iMechanica - Comments for "Stress is defined as the quantity equal to ... what?"
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Comments for "Stress is defined as the quantity equal to ... what?"enRe: Re: stress definition
https://www.imechanica.org/comment/29598#comment-29598
<a id="comment-29598"></a>
<p><em>In reply to <a href="https://www.imechanica.org/comment/29595#comment-29595">Re: stress definition</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Zhigang,</p>
<p>I agree with most of your points, so I will not touch upon them. I will write only where we seem to differ.</p>
<p>===</p>
<p>In your definition, you say that <span>the domain is the vector space of areas. </span></p>
<p><span>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.<br /></span></p>
<p><span>Area is <em>not</em> primarily a vector, but it can be <em>seen</em> 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 <em>primitively</em> 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.)<br /></span></p>
<p><span>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. </span></p>
<p><span>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.<br /></span></p>
<p><span>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 "<em>pseudo</em>-tensor." Forces <em>are</em> proper vectors, and so are tractions (becuase they refer to only the magnitude of the area), but that's not enough. <br /></span></p>
<p><span>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 <em>at one and the same point. </em>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. </span></p>
<p><span>Now, look at Cauchy's formula, viz., </span></p>
<p><span>"{t} = [\sigma]^T{n}". <br /></span></p>
<p><span>This equation is nothing but saying that traction is the result of the dot product given by: </span></p>
<p><span>\vec{t} = \sigma \cdot \vec{n}. </span></p>
<p><span>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}.) </span></p>
<p><span>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. </span></p>
<p><span>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. <br /></span></p>
<p><span>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 <em>surfaces; </em>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 <em>none</em> 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.</span></p>
<p><span>Now the point I want to emphasize is this: </span><span><span>For Cauchy's formula to tumble out, </span>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 <em>the same</em> 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 <em>all</em> 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).</span></p>
<p>OK, enough for now. More, may be later.</p>
<p>====</p>
<p>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.</p>
<p>--Ajit</p>
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</ul>Fri, 23 Feb 2018 09:28:38 +0000Ajit R. Jadhavcomment 29598 at https://www.imechanica.orgRe: stress definition
https://www.imechanica.org/comment/29595#comment-29595
<a id="comment-29595"></a>
<p><em>In reply to <a href="https://www.imechanica.org/comment/29593#comment-29593">Re: Stress Definition</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p dir="ltr"><span><span>In the undergraduate linear algebra class, we introduced </span><a href="https://docs.google.com/document/d/1mDe2xV-tuYqwqNloWixAqiTE-AoHnxMaHgop0N8I-aw/edit?usp=sharing"><span>linear map</span></a><span> about two weeks ago. From past experience, I know students need be reminded of what a </span><a href="https://docs.google.com/document/d/1cqHEgtmp4JhNPv7D0BFZmgMNDYvIMHz17n6z_zP4Zu4/edit?usp=sharing"><span>map</span></a><span> (i.e., a function) is. A map requires three sets: domain, codomain, and graph. </span></span></p>
<p dir="ltr"><span>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.</span></p>
<p dir="ltr"><span>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).</span></p>
<p dir="ltr"><span>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.</span></p>
<p dir="ltr"><span>Thus, in defining stress, I stick to your Equation (3).</span></p>
<p dir="ltr"><span>In teaching the undergraduate course on linear algebra, I try to highlight the importance of the </span><a href="https://docs.google.com/document/d/1GeztAnZIMR8PKNvGLp_Pc8zxf2eKU3FjmlIzj8Kz3-I/edit?usp=sharing"><span>scalar set</span></a><span> (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.</span></p>
<p dir="ltr"><span>Thus, reducing to one component, stress is just force per unit area. The algebra is similar to that each chicken has two feet.</span></p>
<p dir="ltr"><span>In the linear algebra class, I then talked about two vector spaces: chicken-rabbit space and head-foot space. Then I talked about the <a href="https://docs.google.com/document/d/1GeztAnZIMR8PKNvGLp_Pc8zxf2eKU3FjmlIzj8Kz3-I/edit?usp=sharing">vector-vector linear map: (chicken-rabbit space) → (head-foot space)</a>. The linear map is then a table (i.e., matrix) of four facts of life:</span></p>
<ul><li>A chicken has 1 head</li>
<li>A chicken has 2 feet</li>
<li>A rabbit has 1 head</li>
<li>A rabbit has 4 feet</li>
</ul><p> <span>This strategy may be parallel to your way of introducing stress.</span></p>
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</ul>Wed, 21 Feb 2018 13:13:00 +0000Zhigang Suocomment 29595 at https://www.imechanica.orgRe: Defn of stress: A glaring mistake in my reply to Zhigang
https://www.imechanica.org/comment/29594#comment-29594
<a id="comment-29594"></a>
<p><em>In reply to <a href="https://www.imechanica.org/comment/29593#comment-29593">Re: Stress Definition</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Please ignore the second derivation, i.e., starting with Cauchy's formula, and proceeding as follows:</p>
<p> </p>
<p>{t} = [\sigma]^T {n}</p>
<p> </p>
<p>{t}[n] = [\sigma]^T {n}[n] (Post-multiply both sides by {n}-transposed)</p>
<p> </p>
<p>{t}[n] = [\sigma]^T ({n}[n] is the dot product, equal to 1)<strong> <------ THIS PART IS WRONG</strong></p>
<p> </p>
<p>[\sigma] = {n}[t] (Take transposes of both sides, and flip LHS and RHS)"</p>
<p>=======</p>
<p>The error is in saying that {n}[n] is the dot product, equal to 1. Very elementary mistake! </p>
<p>=======</p>
<p>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....</p>
<p>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.</p>
<p>========</p>
<p>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...</p>
<p>And sorry, once again, for bothering you all with the mistake....</p>
<p>--Ajit</p>
<p> </p>
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</ul>Wed, 21 Feb 2018 13:04:11 +0000Ajit R. Jadhavcomment 29594 at https://www.imechanica.orgRe: Stress Definition
https://www.imechanica.org/comment/29593#comment-29593
<a id="comment-29593"></a>
<p><em>In reply to <a href="https://www.imechanica.org/comment/29591#comment-29591">Re: Stress is defined as the quantity equal to ... what?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Zhigang,</p>
<p>Wow, good to know that it's been puzzling for you too...</p>
<p>BTW, yes, I have been aware of your notes, and in fact recently recommended them in a reply at my personal blog, here [<a href="https://ajitjadhav.wordpress.com/2018/02/18/stress-is-defined-as-the-quantity-equal-to-what/#comment-1316" target="_blank">^</a>].</p>
<p>Overall, to cut a long story short, here are a few points that I think we should be explaining to students:</p>
<p>====</p>
<p>We should state and explain the concept of stress as the direct end-result of a tensor product.</p>
<p>Here is one way to bring out the required direct definition. Only the UG-level matrix algebra is used.:</p>
<p>Notation: We adopt the convention that a vector is <em>always</em> denoted via a <em>column</em> vector, <em>never</em> using a row vector, and therefore, that a <em>row</em> vector <em>always</em> represents <em>only</em> 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.)</p>
<p>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.)</p>
<p>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.</p>
<p>Denote the unit normal to surface as {n} and its transpose as [n].</p>
<p>Start with the transpose of the traction vector, denoted as {t}^{T} and also written as [t].</p>
<p>[t] = [t] (and what else?)</p>
<p>= (1) [t]</p>
<p>= ( [n]{n} ) [t]</p>
<p>= [n] ( {n}[t] )</p>
<p>= [n] [\sigma]</p>
<p>In the last step, we have used the direct definition: [\sigma] \equiv {n}[t].</p>
<p>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:</p>
<p>{t} = [\sigma]^T {n} </p>
<p>{t}[n] = [\sigma]^T {n}[n] (Post-multiply both sides by {n}-transposed)</p>
<p>{t}[n] = [\sigma]^T ({n}[n] is the dot product, equal to 1)</p>
<p>[\sigma] = {n}[t] (Take transposes of both sides, and flip LHS and RHS)</p>
<p>The matrix-notation expression {n}[t] stands for a <em>tensor</em> 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}". </p>
<p>This structure of</p>
<p>"{n} \otimes [the surface-intensity of a vector field variable]"</p>
<p>is common to all the <em>flux</em>-tensors of <em>all</em> the <em>vector</em> fields. <em>The direct definition of stress thus brings out the physical idea that the stress is the flux of the traction vector.</em></p>
<p>The idea of the strain is obviously is based on a gradient; it's the symmetrical part of the displacement gradient.</p>
<p>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 <em>all</em> 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.</p>
<p>Speaking in general terms, the flux of a <em>scalar</em> field is a vector field; the gradient of a <em>scalar</em> field is a vector field; and the two vector fields are related via a material (constitutive) law.</p>
<p>Exactly similarly, the flux of a <em>vector</em> field is a tensor field; the gradient of a <em>vector</em> is a tensor field; and the two tensor fields are related via a constitutive law.</p>
<p>===</p>
<p>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.</p>
<p>That's what the indirect equation</p>
<p>{t} = [\sigma]^T {n} (i.e. Cauchy's formula)</p>
<p>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})</p>
<p>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.</p>
<p>===</p>
<p>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.</p>
<p>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 <em>don't</em> need a volume element at all. This is important to realize by the text-book writers. To define stress, you <em>don't</em> 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.</p>
<p>Finally, just one more point.</p>
<p>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.</p>
<p>===</p>
<p>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.</p>
<p>===</p>
<p>But, thanks, once again, for your interest,</p>
<p>Best,</p>
<p>--Ajit</p>
<p> </p>
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</ul>Wed, 21 Feb 2018 11:17:51 +0000Ajit R. Jadhavcomment 29593 at https://www.imechanica.orgRe: Stress is defined as the quantity equal to ... what?
https://www.imechanica.org/comment/29591#comment-29591
<a id="comment-29591"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/22146">Stress is defined as the quantity equal to ... what?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Ajit: Your question has also puzzeled me. I now favor your equation (3). Here is a post called "<a href="http://imechanica.org/node/18249">a state of stress is a linear map</a>". </p>
<p>Incidentally, I am teaching undergarduate linear algebra now. Here are class notes that try to explain <a href="https://docs.google.com/document/d/1GeztAnZIMR8PKNvGLp_Pc8zxf2eKU3FjmlIzj8Kz3-I/edit?usp=sharing">where we get vector spaces and linear maps</a>.</p>
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</ul>Wed, 21 Feb 2018 02:34:00 +0000Zhigang Suocomment 29591 at https://www.imechanica.org