Micromechanics of composites/Proof 10

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 } $$