Hello,
With respect to classic power-law crack growth in brittle materials, the
critical limit is given by
K = Y*S*a^1/2, where Y is a constant, S is the applied stress, and a is the
crack length (which is taken as a square root).
My question is : What is the true independent variable (i.e., stress or
strain) that determines when the material fractures? This equation suggests
that it is the stress, since the strain is not given, but that might be more
for convenience since we tend to measure stress, not strain.
Assuming the stress-strain relationship were perfectly linear it wouldn't
matter, but since materials fracture in a regime in which the stress-strain
is not linear, it changes the interpretation of fracture data.
Thanks,
Jay

Dear seferiad:
stress-strain
Strain is a function of the material, so once you have selected a material
that "degree of freedom" is lost. I would vote for stress to be the
independent variable.
David A. Smith

In article , jbuch
writes
..snip..
...snip...
..snip..
Agreed.
However, in linear elastic fracture mechanics (LEFM)
it is not really necessary to understand the mechanics
of the fracture process. All we need to appreciate
is that the region immediately surrounding the crack
tip is subjected to highly non-linear fields of stress
and strain. However, outside that, we have the unique
K-controlled "square root r" elastic fields of stress
and strain. So, the stress intensity factor, K, tells
you all you need to know about the loading of that crack
region and when it well eventually fail: K_applied=K_mat.
In elastic-plastic fracture mechanics, a similar
'control' of the stresses and strains in the crack
tip region is considered to be achieved by means of the
parameter J (J-integral),
As stated by the previous poster, in reality the material
ahead of the crack tip fails in a manner which depends
on the particular micromechanics of the material.
So, cracks in ductile steels grow by microvoid initiation,
growth and coalescence to give ductile tearing. In a simple
micromechanical model , this requires a combination of
plastic strain and stress triaxiality to achieve.
Cracks in ferritic steels on the lower shelf and in the
ductile-brittle transition are considered to fail by
cleavage. A simple micromechanical model of cleavage
requires a critical stress to be achieved over a
characteristic distance ahead of, or volume
surrounding, the crack tip.
Returning to the original question, it can be argued that
strain is what is imposed on the material and stress is
its response to it. On balance I would prefer to regard
strain as the "independent variable", but engineers
are used to working with stress so this tends to be
more commonly used.
Anyway, just my 'two penny worth' to the thread.
Regards
Martin
--

If you are to use an Instron displacement driven test machine, then you
are indeed imposing strain.
If you are jumping in the air and grab and swing on a vine, then you are
doing a load controlled loading of the structure.
Then there are situations which are fully dynamic, such as falling onto
a matress from 20 feet and the loading in this case is neither simple
strain controlled nor simple load controlled. Inertia and stiffness
both are key parameters.
If you are a tester by orientation, you will generally tend to think in
the way that your test machine operates.
So, an Instron man may believe that the basic material response is
strain controlled.
An MTS or similar pnuematic computerized machine can think of
deformation as either strain controlled (via the fancy computerized
feedback in the MTS machine) or load controlled as old fashioned
pnuematic machines used to be designed.
At one time, the Instron philosophy was more modern when the
electromechanical machines first began replacing the old pnuematic
machines. Now, the Instron philosophy is no longer the more modern one
as the MTS type machines are quite brilliant in the total spectum of
conditions they bring to testing.
As an aside, how do you evaluate impact fracture? [And we will skip the
many different kinds of impact fracture as just extra complications.] It
seems hard to imagine impact fracture as primarily a strain controlled
loading environment.
I still advocate that the oversimplifications we impose on fracture are
one of the reasons it is still technically difficult and seemingly
fragmented.
Jim

While this is a kind of "chicken and egg" question, Over the years
I've found it most instructive to consider strain as the primary
variable when considering mechanics of materials.
There are a number of reasons for this, perhaps the most important
being that if one puts strain on the left hand side of the equation,
the right hand side is a sum of simple influence terms.
Strain = Compliance*Stress + ExpansionCoefficient*Temperature + ..etc.
Another indicator is that the deformation behavior of various
materials varies widely with applied stress and does not vary much
with applied strain; for example, yield of materials occurs in a
fairly small range of strains.
At the atomic level the coherence of a material is dictated by
distances between the atoms. Failure is the consequence of atoms
being drawn so far apart that they rearrange themselves into a new
configuration.
While such considerations cannot assign primacy to intimately tangled
variables like stress and strain, they can sometimes help simplify
certain problems.

Thanks for the detailed responses. I'll pose a more specific question.
Let's say I have a bunch of fibers that perfectly obey Weibull statistics
when I do dynamic pull testing to destruction for the fibers. The
distribution when plotted on Weibull gives a slope = m (it doesn't matter
what m is), but the fiber population is sufficiently strong so that a
significant fraction of the fiber will break at sufficiently high pull force
so that the non-linear stress-strain relationship of glass will be apparent.
I have two "ideal" Instron machines. One machine pulls with a constant
stress rate and the other machine pulls with a constant strain rate.
If I now plot these two Weibull distributions: %Failure vs. Stress (or
Strain), I would expect that for both graphs at low stress (or strain) that
the plots should be perfectly linear (based on the original problem
statement of having ideal fiber). However, at higher Stress (or Strain) one
of the graphs will have a change in slope.
I'm trying to figure out conceptually, which is the plot that is more likely
to be straight -- at least with regards to the equations set forth by
Griffith (for the moment, nevermind that the theory is incomplete I'm trying
to make sure that I am applying that theory correctly). I realize that I
have oversimplified the problem, but this is my simple-minded way of trying
to explain the problem.
Thanks,
Jay

In general, you should think very carefully about what you think is
happening in a specimen and what is actually happening.
Imagine three fibers having identical strengths. Imagine first that each
fiber is exactly the same length as the others and is perfectly
gripped... repeat, perfectly gripped .... in an Instron.
As the specimen is loaded, all three fibers load in exactly the same way
and will fail at the exact same time.... because of the exact identical
strength assumption.
Now imagine these very same fibers. Imagine imperfect gripping of the
specimens so that the length between the perfect grips for each fiber is
slightly different. Imagine looking at a piece of yarn after you have
untwisted is and you wlll see that some fibers are more slack (and
curver) than others).
Now when you load this array of three fibers in an Instron tensile
tester, because of the difference in lengths (Slack Lengths) the
shortest fiber loads first, and then the next and then the next. So, the
load initiation of the fibers reflects the gauge length statistics.
Eventually, the shortest fiber reaches it's failure strength and it
fails. There is a load drop from load redistrubution. Then as the
machine is further run and the specimen continues to elongate, the
remaining fibers will subsequently fail.
Remember the assumptin of a perfect Instron with infinite stiffness.
Because of this pure displacement loading condition, the fibers won't
actually undergo loading increases as the individual fibers fail. For an
Instron represented by a finite spring constant, the second and third
fiber WILL experience a loading increase as a result of the failure of
the first fiber.
But, the result is that you gain evident fiber failure statistics
(incorrect inference) even though the fibers were assumed to have no
failure stress variability.... What you are seeing are geometric
statistics that relate to when fibers begin to be loaded in the perfect
Instron with perfect gripping.
Slack Length statistics.
These slack length statistics will destroy the correlation you are
attempting to draw.
And this work dates to the 1970's.
Even old researchers could and did think.
The basic lesson is that when you are testing fibers, think it through
and understand what the mechanics details really are.
It was a common problem with carbon/graphite fibers that the bare yarn
strengths were typically only about 1/3 of the average fiber strength.
If you understood that the bare yarn test was a lousy way to attempt to
measure filament strengths, then it made sense. Otherwise, folks would
just look at the results and contemplate the mysteries of composites.
I published a paper on this in the "Extended Abstracts of the American
Carbon Society" in about 1975 and discovered some related excellent work
of a similar nature (including twist effects) by an author I can't
remember. There probably have been several prior dissertations on the
detailed effects of similar considerations.
Your best goal to equalizing the load among the fibers is to include a
matrix to transfer load between fibers. However, here you need to
remember to account for the load redistribution.... in two key ways.
1) load redistribution locally from fiber to fiber.
2) load redistribution along the length of the fiber so that an
individual fiber can sustain more than one fracture along its length.
The penalty for the loading equalization of the matrix is to ruin the
assmuption of independent mechanical fracture of the individual fibers.
You can't have your lunch and eat it too.
Remember that the advantage of the matrix is precisely this form of load
redistribution that tends to make each fiber equally share the load.
It helps to think the details through carefully, and even sometimes draw
simple cartoons of what one presupposes the mechanics are.... and then
to actually look at a test sample to guess if your assumptions were a
match to the actual test.
Jim

Don't ignore the effects of stress corrosion. Fibres loaded to less than
their normal breaking stress can eventually fail because of this.
It is arguable that rate of loading could also affect the results.
The same results can be observed in laminates or other composites.

....
"Instron" machines are deflection devices. I take what you say to mean
test type 1 is constant rate of force increase (ie. proportional to
nominal engineering stress rate) & test type 2 is constant rate of
deflection increase(ie. constant nominal engineering strain rate).
Before going on to your question of wiebull distributions for these
two types of tests let's think about the tests themselves.
Consider a large number of non-interacting parallel fibers in a
bundle.
Test type 2: (constant deflection rate)-as deflection increases some
of the fibers break so the total force is for a particular deflection
is less than that expected if no fibers break; ie. the nominal stress
is LESS than predicted by the (initial modulus)*(strain) by the
fraction f of fibers remaining. That is, (actual stress) =
f*(predicted stress). "f" is in effect a fiber's survival probability
at the actual stress on the surviving fibers.
Test type 1: (constant force rate) -as force increases & some of the
fibers break the deflection at a particular nominal stress is MORE
than that expected by the inverse fraction 1/f of fibers broken. That
is, (actual strain)=(predicted strain)/f. Again, f is the fiber's
probability of survival at the actual stress on the surviving fibers.
Note that in either case the fraction of fibers broken can be
estimated by taking the ratio of the observation to the linear
prediction.
....
Now things get a little confusing for me because I don't know exactly
what you intend to plot.
The first test I described (the constant deflection rate test) yields
data that are easily compared with weibull predictions. (force
observed)/(force predicted) is the fraction of surviving fibers and
(engineering strain)*(initial modulus)*time/(initial number of fibers)
is the fiber stress.
I hope this helps. Perhaps your response might clarify your question.
Am I addressing the matter that concerns you?

On Sun, 16 Nov 2003 22:15:02 -0600, Jeff Finlayson =
wrote:
I measure the mean stress using a piezoelectric element inline with the =
sample being stressed. It's really only a measure of the force applied =
to =
the sample, which is then extrapolated to mean stress. This technique h=
as =
other limitations as well.
I'm sure there are other, sexier ways to approxmate (measure) stress. I=
'd =
like to hear about them.

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