# stress vs. strain determines crack limit

• posted
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
• posted
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
• posted
In article , jbuch writes
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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 --
• posted
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
• posted
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.
• posted
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
• posted
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
• posted
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.
• posted
....
"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?
• posted
How is stress MEASURED directly?
• posted
Impossible.
Michael Dahms
f'up2 sci.mat
Michael Dahms
• posted
......
The last equation should be: (fiber stress)=(engineering strain)*(initial modulus)/(initial number of fibers)
I'm sorry for the typo.
• posted
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 =