Micromechanics of composites/Average strain in a RVE

Average Strain in a RVE

The average strain tensor is defined as

$⟨𝜺⟩:=\frac{1}{2}(⟨\mathrm{\nabla }𝐮⟩+⟨\mathrm{\nabla }𝐮{⟩}^T)$

where the average displacement gradient is

$⟨\mathrm{\nabla }𝐮⟩=\frac{1}{V}{\int }_{\Omega }\mathrm{\nabla }𝐮\text{dV}.$

We would like to find the relation between the average strain in a RVE and the applied displacements at the boundary of the RVE. To do that, recall the relation (see Appendix)

${\int }_{\Omega }\mathrm{\nabla }𝐯\text{dV}={\int }_{\partial \Omega }𝐯\otimes 𝐧\text{dA}$

where $𝐯$ is a vector field on $\Omega$ and $𝐧$ is the normal to $\partial \Omega$. Using this relation, we get

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

Hence,

$⟨\mathrm{\nabla }𝐮{⟩}^T=\frac{1}{V}{\int }_{\partial \Omega }𝐧\otimes 𝐮\text{dA}.$

Plugging these into the definition of average strain, we get

$⟨𝜺⟩:=\frac{1}{2V}{\int }_{\partial \Omega }(𝐮\otimes 𝐧+𝐧\otimes 𝐮)\text{dA}.$

This implies that the average strain is completely defined in terms of the applied displacements at the boundary! Also, the average strain tensor is symmetric by virtue of its definition.

We can define the average rotation tensor (which represents an infinitesimal rotation) in an analogous manner. The rotation tensor is given by

$𝝎=\frac{1}{2}(\mathrm{\nabla }𝐮-\mathrm{\nabla }{𝐮}^T).$

Therefore, the average rotation can be defined as

$⟨𝝎⟩:=\frac{1}{2}(⟨\mathrm{\nabla }𝐮⟩-⟨\mathrm{\nabla }𝐮{⟩}^T).$

In terms of the applied boundary displacements,

$⟨𝝎⟩=\frac{1}{2V}{\int }_{\partial \Omega }(𝐮\otimes 𝐧-𝐧\otimes 𝐮)\text{dA}.$

The effect of rigid body motions on the average strain

Let us consider a rigid body displacement given by (see Appendix)

$𝐮(𝐱)=𝐜+𝝎\cdot 𝐱$

where $𝐜$ is a constant translation and $𝝎$ is a second-order skew symmetric tensor representing an infinitesimal rotation. Then,

$⟨\mathrm{\nabla }𝐮⟩=\frac{1}{V}{\int }_{\partial \Omega }𝐮\otimes 𝐧\text{dA}=\frac{1}{V}{\int }_{\partial \Omega }(𝐜+𝝎\cdot 𝐱)\otimes 𝐧\text{dA}=\frac{1}{V}{\int }_{\partial \Omega }𝐜\otimes 𝐧\text{dA}+\frac{1}{V}{\int }_{\partial \Omega }(𝝎\cdot 𝐱)\otimes 𝐧\text{dA}.$

Recall that

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

where $𝑨$ is a second-order tensor and $𝐛$ and $𝐜$ are vectors. Therefore,

$⟨\mathrm{\nabla }𝐮⟩=𝐜\otimes \left(\frac{1}{V}{\int }_{\partial \Omega }𝐧\text{dA}\right)+𝝎\cdot (\frac{1}{V}{\int }_{\partial \Omega }𝐱\otimes 𝐧\text{dA}).$

From the divergence theorem,

${\int }_{\Omega }\mathrm{\nabla }\bullet 𝑨\text{dV}={\int }_{\partial \Omega }𝑨\cdot 𝐧\text{dA}$

where $𝑨$ is a second-order tensor field and $𝐧$ is the unit outward normal vector to $\partial \Omega$. Hence,

${\int }_{\partial \Omega }𝐧\text{dA}={\int }_{\partial \Omega }1\cdot 𝐧\text{dA}={\int }_{\Omega }\mathrm{\nabla }\bullet 1\text{dV}=\mathrm{𝟎}.$

We also have (see appendix),

${\int }_{\Omega }\mathrm{\nabla }𝐯\text{dV}={\int }_{\partial \Omega }𝐯\otimes 𝐧\text{dA}$

where $𝐯$ is a vector and $𝐧$ is the unit outward normal to $\partial \Omega$. Therefore,

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

We then have

$⟨\mathrm{\nabla }𝐮⟩=𝐜\otimes \mathrm{𝟎}+𝝎\cdot 1=𝝎.$

Since $𝝎$ is a skew-symmetric second-order tensor we have

$𝝎=-{𝝎}^T.$

Therefore,

$⟨𝜺⟩=\frac{1}{2}(⟨\mathrm{\nabla }𝐮⟩+⟨\mathrm{\nabla }𝐮{⟩}^T)=\frac{1}{2}(𝝎+{𝝎}^T)=\mathrm{𝟎}.$

Hence, the average strain is not affected by rigid body motions. However, for simplicity, we assume that the displacement field in a RVE does not contain any rigid body motions.

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