iMechanica - Comments for "Johnson Cook plasticity model"
https://imechanica.org/node/11802
Comments for "Johnson Cook plasticity model"enjohnson cook parameter of st 37
https://imechanica.org/comment/25558#comment-25558
<a id="comment-25558"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>is that you can help me to download this thesis, I can not do it in my country.<br />
« Numerical And Experimental Investigation Of Perforation Of St-37 Steel Plates By Oblique Impact » :<br /><a href="http://etd.lib.metu.edu.tr/upload/3/12612045/index.pdf">http://etd.lib.metu.edu.tr/upload/3/12612045/index.pdf</a><br />
here is my email: <a href="mailto:ben2oeb@yahoo.fr">ben2oeb@yahoo.fr</a> <br />
thank you</p>
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</ul>Sat, 01 Feb 2014 16:35:47 +0000junior_2015comment 25558 at https://imechanica.orgHow to Find Parameters for Johnson-Cook Model
https://imechanica.org/comment/23877#comment-23877
<a id="comment-23877"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18205#comment-18205">Finding parameters for Johnson-Cook model</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, Every one i am student doing my masters and i have got the experimental result from tensile test and engineering stress vs strain. but i am going to apply the Johnson cook material model. please can any one give me how to find A,B,C,N and M please i have read it found from the experimental data i have got but i didn't know how to find please help me out. please attach me the steps in detial with my emial: <a href="mailto:bogale.sintayehu@gmail.com">bogale.sintayehu@gmail.com</a>
</p>
<p>
any sort of help is appreciated
</p>
<p>
thank you
</p>
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</ul>Thu, 20 Dec 2012 18:01:21 +0000dawit2012comment 23877 at https://imechanica.orgRe: JC Parameters
https://imechanica.org/comment/23866#comment-23866
<a id="comment-23866"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/23854#comment-23854">JC parameters</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>
Gradient search methods often have problems converging if you start from a point too far from the solution or if the model is not convex. The LM method also suffers from these issues.
</p>
<p>
My approach has been to start with a 2D fit to get good estimates of one parameter. I then use that parameter as the guess for a 3D fit and so on for higher dimensions. If the data are not dense enough, you may get a straight line fit instead of a curve.
</p>
<p>
-- Biswajit
</p>
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</ul>Wed, 19 Dec 2012 04:28:51 +0000Biswajit Banerjeecomment 23866 at https://imechanica.orgJC parameters
https://imechanica.org/comment/23854#comment-23854
<a id="comment-23854"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18205#comment-18205">Finding parameters for Johnson-Cook model</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 Biswajeet Sir,
</p>
<p>
I am using Split Hopkinson Pressure Bar for testing Ti64 under compression. I have my stress strain data and i used that data for fitting using non-linear least squares curve fit in matlab using Levenberg-Marquardt algorithm but its not fitting/i'm not getting theproper values of the constants.
</p>
<p>
I have taken stress strain values after the elastic part of the curve.
</p>
<p>
Could you please elaborate on the 3rd and 4th points of the above post.
</p>
<p>
Thank you.
</p>
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</ul>Mon, 17 Dec 2012 09:28:16 +0000vishal.iisccomment 23854 at https://imechanica.orgJohnson Cook plasticity model for polymer nanocomposie
https://imechanica.org/comment/19478#comment-19478
<a id="comment-19478"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>Can we use Johnson Cook plasticity model for modeling polymer nanocomposites?</p>
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</ul>Mon, 13 Aug 2012 03:10:18 +0000Azadeh Sheidaeicomment 19478 at https://imechanica.orgSame modulus..
https://imechanica.org/comment/18572#comment-18572
<a id="comment-18572"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>
Sir, please read above also....
</p>
<p>
In effect same modulus for quasi static and medium and high rate tests?
</p>
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</ul>Wed, 29 Feb 2012 00:54:43 +0000kajalschopracomment 18572 at https://imechanica.orgStatic modulus of alloy steels..noisy data in dynamic tests
https://imechanica.org/comment/18571#comment-18571
<a id="comment-18571"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Just to confirm- when you say static modulus of alloy steels, do you mean that I should be using the modulus as determined from quasi static tests (i.e. by taking the first few points on the true stress vs true strain curve,fitting a straight line and getting the slope- i.e. 'm' of the straight line equation y = mx + c ).
</p>
<p>
Thanks a lot again Sir.
</p>
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</ul>Wed, 29 Feb 2012 00:52:34 +0000kajalschopracomment 18571 at https://imechanica.orgRe: Noisy data in dynamics tests
https://imechanica.org/comment/18569#comment-18569
<a id="comment-18569"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18554#comment-18554">Quasi static and medium rate tests</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 initial data in dynamic tests is always noisy because of reflections and high frequency mode excitations. One has to be very careful when using that data to determine the modulus.
</p>
<p>
My approach has been to use the static modulus for alloy steels because the modulus doesn't change significantly as you increase the rate of loading. Experimentalists and materials scientists in this forum may be able to tell you more about the physics under those situations.
</p>
<p>
While the yield point may be identical at different rates, fracture usually takes place earlier at high strain rates because of both material properties and stress wave dynamics.
</p>
<p>
-- Biswajit
</p>
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</ul>Tue, 28 Feb 2012 23:05:26 +0000Biswajit Banerjeecomment 18569 at https://imechanica.orgNoisy data...
https://imechanica.org/comment/18556#comment-18556
<a id="comment-18556"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Please read the above also.
</p>
<p>
</p>
<p>
To add: The data in medium rate tests is noisy (wave like pattern)-is the difference in slope/ elastic modulus due to this?
</p>
<p>
</p>
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</ul>Tue, 28 Feb 2012 09:10:00 +0000kajalschopracomment 18556 at https://imechanica.orgQuasi static and medium rate tests
https://imechanica.org/comment/18554#comment-18554
<a id="comment-18554"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Again thanks for continuing the discussion-yes the temperature effect part is clear now.
</p>
<p>
I have observed during quasi static tests and medium rate tests:
</p>
<p>
1)The yield point is almost same in quasi static and medium rate tests
</p>
<p>
2)However when I fit a straight line to the first few points of the quasi static true stress and true strain curve and medium rate true stress starin curve, I observe from the equation of the two lines that the slope (which is the elastic modulus) is lesser in that of medium rate curve than quasi static curve.Should such an effect be noticed? Can the modulus/stiffness comedown as strain rates increase?
</p>
<p>
It is a solid mechanics problem
</p>
<p>
Kajal
</p>
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</ul>Tue, 28 Feb 2012 08:31:53 +0000kajalschopracomment 18554 at https://imechanica.orgRe: Johnson cook temp effect discussion
https://imechanica.org/comment/18552#comment-18552
<a id="comment-18552"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18534#comment-18534">Johnson cook temp effect discussion</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) Yes.
</p>
<p>
2) Depends on the problem you're using your model to solve. If it does not involve solid mechanics or you're interested only in linear elasticity, you don't need any plastic flow stress model. If you don't care about strain rates or temperatures, you just need the hardening model. But yield stress and temperature rise are connected intimately for adiabatic processes. If you're interested in shear banding, you cannot ignore the temperature. If you're interested in anisotropic plastic deformation, you need more information and your isotropic model will not suffice.
</p>
<p>
The issue is: what is the optimum amount of information that's appropriate for your problem. That's for you to determine.
</p>
<p>
-- Biswajit
</p>
<p>
</p>
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</ul>Tue, 28 Feb 2012 04:08:24 +0000Biswajit Banerjeecomment 18552 at https://imechanica.orgJohnson cook temp effect discussion
https://imechanica.org/comment/18534#comment-18534
<a id="comment-18534"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Thanks for the response..
</p>
<p>
1) You said:"<em>unless you measure the temperature of the sample at various points of the test you don't know that the high rate tests are at constant temperature"</em>
</p>
<p>
Can you please clarify what you mean by "various points of the test" .Do you mean at various time instants of the test?
</p>
<p>
2)Sir, yes rho is constant since we are not measring the changed area at every instant of strain measurement (and hence the stress).As well as Cp and Tm.Sir, but anyways we need these values for the concerned material as they need to be applied in the equation for curve fitting.Right Sir? I mean we cannot do without these values??
</p>
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</ul>Mon, 27 Feb 2012 01:02:11 +0000kajalschopracomment 18534 at https://imechanica.orgRe: Johnson Cook temperature effect
https://imechanica.org/comment/18533#comment-18533
<a id="comment-18533"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18528#comment-18528">Johnson Cook temperature effect</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 initial temperature is known but unless you measure the temperature of the sample at various points of the test you don't know that the high rate tests are at constant temperature. So your claim that "Laboratory tests are conducted at constant temperature T and so is our" is not necessarily true.
</p>
<p>
2) You are calculating dT using the energy equation. Make sure you're using it correctly. rho can be assumed to be constant (no volume change). Cp and Tm are needed but those are not necessarily constant. For your purposes you should be able to assume they are constant.
</p>
<p>
-- Biswajit
</p>
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</ul>Sun, 26 Feb 2012 22:16:47 +0000Biswajit Banerjeecomment 18533 at https://imechanica.orgJohnson Cook temperature effect
https://imechanica.org/comment/18528#comment-18528
<a id="comment-18528"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Thank you very much for the response.Yes, we also had a look at this paper:
</p>
<p>
<a href="http://www.dynamore.de/en/training/conferences/past/07-forum/impact/optional-strain-rate-forms-for-the-johnson-cook"><strong>http://www.dynamore.de/en/training/conferences/past/07-forum/impact/optional-strain-rate-forms-for-the-johnson-cook</strong></a>
</p>
<p>
We<br />
have yet to get through the paper by Ravichandran and all in detail.Based on<br />
this I interprate as follows.I shall be grateful if you provide your feedback<br />
to what I interprate;
</p>
<p>
Sir,
</p>
<p>
1)<br />
The equation for the temperature increase is how the J-C model converts plastic<br />
work into temperature. Laboratory tests are conducted at constant temperature T<br />
and so is ours
</p>
<p>
2)Now,<br />
the term in JC model involving temperature effect is;<br />
(1- (TH)^m)
</p>
<p>
<br />
We know,
</p>
<p>
<br />
TH = (T - TR ) / (Tm - TR)
</p>
<p>
where,
</p>
<p>
TR <br />
= reference temperature (or) room temperature i.e. the temperature of the<br />
sorroundings.
</p>
<p>
T= TR + dT
</p>
<p>
dT<br />
is given by the palstic work term which is[ 1/ rho *Cp] * (stress *<br />
plastic strain)
</p>
<p>
where,
</p>
<p>
Cp= specific heat
</p>
<p>
rho= mass density
</p>
<p>
</p>
<p>
Therefore,
</p>
<p>
TH = dT / (TM- TR)
</p>
<p>
=[1/<br />
rho*Cp] * stress*palstic strain / TM-TR
</p>
<p>
So,the data we require is now only: melting temperature and specific heat
</p>
<p>
RightSir?
</p>
<p>
Looking forward for your feedback..
</p>
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</ul>Sun, 26 Feb 2012 17:30:41 +0000kajalschopracomment 18528 at https://imechanica.orgRe: Johnson-Cook temperature dependence
https://imechanica.org/comment/18459#comment-18459
<a id="comment-18459"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18449#comment-18449">Thanks Biswajeet sir..JC</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>
Calibrating the temperature dependence can be a bit tricky in the absence of experimental data at various temperatures.
</p>
<p>
One way of dealing with the issue is to use temperature rise due to plastic deformation to get a temperature dependence.
</p>
<p>
1) Assume that there is a temperature rise during your test due to adiabatic (isentropic) heating. That assumption is reasonably accurate at high strain rates but not for quasistatic tests.
</p>
<p class="ep_block">
2) Calculate the plastic work (energy dissipated ~ stress*plastic_strain) see, for example, "On the Conversion of Plastic Work into Heat During High-Strain-Rate Deformation" by<span class="person_name"> Ravichandran et al. <a href="http://authors.library.caltech.edu/2824/">http://authors.library.caltech.edu/2824/</a></span>
</p>
<p class="ep_block">
3) Convert the plastic work into temperature by solving the energy equation (that is where the specific heat comes in). Assume 90% of the plastic work is converted into heat.
</p>
<p class="ep_block">
4) So you get a value of temperature for each point on your stress-strain curve.
</p>
<p class="ep_block">
That's better than assuming no temperature dependence because you have no temperature data. Specific heat data can be found from the manufacturer's data sheet.
</p>
<p class="ep_block">
-- Biswajit
</p>
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</ul>Tue, 21 Feb 2012 22:22:25 +0000Biswajit Banerjeecomment 18459 at https://imechanica.orgI think if you neglect the
https://imechanica.org/comment/18454#comment-18454
<a id="comment-18454"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18449#comment-18449">Thanks Biswajeet sir..JC</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 think if you neglect the <span>adiabatic temperature rise after the strain reach a critical value，the m could be neglected also.</span></p>
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</ul>Mon, 20 Feb 2012 13:08:13 +0000spacetankcomment 18454 at https://imechanica.orgThanks Biswajeet sir..JC
https://imechanica.org/comment/18449#comment-18449
<a id="comment-18449"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Thanks for the advise and you are right ! I have got the effective plastic straina and efefctive stress data.I just drew a trend line (which is a straight line fit to true stress vs true strain data) in excel got its equation and drew another line parallel to the trend line at 0.002.Since this lien os parallel to the fitted straight line (ternd line) the two slopes are equal and I got its equation by putting x =0 .002 and evaluating constant c.
</p>
<p>
I'm just a bit concerned now that we do not have any data with regards to temperature requirement i.e. we do not have mass density,specifoc heat and for the third term of JC we need these to get the tempertaure increment at every strain increment. In that case (to just continue witht he exercise) can we just neglect the third term and evaluate A,B,n,C?
</p>
<p>
Kajal
</p>
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</ul>Sun, 19 Feb 2012 15:42:40 +0000kajalschopracomment 18449 at https://imechanica.orgRe: Johnson-Cook parameters once again
https://imechanica.org/comment/18428#comment-18428
<a id="comment-18428"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18401#comment-18401">JC evaluation in response</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 Kajal,
</p>
<p>
I think you've reached the stage where you need to start doing your own thinking. As Nike says, "Just do it!".
</p>
<p>
Regards,
</p>
<p>
Biswajit
</p>
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</ul>Sun, 19 Feb 2012 02:12:44 +0000Biswajit Banerjeecomment 18428 at https://imechanica.orgJC evaluation in response
https://imechanica.org/comment/18401#comment-18401
<a id="comment-18401"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>Dear Sir,</span><span>Thanks again for the response.</span><span>A) Regarding your response,</span><em><span>You could calculate the constants for each strain rate and then try to reduce the error overall. That's a lot of effort but may actually give you more reasonable numbers.</span></em><span></span><span>You mean get constants (A,B,n,C,m*) for:</span><span>1)Quasi static tension</span><span>2)Quasi static compression</span><span>3)Quasi static torsion</span><span>4)Medium rate tension</span><span>5)Medium rate comporession</span><span>6)MEdium rate torsion</span><span>7)High rate tension</span><span>8)High rate compression</span><span>9)High rate torsion</span><span>Thus totally 9 sets and then reduce the error? - Just clarifying..</span><span><br />
Sir,regarding the reference on E modulus</span> </p>
<p>
<span>B) That link is inacessable.Sir, we do not have any test as of now for Youngs modulus.In that case, can we use the trend line in excel to get limiting elastic strain.Is it a reasonable start??</span>
</p>
<p>
<span>See attached jpeg file at the first post of this theread which we have just attached.</span>
</p>
<p><span>Kajal</span></p>
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</ul>Thu, 16 Feb 2012 01:45:44 +0000kajalschopracomment 18401 at https://imechanica.orgRe: Johnson-Cook model evaluation
https://imechanica.org/comment/18398#comment-18398
<a id="comment-18398"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18387#comment-18387">JC model evaluation..</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)<strong> Can I use smoorthing algorithm in Matlab?. Having done the smoothing <br />
should I use the smoothened data for curve fir using the LM algo?</strong>
</p>
<p>
There is no need to smooth the data for low rate tests. You should check your testing procedure if you see a lot of noise in a slow test. For high rate tests you have to be careful about the initial part of the stress-strain curve and some smoothing may be necessary.
</p>
<p>
2) <strong>In order to get the elastic part of the stress strain curve, can I use a 'linear trend line' ...</strong>
</p>
<p>
You should use an ASTM or ISO standard procedure to find the Young's modulus. I think ASTM E111 gives you the details for a tension test. See also "Elastic modulus measurement" by JD Lord. <a href="http://iopscience.iop.org/0026-1394/47/2/S05" target="_blank"><cite>iopscience.iop.org/0026-1394/47/S05</cite></a>
</p>
<p>
4 A) <strong>Can we get the JC constants A,B,n the first expression in JC model using quasi static tests alone?</strong>
</p>
<p>
You could calculate the constants for each strain rate and then try to reduce the error overall. That's a lot of effort but may actually give you more reasonable numbers. Just a direct nonlinear fit is easier.
</p>
<p>
You should try out a number of different options to see which leads to the least error (in the least squares sense) for all conditions of interest. You will find that Johnson-Cook cannot be used to represent your material accurately under all situations.
</p>
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</ul>Wed, 15 Feb 2012 21:11:00 +0000Biswajit Banerjeecomment 18398 at https://imechanica.orgJC model evaluation..
https://imechanica.org/comment/18387#comment-18387
<a id="comment-18387"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Thank you very much for the response.Actually, I am planning to use Matlab to do the fit which has the Lavenberg MArquardt algo inbuilt. I just wamnted to clarify a few things
</p>
<p>
1) I observed that there is a little noice in the data obtained through experiments.Can I use smoorthing algorithm in Matlab (Matlab has the moving average filtering for smoothing that which is a low pass filter taking the average of the data points). Having done the smoothing (supposing) should I use the smoothened data for curve fir using the LM algo?
</p>
<p>
2)In order to get the elastic part of the stress strain curve, can I use a 'linear trend line' drawn through excel on my stress strain curve passing through some data points ? Thus, get the proportional limit?Proportional limit indicating the point on the graph where there is deviation from the straight line.Corresponding, to the proportional limit will be the limiting elastic strain and beyonf the proportional limit there will exist a plastic strain = total strain - limiting elastic strain
</p>
<p>
3) Thus now (after 2) we get plastic strain vs stress values.
</p>
<p>
4)Last question, we have experimental data for quasi static, medium strain rate and high strain rate for tension,compresion and torsion tests.
</p>
<p>
A) Can we get the JC constants A,B,n the first expression in JC model using quasi static tests alone? Because, these constants do not have the effect of high strain rate and thermal softening?
</p>
<p>
B) To evaluate C that is constant in the second expression of JC we use the medium and high strain experiments (we substitute the values of A,B,n already evaluated in A above)?
</p>
<p>
Looking forward for the reply.
</p>
<p>
Kajal
</p>
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</ul>Wed, 15 Feb 2012 05:42:00 +0000kajalschopracomment 18387 at https://imechanica.orgRe: Johnson-Cook
https://imechanica.org/comment/18268#comment-18268
<a id="comment-18268"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18247#comment-18247">Dear Sir,
Sorry for being</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>
Answers to some FAQs:
</p>
<p>
1) <strong>Non-linear analysis involves a series of load increments- will the effective plastic strain be different at every load increment?</strong>
</p>
<p>
Non-linear finite element analyis of solids is usually displacement driven (see Chapter 6 of Belytschko et al, Nonlinear Finite Elements on how one such algorithm works and how the displacement increment is estimated from the applied load.)
</p>
<p>
If the displacement increases monotonically (or is constant) and the stress state is on the yield surface, the plastic strain will increase. If the displacement decreases so that the stress state goes back inside the surface, the plastic strain will remain constant.
</p>
<p>
Whether this reflects the physical world is another question altogether.
</p>
<p>
2) <strong>Suppose we do not include the model for fracture then when will the analysis terminate? We mean there has to be limiting strain or limiting stress?</strong>
</p>
<p>
The analysis terminates when your have reached your applied load. You can break your specimen in a finite element calculation only if you explictly specify when it breaks and what happens after that. The JC damage model can be used to decide when something will break. But each element will have a different damage state unless the deformation is homogeneous.
</p>
<p>
3) <strong>We know that true stress involves area of the deformed specimen and nominal stress involves area of the undeformed specimen. Given the nominal stress and strain only (in case of tension specimen as you see in the charts above), how do we get corresponding true stress and true strain for tension specimen?</strong>
</p>
<p>
Assume incompressiblity during plastic deformation to get the deformed area. Ideally you should measure the deformed area if you can. Just because people don't doesn't mean that it is not neceesary to measue the deformed area as a function of load.
</p>
<p>
4) <strong>Elastic indicates that the specimen regains its original shape/length on removal of load, how to get the elastic part from the stress strain curve?</strong>
</p>
<p>
Using the elastic modulus calculate the strain for a given stress.
</p>
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</ul>Sun, 05 Feb 2012 05:33:05 +0000Biswajit Banerjeecomment 18268 at https://imechanica.orgDear Sir,
Sorry for being
https://imechanica.org/comment/18247#comment-18247
<a id="comment-18247"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Sorry for being repetitive..I'm actually from the Structural Engineering stream -not Mechanical and am thus not confortable with questions you asked.
</p>
<p>
Yes,I shall read the book you mentioned.I have read mostly Elements of Strength of Materials by Timoshnko,Mechanics of Structures by Timoshinko and use the e-book by Alan Bower as a reference.
</p>
<p>
Sorry again
</p>
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</ul>Fri, 03 Feb 2012 02:38:36 +0000kajalschopracomment 18247 at https://imechanica.orgRe: yield surface
https://imechanica.org/comment/18246#comment-18246
<a id="comment-18246"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18239#comment-18239">More precisely now...</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 though I had answered that question in my previous comment.
</p>
<p>
Let me ask you a question.
</p>
<p>
When you roll a billet of steel into a sheet, does the billet undergo plastic deformation? Has the material yielded? When you cut a sample out of the rolled sheet for tensile testing is the material still elastic?
</p>
<p>
I think you should read a book on mechanics of materials before you get into all these issues. Boresi et al. "Advanced mechanics of materials" comes to mind.
</p>
<p>
-- Biswajit
</p>
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</ul>Thu, 02 Feb 2012 23:01:56 +0000Biswajit Banerjeecomment 18246 at https://imechanica.orgMore precisely now...
https://imechanica.org/comment/18239#comment-18239
<a id="comment-18239"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>
Thanks again Sir.Again, putting it more precisely
</p>
<p>
1) If the Engineer determines EXPERIMENTALLY the principal stresses on a LOADED STRUCTURE and then goes to his cylinder on the graph which was plotted taking into account the effect of strain rate, strain hardening, thermal softening and locates the point corresponding to sigma_1_load_x, sigma_2_load_x, sigma_3_load_x and then if this point is on the yield surface it indicates the material has yielded and if inside it means material is still elastic. Right?
</p>
<p>
2) I will put it after being clear on 1 Sir..
</p>
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</ul>Thu, 02 Feb 2012 06:00:04 +0000kajalschopracomment 18239 at https://imechanica.orgyield surface
https://imechanica.org/comment/18235#comment-18235
<a id="comment-18235"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18234#comment-18234">Thanks..putting it precisely...</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) Has the engineer modified the size of the yield surface so that hardening, softening, rate effects have been taken into consideration? If the engineer is using the yield surface that corresponds to initial yield (before hardening/softening) and has not taken rate effects into account, the answer is no. Otherwise yes.
</p>
<p>
2) No. The yield surface shape is determined by the yield condition - von Mises, Mohr-Coulomb, Cam-Clay etc.
</p>
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</ul>Thu, 02 Feb 2012 02:39:21 +0000Biswajit Banerjeecomment 18235 at https://imechanica.orgThanks..putting it precisely...
https://imechanica.org/comment/18234#comment-18234
<a id="comment-18234"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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>
Ok Sir, to make it more precise:
</p>
<p>
1) IF the Engineer determines the principal stresses on a LOADED structure - principal stresses being sigma_1_load_x, sigma_2_load_x, sigma_3_load_x and then goes to his graph (on which the cylinder is drawn) and locates the point corresponding to sigma_1_load_x, sigma_2_load_x- the, if this point is on the yield surface it indicates the material has yielded and if inside it means material is still elastic. Right?
</p>
<p>
Another question,Sir,
</p>
<p>
2) As you said that the JC model gives us a way of finding what the size of the yield surface is under different conditions. Can we obtain a specific shape (like cylinder/prism) for the yield surface corresponding to JC model?
</p>
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</ul>Thu, 02 Feb 2012 01:03:53 +0000kajalschopracomment 18234 at https://imechanica.orgVon Mises yield surface
https://imechanica.org/comment/18233#comment-18233
<a id="comment-18233"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18224#comment-18224">Von Mises yield surface</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 we plot the above cylinder on a graph, then calculate the<br />
principal stresses sigma_1,sigma_2,sigma_3 at a point in my structure<br />
,then if we go to the cylinder plotted above and obtain on the graph the<br />
point corresponding to sigma_1,sgma_2,sigma_3 ,then, if this point is<br />
on the yield surface it indicates that the material has yielded and if<br />
inside the yield surface it means that material is elastic.
</p>
<p>
Am I right?"
</p>
<p>
Yes and no. If the material has been unloaded after yield the stress state can be inside the cylinder.
</p>
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</ul>Thu, 02 Feb 2012 00:39:52 +0000Biswajit Banerjeecomment 18233 at https://imechanica.orgVon Mises yield surface
https://imechanica.org/comment/18224#comment-18224
<a id="comment-18224"></a>
<p><em>In reply to <a href="https://imechanica.org/node/11802">Johnson Cook plasticity model</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 Sir,
</p>
<p>
Thanks a lot for the response.
</p>
<p>
A very fundamental question from Von Mises yield surface (not going into Johnson Cooks model at the moment:
</p>
<p>
Now, . Von Mises yield surface is a circular cylinder of infinite length with its axis inclined at equal angles to the three principal stresses.
</p>
<p>
Also, by the definition of the yield surface we can say that:
</p>
<p>
When the stress state lies on the surface the material is said to have reached its yield point and the material is said to have become palstic (Reference: <a href="http://en.wikipedia.org/wiki/Yield_surface">http://en.wikipedia.org/wiki/Yield_surface</a>)
</p>
<p>
Now, my question is that (in undestanding the above definition itself):
</p>
<p>
Regarding;
</p>
<p>
We say above:<em>When the stress state lies on the surface the material is said to have reached its yield point and the material is said to have become palstic </em>
</p>
<p>
Does that mean, if we plot the above cylinder on a graph, then calculate the principal stresses sigma_1,sigma_2,sigma_3 at a point in my structure ,then if we go to the cylinder plotted above and obtain on the graph the point corresponding to sigma_1,sgma_2,sigma_3 ,then, if this point is on the yield surface it indicates that the material has yielded and if inside the yield surface it means that material is elastic.
</p>
<p>
Am I right?
</p>
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</ul>Wed, 01 Feb 2012 08:54:16 +0000kajalschopracomment 18224 at https://imechanica.orgRe: von Mises and Johnson-Cook
https://imechanica.org/comment/18214#comment-18214
<a id="comment-18214"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/18206#comment-18206">Revisiting fundamentals- Von Mises stress defnition and JC model</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 von Mises criterion is a yield condition which says that you cannot have stress states outside the surface defined by that criterion. However, the surface does does not have fixed dimensions except for perfect plasticity. If there is strain hardening or thermal softening the size of the surface can change, allowing you to achieve larger stresses. The material may also be rate dependent, in which case the surface has a different size at different strain rates.
</p>
<p>
The Johnson-Cook model gives you a way of finding what the size of the yield surface is under different conditions.
</p>
<p>
If you take the square of both sides of the von Mises criterion you will notice that you get the equation of an infinite cylinder in 3D (the axes are the principal stresses). That cylinder is the yield surface. The radius of the cylinder is S_y in your notation. The only way you can change the size of the cylinder is by changing the radius. The flow stress given by the Johnson-Cook model gives you a way of calculating S_y under various conditions.
</p>
<p>
-- Biswajit
</p>
<p>
</p>
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</ul>Tue, 31 Jan 2012 23:25:26 +0000Biswajit Banerjeecomment 18214 at https://imechanica.org