Re: stress vs. strain determines crack limit

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


"jbuch" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> > 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
> > --
> > http://www.analysis.demon.co.uk
> > http://www.fracturetraining.co.uk
>
>
> 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
>
> -- 
> ...............................
>
>
> Keepsake gift for young girls.
> Unique and personal one-of-a-kind.
> Builds strong minds 12 ways.
> Guaranteed satisfaction
> - courteous money back
> - keep bonus gifts
>
>   http://www.alicebook.com
>