Nonlinear finite elements/Kinematics

Strain Measures in three dimensions
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.

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
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}$$

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 $$

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} $$

Derivative of J with respect to the deformation gradient
We often need to compute the derivative of $$J = \det{\boldsymbol{F}}$$ with respect to 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,

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, In index notation,

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