# Elasticity/Rigid body motions

## 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.

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

#### Finite Rigid Displacement

If the displacement is rigid we have

{\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

$\mathbf {u} (\mathbf {X} )=\mathbf {u} _{0}+{\boldsymbol {W}}\bullet [\mathbf {X} -\mathbf {X} _{0}]$ where $\textstyle {\boldsymbol {W}}$ is a skew tensor.

## Rigid body displacement field

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

$\mathbf {u} (\mathbf {x} )=\mathbf {c} +{\boldsymbol {\omega }}\cdot \mathbf {x}$ where $\mathbf {c}$ is a constant vector and ${\boldsymbol {\omega }}$ is the infinitesimal rotation tensor.

Proof:

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

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

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

${\boldsymbol {\nabla }}\mathbf {u} ={\frac {\partial \mathbf {u} }{\partial \mathbf {x} }}={\boldsymbol {G}}\qquad \leftarrow \qquad {\text{constant}}~.$ Integrating, we get

$\mathbf {u} (\mathbf {x} )={\boldsymbol {G}}\cdot \mathbf {x} +\mathbf {c} ~.$ Now the strain and rotation tensors are given by

${\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})={\frac {1}{2}}({\boldsymbol {G}}+{\boldsymbol {G}}^{T})~;~~{\boldsymbol {\omega }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})={\frac {1}{2}}({\boldsymbol {G}}-{\boldsymbol {G}}^{T})~.$ For a rigid body motion, the strain ${\boldsymbol {\varepsilon }}=0$ . Therefore,

${\boldsymbol {G}}=-{\boldsymbol {G}}^{T}\qquad \implies \qquad {\boldsymbol {\omega }}={\boldsymbol {G}}~.$ Plugging into the expression for $\mathbf {u}$ for a homogeneous deformation, we have

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