crystallographic Planes (Miller Indices)
Steps to follow1- Move the origin to another corner of the unit cell if necessary. Plane cannot pass through the origin, and plane must intersect the coordinate axes.Determine the intercept of the plane with each axis: x, y, z.If parallel to an axis, it intersects at infinity.p , y
2 - take the reciprocals of the intercepts (1/x, 1/y, 1/z).
3 - multiply or divide by a common factor to obtain the smallest integers.
4 - replace negative integers with a bar over the number,and place in parentheses with no commas (x y z).
-if a plane is parallel to an axis, we say it cuts at ∞ and 1/∞ = 0
-if the plane passes through the origin, we translate the origin to a suitable corner
-round brackets ( ) describe a single plane,
Crystallographic Planes (Miller Indices)
-curly brackets { } to describe a family of planes
-square brackets [ ] describe directions
-NO COMMAS IN FINAL ANSWERS
-for full question credit, you MUST show your work
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Families of Equivalent Planes
The family of {100} planes
The family of {110} planes
Example 3.10, page 65
New origin
For example, if the x, y, and z intercepts are 1/2, 1, and 1/3, the Miller indices are calculated as:
* Take reciprocals: 2, 1, 3* Clear fractions (already there): 2, 1, 3* Reduce to lowest terms (already there)
Thus, the Miller indices are (213).
I separate the numbers with commasin the calculation steps. These commas must be removed for the final answer. (Remember round brackets for planes).
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Crystallographic Planes
z
ya b
c1. Intercepts 1 1 2. Reciprocals 1/1 1/1 1/
1 1 03. Reduction 1 1 0
example a b c
USE THIS FORMAT FOR FULL CREDITON EXAMS AND HOMEWORK
x4. Miller Indices (110)
example a b cz
x
ya b
c
4. Miller Indices (100)
1. Intercepts 1/2 2. Reciprocals 1/½ 1/ 1/
2 0 03. Reduction 2 0 0
Four steps required for full marks – show your work !!!
Crystallographic Planes
z
example
1. Intercepts 1/2 1 3/4
a b c
2. Reciprocals 1/ ½ 1/1 1/ ¾2 1 4/3
3/4z
x
ya b
c
4. Miller Indices (634)
3. Reduction 6 3 4
1/2
1
Some points are better “origins”- easier to see axis intersections
Why is it important to know packing densities?
• It is easiest for crystal lattice deformations to occur in the direction that is close packed (tightest).
• Planes with the highest atomic packing ratios are most likely where crystallographic “slip” occurs
3.11
most likely where crystallographic “slip” occurs.
• Therefore close packed planes are more likely to undergo mechanical deformation.
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Linear and Planar Densities
• LD = (# atoms centered on vector)
length of vectorEquation 3.8
• PD = (# atoms centered on plane)
Area of planeEquation 3.10
Not on plane(atom center above plane)
Not on plane(atom center below plane)
Atom is ½ above plane and ½ below planeatom is centered on the plane
Only count the atom if its exact center in on the plane(or on the line for a linear density)
example: linear density of Al in [110] direction. Atomic radius = 0.143 nm, a=2R(√2), a = 0.405 nm
Linear Density
• Linear Density of Atoms LD =
[110] directionUnit length of direction vector
Number of atoms
a
# atoms
length
3.49 atoms / nma2
2LD
3.49 atoms per nm = 3.49 x 109 atoms/m= 3,490,000 atoms/mm (a dime ~ 1.2 mm thick)
R + (2R) + R = 4R = 2D (2 atoms)
r r
r r
Derive a linear density expression for FCC [100] DIRECTIONFor this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1 atomthat is centered on the direction vector. The length of this direction vector is just the unit cell edge length, [Equation (3.1)]. Therefore, the expression for the linear density of this plane is
LD100 =number of atoms centered on [100] direction vector
l h f [100] di iLD100 length of [100] direction vector
1atom
2R 2
1
2R 2
½ atom + ½ atom = 1
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Unit cell face – dashes lines
Derive a planar density expression for the (001) plane in FCC. For this (001) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this FCC (001) plane. The planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length, [Eq 3.1]; and, thus, the area of
2
2R 2
2this square is just =8R2. Hence, the planar density for this (001) plane is just
2 R 2 2
plane (001) ofarea
plane (001) on centered atoms ofnumber = 001PD
2 atoms
8 R2 1
4R2¼
¼ ¼ 1
¼
Planar Density of (100) IronSolution: At T < 912C iron has the BCC structure.
[note:(100) plane = 1.22 x 1013 atoms/mm2] (~43% less)
2
6
316
2
32
2
1
2
2R
aa
haArea
1 mm = 1x106 nm (nanometers)
• Atoms may assemble into crystalline oramorphous structures.
• We can predict the density of a material,provided we know the atomic weight, atomicradius, and crystal geometry (e.g., FCC,
Summary
radius, and crystal geometry (e.g., FCC,BCC, HCP).
• Material properties generally vary with singlecrystal orientation (i.e., they are anisotropic),but properties are generally non-directional(i.e., they are isotropic) in polycrystals withrandomly oriented grains.
• Be able to determine indices for crystallographic points, directions, and planes
• Be able to calculate linear and planar atomic densities
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EXTRAS
Numbers indicate axis interceptsMiller Indices
Miller IndicesCrystallographic Planes (HCP)
• In hexagonal unit cells the same (Miller) idea is used
Instead of (x y z), we have (a1 a2 a3 c)
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HCP coordinate system (Miller-Bravais)
[ a1 a2 a3 c ] direction
(a1 a2 a3 c ) plane
planesdirections
Crystallographic Planes (HCP)
• In hexagonal unit cells the same idea is used
example a1 a2 a3 c
1 Intercepts 1 1 1
z
HCP planes and directions will not be covered on exams
4. Miller-Bravais Indices (1011)
1. Intercepts 1 -1 1
2. Reciprocals 1 1/1 0
-1-1
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3. Reduction 1 0 -1 1
a2
a3
a1
a1 a2 a3 c
Intercepts 1 -1 ∞ ∞
Reciprocals 1 -1 0 0
Reductions unnecessary
HCP Miller Indices Example – front face of unit cell(Prism planes = {1010} family)
Enclosure (1 1 0 0)
HCP Miller Indices Example – basal plane
a1 a2 a3 c
Intercepts ∞ ∞ ∞ 1
Reciprocals 0 0 0 1
Reductions unnecessary
Enclosure (0 0 0 1)
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More “important” HCP planesRead and understand the chapter summary p. 80-83.
No HCP calculations or questions dealing with section 3.16-x-ray diffraction (but a good read if your interested)
