The strain energy of an incompressible Neo-Hookean solid is given as:

$$ W = C_{10}(I_1 - 3) $$

Implying that at zero deformation $W = 0$, because $F = I \implies C = F^TF = I \implies I_1 = 3$

Strangely though, the second Piola-Kirchoff (2PK) stress is not zero

$$ S_{ij} = 2 \frac{\partial W}{\partial I_1} \delta_{ij} = 2 C_{10} \delta_{ij} $$

Is this an artefact of the 2PK stress not having a physical interpretation, or is $C_{10}$ indicative of residual stress or have I done something wrong?

Following on from this, the 4th order material tensor, given by equation 45 in this article, appears to be always zero for a Neo-Hookean material because all derivatives (first and second order) are zero. This does not add up.

  • $\begingroup$ I don't think it is an artifact of the 2PK stress not having a physical interpretation. The first PK tensor P=FS, which is also not zero. I think this is related to the fact that one can only determine the pressure, which is what $2C_{10}\delta_{ij}$ is upto an arbitrary constant in a displacement boundary value problem (See TJR Hughes 'The Finite Element method: Linear static and dynamic analysis Prentice Hall 1987', Sec 4.2 just after equation 4.2.10. I think If you have traction boundary conditions you will be able to determine $C_{10}$. $\endgroup$
    – NNN
    Commented Dec 20, 2021 at 3:44
  • $\begingroup$ I meant to say "The first PK tensor P=FS which has a physical meaning is also not zero" $\endgroup$
    – NNN
    Commented Dec 20, 2021 at 3:50
  • $\begingroup$ About your follow up question, I'm pretty sure that it's an effect of not including the pressure term. See purdue.edu/biomechanics/wp-content/uploads/2018/10/… equation 3.15 $\endgroup$
    – NNN
    Commented Dec 20, 2021 at 6:04
  • $\begingroup$ @Nachiket The material is incompressible. Does this change anything? Does that justify omitting the term $p(J - 1)$? $\endgroup$
    – Olumide
    Commented Dec 20, 2021 at 14:04

2 Answers 2


The second Piola Stress for incompressible hyperelastic material is expressed as \begin{equation} \mathbf{S}=2 \frac{\partial W(\mathbf{C})}{\partial C}+pJ\mathbf{C}^{-1}=\mathbf{S}^{'}+pJ\mathbf{C}^{-1}, \end{equation} where $\mathbf{C}=\mathbf{F}^\textrm{T}\mathbf{F}$ and $J^2=\textrm{det}(\mathbf{C})$. In order that $p$ can be interpreted as hydrostatic pressure, the following equation has to be true, which is
\begin{equation} p'=\frac{1}{3}J^{-1}\mathbf{S}^{'}:\mathbf{C}=0, \end{equation} since $J>0$, in paricular $J=1$, thus \begin{equation} \mathbf{S}^{'}:\mathbf{C}=2\frac{\partial W(\mathbf{C})}{\partial C}:\mathbf{C}=0. \end{equation} That condition is, in general, true for any valid material model if we restrict our attention only to distortional right Cauchy–Green tensor, i.e. when constrain of incompressibility is somehow imposed on deformation of the material. However, if that is not the case, $W(\mathbf{C})$, i.e. $\mathbf{C}$ is arbitrary, the free energy function has to have an appropriate form, i.e., has to be homogenous of order zero, i.e., $W(\mathbf{C})=W(\alpha\mathbf{C})$ (see note below).

Above is the source of confusion in the question, when we calculate Second Piola Stress, we have to use only the distortional component of the right Cauchy–Green tensor, \begin{equation} \hat{\mathbf{C}}=\textrm{det}(\mathbf{C})^{-1/3}\mathbf{C}. \end{equation} Equivalently we can modify free energy such that is a homogenous function of $\mathbf{C}$. That is done by introducing $\hat{W}(\mathbf{C})$, that is \begin{equation} \hat{W}(\mathbf{C})=W(\hat{\mathbf{C}}). \end{equation} For case of such homogenous function can be showen that $\hat{W} (\alpha\mathbf{C})=\hat{W}(\mathbf{C})$, see [1], thus $\hat{p}^{'}=0$, and $p$ is hydrsotatic pressure.

Applying that reasoning to the case of incompressible Neo-Hookean material, \begin{equation} \hat{W}=\frac{\mu}{2}(\hat{\mathbf{C}}:\mathbf{I}-3) \end{equation} and carefully calulating derivatives, \begin{equation} \frac{\partial (\hat{\mathbf{C}}:\mathbf{I})}{\partial \mathbf{C}} = \textrm{det}(\mathbf{C})^{-1/3}\mathbf{I} - \frac{1}{3}\textrm{det}(\mathbf{C})^{-1/3-1}\textrm{det}(\mathbf{C})\mathbf{C}^{-1}(\mathbf{C}:\mathbf{I}) \\= \textrm{det}(\mathbf{C})^{-1/3} \left( \mathbf{I}-\frac{1}{3}(\mathbf{C}:\mathbf{I})\mathbf{C}^{-1} \right), \end{equation} finally we get \begin{equation} \hat{\mathbf{S}}^{'} = \mu\left[ \textrm{det}(\mathbf{C})^{-1/3} (\mathbf{I}-\frac{1}{3}(\mathbf{C}:I)\mathbf{C}^{-1}) \right], \end{equation} which for zero strain would give zero stress. Having zero stress at zero strain in prinicple is matter of definition of material reference configuration as a stress free, however is not necessity. However, proving that $p^{'}=0$, for $W(\hat{\mathbf{C}})$ is essential for validating hyperelastic model as incompressible.

Note: If we impose incompressibility constrains explicitly of deformation by Lagrange multiplier, e.g. some form of Taylor-Hood elements in case of finite elements discretization, you can use $W(\mathbf{C})$, not $\hat{W}(\mathbf{C})$, since constraints are enforced explicitly on right Cauchy-Green tensor.

Note: For homogenous function (see [2]), we have \begin{equation} W(\alpha\mathbf{C})=\alpha^n W(\mathbf{C}), \end{equation} differntatin both side by $\alpha$, we get \begin{equation} \frac{\partial W(\mathbf{C})}{\partial C}:\mathbf{C} = n\alpha^{n-1}W(\mathbf{C}). \end{equation} Thus for zero oder, when $n=0$, that imples $\frac{\partial W(\mathbf{C})}{\partial \mathbf{C}}:\mathbf{C}=0$, i.e. $p^{'}=0$.

[1] Bonet, Javier, and Richard D. Wood. Nonlinear continuum mechanics for finite element analysis. Cambridge university press, 1997.

\begin{equation} \hat{W}(\alpha\mathbf{C})= \hat{W}(\textrm{det}(\alpha\mathbf{C})^{-1/3}\alpha\mathbf{C}) \\= \hat{W}((\alpha^{3}\textrm{det}(\mathbf{C}))^{-1/3}\alpha\mathbf{C}) \\= \hat{W}((\textrm{det}(\mathbf{C}))^{-1/3}\mathbf{C}) \\= \hat{W}(\mathbf{C}). \end{equation}

[2] https://en.wikipedia.org/wiki/Homogeneous_function

  • 1
    $\begingroup$ Great answer! There might be a $\bf{C}:\bf{I}$ missing in the equation before 'finally we get'. $\endgroup$
    – NNN
    Commented Dec 26, 2021 at 3:34
  • $\begingroup$ Thx @Nachiket. Mistake fixed. $\endgroup$
    – likask
    Commented Dec 26, 2021 at 10:36
  • 1
    $\begingroup$ Thanks for your answer and for pointing me to Bonet and Word. I take it that at zero strain $p = 0$ for a material without residual stress. $\endgroup$
    – Olumide
    Commented Dec 31, 2021 at 2:42
  • $\begingroup$ I'm struggling with the term $\frac{1}{9}$ in the statement of $\mathbb{C}$, equation (6.57a), on page 170 (2nd edition). My workings produce $\frac{1}{3}$. $\endgroup$
    – Olumide
    Commented Jan 1, 2022 at 5:21

From an old set of notes, I think that since your material is incompressible, the stress is determined by the strain energy density function $W$ only upto the hydrostatic pressure. The Cauchy stress is given by

$$ \pmb{\sigma} = -p\pmb{1} + 2\pmb{F}\frac{\partial{W}}{\partial{\pmb{C}}}\pmb{F}^T $$

The second Piola-Kirchhoff stress in incompressible elasticity is given by

$$ \begin{align} \pmb{S} &= \pmb{F}^{-1}\pmb{\sigma}\pmb{F}^{-T} = -p\pmb{F}^{-1}\pmb{F}^{-T} + 2\pmb{F}^{-1}\pmb{F}\frac{\partial{W}}{\partial{\pmb{C}}}\pmb{F}^T\pmb{F}^{-T} = -p\pmb{F}^{-1}\pmb{F}^{-T} + 2\frac{\partial{W}}{\partial{\pmb{C}}}\\ \pmb{S} &= -p\pmb{F}^{-1}\pmb{F}^{-T} + 2\alpha\pmb{1} \end{align} $$

I think this definition of $\pmb{S}$ is what you should be differentiating wrt $\pmb{C}$ in order to get your material tangent modulus $\mathbb{\pmb{C}}$. Since $p$ is an unknown, you would need to differentiate wrt $p$ as well. Maybe you can use your condition that the configuration corresponding to $\pmb{F}=\pmb{1}$ is stress free in order to get a relation between $p$ and $\alpha$. I've changed your strain energy density function to $\alpha(I_1-3)$ in order to avoid the symbol $C_{10}$ clashing with $\pmb{C}$

Edit: Now that I think of it a little bit more, I wouldn't make a relation between $p$ and $\alpha$ using the stress free condition. The pressure should be determined by the equilibrium equation and the material constants by experiments.


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