Chapter 12
Brittle failure
12.1 Stress near a crack tip
We now look at brittle fracture. The theory for brittle fracture is developed in the framework of a plane problem, i.e. we assume that our crack is infinitely extended into one direction. We need to introduce a couple of mathematical tricks to be able to actually compute the stress field, namely the introduction of potential functions.
Note that while our previous applications of the theory of linear elasticity (for beams and plates) lead to solutions that were polynomials in the positions, we will here encounter a solution that is singular, i.e. it diverges as we approach the crack tip.
12.1.1 Airy stress function
First, we invoke Hooke’s law (in plane strain/stress), \begin {equation} \varepsilon _{ij} = \frac {1}{E} [(1+\nu )\sigma _{ij} - \nu \delta _{ij} \sigma _{kk}]. \end {equation} Using the additionalplane condition, \begin {equation} \sigma _{yy} = \nu (\sigma _{xx} + \sigma _{zz}), \end {equation} Hooke’s law itself becomes \begin {align} \label {eq:PlanestrainHookes} \varepsilon _{xx} &= \frac {1+\nu }{E} [(1-\nu )\sigma _{xx} - \nu \sigma _{yy}] \\ \varepsilon _{yy} &= \frac {1+\nu }{E} [(1-\nu )\sigma _{yy} - \nu \sigma _{xx}] \\ \varepsilon _{xy} &= \frac {1}{2G} \sigma _{xy} = \frac {1+\nu }{E} \sigma _{xy} \end {align}
under plane strain conditions. We have just eliminated reference to \(\sigma _{zz}\) and \(\varepsilon _{zz}\) from the equations. Inserting Hooke’s law into the compatibility condition, \begin {equation} \frac {\partial ^2 \varepsilon _{xx}}{\partial z^2} + \frac {\partial ^2 \varepsilon _{zz}}{\partial x^2} = 2\ \frac {\partial ^2 \varepsilon _{xz}}{\partial x \partial z}, \end {equation} yields \begin {equation} \label {eq:Compatibilitystresses} (1-\nu ) \left [\frac {\partial ^2 \sigma _{xx}}{\partial y^2} + \frac {\partial ^2 \sigma _{yy}}{\partial x^2}\right ] - \nu \left [\frac {\partial ^2 \sigma _{xx}}{\partial x^2} + \frac {\partial ^2 \sigma _{yy}}{\partial y^2}\right ] = 2 \frac {\partial ^2 \sigma _{xy}}{\partial x \partial y}, \end {equation} the compatibility condition for the stresses.
We now use a mathematical trick to solve the equations of elastic equilibrium. We define the Airy stress function \(\phi (x,y)\), that gives the stresses (in Cartesian coordinates) as \begin {align} \label {eqn:airyxx} \sigma _{xx} &= \frac {\partial ^2 \phi }{\partial y^2} \equiv [\nabla (\nabla \phi )]_{yy}, \\ \label {eqn:airyzz} \sigma _{yy} &= \frac {\partial ^2 \phi }{\partial x^2} \equiv [\nabla (\nabla \phi )]_{xx},\\ \label {eqn:airyxz} \sigma _{xy} &= -\frac {\partial ^2 \phi }{\partial x \partial y} \equiv [\nabla (\nabla \phi )]_{xy} \end {align}
(Note that the first \(\nabla \) is the gradient of the vector field \(\nabla \phi \), not the divergence! Hence \(\nabla (\nabla \phi )\) is a second order tensor with components \([\nabla (\nabla \phi )]_{ij}=\phi _{,ij}\)!) Equation \begin {equation} \frac {\partial \sigma _{xx}}{\partial x} + \frac {\partial \sigma _{xz}}{\partial z} = 0 \end {equation} becomes \begin {equation} \label {eqn:elastic_equi_airy} \frac {\partial ^3 \phi }{\partial x \partial y^2} - \frac {\partial ^3 \phi }{\partial x \partial y^2} = 0 \end {equation} and is automatically fulfilled! The same holds for the second equation of the elastostatic equilibrium condition. We can now insert Eqs. \eqref{eqn:airyxx} to \eqref{eqn:airyxz} into the compatibility condition Eq. \eqref{eq:Compatibilitystresses} to give \begin {equation} \label {eq:compatibilityairy} (1-\nu )\left [\frac {\partial ^2 }{\partial y^2} \frac {\partial ^2 \phi }{\partial y^2} + \frac {\partial ^2 }{\partial x^2} \frac {\partial ^2 \phi }{\partial x^2}\right ] - \nu \left [\frac {\partial ^2 }{\partial x^2} \frac {\partial ^2 \phi }{\partial y^2} + \frac {\partial ^2}{\partial y^2} \frac {\partial ^2 \phi }{\partial x^2} \right ] + 2 \frac {\partial ^2}{\partial x \partial y} \frac {\partial ^2 \phi }{\partial x \partial y} = 0, \end {equation} which can be rearranged to \begin {equation} \label {eq:compatibilityairyrearr} \frac {\partial ^4 \phi }{\partial x^4} + 2\frac {\partial ^4 \phi }{\partial x^2 \partial y^2} + \frac {\partial ^4 \phi }{\partial y^4} = \left (\frac {\partial ^2}{\partial x^2} + \frac {\partial ^2}{\partial y^2}\right )\left [\left (\frac {\partial ^2}{\partial x^2} + \frac {\partial ^2}{\partial y^2}\right ) \phi \right ] = \nabla ^4 \phi = 0 \end {equation} The conditions of elastostatic equilibrium and the compatibility condition are all simultaneously fulfilled if Eq. \eqref{eq:compatibilityairyrearr} is fulfilled! Equation \eqref{eq:compatibilityairyrearr} is the compatibility condition, but expressed for the Airy stress function.
The operator \(\nabla ^4\) is called the biharmonic operator and any function \(\phi \) satifying \(\nabla ^4 \phi =0\) is called a biharmonic function. (Functions \(\phi \) that fulfill the Laplace equation \(\nabla ^2\phi =0\) are called harmonic functions.) The derivation above tells us, that for isotropic elasticity a biharmonic Airy stress \(\phi (x,y)\) automatically leads to a compatible strain field. This means for any biharmonic function \(\phi (x,y)\), the stress field derived from Eqs. \eqref{eqn:airyxx}-\eqref{eqn:airyxz} will describe a system in static equilibrium. The only thing left to do is to find the function \(\phi (x,y)\) that fulfills a specific boundary condition.
12.1.2 Westergaard stress function
Westergaard (1933) introduced a function \(Z(z)\) that is now known as the Westergaard stress function. The Westergaard stress function builds on top of the Airy function and eliminates the need to even satisfy the biharmonic equation \(\nabla ^4\phi =0\). This means that for any choice of the Westergaard stress function, the conditions for force and moment equilibrium and the compatibility condition are fulfilled automatically. The Westergaard stress function can then be chosen freely as to fulfill the boundary conditions of the problem.
Note that \(z\) is a complex variable that contains the \(x\) and \(y\)-position as its real and imaginary part, \(z=x+iy\), and not the \(z\)-coordinate. Likewise, the function \(Z(z)\) is complex valued. Following common notation, we denote its integrals by \begin {equation} \bar {\bar {Z}}_{,z} = \bar {Z} \quad \text {and}\quad \bar {Z}_{,z} = Z. \end {equation} Westergaard defined the Airy stress function as \begin {equation} \label {eqn:Westergaard} \phi (x, y) = \Re \left \{\bar {\bar {Z}}(x+iy)\right \} + y \Im \left \{\bar {Z}(x+iy)\right \} \end {equation} where \(\Re \) and \(\Im \) denote the real and imaginary part of a complex number, respectively. This Airy stress function fulfills the biharmonic equation, Eq. \eqref{eq:compatibilityairyrearr}, for any arbitrary function \(Z(z)\), as we will show below.
We will now derive the expressions for the stresses. We first note that since \(Z(z)\) is a function of a complex variable, it satisfies the Cauchy-Riemann conditions, \begin {align} \Re Z_{,z} &= \left (\Re Z\right )_{,x} = \left (\Im Z\right )_{,y} \\ \Im Z_{,z} &= \left (\Im Z\right )_{,x} = -\left (\Re Z\right )_{,y}, \end {align}
which implies that \begin {align} \label {eqn:laplaceRe} \nabla ^2 \left (\Re Z\right ) &= \left (\Re Z\right )_{,xx} + \left (\Re Z\right )_{,yy} = 0 \\ \nabla ^2 \left (\Im Z\right ) &= \left (\Im Z\right )_{,xx} + \left (\Im Z\right )_{,yy} = 0, \\ \end {align}
i.e. the real and imaginary parts of the complex function \(Z(z)\) fulfill the Laplace equation.
Note: Cauchy-Riemann equations – The Cauchy-Riemann equations are a cornerstone of complex analysis. They hold for any differential function of a complex variable. Such functions are also called holomorphic. Given \(f(z)=u(z) + iv(z)\) (hence \(u=\Re f\) and \(v=\Im f\)) with \(z=x+iy\), we can formally write the derivative as \begin {equation} f'(z_0) = \lim \limits _{w\in \mathcal {C},w\to 0} \frac {f(z_0+w)-f(z_0)}{w} \end {equation} as the limit of the difference quotient. For real numbers, we can approach \(z_0\) from the left or the right and both limits must equal if the function is differentiable. In the complex plane, we can approach from any direction in two dimensions. For example, we can compute the derivative along the real (\(x\)-)axis, \(w=x\): \begin {equation} f'(z_0) = \lim \limits _{x\in \mathcal {R},x\to 0} \frac {f(z_0+x)-f(z_0)}{x} = \frac {\partial u}{\partial x} + i\frac {\partial v}{\partial x} \end {equation} We can equally well compute the derivative along the imaginary (\(y\)-)axis, \(w=iy\), which gives \begin {equation} f'(z_0) = \lim \limits _{y\in \mathcal {R},y\to 0} \frac {f(z_0+iy)-f(z_0)}{iy} = -i\frac {\partial u}{\partial y} + \frac {\partial v}{\partial y}. \end {equation} Since both expressions have to be equal, we find \begin {equation} \frac {\partial u}{\partial x}=\frac {\partial v}{\partial y} \quad \text {and}\quad \frac {\partial u}{\partial y}=-\frac {\partial v}{\partial x}, \end {equation} the Cauchy-Riemann equations.
Now we differentiate \(\phi \) with respect to \(x\) and \(y\). This yields \begin {align} \phi _{,x} &= \left (\Re \bar {\bar {Z}}\right )_{,x} + y\left (\Im \bar {Z}\right )_{,x} = \Re \bar {\bar {Z}}_{,z} + y \Im \bar {Z}_{,z} = \Re \bar {Z} + y \Im Z \\ \phi _{,y} &= \left (\Re \bar {\bar {Z}}\right )_{,y} + \left (y \Im \bar {Z}\right )_{,y} = -\Im \bar {\bar {Z}}_{,z} + \Im \bar {Z} + y \Re \bar {Z}_{,z} = y \Re Z \end {align}
for the first derivatives and \begin {align} \phi _{,xx} &= \left (\Re \bar {Z}\right )_{,x} + y \left (\Im Z\right )_{,x} = \Re \bar {Z}_{,z} + y \Im Z_{,z} = \Re Z + y \Im Z_{,z} \\ \phi _{,yy} &= \left (y\Re Z\right )_{,y} = \Re Z + y \left (\Re Z\right )_{,y} = \Re Z - y \Im Z_{,z} \\ \phi _{,xy} &= \left (\Re \bar {Z}\right )_{,y} + \left (y \Im Z\right )_{,y} = -\Im \bar {Z}_{,z} + \Im Z + y \Re Z_{,z} = y \Re Z_{,z} \end {align}
for the second derivatives. In summary, we obtain \begin {align} \label {eqn:Westergaardstressxx} \sigma _{xx} &= \phi _{,yy} = \Re Z - y \Im Z_{,z} \\ \label {eqn:Westergaardstressyy} \sigma _{yy} &= \phi _{,xx} = \Re Z + y \Im Z_{,z} \\ \label {eqn:Westergaardstresszz} \sigma _{xy} &= -\phi _{,xy} = -y \Re Z_{,z} \end {align}
for the components of the stress tensor. Note that the Ansatz Eq. \eqref{eqn:Westergaard} fulfills the biharmonic equation. We know from summing Eqs. \eqref{eqn:Westergaardstressxx} and \eqref{eqn:Westergaardstressyy} that \begin {equation} \psi \equiv \left (\frac {\partial ^2}{\partial x^2} + \frac {\partial ^2}{\partial y^2}\right )\phi = 2\Re Z, \end {equation} hence by virtue of Eq. \eqref{eqn:laplaceRe}, \(\nabla ^4\phi =\nabla ^2\psi =0\) is fulfilled for any function \(Z(x+iy)\). The utility of the Westergaard function over the Airy stress function is hence that we have eliminated the need to explicitly fulfill the biharmonic equation.
12.1.3 Stress field
We here discuss only mode I fracture, i.e. crack opening displacement. Consider a plane (strain or stress) situation in which we can derive the stress field from the Westergaard stress function \(Z(x+iy)\), Sec. 12.1.2. Westergaard (1933) made the Ansatz \begin {equation} Z(z) = \frac {\sigma _\infty }{\sqrt {1 - (a/z)^2}} \end {equation} for the stress function of a central crack in a large sheet. Here \(2a\) is the length of the crack and \(\sigma _\infty \) the stress at infinity. The derivative of the stress function is \begin {equation} \label {eqn:Westergaard_derivative} Z_{,z} = \frac {\sigma _\infty }{\sqrt {z^2 - a^2}} \left [ 1 - \frac {1}{1-(a/z)^2} \right ] \end {equation} and its integral is \begin {equation} \label {eqn:Westergaard_integral} \bar {Z} = \sigma _\infty \sqrt {z^2 - a^2}. \end {equation}
Let us first look in the plane of the crack where \(y=0\), \(z=x\) and hence \begin {equation} Z(x) = \frac {\sigma _\infty }{\sqrt {1 - (a/x)^2}}. \end {equation} For \(|x|<a\) (inside the crack) the function is imaginary but for \(|x|>a\) (outside the crack) the function is real. The stresses in the plane of the crack are given by \(\sigma _{xx}=\Re Z\), \(\sigma _{yy}=\Re Z\) and \(\tau _{xy}=0\). Hence they vanish inside the crack. This is the condition for the crack faces which are free surfaces and therefore tractionless. Note that \(\sigma _{xx}\) does not need to vanish from this condition but does here.
Outside the crack (but in the plane of the crack, \(y=0\)), the stress is given by \begin {equation} \sigma _{xx} = \sigma _{yy} = \frac {\sigma _\infty }{\sqrt {1 - (a/x)^2}}. \end {equation} It diverges as \(x\to a\) from above and approaches the hydrostatic state \(\sigma _{xx}=\sigma _{yy}=\sigma _\infty \) as \(x\to \infty \).
We will now focus on the crack line and switch to the variable \(z^*=z-a\). The stress function becomes \begin {equation} Z(z^*) = \frac {\sigma _\infty (z^*+a)}{\sqrt {(z^* + a)^2 - a^2}} = \frac {\sigma _\infty (z^*+a)}{\sqrt {(z^*)^2 + 2az^*}} \approx \sigma _\infty \sqrt {\frac {a}{2z^*}} \end {equation} where the \(\approx \) sign is valid for small \(z^*\). We write this expression as \begin {equation} \label {eqn:stress_intensity} Z(z^*) = \frac {K_I}{\sqrt {2\pi z^*}} \quad \text {with}\quad K_I = \sigma _\infty \sqrt {\pi a}. \end {equation} Note that we have absorbed both the stress at infinity \(\sigma _\infty \) and the crack length \(a\) into a single constant, the stress intensity factor (for mode I fracture), \(K_I\). The stress field near the crack tip depends only on \(K_I\), not on \(\sigma _\infty \) and \(a\) individually and hence the loading condition and geometry, independently.
To derive the component of the stress tensor, we switch to cylindrical coordinates and write \(z^*=r e^{i\theta }\), yielding \begin {equation} Z(r, \theta ) = \frac {K_I}{\sqrt {2\pi r}} e^{-i\theta /2}. \end {equation} In order to obtain the full stress field, we also need the expression for the derivative of the Westergaard stress function near the crack tip. From Eq. \eqref{eqn:Westergaard˙derivative}, we find \begin {equation} Z_{,z} \approx \frac {\sigma _\infty }{\sqrt {2az^*}} \left [ 1-\frac {2az^* + a^2}{2az^*} \right ] = -\frac {\sigma _\infty a^2}{(2az^*)^{3/2}} = -\frac {K_I \pi }{(2\pi z^*)^{3/2}} = -\frac {K_I}{\sqrt {8\pi r}} e^{-3i\theta /2}. \end {equation} We now obtain the individual components of the stress tensor from Eq. \eqref{eqn:Westergaardstressxx}-\eqref{eqn:Westergaardstresszz}: \begin {align} \sigma _{xx} &= \Re Z - y \Im Z_{,z} = \frac {K_I}{\sqrt {2\pi r}} \cos \frac {\theta }{2} \left (1 - \sin \frac {\theta }{2} \sin \frac {3\theta }{2}\right ) \\ \sigma _{yy} &= \Re Z + y \Im Z_{,z} = \frac {K_I}{\sqrt {2\pi r}} \cos \frac {\theta }{2} \left (1 + \sin \frac {\theta }{2} \sin \frac {3\theta }{2}\right ) \\ \sigma _{xy} &= -y \Re Z_{,z} = \frac {K_I}{\sqrt {2\pi r}} \cos \frac {\theta }{2} \sin \frac {\theta }{2} \cos \frac {3\theta }{2} \end {align}
Hence the stress field near the crack tip is entirely described by the stress intensity factor \(K_I\). \(K_I\) is measured in (weird) units of Pa\(\sqrt {\text {m}}\). The \(I\) indicates that this is the stress intensity factor for mode \(I\) fracture. \(K_{II}\) and \(K_{III}\) are related quatities for the two other fracture modes.
Note that the stress intensity factor is the amplitude of the square-root singularity of the stress field at the crack tip. It is often defined from the stress field itself as the limit \begin {align} \label {eq:KI} K_I &= \lim \limits _{r\to 0} \sqrt {2\pi r} \sigma _{yy}(r, 0) \\ \label {eq:KII} K_{II} &= \lim \limits _{r\to 0} \sqrt {2\pi r} \sigma _{xy}(r, 0) \\ \label {eq:KIII} K_{III} &= \lim \limits _{r\to 0} \sqrt {2\pi r} \sigma _{yz}(r, 0) \end {align}
where the singularity has been removed by multiplying with \(\sqrt {2\pi r}\). Note that the limit is taken at the angle \(\theta =0\), i.e. along \(y=0\) within the plane of the crack. From Eqs. \eqref{eq:KI} to \eqref{eq:KIII} is also clear that only \(K_I\) is nonzero for the crack geometry that is discussed in this chapter.
12.2 Fracture toughness
12.2.1 Displacement field at the crack tip
In order to compute the displacement field near the crack, we first need the strain field. As usual, we obtain this from Hooke’s law (in plain stress): \begin {align} \label {eqn:crackstrainxx} \varepsilon _{xx} &\equiv u_{x,x} = \frac {1}{E}\left (\sigma _{xx} - \nu \sigma _{yy}\right ) = \frac {1}{E}\left [(1-\nu ) \Re Z - (1+\nu ) y \Im Z_{,z}\right ] \\ \label {eqn:crackstrainyy} \varepsilon _{yy} &\equiv u_{y,y} = \frac {1}{E}\left (\sigma _{yy} - \nu \sigma _{xx}\right ) = \frac {1}{E}\left [(1-\nu ) \Re Z + (1+\nu ) y \Im Z_{,z}\right ] \\ \label {eqn:crackstrainxy} 2\varepsilon _{xy} &\equiv u_{x,y}+u_{y,x} = 2\frac {1+\nu }{E} \sigma _{xy} = -2\frac {1+\nu }{E} y \Re Z_{,z} \end {align}
Now we use \(Z=\bar {Z}_{,z}\) to express \(\Re Z = \Re \bar {Z}_{,z} = (\Re \bar {Z})_{,x}\) and \(\Im Z_{,z}=(\Im Z)_{,x}\) in Eq. \eqref{eqn:crackstrainxx} to identify \begin {equation} u_x = \frac {1}{E}\left [(1-\nu ) \Re \bar {Z} - (1+\nu ) y \Im Z\right ] =\frac {1}{2\mu }\left [\frac {1-\nu }{1+\nu } \Re \bar {Z} - y \Im Z\right ]. \end {equation} The expression for \(u_y\) is more complicated because of the explicit \(y\) that shows up in the expressions for the strains. It is given by \begin {equation} u_y = \frac {1}{2\mu }\left [\frac {2}{1+\nu } \Im \bar {Z} - y \Re Z\right ]. \end {equation} It is straightforward to verify that from these two expressions we also recover the expression for \(\varepsilon _{xy}\).
It is common to replace \(\nu \) with \(\kappa =(3-\nu )/(1+\nu )\). The displacements can then be written as \begin {align} u_x &= \frac {1}{4\mu }\left [(\kappa -1) \Re \bar {Z} - 2y \Im Z\right ] \\ u_y &= \frac {1}{4\mu }\left [(\kappa +1) \Im \bar {Z} - 2y \Re Z\right ]. \end {align}
Since the integral of the Westergaard function near the crack tip is \begin {equation} \bar {Z}(z^*) \approx \sigma _\infty \sqrt {2 a z^*} = K_I \sqrt {\frac {2 z^*}{\pi }} = 2K_I \sqrt {\frac {r}{2\pi }} e^{i\theta /2}, \end {equation} we can directly write the displacement field in cylindrical coordinates as \begin {align} u_x &= \frac {K_I}{2\mu } \sqrt {\frac {r}{2\pi }} \left [ 2\frac {1-\nu }{1+\nu } \cos \frac {\theta }{2} + \sin ^2 \frac {\theta }{2} \right ]\\ u_y &= \frac {K_I}{2\mu } \sqrt {\frac {r}{2\pi }} \left [ 2\frac {1-\nu }{1+\nu } \cos \frac {\theta }{2} + \sin ^2 \frac {\theta }{2} \right ]. \end {align}
Note that in the plane of the crack (\(y=0\), \(z^*=x\)), the displacement field is given by \begin {align} u_x &= \begin {cases} \frac {K_I}{2\mu } (\kappa -1) \sqrt {\frac {x}{2\pi }} & \text {if}\;x > 0 \\ 0 & \text {if}\;x \leq 0 \end {cases}\\ u_y^+ &= \begin {cases} 0 & \text {if}\;x \geq 0 \\ \frac {K_I}{2\mu } (\kappa +1) \sqrt {-\frac {x}{2\pi }} & \text {if}\;x < 0 \end {cases}. \label {eq:displyat0} \end {align}
The \(y\)-displacement here is denoted with a little \(+\), \(u_y^+\), to indicate that this is the displacement of the top crack face at a position \(y=0^+\), i.e. slightly above \(y=0\). The displacement \(u_y^-\) of the bottom crack face is the negative of this values, \(u_y^-=-u_y^+\), for symmetry reasons. Mathematically, this property emerges because the square-root has a branch cut along the negative real axis.
12.2.2 Strain energy release rate
In order to formulate a fracture criterion we will require an expression for the elastic energy released during propagation of the crack. This will lead to the concept of the strain energy release rate.
We have not yet talked about the concept of energy in these notes. Since elasticity is fully reversible (an elastic object returns to its origin shape when unloaded), the work carried out when deforming an elastic body is conservative. This means we can define an elastic energy that is recoverable by unloading the body. In order to define the energy release rate, we will here focus on the work \(W\) done by a moving crack.
We first note that the crack faces do not contribute to the work because the normal stress is zero by definition, since we are dealing with a free surface. We therefore have to focus on the crack tip. The tip opens by the distance given by Eq. \eqref{eq:displyat0} to both sides. This displacement works against the stress \(\sigma _{yy}\) right at the crack tip. We now assume that the crack moves by a distance \(\Delta a\) which we will later take to zero. We assume that the stress before the crack has moved is taken to zero quasistatically during the crack opening process. The crack faces have opened a distance \begin {equation} u_y^+ - u_y^-=2\frac {K_I}{2\mu } (\kappa + 1) \sqrt {\frac {\Delta a - x^*}{2\pi }} \end {equation} after the crack has moved by \(\Delta a\). The stress was \begin {equation} \sigma _{yy}=\frac {K_I}{\sqrt {2\pi x^*}} \end {equation} before the movement of the crack. The work on the crack faces is then \(W=\int dx^* \sigma _{yy} (u_y^+-u_y^-)/2\). (The factor \(1/2\) enters because we are taking the stress to zero as the crack faces are displacing. Imagine a simple spring with force \(f=kx\) taken from \(x_0\) to \(x=0\). The work is \(kx_0^2/2\), equal to the elastic energy of the string at extension \(x_0\).) This gives \begin {equation} W(\Delta a) = \frac {K_I^2}{4\pi \mu } (\kappa + 1) \int \limits _0^{\Delta a} \dif x^* \sqrt {\frac {\Delta a - x^*}{x^*}} = \frac {K_I^2}{\mu } \frac {\kappa +1}{8} \Delta a = \frac {K_I^2}{E} \Delta a \end {equation} where we have used \(\int _0^1 dx \sqrt {(1-x)/x}=\pi /2\). The energy released per crack length is then \begin {equation} G = \frac {W}{\Delta a} = \frac {K_I^2}{E}. \label {eq:energyreleaserate} \end {equation} \(G\) is called the strain energy release rate and has units of energy per area. Note that Eq. \eqref{eq:energyreleaserate} is valid for a state of plain strain, which has entered through Hooke’s law in Eqs. \eqref{eqn:crackstrainxx} to \eqref{eqn:crackstrainxy}.
12.3 Griffith’s fracture criterion
The value of \(K_I\) uniquely defines the stress field near the crack tip and therefore determines when the crack advances. We define a critical \(K_{Ic}\) beyond which the crack becomes unstable and a new crack opens when \(K_I>K_{Ic}\). \(K_{Ic}\) is called the fracture toughness. Generally, \(K_I\) depends on crack geometry and loading condition. For the example worked above, a straight crack of length \(a\) in an infinite isotropic medium we get Eq. \eqref{eqn:stress˙intensity}. We see that the stress intensity factor \(K_I\) growth with crack length \(a\); hence there is a critical length \(a_c\) beyond which the crack becomes unstable.
As the crack advances it creates new surface area. An advance of the crack from length \(a\) to length \(a+\Delta a\) requires the additional energy \(\Delta E_{\text {surf}}=2\gamma \Delta a\). Here, \(\gamma \) is the surface energy of the pristine crack surface and the factor of \(2\) enters because a crack has two faces. Since we are looking at a plane situation, \(\Delta E_{\text {surf}}\) is an energy per unit length.
Griffith’s fracture criterion now states that the crack advances when the energy released from the elastic field (described by the strain energy release rate \(G\)) is larger than the energy needed to create a new surface, \begin {equation} G>G_c \end {equation} where \(G_c\) is Griffith’s critical energy release rate. For an ideal brittle crack, \begin {equation} G_c=2\gamma , \end {equation} because we only need to “pay” for the new surface with elastic energy. This allows us to define a fracture toughness, \(K_{Ic}=\sqrt {E G_c}\), from Griffith’s theory.