Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

DoITPoMS Teaching & Learning Packages Brittle Fracture When do atomic bonds break?
Brittle Fracture AimsBefore you startIntroductionWhen do atomic bonds break?Why do cracks weaken a material?Inglis and the crack tip stress ideaCan we calculate the energy changes?What about tension?Another way of expressing the energiesAnother way of calculating the energiesWhy bother if they are the same?Coping with a scatter in strengthSimulation of Weibull modulus experimentWhen does the sample fail completely?Sub-critical crack growth and R-curvesSummaryQuestionsGoing further
TLP creditsTLP contentsShow all content
Viewing and downloading resourcesAbout the TLPsTerms of useFeedbackCredits Print this page
PreviousNext

When do atomic bonds break?

Fracture is the separation of atoms. We normally do this by applying a force to the body. How big a force should we need?

We want to know how the energy, U, changes with distance, r, between two atoms. A commonly used expression (known as the Lennard-Jones potential is

$$U(r) = 4\varepsilon \left[ {\left( {{{{\beta ^{12}}} \over {{r^{12}}}}} \right) - \left( {{{{\beta ^6}} \over {{r^6}}}} \right)} \right]$$

Other potentials exist, but the argument is essentially the same.

This contains a short range repulsive term, and a long range attractive term. The parameter ε is a measure of the depth of the potential well, and β is the non-infinite distance where the interparticle potential equals zero.

Use the animation to see how the energy changes as the distance between two atoms is varied.

The force between a pair of atoms is calculated by taking the derivative of the energy function, giving:

$$F = {{{\rm{d}}U{\rm{(r)}}} \over {{\rm{d}}r}} = 24\varepsilon \left[ { - 2\left( {{{{\beta ^{12}}} \over {{r^{13}}}}} \right) + \left( {{{{\beta ^6}} \over {{r^7}}}} \right)} \right]$$

Now see how the force between the atoms changes with distance. What is the force needed to separate the atoms in a crystal?

Because we are interested in the material’s properties we normally think of failure stresses, so in a unit area of crystal we need to know how many bonds in an area of crystal normal to the applied force are carrying the load, say 6.25 × 10−22 m2.

From the animation it is clear that materials should have breaking stresses of the order of 10 GPa, but a piece of glass normally breaks at a stress of 70 MPa.

Where have we gone wrong?

What have we just estimated? It is the stress required to break the bonds on a given plane simultaneously, what becomes the fracture surface. So do all the bonds on a given plane break at once?

Anybody who has opened a packet of crisps or torn a sheet of paper knows the answer is ‘no’ – the bonds break a few at a time – at the tip of a growing crack.

So the analysis above does not describe what actually happens when a material breaks, which is a relief as the predictions were no good anyway.

PreviousNext