Slip in Single Crystals (all content)
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Contents
Main pages
Additional pages
Aims
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 threeindex notation to describe lattice planes and directions. (For simplicity, we have avoided the use of the MillerBravais fourindex 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.
Introduction
When a single crystal is deformed under a tensile stress, it is observed that plastic deformation occurs by slip on welldefined parallel crystal planes. Sections of the crystal slide relative to one another, changing the geometry of the sample as shown in the diagram.
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 closepacked 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 crosssectional area A:
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 τ_{R} at which slip occurs in a given material with specified dislocation density and purity is a constant, known as the critical resolved shear stress τ_{C}. 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 σ_{y}, and corresponds to the quantity F/A in the above equation.
Symbolically, therefore, Schmid's Law can be written:
τ_{C} = σ_{y} 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 τ_{C} 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, τ_{C} 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 l_{0} to length l_{1}, then the angles φ and λ are related as follows:
l_{0} cos φ_{0} = l_{1} cos φ_{1} 
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).
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 closepacked 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.
Geometry of slip in a hexagonal closepacked 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.
The initial angle between the tensile axis and the slip plane normal, φ_{0}, 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, λ_{0}, is
λ_{0} = 90°  φ_{0} = 43.3°
The Schmid factor for the [010](001) slip system is therefore
cos φ_{0} cos λ_{0} = 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 l_{0} = 50 mm, and the crystal is extended by 50 % to l_{1} = 75 mm, then:

The diameter of the crystal will decrease as volume is conserved:
l_{0}d_{0}^{2} = l_{1}d_{1}^{2}
\[{\rm{thus}}\;{d_1} = \sqrt {\frac{{{l_0}{d_0}^2}}{{{l_1}}}} = \sqrt {\frac{{50 \times {3^2}}}{{75}}} = \underline{\underline {2.45{\rm{ mm}}}} \]

The geometry of the slipped crystal changes according to:
l_{0} cos φ_{0} = l_{1} cos φ_{1} and l_{0} sin λ_{0} = l_{1} sin λ_{1}
hence:
φ_{1} = 62.8° and λ_{1} = 27.2°
and so the new Schmid factor for the extended crystal is:
cos φ_{1} cos λ_{1} = 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 closepacked metal (with a nonideal 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 loadextension curves obtained from such an experiment are shown below. Geometric softening may or may not be observed, depending on the geometry of the crystal.
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.
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.



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.
An unusually large slip step can be seen on the lefthand 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.
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 l_{0} cos φ_{0} = l_{1} cos φ_{1} and l_{0} sin λ_{0} = l_{1} sin λ_{1} can be tested.
1 ... in separate window
Slip in CCP metals
Cubic closepacked (c.c.p.) crystals have slip systems consisting of the closepacked directions in the closepacked planes. The shortest lattice vectors are along the face diagonals of the unit cell, as shown below:
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:
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.
Summary
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:
τ_{C} = σ_{y} cos φ cos λ 
where τ_{C} 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 closepacked 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 closepacked metals undergo work hardening due to the simultaneous operation of several slip systems  this mechanism cannot occur in hexagonal closepacked 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 stressstrain 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.
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!

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

Cadmium is hexagonal closepacked 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?

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

In hexagonal and cubic closepacked crystal structures, slip occurs along closepacked directions on the closepacked planes. Bodycentred cubic metals are also ductile through the mechanism of slip, but they have no closepacked planes. What slip systems do b.c.c. crystals slip on?
Deeper questions
The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.

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

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 strain Sample length / mm φ lcosφ / mm λ lsinλ / mm 0 % l_{0} = 18 φ_{0} l_{0}cosφ_{0} λ_{0} l_{0}sinλ_{0} 40 % l_{1} = 25 φ_{1} = 60° l_{1}cosφ_{1} = 12.5 λ_{1} = 30° l_{1}sinλ_{1} = 9.3 100 % l_{2} = 36 φ_{2} = 75° l_{2}cosφ_{2} = 9.3 λ_{2} = 15° l_{2}sinλ_{2} = 12.5 
A polycrystalline sample of a cubic closepacked metal is deformed under tension along a tensile axis parallel to [134]. If the critical resolved shear stress τ_{c} = 95 kPa, estimate the yield stress σ_{y} of the sample, assuming that there are no obstacles to dislocation motion and that the grains have random orientation relative to one another.

Sometimes plastic deformation occurs without slip. Suggest mechanisms by which plastic deformation could occur without slip in the following circumstances:
 At elevated temperature with a very low strain rate.
 In an h.c.p. polycrystalline sample with only 3 independent slip systems.
 In a semicrystalline polymer sample.
Going further
Books
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.
CDROM and websites
The MATTER Project's 'Materials Science on CDROM' 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 timeconsuming, 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 closepacked (c.c.p) or bodycentred cubic (b.c.c.) metals respectively.
The OILS rule: example for a c.c.p. metal
 Write down the indices of the tensile axis [UVW]
 Ignoring the signs, identify the highest (H), intermediate (I) and lowest (L) valued indices.
 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.
 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
 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.
 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 001101111 spherical triangle, which has been shaded.
 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).
 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.
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
l_{0} cos φ_{0} = l_{1} cos φ_{1}
The same can be done for λ:
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
l_{0} sin λ_{0} = l_{1} sin λ_{1}
Hexagonal closepacked crystals: the axial ratio
The ideal axial ratio c/a for a hexagonal closepacked crystal structure can be calculated by considering noninteracting 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:
$$2r = a = {\left( {{{{a^2}} \over 3} + {{{c^2}} \over 4}} \right)^{1/2}}$$
a^{2} = a^{2}/3 + c^{2}/4
4 = 4/3 + c^{2}/a^{2}
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 nonideal 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 ½Gb^{2}, 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 b_{1} and b_{2} to be energetically favourable, the energy of the combined dislocation (with Burgers vector [b_{1} + b_{2}]) must be lower than the sum of the original dislocation energies, i.e. (b_{1} + b_{2})^{2} ≤ b_{1}^{2} + b_{2}^{2}.
When dislocations on the same slip system meet, the reaction depends on their relative signs. If they both have the same sign, (b_{1} + b_{2})^{2} > b_{1}^{2} + b_{2}^{2}, 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.