Elasticity/Rigid body motions

Rigid body motions

Rigid Deformation

A rigid deformation has the form

$\mathit{\phi }(𝐗)={𝐗}_1+𝑸\bullet [𝐗-{𝐗}_0]$

where ${𝐗}_0,{𝐗}_1$ are fixed material points and $𝑸$ is an orthogonal (rotation) tensor.

Therefore

$𝑭=𝑸$

and

$\mathrm{\nabla }𝐮=𝑸-1$.

The strain tensors in this case are given by

$𝑬=0$

but

$\mathit{ϵ}=\frac{1}{2}(𝑸+{𝑸}^T)-1$.

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 $𝐮$ is a rigid displacement field, then the strain field corresponding to $𝐮$ is zero.

Finite Rigid Displacement

If the displacement is rigid we have

$\begin{array}{cc}𝐮(𝐗)& ={𝐗}_1+\mathrm{\nabla }𝐮\bullet [𝐗-{𝐗}_0]+1[𝐗-{𝐗}_0]-𝐗\\ & =({𝐗}_1-{𝐗}_0)+\mathrm{\nabla }𝐮\bullet [𝐗-{𝐗}_0]\\ & ={𝐮}_0+\mathrm{\nabla }𝐮\bullet [𝐗-{𝐗}_0]\end{array}$

Infinitesimal Rigid Displacement

An infinitesimal rigid displacement is given by

$𝐮(𝐗)={𝐮}_0+𝑾\bullet [𝐗-{𝐗}_0]$

where $𝑾$ is a skew tensor.

Rigid body displacement field

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

$𝐮(𝐱)=𝐜+\mathit{\omega }\cdot 𝐱$

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

Proof:

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

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

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

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

$\mathrm{\nabla }𝐮=\frac{\partial 𝐮}{\partial 𝐱}=𝑮\leftarrow \text{constant}.$

Integrating, we get

$𝐮(𝐱)=𝑮\cdot 𝐱+𝐜.$

Now the strain and rotation tensors are given by

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

For a rigid body motion, the strain $\mathit{\epsilon }=0$. Therefore,

$𝑮=-{𝑮}^T⟹\mathit{\omega }=𝑮.$

Plugging into the expression for $𝐮$ for a homogeneous deformation, we have

$𝐮(𝐱)=\mathit{\omega }\cdot 𝐱+𝐜◻$

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