iMechanica - Comments for "1D Plasticity - Isotropic hardening, tutorial with examples."
https://www.imechanica.org/node/11811
Comments for "1D Plasticity - Isotropic hardening, tutorial with examples."enThanks for posting these
https://www.imechanica.org/comment/25505#comment-25505
<a id="comment-25505"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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 for posting these notes up. They helped me get my head around plasticity modelling.</p>
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</ul>Wed, 22 Jan 2014 20:53:57 +0000mohamedmoussacomment 25505 at https://www.imechanica.organswers to your plasticity questions
https://www.imechanica.org/comment/24543#comment-24543
<a id="comment-24543"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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 Weijie,
</p>
<p>
I will answer your questions in the order that you asked them above.
</p>
<p>
1. The elasto-plastic modulus comes about from a purely mathematical(not within an incremental algorithm) derivation. (See Simo and Hughes book page 80, 81 for the general derivation and pages 90 and 91 for particular results for Isotropic hardening and Isotropic with Kinematic hardening, 3dimensionl case). These results are entirely "mathematical" and are NOT derived within the framework of an incremental algorithm that you would implement in a finite element program. The elasto-plastic modulus is basically dsigma/depsilon.
</p>
<p>
The algorithmic tangent modulus comes about from a derivation of dsigma/depsilon within an incremental plasticity algorithm. Such an algorithm is implemented with discrete increments (steps). The algorithmic tangent modulus in this case (3D case) is NOT the same as the purely mathematical elasto plastic modulus. Compare the results with Simo and Hughes on page 91 with the result in box 3.2 on page 124.
</p>
<p>
2. See also remark number 4 on page 125 of Simo and Hughes which answers your question number 2. As delta t and the consistency parameter approach zero the elasto plastic modulus and the algorithmic elasto plastic modulus become the same. But, for large time steps however they are different. Why is this important? The reason this is important is as follows. Suppose you write a finite element analysis program and want to implement a nonlinear material model for plasticity. For each increment of load that causes plastic flow you will need to impose the consistency condition to determine the increase in plastic strain and elastic strain. This is can be done by using Newton Raphson iterations at the local element level. However, you may also do Newton-Raphson iterations at the global level for your finite element model in order to achieve global equilibrium. In order for your global iterations to have the desireable quadratic rate of convergence (that Newton Raphson iterations can achieve if done correctly) you need to use the (algorithmic) elasto plastic tangent modulus that is derived consistently within the frame work of the incremental algorithm. If you do not do this you will not have a quadratic rate of convergence with your global Newton Raphson iterations(your iterations will take longer and may not even converge). So, if you are using Newton Raphson iterations at the global level for equilibrium, you definitely want to use the consistent algorithmic elasto plastic modulus at the local element level.
</p>
<p>
You may also benefit by looking at my paper for the 1D case and seeing how the two are derived (elasto plastic modulus equation 4.11 and the algorithmic elasto plastic modulus equation 4.31). It will help to see how these are derived differently. For the 1D case they end up being the same, but in 3 dimensions they do not turn out the same as I mentioned previously above.
</p>
<p>
One sometimes says that the algorithmic elastoplastic (tangent) modulus is variationally consistent within the plasticity algorithm when Newton Raphson iterations are used at the global level.
</p>
<p>
This is not easy to put into words, but that is the best I can do to explain this. It is a somewhat tricky concept to convey. Understanding this comes about from reading books and literature carefully and even better taking a class where the instructor can explain this.
</p>
<p>
I also want to say that I am still learning these concepts myself and if<br />
some others, here on imechanica, want to add their thoughts on this, I<br />
would be happy for your comments and corrections to what I have said.
</p>
<p>
I hope that helps,
</p>
<p>
Louie
</p>
<p>
</p>
<p>
</p>
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</ul>Sun, 16 Jun 2013 22:12:54 +0000yawloucomment 24543 at https://www.imechanica.orgDear Sir:
This paper is
https://www.imechanica.org/comment/24525#comment-24525
<a id="comment-24525"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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>
This paper is very useful. And I shall be very grateful if you can help me with these questions.
</p>
<p>
1. "For the 1D plasticity case the algorithmic tangent modulus is equivalent to the elasto-plastic modulus. In higher dimensions this is not true."<span> </span>
</p>
<p>
I cann't distinguish the two "<span>algorithmic tangent modulus</span><span>" and "</span><span>the elasto-plastic modulus</span><span>";</span>
</p>
<p>
</p>
<p>
2. For 3D plasticity with general linear isotropic harding. Consider the Mises yiled function.
</p>
<p>
When the SIMO's renturn method converge, the relationship of
</p>
<p>
dσ=[Dsp] * dε should be satisfied? [Dsp] is the elasto-plastic module.
</p>
<p>
σn+1=σn + <span>[Dsp] * dε, should be consistent with stress calculated by SIMO's Return Method (f = 0, flow rules)?</span>
</p>
<p>
</p>
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</ul>Thu, 13 Jun 2013 08:55:13 +0000Weijie Liucomment 24525 at https://www.imechanica.orgThanks to Louie Yaw Sir...
https://www.imechanica.org/comment/19474#comment-19474
<a id="comment-19474"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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>
Many thanks for the reply.My profound gratitude.Thank you for making things clearer for me.
</p>
<p>
Just going a bit further- realting isotropic and Kinematic hardening to Von Mises yield surface:
</p>
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<p>
<strong>Von<br />
Mises yield surface</strong>
</p>
<p>
What I would like to understand is;
</p>
<ul class="unIndentedList"><li>
Let us consider Von Mises<br />
yield criteria. Now, Von Mises yield criteria is a yield criteria which states<br />
that you cannot have stress states outside the surface defined by the criteria.<br />
The surface does not have fixed dimensions except for perfect plasticity.</li>
<li>
If we take the square of<br />
both sides of Von Mises criteria, we notice that we get an equation of infinite<br />
cylinder in 3D (axis being the principal stresses). The cylinder is the yield<br />
surface, the radius of the cylinder is the yield stress.</li>
<li>
Now- for isotropic hardening<br />
(not kinematic hardening) the correct material law is the one (or the correct<br />
material modelling) is the one in which the yield surface increases in size but<br />
remains the same shape as a result of plastic straining.</li>
<li>
<strong>For<br />
kinematic hardening, the correct material law is the one in which allows the<br />
yield surface to translate without changing its shape. So- as one deforms the<br />
material in tension, you drag the yield surface thus increasing the stress thus<br />
modelling strain hardening. This softens the material in compression.</strong></li>
</ul><p>
<strong>Questions</strong>
</p>
<ul class="unIndentedList"><li>
<strong>I<br />
dot follow that for - how is it possible that if the yield surface i.e.<br />
cylinder if dragged (i.e. translated) the material hardens in tension but<br />
softens in compression?</strong></li>
</ul><p>
Now few questions on your paper of 1D plasticity which is turning out to be so useful for me undestand things and concepts:
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<p>
1. <br />
What<br />
does the consistency parameter physically denote? Does it denote amount of<br />
plastic flow? Why is that consistency parameter be always greater than zero?
</p>
<p>
</p>
<p>
On page 4, you have mentioned the Kuhn<br />
Tuckers condition
</p>
<p>
</p>
<p>
2. <br />
Kuhn<br />
Tucker's conditions are conditions (or rules) used to construct mathematical<br />
algorithms which model the process of plastic flow. One of Kuhn Tuckers<br />
condition is : γ >=0.
</p>
<p>
</p>
<p>
Now,<br />
γ is the consistency parameter which allows a means for determining the level<br />
of plastic flow and hardening such that f<=0 is satisfied. What is the<br />
relevance of having : γ >=0. What is the physics/math involved here.
</p>
<p>
</p>
<p>
3. <br />
Again,<br />
what is the relevance of Kuhn Tuckers condition γ f(σ) = 0. How is the<br />
evaluation justified mathematically and physically?
</p>
<p>
</p>
<p>
In box 4.1, which is 1D plasticity<br />
algorithm for general isotropic hardening as well as box 5.1which is 1D<br />
plasticity algorithm with no hardening (perfect plasticity), it has been stated<br />
in step 3 to compute the elastic trial stress,<br />
the trial value for the yield function and test for plastic loading.
</p>
<p>
</p>
<p>
4. What is this trial elastic stress actually? Because, you have mentioned<br />
the expression for trial elastic stress as:
</p>
<p>
sigma_n+1(trial) = E (epsilon_n+1 - epsilon_n^p)
</p>
<p>
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alt="" /></p>
<p>
Also you have marked epsilon_n+1 and epsilon_n^p in Figure 2 of the paper.
</p>
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<p class="MsoNormal">
<span>What does </span><span>εn+1- εnp<br /></span><span>denote? I’m unable to interprate </span><span>εn+1- εnp</span><span> from this graph? </span>
</p>
<p></p>
<p class="MsoNormal">
</p>
<p>
</p>
<p>
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/11811%23comment-form">Log in</a> or <a href="/user/register?destination=node/11811%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 12 Aug 2012 11:40:15 +0000kajalschopracomment 19474 at https://www.imechanica.orgKajal,
[Answers in
https://www.imechanica.org/comment/19466#comment-19466
<a id="comment-19466"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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>
Kajal,
</p>
<p>
[Answers in brackets] to your questions are provided by the questions below.
</p>
<p>
<strong>Isotropic<br />
hardening:</strong>
</p>
<p>
For isotropic hardening, if you plastically<br />
deform a solid, then unload it, then try to reload it again, you will find that<br />
its yield stress (or elastic limit) would have increased compared to what it<br />
was in the first cycle. [YES]
</p>
<p>
Again, when the solid is unloaded and<br />
reloaded, yield stress (or elastic limit) further increases. [YES, if it is reloaded past its previously reached maximum stress.]
</p>
<p>
This continues until a stage (or a cycle) is<br />
reached that the solid deforms elastically throughout. [ I think I understand your statement, eventually if the cycles of load are always to the same level, then after just one cycle your specimen on subsequent cycles will just be loading and unloading along the elastic line of the stress strain curve. Note also that, a steel bar with 1 pound of load on it has deformed elastically thoughout.]
</p>
<p>
This is isotropic hardening. [Isotropic hardening just means if you load something in tension past yield, when you unload it, then load it in compression, it will not yield in compression until it reaches the level past yield that you reached when loading it in tension. In other words if the yield stress in tension increases due to hardening the compression yield stress grows the same amount even though you might not have been loading the speciment in compression. It is a type of hardening used in mathematical models to describe plasticity. It might not be absolutely correct for real materials.]
</p>
<p>
Essentially, in isotropic hardening the yield<br />
stress (or elastic limit) increases whereas ductility keeps decreases. Hence,<br />
in the nth cycle the solid deforms elastically.[Your last sentence here "Hence, in the nth cycle the solid deforms elastically." i'm not sure what you mean. The previous sentences here are ok. As hardening increases, ductility is being used up, or is decreasing. Although, your sentence might make some sense if your cycles are to the same maximum stress each time, in which case after one cycle the material is just being loaded and unloaded along the elastic line of the stress strain curve.]
</p>
<p>
<strong>Is this understanding of isotropic hardening<br />
correct Sir ? [see comments above, you have it mostly right.]<br /></strong>
</p>
<p>
<strong>I shall be grateful if corrected.</strong>
</p>
<p>
<strong>Kinematic<br />
hardening</strong>
</p>
<p>
Isotropic hardening is not useful in<br />
situations where components are subjected to cyclic loading.[I believe real metals exhibit some isotropic hardening AND some kinematic hardening(I'm not sure if this is absolutely correct). Hence Isotropic hardening is useful. But, kinematic hardening is apparently more realistic for cyclic loading of metals.]
</p>
<p>
Isotropic hardening does not account for<br />
Bauschinger effect and predicts that after a few cycles, the material (solid)<br />
just hardens until it responds elastically [Yes, I think what you mean is correct.]
</p>
<p>
To fix this, alternative laws i.e. kinematic<br />
hardening laws have been introduced[Yes]
</p>
<p>
As per these hardening laws, the material<br />
softens in compression and thus can correctly model cyclic behaviour and<br />
Bauschinger effect.
</p>
<p>
<strong>Is this undestanding of kinematic hardening correct Sir?[I think you have it.]</strong>
</p>
<p>
regards,
</p>
<p>
Louie
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/11811%23comment-form">Log in</a> or <a href="/user/register?destination=node/11811%23comment-form">register</a> to post comments</span></li>
</ul>Fri, 10 Aug 2012 22:07:11 +0000yawloucomment 19466 at https://www.imechanica.orgDear Sir,Thank you for
https://www.imechanica.org/comment/19432#comment-19432
<a id="comment-19432"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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 for the response and I have read your paper.I would be grateful if you can correct me if wrong on the following which I feel I should get it right before going into hard math stuff.This is about fundamentals on isotropic and kinematic hardening and my understanding.
</p>
<p>800x600</p>
<p>Normal<br />
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<p>
<strong>Isotropic<br />
hardening:</strong>
</p>
<p>
For isotropic hardening, if you plastically<br />
deform a solid, then unload it, then try to reload it again, you will find that<br />
its yield stress (or elastic limit) would have increased compared to what it<br />
was in the first cycle.
</p>
<p>
Again, when the solid is unloaded and<br />
reloaded, yield stress (or elastic limit) further increases.
</p>
<p>
This continues until a stage (or a cycle) is<br />
reached that the solid deforms elastically throughout.
</p>
<p>
This is isotropic hardening.
</p>
<p>
Essentially, in isotropic hardening the yield<br />
stress (or elastic limit) increases whereas ductility keeps decreases. Hence,<br />
in the nth cycle the solid deforms elastically.
</p>
<p>
<strong>Is this understanding of isotropic hardening<br />
correct Sir ?<br /></strong>
</p>
<p>
<strong>I shall be grateful if corrected.</strong>
</p>
<p>
<strong>Kinematic<br />
hardening</strong>
</p>
<p>
Isotropic hardening is not useful in<br />
situations where components are subjected to cyclic loading.
</p>
<p>
Isotropic hardening does not account for<br />
Bauschinger effect and predicts that after a few cycles, the material (solid)<br />
just hardens until it responds elastically
</p>
<p>
To fix this, alternative laws i.e. kinematic<br />
hardening laws have been introduced
</p>
<p>
As per these hardening laws, the material<br />
softens in compression and thus can correctly model cyclic behaviour and<br />
Bauschinger effect.
</p>
<p>
<strong>Is this undestanding of kinematic hardening correct Sir? </strong>
</p>
<p>
<strong>I shall be grateful if corrected.</strong>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/11811%23comment-form">Log in</a> or <a href="/user/register?destination=node/11811%23comment-form">register</a> to post comments</span></li>
</ul>Sat, 04 Aug 2012 20:25:05 +0000kajalschopracomment 19432 at https://www.imechanica.orgAnswers are provided by the
https://www.imechanica.org/comment/19083#comment-19083
<a id="comment-19083"></a>
<p><em>In reply to <a href="https://www.imechanica.org/comment/19081#comment-19081">Dear Sir,I found this</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 are provided by the questions.
</p>
<li>For<br />
isotropic hardening, hardening accumulates if plastic strain is positive<br />
or negative. How? Material tests show that isotropic hardening happens. If a specimen in tension yields and hardens the same specimen loaded in compression will have a higher yield stress because of the hardening that previously took place when it was loaded in tension. This is due to isotropic hardening. The material has a "memory" (it has been affected by tension yielding) and this affects how it responds when it is loaded in compression.</li>
<ul><li>When<br />
hardening occurs, the value of the yield stress changes. Does it mean that<br />
when hardening occurs, elastic strain increases which means that when<br />
hardening occurs value of yield stress increases? </li>
</ul><p>
yes.
</p>
<p>
Suppose the initial yeild stress was 36 ksi. After hardening has taken place suppose the material stress reaches 40 ksi. Then if you unload the specimen to 0 stress then loaded it again, it will not yield until reaching 40 ksi. Hence the yield stress has increased from 36 initially to 40 ksi due to hardening. As soon as 40 ksi is passed and hardening continues the yield stress will again increase.
</p>
<ul><li>For<br />
isotropic hardening, hardening accumulates if plastic strain is positive<br />
or negative, hence internal hardening variable keep track of the total<br />
change in plastic strain.</li>
</ul><p>
yes
</p>
<p>
Question:
</p>
<p>
Ø <br />
When will the plastic strain become negative?<br />
Is it that the internal hardening variable will become negative because yield<br />
stress (hence elastic strain) increases? If so, does it mean that plastic<br />
strain may be negative when the material has just entered the plastic regime?
</p>
<p>
Based on your choice of coordinate system often tension is positive for plastic strain and compression produces negative plastic strain.
</p>
<p>
No, the internal hardening variable alpha is always positive, it does not ever go negative.
</p>
<ul><li>Hence<br />
internal hardening variable "α" keeps a track of the total change in<br />
plastic strain.</li>
</ul><p>
yes
</p>
<p>
Question:
</p>
<p>
Question: <br />
Yield condition (f) is defined as current<br />
level of stress minus the initial yield stress. When you say-initial yield<br />
stress, do you mean when the yielding began?
</p>
<p>
f is defined as current level of absolute stress minus (initial yield stress plus any change in yield stress due to hardening).
</p>
<p>
Initial yield stress is the stress that material will yield at the first time it is loaded up to yield.
</p>
<p>
Question:Does a negative value of ‘f' always denote<br />
that the material has not started yielding- that is, it is still in the elastic<br />
regime?
</p>
<p>
Negative f means the material is in the elastic range, but it does not mean that it hasn't yielded previously. It might have yielded previously and then it might be unloaded below yield and be in the elastic range and hence have f<0.
</p>
<p>
Question: <br />
Is ‘f' the stress level at a particular<br />
strain?
</p>
<p>
no, it is a function that tries to mathematically describe the boundary between elastic strain causing stresses and plastic strain causing stresses.
</p>
<ul><li>It<br />
has been stated that: if ‘f' is positive it means that current level of<br />
stress is above the yield stress- elastic and plastic strains are<br />
increasing </li>
</ul><p>
No, be careful here. f positive means that plastic flow is taking place. It does not mean stress is above the yield stress. This is a subtle point, but important. Mathematically we may calculate that f is positive, we might be inclined to think then that our stress is above yield, but that isn't quite the right way to think of it. Really what it means is that plastic flow is taking place if f is positive and we need to determine what the amount of plastic flow(strain) has happened such that f is at 0. Hence, f allows us to determine the amount of plastic flow and hardening that has taken place. We do this by using the kuhn tucker conditions, the consistency parameter and the consistency condition (all the equations and derivations indicated in the paper).
</p>
<p>
First<br />
question:
</p>
<p>
Ø <br />
Elastic strain will always increase-correct?
</p>
<p>
No, when the specimen is being unloaded elastic strain is decreasing. When zero stress is applied there may be permanent nonzero plastic strain remaining, but the elastic strain will be zero. Hence, clearly elastic strain can increase or decrease.
</p>
<p>
Second<br />
question:
</p>
<p>
Ø <br />
Does it mean that when f>0, we need to<br />
calculate the flow stress using the material law we are considering?
</p>
<p>
When f>0 we need to find the consistency parameter that makes f=0 again. We don't want f>0, f is only a mathematical device to allow us to calculate the value of the consistency parameter. Then using the consistency parameter we follow our material law for plastic flow and hardening and update internal variables appropriately.
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/11811%23comment-form">Log in</a> or <a href="/user/register?destination=node/11811%23comment-form">register</a> to post comments</span></li>
</ul>Thu, 31 May 2012 19:27:50 +0000yawloucomment 19083 at https://www.imechanica.orgDear Sir,I found this
https://www.imechanica.org/comment/19081#comment-19081
<a id="comment-19081"></a>
<p><em>In reply to <a href="https://www.imechanica.org/node/11811">1D Plasticity - Isotropic hardening, tutorial with examples.</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,<br />
I found this paper very useful and shall be grateful if you can help me with these questions. I have been reading it repeatedly and trying to follow things.The questions are from the first,second and third sections only and address very basic issues</p>
<ul><li>For<br />
isotropic hardening, hardening accumulates if plastic strain is positive<br />
or negative. How?</li>
</ul><p>
</p>
<ul><li>When<br />
hardening occurs, the value of the yield stress changes. Does it mean that<br />
when hardening occurs, elastic strain increases which means that when<br />
hardening occurs value of yield stress increases?</li>
</ul><p>
</p>
<ul><li>For<br />
isotropic hardening, hardening accumulates if plastic strain is positive<br />
or negative, hence internal hardening variable keep track of the total<br />
change in plastic strain.</li>
</ul><p>
</p>
<p>
Question:
</p>
<p>
</p>
<p>
Ø <br />
When will the plastic strain become negative?<br />
Is it that the internal hardening variable will become negative because yield<br />
stress (hence elastic strain) increases? If so, does it mean that plastic<br />
strain may be negative when the material has just entered the plastic regime?
</p>
<p>
</p>
<ul><li>Hence<br />
internal hardening variable "α" keeps a track of the total change in<br />
plastic strain.</li>
</ul><p>
</p>
<p>
Question:
</p>
<p>
</p>
<p>
Question: <br />
Yield condition (f) is defined as current<br />
level of stress minus the initial yield stress. When you say-initial yield<br />
stress, do you mean when the yielding began?
</p>
<p>
Question:Does a negative value of ‘f' always denote<br />
that the material has not started yielding- that is, it is still in the elastic<br />
regime?
</p>
<p>
Question: <br />
Is ‘f' the stress level at a particular<br />
strain?
</p>
<p>
</p>
<p>
</p>
<ul><li>It<br />
has been stated that: if ‘f' is positive it means that current level of<br />
stress is above the yield stress- elastic and plastic strains are<br />
increasing </li>
</ul><p>
</p>
<p>
First<br />
question:
</p>
<p>
</p>
<p>
Ø <br />
Elastic strain will always increase-correct?
</p>
<p>
</p>
<p>
Second<br />
question:
</p>
<p>
</p>
<p>
Ø <br />
Does it mean that when f>0, we need to<br />
calculate the flow stress using the material law we are considering?
</p>
<p>
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/11811%23comment-form">Log in</a> or <a href="/user/register?destination=node/11811%23comment-form">register</a> to post comments</span></li>
</ul>Thu, 31 May 2012 16:18:25 +0000kajalschopracomment 19081 at https://www.imechanica.org