Kinematics

Introduction

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

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

We usually denote a body by the symbol $\textstyle B$ . 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 $\textstyle \boldsymbol{\chi}$ . 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 $\textstyle \boldsymbol{\chi}(\mathbf{X})$ and a current configuration $\textstyle \boldsymbol{\chi}(\mathbf{x})$ .

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

Deformation

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

This means that

\begin{align} \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{align}

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

Deformation Gradient

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

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

In index notation

F_{ij} = \frac{\partial x_i }{\partial X_j}

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

Displacement

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

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

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

In index notation

u_i = x_i - X_i \,

Displacement Gradient

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

The displacement gradient is given by

\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,

\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

Finite Strain Tensor

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

This strain tensor is defined as

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

In index notation,

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 ( $\textstyle \boldsymbol{\epsilon}$ ).

This strain tensor is defined as

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

In index notation,

\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

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

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

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

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

\boldsymbol{E} \approx \boldsymbol{\varepsilon}

Infinitesimal Rotation Tensor

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

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

In index notation,

W_{ij} = \cfrac{1}{2}\left(\frac{\partial u_i}{\partial x_j}- \frac{\partial u_j}{\partial x_i}\right)

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

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

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

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

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

where

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

Volume Change Due To Finite Deformation

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

\delta V = \int_{B} \det (\boldsymbol{F} -\boldsymbol{1}) dv

Volume Change Due To Infinitesimal Deformation

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

\delta V = \int_{B} \boldsymbol{\nabla}\bullet\mathbf{u} dv = \int_{\partial B} \mathbf{u}\bullet\mathbf{n} ds

because

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

A volume change is isochoric (volume preserving) if

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

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

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

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

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

Proof:

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

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

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

W_{ip}~a_p = e_{ijk}~w_j~a_k

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

W_{ip}~a_p = -e_{ikj}~w_j~a_k \equiv -e_{ipq}~w_q~a_p

or,

W_{ip} = -e_{ipq}~w_q ~.

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

\omega_{ij} = -e_{ijk}~\theta_k ~.

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

e_{pij}~\omega_{ij} = -e_{pij}~e_{ijk}~\theta_k = -e_{pij}~e_{kij}~\theta_k~.

Recall the identity

e_{ijk}~e_{pqk} = \delta_{ip}~\delta_{jq} - \delta_{iq}~\delta_{jp}~.

Therefore,

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_{pij}~\omega_{ij} = -2\delta_{pk}~\theta_k = -2\theta_p~.

Rearranging,

\theta_p = -\frac{1}{2}~e_{pij}~\omega_{ij}

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

\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

\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 $\mathbf{v}$ can be written in index notation as

\boldsymbol{\nabla} \times \mathbf{v} = e_{ijk}~\frac{\partial u_k}{\partial x_j}~\mathbf{e}_i

we have

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 $[~]_p$ indicates the $p$ -th component of the vector inside the square brackets.

Hence,

\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,

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

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