Micromechanics of composites/Average stress power in a RVE with finite strain

Average stress power in a RVE
Recall the equation for the balance of energy (with respect to the reference configuration)

\rho_0~\dot{e} = \boldsymbol{P}^T:\dot{\boldsymbol{F}} - \boldsymbol{\nabla}_0 \bullet \mathbf{q} + \rho_0~s ~. $$ The quantity $$\boldsymbol{P}^T:\dot{\boldsymbol{F}}$$ is the stress power.

The average stress power is defined as

{ \langle \boldsymbol{P}^T:\dot{\boldsymbol{F}} \rangle := \cfrac{1}{V_0}\int_{\partial{\Omega}_0}\boldsymbol{P}^T:\dot{\boldsymbol{F}}~\text{dV} ~. } $$ Here $$\boldsymbol{P}^T$$ is an arbitrary self-equilibrating nominal stress field that satisfies the balance of momentum (without any body forces or inertial forces) and $$\dot{\boldsymbol{F}}$$ is the time rate of change of $$\boldsymbol{F}$$. The reference configuration can be arbitrary. Also, the nominal stress and the rate $$\dot{\boldsymbol{F}}$$ need not be related.

Note that in that case

{ \text{tr}(\langle \boldsymbol{P}^T:\dot{\boldsymbol{F}} \rangle) = \cfrac{1}{V_0}\int_{\partial{\Omega}_0}\text{tr}(\boldsymbol{P}\cdot\dot{\boldsymbol{F}})~\text{dV} = \cfrac{1}{V_0}\int_{\partial{\Omega}_0}\text{tr}(\dot{\boldsymbol{F}}\cdot\boldsymbol{P})~\text{dV} = \text{tr}(\langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle) ~. } $$

We can express the stress power in terms of boundary tractions and boundary velocities using the relation (see Appendix)

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} ~. $$ In this case, we have $$\partial{\Omega} \rightarrow \partial{\Omega}_0$$, $$\Omega \rightarrow \Omega_0$$, $$\boldsymbol{\nabla} \rightarrow \boldsymbol{\nabla}_0$$, $$\mathbf{v} \rightarrow \dot{\mathbf{x}}$$, $$\boldsymbol{S} \rightarrow \boldsymbol{P}$$, and $$\mathbf{n} \rightarrow \mathbf{N}$$. Then

\int_{\partial{\Omega}} \dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla}_0~ \dot{\mathbf{x}}\cdot\boldsymbol{P} + \dot{\mathbf{x}}\otimes(\boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T)]~\text{dV} ~. $$ Using the balance of linear momentum (in the absence of body and inertial forces), we get

\int_{\partial{\Omega}} \dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} = \int_{\Omega} \boldsymbol{\nabla}_0~ \dot{\mathbf{x}}\cdot\boldsymbol{P}~\text{dV} ~. $$ Recalling that

\dot{\boldsymbol{F}} = \frac{\partial }{\partial t}\left(\frac{\partial \mathbf{x}}{\partial \mathbf{X}}\right) = \frac{\partial }{\partial \mathbf{X}}\left(\frac{\partial \mathbf{x}}{\partial t}\right) = \boldsymbol{\nabla}_0~ \dot{\mathbf{x}} $$ we then have

\int_{\Omega} \dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} = \int_{\partial{\Omega}} \dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} ~. $$ If $$\bar{\mathbf{T}}$$ is a self equilibrating traction applied on the boundary that leads to the stress field $$\boldsymbol{P}$$, i.e., $$\bar{\mathbf{T}} = \boldsymbol{P}^T\cdot\mathbf{N}$$, then we have

{ \int_{\Omega} \dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} = \int_{\partial{\Omega}} \dot{\mathbf{x}}\otimes\bar{\mathbf{T}}~\text{dA} ~. } $$ Note that the fields $$\dot{\boldsymbol{F}}$$ and $$\boldsymbol{P}$$ need not be related and hence the velocities $$\dot{\mathbf{x}}$$ and the tractions $$\bar{\mathbf{T}}$$ are not related.

If the boundary velocity field $$\dot{\mathbf{x}}$$ leads to the rate $$\dot{\boldsymbol{F}}$$, using the identity (see Appendix)

\langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle = \langle (\dot{\boldsymbol{F}} - \langle \dot{\boldsymbol{F}}\rangle)\cdot(\boldsymbol{P} - \langle \boldsymbol{P} \rangle) \rangle $$ we can show that (see Appendix)

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle\cdot\mathbf{X}]\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N} \right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle\cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N}\right\}~\text{dA} ~. \end{align} $$

Remark
Using similar arguments, if we assume that $$\boldsymbol{F}$$ is a deformation that is compatible with an applied boundary displacement $$\mathbf{u} = \mathbf{x} - \mathbf{X}$$,we can show that

\begin{align} \langle \boldsymbol{F}\cdot\boldsymbol{P} \rangle - \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\mathbf{x} - \langle \boldsymbol{F}\rangle\cdot\mathbf{X}]\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N} \right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\mathbf{x} - \langle \boldsymbol{F}\rangle\cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \mathbf{x}\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N}\right\}~\text{dA} ~. \end{align} $$

We can arrive at $$\langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle = \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle$$ or $$\langle \boldsymbol{F}\cdot\boldsymbol{P} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle$$ if either of the following conditions is satisfied at the boundary:


 * 1) $$\dot{\mathbf{x}} = \langle \dot{\boldsymbol{F}}\rangle\cdot\mathbf{X} $$ or $$\mathbf{x} = \langle \boldsymbol{F}\rangle\cdot\mathbf{X}$$.
 * 2) $$\boldsymbol{P}\cdot\mathbf{N} = \langle \boldsymbol{P} \rangle^T\cdot\mathbf{N}$$.

Linear boundary velocities/displacements
If a linear velocity field is prescribed on the boundary $$\partial{\Omega}_0$$, we can express this field as

\dot{\mathbf{x}}(\mathbf{X},t) = \dot{\boldsymbol{G}}(t)\cdot\mathbf{X} \qquad\qquad \forall \mathbf{X} \in \partial{\Omega}_0~. $$ Now,

\begin{align} \langle \dot{\boldsymbol{F}}\rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes\mathbf{N}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} (\dot{\boldsymbol{G}}\cdot\mathbf{X})\otimes\mathbf{N}~\text{dA} \\ & = \dot{\boldsymbol{G}}\cdot\left(\cfrac{1}{V_0}\int_{\partial{\Omega}_0} \mathbf{X}\otimes\mathbf{N}~\text{dA}\right)~. \end{align} $$ Recall that

\int_{\Omega_0} \boldsymbol{\nabla}_0~ \mathbf{X}~\text{dV} = \int_{\partial{\Omega}_0}\mathbf{X}\otimes\mathbf{N}~\text{dA}~. $$ Therefore,

\begin{align} \langle \dot{\boldsymbol{F}}\rangle & = \dot{\boldsymbol{G}}\cdot\left(\cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{\nabla}_0~ \mathbf{X}~\text{dA}\right) \\ & = \dot{\boldsymbol{G}}\cdot\left(\cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{\mathit{1}}~\text{dA}\right) = \dot{\boldsymbol{G}} ~. \end{align} $$ Hence,

\langle \dot{\boldsymbol{F}}\rangle = \dot{\boldsymbol{G}} \qquad\implies\qquad \dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle\cdot\mathbf{X} = \mathbf{0} ~. $$ Then,

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle\cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \dot{\boldsymbol{G}}\cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA}= \mathbf{0} \end{align} $$ Hence,

{ \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle = \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ Similarly, if a linear displacement field is prescribed on the boundary such that

\mathbf{u}(\mathbf{X},t) = \boldsymbol{G}(t)\cdot\mathbf{X} - \mathbf{X} \qquad\implies\qquad \mathbf{x}(\mathbf{X}) = \boldsymbol{G}(t)\cdot\mathbf{X} \qquad\qquad \forall \mathbf{X} \in \partial{\Omega}_0 $$ we can show that

\langle \boldsymbol{F}\rangle = \boldsymbol{G} \qquad\implies\qquad \mathbf{x} - \langle \boldsymbol{F}\rangle\cdot\mathbf{X} = \mathbf{0} ~. $$ This leads to the equality

{ \langle \boldsymbol{F}\cdot\boldsymbol{P} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ Recall that, the average Kirchhoff stress is given by $$ \langle \overline{\boldsymbol{\tau}} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle $$. Therefore, if a uniform boundary displacement is prescribed, we have

\langle \overline{\boldsymbol{\tau}} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle = \langle \boldsymbol{F}\cdot\boldsymbol{P} \rangle = \langle \boldsymbol{\tau} \rangle $$ or,

{ \langle \overline{\boldsymbol{\tau}} \rangle = \langle \boldsymbol{\tau} \rangle ~. } $$

Uniform boundary tractions
A uniform boundary traction field in the reference configuration can be represented as

\bar{\mathbf{T}}(\mathbf{X},t) = \boldsymbol{G}^T(t)\cdot\mathbf{N}(\mathbf{X}) \qquad\forall~\mathbf{X}\in\partial{\Omega}_0 ~. $$ Now,

\begin{align} \langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \mathbf{X}\otimes\bar{\mathbf{T}}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \mathbf{X}\otimes(\boldsymbol{G}^T\cdot\mathbf{N}~\text{dA} \\ & = \left(\cfrac{1}{V_0}\int_{\partial{\Omega}_0} \mathbf{X}\otimes\mathbf{N}~\text{dA}\right) \cdot\boldsymbol{G} \\ & = \boldsymbol{\mathit{1}}\cdot\boldsymbol{G} = \boldsymbol{G} ~. \end{align} $$ Since the surface tractions are related to the nominal stress by $$\bar{\mathbf{T}}(\mathbf{X},t) = \boldsymbol{P}^T(\mathbf{X},t)\cdot\mathbf{N}(\mathbf{X})$$, we must have

\langle \boldsymbol{P} \rangle = \boldsymbol{P} ~. $$ Therefore,

\langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N}\right\}~\text{dA}= \mathbf{0} $$ or,

{ \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle = \langle \dot{\boldsymbol{F}}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ Similarly,

{ \langle \boldsymbol{F}\cdot\boldsymbol{P} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ Hence, using the same argument as for the previous case, we have

{ \langle \overline{\boldsymbol{\tau}} \rangle = \langle \boldsymbol{\tau} \rangle ~. } $$