Let the average stress in the RVE be defined as
![{\displaystyle {\langle {\boldsymbol {\sigma }}\rangle :={\cfrac {1}{V}}~\int _{\Omega }{\boldsymbol {\sigma }}(\mathbf {x} )~{\text{dV}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ab834d8ce9cfc1c64702c3dd77dedf230aa93c4)
where
is the volume of
.
We would like to find out the relation between the average stress in a RVE and the applied tractions on the boundary of the RVE. To do that, recall the relation (see Appendix)
![{\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aefca55477be1bff6e3f02bf6e0d9167e7f256cd)
If we choose
such that
, we have
![{\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\mathit {1}}}\cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}=\int _{\Omega }{\boldsymbol {S}}~{\text{dV}}+\int _{\Omega }\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c926ecf10587b382e901ae1405b8b4ccd3f1e62)
Therefore,
![{\displaystyle \int _{\Omega }{\boldsymbol {S}}~{\text{dV}}=-\int _{\Omega }\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})~{\text{dV}}+\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/912fe7a723c87d5ba606e1fff73d29b0d1a93bb7)
If we choose
to be the stress tensor
, and involve the symmetry of the stress tensor, we get
![{\displaystyle \int _{\Omega }{\boldsymbol {\sigma }}~{\text{dV}}=-\int _{\Omega }\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})~{\text{dV}}+\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {\sigma }}\bullet \mathbf {n} )~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bee433b9c9128ad76a52756afc6ab825b886247a)
Now, the divergence of the stress is zero (from the conservation of linear momentum). Therefore,
![{\displaystyle \int _{\Omega }{\boldsymbol {\sigma }}~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {\sigma }}\bullet \mathbf {n} )~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e1ca7a5f30a48f0771bc59923add51100d3ae7e)
Using the traction boundary condition, we have
![{\displaystyle \int _{\Omega }{\boldsymbol {\sigma }}~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {v} \otimes {\bar {\mathbf {t} }}~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cc044c9994f31a2bb544f2646a1de415009eb1f)
Now
if
. Therefore, we have
![{\displaystyle \int _{\Omega }{\boldsymbol {\sigma }}~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b89e0269ddbde5ccb4bca581f97d735d9dcbaca)
Hence the average stress is given by
![{\displaystyle {\langle {\boldsymbol {\sigma }}\rangle :={\cfrac {1}{V}}~\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d91522737f10f5bd39da89bd01ac8500db1342d)
This implies that the average stress is completely determined by the applied tractions!
Symmetry of the average stress and the effect of rigid body translation
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Let us now assume that the applied tractions are self equilibrating. Then the resultant forces and moments due to the applied tractions vanish and we have
![{\displaystyle \int _{\partial {\Omega }}{\bar {\mathbf {t} }}~{\text{dA}}=\mathbf {0} \qquad {\text{and}}\qquad \int _{\partial {\Omega }}\mathbf {x} \times {\bar {\mathbf {t} }}~{\text{dA}}=\mathbf {0} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c5a1606bd07c8e9cb91539c2f1cbd885cb5b2b9)
From the moment balance equation above we can show that (see Appendix)
![{\displaystyle \int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}=\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\otimes \mathbf {x} ~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d2a1f396ce291f62b833e3522294beed987d1c0)
Therefore the average stress tensor
is symmetric if the applied tractions are self equilibrated.
Now, if we translate the body by a constant amount
(rigid body
translation), we get
![{\displaystyle {\bar {\langle {\boldsymbol {\sigma }}\rangle }}={\cfrac {1}{V}}~\int _{\partial {\Omega }}(\mathbf {x} +\mathbf {u} _{0})\otimes {\bar {\mathbf {t} }}~{\text{dA}}={\cfrac {1}{V}}~\int _{\partial {\Omega }}[\mathbf {x} \otimes {\bar {\mathbf {t} }}+\mathbf {u} _{0}\otimes {\bar {\mathbf {t} }}]~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/814c2575d1f60be1fb29d68a37d9f20f1b45b60e)
or
![{\displaystyle {\bar {\langle {\boldsymbol {\sigma }}\rangle }}={\cfrac {1}{V}}\left[\int _{\partial {\Omega }}\mathbf {x} \otimes {\bar {\mathbf {t} }}~{\text{dA}}+\mathbf {u} _{0}\otimes \int _{\partial {\Omega }}{\bar {\mathbf {t} }}~{\text{dA}}\right]=\langle {\boldsymbol {\sigma }}\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/814362387ced3a12960fc2aa94d80c2e81cda688)
Therefore, the average stress is not affected by a rigid body translation only if the applied tractions are self equilibrated.
We can conclude that the average stress
is an acceptable measure of stress in a RVE if the applied tractions are self equilibrated.