# Elasticity/Rigid body motions

## Rigid body motions

### Rigid Deformation

A rigid deformation has the form

${\displaystyle {\boldsymbol {\varphi }}(\mathbf {X} )=\mathbf {X} _{1}+{\boldsymbol {Q}}\bullet [\mathbf {X} -\mathbf {X} _{0}]}$

where ${\displaystyle \textstyle \mathbf {X} _{0},\mathbf {X} _{1}}$ are fixed material points and ${\displaystyle \textstyle {\boldsymbol {Q}}}$ is an orthogonal (rotation) tensor.

Therefore

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

and

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

The strain tensors in this case are given by

${\displaystyle {\boldsymbol {E}}=0}$

but

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

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.

 Properties of rigid displacement fields If ${\displaystyle \textstyle \mathbf {u} }$ is a rigid displacement field, then the strain field corresponding to ${\displaystyle \textstyle \mathbf {u} }$ is zero.

#### Finite Rigid Displacement

If the displacement is rigid we have

{\displaystyle {\begin{aligned}\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{aligned}}}

#### Infinitesimal Rigid Displacement

An infinitesimal rigid displacement is given by

${\displaystyle \mathbf {u} (\mathbf {X} )=\mathbf {u} _{0}+{\boldsymbol {W}}\bullet [\mathbf {X} -\mathbf {X} _{0}]}$

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

## Rigid body displacement field

Show that, for a rigid body motion with infinitesimal rotations, the displacement field ${\displaystyle \mathbf {u} (\mathbf {x} )}$ for can be expressed as

${\displaystyle \mathbf {u} (\mathbf {x} )=\mathbf {c} +{\boldsymbol {\omega }}\cdot \mathbf {x} }$

where ${\displaystyle \mathbf {c} }$ is a constant vector and ${\displaystyle {\boldsymbol {\omega }}}$ is the infinitesimal rotation tensor.

Proof:

Note that for a rigid body motion, the strain ${\displaystyle {\boldsymbol {\varepsilon }}}$ is zero. Since

${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}}$

we have a ${\displaystyle {\boldsymbol {\theta }}=}$ constant when ${\displaystyle {\boldsymbol {\varepsilon }}=0}$, i.e., the rotation is homogeneous.

For a homogeneous deformation, the displacement gradient is independent of ${\displaystyle \mathbf {x} }$, i.e.,

${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} ={\frac {\partial \mathbf {u} }{\partial \mathbf {x} }}={\boldsymbol {G}}\qquad \leftarrow \qquad {\text{constant}}~.}$

Integrating, we get

${\displaystyle \mathbf {u} (\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {x} +\mathbf {c} ~.}$

Now the strain and rotation tensors are given by

${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})={\frac {1}{2}}({\boldsymbol {G}}+{\boldsymbol {G}}^{T})~;~~{\boldsymbol {\omega }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})={\frac {1}{2}}({\boldsymbol {G}}-{\boldsymbol {G}}^{T})~.}$

For a rigid body motion, the strain ${\displaystyle {\boldsymbol {\varepsilon }}=0}$. Therefore,

${\displaystyle {\boldsymbol {G}}=-{\boldsymbol {G}}^{T}\qquad \implies \qquad {\boldsymbol {\omega }}={\boldsymbol {G}}~.}$

Plugging into the expression for ${\displaystyle \mathbf {u} }$ for a homogeneous deformation, we have

${\displaystyle {\mathbf {u} (\mathbf {x} )={\boldsymbol {\omega }}\cdot \mathbf {x} +\mathbf {c} \qquad \square }}$