Unformatted text preview: Materials Science and Engineering 206 /196
Imperfections in Crystalline Solids Prof. W.D. Nix
Spring 2008 Homework Assignment #2
SOLUTIONS
Problem 1
In this problem you were asked to find the atomic volumes of Cu and Al in the pure substance
and in solution with a variety of other metals. This was done using the definition of the volume
size factor:
Ω Ω Ω
Ω which can be rearranged to give
ΩΩ Ω 1 where ΩA is the atomic volume of pure solvent A and ΩB is the effective atomic volume of solute
B in A. With the values from the King Tables, we can now tabulate the effective atomic volumes. Solvent
Element Ω Å
Cu
11.81
Al
16.60
Ag
17.06
Au
16.96
Co
11.13
Fe
11.77
Mg
23.23
Mn
12.21
Ni
10.94
Pt
15.10
Zn
15.24
Al
16.60
Cu
11.81
Ag
17.06
Au
16.96
Co
11.13
Fe
11.77
Mg
23.23
Mn
12.21
Ni
10.94
Pt
15.10
Zn
15.24 Element
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Al
Al
Al
Al
Al
Al
Al
Al
Al
Al
Al Solute
Ω%
0.00
37.77
27.75
27.81
6.99
17.53
6.59
21.16
7.18
20.11
54.57
0.00
19.99
9.18
10.17
16.98
12.79
35.80
16.20
14.70
9.25
6.25 Ω Å
11.81
10.33
12.33
12.24
11.91
13.83
21.70
14.79
11.73
12.06
6.92
16.60
14.17
15.49
15.24
13.02
13.28
14.91
14.19
12.55
13.70
14.29 Largest Cu Effective Volume Smallest Cu Effective Volume
Largest Al Effective Volume
(Also acceptable for largest Al volume) Smallest Al Effective Volume Problem 2
In this problem you were asked to find the percent volume change due to Si going from a
precipitate phase into solid solution in a Al + 1 wt.% Si film. We begin by writing expressions for
the average atomic volume of the film, Ω, with the Si in precipitate and then solute form.
Ω Ω 1 Ω (Si precipitates) Ω Ω 1 Ω (Si in solution) is the atomic fraction of Si in the film. We refer to the King Tables to find Ω
where the
20.02 Å . We again use the volume size factor to find Ω ,
16.60 Å and Ω
Ω Ω Ω 1 From the King Tables, Ω
15.78 %, which gives us Ω
13.98 Å . We now need to
convert the weight percent into an atomic fraction. Noting that Si has a molecular weight of 28.09
g/mol, and Al a molecular weight of 26.98 g/mol, for 1 gram of the film, the atomic fraction of Si
can be calculated as:
0.01g / 28.09g/mol
0.01g / 28.09g/mol
0.99g / 26.98g/mol
, where 0.0096 is Avogradro’s Number. Consequently, 1 0.9904. We can now calculate the percent change in volume:
Ω ∆ Ω
Ω Therefore,
∆ Ω
Ω ∆ 0.0096 13.98Å
0.0096 20.02Å Ω
1 Ω 20.02Å
0.9904 16.60Å 0.35% Problem 3
In this problem you need to determine average strain in a AgCu multilayer before and after
the Ag and Cu interdiffuse, where the misfit strain is defined as: Since both Ag and Cu are FCC, we can determine the lattice parameters from the atomic volumes.
Each FCC unit cell, with volume , has 4 atoms, each with volume Ω, so
i.e. 4Ω / From the King Tables, we find that for the pure Ag and Cu, the atomic volumes are Ω
17.06 Å , and Ω 11.81 Å . This gives us lattice parameters (for the pure forms) of
3.61 Å. Calculating the misfit strain before diffusion is now a straightforward 4.09 Å and
exercise, 3.61 Å 4.09 Å 0.130 3.61 Å To calculate the misfit strain after the diffusion occurs, we need to determine the average lattice
parameter for the two AgCu phases. To do this, we first need to know how much Cu is in the Ag
phase and how much Ag is in the Cu phase. This is obtained from the AgCu phase diagram
provided. We see that at 700 , the Ag phase contains ~9 at.% Cu, and the Cu phase contains ~3
at.% Ag. We can use these atomic fractions to calculate the average atomic volume, as was done
in Problem 2.
Ω Ω
Ω Ω Ω
Ω Ω 1 Ω Ω Ω Ω Ω 1 , the volume size factor for Cu in Ag, is –27.75%, and
From the King Table, Ω
, the volume size factor for Ag in Cu, is 43.52%. Thus,
Ω
Ω 0.91 17.06 Å 0.09 17.06 Å
4Ω Ω 0.97 11.81 Å / 1 / 16.63 Å 1 0.4352 11.96 Å 4.05 Å 0.03 11.81 Å
4Ω 0.2775 3.63 Å Thus, the misfit strain after diffusion is
3.63 Å 4.05 Å 3.63 Å 0.116 Problem 4
In this problem you were asked to determine the value of the exponent, n, of the displacement
field,
for a misfitting cylinder. Let’s consider a halfcylindrical shell of radius ,
length, , and infinitesimal thickness, , as shown in fig. 1.
For mechanical equilibrium the forces on the shell in the up direction must exactly balance the
forces in the down direction.
2
2 2 z
Figure 1. Forces on a halfcylindrical shell of infinitesimal thickness For mechanical equilibrium, we must have 2 2 2 If in both sides is removed, the equation of mechanical equilibrium is given by Thus, this equation for mechanical equilibrium does not depend on the length of cylinder. Thus,
the rearranged equation is valid for cylinders with any length (even for the infinitely long cylinder
of this problem). If we rearrange the equilibrium equation again, we have When we ignore terms of order , we have
1 We now consider a displacement field of the form We want to find the value of
strains. They are for which equilibrium is satisfied. First we calculate the principal 0
Now we can directly calculate the stresses associated with the elastic strains using Hooke’s law.
The results are 2
2 2 2
2 2 Thus, the left side of the equilibrium equation (1) is
2 1 2 The right side of the equilibrium equation (1) is 2 2
2 2 2 1 Let’s plug the equation (2) and (3) into the equilibrium equation (1). 2 1 2 2 1 2 2 2
1 Finally, the displacement field is The strains surrounding the misfitting cylinder are 0
The stresses surrounding the misfitting cylinder are
2 2 2 2 0
All shear stresses ( , , ) and shear strains ( , , ) are zero. 3 ...
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 Spring '08
 nix
 Chemistry, Thermodynamics, Solubility, Chemical element

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