iMechanica - Comments for "Why rate equations in Nonlinear FE?"
https://imechanica.org/node/4072
Comments for "Why rate equations in Nonlinear FE?"enObjective rate of stress
https://imechanica.org/comment/26712#comment-26712
<a id="comment-26712"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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 wrote up the <a href="http://imechanica.org/node/17565">notes for my lectures on frame indifference</a> and postes them online. Several people made comments. Both the comments and the notes might contribute to this thread of discussion. </p>
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</ul>Tue, 16 Dec 2014 00:59:14 +0000Zhigang Suocomment 26712 at https://imechanica.orgre
https://imechanica.org/comment/16630#comment-16630
<a id="comment-16630"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/9705#comment-9705">I guess we all know that</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>If you like the rate form of the principle of virtual work "Nonlinear<br />
Finite Elements for Continua And Structures" by Belytschko, Liu, Moran<br />
gives a ton of examples. <a href="http://www.bestessayhelp.com" title="Essay Writing Help">Essay Writing Help</a></p>
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</ul>Mon, 25 Apr 2011 23:49:58 +0000Hanna83comment 16630 at https://imechanica.orgdistinguish that the rate
https://imechanica.org/comment/10828#comment-10828
<a id="comment-10828"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>distinguish that the rate form of constitutive equation is not a same concept as rate dependent.<br />
In case of constitutive equations of nonlinear plasticity, the rate<a href="http://www.seksizleseyret.com/" target="_blank" title="sikis izle,adult film izle"> </a>of loading also doesnt affect the plastic behavior. But the rate form<a href="http://www.sikis-pornoizle.com/" target="_blank" title="sikis izle,porno indir izle"> </a>is still widely used.That is because using the rate form we can get the full stress-strain path from incremental description</p>
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</ul>Fri, 15 May 2009 22:17:39 +0000Scootcomment 10828 at https://imechanica.orgYou don't "need" rate
https://imechanica.org/comment/10798#comment-10798
<a id="comment-10798"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>You don't "need" rate equations unless rates appear in some part of your formulation<a href="http://buy-arimidex.com">.</a> This is especially true of quasistatic problems<a href="http://online--viagra.com">.</a> Once again see the discussion with Rui.</p>
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</ul>Mon, 11 May 2009 23:46:35 +0000simplo345comment 10798 at https://imechanica.org2nd Law of Thermodynamics
https://imechanica.org/comment/10618#comment-10618
<a id="comment-10618"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/9016#comment-9016">Rate Forms in Hyperelasticity</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 Professor Govindjee,
</p>
<p>
I have heard comments which says that the non-assicoate plasticity model may violates the second law of thermodynamcs even for small deformation case. Would you mind to give us some reference about this topic?
</p>
<p>
Thanks,
</p>
<p>
WaiChing
</p>
<p>
</p>
<p>
</p>
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</ul>Wed, 22 Apr 2009 19:36:55 +0000WaiChing Suncomment 10618 at https://imechanica.orgThank you for your site. I
https://imechanica.org/comment/10561#comment-10561
<a id="comment-10561"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8952#comment-8952">Experimental point of view</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>Thank you for your site<a href="http://buy--cialis.net">.</a> I have found here much useful information<a href="http://buyzetia.com">.</a><a href="http://generic--viagra.com">.</a><a href="http://buy-avandia.com">.</a></p>
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</ul>Thu, 16 Apr 2009 19:20:12 +0000werasimosycomment 10561 at https://imechanica.orgI guess we all know that
https://imechanica.org/comment/9705#comment-9705
<a id="comment-9705"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8959#comment-8959">Re: Origin of rate equations</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 guess we all know that when subjected to different loading rates,<br />
most of the engineering materials show <a href="http://www.musthighschool.com/Must/Diploma/index.asp"><span>diploma at home</span></a> significant changes in material<br />
response in the nonlinear portion (i.e., plastic deformation range).<br />
The elastic behavior of the material is not effected due to rate of<br />
loading (unless viscoelastic). This may be due to some microscopic<br />
mechanisms.
</p>
<p>
<a href="http://www.musthighschool.com/Must/GED/index.asp"><span>high school ged</span></a> | <a href="http://www.musthighschool.com/Must/HomeSchooling.asp"><span>diploma for home school</span></a>
</p>
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</ul>Tue, 24 Mar 2009 06:43:03 +0000acotas548comment 9705 at https://imechanica.orgwhy Jaumann rate is popular?
https://imechanica.org/comment/9715#comment-9715
<a id="comment-9715"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8945#comment-8945">Re: Stress rates</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>
<span>I posted a question and just found this thread of<br />
discussions. I post my question again.</span>
</p>
<p>
<span><a href="http://imechanica.org/node/4725">http://imechanica.org/node/4725</a></span>
</p>
<p>
<span>In the above post, we showed that </span><span>Truesdell rate can by<br />
simplified to Green-Naghdi rate by assuming F .=. R </span><span>and can be<br />
further simplified to Jaumann rate by assuming W .=. R(.)R(T),<br /></span><span>where .=. means approximately equal </span>
</p>
<p>
<span>In a stretch dominant deformation, the three rates give different<br />
stress rate. </span><span>This is usually explained that we need a different<br />
tangential modulus for different objective rate. </span><span>However, it is<br />
hard to understand why we need to change "material" modulus when we use a<br />
different "mathematical" form of objective rate as they are all supposed to be<br />
"equivalent." </span>
</p>
<p>
<span>So a simple explanation may be that the Jaumann and Green-Naghdi rates<br />
are "inaccurate" when stretch deformation dominants. </span><span>As we know that<br />
in a shear deformation, Jaumann rate gives a "sin" varying shear stress, while<br />
Truesdell gives the exact answer (linear variation) (See Ted Belytschko, Wing Kam Liu, and Brian<br />
Moran. Nonlinear Finite Elements for Continua and Structures.) In other words,<br />
the three rate forms might not be equivalent, espeically when stretch<br />
deformation dominants</span>
</p>
<p>
<span>So a quick question is why are we often using Jaumann rate, instead of<br />
Truesdell rate in large deformation FE analysis? From the viewpoint of<br />
implementation, there should be no too much difference whichever rate we<br />
choose.</span>
</p>
<p>
<span>Please see </span><a href="http://imechanica.org/files/stress%20rate.pdf">http://imechanica.org/files/stress%20rate.pdf</a> for<br />
the relationship among the three rates.
</p>
<p>
Rong Tian
</p>
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</ul>Sat, 31 Jan 2009 03:47:26 +0000Rong Tiancomment 9715 at https://imechanica.orgGood explanation
https://imechanica.org/comment/9021#comment-9021
<a id="comment-9021"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
Prof. Govindji Sir,
</p>
<p>
Thank you very much for your nice explanation. Thanks to Dr. Biswajitda and all others those who particpated in this thread.
</p>
<p>
With regards,
</p>
<p>
- Ramdas
</p>
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</ul>Sat, 25 Oct 2008 05:50:20 +0000ramdas chennamsetticomment 9021 at https://imechanica.orgRate Forms in Hyperelasticity
https://imechanica.org/comment/9016#comment-9016
<a id="comment-9016"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8966#comment-8966">Re: Rate equations for hyperelasticity</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>
For what is worth, since you asked. Here are my two bits:</p>
<p>1) There is nothing fundamental about using rate forms. One can equally<br />
well use non-rate forms (for the cases being considered). Sometimes, however, people find it easier to think in terms of rates.
</p>
<p>
2) Not all objective rates seen in continuum mechanics are<br />
simple variations of the Lie derivative -- though many can be<br />
expressed as reasonably simple variation of a Lie derivative<br />
(simple being in the eye of the beholder). What really counts is: Can you more<br />
easily express the behavior of your material in terms of one rate or<br />
another?
</p>
<p>
<br />
3) The history of rate forms in mechanics and computational mechanics in<br />
particular is more for convenience than anything else. Written in rate<br />
form the constitutive relations of finite deformation look a lot more like<br />
those of the small strain formulation. This made it easier for people<br />
think about models. Unfortunately when computation came along this also gave people the false impression that they knew how to implement them in finite deformation<br />
simulation codes. Note that I say think! They actually did not do it<br />
correctly from a purist point of view. Even today many commercial code do<br />
not do it correctly (elastic or inelastic); they leave out terms<br />
which results in constitutive evaluations that violate the 2nd Law of Thermodynamics.<br />
Simo and Pister have a nice paper on this. Here too is reference to a small paper on some aspects of this issue:
</p>
<p>
<a href="http://dx.doi.org/10.1108/02644409710157604">Govindjee, S., ``Accuracy<br />
and Stability for Integration of Jaumann Stress Rate Equations in Spinning<br />
Bodies," Engng. Comp., 14, 14-30 (1997).</a>
</p>
<p>
4) The rate forms are only helpful for expressing the tangent matrix needed<br />
in finite element computations when the rate form is EXACTLY integrated in<br />
the constitutive evaluation subroutine. Otherwise what the global Newton-Raphson iteration needs is the algroithmic tangent matrix; i.e. the tangent matrix<br />
associated with the algorithm used to integrate the rate form.
</p>
<p>
</p>
<p>
</p>
<p>
Prof. Dr. Sanjay Govindjee<br />
University of California, Berkeley
</p>
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</ul>Sat, 25 Oct 2008 02:44:36 +0000Sanjay Govindjeecomment 9016 at https://imechanica.orgobjective time derivatives
https://imechanica.org/comment/9013#comment-9013
<a id="comment-9013"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8935#comment-8935">rate equations in nonlinear FE</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 Ramdas and Rui:</p>
<p>I'm not an expert in this topic but have a couple of comments that may be helpful.</p>
<p>1) As has already been mentioned by Biswajit, the existing objective time derivatives in continuum mechanics are all Lie derivatives (with respect to the spatial velocity field). As a matter of fact, the different objective derivatives are nothing but different representations of the same Lie derivative (using the metric tensor for raising and lowering indices). As is already known for a long time, there is no "more" objective stress rate. </p>
<p>One should note that the class of objective time derivatives is larger that just Lie derivatives. The following two papers explain this in some details.</p>
<p>i) F. Bampi and A. Morro, Objectivity and objective time derivatives in continuum physics, Foundations of Physics 10 (1980), 905-920.<br />
ii)Jean-Luc Thiffeault, Covariant time derivative for dynamical systems, Journal of Physics A 34 (2001), 5875–5885.</p>
<p>2)The class of non-dissipative solids is much larger than hyperelastic solids. The following recent paper discusses some of these subtle points.</p>
<p>K.R. Rajagopal and A.R. Srinivasa, On the response of non-dissipative solids, Proceedings of the Royal Society A 463(2007), 357-367.</p>
<p>Regards,<br />
Arash</p>
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</ul>Fri, 24 Oct 2008 21:10:09 +0000Arash_Yavaricomment 9013 at https://imechanica.orgSpatial velocity gradient => strain
https://imechanica.org/comment/9004#comment-9004
<a id="comment-9004"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
</p>
<p>
Thank you!!!
</p>
<p>
Right. I am following Belytschko's book. The approach is virtual power. Do you mean to say that, variation of spatial velocity gradient has to be replaced with variational strain. Then we need to modify constitutive law accordingly. Am I right?
</p>
<p>
Thank you,
</p>
<p>
With regards,
</p>
<p>
- Ramdas
</p>
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</ul>Fri, 24 Oct 2008 06:27:54 +0000ramdas chennamsetticomment 9004 at https://imechanica.orgRe: Rate independent constitutive law
https://imechanica.org/comment/8995#comment-8995
<a id="comment-8995"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8993#comment-8993">Rate independent constitutive 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>
1) Rate equations are not needed even if the constitutive law is rate dependent.
</p>
<p>
e.g. S = A E + B dE/dt
</p>
<p>
is not a true "rate equation" even though dE/dt appears on the right. We can just consider dE/dt to be an independent variable.
</p>
<p>
2) A rate constitutive relation is often of the form
</p>
<p>
dS/dt = A(E) dE/dt
</p>
<p>
Such a relation can have various uses as discussed in earlier comments.
</p>
<p>
3) You don't "need" rate equations unless rates appear in some part of your formulation. This is especially true of quasistatic problems. Once again see the discussion with Rui.
</p>
<p>
4) The easiest way to find out is to generate your own contact formulation and see what constitutive model is needed. Belytschko's book uses a virtual power approach. That is not a must. You can replaced the variation of the velocity with a variation of the displacement and work from there.
</p>
<p>
-- Biswajit
</p>
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</ul>Thu, 23 Oct 2008 05:27:16 +0000Biswajit Banerjeecomment 8995 at https://imechanica.orgRate independent constitutive law
https://imechanica.org/comment/8993#comment-8993
<a id="comment-8993"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
Hi all!
</p>
<p>
Thank you for good discussion. I went through the thread.
</p>
<p>
I understood the use of rate equations if constitutive law is rate dependent. I couldn't really understand why rate equations are required for rate independent constitutive law.
</p>
<p>
Say, I am working on geometric nonlinearities or contact analysis, do I need to use rate equations? Please explain in simple terms.
</p>
<p>
Thank you,
</p>
<p>
Regards,
</p>
<p>
- Ramdas
</p>
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</ul>Thu, 23 Oct 2008 03:05:41 +0000ramdas chennamsetticomment 8993 at https://imechanica.orgThanks for explaining
https://imechanica.org/comment/8989#comment-8989
<a id="comment-8989"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8960#comment-8960">Rate form is not rate of loading</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>
Hello<br />
Xiaoteng Wang,
</p>
<p>
Thanks for your explanation, I guess I didn't get the question properly. Being an experimental person when the rate was mentioned I was looking it from the constitutive behavior point of view. You are absolutely right, now I get what the question is mainly about.
</p>
<p>
Regards,
</p>
<p>
Siva
</p>
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</ul>Wed, 22 Oct 2008 16:36:18 +0000Siva P V Nadimpallicomment 8989 at https://imechanica.orgBe careful about the formulation
https://imechanica.org/comment/8982#comment-8982
<a id="comment-8982"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8977#comment-8977">linear increment for hyperelasticity</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>
Hello RH,
</p>
<p>
In UMAT the default formulation is hypoelastic, not hyperelastic. If you activate the large deformation option, hypoelastic formulation is adopted and the rotation tensor R is automatically calculated and passed by the solver. That means ABAQUS calculates and passes the state variables in a default hypoelastic structure, and automatically rotate the stress rate to the Jaumann rate. The second part which you mentioned before is calculated by ABAQUS but maybe it is incompatible with other formulation....In your case you write your constitutive model based on hyperelastic hypothesis, thus you have to be careful to define your own stress tensor and store it to the state variable matrix and you'd better not use the stress increment tensor given by ABAQUS.
</p>
<p>
</p>
<p>
Regards
</p>
<p>
Xiaoteng Wang
</p>
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</ul>Wed, 22 Oct 2008 01:45:17 +0000Xiaoteng Wangcomment 8982 at https://imechanica.orglinear increment for hyperelasticity
https://imechanica.org/comment/8977#comment-8977
<a id="comment-8977"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8969#comment-8969">Rate form and Incrementally linear form</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>
My thanks to Xu, Xiaoteng, and Biswajit for helping me out with the tangent modulus in hyperelasticity. <a href="http://en.wikiversity.org/wiki/Nonlinear_finite_elements/Rate_form_of_hyperelastic_laws">Biswajit's writeup at wikiversity</a> is particularly helpful with detailed tensor algebra.
</p>
<p>
Starting from an arbitrary strain energy density function for hyperelasticity, I was able to reach the total-form nonlinear equation in the weak form for finite element. To solve this nonlinear equation by iteration, I have to linearize it in an incremental form. Here I ran into the question about the tangent modulus, because the increment of the Kirchhoff stress consists of two parts. One can be written as c:d where c is the fourth elasticity tensor in <a href="http://en.wikiversity.org/wiki/Nonlinear_finite_elements/Rate_form_of_hyperelastic_laws">Biswajit's write up</a>. In addition, the increment also depends on the true stress at the current state as well as the rotation tensor, which is unsymmetric in general. This second part disappears by using the Lie derivative. However I believe both parts are needed to solve the nonlinear equation. In ABAQUS UMAT, it seems that only the first part is needed. My guess is that the second part is taken care of automatically by the nonlinear solver, since it does not really depend on the material. But I am not certain. I would appreciate comments from any ABAQUS experts.
</p>
<p>
RH
</p>
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</ul>Tue, 21 Oct 2008 15:10:00 +0000Rui Huangcomment 8977 at https://imechanica.orgRate form and Incrementally linear form
https://imechanica.org/comment/8969#comment-8969
<a id="comment-8969"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8946#comment-8946">Origin of rate equations!!!</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>
<br />
It seems that the "Rate" we are talking about has two meanings" "Rate" with respect to REAL time and "Rate" with respec to some parameters (for example, to the load increment).
</p>
<p>
If the material behavior is rate-dependent or there is some energy dissipation in the system which is dependent on the velocity (for example velocity dependent friction) of the objects, then time related "Rate form" virtual work principle (virtual power principle) should be used. In this case, the test field is often named as "virtual velocity" and the basic unknow variable is the TRUE VELOCITY field in the structure.
</p>
<p>
For the analysis of hyperelastic material, since this is a conservative system, the equilibrium equation can be expressed in "Total form". But this "Total form" equation(s) is (are) nonlinear in terms of state variables, such as displacement field, etc. Therefore lineariztion and iteration are necessary for the solution of it. At every step of iteration, an "Incrementally linear form" of virtual work principle should be used. In this case, the test field is often named as "virtual diplacement increment " and the basic unknow variable is the TRUE (in the sense of linearization) incremental diplacement field in the structure assocaited with the current load step. The tangent modulus is also obtained in this linearization process.
</p>
<p>
For the palsticity analysis of time rate-independent material, since it is always path-dependent, so "Increment form" must be used. In this case, At every step of iteration, an "Incrementally linear form" of virtual work principle should also be used.
</p>
<p>
From my point of view, for RAEL-time independent problems, the term "Incrementally linear form" is more appropriate than "rate form" Of course, you can also think that "rate" is the rate with respec to the load increment...
</p>
<p>
Hope this helps.<br />
<br />
best regards<br />
X.Guo
</p>
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</ul>Tue, 21 Oct 2008 09:40:04 +0000Xu Guocomment 8969 at https://imechanica.orgRe: Rate equations for hyperelasticity
https://imechanica.org/comment/8966#comment-8966
<a id="comment-8966"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8964#comment-8964">rate equations for total form hyperelasticity</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>
1) The tangent modulus is dsigma/depsilon. If you have sigma(epsilon) in closed form then the tangent can be calculated without resorting to a rate form.
</p>
<p>
2) You're right that a rate equation is often used while deriving a consistent tangent modulus. However, that rate equation can be generated by taking the time derivative of a hyperelastic relation in the material frame and then pushing it forward to the spatial frame if needed. That is equivalent to taking a Lie derivative but consistent with elastic material behavior (without history dependence). You can check out my writeup at <a href="http://en.wikiversity.org/wiki/Nonlinear_finite_elements/Rate_form_of_hyperelastic_laws">http://en.wikiversity.org/wiki/Nonlinear_finite_elements/Rate_form_of_hyperelastic_laws</a>
</p>
<p>
3) You can get an exact relation for the consistent tangent modulus (unless the elastic moduli depend on pressure) after a bit of algebraic trauma.
</p>
<p>
That's not the case for hypoelastic models - no matter what objective stress rate you choose.
</p>
<p>
4) The tangent modulus is often unsymmetric for complicated material models, e.g., soil plasticity with non-associated flow rules. I hope ABAQUS allows for unsymmetric tangent moduli but don't know enough about that software.
</p>
<p>
Prof. Govindjee is an expert on some of these issues as is Prof. Acharya. Let's hope they can find the time to chime in and give their inputs. I've found Simo's 1988 papers in CMAME particularly useful (though very dense).
</p>
<p>
-- Biswajit
</p>
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</ul>Tue, 21 Oct 2008 04:05:00 +0000Biswajit Banerjeecomment 8966 at https://imechanica.orgabout tangent modulus
https://imechanica.org/comment/8965#comment-8965
<a id="comment-8965"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8964#comment-8964">rate equations for total form hyperelasticity</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 tangent modulus (or Jacobian) defined by user is called by ABAQUS to formulate the initial stiffness matrix K.
</p>
<p>
Then ABAQUS solves the nonlinear equation f=Ku using itterative method.
</p>
<p>
A symmetric tangent modulus tensor gives better itterative stability.
</p>
<p>
To formulate Jacobian, you have to write the relationship between stress increment tensor (ds) and total strain increment tensor (de) first. This relationship is not so easy to formulate but from the constitutive equations we can derive it finally.
</p>
<p>
Then the first order derivative d(ds)/d(de) is the material Jacobian J. To achieve that goal you need a bit tensor analysis.
</p>
<p>
Luckily we dont need a super exact J, we just need a J approximately close to the true value to help the solver solving the equation without so many itterations.
</p>
<p>
Once you defined your own constitutive model, you can always formulate the J. Since the relationship between plastic strain increment, total strain increment and stress increment is already defined by yourself.
</p>
<p>
</p>
<p>
</p>
<p>
</p>
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</ul>Tue, 21 Oct 2008 03:15:08 +0000Xiaoteng Wangcomment 8965 at https://imechanica.orgrate equations for total form hyperelasticity
https://imechanica.org/comment/8964#comment-8964
<a id="comment-8964"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8959#comment-8959">Re: Origin of rate equations</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>
It appears to me that even for the total form hyperelasticity the rate equation is needed to define the tangent modulus tensor in the process of solving the nonlinear equilibrium equation. This is the question I have since the definition of tangent modulus depends on which stress rate you use. For example, in ABAQUS UMAT, a symmetric tangent modulus tensor is assumed, but it is not clear to me how the tangent modulus is defined and how it is used to solve the nonlinear equation.
</p>
<p>
RH
</p>
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</ul>Tue, 21 Oct 2008 02:33:00 +0000Rui Huangcomment 8964 at https://imechanica.orgRate form is not rate of loading
https://imechanica.org/comment/8960#comment-8960
<a id="comment-8960"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8952#comment-8952">Experimental point of view</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>
Hello Siva,
</p>
<p>
What you mentioned in your post is about the constitutive behavior of materials.
</p>
<p>
However we should distinguish that the rate form of constitutive equation is not a same concept as rate dependent.
</p>
<p>
In case of constitutive equations of nonlinear plasticity, the rate of loading also doesnt affect the plastic behavior.
</p>
<p>
But the rate form is still widely used.
</p>
<p>
That is because using the rate form we can get the full stress-strain path from incremental description, which is much more convinient.
</p>
<p>
Suppose the materials undergo cyclic loading or ratchetting, in such a case using rate form can easily describe the cyclic stress-strain behavior, especially the plastic loading-elastic unloading transition. But using total form it may be a big problem because at the same strain you may have more than one stress value. You have to make a lot of effort on programming that situation.
</p>
<p>
The rate dependent (viscoplasticity/viscoelasticity) thing is totally another topic. Maybe we can start another thread to talk about it.
</p>
<p>
Regards,
</p>
<p>
Xiaoteng Wang
</p>
<p>
<img src="http://www.iam.rwth-aachen.de/Forschung/Schwabe/Verbundwerkstoffe/altplast.gif" alt="typical ratchetting s-s curve, " width="350" height="300" /></p>
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</ul>Tue, 21 Oct 2008 01:01:03 +0000Xiaoteng Wangcomment 8960 at https://imechanica.orgRe: Origin of rate equations
https://imechanica.org/comment/8959#comment-8959
<a id="comment-8959"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8946#comment-8946">Origin of rate equations!!!</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>
There are two issues here:
</p>
<p>
1) Rate form virtual work (or the principle of virtual power)
</p>
<p>
2) Rate form constitutive equations
</p>
<p>
Neither of these is necessary for nonlinear finite element analysis, though rate forms are widely used (particularly in explicit codes).
</p>
<p>
"Computational Inelasticity" by Simo and Pister takes the approach of not using the rate form of the principle of virtual work. That book discusses the problems with rate form constitutive equations for elasticity and ultimately proposes that total form hyperelasticity is the best way to go. Rate forms are used for history dependent behavior for various reasons (some cited by by another commentor).
</p>
<p>
If you like the rate form of the principle of virtual work "Nonlinear Finite Elements for Continua And Structures" by Belytschko, Liu, Moran gives a ton of examples.
</p>
<p>
Rate form constitutive equations are usually easier to contruct than total forms (and probably easier to express in incrementally linear form).
</p>
<p>
I would like experts on constitutive relations to provide us with some insight into deeper reasons for the use of rate constitutive equations.
</p>
<p>
-- Biswajit
</p>
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</ul>Mon, 20 Oct 2008 23:06:39 +0000Biswajit Banerjeecomment 8959 at https://imechanica.orgExperimental point of view
https://imechanica.org/comment/8952#comment-8952
<a id="comment-8952"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/8946#comment-8946">Origin of rate equations!!!</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>
Hello Ramdas,
</p>
<p>
I had the same questions some time back, but I found some helpful information about it.
</p>
<p>
</p>
<p>
I have not worked extensively on (formulation) coding etc, hence I dont know any reasons from the computational point of view for the above point. However, if we see this from experimental poit of view, it makes some sense.
</p>
<p>
I guess we all know that when subjected to different loading rates, most of the engineering materials show significant changes in material response in the nonlinear portion (i.e., plastic deformation range). The elastic behavior of the material is not effected due to rate of loading (unless viscoelastic). This may be due to some microscopic mechanisms.
</p>
<p>
Hence, the rate equatios may not be required in the elastic region to model the material response.
</p>
<p>
I hope this info is of some help to you.
</p>
<p>
</p>
<p>
Regards,
</p>
<p>
Siva
</p>
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</ul>Mon, 20 Oct 2008 15:23:52 +0000Siva P V Nadimpallicomment 8952 at https://imechanica.orgwithout rate equation you
https://imechanica.org/comment/8947#comment-8947
<a id="comment-8947"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
without rate equation you can not express an exact path.
</p>
<p>
A virtual work W=int(x(y)dy), where x is load, usually an implicit function of y, and dy is virtual displacement.
</p>
<p>
The equation above is an integration process.
</p>
<p>
There are thousands of possible paths to get a same W.
</p>
<p>
The rate equation describes the exact path implicitly.
</p>
<p>
In linear FE formulation, the relation between x and y is simply linear, so we can directly write the virtual work.
</p>
<p>
Also in nonlinear FE, if you can get the integrated function explicitly, you can formulate it without rate form.
</p>
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</ul>Mon, 20 Oct 2008 06:14:08 +0000Xiaoteng Wangcomment 8947 at https://imechanica.orgOrigin of rate equations!!!
https://imechanica.org/comment/8946#comment-8946
<a id="comment-8946"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
Thank you!!! I downloaded your notes.
</p>
<p>
Here, I want to understand why the rate equation came into picture (not which rate equation it is)? What is the origin for this?
</p>
<p>
Why virtual work has to be expressed for per unit time? why can't we formulate non-linear FE frmulations without rate, using virtaul work without rate?
</p>
<p>
Kindly explain.
</p>
<p>
Thanks in advance,
</p>
<p>
Regards,
</p>
<p>
- Ramdas
</p>
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</ul>Mon, 20 Oct 2008 05:33:55 +0000ramdas chennamsetticomment 8946 at https://imechanica.orgRe: Stress rates
https://imechanica.org/comment/8945#comment-8945
<a id="comment-8945"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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>
For more on stress rates see my notes posted on Wikiversity at
</p>
<p>
<a href="http://en.wikiversity.org/wiki/Continuum_mechanics/Objective_stress_rates" target="_blank">http://en.wikiversity.org/wiki/Continuum_mechanics/Objective_stress_rates</a>
</p>
<p>
Which rate you use is not particularly important. What is important is that your material parameters should be such that the rate that you use reflects reality. That is, the parameters should be determined keeping a particular rate in mind and not just be the same as those used in small strain elasticity (for instance). And you have to keep in mind that first-order hypoelasticity is not history-independent.
</p>
<p>
The Jaumann rate is popular because it's easy to implement.
</p>
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</ul>Mon, 20 Oct 2008 04:42:43 +0000Biswajit Banerjeecomment 8945 at https://imechanica.orgrate equations in nonlinear FE
https://imechanica.org/comment/8935#comment-8935
<a id="comment-8935"></a>
<p><em>In reply to <a href="https://imechanica.org/node/4072">Why rate equations in Nonlinear FE?</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 had the same question in mind some time ago. I believe you are right that the rate equation is needed for solving the nonlinear equation iteratively, for example, by the conjugate-gradient method. I am now facing another question about the rate equation. There are definitions of several different rates (Lie rate, Jaumann rate, etc.). Which one shall we use? Why?
</p>
<p>
Thanks.
</p>
<p>
RH
</p>
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</ul>Fri, 17 Oct 2008 14:16:00 +0000Rui Huangcomment 8935 at https://imechanica.org