Micromechanics of composites/Average strain in a RVE

Average Strain in a RVE

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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,

 

The effect of rigid body motions on the average strain

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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.