Plastic e

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PLASTICITY

Professor Khanh Chau Le

Lehrstuhl f ur Allgemeine MechanikRuhr Universit at Bochum

Universit atsstr. 150, D 44780 BochumLecture Notes

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Contents

1 Fundamentals 7

1.1 Phenomenon of plastic deformation . . . . . . . . . . . . . . . 71.2 Mechanical framework . . . . . . . . . . . . . . . . . . . . . . 101.3 Thermodynamical framework . . . . . . . . . . . . . . . . . . 151.4 Constitutive law . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5 Closed system of equations . . . . . . . . . . . . . . . . . . . . 26

2 Elementary theory 292.1 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Torsion of a cylinder . . . . . . . . . . . . . . . . . . . . . . . 392.3 Cylindrical shell under combined load . . . . . . . . . . . . . . 412.4 Simple metal forming processes . . . . . . . . . . . . . . . . . 45

3 Theory of plastic ow 513.1 Governing equations . . . . . . . . . . . . . . . . . . . . . . . 513.2 Torsion of prismatic bars . . . . . . . . . . . . . . . . . . . . . 533.3 Plane strain problems . . . . . . . . . . . . . . . . . . . . . . . 563.4 Plane stress problems . . . . . . . . . . . . . . . . . . . . . . . 70

4 Crystal plasticity 734.1 Physical background . . . . . . . . . . . . . . . . . . . . . . . 734.2 Continuum dislocation theory . . . . . . . . . . . . . . . . . . 79

4.3 Anti plane constrained shear . . . . . . . . . . . . . . . . . . . 844.4 Plane constrained shear . . . . . . . . . . . . . . . . . . . . . 974.5 Single crystals deforming in double slip . . . . . . . . . . . . . 110

3

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4 CONTENTS

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Preliminary remark

In this course we shall restrict ourselves to the deformable solids. In solidmechanics 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 of a material point in terms of the local strain tensor and the localtemperature :

←→( )

Let us rst classify the form of this constitutive relation.As you know from the theory of elasticity, elastic materials are character

ized by a single valued scalar function, called free energy density (per unitvolume) and denoted by ( ), such that

= ∂ ∂

( )

This is the so called state equation for thermoelastic solids. The measuresof strain and stress tensors can still be chosen differently for small and nitedeformations.

For inelastic materials this one to one relation is no longer valid. Thestress strain relation depends now on the history of loading and deformation. We can roughly classify the inelastic material behavior according tothe following features

· rate independent phenomena. The material behavior does not dependon the loading rate. Example: plastic deformation.

5

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6 CONTENTS

· rate dependent phenomena. The material behavior depends on theloading 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 shortdiscussion 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 thecondition of small strains. Within the framework of

i) the so called elementary theory, as well as

ii) the theory of plastic ow

we shall solve some simple problems to show how the elastoplastic deformation of solids can be determined. Finally, we give a short introductionto the modern crystal plasticity incorporating the continuously distributeddislocations.

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Chapter 1

Fundamentals

1.1 Phenomenon of plastic deformation

Simple tension or compression testLet us consider rst the uniaxial tension test with the subsequent unloadingfor two materials: i) pure cooper, and ii) soft annealed carbon steel. Thecorresponding stress strain curves are shown below in Fig. 1.1,

Y Y

O O

AA

BB

i) ii)

C Cp

p e+pe

Figure 1.1: Stress strain curve

where the strain and stress are dened as follows

= Δ

0 =

0

˙ < 10−3

One remark should be made concerning the denition of stress. Since thedeformed cross section at tension shrinks, the true stress should actually bedened as , where is the current cross section area. However, at smallstrains of the order < 1% the error is not so grave.

7

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8 CHAPTER 1. FUNDAMENTALS

Looking at the stress strain curve one can recognize two different typesof material response in the elastic and elasto plastic regions. In the purelyelastic region (within the line OA) no residual strain is observed: the specimen assume its original length after the load is removed. For most of metalsthe stress is proportional to the strain so that the Hooke law is valid. Thepurely elastic region ends at point A corresponding to the yield stress .Beyond this purely elastic region we observe for cooper i) a mild transitionto the elasto plastic region, while for steel ii) a sharp yield stress markedby a nearly horizontal segment. If the specimen is loaded beyond this yieldstress, it begins to deform plastically. The specimen shows a residual strainafter unloading. The total strain is additively decomposed into the elastic

and plastic parts = + =

+

After the unloading the plastic strain remains. In Fig. 1.1 this loading andunloading processes are shown by the stress strain curve (marked with arrows) from O through A to B and nally from B back to C.

Determination of the yield stressIt is not always easy to determine in praxis the yield stress from the stressstrain curve. Normally, one measures the elastic modulus, at which a xedamount of residual strain occurs (say, off = 0 2%). With this modulus of

elasticity 0 one can determine the yield stress (see Fig. 1.2).

off

Y

Figure 1.2: Fixing the yield stress

HardeningIn the elasto plastic region, when the specimen is reloaded again after theunloading, one observes approximately the same stress strain curve (onlyin the opposite direction), apart from a small hysteresis loop and a rathermilder transition to the elasto plastic region. This means that the materialbehaves elastically up to the point B (see Fig. 1.1), and the plastic deformation begins to increase again when the stress achieves its value

∗ at point B

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1.1. PHENOMENON OF PLASTIC DEFORMATION 9

corresponding to the stress at the end of the previous loading process. Thisstress ∗ can be regarded as the new yield stress . Since is higherthan the initial yield stress 0, one speaks of the hardening behavior. Thefollowing power law, which is phenomenological, can be used to describe thehardening behavior

= − 0

≥1

and inversely = √ + 0

Y0

1

1

E

H/(1+H/E)

Figure 1.3: Linear hardening

For = 1 we have a linear hardening (see Fig. 1.3)

= 1

( − 0)

Y

Figure 1.4: Elastic ideal plastic material behavior

If we replace the increasing hardening curve by a horizontal line (seeFig. 1.4), then the material behavior of this idealized material is called elasticideal plastic.

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The same can be said about the hardening behavior for the compressiontest. One needs just to inverse the signs of 0 as well as 0. Thecorresponding inequalities must also be modied appropriately.Bauschinger effectWe consider now a loading process, for which the specimen is rst loadedin tension to attain a certain amount of plastic strain, then is unloaded andimmediately loaded in compression. The corresponding stress strain curve isshown in Fig. 1.5.

Y0

Y+

Y-

Figure 1.5: Process with loading in compression

We observe that + > ∣ −∣

This phenomenon is called Bauschinger effect.

1.2 Mechanical framework

To keep the presentation as simple as possible let us use cartesian coordinatesto describe deformations of solids.Kinematics

At the beginning of the process at time = 0 the body occupies the regionℬ of the three dimensional Euclidean point space. The position vector of anarbitrary material point is denoted by x , and its components by = 1 2 3.The displacement vector of this material point is denoted by u (x ), with

(x ) being its components. The deformation gradient is given by

F = grad( x + u ) = I + grad u

or, in components = +

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some orthogonal transformation of coordinate system. For this purpose oneneeds to nd all eigenvectors n and the corresponding eigenvalues of Afrom the equation

( − ) = 0

This homogeneous equation for the eigenvectors has nontrivial solutions if and only if its determinant vanishes

det( − ) = 0

This is a cubic equation for the eigenvalues which looks in the expanded formas follows

3 − 2 + − = 0

The three coefficients of this cubic equation , , are called principalinvariants of the tensor A . The computations give

= 11 + 22 + 33 =

= 11 12

21 22+ 11 13

31 33+ 22 23

32 33

= det A

Denoting the eigenvalues of A by 1 2 3, we can simply express theseprincipal invariants as

= 1 + 2 + 3

= 1 2 + 1 3 + 2 3

= 1 2 3

According to the above result, we may diagonalize the strain tensor too. Its eigenvalues, called principal strains, will be denoted by 1 2 3.

Balance equationsLet be the mass density, the Cauchy stress tensor, the body force.We formulate the balance of momentum and moment of momentum in theform

=

+ ∂ (1.1)

=

+ ∂

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1.2. MECHANICAL FRAMEWORK 13

for an arbitrary regular volume

⊆ ℬ of the body. Here = 1 2 3

denotes the volume element, the area element,

=⎧⎨⎩

0 when at least two indices coincide1 when is an even permutation

−1 otherwise

is called the permutation symbol, and ˙ corresponds to the time derivativeof . The balance of momentum generalize Newton s second law of theclassical 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 accordancewith Gauss formula. Since is arbitrary and since the integrand is assumedto be continuous, we may derive from (1.1) the balance equations in localform

= + (1.2) =

Thus, the balance of moment of momentum implies the symmetry of thestress tensor .

In case of equilibrium the displacement vector does not depend on time so that the inertial term vanishes. The equation of motion reduces thento the equilibrium condition

+ = 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 correspondinginertial 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 tensorSince the stress tensor is symmetric, it can also be diagonalized. Theeigenvalues of this tensor, 1 2 3, are called principal stresses. The principal invariants of the stress tensor are

= 1 + 2 + 3

= 1 2 + 1 3 + 2 3

= 1 2 3

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The hydrostatic stress is dened as = 3.The stress deviator is dened as follows

= −

The following invariants of the stress deviator are often used in the plasticitytheory

1 = = 0

2 = 12

= 16

[( 1 − 2)2 + ( 2 − 3)2 + ( 3 − 1)2] (1.4)

3 = dets =

1

27[( 1 − 2)2

( 1 − 3 + 2 − 3) + ( 2 − 3)2

( 2 − 1 + 3 − 1) + ( 3 − 1)2( 3 − 2 + 1 − 2)]

One can see that the invariants 2 3 are symmetric functions of − .

Figure 1.7: Mohr s stress circles

The geometric interpretation of 2 can be given in terms of the octahedralshear stress. Consider the normal vector

n = 1√ 3(1 1 1)

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 vectoracting on this area element is given by

t = n = 1√ 3( 1 2 3)

The normal stress equals

oct = n⋅

t = 13

( 1 + 2 + 3) =

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1.3. THERMODYNAMICAL FRAMEWORK 15

The shear stress acting on the side of the octahedron is obtained from theformula

2oct = 13

( 21 + 2

2 + 23) −

19

( 1 + 2 + 3)2 = 23

2

It is interesting to mention that the octahedral shear stress is the averageshear stress over all planes passing through a material point.

With the help of Mohr s stress circles we can also determine the maximumshear stress (see Fig. 1.7)

max = 12

max ∣1 − 2∣∣2 − 3∣∣3 − 1∣

1.3 Thermodynamical frameworkIt is well known from the classical experiment by Taylor and Quinney thatabout 90% of the work done to deform metals plastically will be dissipatedinto heat. This heat suply leads in general to the change of temperature.Thus, if the plastic deformations occur, the process we are dealing withbecomes thermo mechanically coupled. The consequence is that, in plasticity,thermodynamic balance equations should be taken into account.

Energy balanceWe assume that the energy of an arbitrary sub body

is a sum of the kinetic

and internal energies

ℰ =

( + 12

)

with = ˙ being the material velocity. Here corresponds to the internalenergy density. The balance of energy states

˙ℰ = +

where is the power of the external forces, and is the rate at which heatis supplied to the body. The power

of the body and contact forces is given

in the form

=

+ ∂

The heat supply comes from two sources: the body heat supply and the heatow across the boundary; its rate is equal to

=

− ∂

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Here (x ) is the body heat supply per unit mass and unit time, (x ) isthe heat ux vector across the surface per unit time. The heat ow ispositive if q and n are opposite; therefore the minus sign in the last equationagrees with our common sense.

Substituting the above formulas for the power and the heat supply in theright hand side of the balance of energy and transforming the surface integralinto the volume integral, we get

[ ( ˙ + ˙ − − ) −( ) + ] = 0

Since this equation holds true for an arbitrary regular sub body , and sincethe integrand is assumed to be continuous, we obtain the balance of energyin the local form

( ˙ + ˙ − − ) −( ) + = 0

Taking into account the balance of momentum (1.2) we obtain nally

˙ + = ˙ + (1.5)

Second law of thermodynamics

In order to formulate the second law of thermodynamics we need two newquantities. The rst one is the absolute temperature, referred to as an intensive quantity and denoted by (x ). The second one is the entropy, referredto as an extensive quantity, whose density is denoted by (x ). The entropyof the sub body is given by

The second law of thermodynamics states that

≥

−

∂

(1.6)

When the heat supply and the heat ow are absent (adiabatic process with = 0 and = 0), the following inequality holds true

≥0

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1.4. CONSTITUTIVE LAW 17

which means that the entropy of the closed system cannot decrease.With the help of Gauss theorem we obtain

˙ ≥

[ −( ) ]

Since is arbitrary, this inequality leads to

˙ ≥ −( ) = − + 2 (1.7)

We call = ˙− + ( ) the entropy production rate. The inequality(1.7) says that

≥0

There is an alternative form of the entropy production inequality oftenused in plasticity. We introduce the free energy density

= −Provided all other balance equations hold true, then the entropy productioninequality is equivalent to

( ˙ + ˙) − ˙ + ≤0 (1.8)

To prove (1.8) we use the denition of

˙ = ˙ − ˙ − ˙ ⇒ ˙ = ˙ − ˙ − ˙

Substitute this into (1.7) and multiply by

( ˙ − ˙ − ˙) ≥ − +

Combining this equation with the energy balance equation (1.5), we arriveat (1.8).

For isothermal processes with = const the inequality (1.8) reduces to

˙ − ˙ ≤0 (1.9)

1.4 Constitutive law

The formulation of the constitutive laws begins always with the specicationof all quantities characterizing the current state of the material element. Suchquantities are called state variables. Besides, one needs to specify all internalvariables which may inuence the dissipation and the irreversible behaviorof the material element. The constitutive laws for elastoplastic materialsinclude:

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· Specication of the free energy as function of all state variables. Bythis the reversible behavior of the material is xed.

· Evolution law for the internal variables (plastic strains + hardeningparameters)

· A law for the heat ux (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 materialsWe restrict ourselves to isothermal processes with = const. For elasticideal plastic materials we include in the list of variables the following quantities

(1.10)

We assume that the elastic strains characterize completely the currentstate of the deformed material element. This means that the stress tensordepends only on

= ( )

The plastic strains depend on the history of loading and therefore arenot the state variables. They present the internal variables which characterize irreversible behavior of the material element. The total strain tensor isadditively decomposed into the elastic and plastic strain tensors

= + (1.11)

The free energy density assumes the form

= ( )

i. e., it depends only on the state variables. Let us differentiate the freeenergy with respect to time

˙ = ∂∂

˙

We substitute this formula into the dissipation inequality (1.9)

( ∂∂ − ) ˙ − ˙ ≤0 (1.12)

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We rst consider processes with ˙ = 0, i. e. reversible processes. For theseprocesses the second term in (1.12) vanishes, so that

( ∂∂ − ) ˙ ≤0

Since ˙ can be arbitrary, and since the expression in the brackets does notdepend on ˙ , it must be identically equal to zero and thus

= ∂∂

(1.13)

If the free energy density per unit volume = is a quadratic form of

= 12

then (1.13) yield Hooke s law

=

For isotropic elastic material we have

= + 2This equation of state can be decomposed into the volumetric and deviatoricparts

= 3 = 2

where = − 13 is the strain deviator, and = + 2 3. In rate

form we have

˙ = 3 ˙ (1.14)˙ = 2 ˙

With (1.13) we reduce the inequality (1.12) to the following dissipationinequality

˙ ≥0

The left hand side of this equation is called plastic dissipation. The yieldcondition as well as the associate ow rule should satisfy this inequality.One speaks then of thermodynamically consistent constitutive equations. We

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formulate the yield condition in the stress space: the stress tensor mustalways satisfy the condition

( ) ≤0

As long as ( ) < 0, no plastic strain occurs. The surface given by

( ) = 0

is called the yield surface, and function ( ) the yield function. The elasticregion is found inside the yield surface. If the stress tensor lies on the yieldsurface, then the associate ow rule states that ˙ is either zero or shows inthe direction of the gradient of the yield function

= ∂ ∂

(1.15)

with

= 0 for < 0 or = 0 and < 0 (unloading)

> 0 for = = 0 (loading)

If the yield surface has an edge, the above ow rule can still be applied if wereplace the gradient by the sub gradient of . An alternative procedure hasbeen proposed by Koiter: Instead of the product of and the gradient of the

f =0

f =0

f

f

1

2

2

1

p

Figure 1.8: Yield surface with an edge

yield function we take now the linear combination of products

˙ ==1

∂ ∂

(1.16)

with

= 0 for < 0 or = 0 and < 0 (unloading)

> 0 for = = 0 (loading)

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The validity of (1.15) or (1.16) follows from the so called principle of maximum of plastic dissipation , which claims that

( − ∗) ˙ ≥0 (1.17)

for an arbitrary stress state ∗ within the yield surface.

f=0 f=0

p p

*

*

. .

Figure 1.9: Convexity of the yield surface and the normality rule

This principle is equivalent to the requirement of convexity of the yieldsurface, because in the case of non convexity one can always nd the stressstate ∗ which violate the inequality (1.17). Thus, the principle of maximumof plastic dissipation implies the convexity of the yield surface as well as theassociate ow rule. One can show that the plastic dissipation = ˙ isa function of the plastic strain rate ˙ only. When the stress tensor is

found inside the yield surface, then ˙

= 0 and the dissipation vanishes. If the stress tensor during the loading is found on the yield surface, then thedissipation must be a homogeneous function of rst order with respect to˙ . To be consistent with the second law of thermodynamics we require that

≥0.Examples of the yield surfaceFor isotropic materials the yield function must be a symmetric function of three principal stresses

= ( 1 2 3)

Since the principal stresses can be expressed in terms of the principal invariants , the yield function can also be presented in the followingform

= 1( )

Various observations and experiments show that the hydrostatic stress doesnot inuence the plastic yielding. This means that the yield function dependsonly on the principal invariants of the stress tensor, or alternatively,on the invariants 2 3 of the stress deviator

= 2( 2 3)

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Consequently, the yield function must be a symmetric function of

−

= 3( 1 − 2 2 − 3 3 − 1)

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) statesthat the plastic ow occurs when the maximum shear stress achieves somecritical value. According to this criterion the yield function must have theform

= 12

max ∣1 − 2∣∣2 − 3∣∣3 − 1∣ −

or, the equivalent form =

14

(∣1 − 2∣+ ∣2 − 3∣+ ∣3 − 1∣) −

In this equation denotes the yield shear stress. For an uniaxial tension

Figure 1.10: Mohr s stress circle for uniaxial tension

test the plastic ow occurs when (see Fig. 1.10)

1 = = 2

Thus, = 2. The projection of the Tresca yield surface onto the octahedron plane (the so called plane) in 3 D space of principal stresses is ahexagon (Fig. 1.11).

The normality rule then implies that

˙ 1 = 14

[sign( 1 − 2) + sign( 1 − 3)]

where

sign = ∣∣=⎧⎨

⎩

+1 > 0 ∈(−1 1) = 0

−1 < 0

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Figure 1.11: Projection of Tresca s and Mises yield surfaces

Similar equations hold true for ˙ 2 and

3. When all principal stresses aredifferent and are ordered so that 1 > 2 > 3, then ˙ 1 = 2, ˙ 2 = 0,˙ 3 = − 2. When 1 = 2 > 3, then ˙ 1 = (1 + ) 4, ˙ 2 = (1 − ) 4,˙ 3 = − 2, and so on. For each combination of the principal stresses wealways have ∣ 1∣+ ∣

2∣+ ∣ 3∣= . Therefore the dissipation is equal to

= ˙ = = (∣ 1∣+ ∣

2∣+ ∣

3∣)

Von Mises proposed another yield function, which, in terms of the stressinvariants, takes the form

=

2

−With formula (1.4) this yield function can be written in the form

= 16

[( 1 − 2)2 + ( 2 − 3)2 + ( 3 − 1)2]−Mises criterion states that the plastic yielding occurs when the octahedralshear stress achieves a critical value. An alternative form of the Mises yieldfunction reads

= 2 − 2 = 16

[( 1 − 2)2 + ( 2 − 3)2 + ( 3 − 1)2]− 2

For the uniaxial tension test the plastic yielding occurs when

= 13

2 − 2 = 0

Thus, = √ 3. The projection of Mises yield surface onto the octahedronplane in the 3 D space of principal stresses is a circle of radius√ 2 (Fig. 1.11).Using the normality rule we nd that

˙ = (1.18)

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Therefore the plastic dissipation for Mises yield condition is given by

= ˙ = = 2 ˙ ˙

Combining equation (1.18) with Hooke s law in rate form for a linear elasticisotropic material (see equation (1.14)) we obtain the rate of the total strainin the form

˙ = 13

˙

˙ = 12

˙ +

This equation has been obtained by Prandtl and Reuss.Models with hardeningWe again restrict ourselves to isothermal processes with = const. In orderto describe the hardening behavior one needs to include into the list of variables in (1.10) additional internal variables. Different types of hardening canbe described by introducing scalar variable or tensor variable , where

is a traceless tensor of second rank. The hardening parameter is dened ina standard way (Odqvist)

=

2

3 ˙ ˙ (1.19)

If we interpret as the coordinates of the middle point of the yield surfaceand √ 2 as its radius, then various types of hardening can be displayed inthe 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

= 1( − ) − ( )

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which has been proposed by Melan, Prager, Ziegler and Shield. In additionto it the consistence condition = 0 must be fullled for the loading process.Thus

( − )( ˙ − ˙ ) −2 ′ ˙ = 0

With the above evolution equations for the internal variables

˙ = 23

˙ ˙ = 23

( − )( − ) = 2 √ 3( − ) ˙ = ( − )( − ) = 2 2

we can transform the consistence condition to

( − ) ˙ − (2 2 + 4 ′ 2 √ 3) = 0

Therefore =

( − ) ˙2 2( + 2 ′ √ 3)

The ow rule becomes nally

˙ = ( − ) ˙2 2( + 2 ′ √ 3)

( − ) (1.21)

1.5 Closed system of equations

Restricting ourselves to the isothermal processes only, we have altogether thefollowing system of equations

· 3 balance equations of momentum (1.3)

· 6 stress strain relations

· 1 yield condition

· evolution equation for the internal variables.These 10 + equations contain the following unknown functions

· 3 components of displacements (or 3 components of velocity)

· 6 components of the stress tensor

· 1 scalar factor

· internal variables.

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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 valueproblems. However, the ow rule is limited to simple loading situationlike uniaxial strain or pure shear.

ii) theory of plastic ow. This approach is based on the simplied material models (elastic ideal plastic or rigid ideal plastic materials). Apartfrom that no further simplications are made, and the boundary valueproblems 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 fromany simplifying assumption. Due to the mathematical complexity, onlynumerical solutions of boundary value problems based on the niteelement method are available.

Note that, in some special cases solutions based only on the equilibriumequations and on the yield condition can be found without referring to theow rule. We call such problems statically determinate .

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Chapter 2

Elementary theory

The elementary theory uses various simplifying assumptions concerning thekinematics and the stress state. Justication of these assumptions cannot begiven in general, but for particular problems.

2.1 Bending

Pure bending of a beamAs the rst example let us consider the pure bending of a beam having aconstant rectangular cross section

= = 0 = = = = 0

The chosen coordinate system and the sizes of the beam can be seen inFig. 2.1.

Figure 2.1: Straight beam with constant rectangular cross section

According to the elementary theory we assume that the cross sectionsduring bending remain plane and perpendicular to the beam axis, and

= = = = 0

29

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30 CHAPTER 2. ELEMENTARY THEORY

The rst assumption is related to the kinematics of bending, the second tothe stress state. Both coincide with the commonly accepted assumptions of the beam theory. It follows then from the rst assumption

= = 0

= ( ) = 0 +

where is the radius of curvature of the beam axis. Consequently, we havealong the bers parallel to the beam axis an uniaxial stress state

= ( )

The remaining unknown quantities 0 can be found from the equations

= ( ) = 0

= ( )

in which ( ) and ( ) are related to each other by a constitutive law.

Y

Figure 2.2: Stress strain curve

For elastic ideal plastic materials (at loading) we have

= ∣∣<

∣∣≥

The stress distribution over the thickness is shown in Fig. 2.3.Let us consider rst the purely elastic case

= ( ) = ( 0 +

) = 0 ⇒ 0 = 0

= ( ) = 2 =

⇒

1 =

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2.1. BENDING 31

Y

Yz z

z o

z u

i) ii)

el. zone

Figure 2.3: Stress distribution: i) elastic, ii) elastic plastic

Thus, we obtain the well known formula

=

⇒ = ∣∣max = 6

ℎ2

The elastic stress distribution is valid until

∣∣max = ⇒ = ℎ

2

6

1=

2ℎ

The plastic deformation occurs when ≥ . At the boundaries between the elastic and the plastic zone the yield conditions hold true

( 0 + 1

) =

Together with two integral equations for the force and the bending momentthere are four equations to determine four unknowns 0 . With theabove stress distribution the force equation is simplied to

=

( 0 +

) −

−ℎ 2 +

ℎ 2

= 0

0

+

−

−ℎ 2 +

ℎ 2

= 0

0( − ) + 2 ( 2 − 2) − ( +

ℎ2) + (

ℎ

2 − ) = 0

0( − ) + 2

( 2 − 2) − ( + ) = 0

The yield conditions at the boundaries imply

=

− 0 ⇒ + = −2 0

− = 2

2 − 2 = −4 2

0

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and therefore

0 = 02

− 2

4 20

+ 2 0 ⇒ 0 = 0

⇒ =

⇒ =ℎ

2

We compute now the bending moment

=

2 −

−ℎ 2 +

ℎ 2

=

3( 3 − 3) +

2 (ℎ

2

4 − 2 − 2 + ℎ2

4 )

= 23

23

2 +

14 ℎ

2 − 23

2

= (ℎ

2

4 − 2

3

2

2) = (

ℎ2

4 − 2

3ℎ

2

41

2 )

1 2 3 4

0.25

0.5

0.75

1

1.25

1.5

R /Re

M/M e

M/M e

2z /ho

2z /ho

Figure 2.4: Plots of and 2 ℎ

Thus,

= 32

[1− 13

(

)2]

The ultimate moment is achieved when = = 0 or, equivalently, when = 0, and is equal to = 3 2. The plots of and 2∣ ∣ℎ versus

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 rst when = 0, for whichthe strains are innitely large. It must be emphasized, however, that even

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2.1. BENDING 33

for relatively small plastic strains the value of the bending moment is closeto its limit. For example, 98% of this ultimate value is already achieved at

= 4

If we are only interested in the limit state of the plastic bending, we may letthe elastic zone disappear completely so that

= ℎ 2

0 −

0

−ℎ 2 =

ℎ2

4

The angle of bending takes the value = 0 .

Spring back, unloadingWe consider a loading program for which the bending moment is rst increased up to the value

∗, where

< ∗ <

Now, if we unload the beam by decreasing the moment to zero, its curvaturewill decrease from

1

∗

to 1

The difference 1 ∗−1 is the elastic spring back (elastic recovery) of thebeam. is the residual radius of curvature.

The decrease of the moment from ∗ to zero is equivalent to the super

position of the solution found above with the elastic solution correspondingto the moment −

∗, provided, the beam behaves elastically during the un

loading, what we may assume. Thus, for small curvatures we can express theelastic spring back as

1

∗−

1

= ∗

and, accordingly ∗− =

∗ 0

The stress distribution at the end of the loading (corresponding to the moment

∗) is

∗( ) =⎧⎨⎩

∗

∣ ∣≤∗

> ∗

− < − ∗

We have to superimpose this stress distribution with

( ) = − ∗ −

ℎ

2 ≤ ≤ ℎ2

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Y

Yz

M *

Figure 2.5: Stress distribution after the loading

Figure 2.6: Stress distribution due to − ∗

We have thus after the unloading the eigenstress in the cross section of thebeam

( ) =⎧⎨

⎩

( ∗ − ∗

) ∣ ∣≤∗

− ∗ >

∗

− − ∗ <− ∗

where ∗ is determined in accordance with

z

Figure 2.7: Eigenstress after the unloading

∗ =

32[1−

13( ∗)2]

This eigenstress will affect the plastic yielding if we reload the beam in theopposite direction: the absolute value of the moment at which the plasticstrain changes is lower than that of the rst loading. This is the Bauschingereffect due to the eigenstrain. Consider for example the case when the beamis bent up to

= 4 ⇒ ∗

=

4732

∗ =

ℎ

8

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2.1. BENDING 35

The elastic spring back at the unloading is equal to

( 1

∗−

1

) = ∗

= ∗

=

4732

The stress distributions after loading and unloading are shown in Fig. 2.8. At

z

Y

Y

z

M *

-81 128 Y

15 32 Y

i) ii)

Figure 2.8: Stress distributions after loading and unloading

the subsequent reloading in the opposite direction the beam deform elasticallyuntil

= −1732

soΔ

= ∗

−

= 2 0

At the subsequent reloading in the same direction the beam behaves elastically for increasing moment until

= ∗

=

4732

The shakedown of the beam occurs for Δ ≤2 0.

z

Y

Y

z

M *

- 2 Y

i) ii)

Y

Figure 2.9: Reloading: i) in the same direction, ii) in the opposite direction

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Figure 2.10: Hardening and Bauschinger effect at reloading in the same andin the opposite direction

The previous solution can easily be generalized for materials with linearor non linear hardening. One needs just to replace the constant yield stressin the plastic zone by a function ( ).

For linear hardening we have

=⎧

⎨⎩

∣∣ ≤

+ ( + ) >

+ (− + ) < −

If the hardening is symmetric with respect to tension and compression, then

0 = 0 =

⇒ =

As before, the ultimate moment is achieved when → ∞ and →0.These deliberations can further be applied to study:

i) a general case of unsymmetric bending of the beam having rectangularcross section. One has to assume that

( ) = 0 + +

= = = 0

ii) the combination of tension and bending of the beam having rectangularcross section.

iii) similar problems for beams with doubly symmetric cross sections (including thin walled cross section).

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2.1. BENDING 37

iv) the bending of a plate.

Plate bendingFor the bending of a plate we assume that

· = = = = 0,

· but instead of = 0 now = 0 (plane strain state).

In the elastic zone we have

= 0 = 1 ( − )⇒ = =

= = 1

( − ) =

1− 2

Figure 2.11: Plate bending

In the plastic zone we assume the ideal plastic material behavior withMises yield condition, so

˙ = 0 = 1

( ˙ − ˙ ) + 23 ( −

12 )

˙ = ˙ = 1

( ˙ − ˙ ) +

23

( − 12

)

= 2 + 2 − − 2 = 0

We eliminate from the rst two equations

˙ = 1

( ˙ − ˙ ) −

1 ( ˙ − ˙ ) − 2

− 2

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From the yield condition we derive

− 12

= 2 − 34

2 ⇒˙ =

˙ [1− (

12 −

32

2√ ) + (12 −

32

2√ − )(

12 −

32

2√ )]

= ˙

[1 + (

12 −

32

2√ )2 −2 (12 −

32

2√ )]

= ˙

1

4 2 −3 2 [ 2

(5 −4 ) −3 (1 −2 )(√ + 12

)]

Therefore the following differential equation holds true

˙ = ˙ ( )

In the special case = 1 2 we have

˙ = ˙

3 2

4 2 −3 2 =

˙

34

1 − 34

2

where = . Introducing a new variable , with

= 2√ 3 cos

we obtain

˙ = −√ 32

˙

sinThe integration gives

=

−

√ 32

ln 1 −√ 32

1 +√ 32

+

The stress at the boundary of the elastic zone equals

=

√ 1 − + 2and for = 1 2 =

2√ 3

The Ansatz for the solution remains as before

= 0 +

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2.2. TORSION OF A CYLINDER 39

together with the integral equations for the force and bending moment

= = 0 =

Beam with symmetric cross section about one axisIf the cross section of the beam is symmetric only with respect to the axis, then, due to the redistribution of stress during the plastic yielding, theposition of the neutral ber (with = 0 = 0) will change. From thecondition

=

= 0

follows in the case of ultimate moment

= = − = 0 ⇒ = = 2

The ultimate neutral ber will therefore be the line dividing the cross sectioninto equal areas. Take for example the triangle, we have

= 12ℎ

2⇒ = =

14ℎ

2⇒ =

√ 22

ℎ

Figure 2.12: Neutral and ultimately neutral ber

2.2 Torsion of a cylinder

We consider a pure torsion of a cylinder shown in Fig. 2.13, with

= const

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Figure 2.13: Torsion of a cylinder

We assume that cross sections remain plane and perpendicular to the axisof cylinder, and that the straight lines in radial directions remain straight.Besides

= = = = 0

From the rst assumption follows =

with being the twist. Therefore the strain is equal to

( ) = 12

For the elastic torsion we have

= ( ) = 2 =

=

02 ( ) = 2

0

3

= 2

4 bzw. = 0 0 = 2

4

So, the twist is given by

=

0The maximum shear stress is achieved at =

max =

0 =

The threshold value for the plastic strain to occur reads

= 23

where

= 1√ 3 Mises12 Tresca

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2.3. CYLINDRICAL SHELL UNDER COMBINED LOAD 41

When

≥ the plastic strain occurs, and for ideal plastic material

behavior the yield condition at the boundary between the elastic andplastic zone is

= = ⇒ =

The torsion moment is computed as follows

=

02 2 +

2

= 2

4 + 23

( 3 − 3)

= 2 ( )4 + 23 [ 3 −( )3]

= 6

[4 3 −( )3]

With = 0 = we get for

= 6

[4 3 −( )3]

= 23

3[1− 14

( )3]

=

43[1−

14( )

3

]

The ultimate moment is achieved at = 0, i. e. → ∞ and

= 43

This result can be obtained directly if we let the elastic zone disappear

=

02 2 =

23

3 = 43

For cylindrical pipe with the internal radius and external radius theratios decreases with the decreasing ratio , and in the thin walllimit → it tends to 1.

2.3 Cylindrical shell under combined load

We consider a thin walled tank loaded by a longitudinal force , and aninternal pressure (see Fig. 2.14). We assume that the cross section remains

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Figure 2.14: A closed tank under combined loading

plane and perpendicular to the axis. Besides,

= = 2

+ 2

= =

= = = = 0

In addition to this we shall neglect the elastic strains.From these assumptions follows immediately

˙ =˙

˙ = ˙

˙ =˙

where , , and are the length, the radius of the cross section, and thethickness of the cylindrical shell, respectively. Besides,

=⎛⎝

0 0 00 00 0 ⎞⎠

i. e. we have a homogeneous plane stress state. For materials with Misesyield function and isotropic hardening

= 12 − 2( )

The computation of = 12 gives

= 12

= 16

[( − )2 + 2 + 2]

= 13

( 2 − + 2)

The ow rule (1.21) take the form

˙ =√ 3 ˙

4 2 ′

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2.3. CYLINDRICAL SHELL UNDER COMBINED LOAD 43

The loading condition requires that

˙ = ˙ ≥0

Besides, for linear hardening√ 3

12 2 ′ =

1

Therefore

˙ = ˙

(2 − ) =˙

˙ = ˙

(2 − ) = ˙

˙ = − ˙

( + ) =˙

These are three governing differential equations for three unknowns , forwhich the loading paths of and are given.

We can for example load the tank in two steps:

· Step 1: Increase the internal pressure up to the yield stress at = .

· Step 2: Keep = = const and increase the longitudinal force .

p p Y

N

z

Y

Figure 2.15: Loading in two steps

In the rst step

= 2

= 12

=

⇒ =

The yield condition leads to

= 13

2 =

13

( 2 − + 2)

= 13

2 (1 −

12

+ 14

) = 14

2

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Consequently = 2√ 3

In the second step ( = const, ˙ > 0)

˙ = ˙

= ˙

(2 − )

˙ =˙ =

˙(2 − )

with

= 13(

2

− + 2

)˙ =

13

[(2 − ) ˙ + (2 − ) ˙ ]

Now = ⇒ ˙ =

˙( ) = ( ˙ −

˙2 )

Due to the incompressibility

˙ +

˙ +

˙ = 0

we obtain˙ = (

˙ + 2

˙) = ( ˙ + 2 ˙ )

We can therefore express ˙ as follows

˙ = 2 3 ˙

The solution of this equation gives1

+ 3

ln + = 0

At = we have = = 2 3, so

=

11 −

3 ln 32

This is a transcendental equation for . There are still two remaining equations to determine and at a given and , where

˙ =

2 −− 3

22 (2 − )

˙2

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2.4. SIMPLE METAL FORMING PROCESSES 45

2.4 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 beginhere with the simplifying assumptions about the kinematics of the processunder consideration. On the other side the stress states appearing in mostmetal forming processes are much more complicated compared with those of say, simple bending or torsion. Drastic simplications may lead to erroneousresults. 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 plasticcounterpart which may be of the order 1, we shall neglect the former in thecold working problems.

Most metal forming processes such as forging, rolling, drawing and soon can be explained by quite simple models. We restrict ourselves to theplane strain problems. Since plastic deformations in cold worked materialsare normally produced under pressure, we shall regard the latter as positivedespite the traditional convention in mechanics. We denote the pressure by .

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 modelWe consider now a strip obtained by cutting (mentally) the piece of metalat cold rolling along the cuts perpendicular to the plane of motion. Forsimplicity we assume the symmetry with respect to the ( ) plane. Further

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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 axisremain plane, ii) the work done by external forces per unit time is the sameas that of the uni axial compression (the shear strain are neglected), iii) theprocess is quasi static, the body forces are negligibly small. We take for

Figure 2.19: Force equilibrium of a strip

granted that > 0. In the planes = const the shear stress is absent, whilethe normal stress (pressure) is uniformly distributed over the cross section.Thus, the resultant forces in direction acting on the strip is ℎ at andℎ+ ∂ ( ℎ) ∂ at + . The normal force exerted from the roller on the

strip, , causes the friction

=

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Thus, the resultant force from the roller on the strip is decomposed into thevertical component

= = (cos + sin )

and the horizontal component

= (sin − cos )

With

= cos

we obtain

= = (1 + tan )

= (tan − ) = tan −1 + tan

Denoting the coefficient of friction by

= tan

we obtain for the horizontal force =

tan −tan 1 + tan tan

= tan( − )

From the assumption ii) we set the power of external forces equal to thatobtained under the uniaxial compression

˙ = − ℎ = − ℎ + 2 + ℎ −[ ℎ + ∂ ∂

( ℎ ) ]

= − ℎ + 2 − ∂ ∂

( ℎ )

Dividing this equation by ℎ and substituting the formula for into it, weobtain

− ℎ

ℎ = −

ℎ

ℎ + 2

ℎ

tan( − ) − ℎ

∂ ∂

( ℎ) − ∂∂

The incompressibility condition requires that

ℎ = const ⇒ ∂∂

ℎ + ∂ℎ∂

= 0

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Besidesℎ = ∂ℎ∂

Therefore∂∂

= −ℎ

ℎ

In the elementary theory of forming processes we always use Tresca s yieldcondition, for which

− =

It follows from here

ℎ

∂ ∂ +

∂ℎ∂ −2 tan( − ) = 0

Replacing = + and ∂ℎ ∂ = 2 tan , we obtain nally

∂ ∂

+ 2 ℎ

[tan −tan( − )] −2

ℎ tan( − ) = 0

This ordinary differential equation of rst order can be written in the form

+ ( ) + ( ) = 0

so, its general solution reads

( ) = − 0 [ 0 −

0

( 0 ) ]

with = 0 being the value of at = 0. This solution is useful if theintegrals can be computed analytically. Otherwise the numerical solutionturns out to be more effective. After the stress is known, the stress caneasily be determined.The tube modelIf we deal with the axisymmetric forging process, we may imagine to havea tube cut from the cold worked metal at forging which is symmetric aboutthe axis as shown in Fig. 2.20.

We assume i) axisymmetric deformation, i. e. cylindrical surfaces aboutthe axis remain cylindrical, ii) the work done by the external forces is thesame as that in the uni axial compression, iii) the body forces are negligiblysmall. Except that we take for granted that > 0.

As in the previous strip model we obtain for the force acting on the diein the direction

= = (cos − sin ) = (1 −tan tan )

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Figure 2.20: The tube model

and in the direction

= (sin −tan cos ) = (tan −tan ) = tan( − )

From the assumption ii) we set the power of external forces equal to that of the pressure acting on the tube

˙ = − 2 ℎ

=

− 2 ℎ + 2 2 + ℎ 2

−2 [ ℎ +

∂

∂

( ℎ ) ]

= 2 [− ℎ + 2 tan( − ) ]−2 [ ∂ ∂

( ℎ) + ∂ ∂

( ) ℎ]

The incompressibility requires

ℎ

ℎ +

+

∂ ∂

= 0

orℎ

ℎ = −

∂ ∂

( )

We obtain (after dividing by = 2 ℎ )

(− + − )ℎ

ℎ = [2 tan( − ) −

∂ ∂

( ℎ)]ℎ

Using Tresca s yield condition we have

− =

It follows thatℎ

∂ ∂

+ ∂ℎ∂ −2 tan( − ) = 0

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and respectively∂ ∂

+ 2 ℎ

[tan −tan( − )] −2

ℎ tan( − ) = 0

provided ℎ < 0 > 0. Thus, we obtain the same equation as that of thestrip model, with replaced by .

As an example let us consider the compression of a cylinder shown inFig. 2.21). In this case = 0, so the equation reduces to

Figure 2.21: Compression of a cylinder

′ + 2 ℎ

+ 2

ℎ = 0

The boundary condition is

= 0 at = .

This equation yield the following general solution

( ) = − 2 ℎ [ ( ) − 2ℎ

2 ℎ ]

= −2

ℎ ( − )[−

2ℎ

2

ℎ ( − ) ]

= [2

ℎ ( − ) −1]

For ( ) we have ( ) = + =

2

ℎ ( − )

The solution found remains valid as long as the stick zone where = > 2 does not occurs. This means that the shear stress distribution

=

must be controlled and checked, whether its maximum exceed the value 2or not. With this solution we can compute the total compressive force actingon the cold forged material

= 2

0

2

ℎ ( − ) = 2 [

ℎ2

4 2 (2

ℎ −1) − ℎ2

]

We can easily generalize this model to the so called plate model.

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Chapter 3

Theory of plastic ow

3.1 Governing equations

According to the classication given at the end of Chapter 1 we shall assumein the theory of plastic ow the ideal plastic material behavior. Thus,

= const

i. e. any hardening behavior is excluded from consideration.We further admit that the plastic materials obey Mises yield condition

= 2 − 20 = 0 2

0 = 13

20

In most of cases we neglect the elastic strains as small compared with theplastic strains. Thus, as a rule, the material considered in the theory of plasticow is rigid ideal plastic and obey Mises yield condition. The associate owrule reads

= ˙ = ˙ = ∂

∂=

where remains still an unknown parameter. It can be determined only withthe help of the kinematic boundary condition. From this ow rule we derive

= 2 = 2 2 20

Therefore

= 1√ 2 0

51

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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)

˙ = 13

˙

˙ = 12

˙ +

The closed system of governing equations must include the equilibriumconditions, which, in the absence of body forces, read

= 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

· 6 components of the stress tensor

· 1 scalar factor .

The theory of plastic ow is primarily applied for three types of problems:

· onset of plastic ow of a rigid plastic or elastic plastic material,

· non stationary plastic ow, provided the boundary conditions of thestates under consideration are sufficiently known (the so called pseudo

stationary plastic ow),

· stationary plastic ow.

Analytical solutions are available only if some additional assumptions concerning the kinematics of the plastic ow can be made, for examples in

· the torsion of prismatic bars, or

· the plane strain problems.

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3.2. TORSION OF PRISMATIC BARS 53

3.2 Torsion of prismatic bars

Stress distributionFor the torsion of prismatic bars with an arbitrary cross section we use theSt. Venant Ansatz

= − = = ( )

Here is the twist, and the warping. The following components of thestress tensor are zero

= = = = 0

The equilibrium conditions reduce to

+ = 0

Figure 3.1: Cross section of a bar

This equation is automatically fullled if there exists a stress function( ) such that

= = −The differential of

= + = − +

vanishes for =

i. e., when the lines of = const has the same direction as the resultant of (Fig. 3.2). The lines = const will therefore be called stress trajectories.

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Figure 3.2: Lines = const

Since the normal stresses vanishes at the boundary of the cross section,

= const at Γ for simply connected cross section.

The plastic zone in the cross section is determined from the yield condition

2 + 2 = 2

where

=12 Tresca1√ 3 Mises

Thus,( )2 + ( )2 = 2

⇒ ∣∇∣=

Nadai has found from the obtained condition (that the gradient or the maximal slope of remains constant) the sand roof analogy which is constructed by pouring sand on a horizontal sheet of cardboard set out in theshape of a cross section. Due to the constant internal friction of sand, theconstructed roof satises the above equation.

Figure 3.3: Sand roof

The torque is calculated according to

= ( − ) = − ( + ) = 2 so it is equal to twice of the volume under this roof .

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Figure 3.4: Determination of the warping

3.3 Plane strain problems

Governing equationsWe are dealing with a 2 D plastic ow, if the components of the velocity aregiven by

= ( ) = ( ) = 0

and thus,˙ = ˙ = ˙ = 0

Based on the rigid ideal plastic material behavior and the Mises yield condition, the normality rule reads

˙ = > 0

This implies = = = 0

and = =

1

2( + )

and accordingly

=⎛⎝

0 0

0 0 ⎞⎠

So is one of the three principal stresses. The remaining two are obtainedas follows

1 2 = 12 ( − )2 + 4 2

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3.3. PLANE STRAIN PROBLEMS 57

in the coordinate system obtained by an anticlockwise rotation of the originalaxes through an angle (see Fig. 3.5)

= 12

arctan 2

−Thus, the maximum shear stress

max = 12

( 1 − 2) = 12 ( − )2 + 4 2 =

is achieved in the direction inclined at the angle 4 with respect to theprincipal axes. The stress state

1 = + 2 = − 3 = is characterized by superposition of the pure shear stress in the ( ) planeon the hydrostatic pressure.

Figure 3.5: Principal axes and and lines

We denote the directions corresponding to 1 and 2 as rst and second principal direction, respectively, while the directions obtained by theclockwise rotations of the principal axes through the angle 45 0, in which themaximum and minimum of the shear stresses occur as rst 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 sliplines. We denote them as and lines, respectively.

Let be the angle between the line and the axis. Then we obtain

= tan for lines

= −cot for lines

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as differential equations for these two families of slip lines.The stress state must satisfy during the plastic ow the yield condition

16

[( 1 − 2)2 + ( 2 − 3)2 + ( 3 − 1)2]− 2 = 0

For the plane strain problems this condition becomes

=

with = √ 3.The stress state in a particular point is given through . Due to

the yield condition these components of the stress tensor are not independent.Let us introduce two dimensionless quantities, and as follows

= 2

= 12

arctan 2

− − 4

= − 4

Thus, is the angle, through which the axis is rotated in anti clockwisedirection to the rst slip line. Then

= 2 − sin2 = 2 + sin2 (3.1) = cos2

We have in addition the equilibrium equations which reduce to

+ = 0 + = 0

There are altogether three equations to determine three unknown functions.The plane strain problem is therefore static determined .

Slip lines and their propertiesThe stress components can be expressed in the plastic zone in terms of variables and according to (3.1). Substituting formulas (3.1) into the equilibrium conditions, we obtain

− cos 2 − sin2 = 0

− sin 2 + cos 2 = 0

This is the system of homogeneous quasi linear partial differential equationsof rst order. Its solution as well as the suitable method of solution depend

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strongly on the type of these equations. To recognize the latter, we considerthe differential equations of characteristics

= 1

( √ 2 − )

with

= sin 22 = 2 cos 2

= −sin2

The discriminant 2 − = cos2 2 + sin 2 2 = 1

is positive, so the system under consideration is hyperbolic. For the characteristic directions we have

= 1sin2

(−cos2 1)

or

=tancot

Thus, the characteristics of this problem coincide with the slip lines foundabove.

Choosing in an arbitrary point ( ) the directions of the axes suchthat they coincide with the directions of the slip lines ( ), and takinginto account that

sin2 = 0 cos2 = 1

we transform the above equations to

( − ) = 0 ( + ) = 0

These equations imply that − is constant along the rst slip line, while + is constant along the second slip line

= tan − = const line

= −cot + = const line

So, provided the slip lines as well as the values of constants on the and lines are known, then and and the stress state in each point of the

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( ) plane can be determined. However, the directions of characteristicsdepend on the solution of this problem.

The slip lines possess several important properties to be discussed below.Proposition 1 : The pressure changes along a slip line in proportion withthe angle .This property follows directly from the above formulas, according to which

= const along the lines.Proposition 2 : The change in the angle and the pressure is the same fora transition from one slip line of the family to another along any slip lineof the family (Henky s rst theorem).

Figure 3.6: Hencky s rst theorem

From the formula

− = 1 along 1

2 along 2

and respectively,

+ =ℎ1 along 1ℎ2 along 2

it follows

11 =

1

2(ℎ1 − 1) 12 =

1

2(ℎ2 − 1)21 =

12

(ℎ1 − 2) 22 = 12

(ℎ2 − 2)

Therefore

1 = 21 − 11 = 12

( 1 − 2)

2 = 22 − 12 = 12

( 1 − 2) = 1

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AB

C

Figure 3.7: Determination of the pressure

Proposition 3 : If the value of is known at some point of a given slip grid,then it can be found everywhere in the eld.

Since is known everywhere, we have

= 2 (ℎ1 − )

and = 2 ( 1 + )

Proposition 4 : If some segment of a (or ) slip line is straight, then allthe corresponding segments of (or ) lines are also straight and have thesame length.

E

A

BA

B

´

´

Figure 3.8: Straight slip lines

This proposition follows from the second proposition, since the angle betweenthe tangents to any two slip lines remains constant as we move alongthe prescribed line. The evolute (locus of the center of curvature) of anarbitrary curve is the envelope of the normals to the curve (Fig. 3.8). It isevident that the slip line AA ′ and BB′ have the same evolute E. From thedenition 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 intersectionchange by the distance traveled (Hencky s second theorem)Consider neighboring slip lines of the and families, bounding a slip

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A

B

C

D

E

E

ds

ds

d

d

d

´

´

R

R

Figure 3.9: Hencky s second theorem

element shown in Fig. 3.9. It is apparent that

′ = = = =

For the arc length AC we have

= On the other side, from proposition 2 follows

= ( + + )

so = − = −

Proposition 5a : The center of curvature of the lines at point of intersectionwith lines generate the involute PT of the line (Prandtl s theorem).

T S R

Q

PA

A

B

B C

C

DD´ ´ ´

´

d

Figure 3.10: Prandtl s theorem

Since AP=ABQ, PQ must be the involute of the line.

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Proposition 5b : The envelope of the slip lines of one family is the geometriclocus 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 linesas well as the radius of curvature of the lines tend to zero. Therefore: i)T belongs to the envelope of the family of slip lines, ii) the line passingthrough this point has a cusp at T.

Since they have a cusp at T the lines cannot intersect the envelope. Inother words, the envelope is the boundary of an analytic solution.

According to Caratheodory and Schmidt an orthogonal grid of slip linespossessing the above properties is called Hencky Prandtl s grid.

Boundary conditions

The properties of the slip lines discussed in the previous subsection can beused to construct the Hencky Prandtl s grid, provided one slip line is knownfrom each family, and they intersect each other.

In order to construct the grid and the solution of a given boundary valueproblem, the boundary conditions should be taken into account.

Figure 3.11: Static boundary condition

For a static boundary condition we must specify the tractions and along the boundary Γ.

We compute now , and

= 12

( + ) + 12

( − )cos2 + sin 2

= 12( + ) −

12( − )cos2 − sin 2

= −12

( − )sin2 + cos 2

Substitution of and from (3.1) yields

= 2 − sin2( − ) = 2 + sin 2( − )

= cos2( − )

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With the given values of and we get

= 12

arccos

+

= 2

+ 12

sin 2( − )

The presence of two solutions agrees with the quadratic character of theyield condition. To choose the sign we consider the normal stress at theboundary

= 4 −The sign of can sometimes be predetermined, and this enables the correctchoice of the solution.

An important special case occurs when there is no tangential stress atthe boundary Γ ( = 0). Then

= 12

arccos 0 + = 4

+

= 2

+ 12

sin(2 2

) = 2

12

Further

= 4 − = 2Consider for example a traction free straight boundary shown in Fig.3.12,where = 90 = = 0. On this boundary

Figure 3.12: Traction free straight boundary

= 2

4

+

= 2 = = 2

Depending on whether the ber in tangential direction is in tension or compression, one chooses the sign + or − .

In order to study particular plastic ows, we must nd the solution of theabove quasilinear hyperbolic system of equations satisfying certain boundaryconditions. As a rule, the corresponding boundary value problems can only

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AQ

CB

P

P P

x

y ´

´

Figure 3.13: Cauchy s problem

be solved numerically. We now give a brief account of the most importantboundary value problems.

Boundary value problems.1. Cauchy s problem. Let AB be a smooth arc (described by an arc length), which nowhere coincides with characteristics and intersects each charac

teristic once only. Let the functions ( ) and ( ) be continuous togetherwith their derivatives on AB. Then the solution exists and is unique in atriangular region APB, bounded by the arc AB and the and slip linesoriginating at A and B. The solution at poit P depends only on the dataalong 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 pointC on the curve AB, then the jumps propagate along the characteristics CP ′and CP ′′.

If the shear stress is zero at the boundary, the normal to this boundaryis one of the principal directions and the slip lines approach the contourat an angle of 45 . Consequently, the contour coincides nowhere with acharacteristics, and we have Cauchy s problem whose solution is unique.

A

Bx

y

(i,j)

(i+1,j)(i,j-1)

Figure 3.14: Determination of and

If the boundary is traction free, then the stress eld as well as the grid of

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A

B

x

y

O

Figure 3.16: Mixed boundary value problem

Velocity eld.If the stress eld is known, then the velocity eld can also be determined.According to the ow rule we have

˙ = ˙ = ˙ =

From the rst two equations

˙ − ˙ = ( − )

Combine this with the third equation( ˙ − ˙ )

−= ˙

Taking into account the relation

−=

12

tan 2 = 12

tan(2 − 2

) = −12

cot2

and the incompressibility, we obtain

+ = 0( + )tan2 + − = 0

This system of homogeneous quasilinear partial differential equations of rstorder is of hyperbolic type, and its characteristics coincide with the slip lines.

With the transformation

= cos − sin = sin + cos

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we obtain for the derivatives

= cos − sin − sin − cos = sin + cos + cos − sin

Choosing in an arbitrary point ( ) the directions of the axes suchthat they coincide with the directions of the slip lines ( ), and recallingthat

sin = 0 cos = 1

we obtain

= − = 0 ( ) = + = 0 ( )

These equations are called Geiringer s equations for the velocity eld.If the grid of the slip lines is known from the solution of the stress problem,

then the velocity eld can also be determined from these equations for allthree types of the boundary value problems.

Note that the line separating plastic and rigid regions is a slip line orthe envelope of slip lines. To show this we rst assume that the velocitiesare continuous on the line of separation. Then = = 0 on this line.

If the boundary has nowhere a characteristic direction, then the solution of Cauchy s problem gives = = 0 everywhere in the plastic region, whichcontradicts the original assumption. Consider now the case when the velocityis discontinuous on the line of separation. The discontinuity can only be in

, otherwise a crack appears. One can then show that ∣∣ → . Thus, theboundary will be a slip line or an envelope of slip lines.

Using Geiringer s equation and considering the velocity at some nodalpoint ( ) of the grid of slip line, we can easily obtain the following approximate nite difference formulas

( )

−(

−1) = ( )[ ( )

−(

−1)]

( + 1 ) − ( ) = − ( )[ ( + 1 ) − ( )]

For statically determinate problems, i. e. when the static boundary conditions are specied at the boundary Γ, the grid of slip lines as well as thestress eld can be determined numerically, so the velocity eld can be foundfrom these formulas.

For statically undeterminate problems, which have not been consideredup to now, the coupling cannot be removed. To construct the grid of sliplines all four difference equations need to be solve simultaneously.

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For problems with mixed boundary conditions (where the tractions arespecied on one part of the boundary and the velocity on the remaining part)we nd the grid of the slip lines as well as the stress and velocity elds ineach of the zone of inuence as above. The initial data for the velocity can beobtained from the transition conditions at the boundary between the zonesof inuence.

Family of straight slip lines.Many of the above statements become quite simple for the family of straightslip lines.

Due to the Proposition 4 there are only two possible grids with straightslip lines

· grid with two orthogonal families of straight slip lines,

· grid with one family of straight slip lines and the curved slip linesorthogonal to them.

In any case

− = const −line + = const −line

Since remains constant along a straight line, and must also be constantthere.

For the rst case we have a uniform stress state in the whole domain andthe velocity eld of the form

= ( ) = ( )

A domain in which one family of slip lines contains only straight line iscalled a fan. We distinguish between central and non central fans.

The central fan has the lines as concentric circles. Here the envelopedegenerate to a point O which is the middle point of all circles. In this case

= const ⇒ = const on line

and + = const on line

Consequently + = const in the whole domain

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Because =

− is independent of , we obtain for the stresses

= = 2 = 2 ( − ) =

With the help of Geiringer s equation we compute the velocity

= − ( ) = ( ) + ( )

For the non central fan the similar results can also be obtained.

3.4 Plane stress problems

Governing equationsWe are dealing with a plane stress state when, in some cartesian coordinatesystem,

= ( ) = ( ) = ( ) = = = 0

and, accordingly

=⎛⎝

0 0

0 0 0⎞⎠

= 13

( + )

For the principal stresses in the ( ) plane we have as before

1 2 = 12

( + ) 12 ( − )2 + 4 2

These principal stresses occur in a coordinate system which is rotated throughthe angle =

12

arctan 2

−with respect to the original coordinate system.

This plane stress state must fulll in the plastic zone the yield condition,which is either i) Mises

16

[( 1 − 2)2 + ( 2 − 3)2 + ( 3 − 1)2]− 2 = 0

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3.4. PLANE STRESS PROBLEMS 71

with the consequence

21 + 2

2 − 1 2 −3 2 = 0

or ii) Tresca

12

max ∣1 − 2∣∣2 − 3∣∣3 − 1∣ − = 0

with the consequence

∣1 − 2∣−2 = 0 1 2 ≤0

∣1∣−2 = 0 1 2 ≥0 ∣

1∣> ∣

2∣

∣2∣−2 = 0 1 2 ≥0 ∣2∣> ∣1∣

Figure 3.17: Yield surface for the plane stress state

The local stress state at an arbitrary point is xed by three components. For them the yield condition as well as the two remaining

equilibrium conditions

+ = 0

+ = 0hold true. The problem is therefore statically determinate. However, incontrast to the plane strain problems the above two yield conditions lead todifferent results.

Solution based on Mises yield conditionIn addition to the rotation angle let us introduce a new angle

cos = √ 3 2

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so that, with the help of

1 2 = 2 cos( ∓ 6

)

the ow rule is satised identically.Then

= (√ 3cos + sin cos2 )

= (√ 3cos −sin cos2 ) = sin sin2

Substituting these formulas into the equilibrium conditions we obtain nally

(cos −√ 3sin cos2 ) −√ 3sin sin2 + 2 sin = 0√ 3sin sin2 −(cos + √ 3sin cos2 ) + 2 sin = 0

With

= −2sin (cos −√ 3sin cos2 )

= 2 sin 2 √ 3sin2

= −2sin (cos + √ 3sin cos2 )

we derive for the discriminant2

− = 4 sin 2 (3

−4cos2 ) = 4 sin 2 Ω( )

Depending on the sign of Ω( ) we get the hyperbolic, parabolic or ellipticsystem of partial differential equations (see Fig. 3.18).

1 2 3 4

1

0.75

hyperbolic elliptic

Figure 3.18: Graph of function cos 2

For the characteristic directions we have

= −√ 3sin sin2 Ω( )cos −√ 3sin cos2

In the hyperbolic case the problem can be solved with the help of characteristics.

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74 CHAPTER 4. CRYSTAL PLASTICITY

F

-F

Y

a) b)

Figure 4.1: Tensile test and the stress strain curve

slipplane

a b

Figure 4.2: Magnied schematic view of plastic slips: a) front view, b) sideview

planes along the directions of shortest interatomic distances. To recognizethese planes and directions one needs to consider the periodic lattice structureof the crystal. We see for example in Fig. 4.3 the unit cell of a body centeredcubic (bcc) crystal. The unit cell is characterized by three lattice vectors a 1,a 2, and a 3 which, for bcc crystals, are mutually orthogonal and have an equalmagnitude. The translations of the unit cell by 1a 1 + 2a 2 + 3a 3 for allintegers 1 2 3 will develop the regular periodic lattice structure. In eachunit cell of the bcc crystal there are eight corner atoms and one atom locatedat 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 theway to indicate them. We shall use Miller indices for this purpose. Forexample, the Miller indices of the direction in Fig. 4.3 are [111], whilethose of the direction in the same Figure are [101], where the negative signof the 1 component is indicated by a bar over the corresponding index. Wedene the Miller indices of a family of directions as a set of three integers

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4.1. PHYSICAL BACKGROUND 75

1 2 3, enclosed in square brackets, which indicates all directions parallelto the vector 1a 1 + 2a 2 + 3a 3. Since 1 2 3 are dened uniquely upto a common integer factor, we will choose the smallest multiples of themfor the Miller indices. For illustration let us apply this denition to nd theMiller indices of the direction in Fig. 4.3. Since the vector drawn from theorigin parallel to has the smallest integers 1, 1,0 as its components, we getfor the direction the Miller indices [110]. Crystallographic planes are alsoindicated by sets of integers. For example, the Miller indices for the ( 1 2),( 2 3), and ( 3 1) planes are (001), (100), and (010), respectively. Notethat the Miller indices for planes are enclosed in parentheses instead of squarebrackets as for directions. We dene the Miller indices of a family of planes

as a set of three integers 1 2 3, enclosed in parentheses, which indicatesall planes perpendicular to the vector 1a 1 + 2a 2 + 3a 3. Since the normalvector 1a 1 + 2a 2 + 3a 3 is dened uniquely up to a common integer factor,we choose again the smallest multiples of 1 2 3 for the Miller indices. Forillustration let us apply this denition to nd the Miller indices of the planeshown in Fig. 4.3. The normal vector to this plane is obviously a 1 + a 2 + a 3,so the Miller indices of this plane are (111).

x 1

x 2

x 3

s d

ea1

a2

a3

Figure 4.3: Unit cell of bcc crystals

In terms of these Miller indices a precise description of the plastic slipcan be given. For example, the plastic slip may occur in bcc crystals on theslip planes of type (110) along the directions [111]. For fcc crystals, the slipplanes are of type (111), while the slip directions of type [110]. Each slipsystem is characterized by the family of slip planes and slip directions. Forfcc crystals there are 12 slip systems.

The critical shear stress of a perfect crystal at which the plastic slip occurscan roughly be estimated as follows. Consider the glide of the upper half of crystal relative to its lower half under a shear stress . The two adjacent rowsof atoms in the initial state are shown in Fig. 4.4 in which the interplanarspacing is while the interatomic distance in the slip direction is .

Denoting the relative displacement by , the applied shear stress must be

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a

b

u

E

Figure 4.4: Perfect crystal and energy vs. relative displacement

calculated in accordance with = , where is the energy of crystalin terms of . Due to the periodicity of the lattice structure, must be

a periodic function with the period , having a minimum at = 0 and amaximum at = 2 (see Fig. 4.4). The simplest formula for satisfyingthese requirements can be proposed as follows

( ) = 2

1 −cos 2

The derivative of yields the shear stress

=

= sin 2

For small displacement ≪ we have approximately = 2 . Comparing this with the Hooke s law = , with being the shear modulus,we obtain the value of which coincides with the maximum shear stressachieved at = 2

= = 2

In typical situations we have = , so the theoretical shear strength isestimated to be

≈ 2

For most of crystals this theoretically estimated shear strength is three orfour order of magnitude larger than the experimentally observed value of .

In 1934 three scientists, Taylor, Orowan, and Polanyi, simultaneouslycame to the idea that the plastic slip can so easily occur at very low shearstress because of dislocations which are the line defects in crystals. Oneshould mention that, actually, the theory of dislocations in solids was originally developed by Volterra in 1905. But only the discovery by these pioneersgave birth to the dislocation based plasticity. In Fig. 4.5 one can see theso called edge dislocation in a simple cubic lattice structure which can be

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4.1. PHYSICAL BACKGROUND 77

A

D

C

B

Figure 4.5: Edge dislocation in primitive cubic crystal

regarded as the extra half plane of atoms inside the crystal (the half planeABCD). The distortion of the crystal is localized near the edge of this half

plane which is called the dislocation line (the line AB). Far away from thedislocation line the crystal has a nearly perfect lattice structure.

Figure 4.6: Migration of dislocation through crystalFig. 4.6 illustrates how the edge dislocation migrates through the crystal

from left to right under an applied shear stress . This migration, from oneposition to the next, involves only a small rearrangement of the atomic bondsnear the dislocation line. The nal results of these many small steps is theplastic slip considered before. But the shear stress to break the atomic bondand move dislocation in one interatomic distance is much lower that thatrequired for the simultaneous movement of the whole upper half of crystal.

Let us now consider the creation of a single dislocation in a simple cubiccrystal. 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 cutthrough a distance equal to one lattice spacing. Then rejoin the atoms onboth sides of the cut and let the crystal be elastically relaxed. The latticestructure is again almost perfect except near the boundary AB of the cutsurface called dislocation line. If the atoms below the cut are shifted in thedirection parallel to the line AB, a screw dislocation is created; if the shift isperpendicular to AB, an edge dislocation is produced; if the shift is neitherparallel nor perpendicular to AB, the created dislocation is of the mixed type.So, each dislocation is characterized by the dislocation line and the vector

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needs some guiding principles to limit the feasible choices of the availablemodels. Such guiding principles are the laws of thermodynamics.

s

m

p

e

= +e p

Figure 4.9: Additive decomposition of the total strain

Consider rst a crystal deforming in single slip. The strains of an innitesimal element of the crystal is additively decomposed into the plasticstrains and the elastic strains as shown schematically in Fig. 4.9. The plastic strain tensor is the symmetric part of the plastic distortion which mapsthe undeformed crystal to the reference stress free crystal. The subsequentelastic strains deform the crystal in accordance with the Hooke s law. Bothplastic and elastic strains are incompatible so that the sum of them becomescompatible and derivable from a displacement eld. For a large number of loops lying on the parallel slip planes with the mean distance between thembeing much smaller than the characteristic size of the specimen, we propose aspatial average description of plastic distortion produced by this slip systemas follows

= (x )

with s the unit vector pointing in the slip direction, and m the normal vectorto the slip plane. The essential difference to the case of discrete dislocationsis that now function (x ) is assumed to be continuously differentiable. If thecrystal has active slip systems, the plastic distortion is the sum of thoseproduced by these slip systems

==1

(x )

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4.2. CONTINUUM DISLOCATION THEORY 81

The small Gothic index indicating the slip systems runs from 1 to . Onecan see that, in general, = 0, consequently continuous plastic distortionsdo not cause any volume change.

The plastic strains and the plastic rotations are the symmetric andskew symmetric parts of the plastic distortion

= 12

( + ) = 12

( − )

The elastic strains are then the differences between the total compatiblestrains and the incompatible plastic strains

= − = 12( + ) −

with being the components of the displacement vector.Nye introduced an important characteristic of dislocations, the dislocation

density tensor 1

= (4.1)

For a single dislocation loop the dislocation density tensor has the followingphysical meaning: if we take an arbitrary innitesimal surface with theunit normal which is the tangent vector to the dislocation line crossingthis surface, gives the Burgers vector of this dislocation (see Section3.4). Within the continuum dislocation theory, we interpret this quantity asthe resultant Burgers vector of all dislocations whose dislocation lines crossthe surface . For a crystal deforming in single slip = ,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∣ ∣ (4.2)

with 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 conned in the region ℬ reads

Ψ = ℬ ( )

with 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 distortionin form of rotation only. Formula (4.1) was proposed independently by Bilby and Kr oner.

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change of the free energy minus the power of the external forces must benon positive for arbitrary processes

Ψ − = ℬ ( ) − ≤0 (4.3)

The structure of power must be controlled by the form of the energy. In ourcase the power is given by

= ∂ ℬ

( ˙ + ˙ ) (4.4)

where ∂ ℬ is the boundary of ℬ with being the unit outward normal to∂ ℬ . We see that some stresses of higher order enter the theory as a resultof the dependence of the free energy density on the gradient of the plasticdistortion.

Transforming the surface integral in (4.4) into a volume integral by Gausstheorem and requiring that (4.3) is satised for an arbitrary ℬ , we obtainthe inequality

∂∂ − ˙ +

∂∂ − ˙

+ ∂∂ − ˙ − ˙ ≤0 (4.5)

For rigid translations the energy does not change while ˙ , ˙ and ˙ are zero. Since the inequality (4.5) must be fullled for an arbitrary translation, the stress must obey the equilibrium equation

= 0 (4.6)

Similarly, the inequality (4.5) can be satised for arbitrary rigid rotationsonly if the stress tensor is symmetric

= (4.7)

Let us introduce the following notation

= − ∂∂

= − ∂∂

(4.8)

ϰ = − ∂∂

+ (4.9)

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4.2. CONTINUUM DISLOCATION THEORY 83

Then the combined rst and second laws of thermodynamics become

˙ + ϰ ˙ + ˙ ≥0 (4.10)

Equation (4.10) shows that and are those parts of the stresses andthe higher order stresses which cause energy dissipated in heating of thecrystal. Tensor describes heating in a non uniform ow, so it has themeaning of viscous stresses. Tensors ϰ and describe heating caused byhomogeneous and inhomogeneous plastic deformation, respectively.

The widely used closure of non equilibrium thermodynamics assumes thatthere exists a dissipation potential

= ( ˙ ˙ ˙ ) (4.11)such that the tensors , ϰ , and controlling the irreversible processesare linked to ˙ , ˙ , and ˙ by the relations

= ∂∂ ˙

ϰ = ∂∂ ˙

= ∂∂ ˙

(4.12)

The set of equations (4.6),(4.7),(4.8),(4.9), and (4.12) is closed with respectto the unknown functions and .

Thus, each continuum model of dislocations is xed by two functions,

namely, the free energy density and the dissipation potential. For isothermalprocesses we require the free energy density to depend only on the elasticstrain and on the dislocation density :

( ) = 12

+ ( ) (4.13)

where ( ) corresponds to the energy of the dislocation network. In thiscase the stresses are given by

= ∂∂

=

The crucial question is then how the energy of the dislocation network depends on the dislocation density. For a single crystal deforming in single slipwe shall adopt the following formula

= ln 1

1 − (4.14)

where is the saturated dislocation density and a material constant. Thelogarithmic energy stems from two facts: i) for small dislocation densities the

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The free energy density of the crystal with dislocations takes a simple form

= 12

( − )2 + ln 1

1 −∣ ∣ (4.19)

If the resistance to the dislocation motion is negligible (and, hence, thedissipation is zero), the true plastic distortion minimizes the total energy, Ψ,which is a functional of ( ),

Ψ[ ( )] = ℎ

0

12

( − )2 + ln 1

1 −∣ ∣ (4.20)

among all admissible function ( ) satisfying the boundary conditions (4.18).The total strain, , is regarded as a given function of time, so one can studythe evolution of the dislocation network which accompanies the change of the total strain.

If the resistance to the dislocation motion cannot be neglected, then theenergy minimization must be replaced by a ow rule. In case of the rateindependent plasticity, when the dissipation potential is

= ∣˙ ∣

with denoting the critical resolved shear stress, the ow rule reads: for

˙ ⁄= 0 ∂∂ ˙

= −

(4.21)

The right hand side of (4.21) is the negative variational derivative of energywith respect to at xed total strain

ϰ ≡ −

= −

∂∂

+ ∂ ∂

∂∂

With from (4.19) we obtain

ϰ = ( − ) + ( −∣ ∣)

2 (4.22)

According to (4.22), if = const in some sub interval of (0 ), ϰ coincideswith the shear stress = ( − ). For ˙ = 0 the evolution equation (4.21)needs not be satised: it is replaced by the equation ˙ = 0.

So, for ˙ ⁄= 0, the evolution equation for is

sign ˙ = ( − ) +

( −∣ ∣)2 (4.23)

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4.3. ANTI PLANE CONSTRAINED SHEAR 87

which must be subject to the boundary conditions (4.18).According to (4.23) the plastic distortion may evolve only if the yield

condition

∣( − ) +

( −∣ ∣)2∣= (4.24)

is fullled. If ∣ϰ

∣< then is frozen : ˙ = 0. If the yield condition (4.24)holds, function ( ) may evolve or may stay unchanged: this depends onthe time dependence of the control parameter .

We rst analyze the situation when the resistance to the dislocation motion is negligible (and, hence, the dissipation is zero). In this case the determination of ( ) reduces to the minimization problem (4.20). Note that,

since is convex with respect to and , this variational problem has aunique solution. It is convenient to introduce the following dimensionlessquantities

=

( ) =

( ) =

2

= ℎ 3

Ψ (4.25)

The dimensionless variable changes on the interval (0 ), where = .The functional (4.20) reduces to

[ ( )] =

0

12

(1 − )2 + ln 1

1

−∣′∣

(4.26)

where the prime denotes differentiation with respect to . We minimizefunctional (4.26) among functions ( ) satisfying the boundary conditions

(0) = ( ) = 0.It is instructive to analyze rst the variational problem in which the

logarithmic term is replaced by an asymptotic formula

ln 1

1 −∣′∣ ≈ ∣′∣+

12

′2

The functional to be minimized becomes

[ ( )] =

012(1 − )2 + (∣′∣+ 12 ′2) (4.27)

Due to the boundary conditions ′ should change its sign on the interval(0 ). One dimensional theory of dislocation pile ups suggests to seek theminimizer in the form

( ) =⎧⎨

⎩

1( ) for ∈(0 ) for ∈( − )

1( − ) for ∈( − )(4.28)

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where is a constant, an unknown parameter, 0

≤ ≤2, and 1( ) =

at = . We have to nd 1( ) and the constants, and . Since ′ > 0for ∈(0 ), the functional becomes

= 2

0

12

(1 − 1)2 + ′1 + 12

′21 + 12

(1 − )2( −2 ) (4.29)

Function 1( ) is subject to the boundary conditions

1(0) = 0 1( ) = (4.30)

Varying this energy functional with respect to 1( ) we obtain the Euler

equation for 1( ) on the interval (0 )1 − 1 + ′′1 = 0 (4.31)

The variation of (4.29) with respect to and yields the two additionalboundary conditions at =

′1( ) = 0 2 = (1 − )( −2 ) (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

1( ) = 1 −cosh √ + tanh √ sinh √ 0 ≤ ≤ (4.33)

Equations (4.30) 2 and (4.32)2 give the following transcendental equation todetermine in terms of the constants and

( ) ≡2 + 2 cosh √ = (4.34)

According to (4.28) must lie in the segment [0 2]. Since cosh( √ ) ≥1,2

≤ −2 . Thus, equation (4.34) has no positive root if < 2 . Returning

to the original variables according to (4.25) we see that inequality < 2corresponds to the condition < , where

= 2

and for < no dislocations are nucleated. Note that the thresholdvalue, , is inversely proportional to the product of the size of specimentimes the saturated dislocation density (a kind of Hall Petch relation). For

> 2 equation (4.34) has only one root in the interval (0 2). Indeed,

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0.01 0.02 0.03 0.04 0.05

0.002

0.004

0.006

0.008

O

A

B

Figure 4.12: Normalized average stress versus shear strain curve

and the boundary conditions (4.30) and (4.32). Let ( ) be the function

( ) = 1 − ′1

Equation (4.37) and the boundary conditions can be written as the followingboundary value problem

′ = 1 −1 ′1 = 1 −

1(0) = 0 1( ) = ′1( ) = 0 (4.38)

Function ( ) is a decreasing function of . Since ′1( ) = 0, ( ) = .The system of ordinary differential equations (4.38) admits the rst inte

gral12

(1 − 1)2 + ln − = const

The value of the constant can be found from the boundary conditions at thepoint =

12

(1 − 1)2 + ln − = 12

(1 − )2 + ln −Hence

1 − 1( ) = (1 − )2 + 2 ( ( )

−1 −ln ( )

) (4.39)

The expression under the square root is positive because ( ) ≥ and > ln +1 for ≥1. For the function ( ) = ( ) −1 we have the initial

value problem

′ = −1

(1 − )2 + 2 ( −ln(1 + )) (0) = 0 (4.40)

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distortion may evolve only if the yield condition (4.24) is satised. It isconvenient to divide both side of (4.24) by to obtain

∣ − +

( −∣ ∣)2∣= ≡ (4.44)

We regard again as a given function of time (the driving variable) andtry to determine ( ). We consider the following close loading path: isrst increased from zero to some value ∗ > , then decreased to − , andnally increased to zero (Fig. 4.14). The rate of change of ( ) does not affectthe results due to the rate independence of the dissipation. The problem isto determine the evolution of as function of and , provided (0 ) = 0.

Since is initially zero, we see from (4.44) that = 0 as long as < .Thus, the dissipative threshold stress (the yield stress) = in this case.For > the yield condition

− +

( −∣ ∣)2 = (4.45)

takes place everywhere in (0 ). Equation (4.45) follows from (4.44) and theinitial condition (0 ) = 0, since for small ( ) and >

− +

(

−∣

∣

)2 = − +

(

−∣

∣

)2

It is convenient to introduce the deviation of ( ) from the critical shear, , = − > 0, and to dene the following dimensionless quantities

=

( ) =

( )

=

2 (4.46)

They are similar to those of (4.25), with replacing . Equation (4.45)takes the dimensionless form

1 − + ′′

(1 −∣′∣)2 = 0 (4.47)

One dimensional theory of dislocation pile ups with non zero resistancesuggests that the solution of (4.47) is symmetric

( ) = 1( ) for ∈(0 2)1( − ) for ∈( 2 )

(4.48)

where = . Function 1( ) is determined from equation (4.37) andthe boundary conditions

1(0) = 0 ′1( 2) = 0 (4.49)

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t

c

Figure 4.14: A close loading path

The rst condition means that dislocations cannot reach the boundary of theregion because the boundary is clamped. The second condition follows fromthe continuity of plastic distortion and the symmetry property (4.48).

The boundary value problem (4.37), (4.49) can be solved in exactly thesame manner as for the case without dissipation. For the function ( ) = ( ) −1 we have the initial value problem (4.42). Integrating equation(4.42) over from zero to 2 and taking into account the condition ( 2) =0 which is the consequence of (4.49)2, we get

( 0)

≡ 0

0

1 + 2 ( − 0 −ln((1 + ) (1 + 0)))= 2 (4.50)

Thus, 0 is the root of the transcendental equation (4.50). Note that, as longas 0 is smaller than the root of the equation

0 −ln(1 + 0) = 12

= 2

2 (4.51)

the expression under the square root of integral (4.50) is positive and theintegral is well dened. This follows from the inequalities 1 −2 ( 0 −ln(1+

0)) > 0 for 0 smaller than the root of (4.51) (because 1 −2 ( −ln(1 +)) is strictly decreasing function of ) and −ln(1 + ) > 0 for > 0.

Differentiating ( 0) with respect to 0, it is easy to check that 0 > 0,so ( 0) is a strictly increasing function. Since (0) = 0 and ( 0) → ∞ as 0 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 nd 1( ) approximately by solving the initialvalue problem on (0 2)

′ = − 2( −ln(1 + )) (0) = 0

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where 0 is the root of the equation (4.51). The length of the boundarylayer may be dened by the formula

= 2 0

−ln(1 + )

where is a small positive number. We take = 10−5.Fig. 4.15 shows the evolution of ( ) (where = ) as increases.

For the numerical simulation we took as before = 0 04, = = 12.One can see that the dislocation density as well as the length of boundarylayers increase as increases. Note that the expression

− +

( −∣ ∣)2 = − ≡ −( ( ) − )

does not depend on at any instant.

2 4 6 8 10 12

0.0005

0.001

0.0015

0.002

0.0025

0.003

x a

b

c

Figure 4.15: Evolution of : a) = 0 0001, b) = 0 001, c) = 0 003

After reaching some value of shear strain ∗ > we unload the crystalby decreasing . Since ϰ becomes smaller than , does not change ( =

∗( )) until

− ∗ + ∗( −∣ ∗∣)2 = − (4.52)

where ∗( ) is the solution of equation (4.45) for ( ) = ∗:

∗− ∗ + ∗

( −∣ ∗∣)2 = (4.53)

From (4.52) and (4.53) one can see that the ow begins when −( ∗− ) =

− , i.e. for = ∗ = ∗−2 . From that value of the yield condition

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ϰ =

− takes place leading to the decrease of which should be now

determined by the equation

− +

( −∣ ∣)2 = − (4.54)

Since for ∈(− ∗) the deviation = + is positive, equation (4.54)

can again be transformed to equation (4.47) and solved in exactly the samemanner if we replace = ( ) − in all formulas (4.46) (4.51) by =

( ) + . As approaches − , tends to zero because →0. The furtherincrease of from − to zero does not cause change in which remainsidentically zero.

0.5 1 1.5

0.002

0.004

0.006

0.008

0.01

0.012

0.014

x a

b

c

Figure 4.16: Evolution of the normalized dislocation density: a) = 0 0001,b) = 0 001, c) = 0 003

The dislocation density = can be calculated from the solution (4.48).In terms of the dimensionless variable (4.46) we have

( ) =⎧⎨

⎩

( ) (1 + ( )) for ∈(0 )0 for ∈( − )

− (

− ) (1 + (

− )) for

∈(

−)

Since ( ) is decreasing, the maximum dislocation density is achieved at = 0 giving max = 0 (1 + 0). So, the parameter 0 is simply linked

to the maximum dislocation density. Fig. 4.16 shows the distributions of the normalized dislocation density ( ) within the interval (0 ) (where = ) as changes.

As soon as the plastic deformation develops, the shear stress = ( − )becomes inhomogeneous. It is interesting to calculate the average shear stress(4.35) which is a measurable quantity. During the loading, according to

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(4.48), we have for the normalized average shear stress (or the average elasticshear strain)

( ) =

= − 1

−2

0(1 − ( )) = + 2

2

0(1 − ( ))

The integral is evaluated using equation (4.38)

0(1 − ( )) = −

0 ′( ) = −( ( ) − (0)) = 0

Thus( ) = + 2 0 (4.55)

where the second term causing hardening depends on the maximum dislocation density and is inversely proportional to the product of the size of specimen times the saturated dislocation density. Formula (4.55) describesthe size effect in this model.

During the inverse loading, when the yield condition ϰ = − holds true,formula (4.55) changes to

( ) = − + 2

0

where 0 is the root of the equation

0 −ln(1 + 0) = 2

2

0.005 0.01 0.015 0.02

-0.004

-0.002

0.002

0.004

O

AB

CD

cc *

*

Figure 4.17: Normalized average stress versus shear strain curve

Fig. 4.17 shows the normalized average shear stress (or average elasticshear strain) versus shear strain curve for the loading program of Fig. 4.14.

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4.4. PLANE CONSTRAINED SHEAR 97

We took = 12, = 0 04, = 0 004, ∗ = 0 02. The straight lineOA corresponds to the purely elastic loading with increasing from zero to

. The line AB corresponds to the plastic yielding with ϰ = . The yieldbegins at the point A with the yield stress = . The work hardeningdue to the dislocation pile up is observed which is described by the secondterm in (4.55). During the unloading as decreases from ∗ to

∗ = ∗−2

(the line BC) the plastic distortion = ∗ is frozen. As decreases furtherfrom

∗ to − , the plastic yielding occurs with ϰ = − (the line CD). The

yield stress = ∗−2 at the point C, at which the inverse plastic owsets on, is larger than − (because ∗ > ≡ ). Along the line CD,as is decreased, the created dislocations annihilate, and at the point D all

dislocations disappear. Finally, as increases from − to zero, the crystalbehaves elastically with = 0. In this close cycle ABCD dissipation occursonly on the lines AB and CD. It is interesting that the lines DA and BCare parallel and have the same length. In phenomenological plasticity theorythis property is modelled as the translational shift of the yield surface in thestress space, the so called Bauschinger effect.

4.4 Plane constrained shear

0

y

x

h

h

s m

a

L

z

Figure 4.18: Plane constrained shear

As the next boundary value problem we consider a strip made of a singlecrystal undergoing a plane strain shear deformation (see Fig. 4.18). Let thecross section of the strip be a rectangle of width and height ℎ, 0 ≤ ≤ ,0 ≤ ≤ℎ. We realize the shear deformation by placing the strip in a hard

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device with the prescribed displacements at its upper and lower sides

(0) = 0 (0) = 0 (ℎ) = ℎ (ℎ) = 0 (4.56)

where ( ) and ( ) are the longitudinal and transverse displacements, respectively, with being the overall shear strain. We assume that the lengthof the strip is large, and the width is much greater than the height ℎ( ≫≫ℎ) to neglect the end effects and to have the stresses and strainsdepending only on one variable in the central part of the strip.

For the plane strain state the components of the strain tensor are

= 0 = = 1

2

= (4.57)

If the shear strain is sufficiently small, then the crystal deforms elastically and = , = 0 everywhere in the strip. If exceeds some criticalthreshold, then edge dislocations may appear. We admit only the slip directions (or the directions of the Burgers vectors) perpendicular to the axisand inclined at an angle with the axis and the dislocation lines parallel to the axis. Since only one slip system is active, the plastic distortionis given by = , with = (cos sin 0) being the slip direction,and = (−sin cos 0) the normal vector to the slip plane. We assumethat depends only on : = ( ) (translational invariance). Because of

the prescribed boundary conditions (4.56), dislocations cannot penetrate theboundaries = 0 and = ℎ, therefore

(0) = (ℎ) = 0 (4.58)

The in plane components of the plastic strain tensor are

= −12

sin2 = 12

cos 2 = 12

sin2

With these total and plastic strain tensors we obtain the in plane componentsof the elastic strain tensor

= 12 sin2 = 12( − cos2 ) = − 12 sin2

As depends only on , there are two non zero components of Nye s dislocation density tensor (4.1), namely, = sin cos and = sin2 .Thus, the resultant Burgers vector of all dislocations whose dislocation linescut the area perpendicular to the axis is parallel to the slip direction andthe scalar dislocation density equals

= 1 2 + 2 =

1∣ ∣∣sin ∣

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Assuming for simplicity the isotropic elastic property of the crystal, wewrite the free energy per unit volume of the crystal with dislocations as

( ) = 12

( )2 + + ln 1

1 − ∣ ∣∣sin ∣ (4.59)

Thus, the total energy functional becomes

Ψ[ ] = ℎ

0

12

2 + 12

( − cos2 )2 + 14

2 sin2 2 (4.60)

+ ( − 12

sin2 )2 + ln 1

1

− ∣ ∣∣sin ∣

Functional (4.60) can be reduced to a functional depending on ( ) only.Indeed, by rst xing ( ) and taking the variation of (4.60) with respect to

and we derive the equilibrium equations

( − cos2 ) = 0( + 2 ) − sin 2 = 0

Integrating these equations and using the boundary conditions (4.56) we get

= + (

− ⟨

⟩

)cos2

= ( − ⟨ ⟩)sin2 (4.61)

where = +2 , and ⟨⋅⟩ = 1

ℎ ℎ

0 ⋅ . Substituting (4.61) into (4.60) andcollecting common terms we obtain the energy functional in terms of

Ψ[ ] = ℎ

0

12

(1 − ) 2 sin2 2 + 12 ⟨

⟩2 sin2 2 (4.62)

+ 12

( − ⟨ ⟩cos2 )2 + ln 1

1 − ∣ ∣∣sin ∣

If the dissipation is negligible, then the plastic distortion minimizes (4.62)under the constraint (4.58). The overall shear strain is regarded as givenfunction of time (control parameter), so one can study the evolution of thedislocation network which accompanies the change of .

If the resistance to dislocation motion cannot be neglected, then theenergy minimization must be replaced by a ow rule. In case of rateindependent plasticity, the ow rule for ˙ ⁄= 0 reads

∂∂ ˙

= −

(4.63)

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where the dissipation potential for crystals deforming in single slip is

= ∣˙ ∣

with being a positive constant called critical resolved shear stress. Theright hand side of (4.63) is the negative variational derivative of the energywith respect to at xed overall strain

ϰ ≡ −

= −

∂∂

+ ∂ ∂

∂∂

For ˙ = 0, the evolution equation (4.63) does not have to be satised: It isreplaced by the equation ˙ = 0.

For small up to moderate dislocation densities the logarithmic term in(4.62) may be approximated by the formula

ln 1

1 − ∼=

+ 12

2

2 (4.64)

so that

( ) =

ℎ

0

12

(1

−) 2 sin2 2 +

12⟨

⟩

2 sin2 2 (4.65)

+ 12

( − ⟨ ⟩cos2 )2 + ∣ ∣∣sin ∣+

12

2 sin2

( )2

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 ( ) reduces to the minimization of the total energy (4.65).Since is convex with respect to and , the variational problem has aunique solution. It is convenient to introduce the dimensionless quantities

=

Ψ = ℎ = ℎ (4.66)

The dimensionless variable changes on the interval (0 ℎ). Functional (4.65)reduces to

[ ] = ℎ

0

12

(1 − ) 2 sin2 2 + 12 ⟨

⟩2 sin2 2 (4.67)

+ 12

( − ⟨ ⟩cos2 )2 + ∣ ′∣∣sin ∣+ 12

′2 sin2

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where the prime denotes differentiation with respect to , and, for short,the bars over and ℎ are dropped. We minimize functional (4.67) amongfunctions satisfying the boundary conditions

(0) = (ℎ) = 0 (4.68)

For the variational problem of this type, there exists a threshold value such that when < no dislocations are nucleated and = 0. Near thethreshold value the dislocation density must be small so that the last termin (4.67) can be neglected. Besides, the width of the boundary layer tendsto zero as → . This gives us the idea of nding the threshold value byemploying a minimizing sequence of the form

=⎧⎨⎩

for ∈(0 ) for ∈( ℎ− ) (ℎ− ) for ∈(ℎ− ℎ)

(4.69)

where is an unknown constant, and is a small unknown length whichtends to zero as → . Substituting (4.69) into the energy functional(4.67) (with the last term being removed) and neglecting all small terms of order and higher, we obtain

( ) = 12

[( − cos 2 )2 + 2 sin2 2 ]ℎ + 2 ∣ sin ∣ (4.70)

A rather simple analysis shows that the minimum of (4.70) is achieved at ⁄= 0 if and only if

> = 2 ∣sin ∣ℎ∣cos2 ∣

otherwise it is achieved at = 0 (no dislocations are nucleated). Note thatthe sign of depends on the angle : is positive if 0∘ < < 45∘ andnegative if 45∘ < < 90∘. In terms of the original length ℎ the energetic

threshold value reads =

2ℎ∣sin ∣∣cos2 ∣

(4.71)

showing clearly the size effect. Equation (4.71) deviates from the well knownHall Petch relation. The reason for this deviation can be explained by theboundary conditions (4.58) which do not permit the penetration of dislocations through the grain boundaries.

Due to the boundary conditions, ′ should change its sign on the interval(0 ℎ). The one dimensional theory of dislocation pile ups as well as the

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solution of the previous problem suggest to seek the minimizer in the form

( ) =⎧⎨⎩

1( ) for ∈(0 ) for ∈( ℎ− ) 1(ℎ− ) for ∈(ℎ− ℎ)

(4.72)

where is a constant, an unknown length, 0 ≤ ≤ ℎ2 , and 1( ) = .

We have to nd 1( ) and the constants, and . With from (4.72) thetotal energy functional becomes

= 2

0

1

2(1

−) 21 sin2 2 +

∣

′1∣∣

sin

∣

+ 1

2 ′21 sin2

+ 12

(1 − ) 2 sin2 2 (ℎ−2 ) + 12ℎ[ ⟨ ⟩

2 sin2 2 + ( − ⟨ ⟩cos2 )2]

(4.73)

where

⟨ ⟩= 1ℎ

2

0 1 + ( ℎ−2 ) (4.74)

Varying the energy functional (4.73) with respect to 1 we obtain

− ′′1 sin2 +(1

−) 1 sin2 2 +(cos 2 2 + sin2 2 )

⟨

⟩− cos2 = 0 (4.75)

where 1( ) is subject to the boundary conditions

1(0) = 0 1( ) = (4.76)

The variation of (4.73) with respect to gives an additional boundary condition at = ,

′1( ) = 0 (4.77)

which means that the dislocation density must be continuous. Varying theenergy functional with respect to , we obtain a condition for ,

2 ∣sin ∣(sign ′1) + [(cos2 2 + sin2 2 )⟨ ⟩− cos2+ (1 − ) sin2 2 ](ℎ−2 ) = 0 (4.78)

Equations (4.75), (4 76)1, and (4.77) yield the solution

1 = 1 (1 −cosh + tanh sinh ) 0 ≤ ≤ (4.79)

with 1 =

cos 2 −(cos2 2 + sin2 2 )⟨ ⟩(1 − )sin2 2

(4.80)

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and

= 2 1 − ∣cos ∣

With (4.74), (4 76)2, (4.79), and (4.80) we obtain the average of

⟨ ⟩= cos2 2 − tanh + 1 − 1

cosh (ℎ−2 )

( ) (4.81)

where

( ) = ℎ(1 − )sin2

2 + (cos2

2 + sin2

2 )2 −

tanh+ 1 −

1cosh

(ℎ−2 )

and

= cos2 − ⟨ ⟩(cos2 2 + sin2 2 )

(1 − )sin2 21 −

1cosh

(4.82)

Substitution of (4.82) into (4.78) gives the following equation to determine :

( ) ≡2 ∣sin ∣(sign ′1) − cos2 − ⟨ ⟩(cos2

2 + sin2

2 )cosh (ℎ−2 ) = 0

a

b

c

d

e

f

y -

Figure 4.19: Evolution of for single slip constrained shear at zero dissipation: a,d) = 0 0068, b,e) = 0 0118, c,f) = 0 0168

Fig. 4.19 shows the evolution of ( ) as increases, for = 30∘ (continuous lines) and = 60∘ (dashed lines), where = . For the numericalsimulation we took the material parameters from Table 4.1. All materialparameters used in this Chapter (except , , and which is responsible

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for the cross slip interaction) are well known for aluminum. We choose theseadditional parameters to have good agreement of discrete dislocation simulations and the continuum dislocation theory with respect to the yield stressand 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 ℎ = ℎ = 0 349.

It is interesting to plot the shear stress = ( −⟨ ⟩cos2 ) as a functionof the shear strain. As we know, for < no dislocations are nucleatedand = 0, so the shear stress = . For > we take ⟨ ⟩ from (4.81)to compute the shear stress.

A´

A

BB´

0

Figure 4.20: Normalized shear stress versus shear strain curve for single slipconstrained shear at zero dissipation

Fig. 4.20 shows the normalized shear stress versus shear strain curve OABfor = 30∘ and OA B for = 60∘. There is a work hardening sectionAB for > caused by the dislocation pile up. Note, however, thatthere is no residual strain as we unload the crystal by decreasing : thestress strain curve follows the same path BAO, so the plastic deformationis completely reversible and no energy dissipation occurs. In the course of unloading the dislocations nucleated annihilate, and as we approach point Athey all disappear.

If the resistance to dislocation motion (and hence the dissipation) cannotbe neglected, the plastic distortion may evolve only if the yield condition

∣ϰ ∣ = is fullled. If ∣ϰ

∣ < , then is frozen, the dislocation densityremains unchanged and the crystal deforms elastically. Computing the vari

Material (GPa) (A) (m−2) Aluminum 26.3 0.33 2.5 1 3961015 0 000115 0576

Table 4.1: Material characteristics

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ational derivative of (4.65) we derive from (4.63) the yield condition

sin2

2 2 −(1 − ) sin2 2 −(cos2 2 + sin2 2 )⟨ ⟩+ cos2 =

Consider rst the case < 45∘. We divide this equation by and introducethe dimensionless variable = to transform the yield condition to

′′ sin2 −(1 − ) sin2 2 −(cos2 2 + sin2 2 )⟨ ⟩+ cos2= cos 2 (4.83)

with ≡ cos2 and the prime denoting the derivative with respect to. We shall further omit the bar over for short.We regard as a given function of time (the driving variable) and try

to determine ( ). We consider the loading path similar to that shown inFig. 4.14. The problem is to determine the evolution of as a function of and , provided (0 ) = 0 and < 45∘.

Since the plastic distortion, , is initially zero, we see from (4.83) that = 0 as long as < . Thus, the dissipative threshold stress (the yield

stress) = cos2 in this case. For small ( ) and > , the yieldcondition becomes

′′ sin2 −(1− ) sin2 2 −(cos2 2 + sin2 2 )⟨ ⟩+ cos2 = cos2(4.84)

Let us introduce the deviation of ( ) from the critical shear , = − ,and simplify (4.84) to obtain

′′ sin2 −(1 − ) sin2 2 −(cos2 2 + sin2 2 )⟨ ⟩+ cos 2 = 0

Since this equation is linear, is proportional to such that = 1,where 1 is the solution of the equation

′′1 sin

2

−(1− ) 1 sin2

2 −(cos2

2 + sin2

2 )⟨ 1⟩+cos2 = 0 (4.85)The analogous problem of anti plane constrained shear at nonzero resis

tance suggests that the solution of (4.85) is symmetric, i.e.

1( ) = 1(ℎ− ) for ∈(ℎ 2 ℎ) (4.86)

Function 1( ) is determined from equation (4.85) and the boundary conditions

1(0) = 0 ′1(ℎ 2) = 0 (4.87)

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The rst condition means that dislocations cannot reach the boundary of theregion because of the prescribed displacement. The second condition followsfrom the continuity of plastic distortion and the symmetry property (4.86).

Equations (4.85) and (4.87) admit the solution

1 = 1 1 −cosh + tanhℎ

2 sinh 0 ≤ ≤

ℎ2

(4.88)

with 1 =

cos2 −(cos2 2 + sin2 2 )⟨ 1⟩(1 − )sin2 2

(4.89)

and

= 2 1 − ∣cos ∣ (4.90)

The average of 1 is obtained in the form

⟨ 1⟩=cos2 1 −

2 tanh ℎ

2

ℎ

(1 − )sin2 2 + (cos2 2 + sin2 2 ) 1 − 2 tanh ℎ

2

ℎ

(4.91)

y _

1

Figure 4.21: Graphs of 1( ) for single slip constrained shear at non zero

dissipationFig. 4.21 shows the graphs of 1( ) for = 30∘ (continuous line) and

= 60∘ (dashed line). For the numerical simulation we took the materialparameters from Table 4.1, and ℎ = 1 m, so that ℎ = ℎ = 0 349.

After reaching ∗ > , we unload the crystal by decreasing . Since ϰ

becomes smaller than , does not change ( = ∗( )) until

∗′′ sin2 −(1 − ) ∗ sin2 2 −(cos2 2 + sin2 2 )⟨ ∗⟩+ cos2 = − cos2

(4.92)

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where ∗( ) is the solution of (4.84) for ( ) = ∗. From (4.92) we can see thatthe plastic ow begins when −( ∗− ) = − , i.e. for = ∗ = ∗−2 .From that value of , the yield condition ϰ = − holds leading to a decreaseof which should now be determined from

′′ sin2 −(1− ) sin2 2 −(cos2 2 + sin2 2 )⟨ ⟩+ cos2 = − cos2(4.93)

Since for ∈(− ∗), the deviation = + is positive, equation (4.93)

can again be transformed to equation (4.85) and solved in exactly the samemanner if we replace = − in all formulas (4.88) (4.91) by = + .As approaches − , tends to zero because →0. The further increaseof from

− to zero does not cause change in which remains zero.

It is not difficult to modify the construction given above to nd the solution for > 45∘.

y _

1

Figure 4.22: Graphs of 1( ) for single slip constrained shear at non zerodissipation

The normalized dislocation density = sin can be calculated fromthe solution (4.88). Since ( ) is proportional to , ( ) is also proportionalto such that ( ) = 1( ). For ∈(0 ℎ2) we have

1( ) = 1 − sinh + cosh tanh ℎ

2sin

with 1 from (4.89) and from (4.90). For ∈ (ℎ 2 ℎ) we have 1( ) =

− 1(ℎ− ) due to symmetry. Fig. 4.22 shows the graphs of 1 for = 30∘(continuous line) and = 60∘ (dashed line) for ∈(0 ℎ 2).

It is interesting to calculate the shear stress which is a measurablequantity. During loading, we have for the normalized shear stress (or theelastic shear strain)

=

= + − 1 − 2tanh ℎ

2

ℎ 1 cos 2 (4.94)

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∗ >

−cos2 ). Along line CD, as is decreased, the created dislocations

annihilate, and at point D all dislocations have disappeared. Finally, as increases from − to zero, the crystal behaves elastically with = 0. Inthis 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 lengthexhibiting again the Bauschinger effect.

4.5 Single crystals deforming in double slip

r l

y

h

0

z

x a

L

h

s l m l

m r

s r

Figure 4.26: Plane constrained shear of single crystals deforming in doubleslip

Consider the same problem as in the previous Section, but now for crystalsdeforming in double slip. We admit two slip directions (or the directions of the Burgers vectors) perpendicular to the axis and inclined at an angle

(0 ≤ ≤ 2) and ( 2 ≤ ≤ ) with the axis, respectively,and the dislocation lines parallel to the axis. Since two slip systems areactive, the plastic distortion is given by = + , with =(cos sin 0) being the slip directions, and = (−sin cos 0) thenormal vectors to the slip planes ( = ).

We assume that and depend on only: = ( ) and = ( )(translational invariance). Because of the prescribed boundary conditionsdislocations cannot penetrate the boundaries = 0 and = ℎ, therefore

(0) = (0) = (ℎ) = (ℎ) = 0 (4.95)

As previously, the boundaries = 0 and = ℎ serve as obstacles to dislocation motion.

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4.5. SINGLE CRYSTALS DEFORMING IN DOUBLE SLIP 111

Under plane strain state conditions, the non zero components of the plastic strain tensor are

= −12

( sin 2 + sin 2 )

= 12

( cos2 + cos2 )

= 12

( sin 2 + sin 2 )

(4.96)

With (4.57) and (4.96) we obtain the non zero components of the elasticstrain tensor

= 12

( sin 2 + sin 2 )

= 12

( − cos2 − cos 2 )

= − 12

( sin 2 + sin 2 )

(4.97)

As and depend only on , there are two non zero components of Nye s dislocation density tensor

= sin cos + sin cos = sin2 + sin2

These are the components of the resultant Burgers vector of all edge dislocations whose dislocation lines cut the area perpendicular to the axis. Thus,the dislocations produced on two slip systems belong to two different groups:the rst one with the resultant Burgers vector showing in the direction ,the second one with the resultant Burgers vector parallel to . Thereforethe scalar dislocation densities (or the numbers of dislocations per unit area)equal

= 1

∣

sin

∣

= 1

∣

sin

∣

(4.98)

The free energy per unit volume of the crystal with dislocations takes theform

( ) = 12

( )2 + + ln 1

1 − + ln 1

1 − + 2

(4.99)The last term in parentheses corresponds to the energy of the dislocationnetwork which consists of energies of each group of dislocations plus the

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energy of cross slip interaction, with being the interaction factor. With(4.97) and (4.99), the total energy functional becomes

Ψ[ ] = ℎ

0

12

2 + 14

( sin 2 + sin 2 )2

+ 12

( − cos 2 − cos 2 )2 + ( − 12

sin 2 − 12

sin 2 )2

+ ln 1

1 − + ln 1

1 − + 2 (4.100)

Functional (4.100) can be reduced, in exactly the same manner as in the

previous case, to a functional depending on ( ) and ( ) only. The resultis

Ψ[ ] = ℎ

0

12

(1 − )( sin 2 + sin 2 )2

+ 12

(⟨ ⟩sin2 + ⟨ ⟩sin2 )2 + 12

( − ⟨ ⟩cos2 − ⟨ ⟩cos2 )2

+ ln 1

1 − + ln 1

1 − + 2 (4.101)

Using (4.64) for small up to moderate dislocation densities we reduce (4.101)to

Ψ[ ] = ℎ

0

12

(1 − )( sin 2 + sin 2 )2

+ 12

(⟨ ⟩sin2 + ⟨ ⟩sin2 )2 + 12

( − ⟨ ⟩cos2 − ⟨ ⟩cos2 )2

+ + 12

2

+

+ 12

2

+ 2 (4.102)

If the dissipation can be neglected, then and minimize (4.102) underthe constraints (4.95).

If the resistance to dislocation motion cannot be ignored, then the energyminimization must be replaced by the ow rules. For ⁄= 0 and ⁄= 0 wehave

∂∂ ˙

= −

∂

∂ ˙ = −

(4.103)

where the dissipation potential is

= ∣˙ ∣+ ∣

˙ ∣

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with and being positive constants called critical resolved shear stressesof the corresponding slip systems. The right hand sides of (4.103) are thenegative variational derivatives of the energy with respect to and , respectively, at xed overall shear strain

ϰ ≡ −

= −

∂∂

+ ∂ ∂

∂∂

=

The corresponding evolution equation (4.103) does not have to be fullledfor = 0 or = 0. It is replaced by the equation = 0 or = 0.

If the resistance to dislocation motion can be neglected (and hence theenergy dissipation is zero) the determination of ( ) and ( ) reduces tothe minimization of the total energy (4.102). We analyze this variationalproblem rst in the special case = − = − which correspondsto symmetric double slip. It is convenient to introduce the dimensionlessvariables (4.66), in terms of which the functional (4.102) reads

[ ] = ℎ

0

12

(1 − )( − )2 sin2 2 + 12

(⟨ ⟩− ⟨ ⟩)2 sin2 2

+ 12

( −(⟨ ⟩+ ⟨ ⟩)cos2 )2 + ∣sin ∣(∣ ′∣+ ∣ ′∣)

+ 12 sin

2

( ′2

+ ′2

+ 2 ∣ ′∣∣ ′∣) (4.104)

where the prime denotes differentiation with respect to , and, for short,the bars over and ℎ are dropped. We minimize functional (4.104) among

satisfying the boundary conditions (4.95). To guarantee existence anduniqueness of the minimizer we must ensure convexity of the free energydensity with respect to ′. For this purpose let us consider the matrix

′ ′ ′ ′

′ ′ ′ ′= sin2 1 sign ′sign ′

sign ′sign ′ 1

It is easy to see that this matrix is positive denite (and, consequently,the energy is convex with respect to ′) if ⁄= 0 and < 1. For =1 the determinant of the matrix vanishes and there exists an eigenvectorcorresponding to a zero eigenvalue. Thus, for = 1 the energy is no longerstrictly convex and one may expect non uniqueness of the minimizer as wellas some numerical instability. In order to avoid this deciency we will assumethat < 1.

Similarly to the single slip problem, there exists a threshold value such that when < no dislocations are nucleated and = = 0. In

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order to nd the threshold value, we employ minimizing sequences of theform

=⎧⎨⎩

for ∈(0 ) for ∈( ℎ− ) (ℎ− ) for ∈(ℎ− ℎ)

(4.105)

and

=⎧⎨

⎩

for ∈(0 ) for ∈( ℎ− ) (ℎ− ) for ∈(ℎ− ℎ)

(4.106)

where and are unknown constants, and is a small unknown lengthwhich tends to zero as → .

Substituting (4.105) and (4.106) into the energy functional (4.104) (withthe last term being removed) and neglecting all small terms of order andhigher, we obtain the function of two variables

( ) = 12ℎ ( − )2 sin2 2 + ( −( + )cos2 )2

(4.107)+ 2 ∣sin ∣(∣ ∣+ ∣ ∣)

The partial derivatives of (4.107) with respect to and must vanishgiving

∂ ∂

= ℎsin2 2 ( − ) −ℎcos2 ( −( + )cos2 )

+ 2 ∣sin ∣sign = 0

and

∂ ∂

= −ℎsin2 2 ( − ) −ℎcos2 ( −( + )cos2 )

+ 2

∣

sin

∣

sign = 0

It is easy to see that these equations imply = = . Then, a rathersimple analysis shows that the minimum of (4.107) is achieved at ⁄= 0 if and only if (in terms of the original length ℎ)

> = 2ℎ∣sin ∣∣cos2 ∣

otherwise it is achieved at = 0 (no dislocations are nucleated). Note thatthe sign of depends on the angle : is positive if 0∘ < < 45∘ and is

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negative if 45∘ < < 90∘. The energetic threshold value for the symmetricdouble 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 ( ) = ( ) = ( )for ∈(0 ℎ). Due to the boundary conditions, ′ should change its sign onthe interval (0 ℎ). The one dimensional theory of dislocation pile ups as wellas the solution of the previous problem suggest to seek the minimizer in theform

( ) =⎧⎨

⎩

1( ) for ∈(0 ) for ∈( ℎ− ) 1(ℎ

−) for

∈(ℎ

−ℎ)

(4.108)

where is a constant, an unknown length, 0 ≤ ≤ ℎ2 , and 1( ) = .

The total energy functional becomes

[ ] =

0(4 ∣sin ∣∣ ′1∣+4 sin2 ′21 ) +2ℎ

12

− ⟨ ⟩cos22

(4.109)

where

⟨ ⟩= 1ℎ

2

0 1 + (ℎ−2 ) (4.110)

We have to nd 1( ) and the constants, and .This variational problem can be solved in exactly the same manner as in

the previous case. Here, we present the solution:

1 = cos2 ( −2⟨ ⟩cos2 )

(1 + )sin2 − 12

2 0 ≤ ≤ (4.111)

where the average of is given by

⟨ ⟩= (3ℎ−2 ) 2 cos2

6ℎ (1 + )sin2 + 2(3ℎ−2 ) 2 cos2 2 (4.112)

with = cos2 ( −2⟨ ⟩cos2 )2 (1 + )sin2 2 (4.113)

The equation to determine reads

2cos2 −

∣sin ∣(sign ′1)(ℎ−2 )cos2 2 −

(3ℎ−2 ) 2 cos26ℎ (1 + )sin2 + 2(3ℎ−2 ) 2 cos2 2

= 0

Fig. 4.27 shows the evolution of ( ) as increases, for = 30∘ (continuous lines) and = 60∘ (dashed lines), where = . For the numerical

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y _

1

Figure 4.29: Graphs of 1( ) for double slip constrained shear at non zerodissipation

Fig. 4.29 shows the graphs of 1( ) for = 30∘ (continuous line) and = 60∘ (dashed line). For the numerical simulation we took the material

parameters from Table 4.1, and ℎ = 1 m, so that ℎ = ℎ = 0 349.After reaching ∗ > , we unload the crystal by decreasing . Since ϰ

becomes smaller than , does not change ( = ∗( )) until

2cos2 ( −2⟨ ∗⟩cos2 ) + 2 (1 + )sin2 ∗′′ = 2 cos2 (4.120)

where ∗

( ) is the solution of (4.116) for ( ) = ∗

. From (4.120) we can seethat the plastic ow begins when −( ∗− ) = − , i.e. for =

∗ =

∗−2 . From that value of , the yield condition ϰ = − holds leadingto a decrease of which should be now determined from the equation

2cos2 ( −2⟨ ⟩cos2 ) + 2 (1 + )sin2 ′′ = 2 cos2 (4.121)

Since for ∈ (− ∗), the deviation = + is positive, equation

(4.121) can again be transformed to equation (4.117) and solved in exactlythe same manner if we replace = − in all formulas (4.118) and (4.119)by = + . As approaches

− , tends to zero because

→0. The

further increase of from − to zero does not cause change in whichremains zero.

It is not difficult to modify the construction given above to nd the solution for > 45∘.

For symmetric double slip the dislocation densities are the same, = = ′ sin . The resultant Burgers vector of all dislocations has only one

non zero component in the direction: = 2 sin2 . Thus, couples of dislocations near the boundaries form super dislocations with the Burgersvector in the direction. The normalized dislocation density = ′ sin

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y _

1

Figure 4.30: Graphs of 1( ) for double slip constrained shear at non zerodissipation

can be calculated from the solution (4.118). Since ( ) is proportional to ,( ) is also proportional to such that ( ) = 1( ) with

1( ) = cos2 (1 −2⟨ 1⟩cos2 )

2 (1 + )sin (ℎ−2 )

and with ⟨ 1⟩ from (4.119). Fig. 4.30 shows the graphs of 1 for = 30∘(continuous line) and = 60∘ (dashed line) for ∈(0 ℎ).

h/d=40

h/d=160 h/d=80

h/d=240 with dissipationenergy minimization

0

Figure 4.31: Comparison of the normalized shear stress versus shear straincurves obtained from this approach and in [6]

It is interesting to calculate the shear stress which is a measurablequantity. During loading, we have for the normalized shear stress (or theelastic shear strain)

( ) =

= + ( −2⟨ 1⟩cos2 ) (4.122)

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with

⟨ 1

⟩

from (4.119). The second term of (4.122) causes hardening due tothe dislocations piling up. Equation (4.122) describes the size effect in thismodel.

Fig. 4.31 shows the stress strain curves during loading obtained fromenergy minimization and from the ow rule as compared with the averagestress strain curves from discrete dislocation simulations in [6]. In orderto compare with the discrete dislocation simulations we took such that

= = 0 00118, = 60∘, 0 = 1 9 10−3 and let all other materialconstants remain the same as in the previous Section. The average stressstrain curves in the discrete dislocation simulations are provided for fourdifferent ratios ℎ , where is the spacing between the active slip planes. It

is seen that reasonably good agreement of the discrete and the continuumapproaches is observed at ℎ = 240.

h/d=80energy minimizationwith dissipation

y _ h

u ,y

Figure 4.32: Comparison of the total shear strain proles obtained from thepresent approach and in [6]

Fig. 4.32 shows the total shear strain proles obtained from energy min

imization and from the ow rule as compared with the average proles produced by the discrete dislocation simulations in [6]. A somewhat larger discrepancy between the shear strain proles obtained by the discrete and continuum approaches is perhaps due to the rather low ratio ℎ = 80 taken tosimulate these curves in [6].

During inverse loading, when the yield condition ϰ = − holds true,equation (4.122) changes to

= − + ( −2⟨ 1⟩cos2 )

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Find the dislocation density.

4.2 Prove that Nye s dislocation density tensor satises the following equation

= 0

which means that dislocations cannot end in the lattice.

4.3 Prove the following identity

− +

12

( + ) = 0

4.4∗ Given the plastic strain tensor and the dislocation density tensor satisfying the equations in Exercises 4.2 and 4.3. Find the plastic

distortion tensor.

4.5 The plastic distortion tensor is given by

= ( )

where s = (cos sin ) and m = (−sin cos ). Find the dislocationdensity tensor.

x

y M

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 tothe plane = 0 and the dislocation lines parallel to the axis shownin Fig. 4.34 write the energy functional of the crystal beam.

4.7 Develop the numerical procedure for the plane constrained shear of asingle crystal deforming in non symmetric double slip.

4.8 The similar problem for a bicrystal having non symmetric slip system.

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4.9∗ A rigid wedge is pressed into a crystal by a force as shown in Fig. 4.35.Develop a numerical procedure to nd the dislocation density beneaththe wedge and the force versus displacement curve.

4.10∗ Find the dislocation density near a rigid spherical inclusion embeddedin an innite crystal under tension.

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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. Boundarylayers in constrained plastic ow: comparison of nonlocal and discretedislocation plasticity. J. Mech. Phys. Solids , 49:1361 1395, 2001.

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