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# JAcobian interpretation

Thu, 2011-08-04 05:38 - kajalschopra

I was just reading about volumetric locking in Dhondt's book (for Thermomechanical analysis) and came across this (funny) question;

Now, we say the physical interpreation about Jacobian is change in volume, now if a material is incompressible Jacobian = 1, now this incompressibility is a material property, right?

Where is a material parameter attached to Jacobian expression of finite elements (1-d,2-d or 3d finite element)?

Forums:

## If I'm correct, the jacobian

If I'm correct, the jacobian gives the change in integration volume due to a change in integration coordinate system. So it does not have anything to do with incompressibility.

## Jacobian and incompressibility

Hi,

I am not completely clear on the question. So I will just just mention two points for thought.

1. Jacobian is the ratio of final volume to initial volume in a local sense. If the material is incompressible, all the motion (or deformation) the body undergoes should have J=1. In case of compressible materials, J>0; J=0 or negative are NOT acceptable, and J becoming infinity is just meaningless. So this case obvously includes motions with J=1.

2. In case of a linear elastic material (St. Venant-Kirdhhoff material), a specific relation between Lambda and mu (or E and nu) will give incompressibility. Or, in other words, we will have only one of them to be independent in this case. Similar possibilities occur in other material models, and of course, there are material models with built-in incompressibility (or parameters to make sure that compressibility is very small).

Jayadeep

## this may be the reason

There are two things when we talk about Jacobian:

(1) In first case, what Jacobian says is "the mathematical outcome of the deformation happening in the structure and is the ratio of the final material volume to the initial material volume."

Now the material model enforces the deformation to happen in such a way that the Jacobian always remains one. This can be thought of as lets apply some force to a bar made of linear elastic material (lets make it incompressible, eventhough being a wrong one just for an example) , so the material constants will constain the deformation is such a fashion that the deformed volume to the initial volume will always remain one.

(2) And in the second case, If you are talking about an isoparametric element formulation in a finite element modeling then here the jacobian in turn represents the volume ratio of the master element to that of the actual element. Here it does not have anything to do with the deformation happening or the structural material involved, it is just a mathematical representation in the isoparametric formulation and is a representation of the element distortion happening due to this isoparametric idealization.

Hope this may help, if I m not wrong.

## Thanks a lot

Thanks a lot all.

Chandrasekhar- you mentioned that in the first case jacobian is the ratio of the final material volume to initial material volume- where do we have the jacobian determinant in the picture?

## Re: the Jacobian matrix and determinant

Dear Kajal,

Check out Prof. Alan Bower's online book: http://solidmechanics.org/text/Chapter2_1/Chapter2_1.htm#Sect2_1_4 . Notice that in section 2.1.4 he talks of the Jacobian of the _deformation_ gradient, not _displacement_ gradient, and hence he has the additional term of the Kronecker delta.

That's for definitions. Now, returning to your initial queries.

1. I am not sure if incompressibility is, strictly speaking, a material property.

2. Where do material properties get attached to the Jacobian?

Well, first, let's talk in terms of basic continuum mechanics. The considerations carry over to the special case of FEM as well.

First of all, note the general structure of the theory of continnum mechanics. For simplicity and relevance, we consider the static (or time-independent) solid mechanics. The structure is:

displacements <=1=> strains <=2=> stresses <=3=> tractions i.e. forces and moments, where

-- relation 1 denotes the strain-displacment relations, and roughly correspond to (or involve) the compatibility pillar in Shames' description of the three pillars of solid mechanics (see his book on SM).

-- relation 2 denotes the strain-stress relations (e.g. Hooke's law), and correspond to the constitutive equations pillar in Shames' scheme

-- relation 3 denotes the stress-force relations, and roughly correspond (or involve) the equilibrium equations pillar,

and where you can traverse either from left to right or from right to left, even though all the four fields must be consistent to each other.

Let me mention an aside which is important, though not directly relevant to your query: Since the strain tensor represents only a part of the displacement gradient tensor (the other part being the rotation tensor), going from the left to the right automatically satisfies compatibility. Thus, you can always start with an arbitrary displacement field, and then you will always get a set of traction vectors (say forces) that uniquely corresponds with the given displacement field. However, the reverse is not true. For going from the right to the left, compatibility is not automatically satisfied. You cannot impose an arbitrary force field on a solid and expect to get a displacement field that uniquely corresponds to the given force field---you have to additionally also ensure compatibility. The aside over.

The general problem of SM involves boundary conditions involving both displacements and tractions. For the solid as a whole, in physical reality, all the four fields are determined together.

Let's take an example. For simplicity, assume that the displacement field is known. In such a case, none of the remaining three fields is arbitrary. In particular, not only the strain and stress fields, but even the tractions field gets uniquely determined thereby. In other words, you can physically sustain the specified (i.e. the known) displacement field if and only if the implied tractions field also is simultaneously applied to the solid. Not otherwise.

Now, assume that the tractions field is known. There now is an infinite set of displacement fields corresponding to it, but only one that satisfies compatibility for the given solid. Now, this particular displacement field must be allowed to occur physically. Though not directly specified as a boundary condition, it is implied. Thus, again, all the four fields exist simultaneously.

The mechanical state of the solid, thus, always corresponds to a unique combination of all the four fields. A state of deformation can only be physically realized if all the four fields are consistent with each other.

Now, returning to your query.

The nature of the constitutive relations does not alter the above, more fundamental, fact. If you alter the constitutive relation, then its effect propagate through all the four fields in such a way that they become consistent with each other.

For example, take a solid, say a compressible one (loosely speaking :)). Apply a known displacement field to it. As discussed above, a certain unique tractions field is implied thereby. Here, you have to imagine that some agent is actually effecting the implied traction field, for the solid to remain in the given displacement field. That's what the consistency requirement means. Suppose that this traction field is such that at a boundary point P, the traction field comes out to be a force of 10 N acting in +ve x-direction. Next, take an elemental cube somewhere within that solid. You will find that during the deformation, its volume has shrunk. You can use the Jacobian to determine by exactly how much, at that location.

Now, take another, incompressible material. Keep the original geometry of the solid and the applied displacement field the same. In this case, you will find that the traction at point P is not 10 N in the +ve x-direction but something else, say 12 N in +ve x-direction. The traction field corresponding to the same displacement has changed. What changed it? The change of the material. A difference in material implies a difference in that force field which must simultaneously exist in order to physically realize the given displacement field. Now, take the same elemental cube (at the same location). You will find that its volume has not changed. This difference is attributable to the differences in _all_ the three unspecified (but simultaneously existing and unique) fields: strain, stresses, tractions.

The values of the Jacobian in the two cases are different. But they are so only because _all_ of the remaining three fields are different.

So, where precisely do the material properties get attached to the Jacobian? Answer: Only through the constitutive relations, no other place. However, they still are capable of affecting the compressibility property of the associated displacement field because, in general, any change in any one of the four fields must simultaneously affect all of the remaining three fields.

--Ajit

PS: iMechanica was not accessible from Pune, India, for a few days and commenting was not possible for a few more days. Hence the delay in the reply.

PPS: Now that the matter has already been discussed at such a length, any one could come and state it in short-n-sweet words, to let, e.g., the Tweeter founders, feel very very proud. Any one could do that. I promise!

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## My mistake concerning incompressibility and the Jacobian

Hi all,

0. It's somewhat surprising that none has caught a mistake I made in my reply to Kajal above. I noticed it yesterday.

1. The mistake is this. An incompressible deformation cannot be specified

completelyarbitrarily. It must always satisfy the condition that the divergence of the displacement field be zero. (This condition is the same as saying that the mean or spherical strain tensor be zero.)2. The effect of the mistake first begins to show up in my above reply in the paragraph starting with: "For example, take

a solid, say a compressible one (loosely speaking :))...." The mistake then continues through the rest of it.

In my above reply, I take an

arbitrarydisplacement field---first for the compressible case, and then for the incompressible case. The two cases can be madedirectlycomparable only if the displacement field in both cases is taken to be solenoidal (i.e. divergence-free).With that proviso, the rest of the description of the

comparisonis OK (I guess).The displacement field for the compressible case has no constraint by itself---it could very well be non-solenoidal. However, for the two cases to be made directly comparable, we are better off assuming the same, solenoidal, field for both of them.

3. The same mistake also leads to the following somewhat misleading statement. I said: "So, where precisely do the material properties get attached to the Jacobian? Answer: Only through the constitutive relations, no other place."

Well, that seems to be true, but not complete in terms of details.

It indeed is true that material properties get attached to the Jacobian only via the constitutive relations. However, the unstated part is this: they also are such as to constrain the displacement field to be a solenoidal field. After all, the Jacobian refers to the first-order differentials of the displacements, and a constraint that it be exactly 1, does imply that there are some constraints to the displacement field as well.

Going from the left to the right. A displacement field constrained to obey the zero divergence condition implies a zero volumetric strain, and thus, all strain is deviatoric. The corresponding stress and traction fields also reflect the incompressibility constrain, and thus form only a subset of all possible stress and traction fields.

Going from the right to the left is more troublesome (at least to me). The traction field, and therefore the stress field, may be specified in a completely arbitrary manner even for an incompressible material. (I think we can say that. But, do correct me if I am wrong.) However, the constitutive relations for the incompressible material are such that only a deviatoric strain field is to be taken as resulting from them (apart from applyin the constraints due to compatibility considerations). The solenoidal nature of the displacement field is then directly implied.

I am not happy with the above because the two cases seem unsymmetric. Am I making any mistake anywhere? That leads me to the last point.

3. The better I seem to understand QM, the dumber I seem to be getting in SM (and everything else). My intellectual capacities thus seem to be incompressible in nature---a stretch in one direction is exactly compensated for by a squeeze in the other directions. ...

So, it's easily possible that I still don't understand this issue the way it should be. (For that matter, I have never studied nonlinear/finite strain/large deformation theory much---never had a course, and my reading through self-studies has been rather cursory/superficial.)

So, I will appreciate if someone could go through this issue and point out mistakes. (Even minor ones---they help solidify understanding.) Thanks in advance.

--Ajit

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## Hydrostatic stress is indeterminate in incompressible material

Ajit,

I don't think I am fully qualified to answer your question. Still I would express my understanding on the matter.

As far as I understand, what you are getting into in item 2 above is the fact that the hydrostatic component of stress is indeterminate (from constitutive relations) in an incompressible material. So, whether you start from the displacement and then compute strain and stress from it or start from stress and compute strain and displacement from it, the hydrostatic (spherical) part of the stress tensor can be specified arbitrarily, or it is indeterminate from constitutive relations. Of course, the other specifications in the problem should allow us to compute it (possibly the specified tractions). So, in my opinion, there in no loss of symmetry...

Regards,

Jayadeep

## Reply to Jayadeep. One more clarification! Still open questions

Hi Jayadeep,

0. Thanks for joining the discussion. This incompressibility animal

iscrazy! Though it hasn't exactly driven me nuts yet, it's getting pretty close. I am shaking my head in disbelief for a third day in row! ... So, any help, any second thought, is only welcome. And, then, you do point out some very relevant things.1. OK. A clarification (and a correction to my very first reply above).

In the context of making the intended comparisons: if the displacement field is the same, then obviously the Jacobian will be the same, and therefore volume change will also remain the same---

regardlessof the nature of the constitutive relations.So, indeed, we should talk in terms of (in)compressible

deformationrather than (in)compressiblematerial.And, therefore, the (in)compressibility issue should be kept separate from that of the material properties like E & nu and the singularities associated with nu = 0.5. The two are separate issues.

All materials, regardless of their nu values, will show

incompressible deformationprovided that the displacement field is solenoidal.Therefore, for identical displacements, the only differences in the compared solids would be in their stress and traction fields.

2. As you point out, and also quoting from Bower's book [^]: "Because you can apply any pressure to an incompressible solid without changing its shape, the stress cannot be uniquely determined from the strains." And, further: "The hydrostatic stress p is an unknown variable, which must be calculated by solving the boundary value problem."

For a solenoidal displacement field (i.e. for incompressible deformation), the strain is purely deviatoric, which can be used to uniquely determine the deviatoric stress. (Right?) The pressure is to be separately determined from the tractions.

3. For the comparison of the compressible vs. incompressible

deformation, it now seems to me, we don't have two options. We can begin only from the tractions side. (For simplicity, I will not consider the mixed case of some tractionsandsome displacements being specified as constraints.) If we begin from displacements side, they both will be either compressible or incompressible, and so, comparisons in terms of volume changes won't be possible; only the differences in stresses/tractions will show up.So, to see a change of volume with a same initally applied field, let's assume that we begin from the tractions side.

Now, tractions always uniquely determine stresses. (So, it could just be a stress boundary-condition problem as well.) The trouble of non-unique determinations comes up only for the constitutive relations, and then for the compatibility relations.

Here, the question is: for an

incompressibledeformation, how would the constitutive relations "know" that they have to output only a deviatoric strain (i.e. a zero hydrostatic strain) even if the input stress is arbitrary (i.e. it in general carries also nonzero pressure)? I have no idea!! Using Hooke's law, a non-zero hydrostatic stress would necessarily give a non-zero hydrostatic strain.So, is it the case that for incompressible deformation, even the traction and stress fields cannot be specified arbitrarily? Now, it does seem so to me! But I will leave it open for further comments/discussions.

4. For the comparison of a material with nu = 0.5 (for convenience, let's call it rubber) vs. that with nu = 0.3 ("metal").

Is it possible that the metal can be deformed (at least theoretically) to show up compressible deformation as well as incompressible deformation? Seems possible---just tweek around the applied displacement field (and the implied traction field) until you have a solenoidal displacement field.

Is it possible that the rubber can be deformed in both compressible and incompressible modes? ... Hmm... I am thinking!!

5. For convenience, let me collect together the

questionable assumptions and open questions:5.1Will the deviatoric strain always uniquely determine the deviatoric stress, for all materials and constitutive models (esp. nu = 0.5 or not), and under all deformation models (esp. compressible vs. incompressible)?5.2For anincompressible deformation, for atraction/stress BV problem, is it true that thetraction or stress fields cannot be specified arbitrarily?5.3Can a material with nu other than 0.5, say ametal, be made to show? Can a material with nu = 0.5, say aincompressibledeformationrubber, be made to showcompressibledeformation?--Ajit

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## Volume preserving deformation vs Incompressibility

Hi Ajith,

I think we need to differentiate between "Volume preserving deformation"

and "Incompressibility". Merriam-Webster (.com) defines incompressible

to be "incapable of or resistant to compression",

and hence it must be a material property (obviously nu=0.5 also says the

same; though only for the case of infitesimal strain). The deformation

could be volume-preserving or not; but it is misleading to call it

"incompressible". compressible materials can undergo both

volume-preserving and non-volume preserving deformations, and the stress

tensor is uniquely determined by the displacement field in all cases.

Hypothetical incompressible materials can undergo only volume-preserving

deformations, and the displacment field can not determine the

hydrostatic part of stress tensor. Please refer to an earlier discussion on imechanica, to which you also contributed...

From an energy perspective, deviatoric parts of stress & strain

(assuming a choice of proper measures for both) and spherical part of

stress & strain are energy conjugates. For compressible materials

both parts could be present for both stress and strain, thus each

uniquely determines the other. For incompressible materials, spherical

part of strain is never present, and hence there is no energy

contribution from the hydrostatic stress, which makes it indeterminate

from the strain. On the other hand, the material model will be so as to

make the strain energy infinite for any non-zero spherical strain (bulk

modulus = infinity), and hence the strain tensor cannot have a non-zero

spherical part, whatever be the spherical part of the stress tensor.

Let me attempt to answer you questions in section 5:

5.1 I believe that deviatoric strain should always be capable of

determining the deviatoric stress uniquely, unless we can come up with a

material which can undergo only volumetric deformations (no shape

changes)...?!! I think such a material would be really interesting.

Other way around, we have enough examples, where deviatoric stress

cannot uniquely determine deviatoric strains... all the fluids for

example.

5.2 We can specify tractions arbitrarily -- stress field should be an

outcome -- whatever be the material. If the material is truly

incompressible, there will not be any volumetric deformation.

5.3 I assume you mean "volume-preserving" by the saying "incompressible

deformation" here. As mentioned above, compressible materials can show

both volume-preserving as well as non-volume-preserving deformations,

while all the deformation undergone by an incompressible material should

be volume-preserving.

Regards,

Jayadeep

P.S.: I hope I am not making any mistake by interchangeably using the

words "spherical" and "hydrostatic" in connection with stresses and

strains.

## Reply to Jayadeep: volume- preserving vs. incompressibility etc.

Hi Jayadeep,

0. I have to rush for an upcoming meeting, but would like to squeeze in a few points right away. I will come back tomorrow once again.

1. I, too, have used the terms: mean, hydrostatic, spherical, etc., interchangeably. Don't at all know if the terms: spherical and hydrostatic, carry over to more general definitions of stresses and strains. Hope someone could clarify.

2. You say that the mean and the deviatoric parts are "energy conjugate." I am not sure what you mean by that. In the usual usage, pairs of conjugate variables have different dimensions/physical units. For example, for action, momentum and position is a conjugate pair, and so is the pair of energy and time. For energy, force and distance are conjugate and so are mass and velocity-squared.

However, here, both the variables are of the same kind. The devioatoric and mean quantities are either both stresses, or both strains, or both strain energy densities. They are additive, not multiplicative. So, perhaps you meant independent? (Complementary quantities are again of a different kind, in SM. Energy and complementary energy are the different areas under the stress-strain curve. So, we have to look for another word.)

For the simplest linear elasticity, of course, energy-wise, they are independent; deviatoric stress produces only deviatoric strain and vice versa; the energy does not move from deviatoric mode to hydrostatic mode. The question is, do they remain that way also in the more general descriptions? Does a deviatoric stress always produce only deviatoric strain and no mean strain, and vice versa, also in the more general descriptions? Can it be rigorously proved? I have no idea. ... As to fluids, I will have to think a bit about it from this angle, say. e.g., for the Navier Stokes equations. (Never did it, so far!) Later... (There was a paper from someone at IISc a few years ago that derived equations of SM from FM.) But, no, my main point wasn't about indeterminancy or multiplicity---whether deviatoric stress produces unique deviatoric strain isn't the main point. It is: whether deviatoric stress produces mean strain or not, and whether pressure produces deviatoric strain or not, in general. Thus, it was about independence. (Sorry for the repetition of the same point. Am in too much hurry to edit it.)

But there, in fluid mechanics, they use the word "incompressible" very routinely, not "volume-preserving," don't they?

More, later.

Thanks, and best,

--Ajit

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## On energy conjugates

Hi Ajit,

On item 2 in your comment above: I am sorry that I was not sufficiently clear about what I meant while using the term energy conjugate. What I meant to say was that the deviatoric part of stress is energy conjugate to deviatoric strain and similarly the spherical components, which is strictly true in case of isotropic materials, undergoing small strains. (I had a feeling that such a confusion could arise, but I didn't want to spend too much time to make my statements very precise, devoid of all chances of misunderstanding. May be I need to improve my writing...!!!)

The hydrostatic stress is sure to cause deviatoric components of strains in anisotropic materials, but I think we are not much concerned about that issue in the present discussion.

I also would like to hear the opinions of others on imechanica (or need to read/think/work more) about these concepts in a more general setting (finite strains, anisotropy, plastic deformations etc.).

Regards,

Jayadeep

## Reply to Jayadeep

Hi Jayadeep,

Cool. Thanks for clarifying the matter with the example of anisotropic materials.

Yes, I also would like to hear from other iMechanicians, as to how all these concepts/principles/ideas pan out in more general settings. ... With the emphasis of linear elasticity in all practical design methods (and courses), these often are neglected topics. That's why, I think, we all (i.e. those trained in engg. and without a direct exposure to advanced courses/research experience in these areas) can certainly allow ourselves a bit of fumbling here.

Best,

--Ajit

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## Re: Incompressibility etc. Knocking off (?) the second question

Hi all,

1. I think I can knock off the second question, i.e. question 5.2 above.

The crucial thing is this: Tractions uniquely determine stresses

onlyin the infinitesimal neighbourhood of the point(s) of their application---but, in general, not at any other points of the contiuum.2. Thus, we have to introduce some further distinctions: the internal vs. the external, the applied field vs. the developed field.

The tensor fields (stresses and strains) exist only internal to the continuum. The vector fields (displacements and tractions) may exist as either internal or externally applied fields.

Displacements can be specified at points. Tractions require at least differential surface elements, which, for calculation purposes, may be imaginary.

If a field is not specified everywhere in a region, then we have to distinguish the developed field from the applied field.

Any asymmetry, if at all existing, must arise due to differences like these.

3. Consider an identical and arbitrary traction field being applied to the external surfaces of two solids of identical geometry but different materials: rubber, and metal.

The "applied" stresses---i.e. the stresses in the infinitesimal region at the points and on the surfaces of application---will be uniquely determined by the tractions. Hence, the

"applied"stresses will be the same for both materials.However, the continuum as a whole will respond differently to the applied stresses. The problem is structurally indeterminate and so you have to invoke an additional principle, say a variational or energy principle, say the principle of minimization of the total potential energy (TPE for short).

The TPE field will be different in the two solids because of the differences in their respective constitutive relationships.

The difference in the values of material properties will in any case (even for two different

metals) make the two TPE fields hit their respective minima with a different combination of stress and strain fields---their values will be different.In the case of rubber (nu = 0.5), a further constraint will be enforced by the constitutive relation, viz. that the strains must remain deviatoric. This constraint will affect the values of all of the equilibrium fields, apart from also ensuring that the displacements remain only solenoidal.

Thus, for the rubber, the

appliedstress field may indeed be arbitrary. However, the resulting displacements will still be divergence-free.In the case of the metal, the constraint of zero hydrostatic strains will be absent, and the minimum of TPE will be reached with a general state of strain (i.e. nonzero values for both deviatoric and hydrostatic parts).

So, the answer is, yes, the applied traction field can be arbitrary even for an incompressible deformation. The constraint that I was talking about, gets placed only on the developed fields.

4. In view of this discussion, I guess that for the stress BV problem to be well-posed for the incompressible deformation, you have to choose. (i) You can either specify an arbitrary stress field but at only

somepoints of the solid (say at its bounding surfaces) and not all, or (ii) if you choose to specify the applied stress field at all points of the solid, then it must be a deviatoric-only field.Caveat: The truthhood of the second option depends on whether a deviatoric stress produces only deviatoric strains in any material or not, i.e. question 5.1.

5. A question now: Does any one know of any good book that explicitly discusses issues like these... incompressible materials vs. incompressible deformation (in the context of solid mechanics, not fluid mechanics), incompressible deformation of metals (at nu not equal to 0.5), and the other questions? Thanks in advance.

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An aside: Hmmm.... Time for the score-card: Ajit Jadhav: 1 point. Solid Mechanics: 2 points. ... Not bad...

Looks like this is a good set of questions to ask on the PhD qualifiers---what da you think, UAB?

--Dr. Ajit R. Jadhav

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## Volume preserving deformation vs Incompressibility

In order to avoid confusion, moved this comment up as a reply for the earlier comment by Ajith. Sorry...

Jayadeep

## Re: Incompressible Nonlinear Elasticity

I haven't read all the above discussions and this is just a quick note. Perhaps it would be easier if we discuss the issues one by one.

Incompressibility in nonlinear elasticity means the (time-dependent) motion is volume preserving. In other words Jacobian J=1 (note that dv = J dV). Given a reference motion with Jacobian J_0, linearization of Jacobian for a nearby motion reads \delta J = J_0 + J_0 div U, where U is the variation field (or displacement field). Now Assuming that J_0=1, incompressibility in the linearized setting implies div U=0.

Incompressibility is an internal constraint. Incompressible motions can be modeled in an action principle setting using two methods: 1) One can use a Lagrange multiplier (pressure field) to enforce J=1. For an incompressible motion stress has two parts; one that can be determined constitutively and the other one is not constitutive (pressure field). 2) One can restrict the action to the manifold of volume-preserving motions and then extremize it over this manifold. This will give the same governing equations (with a bit more work). The unknown pressure field will be determined using the governing equations.

An example may help. Consider an incompressible Neo-Hookean solid (energy is \mu/2 (tr C -3)). If one is looking at spherically symmetric deformations (for example a point defect or a hollow thick sphere under internal pressure) incompressibility of motion determines the kinematics. Now the only unknown is the pressure field p(R), which will be determined from the only nontrivial balance of linear momentum P^{rA}{}_{|A}=0.

Regards,

Arash

## Re: incompressible nonlinear elasticity

Dear Arash,

Thanks for joining the discussion. Your reply is not just informative but also greatly clarifying. Further, the example of the Neo-Hookean solid that you introduce in the discussion is very relevant, too.

Of course, my own preparation of these topics (topology, differentiable manifolds, applications to nonlinear elasticity) is just about at the level of a vague qualitative grasp---if at all that! (All of it, while at UAB, some 2 decades ago, and never refreshed later.) I think I ceased pursuing reading/browsing on these topics at the point where I found someone mentioning that while the mathematical framework itself is quite general, most of the particular theory deals only with the quadratic forms. I think I also did not realize the relevance of the theory for solid mechanics. I just vaguely got the idea (probably right) that no particularly new physical light is shed on the problem of singularity, e.g. one at the crack tip. At that point, I decided not to pursue these topics any further. And, with that decision, I now realize, I seem to have also lost access to some great mathematical techniques, some really wonderful insights!

Now, a request: If someone with a typical engineering background wishes to pursue these topics, what books would you recommend? Is there any easy-to-understand book concerning manifolds, action principles and solid mechanics problems, accessible via self-studies to, say, a beginning graduate student? Any tips? (Personally speaking, I don't think I could possibly find the time in the near future to pursue those studies. However, I would definitely at least browse through the book(s).)

Ideally, the book should deal with even simple specifics of SM such as, e.g., the following: With regard to extremization of action over a specific manifold (e.g, a volume-preserving one) just how does the method handle the compatibility constraints for a traction BV problem? Where do these enter the theory?

Thanks in advance for any tips for such book(s)/study material...

Best,

--Ajit

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## Re: incompressible nonlinear elasticity

Dear Ajit:

Unfortunately, there is no single book to learn these topics. I would say to get started with geometric ideas one could look at the following book: "The Geometry of Physics: An Introduction" by Theodore Frankel. I'm writing a paper on geometric discritization of incompressible elasticity and will be happy to share it with you as soon as it is completed.

Regards,

Arash

## Thanks, Arash! I have

Thanks, Arash!

I have already ordered Frankel's book. (Also, Schutz', just to help me juggle my memories---after a rapid Google book preview, I think that is the book I

mighthavebrowsedthrough, in those days.) I may not find the time to master the topics, but I will make sure to read through at least the more relevant portions.It would be great to receive your paper. Please do remember me once you complete it: aj175tp {at) yahoo (dot] co [dot} in .

Best,

--Ajit

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## Dear Kajal, If one

Dear Kajal,

If one considers for an example, lets say a bar is loaded in extension then the longitudinal strain, lateral strain and shear strain developed in the bar are outcome of the material property of the bar. Now if we see that the strains are simply some del(x_i)/del(X_j), then also the x_i and X_j are related by means of material properties only. That's why force-stress-strain relations only raise a well defined problem and we require E's (generalized material modulus) to find u_i so that x_i = X_i + u_i for any material configuration with some applied load. For our bar only, linear extension and lateral contraction are constrained with materail property itself and hence initial volume and final volume ratio is in turn a material property. Now for the ratio you can evaluate the determinant (del(x_i)/del(X_j)) or else the divergence of the resultant displacement field only after the equilibrium, constitutive, and kinematic relations are satisfied.

And if your material gives rise to incompressibility then you need some means to counter this problem in computational mechanics, such as mixed formulations etc. This has already been mentioned in the course of this discussion by some of our friend here.