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 $\partial{\Omega}$:

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.

We can also show in an identical manner that

${ \langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\mathbf{u}~\text{dV} ~. }$

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.