iMechanica - Comments for "New theory of elasticity & deformation"
https://imechanica.org/node/5014
Comments for "New theory of elasticity & deformation"enAs a good painkiller...
https://imechanica.org/comment/24966#comment-24966
<a id="comment-24966"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
I would recommend this beautiful<br />
article:
</p>
<p>
</p>
<p>
<em>The Existence of the Flux Vector and<br />
the Divergence Theorem for General Cauchy Fluxes</em>.
</p>
<p>
Silhavy, M., 1985. Archive for Rational<br />
Mechanics and Analysis, 90, 3, 195-212.
</p>
<p>
</p>
<p>
I hope it will be a relief for much of<br />
the torment, if not all...
</p>
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</ul>Sun, 22 Sep 2013 18:44:43 +0000Stefan C. Soarecomment 24966 at https://imechanica.org?
https://imechanica.org/comment/12673#comment-12673
<a id="comment-12673"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>why there isn't more comments? it was very funny reading this. all the best to everyone:)</p>
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</ul>Wed, 28 Oct 2009 19:23:43 +0000sasaborgcomment 12673 at https://imechanica.orgthe virial law
https://imechanica.org/comment/10745#comment-10745
<a id="comment-10745"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>Craig, </p>
<p>elastic behavior is time-independent. The situations you refer to are clearly relating to heat, not mechanics. Having said that, I do not question what you wrote. I just think there is more to it than is apparent so far. I would have to second-guess your words, and I'd rather not; but I am not talking about heat effects and/or diffusion. I am talking about the loaded state below the threshold temp at which diffusion becomes important, such that the elastically loaded state can be maintained, indefinitely if necessary. (A few billion years should suffice; I have seen recent cracks in rocks that developed due to an elastic potential that had waited 2,4by for its release.) My theory has nothing to do whatsoever with rigid-body motion, it has nothing to do with motion (in the Newtonian sense) at all, and I cannot see a reason why it should not translate. My theory is a generalization of thermodynamics; the theory you know to be written in P and V, I have rewritten in f and r, with the condition (always observed) that the surface integral of the f-r calculations must be compatible with the P-V calculations for identical situations. It does work. </p>
<p>I understand elastic deformation as a change of state. The simplest deformation conceivable is isotropic compression of an ideal gas. If you take that theory, transform it into vector form, and spend a few thougths on boundary conditions, you arrive at my theory. But you will not get into conventional CM because the Cauchy theory cannot consider the distinction of system and surrounding. It is simply wrong to start a theory of stress – implying a change of state – with an equation of motion (implying Newtonian mechanics, E_kin + E_pot = const, and therefore by definition excluding changes of state from consideration). </p>
<p>Falk</p>
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</ul>Wed, 06 May 2009 23:24:03 +0000Falk H. Koenemanncomment 10745 at https://imechanica.orgPlease, enough of this
https://imechanica.org/comment/10743#comment-10743
<a id="comment-10743"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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 Mr Koenemann,
</p>
<p>
I am certain I will regret re-engaging with this discussion, but I cannot let your comments pass unchallenged.
</p>
<p>
Several posts above you make a serious mistake in your interpretation of the virial theorem. Virial-type relations are a consequence of the vanishing of the time-average of bounded functions as time increases. An arbitrary deformation is not a bounded function. Furthermore, a second requirement of the virial theorem is that the time average of the external force resultants be zero, which is not the case on the time scale of a deformation. One of the consequences of your errors is that your theory is not invariant under translations. This means that a rigid-body motion will change the stress-state (or whatever you want to call it in your theory). Clearly this is absurd. This is also one of the foundations of your theory and, by itself, this mistake invalidates your claims.
</p>
<p>
I believe Schweitz did some work on this topic back in the 1970s, which may provide you with some greater insight. I expect that you will find that your theory is incorrect and you may wish to recant.
</p>
<p>
Sincerely,
</p>
<p>
Craig
</p>
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</ul>Wed, 06 May 2009 16:49:33 +0000Craig A Steevescomment 10743 at https://imechanica.orgTrace of stress s_ii: physical significance
https://imechanica.org/comment/10702#comment-10702
<a id="comment-10702"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>With one recent correspondent I had the following exchange: </p>
<p>Just answer: f is a flux. </p>
<p>(1) Consider heat flow. For a typical region the condition is div f = 0. I conclude that the temp of the region is constant. Agree or disagree?</p>
<p>(2) Consider mass flow, e.g. water. For a typical region the flow is characterized by div f = 0. I conclude that no water is stored in the region, and no stored water is released from the region. Agree or disagree?</p>
<p>(3) Consider energy. A volume of gas is deformed (equilibrium flow) such that its shape changes, but its volume is constant. The flow is characterized by div f = 0. Has work been done, has a change of state occurred?</p>
<p>To which he replied: </p>
<p>to (1): This is not even a relevant question. But I will answer briefly. If f is the heat flux tensor, which is typically defined as proportional to the gradient of temperature, div f = 0 implies no heat source in the region and leads to a Laplace's equation for temperature.</p>
<p>to (2) Again, this is an irrelevant question. If f is the mass flux vector, div f = 0 is the simple mass conservation rule. Water comes in and out, but the total mass of water in the region does not change. </p>
<p>to (3): What is f in this case? If you are thinking about ideal gas, only pressure does work. Then, if the volume of the gas does not change, no work is done and no change of state occurs (assuming isothermal condition). However, we cannot extend this to solids […]. </p>
<p>I wrote: </p>
<p>I agree with your last sentence. Question now: why not? There is no mathematical reason. Of course I do not maintain that no work is done in deforming a solid. But if the condition s_ii = 0 is assigned the importance it has in conventional CM, it must have a meaning. It cannot have another meaning than "what goes in, goes out". </p>
<p>Or do you see an alternative? What does the condition mean physically in your view? </p>
<p>Why is the statement correct for a gas, but not correct for a solid, when the external boundary conditions are the same? <br />
If you consider the situation that you have two pistons with only air in between, and you close the pistons slowly, the gas will deform. It will spread laterally, just as a solid. It will maintain its volume. Thus you can say that a gas has Poisson's ratio 0.5. As long as you are under the rule of dw = d(E_kin – E_pot), you deal with acceleration work only. For a gas with n particles it works: the work done on all particles sums to zero. </p>
<p>Since then he does not answer any more. That's just the easy way out. </p>
<p>The idea that the trace of a square matrix is meaningful comes from linear algebra and from the theory of potentials; and for heat, mass flow and for the volume-neutral 'deformation' of a gas the zero result is correct, showing the accuracy of the concept. If this is not applicable to elastic deformation of solids, the reason is that work is done which is stored in the system, and an elastic potential builds up. But this cannot be considered under the Laplace condition div f = 0: there are no excess forces left that can be used to define a non-zero work term. </p>
<p>Sharper: if a physical situation is described involving the Laplace condition div f = 0, it is the most stringent description of the fact that no work is done upon a system. If a theory containing this condition comes to a non-zero result anyway, by applying proportionality factors etc. (such as Poisson's ratio) it is not a sign that the theory is reliable nonetheless. Rather, it is a classic perpetuum mobile of the second kind: more energy comes out than goes in. </p>
<p>However, if the contention is dropped that the condition div f = 0 (or tr s = 0) is applicable to solids, if it is acknowledged that the divergence (or an equivalent expression) must be non-zero if work is done upon a system, it would be the proper step to understand elasticity better. In the light of potential theory this is the only way to go. But Euler-Cauchy will fall. </p>
<p>I could not get the correspondent to address my question what s_ii = 0 physically means in his opinion. This is a pity, this is the core of the entire dispute. </p>
<p>As long as it is maintained that the condition tr s = 0 is meaningful in deformation of solids, and correct to use, continuum mechanics remains out of touch with the rest of physics. <strong>There are no sanctuaries for endangered theories. </strong></p>
<p>
Falk H. Koenemann</p>
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</ul>Mon, 04 May 2009 03:13:30 +0000Falk H. Koenemanncomment 10702 at https://imechanica.orgFramework of classical physics
https://imechanica.org/comment/10700#comment-10700
<a id="comment-10700"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Classical physics (pre-1900) began to find its first reliable foundations in the 17th C. Newton's three laws are indispensable for the mechanics of discrete bodies in freespace. Originally one used a force conservation law; but it soon became apparent that the entity to be conserved is some thing else for which the term 'energy' was proposed by Thomas Young in the 1790s, but the term did not come into common use before the late 19th C. It was certainly considered, under the name 'vis mortua'. The first energy conservation law was found by J.G. Leibnitz in the 1670s and Johann Bernoulli 1735 who concluded that E_kin + E_pot = const. Bernoulli had a student who would gain the stature of a giant: Leonhard Euler. </p>
<p>In 1776 two entirely different steps of importance for the discussion of deformation were made: Euler proposed the stress tensor; and James Watt invented the steam engine. Clearly the connection was not seen for another 70 years. </p>
<p>Watt's step culminated in the First Law of Thermodynamics in 1847 by Hermann Helmholtz. Since then the framework of classical physics has been completed: </p>
<p><strong>1. </strong>There is the field of <strong>conservative</strong> processes under the rule of E_kin + E_pot = const. The 'const' is called the entire energy H of a kinetic system. This is Newton's mechanics.
</p>
<p>
<br />
<strong>2. </strong>If a process is <strong>non-conservative</strong> it means that H is 'not conserved' i.e. it is a variable. The energy conservation law for such processes is the First Law dU = dw + dq. </p>
<p>Two different types of non-conservative processes were soon distinguished:</p>
<p><strong>2a.</strong> The process may be <strong>reversible</strong>. </p>
<p><strong>2b.</strong> The process may be <strong>irreversible</strong>, and entropy is produced. </p>
<p>If a process is conservative, the energy of the system is invariant. The system is a kinetic system consisting of n bodies in freespace, and the system does not exchange energy with its surrounding. It is isolated. A conservative theory may be started with an equation of motion; it does not need an equation of state. </p>
<p>If a process is non-conservative, it must start with an equation of state. </p>
<p>The connection from Bernoulli's energy conservation law of conservative mechanics and the First Law of non-conservative mechanics was provided by Rudolf Clausius 1870. He considered the free vibrations of atoms in a solid or fluid and their kinetic energy E_kin and correlated them with the heat. He correlated the potential energy term with the bonds in a solid, or – my liberty – other, less permanent forms of atomic interactions. Thus he arrived at the <strong>virial law</strong>,</p>
<p>E_kin + E_pot = PV</p>
<p>Clausius interpreted E_kin as the heat. Thus if the internal energy PV = U of a system is a variable, we can consider it by virtue of the equation of state </p>
<p>PV = nRT, </p>
<p>and the energy conservation law for non-conservative physics, i.e. the physics of changes of the energetic state, is, of course, </p>
<p>dU = dw + dq.</p>
<p>Considering an elastic process, it is clearly non-conservative-yet-reversible. Thus it must start with an equation of state. I have not seen a textbook in which the attempt was made to derive the theory of stress from an equation of state. Rather, I have quoted enough textbooks in which an elastic deformation is understood as a variation. But a variation is a concept that was invented by Euler, it is strictly under the rule of E_kin + E_pot = const. Maybe the theory of variations has its benefits in other parts of physics, but this energy conservation law is the wrong one for elasticity and deformation of solids. Work in conventional CM is defined as the negative of the kinetic potential, </p>
<p>dw = d(E_pot – E_kin)</p>
<p>which makes it clear that this work is acceleration work, but not change-of-state work.
</p>
<p>
The difference between conservative and non-conservative physics is so profound that they are commonly explained in different textbooks. Mixing them up is arguably the worst mistake one can make in classical physics. But the difference between conservative and nonconservative physics is systematically blurred in continuum mechanics. </p>
<p>The form of the "First Law" given in the textbooks is</p>
<p>int (d E_kin + dU) dV = dw + dq</p>
<p>where U is the internal energy. (the d is a lower case greek delta, indicating a very small, but finite quantity.) I cannot see how E_kin and U can be reasonably summed. This "law" and the virial law above cannot be reconciled. </p>
<p>The fire I have drawn here on iMechanica so far did not address these points. I have some human understanding for those who have difficulties to accept this difference because an academic education, and possibly a decade or three of professional work and teaching, form the mind very strongly, and it is not easy to find out of that. However, the mixup must be cleared up. The form of the First Law as used in CM is invalid, it is not the First Law. </p>
<p>Falk H. Koenemann
</p>
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</ul>Fri, 01 May 2009 10:17:35 +0000Falk H. Koenemanncomment 10700 at https://imechanica.orgPoisson's ratio
https://imechanica.org/comment/10620#comment-10620
<a id="comment-10620"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Rui, </p>
<p>No, I have not given in. I was disappointed that you insisted that I accept a term "because it is part of the conventional theory" which I question because of its purely phenomenological nature. I was disappointed because I thought I had given you enough insight in my arguments, which you would not reply to. All you got from me was the admission that if you blindly follow the cookbook you get what you want; but I don't like the recipe at all, I consider it thoughtless, and I didn't eat it. </p>
<p>My critique of Poisson's ratio is this:
</p>
<p>
(1) It can be observed only if free surfaces are nearby; it is therefore dependent on the shape and the dimensions of the sample. There is no basis for a Poisson's ratio in an infinite continuum. Such an infinite continuum may be hard to find in engineering, but it is the standard situation when considering deformation in the Earth's crust & mantle, that is: for a geologist the constraints are tighter.
</p>
<p>
(2) Step 1: if a sample is stretched in X, and no attenuation is permitted in Y & Z, the work done per stretch in X is substantial because the sample is in effect expanded. Step 2: if then the boundary conditions are released in Y & Z, the sample will contract in Y & Z, and the energetic state will relax. Let's assume that now the deformation in the center of the bar is isochoric. Then it is obvious that in step 1 work was done in Y & Z _although_nothing_happened_ in these directions: without the constraints provided by the surrounding the sample would have contracted; and in step 2 the sample did work on the surrounding by pulling it inward. (I did not imply free surfaces, I implied external conditions in an infinite continuum that resemble the situation observed in a bar. If there is an energetic relaxation in step 2, it is wrong to assume that for a bar in freespace the attenuation comes free of work. )
</p>
<p>
(3) Consider an infinite continuum, and a subvolume in it (the system; 2D conditions – XZ – for simplicity, it works in 3D too). Boundary conditions are such that the sample is compressed in Z, and may expand in X. What we observe is: the sample is shortened in Z. The sample will also expand by itself in X. But since the system is part of a bonded continuum, such that system and surrounding are solidly bonded, the surface point on the X axis is under tension by the surrounding. Thus there are two independent stretch components in X: one provided by the system itself, the other one caused by the surrounding on the system. Without the second stretch component the surface point of the system on X would be in equilibrium after the system-caused stretch in X. </p>
<p>The latter situation must be expected on free surfaces. Thus there must be a strain gradient near free surfaces: far inside a solid there will be two stretch components in X, but one of them must reach zero on an unloaded surface. </p>
<p>Thus the question: if you talk about strain in a sample, do you mean a situation deep inside, on the surface, or someplace in between?</p>
<p>All I want to show at this point is this: point (1) shows that a very careful analysis of the boundary conditions is required. In order to write a general approch one must take care not to imply specific boundary conditions. (2) The attenuation is indeed work saved. The real significance here is not Poisson's ratio; but it is evident that the attenuation is caused by the principle of least work. Thus we must talk about work as a function of the boundary conditions. (3) The fact that there is only one shortening component in Z, but two possible stretch components in X, one of which must be zero on free surfaces, such that inevitably there must be a strain gradient from the interior towards a free surface, is probably new to most readers. It is nonetheless a fact. </p>
<p>Summary. If you consider Poisson's ratio as a material property (which is the standard way) you cannot consider the implications I outlined. Rather, we must very carefully analyze the possible material reactions as a function of the boundary conditions. A material property called "Poisson's ratio" does not exist. The principle of least work certainly does exist. </p>
<p>Falk
</p>
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</ul>Wed, 22 Apr 2009 22:01:23 +0000Falk H. Koenemanncomment 10620 at https://imechanica.orgSee there. Falk
https://imechanica.org/comment/10615#comment-10615
<a id="comment-10615"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10608#comment-10608">Just do a real physical test</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>See there. Falk</p>
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</ul>Wed, 22 Apr 2009 13:40:47 +0000Falk H. Koenemanncomment 10615 at https://imechanica.orgJust do a real physical test
https://imechanica.org/comment/10608#comment-10608
<a id="comment-10608"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Falk,
</p>
<p>
See Chad's challenge here:
</p>
<p>
<a href="http://imechanica.org/node/5321">http://imechanica.org/node/5321</a>
</p>
<p>
</p>
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</ul>Wed, 22 Apr 2009 12:25:53 +0000kwlimcomment 10608 at https://imechanica.orgRe: Unjustified Poisson's ratio
https://imechanica.org/comment/10611#comment-10611
<a id="comment-10611"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10607#comment-10607">Re: deformation work</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>
Falk,
</p>
<p>
What I said is that the sum of external forces acting on a system is zero while the pressure in the system is not necessarily zero. I will leave this for the public to judge.
</p>
<p>
I used Poisson's ratio because we were discussing how the conventional theory calculates the strain and work during isochoric deformation. You refused Poisson's ratio but claimed that the conventional theory is wrong because it leads to zero work. Do you still think you have applied the conventional theory correctly (without using Poisson's ratio) to reach the wrong conclusion? I thought I have convinced you what is the right way to apply the conventional theory. You agreed in your email.
</p>
<p>
Gibbs' claim has no contradiction with the conventional theory. In a direction in which nothing happens, no displacement and thus no work done. Remember here we consider work to be force times displacement in the conventional theory. On the other hand, in a direction in which there is no force, no work is done either even when there is displacement (contraction or attenuation). The two cases are not equivalent however. The first case is uniaxial strain, and the second is uniaxial stress. If you do not understand the difference, you do not understand any mechanics, conventional or not.
</p>
<p>
Indeed there are many contradictions that you have claimed and we have not addressed. The reason however is not that we intentionally remain silent on any of those. Instead, we could not make any progress with you on any single point that we wish to get started. You simply refuse to correctly apply the conventional theory and yet claim the conventional theory is wrong. I hope you understand the simple logic: (1) the theory is right; (2) you applied it incorrectly (or simply did not follow the theory); (3) you reached an obviously wrong conclusion; (4) you think the theory is wrong. You may apply this to any theory.
</p>
<p>
RH
</p>
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</ul>Tue, 21 Apr 2009 23:37:00 +0000Rui Huangcomment 10611 at https://imechanica.orgUnjustified Poisson's ratio
https://imechanica.org/comment/10609#comment-10609
<a id="comment-10609"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10607#comment-10607">Re: deformation work</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>Rui, </p>
<p>you claimed that a zero sum of forces can nevertheless result in a non-zero pressure. If I misunderstood you, you have a highly misleading way of choosing your words. As of this point you are by no means on the safe side. And I do not accept calculations involving fudge factors, ok? I gave you my reasons why I think Poisson's ratio is not a physical term; you insisted, but without rationale other than that it is commonly accepted. I still think the energetic implications of this ratio need to be explored. </p>
<p>Plus, neither you nor anyone else has addressed the obvious contradiction between common understanding and Gibbs: common understanding has it that a bar may attenuate, and no work is done in this; Gibbs claimed that no work is done in a direction in which nothing happens. So either the thing attenuates with work (or relaxation) being done, or withhout, so which one is it? Just wrapping this in silence is not an answer. </p>
<p>Falk</p>
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</ul>Tue, 21 Apr 2009 22:46:35 +0000Falk H. Koenemanncomment 10609 at https://imechanica.orgRe: deformation work
https://imechanica.org/comment/10607#comment-10607
<a id="comment-10607"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10600#comment-10600">Deformation work</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>
Falk,
</p>
<p>
If your arguments make any sense, we would have gone much further into those questions. The fact is, you do not understand the conventional continuum mechanics theory, or you have intentionally applied it incorrectly to get all kinds of wrong/strange conclusions, for which you simply balme on the theory rather than your own intelligenece or honesty. Through all these discussions I have followed on imechanica and via private emails, you have shown clearly how badly you understand the conventional continuum mechanics theory. You have used thermodynamics as a weapon to attack continuum mechanics, but really you do not understand thermodynamics that well either. You have also shown your weakness in basic calculus several times regarding integrals and vectors. It is unbelievable that you can still maintain yourself with all the errors and misinterpretations of math, physics, and continuum mechanics.
</p>
<p>
To all other readers: I am one of the correspondents trying to understand Falk. Unfortunately, I am afraid it was a bad idea to waste time on this.
</p>
<p>
RH
</p>
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</ul>Tue, 21 Apr 2009 17:53:00 +0000Rui Huangcomment 10607 at https://imechanica.orgDeformation work
https://imechanica.org/comment/10600#comment-10600
<a id="comment-10600"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
The discussion has continued with a few correspondents by private emails. Two subjects came up with two correspondents simulaneously: the sign of deformation work, and the way to integrate the stress-strain equation. </p>
<p>Work: it was claimed that the equation for elastic work can only be positive: if s = [1 -1/2 -1/2], and e = [1 -1/2 -1/2], then necessarily the trace of the result is 1 + 1/4 + 1/4. I think this cannot be right. An elastic expansion and a contraction must produce work with opposite sign, for two reasons: (1) the relaxation of a compressional loading condition is an expansion, and the work of loading and unloading must cancel; if they were both positive, they should add. (2) The work done in an expansion or compression, starting from some standard state, is different in thermodynamics, as everybody can check for himself. </p>
<p>I have pointed out that there are two ways to interpret the trace of two tensors A and B: it may be tr (A.B), or it may be tr A . tr B. The product above would be tr (A.B), and I think it is physically meaningless, for the reasons given. The product tr A . tr B can only be zero if tr A is zero. </p>
<p>I maintain that the condition tr s = 0 is a statement to the effect that no work is done in a process (s = stress, tr s = s_11 + s_22 + s_33). Consider a region V, consider fluxes f in and out of the region. If the fluxes balance such that div f = 0 holds, the state of the region is unchanged. If the fluxes are flow of water, the region is neither a source nor a sink of water. </p>
<p>If the flux is heat, the temp of the region is constant. And if f are forces, no work is done upon the region. Since no one would maintain that indeed no work is done in an elastic deformation, it is clear that something has so far been overlooked. I think the power of this argument is so far vastly underestimated. </p>
<p>One correspondent claimed that the involvement of Poisson's ratio changes the zero result. I disagree. Poisson's ratio is the attempt to explain an observation by itself, it is phenomenological. Rather, one should ask how much work is saved by allowing a stretched bar to attenuate, and then focus on very carefully analyzing the energetics of deformation. </p>
<p>Two correspondents took issue with my claim that in the integration of s de, de is isotropic. I think that the result of int f(x) dx should be fully determined by the integrand f(x), and the magnitude by the limits of integration. Thus if the result is a tensor, it is not appropriate to let the limits of integration be anisotropic unless there are compelling reasons which have not been offered so far. </p>
<p>Characterizing the discussion so far, it is still completely from the basis of conventional CM. While the correspondence with some partners continues, and it is too early to come to a final <br />
opinion, the actual desire to simply explore what else is offered, and where it may lead to, is nil. People would not admit it, but it is plain orthodoxy. Questions that do not fit into the known canon will not be discussed. Examples: </p>
<p>(1) the obvious conflict between standard elasticity (which holds that the attenuation of a bar happens without work being done) and Gibbs (who said that no work is done in a direction in which nothing happens) has not been approached by anyone. </p>
<p>(2) What is so wrong about "Why does the theory of elasticity start with an equation of motion, and not with an equation of state?" </p>
<p>(3) My claim that Clausius' virial law must be taken into consideration, has not found any response. </p>
<p>Falk H. Koenemann
</p>
<p>
</p>
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</ul>Tue, 21 Apr 2009 08:12:56 +0000Falk H. Koenemanncomment 10600 at https://imechanica.orgstress, strain, and material
https://imechanica.org/comment/10553#comment-10553
<a id="comment-10553"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Craig,
</p>
<p>
I did not say that e_11 = e_22 = e_33, but I said that de_11 = de_22 = de_33. The numerical values of the e_ii are yet to be determined.
</p>
<p>
Since I had just had nearly the same discussion with a private correspondent, I repeat the entire argument here:
</p>
<p>
Tensor products do not communicate, AB ≠ BA
</p>
<p>
but the trace of a tensor is an invariant, and tr AB = tr BA communicates.
</p>
<p>
Also, tr A . tr B = tr B . tr A
</p>
<p>
but tr A . tr B =/= tr AB.
</p>
<p>
Let the off-diagonal terms be zero to make it simpler, then <br />
tr AB = a_11 b_11 + a_22 b_22 + a_33 b_33
</p>
<p>
but tr A . tr B = (a_11 + a_22 + a_33) (b_11 + b_22 + b_33)
</p>
<p>
If A = B such that a_11 =/= a_22 =/= a_33 but tr A = 0, (at least one term must be negative)
</p>
<p>
tr AB = tr AA = (a_11)^2 + (a_22)^2 + (a_33)^2 > 0 (it cannot contain negative terms)
</p>
<p>
whereas tr A . tr A = 0 . 0 = 0
</p>
<p>
So, the question is, do I pick my results as they please me? No:
</p>
<p>
Let s_ij = [1 -1/2 -1/2]
</p>
<p>
you can write s de as
</p>
<p>
s [1 -1/2 -1/2] de [1 1 1]
</p>
<p>
Why so?
</p>
<p>
The s matrix determines the properties of stress, the de matrix must be an identity matrix I for an isotropic material. If you let s be isotropic, but de is not isotropic, the result is necessarily anisotropic. This is what you get if you subject a single crystal of some non-cubic material to a high hydrostatic pressure. So, if de = I, and s is as above, the properties of the finite e are those of s. Thus
</p>
<p>
tr se = (s_11)^2 + (s_22)^2 + (s_33)^2
</p>
<p>
which is sign-insensitive and physically meaningless, it no longer reflects the properties of s itself;
</p>
<p>
but tr s tr e = 0. </p>
<p>Regarding my own theory, you haven't even looked at it yet. All of the above is in terms of the conventional theory.
</p>
<p>
Greetings, <br />
Falk
</p>
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</ul>Thu, 16 Apr 2009 10:33:38 +0000Falk H. Koenemanncomment 10553 at https://imechanica.orgWork-free isochoric deformation?
https://imechanica.org/comment/10542#comment-10542
<a id="comment-10542"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10535#comment-10535">signs?</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 Mr Koenemann,
</p>
<p>
This is, at best, bizarre and laughable. If you integrate de_ij and get a constant e for e_11, e_22 and e_33, you are explicitly stating that e_11 = e_22 = e_33 = e ie isotropic expansion. Moreover, if e_11 does not equal e_22 and e_33, then you cannot, algebraically, remove s_11, s_22 and s_33 from the equation s_11 e_11 + s_22 e_22 + s_33 e_33. Furthermore, you are also ignoring the functional dependence of s_ij on e_ij: what is your constitutive relation? We are now reduced to discussing what is possible and not possible in basic algebra and calculus. To me, thus far, it appears that every step of your "theory" is based upon similar examples of mathematical or logical mistakes and misunderstandings. At each stage, a variety of errors in this "theory" have been pointed out by a series of different people, all of whom seem to have reached the same two conclusions:
</p>
<p>
(1) your theory is rubbish.
</p>
<p>
(2) it is futile trying to point out the errors in your thinking.
</p>
<p>
With a certain amount of relief, I will have now to abandon this thread as I am going to be away for several days, but I am left wondering if this is a strange piece of modernist performance art in which I have been foolish enough to participate. Most importantly, though, I do not want any future readers to stumble across this thread and be led to believe that the Koenemann theory is valid or uncontested. It is not. I don't recommend that anyone else take up this Quixotic discussion, but I will certainly be amused to read any further posts.
</p>
<p>
Regards,
</p>
<p>
Craig
</p>
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</ul>Wed, 15 Apr 2009 15:12:08 +0000Craig A Steevescomment 10542 at https://imechanica.orgsigns?
https://imechanica.org/comment/10535#comment-10535
<a id="comment-10535"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Craig,
</p>
<p>
sorry, there is a sign error in your point 2, line 1: if tr s = s_11 + s_22 + s_33 = 1 -1/2 -1/2 then tr s = 0. The s_ii (no sum) are the algebraic terms; only their value is positive or negative. The relation s de (s = sigma = stress, e = epsilon = strain) is the formula by which the strain is calculated as a function of stress. – That's how I learned it. I don't agree with it. But right now you ask me to argue in terms of the conventional theory.
</p>
<p>
If we calculate s de, the sign of the result comes from the sign of the s term. Hence if we integrate over unit magnitude of e, the result is s_11 e + s_22 e + s_33 e = 1 e – 1/2 e – 1/2 e. so this is not an isotropic expansion, but an isochoric deformation.
</p>
<p>
Again, please check the signs. I did not indicate an isotropic expansion. And, no, I do not think we can arbitrarily assign stress and strain fields. But if we choose a stress with magnitudes as I proposed such that tr s = 0, by means of s de we cannot come to another result than that the total is zero too. For strain e it means that it is isochoric; a non-zero value would indicate a volume change. For work s de it means that it is zero. </p>
<p>Thank you,
</p>
<p>
Falk
</p>
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</ul>Tue, 14 Apr 2009 20:05:42 +0000Falk H. Koenemanncomment 10535 at https://imechanica.orgWork-free isochoric deformation?
https://imechanica.org/comment/10508#comment-10508
<a id="comment-10508"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10470#comment-10470">isochoric</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 Mr Koenemann,
</p>
<p>
I appreciate your response, but I am afraid this raises further questions.
</p>
<p>
(1) Your original claim was, effectively, that conventional continuum mechanics was so flawed that it will produce absurd results for simple cases. In particular you claimed that for all cases of isochoric deformation, conventional continuum mechanics will calculate zero work. You then demonstrate this claim by using your own unusual, perhaps unique, interpretation of mechanics, with which most if not all workers in the field would disagree, to generate absurd results. Surely if you are going to criticise conventional continuum mechanics for generating absurd results, you should at least use conventional continuum mechanics to generate the absurd results.
</p>
<p>
(2) After assigning the stress field, your equation would be correct provided de_11 = 1/2 (de_22 + de_33). This could be, most obviously, an isotropic expansion or contraction such that de_11 = de_22 = de_33, but it is most certainly not isochoric, nor is e_ij proportional to the s_ij you set. What exactly is s, and why are you multiplying by it? And how is this connected to the stress field you propose? I am not familiar with stress fields of the sort you suggest which would produce isotropic expansions. Are you suggesting that you can arbitrarily assign stress and strain fields? This makes no sense at all to me, physically, mathematically or logically.
</p>
<p>
Regards,
</p>
<p>
Craig
</p>
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</ul>Mon, 13 Apr 2009 16:34:32 +0000Craig A Steevescomment 10508 at https://imechanica.orgisochoric
https://imechanica.org/comment/10470#comment-10470
<a id="comment-10470"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Craig,
</p>
<p>
because I do not believe in the necessity of Poisson's ratio as a physical term, because I have made it clear that free surfaces nearby change the end result, and because I reject the proposition that the attenuation in y and z comes free, I prefer a perfect infinite continuum simply so we are both in unambiguous terrain. </p>
<p>Assume a situation in which stress s_11 = a (a is a free numerical, not an area), </p>
<p>s_22 = -a/2, s_33 = -a/3 such that tr s = 0; in the following the brackets are 3x3 matrices with zero terms left out</p>
<p>then s_ij de_ij = s [a -a/2 -a/2] [de_11 de_22 de_33]</p>
<p>= as de_11 - (as/2) de_22 - (as/2) de_33</p>
<p>= as (de_11 – 1/2 de_22 – 1/2 de_33/2)</p>
<p>= 0</p>
<p>which means you do not have to do the integration any more. Integrated over unit length the strains e will be proportional to s, so the deformation is isochoric. But since the paths cancel, the work cancels. </p>
<p>Falk
</p>
<p>
I shall be absent for a day or two.
</p>
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</ul>Fri, 10 Apr 2009 22:51:13 +0000Falk H. Koenemanncomment 10470 at https://imechanica.orgWork-free isochoric deformation
https://imechanica.org/comment/10455#comment-10455
<a id="comment-10455"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10454#comment-10454">work again</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 Mr Koenemann,
</p>
<p>
Yet again, thank you for your reply. I feel that it is very important to be completely clear about these preliminary points, otherwise we will end up speaking past each other due to differences in terminology or in the way that we are defining our concepts. So, what you are stating is that:
</p>
<p>
(1) Conventional continuum mechanics will calculate non-zero work for an isochoric deformation such as the uniaxial stretching of a material with Poisson's ratio 0.5 (as I have done above, using conventional continuum mechanics).
</p>
<p>
(2) However, you believe that conventional continuum mechanics is incorrect; as an example, you believe that conventional continuum mechanics wrongly allows the trace of the stress tensor to be non-zero during an isochoric deformation.
</p>
<p>
(3) When you apply your interpretation of continuum mechanics to an isochoric deformation, you produce absurd results, such as uniformly zero work for volume-neutral deformations.
</p>
<p>
(4) Therefore conventional continuum mechanics is invalid.
</p>
<p>
Have we agreed on this much, at least?
</p>
<p>
Regards,
</p>
<p>
Craig
</p>
<p>
</p>
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</ul>Thu, 09 Apr 2009 23:25:12 +0000Craig A Steevescomment 10455 at https://imechanica.orgwork again
https://imechanica.org/comment/10454#comment-10454
<a id="comment-10454"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Craig,
</p>
<p>
the condition tr sigma = 0 is the only possible condition for a volume-neutral deformation, pure or simple shear or other. Since that condtion is compatible – in the light of the principles of potential theory – only with the conclusion that no work has been done, I must conclude that a volume-neutral deformation can be achieved without doing work, according to the conventional theory. This is the perpetuum mobile condition of the first kind, namely that we get an effect (a deformation) without doing work. I hope this is as unambiguous as words can be.
</p>
<p>
I think I have made it clear that I have a sufficient number of systematic objections to the conventional theory in non-straightforward cases to be able to deal with them as well. Among others, the L&L-textbook example demonstrates that the error is solidly built into the conventional theory.
</p>
<p>
This is as focussed as possible. I have given one example where the zero work condition for a deformation is actually correct, namely for the volume-neutral deformation of a gas when no elastic potential builds up. The point is, from the entire mathematical & philosophical structure of conventional CM, one can only conclude that the work done in a deformation is work done under the conservative energy conservation law E_kin + E_pot = const; whereas it has never been properly considered that elastic deformation work must be akin to change-of-state work, i.e. work done under the First Law, in the isotropic case this is PdV-work.
</p>
<p>
Falk
</p>
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</ul>Thu, 09 Apr 2009 22:41:25 +0000Falk H. Koenemanncomment 10454 at https://imechanica.orgWork-free isochoric deformation
https://imechanica.org/comment/10446#comment-10446
<a id="comment-10446"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10442#comment-10442">condition tr s = 0</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 Mr Koenemann,
</p>
<p>
Thank you again for your detailed explanation. I am, however, still confused about the very first point which we are discussing. Consequently our conversations cannot be very productive as I am not certain that we are in fact discussing the same thing. I return to your original statement: "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". We have not yet managed to agree even on the meaning of this (or if we have, I missed that, since I don't yet know quite what this means). Are you stating that, for all volume-neutral deformations, the standard application of continuum mechanics (with which you, admittedly, disagree) will calculate zero work?</p>
<p>Let's come to an agreement on what this means. Once we have cleared up this point then I will at least know what we are discussing. Until then, it is very difficult for me to make rigorous statements about something which seems rather undefined. If we work on one thing at a time, we will at least make that much progress. By discussing dozens of topics simultaneously we seem to be foundering on the complex interrelationships between these things.
</p>
<p>
Regards,<br />
Craig<br />
</p>
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</ul>Thu, 09 Apr 2009 18:16:40 +0000Craig A Steevescomment 10446 at https://imechanica.orgcondition tr s = 0
https://imechanica.org/comment/10442#comment-10442
<a id="comment-10442"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Craig,
</p>
<p>
Yes, I imply that the condition "trace of stress tensor = 0" implies that no work is done. In fact, it is not an implication, but there cannot be a more explicit statement to that effect. And yes, I know it flies into the face of everything you and I have learned in grad school. Potential theory does not leave you another option.
</p>
<p>
If I am incredibly glad for having found iMechanica, it is because I can now make such statements in public, and they cannot be drowned in silence any more. This blog is intensely watched. The condition "trace of the tensor = 0" is an expression of the Laplace condition div v = 0 (v = general vector field), indicating that the net flux of the variable in question is zero (in this case energy flux). Of course I know that most of the people out there are unfamiliar with potential theory, and they are as incredulous as you are, but this is why I create all this noise.
</p>
<p>
The question as to what source density is, has come up a week ago (31 March). Assume that the matrix [-1 0 | 0 1] describes the flow of water in and out of a region. The trace is zero; as much water flows out as flows in, such that no extra water flows out (the region would be a source) or remains there (it would be a sink). With mass flows this appears natural, but the same matrix describing a heat flow tells you that the region does not contain an oven. Now think of energy U and its flux dU/dx = f. If the trace is zero, it means that the region remains energetically unchanged, i.e. no work is done upon the system. – To be fair to Euler & Cauchy, these systematics were understood only way after 1830. </p>
<p>Second question: my first answer to you was a bit long because I did my best to discredit the Poisson ratio as a physical term. Its observation depends on the existence of free surfaces nearby. Your statement that the principal stresses are [sigma 0 0] is in my view wrong too. – You force me to another long answer. Please download</p>
<p><a href="http://www.elastic-plastic.de/slide25.jpg">http://www.elastic-plastic.de/slide25.jpg</a></p>
<p>This is a matrix of states. First column: the unloaded state; second column: the loaded state with the condition that system and surrounding are solidly bonded; third column: the loaded state with the condition that system and surrounding are not bonded. First row: isotropic compression; second row: isotropic dilation; third row: anisotropic deformation. In isotropic compression, it does not matter if system and surrounding are bonded; the system must contract. Conclusion: normal compressive forces can always do work. </p>
<p>In dilation the bonding condition matters; if system and surrounding are not bonded, a cavity will be created in which the unloaded system rests as a discrete body. </p>
<p>In an anisotropic loading configuration normal compressive forces can do work (in y), and as a result the system bulges (in x). It does so in order to minimize the change of state. But it depends on the bonding condition whether the surface point of the system on the x axis is under tension by the surrounding, or unloaded. If the system is under tension in x, there will be two stretch components in x, one supplied by the system itself, another by the surrounding. Of necessity, the first stretch component must exist from the interior of the sample all the way to a free surface; but the second stretch component must reach zero on free surfaces. </p>
<p>This example shows a bunch of things:</p>
<p>- Inevitably, there is a strain gradient in loaded bodies with free surfaces that fades towards the interior. </p>
<p>- The state of loading (I avoid "stress" because I do not want to refer to Cauchy's tensor) must also vary as a function of position within the sample. Inside a sample the stored energy must be larger than near free surfaces. (See the model calculations in Approach, last chapter.)</p>
<p>- It is impossible to ignore bonds. The bonding forces are constraint forces (see my answer to Rui Huang on April 7).</p>
<p>Altogether, the description you gave above – [sigma 0 0 ] – becomes too crude. And what does Poisson's ratio mean here? </p>
<p>To an engineer, it is all too natural to start with stretching a bar, i.e. a discrete body with free surfaces. However, I am a geologist, and the infinite continuum came very natural to me. Free surfaces do not exist at metamorphic depths, and I missed the second stretch component from the beginning. I am saying this because it shows that different backgrounds produce different perceptions, and we must be careful with what we take for granted. </p>
<p>Falk
</p>
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</ul>Thu, 09 Apr 2009 14:35:20 +0000Falk H. Koenemanncomment 10442 at https://imechanica.orgWork-free isochoric deformation
https://imechanica.org/comment/10435#comment-10435
<a id="comment-10435"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10433#comment-10433">Work</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 Mr Koenemann,
</p>
<p>
I remain confused by this, and I am afraid that we still haven't quite agreed upon what exactly you mean by your statement: "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". Are you stating that by applying standard continuum mechanics to an isochoric deformation, then it is impossible to calculate non-zero work? Or am I misunderstanding this? We are both reasonable people, so we should try to come to an agreement on this point before complicating matters further.
</p>
<p>
If I am understanding you correctly, then my first example of an isochoric deformation (I apologise for being insufficiently specific in my previous post; sometimes my writing lacks rigour. I meant the one involving the stretching of a material with 0.5 Poisson's ratio by an axial force.) has principal stresses [sigma 0 0], and the corresponding tensor does not have zero trace. (Parenthetically, I disagree with your claim that an isochoric deformation must be accompanied by a stress tensor with trace zero, and further with your claim that the traces of the stress and strain tensors are related to work. Work is related to sigma_ij epsilon_ij; the products of the elements of the two tensors are taken before the summation, but to keep life simple I don't think we should consider that yet.) Standard continuum mechanics clearly calculates non-zero work (for a linear elastic material it will be something akin to sigma^2/2E) for the isochoric deformation in this example, and hence if your statement means what it appears to mean, then your statement is incorrect. Am I interpreting your statement accurately? If not, how ought it be interpreted?
</p>
<p>
Thanks again for your help with this.
</p>
<p>
Regards,
</p>
<p>
Craig
</p>
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</ul>Wed, 08 Apr 2009 22:37:18 +0000Craig A Steevescomment 10435 at https://imechanica.orgWork
https://imechanica.org/comment/10433#comment-10433
<a id="comment-10433"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10432#comment-10432">Work-free isochoric deformation</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>
Craig,
</p>
<p>
work: I thought I had answered this. The trace of the stress tensor is sigma_ii = 1 – 1/2 – 1/2 = 0, and ditto for epsilon_ii. Hence the product of the terms relating to work – the traces of the respective tensors – is 0 x 0 = 0.
</p>
<p>
It is entirely sufficient to consider sigma_ii only. sigma_ii = 0 is a tatement to the effect that the inward-directed and the outward-directed forces balance. In the conventional theory work done by shear forces is not considered.</p>
<p>Let me show you how this is done in my theory (a little simplified): the total force field was split into (1) an isotropic component, and the deviatoric field; the latter was split again into (2) the normal force field and (3) the shear force field. I assumed that the ideal deformation is the one that requires the most work, which is an isotropic contraction, caused by the isotropic inward-directed force field of step 1. Then I considered the deviatoric force field, and here normal forces and shear forces separately. The volumetric and energetic net effect of the normal force field is indeed zero. The work done by he shear forces is a geometric expansion which indeed balanced the volume loss by the first step; and an energetic relaxation, but here the work done in the first step is only partly relaxed, a rest remains, so the net work is non-zero. </p>
<p>E_kin = m v^2</p>
<p>Falk
</p>
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</ul>Wed, 08 Apr 2009 21:14:34 +0000Falk H. Koenemanncomment 10433 at https://imechanica.orgWork-free isochoric deformation
https://imechanica.org/comment/10432#comment-10432
<a id="comment-10432"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10418#comment-10418">work, and the First Law</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 Mr Koenemann,</p>
<p>Thank you for your detailed response. I would like to suggest two things. First, since you have written a response twice as long as my original post, and because each of us could easily continue to do the same until we have used all of the electrons on the internet, I'd like to reply only to one aspect of your post: the original assertion from your Gibbs paper (which you have rephrased earlier in this thread). The second suggestion is that sometimes, in cases of disagreement such as this, it is helpful first to determine on which points we are in agreement, and then progress to where our views diverge.</p>
<p>The original claim you made in your Gibbs paper is: "The consequence of the attempt to interpret elastic deformation as a conservative process in the sense of eqn.1 is that it is not possible to derive a non-zero work term for a volume-constant elastic deformation." In this thread you have rephrased this in your first, summary post of 9 March 2009 as "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". I interpret these to mean what they say on their faces: that if continuum mechanics is applied as it is currently understood, then zero work will be calculated for all isochoric deformations. In my previous post I gave a trivial example of stretching a material with Poisson's ratio 0.5 for which continuum mechanics calculates net positive work (or negative, if that is your preferred sign convention, but in either case non-zero). Either this contradicts your original assertion, or your original assertion does not mean what it seems to state. Before taking this further, could we come to an agreement on what your original statement means? It would help very much to know exactly what we are discussing.</p>
<p>By the way (violating my intention above to address only one issue), how are you defining E_kin?</p>
<p>Craig</p>
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</ul>Wed, 08 Apr 2009 18:42:37 +0000Craig A Steevescomment 10432 at https://imechanica.orgwork, and the First Law
https://imechanica.org/comment/10418#comment-10418
<a id="comment-10418"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10406#comment-10406">Work-free isochoric deformation</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>
First of all, to general audience: sorry for posting the same twice, a technical misunderstanding. </p>
<p>Craig, </p>
<p>1. I separate your point 1 into two. But I must discuss the second point first. </p>
<p>Your argument contains a sign error. In thermodynamics, the sign convention is this: a contraction is due to a pressure increase and perceived as positive work, so that an expansion requires negative work. If you perform an isotropic contraction and then let go, the system will expand; the work of step 1 and 2 cancel such that U_2 = U_0. Your logic would imply that contraction work done by the surrounding and contraction work done by the system have opposite sign, or that the work done in a contraction starting from U_0 and an expansion starting from U_0 have the same sign. Not so.</p>
<p>Now the hard part. If you stretch a bar of length d in x by Delta d_x, it will attenuate in y and z, but only if you let it. If you prevent the system from contracting in y and z, this Delta d_x will cost you far more Joule because in effect you perform an expansion. Gibbs (explained at the end of my paper) concluded that if nothing happens in y and z, no work is done in these directions. This is to show that Gibbs' interpretation and yours are already in conflict: either no work is done when nothing happens, or no work is done when the thing attenuates. I think both views are too simplistic. </p>
<p>If you stretch in x while keeping y and z unchanged, and then relax the boundary conditions by making the walls soft, the material will relax energetically by pulling the walls in along y and z, and the energy released is the work done in y and z in the first place. What we have now is this:
</p>
<p>
- if the substance is a gas, and V = const, indeed no work would be done in the deformation. But it would not snap back upon release; no elastic potential has built up.
</p>
<p>
- If the substance is a solid, and V = const, a finite amount of work is observed: negative stretch work is done by the surrounding in x, and positive work is done by the system upon the surrounding in y and z – and there is an elastic potential. </p>
<p>Now you can conclude: the material contracts because it has this and this Poisson's ratio. Or you can conclude that the material attenuates because of the law of least work: the change of state is kept to a minimum. The former is phenomenological. The latter is energetic, a physical thought.</p>
<p>I think a proper deformation theory should not need the Poisson term. It is not a material property, but a result of the physical setup, subject to specific boundary conditions, the presence of the free surfaces. It is dependent on the shape of the bar, it matters if the bar has square or rectangular cross section. </p>
<p>This example, the stretching of a bar, is a nice example to show that the boundary conditions were never sufficiently examined in the beginning: it matters a lot if the bar is allowed to attenuate, i.e. if there are free surfaces nearby. But if you think of deformation in the Earth's mantle, the nearest free surface is very far away, and talking about Poisson's ratio down there is at best funny. </p>
<p>All this is to show that a change of state in the sense of the First Law is always work done upon a volume, not work done in a particular direction. Whatever happens, it is spatial, and the material reactions in the various directions are never independent of one another. A proper theory must take this into account. </p>
<p>2. In the light of the above, your sentence "The latter is true, the former is not" no longer holds. Besides, you can find the statement – that for an isochoric deformation the trace of the stress tensor is zero – in textbooks, e.g. Gurtin 1972, p.53, and I have seen it in Gurtin 1981, but there I cannot give you the page. Given the conservative mathematical structure of conventional CM, the zero condition for the stress tensor in isochoric deformation is in fact unavoidable: if the material paths cancel, the work done in the deformation must cancel. And if epsilon_ii = 0 and sigma_ii = 0: it is true, tensor products do not communicate; but the products of the invariants do communicate. </p>
<p>Bachelor: I refer to the entire chapter, specifically to the eqn. without number above eqn.3.3.8, p.144. But I disregard parts where Bachelor explicitly refers to viscous effects, because viscous flow is irreversible, and thus the Second Law is involved. Before we get into the irreversible field we should first clear the reversible field. Bachelor must have been aware of Poisson's partition of a viscous flow step into an elastic, time-independent, reversible loading step and a diffusion-controlled, time-dependent, irreversible relaxation step. He must have learned about this from the same source as myself: from Stokes' paper. I refer to the elastic part only. </p>
<p>3. There is no typo in my eqn.10, and the pages of L&L referring to stress are available at <br />
<a href="http://www.elastic-plastic.de/landau-stress.pdf">www.elastic-plastic.de/landau-stress.pdf</a></p>
<p>Love, Fung et al: you must make a decision which process you wish to consider. If it is the acceleration of a body as a whole, it is done under E_kin + E_pot = const = H where H is the entire mechanical energy of the system. Note that up to here the system is isolated. If a change of state is to be considered, be it isotropic leading to PdV-work, or anisotropic as in elasticity, interaction of system and surrounding is required, and the energy conservation law is dU = dw + dq. U_0 and H have different names in CM and thermodynamics for historical conventions, but they are identical terms. If a change of state is to be studied, U = H is a variable, and Delta U is the elastic potential. In all the quoted books an attempt is made to treat E_kin and U as separate variables. This is physically impossible, E_kin is a subset of U, they cannot be summed. </p>
<p>An external acceleration of a body cannot change its internal energy. If a change of state is the subject, any possibly existing external acceleration is simply irrelevant, besides I think it is impossible, because in order to perform a change of state, external equilibrium must exist. </p>
<p>Clausius outlined it quite well in the virial law: in <br />
E_kin + E_pot = PV<br />
the first term is the heat, the random motion of the atoms which accelerate one another. If so, an equation of motion causing an acceleration is no longer available for further use. The second term LHS is the bonds. I have always missed them in conventional CM. The term RHS is the state of the system in the standard state. If that term is a variable, RHS = PV = nRT, we have the equation of state and get into standard thermodynamics, and the LHS is no longer of interest; in effect, E_pot interacts with the external forces. But it is clear that Clausius has been completely overlooked in CM.</p>
<p>I have not found out who wrote the form of the First Law as used in Love etc., I suspect it must have happened already in the 1850s. But the attempt to treat U and E_kin as variables that can be summed amounts to making the First Law, the energy conservation law of non-conservative physics, subordinate to E_kin + E_pot = const. This is the First Law upside-down. </p>
<p>Falk H. Koenemann
</p>
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</ul>Tue, 07 Apr 2009 20:03:27 +0000Falk H. Koenemanncomment 10418 at https://imechanica.orgpoints & distributions
https://imechanica.org/comment/10413#comment-10413
<a id="comment-10413"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10399#comment-10399">Re: discrete system</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>
Rui: </p>
<p>point 0: agreed.<br />
point 1: agreed.<br />
point 2: agreed.</p>
<p>point 3: sentence 1 & 2 agreed. Sentence 3 is in conflict with the thermodynamic theory because the P-terms in the thermodynamic equilibrium condition P_syst + P_surr = 0 are not zero in the loaded state. It is time to consider the nature of pressure itself. U/V is the energy density, a scalar by definition. f/A is a force density on a surface. They are not equivalent, and U/V is the more fundamental one. </p>
<p>point 4: disagree. You avoid the thermodynamic definition P = U/V at all cost. Area does not have vector nature, but the surface normal vector is a representative of the plane. If you want to use it, you can do it for planes at any point except the chosen reference point Q, because there is no notation for these planes. The use of planes in space in the Cauchy theory is incompatible with basic vector space properties, see <strong>Unorthodox Thoughts</strong>, chapter 8. </p>
<p>point 5: no contest, but irrelevant in the present context. Thermodynamic work is time-independent, and not an acceleration. </p>
<p>point 6: Sentence 1 & 2 agreed. Last sentence is foggy. </p>
<p>In my theory, bonds occur in two different ways:
</p>
<p>
<br />
(1) Bonded substances have an internal pressure. This pressure is internally balanced in the unloaded solid/fluid. Its nature is compatible with your sentences 6.1 & 2. See <strong>Grueneisen 1908</strong>, and farther back <strong>Clausius 1870</strong>. <br />
(2) The bonds between system and surrounding must be taken into account because without them the surrounding cannot do tensional work on the system. This has implications for the final deformation. Surface-bonding forces are constraint forces, i.e. they do not do work themselves, but without them the acting forces (here the tensional forces exerted by the surrounding) cannot do work. </p>
<p>Unorthodox Thoughts <a href="http://www.elastic-plastic.de/thoughts.pdf">http://www.elastic-plastic.de/thoughts.pdf</a><br />
Clausius <a href="http://www.elastic-plastic.de/clausius1870.pdf">http://www.elastic-plastic.de/clausius1870.pdf</a> <br />
Grueneisen <a href="http://www.elastic-plastic.de/grueneisen1908.pdf">http://www.elastic-plastic.de/grueneisen1908.pdf</a></p>
<p>
CM suffers from the heritage it carries around from Newtonian mechanics of discrete bodies. The 'part' or 'particle' is required, among other reasons, for the definition of the force. The limit operation in Cauchy's theory is called the continuity approach because it is thought to dissolve the continuously distributed mass into mass points, mass differentials, parts or particles which move past another. The concept is in part influenced by the recognition that all matter is discrete, but the two concepts do not harmonize. </p>
<p>In potential theory a very different view is offered. A mass point is something entirely different: a body of mass in freespace can be considered a mass point with the condition that its mass is conserved, but concentrated in one point. In effect, only the volume of the mass is ignored. One condition for this assumption to be valid is that the mass distribution in the system is scale-independent, in clear English it is constant. The concept is of use in astronomy, for example. It must be given up whenever the dimension of the mass can no longer be ignored, e.g. if the shape of the body matters. </p>
<p>This concept of the mass point cannot be carried into continuum physics, however, because the mass considered in a continuum is necessarily scale-dependent. The logical alternative is then to consider an unit mass of finite extent. </p>
<p>This is what is done in thermodynamics, and this is what I propose. It would mean to give up the concept of the 'part' or 'particle' entirely and treat a solid as a perfect continuum of mass. </p>
<p>I submit this to your consideration. </p>
<p>Falk
</p>
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</ul>Tue, 07 Apr 2009 12:41:04 +0000Falk H. Koenemanncomment 10413 at https://imechanica.orgWork-free isochoric deformation
https://imechanica.org/comment/10406#comment-10406
<a id="comment-10406"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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 Mr Koenemann,</p>
<p>I am fascinated by this exchange, but I too suspect that there are serious problems with your new theory. I will nevertheless try to keep an open mind and give you an opportunity to respond to queries. Since no one else has addressed your contention that continuum mechanics will always predict zero work for an isochoric elastic deformation, I will make an attempt to understand it. I have read your paper entitled "ON THE SYSTEMATICS OF ENERGETIC TERMS IN CONTINUUM MECHANICS, AND A NOTE ON GIBBS (1877)", in which this claim is detailed. You have asserted that you demonstrate three times in this paper that standard continuum mechanics erroneously evaluates the work necessary for a volume-neutral deformation, and hence I will comment on the three examples you provide.</p>
<p>1. You use the standard continuity equation, and make the statement (a) "that stretch displacements and contraction displacements sum to zero" and hence (b) work must be equal to zero. In a sense (a) is true: if one takes a bar of material with Poisson's ratio 0.5 and stretches it, volume is conserved because radial contractions occur in directions perpendicular to the stretching. In this case, using standard notation and stretching in the 11 direction, sigma_11 = sigma, while sigma_22 = sigma_33 = 0; the corresponding strains will be something like epsilon_11 = sigma_11 / E; epsilon_22 = epsilon_33 = - sigma_11 / 2E. Continuum mechanics would find that work has been performed because the extension is accompanied by a force in the same direction, while the radial contractions are work-free because no forces are associated with the direction of these displacements. Hence there is a net positive work, and your statement (b) that continuum mechanics would find zero net work is incorrect. Then you make the remarkable claim that "stretch work and contraction work have opposite sign". Typically, positive work is that which is done upon a system while negative work is done by a system (or vice versa, if that suits your tastes). If your statement were true I would be able to extract energy from a system merely by compressing it (or stretching it, depending upon your sign convention). </p>
<p>2. You note that for isochoric, isentropic deformations, continuum mechanics would say that the increment in internal energy dU is equal to the product of sigma_ij (the stress) and d epsilon_ij (the strain increment). You then state that there are two conditions for an isochoric deformation: sigma_ii = 0 and epsilon_ii = 0. The latter is true; the former is not (see the example in 1. above, which has an isochoric deformation). In any case, even if this were true, it would not make the product sigma_ij d epsilon_ij identical to zero. Take, for example, the tensors associated with the principal stresses sigma_i = [1 -1/2 -1/2] and the principal strains epsilon_i = [1 -1/2 -1/2]: the traces of both tensors would be zero but the product would not. Moreover, you state that Bachelor notes that "the deviatoric stress tensor [...] does not contribute energetically to the deformation". Bachelor, on page 142 in my edition (section 3.3), states that the deviatoric stress tensor is directly associated with the motion of a viscous fluid, which is a disspiative process requiring energy and hence work. This is the opposite to what you claim Bachelor wrote.</p>
<p>3. I don't have Landau & Lifshitz on my desk so I cannot adequately comment on point 3. There is also, I suspect, a typographical error in (10), which would further complicate discussion until clarified.</p>
<p>I also found curious your critiques of Love and Fung earlier in this paper. Are you objecting to the partitioning of the energy of an object into the kinetic energy of the object as a whole and the internal energy due to temperature, strain, etc? Or are you objecting to the integration of the kinetic energy of the differential elements composing the object to determine the kinetic energy of the entire object?</p>
<p>I appreciate your assistance with these issues. I am, however, increasingly convinced that the theory you propound is incorrect.</p>
<p>Regards,<br />
Craig</p>
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</ul>Mon, 06 Apr 2009 22:19:08 +0000Craig A Steevescomment 10406 at https://imechanica.orgRe: discrete system
https://imechanica.org/comment/10399#comment-10399
<a id="comment-10399"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
In my language, a discrete system consists of a finite number of particles (e.g., atoms, molecules, etc.) enclosed by a closed surface. The particles interact with each other (not necessarily bonded) and the system interacts with its environment across its surface (e.g., mass transport, heat conduction, energy flux). Please see my post on "<a href="http://www.imechanica.org/node/5214">a point and a particle</a> " for clarification.
</p>
<p>
RH
</p>
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</ul>Mon, 06 Apr 2009 17:44:55 +0000Rui Huangcomment 10399 at https://imechanica.orgdiscrete system, and others
https://imechanica.org/comment/10398#comment-10398
<a id="comment-10398"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5014">New theory of elasticity & deformation</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>
Rui,
</p>
<p>
(1) Mathematically, the equality between a surface integral and a volume integral can be established for an arbitrary vector field f. The mathematic conditions for the divergence theorem to hold may be further discussed, with specific physical problems in mind.
</p>
<p>
<br />
OK. - The point of importance is this: the validity of the divergence theorem is unlimited in freespace, it is certainly limited in distributed mass, and it is further limited – this is not in Kellogg, but it is my conviction – in bonded distributed mass.
</p>
<p>
(2) If the vector f is the external force acting on the surface of the system, the surface integral is not a force quantity. Its unit is [force(N)][length(m)]^2.
</p>
<p>
Yes. As far as I could find out, [Jm] is the unit in which Coulomb's law was originally perceived. The definition was changed later when the Ampere was chosen to be the most basic unit in electricity.
</p>
<p>
The explanation above by Mr. Koenemann as to how the unit of this integral remains in [N] does not make sense to me. One may convert an integral over a spherical surface area to an integral over spherical angles, but the conversion does not change the unit.
</p>
<p>
<br />
int f.n dA is in [Jm] because it is an integration over a surface term. If you integrate over the space angles, you need a scaling factor to indicate what theta means in real distances. That scaling factor is the magnitude of the radius in question. It is always involved (e.g. in fxr). The unit is in [J]. This is convenient because [J]-terms are scalars by definition, so there's no ambiguity.
</p>
<p>
(3) In vector form, summation of all external forces acting on a system equals zero at equilibrium, and this is the only correct way to sum up the forces.
</p>
<p>
f.r and |fxr| are [J] terms, and scalars which you can integrate around 2pi.
</p>
<p>
(4) With (2) and (3), I do not understand the physical meaning of the surface integral of f.n (normal component of external forces) in the divergence theorem.
</p>
<p>
I can only repeat what I wrote further up: I accepted Gauss' view that [Jm] is a meaningful unit. In the first definition of Coulomb's law this unit was clearly implied. (It has been changed since, see above.) But I preferred integrals which led to [N] or [J], and integrating over the angles satisfied me.
</p>
<p>
(5) It remains puzzling to me what f is in the system (V) and why the divergence of f has to be a constant everywhere in V.
</p>
<p>
Excluding external gradients, the constancy of div f within distributed mass is not my contention, but that of Gauss. But I do not wish to defer responsibility to others. Treating del.f as a constant made sense to me. There must be some term that scales a force to a length term of the system involved (e.g. a spring) to make a relation of force to work possible, see answer to Chad, April 3, 23.49h. – del.f is not a constant if V varies in scale, but mass does not. This is the case of a system with a finite body in freespace.
</p>
<p>
Within distributed mass, the constancy of del.f follows in my view from the fact that int del.f dV must be linearly proportional to the limits of integration, or else the result is not proportional to mass. The latter would bring us into a real mess. consider k an extensive parameter, and del.f an intensive parameter.
</p>
<p>
Mr. Koenemann’s answer, “f is either f_syst or f_surr”, seems to apply on the surface, but not in the volume. Say, at an arbitrary point in V, what is f_syst and what is f_surr? Do we need a surface passing that particular point to separate system from surrounding? If so, there are infinite choices for such a surface at each point, and which one should be used?
</p>
<p>
<br />
On the surface of the system, two forces interact, f_syst and f_surr. The question what happens inside is not the subject of thermodynamics; this theory only considers exchanges between system and surrounding. – I am not trying to dodge your question. If you find that unsatisfying, we can only change the scale of the system, i.e. chose systems at varying scale.
</p>
<p>
In principle we do not need the surface as a planar or curviplanar geometrical element. But the sum of all radius vectors emanating from Q, indicating all points of action of external forces f_surr, necessarily forms a surface.
</p>
<p>
(6) The purpose of the divergence theorem in Mr. Koenemann’s theory seems to show that f is proportional to r and thus f/A does not exist as r -> 0 (i.e., non-existence of the Cauchy stress). This would only make sense to me if I understand (5).
</p>
<p>
The purpose of the div-theorem argument is (a) to show that the Cauchy theory does not work, (b) to show that we must establish an unit distance between Q and P, the radius, which can then serve as a zero potential distance. The latter can be used to define work; e.g., if V_0 is changed to V_1 by a pressure increase, the same holds for the radius r_0 to r_1.
</p>
<p>
<br />
(7) In Mr. Koenemann’s paper (Approach), the force is also defined as the energy flux, f_I = dU/dx_I (partial differentiation). Such a definition for a force field seems to make sense in a discrete system (particles, atoms), but in a continuum the partial differentiation with respect to the coordinate of a particular point would inevitably lead to a singularity.
</p>
<p>
Surprised, explain. The particular point is in my case always the point Q, the center of mass of the system, in the ideal case removed from the surface by one unit distance. </p>
<p>Perhaps the confusion is in the expression 'discrete system'. I have seen this expression used with a different meaning. A discrete body is a body bounded by sharp surfaces against some other medium or the vacuum such that a surface can be wrapped around it that does not run through mass. But a discrete system need not be a discrete body; a discrete system is a system defined by a given mass the identity of which is fixed. This system can be separated from its surrounding (which may consist of the same substance) by a sharp surface. This condition is implied here. Thermodynamicists use other types of systems, e.g. in the case of diffusion: a system of water may be open to mass exchange provided the mass in the system remains constant (in the same way as heat is exchanged between system and surrounding at constant temperature). <br />
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Falk
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</ul>Mon, 06 Apr 2009 17:08:20 +0000Falk H. Koenemanncomment 10398 at https://imechanica.org