# Micromechanics of composites/Average stress power in a RVE
## Average Stress Power in a RVE

The equation for the balance of energy is
    $\rho ~{\dot {e}}-{\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )+{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.$
If the absence of heat flux or heat sources in the RVE, the equation reduces to
    $\rho ~{\dot {e}}={\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )~.$
The quantity on the right is the stress power density and is a measure of the internal energy density of the material.
The average stress power in a RVE is defined as
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}~.}$
Note that the quantities ${\boldsymbol {\sigma }}$ and ${\boldsymbol {\nabla }}\mathbf {v}$ need not be related in the general case.
The average velocity gradient $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle$ is defined as
    ${\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}~.}$
To get an expression for the average stress power in terms of the boundary conditions, we use the identity
    ${\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\cdot \mathbf {v} )={\boldsymbol {S}}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})\cdot \mathbf {v}$
to get
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}={\cfrac {1}{V}}\int _{\Omega }\left[{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )-({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\cdot \mathbf {v} \right]~{\text{dV}}~.$
Using the balance of linear momentum (${\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}=0$), we get
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )~{\text{dV}}~.$
Using the divergence theorem, we have
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )\cdot \mathbf {n} ~{\text{dV}}={\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )\cdot \mathbf {n} ~{\text{dV}}={\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {n} )\cdot \mathbf {v} ~{\text{dV}}~.$
Now, the surface traction is given by ${\bar {\mathbf {t} }}={\boldsymbol {\sigma }}\cdot \mathbf {n}$. Therefore,
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot \mathbf {v} ~{\text{dV}}~.}$
{\scriptsize } In micromechanics, it is of interest to see how the average stress power of a RVE is related to the product of the average stress $\langle {\boldsymbol {\sigma }}\rangle$ and the average velocity gradient $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle$. While homogenizing a RVE, we would ideally like to have
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ~.$
However, this is not true in general. We can show that if the gradient of the velocity is a symmetric tensor (i.e., there is no spin), then (see Appendix for proof)
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}[\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot [({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]~{\text{dA}}~.}$
We can arrive at $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle$ if either of the following conditions is met on the boundary $\partial {\Omega }$:
* $\mathbf {v} =\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x}$ ~.
* ${\boldsymbol {\sigma }}\cdot \mathbf {n} =\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n}$ ~.

#### Linear boundary velocities

If the prescribed velocities on $\partial {\Omega }$ are a linear function of $\mathbf {x}$, then we can write
    $\mathbf {v} (\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {x} \qquad \qquad \forall ~\mathbf {x} \in \partial {\Omega }$
where ${\boldsymbol {G}}$ is a constant second-order tensor.
From the divergence theorem
    $\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {a} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {a} \otimes \mathbf {n} ~{\text{dA}}~.$
Therefore,
    $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}={\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {v} \otimes \mathbf {n} ~{\text{dA}}~.$
Hence, on the boundary
    $\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ={\boldsymbol {G}}\cdot \mathbf {x} -\left[{\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {G}}\cdot \mathbf {x} )\otimes \mathbf {n} ~{\text{dA}}\right]\cdot \mathbf {x}$
Using the identity (see Appendix)
    $({\boldsymbol {A}}\cdot \mathbf {a} )\otimes \mathbf {b} ={\boldsymbol {A}}\cdot (\mathbf {a} \otimes \mathbf {b} )$
and since ${\boldsymbol {G}}$ is constant, we get
    $\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \mathbf {x} ={\boldsymbol {G}}\cdot \mathbf {x} -\left[{\boldsymbol {G}}\cdot \left({\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}\right)\right]\cdot \mathbf {x} ~.$
From the divergence theorem,
    $\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}=\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {x} ~{\text{dV}}=\int _{\Omega }{\boldsymbol {\mathit {1}}}~{\text{dV}}=V~{\boldsymbol {\mathit {1}}}~.$
Therefore,
    $\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ={\boldsymbol {G}}\cdot \mathbf {x} -({\boldsymbol {G}}\cdot {\boldsymbol {\mathit {1}}})\cdot \mathbf {x} ={\boldsymbol {G}}\cdot \mathbf {x} -{\boldsymbol {G}}\cdot \mathbf {x} =\mathbf {0} \qquad \implies \qquad {\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle }~.$

#### Uniform boundary tractions

If the prescribed tractions on the boundary $\partial {\Omega }$ are uniform, they can be expressed in terms of a constant symmetric second-order tensor ${\boldsymbol {G}}$ through the relation
    ${\bar {t}}(\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {n} (\mathbf {x} )\qquad \qquad \forall ~\mathbf {x} \in \partial {\Omega }~.$
The tractions are related to the stresses at the boundary of the RVE by ${\bar {t}}={\boldsymbol {\sigma }}\cdot \mathbf {n}$.
The average stress in the RVE is given by
    $\langle {\boldsymbol {\sigma }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {t}}~{\text{dA}}={\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes ({\boldsymbol {G}}\cdot \mathbf {n} )~{\text{dA}}~.$
Using the identity $\mathbf {a} \otimes ({\boldsymbol {A}}\cdot \mathbf {b} )=(\mathbf {a} \otimes \mathbf {b} )\cdot {\boldsymbol {A}}^{T}$ (see Appendix), we have
    $\langle {\boldsymbol {\sigma }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}(\mathbf {x} \otimes \mathbf {n} )\cdot {\boldsymbol {G}}^{T}~{\text{dA}}~.$
Since ${\boldsymbol {G}}$ is constant and symmetric, we have
    $\langle {\boldsymbol {\sigma }}\rangle =\left({\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}\right)\cdot {\boldsymbol {G}}~.$
Applying the divergence theorem,
    $\langle {\boldsymbol {\sigma }}\rangle =\left({\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {x} ~{\text{dV}}\right)\cdot {\boldsymbol {G}}={\boldsymbol {\mathit {1}}}\cdot {\boldsymbol {G}}={\boldsymbol {G}}~.$
Therefore,
    ${\boldsymbol {\sigma }}\cdot \mathbf {n} -\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} ={\bar {t}}-{\boldsymbol {G}}\cdot \mathbf {n} =\mathbf {0} \qquad \implies \qquad {\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle }~.$

#### Remark

Recall that for small deformations, the displacement gradient ${\boldsymbol {\nabla }}\mathbf {u}$ can be expressed as
    ${\boldsymbol {\nabla }}\mathbf {u} ={\boldsymbol {\varepsilon }}+{\boldsymbol {\omega }}~.$
For small deformations, the time derivative of ${\boldsymbol {\nabla }}\mathbf {u}$ gives us the velocity gradient ${\boldsymbol {\nabla }}\mathbf {v}$, i.e.,
    ${\boldsymbol {\nabla }}\mathbf {v} ={\dot {\boldsymbol {\varepsilon }}}+{\dot {\boldsymbol {\omega }}}~.$
If ${\boldsymbol {\omega }}=0$, we get
    ${\boldsymbol {\nabla }}\mathbf {v} ={\dot {\boldsymbol {\varepsilon }}}~.$
Hence, for small strains and in the absence of rigid body rotations, the stress power density is given by ${\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}$. Then the average stress power is defined as
    ${\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}~{\text{dV}}~.}$
and the average strain rate is defined as
    ${\langle {\dot {\boldsymbol {\varepsilon }}}\rangle :={\cfrac {1}{V}}\int _{\Omega }{\dot {\boldsymbol {\varepsilon }}}~{\text{dV}}~.}$
In terms of the surface tractions and the applied boundary velocities, we have
    ${\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot {\dot {\mathbf {u} }}~{\text{dV}}~.}$
For small strains and no rotation, the stress-power difference relation becomes
    ${\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\dot {\boldsymbol {\varepsilon }}}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}[{\dot {\mathbf {u} }}-\langle {\boldsymbol {\nabla }}{\dot {\mathbf {u} }}\rangle \cdot \mathbf {x} ]\cdot [({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]~{\text{dA}}~.}$
We can arrive at $\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\dot {\boldsymbol {\varepsilon }}}\rangle$ if either of the following conditions is met on the boundary $\partial {\Omega }$:
* ${\dot {\mathbf {u} }}=\langle {\boldsymbol {\nabla }}{\dot {\mathbf {u} }}\rangle \cdot \mathbf {x} \qquad \implies \qquad$ Linear boundary velocity field.
* ${\boldsymbol {\sigma }}\cdot \mathbf {n} =\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} \qquad \implies \qquad$ Uniform boundary tractions.
We can also show in an identical manner that
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot \mathbf {u} ~{\text{dV}}~.}$
and that, when ${\boldsymbol {\nabla }}\mathbf {u}$ is symmetric,
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}\rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\varepsilon }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}[\mathbf {u} -\langle {\boldsymbol {\nabla }}\mathbf {u} \rangle \cdot \mathbf {x} ]\cdot [({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]~{\text{dA}}~.}$
In this case, we can arrive at the relation $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}\rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\varepsilon }}\rangle$ if either of the following conditions is met at the boundary:
* $\mathbf {u} =\langle {\boldsymbol {\nabla }}\mathbf {u} \rangle \cdot \mathbf {x} \qquad \implies \qquad$ Linear boundary displacement field.
* ${\boldsymbol {\sigma }}\cdot \mathbf {n} =\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} \qquad \implies \qquad$ Uniform boundary tractions.