Micromechanics of composites/Average stress power in a RVE with finite strain

Average stress power in a RVE

Recall the equation for the balance of energy (with respect to the reference configuration)

${\rho }_0\dot{e}={𝑷}^T:\dot{𝑭}-{\mathrm{\nabla }}_0\bullet 𝐪+{\rho }_{0}s.$

The quantity ${𝑷}^T:\dot{𝑭}$ is the stress power.

The average stress power is defined as

$⟨{𝑷}^T:\dot{𝑭}⟩:=\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }{𝑷}^T:\dot{𝑭}\text{dV}.$

Here ${𝑷}^T$ is an arbitrary self-equilibrating nominal stress field that satisfies the balance of momentum (without any body forces or inertial forces) and $\dot{𝑭}$ is the time rate of change of $𝑭$. The reference configuration can be arbitrary. Also, the nominal stress and the rate $\dot{𝑭}$ need not be related.

Note that in that case

$\text{tr}(⟨{𝑷}^T:\dot{𝑭}⟩)=\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }\text{tr}(𝑷\cdot \dot{𝑭})\text{dV}=\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }\text{tr}(\dot{𝑭}\cdot 𝑷)\text{dV}=\text{tr}(⟨\dot{𝑭}\cdot 𝑷⟩).$

We can express the stress power in terms of boundary tractions and boundary velocities using the relation (see Appendix)

$\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\cdot 𝐧)\text{dA}=\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}.$

In this case, we have $\partial \mathrm{\Omega }\to \partial {\mathrm{\Omega }}_0$, $\mathrm{\Omega }\to {\mathrm{\Omega }}_0$, $\mathrm{\nabla }\to {\mathrm{\nabla }}_0$, $𝐯\to \dot{𝐱}$, $𝑺\to 𝑷$, and $𝐧\to 𝐍$. Then

$\underset{\partial \mathrm{\Omega }}{\int }\dot{𝐱}\otimes ({𝑷}^T\cdot 𝐍)\text{dA}=\underset{\mathrm{\Omega }}{\int }[{\mathrm{\nabla }}_0\dot{𝐱}\cdot 𝑷+\dot{𝐱}\otimes ({\mathrm{\nabla }}_0\bullet {𝑷}^T)]\text{dV}.$

Using the balance of linear momentum (in the absence of body and inertial forces), we get

$\underset{\partial \mathrm{\Omega }}{\int }\dot{𝐱}\otimes ({𝑷}^T\cdot 𝐍)\text{dA}=\underset{\mathrm{\Omega }}{\int }{\mathrm{\nabla }}_0\dot{𝐱}\cdot 𝑷\text{dV}.$

Recalling that

$\dot{𝑭}=\frac{\partial }{\partial t}\left(\frac{\partial 𝐱}{\partial 𝐗}\right)=\frac{\partial }{\partial 𝐗}\left(\frac{\partial 𝐱}{\partial t}\right)={\mathrm{\nabla }}_0\dot{𝐱}$

we then have

$\underset{\mathrm{\Omega }}{\int }\dot{𝑭}\cdot 𝑷\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }\dot{𝐱}\otimes ({𝑷}^T\cdot 𝐍)\text{dA}.$

If $\overline{𝐓}$ is a self equilibrating traction applied on the boundary that leads to the stress field $𝑷$, i.e., $\overline{𝐓}={𝑷}^T\cdot 𝐍$, then we have

$\underset{\mathrm{\Omega }}{\int }\dot{𝑭}\cdot 𝑷\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }\dot{𝐱}\otimes \overline{𝐓}\text{dA}.$

Note that the fields $\dot{𝑭}$ and $𝑷$ need not be related and hence the velocities $\dot{𝐱}$ and the tractions $\overline{𝐓}$ are not related.

If the boundary velocity field $\dot{𝐱}$ leads to the rate $\dot{𝑭}$, using the identity (see Appendix)

$⟨\dot{𝑭}\cdot 𝑷⟩-⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩=⟨(\dot{𝑭}-⟨\dot{𝑭}⟩)\cdot (𝑷-⟨𝑷⟩)⟩$

we can show that (see Appendix)

$\begin{array}{cc}⟨\dot{𝑭}\cdot 𝑷⟩-⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩& =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[\dot{𝐱}-⟨\dot{𝑭}⟩\cdot 𝐗]\otimes \{[𝑷-⟨𝑷⟩{]}^T\cdot 𝐍\}\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[\dot{𝐱}-⟨\dot{𝑭}⟩\cdot 𝐗]\otimes ({𝑷}^T\cdot 𝐍)\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }\dot{𝐱}\otimes \{[𝑷-⟨𝑷⟩{]}^T\cdot 𝐍\}\text{dA}.\end{array}$

Remark

Using similar arguments, if we assume that $𝑭$ is a deformation that is compatible with an applied boundary displacement $𝐮=𝐱-𝐗$,we can show that

$\begin{array}{cc}⟨𝑭\cdot 𝑷⟩-⟨𝑭⟩\cdot ⟨𝑷⟩& =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[𝐱-⟨𝑭⟩\cdot 𝐗]\otimes \{[𝑷-⟨𝑷⟩{]}^T\cdot 𝐍\}\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[𝐱-⟨𝑭⟩\cdot 𝐗]\otimes ({𝑷}^T\cdot 𝐍)\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐱\otimes \{[𝑷-⟨𝑷⟩{]}^T\cdot 𝐍\}\text{dA}.\end{array}$

We can arrive at $⟨\dot{𝑭}\cdot 𝑷⟩=⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩$ or $⟨𝑭\cdot 𝑷⟩=⟨𝑭⟩\cdot ⟨𝑷⟩$ if either of the following conditions is satisfied at the boundary:

1. $\dot{𝐱}=⟨\dot{𝑭}⟩\cdot 𝐗$ or $𝐱=⟨𝑭⟩\cdot 𝐗$.
2. $𝑷\cdot 𝐍=⟨𝑷{⟩}^T\cdot 𝐍$.

Linear boundary velocities/displacements

If a linear velocity field is prescribed on the boundary $\partial {\mathrm{\Omega }}_0$, we can express this field as

$\dot{𝐱}(𝐗,t)=\dot{𝑮}(t)\cdot 𝐗\mathrm{\forall }𝐗\in \partial {\mathrm{\Omega }}_0.$

Now,

$\begin{array}{cc}⟨\dot{𝑭}⟩& =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }\dot{𝐱}\otimes 𝐍\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }(\dot{𝑮}\cdot 𝐗)\otimes 𝐍\text{dA}\\ & =\dot{𝑮}\cdot (\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐗\otimes 𝐍\text{dA}).\end{array}$

Recall that

$\underset{{\mathrm{\Omega }}_0}{\int }{\mathrm{\nabla }}_0𝐗\text{dV}=\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐗\otimes 𝐍\text{dA}.$

Therefore,

$\begin{array}{cc}⟨\dot{𝑭}⟩& =\dot{𝑮}\cdot \left(\frac{1}{V_0}\underset{{\mathrm{\Omega }}_0}{\int }{\mathrm{\nabla }}_0𝐗\text{dA}\right)\\ & =\dot{𝑮}\cdot \left(\frac{1}{V_0}\underset{{\mathrm{\Omega }}_0}{\int }1\text{dA}\right)=\dot{𝑮}.\end{array}$

Hence,

$⟨\dot{𝑭}⟩=\dot{𝑮}⟹\dot{𝐱}-⟨\dot{𝑭}⟩\cdot 𝐗=\mathrm{𝟎}.$

Then,

$\begin{array}{cc}⟨\dot{𝑭}\cdot 𝑷⟩-⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩& =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[\dot{𝐱}-⟨\dot{𝑭}⟩\cdot 𝐗]\otimes ({𝑷}^T\cdot 𝐍)\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }[\dot{𝐱}-\dot{𝑮}\cdot 𝐗]\otimes ({𝑷}^T\cdot 𝐍)\text{dA}=\mathrm{𝟎}\end{array}$

Hence,

$⟨\dot{𝑭}\cdot 𝑷⟩=⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩.$

Similarly, if a linear displacement field is prescribed on the boundary such that

$𝐮(𝐗,t)=𝑮(t)\cdot 𝐗-𝐗⟹𝐱(𝐗)=𝑮(t)\cdot 𝐗\mathrm{\forall }𝐗\in \partial {\mathrm{\Omega }}_0$

we can show that

$⟨𝑭⟩=𝑮⟹𝐱-⟨𝑭⟩\cdot 𝐗=\mathrm{𝟎}.$

This leads to the equality

$⟨𝑭\cdot 𝑷⟩=⟨𝑭⟩\cdot ⟨𝑷⟩.$

Recall that, the average Kirchhoff stress is given by $⟨\bar{\mathit{\tau }}⟩=⟨𝑭⟩\cdot ⟨𝑷⟩$. Therefore, if a uniform boundary displacement is prescribed, we have

$⟨\bar{\mathit{\tau }}⟩=⟨𝑭⟩\cdot ⟨𝑷⟩=⟨𝑭\cdot 𝑷⟩=⟨\mathit{\tau }⟩$

or,

$⟨\bar{\mathit{\tau }}⟩=⟨\mathit{\tau }⟩.$

Uniform boundary tractions

A uniform boundary traction field in the reference configuration can be represented as

$\overline{𝐓}(𝐗,t)={𝑮}^T(t)\cdot 𝐍(𝐗)\mathrm{\forall }𝐗\in \partial {\mathrm{\Omega }}_0.$

Now,

$\begin{array}{cc}⟨𝑷⟩& =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐗\otimes \overline{𝐓}\text{dA}\\ & =\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐗\otimes ({𝑮}^T\cdot 𝐍\text{dA}\\ & =(\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }𝐗\otimes 𝐍\text{dA})\cdot 𝑮\\ & =1\cdot 𝑮=𝑮.\end{array}$

Since the surface tractions are related to the nominal stress by $\overline{𝐓}(𝐗,t)={𝑷}^T(𝐗,t)\cdot 𝐍(𝐗)$, we must have

$⟨𝑷⟩=𝑷.$

Therefore,

$⟨\dot{𝑭}\cdot 𝑷⟩-⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩=\frac{1}{V_0}\underset{\partial {\mathrm{\Omega }}_0}{\int }\dot{𝐱}\otimes \{[𝑷-⟨𝑷⟩{]}^T\cdot 𝐍\}\text{dA}=\mathrm{𝟎}$

or,

$⟨\dot{𝑭}\cdot 𝑷⟩=⟨\dot{𝑭}⟩\cdot ⟨𝑷⟩.$

Similarly,

$⟨𝑭\cdot 𝑷⟩=⟨𝑭⟩\cdot ⟨𝑷⟩.$

Hence, using the same argument as for the previous case, we have

$⟨\bar{\mathit{\tau }}⟩=⟨\mathit{\tau }⟩.$

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