AAE 590, CRN 65011, Spring 2013
Homework #6; Due on Thursday, March 21
Solve the following problems. This HW is for your own benefit and is necessary to properly learn the material. It is
expected that each student put forward an honest effort in solving each problem and contributes to every part of the
HW. Please place all names on the HW. Please ask any questions during class or office hours.
Problem 1: Precipitation Hardening
Consider a material strengthened by precipitation of fine particles. Suppose the shear yield
strength of the particle is µ/10 (theoretical) and the strength of the matrix is µ/500. Also, assume
that the volume fraction of particles is 1 vol. % and they have a diameter of 0.010 µm. Assume
that b = 0.2 nm.
a.) Calculate the applied shear stress, τ (relative to µ), which is necessary to bow the dislocations
between particles. Hint: First calculate the number of particles (Vf=na* πd2/4).
b.) Would dislocations bow between the particles or shear them?
Solution:
a.) the volume fraction, Vf = naπd2/4, na = 4Vf /(πd2) = 0.04/(π*(0.010x10-6)2) = 1.27x1014 so
the distance, L, between particles is L = 1/√na = 0.0886 µm.
τ = µ b/L = µ (.02nm)/(0.157 µm) = 2.256x10-4*µ
b.) To shear the particles, τ = (shear strength)*(area fraction) = (µ /10)(.01) = 10-3*µ.
The dislocations will bow between the particles.
Problem 2: Precipitation Hardening
The first figure below shows the yield strength of aluminum alloy 2014, which contains about
4.5% Cu by volume, as a function of aging conditions. The shear modulus for 2014 aluminum is
28.0 GPa.
a.) For optimal properties, using both figures below, describe the ideal heat treatment of
2014 aluminum. What is your target for the particle radius, r (use your best engineering
knowledge to estimate a value for r)?
b.) Al is an FCC material with lattice constant aAl = 0.405 nm. Also, Al-Cu has an FCC
structure with lattice constant aAl-Cu = 0.404 nm. Hence, the Al-Cu precipitate is coherent
within the Al matrix. Now, assume the Al-Cu precipitate particles are hard and
impenetrable. What is the strengthening benefit (in terms of shear stress, i.e. Δτ) of the
precipitation strengthening (using your value of r from part a)?
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Solution:
(a.) Considering weight fraction of Cu equal to volume fraction (for ease of using the charts
particularly for this problem). Now, for 4.5 % Cu by weight, from the 2nd figure, it can be
seen that homogenization temperature should be around 540 o C. The material should be
heated to 540 o C and held there to allow uniform distribution of the Cu during the
homogenization heat treatment. At that point, the material should be quenched to create a
nonequilibrium structure of Cu supersaturated within the Al. From 1st figure, the aging
temperature can be chosen based on desire for strength/ductility and economic consideration
of energy to heat the oven plus time it takes to age the material. The highest strength can be
achieved at 175o C. Hence the material should then be heated to 175 o C during the aging
process and held there for 9 hours to reach the maximum possible yield stress attainable (415
MPa).
The precipitate particles ought to be very small, as for a given volume fraction of strong Cu
precipitates – the best possible strategy for strengthening is small evenly spaced particles. A
good estimate of those particles would be ranging from a few nanometers to several
micrometers.
(b.)
Δτ =
µbf
r
Let, r = 10 nm. Given, µ = 28 GPa and f = 0.045
a
b = Al
2
Δτ =
µbf
r
=
28000 × 0.405 × 0.045
= 36 MPa
2 × 10
2
Problem 3: Grain Size Effects
The following data were obtained for a carbon steel and an aluminum alloy.
Carbon Steel
d (µm)
406
106
75
43
30
16
σy (MPa)
93
129
145
158
189
233
d (µm)
42
16
11
8.5
5.0
3.1
Aluminum alloy
σy (MPa)
223
225
225
226
231
238
a.) Show that the yield strengths of this steel and aluminum alloy obey the Hall-Petch
relationship. Determine σo and ky for each material.
b.) Certain micro-alloyed steels contain small additions of vanadium or niobium that permit the
grain size to be reduced to about 2 µm if the processing of the steel is carefully controlled.
Likewise, advanced aluminum alloys containing special types of particles can be processed to
yield a grain size of about 2 µm. Suppose we reduce the grain size of steel and aluminum from
150 µm to 2 µm by such processing. What is the relative increase in strength in each material?
Comment on the significance of your answer.
Solution:
a) Hall-Petch relationship is given as :
K
σ y =σ0 +
d
From given data, if yield stress is plotted against
Hall-Petch equation.
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1
, two linear plots are obtained indicating
d
From linear fit ,
For carbon steel,
K = 689.49 MPa µm and σ 0 = 60.466 MPa
For Aluminum alloys,
K = 36.385 MPa µm and σ 0 = 215.48 MPa
b.) Rewriting the Hall-Petch equation, we get the change yield stress that would incur from
the change in grain diameter (d).
1
1
Δσ y = K (
−
)
d1
d2
For steel,
1
1
1
1
−
) = 757.9(
−
) = 474.03 MPa
d1
d2
2
150
For Aluminum alloys,
Δσ y = K (
Δσ y = K (
1
1
1
1
−
) = 20.8(
−
) = 13.01 MPa
d1
d2
2
150
From these results, it can be said that strength in steel is more sensitive to grain size than
aluminum alloys.
Problem 4: Grain Size Distributions
Many engineering alloys do not have a uniform grain size. Rather the grain size is a distribution
as shown below for a Ni-based superalloy. Comment how this would affect the overall strength
of the material in terms of the average and standard deviation of the distribution. Extra Credit:
How would this affect other properties of the material?
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Solution: The average grain size from the distribution will still follow the Hall-Petch
relationship. For instance, a smaller average grain size will be stronger than a similar alloy with
a larger average grain size. In general, increasing the standard deviation of the grain size will
reduce the strengthening benefit. Ideally, we would like the grain size to be uniform to obtain
maximum strengthening and hardening behavior.
Extra: Having larger standard deviations in grain size can be detrimental to the overall
performance of the material. Fatigue fails by finding the weakest link in the material. Hence,
having a large standard deviation will result in large grains within the material, which are weak
and prone to fatigue failure. Additionally, the various grain sizes may change the failure
mechanism in the material in cases of creep, fracture, fatigue crack growth, and environmental
assisted cracking (such as corrosion or hydrogen environments). Further, thermal, electrical,
diffusivity properties vary as a function of grain size, hence having large distributions can result
in spatial differences in these properties throughout the material.
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