Modeling elastomer in compression

[samuel]

Hello,

This is my first post here, and I don't have any experience with polymer, so please, go easy on me 8O ...

I am trying to model compression of an elastomer membrane, and I would like to have a simplified analytical model (to start with). The idea, would be of course to make some FEM simulations, but I would like to understand the theory behind it.

I don't know if it is a good way to start, but at first I would consider the equilibrium state when t->infinity to get rid of the viscoelasticity. So let's use hooke law. But with an elastomer, if I got it right, Young's modulus is not constant for large strain, so how can I take this into account? Is there a way to model the behaviour of the Young's modulus as a function of strain? I assume my elastomer has a poisson coeff of 0.5.

Thank you in advance for your replies

[Jorgen]

Hello Samuel,

Welcome to the world of polymers :wink:
Your approach to only consider the equilibrium response (t -> infinity) is very good. As you mentioned, for elastomers, Hooke's law is not the best approach. Instead I recommend that you use a hyperelastic model, such as the Neo-Hookean model. This is a very simple model that captures the change in modulus with strain and is quite suitable for moderately strained elastomers.

Best of luck,
Jorgen

[samuel]

Thanks for the info!

Now I notice that I lack some basic knowledge on the suject. For example the term "invariant" is new to me, and all the information I have found on the web about modeling polymers require knowledge I unfortunately do not have at this time.

So do you have any suggestion about a good book on the subject which could help me get the basic knowledge necessary to understand the different models.

Thanks in advance

Samuel

[Jorgen]

Hello Samuel,

There are a number different books that are useful for learning more about how to model polymers. Two books that are often recommended are:

:arrow: "An Introduction to the Mechanical Properties of Solid Polymers", but I.M. Ward.

:arrow: "Nonlinear Solid Mechanics" by Gerhard Holzapfel.

Jorgen

[samuel]

Thanks, I will have a look at those books. As a matter of fact, I had already booked the second one at the library and I will get it at the end of the week. I hope that I will then have a clearer idea on the subject.

[samuel]

Ok, I'm back on track thanks to the book you recommanded to me. (I'm reading Holzapfel's)

I have some questions though, and I hope you can help me...
According to your first reply, I first concentrated on the neo-Hookean model. It has the advantage of using a single constant C1 which can be readily identified with mu/2, Half of the shear modulus. But I wanted to apply the model to uniaxial loading on a polymer membrane, with free boundary conditions on the x and y axis and a deformation Sz=(Z-Z0)/Z0 on the Z axis (The simpliest problem I could think of...).
I also assume that the polymer is incompressible and therefore Sx=Sy=-0.5Sz. By using the free boundary condition, I can find the lagrange multiplier and my Cauchy-Stress tensor T is null everywhere, except for T33 (=Tz).
With T33=3*mu*Sz(1+0.25*Sz)
The problem I see with this equation is that T33(-Sz) is not equal to -T33(Sz), and therefore, the model does not predict the same stress for a compressive than for a tensil strain. (not to mention the fact of a second zero for negative Sz)
What did I get wrong?

I thank you in advance for your answer, and I apologize for asking such trivial questions...

[Jorgen]

Hi Samuel,

I am not sure I understand your question. Can you define the variables that you use (e.g. Sx, Sy, Sz, Tz, T33, etc), and the loading conditions.

- Jorgen

[samuel]

Hello and thank you for trying to help me. I realized my explanations were quite obscure, and I will try to explain what I was trying to do in a clearer manner.

I take the easiest possible case : a cube of polymer with side length a whose principal axes are aligned with the XYZ axes. I choose uniaxial loading by applying a pressure on the two faces perpendicular to Z axis The faces perpendicular to X and Y have free boundaries condition and can move freely, implying that the stresses SigmaX and SigmaY are zero. With a neo-hookean model, I want to model the deformation of the cube along the three axes.
From Holzapfel’s book, the Neo-hookean model linking the material stress with the left Cauchy strain tensor B is : Sigma=-p*I+2c1*B.
The deformation Gradient for this simple case is :

.....|l ,0 ,0 |
F= |0 ,l ,0 |
.....|0 ,0 ,lz |

(the dots are only there to display the tensor correctly)
With l the stretch in the X and Y direction and lz, the stretch in the Z direction.
I then calculate B=F*F’ (’ = transposed) which gives B11=B22=l^2 B33=lz^2 and the 6 other elements = 0. Using the fact that for the neo-Hookean model, the constant c1=mu/2, (mu, shear modulus (initial mu I suppose)) I write :

............|(mu*l^2)-p , 0 , 0 |
Sigma= | 0 ,(mu*l^2)-p , 0 |
............| 0 , 0 ,(mu*lz^2)-p|

Using the boundary conditions, I know that the stress is null on X and Y, and therefore p=mu*l^2. The entire Sigma tensor is null, except Sigma33=mu(lz^2-l^2)
Assuming that the polymer’s poisson ration is 0.5, I can express the relation between the stretch in Z and in X,Y : l=-0.5*lz+1.5
I finally get : SigmaZ=4*mu(3*lz^2+6*lz-9)
Or, in term of relative stretch Sz=(Z-Z0)/Z :
SigmaZ=4*mu(3*Sz^2+12*Sz)

And I do not understand the result, because if I plot SigmaZ as a function of strain Sz, I do not get a similar behavior for traction than for compression (of course, the relation is quadratic). I would have expected a relation with an un-even power of Sz to get an anti-symmetric behavior. So I probably missed something somewhere ang got something wrong.

I thank you in advance for your help.

[Jorgen]

Samual,

It sounds like you are applying a biaxial compressive stress:

I take the easiest possible case : a cube of polymer with side length a whose principal axes are aligned with the XYZ axes. I choose uniaxial loading by applying a pressure on the two faces perpendicular to Z axis The faces perpendicular to X and Y have free boundaries condition and can move freely, implying that the stresses SigmaX and SigmaY are zero.

I don't see why you state that SigmaX and SigmaY are zero, shouldn't it be SigmaZ that is zero?

- Jorgen

[samuel]

Jörgen,

I said it was uniaxial loading because I am only applying a pressure along Z-axis, on two opposite sides of the cube. Along X and Y I said that Sigma was 0 because I chose free boundary conditions for these directions, meaning the polymer can move freely along X and Y. Let’s say I put my cube on a frictionless table and that I apply a pressure on top of the cube: this is the situation I was trying to describe.

If the cube was in steel, and the pressure applied was small, with Hooke law it would give : SigmaX=SigmaY=0, SigmaZ=press
And for the strain : Sx=Sy=-mu/E*(SigmaZ) and Sz=press/E
With press=pressure applied along Z, E=Young Modulus, Sx=(X-X0)/X0 etc…

Now in my previous post, I tried to apply the same loading case to a polymer, using the Neo-Hookean model. As I wrote, I do not understand the result being a quadratic function of strain.

As for your question, can SigmaX and SigmaY not be zero if the faces of the cube perpendicular to these direction are free to move when I push on top of the cube?

Thanks for your help,

Samuel

[samuel]

Well, I found what I did wrong in my reasoning :
Using Sx=Sy=-0.5*Sz for incompressible material is an approximation made by neglecting higher order terms. Indeed in terms of stretch, it gives l=-0.5*lz+3/2.
But the exact relation between lx=ly=l and lz for an incompressible material is simply l=lz^-0.5. (The other equation is a good approximation only around l=1).
By using this new relation, I get :
SigmaZ=mu*(lz^2-lz^-1), which looks better.

But I still have trouble to interpret the data obtained with this model. As I expected, I have SigmaZ->-infinity for Sz=-1, meaning that it is impossible to fully compress the polymer with a finite pressure, and symmetrically, I have SigmaZ->infinity for Sz->infinity, meaning that extending the polymer to an infinite length is impossible with a finite force (of course, without taking into account the braking of the material).
What I don’t understand is that if I plot on the same graph the linear case (i.e. SigmaZ=E*true_strain=E*(1+Sz)) I do not get a symmetric deviation of the neo-hookean model from the linear law in compression and traction. To make things clearer, I attached a picture :
http://img215.exs.cx/img215/7753/neo7iw.jpg
You can see that in the traction case, the Neo-Hookean model is always “stiffer” than the linear law, which is not the case for the compressive part of the graph. I do not understand why the results in compression and traction would not be the same.
I was thinking than if I get SigmaZ=x for a true strain of y, then I would get SigmaZ=y for true strain –y…

Thanks in advance for the light you can bring me…

Samuel

[Jorgen]

Samuel,

Here are my comments:
But the exact relation between lx=ly=l and lz for an incompressible material is simply l=lz^-0.5. (The other equation is a good approximation only around l=1).
By using this new relation, I get :
SigmaZ=mu*(lz^2-lz^-1), which looks better.
Sounds OK to me.

What I don’t understand is that if I plot on the same graph the linear case (i.e. SigmaZ=E*true_strain=E*(1+Sz)) I do not get a symmetric deviation of the neo-hookean model from the linear law in compression and traction.
This equation is incorrect, I believe it should be: SigmaZ = E*log(1+Sz).

You can see that in the traction case, the Neo-Hookean model is always “stiffer” than the linear law, which is not the case for the compressive part of the graph. I do not understand why the results in compression and traction would not be the same.
Try plotting true stress vs. true strain

]I was thinking than if I get SigmaZ=x for a true strain of y, then I would get SigmaZ=y for true strain –y…
Why do you think the stress response should be symmetric in strain? Most hyperelastic models do not have that symmetry.

-Jorgen

[samuel]

Jörgen,

For the equation, it was a typo. I indeed forgot the natural logarithm, but the brown curve in the graph was plotted with the right equation (i.e. it is linear versus true strain, but not versus engineering strain, as it is plotted in the graph).

The important point of your message is that most hyperelastic models do not show symmetry for the behavior in traction and compression. I am still surprised to see that in traction, the hyperelastic gives always a higher stress in traction than the linear law (hence the "hyperelastic" word?), but in compression there is a region for small compressive Sz, where the material will be less stiff than the linear law, and then for an engineering strain of -0.79 (for the Neo-Hookean model) it crosses the linear law and gets stiffer.

I didn’t know there were such drastic differences between traction and compression.
I was planning to do some traction tests to study the behavior of my polymer, and then, use the true strain-true stress data in traction multiplied by -1 to model the compressive behavior, but I understand now that this is not the way to go if the behavior is not symmetric.

Two ideas come to my mind to model the compressive strain-stress behavior of the polymer I’d like to study (a PDMS) :
1) doing compressive stress test, but this seems complicated due to the fact that to get truly uniaxial compressive stress, the polymer should be able to expand freely between the compression plates, which seems difficult to achieve.
2) Stick with tensile test, but fit Mooney-Rivlin coefficient to the data, and use that theoretical law to model the compressive behavior of the PDMS
Does one of the method seems better than the other one? Do you have any advice?

…and thank you very much for the help you provided

Samuel

[Jorgen]

Hello Samuel,

Both of those 2 methods are fine. If the compressive strains are not too large and if you lubricate the compression specimen, then you should be able to get quite accurate data from simple compression tests.

Also, if you want to fit a constitutive model to tensile data, then I wouldn't use the Mooney-Rivlin model but a hyperelastic model that is only dependent on I1, such as the 8-chain model, or the Yeoh model.

Jorgen

[samuel]

Many thanks for the answers and advices. I'll see what I can do.
The reason I mentioned the use of the Mooney-Rivlin model was that it was readily available with Ansys

Samuel