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The distribution of stress near the tip of a radial crack at the edge of a circular hole
Inr. .I. Engng Sci.. 1973, Vol. 11, pp. I I@-
1195.
Printed in Great Britain
Pergamon Press.
THE DISTRIBUTION OF STRESS NEAR THE TIP OF A RADIAL CRACK AT THE EDGE OF A CIRCULAR HOLE? J. TWEED The University of Glasgow, Glasgow, Scotland and D. P. ROOKE The Royal Aircraft Establishment, Famborough,
Hants
(Communicated by I. N. SNEDDON) Abstract- A Mellin transform technique is used to find an integral equation the solution of which is related to the stress intensity factor and the formation energy of a crack at the edge of a circular hole in an infinite elastic solid. The biaxial loading case is considered in detail and numerical results given.
I. INTRODUCTION
of determining the distribution of stress near the tip of a crack which originates at the edge of a circular hole in an infinite elastic solid appears to have been considered first by Bowie[ 11 who solves it by using a complex mapping technique. The results given by Bowie are not very accurate, so in this paper we show that the stress intensity factor and crack energy are related to the solution of a Fredholm equation and may therefore be calculated to a high degree of accuracy. We shall assume that the problem is to be solved under the conditions of plane strain and that the crack and the hole are defined, in plane polar coordinates (r, 8), by the relations R s r s Rb, 19= 0 and 0 S r c R, 0 s 8 s 2 7~respectively. THE PROBLEM
-------
V=O
Y--K
Fig. 1.
If the loading is symmetric about the plane of the crack the problem may be reduced to that of finding a solution of the equations of elasticity for the region R < r < w, 0 < d < W,which is such that (I) at infinity the stresses u,,(r, a), u,&, ments ur(r, 6), u$(T, 29)areO(r-‘),
6), cr&r, 8) are O(r+) and the displace-
tThis paper was prepared as a part of the work of the Applied Mathamatics Research Group at North Carolina State University through the grant AF-AFOSR-69-1779 and is under the joint sponsorship of AFOSR, AR0 and ONR through the Joint Services Advisory Group. 1185
1186
(2) (3) (4) (5) (6)
J. TWEED
and D. P. ROOKE
a$%3(r,O)=O,R < F-c a, V.,$(F-,?t) = u+(r, 77) = 0, R < r < co, arr(R, 8) = 0,o < 3 < lr, cr@(R, 4 = 0,o < 6 < 7r, o,s,j(r, 0) =-f(r), R < Y < Rb,
(7) u*(r. 0) = 0, Rb < r < m,
and (8)
limit
r-R+
au89(r’ O)< ar
W
*
2. REDUCTION
OF THE
PROBLEM
TO AN INTEGRAL
EQUATION
In order to find a suitable representation for the stresses and displacements in the problem set out above we shall begin by superimposing the solutions of problems 1 and 2 below. Problem I. Find a solution of the equations of elasticity for the region R < r < ~0, 0 < 19 < P. which is such that (a) at infinity the stresses are O(@) and the displacements are O(r-‘), (b) c,.+(r, 0) = u*(r, 0) = 0, R < r < ~0, and (c) cT,$(r, n) = uB(rr z) =O,R
zY+~$~ [cnr-n+d,r-“+2]cosn8,
and the corresponding
stresses and displacements
o,, (r, 6) = coF2 - i,
[n(n+l)c,r-n-2+
o,~(r, 6) =-i
[n(n+ VI=1
vh9(r, 7%=-car++
5
[n(nt
(2.2)
l)d,r-n]sinnS,
l)~,r-~-~+
I$-?) -ccor~1+-c1r-2cos U,(Y,6) = ---jy-
by
(n+2)(n-l)d,r~n]c0sn~
l)c,r-n-2+n(n-
f?=l
(2.1)
(n-2)(n-
(2.3)
(2.4)
l)d,r-n]cosk?,
6-i-E[r~c,P-~+
(n+2-4~)dnr-n+‘]cos
n6 , I
It=?2
(2.5)
and u19(r, 8) = 9
[nc,r-n-1+
cIr-zsin 8+i
1
n=2
(n-4+4r))d,r-n+1]sinn*
I
,
(2.6)
where E is the Young’s modulus and q is the Poisson’s ratio of the material. problem 2. Find a solution of the equations of elasticity, for the half-plane 0 s r < m, 0 < 6 < T, which is such that (a) at infinity the stresses are 0(r-2) and the disptacements are O(r-l), (b) at the origin the stresses and displacements are bounded, (c) ~,.~(r, 0) = 0,O G r < 03,and(d) u,+(r, V) = ub(r, 7~) = 0,O s r < ~0. By utilising the properties of the Me&n transform (e.g. see Tranter [3]) it can be
1187
Stress distribution at tip of crack at edge of a circular hole
shown that the solution of this problem may be written in the form urr
(r, 6) = re2A-’
cos (a-P)(s+~)-((s+~)cos
cr,*
(r, 8) = r-2dZ-1
&ps+2)
ugo (r, 8) = r-2.K1
h(r,
(sin (I%GT)(s+~)-sin
1+7&l 6) = 7
[
(zY-7r)s);r
2(S+A1(~~in,,((s+2)cos
1,
(t+-n)(s+2));r
COS(6-7r)s-sC0S
(~-7T)s
-((s+4-47))cos
1+9&l [
2(s+A1(fs)in?rs((S+2)sin
, (2.7)
(2.8)
1, 1,
(S-7r)(s+2));r
and ufi(r, 6) =x
1
(S-7~)s);r
(2.10)
(G-r)S
(8-7r)(s+2)):r].
- (s-224q)sin
(2.9)
(2.11)
where AZ’-’is the inverse Mellin transform and - 1 < Re(s) < 0. Superimposing the solutions of these two problems, we obtain a solution of the equations of elasticity for the region R < r < 03, 0 < 8 < v which automatically satisfies conditions (l), (2) and (3) and which is such that u,,(r,
8) =cgrp2-5
[n(n+l)~,r-~-~+(rz+2)(n--l)d,r-“]cosn29 n=1
+ r-2dh-1
[
(6-7F)(S+2)-
&((sf4)COS
(st2)COS
(a-r)s);r],
(2.12) a,,(r,
9) = - 5
[n(n+
1)c,r-n-2+rz(rz-
1)&r-“Isinn
n=1
r-2di-1
+
[
(“2fs~~~~‘)(sin
o60(r, S) =-cOrm2+~
(8-?r)s);r],
(8-T)(s+~)-sin
(2.13)
[n(n+1)c,r-“-2+(~-2)(rz-l)d,r-“]cosn8 ?l=l
+ r-2A?-1 &@+a
b(r, 6)
=E
1+rl -ccor~‘+clr-2c0S [ 1
1+rl&&1 rE
Cos (8--T)S--SCOS
(8-v)(S+2));r
a
S+x
(nc,r-n-1+(n+2-4)d,r-“-1)cosn9It=2
2(S+A~~~in,,((s+2)cos - (s+4-4v)cos
1 ,
(2.14)
1
(6-7T)s (79-r)(s+2)):r],
(2.15)
1188
J. TWEED
and D. P. ROOKE
and
1+7) clr-* sin 6+ 2 (nc,r-“-‘+ 49@,9) = -jy[ II=2
+i3&1
(n-4+4q)d,r-~+1)sin
2(S$l(yjin,,((“+2)sin
rE
- (s-224q)sin
(S-r)(s+2)
n6
1
(@-a)s
):r],
(2.16)
where - 1 < Re(s) < 0. The complete solution of the problem may now be obtained by choosing the unknown function A (s) and the unknown sequences (c,) and (d,) in such a way that the remaining boundary conditions are satisfied. From (2.14) and (2.16) we see that conditions (6) and (7) will be satisfied if A (s) is a solution of the dual equations dwl[A(s)cot
m; r] = -r”f(r)
AP[(l-ts)-‘A(s);r] where-
1 < Re(s)
-?F(r),
R < r < Rb Rb
=o.
(2.17)
< 0 and
F(r) =-c&2+$
Il=l
[n(n+1)C,rn-2+(,-2)(n-l)d,r-“].
(2.18)
If we now assume that A (s) may be written in the form A (s) = I,“” p(t)P+‘dt
(2.19)
we find that (see Tweed [4]) “&‘[A(s)(l+s)-‘:r]
O.RbGrCm
=r
(2.20) / ,? p(t) dt,
R SrSRb
and AC’[A (s)cot 7~s;r] = i
Rbtp(t> & I R t-r
(2.21)
and hence that the equations (2.17) will be satisfied if tp(t) dt -1 Rbp=-rf(r) t-r = IR
-r&‘(r),
R < r < Rb.
(2.22)
The equation (2.22) is well known and Tricomi[5] has shown that its solution is given by @(t) =
f(&$)“’ r
~~)““yf(y)(y)
where C is an arbitrary constant.
dy+ [(Rb_t)&),~,”
In order to determine C, we make use of condition
Stress dist~bution at tip of crack at edge of a circular hole
1189
(8) which together with (2.16) and (2.20) implies that hm,t p(r) exists and hence that C = 0. It follows that p (t) is given by the expression (2.23) Similarly, on applying conditions (4) and (5) we find that
=dd-’ i
[n(n+
I
cos (79-v)(s+2)-(s+2)cos
&((s+4)
l)c,R-n+n(n-
(z?-r)s);R
I
l)d,,R-n*2]sinn6
fl=l
(sin (+-n)(s+2)-sin where0
(Ir,-n)s>;R
1
< 8 < P, and hence that
1 (s+z2)(S+2+n);R A (s) (8+ 2)
1 1
;R
,
(2.24)
vn 3 1
(2.25)
and d -_ 2R”-2M-I n P
A(s) @+2) (s+n)([s+2]“-n”);R
1
Fn a 2
(2.26)
where-l < Re(s) < 0. Substituting from (2.19) into (2.24) through (2.26) and working out the inverse Mellin transforms, we find that t-‘p(t)
c, =
dt,
(2.27)
E 6”rp(r~[~(~~+2-~(~)n}dt, Iz 2
1
(2.28)
and &=s
l
Ip(j){s($)‘-(;!)nH)
dt, n 3 2
(2.29)
and hence that p(j)K(r,
(2.30)
8) dt,
where K(r
,
j) = R2(R2-f2)2_f(R2-t2) t(R2-
rt)”
At this point we find it convenient
(RZ-a)”
R2-t2 Rzt_rt+F-;+
1
(2.31)
to introduce the function P(r) which is defined
1190
J. TWEED
and D. P. ROOKE
by the equation P(t) = [(r-R)(Rb-~tf]“~p(r).
(2.32)
On substituting from (2.30) into (2.23) and taking account of (2.32) we see that P(t) must satisfy the integral equation lib
P(f) -
Pf )~(t,P) ~(p_~)(Rb-p)]l,2dp=~(~).
R
i
(2.33)
where S(t)
+$”
*‘2xfbY& y-t
R
(2.34)
’
and Rh
(2.35)
M(t,p) =y i
R
If we now substitute from (2.3 1) into (2.35) we find thatM(t, form RZ/p) +p-f(R2-p2)
M(t, p) = ~(p-4Rw?‘-p2)2J&. +J,(t,
0) -J,(t,
p) may be written in the
[52(&O) -.&(t,
R21p)]
WP)),
(2.36)
where (2.37) Osx
By using the result
-l,Rb
RCxcRb
and the fact that Jl(f, x) = (x-t)-‘[tJ(t) J~(t,x)
=-&J1(~,x)
--xl(x)], and J3(t,x)
=-+$Jz(t,x)
it follows that M (t, p) is given by M(t
p>
3
=
(t-R)R2Wlrt
p2Y
(b- l)t 2R(R2-pf)2(bp-R)*‘2(p-R)5’2
t(bp-R)“” (b-1)[(1+3b)p-4R] -8p(R2-pt)(bp-R)3’2(p-R)5’2-p(R2-pt)3(p-R)”2
I
1191
Stress distribution at tip of crack at edge of a circular hole
+
(t--R)W-P7
1
pt(bp-R)“2
(R2-pt)Q-R)l’2
Irt
/g/2
R(b- l)P
--
-2(R2-pt)(bp-R)“2(p-R)3’2
pt I
(t-R)RZ(bp--R)l’2 + m(R2-pt) (p-R)“2’ 3. THE
STRESS
(2.38)
INTENSITY
FACTOR
AND
THE
CRACK
ENERGY
We now show that the stress intensity factor K and the crack energy W, defined by K=-liliiJ
[2(Rb-r)]“22(1~112)
blip),
(3.1)
and w-j-,”
o+(r, O)uti(r, 0) dr
(3.2)
respectively, are simply related to the function P(t) introduced in the last section. By substituting from (2.32) into (2.20) and taking account of (2.16) we see that &(r,O)
=-2(1E’2)
1:” [(r_$(($~f-t)]l,2,R
s r s Rb.
(3.3)
It now follows that (3.4) and
w=_-2(l-~2)
P(t) dt [(Rb-t)(t-R)]“2
E 4. SPECIAL
CASE-
BIAXIAL
(3.5) TENSION
In this section we consider the special case in which the loading consists of a uniform tension T perpendicular to the plane of the crack and a uniform tension CYT parallel to the plane of the crack. The loading functionf(r) appropriate to this case is given (Sokolnikoff[6], p. 29 1) by f(r) = TM(r/R)
+ofi(rlR)l
(4.1)
wheref, (r) = i(2 + rp2 + 3rm4) andf, (r) = f(r-” - 3rm4). It follows that we can put P(t) = PI(t) +aP,(t)
(4.2)
where Pi(t) is the solution of the equation Rb pi(t)-
I
Pibw(t, 1 [(Rb_p)(p_PR)I"Zdp=Si(t),i=172
R
(4.3)
with S.(t) 1
=
T(t--RI 1:
and M(t, p) given by (2.38).
it
ys)1'2yedy,
i=
1, 2
(4.4)
1192
J. TWEED
and D. P. ROOKE
By making the change of variables I = RT and p = RCT,and putting qi(T) zf’i(R~)/RT,i= we find that
(fl)Ml(T, I * qi
1, 2
(4.5)
a)
Q(T)- 1 [(b_cr)(~_l)]1,2d~=Si(T) where
(4.6)
(4.7) and M, (7, o) = M(R7, Rv). Performing the integrations in (4.7) gives (17b”-6b-3)_3(b8rb312
1) -- 3b”2 2r2b112 73
(4.9)
and 3b1’2+3(b- 1) + (b2-6b-3) 2T2b’12
(4.9)
Jr
If K,, and W,, are the stress intensity factor and crack energy respectively of a Griffith crack of length 2R(b- 1) in an infinite elastic sheet which is subject to a uniform all-round tension T, then K0 = T[R(b1)]1’2 (4.10) and w = n(l-v2)T2R2(b1)2 (4.11) 0 E The stress intensity factor can be written in the form K/K, = KJKo + aK21Ko where, from (3.4), (3.5) and (4. lo), we have KJKo =-21’2qi(b)/(b-
l), i= 1,2.
Similarly, on substituting the expressions forJ(r) we discover that the energy can be written as
(4.12)
into (4.1) and (3.5) and integrating,
(4.13)
wlwo=w,/wO+a(W2/WO+W3/WO)+a2W~/WO
where W,lW,=-
w,/w,=-
l
a(b-I)”
m(b-
1
’ q1(~)(27-v’-~-3)
I
1
(v
* g1(~)(~-3-~-1)
dr 1)]“2
'
dr
1)2 I 1 [(b-7)(7-l)]1’2r
1 w31wO=-n(b_1)2
[(b--7)
I
* q2(~)(2~-~-1-~-3) [(b-7)(7-
1
d,r 1)]1’2 ’
(4.14)
(4.15)
(4.16)
Stress distribution at tip of crack at edge of a circular hole
1193
and ’ g2(7) (r-“-~-l)
dr
w4~wo=-7T(bL)~ I [(b-7)(7-1)]1’2’
(4.17)
1
it
From Betti’s Integral Theorem
follows (see e.g. Sokolnikoff [6], p. 39 1) that W2 = W3.
5. NUMERICAL
PROCEDURE
By using Gauss-Chebyshev quadrature by the simultaneous linear equations
%(jk)
AND
RESULTS
we can replace the integral equations (4.6)
-a m=1 i 41(hd~l(fk, jrn) = Si(fk),
(5.1)
where
(5.2) k= 1, 2,3,. . . , n; i = 1,2. These equations may be solved for the qi (tk) and the quantities Ki/Ko calculated [see (4.2 l)] from the formula Ki/KO=-
(fTlj -[q(b) +: nl=l e qi(jm)Ml(b, I,)),i=192. Table
1.The variation of KJK, and W,/W, with a/R
a/R
K,I&
0.01 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.25 0130 040 0.50 0.75 1.00 1.50 2.00 3.00 4.00 5.00 7.00 90-I
3.291 3.223 3.095 2.978 2.870 2.771 2.679 2.594 2.515 2442 2.373 2.221 2.092 1.884 1.727 1,464 1.306 1.127 1.030 0.930 0.877 0.845 0.808 0.787
-
&I& - 1.079 - 1.040 - 0.966 - 0.900 - 0.839 - 0.783 - 0.732 - 0.685 - 0642 - 0603 - 0.566 - 0.487 -0.421 - 0.320 - 0.247 -0.137 -0.080 -0.030 -0.010 0.002 0.005 0.005 om4 0@04
WI/W0 5.494 5.342 5.060 4.803 4.568 4.354 4.157 3.975 3.808 3.653 3.509 3.192 2.925 2.502 2.185 1.662 1.349 1mO 0.814 0.624 0.528 0.471 0407 0.371
w2wcl - 1.812 - 1.743 - 1.616 - 1.502 - 1.399 - 1.306 - 1.222 - 1.145 - 1.075 - 1.011 - 0.953 - 0.826 - 0.723 - 0.565 - 0.452 -0.281 -0.190 -0.101 -0.061 -0.028 -0.015 -o&W -0m4 - 0.002
w4wcl 0.597 0.568 0.516 0.470 0.429 0.392 0.360 0.331 0.305 0.282 0.260 0.216 0.181 0.131 0.098 0.053 0.032 0.015 OW8 0.004 0*002 O@Ol om1 0.000
1194
J. TWEED
Fig. 2. The variation
and
D. P. ROOKE
of K/K,, with a/R for several
values of the parameter
a.
Similarly, the energies Wj/ W, may be calculated from
Wl/WO = -,g
I)2
E 41(r,)(an-r~1-t,3),
m=1
i 41(bz)(Gt3-G1)7 w2/wo=-n(bi*)2 T7l=l
w3/wo=
-n(b! I)2
and
i
92(fm)m?n-Gi1-G13)
m=1
w’/wO=-,(b!*)~~
%(fm)(f-z-f>). WI=1
Table 1 shows the variation of KJK,, and WJW,, i = 1, 2; j = 1, 2, 3, 4, with a/R and Fig. 2 shows the variation of K/K0 with a/R for several values of the parameter CL REFERENCES
[ 11 0. L. B0WIE.J.
Mark Phys. 35,60 E. A. PHILLIPS
(1956). and C. H. TSAO.
[2] A. J. DURELLI. Analysis of Stress and Strain. (1958). [3] C. J. TRANTER, Q. J. Math. appl. Marh 1, 125 ( 1948). [4] J. TWEED, Glasgow Math. J. 14,65 (1973). [5] F. G. TRICOMI, Q.J. Math. 2, 199 (195 I). [6] I. S. SOKOLNIKOFF,Mathematical Theory ofElasticity. McGraw-Hill ( 1956). [7] H. K. KUTTER,fnt.J. Fract. Mech. 6,233 (1970). [8] J. TWEED, Tech Rpr. PSR-9919, N.C. State University, Raleigh. (Received
9 February
1972)
McGraw-Hill
Stress distribution at tip of crack at edge of a circular hole
1195
Rksumk-
Une technique de transformation de Mellin est utiliste pour trouver une equation integrale dont la solution est lide au facteur d’intensitt de contrainte et a l’energie de formation d’une crique sur le bord d’un trou circulaire dans un solide elastique infini Le cas d’une charge bi-axiale est considere en detail, et des resultats numeriques sont donnes. ZusammenfassungEs wird eine Mellin-Transformethode verwendet, urn eine Integralgleichung zu finden. deren Losung auf den Spannungsintensitatsfaktor und die Bildungsenergie eines Risses an der Kante eines kreisformigen Loches in einem unendlichen elastischen Festkorper Bezug hat. Der Fall zweiachsiger Last wird in Einzelheiten behandelt un numerische Resultate werden gegeben.
Sommario-La tecnica di trasformazione Mellin viene usata per trovare un’equazione integrale la cui soluzione sia relativa al fattore d’intensita della sollecitazione e all’energia di formazione di una incrinatura sull’orlo di un foro circolare in un solid0 elastic0 infinito. II case di carico biassiale viene considerato in dettaglio e vengono offerti risultati numerici. A~CTWKT peIIIUikiC TpeWiHbI CMOTPtX
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1195.
Printed in Great Britain
Pergamon Press.
THE DISTRIBUTION OF STRESS NEAR THE TIP OF A RADIAL CRACK AT THE EDGE OF A CIRCULAR HOLE? J. TWEED The University of Glasgow, Glasgow, Scotland and D. P. ROOKE The Royal Aircraft Establishment, Famborough,
Hants
(Communicated by I. N. SNEDDON) Abstract- A Mellin transform technique is used to find an integral equation the solution of which is related to the stress intensity factor and the formation energy of a crack at the edge of a circular hole in an infinite elastic solid. The biaxial loading case is considered in detail and numerical results given.
I. INTRODUCTION
of determining the distribution of stress near the tip of a crack which originates at the edge of a circular hole in an infinite elastic solid appears to have been considered first by Bowie[ 11 who solves it by using a complex mapping technique. The results given by Bowie are not very accurate, so in this paper we show that the stress intensity factor and crack energy are related to the solution of a Fredholm equation and may therefore be calculated to a high degree of accuracy. We shall assume that the problem is to be solved under the conditions of plane strain and that the crack and the hole are defined, in plane polar coordinates (r, 8), by the relations R s r s Rb, 19= 0 and 0 S r c R, 0 s 8 s 2 7~respectively. THE PROBLEM
-------
V=O
Y--K
Fig. 1.
If the loading is symmetric about the plane of the crack the problem may be reduced to that of finding a solution of the equations of elasticity for the region R < r < w, 0 < d < W,which is such that (I) at infinity the stresses u,,(r, a), u,&, ments ur(r, 6), u$(T, 29)areO(r-‘),
6), cr&r, 8) are O(r+) and the displace-
tThis paper was prepared as a part of the work of the Applied Mathamatics Research Group at North Carolina State University through the grant AF-AFOSR-69-1779 and is under the joint sponsorship of AFOSR, AR0 and ONR through the Joint Services Advisory Group. 1185
1186
(2) (3) (4) (5) (6)
J. TWEED
and D. P. ROOKE
a$%3(r,O)=O,R < F-c a, V.,$(F-,?t) = u+(r, 77) = 0, R < r < co, arr(R, 8) = 0,o < 3 < lr, cr@(R, 4 = 0,o < 6 < 7r, o,s,j(r, 0) =-f(r), R < Y < Rb,
(7) u*(r. 0) = 0, Rb < r < m,
and (8)
limit
r-R+
au89(r’ O)< ar
W
*
2. REDUCTION
OF THE
PROBLEM
TO AN INTEGRAL
EQUATION
In order to find a suitable representation for the stresses and displacements in the problem set out above we shall begin by superimposing the solutions of problems 1 and 2 below. Problem I. Find a solution of the equations of elasticity for the region R < r < ~0, 0 < 19 < P. which is such that (a) at infinity the stresses are O(@) and the displacements are O(r-‘), (b) c,.+(r, 0) = u*(r, 0) = 0, R < r < ~0, and (c) cT,$(r, n) = uB(rr z) =O,R
zY+~$~ [cnr-n+d,r-“+2]cosn8,
and the corresponding
stresses and displacements
o,, (r, 6) = coF2 - i,
[n(n+l)c,r-n-2+
o,~(r, 6) =-i
[n(n+ VI=1
vh9(r, 7%=-car++
5
[n(nt
(2.2)
l)d,r-n]sinnS,
l)~,r-~-~+
I$-?) -ccor~1+-c1r-2cos U,(Y,6) = ---jy-
by
(n+2)(n-l)d,r~n]c0sn~
l)c,r-n-2+n(n-
f?=l
(2.1)
(n-2)(n-
(2.3)
(2.4)
l)d,r-n]cosk?,
6-i-E[r~c,P-~+
(n+2-4~)dnr-n+‘]cos
n6 , I
It=?2
(2.5)
and u19(r, 8) = 9
[nc,r-n-1+
cIr-zsin 8+i
1
n=2
(n-4+4r))d,r-n+1]sinn*
I
,
(2.6)
where E is the Young’s modulus and q is the Poisson’s ratio of the material. problem 2. Find a solution of the equations of elasticity, for the half-plane 0 s r < m, 0 < 6 < T, which is such that (a) at infinity the stresses are 0(r-2) and the disptacements are O(r-l), (b) at the origin the stresses and displacements are bounded, (c) ~,.~(r, 0) = 0,O G r < 03,and(d) u,+(r, V) = ub(r, 7~) = 0,O s r < ~0. By utilising the properties of the Me&n transform (e.g. see Tranter [3]) it can be
1187
Stress distribution at tip of crack at edge of a circular hole
shown that the solution of this problem may be written in the form urr
(r, 6) = re2A-’
cos (a-P)(s+~)-((s+~)cos
cr,*
(r, 8) = r-2dZ-1
&ps+2)
ugo (r, 8) = r-2.K1
h(r,
(sin (I%GT)(s+~)-sin
1+7&l 6) = 7
[
(zY-7r)s);r
2(S+A1(~~in,,((s+2)cos
1,
(t+-n)(s+2));r
COS(6-7r)s-sC0S
(~-7T)s
-((s+4-47))cos
1+9&l [
2(s+A1(fs)in?rs((S+2)sin
, (2.7)
(2.8)
1, 1,
(S-7r)(s+2));r
and ufi(r, 6) =x
1
(S-7~)s);r
(2.10)
(G-r)S
(8-7r)(s+2)):r].
- (s-224q)sin
(2.9)
(2.11)
where AZ’-’is the inverse Mellin transform and - 1 < Re(s) < 0. Superimposing the solutions of these two problems, we obtain a solution of the equations of elasticity for the region R < r < 03, 0 < 8 < v which automatically satisfies conditions (l), (2) and (3) and which is such that u,,(r,
8) =cgrp2-5
[n(n+l)~,r-~-~+(rz+2)(n--l)d,r-“]cosn29 n=1
+ r-2dh-1
[
(6-7F)(S+2)-
&((sf4)COS
(st2)COS
(a-r)s);r],
(2.12) a,,(r,
9) = - 5
[n(n+
1)c,r-n-2+rz(rz-
1)&r-“Isinn
n=1
r-2di-1
+
[
(“2fs~~~~‘)(sin
o60(r, S) =-cOrm2+~
(8-?r)s);r],
(8-T)(s+~)-sin
(2.13)
[n(n+1)c,r-“-2+(~-2)(rz-l)d,r-“]cosn8 ?l=l
+ r-2A?-1 &@+a
b(r, 6)
=E
1+rl -ccor~‘+clr-2c0S [ 1
1+rl&&1 rE
Cos (8--T)S--SCOS
(8-v)(S+2));r
a
S+x
(nc,r-n-1+(n+2-4)d,r-“-1)cosn9It=2
2(S+A~~~in,,((s+2)cos - (s+4-4v)cos
1 ,
(2.14)
1
(6-7T)s (79-r)(s+2)):r],
(2.15)
1188
J. TWEED
and D. P. ROOKE
and
1+7) clr-* sin 6+ 2 (nc,r-“-‘+ 49@,9) = -jy[ II=2
+i3&1
(n-4+4q)d,r-~+1)sin
2(S$l(yjin,,((“+2)sin
rE
- (s-224q)sin
(S-r)(s+2)
n6
1
(@-a)s
):r],
(2.16)
where - 1 < Re(s) < 0. The complete solution of the problem may now be obtained by choosing the unknown function A (s) and the unknown sequences (c,) and (d,) in such a way that the remaining boundary conditions are satisfied. From (2.14) and (2.16) we see that conditions (6) and (7) will be satisfied if A (s) is a solution of the dual equations dwl[A(s)cot
m; r] = -r”f(r)
AP[(l-ts)-‘A(s);r] where-
1 < Re(s)
-?F(r),
R < r < Rb Rb
=o.
(2.17)
< 0 and
F(r) =-c&2+$
Il=l
[n(n+1)C,rn-2+(,-2)(n-l)d,r-“].
(2.18)
If we now assume that A (s) may be written in the form A (s) = I,“” p(t)P+‘dt
(2.19)
we find that (see Tweed [4]) “&‘[A(s)(l+s)-‘:r]
O.RbGrCm
=r
(2.20) / ,? p(t) dt,
R SrSRb
and AC’[A (s)cot 7~s;r] = i
Rbtp(t> & I R t-r
(2.21)
and hence that the equations (2.17) will be satisfied if tp(t) dt -1 Rbp=-rf(r) t-r = IR
-r&‘(r),
R < r < Rb.
(2.22)
The equation (2.22) is well known and Tricomi[5] has shown that its solution is given by @(t) =
f(&$)“’ r
~~)““yf(y)(y)
where C is an arbitrary constant.
dy+ [(Rb_t)&),~,”
In order to determine C, we make use of condition
Stress dist~bution at tip of crack at edge of a circular hole
1189
(8) which together with (2.16) and (2.20) implies that hm,t p(r) exists and hence that C = 0. It follows that p (t) is given by the expression (2.23) Similarly, on applying conditions (4) and (5) we find that
=dd-’ i
[n(n+
I
cos (79-v)(s+2)-(s+2)cos
&((s+4)
l)c,R-n+n(n-
(z?-r)s);R
I
l)d,,R-n*2]sinn6
fl=l
(sin (+-n)(s+2)-sin where0
(Ir,-n)s>;R
1
< 8 < P, and hence that
1 (s+z2)(S+2+n);R A (s) (8+ 2)
1 1
;R
,
(2.24)
vn 3 1
(2.25)
and d -_ 2R”-2M-I n P
A(s) @+2) (s+n)([s+2]“-n”);R
1
Fn a 2
(2.26)
where-l < Re(s) < 0. Substituting from (2.19) into (2.24) through (2.26) and working out the inverse Mellin transforms, we find that t-‘p(t)
c, =
dt,
(2.27)
E 6”rp(r~[~(~~+2-~(~)n}dt, Iz 2
1
(2.28)
and &=s
l
Ip(j){s($)‘-(;!)nH)
dt, n 3 2
(2.29)
and hence that p(j)K(r,
(2.30)
8) dt,
where K(r
,
j) = R2(R2-f2)2_f(R2-t2) t(R2-
rt)”
At this point we find it convenient
(RZ-a)”
R2-t2 Rzt_rt+F-;+
1
(2.31)
to introduce the function P(r) which is defined
1190
J. TWEED
and D. P. ROOKE
by the equation P(t) = [(r-R)(Rb-~tf]“~p(r).
(2.32)
On substituting from (2.30) into (2.23) and taking account of (2.32) we see that P(t) must satisfy the integral equation lib
P(f) -
Pf )~(t,P) ~(p_~)(Rb-p)]l,2dp=~(~).
R
i
(2.33)
where S(t)
+$”
*‘2xfbY& y-t
R
(2.34)
’
and Rh
(2.35)
M(t,p) =y i
R
If we now substitute from (2.3 1) into (2.35) we find thatM(t, form RZ/p) +p-f(R2-p2)
M(t, p) = ~(p-4Rw?‘-p2)2J&. +J,(t,
0) -J,(t,
p) may be written in the
[52(&O) -.&(t,
R21p)]
WP)),
(2.36)
where (2.37) Osx
By using the result
-l,Rb
RCxcRb
and the fact that Jl(f, x) = (x-t)-‘[tJ(t) J~(t,x)
=-&J1(~,x)
--xl(x)], and J3(t,x)
=-+$Jz(t,x)
it follows that M (t, p) is given by M(t
p>
3
=
(t-R)R2Wlrt
p2Y
(b- l)t 2R(R2-pf)2(bp-R)*‘2(p-R)5’2
t(bp-R)“” (b-1)[(1+3b)p-4R] -8p(R2-pt)(bp-R)3’2(p-R)5’2-p(R2-pt)3(p-R)”2
I
1191
Stress distribution at tip of crack at edge of a circular hole
+
(t--R)W-P7
1
pt(bp-R)“2
(R2-pt)Q-R)l’2
Irt
/g/2
R(b- l)P
--
-2(R2-pt)(bp-R)“2(p-R)3’2
pt I
(t-R)RZ(bp--R)l’2 + m(R2-pt) (p-R)“2’ 3. THE
STRESS
(2.38)
INTENSITY
FACTOR
AND
THE
CRACK
ENERGY
We now show that the stress intensity factor K and the crack energy W, defined by K=-liliiJ
[2(Rb-r)]“22(1~112)
blip),
(3.1)
and w-j-,”
o+(r, O)uti(r, 0) dr
(3.2)
respectively, are simply related to the function P(t) introduced in the last section. By substituting from (2.32) into (2.20) and taking account of (2.16) we see that &(r,O)
=-2(1E’2)
1:” [(r_$(($~f-t)]l,2,R
s r s Rb.
(3.3)
It now follows that (3.4) and
w=_-2(l-~2)
P(t) dt [(Rb-t)(t-R)]“2
E 4. SPECIAL
CASE-
BIAXIAL
(3.5) TENSION
In this section we consider the special case in which the loading consists of a uniform tension T perpendicular to the plane of the crack and a uniform tension CYT parallel to the plane of the crack. The loading functionf(r) appropriate to this case is given (Sokolnikoff[6], p. 29 1) by f(r) = TM(r/R)
+ofi(rlR)l
(4.1)
wheref, (r) = i(2 + rp2 + 3rm4) andf, (r) = f(r-” - 3rm4). It follows that we can put P(t) = PI(t) +aP,(t)
(4.2)
where Pi(t) is the solution of the equation Rb pi(t)-
I
Pibw(t, 1 [(Rb_p)(p_PR)I"Zdp=Si(t),i=172
R
(4.3)
with S.(t) 1
=
T(t--RI 1:
and M(t, p) given by (2.38).
it
ys)1'2yedy,
i=
1, 2
(4.4)
1192
J. TWEED
and D. P. ROOKE
By making the change of variables I = RT and p = RCT,and putting qi(T) zf’i(R~)/RT,i= we find that
(fl)Ml(T, I * qi
1, 2
(4.5)
a)
Q(T)- 1 [(b_cr)(~_l)]1,2d~=Si(T) where
(4.6)
(4.7) and M, (7, o) = M(R7, Rv). Performing the integrations in (4.7) gives (17b”-6b-3)_3(b8rb312
1) -- 3b”2 2r2b112 73
(4.9)
and 3b1’2+3(b- 1) + (b2-6b-3) 2T2b’12
(4.9)
Jr
If K,, and W,, are the stress intensity factor and crack energy respectively of a Griffith crack of length 2R(b- 1) in an infinite elastic sheet which is subject to a uniform all-round tension T, then K0 = T[R(b1)]1’2 (4.10) and w = n(l-v2)T2R2(b1)2 (4.11) 0 E The stress intensity factor can be written in the form K/K, = KJKo + aK21Ko where, from (3.4), (3.5) and (4. lo), we have KJKo =-21’2qi(b)/(b-
l), i= 1,2.
Similarly, on substituting the expressions forJ(r) we discover that the energy can be written as
(4.12)
into (4.1) and (3.5) and integrating,
(4.13)
wlwo=w,/wO+a(W2/WO+W3/WO)+a2W~/WO
where W,lW,=-
w,/w,=-
l
a(b-I)”
m(b-
1
’ q1(~)(27-v’-~-3)
I
1
(v
* g1(~)(~-3-~-1)
dr 1)]“2
'
dr
1)2 I 1 [(b-7)(7-l)]1’2r
1 w31wO=-n(b_1)2
[(b--7)
I
* q2(~)(2~-~-1-~-3) [(b-7)(7-
1
d,r 1)]1’2 ’
(4.14)
(4.15)
(4.16)
Stress distribution at tip of crack at edge of a circular hole
1193
and ’ g2(7) (r-“-~-l)
dr
w4~wo=-7T(bL)~ I [(b-7)(7-1)]1’2’
(4.17)
1
it
From Betti’s Integral Theorem
follows (see e.g. Sokolnikoff [6], p. 39 1) that W2 = W3.
5. NUMERICAL
PROCEDURE
By using Gauss-Chebyshev quadrature by the simultaneous linear equations
%(jk)
AND
RESULTS
we can replace the integral equations (4.6)
-a m=1 i 41(hd~l(fk, jrn) = Si(fk),
(5.1)
where
(5.2) k= 1, 2,3,. . . , n; i = 1,2. These equations may be solved for the qi (tk) and the quantities Ki/Ko calculated [see (4.2 l)] from the formula Ki/KO=-
(fTlj -[q(b) +: nl=l e qi(jm)Ml(b, I,)),i=192. Table
1.The variation of KJK, and W,/W, with a/R
a/R
K,I&
0.01 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.25 0130 040 0.50 0.75 1.00 1.50 2.00 3.00 4.00 5.00 7.00 90-I
3.291 3.223 3.095 2.978 2.870 2.771 2.679 2.594 2.515 2442 2.373 2.221 2.092 1.884 1.727 1,464 1.306 1.127 1.030 0.930 0.877 0.845 0.808 0.787
-
&I& - 1.079 - 1.040 - 0.966 - 0.900 - 0.839 - 0.783 - 0.732 - 0.685 - 0642 - 0603 - 0.566 - 0.487 -0.421 - 0.320 - 0.247 -0.137 -0.080 -0.030 -0.010 0.002 0.005 0.005 om4 0@04
WI/W0 5.494 5.342 5.060 4.803 4.568 4.354 4.157 3.975 3.808 3.653 3.509 3.192 2.925 2.502 2.185 1.662 1.349 1mO 0.814 0.624 0.528 0.471 0407 0.371
w2wcl - 1.812 - 1.743 - 1.616 - 1.502 - 1.399 - 1.306 - 1.222 - 1.145 - 1.075 - 1.011 - 0.953 - 0.826 - 0.723 - 0.565 - 0.452 -0.281 -0.190 -0.101 -0.061 -0.028 -0.015 -o&W -0m4 - 0.002
w4wcl 0.597 0.568 0.516 0.470 0.429 0.392 0.360 0.331 0.305 0.282 0.260 0.216 0.181 0.131 0.098 0.053 0.032 0.015 OW8 0.004 0*002 O@Ol om1 0.000
1194
J. TWEED
Fig. 2. The variation
and
D. P. ROOKE
of K/K,, with a/R for several
values of the parameter
a.
Similarly, the energies Wj/ W, may be calculated from
Wl/WO = -,g
I)2
E 41(r,)(an-r~1-t,3),
m=1
i 41(bz)(Gt3-G1)7 w2/wo=-n(bi*)2 T7l=l
w3/wo=
-n(b! I)2
and
i
92(fm)m?n-Gi1-G13)
m=1
w’/wO=-,(b!*)~~
%(fm)(f-z-f>). WI=1
Table 1 shows the variation of KJK,, and WJW,, i = 1, 2; j = 1, 2, 3, 4, with a/R and Fig. 2 shows the variation of K/K0 with a/R for several values of the parameter CL REFERENCES
[ 11 0. L. B0WIE.J.
Mark Phys. 35,60 E. A. PHILLIPS
(1956). and C. H. TSAO.
[2] A. J. DURELLI. Analysis of Stress and Strain. (1958). [3] C. J. TRANTER, Q. J. Math. appl. Marh 1, 125 ( 1948). [4] J. TWEED, Glasgow Math. J. 14,65 (1973). [5] F. G. TRICOMI, Q.J. Math. 2, 199 (195 I). [6] I. S. SOKOLNIKOFF,Mathematical Theory ofElasticity. McGraw-Hill ( 1956). [7] H. K. KUTTER,fnt.J. Fract. Mech. 6,233 (1970). [8] J. TWEED, Tech Rpr. PSR-9919, N.C. State University, Raleigh. (Received
9 February
1972)
McGraw-Hill
Stress distribution at tip of crack at edge of a circular hole
1195
Rksumk-
Une technique de transformation de Mellin est utiliste pour trouver une equation integrale dont la solution est lide au facteur d’intensitt de contrainte et a l’energie de formation d’une crique sur le bord d’un trou circulaire dans un solide elastique infini Le cas d’une charge bi-axiale est considere en detail, et des resultats numeriques sont donnes. ZusammenfassungEs wird eine Mellin-Transformethode verwendet, urn eine Integralgleichung zu finden. deren Losung auf den Spannungsintensitatsfaktor und die Bildungsenergie eines Risses an der Kante eines kreisformigen Loches in einem unendlichen elastischen Festkorper Bezug hat. Der Fall zweiachsiger Last wird in Einzelheiten behandelt un numerische Resultate werden gegeben.
Sommario-La tecnica di trasformazione Mellin viene usata per trovare un’equazione integrale la cui soluzione sia relativa al fattore d’intensita della sollecitazione e all’energia di formazione di una incrinatura sull’orlo di un foro circolare in un solid0 elastic0 infinito. II case di carico biassiale viene considerato in dettaglio e vengono offerti risultati numerici. A~CTWKT peIIIUikiC TpeWiHbI CMOTPtX
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