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Stress distribution at the edge of an equilibrium crack
J. Mech. Phys. Solids, 1964, Vol. 12, pp. 149 to 163. Perpamon Press Ltd. Printed in Great Britain
STRESS
DISTRIBUTION
AT THE
EQUILIBRIUM By Department
of Civil Engineering
L.
M.
EDGE
OF AN
CRACK KEER
and Engineering
(Received 28ti January,
Mechanics, Columbia Clniversity
1984)
SUMMARY
USING the assumptions
of G. I. Barenblatt the problem of determining the stresses over the entire surface of an equilibrium crack is formulated as a mixed-mixed boundary value problem in the Solutions are obtained for a penny-shaped crack in an infinite, classical theory of elasticity. elastic medium under conditions of uniform tension parallel to the axis of the crack and uniform The analogous two dimensional problem for simple shear parallel to the plane of the crack. The results show that the stress distribution at the edge of the crack tension is also investigated. depends only slightly upon the applied stresses when compared with the stresses associated with cohesive forces. The distance over which the cohesive forces act is computed in terms of Barenblatt’s modulus of cohesion.
1.
THIS paper
deals
with
opened in an infinite,
stress
isotropic,
INTRODUCTION
distributions
elastic
arising
medium
from
an isolated
under isothermal
crack
conditions,
a crack will be defined as a plane cut in the material. In three boundary of the cut is a circle and in two dimensions a line.
being where
dimensions the The theory of
BARENBLATT (1962) assumes that forces of cohesion exist at the edges of the crack causing the faces to be smoothly joined together. Thus all stresses are finite everywhere. stresses
The
purpose
within
of this
the framework
paper
is to determine
of classical
region of the crack over which the cohesive for such
a theory
elasticity stresses
as well as a comprehensive
the
distribution
of cohesive
theory and to calculate the act. The physical justifications
review
of theoretical
and experi-
mental work concerning the solution of crack problems are summarized blatt’s recent article. His main assumptions are as follows :
in Baren-
(a)
The longitudinal dimensions of the end region where cohesive are small when compared with the entire crack dimension.
(b)
The opposite faces of the crack are smoothly joined with each other, what is the same, the stress is finite at the tip of the crack.
Assumptions (a) and (b) will be used in the sequel stress distributions to a simple form.
to reduce
forces
expressions
act or
for the
For the three-dimensional case the solid contains a penny-shaped crack of radius a, given by z = 0 (0 < T f a), where (r, 0, z) are cylindrical polar coordinates with the centre and axis of the crack as origin and z-axis respectively. The 149
130
1,. hi.
displacement~s
ad
stresses
I
rorrespording
to
the
respecti\-c
c~ordiriates
arc’
(trr. z(& ur) and (OZT. cr& DIzz), where the stresses arc comput.ed on an elemcsnt whose normal is in the direction of the z-axis. In the sequel it will hc ran\-enient to use a rectangular. coincide
with
Cartesian coordinate
the z-axis
and
origin
system
(s. JJ,-_) M’ILOSC z-axis
of the cylindrical
system.
origirl
mtl
The
associated
displacements
and stresses are (u%. uy. u,) and (uzz, azy, azr). The solutiorl will (bonsider separately the cases where ozz and ozz are the only nonzero stress houndar\, conditions on the face of the crack. For two-dimensional
problems
where Cartesian coordinates k,v _
I/
-o(~
us
,r
::
(I).
a crack is a cut in the material
along the S-ISIS.
(z, y) are used.
The crack has length ‘La and is
The displarenunts
and stresses corrcspondinp
detirletl
to (s. !,,
arc (u, 71) ant1 (~7~~.uey). where the stresses are computed 011 ;II~ element whose. rlormal is in the direction of the !I-axis. The solrrtion consiclerc~tl for th;s (use w11l he for ozy non-zero OII the face of the crack. ‘l’ltc results for penny-shaped cracks are hasetl upor~ work 01’ C’OI.I.I>S ( I!M+ZJ and
I~11sur.r~
presentation
(1955a),
and for the two-dimensional
for complex
potentials
as given
b!.
crack
problems
E:s(;r..\sl)
on the rc-
anti GHI~:K~ (lM3).
W’itll the assumptions of Barenblatt the distribution of strtxss at the edge of’ tllc, crack can be easily found by a semi-inr,erse method. ‘I’hcsc stresses are the
’ stresses that ensure the crack will be smoothI>. joined.
‘I’hc magnitrtde
of these stresses will he seen to be of the order (/,/CL)*as compared
’ cohesive
with the normal
stresses that are applied to the crack, where d is the radial distance (length) ol.cr which the cohesi\-c stresses are assumed to a& for a t~irc,c,-clirliensiorl;ll (twodimensional)
2. In
crack.
PENNY-SHAPI~D
this section
CRACK
a penny-shaped
OPENED
crack
ISY
opened
NORMAI,
by
axially
PI~E:SSI:HE:
symmetric
pressure is considered. With tl 1e i n t en t’Ion of using the assumptions gi\-en earlier the boundary conditions are formulated as follows : UZE
:-f(r),
the =
g
z =
(r),
0
z = 0
(0 .s: r SC 1,) (6 < r
&
a)
normal
(a) and (h)
1
(2.1)
where vpz is finite at r --= b. In this way the problem of finding the stress distribution due to cohesive stresses at the tip of the crack is sought by the following seminverse method : assume a given distribution of displacements consistent with Harenblatt’s assumptions near the edge of the crack, solve the resulting elasticity problem, and then determine the stresses necessary to produce this displacement. To solve the problem given by (2.1) we first solve the following two boundary. \.alue problems which are analogous to equations (2.1) :
where
all quantities
at the edge of an equilibrium
uz2 = g (r),
(6 < r d a)
u22 = 0.
(a < T <
are computed
on the plane,
and ZERKA
this section. The boundary
(1954). The solution
value problems
means of the following
z = 0.
The problems
given
by
value problems in the classical are given by SICEDDON (1946)
of COLLINS (1962) will be used for
are reduced
representation
(2.3)
1
00)
(2.2) and (2.3) are well-set mixed-mixed boundary theory of elasticity. In particular the solutions and GREEK
151
crack
to problems
of displacement
of potential
theory
by
:
(2.4) c2 I#’ = 0 2p u’ = (1 -
where TVis the modulus ment
and stress
= 1, 2Y) (i v 9’ $ 2) 2 v z
-
L0, 0, (3 -
of shear and Y is Poisson’s
significant
to this problem
ratio.
4 v) ;$
From
I> (2.4)
the displace-
are
(2.5)
Following
Collins,
3+‘!3z
3$’
are given
1
h, 0) dt
;
=--J 2i*
3z
as follows
[ r2 +
(2 t
ity]t
’
:
h, (t) = - h, (-
t)
(2.6)
(I &#J2
1 dz = ", s p
where
h, (1)dt + (z + q2y
6, (t) = o for 0 < r < 6.
the first of (2.3). the following
This
On the surface
values
'
latter
z = 0, 3$*/3x
h2 (t)=
- h2 (- t,
condition
will automatically
and their
normal
derivatives
(2-V satisfy have
: b
s
N’ -=-
h, (4 dt
(P -
32
r2)i
(0 < r d 6)
(2.8)
r
0
zzz
3” +’
-_=_-32~
1 d
(6 < )’ < ’ th, (t) dt
rdro s (r2-t2)*
(0 <7.
co)
b
1 d * th, (t) dt = .~ -_ r droJ (re - te)i
\
(6 < r <
co) J
(2.9)
1,. M.
162
KEER
(0 < T < CO)
0
The solution to t.he problems the followin,n values :
given
J
by (2.2) and (8.3) is completed
by giving /II (t)
0 except within the limits specified by (?.lt,j anti (‘L.13). ‘I’he ardysia where hi (t) in\-ol\,ecl in sol\ing similar integral equations has been given h>. Gr{xx anti %I.:its I (195-b) and Gool~ai~s (1962). .Issuruirrga (7) has a sectionall\. continuous deri\ati\~e.
h, (2) -~ --
“t - (i2g’ i 7?
(rj
,
dr
(2.14)
tyt
I
since g (a) = 0 by the requirement that the displacements are continuous. Stresses using the functions h, (t) and 1~~(t) on the plant 2 := 0 may now be computed as given by (2.12) ant1 (2.1-l). Tlir!, arc as follows :
tr sf(s)
‘J**l
(b2 -
s”)t ds
0
(6 -r: r
<‘.
cc)
(2.15)
Stressdistributionat the edge of an equilibriumcrack
158
where (2.15) is obtained from (2.12), (2.5), and the second of (2.9) by evaluation of the result from reversing the order of integration. It is clear that a singularity will arise as r --f .?J+due to the first terms in the right-hand side of (2.16) and that, the second term does not have a singularity. We require that (4
+ 4(
< 00,
2 = 0 (0 < r < co).
(2.17)
This requirement is met if b
a
l-v
sf(s) h g’ (4 as o (P Sa)* = b (s2 - P)’ s s tLb
Using (2.18) one writes the sum of the stresses as follows
(2.18)
:
b
uzz
=
(9 -
f
bz)’
s
cf(s) ds
n (r* -
se) (ba -
se)4 -
[One observes that unless the crack is smoothly joined, the second integral on the right-hand side of (2.19) is not finite, i.e. g’ (a) = O]. At this stage the problem has been solved with certain restrictions on .f (r) and g (T) that will ensure that displacements are physically reasonable. Now assumption (a) of Barenblatt’s theory is used by writing out the displacement near the tip of the crack as follows : ut = uz (a) + I& (a) (a -
T) + 4 uz” (a) (a - ry + . . .
(2.20)
where terms higher than (a - r)e have been ignored. By continuity of displacements U, (a) = 0, and by assumption (b) uz’ (a) = 0. Therefore g (r) = I.& = $ where A1 = uz” (a).
(2.21)
(a - r)B
From (2.18) b
sf(s) ds
s*)* = ,8bK’
(a’ -
bs)’
where K’ = a log
The stress distribution
a +
b
is therefore
b uzz
=
?.
r
(r* -
ba)*
os (r* -
q-(4 ds &‘) (b4 -
1 -
(62 -
ss)* +
Y BI
1- (a*
b*)*.
(2.22)
I54
1,. 11. Km%
*4t r :z 0 the stress is continuous since the first integral equals.f‘(O) and t,hc second vanishes. The serond integral on the right-hand side of (?.l!)) may 1~ evaluated by the following
rhange
of variables
:
+ “6) stxtus that the ~iistrib~~ti~)r~of stress in the reFiort The resltlt giycn ~if I_._ wliere rohcsivc forces act dcyends only slightly upon the applied loads. ‘I’hr sign of the stress changes as r = /I-, and the absolute \ralue of the stress increases rapidly and is of order (G/C_@as compared with the value 01’ stress on the snrfact, ijO. where fjO is a constant of the crack. A particulari>, interesting case is wIicnJ‘(r) normal pressure.
3.
For this special case (2.26) becomes
!'EK;SS-SIIAPEI)
C'I(,~C'li
OI’EN1.X)
StMP1.l~~ SIIE.%H
111~
The problem for this section is concerned with a crack acted upon by a shear The applied shear stress is assumed to be &ally s~~~etr~c~. stress 6rzz only. Boundary conditions are as follows :
flzv
As in the preceding
o*z
0.
section the problem
: -. 0
(0 I< t
is broken
x;).
Ij
up into two separate
parts.
The cancelling of the singularity will give a relation each producing a singldarity. between 12(T) and I (r). Equations (3.1) are written as follows :
nzI’
k (r).
(0 I
r Y> It).
/frl
0.
(h ”
)‘ %I 33),
~zz2 1Lz2
0.
(0 “- r & cr?).
1 (r).
(h i
r < a),
(i-2 4
r -’
Ii,” where
O&
-- 0.
=- cTzzi := o and all quantities
a-2).
are cotnputed
I !. i
(3.2)
(323) !
on the plane z = O.
1.55
Stress distribution at the edge of an equilibrium crack To develop
solutions
are considered
(3.2) and (3.3) PAPKOVITCF~(1332) functions
to equations
in the following
2/l llL: = -
(3 -
form : -cY) YZ + X
2/Luu, =
2puz
=
The representation
(3.5)
X
-
(3 -
au) Yy, + E
of displacement
given
by (3..$) lead to the following
system
of stresses :
uzz=
-(l
-2”)
(3.6) (3.7)
MINDLIK (1355s) established
observes
that. if gt5 is known. then an equation can be on the boundary z = 0. This is done by differentiating the results. to y and (3.7) with respect to x and subtracting
for 3Y&z
(3.6) with respect The following
relation
and an integration
is developed
:
gives
where an arbitrary inspection
constant
2p.U
Z =
of
-
integration
(3 -
has
been
3Yz 4v) YZ + x T +
set equal
to
zero.
b@ -3s
By
(3.10)
on z = 0. Equation (3.16) as it stands is not harmonic and does not lead to the same type of boundary conditions as found in Section 2. However, appropriate solutions
to the two problems
selection
of YZ.
MINDLIN (1955b)
posed by (3.2) and (3.3) can be obtained
has shown
that if
~,--2(l-v)~-2v~~+~~+x~~~_
v2==
-(I
by proper
LI@ -2v)Y~I~+x~-zJ-~0)
3’ Yz 5= 0. 2s bz
‘ 3\y,
av,
(3.11)
(3.12)
where V2 V, = V2 F12 = 0, then the last of equations (3.2) and (3.3) are identically satisfied on z = 0. If YZ is assumed to be a known harmonic function, then the following relations for ~Y+z and b2 a/&z2 are obtained :
Stress distribution at the edge of an equilibrium crack
156
(3.13)
(3.14)
It should and
be remarked
@ to within
that
Y, can be determined
& (z, y) + z dS (z, y).
Further
to within a function clarification
will be made later but only for the case where ul, is an axially symmetric The problem also be axially
given
by (3.2)
symmetric.
is axially
symmetric
& (z, y)
of these functions
and consequently
function. YZ1 must
We th ere f ore assume that the form for YZ1 is given as b
~~
yzl==-!.
2i_b I p
where YZ1 is antisymmetric given bv (3.2) is sought. derivative
Pl
+
(t)dt
(z +
(3.15)
itp]i
about the plane z = O, and the solution
to the problem
With YZ1 as in (3.15) its value and the value of its normal
on the plane c = 0 are as follows
:
b
Yr’
;
I
-
r
~1
(0 zg r < 21)
(1)dt
(3.16)
(t2 - r2)t
T
(0 ,Cr
b
Y, l is a function
‘l‘herefore, by assumption plane z =- 0. From
(3.13)
substitution.
and
(3.14)
values
Using equation
for
(X13),
that IS zero exterior
3YZ1/az and 32 @l/bz2
aY,l/bz
to v =: h on the
are found
by direct
is
b
3Yf/,' -__
iJZ
where
= -*-
is antisymmetric
2‘YZ1/&
(t) dt cos B 22 3Cbs [Y” + (z + itpp 1
2-‘,
3
about
the plane
z = 0.
(3.18)
One integration
gives
b
Y,’ ::
T$f ;; I p,(t) log (z
+ it + [ r2 + (z + it)“]‘}
dt +- D
(3.19)
-b
nherc
1) is an arbitrary
constant.
From
the requirement
\-anishes as r -t 00, D must be zero. The problem is completed when @’ is specified. the following
requirements
that the displacement
The function
@
must satisfy
as r 3 co :
@’ -= *= (T, z)cosf?,
a@’ 7
=
0
t1r 1,
(3.20)
Stress distribution at the edge of an equilibrium crack where the order relationship is used in the usual manner. function satisfying these relations is
157
A suitable
potential
b
@’ = ;;
GrJ-q1 (t) dt { (z
+ it) log([ rt + (2 + it)z]+ + (23 + it,} -
-b
-
where q (t) =
-
t) and a constant
q (-
[ ra + (z + i2)z]*} cos B
and logarithmic
The functions @I and b2 4p1/3zz are antisymmetric symmetric. On the plane z = 0
(3.21)
term have been omitted.
about
z = 0 while
B@/dz
is
b
32 @’
s
a!
g=-xr
(0 < r < 6)
(3.22)
r =O Therefore
(3.14)
(b < r <
may be expressed
b
co)
as
b
b
p1 (t) dt + E, (t” - r2)) b
b
Or
s
ql (4 dt (P - r2)t
==
r
+
b
d
Pl(4
dt
(t2 -
P)+
~
dr
I
+
2
(1
_
I’1
“)
where E, is an arbitrary
dt
+
E’,
r2)+
(3.24)
r
constant
is integrated
sequently performed. p, (t) with the result
(t)
(t2 -
s
7
hand side of (3.24)
(3.23)
of integration.
The first integral
by parts with the indicated
The resultant
equation
on the right-
differentiation
suh-
is then solved for q1 (t) in terms
of
b
!A@) = P,(b)
+ t
(3.25)
s
1
Equation
(3.25) determines
differentiation indicated value for @ :
@l to within an arbitrary
by (3.19)
constant,
E,.
Performing
the
and taking limits as z --f 0, we find the following
b -
To remove
all displacements
for 6 < r <
+
case.
itq*(t)dt]
o
(3.26)
03 requires
b
(3.27)
tq, (t) dt = o s 0
and it is this restriction
on q (t) that
determines
b
f.js =
! r
z
s
t2 q1 (t) dt cos
7
(t2 -
0,
r2)*
e ’
the value for E,.
(3.23)
z=O(O
Hence
< CO)
(3.29)
I,. 11. KICIIR
158 and the boundary WC observr that
condit.ions
given
by equations
(3.2)
are rompletely
satislicd.
Thcrcforc displaccmcnts YG1 and !Pb’ are axially symmrtric I’unctions. arising from an axially symmetric shear stress, ozz. will not be axially symmetric but will instead have nonsymmetrical contributions of the form given by (3.30).
where
If
one
is primarily
interested
in stresses.
(3 ~~ 4”) Y=’
then the problem
:
0.
0 (I/>;
(3.32
r/ )
r
I
is in a sense cqui\-alent to the problem dcfirrcd by (3.2) pro\.idcd that a displaccmcnt This completes the problem defined h> field given h\r (3.30) is added to thr solution. (3.2). ‘I‘o soh~ the problem satisfied :
given bv (3.3) the following
J~$’ _ {, dz ’
z
boundar\.
-.
(3.33 )
0 (0 :-_ ?’ > b)
z=
0.
c~onclitions must I,(,
0 (/I
x
r
0 (/I
ml
5.
r i.. n)
(8.3 t)
(3.3.;)
wllr~e 1 (1.) plays the role of the similar frlllction given in Se&on L’. ‘I’lic rcnutiriin~ functions Ya2 and Yb2 are given by analogy with (3.30) and are determined aft& the following
boundary
value problem
is solved
for Yz2 :
‘I’hr problems given by (3.31 ) a11t1(3.32) and those gi\.en by (3.36) and (3.37) are identical in form to the prohlrm solved in Section 2. If we d&tic
(list ribrrtion
Stress
at the edge of an
cquilibriuu~ crack
IS!)
a
s .r
sl (s) ds
L’ d
(3 - Ay)ZJa(4 = - ; -y
-___ (s2
_
t2)t
(3.4))
n
2
‘I’heac fnnctions grntial
result
shearing
-t,
7r
in the determination
stresses
in the region
1’ (s) ds
(R2-P)i
of the following
distribution
for tan-
1) < Y < u :
(3.41 )
The singularity and
I (T,
:
at 1‘ -
I/ is removed
by the following
h - sh (5) ds 2(1 --I$, .Ii (/P - sy+ = (3 - 4Y)
1)
relation
hctwccn
/i (r)
u I’ (s) ds bs (9
-
(3. 43)
t2)+ .
If ISarcnMatt’s assumptions are now applied to the stresses occurring at the edges of the crack, it 3 then seen that the same requirements arc necessary for 1 (v) as WC~Cnecessary for g (r) in the preceding section. Therefore
is determined
from (3.43).
P ” (1 -
77 (3 --llY) 1Vhrn
v)
The stress
distribution
in the region b < r bC n is
_.~~j:,e~~2)i{blogrU+(u~-il)lj
_qa’-L.‘)+
(3:L6)
i-
czl: = r,,, where
TV is a constant,
the resulting
stress
distribution
is
160
For this section the t,wo dimensional formulation as given by ~~:NGLANLI ar~l GREEN (1963) is chosen, and the solution will be brief since the development so closely parallels that of the previous sections. The basic equations for two dimensional isotropic elasticity Kivcn in GHEHN and ZERNA (1954) are
\
/Lx1-- p (UT t j/Au), =: K J2(2) -~ .O (Sj - (2 -. 2) D’ (z) u.vy ~- i o*y
2: a’ (2) -~ a’(z)
. (2)
(Z f>,,i)
tJ (2),
(4.1 j
~. 4’(z).
1
where K == 3 ~. 4v for plane strain and K ~~ (3 r,);(1 8,)for gcnrralizc71 fblanc stress. We suppose that there is D single line crack between .I* Ym-1 rr and J 0 : ihr region 15 otherwise unbounded. The crack is being opened by equal and oppositr norlnal pressure’. on g 0 the following boundary conditions must bc satisfied :
In the last of (4.1) if IJ (z) = 0, then on ~1 ”
~!a
zi P
uy
0 XI,’ have
[S”’
(2)
f?
(M : l)[Q(Z)
(;,I.
i
.ct (:,I.
(4.3)
13~ using the function
the problem defined b> 1_ OYY
,,, (T), !, -~ 0 (0 I /,(‘j s_h)
IL,,1= 0,
!I
(4.5)
0 (b . j.t.(. . 7 ,
is solved by reduction to the following solvable Abel typv t(luation :
which has for its solution
Similarly, the problem defined b>
(4.8,
Q2(z)
_ I u
P2 (t) dt -__(22 -. 12)$
This second problem is reduced to the following
Abel type ecluation
(4.9)
:
Stress distribution at the edge of an equilibrium crack
161
a
-~ K+l P
s
z- (0 dJ
(_
n(x)
=
(4.10)
(E
which has for its solution K+l -yF+5
s ~n’
(9
n
(8) ds
(4.11)
la)+
-
t The stresses on y = 0 (a -
(~1 < d) are given as m (9) (U - sz) ~2%
(4.12)
(x2 - 81)
0 a
(4.13)
Removal
of the stress singularity at x = b requires
that
b
a
s
m (4 d.3
n’
(bB=-
ds
(4.14) ’
b
0
Using Barenblatt’s
(8)
(P--
assumptions on uy as before, we take
n(x) z
$(a -
(4.15)
ISI,”
where with the use of (4.14) the constant As becomes b
AS
=
(* + ‘) 4,~ bK
s
m (4 d8
(4.16)
(bB_ga)t *
0
These values when used in (4.12) and (4.13) give the following distribution for the cohesive stresses on the tip of the crack : b
(x2-
w (aa- w*+
(& @‘d _
When a constant normal pressure is assumed, the distribution
b8)t -
x’)i (x2 -
bz)’
I> .
(4.17)
of cohesive stresses is
u”*=-~po{;-3(~)+[(~~+(l-~)logp&f..q}. (4.13)
Therefore, the distribution of the cohesive stresses for the two dimensional crack is inherently the same as that for the penny-shaped crack.
h_
/1,:ytrlf
_
(.i.i!
1 0
where G (t) is the intcnsit>. of tllc forces of cohesion Il<‘i\r tllc, c&c” of‘ thch c’racah (11 < t’ :: (I). and t is the dist,ante from the edge of the c~:tc~k. A- is also writtrtlx :I\
(.i..i)
The distribution cat’ cohesive stresses for isolated equilibrittrn cracks determined by a semi-inyerst method within the framework of classical
has I~WI elasticit!.
are used to obtain the distribution of cohesi1.c theory. li?-:-:.‘,~L~I t’s hypotheses stresses . These stresses. which are assumed to act over a specified region near ttic cdgc of the crack. arc of’ order (u/d) w 1&en compared with the applied stress3 For the three problems consideret 1 fix both two and three dimensiona problems.
tlw stress distrihutidns IIowe\cr, upon the
near the edge of the crack
(/I .-- 1’ -- (I) are nearly
identical.
thcb joining of the cracks depends different for cac4r of the three profor the blems. Using Rarenblatt’s modulus of cohesion hF. \-alucs are obtained stresses act in terms of K and radial distance (length) over which the cohesive the applied
THE author
the constant di associated with elastic constants and is therefore
load.
wishes to thank Professor Ii. 1). Mindlin for his many helpful discussions and partieularlp for pointing out (3.1 I) and (3.12) which allowed an ea..y solution to the problem in Stcztion 3. The ORire of Naval Iteseareh supported this work.
Strew distribution at the edge of an equilibrium crack
163
REFERENCES BAEENBLA~, G. I. COLLINS, w. D., ENOLAND, A. H. and G-N, A. E. GOODXAN, L. E. GEEEN, A. E. and <NA, KEEX,
w.
L. M.
bbDLIN, R. D. MINDLIN, R. D. PAFXOVITC~~,P. F. SNEDDON, I. N.
1962 1962
Advanca in Applied Mechanics, Vol. 2 Academic Press. Rae. ZZoy.Sot. A266,859
lQ68 1962
Proc. Camb. Phil. Sot. 59, 489. J. A@. Mcch. 29, 515.
1954 lQ64 1955a 1955b 1982 1946
Theoretical Elaatkity. Oxford University Press. QJMAM (in print). Z’rac. 1st i%&u&m Conferenceon Solid Mechanics, p. 56. Unpublished. Zao. Akud. Nauk SSR, Phys.-Maih. Ser. 10, 1425. PToe. Roy sot. A187,22Q.
STRESS
DISTRIBUTION
AT THE
EQUILIBRIUM By Department
of Civil Engineering
L.
M.
EDGE
OF AN
CRACK KEER
and Engineering
(Received 28ti January,
Mechanics, Columbia Clniversity
1984)
SUMMARY
USING the assumptions
of G. I. Barenblatt the problem of determining the stresses over the entire surface of an equilibrium crack is formulated as a mixed-mixed boundary value problem in the Solutions are obtained for a penny-shaped crack in an infinite, classical theory of elasticity. elastic medium under conditions of uniform tension parallel to the axis of the crack and uniform The analogous two dimensional problem for simple shear parallel to the plane of the crack. The results show that the stress distribution at the edge of the crack tension is also investigated. depends only slightly upon the applied stresses when compared with the stresses associated with cohesive forces. The distance over which the cohesive forces act is computed in terms of Barenblatt’s modulus of cohesion.
1.
THIS paper
deals
with
opened in an infinite,
stress
isotropic,
INTRODUCTION
distributions
elastic
arising
medium
from
an isolated
under isothermal
crack
conditions,
a crack will be defined as a plane cut in the material. In three boundary of the cut is a circle and in two dimensions a line.
being where
dimensions the The theory of
BARENBLATT (1962) assumes that forces of cohesion exist at the edges of the crack causing the faces to be smoothly joined together. Thus all stresses are finite everywhere. stresses
The
purpose
within
of this
the framework
paper
is to determine
of classical
region of the crack over which the cohesive for such
a theory
elasticity stresses
as well as a comprehensive
the
distribution
of cohesive
theory and to calculate the act. The physical justifications
review
of theoretical
and experi-
mental work concerning the solution of crack problems are summarized blatt’s recent article. His main assumptions are as follows :
in Baren-
(a)
The longitudinal dimensions of the end region where cohesive are small when compared with the entire crack dimension.
(b)
The opposite faces of the crack are smoothly joined with each other, what is the same, the stress is finite at the tip of the crack.
Assumptions (a) and (b) will be used in the sequel stress distributions to a simple form.
to reduce
forces
expressions
act or
for the
For the three-dimensional case the solid contains a penny-shaped crack of radius a, given by z = 0 (0 < T f a), where (r, 0, z) are cylindrical polar coordinates with the centre and axis of the crack as origin and z-axis respectively. The 149
130
1,. hi.
displacement~s
ad
stresses
I
rorrespording
to
the
respecti\-c
c~ordiriates
arc’
(trr. z(& ur) and (OZT. cr& DIzz), where the stresses arc comput.ed on an elemcsnt whose normal is in the direction of the z-axis. In the sequel it will hc ran\-enient to use a rectangular. coincide
with
Cartesian coordinate
the z-axis
and
origin
system
(s. JJ,-_) M’ILOSC z-axis
of the cylindrical
system.
origirl
mtl
The
associated
displacements
and stresses are (u%. uy. u,) and (uzz, azy, azr). The solutiorl will (bonsider separately the cases where ozz and ozz are the only nonzero stress houndar\, conditions on the face of the crack. For two-dimensional
problems
where Cartesian coordinates k,v _
I/
-o(~
us
,r
::
(I).
a crack is a cut in the material
along the S-ISIS.
(z, y) are used.
The crack has length ‘La and is
The displarenunts
and stresses corrcspondinp
detirletl
to (s. !,,
arc (u, 71) ant1 (~7~~.uey). where the stresses are computed 011 ;II~ element whose. rlormal is in the direction of the !I-axis. The solrrtion consiclerc~tl for th;s (use w11l he for ozy non-zero OII the face of the crack. ‘l’ltc results for penny-shaped cracks are hasetl upor~ work 01’ C’OI.I.I>S ( I!M+ZJ and
I~11sur.r~
presentation
(1955a),
and for the two-dimensional
for complex
potentials
as given
b!.
crack
problems
E:s(;r..\sl)
on the rc-
anti GHI~:K~ (lM3).
W’itll the assumptions of Barenblatt the distribution of strtxss at the edge of’ tllc, crack can be easily found by a semi-inr,erse method. ‘I’hcsc stresses are the
’ stresses that ensure the crack will be smoothI>. joined.
‘I’hc magnitrtde
of these stresses will he seen to be of the order (/,/CL)*as compared
’ cohesive
with the normal
stresses that are applied to the crack, where d is the radial distance (length) ol.cr which the cohesi\-c stresses are assumed to a& for a t~irc,c,-clirliensiorl;ll (twodimensional)
2. In
crack.
PENNY-SHAPI~D
this section
CRACK
a penny-shaped
OPENED
crack
ISY
opened
NORMAI,
by
axially
PI~E:SSI:HE:
symmetric
pressure is considered. With tl 1e i n t en t’Ion of using the assumptions gi\-en earlier the boundary conditions are formulated as follows : UZE
:-f(r),
the =
g
z =
(r),
0
z = 0
(0 .s: r SC 1,) (6 < r
&
a)
normal
(a) and (h)
1
(2.1)
where vpz is finite at r --= b. In this way the problem of finding the stress distribution due to cohesive stresses at the tip of the crack is sought by the following seminverse method : assume a given distribution of displacements consistent with Harenblatt’s assumptions near the edge of the crack, solve the resulting elasticity problem, and then determine the stresses necessary to produce this displacement. To solve the problem given by (2.1) we first solve the following two boundary. \.alue problems which are analogous to equations (2.1) :
where
all quantities
at the edge of an equilibrium
uz2 = g (r),
(6 < r d a)
u22 = 0.
(a < T <
are computed
on the plane,
and ZERKA
this section. The boundary
(1954). The solution
value problems
means of the following
z = 0.
The problems
given
by
value problems in the classical are given by SICEDDON (1946)
of COLLINS (1962) will be used for
are reduced
representation
(2.3)
1
00)
(2.2) and (2.3) are well-set mixed-mixed boundary theory of elasticity. In particular the solutions and GREEK
151
crack
to problems
of displacement
of potential
theory
by
:
(2.4) c2 I#’ = 0 2p u’ = (1 -
where TVis the modulus ment
and stress
= 1, 2Y) (i v 9’ $ 2) 2 v z
-
L0, 0, (3 -
of shear and Y is Poisson’s
significant
to this problem
ratio.
4 v) ;$
From
I> (2.4)
the displace-
are
(2.5)
Following
Collins,
3+‘!3z
3$’
are given
1
h, 0) dt
;
=--J 2i*
3z
as follows
[ r2 +
(2 t
ity]t
’
:
h, (t) = - h, (-
t)
(2.6)
(I &#J2
1 dz = ", s p
where
h, (1)dt + (z + q2y
6, (t) = o for 0 < r < 6.
the first of (2.3). the following
This
On the surface
values
'
latter
z = 0, 3$*/3x
h2 (t)=
- h2 (- t,
condition
will automatically
and their
normal
derivatives
(2-V satisfy have
: b
s
N’ -=-
h, (4 dt
(P -
32
r2)i
(0 < r d 6)
(2.8)
r
0
zzz
3” +’
-_=_-32~
1 d
(6 < )’ < ’ th, (t) dt
rdro s (r2-t2)*
(0 <7.
co)
b
1 d * th, (t) dt = .~ -_ r droJ (re - te)i
\
(6 < r <
co) J
(2.9)
1,. M.
162
KEER
(0 < T < CO)
0
The solution to t.he problems the followin,n values :
given
J
by (2.2) and (8.3) is completed
by giving /II (t)
0 except within the limits specified by (?.lt,j anti (‘L.13). ‘I’he ardysia where hi (t) in\-ol\,ecl in sol\ing similar integral equations has been given h>. Gr{xx anti %I.:its I (195-b) and Gool~ai~s (1962). .Issuruirrga (7) has a sectionall\. continuous deri\ati\~e.
h, (2) -~ --
“t - (i2g’ i 7?
(rj
,
dr
(2.14)
tyt
I
since g (a) = 0 by the requirement that the displacements are continuous. Stresses using the functions h, (t) and 1~~(t) on the plant 2 := 0 may now be computed as given by (2.12) ant1 (2.1-l). Tlir!, arc as follows :
tr sf(s)
‘J**l
(b2 -
s”)t ds
0
(6 -r: r
<‘.
cc)
(2.15)
Stressdistributionat the edge of an equilibriumcrack
158
where (2.15) is obtained from (2.12), (2.5), and the second of (2.9) by evaluation of the result from reversing the order of integration. It is clear that a singularity will arise as r --f .?J+due to the first terms in the right-hand side of (2.16) and that, the second term does not have a singularity. We require that (4
+ 4(
< 00,
2 = 0 (0 < r < co).
(2.17)
This requirement is met if b
a
l-v
sf(s) h g’ (4 as o (P Sa)* = b (s2 - P)’ s s tLb
Using (2.18) one writes the sum of the stresses as follows
(2.18)
:
b
uzz
=
(9 -
f
bz)’
s
cf(s) ds
n (r* -
se) (ba -
se)4 -
[One observes that unless the crack is smoothly joined, the second integral on the right-hand side of (2.19) is not finite, i.e. g’ (a) = O]. At this stage the problem has been solved with certain restrictions on .f (r) and g (T) that will ensure that displacements are physically reasonable. Now assumption (a) of Barenblatt’s theory is used by writing out the displacement near the tip of the crack as follows : ut = uz (a) + I& (a) (a -
T) + 4 uz” (a) (a - ry + . . .
(2.20)
where terms higher than (a - r)e have been ignored. By continuity of displacements U, (a) = 0, and by assumption (b) uz’ (a) = 0. Therefore g (r) = I.& = $ where A1 = uz” (a).
(2.21)
(a - r)B
From (2.18) b
sf(s) ds
s*)* = ,8bK’
(a’ -
bs)’
where K’ = a log
The stress distribution
a +
b
is therefore
b uzz
=
?.
r
(r* -
ba)*
os (r* -
q-(4 ds &‘) (b4 -
1 -
(62 -
ss)* +
Y BI
1- (a*
b*)*.
(2.22)
I54
1,. 11. Km%
*4t r :z 0 the stress is continuous since the first integral equals.f‘(O) and t,hc second vanishes. The serond integral on the right-hand side of (?.l!)) may 1~ evaluated by the following
rhange
of variables
:
+ “6) stxtus that the ~iistrib~~ti~)r~of stress in the reFiort The resltlt giycn ~if I_._ wliere rohcsivc forces act dcyends only slightly upon the applied loads. ‘I’hr sign of the stress changes as r = /I-, and the absolute \ralue of the stress increases rapidly and is of order (G/C_@as compared with the value 01’ stress on the snrfact, ijO. where fjO is a constant of the crack. A particulari>, interesting case is wIicnJ‘(r) normal pressure.
3.
For this special case (2.26) becomes
!'EK;SS-SIIAPEI)
C'I(,~C'li
OI’EN1.X)
StMP1.l~~ SIIE.%H
111~
The problem for this section is concerned with a crack acted upon by a shear The applied shear stress is assumed to be &ally s~~~etr~c~. stress 6rzz only. Boundary conditions are as follows :
flzv
As in the preceding
o*z
0.
section the problem
: -. 0
(0 I< t
is broken
x;).
Ij
up into two separate
parts.
The cancelling of the singularity will give a relation each producing a singldarity. between 12(T) and I (r). Equations (3.1) are written as follows :
nzI’
k (r).
(0 I
r Y> It).
/frl
0.
(h ”
)‘ %I 33),
~zz2 1Lz2
0.
(0 “- r & cr?).
1 (r).
(h i
r < a),
(i-2 4
r -’
Ii,” where
O&
-- 0.
=- cTzzi := o and all quantities
a-2).
are cotnputed
I !. i
(3.2)
(323) !
on the plane z = O.
1.55
Stress distribution at the edge of an equilibrium crack To develop
solutions
are considered
(3.2) and (3.3) PAPKOVITCF~(1332) functions
to equations
in the following
2/l llL: = -
(3 -
form : -cY) YZ + X
2/Luu, =
2puz
=
The representation
(3.5)
X
-
(3 -
au) Yy, + E
of displacement
given
by (3..$) lead to the following
system
of stresses :
uzz=
-(l
-2”)
(3.6) (3.7)
MINDLIK (1355s) established
observes
that. if gt5 is known. then an equation can be on the boundary z = 0. This is done by differentiating the results. to y and (3.7) with respect to x and subtracting
for 3Y&z
(3.6) with respect The following
relation
and an integration
is developed
:
gives
where an arbitrary inspection
constant
2p.U
Z =
of
-
integration
(3 -
has
been
3Yz 4v) YZ + x T +
set equal
to
zero.
b@ -3s
By
(3.10)
on z = 0. Equation (3.16) as it stands is not harmonic and does not lead to the same type of boundary conditions as found in Section 2. However, appropriate solutions
to the two problems
selection
of YZ.
MINDLIN (1955b)
posed by (3.2) and (3.3) can be obtained
has shown
that if
~,--2(l-v)~-2v~~+~~+x~~~_
v2==
-(I
by proper
LI@ -2v)Y~I~+x~-zJ-~0)
3’ Yz 5= 0. 2s bz
‘ 3\y,
av,
(3.11)
(3.12)
where V2 V, = V2 F12 = 0, then the last of equations (3.2) and (3.3) are identically satisfied on z = 0. If YZ is assumed to be a known harmonic function, then the following relations for ~Y+z and b2 a/&z2 are obtained :
Stress distribution at the edge of an equilibrium crack
156
(3.13)
(3.14)
It should and
be remarked
@ to within
that
Y, can be determined
& (z, y) + z dS (z, y).
Further
to within a function clarification
will be made later but only for the case where ul, is an axially symmetric The problem also be axially
given
by (3.2)
symmetric.
is axially
symmetric
& (z, y)
of these functions
and consequently
function. YZ1 must
We th ere f ore assume that the form for YZ1 is given as b
~~
yzl==-!.
2i_b I p
where YZ1 is antisymmetric given bv (3.2) is sought. derivative
Pl
+
(t)dt
(z +
(3.15)
itp]i
about the plane z = O, and the solution
to the problem
With YZ1 as in (3.15) its value and the value of its normal
on the plane c = 0 are as follows
:
b
Yr’
;
I
-
r
~1
(0 zg r < 21)
(1)dt
(3.16)
(t2 - r2)t
T
(0 ,Cr
b
Y, l is a function
‘l‘herefore, by assumption plane z =- 0. From
(3.13)
substitution.
and
(3.14)
values
Using equation
for
(X13),
that IS zero exterior
3YZ1/az and 32 @l/bz2
aY,l/bz
to v =: h on the
are found
by direct
is
b
3Yf/,' -__
iJZ
where
= -*-
is antisymmetric
2‘YZ1/&
(t) dt cos B 22 3Cbs [Y” + (z + itpp 1
2-‘,
3
about
the plane
z = 0.
(3.18)
One integration
gives
b
Y,’ ::
T$f ;; I p,(t) log (z
+ it + [ r2 + (z + it)“]‘}
dt +- D
(3.19)
-b
nherc
1) is an arbitrary
constant.
From
the requirement
\-anishes as r -t 00, D must be zero. The problem is completed when @’ is specified. the following
requirements
that the displacement
The function
@
must satisfy
as r 3 co :
@’ -= *= (T, z)cosf?,
a@’ 7
=
0
t1r 1,
(3.20)
Stress distribution at the edge of an equilibrium crack where the order relationship is used in the usual manner. function satisfying these relations is
157
A suitable
potential
b
@’ = ;;
GrJ-q1 (t) dt { (z
+ it) log([ rt + (2 + it)z]+ + (23 + it,} -
-b
-
where q (t) =
-
t) and a constant
q (-
[ ra + (z + i2)z]*} cos B
and logarithmic
The functions @I and b2 4p1/3zz are antisymmetric symmetric. On the plane z = 0
(3.21)
term have been omitted.
about
z = 0 while
B@/dz
is
b
32 @’
s
a!
g=-xr
(0 < r < 6)
(3.22)
r =O Therefore
(3.14)
(b < r <
may be expressed
b
co)
as
b
b
p1 (t) dt + E, (t” - r2)) b
b
Or
s
ql (4 dt (P - r2)t
==
r
+
b
d
Pl(4
dt
(t2 -
P)+
~
dr
I
+
2
(1
_
I’1
“)
where E, is an arbitrary
dt
+
E’,
r2)+
(3.24)
r
constant
is integrated
sequently performed. p, (t) with the result
(t)
(t2 -
s
7
hand side of (3.24)
(3.23)
of integration.
The first integral
by parts with the indicated
The resultant
equation
on the right-
differentiation
suh-
is then solved for q1 (t) in terms
of
b
!A@) = P,(b)
+ t
(3.25)
s
1
Equation
(3.25) determines
differentiation indicated value for @ :
@l to within an arbitrary
by (3.19)
constant,
E,.
Performing
the
and taking limits as z --f 0, we find the following
b -
To remove
all displacements
for 6 < r <
+
case.
itq*(t)dt]
o
(3.26)
03 requires
b
(3.27)
tq, (t) dt = o s 0
and it is this restriction
on q (t) that
determines
b
f.js =
! r
z
s
t2 q1 (t) dt cos
7
(t2 -
0,
r2)*
e ’
the value for E,.
(3.23)
z=O(O
Hence
< CO)
(3.29)
I,. 11. KICIIR
158 and the boundary WC observr that
condit.ions
given
by equations
(3.2)
are rompletely
satislicd.
Thcrcforc displaccmcnts YG1 and !Pb’ are axially symmrtric I’unctions. arising from an axially symmetric shear stress, ozz. will not be axially symmetric but will instead have nonsymmetrical contributions of the form given by (3.30).
where
If
one
is primarily
interested
in stresses.
(3 ~~ 4”) Y=’
then the problem
:
0.
0 (I/>;
(3.32
r/ )
r
I
is in a sense cqui\-alent to the problem dcfirrcd by (3.2) pro\.idcd that a displaccmcnt This completes the problem defined h> field given h\r (3.30) is added to thr solution. (3.2). ‘I‘o soh~ the problem satisfied :
given bv (3.3) the following
J~$’ _ {, dz ’
z
boundar\.
-.
(3.33 )
0 (0 :-_ ?’ > b)
z=
0.
c~onclitions must I,(,
0 (/I
x
r
0 (/I
ml
5.
r i.. n)
(8.3 t)
(3.3.;)
wllr~e 1 (1.) plays the role of the similar frlllction given in Se&on L’. ‘I’lic rcnutiriin~ functions Ya2 and Yb2 are given by analogy with (3.30) and are determined aft& the following
boundary
value problem
is solved
for Yz2 :
‘I’hr problems given by (3.31 ) a11t1(3.32) and those gi\.en by (3.36) and (3.37) are identical in form to the prohlrm solved in Section 2. If we d&tic
(list ribrrtion
Stress
at the edge of an
cquilibriuu~ crack
IS!)
a
s .r
sl (s) ds
L’ d
(3 - Ay)ZJa(4 = - ; -y
-___ (s2
_
t2)t
(3.4))
n
2
‘I’heac fnnctions grntial
result
shearing
-t,
7r
in the determination
stresses
in the region
1’ (s) ds
(R2-P)i
of the following
distribution
for tan-
1) < Y < u :
(3.41 )
The singularity and
I (T,
:
at 1‘ -
I/ is removed
by the following
h - sh (5) ds 2(1 --I$, .Ii (/P - sy+ = (3 - 4Y)
1)
relation
hctwccn
/i (r)
u I’ (s) ds bs (9
-
(3. 43)
t2)+ .
If ISarcnMatt’s assumptions are now applied to the stresses occurring at the edges of the crack, it 3 then seen that the same requirements arc necessary for 1 (v) as WC~Cnecessary for g (r) in the preceding section. Therefore
is determined
from (3.43).
P ” (1 -
77 (3 --llY) 1Vhrn
v)
The stress
distribution
in the region b < r bC n is
_.~~j:,e~~2)i{blogrU+(u~-il)lj
_qa’-L.‘)+
(3:L6)
i-
czl: = r,,, where
TV is a constant,
the resulting
stress
distribution
is
160
For this section the t,wo dimensional formulation as given by ~~:NGLANLI ar~l GREEN (1963) is chosen, and the solution will be brief since the development so closely parallels that of the previous sections. The basic equations for two dimensional isotropic elasticity Kivcn in GHEHN and ZERNA (1954) are
\
/Lx1-- p (UT t j/Au), =: K J2(2) -~ .O (Sj - (2 -. 2) D’ (z) u.vy ~- i o*y
2: a’ (2) -~ a’(z)
. (2)
(Z f>,,i)
tJ (2),
(4.1 j
~. 4’(z).
1
where K == 3 ~. 4v for plane strain and K ~~ (3 r,);(1 8,)for gcnrralizc71 fblanc stress. We suppose that there is D single line crack between .I* Ym-1 rr and J 0 : ihr region 15 otherwise unbounded. The crack is being opened by equal and oppositr norlnal pressure’. on g 0 the following boundary conditions must bc satisfied :
In the last of (4.1) if IJ (z) = 0, then on ~1 ”
~!a
zi P
uy
0 XI,’ have
[S”’
(2)
f?
(M : l)[Q(Z)
(;,I.
i
.ct (:,I.
(4.3)
13~ using the function
the problem defined b> 1_ OYY
,,, (T), !, -~ 0 (0 I /,(‘j s_h)
IL,,1= 0,
!I
(4.5)
0 (b . j.t.(. . 7 ,
is solved by reduction to the following solvable Abel typv t(luation :
which has for its solution
Similarly, the problem defined b>
(4.8,
Q2(z)
_ I u
P2 (t) dt -__(22 -. 12)$
This second problem is reduced to the following
Abel type ecluation
(4.9)
:
Stress distribution at the edge of an equilibrium crack
161
a
-~ K+l P
s
z- (0 dJ
(_
n(x)
=
(4.10)
(E
which has for its solution K+l -yF+5
s ~n’
(9
n
(8) ds
(4.11)
la)+
-
t The stresses on y = 0 (a -
(~1 < d) are given as m (9) (U - sz) ~2%
(4.12)
(x2 - 81)
0 a
(4.13)
Removal
of the stress singularity at x = b requires
that
b
a
s
m (4 d.3
n’
(bB=-
ds
(4.14) ’
b
0
Using Barenblatt’s
(8)
(P--
assumptions on uy as before, we take
n(x) z
$(a -
(4.15)
ISI,”
where with the use of (4.14) the constant As becomes b
AS
=
(* + ‘) 4,~ bK
s
m (4 d8
(4.16)
(bB_ga)t *
0
These values when used in (4.12) and (4.13) give the following distribution for the cohesive stresses on the tip of the crack : b
(x2-
w (aa- w*+
(& @‘d _
When a constant normal pressure is assumed, the distribution
b8)t -
x’)i (x2 -
bz)’
I> .
(4.17)
of cohesive stresses is
u”*=-~po{;-3(~)+[(~~+(l-~)logp&f..q}. (4.13)
Therefore, the distribution of the cohesive stresses for the two dimensional crack is inherently the same as that for the penny-shaped crack.
h_
/1,:ytrlf
_
(.i.i!
1 0
where G (t) is the intcnsit>. of tllc forces of cohesion Il<‘i\r tllc, c&c” of‘ thch c’racah (11 < t’ :: (I). and t is the dist,ante from the edge of the c~:tc~k. A- is also writtrtlx :I\
(.i..i)
The distribution cat’ cohesive stresses for isolated equilibrittrn cracks determined by a semi-inyerst method within the framework of classical
has I~WI elasticit!.
are used to obtain the distribution of cohesi1.c theory. li?-:-:.‘,~L~I t’s hypotheses stresses . These stresses. which are assumed to act over a specified region near ttic cdgc of the crack. arc of’ order (u/d) w 1&en compared with the applied stress3 For the three problems consideret 1 fix both two and three dimensiona problems.
tlw stress distrihutidns IIowe\cr, upon the
near the edge of the crack
(/I .-- 1’ -- (I) are nearly
identical.
thcb joining of the cracks depends different for cac4r of the three profor the blems. Using Rarenblatt’s modulus of cohesion hF. \-alucs are obtained stresses act in terms of K and radial distance (length) over which the cohesive the applied
THE author
the constant di associated with elastic constants and is therefore
load.
wishes to thank Professor Ii. 1). Mindlin for his many helpful discussions and partieularlp for pointing out (3.1 I) and (3.12) which allowed an ea..y solution to the problem in Stcztion 3. The ORire of Naval Iteseareh supported this work.
Strew distribution at the edge of an equilibrium crack
163
REFERENCES BAEENBLA~, G. I. COLLINS, w. D., ENOLAND, A. H. and G-N, A. E. GOODXAN, L. E. GEEEN, A. E. and <NA, KEEX,
w.
L. M.
bbDLIN, R. D. MINDLIN, R. D. PAFXOVITC~~,P. F. SNEDDON, I. N.
1962 1962
Advanca in Applied Mechanics, Vol. 2 Academic Press. Rae. ZZoy.Sot. A266,859
lQ68 1962
Proc. Camb. Phil. Sot. 59, 489. J. A@. Mcch. 29, 515.
1954 lQ64 1955a 1955b 1982 1946
Theoretical Elaatkity. Oxford University Press. QJMAM (in print). Z’rac. 1st i%&u&m Conferenceon Solid Mechanics, p. 56. Unpublished. Zao. Akud. Nauk SSR, Phys.-Maih. Ser. 10, 1425. PToe. Roy sot. A187,22Q.