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Stress intensity factors in a reinforced thick-walled cylinder
Inr. J. Engng So Vol. 22, No. 7, pp. 867-879, tinted in the U.S.A.
002&7225/84 Pergamon
1984
$3.00 + .oO Press Ltd.
STRESS INTENSITY FACTORS IN A REINFORCED THICK-WALLED CYLINDER? RENJI TANGS and F. ERDOGAN Department of Mechanical Engineering and Mechanism, Lehigh University, Bethlehem, PA 18015, U.S.A. Ahstraet-In this paper an elastic (hick-walled cylinder containing a radial crack is considered. It is assumed that the cylinder is reinforced by an elastic membrane on its inner surface. The model is intended to simulate pressure vessels with cladding. The formulation of the problem is reduced to a singular integral equation. Various special cases including that of a crack terminating at the cylinder-reinforcement interface are investigated and numerical examples are given. Among the interesting results found one may mention the following: In the case of the crack touching the interface the crack surface displacement derivative is finite and cons~uently the stress state around the corresponding crack tip is bounded, and generally, for realistic values of the stiffness parameter, the effect of the reinforcement is not very significant. 1. INTRODUCTION
its practical importance the problem of a thick-walled cylinder containing a radial crack has been studied rather extensively (see [I] for review and references). Considered as a plane problem the results obtamed from such a study provide an upper bound for the stress intensity factors in pressure vessels, pipes and disks containing an axial part-through crack. In all these previous studies it is assumed that the cylinder is a homogeneous elastic solid. However, there is at least one group of technologically important problems in which the material may not be considered homogeneous. These are the problems which may generally be classified under “multi-1ayere~’ cylinders. Variety of reinforced disks and cylinders and pressure vessels with cladding may be cited as some examples. In this paper we are primarily interested in thick-walled cylinders with cladding containing a radial crack. This is a nonhomogeneous cylinder problem the distinguishing feature of which is that the thickness of the clad (that is, the inner cylinder) is very small compared to that of the main cylinder. The crack problem may be formulated and solved directly as a nonhomogeneous cylinder problem. However, there are at least two reasons for not following this procedure. First, the analysis of the nonhomogeneous problem is highly complicated and secondly if, as in the cladded pressure vessels, the thickness ratio of the two cylinders is very small, past experiences indicate that there would be severe convergence problems in the related numerical calculations. In this paper the cladding will instead be approximated by a membrane and the problem will be treated as a reinforced thick-walled cylinder. A similar problem for a half plane and an infinite strip is considered in[2, 31. Introducing a dislocation into Mitchell’s solution the problem is formulated in terms of a singular integral equation. The case of the crack terminating at the interface is separately studied and the stress intensity factors for various values of the stiffness constant characterizing the reinfor~ment are given. BECAUSE OF
2. THE BASIC
FORMULATION
Except for the application of the boundary conditions, the formulation of the problem is quite similar to that given in[l].Q The stress state in the reinforced cylinder will again be expressed as the sum of two solutions containing the main features of the crack and the hollow cylinder, respectively. hitting the details and referring to Fig. 1 these two solutions may be written as follows (see [I] and [4-6]): tThis work was supported by NSF under the Grant MEA-8209083 and by NASA-Langley under the Grant NGR 39-007-011. $Visiting scholar. Permanent address: Associate Professor, Department of Mathematics and Mechanics, Lanzbou University, China. §In this paper certain expressions have been simplified through further manipulations and by evaluating some of the series in closed form. 867
868
R. TANG and F. ERDOGAN
I
(a)
Fig. 1. Geometry of a reinforced thick-walled cylinder containing a radial crack.
(a) The crack solution g,,,(r, e) = -
adr, e) =
~(~ +1) e S[ e+1)e 2P
2P
g r cos e - t - 2t sin’e 2t2 sin’ e(r cos e - t) r2+ t2 - 2rt cos e - (r2 + t2 - 2rt cos e)2
dt,(2.1) 1j-0) dt, 1f(t)
g sin e(2t cos e - r) 2t sin tl(r cos e - t)(r - t cos (r2 + t2 - 2rt cos e)2 f[ r2 + t2 - 2rt cos e -
e)
(2.2)
2c0se(r-tc0se)+rc0se-t
fldr, 8) = -
r2 + t2 - 2rt cos e
1
- 2(r;:“,;1;;tr-;s;;2e)2
f(t) dt,
(2.3)
where p is the shear modulus, JC= 3 - 4v for plane strain, K = (3 - v)/(l + v) for plane stress, v the Poisson’s ratio and the function f is defined by
f(r) = i b4(r,+O) - udr, -
0)],
(e < r < g).
(2.4)
In order to facilitate the application of the boundary conditions, it is convenient to express the stress components on the boundary in terms of the following Fourier series
ada, 0) =
~&,
0) = -
’
71(Ic+ 1)
n$ seg
A,(t)
sinntf0)
dt,
’ CgnzOR(t) cos nUQ) dt, A(K + 1) s
(2.5) (2.6)
(2.7)
(28
4b,
0) = -
p
II(K + 1)
nzo MO W(t) dt. ~0s
(2.9)
Stress intensity factors in a reinforced thick-walled cylinder
869
From (2.1~(2.3) and (2.5 j-(2.9) the Fourier coefficients may be obtained as follows (2.10)
-E&)=$3($-2],
D,(t)=
(2.11) and
C&Q
-(~+2)(~~+‘~~~-2)(~~-‘],
.,I)=;[
,,I~=~[,+2)(~)l+‘-n(~~-‘],
-(n+2)($+‘+(n
+2)@-‘1,
(n =2,3,...).
(2.12)
(b) Michell’s Solution If the loads satisfy the symmetry condition about the polar axis ox shown in Fig. 1, and the resultants of the tractions on the inner circular bounda~ r = a are zero, then the Mitchell’s Solution for the hollow cylinder gives the following stress components[4, l]
-
!2[u,r”-2 + b&t -
2)~ + c,r -n-2 + d,(n + 2)r -“I cos n0 dt, (2.13)
~2&,
0)
=
Je gf(o
{(
-~+24r)sins+~~[4rl-2+blnln
- c,r -n-2 - d,,nr -“I sin nt? dt,
(2.14)
+ b,(n + 2)r” + c,r -n-2 + d,(n - 2)7
cos &I dt,
(2.15)
where a,, b,, c,, and d,, (n = 0, 1,. . .) are functions of t and the constants
sgf(tb,,(Osg~WMt) dt,
e
dt,. . .
e
replace the regular coefficients in the ~itchel~s 3. BOUNDARY
solution.
CONDITIONS
Consider now a hollow elastic cylinder having radii a and b and containing a radial crack along 8 = 0, e < r < g as shown in Fig. 1. Assume that the inner boundary of the cylinder is reinforced by a thin membrane of thickness k and the self-equilibrating normal tractions on the crack surfaces are the only nonvanishing external loads.
870
R. TANG and F. ERDOGAN
Let the stress state in the cracked hollow cylinder be the sum of the two fundamental solutions given by (2.1k(2.3) and (2.13)-(2.15), that is, let
Qr, 0) = gl,,(r,0) + c2&, e), (i,j = r, 0).
(3.1)
Then the boundary conditions of the problem may be expressed as
4b,
e) + c2rrtb,e) = 0, 6db, 0) + o,db, e) = 0,
~(a,
0) + ~(a,
toI
8
c 274
(3.2)
~,&(a, e) + o*~(~, e) = P@(e),(0 I e < 2a) (3.3)
e) = p,(e),
and
~dr, f 0)+ ~200tr, T 0) = q(r), (e -cr -cg),
(3.4)
where pr and pe are the normal and shear tractions at the cylinder-reinforcement interface (Fig. lb). From the equilibrium of the membrane it may easily be shown that p,(e) = gr,(a, e) = k,,(~,
p,(e)=ati(a,e)=
e),
(0 I e < 274
-+$.O(u,e),
(010
(3.5)
~271)
(3.6)
where hE, for plane stress, aE (3.7)
E,,,h(l - v’)
for plane strain.
uE(1 - v,‘) + hE,v(l + v)
In (3.7) E and v are the elastic constants of the cylinder and E,,, and v, of the reinforcing membrane. By defining now
G = -+
PfG)
d, = -
+ ,p
pftt) 6,,
7+c + 1)
and by substituting the stresses into the boundary obtain $+2,,,=
-%+2&b
-$+2&b
E,,
= -E,,
= D,,
(1+@+2(1
(n = 0, I,...)
(3.8)
conditions (3.2), (3.5) and (3.6) we
-A)y,=lCO-B,,
(3.9)
2(1 + A) $ + (61 - 2)&u = B, - AC,,
2(1 + A)$+
(61 - 2)&z = -A, - Xi,
(3.10)
(3.11)
and a,b “-‘+/?,(n
-2)b”+y,b-‘“+2’+6,(n
+2)b-“=
En
(3.12)
Stress intensity factors in a reinforced thick-walled cylinder
871
a,b”-2+~,nb”-y,b-(“+2~-~6,nb-“=D,
(3.13)
a,une2( 1 + 1) + B,d[(n - 2) + I(n + 2)] + ?,a - (n+2)(1+1)+6,a-“[(n
+2)+I(n
-2)]
= B,, - AC,, a,a”-2(1 -d)
(3.14)
+ S,d[n - hz(n + 2)] - y,~-(“+~)(l + d) - &a-“[n + hz(n - 2)] = A, + A,$,
(3.15)
for n = 2,3,. . . Note that Ai, Bi,. . . are known functions and are given by (2.10)-(2.12). Equations (3.10) and (3.11) imply that the constants A,, B,, D, and E, must satisfy A, + B, = 0,
(3.16)
D, + E, = 0.
It can be shown that eqns (3.16) indeed follow from the fact that in the “crack solution” u,~ the self-equilibrating crack surface tractions are the only non-zero external loads and consequently the stress vector acting on r = a has a zero resultant[l]. With (3.16), the solution of (3.9)-(3.11) is found to be 2(1 - A.)u2
PO= u’(l--A)-b2(1+1)
(3.17)
1 -(1+A)t+ ” = a’(1 - A) - b2(1 + A)
2 u4(1 - 32) -$+b2+&, ” = u4(1 - 31) - b4(1 + A)
6, =
1 u4(1-31)-b4(1+IZ)
1 1
u2(1 - 1) t ,
(3.18)
(3.19)
1
+ b2(1 + A) + u4(12;231) .
(3.20)
Similarly, the solution of the infinite system (3.12~(3.15) can be obtained as follows an=b-2(“-I)tn-l_~,(n
_ l)b2_~,b-2(“-‘),
(3.21)
A=$
(3.22)
0
~*(~+‘)t-(“+‘)(l - 2nl - 1) - 6,u*(l + n -nL
-A)
+ &z2(‘+‘)(l - 2nl -A)],
(3.23)
iin=+,
(3.24)
0
where the determinants
A,, A, and A, are given by
A, = n2(1 + 21 + ,12)um2b2- 2(n2 - l)(l - A’) - [I + 2nl + (2n - 1)12]u-2”b2” - [l - 2nl - (2n + 1)A2]aZ”b-2n + n2(1 - 2A - 31’)a’b -‘,
(3.25)
A, = ((1 - n2)(l - 2’) - [l - 2d - (2n + 1)12]uhb-2” + n2(1 - 21 - 31.2)u2b-2)t-(“f1) +((2+n
-n’)(l
-A2)bm2-(2+n
-n’)(l
+21 +122)u-2)t-(n-1)
+ ((1 + n)[l + 2nl - (1 - 2n)12]ueh - (1 + n)(l - ~2)b-h}tn-1 + ((2 + n)(l + 21 + L2)u-2b-Zn - (2 + n)[l + 2nl - (1 - 2n)L2]u-hb-2}t”+1,
(3.26)
872 da=
R. TANG and F. ERDOGAN {
-(l
-n)[l
-2&l
-(1+2n)12]a2”+(1
-n)(l
-12z)tP}f-(“+1)-{(2-?z)
x(1+21+~~)a-~b2”-(2-n)[l-22n~-(1+2n)~*]a2”b-*}t-“-” +{n~(l+2~+P)a-W+(l-n2)(1-12)-[1-2n~-(1+2n)12] x &-~)&-U
+((2-n-n2)(1+2~+1*)a-2-(2-n-n2)
x (1 - 12)b -*}t”+‘.
(3.27)
4. THE INTEGRAL
EQUATION
Having determined c1,,/I,,, y,, and a,, from (3.8) and from the expressions for stresses it is seen that the solution of the cracked cylinder problem is completely known in terms of the crack surface displacement derivativef(r). This function is as yet unknown and may be determined by using the remaining boundary condition (3.4). Thus, by substituting the stresses crleeand crzoO as expressed in terms off(r) into (3.4), after some manipulations and separating the singular part of the kernel we obtain the following integral equation k(r, t)f(t)dt legsdt +Jeg
= -
n(l + Jc)
q(r), e
2p
(4.1)
where the bounded kernel k(r, t) is given by k(r, t)=
--i
x
r
-$+2y,,+$!+fS,r
+ f
[rx,F2+/?,,(n
+2)
n=2
r”+y,r-(“+2)+6,(n
-2)r-“1
.
From the definition off(r) as given by (2.4) it is seen that the following singlevaluedness condition must be satisfied
s pf(r)
dr = 0.
(4.3)
The singular integral eqn (4.1) may easily be solved numerically by using the procedure outlined, for example, in [7]. Noting that the solution of (4.1) is of the form[8]
f(r) = W&/k -elk - r),
(4.4)
where F(r) is a bounded function, after determiningf (r) the stress intensity factors at the crack tips may be defined and evaluated as follows k(e) = lim ,/m
r-t?
o,(r,
0) = lim
k(g)=lim,/2(r--g)aee(r,0)=
(4.5)
-iii-$--J-f(r).
r-g
5. SPECIAL
(4.6)
CASES
The first special case to be considered is an infinite plane which contains a reinforced circular hole and a finite radial crack. In this case by letting b + co the Fredholm kernel k(r, t) in (4.1) may be simplified as lim k(r, t) = k,(r, t) = -i b-m
- $+
F+
5 n=2
[y,r -(“+‘)+6,(n
-2)r-“1
,
(5.1)
873
Stress intensity factors in a reinforced thick-walled cylinder
(5.2) a*(n+‘)(n* - 31n2)
yn=-
2An+l_1
6 =a”(n
n
--An - 1
1 *t”f’+
-2-2d+In+21)
a”(??-n
.- 1
21n + 1 - A
f-l
1 1 21) +A) .-_ i + In ‘t”-” 21n+l-1 [II+1 u2(“-“(n
2iln + 1 - 1
(5.3)
(5.4)
Note that the new kernel given by (5.1) can be evaluated partly in closed form by using the following infinite series Ax Ax+(A
+B)x2+(A
+2B)x3+
... =
+ (B
-
A)X2
(5.5)
(1 -x)2
where A and B are arbitrary constants. After some obvious transformations it can be shown that k
( 1
6
t)
=
_’ 21
A 2Yl; 1 1w 3a4 a4 ; 3a2 I
r2
I
r3
4 ( r3t2
r’
rt2
1
r > 2rt - u
(5.6) Now observing that
z22in il g>“‘=” f yf)“-q$~2n(2ni: _A)(;~-‘7 2Lu2,,2n rt 4
1
(5.7)
and
( >
1% &
=TZ,
-llx
(5.8)
the kernel k,(r, t) may be decomposed as follows
kdr, t) = kdr, t) + h&,
t>+ Mr, t)
(5.9)
where 3u4 u4 3u2 -;r+~+-$--~;(~-$$-;)(~~
kr,(r, t, = -&
fuz)2~ (5.1o) log(rt
_a’)
(5.11)
874
R. TANG
k&.1)=;
and F. ERDOGAN
J$+y=$+y (
i
a2
1 3a4 -5
(1 -A) ; n-’ 0 f “-2 2Ln(2Ln + 1 - 1)
&logrr-&
a2
5a2
( r3t2+7+7-T
9
m
0
n-l
n It
) n:221n
+ 1 --il.
(5.12)
The subscripts in k,,, ki, and klb refer to the fact that as r and t both approach e = a (in the case of the crack touching the interface), these kernels become “singular”, “weakly singular” and “bounded”, respectively. The second special case is that of an infinite plane containing a radial crack and a circular hole without reinforcement. In this case by letting A = 0 (see (3.7)) it can be shown that the kernel k(r, t) in (4.1) may be evaluated in closed form as follows
lim k(r, t) = k2(r, t) = -
b-m 1-O
(t2 - a2)2a2 t(rt-a2)3
t(t2 - a2)
-
+(rt-a2)2+rf1a2+
-a2-t2
a2
r2t
1
-p-;‘(5’13)
The third special case refers to the infinite plane with a reinforced circular hole and a radial crack in which the radius a is infinitely large. This is the problem of a cracked half plane reinforced by a membrane along its boundary (Fig. 2). In this case if we define r=a+y,
t=a+L
e=a+e,,
g=a+g,,
(5.14)
and assume that y and c remains finite as a tends to infinity, the kernel k(r, t) (4.1) becomes
W, t)
= lim k,(a +y, a + 0 = k&y, C) (e. < CY,C)
(5.15)
Again, by evaluating the asymptotically dominant terms in the infinite series in closed form, after somewhat lengthy manipulations we obtain
X
Fig. 2. Notation
for an elastic half plane which is reinforced a finite crack.
by a membrane
and which contains
875
Stress intensity factors in a reinforced thick-walled cylinder 25$-Y YY -J- cph(i;+Y)-2~Zh2(1+Y)+bjh(i+Y)Z 5+v
k,(_v,lJ=
5Y
(5.16)
where 4=
E,,,/E for plane stress, E,( 1 - v2)/E( 1 - vnt2) for plane strain.
(5.17)
Of course, the integral equation for the half plane problem may also be derived directly by using the method of Fourier transforms. By following this technique and by omitting the details, after evaluating the as~ptotically dominant infinite integrals, the kernel k&y, [) may be obtained as foflows
26+Y CY TV -J- &X5 5 +Y +Y)- W2h2t5 +y)+#K+y12
k3(y,[)=
(5.18)
dir.
Finally, for the half plane without reinforcement it can be shown that for a --)m (5.13) would reduce to the following known kernel[7] 1 k4hO=
6. THE CASE
6y
4y2
-i+m-(i+
OF THE CRACK
(5.19)
+y)3’
TOUCHING
THE INTERFACE
(e = a)
In the infinite plane containing a reinforced circular hole considered in the previous section, let us assume that the inner crack tip touches the interface (i.e. e = a). In this case, from (5.9)-(5.12) the kernel k(r, I) of the integral eqn (4.1) may be shown to reduce to
lim k(r, t) =
h&r,t) =kdr, I)+
kdr,
t) + &(r, t)
(6.1)
~,~(~,~)~~~-~-5~~+a2(~+~~-a’2.~:;!,
(6.2) r
r GIws(r,t) = -&
Ir,,(r, t) = kdr,
>
log (rt - a2)
t) + i
(6.3)
(6.4)
Note that as both r and t approach the interface r = a, the kernel &, become unbounded and hence influence the singular behavior of the solution f (E). This behavior may be investigated by letting f(t)
= (g - t)-“(t
- a)-B,
(0 < Re(ct, 8) x 1)
(6.5)
and by using the function theoretic method described in[7]. Omitting the details, the
876
R. TANG and F. ERDOGAN
characteristic equations giving a and /I may thus be obtained as cosna =O,
cot7$ --
1 = 0, sin n/I
giving a = l/2 and B = 0 as the acceptable roots. The root /I = 0 indicate tip r = a the crack surface derivative is finite. Since the bending reinforcement is zero, this is the physically expected result. In the limiting case of a + cc (Fig. 2), the kernel given by (5.18) which the method of Fourier transforms is still valid. In this case the integral evaluated as
s m
0
that at the crack stiffness of the is obtained from in (5.18) may be
e-K+Yb
24ha + 1
da = -~[Eo+log(!$f)+~~],
(6.7)
where 8 = - (c +y)/2+h and E, = 0.5772157 is the Euler’s constant. Hence the infinite integral in (5.18) has the following order
s m
0
,-aci+Y)
24ha + 1
da - Wog(!: +
~11
(6.8)
and the kernel of the integral eqn (4.1) may be arranged as follows
s
,-a+Y)
m
LsCY, 5) = 4
24ha + 1
0
lC&Y,o=
-
2r+y MC +Y)-
CY W*h%
21 +Y 4h
(6.11)
da, iY
+~)+&(5 m
+Y)*
e-a(i+Y)
>S
o 24ha + 1
da.
(6.12)
It is seen that as 5 and y both approach zero & becomes bounded, &, exhibits only a weak singularity and & remains unbounded. Thus, if we express f(t) again as in (6.5) the function theoretic method gives the characteristic equations for a and /I as cos xa = 0,
(cos 7$ - l)/sin 7$ = 0
(6.13)
which are the same as (6.6) and give a = l/2 and j3 = 0. This result could also be obtained from the circular hole problem by substituting T = a + y, t = a + (I, and by letting a + CC in (6.2). After some simple algebra one may then observe that the principal part of the kernel E,;, is - l/(4’ + y) which is identical to (6.10).
7. SOLUTION
AND
RESULTS
Excepting the problem of unreinforced edge crack, in all cases considered the displacement u, at the crack tips is zero and consequently the integral eqn (4.1) and all its special cases must be solved under the single-valuedness condition (4.3). For the numerical solution following normalized quantities are introduced
Stress intensity factors in a reinforced thick-walled cylinder
&--e-
2
g+e
g-e T+g+e r=Tp+T 2 ’ k(r, f),
lqp, 7) = g+
4(p)
m =&f(r) =P(z)w(z),
877
(7.1)
= 4(r),
(-
(7.2)
1
(7.3)
In the embedded crack problem (i.e. for e > a) w(r) = (1 -r2)-1/2 and a Gauss-Chebshev integration technique is used to determine F”(r)[7]. The stress intensity factors are then evaluated from (see (4.5) and (4.6)) k(e) =
JaF”(
-
l),
(7.4)
k(g) = -J-P(l).
For e = a the function fir) is bounded at r = a (i.e. in (6.5) p = 0) and the stress intensity factor at r = g is obtained from
k(g)= -&Z&l).
(7.5)
The calculated stress intensity factors are given in Tables l-6. In the case of an embedded crack (i.e. for e > a or e, > 0) the stress intensity factors are normalized as follows (7.6) Table 1 gives an example showing the effect of the stiffness parameter I defined by (3.7) on the stress intensity factors. Similar results for a reinforced half plane are shown in Table 2. The stiffness constant 4 shown in Table 2 is defined by (5.17). The results for somewhat more realistic values of the stiffness parameter 1 and the radius ratio a/b which may simulate a pressure vessel with internal claddding are shown in Tables 3-6. The normalized stress intensity factors are calculated for two types of external loads. In the first it is assumed that a uniform crack surface pressure a,&, 0) = q(r) = -q,, is the only nonzero applied load. In the second the cylinder is assumed to be under internal pressure a,,(~ - h, 0) = -p,,. Generally, for the embedded crack the influence of the membrane reinforcement or the cladding on the stress intensity factors does not seem to be very significant. Since the properties of the clad material cannot be very different than that of the cylinder, this result is not surprising. However, in the case of the crack terminating at the interface (i.e. for e = a, Fig. 1) the stress intensity factors k(g) calculated for 1, # 0 and 1 = 0 are very Table 1. Stress intensity factors in a reinforced hollow circular cylinder containing a radial crack (Fig. 1). External load: uniform crack surface pressure q(r) = - qo, II = 0.56, g -e = 0.5(b -a), k(e)
e-a 6z
= &)lqoJE~,
Mg) = U)lqoJFFP
A=0
A = 0.2
x = 0.5
k,(e)
h(S)
G(e)
G(9)
0.15
1.1968
1.1754
1.1085
1.1255
1.0549
1.0988
0.20
1.1517
1.1769
1.1123
1.1436
1.0740
1.1223
0.25
1.1331
1.1951
1.1215
1.1706
1.0926
1:1523
0.30
1.1727
1.2303
1.1379
1.2100
1.1148
1.1930
0.35
1.1893
1.2912
1.1637
1.2702
1.1439
1.2526
0.40
1.2263
1.4033
1.2042
1.3764
1.1854
1.3559
t(e)
S(g)
878
R. TANG and F. ERDOGAN
Table 2. Stress intensity factors in a reinforced half plane containing a crack perpendicular to the boundary (Fig. 2). External load: uniform crack surface pressure q(y) = -q,,, & = h+/g, kAg0) = k ~~)/q~~~~, k&t3 = k(e~)~q~~~~~
xo=0
r
e”‘go k,(e,)
x* = 0.0375
k,(s,)
x* = 0.075
xo = 0.15
xo = 0.3
k,(eo)
k&g,)
k,(e,)
k&9,)
k,(eo)
k&9,)
k,(ec)
k&,1
0.1
1.42667
1.15500
1.41438
1.15234
1.40300
1.14983
1.38222
1.14510
1.34633
1.13661
0.2
1.20354
1.09670
1.19970
1.09543
1.19604
1.09422
1.18921
1.09190
1.17698
1.08765
0.3
1.11075
1.06242
~.?0909
1.06171
1.10750
1.06103
1.10449
1.05972
1.09902
1.05730
0.4
1.06224
1.03990
1.06145
1.03949
1.06068
1.03909
1.05923
1.03833
1.05655
1.03697
Table 3. Stress intensity factors in a reinforced hollow circular cylinder containing a radial crack (Fig. load: uniform crack surface pressure q(r) = - qo, a/b = 0.8, e/a = 1.05, 1). External k&z) = &~)~q~~~, k,(e) = k(e)lq~~~~ x=0 zz
A = 0.001
A = 0.002
k,(a)
k,(9)
k,(e)
k”(9)
k”(e)
a = 0.004
k,(g)
k,(e)
k,,(9)
0.05
1.0044
1.0041
1.0043
1.0039
1.0042
1.0038
1.0039
1.0036
0.10 0.20
1.0157 1.0511
1.0133 1.0382
1.0153 1.0497
1.0129 1.0373
I.0148 1.0484
1.0126 1.0364
1.0140 1.0459
1.0119 1.0349
0.30
7.0962
1.0668
t -0938
1.0656
1.0915
1.0644
1.0871
1.0621
0.40
1.1466
1.1023
1.1432
1 .lOll
1.1400
1.0998
1.1339
1.0974
0.50
1.2021
1.1596
1.1979
1.1584
1.1940
1.1573
1.1866
1.1552
Table 4. Stress intensity factors in a reinforced pressurized cylinder containing a radial crack. External e/a = 1.05, a/b = 0.8, pressure Ue) = W)bo$?, load: internal 6, = -Pa, k.(g) = W/&$~ )i = 0.004
x = 0.002
x=0 E
k,,(e)
k,(g)
k,(e)
k,(9)
k,,(a)
k,,(g)
0.05 0.10
4.3014 4.3346
4.2703 4.2662
4.3003 4.3307
4.2693 4.2633
4.2992 4.3273
4.2684 4.2606
0.20
4.4534
4.2872
4.4420
4.2800
4.4316
4.2734
0.30 0.40
4.6101 4.7854
4.3256 4.3935
4.5903 4.7581
4.3156 4.3828
4.5723 4.7332
4.3064 4.3730
0.50
4.9779
4.5473
4.945t
4.5382
4.9150
4.5297.
a in a reinforced cylinder containing a radial Table 5. The stress intensity factor k,(g) = k(g)/qoF crack which terminates at the interface (e = a, Fig. 1). External load: crack surface pressure q(r) = -qo, a/b = 0.8
is
x-o
0.25
1.3380
0.8894
0.8701
0.30
1.4477
0.9256
0.9023
0.40
1.7159
0.9886
0.95eo
0.50
2.0489
1.0375
1.0013
0.60
2.4352
1.0735
1.0345
0.70
2.8271
1.1065
1.0681
7
x = 0.002
x = 0.004
different. For A = 0 there is no reinforcement and the crack is an edge crack with a non-zero opening at r = a (i.e. uO(cz,+ 0) - ~+(a, - 0) > 0). On the other hand, for I # 0 since ~,(a, 0) = 0 there is a great deal of constraint tending to reduce the stress intensity factor. One may note that because of the membrane assumption made for the reinforcement at r = a the derivative of the crack surface displa~ment is finite giving a finite angle between the upper and lower crack surfaces.
Stress intensity factors in a reinforced thick-walled cylinder
879
Table 6. Stress intensity factor k.(g) = k(g)/poG in a pressurized reinforced cylinder containing a radial crack which terminates at the interface (e = a, Fig. 1). External load: a,,(a, 0) = -p,,, a/b = 0.8. (For the edge crack case, that is, for X = 0 the effect of the pressure q&r, 0) = -pO on the crack surfaces is not included.) (k,*(g) = k,,(g)/oee(a, O)\/g--a) k=o
x=0
k,(g)
k,*(g)
A = 0.002 k,(g)
0.25
5.8477
1.2836
3.8533
3.7684
0.30
6.2838
1.3794
3.9755
3.8734
0.40
7.3578
1.6151
4.1768
0.50
8.6933 10.2352
4.3154 4.3977
4.0443 4.1603
0.60 0.70
1.9083 2.2467
11.7739
2.5845
4.4642
x = 0.004 k,(g)
4.2320 4.3026
REFERENCES F. DELALE and F. ERDOGAN, Int. J. Frac. 20, 251 (1982). F. DELALE and F. ERDGGAN, Int. J. Solid Structures 5, 381 (1982). S. KRENK and M. BAKIOGLU, Int. J. Frac. 3, 441 (1975). J. DUNDURS, Mathematical Theory of Dislocations (Edited by T. Mura), p. 70 (1969). R. W. LITTLE, Elasticity. Prentice-Hall, New York (1973). R. J. TANG and K. WANG, Engng Frac. Mech. 16, 47 (1982). F. ERDOGAN, Mechanics Today (Edited by S. Nemat-Nasser), Vol. 4, pp. 1-81 (1978) and Vol. 6, pp. 199-202 (1981). [8] N. I. MUSKHELISHVILI, Singular Integral Equation. Noordhoff, Groningen, Holland (1953).
[l] [2] [3] [4] [5] [6] [7]
(Received 5 August 1983)
002&7225/84 Pergamon
1984
$3.00 + .oO Press Ltd.
STRESS INTENSITY FACTORS IN A REINFORCED THICK-WALLED CYLINDER? RENJI TANGS and F. ERDOGAN Department of Mechanical Engineering and Mechanism, Lehigh University, Bethlehem, PA 18015, U.S.A. Ahstraet-In this paper an elastic (hick-walled cylinder containing a radial crack is considered. It is assumed that the cylinder is reinforced by an elastic membrane on its inner surface. The model is intended to simulate pressure vessels with cladding. The formulation of the problem is reduced to a singular integral equation. Various special cases including that of a crack terminating at the cylinder-reinforcement interface are investigated and numerical examples are given. Among the interesting results found one may mention the following: In the case of the crack touching the interface the crack surface displacement derivative is finite and cons~uently the stress state around the corresponding crack tip is bounded, and generally, for realistic values of the stiffness parameter, the effect of the reinforcement is not very significant. 1. INTRODUCTION
its practical importance the problem of a thick-walled cylinder containing a radial crack has been studied rather extensively (see [I] for review and references). Considered as a plane problem the results obtamed from such a study provide an upper bound for the stress intensity factors in pressure vessels, pipes and disks containing an axial part-through crack. In all these previous studies it is assumed that the cylinder is a homogeneous elastic solid. However, there is at least one group of technologically important problems in which the material may not be considered homogeneous. These are the problems which may generally be classified under “multi-1ayere~’ cylinders. Variety of reinforced disks and cylinders and pressure vessels with cladding may be cited as some examples. In this paper we are primarily interested in thick-walled cylinders with cladding containing a radial crack. This is a nonhomogeneous cylinder problem the distinguishing feature of which is that the thickness of the clad (that is, the inner cylinder) is very small compared to that of the main cylinder. The crack problem may be formulated and solved directly as a nonhomogeneous cylinder problem. However, there are at least two reasons for not following this procedure. First, the analysis of the nonhomogeneous problem is highly complicated and secondly if, as in the cladded pressure vessels, the thickness ratio of the two cylinders is very small, past experiences indicate that there would be severe convergence problems in the related numerical calculations. In this paper the cladding will instead be approximated by a membrane and the problem will be treated as a reinforced thick-walled cylinder. A similar problem for a half plane and an infinite strip is considered in[2, 31. Introducing a dislocation into Mitchell’s solution the problem is formulated in terms of a singular integral equation. The case of the crack terminating at the interface is separately studied and the stress intensity factors for various values of the stiffness constant characterizing the reinfor~ment are given. BECAUSE OF
2. THE BASIC
FORMULATION
Except for the application of the boundary conditions, the formulation of the problem is quite similar to that given in[l].Q The stress state in the reinforced cylinder will again be expressed as the sum of two solutions containing the main features of the crack and the hollow cylinder, respectively. hitting the details and referring to Fig. 1 these two solutions may be written as follows (see [I] and [4-6]): tThis work was supported by NSF under the Grant MEA-8209083 and by NASA-Langley under the Grant NGR 39-007-011. $Visiting scholar. Permanent address: Associate Professor, Department of Mathematics and Mechanics, Lanzbou University, China. §In this paper certain expressions have been simplified through further manipulations and by evaluating some of the series in closed form. 867
868
R. TANG and F. ERDOGAN
I
(a)
Fig. 1. Geometry of a reinforced thick-walled cylinder containing a radial crack.
(a) The crack solution g,,,(r, e) = -
adr, e) =
~(~ +1) e S[ e+1)e 2P
2P
g r cos e - t - 2t sin’e 2t2 sin’ e(r cos e - t) r2+ t2 - 2rt cos e - (r2 + t2 - 2rt cos e)2
dt,(2.1) 1j-0) dt, 1f(t)
g sin e(2t cos e - r) 2t sin tl(r cos e - t)(r - t cos (r2 + t2 - 2rt cos e)2 f[ r2 + t2 - 2rt cos e -
e)
(2.2)
2c0se(r-tc0se)+rc0se-t
fldr, 8) = -
r2 + t2 - 2rt cos e
1
- 2(r;:“,;1;;tr-;s;;2e)2
f(t) dt,
(2.3)
where p is the shear modulus, JC= 3 - 4v for plane strain, K = (3 - v)/(l + v) for plane stress, v the Poisson’s ratio and the function f is defined by
f(r) = i b4(r,+O) - udr, -
0)],
(e < r < g).
(2.4)
In order to facilitate the application of the boundary conditions, it is convenient to express the stress components on the boundary in terms of the following Fourier series
ada, 0) =
~&,
0) = -
’
71(Ic+ 1)
n$ seg
A,(t)
sinntf0)
dt,
’ CgnzOR(t) cos nUQ) dt, A(K + 1) s
(2.5) (2.6)
(2.7)
(28
4b,
0) = -
p
II(K + 1)
nzo MO W(t) dt. ~0s
(2.9)
Stress intensity factors in a reinforced thick-walled cylinder
869
From (2.1~(2.3) and (2.5 j-(2.9) the Fourier coefficients may be obtained as follows (2.10)
-E&)=$3($-2],
D,(t)=
(2.11) and
C&Q
-(~+2)(~~+‘~~~-2)(~~-‘],
.,I)=;[
,,I~=~[,+2)(~)l+‘-n(~~-‘],
-(n+2)($+‘+(n
+2)@-‘1,
(n =2,3,...).
(2.12)
(b) Michell’s Solution If the loads satisfy the symmetry condition about the polar axis ox shown in Fig. 1, and the resultants of the tractions on the inner circular bounda~ r = a are zero, then the Mitchell’s Solution for the hollow cylinder gives the following stress components[4, l]
-
!2[u,r”-2 + b&t -
2)~ + c,r -n-2 + d,(n + 2)r -“I cos n0 dt, (2.13)
~2&,
0)
=
Je gf(o
{(
-~+24r)sins+~~[4rl-2+blnln
- c,r -n-2 - d,,nr -“I sin nt? dt,
(2.14)
+ b,(n + 2)r” + c,r -n-2 + d,(n - 2)7
cos &I dt,
(2.15)
where a,, b,, c,, and d,, (n = 0, 1,. . .) are functions of t and the constants
sgf(tb,,(Osg~WMt) dt,
e
dt,. . .
e
replace the regular coefficients in the ~itchel~s 3. BOUNDARY
solution.
CONDITIONS
Consider now a hollow elastic cylinder having radii a and b and containing a radial crack along 8 = 0, e < r < g as shown in Fig. 1. Assume that the inner boundary of the cylinder is reinforced by a thin membrane of thickness k and the self-equilibrating normal tractions on the crack surfaces are the only nonvanishing external loads.
870
R. TANG and F. ERDOGAN
Let the stress state in the cracked hollow cylinder be the sum of the two fundamental solutions given by (2.1k(2.3) and (2.13)-(2.15), that is, let
Qr, 0) = gl,,(r,0) + c2&, e), (i,j = r, 0).
(3.1)
Then the boundary conditions of the problem may be expressed as
4b,
e) + c2rrtb,e) = 0, 6db, 0) + o,db, e) = 0,
~(a,
0) + ~(a,
toI
8
c 274
(3.2)
~,&(a, e) + o*~(~, e) = P@(e),(0 I e < 2a) (3.3)
e) = p,(e),
and
~dr, f 0)+ ~200tr, T 0) = q(r), (e -cr -cg),
(3.4)
where pr and pe are the normal and shear tractions at the cylinder-reinforcement interface (Fig. lb). From the equilibrium of the membrane it may easily be shown that p,(e) = gr,(a, e) = k,,(~,
p,(e)=ati(a,e)=
e),
(0 I e < 274
-+$.O(u,e),
(010
(3.5)
~271)
(3.6)
where hE, for plane stress, aE (3.7)
E,,,h(l - v’)
for plane strain.
uE(1 - v,‘) + hE,v(l + v)
In (3.7) E and v are the elastic constants of the cylinder and E,,, and v, of the reinforcing membrane. By defining now
G = -+
PfG)
d, = -
+ ,p
pftt) 6,,
7+c + 1)
and by substituting the stresses into the boundary obtain $+2,,,=
-%+2&b
-$+2&b
E,,
= -E,,
= D,,
(1+@+2(1
(n = 0, I,...)
(3.8)
conditions (3.2), (3.5) and (3.6) we
-A)y,=lCO-B,,
(3.9)
2(1 + A) $ + (61 - 2)&u = B, - AC,,
2(1 + A)$+
(61 - 2)&z = -A, - Xi,
(3.10)
(3.11)
and a,b “-‘+/?,(n
-2)b”+y,b-‘“+2’+6,(n
+2)b-“=
En
(3.12)
Stress intensity factors in a reinforced thick-walled cylinder
871
a,b”-2+~,nb”-y,b-(“+2~-~6,nb-“=D,
(3.13)
a,une2( 1 + 1) + B,d[(n - 2) + I(n + 2)] + ?,a - (n+2)(1+1)+6,a-“[(n
+2)+I(n
-2)]
= B,, - AC,, a,a”-2(1 -d)
(3.14)
+ S,d[n - hz(n + 2)] - y,~-(“+~)(l + d) - &a-“[n + hz(n - 2)] = A, + A,$,
(3.15)
for n = 2,3,. . . Note that Ai, Bi,. . . are known functions and are given by (2.10)-(2.12). Equations (3.10) and (3.11) imply that the constants A,, B,, D, and E, must satisfy A, + B, = 0,
(3.16)
D, + E, = 0.
It can be shown that eqns (3.16) indeed follow from the fact that in the “crack solution” u,~ the self-equilibrating crack surface tractions are the only non-zero external loads and consequently the stress vector acting on r = a has a zero resultant[l]. With (3.16), the solution of (3.9)-(3.11) is found to be 2(1 - A.)u2
PO= u’(l--A)-b2(1+1)
(3.17)
1 -(1+A)t+ ” = a’(1 - A) - b2(1 + A)
2 u4(1 - 32) -$+b2+&, ” = u4(1 - 31) - b4(1 + A)
6, =
1 u4(1-31)-b4(1+IZ)
1 1
u2(1 - 1) t ,
(3.18)
(3.19)
1
+ b2(1 + A) + u4(12;231) .
(3.20)
Similarly, the solution of the infinite system (3.12~(3.15) can be obtained as follows an=b-2(“-I)tn-l_~,(n
_ l)b2_~,b-2(“-‘),
(3.21)
A=$
(3.22)
0
~*(~+‘)t-(“+‘)(l - 2nl - 1) - 6,u*(l + n -nL
-A)
+ &z2(‘+‘)(l - 2nl -A)],
(3.23)
iin=+,
(3.24)
0
where the determinants
A,, A, and A, are given by
A, = n2(1 + 21 + ,12)um2b2- 2(n2 - l)(l - A’) - [I + 2nl + (2n - 1)12]u-2”b2” - [l - 2nl - (2n + 1)A2]aZ”b-2n + n2(1 - 2A - 31’)a’b -‘,
(3.25)
A, = ((1 - n2)(l - 2’) - [l - 2d - (2n + 1)12]uhb-2” + n2(1 - 21 - 31.2)u2b-2)t-(“f1) +((2+n
-n’)(l
-A2)bm2-(2+n
-n’)(l
+21 +122)u-2)t-(n-1)
+ ((1 + n)[l + 2nl - (1 - 2n)12]ueh - (1 + n)(l - ~2)b-h}tn-1 + ((2 + n)(l + 21 + L2)u-2b-Zn - (2 + n)[l + 2nl - (1 - 2n)L2]u-hb-2}t”+1,
(3.26)
872 da=
R. TANG and F. ERDOGAN {
-(l
-n)[l
-2&l
-(1+2n)12]a2”+(1
-n)(l
-12z)tP}f-(“+1)-{(2-?z)
x(1+21+~~)a-~b2”-(2-n)[l-22n~-(1+2n)~*]a2”b-*}t-“-” +{n~(l+2~+P)a-W+(l-n2)(1-12)-[1-2n~-(1+2n)12] x &-~)&-U
+((2-n-n2)(1+2~+1*)a-2-(2-n-n2)
x (1 - 12)b -*}t”+‘.
(3.27)
4. THE INTEGRAL
EQUATION
Having determined c1,,/I,,, y,, and a,, from (3.8) and from the expressions for stresses it is seen that the solution of the cracked cylinder problem is completely known in terms of the crack surface displacement derivativef(r). This function is as yet unknown and may be determined by using the remaining boundary condition (3.4). Thus, by substituting the stresses crleeand crzoO as expressed in terms off(r) into (3.4), after some manipulations and separating the singular part of the kernel we obtain the following integral equation k(r, t)f(t)dt legsdt +Jeg
= -
n(l + Jc)
q(r), e
2p
(4.1)
where the bounded kernel k(r, t) is given by k(r, t)=
--i
x
r
-$+2y,,+$!+fS,r
+ f
[rx,F2+/?,,(n
+2)
n=2
r”+y,r-(“+2)+6,(n
-2)r-“1
.
From the definition off(r) as given by (2.4) it is seen that the following singlevaluedness condition must be satisfied
s pf(r)
dr = 0.
(4.3)
The singular integral eqn (4.1) may easily be solved numerically by using the procedure outlined, for example, in [7]. Noting that the solution of (4.1) is of the form[8]
f(r) = W&/k -elk - r),
(4.4)
where F(r) is a bounded function, after determiningf (r) the stress intensity factors at the crack tips may be defined and evaluated as follows k(e) = lim ,/m
r-t?
o,(r,
0) = lim
k(g)=lim,/2(r--g)aee(r,0)=
(4.5)
-iii-$--J-f(r).
r-g
5. SPECIAL
(4.6)
CASES
The first special case to be considered is an infinite plane which contains a reinforced circular hole and a finite radial crack. In this case by letting b + co the Fredholm kernel k(r, t) in (4.1) may be simplified as lim k(r, t) = k,(r, t) = -i b-m
- $+
F+
5 n=2
[y,r -(“+‘)+6,(n
-2)r-“1
,
(5.1)
873
Stress intensity factors in a reinforced thick-walled cylinder
(5.2) a*(n+‘)(n* - 31n2)
yn=-
2An+l_1
6 =a”(n
n
--An - 1
1 *t”f’+
-2-2d+In+21)
a”(??-n
.- 1
21n + 1 - A
f-l
1 1 21) +A) .-_ i + In ‘t”-” 21n+l-1 [II+1 u2(“-“(n
2iln + 1 - 1
(5.3)
(5.4)
Note that the new kernel given by (5.1) can be evaluated partly in closed form by using the following infinite series Ax Ax+(A
+B)x2+(A
+2B)x3+
... =
+ (B
-
A)X2
(5.5)
(1 -x)2
where A and B are arbitrary constants. After some obvious transformations it can be shown that k
( 1
6
t)
=
_’ 21
A 2Yl; 1 1w 3a4 a4 ; 3a2 I
r2
I
r3
4 ( r3t2
r’
rt2
1
r > 2rt - u
(5.6) Now observing that
z22in il g>“‘=” f yf)“-q$~2n(2ni: _A)(;~-‘7 2Lu2,,2n rt 4
1
(5.7)
and
( >
1% &
=TZ,
-llx
(5.8)
the kernel k,(r, t) may be decomposed as follows
kdr, t) = kdr, t) + h&,
t>+ Mr, t)
(5.9)
where 3u4 u4 3u2 -;r+~+-$--~;(~-$$-;)(~~
kr,(r, t, = -&
fuz)2~ (5.1o) log(rt
_a’)
(5.11)
874
R. TANG
k&.1)=;
and F. ERDOGAN
J$+y=$+y (
i
a2
1 3a4 -5
(1 -A) ; n-’ 0 f “-2 2Ln(2Ln + 1 - 1)
&logrr-&
a2
5a2
( r3t2+7+7-T
9
m
0
n-l
n It
) n:221n
+ 1 --il.
(5.12)
The subscripts in k,,, ki, and klb refer to the fact that as r and t both approach e = a (in the case of the crack touching the interface), these kernels become “singular”, “weakly singular” and “bounded”, respectively. The second special case is that of an infinite plane containing a radial crack and a circular hole without reinforcement. In this case by letting A = 0 (see (3.7)) it can be shown that the kernel k(r, t) in (4.1) may be evaluated in closed form as follows
lim k(r, t) = k2(r, t) = -
b-m 1-O
(t2 - a2)2a2 t(rt-a2)3
t(t2 - a2)
-
+(rt-a2)2+rf1a2+
-a2-t2
a2
r2t
1
-p-;‘(5’13)
The third special case refers to the infinite plane with a reinforced circular hole and a radial crack in which the radius a is infinitely large. This is the problem of a cracked half plane reinforced by a membrane along its boundary (Fig. 2). In this case if we define r=a+y,
t=a+L
e=a+e,,
g=a+g,,
(5.14)
and assume that y and c remains finite as a tends to infinity, the kernel k(r, t) (4.1) becomes
W, t)
= lim k,(a +y, a + 0 = k&y, C) (e. < CY,C)
(5.15)
Again, by evaluating the asymptotically dominant terms in the infinite series in closed form, after somewhat lengthy manipulations we obtain
X
Fig. 2. Notation
for an elastic half plane which is reinforced a finite crack.
by a membrane
and which contains
875
Stress intensity factors in a reinforced thick-walled cylinder 25$-Y YY -J- cph(i;+Y)-2~Zh2(1+Y)+bjh(i+Y)Z 5+v
k,(_v,lJ=
5Y
(5.16)
where 4=
E,,,/E for plane stress, E,( 1 - v2)/E( 1 - vnt2) for plane strain.
(5.17)
Of course, the integral equation for the half plane problem may also be derived directly by using the method of Fourier transforms. By following this technique and by omitting the details, after evaluating the as~ptotically dominant infinite integrals, the kernel k&y, [) may be obtained as foflows
26+Y CY TV -J- &X5 5 +Y +Y)- W2h2t5 +y)+#K+y12
k3(y,[)=
(5.18)
dir.
Finally, for the half plane without reinforcement it can be shown that for a --)m (5.13) would reduce to the following known kernel[7] 1 k4hO=
6. THE CASE
6y
4y2
-i+m-(i+
OF THE CRACK
(5.19)
+y)3’
TOUCHING
THE INTERFACE
(e = a)
In the infinite plane containing a reinforced circular hole considered in the previous section, let us assume that the inner crack tip touches the interface (i.e. e = a). In this case, from (5.9)-(5.12) the kernel k(r, I) of the integral eqn (4.1) may be shown to reduce to
lim k(r, t) =
h&r,t) =kdr, I)+
kdr,
t) + &(r, t)
(6.1)
~,~(~,~)~~~-~-5~~+a2(~+~~-a’2.~:;!,
(6.2) r
r GIws(r,t) = -&
Ir,,(r, t) = kdr,
>
log (rt - a2)
t) + i
(6.3)
(6.4)
Note that as both r and t approach the interface r = a, the kernel &, become unbounded and hence influence the singular behavior of the solution f (E). This behavior may be investigated by letting f(t)
= (g - t)-“(t
- a)-B,
(0 < Re(ct, 8) x 1)
(6.5)
and by using the function theoretic method described in[7]. Omitting the details, the
876
R. TANG and F. ERDOGAN
characteristic equations giving a and /I may thus be obtained as cosna =O,
cot7$ --
1 = 0, sin n/I
giving a = l/2 and B = 0 as the acceptable roots. The root /I = 0 indicate tip r = a the crack surface derivative is finite. Since the bending reinforcement is zero, this is the physically expected result. In the limiting case of a + cc (Fig. 2), the kernel given by (5.18) which the method of Fourier transforms is still valid. In this case the integral evaluated as
s m
0
that at the crack stiffness of the is obtained from in (5.18) may be
e-K+Yb
24ha + 1
da = -~[Eo+log(!$f)+~~],
(6.7)
where 8 = - (c +y)/2+h and E, = 0.5772157 is the Euler’s constant. Hence the infinite integral in (5.18) has the following order
s m
0
,-aci+Y)
24ha + 1
da - Wog(!: +
~11
(6.8)
and the kernel of the integral eqn (4.1) may be arranged as follows
s
,-a+Y)
m
LsCY, 5) = 4
24ha + 1
0
lC&Y,o=
-
2r+y MC +Y)-
CY W*h%
21 +Y 4h
(6.11)
da, iY
+~)+&(5 m
+Y)*
e-a(i+Y)
>S
o 24ha + 1
da.
(6.12)
It is seen that as 5 and y both approach zero & becomes bounded, &, exhibits only a weak singularity and & remains unbounded. Thus, if we express f(t) again as in (6.5) the function theoretic method gives the characteristic equations for a and /I as cos xa = 0,
(cos 7$ - l)/sin 7$ = 0
(6.13)
which are the same as (6.6) and give a = l/2 and j3 = 0. This result could also be obtained from the circular hole problem by substituting T = a + y, t = a + (I, and by letting a + CC in (6.2). After some simple algebra one may then observe that the principal part of the kernel E,;, is - l/(4’ + y) which is identical to (6.10).
7. SOLUTION
AND
RESULTS
Excepting the problem of unreinforced edge crack, in all cases considered the displacement u, at the crack tips is zero and consequently the integral eqn (4.1) and all its special cases must be solved under the single-valuedness condition (4.3). For the numerical solution following normalized quantities are introduced
Stress intensity factors in a reinforced thick-walled cylinder
&--e-
2
g+e
g-e T+g+e r=Tp+T 2 ’ k(r, f),
lqp, 7) = g+
4(p)
m =&f(r) =P(z)w(z),
877
(7.1)
= 4(r),
(-
(7.2)
1
(7.3)
In the embedded crack problem (i.e. for e > a) w(r) = (1 -r2)-1/2 and a Gauss-Chebshev integration technique is used to determine F”(r)[7]. The stress intensity factors are then evaluated from (see (4.5) and (4.6)) k(e) =
JaF”(
-
l),
(7.4)
k(g) = -J-P(l).
For e = a the function fir) is bounded at r = a (i.e. in (6.5) p = 0) and the stress intensity factor at r = g is obtained from
k(g)= -&Z&l).
(7.5)
The calculated stress intensity factors are given in Tables l-6. In the case of an embedded crack (i.e. for e > a or e, > 0) the stress intensity factors are normalized as follows (7.6) Table 1 gives an example showing the effect of the stiffness parameter I defined by (3.7) on the stress intensity factors. Similar results for a reinforced half plane are shown in Table 2. The stiffness constant 4 shown in Table 2 is defined by (5.17). The results for somewhat more realistic values of the stiffness parameter 1 and the radius ratio a/b which may simulate a pressure vessel with internal claddding are shown in Tables 3-6. The normalized stress intensity factors are calculated for two types of external loads. In the first it is assumed that a uniform crack surface pressure a,&, 0) = q(r) = -q,, is the only nonzero applied load. In the second the cylinder is assumed to be under internal pressure a,,(~ - h, 0) = -p,,. Generally, for the embedded crack the influence of the membrane reinforcement or the cladding on the stress intensity factors does not seem to be very significant. Since the properties of the clad material cannot be very different than that of the cylinder, this result is not surprising. However, in the case of the crack terminating at the interface (i.e. for e = a, Fig. 1) the stress intensity factors k(g) calculated for 1, # 0 and 1 = 0 are very Table 1. Stress intensity factors in a reinforced hollow circular cylinder containing a radial crack (Fig. 1). External load: uniform crack surface pressure q(r) = - qo, II = 0.56, g -e = 0.5(b -a), k(e)
e-a 6z
= &)lqoJE~,
Mg) = U)lqoJFFP
A=0
A = 0.2
x = 0.5
k,(e)
h(S)
G(e)
G(9)
0.15
1.1968
1.1754
1.1085
1.1255
1.0549
1.0988
0.20
1.1517
1.1769
1.1123
1.1436
1.0740
1.1223
0.25
1.1331
1.1951
1.1215
1.1706
1.0926
1:1523
0.30
1.1727
1.2303
1.1379
1.2100
1.1148
1.1930
0.35
1.1893
1.2912
1.1637
1.2702
1.1439
1.2526
0.40
1.2263
1.4033
1.2042
1.3764
1.1854
1.3559
t(e)
S(g)
878
R. TANG and F. ERDOGAN
Table 2. Stress intensity factors in a reinforced half plane containing a crack perpendicular to the boundary (Fig. 2). External load: uniform crack surface pressure q(y) = -q,,, & = h+/g, kAg0) = k ~~)/q~~~~, k&t3 = k(e~)~q~~~~~
xo=0
r
e”‘go k,(e,)
x* = 0.0375
k,(s,)
x* = 0.075
xo = 0.15
xo = 0.3
k,(eo)
k&g,)
k,(e,)
k&9,)
k,(eo)
k&9,)
k,(ec)
k&,1
0.1
1.42667
1.15500
1.41438
1.15234
1.40300
1.14983
1.38222
1.14510
1.34633
1.13661
0.2
1.20354
1.09670
1.19970
1.09543
1.19604
1.09422
1.18921
1.09190
1.17698
1.08765
0.3
1.11075
1.06242
~.?0909
1.06171
1.10750
1.06103
1.10449
1.05972
1.09902
1.05730
0.4
1.06224
1.03990
1.06145
1.03949
1.06068
1.03909
1.05923
1.03833
1.05655
1.03697
Table 3. Stress intensity factors in a reinforced hollow circular cylinder containing a radial crack (Fig. load: uniform crack surface pressure q(r) = - qo, a/b = 0.8, e/a = 1.05, 1). External k&z) = &~)~q~~~, k,(e) = k(e)lq~~~~ x=0 zz
A = 0.001
A = 0.002
k,(a)
k,(9)
k,(e)
k”(9)
k”(e)
a = 0.004
k,(g)
k,(e)
k,,(9)
0.05
1.0044
1.0041
1.0043
1.0039
1.0042
1.0038
1.0039
1.0036
0.10 0.20
1.0157 1.0511
1.0133 1.0382
1.0153 1.0497
1.0129 1.0373
I.0148 1.0484
1.0126 1.0364
1.0140 1.0459
1.0119 1.0349
0.30
7.0962
1.0668
t -0938
1.0656
1.0915
1.0644
1.0871
1.0621
0.40
1.1466
1.1023
1.1432
1 .lOll
1.1400
1.0998
1.1339
1.0974
0.50
1.2021
1.1596
1.1979
1.1584
1.1940
1.1573
1.1866
1.1552
Table 4. Stress intensity factors in a reinforced pressurized cylinder containing a radial crack. External e/a = 1.05, a/b = 0.8, pressure Ue) = W)bo$?, load: internal 6, = -Pa, k.(g) = W/&$~ )i = 0.004
x = 0.002
x=0 E
k,,(e)
k,(g)
k,(e)
k,(9)
k,,(a)
k,,(g)
0.05 0.10
4.3014 4.3346
4.2703 4.2662
4.3003 4.3307
4.2693 4.2633
4.2992 4.3273
4.2684 4.2606
0.20
4.4534
4.2872
4.4420
4.2800
4.4316
4.2734
0.30 0.40
4.6101 4.7854
4.3256 4.3935
4.5903 4.7581
4.3156 4.3828
4.5723 4.7332
4.3064 4.3730
0.50
4.9779
4.5473
4.945t
4.5382
4.9150
4.5297.
a in a reinforced cylinder containing a radial Table 5. The stress intensity factor k,(g) = k(g)/qoF crack which terminates at the interface (e = a, Fig. 1). External load: crack surface pressure q(r) = -qo, a/b = 0.8
is
x-o
0.25
1.3380
0.8894
0.8701
0.30
1.4477
0.9256
0.9023
0.40
1.7159
0.9886
0.95eo
0.50
2.0489
1.0375
1.0013
0.60
2.4352
1.0735
1.0345
0.70
2.8271
1.1065
1.0681
7
x = 0.002
x = 0.004
different. For A = 0 there is no reinforcement and the crack is an edge crack with a non-zero opening at r = a (i.e. uO(cz,+ 0) - ~+(a, - 0) > 0). On the other hand, for I # 0 since ~,(a, 0) = 0 there is a great deal of constraint tending to reduce the stress intensity factor. One may note that because of the membrane assumption made for the reinforcement at r = a the derivative of the crack surface displa~ment is finite giving a finite angle between the upper and lower crack surfaces.
Stress intensity factors in a reinforced thick-walled cylinder
879
Table 6. Stress intensity factor k.(g) = k(g)/poG in a pressurized reinforced cylinder containing a radial crack which terminates at the interface (e = a, Fig. 1). External load: a,,(a, 0) = -p,,, a/b = 0.8. (For the edge crack case, that is, for X = 0 the effect of the pressure q&r, 0) = -pO on the crack surfaces is not included.) (k,*(g) = k,,(g)/oee(a, O)\/g--a) k=o
x=0
k,(g)
k,*(g)
A = 0.002 k,(g)
0.25
5.8477
1.2836
3.8533
3.7684
0.30
6.2838
1.3794
3.9755
3.8734
0.40
7.3578
1.6151
4.1768
0.50
8.6933 10.2352
4.3154 4.3977
4.0443 4.1603
0.60 0.70
1.9083 2.2467
11.7739
2.5845
4.4642
x = 0.004 k,(g)
4.2320 4.3026
REFERENCES F. DELALE and F. ERDOGAN, Int. J. Frac. 20, 251 (1982). F. DELALE and F. ERDGGAN, Int. J. Solid Structures 5, 381 (1982). S. KRENK and M. BAKIOGLU, Int. J. Frac. 3, 441 (1975). J. DUNDURS, Mathematical Theory of Dislocations (Edited by T. Mura), p. 70 (1969). R. W. LITTLE, Elasticity. Prentice-Hall, New York (1973). R. J. TANG and K. WANG, Engng Frac. Mech. 16, 47 (1982). F. ERDOGAN, Mechanics Today (Edited by S. Nemat-Nasser), Vol. 4, pp. 1-81 (1978) and Vol. 6, pp. 199-202 (1981). [8] N. I. MUSKHELISHVILI, Singular Integral Equation. Noordhoff, Groningen, Holland (1953).
[l] [2] [3] [4] [5] [6] [7]
(Received 5 August 1983)