Lecture #5
Materials science I
(MAE2001)
March 19, 2025
Materials Scieince 1
Sunyoung Lee
Crystals as Building Blocks
• Some engineering applications require single crystals:
--diamond single
crystals for abrasives
(Courtesy Martin Deakins,
GE Superabrasives,
Worthington, OH. Used with
permission.)
--turbine blades
Fig. 8.33(c), Callister 7e.
(Fig. 8.33(c) courtesy
of Pratt and Whitney).
• Properties of crystalline materials
often related to crystal structure.
--Ex: Quartz fractures more easily
along some crystal planes than
others.
(Courtesy P.M. Anderson)
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Polycrystals
• Most engineering materials are polycrystals.
Anisotropic
Adapted from Fig. K,
color inset pages of
Callister 5e.
(Fig. K is courtesy of
Paul E. Danielson,
Teledyne Wah Chang
Albany)
1 mm
• Nb-Hf-W plate with an electron beam weld.
• Each "grain" is a single crystal.
• If grains are randomly oriented,
Isotropic
overall component properties are not directional.
• Grain sizes typ. range from 1 nm to 2 cm
(i.e.,
from1 a few to millions of atomic layers).
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Single vs Polycrystals
• Single Crystals
E (diagonal) = 273 GPa
Data from Table 3.3,
Callister 7e.
(Source of data is R.W.
Hertzberg, Deformation
and Fracture Mechanics
of Engineering
Materials, 3rd ed., John
Wiley and Sons, 1989.)
-Properties vary with
direction: anisotropic.
-Example: the modulus
of elasticity (E) in BCC iron:
• Polycrystals
-Properties may/may not
vary with direction.
-If grains are randomly
oriented: isotropic.
(Epoly iron = 210 GPa)
-If grains are textured,
anisotropic.
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E (edge) = 125 GPa
200 m
Adapted from Fig.
4.14(b), Callister 7e.
(Fig. 4.14(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].)
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Section 3.6 – Polymorphism
• Two or more distinct crystal structures for the
same material (allotropy/polymorphism)
iron system
titanium
liquid
, -Ti
1538ºC
-Fe
BCC
carbon
1394ºC
diamond, graphite
-Fe
FCC
912ºC
BCC
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-Fe
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Section 3.8 Point Coordinates
z
Point coordinates for unit cell
center are
111
c
a/2, b/2, c/2
000
a
x
y
b
Point coordinates for unit cell
corner are 111
•
z
½ ½ ½
2c
•
•
•
b
b
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y
Translation: integer multiple of
lattice constants →
identical position in another
unit cell
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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
[uvw]
x
ex: 1, 0, ½
=> 2, 0, 1 => [ 201 ]
-1, 1, 1 => [ 111 ]
where overbar represents a
negative index
families of directions <uvw>
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Linear Density
• Linear Density of Atoms  LD =
Number of atoms
Unit length of direction vector
[110]
ex: linear density of Al in [110]
direction
a = 0.405 nm
# atoms
a
LD =
length
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2
= 3.5 nm −1
2a
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HCP Crystallographic Directions
z
Algorithm
a2
-
a3
a1
1. Vector repositioned (if necessary) to pass
through origin.
2. Read off projections in terms of unit
cell dimensions a1, a2, a3, or c
3. Adjust to smallest integer values
4. Enclose in square brackets, no commas
[uvtw]
a
2
-a3
a2
2
ex:
½ , ½ , -1, 0
=>
[ 1120 ]
a3
dashed red lines indicate
projections onto a1 and a2 axes
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a1
2
a1
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HCP Crystallographic Directions
• Hexagonal Crystals
– 4 parameter Miller-Bravais lattice coordinates
are related to the direction indices (i.e., u'v'w')
as follows.
z
[ u 'v 'w ' ] → [ uvtw ]
a2
-
a3
a1
1
u = (2 u ' - v ')
3
1
v = (2 v ' - u ')
3
t = - (u +v )
w = w'
Fig. 3.8(a), Callister 7e.
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https://www.youtube.com/watch?v=6KwH1WTpiPM
Crystallographic Planes
Adapted from Fig. 3.9, Callister 7e.
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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
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
2
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
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Crystallographic Planes
z
example
1. Intercepts
2. Reciprocals
3.
Reduction
a
1/2
1/½
2
6
4.
Miller Indices
(634)
b
1
1/1
1
3
c
c
3/4
•
1/¾
4/3
•
4 a
x
•
y
b
Family of Planes {hkl}
Ex: {100} = (100), (010), (001), (100), (010), (001)
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Crystallographic Planes (HCP)
• In hexagonal unit cells the same idea is
z
used
example
1. Intercepts
2. Reciprocals
3.
Reduction
a1
1
1
1
1
a2

1/
0
0
a3
-1
-1
-1
-1
c
1
1
1
1
a2
a3
4.
Miller-Bravais Indices
(1011)
a1
Adapted from Fig. 3.8(a), Callister 7e.
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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.
2D repeat unit
(100)
Planar Density =
area
2D repeat unit
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1
a2
=
4 3
R
3
Radius of iron R = 0.1241 nm
Adapted from Fig. 3.2(c), Callister 7e.
atoms
2D repeat unit
a=
1
4 3
R
3
atoms
atoms
19
= 1.2 x 10
2 = 12.1
2
nm
m2
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Atomic Packing Factor: BCC
• APF for a body-centered cubic structure = 0.68
3a
a
2a
Adapted from
Fig. 3.2(a), Callister 7e.
atoms
R
a
4
Close-packed directions:
length = 4R = 3 a
volume
atom
 ( 3a/4) 3
2
unit cell
3
APF =
volume
3
a
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unit cell
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Planar Density of (111) Iron
Solution (cont): (111) plane 1 atom in plane/ unit surface cell
2a
atoms in plane
atoms above plane
atoms below plane
h=
3
a
2
2
atoms
2D repeat unit
 4 3  16 3 2
2
=
area = 2 ah = 3 a = 3 
R÷
R
÷
3
 3

1
= 7.0
Planar Density =
area
2D repeat unit
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16 3
3
R2
atoms =
nm2
0.70 x 1019
atoms
m2
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See You next time!!
Materials Scieince 1
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