The equation for the balance of energy is
![{\displaystyle \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )+{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc70611bc17203808b8f11583f8fdeb33b33f467)
If the absence of heat flux or heat sources in the RVE, the equation reduces to
![{\displaystyle \rho ~{\dot {e}}={\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5026df323c16ac2c50769b6ebf3540287d016f)
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
![{\displaystyle {\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e29bd6ac8737cfb144448e7792e909f6b500e45)
Note that the quantities
and
need not be related in the general case.
The average velocity gradient
is defined as
![{\displaystyle {\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\mathbf {v} ~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef7074cf0eb17c259f0f6198ec9faa4b21e523c3)
To get an expression for the average stress power in terms of the boundary conditions, we use the identity
![{\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\cdot \mathbf {v} )={\boldsymbol {S}}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})\cdot \mathbf {v} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fa02849d1fe02323a9ee291ed7f88e3aa2a87ba)
to get
![{\displaystyle \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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d22cf3137b65175518e8a533278573492809d469)
Using the balance of linear momentum (
), we get
![{\displaystyle \langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97abeb1a400dc2a5161640c424fee89d4844758c)
Using the divergence theorem, we have
![{\displaystyle \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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1019a796c5f9505a8d3e231b77d800b1a7ae9e9)
Now, the surface traction is given by
. Therefore,
![{\displaystyle {\langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot \mathbf {v} ~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4cbde2391f92f9f9d56d77671892f6f5be099c)
{\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
and the average velocity gradient
. While homogenizing a RVE, we would ideally like to have
![{\displaystyle \langle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \rangle =\langle {\boldsymbol {\sigma }}\rangle :\langle {\boldsymbol {\nabla }}\mathbf {v} \rangle ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b4301385081ff02ac81ab327bc3a058d416d386)
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)
![{\displaystyle {\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}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1440a30417998062deb25a8876f186cdfa719968)
We can arrive at
if either of the following
conditions is met on the boundary
:
~.
~.
If the prescribed velocities on
are a linear function of
, then we can write
![{\displaystyle \mathbf {v} (\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {x} \qquad \qquad \forall ~\mathbf {x} \in \partial {\Omega }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6acf278808a173302eb891e1443b12a2de43bb8f)
where
is a constant second-order tensor.
From the divergence theorem
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\mathbf {a} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {a} \otimes \mathbf {n} ~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01bb5971444ab2efe8b1a657aecd2c6a74958541)
Therefore,
![{\displaystyle \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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/352f4d3b2f6393b20fb33739263ccd6059e4b8a2)
Hence, on the boundary
![{\displaystyle \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} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f0551303921fd242b12f53cf5aedda1bb5693c4)
Using the identity (see Appendix)
![{\displaystyle ({\boldsymbol {A}}\cdot \mathbf {a} )\otimes \mathbf {b} ={\boldsymbol {A}}\cdot (\mathbf {a} \otimes \mathbf {b} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10ff9ec7f0e010fc900847bd48c340d97f1c6a83)
and since
is constant, we get
![{\displaystyle \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} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93c7fe9e1e1b7bdd7e2ca9f632d2e0fc4bc0da38)
From the divergence theorem,
![{\displaystyle \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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2df7188712d0523151345ac849dda0b277fa8716)
Therefore,
![{\displaystyle \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 }~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ef316cc6924e11ab9deb3e12ea81e0f52efaa67)
If the prescribed tractions on the boundary
are uniform,
they can be expressed in terms of a constant symmetric second-order tensor
through the relation
![{\displaystyle {\bar {t}}(\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {n} (\mathbf {x} )\qquad \qquad \forall ~\mathbf {x} \in \partial {\Omega }~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5684f4bfd47703ced404399a5d481ae6557d0c48)
The tractions are related to the stresses at the boundary of the RVE by
.
The average stress in the RVE is given by
![{\displaystyle \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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef056ebb1cacb7d2f2ff9242cfebd8775a337f4d)
Using the identity
(see Appendix),
we have
![{\displaystyle \langle {\boldsymbol {\sigma }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}(\mathbf {x} \otimes \mathbf {n} )\cdot {\boldsymbol {G}}^{T}~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9e9f18c415cbe0367fe8fba4e13a4cf6071a102)
Since
is constant and symmetric, we have
![{\displaystyle \langle {\boldsymbol {\sigma }}\rangle =\left({\cfrac {1}{V}}\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {n} ~{\text{dA}}\right)\cdot {\boldsymbol {G}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24d8866c4afcf7a0a6ac4e32092fa302f5754e55)
Applying the divergence theorem,
![{\displaystyle \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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/284410ebe0138ecb2b8508925fff177ff09e03d0)
Therefore,
![{\displaystyle {\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 }~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8459a0ded792fe1df2e2b49b9b41758bba35f9d)
Recall that for small deformations, the displacement gradient
can be expressed as
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {u} ={\boldsymbol {\varepsilon }}+{\boldsymbol {\omega }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e24a6b40236ec2681f595faae8ebaa8317c78202)
For small deformations, the time derivative of
gives us the velocity gradient
, i.e.,
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\dot {\boldsymbol {\varepsilon }}}+{\dot {\boldsymbol {\omega }}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be1bca15e5b8c2fc5ab72e12bc49717afc9c6a16)
If
, we get
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\dot {\boldsymbol {\varepsilon }}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c174a949ce26e657850a0dc935c6ddde6bb645a7)
Hence, for small strains and in the absence of rigid body rotations, the stress power density is given by
. Then the average stress power is defined as
![{\displaystyle {\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50cbe9304a953d59aaa8c285a4c33dae936e019b)
and the average strain rate is defined as
![{\displaystyle {\langle {\dot {\boldsymbol {\varepsilon }}}\rangle :={\cfrac {1}{V}}\int _{\Omega }{\dot {\boldsymbol {\varepsilon }}}~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b43fb2d9f1da3e16d6e56e620739169da956e28)
In terms of the surface tractions and the applied boundary velocities, we have
![{\displaystyle {\langle {\boldsymbol {\sigma }}:{\dot {\boldsymbol {\varepsilon }}}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot {\dot {\mathbf {u} }}~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ef46ad1d81795c938f5c1850c612183e2b2fc1)
For small strains and no rotation, the stress-power difference relation becomes
![{\displaystyle {\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}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7aa0ed2e928f825db24d9631681911d1cbe448c)
We can arrive at
if either of the following conditions is met on the boundary
:
Linear boundary velocity field.
Uniform boundary tractions.
We can also show in an identical manner that
![{\displaystyle {\langle {\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}\rangle ={\cfrac {1}{V}}\int _{\partial {\Omega }}{\bar {\mathbf {t} }}\cdot \mathbf {u} ~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1074d878c32627e42175e6568c9f6acf2c068d96)
and that, when
is symmetric,
![{\displaystyle {\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}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7ae7768c9861cfedcfcbed92e15284113b4a6fc)
In this case, we can arrive at the relation
if either of the following conditions is met at the boundary:
Linear boundary displacement field.
Uniform boundary tractions.