Micromechanics of composites/Average stress power in a RVE

Average Stress Power in a RVE

The equation for the balance of energy is

$\rho \dot{e}-\mathit{\sigma }:(\mathrm{\nabla }𝐯)+\mathrm{\nabla }\bullet 𝐪-\rho s=0.$

If the absence of heat flux or heat sources in the RVE, the equation reduces to

$\rho \dot{e}=\mathit{\sigma }:(\mathrm{\nabla }𝐯).$

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

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }:\mathrm{\nabla }𝐯\text{dV}.$

Note that the quantities $\mathit{\sigma }$ and $\mathrm{\nabla }𝐯$ need not be related in the general case.

The average velocity gradient $⟨\mathrm{\nabla }𝐯⟩$ is defined as

$⟨\mathrm{\nabla }𝐯⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }𝐯\text{dV}.$

To get an expression for the average stress power in terms of the boundary conditions, we use the identity

$\mathrm{\nabla }\bullet ({𝑺}^T\cdot 𝐯)=𝑺:\mathrm{\nabla }𝐯+(\mathrm{\nabla }\bullet 𝑺)\cdot 𝐯$

to get

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }:\mathrm{\nabla }𝐯\text{dV}=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }\bullet ({\mathit{\sigma }}^T\cdot 𝐯)-(\mathrm{\nabla }\bullet \mathit{\sigma })\cdot 𝐯]\text{dV}.$

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

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet ({\mathit{\sigma }}^T\cdot 𝐯)\text{dV}.$

Using the divergence theorem, we have

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }({\mathit{\sigma }}^T\cdot 𝐯)\cdot 𝐧\text{dV}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }({\mathit{\sigma }}^T\cdot 𝐯)\cdot 𝐧\text{dV}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(\mathit{\sigma }\cdot 𝐧)\cdot 𝐯\text{dV}.$

Now, the surface traction is given by $\overline{𝐭}=\mathit{\sigma }\cdot 𝐧$. Therefore,

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\cdot 𝐯\text{dV}.$

{\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 $⟨\mathit{\sigma }⟩$ and the average velocity gradient $⟨\mathrm{\nabla }𝐯⟩$. While homogenizing a RVE, we would ideally like to have

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=⟨\mathit{\sigma }⟩:⟨\mathrm{\nabla }𝐯⟩.$

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)

$⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩-⟨\mathit{\sigma }⟩:⟨\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }[𝐯-⟨\mathrm{\nabla }𝐯⟩\cdot 𝐱]\cdot [(\mathit{\sigma }-⟨\mathit{\sigma }⟩)\cdot 𝐧]\text{dA}.$

We can arrive at $⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=⟨\mathit{\sigma }⟩:⟨\mathrm{\nabla }𝐯⟩$ if either of the following conditions is met on the boundary $\partial \mathrm{\Omega }$:

1. $𝐯=⟨\mathrm{\nabla }𝐯⟩\cdot 𝐱$ ~.
2. $\mathit{\sigma }\cdot 𝐧=⟨\mathit{\sigma }⟩\cdot 𝐧$ ~.

Linear boundary velocities

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

$𝐯(𝐱)=𝑮\cdot 𝐱\mathrm{\forall }𝐱\in \partial \mathrm{\Omega }$

where $𝑮$ is a constant second-order tensor.

From the divergence theorem

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }𝐚\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐚\otimes 𝐧\text{dA}.$

Therefore,

$⟨\mathrm{\nabla }𝐯⟩=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }𝐯\text{dV}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes 𝐧\text{dA}.$

Hence, on the boundary

$𝐯-⟨\mathrm{\nabla }𝐯⟩\cdot 𝐱=𝑮\cdot 𝐱-[\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(𝑮\cdot 𝐱)\otimes 𝐧\text{dA}]\cdot 𝐱$

Using the identity (see Appendix)

$(𝑨\cdot 𝐚)\otimes 𝐛=𝑨\cdot (𝐚\otimes 𝐛)$

and since $𝑮$ is constant, we get

$𝐯-⟨\mathrm{\nabla }𝐯{⟩}^T\cdot 𝐱=𝑮\cdot 𝐱-[𝑮\cdot (\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes 𝐧\text{dA})]\cdot 𝐱.$

From the divergence theorem,

$\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes 𝐧\text{dA}=\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }𝐱\text{dV}=\underset{\mathrm{\Omega }}{\int }1\text{dV}=V1.$

Therefore,

$𝐯-⟨\mathrm{\nabla }𝐯⟩\cdot 𝐱=𝑮\cdot 𝐱-(𝑮\cdot 1)\cdot 𝐱=𝑮\cdot 𝐱-𝑮\cdot 𝐱=\mathrm{𝟎}⟹⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=⟨\mathit{\sigma }⟩:⟨\mathrm{\nabla }𝐯⟩.$

Uniform boundary tractions

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

$\overline{t}(𝐱)=𝑮\cdot 𝐧(𝐱)\mathrm{\forall }𝐱\in \partial \mathrm{\Omega }.$

The tractions are related to the stresses at the boundary of the RVE by $\overline{t}=\mathit{\sigma }\cdot 𝐧$.

The average stress in the RVE is given by

$⟨\mathit{\sigma }⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes \overline{t}\text{dA}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes (𝑮\cdot 𝐧)\text{dA}.$

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

$⟨\mathit{\sigma }⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(𝐱\otimes 𝐧)\cdot {𝑮}^T\text{dA}.$

Since $𝑮$ is constant and symmetric, we have

$⟨\mathit{\sigma }⟩=(\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes 𝐧\text{dA})\cdot 𝑮.$

Applying the divergence theorem,

$⟨\mathit{\sigma }⟩=\left(\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }𝐱\text{dV}\right)\cdot 𝑮=1\cdot 𝑮=𝑮.$

Therefore,

$\mathit{\sigma }\cdot 𝐧-⟨\mathit{\sigma }⟩\cdot 𝐧=\overline{t}-𝑮\cdot 𝐧=\mathrm{𝟎}⟹⟨\mathit{\sigma }:\mathrm{\nabla }𝐯⟩=⟨\mathit{\sigma }⟩:⟨\mathrm{\nabla }𝐯⟩.$

Remark

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

$\mathrm{\nabla }𝐮=\mathit{\epsilon }+\mathit{\omega }.$

For small deformations, the time derivative of $\mathrm{\nabla }𝐮$ gives us the velocity gradient $\mathrm{\nabla }𝐯$, i.e.,

$\mathrm{\nabla }𝐯=\dot{\mathit{\epsilon }}+\dot{\mathit{\omega }}.$

If $\mathit{\omega }=0$, we get

$\mathrm{\nabla }𝐯=\dot{\mathit{\epsilon }}.$

Hence, for small strains and in the absence of rigid body rotations, the stress power density is given by $\mathit{\sigma }:\dot{\mathit{\epsilon }}$. Then the average stress power is defined as

$⟨\mathit{\sigma }:\dot{\mathit{\epsilon }}⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }:\dot{\mathit{\epsilon }}\text{dV}.$

and the average strain rate is defined as

$⟨\dot{\mathit{\epsilon }}⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\dot{\mathit{\epsilon }}\text{dV}.$

In terms of the surface tractions and the applied boundary velocities, we have

$⟨\mathit{\sigma }:\dot{\mathit{\epsilon }}⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\cdot \dot{𝐮}\text{dV}.$

For small strains and no rotation, the stress-power difference relation becomes

$⟨\mathit{\sigma }:\dot{\mathit{\epsilon }}⟩-⟨\mathit{\sigma }⟩:⟨\dot{\mathit{\epsilon }}⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }[\dot{𝐮}-⟨\mathrm{\nabla }\dot{𝐮}⟩\cdot 𝐱]\cdot [(\mathit{\sigma }-⟨\mathit{\sigma }⟩)\cdot 𝐧]\text{dA}.$

We can arrive at $⟨\mathit{\sigma }:\dot{\mathit{\epsilon }}⟩=⟨\mathit{\sigma }⟩:⟨\dot{\mathit{\epsilon }}⟩$ if either of the following conditions is met on the boundary $\partial \mathrm{\Omega }$:

1. $\dot{𝐮}=⟨\mathrm{\nabla }\dot{𝐮}⟩\cdot 𝐱⟹$ Linear boundary velocity field.
2. $\mathit{\sigma }\cdot 𝐧=⟨\mathit{\sigma }⟩\cdot 𝐧⟹$ Uniform boundary tractions.

We can also show in an identical manner that

$⟨\mathit{\sigma }:\mathit{\epsilon }⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\cdot 𝐮\text{dV}.$

and that, when $\mathrm{\nabla }𝐮$ is symmetric,

$⟨\mathit{\sigma }:\mathit{\epsilon }⟩-⟨\mathit{\sigma }⟩:⟨\mathit{\epsilon }⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }[𝐮-⟨\mathrm{\nabla }𝐮⟩\cdot 𝐱]\cdot [(\mathit{\sigma }-⟨\mathit{\sigma }⟩)\cdot 𝐧]\text{dA}.$

In this case, we can arrive at the relation $⟨\mathit{\sigma }:\mathit{\epsilon }⟩=⟨\mathit{\sigma }⟩:⟨\mathit{\epsilon }⟩$ if either of the following conditions is met at the boundary:

1. $𝐮=⟨\mathrm{\nabla }𝐮⟩\cdot 𝐱⟹$ Linear boundary displacement field.
2. $\mathit{\sigma }\cdot 𝐧=⟨\mathit{\sigma }⟩\cdot 𝐧⟹$ Uniform boundary tractions.