# Computational & Multiscale Mechanics of Materials

## Fracture Mechanics Online Class

Other online classes : Aircraft structures

# Energetic Approach > J-Integral

As seen in the previous pages, the crack closure integral has some limitations: It can only be "easily" computed for linear and elastic materials, and is useful for cracks growing straight ahead only. The following introduces a more general energy-related concept called the J-Integral.

## The J-integral concept

### Homogeneous bodies

In 1968, Rice suggested to compute the energy that flows toward the crack tip. To do so he considered a homogeneous un-cracked body $B$ with the following conditions:

• $D$ is a sub-volume of boundary $\partial D$;
• The stress tensor derives from a potential $U$;
• On the boundary $\partial D$, traction forces $\mathbf{T}$ are defined as $\mathbf{\sigma} \cdot \mathbf{n}$;
• The problem is static, so $\nabla \cdot \mathbf{\sigma} = 0$.

Using these notations the J-integral is a vector defined as:

$$\mathbf{J} = \int_{\partial D} \left[U\left(\mathbf{\nabla}\mathbf{u}\right) \mathbf{n} - \left(\mathbf{\nabla}\mathbf{u}\right)^T \mathbf{T}\right]dA,\label{eq:Jintegral}$$

or, in the indicial form:

\mathbf{J}_i = \int_{\partial D} \left[U\left(\mathbf{\varepsilon}\right) \mathbf{n}_i - \mathbf{\nabla}_i \mathbf{u}_k \mathbf{\sigma}_{km} \mathbf{n}_m\right] dA.\label{eq:JintegralIndices}

As $\mathbf{\sigma}^T=\mathbf{\sigma}$ and $\nabla \cdot \mathbf{\sigma} = 0$, applying the Gauss theorem on this last expression leads to

\begin{eqnarray} \mathbf{J}_i &=& \int_{D} \left[\underbrace{\frac{\partial U}{\partial \mathbf{\varepsilon}_{km}}}_{\mathbf{\sigma}_{km}}\frac{\mathbf{u}_{k,mi}+\mathbf{u}_{m,ki}}{2} - \mathbf{\sigma}_{km}\mathbf{u}_{k,im}-\underbrace{\mathbf{\sigma}_{km,m}}_{=0}\mathbf{u}_{k,i} \right] dD=\mathbf{\sigma}_{km}\mathbf{u}_{k,mi}- \mathbf{\sigma}_{km}\mathbf{u}_{k,im}= 0. \label{eq:Jhomog}\end{eqnarray}

This relation means that the energy flowing through a closed surface in an homogeneous medium is equal to zero. But what happens if the body is heterogeneous or cracked?

### Heterogeneous bodies

For heterogeneous bodies the internal material potential $U(\mathbf{\nabla} \mathbf{u},\, \mathbf{X})$ depends on the position $\mathbf{X}$, so when applying the Gauss theorem we should consider that

$$\frac{D U\left(\mathbf{\nabla}\mathbf{u},\mathbf{X}\right)}{D\mathbf{X}_i} = \frac{\partial U}{\partial \mathbf{\varepsilon}}:\mathbf{\varepsilon}_{,i} +\frac{\partial U}{\partial \mathbf{X}_i}.$$

As a result $\mathbf{J} \neq 0$ even for un-cracked bodies as the second term remains in (\ref{eq:Jhomog}).

### Cracked homogeneous bodies (2D)

Let us now consider a cracked homogeneous material. We can still use (\ref{eq:Jhomog}) if the contour $\Gamma$ embeds a homogeneous part, meaning if it does not intercept the crack. Considering a crack parallel to the x-axis with the contour $\Gamma=\Gamma_1+\Gamma^++\Gamma^--\Gamma_2$ going around the crack, see Picture III.26, then (\ref{eq:Jhomog}) is satisfied and

$$\mathbf{J}_x = \oint_\Gamma \left[U\left(\mathbf{\varepsilon}\right) \mathbf{n}_x - \mathbf{\mathbf{u}}_{,x} \cdot \mathbf{T}\right] dl = 0, \label{eq:JCrackedFull}$$

with $\oint_\Gamma = \int_{\Gamma_1} + \int_{\Gamma^+}+ \int_{\Gamma^-} - \int_{\Gamma_2}$. We can now analyze the contribution on the different curves. Along $\Gamma^+$ and $\Gamma^-$ we have

• $\mathbf{n}_x = 0$ and $\mathbf{n}_y \pm 1$;
• The crack is stress-free (no friction): $\mathbf{T}_\alpha = \mathbf{\sigma}_{\alpha y} \mathbf{n}_y = 0$;
• Hence $\int_{\Gamma^-} =\int_{\Gamma^+} = 0$.

So (\ref{eq:JCrackedFull}) simplifies into the so-called J-integral, which is the energy that flows toward the crack tip:

$$J = \int_{\Gamma_1} \left[U\left(\mathbf{\varepsilon}\right) \mathbf{n}_x - \mathbf{\mathbf{u}}_{,x} \cdot \mathbf{T}\right] dl = \int_ {\Gamma_2} \left[U\left(\mathbf{\varepsilon}\right) \mathbf{n}_x - \mathbf{\mathbf{u}}_{,x} \cdot \mathbf{T}\right] dl.\label{eq:J}$$

As it can be deduced from (\ref{eq:J}), this integral

• Is path independent (as long as the contour embeds the crack tip);
• Does not request linearity of the material (only the existence of an internal potential has been assumed);
• Does not depend on subsequent crack growth direction.

This concept is thus particularly general and is widely used in analytical and numerical analyzes, as it will be shown later.

## Particular cases and the J-integral

Although the concept is general as long as an internal potential exists, the J-Integral can be specialized and in particular can be related to the concepts of energy release rate and of SIFs studied before.

### For cracks growing straight ahead

Heading back to the crack closure integral, we have found the following potential energy variation for a growing cavity, see Picture III.27:

\begin{eqnarray} \Delta \Pi_T &=& \int_{B-\Delta B} U\left(\mathbf{\nabla}\mathbf{u}+\mathbf{\nabla}\Delta\mathbf{u}\right)-U\left(\mathbf{\nabla}\mathbf{u}\right) dB-\nonumber\\&&\int_{\Delta B} U\left(\mathbf{\nabla}\mathbf{u}\right) dB - \int_{\partial_N B} \bar{\mathbf{T}} \cdot\Delta\mathbf{u}d\partial B.\label{eq:DpiCCI1} \end{eqnarray}

The energy release rate for a crack (no change of volume) can thus be evaluated as

\begin{eqnarray} G = \lim_{\Delta a \rightarrow 0} - \frac{\Delta \Pi_T}{ t \Delta a} = \lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\left\{ \int_{B}
U\left(\mathbf{\nabla}\mathbf{u}\right) -U\left(\mathbf{\nabla}\mathbf{u}+\mathbf{\nabla}\Delta\mathbf{u}\right) d B + \int_{\partial_N B} \bar{\mathbf{T}} \cdot \Delta\mathbf{u} d \partial B\right\}.\label{eq:JToG1} \end{eqnarray}

As the crack is stress-free, as $\Delta \mathbf{u}=0$ on $\partial_DB$, and as $\Delta\mathbf{\sigma}=0$ on $\partial_N B$, the last term of this equation can be simplified into

$$\int_{\partial_N B}\bar{\mathbf{T}} \cdot \Delta \mathbf{u} d \partial B = \int_{\partial_N B+\partial_D B + S+\Delta S} \Delta \mathbf{u}\cdot \left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right)\cdot \mathbf{n} d \partial B = \int_{B} \mathbf{\nabla}\cdot\left( \Delta \mathbf{u}\cdot \left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right)\right) d B.\label{eq:JToG2},$$

or, as $\mathbf{\nabla}\cdot\mathbf{\sigma}=\mathbf{\nabla}\cdot\left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right)=0$, into

$$\int_{\partial_N B}\bar{\mathbf{T}} \cdot \Delta \mathbf{u} d \partial B = \int_{B} \left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right):\left(\mathbf{\nabla} \Delta \mathbf{u}\right) d B\label{eq:JToG3}.$$

Using this last result in (\ref{eq:JToG1}) yields

G = \lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\left\{ \int_{B} U\left(\mathbf{\varepsilon}\right) - U\left(\mathbf{\varepsilon}+\Delta\mathbf{\varepsilon}\right) +
\left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right):\Delta\mathbf{\varepsilon} d B\right\}.\label{eq:JToG4}

Assuming the crack grows straight ahead we can consider the following change of variables, see Picture III.28,

$$\begin{cases} x' = x- a\\ y' = y \end{cases} \rightarrow \begin{cases}f(x',y',a) = f(x-a,y,a) \\ \frac{d f}{d a} = - \partial_{x'} f +\partial_a f\end{cases}. \label{eq:JToGCV}$$

The energy release rate (\ref{eq:JToG4}) involves the whole body $B$. However as $\Delta a\rightarrow 0$, the non-vanishing contributions are around the crack tip. So we can limit the integral to a fixed region $D$ of boundary $\Gamma$, see Picture III.28, and this equation becomes

\begin{eqnarray} G = \lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}&&\left\{ \int_{D} U\left(\mathbf{\varepsilon}\right) - U\left(\mathbf{\varepsilon}+\Delta\mathbf{\varepsilon}\right) +\right.\nonumber\\&&\left.\left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right):\Delta\mathbf{\varepsilon} d D\right\}.\label{eq:JToG5} \end{eqnarray}

We can integrate by parts and apply the Gauss theorem on the last term of (\ref{eq:JToG5}), and as $\mathbf{\nabla}\cdot\left(\mathbf{\sigma}+\Delta\mathbf{\sigma}\right)=0$, it yields

$$\int_D \left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right): \Delta \mathbf{\varepsilon} dD = \int_\Gamma \Delta \mathbf{u}\cdot\left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right)\cdot\mathbf{n} dl ,\label{eq:JToG6}$$

or

$$\lim_{\Delta a\rightarrow 0}\frac{1}{\Delta a} \int_D \left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right): \Delta \mathbf{\varepsilon} dD =\lim_{\Delta a\rightarrow 0}\frac{1}{\Delta a} \int_\Gamma \Delta \mathbf{u}\cdot\left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right)\cdot\mathbf{n} dl .\label{eq:JToG8}$$

Considering the change of variables (\ref{eq:JToGCV}), this last equation reads

\lim_{\Delta a\rightarrow 0}\frac{1}{\Delta a} \int_D \left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right): \Delta \mathbf{\varepsilon} dD =\int_\Gamma \frac{d
\mathbf{u}}{d a} \cdot \mathbf{T} dl = \int_\Gamma \left( - \partial_{x'} \mathbf{u} +\partial_a \mathbf{u} \right) \cdot \mathbf{T} dl. \label{eq:JToG9}

One more time using $\mathbf{\nabla}\cdot\mathbf{\sigma}=\mathbf{\nabla}\cdot\partial_{\mathbf{\varepsilon}}U=0$ we have

\int_\Gamma \partial_a \mathbf{u} \cdot \mathbf{T} dl = \int_\Gamma \partial_a\mathbf{u} \cdot \partial_{\mathbf{\varepsilon}} U \cdot \mathbf{n} dl = \int_D
\partial_a\mathbf{\varepsilon} : \partial_{\mathbf{\varepsilon}} U d D = \int_D \partial_a U d D,\label{eq:JToG10}

\lim_{\Delta a\rightarrow 0}\frac{1}{\Delta a} \int_D \left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right): \Delta \mathbf{\varepsilon} dD = - \int_\Gamma \partial_{x'}
\mathbf{u} \cdot \mathbf{T} dl + \int_D \partial_a U d D. \label{eq:JToG11}

We can now consider the first term of (\ref{eq:JToG5}). The domain $D$ is fixed to the initial crack tip, but we can define a domain $D^\star$ moving with the crack tip such that $D=D^\star+\Delta D_L-\Delta D_R$, see Picture III.29. The first term of (\ref{eq:JToG5}) reads

\lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\left\{ \int_{D}
\left. \int_{D*+\Delta D_L -
\Delta D_R} U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) d D\right\}.\label{eq:JToG12} \end{eqnarray}

As $D^\star \rightarrow D$ for infinitesimal crack growth, this relation becomes (formally, one should use derivatives & limits of integrals with non-constant intervals)

\begin{eqnarray} \lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\int_{D} U\left(\mathbf{\varepsilon}\right) - U\left(\mathbf{\varepsilon}+\Delta\mathbf{\varepsilon}\right) d D &=&
\lim_{\Delta a \rightarrow 0} \left\{ \int_{D} \frac{U\left(\mathbf{\varepsilon}\left(a\right)\right)-U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) }{\Delta a} dD -\right.\nonumber\\
&&\left. \frac{1}{\Delta a} \int_{\Delta D_L - \Delta D_R} U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) d D\right\} .\label{eq:JToG13}\end{eqnarray}

These integrals can be successively computed:

• As we have an homogeneous material, one has $\partial_{x'}U=0$ and $$\lim_{\Delta a \rightarrow 0} \int_{D} \frac{U\left(\mathbf{\varepsilon}\left(a\right)\right)-U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) }{\Delta a} dD = -\int_{D} \partial_a U dD.\label{eq:JtoGtmp1}$$
• Considering the left increment $\Delta D_L$ and the open curve $\Gamma_L$, see Picture III.30, for an infinitesimal crack advance we have
\int_{\Delta D_L} U d D = \int_{-\Gamma_L} U \mathbf{n}_x \Delta a dl, so that
$$\lim_{\Delta a \rightarrow 0}\frac{1}{\Delta a} \int_{\Delta D_L} U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) d D = -\int_{ \Gamma_L} U\left(\mathbf{\varepsilon}\left(a\right)\right)\mathbf{n}_x d l\label{eq:JtoGtmp2}.$$

• Considering the right increment $\Delta D_R$ and the open curve $\Gamma_R^\star$, see Picture III.31, for infinitesimal crack advance we have
\int_{\Delta D_R} U d D = \int_{\Gamma_R} U \mathbf{n}_x \Delta a dl, so that
$$\lim_{\Delta a \rightarrow 0}\frac{1}{\Delta a} \int_{\Delta D_R} U\left(\mathbf{\varepsilon}\left(a+\Delta a\right)\right) d D = \int_{ \Gamma_R^\star} U\left(\mathbf{\varepsilon}\left(a\right)\right)\mathbf{n}_x d l\label{eq:JtoGtmp3}.$$

Combining (\ref{eq:JtoGtmp1}-\ref{eq:JtoGtmp3}), and as $\Gamma_L + \Gamma_R^\star\rightarrow$\Gamma$, the relation (\ref{eq:JToG13}) becomes $$\lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\int_{D} U\left(\mathbf{\varepsilon}\right) - U\left(\mathbf{\varepsilon}+\Delta\mathbf{\varepsilon}\right) d D=-\int_{D} \partial_a U dD +\int_{ \Gamma} U\left(\mathbf{\varepsilon}\left(a\right)\right)\mathbf{n}_x d l .\label{eq:JToG14}$$ If the crack grows straight ahead, for$\Delta a \rightarrow 0$, we have found (\ref{eq:JToG11}) and (\ref{eq:JToG14}) which are summarized as \begin{cases} \lim_{\Delta a \rightarrow 0} \frac{1}{\Delta a}\int_{D} U\left(\mathbf{\varepsilon}\right) - \left(\mathbf{\varepsilon}+\Delta\mathbf{\varepsilon}\right) d D = -\int_{D} \partial_a U dD +\int_{ \Gamma} U\left(\mathbf{\varepsilon}\left(a\right)\right)\mathbf{n}_x dl \\ \lim_{\Delta a\rightarrow 0}\frac{1}{\Delta a} \int_D \left(\mathbf{\sigma}+\Delta \mathbf{\sigma}\right): \Delta \mathbf{\varepsilon} dD = - \int_\Gamma \partial_{x'} \mathbf{u} \cdot \mathbf{T} dl + \int_D \partial_a U d D \end{cases}, so that (\ref{eq:JToG5}) finally reads G = \int_{ \Gamma} U\left(\mathbf{\varepsilon}\left(a\right)\right)\mathbf{n}_x d l- \int_\Gamma \partial_{x} \mathbf{u} \cdot \mathbf{T} dl = J. So, if the material is defined by an internal potential and if the crack grows straight ahead,$G=J$. ### Linear Elasticity Let us now consider the case of linear elastic materials. The general expression of$J$reads $$J = \int_{\Gamma} \left[U\left(\mathbf{\varepsilon}\right) \mathbf{n}_x - \mathbf{u}_{,x} \cdot \mathbf{T}\right] dl, \label{eq:Jelast1}$$ and can be particularized for linear elasticity ($U=\frac{\mathbf{\sigma}\mathbf{\varepsilon}}{2}$) as $$J = \int_{\Gamma} \left[\frac{\mathbf{\sigma}_{ij}\mathbf{u}_{i,j}}{2}\delta_{xk} - \mathbf{u}_{i,x} \mathbf{\sigma}_{ik}\right]\mathbf{n}_k dl .\label{eq:Jelast2}$$ Let the contour$\Gamma$be a circle as shown in Figure III.32. Thus as $$\begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix} = \begin{pmatrix} \cos \theta & -\frac{\sin \theta}{r} \\ \sin \theta & \frac{\cos \theta}{r} \end{pmatrix} \begin{pmatrix} \partial_r \\ \partial_\theta \end{pmatrix},\label{eq:Jelast3}$$ the J-Integral (\ref{eq:Jelast2}) becomes \begin{eqnarray} J &=& \int_{-\pi}^{\pi} \frac{r\mathbf{\sigma}_{ix}\cos{\theta}u_{i,r}-\mathbf{\sigma}_{ix}\sin{\theta}u_{i,\theta}+ r \mathbf{\sigma}_{iy} \sin{\theta}u_{i,r} + \mathbf{\sigma}_{iy}\cos{\theta}u_{i,\theta}}{2}\cos{\theta} d\theta-\nonumber\\ && \int_{-\pi}^{\pi} \left[r\mathbf{\sigma}_{ix}\cos{\theta}u_{i,r}-\mathbf{\sigma}_{ix}\sin{\theta}u_{i,\theta}\right]\cos{\theta} d\theta-\nonumber\\&& \int_{-\pi}^{\pi} \left[r\mathbf{\sigma}_{iy}\cos{\theta}u_{i,r}-\mathbf{\sigma}_{iy}\sin{\theta}u_{i,\theta}\right]\sin{\theta} d\theta \label{eq:Jelast4}.\end{eqnarray} As J is path-independent we can take$r$tending to 0 and can thus use the asymptotic solution $$\begin{cases} \mathbf{\sigma}^\text{mode i} = \frac{K_i}{\sqrt{2 \pi r}} \mathbf{f}^\text{mode i}(\theta) \\ \mathbf{u}^\text{mode i} = K_i \sqrt{\frac{r}{2 \pi}} \mathbf{g}^\text{mode i}(\theta) \end{cases}.$$ Superposing the fracture modes yields $$\begin{cases} \mathbf{u} = \sqrt{\frac{r}{2\pi}}\sum_{i=I}^{III} K_i \mathbf{g}^{\text{mode i}}\left(\theta\right) \\ \mathbf{\sigma} = \frac{1}{\sqrt{2\pi r}}\sum_{i=I}^{III} K_i \mathbf{f}^{\text{mode i}}\left(\theta\right) \end{cases},$$ and after some manipulations the J-integral (\ref{eq:Jelast4}) becomes $$J = \frac{K^2_I}{E'}+\frac{K^2_{II}}{E'}+\frac{K^2_{III}}{2\mu}.\label{eq:Jelast5}$$ So, if the material is linear and elastic for any direction of the crack growth$J=\frac{K^2_I}{E'}+\frac{K^2_{II}}{E'}+\frac{K^2_{III}}{2\mu}\$.

 Last update: September 02, 2015