iMechanica - fundamentality
https://imechanica.org/taxonomy/term/703
enStress or strain: which one is more fundamental?
https://imechanica.org/node/1001
<div class="field field-name-taxonomy-vocabulary-6 field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/taxonomy/term/128">education</a></div></div></div><div class="field field-name-taxonomy-vocabulary-8 field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/taxonomy/term/131">stress</a></div><div class="field-item odd"><a href="/taxonomy/term/132">strain</a></div><div class="field-item even"><a href="/taxonomy/term/157">students</a></div><div class="field-item odd"><a href="/taxonomy/term/703">fundamentality</a></div><div class="field-item even"><a href="/taxonomy/term/704">measurement</a></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>
In between stress and strain, which one is the more fundamental physical quantity? Or is it the case that each is defined independent of the other and so nothing can be said about their order? Is this the case?
</p>
<p>
To begin with these questions, consider the fact that first we have to apply a force to an object and it is only then that the object is observed to have been deformed or strained. Accordingly, one may say that forces produce strains, and therefore, it <em>seems</em> that stress has to be more fundamental. If so, how come stress cannot be measured directly? This is the paradox I would like to address here.
</p>
<p>
Of course, to begin with, my position is that you can never directly measure stress.
</p>
<p>
I have read somewhere an argument (and forgot exactly where!) that even in photoelasticity what you really measure is strain. The argument, essentially, is this: Birefringence arises because the molecular chains in the photoelastic polymer get stretched. (In case of crystals, "stress-induced" birefringence arises if the deformation is inherently anisotropic.) So, what is important here is the relative positions of atoms in the chain--not whether the atoms were carrying any load or not.
</p>
<p>
The above argument, of course, is sound. Yet, it does not quite settle the issue by itself. This issue is <em>somewhat</em> (but not fully) similar to the hen and eggs situation. To settle the issue, we have to go beyond photoelasticity mechanisms and examine it from the viewpoint of the theoretical structure of the mechanics of solids/fluids.
</p>
<p>
What is clearly undisputed is the primacy of displacements. (Definition: Displacement is the total movement of a point with respect to a fixed coordinate frame.) Displacements can be measured directly and do not need other physical quantities to be measured. Hence, they are primary.
</p>
<p>
Further, what can also be thought of directly is deformation. (Definition: Deformation is the relative movement of a point with respect to another point in the body.)
</p>
<p>
Now, displacement can be related to deformation via the relative deformation tensor, a tensor of second order. (As usual, in the simplest analysis, one assumes infinitesimally small deformations.) Now, if you split the relative deformation tensor into its symmetric and anti-symmetric parts, and ignore the anti-symmetric part (representing rotations), what you get is the strain tensor. This is the primary way strain is defined.
</p>
<p>
For homogeneous linear elastic isotropic materials, strains and stresses are directly related. So, we should expect to find a similar theoretical structure for the concept of stress too. In a way, this does turn out to be the case--but not quite fully. Let's see how.
</p>
<p>
Stress also is a second rank tensor and it also is symmetrical--it drops out the torques/couples part. (This is in analogy with the dropping of the rotation part while defining strain.) This fact about the stress tensor is usually taught in the introductory courses as the result of having moments balance out over the infinitesimal element. But the real reason is that this way we can maintain the similarity of the theoretical structure. The fact is, one could keep moment-balance (as required for static equilibrium) and yet choose <em>not</em> to drop out the torques-related part. However, a discussion on the so-called couple stresses would be a digression here.
</p>
<p>
In short, since both are second order symmetric tensors, stress and strain tensors do seem completely similar.
</p>
<p>
But are they?
</p>
<p>
There is that "displacement<->the gradient tensor<->deformation" relation on the strain side. What is its parallel on the stress side?
</p>
<p>
Here, even in the simplest case of the linear elastic (etc.) solids, it is difficult to believe that a conceptual parallel could be derived independently. One could, of course, argue from an abstract viewpoint that such derivation is possible "mathematically". But remember: deformation is by definition a point phenomenon, whereas force is by definition related to the momentum of an object. For <em>field</em> quantities, force necessarily arises only in the context of a geometric element of nonzero side--e.g. area (as in stress components), or line (as in surface tension). You always need a geometric entity like area or line element (even if it is infinitesimally small) before quantities like stress or flux can at all be defined.
</p>
<p>
Deformation, in contrast, can be defined <em>at</em> a point. We don't have to refer to a geometric element in order to define what this concept means.
</p>
<p>
It is this particular difference which makes it impossible to have a direct analog of deformation on the force/stress side.
</p>
<p>
Consequently, any quantity on the force/stress side must always be defined in reference to some or the other <em>external</em> assumptions as to how the quantity ought to vary across the relevant geometrical element. The simplest assumption of this nature is to say that stress must conceptually remain analogous to strain.
</p>
<p>
The rest then follows.
</p>
<p>
-----
</p>
<p>
Sometimes, it is said that it is very <em>obvious</em> that strain must be more fundamental because in the usual stress-strain diagrams, it is strain that is taken on the x-axis, i.e. as the independent variable.
</p>
<p>
However, note that this "argument" is very superfluous. To plot stress-strain in the usual manner does not need all the above kind of thought. One need only observe that the constitutive law of metals is nonlinear and metal specimens experience necking so that there is a drop in the graph of engineering stress once the point of ultimate tensile strength is reached. In such a situation, taking strain on the x-axis avoids the possibility of having a multi-valued "function." Thus, the choice to take strains on the x-axis is more of a simpler convenience that happens to be in accord with the more fundamental reasoning discussed above.
</p>
<p>
-----
</p>
<p>
As an aside: the difference about a quantity defined <em>at</em> at point by itself (e.g. displacement of a point) and a quantity that requires an infinitesimal volume, area, or line for its definition (e.g. stress, strain, electric field vector) is a fundamental one. It marks the conceptual difference between particles and fields. It therefore plays a crucial role in many other matters such as flux-conservative laws and wave-particle duality.)
</p>
<p>
-----
</p>
<p>
Acknowledgment: It was the discussion on Henry Tan's blog about whether stresses can ever be measured directly or not that provided the spark to write this post.
</p>
<p>
-----
</p>
<p>
If there are deeply thought out and interesting possibilities running counter to the arguments presented above, I would like to know about them.
</p>
<p>
</p>
</div></div></div>Thu, 08 Mar 2007 10:02:32 +0000Ajit R. Jadhav1001 at https://imechanica.orghttps://imechanica.org/node/1001#commentshttps://imechanica.org/crss/node/1001