Shear force

If a material is stressed under tension, it will eventually deform if the tensile forces F₀ are too high. As  explained in the article Fundamentals of deformation section, lattice planes must be slipped for this purpose.

For sliding, forces must act in a suitable manner in a slip plane, so that they can be sheared off. However, a force in the slip plane alone is not enough. Because within the slip plane, the force must also be oriented in the slip direction. If, for example, a shearing force acts in a slip plane, which, however, is oriented perpendicular to the slip direction, then the slip plane will not be easily shorn off.

Only forces in slip direction are relevant for shifting of lattice planes!

Since the combination of slip plane and slip direction forms a slip system, the following statement applies:

In order to trigger a deformation process, forces must act in a slip system!

Therefore, the question arises how the force F can be determined in a slip system according to its spatial orientation on the basis of the external force F₀.

For this purpose, on the one hand, the spatial orientation of the slip plane is relevant and, on the other hand, the orientation of the slip direction within the slip plane. Both orientations are defined by an angle.

The angle \(\alpha\) describes the angle between external tensile force F₀ and surface normal of the slip plane. The orientation of the slip direction is analogously defined as angle \(\beta\) between the force’s axis and the slip direction. Note that both angles are only independent within a certain range.

First, the external force F₀ is broken down into a component acting in slip direction F_||. This is done via the cosine of the angle \(\beta\):

\begin{equation}
\label{kraft}
F_{\parallel} = F_0 \cdot \cos(\beta)
\end{equation}

Obviously, one obtains a maximum force in the slip direction, if the force is oriented at a small angle \(\beta\). The slip plane is then inevitably very steeply inclined (large angle \(\alpha\)). With such a steep orientation of the slip plane, however, its area also increases over the cross section.

This in turn means that now significantly more atomic bonds between the lattice planes must be “sheared off”. The lattice planes to be shifted “stick” more closely together, so to speak. In this case, then even the greater force due to the disproportionately increased atomic surface (“adhesion”) is not enough to bring the plane to shearing.

So it is not the force alone one have to take into account for a sliding process but the best possible ratio of force and area (highest force per area)! It is therefore always the shear stress that has to be considered for a possible sliding process.

Shear stress

The quotient of shear force F_|| (in slip direction) and slip plane surface \(A\) yields the shear stress relevant for the sliding process \(\tau\):

\begin{equation}
\label{schubspannung}
\tau=\frac{F_{\parallel}}{A}
\end{equation}

Thus, in addition to the force F_|| (dependent on the angle \(\beta\)), the dependence of the slip plane surface on the angle \(\alpha \) must be known. If the cross-sectional area of ​​the sample is denoted by A₀, then the slip plane area A is given by the angle \(\alpha\) as follows:

\begin{equation}
\label{flaeche}
A=\frac{A_0}{\cos(\alpha)}
\end{equation}

If equation (\ref{flaeche}) and equation (\ref{kraft}) are used in equation (\ref{schubspannung}), then the shear stress \(\tau\) in slip direction of an arbitrarily inclined slip plane is calculated as follows:

\begin{align}
\label{a}
&\tau=\frac{F_{\parallel}}{A} = \frac{ F_0 \cdot \cos(\beta) }{ \tfrac{A_0}{\cos(\alpha)}} =  \frac{F_0}{A_0}\cdot \cos(\alpha) \cdot \cos(\beta) \\[5px]
\end{align}

The expression \(\frac{F_0}{A_0}\) in equation (\ref{a}) corresponds to the externally applied normal stress \(\sigma_0\). In this way, the following relationship between the externally applied normal stress \(\sigma_0\) and the shear stress induced in a slip system \(\tau\) can finally be established:

\begin{align}
\label{s}
&\boxed {\tau = \sigma_0 \cdot \cos (\alpha) \cos (\beta)} ~~~~~ \text {Schmid’s law} \\[5px]
\end{align}

Equation (\ref{s}) is also called Schmid’s law. The geometry factor \(\cos(\alpha)\cos(\beta)\) is referred to as Schmid factor \(m\):

\begin{align}
\label{schmid-faktor}
&\boxed{\tau = \sigma_0 \cdot m} ~~~~~\text{mit}~~~~~\boxed{m=\cos(\alpha) \cos(\beta)} \\[5px]
\end{align}

The Schmid factor describes the relationship between external normal stresses and the shear stresses caused in a slip system!

Critical resolved shear stress

If a certain critical shear stress is exceeded in a slip system (slip system with the largest Schmid factor), then the lattice planes begin to shear off or dislocations begin to move. The deformation process begins.

Note that the critical shear stresses do not refer to the mere shear stress in a slip plane but to that component acting in slip direction. Therefore, these critical shear stresses are referred to as  critical resolved shear stresses (shear stress resolved into the component acting in slip direction).

In order to shear off slip planes, certain critical resolved shear stresses must be exceeded in a slip system (especially in slip direction)!

The difference between the mere shear stress in a slip plane and the resolved shear stress in the slip direction will be explained by a short numerical example. For this purpose, a slip plane is considered which is inclined at an angle of \(\alpha\) = 15°.

If a normal stress of \(\sigma_0\) = 400 N/mm² is applied from the outside, then a shear stress of \(\tau_{p}\) = 100 N/mm² acts within the slip plane. Only if the slip direction were oriented at an angle of \(\beta\) = 75°, 100 N/mm² of (resolved) shear stress would also act in slip direction.

If, however, the slip  direction in the slip plane is oriented at an angle of \(\beta\) = 80, then only a shear stress of \(\tau_{R}\) = 67 N/mm² will result in this slip direction. Only this shear stress, which is resolved in slip direction, is relevant for a possible slipping process!

Maximum resolved shear stress

This raises the question of how a slip system must be spatially oriented in order to obtain the greatest possible shear stress there (maximum Schmid factor). Is it perhaps even possible to orient a slip system in such a way that the entire external normal stress in the slip direction is active?

For this equation (\ref{schmid-faktor}) must be considered closer. First, the expression \(\cos(\alpha)\) reaches the maximum value 1 for  \(\alpha\) = 0, but in this case \(\beta\) inevitably reaches 90° and thus \(\cos(\beta)\) = 0. The Schmid factor is thus zero. In slip planes that are aligned perpendicular to the pulling axis, therefore, no shear stress acts!

Even for the reverse extreme case, when the slip planes are aligned parallel to the tensile axis (\(\alpha\) = 90°), there is no stress in the slip plane, because then inevitably applies \(\cos(\alpha)\) = 0!

In fact, the largest shear stresses in a slip plane will be found at an angle of \(\alpha\) = 45°. The maximum resolved shear stress results when the slip direction is directed “upwards” in this 45° inclined sliding plane. The angle \(\beta\) in this case is also 45° (note that both angles can not be chosen independently). The Schmid factor reaches 0.5 in this ideal case. This corresponds to the maximum possible value of Schmid’s orientation factor:

\begin{align}
&m_{max}=\cos(45°) \cos(45°) = 0,5 \\[5px]
&\boxed{m \le 0,5} \\[5px]
\end{align}

Thus, the maximum resolved shear stresses are at most half the value of the external normal stresses.

In slip systems inclined at an angle of 45°, the greatest shear stresses occur. They amount to half of the externally applied normal stresses!

This is also the reason why special oriented single crystals in the tensile test have slip steps at an angle of 45° to the tensile axis. Further information can be found in the article deformation of single crystals again.

Introduction

Even if single crystals are used far less than polycrystals, one can understand very well the processes of deformation. In particular, the stress-strain curve of a single crystal sample will be discussed in more detail below.

Stage 0 – elastic strain

As explained in the article deformation of single crystals, at first the slip plane which is aligned at an angle of about 45° to the tensile axis will be prefarably activated, since in this slip plane the critical resolved shear stress (CRSS) is reached first.

On the other hand, if the stress is below the CRSS, the deformation vanishes completely after removing the external force (reversible, elastic deformation).

If the shear stress value \(\tau\) ist represented in a graph against the strain value \(\epsilon\), one will see at first a linear relationship between both values (stage 0). This linear relationship is also referred to Hooke’s law. Within the straight line in the stress-strain-curve only elastic deformation will occur. This applies only as long as the shear stress does not exceed the CRSS.

If the shear stress is below the CRSS the material will only be exposed to elastic deformation. Shear stress and strain will then show a linear behavior (Hooke’s law)!

Stage I – single gliding

If the single crystal is stretched beyond the elastic strain, then the critical resolved shear stress is exceeded in a favorable oriented slip plane. The dislocations begin to migrate and the irreversible plastic deformation process starts. In order to maintain the deformation process, the dislocations must then only be kept “running”. The deformation process therefore takes place without the need for an excessive increase in tension. The material behaves relatively soft in this state.

In addition, new dislocations are introduced in the material due to the deformation process. Many moving dislocations ultimately mean good ductility, so that the deformation is practically multiplied. Therefore, the curve in the stress-strain diagram becomes increasingly flatter (stage I). Since only one a slip system is active, this stage is also called single gliding or easy gliding.

At moderate stresses, only one slip system is active (preferably oriented at 45°). This is also called single gliding or easy gliding!

Note: In extreme cases, the dislocation multiplication even leads to the fact that in dislocation-free single crystals the stress even decreases after starting the plastic deformation process. Because in such a case, dislocations must be generated first to a significant extent under increased force. Only when sufficient dislocations are present, the deformation process can be continued with lower shear stress.

Stage II – multiple gliding

If the deformation process is continued and the stress is therefore further increased, the critical shear stress will eventually be exceeded in other slip planes (which are slightly less favorable).

In addition, due to the deformation of the single crystal, a spatial reorientation of the slip planes will occur, so that previously unfavorably located slip planes can now be displaced into more favorable positions. The movement of dislocations then takes place in several differently oriented slip systems. This is the stage of multiple gliding (stage II).

In the stage of multiple gliding several slip systems are active due to higher stresses!

Inevitably, the moving dislocations will meet at the different slip planes and influence each other. Because every dislocation causes compressive and tensile stresses in the immediate vicinity of their dislocation line, which in turn influence other mofing dislocations.

This becomes particularly significant when two dislocations intersect. Energetically, this is possible only with great effort. However, such situations will inevitably occur when several different slip systems become active at the same time. At this stage of multiple gliding, the dislocations therefore hinder each other from moving.

As a result, however, the deformation process is made considerably more difficult, because ultimately a good deformation process is based precisely on the free displacement movement. Thus, now a larger stress is required to further drive the deformation process. At the same time, however, more dislocations are also introduced into the material as a result, which leads to a further increase in the mutual hindrance of the dislocation movement.

It comes in the material to a “dislocation jam”, so to speak. The stress must therefore be greatly increased to ensure further deformation. Stress and strain behave almost linearly in this stage.

If two dislocations are perpendicular to one another on different slip planes, this is also referred to as a forest dislocation. As well as other dislocations, forest dislocations remain in the material after removal of the force. If the deformation process is repeated again, it is mainly these forest dislocations that make the deformation process more difficult from the start. The materiel is now much harder to deform than before. This principle is referred to as work hardening to achieve a strength-increasing effect (also called strain hardening).

Forest dislocations are perpendicular dislocations that block each other! Forest dislocations significantly contribute to strain hardening!

Stage III – crystal recovery

The dislocations that are dammed up or held on crystallographic defects can only be torn loose by larger stresses. The dislocations then spread on other slip planes by diffusion processes. In the case of screw dislocations, this phenomenon is called cross-slipping or, in the case of step dislocations, climbing.

The dislocations thus open up new opportunities to move. Due to the newly gained mobility, the deformation process becomes easier. Therefore, now only a lower increase of stress is needed to further deform the single crystal (stage III). The crystal has, so to speak, “recovered” from the blocking of the dislocation movement, which is why this stage is also referred to as a crystal recovery.

Dislocations can cross-slip oder climb by diffusion processes to other slip planes. The dislocations gain new mobility and the material becomes more ductile (“crystal recovery”).

At some point, however, the atomic cohesion can no longer withstand the stresses in the material. The sample finally breaks.

Note

If the considered monocrystal is spatially aligned in such a manner that under load no slip plane is preferably oriented, the critical shear stress is exceeded simultaneously in several different slip planes. Thus, already at the beginning of the plastic deformation multiple gliding will occur. In this case, the stress-strain curve omits stage I.

Such multiple gliding will generally be present in polycrystalline materials, since the individual crystallites (grains) are randomly oriented. In this way, the stress-strain curve of these materials can be explained, too.

Introduction

If a material on the atomic level is characterized by a uniform lattice orientation (no grain boundaries), it is also called a single crystal.

If such a single crystal is deformed under tension, inclined rings are formed in the deformation region. This can be clearly seen from the copper single crystal shown in the figure below. These rings are so-called slip steps.

Why these slip steps preferentially aligned at an angle of 45°, will be clarified in the following sections.

Shear force

The single crystal with a cross-sectional area A₀is externally loaded with the tensile force F₀. In order to initiate the deformation process and to allow the atomic planes to shear off, a force F_|| must act in the slip planes. Only in this way can the plane actually be moved in the slip direction. On the other hand, a normal force acting perpendicular to the slip planes has no influence on the deformation. As a result, the layers would merely be compressed but not moved.

If a slip plane lies at an angle \(\alpha\) to the tensile axis, then the external force F₀ can be resloved into a parallel component (shear force F_||) and a perpendicular component (normal force F_⊥) to the slip plane. Mathematically, the shear force and the normal force can be determined as a function of the angle \(\alpha\) as follows:

\begin{align}
\label{kraft}
F_{\parallel}&=F_0 \cdot \cos(\alpha) ~~~~~\text{shear force}  \\[5px]
F_{\perp}&=F_0 \cdot \sin(\alpha)~~~~~\text{normal force}  \\[5px]
\end{align}

If the angle \(\alpha\) between the surface normal and the tensile axis is relatively small, the shear force in the slip plane is also relatively low. The force may not be sufficient to activate the slip plane and initiate the deformation process.

If, on the other hand, the slip plane normal lies at a relatively large angle (\alpha\) to the tensile axis, the shear force in the slip plane is relatively large at first. However, at the same time the slip plane area and with it the binding force between the planes increases. Even in this case, the greater shear force may still not be enough to shear off the plane and start the deformation process.

The situation can be compared with a hook-and-loop fastener, wherein the mutual entanglements between the closure materials correspond to the bonds between the atomic planes. If the shear forces to open the hook-and-loop fastener are too low, the closure materials do not slide off each other. However, a multiplication of the force does not lead to slippage, if at the same time the overlap area of ​​the two closure materials increases disproportionately.

There must therefore be a favorable ratio of force and area in order to reach the optimum state where most force per area is obtained. That means where the greatest shear stress is achieved.

Shear stress

Decisive for a slipping of the atomic planes is therefore not the force in the slip plane alone but the acting force per area, i.e. the shear stress (see also fundamentals of deformation):

\begin{equation}
\label{schubspannung}
\tau=\frac{F_{\parallel}}{A}
\end{equation}

While the force F_|| depends on the angle \(\alpha\) of the slip plane accordingly to the equation (\ref{kraft}), the slip plane area A is determined as follows by the sample’s cross section F₀:

\begin{equation}
\label{flaeche}
A=\frac{A_0}{\sin(\alpha)}
\end{equation}

The increase of the shear force (blue curve) and the increase of the area (black curve) with increasing angle schematically shows the diagram below. In addition, the resulting curve of the shear stress is shown (red curve). Obviously, the largest force per area acts at an angle of 45°. This will be shown mathematically in the following section.

Maximum shear stress

The shear stress curve shown in the figure above is obtained by inserting the equation (\ref{kraft}) and the equation (\ref{flaeche}) in the shear stress equation (\ref{schubspannung}):

\begin{equation}
\label{scherspannung}
\tau=\frac{F_{\parallel}}{A}=\frac{F_0 \cdot \cos(\alpha)}{\frac{A_0}{\sin(\alpha)}} =\frac{F_0}{A_0} \cdot \cos(\alpha) \cdot \sin(\alpha)
\end{equation}

The quotient \(\frac{F_0}{A_0}\) appearing in equation (\ref{scherspannung}) corresponds to the normal stress \(\sigma_0\) applied from the outside (“force per cross-sectional area”). For the angle-dependent shear stress \(\tau(\alpha)\), therefore, the following equation applies:

\begin{equation}
\tau(\alpha)=\sigma_0 \cdot \cos(\alpha) \cdot \sin(\alpha)
\end{equation}

The term \(\cos(\alpha)\cdot \sin(\alpha)\) can further be replaced by the expression \(\frac{\sin(2\alpha)}{2}\). Thus, the effective shear stress \(\tau\) in a slip plane which is aligned at the angle \(\alpha\) to the tensile axis, can be calculated as follows:

\begin{equation}
\label{tau}
\boxed{\tau(\alpha)={\frac{\sin(2 \alpha)}{2}  \cdot \sigma_0}  }
\end{equation}

The internal shear stress \(\tau\) thus depends on the applied external stress \(\sigma_0\) and in particular on the angular position \(\alpha\). The term \(\sin(2 \alpha)\) reaches the maximum value 1 for an angle of \(\alpha\) = 45°.

It is now also shown mathematically that the maximum shear stress \(\tau_ {max}\) is thus reached at an angle of \(\alpha\) = 45°. For this case, the maximum shear stress corresponds to exactly half of the outer normal stresses \(\sigma_0\)!

\begin{equation}
\label{max}
\boxed{\tau_{max}=\frac{\sigma_0}{2}} ~~~ \text{for} ~~~ \alpha=45°
\end{equation}

External normal stresses induce shear stresses in the material interior, which become maximum at an angle of 45°. In those favorably lying slip planes, the atomic layers therefore prefer to shear off!

This is the reason why the copper single crystal shown above has slip steps at an angle of about 45° to the tensile axis. The prerequisite for this is that the lattice structure is spatially oriented in such a way that a slip plane is aligned at an angle of about 45°. The effects of the deformation process when the slip plane is not at 45 ° will be discussed in the next section.

Note: In addition to the optimal orientation of the slip plane, the slip direction must also be oriented accordingly. This optimal orientation has been assumed tacitly. The article Schmid’s law deals with this topic in more detail.

Influence of lattice orientation

What does the deformation of a single crystal look like if its slip planes are not at an angle of 45°? Due to the less favorable position, larger external normal stresses are required in this case, so that the critical resolved shear stress in the slip plane (and in the slip direction!) will be exceeded.

Multiple gliding

In the case of a very unfavorable position of the (primary) slip planes, this can even lead to an activation of other slip planes, which normally only come into account at higher stresses, because in them the critical shear stress is exceeded first. Slipping then probably takes place in several different slip systems.

This will be the case if the different slip planes are relatively symmetrical arranged to the tensile axis. Then the critical shear stress in all slip systems is reached approximately simultaneously. For face-centered cubic monocrystals, the unit cells must be aligned nearly parallel to the tensile axis (see figure after next).

All four slip planes are then activated equally. Due to the fact that the shearing takes place on several different slip planes, one also speaks of multiple gliding. By such multiple gliding, the crystal is then deformed in different directions. The slip steps exit the material at different angles and spatial directions. The superposition of these slip steps makes it impossible to distingish different slip steps from one another. Therefore they are not visible with the naked eye.

Figure: Multiple gliding in a single crystal

Easy gliding

On the other hand, in the image of the copper single crystal obviously shows distinguished uniform slip steps. The reason for this is that the fcc lattice was specially aligned to the tensile axis. The single crystal was so designed that only one slip plane was oriented at an angle of approximately 45° to the tensile axis.

Thus, upon reaching the limit stress, only this slip plane was activated, while the shear stresses in the remaining slip planes (or in the slip direction) remained below the critical resolved shear stress. Since in such a case, the sliding takes place only within one slip plane, this process is also referred to as a single gliding or easy gliding.

The external normal stress that has to be applied \(\sigma_{crit}\) to a material in order to exceed the critical resolved shear stress in a slip plane, is referred to as yield strength or elastic limit.

This characteristic value thus refers to the (tensile) stress at which an irreversible deformation process begins. As explained, the yield strength for single crystals depends on the spatial orientation of the lattice! The lowest value of the yield strength is therefore obtained for single crystals when their slip planes are aligned at an angle of 45 ° to the tensile axis.

The yield strength marks the stress limit from which on a material deforms plastically!

Polycrystals

The annular depiction of the slip steps at an angle of 45° is only evident in specially oriented single crystals. However, single crystals can only be produced with great technical effort. They are therefore limited to special applications such as nickel-based single-crystal turbine blades or silicon chip production.

In general metals do not have a uniform lattice orientation (note: the term lattice orientation must not be confused with the term lattice structure!). Rather, many small microscopic areas (grains) are found in a metal, each with a different lattice orientation. Such crystals are also referred to as polycrystals.

Thus, in such polycrystals several slip systems with different slip plane orientation are always simultaneously activated (multiple gliding). A uniform exit of slip steps is therefore not visible in such materials anyway. How it comes to such a nonuniform lattice orientation is explained in more detail in the chapter microstructure.

In this article, only the processes that lead to the initiation of a deformation were discussed. The actual deformation process in single crystals is therefore discussed in more detail in the article deformation process in single crystals.

Introduction

To initiate a deformation process, a certain critical resolved shear stress (CRSS) is required at the atomic level. As described in the article Fundamentals of Deformation, the theoretical values ​​for an ideal crystal are about a factor of 1000 higher than those measured in reality.

The reason for this discrepancy are certain crystallographic defects, which occur in real metals. These are the dislocations, whereby only the edge dislocation will be discussed in this article.

Role of the dislocations in the deformation process

Due to dislocations, it is no longer necessary to break all bonds between two atomic planes at once in order to shear off a lattice planes. Rather, it is enough to overcome only one binding series at a time. The dislocation line jumps step-by-step from atomic row to atomic row with little effort and finally emerges as a slip step on the surface of the material.

Due to the low-energy migration of the dislocations, the deformation process starts at already much lower critical shear stresses than the theory predicts without considering dislocations! This critical shear stress is referred to as Peierls stress.

Dislocations allow low-energy slippage of atomic blocks, so that deformation processes in real crystals already occur at lower critical shear stresses than in ideal crystals!

The low-energy sliding of the atomic plane by a dislocation can be illustrated by moving a carpet. Moving a large and heavy carpet usually requires a very large force due to the friction between the carpet and the floor. If, however, wrinkles are struck in the carpet and these are then moved through the carpet, then one achieves the same result with less effort. The carpet can move in stages like a caterpillar.

By understanding the atomic mechanism of deformation and the central role of dislocations, specific measures can now be taken to prevent deformation. After all, many materials that are used under high load should not deform so easily. Therefore they should be high-strength. How measures for increasing the strength of metals can look like is explained in the following section.

Strengthening mechanisms

By the term strengthening mechanisms one understands measures which aim to prevent the deformation of a metal purposefully. Thus, the highest possible strength of the corresponding material is achieved. Since the primary cause of a deformation process is the migration of dislocations, ultimately all strengthening mechanisms are based on blocking of this movement.

All measures with the aim of blocking dislocation movement are summerazied under the term strengthening mechanisms!

The most important strengthening mechanisms are described in the following sections:

• solid solution strengthening
• precipitation hardening
• grain boundary strengthening (Hall–Petch strengthening)
• work hardening (strain hardening)

Solid solution strengthening

The principle of solid solution hardening is based on the distortion of the lattice by foreign atoms. These can be either substitutional atoms or interstitial atoms. Due to their blocking of the dislocation movement, the lattice planes can consequently no longer slide off so easily. A deformation of the lattice thus occurs only at significantly higher critical shear stresses, since the lattice distortion must also be overcome. In this way, an increase in the strength of the material is ultimately achieved.

With solid solution hardening, foreign atoms block the dislocation movement!

Precipitation strengthening

Not only single foreign atoms can impede the dislocation movement, but precipitates can also block the movement of dislocations. This is called precipitation hardening or precipitation strengthening.

In a precipitation hardening precipitates block the dislocation movement!

This principle of precipitation strengthening is used in so-called hardenable aluminum alloys. The aluminum alloy is first heated to a relatively high temperature, so that the foreign atoms contained therein can completely dissolve in the aluminum lattice structure. Note that solubility generally decreases with decreasing temperature, so high temperatures are required for complete solubility.

If it is cooled rapidly (called quenching), then the foreign atoms, despite the lower solubility, remain forcibly dissolved in the lattice. Since the concentration of dissolved atoms in this state is above the actual solubility limit, one speaks of a so-called supersaturated solid solution.

This state is not thermodynamically stable, so that the forcibly solved foreign atoms begin to segregate from the lattice and form their own compounds (precipitates) within the metal. To accelerate this process of so-called aging, the alloy is heated slightly, so that the diffusion processes can proceed more quickly.

Grain boundary hardening (grain refining)

The principle of grain boundary hardening is based on the complicated dislocation movement across grain boundaries. Grain boundaries are therefore no weak points in the material in this context but contribute to a particular extent to the increase in strength!

A high number of grain boundaries can be achieved by fine grains in the material. Therefore, grain boundary strengthening is also called grain refinement (sometimes referred to as Hall–Petch strengthening). A small grain size can be achieved by targeted influencing of the melt during cooling (for example by seeding or supercooling of the melt).

With a grain refining, grain boundaries block the dislocation movement!

The principle of grain refining is applied to so-called weldable fine grain steels in steel building. Although carbon has a strength-increasing effect in steel, it is undesirable in terms of good weldability. Carbon makes the steel hard and brittle due to the rapid cooling after welding. Therefore, due to a good weldability (as is often required in steel building), it is necessary to keep the carbon content as low as possible. Nevertheless, in order to ensure a high strength, one is dependent on grain refining. The diameters of the individual grains in fine grain steels are in the range of approximately 20 μm.

Work hardening (strain strengthening)

The principle of work hardening is based on the introduction of additional dislocations during plastic deformation. In every deformation process, new dislocations are always introduced into the material. The dislocations thus hinder each other from moving, which results in a strength-increasing effect.

With work hardening additionally introduced dislocations block each other from moving!

The increase in strength during plastic deformation can also be comprehended using the stress-strain diagram. For this purpose, a tensile specimen is first stretched out beyond the yield strength \(R_p\) (red line to point A) and therefore suffers a plastic deformation.

If the force is subsequently removed, the operating point returns to zero stress in a direction parallel to Hooke’s straight line (point B). Accordingly, the material has a permanent elongation.

If the tensile test is repeated again, the operating point first goes up the elastic straight line again and only reaches plastic deformation at higher stress ​​(blue curve)!

The comparison of both stress–strain curves makes it clear that the plastic deformation now only occurs at higher stress. This means that the work-hardened sample obviously has an increased yield strength \(R_p\)! The tensile strength also increases accordingly. Due to the strain a workpiece has suffered during plastic deformation, work hardening is also referred to as strain strengthening.

Work hardening is specifically brought about, for example, in the manufacture of cold-rolled sheets in order to achieve a significantly higher strength compared to the hot-rolled condition.

Note that work hardening can not be done to any degree. If too many dislocations are introduced by plastic deformation, the material is thereby locally destroyed and ruptured.

This behavior is evident, for example, in the repeated bending back and forth of a wire. This only works well until too many dislocations have been introduced and the wire eventually breaks.

Slip system

As explained in the article on Fundamentals of Deformation, a plastic deformation processes in metals can be attributed to slippage of atomic blocks on lattice planes. A lattice plane is an arbitrarily imaginary plane which is regularly occupied with atoms. Not only those planes are considered as lattice planes with which the structure of the different lattice types was explained in the chapter Important types of lattice structures!

A lattice plane refers to a plane in a lattice structure that is regularly covered with atoms!

The figure above shows schematically possible lattice planes in a crystal. At this point, it should be expressly mentioned that the “bars” represent no bonds in the imaging of unit cells. Just because two atoms are connected by a bar does not necessarily mean that there is a particularly strong bond between these atoms! Thus, for example, in cubic lattice types not only the outer surfaces of the cubes form lattice planes, but also arbitrarily inclined planes. One should imagine the lattice structure just without the “bars” when it comes to lattice planes.

If a shearing of entire atomic layers can take place on such a lattice plane, it is referred to as slip plane.

Slip planes are lattice planes on which entire atomic layers can slide!

Although there are  countless lattice planes in a lattice structure, not all are suitable as slip planes. The reason for this is the electrostatic forces acting between the atoms, which bind the atomic planes more or less strongly to each other. In a lattice structure, those lattice planes preferably serve as slip planes, which have the greatest possible distance from each other (“low attraction forces between the planes”). In addition, the lattice planes should contain as many atoms as possible (“gentle sliding of the planes”).

The displaceability of a slip plane is basically not given in all directions for energetic reasons. Thus, a slip plane is very difficult to move in the direction in which the stacking sequence would change. For then, for example, a face-centered cubic lattice would create a hexagonal one. But finally there are energetic reasons why a lattice crystallizes in a certain lattice type.

For a good ductility, therefore, a metal must, in addition to a large number of different slip planes, also have the highest possible number of so-called  slip directions. It is this combination of slip plane number and number of slip directions that determines the overall slip options and thus the deformability. The combination of slip plane and slip direction is referred to as a slip system.

The number of slip systems describes the sliding possibilities of a lattice structure and is therefore a measure of their ductility.

Note that good deformability does not necessarily mean that a lattice structure must already be deformed by small forces. Good ductility rather means, that the lattice structure should deform without damage (“cracking”). Therefore a lattice structure should have as many slip systems as possible.

Metals and their lattice structures

The different lattice structures (fcc, bcc, hcp, etc.) each have different numbers of slip systems both in quantity and in quality. This is primarily the reason why metals can be deformed to different degrees depending on the type of lattice.

Metals such as magnesium, cobalt, zinc or titanium can hardly be plastically deformed under normal conditions. They would rather crack. All these metals have in common that they are present in a hexagonal closest packed lattice structure (hcp-lattice). This type of lattice obviously offers only a few slip systems.

By contrast, metals such as aluminum, lead, copper and nickel have very good deformability. This is obviously due to their face-centered cubic structure (fcc-lattice), which offers many slip systems.

The ductility of the body-centered cubic lattice (bcc-lattice) lies between the lattice types mentioned above. Typical representatives of this structure are metals such as iron, chromium, molybdenum and vanadium.

The ductility steadily decreases starting from the fcc lattice via the bcc lattice all the way to the hcp-lattice!

Due to the good ductility of the face-centered cubic lattice structure of copper and aluminum (or alloys thereof), these metals are often used as so-called wrought metals. On the other hand, the metals magnesium and zinc are mainly used as cast metals due to their hardly deformable lattice.

Body-centered cubic structure

The body-centered cubic lattice (bcc) has mainly 6 slip planes on which the lattice blocks preferably slide off. This corresponds to the levels which in addition to two atoms on a cube edge also include the body-centered atom as well as the diagonally opposite atoms.

There are 2 possible slip directions for each slip plane. This corresponds to the area diagonals of the defined plane (marked in yellow). Other directions of sliding are out of the question due to the strong forces of attraction. The bcc-lattice thus has mainly 12 sliding possibilities (= 6 slip planes x 2 slip directions), how the atomic blocks can shear one another.

The body-centered cubic lattice structure has mainly 12 slip systems!

Figure: Slip planes and directions in the bcc-lattice

Not only the diagonal planes but also other atomic planes can serve as slip planes if greater forces are applied. However, since these slip planes are activated only with higher forces due to the larger electrostatic attraction, they do not decisively determine the deformability for small forces. So it remains at the 12 mainly operated slip planes in the body-centered cubic lattice. If all sliding options are taken into account, then one would come to a total of 48 slip systems.

At high forces, up to 48 slip systems can be activated in the body-centered cubic lattice!

Face-centered cubic structure

In contrast to the bcc-lattice, a total of 4 four slip planes can be found in the face-centered cubic lattice (fcc), on which atomic blocks preferably slide off. These are the closest packed layers that run diagonally through the unit cell. There are 3 slip directions for each of these slip planes.

Note that shearing only occurs in the direction in which the stacking sequence (ABCABC …) is maintained. Because by changing the stacking sequence (e.g. in ABAB …), for example, a fcc-lattice would become partially a hexagonal closest packed structure. However, this condition is hardly realizable due to the high forces required. After all, there is an (energetic) reason why the lattice structure is face-centered cubic and not hexagonal.

Like the bcc-lattice, the fcc-lattice therefore has a total of 12 slip systems (= 4 slip planes x 3 slip directions).

The face-centered cubic lattice structure has a total of 12 slip systems!

Due to the same number of slip systems, one could prematurely conclude that both lattices (fcc and bcc) are equally deformable. In practice, however, it is found that the fcc-lattice shows a significantly higher ductility.

The reason for this is that the slip planes in the fcc-structure are closest packed compared to the bcc-lattice. The “lifting out” of the atomic blocks from the “hollows” thus requires a significantly lower energy expenditure. Shearing processes therefore take place with less effort.

In addition to the main criterion of the number of slip systems (quantity), the quality of the slip planes also plays a role in ductility.

Due to the closest packed slip planes, the face-centered cubic lattice is more ductile despite the same number of slip systems as the body centered cubic lattice!

Note: Slip planes on which the atoms are closest packed are also called primary slip planes. Such closest packed levels are not present in the bcc-latticed (see article Important types of lattice structures)!

Hexagonal closest-packed lattice structure

In contrast to the fcc-lattice with a total of 4 primary slip planes, the hexagonal closest packed lattice (hcp) has only 1 primary slip plane. It is the hexagonal base of the unit cell. There are 3 slip directions for this slip plane, so that the hcp-lattice has a total of 3 slip systems.

The hexagonal closest lattice structure has only 3 slip systems!

Note that for the determination of the number of slip planes, only non-parallel atomic layers are important, which can thus be activated independently of each other. The atomic levels lying above the base are ultimately identical; they are only shifted against each other. The hcp-lattice thus has only a small ductility compared to the fcc-lattice and the bcc-lattice.

Even in the hcp-lattice, additional slip planes can be activated by greater force. Thus, for example, the outer surfaces of the unit cell can also serve as slip planes. However, this requires very high forces, which is why the deformability of metals with hcp-lattice is very low under normal conditions.

Summary

The most important lattice structures are compared with respect to their ductility in the table below.

	face-centered cubic (fcc)	body-centered cubic (bcc)	hexagonal closest packed (hcp)
lattice structure
number of preferred slip planes	4	6	1
number of slip directions	3	2	3
number of preferred slip systems	12	12	3
packing density	74 % (densest packed)	68 % (loosely packed)	74 % (densest packed)
ductility	high	medium	low
metals	γ-iron (<1392°C), aluminium, β-cobalt (>417°C), lead, copper, nickel	α-iron (<911°C), chrome, molybdenum, β-titanium (>882°C), vanadium, tungsten	α-cobalt(<417°C), α-titanium (<882°C), zinc, magnesium

For good deformability, a lattice structure should have at least 5 slip systems to be activated during a deformation process, so that the workpiece can deform in any direction as desired.

For good deformability, a grid structure should have at least 5 slip systems!

Polymorphism (allotropy)

Some metals can change their lattice structure by external thermodynamic influences such as pressure and temperature, whereby only the influence of temperature is relevant for mechanical engineering.

Although iron has a body-centered cubic lattice at room temperature, it is face-centered cubic above 911 °C. At a temperature of 1392 °C, iron reverts to the body-centered cubic lattice modification, before the melting temperature is reached at 1536 °C and the lattice structure dissolves. This transition from the bcc-lattice, which is only moderately deformable, to the much better deformable fcc-lattice is also the reason why iron or steel is heated during forging.

Titanium, too, possesses such a property of changing the lattice structure. At a temperature of 882 °C titanium passes from the hexagonal structure into the body-centered cubic lattice structure.

The property of metals to change their lattice structures, depending on temperature or pressure, is called allotropy or polymorphism.

The cause of allotropy lies in the more favorable energetic conditions at the corresponding temperature, so that the unstable lattice structure “decomposes” into the more stable modification.

The different lattice modification of a polymorphic metal is replaced by Greek letters in front of the element symbol. For example, the hexagonal lattice structure of titanium at room temperature is also known as α-titanium (or \(\alpha\)-Ti). Finally, after the first lattice transformation, \(\beta\)-titanium (or \(\beta\)-Ti). A Further transformation at next higher temperature would be called \(\gamma\) (however, such further lattice conversion does not occur with titanium).

Introduction

The relatively good deformability of metals (also referred to as ductility) compared to other materials is a significant feature. The reason for this lies in the special metallic bond. The good formability is the basis for many manufacturing processes such as bending, deep drawing, forging, etc.

Not every metal can be deformed equally well. The different degrees of ductility can be attributed mainly to the different lattice structures. In order to understand this, basic knowledge about the atomic processes during deformation is necessary.

In principle, a distinction can be made between elastic deformation and plastic deformation.

Elastic deformation

One speaks of an elastic deformation when only a relatively low force ist action on the atoms in the respective material and therefore the atoms are only moved slightly. After removing the force, the atoms regain their initial positions. The deformed workpiece recovers completely back to its original shape after the elastic deformation.

Elastic deformation is a non-permanent deformation. The deformed material returns to its original shape after the force has been removed!

Mechanically loaded components in machines (e.g. cylinder head bolts in engines) should only be subjected to elastic deformations in order not to be permanently deformed.

Plastic deformation

In contrast to the elastic deformation, the applied force during a plastic deformation is relatively large. This leads to a sliding of individual atomic planes. The resulting atomic shifts are retained after removal of the force. The individual atomic planes no longer return to their original positions but have moved on by one or more atomic distances. The workpiece remains permanently deformed after removing the force.

A plastic deformation is a permanent deformation. The deformed material does not return to its original shape after the force has been removed!

In some manufacturing processes (e.g. forging, bending or deep drawing) such a plastic deformation is desired by means of which the corresponding components permanently obtain their desired shape.

Note that with every plastic deformation, the material is always elastically deformed to a certain extent (see animation above). Thus, the material springs back a little after removing the force, even at a plastic deformation. This is also called springback.

Springback refers to the elastic portion which a deformed material recovers when the force is removed!

Such springback must be taken into account, for example, during bending. This makes it necessary to bend the component beyond the desired bending angle in order to compensate the springback.

Slip system

The atomic planes at which the atomic blocks shear during the plastic deformation are also called slip planes. After the atomic blocks have emerged from the material by one or more atomic distances, they are visible under a microscope as slip steps.

Since the reflection behavior changes with the formation of the slip steps, this manifests in a matting of the surface. This is the reason why the bending point of polished pipes often appears dull.

Note that ultimately any plastic deformation process, regardless of the type of stress (whether tensile, compressive, bending, torsional or shear stress) can be attributed to gliding of atomic blocks. Due to the strong electrostatic forces between the individual atoms and the associated stability, however, the shape of the unit cell does not change (permanently) during the deformation processes!

The cause of plastic deformation is the shearing off of atomic blocks on slip planes!

A metal is well deformable, if there are many slip planes with as many different directions of sliding as possible. This means that a deformation process can take place in many directions at the same time, without rupturing the atomic structure irreparably. The combination of slip plane and slip direction is also referred to as a slip system. For high ductility a lattice structure should therefore have as many slip systems as possible.

A slip system is a combination of a slip plane and a slip direction. The more slip systems a lattice structure has, the more deformable is the respective metal.

The different types of lattice structures, such as face-centered cubic, body-centered cubic and hexagonal closest packed, each have different numbers of slip systems. This is primarily the cause of the different deformability of the lattice structures or the corresponding metals.

Tension

As discussed in the previous section, metal deformation processes are based on slipping of atomic planes. This is only possible if a force acts in a proper way. A mere “squeezing” of the atomic structure would only cause the atomic blocks to compress (normal strain).

Slipping will only happen if the force acts in such a way that a lateral “shift” of the atomic structure occurs (shear strain). It is therefore useful to divide forces according to their direction of action on surfaces. Forces acting perpendicular to surfaces or a cross section are referred to as normal forces (normal forces can in principle be further divided into tensile forces and compressive forces). Forces acting parallel to a surface or a cross section are called shear forces.

Only shear forces which are directed parallel to atomic planes (shear stresses) lead to a slipping of lattice planes and thus initiate a deformation process!

Whether a force is capable of causing an atomic layer to slip does not depend solely on the force alone. In addition, of course, it is still crucial how big the atomic plane is that should be sheared off. Because the larger the surface of the atomic layer, the more “bonding points” between two atomic levels arise and must be broken up to slide off. The force per bond or the force per area is of importance!

Such area-related forces are then also referred to as stresses. For normal forces, these stresses are therefore called normal stresses. In case of shear forces the stresses are called shear stresses. The distinction between these stresses also becomes clear in the symbolism. Normal stresses are denoted by the Greek letter σ (sigma), shear stresses are given the Greek letter τ (tau):

\begin{equation}
\label{spannung}
\text{normal stress: }\boxed{\sigma=\frac{F_{\perp}}{A}} \;\;\;\;\;\; \text{shear stress: }\boxed{\tau=\frac{F_{\parallel}}{A}}
\end{equation}

Normal stress acts on a cross section and shear stress in a cross section!

However, the fact that only shear stresses lead to a slipping of atomic planes does not mean that normal stresses acting on a material would not lead to deformation! The animation below shows that the externally applied normal stress (compressive stress) causes shear stress inside the material and atomic blocks to shear off.

By a resolution of forces, that can quickly be comprehended. The force, which has been applied from the outside, is broken down into a vertical and a parallel component regarding the slip plane. Although only normal stresses are applied from the outside, shear stresses are induced in the slip plane.

Normal stresses that are applied from the outside of a material induce shear stresses inside the material!

One must therefore always distinguish: While on a macroscopic level shear stresses as well as normal stresses can lead to deformations, the deformation process on a microscopic level can always be attributed to shear stresses.

To initiate a deformation process, a critical resolved shear stress (CRSS) must be exceeded in a slip plane (and in particular in slip direction) in order to shear off the lattice plane. Due to the binding forces between the atoms, one can make theoretical predictions what critical shear stress is necessary. For metals, the theoretical CRSS ​​are in the range of 1000 to 3000 N/mm² (1 to 3 GPa). Theoretically, therefore, a force of 1000 to 3000 newton per square millimeter must act in a slip plane in order to shear it off.

However, in reality only a fraction of this theoretical shear stress is needed to actually deform a material! The experimental values ​​are in a single-digit range between 1 and 30 N/mm²! In practice, the deformation already starts at much lower shear stress ​​than calculated theoretically. The article on deformation process in real crystal structures deals with this phenomenon more closely.

Note: The word “resolved” in the term “critical resolved shear stress” means that the force which acts in the slip plane must be resolved in slip direction! The critical resolved shear stress is then calculated by this force! Only this dissolved stress, is decisive for the deformation process, since the CRSS is not only acting in the slip plane but also in slip direction! For more details see Schmid’s law.