 Chapter 9

Modeling Material Failure

Energy methods provide additional insight into fracture, and also provide a foundation for a range of analytical and numerical methods in fracture mechanics.  In this section, we outline some of the most important results.

9.4.1 Definition of crack tip energy release rate for cracks in linear elastic solids The crack tip energy release rate quantifies the rate of change of the potential energy of a cracked elastic solid as the crack grows.

To make this precise, consider an ideally elastic solid, subjected to some loading (applied tractions, displacements, or body forces).  Suppose the solid contains a crack (the figure shows a circular crack with radius a as a representative example).  Define the potential energy of the solid in the usual way (Sect 5.6.1) as

$V\left(a\right)=\underset{V}{\int }U\left(u\right)dV-\underset{V}{\int }{\rho }_{0}{b}_{i}{u}_{i}dV-\underset{{\partial }_{2}R}{\int }{t}_{i}{u}_{i}dA$

Suppose the crack increases in size, so that the crack advances a distance $\alpha \delta a\left(s\right)$ with loading kept fixed, where s measures position around the crack front. The principle of minimum potential energy (sect 5.6.2) shows that $V\left(a+\alpha \delta a\right)\le V\left(a\right)$, since the displacement field associated with $V\left(a\right)$ is a kinematically admissible field for the solid with a longer crack.  The energy release rate $G\left(s\right)$ around the crack front is defined so that

$\underset{C}{\int }G\left(s\right)\delta a\left(s\right)ds=-\underset{\alpha \to 0}{\mathrm{lim}}\frac{\partial V\left(a+\alpha \delta a\left(s\right)\right)}{\partial \alpha }$

Energy release rate has units of $N{m}^{-1}$ (energy per unit area).

For the special case of a 2D slit crack with length a, the energy release rate is

$G=-\frac{\partial \overline{V}\left(a\right)}{\partial a}$

where $\overline{V}$ is now the potential energy per unit out-of-plane distance.

9.4.2 Energy release rate as a fracture criterion

Phenomenological fracture (or fatigue) criteria can be based on energy release rate arguments as an alternative to the K based fracture criteria discussed earlier.

The argument is as follows.  Regardless of the actual mechanisms involved, crack propagation involves dissipation (or conversion) of energy.  A small amount of energy is required to create two new free surfaces (twice the surface energy per unit area of crack advance, to be precise).  In addition, there may be a complex process zone at the crack tip, where the material is plastically deformed; voids may be nucleated; there may be chemical reactions; and generally all hell breaks loose.  All these processes involve dissipation of energy.  We postulate, however, that the process zone remains self-similar during crack growth.  If this is the case, energy will be dissipated at a constant rate during crack growth.  The crack can only grow if the rate of change of potential energy is sufficient to provide this energy.

This leads to a fracture criterion of the form

$G\ge {G}_{C}$

for crack growth, where ${G}_{C}$ is a property of the material. Unfortunately ${G}_{C}$ is often referred to as the fracture toughness of a solid, just like ${K}_{IC}$ defined earlier.  It is usually obvious from dimensional considerations which one is being used, but its an annoying source of confusion.

9.4.3 Relation between energy release rate and stress intensity factor

The energy release rate G is closely related to the stress intensity factors defined in Sect 9.3.  Specifically, for an isotropic, linear elastic solid with Young’s modulus $E$ and Poisson’s ratio $\nu$ the energy release rate is related to stress intensity factors by

$G=\frac{1-{\nu }^{2}}{E}\left({K}_{I}^{2}+{K}_{II}^{2}\right)+\frac{1+\nu }{E}{K}_{III}^{2}$

HEALTH WARNING: The result relating G to ${K}_{I}$ and ${K}_{II}$ is valid only for plane strain deformation at the crack tip. Derivation A neat argument due to Irwin provides the connection.

A crack of length a can be regarded as a crack with $a+\delta a$ which is being pinched closed by an appropriate distribution of traction acting on the crack faces between ${x}_{1}=-\delta a$ and ${x}_{1}=0$.  We can therefore calculate the change in potential energy as the crack propagates by distance $\delta a$ by computing the work done as these tractions are progressively relaxed to zero.   To this end, note that

1.      The tractions that pinch the crack tip closed can be calculated from the asymptotic crack tip field (Sect 9.3.1)

${t}_{1}=\frac{{K}_{II}}{\sqrt{2\pi \left(\delta a+{x}_{1}\right)}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{t}_{2}=\frac{{K}_{I}}{\sqrt{2\pi \left(\delta a+{x}_{1}\right)}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{t}_{3}=\frac{{K}_{III}}{\sqrt{2\pi \left(\delta a+{x}_{1}\right)}}$

(equal and opposite tractions must act on the lower crack face).

2.      As the crack is allowed to open, the upper crack face displaces by

$\Delta {u}_{1}=\frac{2\left(1-{\nu }^{2}\right)}{E}{K}_{II}\sqrt{\frac{-2{x}_{1}}{\pi }\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta {u}_{2}=\frac{2\left(1-{\nu }^{2}\right)}{E}{K}_{I}\sqrt{\frac{-2{x}_{1}}{\pi }}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta {u}_{3}=\frac{2\left(1+\nu \right)}{E}{K}_{III}\sqrt{\frac{-2{x}_{1}}{\pi }}$

where we have assumed plane strain deformation.

3.      The total work done as the tractions are relaxed quasi-statically to zero is

$G\delta a=\underset{-\delta a}{\overset{0}{\int }}\left({t}_{1}\Delta {u}_{1}+{t}_{2}\Delta {u}_{2}+{t}_{3}\Delta {u}_{3}\right)d{x}_{1}$

(the work done by tractions acting on the upper crack face per unit length is ${t}_{i}\Delta {u}_{i}/2$, and there are two crack faces).

4.      Evaluating the integrals gives

$G=\frac{1-{\nu }^{2}}{E}\left({K}_{I}^{2}+{K}_{II}^{2}\right)+\frac{1+\nu }{E}{K}_{III}^{2}$

The same result can be obtained by applying crack tip energy flux integrals, to be discussed below.

9.4.4 Relation between energy release rate and compliance Energy release rate is related to the compliance of a structure or specimen, as follows.  Consider the compact tension specimen shown in the picture.  Suppose that the specimen is subjected to a load P, which causes the point of application of the load to displace by a distance ${x}_{0}$ in a direction parallel to the load. The compliance of the specimen is defined as

$C=\frac{{x}_{0}}{P}$

As the crack grows, the compliance of the specimen always increases, so C is a function of crack length.  The energy release rate is related to compliance C by

$G=\frac{1}{2}\frac{{P}^{2}}{B}\frac{dC}{da}$

This formula applies to any structure or component, not just to compact tension specimens.  The formula is useful for two reasons:

(i) It can be used to measure energy release rate in an experiment.  All you need to do is to measure the crack length as it grows, and at the same time measure the compliance of your specimen.

(ii) It can be used to calculate stress intensity factors, as outlined in the next section.

Derivation: This result can be derived by calculating the change in energy of the system as the crack grows.  Note that

1.      The load P induces a total strain energy $\Phi =\frac{1}{2}{x}_{0}P=\frac{1}{2}C{P}^{2}$ in the specimen.  To see this, note that the the solid is elastic and so behaves like a linear spring $–$ this is just the formula for the energy in a spring.

2.      Now, suppose that the crack extends by a distance $\delta a$.  During crack growth, the load increases to $P+\delta P$ and displaces to ${x}_{0}+\delta x$.  In addition, the strain energy changes to $\Phi +\delta \Phi$, while the compliance increases to $C+\delta C$.

3.      The energy released during crack advance is equal to the decrease in potential energy of the system, so that

$G\text{\hspace{0.17em}}B\text{\hspace{0.17em}}\delta a=-\delta V=-\left[\left(\Phi +\delta \Phi \right)-\Phi -P\delta x\right]$

4.      Note that

$\begin{array}{l}\Phi +\delta \Phi =\frac{1}{2}\left(C+\delta C\right){\left(P+\delta P\right)}^{2}\approx \frac{1}{2}C{P}^{2}+CP\delta P+\frac{1}{2}\delta C{P}^{2}\\ \delta x=\left(C+\delta C\right)\left(P+\delta P\right)-CP\approx C\delta P+P\delta C\end{array}$

5.      Substituting these results into the expression in step (3) and simplifying shows that

$G\text{\hspace{0.17em}}B\text{\hspace{0.17em}}\delta a=\frac{1}{2}{P}^{2}\delta C=\frac{1}{2}{P}^{2}\frac{dC}{da}\delta a$

The energy release rate therefore is related to compliance by

$G=\frac{1}{2}\frac{{P}^{2}}{B}\frac{dC}{da}$

9.4.5 Calculating stress intensity factors using compliance The relation between compliance and energy release rate can be used to determine energy release rates, and sometimes also stress intensity factors, for structures whose rate of change of compliance with crack length can be easily determined.  One example is the cantilever beam specimen shown in the figure.  The mode I stress intensity factor for this specimen can be derived as

${K}_{I}=\frac{2\sqrt{3}}{\sqrt{1-{\nu }^{2}}}\frac{Pa}{B{h}^{3/2}}$ Derivation This result is derived by first calculating the compliance of the solid; then using the formula to deduce the energy release rate, and finally using the relationship between stress intensity factor and energy release rate.  To proceed,

1.      Note that the deflection d of the loaded point can be calculated by visualizing the specimen as two cantilever beams, length a, width B and height h, clamped on their right hand end and subjected to a load P at their left hand ends.  From elementary beam theory, the deflection is

$d=2\frac{{a}^{3}P}{3E\left(B{h}^{3}/12\right)}=8\frac{{a}^{3}P}{EB{h}^{3}}$

where E is the Young’s modulus of the specimen.

2.      The compliance follows as

$C=\frac{d}{P}=8\frac{{a}^{3}}{EB{h}^{3}}$

3.      The energy release rate formula in Sect 9.4.4 gives

$G=\frac{1}{2}\frac{{P}^{2}}{B}\frac{dC}{da}=12\frac{{P}^{2}{a}^{2}}{E{B}^{2}{h}^{3}}$

4.      By symmetry, the crack must be loaded in pure mode I.  We can therefore deduce the stress intensity factor using the relation

$G=\frac{1-{\nu }^{2}}{E}{K}_{I}^{2}⇒{K}_{I}=\frac{2\sqrt{3}}{\sqrt{1-{\nu }^{2}}}\frac{Pa}{B{h}^{3/2}}$

9.4.6 Integral expressions for energy flux to a crack tip

In this section we outline a way to compute the energy release rate for a crack, which applies not only to linear elastic solids under quasi-static loading conditions, but is completely independent of the constitutive response of the solid, and also applies under dynamic loading (it is restricted to small strains, however).  The approach will be to find an expression for the flux of energy through a cylindrical surface $\Gamma$ enclosing the crack tip, which moves with the crack.  We will get the energy release rate by shrinking the surface down onto the crack tip. Energy flux across a surface in a solid: We first derive a formula that can be used to calculate the flux of kinetic and potential energy across a surface in a deformable solid.  To this end, Consider an arbitrary surface S, which encloses some volume V in a solid.  The surface need not necessarily be a material surface $–$ it could move with respect to the solid.  We will denote the velocity of S  (with respect to a fixed origin) by ${v}_{j}$ Assume that the solid is free of body forces, for simplicity. Let $\left[{u}_{i},{\epsilon }_{ij},{\sigma }_{ij}\right]$ denote the displacement, (infinitesimal) strain and stress field in the solid, and let ${\stackrel{˙}{u}}_{i}$ denote the velocity of a material point with respect to a fixed origin. Let $T=\rho {\stackrel{˙}{u}}_{i}{\stackrel{˙}{u}}_{i}/2$ denote the kinetic energy of a material particle in the solid Let $\stackrel{˙}{W}={\sigma }_{ij}{\stackrel{˙}{\epsilon }}_{ij}={\sigma }_{ij}\frac{\partial {\stackrel{˙}{u}}_{i}}{\partial {x}_{j}}$ denote the rate of work done by stresses at a point in the solid Define the rate of change of mechanical energy density at an arbitrary point in the solid as $\stackrel{˙}{\psi }=\stackrel{˙}{W}+\stackrel{˙}{T}$, and let $\psi =\underset{-\infty }{\overset{t}{\int }}\stackrel{˙}{\psi }dt$ Denote the total energy within V as $\Psi =\underset{V}{\int }\psi dV$ Define the work flux vector as ${\omega }_{j}={\sigma }_{ij}{\stackrel{˙}{u}}_{i}$

The energy flux across S can be calculated in terms of these quantities as follows:

$\frac{d\Psi }{dt}=\frac{d}{dt}\underset{V}{\int }\psi dV=\underset{S}{\int }\left({\omega }_{j}+\psi {v}_{j}\right){m}_{j}dA$

The right hand side of this expression denotes the energy flux across the surface; the left hand side is the rate of change of the total energy within V.  The two are equal by energy conservation, as shown below.

Derivation:

1.      Begin by showing that the energy flux vector and the rate of change of mechanical energy density are related by

$\partial {\omega }_{j}/\partial {x}_{j}=\stackrel{˙}{\psi }$

To see this, note that

$\stackrel{˙}{\psi }=\stackrel{˙}{W}+\stackrel{˙}{T}={\sigma }_{ij}\frac{\partial {\stackrel{˙}{u}}_{i}}{\partial {x}_{j}}+\rho {\stackrel{¨}{u}}_{i}{\stackrel{˙}{u}}_{i}={\sigma }_{ij}\frac{\partial {\stackrel{˙}{u}}_{i}}{\partial {x}_{j}}+\frac{\partial {\sigma }_{ij}}{\partial {x}_{j}}{\stackrel{˙}{u}}_{i}=\frac{\partial }{\partial {x}_{j}}\left({\sigma }_{ij}{\stackrel{˙}{u}}_{i}\right)=\frac{\partial {\omega }_{j}}{\partial {x}_{j}}$

where we have used the linear and angular momentum balance equations $\partial {\sigma }_{ij}/\partial {x}_{i}=\rho \stackrel{¨}{u}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{ij}={\sigma }_{ji}$.

2.      Now, integrate both sides of this equation over the volume V and apply the divergence theorem to see that

$\underset{V}{\int }\frac{\partial {\omega }_{j}}{\partial {x}_{j}}dV=\underset{V}{\int }\stackrel{˙}{\psi }dV\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{S}{\int }{\omega }_{j}{m}_{j}dA=\underset{V}{\int }\stackrel{˙}{\psi }dV$

3.      Next note that the total rate of change of $\psi$ within the volume V bounded by S can be expressed as

$\frac{d}{dt}\underset{V}{\int }\psi dV=\underset{V}{\int }\stackrel{˙}{\psi }dV+\underset{S}{\int }\psi {v}_{j}{m}_{j}dA$

Here, the first term on the right represents the rate of change due to the time derivative of $\psi$ within V, while the second term represents the flux of energy crossing S as the surface moves with velocity ${v}_{j}$.

4.      Combining (2) and (3) shows that

$\underset{S}{\int }{\omega }_{j}{m}_{j}dA=\frac{d}{dt}\underset{V}{\int }\psi dV-\underset{S}{\int }\psi {v}_{j}{m}_{j}dA\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{S}{\int }\left({\omega }_{j}+\psi {v}_{j}\right){m}_{j}dA=\frac{d}{dt}\underset{V}{\int }\psi dV$

The term on the right hand side clearly represents the total rate of change of mechanical energy in V.  Consequently, the term on the left hand side must represent the mechanical energy flux across $S$.  This is the result we need. Energy flux to a crack tip. We can use the energy flux integral to obtain an expression for the energy flux to a crack tip.  Suppose the crack tip runs with steady speed v in the ${x}_{1}$ direction.  Let $\Gamma$ denote a cylindrical surface enclosing the crack tip, which moves with the crack tip.  The energy flux through $\Gamma$ follows as

$\frac{d\Psi }{dt}=\underset{\Gamma }{\int }\left({\omega }_{j}+\psi v{\delta }_{j1}\right){m}_{j}dA=\underset{\Gamma }{\int }\left({\omega }_{j}+\left(T+W\right)v{\delta }_{j1}\right){m}_{j}dA$

where

$W=\underset{-\infty }{\overset{t}{\int }}\stackrel{˙}{W}dt=\underset{-\infty }{\overset{t}{\int }}{\sigma }_{ij}{\stackrel{˙}{\epsilon }}_{ij}dt=\underset{0}{\overset{{\epsilon }_{ij}}{\int }}{\sigma }_{ij}d{\epsilon }_{ij}$

is the net work done on the solid per unit volume by stresses, and $T=\rho {\stackrel{˙}{u}}_{i}{\stackrel{˙}{u}}_{i}/2$ is the kinetic energy density. The energy flux to the crack tip follows by taking the limit as $\Gamma$ shrinks down onto the crack tip.

Contour integral formula for energy release rate. To obtain an expression for the energy release rate, assume that the crack tip fields remain self-similar (i.e. an observer traveling with the crack tip sees a fixed state of strain and stress).  In addition, assume that the crack front is straight, and has length L in direction perpendicular to the plane of the figure.  Under these conditions ${\stackrel{˙}{u}}_{i}=-v\partial {u}_{i}/\partial {x}_{1}$, and $d\Psi /dt=GLv$.  Consequently

$G=\frac{1}{L}\underset{\Gamma \to 0}{\mathrm{lim}}\underset{\Gamma }{\int }\left(\left(T+W\right){\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}dA=\underset{C\to 0}{\mathrm{lim}}\underset{C}{\int }\left(\left(T+W\right){\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds$

where C is a contour enclosing the crack tip. (Equivalent results can be derived for general 3D cracks, but these details are omitted here).

This result is valid for any material response (including plastic materials), and applies to both static and dynamic conditions.

9.4.7 Rice’s J integral The result derived in the preceding section becomes particularly useful if we make two further assumptions:

2.      The material is elastic.

In this case T=0 and $W$ is simply the strain energy density in the solid -  e.g. for a linear elastic solid with no thermal stress,

$W=\frac{E}{2\left(1+\nu \right)}{\epsilon }_{ij}^{}{\epsilon }_{ij}^{}+\frac{E\nu }{2\left(1+\nu \right)\left(1-2\nu \right)}{\epsilon }_{jj}^{}{\epsilon }_{kk}^{}$

The expression for energy flux through a surface surrounding the crack tip reduces to

$J=\underset{\Gamma }{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds$

This is the famous J integral.  It has the following properties:

1.      The crack tip energy integral is path independent, as long as the material enclosed by the contour is homogeneous.  There is no need then to shrink the contour down onto the crack tip $–$ we get the same answer for any contour that encloses the crack tip.

2.      J=G  for an elastic solid - so the contour integral gives an elegant way to calculate the crack tip energy release rate. Path independence of J: To show this, we first show that if the J integral is evaluated around any closed contour that does not enclose the crack tip, it is zero.  To see this, apply the divergence theorem

$J=\underset{\Gamma }{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds=\underset{A}{\int }\frac{\partial }{\partial {x}_{j}}\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right)dA=0$

where A is the area enclosed by $\Gamma$.  To see that the area integral on the right hand side is zero, note that

$\begin{array}{l}\frac{\partial W}{\partial {x}_{j}}{\delta }_{j1}=\frac{\partial W}{\partial {\epsilon }_{kl}}\frac{\partial {\epsilon }_{kl}}{\partial {x}_{1}}={\sigma }_{kl}\frac{\partial {u}_{k}}{\partial {x}_{l}\partial {x}_{1}}\\ \frac{\partial }{\partial {x}_{j}}\left({\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right)=\frac{\partial {\sigma }_{ij}}{\partial {x}_{j}}\frac{\partial {u}_{i}}{\partial {x}_{1}}+{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{j}\partial {x}_{1}}={\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{j}\partial {x}_{1}}\end{array}$

where we have used the equilibrium equation $\partial {\sigma }_{ij}/d{x}_{j}=0$. Now, evaluate the integral around the closed contour shown on the right. Note that the integrand vanishes on ${C}_{2}$ and ${C}_{4}$ so that

$\underset{{C}_{1}}{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds+\underset{{C}_{3}}{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds=0$

Now reverse the direction of integration around ${C}_{3}$ (note that m = -n) to get

$\underset{{C}_{1}}{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){m}_{j}ds=\underset{{C}_{3}}{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){n}_{j}ds$

showing that the integral is equal for any two contours that start and end on the two crack faces.

9.4.8 Calculating energy release rates using the J integral The J integral has many applications.  In some cases it can be used to compute energy release rates.  For example, consider the problem shown below.  A cracked linear elastic cracked sheet is clamped between rigid boundaries.  The bottom boundary is held fixed; the top is displaced vertically by a distance $\Delta$.  Calculate the energy release rate for the crack.

For this case G=J, and we can easily evaluate the J integral around the contour shown. To do so, note that

1.      Far behind the crack tip ( ${x}_{1}\to -\infty$ ) the solid is stress free.  The J integral vanishes on ${\Gamma }_{1}$ and ${\Gamma }_{5}$

2.      The displacement field is constant on ${x}_{2}=±h/2$ so that $\partial {u}_{i}/\partial {x}_{1}=0$ there.  In addition ${m}_{1}=0$ on ${\Gamma }_{2}$ and ${\Gamma }_{4}$.   The J integral vanishes on ${\Gamma }_{2}$ and ${\Gamma }_{4}$, therefore.

3.      Far ahead of the crack tip ${x}_{1}\to \infty$, the displacement, stress and strain energy density can easily be calculated as

$\begin{array}{l}{u}_{2}={x}_{2}\Delta /h,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{1}={u}_{3}=0\\ {\sigma }_{22}=E\left(1-\nu \right)\Delta /\left(1+\nu \right)\left(1-2\nu \right)h\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{11}={\sigma }_{33}=E\nu \Delta /\left(1+\nu \right)\left(1-2\nu \right)h\\ W=E\left(1-\nu \right){\Delta }^{2}/2\left(1+\nu \right)\left(1-2\nu \right){h}^{2}\end{array}$

The contribution to the J integral from ${\Gamma }_{3}$ follows as

$\underset{{\Gamma }_{3}}{\int }\left(W{\delta }_{j1}-{\sigma }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{1}}\right){n}_{j}ds=\frac{E\left(1-\nu \right){\Delta }^{2}}{2\left(1+\nu \right)\left(1-2\nu \right)h}$

4.      The energy release rate is therefore

$G=\frac{E\left(1-\nu \right){\Delta }^{2}}{2\left(1+\nu \right)\left(1-2\nu \right)h}$

Symmetry conditions show that the crack must be loaded in pure mode I, so the stress intensity factor can also be computed.