Average stress power in a RVE
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Recall the equation for the balance of energy (with respect to the reference
configuration)
ρ 0 e ˙ = P T : F ˙ − ∇ 0 ∙ q + ρ 0 s . {\displaystyle \rho _{0}~{\dot {e}}={\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}-{\boldsymbol {\nabla }}_{0}\bullet \mathbf {q} +\rho _{0}~s~.} The quantity P T : F ˙ {\displaystyle {\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}}
The average stress power is defined as
⟨ P T : F ˙ ⟩ := 1 V 0 ∫ ∂ Ω 0 P T : F ˙ dV . {\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}}~.}} Here P T {\displaystyle {\boldsymbol {P}}^{T}} F ˙ {\displaystyle {\dot {\boldsymbol {F}}}} F {\displaystyle {\boldsymbol {F}}} F ˙ {\displaystyle {\dot {\boldsymbol {F}}}}
Note that in that case
tr ( ⟨ P T : F ˙ ⟩ ) = 1 V 0 ∫ ∂ Ω 0 tr ( P ⋅ F ˙ ) dV = 1 V 0 ∫ ∂ Ω 0 tr ( F ˙ ⋅ P ) dV = tr ( ⟨ F ˙ ⋅ P ⟩ ) . {\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 )~.}} We can express the stress power in terms of boundary tractions and
boundary velocities using the relation (see Appendix)
∫ ∂ Ω v ⊗ ( S T ⋅ n ) dA = ∫ Ω [ ∇ v ⋅ S + v ⊗ ( ∇ ∙ S T ) ] dV . {\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}}~.} In this case, we have ∂ Ω → ∂ Ω 0 {\displaystyle \partial {\Omega }\rightarrow \partial {\Omega }_{0}} Ω → Ω 0 {\displaystyle \Omega \rightarrow \Omega _{0}} ∇ → ∇ 0 {\displaystyle {\boldsymbol {\nabla }}\rightarrow {\boldsymbol {\nabla }}_{0}} v → x ˙ {\displaystyle \mathbf {v} \rightarrow {\dot {\mathbf {x} }}} S → P {\displaystyle {\boldsymbol {S}}\rightarrow {\boldsymbol {P}}} n → N {\displaystyle \mathbf {n} \rightarrow \mathbf {N} }
∫ ∂ Ω x ˙ ⊗ ( P T ⋅ N ) dA = ∫ Ω [ ∇ 0 x ˙ ⋅ P + x ˙ ⊗ ( ∇ 0 ∙ P T ) ] dV . {\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}}~.} Using the balance of linear momentum (in the absence of body and inertial
forces), we get
∫ ∂ Ω x ˙ ⊗ ( P T ⋅ N ) dA = ∫ Ω ∇ 0 x ˙ ⋅ P dV . {\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}}~.} Recalling that
F ˙ = ∂ ∂ t ( ∂ x ∂ X ) = ∂ ∂ X ( ∂ x ∂ t ) = ∇ 0 x ˙ {\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} }}} we then have
∫ Ω F ˙ ⋅ P dV = ∫ ∂ Ω x ˙ ⊗ ( P T ⋅ N ) dA . {\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}}~.} If T ¯ {\displaystyle {\bar {\mathbf {T} }}} P {\displaystyle {\boldsymbol {P}}} T ¯ = P T ⋅ N {\displaystyle {\bar {\mathbf {T} }}={\boldsymbol {P}}^{T}\cdot \mathbf {N} }
∫ Ω F ˙ ⋅ P dV = ∫ ∂ Ω x ˙ ⊗ T ¯ dA . {\displaystyle {\int _{\Omega }{\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}~{\text{dV}}=\int _{\partial {\Omega }}{\dot {\mathbf {x} }}\otimes {\bar {\mathbf {T} }}~{\text{dA}}~.}} Note that the fields F ˙ {\displaystyle {\dot {\boldsymbol {F}}}} P {\displaystyle {\boldsymbol {P}}} x ˙ {\displaystyle {\dot {\mathbf {x} }}} T ¯ {\displaystyle {\bar {\mathbf {T} }}}
If the boundary velocity field x ˙ {\displaystyle {\dot {\mathbf {x} }}} F ˙ {\displaystyle {\dot {\boldsymbol {F}}}}
⟨ F ˙ ⋅ P ⟩ − ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ = ⟨ ( F ˙ − ⟨ F ˙ ⟩ ) ⋅ ( P − ⟨ P ⟩ ) ⟩ {\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 } we can show that (see Appendix)
⟨ F ˙ ⋅ P ⟩ − ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ = 1 V 0 ∫ ∂ Ω 0 [ x ˙ − ⟨ F ˙ ⟩ ⋅ X ] ⊗ { [ P − ⟨ P ⟩ ] T ⋅ N } dA = 1 V 0 ∫ ∂ Ω 0 [ x ˙ − ⟨ F ˙ ⟩ ⋅ X ] ⊗ ( P T ⋅ N ) dA = 1 V 0 ∫ ∂ Ω 0 x ˙ ⊗ { [ P − ⟨ P ⟩ ] T ⋅ N } dA . {\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}}} Linear boundary velocities/displacements
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If a linear velocity field is prescribed on the boundary ∂ Ω 0 {\displaystyle \partial {\Omega }_{0}}
x ˙ ( X , t ) = G ˙ ( t ) ⋅ X ∀ X ∈ ∂ Ω 0 . {\displaystyle {\dot {\mathbf {x} }}(\mathbf {X} ,t)={\dot {\boldsymbol {G}}}(t)\cdot \mathbf {X} \qquad \qquad \forall \mathbf {X} \in \partial {\Omega }_{0}~.} Now,
⟨ F ˙ ⟩ = 1 V 0 ∫ ∂ Ω 0 x ˙ ⊗ N dA = 1 V 0 ∫ ∂ Ω 0 ( G ˙ ⋅ X ) ⊗ N dA = G ˙ ⋅ ( 1 V 0 ∫ ∂ Ω 0 X ⊗ N dA ) . {\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}}} Recall that
∫ Ω 0 ∇ 0 X dV = ∫ ∂ Ω 0 X ⊗ N dA . {\displaystyle \int _{\Omega _{0}}{\boldsymbol {\nabla }}_{0}~\mathbf {X} ~{\text{dV}}=\int _{\partial {\Omega }_{0}}\mathbf {X} \otimes \mathbf {N} ~{\text{dA}}~.} Therefore,
⟨ F ˙ ⟩ = G ˙ ⋅ ( 1 V 0 ∫ Ω 0 ∇ 0 X dA ) = G ˙ ⋅ ( 1 V 0 ∫ Ω 0 1 dA ) = G ˙ . {\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}}} Hence,
⟨ F ˙ ⟩ = G ˙ ⟹ x ˙ − ⟨ F ˙ ⟩ ⋅ X = 0 . {\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} ~.} Then,
⟨ F ˙ ⋅ P ⟩ − ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ = 1 V 0 ∫ ∂ Ω 0 [ x ˙ − ⟨ F ˙ ⟩ ⋅ X ] ⊗ ( P T ⋅ N ) dA = 1 V 0 ∫ ∂ Ω 0 [ x ˙ − G ˙ ⋅ X ] ⊗ ( P T ⋅ N ) dA = 0 {\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}}} Hence,
⟨ F ˙ ⋅ P ⟩ = ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ . {\displaystyle {\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle =\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}} Similarly, if a linear displacement field is prescribed on the
boundary such that
u ( X , t ) = G ( t ) ⋅ X − X ⟹ x ( X ) = G ( t ) ⋅ X ∀ X ∈ ∂ Ω 0 {\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}} we can show that
⟨ F ⟩ = G ⟹ x − ⟨ F ⟩ ⋅ X = 0 . {\displaystyle \langle {\boldsymbol {F}}\rangle ={\boldsymbol {G}}\qquad \implies \qquad \mathbf {x} -\langle {\boldsymbol {F}}\rangle \cdot \mathbf {X} =\mathbf {0} ~.} This leads to the equality
⟨ F ⋅ P ⟩ = ⟨ F ⟩ ⋅ ⟨ P ⟩ . {\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}} Recall that, the average Kirchhoff stress is given by
⟨ τ ¯ ⟩ = ⟨ F ⟩ ⋅ ⟨ P ⟩ {\displaystyle \langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle }
⟨ τ ¯ ⟩ = ⟨ F ⟩ ⋅ ⟨ P ⟩ = ⟨ F ⋅ P ⟩ = ⟨ τ ⟩ {\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 } or,
⟨ τ ¯ ⟩ = ⟨ τ ⟩ . {\displaystyle {\langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {\tau }}\rangle ~.}} Uniform boundary tractions
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A uniform boundary traction field in the reference configuration can be represented as
T ¯ ( X , t ) = G T ( t ) ⋅ N ( X ) ∀ X ∈ ∂ Ω 0 . {\displaystyle {\bar {\mathbf {T} }}(\mathbf {X} ,t)={\boldsymbol {G}}^{T}(t)\cdot \mathbf {N} (\mathbf {X} )\qquad \forall ~\mathbf {X} \in \partial {\Omega }_{0}~.} Now,
⟨ P ⟩ = 1 V 0 ∫ ∂ Ω 0 X ⊗ T ¯ dA = 1 V 0 ∫ ∂ Ω 0 X ⊗ ( G T ⋅ N dA = ( 1 V 0 ∫ ∂ Ω 0 X ⊗ N dA ) ⋅ G = 1 ⋅ G = G . {\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}}} Since the surface tractions are related to the nominal stress by T ¯ ( X , t ) = P T ( X , t ) ⋅ N ( X ) {\displaystyle {\bar {\mathbf {T} }}(\mathbf {X} ,t)={\boldsymbol {P}}^{T}(\mathbf {X} ,t)\cdot \mathbf {N} (\mathbf {X} )}
⟨ P ⟩ = P . {\displaystyle \langle {\boldsymbol {P}}\rangle ={\boldsymbol {P}}~.} Therefore,
⟨ F ˙ ⋅ P ⟩ − ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ = 1 V 0 ∫ ∂ Ω 0 x ˙ ⊗ { [ P − ⟨ P ⟩ ] T ⋅ N } dA = 0 {\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} } or,
⟨ F ˙ ⋅ P ⟩ = ⟨ F ˙ ⟩ ⋅ ⟨ P ⟩ . {\displaystyle {\langle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {P}}\rangle =\langle {\dot {\boldsymbol {F}}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}} Similarly,
⟨ F ⋅ P ⟩ = ⟨ F ⟩ ⋅ ⟨ P ⟩ . {\displaystyle {\langle {\boldsymbol {F}}\cdot {\boldsymbol {P}}\rangle =\langle {\boldsymbol {F}}\rangle \cdot \langle {\boldsymbol {P}}\rangle ~.}} Hence, using the same argument as for the previous case, we have
⟨ τ ¯ ⟩ = ⟨ τ ⟩ . {\displaystyle {\langle {\overline {\boldsymbol {\tau }}}\rangle =\langle {\boldsymbol {\tau }}\rangle ~.}}
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)