Kinematics

• Kinematics
  • Displacements, strains and the relations between displacements and strains.
  • We view the theory of infinitesimal linear deformations as a first-order approximation of the theory of finite deformations.
  • Note that the linear theory can be derived independently of the finite theory and is completely self-consistent on its own.

The following concepts and definitions are based on Gurtin (1972) and Truesdell and Noll (1992). These definitions are useful both for the linear and the nonlinear theory of elasticity.

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)

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.

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., ${\displaystyle \textstyle J=\det({\boldsymbol {\nabla }}{\boldsymbol {\varphi }})>0}$.

The deformation gradient is usually denoted by ${\displaystyle \textstyle {\boldsymbol {F}}}$ and is defined as

${\displaystyle {\boldsymbol {F}}:={\boldsymbol {\nabla }}{\boldsymbol {\varphi }}}$

In index notation

${\displaystyle F_{ij}={\frac {\partial x_{i}}{\partial X_{j}}}}$

For a deformation to be allowable, we must be able to invert ${\displaystyle \textstyle {\boldsymbol {F}}}$. That is why we require that ${\displaystyle \textstyle J=\det {\boldsymbol {F}}>0}$. Otherwise, the body may undergo deformations that are unphysical.

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

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

The gradient of the displacement is denoted by ${\displaystyle \textstyle {\boldsymbol {\nabla }}\mathbf {u} }$.

The displacement gradient is given by

${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} ={\boldsymbol {\nabla }}{(\mathbf {x} -\mathbf {X} )}={\boldsymbol {\nabla }}{\mathbf {x} }-{\boldsymbol {\nabla }}{\mathbf {X} }={\boldsymbol {\nabla }}{({\boldsymbol {\varphi }}(\mathbf {X} )-\mathbf {X} )}={\boldsymbol {F}}-{\boldsymbol {\it {1}}}}$

In index notation,

${\displaystyle {\frac {\partial u_{i}}{\partial X_{j}}}={\frac {\partial x_{i}}{\partial X_{j}}}-{\frac {\partial X_{i}}{\partial X_{j}}}=F_{ij}-\delta _{ij}}$

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,

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

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

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

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

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

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

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

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

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

Relation between axial vector and displacement [edit | edit source] Let ${\displaystyle \mathbf {u} }$ be a displacement field. The displacement gradient tensor is given by ${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} }$. Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be ${\displaystyle {\boldsymbol {\omega }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})~.}$ Let ${\displaystyle {\boldsymbol {\theta }}}$ be the axial vector associated with the skew symmetric tensor ${\displaystyle {\boldsymbol {\omega }}}$. Show that ${\displaystyle {\boldsymbol {\theta }}={\frac {1}{2}}~{\boldsymbol {\nabla }}\times \mathbf {u} ~.}$

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,

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

Relation between axial vector and strain [edit | edit source] Let ${\displaystyle \mathbf {u} }$ be a displacement field. Let ${\displaystyle {\boldsymbol {\varepsilon }}}$ be the strain field (infinitesimal) corresponding to the displacement field and let ${\displaystyle {\boldsymbol {\theta }}}$ be the corresponding infinitesimal rotation vector. Show that ${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}~.}$

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,

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

• Introduction to Elasticity
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