Micromechanics of composites/Average stress power in a RVE

Average Stress Power in a RVE
The equation for the balance of energy is

\rho~\dot{e} - \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) + \boldsymbol{\nabla} \bullet \mathbf{q} - \rho~s = 0~. $$ If the absence of heat flux or heat sources in the RVE, the equation reduces to

\rho~\dot{e} = \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v})~. $$ 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

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}~\text{dV} ~. } $$ Note that the quantities $$\boldsymbol{\sigma}$$ and $$\boldsymbol{\nabla}\mathbf{v}$$ need not be related in the general case.

The average velocity gradient $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ is defined as

{ \langle\boldsymbol{\nabla}\mathbf{v} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\nabla}\mathbf{v}~\text{dV} ~. } $$ To get an expression for the average stress power in terms of the boundary conditions, we use the identity

\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\cdot\mathbf{v}) = \boldsymbol{S}:\boldsymbol{\nabla}\mathbf{v} + (\boldsymbol{\nabla} \bullet \boldsymbol{S})\cdot\mathbf{v} $$ to get

\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} ~. $$ Using the balance of linear momentum ($$\boldsymbol{\nabla} \bullet \boldsymbol{\sigma} = 0$$), we get

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\Omega}\boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}^T\cdot\mathbf{v})~\text{dV} ~. $$ Using the divergence theorem, we have

\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} ~. $$ Now, the surface traction is given by $$\bar{\mathbf{t}} = \boldsymbol{\sigma}\cdot\mathbf{n}$$. Therefore,

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\mathbf{v}~\text{dV} ~. } $$ {\scriptsize } In micromechanics, it is of interest to see how the average stress power of a RVE is related to the product of the average stress $$\langle \boldsymbol{\sigma} \rangle$$ and the average velocity gradient $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$. While homogenizing a RVE, we would ideally like to have

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle ~. $$ 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)

{ \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} ~. } $$

We can arrive at $$\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ if either of the following conditions is met on the boundary $$\partial{\Omega}$$:


 * 1) $$\mathbf{v} = \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}$$ ~.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}$$ ~.

Linear boundary velocities
If the prescribed velocities on $$\partial{\Omega}$$ are a linear function of $$\mathbf{x}$$, then we can write

\mathbf{v}(\mathbf{x}) = \boldsymbol{G}\cdot\mathbf{x} \qquad \qquad \forall ~\mathbf{x} \in \partial{\Omega} $$ where $$\boldsymbol{G}$$ is a constant second-order tensor.

From the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \mathbf{a}~\text{dV} = \int_{\partial{\Omega}} \mathbf{a}\otimes\mathbf{n}~\text{dA} ~. $$ Therefore,

\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}~. $$ Hence, on the boundary

\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} $$ Using the identity (see Appendix)

(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) $$ and since $$\boldsymbol{G}$$ is constant, we get

\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} ~. $$ From the divergence theorem,

\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}}~. $$ Therefore,

\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}~. $$

Uniform boundary tractions
If the prescribed tractions on the boundary $$\partial{\Omega}$$ are uniform, they can be expressed in terms of a constant symmetric second-order tensor $$\boldsymbol{G}$$ through the relation

\bar{t}(\mathbf{x}) = \boldsymbol{G}\cdot\mathbf{n}(\mathbf{x}) \qquad\qquad \forall~\mathbf{x} \in \partial{\Omega}~. $$ The tractions are related to the stresses at the boundary of the RVE by $$ \bar{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$.

The average stress in the RVE is given by

\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} ~. $$ Using the identity $$\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T$$ (see Appendix), we have

\langle \boldsymbol{\sigma} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} (\mathbf{x}\otimes\mathbf{n})\cdot\boldsymbol{G}^T~\text{dA} ~. $$ Since $$\boldsymbol{G}$$ is constant and symmetric, we have

\langle \boldsymbol{\sigma} \rangle = \left(\cfrac{1}{V}\int_{\partial{\Omega}}\mathbf{x}\otimes\mathbf{n}~\text{dA}\right) \cdot\boldsymbol{G}~. $$ Applying the divergence theorem,

\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} ~. $$ Therefore,

\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}~. $$

Remark
Recall that for small deformations, the displacement gradient $$\boldsymbol{\nabla}\mathbf{u}$$ can be expressed as

\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{\omega} ~. $$ For small deformations, the time derivative of $$\boldsymbol{\nabla}\mathbf{u}$$ gives us the velocity gradient $$\boldsymbol{\nabla}\mathbf{v}$$, i.e.,

\boldsymbol{\nabla}\mathbf{v} = \dot{\boldsymbol{\varepsilon}} + \dot{\boldsymbol{\omega}} ~. $$ If $$\boldsymbol{\omega} = 0$$, we get

\boldsymbol{\nabla}\mathbf{v} = \dot{\boldsymbol{\varepsilon}} ~. $$ Hence, for small strains and in the absence of rigid body rotations, the stress power density is given by $$\boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}$$. Then the average stress power is defined as

{     \langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}~\text{dV} ~. }   $$ and the average strain rate is defined as

{     \langle \dot{\boldsymbol{\varepsilon}} \rangle := \cfrac{1}{V}\int_{\Omega} \dot{\boldsymbol{\varepsilon}}~\text{dV} ~. }   $$ In terms of the surface tractions and the applied boundary velocities, we have

{     \langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\dot{\mathbf{u}}~\text{dV} ~. }   $$ For small strains and no rotation, the stress-power difference relation becomes

{     \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} ~. }   $$ We can arrive at $$\langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle = \langle \boldsymbol{\sigma} \rangle:\langle \dot{\boldsymbol{\varepsilon}} \rangle$$ if either of the following conditions is met on the boundary $$\partial{\Omega}$$: We can also show in an identical manner that
 * 1) $$\dot{\mathbf{u}} = \langle\boldsymbol{\nabla} \dot{\mathbf{u}} \rangle\cdot\mathbf{x} \qquad\implies\qquad$$ Linear boundary velocity field.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n} \qquad\implies\qquad$$ Uniform boundary tractions.

{     \langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\mathbf{u}~\text{dV} ~. }   $$ and that, when $$\boldsymbol{\nabla}\mathbf{u}$$ is symmetric,

{     \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} ~. }   $$ In this case, we can arrive at the relation $$\langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle = \langle \boldsymbol{\sigma} \rangle:\langle \boldsymbol{\varepsilon} \rangle$$ if either of the following conditions is met at the boundary:
 * 1) $$\mathbf{u} = \langle \boldsymbol{\nabla} \mathbf{u} \rangle\cdot\mathbf{x} \qquad\implies\qquad$$ Linear boundary displacement field.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n} \qquad\implies\qquad$$ Uniform boundary tractions.