Elasticity/Rigid body motions

Rigid Deformation
A rigid deformation has the form



\boldsymbol{\varphi}(\mathbf{X}) = \mathbf{X}_1 + \boldsymbol{Q}\bullet[\mathbf{X}-\mathbf{X}_0] $$ where $$\textstyle \mathbf{X}_0, \mathbf{X}_1$$ are fixed material points and $$\textstyle \boldsymbol{Q}$$ is an orthogonal (rotation) tensor.

Therefore

\boldsymbol{F} = \boldsymbol{Q} $$ and

\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{Q} - \boldsymbol{1} $$.

The strain tensors in this case are given by
 * $$ \boldsymbol{E} = 0

$$ but
 * $$ \boldsymbol{\epsilon} = \cfrac{1}{2}(\boldsymbol{Q}+\boldsymbol{Q}^T)-\boldsymbol{1} $$.

Hence the infinitesimal strain tensor does not measure the correct strain when there are large rotations though the finite strain tensor can.

Rigid Displacement
Rigid displacements involve motions in which there are no strains.

Finite Rigid Displacement
If the displacement is rigid we have

\begin{align} \mathbf{u}(\mathbf{X}) &= \mathbf{X}_1 + \boldsymbol{\nabla}\mathbf{u}\bullet[\mathbf{X}-\mathbf{X}_0] + \boldsymbol{1}[\mathbf{X}-\mathbf{X}_0] - \mathbf{X} \\ & = (\mathbf{X}_1-\mathbf{X}_0) + \boldsymbol{\nabla}\mathbf{u}\bullet[\mathbf{X}-\mathbf{X}_0] \\ &= \mathbf{u}_0 + \boldsymbol{\nabla}\mathbf{u}\bullet[\mathbf{X}-\mathbf{X}_0] \end{align} $$

Infinitesimal Rigid Displacement
An infinitesimal rigid displacement is given by

\mathbf{u}(\mathbf{X}) = \mathbf{u}_0 + \boldsymbol{W}\bullet[\mathbf{X}-\mathbf{X}_0] $$

where $$\textstyle \boldsymbol{W}$$ is a skew tensor.