A
SIMPLE
MODEL
OF
RELAXATION
C. ATKINSONt$
and
T.
AT R.
A
CRACK
TIP*
KAY-f
A model is proposed whereby plastic flow at a crack tip is representedby a single dislocation of unknown Burgers vector emitting from the crack tip along each slip direction. The magnitude of the Burgers vector and the distance of each dislocation from the crack tip are determined by two conditions (i) that the singularity in the stress at the crack tip should be zero and (ii) that the total force on the dislocation in its slip plane is zero (assuming it has a resistance to motion oil the friction stress). The results of the model are compared with other models when the dislocations are collinear with the crack, and there is shown to be good agreement for the predicted displacement at the crack tip. The virtue of this model is that it oan be extended relatively simply to situations where the relaxation is not collinear with the crack tip and one such example is considered. MODELE
SIMPLE
DE RELAXATION
A LA POINTE
DUNE
FISSURE
Les auteurs proposent un modele ou la deformation plastique B la pointe d’une fissure est represent&e par une seule dislocation de vecteur de Burgers inconnu emettant B partir de l’extremite de la fissure le long de chaque glissement. L’amplitude du vecteur de Burgers et la distance de la pointe de la fissure a chaque dislocation sont d&ermin&ss par deux conditions: (i) que la singularite dans la contrainte a l’extremite de la fissure soit nulle et (ii) que la force totale SUPla dislocation dans son plan de glissement soit nulle (en supposant qu’il existe une resistance au mouvement, ui, contra&e de frottement). Les rbultats du modele sont compares a d’autres modeles quand les dislocations sont colineaires aveo la fissure, et les auteurs montrent qu’il existe un bon accord pour le d&placement pr&u B l’extremit6 de la fissure. Le merite de ce modele est qu’il peut 6tre Btendu relativement simplement aux situations pour lesquelles la relaxation n’est pas oolineaire aveo la pointe de la fissure, et les auteurs Btudient un tel exemple. EIN
EINFACHES
MODELL
FUR
DIE
RELAXATION
AN DER
RIDSPITZE
Ein Model1 wird vorgeschlagen, bei dem plastisches FlieBen an einer RiBspitze durch eine einzige Versetzung mit unbekanntem Burgersvektor beschrieben wird, die von der RiBspitze aus entlang jeder Gleitrichtung verlauft. Der Betrag des Burgersvektors und der Abstand jeder Versetzung van der RiBspitze werden durch zwei Bedingungen bestimmt: (i) die Singularitat der Spannung an der RiBspitze mu13 Null sein und (ii) die Gesamtkraft in der Gleitebene der Versetzung auf die Versetzung ist Null (unter der Annahme, da13sie der Bewegung die Reibungsspannung ai entgegensetzt). Die Ergebnisse des Modells werden mit denen anderer Modelle verglichen fur den Fall, da6 die Versetzungen mit dem Rifi kolinear sind und es wird gezeigt, dalj die vorhergesagten Verschiebungen an der RiDspitze gut iibereinstimmen. Der Vorteil dieses Modells ist, daB es relativ einfach auf Situationen angewandt werden kann, in denen die Relaxation nicht kolinear mit der RiDspitze ist; rin solches Beispiel wird betrachtet.
INTRODUCTION
In recent years a number have
root
been proposed of a notch.
models the
is that
plastic
tribution
field of
to represent
Perhaps of
Bilby
at the
edge
In this note we consider
of approximate
or
plastic
models
flow at the
the best known et .1.(l) tip of
screw
who,
a crack
of these represent by
dislocations
a dis-
collinear
with the crack and subject to a constant friction stress oi (this stress is associated material). applied
The crack shear
zone length
stress
required
with the yield stress of the
is acted at
upon
infinity
to relax
by a constant
and
He represented
a single
above.
dislocation
superdislocation) the
crack
of arbitrary
has been emitted
and
has travelled
plane until it has reached the
total
force
coefficient
on the
between
(a
along
a suitable
slip
and the
These two conditions
the Burgers
vector
give two
of the super-
the
plastic
dislocation,
at the
crack length and the distance of the dislocation from the crack tip. So for a crack of given length in a mate-
the plastic zone by a
the applied stress, the friction
applied
stress, these two equations
and
(and hence the displacement
stress of the material)
solved the problem as a boundary value problem in the classical theory of elasticity without any reference
stress, the
rial with given yield stress and acted upon by a given
mal stress
of the Burgers
vector
will give us the of the dislocation
at the crack tip) and
also its distance away from the crack tip. To test the model we first consider anti-plane strain deformation with the dislocations lying in the plane of the
crack
and
compare
the results
with
those
obtained by Bilby et o1.o) and by Dugdale.t2) We then consider the problem when the dislocations lie
t Department of the Theory of Materials, University of Sheffield, Sheffield, England. $, Now at: Department of Mathematics, Imperial College, on slip planes which intersect London, England. 679 ACTA METALLURGICA, VOL. 19, JULY 1971 8
vector
from each end of
in the stress at the tip
magnitude
* Received November 19, 1970.
Burgers
“superdislocation”
nor-
to dislocations.
to
that both
thin strip collinear with the crack at a constant Y (the yield
model
such a position
of the singularity
of the crack are zero. relations
an alternative
In our model we imagine that
the singularity
crack tip is determined. Dugdalec2) considered a similar model for a crack under tension in a plane stress geometry.
those described
the crack at an angle,
ACTA
680
METALLURGICA,
VOL.
19,
1971
where A = ,~4Bb/Brr in this case since we are considering anti-plane plane
strain.
strain
The relevant
or plane
corresponding
stres
to those
respectively.
of A for
definition
will
of Bilby
reproduce
results
et al. or Dugdale and b is the
,u is the shear modulus
Burgers vector of a unit dislocation. A between
Eliminating
equations
(1) and (2) gives
the relation
-e-y
-C
-0
FIG. 1 A crack lying on the z axis between z = +c and z = - c with positive and negative sorew “superdislocations” of unknown but equal magnitude placed at z = +a, y = 0 and z = --a, y = 0 respectively. The infinite body containing the crack and dislocations is subject to the applied shear stress Pva = u.
again
assuming
anti-plane
problem
has been
similar
where he represents distributions
of
strain
deformation.
considered
The corresponding
dislocations.
complicated
(3)
result for the Dugdalec2) model is ?T
2 cos-l
(4)
(c/a)
A
by Lardnerc3)
the plastic zones by continuous
screw
c2
4a(a2 - c2)1’2
q/u =
However,
his
approach
is rather
numerical
results may not be reliable since in the one
Rearranging
these
two
expressions
and
expanding
c/a in terms of o/gi for small scale yielding
(alai < 1)
we find
and his subsequent
case where he considers higher terms in his successive approximation
3a2 -
ai/0=
scheme he does not get convergence.
c/a = 1 -
+(a/ui)2
(5)
for our model and
ANALYSIS
c/a =
1 -
f
(a/bi)2
(6)
1. Stress relaxation in the plane of the crack The model is illustrated of length x =
2c on the
in Fig. 1. There is a crack
x axis
between
+c with large dislocations
of opposite
sign placed
x = -c
of magnitude
and
Bb and
on the x axis at x = fa.
for the model of Bilby et al. and Dugdale. the ratio of plastic function
of a/oi from equations
Fig.
In the
2.
associated
shear stress Pug = a at infinity
distance of the dislocation
opposing
conditions
the
motion
of
and a friction
the
The
a and B are:
from which we determine
(i) l!he stress singularity
stress
dislocations.
at the tip of the crack
should be zero. (ii) ‘llhe total force on each of the dislocations their slip direction The stress field produced be deduced
methods
deduce
the
for crack problems
(Kayc4)).
relevant
results
and (ii) from a paper by Smith.c5)
model
we have
zone with the
from the tip of the crack.
The main difference
between
the two models
flected by equations
(3) and (4) is that as a +
predicts oJa +
4 whereas (4) predicts GJO +
ever, it should
be noted
as reco (3)
1. How-
that since the model is to
represent plastic relaxation at a crack tip a < oi and so the results to the right of the dotted
line in Fig. 2 will
To represent a/c in terms of the stresses
we can solve equation
(3) as a quadratic
equation
in
a/c and find
Alternatively for
conditions
one (i)
a/c =
We find that
(aa -
(1)
cz)iP-
1.1 The displacement at the crack tip. “superdislocation” model the displacement
and -Ac2 a(a2 -
II2
2
a 2 -
f_Ti==
of the plastic
this has been done in a thesis
by one of the authors can
the extent
not be used.
by the model of Fig. 1 can
by standard
in linear elasticity;
in
should also be zero.
(3) and (4) is given in
“superdislocation”
The problem is to find B and a when there is an applied oi
A plot of
zone length to crack length as a
A c”) + (a2 :C2,1,2
-
g
(2)
In
the
at the
tip of the crack will be that due to the dislocations alone and can be calculated easily using equation (1).
riTKINSON
KAY:
AND
MODEL
OF
RELAXATION
ST
A CRACK
TIP
681
3
2, This
Model
Lt
1,
I
0.3
I.2
0.9
Fro. 2. A plot of L+ = (a - c)/c against P+ = O/U~ as predicted by the model of Fig. 1 and the model of B.C.S. and Dugdale for relaxation by dislocations collinear with a crack. 3.0 t
2.0
-
0.4
o-2
FIG. 3. A
0.6
I.0
plot of W+ = Aw~/oic against P+ = o/u+ a8 predicted by the model of Fig. 1 and that of B.C.S. and Dugdale.
The result is
an approximate Ahw = “0 (a‘2 _ +/2 P
where
0.8
Aw is the
discontinuity
in displacement
(8)
scale yielding
Aw = 2
at
x = c the crack tip and a is given in terms of hi/o and c from equation (3). Using equation (3) we can derive
expression
for Aw from (8) for small
(~/a~ < 1). The result is (1 + &((~/a~)~}
2 The
corresponding
result
using
the
B.C.S.
(9) model
ACT-4
682
METALLURGICA,
VOL.
19,
1971
As can be seen from expressions (9) and (1 l), for small scale yielding the crack tip displacement
predicted
by the two models are quite close. Figure 3 shows the displacements predicted by the two models as a function of g/aj for a fixed crack length c. 1.2 S&es8 reZ&xcztionat an angle to the plane of the crack.
The model is as shown in Fig. 4. The crack is
now assumed to be relaxed by dislocations lying along planes which intersect the crack at an angle 0 with the x axis. The geometry is again anti-plane strain and there is an applied stress Puz = o at infinity and a resistance to motion cri of the dislocations along FIG. 4. A crack lying on the z axis between z = -kc and z = -c with equal positive screw “superdisloclstions” at A and D and negative screw “su~rdislocations” at B and C of equal magnitude t,o those at A and D.
(Bilby et ~1.‘~)) is Aw = 3
log (a/c)
(10)
where a/e is given in terms of d/o2 from equation (4). An expression for ajdi small gives A,=g{I+$)
the dip planes. Again the solution can be calculated using standard methods in elasticity or written down in terms of a complex variable using results given in smith.‘5 For brevity we quote here the results in real variable form as they are derived in fuIl in Kay.c4) The condition of no singularity at the crack tip gives 4A CJ= -= cos y’ f.lf.2
where rr, rs, 0, and 0, are as illustrated in Fig. 4. The condition that the force on the dislocation at A,
O-07-
0-06 -
0.05-
I?
(12)
0-04-
Fra. 5. A plot of L+ = rJcagainst P+ = a/o,for different values of 0, as predicted by the model illustrated in Fig. 4.
ATKINSON
FIG.
AND
KAY:
MODEL
6. A plot of W+ == Awp12cricagainst P + =
OF
rr,r2
A CRACK
TIP
683
given 8, we merely fix r1 and calculate the corresponding b/A from (12) since rz and 0, are correspondingly fixed by simple geometry. Substituting for A from (12) in equation (13) we can then evaluate C/G, for the given value of r1 and f&. The displacement at the crack tip is then 2A.
2A co.9 81 + ____ r sin 28
AC2cos (8 + e1 + 0,)
AT
o/ui for different values of 6, as predioted by the model illus. trated in Fig. 4.
acting along the slip band CA, be zero gives -
RELAXATION
CONCLUSION
AC2 - -_- sin
rr,rs
(e + e1 + e,)
sin
In this note a simple model has been suggested which can be used w an approximate model of plastic relaxation just EMcan those of Bilby et al. and Dugdale. The virtue of the model is that it can be extended to more eornp~c~~d situations without too much more difficulty, for example it can be used to represent oblique slip bands without the need for numerically inverting oomplicated integral equations. The disp’lacement at the crack tip can also be worked out much more simply.
8, + _ZfP!!L r sin 28
x ie-“I.fB,isin(~~)sine~--~
X sin 0 (
%+
2
1 +
sin 0
1
ACKNOWLEDGEMENTS
A ~05%8 sin 0, 2r sin 8
- (Ti = 0
(13)
where r and z9are again defined in Fig. 4 and equation (13) is true provided 8, # 0. Equations (12) and (13) have been used to calculate the displacement at the crack tip and the length rz of the plastic zone as a function of d/bd for a fixed &. The results are shown graphically in Figs. 5 and 6. It is a simple matter to calculate these results, for a
We would like to thank Professor B. A. Bilby for his comments on the manuscript, and one of us (T. R. K.) would like to thank the Science Research Council for financial support. REFERENCES 1. B.A. 2. 3. 4. 5.
R. D.
R. T. E.
BILBY, A. H. COTTRELL and K.H. SWINDEN,PTOC. SOC. A272, 304 (1963). S. D~a~~~~,J.Mech.Pfiys.Sfflids 8, 100 (1960). W. LARDNER, 1st. J. Fmcture Neck 4, 299 (1968). R. KAY, M.Sc. Thesis, University of Sheffield (1970). SMITH, Proc. R. Sm. AS05, 387 (1968).