Recall the equation for the balance of energy (with respect to the reference
configuration)
![{\displaystyle \rho _{0}~{\dot {e}}={\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}-{\boldsymbol {\nabla }}_{0}\bullet \mathbf {q} +\rho _{0}~s~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9be15e8447b04c6f9c200e6fb980f81652c1c24)
The quantity
is the stress power.
The average stress power is defined as
![{\displaystyle {\langle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\rangle :={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}~{\text{dV}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/286f647be52686f5a32c38a84a3835e32f5ada61)
Here
is an arbitrary self-equilibrating nominal stress field
that satisfies the balance of momentum (without any body forces or
inertial forces) and
is the time rate of change of
.
The reference configuration can be arbitrary. Also, the nominal stress
and the rate
need not be related.
Note that in that case
![{\displaystyle {{\text{tr}}(\langle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\rangle )={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\text{tr}}({\boldsymbol {P}}\cdot {\dot {\boldsymbol {F}}})~{\text{dV}}={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\text{tr}}({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}})~{\text{dV}}={\text{tr}}(\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle )~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a7bf102228f5e1a1bcb114b538133f470855989)
We can express the stress power in terms of boundary tractions and
boundary velocities using the relation (see Appendix)
![{\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \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/c526aaf5e74863755e2cc1709c003642c21894e7)
In this case, we have
,
,
,
,
, and
. Then
![{\displaystyle \int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}\cdot {\boldsymbol {P}}+{\dot {\mathbf {x} }}\otimes ({\boldsymbol {\nabla }}_{0}\bullet {\boldsymbol {P}}^{T})]~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b8c328c31c0f60cc0b4d43fe87d088325563393)
Using the balance of linear momentum (in the absence of body and inertial
forces), we get
![{\displaystyle \int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}=\int _{\Omega }{\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}\cdot {\boldsymbol {P}}~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/189c8cc421879b85d524959713e3879a6fc809bd)
Recalling that
![{\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial }{\partial t}}\left({\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}\right)={\frac {\partial }{\partial \mathbf {X} }}\left({\frac {\partial \mathbf {x} }{\partial t}}\right)={\boldsymbol {\nabla }}_{0}~{\dot {\mathbf {x} }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4334a7ea2cc9bb7a55d98b9792ef3381c58f853e)
we then have
![{\displaystyle \int _{\Omega }{\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}~{\text{dV}}=\int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/770125a9f589731c3a7db5853d4569f509407b58)
If
is a self equilibrating traction applied on the boundary
that leads to the stress field
, i.e.,
,
then we have
![{\displaystyle {\int _{\Omega }{\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}~{\text{dV}}=\int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes {\bar {\mathbf {T} }}~{\text{dA}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e58b0e25a33ab6341b6fc5416357573b29ae5393)
Note that the fields
and
need not be related and hence the velocities
and the tractions
are not related.
If the boundary velocity field
leads to the rate
, using the identity (see Appendix)
![{\displaystyle \langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle =\langle ({\dot {\boldsymbol {F}}}-\langle {\dot {\boldsymbol {F}}}\rangle )\cdot ({\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle )\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4dd3d4b5386dc142305cedfd19dddd91aa6ad78)
we can show that (see Appendix)
![{\displaystyle {\begin{aligned}\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle &={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[{\dot {\mathbf {x} }}-\langle {\dot {\boldsymbol {F}}}\rangle \cdot \mathbf {X} ]\otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[{\dot {\mathbf {x} }}-\langle {\dot {\boldsymbol {F}}}\rangle \cdot \mathbf {X} ]\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\dot {\mathbf {x} }}\otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e391e8782b16ba0291b257e0fa3f733ec1542284)
Using similar arguments, if we assume that
is a deformation that is compatible with an applied boundary displacement
,we can show that
![{\displaystyle {\begin{aligned}\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle -\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle &={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[\mathbf {x} -\langle {\boldsymbol {F}}\rangle \cdot \mathbf {X} ]\otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[\mathbf {x} -\langle {\boldsymbol {F}}\rangle \cdot \mathbf {X} ]\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}\mathbf {x} \otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/942ce4eae345ada758edfe58c4fd8186773f0794)
We can arrive at
or
if either of
the following conditions is satisfied at the boundary:
or
.
.
If a linear velocity field is prescribed on the boundary
,
we can express this field as
![{\displaystyle {\dot {\mathbf {x} }}(\mathbf {X} ,t)={\dot {\boldsymbol {G}}}(t)\cdot \mathbf {X} \qquad \qquad \forall \mathbf {X} \in \partial {\Omega }_{0}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c98709e08d5dec80a3facc762b15e2a6e28beca6)
Now,
![{\displaystyle {\begin{aligned}\langle {\dot {\boldsymbol {F}}}\rangle &={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\dot {\mathbf {x} }}\otimes \mathbf {N} ~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}({\dot {\boldsymbol {G}}}\cdot \mathbf {X} )\otimes \mathbf {N} ~{\text{dA}}\\&={\dot {\boldsymbol {G}}}\cdot \left({\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes \mathbf {N} ~{\text{dA}}\right)~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b034d31a5935a95a616036d4be420d183852f79d)
Recall that
![{\displaystyle \int _{\Omega _{0}}{\boldsymbol {\nabla }}_{0}~\mathbf {X} ~{\text{dV}}=\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes \mathbf {N} ~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42f09a255783b6ab09f7512af4552b6111e173fa)
Therefore,
![{\displaystyle {\begin{aligned}\langle {\dot {\boldsymbol {F}}}\rangle &={\dot {\boldsymbol {G}}}\cdot \left({\cfrac {1}{V_{0}}}\int _{\Omega _{0}}{\boldsymbol {\nabla }}_{0}~\mathbf {X} ~{\text{dA}}\right)\\&={\dot {\boldsymbol {G}}}\cdot \left({\cfrac {1}{V_{0}}}\int _{\Omega _{0}}{\boldsymbol {\mathit {1}}}~{\text{dA}}\right)={\dot {\boldsymbol {G}}}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cf1f60778d4d33193b9d3fcb830f7c434b6ef67)
Hence,
![{\displaystyle \langle {\dot {\boldsymbol {F}}}\rangle ={\dot {\boldsymbol {G}}}\qquad \implies \qquad {\dot {\mathbf {x} }}-\langle {\dot {\boldsymbol {F}}}\rangle \cdot \mathbf {X} =\mathbf {0} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2bb78b2333422bb2e05a58d723ebd60e9463e04)
Then,
![{\displaystyle {\begin{aligned}\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle &={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[{\dot {\mathbf {x} }}-\langle {\dot {\boldsymbol {F}}}\rangle \cdot \mathbf {X} ]\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}[{\dot {\mathbf {x} }}-{\dot {\boldsymbol {G}}}\cdot \mathbf {X} ]\otimes ({\boldsymbol {P}}^{T}\cdot \mathbf {N} )~{\text{dA}}=\mathbf {0} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b63f2b49a3c0d48e5469b9ed70a4b990be83557a)
Hence,
![{\displaystyle {\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle =\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7318b429c7b17e64d4a846adf1456f44b92b5422)
Similarly, if a linear displacement field is prescribed on the
boundary such that
![{\displaystyle \mathbf {u} (\mathbf {X} ,t)={\boldsymbol {G}}(t)\cdot \mathbf {X} -\mathbf {X} \qquad \implies \qquad \mathbf {x} (\mathbf {X} )={\boldsymbol {G}}(t)\cdot \mathbf {X} \qquad \qquad \forall \mathbf {X} \in \partial {\Omega }_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ce5392bce7e63a57d8583839c2af7d205c5fd3a)
we can show that
![{\displaystyle \langle {\boldsymbol {F}}\rangle ={\boldsymbol {G}}\qquad \implies \qquad \mathbf {x} -\langle {\boldsymbol {F}}\rangle \cdot \mathbf {X} =\mathbf {0} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f514b8b4017d39f90ed8a4a8a609c1a4e605034)
This leads to the equality
![{\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0df70eaebb85dcc2c55bd81a04de2d8cb2476a31)
Recall that, the average Kirchhoff stress is given by
.
Therefore, if a uniform boundary displacement is prescribed, we
have
![{\displaystyle \langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {\tau }}\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/64eb99e30b9a8fcee84e72ca08ebd074f483a8b1)
or,
![{\displaystyle {\langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {\tau }}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f9b10f2f88451c87fcc56a911ecebc3a6817123)
A uniform boundary traction field in the reference configuration can be represented as
![{\displaystyle {\bar {\mathbf {T} }}(\mathbf {X} ,t)={\boldsymbol {G}}^{T}(t)\cdot \mathbf {N} (\mathbf {X} )\qquad \forall ~\mathbf {X} \in \partial {\Omega }_{0}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b0626ca132c6cac90472adfce41971949170bdb)
Now,
![{\displaystyle {\begin{aligned}\langle {\boldsymbol {P}}\rangle &={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes {\bar {\mathbf {T} }}~{\text{dA}}\\&={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes ({\boldsymbol {G}}^{T}\cdot \mathbf {N} ~{\text{dA}}\\&=\left({\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes \mathbf {N} ~{\text{dA}}\right)\cdot {\boldsymbol {G}}\\&={\boldsymbol {\mathit {1}}}\cdot {\boldsymbol {G}}={\boldsymbol {G}}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/340a834298088cec2d5f7d8ea6d0001684b61ea2)
Since the surface tractions are related to the nominal stress by
, we must have
![{\displaystyle \langle {\boldsymbol {P}}\rangle ={\boldsymbol {P}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee02074293eb665e438acfc38dda8956b850d6ef)
Therefore,
![{\displaystyle \langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle -\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ={\cfrac {1}{V_{0}}}\int _{\partial {\Omega }_{0}}{\dot {\mathbf {x} }}\otimes \left\{[{\boldsymbol {P}}-\langle {\boldsymbol {P}}\rangle ]^{T}\cdot \mathbf {N} \right\}~{\text{dA}}=\mathbf {0} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/966ca46809534424457f51b9e940b67140551862)
or,
![{\displaystyle {\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle =\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7318b429c7b17e64d4a846adf1456f44b92b5422)
Similarly,
![{\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0df70eaebb85dcc2c55bd81a04de2d8cb2476a31)
Hence, using the same argument as for the previous case, we have
![{\displaystyle {\langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {\tau }}\rangle ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f9b10f2f88451c87fcc56a911ecebc3a6817123)