# Micromechanics of composites/Proof 13
## Question

Let ${\boldsymbol {\sigma }}$ be the Cauchy stress and let ${\boldsymbol {\nabla }}\mathbf {v}$ be the velocity gradient in a body $\Omega$ with boundary $\partial {\Omega }$. Let $\mathbf {n}$ be the normal to the boundary. Let $V$ be the volume of the body. If the skew-symmetric part of the velocity gradient is zero, i.e., ${\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {\nabla }}\mathbf {v} ^{T}$, or if the stress field is self equilibrated, i.e., $\langle {\boldsymbol {\sigma }}\rangle =\langle {\boldsymbol {\sigma }}\rangle ^{T}$, show that
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}[\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot [({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]~{\text{dA}}~.$

## Proof

Taking the trace of each term in the identity
    $\langle {\boldsymbol {A}}\cdot {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle =\langle [{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]\rangle$
the difference between the average stress power and the product of the average stress and the average velocity gradient can be written as (using either the symmetry of the stress or of the velocity gradient)
    ${\begin{aligned}\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle &=\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :\langle {\boldsymbol {\sigma }}\rangle +\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :\langle {\boldsymbol {\sigma }}\rangle \\&=\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :\langle {\boldsymbol {\sigma }}\rangle +[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \langle {\boldsymbol {\sigma }}\rangle ]:{\boldsymbol {\mathit {1}}}\end{aligned}}$
Recall that
    $\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {n} )\cdot \mathbf {v} ~{\text{dV}}~;~~\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}~;~~\langle {\boldsymbol {\sigma }}\rangle ={\cfrac {1}{V}}~\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}~;~~{\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {x} ~{\text{dV}}={\boldsymbol {\mathit {1}}}~.$
Also, from the divergence theorem
    $\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {v} \otimes \mathbf {n} ~{\text{dA}}~;~~\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {x} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}~.$
Therefore,
    ${\begin{aligned}\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle &={\cfrac {1}{V}}\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {n} )\cdot \mathbf {v} ~{\text{dV}}-\langle {\boldsymbol {\sigma }}\rangle :\left[{\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {v} \otimes \mathbf {n} ~{\text{dA}}\right]-\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :\left[{\cfrac {1}{V}}~\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}\right]\\&\qquad \qquad +[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \langle {\boldsymbol {\sigma }}\rangle ]:\left[{\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}\right]~.\end{aligned}}$
Since $\langle {\boldsymbol {\sigma }}\rangle$ and $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle$ are independent of $\mathbf {x}$, we can take these inside the integrals to get
    ${\begin{aligned}\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle &={\cfrac {1}{V}}\int _{\partial {\Omega }}\left[({\boldsymbol {\sigma }}\cdot \mathbf {n} )\cdot \mathbf {v} -\langle {\boldsymbol {\sigma }}\rangle :(\mathbf {v} \otimes \mathbf {n} )-\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :(\mathbf {x} \otimes {\bar {\mathbf {t} }})+[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \langle {\boldsymbol {\sigma }}\rangle ]:(\mathbf {x} \otimes \mathbf {n} )\right]~{\text{dA}}\end{aligned}}$
Using the identity
    $({\boldsymbol {A}}\cdot \mathbf {a} )\cdot ({\boldsymbol {B}}\cdot \mathbf {b} )=({\boldsymbol {A}}^{T}\cdot {\boldsymbol {B}}):(\mathbf {a} \otimes \mathbf {b} )$
we get
    $[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \langle {\boldsymbol {\sigma }}\rangle ]:(\mathbf {x} \otimes \mathbf {n} )=[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot [\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} ]~.$
Also, using the identity
    ${\boldsymbol {A}}:(\mathbf {a} \otimes \mathbf {b} )=({\boldsymbol {A}}\cdot \mathbf {b} )\cdot \mathbf {a}$
we get
    $\langle {\boldsymbol {\sigma }}\rangle :(\mathbf {v} \otimes \mathbf {n} )=[\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} ]\cdot \mathbf {v} ~;~~\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :(\mathbf {x} \otimes {\bar {\mathbf {t} }})=[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot {\bar {\mathbf {t} }}]\cdot \mathbf {x} =[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \mathbf {x} ]\cdot {\bar {\mathbf {t} }}=[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}\cdot \mathbf {x} ]\cdot ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~.$
Since ${\boldsymbol {\nabla }}\mathbf {v} ^{T}={\boldsymbol {\nabla }}\mathbf {v}$, we have $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ^{T}=\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle$ (we could alternatively use the symmetry of $\langle {\boldsymbol {\sigma }}\rangle$ to arrive at the following equation). Hence,
    $\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :(\mathbf {x} \otimes {\bar {\mathbf {t} }})=[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~.$
Plugging these back into the original equation, we have
    ${\begin{aligned}\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle &={\cfrac {1}{V}}\int _{\partial {\Omega }}\left\{({\boldsymbol {\sigma }}\cdot \mathbf {n} )\cdot \mathbf {v} -[\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} ]\cdot \mathbf {v} -[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot ({\boldsymbol {\sigma }}\cdot \mathbf {n} )+[\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot [\langle {\boldsymbol {\sigma }}\rangle \cdot \mathbf {n} ]\right\}~{\text{dA}}\\&={\cfrac {1}{V}}\int _{\partial {\Omega }}\left\{[({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]\cdot \mathbf {v} -[({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]\cdot (\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} )\right\}~{\text{dA}}\\&={\cfrac {1}{V}}\int _{\partial {\Omega }}\left\{[({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]\cdot [\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\right\}~{\text{dA}}~.\end{aligned}}$
Hence
    ${\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle -\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}\left[[\mathbf {v} -\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle \cdot \mathbf {x} ]\cdot [({\boldsymbol {\sigma }}-\langle {\boldsymbol {\sigma }}\rangle )\cdot \mathbf {n} ]\right]~{\text{dA}}~.}\qquad \qquad \qquad \qquad \square$