Continuum mechanics/Strains and deformations

Strain Measures in three dimensions

The motion of a body

Initial orthonormal basis:

$({𝑬}_1,{𝑬}_2,{𝑬}_3)$

Deformed orthonormal basis:

$({𝐞}_1,{𝐞}_2,{𝐞}_3)$

We assume that these coincide.

Motion

$𝐱=𝝋(𝐗,t)=𝐱(𝐗,t)$

Deformation Gradient

$\begin{array}{cc}𝑭& =\frac{\partial 𝝋}{\partial 𝐗}={𝜵}_o𝝋\\ & =\frac{\partial 𝐱}{\partial 𝐗}={𝜵}_X𝝋\end{array}$

Effect of $𝑭$:

$\begin{array}{cccc}d{𝐱}_1& =𝑭\bullet d{𝐗}_1;& d{𝐱}_2& =𝑭\bullet d{𝐗}_2\end{array}$

Dyadic notation:

$𝑭=F_{iJ}{𝐞}_i\otimes {𝑬}_J$

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.,

$J=det𝑭$

Push Forward and Pull Back

Forward Map:

$𝐱=𝝋(𝐗,t)$

Forward deformation gradient:

$𝑭=\frac{\partial 𝐱}{\partial 𝐗}={𝜵}_o𝝋$

Dyadic notation:

$𝑭=\sum _{i,J=1}^3\frac{\partial x_i}{\partial X_J}{𝐞}_i\otimes {𝑬}_J$

Effect of deformation gradient:

$d𝐱=𝑭\bullet d𝐗={𝝋}_{\ast }[d𝐗]$

Push Forward operation:

${𝝋}_{\ast }[\bullet ]$

• $d𝐗$ = material vector.
• $d𝐱$ = spatial vector.

Inverse map:

$𝐗={𝝋}^{-1}(𝐱,t)$

Inverse deformation gradient:

${𝑭}^{-1}=\frac{\partial 𝐗}{\partial 𝐱}=𝜵{𝝋}^{-1}$

Dyadic notation:

${𝑭}^{-1}=\sum _{i,J=1}^3\frac{\partial X_I}{\partial x_j}{𝑬}_I\otimes {𝐞}_j$

Effect of inverse deformation gradient:

$d𝐗={𝑭}^{-1}\bullet d𝐱={𝝋}^{\ast }[d𝐱]$

Pull Back operation:

${𝝋}^{\ast }[\bullet ]$

• $d𝐗$ = material vector.
• $d𝐱$ = spatial vector.

Example

Push forward and pull back

Motion:

$\begin{array}{cc}{x}_1& =\frac{1}{4}(18+4X_1+6X_2)\\ x_2& =\frac{1}{4}(14+6X_2)\end{array}$

Deformation Gradient:

$F_{ij}=\frac{\partial x_i}{\partial X_j}⟹𝐅=\frac{1}{2}\left[\begin{array}{cc}2& 3\\ 0& 3\end{array}\right]$

Inverse Deformation Gradient:

${𝐅}^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& -3\\ 0& 2\end{array}\right]$

Push Forward:

$\begin{array}{cc}{𝝋}_{\ast }[{𝑬}_1]& =𝐅\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\\ {𝝋}_{\ast }[{𝑬}_2]& =𝐅\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1.5\\ 1.5\end{array}\right]\end{array}$

Pull Back:

$\begin{array}{cc}{𝝋}^{\ast }[{𝐞}_1]& ={𝐅}^{-1}\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\\ {𝝋}^{\ast }[{𝐞}_2]& ={𝐅}^{-1}\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}-1\\ 2/3\end{array}\right]\end{array}$

Cauchy-Green Deformation Tensors

Right Cauchy-Green Deformation Tensor

Recall:

$d{𝐱}_1=𝑭\bullet d{𝐗}_1;d{𝐱}_2=𝑭\bullet d{𝐗}_2$

Therefore,

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

Using index notation:

$\begin{array}{cc}d{𝐱}_1\bullet d{𝐱}_2& =(F_{ij}d{X}_j^1)(F_{ik}d{X}_k^2)\\ & =d{X}_j^1(F_{ij}{F}_{ik})d{X}_k^2\\ & =d{𝐗}_1\bullet ({𝑭}^T\bullet 𝑭)\bullet d{𝐗}_2\\ & =d{𝐗}_1\bullet 𝑪\bullet d{𝐗}_2\end{array}$

Right Cauchy-Green tensor:

$𝑪={𝑭}^T\bullet 𝑭$

Left Cauchy-Green Deformation Tensor

Recall:

$d{𝐗}_1={𝑭}^{-1}\bullet d{𝐱}_1;d{𝐗}_2={𝑭}^{-1}\bullet d{𝐱}_2$

Therefore,

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

Using index notation:

$\begin{array}{cc}d{𝐗}_1\bullet d{𝐗}_2& =(F_{ij}^{-1}d{x}_j^1)(F_{ik}^{-1}d{x}_k^2)\\ & =d{x}_j^1(F_{ij}^{-1}{F}_{ik}^{-1})d{x}_k^2\\ & =d{𝐱}_1\bullet ({𝑭}^{-T}\bullet {𝑭}^{-1})\bullet d{𝐱}_2\\ & =d{𝐱}_1\bullet (𝑭\bullet {𝑭}^T{)}^{-1}\bullet d{𝐱}_2\\ & =d{𝐱}_1\bullet {𝐛}^{-1}\bullet d{𝐱}_2\end{array}$

Left Cauchy-Green (Finger) tensor:

$𝐛=𝑭\bullet {𝑭}^T$

Strain Measures

Green (Lagrangian) Strain

$\begin{array}{cc}\frac{1}{2}(d{𝐱}_1\bullet d{𝐱}_2& -d{𝐗}_1\bullet d{𝐗}_2)\\ & =\frac{1}{2}d{𝐗}_1\bullet (𝑪-𝑰)\bullet d{𝐗}_2\\ & =d{𝐗}_1\bullet 𝑬\bullet d{𝐗}_2\end{array}$

Green strain tensor:

$\begin{array}{cc}𝑬& =\frac{1}{2}(𝑪-𝑰)\\ & =\frac{1}{2}({𝑭}^T\bullet 𝑭-𝑰)\\ & =\frac{1}{2}[{\mathrm{\nabla }}_o𝐮+({\mathrm{\nabla }}_o𝐮{)}^T+{\mathrm{\nabla }}_o𝐮\bullet ({\mathrm{\nabla }}_{𝒐}𝐮{\mathit{)}}^{𝑻}]\end{array}$

Index notation:

$\begin{array}{cc}{E}_{ij}& =\frac{1}{2}(F_{ki}{F}_{kj}-{\delta }_{ij})\\ & =\frac{1}{2}(\frac{\partial u_i}{\partial X_j}+\frac{\partial u_j}{\partial X_i}+\frac{\partial u_k}{\partial X_i}\frac{\partial u_k}{\partial X_j})\end{array}$

Almansi (Eulerian) Strain

$\begin{array}{cc}\frac{1}{2}(d{𝐱}_1\bullet d{𝐱}_2& -d{𝐗}_1\bullet d{𝐗}_2)\\ & =\frac{1}{2}d{𝐱}_1\bullet (𝑰-{𝐛}^{-1})\bullet d{𝐱}_2\\ & =d{𝐱}_1\bullet 𝐞\bullet d{𝐱}_2\end{array}$

Almansi strain tensor:

$\begin{array}{cc}𝐞& =\frac{1}{2}(𝑰-{𝐛}^{-1})\\ & =\frac{1}{2}(𝑰-{𝑭}^{-T}\bullet {𝑭}^{-1})\end{array}$

Index notation:

$\begin{array}{cc}{e}_{ij}& =\frac{1}{2}({\delta }_{ij}-F_{ki}^{-1}{F}_{kj}^{-1})\end{array}$

Push Forward and Pull Back

Recall:

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

Now,

$\begin{array}{cc}d{𝐱}_1\bullet 𝐞\bullet d{𝐱}_2& =(𝑭\bullet d{𝐗}_1)\bullet 𝐞\bullet (𝑭\bullet d{𝐗}_2)\\ & =d{𝐗}_1\bullet ({𝑭}^T\bullet 𝐞\bullet 𝑭)\bullet d{𝐗}_2\\ & =d{𝐗}_1\bullet 𝑬\bullet d{𝐗}_2\end{array}$

Therefore,

$\begin{array}{cc}𝑬& ={𝑭}^T\bullet 𝐞\bullet 𝑭\\ ⟹𝐞& ={𝑭}^{-T}\bullet 𝑬\bullet {𝑭}^{-1}\end{array}$

Push Forward:

$𝐞={𝝋}_{\ast }[𝑬]={𝑭}^{-T}\bullet 𝑬\bullet {𝑭}^{-1}$

Pull Back:

$𝑬={𝝋}^{\ast }[𝐞]={𝑭}^T\bullet 𝐞\bullet 𝑭$

Some useful results

Derivative of J with respect to the deformation gradient

We often need to compute the derivative of $J=det𝑭$ with respect to the deformation gradient $𝑭$. From tensor calculus we have, for any second order tensor $𝑨$

$\frac{\partial }{\partial 𝑨}(det𝑨)=det𝑨{𝑨}^{-T}$

Therefore,

$\frac{\partial J}{\partial 𝑭}=J{𝑭}^{-T}$

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 ($𝑪$) is also often encountered in continuum mechanics.

To calculate the derivative of $J=det𝑭$ with respect to $𝑪$, we recall that (for any second order tensor $𝑻$)

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

Also,

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

From the symmetry of $𝑪$ we have

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

Therefore, involving the arbitrariness of $𝑻$, we have

$\frac{\partial J}{\partial 𝑭}=2𝑭\cdot \frac{\partial J}{\partial 𝑪}$

Hence,

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

Also recall that

$\frac{\partial J}{\partial 𝑭}=J{𝑭}^{-T}$

Therefore,

$\frac{\partial J}{\partial 𝑪}=\frac{1}{2}J{𝑭}^{-1}\cdot {𝑭}^{-T}=\frac{J}{2}{𝑪}^{-1}$

In index notation,

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

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 ($𝑪$).

Recall that, for a second order tensor $𝑨$,

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

In index notation

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

or,

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

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

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

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