New theory of elasticity & deformation

Dear Ajit,

to other readers: this is a discussion that started on Ajit's blog . 

Regarding your questions: 

(2.1.) He says in his paper that "div f is a measure of the work done by/upon a system."

See here, p.52 in Kellogg (1929) Foundations of potential theory, Springer Verlag. I understand your reservations farther down, but I did not say, and Kellogg did not say that div f is the work term itself, just a measure of work. div f is the source density or charge density, and int div f dV is the charge (int = integral sign). 

Having said that, I think Kellogg's statement must be taken with caution. To make things simple, I prefer to restrict the discussion for the moment to isotropic conditions. I think that Kellogg's statement is not universally valid within bonded mass distributions, but this is not the right moment yet to discuss details. And do not take me as simple-minded as you appear to have taken me in the July 08 discussion (which I became aware of only in Feb 09 because no one bothered to contact me, including you). 

(3.0) In another paper: "General solutions for the Poisson equation exist only for reversible processes, e.g. the Helmholtz equation." [sic]

That's what I learned. Apart from that I have my own opinion. The thermodynamic equilibrium condition states that two sources of forces interact, the external and the internal one, such that P_syst + P_surr = 0. This implies in my view that both system and surrounding act as a sources of forces, and they serve to one another as sinks; hence there are two Poisson equations in one equilibrium equation, because this eqn can also be written div f_syst + div f_surr = 0 (for isotropic conditions, n.b.). They balance only if there are no inequality terms, i.e. Second-Law effects, a.k.a. entropy. Hence the eqn is reversible. That's elasticity. 

Very much apart from that, your distaste for my writing is misplaced, whether I am right or wrong. I have fought for 20 years for discussion, and have experienced very literally the most massive and long-lasting excommunication from the moment on I let it be known that I differ. The reasons never mattered, it has simply been impossible to discuss the foundations of CM. In my experience, CM is like a faith or a church. While I disagree with your assessment of my work in the discussion on Biswajit's blog, I gathered that you are a thoughtful man. That's why I placed my comment on your blog. 

(i) Why do you say that a characterization of mechanical response of solids from a thermodynamic perspective is absent in the traditional theory (i.e. all solid mechanics than what you say of it)? I can clearly see that such a description exists. What makes you say that it doesn't?

I cannot see the faintest trace of thermodynamic thinking in continuum mechanics, starting with the assumption that elasticity is linear (Hookean). Thermodynamic potentials are logarithmic. Thermodynamics distinguishes system and surrounding, Cauchy let the system vanish identically. I have never ever seen any mention of the thermodynamic equilibrium condition in CM. Just the idea to start a theory of elasticity with an equation of motion is un-thermodynamic. I could go on ranting, but it is all published . What I do see in the literature is a mix of conservative and nonconservative terms as if they belong together. They don't. They are wolds apart. 

I have shown - in Systematics & Gibbs , Chapter 2 - that the form of the First Law of thermodynamics as it is used in CM is invalid, and that as a consequence, the work done in a volume-neutral elastic deformation is always zero (Chapter 4). The path from E_kin to E_pot = const to dU = dw + dq is the virial law of Clausius who interpreted

E_kin + E_pot = const

as 

mv^2 + fr = PV

Note that only the first term LHS is accessible for an equation of motion, but Clausius interpreted this as the _heat_: it is the free vibration of the individual atoms. The second term LHS are the bonds in a solid. The term RHS is the total energy of the system in the standard state. In case of a change of state, evidently the RHS becomes a variable, and the LHS is no longer interesting because now we consider exchange of energy between system and surrounding. Thus I see no way to start an elastic deformation with an equation of motion under the rule of E_kin + E_pot = const. The key to elasticity is PV = nRT and dU = dq + dq. For the rest, see my theory.

(ii) What part of Cauchy's definition lacks thermodynamic implications? 

Cauchy's tetrahedron is effectively a thermodynamic system. But then he lets it vanish identically. There was no mention ever of the thermodynamic equilibrium condition in any of the many derivations of the Cauchy tensor I have seen; it was never considered. In potential theory it is stated that for logarithmic problems (such as thermodynamics) one must define a zero potential state by convention because they do not have a natural zero point. That means - as it is done in thermodynamics - that a system with unit mass and unit size is chosen (n and V are finite in PV = nRT). It cannot vanish. Its dimension can be taken as the zero potential distance d_0 of potential theory. It was Cauchy's error to let d_0 vanish; it is required for the definition of work. - In fact, this d_0 is in Hooke's law the length of the spring. Cauchy let it vanish. 

(iii) In your theory, do you believe that stress remains a 2nd rank tensor quantity? 

No. 

If not, what are the physical units or dimensions of stress according to your? 

I have written a force field theory (in [Newton], and not in [Pascal]). There are two fields: the external one that is controlled by the external boundary conditions; and the internal one which is subject to the material properties (compressibility, anisotropy, orientation thereof, etc.). The two fields are independent of one another (in origin, and in properties, but yes, they are controlled by field property tensors); but they are brought into equilibrium by the condition that system and surrounding are solidly bonded to one another, hence disequilibrium within a solid (below the yield point) is impossible. The two independent force fields result in a third one that combines their properties. To calculate the displacement the work function fdr is used in the same way as PdV is used in standard thermodynamics (f and r as in the virial law above). The details of that are explained in my paper, but be assured that no Joules vanish. 

Can you give component-wise derivation as to how such a quantity is defined after starting with the thermodynamic laws?

Very simple:  f = dU/dx. That's the only definition of a force field known to me. The fact that conventional CM does _not_ use this definition is enough evidence to show that conventional CM is not a proper field theory. (in fact, it uses f = ma exclusively, which is by nature _not_ a field force.)

And, why should anyone begin using such a "stress"?

I do not use the term "stress", simply because everybody would instantly take it as an orthogonal tensor in the sense of Cauchy. My theory is a vector field theory. I use the term "properties of the loaded state". Its eigendirections may be non-orthogonal. The reason for using it is: the rationale of Cauchy's tensor is simply invalid, the term does not exist; if you disagree I expect you to refute my argument in Systematics & Gibbs , Chapter 5. Or better, everybody is invited. Hint: no one, from my first discussion partners in 1986 to the last one two weeks ago , could do it. The argument is of a quality that cannot be refuted. It is time to acknowledge this. Thus, whatever you do, you are in need of a new theory. 

Apart from that, I suggest you read my Approach paper. There are enough demonstrations that my theory delivers excellent results. - Of course, satisfactory predictions do not prove that a theory is right. But I offer solutions to problems that have been considered intractable using the conventional theory, - such as turbulence in viscous flow - so it might be worthwhile looking at it.

Falk H. Koenemann

The standard understanding of simple shear is strongly based on the Helmholtz partition: "a simple shear can be decomposed into a strain and an external rotation". Sorry, no – neither geometrically, nor energetically. Nor would Helmholtz agree to the way the partition is used today.

The partition works only once. It cannot be used in an iterative way. If a strain step is chosen, and the required rotation is found, ok that's a simple shear. But if the new points are fed back into the same equations it does not work any more. If one plays with other points iteratively, it becomes clear that the points follow a transformation matrix in the sense of Ax = b. It is possible to find A from the chosen strain and rotation, hence it is possible to find its eigenvectors. The field is incompatible with simple shear.

The partition works only if one starts all over again – again and again and again. This is in incompatible with any attempt to develop a physical relation between energy spent and effect achieved.

Furthermore, if other strain steps are chosen, the eigenvectors vary. To anyone who is familiar with eigenvector problems this is an absolute No-No; one cannot manipulate the eigendirections as one pleases – except that it is still impossible to get the eigendirections for simple shear. Eigendirections are non-rotating directions, they are highly characteristic for the field to which they belong.

The eigendirections are non-orthogonal. The smaller the strain step, the more orthogonal the eigendirections become – and the closer they are to the eigendirections of the initial strain step. In the limit the non-orthogonality should vanish – great for those who have
faith in this concept, impossible from a physical point of view. A rotating eigendirection – that's like a roasted icicle.

The partition cannot be expressed in analytic form. The partition of the influence of strain and rotation is non-unique. It is therefore not possible to write a differential approach for simple shear that can be integrated to get the finite deformation. The  concept is
entirely phenomenological.

The partition is energetically wrong. A rigid-body rotation should not cost extra energy, hence all identical strain states, achieved through pure or simple shear, should cost the same amount of Joules. From the rubber experiments done in the mid-20th C it is perfectly clear that this is not the case – simple shear costs more energy than pure shear.

Strong energetic differences are also observed in the plastic field, but there simple shear costs substantially less than pure shear.

This finding has the most profound significance: strain cannot be a thermodynamic state function.

(Elastic field: the data by Rivlin & Treloar are hard to evaluate because they are not given in raw form, but worked through a "stored energy function" of which it is not clear what it really means. But the fact that simple shear costs more energy is clearly apparent. Plastic field: thesis Franssen, Utrecht 1993.)

Helmholtz wrote the paper explicitly for an ideal Newtonian fluid – friction-free, conservative in the sense of E_kin + E_pot = const. Whether this is an acceptable thought today is not the question here  (viscous flow is irreversible); but carrying this partition out of blind faith into the mechanics of solids is preposterous. Helmholtz never thought of a solid with strong internal bonds.

Apart from that: would please anyone explain to me how a free rotation is supposed to come about in a continuously bonded solid? Isn't this so ridiculously out of the question that it is not worth thinking twice about this partition? A free rotation requires free surfaces around the volume to be rotated. Why is such nonsense being taught to this day?

All these inconsistencies are a clear sign that elastic deformation as a physical process is simply not understood. In Approach Chapter 7 I give more reasons why the current ideas about simple shear cannot be right, and I develop a theory which delivers far more realistic results.

To those who can read German: Helmholtz' original article can be downloaded here , the ref is:

Helmholtz H (1858) Über die Integrale der hydrodynamischen Gleichungen. Crelles Journal/Journal für die reine und angewandte Mathematik 55, 25-55

Falk H. Koenemann

Chad,

I did not make a mistake, you can't read. The quoted paper is entirely correct.

The data fully support my predictions, both in the elastic and in the plastic field: download the experimental data to compare for yourself.

Alright. Please proceed to shoot down my refutation of the Cauchy stress tensor, including a reason why in your opinion Kellogg's Lemma 1 does not apply. Or do you need any help to remember the other points you dodged? I'd be happy to assist you.

Cheers,
Falk

Yes, k has unit C^2. The Coulomb is kg^(1/2) m^(3/2) / sec, and not what you think. k is the charge, see e.g. the Feynman lectures. I hope this furthers your enlightenment.

I don't know why you keep asking questions that have nothing to do with the subject here. I explained on March 12 to Louie – who had a decent way of asking questions – that f ~ A on free planes (dividing the universe into left & right) and on closed planes (around a volume V) if Delta A approaches zero while the dimension of V is not itself subject to the limit operation. But in Cauchy's limit operation V approaches zero.

F.

Mike,

I am perfectly aware of the long history of Cauchy stress, and of the long tradition of design. And to say this right away, I do not think that all these engineers are brain-amputated. They are very careful and conscious of the fact that most of their work has safety implications, and before they trust a prediction they do lots of testing to be sure it works. On the other hand, I have observed a most profound and massive distrust in theory especially among engineers, to the extent that they are not aware of it any more, but it exists nonetheless. I live in Aachen with its large technical university. People never admit they don't trust theory - in public. In private, a large number of profs told me they don't care about theory at all, "We don't work with thermodynamics, we run our projects" is a very recent quip. A Rektor of this RWTH told me once "Nobody wants new theories, this would create far too much unrest in the universities." A mechanics prof here sent my work to a colleague to evaluate it. When the answer was straightforwardly positive and endorsing, the prof here ceased to communicate with me. That was 15 years ago. The colleague helped me get a paper in print.

I am not at all worried about people being sceptic, that is their natural right. But I find it disconcerting if you or others ask for examples of errors - I have done that, I have shown in three different ways that the conventional theory always predicts that the work done in an isochoric elastic deformation is zero. At the same time, I do consider my published work to be part of this discussion, I don't have to repeat it here. The predictions for simple shear cannot be better.

I am not concerned about the limit of Cauchy stress, but about the fact that the Cauchy theory does not distinguish between system and surrounding, the most basic theoretical concept of thermodynamics. Also, I consider it incompatible to approach a change of state with an equation of motion. A theory of deformation of solids that does not take the existence of bonds in solids into account is nonsense, excuse me, no matter how well it works. I cannot agree with the way the First Law of thermodynamics has been turned upside down in CM. And I cannot live with the fact that it was so far simply impossible to discuss these matters. As said someplace above: potential theory says that for logarithmic problems in distributed mass one needs to take a unit volume in some standard state (to be defined by convention), and work is done if the state of that unit volume is changed (at constant mass). This is precisely as it is done in thermodynamics. I see no way to reconcile Cauchy with the basics of thermodynamics. Cauchy let the system vanish identically.

I have found out why the Cauchy theory works to some degree, but at the same time I have found why I ran into such massive inconsistencies between prediction and observation in the first place. The explanation may be a bit out of context for you, but anyway:

when Euler reduced Newton's fxr to fxn, he effectively reduced the shape of the system to being spherical. This is an unrecognized boundary condition. In my theory, the shape of the system is a function of the material properties and the equilibrium conditions. The conventional theory is not completely out of whack with reality as long as this unrecognized boundary condition is in fact observed by the physical setup, namely that the system has spherical shape; if not, the disparity between prediction and observation is massive and profound. For simple shear the system must have elliptic shape, a sphere does not work.

Another reason why the error is not obvious in most cases is this: in the elastic field, Nature prefers pure shear deformation for energetic reasons. This is - independent of one another - my repeated observation in the field, and my prediction. But for pure shear (or higher symmetric situations) the shape of the system is indeed a sphere . The flipside is: simple shear is by far the most common deformation type in the plastic field (observation). I found out that it is strongly energetically favored (prediction). It cannot be understood by conventional means. The list of phenomena which I think to have given an explanation for, relating to simple shear, is quite long. - To be sure, it is at this point still a matter of contention if my theory is acceptable. A student has written a thesis on it, and enough people have checked it, so it is a serious proposition. The rest is up to public discussion. But this we need badly.

Falk

I have not seen hard arguments so far from anyone, disregarding Chad's various attempts to willfully misunderstand the divergence theorem. By hard argument I mean either mathematical or physical logic. So far, it has survived. The best contributions, by tone, attitude and quality, came from Rui.

I have far more reasons to question the Euler-Cauchy theory than just the divergence theorem argument. No one addressed them.

No one commented on the distinction "conservative – non-conservative", the most profound distinction in classical physics. If you do not understand the difference between E_kin + E_pot = const and dU = dw + dq, you cannot claim to know much about physics altogether.

No one objected to the zero work predictions of the Euler-Cauchy theory. This is simply ignoring straightforward proof.

No one seriously replied to my claim that bonds in solids are not contained in the theory. Instead, once again I am given reasons why bonds are not important. This is voodoo science.

No one challenged my claim that the foundations of CM are clinically free of any thought compatible with thermodynamics. Quite the opposite, Louie's references have confirmed this.

And, as far as I can tell, none of those who contributed, bothered to read my papers. What do you think they are there for? Louie asked for the equation of state (EOS) that works for for gas and solids. I gave an illustration for the EOS I propose – it is possible to show that one and the same EOS can be used for different materials. So, Louie, if you withdraw, it is not because I did not answer your question. It is because you feel better in the crowd. Or do you have any objections to my discussion regarding the application and validity of the divergence theorem? Let me guess: you don't know enough, but you are not going to admit it.

Never mind. The gun is loose now. Among 7000 reads there must be a few thoughtful readers, and that's enough. 

Falk H. Koenemann
http:/www.elastic-plastic.de

Rui, 

(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.

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.

(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.

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.

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.

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.

(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.

f.r and |fxr| are [J] terms, and scalars which you can integrate around 2pi. 

(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.

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.

(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.

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.

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.

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?

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.

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.

(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).

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.

 (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.

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.

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).
 
Falk

Craig,

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.

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.

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.

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

http://www.elastic-plastic.de/slide25.jpg

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.

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.

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.

This example shows a bunch of things:

-    Inevitably, there is a strain gradient in loaded bodies with free surfaces that fades towards the interior.

-    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.)

-    It is impossible to ignore bonds. The bonding forces are constraint forces (see my answer to Rui Huang on April 7).

Altogether, the description you gave above – [sigma  0  0 ] – becomes too crude. And what does Poisson's ratio mean here?

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.

Falk

Craig,

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.

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.

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.

Falk

Craig,

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.

Assume a situation in which stress s_11 = a (a is a free numerical, not an area),

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

then s_ij de_ij = s [a   -a/2   -a/2] [de_11   de_22   de_33]

= as de_11  - (as/2) de_22 - (as/2) de_33

= as (de_11 – 1/2 de_22 – 1/2 de_33/2)

= 0

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.

Falk

I shall be absent for a day or two.

Craig,

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.

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.

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.

Thank you,

Falk

Craig,

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.

Since I had just had nearly the same discussion with a private correspondent, I repeat the entire argument here:

Tensor products do not communicate, AB ≠ BA

but the trace of a tensor is an invariant, and tr AB = tr BA communicates.

Also, tr A . tr B = tr B . tr A

but tr A . tr B =/= tr AB.

Let the off-diagonal terms be zero to make it simpler, then
tr AB = a_11 b_11 + a_22 b_22 + a_33 b_33

but tr A . tr B = (a_11 + a_22 + a_33) (b_11 + b_22 + b_33)

If A = B such that a_11 =/= a_22 =/= a_33 but tr A = 0, (at least one term must be negative)

tr AB = tr AA = (a_11)^2 + (a_22)^2 + (a_33)^2 > 0 (it cannot contain negative terms)

whereas tr A . tr A = 0 . 0 = 0

So, the question is, do I pick my results as they please me? No:

Let s_ij = [1  -1/2  -1/2]

you can write s de as

s [1  -1/2  -1/2] de [1  1  1]

Why so?

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

tr se = (s_11)^2 + (s_22)^2 + (s_33)^2

which is sign-insensitive and physically meaningless, it no longer reflects the properties of s itself;

but tr s tr e = 0.

Regarding my own theory, you haven't even looked at it yet. All of the above is in terms of the conventional theory.

Greetings,
Falk

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.

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.

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.

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.

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.

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.

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.

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
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:

(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.

(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?"

(3) My claim that Clausius' virial law must be taken into consideration, has not found any response.

Falk H. Koenemann

Rui,

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.

My critique of Poisson's ratio is this:

(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.

(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. )

(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.

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.

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?

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.

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.

Falk

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.

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.

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:

1. There is the field of conservative 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.

2. If a process is non-conservative 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.

Two different types of non-conservative processes were soon distinguished:

2a. The process may be reversible.

2b. The process may be irreversible, and entropy is produced.

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.

If a process is non-conservative, it must start with an equation of state.

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 virial law,

E_kin + E_pot = PV

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

PV = nRT,

and the energy conservation law for non-conservative physics, i.e. the physics of changes of the energetic state, is, of course,

dU = dw + dq.

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,

dw = d(E_pot – E_kin)

which makes it clear that this work is acceleration work, but not change-of-state work.

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.

The form of the "First Law" given in the textbooks is

int (d E_kin + dU) dV = dw + dq

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.

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.

Falk H. Koenemann

With one recent correspondent I had the following exchange:

Just answer: f is a flux.

(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?

(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?

(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?

To which he replied:

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.

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.

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 […].

I wrote:

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".

Or do you see an alternative? What does the condition mean physically in your view?

Why is the statement correct for a gas, but not correct for a solid, when the external boundary conditions are the same?
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.

Since then he does not answer any more. That's just the easy way out.

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.

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.

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.

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.

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. There are no sanctuaries for endangered theories.

Falk H. Koenemann

Craig,

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.

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).

Falk