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DoITPoMS Teaching & Learning Packages Slip in Single Crystals Slip in Single Crystals (all content)

Slip in Single Crystals (all content)

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On completion of this TLP you should:

  • appreciate that slip occurs on a given slip system when the resolved shear stress on that system reaches a critical value
  • be able to calculate and predict slip behaviour in simple situations
  • understand how slip geometry changes as slip proceeds
  • understand the phenomena of geometric softening and work hardening and their effect on slip in hexagonal close packed crystals, cubic close packed crystals and polycrystalline materials

Before you start

In this package, we use the Miller three-index notation to describe lattice planes and directions. (For simplicity, we have avoided the use of the Miller-Bravais four-index notation for the description of hexagonal crystal systems).

It is assumed that you are familiar with the concept of dislocations, including their structure and movement. It might be useful to look at the Introduction to dislocations teaching and learning package.

It would be useful to be familiar with some basic crystal structures and crystallography.


When a single crystal is deformed under a tensile stress, it is observed that plastic deformation occurs by slip on well-defined parallel crystal planes. Sections of the crystal slide relative to one another, changing the geometry of the sample as shown in the diagram.

Diagram illustrating slip in a single crystal

By observing slip on a number of specimens of the same material, it is possible to determine that slip always occurs on a particular set of crystallographic planes, known as slip planes. In addition, slip always takes place along a consistent set of directions within these planes – these are called slip directions. The combination of slip plane and slip direction together makes up a slip system. Slip systems are usually specified using the Miller index notation. For example, cubic close-packed metals slip on <1 bar1 0>{111}: that is, in directions related to [1 bar1 0] by symmetry and on planes related to (111) by symmetry. The slip direction must lie in the slip plane.

Generally, one set of crystallographically equivalent slip systems dominates the plastic deformation of a given material. However, other slip systems might operate at high temperature or under high applied stress. The crystal structure and the nature of the interatomic bonding determine the slip systems that operate in a material.

Slip geometry: the critical resolved shear stress

Slip occurs by dislocation motion. To move dislocations, a certain stress must be applied to overcome the resistance to dislocation motion. This is discussed further in the Introduction to dislocations package on this site.

It is observed experimentally that slip occurs when the shear stress acting in the slip direction on the slip plane reaches some critical value. This critical shear stress is related to the stress required to move dislocations across the slip plane.

The tensile yield stress of a material is the applied stress required to start plastic deformation of the material under a tensile load. We want to relate the tensile stress applied to a sample to the shear stress that acts along the slip direction. This can be done as follows. Consider applying a tensile stress along the long axis of a cylindrical single crystal sample with cross-sectional area A:

Diagram illustrating application of tensile stress along the long axis of a cylindrical single crystal sample

The applied force along the tensile axis is F = σA. If slip occurs on the slip plane shown in the diagram, with plane normal n, then the slip direction will lie in this plane. We can calculate the resolved shear stress acting parallel to the slip direction on the slip plane as follows.

The area of the slip plane is A/cosφ, where φ is the angle between the tensile axis and the slip plane normal.

The component of the axial force F that lies parallel to the slip direction is F cos λ. The resolved shear stress on the slip plane parallel to the slip direction is therefore given by:

\[{\tau _{\rm{R}}} = \frac{{{\rm{resolved}}\;{\rm{force}}\;{\rm{acting}}\;{\rm{on}}\;{\rm{slip}}\;{\rm{plane}}}}{{{\rm{area}}\;{\rm{of}}\;{\rm{slip}}\;{\rm{plane}}}} = \frac{{F\cos \lambda }}{{A/\cos \varphi }} = \frac{F}{A}\cos \varphi \cos \lambda \]

It is found that the value of τR at which slip occurs in a given material with specified dislocation density and purity is a constant, known as the critical resolved shear stress τC. This is Schmid's Law.

The quantity cos φ cos λ is called the Schmid factor. The tensile stress at which the crystal starts to slip is known as the yield stress σy, and corresponds to the quantity F/A in the above equation.

Symbolically, therefore, Schmid's Law can be written:

τC = σy cos φ cos λ

In a given crystal, there may be many available slip systems. As the tensile load is increased, the resolved shear stress on each system increases until eventually τC is reached on one system. The crystal begins to plastically deform by slip on this system, known as the primary slip system. The stress required to cause slip on the primary slip system is the yield stress of the single crystal. As the load is increased further, τC may be reached on other slip systems; these then begin to operate.

From Schmid's Law, it is apparent that the primary slip system will be the system with the greatest Schmid factor. It is possible to calculate the values of cos φ cos λ for every slip system and subsequently determine which slip system operates first. This can be time consuming, but for cubic crystal systems, the OILS rule and Diehl's rule provide quick routes to identifying the primary slip system.

Geometry during slip

Two conditions restrict the geometry of a crystal as slip proceeds:

  • the spacing of the planes remains constant;
  • the number of planes in the specimen is conserved.

These give rise to two important relationships that describe the way that the orientation of slip planes and slip directions changes as slip proceeds:

  • l cos φ is constant, so that as the specimen length l increases, the angle between the slip plane normal and the tensile axis approaches 90°
  • l sin λ is constant, so that as l increases, the angle between the slip direction and the tensile axis approaches zero.

If a crystal is extended from length l0 to length l1, then the angles φ and λ are related as follows:

l0 cos φ0 = l1 cos φ1
l0 sin λ0 = l1 sin λ1


Slip in HCP metals 1: slip systems

In hexagonal close packed (h.c.p.) metals, such as cadmium, slip occurs in <100> type directions on {001} type planes. These correspond to the close packed directions in the close packed planes.

Examination of the crystal structure (see the diagram below) shows that there is only one distinct lattice plane of the {001} type, i.e. (001). There are three distinct <100> directions lying in this (001) plane: [100], [010] and [110]. Hence, the h.c.p. structure exhibits three distinct slip systems. The h.c.p structure has only two independent slip systems, since any slip on [110](001) can be described entirely as a combination of slip on [100](001) and [010](001).

Diagram of h.c.p. crystal structure

Hexagonal close packed crystals slip on <100>{001} slip systems. This diagram shows a 2x2 array of unit cells projected onto the (001) plane. The three slip directions lying in the plane are shown as blue arrows.

Slip in HCP metals 2: application of Schmid's Law

The analysis of slip in h.c.p. crystals can be demonstrated by considering the special case in which the slip direction, the tensile axis and the normal of the slip plane are all coplanar. In this special case, φ + λ = 90°. This condition is not generally true.

For example, consider a cadmium single crystal strained with the tensile axis along the [021] direction. Cadmium is (approximately) hexagonal close-packed with a = b = 2.98 Å and c = 5.62 Å, and it slips on <100>{001}. The slip plane must be (001), since that is the only plane of the {001} type. Hence, the slip plane normal is parallel to [001]. The tensile axis [021] and the slip plane normal [001] both lie in the (100) plane.

The operating slip system will be that with the highest resolved shear stress acting upon it. By considering the geometry, shown in the diagram below, it is apparent that the primary slip system (i.e. the system with the greatest Schmid factor) in h.c.p. crystals will be [010](001) for all tensile axes of the [0VW] type.

Diagram illustrating slip geometry

Geometry of slip in a hexagonal close-packed crystal system. The angle between the [100] and [010] directions is 120°. The three possible slip systems are [100](001), [010](001) and [110](001). For a tensile axis lying in the (100) plane of the type [0VW], the angle between the slip direction and the tensile axis is smallest for the [010] slip direction. cos φ cos λ is largest when λ = λ[010], hence the operating slip system is [010](001).

Slip in HCP metals 3: calculation of forces

Having determined which slip system will operate first, it should now be possible to calculate the minimum force needed to cause plastic flow during the application of a stress to the crystal. The calculation proceeds as follows.

The following diagram shows the orientation of the [021] tensile axis with respect to the unit cell vectors b and c (parallel to [010] and [001] respectively). These three vectors all lie in the (100) plane.

Diagram showing orientation of the [021] tensile axis with respect to the unit cell vectors b and c

Identification of the angles λ and φ. The diagram shows coplanar vectors on the (100) plane. The (001) slip plane lies horizontal and extends perpendicular to the screen.

The initial angle between the tensile axis and the slip plane normal, φ0, is

\[{\varphi _0} = {\tan ^{ - 1}}\left( {\frac{{2a}}{c}} \right) = {\tan ^{ - 1}}\left( {\frac{{2 \times 2.98}}{{5.62}}} \right) = 46.7^\circ \]

and the angle between the tensile axis and the slip direction, λ0, is

λ0 = 90° - φ0 = 43.3°

The Schmid factor for the [010](001) slip system is therefore

cos φ0 cos λ0 = cos 46.7° × cos 43.3° = 0.499

If the critical resolved shear stress for cadmium is 0.15 MPa, and the initial crystal diameter is 3 mm, then the force required to cause slip can be calculated:

\[{\tau _c} = {\sigma _y}\cos {\varphi _0}\cos {\lambda _0} = \frac{F}{A}\cos {\varphi _0}\cos {\lambda _0}\]

\[F = \frac{{{\tau _c}A}}{{\cos {\varphi _0}\cos {\lambda _0}}} = \frac{{0.15 \times {{10}^6} \times \pi \times {{(1.5 \times {{10}^{ - 3}})}^2}}}{{0.499}} = \underline{\underline {2.12{\rm{ N}}}} \]

Now consider what happens when the crystal is plastically extended. If the original length of crystal was l0 = 50 mm, and the crystal is extended by 50 % to l1 = 75 mm, then:

  • The diameter of the crystal will decrease as volume is conserved:

    l0d02 = l1d12

    \[{\rm{thus}}\;{d_1} = \sqrt {\frac{{{l_0}{d_0}^2}}{{{l_1}}}} = \sqrt {\frac{{50 \times {3^2}}}{{75}}} = \underline{\underline {2.45{\rm{ mm}}}} \]

  • The geometry of the slipped crystal changes according to:

    l0 cos φ0 = l1 cos φ1 and l0 sin λ0 = l1 sin λ1


    φ1 = 62.8° and λ1 = 27.2°

    and so the new Schmid factor for the extended crystal is:

    cos φ1 cos λ1 = 0.407

The force required to cause further deformation of the crystal in this condition can be calculated as before:

\[F = \frac{{{\tau _c}A}}{{\cos {\varphi _1}\cos {\lambda _1}}} = \frac{{0.15 \times {{10}^6} \times \pi \times {{(1.225 \times {{10}^{ - 3}})}^2}}}{{0.407}} = \underline{\underline {1.{\rm{74 N}}}} \]

The force required to cause slip is lower after the crystal has been deformed. This phenomenon is known as geometric softening - once deformation has started, less load is required to further deform the crystal. Geometric softening depends heavily on the orientation of the tensile axis within the crystal. For some orientations, no geometric softening is observed. There are other factors that control slip, which will not be discussed here, and these can dominate over the geometric factor.

Slip in HCP metals 4: observing slip in cadmium

Cadmium is a hexagonal close-packed metal (with a non-ideal axial ratio c/a: more ). We have seen that it slips on <100>{001}.

Cadmium single crystals can be grown in the form of long cylinders. The crystal orientation of each crystal is random - i.e. no particular orientation is favoured. The specimen can be deformed under extension control, where the crystal is extended by a given amount and the load required to achieve that extension is recorded.

Typical load-extension curves obtained from such an experiment are shown below. Geometric softening may or may not be observed, depending on the geometry of the crystal.

Graph of load against extension

Schematic load-extension curves for tensile deformation of two cylindrical cadmium crystals with the same initial diameter but different crystallographic orientations. After an initial period of elastic deformation, both begin to plastically deform. The solid line shows the behaviour of a crystal showing geometric softening, where the load decreases over part of the extension. The broken line shows a sample where no geometric softening occurs. The two crystals yield at different loads because the Schmid factors for the primary slip systems are different but τc is constant for the material. Both crystals have the same Young's modulus.

As the crystal deforms, the geometry of the crystal changes according to the relationships given in Geometry during slip. The intersection of the slip planes with the surface of the crystal gives rise to steps and bands on the surface, which can be seen in the optical and scanning electron microscopes.

Scanning electron micrograph of undeformed cadmium single crystal

Scanning electron micrograph of undeformed cadmium single crystal. (Click on image to view larger version.)

Scanning electron micrograph of cadmium single crystal after deformation to 100% strain

Cadmium crystal after deformation to 100 % strain. (Click on image to view larger version.)

Scanning electron micrograph of cadmium single crystal after deformation to 200% strain

High magnification image of slip steps in cadmium crystal after deformation to 200 % strain. The tensile axis runs approximately from the bottom left corner to the top right corner of the image. (Click on image to view larger version.)

Video clips of a crystal undergoing deformation in the SEM are available on the following page.

Video clips of slip in a single cadmium crystal

The following video clips were made as a single crystal of cadmium underwent deformation in a scanning electron microscope (SEM) chamber.

A single cadmium crystal undergoing deformation in an SEM:
appearance of slip steps as slip begins on a previously
undeformed crystal

A single cadmium crystal undergoing deformation in an SEM:
appearance of slip steps as slip begins on a previously
undeformed crystal

A single cadmium crystal undergoing deformation in the SEM:
slip proceeding on previously deformed crystal

The trace of the slip planes can be seen. Markers have been dropped onto the surface of the crystal to allow the relative movement of parts of the crystal to be clearly seen.

A single cadmium crystal undergoing deformation in the SEM: slip proceeding on previously deformed crystal

An unusually large slip step can be seen on the left-hand edge of the specimen. As slip proceeds, the slip planes rotate towards the tensile axis, but the large slip step does not increase in size.

A single cadmium crystal undergoing deformation in the SEM: the crystal fractures under tensile load

The mode of fracture is ductile tearing. Note how the angle of the fracture surfaces mimics the angle of the slip planes in the deformed crystal.

Exercise: Determination of the critical resolved shear stress for slip in cadmium

The following sequence of images shows slip in a cadmium crystal after deformation to 40 % and 100 % of its original length. By measuring the angles φ and λ at each stage of deformation, the slip geometry relationships l0 cos φ0 = l1 cos φ1 and l0 sin λ0 = l1 sin λ1 can be tested.

1 ... in separate window

Slip in CCP metals

Cubic close-packed (c.c.p.) crystals have slip systems consisting of the close-packed directions in the close-packed planes. The shortest lattice vectors are along the face diagonals of the unit cell, as shown below:

Diagram of slip directions in a c.c.p. unit cell

Slip systems in c.c.p. on the (111) plane. There are 3 distinct slip directions lying in this plane, and there are 3 other planes of the {111} type, making 12 distinct slip systems of the <1 bar1 0>{111} type.

The cubic symmetry requires that there be many distinct slip systems, using all <1 bar1 0> directions and {111} planes. There are 12 such <1 bar1 0>{111} systems, five of which are independent. Note that on a given slip system, slip may occur in either direction along the specified slip vector. The c.c.p crystal therefore has many more slip systems than an h.c.p. crystal, and slip progresses through three stages:

Graph of resolved shear stress against shear strain during deformation of a c.c.p. single crystal

Plot of the resolved shear stress τR acting on one slip system against the shear strain γ during deformation of a c.c.p. single crystal.

The initial elastic strain is caused by the simple stretching of bonds. Hooke's Law applies to this region.

At the yield point, stage I begins. The crystal will extend considerably at almost constant stress. This is called easy glide, and is caused by slip on one slip system (the primary slip system).

The geometry of the crystal changes as slip proceeds. The Schmid factor changes for each slip system, and slip may begin on a second slip system when its Schmid factor is equal to that of the primary slip system. In this stage of deformation, known as stage II, dislocations are gliding on two slip systems, and they can interact in ways that inhibit further glide. Consequently, the crystal becomes more difficult to extend. This phenomenon is called work hardening. The stress / strain ratio in stage II may be constant.

Further details of work hardening.

Stage III corresponds to extension at high stresses, where the applied force becomes sufficient to overcome the obstacles, so the slope of the graph becomes progressively less steep. The work hardening saturates. Stage III ends with the failure of the crystal.


In this teaching and learning package, we have seen how the phenomenon of plastic deformation proceeds by slip. This involves dislocation motion in specific directions on specific planes, which in combination are known as slip systems.

The observed yield stress of a single crystal is related to the geometry of the crystal structure via Schmid's Law:

τC = σy cos φ cos λ

where τC is the critical resolved shear stress which is related to the stress required to move dislocations across the slip plane.

The macroscopic behaviour of cadmium crystals was examined as an example of slip in a hexagonal close-packed metal. It was demonstrated that the orientation of the crystal with respect to the tensile axis is crucial in determining the behaviour of a single crystal undergoing deformation. Microscopic slip steps were observed on the crystals, which confirm the geometry of slipand show that certain geometrical relationships are obeyed as slip proceeds.

The crystal structure of the material can affect the nature of slip. We have seen how cubic close-packed metals undergo work hardening due to the simultaneous operation of several slip systems - this mechanism cannot occur in hexagonal close-packed crystals unless unusual slip systems operate.

In polycrystalline materials, the distribution of grain orientations and the constraint to deformation offered by neighbouring grains gives rise to a simplified overall stress-strain curve in comparison to the curve from a single crystal sample. Crystal structure is also important in polycrystalline samples - the von Mises criterion states that a minimum of five independent slip systems must exist for general yielding.


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. A cubic close-packed (cubic F) metal is deformed under tension. The tensile axis lies along [2 -3 1]. What is the primary slip system?

    a [0 -1 1] (1 -1 -1)
    b [0 1 1] (1 -1 1)
    c [1 -1 0] (1 1 -1)
    d [1 1 0] (1 -1 1)

  2. Cadmium is hexagonal close-packed and slips on <100>{001} slip systems. As dislocations glide in a cadmium single crystal during plastic flow, steps form at the edges of the crystal where the dislocations reach the surface. What height will the slip step arising from the arrival of one single dislocation at the surface of the crystal be, in terms of the lattice parameters a and c?

    a a ∗√(3/4)
    b a
    c c
    d a / 2

  3. An amorphous solid is deformed under tension. Which of the following statements describes its behaviour best?

    a Dislocations glide through the material, resulting in bulk plastic deformation.
    b There is virtually no plastic deformation by slip, since dislocation movement through the structure is very difficult.
    c There is no plastic deformation by slip, since dislocations cannot exist in an amorphous material.
    d Dislocations can move through the solid, but there are no defined slip planes so plastic deformation tends to occur by other mechanisms.

  4. In hexagonal and cubic close-packed crystal structures, slip occurs along close-packed directions on the close-packed planes. Body-centred cubic metals are also ductile through the mechanism of slip, but they have no close-packed planes. What slip systems do b.c.c. crystals slip on?

    a <1 -1 1>{110}
    b <1 -1 1>{111}
    c <1 -1 0>{110}
    d <001>{110}

Deeper questions

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

  1. Find the Schmid factor for the primary slip system in a cubic close-packed single crystal when the tensile axis is parallel to [3 4 -1].

  2. This question refers to the exercise involving deformation of a cadmium single crystal.

    The experimental data from the experiment are given in the table below. The angles φ and λ at the two stages of elongation were measured from the diagrams in the exercise.

    Determine what the values of φ and λ must have been in the crystal before deformation. (Assume that only one slip system operated.) The values of lcosφ and lsinλ have been calculated for you.

    Percent strainSample length / mmφ lcosφ / mmλlsinλ / mm
    0 %l0 = 18φ0 l0cosφ0λ0l0sinλ0
    40 %l1 = 25φ1 = 60°l1cosφ1 = 12.5λ1 = 30°l1sinλ1 = 9.3
    100 %l2 = 36φ2 = 75°l2cosφ2 = 9.3λ2 = 15°l2sinλ2 = 12.5

  3. A polycrystalline sample of a cubic close-packed metal is deformed under tension along a tensile axis parallel to [134]. If the critical resolved shear stress τc = 95 kPa, estimate the yield stress σy of the sample, assuming that there are no obstacles to dislocation motion and that the grains have random orientation relative to one another.

  4. Sometimes plastic deformation occurs without slip. Suggest mechanisms by which plastic deformation could occur without slip in the following circumstances:

    1. At elevated temperature with a very low strain rate.
    2. In an h.c.p. polycrystalline sample with only 3 independent slip systems.
    3. In a semi-crystalline polymer sample.

Going further


Most general 'engineering materials' books cover relevant topics. For example, try:

  • Newey C and Weaver G, Materials: Principles and Practice, Open University and Butterworths, 1990.
  • Weidmann G, Lewis P and Reid N, Structural Materials, Open University and Butterworths, 1990.

Also consider looking at: Kelly A and Knowles K M, Crystallography and Crystal Defects (second edition), John Wiley, 2012 -- a comprehensive (and mathematically detailed) exploration of the relationship between the crystal structure and properties of solids.

CD-ROM and websites

The MATTER Project's 'Materials Science on CD-ROM' includes relevant modules on:

  • Introduction to Crystallography
  • Dislocations

See the MATTER website for details of availability.

Which slip system will operate?

Calculating the Schmid factor for each slip system is possible but can be complicated and time-consuming, depending on the crystal structure and the orientation of the tensile axis. For some cubic structures, there are two short cuts for finding the slip system that will operate. These methods work only for cubic crystals that slip on <1 bar1 0> {111} or <1 bar1 1> {110}, i.e. cubic close-packed (c.c.p) or body-centred cubic (b.c.c.) metals respectively.

The OILS rule: example for a c.c.p. metal

  1. Write down the indices of the tensile axis [UVW]
  2. Ignoring the signs, identify the highest (H), intermediate (I) and lowest (L) valued indices.
  3. The slip direction is the <110> direction with zero in the position of the Intermediate index and the signs of the other two indices preserved.
  4. The slip plane is the {111} plane with the signs of the Highest and Intermediate indices preserved, but with the sign of the Lowest index reversed.

OILS stands for zerO Intermediate, Lowest Sign. The method works equally well for bcc metals in which the directions are of the form <111> and the planes {110}.

Example: In a c.c.p metal, if the tensile axis is along [214] the indices are identified as [ILH].

So zerO Intermediate gives the slip direction as [011], and Lowest Sign tells us that the slip plane is (1 bar1 1). Hence the slip system with the greatest Schmid factor is [011](1 bar1 1).

Diehl's Rule

This method also works for c.c.p. and b.c.c. crystals, but requires some knowledge of stereograms. It can be shown that Diehl's rule is exactly equivalent to the OILS rule, so if you have not studied stereograms, there is no need to learn Diehl's method.

Diehl's Rule: example for a c.c.p. metal

  1. Draw a cubic stereogram displaying all poles of the forms {100}, {110} and {111}, and the great circles which connect them, as shown above. NB: In the above sketch, only relevant poles have been labelled, to preserve clarity.
  2. Your diagram will show 48 standard triangles. Locate the triangle that contains the pole of the tensile axis (TA). For example, in the above diagram the TA lies in the 001-101-111 spherical triangle, which has been shaded.
  3. The slip plane in a c.c.p. crystal must be of the form {111}. To identify the operational slip plane, take the {111}-type pole in the shaded triangle and form its reflection in the side opposite to it, i.e. follow the purple arrow above. In this case the slip plane is (1 bar1 1).
  4. Similarly, the slip direction must be of the <110> type. Take the reflection of the <110>-type pole in the side opposite to it to find the actual slip direction, i.e. the blue arrow. Hence for this example, the slip direction is [011].

For a b.c.c. metal, the same method can be used by interchanging the slip plane and slip direction indices, so that directions are of the form <111> and planes are of the form {110}. If the tensile axis lies at the boundary of two or more triangles, then the Schmid factors of each corresponding slip systems are equal.

Geometry as slip proceeds

Let N = number of slip planes in the sample, d = perpendicular spacing of slip planes. During slip, both N and d remain constant.

Diagram illustrating geometry as slip proceeds

d = (l/N) cos φ \ N d = l cos φ

N and d are constant, so lcos φ must also be constant throughout slip. Hence

l0 cos φ0 = l1 cos φ1

The same can be done for λ:

Diagram illustrating geometry as slip proceeds

d= (l/N) cos (90º - λ) \ N d = l sin λ

N and d are constant, so l sin λ must also be constant throughout slip. Hence

l0 sin λ0 = l1 sin λ1

Hexagonal close-packed crystals: the axial ratio

The ideal axial ratio c/a for a hexagonal close-packed crystal structure can be calculated by considering non-interacting identical hard spheres packed together in the h.c.p. crystal structure.

If the sphere radius is r, then the lattice parameters a (=b) and c can be written in terms of r:

These two relationships can be solved for the ideal axial ratio c/a:

$$2r = a = {\left( {{{{a^2}} \over 3} + {{{c^2}} \over 4}} \right)^{1/2}}$$

a2 = a2/3 + c2/4

4 = 4/3 + c2/a2

c/a = 1.633

Many metallic materials have the h.c.p. crystal structure, but the axial ratio is rarely ideal. Cadmium, for example, has an axial ratio of c/a = 1.886. This non-ideal structure has implications for the behaviour of the material, for example in slip.

Further details of work hardening

Work hardening is the result of many contributing factors. It arises because of interactions between dislocations. As a material is plastically deformed, dislocations move extensively throughout the crystal, and in addition the dislocation density increases. The effect of this is to increase the number of entanglements - these are points where dislocations interact in such a way that their further motion is hindered.

Dislocations may combine with each other, but only if the reaction is energetically favourable. The energy (per unit length) of a dislocation, U, is ½Gb2, where b is the magnitude of the Burgers vector b and G is the shear modulus. For a dislocation reaction between two dislocations with Burgers vectors b1 and b2 to be energetically favourable, the energy of the combined dislocation (with Burgers vector [b1 + b2]) must be lower than the sum of the original dislocation energies, i.e. (b1 + b2)2 ≤ b12 + b22.

When dislocations on the same slip system meet, the reaction depends on their relative signs. If they both have the same sign, (b1 + b2)2 > b12 + b22, so they repel each other. This repulsion can inhibit further glide. Conversely, if they have opposite signs, they will attract each other, again reducing the mobility of the dislocations.

Dislocations on different slip systems may also react. If the "product" dislocation is not on a valid slip system, it is sessile, and blocks further slip on both systems. This is called a Lomer lock.

c.c.p crystals have many slip systems, so it is likely that many dislocations will interact to inhibit slip. This leads to the increased resistance to deformation in Stage II of the slip process.

Academic consultant: Kevin Knowles (University of Cambridge)
Content development: Derek Holmes, Ashleigh Bridges and Heather Scott
Photography and video: Brian Barber and Carol Best
Web development: Dave Hudson

This TLP was prepared when DoITPoMS was funded by the Higher Education Funding Council for England (HEFCE) and the Department for Employment and Learning (DEL) under the Fund for the Development of Teaching and Learning (FDTL). Additional support for the development of this TLP came from the Armourers and Brasiers' Company and Alcan.