Introduction to Elasticity/Rigid body motions

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[edit] Rigid body motions

[edit] 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.

[edit] 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.

[edit] 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}

[edit] 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.

[edit] 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 }
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