You are here
Stress is defined as the quantity equal to ... what?
In this post, I am going to note a bit from my personal learning history. I am going to note what had happened when a clueless young engineering student that was me, was trying hard to understand the idea of tensors during my UG years, and then for quite some time even after my UG days. May be for a decade or even more....
There certainly were, and are likely to be even today, many students like [the past] me. So, in the further description, I will use the term ``we.'' Obviously, the ``we'' here is the collegial ``we,'' perhaps even the pedagogical ``we,'' but certainly neither the pedestrian nor the royal ``we.''
========================
What we would like to understand is the idea of tensors; the question of what these beasts are really, really like.
As with developing an understanding of any new concept, we first go over some usage examples involving that idea, some instances of that concept.
Here, there is not much of a problem; our mind easily picks up the stress as a ``simple'' and familiar example of a tensor. So, we try to understand the idea of tensors via the example of the stress tensor. [Turns out that it becomes far more difficult this way... But read on, anyway!]
Not a bad decision, we think.
After all, even if the tensor algebra (and tensor calculus) was an achievement wrought only in the closing decade(s) of the 19th century, Cauchy was already up and running with the essential idea of the stress tensor right by 1822---i.e., more than half a century earlier. We come to know of this fact, say via James Rice's article on the history of solid mechanics [(.PDF) ^].
Given this bit of history, we become confident that we are on the right track. After all, if the stress tensor could not only be conceived of, but even a divergence theorem for it could be spelt out, and the theorem could even be used in applications of engineering importance, all for decades before any other tensors were even abstractly conceived of, then, of course, developing a good understanding of the stress tensor ought to provide for a sound pathway to understanding tensors in general.
So, we begin with the stress tensor, and try [very hard] to understand it.
========================
We recall what we have already been taught: stress is defined as force per unit area. In symbolic terms, read for the very first time in our XI standard physics texts, the equation reads:
... Eq. (1)
Admittedly, we had been made aware, that Eq. (1) holds only for the 1D case.
But given this way of putting things as the starting point, the only direction which we could at all possibly be pursuing, would be nothing but the following:
The 3D representation ought to be just a simple generalization of Eq. (1), i.e., it must look something like this:
... Eq. (2)
where the two overlines over represents the idea that it is to be taken as a tensor quantity.
But obviously, there is some trouble with the Eq. (2). This way of putting things can only be wrong, we suspect.
The reason behind our suspicion, well-founded in our knowledge, is this:
The operation of a division by a vector is not well-defined, at least, it is not at all noted in the UG vector-algebra texts. [And, our UG maths teachers would happily fail us if we were to try an expression of that sort in our exams!]
For that matter, from what we already know, even the idea of ``multiplication'' of two vectors is not uniquely defined: We have at least two ``product''s: the dot product [or the inner product], and the cross product [a case of the outer or the tensor product]. The absence of divisions and unique multiplications is what distinguishes vectors from complex numbers (including phasors, which are often noted as ``vectors'' in the EE texts).
Now, even if you attempt to ``generalize'' the idea of divisions, just the way you have ``generalized'' the idea of multiplications, it still doesn't help a lot.
[To speak of a tensor object as representing the result of a division is nothing but to make an indirect reference to the very operation [viz. that of taking a tensor product], and the very mathematical structure [viz. the tensor structure] which itself is what we are trying to understand. ... ``Circles in the sand, round and round... .'' In any case, at this stage, the student is just as clueless about divisions by vectors, as he is about tensor products.]
But, still being under the spell of what had been taught to us during our XI-XII physics courses, and later on, also in the UG engineering courses--- their line and method of developing these concepts---we then make the following valiant attempt. We courageously rearrange the same equation, obtain the following, and try to base our ``thinking'' in reference to the rearrangement it represents:
... Eq (3)
It takes a bit of time and energy, but then, very soon, we come to suspect that this too could be a wrong way of understanding the stress tensor. How can a mere rearrangement lead from an invalid equation to a valid equation? That's for the starters.
But a more important consideration is this one: Any quantity must be definable directly, i.e., via an equation that follows the following format:
the quantiy being defined, and nothing else but only that quantity, appearing on the left hand-side
=
some expression involving some other quantities, appearing on the right hand-side.
Let's call this format Eq. (4).
Clearly, Eq. (3) does not follow the format of Eq. (4).
So, despite the rearrangement from Eq. (2) to Eq. (3), the question remains:
How can we define the stress tensor (or for that matter, any tensors of similar kind, say the second-order tensors of strain, conductivity, etc.) such that its defining expression follows the format given in Eq. (4)?
========================
A few more words:
It would be easy enough to abstractly do just a bit of algebraic manipulation and arrive at the solution. The point isn't that. The point is to understand the physical implications of that manipulation.
And, further, the point is this: If it were that obvious or simple, why is it that not even a single text-book/class-notes ever anticipates the above-mentioned possible line of thought on the part of a beginning student, and therefore, proceeds to provide him with the required definition in direct terms?
And then, as I might note later on, there are a few other conceptual advantages with a direct defintion, too. But more on it, later, if there is enough interest in this topic.
--Ajit
- Ajit R. Jadhav's blog
- Log in or register to post comments
- 15088 reads
Comments
Re: Stress is defined as the quantity equal to ... what?
Ajit: Your question has also puzzeled me. I now favor your equation (3). Here is a post called "a state of stress is a linear map".
Incidentally, I am teaching undergarduate linear algebra now. Here are class notes that try to explain where we get vector spaces and linear maps.
Re: Stress Definition
Dear Zhigang,
Wow, good to know that it's been puzzling for you too...
BTW, yes, I have been aware of your notes, and in fact recently recommended them in a reply at my personal blog, here [^].
Overall, to cut a long story short, here are a few points that I think we should be explaining to students:
====
We should state and explain the concept of stress as the direct end-result of a tensor product.
Here is one way to bring out the required direct definition. Only the UG-level matrix algebra is used.:
Notation: We adopt the convention that a vector is always denoted via a column vector, never using a row vector, and therefore, that a row vector always represents only a transposed vector. Column vectors are denoted using braces {c}, and row vectors and square matrices are denoted with square brackets [r] or [M]. (When using LaTeX, we use the left- and right-floor symbols for the row vectors, and the square brackets for matrices.)
First, define the local field of traction-vector acting on a given surface (whether the surface is internal or external, whether it is used in specifying BCs or not) in the usual way, viz., as the ratio \vec{F} / |\vec{A}| in the limit that |\vec{A}| vanishes. (Or, if you want to avoid limits at least initially, consider a uniform \vec{F} field.)
The denominator here is a scalar (it's only the magnitude of the area vector), and so, the division is not a problem. Basically, before using the area vector further, we isolate apart its two aspects, viz., the scalar magnitude and the direction.
Denote the unit normal to surface as {n} and its transpose as [n].
Start with the transpose of the traction vector, denoted as {t}^{T} and also written as [t].
[t] = [t] (and what else?)
= (1) [t]
= ( [n]{n} ) [t]
= [n] ( {n}[t] )
= [n] [\sigma]
In the last step, we have used the direct definition: [\sigma] \equiv {n}[t].
The above was the round-about way in which I had got to the defintion for the first time in my life. However, another, perhaps simpler, way is this. Start with Cauchy's formula:
{t} = [\sigma]^T {n}
{t}[n] = [\sigma]^T {n}[n] (Post-multiply both sides by {n}-transposed)
{t}[n] = [\sigma]^T ({n}[n] is the dot product, equal to 1)
[\sigma] = {n}[t] (Take transposes of both sides, and flip LHS and RHS)
The matrix-notation expression {n}[t] stands for a tensor product taken between the unit surface-normal vector on the left hand-side of the operator, and the traction vector on the right hand-side. Using the vector language, this tensor product is the same as what is expressed by: "\hat{n}\otimes\vec{t}".
This structure of
"{n} \otimes [the surface-intensity of a vector field variable]"
is common to all the flux-tensors of all the vector fields. The direct definition of stress thus brings out the physical idea that the stress is the flux of the traction vector.
The idea of the strain is obviously is based on a gradient; it's the symmetrical part of the displacement gradient.
The student can now appreciate that the stress-strain relation is nothing but just a case of a flux-gradient relation. The flux-gradient relations occur in almost all other areas of physics; as just one example, consider Fourier's law of heat conduction. So, hopefully, he can see the commonality of analysis with other branches of physics and engineering.
Speaking in general terms, the flux of a scalar field is a vector field; the gradient of a scalar field is a vector field; and the two vector fields are related via a material (constitutive) law.
Exactly similarly, the flux of a vector field is a tensor field; the gradient of a vector is a tensor field; and the two tensor fields are related via a constitutive law.
===
The idea that the stress is a linear map, as emphasized by you, is indeed very helpful. It throws light on what kind of purpose it serves. So, it is very valuable.
That's what the indirect equation
{t} = [\sigma]^T {n} (i.e. Cauchy's formula)
shows. It brings out, very clearly, the fact that the stress as a mathematical object is, from the external viewpoint, a linear map (from {n} to {t})
However, IMHO, this idea also should be complemented by pointing out that the stress is a tensor product (which highlights its internal structure) and that it is a flux (which identifies its physical meaning). The direct definition allows one to see both these latter facts.
===
BTW, the idea of a flux does not always refer to something that flows; it can refer to something that simply exists (rather than flows across) a surface. The idea of flux, however, necessarily refers to a reference (planar) surface.
Therefore, the flux nature of stress also helps reinforce the idea that it is purely a surface phenomenon. To define stress, all that you need is a planar element (whose orientation can be made to change) around the point of interest. In particular, you don't need a volume element at all. This is important to realize by the text-book writers. To define stress, you don't need either Cauchy's tetrahedron or the hexahedral element (as used in deriving the stress-divergence theorem). Both these are volume elements. But stress can be completely defined with just an orientable surface element. This point becomes clear only when you understand the stress as a flux, which itself becomes clear only after you consider the direct definition via tensor product of two vectors.
Finally, just one more point.
Stress analysis is based on the differential equation paradigm. For a problem involving only stresses (i.e., no equations or calculations of strains or displacements being involved), the auxiliary data for such a problem is fully stated in terms of just the unit normals and the traction vectors. The point to note is this: the auxiliary data in fact cannot be stated in any others terms. Specifically, it cannot be stated in terms of forces and areas. When in engineering we say that a force is being specified as a BC for a stress analysis problem, what we are actually doing is to mentally translate the force vectors into the area-intensity vectors, before importing them into the stress analysis problem as traction vectors. The direct definition simply helps makes this part more direct and explicit.
===
I have grown so convinced about the necessity to have a direct definition that I am seriously thinking of writing a journal paper on the topic, just to let the argument have a permanent place in the archives.
===
But, thanks, once again, for your interest,
Best,
--Ajit
Re: Defn of stress: A glaring mistake in my reply to Zhigang
Please ignore the second derivation, i.e., starting with Cauchy's formula, and proceeding as follows:
{t} = [\sigma]^T {n}
{t}[n] = [\sigma]^T {n}[n] (Post-multiply both sides by {n}-transposed)
{t}[n] = [\sigma]^T ({n}[n] is the dot product, equal to 1) <------ THIS PART IS WRONG
[\sigma] = {n}[t] (Take transposes of both sides, and flip LHS and RHS)"
=======
The error is in saying that {n}[n] is the dot product, equal to 1. Very elementary mistake!
=======
I wrote the reply while in office, interrupted by an inaugural function, some 10+ visitors, two meetings, quite a few phone-calls, and a skipped lunch... All simply because I am so enthusiastic about fundamental topics like these, that I can overboard.... But no excuses, so let me say sorry....
I also realize that the characterization of the stress-strain relation as a flux-gradient relation has a very definite merit to it, but the fact is, the strain tensor is just a part of the gradient tensor, not the entire one. I need to think more compreshensively about it.
========
Bye for now, I will go home, think a bit about it, and come back either tomorrow or the day after or so. In the meanwhile, everyone, feel absolutely free to leave your ideas for going from Cauchy's formula to a direct definition of stress, and please see if there is any other error in what I say. Thanks in advance for your feedback...
And sorry, once again, for bothering you all with the mistake....
--Ajit
Re: stress definition
In the undergraduate linear algebra class, we introduced linear map about two weeks ago. From past experience, I know students need be reminded of what a map (i.e., a function) is. A map requires three sets: domain, codomain, and graph.
For a linear map, the domain must be a vector space, the codomain must be also a vector space, and the graph must be linear.
In saying the stress is a linear map, I stick to this prescription. Thus, the domain is the vector space of areas, the codomain is the vector space of forces, and the graph is the stress. That is your equation (3).
The Cauchy formula traction = (stress)(unit normal) does not follow this prescription. The collection of all unit normal vectors is not a vector space, because the addition of two unit normal vectors do not give another unit normal vector. Also, the collection of all traction vectors is not a vector space, because tractions on different planes are not additive.
Thus, in defining stress, I stick to your Equation (3).
In teaching the undergraduate course on linear algebra, I try to highlight the importance of the scalar set (i.e., the one-directional vector space). Then an n-dimension vector space is the Cartesian product of n scalar sets. A vector-vector linear map is a table of scalar-scalar linear map. Thus all linear map is similar to the fact of nature: a chicken has two feet. This fact is a linear map between two sets: the chicken set and the foot set.
Thus, reducing to one component, stress is just force per unit area. The algebra is similar to that each chicken has two feet.
In the linear algebra class, I then talked about two vector spaces: chicken-rabbit space and head-foot space. Then I talked about the vector-vector linear map: (chicken-rabbit space) → (head-foot space). The linear map is then a table (i.e., matrix) of four facts of life:
This strategy may be parallel to your way of introducing stress.
Re: Re: stress definition
Dear Zhigang,
I agree with most of your points, so I will not touch upon them. I will write only where we seem to differ.
===
In your definition, you say that the domain is the vector space of areas.
However, technically speaking, areas also face exactly the same technical objection as that for the unit normals. ... After all, a unit normal is nothing but only the area vector, as scaled by the reciprocal of its magnitude. Mere scaling doesn't affect the question of whether something is indeed a vector or not.
Area is not primarily a vector, but it can be seen as a vector, at least for the infinitesimally small parts of arbitrary surfaces, or for planar (orientable) surfaces. Even for such surfaces, the area is not a primitively defined vector. It is actually the result-object of taking the cross product of two primitively defined vectors---two non-colinear vectors lying in the plane of the area. (By primitively defined vector, we mean a member of a properly defined vector space.)
Exactly the same thing goes for the unit normal vector. For example, \hat{k} is a basis vector, and thus, of course, a proper vector. However, if we want it to represent a unit surface area in the xy-plane, then something defining about that area has to be associated with this object. This we do via the relation \hat{k}= \hat{i}\times\hat{j} where "\times" denotes the cross product.
However, this is a bit of notation abuse, because we know that a cross-product is not a "true" vector; it's a pseudo-vector. The same object, viz. \hat{k}, appears in our equations sometimes as a proper vector and at other times as a pseudo-vector.
Since areas are pseudo-vectors, so must be stress, whether you define it as {n}[t] or as {A}[F], or even as a linear map. So long as areas are involved, it should turn out to be a "pseudo-tensor." Forces are proper vectors, and so are tractions (becuase they refer to only the magnitude of the area), but that's not enough.
We manage to get by treating the stress as a tensor (and the areas or the unit normals as vectors), primarily because all the relevant quantities are defined and used at one and the same point. The basic trouble with a pseudovector is that it cannot be freely transported. It has the meaning of a vector only at the point of its definition.
Now, look at Cauchy's formula, viz.,
"{t} = [\sigma]^T{n}".
This equation is nothing but saying that traction is the result of the dot product given by:
\vec{t} = \sigma \cdot \vec{n}.
In the first representation, the transpose-sign for the stress matrix appears simply because we are translating from the vector algebraic notation to the matrix notation. (This is exactly like saying: \vec{a} \cdot \vec{b} = {a}^T{b}.)
In either representation, what the equation basically says is that the projection of the stress (tensor) on to the specified unit normal (vector) is the traction (vector) acting on the specified side of the specified surface---at the specified point.
We can get by because all the quantities used in this equation, viz. {t}, [\sigma] and {n}, (or, if you prefer, {F}, [\sigma] and {A}) are defined, related and used at exactly one and the same point; they are never transported to (or used at) any other point. The projection occurs and is meaningful only at the single, distinguished, point.
This incidentally is the point I wanted to bring out and emphasize. Text-book derivations keep Cauchy's original line of thought completely intact. So, they make reference to the tetrahedral element. But as I said in the earlier reply, the tetrahedral element is a volume element, sort of. Actually, Cauchy's argument makes use of only the external, bounding surfaces; it does not refer to the interior or the volume region. Yet my objection remains: Cauchy's derivation of the above formula refers to four different planes none of which pass through the centroid of the tetrahedron. It's only because the derivation is for the limiting case of the vanishing region of space that the different planes can be seen as virtually crossing at the centroid point---only in the limiting case.
Now the point I want to emphasize is this: For Cauchy's formula to tumble out, you don't need to erect four different planes in the first place! Just take one plane passing through the point of interest (the single distinguished point), and rotate the same plane around that point any which way you wish. For example, rotate it so as to align along the xy plane, yz plane, etc. Since the plane always remains tucked on to the same distinguished point during any rotations, any and every quantity you define w.r.t. that plane and at that point, whether with this orientation of plane or that, escapes the trouble which the pseudovector faces due to "transportation." The weakness of text-book derivations (common to all text-books) is that, since all the four planes don't pass through a single point (say the centroid of the tetrahdron), the area vectors are implicitly assumed in their derivations as transported---which brings in the trouble of the pseudovector (and which trouble text-books summarily skip or ignore).
OK, enough for now. More, may be later.
====
Sorry for keeping you all waiting, but have been too busy (and will be, until Feb-end.) I will try to check out this thread before March arrives, but practically speaking, it may not be possible. However, I will definitely come back in early March, and so, please feel free to post your ideas, objections etc. in the meanwhile... Bye for now.
--Ajit