J Integral - Two-dimensional J-integral

Two-dimensional J-integral

The two-dimensional J-integral was originally defined as (see Figure 1 for an illustration)

$J := \int_\Gamma \left(W~dx_2 - \mathbf{t}\cdot\cfrac{\partial\mathbf{u}}{\partial x_1}~ds\right)$

where is the strain energy density, are the coordinate directions, is the surface traction vector, is the normal to the curve, is the Cauchy stress tensor, and is the displacement vector. The strain energy density is given by

$W = \int_0^\epsilon \boldsymbol{\sigma}:d\boldsymbol{\epsilon} ~;~~ \boldsymbol{\epsilon} = \tfrac{1}{2}\left ~.$

The J-Integral around a crack tip is frequently expressed in a more general form (and in index notation) as

$J_i := \lim_{\epsilon\rightarrow 0} \int_{\Gamma_\epsilon} \left(W n_i - n_j\sigma_{jk}~\cfrac{\partial u_k}{\partial x_i}\right) d\Gamma$

where is the component of the J-integral for crack opening in the direction and is a small region around the crack tip. Using Green's theorem we can show that this integral is zero when the boundary is closed and encloses a region that contains no singularities and is simply connected. If the faces of the crack do not have any surface tractions on them then the J-integral is also path independent.

Rice also showed that the value of the J-integral represents the energy release rate for planar crack growth. The J-integral was developed because of the difficulties involved in computing the stress close to a crack in a nonlinear elastic or elastic-plastic material. Rice showed that if monotonic loading was assumed (without any plastic unloading) then the J-integral could be used to compute the energy release rate of plastic materials too.

Proof that the J-integral is zero over a closed path
To show the path independence of the J-integral, we first have to show that the value of is zero over a closed contour in a simply connected domain. Let us just consider the expression for which is $J_1 := \int_{\Gamma} \left(W n_1 - n_j\sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right) d\Gamma$ We can write this as $J_1 = \int_{\Gamma} \left(W \delta_{1j} - \sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right)n_j d\Gamma$ From Green's theorem (or the two-dimensional divergence theorem) we have $\int_{\Gamma} f_j~n_j~d\Gamma = \int_A \cfrac{\partial f_j}{\partial x_j}~dA$ Using this result we can express as $\begin{align} J_1 & = \int_{A} \cfrac{\partial}{\partial x_j}\left(W \delta_{1j} - \sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right) dA \\ & = \int_A \left[\cfrac{\partial W}{\partial x_1} - \cfrac{\partial\sigma_{jk}}{\partial x_j}~\cfrac{\partial u_k}{\partial x_1} - \sigma_{jk}~\cfrac{\partial^2 u_k}{\partial x_1 \partial x_j}\right]~dA \end{align}$ where is the area enclosed by the contour . Now, if there are no body forces present, equilibrium (conservation of linear momentum) requires that $\boldsymbol{\nabla}\cdot\boldsymbol{\sigma} = \mathbf{0} \qquad \implies \qquad \cfrac{\partial\sigma_{jk}}{\partial x_j} = 0 ~.$ Also, $\boldsymbol{\epsilon} = \tfrac{1}{2}\left \qquad \implies \qquad \epsilon_{jk} = \tfrac{1}{2}\left(\cfrac{\partial u_k}{\partial x_j} + \cfrac{\partial u_j}{\partial x_k}\right) ~.$ Therefore, $\sigma_{jk}\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \tfrac{1}{2}\left(\sigma_{jk}\cfrac{\partial^2 u_k}{\partial x_1 \partial x_j} + \sigma_{jk}\cfrac{\partial^2 u_j}{\partial x_1 \partial x_k}\right)$ From the balance of angular momentum we have . Hence, $\sigma_{jk}\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \sigma_{jk}\cfrac{\partial^2 u_j}{\partial x_1 \partial x_k}$ The J-integral may then be written as $J_1 = \int_A \left[\cfrac{\partial W}{\partial x_1} - \sigma_{jk}~\cfrac{\partial\epsilon_{jk}}{\partial x_1}\right]~dA$ Now, for an elastic material the stress can be derived from the stored energy function using $\sigma_{jk} = \cfrac{\partial W}{\partial\epsilon_{jk}}$ Then, using the chain rule of differentiation, $\sigma_{jk}~\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \cfrac{\partial W}{\partial\epsilon_{jk}}~\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \cfrac{\partial W}{\partial x_1}$ Therefore we have for a closed contour enclosing a simply connected region without any stress singularities.

Proof that the J-integral is zero over a closed path

To show the path independence of the J-integral, we first have to show that the value of is zero over a closed contour in a simply connected domain. Let us just consider the expression for which is

$J_1 := \int_{\Gamma} \left(W n_1 - n_j\sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right) d\Gamma$

We can write this as

$J_1 = \int_{\Gamma} \left(W \delta_{1j} - \sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right)n_j d\Gamma$

From Green's theorem (or the two-dimensional divergence theorem) we have

$\int_{\Gamma} f_j~n_j~d\Gamma = \int_A \cfrac{\partial f_j}{\partial x_j}~dA$

Using this result we can express as

$\begin{align} J_1 & = \int_{A} \cfrac{\partial}{\partial x_j}\left(W \delta_{1j} - \sigma_{jk}~\cfrac{\partial u_k}{\partial x_1}\right) dA \\ & = \int_A \left[\cfrac{\partial W}{\partial x_1} - \cfrac{\partial\sigma_{jk}}{\partial x_j}~\cfrac{\partial u_k}{\partial x_1} - \sigma_{jk}~\cfrac{\partial^2 u_k}{\partial x_1 \partial x_j}\right]~dA \end{align}$

where is the area enclosed by the contour . Now, if there are no body forces present, equilibrium (conservation of linear momentum) requires that

$\boldsymbol{\nabla}\cdot\boldsymbol{\sigma} = \mathbf{0} \qquad \implies \qquad \cfrac{\partial\sigma_{jk}}{\partial x_j} = 0 ~.$

Also,

$\boldsymbol{\epsilon} = \tfrac{1}{2}\left \qquad \implies \qquad \epsilon_{jk} = \tfrac{1}{2}\left(\cfrac{\partial u_k}{\partial x_j} + \cfrac{\partial u_j}{\partial x_k}\right) ~.$

Therefore,

$\sigma_{jk}\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \tfrac{1}{2}\left(\sigma_{jk}\cfrac{\partial^2 u_k}{\partial x_1 \partial x_j} + \sigma_{jk}\cfrac{\partial^2 u_j}{\partial x_1 \partial x_k}\right)$

From the balance of angular momentum we have . Hence,

$\sigma_{jk}\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \sigma_{jk}\cfrac{\partial^2 u_j}{\partial x_1 \partial x_k}$

The J-integral may then be written as

$J_1 = \int_A \left[\cfrac{\partial W}{\partial x_1} - \sigma_{jk}~\cfrac{\partial\epsilon_{jk}}{\partial x_1}\right]~dA$

Now, for an elastic material the stress can be derived from the stored energy function using

$\sigma_{jk} = \cfrac{\partial W}{\partial\epsilon_{jk}}$

Then, using the chain rule of differentiation,

$\sigma_{jk}~\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \cfrac{\partial W}{\partial\epsilon_{jk}}~\cfrac{\partial\epsilon_{jk}}{\partial x_1} = \cfrac{\partial W}{\partial x_1}$

Therefore we have for a closed contour enclosing a simply connected region without any stress singularities.

Proof that the J-integral is path-independent
Consider the contour . Since this contour is closed and encloses a simply connected region, the J-integral around the contour is zero, i.e. $J = J_{(1)} + J^{+} - J_{(2)} - J^{-} = 0$ assuming that counterclockwise integrals around the crack tip have positive sign. Now, since the crack surfaces are parallel to the axis, the normal component on these surfaces. Also, since the crack surfaces are traction free, . Therefore, $J^{+} = J^{-} = \int_{\Gamma} \left(W n_1 - t_k~\cfrac{\partial u_k}{\partial x_1}\right) d\Gamma = 0$ Therefore, $J_{(1)} = J_{(2)}$ and the J-integral is path independent.

Proof that the J-integral is path-independent

Consider the contour . Since this contour is closed and encloses a simply connected region, the J-integral around the contour is zero, i.e.

$J = J_{(1)} + J^{+} - J_{(2)} - J^{-} = 0$

assuming that counterclockwise integrals around the crack tip have positive sign. Now, since the crack surfaces are parallel to the axis, the normal component on these surfaces. Also, since the crack surfaces are traction free, . Therefore,

$J^{+} = J^{-} = \int_{\Gamma} \left(W n_1 - t_k~\cfrac{\partial u_k}{\partial x_1}\right) d\Gamma = 0$

Therefore,

$J_{(1)} = J_{(2)}$

and the J-integral is path independent.