Composite material yielding question

A question on a recent test asked me to determine what the stress inside a uniaxial composite was at a given strain value. The given
strain value was below the failure strain for the fibers and matrix material, but above the yield strain for the matrix. I was docked points for assuming that the matrix material would still be bearing load once it was beyond yielding, and my professor says that once the matrix has begun yeilding, it won't carry any of the load. Why is this so? I'm still not 100% sure about my professor's explanation.
Thanks in advance, Dave
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snipped-for-privacy@gmail.com wrote:

Are you sure that you accurately described the professor's remarks?
Was there any mention made of the type of material in the fiber and the matrix? Ductileness or brittleness - were these specified?
Was the reinforcement continuous or discontinuous?
Was the loading tensile, not compressive?
The most common assumption is that the strain will distribute more or less uniformly between the fiber and the matrix if the fibers are continuous and parallel to the loading direction. If not, then the fibers would likely be slipping in the matrix. In other words, the total length of the fibers would be different from the total length of the matrix, as obtained by the integration of the axial strain along the specimen length.
If the matrix is carrying no load, then it doesn't elastically stretch and there must be slippage between the fiber and the matrix, again by integrating the strain along the axis of the fiber. Of course, a material with zero post yield load bearing capability would, by definition satisfy such a criterion, but with the exception of quite brittle materials, such a constitutive equation is unexpected. The sketchy description you gave would imply that the matrix isn't brittle in this sense.
Hit the library and look in other introductory textbooks. You should find the printed answer there, and you can take that back to your professor for further discussion. The old text by R M Jones may have it.
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jbuch wrote:

the
this
Yes, his solution only accounted for the fiber's loadbearing, not the matrix (which he said did not carry any load under yielding).

the
The values given to us were:
1. ultimate tensile stress of the fiber and matrix 2. ultimate tensile strain of the fiber and matrix 3. yield stress and strain of the matrix material 4. volume fraction
The question was along the lines of "what is the elastic modulus of the composite (a) prior to matrix yielding and (b) after matrix yielding?"
For (a), he included the loadbearing of the matrix into the solution, but not for (b) as he said it was no longer bearing load...???

Continuous
Yes, tensile.

of
the
Yes, I'm assuming that under yeilding, strain(fiber)=strain(matrix).

stretch
brittle
it.
Well, I (and some of my fellow classmates) are having trouble with this. I'm not too sure the professor hasn't made an incorrect assumption about matrix yielding.
Thanks for your input, and any additional advise would be appreciated!
Dave
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snipped-for-privacy@gmail.com wrote:

You might make discrete inquiries as to any formal classes in continuum mechanics that may be in the professor's background, or ABSENT from it.
Without a decent background in solid mechanics (elasticity, continuum mechanics, plasticity....) it is easy to make absurd statements on loads and stresses and strains.
Being ignorant of mechanics is not a forbidding variable to claim competence in composite materials. Unfortunately, understanding just mechanics is sometimes considered adequate to be an expert in composite materials.
Conventional Chemistry professors would distain solid mechanics because it was so "elementary" (just the same old Newton's Laws). These chemistry profs knew all about stress and strain, except that they couldn't actually accurately distinguish between stress and strain when asked.
Frankly, assuming the correctness of what you describe, your professor is using mechanics as taught on Venus or some other planet. So, if the professor is a Venusian, then it would make sense to allow him or her to conform to the physics of his or her native planet.
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snipped-for-privacy@gmail.com wrote:

given
matrix
docked
bearing
is
If the matrix does not work harden, ie. is elastic-perfectly plastic, the stress it supports above its yield point is constant. Since a perfectly plastic material's resistance to deformation does not change with stress, it does not contribute to modulus even though it does provide a constant resistance to deformation. (modulus is the *change* in stress with deformation.)
Elastic-perfectly plastic is not a bad description for many cases; it just means that the work hardening slope is much, much less than the pre-yield modulus.
Are you sure your professor said "once the matrix has begun yeilding, it won't carry any of the load?" Might s/he have said that "above the yield point the matrix contribution to the modulus is very small."
The point is "it won't carry any of the load" is not the same as "it doesn't contribute to the modulus".
It might have been a poorly posed question about a subtle point.
Dave
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Ok, I have my test in front of me. The actual question was:
"Titanium alloys are reinforced with longitudinal continuous silicon carbide fiber. The following are material properties for both matrices and fiber:
Ef@0Gpa Ultimate stress (fiber)440Mpa Volume fraction (fiber/total)=.36 Em0Gpa Yield stress (matrix)0Mpa Ultimate stress (matrix)00Mpa Ultimate strain (matrix)=1.0%
a) Calculate the longitudinal modulus of the lamina (with matrix 1) before and after matrix 1 yielding."
Any insight?
Thanks, Dave
snipped-for-privacy@newarts.com wrote:

stress
once
change
*change*
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You said... " Em0Gpa {I take this to refer to "matrix 1" in the problem} Yield stress (matrix)0Mpa Ultimate stress (matrix)00Mpa Ultimate strain (matrix)=1.0% "
The matrix yield point was .00818 strain = 900mpa/110,000mpa The average effective modulus of the matrix above its yield point is (1000-900)mpa/(.01-.00818)T.945 gpa.
While not exactly zero it is much smaller than the modulus of the fibers.
A simple parallel model for the composite above the yield point has an effective modulus of 0.36x400+0.64x54.99 gpa
If the matrix had zero modulus the effective modulus would be 0.36*400+0.04 gpa
Dave
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If the matrix ruptures; ie. the strain is taken above the matrix's UTS it stops contributing anything.
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snipped-for-privacy@newarts.com wrote:

an
<snip>
..which is what he's saying. He says the answer is 144Gpa, and says the modulus doesn't contribute during yielding...?
Dave
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snipped-for-privacy@gmail.com wrote:

is
Well, he may be correct above a total strain greater than 1% (ie. *if* the matrix truly ruptures) but he's incorrect between 0.818% and 1% because the problem statement implies work hardening plastic behavior from 0.818-1.00% and continuity requires the matrix' strain to increase in that range. The work to cause that strain must come from somewhere.
You'll have to decide if it is in your best interests to argue about a range of 0.182%.
Also be sure he's refering to the first strain excursion above the yield point. While subsequent lower strain modulii may the same as in the original case, the effects of residual stresses and strains make it harder to understand.
If you are going to pursue the matter, you'd better think through the residual strain case so you can defend your position.
Dave
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The question doesn't specifiy which region above yeilding, leading me to think that if I can proove that the matrix's modulus in the area immediately beyond yeilding is not zero, I can make a good case for my solution. On the other hand, getting on the professor's bad side could proove more harmful than taking the 4 point hit.
Dave
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snipped-for-privacy@gmail.com wrote:

my
could
Be sure you say "the matrix' effective modulus", ie. the slope of the workhardening curve, for the first excursion.
And you better understand why.
Dave
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