FEM: Material tangent of neo-Hookean material

Let’s calculate material tangent of hyperelastic material, taking neo-Hookean for instance.

The elastic energy of neo-Hookean material is written as

\psi^e=\frac{\mu}2tr(\bold{C})-\mu\ln J

A penalty method is applied to constrain volume unchanged

\psi^b=\frac{\kappa}2(J^2-2\ln J-1)

The total free energy density is therefore written into

\psi=\psi^e+\psi^b

The stress is obtained by derivative

P_{iA}=\frac{\partial\psi}{\partial F_{iA}}=\frac{\partial\psi^e}{\partial F_{iA}}+\frac{\partial\psi^b}{\partial F_{iA}}

The elastic stress

P^e_{iA}=\frac{\mu}{2}\frac{\partial I_1}{\partial C_{MN}}\frac{\partial C_{MN}}{\partial F_{iA}}-\frac{\mu}{J}\frac{\partial J}{\partial F_{iA}}

and bulk stress

P^b_{iA}=\kappa(J-J^{-1})\frac{\partial J}{\partial F_{iA}}

The math

\frac{\partial J}{\partial F_{iA}}=JF^{-1}_{Ai}

and

\frac{\partial C_{MN}}{\partial F_{iA}}=\frac{\partial F_{kM}}{\partial F_{iA}}F_{kN}+\frac{\partial F_{kN}}{\partial F_{iA}}F_{kM}=\delta_{AN}F_{iM}+\delta_{AM}F_{iN}

Material tangent is defined as

C_{iAjB}=\frac{\partial P_{iA}}{\partial F_{jB}}=C^e_{iAjB}+C^b_{iAjB}

We calculate it by part

C^e_{iAjB}=\frac{\partial P^e_{iA}}{\partial F_{jB}}=\frac{\partial}{\partial F_{jB}}\left[\frac{\mu}2\delta_{MN}(\delta_{AN}F_{iM}+\delta_{AM}F_{iN})-\mu F^{-1}_{Ai}\right]

Do math

\frac{\partial F^{-1}_{Ai}}{\partial F_{jB}}=F^{-1}_{Aj}F^{-1}_{Bi}

Now we can rewrite it into

C^e_{iAjB}=\frac{\mu}2\delta_{MN}(\delta_{AN}\delta_{ij}\delta_{MB}+\delta_{AM}\delta_{ij}\delta_{NB})-\mu F^{-1}_{Aj}F^{-1}_{Bi}=\mu\delta_{ij}\delta_{MA}\delta_{MN}-\mu F^{-1}_{Aj}F^{-1}_{Bi}

The bulk part of material tangent is

\begin{align*}
C^b_{iAjB}&=\kappa\frac{\partial(J^2-1)}{\partial J}\frac{\partial J}{\partial F_{jB}}F^{-1}_{Ai}+\kappa(J-J^{-1})\frac{\partial F^{-1}_{Ai}}{\partial F_{jB}}\\
&=2\kappa J^2F^{-1}_{Ai}F^{-1}_{Bj}+\kappa(J-J^{-1})F^{-1}_{Aj}F^{-1}_{Bi}
\end{align*}