John (Waldron) raises a few points that I think I may be able to help
with. If I am incorrect (from a mechanics standpoint), I hope that
someone will politely point out any mistakes. I know there are lots of
folks on this list with considerably more mechanics knowledge than I.
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"Then comes the part that always leaves me with nagging doubts. There
is an argument in all the texts that the shear stresses sigma-x-y and
sigma-y-x are identical, based on the case that there is no net moment
about the z axis in this vanishingly small cube. When applied to all
the off-diagonal elements, this leads to a symmetrical stress tensor
with 6 independent terms, in contrast to the asymmetric deformation
gradient tensor with 9 terms. I am uncomfortable with this contrast,
which seems counter-intuitive"
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I, too, struggled with this because I was always taught that stress
causes strain, which in turn causes displacements. Whether this is true
or not makes no difference to me because they are all related
mathematically. Also, stress and strain cannot be directly measured;
only displacements can. So, I think if one looks at this from the
perspective of what can be measured (displacements) and then discuss how
the stress tensor comes from measurements of displacements things may
make a bit more sense.
The displacement gradient tensor (which can be directly measured) can
have 9 independent components while the stress/strain tensor only has 6.
This is because the displacement gradient tensor is the sum of the
strain tensor and the rotation tensor. The strain tensor is symmetric
like the stress tensor and the rotation tensor is antisymmetric. This is
nicely described in Jaeger, Cook, and Zimmerman's Rock Mechanics text
(pg. 43-54). In the case of equilibrium, the rotation tensor is zero.
Conceptually speaking, a rigid block rotation does not cause fractures,
faults, or bodies of rock that are completely contained within a block
to be strained because rigid block rotations, by definition, do not
cause strain...i.e. rotations do not change the size/shape of a body.
My understanding is that the stress and strain tensors must be
symmetrical because asymmetry in the stress/strain tensors would imply
that (infinitesimally) small parts of the body are rotating while others
are not. This means that the body does not satisfy the definition of a
continuum because the displacements would not be continuous throughout.
I hope this helps.
-Scott
On 3/26/2011 5:50 PM, John Waldron wrote:
> I have a plea and a question.
>
> I have too found the discussion on this list very informative over time, and I would like it to remain so. However, the last 18 posts have been about the behaviour of people, not rocks. Public statements of opinion, or announcements of intent to leave the list, however well-intentioned, may contribute to the problem; more list members will be tempted to leave because they don't want to read this stuff. So, I would make a plea based on my experience on the Canadian list mentioned by Jürgen. If you intend to leave the list, I would urge you to leave quietly, or to make your opinions on individuals (on whichever side of the argument) known in private emails or to the list owner (he may not thank me for this), rather than to the whole list. I intend to stay on, and hope there will be enough expertise left in the list to make it as informative in the future as it has been in the past.
>
> In that spirit (and lest I contribute to the same problem) I would like to ask a question, that has been raised in my head by some of Dr. Koenemann's comments. Like many members, I work in general field-based structural geology, and am not an expert in continuum mechanics. However, I do teach the basics of stress and strain in my undergraduate and graduate classes, typically to students with even less background in physics and mathematics than mine. Like most of us who teach this stuff, I take my students through the hypothetical vanishingly small cubic element of a solid under stress, and represent the three components of stress (or more properly traction) on each surface so as to fill out the 9 components of the stress tensor.
>
> Then comes the part that always leaves me with nagging doubts. There is an argument in all the texts that the shear stresses sigma-x-y and sigma-y-x are identical, based on the case that there is no net moment about the z axis in this vanishingly small cube. When applied to all the off-diagonal elements, this leads to a symmetrical stress tensor with 6 independent terms, in contrast to the asymmetric deformation gradient tensor with 9 terms. I am uncomfortable with this contrast, which seems counter-intuitive. If deformation is driven by stress, and the stress tensor only controls the six terms that describe distortion (or distortion rate) then how is the rotational part of deformation controlled? I realize that rotation can be constrained by setting appropriate boundary conditions, but my discomfort is that that vanishingly small cube doesn't 'know' about the boundary conditions of the system in which it sits, so what controls its rotation if not the state of stress?
So I always end my lecture with the feeling that the argument is sleight of hand - I have used phrases like 'arguments beyond the scope of this course lead to...', without feeling that I actually have a proper grasp of those arguments.
>
> This may be something that can be very simply answered, and that I simply missed out on in my own education. However, Dr. Koenemann's discourses raised the idea that we should be able to explain stress-strain relationships in terms of forces that act along bonds between atoms, not infinite imaginary surfaces within continua, so I am tempted to wonder whether there are elements of his argument that might lead to a resolution of my question, perhaps by including a rotational element into the description of stress. If anyone has any suggestions or explanations that help to make this make sense, and help me make sense of this to my students, it would be most welcome.
>
> John Waldron
>
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Scott T. Marshall
Department Of Geology
Appalachian State University
572 Rivers St.
Boone, NC 28608
http://www.appstate.edu/~marshallst/
ftp://pm.appstate.edu/pub/prog/marshallst/
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