freund_calculation

Freund (1998) $\theta$ ), and following schematics: Annotation 2020-04-03 092325

The starting point is the following expression of the total free energy of the body:

\mathscr{E}(t)=\int_{R} U\left(F_{i j}\right) \mathrm{d} R+\int_{S} \tilde{U}\left(\tilde{F}_{i}\right) \mathrm{d} S,\label{eq:1}

$U$ $\widetilde U$ is the surface energy, the latter being dependent on the restriction

\tilde{F}_{i}=F_{i j} m_{j}

$F$ $m$ $\widetilde F$ , in particular, represents the tangential stretch. The goal of the calculation is to obtain the form

\dot{\mathscr{E}}=\int_{S} \Phi(\text { deformation shape }) v_{n} \mathrm{d} S+\textrm{boundary terms}.\label{eq:3}

$\eqref{eq:3}$ , one follows two steps.

$\eqref{eq:1}$ , one obtains:

\dot{\mathscr{E}}=\int_{R} \frac{\partial U}{\partial F_{i j}} \dot{F}_{i j} \mathrm{d} R+\int_{S} U v_{n} \mathrm{d} S+\int_{S}\left[\frac{\partial \tilde{U}}{\partial \tilde{F}_{i}} \frac{\mathrm{d} \tilde{F}_{i}}{\mathrm{d} t}-\tilde{U} v_{n} \kappa\right] \mathrm{d} S

$v_n$ is the normal velocity, and where

\frac{\mathrm{d} \tilde{F}_{i}}{\mathrm{d} t}=\left(\dot{F}_{i j}+v_{n} \frac{\partial F_{i j}}{\partial p_{k}} n_{k}\right) m_{j}+F_{i j} \dot{m}_{j}

$\widetilde F$ $\Sigma$ $\widetilde\Sigma$ is defined by

\Sigma_{ij}=\frac{\partial U}{\partial F_{ij}},\qquad \widetilde{\Sigma}_{i}=\frac{\partial \tilde{U}}{\partial \tilde{F}_{i}}

we obtain

\dot{\mathscr{E}}=\int_{R} \Sigma_{i j} \dot{F}_{i j} \mathrm{d} R+\int_{S} U v_{n} \mathrm{d} S+\int_{S}\left[\widetilde \Sigma_{i}\left(\dot{F}_{i j} m_{j}+v_{n} \frac{\partial F_{i j}}{\partial n} m_{j}+\frac{\partial v_{n}}{\partial S} F_{i j} n_{j}\right)-\tilde{U} v_{n} \kappa\right] \mathrm{d} S

$\kappa$ is the curvature in the current configuration. Integrating by parts and using the divergence theorem, along with the boundary condition

\Sigma_{i j} n_{j}=\frac{\partial \widetilde{\Sigma}_{i}}{\partial S} \quad \text { on } S

$\eqref{eq:3}$ with ¹

\Phi\left(F_{i j}, \theta\right)=U+\widetilde{\Sigma}_{i} \frac{\partial{F}_{ij}}{\partial n} m_j -\frac{\partial}{\partial S}\left(\tilde{\Sigma}_{i} F_{i j} n_{j}\right)-\kappa\tilde{U}.

\int_{R} \Sigma \cdot \dot{F}=\int_{S} \frac{\partial \tilde{\Sigma}}{\partial S} \cdot \dot{\chi},

and that

\tilde{\Sigma}_i \cdot \dot{F_{ij}}{m_j}=\tilde{\Sigma}_i \cdot \frac{\partial \dot{\chi_i}}{\partial S}

Thus, we have

\begin{multline} \dot{\mathscr{E}}=\int_{R} \Sigma_{i j} \dot{F}_{i j} \mathrm{d} R+\int_{S} U v_{n} \mathrm{d} S+\int_{S}\left[\widetilde \Sigma_{i}\left(\dot{F}_{i j} m_{j}+v_{n} \frac{\partial F_{i j}}{\partial n} m_{j}+\frac{\partial v_{n}}{\partial S} F_{i j} n_{j}\right)-\tilde{U} v_{n} \kappa\right] \mathrm{d} S\\ % =\int_{S} \left[\frac{\partial \tilde{\Sigma}}{\partial S} \cdot \dot{\chi} +\tilde{\Sigma}_i \cdot \frac{\partial \dot{\chi_i}}{\partial S}\right]\mathrm{d} S+\int_{S} U v_{n} \mathrm{d} S+\int_{S}\left[\widetilde \Sigma_{i}\left(v_{n} \frac{\partial F_{i j}}{\partial n} m_{j}+\frac{\partial v_{n}}{\partial S} F_{i j} n_{j}\right)-\tilde{U} v_{n} \kappa\right] \mathrm{d} S\\ \end{multline}

$S$ $\Phi$ .

1 Observe that using force balance ↩