Continuum mechanics/Strains and deformations: Wikis

Advertisements

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

Study guide

Up to date as of January 14, 2010

From Wikiversity

Contents

Strain Measures in three dimensions

The motion of a body

Initial orthonormal basis:

 (\boldsymbol{E}_1, \boldsymbol{E}_2, \boldsymbol{E}_3)

Deformed orthonormal basis:

 (\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)

We assume that these coincide.

Advertisements

Motion

 \mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t) = \mathbf{x}(\mathbf{X}, t)

Deformation Gradient

\begin{align} \boldsymbol{F} & = \frac{\partial \boldsymbol{\varphi}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_o \boldsymbol{\varphi} \ & = \frac{\partial \mathbf{x}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_X \boldsymbol{\varphi} \end{align}

Effect of \boldsymbol{F}:

\begin{align} d\mathbf{x}_1 & = \boldsymbol{F}\bullet d\mathbf{X}_1 ~;& d\mathbf{x}_2 & = \boldsymbol{F}\bullet d\mathbf{X}_2 \end{align}

Dyadic notation:

 \boldsymbol{F} = F_{iJ}~\mathbf{e}_i\otimes\boldsymbol{E}_J

Index notation:

 F_{iJ} = \frac{\partial x_i}{\partial X_J}

The determinant of the deformation gradient is usually denoted by J and is a measure of the change in volume, i.e.,

 J = \det{\boldsymbol{F}}

Push Forward and Pull Back

Forward Map:

 \mathbf{x} = \boldsymbol{\varphi}(\mathbf{X},t)

Forward deformation gradient:

 \boldsymbol{F} = \frac{\partial \mathbf{x}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_o \boldsymbol{\varphi}

Dyadic notation:

 \boldsymbol{F} = \sum_{i,J=1}^3 \frac{\partial x_i}{\partial X_J}~\mathbf{e}_i\otimes\boldsymbol{E}_J

Effect of deformation gradient:

 d\mathbf{x} = \boldsymbol{F}\bullet d\mathbf{X} = \boldsymbol{\varphi}_{*}[d\mathbf{X}]

Push Forward operation:

 \boldsymbol{\varphi}_{*}[\bullet]
  • d\mathbf{X} = material vector.
  • d\mathbf{x} = spatial vector.

Inverse map:

 \mathbf{X} = \boldsymbol{\varphi}^{-1}(\mathbf{x},t)

Inverse deformation gradient:

 \boldsymbol{F}^{-1} = \frac{\partial \mathbf{X}}{\partial \mathbf{x}} = \boldsymbol{\nabla} \boldsymbol{\varphi}^{-1}

Dyadic notation:

 \boldsymbol{F}^{-1} = \sum_{i,J=1}^3 \frac{\partial X_I}{\partial x_j}~\boldsymbol{E}_I\otimes\mathbf{e}_j

Effect of inverse deformation gradient:

 d\mathbf{X} = \boldsymbol{F}^{-1}\bullet d\mathbf{x} = \boldsymbol{\varphi}^{*}[d\mathbf{x}]

Pull Back operation:

 \boldsymbol{\varphi}^{*}[\bullet]
  • d\mathbf{X} = material vector.
  • d\mathbf{x} = spatial vector.
Example
Push forward and pull back

Motion:

\begin{align} x_1 & = \cfrac{1}{4}(18 + 4X_1 + 6X_2) \ x_2 & = \cfrac{1}{4}(14 + 6X_2) \end{align}

Deformation Gradient:

 F_{ij} = \frac{\partial x_i}{\partial X_j} \implies \mathbf{F} = \frac{1}{2}\begin{bmatrix} 2 & 3 \\ 0 & 3 \end{bmatrix}

Inverse Deformation Gradient:

 \mathbf{F}^{-1} = \cfrac{1}{3}\begin{bmatrix} 3 & -3 \\ 0 & 2 \end{bmatrix}

Push Forward:

\begin{align} \boldsymbol{\varphi}_{*}[\boldsymbol{E}_1] & = \mathbf{F}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \boldsymbol{\varphi}_{*}[\boldsymbol{E}_2] & = \mathbf{F}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1.5 \\ 1.5 \end{bmatrix} \end{align}

Pull Back:

\begin{align} \boldsymbol{\varphi}^{*}[\mathbf{e}_1] & = \mathbf{F}^{-1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \boldsymbol{\varphi}^{*}[\mathbf{e}_2] & = \mathbf{F}^{-1}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 2/3 \end{bmatrix} \end{align}

Cauchy-Green Deformation Tensors

Right Cauchy-Green Deformation Tensor

Recall:

 d\mathbf{x}_1 = \boldsymbol{F}\bullet d\mathbf{X}_1 ~;~~ d\mathbf{x}_2 = \boldsymbol{F}\bullet d\mathbf{X}_2

Therefore,

 d\mathbf{x}_1\bullet d\mathbf{x}_2 = (\boldsymbol{F}\bullet d\mathbf{X}_1)\bullet(\boldsymbol{F}\bullet d\mathbf{X}_2)

Using index notation:

\begin{align} d\mathbf{x}_1\bullet d\mathbf{x}_2 & = (F_{ij}~dX^1_j)(F_{ik}~dX^2_k)\ & = dX^1_j~(F_{ij}~F_{ik})~dX^2_k \ & = d\mathbf{X}_1\bullet(\boldsymbol{F}^T\bullet\boldsymbol{F})\bullet d\mathbf{X}_2 \ & = d\mathbf{X}_1\bullet\boldsymbol{C} \bullet d\mathbf{X}_2 \end{align}

Right Cauchy-Green tensor:

 \boldsymbol{C} = \boldsymbol{F}^T\bullet\boldsymbol{F}

Left Cauchy-Green Deformation Tensor

Recall:

 d\mathbf{X}_1 = \boldsymbol{F}^{-1}\bullet d\mathbf{x}_1 ~;~~ d\mathbf{X}_2 = \boldsymbol{F}^{-1}\bullet d\mathbf{x}_2

Therefore,

 d\mathbf{X}_1\bullet d\mathbf{X}_2 = (\boldsymbol{F}^{-1}\bullet d\mathbf{x}_1)\bullet(\boldsymbol{F}^{-1}\bullet d\mathbf{x}_2)

Using index notation:

\begin{align} d\mathbf{X}_1\bullet d\mathbf{X}_2 & = (F^{-1}_{ij}~dx^1_j)(F^{-1}_{ik}~dx^2_k)\ & = dx^1_j~(F^{-1}_{ij}~F^{-1}_{ik})~dx^2_k \ & = d\mathbf{x}_1\bullet(\boldsymbol{F}^{-T}\bullet\boldsymbol{F}^{-1})\bullet d\mathbf{x}_2 \ & = d\mathbf{x}_1\bullet(\boldsymbol{F}\bullet\boldsymbol{F}^T)^{-1}\bullet d\mathbf{x}_2 \ & = d\mathbf{x}_1\bullet\mathbf{b}^{-1}\bullet d\mathbf{x}_2 \end{align}

Left Cauchy-Green (Finger) tensor:

 \mathbf{b} = \boldsymbol{F}\bullet\boldsymbol{F}^T

Strain Measures

Green (Lagrangian) Strain

\begin{align} \frac{1}{2}(d\mathbf{x}_1\bullet d\mathbf{x}_2 & - d\mathbf{X}_1\bullet d\mathbf{X}_2) \ & = \frac{1}{2}d\mathbf{X}_1\bullet(\boldsymbol{C} - \boldsymbol{I}) \bullet d\mathbf{X}_2 \ & = d\mathbf{X}_1\bullet\boldsymbol{E} \bullet d\mathbf{X}_2 \end{align}

Green strain tensor:

 \begin{align} \boldsymbol{E} & = \frac{1}{2}(\boldsymbol{C} - \boldsymbol{I}) \ & = \frac{1}{2}(\boldsymbol{F}^T\bullet\boldsymbol{F} - \boldsymbol{I}) \ & = \frac{1}{2}\left[\boldsymbol{\nabla}_o \mathbf{u} + (\boldsymbol{\nabla}_o \mathbf{u})^T + \boldsymbol{\nabla}_o \mathbf{u}\bullet(\boldsymbol{\nabla_o \mathbf{u})^T}\right] \end{align}

Index notation:

 \begin{align} E_{ij} & = \frac{1}{2}(F_{ki}~F_{kj} - \delta_{ij})\ & = \frac{1}{2}\left(\frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} + \frac{\partial u_k}{\partial X_i}\frac{\partial u_k}{\partial X_j}\right) \end{align}

Almansi (Eulerian) Strain

\begin{align} \frac{1}{2}(d\mathbf{x}_1\bullet d\mathbf{x}_2 & - d\mathbf{X}_1\bullet d\mathbf{X}_2) \ & = \frac{1}{2}d\mathbf{x}_1\bullet(\boldsymbol{I} - \mathbf{b}^{-1})\bullet d\mathbf{x}_2 \ & = d\mathbf{x}_1\bullet\mathbf{e} \bullet d\mathbf{x}_2 \end{align}

Almansi strain tensor:

 \begin{align} \mathbf{e} & = \frac{1}{2}(\boldsymbol{I} - \mathbf{b}^{-1}) \ & = \frac{1}{2}(\boldsymbol{I} - \boldsymbol{F}^{-T}\bullet\boldsymbol{F}^{-1}) \end{align}

Index notation:

 \begin{align} e_{ij} & = \frac{1}{2}(\delta_{ij} - F^{-1}_{ki}~F^{-1}_{kj}) \end{align}

Push Forward and Pull Back

Recall:

 d\mathbf{x}_1\bullet\mathbf{e} \bullet d\mathbf{x}_2 = d\mathbf{X}_1\bullet\boldsymbol{E} \bullet d\mathbf{X}_2

Now,

\begin{align} d\mathbf{x}_1\bullet\mathbf{e} \bullet d\mathbf{x}_2 & =(\boldsymbol{F}\bullet d\mathbf{X}_1)\bullet\mathbf{e}\bullet(\boldsymbol{F}\bullet d\mathbf{X}_2) \ & =d\mathbf{X}_1\bullet(\boldsymbol{F}^T\bullet\mathbf{e}\bullet\boldsymbol{F})\bullet d\mathbf{X}_2 \ & =d\mathbf{X}_1\bullet\boldsymbol{E}\bullet d\mathbf{X}_2 \end{align}

Therefore,

\begin{align} \boldsymbol{E} & = \boldsymbol{F}^T\bullet\mathbf{e}\bullet\boldsymbol{F} \ \implies \mathbf{e} & = \boldsymbol{F}^{-T}\bullet\boldsymbol{E}\bullet\boldsymbol{F}^{-1} \end{align}

Push Forward:

 \mathbf{e} = \boldsymbol{\varphi}_{*}[\boldsymbol{E}] =\boldsymbol{F}^{-T}\bullet\boldsymbol{E}\bullet\boldsymbol{F}^{-1}

Pull Back:

 \boldsymbol{E} = \boldsymbol{\varphi}^{*}[\mathbf{e}] =\boldsymbol{F}^T\bullet\mathbf{e}\bullet\boldsymbol{F}

Some useful results

Derivative of J with respect to the deformation gradient

We often need to compute the derivative of J = \det{\boldsymbol{F}} with respect the the deformation gradient \boldsymbol{F}. From tensor calculus we have, for any second order tensor \boldsymbol{A}

 \cfrac{\partial}{\partial \boldsymbol{A}}( \det{\boldsymbol{A}}) = \det{\boldsymbol{A}}~\boldsymbol{A}^{-T}

Therefore,

 \cfrac{\partial J}{\partial \boldsymbol{F}} = J~\boldsymbol{F}^{-T}

Derivative of J with respect to the right Cauchy-Green deformation tensor

The derivative of J with respect to the right Cauchy-Green deformation tensor (\boldsymbol{C}) is also often encountered in continuum mechanics.

To calculate the derivative of J = \det{\boldsymbol{F}} with respect to \boldsymbol{C}, we recall that (for any second order tensor \boldsymbol{T})

 \frac{\partial \boldsymbol{C}}{\partial \boldsymbol{F}}:\boldsymbol{T} = \frac{\partial }{\partial \boldsymbol{F}}(\boldsymbol{F}^T\cdot\boldsymbol{F}):\boldsymbol{T} = (\boldsymbol{\mathsf{I}}^T:\boldsymbol{T})\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot(\boldsymbol{\mathsf{I}}:\boldsymbol{T}) = \boldsymbol{T}^T\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\boldsymbol{T}

Also,

 \frac{\partial J}{\partial \boldsymbol{F}}:\boldsymbol{T} = \frac{\partial J}{\partial \boldsymbol{C}}:(\frac{\partial \boldsymbol{C}}{\partial \boldsymbol{F}}:\boldsymbol{T}) = \frac{\partial J}{\partial \boldsymbol{C}}:(\boldsymbol{T}^T\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\boldsymbol{T}) = \left[\boldsymbol{F}\cdot\frac{\partial J}{\partial \boldsymbol{C}}\right]:\boldsymbol{T} + \left[\boldsymbol{F}\cdot\left(\frac{\partial J}{\partial \boldsymbol{C}}\right)^T\right]:\boldsymbol{T}

From the symmetry of \boldsymbol{C} we have

 \frac{\partial J}{\partial \boldsymbol{C}} = \left(\frac{\partial J}{\partial \boldsymbol{C}}\right)^T

Therefore, involving the arbitrariness of \boldsymbol{T}, we have

 \frac{\partial J}{\partial \boldsymbol{F}} = 2~\boldsymbol{F}\cdot\frac{\partial J}{\partial \boldsymbol{C}}

Hence,

 \frac{\partial J}{\partial \boldsymbol{C}} = \frac{1}{2}~\boldsymbol{F}^{-1}\cdot\frac{\partial J}{\partial \boldsymbol{F}} ~.

Also recall that

 \frac{\partial J}{\partial \boldsymbol{F}} = J~\boldsymbol{F}^{-T}

Therefore,

 \frac{\partial J}{\partial \boldsymbol{C}} = \frac{1}{2}~J~\boldsymbol{F}^{-1}\cdot\boldsymbol{F}^{-T} = \cfrac{J}{2}~\boldsymbol{C}^{-1}

In index notation,

 \frac{\partial J}{\partial C_{IJ}} = \cfrac{J}{2}~C^{-1}_{IJ}

Derivative of the inverse of the right Cauchy-Green tensor

Another result that is often useful is that for the derivative of the inverse of the right Cauchy-Green tensor (\boldsymbol{C}).

Recall that, for a second order tensor \boldsymbol{A},

 \frac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}}:\boldsymbol{T} = -\boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}

In index notation

 \frac{\partial A^{-1}_{ij}}{\partial A_{kl}}~T_{kl} = B_{ijkl}~T_{kl} = -A^{-1}_{ik}~T_{kl}~A^{-1}_{lj}

or,

 \frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = B_{ijkl} = -A^{-1}_{ik}~A^{-1}_{lj}

Using this formula and noting that since \boldsymbol{C} is a symmetric second order tensor, the derivative of its inverse is a symmetric fourth order tensor we have

 \frac{\partial C^{-1}_{IJ}}{\partial C_{KL}} = -\frac{1}{2}~(C^{-1}_{IK}~C^{-1}_{JL} + C^{-1}_{JK}~C^{-1}_{IL})

Advertisements






Got something to say? Make a comment.
Your name
Your email address
Message