Yazhuo Liu

Decouple the entropic and enthalpic contributions to the elastic energy of a polymer network

Assuming that the deformation of the hyperelastic solid can be divided into two steps: The first step is purely entropic, in which the length of each atomic bond is fixed and the volume taken by each atom is unchanged. In other words, the deformation only associates with the rearrangement of the polymer chains modelled as rigid links. And the second step is purely enthalpic, which means the relative positions of all atoms are frozen and only the atomic bond is stretched.

Mathematically, the total deformation gradient can be decomposed into two parts: $$F_{iK}=F_{ij^{\prime}}^H F_{j^{\prime}K}^S,$$ where $\boldsymbol{F}^H$ can be any second-order tensor and $\operatorname{det}\boldsymbol{F}^S=1$.

As the enthalpic deformation is relatively small, we adopt linear elastic assumption. The enthalpic strain energy can be represented by $$W_{\mathrm{H}}\left(\boldsymbol{\epsilon}^{\mathrm{H}}\right)=\frac{1}{2} C_{i j k l}^{\mathrm{H}} \epsilon_{i j}^{\mathrm{H}} \epsilon_{k l}^{\mathrm{H}}$$, in which the small enthalpic strain tensor \(\epsilon_{i j}^{\mathrm{H}}=\frac{1}{2}\left(F_{i j}^{\mathrm{H}}+F_{j i}^{\mathrm{H}}\right)-\delta_{i j}\) is used as internal variables

The change of bond length is only reflected in the second enthalpic step, and fracture usually occurs after a covalent bond has been stretched to its limit. In our model, the stretch of the covalent bond can be represented by the enthalpic strain energy $W_{\mathrm{H}}\left(\boldsymbol{\epsilon}^{\mathrm{H}}\right)$. Note that without a trustworthy stress/strain-based fracture criterion for soft solids, the energy-based fracture criteria is somehow acceptable. we then set the fracture criteria as the enthalpic strain energy reaches a certain threshold.

Implemented the above discussion with the Phase-field method, then we can get the results shown at the beginning.