PLASTICITY
Professor Khanh Chau Le
Lehrstuhl für Allgemeine Mechanik
Ruhr-Universität Bochum
Universitätsstr. 150, D-44780 Bochum
Lecture Notes
2
Contents
1 Fundamentals
1.1 Phenomenon of plastic deformation
1.2 Mechanical framework . . . . . . .
1.3 Thermodynamical framework . . .
1.4 Constitutive law . . . . . . . . . . .
1.5 Closed system of equations . . . . .
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2 Elementary theory
2.1 Bending . . . . . . . . . . . . . . . .
2.2 Torsion of a cylinder . . . . . . . . .
2.3 Cylindrical shell under combined load
2.4 Simple metal forming processes . . .
3 Theory of plastic flow
3.1 Governing equations . .
3.2 Torsion of prismatic bars
3.3 Plane strain problems . .
3.4 Plane stress problems . .
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4 Crystal plasticity
4.1 Physical background . . . . . . . . . .
4.2 Continuum dislocation theory . . . . .
4.3 Anti-plane constrained shear . . . . . .
4.4 Plane constrained shear . . . . . . . .
4.5 Single crystals deforming in double slip
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4
CONTENTS
Preliminary remark
In this course we shall restrict ourselves to the deformable solids. In solid
mechanics we distinguish
1. material independent universal relations such as
∙ kinematic relations,
∙ mechanical balance equations, as well as
∙ thermodynamical balance equations
2. from the constitutive relation, which expresses the stress tensor 𝝈 of
a material point in terms of the local strain tensor 𝜺 and the local
temperature 𝑇 :
𝝈 ←→ (𝜺, 𝑇 ).
Let us first classify the form of this constitutive relation.
As you know from the theory of elasticity, elastic materials are characterized by a single-valued scalar function, called free energy density (per unit
volume) and denoted by 𝜙(𝜺, 𝑇 ), such that
𝝈=
∂
𝜙(𝜺, 𝑇 ).
∂𝜺
This is the so-called state equation for thermoelastic solids. The measures
of strain and stress tensors can still be chosen differently for small and finite
deformations.
For inelastic materials this one-to-one relation is no longer valid. The
stress-strain relation depends now on the history of loading and deformation. We can roughly classify the inelastic material behavior according to
the following features
∙ rate-independent phenomena. The material behavior does not depend
on the loading rate. Example: plastic deformation.
5
6
CONTENTS
∙ rate-dependent phenomena. The material behavior depends on the
loading rate. Examples: visco-plastic deformation, creep, relaxation.
In this course we shall study isothermal deformation processes of elastoplastic bodies, where we sometimes even neglect the contribution of the elastic strains as small compared with its plastic counterpart. After a short
discussion about the phenomenon of plastic deformation on the example of
the uniaxial tension or compression test we shall propose thermodynamically consistent constitutive equations for elastoplastic materials under the
condition of small strains. Within the framework of
i) the so-called elementary theory, as well as
ii) the theory of plastic flow
we shall solve some simple problems to show how the elastoplastic deformation of solids can be determined. Finally, we give a short introduction
to the modern crystal plasticity incorporating the continuously distributed
dislocations.
Chapter 1
Fundamentals
1.1
Phenomenon of plastic deformation
Simple tension or compression test
Let us consider first the uniaxial tension test with the subsequent unloading
for two materials: i) pure cooper, and ii) soft-annealed carbon steel. The
corresponding stress-strain curves are shown below in Fig. 1.1,
s
s
B
B
sY
sY A
O
e
p
C
e
e
A
O
e
i)
C
p
ep e
e +e
e
ii)
Figure 1.1: Stress-strain curve
where the strain and stress are defined as follows
𝜀=
Δ𝑙
,
𝑙0
𝜎=
𝐹
,
𝐴0
𝜀˙ < 10−3 /𝑠.
One remark should be made concerning the definition of stress. Since the
deformed cross-section at tension shrinks, the true stress should actually be
defined as 𝐹/𝐴, where 𝐴 is the current cross-section area. However, at small
strains of the order 𝜀 < 1% the error is not so grave.
7
8
CHAPTER 1. FUNDAMENTALS
Looking at the stress-strain curve one can recognize two different types
of material response in the elastic and elasto-plastic regions. In the purely
elastic region (within the line OA) no residual strain is observed: the specimen assume its original length after the load is removed. For most of metals
the stress is proportional to the strain so that the Hooke law is valid. The
purely elastic region ends at point A corresponding to the yield stress 𝜎𝑌 .
Beyond this purely elastic region we observe for cooper i) a “mild” transition
to the elasto-plastic region, while for steel ii) a sharp yield stress marked
by a nearly horizontal segment. If the specimen is loaded beyond this yield
stress, it begins to deform plastically. The specimen shows a residual strain
after unloading. The total strain is additively decomposed into the elastic
and plastic parts
𝜎
𝜀 = 𝜀𝑒 + 𝜀𝑝 = + 𝜀𝑝 .
𝐸
After the unloading the plastic strain remains. In Fig. 1.1 this loading and
unloading processes are shown by the stress-strain curve (marked with arrows) from O through A to B and finally from B back to C.
Determination of the yield stress
It is not always easy to determine in praxis the yield stress from the stressstrain curve. Normally, one measures the elastic modulus, at which a fixed
amount of residual strain occurs (say, 𝜀off = 0.2%). With this modulus of
elasticity 𝐸0 one can determine the yield stress 𝜎𝑌 (see Fig. 1.2).
s
sY
eoff
e
Figure 1.2: Fixing the yield stress
Hardening
In the elasto-plastic region, when the specimen is reloaded again after the
unloading, one observes approximately the same stress-strain curve (only
in the opposite direction), apart from a small hysteresis loop and a rather
milder transition to the elasto-plastic region. This means that the material
behaves elastically up to the point B (see Fig. 1.1), and the plastic deformation begins to increase again when the stress achieves its value 𝜎∗ at point B
1.1. PHENOMENON OF PLASTIC DEFORMATION
9
corresponding to the stress at the end of the previous loading process. This
stress 𝜎∗ can be regarded as the new yield stress 𝜎𝑌 . Since 𝜎𝑌 is higher
than the initial yield stress 𝜎𝑌 0 , one speaks of the hardening behavior. The
following power law, which is phenomenological, can be used to describe the
hardening behavior
(
)𝑛
𝜎𝑌 − 𝜎𝑌 0
𝑝
𝜀 =
, 𝑛 ≥ 1,
𝐻
and inversely
𝜎𝑌 = 𝐻
√
𝑛
𝜀 𝑝 + 𝜎𝑌 0 .
s
1
sY0
H/(1+H/E)
E
1
e
Figure 1.3: Linear hardening
For 𝑛 = 1 we have a linear hardening (see Fig. 1.3)
𝜀𝑝 =
1
(𝜎𝑌 − 𝜎𝑌 0 ).
𝐻
s
sY
e
Figure 1.4: Elastic-ideal-plastic material behavior
If we replace the increasing hardening curve by a horizontal line (see
Fig. 1.4), then the material behavior of this idealized material is called elasticideal-plastic.
10
CHAPTER 1. FUNDAMENTALS
The same can be said about the hardening behavior for the compression
test. One needs just to inverse the signs of 𝜎, 𝜎𝑌 0 as well as 𝜀, 𝜀𝑌 0 . The
corresponding inequalities must also be modified appropriately.
Bauschinger effect
We consider now a loading process, for which the specimen is first loaded
in tension to attain a certain amount of plastic strain, then is unloaded and
immediately loaded in compression. The corresponding stress-strain curve is
shown in Fig. 1.5.
s
sY+
sY0
e
sY-
Figure 1.5: Process with loading in compression
We observe that
𝜎𝑌 + > ∣𝜎𝑌 − ∣.
This phenomenon is called Bauschinger effect.
1.2
Mechanical framework
To keep the presentation as simple as possible let us use cartesian coordinates
to describe deformations of solids.
Kinematics
At the beginning of the process at time 𝑡 = 0 the body occupies the region
ℬ of the three-dimensional Euclidean point space. The position vector of an
arbitrary material point is denoted by x, and its components by 𝑥𝑖 , 𝑖 = 1, 2, 3.
The displacement vector of this material point is denoted by u(x, 𝑡), with
𝑢𝑖 (x, 𝑡) being its components. The deformation gradient is given by
F = grad(x + u) = I + gradu
or, in components
𝐹𝑖𝑗 = 𝛿𝑖𝑗 + 𝑢𝑖,𝑗 ,
11
1.2. MECHANICAL FRAMEWORK
u
B
x
Figure 1.6: Motion of a solid in the Euclidean space
where the comma before an index denotes the partial derivative with respect
to the corresponding co-ordinate. The Green strain tensor is defined as E =
1
(F𝑇 F − I), or in components
2
1
𝐸𝑖𝑗 = [(𝛿𝑖𝑘 + 𝑢𝑖,𝑘 )(𝛿𝑗𝑘 + 𝑢𝑗,𝑘 ) − 𝛿𝑖𝑗 ].
2
Unless otherwise specified we always use the Einstein summation convention:
summation on repeated indices from 1 to 3 is understood. For small displacement gradients the quadratic term 𝑢𝑖,𝑘 𝑢𝑗,𝑘 in this formula can be neglected
giving
1
𝜀𝑖𝑗 = (𝑢𝑖,𝑗 + 𝑢𝑗,𝑖 ).
2
Like the Green strain tensor, this small strain tensor 𝜀𝑖𝑗 is also symmetric.
The trace of the strain tensor 𝜀𝑖𝑗
𝜀11 + 𝜀22 + 𝜀33 = 𝜀𝑖𝑖 = 𝑢𝑖,𝑖
means nothing else but the volume change Δ𝑉 /𝑉0 . To show this we use the
well-known Euler formula
1 + 𝑢1,1
𝑢
𝑢
1,2
1,3
1 + 𝑢2,2
𝑢2,3 .
𝑉 /𝑉0 = det F = 𝑢2,1
𝑢3,1
𝑢3,2
1 + 𝑢3,3 Since the displacement gradients 𝑢𝑖,𝑗 are small compared with 1, we can
neglect the quadratic and cubic terms of 𝑢𝑖,𝑗 in this determinant to obtain
𝑉 /𝑉0 = 1 + 𝑢𝑖,𝑖
⇒
Δ𝑉 /𝑉0 = 𝑢𝑖,𝑖 .
We recall the following preliminary result from linear algebra: any symmetric tensor of second rank A can be brought into the diagonal form by
12
CHAPTER 1. FUNDAMENTALS
some orthogonal transformation of coordinate system. For this purpose one
needs to find all eigenvectors n and the corresponding eigenvalues 𝜆 of A
from the equation
(𝐴𝑖𝑗 − 𝜆𝛿𝑖𝑗 )𝑛𝑗 = 0
This homogeneous equation for the eigenvectors has nontrivial solutions if
and only if its determinant vanishes
det(𝐴𝑖𝑗 − 𝜆𝛿𝑖𝑗 ) = 0.
This is a cubic equation for the eigenvalues which looks in the expanded form
as follows
𝜆3 − 𝐼𝐴 𝜆2 + 𝐼𝐼𝐴 𝜆 − 𝐼𝐼𝐼𝐴 = 0.
The three coefficients of this cubic equation 𝐼𝐴 , 𝐼𝐼𝐴 , 𝐼𝐼𝐼𝐴 are called principal
invariants of the tensor A. The computations give
𝐼𝐴 = 𝐴11 + 𝐴22 + 𝐴33 = 𝐴𝑖𝑖 ,
𝐴11 𝐴12 𝐴11 𝐴13 𝐴22 𝐴23 ,
+
+
𝐼𝐼𝐴 = 𝐴21 𝐴22 𝐴31 𝐴33 𝐴32 𝐴33 𝐼𝐼𝐼𝐴 = det A.
Denoting the eigenvalues of A by 𝜆1 , 𝜆2 , 𝜆3 , we can simply express these
principal invariants as
𝐼 𝐴 = 𝜆1 + 𝜆2 + 𝜆3 ,
𝐼𝐼𝐴 = 𝜆1 𝜆2 + 𝜆1 𝜆3 + 𝜆2 𝜆3 ,
𝐼𝐼𝐼𝐴 = 𝜆1 𝜆2 𝜆3 .
According to the above result, we may diagonalize the strain tensor 𝜀𝑖𝑗
too. Its eigenvalues, called principal strains, will be denoted by 𝜀1 , 𝜀2 , 𝜀3 .
Balance equations
Let 𝜌 be the mass density, 𝜎𝑖𝑗 the Cauchy stress tensor, 𝑏𝑖 the body force.
We formulate the balance of momentum and moment of momentum in the
form
∫
∫
∫
𝜌¨
𝑢𝑖 𝑑𝑥 =
𝜌𝑏𝑖 𝑑𝑥 +
𝜎𝑖𝑗 𝑛𝑖 𝑑𝑎,
(1.1)
𝒰
𝒰
∂𝒰
∫
∫
∫
𝜖𝑖𝑗𝑘 𝑥𝑗 𝜌¨
𝑢𝑘 𝑑𝑥 =
𝜖𝑖𝑗𝑘 𝑥𝑗 𝜌𝑏𝑘 𝑑𝑥 +
𝜖𝑖𝑗𝑘 𝑥𝑗 𝜎𝑘𝑙 𝑛𝑙 𝑑𝑎
𝒰
𝒰
∂𝒰
1.2. MECHANICAL FRAMEWORK
13
for an arbitrary regular volume 𝒰 ⊆ ℬ of the body. Here 𝑑𝑥 = 𝑑𝑥1 𝑑𝑥2 𝑑𝑥3
denotes the volume element, 𝑑𝑎 the area element,
⎧

when at least two indices coincide
⎨0
𝜖𝑖𝑗𝑘 = 1
when 𝑖𝑗𝑘 is an even permutation

⎩
−1 otherwise
is called the permutation symbol, and 𝑢˙ 𝑖 corresponds to the time derivative
of 𝑢𝑖 . The balance of momentum generalize Newton’s second law of the
classical mechanics to continua; together with the balance of moment of
momentum they present the most general laws of mechanics. The surface
integrals in (1.1) can be transformed into the volume integrals in accordance
with Gauss’ formula. Since 𝒰 is arbitrary and since the integrand is assumed
to be continuous, we may derive from (1.1) the balance equations in local
form
𝜌¨
𝑢𝑖 = 𝜌𝑏𝑖 + 𝜎𝑖𝑗,𝑗 ,
𝜎𝑗𝑖 = 𝜎𝑖𝑗 .
(1.2)
Thus, the balance of moment of momentum implies the symmetry of the
stress tensor 𝜎𝑖𝑗 .
In case of equilibrium the displacement vector 𝑢𝑖 does not depend on time
𝑡 so that the inertial term 𝜌¨
𝑢𝑖 vanishes. The equation of motion reduces then
to the equilibrium condition
𝜎𝑖𝑗,𝑗 + 𝜌𝑏𝑖 = 0.
(1.3)
In plasticity we often have a very slow loading process. Therefore the deformation process runs quite slowly, and the acceleration and the corresponding
inertial term turns out to be small compared with other terms. Such processes are called quasi-static, and for them the equilibrium equation (1.3)
presents a good approximation.
Stress tensor
Since the stress tensor 𝜎𝑖𝑗 is symmetric, it can also be diagonalized. The
eigenvalues of this tensor, 𝜎1 , 𝜎2 , 𝜎3 , are called principal stresses. The principal invariants of the stress tensor are
𝐼 𝜎 = 𝜎1 + 𝜎2 + 𝜎3 ,
𝐼𝐼𝜎 = 𝜎1 𝜎2 + 𝜎1 𝜎3 + 𝜎2 𝜎3 ,
𝐼𝐼𝐼𝜎 = 𝜎1 𝜎2 𝜎3 .
14
CHAPTER 1. FUNDAMENTALS
The hydrostatic stress is defined as 𝜎𝑚 = 𝐼𝜎 /3.
The stress deviator is defined as follows
𝑠𝑖𝑗 = 𝜎𝑖𝑗 − 𝜎𝑚 𝛿𝑖𝑗 .
The following invariants of the stress deviator are often used in the plasticity
theory
𝐽1 = 𝑠𝑖𝑖 = 0,
1
1
𝐽2 = 𝑠𝑖𝑗 𝑠𝑖𝑗 = [(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ],
2
6
1
𝐽3 = dets = [(𝜎1 − 𝜎2 )2 (𝜎1 − 𝜎3 + 𝜎2 − 𝜎3 ) + (𝜎2 − 𝜎3 )2
27
(𝜎2 − 𝜎1 + 𝜎3 − 𝜎1 ) + (𝜎3 − 𝜎1 )2 (𝜎3 − 𝜎2 + 𝜎1 − 𝜎2 )].
(1.4)
One can see that the invariants 𝐽2 , 𝐽3 are symmetric functions of 𝜎𝑖 − 𝜎𝑗 .
Figure 1.7: Mohr’s stress circles
The geometric interpretation of 𝐽2 can be given in terms of the octahedral
shear stress. Consider the normal vector
1
n = √ (1, 1, 1)
3
in the principal coordinates of the stress tensor and an area element perpendicular to it which lies on the side of the octahedron. The stress vector
acting on this area element is given by
1
t = 𝝈n = √ (𝜎1 , 𝜎2 , 𝜎3 ).
3
The normal stress equals
1
𝜎oct = n ⋅ t = (𝜎1 + 𝜎2 + 𝜎3 ) = 𝜎𝑚 .
3
1.3. THERMODYNAMICAL FRAMEWORK
15
The shear stress acting on the side of the octahedron is obtained from the
formula
1
1
2
2
𝜏oct
= (𝜎12 + 𝜎22 + 𝜎32 ) − (𝜎1 + 𝜎2 + 𝜎3 )2 = 𝐽2 .
3
9
3
It is interesting to mention that the octahedral shear stress is the average
shear stress over all planes passing through a material point.
With the help of Mohr’s stress circles we can also determine the maximum
shear stress (see Fig. 1.7)
𝜏max =
1.3
1
max{∣𝜎1 − 𝜎2 ∣, ∣𝜎2 − 𝜎3 ∣, ∣𝜎3 − 𝜎1 ∣}.
2
Thermodynamical framework
It is well known from the classical experiment by Taylor and Quinney that
about 90% of the work done to deform metals plastically will be dissipated
into heat. This heat suply leads in general to the change of temperature.
Thus, if the plastic deformations occur, the process we are dealing with
becomes thermo-mechanically coupled. The consequence is that, in plasticity,
thermodynamic balance equations should be taken into account.
Energy balance
We assume that the energy of an arbitrary sub-body 𝒰 is a sum of the kinetic
and internal energies
∫
1
ℰ = 𝜌(𝑒 + 𝑣𝑖 𝑣𝑖 ) 𝑑𝑥,
2
𝒰
with 𝑣𝑖 = 𝑢˙ 𝑖 being the material velocity. Here 𝑒 corresponds to the internal
energy density. The balance of energy states
ℰ̇ = 𝒫 + 𝒬.
where 𝒫 is the power of the external forces, and 𝒬 is the rate at which heat
is supplied to the body. The power 𝒫 of the body and contact forces is given
in the form
∫
∫
𝒫=
𝜌𝑏𝑖 𝑣𝑖 𝑑𝑥 +
𝒰
𝜎𝑖𝑗 𝑛𝑗 𝑣𝑖 𝑑𝑎.
∂𝒰
The heat supply comes from two sources: the body heat supply and the heat
flow across the boundary; its rate is equal to
∫
∫
𝒬 = 𝜌𝑟 𝑑𝑥 − 𝑞𝑖 𝑛𝑖 𝑑𝑎,
𝒰
∂𝒰
16
CHAPTER 1. FUNDAMENTALS
Here 𝑟(x, 𝑡) is the body heat supply per unit mass and unit time, 𝑞𝑖 (x, 𝑡) is
the heat flux vector across the surface 𝑑𝑎 per unit time. The heat flow is
positive if q and n are opposite; therefore the minus sign in the last equation
agrees with our common sense.
Substituting the above formulas for the power and the heat supply in the
right-hand side of the balance of energy and transforming the surface integral
into the volume integral, we get
∫
[𝜌(𝑒˙ + 𝑣𝑖 𝑣˙ 𝑖 − 𝑏𝑖 𝑣𝑖 − 𝑟) − (𝜎𝑖𝑗 𝑣𝑖 ),𝑗 + 𝑞𝑖,𝑖 ] 𝑑𝑥 = 0.
𝒰
Since this equation holds true for an arbitrary regular sub-body 𝒰, and since
the integrand is assumed to be continuous, we obtain the balance of energy
in the local form
𝜌(𝑒˙ + 𝑣𝑖 𝑣˙ 𝑖 − 𝑏𝑖 𝑣𝑖 − 𝑟) − (𝜎𝑖𝑗 𝑣𝑖 ),𝑗 + 𝑞𝑖,𝑖 = 0.
Taking into account the balance of momentum (1.2) we obtain finally
𝜌𝑒˙ + 𝑞𝑖,𝑖 = 𝜎𝑖𝑗 𝜀˙𝑖𝑗 + 𝜌𝑟.
(1.5)
Second law of thermodynamics
In order to formulate the second law of thermodynamics we need two new
quantities. The first one is the absolute temperature, referred to as an intensive quantity and denoted by 𝑇 (x, 𝑡). The second one is the entropy, referred
to as an extensive quantity, whose density is denoted by 𝑠(x, 𝑡). The entropy
of the sub-body 𝒰 is given by
∫
𝜌𝑠 𝑑𝑥.
𝒰
The second law of thermodynamics states that
∫
∫
∫
𝑑
𝜌𝑟
𝑞 𝑖 𝑛𝑖
𝜌𝑠 𝑑𝑥 ≥
𝑑𝑥 −
𝑑𝑎.
𝑑𝑡
𝑇
𝑇
𝒰
𝒰
(1.6)
∂𝒰
When the heat supply and the heat flow are absent (adiabatic process with
𝑟 = 0 and 𝑞𝑖 = 0), the following inequality holds true
𝑑
𝑑𝑡
∫
𝒰
𝜌𝑠 𝑑𝑥 ≥ 0
17
1.4. CONSTITUTIVE LAW
which means that the entropy of the closed system cannot decrease.
With the help of Gauss’ theorem we obtain
∫
∫
𝜌𝑟
𝜌𝑠˙ 𝑑𝑥 ≥ [ − (𝑞𝑖 /𝑇 ),𝑖 ] 𝑑𝑥.
𝑇
𝒰
𝒰
Since 𝒰 is arbitrary, this inequality leads to
𝜌𝑠˙ ≥ 𝜌𝑟/𝑇 − (𝑞𝑖 /𝑇 ),𝑖 = 𝜌𝑟/𝑇 − 𝑞𝑖,𝑖 /𝑇 + 𝑞𝑖 𝑇,𝑖 /𝑇 2 .
(1.7)
We call 𝛾 = 𝜌𝑠˙ − 𝜌𝑟/𝑇 + (𝑞𝑖 /𝑇 ),𝑖 the entropy production rate. The inequality
(1.7) says that 𝛾 ≥ 0.
There is an alternative form of the entropy production inequality often
used in plasticity. We introduce the free energy density
𝜓 = 𝑒 − 𝑇 𝑠.
Provided all other balance equations hold true, then the entropy production
inequality is equivalent to
˙ − 𝜎𝑖𝑗 𝜀˙𝑖𝑗 + 𝑞𝑖 𝑇,𝑖 /𝑇 ≤ 0.
𝜌(𝑠𝑇˙ + 𝜓)
(1.8)
To prove (1.8) we use the definition of 𝜓
𝜓˙ = 𝑒˙ − 𝑇˙ 𝑠 − 𝑇 𝑠˙
⇒
˙
𝑇 𝑠˙ = 𝑒˙ − 𝑠𝑇˙ − 𝜓.
Substitute this into (1.7) and multiply by 𝑇
˙ ≥ 𝜌𝑟 − 𝑞𝑖,𝑖 + 𝑞𝑖 𝑇,𝑖 /𝑇.
𝜌(𝑒˙ − 𝑠𝑇˙ − 𝜓)
Combining this equation with the energy balance equation (1.5), we arrive
at (1.8).
For isothermal processes with 𝑇 = const the inequality (1.8) reduces to
𝜌𝜓˙ − 𝜎𝑖𝑗 𝜀˙𝑖𝑗 ≤ 0.
1.4
(1.9)
Constitutive law
The formulation of the constitutive laws begins always with the specification
of all quantities characterizing the current state of the material element. Such
quantities are called state variables. Besides, one needs to specify all internal
variables which may influence the dissipation and the irreversible behavior
of the material element. The constitutive laws for elastoplastic materials
include:
18
CHAPTER 1. FUNDAMENTALS
∙ Specification of the free energy as function of all state variables. By
this the reversible behavior of the material is fixed.
∙ Evolution law for the internal variables (plastic strains + hardening
parameters)
∙ A law for the heat flux (if the process under consideration is thermomechanically coupled)
Different models of elasto-plastic materials can be proposed. Below we consider some of them.
Elastic-ideal-plastic materials
We restrict ourselves to isothermal processes with 𝑇 = const. For elasticideal-plastic materials we include in the list of variables the following quantities
(1.10)
𝜀𝑒𝑖𝑗 , 𝜀𝑝𝑖𝑗 .
We assume that the elastic strains 𝜀𝑒𝑖𝑗 characterize completely the current
state of the deformed material element. This means that the stress tensor
depends only on 𝜀𝑒𝑖𝑗
𝜎𝑖𝑗 = 𝜎𝑖𝑗 (𝜀𝑒𝑖𝑗 ).
The plastic strains 𝜀𝑝𝑖𝑗 depend on the history of loading and therefore are
not the state variables. They present the internal variables which characterize irreversible behavior of the material element. The total strain tensor is
additively decomposed into the elastic and plastic strain tensors
𝜀𝑖𝑗 = 𝜀𝑒𝑖𝑗 + 𝜀𝑝𝑖𝑗 .
(1.11)
The free energy density assumes the form
𝜓 = 𝜓(𝜀𝑒𝑖𝑗 ),
i. e., it depends only on the state variables. Let us differentiate the free
energy with respect to time 𝑡
∂𝜓
𝜓˙ = 𝑒 𝜀˙𝑒𝑖𝑗 .
∂𝜀𝑖𝑗
We substitute this formula into the dissipation inequality (1.9)
(𝜌
∂𝜓
− 𝜎𝑖𝑗 )𝜀˙𝑒𝑖𝑗 − 𝜎𝑖𝑗 𝜀˙𝑝𝑖𝑗 ≤ 0.
𝑒
∂𝜀𝑖𝑗
(1.12)
19
1.4. CONSTITUTIVE LAW
We first consider processes with 𝜀˙𝑝𝑖𝑗 = 0, i. e. reversible processes. For these
processes the second term in (1.12) vanishes, so that
(𝜌
∂𝜓
− 𝜎𝑖𝑗 )𝜀˙𝑒𝑖𝑗 ≤ 0.
∂𝜀𝑒𝑖𝑗
Since 𝜀˙𝑒𝑖𝑗 can be arbitrary, and since the expression in the brackets does not
depend on 𝜀˙𝑒𝑖𝑗 , it must be identically equal to zero and thus
𝜎𝑖𝑗 = 𝜌
∂𝜓
.
∂𝜀𝑒𝑖𝑗
(1.13)
If the free energy density per unit volume 𝜙 = 𝜌𝜓 is a quadratic form of 𝜀𝑒𝑖𝑗
1
𝜌𝜓 = 𝐶𝑖𝑗𝑘𝑙 𝜀𝑒𝑖𝑗 𝜀𝑒𝑘𝑙 ,
2
then (1.13) yield Hooke’s law
𝜎𝑖𝑗 = 𝐶𝑖𝑗𝑘𝑙 𝜀𝑒𝑘𝑙 .
For isotropic elastic material we have
𝜎𝑖𝑗 = 𝜆𝜀𝑒𝑘𝑘 𝛿𝑖𝑗 + 2𝜇𝜀𝑒𝑖𝑗 .
This equation of state can be decomposed into the volumetric and deviatoric
parts
𝜎𝑘𝑘 = 3𝐾𝜀𝑒𝑘𝑘 ,
𝑠𝑖𝑗 = 2𝜇𝑒𝑒𝑖𝑗 ,
where 𝑒𝑒𝑖𝑗 = 𝜀𝑒𝑖𝑗 − 31 𝜀𝑒𝑘𝑘 𝛿𝑖𝑗 is the strain deviator, and 𝐾 = 𝜆 + 2𝜇/3. In rate
form we have
𝜎˙ 𝑘𝑘 = 3𝐾 𝜀˙𝑒𝑘𝑘 ,
𝑠˙ 𝑖𝑗 = 2𝜇𝑒˙ 𝑒𝑖𝑗 .
(1.14)
With (1.13) we reduce the inequality (1.12) to the following dissipation
inequality
𝜎𝑖𝑗 𝜀˙𝑝𝑖𝑗 ≥ 0.
The left-hand side of this equation is called plastic dissipation. The yield
condition as well as the associate flow rule should satisfy this inequality.
One speaks then of thermodynamically consistent constitutive equations. We
20
CHAPTER 1. FUNDAMENTALS
formulate the yield condition in the stress space: the stress tensor must
always satisfy the condition
𝑓 (𝜎𝑖𝑗 ) ≤ 0
As long as 𝑓 (𝜎𝑖𝑗 ) < 0, no plastic strain occurs. The surface given by
𝑓 (𝜎𝑖𝑗 ) = 0
is called the yield surface, and function 𝑓 (𝜎𝑖𝑗 ) the yield function. The elastic
region is found inside the yield surface. If the stress tensor lies on the yield
surface, then the associate flow rule states that 𝜀˙𝑝𝑖𝑗 is either zero or shows in
the direction of the gradient of the yield function
𝜀˙𝑝𝑖𝑗 = 𝜆
∂𝑓
∂𝜎𝑖𝑗
(1.15)
with
𝜆 = 0 for 𝑓 < 0 or 𝑓 = 0 and 𝑓˙ < 0 (unloading),
𝜆 > 0 for 𝑓 = 𝑓˙ = 0 (loading).
If the yield surface has an edge, the above flow rule can still be applied if we
replace the gradient by the sub-gradient of 𝑓 . An alternative procedure has
been proposed by Koiter: Instead of the product of 𝜆 and the gradient of the
¶f2
¶s
.p
e
f2 =0
¶f1
¶s
f1 =0
Figure 1.8: Yield surface with an edge
yield function we take now the linear combination of 𝑛 products
𝜀˙𝑝𝑖𝑗
=
𝑛
∑
𝛼=1
𝜆𝛼
∂𝑓𝛼
∂𝜎𝑖𝑗
with
𝜆𝛼 = 0 for 𝑓𝛼 < 0 or 𝑓𝛼 = 0 and 𝑓˙𝛼 < 0 (unloading)
𝜆𝛼 > 0 for 𝑓𝛼 = 𝑓˙𝛼 = 0 (loading).
(1.16)
21
1.4. CONSTITUTIVE LAW
The validity of (1.15) or (1.16) follows from the so-called “principle of
maximum of plastic dissipation”, which claims that
(𝜎𝑖𝑗 − 𝜎𝑖𝑗∗ )𝜀˙𝑝𝑖𝑗 ≥ 0
(1.17)
for an arbitrary stress state 𝜎𝑖𝑗∗ within the yield surface.
.p
e
.p
e
s
s
s*
s*
f=0
f=0
Figure 1.9: Convexity of the yield surface and the normality rule
This principle is equivalent to the requirement of convexity of the yield
surface, because in the case of non-convexity one can always find the stress
state 𝜎𝑖𝑗∗ which violate the inequality (1.17). Thus, the principle of maximum
of plastic dissipation implies the convexity of the yield surface as well as the
associate flow rule. One can show that the plastic dissipation 𝐷 = 𝜎𝑖𝑗 𝜀˙𝑝𝑖𝑗 is
a function of the plastic strain rate 𝜀˙𝑝𝑖𝑗 only. When the stress tensor 𝜎𝑖𝑗 is
found inside the yield surface, then 𝜀˙𝑝𝑖𝑗 = 0 and the dissipation vanishes. If
the stress tensor during the loading is found on the yield surface, then the
dissipation must be a homogeneous function of first order with respect to
𝜀˙𝑝𝑖𝑗 . To be consistent with the second law of thermodynamics we require that
𝐷 ≥ 0.
Examples of the yield surface
For isotropic materials the yield function must be a symmetric function of
three principal stresses
𝑓 = 𝑓 (𝜎1 , 𝜎2 , 𝜎3 ).
Since the principal stresses can be expressed in terms of the principal invariants 𝐼𝜎 , 𝐼𝐼𝜎 , 𝐼𝐼𝐼𝜎 , the yield function can also be presented in the following
form
𝑓 = 𝑓1 (𝐼𝜎 , 𝐼𝐼𝜎 , 𝐼𝐼𝐼𝜎 ).
Various observations and experiments show that the hydrostatic stress does
not influence the plastic yielding. This means that the yield function depends
only on the principal invariants 𝐼𝐼𝜎 , 𝐼𝐼𝐼𝜎 of the stress tensor, or alternatively,
on the invariants 𝐽2 , 𝐽3 of the stress deviator
𝑓 = 𝑓2 (𝐽2 , 𝐽3 ).
22
CHAPTER 1. FUNDAMENTALS
Consequently, the yield function must be a symmetric function of 𝜎𝑖 − 𝜎𝑗
𝑓 = 𝑓3 (𝜎1 − 𝜎2 , 𝜎2 − 𝜎3 , 𝜎3 − 𝜎1 )
√
and any parallel translation in the direction (1, 1, 1)/ 3 in the 3-D space of
principal stresses does not change the yield surface.
The criterion of maximum shear stress (Tresca’s yield condition) states
that the plastic flow occurs when the maximum shear stress achieves some
critical value. According to this criterion the yield function must have the
form
1
𝑓 = max{∣𝜎1 − 𝜎2 ∣, ∣𝜎2 − 𝜎3 ∣, ∣𝜎3 − 𝜎1 ∣} − 𝜏𝑌 ,
2
or, the equivalent form
1
𝑓 = (∣𝜎1 − 𝜎2 ∣ + ∣𝜎2 − 𝜎3 ∣ + ∣𝜎3 − 𝜎1 ∣) − 𝜏𝑌 .
4
In this equation 𝜏𝑌 denotes the yield shear stress. For an uniaxial tension
Figure 1.10: Mohr’s stress circle for uniaxial tension
test the plastic flow occurs when (see Fig. 1.10)
𝜎1 = 𝜎𝑌 = 2𝜏𝑌 .
Thus, 𝜏𝑌 = 𝜎𝑌 /2. The projection of the Tresca yield surface onto the octahedron plane (the so-called 𝜋-plane) in 3-D space of principal stresses is a
hexagon (Fig. 1.11).
The normality rule then implies that
1
𝜀˙𝑝1 = 𝜆[sign(𝜎1 − 𝜎2 ) + sign(𝜎1 − 𝜎3 )],
4
where
⎧

𝑥 > 0,
⎨+1
𝑑
∣𝑥∣ = 𝛽, 𝛽 ∈ (−1, 1) 𝑥 = 0,
sign𝑥 =

𝑑𝑥
⎩
−1
𝑥 < 0.
23
1.4. CONSTITUTIVE LAW
Figure 1.11: Projection of Tresca’s and Mises’ yield surfaces
Similar equations hold true for 𝜀˙𝑝2 and 𝜀˙𝑝3 . When all principal stresses are
different and are ordered so that 𝜎1 > 𝜎2 > 𝜎3 , then 𝜀˙𝑝1 = 𝜆/2, 𝜀˙𝑝2 = 0,
𝜀˙𝑝3 = −𝜆/2. When 𝜎1 = 𝜎2 > 𝜎3 , then 𝜀˙𝑝1 = 𝜆(1 + 𝛽)/4, 𝜀˙𝑝2 = 𝜆(1 − 𝛽)/4,
𝜀˙𝑝3 = −𝜆/2, and so on. For each combination of the principal stresses we
always have ∣𝜀˙𝑝1 ∣ + ∣𝜀˙𝑝2 ∣ + ∣𝜀˙𝑝3 ∣ = 𝜆. Therefore the dissipation is equal to
𝐷 = 𝜎𝑖 𝜀˙𝑝𝑖 = 𝜆𝜏𝑌 = 𝜏𝑌 (∣𝜀˙𝑝1 ∣ + ∣𝜀˙𝑝2 ∣ + ∣𝜀˙𝑝3 ∣).
Von Mises proposed another yield function, which, in terms of the stress
invariants, takes the form
√
𝑓 = 𝐽2 − 𝑘
With formula (1.4) this yield function can be written in the form
√
1
[(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ] − 𝑘.
𝑓=
6
Mises’ criterion states that the plastic yielding occurs when the octahedral
shear stress achieves a critical value. An alternative form of the Mises yield
function reads
1
𝑓 = 𝐽2 − 𝑘 2 = [(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ] − 𝑘 2 .
6
For the uniaxial tension test the plastic yielding occurs when
1
𝑓 = 𝜎𝑌2 − 𝑘 2 = 0.
3
√
Thus, 𝑘 = 𝜎𝑌 / 3. The projection of Mises’ yield surface onto√the octahedron
plane in the 3-D space of principal stresses is a circle of radius 2𝑘 (Fig. 1.11).
Using the normality rule we find that
𝜀˙𝑝𝑖𝑗 = 𝜆𝑠𝑖𝑗 .
(1.18)
24
CHAPTER 1. FUNDAMENTALS
Therefore the plastic dissipation for Mises’ yield condition is given by
√
𝑝
𝐷 = 𝜎𝑖𝑗 𝜀˙𝑖𝑗 = 𝑠𝑖𝑗 𝜆𝑠𝑖𝑗 = 𝑘 2𝜀˙𝑝𝑖𝑗 𝜀˙𝑝𝑖𝑗 .
Combining equation (1.18) with Hooke’s law in rate form for a linear elastic
isotropic material (see equation (1.14)) we obtain the rate of the total strain
in the form
1
𝜎˙ 𝑘𝑘 ,
𝜀˙𝑘𝑘 =
3𝐾
1
𝑒˙ 𝑖𝑗 =
𝑠˙ 𝑖𝑗 + 𝜆𝑠𝑖𝑗 .
2𝜇
This equation has been obtained by Prandtl and Reuss.
Models with hardening
We again restrict ourselves to isothermal processes with 𝑇 = const. In order
to describe the hardening behavior one needs to include into the list of variables in (1.10) additional internal variables. Different types of hardening can
be described by introducing scalar variable 𝜅 or tensor variable 𝜶, where 𝜶
is a traceless tensor of second rank. The hardening parameter 𝜅 is defined in
a standard way (Odqvist)
∫ √
2 𝑝 𝑝
𝜅=
𝜀˙ 𝜀˙ 𝑑𝑡.
(1.19)
3 𝑖𝑗 𝑖𝑗
If we√interpret 𝜶 as the coordinates of the middle point of the yield surface
and 2𝑘 as its radius, then various types of hardening can be displayed in
the 𝜋-plane as shown in Fig. 1.12. Choosing the yield function in the form
Figure 1.12: Hardening: i) purely isotropic, ii) purely kinematic iii) combined
𝑓 = 𝑓1 (𝝈 − 𝜶) − 𝑘(𝜅),
25
1.4. CONSTITUTIVE LAW
we can describe both isotropic and kinematic hardening.
For the plastic strain rate the normality rule in the form (1.15) remains
valid. Its validity for the hardening material behavior follows directly from
Drucker’s postulate, which requires that
1. additional stresses produce non-negative work during a loading process
(material stability)
𝑑𝜎𝑖𝑗 𝑑𝜀𝑖𝑗 ≥ 0,
2. for a complete cycle of additional loading and unloading, additional
stresses do positive work if plastic strains occur, and zero when the
strains are purely elastic
∮
(𝜎𝑖𝑗 − 𝜎𝑖𝑗∗ )𝑑𝜀𝑖𝑗 ≥ 0.
Here 𝜎𝑖𝑗∗ is an arbitrary starting point in the stress space. The elastic part
in the last inequality can be removed because it does not contribute to the
work done. Consider a cycle ABCA in the stress space. Since the plastic
strains occur only along the path BC (see Fig. 1.13), this inequality can also
be written in the form (1.17).
C
B
s
A
s*
Figure 1.13: Normality rule and material stability
Drucker’s postulate is an additional requirement, that cannot be derived
from the second law of thermodynamics.
From the definition (1.19) follows that the hardening parameter 𝜅 satisfies
the following equation
√
2 𝑝 𝑝
𝜀˙ 𝜀˙ .
𝜅˙ =
3 𝑖𝑗 𝑖𝑗
For the evolution of the internal variables 𝜶 we propose a very simple equation in the form
𝛼˙ 𝑖𝑗 = 𝑐𝜀˙𝑝𝑖𝑗 .
Then the unknown factor 𝜆 in the flow rule for materials with hardening can
be determined. We consider for example the yield function in the form
1
𝑓 = (𝑠𝑖𝑗 − 𝛼𝑖𝑗 )(𝑠𝑖𝑗 − 𝛼𝑖𝑗 ) − 𝑘 2 (𝜅),
(1.20)
2
26
CHAPTER 1. FUNDAMENTALS
which has been proposed by Melan, Prager, Ziegler and Shield. In addition
to it the consistence condition 𝑓˙ = 0 must be fulfilled for the loading process.
Thus
(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )(𝑠˙ 𝑖𝑗 − 𝛼˙ 𝑖𝑗 ) − 2𝑘𝑘 ′ 𝜅˙ = 0.
With the above evolution equations for the internal variables
√
√
√
2 𝑝 𝑝
2
𝜅˙ =
𝜀˙𝑖𝑗 𝜀˙𝑖𝑗 = 𝜆
(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )(𝑠𝑖𝑗 − 𝛼𝑖𝑗 ) = 2𝜆𝑘/ 3,
3
3
(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )𝛼˙ 𝑖𝑗 = 𝑐𝜆(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )(𝑠𝑖𝑗 − 𝛼𝑖𝑗 ) = 2𝑐𝜆𝑘 2 ,
we can transform the consistence condition to
√
(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )𝑠˙ 𝑖𝑗 − 𝜆(2𝑐𝑘 2 + 4𝑘 ′ 𝑘 2 / 3) = 0.
Therefore
𝜆=
(𝑠𝑖𝑗 − 𝛼𝑖𝑗 )𝑠˙ 𝑖𝑗
√ .
2𝑘 2 (𝑐 + 2𝑘 ′ / 3)
The flow rule becomes finally
𝜀˙𝑝𝑖𝑗 =
1.5
(𝑠𝑘𝑙 − 𝛼𝑘𝑙 )𝑠˙ 𝑘𝑙
√ (𝑠𝑖𝑗 − 𝛼𝑖𝑗 ).
2𝑘 2 (𝑐 + 2𝑘 ′ / 3)
(1.21)
Closed system of equations
Restricting ourselves to the isothermal processes only, we have altogether the
following system of equations
∙ 3 balance equations of momentum (1.3)
∙ 6 stress-strain relations
∙ 1 yield condition
∙ 𝑛 evolution equation for the internal variables.
These 10 + 𝑛 equations contain the following unknown functions
∙ 3 components of displacements (or 3 components of velocity)
∙ 6 components of the stress tensor 𝜎𝑖𝑗
∙ 1 scalar factor
∙ 𝑛 internal variables.
1.5. CLOSED SYSTEM OF EQUATIONS
27
Thus, the system of equations is closed. To solve this system we may develop,
depending on the particular problems, different methods and approaches:
i) elementary theory of elasto-plastic deformation. This approach is characterized by hypotheses which strongly simplify the boundary-value
problems. However, the flow rule is limited to simple loading situation
like uniaxial strain or pure shear.
ii) theory of plastic flow. This approach is based on the simplified material models (elastic-ideal-plastic or rigid-ideal-plastic materials). Apart
from that no further simplifications are made, and the boundary-value
problems will be solved exactly. Due to the mathematical complexity,
analytical solutions may be obtained only in exceptional cases.
iii) general theory of elasto-plastic deformation. This approach is free from
any simplifying assumption. Due to the mathematical complexity, only
numerical solutions of boundary-value problems based on the finite
element method are available.
Note that, in some special cases solutions based only on the equilibrium
equations and on the yield condition can be found without referring to the
flow rule. We call such problems “statically determinate”.
28
CHAPTER 1. FUNDAMENTALS
Chapter 2
Elementary theory
The elementary theory uses various simplifying assumptions concerning the
kinematics and the stress state. Justification of these assumptions cannot be
given in general, but for particular problems.
2.1
Bending
Pure bending of a beam
As the first example let us consider the pure bending of a beam having a
constant rectangular cross-section
𝑀𝑧 = 𝑀𝑥 = 0, 𝑀𝑦 = 𝑀,
𝑄𝑦 = 𝑄𝑧 = 𝑁 = 0.
The chosen coordinate system and the sizes of the beam can be seen in
Fig. 2.1.
Figure 2.1: Straight beam with constant rectangular cross-section
According to the elementary theory we assume that the cross-sections
during bending remain plane and perpendicular to the beam axis, and
𝜎𝑦𝑦 = 𝜎𝑧𝑧 = 𝜎𝑦𝑧 = 𝜎𝑥𝑧 = 0.
29
30
CHAPTER 2. ELEMENTARY THEORY
The first assumption is related to the kinematics of bending, the second to
the stress state. Both coincide with the commonly accepted assumptions of
the beam theory. It follows then from the first assumption
𝜀𝑥𝑦 = 𝜀𝑥𝑧 = 0,
𝜀𝑥𝑥 = 𝜀(𝑧) = 𝜀0 +
𝑧
,
𝑅
where 𝑅 is the radius of curvature of the beam axis. Consequently, we have
along the fibers parallel to the beam axis an uniaxial stress state
𝜎𝑥𝑥 = 𝜎(𝑧).
The remaining unknown quantities 𝜀0 , 𝑅 can be found from the equations
∫
𝑁=
𝜎(𝑧) 𝑑𝑎 = 0,
𝐴
∫
𝑀=
𝜎(𝑧)𝑧 𝑑𝑎
𝐴
in which 𝜎(𝑧) and 𝜀(𝑧) are related to each other by a constitutive law.
s
sY
e
Figure 2.2: Stress-strain curve
For elastic-ideal-plastic materials (at loading) we have
{
𝐸𝜀
∣𝜀∣ < 𝜎𝐸𝑌 ,
𝜎=
±𝜎𝑌 ∣𝜀∣ ≥ 𝜎𝐸𝑌 .
The stress distribution over the thickness is shown in Fig. 2.3.
Let us consider first the purely elastic case
∫
∫
𝑧
𝑁=
𝜎(𝑧) 𝑑𝑎 = 𝐸 (𝜀0 + ) 𝑑𝑎 = 0 ⇒ 𝜀0 = 0,
𝑅
𝐴
∫𝐴
∫
𝐸
1
𝑀
𝐸
𝑀 =𝐸
𝜀(𝑧)𝑧 𝑑𝑎 =
=
.
𝑧 2 𝑑𝑎 = 𝐽𝑧 ⇒
𝑅 𝐴
𝑅
𝑅
𝐸𝐽𝑧
𝐴
31
2.1. BENDING
-sY
zu
el. zone
zo
sY
z
z
ii)
i)
Figure 2.3: Stress distribution: i) elastic, ii) elastic-plastic
Thus, we obtain the well-known formula
𝜀=
𝑀
𝑧
𝐸𝐽𝑧
⇒𝜎=
𝑀
𝑧,
𝐽𝑧
∣𝜎∣max =
6𝑀
.
𝑏ℎ2
The elastic stress distribution is valid until
∣𝜎∣max = 𝜎𝑌
⇒ 𝑀𝑒 = 𝜎 𝑌
𝑏ℎ2
,
6
1
2 𝜎𝑌
.
=
𝑅𝑒
ℎ𝐸
The plastic deformation occurs when 𝑀 ≥ 𝑀𝑒 . At the boundaries between the elastic and the plastic zone 𝑧𝑜 , 𝑧𝑢 the yield conditions hold true
𝐸(𝜀0 +
1
𝑧𝑜,𝑢 ) = ±𝜎𝑌 .
𝑅
Together with two integral equations for the force and the bending moment
there are four equations to determine four unknowns 𝜀0 , 𝑅, 𝑧𝑜 , 𝑧𝑢 . With the
above stress distribution the force equation is simplified to
∫ 𝑧𝑢
∫ ℎ/2
∫ 𝑧𝑜
𝑧
𝑏 𝑑𝑧 + 𝜎𝑌
𝑏 𝑑𝑧 = 0,
𝑁=
𝐸(𝜀0 + )𝑏 𝑑𝑧 − 𝜎𝑌
𝑅
−ℎ/2
𝑧𝑜
𝑧𝑢
∫ 𝑧𝑜
∫
∫ 𝑧𝑢
∫ ℎ/2
𝐸𝑏 𝑧𝑜
𝐸𝑏𝜀0
𝑑𝑧 +
𝑧 𝑑𝑧 − 𝜎𝑌
𝑏 𝑑𝑧 + 𝜎𝑌
𝑏 𝑑𝑧 = 0,
𝑅 𝑧𝑢
𝑧𝑢
−ℎ/2
𝑧𝑜
ℎ
ℎ
𝐸 2
(𝑧𝑜 − 𝑧𝑢2 ) − 𝜎𝑌 (𝑧𝑢 + ) + 𝜎𝑌 ( − 𝑧𝑜 ) = 0,
𝐸𝜀0 (𝑧𝑜 − 𝑧𝑢 ) +
2𝑅
2
2
𝐸 2
𝐸𝜀0 (𝑧𝑜 − 𝑧𝑢 ) +
(𝑧 − 𝑧𝑢2 ) − 𝜎𝑌 (𝑧𝑜 + 𝑧𝑢 ) = 0.
2𝑅 𝑜
The yield conditions at the boundaries 𝑧𝑜 , 𝑧𝑢 imply
𝜎𝑌
− 𝑅𝜀0 ⇒ 𝑧𝑜 + 𝑧𝑢 = −2𝑅𝜀0 ,
𝐸
𝜎𝑌
𝜎𝑌
𝑧𝑜 − 𝑧𝑢 = 2𝑅 , 𝑧𝑜2 − 𝑧𝑢2 = −4𝑅2 𝜀0 ,
𝐸
𝐸
𝑧𝑜,𝑢 = ±𝑅
32
CHAPTER 2. ELEMENTARY THEORY
and therefore
𝜎𝑌
𝐸
𝜎𝑌
−
4𝑅2 𝜀0
+ 𝜎𝑌 2𝑅𝜀0 ⇒ 𝜀0 = 0,
𝐸
2𝑅
𝐸
𝜎𝑌
ℎ𝑅
⇒ 𝑧𝑜,𝑢 = ±𝑅
⇒ 𝑧𝑜,𝑢 = ±
.
𝐸
2 𝑅𝑒
0 = 𝐸𝜀0 2𝑅
We compute now the bending moment
𝐸
𝑀=
𝑅
∫
𝑧𝑜
𝑧𝑢
2
𝑏𝑧 𝑑𝑧 − 𝜎𝑌
∫
𝑧𝑢
𝑏𝑧 𝑑𝑧 + 𝜎𝑌
−ℎ/2
∫
ℎ/2
𝑏𝑧 𝑑𝑧,
𝑧𝑜
𝐸𝑏 3
𝜎𝑌 𝑏 ℎ 2
ℎ2
(𝑧𝑜 − 𝑧𝑢3 ) +
( − 𝑧𝑜2 − 𝑧𝑢2 + ),
𝑅3
2 4
4
3
1
𝜎
2 2 𝜎𝑌3
𝑀 = 𝑏𝑅 2 + 𝜎𝑌 𝑏ℎ2 − 𝑏𝑅2 𝑌2 ,
3
𝐸
4
𝐸
2
2 2
2
ℎ
𝑅 𝜎𝑌
ℎ
𝑅 2 ℎ2 1
𝑀 = 𝜎𝑌 𝑏( −
).
) = 𝜎𝑌 𝑏( −
4
3 𝐸2
4
3 4 𝑅𝑒2
𝑀=
M/Me
2zo/h
1.5
M/Me
1.25
1
0.75
0.5
2zo/h
0.25
1
2
3
4
Re/R
Figure 2.4: Plots of 𝑀/𝑀𝑒 and 2𝑧𝑜 /ℎ
Thus,
𝑀
1 𝑅
3
= [1 − ( )2 ].
𝑀𝑒
2
3 𝑅𝑒
The ultimate moment is achieved when 𝑧𝑜 = 𝑧𝑢 = 0 or, equivalently, when
𝑅 = 0, and is equal to 𝑀𝑢 /𝑀𝑒 = 3/2. The plots of 𝑀/𝑀𝑒 and 2∣𝑧𝑜 ∣/ℎ versus
𝑅/𝑅𝑒 are shown in Fig. 2.4.
Mention that the above formula for the moment is valid only for small
strains, while the ultimate moment is achieved first when 𝑅 = 0, for which
the strains are infinitely large. It must be emphasized, however, that even
33
2.1. BENDING
for relatively small plastic strains the value of the bending moment is close
to its limit. For example, 98% of this ultimate value is already achieved at
𝑅𝑒
= 4.
𝑅
If we are only interested in the limit state of the plastic bending, we may let
the elastic zone disappear completely so that
∫ ℎ/2
∫ 0
𝑏ℎ2
𝑀𝑢 =
𝜎𝑌 𝑏𝑧 𝑑𝑧 −
𝜎𝑌 𝑏𝑧 𝑑𝑧 = 𝜎𝑌
.
4
0
−ℎ/2
The angle of bending 𝛼 takes the value 𝛼 = 𝑙0 /𝑅.
Spring-back, unloading
We consider a loading program for which the bending moment is first increased up to the value 𝑀∗ , where
𝑀𝑒 < 𝑀 ∗ < 𝑀𝑢 .
Now, if we unload the beam by decreasing the moment to zero, its curvature
will decrease from
1
1
to
.
𝑅∗
𝑅𝑝
The difference 1/𝑅∗ − 1/𝑅𝑝 is the elastic spring-back (elastic recovery) of the
beam. 𝑅𝑝 is the residual radius of curvature.
The decrease of the moment from 𝑀∗ to zero is equivalent to the superposition of the solution found above with the elastic solution corresponding
to the moment −𝑀∗ , provided, the beam behaves elastically during the unloading, what we may assume. Thus, for small curvatures we can express the
elastic spring-back as
1
1
𝑀∗
−
=
𝑅∗ 𝑅𝑝
𝐸𝐽𝑧
and, accordingly 𝛼∗ − 𝛼𝑝 =
𝑀∗
𝑙0
𝐸𝐽𝑧
The stress distribution at the end of the loading (corresponding to the moment 𝑀∗ ) is
⎧
𝜎𝑌
𝐸

⎨ 𝑅∗ 𝑧 ∣𝑧∣ ≤ 𝑅∗ 𝐸 ,
𝜎∗ (𝑧) = 𝜎𝑌
𝑧 > 𝑅∗ 𝜎𝐸𝑌 ,

⎩
−𝜎𝑌 𝑧 < −𝑅∗ 𝜎𝐸𝑌 .
We have to superimpose this stress distribution with
𝜎(𝑧) = −
𝑀∗
𝑧,
𝐽
−
ℎ
ℎ
≤𝑧≤ .
2
2
34
CHAPTER 2. ELEMENTARY THEORY
-sY
M*
sY
z
Figure 2.5: Stress distribution after the loading
Figure 2.6: Stress distribution due to −𝑀∗
We have thus after the unloading the eigenstress in the cross section of the
beam
⎧
𝜎𝑌
𝑀∗
𝐸

⎨( 𝑅∗ − 𝐽𝑧 )𝑧 ∣𝑧∣ ≤ 𝑅∗ 𝐸 ,
∗
𝜎
ˆ (𝑧) = 𝜎𝑌 − 𝑀
𝑧
𝑧 > 𝑅∗ 𝜎𝐸𝑌 ,
𝐽𝑧

⎩
∗
𝑧 𝑧 < −𝑅∗ 𝜎𝐸𝑌 .
−𝜎𝑌 − 𝑀
𝐽𝑧
where 𝑅∗ is determined in accordance with
z
Figure 2.7: Eigenstress after the unloading
𝑀∗
1 𝑅∗
3
= [1 − ( )2 ].
𝑀𝑒
2
3 𝑅𝑒
This eigenstress will affect the plastic yielding if we reload the beam in the
opposite direction: the absolute value of the moment at which the plastic
strain changes is lower than that of the first loading. This is the Bauschinger
effect due to the eigenstrain. Consider for example the case when the beam
is bent up to
ℎ
𝑀∗
47
𝑅
=4 ⇒
= , 𝑧𝑜,𝑢∗ = ± .
𝑅𝑒
𝑀𝑒
32
8
35
2.1. BENDING
The elastic spring-back at the unloading is equal to
𝑅𝑒 (
1
1
𝑀∗ 𝑀𝑒
𝑀∗
47
−
)=
=
= .
𝑅∗ 𝑅𝑝
𝑀𝑒 𝐸𝐽𝑧
𝑀𝑒
32
The stress distributions after loading and unloading are shown in Fig. 2.8. At
-sY
15sY/32
-81sY/128
M
*
sY
z
z
i)
ii)
Figure 2.8: Stress distributions after loading and unloading
the subsequent reloading in the opposite direction the beam deform elastically
until
¯
𝑀
17
=− ,
𝑀𝑒
32
so
¯
Δ𝑀
𝑀
𝑀∗
=
−
= 2.0.
𝑀𝑒
𝑀 𝑒 𝑀𝑒
At the subsequent reloading in the same direction the beam behaves elastically for increasing moment until
𝑀∗
47
𝑀
=
= .
𝑀𝑒
𝑀𝑒
32
The shakedown of the beam occurs for Δ𝑀/𝑀𝑒 ≤ 2.0.
-sY
sY
- s Y/ 2
M
*
sY
z
z
i)
ii)
Figure 2.9: Reloading: i) in the same direction, ii) in the opposite direction
36
CHAPTER 2. ELEMENTARY THEORY
Figure 2.10: Hardening and Bauschinger effect at reloading in the same and
in the opposite direction
The previous solution can easily be generalized for materials with linear
or non-linear hardening. One needs just to replace the constant yield stress
in the plastic zone by a function 𝜎𝑌 (𝜀).
For linear hardening we have
⎧

∣𝜀∣ ≤ 𝜎𝐸𝑌 ,
⎨𝐸𝜀
𝐸
𝜎 = 𝐸+𝐻
(𝜎𝑌 + 𝐻𝜀)
𝜀 > 𝜎𝐸𝑌 ,

⎩ 𝐸
(−𝜎𝑌 + 𝐻𝜀) 𝜀 < − 𝜎𝐸𝑌 .
𝐸+𝐻
If the hardening is symmetric with respect to tension and compression, then
𝜀0 = 0,
𝜀=
𝑧
𝑅
⇒ 𝑧𝑜,𝑢 = ±𝑅
𝜎𝑌
.
𝐸
As before, the ultimate moment is achieved when 𝜀 → ∞ and 𝑅 → 0.
These deliberations can further be applied to study:
i) a general case of unsymmetric bending of the beam having rectangular
cross section. One has to assume that
𝑧
𝑦
𝜀(𝑦, 𝑧) = 𝜀0 +
+
,
𝑅𝑧 𝑅𝑦
𝜎𝑦𝑦 = 𝜎𝑧𝑧 = 𝜎𝑦𝑧 = 0.
ii) the combination of tension and bending of the beam having rectangular
cross section.
iii) similar problems for beams with doubly-symmetric cross sections (including thin-walled cross section).
37
2.1. BENDING
iv) the bending of a plate.
Plate bending
For the bending of a plate we assume that
∙ 𝜎𝑧𝑧 = 𝜎𝑧𝑦 = 𝜎𝑧𝑥 = 𝜎𝑥𝑦 = 0,
∙ but instead of 𝜎𝑦𝑦 = 0 now 𝜀𝑦𝑦 = 0 (plane strain state).
In the elastic zone we have
1
(𝜎𝑦𝑦 − 𝜈𝜎𝑥𝑥 ) ⇒ 𝜎𝑦𝑦 = 𝜈𝜎𝑥𝑥 = 𝜈𝜎,
𝐸
1
1 − 𝜈2
= 𝜀 = (𝜎𝑥𝑥 − 𝜈𝜎𝑦𝑦 ) =
𝜎.
𝐸
𝐸
𝜀𝑦𝑦 = 0 =
𝜀𝑥𝑥
Figure 2.11: Plate bending
In the plastic zone we assume the ideal-plastic material behavior with
Mises’ yield condition, so
1
2
1
(𝜎˙ 𝑦𝑦 − 𝜈 𝜎˙ 𝑥𝑥 ) + 𝜆(𝜎𝑦𝑦 − 𝜎𝑥𝑥 ),
𝐸
3
2
2
1
1
= 𝜀˙ = (𝜎˙ 𝑥𝑥 − 𝜈 𝜎˙ 𝑦𝑦 ) + 𝜆(𝜎𝑥𝑥 − 𝜎𝑦𝑦 ),
𝐸
3
2
2
2
2
𝑓 = 𝜎𝑥𝑥 + 𝜎𝑦𝑦 − 𝜎𝑥𝑥 𝜎𝑦𝑦 − 𝜎𝑌 = 0.
𝜀˙𝑦𝑦 = 0 =
𝜀˙𝑥𝑥
We eliminate 𝜆 from the first two equations
𝜀˙ =
1
1
𝜎𝑥𝑥 − 𝜎𝑦𝑦 /2
(𝜎˙ 𝑥𝑥 − 𝜈 𝜎˙ 𝑦𝑦 ) − (𝜎˙ 𝑦𝑦 − 𝜈 𝜎˙ 𝑥𝑥 )
.
𝐸
𝐸
𝜎𝑦𝑦 − 𝜎𝑥𝑥 /2
38
CHAPTER 2. ELEMENTARY THEORY
From the yield condition we derive
√
1
3 2
𝜎𝑦𝑦 − 𝜎𝑥𝑥 = 𝜎𝑌2 − 𝜎𝑥𝑥
⇒
2
4
3
3
3
𝜎
𝜎
𝜎
1
𝜎˙
1
1
2
2
𝜀˙ = [1 − 𝜈( − √
) + ( − √2
)( − √
)]
𝐸
2 2 ...
2 2 ... − 𝜈 2 2 ...
3
3
𝜎 2
𝜎
1
1
𝜎˙
2
2
[1 + ( − √
) − 2𝜈( − √
)]
𝐸
2 2 ...
2 2 ...
1
𝜎˙
1
√
[𝜎𝑌2 (5 − 4𝜈) − 3𝜎(1 − 2𝜈)( ... + 𝜎)].
=
2
2
𝐸 4𝜎𝑌 − 3𝜎
2
=
Therefore the following differential equation holds true
𝜀˙ = 𝜎𝑔(𝜎).
˙
In the special case 𝜈 = 1/2 we have
3
𝜎˙
𝜎˙
3𝜎𝑌2
4
𝜀˙ =
,
=
𝐸 4𝜎𝑌2 − 3𝜎 2
𝐸 1 − 43 𝜎
¯2
where 𝜎
¯ = 𝜎/𝜎𝑌 . Introducing a new variable 𝑢, with
2
𝜎
¯ = √ cos 𝑢,
3
we obtain
𝜀˙ = −
√
3 𝜎𝑌 𝑢˙
.
2 𝐸 sin 𝑢
The integration gives
√
v
u
1−
3 𝜎𝑌 u
ln ⎷
𝜀=−
2 𝐸
1+
√
3
𝜎
¯
2
√
3
𝜎
¯
2
+ 𝑐.
The stress at the boundary of the elastic zone equals
𝜎𝑒 = √
𝜎𝑌
1 − 𝜈 + 𝜈2
2
and for 𝜈 = 1/2 𝜎𝑒 = √ 𝜎𝑌 .
3
The Ansatz for the solution remains as before
𝜀 = 𝜀0 +
𝑧
,
𝑅
39
2.2. TORSION OF A CYLINDER
together with the integral equations for the force and bending moment
∫
∫
𝑛 = 𝜎 𝑑𝑧 = 0, 𝑚 = 𝜎𝑧 𝑑𝑧.
Beam with symmetric cross section about one axis
If the cross section of the beam is symmetric only with respect to the 𝑧axis, then, due to the redistribution of stress during the plastic yielding, the
position of the neutral fiber (with 𝜀 = 0, 𝜎 = 0) will change. From the
condition
∫
𝑁=
𝜎 𝑑𝑎 = 0
𝐴
follows in the case of ultimate moment
∫
𝑁=
𝜎 𝑑𝑎 = 𝜎𝑌 𝐴𝑢 − 𝜎𝑌 𝐴𝑜 = 0
𝐴
⇒ 𝐴𝑜 = 𝐴𝑢 = 𝐴/2.
The ultimate neutral fiber will therefore be the line dividing the cross section
into equal areas. Take for example the triangle, we have
√
1 2
2
1 2
𝐴= ℎ
⇒ 𝐴𝑜 = 𝐴𝑢 = ℎ
⇒𝑧=
ℎ.
2
4
2
Figure 2.12: Neutral and ultimately neutral fiber
2.2
Torsion of a cylinder
We consider a pure torsion of a cylinder shown in Fig. 2.13, with
𝑀𝑡 = const.
40
CHAPTER 2. ELEMENTARY THEORY
Figure 2.13: Torsion of a cylinder
We assume that cross sections remain plane and perpendicular to the axis
of cylinder, and that the straight lines in radial directions remain straight.
Besides
𝜎𝑥𝑥 = 𝜎𝑦𝑦 = 𝜎𝑧𝑧 = 𝜎𝑦𝑧 = 0.
From the first assumption follows
𝑤𝜑 = 𝑥𝑟𝜗,
with 𝜗 being the twist. Therefore the strain is equal to
1
𝜀𝑥𝜑 (𝑟) = 𝑟𝜗.
2
For the elastic torsion we have
𝜎𝑥𝜑 = 𝜏 (𝑟) = 2𝐺𝜀𝑥𝜑 = 𝐺𝜗𝑟,
∫
∫ 𝑅
2𝜋𝑟𝜏 (𝑟)𝑟 𝑑𝑟 = 2𝜋𝐺𝜗
𝑀𝑡 =
0
𝜋
𝑀𝑡 = 𝐺𝜗𝑅4
2
𝑅
𝑟3 𝑑𝑟,
0
bzw. 𝑀𝑡 = 𝐺𝐽0 𝜗,
𝐽0 =
So, the twist 𝜗 is given by
𝑀𝑡
.
𝐺𝐽0
The maximum shear stress is achieved at 𝑟 = 𝑅
𝜗=
𝜏max =
𝑀𝑡
𝑀𝑡
𝑅=
.
𝐽0
𝑊𝑡
The threshold value for the plastic strain to occur reads
𝜋
𝑀𝑡𝑒 = 𝜏𝑌 𝑅3 ,
2
where
𝜏𝑌 =
{
√1 𝜎𝑌
3
1
𝜎
2 𝑌
Mises,
Tresca.
𝜋 4
𝑅 .
2
2.3. CYLINDRICAL SHELL UNDER COMBINED LOAD
41
When 𝑀𝑡 ≥ 𝑀𝑡𝑒 the plastic strain occurs, and for ideal-plastic material
behavior the yield condition at the boundary 𝑟𝑔 between the elastic and
plastic zone is
𝜏𝑌
.
𝜏 = 𝜏𝑌 = 𝐺𝜗𝑟𝑔 ⇒ 𝑟𝑔 =
𝐺𝜗
The torsion moment 𝑀𝑡 is computed as follows
∫ 𝑟𝑔
∫ 𝑅
2
𝑀𝑡 =
2𝜋𝑟𝐺𝜗𝑟 𝑑𝑟 +
2𝜋𝑟𝜏𝑌 𝑟 𝑑𝑟
0
𝑟𝑔
𝜋
2
= 𝐺𝜗𝑟𝑔4 + 𝜋𝜏𝑌 (𝑅3 − 𝑟𝑔3 )
2
3
𝜋
𝜏𝑌 4 2
𝜏𝑌 3
= 𝐺𝜗(
) + 𝜋𝜏𝑌 [𝑅3 − (
)]
2
𝐺𝜗
3
𝐺𝜗
𝜋
𝜏𝑌 3
) ].
𝑀𝑡 = 𝜏𝑌 [4𝑅3 − (
6
𝐺𝜗
With 𝜗𝑒 = 𝑀𝑡𝑒 /𝐺𝐽0 = 𝜏𝑌 /𝐺𝑅 we get for 𝑀𝑡
𝑅𝜗𝑒 3
𝜋
𝜏𝑌 [4𝑅3 − (
)]
6
𝜗
2
1 𝜗𝑒
𝑀𝑡 = 𝜋𝜏𝑌 𝑅3 [1 − ( )3 ],
3
4 𝜗
1 𝜗𝑒 3
4
𝑀𝑡
= [1 − ( ) ].
𝑀𝑡𝑒
3
4 𝜗
𝑀𝑡 =
The ultimate moment is achieved at 𝑟𝑔 = 0, i. e. 𝜗 → ∞ and
4
𝑀𝑡𝑢
= .
𝑀𝑡𝑒
3
This result can be obtained directly if we let the elastic zone disappear
∫ 𝑅
2
4
2𝜋𝜏𝑌 𝑟2 𝑑𝑟 = 𝜋𝜏𝑌 𝑅3 = 𝑀𝑡𝑒 .
𝑀𝑡𝑢 =
3
3
0
For cylindrical pipe with the internal radius 𝑑𝑖 and external radius 𝑑𝑎 the
ratios 𝑀𝑡𝑢 /𝑀𝑡𝑒 decreases with the decreasing ratio 𝑑𝑖 /𝑑𝑎 , and in the thin-wall
limit 𝑑𝑎 → 𝑑𝑖 it tends to 1.
2.3
Cylindrical shell under combined load
We consider a thin-walled tank loaded by a longitudinal force 𝑁 , and an
internal pressure 𝑝 (see Fig. 2.14). We assume that the cross-section remains
42
CHAPTER 2. ELEMENTARY THEORY
Figure 2.14: A closed tank under combined loading
plane and perpendicular to the 𝑧-axis. Besides,
𝑝𝑐
𝑁
+
,
2𝑡 2𝜋𝑡𝑐
𝑝𝑐
= 𝜎𝜑 = ,
𝑡
= 𝜎𝑟𝑟 = 𝜎𝑧𝑟 = 𝜎𝑟𝜑 = 0.
𝜎𝑧𝑧 = 𝜎𝑧 =
𝜎𝜑𝜑
𝜎𝜑𝑧
In addition to this we shall neglect the elastic strains.
From these assumptions follows immediately
𝑙˙
𝜀˙𝑧 = ,
𝑙
𝑐˙
𝜀˙𝜑 = ,
𝑐
𝑡˙
𝜀˙𝑟 = ,
𝑡
where 𝑙, 𝑐, and 𝑡 are the length, the radius of the cross section, and the
thickness of the cylindrical shell, respectively. Besides,
⎞
⎛
0 0 0
𝜎𝑖𝑗 = ⎝0 𝜎𝜑 0 ⎠ ,
0 0 𝜎𝑧
i. e. we have a homogeneous plane stress state. For materials with Mises’
yield function and isotropic hardening
1
𝑓 = 𝑠𝑖𝑗 𝑠𝑖𝑗 − 𝑘 2 (𝜅).
2
The computation of 𝑔 = 12 𝑠𝑖𝑗 𝑠𝑖𝑗 gives
1
1
𝑔 = 𝑠𝑖𝑗 𝑠𝑖𝑗 = [(𝜎𝜑 − 𝜎𝑧 )2 + 𝜎𝜑2 + 𝜎𝑧2 ]
2
6
1
= (𝜎𝜑2 − 𝜎𝜑 𝜎𝑧 + 𝜎𝑧2 ).
3
The flow rule (1.21) take the form
√
3𝑠𝑘𝑙 𝑠˙ 𝑘𝑙
𝑝
𝑠𝑖𝑗 .
𝜀˙𝑖𝑗 =
4𝑘 2 𝑘 ′
43
2.3. CYLINDRICAL SHELL UNDER COMBINED LOAD
The loading condition requires that
𝑠𝑘𝑙 𝑠˙ 𝑘𝑙 = 𝑔˙ ≥ 0.
Besides, for linear hardening
√
3
1
=
12𝑘 2 𝑘 ′
𝑏𝑔
Therefore
𝑙˙
𝑔˙
(2𝜎𝑧 − 𝜎𝜑 ) = ,
𝑏𝑔
𝑙
𝑐˙
𝑔˙
𝜀˙𝜑 = (2𝜎𝜑 − 𝜎𝑧 ) = ,
𝑏𝑔
𝑐
𝑔˙
𝑡˙
𝜀˙𝑟 = − (𝜎𝑧 + 𝜎𝜑 ) = .
𝑏𝑔
𝑡
𝜀˙𝑧 =
These are three governing differential equations for three unknowns 𝑐, 𝑡, 𝑙, for
which the loading paths of 𝜎𝜑 and 𝜎𝑧 are given.
We can for example load the tank in two steps:
∙ Step 1: Increase the internal pressure up to the yield stress at 𝑝 = 𝑝𝑌 .
∙ Step 2: Keep 𝑝 = 𝑝𝑌 = const and increase the longitudinal force 𝑁 .
sz
N
pY
p
sjY
Figure 2.15: Loading in two steps
In the first step
𝜎𝑧 =
𝑝𝑐
1
= 𝜎𝜑 ,
2𝑡
2
𝜎𝜑 =
𝑝𝑐
𝑡
⇒ 𝜎𝜑𝑌 = 𝑝𝑌
The yield condition leads to
1
1
𝑔𝑌 = 𝜎𝑌2 = (𝜎𝜑2 − 𝜎𝜑 𝜎𝑧 + 𝜎𝑧2 )
3
3
1 2
1 1
1 2
= 𝜎𝜑𝑌 (1 − + ) = 𝜎𝜑𝑌
.
3
2 4
4
𝑐𝑌
.
𝑡𝑌
sj
44
CHAPTER 2. ELEMENTARY THEORY
Consequently
2 𝑡𝑌
𝜎𝑌 .
𝑝𝑌 = √
3 𝑟𝑌
In the second step (𝑝𝑌 = const, 𝜎˙ 𝑧 > 0)
𝑐˙
𝑔˙
= (2𝜎𝜑 − 𝜎𝑧 ),
𝑐
𝑏𝑔
𝑔˙
𝑙˙
𝜀˙𝑧 = = (2𝜎𝑧 − 𝜎𝜑 )
𝑙
𝑏𝑔
𝜀˙𝜑 =
with
1
𝑔 = (𝜎𝜑2 − 𝜎𝜑 𝜎𝑧 + 𝜎𝑧2 ),
3
1
𝑔˙ = [(2𝜎𝜑 − 𝜎𝑧 )𝜎˙ 𝜑 + (2𝜎𝑧 − 𝜎𝜑 )𝜎˙ 𝑧 ].
3
Now
𝑐˙ 𝑐𝑡˙
𝑐
˙
⇒ 𝜎˙ 𝜑 = 𝑝𝑌 (𝑐/𝑡) = 𝑝𝑌 ( − 2 ).
𝑡
𝑡 𝑡
Due to the incompressibility
𝜎𝜑 = 𝑝 𝑌
𝑡˙ 𝑐˙ 𝑙˙
+ + =0
𝑡 𝑐 𝑙
we obtain
𝑐 𝑙˙
𝑐˙
𝑐
𝜎˙ 𝜑 = 𝑝𝑌 ( + 2 ) = 𝑝𝑌 (𝜀˙𝑧 + 2𝜀˙𝜑 ).
𝑡 𝑙
𝑐
𝑡
We can therefore express 𝜎˙ 𝜑 as follows
𝜎˙ 𝜑 = 𝜎𝜑2
3 𝑔˙
.
𝑏𝑔
The solution of this equation gives
1
3
+ ln 𝑔 + 𝑎 = 0.
𝜎𝜑 𝑏
At 𝜎𝜑 = 𝜎𝜑𝑌 we have 𝑔 = 𝑔𝑌 = 𝜎𝑌2 /3, so
𝜎𝜑
=
𝜎𝜑𝑌
1−
1
3𝜎𝜑𝑌
𝑏
ln 𝜎3𝑔2
.
𝑌
This is a transcendental equation for 𝜎𝜑 . There are still two remaining equations to determine 𝑟 and 𝑙 at a given 𝜎𝑧 and 𝜎𝜑 , where
2𝜎𝑧 − 𝜎𝜑
𝑔˙
𝜎˙ 𝑧
=
.
3 2
𝑏𝑔
𝑔 − 2𝑏 𝜎𝜑 (2𝜎𝜑 − 𝜎𝑧 ) 2𝑏
2.4. SIMPLE METAL FORMING PROCESSES
2.4
45
Simple metal forming processes
The elementary theory of metal forming processes (cold work) is quite similar to the elementary theory of elasto-plastic deformation. We also begin
here with the simplifying assumptions about the kinematics of the process
under consideration. On the other side the stress states appearing in most
metal forming processes are much more complicated compared with those of
say, simple bending or torsion. Drastic simplifications may lead to erroneous
results. That is why we normally made in this type of problems only assumptions about some integral characteristics (for example the plastic work)
rather than assumptions about the stress state.
Since the elastic strains are normally quite small compared with its plastic
counterpart which may be of the order 1, we shall neglect the former in the
cold working problems.
Most metal forming processes such as forging, rolling, drawing and so
on can be explained by quite simple models. We restrict ourselves to the
plane strain problems. Since plastic deformations in cold-worked materials
are normally produced under pressure, we shall regard the latter as positive
despite the traditional convention in mechanics. We denote the pressure by
𝑝, 𝑞.
We consider for instance:
i) rolling (Fig. 2.16),
ii) drawing, extrusion (Fig. 2.17),
iii) pressing, forging (Fig. 2.18).
Figure 2.16: Rolling: Boundary conditions are time-independent
The strip model
We consider now a strip obtained by cutting (mentally) the piece of metal
at cold rolling along the cuts perpendicular to the plane of motion. For
simplicity we assume the symmetry with respect to the (𝑥, 𝑧)-plane. Further
46
CHAPTER 2. ELEMENTARY THEORY
Figure 2.17: Drawing: Boundary conditions are time-independent
Figure 2.18: Forging: Boundary conditions are time-dependent
we assume i) the plane strain state, i. e. the planes perpendicular to 𝑥-axis
remain plane, ii) the work done by external forces per unit time is the same
as that of the uni-axial compression (the shear strain are neglected), iii) the
process is quasi-static, the body forces are negligibly small. We take for
Figure 2.19: Force equilibrium of a strip
granted that 𝑣𝑥 > 0. In the planes 𝑥 = const the shear stress is absent, while
the normal stress (pressure) 𝑞 is uniformly distributed over the cross section.
Thus, the resultant forces in 𝑥-direction acting on the strip is 𝑞ℎ at 𝑥 and
𝑞ℎ + ∂(𝑞ℎ)/∂𝑥𝑑𝑥 at 𝑥 + 𝑑𝑥. The normal force exerted from the roller on the
strip, 𝑝𝑛 𝑑𝑠, causes the friction
𝜏 𝑑𝑠 = 𝜇𝑝𝑛 𝑑𝑠
47
2.4. SIMPLE METAL FORMING PROCESSES
Thus, the resultant force from the roller on the strip is decomposed into the
vertical component
𝑝𝑦 𝑑𝑥 = 𝑝𝑑𝑥 = 𝑝𝑛 𝑑𝑠(cos 𝛼 + 𝜇 sin 𝛼)
and the horizontal component
𝑝𝑥 𝑑𝑥 = 𝑝𝑛 𝑑𝑠(sin 𝛼 − 𝜇 cos 𝛼).
With
𝑑𝑥 = cos 𝛼𝑑𝑠,
we obtain
𝑝𝑦 𝑑𝑥 = 𝑝𝑑𝑥 = 𝑝𝑛 𝑑𝑥(1 + 𝜇 tan 𝛼),
tan 𝛼 − 𝜇
𝑝𝑥 𝑑𝑥 = 𝑝𝑛 𝑑𝑥(tan 𝛼 − 𝜇) = 𝑝𝑑𝑥
.
1 + 𝜇 tan 𝛼
Denoting the coefficient of friction by
𝜇 = tan 𝜈,
we obtain for the horizontal force
𝑝𝑥 𝑑𝑥 = 𝑝𝑑𝑥
tan 𝛼 − tan 𝜈
= 𝑝𝑑𝑥 tan(𝛼 − 𝜈).
1 + tan 𝜈 tan 𝛼
From the assumption ii) we set the power of external forces equal to that
obtained under the uniaxial compression
𝑤˙ = −𝜎𝑌 𝑑𝑥ℎ̇ = −𝑝𝑑𝑥ℎ̇ + 2𝑝𝑥 𝑑𝑥𝑣𝑥 + 𝑞ℎ𝑣𝑥 − [𝑞ℎ𝑣𝑥 +
= −𝑝𝑑𝑥ℎ̇ + 2𝑝𝑥 𝑑𝑥𝑣𝑥 −
∂
(𝑞ℎ𝑣𝑥 )𝑑𝑥]
∂𝑥
∂
(𝑞ℎ𝑣𝑥 )𝑑𝑥.
∂𝑥
Dividing this equation by ℎ𝑑𝑥 and substituting the formula for 𝑝𝑥 into it, we
obtain
−𝜎𝑌
ℎ̇
ℎ̇
𝑝
𝑣𝑥 ∂
∂𝑣𝑥
= −𝑝 + 2 tan(𝛼 − 𝜈)𝑣𝑥 −
(𝑞ℎ) − 𝑞
.
ℎ
ℎ
ℎ
ℎ ∂𝑥
∂𝑥
The incompressibility condition requires that
𝑣𝑥 ℎ = const
⇒
∂𝑣𝑥
∂ℎ
ℎ + 𝑣𝑥
= 0.
∂𝑥
∂𝑥
48
CHAPTER 2. ELEMENTARY THEORY
Besides
ℎ̇ =
∂ℎ
𝑣𝑥 .
∂𝑥
Therefore
∂𝑣𝑥
ℎ̇
=− .
∂𝑥
ℎ
In the elementary theory of forming processes we always use Tresca’s yield
condition, for which
𝑝 − 𝑞 = 𝜎𝑌 .
It follows from here
ℎ
∂𝑞
∂ℎ
+𝑞
− 2𝑝 tan(𝛼 − 𝜈) = 0.
∂𝑥
∂𝑥
Replacing 𝑝 = 𝑞 + 𝜎𝑌 and ∂ℎ/∂𝑥 = 2 tan 𝛼, we obtain finally
𝑞
𝜎𝑌
∂𝑞
+ 2 [tan 𝛼 − tan(𝛼 − 𝜈)] − 2
tan(𝛼 − 𝜈) = 0.
∂𝑥
ℎ
ℎ
This ordinary differential equation of first order can be written in the form
𝑑𝑞
+ 𝑓 (𝑥)𝑞 + 𝑔(𝑥) = 0,
𝑑𝑥
so, its general solution reads
𝑞(𝑥) = 𝑒
−
∫𝑥
𝑥0
𝑓 𝑑𝑥
[𝑞0 −
∫
𝑥
(𝑔𝑒
∫𝑥
𝑥0
𝑓 𝑑𝑥
)𝑑𝑥],
𝑥0
with 𝑞 = 𝑞0 being the value of 𝑞 at 𝑥 = 𝑥0 . This solution is useful if the
integrals can be computed analytically. Otherwise the numerical solution
turns out to be more effective. After the stress 𝑞 is known, the stress 𝑝 can
easily be determined.
The tube model
If we deal with the axisymmetric forging process, we may imagine to have
a tube cut from the cold worked metal at forging which is symmetric about
the 𝑧-axis as shown in Fig. 2.20.
We assume i) axisymmetric deformation, i. e. cylindrical surfaces about
the 𝑧-axis remain cylindrical, ii) the work done by the external forces is the
same as that in the uni-axial compression, iii) the body forces are negligibly
small. Except that we take for granted that 𝑣𝑟 > 0.
As in the previous strip model we obtain for the force acting on the die
in the 𝑧-direction
𝑝𝑧 𝑑𝑟 = 𝑝𝑑𝑟 = 𝑝𝑛 𝑑𝑠(cos 𝛼 − 𝜇 sin 𝛼) = 𝑝𝑛 𝑑𝑟(1 − tan 𝜈 tan 𝛼),
2.4. SIMPLE METAL FORMING PROCESSES
49
Figure 2.20: The tube model
and in the 𝑟-direction
𝑝𝑟 𝑑𝑟 = 𝑝𝑛 𝑑𝑠(sin 𝛼 − tan 𝜈 cos 𝛼) = 𝑝𝑛 𝑑𝑟(tan 𝛼 − tan 𝜈) = 𝑝𝑑𝑟 tan(𝛼 − 𝜈).
From the assumption ii) we set the power of external forces equal to that of
the pressure acting on the tube
𝑤˙ = −𝜎𝑌 2𝜋𝑟𝑑𝑟ℎ̇
∂
(𝑞𝑟ℎ𝑣𝑟 )𝑑𝑟]
∂𝑟
∂
∂
= 2𝜋𝑟𝑑𝑟[−𝑝ℎ̇ + 2𝑝 tan(𝛼 − 𝜈)𝑣𝑟 ] − 2𝜋𝑑𝑟[ (𝑞ℎ)𝑟𝑣𝑟 + (𝑟𝑣𝑟 )𝑞ℎ].
∂𝑟
∂𝑟
= −𝑝2𝜋𝑟𝑑𝑟ℎ̇ + 2𝑝𝑟 2𝜋𝑟𝑑𝑟𝑣𝑟 + 𝑞ℎ𝑣𝑟 2𝜋𝑟 − 2𝜋[𝑞𝑟ℎ𝑣𝑟 +
The incompressibility requires
∂
ℎ̇ 𝑣𝑟
+ + 𝑣𝑟 = 0,
ℎ
𝑟
∂𝑟
or
ℎ̇
∂
= − (𝑟𝑣𝑟 ).
ℎ
∂𝑟
We obtain (after dividing by 𝑑𝑉 = 2𝜋𝑟ℎ𝑑𝑟)
𝑟
(−𝜎𝑌 + 𝑝 − 𝑞)
∂
𝑣𝑟
ℎ̇
= [2𝑝 tan(𝛼 − 𝜈) − (𝑞ℎ)] .
ℎ
∂𝑟
ℎ
Using Tresca’s yield condition we have
𝑝 − 𝑞 = 𝜎𝑌 .
It follows that
ℎ
∂𝑞
∂ℎ
+𝑞
− 2𝑝 tan(𝛼 − 𝜈) = 0,
∂𝑟
∂𝑟
50
CHAPTER 2. ELEMENTARY THEORY
and respectively
∂𝑞
𝑞
𝜎𝑌
+ 2 [tan 𝛼 − tan(𝛼 − 𝜈)] − 2
tan(𝛼 − 𝜈) = 0,
∂𝑟
ℎ
ℎ
provided ℎ̇ < 0, 𝑣𝑟 > 0. Thus, we obtain the same equation as that of the
strip model, with 𝑥 replaced by 𝑟.
As an example let us consider the compression of a cylinder shown in
Fig. 2.21). In this case 𝛼 = 0, so the equation reduces to
Figure 2.21: Compression of a cylinder
𝑞
𝜎𝑌
𝑞 ′ + 2𝜇 + 2𝜇
= 0.
ℎ
ℎ
The boundary condition is
𝑞 = 0 at 𝑟 = 𝑅.
This equation yield the following general solution
∫ 𝑟 ∫
∫
𝑟
2𝜇
− 𝑅𝑟 2𝜇/ℎ𝑑𝑟
𝑞(𝑟) = 𝑒
[𝑞(𝑅) −
𝜎𝑌
𝑒 𝑅 2𝜇/ℎ𝑑𝑟 𝑑𝑟]
ℎ
𝑅
∫ 𝑟
2𝜇
2𝜇
2𝜇
= 𝑒− ℎ (𝑟−𝑅) [− 𝜎𝑌
𝑒 ℎ (𝑟−𝑅) 𝑑𝑟]
ℎ
𝑅
2𝜇
= 𝜎𝑌 [𝑒 ℎ (𝑅−𝑟) − 1].
For 𝑝(𝑟) we have
2𝜇
𝑝(𝑟) = 𝜎𝑌 + 𝑞 = 𝜎𝑌 𝑒 ℎ (𝑅−𝑟)
The solution found remains valid as long as the stick zone where 𝜏 = 𝜇𝑝 >
𝜎𝑌 /2 does not occurs. This means that the shear stress distribution
𝜏 = 𝜇𝑝
must be controlled and checked, whether its maximum exceed the value 𝜎𝑌 /2
or not. With this solution we can compute the total compressive force acting
on the cold forged material
∫ 𝑅
2𝜇
ℎ𝑅
ℎ2 2𝜇𝑅
].
𝐹 = 2𝜋𝜎𝑌
𝑒 ℎ (𝑅−𝑟) 𝑟𝑑𝑟 = 2𝜋𝜎𝑌 [ 2 (𝑒 ℎ − 1) −
4𝜇
2𝜇
0
We can easily generalize this model to the so-called plate model.
Chapter 3
Theory of plastic flow
3.1
Governing equations
According to the classification given at the end of Chapter 1 we shall assume
in the theory of plastic flow the ideal plastic material behavior. Thus,
𝜎𝑌 = const,
i. e. any hardening behavior is excluded from consideration.
We further admit that the plastic materials obey Mises’ yield condition
𝑓 = 𝐽2 − 𝑘02 = 0,
1
𝑘02 = 𝜎02 .
3
In most of cases we neglect the elastic strains as small compared with the
plastic strains. Thus, as a rule, the material considered in the theory of plastic
flow is rigid-ideal-plastic and obey Mises’ yield condition. The associate flow
rule reads
∂𝑓
𝑑𝑖𝑗 = 𝜀˙𝑖𝑗 = 𝜀˙𝑝𝑖𝑗 = 𝜆
= 𝜆𝑠𝑖𝑗
∂𝜎𝑖𝑗
where 𝜆 remains still an unknown parameter. It can be determined only with
the help of the kinematic boundary condition. From this flow rule we derive
𝑑𝑖𝑗 𝑑𝑖𝑗 = 𝜆2 𝑠𝑖𝑗 𝑠𝑖𝑗 = 2𝜆2 𝑘02 .
Therefore
𝜆= √
1 √
𝑑𝑖𝑗 𝑑𝑖𝑗 .
2𝑘0
51
52
CHAPTER 3. THEORY OF PLASTIC FLOW
If the material behaves as elastic-plastic, then the Prandtl-Reuss equations hold true (see Section 1.4)
1
𝜎˙ 𝑘𝑘 ,
3𝐾
1
𝑒˙ 𝑖𝑗 =
𝑠˙ 𝑖𝑗 + 𝜆𝑠𝑖𝑗 .
2𝜇
𝜀˙𝑘𝑘 =
The closed system of governing equations must include the equilibrium
conditions, which, in the absence of body forces, read
𝜎𝑖𝑗,𝑗 = 0.
Thus, the whole system of governing equations consists of
∙ 3 equilibrium conditions
∙ 6 stress-strain relations
∙ 1 yield condition
These 10 equations contain 10 unknowns, among which
∙ 3 components of velocity 𝑣𝑖
∙ 6 components of the stress tensor 𝜎𝑖𝑗
∙ 1 scalar factor 𝜆.
The theory of plastic flow is primarily applied for three types of problems:
∙ onset of plastic flow of a rigid-plastic or elastic-plastic material,
∙ non-stationary plastic flow, provided the boundary conditions of the
states under consideration are sufficiently known (the so-called pseudostationary plastic flow),
∙ stationary plastic flow.
Analytical solutions are available only if some additional assumptions concerning the kinematics of the plastic flow can be made, for examples in
∙ the torsion of prismatic bars, or
∙ the plane strain problems.
53
3.2. TORSION OF PRISMATIC BARS
3.2
Torsion of prismatic bars
Stress distribution
For the torsion of prismatic bars with an arbitrary cross-section we use the
St. Venant Ansatz
𝑢𝑦 = −𝜗𝑧𝑥,
𝑢𝑧 = 𝜗𝑦𝑥,
𝑢𝑥 = 𝑢(𝑦, 𝑧).
Here 𝜗 is the twist, and 𝑢 the warping. The following components of the
stress tensor are zero
𝜎𝑥𝑥 = 𝜎𝑦𝑦 = 𝜎𝑧𝑧 = 𝜎𝑦𝑧 = 0.
The equilibrium conditions reduce to
𝜎𝑥𝑦,𝑦 + 𝜎𝑥𝑧,𝑧 = 0.
Figure 3.1: Cross section of a bar
This equation is automatically fulfilled if there exists a stress function
𝜓(𝑦, 𝑧) such that
𝜎𝑥𝑦 = 𝜓,𝑧 , 𝜎𝑥𝑧 = −𝜓,𝑦 .
The differential of 𝜓
𝑑𝜓 = 𝜓,𝑦 𝑑𝑦 + 𝜓,𝑧 𝑑𝑧 = −𝜎𝑥𝑧 𝑑𝑦 + 𝜎𝑥𝑦 𝑑𝑧
vanishes for
𝑑𝑧
𝜎𝑥𝑧
=
𝑑𝑦
𝜎𝑥𝑦
i. e., when the lines of 𝜓 = const has the same direction as the resultant of
𝜏 (Fig. 3.2). The lines 𝜓 = const will therefore be called stress trajectories.
54
CHAPTER 3. THEORY OF PLASTIC FLOW
Figure 3.2: Lines 𝜓 = const
Since the normal stresses vanishes at the boundary of the cross section,
𝜓 = const at Γ for simply connected cross section.
The plastic zone in the cross section is determined from the yield condition
2
2
𝜎𝑥𝑦
+ 𝜎𝑥𝑧
= 𝜏𝑌2 ,
where
𝜏𝑌 =
{
1
𝜎
2 𝑌
1
√ 𝜎𝑌
3
Tresca,
Mises.
Thus,
(𝜓,𝑦 )2 + (𝜓,𝑧 )2 = 𝜏𝑌2
⇒
∣∇𝜓∣ = 𝜏𝑌 .
Nadai has found from the obtained condition (that the gradient or the maximal slope of 𝜓 remains constant) the “sand roof analogy” which is constructed by pouring sand on a horizontal sheet of cardboard set out in the
shape of a cross section. Due to the constant internal friction of sand, the
constructed roof satisfies the above equation.
Figure 3.3: Sand roof
The torque is calculated according to
∫
∫
∫
𝑀𝑡 = (𝜎𝑥𝑧 𝑦 − 𝜎𝑥𝑦 𝑧) 𝑑𝑎 = − (𝜓,𝑦 𝑦 + 𝜓,𝑧 𝑧) 𝑑𝑎 = 2 𝜓 𝑑𝑎,
𝐴
𝐴
so it is equal to twice of the volume under this “roof”.
𝐴
3.2. TORSION OF PRISMATIC BARS
55
The determination of the stress distribution 𝜎𝑖𝑗 within the cross section
does not requires the stress-strain relation. Thus, this is the statically determinate problem.
Determination of the warping
According to the flow rule we have in the plastic zone
𝜀˙𝑥𝑧 = 𝜆𝜎𝑥𝑧 ,
𝜀˙𝑥𝑦 = 𝜆𝜎𝑥𝑦 .
Eliminating 𝜆 we derive
𝑑𝜀𝑥𝑧
𝜎𝑥𝑧
=
.
𝑑𝜀𝑥𝑦
𝜎𝑥𝑦
Since the stress components in the plastic zone are time-independent, we can
integrate this equation to get
𝜀𝑥𝑧
𝜎𝑥𝑧
=
+ const.
𝜀𝑥𝑦
𝜎𝑥𝑦
The constant of integration must be equal to zero because it vanishes identically at the boundary between the elastic and plastic zones. Based on the
St. Venant’s Ansatz we have
1
𝜀𝑥𝑧 = (𝑢,𝑧 + 𝜗𝑦),
2
1
𝜀𝑥𝑦 = (𝑢,𝑦 − 𝜗𝑧).
2
Substitution in the above equation gives
𝑢,𝑧 + 𝜗𝑦
𝜎𝑥𝑧
= tan 𝜑,
=
𝜎𝑥𝑦
𝑢,𝑦 − 𝜗𝑧
what is completely analogous to the purely elastic torsion, where 𝜑 denotes
as before the angle between the 𝑦-Achse and the direction of 𝜏 . A simple
transformation leads to
𝑢,𝑧 cos 𝜑 − 𝑢,𝑦 sin 𝜑 = −𝜗(𝑧 sin 𝜑 + 𝑦 cos 𝜑),
or
𝑢,𝑛 = −𝜗𝑑,
where 𝑛 is the direction perpendicular to 𝜏 . With the known stress distribution and the known direction of 𝜏 we can find the warping from this
equation.
56
CHAPTER 3. THEORY OF PLASTIC FLOW
Figure 3.4: Determination of the warping
3.3
Plane strain problems
Governing equations
We are dealing with a 2-D plastic flow, if the components of the velocity are
given by
𝑣𝑥 = 𝑣𝑥 (𝑥, 𝑦), 𝑣𝑦 = 𝑣𝑦 (𝑥, 𝑦), 𝑣𝑧 = 0,
and thus,
𝜀˙𝑧𝑧 = 𝜀˙𝑦𝑧 = 𝜀˙𝑥𝑧 = 0.
Based on the rigid-ideal-plastic material behavior and the Mises yield condition, the normality rule reads
𝜀˙𝑖𝑗 = 𝜆𝑠𝑖𝑗 ,
𝜆 > 0.
This implies
𝑠𝑧𝑧 = 𝜎𝑦𝑧 = 𝜎𝑥𝑧 = 0,
and
1
𝜎𝑧𝑧 = 𝜎 = (𝜎𝑥𝑥 + 𝜎𝑦𝑦 ),
2
and accordingly
⎞
𝜎𝑥𝑥 𝜎𝑥𝑦 0
𝜎𝑖𝑗 = ⎝𝜎𝑥𝑦 𝜎𝑦𝑦 0 ⎠ .
0
0 𝜎
⎛
So 𝜎 is one of the three principal stresses. The remaining two are obtained
as follows
1√
2
𝜎1,2 = 𝜎 ±
(𝜎𝑥𝑥 − 𝜎𝑦𝑦 )2 + 4𝜎𝑥𝑦
2
57
3.3. PLANE STRAIN PROBLEMS
in the coordinate system obtained by an anticlockwise rotation of the original
axes through an angle 𝜑 (see Fig. 3.5)
𝜑=
1
2𝜎𝑥𝑦
arctan
.
2
𝜎𝑥𝑥 − 𝜎𝑦𝑦
Thus, the maximum shear stress
𝜏max
1
1√
2 = 𝜏
(𝜎𝑥𝑥 − 𝜎𝑦𝑦 )2 + 4𝜎𝑥𝑦
= (𝜎1 − 𝜎2 ) =
2
2
is achieved in the direction inclined at the angle 𝜋/4 with respect to the
principal axes. The stress state
𝜎1 = 𝜎 + 𝜏,
𝜎2 = 𝜎 − 𝜏,
𝜎3 = 𝜎
is characterized by superposition of the pure shear stress in the (𝑥, 𝑦)-plane
on the hydrostatic pressure.
Figure 3.5: Principal axes and 𝛼- and 𝛽-lines
We denote the directions corresponding to 𝜎1 and 𝜎2 as first and second principal direction, respectively, while the directions obtained by the
clockwise rotations of the principal axes through the angle 450 , in which the
maximum and minimum of the shear stresses occur as first and second slip
direction.
The curve, whose tangent coincides with the slip direction in each point,
is called slip line. It is obvious that there are two orthogonal families of slip
lines. We denote them as 𝛼- and 𝛽-lines, respectively.
Let 𝜗 be the angle between the 𝛼-line and the 𝑥-axis. Then we obtain
𝑑𝑦
= tan 𝜗, for 𝛼-lines,
𝑑𝑥
𝑑𝑦
= − cot 𝜗, for 𝛽-lines
𝑑𝑥
58
CHAPTER 3. THEORY OF PLASTIC FLOW
as differential equations for these two families of slip lines.
The stress state must satisfy during the plastic flow the yield condition
1
[(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ] − 𝑘 2 = 0
6
For the plane strain problems this condition becomes
𝜏 = ±𝑘,
√
with 𝑘 = 𝜎𝑌 / 3.
The stress state in a particular point is given through 𝜎𝑥𝑥 , 𝜎𝑦𝑦 , 𝜎𝑥𝑦 . Due to
the yield condition these components of the stress tensor are not independent.
Let us introduce two dimensionless quantities, 𝜔 and 𝜗 as follows
𝜔=
𝜎
,
2𝑘
𝜗=
1
2𝜎𝑥𝑦
𝜋
𝜋
arctan
− =𝜑− .
2
𝜎𝑥𝑥 − 𝜎𝑦𝑦
4
4
Thus, 𝜗 is the angle, through which the 𝑥-axis is rotated in anti-clockwise
direction to the first slip line. Then
𝜎𝑥𝑥 = 2𝑘𝜔 − 𝑘 sin 2𝜗,
𝜎𝑦𝑦 = 2𝑘𝜔 + 𝑘 sin 2𝜗,
𝜎𝑥𝑦 = 𝑘 cos 2𝜗.
(3.1)
We have in addition the equilibrium equations which reduce to
𝜎𝑥𝑥,𝑥 + 𝜎𝑥𝑦,𝑦 = 0,
𝜎𝑥𝑦,𝑥 + 𝜎𝑦𝑦,𝑦 = 0.
There are altogether three equations to determine three unknown functions.
The plane strain problem is therefore “static determined”.
Slip lines and their properties
The stress components can be expressed in the plastic zone in terms of variables 𝜔 and 𝜗 according to (3.1). Substituting formulas (3.1) into the equilibrium conditions, we obtain
𝜔,𝑥 − 𝜗,𝑥 cos 2𝜗 − 𝜗,𝑦 sin 2𝜗 = 0,
𝜔,𝑦 − 𝜗,𝑥 sin 2𝜗 + 𝜗,𝑦 cos 2𝜗 = 0.
This is the system of homogeneous quasi-linear partial differential equations
of first order. Its solution as well as the suitable method of solution depend
59
3.3. PLANE STRAIN PROBLEMS
strongly on the type of these equations. To recognize the latter, we consider
the differential equations of characteristics
√
𝑑𝑦
1
= (𝑏 ± 𝑏2 − 𝑎𝑐)
𝑑𝑥
𝑎
with
𝑎 = sin 2𝜗,
2𝑏 = 2 cos 2𝜗,
𝑐 = − sin 2𝜗.
The discriminant
𝑏2 − 𝑎𝑐 = cos2 2𝜗 + sin2 2𝜗 = 1
is positive, so the system under consideration is hyperbolic. For the characteristic directions we have
𝑑𝑦
1
=
(− cos 2𝜗 ± 1),
𝑑𝑥
sin 2𝜗
or
𝑑𝑦
=
𝑑𝑥
{
tan 𝜗,
cot 𝜗.
Thus, the characteristics of this problem coincide with the slip lines found
above.
Choosing in an arbitrary point 𝑃 (𝑥, 𝑦) the directions of the 𝑥-𝑦-axes such
that they coincide with the directions of the slip lines (𝑠𝛼 , 𝑠𝛽 ), and taking
into account that
sin 2𝜗 = 0, cos 2𝜗 = 1
we transform the above equations to
(𝜔 − 𝜗),𝑠𝛼 = 0,
(𝜔 + 𝜗),𝑠𝛽 = 0.
These equations imply that 𝜔 − 𝜗 is constant along the first slip line, while
𝜔 + 𝜗 is constant along the second slip line
𝑑𝑦
= tan 𝜗, 𝜔 − 𝜗 = const, 𝛼-line,
𝑑𝑥
𝑑𝑦
= − cot 𝜗, 𝜔 + 𝜗 = const, 𝛽-line.
𝑑𝑥
So, provided the slip lines as well as the values of constants on the 𝛼- and
𝛽-lines are known, then 𝜔 and 𝜗 and the stress state in each point of the
60
CHAPTER 3. THEORY OF PLASTIC FLOW
(𝑥, 𝑦)-plane can be determined. However, the directions of characteristics
depend on the solution of this problem.
The slip lines possess several important properties to be discussed below.
Proposition 1 : The pressure 𝜎 changes along a slip line in proportion with
the angle 𝜗.
This property follows directly from the above formulas, according to which
𝜔 ± 𝜗 = const along the 𝛼, 𝛽-lines.
Proposition 2 : The change in the angle 𝜗 and the pressure 𝜎 is the same for
a transition from one slip line of the 𝛽-family to another along any slip line
of the 𝛼-family (Henky’s first theorem).
Figure 3.6: Hencky’s first theorem
From the formula
𝜔−𝜗=
{
𝑘1
𝑘2
along 𝛼1 ,
along 𝛼2 ,
𝜔+𝜗=
{
ℎ1
ℎ2
along 𝛽1 ,
along 𝛽2 ,
and respectively,
it follows
1
𝜗𝑎11 = (ℎ1 − 𝑘1 ),
2
1
𝜗𝑎21 = (ℎ1 − 𝑘2 ),
2
1
𝜗𝑎12 = (ℎ2 − 𝑘1 ),
2
1
𝜗𝑎22 = (ℎ2 − 𝑘2 ).
2
Therefore
1
𝜑1 = 𝜗𝑎21 − 𝜗𝑎11 = (𝑘1 − 𝑘2 ),
2
1
𝜑2 = 𝜗𝑎22 − 𝜗𝑎12 = (𝑘1 − 𝑘2 ) = 𝜑1 .
2
61
3.3. PLANE STRAIN PROBLEMS
a
b
C
A
B
Figure 3.7: Determination of the pressure
Proposition 3 : If the value of 𝜎 is known at some point of a given slip grid,
then it can be found everywhere in the field.
Since 𝜗 is known everywhere, we have
𝜎𝐵 = 2𝑘(ℎ1 − 𝜗𝐵 ),
and
𝜎𝐶 = 2𝑘(𝑘1 + 𝜗𝐶 ).
Proposition 4 : If some segment of a 𝛽- (or 𝛼)-slip line is straight, then all
the corresponding segments of 𝛽- (or 𝛼)-lines are also straight and have the
same length.
E
B´ a
B
A
A´
b
Figure 3.8: Straight slip lines
This proposition follows from the second proposition, since the angle between
the tangents to any two 𝛽-slip lines remains constant as we move along
the prescribed 𝛽-line. The evolute (locus of the center of curvature) of an
arbitrary curve is the envelope of the normals to the curve (Fig. 3.8). It is
evident that the slip line AA′ and BB′ have the same evolute E. From the
definition of the evolute follows that AB=A′ B′ .
Proposition 5 : Suppose that we move along some slip line; then the radii of
curvature of the slip lines of the other family at the points of intersection
change by the distance traveled (Hencky’s second theorem)
Consider neighboring slip lines of the 𝛼 and 𝛽 families, bounding a slip
62
CHAPTER 3. THEORY OF PLASTIC FLOW
a ds
b
C
A
dsa
B
Rb
Ea
d g´
Ra
b
d g´
D
dg
Eb
Figure 3.9: Hencky’s second theorem
element 𝑑𝑠𝛼 𝑑𝑠𝛽 shown in Fig. 3.9. It is apparent that
𝑅𝛼 𝑑𝛾 ′ = 𝑑𝑠𝛼 = 𝐴𝐵,
𝑅𝛽 𝑑𝛾 = 𝑑𝑠𝛽 = 𝐴𝐶.
For the arc-length AC we have
𝐴𝐶 = 𝑅𝛽 𝑑𝛾.
On the other side, from proposition 2 follows
𝐴𝐶 = (𝑅𝛽 + 𝑑𝑅𝛽 + 𝑑𝑠𝛼 )𝑑𝛾,
so
𝑑𝑅𝛽 = −𝑑𝑠𝛼 ,
𝑑𝑅𝛼 = −𝑑𝑠𝛽 .
Proposition 5a: The center of curvature of the 𝛽-lines at point of intersection
with 𝛼-lines generate the involute PT of the 𝛼-line (Prandtl’s theorem).
T S
R
´
A´ B´ C
b
C
a
A B
D´
Q
D
dg
P
Figure 3.10: Prandtl’s theorem
Since AP=ABQ, PQ must be the involute of the 𝛼-line.
3.3. PLANE STRAIN PROBLEMS
63
Proposition 5b: The envelope of the slip lines of one family is the geometric
locus of the cusps of the slip lines of the other family.
It can be seen from Fig. 3.10 that in point T the distance between the 𝛼-lines
as well as the radius of curvature of the 𝛽-lines tend to zero. Therefore: i)
T belongs to the envelope of the family of slip lines, ii) the 𝛽-line passing
through this point has a cusp at T.
Since they have a cusp at T the 𝛽-lines cannot intersect the envelope. In
other words, the envelope is the boundary of an analytic solution.
According to Caratheodory and Schmidt an orthogonal grid of slip lines
possessing the above properties is called Hencky-Prandtl’s grid.
Boundary conditions
The properties of the slip lines discussed in the previous subsection can be
used to construct the Hencky-Prandtl’s grid, provided one slip line is known
from each family, and they intersect each other.
In order to construct the grid and the solution of a given boundary-value
problem, the boundary conditions should be taken into account.
Figure 3.11: Static boundary condition
For a static boundary condition we must specify the tractions 𝜎𝑛 and 𝜏𝑛
along the boundary Γ.
We compute now 𝜎𝑛 , 𝜎𝑡 and 𝜏𝑛
1
1
𝜎𝑛 = (𝜎𝑥𝑥 + 𝜎𝑦𝑦 ) + (𝜎𝑥𝑥 − 𝜎𝑦𝑦 ) cos 2𝜓 + 𝜎𝑥𝑦 sin 2𝜓,
2
2
1
1
𝜎𝑡 = (𝜎𝑥𝑥 + 𝜎𝑦𝑦 ) − (𝜎𝑥𝑥 − 𝜎𝑦𝑦 ) cos 2𝜓 − 𝜎𝑥𝑦 sin 2𝜓,
2
2
1
𝜏𝑛 = − (𝜎𝑥𝑥 − 𝜎𝑦𝑦 ) sin 2𝜓 + 𝜎𝑥𝑦 cos 2𝜓.
2
Substitution of 𝜎𝑥𝑥 , 𝜎𝑥𝑦 and 𝜎𝑦𝑦 from (3.1) yields
𝜎𝑛 = 2𝑘𝜔 − 𝑘 sin 2(𝜗 − 𝜓),
𝜎𝑡 = 2𝑘𝜔 + 𝑘 sin 2(𝜗 − 𝜓),
𝜏𝑛 = 𝑘 cos 2(𝜗 − 𝜓).
64
CHAPTER 3. THEORY OF PLASTIC FLOW
With the given values of 𝜎𝑛 , 𝜏𝑛 and 𝜓 we get
𝜏𝑛
1
arccos + 𝑚𝜋,
2
𝑘
𝜎𝑛 1
𝜔=
+ sin 2(𝜗 − 𝜓).
2𝑘 2
𝜗=𝜓±
The presence of two solutions agrees with the quadratic character of the
yield condition. To choose the sign we consider the normal stress 𝜎𝑡 at the
boundary
𝜎𝑡 = 4𝑘𝜔 − 𝜎𝑛 .
The sign of 𝜎𝑡 can sometimes be predetermined, and this enables the correct
choice of the solution.
An important special case occurs when there is no tangential stress at
the boundary Γ (𝜏𝑛 = 0). Then
1
𝜋
arccos 0 + 𝑚𝜋 = 𝜓 ± + 𝑚𝜋,
2
4
𝜋
𝜎𝑛 1
𝜎𝑛 1
+ sin(2𝑚𝜋 ± ) =
± .
𝜔=
2𝑘 2
2
2𝑘 2
𝜗=𝜓±
Further
𝜎𝑡 = 4𝑘𝜔 − 𝜎𝑛 = 𝜎𝑛 ± 2𝑘.
Consider for example a traction free straight boundary shown in Fig.3.12,
where 𝜓 = 90𝑜 , 𝜎𝑛 = 𝜏𝑛 = 0. On this boundary
Figure 3.12: Traction free straight boundary
𝜋 𝜋
± + 𝑚𝜋,
2
4
𝜎 = 2𝑘𝜔 = ±𝑘, 𝜎𝑡 = ±2𝑘.
𝜗=
Depending on whether the fiber in tangential direction is in tension or compression, one chooses the sign “+” or “−”.
In order to study particular plastic flows, we must find the solution of the
above quasilinear hyperbolic system of equations satisfying certain boundary
conditions. As a rule, the corresponding boundary value problems can only
65
3.3. PLANE STRAIN PROBLEMS
A
P´´
P
C
P´
Q
y
B
x
Figure 3.13: Cauchy’s problem
be solved numerically. We now give a brief account of the most important
boundary-value problems.
Boundary-value problems.
1. Cauchy’s problem. Let AB be a smooth arc (described by an arc-length
𝑠), which nowhere coincides with characteristics and intersects each characteristic once only. Let the functions 𝜔(𝑠) and 𝜗(𝑠) be continuous together
with their derivatives on AB. Then the solution exists and is unique in a
triangular region APB, bounded by the arc AB and the 𝛼- and 𝛽-slip lines
originating at A and B. The solution at poit P depends only on the data
along AB, therefore the region APB is called the domain of dependence for
point P.
Note that if derivatives of the initial data are discontinuous at some point
C on the curve AB, then the jumps propagate along the characteristics CP′
and CP′′ .
If the shear stress 𝜏𝑛 is zero at the boundary, the normal to this boundary
is one of the principal directions and the slip lines approach the contour
at an angle of 45𝑜 . Consequently, the contour coincides nowhere with a
characteristics, and we have Cauchy’s problem whose solution is unique.
a
(i,j)
b
A
y
(i+1,j)
(i,j-1)
x
B
Figure 3.14: Determination of 𝜔 and 𝜗
If the boundary is traction-free, then the stress field as well as the grid of
66
CHAPTER 3. THEORY OF PLASTIC FLOW
slip lines in the influence domain depend only on the shape of the boundary.
Provided the values of 𝜎𝑛 and 𝜏𝑛 are given at the boundary 𝐴𝐵, then
we can compute 𝜔 and 𝜗 in each point of this curve. The proper choice of
solution is based on the predetermined sign of 𝜎𝑡 .
Let us denote the nodal points of the grid of characteristics such that
on 𝛼-lines the first index 𝑖 is constant and on 𝛽-lines the second index 𝑗 is
constant. Then according to proposition 2 the quantities 𝜔 and 𝜗 at some
point (𝑖, 𝑗) are given by
1
𝜔(𝑖, 𝑗) = [𝜔(𝑖, 𝑗 − 1) + 𝜔(𝑖 + 1, 𝑗) + 𝜗(𝑖 + 1, 𝑗) − 𝜗(𝑖, 𝑗 − 1)],
2
1
𝜗(𝑖, 𝑗) = [𝜔(𝑖 + 1, 𝑗) − 𝜔(𝑖, 𝑗 − 1) + 𝜗(𝑖 + 1, 𝑗) + 𝜗(𝑖, 𝑗 − 1)].
2
Thus, the grid as well as the stress field can be determined by these equations in finite difference. Note that such the grid of slip lines satisfies the
orthogonality condition only approximately, since the small arcs connecting
nodal points are replaced by the chords.
b
A
y
B
a
x
Figure 3.15: Riemann’s problem
2. Riemann’s problem. Suppose the initial data are known on the segment
AB of the slip line. In this case the domain of influence of AB covers the whole
strip between the slip lines of the other family passing through A and B. In
this case the unique solution must be found by the following requirements:
i) the second slip lines are orthogonal to AB, and ii) 𝜔 + 𝜗 = const along
these slip lines.
3. Mixed boundary value problem. Let us consider now the combination of
the above problems. Namely, assume that 𝜔 and 𝜗 are known on a segment
of the slip line OB. This segment joints a chracteristic curve OA, along which
the angle 𝜗 is prescribed. The solution to the mixed boundary value problem
is determined in the triangle 𝐴𝑂𝐵. The approximate solutions to 2. and 3.
are similar to 1.
67
3.3. PLANE STRAIN PROBLEMS
A
b
B
y
a
x
O
Figure 3.16: Mixed boundary value problem
Velocity field.
If the stress field is known, then the velocity field can also be determined.
According to the flow rule we have
𝜀˙𝑥𝑥 = 𝜆𝑠𝑥𝑥 ,
𝜀˙𝑦𝑦 = 𝜆𝑠𝑦𝑦 ,
𝜀˙𝑥𝑦 = 𝜆𝜎𝑥𝑦 .
From the first two equations
𝜀˙𝑥𝑥 − 𝜀˙𝑦𝑦 = 𝜆(𝜎𝑥𝑥 − 𝜎𝑦𝑦 ).
Combine this with the third equation
(𝜀˙𝑥𝑥 − 𝜀˙𝑦𝑦 )
𝜎𝑥𝑦
= 𝜀˙𝑥𝑦 .
𝜎𝑥𝑥 − 𝜎𝑦𝑦
Taking into account the relation
𝜎𝑥𝑦
1
𝜋
1
1
= tan 2𝜑 = tan(2𝜗 − ) = − cot 2𝜗
𝜎𝑥𝑥 − 𝜎𝑦𝑦
2
2
2
2
and the incompressibility, we obtain
𝑣𝑥,𝑥 + 𝑣𝑦,𝑦 = 0,
(𝑣𝑥,𝑦 + 𝑣𝑦,𝑥 ) tan 2𝜗 + 𝑣𝑥,𝑥 − 𝑣𝑦,𝑦 = 0.
This system of homogeneous quasilinear partial differential equations of first
order is of hyperbolic type, and its characteristics coincide with the slip lines.
With the transformation
𝑣𝑥 = 𝑣𝛼 cos 𝜗 − 𝑣𝛽 sin 𝜗,
𝑣𝑦 = 𝑣𝛼 sin 𝜗 + 𝑣𝛽 cos 𝜗,
68
CHAPTER 3. THEORY OF PLASTIC FLOW
we obtain for the derivatives
𝑣𝑥,𝑥 = 𝑣𝛼,𝑥 cos 𝜗 − 𝑣𝛼 sin 𝜗 𝜗,𝑥 − 𝑣𝛽,𝑥 sin 𝜗 − 𝑣𝛽 cos 𝜗 𝜗,𝑥 ,
𝑣𝑦,𝑦 = 𝑣𝛼,𝑦 sin 𝜗 + 𝑣𝛼 cos 𝜗 𝜗,𝑦 + 𝑣𝛽,𝑦 cos 𝜗 − 𝑣𝛽 sin 𝜗 𝜗,𝑦 .
Choosing in an arbitrary point 𝑃 (𝑥, 𝑦) the directions of the 𝑥-𝑦-axes such
that they coincide with the directions of the slip lines (𝑠𝛼 , 𝑠𝛽 ), and recalling
that
sin 𝜗 = 0, cos 𝜗 = 1
we obtain
𝑣𝑥,𝑥 = 𝑣𝛼,𝑥 − 𝑣𝛽 𝜗,𝑥 = 0 (𝛼),
𝑣𝑦,𝑦 = 𝑣𝛼 𝜗,𝑦 + 𝑣𝛽,𝑦 = 0 (𝛽).
These equations are called Geiringer’s equations for the velocity field.
If the grid of the slip lines is known from the solution of the stress problem,
then the velocity field can also be determined from these equations for all
three types of the boundary value problems.
Note that the line separating plastic and rigid regions is a slip line or
the envelope of slip lines. To show this we first assume that the velocities
are continuous on the line of separation. Then 𝑣𝑥 = 𝑣𝑦 = 0 on this line.
If the boundary has nowhere a characteristic direction, then the solution of
Cauchy’s problem gives 𝑣𝑥 = 𝑣𝑦 = 0 everywhere in the plastic region, which
contradicts the original assumption. Consider now the case when the velocity
is discontinuous on the line of separation. The discontinuity can only be in
𝑣𝑡 , otherwise a crack appears. One can then show that ∣𝜎𝑡 ∣ → 𝑘. Thus, the
boundary will be a slip line or an envelope of slip lines.
Using Geiringer’s equation and considering the velocity at some nodal
point (𝑖, 𝑗) of the grid of slip line, we can easily obtain the following approximate finite difference formulas
𝑣𝛼 (𝑖, 𝑗) − 𝑣𝛼 (𝑖, 𝑗 − 1) = 𝑣𝛽 (𝑖, 𝑗)[𝜗(𝑖, 𝑗) − 𝜗(𝑖, 𝑗 − 1)],
𝑣𝛽 (𝑖 + 1, 𝑗) − 𝑣𝛽 (𝑖, 𝑗) = −𝑣𝛼 (𝑖, 𝑗)[𝜗(𝑖 + 1, 𝑗) − 𝜗(𝑖, 𝑗)].
For statically determinate problems, i. e. when the static boundary conditions are specified at the boundary Γ, the grid of slip lines as well as the
stress field can be determined numerically, so the velocity field can be found
from these formulas.
For statically undeterminate problems, which have not been considered
up to now, the coupling cannot be removed. To construct the grid of slip
lines all four difference equations need to be solve simultaneously.
69
3.3. PLANE STRAIN PROBLEMS
For problems with mixed boundary conditions (where the tractions are
specified on one part of the boundary and the velocity on the remaining part)
we find the grid of the slip lines as well as the stress and velocity fields in
each of the zone of influence as above. The initial data for the velocity can be
obtained from the transition conditions at the boundary between the zones
of influence.
Family of straight slip lines.
Many of the above statements become quite simple for the family of straight
slip lines.
Due to the Proposition 4 there are only two possible grids with straight
slip lines
∙ grid with two orthogonal families of straight slip lines,
∙ grid with one family of straight slip lines and the curved slip lines
orthogonal to them.
In any case
𝜔 − 𝜗 = const,
𝜔 + 𝜗 = const,
𝛼 − line,
𝛽 − line.
Since 𝜗 remains constant along a straight line, 𝜔 and 𝜎 must also be constant
there.
For the first case we have a uniform stress state in the whole domain and
the velocity field of the form
𝑣𝑥 = 𝑓 (𝑦),
𝑣𝑦 = 𝑔(𝑥).
A domain in which one family of slip lines contains only straight line is
called a fan. We distinguish between central and non-central fans.
The central fan has the 𝛽-lines as concentric circles. Here the envelope
degenerate to a point O which is the middle point of all circles. In this case
𝜗 = const ⇒ 𝜔 = const on 𝛼-line,
and
𝜔 + 𝜗 = const on 𝛽-line.
Consequently
𝜔 + 𝜗 = const in the whole domain.
70
CHAPTER 3. THEORY OF PLASTIC FLOW
Because 𝜔 = 𝑐 − 𝜗 is independent of 𝑟, we obtain for the stresses
𝜎𝑟 = 𝜎𝜑 = 2𝑘𝜔 = 2𝑘(𝑐 − 𝜗),
𝜎𝑟𝜑 = 𝑘.
With the help of Geiringer’s equation we compute the velocity
𝑣𝑟 = −𝑓 (𝜗),𝜗 ,
𝑣𝜑 = 𝑓 (𝜗) + 𝑔(𝑟).
For the non-central fan the similar results can also be obtained.
3.4
Plane stress problems
Governing equations
We are dealing with a plane stress state when, in some cartesian coordinate
system,
𝜎𝑥𝑥
𝜎𝑦𝑦
𝜎𝑥𝑦
𝜎𝑧𝑧
= 𝜎𝑥𝑥 (𝑥, 𝑦),
= 𝜎𝑦𝑦 (𝑥, 𝑦),
= 𝜎𝑥𝑦 (𝑥, 𝑦),
= 𝜎𝑥𝑧 = 𝜎𝑦𝑧 = 0,
and, accordingly
⎞
⎛
𝜎𝑥𝑥 𝜎𝑥𝑦 0
𝜎𝑖𝑗 = ⎝𝜎𝑥𝑦 𝜎𝑦𝑦 0⎠ ,
0
0 0
1
𝜎 = (𝜎𝑥𝑥 + 𝜎𝑦𝑦 ).
3
For the principal stresses in the (𝑥, 𝑦)-plane we have as before
1√
1
2 .
(𝜎𝑥𝑥 − 𝜎𝑦𝑦 )2 + 4𝜎𝑥𝑦
𝜎1,2 = (𝜎𝑥𝑥 + 𝜎𝑦𝑦 ) ±
2
2
These principal stresses occur in a coordinate system which is rotated through
the angle
1
2𝜎𝑥𝑦
𝜑 = arctan
2
𝜎𝑥𝑥 − 𝜎𝑦𝑦
with respect to the original coordinate system.
This plane stress state must fulfill in the plastic zone the yield condition,
which is either i) Mises
1
[(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ] − 𝑘 2 = 0,
6
71
3.4. PLANE STRESS PROBLEMS
with the consequence
𝜎12 + 𝜎22 − 𝜎1 𝜎2 − 3𝑘 2 = 0,
or ii) Tresca
1
max{∣𝜎1 − 𝜎2 ∣, ∣𝜎2 − 𝜎3 ∣, ∣𝜎3 − 𝜎1 ∣} − 𝜏𝑌 = 0,
2
with the consequence
∣𝜎1 − 𝜎2 ∣ − 2𝜏𝑌 = 0, 𝜎1 𝜎2 ≤ 0,
∣𝜎1 ∣ − 2𝜏𝑌 = 0, 𝜎1 𝜎2 ≥ 0, ∣𝜎1 ∣ > ∣𝜎2 ∣,
∣𝜎2 ∣ − 2𝜏𝑌 = 0, 𝜎1 𝜎2 ≥ 0, ∣𝜎2 ∣ > ∣𝜎1 ∣.
Figure 3.17: Yield surface for the plane stress state
The local stress state at an arbitrary point is fixed by three components
𝜎𝑥𝑥 , 𝜎𝑦𝑦 , 𝜎𝑥𝑦 . For them the yield condition as well as the two remaining
equilibrium conditions
𝜎𝑥𝑥,𝑥 + 𝜎𝑥𝑦,𝑦 = 0,
𝜎𝑥𝑦,𝑥 + 𝜎𝑦𝑦,𝑦 = 0,
hold true. The problem is therefore statically determinate. However, in
contrast to the plane strain problems the above two yield conditions lead to
different results.
Solution based on Mises’ yield condition
In addition to the rotation angle 𝜑 let us introduce a new angle 𝜔
cos 𝜔 =
√ 𝜎
3 ,
2𝑘
72
CHAPTER 3. THEORY OF PLASTIC FLOW
so that, with the help of
𝜎1,2 = 2𝑘 cos(𝜔 ∓
𝜋
)
6
the flow rule is satisfied identically.
Then
√
𝜎𝑥𝑥 = 𝑘( 3 cos 𝜔 + sin 𝜔 cos 2𝜑),
√
𝜎𝑦𝑦 = 𝑘( 3 cos 𝜔 − sin 𝜔 cos 2𝜑),
𝜎𝑥𝑦 = 𝑘 sin 𝜔 sin 2𝜑.
Substituting these formulas into the equilibrium conditions we obtain finally
√
√
(cos 𝜔 − 3 sin 𝜔 cos 2𝜑)𝜔,𝑥 − 3 sin 𝜔 sin 2𝜑𝜔,𝑦 + 2 sin 𝜔𝜑,𝑦 = 0,
√
√
3 sin 𝜔 sin 2𝜑𝜔,𝑥 − (cos 𝜔 + 3 sin 𝜔 cos 2𝜑)𝜔,𝑦 + 2 sin 𝜔𝜑,𝑥 = 0.
With
√
𝑎 = −2 sin 𝜔(cos 𝜔 − 3 sin 𝜔 cos 2𝜑),
√
𝑏 = 2 sin2 𝜔 3 sin 2𝜑,
√
𝑐 = −2 sin 𝜔(cos 𝜔 + 3 sin 𝜔 cos 2𝜑),
we derive for the discriminant
𝑏2 − 𝑎𝑐 = 4 sin2 𝜔(3 − 4 cos2 𝜔) = 4 sin2 𝜔Ω(𝜔).
Depending on the sign of Ω(𝜔) we get the hyperbolic, parabolic or elliptic
system of partial differential equations (see Fig. 3.18).
1
0.75
1
2
3
4
w
hyperbolic elliptic
Figure 3.18: Graph of function cos2 𝜔
For the characteristic directions we have
√
√
3 sin 𝜔 sin 2𝜑 ± Ω(𝜔)
𝑑𝑦
√
.
=−
𝑑𝑥
cos 𝜔 − 3 sin 𝜔 cos 2𝜑
In the hyperbolic case the problem can be solved with the help of characteristics.
Chapter 4
Crystal plasticity
This Chapter discusses crystal plasticity. We aim at studying the dislocation
nucleation and accumulation, the resulting work hardening and the influence
of the resistance to dislocation motion. Among various problems we select
antiplane shear and plane constrained shear of single crystals which admit
analytical solutions. The interesting features of these solutions are the energetic and dissipative thresholds for dislocation nucleation, the Bauschinger
translational work hardening, and the size effects.
4.1
Physical background
One of the remarkable properties of crystals is their ability to glide easily
on certain crystallographic planes along certain crystallographic directions.
The simplest way to see this is to do a tension test for a bar made of a single
crystal whose slip planes are inclined at some angle, say, 45∘ to the bar axis
(see Fig. 4.1a). By measuring the load 𝐹 and the corresponding elongation
Δ𝑙 of the bar, we can plot the stress 𝜎 = 𝐹/𝐴 versus strain 𝜀 = Δ𝑙/𝑙 curve,
which is shown (schematically) in Fig. 4.1b.
At small stresses the bar deforms elastically, and the stress-strain curve
is nearly linear. If the force is removed, the bar length returns to its initial
value 𝑙 so that Δ𝑙 = 0. Beginning from a critical value 𝜎𝑌 , called yield stress,
the bar deforms plastically and then, at larger stresses, easily stretches out
to a narrow ribbon (necking) which may be four or five times longer than the
original length. If one examines the surface of the plastically deformed bar,
steps can be seen which run more or less continuously around the boundary
in the form of ellipses. These steps appear as witnesses of the plastic slips
on the active slip planes (see Fig. 4.2).
It is observed experimentally that plastic slips occur on the close packed
73
74
CHAPTER 4. CRYSTAL PLASTICITY
F
s
sY
e
-F
a)
b)
Figure 4.1: Tensile test and the stress-strain curve
slip
plane
a
b
Figure 4.2: Magnified schematic view of plastic slips: a) front view, b) side
view
planes along the directions of shortest interatomic distances. To recognize
these planes and directions one needs to consider the periodic lattice structure
of the crystal. We see for example in Fig. 4.3 the unit cell of a body-centered
cubic (bcc) crystal. The unit cell is characterized by three lattice vectors a1 ,
a2 , and a3 which, for bcc crystals, are mutually orthogonal and have an equal
magnitude. The translations of the unit cell by 𝑛1 a1 + 𝑛2 a2 + 𝑛3 a3 for all
integers 𝑛1 , 𝑛2 , 𝑛3 will develop the regular periodic lattice structure. In each
unit cell of the bcc crystal there are eight corner atoms and one atom located
at its center. Crystals may have also other lattice structures, for instance,
face-centered cubic (fcc) or hexagonal closed-packed (hcp) lattice structures.
When referring to crystallographic directions and planes, we need the
way to indicate them. We shall use Miller indices for this purpose. For
example, the Miller indices of the direction 𝑑 in Fig. 4.3 are [111], while
those of the direction 𝑠 in the same Figure are [101], where the negative sign
of the 𝑥1 -component is indicated by a bar over the corresponding index. We
define the Miller indices of a family of directions as a set of three integers
75
4.1. PHYSICAL BACKGROUND
𝑛1 , 𝑛2 , 𝑛3 , enclosed in square brackets, which indicates all directions parallel
to the vector 𝑛1 a1 + 𝑛2 a2 + 𝑛3 a3 . Since 𝑛1 , 𝑛2 , 𝑛3 are defined uniquely up
to a common integer factor, we will choose the smallest multiples of them
for the Miller indices. For illustration let us apply this definition to find the
Miller indices of the direction 𝑒 in Fig. 4.3. Since the vector drawn from the
origin parallel to 𝑒 has the smallest integers 1,-1,0 as its components, we get
for the direction 𝑒 the Miller indices [110]. Crystallographic planes are also
indicated by sets of integers. For example, the Miller indices for the (𝑥1 , 𝑥2 ),
(𝑥2 , 𝑥3 ), and (𝑥3 , 𝑥1 )-planes are (001), (100), and (010), respectively. Note
that the Miller indices for planes are enclosed in parentheses instead of square
brackets as for directions. We define the Miller indices of a family of planes
as a set of three integers 𝑛1 , 𝑛2 , 𝑛3 , enclosed in parentheses, which indicates
all planes perpendicular to the vector 𝑛1 a1 + 𝑛2 a2 + 𝑛3 a3 . Since the normal
vector 𝑛1 a1 + 𝑛2 a2 + 𝑛3 a3 is defined uniquely up to a common integer factor,
we choose again the smallest multiples of 𝑛1 , 𝑛2 , 𝑛3 for the Miller indices. For
illustration let us apply this definition to find the Miller indices of the plane
shown in Fig. 4.3. The normal vector to this plane is obviously a1 + a2 + a3 ,
so the Miller indices of this plane are (111).
x3
a3
s
d
a2
a1
e
x2
x1
Figure 4.3: Unit cell of bcc crystals
In terms of these Miller indices a precise description of the plastic slip
can be given. For example, the plastic slip may occur in bcc crystals on the
slip planes of type (110) along the directions [111]. For fcc crystals, the slip
planes are of type (111), while the slip directions of type [110]. Each slip
system is characterized by the family of slip planes and slip directions. For
fcc crystals there are 12 slip systems.
The critical shear stress of a perfect crystal at which the plastic slip occurs
can roughly be estimated as follows. Consider the glide of the upper half of
crystal relative to its lower half under a shear stress 𝜏 . The two adjacent rows
of atoms in the initial state are shown in Fig. 4.4 in which the interplanar
spacing is 𝑎 while the interatomic distance in the slip direction is 𝑏.
Denoting the relative displacement by 𝑢, the applied shear stress must be
76
CHAPTER 4. CRYSTAL PLASTICITY
b
t
a
E
u
Figure 4.4: Perfect crystal and energy vs. relative displacement
calculated in accordance with 𝜏 = 𝑑𝐸/𝑑𝑢, where 𝐸 is the energy of crystal
in terms of 𝑢. Due to the periodicity of the lattice structure, 𝐸 must be
a periodic function with the period 𝑏, having a minimum at 𝑢 = 0 and a
maximum at 𝑢 = 𝑏/2 (see Fig. 4.4). The simplest formula for 𝐸 satisfying
these requirements can be proposed as follows
(
)
2𝜋𝑢
𝐾𝑏
1 − cos
.
𝐸(𝑢) =
2𝜋
𝑏
The derivative of 𝐸 yields the shear stress
𝜏=
𝑑𝐸
2𝜋𝑢
= 𝐾 sin
.
𝑑𝑢
𝑏
For small displacement 𝑢 ≪ 𝑏 we have approximately 𝜏 = 𝐾2𝜋𝑢/𝑏. Comparing this with the Hooke’s law 𝜏 = 𝜇𝑢/𝑎, with 𝜇 being the shear modulus,
we obtain the value of 𝐾 which coincides with the maximum shear stress
achieved at 𝑢 = 𝑏/2
𝜇𝑏
𝜏𝑡 = 𝐾 =
.
2𝜋𝑎
In typical situations we have 𝑏 = 𝑎, so the theoretical shear strength is
estimated to be
𝜇
.
𝜏𝑡 ≈
2𝜋
For most of crystals this theoretically estimated shear strength is three or
four order of magnitude larger than the experimentally observed value of 𝜎𝑌 .
In 1934 three scientists, Taylor, Orowan, and Polanyi, simultaneously
came to the idea that the plastic slip can so easily occur at very low shear
stress because of dislocations which are the line defects in crystals. One
should mention that, actually, the theory of dislocations in solids was originally developed by Volterra in 1905. But only the discovery by these pioneers
gave birth to the dislocation based plasticity. In Fig. 4.5 one can see the
so-called edge dislocation in a simple cubic lattice structure which can be
77
4.1. PHYSICAL BACKGROUND
C
D
B
A
Figure 4.5: Edge dislocation in primitive cubic crystal
regarded as the extra half-plane of atoms inside the crystal (the half-plane
ABCD). The distortion of the crystal is localized near the edge of this halfplane which is called the dislocation line (the line AB). Far away from the
dislocation line the crystal has a nearly perfect lattice structure.
t
t
t
t
Figure 4.6: Migration of dislocation through crystal
Fig. 4.6 illustrates how the edge dislocation migrates through the crystal
from left to right under an applied shear stress 𝜏 . This migration, from one
position to the next, involves only a small rearrangement of the atomic bonds
near the dislocation line. The final results of these many small steps is the
plastic slip considered before. But the shear stress to break the atomic bond
and move dislocation in one interatomic distance is much lower that that
required for the simultaneous movement of the whole upper half of crystal.
Let us now consider the creation of a single dislocation in a simple cubic
crystal. Cut this crystal along any of the surfaces indicated in Fig. 4.7a,b,c.
Let the atoms on one side of the cut shift in a direction parallel to the cut
through a distance equal to one lattice spacing. Then rejoin the atoms on
both sides of the cut and let the crystal be elastically relaxed. The lattice
structure is again almost perfect except near the boundary AB of the cut
surface called dislocation line. If the atoms below the cut are shifted in the
direction parallel to the line AB, a screw dislocation is created; if the shift is
perpendicular to AB, an edge dislocation is produced; if the shift is neither
parallel nor perpendicular to AB, the created dislocation is of the mixed type.
So, each dislocation is characterized by the dislocation line and the vector
78
CHAPTER 4. CRYSTAL PLASTICITY
pointing in the direction of shift whose magnitude is equal to one lattice
spacing. This vector is called Burgers’ vector.
B
A
D
C
a)Screw
B
D
A
C
b)Edge
B
A
C
c)Mixed
Figure 4.7: Creation of dislocation in a simple cubic crystal
11
12
10
7
14
6
2
3
a)
4
9
8
13
1
11 10
12
9
5
8
7
13
14
1
b
6
2
3
5
4
b)
Figure 4.8: Burgers’ circuit around an edge dislocation
The definition of Burgers’ vector depends on the sense of a dislocation
line. We define the sense of the dislocation line by assigning a unit vector 𝝉
tangent to the dislocation line and taking the positive sense in the positive
direction of 𝝉 . Burgers’ vector can be determined by the following geometric
construction. Consider two circuits drawn in Fig. 4.8a,b. The left close circuit
is drawn in the reference dislocation-free crystal; the right is drawn around
an edge dislocation in real crystal. When moving along each circuit in the
positive direction agreeing with the sense of the dislocation line (anticlockwise
4.2. CONTINUUM DISLOCATION THEORY
79
if the positive sense of the dislocation line is taken to be out of the paper)
the same number of jumps from atom to atom are made to the right as to
the left, up as down. The starting and the end points correspond to the
same atom in the case of the circuit in the reference crystal. However the
starting and end points are not the same atom for the circuit that includes
the dislocation in the real crystal. Thus, there is a closure failure (or a misfit)
in this circuit defined as Burgers’ vector.
4.2
Continuum dislocation theory
Macroscopically observable plastic deformation of single and polycrystals is
produced essentially by the motion of a large number of dislocations. On the
other side, these newly formed dislocations in crystals pile up near various
obstacles like grain or phase boundaries, or particulate inclusions, giving rise
to size dependent hardening of the material. Dislocations appear in the deformed crystal lattice to reduce its energy. Motion of dislocations generates
the dissipation of energy which, in turn, results in a resistance to dislocation motion. The understanding of nucleation mechanism and the motion
of dislocations is therefore a cornerstone for describing plastic yielding, work
hardening, and hysteresis effects in crystal plasticity.
Furthermore, dislocations are not only a key microstructural defect for
plastic slip but also the core ingredient for forming microstructural patterns
and substructures. There are numerous examples illustrating this. The first
one is the formation of lamellar twin patterns in manganese steels and other
twinning induced plasticity (TWIP)-alloys, which has significant impact on
the macroscopic stress-strain response. The formation of twins provides
TWIP-alloys with excellent hardening behavior, allowing for higher stresses
and larger strains than in common fcc or bcc metals. The other example
is the recrystallization produced by severe plastic deformation during equal
channel angular extrusion which leads to almost dislocation-free grains of
an average diameter of a few hundred nanometers, yielding materials with
exceptional room-temperature strength.
From the above examples, one can clearly see the importance of being
able to predict the nucleation and motion of dislocations in crystals. The
high dislocation densities accompanying plastic deformation, in the range of
108 -1015 m−2 , along with the complexity of the dislocation network necessitate
the continuum approach to dislocations, which is the subject of this Chapter.
Continuum dislocation theory deals with the ensembles of a huge number of
dislocations by the methods of continuum mechanics. The complexity of the
system makes the phenomenological approach unavoidable. However, one
80
CHAPTER 4. CRYSTAL PLASTICITY
needs some guiding principles to limit the feasible choices of the available
models. Such guiding principles are the laws of thermodynamics.
e
e
e
p
e =e +e
m
s
p
e
Figure 4.9: Additive decomposition of the total strain
Consider first a crystal deforming in single slip. The strains of an infinitesimal element of the crystal is additively decomposed into the plastic
strains and the elastic strains as shown schematically in Fig. 4.9. The plastic strain tensor is the symmetric part of the plastic distortion which maps
the undeformed crystal to the reference stress-free crystal. The subsequent
elastic strains deform the crystal in accordance with the Hooke’s law. Both
plastic and elastic strains are incompatible so that the sum of them becomes
compatible and derivable from a displacement field. For a large number of
loops lying on the parallel slip planes with the mean distance between them
being much smaller than the characteristic size of the specimen, we propose a
spatial average description of plastic distortion produced by this slip system
as follows
𝛽𝑖𝑗 = 𝛽(x)𝑠𝑖 𝑚𝑗 ,
with s the unit vector pointing in the slip direction, and m the normal vector
to the slip plane. The essential difference to the case of discrete dislocations
is that now function 𝛽(x) is assumed to be continuously differentiable. If the
crystal has 𝑛 active slip systems, the plastic distortion is the sum of those
produced by these slip systems
𝛽𝑖𝑗 =
𝑛
∑
𝔞=1
𝛽𝔞 (x)𝑠𝔞𝑖 𝑚𝔞𝑗 .
4.2. CONTINUUM DISLOCATION THEORY
81
The small Gothic index 𝔞 indicating the slip systems runs from 1 to 𝑛. One
can see that, in general, 𝛽𝑖𝑖 = 0, consequently continuous plastic distortions
do not cause any volume change.
The plastic strains 𝜀𝑝𝑖𝑗 and the plastic rotations 𝜔𝑖𝑗 are the symmetric and
skew-symmetric parts of the plastic distortion
1
𝜀𝑝𝑖𝑗 = (𝛽𝑖𝑗 + 𝛽𝑗𝑖 ),
2
1
𝜔𝑖𝑗 = (𝛽𝑖𝑗 − 𝛽𝑗𝑖 ).
2
The elastic strains are then the differences between the total compatible
strains and the incompatible plastic strains
1
𝜀𝑒𝑖𝑗 = 𝜀𝑖𝑗 − 𝜀𝑝𝑖𝑗 = (𝑢𝑖,𝑗 + 𝑢𝑗,𝑖 ) − 𝜀𝑝𝑖𝑗 ,
2
with 𝑢𝑖 being the components of the displacement vector.
Nye introduced an important characteristic of dislocations, the dislocation
density tensor1
𝛼𝑖𝑗 = 𝜖𝑗𝑘𝑙 𝛽𝑖𝑙,𝑘 .
(4.1)
For a single dislocation loop the dislocation density tensor has the following
physical meaning: if we take an arbitrary infinitesimal surface 𝑑𝑎 with the
unit normal 𝝉 which is the tangent vector to the dislocation line crossing
this surface, 𝛼𝑖𝑗 𝑛𝑗 𝑑𝑎 gives the Burgers’ vector of this dislocation (see Section
3.4). Within the continuum dislocation theory, we interpret this quantity as
the resultant Burgers’ vector of all dislocations whose dislocation lines cross
the surface 𝑑𝑎. For a crystal deforming in single slip 𝛼𝑖𝑗 𝑛𝑗 = 𝑠𝑖 𝜖𝑗𝑘𝑙 𝛽,𝑘 𝑚𝑙 𝑛𝑗 ,
so the resultant Burgers’ vector turns out to be parallel to the slip direction.
The number of dislocations per unit area can then be computed as
1
𝜌 = ∣𝜖𝑗𝑘𝑙 𝛽,𝑘 𝑚𝑙 𝑛𝑗 ∣,
𝑏
(4.2)
with 𝑏 the magnitude of Burgers’ vector.
Let ℬ be any regular sub-region of the crystal in its initial state. The free
energy of the crystal confined in the region ℬ reads
∫
Ψ=
𝜙(𝜀𝑖𝑗 , 𝛽𝑖𝑗 , 𝛼𝑖𝑗 , 𝜃) 𝑑𝑥,
ℬ
with 𝜃 being the absolute temperature which is assumed to be constant.
Under this assumption the laws of thermodynamics state that the rate of
1
In fact, Nye introduced the dislocation density tensor produced by a plastic distortion
in form of rotation only. Formula (4.1) was proposed independently by Bilby and Kröner.
82
CHAPTER 4. CRYSTAL PLASTICITY
change of the free energy minus the power of the external forces must be
non-positive for arbitrary processes
∫
𝑑
𝜙(𝜀𝑖𝑗 , 𝛽𝑖𝑗 , 𝛼𝑖𝑗 , 𝜃) 𝑑𝑥 − 𝑃 ≤ 0.
(4.3)
Ψ̇ − 𝑃 =
𝑑𝑡 ℬ
The structure of power must be controlled by the form of the energy. In our
case the power is given by
∫
𝑃 =
(𝜎𝑖𝑗 𝑛𝑗 𝑢˙ 𝑖 + 𝜎𝑖𝑗𝑘 𝑛𝑘 𝛽˙ 𝑖𝑗 ) 𝑑𝑎,
(4.4)
∂ℬ
where ∂ℬ is the boundary of ℬ with 𝑛𝑖 being the unit outward normal to
∂ℬ. We see that some stresses of higher order enter the theory as a result
of the dependence of the free energy density on the gradient of the plastic
distortion.
Transforming the surface integral in (4.4) into a volume integral by Gauss’
theorem and requiring that (4.3) is satisfied for an arbitrary ℬ, we obtain
the inequality
(
)
(
)
∂𝜙
∂𝜙
− 𝜎𝑖𝑗 𝑢˙ 𝑖,𝑗 +
− 𝜎𝑖𝑗𝑘,𝑘 𝛽˙ 𝑖𝑗
∂𝜀𝑖𝑗
∂𝛽𝑖𝑗
(
)
∂𝜙
+
𝜖𝑚𝑘𝑗 − 𝜎𝑖𝑗𝑘 𝛽˙ 𝑖𝑗,𝑘 − 𝜎𝑖𝑗,𝑗 𝑢˙ 𝑖 ≤ 0. (4.5)
∂𝛼𝑖𝑚
For rigid translations the energy does not change while 𝑢˙ 𝑖,𝑗 , 𝛽˙ 𝑖𝑗 and 𝛽˙ 𝑖𝑗,𝑘
are zero. Since the inequality (4.5) must be fulfilled for an arbitrary translation, the stress must obey the equilibrium equation
𝜎𝑖𝑗,𝑗 = 0.
(4.6)
Similarly, the inequality (4.5) can be satisfied for arbitrary rigid rotations
only if the stress tensor is symmetric
𝜎𝑖𝑗 = 𝜎𝑗𝑖 .
(4.7)
Let us introduce the following notation
𝜏𝑖𝑗 = 𝜎𝑖𝑗 −
∂𝜙
,
∂𝜀𝑖𝑗
ϰ𝑖𝑗 = −
𝜏𝑖𝑗𝑘 = 𝜎𝑖𝑗𝑘 −
∂𝜙
+ 𝜎𝑖𝑗𝑘,𝑘 .
∂𝛽𝑖𝑗
∂𝜙
𝜖𝑚𝑘𝑗 ,
∂𝛼𝑖𝑚
(4.8)
(4.9)
83
4.2. CONTINUUM DISLOCATION THEORY
Then the combined first and second laws of thermodynamics become
𝜏𝑖𝑗 𝑢˙ 𝑖,𝑗 + ϰ𝑖𝑗 𝛽˙ 𝑖𝑗 + 𝜏𝑖𝑗𝑘 𝛽˙ 𝑖𝑗,𝑘 ≥ 0.
(4.10)
Equation (4.10) shows that 𝜏𝑖𝑗 and 𝜏𝑖𝑗𝑘 are those parts of the stresses and
the higher order stresses which cause energy dissipated in heating of the
crystal. Tensor 𝜏𝑖𝑗 describes heating in a non-uniform flow, so it has the
meaning of viscous stresses. Tensors ϰ𝑖𝑗 and 𝜏𝑖𝑗𝑘 describe heating caused by
homogeneous and inhomogeneous plastic deformation, respectively.
The widely used closure of non-equilibrium thermodynamics assumes that
there exists a dissipation potential
𝐷 = 𝐷(𝑢˙ 𝑖,𝑗 , 𝛽˙ 𝑖𝑗 , 𝛽˙ 𝑖𝑗,𝑘 )
(4.11)
such that the tensors 𝜏𝑖𝑗 , ϰ𝑖𝑗 , and 𝜏𝑖𝑗𝑘 controlling the irreversible processes
are linked to 𝑢˙ 𝑖,𝑗 , 𝛽˙ 𝑖𝑗 , and 𝛽˙ 𝑖𝑗,𝑘 by the relations
𝜏𝑖𝑗 =
∂𝐷
,
∂ 𝑢˙ 𝑖,𝑗
ϰ𝑖𝑗 =
∂𝐷
,
∂ 𝛽˙ 𝑖𝑗
𝜏𝑖𝑗𝑘 =
∂𝐷
.
∂ 𝛽˙ 𝑖𝑗,𝑘
(4.12)
The set of equations (4.6),(4.7),(4.8),(4.9), and (4.12) is closed with respect
to the unknown functions 𝑢𝑖 and 𝛽𝑖𝑗 .
Thus, each continuum model of dislocations is fixed by two functions,
namely, the free energy density and the dissipation potential. For isothermal
processes we require the free energy density to depend only on the elastic
strain 𝜀𝑒𝑖𝑗 and on the dislocation density 𝛼𝑖𝑗 :
1
𝜙(𝜀𝑖𝑗 , 𝛽𝑖𝑗 , 𝛼𝑖𝑗 ) = 𝐶𝑖𝑗𝑘𝑙 𝜀𝑒𝑖𝑗 𝜀𝑒𝑘𝑙 + 𝜙𝑚 (𝛼𝑖𝑗 ),
2
(4.13)
where 𝜙𝑚 (𝛼𝑖𝑗 ) corresponds to the energy of the dislocation network. In this
case the stresses are given by
𝜎𝑖𝑗 =
∂𝜙
= 𝐶𝑖𝑗𝑘𝑙 𝜀𝑒𝑘𝑙 .
𝑒
∂𝜀𝑖𝑗
The crucial question is then how the energy of the dislocation network depends on the dislocation density. For a single crystal deforming in single slip
we shall adopt the following formula
𝜙𝑚 = 𝑘𝜇 ln
1
,
1 − 𝜌/𝜌𝑠
(4.14)
where 𝜌𝑠 is the saturated dislocation density and 𝑘 a material constant. The
logarithmic energy stems from two facts: i) for small dislocation densities the
84
CHAPTER 4. CRYSTAL PLASTICITY
energy of the dislocation network must be proportional to the dislocation density, and ii) there exists a saturated dislocation density which characterizes
the closest packing of dislocations of equal signs admissible in the discrete
crystal lattice. The logarithmic term ensures a linear increase of the energy
for small dislocation density 𝜌 and tends to infinity as 𝜌 approaches the saturated dislocation density 𝜌𝑠 hence providing an energetic barrier against
over-saturation.
Concerning the dissipation potential several models can be considered.
The simplest model assumes that the dissipation is zero. In this case all
tensors 𝜏𝑖𝑗 , ϰ𝑖𝑗 , and 𝜏𝑖𝑗𝑘 vanish, and functions 𝑢𝑖 and 𝛽𝑖𝑗 should be found
from energy minimization. The next model, also quite simple, neglects the
viscous effect as well as the dissipation caused by 𝛽˙ 𝑖𝑗,𝑘 . In this model 𝐷 is
assumed to depend only on 𝛽˙ 𝑖𝑗 so that 𝜏𝑖𝑗 = 0 and 𝜏𝑖𝑗𝑘 = 0 and
𝜎𝑖𝑗 =
∂𝜙
,
∂𝜀𝑖𝑗
𝜎𝑖𝑗𝑘 =
∂𝜙
𝜖𝑚𝑘𝑗 .
∂𝛼𝑖𝑚
(4.15)
If, furthermore, 𝐷 is a homogeneous function of first order with respect to
𝛽˙ 𝑖𝑗 , then the evolution equation for 𝛽𝑖𝑗 becomes
ϰ𝑖𝑗 =
∂𝐷
.
∂ 𝛽˙ 𝑖𝑗
(4.16)
Note that, for the free energy density in form (4.13), equation (4.16) can be
written in a “variational” form
∂𝜙 𝛿𝜀 𝜙
∂𝐷
∂ ∂𝜙
≡−
.
(4.17)
=−
+
𝛿𝛽𝑖𝑗
∂𝛽𝑖𝑗 𝜀𝑖𝑗 ∂𝑥𝑘 ∂𝛽𝑖𝑗,𝑘
∂ 𝛽˙ 𝑖𝑗
We will consider in what follows only these simplified dissipation potentials.
4.3
Anti-plane constrained shear
We start as always with the simplest problem of a beam made of a single
crystal undergoing an anti-plane shear deformation. Let 𝒞 be the cross section of the beam by planes 𝑧 = const. For simplicity, we consider 𝒞 to be
a rectangle of width 𝑎 and height ℎ, 0 < 𝑥 ≤ 𝑎, 0 < 𝑦 ≤ ℎ. We place the
crystal in a “hard” device with the prescribed displacement at the boundary
∂𝒞 × [0, 𝐿] (see Fig. 4.10)
𝑤 = 𝛾𝑦
at ∂𝒞 × [0, 𝐿],
85
4.3. ANTI-PLANE CONSTRAINED SHEAR
y
x
h
L
a
z
Figure 4.10: Anti-plane constrained shear
where 𝑤(𝑥, 𝑦) is the 𝑧-component of the displacement vector and 𝛾 corresponds to the overall shear strain. The height of the cross section, ℎ, and
the length of the beam, 𝐿, are assumed to be much larger than the width
𝑎 (𝑎 ≪ ℎ, 𝑎 ≪ 𝐿) to neglect the end effects and to have the stresses and
strains depending only on one variable 𝑥 in the central part of the beam. If
the shear strain is sufficiently small, then the crystal deforms elastically and
𝑤 = 𝛾𝑦 everywhere in the beam. If 𝛾 exceeds some critical value, then the
screw dislocations may appear. We allow only the slip planes parallel to the
plane 𝑦 = 0 and the dislocation lines parallel to the 𝑧-axis. Our aim is to
determine the distribution of dislocations as function of 𝛾 within the framework of continuum theory of dislocations proposed in Section 4.2. For screw
dislocations with the slip planes parallel to the plane 𝑦 = 0, the tensor of
plastic distortion, 𝛽𝑖𝑗 , has only one non-zero component 𝛽𝑧𝑦 ≡ 𝛽. We assume
that 𝛽 depends only on 𝑥-coordinate: 𝛽 = 𝛽(𝑥). Since the displacements are
prescribed at the boundary of the crystal, dislocations cannot penetrate the
boundaries 𝑥 = 0 and 𝑥 = 𝑎, therefore
𝛽(0) = 𝛽(𝑎) = 0.
The plastic strains are given by
1
𝜀𝑝𝑦𝑧 = 𝜀𝑝𝑧𝑦 = 𝛽(𝑥).
2
The only non-zero component of Nye’s tensor of dislocation density is
𝛼𝑧𝑧 = 𝛽,𝑥 .
(4.18)
86
CHAPTER 4. CRYSTAL PLASTICITY
The free energy density of the crystal with dislocations takes a simple form
1
1
,
𝜙 = 𝜇(𝛾 − 𝛽)2 + 𝜇𝑘 ln
2
1 − ∣𝛽,𝑥 ∣/𝜌𝑠 𝑏
(4.19)
If the resistance to the dislocation motion is negligible (and, hence, the
dissipation is zero), the true plastic distortion minimizes the total energy, Ψ,
which is a functional of 𝛽(𝑥),
]
∫ 𝑎[
1
1
2
Ψ[𝛽(𝑥)] = ℎ𝐿
𝑑𝑥
(4.20)
𝜇(𝛾 − 𝛽) + 𝜇𝑘 ln
2
1 − ∣𝛽,𝑥 ∣/𝜌𝑠 𝑏
0
among all admissible function 𝛽(𝑥) satisfying the boundary conditions (4.18).
The total strain, 𝛾, is regarded as a given function of time, so one can study
the evolution of the dislocation network which accompanies the change of
the total strain.
If the resistance to the dislocation motion cannot be neglected, then the
energy minimization must be replaced by a flow rule. In case of the rateindependent plasticity, when the dissipation potential is
˙
𝐷 = 𝐾∣𝛽∣,
with 𝐾 denoting the critical resolved shear stress, the flow rule reads: for
𝛽˙ ∕= 0
𝛿𝛾 𝜙
∂𝐷
.
(4.21)
=−
𝛿𝛽
∂ 𝛽˙
The right-hand side of (4.21) is the negative variational derivative of energy
with respect to 𝛽 at fixed total strain 𝛾
ϰ≡−
𝛿𝛾 𝜙
∂𝜙
∂ ∂𝜙
=−
+
.
𝛿𝛽
∂𝛽 ∂𝑥 ∂𝛽,𝑥
With 𝜙 from (4.19) we obtain
ϰ = 𝜇(𝛾 − 𝛽) + 𝜇𝑘
𝛽,𝑥𝑥
.
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
(4.22)
According to (4.22), if 𝛽 = const in some sub-interval of (0, 𝑎), ϰ coincides
with the shear stress 𝜎 = 𝜇(𝛾 − 𝛽). For 𝛽˙ = 0 the evolution equation (4.21)
needs not be satisfied: it is replaced by the equation 𝛽˙ = 0.
So, for 𝛽˙ ∕= 0, the evolution equation for 𝛽 is
𝐾sign𝛽˙ = 𝜇(𝛾 − 𝛽) + 𝜇𝑘
𝛽,𝑥𝑥
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
(4.23)
87
4.3. ANTI-PLANE CONSTRAINED SHEAR
which must be subject to the boundary conditions (4.18).
According to (4.23) the plastic distortion may evolve only if the yield
condition
𝛽,𝑥𝑥
∣𝜇(𝛾 − 𝛽) + 𝜇𝑘
∣=𝐾
(4.24)
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
is fulfilled. If ∣ϰ∣ < 𝐾 then 𝛽 is “frozen”: 𝛽˙ = 0. If the yield condition (4.24)
holds, function 𝛽(𝑡, 𝑥) may evolve or may stay unchanged: this depends on
the time dependence of the control parameter 𝛾.
We first analyze the situation when the resistance to the dislocation motion is negligible (and, hence, the dissipation is zero). In this case the determination of 𝛽(𝑥) reduces to the minimization problem (4.20). Note that,
since 𝜙 is convex with respect to 𝛽 and 𝛽,𝑥 , this variational problem has a
unique solution. It is convenient to introduce the following dimensionless
quantities
𝜉=
𝑥𝑏𝜌𝑠
,
𝛾
𝑢(𝜉) =
𝛽(𝑥)
,
𝛾
𝑚=
𝑘
,
𝛾2
𝐸=
𝑏𝜌𝑠
Ψ.
𝜇ℎ𝐿𝛾 3
(4.25)
The dimensionless variable 𝜉 changes on the interval (0, 𝑐), where 𝑐 = 𝑎𝑏𝜌𝑠 /𝛾.
The functional (4.20) reduces to
]
∫ 𝑐[
1
1
2
𝐸[𝑢(𝜉)] =
𝑑𝜉,
(4.26)
(1 − 𝑢) + 𝑚 ln
2
1 − ∣𝑢′ ∣
0
where the prime denotes differentiation with respect to 𝜉. We minimize
functional (4.26) among functions 𝑢(𝜉) satisfying the boundary conditions
𝑢(0) = 𝑢(𝑐) = 0.
It is instructive to analyze first the variational problem in which the
logarithmic term is replaced by an asymptotic formula
ln
1
1
≈ ∣𝑢′ ∣ + 𝑢′2 .
′
1 − ∣𝑢 ∣
2
The functional to be minimized becomes
]
∫ 𝑐[
1
1 ′2
2
′
𝐸[𝑢(𝜉)] =
(1 − 𝑢) + 𝑚(∣𝑢 ∣ + 𝑢 ) 𝑑𝜉,
2
2
0
(4.27)
Due to the boundary conditions 𝑢′ should change its sign on the interval
(0, 𝑐). One-dimensional theory of dislocation pile-ups suggests to seek the
minimizer in the form
⎧

for 𝜉 ∈ (0, 𝑙),
⎨𝑢1 (𝜉)
𝑢(𝜉) = 𝑢𝑚
(4.28)
for 𝜉 ∈ (𝑙, 𝑐 − 𝑙),

⎩
𝑢1 (𝑐 − 𝜉) for 𝜉 ∈ (𝑐 − 𝑙, 𝑐),
88
CHAPTER 4. CRYSTAL PLASTICITY
where 𝑢𝑚 is a constant, 𝑙 an unknown parameter, 0 ≤ 𝑙 ≤ 𝑐/2, and 𝑢1 (𝑙) = 𝑢𝑚
at 𝜉 = 𝑙. We have to find 𝑢1 (𝜉) and the constants, 𝑢𝑚 and 𝑙. Since 𝑢′ > 0
for 𝜉 ∈ (0, 𝑙), the functional becomes
)]
(
∫ 𝑙[
1
1
1 ′2
2
′
𝐸=2
𝑑𝜉 + (1 − 𝑢𝑚 )2 (𝑐 − 2𝑙). (4.29)
(1 − 𝑢1 ) + 𝑚 𝑢1 + 𝑢1
2
2
2
0
Function 𝑢1 (𝜉) is subject to the boundary conditions
𝑢1 (0) = 0,
𝑢1 (𝑙) = 𝑢𝑚 .
(4.30)
Varying this energy functional with respect to 𝑢1 (𝜉) we obtain the Euler
equation for 𝑢1 (𝜉) on the interval (0, 𝑙)
1 − 𝑢1 + 𝑚𝑢′′1 = 0.
(4.31)
The variation of (4.29) with respect to 𝑢𝑚 and 𝑙 yields the two additional
boundary conditions at 𝜉 = 𝑙
𝑢′1 (𝑙) = 0,
2𝑚 = (1 − 𝑢𝑚 )(𝑐 − 2𝑙).
(4.32)
Condition (4.32)1 means that the dislocation density must be continuous.
Equations (4.31), (4.30)1 , and (4.32)1 have the solution
𝜉
𝑙
𝜉
𝑢1 (𝜉) = 1 − cosh √ + tanh √ sinh √ ,
𝑚
𝑚
𝑚
0 ≤ 𝜉 ≤ 𝑙.
(4.33)
Equations (4.30)2 and (4.32)2 give the following transcendental equation to
determine 𝑙 in terms of the constants 𝑚 and 𝑐
𝑙
𝑓 (𝑙) ≡ 2𝑙 + 2𝑚 cosh √ = 𝑐.
𝑚
(4.34)
√
According to (4.28) 𝑙 must lie in the segment [0, 𝑐/2]. Since cosh(𝑙/ 𝑚) ≥ 1,
2𝑙 ≤ 𝑐 − 2𝑚. Thus, equation (4.34) has no positive root if 𝑐 < 2𝑚. Returning
to the original variables according to (4.25) we see that inequality 𝑐 < 2𝑚
corresponds to the condition 𝛾 < 𝛾𝑒𝑛 , where
𝛾𝑒𝑛 =
2𝑘
,
𝑎𝑏𝜌𝑠
and for 𝛾 < 𝛾𝑒𝑛 no dislocations are nucleated. Note that the threshold
value, 𝛾𝑒𝑛 , is inversely proportional to the product of the size 𝑎 of specimen
times the saturated dislocation density (a kind of Hall-Petch relation). For
𝑐 > 2𝑚 equation (4.34) has only one root in the interval (0, 𝑐/2). Indeed,
89
4.3. ANTI-PLANE CONSTRAINED SHEAR
since 𝑓 (0) = 2𝑚 < 𝑐 and 𝑓 (𝑐/2) > 𝑐, the curve 𝑦 = 𝑓 (𝑙) crosses the line
𝑦 = 𝑐, so, there are roots of the equation (4.34) in (0, 𝑐/2). In fact, there is
only one root because 𝑓 (𝑙) is strictly increasing function of 𝑙.
Fig. 4.11 shows the evolution of 𝛽(¯
𝑥) (where 𝑥¯ = 𝑥𝑏𝜌𝑠 ) as 𝛾 increases. For
the numerical simulation we took 𝑘 = 0.04, 𝜌𝑠 = 3 .1014 m−2 , 𝑏 = 4 .10−10 m,
𝑎 = 10−4 m, so that 𝑎
¯ = 𝑎𝑏𝜌𝑠 = 12.
b
c
0.06
0.05
b
0.04
0.03
0.02
0.01
a
2
4
6
8
10
12
x
Figure 4.11: Evolution of 𝛽: a) 𝛾 = 0.01, b) 𝛾 = 0.05, c) 𝛾 = 0.07
It is interesting to plot the average shear stress
∫
1 𝑎
𝜎
¯=
𝜇(𝛾 − 𝛽(𝑥)) 𝑑𝑥
𝑎 0
(4.35)
as function of the shear strain. For 𝛾 < 𝛾𝑒𝑛 the plastic distortion 𝛽 = 0, so
𝜎
¯ = 𝜇𝛾. For 𝛾 > 𝛾𝑒𝑛 from (4.32)2 and (4.33) we obtain
√
𝜎
¯
2𝑘 2 𝑘𝛾
𝑙
=
+
tanh √ .
(4.36)
𝜇
𝑎
¯
𝑎
¯
𝑚
Fig. 4.12 shows the normalized average shear stress versus shear strain curve
OAB. There is a “work hardening” section AB for 𝛾 > 𝛾𝑒𝑛 due to the second
term in (4.36) caused by the dislocation pile-up. Mention, however, that
there is no residual strain as we unload the crystal by decreasing 𝛾: the
stress-strain curve follows the same path BAO, so the plastic deformation
is completely reversible, and no energy dissipation occurs. In the course
of unloading the dislocations nucleated annihilate, and as we approach the
point A they all disappear.
Consider now the functional (4.26) with the logarithmic energy of dislocation network. Assuming 𝑢(𝜉) as before in the form (4.28), we obtain for
𝑢1 (𝜉) the nonlinear equation
1 − 𝑢1 + 𝑚
𝑢′′1
= 0,
(1 − 𝑢′1 )2
(4.37)
90
CHAPTER 4. CRYSTAL PLASTICITY
_
s/m
B
0.008
A
0.006
0.004
0.002
O
0.01
0.02
0.03
0.04
0.05
g
Figure 4.12: Normalized average stress versus shear strain curve
and the boundary conditions (4.30) and (4.32). Let 𝑝(𝜉) be the function
𝑝(𝜉) =
𝑚
.
1 − 𝑢′1
Equation (4.37) and the boundary conditions can be written as the following
boundary-value problem
𝑝′ = 𝑢1 − 1,
𝑢′1 = 1 −
𝑚
,
𝑝
𝑢1 (0) = 0,
𝑢1 (𝑙) = 𝑢𝑚 ,
𝑢′1 (𝑙) = 0. (4.38)
Function 𝑝(𝜉) is a decreasing function of 𝜉. Since 𝑢′1 (𝑙) = 0, 𝑝(𝑙) = 𝑚.
The system of ordinary differential equations (4.38) admits the first integral
1
(1 − 𝑢1 )2 + 𝑚 ln 𝑝 − 𝑝 = const.
2
The value of the constant can be found from the boundary conditions at the
point 𝜉 = 𝑙
1
1
(1 − 𝑢1 )2 + 𝑚 ln 𝑝 − 𝑝 = (1 − 𝑢𝑚 )2 + 𝑚 ln 𝑚 − 𝑚.
2
2
Hence
1 − 𝑢1 (𝜉) =
√
(1 − 𝑢𝑚 )2 + 2𝑚(
𝑝(𝜉)
𝑝(𝜉)
− 1 − ln
).
𝑚
𝑚
(4.39)
The expression under the square root is positive because 𝑝(𝜉) ≥ 𝑚 and
𝑥 > ln 𝑥 + 1 for 𝑥 ≥ 1. For the function 𝑞(𝜉) = 𝑝(𝜉)/𝑚 − 1 we have the initial
value problem
𝑞′ = −
1√
(1 − 𝑢𝑚 )2 + 2𝑚(𝑞 − ln(1 + 𝑞)),
𝑚
𝑞(0) = 𝑞0 ,
(4.40)
91
4.3. ANTI-PLANE CONSTRAINED SHEAR
where, according to (4.39), 𝑞0 is related to 𝑢𝑚 by a transcendental equation
(1 − 𝑢𝑚 )2 + 2𝑚(𝑞0 − ln(1 + 𝑞0 )) = 1.
(4.41)
Expressing 𝑢𝑚 in (4.41) by 𝑞0 and putting the result in (4.40), we obtain the
initial value problem
√
1
1+𝑞
′
), 𝑞(0) = 𝑞0 .
(4.42)
𝑞 =−
1 + 2𝑚(𝑞 − 𝑞0 − ln
𝑚
1 + 𝑞0
The length 𝑙 is determined by the formula
∫ 𝑞0
𝑚𝑑𝑞
√
𝑙=
.
1 + 2𝑚(𝑞 − 𝑞0 − ln((1 + 𝑞)/(1 + 𝑞0 )))
0
Together with the condition (4.32)2 , we get the following equation for 𝑞0
)
(
∫ 𝑞0
𝑚𝑑𝑞
√
𝑐−2
1 + 2𝑚(𝑞 − 𝑞0 − ln((1 + 𝑞)/(1 + 𝑞0 )))
(4.43)
0
√
× 1 + 2𝑚(ln(𝑞0 + 1) − 𝑞0 ) = 2𝑚.
Having found 𝑞0 from (4.43) we can solve the initial value problem (4.42) to
determine 𝑢1 (𝜉). The solution of this problem shows that the minimizer of
(4.26) does not differ much from that of (4.27) for the values of the parameters
chosen. Fig. 4.13 shows the plot of 𝑢1 (𝜉) for 𝜉 ∈ (0, 𝑙) obtained by minimizing
(4.26) and (4.27) for 𝛾 = 0.01.
u1
0.3
0.25
0.2
0.15
0.1
0.05
2.5
5
7.5
10 12.5 15 17.5
x
Figure 4.13: The graph of 𝑢1 (𝜉) for 𝛾 = 0.01: a) bold line: minimizer of
(4.26), b) dash line: minimizer of (4.27)
We turn now to the case when the resistance to the dislocation motion
(and hence the dissipation) cannot be neglected. In this case the plastic
92
CHAPTER 4. CRYSTAL PLASTICITY
distortion may evolve only if the yield condition (4.24) is satisfied. It is
convenient to divide both side of (4.24) by 𝜇 to obtain
∣𝛾 − 𝛽 + 𝑘
𝛽,𝑥𝑥
∣ = 𝛾𝑐 ≡ 𝐾/𝜇.
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
(4.44)
We regard 𝛾 again as a given function of time (the “driving” variable) and
try to determine 𝛽(𝑡, 𝑥). We consider the following close loading path: 𝛾 is
first increased from zero to some value 𝛾 ∗ > 𝛾𝑐 , then decreased to −𝛾𝑐 , and
finally increased to zero (Fig. 4.14). The rate of change of 𝛾(𝑡) does not affect
the results due to the rate independence of the dissipation. The problem is
to determine the evolution of 𝛽 as function of 𝑡 and 𝑥, provided 𝛽(0, 𝑥) = 0.
Since 𝛽 is initially zero, we see from (4.44) that 𝛽 = 0 as long as 𝛾 < 𝛾𝑐 .
Thus, the dissipative threshold stress (the yield stress) 𝜎𝑦 = 𝐾 in this case.
For 𝛾 > 𝛾𝑐 the yield condition
𝛾−𝛽+𝑘
𝛽,𝑥𝑥
= 𝛾𝑐
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
(4.45)
takes place everywhere in (0, 𝑎). Equation (4.45) follows from (4.44) and the
initial condition 𝛽(0, 𝑥) = 0, since for small 𝛽(𝑡, 𝑥) and 𝛾 > 𝛾𝑐
𝛽,𝑥𝑥
𝛽
,𝑥𝑥
=𝛾−𝛽+𝑘
𝛾 − 𝛽 + 𝑘
.
2
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)
(𝜌𝑠𝑡 𝑏 − ∣𝛽,𝑥 ∣)2
It is convenient to introduce the deviation of 𝛾(𝑡) from the critical shear, 𝛾𝑐 ,
𝛾𝑟 = 𝛾 − 𝛾𝑐 > 0, and to define the following dimensionless quantities
𝜉=
𝑥𝑏𝜌𝑠
,
𝛾𝑟
𝑢(𝜉) =
𝛽(𝑥)
,
𝛾𝑟
𝑚=
𝑘
.
𝛾𝑟2
(4.46)
They are similar to those of (4.25), with 𝛾𝑟 replacing 𝛾. Equation (4.45)
takes the dimensionless form
𝑢′′
1−𝑢+𝑚
= 0.
(4.47)
(1 − ∣𝑢′ ∣)2
One-dimensional theory of dislocation pile-ups with non-zero resistance
suggests that the solution of (4.47) is symmetric
{
𝑢1 (𝜉)
for 𝜉 ∈ (0, 𝑐/2),
(4.48)
𝑢(𝜉) =
𝑢1 (𝑐 − 𝜉) for 𝜉 ∈ (𝑐/2, 𝑐),
where 𝑐 = 𝑎𝑏𝜌𝑠 /𝛾𝑟 . Function 𝑢1 (𝜉) is determined from equation (4.37) and
the boundary conditions
𝑢1 (0) = 0,
𝑢′1 (𝑐/2) = 0.
(4.49)
93
4.3. ANTI-PLANE CONSTRAINED SHEAR
g
g*
t
-gc
Figure 4.14: A close loading path
The first condition means that dislocations cannot reach the boundary of the
region because the boundary is clamped. The second condition follows from
the continuity of plastic distortion and the symmetry property (4.48).
The boundary-value problem (4.37), (4.49) can be solved in exactly the
same manner as for the case without dissipation. For the function 𝑞(𝜉) =
𝑝(𝜉)/𝑚 − 1 we have the initial value problem (4.42). Integrating equation
(4.42) over 𝜉 from zero to 𝑐/2 and taking into account the condition 𝑞(𝑐/2) =
0 which is the consequence of (4.49)2 , we get
∫ 𝑞0
𝑚𝑑𝑞
√
𝑔(𝑞0 ) ≡
= 𝑐/2.
(4.50)
1 + 2𝑚(𝑞 − 𝑞0 − ln((1 + 𝑞)/(1 + 𝑞0 )))
0
Thus, 𝑞0 is the root of the transcendental equation (4.50). Note that, as long
as 𝑞0 is smaller than the root of the equation
𝛾𝑟2
1
=
,
(4.51)
𝑞0 − ln(1 + 𝑞0 ) =
2𝑚
2𝑘
the expression under the square root of integral (4.50) is positive and the
integral is well defined. This follows from the inequalities 1 − 2𝑚(𝑞0 − ln(1 +
𝑞0 )) > 0 for 𝑞0 smaller than the root of (4.51) (because 1 − 2𝑚(𝑥 − ln(1 +
𝑥)) is strictly decreasing function of 𝑥) and 𝑞 − ln(1 + 𝑞) > 0 for 𝑞 > 0.
Differentiating 𝑔(𝑞0 ) with respect to 𝑞0 , it is easy to check that 𝑑𝑔/𝑑𝑞0 > 0,
so 𝑔(𝑞0 ) is a strictly increasing function. Since 𝑔(0) = 0 and 𝑔(𝑞0 ) → ∞ as
𝑞0 approaches the root of (4.51), equation (4.50) has only one root which is
smaller than the root of (4.51).
Numerical calculations show that the root of (4.50) is very close to that
of (4.51). Therefore we may find 𝑢1 (𝜉) approximately by solving the initial
value problem on (0, 𝑐/2)
√
2
𝑞′ = −
(𝑞 − ln(1 + 𝑞)), 𝑞(0) = 𝑞0 ,
𝑚
94
CHAPTER 4. CRYSTAL PLASTICITY
where 𝑞0 is the root of the equation (4.51). The length 𝑙 of the boundary
layer may be defined by the formula
√ ∫ 𝑞0
𝑑𝑞
𝑚
√
𝑙=
,
2 𝛿
𝑞 − ln(1 + 𝑞)
where 𝛿 is a small positive number. We take 𝛿 = 10−5 .
Fig. 4.15 shows the evolution of 𝛽(¯
𝑥) (where 𝑥¯ = 𝑥𝑏𝜌𝑠 ) as 𝛾𝑟 increases.
For the numerical simulation we took as before 𝑘 = 0.04, 𝑎
¯ = 𝑎𝑏𝜌𝑠 = 12.
One can see that the dislocation density as well as the length of boundary
layers increase as 𝛾𝑟 increases. Note that the expression
−𝛽 + 𝑘
𝛽,𝑥𝑥
= −𝛾𝑟 ≡ −(𝛾(𝑡) − 𝛾𝑐 )
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
does not depend on 𝑥 at any instant.
b
c
0.003
0.0025
0.002
0.0015
b
0.001
0.0005
a
2
4
6
8
10
12
x
Figure 4.15: Evolution of 𝛽: a) 𝛾𝑟 = 0.0001, b) 𝛾𝑟 = 0.001, c) 𝛾𝑟 = 0.003
After reaching some value of shear strain 𝛾 ∗ > 𝛾𝑐 we unload the crystal
by decreasing 𝛾. Since ϰ becomes smaller than 𝐾, 𝛽 does not change (𝛽 =
𝛽 ∗ (𝑥)) until
∗
𝛽,𝑥𝑥
∗
= −𝛾𝑐 ,
(4.52)
𝛾−𝛽 +𝑘
∗ ∣)2
(𝜌𝑠 𝑏 − ∣𝛽,𝑥
where 𝛽 ∗ (𝑥) is the solution of equation (4.45) for 𝛾(𝑡) = 𝛾 ∗ :
𝛾∗ − 𝛽∗ + 𝑘
∗
𝛽,𝑥𝑥
= 𝛾𝑐 .
∗ ∣)2
(𝜌𝑠 𝑏 − ∣𝛽,𝑥
(4.53)
From (4.52) and (4.53) one can see that the flow begins when 𝛾 − (𝛾 ∗ − 𝛾𝑐 ) =
−𝛾𝑐 , i.e. for 𝛾 = 𝛾∗ = 𝛾 ∗ − 2𝛾𝑐 . From that value of 𝛾 the yield condition
95
4.3. ANTI-PLANE CONSTRAINED SHEAR
ϰ = −𝐾 takes place leading to the decrease of 𝛽 which should be now
determined by the equation
𝛾−𝛽+𝑘
𝛽,𝑥𝑥
= −𝛾𝑐 .
(𝜌𝑠 𝑏 − ∣𝛽,𝑥 ∣)2
(4.54)
Since for 𝛾 ∈ (−𝛾𝑐 , 𝛾∗ ) the deviation 𝛾𝑙 = 𝛾 + 𝛾𝑐 is positive, equation (4.54)
can again be transformed to equation (4.47) and solved in exactly the same
manner if we replace 𝛾𝑟 = 𝛾(𝑡) − 𝛾𝑐 in all formulas (4.46)–(4.51) by 𝛾𝑙 =
𝛾(𝑡) + 𝛾𝑐 . As 𝛾 approaches −𝛾𝑐 , 𝛽 tends to zero because 𝛾𝑙 → 0. The further
increase of 𝛾 from −𝛾𝑐 to zero does not cause change in 𝛽 which remains
identically zero.
a
0.014
0.012
0.01
0.008
c
0.006
0.004
0.002
b
a
0.5
1
1.5
x
Figure 4.16: Evolution of the normalized dislocation density: a) 𝛾𝑟 = 0.0001,
b) 𝛾𝑟 = 0.001, c) 𝛾𝑟 = 0.003
The dislocation density 𝛼 = 𝛽,𝑥 can be calculated from the solution (4.48).
In terms of the dimensionless variable (4.46) we have
⎧

for 𝜉 ∈ (0, 𝑙),
⎨𝑏𝜌𝑠 𝑞(𝜉)/(1 + 𝑞(𝜉))
𝛼(𝜉) = 0
for 𝜉 ∈ (𝑙, 𝑐 − 𝑙),

⎩
−𝑏𝜌𝑠 𝑞(𝑐 − 𝜉)/(1 + 𝑞(𝑐 − 𝜉)) for 𝜉 ∈ (𝑐 − 𝑙, 𝑐).
Since 𝑞(𝜉) is decreasing, the maximum dislocation density is achieved at
𝜉 = 0 giving 𝛼max = 𝑏𝜌𝑠 𝑞0 /(1 + 𝑞0 ). So, the parameter 𝑞0 is simply linked
to the maximum dislocation density. Fig. 4.16 shows the distributions of
the normalized dislocation density 𝛼(¯
𝑥)/𝑏𝜌𝑠 within the interval (0, ¯𝑙) (where
¯𝑙 = 𝑙𝛾𝑟 ) as 𝛾𝑟 changes.
As soon as the plastic deformation develops, the shear stress 𝜎 = 𝜇(𝛾 −𝛽)
becomes inhomogeneous. It is interesting to calculate the average shear stress
(4.35) which is a measurable quantity. During the loading, according to
96
CHAPTER 4. CRYSTAL PLASTICITY
(4.48), we have for the normalized average shear stress (or the average elastic
shear strain)
[
]
∫ 𝑙
∫ 𝑙
1
𝜎
¯
𝛾𝑟2
(𝑒)
𝜀¯ = = 𝛾 −
𝛾𝑟 𝑎 − 2𝛾𝑟 (1 − 𝑢(𝑥)) 𝑑𝑥 = 𝛾𝑐 + 2
(1 − 𝑢(𝜉)) 𝑑𝜉.
𝜇
𝑎
𝑎𝑏𝜌𝑠 0
0
The integral is evaluated using equation (4.38)
∫ 𝑙
∫ 𝑙
𝑝′ (𝜉) 𝑑𝜉 = −(𝑝(𝑙) − 𝑝(0)) = 𝑚𝑞0 .
(1 − 𝑢(𝜉)) 𝑑𝜉 = −
0
0
Thus
2𝑘
𝑞0 ,
(4.55)
𝑎𝑏𝜌𝑠
where the second term causing hardening depends on the maximum dislocation density and is inversely proportional to the product of the size 𝑎 of
specimen times the saturated dislocation density. Formula (4.55) describes
the size effect in this model.
During the inverse loading, when the yield condition ϰ = −𝐾 holds true,
formula (4.55) changes to
𝜀¯(𝑒) = 𝛾𝑐 +
𝜀¯(𝑒) = −𝛾𝑐 +
2𝑘
𝑞0 ,
𝑎𝑏𝜌𝑠
where 𝑞0 is the root of the equation
𝑞0 − ln(1 + 𝑞0 ) =
𝛾𝑙2
.
2𝑘
s/m
0.004
B
A
0.002
-gc
O
gc
0.005
g*
0.01
g*
0.015
0.02
g
-0.002
-0.004
D
C
Figure 4.17: Normalized average stress versus shear strain curve
Fig. 4.17 shows the normalized average shear stress (or average elastic
shear strain) versus shear strain curve for the loading program of Fig. 4.14.
97
4.4. PLANE CONSTRAINED SHEAR
We took 𝑎𝑏𝜌𝑠 = 12, 𝑘 = 0.04, 𝛾𝑐 = 0.004, 𝛾 ∗ = 0.02. The straight line
OA corresponds to the purely elastic loading with 𝛾 increasing from zero to
𝛾𝑐 . The line AB corresponds to the plastic yielding with ϰ = 𝐾. The yield
begins at the point A with the yield stress 𝜎𝑦 = 𝐾. The work hardening
due to the dislocation pile-up is observed which is described by the second
term in (4.55). During the unloading as 𝛾 decreases from 𝛾 ∗ to 𝛾∗ = 𝛾 ∗ − 2𝛾𝑐
(the line BC) the plastic distortion 𝛽 = 𝛽 ∗ is frozen. As 𝛾 decreases further
from 𝛾∗ to −𝛾𝑐 , the plastic yielding occurs with ϰ = −𝐾 (the line CD). The
yield stress 𝜎𝑦 = 𝜇𝛾 ∗ − 2𝐾 at the point C, at which the inverse plastic flow
sets on, is larger than −𝐾 (because 𝛾 ∗ > 𝛾𝑐 ≡ 𝐾/𝜇). Along the line CD,
as 𝛾 is decreased, the created dislocations annihilate, and at the point D all
dislocations disappear. Finally, as 𝛾 increases from −𝛾𝑐 to zero, the crystal
behaves elastically with 𝛽 = 0. In this close cycle ABCD dissipation occurs
only on the lines AB and CD. It is interesting that the lines DA and BC
are parallel and have the same length. In phenomenological plasticity theory
this property is modelled as the translational shift of the yield surface in the
stress space, the so-called Bauschinger effect.
4.4
Plane constrained shear
gh
y
h
0
m
s
L
j
a
x
z
Figure 4.18: Plane constrained shear
As the next boundary-value problem we consider a strip made of a single
crystal undergoing a plane strain shear deformation (see Fig. 4.18). Let the
cross-section of the strip be a rectangle of width 𝑎 and height ℎ, 0 ≤ 𝑥 ≤ 𝑎,
0 ≤ 𝑦 ≤ ℎ. We realize the shear deformation by placing the strip in a “hard”
98
CHAPTER 4. CRYSTAL PLASTICITY
device with the prescribed displacements at its upper and lower sides
𝑢(0) = 0,
𝑣(0) = 0,
𝑢(ℎ) = 𝛾ℎ,
𝑣(ℎ) = 0,
(4.56)
where 𝑢(𝑦) and 𝑣(𝑦) are the longitudinal and transverse displacements, respectively, with 𝛾 being the overall shear strain. We assume that the length
of the strip 𝐿 is large, and the width 𝑎 is much greater than the height ℎ
(𝐿 ≫ 𝑎 ≫ ℎ) to neglect the end effects and to have the stresses and strains
depending only on one variable 𝑦 in the central part of the strip.
For the plane strain state the components of the strain tensor are
𝜀𝑥𝑥 = 0,
1
𝜀𝑥𝑦 = 𝜀𝑦𝑥 = 𝑢,𝑦 ,
2
𝜀𝑦𝑦 = 𝑣,𝑦 .
(4.57)
If the shear strain 𝛾 is sufficiently small, then the crystal deforms elastically and 𝑢 = 𝛾𝑦, 𝑣 = 0 everywhere in the strip. If 𝛾 exceeds some critical
threshold, then edge dislocations may appear. We admit only the slip directions (or the directions of the Burgers’ vectors) perpendicular to the 𝑧-axis
and inclined at an angle 𝜑 with the 𝑥-axis and the dislocation lines parallel to the 𝑧-axis. Since only one slip system is active, the plastic distortion
is given by 𝛽𝑖𝑗 = 𝛽𝑠𝑖 𝑚𝑗 , with 𝑠 = (cos 𝜑, sin 𝜑, 0) being the slip direction,
and 𝑚 = (− sin 𝜑, cos 𝜑, 0) the normal vector to the slip plane. We assume
that 𝛽 depends only on 𝑦: 𝛽 = 𝛽(𝑦) (translational invariance). Because of
the prescribed boundary conditions (4.56), dislocations cannot penetrate the
boundaries 𝑦 = 0 and 𝑦 = ℎ, therefore
𝛽(0) = 𝛽(ℎ) = 0.
(4.58)
The in-plane components of the plastic strain tensor are
1
𝜀𝑝𝑥𝑥 = − 𝛽 sin 2𝜑,
2
1
𝜀𝑝𝑥𝑦 = 𝛽 cos 2𝜑,
2
1
𝜀𝑝𝑦𝑦 = 𝛽 sin 2𝜑.
2
With these total and plastic strain tensors we obtain the in-plane components
of the elastic strain tensor
1
𝜀𝑒𝑥𝑥 = 𝛽 sin 2𝜑,
2
1
𝜀𝑒𝑥𝑦 = (𝑢,𝑦 − 𝛽 cos 2𝜑),
2
1
𝜀𝑒𝑦𝑦 = 𝑣,𝑦 − 𝛽 sin 2𝜑.
2
As 𝛽 depends only on 𝑦, there are two non-zero components of Nye’s dislocation density tensor (4.1), namely, 𝛼𝑥𝑧 = 𝛽,𝑦 sin 𝜑 cos 𝜑 and 𝛼𝑦𝑧 = 𝛽,𝑦 sin2 𝜑.
Thus, the resultant Burgers’ vector of all dislocations whose dislocation lines
cut the area perpendicular to the 𝑧-axis is parallel to the slip direction 𝑠 and
the scalar dislocation density equals
1√ 2
1
2 = ∣𝛽 ∣∣ sin 𝜑∣.
𝜌=
𝛼𝑥𝑧 + 𝛼𝑦𝑧
,𝑦
𝑏
𝑏
99
4.4. PLANE CONSTRAINED SHEAR
Assuming for simplicity the isotropic elastic property of the crystal, we
write the free energy per unit volume of the crystal with dislocations as
1
1
.
𝜙(𝜀𝑒𝑖𝑗 , 𝛼𝑖𝑗 ) = 𝜆 (𝜀𝑒𝑖𝑖 )2 + 𝜇𝜀𝑒𝑖𝑗 𝜀𝑒𝑖𝑗 + 𝜇𝑘 ln
∣𝛽
2
1 − ,𝑦 ∣∣ sin 𝜑∣
(4.59)
𝑏𝜌𝑠
Thus, the total energy functional becomes
∫ ℎ[
1 2
1
1
Ψ[𝑢, 𝑣, 𝛽] = 𝑎𝐿
𝜆𝑣,𝑦 + 𝜇(𝑢,𝑦 − 𝛽 cos 2𝜑)2 + 𝜇𝛽 2 sin2 2𝜑
2
2
4
0
]
1
1
2
𝑑𝑦.
+ 𝜇(𝑣,𝑦 − 𝛽 sin 2𝜑) + 𝜇𝑘 ln
2
1 − ∣𝛽,𝑦 ∣∣ sin 𝜑∣
(4.60)
𝑏𝜌𝑠
Functional (4.60) can be reduced to a functional depending on 𝛽(𝑦) only.
Indeed, by first fixing 𝛽(𝑦) and taking the variation of (4.60) with respect to
𝑢 and 𝑣 we derive the equilibrium equations
𝜇(𝑢,𝑦𝑦 − 𝛽,𝑦 cos 2𝜑) = 0,
(𝜆 + 2𝜇)𝑣,𝑦𝑦 − 𝜇𝛽,𝑦 sin 2𝜑 = 0.
Integrating these equations and using the boundary conditions (4.56) we get
𝑢,𝑦 = 𝛾 + (𝛽 − ⟨𝛽⟩) cos 2𝜑,
(4.61)
𝑣,𝑦 = 𝜅(𝛽 − ⟨𝛽⟩) sin 2𝜑,
∫ℎ
𝜇
where 𝜅 = 𝜆+2𝜇
, and ⟨⋅⟩ = ℎ1 0 ⋅ 𝑑𝑦. Substituting (4.61) into (4.60) and
collecting common terms we obtain the energy functional in terms of 𝛽
∫ ℎ [
1
1
Ψ[𝛽] =𝑎𝐿
𝜇 (1 − 𝜅)𝛽 2 sin2 2𝜑 + 𝜅⟨𝛽⟩2 sin2 2𝜑
(4.62)
2
2
0
]
1
1
2
𝑑𝑦.
+ (𝛾 − ⟨𝛽⟩ cos 2𝜑) + 𝑘 ln
∣∣ sin 𝜑∣
2
1 − ∣𝛽,𝑦 𝑏𝜌
𝑠
If the dissipation is negligible, then the plastic distortion 𝛽 minimizes (4.62)
under the constraint (4.58). The overall shear strain 𝛾 is regarded as given
function of time (control parameter), so one can study the evolution of the
dislocation network which accompanies the change of 𝛾.
If the resistance to dislocation motion cannot be neglected, then the
energy minimization must be replaced by a flow rule. In case of rateindependent plasticity, the flow rule for 𝛽˙ ∕= 0 reads
∂𝐷
𝛿𝛾 𝜙
=−
,
𝛿𝛽
∂ 𝛽˙
(4.63)
100
CHAPTER 4. CRYSTAL PLASTICITY
where the dissipation potential for crystals deforming in single slip is
˙
𝐷 = 𝐾∣𝛽∣,
with 𝐾 being a positive constant called critical resolved shear stress. The
right hand side of (4.63) is the negative variational derivative of the energy
with respect to 𝛽 at fixed overall strain 𝛾
ϰ≡−
𝛿𝛾 𝜙
∂𝜙
∂ ∂𝜙
=−
+
.
𝛿𝛽
∂𝛽 ∂𝑦 ∂𝛽,𝑦
For 𝛽˙ = 0, the evolution equation (4.63) does not have to be satisfied: It is
replaced by the equation 𝛽˙ = 0.
For small up to moderate dislocation densities the logarithmic term in
(4.62) may be approximated by the formula
ln
1 𝜌2
1 ∼ 𝜌
,
+
=
1 − 𝜌𝜌𝑠
𝜌𝑠 2 𝜌2𝑠
(4.64)
so that
ℎ
[
1
1
(1 − 𝜅)𝛽 2 sin2 2𝜑 + 𝜅⟨𝛽⟩2 sin2 2𝜑
(4.65)
2
2
0
)]
(
1
∣𝛽,𝑦 ∣∣ sin 𝜑∣ 1 𝛽,𝑦2 sin2 𝜑
2
𝑑𝑦.
+
+ (𝛾 − ⟨𝛽⟩ cos 2𝜑) + 𝑘
2
𝑏𝜌𝑠
2 (𝑏𝜌𝑠 )2
𝐸(𝛽) =𝑎𝐿
∫
𝜇
For simplicity of the analysis, we shall further deal with this functional only.
In the case of zero resistance (and hence the energy dissipation is zero) the
determination of 𝛽(𝑦) reduces to the minimization of the total energy (4.65).
Since 𝑈 is convex with respect to 𝛽 and 𝛽,𝑦 , the variational problem has a
unique solution. It is convenient to introduce the dimensionless quantities
𝐸=
𝑏𝜌𝑠
Ψ,
𝑎𝐿𝜇
𝑦¯ = 𝑦𝑏𝜌𝑠 ,
ℎ̄ = ℎ𝑏𝜌𝑠 .
(4.66)
The dimensionless variable 𝑦¯ changes on the interval (0, ℎ̄). Functional (4.65)
reduces to
∫ ℎ[
1
1
𝐸[𝛽] =
(1 − 𝜅)𝛽 2 sin2 2𝜑 + 𝜅⟨𝛽⟩2 sin2 2𝜑
(4.67)
2
2
0
]
1
1 ′2 2
2
′
+ (𝛾 − ⟨𝛽⟩ cos 2𝜑) + 𝑘∣𝛽 ∣∣ sin 𝜑∣ + 𝑘𝛽 sin 𝜑 𝑑𝑦,
2
2
4.4. PLANE CONSTRAINED SHEAR
101
where the prime denotes differentiation with respect to 𝑦¯, and, for short,
the bars over 𝑦 and ℎ are dropped. We minimize functional (4.67) among
functions satisfying the boundary conditions
𝛽(0) = 𝛽(ℎ) = 0.
(4.68)
For the variational problem of this type, there exists a threshold value 𝛾𝑒𝑛
such that when 𝛾 < 𝛾𝑒𝑛 no dislocations are nucleated and 𝛽 = 0. Near the
threshold value the dislocation density must be small so that the last term
in (4.67) can be neglected. Besides, the width of the boundary layer tends
to zero as 𝛾 → 𝛾𝑒𝑛 . This gives us the idea of finding the threshold value by
employing a minimizing sequence of the form
⎧
𝛽𝑚

for 𝑦 ∈ (0, 𝜖),
⎨ 𝜖 𝑦,
𝛽 = 𝛽𝑚 ,
(4.69)
for 𝑦 ∈ (𝜖, ℎ − 𝜖),

⎩ 𝛽𝑚
(ℎ − 𝑦), for 𝑦 ∈ (ℎ − 𝜖, ℎ),
𝜖
where 𝛽𝑚 is an unknown constant, and 𝜖 is a small unknown length which
tends to zero as 𝛾 → 𝛾𝑒𝑛 . Substituting (4.69) into the energy functional
(4.67) (with the last term being removed) and neglecting all small terms of
order 𝜖 and higher, we obtain
1
2
𝐸(𝛽𝑚 ) = [(𝛾 − 𝛽𝑚 cos 2𝜑)2 + 𝛽𝑚
sin2 2𝜑]ℎ + 2𝑘 ∣𝛽𝑚 sin 𝜑∣ .
2
(4.70)
A rather simple analysis shows that the minimum of (4.70) is achieved at
𝛽𝑚 ∕= 0 if and only if
2𝑘∣ sin 𝜑∣
,
𝛾 > 𝛾𝑒𝑛 =
ℎ∣ cos 2𝜑∣
otherwise it is achieved at 𝛽𝑚 = 0 (no dislocations are nucleated). Note that
the sign of 𝛽𝑚 depends on the angle 𝜑: 𝛽𝑚 is positive if 0∘ < 𝜑 < 45∘ and
negative if 45∘ < 𝜑 < 90∘ . In terms of the original length ℎ the energetic
threshold value reads
2𝑘 ∣ sin 𝜑∣
𝛾𝑒𝑛 =
,
(4.71)
ℎ𝑏𝜌𝑠 ∣ cos 2𝜑∣
showing clearly the size effect. Equation (4.71) deviates from the well-known
Hall-Petch relation. The reason for this deviation can be explained by the
boundary conditions (4.58) which do not permit the penetration of dislocations through the grain boundaries.
Due to the boundary conditions, 𝛽 ′ should change its sign on the interval
(0, ℎ). The one-dimensional theory of dislocation pile-ups as well as the
102
CHAPTER 4. CRYSTAL PLASTICITY
solution of the previous problem suggest to
⎧

for
⎨𝛽1 (𝑦),
𝛽(𝑦) = 𝛽𝑚 ,
for

⎩
𝛽1 (ℎ − 𝑦), for
seek the minimizer in the form
𝑦 ∈ (0, 𝑙),
𝑦 ∈ (𝑙, ℎ − 𝑙),
𝑦 ∈ (ℎ − 𝑙, ℎ),
(4.72)
where 𝛽𝑚 is a constant, 𝑙 an unknown length, 0 ≤ 𝑙 ≤ ℎ2 , and 𝛽1 (𝑙) = 𝛽𝑚 .
We have to find 𝛽1 (𝑦) and the constants, 𝛽𝑚 and 𝑙. With 𝛽 from (4.72) the
total energy functional becomes
∫ 𝑙[
]
1 ′2 2
1
2
2
′
(1 − 𝜅)𝛽1 sin 2𝜑 + 𝑘∣𝛽1 ∣∣ sin 𝜑∣ + 𝑘𝛽1 sin 𝜑 𝑑𝑦
𝐸=2
2
0 2
1
1
2
+ (1 − 𝜅)𝛽𝑚
sin2 2𝜑(ℎ − 2𝑙) + ℎ[𝜅⟨𝛽⟩2 sin2 2𝜑 + (𝛾 − ⟨𝛽⟩ cos 2𝜑)2 ],
2
2
(4.73)
where
1
⟨𝛽⟩ =
ℎ
( ∫ 𝑙
)
2
𝛽1 𝑑𝑦 + (ℎ − 2𝑙)𝛽𝑚 .
(4.74)
0
Varying the energy functional (4.73) with respect to 𝛽1 we obtain
−𝑘𝛽1′′ sin2 𝜑+(1−𝜅)𝛽1 sin2 2𝜑+(cos2 2𝜑+𝜅 sin2 2𝜑)⟨𝛽⟩−𝛾 cos 2𝜑 = 0, (4.75)
where 𝛽1 (𝑦) is subject to the boundary conditions
𝛽1 (0) = 0,
𝛽1 (𝑙) = 𝛽𝑚 .
(4.76)
The variation of (4.73) with respect to 𝑙 gives an additional boundary condition at 𝑦 = 𝑙,
𝛽1′ (𝑙) = 0,
(4.77)
which means that the dislocation density must be continuous. Varying the
energy functional with respect to 𝛽𝑚 , we obtain a condition for 𝛽𝑚 ,
2𝑘∣ sin 𝜑∣(sign𝛽1′ ) + [(cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽⟩ − 𝛾 cos 2𝜑
+ (1 − 𝜅)𝛽𝑚 sin2 2𝜑](ℎ − 2𝑙) = 0.
(4.78)
Equations (4.75), (4.76)1 , and (4.77) yield the solution
𝛽1 = 𝛽1𝑝 (1 − cosh 𝜂𝑦 + tanh 𝜂𝑙 sinh 𝜂𝑦) ,
with
𝛽1𝑝 =
0≤𝑦≤𝑙
𝛾 cos 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽⟩
,
(1 − 𝜅) sin2 2𝜑
(4.79)
(4.80)
103
4.4. PLANE CONSTRAINED SHEAR
and
𝜂=2
√
1−𝜅
∣ cos 𝜑∣.
𝑘
With (4.74), (4.76)2 , (4.79), and (4.80) we obtain the average of 𝛽
[ (
) (
)
]
𝜂𝑙
1
𝛾 cos 2𝜑 2 𝑙 − tanh
+
1
−
(ℎ
−
2𝑙)
𝜂
cosh 𝜂𝑙
⟨𝛽⟩ =
,
𝑔(𝑙)
(4.81)
where
𝑔(𝑙) = ℎ(1 − 𝜅) sin2 2𝜑 + (cos2 2𝜑 + 𝜅 sin2 2𝜑)
[ (
) (
)
]
tanh 𝜂𝑙
1
× 2 𝑙−
+ 1−
(ℎ − 2𝑙) ,
𝜂
cosh 𝜂𝑙
and
𝛾 cos 2𝜑 − ⟨𝛽⟩(cos2 2𝜑 + 𝜅 sin2 2𝜑)
𝛽𝑚 =
(1 − 𝜅) sin2 2𝜑
(
1
1−
cosh 𝜂𝑙
)
.
(4.82)
Substitution of (4.82) into (4.78) gives the following equation to determine 𝑙:
𝑓 (𝑙) ≡ 2𝑘∣ sin 𝜑∣(sign𝛽1′ ) −
b
𝛾 cos 2𝜑 − ⟨𝛽⟩(cos2 2𝜑 + 𝜅 sin2 2𝜑)
(ℎ − 2𝑙) = 0.
cosh 𝜂𝑙
j=30°
j=60°
c
b
a
y
d
e
f
Figure 4.19: Evolution of 𝛽 for single-slip constrained shear at zero dissipation: a,d) 𝛾 = 0.0068, b,e) 𝛾 = 0.0118, c,f) 𝛾 = 0.0168
Fig. 4.19 shows the evolution of 𝛽(¯
𝑦 ) as 𝛾 increases, for 𝜑 = 30∘ (contin∘
uous lines) and 𝜑 = 60 (dashed lines), where 𝑦¯ = 𝑦𝑏𝜌𝑠 . For the numerical
simulation we took the material parameters from Table 4.1. All material
parameters used in this Chapter (except 𝑘, 𝜌𝑠 , and 𝜒 which is responsible
104
CHAPTER 4. CRYSTAL PLASTICITY
for the cross-slip interaction) are well-known for aluminum. We choose these
additional parameters to have good agreement of discrete dislocation simulations and the continuum dislocation theory with respect to the yield stress
and the hardening rate for both single and double slip (see also Section 4.3).
In all numerical simulations we take ℎ = 10−6 m so that ℎ̄ = ℎ𝑏𝜌𝑠 = 0.349.
It is interesting to plot the shear stress 𝜏 = 𝜇(𝛾 −⟨𝛽⟩ cos 2𝜑) as a function
of the shear strain. As we know, for 𝛾 < 𝛾𝑒𝑛 no dislocations are nucleated
and 𝛽 = 0, so the shear stress 𝜏 = 𝜇𝛾. For 𝛾 > 𝛾𝑒𝑛 we take ⟨𝛽⟩ from (4.81)
to compute the shear stress.
t/m
j=30°
j=60°
B´
B
A´
A
g
0
Figure 4.20: Normalized shear stress versus shear strain curve for single-slip
constrained shear at zero dissipation
Fig. 4.20 shows the normalized shear stress versus shear strain curve OAB
for 𝜑 = 30∘ and OA’B’ for 𝜑 = 60∘ . There is a work hardening section
AB for 𝛾 > 𝛾𝑒𝑛 caused by the dislocation pile-up. Note, however, that
there is no residual strain as we unload the crystal by decreasing 𝛾: the
stress-strain curve follows the same path BAO, so the plastic deformation
is completely reversible and no energy dissipation occurs. In the course of
unloading the dislocations nucleated annihilate, and as we approach point A
they all disappear.
If the resistance to dislocation motion (and hence the dissipation) cannot
be neglected, the plastic distortion may evolve only if the yield condition
∣ϰ∣ = 𝐾 is fulfilled. If ∣ϰ∣ < 𝐾, then 𝛽 is frozen, the dislocation density
remains unchanged and the crystal deforms elastically. Computing the variMaterial
Aluminum
𝜇 (GPa)
𝜈
26.3
0.33
𝑏 (Å)
2.5
𝜌𝑠 (m−2 )
1.396 1015
𝑘
0.000115
Table 4.1: Material characteristics
𝜒
0.576
105
4.4. PLANE CONSTRAINED SHEAR
ational derivative of (4.65) we derive from (4.63) the yield condition
𝛽,𝑦𝑦 sin2 𝜑
2
2
2
= 𝐾.
𝜇 𝑘
−
(1
−
𝜅)𝛽
sin
2𝜑
−
(cos
2𝜑
+
𝜅
sin
2𝜑)⟨𝛽⟩
+
𝛾
cos
2𝜑
𝑏2 𝜌2𝑠
Consider first the case 𝜑 < 45∘ . We divide this equation by 𝜇 and introduce
the dimensionless variable 𝑦¯ = 𝑦𝑏𝜌𝑠 to transform the yield condition to
′′ 2
𝑘𝛽 sin 𝜑 − (1 − 𝜅)𝛽 sin2 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽⟩ + 𝛾 cos 2𝜑
= 𝛾𝑐𝑟 cos 2𝜑, (4.83)
with 𝛾𝑐𝑟 ≡ 𝐾/𝜇 cos 2𝜑 and the prime denoting the derivative with respect to
𝑦¯. We shall further omit the bar over 𝑦 for short.
We regard 𝛾 as a given function of time (the driving variable) and try
to determine 𝛽(𝑡, 𝑦). We consider the loading path similar to that shown in
Fig. 4.14. The problem is to determine the evolution of 𝛽 as a function of 𝑡
and 𝑦, provided 𝛽(0, 𝑦) = 0 and 𝜑 < 45∘ .
Since the plastic distortion, 𝛽, is initially zero, we see from (4.83) that
𝛽 = 0 as long as 𝛾 < 𝛾𝑐𝑟 . Thus, the dissipative threshold stress (the yield
stress) 𝜏𝑦 = 𝐾/ cos 2𝜑 in this case. For small 𝛽(𝑡, 𝑥) and 𝛾 > 𝛾𝑐𝑟 , the yield
condition becomes
𝑘𝛽 ′′ sin2 𝜑 − (1 − 𝜅)𝛽 sin2 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽⟩ + 𝛾 cos 2𝜑 = 𝛾𝑐𝑟 cos 2𝜑.
(4.84)
Let us introduce the deviation of 𝛾(𝑡) from the critical shear 𝛾𝑐𝑟 , 𝛾𝑟 = 𝛾 −𝛾𝑐𝑟 ,
and simplify (4.84) to obtain
𝑘𝛽 ′′ sin2 𝜑 − (1 − 𝜅)𝛽 sin2 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽⟩ + 𝛾𝑟 cos 2𝜑 = 0.
Since this equation is linear, 𝛽 is proportional to 𝛾𝑟 such that 𝛽 = 𝛾𝑟 𝛽1 ,
where 𝛽1 is the solution of the equation
𝑘𝛽1′′ sin2 𝜑 − (1 − 𝜅)𝛽1 sin2 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽1 ⟩ + cos 2𝜑 = 0. (4.85)
The analogous problem of anti-plane constrained shear at nonzero resistance suggests that the solution of (4.85) is symmetric, i.e.
𝛽1 (𝑦) = 𝛽1 (ℎ − 𝑦) for 𝑦 ∈ (ℎ/2, ℎ).
(4.86)
Function 𝛽1 (𝑦) is determined from equation (4.85) and the boundary conditions
(4.87)
𝛽1 (0) = 0,
𝛽1′ (ℎ/2) = 0.
106
CHAPTER 4. CRYSTAL PLASTICITY
The first condition means that dislocations cannot reach the boundary of the
region because of the prescribed displacement. The second condition follows
from the continuity of plastic distortion and the symmetry property (4.86).
Equations (4.85) and (4.87) admit the solution
)
(
ℎ
ℎ
0≤𝑦≤ ,
(4.88)
𝛽1 = 𝛽1𝑝 1 − cosh 𝜂𝑦 + tanh 𝜂 sinh 𝜂𝑦 ,
2
2
with
𝛽1𝑝
and
cos 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽1 ⟩
=
,
(1 − 𝜅) sin2 2𝜑
√
1−𝜅
∣ cos 𝜑∣.
𝑘
The average of 𝛽1 is obtained in the form
)
(
2 tanh 𝜂 ℎ
2
cos 2𝜑 1 − 𝜂ℎ
(
⟨𝛽1 ⟩ =
2
2
2
(1 − 𝜅) sin 2𝜑 + (cos 2𝜑 + 𝜅 sin 2𝜑) 1 −
𝜂=2
(4.89)
(4.90)
2 tanh 𝜂 ℎ
2
𝜂ℎ
).
(4.91)
j=30°
j=60°
b1
_
y
Figure 4.21: Graphs of 𝛽1 (¯
𝑦 ) for single-slip constrained shear at non-zero
dissipation
Fig. 4.21 shows the graphs of 𝛽1 (¯
𝑦 ) for 𝜑 = 30∘ (continuous line) and
𝜑 = 60∘ (dashed line). For the numerical simulation we took the material
parameters from Table 4.1, and ℎ = 1𝜇m, so that ℎ̄ = ℎ𝑏𝜌𝑠 = 0.349.
After reaching 𝛾 ∗ > 𝛾𝑐𝑟 , we unload the crystal by decreasing 𝛾. Since ϰ
becomes smaller than 𝐾, 𝛽 does not change (𝛽 = 𝛽 ∗ (𝑦)) until
𝑘𝛽 ∗′′ sin2 𝜑 − (1 − 𝜅)𝛽 ∗ sin2 2𝜑 − (cos2 2𝜑 + 𝜅 sin2 2𝜑)⟨𝛽 ∗ ⟩
+𝛾 cos 2𝜑 = −𝛾𝑐𝑟 cos 2𝜑,
(4.92)
107
4.4. PLANE CONSTRAINED SHEAR
where 𝛽 ∗ (𝑦) is the solution of (4.84) for 𝛾(𝑡) = 𝛾 ∗ . From (4.92) we can see that
the plastic flow begins when 𝛾 − (𝛾 ∗ − 𝛾𝑐𝑟 ) = −𝛾𝑐𝑟 , i.e. for 𝛾 = 𝛾∗ = 𝛾 ∗ − 2𝛾𝑐𝑟 .
From that value of 𝛾, the yield condition ϰ = −𝐾 holds leading to a decrease
of 𝛽 which should now be determined from
𝑘𝛽 ′′ sin2 𝜑−(1−𝜅)𝛽 sin2 2𝜑−(cos2 2𝜑+𝜅 sin2 2𝜑)⟨𝛽⟩+𝛾 cos 2𝜑 = −𝛾𝑐𝑟 cos 2𝜑,
(4.93)
Since for 𝛾 ∈ (−𝛾𝑐𝑟 , 𝛾∗ ), the deviation 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 is positive, equation (4.93)
can again be transformed to equation (4.85) and solved in exactly the same
manner if we replace 𝛾𝑟 = 𝛾 − 𝛾𝑐𝑟 in all formulas (4.88)-(4.91) by 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 .
As 𝛾 approaches −𝛾𝑐𝑟 , 𝛽 tends to zero because 𝛾𝑙 → 0. The further increase
of 𝛾 from −𝛾𝑐𝑟 to zero does not cause change in 𝛽 which remains zero.
It is not difficult to modify the construction given above to find the solution for 𝜑 > 45∘ .
a1
j=30°
j=60°
_
y
Figure 4.22: Graphs of 𝛼1 (¯
𝑦 ) for single-slip constrained shear at non-zero
dissipation
The normalized dislocation density 𝛼 = 𝛽,𝑦 sin 𝜑 can be calculated from
the solution (4.88). Since 𝛽(𝑦) is proportional to 𝛾𝑟 , 𝛼(𝑦) is also proportional
to 𝛾𝑟 such that 𝛼(𝑦) = 𝛾𝑟 𝛼1 (𝑦). For 𝑦 ∈ (0, ℎ/2) we have
)
(
𝜂ℎ
sin 𝜑,
𝛼1 (𝑦) = 𝛽1𝑝 −𝜂 sinh 𝜂𝑦 + 𝜂 cosh 𝜂𝑦 tanh
2
with 𝛽1𝑝 from (4.89) and 𝜂 from (4.90). For 𝑦 ∈ (ℎ/2, ℎ) we have 𝛼1 (𝑦) =
−𝛼1 (ℎ − 𝑦) due to symmetry. Fig. 4.22 shows the graphs of 𝛼1 for 𝜑 = 30∘
(continuous line) and 𝜑 = 60∘ (dashed line) for 𝑦¯ ∈ (0, ℎ̄/2).
It is interesting to calculate the shear stress 𝜏 which is a measurable
quantity. During loading, we have for the normalized shear stress (or the
elastic shear strain)
(
(
)
)
𝜂ℎ
2
tanh
𝜏
2
𝛾 𝑒 = = 𝛾𝑐𝑟 + 𝛾𝑟 − 1 −
𝛽1𝑝 cos 2𝜑 ,
(4.94)
𝜇
𝜂ℎ
108
CHAPTER 4. CRYSTAL PLASTICITY
t/t0
h/d=1.15
h/d=2.3
h/d=80
with dissipation
energy minimization
g
Figure 4.23: The normalized shear stress versus shear strain curve
with 𝛽1𝑝 from (4.89). The second term of (4.94) causes hardening due to the
dislocation pile-ups. Equation (4.94) describes the size effect in this model.
u,y
h/d=2.3
h/d=80
energy minimization
with dissipation
g=0.0218
g=0.0168
g=0.0118
g=0.0068
y
_
h
Figure 4.24: Comparison of the total shear strain profile obtained from this
approach and from [6]
Fig. 4.23 shows a comparison of the stress-strain curves during loading
obtained from energy minimization, from (4.94), and from the discrete dislocation simulations reported in [6]. In order to compare with the discrete
dislocation simulations we took 𝜑 = 60∘ , 𝜏0 = 1.9 10−3 𝜇 and let all other material constants remain the same as in previous simulations. The stress-strain
curves from the discrete dislocation simulations are provided for three different ratios ℎ/𝑑, where 𝑑 is the spacing between the active slip planes. Both
109
4.4. PLANE CONSTRAINED SHEAR
curves obtained from energy minimization as well as from the flow rule nearly
coincide and show good agreement with the discrete dislocation simulations
for ℎ/𝑑 = 80.
Fig. 4.24 shows a comparison of the total shear strain profiles obtained
from energy minimization, from the flow rule, and from the discrete dislocation simulations reported in [6]. Both profiles obtained from energy minimization and from the flow rule again show good agreement with the discrete
dislocation simulations for ℎ/𝑑 = 80.
During inverse loading, when the yield condition ϰ = −𝐾 holds true,
equation (4.94) changes to
(
(
)
)
2 tanh 𝜂ℎ
𝜏
2
= −𝛾𝑐𝑟 + 𝛾𝑙 − 1 −
𝛽1𝑝 cos 2𝜑 ,
𝜇
𝜂ℎ
with the deviation 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 used instead of 𝛾𝑟 = 𝛾 − 𝛾𝑐𝑟 for the average
of 𝛽.
B
t/m
C
A
-gcr 0
gcr
g*
g* g
D
Figure 4.25: Normalized shear stress versus shear strain curve for single-slip
constrained shear at non-zero dissipation, for 𝜑 = 60∘
Fig. 4.25 shows the normalized shear stress (or elastic shear strain) versus shear strain curve for the loading path of Fig. 4.14, with 𝛾 ∗ = 0.006,
𝜑 = 60∘ , while all other parameters remain the same. The straight line OA
corresponds to purely elastic loading with 𝛾 increasing from zero to 𝛾𝑐𝑟 . Line
AB corresponds to plastic yielding with ϰ = 𝐾. Yielding begins at point A
with the yield stress 𝜏𝑦 = −𝐾/ cos 2𝜑, and we can observe the work hardening due to the dislocation pile-up, which is described by the second term
of (4.94). During unloading, as 𝛾 decreases from 𝛾 ∗ to 𝛾∗ = 𝛾 ∗ − 2𝛾𝑐𝑟 (line
BC), the plastic distortion 𝛽 = 𝛽 ∗ is frozen. As 𝛾 decreases further from
𝛾∗ to −𝛾𝑐𝑟 , plastic yielding occurs with ϰ = −𝐾 (line CD). From Fig. 4.25
it becomes obvious that the yield stress 𝜏𝑦 = 𝜏 ∗ + 2𝐾/ cos 2𝜑 at point C,
at which the inverse plastic flow sets on, is larger than −𝐾/ cos 2𝜑 (because
110
CHAPTER 4. CRYSTAL PLASTICITY
𝜏 ∗ > −𝐾/ cos 2𝜑). Along line CD, as 𝛾 is decreased, the created dislocations
annihilate, and at point D all dislocations have disappeared. Finally, as 𝛾
increases from −𝛾𝑐𝑟 to zero, the crystal behaves elastically with 𝛽 = 0. In
this closed cycle OABCDO dissipation occurs only along lines AB and CD.
It is interesting that lines DA and BC are parallel and have the same length
exhibiting again the Bauschinger effect.
4.5
Single crystals deforming in double slip
gh
y
h
m
sr
l
s
l
L
r
0
m
jl
jr
a
x
z
Figure 4.26: Plane constrained shear of single crystals deforming in double
slip
Consider the same problem as in the previous Section, but now for crystals
deforming in double slip. We admit two slip directions (or the directions of
the Burgers’ vectors) perpendicular to the 𝑧-axis and inclined at an angle
𝜑𝑙 (0 ≤ 𝜑𝑙 ≤ 𝜋/2) and 𝜑𝑟 (𝜋/2 ≤ 𝜑𝑟 ≤ 𝜋) with the 𝑥-axis, respectively,
and the dislocation lines parallel to the 𝑧-axis. Since two slip systems are
active, the plastic distortion is given by 𝛽𝑖𝑗 = 𝛽𝑙 𝑠𝑙𝑖 𝑚𝑙𝑗 + 𝛽𝑟 𝑠𝑟𝑖 𝑚𝑟𝑗 , with 𝑠𝔞𝑖 =
(cos 𝜑𝔞 , sin 𝜑𝔞 , 0) being the slip directions, and 𝑚𝔞𝑗 = (− sin 𝜑𝔞 , cos 𝜑𝔞 , 0) the
normal vectors to the slip planes (𝔞 = 𝑙, 𝑟).
We assume that 𝛽𝑙 and 𝛽𝑟 depend on 𝑦 only: 𝛽𝑙 = 𝛽𝑙 (𝑦) and 𝛽𝑟 = 𝛽𝑟 (𝑦)
(translational invariance). Because of the prescribed boundary conditions
dislocations cannot penetrate the boundaries 𝑦 = 0 and 𝑦 = ℎ, therefore
𝛽𝑙 (0) = 𝛽𝑟 (0) = 𝛽𝑙 (ℎ) = 𝛽𝑟 (ℎ) = 0.
(4.95)
As previously, the boundaries 𝑦 = 0 and 𝑦 = ℎ serve as obstacles to dislocation motion.
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
111
Under plane strain state conditions, the non-zero components of the plastic strain tensor are
1
𝜀𝑝𝑥𝑥 = − (𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 ),
2
1
𝜀𝑝𝑥𝑦 = (𝛽𝑙 cos 2𝜑𝑙 + 𝛽𝑟 cos 2𝜑𝑟 ),
2
1
𝜀𝑝𝑦𝑦 = (𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 ).
2
(4.96)
With (4.57) and (4.96) we obtain the non-zero components of the elastic
strain tensor
1
𝜀𝑒𝑥𝑥 = (𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 ),
2
1
𝑒
𝜀𝑥𝑦 = (𝑢,𝑦 − 𝛽𝑙 cos 2𝜑𝑙 − 𝛽𝑟 cos 2𝜑𝑟 ),
2
1
𝑒
𝜀𝑦𝑦 = 𝑣,𝑦 − (𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 ).
2
(4.97)
As 𝛽𝑙 and 𝛽𝑟 depend only on 𝑦, there are two non-zero components of
Nye’s dislocation density tensor
𝛼𝑥𝑧 = 𝛽𝑙,𝑦 sin 𝜑𝑙 cos 𝜑𝑙 + 𝛽𝑟,𝑦 sin 𝜑𝑟 cos 𝜑𝑟 ,
𝛼𝑦𝑧 = 𝛽𝑙,𝑦 sin2 𝜑𝑙 + 𝛽𝑟,𝑦 sin2 𝜑𝑟 .
These are the components of the resultant Burgers’ vector of all edge dislocations whose dislocation lines cut the area perpendicular to the 𝑧-axis. Thus,
the dislocations produced on two slip systems belong to two different groups:
the first one with the resultant Burgers’ vector showing in the direction 𝑠𝑙𝑖 ,
the second one with the resultant Burgers’ vector parallel to 𝑠𝑟𝑖 . Therefore
the scalar dislocation densities (or the numbers of dislocations per unit area)
equal
1
1
(4.98)
𝜌𝑙 = ∣𝛽𝑙,𝑦 sin 𝜑𝑙 ∣, 𝜌𝑟 = ∣𝛽𝑟,𝑦 sin 𝜑𝑟 ∣.
𝑏
𝑏
The free energy per unit volume of the crystal with dislocations takes the
form
)
(
1
𝜌
𝜌
1
1
𝑙 𝑟
+ ln
+𝜒 2 .
𝜙(𝜀𝑒𝑖𝑗 , 𝛼𝑖𝑗 ) = 𝜆 (𝜀𝑒𝑖𝑖 )2 + 𝜇𝜀𝑒𝑖𝑗 𝜀𝑒𝑖𝑗 + 𝜇𝑘 ln
2
1 − 𝜌𝜌𝑠𝑙
1 − 𝜌𝜌𝑟𝑠
𝜌𝑠
(4.99)
The last term in parentheses corresponds to the energy of the dislocation
network which consists of energies of each group of dislocations plus the
112
CHAPTER 4. CRYSTAL PLASTICITY
energy of cross-slip interaction, with 𝜒 being the interaction factor. With
(4.97) and (4.99), the total energy functional becomes
∫ ℎ[
1 2
1
𝜆𝑣,𝑦 + 𝜇(𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 )2
2
4
0
1
1
1
+ 𝜇(𝑢,𝑦 − 𝛽𝑙 cos 2𝜑𝑙 − 𝛽𝑟 cos 2𝜑𝑟 )2 + 𝜇(𝑣,𝑦 − 𝛽𝑙 sin 2𝜑𝑙 − 𝛽𝑟 sin 2𝜑𝑟 )2
2
2
)2]
(
1
𝜌𝑙 𝜌𝑟
1
𝑑𝑦. (4.100)
+ 𝜇𝑘 ln
𝜌𝑙 + ln
𝜌𝑟 + 𝜒 2
1 − 𝜌𝑠
1 − 𝜌𝑠
𝜌𝑠
Ψ[𝑢, 𝑣, 𝛽𝑙 , 𝛽𝑟 ] = 𝑎𝐿
Functional (4.100) can be reduced, in exactly the same manner as in the
previous case, to a functional depending on 𝛽𝑙 (𝑦) and 𝛽𝑟 (𝑦) only. The result
is
∫ ℎ [
1
𝜇 (1 − 𝜅)(𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 )2
Ψ[𝛽𝑙 , 𝛽𝑟 ] = 𝑎𝐿
2
0
1
1
+ 𝜅(⟨𝛽𝑙 ⟩ sin 2𝜑𝑙 + ⟨𝛽𝑟 ⟩ sin 2𝜑𝑟 )2 + (𝛾 − ⟨𝛽𝑙 ⟩ cos 2𝜑𝑙 − ⟨𝛽𝑟 ⟩ cos 2𝜑𝑟 )2
2
2
(
)]
1
1
𝜌𝑙 𝜌𝑟
+ 𝑘 ln
𝑑𝑦. (4.101)
+ ln
+𝜒 2
1 − 𝜌𝜌𝑠𝑙
1 − 𝜌𝜌𝑟𝑠
𝜌𝑠
Using (4.64) for small up to moderate dislocation densities we reduce (4.101)
to
∫ ℎ [
1
𝜇 (1 − 𝜅)(𝛽𝑙 sin 2𝜑𝑙 + 𝛽𝑟 sin 2𝜑𝑟 )2
Ψ[𝛽𝑙 , 𝛽𝑟 ] = 𝑎𝐿
2
0
1
1
+ 𝜅(⟨𝛽𝑙 ⟩ sin 2𝜑𝑙 + ⟨𝛽𝑟 ⟩ sin 2𝜑𝑟 )2 + (𝛾 − ⟨𝛽𝑙 ⟩ cos 2𝜑𝑙 − ⟨𝛽𝑟 ⟩ cos 2𝜑𝑟 )2
2
2
(
)]
( )2
( )2
𝜌𝑙
𝜌𝑟 1 𝜌𝑟
𝜌𝑙 𝜌𝑟
1 𝜌𝑟
+
+𝜒 2
+𝑘
𝑑𝑦. (4.102)
+
+
𝜌𝑠 2 𝜌𝑠
𝜌𝑠 2 𝜌𝑠
𝜌𝑠
If the dissipation can be neglected, then 𝛽𝑙 and 𝛽𝑟 minimize (4.102) under
the constraints (4.95).
If the resistance to dislocation motion cannot be ignored, then the energy
minimization must be replaced by the flow rules. For 𝛽˙𝑙 ∕= 0 and 𝛽˙𝑟 ∕= 0 we
have
∂𝐷
∂𝐷
𝛿𝛾 𝜙
𝛿𝛾 𝜙
,
,
(4.103)
=−
=−
𝛿𝛽𝑙
𝛿𝛽𝑟
∂ 𝛽˙ 𝑙
∂ 𝛽˙ 𝑟
where the dissipation potential is
𝐷 = 𝐾𝑙 ∣𝛽˙ 𝑙 ∣ + 𝐾𝑟 ∣𝛽˙ 𝑟 ∣,
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
113
with 𝐾𝑙 and 𝐾𝑟 being positive constants called critical resolved shear stresses
of the corresponding slip systems. The right hand sides of (4.103) are the
negative variational derivatives of the energy with respect to 𝛽𝑙 and 𝛽𝑟 , respectively, at fixed overall shear strain 𝛾
ϰ𝔞 ≡ −
𝛿𝛾 𝜙
∂𝜙
∂ ∂𝜙
=−
+
,
𝛿𝛽𝔞
∂𝛽𝔞 ∂𝑦 ∂𝛽𝔞,𝑦
𝔞 = 𝑙, 𝑟.
The corresponding evolution equation (4.103) does not have to be fulfilled
for 𝛽˙𝑙 = 0 or 𝛽˙𝑟 = 0. It is replaced by the equation 𝛽˙𝑙 = 0 or 𝛽˙𝑟 = 0.
If the resistance to dislocation motion can be neglected (and hence the
energy dissipation is zero) the determination of 𝛽𝑙 (𝑦) and 𝛽𝑟 (𝑦) reduces to
the minimization of the total energy (4.102). We analyze this variational
problem first in the special case 𝜑𝑟 = 𝜋 − 𝜑𝑙 = 𝜋 − 𝜑 which corresponds
to symmetric double slip. It is convenient to introduce the dimensionless
variables (4.66), in terms of which the functional (4.102) reads
𝐸[𝛽𝑙 , 𝛽𝑟 ] =
∫ ℎ[
0
1
1
(1 − 𝜅)(𝛽𝑙 − 𝛽𝑟 )2 sin2 2𝜑 + 𝜅(⟨𝛽𝑙 ⟩ − ⟨𝛽𝑟 ⟩)2 sin2 2𝜑
2
2
1
+ (𝛾 − (⟨𝛽𝑙 ⟩ + ⟨𝛽𝑟 ⟩) cos 2𝜑)2 + 𝑘∣ sin 𝜑∣(∣𝛽𝑙′ ∣ + ∣𝛽𝑟′ ∣)
2
]
1
2
′2
′2
′
′
+ 𝑘 sin 𝜑(𝛽𝑙 + 𝛽𝑟 + 2𝜒∣𝛽𝑙 ∣∣𝛽𝑟 ∣) 𝑑𝑦, (4.104)
2
where the prime denotes differentiation with respect to 𝑦¯, and, for short,
the bars over 𝑦 and ℎ are dropped. We minimize functional (4.104) among
𝛽𝔞 satisfying the boundary conditions (4.95). To guarantee existence and
uniqueness of the minimizer we must ensure convexity of the free energy
density 𝜙 with respect to 𝛽𝔞′ . For this purpose let us consider the matrix
(
)
(
)
𝜙,𝛽𝑙′ 𝛽𝑙′ 𝜙,𝛽𝑙′ 𝛽𝑟′
1
𝜒sign𝛽𝑙′ sign𝛽𝑟′
2
= 𝑘 sin 𝜑
𝜙,𝛽𝑟′ 𝛽𝑙′ 𝜙,𝛽𝑟′ 𝛽𝑟′
𝜒sign𝛽𝑙′ sign𝛽𝑟′
1
It is easy to see that this matrix is positive definite (and, consequently,
the energy is convex with respect to 𝛽𝔞′ ) if 𝜑 ∕= 0 and 𝜒 < 1. For 𝜒 =
1 the determinant of the matrix vanishes and there exists an eigenvector
corresponding to a zero eigenvalue. Thus, for 𝜒 = 1 the energy is no longer
strictly convex and one may expect non-uniqueness of the minimizer as well
as some numerical instability. In order to avoid this deficiency we will assume
that 𝜒 < 1.
Similarly to the single slip problem, there exists a threshold value 𝛾𝑒𝑛
such that when 𝛾 < 𝛾𝑒𝑛 no dislocations are nucleated and 𝛽𝑙 = 𝛽𝑟 = 0. In
114
CHAPTER 4. CRYSTAL PLASTICITY
order to find the threshold value, we
form
⎧
𝛽𝑙𝑚

⎨ 𝜖 𝑦,
𝛽𝑙 = 𝛽𝑙𝑚 ,

⎩ 𝛽𝑙𝑚
(ℎ − 𝑦),
𝜖
and
employ minimizing sequences of the
for 𝑦 ∈ (0, 𝜖),
for 𝑦 ∈ (𝜖, ℎ − 𝜖),
for 𝑦 ∈ (ℎ − 𝜖, ℎ),
⎧
𝛽𝑟𝑚

for 𝑦 ∈ (0, 𝜖),
⎨ 𝜖 𝑦,
𝛽𝑟 = 𝛽𝑟𝑚 ,
for 𝑦 ∈ (𝜖, ℎ − 𝜖),

⎩ 𝛽𝑟𝑚
(ℎ − 𝑦), for 𝑦 ∈ (ℎ − 𝜖, ℎ),
𝜖
(4.105)
(4.106)
where 𝛽𝑙𝑚 and 𝛽𝑟𝑚 are unknown constants, and 𝜖 is a small unknown length
which tends to zero as 𝛾 → 𝛾𝑒𝑛 .
Substituting (4.105) and (4.106) into the energy functional (4.104) (with
the last term being removed) and neglecting all small terms of order 𝜖 and
higher, we obtain the function of two variables
)
1 (
𝐸(𝛽𝑙𝑚 , 𝛽𝑟𝑚 ) = ℎ (𝛽𝑙𝑚 − 𝛽𝑟𝑚 )2 sin2 2𝜑 + (𝛾 − (𝛽𝑙𝑚 + 𝛽𝑟𝑚 ) cos 2𝜑)2
2
(4.107)
+ 2𝑘∣ sin 𝜑∣(∣𝛽𝑙𝑚 ∣ + ∣𝛽𝑟𝑚 ∣).
The partial derivatives of (4.107) with respect to 𝛽𝑙𝑚 and 𝛽𝑟𝑚 must vanish
giving
∂𝐸
= ℎ sin2 2𝜑(𝛽𝑙𝑚 − 𝛽𝑟𝑚 ) − ℎ cos 2𝜑(𝛾 − (𝛽𝑙𝑚 + 𝛽𝑟𝑚 ) cos 2𝜑)
∂𝛽𝑙𝑚
+ 2𝑘∣ sin 𝜑∣sign𝛽𝑙𝑚 = 0,
and
∂𝐸
= −ℎ sin2 2𝜑(𝛽𝑙𝑚 − 𝛽𝑟𝑚 ) − ℎ cos 2𝜑(𝛾 − (𝛽𝑙𝑚 + 𝛽𝑟𝑚 ) cos 2𝜑)
∂𝛽𝑟𝑚
+ 2𝑘∣ sin 𝜑∣sign𝛽𝑟𝑚 = 0.
It is easy to see that these equations imply 𝛽𝑙𝑚 = 𝛽𝑟𝑚 = 𝛽𝑚 . Then, a rather
simple analysis shows that the minimum of (4.107) is achieved at 𝛽𝑚 ∕= 0 if
and only if (in terms of the original length ℎ)
𝛾 > 𝛾𝑒𝑛 =
2𝑘 ∣ sin 𝜑∣
,
ℎ𝑏𝜌𝑠 ∣ cos 2𝜑∣
otherwise it is achieved at 𝛽𝑚 = 0 (no dislocations are nucleated). Note that
the sign of 𝛽𝑚 depends on the angle 𝜑: 𝛽𝑚 is positive if 0∘ < 𝜑 < 45∘ and is
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
115
negative if 45∘ < 𝜑 < 90∘ . The energetic threshold value for the symmetric
double slip turns out to be exactly the same as that of the single slip problem,
see equation (4.71).
Based on the previous analysis, we now assume that 𝛽𝑙 (𝑦) = 𝛽𝑟 (𝑦) = 𝛽(𝑦)
for 𝑦 ∈ (0, ℎ). Due to the boundary conditions, 𝛽 ′ should change its sign on
the interval (0, ℎ). The one-dimensional theory of dislocation pile-ups as well
as the solution of the previous problem suggest to seek the minimizer in the
form
⎧

for 𝑦 ∈ (0, 𝑙),
⎨𝛽1 (𝑦),
𝛽(𝑦) = 𝛽𝑚 ,
(4.108)
for 𝑦 ∈ (𝑙, ℎ − 𝑙),

⎩
𝛽1 (ℎ − 𝑦), for 𝑦 ∈ (ℎ − 𝑙, ℎ),
where 𝛽𝑚 is a constant, 𝑙 an unknown length, 0 ≤ 𝑙 ≤ ℎ2 , and 𝛽1 (𝑙) = 𝛽𝑚 .
The total energy functional becomes
)2
(
∫ 𝑙
1
′
2
′2
𝛾 − ⟨𝛽⟩ cos 2𝜑 , (4.109)
𝐸[𝛽] =
(4𝑘∣ sin 𝜑∣∣𝛽1 ∣+4𝑘 sin 𝜑𝛽1 ) 𝑑𝑦 +2ℎ
2
0
where
1
⟨𝛽⟩ =
ℎ
( ∫ 𝑙
)
2
𝛽1 𝑑𝑦 + (ℎ − 2𝑙)𝛽𝑚 .
(4.110)
0
We have to find 𝛽1 (𝑦) and the constants, 𝛽𝑚 and 𝑙.
This variational problem can be solved in exactly the same manner as in
the previous case. Here, we present the solution:
(
)
cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑)
1 2
𝛽1 =
𝑙𝑦 − 𝑦 , 0 ≤ 𝑦 ≤ 𝑙,
(4.111)
2
𝑘(1 + 𝜒) sin2 𝜑
where the average of 𝛽 is given by
⟨𝛽⟩ =
𝛾(3ℎ − 2𝑙)𝑙2 cos 2𝜑
,
6ℎ𝑘(1 + 𝜒) sin2 𝜑 + 2(3ℎ − 2𝑙)𝑙2 cos2 2𝜑
(4.112)
cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑) 2
𝑙 .
2𝑘(1 + 𝜒) sin2 𝜑
(4.113)
with
𝛽𝑚 =
The equation to determine 𝑙 reads
(
)
𝛾
𝛾(3ℎ − 2𝑙)𝑙2 cos 2𝜑
𝑘∣ sin 𝜑∣(sign𝛽1′ )
=0
−
−
2 cos 2𝜑
(ℎ − 2𝑙) cos2 2𝜑
6ℎ𝑘(1 + 𝜒) sin2 𝜑 + 2(3ℎ − 2𝑙)𝑙2 cos2 2𝜑
Fig. 4.27 shows the evolution of 𝛽(¯
𝑦 ) as 𝛾 increases, for 𝜑 = 30∘ (continuous lines) and 𝜑 = 60∘ (dashed lines), where 𝑦¯ = 𝑦𝑏𝜌𝑠 . For the numerical
116
CHAPTER 4. CRYSTAL PLASTICITY
j=30°
j=60°
b
c
b
a
_
y
d
e
f
Figure 4.27: Evolution of 𝛽 for double-slip constrained shear at zero dissipation: a,d) 𝛾 = 0.02, b,e) 𝛾 = 0.04, c,f) 𝛾 = 0.06
simulation we took the material parameters from Table 4.1, and ℎ = 1𝜇m,
so that ℎ̄ = ℎ𝑏𝜌𝑠 = 0.349.
It is interesting to plot the shear stress 𝜏 = 𝜇(𝛾−2⟨𝛽⟩ cos 2𝜑) as a function
of the shear strain. As we know, for 𝛾 < 𝛾𝑒𝑛 no dislocations are nucleated
and 𝛽 = 0, so the shear stress 𝜏 = 𝜇𝛾. For 𝛾 > 𝛾𝑒𝑛 we take ⟨𝛽⟩ from (4.112)
to compute the shear stress.
t/m
j=30°
j=60°
B´
A´
B
A
0
g
Figure 4.28: Normalized shear stress versus shear strain curve for double-slip
constrained shear at zero dissipation
Fig. 4.28 shows the normalized shear stress versus shear strain curve OAB
for 𝜑 = 30∘ and OA’B’ for 𝜑 = 60∘ . The behavior is quite similar to that of
a crystal deforming in single slip.
If the resistance to dislocation motion (and hence the dissipation) cannot be neglected, the plastic distortion may evolve only if one of the yield
conditions ∣ϰ𝑙 ∣ = 𝐾𝑙 and ∣ϰ𝑟 ∣ = 𝐾𝑟 is fulfilled. If ∣ϰ𝔞 ∣ < 𝐾𝔞 , then 𝛽𝔞 is
frozen, the dislocation density remains unchanged, and the crystal deforms
elastically. Again, it is useful to analyze the evolution of 𝛽𝔞 in the special
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
117
case of symmetric double slip for which we assume that 𝛽𝑙 (𝑦) = 𝛽𝑟 (𝑦) = 𝛽(𝑦)
for 𝑦 ∈ (0, ℎ) and 𝐾𝑙 = 𝐾𝑟 = 𝐾. Introducing the dimensionless variable
𝑦¯ = 𝑦𝑏𝜌𝑠 , we write the total energy functional as
)2
]
∫ ℎ [ (
1
′
2
′2
𝜇 2
𝐸(𝛽) =
𝛾 − ⟨𝛽⟩ cos 2𝜑 + 2𝑘∣ sin 𝜑∣∣𝛽 ∣ + 𝑘(1 + 𝜒) sin 𝜑𝛽 𝑑𝑦.
2
0
(4.114)
Computing the variational derivative of (4.114), we derive from (4.103) the
yield condition
𝜇 2 cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑) + 2𝑘(1 + 𝜒) sin2 𝜑𝛽 ′′ = 𝐾.
Consider first the case 𝜑 < 45∘ . We divide this equation by 𝜇 to transform
the yield condition to
2 cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑) + 2𝑘(1 + 𝜒) sin2 𝜑𝛽 ′′ = 2𝛾𝑐𝑟 cos 2𝜑,
(4.115)
with 𝛾𝑐𝑟 ≡ 𝐾/2𝜇 cos 2𝜑 and the prime denoting differentiation with respect
to 𝑦¯. We shall further omit the bar over 𝑦 for short.
We regard 𝛾 as a given function of time (the driving variable) and try
to determine 𝛽(𝑡, 𝑦). We consider the loading path shown in Fig. 4.14. The
problem is to determine the evolution of 𝛽 as a function of 𝑡 and 𝑦, provided
𝛽(0, 𝑦) = 0 and 𝜑 < 45∘ .
Since the plastic distortion, 𝛽, is initially zero, we see from (4.115) that
𝛽 = 0 as long as 𝛾 < 𝛾𝑐𝑟 . Thus, the dissipative threshold stress (the yield
stress) 𝜏𝑦 = 𝐾/2 cos 2𝜑 in this case. For small 𝛽(𝑡, 𝑥) and 𝛾 > 𝛾𝑐𝑟 , the yield
condition becomes
2 cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑) + 4𝑘 sin2 𝜑𝛽 ′′ = 2𝛾𝑐𝑟 cos 2𝜑,
(4.116)
Let us introduce the deviation of 𝛾(𝑡) from the critical shear 𝛾𝑐𝑟 , 𝛾𝑟 = 𝛾 − 𝛾𝑐𝑟
and simplify (4.116) to obtain
2 cos 2𝜑(𝛾𝑟 − 2⟨𝛽⟩ cos 2𝜑) + 4𝑘 sin2 𝜑𝛽 ′′ = 0.
(4.117)
Since this equation is linear, 𝛽 is proportional to 𝛾𝑟 such that 𝛽 = 𝛾𝑟 𝛽1 ,
where 𝛽1 is
cos 2𝜑(1 − 2⟨𝛽1 ⟩ cos 2𝜑)
𝛽1 =
(ℎ − 𝑦)𝑦.
(4.118)
2𝑘(1 + 𝜒) sin2 𝜑
The average of 𝛽1 is obtained in the form
⟨𝛽1 ⟩ =
ℎ2 cos 2𝜑
.
12𝑘(1 + 𝜒) sin2 𝜑 + 2ℎ2 cos2 2𝜑
(4.119)
118
CHAPTER 4. CRYSTAL PLASTICITY
b1
j=30°
j=60°
_
y
Figure 4.29: Graphs of 𝛽1 (¯
𝑦 ) for double-slip constrained shear at non-zero
dissipation
Fig. 4.29 shows the graphs of 𝛽1 (¯
𝑦 ) for 𝜑 = 30∘ (continuous line) and
𝜑 = 60∘ (dashed line). For the numerical simulation we took the material
parameters from Table 4.1, and ℎ = 1𝜇m, so that ℎ̄ = ℎ𝑏𝜌𝑠 = 0.349.
After reaching 𝛾 ∗ > 𝛾𝑐𝑟 , we unload the crystal by decreasing 𝛾. Since ϰ
becomes smaller than 𝐾, 𝛽 does not change (𝛽 = 𝛽 ∗ (𝑦)) until
2 cos 2𝜑(𝛾 − 2⟨𝛽 ∗ ⟩ cos 2𝜑) + 2𝑘(1 + 𝜒) sin2 𝜑𝛽 ∗′′ = 2𝛾𝑐𝑟 cos 2𝜑,
(4.120)
where 𝛽 ∗ (𝑦) is the solution of (4.116) for 𝛾(𝑡) = 𝛾 ∗ . From (4.120) we can see
that the plastic flow begins when 𝛾 − (𝛾 ∗ − 𝛾𝑐𝑟 ) = −𝛾𝑐𝑟 , i.e. for 𝛾 = 𝛾∗ =
𝛾 ∗ − 2𝛾𝑐𝑟 . From that value of 𝛾, the yield condition ϰ = −𝐾 holds leading
to a decrease of 𝛽 which should be now determined from the equation
2 cos 2𝜑(𝛾 − 2⟨𝛽⟩ cos 2𝜑) + 2𝑘(1 + 𝜒) sin2 𝜑𝛽 ′′ = 2𝛾𝑐𝑟 cos 2𝜑.
(4.121)
Since for 𝛾 ∈ (−𝛾𝑐𝑟 , 𝛾∗ ), the deviation 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 is positive, equation
(4.121) can again be transformed to equation (4.117) and solved in exactly
the same manner if we replace 𝛾𝑟 = 𝛾 −𝛾𝑐𝑟 in all formulas (4.118) and (4.119)
by 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 . As 𝛾 approaches −𝛾𝑐𝑟 , 𝛽 tends to zero because 𝛾𝑙 → 0. The
further increase of 𝛾 from −𝛾𝑐𝑟 to zero does not cause change in 𝛽 which
remains zero.
It is not difficult to modify the construction given above to find the solution for 𝜑 > 45∘ .
For symmetric double slip the dislocation densities are the same, 𝛼𝑙 =
𝛼𝑟 = 𝛽 ′ sin 𝜑. The resultant Burgers’ vector of all dislocations has only one
non-zero component in the 𝑦-direction: 𝛼𝑦𝑧 = 2𝛽,𝑦 sin2 𝜑. Thus, couples of
dislocations near the boundaries form “super” dislocations with the Burgers’
vector in the 𝑦-direction. The normalized dislocation density 𝛼 = 𝛽 ′ sin 𝜑
119
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
a1
j=30°
j=60°
_
y
Figure 4.30: Graphs of 𝛼1 (¯
𝑦 ) for double-slip constrained shear at non-zero
dissipation
can be calculated from the solution (4.118). Since 𝛽(𝑦) is proportional to 𝛾𝑟 ,
𝛼(𝑦) is also proportional to 𝛾𝑟 such that 𝛼(𝑦) = 𝛾𝑟 𝛼1 (𝑦) with
𝛼1 (𝑦) =
cos 2𝜑(1 − 2⟨𝛽1 ⟩ cos 2𝜑)
(ℎ − 2𝑦),
2𝑘(1 + 𝜒) sin 𝜑
and with ⟨𝛽1 ⟩ from (4.119). Fig. 4.30 shows the graphs of 𝛼1 for 𝜑 = 30∘
(continuous line) and 𝜑 = 60∘ (dashed line) for 𝑦¯ ∈ (0, ℎ̄).
t/t0
h/d=40
h/d=80
h/d=160
h/d=240
with dissipation
energy minimization
g
Figure 4.31: Comparison of the normalized shear stress versus shear strain
curves obtained from this approach and in [6]
It is interesting to calculate the shear stress 𝜏 which is a measurable
quantity. During loading, we have for the normalized shear stress (or the
elastic shear strain)
𝜖(𝑒) =
𝜏
= 𝛾𝑐𝑟 + (𝛾𝑟 − 2⟨𝛽1 ⟩ cos 2𝜑) ,
𝜇
(4.122)
120
CHAPTER 4. CRYSTAL PLASTICITY
with ⟨𝛽1 ⟩ from (4.119). The second term of (4.122) causes hardening due to
the dislocations piling up. Equation (4.122) describes the size effect in this
model.
Fig. 4.31 shows the stress-strain curves during loading obtained from
energy minimization and from the flow rule as compared with the average
stress-strain curves from discrete dislocation simulations in [6]. In order
to compare with the discrete dislocation simulations we took 𝐾 such that
𝛾𝑐𝑟 = 𝛾𝑒𝑛 = 0.00118, 𝜑 = 60∘ , 𝜏0 = 1.9 10−3 𝜇 and let all other material
constants remain the same as in the previous Section. The average stressstrain curves in the discrete dislocation simulations are provided for four
different ratios ℎ/𝑑, where 𝑑 is the spacing between the active slip planes. It
is seen that reasonably good agreement of the discrete and the continuum
approaches is observed at ℎ/𝑑 = 240.
u,y
h/d=80
energy minimization
with dissipation
g=0.0218
g=0.0168
g=0.0118
g=0.0068
y
_
h
Figure 4.32: Comparison of the total shear strain profiles obtained from the
present approach and in [6]
Fig. 4.32 shows the total shear strain profiles obtained from energy minimization and from the flow rule as compared with the average profiles produced by the discrete dislocation simulations in [6]. A somewhat larger discrepancy between the shear strain profiles obtained by the discrete and continuum approaches is perhaps due to the rather low ratio ℎ/𝑑 = 80 taken to
simulate these curves in [6].
During inverse loading, when the yield condition ϰ = −𝐾 holds true,
equation (4.122) changes to
𝜏
= −𝛾𝑐𝑟 + (𝛾𝑙 − 2⟨𝛽1 ⟩ cos 2𝜑) ,
𝜇
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
121
where
ℎ2 cos 2𝜑
,
12𝑘(1 + 𝜒) sin2 𝜑 + 2ℎ2 cos2 2𝜑
with the deviation 𝛾𝑙 = 𝛾 + 𝛾𝑐𝑟 used instead of 𝛾𝑟 = 𝛾 − 𝛾𝑐𝑟 .
⟨𝛽1 ⟩ =
t/m
B
A
g*
-gcr 0 gcr
g*
g
C
D
Figure 4.33: Normalized shear stress versus shear strain curve for double-slip
constrained shear at non-zero dissipation for 𝜑 = 60∘
Fig. 4.33 shows the normalized shear stress (or elastic shear strain) versus
shear strain curve for the loading path of Fig. 4.14, with 𝛾 ∗ = 0.025, 𝜑 = 60∘ ,
ℎ = 1𝜇m, while all other parameters remain the same. The straight line OA
corresponds to purely elastic loading with 𝛾 increasing from zero to 𝛾𝑐𝑟 . Line
AB corresponds to plastic yielding with ϰ = 𝐾. Yielding begins at point
A with the yield stress 𝜏𝑦 = −𝐾/2 cos 2𝜑, and we can observe the work
hardening due to the dislocation pile-up which is described by the second
term of (4.122). During unloading, as 𝛾 decreases from 𝛾 ∗ to 𝛾∗ = 𝛾 ∗ − 2𝛾𝑐𝑟
(line BC), the plastic distortion 𝛽 = 𝛽 ∗ is frozen. As 𝛾 decreases further
from 𝛾∗ to −𝛾𝑐𝑟 , plastic yielding occurs with ϰ = −𝐾 (line CD). From
Fig. 4.33 it is seen that the yield stress 𝜏𝑦 = 𝜏 ∗ + 2𝐾/2 cos 2𝜑 at point C, at
which the inverse plastic flow sets on, is larger than −𝐾/2 cos 2𝜑 (because
𝜏 ∗ > −𝐾/2 cos 2𝜑). Along line CD, as 𝛾 is decreased, the created dislocations
annihilate, and at point D all dislocations have disappeared. Finally, as 𝛾
increases from −𝛾𝑐𝑟 to zero, the crystal behaves elastically with 𝛽 = 0. In
this closed cycle OABCDO dissipation occurs only along lines AB and CD.
It is interesting that lines DA and BC are parallel and have the same length.
This corresponds again to the Bauschinger effect as in the case of single slip.
Exercises
4.1 The only non-zero component of the plastic distortion is
𝛽𝑦𝑧 = 𝑥2 (1 + 𝑦 4 ).
122
CHAPTER 4. CRYSTAL PLASTICITY
Find the dislocation density.
4.2 Prove that Nye’s dislocation density tensor satisfies the following equation
𝛼𝑖𝑗,𝑗 = 0,
which means that dislocations cannot end in the lattice.
4.3 Prove the following identity
1
−𝜖𝑖𝑎𝑏 𝜖𝑗𝑐𝑑 𝜀𝑝𝑑𝑏,𝑎𝑐 + (𝜖𝑖𝑎𝑏 𝛼𝑏𝑗,𝑎 + 𝜖𝑗𝑎𝑏 𝛼𝑏𝑖,𝑎 ) = 0.
2
4.4∗ Given the plastic strain tensor 𝜀𝑝𝑖𝑗 and the dislocation density tensor
𝛼𝑖𝑗 satisfying the equations in Exercises 4.2 and 4.3. Find the plastic
distortion tensor.
4.5 The plastic distortion tensor is given by
𝛽𝛼𝛽 = 𝛽(𝑥, 𝑦)𝑠𝛼 𝑚𝛽 ,
where s = (cos 𝜑, sin 𝜑) and m = (− sin 𝜑, cos 𝜑). Find the dislocation
density tensor.
y
M
x
Figure 4.34: Bending of a beam
F
Figure 4.35: Indentation
4.6∗ For the plane strain bending of a beam with the slip planes parallel to
the plane 𝑦 = 0 and the dislocation lines parallel to the 𝑧-axis shown
in Fig. 4.34 write the energy functional of the crystal beam.
4.7 Develop the numerical procedure for the plane constrained shear of a
single crystal deforming in non-symmetric double slip.
4.8 The similar problem for a bicrystal having non-symmetric slip system.
4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP
123
4.9∗ A rigid wedge is pressed into a crystal by a force 𝐹 as shown in Fig. 4.35.
Develop a numerical procedure to find the dislocation density beneath
the wedge and the force versus displacement curve.
4.10∗ Find the dislocation density near a rigid spherical inclusion embedded
in an infinite crystal under tension.
124
CHAPTER 4. CRYSTAL PLASTICITY
Bibliography
[1] Calladine, C. R., Engineering Plasticity, Pergamon 1969.
[2] Hill, R., The Mathematical Theory of Plasticity, Clarendon Press 1983.
[3] Hull, D., Introduction to Dislocations, Pergamon Press 1975.
[4] Le, K. C., Introduction to Micromechanics, Nova Science 2010.
[5] Lubliner, J., Plasticity theory, Macmillan Publishing Company 1990.
[6] J. Y. Shu, N. A. Fleck, E. Van der Giessen, and A. Needleman. Boundary
layers in constrained plastic flow: comparison of nonlocal and discrete
dislocation plasticity. J. Mech. Phys. Solids, 49:1361–1395, 2001.
125