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6 N
ov 20
12The Yield-Strain in Shear Banding Amorphous Solids
Ratul Dasgupta, H. George E. Hentschel and Itamar
ProcacciaDepartment of Chemical Physics, The Weizmann Institute of
Science, Rehovot 76100, Israel.
Dept of Physics, Emory University, Atlanta GA. 30322
(Dated: November 7, 2012)
In recent research it was found that the fundamental
shear-localizing instability of amorphoussolids under external
strain, which eventually results in a shear band and failure,
consists of ahighly correlated array of Eshelby quadrupoles all
having the same orientation and some density. In this paper we
calculate analytically the energy E(, ) associated with such highly
correlatedstructures as a function of the density and the external
strain . We show that for strains smallerthan a characteristic
strain Y the total strain energy initially increases as the
quadrupole densityincreases, but that for strains larger than Y the
energy monotonically decreases with quadrupoledensity. We identify
Y as the yield strain. Its value, derived from values of the
qudrupole strengthbased on the atomistic model, agrees with that
from the computed stress-strain curves and broadlywith experimental
results.
I. INTRODUCTION
Amorphous solids are obtained when a glass-former iscooled below
the glass transition [13] to a state whichon the one hand is
amorphous, exhibiting liquid like or-ganization of the constituents
(atoms, molecules or poly-mers), and on the other hand is a solid,
reacting elasti-cally (reversibly) to small strains. There is a
large vari-ety of experimental examples of such glassy systems,
andtheoretically there are many well studied models [46]based on
point particles with a variety of inter-particlepotentials that
exhibit stable supercooled liquids phaseswhich then solidify to an
amorphous solid when cooledbelow the glass transition. Typically
all these materials,both in the lab and on the computer, exhibit a
so-calledyield-stress above which the material fails to a
plasticflow. In previous research [710] it was pointed out
thatdepending on the protocol of cooling to the glass state,the
plastic response of the system can be either via ho-mogeneous flow
or via shear bands. The former obtainstypically when the quenching
to the glass state is fast,whereas the latter when the quench is
slow. In thelatter case when the stress exceeds some yield stress,
thesample, rather than flowing homogeneously in a plasticflow,
localizes all the shear in a plane that is at 45 degreesto the
compressive stress axis, and then breaks along thisplane [1].
In recent work [11] it was argued that the
fundamentalinstability that gives rise to shear bands is the
appearanceof highly correlated lines of Eshelby quadrupoles (andsee
below for precise definition) which organize the non-affine
displacement field of an amorphous solid such thatthe shear is
highly localized along a narrow band. Howthis fundamental
instability results in macroscopic shearbands, why these appear in
45 degrees to the principalstress axis, and what determines the
difference in plasticresponse between fastly and slowly quenched
glasses areall subjects of this paper. We will also present an
ab-initio calculation of the yield stress at which an amor-phous
solid is expected to response plastically with
shearlocalization.
In Sect. II we review briefly the type of numericalsimulations
that we do and explain the basic facts aboutplasticity of amorphous
solids. Section III exhibits thefundamental solution of an Eshelby
quadrupolar plasticevent. This section is not particularly new but
is im-portant for our purposes in setting the stage and thenotation
for the next Sect. IV in which we computethe energy of N such
Eshelby quadrupoles in the elasticmedium. We show explicitly that
as a function of the ex-ternal strain (or the resulting stress)
there is a thresholdvalue at which a bifurcation occurs. Below this
value onlyisolated Eshelby quadrupoles can appear in the
system,leading to localized plastic events. Above this thresh-old a
density of such quadrupoles can appear, and whenthey do appear they
are highly correlated, preferring toorganize on a line at 45
degrees to the principal sheardirection. In Sect. V we present the
analytic estimateof the yield strain, and demonstrate a
satisfactory agree-ment with the numerical simulations. Finally, in
Sect.VI we provide a summary of the most important resultsof the
paper and offer a discussion of the road ahead.
II. PLASTICITY IN AMORPHOUS SOLIDS
AND SIMULATIONS
As a background to the calculations in this paper weneed to
briefly review recent progress in understandingplasticity in
amorphous solids [5, 6, 1214]. Below wedeal with 2-dimensional
systems composed of N pointparticles in an area A, characterized by
a total energyU(r1, r2, rn) where ri is the position of the ith
parti-cle. Generalization to 3-dimensional systems is
straight-forward if somewhat technical. The fundamental
plasticinstability is most cleanly described in athermal (T = 0)and
quasi-static (AQS) conditions when an amorphoussolid is subjected
to quasi-static strain, allowing the sys-tem to regain mechanical
equilibrium after every differen-tial strain increase. Higher
temperatures and finite strainrates introduce fluctuations and lack
of mechanical equi-librium which cloud the fundamental physics of
plastic
-
2instabilities with unnecessary details [12].In our AQS
numerical simulations we use a 50 50
binary Lennard-Jones mixture to simulate the shear lo-calization
discussed in this work. The potential energyfor a pair of particles
labeled i and j has the form
Uij(rij) = 4ij
[(ijrij
)12(ijrij
)6+A0
+ A1
( rijij
)+A2
( rijij
)2], (1)
where the parametersA0, A1 andA2 are added to smooththe
potential at a scaled cut-off of r/ = 2.5 (up to thesecond
derivative). The parameters AA, BB and ABwere chosen to be 2
sin(/10), 2 sin(/5) and 1 respec-tively and AA = BB = 0.5, AB =
1(see [15]). Theparticle masses were taken to be equal. The
sampleswere prepared using high-temperature equilibration fol-lowed
by a quench to zero temperature (T = 0.001) (see[16]). For
shearing, the usual athermal-quasistatic shearprotocol was followed
where each step comprises of anaffine shift followed by an
non-affine displacement usingconjugate gradient minimization. The
simulations wereconducted in two dimensions (2d) and employed
Lees-Edwards periodic boundary conditions. This implies thata
square sample of size L2 remains so also after strain.Samples were
generated with quench rates ranging from3.2106 to 3.2102 ( in LJ
units ), and were strainedto greater than 100 percent. Simulations
were performedon system-sizes ranging from 5000 to 20000 particles
witha fixed density of = 0.976 (in LJ units). The simula-tions
reported in the paper have 10000 particles and aquench-rate of 6.4
106 (in LJ units).We choose to develop the theory for the case of
exter-
nal simple shear since then the strain tensor is
traceless,simplifying some of the theoretical expressions.
Apply-ing an external shear, one discovers that the response ofan
amorphous solids to a small increase in the externalshear strain
(we drop tensorial indices for simplicity)is composed of two
contributions. The first is the affineresponse which simply follows
the imposed shear, suchthat the particles positions ri = xi, yi
change via
xi xi + yi xiyi yi yi. (2)
This affine response results in nonzero forces between
theparticles (in an amorphous solid) and these are relaxedby the
non-affine response ui which returns the systemto mechanical
equilibrium. Thus in total ri ri + ui.The nonaffine response ui
solves an exact (and modelindependent) differential equation of the
form [5, 17]
duid
= H1ij j (3)
where Hij 2U(r1,r2,rn)
rirjis the so-called Hessian ma-
trix and i 2U(r1,r2,rn)
riis known as the non-affine
FIG. 1: (Color Online). Left panel: the localization of
thenon-affine displacement onto a quadrupolar structure whichis
modeled by an Eshelby inclusion, see right panel. Rightpanel: the
displacement field associated with a single Eshelbycircular
inclusion of radius a, see text. The best fit parametersare a 2.5
and 0.1. To remove the effect of boundaryconditions, the best fit
is generated on a smaller box of size(x, y) [25.30, 75.92]
force. The inverse of the Hessian matrix is evaluated af-ter the
removal of any Goldstone modes (if they exist). Aplastic event
occurs when a nonzero eigenvalue P of Htends to zero at some strain
value P . It was proven thatthis occurs universally via a saddle
node bifurcation suchthat P tends to zero like P P [14]. For
valuesof the stress which are below the yield stress the
plasticinstability is seen [5] as a localization of the
eigenfunctionof H denoted as P which is associated with the
eigen-value P , (see Fig. 1 left panel). While at = 0 all
theeigenfunctions associated with low-lying eigenvalues
aredelocalized, P localizes as P (when P 0) on aquadrupolar
structure as seen in Fig. 1 left panel for thenon-affine
displacement field when the plastic instabilityis approached. These
simple plastic instabilities involvethe motion of a relatively
small number of particles (say20 to 30 particles) but the stress
field that is released hasa long tail.When the strain increases
beyond some yield strain,
the nature of the plastic instabilities can change in a
fun-damental way [10]. The main analytic calculation thatis
reported in Sect. IV shows that when the stress builtin the system
is sufficiently large, instead of the eigen-function localizing on
a single quadrupolar structure, itcan now localize on a series of N
such structures,which are organized on a line that is at 45
degreesto the principal stress axis, with the quadrupolar
structures having a fixed orientation relative to
the applied shear. Fig. 2 shows the non-affine fieldthat is
identical to the eigenfunction which is associatedwith this
instability, clearly demonstrating the series ofquadrupolar
structures that are now organizing the flowsuch as to localize the
shear in a narrow strip aroundthem. This is the fundamental shear
banding instability.Note that this instability is reminiscent of
some chainlikestructure seen in liquid crystals, arising from the
orien-tational elastic energy of the anisotropic host fluid
[18],and ferromagnetic chains of particles in strong magneticfields
[19]. The reader should note that the event shown
-
3FIG. 2: (Color Online). Left panel: The nonaffine displace-ment
field associated with a plastic instability that results ina shear
band. Right panel: the displacement field associatedwith 7 Eshelby
inclusions on a line with equal orientation.Note that in the left
panel the quadrupoles are not preciselyon a line as a result of the
finite boundary conditions and therandomness. In the right panel
the series of N = 7 Eshelbyinclusions, each given by Eq. (18) and
separated by a distanceof 13.158, using the best fit parameters of
Fig. 1, have beensuperimposed to generate the displacement field
shown.
in Fig. 2 will move the particles only a tiny amount, andit is
the repeated instability where many such event hitat the same
region which results in the catastrophic eventthat is seen as a
shear band in experiments. Neverthelessthis is the fundamental
shear localization instability andin the sequel we will have to
understand why repeatedinstabilities hit again and again in the
same region. Weshould stress that this is not inevitable, for
samples thatare prepared by a fast quench these instabilities
appearat random places adding up to what seems to be a ho-mogeneous
flow. For further discussion of this point seeSect. VI.
III. DISPLACEMENT IN 2D FOR A CIRCULAR
INCLUSION
As said above, the shear localizing instability appearsonly when
the stress exceeds a threshold. To explainwhy, we turn now to
analysis. As a first step we modelthe quadrupolar stress field
which is associated with thesimple plastic instability as a
circular Eshelby inclusion[5].
A. Circular Inclusion
We consider a 2d circular inclusion that has beenstrained into
an ellipse using an eigenstrain or a stress-free strain which we
take to be traceless ie
= 0
[20]. Here and below repeated indices imply a summationin
2-dimensions. A general expression for such a tracelesstensor can
be written in terms of a unit vector n and ascalar as
= (2nn ) . (4)
We also assume that a homogenous strain acts glob-
ally (which in our case also triggers the local transforma-tion
of the inclusion). This strained ellipsoidal inclusionboth feels a
traction exerted by the surrounding elas-tic medium resulting in a
constrained strain c in theinclusion, and itself exerts a traction
at the inclusion-elastic medium interface resulting in the
originally un-strained surroundings developing a constrained
strain
field c(~X). Here and below ~X stands for an arbitrary
cartesian point in the material which for this purpose
isapproximated as a continuum.A fourth-order Eshelby tensor S can
be defined
which relates the constrained strain in the inclusion cto the
eigenstrain viz.
c = S. (5)
Now for an inclusion of arbitrary shape the constrainedstrain c,
stress
c , and displacement field u inside
the inclusion are in general functions of space. For
ellip-soidal inclusions, however, it was shown by Eshelby [2123]
that the Eshelby tensor and the constrained stressand strain fields
inside the inclusion become indepen-dent of space. We work here
with a circular inclusionwhich is a special case of an ellipse and
hence for such aninclusion, the Eshelby tensor is [2123]
S =4 18 (1 ) +
3 48 (1 ) ( + ) ,(6)
where is the Poissons ratio. Note that this is the Es-helby
tensor for an inclusion in 2-dimensions. It is thesame as a
cylindrical inclusion in 3-dimensions under aplane strain [23].
From Eqs. (5) and (6), we obtain
c =
[4 18 (1 ) +
3 48 (1 ) ( + )
]
=3 44 (1 )
For a traceless eigenstrain. (7)
The total stress, strain and displacement field inside
thecircular inclusion is then given by
I = c +
=
3 44 (1 )
+
I = c + C
(c +
)uI = u
c + u
=
[3 44 (1 )
+
]X .
(8)
Here the super-script I indicates the inclusion and the denotes
the eigenstress which is linearly related tothe eigenstrain by C,
and which for anisotropic elastic medium simplifies further
using
C + ( + ) , (9)to:
= 2 +
(10)
=E
1 + +
E(1 + ) (1 2)
,
-
4where and are the Lames parameters. One can eitherchoose the
two Lames coefficients or E and as the twoindependent material
parameters. The relations betweenthem are given by
= E2(1+) , =E
(1+)(12) (11)
E = (3+2)+ , =
2(+) (12)
These relations are correct in 3-dimensions with planestrain
conditions and therefore also in 2-dimensions [23].The stress in
the inclusion can now be written down
in terms of independent variables using Eq. (10) by
I = C(c +
)=
E1 +
c +E
(1 + ) (1 2)c
E1 +
+E
1 + (13)
as and are traceless. Note that Eq. (7) implies
that for a traceless eigenstrain, the constrained straininside
the inclusion viz. c is also traceless and thus we
obtain from Eq. (13),
I =E
1 + c
E1 +
+E
1 +
=E
1 +
3 44 (1 )
E1 +
+E
1 +
=E
4 (1 + ) (1 ) +
E1 +
(14)
B. Constrained Fields in the Elastic Medium
In the surrounding elastic medium the stress, strain
anddisplacement fields are all explicit function of space andcan be
written
m(~X) = c(
~X) +
m( ~X) = c( ~X) +
um ( ~X) = uc( ~X) + u
( ~X) .
(15)
In order to compute the displacement field uc(~X) in
the isotropic elastic medium we need to solve the Lame-Navier
equation
E2 (1 + ) (1 2)
2ucXX
+E
2 (1 + )
2ucXX
= 0,(16)
as there are no body forces present in our calculation.The
constrained fields in the inclusion will supply theboundary
conditions for the fields in the elastic mediumat the inclusion
boundary. Also as r the con-strained displacement field will
vanish.All solutions of equation Eq. (16) also obey the higher
order bi-harmonic equation
4ucXXXX
= 22uc = 0. (17)
Thus our objective is to construct from the radial so-lutions of
the bi-laplacian equation Eq. (17) derivativeswhich also satisfy
Eq. (16). Note that Eq. (17) is only anecessary (but not a
sufficient) condition for the solutionsand Eq. (16) still needs to
be satisfied. The calculationis presented in Appendix A, with the
final result
uc( ~X) = (18)
4(1 )(ar
)2 [2(1 2) +
(ar
)2 ][2nn ~X X
]
+
2(1 )(ar
)2 [1
(ar
)2 ][2(n ~X)2r2
1]X .
C. Fit to the data
Armed with this analytic expression we return now toour
numerics, cf. Fig. 1, and fit the two parametersin Eq. (18) to the
data of the displacement exhibitedby a single localized plastic
event. The result of thisprocedure is a 2.5 and 0.1. The quality of
thisfit can be judged from the right panel of Fig. 1 wherewe
exhibit the form of Eq. (18) with the parametersfitted to the
displacement field in the left panel. Alsothe value of a appears
reasonable since it means thatabout a2 20 particles are involved in
the core of therelaxation event. On physical grounds this is about
theright order of magnitude.We will keep these parameters fixed in
all our calcula-
tions below. The reader should note that this is an
ap-proximation when there are multiple quadrupoles in thesystem,
since they influence each other and the solutionleading to Eq. (18)
should be repeated in the presence ofmany quadrupoles. We expect
however that the changesin the parameters should not be large when
the densityof the quadrupoles is small. We will always work in
thesmall density limit a2 1 where is the area densityof quadrupoles
N/L2.
IV. THE ENERGY OF N ESHELBY
INCLUSIONS EMBEDDED IN A MATRIX
A. Notation
Having the form of a single quadrupole, Eq. (18),we turn now to
the calculation of the energy associ-ated with N quadrupoles
embedded in an elastic ma-trix. Once computed, we will show later
that for largestrains the minimum of this energy is obtained for a
lineof quadrupoles all having the same orientation. Fromnow on we
use the notation that u, u . The en-ergy of N Eshelby inclusions
embedded in a linear elasticmedium (or matrix) N , is given by the
expression
E=1
2
Ni=1
V
(i)0
(i)
(i)dV +
1
2
V
Ni=1 V
(i)0
(m)
(m) dV(19)
-
5FIG. 3: Schematic of four Eshelby inclusions embedded in a
matrix m
where the superscript i indicates the index of the inclu-sion
and m indicates the matrix. We evaluate Eq. (19)in Appendix B. The
result can be expressed in termsof four contributions: Emat which
is the contribution ofthe strained matrix, E which is the energy of
the N
qudrupoles in the external strain, Eesh which representsthe self
energy of the N quadrupoles (their cost of cre-ation) and lastly
Einc represents the energy of interactionbetween the inclusions.
Explicitly
Emat 12()
() V = V
()xy
()xy =
V E22 (1 + )
(20)
E 12()
(Ni=1
(,i) V
(i)0
)= a2()xy
Ni=1
(,i)yx = a2E(1 + )
Ni=1
n(i)x n(i)y (21)
Eesh 12
Ni=1
(,i)
(c,i) V
(i)0 +
1
2
Ni=1
(,i)
(,i) V
(i)0 =
a2
2
Ni=1
((,i) (c,i)
)(,i) (22)
Einc 12
Ni=1
(,i)V(i)0
j 6=i
(c,j) (Rij)
= a2
2
ij
[(,i)
(c,j) (Rij) +
(,j)
(c,i) (Rij)
](23)
The above expressions are specific to 2D, for a global strain
corresponding to simple shear under the linear
approximation. Thus xy =2 here, and the traceless eigenstrain
takes the form
(,i)yx = 2n
(i)x n
(i)y .
The form of Einc is not final, and we bring it to its final form
in Appendix C. The final result is
Einc = E()2a2
8(1 2)ij
(a
Rij
)2
[ 8
{(1 2) +
(a
Rij
)2}(4(n(i) n(j)
)(n(i) rij
)(n(j) rij
)
2(n(i) rij
)2 2
(n(j) rij
)2+ 1
)+ 4
(2 (1 2) +
(a
Rij
)2)(2(n(i) n(j)
)2 1
)
8(1 2
(a
Rij
)2)(2(n(i) rij
)2 1
)(2(n(j) rij
)2 1
)
+ 32
(1
(a
Rij
)2)((n(i) rij
)(n(j) rij
)(n(i) n(j)
)(n(i) rij
)2 (n(j) rij
)2)](24)
-
6where rij ~XijRij .Our task is now to find the configuration of
N Eshelby
quadrupoles that minimize the total energy. Obviously,if the
external strain is sufficiently large, we need tominimize E
separately, since it is proportional to .The minimum of (21) is
obtained for
n(i)x = n(i)y =
12. (25)
Substituting this result in Eq. (24) simplifies it
consid-erably. We find
Einc = a2()2E
8(1 2)
(a
Rij)2{ 8[(1 2) + ( a
Rij)2] + 4[2(1 2) + ( a
Rij)2] 8[1 2( a
Rij)2][2(n r)2 1]2
+32[1 ( aRij
)2][(n r)2 (n r)4]}. (26)
We can find the minimum energy very easily. Denotex (n r)2, and
minimize the expression A[2x 1]2 B[x x2]. The minimum is obtained
at x = 1/2, orcos =
1/2. We thus conclude that when the line of
correlated quadrupole forms under shear, this line is in45
degrees to the compressive axis, as is indeed seen inexperiments.
Of course there are two solutions for this,perpendicular to each
other. Note that for other externalstrains which are not consistent
with a traceless straintensor (or in 3-dimensions) this conclusion
may change.
The physical meaning of this analytic result is thatit is
cheaper (in energy) for the material to organize Nquadrupolar
structures on a line of 45 degrees with thecompressive stress, all
having the same orientation, thanany other arrangement of these N
quadrupoles, includ-ing any random distribution. This explains why
sucha highly correlated distribution appears in the
strainedamorphous solid, and why it can only appear when
theexternal strain (or the built-up stress) are high enough.This
fact, in addition to the observation that such anarrangement of
Eshelby quadrupoles organizes the dis-placement field into a
localized shear, explains the originof this fundamental
instability.
V. ESTIMATE OF YIELD-STRESS AND
NUMBER OF ESHELBY QUADRUPOLES
In this section we turn to estimate the yield stress andthe
associated density of Eshelby quadrupoles. To thisaim we need to
compute one other energy term that wasnot needed until now, namely
Eesh which was the samefor all the configurations of the
quadrupoles.
A. Expression for Eesh
The energy term Eesh was given by Eq. (22) which canbe
re-written as
=a2
2
Ni=1
(,i) C
((,i) (c,i)
)(27)
=a2
2
Ni=1
(,i) C
((,i)
3 44(1 )
(,i)
), Cf. Eq. 7
=a2
2
Ni=1
(,i) C
(,i)
4(1 ) .
For an isotropic matrix, using Eq. (9), we have
C(,i) =
(,i) + ( + )
(,i)
= 2(,i) =
E1 +
(,i) (28)
for a symmetric traceless eigenstrain. Using Eq. (28)
forcircular inclusions each of radius a in 2D, we obtain
Eesh =a2
2
E4(1 2)
Ni=1
(,i)
(,i) (29)
Now,
(,i)
(,i) = (
(,i))2(2n(i) n
(i)
)(2n
(i) n
(i)
)= 2((,i))2 (30)
For all eigenstrains equal (ie. (,i) = ), we have
Eesh =Ea2N ()24(1 2) (31)
-
7B. Einc for a line of equidistant quadrupoles with
the same polarization
At this point we need to compute the energy term Eincfor the
special configurations of N quadrupoles that areequi-distant and
with the same polarization, organized
in a line. In this case we have
n(i) = n, rij = r, and Rij = |j i|R (32)
Starting from Eq. (26) we specialize to the present
situ-ation
Einc = E()2a2
8(1 2)N1i=1
Nj,j>i
( aR
)2 1(j i)2
{ 8
[(1 2) +
( aR
)2 1(j i)2
]
+ 4
[2 (1 2) +
( aR
)2 1(j i)2
] 8
[1 2
( aR
)2 1(j i)2
] [2(n r)2 1
]2+ 32
[1
( aR
)2 1(j i)2
])[(n r)2 (n r)4
]}(33)
For (n r)2 = 1/2, the above expression reduces to
Einc = E()2a2
2(1 2)N1i=1
Nj,j>i
[2
(j i)2( aR
)2 3
( aR
)4 1(j i)4
](34)
We have
N1i=1
Nj=i+1
1
(j i)s =N1i=1
Nin=1
1
ns N (s) for N >> 1 , (35)
where (s) is the Riemann zeta function. Thus we obtain
Einc = E()2a2N
2(1 2)[2( aR
)2(2) 3
( aR
)4(4)
]. (36)
At this point we realize that the distance R betweenthe
quadrupoles is not determined. We will choose Rby demanding that
the line density of the quadrupolesremains invariant in the
thermodynamic limit, or R L/N . Thus, with N/L, the energy density
in thestrip L a reads
EincLa
= E()2a
2(1 2)[2(a)2(2) 3(a)4(4)
](37)
Similarly from equation 31, we obtain
EeshLa
=Ea()24(1 2) (38)
From equation (25) we have
E
La= aE
2 (1 + ) (39)
Thus the plastic energy density is given by
E(, )
La E
+ Eesh + EincLa
(40)
=E()24(1 2)
[A
(1
Y
)aB(a)3 + C(a)5
],
where A = 1, B = 4(2) and C = 6(4). Yis defined as
Y
2(1 ) . (41)
Eq. (40) is plotted, using the numerical values of theparameters
found in our numerical simulations in Fig.VB for various values of
. For < Y the minimumof the expression is attained always at =
0, and theshear localization cannot occur. Only at Y a newsolution
opens up to allow a finite density of the Es-helby quadrupoles.
Therefore Y is by definition the yieldstrain.
-
8FIG. 4: (Color Online). The total plastic energy for the
cre-ation of an array of quadrupoles with line density for
threevalues of : = Y 0.1, = Y 0.05, and = Y
VI. SUMMARY AND CONCLUSIONS
We have presented a theory of the fundamental insta-bility that
leads to shear localization and eventually toshear bands. One
remarkable observation is that the nat-ural plastic instability
that occurs spontaneously in oursimulations results in a
displacement field that is sur-prisingly close to the one made by
an Eshelby circularinclusion, see Fig. 1. The best fit for the
parameter a, ofthe order of 2.5 is in agreement with the intuitive
beliefthat shear transformation zones involve 20-30 particles,as a2
would predict. Basing our analysis on this simi-larity we could
develop an analytic theory of the energyneeded to create N such
inclusions, whether scattered inthe system randomly or aligned and
organized in highlycorrelated shear localized structure. We
discover thatthe latter becomes energetically favorable when
exceedsY 2(1) . In our system 0.215, and with our best
fit 0.1 we predict Y 0.07 which is right on the
mark as one can seen from Ref. [11].While we believe that our
calculation of the energy of
N quadrupoles is accurate for densities such that a2 1, the
interaction between the quadrupoles become muchmore involved for
higher densities, and we avoided thiscomplication. The consequence
is that we cannot pre-dict a-priory the critical density of our
quadrupoles, andwe leave this interesting issue for future
research. An-other issue that warrants further study is whether
theparameters a and are material parameters which aredetermined by
the small scale structure of the glass, andif so, how to estimate
them a-priori. Also, are these pa-rameters dependent on the way the
system is stressed,i.e. via shear or via tensile compression
etc..Next we need to discuss the difference in behavior of
glasses that were quenched relatively quickly and thosethat were
quenched relatively slowly. In our simulationswe find that the
latter exhibit the fundamental shearlocalization instability again
and again along approxi-mately the same line, accumulating
displacement that
FIG. 5: (Color Online). Repeated shear localization
insta-bilities in the case of slow quench. Upper panel: the
energyvs. strain and the instabilities that were chosen for
displayin the lower panel. The average non affine displacement
fieldafter the instabilities that are marked in the upper panel.
Thepoint to notice is that the instability falls repeatedly on
thesame band, accumulating to a shear band.
develops into a shear band. On the other hand theformer tends
every time to have plastic instabilities atdifferent places,
sometime localized and sometime morecorrelated, and on the average
this then appears like ahomogenous flow. This difference is
stressed in Fig. 5where the case of slow quench is exhibited. It
appearsthat fastly quenched systems may have plastic instabili-ties
almost everywhere, and there is no special preferencefor one line
or another. Slowly quenched systems are ini-tially harder to shear
localize, but once it happens in aparticular (random) line it is
likely to repeat again andagain along the same line. Making these
words quantita-tive is again an issue left to future research. In
particularit is interesting to find what is the precise meaning of
fastand slow quenches, fast and slow compared to what, andhow this
changes the microscopic local structure.Finally the effects of
temperature and finite strain rates
on the present mechanism constitute a separate piece ofwork
which of course is of the utmost importance. Likethe other open
subject mentioned above, it will be takenup in future research.
Acknowledgments
This work had been supported in part by the IsraelScience
Foundation, the German-Israeli Foundation andby the European
Research Council under an ideasgrant.
-
9Appendix A: The displacement field of an Eshelby
circular inclusion
1. Solutions of the Lame-Navier equation
We look for linear combinations of derivatives of the ra-dial
solutions of Eq. 17 which are linear in the eigenstrain and go to
zero at large radii. In addition the termsmust transform as
components of a vector field. Such asolution can be written down
as
uc = A
ln r
X+B
3 ln r
XXX
+ C3(r2 ln r
)XXX
, (A1)
where X is the component of the position vector withthe origin
at the center of the Eshelby quadrupole, and
r | ~X|. It can be checked that any other terms areeither zero,
do not go to zero as r , or do not trans-form as components of a
vector. We re-write equationEq. (16) as
(1
1 2)
2ucXX
+2uc
XX= 0 (A2)
which is the equation for the constrained displacementfield in
the elastic matrix subject to appropriate bound-ary conditions.
From Eq. (A1), we obtain
2ucXX
=A
X
[2
XXln r
]+B
3
XXX
[2
XXln r
]+C
3
XXX
[2
XXr2 ln r
].(A3)
We need the following identities
2
XX(ln r) = 0
2
XX
(r2 ln r
)= 4 ln r + 4
(A4)
Thus we obtain from equation A3,
2ucXX
= 4C3 ln r
XXX(A5)
Similarly, the expression for
2ucXX
=2
XX
[A
ln r
X+B
3 ln r
XXX+ C
3(r2 ln r
)XXX
]
=
X
[A
2 ln r
XX+B
4 ln r
XXXX+ C
4(r2 ln r
)XXXX
]
(A6)
which can be re-written (noting that the second and thirdterms
involve the laplacian for which we have identities
from Eqs. (A4) as
2ucXX
=
X
[A
2 ln r
XX+ C
2 (4 ln r + 4)
XX
]
= (A+ 4C) 3 ln r
XXX(A7)
-
10
Plugging expressions (A5) and (A7) in Eq. (A2), we
thusobtain
(A+4C)12
3 ln rXXX
+ 4C3 ln r
XXX= 0
[A+4C12 + 4C
]
3 ln rXXX
= 0 C = A8(1) (A8)
We can thus re-write Eq. (A1) as
uc = A
ln r
X+B
3 ln r
XXX A
8 (1 )
3(r2 ln r
)XXX
(A9)
The following identities are now required:
ln r
X=
Xr2
3 ln r
XXX=
2r2 (X +X +X) + 8XXXr6
3(r2 ln r
)XXX
=2r2 (X +X +X) 4XXX
r4(A10)
Using these relations we can re-write Eq. A9 as
uc = A
Xr2
+B
[2r2 (X +X +X) + 8XXXr6
]
A8 (1 )
[2r2 (X +X +X) 4XXX
r4
]
= AXr2
[2B
r4+
A
4 (1 ) r2] (X +X +X) +
[8B
r6+
A
2 (1 ) r4]XXX(A11)
Remembering that the eigenstrain is traceless,we find that (X +X
+X) = 2X using which we
can simplify Eq. (A11) to obtain
uc =[Ar2 4B
r4 A2(1)r2
]X +
[8Br6
+ A2(1)r4
]X
XX
=[Ar2
122(1) 4Br4
]X +
[8Br6
+ A2(1)r4
]X
XX (A12)
At r = a (the radius of the circular inclusion), the form of
expression Eq. (A12) must match the form of the
constraineddisplacement field of the inclusion which from Eq. (7)
is 344(1)
X . Thus the co-efficient of the second term in
expression (A12) must go to zero at the inclusion boundary,
which gives us
8Ba6
+ A2(1)a4 = 0
B = a2A16(1) (A13)
Thus we have
uc =[Ar2
122(1) 4r4 a
2A16(1)
]X +
[8r6
a2A16(1) +
A2(1)r4
]X
XX
= A4r2(1)
[2 (1 2) + a2
r2
]X +
A2r4(1)
[1 a2
r2
]X
XX
(A14)
And the value of uc at r = a should match the value obtained
from Eq. (7), implying
3 44 (1 ) =
A
4 (1 ) a2 [2 (1 2) + 1]3 44 (1 ) =
A (3 4)4 (1 ) a2 A = a
2 (A15)
-
11
The expression for uc becomes:
uc =1
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)]X +
1
2 (1 )(a2
r2
)[1
(a2
r2
)]
XXXr2
(A16)
From Eq. 4, we have = (2nn ) and thus
X = [2n
(n ~X
)X
]X
X =
X (2nn )X = [2(n ~X
)2 r2
](A17)
allowing us to write the final vectorial expression for the
displacement field:
~uc(~X)=
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)] [2n(n ~X
) ~X
]
+
2 (1 )(a2
r2
)[1
(a2
r2
)]2(n ~X
)2r2
1
~X
(A18)
We can also derive expressions for the constrained stress and
strain fields. Noting that
f(r)
X= f (r)
r
X= f (r)
Xr
(n ~X
)X
= n (A19)
-
12
we obtain
ucX
=
X
[
4 (1 )(a2
r2
){2 (1 2) +
(a2
r2
)}{2n
(n ~X
)X
}
+
2 (1 )(a2
r2
){1
(a2
r2
)}2(n ~X
)2r2
1
X
]
=
4 (1 ){4 (1 2)
(a2
r4
) 4
(a4
r6
)}{2n
(n ~X
)X
}X
+
4 (1 ){2 (1 2)
(a2
r2
)+
(a4
r4
)}{2nn }
+
2 (1 ){(
2a2
r4
)+
(4a4
r6
)}2(n ~X
)2r2
1
XX
+
2 (1 ){(
a2
r2
)(a4
r4
)}4(n ~X
)n
r2
4(n ~X
)2X
r4
X
+
2 (1 ){(
a2
r2
)(a4
r4
)}2(n ~X
)2r2
1
uc
X=
4 (1 )[ 4
(a2
r2
){(1 2) +
(a2
r2
)}2n
(n ~X
)r
Xr
Xr
+
(a2
r2
){2 (1 2) +
(a2
r2
)}{2nn }
4(a2
r2
){1 2
(a2
r2
)}2(n ~X
)2r2
1
XXr2
+8
(a2
r2
){1
(a2
r2
)}(n ~X
)n
r(n ~X
)2r2
Xr
Xr
+2
(a2
r2
){1
(a2
r2
)}2(n ~X
)2r2
1
]
(A20)
-
13
Thus the expression for c 12(ucX
+ucX
)is
c( ~X) =
8 (1 )[ 4
(a2
r2
){(1 2) +
(a2
r2
)}{2
(n ~Xr
)(nXr
+nXr
) 2XX
r2
}
+2
(a2
r2
){2 (1 2) +
(a2
r2
)}{2nn }
8(a2
r2
){1 2
(a2
r2
)}2(n ~X
)2r2
1
XXr2
+8
(a2
r2
){1
(a2
r2
)}(n ~Xr
)(Xnr
+Xnr
) 2
(n ~X
)2r2
XXr2
+4
(a2
r2
){1
(a2
r2
)}2(n ~X
)2r2
1
]
(A21)
allowing us to write the final expression for the constrained
strain in the matrix
c(~X) =
4 (1 )[ 4
(a2
r2
){(1 2) +
(a2
r2
)}{(n ~Xr
)(nXr
+nXr
) XX
r2
}
+
(a2
r2
){2 (1 2) +
(a2
r2
)}{2nn }
4(a2
r2
){1 2
(a2
r2
)}2(n ~X
)2r2
1
XXr2
+4
(a2
r2
){1
(a2
r2
)}(n ~Xr
)(Xnr
+Xnr
) 2
(n ~X
)2r2
XXr2
+2
(a2
r2
){1
(a2
r2
)}2(n ~X
)2r2
1
]
(A22)
-
14
The trace of c is not zero in the elastic medium and is given
by
c =
4 (1 )[ 4
(a2
r2
){(1 2) +
(a2
r2
)}2
(n ~Xr
)2 1
+ 0
4(a2
r2
){1 2
(a2
r2
)}2(n ~X
)2r2
1
+ 0
+ 4
(a2
r2
){1
(a2
r2
)}2(n ~X
)2r2
1
]
(A23)
c =
4 (1 )[ 4
(a2
r2
){(1 2) +
(a2
r2
)}2
(n ~Xr
)2 1
+ 4
(a2
r2
)(a2
r2
)2(n ~X
)2r2
1
]
c = (1 21
)(a2
r2
)2(n ~X
)2r2
1
(A24)
We are now in a position to calculate the constrained stress in
the elastic medium due to the deformed inclusion.It is given by the
expression
cij =E
1 + c +
E(1 + ) (1 2)
c =
E4 (1 2)
[.....
] E
(1 2)(a2
r2
)2(n ~X
)2r2
1
(A25)
where
[.....
]is the expression inside the square brackets in Eq. A22.
2. Constrained displacement field - Cartesian components
It proves useful to have the explicit cartesian components of
the displacement field for computational and graphicalpurposes. If
we consider the unit-vector n making an angle of with the positive
direction of the x-axis, then the
-
15
cartesian components of equation A18 are :
ucx =
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)][2 cos (x cos+ y sin) x]
+
2 (1 )(a2
r2
)[1
(a2
r2
)][2 (x cos+ y sin)
2
r2 1
]x
ucx =
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)][x cos 2+ y sin 2]
+
2 (1 )(a2
r2
)[1
(a2
r2
)][(x2 y2) cos 2+ 2xy sin 2
r2
]x
ucy =
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)][2 sin (x cos+ y sin) y]
+
2 (1 )(a2
r2
)[1
(a2
r2
)][2 (x cos+ y sin)
2
r2 1
]y
ucy =
4 (1 )(a2
r2
)[2 (1 2) +
(a2
r2
)][x sin 2 y cos 2]
+
2 (1 )(a2
r2
)[1
(a2
r2
)][(x2 y2) cos 2+ 2xy sin 2
r2
]y
(A26)
Appendix B: Calculation of the energy of N quadrupoles
Eq. (19) can be re-written using 1/2 (u, + u,) as
E =1
4
Ni=1
V i0
(i)
(u(i), + u
(i),
)dV +
1
4
V
Ni=1 V
(i)0
(m)
(u(m), + u
(m),
)dV (B1)
Using the symmetry of the stress tensor, we obtain
E =1
2
Ni=1
V i0
(i)u
(i),dV +
1
2
V
Ni=1 V
(i)0
(m) u
(m),dV (B2)
We also have the identity
u, = (u), ,u = (u), (B3)
if there are no body forces. Thus we can write Eq. B2 as
E =1
2
Ni=1
V
(i)0
((i)u
(i)
),dV +
1
2
V
Ni=1 V
(i)0
((m) u
(m)
),dV (B4)
Using Gausss theorem to convert these volume integrals into area
integrals, we obtain
E =1
2
Ni=1
Si0
(i)u
(i) n
(i) dS
Ni=1
1
2
Si0
(m) u
(m) n
(i) dS +
1
2
S
(m) u
(m) n
() dS (B5)
-
16
where n(i) and n() are unit normal vectors both pointing
outwards respectively from the inclusion volume V i0 andthe matrix
boundary. Eq. B5 can be rewritten as follows
E =1
2
S()
(m) u
(m) n
() dS +
1
2
Ni=1
S
(i)0
((i)u
(i) (m) u(m)
)n(i) dS
E = 12()
()
S()
Xn() dS +
1
2
Ni=1
S
(i)0
((i)u
(i) (m) u(m)
)n(i) dS
E = 12
V +
1
2
Ni=1
S
(i)0
((i)u
(i) (m) u(m)
)n(i) dS
(B6)
Thus we can write using the expressions earlier writtendown in
Eq. (15)
(m)
(~X)=
() +
Ni=1
(c,i)
(~X)
(m)
(~X)=
() +
Ni=1
(c,i)
(~X)
u(m)
(~X)= u() (
~X) +
Ni=1
u(c,i)
(~X)
(B7)
where (c,i)
(~X)indicates the constrained strain at lo-
cation ~X in the matrix due to the eshelby labeled with
the index i etc. We also have for locations ~X inside
theinclusions
(i)
(~X)=
() +
j 6=i
(c,j)(~X)+
(c,i) (,i)
(i)
(~X)=
() +
j 6=i
(c,j)(~X)+
(c,i) (,i)
u(i)
(~X)= u() +
j 6=i
u(c,j)
(~X)+ u(c,i) (,i) X
(B8)
where (,i) is the eigenstrain of the ith Eshelby and so on.Note
that in the expression for the strain in the inclusiongiven by Eq.
(B8) we have removed the eigenstrain from
the constrained strain (c,i) (,i) leaving only the elastic
contribution in order to calculate correctly the
elasticcontribution to the energy. Using these expressions,
theelastic energy of the system can be written from Eq. (B6)
E =1
2()
() V (B9)
+1
2
Ni=1
S
(i)0
((i)u
(i) (m) u(m)
)n(i) dS .
Since the traction force has to be continuous at the in-clusion
boundary (Newtons third law), we have
(i) n
(i) =
(m) n
(i) at the inclusion boundary (B10)
which gives us from Eq. (B9),
E=1
2()
() V +
1
2
Ni=1
S
(i)0
(i)n
(i)
(u(i) u(m)
)dS(B11)
We also have from Eqs. (B7) and (B8),
u(i) u(m) = (,i) X . (B12)
On plugging this expression into Eq. (B11) gives us finally
E =1
2()
() V
1
2
Ni=1
S
(i)0
(i) n
(i)
(,i) XdS
= 12()
() V
1
2
Ni=1
(,i)
V
(i)0
((i)X
),dV
= 12()
() V
1
2
Ni=1
(,i)
V
(i)0
(i)dV
E = 12()
() V
1
2
Ni=1
V(i)0
(,i)
(i)
(B13)
where i (1/V i0 )V i0
(i)dV . Using the expression for
(i) from Eq. (B8), we obtain
(i)(
~X) () +j 6=i
(c,j) (Rij) + (c,i) (,i).(B14)
Eq. (B14) is a far field approximation that assumes thatRij a.
As Rij a clearly the spatial integrals con-tibuting to c,i must be
computed explicitly and cannotbe replaced by the single distance
Rij between the centersof the Eshelby inclusions i and j.
Using expression (B13) we obtain
-
17
E =1
2()
() V
1
2()
(Ni=1
(,i) V
(i)0
) 1
2
Ni=1
(,i)
(c,i) V
(i)0 +
1
2
Ni=1
(,i)
(,i) V
(i)0
12
Ni=1
(,i)V(i)0
j 6=i
(c,j) (Rij)
E = Emat + E + Eesh + Einc. (B15)
where all these terms are defined in Eqs. (20)-(23).
Appendix C: The final form of Einc
We can also explicitly write Einc showing its linear dependence
on the eigenstrain by using Eq. (23). This gives us
Einc = a2
2ij
[(2n(i) n
(i)
)(c,j) (Rij) +
(2n(j) n
(j)
)(c,i) (Rij)
]. (C1)
Plugging Eq. (A25) into the above equation, we find that the
term inside the square braces in Eq. (C1) can be writtenas:
=E
(2n
(j) n
(j)
)4 (1 2)
[ 4
(a
Rij
)2{(1 2) +
(a
Rij
)2}
{(
ni ~X ijRij
)(n(i) X
(ij)
Rij+n(i) X
(ij)
Rij
) X
(ij) X
(ij)
(Rij)2
}
+
(a
Rij
)2{2 (1 2) +
(a
Rij
)2}{2n(i) n
(i)
}
4(
a
Rij
)2{1 2
(a
Rij
)2}2(n(i) ~X(ij)
)2(Rij)2
1
X(ij) X
(ij)
(Rij)2
+ 4
(a
Rij
)2{1
(a
Rij
)2}
(n(i) ~X(ij)
Rij
)(X
(ij) n
(i)
Rij+X
(ij) n
(i)
Rij
) 2
(n(i) ~X(ij)
)2(Rij)2
X(ij) X
(ij)
(Rij)2
+ 2
(a
Rij
)2{1
(a
Rij
)2}2(n(i) ~X(ij)
)2(Rij)2
1
]
E
1 2(a2
r2
)(2(n(i) ~X(ij))2
r2 1
)(2n(j) n
(j)
)
+ i j (C2)
where ~X(ij) indicates the vector joining the centers of the
eshelby pair labeled as i and j, and i j in Eq. (C2)represents the
term obtained by exchanging i and j.
-
18
In order to simplify Eq. (C2), we need the following
identities
(2n(j) n
(j)
){( n(i) ~X(ij)Rij
)(n(i) X
(ij)
Rij+n(i) X
(ij)
Rij
) X
(ij) X
(ij)
(Rij)2
}+ i j
= 8n(i) n(j)(n(i) ~X(ij)
Rij
)(n(j) ~X(ij)
Rij
) 4
(n(i) ~X(ij)
Rij
)2 4
(n(j) ~X(ij)
Rij
)2+ 2 (C3)
(2n(j) n
(j)
)(2n(i) n
(i)
)+ i j = 4
[2(n(i) n(j)
)2 1
](C4)
(2n(j) n
(j)
) X(ij) X(ij)(Rij)2
+ i j = 22(n(j) ~X(ij)
Rij
)2 1
(C5)
(2n(j) n
(j)
)(n(i) ~X(ij)
Rij
)(X
(ij) n
(i)
Rij+X
(ij) n
(i)
Rij
) 2
(n(i) ~X(ij)
)2(Rij)2
X(ij) X
(ij)
(Rij)2
+ i j
= 8
(n(i) ~X(ij)
Rij
)(n(j) ~X(ij)
Rij
)(n(i) n(j)
) 8
(n(i) ~X(ij)
Rij
)2(n(j) ~X(ij)
Rij
)2(C6)
(2n(j) n
(j)
) + i j = 0 (C7)
Using these identities, we can write the final expression for
the interaction energy in the form shown in Eq. (24).
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(2008);http://www.wpi-aimr.tohoku.ac.jp/en/modules/chengroup/.
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