Micromechanics of composites/Proof 10

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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