# 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)

${\displaystyle \rho _{0}~{\dot {e}}={\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}-{\boldsymbol {\nabla }}_{0}\bullet \mathbf {q} +\rho _{0}~s~.}$

The quantity ${\displaystyle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}}$ is the stress power.

The average stress power is defined as

${\displaystyle {\langle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\rangle :={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}~{\text{dV}}~.}}$

Here ${\displaystyle {\boldsymbol {P}}^{T}}$ is an arbitrary self-equilibrating nominal stress field that satisfies the balance of momentum (without any body forces or inertial forces) and ${\displaystyle {\dot {\boldsymbol {F}}}}$ is the time rate of change of ${\displaystyle {\boldsymbol {F}}}$. The reference configuration can be arbitrary. Also, the nominal stress and the rate ${\displaystyle {\dot {\boldsymbol {F}}}}$ need not be related.

Note that in that case

${\displaystyle {{\text{tr}}(\langle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\rangle )={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\text{tr}}({\boldsymbol {P}}\cdot {\dot {\boldsymbol {F}}})~{\text{dV}}={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\text{tr}}({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}})~{\text{dV}}={\text{tr}}(\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle )~.}}$

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

${\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}~.}$

In this case, we have ${\displaystyle \partial {\Omega }\rightarrow \partial {\Omega }_{0}}$, ${\displaystyle \Omega \rightarrow \Omega _{0}}$, ${\displaystyle {\boldsymbol {\nabla }}\rightarrow {\boldsymbol {\nabla }}_{0}}$, ${\displaystyle \mathbf {v} \rightarrow {\dot {\mathbf {x} }}}$, ${\displaystyle {\boldsymbol {S}}\rightarrow {\boldsymbol {P}}}$, and ${\displaystyle \mathbf {n} \rightarrow \mathbf {N} }$. Then

${\displaystyle \int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}\cdot {\boldsymbol {P}}+{\dot {\mathbf {x} }}\otimes ({\boldsymbol {\nabla }}_{0}\bullet {\boldsymbol {P}}^{T})]~{\text{dV}}~.}$

Using the balance of linear momentum (in the absence of body and inertial forces), we get

${\displaystyle \int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}=\int _{\Omega }{\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}\cdot {\boldsymbol {P}}~{\text{dV}}~.}$

Recalling that

${\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial }{\partial t}}\left({\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}\right)={\frac {\partial }{\partial \mathbf {X} }}\left({\frac {\partial \mathbf {x} }{\partial t}}\right)={\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}}$

we then have

${\displaystyle \int _{\Omega }{\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}~{\text{dV}}=\int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}~.}$

If ${\displaystyle {\bar {\mathbf {T} }}}$ is a self equilibrating traction applied on the boundary that leads to the stress field ${\displaystyle {\boldsymbol {P}}}$, i.e., ${\displaystyle {\bar {\mathbf {T} }}={\boldsymbol {P}}^{T}\cdot \mathbf {N} }$, then we have

${\displaystyle {\int _{\Omega }{\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}~{\text{dV}}=\int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes {\bar {\mathbf {T} }}~{\text{dA}}~.}}$

Note that the fields ${\displaystyle {\dot {\boldsymbol {F}}}}$ and ${\displaystyle {\boldsymbol {P}}}$ need not be related and hence the velocities ${\displaystyle {\dot {\mathbf {x} }}}$ and the tractions ${\displaystyle {\bar {\mathbf {T} }}}$ are not related.

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

${\displaystyle \langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle =\langle ({\dot {\boldsymbol {F}}}-\langle {\dot {\boldsymbol {F}}}\rangle )\cdot ({\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle )\rangle }$

we can show that (see Appendix)

{\displaystyle {\begin{aligned}\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{aligned}}}

## Remark

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

{\displaystyle {\begin{aligned}\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{aligned}}}

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

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

## Linear boundary velocities/displacements

If a linear velocity field is prescribed on the boundary ${\displaystyle \partial {\Omega }_{0}}$, we can express this field as

${\displaystyle {\dot {\mathbf {x} }}(\mathbf {X} ,t)={\dot {\boldsymbol {G}}}(t)\cdot \mathbf {X} \qquad \qquad \forall \mathbf {X} \in \partial {\Omega }_{0}~.}$

Now,

{\displaystyle {\begin{aligned}\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{aligned}}}

Recall that

${\displaystyle \int _{\Omega _{0}}{\boldsymbol {\nabla }}_{0}~\mathbf {X} ~{\text{dV}}=\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes \mathbf {N} ~{\text{dA}}~.}$

Therefore,

{\displaystyle {\begin{aligned}\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{aligned}}}

Hence,

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

Then,

{\displaystyle {\begin{aligned}\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{aligned}}}

Hence,

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

Similarly, if a linear displacement field is prescribed on the boundary such that

${\displaystyle \mathbf {u} (\mathbf {X} ,t)={\boldsymbol {G}}(t)\cdot \mathbf {X} -\mathbf {X} \qquad \implies \qquad \mathbf {x} (\mathbf {X} )={\boldsymbol {G}}(t)\cdot \mathbf {X} \qquad \qquad \forall \mathbf {X} \in \partial {\Omega }_{0}}$

we can show that

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

${\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}$

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

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

or,

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

## Uniform boundary tractions

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

${\displaystyle {\bar {\mathbf {T} }}(\mathbf {X} ,t)={\boldsymbol {G}}^{T}(t)\cdot \mathbf {N} (\mathbf {X} )\qquad \forall ~\mathbf {X} \in \partial {\Omega }_{0}~.}$

Now,

{\displaystyle {\begin{aligned}\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{aligned}}}

Since the surface tractions are related to the nominal stress by ${\displaystyle {\bar {\mathbf {T} }}(\mathbf {X} ,t)={\boldsymbol {P}}^{T}(\mathbf {X} ,t)\cdot \mathbf {N} (\mathbf {X} )}$, we must have

${\displaystyle \langle {\boldsymbol {P}}\rangle ={\boldsymbol {P}}~.}$

Therefore,

${\displaystyle \langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\dot {\mathbf {x} }}\otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}=\mathbf {0} }$

or,

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

Similarly,

${\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}$

Hence, using the same argument as for the previous case, we have

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