# Exponential model W (strain energy density function)- Zero stress zero strain condition

Hi,

I am new here and have a very basic question on the Strain Energy Density (SED) function of exponential model.

This is the exponential model SED form : W=a(e(b(I1-3))-1)

Now we have certain postulates such as that

1.W=0 when there is no deformation and

2. stress=0 when strain =0.

When there is no deformation I1=3 and hence W=0, which is correct.

However, when we take the 1st derivative of W with respect to I1 to get our stress and then substitute I1=3 we don't get stress=0.

I know this is a standard model used very often, so I am obviously wrong in my understanding of how the 2nd postulate works for SED functions.

Can some one clarify as to how to work the 2nd postulate (stress=0 when strain=0) for this particular constitutive model ?

### What is derevative of W

What is derevative of W with respect to I1?

Stresses is derevative of W with respect strains.

### Re. exponential strain energy density function

The strain energy density function is:
$W=a\left[\exp[b(I_1-3)]-1\right]$.
From Wiki Hyperelastic material we see that
$\boldsymbol{\sigma} = -p~\boldsymbol{\mathit{1}} + 2~\boldsymbol{F}\cdot\frac{\partial W}{\partial \boldsymbol{C}}\cdot\boldsymbol{F}^T = -p~\boldsymbol{\mathit{1}} + 2~\boldsymbol{F}\cdot\left(\frac{\partial W}{\partial I_1}\,\frac{\partial I_1}{\partial \boldsymbol{C}}\right)\cdot\boldsymbol{F}^T$. The derivative of W wrt I1 is
$\frac{\partial W}{\partial I_1}=a\,b\,\exp[b(I_1-3)]$ and from Wiki Tensor derivatives $\frac{\partial I_1}{\partial \boldsymbol{C}} & = \boldsymbol{\mathit{1}}$. Hence the stress is not-zero at zero strain. Your model is not hyperelastic.

A better model is $W=a\left[\exp[b(I_1-3)^2]-1\right]$ which is similar to Fung's exponential model and does give zero stress at zero strain.

### Thank you

You consider the incompressible material. In this case we have hydrostatic pressure which is not determined by strains.

If I understand right then for Fung material sresses is not zero under simplification for big stretches.

### Re: Fung material and undetermined mean stress

The model and equations discussed in this blog describes isochoric deformations. The stress computed using these equations may not be able to balance the applied external forces.  The unbalanced forces will have to be balanced by the mean stress (p).   In other words, the mean stress  will depend on the applied external forces in such a way that internal and external forces are balanced.

At zero strain the external force is (presumably) zero and hence the pressure predicted by the Fung model will be zero.

-- Biswajit

### Undetermined mean stress

Yes, typically the pressure field will have to be solved for separately in order to satisfy equilibrium and boundary conditions. This is one of the blessings and curses of being literal about incompressibility.

Matt Lewis
Los Alamos, New Mexico

Null

### How is the proposed model better than the initial one?

Thanks for the explanation.

Is'nt it possible that W could still be zero based on what value our lagrange multiplier p might turn out to be say in an uniaxial condition?

But I don't understand as to how is the proposed one any different from the initial one I mentioned in my post W=a[exp[b(I1-3)]-1]

And a follow up question would be as to what will happen if such models are used to fit our experimental data? It would still be able to get the constants, so my qn is as to what is the implication of using such models to fit our experimental data?

### Re: Incompressible hyperelasticity and exponential models

Consider a ball made of an incompressible solid.  Apply a uniform external pressure to the ball.  Under this condition the ball is effectively rigid and does not have any internal energy.  What is the stress in the ball?

Hyperelasticity is a model for material behavior.  Such models should be independent of external forces, body forces, and other boundary conditions.  Your model depends on the boundary conditions for predicting the expected elastic response.  The modified model does not.  Both models are less than realistic.  That is why we prefer to use compressible hyperelastic models when possible.

-- Biswajit

### Re: Incompressible hyperelasticity and exponential models

Mathematically if there is no stored energy then the stress should be zero as well for a hyperelastic material. So for the example you have given also this will hold good right?

However, I still don't quite understand how the modified model response function will make the stress=0 when strain =0.

So according to the modified equation, we have $W=a\left[\exp[b(I_1-3)^2]-1\right]$ so
mathematically our
∂W/(∂I_1 )=-4abe^(b〖(I_1-3)〗^2 ) and we know that  $\frac{\partial I_1}{\partial \boldsymbol{C}} & = \boldsymbol{\mathit{1}}$, so still we donot satisfy the stress=0, when strain=0 condition here. What am i missing here?

Considering that we do not need the p to be solved to get our zero-stress-strain configuration..

### More on exponential hyperelasticity

A good way to understand incompressible behavior in rubbers is to start with the ball image and vary boundary conditions.  In the previous example you could substitute the applied pressue with zero pressure and not observe any change.  Now take the ball and put it between two perfectly smooth plates and apply a clamping force on the plates.  That is equivalent to having a slightly more complex  pressure boundary condition on the ball.  What is the stress in the ball? And so on.

Regarding the formula above see  http://www.wolframalpha.com/input/?i=A+%28exp%28b%28x-3%29^2%29-1%29

-- BIswajit

### Thanks Dr.Biswajit

I understand now, I think.

Can you please provide me with any book references that discuss this concept of zero stress-zero strain with an example like you have said to better understand how this works for the various phenomenological models that exist out in the literature?

Regards

Nisha

Thank you