Body

We usually denote a body by the symbol ${\displaystyle \textstyle B}$. A body is essentially a set of points in Euclidean space. For mathematical definition see Truesdell and Noll (1992)

Configuration

A configuration of a body is denoted by the symbol ${\displaystyle \textstyle {\boldsymbol {\chi }}}$. A configuration of a body is just what the name suggests. Sometimes a configuration is also referred to as a placement.

Mathematically, we can think of a configuration as a smooth one-to-one mapping of a body into a region of three-dimensional Euclidean space.

Thus, we can have a reference configuration ${\displaystyle \textstyle {\boldsymbol {\chi }}(\mathbf {X} )}$ and a current configuration ${\displaystyle \textstyle {\boldsymbol {\chi }}(\mathbf {x} )}$.

A one-to-one mapping is also called a homeomorphism.

Deformation

A deformation is the relationship between two configurations and is usually denoted by ${\displaystyle \textstyle {\boldsymbol {\varphi }}}$. Deformations include both volume and shape changes and rigid body motions.

For a continuous body, a deformation can be thought of as a smooth mapping from one configuration (${\displaystyle \textstyle \mathbf {X} }$) to another (${\displaystyle \textstyle \mathbf {x} }$). The inverse mapping should be possible.

This means that

${\displaystyle {\begin{aligned}\mathbf {x} &\equiv \{x_{1},x_{2},x_{3}\}={\boldsymbol {\varphi }}(\mathbf {X} )\\\mathbf {X} &\equiv \{X_{1},X_{2},X_{3}\}={\boldsymbol {\varphi }}^{-1}(\mathbf {x} )\end{aligned}}}$

For the inverse mapping to exist, we require that the Jacobian of the deformation is positive, i.e., ${\displaystyle \textstyle J=\det({\boldsymbol {\nabla }}{\boldsymbol {\varphi }})>0}$.

Displacement

The displacement is usually denoted by the symbol ${\displaystyle \textstyle \mathbf {u} }$.

The displacement is defined as a vector from the location of a material point in one configuration to the location of the same material point in another configuration.

The definition is

${\displaystyle \mathbf {u} (\mathbf {X} ):={\boldsymbol {\varphi }}(\mathbf {X} )-\mathbf {X} =\mathbf {x} -\mathbf {X} \,}$

In index notation

${\displaystyle u_{i}=x_{i}-X_{i}\,}$

Finite Strain Tensor

The finite strain tensor (${\displaystyle \textstyle {\boldsymbol {E}}}$) is also called the Green-St. Venant Strain Tensor or the Lagrangian Strain Tensor.

This strain tensor is defined as

${\displaystyle {\boldsymbol {E}}:={\cfrac {1}{2}}({\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}}-{\boldsymbol {1}})}$

In index notation,

${\displaystyle E_{ij}={\cfrac {1}{2}}(F_{ki}F_{kj}-\delta _{ij})}$

Infinitesimal Strain Tensor

In the limit of small strains, the Lagrangian finite strain tensor reduces to the infinitesimal strain tensor (${\displaystyle \textstyle {\boldsymbol {\epsilon }}}$).

This strain tensor is defined as

${\displaystyle {\boldsymbol {\varepsilon }}:={\cfrac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})}$

In index notation,

${\displaystyle \varepsilon _{ij}={\cfrac {1}{2}}\left({\cfrac {\partial u_{i}}{\partial x_{j}}}+{\cfrac {\partial u_{j}}{\partial x_{i}}}\right)}$

Therefore we can see that the finite strain tensor and the infinitesimal strain tensor are related by

${\displaystyle {\boldsymbol {E}}={\boldsymbol {\varepsilon }}+{\cfrac {1}{2}}{\boldsymbol {\nabla }}\mathbf {u} ^{T}\bullet {\boldsymbol {\nabla }}\mathbf {u} }$

If ${\displaystyle \textstyle \epsilon =|{\boldsymbol {\nabla }}\mathbf {u} |}$, then

${\displaystyle {\boldsymbol {E}}={\boldsymbol {\varepsilon }}+O(\epsilon ^{2})}$

For small strains, ${\displaystyle \textstyle \epsilon ^{2}\rightarrow 0}$ and

${\displaystyle {\boldsymbol {E}}\approx {\boldsymbol {\varepsilon }}}$.

Infinitesimal Rotation Tensor

For small deformation problems, in addition to small strains we can also have small rotations (${\displaystyle \textstyle {\boldsymbol {W}}}$). The infinitesimal rotation tensor is defined as

${\displaystyle {\boldsymbol {W}}:={\cfrac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})}$

In index notation,

${\displaystyle W_{ij}={\cfrac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}$

If ${\displaystyle {\boldsymbol {A}}}$ is a skew-symmetric tensor, then for any vector ${\displaystyle \textstyle \mathbf {a} }$ we have

${\displaystyle {\boldsymbol {A}}\bullet \mathbf {a} ={\boldsymbol {\omega }}\times {\mathbf {a} }~.}$

The vector ${\displaystyle {\boldsymbol {\omega }}}$ is called the axial vector of the skew-symmetric tensor.

In our case, ${\displaystyle {\boldsymbol {W}}}$ is the skew-symmetric infinitesimal rotation tensor. The corresponding axial vector is the rotation vector ${\displaystyle {\boldsymbol {\theta }}}$ defined as

${\displaystyle {\boldsymbol {W}}\bullet \mathbf {a} ={\boldsymbol {\theta }}\times {\mathbf {a} }~.}$

where

${\displaystyle {\boldsymbol {\theta }}={\cfrac {1}{2}}{\boldsymbol {\nabla }}\times {\boldsymbol {u}}}$

Volume Change Due To Finite Deformation

The change in volume (${\displaystyle \textstyle \delta V}$) during a finite deformation is given by

${\displaystyle \delta V=\int _{B}\det({\boldsymbol {F}}-{\boldsymbol {1}})dv}$

Volume Change Due To Infinitesimal Deformation

The volume change during an infinitesimal deformation (${\displaystyle \textstyle \delta V}$) is given by

${\displaystyle \delta V=\int _{B}{\boldsymbol {\nabla }}\bullet \mathbf {u} dv=\int _{\partial B}\mathbf {u} \bullet \mathbf {n} ds}$

because

${\displaystyle \det {\boldsymbol {F}}=\det({\boldsymbol {1}}+{\boldsymbol {\nabla }}\mathbf {u} )=1+{\text{Tr}}{\boldsymbol {\nabla }}\mathbf {u} +O(\epsilon ^{2})=1+{\boldsymbol {\nabla }}\bullet \mathbf {u} +O(\epsilon ^{2})}$

The quantity ${\displaystyle \textstyle {\boldsymbol {\nabla }}\bullet \mathbf {u} ={\text{Tr}}({\boldsymbol {\varepsilon }})}$ is called the dilatation.

A volume change is isochoric (volume preserving) if

${\displaystyle \textstyle {\text{Tr}}({\boldsymbol {\varepsilon }})=0}$.

Proof:

The axial vector ${\displaystyle \mathbf {w} }$ of a skew-symmetric tensor ${\displaystyle {\boldsymbol {W}}}$ satisfies the condition

${\displaystyle {\boldsymbol {W}}\cdot \mathbf {a} =\mathbf {w} \times \mathbf {a} }$

for all vectors ${\displaystyle \mathbf {a} }$. In index notation (with respect to a Cartesian basis), we have

${\displaystyle W_{ip}~a_{p}=e_{ijk}~w_{j}~a_{k}}$

Since ${\displaystyle e_{ijk}=-e_{ikj}}$, we can write

${\displaystyle W_{ip}~a_{p}=-e_{ikj}~w_{j}~a_{k}\equiv -e_{ipq}~w_{q}~a_{p}}$

or,

${\displaystyle W_{ip}=-e_{ipq}~w_{q}~.}$

Therefore, the relation between the components of ${\displaystyle {\boldsymbol {\omega }}}$ and ${\displaystyle {\boldsymbol {\theta }}}$ is

${\displaystyle \omega _{ij}=-e_{ijk}~\theta _{k}~.}$

Multiplying both sides by ${\displaystyle e_{pij}}$, we get

${\displaystyle e_{pij}~\omega _{ij}=-e_{pij}~e_{ijk}~\theta _{k}=-e_{pij}~e_{kij}~\theta _{k}~.}$

Recall the identity

${\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}~\delta _{jq}-\delta _{iq}~\delta _{jp}~.}$

Therefore,

${\displaystyle e_{ijk}~e_{pjk}=\delta _{ip}~\delta _{jj}-\delta _{ij}~\delta _{jp}=3\delta _{ip}-\delta _{ip}=2\delta _{ip}}$

Using the above identity, we get

${\displaystyle e_{pij}~\omega _{ij}=-2\delta _{pk}~\theta _{k}=-2\theta _{p}~.}$

Rearranging,

${\displaystyle \theta _{p}=-{\frac {1}{2}}~e_{pij}~\omega _{ij}}$

Now, the components of the tensor ${\displaystyle {\boldsymbol {\omega }}}$ with respect to a Cartesian basis are given by

${\displaystyle \omega _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}$

Therefore, we may write

${\displaystyle \theta _{p}=-{\cfrac {1}{4}}~e_{pij}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}$

Since the curl of a vector ${\displaystyle \mathbf {v} }$ can be written in index notation as

${\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =e_{ijk}~{\frac {\partial u_{k}}{\partial x_{j}}}~\mathbf {e} _{i}}$

we have

${\displaystyle e_{pij}~{\frac {\partial u_{j}}{\partial x_{i}}}~=[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\qquad {\text{and}}\qquad e_{pij}~{\frac {\partial u_{i}}{\partial x_{j}}}~=-e_{pji}{\frac {\partial u_{i}}{\partial x_{j}}}=-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}}$

where ${\displaystyle [~]_{p}}$ indicates the ${\displaystyle p}$-th component of the vector inside the square brackets.

Hence,

${\displaystyle \theta _{p}=-{\cfrac {1}{4}}~\left(-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\right)={\frac {1}{2}}~[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}~.}$

Therefore,

${\displaystyle {{\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} \qquad \square }}$

Proof:

The infinitesimal strain tensor is given by

${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})~.}$

Therefore,

${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\frac {1}{2}}[{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} )+{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} ^{T})]~.}$

Recall that

${\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} )=\mathbf {0} \qquad {\text{and}}\qquad {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} ^{T})={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {u} )~.}$

Hence,

${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\frac {1}{2}}[{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {u} )]~.}$

Also recall that

${\displaystyle {\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} ~.}$

Therefore,

${\displaystyle {{\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}\qquad \square }}$