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,