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 }