Micromechanics of composites/Proof 9

Relation between axial vector and strain Let $𝐮$ be a displacement field. Let $\mathit{\epsilon }$ be the strain field (infinitesimal) corresponding to the displacement field and let $\mathit{\theta }$ be the corresponding infinitesimal rotation vector. Show that $\mathrm{\nabla }\times \mathit{\epsilon }=\mathrm{\nabla }\mathit{\theta }.$

Proof:

The infinitesimal strain tensor is given by

$\mathit{\epsilon }=\frac{1}{2}(\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T).$

Therefore,

$\mathrm{\nabla }\times \mathit{\epsilon }=\frac{1}{2}[\mathrm{\nabla }\times (\mathrm{\nabla }𝐮)+\mathrm{\nabla }\times (\mathrm{\nabla }{𝐮}^T)].$

Recall that

$\mathrm{\nabla }\times (\mathrm{\nabla }𝐮)=\mathrm{𝟎}\text{and}\mathrm{\nabla }\times (\mathrm{\nabla }{𝐮}^T)=\mathrm{\nabla }(\mathrm{\nabla }\times 𝐮).$

Hence,

$\mathrm{\nabla }\times \mathit{\epsilon }=\frac{1}{2}[\mathrm{\nabla }(\mathrm{\nabla }\times 𝐮)].$

Also recall that

$\mathit{\theta }=\frac{1}{2}\mathrm{\nabla }\times 𝐮.$

Therefore,

$\mathrm{\nabla }\times \mathit{\epsilon }=\mathrm{\nabla }\mathit{\theta }◻$

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