Penny-shaped interface crack between two dissimilar transversely isotropic layers

PEW-SHAPED ~~SS~~~~AR ~~~~t

INTERFACE CRACK BETWEEN TWO T~~SVERSE~Y fSUTRCWB2 LAYERS

WENHUA WE, R. S. DHALIWAL and H. S. SAXENA of Mat~~ati~ and Statistks, The Un~~~t~ of Calgary, CItEgary A&&s, Cam& T2N IN4

Abstract-TItt? problem of a ~~~~~~~ crack at tk? interface of two ~j~jrnj~ar traosmly isotropic Iayers has bexmconsidered. The m&od of Hankei trattsfm-ms&asbeen empliayedto reduce the problem to the solution of a system of singular integra1 equations, These equations have been further reduced to a system of algebraic equations by means of Jacobi pofynomiais.

IN TBEMA~~~~~~ and design of engineering structures and machines, acxxwdhg ta modem principles, flaws at the interfaces of dissimilar bonded materials are considered af great irn~r~~~~ In general, a complex ~s~~lat~~ stress field develops in the ~e~~~urh~~ of mechanical defmts such as cracks. Sneddon fl] has successfully introduced Hankel transforms to salve the pennyshtqd crack problem for isotropic materials when the pressure, which opens the surface of the cock, is constant. Later Sneddon f2] solved the problem of a ~~~y~sha~ crack under torsion, Lowengrub p], Sneddon and Tait [4] and Sneddan and Welch [Sj also investigated the pennyshaped crack in an isotropic medium for various other cases, M~s~k~~ki~ and Rybka f6] intros a method of f~~ulat~n~ the axially s~rnet~c penny-shaped crack problem in dissimilar media. Kassir and Bregman [7j and WiIlis[gl also ~~~~t~~at~ the stress intensity factors for a ~n~~~~~~d crack between two dissimilar materials, Etiogan [!I]considered the interface deny-shah crack problem by reducing it to singular integral equations. Kassir and Sib [I Of investi ted an e~~jp~~~crack problem in a transve~e~y isotropic solid md Parhi and Audi ElI] dads the d~st~b~t~~n of stress in a transversely isotropic cyfindcr containing a penny-shaped crack, Recently, Sxetia and Dhafiwal[12] considered the problem of a deny-shah crack at the interface of two t~~s~e~ly isotropic ha&spaces. In this paper we consider a penny-shaped inte$ace crack between two dissimilar transverse& isotrczpictayers (Fig. 1) and by means af Hankel transforms and Fourier tmnsfo~s the problem is reduced to the s&&n of a system of sin~ar integral stations. These equations are fu~er reduced to a system of sirn~lt~e~~ aIgeb&c equations by using the Jacobi ~~ly~~rn~a~ approximation. Numerical methods are empiayed to determine the stress intensity factors, which IIZEVCbeen displayed ~~~hica~ly,

84

HE et al.

WENHUA

-a

a

j

z=-h,

Fig. 1. A penny-shaped

interface

crack

between

two dissimilar

transversely

isotropic

layers

(2) where cijs are the elastic moduli of the transversely isotropic medium. In the absence of body forces, the equations of equilibrium may be written as follows: Cl1

ak = 0, $+pg-: +c&q~+(C,3+q.$)2 araz [ 1 (3)

Let ciiksbe the elastic moduli for the transversely isotropic medium in the region R, (k = 1,2) where R, is 0 < z 6 hz and R2 is --A, < z d 0. By means of the Hankel transforms, eqs (3) lead to the following solution: urj(r, z) = 2, [5 -‘{Aje 8,5;+Bie-.he.-+Cje6,5r+Dje-6,;1};5~r], uzj(r,z) = Xo[~ -‘{A,ccjeBJ” - Bjcr,e p~c’+ C,y,e6Jt’ -D,yje-6~eZ}; t-t-1,

(4)

where urj and uzj denote the displacements for the region R, (j = 1,2), C.&, - cllj6,-

~ = c4& - cn,P,-’ I

’ Cl3,

+

(8,Y4) =

‘i,=

c44,

VI

i

I!I

J

?: -

’

cl,,+%,

4clI jc33jcLj

2c33jc44j

’

“’

I’

and Aj, 4, C, and 0, (j = 1,2) are unknown functions of 5 to be determined by the boundary and continuity conditions.

85

Penny-shaped interface crack between two isotropic layers

Substituting from eqs (4) into (2), we have crzj = %,[((c,~j+

c,,4&)(Aje”jtz

+

Bje-‘~“)

+ (C,Jj+

~jdjC~~j)(Cje6jtz

B,e+@) + (6, - y,)(Ciedi”-

a,zj=c,j~,[((/lj-aj)(Aj&tz-

+ Dje-“jtz)};

< +r],

Die-gltz)}; {+r],

(6a) (6b)

where trzzjand CJ,=~ denote stresses for the region R, (j = 1,2). 3. STATEMENT

OF THE PROBLEM

AND BOUNDARY CONDITIONS

We assume that two dissimilar transversely isotropic elastic layers, which occupy the regions < z < 0) respectively, are perfectly bonded except that there is a crack in the region 0 < r G a, z = 0. In the first case we also assume that the surfaces z = -h, and z = h, are stress-free. It is assumed that the surfaces of the crack are subjected to prescribed normal and shear stresses p,(r) and p2(r) respectively. Hence the problem of determining the stress and displacement field is subjected to the following boundary and continuity conditions: R, (0 < z c h2) and R, (-h,

orrj(r, O) =PI Cr)9 arzj(r, 0) =P2(r),

&I (r, O+) = &, b,,, (r, O+) = &r,

0< r < 4

(7)

r > a,

(8)

0%

u,, (r, O+) = n,z(r, o-),

O-),

a,,, (r, O+) = o,&, O-),

G (r, &) = 0, ozzz(r, -h,)

i = 19%

=0,

Q,~,(r, h,) = 0, cr,*(r, -h,)

r > a,

r 2 0,

=O,

(9) (10)

r 2 0.

(11)

With the help of conditions (7), the conditions (9) may be replaced by the following: a,,, (r, O+) = nzz2(r, O-),

crz, (r, O+) = 6,:?(r, O-),

r 3 0.

(W

4. ANALYSIS Let us denote bj = Cuj(fij - ai),

aj = c,sj + ajfijcjjj,

dj= c,U+ Giyjcsjj,ej= cMj(dj- yj), j = 1, 2.

(12)

Applying conditions (9a), eqs (6) give a#,

+B,)+d,(C,

+o,)=a*(A,+B,)+d*(C,+D,),

b,(&-~,)+e,(C,-~,)=b~(~~-~~)+ez(C,-~z);

(134 (13b)

and the conditions (l(y) and (11) yield a, A, eSIChz + a, B, e -hh

+ d, C, &h + d, D, e-4& = 0,

VW

b,A, eflICh2 - 6, B, e-hth + e, C, e4% - e, D, e-&P2 = 0 7

U4W

a, A2 e-s2th, -I-a2B, es2B 1+ d2C, ee62thl+ d2& &thl = 0 >

(14c)

b, A2 e-hthl - b2 B2 &thl + e, C, em62thl- e2D, e6,@,= 0.

(14d)

Solving eqs (13) and (14) we obtain C, = -1,,A, e@,-“,Kh2-/,2B, e-U,+J,Khz, D, = -luA,

e(B,+J,Kh2 - l,,B, e(-B,+S,Khz ,

C, = - l,, A2 &h

+ Qth, _ I,, B2 ebb+ Wh,

D, = -122A2e-@2+62’h _&,B2e(82-Qth~, 4=4,4+1x4,

B,=l,,A,+l,,B,,

(1%

86

WENHUA HE et al.

where l,j

=

[ale, - (- l)%,d,]/(2d,e,),

L = (a,,& - a&,)/A,

12,= [a2e2 - (- l)jb2d2]/(2dZe2); j = 1, 2,

132= (a,A

- a22b12)lA,

l,, = (a~, b,, - a,, bz, )/A, 142= (az, b,, - a,zb,, )/A> alI = a, _ d, I,, e(B,-61,ihZ_ d, I,, e-(81+61,Sh2, aI2 =a, _~,~,2e-,81+61)t*~_d,~,,e(-B~+61)C*z, _ d21z2e-(82+&X*,, &I = a, _ d2&, e(-82+a>,C*, - d21z,eW+%,th,, a2, = a, _ dzlu e(B2+~z,C*, b , , = b, - q I,, e’JI - 61)Ch+ e, I,, &‘I + 61X*2, b ,2= -b,

-ee,l,,e-(B~+s,)Ch2+e,l,,

~(-PI+~I)%,

b *, = b2 - e212, e(-i%+h)t*l + e21z2e-(82+*2)S*t, b22 = _b2 _ e21z2&Jz+h)
(16)

The conditions (7) and (8) lead to the following dual integral equations:

~~[(a,A,+a,B,+d,C,+d,D,);5~rl=p,(r);

r
(17)

~,[(b,A,-b,B,+e,C,-e,D,);5-*rl=P*(r);

r
(18) r>a,

~,[5-‘(A,+B,+C,+D,-A*-B,-C,-D,);5--*r]=O;

S’,,[<-‘(al A, - a,B, + y, C, - y,D, - cc,A, + u,B, - y2C2 + yzDz); 5 -+r] = 0;

(19) r > a.

(20)

If we introduce two new unknown functions 4, (0 and (bz(5) by the relations A, + B, + C, + D, - AZ - B, - Cz - D, = 4, (t), a,A,-a,B,+y,C,-~,D,-cc~A,+cr,B,-y,C,+y,D,=~,(r), then eqs (17x20)

(21)

may be rewritten in the form ~OKA,A

+ LA);

5 +I

=P,(r),

r
(22)

~,I(&&

+&42);

5--l =Mh

r
(23)

X,[5-‘6,(~);5+rl=O. Ho]5 -‘&(5); 5 -1

r >a,

(24)

= 0, r > a,

(25)

where A# are functions of < and are given by I.,, = (a,,~,, - a,,y,,)lA*,

&z = (a,,~,, - a,,y,,)lA*,

A,,

=

(b,,

A22 =

A*

=

~12~2,

y,z

-

-

b,,y,,

YI,

)/A*,

(h2Y2,

-

h,

722)/A*, (26)

722

and yik are defined as follows: y,, =a,-~,l,,ecB~-6~)Ch~+~,l,2e(B~+‘1)e*2-a2l~,+a2l,,

-y21s,+y216,,

y12= -a, - y,l,2e-(PI+s,)eh2+ y,l,, e(-*~+a~)eh2-a2132+a21~2- y21s2+y2162, y2, = 1 _ I,, e(BI-~Oh _ I,, @t+6t)C”2- 13,- 14,- 15,- &,, YE= 1 _1,2e-(B1+“~)C*~-~,,e(-81+~~)~*~_1~2-1~2-152-1~2r 1~, = _12,13, e(-82+4)C*l _ lz214,e(h+62)h, ls2= _12, 132e,-L%+%K*l- lz21d2 $%+hJthl

s’@(r); y] = (2/1r)“* wP(r)sin(ry) dr,

s0

y3 = (2/7G’” mF@)cos(ry) dr. (33) 50 Applying the operators J$l r, dr, .d&r and d#, to eqs (22), (231,(24) and f25) respectively, we obtain ~,W);

(34) (35) (361 (37) where f; W = dt ivt W; yl, My) = y4 IMr); y] + cT c = *,ifn,, Qit+ &&); 01.

138)

By taking y = ax, eqs (34)-(37) can be written in the foliowing form: @x&,6, +&&);4=m~

OLaEx Q 1;

(34’)

~cE@i,16,+ &;2$6); 4 =Aw,

0 d x G 1;

(35’1

gFfp.&;x]=O, x > I;

(351

3q&;n]=O,

(371

x > 1;

where &W =&Ma),

&p;C@ = rbsW),

_&x1= a@x),

j, k = I, 2.

(38’)

If we introduce two new unknown functions er(t) and e2(t) such that

gPc f&(x1; 4 = $4ww

- Q, t 2 0;

(39)

~J&Cx>; fl = MW(1

- t), t 3 0;

(@I

WENHUA

88

HE et al

where H(x) denotes the Heaviside unit function, eqs (36’) and (37’) are identically satisfied. Let #Jd) and e@(t) be the even extension of $r(r) and the odd extension of &(t) on I-- 1, $3 respectively, then we get [I 31

Now if we denote by F,(x) the odd extension of j; (x) and by F,(x) the even extension of j;(x) on [- 1, I], eqs (34’) and (35’) become ~~~~~)~~~~~-F

‘, g I

h(oQMA~~

&2(~) + ~ 71

n<,(x) -

9

dt +

’ ~~~~~~~~ G, fl df s --I I + Jld~Wrzfx, s I

f) dt = F,(x),

Ix I -= 1;

(43)

’ ’ tioU) & + $,(~)~a h, fl dt s -i s _it -.x

-$ ;I F;

dt +

S’-I

i, (t)Mz, (A t) dt i-

S’-I

Mf)W!z(x>t)dt

=gz(x),

(47)

Penny-shaped interface crack between two isotropic layers

Y*k t) +

J;i;

89

K*,(x,t)+i

.

~**W&l

(48)

The analytic solution of eqs (46) and (47) has been extensively studied (see, for example, refs [14] and [IS]) by using a regularization method, which, in this case, however, becomes cumbersome. Here we use an approximation method described by Erdogan [16] to find the stress intensity factors. Since the kernels Miks are bounded, aside from a multiplication constant, the singular behaviour of the functions cj (j = 1,2) at the points x = + 1 is determined by the dominant part of the singular integral equations. Equations (46) and (47) will be solved under the assumption that iyezdls, 2) satisfy a Holder condition on every closed part of the interval (- 1, 1) not containing The solution of eqs (46) and (47) may be assumed in the form of Jacobi polynomials Pfi*‘Ik)(x) by [161

(49) where

k = 1,2; and C,, are unknown coefficients. Substituting from eq. (49) into eqs (46) and (47) we obtain f

n=I

G”

(1 -A*)“*

P(-“I*-rqx) + ‘, n-l S[

2i

$, (C,,M,,(x, + C&,2(x,

~‘, 2i

n=l

(1 - A*)“* P::“;*-‘2’(x) -

1, S[ +

t)W,(t)PpJ”(t)

1

t)W2(t)P$-(t))

dt = g, (x),

Ix I < 1; (51)

$, (C,,M*,(x, t)W,(t)P$“Iyt)

1

~)W~(C)P~~~~~(~))dt = -g2(x),

C2,,~22(4

Ix 1< 1; (52)

where we have used the following result of Karpenko [17]: -$

; W&)P$*‘k)(t) I I

&

+(-l)~nW,(x)Ppqx)=

-(l

-~*)“*P~-et,-9)(X),

Ix/< 1. (53)

Multiplying eqs (51) and (52) by W;‘(x)Pj!-“l~-‘l)(x) and W,-l(x)P~-Q~-rz)(x) and using the orthogonality relations of Jacobi polynomials

s I

W(t)P~)(t)P~)(t)

dt =

-I

@.e =

’ 02)

respectively,

n #m, n=m,

2"+'+'f(m+ d + I)f(m+ T + 1) m!(2m + 0 + t + l)r(m + 0 + T + 1) ’

(54)

WENAUA

90

HE et al.

we

get the following infinite system of simultaneous algebraic equations for the determination C kS4l: -(l

-1’)‘nCrM6”,-“~~-‘~~+ 2 (L&,,,,C,,+L&,,,C,)=G,,, n-1

of

(551

where

I Gkm

gk(x)w,-‘(x)P’-~~,-‘~) M

=

I k,j=

h)dx,

-I

t,2;

n,m=l,2,3,4,

,_=.

(57)

After we find the coefkients C,, by solving the system of linear algebraic stations (55) and (56), we can determine the stress intensity factors. To do so, we make use of the following

91

Penny-shaped interface crack between two isotropic layers

which by using eq. (61) gives ~&,,

(aMI (r);

rl = 0, r > a.

(65)

Also,

which by using eq. (59) gives ~rlM~M*(t);

J

= -“‘:p”’ (2/~)“~ f 2 r’ +2(t)t/(s2 - t2)“2 dt,

rl

r

(67)

>a.

0

Similarly by means of eqs (58) and (60) we get

rl = 0, r > a;

XI [n22(~)#2(0;

(2/~)“~ ii

(68)

’ $, (t)/(s2 - t2)“’ dt,

r

>

a.

(69)

f 0

Now

I

I‘Sd~)CWW

m5[IZ;,(r)-1,,(~)IJo(~S)d5

JI",I(n,,(S)-~,,(m))~~(~);r1=~(2/n)'"

=~(2,~)“2~~,(Ildf~~~~,(~)i,,LalY,od(

=

and when r+u+,

&‘,[(A,,(t)

- A,,)&;

-Jf0[(42(5)

3f1[@2,

as r+a+,

O(l),

(0

9f1U22W

has no singularity. Similarly we have

r]

-

42(~N&2(0;

-

R2,

-

~22(=3))42(0;

(a~)%

(70)

rl =

O(l),

as

r+a+;

rl = O(1), as r-a+;

(0;

rl= O(1), as r--w+.

(71)

So as r+a+, azzl(r, 0+)a2&A,,(co)

=

lim $ ’ ,/&h2(t)t/(s2 s-it+ s 0

s’ 0

J&II 0)l(s2-

- t2)“2dr + O(l),

(72)

t2)“2dr + O(1).

(73)

But

$ I’ ,&b2(t)t/(s2 -

t2)“2dt = g

I’ [C,(t) - [&)]r/(s2 - t2)“2 dt

0

=+..g,k$,

(-l)*+‘C+

I’r(~)~‘~~,r&)(~)(~2-rt)-“2dr], 0

(74)

92

WENHUA HE er al.

and

I ;

[

,/&

(t)t/(s* - t*)“* dt = 2

j’ [i,(t) +

0

idt)l/(s* - t2F2df

0

If we define the stress intensity factors K, and K2 by mxl% 12

K, f (- l)‘+‘im fi

K, = lim (7c/2)“*[(r - a)“2-iwk(r + u)“~+~“‘,] f-W+

21

-~z,l(r,O+)+(-l)k+‘i

-

J*,(cf3)

then using the substitution s = r/a, Vainshelbaum [ 191 by taking

1 k=

v%orr,(r, O+)

I

1,2;

(76)

eqs (72)--(75) and following the method of Goldstein and

x=1---.--

s-l S--t’

Fig. 2. Numerical values of the stress intensity factor K, against h/a (0.2-1.0) for a fixed crack length a and equal thickness (h, = h, = h).

Penny-shaped interface crack between two isotropic layers

93

we get

n$)K

I +(-lri’i&K

2-

-(-l)k+‘iu[7r(1

-~*)]“*

21

12

r(l + io,) x f C,,Pj,Q~Q’(l), ’ r(1/2 + iok) II=1

k = 1,2.

(77)

5. OTHER CASES In the previous two sections we have considered the case when the surfaces z = -h, and z = h2 are stress-free. In this section we will consider other kinds of boundary conditions. 5.1. One face jxed

and the other stress-free

In this case, we assume that all the conditions (7)-(10) remain the same and the conditions (11) are replaced by ~~~(r,-h,)=O,

u,~(T,-h,)=O,

r20.

(78)

In this case conditions (78) yield the following algebraic equations: A2 e-b2ghl+ B, e82ghl+ C, e-J2thl + D2 e4Chi= 0, A2a2 e-@2ghl - B,a, efi2chl+ C, y2eP62ghi- D2y2 es2chl=

(79) 0,

z P. d B. 6

f-

Ei6.

ri

!i, 1.0

a.0

a.0

GO

s.0

a.0

7.0

0.0

LO

h/o

Fig. 3. Numerical values of the stress intensity factor K, against h/a (I&10.0) a and equal layer thickness (/I, = h, = h). EFM U/I-G

for a fixed crack length

(80)

WENHUA

94

HE et al.

which replace the algebraic equations (14a) and (14b) respectively. Comparing equations, we find that if we take a2=

1, d,= 1,

bz=a2,

the two

e,=yZ,

sets

of

(811

in the results of Section 4, we get the solution for this case.

In this case all the conditions (7)-(9) remain the same but the conditions fl0) are replaced by the following: utl (r,@=O,

z+(r,h2)-0,

r 2 0;

W

and the conditions (11) are replaced by f7gf. In this case, eqs (14) will be replaced by eqs (79}, (80) and the following equations: A, eSl& + &, e-81th2+ C, e”1@2 + Dz e-61& = 0,

(8%

~~~~ $I& _ &,a, e-#%& + Cry, e-&chz_ ~,y$ e-&t% = @_

(84)

Comparing the two sets of equations, we find that if we take ai==di=

1, bi=ai,

e,=y,

(i =

1,2),

(85)

in the results of Section 4, we get the solution for this case.

6

Penny-shaped interface crack between two isotropic layers

95

5.3. One face rigidly restrained and the other fixed In this case, we assume that the boundary conditions (11) are replaced by (78) and conditions (10) are replaced by uz,(z,hz)=O,

g,z,(z,h,)=O;

(86)

and all the other conditions remain the same. So eqs (14) should be replaced by (84), (14b), (79) and (80). Solving eqs (13), (84), (14b), (79) and (80) we can express B, , D,,A*, B,, C, and D2 in terms of A, and C, by the following equations: B, = A,

e2dleh2, D, = C, ealCh2,

C, = _12,A2 e(-B2+Wh~_ /,,B, &%+*2Xhl , - /,,B,&b-&k?‘~9 D2= -~,,A2e-&2+~2Xh~ A2

=

4,

A,

+

1,2G,

(72

+

a2 WY2

B2

=

4,

A,

+

(87)

42C,,

where I,, and l,, are given by

12, =

3

42

=

672 -

a2 m,

,

Fig. 5. Numerical values of the stress intensity factor K2 against h/a (1.0-10.0) for a fixed crack length a and equal layer thickness (h, = h2 = h).

(88)

96

WENHUA HE et al.

and ifi (j = 3,4; k = 1,2) are still given by (16) while the expressions for a$ and bik (j, k = 1,2) are given by the following:

b

Consequently

2,

=

b, -e212, ec-@2+s2)chf,& = -b, + e2iue@2e+)thr.

(89)

the expressions for yjk (j, k = 1,2) for this case are given by - @z&r+ Y~~=~~-Q~~ *@lCh2 y12 = y1 -

yI e2’iih2 -

h&2 + a2b2 + ~~42 + ~~~~~~

~2~ = 1 + e2BiTh2 - i,, y22 =

=2&t + Yd,, + Yzf,, ,

I,, -

is, -

&,

I + e261th* - I,, - l,, - is2- la2,

(90)

where bk (j = 5,6; k = 1,2) are the same as given in (27). The solution for this case is given by the results of Section 4 after we have made the above replacements. 5.4. One face rigidly restrained and the other stress-free

In this case, the boundary conditions (10) are replaced by (86) and all the other boundary conditions remain the same as in Section 4. So in eqs (13) and (14) we only have to replace eq.

Fig. 6. Numerical values of the stress intensity factor KI against the crack length a (1 B-10.0) for the layer thickness h, = hpco.

Penny-shaped interface crack between two isotropic layers

97

(14a) by (84). Now we find that for this case b (j = 2,3,4; k = 1,2) are all given by (16), uik and bfi (j, k = 1,2) are given by (89) and the solution for this case is given by the results of Section 4 after we have made the above modifications. 5.5. Both faces rigidly restrained In this case, we are assuming that the boundary conditions (10) and (11) are replaced by conditions (86) and a,&, -h,)

= 0,

G&,

-h,)

= 0.

(91)

Equations (14a) and (14c) will be replaced by eqs (84) and (80) while all the other equations remain the same. Solving eqs (13), (84), (14b), (80) and (14d) we can write B, = A, ezfil@l, D, = C, e261 th2, B2 = A, e-%thl,

D2 = C2 e-%Ehl,

A2

,

=

4,

A,

+

132 C,

C2

=

14, A I +

42 G

,

(92)

where 4 (j = 3,4; k = 1,2) are given by (16) but the expressions for ajkand bjk(j, k = 1,2) should be replaced by the following: alI =

a, (1 + ez81th2), a,, = d, (1 + ezalChz),

azl = a2( 1 + e-‘fi2thl), az2= d2( 1 + eVza2thl),

Fig. 7. Numerical values of the stress intensity factor K2against the crack length a (1.0-10.0) for the layer thickness h, = /I~-CCI.

98

WENHUA HE er al. b,, =

b,(l - e2Bl(ih2),b,, = e,(l - e2slE*2),

b,, = 6,(1 - e-2g2Chl), b,* = e,(l - eez62thl). Consequently,

(93)

for this case we should replace yjk (j, k = 1,2) by yII = a, (1 - e2BICh2) - a2i,, (1 - e-282chl)- y21d,(1 - e-262thl), y12= ~~(1 - e2’lC*2) - ~r,l~~(l- e-2a2Chl)- y21d2(1- e-262’hl), y2, = 1 + t+‘lCh - I,, (1 + e-*h~h) _ l,, (1 + e-*bMI) , y2* = 1 + e2bch2- [,,(I + e-*62%) _ /,,(I + e-%thl)

(94)

Then the solution for this case is given by the results of Section 4.

6. NUMERICAL RESULTS AND DISCUSSION To evaluate the stress intensity factors, we truncate the infinite system of simultaneous algebraic equations (55) and (56) at n = 10, and use a Gaussian quadrature formula to perform the numerical integrations involved in the solution. Numerical results for the stress intensity factors K, and K2 are obtained for the case when the crack is subjected to a constant pressure p, (r) = p0 and p*(r) = 0, the thickness of the layers is kept the same (i.e. h, = h, = h) and the surfaces z = -h

Fig. 8. Numerical values of the stress intensity factor K, against the crack length a ([email protected]) for the layer thickness h, = h, = 20.

99

Penny-shaped interface crack between two isotropic layers

and z = h are stress-free. For the two transversely isotropic materials considered here, the numerical values of the elastic moduli are taken as follows (10” dynes/cm*):

Cadmium Beryl

CII

Cl2

Cl3

c33

11.00 26.94

4.04 9.61

3.83 6.61

4.69 23.63

126 6.53

(layer 1) (layer 2).

Numerical values of the stress intensity factors have been calculated for the following three particular cases. Case 1. Fix the length of crack a = 1.0 and let h, /a = h2/a = h/a = 0.2(0.2), 1.0,2.0(2~0)10.0; the stress intensity factor K, against h/u is displayed in Figs 2 and 3, and the stress intensity factor K2 against h/u is displayed in Figs 4 and 5. Case 2. Let h, = h2-*c0, and the length of the crack a = l.O( 1.O)lO.O;the stress intensity factor K, against a is displayed in Fig. 6, and K2 against a is displayed in Fig. 7. Case 3. Let h, = h2 = 20.00 and the length of the crack a = l.O(l.0)lO.O; the stress intensity factor K, against a is displayed in Fig. 8, and K2 against a is displayed in Fig. 9. We make the following observations from the graphs: when the length of crack a is fixed at a = 1.0, the stress intensity factor K2 increases as the thickness of the layers increases, while the stress intensity factor K, decreases as the thickness of the layers increases from 0.1 to 1.0, but increases when the thickness of the layers increases from 1.0 to 10.0. If the thickness of the layers is fixed at 20.0 or tends to co, the stress intensity factors K, and K2 decrease as the length of the crack increases.

Fig. 9. Numerical values of the stress intensity factor K2against the crack length a (1.0-l 0.0) for the layer thickness /I, = h, = 20.

100

WENHUA HE et al.

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HI

I. N. Sneddon, The distribution of stress in the neighborhood of a crack in an elastic solid.

32, 829 (1965).

stress around elliptical crack in transversely isotropic solid. Engng Fracture Mech. 1, 327 (1968). [Ill l-l. Parhi and A. Atsumi, The distribution of stress in a transversely isotropic cylinder containing a penny-shaped crack. In!. J. Engng Sci. 13, 675 (1975). WI H. S. Saxena and R. S. Dhaliwal, A penny-shaped crack at the interface of two bonded dissimilar transversely isotropic elastic half-spaces. Engng Fracture Mech. 37, 891 (1990). [lOI M. K. Kassir and G. C. Sih, Three-dimensional

P31 M. Lowengrub and I. N. Sneddon, The stress field near a Griffith crack at the interface of two bonded dissimilar elastic half-plane; Int. J. Engng Sci. 11, 1025 (1973). iI41 N. I. Muskhelishvili, Singular Integral Equations. Noordhoff, Groningen, Holland (1953). F. D. Gakhov, Boundary Value Problems. Pergamon Press, Oxford (1966). iti; F. Erdogan, Approximate solutions of systems of singular integral equations. SIAM J. appl. Math. 17, 1041 (1969). (171 L. N. Karpenko, Approximate solution of singular integral equation by means of Jacobi polynomials. PMM 30, 668 (1966). A. Erdelyi, Tables of Integral Transforms, Vols I and Il. McGraw-Hill, New York (1954). iti; R. V. Goldstein and V. M. Vainshelbaum, Axisymmetric problem of a crack at the interface of layers in a multi-layered medium. In!. J. Engng Sci. 14, 335 (1976). (Received 24 April 199 1)