Kinematics

## Introduction

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* 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.

## Concepts and Definitions

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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.

### Body

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

### Configuration

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A configuration of a body is denoted by the symbol  χ {\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  χ ( X ) {\displaystyle \textstyle {\boldsymbol {\chi }}(\mathbf {X} )} and a current configuration  χ ( x ) {\displaystyle \textstyle {\boldsymbol {\chi }}(\mathbf {x} )} .

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

### Deformation

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A deformation is the relationship between two configurations and is usually denoted by  φ {\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 ( X {\displaystyle \textstyle \mathbf {X} } ) to another ( x {\displaystyle \textstyle \mathbf {x} } ). The inverse mapping should be possible.

This means that

x ≡ { x 1 , x 2 , x 3 } = φ ( X ) X ≡ { X 1 , X 2 , X 3 } = φ − 1 ( x ) {\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.,  J = det ( ∇ φ ) > 0 {\displaystyle \textstyle J=\det({\boldsymbol {\nabla }}{\boldsymbol {\varphi }})>0} .

#### Deformation Gradient

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

F := ∇ φ {\displaystyle {\boldsymbol {F}}:={\boldsymbol {\nabla }}{\boldsymbol {\varphi }}}

In index notation

F i j = ∂ x i ∂ X j {\displaystyle F_{ij}={\frac {\partial x_{i}}{\partial X_{j}}}}

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

### Displacement

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The displacement is usually denoted by the symbol  u {\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

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

In index notation

u i = x i − X i {\displaystyle u_{i}=x_{i}-X_{i}\,}

#### Displacement Gradient

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

The displacement gradient is given by

∇ u = ∇ ( x − X ) = ∇ x − ∇ X = ∇ ( φ ( X ) − X ) = F − 1 {\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,

∂ u i ∂ X j = ∂ x i ∂ X j − ∂ X i ∂ X j = F i j − δ i j {\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}}

## Strains

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### Finite Strain Tensor

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

This strain tensor is defined as

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

In index notation,

E i j = 1 2 ( F k i F k j − δ i j ) {\displaystyle E_{ij}={\cfrac {1}{2}}(F_{ki}F_{kj}-\delta _{ij})}

### Infinitesimal Strain Tensor

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

This strain tensor is defined as

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

In index notation,

ε i j = 1 2 ( ∂ u i ∂ x j + ∂ u j ∂ x i ) {\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

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

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

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

For small strains,  ϵ 2 -> 0 {\displaystyle \textstyle \epsilon ^{2}\rightarrow 0} and

E ≈ ε {\displaystyle {\boldsymbol {E}}\approx {\boldsymbol {\varepsilon }}} .

### Infinitesimal Rotation Tensor

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

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

In index notation,

W i j = 1 2 ( ∂ u i ∂ x j − ∂ u j ∂ x i ) {\displaystyle W_{ij}={\cfrac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}

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

A ∙ a = ω × a . {\displaystyle {\boldsymbol {A}}\bullet \mathbf {a} ={\boldsymbol {\omega }}\times {\mathbf {a} }~.}

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

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

W ∙ a = θ × a . {\displaystyle {\boldsymbol {W}}\bullet \mathbf {a} ={\boldsymbol {\theta }}\times {\mathbf {a} }~.}

where

θ = 1 2 ∇ × u {\displaystyle {\boldsymbol {\theta }}={\cfrac {1}{2}}{\boldsymbol {\nabla }}\times {\boldsymbol {u}}}

### Volume Change Due To Finite Deformation

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

δ V = ∫ B det ( F − 1 ) d v {\displaystyle \delta V=\int _{B}\det({\boldsymbol {F}}-{\boldsymbol {1}})dv}

### Volume Change Due To Infinitesimal Deformation

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

δ V = ∫ B ∇ ∙ u d v = ∫ ∂ B u ∙ n d s {\displaystyle \delta V=\int _{B}{\boldsymbol {\nabla }}\bullet \mathbf {u} dv=\int _{\partial B}\mathbf {u} \bullet \mathbf {n} ds}

because

det F = det ( 1 + ∇ u ) = 1 + Tr ∇ u + O ( ϵ 2 ) = 1 + ∇ ∙ u + O ( ϵ 2 ) {\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  ∇ ∙ u = Tr ( ε ) {\displaystyle \textstyle {\boldsymbol {\nabla }}\bullet \mathbf {u} ={\text{Tr}}({\boldsymbol {\varepsilon }})} is called the dilatation.

A volume change is isochoric (volume preserving) if

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

## Relation between axial vector and displacement

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Let  u {\displaystyle \mathbf {u} } be a displacement field. The displacement gradient tensor is given by  ∇ u {\displaystyle {\boldsymbol {\nabla }}\mathbf {u} } . Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be

ω = 1 2 ( ∇ u − ∇ u T ) . {\displaystyle {\boldsymbol {\omega }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})~.}

Let  θ {\displaystyle {\boldsymbol {\theta }}} be the axial vector associated with the skew symmetric tensor  ω {\displaystyle {\boldsymbol {\omega }}} . Show that

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

Proof:

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

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

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

W i p a p = e i j k w j a k {\displaystyle W_{ip}~a_{p}=e_{ijk}~w_{j}~a_{k}}

Since  e i j k = − e i k j {\displaystyle e_{ijk}=-e_{ikj}} , we can write

W i p a p = − e i k j w j a k ≡ − e i p q w q a p {\displaystyle W_{ip}~a_{p}=-e_{ikj}~w_{j}~a_{k}\equiv -e_{ipq}~w_{q}~a_{p}}

or,

W i p = − e i p q w q . {\displaystyle W_{ip}=-e_{ipq}~w_{q}~.}

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

ω i j = − e i j k θ k . {\displaystyle \omega _{ij}=-e_{ijk}~\theta _{k}~.}

Multiplying both sides by  e p i j {\displaystyle e_{pij}} , we get

e p i j ω i j = − e p i j e i j k θ k = − e p i j e k i j θ k . {\displaystyle e_{pij}~\omega _{ij}=-e_{pij}~e_{ijk}~\theta _{k}=-e_{pij}~e_{kij}~\theta _{k}~.}

Recall the identity

e i j k e p q k = δ i p δ j q − δ i q δ j p . {\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}~\delta _{jq}-\delta _{iq}~\delta _{jp}~.}

Therefore,

e i j k e p j k = δ i p δ j j − δ i j δ j p = 3 δ i p − δ i p = 2 δ i p {\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

e p i j ω i j = − 2 δ p k θ k = − 2 θ p . {\displaystyle e_{pij}~\omega _{ij}=-2\delta _{pk}~\theta _{k}=-2\theta _{p}~.}

Rearranging,

θ p = − 1 2 e p i j ω i j {\displaystyle \theta _{p}=-{\frac {1}{2}}~e_{pij}~\omega _{ij}}

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

ω i j = 1 2 ( ∂ u i ∂ x j − ∂ u j ∂ x i ) {\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

θ p = − 1 4 e p i j ( ∂ u i ∂ x j − ∂ u j ∂ x i ) {\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  v {\displaystyle \mathbf {v} } can be written in index notation as

∇ × v = e i j k ∂ u k ∂ x j e i {\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =e_{ijk}~{\frac {\partial u_{k}}{\partial x_{j}}}~\mathbf {e} _{i}}

we have

e p i j ∂ u j ∂ x i = [ ∇ × u ] p and e p i j ∂ u i ∂ x j = − e p j i ∂ u i ∂ x j = − [ ∇ × u ] p {\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  [ ] p {\displaystyle [~]_{p}} indicates the  p {\displaystyle p} -th component of the vector inside the square brackets.

Hence,

θ p = − 1 4 ( − [ ∇ × u ] p − [ ∇ × u ] p ) = 1 2 [ ∇ × u ] p . {\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,

θ = 1 2 ∇ × u ◻ {\displaystyle {{\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} \qquad \square }}

Kinematics

## Relation between axial vector and strain

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Let  u {\displaystyle \mathbf {u} } be a displacement field. Let  ε {\displaystyle {\boldsymbol {\varepsilon }}} be the strain field (infinitesimal) corresponding to the displacement field and let  θ {\displaystyle {\boldsymbol {\theta }}} be the corresponding infinitesimal rotation vector. Show that

∇ × ε = ∇ θ . {\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}~.}

Proof:

The infinitesimal strain tensor is given by

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

Therefore,

∇ × ε = 1 2 [ ∇ × ( ∇ u ) + ∇ × ( ∇ u T ) ] . {\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

∇ × ( ∇ u ) = 0 and ∇ × ( ∇ u T ) = ∇ ( ∇ × u ) . {\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,

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

Also recall that

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

Therefore,

∇ × ε = ∇ θ ◻ {\displaystyle {{\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}\qquad \square }}

Kinematics

## Related Content

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* Introduction to Elasticity
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