Material Engineering
MEEN :2311
Structure of Crystalline Solids
Structure of Crystalline Solids
• How do atoms assemble into solid structures?
• How does the density of a material depend on
its structure?
• When do material properties vary with the
sample (i.e., part) orientation?
2
Materials and Packing
Solid materials may be classified according to the regularity with which
atoms or ions are arranged with respect to one another
Crystalline materials...
• atoms pack in periodic, 3D arrays
• typical of: -metals
-many ceramics
-some polymers
Noncrystalline materials...
• atoms have no periodic packing
• occurs for: -complex structures
-rapid cooling
"Amorphous" = Noncrystalline
crystalline SiO2
Si
Oxygen
Non crystalline SiO2
3
Materials and Packing
Crystalline materials...
• atoms pack in periodic, 3D arrays
• typical of: -metals
-many ceramics
-some polymers
crystalline SiO2
Single crystal: atoms are in a repeating or periodic
array over the entire extent of the material
Polycrystalline material: comprised of many small
crystals or grains
4
Crystal Structures: Energy and Packing
Why do atoms assemble into ordered structures (crystals)?
• Non dense, random packing
Energy of interatomic bond
• Dense, ordered packing
Energy
typical neighbor
bond energy
r
typical neighbor
bond energy
Energy
r
Nearest neighbor atoms distances tend to be small in order to lower
bond energy.
Dense, ordered packed structures tend to have lower energies
Crystal Structures
Atoms are considered as hard spheres with well-defined radii.
The shortest distance between two like atoms is one diameter of
the hard sphere.
We can consider crystalline structure as a lattice of points at
atom/sphere centers
Unit Cell
The unit cell is a structural unit or building block that describe the
crystal structure. Repetition of the unit cell generates the entire
crystal.
CRYSTAL STRUCTURES
Metallic Crystal Structures
Metals are usually (poly) crystalline; although formation of
amorphous metals is possible by rapid cooling
Atomic bonding in metals (Metallic Bonds) is non-directional
⇒ no restriction on numbers or positions of nearest-neighbor
atoms ⇒ large number of nearest neighbors and dense atomic
packing
Atomic (hard sphere) radius, R, defined by ion core radius typically 0.1 - 0.2 nm
The most common types of unit cells are
faced-centered cubic (FCC)
body-centered cubic (BCC)
hexagonal close-packed (HCP)
Face Centered Cubic Crystal Structure (FCC)
Atoms are located at each of the corners and on the centers of all the
faces of cubic unit cell
Cu, Al, Ag, Au have this crystal structure
Atoms touch each other along face diagonals
Cube edge length, a
⇒
Face Centered Cubic Crystal Structure (FCC)
Number of atoms per unit cell, n = 4.
(For an atom that is shared with m adjacent unit
cells, we only count a fraction of the atom, 1/m).
In FCC unit cell we have:
Each face atom(6) shared by two cells: 6×1/2 = 3
Each corner atom(8) shared by eight cells: 8×1/8 = 1
Coordination number, CN =12
Number of closest neighbors to which an atom is bonded
= number of touching atoms,
ABCABC... Stacking Sequence
Face Centered Cubic Crystal Structure (FCC)
Atomic packing factor, APF
fraction of volume occupied by hard spheres = (Sum of atomic
volumes)/(Volume of cell) = 0.74 (maximum possible)
APF= (Sum of atomic volumes)/(Volume of
unit cell)
Volume of 4 hard spheres in the unit cell
= 4 x 4/3 π R3
Volume of the unit cell = a3 =
FCC is a close-packed structure with highest density of atoms
Body Centered Cubic Crystal Structure (BCC)
Atom at each corner and at center of cubic unit cell
Cr, α-Fe, Mo, Tantalum have this crystal structure
The coordination number, CN = 8
Number of atoms per unit cell, n = 2
Center atom (1) shared by no other cells: 1 x 1 = 1
8 corner atoms shared by eight cells: 8 x 1/8 = 1
Body Centered Cubic Crystal Structure (BCC)
The hard spheres touch one another along cube diagonal
3
3a
a
3
a
2
2a
2
1
a
atoms
R
Cube diagonal length = 4R = 3 a
a
4
1
a= 4R/√3
volume
atom
p ( 3a/4) 3
2
unit cell
3
APF =
volume
3
a
APF= 0.68
unit cell
Hexagonal Close-Packed Crystal Structure (HCP)
HCP is one more common structure of metallic crystals
Six atoms form regular hexagon, surrounding one atom in center.
Another plane is situated halfway up unit cell (c-axis), with 3 additional
atoms situated at interstices of hexagonal (close-packed) planes
Cd, Mg, Zn, Ti have this crystal structure
Number of atoms per unit cell,
n = 6.
• 3 mid-plane atoms shared by
no other cells: 3 x 1 = 3
• 12 hexagonal corner atoms
shared by 6 cells: 12 x 1/6 = 2
• 2 top/bottom plane center
atoms shared by 2 cells: 2 x
1/2 = 1
Unit cell has two lattice parameters
a and c. Ideal ratio c/a = 1.633
Hexagonal Close-Packed Crystal Structure (HCP)
The coordination number, CN =12 (same as in FCC)
Atomic packing factor, APF = 0.74 (same as in FCC)
a = 2R, c = 1.633a
Volume of one HCP unit Cell = [Area of triangle ABC)x 6] x c
Area of triangle = ½ Base x height
The difference between the FCC and HCP structures is in the stacking
sequence.
Cubic lattice structures allow slippage to occur more easily than noncubic lattices, so HCP metals are not as ductile as the FCC metals
Density, r Computation
Entire crystal can be generated by the repetition
of the unit cell.
Density of a crystalline material, ρ
Mass of total atoms in a unit cell
Density = r =
Total Volume of Unit Cell
r =
where
nA
VC NA
n = number of atoms/unit cell
A = atomic weight (g/mol)
VC = Volume of unit cell (cm3)= a3 for cubic
NA = Avogadro’s number = 6.022 x 1023 atoms/mol
Example: Copper
• crystal structure = FCC: 4 atoms/unit cell
rCu = 8.89 g/cm3
• atomic weight = 63.55 g/molResult:
(1 amu =theoretical
1 g/mol)
• atomic radius R = 0.128 nm (1 nm = 10-7 cm)
Characteristics of Selected Elements at 20°C
At. Weight
Element
Symbol (amu)
Aluminum Al
26.98
Argon
Ar
39.95
Barium
Ba
137.33
Beryllium
Be
9.012
Boron
B
10.81
Bromine
Br
79.90
Cadmium
Cd
112.41
Calcium
Ca
40.08
Carbon
C
12.011
Cesium
Cs
132.91
Chlorine
Cl
35.45
Chromium Cr
52.00
Cobalt
Co
58.93
Copper
Cu
63.55
Flourine
F
19.00
Gallium
Ga
69.72
Germanium Ge
72.59
Gold
Au
196.97
Helium
He
4.003
Hydrogen
H
1.008
Density
(g/cm3)
2.71
-----3.5
1.85
2.34
-----8.65
1.55
2.25
1.87
-----7.19
8.9
8.94
-----5.90
5.32
19.32
-----------
Atomic radius
(nm)
0.143
-----0.217
0.114
----------0.149
0.197
0.071
0.265
-----0.125
0.125
0.128
-----0.122
0.122
0.144
----------12
Densities of Material Classes
In general
rmetals > rceramics > rpolymers
Why?
Polymers have...
r (g/cm3 )
• close-packing
(metallic bonding)
• often large atomic masses
• less dense packing
• often lighter elements
• low packing density
(often amorphous)
• lighter elements (C,H,O)
Composites have...
• intermediate values
Graphite/
Ceramics/
Semicond
Polymers
Composites/
fibers
30
Metals have...
Ceramics have...
Metals/
Alloys
20
Platinum
Gold, W
Tantalum
10
Silver, Mo
Cu,Ni
Steels
Tin, Zinc
5
4
3
2
1
0.5
0.4
0.3
Titanium
Aluminum
Magnesium
*GFRE, CFRE, & AFRE are Glass,
Carbon, & Aramid Fiber-Reinforced
Epoxy composites (values based on
60% volume fraction of aligned fibers
in an epoxy matrix).
Zirconia
Al oxide
Diamond
Si nitride
Glass -soda
Concrete
Silicon
Graphite
PTFE
Silicone
PVC
PET
PC
HDPE, PS
PP, LDPE
Glass fibers
GFRE*
Carbon fibers
CFRE*
Aramid fibers
AFRE*
Wood
20
CRYSTAL SYSTEMS
• Unit cell = 3-dimensional unit that repeats in space
• Unit cell geometry completely specified by a, b, c & a, b, g
(lattice parameters or lattice constants)
CRYSTALLOGRAPHIC POINTS, DIRECTIONS & PLANES
• In crystalline materials, often necessary to specify points,
directions and planes within unit cell and in crystal lattice
• Three numbers (or indices) used to designate points,
directions (lines) or planes, based on basic geometric notions
• The three indices are determined by placing the origin at one
of the corners of the unit cell, and the coordinate axes along
the unit cell edges
POINT COORDINATES
 Any point within a unit cell specified as fractional
multiples of the unit cell edge lengths
 Position P specified as q r s; convention: coordinates
not separated by commas or punctuation marks
EXAMPLE: POINT COORDINATES
• Locate the point (1/4 1 ½)
• Specify point coordinates for all atom positions for a
BCC unit cell
– Answer: 0 0 0, 1 0 0, 1 1 0, 0 1 0, ½ ½ ½, 0 0 1, 1
0 1, 1 1 1, 0 1 1
Notation for lattice positions
CRYSTALLOGRAPHIC DIRECTIONS
• Defined as line between two points: a vector
• Steps for finding the 3 indices denoting a direction
– Determine the point positions of a beginning point (X1 Y1 Z1) and a
ending point (X2 Y2 Z2) for direction, in terms of unit cell edges
– Calculate difference between ending and starting point
– Multiply the differences by a common constant to convert them to the
smallest possible integers u, v, w
– The three indices are not separated by commas and are enclosed in
square brackets: [uvw]
– If any of the indices is negative, a bar is placed in top of that index
Notation for lattice directions
COMMON DIRECTIONS
EXAMPLES: DIRECTIONS
 Draw a [1,-1,0] direction within a cubic unit cell
• Determine the indices for this direction
– Answer: [120]
CRYSTALLOGRAPHIC PLANES
• Crystallographic planes specified by 3 Miller indices as (hkl)
• Procedure for determining h, k and l:
Z
1.
2.
3.
4.
5.
6.
If plane passes through origin, translate plane or
choose new origin
Determine intercepts of planes on each of the axes in
terms of unit cell edge lengths (lattice parameters) (½
¼ ½) . Note: if plane has no intercept to an axis (i.e., it
is parallel to that axis), intercept is infinity
Determine reciprocal of the three intercepts (2 4 2)
If necessary, multiply these three numbers by a
common factor which converts all the reciprocals to
small integers (1 2 1)
The three indices are not separated by commas and
are enclosed in curved brackets: (hkl) (121)
If any of the indices is negative, a bar is placed in top of
that index
X
1/2
1/4
Y
1/2
(1 2 1)
Crystallographic Planes
example
1. Intercepts
2. Reciprocals
3.
Reduction
4.
Miller Indices
example
a
b
z
c
c
b
a
x
a
b
y
z
c
c
1.
2.
Intercepts
Reciprocals
3.
Reduction
4.
Miller Indices
y
a
b
x
31
Crystallographic Planes
z
example
1. Intercepts
2. Reciprocals
3.
Reduction
a
1
1/1
1
1
4.
Miller Indices
(110)
example
1. Intercepts
2. Reciprocals
3.
Reduction
a
1/2
1/½
2
1
4.
Miller Indices
(100)
b
1
1/1
1
1
c

1/
0
0
c
y
b
a
x
b

1/
0
0
c

1/
0
0
z
c
y
a
b
x
32
Crystallographic Planes
• Construct a (0,-1,1) plane
Notation for lattice planes.
THREE IMPORTANT CRYSTAL PLANES
• Parallel
planes are
equivalent
For further practice https://www.doitpoms.ac.uk/tlplib/miller_indices/lattice_examples.php
Polymorphism and Allotropy
Some materials (compounds) may exist in more than one crystal
structure, this is called polymorphism.
If the material is an elemental solid, it is called allotropy.
Elements exhibiting allotropy include tin, carbon, titanium, sulfur,
phosphorus, and oxygen.
Carbon, which can exist as diamond, graphite, and amorphous carbon.
Polymorphism and Allotropy
Iron (Fe) system
liquid
BCC
1538ºC
-Fe
FCC
1394ºC
g-Fe
912ºC
BCC
a-Fe
Single Crystals & Polycrystalline Materials
Single crystal: atoms are in a repeating or periodic array over the entire
extent of the material
Polycrystalline material: comprised of many small crystals or grains.
The grains have different crystallographic orientation. There exist
atomic mismatch within the regions where grains meet. These regions
are called grain boundaries.
Anisotropy
Different directions in a crystal have different packing. For instance,
atoms along the edge of FCC unit cell are more separated than along
the face diagonal. This causes Anisotropy in the properties of crystals,
for instance, the deformation depends on the direction in which a stress
is applied
In polycrystalline materials, grain
orientations are random, so bulk material
properties are isotropic
E (diagonal) = 273 GPa
E (edge) = 125 GPa
Non Crystalline Amorphous Solids
 Amorphous materials - Materials, including glasses, that have no long-range
order, or crystal structure.
 But amorphous does not mean random, in many cases there is some form of
short-range order
amorphous SiO2 structure