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DoITPoMS Teaching & Learning Packages Dislocation Energetics Dislocation Energetics (all content)

Dislocation Energetics (all content)

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Contents

Aims

On completion of this TLP you should:

  • Understand why a dislocation forms and how the dislocation width can be determined using an atomistic model, by minimisation of misfit energy due to in-plane stress and misalignment.
  • Appreciate how the energy changes as a dislocation moves and how the parameters of atomic spacing, Shear modulus and Poisson ratio affect the total misfit energy.
  • Be familiar with the concepts of Peierls stress and lattice resistance and compare theoretical and experimental values. Also understand the difference between the linear elastic analytic solution used by Peierls and the atomistic model.

Before you start

 

Introduction

Plastic deformation of crystals generally proceeds by the propagation of dislocations along the slip planes. In most brittle crystals and some metals, it is the changes in misfit energy as the dislocation moves that are the predominant obstacle to dislocation motion. These arise from atoms in the crystal lattice being displaced from their equilibrium positions, so this effect is known as lattice resistance.

Why do dislocations ever form? Deformation can be thought of a kinetic process since dislocation motion (the most common cause of plastic flow) requires the breaking and reforming of bonds, like in a chemical reaction.

In this TLP we will be setting up an atomistic model, rather than using continuum elasticity as in the Peierls Nabarro model, where it is assumed that in-plane strains do not change as the dislocation moves.

It can be shown that despite the general form of the expression it is the same as that derived by Peierls, but with a slightly lower magnitude.

The misfit energy is made up of in-plane strains and misalignment strains (across the slip plane). This energy changes as the dislocation moves. The misalignment energy increases as the dislocation width increases, so act to localise the misfit strains. Whereas the in-plane strain energy decreases as dislocation width increases, so acts to spread the misfit strain over a larger region. The final arrangement of atoms results in a minimisation of the overall misfit energy for a given dislocation width.

Considering the energy changes as the dislocation moves allows us to calculate the Peierls stress, required to move the dislocation, and see that it is exponentially dependent on the dislocation width. The Peierls stress depends on the atomic spacing both normal to the slip plane and parallel to it – in fact it is extremely sensitive to the ratio of lattice parameters b/d. 

Making a dislocation

We will consider the changes in atomic positions in a given plane of atoms when an edge dislocation is formed in a simple cubic lattice. If the extension of the dislocation is large compared with the atomic spacing b, the horizontal displacement u varies slowly from atom to atom. The relative displacement of neighbouring atoms within each half crystal is much smaller than b.

We start out by considering the initial atomic positions in just two planes A and B.

form of a dislocation

The variation of the displacement u with distance from the dislocation line

It is clear from the above figure that a tan function would be consistent with the displacement. Peierls showed this was the case.

    • The following animation shows how the model is set up using the initial atomic positions in just two planes; here we are making the approximation that the displacements of atoms in the planes on either side (in the y direction) are small enough to be neglected from the overall energy. We are able to show that the final displacement (in the x direction) varies as an arctan function, as demonstrated by Peierls.

Join the crystals to form the dislocation

We have determined that the displacements of the atoms are:

\[{u_A}\left( x \right) = - \frac{b}{{2\pi }}ta{n^{ - 1}}\left( {\frac{{{x_A}}}{w}} \right)\]

\[{u_B}\left( x \right) = - \frac{b}{{2\pi }}ta{n^{ - 1}}\left( {\frac{{{x_B}}}{w}} \right)\]

From below, the displacements of the atoms on the A-plane are symmetrical on either side of the dislocation line, and are zero at the centre. It is clear that the misfit around the dislocation is of two types: a strain in the planes above and below the dislocation and a misalignment of the atoms across the slip plane.

The process of determining the energies is shown in the following animation:

Dislocation width

For a given atomic configuration, we can work out the total energy and then try different values of w until we find the minimum. Practically (see graph below), this can be done by summing the energy over planes from n=-1000 to n=+1000 either side of the “extra plane of atoms”, since the effects of increasing the number of planes beyond this is negligible.

As w increases, the decrease in energy associated with localising the in-plane strains (which is why we said a dislocation should never form) is offset by the increase in misalignment energy, which increases with w. There is therefore a minimum in the misfit energy, which gives the dislocation width.

The following graph shows the sums of the misfit energies (in-plane, misalignment and total) resulting from the displacements of planes either side of the initial “extra half plane of atoms”. We can determine the dislocation width (w/b) for given parameters by considering the minimum of the total misfit energy curve. 

If we look at the value of 0.25Gb2 for the four materials given, it is similar to the energy value calculated by the model. This shows an approximate agreement with the linear elastic solution.

What determines w/b?

  • What we are interested in is the ratio of d/b. As d/b increases, w/b increases. d/b is determined by crystal structure.
    • b is the Burger's vector
    • d is the spacing of the close packed planes
    • We also need to take into consideration partial dislocations. For example, copper is ccp so the slip system is {111} <110>. But the dislocations dissociate into two partial dislocations \( \frac{a}{6}\) <112>; hence the magnitude of b will be  \( \frac{a}{{\sqrt 6 }}\; \)  where a is the lattice parameter. 
    • Therefore, the ratio of d/b in copper will be \( \frac{a}{{\sqrt 3 }} \div \frac{a}{{\sqrt 6 }}\; = \sqrt 2 \) 
  • Poisson ratio (ν) affects the in-plane component of the misfit energy. A higher Poisson ratio means the width of the dislocation is greater.

What determines the total energy?

  • The total energy is affected by the magnitude of Gb2, where G is the shear modulus and b is the Burger's vector.

The following clip shows the free surface of a Cd single crystal subject to tensile testing. Slip is occurring on a particular set of planes, and the set of ridges that form on the free surface are created by the arrival of sets of dislocations.

Form of the displacement

We now need to modify the stationary dislocation model to estimate the change in misfit energy of a dislocation as it moves.

The half-plane of atoms moves by much less than the dislocation. The half-plane of atoms was used as the origin, so we think of the dislocations as a moving origin from which we can estimate the displacements. We use the parameter alpha to describe the fraction across the unit cell across which the dislocation has moved.  

As shown in the following animation, when the dislocation has moved through b (where α = 1) the extra half plane of atoms is now moved to the plane adjacent to the new half plane.

The dislocation has moved about twice as far as the extra half plane.

Change in the misfit energy of a dislocation as it moves

The dislocation remains fixed (at the origin) and the lattice moves around it, therefore the initial positions of atoms xA and xB are given by

\[{x_A}{\rm{ }} = {\rm{ }}nb{\rm{ }}-\alpha b\]

\[{x_B}{\rm{ }} = {\rm{ }}nb{\rm{ }}-b/2{\rm{ }}-\alpha b\]

We use the same method as for the α = 0 case to determine the in-plane displacement and misalignment. As a result, the potentials for different α values can be determined. The following animation (which is a magnified version of the strain energy graph on the previous page) demonstrates how changing the parameter alpha affects the position of the energy minimum and the dislocation width.

By calculating the misfit energy for each value of alpha, we can determine the change in misfit energy \(\Delta {U_T}(\alpha )\) as the dislocation moves.

  • ΔUT(x) is the difference in total energy from α = x and α = 0. 
  • At first, as the dislocation moves across the unit cell 0<|α|<0.25, ΔUT increases to a maximum. 
  • At the maximum (α = 0.25), the value of ΔUT is the Peierls energy, ΔUP
  • Then for 0.25<|α|<0.5, ΔUT decreases. At α = 0.5, ΔUT = 0.
  • w also changes as the dislocation moves across the cell as seen. This is because the configuration of the dislocation changes as the dislocation moves across the unit cell. For the energy maxima, the dislocation width is also a maximum. 

Peierls energy

The maximum change in the total misfit energy is called the Peierls energy \(\Delta {U_P}\). It is the energy required to move unit length of dislocation over the resistance of the lattice. The following graph shows how the fraction of the unit cell across which the dislocation has moved affects the misfit energy. Adjusting parameters demonstrates how these affect the Peierls energy.

The misfit energy associated with in plane strains varies sinusoidally, as does the total energy. Whereas the misfit energy associated with the misalignments is greatest at the position of lowest overall energy. The period is b/2. Hence:

\[ \Delta {U_T}\left( \alpha \right) = \frac{1}{2}\Delta {U_P}\left( {1 - cos4\pi \alpha } \right)       (1)\]

This is shown in the following interactive graph.

As can be seen in the magnified strain energy graph, the dislocation width also changes as the dislocation moves, although for simplicity simulations often assume it is constant.

The magnitude of the Peierl's energy (and hence the misfit energy) scales with Gb2. The Peierl's energy varies exponentially with w/b.  

\[\frac{{{\rm{\Delta }}{{U}_{\rm{P}}}}}{{{G}{{\rm{b}}^2}}} ∝ \exp ( - \frac{w}{b})       (2)\]

dislocation width

As can be seen in the misfit energy graph animation, the dislocation width also changes as the dislocation moves, although for simplicity, simulations often assume it is constant. For example, in the energy calculation, we assumed that w/b was a constant for any value of α.

What is the Peierls stress?

The Peierls stress is the minimum shear stress required to move a single dislocation of unit length in a perfect crystal. The magnitude of the Peierls stress determines the ability of the lattice to resist dislocation motion – the lattice resistance in the absence of thermal activation.

Another way is to think about the resistive force to dislocation motion as the magnitude of the gradient of strain energy curve. This means that the Peierls stress is proportional to the maximum gradient of the misfit energy curve. If we use the continuum model, the energy of the dislocation is independent of position. Any stress, however small, would set the dislocation into motion, because the dislocation is always in neutral equilibrium and so will move under any force. We know this is not right. We need to take into account the crystal structure on an atomic level such that the energy of the dislocation depends on its exact position. The dislocation will have several stable equilibrium positions which persist up until a stress is applied that exceeds a certain magnitude (the Peierls stress), at which point the dislocation can move.

Determining the Peierls stress

The force required to move the dislocation is \(F = \;\frac{{\delta {\rm{\Delta }}{U_T}\left( \alpha \right)}}{{\delta \left( {\alpha b} \right)}}           (3)\)

Plotting the derivative of the energy-alpha graph gives a plot of the force required to move a dislocation.  The corresponding graph gives the stress required to move the dislocation per unit length of dislocation.

\[\tau = \;\frac{1}{{{b^2}}}\frac{{\delta {\rm{\Delta }}{U_T}\left( \alpha \right)}}{{\delta \left( {\alpha b} \right)}}           (4)\]

As we saw in expression (1) on the “Peierls energy” page.

\[\;\Delta {U_T}\left( \alpha \right) = \;\frac{1}{2}\Delta {U_P}\left( {1 - cos4\pi \alpha } \right)           (5)\]

Hence,, differentiating, we get \({\tau _P} = \)\(\frac{{2\pi }}{{{b^2}}} \)\(\;\Delta {U_P}\;{\rm{sin}}\left( {4\pi \alpha } \right)           (6)\)

The maximum stress is where \(\;\sin \left( {4\pi \alpha } \right) = \;1 \). The expression is often written as

\[\frac{{{\tau _P}}}{G} = \;\frac{{2\pi }}{{G{b^2}}}\;\Delta {U_P}          (7)\]

A note on this graph: It's important to remember that even though we are multiplying by Gb2, d/b also affects the dislocation width - increasing d/b increases the dislocation width as seen on the "determining w" graph. So for this graph if we impose a w/b, then we can see the effects of changing the magnitude of Gb2 by adjusting the parameters G and b/d. Likewise if we impose a b/d we can see the effects of altering w/b through changing something else like the Poisson ratio.

The graph shows the stress calculated using the maximum of the graph (expression 4) and also the differentiated expression for ΔU (equation 7). They are very similar, which is expected.

Therefore, using expression (2) from the “Peierls energy” page.

\[\frac{{{\tau _P}}}{G} ∝ \;2\pi \exp ( - \frac{w}{b})          (8)\]

The analytical solution provided by the “continuum elasticity” model (/tlplib/dislocations/slip_via_dislocation.php) gives the constants as:

\[ {\tau _P} = \;3G\exp ( - 2\pi \frac{w}{b})           (9)\]

which is sometimes simplified to \( \frac{G}{{180}}\) if we assume that w = b (which is not in fact true as we have seen!). This solution gives us a theoretical shear stress – i.e. the stress required for uniform slip – but as can be seen from the above graph, it is several orders of magnitude higher than the actual Peierls stress.

The analytical solution is from Peierls 1940 and Nabarro 1947. In essence, they assume that the in-plane strain is the surface of a semi-infinite elastic continuum and use that and the sum of misalignment energies to find the width at the equilibrium point. The elastic continuum (i.e. in-plane energy) is assumed to be fixed as is the width; therefore, the changes in energy are due only to the misalignment summation. This gives a wrong answer and indeed the maximum and minimum positions swap (U-tot Vs U-misalignment are out of phase). Since both the in-plane strain energy and the width change as the dislocation is displaced these should be included.

Compared with real values, the atomistic model is actually a small underestimate of the Peierls stress, but the analytical solution is a vast overestimate, which renders it is not very useful.  

Lattice resistance

The higher the Peierls stress the higher the lattice resistance. The model showed an exponential dependence of Peierls stress on w/b (and hence d/b since we found that was the key factor affecting dislocation width).This exponential dependence is very important and is seen in materials experimentally.

d over b versus tau graph

That τP can be predicted for both metallic and covalent materials using the atomistic model; there is no obvious effect of bonding. d/b is the important factor. We often think of covalent materials as the "strongest", but this graph shows otherwise. Experimental determination of the Peierls stress is done by considering the yield stress at zero kelvin. It is seen that the materials tend to follow the trend for giant covalent crystals (diamond, TiC); but it is more complicated when there are coulombic forces coming in to play, for example in ionic crystals.

It is important to consider what slip system we are using. For example, halides (NaCl, LiF) and MgO have two slip systems – {100} <110> (for which there is good agreement) and {110} <110> (for which the model predicts too high a Peierls stress by ≈102).

Uses and limitations of the atomistic model

The atomistic model improves substantially upon the continuum elasticity model in providing an accurate estimate for the Peierls stress.  This is because it takes into account the change in the in-plane strain energy as the dislocation moves, as well as the misalignment energy.

However the model assumes that lattice resistance is dominating plastic flow and so cannot be used to predict τY. In materials (especially metals such as Al, Cu, Ni), the yield stress is substantially higher than τP. This is due to interactions with other obstacles and dislocation interactions, which this model does not consider. It is important to remember that we are considering only independent dislocations in this model; the calculations would become much more complicated if we consider dislocations interacting.

 

The model also does not take into account the anisotropy of crystals; the critical shear stress would depend on some combination of elastic constants, different for each plane. It would also be influenced by the anharmonic forces between atoms, which are here neglected, since the displacements near the dislocation line are large.

One application is in the toughening of non-metallic materials. Increasing toughness requires that the stress required for a dislocation to move must be substantially reduced. In this case we wish to decrease the energy associated with the misalignments across the slip plane, which act to reduce the dislocation width, w. These energies scale with the shear modulus.

 

The data from which the graphs were devised can be seen here.

 

 

 

 

Summary

There are two types of energy associated with a dislocation

  • In-plane energy – decreases as dislocation width increases, so acts to spread the misfit strain over a larger region
  • Misalignment energy – increases as the dislocation width increases, so acts to localise the misfit strains
    • The dislocation width will be the value for which the sum of the two types of energy is a minimum
    • w/b is strongly dependent on d/b, where b is the atom spacing parallel to the slip plane and d normal to it.
    • Changes in misfit energy are the primary obstacle to dislocation motion.
    • Using the atomistic model with a moving origin allows us to estimate the energy as the dislocation moves, hence we can determine the Peierls energy and the Peierls stress.
    • Peierls stress increases exponentially as the dislocation width w/b decreases.

Questions

Quick questions

You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

  1. What is the main cause of lattice resistance?

    a Change in the dislocation width as the dislocation moves
    b Changes in misfit energy as the dislocation moves
    c Change in the Peierls stress as the dislocation moves
    d Dislocations interacting and causing work hardening

  2. What is the main problem with the Peierls continuum elasticity model?

    a It only includes the nearest-neighbour misalignment energy and does not include the next-nearest-neighbour term.
    b It assumes the in-plane energy is negligible and can be ignored
    c It assumes the total energy remains constant as the dislocation moves
    d It assumes that in-plane energy is constant and does not vary as the dislocation moves

  3. How do the misalignment and in-plane energies change with w/b?

    a As w/b increases, both misalignment and in-plane energies increase
    b As w/b increase, both misalignment and in-plane energies decreases
    c As w/b increases, misalignment energy increases; in-plane energy decreases
    d As w/b increases, misalignment energy decreases; in-plane energy increases

  4. What is the relationship between dislocation width and Peierls stress?

    a Peierls stress increases linearly as the dislocation width increases.
    b Peierls stress decreases exponentially as the dislocation width increases.
    c Peierls stress increases exponentially as the dislocation width increases.
    d Peierls stress decreases linearly as the dislocation width increases.

  5. What is the Peierls stress?

    a The minimum stress for which fracture occurs
    b The tensile stress at which slip starts to occur
    c The minimum shear stress required to move a dislocation across the slip plane
    d The minimum stress at which plastic deformation begins

Deeper questions

The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.

  1. For what value(s) of alpha (fraction the dislocation has moved across the unit cell) is the mifit energy maximum?

    Yes No a Alpha = 0
    Yes No b Alpha = 0.25
    Yes No c Alpha = 0.5
    Yes No d Alpha = 0.75

Going further

Books

Derek Hull and, D.J.Bacon, Introduction to Dislocatioins (Volume 3 of Materials Science and technology, 5th Edition), Elsevier, 2011, ISBN: 008096673X, 9780080966731

H.J.Frost and M.F.Ashby, Deformation-mechanism maps: the plasticity and creep of metals and ceramics, First Edidtion, Pergamon Press, 1982, ISBN: 0080293379, 9780080293379, http://engineering.dartmouth.edu/defmech/

Ulrich Messerschmidt, Dislocation Dynamics During Plastic Deformation, Springer-Berlin, Heidelburg, 2010, ISBN: 978-3-642-03176-2, 978-3-642-03177-9, https://doi.org/10.1007/978-3-642-03177-9

Paper

Howie,P.R., Thompson, R.P., Korte-Kerzel,.S., & Clegg,W.J. (2017), Softening non-metallic crystals by inhomogenenous elasticity. Scientific Reports, 7(1), 11602. https://doi.org/10.1038/s41598-017-09453-1

Academic consultant: Bill Clegg (University of Cambridge)
Content development: Emma Bryan
Photography and video:
Web development: Lianne Sallows and David Brook

This DoITPoMS TLP was funded by the Department of Materials Science and Metallurgy, University of Cambridge.