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# Griffith's Energy Release Rate

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## Introduction

Alan Arnold Griffith's energy-based analysis of cracks in 1920 is considered to be the birth of the field of fracture mechanics [1]. A copy of his paper can be found here. He was motivated by Inglis's linear elastic solution for stresses around an elliptical hole [2], which predicted that the stress level approached infinity as the ellipse flattened to form a crack.

Inglis's result had prompted much discussion concerning the fact that it could not be "correct" because no material can support an infinite stress without yielding and failing. Therefore, everything should immediately fail under even the smallest load if a crack were present. (But this obviously doesn't happen.) So Griffith looked to propose an energy-based failure criterion that effectively side-stepped Inglis's infinite-stress prediction, while nevertheless making direct use of his linear elastic solution.

Griffith compared the work required to break atomic bonds to the strain energy released as a crack grows. This page begins by covering these topics separately, then combines them to form Griffith's energy release rate criterion.

## Energy Principles

In order to understand Griffith's work, one must understand the basic principles of work and energy, and especially strain energy density. Work, $$W$$, is the mechanical form of energy and is given by

$W = \int {\bf F} \cdot {\bf dx}$
For mechanics calculations, it is often convenient to calculate work (or energy) in terms of stress and strain rather than force and displacement. To do so, multiply and divide the above equation through by the volume, $$V$$, but express $$V$$ in the denominator as $$A L$$, the area times length, and group them as follows

$W \; \; = \; \; \int {\bf F} \cdot {\bf dx} \left( V \over A L \right) \; \; = \; \; \int \left ( { {\bf F} \over A} \right ) \cdot \left ( { d{\bf x} \over L} \right) \; V$
But $$({\bf F} / A)$$ is stress, $$\boldsymbol{\sigma}$$, and $$d{\bf x} / L$$ is strain, $$d \boldsymbol{\epsilon}$$. This gives

$U = \int \boldsymbol{\sigma} : d{\boldsymbol{\epsilon}} \; V$
In this case, $$U$$ has been used instead of $$W$$ to indicate that strain energy is the specific form of work (and energy) being considered. The integral $$\int \boldsymbol{\sigma} : d{\boldsymbol{\epsilon}}$$ is strain energy density. It is strain energy per unit volume and is represented by $$U'$$ as

$U' = \int \boldsymbol{\sigma} : d{\boldsymbol{\epsilon}} \qquad \text{and} \qquad U = U' * V$
For a linear elastic material in uniaxial tension, Hooke's Law reduces to $$\sigma = E \epsilon$$. Inserting this into the integral for strain energy density gives

$U' = \int \boldsymbol{\sigma} : d{\boldsymbol{\epsilon}} = \int E \, \epsilon \, d\epsilon = {1 \over 2} E \, \epsilon^2$
And then the $$\sigma = E \epsilon$$ relation can be substituted to form several different equivalent expressions

$U' \; = \; {1 \over 2} E \, \epsilon^2 \; = \; {1 \over 2} \sigma \epsilon \; = \; {\sigma^2 \over 2 E}$
We will return to the last expression shortly because it appears in the derivation of Griffith's energy criterion.

## Atomic Bonds and Surface Energy

Solids have an equilibrium spacing distance between pairs of atoms making up a material. Too close, and the atoms repell each other. Too far, and the atoms attract each other. And at an intermediate distance, there is no net force between them. They are in equilibrium.

Imagine an atom being pulled away from its neighbor. The x-axis below represents the displacement of an atom from its equilibrium position. The graph shows that the force required to move the atom initially increases as the atoms are separated. (The more the atoms separate, the more they attract each other.) But as the distance increases, the atoms eventually become separated so far that they no longer attract each other. By this point, the force will have leveled out, then begun to decrease, before eventually returning to zero.

The key principle here is that the area under the force-displacement curve represents the energy of an atomic bond.

The above discussion presented the concept of atomic bond energy. In order to break a bond, an amount of work equal to the bond energy must be performed on the system. The next figure extends this to the case of a crack growing in a solid.

The figure shows a crack that has grown to length, $$a$$, and in the process, has broken several atomic bonds along the way, each requiring a certain amount of work to overcome the atomic bond energy. The total energy is expressed as

$E_{bond} = 2 \, \gamma_s a \, B$
where $$\gamma_s$$ represents the energy required to break atomic bonds per unit surface area created by the crack. (It does not stand for shear strain here.) The surface area is $$a * B$$ where $$a$$ is the crack length and $$B$$ is the part thickness. Curiously, it is convention to measure $$\gamma_s$$ relative to free surface area (the $$s$$ stands for surface), and since there are two free surfaces created by a crack, a top and a bottom, the 2 is necessary in the equation.

The bottom line is that $$E_{bond} = 2 \, \gamma_s a \, B$$ gives the amount of work (energy) that must be input into an object to overcome the atomic bond energy of the atoms forming a crack of length, $$a$$.

### Conventions

It is common in fracture mechanics to use $$B$$ for thickness, perhaps because $$t$$ and $$T$$ are usually assigned to time and temperature, respectively. And $$a$$ is used for crack length, probably a carry-over from using $$a$$ and $$b$$ for the major and minor axes of an ellipse.

## Strain Energy Release

The concept of Strain Energy Release with crack growth is simple. Consider a bar pulled in tension by a stress, $$\sigma$$. Its strain energy will be

$U = {\sigma^2 \over 2 E} \, V$
according to the Energy Principles discussed above. Now assume that a crack propagates all the way across the bar, causing it to snap in two halves. Both halves will now be unloaded following the break, so they can have no strain energy. The original strain energy, $$U$$, has been released as a result of the propagation of the crack across the part.

Griffith was the first to compute the strain energy release associated with crack growth. He did so for the case of an infinite plate in uniaxial tension. Why this scenario? Easy, because the infinite plate in uniaxial tension was the very case that Inglis had just solved seven years earlier in 1913. Griffith used Ingils' limiting case of an ellipse flattened to form a crack and integrated the stress and strain fields to obtain the strain energy as a function of crack length, $$a$$. He obtained the following result (for one-half of the infinite plate).

$U = {\sigma^2 \over 2 E} \, V - {\sigma^2 \over 2 E} \, B \, \pi \, a^2$
The result is surprisingly simple. It is a combination of a baseline value, $${\sigma^2 \over 2 E} \, V$$, corresponding to zero crack length, and a second term that subtracts strain energy away as the crack length increases according to $$a^2$$. It is interesting that the quadratic dependence on crack length means that the change in strain energy with crack length at short lengths is very small, but then becomes very sensitive to crack length at longer lengths.

One might wonder about the utility of this result given that the plate is infinite in size and therefore, the strain energy is infinite because $$V$$ is infinite. Nevertheless, this equation will proove to be useful because we will eventually differentiate it, thereby eliminating the infinite volume term because it is constant.

### Quick Geometric Interpretation

There is an easy way to remember Griffith's result, and it is shown in the figure below. It is to think of an unloaded portion of material above and below the crack that is triangular in nature, having width $$a$$ and height $$\pi a$$. This gives an area of $$\pi a^2 / 2$$, and since there are two triangles, one above and one below, the area of both triangles is $$\pi a^2$$. Finally since the thickness is $$B$$, the volume is $$B \pi a^2$$. This matches the volume term in Griffith's equation.

So the easy way to think of this is there is a volume of material near the crack equal to $$B \pi a^2$$ that subtracts out the strain energy density, $$\sigma^2 / 2 E$$. Keep in mind that this is just an easy way to remember the result. It is not meant to imply that the triangular region is in fact a hard boundary separating nonzero strain energy regions from zero strain energy regions.

## Griffith's Failure Criterion

Recall that the energy input required to break the atomic bonds and grow the crack is given by

$E_{bond} = 2 \, \gamma_s a \, B$
whereas as the crack grows, the stored mechanical strain energy decreases according to

$U = {\sigma^2 \over 2 E} \, V - {\sigma^2 \over 2 E} \, B \, \pi \, a^2$
The total energy in the system is simply the sum of the two expressions. This is reflected by the red curve in the figure below.

$E_\text{total} = 2 \, \gamma_s a \, B + {\sigma^2 \over 2 E} \, V - {\sigma^2 \over 2 E} \, B \, \pi \, a^2$

The key insight is to recognize that for short crack lengths (left of the max value on the graph), the total energy of the object increases with increasing crack length. Therefore, additional energy must be input into the material in order to cause the crack to grow. It is stable.

However, at longer crack lengths (right of the max curve value), an increase in crack length leads to a decrease in total energy. This means the crack can grow without any additional external input. This is an unstable situation that can lead to catastrophic failure as a crack suddenly propagates completely through a part. In technical lingo, it is said that the crack can grow spontaneously.

We need to differentiate the total energy curve with respect to crack length, $$a$$, and then set the derivative to zero, in order to find the length at which the unstable crack growth (and part failure) can occur.

${dE_\text{total} \over da} \, = \, 2 \, \gamma_s \, B \, - \, {\sigma^2 \over E} \, B \, \pi \, a \, = \, 0$
Cancelling out $$B$$ from the terms and solving for $$\sigma$$ gives

$\sigma_f = \sqrt{2 \, \gamma_s E \over \pi \, a}$
An $$_f$$ subscript has been added to $$\sigma$$ to signify failure. The equation shows that the failure stress increases with the square root of bond energy, $$\gamma_s$$, but decreases with the square root of crack length, $$a$$. Finally, the quantity $$2 \gamma_s$$ is combined into the Griffith Critical Energy Release Rate, $$G_c$$.

$\sigma_f = \sqrt{G_c E \over \pi \, a}$

### Example of a Crack in Glass

It is well known that glass panes are brittle, and are especially susceptible to shattering when cracks are present. Griffith's criterion can be used to estimate the critical failure stress for a given crack length. Let's calculate the critical stress of a typical glass pane when a 1" (25.4 mm) crack is present.

The modulus of glass is $$E = 70,000\;\text{MPa}$$ and the critical energy release rate is about $$G_c = 7\;\text{J/m}^2$$.

$\sigma_f \; = \; \sqrt{G_c E \over \pi \, a} \; = \; \sqrt{(7\;\text{J/m}^2) (7\text{E}10\;\text{N/m}^2) \over (\pi) (0.025\;\text{m})} \; = \; 2.5\text{E}6\;\text{N/m}^2 \; = \; 2.5\;\text{MPa}$
This is a very low failure stress. It can be easily exceeded when bending loads are imposed on a pane of glass. For comparison, the yield strength of aircraft-grade aluminum (e.g., Al 7075-T73) is approximately $$400\;\text{MPa}$$.

### Metal Plasticity and Griffith's Criterion

It is important to recognize that the equation $$E_{bond} = 2 \gamma_s B a$$ neglects any energy associated with metal plasticity. It applies only to brittle materials, such as glass, because it only accounts for energy associated with atomic bond breaking. Therefore, Griffith's energy-based failure criterion only applies to brittle materials as well. In fact, it was glass that Griffith used in tests to experimentally confirm his failure criterion.

It was soon discovered that his criterion greatly underestimated the critical failure strength of many metals. It turns out that in metallic materials, there is a great deal more energy dissipation associated with plastic deformation near the crack tip than with atomic bond breaking. This was addressed by Irwin [3] and Orowan [4], who modified Griffith's initial development by adding $$\gamma_p$$, the energy due to plastic deformation at the crack tip per unit surface area created by crack propagation.

$E_{bond} = 2 (\gamma_s + \gamma_p) B a$
Therefore, Griffith's critical energy release rate becomes

$G_c = 2 (\gamma_s + \gamma_p)$
Nevertheless, the equation relating crack length, modulus, critical stress, and critical release rate remains unchanged. It's just that $$G_c$$ now includes plastic energy dissipation.

$\sigma_f \; = \; \sqrt{G_c E \over \pi \, a}$

### Stress Intensity Factors

Solving Griffith's equation for $$G_c$$ gives

$G_c \; = \; { \sigma_f^2 \, \pi \, a \over E}$
which in itself is nothing special. But in 1957, Irwin introduced the critical Stress Intensity Factor, $$K_c$$, defined as [5]

$K_c = \sigma_f \sqrt{\pi \, a}$
and this means that Griffith's equation can also be written as

$G_c = { K_c^2 \over E}$
This has served as a "teaser" to the complete discussion of stress intensity factors, which comes later. It turns out that over time, the use of stress intensity factors became much more popular than energy release rates for linear elastic problems.

## References

1. Griffith, A.A., "The Phenomena of Rupture and Flow in Solids," Philosophical Transactions, Series A, Vol. 221, pp. 163-198, 1920.
2. Inglis, C.E., "Stresses in Plates Due to the Presence of Cracks and Sharp Corners," Transactions of the Institute of Naval Architects, Vol. 55, pp. 219-241, 1913.
3. Irwin, G.R., "Fracture Dynamics," Fracturing of Metals, American Society for Metals, Cleveland, OH, pp. 147-166, 1948.
4. Orowan, E., "Fracture and Strength of Solids," Reports on Progress in Physics, Vol. XII, p. 185, 1948.
5. Irwin, G.R., "Analysis of Stresses and Strains near the End of a Crack Traversing a Plate," Journal of Applied Mechanics, Vol. 24, pp. 361-364, 1957.