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

${\mathbf {u} (\mathbf {x} )={\boldsymbol {\omega }}\cdot \mathbf {x} +\mathbf {c} \qquad \square }$

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Micromechanics of composites

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