Chapter 3

Constitutive Models $–$ Relations between Stress and Strain

3.12 Constitutive models for metal single crystals

Plastic flow in a single crystal is anisotropic, and so cannot be modeled using the simple constitutive equations described in Section 3.6.  Instead, a more complicated constitutive law is used, which considers the slip activity in the crystal directly. The main application of the constitutive equation is to model the rotations of individual grains in a polycrystal, and hence to predict the evolution of texture, and to account for the effects of texture on the development of anisotropy in the solid.

3.12.1 Review of some important concepts from crystallography

Common crystal structures

Most metal crystals of practical interest have either face centered cubic, body centered cubic, or hexagonal crystal structures.  These are illustrated in the picture below.  Crystal plasticity models exist for all three crystal structures, but fcc materials are the most extensively studied.

Miller index notation for crystallographic planes and directions

Planes and directions in a single crystal are referred to as follows.   For a cubic crystal, we choose basis vectors $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ perpendicular to the faces of the basic cubic unit cell, as illustrated in the picture.  Then

1.      The symbol [l,m,n], where l,m,n are positive integers, denotes a direction parallel to a unit vector with components $\left(l{e}_{1}+m{e}_{2}+n{e}_{3}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}}$

2.      The symbol $\left[l,\overline{m},n\right]$ denotes a direction parallel to the unit vector with components $\left(l{e}_{1}-m{e}_{2}+n{e}_{3}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}}$

3.      The symbol <l,m,n> denotes the family of [l,m,n] directions that are identical due to the symmetry of the crystal.  For example $<111>$ in a cubic crystal includes all of

$\left[111\right],\left[\overline{1}11\right],\left[1\overline{1}1\right],\left[11\overline{1}\right],\left[\overline{1}\overline{1}1\right],\left[\overline{1}1\overline{1}\right],\left[1\overline{1}\overline{1}\right],\left[\overline{1}\overline{1}\overline{1}\right]$.

4.      The symbol $\left(l,m,n\right)$ denotes a plane that is perpendicular to a unit vector with components $\left(l{e}_{1}+m{e}_{2}+n{e}_{3}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}}$

5.      The symbol $\left(l,\overline{m},n\right)$ denotes a plane that is perpendicular to a unit vector with components $\left(l{e}_{1}-m{e}_{2}+n{e}_{3}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}}$

6.      The symbol $\left\{l,m,n\right\}$ denotes the family of (l,m,n) planes that are crystallographically identical by symmetry.

For a hexagonal crystal, planes and directions are defined by introducing four auxiliary unit vectors ${m}_{i}$, i=1..4 as shown in the picture above.  The first three unit vectors lie in the basal plane and are oriented parallel to the three shortest distances between neighboring atoms.  The fourth vector is perpendicular to the basal plane.  These vectors are related to a Cartesian basis $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ by

${m}_{1}={e}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{2}=\left(-{e}_{1}+\sqrt{3}{e}_{2}\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{3}=\left(-{e}_{1}-\sqrt{3}{e}_{2}\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{4}={e}_{3}$

Then

1.      The symbol [l,m,n,p], where l,m,n,p are positive integers, denotes a direction parallel to a unit vector  $\left(l{m}_{1}+m{m}_{2}+n{m}_{3}+p{m}_{4}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}-lm-ln-mn+{p}^{2}}$

2.      A bar over one of the indices $\left[\overline{l},m,n,p\right]$ changes the sign of the index, exactly as for cubic crystals;

3.      The symbol <l,m,n,p> denotes the family of [l,m,n,p] directions that are crystallographically identical by symmetry.

4.      The symbol $\left(l,m,n,p\right)$ denotes a plane that is perpendicular to a unit vector with components $\left(l{m}_{1}+m{m}_{2}+n{m}_{3}+p{m}_{4}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}-lm-ln-mn+{p}^{2}}$

5.      The symbol $\left(l,\overline{m},n,p\right)$ denotes a plane that is perpendicular to a unit vector with components $\left(l{m}_{1}-m{m}_{2}+n{m}_{3}+p{m}_{4}\right)/\sqrt{{l}^{2}+{m}^{2}+{n}^{2}-lm-ln-mn+{p}^{2}}$

6.      The symbol $\left\{l,m,n,p\right\}$ denotes the family of (l,m,n,p) planes that are crystallographically identical by symmetry.

Representing crystallographic directions and orientations using stereographic projections

Stereographic projection is a way to represent 3D orientations on a 2D plane.  The figure shows one way to interpret the projection:

1.      The specimen is placed in the center of an imaginary sphere with unit radius, with some convenient material directions aligned with the $\left\{i,j,k\right\}$ basis.

2.      A direction of interest is represented by a unit vector $m=xi+yj+zk$, which intersects the sphere at some point P on its surface.

3.      A line is then drawn from P to the point where the k axis intersects the sphere at Q.

4.      The line PQ cuts through the equatorial plane of the sphere at some point R

5.      The vector $\stackrel{\to }{OR}\equiv \rho$ is the stereographic projection of the orientation m.

6.      The general conversion between the 2D projection and the 3D unit vector is easily shown to be

$\rho =\frac{m-\left(m\cdot k\right)k}{1+m\cdot k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m=\frac{2\rho +k\left(1-\rho \cdot \rho \right)}{1+\rho \cdot \rho }$

7.      The symmetry of the crystal makes the $±m$ directions equivalent.  For this reason, projections usually only show vectors with a positive k component (the projection of a vector with a negative k component lies outside the sphere)

Pole figures and inverse pole figures

In crystal plasticity, the projection is used in two ways. In one approach, specific crystallographic directions are chosen to be aligned with the $\left\{i,j,k\right\}$ directions, and other directions of interest (which could be other crystallographic directions, or the direction of the loading axis in a tensile test, for example) are projected. This is known as an inverse pole figure.

For example, the figure shows the standard projection for a cubic crystal. To interpret the figure, note that

1.      The [100],[010] and [001] directions are parallel to i, j, k, respectively. HEALTH WARNING: this is not the only choice of crystal orientation $–$ you will often see the inverse pole figure with [100] parallel to i, for example.

2.      The points mark the projections of the specified crystallographic directions.

3.      The lines mark the traces of the planes specified: that is to say, the projection of the line where the plane intersects the surface of the unit sphere.

4.      The figure also shows the standard triangle for a cubic crystal.  Notice that the traces of the planes divide the plane into a set of 24 curvilinear triangles, each of which has  <100>, <110> and <111> directions at its corners.  These triangles are indistinguishable because of the symmetry of the crystal $–$ you could interchange any two triangles by applying an appropriate rigid rotation to the crystal, without influencing the arrangement of atoms.   This has important consequences: for example, when testing the response of a cubic crystal to uniaxial tension, you only need to run tests with the loading direction inside the standard triangle.  For this reason, inverse pole figures often only display the standard triangle.

A second application of the stereographic projection in crystal plasticity is to display pole figures.  In this approach, specific physical directions are chosen to be aligned with the $\left\{i,j,k\right\}$ basis, and the orientations of crystallographic directions of interest are displayed on the projection.

For example, when a pole figure is plotted for a rolled sheet specimen, the rolling direction (denoted RD) is often chosen to be parallel to j, the direction transverse to the rolling direction in the plane of the sheet (denoted TD) is chosen to be parallel to i, and the direction perpendicular to the sheet (denoted ND) is chosen to be parallel to k.  The sheet generally contains many grains, and each grain is a single crystal.  The orientations the grains are displayed on a pole figure by choosing some convenient crystallographic direction (<100> or <111> are common) and plotting the stereographic projection for each member of this family of crystallographic directions.  In a cubic crystal, each grain contributes four points to the projection (there are 8 <100> directions but only 4 of them have positive k component).  A typical <100> pole figure for a rolled aluminum sheet after a 40% reduction is shown in the picture.   The pole figure shows that grain orientations tend to cluster together, indicating that the sheet has developed a texture.

3.12.2 Features of plastic flow in single crystals

Plastic flow in a crystal is most often measured by conducting a tensile test with the loading axis parallel to a chosen crystallographic direction.  The main results of these experiments are:

1.      For most orientations of the loading axis, the plastic flow initially consists of shearing parallel to one member of a family crystallographic planes in the crystal, in the direction of a vector s lying in that plane, as illustrated in the picture above.  The crystallographic plane on which shear occurs is called a slip plane.  The shearing direction is known as the slip Direction.  Slip planes and directions for common crystals are listed in the table.

 Structure Slip Direction Slip plane fcc <110> {111} bcc <111> {110},{112} hcp $\begin{array}{l}<11\overline{2}0>\\ <11\overline{2}3>\end{array}$ $\begin{array}{l}\left(0001\right),\left\{1\overline{1}11\right\}\\ \left\{1\overline{1}01\right\},\left\{1\overline{1}0\overline{1}\right\}\end{array}$

2.      Crystals contain a large number of candidate slip systems.  For example, an fcc crystal contains 12 possible slip systems.  These are conventionally designated as listed in the table below.

 Slip plane Slip direction (111) $\left[0\overline{1}1\right]$ ${a}_{1}$ $\left[10\overline{1}\right]$ ${a}_{2}$ $\left[\overline{1}10\right]$ ${a}_{3}$ $\left(\overline{1}\overline{1}1\right)$ $\left[011\right]$ ${b}_{1}$ $\left[\overline{1}0\overline{1}\right]$ ${b}_{2}$ $\left[1\overline{1}0\right]$ ${b}_{3}$ $\left(\overline{1}11\right)$ $\left[0\overline{1}1\right]$ ${c}_{1}$ $\left[\overline{1}0\overline{1}\right]$ ${c}_{2}$ $\left[110\right]$ ${c}_{3}$ $\left(1\overline{1}1\right)$ $\left[011\right]$ ${d}_{1}$ $\left[10\overline{1}\right]$ ${d}_{2}$ $\left[\overline{1}\overline{1}0\right]$ ${d}_{3}$

3.      The slip systems in the undeformed crystal are identified by unit vectors ${m}^{\alpha }$ normal to the slip plane, together with unit vectors ${s}^{\alpha }$ parallel to the slip direction. Here $\alpha =1...N$ and N denotes the total number of slip systems (eg N=12 for an fcc crystal).  The crystal can rotate during deformation.  In the deformed solid the slip plane normals and slip directions are denoted ${m}^{*\alpha }$, ${s}^{*\alpha }$.

4.      In a tensile test on an annealed fcc single crystal, shearing occurs on the slip system that is subjected to the largest resolved shear stress.  The resolved shear stress on the $\alpha$ th system can be computed from the Cauchy stress ${\sigma }_{ij}$ acting on the solid as

${\tau }^{\alpha }=J{\sigma }_{ij}{m}_{i}^{*\alpha }{s}_{j}^{*\alpha }$

where $J=dV/d{V}_{0}$ is the ratio of deformed to undeformed volume of the specimen ( $J\approx 1$ ). For example, the inverse pole figure shows the active slip system for all possible orientations of the tensile axis with respect to an fcc crystal (a bar over a slip system indicates a negative resolved shear stress).  In materials with other crystal structures, some slip systems may be inherently stronger than others.  In this case slip occurs on the system with highest value of ${\tau }^{\alpha }/{g}^{\alpha }$, where ${g}^{\alpha }$ denotes the strengths of the slip systems

5.      Slip on the critical system initiates when the resolved shear stress exceeds a critical magnitude (the strength of the slip system) ${\tau }^{\alpha }>{g}^{\alpha }$.  The strength of the slip systems increases with plastic straining: this behavior will be discussed in more detail later.

6.      For special orientations of the tensile axis, more than one slip system may be activated.  For example, if an fcc crystal is loaded parallel to a <100> direction, 8 slip systems are subjected to the same resolved shear stress, and so are active at the same time (the inverse pole figure above shows the active systems)

7.      The deformation gradient resulting from a shear strain ${\gamma }^{\alpha }$ on the $\alpha$ th system is

${F}_{ij}={R}_{ik}\left({\delta }_{kj}+{\gamma }^{\alpha }{s}_{k}^{\alpha }{m}_{j}^{\alpha }\right)$

where ${R}_{ij}$ is a proper orthogonal tensor (i.e. det(R)=1, ${R}_{ik}{R}_{jk}={\delta }_{ij}$ ), representing a rigid rotation.  The rotation is determined by the way the solid is loaded.   For example, in a tensile test, ${R}_{ij}$ is often calculated from the condition that the material fiber parallel to the loading axis (specified by a unit vector p) does not rotate during deformation.  This gives

${R}_{ij}={\delta }_{ij}\mathrm{cos}\theta +\left(1-\mathrm{cos}\theta \right){n}_{i}{n}_{j}+\mathrm{sin}\theta {\in }_{ikj}{n}_{k}$

where ${n}_{i}={\in }_{ijk}{s}_{j}{p}_{k}$ $\mathrm{cos}\theta =\left(1+{\gamma }^{\alpha }{p}_{i}{s}_{i}^{\alpha }{p}_{k}{m}_{k}^{\alpha }\right)/C$, $\mathrm{sin}\theta ={\gamma }^{\alpha }\left({p}_{i}{m}_{i}^{\alpha }\right)\sqrt{\left(1-{\left({p}_{i}{s}_{i}^{\alpha }\right)}^{2}\right)}/C$, and $C=\sqrt{1+{\gamma }^{\alpha 2}{\left({p}_{i}{m}_{i}^{\alpha }\right)}^{2}+2{\gamma }^{\alpha }{p}_{i}{s}_{i}^{\alpha }{p}_{k}{m}_{k}^{\alpha }}$Other assumptions are also used to calculate R.

8.      The crystal lattice is rotated by R, so that after deformation ${s}_{i}^{*}={R}_{ij}{s}_{j}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{i}^{*}={R}_{ij}{m}_{j}$ for all the slip systems.

9.      The rate of deformation resulting from a shearing rate ${\stackrel{˙}{\gamma }}^{\alpha }$ on the $\alpha$ th system is

${L}_{ij}={\stackrel{˙}{F}}_{ip}{F}_{pj}^{-1}={\stackrel{˙}{R}}_{ik}{R}_{jk}+{R}_{ik}{\stackrel{˙}{\gamma }}^{\alpha }{s}_{k}^{\alpha }{m}_{p}^{\alpha }{R}_{jp}={\stackrel{˙}{R}}_{ik}{R}_{jk}+{\stackrel{˙}{\gamma }}^{\alpha }{s}_{i}^{*\alpha }{m}_{j}^{*\alpha }$

This can be decomposed into a symmetric part, representing a stretching, together with a skew part, representing a spin, as

${L}_{ij}={D}_{ij}+{W}_{ij}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{W}_{ij}={\stackrel{˙}{R}}_{ik}{R}_{jk}+{\stackrel{˙}{\gamma }}^{\alpha }\left({s}_{i}^{*\alpha }{m}_{j}^{*\alpha }-{s}_{j}^{*\alpha }{m}_{i}^{*\alpha }\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{ij}={\stackrel{˙}{\gamma }}^{\alpha }\left({s}_{i}^{*\alpha }{m}_{j}^{*\alpha }+{s}_{j}^{*\alpha }{m}_{i}^{*\alpha }\right)/2$

Here, the first term in ${W}_{ij}$ represents the rotation of the lattice, while the second term is the spin due to lattice shearing.

10.  In a tensile test oriented for single slip, the crystal rotates so as to align the slip direction with the loading axis.  This rotation is illustrated for an fcc crystal on the inverse pole figure on the right.  Eventually, the crystal rotates far enough to activate a second slip system.  The exact point where this occurs depends on how the crystal hardens; it usually occurs shortly before the loading axis moves out of the standard triangle.  The rotation direction changes after the second slip system becomes active: eventually, the loading axis aligns with the [112] direction. This is a stable orientation, and the crystal continues to deform in double slip without further rotation.

11.  The resistance of each slip plane to shearing increases with plastic strain, due to strain hardening.   A typical stress-strain curve for a single crystal that is initially oriented for single slip is illustrated in the figure. The curve is divided into three characteristic regions. Stage I corresponds to the period while a single slip system is active, and has a low hardening rate (due to self hardening); Stage II begins when a second slip system activates, and has a higher hardening rate (due to both self and latent hardening); while Stage III occurs at large strains, and the hardening rate decreases due to dynamic recovery.  The hardening rates in Stages I and II are insensitive to temperature; but the Stage III hardening rate decreases with temperature.

12.  Shearing on the $\alpha$ th system increases its own strength ${g}^{\alpha }$: this is known as self-hardening.  Shearing on the $\alpha$ th system also increases the strength of all the other slip systems ${g}^{\beta }$, $\beta \ne \alpha$: this is known as latent hardening.  Self-hardening can be measured using single-slip tests.  Latent hardening is often measured by first deforming the material in single slip, then re-loading the specimen to activate a second slip system.   Latent hardening is often quantified by the Latent Hardening Ratio, which specifies the ratio of the strength of the second system to that of the first ${q}^{\alpha \beta }={g}^{\beta }/{g}^{\alpha }$.  The details of the hardening behavior of single crystals are very complex, and at present there is no consensus on how best to measure or characterize hardening.

13.  Lattice rotation during a tensile test gives rise to a phenomenon known as geometric softening,’ which plays an important role in shear localization in single crystals.  The term geometric softening’ refers to the fact that the crystal may rotate so as to increase the resolved shear stress on its active slip system, and therefore lead to a decrease in the tensile flow stress of the crystal.

3.12.3 Kinematic descriptions used in constitutive models of single crystals

Let ${x}_{i}$ be the position of a material particle in the undeformed crystal. Suppose that the solid is subjected to a displacement field ${u}_{i}\left({x}_{k}\right)$, so that the point at ${x}_{i}$ moves to ${y}_{i}={x}_{i}+{u}_{i}$.  Define

The deformation gradient and its jacobian

${F}_{ij}={\delta }_{ij}+\frac{\partial {u}_{i}}{\partial {x}_{j}}$       $J=\mathrm{det}\left(F\right)$

${L}_{ij}=\frac{\partial {\stackrel{˙}{u}}_{i}}{\partial {y}_{j}}={\stackrel{˙}{F}}_{ik}{F}_{kj}^{-1}$

The stretch rate and spin

${D}_{ij}=\left({L}_{ij}+{L}_{ji}\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{W}_{ij}=\left({L}_{ij}-{L}_{ji}\right)/2$

Recall that ${F}_{ij}$ relates infinitesimal material fibers $d{y}_{i}$ and $d{x}_{i}$ in the deformed and undeformed solid, respectively, as

$d{y}_{i}={F}_{ij}d{x}_{j}$

To decompose the deformation gradient into elastic and plastic parts, we assume that deformation takes place in two stages.  The plastic strain is assumed to shear the lattice, without stretching or rotating it.  The elastic deformation rotates and stretches the lattice. We think of these two events occurring in sequence, with the plastic deformation first, and the stretch and rotation second, giving

$d{y}_{i}={F}_{ij}d{x}_{j}={F}_{ik}^{e}{F}_{kj}^{p}d{x}_{j}$

To decompose the velocity gradient into elastic and plastic parts, note that

${L}_{ij}={\stackrel{˙}{F}}_{ik}{F}_{kj}^{-1}=\left({\stackrel{˙}{F}}_{ik}^{e}{F}_{kl}^{p}+{F}_{ik}^{e}{\stackrel{˙}{F}}_{kl}^{p}\right)\left({F}_{lm}^{p-1}{F}_{mj}^{e-1}\right)={\stackrel{˙}{F}}_{ik}^{e}{F}_{kj}^{e-1}+{F}_{ik}^{e}{\stackrel{˙}{F}}_{kl}^{p}{F}_{lm}^{p-1}{F}_{mj}^{e-1}$

Thus the velocity gradient contains two terms, one of which involves only measures of elastic deformation, while the other contains measures of plastic deformation.  We use this to decompose L into elastic and plastic parts

${L}_{ij}={L}_{ij}^{e}+{L}_{ij}^{p}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{L}_{ij}^{e}={\stackrel{˙}{F}}_{ik}^{e}{F}_{kj}^{e-1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{L}_{ij}^{p}={F}_{ik}^{e}{\stackrel{˙}{F}}_{kl}^{p}{F}_{lm}^{p-1}{F}_{mj}^{e-1}$

Plastic flow in the crystal occurs by shearing a set of N slip systems.  The slip systems are characterized by unit vectors parallel to slip directions ${s}_{i}^{\alpha }$ and slip plane normals ${m}_{i}^{\alpha }$ in the undeformed solid.  The rate of shear on the $\alpha$ th system is denoted by ${\stackrel{˙}{\gamma }}^{\alpha }$.  The velocity gradient due to this shearing is

${\stackrel{˙}{F}}_{ik}^{p}{F}_{kj}^{p-1}=\sum _{\alpha =1}^{N}{\stackrel{˙}{\gamma }}^{\alpha }{s}_{i}^{\alpha }{m}_{j}^{\alpha }$

It is convenient to define vectors that describe plastic shearing in the current configuration, as

${s}_{i}^{*\alpha }={F}_{ik}^{e}{s}_{k}^{\alpha }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{i}^{*\alpha }={m}_{k}^{\alpha }{F}_{ki}^{e-1}$

The former can be interpreted as the slip direction in the deformed solid (note that it is not a unit vector, however), while ${m}_{i}^{*\alpha }$ can be interpreted as the slip plane normal in the deformed solid.

The plastic part of the velocity gradient can then be expressed in terms of the shearing rates as

${L}_{ij}^{p}=\sum _{\alpha =1}^{N}{\stackrel{˙}{\gamma }}^{\alpha }{s}_{i}^{*\alpha }{m}_{j}^{*\alpha }$

The elastic and plastic parts of the velocity gradient can be decomposed in to symmetric and skew symmetric parts, representing stretching and spin, respectively as

$\begin{array}{l}{D}_{ij}^{e}=\left({L}_{ij}^{e}+{L}_{ji}^{e}\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{W}_{ij}^{e}=\left({L}_{ij}^{e}-{L}_{ji}^{e}\right)/2\\ {D}_{ij}^{p}=\left({L}_{ij}^{p}+{L}_{ji}^{p}\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{W}_{ij}^{p}=\left({L}_{ij}^{p}-{L}_{ji}^{p}\right)/2\end{array}$

The plastic stretching and spin can be expressed in terms of the lattice shearing as

${D}_{ij}^{p}=\sum _{\alpha =1}^{N}{\stackrel{˙}{\gamma }}^{\alpha }\left({s}_{i}^{*\alpha }{m}_{j}^{*\alpha }+{s}_{j}^{*\alpha }{m}_{i}^{*\alpha }\right)/2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{W}_{ij}^{p}=\sum _{\alpha =1}^{N}{\stackrel{˙}{\gamma }}^{\alpha }\left({s}_{i}^{*\alpha }{m}_{j}^{*\alpha }-{s}_{j}^{*\alpha }{m}_{i}^{*\alpha }\right)/2$

3.12.4 Stress measures used crystal plasticity

Stress measures that appear in descriptions of crystal plasticity are summarized below:

The Cauchy (“true”) stress represents the force per unit deformed area in the solid and is defined by

${n}_{i}{\sigma }_{ij}=\underset{dA\to 0}{Lim}\frac{d{P}_{j}^{\left(n\right)}}{dA}$

Kirchhoff stress  $\text{\hspace{0.17em}}{\tau }_{ij}=J{\sigma }_{ij}$

Material stress for the intermediate configuration  ${\Sigma }_{ij}=J{F}_{ik}^{e-1}{\sigma }_{kl}{F}_{jl}^{e-1}$

Resolved shear stress on a slip system  ${\tau }^{\alpha }=J{m}_{i}^{*\alpha }{\sigma }_{ij}{s}_{j}^{*\alpha }$

Lattice Jaumann rate of Kirchoff stress  $\stackrel{\nabla e}{{\tau }_{ij}}=\frac{d{\tau }_{ij}}{dt}-{W}_{ik}^{e}{\tau }_{kj}+{\tau }_{ik}{W}_{kj}^{e}$

The constitutive equations must specify relationships between these stress measures, and the deformation measures outlined in the preceding section.  In particular, the constitutive equations must relate:

1.      The elastic part of the deformation gradient to stress;

2.      The rate of shearing on each slip system to the resolved shear stress.

3.12.5 Elastic stress-strain relation used in crystal plasticity

The relations between stress and the elastic part of the deformation gradient follow the procedure developed for finite strain plasticity in Section 3.8.3.  Only the final results will be repeated here

1.      Define the Lagrangean elastic strain as ${E}_{ij}^{e}=\left({F}_{ki}^{e}{F}_{kj}^{e}-{\delta }_{ij}\right)/2$

2.      Assume that the material stress is proportional to Lagrange strain, as ${\Sigma }_{ij}={C}_{ijkl}{E}_{kl}^{e}$, where ${C}_{ijkl}$ are the components of the elastic stiffness tensor (as defined and tabulated in Section 3.1), for the material with orientation in the undeformed configuration.

3.      The elastic stress-strain law is often expressed in rate form, as follows

$\stackrel{\nabla e}{{\tau }_{ij}}\approx {C}_{ijkl}^{e}{D}_{kl}^{e}$

where $\stackrel{\nabla e}{{\tau }_{ij}}$ is the Jaumann rate of Kirchhoff stress with respect to axes that rotate with the crystal lattice; ${C}_{ijkl}^{e}={F}_{in}^{e}{F}_{jm}^{e}{C}_{nmpq}{F}_{kp}^{e}{F}_{lq}^{e}$ can be thought of as the components of the elastic compliance tensor for material with orientation in the deformed configuration, and ${D}_{ij}^{e}$ is the elastic stretch rate.

3.12.6 Plastic stress-strain relation used in crystal plasticity

The plastic constitutive equations specify the relationship between the stress on the crystal and slip rates ${\stackrel{˙}{\gamma }}^{\alpha }$ on each slip system.   Here, we outline a simple and widely used approach to doing this, based on the work of Pierce, Asaro and Needleman Acta Metall. 31 1951-1976 (1983).  This model is not the best fit to experimental observations, however $–$ in particular more sophisticated equations are required to accurately describe latent hardening behavior.

Flow Rule

There are many advantages to using a viscoplastic flow rule to predict the slip rates in a single crystal: this avoids having to use an iterative procedure to identify active slip systems, and also helps to stabilize material behavior.  The simplest such flow rule is

${\stackrel{˙}{\gamma }}^{\alpha }={\stackrel{˙}{\gamma }}_{0}sign\left({\tau }^{\alpha }\right){\left(\frac{|{\tau }^{\alpha }|}{{g}^{\alpha }}\right)}^{m}$

where ${\tau }^{\alpha }$ is the resolved shear stress on the slip system, ${g}^{\alpha }$ is its current strength (which evolves with plastic straining), and ${\stackrel{˙}{\gamma }}_{0},m$ are material properties.

Hardening rule

The hardening rule must specify the relationship between the slip system strengths ${g}^{\alpha }$ and the plastic strain.  At time t=0 each slip system has the same initial strength ${g}_{0}$.  Thereafter, the slip systems increase in strength as a result of the plastic shearing according to

${\stackrel{˙}{g}}^{\alpha }=\sum _{\beta =1}^{N}{h}_{\alpha \beta }|{\stackrel{˙}{\gamma }}^{\beta }|$

where ${h}_{\alpha \beta }$ are strain dependent hardening rates.  The hardening rate is approximated as

$\begin{array}{l}{h}_{\alpha \beta }={q}_{\alpha \beta }h\left(\overline{\gamma }\right)\\ h\left(\overline{\gamma }\right)={h}_{s}+\left({h}_{s}-{h}_{0}\right){\text{sech}}^{\text{2}}\left\{\left(\frac{{h}_{0}-{h}_{s}}{{g}_{s}-{g}_{0}}\right)\overline{\gamma }\right\}\end{array}$

where ${h}_{s},{h}_{0},{g}_{s},{q}_{\alpha \beta }$ are material properties, and $\overline{\gamma }$ is the total accumulated slip on all slip systems

$\overline{\gamma }=\underset{0}{\overset{t}{\int }}\sum _{\alpha =1}^{N}|{\stackrel{˙}{\gamma }}^{\alpha }|dt$

The matrix ${q}_{\alpha \beta }$ controls the latent hardening rate: for an fcc crystal, it is usually taken to have the form

where q is a material property. The slip systems for an fcc crystal are listed in Section 3.12.2: for example, slip systems ${a}_{1},{a}_{2},{a}_{3}$ are coplanar, while ${a}_{1},{b}_{1}$ non-coplanar.

3.12.7 Representative values for plastic properties of single crystals

Elastic properties of single crystals are listed in Sections 3.1.15 and 3.1.16. The plastic properties of single crystals are strongly sensitive to the material’s crystal structure and composition.  For accurate predictions you will need to test the actual material you plan to use.  As a rough guide, representative parameters for a copper single crystal (taken from Wu, Neale and Van der Giessen, Int J plasticity, 12, p.1199, 1996) are listed in the table.

 Properties of Cu single crystal ${\stackrel{˙}{\gamma }}_{0}$ $m$ ${g}_{0}$ ${g}_{s}$ ${h}_{0}$ ${h}_{s}$ $q$ 0.001 ${\text{s}}^{-1}$ 10-20 16 MPa 70.4 MPa 132 MPa 8 MPa 1.4