Chapter 3: The Structure of Crystalline Solids
Theoretical Density, r
Density = r =
r =
where
Mass of Atoms in Unit Cell
Total Volume of Unit Cell
nA
VC NA
n = number of atoms/unit cell
A = atomic weight
VC = Volume of unit cell = a3 for cubic
NA = Avogadro’s number
= 6.022 x 1023 atoms/mol
Chapter 3 - 1
Theoretical Density, r
• Ex: Cr (BCC)
A = 52.00 g/mol
R = 0.125 nm
n = 2 atoms/unit cell
Adapted from
Fig. 3.2(a), Callister &
Rethwisch 8e.
atoms
unit cell
r=
volume
unit cell
R
a
2 52.00
a3 6.022 x 1023
a = 4R/ 3 = 0.2887 nm
g
mol
rtheoretical = 7.18 g/cm3
ractual
atoms
mol
= 7.19 g/cm3
Chapter 3 - 2
Densities of Material Classes
In general
rmetals > rceramics > rpolymers
30
Why?
Metals have...
Ceramics have...
• less dense packing
• often lighter elements
Polymers have...
r (g/cm3 )
• close-packing
(metallic bonding)
• often large atomic masses
• low packing density
(often amorphous)
• lighter elements (C,H,O)
Composites have...
• intermediate values
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
Graphite/
Ceramics/
Semicond
Composites/
fibers
Polymers
Based 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).
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
Data from Table B.1, Callister & Rethwisch, 8e.
Chapter 3 - 3
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 &
Rethwisch 8e. (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)
Chapter 3 - 4
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 typically range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Chapter 3 - 5
Single vs Polycrystals
• Single Crystals
E (diagonal) = 273 GPa
Data from Table 3.3,
Callister & Rethwisch
8e. (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.
E (edge) = 125 GPa
200 mm
Adapted from Fig.
4.14(b), Callister &
Rethwisch 8e.
(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].)
Chapter 3 - 6
Polymorphism
• Two or more distinct crystal structures for the same
material.
• Allotropy-when found in elemental solids.
titanium
, -Ti
carbon
diamond, graphite
iron system
liquid
BCC
1538ºC
-Fe
FCC
1394ºC
-Fe
912ºC
BCC
-Fe
Chapter 3 - 7
Tin (Its Allotropic Transformation)
White (or beta) tin, having a body-centered tetragonal crystal structure at room
temperature, transforms, at 13.2°C (55.8°F), to gray (or alpha) tin, which has a
crystal structure similar to diamond.
The rate at which this change takes place is extremely slow; however, the lower the
temperature (below 13.2°C) the faster the rate.
This transformation results in:
Increase in volume (27 %), and, a decrease in density (from 7.30 g/cm3 to 5.77 g/cm3).
Consequently, this volume expansion results in the disintegration of the white tin metal
into a coarse powder of the gray allotrope.
This produced some rather dramatic results in 1850 in Russia. The winter that year
was particularly cold, and record low temperatures persisted for extended periods of
time. The uniforms of some Russian soldiers had tin buttons, many of which crumbled
because of these extreme cold conditions, as did also many of the tin church organ
Chapter 3 - 8
pipes. This problem came to be known as the “tin disease.”
3.7 Iron has a BCC crystal structure, an atomic radius of
0.124 nm, and an atomic weight of 55.85 g/mol. Compute and
compare its theoretical density with the experimental value
found inside the front cover.
3.8 Calculate the radius of an iridium atom, given that Ir has
an FCC crystal structure, a density of 22.4 g/cm3, and an
atomic weight of 192.2 g/mol.
Chapter 3 - 9
Crystal Systems
Unit cell: smallest repetitive
volume which contains the
complete lattice pattern of a
crystal.
Fig. 3.4, Callister & Rethwisch 8e.
a, b, and c are the lattice
constant
Chapter 3 - 10
Crystal Structure and Crystal System
Q. What is the difference between crystal structure and crystal
system?
A: A crystal structure is described by both the geometry of,
and atomic arrangements within, the unit cell, whereas a
crystal system is described only in terms of the unit cell
geometry. For example, face-centered cubic and bodycentered cubic are crystal structures that belong to the cubic
crystal system.
Chapter 3 - 11
Point Coordinates
z
Point coordinates for unit cell
center are
111
c
a
a/2, b/2, c/2
000
y
b
½½½
Point coordinates for unit cell
corner are 111
x
Chapter 3 -
Point coordinates for all atom positions
for a BCC unit cell
Chapter 3 - 13
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>
Chapter 3 - 14
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
Adapted from Fig. 3.8(a),
Callister & Rethwisch 8e.
ex:
½, ½, -1, 0
-a3
a2
2
=>
[ 1120 ]
a3
dashed red lines indicate
projections onto a1 and a2 axes
a1
2
a1
Chapter 3 - 15
Problem 3.32
3.32 Determine the indices for
the directions shown in the
following cubic unit cell:
Chapter 3 - 16