Micromechanics of composites/Average displacement in a RVE

Average Displacement in a RVE

The average displacement in a RVE may be defined as

$⟨𝐮⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝐮(𝐱)\text{dV}.$

We would like to find the relation between the average displacement in a RVE and the applied displacements at the boundary of the RVE. To do that, recall the identity

$\mathrm{\nabla }\bullet (𝐯\otimes 𝒘)=𝐯\cdot (\mathrm{\nabla }\bullet 𝒘)+(\mathrm{\nabla }𝐯)\cdot 𝒘$

where $𝐯$ and $𝒘$ are two vector fields.

If we choose $𝐯$ such that $\mathrm{\nabla }𝐯=1$ in the above identity, then we can get an equation for $𝒘$, i.e.,

$\mathrm{\nabla }\bullet (𝐯\otimes 𝒘)=𝐯\cdot (\mathrm{\nabla }\bullet 𝒘)+1\cdot 𝒘=𝐯\cdot (\mathrm{\nabla }\bullet 𝒘)+𝒘.$

Now, $\mathrm{\nabla }𝐯=1$ if $𝐯=𝐱$. Therefore,

$\mathrm{\nabla }\bullet (𝐱\otimes 𝒘)=𝐱\cdot (\mathrm{\nabla }\bullet 𝒘)+𝒘⟹𝒘=\mathrm{\nabla }\bullet (𝐱\otimes 𝒘)-𝐱\cdot (\mathrm{\nabla }\bullet 𝒘).$

Using the above in the expression for the average displacement, we have

$⟨𝐮⟩=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }\bullet (𝐱\otimes 𝐮)-𝐱\cdot (\mathrm{\nabla }\bullet 𝐮)]\text{dV}.$

Applying the divergence theorem to the first term on the right, we get

$⟨𝐮⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(𝐱\otimes 𝐮)\cdot 𝐧\text{dV}-\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝐱\cdot (\mathrm{\nabla }\bullet 𝐮)\text{dV}.$

There are two terms in the above expression: the first is a boundary term while the second requires information from the interior of the body. Hence, in general, the average displacement of a RVE cannot be determined solely on the basis of boundary displacements.

Incompressible materials

In the material is incompressible, the balance of mass gives us

$\mathrm{\nabla }\bullet 𝐮=0.$

In that case,

$⟨𝐮⟩=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(𝐱\otimes 𝐮)\cdot 𝐧\text{dV}.$

It's only in this special case that the average displacement in the RVE can be expressed in terms of boundary displacements.

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