The average strain tensor is defined as
where the average displacement gradient is
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)
where is a vector field on and is the normal to . Using this relation, we get
Hence,
Plugging these into the definition of average strain, we get
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
Therefore, the average rotation can be defined as
In terms of the applied boundary displacements,
Let us consider a rigid body displacement given by (see Appendix)
where is a constant translation and is a second-order skew symmetric tensor representing an infinitesimal rotation. Then,
Recall that
where is a second-order tensor and and are vectors. Therefore,
From the divergence theorem,
where is a second-order tensor field and is the unit outward normal vector to . Hence,
We also have (see appendix),
where is a vector and is the unit outward normal to . Therefore,
We then have
Since is a skew-symmetric second-order tensor we have
Therefore,
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.