Chapter 2
2.1 Number of Atoms
2.2 Atoms and Density
2.3 Mass and Volume
2.4 Plating Thickness
2.4 Electronegativity and Bonding
2.1 Number of Atoms
Calculate the number of atoms in 100 grams of silver.
This problem is a reminder of the definition of Avogadro's number, and also a reminder of the very
large number of atoms present in macroscopic objects.
We start, as usual with Mathcad problems, by defining what we know.
100. gm
M Ag
Atomic weight of silver:
AW Ag
Avogadro's number:
A0
gm
107.87.
mole
from
Appendix A
23 atoms
6.023. 10 .
mole
Now let us write down the definition of Avogadro's number. It is the number of atoms (or molecules)
present in a gram mole of any substance, so
It is usually easier to remember the basic form of a definition, and then solve it for whatever is the
unknown term in any particular problem, than to try to memorize all of the different ways an
equation may appear. In this case, we can rearrange and solve for the unknown number of atoms:
N Ag
A 0.
M Ag
AW Ag
N Ag = 5.5836 1023 atoms
This is the number of atoms of silver. Double clicking on bold, underlined text above takes you to a
table of atomic values. Try substituting in values for other elements, and watch how N Ag changes.
In Askeland Homework Problem Number 2-1, the same relationship is used to solve for a different
element, aluminum.
We are told that Aluminum foil used for food storage weighs 0.3 grams per square inch (a mixture of
units that is unfortunate but common in engineering problems). How many atoms of aluminum are
contained in this sample? The method of solution is identical to that above.
Weight per area of Aluminum:
wpa Al
gm
0.3.
2
cm
A Al
1. cm2
Mass of Aluminum:
M Al
wpa Al. A Al
Atomic weight of Aluminum:
AW Al
gm
26.98.
mole
from
Appendix A
The number of atoms present in one square inch of foil is then
N Al = 6.6972 1021 atoms
Looking at the definition of the number of atoms/molecules present in a given sample, do you
expect the number to vary linearly with mass? That is, if you double the mass, will you double the
number of atoms/molecules? Let's look at a plot.
N M Al
A 0.
M Al
AW Al
M Al
1. gm, 2. gm.. 1. kg
25
3 10
25
2 10
N M Al
25
1 10
0
0
0.5
1
M Al
Number of molecules vs. Mass
1.5
You can see here, as you
can in the definition, that the
number of atoms/molecules
in a sample is linearly
dependent on mass.
Although our plot uses
Aluminum as a sample, this
will remain true for any
material.
2.2 Atoms and Density
Compare the number of atoms in one gram of Uranium with the number in one gram of Boron. Then,
using the densities of each, calculate the number of atoms per cubic centimeter in Uranium and
Boron.
Table values of atomic weights and densities for these two elements are:
Boron (Z = 5):
AW B
gm
10.81.
mole
ρB
gm
2.3.
3
cm
Uranium (Z = 92):
AW U
gm
238.03.
mole
ρU
gm
19.05.
3
cm
And Avogadro's number is defined as
A0
23 atoms
6.02. 10 .
mole
The atomic weight of an element is the number of grams per gram mole, and one gram mole contains
Avogadro's number of atoms. Therefore, the number of atoms per gram (apg) of each element is
determined by
apg B
apg U
A0
AW B
A0
AW U
22
apg B = 5.5689 10
atoms
gm
21
apg U = 2.5291 10
atoms
gm
The definition of density is mass per volume ( ρ = M / V ). We know the number of atoms per mass,
and we know the mass per volume. Multiplying these two together will give us the number of atoms
per volume:
NB
apg B. ρ B
23
N B = 1.2809 10
atoms
3
cm
NU
apg U. ρ U
22
N U = 4.8179 10
atoms
3
cm
2.3 Mass and Volume
Suppose you collect 5 x 10 26 atoms of nickel. Calculate the mass in grams and the volume in cubic
centimeters represented by this number of atoms.
The first point of this problem is to remind students of the large numbers of atoms in macroscopic
objects, of the exponential notation used, and of the use and meaning of Avogadro's number. The
second point is to emphasize the concept of density, which is simply mass/volume.
We begin by defining our known terms:
Number of atoms:
N Ni
26
5. 10 . atoms
Density of nickel:
ρ Ni
gm
8.902.
3
cm
Atomic weight of nickel:
AW Ni
Avogadro's number:
A0
from
Appendix A
gm
58.71.
mole
23 atoms
6.02. 10 .
mole
The mass of the atoms is solved by dragging the numeric values into the expression
M Ni
N Ni. AW Ni
A0
M Ni = 48.762 kg
Density is defined as mass/volume ( ρ = M / V ), so we can rearrange this expression to solve for
volume:
V Ni
M Ni
ρ Ni
3
V Ni = 5477.7 cm
As you can see in the definition of volume, it is inversely proportional to the density. This means that
as the density of a substance gets larger, it's corresponding volume will get smaller (IF you keep the
mass the same). This is the reason why one pound of bread takes up considerably more space than
one pound of gold.
2.4 Plating Thickness
In order to plate a steel part having a surface area of 200 square inches with a 0.002 inch thick layer
of nickel, how many atoms of nickel are required, and how many moles?
As usual, we begin by defining what we know.
2
200.in
Surface area of plating:
A
Thickness of plating:
t
0.002.in
Density of nickel:
ρ
gm
8.902.
3
cm
Atomic weight of nickel:
AW
Avogadro's number:
A0
from Appendix A
gm
58.71.
mole
23 atoms
6.023. 10 .
mole
This problem is similar to the number of atoms problems (see Example 2.1), except that the known
quantity is volume instead of mass, so we must use the density of nickel to convert.
The first part of the solution is to compute the volume of the nickel plating:
V
3
A.t
V = 6.5548 cm
Density is defined as mass/volume, so we can rearrange this expression and solve it for mass:
M
ρ. V
M = 58.351 gm
To convert this mass to the number of atoms, we recall that Avogadro's number is the number of
atoms in one mole, and the atomic weight is the number of grams in one mole:
N atom
M.
A0
AW
23
N atom = 5.9862 10
atoms
Since one mole of any material contains Avogadro's number of atoms, we can use the following ratio
to determine the number of moles of nickel present:
N mol
N atom
A0
N mol = 0.994 mole
This is exactly equivalent to using the ratio of the mass of nickel present to its atomic mass:
N mol
M
AW
N mol = 0.994 mole
2.5 Electronegativity and Bonding
What fraction of the bonding in SiO 2 is covalent?
The calculation here is straightforward, using the following equation to determine the
fraction of the bond that is covalent in terms of the difference in negativities of the atoms.
Covalent( ∆ E)
exp
2
0.25. ∆ E
The electronegativity values can be taken from tables or from a graph such as the one shown below.
Notice that the electronegativity varies in a more-or-less linear way according to the number of
"valence" electrons, those in the outermost shell (for the elements shown in the graph, these are
the number of s and p electrons in the outermost shell - electronegativities are a bit more
complicated for the transition metals that are filling the d shell). Generally speaking, the
electronegativity is highest at the upper right corner of the periodic table, and drops as you go to the
left, or down.
The key to understanding the
differences between the
tendencies for particular
combinations of atoms to form
ionic, covalent or metallic bonds
is the electronegativity. If the
atoms have a low
electronegativity, then their hold
on outer electrons is weak, and
they can all donate them to the
electron sea forming a metallic
bond. If one atom has a low
electronegativity and the other a
high value, the atom with the low
electronegativity value will transfer
electron(s) to the other atom
forming ions, and resulting in an
ionic bond. If both atoms have a
high electronegativity, neither
wants to give up an electron and
From this graph, we can estimate the electronegativity
so they share them forming a
of silicon to be about 1.8, and that of oxygen to be
covalent bond.
about 3.5.
E Si
1.8
∆E
E Si
EO
3.5
∆ E = 1.7
EO
Note that the capital greek Delta is made by
typing a capital D, followed by Ctrl-G
Covalent( ∆ E) = 0.486
This says that about half of the bond formed between Si and O is covalent. As we will see
when we get to a later chapter on ceramic structures, this is enough to cause the bond to
be highly directional. The physical arrangement of Si and O ions in glasses and ceramics is
strongly dependent on the covalent nature of the bond.
On the other hand, when bonds are predominantly ionic in nature, there is little inherent
directionality to the bond and it is the desire for each ion to surround itself with the
maximum number of ions of the opposite sign that dominates the structure. This in turn
depends on the ratio of ionic radii, and gives rise to structures with 4, 6 and 8 coordination
as discussed in Chapter 3.
For example, for sodium chloride (NaCl), the electronegativies are about 0.9 for Na and 3 for
Cl., and the amount of covalency in the bond drops to about 33%. This is enough less than
the value for SiO2 to make the bond predominatly ionic and allow dense packing of the
atoms.
E Na
∆E
0.9
E Na
E Cl
E Cl
Covalent( ∆ E) = 0.332
3
∆ E = 2.1
Try this for other combinations of elements
which you know form viable compounds.
Chapter 3
3.1 Unit Cell Dimensions, Packing Factor and Density
3.2 Packing factor in ionic unit cell
3.3 Crystal Structure
3.4 Packing Factor in Diamond Cube Structure
3.5 Lattice Parameter
3.6 Alloy Proportions
3.7 BCC to FCC Iron
3.8 Calculation of Linear Atom Density
3.9 Calculation of Planar Density
3.10 Calculation of Planar Spacing
3.11 Interstitial Site Size
3.12 Radius Ratio Calculations
3.13 Ionic Unit Cell Geometry
3.14 Basis for a Complex Unit Cell
3.1 Unit Cell Dimensions, Packing Factor and Density
In Example 3.2, the relationship between atomic radius and lattice parameters is derived for
cubic unit cells. This relationship is obtained from simple geometry. In this example, we use that
formula to calculate the lattice parameters of BCC Fe and FCC Fe.
In the SIMPLE CUBIC cell, the atoms of radius
r touch along each edge. Consequently the
length of the edge a is equal to two times r, or
a0 = 2r.
In the BODY CENTERED CUBIC cell,
the atoms of radius r touch along the
body diagonal. There is an atom in the
center of the cube, so the total length
of this line must be four times r. But for
a cube, the body diagonal is equal to
the square root of three times the
edge, so
3. a 4. r or
a BCC( r )
4. r
3
In the FACE CENTERED CUBIC cell, the atoms
of radius r touch along a face diagonal. There is
an atom in the center of the face, so the total
length of this line must be four times r. But for a
cube, the face diagonal is equal to the square
root of two times the edge, so
2. a 4. r
or
a FCC( r )
4. r
2
Use these rules to calculate the length of the edge of the unit cell for body-centered and
face-centered iron. BCC is the structure that exists at room temperature, FCC is the structure
that exists at higher temperature (e.g., 1000C). The change from one structure to another on
heating and cooling (called an allotropic transformation) is the basis for much of the control over
iron and steel microstructure and properties.
We are given the atomic radius of iron:
8
1.24. 10 . cm
r Fe
The formula for BCC iron is:
a BCC r Fe = 2.86 10
The formula for FCC iron is:
a FCC r Fe = 3.51 10
8
8
cm
cm
To calculate the packing factor of a FCC cell, we take the number of atoms per cell multiplied by
the volume of the atoms, and divided by the volume of the unit cell. In a FCC structure, there are
4 lattice points and for a metal structure with 1 atom per lattice point, there are thus 4 atoms.
We also know that the volume of the atom is 4/3 π r3 and the volume of the unit cell is the cube of
the lattice parameter, so:
Number of atoms:
N FCC
Packing factor:
4
4. .
3
π r Fe
3
N FCC.
P
a FCC r Fe
P = 0.74
3
Calculate the density of iron.
Atoms:
N BCC
2
Atomic weight:
AW Fe
gm
55.8.
mole
Avogadro's number:
A0
23 atoms
6.023.10 .
mole
According to the book, and as calculated in the problem above, BCC iron has a lattice
parameter of 2.866. The volume of a unit cell for BCC iron is then
V
a BCC r Fe
3
V = 2.35 10
23
3
cm
And the density is
ρ
N BCC. AW Fe
A 0. V
ρ = 7.89
gm
3
cm
There are several homework problems that solve the density equation for other parameters. For
instance, suppose we were given the density value but asked for the lattice parameter and atom
radius for lead.
gm
11.4.
3
cm
Density:
ρ Pb
Atomic weight:
AW Pb
Atoms per unit cell:
N FCC
gm
207.
mole
4
The number of atoms per unit cell comes from the statement in the problem that lead has an fcc
structure.
We know that
ρ
N FCC. AW
A 0. V
Solving this for volume gives us
V
N FCC. AW Pb
A 0. ρ Pb
V = 1.21 10
22
3
cm
Now the lattice parameter is obtained from the unit cell volume.
1
3
a 0.Pb
V
a 0.Pb = 4.94 10
8
cm
And the relationship derived above between the lattice parameter and atom radius for an fcc
structure is used to solve for the atom size:
a0
4. r
2
Solving for the radius gives us
r Pb
2. a 0.Pb
4
r Pb = 1.75 10
8
cm
3.2 Packing factor in ionic unit cell
For KCl (a) verify that the compound may have the cesium chloride structure and (b) calculate
the packing factor for the compound.
This is an example of how to handle unit cell calculations when more than one type of atom is
involved.
The structure of KCl is shown in
the figure. The large central ion
is the Cl -- and the corner ions are
the K+.
There are a few important
points to remember about
this structure.
First, although 8 potassium ions are shown and only a single chorine, the unit cell contains just
one of each. As discussed in the section on packing factors and density calculations, only
one-eighth of each of the corner atoms actually lies within the cubic unit cell (the other parts lie
in other, adjacent unit cells), so there is just 8 = 1 potassium ion, and the stoichiometric ratio of
1 K : 1 Cl is maintained.
Second, although there is an atom at the center of the cubic unit cell, this is NOT a
body-centered cubic structure. That is because the definition of a lattice point requires that the
same atom or group of atoms lie on each point. In this case, the atom at the center (Cl) is not
the same as the ones at the corners (K) so the center point cannot be a lattice point. In fact,
this is an example of the CsCl structure which is simple cubic.
But unlike many of the unit cells discussed earlier in this chapter, there is not just one atom per
lattice point. Now there are two. If you group together the K ion at the 0,0,0 point and the Cl ion
at 1/2, 1/2, 1/2, this forms the BASIS for the structure (2 atoms - some structures have many
more in the group). Placing this group of atoms at each lattice point (for simple cubic, the corner
of the cube) produces the entire structure.
Now we will calculate the radius ratio for the structure. From Appendix B:
rK
1.33. 10
10.
m
r Cl
1.81. 10
10.
m
So, the radius ratio is:
r rat
rK
r rat = 0.73
r Cl
Since this is between 0.732 and 1.000, the coordination number is 8 and the CsCl structure is
likely.
The ions touch along the body diagonal of the unit cell, so:
3. a 0 2. r K
a0
2. r K
2. r Cl
2. r Cl
a 0 = 3.63 10
10
m
3
If you are in doubt about how the ions touch in a unit cell, remember that the positive ions want
to touch the negative ones, and not the other positive ones (and vice versa). In this example it is
pretty simple. In more complicated structures like NaCl (the next problem), which is face
centered cubic with a basis of 2 and has octahedral coordination with six neighbors around each
ion, it is not so obvious. The touching occurs between (+) and (_ ) ions along the cube edges in
that case.
Knowing the length of the side of the unit cell allows calculating its volume, which is needed to
compute the packing factor.
PF
4. .
3
π rK
3
3
r Cl
PF = 0.73
3
a0
The numerator counts up the atoms and their volumes. In this case there is one of each atom in
the unit cell, so the expression for the volume of a sphere is used with each radius to compute
the volume occupied by the atoms. The denominator is just the volume of the unit cell.
The density of the material would be calculated in much the same way, adding up the atomic
weights of the atoms in the unit cell, dividing by the volume and Avogadro's number.
3.3 Crystal Structure
Show that MgO can have the NaCl crystal structure and calculate the density of MgO.
This problem first asks that you recall the NaCl crystal structure. The figure shows this, for the
specific case of Mg and O ions (remember that the ion radii are not the same as the atom radii).
The large ions are the oxygen.
In this perspective view with space-filling atoms, the
overall structure is evident and you should be able
to see that each ion is surrounded by six ions of
the opposite kind, corresponding to octahedral
coordination. This is what would be predicted from
the radius ratio.
Radius of Mg:
Radius ratio:
r Mg
6.6. 10
11.
r ratio
Radius of oxygen:
m
r Mg
rO
rO
1.32. 10
10.
m
r ratio = 0.5
Since this result of 0.5 lies between 0.414 and 0.732, we would expect a coordination number of 6.
Now we will calculate the unit cell size. In this structure, the atoms touch along an edge.
However, that is not easy to see in the perspective figure of the unit cell, and may be unexpected
since in the case of an fcc metal in earlier problems, we used the fact that the atoms touch
along a face diagonal, and NOT along the edge, to determine the unit cell size.
The key here is that MgO and NaCl are ionic compounds. The (+) and (-) ions try to surround and
touch each other, but try to avoid touching other ions of the same charge. Looking at the (100)
face plane of the unit cell and at the (200) plane across the center of the unit cell shows the
atom arrangement.
The key here is that MgO and NaCl are ionic compounds. The (+) and (-) ions try to surround and
touch each other, but try to avoid touching other ions of the same charge. Looking at the (100)
face plane of the unit cell and at the (200) plane across the center of the unit cell shows the
atom arrangement.
The (100) face plane
The (200) midplane
Notice in both of these illustrations that the Mg ions do not touch other Mg ions, and neither do
the larger O ions. But they do touch the ions of the opposite sign. This means that the ions
touch along the cube edge, whose length (the a0 for the unit cell) is just the sum of twice the
radius of Mg and twice the radius of O.
Lattice parameter:
a0
2. r Mg
2. r O
a 0 = 3.96 10
10
m
The number of ions per unit cell is equal to the number of lattice points per unit cell since the
NaCl structure is fcc with a basis of 2. In order to maintain charge neutrality, the number of ions
per unit cell is the same for Mg and O:
N Mg
4
NO
4
In this case, we determined the number of atoms per unit cell directly by using the known
number of lattice points in the FCC unit cell (4), and the knowledge that the basis of two
assigned one Mg ion and one O ion to each lattice point. There are other ways to get the same
information:
1. We could just count all of the atoms, taking into account the parts that lie within the cube. It
is easy to make a "silly" error this way, but at least the concept is clear. There are 8 oxygen
atoms on the corners of the cube, but only 1/8 of each lies within the cube, so 8(1/8) amounts to
1 atom. There is also an oxygen atom in the center of each of the six faces, but only one half
lies within the cube. 6(1/2) = 3. So 1 + 3 makes four oxygen atoms. In the same way, there are
12 cube edges and there is a magnesium ion on each, but only one-quarter of these atoms lie
within the cube. 12(1/4) = 3 atoms. Added to the Mg ion at the center of the cube, this gives a
total of 4 Mg atoms. The ratio of Mg to O is the correct 1:1.
2. We could count atoms based on whether their center coordinates lie within the range
0 ≤ x,y,z < 1. This means that all three of the coordinates must be greater than or
equal to zero, but strictly less than 1. Any atom with a center coordinate in x, y, or z
that is 1 actually lies in the next unit cell. If we do this, then we would count the oxygen
atoms at 0,0,0; 1/2, 1/2, 0; 1/2, 0, 1/2; and 0, 1/2, 1/2 (4 atoms) and the magnesium
atoms at 1/2, 0, 0; 0, 1/2, 0; 0, 0, 1/2; and 1/2, 1/2, 1/2 (4 atoms). All of the other
atoms have one or more of their x, y, z coordinates equal to 1.
Combining the number of ions, their masses, the volume of the unit cell, and Avogadro's number
lets us calculate the density.
Atomic weight of magnesium:
AW Mg
Atomic weight of oxygen:
AW O
Avogadro's number:
A0
gm
24.3.
mole
gm
16.
mole
23 atoms
6.023. 10 .
mole
The density is now determined by
ρ
N Mg . AW Mg
3.
N O. AW O
a0 A0
6
ρ = 4.31 10
gm
3
m
3.4 Packing Factor in Diamond Cube Structure
Determine the packing factor in DC silicon.
This is often one of the more difficult packing factor calculations. The geometry is more difficult to
visualize because the atoms do not line up in a continuous touching line in the body diagonal
direction. The image below shows the diamond-cubic unit cell with space filling atoms. This unit
cell has a face-centered cubic lattice, with two atoms as the basis. If you place a pair of touching
atoms, one at coordinates (0, 0, 0) and the other at (1/4, 1/4, 1/4) at each of the lattice points it
generates the diamond cubic structure. Each of these atoms touches four others, which are
arranged around the atom in a tetrahedral configuration.
Since the silicon atoms are covalently
bonded to each other, the tetrahedral
arrangement allows the bonds (where the
electrons are localized) to be maximally
separated from each other. To see the
tetrahedral arrangement of atoms in the
image below, consider one of the atoms
at the internal sites (with coordinates
1/4, 1/4, 1/4). This atom touches four
others - the one at the nearest corner
and three atoms in the center of the
nearest faces.
It is somewhat easier to see the atom
relationships if we shrink down the atoms,
and add a few construction lines. In the
sketch below, only the atom centers are
shown. Also, in addition to the edges of the
cubic unit cell, there is a line shown along
the body diagonal of the cell that passes
through one of the atoms at the tetrahedral
site inside the cell. This atom is at
coordinates of (1/4, 1/4, 1/4), as shown by
the construction lines parallel to the edges
that show its coordinates.
The textbook (Askeland) suggests
one way to relate the atom
dimension to the unit cell. It notes
that IF additional atoms were
inserted along the diagonal, they
would just fit. Then the diagonal
line through the cubic unit cell
would have a length exactly equal
to 4 times the atom diameter.
This line has a length equal to 8 times the radius of the atom, but it is also equal to the body
diagonal of the cube, or √ 3 a0 (the edge of the cube). Hence
8. r
3. a 0
a0
8. r
3
There is actually an easier way to reach the same conclusion. Refer back to the image showing
the atoms reduced in size, with the diagonal line. Notice that the line passes through the atom at
the 1/4, 1/4, 1/4 point. The distance from this point to the corner is just twice the radius of the
atoms. But in terms of the coordinates, the Pythagorean theorem tells us that the length of the
line segment is
2. r
a0
4
2
a0
4
2
a0
2
4
3.
a0
2
4
There are two solutions for this, one positive and one negative, but only the positive answer has
physical meaning:
a0
8. r
3
This is, of course, the same answer as we obtained with the first method. The reason for showing
two different approaches to solving for the relationship between a0 and r is to help you
understand the geometry of the unit cell.
Now we can proceed to calculate the packing factor. This is defined as the volume occupied by
the atoms divided by the volume of the unit cell.
4. . 3
π r
3
.
P N atom
3
a0
N is the number of atoms per unit cell. We can determine this in three different ways, all of which
give the same result. You may either select the method that you visualize best, or by
understanding them all and how they relate, further develop your familiarity with the idea of a unit
cell.
A. We can count the fractions of atoms within the bounds of the unit cell, referring to the
first image shown above. This would give
8 corner atoms * 1/8 + 6 face-centered atoms * 1/2 + 4 internal atoms = 8 total atoms per
unit cell.
B. We can count those atoms whose coordinates (x,y,z) have values of x, y and z that
are greater than or equal to zero but strictly less than one. A complete list of such atoms
is (0,0,0) - the corner atom at the origin. Other corner atoms have at least one coordinate
of 1 and are therefore not counted:
(1/2, 1/2, 0); (1/2, 0, 1/2); (0, 1/2, 1/2) - three face atoms. Other face atoms
have at least one coordinate of 1 and are therefore not counted.
(1/4, 1/4, 1/4); (3/4, 3/4, 1/4); (1/4, 3/4, 3/4); (3/4, 1/4, 3/4) - the four internal
atoms at tetrahedral sites. This gives a total of eight atoms.
C. We can use the fact that the unit cell has an FCC lattice and a basis of two atoms per
lattice point. The FCC lattice has four points per unit cell, so two times four equals 8.
Each of these methods gives us N = 8 as the number of atoms per unit cell.
N
8
We now substitute our definition of a 0 into our definition of P in order to cancel the r3 factor:
a0
8. r
3
4. . 3
π r
3
.
P N atom
3
a0
4. . 3
π r
3
.
P N atom
3
8. r
3
This leaves us with
P
N. 3. π
P = 0.34
128
The final answer is a packing factor of 0.34, which is much lower than metallic unit cells (BCC =
0.68, FCC = 0.74), or most ionic structures. This is because of the strong covalent bonding in
diamond cubic materials.
3.5 Lattice Parameter
The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855
grams per cubic centimeter. The atomic weight of potassium is 39.09 grams per gram-mole.
Calculate the lattice parameter and atomic radius of potassium.
This problem requires us to use the density equation for the unit cell. We know that the three
types of cubic unit cells are simple cubic with one lattice point (and hence one atom) per unit
cell, body centered cubic with two lattice points per unit cell, and face centered cubic with four
lattice points per unit cell.
First, we enter the constants that we have been given.
gm
0.855.
3
cm
Density:
ρK
Atomic weight:
AW K
gm
39.09.
mole
Atoms per cell (BCC structure):
Avogadro's number:
A0
NK
2
23 atoms
6.023. 10 .
mole
We know that
ρ
N. AW
V. A 0
3
and
V a0
Solving for volume, we can substitute in the relationship between volume and lattice parameter,
solve for a0 and calculate the answer
V
N. AW
A 0. ρ
3
a0
N. AW
A 0. ρ
1
a 0.K
N K. AW K
A 0. ρ K
3
a 0.K = 5.33 10
8
cm
This is the lattice parameter for potassium. (The value in the book's table is 5.344 x 10 -8 cm.)
Since the structure is BCC, we know that the atoms touch along the body diagonal of the unit
cell. The diagonal must therefore have a length that is equal to four times the atom radius. This is
the <111> direction, and the relationship between the length of the body diagonal and the edge
was covered in detail in example problems in the chapter.
Diag 4. r
rK
Diag
3. a 0.K
3. a 0
r K = 2.31 10
4
4. r
8
3. a 0
cm
This is the radius of the potassium atom. (The book value = 2.314 x 10 -8 cm.)
Problem 3.4 is identical except that the element is thorium (a much heavier element) and the
structure is FCC, which changes the geometric relationship between the lattice parameter and
atom radius.
The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72
gm/cm3 . The atomic weight of Thorium is 232 gm/mole. Calculate the lattice parameter and the
atomic radius of Thorium.
gm
11.72.
3
cm
ρ Th
Density:
Atomic weight:
AW Th
gm
232.
mole
Atoms per unit cell (FCC structure):
N Th
4
1
N Th. AW Th
a 0.Th
3
a 0.Th = 5.08 10
A 0. ρ Th
8
cm
This is the lattice parameter for thorium. (The value in the book's table is 5.086 x 10 -8 cm.)
Since this structure is FCC, we know from example problems in the chapter that the atoms
touch along a face diagonal. This distance is has a length of four times the atom radius. It is a
<110> direction, and has a length √ 2 times the edge of the cube.
Diag 4. r
Diag
2. a 0
4. r
2. a 0
We solve this as before to get the radius of the Thorium atom. (The book value = 1.798 x 10 -8
cm.)
r Th
2. a 0.Th
4
r Th = 1.8 10
8
cm
Another related type of problem performs the same calculations in reverse, in which the crystal
structure is the unknown. For instance:
A material with a cubic structure has a density of 0.855 gm/cm 3 , an atomic mass of 39.09
gm/mole and a lattice parameter of 5.344 x 10 -8 cm. If one atom is located at each lattice point,
determine the type of unit cell.
This problem requires us to use the density equation, but to solve for the number of atoms per
unit cell. We know that the three types of cubic unit cells are simple cubic with one lattice point
(and hence one atom) per unit cell, body centered cubic with two lattice points per unit cell, and
face centered cubic with four lattice points per unit cell.
First, we enter the constants that we have been given.
gm
0.855.
3
cm
Density of material:
ρm
Atomic weight of material:
AW m
Lattice parameter:
a 0.m
gm
39.09.
mole
8
5.344. 10 . cm
We know that
ρ
N. AW
3
a 0 .A 0
which allows us to rearrange and solve for our unknown, N:
Nm
3
ρ m. a 0.m . A 0
AW m
N m = 2.01
Since it has two atoms/lattice, it must be BCC.
3.6 Alloy Proportions
Most real materials are not one pure element, but a mixture of several. We should become
familiar with the rule for calculating density for a mixed alloy, also.
A BCC alloy of tungsten containing substitutional atoms of vanadium has the following properties:
Density:
ρ WV
Lattice parameter:
a 0.WV
gm
16.912.
3
cm
8
3.1378. 10 . cm
Atoms per cell (BCC structure):
N WV
2
Calculate the fraction of vanadium atoms in the alloy.
Since we are given the lattice information (the size of the unit cell and the fact that it has BCC
geometry with 2 atoms per cubic unit cell), the first step is to calculate the number of atom sites
per cubic centimeter.
atom
N WV
1
22
atom = 6.47 10
3
3
a 0.WV
cm
The density of the material is just mass/volume, which is the number of tungsten atoms times
their weight, plus the number of vanadium atoms times their weight. Further, we know that the
amount of tungsten is just (1 - F) where F is the fraction of atoms that are vanadium.
Atomic weight (vanadium):
AW V
Atomic weight (tungsten):
AW W
Avogadro's number:
A0
gm
50.941.
mole
gm
183.85.
mole
23 atoms
6.023. 10 .
mole
We know that
ρ atom.
F. AW V ( 1
F ) . AW W
A0
which allows us to solve for the fraction of vanadium atoms as
ρ WV. A 0
atom
F
AW W
F = 19.94 %
AW V AW W
So approximately 20% of the atoms are vanadium atoms, the remainder are tungsten atoms.
3.7 Volume Change in Allotropic Transformation
Calculate the change in volume that occurs when BCC iron is heated and changes to FCC iron.
In this case, the lattice parameters, a0 , for the bcc and fcc structures are given, so the unit cell
volume can be calculated directly, and from that the change in volume.
a 0.B
NB
8
2.86. 10 . cm
2
a 0.F
NF
8
3.59. 10 . cm
4
The factors of 2 and 4 are needed here because the FCC unit cell contains 4 atoms while the
BCC unit cell contains 2. Hence it is meaningful to compare the change in volume per atom,
since a real chunk of material will contain a constant number of atoms, and hence will
experience an actual change in volume as calculated in this way.
VB
a 0.B
3
V B = 2.34 10
23
cm
VF
a 0.F
3
V F = 4.63 10
23
cm
∆V
VF
VB
NF
NB
VB
3
3
∆ V = 1.11 %
NB
So the net change is -1.11%, indicating shrinkage when iron is heated. This is expected since
the FCC structure is more dense (has a higher atomic packing factor) than the BCC structure.
Notice that the change does not depend on the size of the Fe atom.
3.8 Linear Density
Calculate the repeat distance, linear density, and packing fraction for the [111] direction in FCC
copper.
This question is important because the directions in which the atom density is high, and the
planes in which the atom density is high, are involved in the deformation of materials due to slip
by the motion of dislocations, as discussed in the next chapter. The Burgers' vector of the
dislocations will be shortest (and thus more likely to occur) when atoms are close together.
It will help to treat each atom as a point at its center and to ignore the finite size of the atom. An
atom whose center lies on the line or plane is included in the density calculation. An atom
whose center does not lie on the line or plane, even if some part of its spherical shape is
intersected by the line or plane, is ignored. This convention is the same as the way we
calculated the packing factor and density of unit cells in earlier problems in this chapter.
One method is to painstakingly keep track of the eighths, quarters and halves of atoms at
corners, edges and faces of the unit cell that lie inside the cube. This method is prone to errors
in counting.
A much simpler way is to just use the center of each atom, and to count the atom as lying
entirely inside the cell if the center has x, y, z coordinates that are greater than or equal to zero
and strictly less than 1. This works because the unit cells repeat in all directions, and any part of
an atom that sticks out into another cell is matched by a part of one sticking in on the opposite
side.
For the linear and planar density problem, we just have to count the center points that lie along the
directions or in the planes. First, for the linear density, if we start at 0,0,0, we do not pass another
lattice point in the FCC structure until 1,1,1. The drawing shows an FCC unit cell with the lattice
sites marked. A plane (of type 110) is drawn in showing the orientation of the [111] direction line.
Geometry tells us then that this distance is just the length of the body diagonal of the unit cell.
Lattice parameter:
a0
3.62. 10
The repeat distance is then
D repeat
3. a 0
D repeat = 6.27 10
10
m
10.
m
And the linear density (the number of lattice points per length) is the reciprocal of the repeat
distance, or
λ
1
9
λ = 1.59 10
D repeat
m
1
The calculation of planar density is illustrated in Example 3.10.
3.9 Calculation of Planar Density
Calculate the planar density and planar packing fraction for the (010) and (020) planes in simple
cubic polonium, which has a lattice parameter of 3.34 x 10 -10 m.
Lattice parameter:
a0
3.34. 10
10.
m
The figure below shows the geometry of the simple cubic unit cell with the lattice points where
the polonium atoms are located. The (010) and (020) planes are shown. The (010) plane is a face
plane of the cube, and so its area is just a0 2 . The (020) plane is identical to the (010) in area,
and parallel to it, but passes through the center of the cube.
The (010) plane passes through the four corner
atoms in the cube, so it may at first appear that
the density would be four atoms divided by the
area of the face. This is NOT the case. As
discussed in Example 3.8 for linear density,
and in the calculation of atomic packing factors,
there are two equivalent ways to correctly
determine how to count atoms. One is to keep
track of fractions. Looking at the face
straight-on, it appears as shown in the figure
below (where now the atoms are drawn full size).
Notice that exactly one-quarter of each
atom lies within the square area of the
face. Four times one-quarter gives one
atom per face. The equivalent (and
easier) way to get the same result is to
count those atoms whose center
coordinates are greater than or equal to
0, and strictly less than 1. This counts
the atom at x = 0, z = 0 but not those
at x = 1, z = 0 or x = 0, z = 1 or x = 1,
z = 1. In either case, we can use the
lattice parameter and the fact that there
is one atom per face to get the planar
density of the (010) plane.
Planar area:
area p
Atoms per face:
atom
2
a0
area p = 1.12 10
19
2
m
1
And the planar density is
σ
atom
area p
18
σ = 8.96 10
m
2
The planar packing fraction is given by the planar area times the number of atoms per face
divided by the area of one face:
P
area p . atom
area face
P
4. atom
π
P
2
( 2. r ) . atom
2
π .r
P = 1.27
For the (020) plane, the area of the square remains the same as the (010), but notice in the top
figure that the plane does not pass through ANY atoms. Hence, the number of atoms per face is
zero, and the planar density and planar packing fraction are both exactly zero.
3.10 Calculation of Planar Spacing
Calculate the distance between adjacent (111) planes in gold, which has a lattice parameter of
4.079 x 10-10 m.
Lattice parameter:
a0
x, y and z values:
x
4.079. 10
1
10.
m
y
1
z
1
The equation for the distance between adjacent (111) planes is given in the text:
a0
d 111
2
x
y
2
d 111 = 2.36 10
10
m
2
z
It is not difficult to derive this expression for the case of a cubic unit cell. The simple geometry
and high symmetry of this unit cell is such that a direction [hkl] is always perpendicular to the
plane (hkl). This allows using a little geometry with the Pythagorean theorem to get the equation.
For the purposes of this example, we will accept it as a given.
As an aside, it is interesting to know how the seemingly esoteric fact of distance between
adjacent planes is used. For one thing, the most common technique for studying crystal
structures depends on it. X-ray diffraction directly measures the spacing between planes of
atoms in the crystal structure.
A beam of X-rays of known wavelength is directed toward a sample.
When the beam strikes a set of atomic planes at a critical angle (the Bragg angle) that
depends upon the plane spacing, it is diffracted. The geometry is much the same as
reflecting light from a mirror, except that it depends on constructive interference in the
wave motion of the electromagnetic radiation, and only occurs when the angle is exactly
correct.
Measuring the angle then gives the spacing.
Diffraction is the same phenomenon that causes the appearance of color fringes when light (like
x-rays but with a longer wavelength) if reflected from a thin layer of oil on a puddle. The thickness
of the layer interacts with the various colors in the white light, and each color (wavelength) is
diffracted only at the particular angle that matches the thickness.
3.11 Interstitial Site Size
What is the radius of an atom that will just fit into the octahedral site in FCC copper without
disturbing the lattice?
There are two different ways that we
can use to approach this problem. If
we start from basic geometric
principles, we can draw the
appearance of a face of the FCC unit
cell, as shown below.
The atoms in an FCC structure like copper touch along the face diagonal as shown. The location
of an interstitial atom (radius r) in an octahedral site is shown.
There are several geometric relationships that can be written down from this diagram.
Because the atoms touch along diagonal of the square, its length is four times the radius R
diag 4. R
Since the face is a square, the diagonal is √ 2 times the edge.
diag
2. a 0
Since the interstitial atom touches the corner atoms, the length of the edge is equal to 2R+ 2r:
a 0 2. R
2. r
According to the appendix in the text:
Radius of copper atom:
R Cu
1.278. 10
10.
m
from Appendix B
Now we can combine these statements to solve for the interstitial atom radius, r. This can be
done in many different sequences. The first option is as follows:
4. R
2. a 0
These are the two definitions of the diagonal.
4. R
2. ( 2. R
2. r )
Here we have substituted in the definition of a0 .
Now we solve for r to get the interstitial atom radius:
r1
2. R Cu
R Cu
11
r 1 = 5.2936 10
m
2
An alternative approach would be to use the size of the unit cell in copper, instead of the atom
radius. In that case, we would write down the following statements as our starting point. The first
three are identical to those used above, but instead of the table value for R, we use the table
(inside the cover) to get the value for a 0 .
diag 4. R
diag
2. a 0
Lattice parameter:
a 0 2. R
a0
3.6151. 10
2. r
10.
m
from Appendix A
Again, there are many ways to combine these to get the final answer. The one shown here
begins by setting our two definitions of the diagonal equal to each other and solving for R:
4. R
2. a 0
R
2.
4
a0
Now we solve our third expression, which is in terms of our two knowns, R and a 0 , for r:
r2
2. R
a0
2
r 2 = 5.2942 10
11
m
Notice that this value is ALMOST exactly the same as the one from above:
r 1 = 5.2936 10
11
m
The difference is due to the different starting points, and the different precision of the two table
values.
There is a third way to answer the problem, which is by far the easiest in terms of calculation,
although is presumes that we understand the principles behind the radius ratio for different
coordination numbers. The minimum radius ratio (r/R) for octahedral (6-neighbor) coordination is
√ 2 - 1. This corresponds to the case when the central atom just touches the six neighbors, in the
limit when they also touch each other. Recognizing that this applies to the case given in the
problem, and using the table value for the copper atom radius, gives a direct solution.
Radius of a copper atom
(according to the
appendix in the text):
R
The minimum radius ratio (r/R):
rad rat
from Table 3-6,
Askeland
r3
rad rat. R
1.278. 10
10.
2
r 3 = 5.2936 10
m
1
11
m
Here we see that the answer is the same as our first method of computation:
r 1 = 5.2936 10
11
m
r 2 = 5.2942 10
11
m
In principle this method is identical to the first method shown above (the radius ratio values in
Table 3-6 came from the same geometric reasoning). Because Mathcad carries precision to 15
places in its internal calculations, the difference in results between the two methods is
imperceivable. If, however, your approximate value of √ 2 - 1 were less precise, there would be a
noticeable discrepancy in the results.
2
1 = 0.414
r3
0.414. R
r 3 = 5.2909 10
11
m
3.12 Radius Ratio Calculations
Calculate the radius of an atom that will just fit into the cubic site and the octahedral site. The
radius of the atoms in the normal lattice positions is R.
For the cubic hole (8 neighbors), the central atom (radius r) touches two of its neighbors (radius
R) along a body diagonal of the cube. The edge of the cube has length 2R, so the diagonal has
length √ 3 2R. But this must equal the sum of the atom radii, or 2R+2r.
2. R
2. r
3. 2. R
Solve for r:
3. 2. R
r
2. R
2
Simplify:
3. R
r
R
Collect the R terms:
r R.
3
1
In terms of solutions, the R = 0 case is not physically significant since no self-respecting atom
would have a radius of zero. Looking at the R ≠ 0 case allows us to solve for the radius ratio, r/R:
rad rat
3
1
rad rat = 0.732
The same sort of computation applies to the octahedral case. Only the geometry changes.
2. R
r
2. r
2. 2. R
2. 2. R
2. R
2
r
2. R
r R.
2
R
1
For the octahedral coordination
case, the central atom (radius r)
lies in a plane surrounded by four
neighbor atoms (radius R). These
form a square of side 2R, and
hence diagonal √ 2 2R But the
diagonal is also of length 2R+2r.
Hence, following the same algebraic
steps as before, we can determine:
Again, the R=0 case is physically meaningless, and the R ≠ 0 gives us a radius ratio of
rad rat
2
1
rad rat = 0.414
If you are having difficulty in "seeing" the geometric relationships between the atoms in these
packings, or understanding why it is important, try the following experiments.
Take some ping-pong balls and some oranges. Imagine that these are two
different kinds of atoms, held together by an ionic bond.
For the ionic bond, each positive ion wants to surround itself with as many
negative ions as possible, and vice versa. In addition, the ions never want to
touch other ions of the same charge.
Now try to pack the oranges around the ping-pong ball. For a typical
orange, the diameter will typically be a little bit more than twice the
diameter of the ping-pong ball, so you will have a radius ratio close to 0.414
(√ 2 - 1). This is the critical ratio that separates tetrahedral from octahedral
packing.
Notice that if you put four oranges around the ball, corresponding to the
corners of a tetrahedron, they are well separated. In fact, there is so much
space it looks like you could fit in a few more oranges.
Now try to pack together six oranges around the ball, and it is likely that
instead of all touching the ball they will instead touch each other and let the
ping pong ball rattle around inside. This would not be a stable arrangement
since the like charges touch.
Shifting over from 4 to 6 to 8 to 12 neighbors are radius ratios that allow enough space around
the central ion to pack the neighbors so they can all touch the central ion and not touch each
other. Only 4, 6, 8 and 12 neighbor coordination occurs because other packing arrangements (5,
7, etc.) cannot be repeated to fill three-dimensional space.
3.13 Ionic Unit Cell Geometry
Would you expect NiO to have the cesium chloride, sodium chloride, or zinc blende structure?
Based on you answer, determine the lattice parameter, density, and packing factor.
This problem assumes that the material is held together by purely ionic bonds, with no covalent
component that might reduce the coordination number or dictate specific geometries for the atom
arrangement. For transition metal ions, this is usually a good assumption. Consequently, the
determining factor is the stoichiometry and the radius ratio.
NiO has equal numbers of Ni (++) and O (--) ions, so the structure must be one that has equal
numbers of the two ions. This would, for instance, rule out structures such as the hexagonal
Al2 O3 structure (which has a 3:2 ratio of atoms), or the fluorite structure (which has a 2:1 ratio).
The difference between the cesium chloride, sodium chloride and zinc blende structures is the
geometry of the atoms.
The cesium chloride structure is
simple cubic, with a basis of two.
The central (+) ion has eight-fold
coordination to the corner (-) ions
and vice versa (cubic
coordination). A radius ratio
between 0.732 and 1.000 can
support this type of packing.
The sodium chloride structure is
face-centered cubic with a basis of 2. Each
(+) ion is surrounded by six (-) ions and vice
versa (octahedral coordination). A radius
ratio between 0.414 and 0.732 can support
this type of packing.
The zinc blende structure is
face-centered cubic with a basis of
2. Each (+) ion is surrounded by
four (-) ions and vice versa
(tetrahedral coordination). A radius
ratio between 0.225 and 0.414 can
support this type of packing.
The radii of nickel (2+) and oxygen (2-) ions are taken from the table in Appendix B. Note that it
is important to use the ionic radii and not the atomic radii.
Nickel radius:
R Ni
0.69. 10
Oxygen radius:
RO
1.32. 10
10.
10.
m
m
Substituting the above values lets us calculate the radius ratio.
rad rat
R Ni
rad rat = 0.52
RO
Since the radius ratio falls into the range for octahedral (6-neighbor) packing, the structure of NiO
has the sodium-chloride geometry.
Now we can proceed to calculate the requested parameters.
For the NaCl structure, the positive and negative ions touch along the cube edge. It is always
true in ionic compounds that the oppositely charged ions touch each other and try to avoid
touching ions of like charge. This helps us to rule out any other directions in which the ions in
the structure might touch. This lets us directly determine the lattice parameter, a 0 .
The edge length (a.k.a., a 0 ) is the sum of twice the radii of the touching ions.
2. R Ni 2. R O
a0
a 0 = 4.02 10
10
m
We can now calculate the density using the standard equation. The FCC structure has four
lattice points, and the basis of two means that each point has one Nickel and one Oxygen atom.
Atomic weight of nickel:
AW Ni
gm
58.71.
mole
Atomic weight of oxygen:
AW O
gm
15.999.
mole
Avogadro's number:
A0
23 atoms
6.023. 10 .
mole
The basic density equation then yields
ρ
4. AW Ni
AW O
3.
ρ = 7.64
a0 A0
gm
cm3
Next, we calculate the packing factor. This is defined as the volume of the atoms divided by the
volume of the unit cell. First we calculate the volume of the atoms using the equation for a
sphere:
V Ni
4. .
3
π R Ni
3
V Ni = 1.38 10
VO
4. .
3
π RO
3
V O = 9.63 10
Now the packing factor for NiO is calculated as
PF
4. V Ni
VO
3
a0
PF = 67.79 %
30
30
3
m
3
m
3.14 Basis in Complex Unit Cells
Determine the number of carbon and hydrogen atoms in each unit cell of crystalline polyethylene.
The density of polyethylene is 0.9972 gm/cm 3.
The constants we know are the atomic weights of carbon and hydrogen, the dimensions of the
unit cell (which are given in Figure 3-34 in the text) and the fact that the ethylene mer that is the
basis of polyethylene has the formula C2H4 , so that the ratio of hydrogen to carbon is exactly
2:1.
gm
0.9972.
3
cm
Density of polyethylene:
ρ
Atomic weight of hydrogen:
AW H
gm
1.
mole
Atomic weight of carbon:
AW C
12.
Dimensions of unit cell:
Avogadro's number:
gm
mole
a
8
7.41. 10 . cm
b
8
4.94. 10 . cm
c
8
2.55. 10 . cm
A0
23 atoms
6.023 .10 .
mole
It is usually easiest to write down the expressions we know because of their basic definition, and
then solve for the unknown. In this example, we use the density equation, with the variable x to
get the number of atoms.
ρ
x. AW C
2.x. AW H
a. b. c. A 0
Solving for x gives us
x
ρ. a. b. c. A 0
AW C
2. AW H
x =4
Which indicates that there are 4 C atoms and 8 H atoms
This analysis is a little bit simplistic, and the density of real polyethylene is a bit lower than this.
Actually, most polymers do not form such perfect crystalline arrays of atoms that we can expect
this type of density equation to give exact answers. In Chapter 15, we will see that most
polymers are, at most, only partially crystalline.
For a metal or ceramic material that is more ideally crystalline, this method should give the
exact answer. However, we will see in the next chapter that even for such materials, the number
of atoms per unit cell may be slightly less than the theoretical integer values. This is due to the
presence of vacancies (lattice positions that are missing an atom).
Chapter 4
4.1 Dislocation Density in Aluminum
4.2 Dislocation Density in Copper
4.3 Burgers' Vectors
4.4 Density Slip Planes
4.5 Resolved Shear Stress
4.6 Critical Resolved Shear Stress
4.7 Vacancy Density - BCC
4.8 Vacancy Density - FCC
4.9 Vacancy Density and Activation Energy
4.10 Interstitial C in Iron
4.11 Grain Size
4.1 Dislocation Density in Aluminum
How many grams of aluminum, with a dislocation density of 10 10 cm/cm3 are required to give a
total dislocation length that would stretch from New York City to Los Angeles (3000 miles).
Dislocation density:
Dis ρ
Dislocation length:
L
10 cm
1. 10 .
3
cm
3000. mi
The point of the problem is to help the student visualize the amount of dislocation length that can
be packed into small volumes of metals. Of course, since the dislocation core is essentially one
atom wide, the total volume occupied by the dislocations is still small, and most atoms in the
structure are still in their ideal lattice configuration.
We know that
Dis ρ
L
V
Solving this for V gives
L
V
V = 4.83 10
8
3
m
Dis ρ
Now, we can convert the volume to mass using the table value for the density of aluminum, and
the definition of mass (mass = ρ V).
Density of aluminum:
ρ
gm
2.699.
3
cm
The mass is then
mass
ρ. V
Pretty small piece of metal!
mass = 0.13 gm
from Appendix A
4.2 Dislocation Density in Copper
A typical dislocation density in soft copper is 10 6 cm/cm3 . If the dislocations in 1000 gm of
copper were placed end to end, how many miles of dislocation would be available?
Dislocation density:
Dis ρ
Density of material:
ρ
Mass:
mass
6 cm
10 .
3
cm
gm
8.93.
3
cm
1000. gm
We can now compute the volume of the sample as
V
mass
3
V = 111.98 cm
ρ
And the dislocation length is simply volume times the dislocation density:
L
V. Dis ρ
L = 695.82 mi
The point is that an enormous length of dislocation lines are present in typical material. One
kilogram of copper (2.2 pounds) is a piece that can easily be held in one hand. It may seem
difficult to imagine how nearly 700 miles of dislocation can be crumpled and curled up inside. But
dislocations are very narrow - only a few atoms near the immediate core of the dislocation line
are affected. Most of the atoms in the material are unaffected, and lie in the ideal crystalline
arrangement.
V( mass )
mass
L( mass )
ρ
V( mass ) . Dis ρ
mass
Let's look at a plot of length as a function of mass.
4
1 10
L( mass )
mi
5000
0
0
2
4
6
8
mass
Dislocation length (mi) vs. Mass (kg)
10
1. gm, 5. gm.. 10. kg
4.3 Burgers' Vectors
Given a BCC structure with a lattice parameter of 4 x 10 -10 m and a dislocation as shown in the
sketch, determine the Burgers vector.
The figure is easier to understand if we separate it into three sketches, showing the dislocation,
a loop around the dislocation, and the Burgers' vector.
A loop consisting of a number of steps in the east, south, west and north directions would close
exactly in a perfect lattice, but does not close when it surrounds a dislocation. The clockwise
loop around the dislocation that is shown in the sketch does not exactly close. The vector
needed to close it is shown. Since this vector is perpendicular to the (222) plane, the Miller
indices of the direction b must be [111]. The length is determined by the distance between the
adjacent (222) planes.
Lattice parameter:
a0
Dislocation:
h
4. 10
2
10.
m
k
2
l
2
The length of the Burger's vector is then
a0
d 222
h
2
2
k
d 222 = 1.15 10
2
l
The Burgers vector has a [111] direction and a length of d222 .
10
m
4.4 Density of Slip Planes
The planar density of the (112) plane in BCC iron is 9.94 atoms/cm 2 . Calculate the planar
density of the (110) plane and the interplanar spacings for both the (112) and the (110) planes.
On which type of plane would slip normally occur.
(112) planar density:
ρ 112
atoms
9.94.
2
cm
The point of this problem is that slip generally occurs in high density directions and on high
density planes. The high density directions are directions in which the Burgers' vector is short,
and the high density planes are the "smoothest" for slip.
It will help to visualize these two planes as we calculate the atom density.
The (110) plane passes through the atom on the lattice point in the center of the unit cell. The
plane is rectangular, with a height equal to the lattice parameter a0 and a width equal to the
diagonal of the cube face, which is √ 2 a0 .
Lattice parameter (height):
a0
8
2.866. 10 . cm
Width:
w
2. a 0
Thus, according to the geometry, the area of a (110) plane would be
A 110
a 0 .w
A 110 = 1.16 10
15
2
cm
There are two atoms in this area. We can determine that by counting the piece of atoms that lie
within the circle (1 for the atom in the middle and 4 times 1/4 for the corners), or using atom
coordinates as discussed in Chapter 3. Then the planar density is
ρ 110
2
15
ρ 110 = 1.72 10
A 110
atoms
2
cm
The interplanar spacing for the (110) planes is
a0
d 110
1
2
d 110 = 2.03 10
2
1
0
8
cm
2
For the (112) plane, the planar density is not quite so easy to determine. Let us draw a larger
array of four unit cells, showing the plane and the atoms it passes through.
This plane is also rectangular, with a base width of √2 a0 (the diagonal of a cube face), and a
height of √3 a0 (the body diagonal of a cube). It has four atoms at corners, which are counted as
1/4 for the portion inside the rectangle (4 x 1/4) and two atoms on the edges, counted as 1/2 for
the portion inside the rectangle (2 x 1/2). This is a total of 2 atoms.
Base width:
w
2. a 0
Height:
h
3. a 0
Hence, we can calculate the area and density as for the (110) plane.
area 112
w. h
area 112 = 2.01 10
ρ 112
2
ρ 112 = 9.94 10
14
area 112
1
2
2
1
2
cm
atoms
2
cm
a0
d 112
15
d 112 = 1.17 10
2
8
cm
2
The planar density and interplanar spacing of the (110) plane are larger than that of the (112)
plane, thus the (110) plane would be the preferred slip plane.
4.5 Resolved Shear Stress
Calculate the resolved shear stress on the (111) [-1 0 1] slip system if a stress of 10,000 psi is
applied in the [001] direction of an FCC unit cell.
Stress on system:
σ
10000. psi
The geometry for this problem is given on pg. 99 and summarized in the figure below. The point
of the problem is to understand how the resolved shear stress depends on geometry. Slip occurs
on the high-atom-density planes and in high-atom-density directions in unit cells. In the FCC unit
cell, the highest density planes are the {111} family of planes, and the highest density directions
in those planes are the <110> directions.
Since the face of the cubic unit cell is a square, and the [-1 0 1] direction is a diagonal, the angle
λ is 45 degrees. We also know that in any cubic unit cell, the normal to a plane (hkl) is the
direction [hkl). Hence, the normal to the (111) plane is the [111] direction. So,
Angle ([-1 0 1] direction):
λ
45. deg
Angle ([1 1 1] direction):
φ
acos
1
φ = 54.74 deg
3
The equation for resolved shear stress is given by
τr
σ . cos( λ ) . cos( φ )
3
τ r = 4.08 10
psi
This is value is applied in the direction that slip would actually occur. If this stress exceeds the
critical value for slip to occur, then dislocations will move on these slip planes in this direction,
resulting in deformation.
4.6 Critical Resolved Shear Stress
Calculate the critical resolved shear stress when λ = 70° and φ = 30°, where slip begins at σ =
5000 psi.
λ
70. deg
φ
30. deg
σ
5000. psi
This is basically the same type of problem as the preceding (Example 4.5), except that the idea
of the critical resolved shear stress is introduced, and we are given the angles instead of having
to figure them out from the geometry of the unit cell.
τr
σ . cos( λ ) . cos( φ )
3
τ r = 1.48 10
psi
We are told that slip is observed to occur when the applied force equals 5000 psi. We calculate
that the resolved shear stress on the slip planes for that applied force is 1480 psi. Hence, this
value is the critical resolved shear stress.
4.7 Vacancy Density in BCC
Determine the number of vacancies, n, needed for a BCC iron lattice to have a density of 7.87
gm/cm3 . The lattice parameter of the iron is 2.866 x 10 -8 cm.
gm
7.87.
cm3
Density:
ρ
Lattice parameter:
a0
Atomic weight:
AW
Avogadro's number:
A0
Volume:
V
8
2.866. 10 . cm
gm
55.85.
mole
23 atoms
6.023. 10 .
mole
3
a0
The equation for density of a material based on its unit cell includes the number of atoms per unit
cell. Instead of setting this value equal to 2 (for BCC) we will treat it as an unknown. The amount
by which it is less than 2.0 will tell us how many of the atom sites are empty, giving the vacancy
density. The density is defined as
N. AW
V. A 0
ρ
Solving for N gives us
N
ρ . V. A 0
N = 1.998
AW
This is less than 2, as expected, indicating that there are vacancies at some of the sites. The
vacancy fraction is
2
F
N
F = 0.0010058
2
The number of vacancies per cubic centimeter is just the product of this number and the number
of sites per cm3. We are given the lattice parameter of the iron, and there are two lattice points
per unit cell in bcc, thus the number of lattice points per unit volume is calculated as shown
below, and multiplied by F to get the density of vacancies in the iron, n.
n
2.
V
F
19
n = 8.54 10
1
3
cm
4.8 Vacancy Density in FCC
The density of a sample of FCC palladium is 11.98 gm/cm 3 , and the lattice parameter is 3.8902
x 10-8 cm. Calculate the fraction of lattice points that are vacancies, F, and the total number of
vacancies in a cubic centimeter, n, of palladium.
gm
11.98.
3
cm
Density:
ρ
Lattice parameter:
a0
Atoms per unit cell:
A fcc
Atomic weight:
AW
Avogadro's number:
A0
Volume:
V
8
3.8902. 10 . cm
4
gm
106.4.
mole
23 atoms
6.023. 10 .
mole
3
a0
This is very similar to Example 4.7, except that the element is different and has an FCC rather
than a BCC lattice. The equation for density of a material based on its unit cell includes the
number of atoms per unit cell. Instead of setting this value equal to 4 (for FCC) we will treat it as
an unknown. The amount by which it is less than 4.0 will tell us how many of the atom sites are
empty, giving the vacancy density.
The density of a material is defined as
ρ
N. AW
V. A 0
Solving this for N gives us
N
ρ . V. A 0
N = 3.99
AW
This is less than 4, as expected, indicating that there are vacancies at some of the sites.
The vacancy fraction is
A fcc
F
N
A fcc
F = 0.0018775
The number of vacancies per cubic centimeter is just the product of this number and the number
of sites per cm3 . We are given the lattice parameter of the palladium, and there are four lattice
points per unit cell in fcc, thus the number of lattice points per unit volume is calculated as
shown below, and multiplied by the fraction of vacancies to determine the number of vacancies
per unit volume, n.
A fcc
n
V
20
.F
n = 1.28 10
1
cm3
BCC Lithium has a lattice parameter of 3.5089 x 10-8 cm and contains one vacancy per 200 unit
cells. Calculate the number of vacancies per cubic centimeter and the density of the Li.
8
3.5089 . 10 . cm
Lattice parameter:
a0
Vacancy rate:
r
Atoms per unit cell:
A bcc
Atomic weight:
AW
Volume:
V
1
200
2
gm
6.939.
mole
3
a0
This problem uses the same relationships as the previous one, but in a different combination.
This time the problem gives the vacancy fraction and asks for the density.
We begin by determining the vacancy fraction:
r
F
F = 0.0025
A bcc
Now we can use the expression for number of vacancies per unit volume from above, and
substitute in the numeric values for lithium, to get the vacancy density, n.
A bcc
n
V
.F
20
n = 1.16 10
1
cm3
To solve for the density of the lithium, we use the same expressions that were used before, but
with the numeric values for lithium, and solving for a different unknown. We know that the density
fraction is defined as
F
Ac
N
Ac
which allows us to solve for N:
N
(1
F ) . A bcc
N = 1.995
And now we can solve for the density as
ρ
N. AW
V. A 0
Note that Lithium is pretty light stuff!
ρ = 0.53
gm
3
cm
4.9 Vacancy Density and Activation Energy
Calculate the number of vacancies per cm 3 and per copper atom when copper is at room
temperature 298 K and at 1357 (just below the melting temperature). The activation energy Q
required to produce a vacancy is 20,000 calories per mole. The lattice parameter of FCC copper
is 3.62 x 10-8 cm, and there are four lattice points per unit cell in fcc.
298. K
T1
Q
cal
20000.
mole
pt fcc
4
T2
1357. K
a0
8
3.62. 10 . cm
R
cal
1.99.
mole . K
The number of lattice points per unit volume is calculated as shown below:
n
pt fcc
atoms
22
n = 8.43 10
3
3
a0
cm
The equation for number of vacancies is
Q
n. e
n v1
R. T 1
1
8
n v1 = 1.9 10
3
cm
Q
n. e
n v2
R. T 2
1
19
n v2 = 5.12 10
3
cm
The vacancy density per copper atom (fraction of copper atom sites that are empty) is then
ρ1
ρ2
n v1
n
n v2
n
ρ 1 = 2.25 10
15
ρ 2 = 6.07 10
4
Notice the increase in vacancies as the temperature increases. This can be shown graphically.
Q
n v( T )
n. e
R. T
T
250. K .. 1500. K
20
2 10
298
1357
20
1 10
3
n v ( T ) . cm
19
5 10
0
500
1000
1500
T
The exponential increase in vacancy concentration as a function of temperature is difficult to
graph on these conventional linear axes. Equations of the same form as this one, for vacancies,
occur throughout materials science and chemistry, and are called Arrhenius relationships. It is
very instructive to plot them on different axes so that the relationship is reduced to a straight
line. The slope of the line will be the activation energy, an important material property. The
straight line relationship also makes it easier to solve for values, such as in this problem.
Let's go back to the original equation. If we take the logarithm of both sides, it reduces to a
linear form like y = ax + b. The resulting equation is linear, like y = ax + b, if we treat log(nv) as
y and 1/T as x. Then the slope is proportional to Q (since -0.43429/R is a constant).
log n v( T )
x( T )
0.434.
1
Q
R. T
log( n )
a
0.434.
T
y( T )
Q
b
log n. cm3
R
a .x( T )
b
30
20
y( T )
10
0
0.001
0.002
0.003
0.004
x( T )
These plots can be further manipulated, and we will see them again often. Arrhenius
relationships are encountered in almost every situation in which we need to describe how some
property varies with temperature - the creep rate of a metal, the viscosity of a liquid, the
conductivity of a semiconductor, and so forth. Solving these equations will be much easier if you
understand the transformation that makes them linear, and the meaning of the axes. The
horizontal axis is 1/T (temperature) so high temperatures are to the left side. The vertical axis is
the log of the property, in this example the number of vacancies. Logarithmic axes have the
characteristic that equal distances are equal ratios of values. If you are not already familiar with
log scales, you will want to become so. Refer to example notebooks 5-G-1 and 5-G-2 on
Diffusion (another Arrhenius relationship) in Chapter 5.
4.10 Interstitial C in Iron
In FCC iron, carbon atoms are located at octahedral interstitial sites with coordinates (1/2, 0, 0),
whereas carbon atoms enter sites at (0,0.5,0.25) in BCC iron. The lattice parameter is 3.571 x
10-10 m for FCC iron and 2.866 x 10 -10 m for BCC iron. Carbon atoms have a radius of 0.71 x
10-10 m. Would there be a greater distortion of the lattice by an interstitial carbon atom in FCC or
BCC iron?
0.71. 10
10.
Radius of carbon atom:
RC
Radius of iron atom:
R Fe
1.241. 10
10.
Lattice parameter (FCC):
a 0.f
3.571. 10
10.
Lattice parameter (BCC):
a 0.b
2.866. 10
10.
Atoms per cell (FCC):
A fcc
4
Atoms per cell (BCC):
A bcc
2
m
m
m
m
We will also compare these sites to the location at (1/4, 1/4, 1/4) in the FCC structure, to verify
that it is smaller than the site noted above.
There site available at 1/2, 1/2, 1/2 is octahedrally coordinated (6 neighbors). This site is
identical to the ones in the center of each edge (1/2,0,0), which is shown in the sketch. Let us
calculate the size of those sites.
rf
2. R Fe
a 0.f
2
r f = 5.45 10
11
m
We know that in the FCC structure the (0.25,0.25,0.25) site is surrounded by four Fe atoms in a
tetrahedral arrangement. We also know that the minimum radius ratio for fourfold coordination is
0.225.
Minimum radius ratio (FCC):
rad rat
0.225
We can use this to directly calculate the size of the largest sphere that would fit at this location
so that it can just touch the four surrounding Fe atoms. The figure shows the FCC unit cell with
one of these sites, and the tetrahedron formed by the neighboring Fe atoms.
r inter
rad rat.
r inter = 2.84 10
2. a 0.f
A fcc
11
m
This is a much smaller site than the (1/2, 0 , 0) site calculated above, so it does not act as a site
for interstitial C.
In BCC iron, the size of the (0,0.5,0.25) site is calculated from the geometry. First, it is important
to visualize the interstitial site. This point lies in the face plane of the cube. It is a site with
tetrahedral coordination, since the carbon atom touches the two corner atoms in the face plane,
the atom at the center of the cube, and the atom at the center of the next unit cell. However,
these atoms at not at the corners of a REGULAR tetrahedron, so we can not use the radius ratio
of atoms with tetrahedral coordination to solve this case.
What we do know is that the radius of the iron atom can be calculated from the unit cell
geometry. Then the distance from the corner of the unit cell to the center of the interstitial site
must be equal to the sum of the radii of the iron atom and the size sphere that would fit into the
site, and this must equal the distance from the 0,0,0 point to the 1/2,1/4,0 point determined by
the Pythagorean theorem:
r
R Fe
h. a 0
2
k. a 0
2
l. a 0
2
Distances:
h
1
1
k
2
l
0
4
Calculated value of the radius of the iron atom:
R Fe
3. a 0.b
R Fe = 1.24 10
4
10
m
Now the radius of the BCC site is
rb
h . a 0.b
2
k . a 0.b
2
l. a 0.b
2
R Fe
r b = 3.61 10
11
m
Since the spacing in the BCC is smaller, the large carbon atom would disrupt the lattice more.
The FCC site is a LARGER site than available in BCC, although still smaller than the 0.71
angstrom size of the C atom. This is why the solubility of carbon is much greater in FCC iron
than it is in BCC iron (about 2% maximum vs. about 0.02% as will be seen in Chapter 11).
4.11 Grain Size
If we count 16 grains/in2 at magnification x250 in a photomicrograph, determine the ASTM grain
size, n.
Count:
c
16
Magnification:
m
250
This is a trivial calculation, but it offers an opportunity to discuss the importance of
measurements of microstructure, and the fact that the grains that comprise the microstructure
exist in three dimensions.
We know that
n
N 2
1
where N is the number per square inch at a magnification of 100x and n is the ASTM grain size.
To convert the number per square inch at 250x to the requisite 100x, we perform the following
calculation:
N
m
2
.c
N = 100
100
Now we can solve our first expression for n:
n
ln( N)
1
n = 7.64
ln( 2)
There is a problem with this definition of SIZE for the grains. Not only is the number of grains
counted (16) too small to provide useful measurement precision (since grains vary in size in the
microstructure, you must count enough to get a good average). There is also a flawed
assumption about the three-dimensional nature of the microstructure.
It has to do with the fact that the average area (which is 10-6 square inches per grain) calculated
here is actually the size of a random slice through the grain - not a cut across its midsection.
Whether this "SIZE" is related to the actual three-dimensional size of the grains, even in the
average of many measurements, depends on the shape of the grains, which can be quite
variable, and on the assumption that all of the grains in the material are the SAME in size and
shape, which they clearly are not. In fact, the ASTM measurement performed in this way is
actually a measure of the length of the contact lines where three grains meet in the structure, an
esoteric parameter indeed.
There is another definition of grain size, based on drawing lines on the image and counting the
number of grain boundaries that are crossed per unit length of the line. This gives an average
length of intercept across the grains, which seems at first to be a "SIZE" but is also related only
indirectly at best to the 3D size of the grains. This intercept length is also converted to a number
using some adjustment constants that make the result agree with the "ASTM Size" value defined
above for the case of equiaxed, fully recrystallized steels (this idea is discussed more in Chapter
9). But again, it is not really a measure of grain size but, in this case, of the amount of contact
area between grains - the total area of the grain boundary surfaces. This correlates quite well
with many mechanical properties, because grain boundaries can act as sources and final
destinations for dislocations.
Chapter 5
5.1 Activation Energy
5.2 Activation Energy/Diffusion: Cr in Cr2O3
5.3 Activation Energy/Diffusion: Al in Cu
5.4 Activation Energy/Diffusion: Ni in MgO
5.5 Concentration Gradient
5.6 Fick's First Law: Solids
5.7 Fick's First Law: Hydrogen Gas
5.8 Fick's First Law: Nitrogen Gas
5.9 The Arrhenius Equation: Plotting
5.10 The Arrhenius Equation: Solving
5.11 Fick's Second Law: Time vs. Temp.
5.12 Solving for Time with The Error Function
5.13 Fick's Second Law: Solving for Concentration
5.14 Fick's Second Law: Solving for Concentration II
5.15 Fick's Second Law: Solving for Temperature
5.16 Fick's Second Law: Solving for Time
5.17 Fick's Second Law: Time vs. Temp. II
5.18 Fick's Second Law: Temp. vs. Two Times
5.2 Activation Energy/Diffusion: Cr in Cr2O3
The diffusion coefficient for Cr in Cr 2 O3 is found to be 6 x 10-15 cm2 /sec at 1000 K, and 1 x 10 -9
cm2 /sec at 1673 K. Calculate the activation energy and D0 .
This is an example of solving for the activation energy as the slope of a straight line Arrhenius
plot, in this case of log D(T) vs. 1/T. But basically, it is just a problem of solving two
simultaneous equations in two unknowns.
We start by entering all of the things we know.
T1
1000. K
D1
6. 10
2
15. cm
sec
Gas constant:
T2
1673. K
D2
9 cm
1. 10 .
sec
2
R
cal
1.987.
mole . K
This example serves as a prototype for solution of simultaneous equations of the Arrhenius form,
which are used in several of the homework problems in this and subsequent chapters as well.
Q
D D 0. e
R. T
so we can set up a system of two equations and two unknowns (D 0 and Q) to find Q. First we
define
D 1 D 0. e
Q
Q
R. T 1
R. T 2
D 2 D 0. e
Now we solve for D0 in terms of rate1 and plug the result into the definition of D2 :
Q
D1
D0
e
D1
D2
Q
R. T 1
e
Q
R. T
. e
R. T 2
1
Now we engage in some algebraic manipulation... Knowing that ex/ey = ex - y, we have
D2
D1
e
Q
Q
R. T 2
R. T 1
D2
D1
e
Q.
1
1
R
T1
T2
Now using the detail that ln(ez) = z, we can write
ln
rate 2
Q. 1
1
rate 1
R T1
T2
And FINALLY, we can solve for Q:
Q
R. ln
D2
. 1
D1 T1
1
1
cal
Q = 59391
T2
mole
To get D0 , we can substitute into either of the original expressions for D1 or D2 .
D1
D0
Q
e
D2
D0
R. T 1
Q
e
R. T 2
2
D 0 = 0.0574
cm
sec
Both expressions give the same answer, of course.
2
D 0 = 0.0574
cm
sec
5.3 Activation Energy/Diffusion: Al in Cu
The diffusion coefficient for aluminum in copper is found to be 2.5 x 10 -20 cm2 /sec at 473 K, and
3.1 x 10-13 cm2 /sec at 773 K. Calculate the activation energy for the diffusion of aluminum in
copper.
T1
473. K
D1
2.5. 10
R
2
20. cm
sec
T2
773. K
D2
3.1. 10
2
13. cm
sec
cal
1.98.
mole . K
This is another example of solving for the activation energy as the slope of a straight line
Arrhenius plot, in this case of log D vs. 1/T. But basically, it is just a problem of solving two
simultaneous equations in two unknowns, one of which is the Q value asked for.
Q
D D 0. e
R. T
First we write our two equations:
D 1 D 0. e
Q
Q
R. T 1
R. T 2
D 2 D 0. e
Now we define a ratio and cancel out terms:
Q
D 2 D 0. e
D1
Q
R. T 2
D2 e
D1
Q
D 0. e
R. T 1
R. T 2
Q
e
R. T 1
Now we use the fact that ex/ey = ex - y
D2
Q.
e
1
1
R. T 2
R. T 1
D1
Now we take the natural log of both sides bearing in mind that ln(e z) = z:
ln
D2
Q.
D1
1
R. T 2
1
R. T 1
Finally, we solve for Q:
D2
ln
D1
Q
1
R. T
2
1
R. T
4
Q = 3.94 10
cal
mole
1
This example serves as a prototype for solution of simultaneous equations of the Arrhenius form,
which are used in several of the homework problems in this and subsequent chapters as well.
If you still don't feel comfortable with solving simultaneous equations of the form D = D 0 e-Q/RT ,
this is a good place to overcome your reluctance and learn to handle them, because they come
up over and over in this course and others. You may find it helpful to "linearize" the equations so
they are log D = log D0 -Q/RT. This is equivalent to y = ax + b were y = Log D, b = Log D0 , a =
-Q/R and x = 1/T. You may also find it useful to graph the function using the two points and then
measure the slope of the line.
5.4 Activation Energy/Diffusion: Ni in MgO
The diffusion coefficient for Ni in MgO is 1.23 x 10 -12 cm2 /sec at 1473 K and 1.45 x 10 -10
cm2 /sec at 2073 K. Calculate Q and D0 .
T1
1473. K
D1
1.23. 10
2
12. cm
sec
Gas constant (R):
T2
2073. K
D2
1.45. 10
R
2
10. cm
sec
cal
1.987.
mole . K
This is a classic example of the calculations necessary to solve many problems that depend
upon an Arrhenius plot. We are given two points: the diffusion constant is D 1 at T1 and D2 at T2 .
If we plot these values on an Arrhenius graph, the vertical axis is the logarithm of D and the
horizontal axis is 1/T. The straight line that connects the two points is the solution. It shows how
D varies with temperature, according to the relationship:
Q
D D 0. e
R. T
1.5 10
1.122 10
12
12
D1
cm
sec
7.439 10
13
D2
cm
sec
3.658 10
1.23 10
13
14
0.000450.0005
0.00055
0.0006
0.000650.0007
1
1
,
T1 T2
Our task is now to find the values of Q and D0 , which is simplified if we take the natural log of
both sides of our expression to give it a linear form.
Our task is now to find the values of Q and D0 , which is simplified if we take the natural log of
both sides of our expression to give it a linear form.
Q
R. T
D D 0. e
Q
R. T
ln( D ) ln D 0
Now Q/R is seen to be the slope of the plot of ln D vs. 1/T, and ln D0 is the intercept.
We can solve the two simultaneous equations (one for each point) to get the values of Q and D 0
without necessarily visualizing the relationship of the data to the slope of the Arrhenius plot.
However, it is very important to having a full understanding of what this type of equation is all
about to keep this graphical presentation in mind.
We know that
Q
R. T
ln( D ) ln D 0
so we can set up a system of two equations and two unknowns (C and Q) to find Q. First we
define
ln D 1
Q
.
RT1
ln D 0
ln D 2
ln D 0
Q
.
RT2
Now we solve both of these expressions for Q and set them equal to each other (remember that
ln(x) - ln(y) = ln(x/y)):
Q
R. T 1. ln
D1
Q
D0
R. T 1 . ln
D1
D0
R. T 2. ln
R. T 2. ln
D2
D0
D2
D0
Now we can do some rearranging (remember that ln(x)/ln(y) = ln(x-y)):
T1
T2
ln
D2
D1
D0
D0
Now we use the fact that eln(x) = x:
T1
e
T2
D2
D1
D0
And, at long last, we can solve for D 0 :
D0
D2
D1
D 0 = 7.06 10
T1
e
2
cm
11
sec
T2
Now we can substitute this value of D 0 back into either of the two the original equations to get
the value of Q.
Q
R. T 1. ln
3
Q = 2.96 10
D1
D0
cal
mole
Q
R. T 2. ln
3
Q = 2.96 10
D2
D0
cal
mole
Notice that the method for solving simultaneous equations is the same one you are used to from
algebra.
5.5 Concentration Gradient
Suppose a silicon wafer 0.1 cm thick, which originally contains 1 phosphorus atom for every 10 7
silicon atoms, is treated so that there are 400 P atoms for every 10 7 Si atoms at the surface.
Calculate the concentration gradient (a) in atomic percent/cm and (b) in atoms/cm 3 cm. The
lattice parameter of silicon is 5.4307 x 10 -10 m.
0.1. cm
Thickness of wafer:
d
Lattice parameter:
a0
Initial concentration of P:
ci
10.
5.4307. 10
m
1
c i = 1 10
7
5
%
10
Surface concentration of P:
cs
400
c s = 0.004 %
7
10
Now we can compute the gradient as
ci
∆c∆x
cs
d
%
∆ c ∆ x = 0.0399
cm
To find the gradient in terms of atoms/cm 3 , we must first find the volume of the unit cell:
3
V cell
a0
V cell = 1.6 10
22
3
cm
The total volume of 107 Si atoms which are arranged in the DC structure (8 atoms/cell) is
7
V
10 .
V cell
8
V = 2 10
16
3
cm
Thus, the compositions are
ci
cs
1
15
c i = 4.99 10
V
3
cm
400
∆c∆x
atoms
18
c s = 2 10
V
atoms
3
cm
ci
cs
d
19
∆ c ∆ x = 1.99 10
atoms
3
cm . cm
A 0.05 cm. layer of MgO is deposited between layers of nickel and tantalum to provide a diffusion
barrier that prevents reactions between the two metals. At 1673 K, nickel ions are created and
diffuse through the MgO ceramic to the tantalum. Determine the number of nickel ions that pass
through the MgO per second. The diffusion coefficient of nickel in MgO is 9 x 10 -12 cm2 /sec, and
the lattice parameter of nickel at 1673 K is 3.6 x 10 -8 cm.
First we calculate the initial concentration of nickel on each side of the barrier. Clearly, on the
tantalum side it is zero. On the nickel side, we use the fact the nickel is FCC and has four
atoms per unit cell, plus the size of the unit cell to determine the concentration.
Barrier thickness:
d
0.05. cm
Diffusion coefficient:
D
9. 10
2
12. cm
sec
Lattice parameter:
a0
Atoms per cell:
A fcc
8
3.6. 10 . cm
4
So the concentration of nickel atoms is
A fcc
c Ni
3
atoms
22
c Ni = 8.57 10
3
a0
cm
and the gradient is
∆c∆x
c Ni
d
24
∆ c ∆ x = 1.71 10
atoms
3
cm . cm
Fick's first law relates the flux of atoms to the gradient and the diffusion coefficient:
J
D. ∆ c ∆ x
atoms
13
J = 1.54 10
2
cm . sec
The - sign indicates that the direction of the flux is out of the nickel toward the tantalum.
Since the actual sample area is 2x2 cm, the total number of Ni atoms crossing the barrier is the
flux times the area, or
w
total
2. cm
2
J. w
13
total = 6.17 10
atoms
sec
5.7 Fick's First Law: Hydrogen Gas
A BCC iron structure is to be manufactured that will allow no more than 50 grams of hydrogen to
be lost per year through each cm 2 of iron, at 673 K. If the concentration at one surface is 0.05 H
atoms per unit cell and 0.001 H atoms per unit cell at the second surface, determine the
minimum thickness of the iron.
Fick's first law tells us that the flux is proportional to the concentration gradient, which is simply
the difference in concentration divided by the distance, and to the diffusion coefficient. As usual,
it is wise to write down what we know. We cannot get the diffusion coefficient for H in BCC iron
from figure 5-0 in the text, because the graph does not extend to 673 K. Rather than try to
graphically extrapolate, we will use the values for Q and D0 given in Table 5-1.
gm
50.
2
yr. cm
Allowed rate of loss:
loss
First concentration:
c1
0.05. atoms
Second concentration:
c2
0.001. atoms
Diffusion constant:
D0
cm
0.0012.
sec
2
Activation energy:
Q
cal
3600.
mole
Gas constant:
R
cal
1.987.
mole . K
Temperature:
T
673. K
Lattice parameter:
a0
Atomic weight of H:
AW
Avogadro's number:
A0
from Table 5.1
8
2.866. 10 . cm
gm
1.008.
mole
23 atoms
6.023. 10 .
mole
Now we can calculate the diffusion coefficient for H in BCC iron at T:
Q
D
D 0. e
R. T
D = 8.13 10
5
2
cm
sec
The gradient is given in terms of atoms per unit cell. The volume of the BCC iron unit cell is the
lattice parameter cubed, so in terms of atoms/cm 3 this becomes
grad
c2
c1
3
a 0 .d
We also know that the flux rate, F, is defined by Fick's first law as
D. grad
F
D.
c2
c1
Let's remember that we're looking to solve for d.
3.
a0 d
So now we have to determine the flux rate:
loss .
F
A0
17
F = 9.47 10
AW
atoms
2
cm . sec
So now, we are at long last ready to determine the minimum thickness of the iron:
d
D .
c2
3
.
Fa0
c1
d = 0.1787 cm
5.8 Fick's First Law: Nitrogen Gas
Nitrogen gas is held in a pressure chamber by an iron foil (atomic structure BCC) only 0.02 cm.
thick. The concentration of nitrogen atoms on one side is 3 x 10 20 atoms/cm3 , and on the other
side it is 5 x 1010 atoms/cm3 . Calculate the flux of nitrogen atoms through the foil at 1023 K.
0.02. cm
Thickness:
d
Concentration of atoms:
c1
20 atoms
3. 10 .
3
cm
c2
10 atoms
5. 10 .
3
cm
1023. K
Temperature:
T
Diffusion constant for N in BCC Fe:
D0
Activation energy for N in BCC Fe:
Q
cal
18300.
mole
Gas constant:
R
cal
1.987.
mole . K
2
cm
0.0047.
sec
from Table 5.1
Fick's first law tells us that the flux is proportional to the concentration gradient, which is simply
the difference in concentration divided by the distance, and to the diffusion coefficient which we
can calculate as a function of temperature.
grad
c2
c1
22
grad = 1.5 10
d
atoms
3
cm . cm
The (-) sign indicates the concentration drops from the inside to the outside of the container.
The expression for diffusion coefficient depends on two constants D0 and Q, which we obtained
from Table 5-1 above. This allows us to calculate the diffusion coefficient for nitrogen in BCC iron:
Q
D
D 0. e
R. T
D = 5.78 10
7
2
cm
sec
Now we get the flux from Fick's First law:
F
D. grad
15
F = 8.68 10
atoms
cm2. sec
So this is the number of nitrogen atoms per second per square centimeter diffusing through the
container wall.
5.9 The Arrhenius Equation: Plotting
Temperature Dependence of the Diffusion Coefficient
Created by John Chun.
In much of chemistry and materials science, the dependence of properties and reaction rates are
described by an equation of the form
Q
D D 0. e
R. T
R
cal
1.986.
mole . K
the so-called Arrhenius equation. Here we are using diffusion as such an example where
D is the diffusion coefficient, given in units of cm 2 /sec.
D0 is the diffusion parameter which is characteristic for each material.
Q is the activation energy for diffusion which is also characteristic for each
material. This has units of cal/mole.
R is the gas constant.
T is the absolute temperature.
For each material, D0 and Q are fixed, so the temperature dependence of D is exponential.
Example 1: Linear Plot of the Diffusion Coefficient
Let's take diffusion of carbon in α -iron, for instance, which has the properties
2
D0
8 cm
2.2. 10 .
sec
Q
cal
29300.
mole
We can plot the Arrhenius equation to see its temperature dependence graphically:
Q
D( T )
D 0. e
1.007 10
7.553 10
D( T )
cm
R. T
2 5.035 10
2.518 10
1.086 10
T
350. K .. 1200. K
13
14
14
14
26
200
450
700
950
1200
T
Well, this doesn't convey much information except it shows that D increases exponentially as T
increases. In fact, D increases so fast that we can see its values only from 700 K -- below 700 K,
D becomes too small.
Example 2: Log Plot of the Diffusion Coefficient
What can we do to overcome this limitation? We can plot it as a log plot by double-clicking on
the plot and selecting the Log Scale option in the dialog box. We also plot against 1/T.
1
1
1
1
1
1
D( T ) 1
2 1
cm 1
1
1
1
1
1
1
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
0.001
0.001
0.002
1
T
0.002
0.003
The exponential function has now become a linear function with a straight line of the form y = ax
+ b where x = 1/T and y = log(D). The slope of the line is given by -0.43429 x Q/R. Here, what's
important is that the slope is proportional to the activation energy Q. The numeric coefficient
0.43429 is an artifact from the fact that we took the log of ex. Had we taken the natural log (ln) of
this instead, this coefficient would've been 1. We notice that the diffusion coefficient becomes
very small at room temperature. This log plot conveys much more information than the linear plot
we made above.
Example 3: Many Curves On One Graph
Now we've seen how to make a diffusion coefficient plot for one material, let's make a plot that
has diffusion coefficient curves for many materials superimposed on one graph.
We begin by defining the dependent terms for each of the six chosen materials:
Carbon in FCC iron:
Carbon in BCC iron:
Iron in FCC iron:
Iron in BCC iron:
Nickel in FCC iron:
Manganese in FCC iron:
2
10. cm
D0 CF
20. 10
D0 CB
220. 10
D0 FF
22. 10
D0 FB
200. 10
D0 NF
77. 10
D0 MF
35. 10
sec
2
10. cm
sec
2
10. cm
sec
2
10. cm
sec
2
10. cm
sec
2
10. cm
sec
Q CF
3 cal
34.0. 10 .
mole
Q CB
3 cal
29.3. 10 .
mole
Q FF
3 cal
64. 10 .
mole
Q FB
3 cal
57.5. 10 .
mole
Q NF
3 cal
67. 10 .
mole
Q MF
3 cal
67.5. 10 .
mole
Now we define our expression as a function of three variables, D0 , Q and T:
Q
D D 0, Q, T
D 0. e
R. T
T
350. K .. 1200. K
Now we create our plot (remember that the y-axis is log scale and the x-axis is 1/T):
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
0.0005
0.001
Carbon in FCC iron
Carbon in BCC iron
Iron in FCC iron
Iron in BCC iron
Nickel in FCC iron
Manganese in FCC iron
0.0015
0.002
0.0025
0.003
5.10 The Arrhenius Equation - Part 2 (Solving)
Temperature Dependence of Diffusion Coefficient
Created by John Chun.
Introduction
In Lesson 1, we saw how diffusion coefficients change with temperature through the Arrhenius
equation. By seeing examples for several different species we saw how the plot of the Arrhenius
equation could be characterized by two parameters: the activation energy Q and diffusion
parameter D0 . From these two parameters we could figure out the diffusion coefficient D at any
temperature T.
Now we ask ourselves; how do we know Q and D0 ? In Lesson 1 these values were given for
several materials and that was easy. But how do we get these values in real life? The answer is
we get these values by running experiments. We can run an experiment where we control the
temperature and measure the extent of diffusion using a certain technique. From such an
experiment, we directly obtain D and T. Theoretically if we run this experiment at two different
temperatures, T1 and T2 , and measure D1 and D2 , then we can solve two equations for two
unknowns and obtain Q and D0 . We will see how this is done in Example 1 below. However, in
practice, many such experiments are performed at different temperatures to minimize
experimental errors.
Example 1: Solving Simultaneous Equations
We saw in Lesson 1 that diffusion coefficients can be described by the Arrhenius equation:
Q
D D 0. e
R. T
where
D is the diffusion coefficient.
D0 is the diffusion parameter which is characteristic for each material.
Q is the activation energy for diffusion which is also characteristic for each material.
R is the gas constant.
T is the absolute temperature in Kelvin.
Let us suppose that someone in the lab ran some experiments and reported the following results:
T1
1000. K
T2
1500. K
D1
5.7. 10
D2
2.3. 10
2
15. cm
sec
2
12. cm
sec
We also know that the gas constant is
R
cal
1.986.
mole . K
From these we have been asked to find the activation energy Q and diffusion parameter D 0 for
this material.
Rewriting the Arrhenius equation for the two sets of data gives:
D 1 D 0. e
Q
Q
R. T 1
R. T 2
D 2 D 0. e
Now we manipulate these equations symbolically to come up with the equations for Q and D 0 in
terms of the reported data. First we solve for Q (to see the steps involved, double-click on the
icon to the left):
POPUP
05_10p
Q
T 2. T 1
T2
. R. ln
T1
D2
cal
4
Q = 3.57 10
D1
mole
This is the equation for Q that is expressed by the four reported data T1 , D1 , T2 , and D2 .
Now we can solve either of our expressions of D for D0 and use our definition of Q:
D1
D0
Q
e
Q
e
7
D 0 = 3.74 10
7
2
cm
sec
R. T 1
D2
D0
D 0 = 3.74 10
2
cm
sec
R. T 2
Both forms, which are expressed in terms of the reported data, give the same result.
Now that we've found these values, let's make a plot as we did in Lesson 1 for this data set.
Q
D( T )
D 0. e
R. T
T
750. K .. 1750. K
2 10
15
D1
D2
1 10
15
D( T )
0
1000
1500
T 1, T 2, T
Notice how the line nicely passes through the two data points.
5.11 Fick's Second Law
We find that 10 hours are required to cause carbon to diffuse 0.1 cm into the surface of a steel
gear at 800 C (1073 K). What time is required to achieve the same carbon depth at 900 C (1173
K)? The activation energy Q for diffusion of C atoms in FCC iron is 32900 cal/mole.
As usual, we start be entering what we know.
cal
32900.
mole
Activation energy:
Q
First temperature:
T1
1073. K
Second temperature:
T2
1173. K
Time:
t1
Gas constant:
R
10. hr
cal
1.99.
mole . K
There is a great simplification available to solve problems like this, so that we don't have to work
out the entire Arrhenius equation solution or use the error function values. The expression for
concentration at any depth x (equation 5-5) is
cs
cx
x
cs
co
2. D. t
Notice that in this problem, we are told to achieve the same depth of diffusion of carbon. The left
side of this equation will be constant, so the right side must also be constant. Furthermore,
whatever the actual values of the error function may be, it is a single-valued function. For the
value of erf(z) to be constant, z itself must be constant. This means that the product of D, the
diffusion coefficient, times t, the time, must be a constant. We can use that to solve for the
change in t for a given change in D, or vice versa.
All we need to do is find the ratio of the D values at the two temperatures, and that will give us
the ratio of times.
D1
e
Q
T 1. R
We know that
D 1. t 1 D 2. t 2
D2
e
Q
T 2. R
so solving for t2 gives
t2
D 1. t 1
t 2 = 2.69 hr
D2
Raising the temperature from 800 to 900 °C shortens the time required from 10 to 2.7 hours.
5.12 Solving for Time with The Error Function
The surface of a 0.1% C steel is to be strengthened by carburizing. In carburizing the steel is
placed in an atmosphere that provides a maximum of 1.2% C at the surface of the steel. At a
high temperature, carbon then diffuses into the surface of the steel. For optimum properties for a
particular application, the steel must contain 0.45% C at the depth of 0.2 cm below the surface.
How long will carburizing take if the diffusion coefficient is 2 x 10 -7 cm2 /sec?
Concentrations:
cs
Depth:
x
Activation energy:
Q
Diffusion constant:
D0
Gas constant:
R
1.2. %
0.1. %
co
cx
0.45. %
0.2. cm
cal
32900.
mole
2
cm
0.115.
sec
cal
1.987.
mole . K
First we define the error function. (The error function is actually a built-in function in Mathcad, so
the definition below is not active.) This is used below instead of requiring the value to be read
from the graph in the textbook. However, it is always wise to confirm the calculated value by
comparing it to the graph. This will serve both as a check on the entered expression, and (more
importantly) on your ability to read the graph.
x
erf( x)
2 .
π
e
2
u
du
z
2 , 1.99 .. 2
0
1
erf( z ) 0
1
2
1
0
z
1
2
The form of Fick's second law given in the text relates the concentration at any depth c x to the
original and surface concentrations c S and co . It assumes that diffusion is one-dimensional, and
that D does not change with concentration. The relationship then becomes just the "error
function." This is a mathematical function (defined above in terms of the Gaussian curve) that we
can treat just as we would any other expression. It provides a value between 0 and 1 for each
positive value of the argument z.
erf z
cs
cx
cs
co
erf z = 0.68
So now we define and plot
y( z)
erf( z)
erf z
1
0
y( z )
1
2
2
1
0
1
2
z
From the plot, we can make a rough estimate of the solution for z (where the trace crosses the y
axis). In Mathcad, we need this rough guess in order to use the root function to find the solution.
First we enter the rough estimate:
z
1
Now we define our root function:
sol
root( y( z) , z)
erf( sol ) = 0.68
sol = 0.7072
erf z = 0.68
Now we can equate this to a relationship involving the depth x, the time t, and the diffusion
coefficient D:
z
x
2. D. t
We can solve this for t as a function of D and graph it:
2
2
x
t( D )
D
x
0.1 , 0.2 .. 3
2
4. D. z
2
= 0.01 cm
2
4. z
1 10
5
t( D ) 5 10 6
0
0
1
2
3
4
D
But we also know that D varies with temperature for the diffusion of C in FCC iron.
Q
D( T )
D 0. e
R .T
Substituting the expression for D gives a relationship between time t and temperature T
2
x
t( T )
2
4. D( T ) . z
We can now substitute in specific T values to evaluate this:
t( 1173. K ) = 32.61 hr
t( 1273. K ) = 10.76 hr
t( 1373. K ) = 4.17 hr
t( 1473. K ) = 1.84 hr
But it is even more useful to make a plot of t(T) vs T:
T
800. K , 810. K .. 1600. K
4
3 10
4
2 10
t( T )
hr
4
1 10
0
800
1000
1200
T
1400
1600
Plotting the vertical scale logarithmically lets us see the relationship better
5
1 10
4
1 10
1000
t( T )
hr
100
10
1
0.1
800
1000
1200
T
1400
1600
5.13 Fick's Second Law: Concentration I
Iron containing 0.05% C is heated to 912 °C (1185 K) in an atmosphere that produces 1.20 % C
at the surface, and is held for 24 hours. Calculate the carbon content at a depth of 0.05 cm.
beneath the surface, cx, if a) the iron is BCC and b) the iron is FCC. Explain the difference
Surface composition:
cs
1.2
Initial material composition:
c0
0.05
Distance beneath surface:
x
0.05. cm
Temperature:
T
1185. K
Time:
t
Gas constant:
R
Diffusion constant (FCC):
D 0.f
24. hr
cal
1.987.
mole . K
2
Activation energy (FCC):
Qf
Diffusion constant (BCC):
D 0.b
Activation energy (BCC):
Qb
cm
0.23.
sec
cal
32900.
mole
from Table 5.1,
Askeland
2
cm
0.011.
sec
cal
20900.
mole
The form of Fick's second law given in the text relates the concentration at any depth c x to the
original and surface concentrations c S and co . It assumes that diffusion is one-dimensional, and
that D does not change with concentration.
cs
cx
cs
c0
erf( z)
z
x
2. D. t
erf(z) is the "error function." This is a mathematical function that we can treat just as we would
any other expression, such as exp(z) or sin(z). It provides a value between 0 and 1 for each
positive value of the argument z and is defined as
x
erf( x)
2 .
π
e
0
2
u
du
So now we have to find z. We begin by finding D for both the FCC and BCC forms of irons.
The FCC structure has a higher atomic packing factor than BCC, and it is therefore more difficult
for interstitial atoms such as carbon to diffuse through it. We can calculate the diffusion
coefficient for each using the values for D0 and Q given in the text. The temperature of 912 °C
was selected for this problem, because that is the temperature at which the structure of iron
changes from FCC to BCC, and hence both structures can exist at that temperature.
Qf
Df
R. T
D 0.f. e
D f = 1.97 10
7
D b = 1.54 10
6
sec
Qb
Db
R. T
D 0.b . e
2
cm
2
cm
sec
Fick's second law gives us the relationship between the concentration at any depth and z that in
turn depends on the diffusion coefficient D and time t.
zf
zb
x
z f = 0.19
2. D f. t
x
z b = 0.07
2. D b . t
Now we can find cx:
cs
cx
cs
c0
erf( z)
c x.f
cs
cs
c 0 . erf z f
c x.f = 0.9541
c x.b
cs
cs
c 0 . erf z b
c x.b = 1.1111
There is a difference in concentration of carbon at the same depth in the material, and under
identical time and temperature conditions, for the case of FCC and BCC iron. This is because
the diffusion coefficient is different, and that in turn is the consequence of the different crystal
structures. In general, the higher the atomic packing factor, the more difficult it is for interstitial
atoms to diffuse through the lattice.
5.14 Fick's Second Law: Concentration II
Carbon steel by introducing 1.0%C at the surface. Calculate the carbon content at a depth 0.1
cm. beneath the surface after holding at 912 °C (1185 K) for one hour, if a) the iron is BCC and b)
the iron is FCC. Explain the difference that the structure makes.
Surface composition:
cs
1
Initial material composition:
c0
0
Distance beneath surface:
x
0.1. cm
Temperature:
T
1185. K
Time:
t
Gas constant:
R
Diffusion constant (FCC):
D 0.f
1. hr
cal
1.987.
mole . K
2
Activation energy (FCC):
Qf
Diffusion constant (BCC):
D 0.b
Activation energy (BCC):
Qb
cm
0.23.
sec
cal
32900.
mole
from Table 5.1,
Askeland
2
cm
0.011.
sec
cal
20900.
mole
The form of Fick's second law given in the text relates the concentration at any depth c x to the
original and surface concentrations c S and co . It assumes that diffusion is one-dimensional, and
that D does not change with concentration.
cs
cx
cs
c0
erf( z)
z
x
2. D. t
erf(z) is the "error function." This is a mathematical function that we can treat just as we would
any other expression, such as exp(z) or sin(z). It provides a value between 0 and 1 for each
positive value of the argument z and is defined as
x
erf( x)
2 .
π
e
0
2
u
du
However, erf(x) is a built-in function in Mathcad, just as sin(x) and exp(z) are, so it is not
necessary to define this as an active definition.
So now we have to find z. We begin by finding D for both the FCC and BCC forms of irons.
The FCC structure has a higher atomic packing factor than BCC, and it is therefore more difficult
for interstitial atoms such as carbon to diffuse through it. We can calculate the diffusion
coefficient for each using the values for D0 and Q given in the text. The temperature of 912 °C
was selected for this problem, because that is the temperature at which the structure of iron
changes from FCC to BCC, and hence both structures can exist at that temperature.
Qf
Df
R. T
D 0.f. e
D f = 1.97 10
7
sec
Qb
Db
R. T
D 0.b . e
D b = 1.54 10
2
cm
6
2
cm
sec
Fick's second law gives us the relationship between the concentration at any depth and z that in
turn depends on the diffusion coefficient D and time t.
zf
zb
x
z f = 1.88
2. D f. t
x
z b = 0.67
2. D b . t
Now we can find cx:
cs
cx
cs
c0
erf( z)
c x.f
cs
cs
c 0 . erf z f
c x.f = 0.0079
c x.b
cs
cs
c 0 . erf z b
c x.b = 0.3417
Notice that these answers are somewhat different from those in the textbook. That is because it
is very difficult to read the graph of the error function with enough accuracy to get precise
calculations, while in Mathcad, the value for the error function is built in to the math engine.
There is a large difference in concentration of carbon at the same depth in the material, and
under identical time and temperature conditions, for the case of FCC and BCC iron. This is
because the diffusion coefficient is very different, and that in turn is the consequence of the
different crystal structures. In general, the higher the atomic packing factor, the more difficult it is
for interstitial atoms to diffuse through the lattice.
5.15 Fick's Second Law: Temperature
We can successfully perform a carburizing treatment at 1200 °C (1473 K) in 1 hour. In an effort
to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing
temperature to 950 °C (1223 K). What time will be required to give the same carburizing
treatment?
T1
1473. K
T2
1223. K
t1
Gas constant:
R
Diffusion constant:
D0
1. hr
cal
1.987.
mole . K
2
Activation energy:
Q
cm
0.23.
sec
cal
32900.
mole
from Table 5.1,
Askeland
In several preceding problems in this chapter, a solution involving the error function (or inverse
error function) has been required in order to obtain time, temperature or distance results. This
problem does not require that. The solution to Fick's second law that we have been using states
that
cs
cx
cs
c0
erf( z)
z
x
2. D. t
Regardless of what the value of the error function, we know that if the left side of the equation
does not change (the concentration profile in the sample remains constant), then the error
function value must also remain constant, and hence the argument (z) of the error function
cannot change.
But z is itself a function of the diffusion coefficient D, the time t and the distance x. These must
therefore be interdependent in such a way that the value of z remains fixed. This means that
x1
x2
2. D 1. t 1 2. D 2. t 2
In this example, x, the depth of carbon diffusion, is also constant. Hence this (and the constant
2) can be pulled out, leaving the simplified relationship
D 1. t 1 D 2. t 2
The Diffusion Coefficient D can be calculated for FCC iron at each of the temperatures specified
in the problem.
Q
D1
D 0. e
R. T 1
D 1 = 3.02 10
6
D 2 = 3.03 10
7
2
cm
sec
Q
D2
D 0. e
R. T 2
2
cm
sec
We can now use these in the expression relating D and t from above:
t2
D 1. t 1
D2
t 2 = 9.95 hr
This is the time required to diffuse carbon into the steel with the identical concentration profile,
but at the lower temperature.
5.16 Fick's Second Law: Time
A 0.15% C steel is to be carburized at 1100C giving 0.35% C at a distance 1 mm. beneath the
surface. If the surface composition is maintained at 0.90% C, what time is required.
surface concentration:
cs
0.90. %
initial metal composition:
c0
required carbon concentration:
cx
0.35. %
depth beneath the surface
x
Gas constant:
R
cal
1.987.
mole . K
Temperature (use Kelvin):
T
( 1100
0.15. %
1. mm
273 ) . K
The form of Fick's second law we've been using in previous problems relates the concentration at
any depth Cx to the original and surface concentrations Cs and Co . It assumes that diffusion is
one-dimensional, and that D does not change with concentration. The relationship is equal to the
error function, and we can treat that just as we would any other expression, such as exp(z) or
sin(z). It provides a value between 0 and 1 for each positive value of the argument z.
ERFZ
cs
cx
cs
c0
ERFZ = 0.7333
In this problem, we use the error function in the "opposite direction" to that in problem number
18. Instead of calculating z, the argument of the function, and proceeding from that to the erf
value, we calculate the value of the function and then must determine the value of the argument.
In order to solve this expression for z, construct a function
y( z)
erf( z)
ERFZ
z
0 , 0.1 .. 1
and solve it for the case of y = 0. The solution is the zero-crossing of the function y. This is done
numerically and shown graphically below. The graph helps us pick a guess for Z.
guess
0.8
z0
root( y( guess ) , guess )
z 0 = 0.7854
y z 0 = 1.2093 10
0.5
z0
y z0
0
y( z )
0.5
1
0
0.2
0.4
0.6
z
0.8
1
6
Now we can solve for the time t, since the argument of the error function, z 0 , is related to x, the time
t, and the diffusion coefficient D. Calculate it from the D0 and Q values given in Table 5-1.Since the
temperature is above 912C, the iron is FCC.
2
Diffusion constant:
cm
0.23.
sec
D0
Activation energy:
Q
from Table 5.1,
Askeland
cal
32900.
mole
The expression for the diffusion coefficient. :
Q
D
D 0. e
R. T
D = 1.3316 10
6
2
cm
sec
The expression for z, the argument of the error function, is
z0
x
2. D. t
2
so
t
x
2
4. z 0 . D
t = 3043.2871 sec
or
t = 0.8454 hr
5.17 Fick's Second Law: Time vs. Temp. II
We can successfully perform a carburizing treatment at 1200C in 1 hour. In an effort to reduce
the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to
950C. What time will be required to give the same carburizing treatment.
We need (as usual) the values of R, D0 and Q:
cal
1.987.
mole . K
Gas constant:
R
Diffusion constant:
D0
2
cm
0.23.
sec
from Table 5.1,
Askeland
cal
32900.
mole
Activation energy:
Q
Temperature 1 and 2:
T1
( 1200
T2
( 950
273 ) . K
as always, don't
forget to use K!
273 ) . K
As explained in Problem 5.16,
D 1. t 1 D 2. t 2
Substituting these values into the expression for D gives
Q
D1
D 0. e
R. T 1
D 1 = 3.019 10
6
D 2 = 3.034 10
7
2
cm
sec
value at T 1 = 1473 K
Q
D2
D 0. e
R. T 2
2
cm
sec
value at T 2 = 1223 K
we can now use these in the expression relating D and t from above
t1
t2
1. hr the initial time
D 1. t 1
D2
t 2 = 9.952 hr
this is the time required to diffuse carbon into the steel with the identical concentration
profile, but at the lower temperature.
5.18 Fick's Second Law: Temp
A nitriding treatment of an BCC steel normally requires 2 hours at 600 °C (873 K). What
temperature would be required to reduce the heat treatment time to 1 hour?
T1
873. K
t1
2. hr
t2
1. hr
Gas constant:
R
Diffusion constant:
D0
cal
1.987.
mole . K
2
Activation energy:
Q
cm
0.0047.
sec
cal
18300.
mole
from Table 5.1,
Askeland
In several preceding problems in this chapter, a solution involving the error function (or inverse
error function) has been required in order to obtain time, temperature or distance results. This
problem does not require that. The solution to Fick's second law that we have been using states
that
cs
cx
cs
c0
erf( z)
z
x
2. D. t
Regardless of what the value of the error function, we know that if the left side of the equation
does not change (the concentration profile in the sample remains constant), then the error
function value must also remain constant, and hence the argument (z) of the error function
cannot change.
But z is itself a function of the diffusion coefficient D, the time t and the distance x. These must
therefore be interdependent in such a way that the value of z remains fixed. This means that
x1
x2
2. D 1. t 1 2. D 2. t 2
In this example, x, the depth of carbon diffusion, is also constant. Hence this (and the constant
2) can be pulled out, leaving the simplified relationship
D 1. t 1 D 2. t 2
The diffusion coefficient D can be calculated for BCC iron at the known temperature:
Q
D1
R. T 1
D 0. e
D 1 = 1.23 10
7
2
cm
sec
We can now use these in the expression relating D and t from above:
D2
D 1. t 1
D 2 = 2.46 10
t2
7
2
cm
sec
We can now use the definition of D2 to solve for T2 :
Q
D 2 D 0. e
R. T 2
R. ln
---->
e
R. T 2
---->
D0
Q
T2
Q
D2
D2
D0
ln
D2
D0
Q
.
RT2
T 2 = 934.39 K
T 1 = 873 K
Raising the temperature from T1 to T2 would reduce the nitriding time from 2 hours to 1 hour.
Chapter 6
6.1 Tensile Properties
6.2 Stress Calculation
6.3 Modulus of Elasticity
6.4 Modulus and Strain
6.5 Stress and Strain
6.6 Elastic Deformation
6.7 True Stress-Strain
6.8 Hardness
6.9 Fracture Toughness: Crack Size
6.10 Fracture Toughness: Failure Testing
6.11 Fracture Toughness: Failure Stress
6.12 Fatigue in Bending
6.13 Fatigue in Rotating Beam
6.1 Tensile Properties
A force of 100,000 N is applied to a 10 mm x 20 mm iron bar having a yield strength of 400 MPa
and a tensile strength of 480 MPa. Determine whether the bar will plastically deform, and
whether the bar will experience necking.
100000. newton
F
A
10. mm. 20. mm
It helps to know the definition of the units used here. 1 MPa (megaPascal) is equal to one
Newton per square mm. Hence we can directly calculate the stress applied to the bar.
σ
F
A
σ = 500 MPa
The applied stress is greater than the yield strength, so there will be plastic deformation. It is
also greater than the tensile strength, so there will be necking.
6.2 Stress Calculation
Calculate the minimum diameter of the cable required to support an elevator cab weighing 10000
lbs. The yield strength of the alloy is 35000 psi (pounds per square inch).
10000. lbf
F
σ
35000. psi
We know the definitions of σ and A:
σ
F
A
A
π.
d
2
4
Which allows us to substitute and solve for d:
σ
d
4. F
π .d
2
4. F
π .σ
d = 0.6 in
In actual practice, you would increase the expected load to take into account the load carried by
the elevator, plus additional forces due to starting and stopping, plus the weight of the cable
itself, and then include a safety factor (or factor of ignorance) to allow some margin in case you
underestimated the load, overestimated the strength of the material, or had some defects
somewhere in the cable that might locally reduce its yield strength.
6.3 Modulus of Elasticity
A force of 20000 Newtons will cause a 1 cm. x 1 cm. bar of magnesium to stretch from 10 cm. to
10.045 cm. Calculate the modulus of elasticity.
F
20000. newton
10. cm
L0
A
2
1. cm
L
10.045. cm
We start with Hooke's Law, which relates stress and strain:
σ E. ε
where σ is the stress, ε is the strain and E is what we're looking for, the modulus of elasticity.
We solve for σ as
σ
ε
F
4
σ = 2.9 10
A
L L0
L
psi
ε = 0.0045
Now we solve for E:
E
σ
ε
6
psi
4
MPa
E = 6.48 10
E = 4.46 10
So the calculated modulus in psi is close to the value of 6.5 x 10 6 given in Table 6.3.
6.4 Modulus and Strain
(1) Calculate the modulus of elasticity of an aluminum alloy that produces a strain of 0.0035 in/in
under a stress of 35000 psi. (2) Calculate the length of a 50 in bar to which a stress of 30000 psi
is applied.
in
0.0035.
in
ε1
σ1
35000. psi
Insert these values into Hooke's law for elastic deformation.
σ1
E
7
E = 1 10
ε1
psi
E is the modulus of elasticity.
Second part - how much longer is the bar when an applied force of 30,000 psi is present?
σ2
30000. psi
L0
50. in
Strain is defined as
ε
L L0
L
so if we compute the strain, we can solve for the new length, L.
ε2
L
σ2
ε 2 = 0.003
E
L0
ε2
L = 50.15 in
1
6.5 Stress and Strain
The modulus of elasticity of Nickel is 29.9 x 10 6 psi. Determine the length L of the bar when a
force of 1550 lb is applied to a bar with dimensions of 0.5 in x 0.3 in. that is originally 36 in.
long, without causing plastic deformation.
E
6
29.9. 10 . psi
A
F
1550. lbf
L0
0.5. in . 0.3. in
36. in
Since we are told that there is no plastic deformation of the bar, the deformation must be entirely
elastic. Hence, the relationship σ = Eapplied , and can be used to determine the strain. The
problem does require calculating the strain in the bar, given the applied force and cross sectional
area.
With the knowledge that
σ
E
and
ε
ε
L L0
L
we can solve for σ and find L.
σ
F
4
σ = 1.03 10
A
Now we have
E
σ.L
L L0
which, when rearranged, yields a solution for L of
L
E. L 0
σ
E
L = 36.01 in
psi
6.6 Elastic Deformation
A steel cable 1.25 inches in diameter and 50 feet long is to lift a 20 ton load (40,000 lb). What is
the length of the cable during lifting? The modulus of elasticity of he steel is 30 x 10 6 psi.
F
40000. lbf
d
1.25. in
50. ft
E
6
30. 10 . psi
L0
A
π.
d
2
4
We know that
ε
L L0
L0
Hooke's law gives us ε (the strain in the table)
σ E. ε
or
ε
σ
E
if we can find σ (the stress in the cable). And, as fate would have it, we can calculate σ quite
easily:
σ
F
4
σ = 3.26 10
A
psi
now we have
σ L L0
E
L0
which allows us to solve for L (the length of the cable during lifting):
L
σ
L 0.
E
1
L = 50.054 ft
6.7 True Stress and Strain
Compare the engineering stress and strain to the true stress and strain for the aluminum alloy in
Example 6.1 in the main text at the maximum load and at the fracture point.
The diameter at maximum load is 0.497 in. and at fracture is 0.398 in. The explanation
accompanying Example 6.1 tells us that the tensile specimen has an original diameter of 0.505
inches, and an original gauge length of 2 inches. Figure 6.2 shows the maximum applied load to
be 8000 pounds, and the load at fracture to be 7600 pounds.
Final length:
L0
2. in
L
2.12. in
d0
F max
8000. lbf
F frac
7600. lbf
0.505. in
A0
d max
0.497. in
d frac
0.398. in
Definition of engineering stress and strain:
σ eng
ε eng
F max
A0
L L0
L
σ eng = 39941 psi
ε eng = 0.0566
in
in
Definition of true stress and strain:
σ true
ε true
F max
A max
ln
L
L0
σ true = 41237 psi
ε true = 0.0583
2.205. in
L frac
in
in
At fracture, the engineering and true stresses are calculated as
A max
A frac
π.
4
2
d0
π.
4
π.
4
2
d max
2
d frac
σ eng.f
σ true.f
F frac
σ eng.f = 37944 psi
A0
F frac
σ true.f = 61088 psi
A frac
and the engineering and true strains are calculated as
ε eng.f
ε true.f
L frac
L0
L0
ln
A0
A frac
ε eng.f = 0.1025
in
ε true.f = 0.4762
in
in
in
It is only after necking begins that the reduction of area causes the true stress to become
significantly larger than the engineering stress. In this material, the true stress and true strain at
failure are significantly greater than the engineering values.
6.8 Hardness
A Brinell hardness test is performed on a steel using a 10 mm indentor with a load of 3000 kg. A
3.2-mm impression is measured on the surface of the steel. Calculate the BHN, the tensile
strength and the endurance limit.
10. mm
D
Di
3.2. mm
F
3000. kg
The relationship between the applied load and the shape and depth of the indentation is different
for each of the various types of hardness tests. For the Brinell test, the indentor is a sphere
which is assumed to be sufficiently harder than the material being tested that it does not
significantly deform, so that the indentation is spherical. Applying a known force to produce a
measurable indentation, allows calculating the hardness number.
F
BHN
π . .
D D
2
BHN = 363.21
2
D
kg
2
mm
2
Di
The empirical relationships given to estimate other properties from the hardness are specific to a
particular type of material (in this case steels). The tensile strength is:
σt
500. BHN
σ t = 181606
kg
2
mm
And the endurance limit is:
E lim
0.5. σ t
E lim = 90803
kg
2
mm
For different materials, the choice of load to produce a measurable indentation might be different,
and the relationship between the hardness number and the tensile strength would certainly be
different. Since the indentation method of measurement involves a complex combination of
elastic and plastic deformation, and work hardening, there is no fundamental expression that can
be used to relate the two. However, for a given class of materials, an empirical relationship may
be useful to allow this quick and non-destructive test to be used.
6.9 Fracture Toughness: Crack Size
For a large plate, the geometry factor f is one. Suppose a steel plate has a critical fracture
toughness of 80000 psi in 1/2 . The steel will be exposed to a stress of 45000 psi during service.
Calculate the minimum size of a crack, a, at the surface and an internal crack, x, that will grow.
1
f
1
K Ic
2
80000. psi. in
σ
45000. psi
This is an example of the calculation of critical crack size that must be able to be detected by
inspection, to prevent catastrophic failure of a part at an applied stress well below the measured
failure stress in a typical tensile test.
Usually, the calculation solves the fracture toughness relationship in one of two ways:
The maximum applied stress is calculated from the size of the
largest crack that can be detected by inspection.
The crack size is calculated from the design maximum applied
stress (this is our method here).
Notice that in all cases, the KIc value is a fixed constant that depends only on the material.
K Ic f. σ . π . a
A common error is to assume that it somehow depends on the parameters on the right side of
the equation. This is due to the usual convention for writing algebraic relationships, with the
dependent variable on the left, but it does not apply here. With K Ic determined by the material,
the important interdependence is between the failure stress and the crack size.
We can now solve for the size of the crack at the surface, a, as
a
K Ic
2
2 2
π .f .σ
a = 1.01 in
For a internal crack
x
2.a
x = 2.01 in
It is worth noting that this example uses a value of fracture toughness that is quite high. Many
brittle materials, metals and particularly ceramics, have much smaller values of K Ic, so that in
order to have the failure stress approach the value observed in a pure tensile test, the critical
crack size becomes very small, on the order of micrometers or even nanometers (about the size
of the unit cell of the atomic lattice). Such cracks are always present and may be impossible to
detect, so that other techniques are needed to avoid catastrophic failure. One such method is to
subject the material (or at least its surface) only to compressive forces, rather than tensile
forces.
6.10 Fracture Toughness: Failure Testing
A ceramic matrix composite contains internal flaws as large as 0.001 cm. in length. The plane
strain fracture toughness of the composite is 45 MPa m 1/2, and the tensile strength is 550 MPa.
Will stress cause the composite to fail before the tensile strength is reached? Assume that f = 1.
a
σt
0.001. cm
550. MPa
K Ic
f
45. MPa. m
1
This is an example of the calculation of critical crack size that must be able to be detected by
inspection, to prevent catastrophic failure of a part at an applied stress well below the measured
failure stress in a typical tensile test.
Usually, the calculation solves the fracture toughness relationship in one of two ways:
The maximum applied stress is calculated from the size of the
largest crack that can be detected by inspection.
The crack size is calculated from the design maximum applied
stress (this is our method here).
Notice that in all cases, the KIc value is a fixed constant that depends only on the material.
K Ic f. σ . π . a
A common error is to assume that it somehow depends on the parameters on the right side of
the equation. This is due to the usual convention for writing algebraic relationships, with the
dependent variable on the left, but it does not apply here. With K Ic determined by the material,
the important interdependence is between the failure stress and the crack size.
We can now use our expression to solve for the stress:
σ
K Ic
σ = 8029 MPa
f. π . a
σ t = 550 MPa
The applied stress that would be needed to cause the internal cracks to cause failure is far larger
than the tensile strength of the material. The cracks will not limit the service of the material.
It is worth noting that this example uses a value of fracture toughness that is quite high. Many
brittle materials, metals and particularly ceramics, have much smaller values of K Ic , so that in
order to have the failure stress approach the value observed in a pure tensile test, the critical
crack size becomes very small, on the order of micrometers or even nanometers (about the size
of the unit cell of the atomic lattice). Such cracks are always present and may be impossible to
detect, so that other techniques are needed to avoid catastrophic failure. One such method is to
subject the material (or at least its surface) only to compressive forces, rather than tensile
forces.
6.11 Fracture Toughness: Failure Stress
An aluminum-copper alloy with a yield strength of 47000 psi and a critical fracture toughness of
33000 psi in1/2 contains surface flaws that are 0.0003 in. deep. What is the maximum allowable
applied stress if the flaws are not to propagate? How does this compare to the yield strength?
33000. psi. in
K Ic
0.0003. in
a
f
1
This problem is typical of fracture toughness calculations. Since K is a material property, and in
all problems in this course we assume that the geometrical factor f is exactly 1, the relationship
we are calculating is between failure stress σ and crack size a. In this case, we are given the
maximum flaw size, so we solve the expression for the failure stress.
K Ic σ . f. π . a
K Ic
σ
6
σ = 1.07 10
psi
f. π . a
This is the failure stress that corresponds to the propagation of the crack. Since it is greater than
the yield strength of the material, this material is not crack sensitive. Instead, the material is
sufficiently ductile that the stress concentration at the crack tip causes the material there to
deform plastically (exceed the yield strength), which blunts the crack tip and reduces the stress
concentration. If the failure stress calculated according to the fracture toughness equation is less
than the yield strength, then the material is crack sensitive and may fail because of the presence
of cracks.
It is useful to use a graph to see how the critical stress changes with flaw depth in this material
K Ic
σ( a )
a
0.005. in , 0.01. in .. 3. in
f. π . a
5
3 10
5
2 10
σ ( a)
psi
5
1 10
0
0
1
2
a
in
3
6.12 Fatigue in Bending
A cyclical load of 1500 pounds is to be exerted at the end of a 10 inch long aluminum beam. The
bar must survive for at least 10 6 cycles. What is the minimum diameter d of the bar?
F
1500. lbf
L
10. in
cycles
6
10
The problem is not explicit about the way the bar is stressed, but from the example given in the
text, it is reasonable to assume that the bar is a rotating cantilever to which a bending stress is
applied.
Data giving the fatigue properties of this aluminum are summarized below.
80
70
60
σi
psi
50
40
30
20
4
1 10
5
1 10
6
1 10
7
1 10
8
1 10
c
i
Applied stress (kpsi) vs Number of cycles
At 106 cycles, the failure stress is approximately 34000 psi.
σ
34000. psi
The relationship for a rotating beam is given below.
σ
10.18. L. F
d
3
where d is the bar diameter (our desired value).
We can now solve this for d:
1
d
10.18. L. F
σ
3
d = 1.65 in
Of course, in real design cases, we would then apply a margin of safety to increase the diameter.
6.13 Fatigue in Rotating Beam
A 650-lb force is applied to a tool steel bar rotating at 3000 cycles/min. The bar is 1 inch in
diameter and 12 inches long. Estimate time to failure and calculate the diameter of the shaft
necessary to prevent fatigue failure.
F
650. lbf
r
3000. rpm
d
1. in
L
12. in
This problem requires reading a graph, which is reproduced in simplified form below. One of the
points raised in this problem is how to read a graph with a logarithmic scale. With the advent of
electronic calculators and the demise of the once-common slide rule for performing arithmetic,
students have lost familiarity with logarithmic scales, and may be confused by the unequal
spacing of the tick marks, or the exponential notation, so that they cannot easily read values
from the scales.
120
110
100
σi
90
80
70
60
4
1 10
5
1 10
6
1 10
7
1 10
c
i
Applied stress (k psi) vs Number of cucles
First, we calculate the relationship between the maximum stress in the part and its dimensions
and the applied force.
σ
L. F
10.18.
3
d
σ = 79404 psi
According to the graph shown above, the specimen will fail after about 250000 cycles.
r fail
250000
For a discussion on interpreting log scale plots, double-click here.
So now we can compute
POPUP (06_13p)
t
r fail
t = 83.33 min
r
The endurance limit is the horizontal portion of the curve, representing the applied stress below
which failure will not occur due to cyclic loading. From the graph, this value is 60,000 psi.
σ limit
L. F
σ limit 10.18.
3
d limit
60000. psi
1
d limit
10.18. L. F
3
d limit = 1.1 in
σ limit
One of the capabilities of Mathcad that can be exploited by the interested student to better
understand complex relationships is making graphs. The following graphs show the surface of
stresses vs. F and L, indicating how the maximum stress in the rotating bar test specimen
depends on the applied load and bar length.
i
0 .. 19
σ2
i,j
j
0 .. 19
Fi
100. ( i
10.18. Lj. Fi
2
d2
7
1.765 10
7
1.622 10
σ2
9.1 ) . lbf
Lj
(j
9 ) . in
d2
1. in
Chapter 7
7.1 Strain Hardening
7.2 Cold Work and Properties
7.3 Cold Work and Deformation
7.4 Cold Work and Thickness
7.5 Wire Drawing: Diameter
7.6 Wire Drawing: Forces
7.7 Annealing Temperature
7.1 Strain Hardening
A 0.505 inch diameter bar with a gage length of 2 inches is subjected to a tensile test. When a
force of 12,000 lb. is applied, the specimen has a diameter of 0.497 inches and a gage length of
2.007 inches. When a force of 23,000 lb. is applied, the diameter is 0.426 inches and the gage
length is 2.354 inches. Determine the strain hardening coefficient, n. Would you expect the
metal to have an FCC structure?
First, we enter the numeric values for the two points specified on the stress-strain curve.
d
0.505. in
L
2. in
d1
0.497. in
d2
0.426. in
F1
12000. lbf
F2
23000. lbf
L1
2.007. in
L2
2.354. in
Now we can compute the following:
Cross-sectional area:
A1
π.
4
2
d1
A2
π.
4
2
d2
True stress:
σ1
F1
A1
σ2
F2
A2
True strain:
ε1
ln
L1
L
ε2
ln
L2
L
Now we must solve the simultaneous equations. We will use Mathcad's Solve Block feature to
do this.
Because the solve block is an iterative process, we must provide Mathcad with a starting point
for each of the unknowns:
K
25000
n
1
Now we enter the word "Given" followed by our two expressions:
Given
ln
σ1
ln( K )
psi
n . ln ε 1
ln
σ2
ln( K )
psi
n . ln ε 2
Now we can determine the solution using "Find":
K
n
Find( K , n )
5
K = 2.5377 10
n = 0.2495
n is the strain hardening coefficient. Most FCC materials have somewhat higher coefficients than
this, so it is not likely that this material is FCC.
7.2 Cold Work and Properties
A Copper-30% Zinc brass bar is reduced from 1 inch diameter to 0.45 inch diameter. Determine
the final properties of the bar.
di
1. in
df
0.45. in
The properties
of cold-worked
70:30 brass
are shown. But
before we can
read the graph,
we must first
calculate the
amount of cold
work involved
in reducing the
bar diameter.
The formula for percent cold work is based on the cross-sectional area of the bar.
The areas are of course calculated from the diameters as
Ai
Af
π.
2
2
di
A i = 0.79 in
π. 2
d
4 f
A f = 0.16 in2
4
Then the equation for the percentage cold work is
CW
Ai
Af
CW = 79.75 %
Ai
The result shows nearly 80% cold work
Now the properties can be read directly from the graph, which gives
Tensile strength:
105000. psi
Yield strength:
70000.p s i
Elongation:
2. %
Note that the graph shown in the textbook does not plot tensile strength to 80% elongation, so
some extrapolation is required. In general, reading small graphs from the textbook will not give
better than about 5% accuracy, so if your answer disagrees slightly with these, that is no cause
for alarm.
7.3 Cold Work and Deformation
We want a Copper-30% Zinc plate originally 1.2 inches thick to have a yield strength greater than
50,000 psi and an elongation of at least 10%. What range of final thicknesses must be obtained?
ti
1.2. in
The properties of cold-worked 70:30 brass are shown.
In order to have a yield strength of at least 50,000 psi, a minimum of 20% cold work is required
(for larger amounts of cold work, the curve of yield strength lies at values greater than 50,000
psi).
CW min
20. %
Likewise, in order to have an elongation of at least 10%, the amount of cold work cannot be
greater than about 36% (for larger amounts of cold work, the curve of elongation lies at values
less than 10%).
CW max
36. %
Actually, the ability to read numeric values from these small graphs is limited, and if your
answers are slightly different from these, it is OK. The point is to learn how to read the graph to
get the limits, and use them properly in the calculation below.
The equation for the percentage cold work in a plate is shown below. This is actually the same
as the equation used in Example 7.2, where we used the cross sectional area of a bar. But for
the case of rolling a plate, the width does not change (the plate becomes longer in the rolling
process, of course), so the area is just the product of the thickness and width, and the width
cancels out to give the equation shown.
CW
ti
tf
ti
where ti is the original thickness and t f the final thickness. We can now solve this expression for
tf and determine the minimum and maximum values using our values of CW:
t f.min
t i. 1
CW min
t f.min = 0.96 in
t f.max
t i. 1
CW max
t f.max = 0.77 in
Consequently, the rolling process must result in a final plate thickness between 0.77 and 0.96
inches thick in order to achieve the required minimum yield strength of 50,000 psi and elongation
no less than 10%.
7.4 Cold Work and Thickness
Suppose we want to reduce a copper plate to a final thickness of 0.1 cm with at least 65000 psi
tensile strength, 60000 psi yield strength and 5% elongation. How much cold work is required?
Calculate the original plate thickness.
tf
0.1. cm
From Figure 7-6 in
Askeland (reproduced here),
we need at least 35% cold
work to meet the 65000 psi
tensile strength requirement
and 40% cold work to meet
the 60000 psi yield strength
requirement. However, we
must not exceed 45% cold
work, lest we fail to satisfy
the 5% elongation
requirement.
40. %
CW min
CW max
45. %
We know that
CW
ti
tf
ti
so we can solve this for t i (the initial thickness) and plug in our min and max values of CW to find
the range:
t i.min
t i.max
tf
1
CW min
tf
1
CW max
t i.min = 0.167 cm
t i.max = 0.182 cm
The initial thickness must be at least 0.167 cm. but not more than 0.181 cm.
7.5 Wire Drawing: Diameter
A 0.20 inch titanium wire is passed through a 0.18 inch diameter die in a wire-drawing process,
producing a wire with 60,000 psi yield strength and an 85,000 psi tensile strength. If the modulus
of elasticity of the titanium is 16 x 10 6 psi, calculate the diameter of the final wire product.
di
σ
0.18. in
60000. psi
E
6
16. 10 . psi
The process of wire drawing produces yielding in the material, hence the stress on the titanium
as it passes through the die is 60,000 psi. The strain is then given by Hooke's law for elastic
behavior σ = Eε . Rearranging to solve for ε gives us
ε
σ
ε = 0.00375
E
This is the magnitude of the strain in the wire. Since the stress is compressive as the wire
passes through the die, the diameter of the wire increases elastically after it has been deformed.
The strain is defined in terms of the original and final diameters, which allows us to find the final
diameter:
ε
df
df
di
di
d i. ( ε
1)
d f = 0.18068 in
The final diameter of the wire is slightly larger than the die.
7.6 Wire Drawing: Forces
A copper rod 0.25 in. in diameter is to be drawn through a 0.20 in. diameter die in a wire drawing
process. What is the force required to deform the metal? Is the force sufficient to break the wire
after it has been drawn?
di
Ai
0.25.in
π.
4
df
2
di
Af
0.20.in
π.
4
2
df
The percent cold work is calculated from the initial and final cross-sectional areas of the wire.
Ai
CW
Af
CW = 36 %
Ai
This is the % cold work produced when drawing the wire through the die.
From Figure 7.6 in
Askeland (reproduced),
the initial yield strength is
22000 psi and the yield
strength at 36% cold work
is 58000 psi.
σy
22000. psi
σt
58000. psi
The force required to deform the initial wire is:
F
σ y. A i
F = 1079.9 lbf
And this same force produces a stress acting on the wire after it passes through the die that is
increased because of the smaller cross-sectional area:
σ
F
Af
σ = 34375 psi
Since 34375 < 58000 (the tensile strength), the wire will not break.
This problem can be extended as in Example 7.5 to find the maximum amount of reduction
drawing for which the wire will not break.
7.7 Annealing Temperature
The following data were obtained when a cold-worked metal was annealed:
Number of data points:
Annealing
temperature
n
8
i
0 .. n
Residual
stresses
Tensile
strength
σR
i
Ti
523. K
548. K
573. K
598. K
623. K
648. K
673. K
698. K
1
Grain
size
σt
i
21000. psi
21000. psi
5000. psi
0. psi
0. psi
0. psi
0. psi
0. psi
grain i
52000. psi
52000. psi
52000. psi
52000. psi
34000. psi
30000. psi
27000. psi
25000. psi
0.003. in
0.003. in
0.003. in
0.003. in
0.001. in
0.001. in
0.0035. in
0.0072. in
a) Estimate the recovery, recrystallization and grain growth temperatures.
b) Recommend a suitable temperature for obtaining a high-strength, high-electrical conductivity
wire.
c) Recommend a suitable temperature for a hot-working process.
d) Estimate the melting temperature of the alloy.
Most people have some difficulty in scanning tables of numbers, and find it easier to visualize the
relationships by first plotting the data.
4
6 10
0.01
4
σ R 4.5 10
i
psi
σt
i
grain
i
4
3 10
0.005
in
psi 1.5 104
0
0
500
550
600
T
650
700
i
Residual stress vs Temperature
Tensile strength vs Temperature
500
600
T
700
i
Grain size vs Temperature
From the plot, and recalling the definition of the recovery, recrystallization, and grain growth
processes, we can estimate the temperatures.
Recovery is marked by a drop in (or elimination of) residual stresses, but without any decrease in
mechanical strength. On the graph, this happens in the temperature range from 573 to 598 K.
Since we cannot read the graph to any finer precision than the given data, any answer in that
range would be correct. We cannot select a temperature greater than 598 K because the tensile
strength would begin to decrease.
T Recov
598.K
Recrystallization is marked by a sharp decrease in grain size as new, equiaxed grains form. On
the plot, this happens somewhere in the temperature range between 598 and 623 K. Again, we
cannot specify it more closely than this because of the finite spacing of the data points. We also
note that the strength begins to drop in this temperature range.
T Recrys
623. K
Grain growth is indicated by an abrupt increase in grain size, which is evident on the plot at a
temperature between 648 and 673 K. Because of the finite spacing of the points, the line labeled
grain size appears to begin rising at 648 K, but in fact it may actually start at any temperature
between 648 and 673 K.
673.K
T GG
We are asked to recommend an annealing temperature for the production of high-strength,
high-electrical conductivity wire. This plot does not give the electrical conductivity data, but in the
chapter it is shown that electrical conductivity improves in the recovery stage of annealing, as the
dislocations are rearranged. We also note that this happens without the strength of the material
being affected. Consequently an anneal that heats the wire to the recovery temperature, 573 598 K, would be appropriate for this purpose.
TA
598.K
Next, we are asked to recommend a temperature for hot working the same material. Hot working
is defined as deforming the material at a temperature above the recrystallization temperature, so
that continuous grain refinement is possible without strain hardening. For this material, a
temperature above 623 K would be suitable for hot-working.
T H.min
623.K
Finally, we are asked to estimate the melting temperature of the alloy. The textbook suggests
the rather approximate relationship that the recrystallization temperature is about 0.4 times the
melting temperature, on an absolute temperature scale. Since we have estimated the
recrystallization temperature as about 623 K, the estimated melting temperature would be
calculated as shown.
T Recrys 0.4 .T melt
T melt
T Recrys
T melt = 1557.5 K
0.4
Of course, this is in Kelvin, so we may want to convert it to Celsius:
T melt.c
T melt
K
273
T melt.c = 1284.5
The melting temperature is probably between 1250 and 1300 °C.
Chapter 8
8.1 Critical Size for Nucleation in Cu
8.2 Critical Size for Nucleation in Ni
8.3 Undercooling for Homogeneity
8.1 Critical Size for Nucleation in Cu
Calculate the size of the critical radius and the number of atoms in the critical nucleus when
solid copper forms by homogenous nucleation.
8
3.615. 10 . cm
Lattice parameter for Cu (FCC):
a0
Melting temperature:
Tm
( 1085
Required undercooling:
∆T
236. K
Latent heat of fusion:
∆Hf
Surface energy:
σ
273 ) . K
joule
1628.
3
cm
from Table 8.1,
Askeland
5 joule
1.77. 10 .
2
cm
Now we can define the radius of the critical nucleus:
rc
2. σ . T m
∆ H f. ∆ T
r c = 1.25 10
7
cm
To determine the number of nickel atoms in the nucleus, we must determine the volume of the
critical size nucleus and the volume of the unit cell:
Vc
4. . 3
π rc
3
Vu
a0
3
V c = 8.21 10
21
cm
3
volume of the critical nucleus
V u = 4.72 10
23
cm
3
volume of the copper unit cell
We can now compute the number of unit cells in the nucleus as
n cell
Vc
Vu
n cell = 174
And since FCC has 4 atoms per cell, the number of atoms per nucleus is
n atom
4. n cell
n atom = 695
Problem 8-2 goes through the same series of calculations for nickel, comparing the size of the
critical nucleus for normal homogeneous nucleation to the size that would be needed with a
smaller amount of undercooling. The size of the nucleus increases rapidly to many thousands of
atoms, which are not likely to congregate in the crystallographic arrangement. This is why so
much undercooling is needed to cause homogeneous nucleation.
8.2 Critical Size for Nucleation in Ni
Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate
a) The critical radius of the nucleus requires.
b) The number of nickel atoms in the nucleus.
Assume that the lattice parameter of solid FCC nickel is 3.56 x 10 -8 cm. We begin to solve this
problem by using the observation that homogeneous nucleation typically requires an
undercooling (∆ T) of about 0.2 times the absolute melting temperature. And as usual with these
solutions, we start out by writing down the things we know.
8
3.56. 10 . cm
Lattice parameter:
a0
Melting temperature:
Tm
1726. K
Required undercooling:
∆T
0.2. T m
Latent heat of fusion:
∆Hf
Surface energy:
σ
joule
2756.
3
cm
from Table 8.1,
Askeland
5 joule
2.55. 10 .
2
cm
Now we can define the radius of the critical nucleus:
rc
2. σ . T m
∆ H f. ∆ T
r c = 9.25 10
8
cm
To determine the number of nickel atoms in the nucleus, we must determine the volume of the
critical size nucleus and the volume of the unit cell:
Vc
4. . 3
π rc
3
Vu
a0
3
V c = 3.32 10
21
cm
V u = 4.51 10
23
cm
We can now compute the number of unit cells in the nucleus as
3
3
Vc
n cell
n cell = 74
Vu
And since FCC has 4 atoms per cell, the number of atoms per nucleus is
4. n cell
n atom
n atom = 294
Compare this to the size of the nucleus that would have to form to become stable at a smaller
degree of undercooling. The size of the nucleus increases rapidly to many thousands of atoms,
which are not likely to congregate in the crystallographic arrangement. This is why so much
undercooling is needed to cause homogeneous nucleation.
Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22
K. How many atoms would have to group together spontaneously for this to occur?
Some of the constants from above are still useful, but are repeated here for convenience.
a 0 = 3.56 10
8
3
∆ H f = 2.76 10
∆T
cm
T m = 1726 K
joule
3
σ = 2.55 10
joule
5
2
cm
cm
22.K
We use t he same expression for critical radius from above, but now solved for a smaller value of
∆ T:
rc
2 .σ . T m
∆ H f .∆ T
r c = 1.45 10
6
cm
Note that the critical radius is larger with a smaller ∆ T.
Now we carry out the same calculation for the size of the nucleus and the number of atoms in it,
as was done above.
Vc
4. . 3
π rc
3
Vu
a0
n cell
3
Vc
Vu
V c = 1.28 10
17
cm
3
V u = 4.51 10
23
cm
3
5
n cell = 3 10
And since FCC has 4 atoms per cell, the number of atoms per nucleus is
n atom
4. n cell
6
n atom = 1.14 10
Compare this to the value of 294 for the case of normal undercooling, above, and you can see
how unlikely it would be for more than a million atoms to spontaneously line up in an FCC
arrangement and form a stable particle that would allow homogeneous nucleation in Nickel with
only 22 K undercooling.
8.3 Critical Size for Homogeneity
This series of problems involves the homogeneous nucleation of silver.
a) Calculate the temperature and number of degrees of undercooling at which silver should
nucleate homogeneously.
b) Calculate the size of the critical radius when silver nucleates homogeneously.
c) Calculate the number of atoms in the nucleus having the critical radius when silver nucleates
homogeneously. The lattice parameter of FCC silver is about 4.0862.
d) Suppose that a nucleus of silver is formed homogeneously with an undercooling of only 25C.
How many atoms would have to spontaneously group together in order for this to occur?
8
4.0862. 10 . cm
Lattice parameter:
a0
Melting temperature:
Tm
Latent heat of fusion:
∆Hf
Surface energy:
σ
273 ) . K
( 962
joule
965.
3
cm
from Table 8.1,
Askeland
7 joule
126. 10 .
2
cm
a) Askeland gives the rule of thumb that the ∆ T of undercooling for homogeneous nucleation is
about 0.2 times the absolute melting temperature.
Required undercooling:
∆T
∆ T = 247 K
0.2. T m
247
273 = 26 in Celcius
The temperature at which homogeneous nucleation should occur.
T homo
Tm ∆T
T homo = 988 K
You could go back to Celcius as well.
b) To calculate the critical nucleus:
rc
2. σ . T m
∆ H f. ∆ T
r c = 1.3057 10
7
cm
This is the critical radius of the nucleus that must form in order to become stable and to grow for
homogeneous nucleation to take place.
c) Using this value, how many atoms does that particle include? This is the number of atoms
that must "spontaneously" assemble into a particle for it to reach the critical radius, and grow to
produce homogeneous nucleation. First, calculate the volume of the spherical particle with
radius r:
Vc
4. . 3
π rc
3
V c = 9.3243 10
21
cm3
Now calculate the size of the unit cell for silver
Vu
3
a0
23
V u = 6.8227 10
3
cm
The number of cells is
Vc
n cell
Vu
n cell = 137
Since each unit cell contains 4 atoms (we were told above that it is FCC), the number of atoms
per particle is simply:
4.n cell
n atom
n atom = 547
This result says that more than 500 atoms must spontaneously collect together into a solid
particle with the proper crystal structure, in order to form a nucleus that is stable and will grow for
homogeneous nucleation. The reason that so much undercooling is required for homogeneous
nucleation is that the critical particle radius depends on the temperature difference, and at higher
temperatures the particle size is much larger. This requires that more atoms assemble
spontaneously to form the stable nucleus, and of course this is much less likely to occur.
d) The expression for critical radius from above can be solved again with the same values of
surface energy and latent heat, but with a ∆ T of 25C.
∆T
25.K
We use the same expression for critical radius from above, but now solved for a smaller value of
∆ T:
2.σ . T m
r cnew
∆ H f. ∆ T
6
r cnew = 1.29 10
cm
r c = 1.3057 10
7
cm
Note that the critical radius is larger with a smaller ∆ T.
Now we carry out the same calculation for the size of the nucleus and the number of atoms in it,
as was done above.
Vc
4. .
3
π r cnew
3
Vu
a0
n cell
3
Vc
Vu
V c = 8.9927 10
V u = 6.8227 10
18
23
3
cm
3
cm
5
n cell = 1 10
And since FCC has 4 atoms per cell, the number of atoms per nucleus is
n atom
4.n cell
5
n atom = 5.27 10
This is the number of atoms that would have to spontaneously assemble to form a stable
nucleus if homogeneous nucleation were to occur with an undercooling of only 25K (a
temperature of 937C). Comparing this to the 500 atoms needed at 715C, it is easy to see why
large amounts of undercooling are needed to produce homogeneous nucleation.