Homework 4 on Dislocations, Yield Stress, Hardness, Creep, Grain Size
27-301, A. D. Rollett, Fall 2002
Chemical processing plant sometimes uses nickel or nickel-based alloys because of their
corrosion resistance. It is important nonetheless to understand their strength in order to
select the correct microstructural state. Taking the uncrystallized Nickel of homework 1
as our example, we are going to estimate its strength and suitability for service.
1a. [10 points] Based on a dislocation density of 1014m-2, estimate the mean spacing
between dislocations. Explain how you made your estimate and, if you can, relate it to
what has been discussed on stereology in class.
Notes: you should first notice that dislocation density can be defined in two ways.
The first is to consider the density as an amount of dislocation line length per unit
volume (with units m/m3, = m-2). You need to relate this to a density of points
where dislocations intersect with a slip plane (with units 1/m2 = m-2, again). If
you consider dislocations arranged randomly in the material (which is the picture
in Porter & Easterling) you may apply the standard stereological relationships that
I gave you, table 2.1 from Underwood in lecture 1). If you consider a set of
parallel dislocations, however, which is often done in textbooks for simplicity,
and work it out directly, you will obtain an answer that is different by a factor of
two. Do not be alarmed by this! This is a nice illustration of the importance of
stereology. Finally, once you know the density of points that intersect a slip
plane, you can estimate the spacing between them based, for example, on the area
per dislocation (think of arranging the dislocation intersection points on a regular
lattice). There is also a stereological relationship between the density of points on
a plane (PA) and the mean spacing between those points; we did not have time to
go into this in class. I will make some supplemental slides available on this topic.
Answer: For randomly arranged dislocations, the relationship PA=LV/2 applies.
For straight and parallel dislocations, PA=LV applies since all the dislocations
thread through the slip plane. Then we need an estimate of the mean spacing.
Not discussed in class is the case for random dislocations, for which ∆2=(2√PA)-1;
for straight and parallel dislocations, ∆2=(√PA)-1. So for the former case (random),
l=(2√{LV/2})-1and for the latter case (straight), l=1/√LV. Based on LV=1014.m-2, l=
71 nm (random) or 100 nm (straight).
1b. [10 points] Now estimate the critical resolved shear stress of the nickel based on the
relations given in class, given that the shear modulus is 76 GPa and the Burgers vector is
0.25 nm.
Using t= Gb/l, and assuming that we have straight and parallel dislocations, t = 190
MPa.
Alternate answer #1:Using t= Gb/l, and assuming that we have randomly oriented
dislocations, t = 271 MPa.
Alternate answer #2: use t= aGb√r, which gives t= 95 MPa. Note the difference in
result!
1c. [10 points] What effect will raising the temperature have on this estimate of the flow
stress? Hint: consider what happens to the shear modulus.
The flow stress will tend to decrease with increasing temperature because the shear
modulus decreases with temperature (in almost all materials). Thermal activation of
dislocation motion (both conservative and non-conservative) will also lower the flow
stress. Either of these is an acceptable answer.
1d. [20 points] In a single crystal tensile test, the orientation is given as [259]//tensile
axis. Calculate the tensile yield stress based on the critical resolved shear stress that you
obtained in part b. Note: Ni is fcc and can slip on any {111} plane in any <110>
direction. You will have to find out which is the most highly stressed slip system, i.e.
find the largest Schmid factor. This value of the Schmid factor is what you should use to
determine the tensile yield stress because it determines which slip system will be
activated first. The direction [259] is approximately in the center of the standard
stereographic triangle (i.e. 001-011-111).
A suggestion for how to proceed is to use a spreadsheet (e.g. excel) and make a
list of all possible combinations of slip plane (111, -111, 1-11, -1-11) and slip
direction (e.g. 111 is orthogonal to –110, -101 and 0-11), taking only positive
versions of each (unit) vector. This will give you a table with 12 rows, one for
each slip system (and 2 columns). Then calculate the dot products of the tensile
axis with each pair of plane+direction in turn in order to obtain cosf and
cosl, respectively (2 more columns). Then you can calculate the Schmid factor as
cosf*cosl (1 more column). Finally, identify the row with the largest absolute
value of the Schmid factor in it (i.e. positive or negative). {You can expand the
table to include the negatives of each slip direction in addition: this will give you
24 rows (e.g. 111 is orthogonal to –110, -101, 0-11, 1-10, 10-1 and 01-1). If you
go with the 24 row version, you will find that you obtain a pair of positive and
negative Schmid factors for each pair of positive and negative slip directions.
This positive/negative pairing corresponds to positive and negative directions of
slip.}
Answer: The largest Schmid factor is 0.49. If you used only 12 slip systems (no
negatives) then you would have had to take the largest absolute value of the
Schmid factor. Negative Schmid factors simply mean slip in the opposite
direction on the same slip system. If you used 24 systems, then you can take the
largest positive value. Based on this analysis, the tensile yield stress is 190/0.49,
or, s = 388 MPa. [20]
1e. [15 points] The orientation of several grains has been characterized in a polycrystal:
their orientations with respect to the tensile axis are [259], [001], [011] and [111].
Calculate their Schmid factors and calculate their average in order to estimate the yield
stress of the polycrystal. You may find it helpful to draw a stereographic projection and
note the locations of the slip direction and plane on it that satisfy the maximum Schmid
factor criterion. Obviously the symmetry axes give an ambiguous result in that several
slip systems satisfy the criterion with equal Schmid factors.
[Your answers should be close to 0.49, 0.43, 0.425, and 0.275, respectively; you should
be able to use the same procedure to solve the problem as in part (e) above].
Answer: the average Schmid factor = (0.49+0.43+0.425+0.275)/4=0.405. Therefore
the polycrystal should yield at 190/0.405 = 469 MPa. The other Schmid factors can
be calculated by replacing [259] as the tensile axis direction with the other three
directions. Note that these more symmetric directions give more than one system
with the same factor, indicating that more than one slip system is active even when
only single slip is considered.
1f. [10 points] Unfortunately the above estimate is not very accurate because multiple slip
is necessary for a polycrystal to yield. The Taylor factors of the above orientations are
3.010, 2.449, 3.674, and 3.674, respectively. Compute the yield strength again from the
average of the Taylor factors, and compare to the previous result. [In the graduate
course, we would use a computer code to compute the Taylor factors, after having
computed one by hand as an example.]
Answer: the average Taylor factor, <M> = 3.20; therefore the yield stress would be
608 MPa.
1g. [10 points] If you were to measure the hardness of this nickel, what result would you
expect in terms of a Vickers hardness (as you might obtain from a microhardness test)?
A reasonable rule-of-thumb is that the hardness is approximately 3 times the yield
stress. Thus HV ~ 3 x 608= 1825 MPa! This is admittedly a very high hardness and
unlikely to be observed in practice.
1h. [5 points] Finally, what is your opinion of the suitability of this material for use in
structural applications? Would you use this, or a nickel alloy that had been strengthened
through some other means? Why?
Answer: on the whole, this particular batch of nickel is a bit too hard to be useful. It
would not fail gracefully if strained in tension, i.e. it would behave in a somewhat
brittle manner. It would be likely to recrystallize if heated and so could not be
exposed to elevated temperatures without losing strength. [5]
2a. [40 points] Nickel has a Hall-Petch coefficient of 0.13 MPa(m3/2) and a friction stress
of 60 MPa. Its shear modulus is 76 GPa and its Burgers vector is 0.25 nm. Assuming
that the grain size (diameter) determines a minimum Orowan bowing stress (ignoring
other dislocations that may be present within the grains), plot both the flow stress
determined from the Hall-Petch equation, and the Orowan bowing stress (on the same
graph) as a function of grain size, d. Use logarithmic axes in order to cover the range
1nm ≤ d ≤ 1mm. Assume an average Taylor factor appropriate for randomly oriented
polycrystals, <M>=3.1. Below what grain size does the Orowan bowing stress become
dominant?
Answer: using stresses calculated in the standard manner (see formulae on graph), the
Orowan stress becomes dominant below about 140 nm. Note the Taylor factor is needed
to convert a critical resolved shear stress based on the Orowan dowing stress to a yield
stress (whereas the Hall-Petch formula is already in terms of macroscopic yield stress).
creep.strength.hwk4.F02.Kdata
11
10
Orowan stress = <M>Gb/l
Hall-Petch = sfriction+kd-1/2
10
Flow Stress (Pa)
10
109
108
107
106
Cross-over grain size = 144 nm
105
104
1
10
100
1000
104
105
106
Grain Size (nm)
2b. [30 points] Explaining the flow stress of very fine-grained materials is challenging.
Based, however, on what you have learned about creep, it is useful to calculate the
expected creep rates based on diffusion mechanisms. Use the following equations and
data for Nabarro-Herring and Coble creep at room temperature (300K) in Cu. NabarroHerring creep is the bulk diffusion mechanism discussed in class for creep, and Coble
creep is the equivalent diffusion-based creep mechanism except that the diffusion takes
place along grain boundaries (instead of the bulk).
Nabarro-Herring Creep:
Ê D ˆ Ê sW ˆ
e˙ = ANH Á bulk
˜Á
˜
Ë d 2 ¯ Ë kT ¯
Coble Creep:
Ê D d ˆ Ê sW ˆ
e˙ = ACoble ÁÁ gb3 ˜˜ Á
˜
Ë d ¯ Ë kT ¯
For bulk diffusion, Qbulk = 200,000 J/mole, D0 = 31 mm2/s (be careful of units) and ANH =
10. For grain boundary diffusion, Qgb = 100,000 J/mole, D0 = 50 mm2/s, d = 0.25nm,
ACoble = 40. Plot the creep rate for both mechanisms against grain size, d, over the same
range of grain size and flow stress as calculated in part (a); use the larger of the two flow
stresses at each grain size. You should find that the creep rate (in units of "per second")
becomes appreciable at small enough grain sizes for one of the mechanisms despite the
low temperature. Based on this result, comment on what this means for the chances of
observing high strength in nanocrystalline (pure) copper. Try repeating your calculations
(in a spreadsheet, I trust!) for 400K. What hope is there for high strength at (slightly)
elevated temperature?!
Nabarro-Herring, room T
Coble, room T
NH (400K)
Coble (400K)
1000
1
0.001
Creep Rate
1 0-6
1 0-9
1 0-12
1 0-15
1 0-18
1 0-21
1 0-24
1 0-27
1 0-30
1 0-33
1 0-36
1
10
100
1000
104
Grain Size (nm)
105
106
Clearly, at small enough grain sizes, the creep rate (in Coble creep, but not N-H creep)
becomes appreciable for stresses close to the nominal yield stress. Raising the
temperature only makes things worse. Moreover one could reasonably expect grain
growth to be very rapid for nanocrystalline materials since the driving force rises to
significant levels for nanocrystalline materials. Copper is also prone to dynamic
recrystallization at temperatures not far above room temperature which would also tend
to coarsen the microstructure. Generally, the only hope for high strengths from
nanocrystalline materials is to stabilize the microstructure with second phase particles.
One final comment is that it is, obviously, a bit artificial to take stress values from a
different material, but nickel is very similar to copper and the creep rate is more sensitive
to grain size than to stress, so the approach is reasonable.