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Two bonded strips containing an infinite row of interface cracks
hf. 3. finmg ScL, 1976, Vol. 14, pp. 831-843.
Perpunon Press.
Printed in Great Britain
TWO BONDED STRIPS CONTAINING AN INFINITE ROW OF INTERFACE CRACKS K. S. PARH-IAR and A. C. GARG Departments of Mathematics and Aeronautical Engineering, Indian Institute of Technology, Bombay, India Abstract-The plane strain problem of two bonded dissimilar isotropic elastic strips is considered. It is assumed that the composite strip contains an infinite row of interface cracks located symmetrically on the centerline. The case in which each of the cracks is opened out by the same constant pressure is discussed in detail and numerical results are reported for quantities of practical interest.
I. INTRODUCTION THE STRUCTURAL strength
of composite materials is controlled to a considerable extent by the size, shape and distribution of the flaws which exist in the material, usually, in the form of weak impurities on the interface, ruptured bonds and cracks. From the viewpoint of fracture initiation and propagation in the medium particularly important are the manufacturing flaws such as flat cavities which develop during bonding and casting. The problem of a multilayered composite containing a single crack on an interface has been studied by Erdogan and Gupta [ 11.In particular, they solve the problem of two bonded layers containing an interface crack opened by a uniform internal pressure. In this paper we consider the plane strain problem of stresses near an infinite row of cracks at the interface of two bonded dissimilar strips (see Fig. 1). The cracks are assumed to be of equal length, spaced at equal distance and opened by the same internal pressure. Even though the details are presented only for the case in which the edges of the composite strip are free from tractions, the results for some other types of boundary conditions (for example, smoothly restrained edge conditions) can be easily reproduced following a similar procedure. By simple symmetry considerations we see that the stress distribution in the entire strip can be deduced from that in the rectangle (xl5 a, ly] 5 S with the interface crack 1x1I c (y = 0). The edges y = f S are free from tractions and on the edges x = 2 a the symmetry conditions are equivalent to the smoothly restrained edge conditions.
Fig. I. Strip geometry.
The method of solution has affinities with those of Sneddon and Srivastav [2], Lowengrub and Sneddon[3]. We take a finite Fourier transform representation for the displacement field in the composite rectangle and derive a set of simultaneous dual series relations. By using a technique similar to that of Parihar [4] we reduce the problem of solving the dual series to the solution of a pair of singular integral equations which in turn is reduced to the Hilbert problem[5] with the line of discontinuity as an arc of a unit circle. When exp (4s) s 1 (4 = w/a), an iterative solution of the Hilbert problem can be obtained. The corresponding expressions, up to the order of exp (-4qS), for the stress intensity factors are derived and their dependence on various geometry parameters is shown graphically. 2. FORMULATION
OF THE PROBLEM
Consider the plane strain problem for the composite rectangle 1x1I a, (y I 5 S. Let a medium with elastic properties CL,and k, = 3-4~~ occupy the upper rectangle y > 0 and a medium with 831
K. S. PARIHAR and A. C. GARG
832
elastic properties p2 and k2 = 3 - 4v2 occupy the lower rectangle y < 0, where p ,, pz denote the rigidity moduli and yl, yz the Poisson ratios. Let [UP’, u:‘], k = 1,2, denote the displacement vectors for the upper and lower rectangles respectively and there is a corresponding notation for the stress components. On the edges x = 2 a and y = 2 6 we have u:k’(+a, y) = 0,
u:y(ka, y) = 0,
k=1,2
(2.1)
aE’(x, 5s) = 0,
a:;‘(x, 55) = 0,
k = 1,2
(2.2)
and the mixed boundary conditions at the interface are given by ai:‘(x, 0) = ug)(x, O), 1x15 a,
u::)(x, 0) = u:;(x, O),
(2.3)
u:“(x, 0) = uY’(x,O),u’,“(x,0) = u:*‘(x,O),c < 1x1I a, 6%
(2.4)
&)(x, 0) = 0, Ix) I c,
0) = - P (Ix/),
(2.5)
where p(x) is the internal pressure applied to the crack faces. To solve the mixed boundary value problem for the composite rectangle we assume a displacement field in the form u:‘(x,
y) = 2 [A, “=I +
+ 2~0,
+
{A,(k, + Z&Y - 2~5)
2cz,,(2a,ys t k,y - 2a,6’)B, - (1 - k:)BJ
x exp {~CL(Y- @)I exp (my) ul”(x, y) = ud”+ s, [A, + 2(a,y
t k@,
t
{(k, t 24
x
exp {2a.(y - S))l exp (-KY) cos (u),
ur’(x, y) = 2 [C, + 2(~,yD, + {(kz - 2~ n=l x exp {-2~(y
+ kJ(2h
- 2cw,y)A, t (24
t2cr,(2a,S2t2a.Sy
tk2y)Dn
+ k, - ky
Uk + B. 1
(2.7)
- 24)C, +(l-k:)D.}
(2.8)
+ S)}l exp (Amy)sin(u),
u:“(x, y) = us”-- z, [C,, t 2&y - k,)D, + {(kz + &Y -
(2.6)
sin (ML
t 2&W’”
(2~~~6t k2)(2a,S t k? t 2cy,y)D, - Dn) exp
{-2a,(y
t
@)I (2.9)
x exp (a.y) cm (w 1, where A., B,, C., D, are arbitrary constants and a” =
nq = nala.
(2.10)
By using the stress displacement relations we find that the corresponding stress components for the upper rectangle are given by
uZ(x, y) = 2/.~,2 (Y,[A, + {k, - 3 + 2cu,y}B, n=I + ((3 t 2a.y - 2a,6)A, - [2a. (6 - y)(2a,6 - k,) + I- 3kl - 6a,SlB,} exp {2cr.(y - S)}] exp (-a,~) cos (a,~), a::(~, y) = - 2/~, 2 cz,[A, + (2a,y t k, - l)B, n=, t ((2~3 - 2cz,y - l)A, t (2~x,(S- y)(2&
t k,)
(2.11)
TWObonded stripscontaining an infinite row of interface cracks f I - k, -
2cd>B,} exp {2cu,(y - &)}Iexp (-fky) sin (ax),
833
(2.12)
a:;(~, y) = - 2p, 2 a,, [A, t (2amy+ kl -I-I)&, - {(2ad - 2cr,y t l)A, II=1 t (2a,(S - y)(2a$ i- &,If 1+ k, f 2ad)B,) exp {2~y,(y- S)}] (2.13)
x exp (-any 1cos frullx 1, and for the tower rectangle we have
t ((3 - 2&y - 2anS)C, + (2a, (S t y)(2ru,St k*) t 1- 3k2 - 6a,S)D,} exp {-2ff,(S f y)}] exp (cu,y) cos (cu,x),
(2.14)
&Xx, y) = 21.122 cu,lC, + &K,Y - kz + W. n=l
t
{(2a,6 t 2amy- l)C, - (2tu,(S t y)(2& t kz) + 1- kz - 2a,&>Dm)
x exp {---2cy,(y+ S))l exp (any) sin (a~),
(2.15)
- ((2anS t 2any t I K’,, - (2~ (S t y N2ad t kd t 1 t kz t 2a,,S)D,} exp {-2a,(y
t S)}] exp (~,y) cos (a,~).
(2.16)
It is easily verified that the displacements and stresses (2.6)-(2.f6) satisfy the boundary conditions (2.1)-(2.2). The conditions (2.3) at the interface are also fulfilled if
cm= - ~~lAtff~~)~[A”{k*+ Gb-nS)l+ RVcrkz- I+ D, = --{P/A(QI,S)}~A~~~ t GdwVlt
GhMlt
Bnk t G.&nWl,
(2.17) (2.18)
where A(X) and G&X) (i = 1, 2, 3, 4) are defined in the Appendix and we have CL =
cLll/.h
(2.19)
The arbitrary constants A,, B, are to be determined by the mixed boundary conditions (2.4)-(2.5). We might note here that the resultant force on any cross section normal to the bond line is zero. Indeed, by means of the relations (2.11)-(2.19) we have 6 I0
a%,
Y) dy +
a:?@, y) dy = 0, 1x1~ a,
(2.20) (2.21)
The stress components at the interface may be written as a::‘&, 0) = u:(x, 0) = - g ,{( 1 + pkz)(p t k,))-’ (2.22) a$yn, 0) = uZz:Cn, 0) = - p,ifl + /.&)(1.L + Ml-’ (2.23)
where Hi, (i = 1,2,3,4) are defined in the Appendix and
834
K. S. PARfHAR and A. C. GARG
a,=Cr,+k,-l-Nkzt
(2.24)
az=Ir.fkt+l+~kz.
The constants S,, T, in (2.22)-(2.23) are given by S,, = [An{1 f yk,t
N,(rw,S)l+ B,(/.&,kz-- iI+
T, = [A,(1 t kkz + N&,,S)). f &(2kr + plk,k,+
2ru2(a,s))l{A(cr,S)l-‘,
(2.25)
1) + ~N~(~~,S)~I~A~LU~S))-‘~ (2.26)
where Ni(x) (i = 1,2,3,4) are defined in the Appendix. In view of the relations (2.17)-(2.18) and (2.25)-(2.26), the displacement and stress fields (2.6)-(2.16) can be expressed in terms of the undetermined constants S,, T,. The mixed boundary conditions (2.4H2.5) are also satisfied if S,, T, are solutions of the simultaneous dual series relations $BS,sin(*mx)=O,
To+
(2.27)
c
2 T, cos(a,x)=o,
(2.28)
C
tl=l
m
x Pft,S, + H2,T,la,
+
cos (c&x) = P(x),
0 < x < c,
(2.29)
“-8
-.fg,
[a,T,
-
ta2S~lm(GJ)
s,[H3”T, t H4,,S,,]a, sin (GX) = 0,
0 < x < c,
(2.30)
where H;:, (i = l, 2, 3, 4) are defined in the Appendix, a,, a2 are given by (2.24) and P(x)=p(x)(l”rCLk2)(~Lkki)/lU,,
To=u;‘--t@.
(2.31)
If the material constituting the rectangle is homogeneous, the dual series relations (2.27X2.30) reduce to the form Tot
5 T, cos (a,~) = 0,
c < x < a,
(2.32)
PI=,
$2,
T*[a~ + K(~Y,s)I sin (a,x) = P(X),
o < x c c,
(2.33)
where a2, P(x) are given by (2.24), (2.31) with lu.= 1, k, = k2 and Hz is defined by Hz(x) = - &2x2+ 2x t 1 -e-2x)(2x f sinh 2x)‘.
(2.34)
The dual series (2.32)-(2,33) are the same as obtained by Srivastav[6]. In the next section we consider the problem of determining S,, Tn satisfying the dual series relations (2.27)-(2.30). 3. SOLUTION
OF THE DUAL SERIES RELATIONS
We first reduce the dual series relations (2.27H2.30) to a pair of simultaneous singular integral equations. Follo~ng Parihar[4] we set
S, =;
I
f h(t) cos (ant) dt,
(3.1)
Two bonded strips containing an infinite row of interface cracks
sin (a~) dt,
T. = $ \‘g(l)
cl
To = (q/2) [
835
tg(t) dt.
(3.2)
This trial solution satisfies (2.27H2.28) for all choices of g(t) and h(t) provided ch(t)dt I cl
=O.
(3.3)
The remaining eqns (2.29)-(2.30) are also satisfied when the functions g(t) and h(t) are solutions of the pair of singular integral equations
(2m)h(x)
- qa2
(2ruJg(x)
q
g(t) cot
+ qa2
I
dt = 4[P(X) - F,(x) - &(x)1,
-c
(3.4)
-c
(3.5)
e
--c
h(l)cos
(
y
>
q dt = - d[&(x)
t F,(x)],
where h(x) and g(x) have been extended to the range -c < x < 0 as even and odd functions respectively and we have F,(x) = q[
(3.6)
h(t) dt “$, HI, cos (CYJ)cos (ant),
(3.7)
F,(x) = q 1’ g(t) dt “2, 0 F~(~) = 4 [
(3.8)
H3,sin ((Y.x)sin (a.t),
h(t) dt “$, H,.sin (cu.x) cos (oJ).
(3.9)
If we set G(x) = g(x) + ih(x),
(3.10)
the eqns (3.4)-(3.5) can be combined to yield the single integral equation -c
(3.11)
where Q(x) = (2i/7r)[P(x) - F,(x)- F2(x) + iW(x)
+
F4(xNl.
(3.12)
Noting that Q(x) involves the unknown function G(x), it seems eqn (3.11) cannot be solved in closed form. When exp (qS) s 1, the following approximate method gives useful results. It is easily seen that eqn (3.11) has the solution G(X)=
2 Gdx),
k-0
Gdx)
=g&)+
ik(x),
(3.13)
where Gk are determined successively from the sequence of singular integral equations
a,G,(x) t2
qdt=R,(x),-cCxXc,
k=0,1,2...
(3.14)
K. S. PARIHAR and A. C, GARG
836
(3.15)
fk4-x) = C?i/rVYx),
f&(x) = - (2iq/v)[\oc
f
hdt)
dt 2 H,, cos (QJ) cos ((Y,J) n=,
m C gAt)dt”z, Hzncm kw) sin (cu,t) I cl
m (3.16)
- i ’ h_,(t) dt c &, sin (ru,x) cos (a,t) , I0
n=1
I
k = 1, 2, 3, . . .
Since C&,is known the integral eqn (3.14) with k = 0 determines G, which in turn determines s1, and so on. It is obvious that, in practice, this process may not be continued indefiniteIy. We might note that Hi, (i = 1,2,3,4) are of the order exp (-2nq8) and the functions g,, k, will be of the order exp (-2kq6) for large exp (q6). The exponential decay of gk, hk suggests that a fairly good accuracy can be achieved even by retaining the first few terms in the solution (3.13). To solve the singular integral eqns (3.14) we consider the equations with Cauchy kernel given by
(3.17) k=0,1,2,... where M is an arc C&C& of the unit circle with center at the origin and the positive direction is from d, to dZ when passing round the circle in the counterclockwise direction. If we set (3.18) (3.19) and make use of the conditions
I
M
k =0, 1,2,.. .,
r-‘&(r)dr=O,
(3.20)
then the integral eqns (3.17) reduce to (3.14). In view of the relations (3.18)-(3.19), the conditions (3.20) are equivalent to
Ic -r
G,(X)dx=O,
k=O,1,2 ,...
(3.21)
If Gk satisfy the above conditions, the solution (3.13) will satisfy the condition (3.3) Thus, we shall have solved the dual series equations ~2.27~2.30) if we can find RI,(r) satisfying the integral eqns (3.17) and conditions (3.20). If we set k(w)=&I,
+$-$,
k =0,1,2 ,...
(3.22)
and make use of the Plemelj formulae (see Muskhelishvili[S~) (3.23) @Z(r) + @G(r) = -$ I,
$$-$,
rEM
(3.24)
Twobonded strips containing an infinite row of interface cracks
a37
then from (3.17) we get the relations 4;(r) + m+;(r) = - (2 + &kz)-‘Q(r),
r E M
(3.25)
M = (a, + &(a* - a,)-’ = (CL+ k,)(l + pkz)-’ > 0.
(3.26)
k=0,1,2,...
The Hilbert problems (3.25) are known to have the solutions (see [5], pp. 446-451) k=0,1,2...
(3.27)
where C, are arbitrary complex constants which may be determined by conditions (3.20) and Y(w) defined by
Y(w)= (w
- dJ”2”@
(w - dJ”2-‘B,
/3 = (2ff)-’ In m,
(3.28)
is particular solution of the corresponding homogeneous problem. If c3, are polynomials or rational functions, the integrals in (3.27) can be evaluated in finite form. To illustrate the use of the general formulae, a particular case is discussed below in some more detail. 4.THECASEOFCONSTANTINTERNALPRESSURE
Let us consider the case in which the prescribed pressure p(x) in (2.5) is a constant, say po, then Q. defined by relations (2.31), (3.15) and (3.19) becomes @o(r) = (2il7r)(~ + k,)(l + &ua,
0, = PO/PI.
(4.1)
In this case eqns (3.27), (3.23) and (3.20) with k = 0 yield t+&(w)= - i(p + k,)+(r
+ rm)-‘11 -(w - n’)Y(w)l,
rl = exp (-c#),
(4.2) (4.3)
where M, /3 and Y are given by (3.26) and (3.28). Now making use of relations (4.2), (3.23), (3.19) and (3.13) with k = 0 we find (4.4) h,(x)=u&
+ k,)[Vo(x)sinP8(x)-
UO(X)CO~~B(X)I{~P~~/(~)S(~)}-‘,
(4.5)
where a, is given by (4.1) and we have Uo(x) = (1 - 7’) cos (4x/2), e(x) = In Jsin (y)q
V,(x) = (1 + 772)sin (4x/2),
(4.6)
cosec (y)q(.
(4.7)
S(X) = [Isin’(qc/2) - sin2(qx/2)ll”‘.
(4.8)
Further calculations can be carried out by restricting the number of terms in the solution (3.13). We derive here the solution which gives the stress intensity factors upto the order exp (-4qS) for large exp (96). Neglecting small quantities of higher order, from (3.16) and (3.19) we get C&+,(r)= -(iq/?r)[b,g
tb21r~-'tbjLr2+b4~r-21+O(e-~4q,
b,, = L,, + L2lr - L31r-La,
DESVd. 14.No. 9-D
ba = i: L,k I-1
(4.9) (4.10)
838
K.S.PARIHARand b 3k = El, + E*k - EJ~ - Edto
b4k = i
A. C. GARG
i-1
L ,k = HI, ’ h,(t) cos (@) dt, f0
Ejk
L = IL
(4.11) ‘g&) sin (qt) dt, I0
Lk = H,, oc&(t)sin (@) dt, L4k= H,, 'h,(t) I I II
cos
(qt)
dt,
E,, = Hz2 ’ ~~(~)cos(2~~~d~ Et, = Hz 'gk(~)Sin(2~~) de I0 I0 &k
=
sin(2qt)dt,
‘gk(t)
Hz I
0
Ed.+ = HA2 chk(t)cos(2qt) dt, I0
(4.12) (4.13) (4.14) (4.15)
k =O,l
where H1,(i= 1,2,3,4; n = f, 2) are defined in the Appendix. In (4.91,Qt will be known only when g,, h, have been determined by using Q,. Substituting Qk from (4.9) into (3.27) and evaluating ck using conditions (3.20)we have @k+‘(w)
- iq{2r(1+ Cl.kz)(l+ m)}-‘[bt,W+ balrw-’ + hw2 + baw-’ - #bmw* - b2irq2w-‘) + 2~(b,~w + bZkq21 sin (4~) -(bgkw - tf*&_)+ b3icW3-b4k~2w-2fb3,(2fl sin(qc)-cos(qc))wZ t b4kT2(COS(qc) + 2/9sin (@))W_ + $(1+ 4@*)(blkW - b&T*) Sin2(qc)}Y(w)] + o(e-“7, k=O,l.
=
(4.16)
The above relation together with the eqns (3.23),(3.19)and (3.13)yields gk+dx)
=
-
hii+&)= -
u4[Uk+1(x)
sin/38(x)+ vk+dx)cospe(x)l{6(x))-‘7
(4.17)
a4[Vk+l(x)
sin @e(x)- &+l(x) cos fi@(x)]{8(x)}-‘,
(4.18)
where k = 0, 1 and uk+,(x) = (d3k- d4kq2)cos (5qx/2)+ [d,k
t ${d4r -(dtk
(COS(qC) t
+ d3kt28 sin (w)-cos (@))
33Sin (qC)) - d,k}l
COS&IX/~)
- q2dzrr) COS(~C) ~0~~~~~2) t $(1+4fi*)(&
- T2&J sin2(~c}Cos(~x~2)y (4.19)
vk+,(x)= (dsr, -t d&)*) Sin(5qX/2)+ t&n -t &,(28 sin (4c) - cos (4C)) - q*{d41:(COS(qc)+ 2/3 sin (qC))- da)] Sin (3qx/2)- (& + ?j2&) x cos (qc) sin (~~~2)t $1 t dik = bfk/u3, a4
j =
1,2,3,4;
4P2)tdsk f q2da)
sin’(qc) sin (4x/2),
k =0, 1
(4.21)
qa3{4~?70 Xl -t ok&-’
=
(4.20)
(4.22)
Since bli; in (4.16)contain u3 as a multip~er, d& are independent of as. Thus, an approbate solution of the integraI equations (3.3)-(3.5)is given by g(x) = a& + hXStx))-‘L&(x) - q&(2(1 + pk&qx))-’
sinJWx) + V&x) cos @l(x)1 2
[ uk(x)
Sinp@(X)+ v&(x)cos j%?(X)]
k=-I +
O(e+%
(4.23)
839
Two bonded strips containing an infinite row of interface cracks
h(x) = a& -
t k,){S(x)}-‘[V&x) sin /30(x) - L&(x)cos /Wx)l
4431+ b&M(x))-
k$, [Vktx) sin /38(x)- U,(x) cm petx)l
t O(e+@),
(4.24)
as = Po{2?r.llp,v/m}-‘.
(4.25)
Bonding stresses By using the relations (2.22)-(2.23), (3.1H3.2) and (3.10) we have u~;(x, 0)- iul:‘(x, 0) = u%(x, 0) - i&Yx, 0)
= pI{(p t k,)(l t /&)j-‘[(a,q/4)
I_: G(t) cot (+
-F,(x)-F,(x)tiIF,(x)+F,(x)}],
dt
c
(4.26)
Further, in view of the relations (3.13), (3.18)-(3.19) and (3.22) we have u~~(x, 0)- bXy(x, 0) = p,{(p + h)(l + rkJ}-‘[-im
k$O$W+)
- F,(x) - K(x)
c
+ @Xx ) + F.0 )I] + W-*%
(4.27)
where I,Q~ (k = 0, 1, 2) are given by (4.2), (4.16) and E (i = 1, 2, 3, 4) are defined by WW9). Thus we can write qv (4 0) = PO[
-
It q{(,.t t k,)(l t pk&-l(k$o{(d,~ + dz,) cm (qx)+(d,t + d.,i) ~0s (24x)}
-{r~S(x)}-‘~,{U,(x)sinpB(x)+ t {2+(x)}-'{Uo(x) sin Be(x)
Vk(x)cosBB(x)l) +
Vo(x) cbs pe(x)}
1
~W~[j-ock~o h,(t) dt “2, H,: cos((YJ)cos(ant) •t Ioc$o gk(t)dt$, HZncos(cu,x)sin(cy,t) I tO(e-6'"),c
(ICLI~~LL + kdtl +
+ (d,, t ddk) Sh (2qX)}+{4@(X))-’ -
vk(x)
sin@(x)])
i
{uk(x)
(4.28)
COS be(x)
+~2~~(~)~-'~~O~~~co~~~(~)-
vO(x)
sin@(x))] J
(4.29)
where the notation of (4.4H4.8) and (4.17)-(4.22) is employed. If we let 6 + m, eqns (4.28H4.29) yield G(x, 0) = - P0 +P0{2$(x)}-‘[Uo(x) sin Be(x) t ~.dx,0)= ~o{2~1s(x)~-'wo(x) c0S Be(x)
- Vo(x)
Vo(x)
c <
cos /M(X)],
sin Be(x)],
c <
x <
U,
x <
U,
(4.30)
(4.31)
where Uo, V. are given by (4.6). The bonding stresses (4.3OH4.31) are the same as one would obtain from the Goursat functions given by Rice and Sih ([7], Appendix 2) for the problem of an infinite row of collinear cracks at the interface of two bonded dissimilar elastic half-planes.
K. S. PARIHAR
840
and A. C. GARG
Furthermore, in the limiting case of a + a, the results (4.30)-(4.31) reduce to those of Lowengrub and Sneddon [3] for a single crack at the interface. On the other hand, if elastic properties of both the media are same (CL= l), then the shearing stress (4.31) vanishes snd the normal stress (4.30) on the crack axis reduces to that obtained by Sneddon and Srivastav[2].
-90I
-35 f5
-30
-25
-20
-15
-10
-5
01-2-O
loglo r/c-
Fig. 2. Contact stress go = a,,(~. O)lpO with r = x - c in the vicinity of crack/tip x = c for p = 0, yI = 0.3, c/a = 0.9and 6/a = 0.9, m. -, s/a = 0.9; ---, S/a = m.
From eqns (4.2&o-(.29) we see that as x -+ c the stresses change in sign an infinite number of times and their magnitude diverges, indicating that large tensile as well as compressive forces occur in the region near the tip of the crack. To study the oscillating nature of stresses as the points of singularity -c and c are approached, d(r/c)ao and In ]w,,]are calculated as functions of log,, (r/c) where a0 = a,,(~, O)/pOand r = x - c. The numerical results for s/a = 0.9 and ~4when c/a = 0.9, vI = 0.3 and I_L= 0 are shown in Fig. 2. Before gYrv(x, 0), undergoes a change in sign, m. takes its maximum value around r/c = 10m7.25 and lo-’ for S/a = 0.9 and m respectively. The values of a0 at these points are over 1175and 794 respectively which may be considered as stress concentration factors. This suggests that the stress concentration decreases with the increase in ala.
The phenomenon of stress oscillation has been studied also by Erdogan[S] for the external crack problem of two dissimilar half-planes bonded along a straight line segment and subjected to a wedge loading. Even though the crack geometry and the loading therein are entirely different from those considered in the present paper, the trend of curves in Fig. 8 of Erdogan[8] for the normal component of contact stress is similar to that of Fig. 2 here. In fact, the magnitudes of distances and stresses involved are also comparable and lead to the conclusion that in practical applications the phenomenon of stress oscillation, which is a peculiar characteristic of interfacial crack problems, may be ignored. Stress intensity factors
The stress intensity factors K, and K, defined by K, + iKz = lim V(X - c)(eiqx - eisc)‘B(eiqx - e-iqC)-ie {a,,(~, 0) - iuX,(x,0)},
X-l-*
may be computed directly from the formula (4.27). By using (4.2) and (4.16) we have
(4.32)
841
Two bonded strips containing an infinite row of interface cracks
K, = p,(2qq)-‘[(l
x i Vk(c) ll=l
t n*)d/[tan (qc/2)1- &.A +
M(l + A)dI[2 sin (~)l]-’
1
K2 = -~.(2\/q)-‘[(l-
(4.33)
tO(e+‘), n’)q/[cot(9c/2)]
(4.34)
t q{(p t k,)(l t j.&)d/[2 sin (qc)l)-’ $, Ll,(c)] + O(e-64s)7
~ fizO.0 ----)A=,_0
4'3j.O
20
3-o a/C -
40
5
Fig. 3. Dependence on a/c of the stress intensity factor (p,~c)-’ K, and (pO~c)-’ K2 for various S/c when /A= 0, Y,= 0.3; /L = I, Y,= Y*= 0.3.----, p = 0.0; ---,/I = 1.0.
where IX and V, are given by (4.19)-(4.21). Letting 6 + cQin the above equations we find that K,, K2 reduce to KT, KT given by
KT = po(2v’4)-‘(1+ q’W/[tan(4cDk
(4.35)
KT = - p&/‘/q)-‘(1 - nW/[cof (w/2)1,
(4.36)
which are in agreement with the stress intensity factors obtained by Sneddon and Srivastav [2] for the case where elastic properties of both the media are same (CL= n = 1). The variation of (pod/c)-‘K, and (po~c)~‘K2 with a/c for 6/c = 1, 2, 4, m when Y,= 0.3, p = 0 and yl = v2= 0.3, p = 1 which corresponds to the homogeneous case is shown in Fig. 3. The curves are drawn with the restriction 6/a > 0.8 (i.e. exp (qS) > 12.30approx.) so that the results are fairly accurate.
Strip with a crack on the centerline
Let us consider the problem of stress distribution in the neighbourhood of a single crack 1x1I c (y = 0) on the centerline of a strip with stress free edges y = 2 S. This is a limiting case (F = 1, a + m) of the problem discussed above. Indeed, if we set p = 1 and take the limit as a + m in eqns (3.4) and (4.26), the normal stress on the crack axis is given by CT:,:(X,O)= -2p,(k,tl)[[t(t2-x2)-‘g’“‘(t)dt-F,(x)],
where g’“’ is the solution of the integral equation
x>c,
(4.37)
K. S. PARIHAR and A. C. GARG
842 c I0
,(,2-.z)-1g(m’lt)dt-Fs(~)=(2~,)-’(k,+1)p(~),
O
(4.38)
in which p is the internal pressure applied to the crack faces and we have Fs(x) = [ K,(x, t)g’“‘(t) dt, K,(x, t) = [mH(@) cos (6x) sin (Tt) d5,
(4.39)
H(x)=(2~*+2x+l-e-~“)(2x+sinh2x)-’.
(4.40)
Inverting[4] the singular integral equation f4.38) we can write g’“(x) + l ~(~~~~)~~x,t) dr = Q(x),
o
(4.41)
where K(X, t) = (4xlrrz)(cz - x2)-“* [(y’Q(x) = - 2x(k, -t i)(n’~,)-‘(cZ-
xWc2
-
y2Y2K,(y,
t)
dy,
Xz)-“2 ~=(yz-xz)-1(~~-yz)li2P(y)dy. f
(4.42) (4.43)
The expression (4.37) for the normal stress on the crack axis along with the Fr~holm integral equation (4.41) can be obtained also by the method used by Loweng~b and S~vas~va~9] for the corresponding two crack problem. Also, their procedure in 181can be used to derive an iterative solution of the single crack problem for large values of S/c,
REFERENCES [l] F. ERDOGAN and G. D. GUPTA, Inr, J. Solids 9ruct. 7, 1089 (1971). [Z] I. N. SNEDDON and R. P. SRIVASTAV, Pmt. Roy. Sot. Edithugh A67, 39 (1965). [3] M. LOWENGRUB and I. N. SNEDDON, fnt. J. Engng Sci. 11, 1025 (1973). [4] K, S. PARIHAR, Pm. Roy. Sot. ~di~b~~~ A@, 255 f1971). 151 N~~~o~, The Ne~e~a~s . N. 1. MUSKHELISHVl~i, Some Basic P~b~e~s in tke ~~~e~atj~uf Them-yof ~~usr~c~~. (1963). [6] R. P. SRIVASTAV, Ph.D. Thesis, University of GIasgow (1%3). [?] J. R. RICE and G. C. SIH, 1. Appl. Me& 32,418 (1%5). [SJ F. ER~AN, I. Appt. Mech. 32, 403 (1965). f91 M. ~OWENGRUB and K. N. SRIVASTAVA, Int. J. Engng Sci. 6,425 (f968).
APPENDIX The functions A(x), G,(x), N,(x) and &(i
= 1, 2, 3. 4)
usedin the text are givenby:
A(x)= (1 -e-2x))2- 4x* e-**. G,fxf=RIe-“(e-2”
- 2) -4x2(2x f k2feFxx,
G,(x)= (k,& - lf(e-2” - 2je-2” i-8.x2 e-‘” -4x*e-*‘{f
i-(2x f k&2x + k&,
G,(x) = - 2 e-” + (1-4~~) e- Ir, G,(x) = k,(e- 2x- 2) ewzX- 4x2(2x f k,) e-“, H, = H&x?&.
(2)
i = I. 2, 3. 4
(31
~,(x)=l~fk,-l-~~,Gfx)+IC[,(xXl+G6U. ~,(~)=(~+k~~l+k~~G(x~-t~,~x~Il+G(x~~,
i=1,3 j=2,4
M,(x) = 2N,(x) - (k, + l)Ns(x) t [(2x f 1){-2N,(x) + 1- ~1f&z-k, t (1 + k,)N,(x)} f 4x2(1 t pk, + NJ(x))1e-“, b&(x)= -2N,(xf+(lfk,)N,(z)+[(2x~lX2N,(x~-p-l-~kt-~, -N,(x)(1~klf~-4x2(1t~k2+Nl(~ffle~**~
843 ~~txj= -2~2(x)+(k,- im'ttx)+~21 -i)~-2~*(~~~~
- 1 +ka-/A
+N,(x)(kr-1)~+4x1(1+~z+Nr(x)lle-~, M,(x) = 2N,(x) - (k, - l)N,(x) + [(2x - 1){2N,(x)
- N,(x)(k, - I)} - 4x*(1 t #kr + N,(x))] G(x) = {1+ f&(x)}-‘fe-”
-2 eF2” -4x*P
N,(x) = [(k, - 2x) V,(x)
NAx)l-
(4)
-f&(x)},
G&f = {(I + k,)(f + ~k*)~-‘IN~(x~k, + (d2)fl +(I t sk,NWxl-
t fi - 1t k, - pk,
e-” ;
t ktk,) + N,(x)1
NhW,(x)+
WM,k~-
#I,
(5)
t (1t pk2)(ev2’ - 2) -4x* p V8(x)(4x2 - 2xk, t I)] es”,
- p V&)(2x t k, + ee2=) t
N2(~)=(1/2)[-(4x2-k:+l)V,(x)-p(k,k,-
1)(2x-t)
+p(l-(Zr-k3e-“~{2-e-z”X4x2-k,k~+1)-V~(xXZr+klNZrtk,)~ +(4x’-k:+lXk,V,(x)-
V2(x)(2xt kt)e-tijlesti,
N,(x)= [(2x t k,) V3(x)t (eezX- 2)(1+ pk,) - 4x2
t p {(4x* + 2xk, t 1)V,(x) t (2x - k2 - em- ) V,(x)}] e-l*, N,(x)=(1/2)e-*“[{1+(kt+2x)‘}V~(x)+2kr(e-”-2-4x3 -~~1t(2rtkt)e-~~(2-e-Z"X4x2-ktk,+1) - V~(X)(~X + k&2x t kz)} + (2x t kzXk,kz- 1)
-{I t(2x tk,)*Kk,V,(X)-Ox tk,)V2(x)e-*‘l - 2k,{kl(e-” V,(x) = (1 -e-*‘)‘,
- 2) - V2(x)(2x t k,))], Vz(x) = 4x’e+,
V,(x) = V,(x) - V,(x).
(6) (7)
Perpunon Press.
Printed in Great Britain
TWO BONDED STRIPS CONTAINING AN INFINITE ROW OF INTERFACE CRACKS K. S. PARH-IAR and A. C. GARG Departments of Mathematics and Aeronautical Engineering, Indian Institute of Technology, Bombay, India Abstract-The plane strain problem of two bonded dissimilar isotropic elastic strips is considered. It is assumed that the composite strip contains an infinite row of interface cracks located symmetrically on the centerline. The case in which each of the cracks is opened out by the same constant pressure is discussed in detail and numerical results are reported for quantities of practical interest.
I. INTRODUCTION THE STRUCTURAL strength
of composite materials is controlled to a considerable extent by the size, shape and distribution of the flaws which exist in the material, usually, in the form of weak impurities on the interface, ruptured bonds and cracks. From the viewpoint of fracture initiation and propagation in the medium particularly important are the manufacturing flaws such as flat cavities which develop during bonding and casting. The problem of a multilayered composite containing a single crack on an interface has been studied by Erdogan and Gupta [ 11.In particular, they solve the problem of two bonded layers containing an interface crack opened by a uniform internal pressure. In this paper we consider the plane strain problem of stresses near an infinite row of cracks at the interface of two bonded dissimilar strips (see Fig. 1). The cracks are assumed to be of equal length, spaced at equal distance and opened by the same internal pressure. Even though the details are presented only for the case in which the edges of the composite strip are free from tractions, the results for some other types of boundary conditions (for example, smoothly restrained edge conditions) can be easily reproduced following a similar procedure. By simple symmetry considerations we see that the stress distribution in the entire strip can be deduced from that in the rectangle (xl5 a, ly] 5 S with the interface crack 1x1I c (y = 0). The edges y = f S are free from tractions and on the edges x = 2 a the symmetry conditions are equivalent to the smoothly restrained edge conditions.
Fig. I. Strip geometry.
The method of solution has affinities with those of Sneddon and Srivastav [2], Lowengrub and Sneddon[3]. We take a finite Fourier transform representation for the displacement field in the composite rectangle and derive a set of simultaneous dual series relations. By using a technique similar to that of Parihar [4] we reduce the problem of solving the dual series to the solution of a pair of singular integral equations which in turn is reduced to the Hilbert problem[5] with the line of discontinuity as an arc of a unit circle. When exp (4s) s 1 (4 = w/a), an iterative solution of the Hilbert problem can be obtained. The corresponding expressions, up to the order of exp (-4qS), for the stress intensity factors are derived and their dependence on various geometry parameters is shown graphically. 2. FORMULATION
OF THE PROBLEM
Consider the plane strain problem for the composite rectangle 1x1I a, (y I 5 S. Let a medium with elastic properties CL,and k, = 3-4~~ occupy the upper rectangle y > 0 and a medium with 831
K. S. PARIHAR and A. C. GARG
832
elastic properties p2 and k2 = 3 - 4v2 occupy the lower rectangle y < 0, where p ,, pz denote the rigidity moduli and yl, yz the Poisson ratios. Let [UP’, u:‘], k = 1,2, denote the displacement vectors for the upper and lower rectangles respectively and there is a corresponding notation for the stress components. On the edges x = 2 a and y = 2 6 we have u:k’(+a, y) = 0,
u:y(ka, y) = 0,
k=1,2
(2.1)
aE’(x, 5s) = 0,
a:;‘(x, 55) = 0,
k = 1,2
(2.2)
and the mixed boundary conditions at the interface are given by ai:‘(x, 0) = ug)(x, O), 1x15 a,
u::)(x, 0) = u:;(x, O),
(2.3)
u:“(x, 0) = uY’(x,O),u’,“(x,0) = u:*‘(x,O),c < 1x1I a, 6%
(2.4)
&)(x, 0) = 0, Ix) I c,
0) = - P (Ix/),
(2.5)
where p(x) is the internal pressure applied to the crack faces. To solve the mixed boundary value problem for the composite rectangle we assume a displacement field in the form u:‘(x,
y) = 2 [A, “=I +
+ 2~0,
+
{A,(k, + Z&Y - 2~5)
2cz,,(2a,ys t k,y - 2a,6’)B, - (1 - k:)BJ
x exp {~CL(Y- @)I exp (my) ul”(x, y) = ud”+ s, [A, + 2(a,y
t k@,
t
{(k, t 24
x
exp {2a.(y - S))l exp (-KY) cos (u),
ur’(x, y) = 2 [C, + 2(~,yD, + {(kz - 2~ n=l x exp {-2~(y
+ kJ(2h
- 2cw,y)A, t (24
t2cr,(2a,S2t2a.Sy
tk2y)Dn
+ k, - ky
Uk + B. 1
(2.7)
- 24)C, +(l-k:)D.}
(2.8)
+ S)}l exp (Amy)sin(u),
u:“(x, y) = us”-- z, [C,, t 2&y - k,)D, + {(kz + &Y -
(2.6)
sin (ML
t 2&W’”
(2~~~6t k2)(2a,S t k? t 2cy,y)D, - Dn) exp
{-2a,(y
t
@)I (2.9)
x exp (a.y) cm (w 1, where A., B,, C., D, are arbitrary constants and a” =
nq = nala.
(2.10)
By using the stress displacement relations we find that the corresponding stress components for the upper rectangle are given by
uZ(x, y) = 2/.~,2 (Y,[A, + {k, - 3 + 2cu,y}B, n=I + ((3 t 2a.y - 2a,6)A, - [2a. (6 - y)(2a,6 - k,) + I- 3kl - 6a,SlB,} exp {2cr.(y - S)}] exp (-a,~) cos (a,~), a::(~, y) = - 2/~, 2 cz,[A, + (2a,y t k, - l)B, n=, t ((2~3 - 2cz,y - l)A, t (2~x,(S- y)(2&
t k,)
(2.11)
TWObonded stripscontaining an infinite row of interface cracks f I - k, -
2cd>B,} exp {2cu,(y - &)}Iexp (-fky) sin (ax),
833
(2.12)
a:;(~, y) = - 2p, 2 a,, [A, t (2amy+ kl -I-I)&, - {(2ad - 2cr,y t l)A, II=1 t (2a,(S - y)(2a$ i- &,If 1+ k, f 2ad)B,) exp {2~y,(y- S)}] (2.13)
x exp (-any 1cos frullx 1, and for the tower rectangle we have
t ((3 - 2&y - 2anS)C, + (2a, (S t y)(2ru,St k*) t 1- 3k2 - 6a,S)D,} exp {-2ff,(S f y)}] exp (cu,y) cos (cu,x),
(2.14)
&Xx, y) = 21.122 cu,lC, + &K,Y - kz + W. n=l
t
{(2a,6 t 2amy- l)C, - (2tu,(S t y)(2& t kz) + 1- kz - 2a,&>Dm)
x exp {---2cy,(y+ S))l exp (any) sin (a~),
(2.15)
- ((2anS t 2any t I K’,, - (2~ (S t y N2ad t kd t 1 t kz t 2a,,S)D,} exp {-2a,(y
t S)}] exp (~,y) cos (a,~).
(2.16)
It is easily verified that the displacements and stresses (2.6)-(2.f6) satisfy the boundary conditions (2.1)-(2.2). The conditions (2.3) at the interface are also fulfilled if
cm= - ~~lAtff~~)~[A”{k*+ Gb-nS)l+ RVcrkz- I+ D, = --{P/A(QI,S)}~A~~~ t GdwVlt
GhMlt
Bnk t G.&nWl,
(2.17) (2.18)
where A(X) and G&X) (i = 1, 2, 3, 4) are defined in the Appendix and we have CL =
cLll/.h
(2.19)
The arbitrary constants A,, B, are to be determined by the mixed boundary conditions (2.4)-(2.5). We might note here that the resultant force on any cross section normal to the bond line is zero. Indeed, by means of the relations (2.11)-(2.19) we have 6 I0
a%,
Y) dy +
a:?@, y) dy = 0, 1x1~ a,
(2.20) (2.21)
The stress components at the interface may be written as a::‘&, 0) = u:(x, 0) = - g ,{( 1 + pkz)(p t k,))-’ (2.22) a$yn, 0) = uZz:Cn, 0) = - p,ifl + /.&)(1.L + Ml-’ (2.23)
where Hi, (i = 1,2,3,4) are defined in the Appendix and
834
K. S. PARfHAR and A. C. GARG
a,=Cr,+k,-l-Nkzt
(2.24)
az=Ir.fkt+l+~kz.
The constants S,, T, in (2.22)-(2.23) are given by S,, = [An{1 f yk,t
N,(rw,S)l+ B,(/.&,kz-- iI+
T, = [A,(1 t kkz + N&,,S)). f &(2kr + plk,k,+
2ru2(a,s))l{A(cr,S)l-‘,
(2.25)
1) + ~N~(~~,S)~I~A~LU~S))-‘~ (2.26)
where Ni(x) (i = 1,2,3,4) are defined in the Appendix. In view of the relations (2.17)-(2.18) and (2.25)-(2.26), the displacement and stress fields (2.6)-(2.16) can be expressed in terms of the undetermined constants S,, T,. The mixed boundary conditions (2.4H2.5) are also satisfied if S,, T, are solutions of the simultaneous dual series relations $BS,sin(*mx)=O,
To+
(2.27)
c
2 T, cos(a,x)=o,
(2.28)
C
tl=l
m
x Pft,S, + H2,T,la,
+
cos (c&x) = P(x),
0 < x < c,
(2.29)
“-8
-.fg,
[a,T,
-
ta2S~lm(GJ)
s,[H3”T, t H4,,S,,]a, sin (GX) = 0,
0 < x < c,
(2.30)
where H;:, (i = l, 2, 3, 4) are defined in the Appendix, a,, a2 are given by (2.24) and P(x)=p(x)(l”rCLk2)(~Lkki)/lU,,
To=u;‘--t@.
(2.31)
If the material constituting the rectangle is homogeneous, the dual series relations (2.27X2.30) reduce to the form Tot
5 T, cos (a,~) = 0,
c < x < a,
(2.32)
PI=,
$2,
T*[a~ + K(~Y,s)I sin (a,x) = P(X),
o < x c c,
(2.33)
where a2, P(x) are given by (2.24), (2.31) with lu.= 1, k, = k2 and Hz is defined by Hz(x) = - &2x2+ 2x t 1 -e-2x)(2x f sinh 2x)‘.
(2.34)
The dual series (2.32)-(2,33) are the same as obtained by Srivastav[6]. In the next section we consider the problem of determining S,, Tn satisfying the dual series relations (2.27)-(2.30). 3. SOLUTION
OF THE DUAL SERIES RELATIONS
We first reduce the dual series relations (2.27H2.30) to a pair of simultaneous singular integral equations. Follo~ng Parihar[4] we set
S, =;
I
f h(t) cos (ant) dt,
(3.1)
Two bonded strips containing an infinite row of interface cracks
sin (a~) dt,
T. = $ \‘g(l)
cl
To = (q/2) [
835
tg(t) dt.
(3.2)
This trial solution satisfies (2.27H2.28) for all choices of g(t) and h(t) provided ch(t)dt I cl
=O.
(3.3)
The remaining eqns (2.29)-(2.30) are also satisfied when the functions g(t) and h(t) are solutions of the pair of singular integral equations
(2m)h(x)
- qa2
(2ruJg(x)
q
g(t) cot
+ qa2
I
dt = 4[P(X) - F,(x) - &(x)1,
-c
(3.4)
-c
(3.5)
e
--c
h(l)cos
(
y
>
q dt = - d[&(x)
t F,(x)],
where h(x) and g(x) have been extended to the range -c < x < 0 as even and odd functions respectively and we have F,(x) = q[
(3.6)
h(t) dt “$, HI, cos (CYJ)cos (ant),
(3.7)
F,(x) = q 1’ g(t) dt “2, 0 F~(~) = 4 [
(3.8)
H3,sin ((Y.x)sin (a.t),
h(t) dt “$, H,.sin (cu.x) cos (oJ).
(3.9)
If we set G(x) = g(x) + ih(x),
(3.10)
the eqns (3.4)-(3.5) can be combined to yield the single integral equation -c
(3.11)
where Q(x) = (2i/7r)[P(x) - F,(x)- F2(x) + iW(x)
+
F4(xNl.
(3.12)
Noting that Q(x) involves the unknown function G(x), it seems eqn (3.11) cannot be solved in closed form. When exp (qS) s 1, the following approximate method gives useful results. It is easily seen that eqn (3.11) has the solution G(X)=
2 Gdx),
k-0
Gdx)
=g&)+
ik(x),
(3.13)
where Gk are determined successively from the sequence of singular integral equations
a,G,(x) t2
qdt=R,(x),-cCxXc,
k=0,1,2...
(3.14)
K. S. PARIHAR and A. C, GARG
836
(3.15)
fk4-x) = C?i/rVYx),
f&(x) = - (2iq/v)[\oc
f
hdt)
dt 2 H,, cos (QJ) cos ((Y,J) n=,
m C gAt)dt”z, Hzncm kw) sin (cu,t) I cl
m (3.16)
- i ’ h_,(t) dt c &, sin (ru,x) cos (a,t) , I0
n=1
I
k = 1, 2, 3, . . .
Since C&,is known the integral eqn (3.14) with k = 0 determines G, which in turn determines s1, and so on. It is obvious that, in practice, this process may not be continued indefiniteIy. We might note that Hi, (i = 1,2,3,4) are of the order exp (-2nq8) and the functions g,, k, will be of the order exp (-2kq6) for large exp (q6). The exponential decay of gk, hk suggests that a fairly good accuracy can be achieved even by retaining the first few terms in the solution (3.13). To solve the singular integral eqns (3.14) we consider the equations with Cauchy kernel given by
(3.17) k=0,1,2,... where M is an arc C&C& of the unit circle with center at the origin and the positive direction is from d, to dZ when passing round the circle in the counterclockwise direction. If we set (3.18) (3.19) and make use of the conditions
I
M
k =0, 1,2,.. .,
r-‘&(r)dr=O,
(3.20)
then the integral eqns (3.17) reduce to (3.14). In view of the relations (3.18)-(3.19), the conditions (3.20) are equivalent to
Ic -r
G,(X)dx=O,
k=O,1,2 ,...
(3.21)
If Gk satisfy the above conditions, the solution (3.13) will satisfy the condition (3.3) Thus, we shall have solved the dual series equations ~2.27~2.30) if we can find RI,(r) satisfying the integral eqns (3.17) and conditions (3.20). If we set k(w)=&I,
+$-$,
k =0,1,2 ,...
(3.22)
and make use of the Plemelj formulae (see Muskhelishvili[S~) (3.23) @Z(r) + @G(r) = -$ I,
$$-$,
rEM
(3.24)
Twobonded strips containing an infinite row of interface cracks
a37
then from (3.17) we get the relations 4;(r) + m+;(r) = - (2 + &kz)-‘Q(r),
r E M
(3.25)
M = (a, + &(a* - a,)-’ = (CL+ k,)(l + pkz)-’ > 0.
(3.26)
k=0,1,2,...
The Hilbert problems (3.25) are known to have the solutions (see [5], pp. 446-451) k=0,1,2...
(3.27)
where C, are arbitrary complex constants which may be determined by conditions (3.20) and Y(w) defined by
Y(w)= (w
- dJ”2”@
(w - dJ”2-‘B,
/3 = (2ff)-’ In m,
(3.28)
is particular solution of the corresponding homogeneous problem. If c3, are polynomials or rational functions, the integrals in (3.27) can be evaluated in finite form. To illustrate the use of the general formulae, a particular case is discussed below in some more detail. 4.THECASEOFCONSTANTINTERNALPRESSURE
Let us consider the case in which the prescribed pressure p(x) in (2.5) is a constant, say po, then Q. defined by relations (2.31), (3.15) and (3.19) becomes @o(r) = (2il7r)(~ + k,)(l + &ua,
0, = PO/PI.
(4.1)
In this case eqns (3.27), (3.23) and (3.20) with k = 0 yield t+&(w)= - i(p + k,)+(r
+ rm)-‘11 -(w - n’)Y(w)l,
rl = exp (-c#),
(4.2) (4.3)
where M, /3 and Y are given by (3.26) and (3.28). Now making use of relations (4.2), (3.23), (3.19) and (3.13) with k = 0 we find (4.4) h,(x)=u&
+ k,)[Vo(x)sinP8(x)-
UO(X)CO~~B(X)I{~P~~/(~)S(~)}-‘,
(4.5)
where a, is given by (4.1) and we have Uo(x) = (1 - 7’) cos (4x/2), e(x) = In Jsin (y)q
V,(x) = (1 + 772)sin (4x/2),
(4.6)
cosec (y)q(.
(4.7)
S(X) = [Isin’(qc/2) - sin2(qx/2)ll”‘.
(4.8)
Further calculations can be carried out by restricting the number of terms in the solution (3.13). We derive here the solution which gives the stress intensity factors upto the order exp (-4qS) for large exp (96). Neglecting small quantities of higher order, from (3.16) and (3.19) we get C&+,(r)= -(iq/?r)[b,g
tb21r~-'tbjLr2+b4~r-21+O(e-~4q,
b,, = L,, + L2lr - L31r-La,
DESVd. 14.No. 9-D
ba = i: L,k I-1
(4.9) (4.10)
838
K.S.PARIHARand b 3k = El, + E*k - EJ~ - Edto
b4k = i
A. C. GARG
i-1
L ,k = HI, ’ h,(t) cos (@) dt, f0
Ejk
L = IL
(4.11) ‘g&) sin (qt) dt, I0
Lk = H,, oc&(t)sin (@) dt, L4k= H,, 'h,(t) I I II
cos
(qt)
dt,
E,, = Hz2 ’ ~~(~)cos(2~~~d~ Et, = Hz 'gk(~)Sin(2~~) de I0 I0 &k
=
sin(2qt)dt,
‘gk(t)
Hz I
0
Ed.+ = HA2 chk(t)cos(2qt) dt, I0
(4.12) (4.13) (4.14) (4.15)
k =O,l
where H1,(i= 1,2,3,4; n = f, 2) are defined in the Appendix. In (4.91,Qt will be known only when g,, h, have been determined by using Q,. Substituting Qk from (4.9) into (3.27) and evaluating ck using conditions (3.20)we have @k+‘(w)
- iq{2r(1+ Cl.kz)(l+ m)}-‘[bt,W+ balrw-’ + hw2 + baw-’ - #bmw* - b2irq2w-‘) + 2~(b,~w + bZkq21 sin (4~) -(bgkw - tf*&_)+ b3icW3-b4k~2w-2fb3,(2fl sin(qc)-cos(qc))wZ t b4kT2(COS(qc) + 2/9sin (@))W_ + $(1+ 4@*)(blkW - b&T*) Sin2(qc)}Y(w)] + o(e-“7, k=O,l.
=
(4.16)
The above relation together with the eqns (3.23),(3.19)and (3.13)yields gk+dx)
=
-
hii+&)= -
u4[Uk+1(x)
sin/38(x)+ vk+dx)cospe(x)l{6(x))-‘7
(4.17)
a4[Vk+l(x)
sin @e(x)- &+l(x) cos fi@(x)]{8(x)}-‘,
(4.18)
where k = 0, 1 and uk+,(x) = (d3k- d4kq2)cos (5qx/2)+ [d,k
t ${d4r -(dtk
(COS(qC) t
+ d3kt28 sin (w)-cos (@))
33Sin (qC)) - d,k}l
COS&IX/~)
- q2dzrr) COS(~C) ~0~~~~~2) t $(1+4fi*)(&
- T2&J sin2(~c}Cos(~x~2)y (4.19)
vk+,(x)= (dsr, -t d&)*) Sin(5qX/2)+ t&n -t &,(28 sin (4c) - cos (4C)) - q*{d41:(COS(qc)+ 2/3 sin (qC))- da)] Sin (3qx/2)- (& + ?j2&) x cos (qc) sin (~~~2)t $1 t dik = bfk/u3, a4
j =
1,2,3,4;
4P2)tdsk f q2da)
sin’(qc) sin (4x/2),
k =0, 1
(4.21)
qa3{4~?70 Xl -t ok&-’
=
(4.20)
(4.22)
Since bli; in (4.16)contain u3 as a multip~er, d& are independent of as. Thus, an approbate solution of the integraI equations (3.3)-(3.5)is given by g(x) = a& + hXStx))-‘L&(x) - q&(2(1 + pk&qx))-’
sinJWx) + V&x) cos @l(x)1 2
[ uk(x)
Sinp@(X)+ v&(x)cos j%?(X)]
k=-I +
O(e+%
(4.23)
839
Two bonded strips containing an infinite row of interface cracks
h(x) = a& -
t k,){S(x)}-‘[V&x) sin /30(x) - L&(x)cos /Wx)l
4431+ b&M(x))-
k$, [Vktx) sin /38(x)- U,(x) cm petx)l
t O(e+@),
(4.24)
as = Po{2?r.llp,v/m}-‘.
(4.25)
Bonding stresses By using the relations (2.22)-(2.23), (3.1H3.2) and (3.10) we have u~;(x, 0)- iul:‘(x, 0) = u%(x, 0) - i&Yx, 0)
= pI{(p t k,)(l t /&)j-‘[(a,q/4)
I_: G(t) cot (+
-F,(x)-F,(x)tiIF,(x)+F,(x)}],
dt
c
(4.26)
Further, in view of the relations (3.13), (3.18)-(3.19) and (3.22) we have u~~(x, 0)- bXy(x, 0) = p,{(p + h)(l + rkJ}-‘[-im
k$O$W+)
- F,(x) - K(x)
c
+ @Xx ) + F.0 )I] + W-*%
(4.27)
where I,Q~ (k = 0, 1, 2) are given by (4.2), (4.16) and E (i = 1, 2, 3, 4) are defined by WW9). Thus we can write qv (4 0) = PO[
-
It q{(,.t t k,)(l t pk&-l(k$o{(d,~ + dz,) cm (qx)+(d,t + d.,i) ~0s (24x)}
-{r~S(x)}-‘~,{U,(x)sinpB(x)+ t {2+(x)}-'{Uo(x) sin Be(x)
Vk(x)cosBB(x)l) +
Vo(x) cbs pe(x)}
1
~W~[j-ock~o h,(t) dt “2, H,: cos((YJ)cos(ant) •t Ioc$o gk(t)dt$, HZncos(cu,x)sin(cy,t) I tO(e-6'"),c
(ICLI~~LL + kdtl +
+ (d,, t ddk) Sh (2qX)}+{4@(X))-’ -
vk(x)
sin@(x)])
i
{uk(x)
(4.28)
COS be(x)
+~2~~(~)~-'~~O~~~co~~~(~)-
vO(x)
sin@(x))] J
(4.29)
where the notation of (4.4H4.8) and (4.17)-(4.22) is employed. If we let 6 + m, eqns (4.28H4.29) yield G(x, 0) = - P0 +P0{2$(x)}-‘[Uo(x) sin Be(x) t ~.dx,0)= ~o{2~1s(x)~-'wo(x) c0S Be(x)
- Vo(x)
Vo(x)
c <
cos /M(X)],
sin Be(x)],
c <
x <
U,
x <
U,
(4.30)
(4.31)
where Uo, V. are given by (4.6). The bonding stresses (4.3OH4.31) are the same as one would obtain from the Goursat functions given by Rice and Sih ([7], Appendix 2) for the problem of an infinite row of collinear cracks at the interface of two bonded dissimilar elastic half-planes.
K. S. PARIHAR
840
and A. C. GARG
Furthermore, in the limiting case of a + a, the results (4.30)-(4.31) reduce to those of Lowengrub and Sneddon [3] for a single crack at the interface. On the other hand, if elastic properties of both the media are same (CL= l), then the shearing stress (4.31) vanishes snd the normal stress (4.30) on the crack axis reduces to that obtained by Sneddon and Srivastav[2].
-90I
-35 f5
-30
-25
-20
-15
-10
-5
01-2-O
loglo r/c-
Fig. 2. Contact stress go = a,,(~. O)lpO with r = x - c in the vicinity of crack/tip x = c for p = 0, yI = 0.3, c/a = 0.9and 6/a = 0.9, m. -, s/a = 0.9; ---, S/a = m.
From eqns (4.2&o-(.29) we see that as x -+ c the stresses change in sign an infinite number of times and their magnitude diverges, indicating that large tensile as well as compressive forces occur in the region near the tip of the crack. To study the oscillating nature of stresses as the points of singularity -c and c are approached, d(r/c)ao and In ]w,,]are calculated as functions of log,, (r/c) where a0 = a,,(~, O)/pOand r = x - c. The numerical results for s/a = 0.9 and ~4when c/a = 0.9, vI = 0.3 and I_L= 0 are shown in Fig. 2. Before gYrv(x, 0), undergoes a change in sign, m. takes its maximum value around r/c = 10m7.25 and lo-’ for S/a = 0.9 and m respectively. The values of a0 at these points are over 1175and 794 respectively which may be considered as stress concentration factors. This suggests that the stress concentration decreases with the increase in ala.
The phenomenon of stress oscillation has been studied also by Erdogan[S] for the external crack problem of two dissimilar half-planes bonded along a straight line segment and subjected to a wedge loading. Even though the crack geometry and the loading therein are entirely different from those considered in the present paper, the trend of curves in Fig. 8 of Erdogan[8] for the normal component of contact stress is similar to that of Fig. 2 here. In fact, the magnitudes of distances and stresses involved are also comparable and lead to the conclusion that in practical applications the phenomenon of stress oscillation, which is a peculiar characteristic of interfacial crack problems, may be ignored. Stress intensity factors
The stress intensity factors K, and K, defined by K, + iKz = lim V(X - c)(eiqx - eisc)‘B(eiqx - e-iqC)-ie {a,,(~, 0) - iuX,(x,0)},
X-l-*
may be computed directly from the formula (4.27). By using (4.2) and (4.16) we have
(4.32)
841
Two bonded strips containing an infinite row of interface cracks
K, = p,(2qq)-‘[(l
x i Vk(c) ll=l
t n*)d/[tan (qc/2)1- &.A +
M(l + A)dI[2 sin (~)l]-’
1
K2 = -~.(2\/q)-‘[(l-
(4.33)
tO(e+‘), n’)q/[cot(9c/2)]
(4.34)
t q{(p t k,)(l t j.&)d/[2 sin (qc)l)-’ $, Ll,(c)] + O(e-64s)7
~ fizO.0 ----)A=,_0
4'3j.O
20
3-o a/C -
40
5
Fig. 3. Dependence on a/c of the stress intensity factor (p,~c)-’ K, and (pO~c)-’ K2 for various S/c when /A= 0, Y,= 0.3; /L = I, Y,= Y*= 0.3.----, p = 0.0; ---,/I = 1.0.
where IX and V, are given by (4.19)-(4.21). Letting 6 + cQin the above equations we find that K,, K2 reduce to KT, KT given by
KT = po(2v’4)-‘(1+ q’W/[tan(4cDk
(4.35)
KT = - p&/‘/q)-‘(1 - nW/[cof (w/2)1,
(4.36)
which are in agreement with the stress intensity factors obtained by Sneddon and Srivastav [2] for the case where elastic properties of both the media are same (CL= n = 1). The variation of (pod/c)-‘K, and (po~c)~‘K2 with a/c for 6/c = 1, 2, 4, m when Y,= 0.3, p = 0 and yl = v2= 0.3, p = 1 which corresponds to the homogeneous case is shown in Fig. 3. The curves are drawn with the restriction 6/a > 0.8 (i.e. exp (qS) > 12.30approx.) so that the results are fairly accurate.
Strip with a crack on the centerline
Let us consider the problem of stress distribution in the neighbourhood of a single crack 1x1I c (y = 0) on the centerline of a strip with stress free edges y = 2 S. This is a limiting case (F = 1, a + m) of the problem discussed above. Indeed, if we set p = 1 and take the limit as a + m in eqns (3.4) and (4.26), the normal stress on the crack axis is given by CT:,:(X,O)= -2p,(k,tl)[[t(t2-x2)-‘g’“‘(t)dt-F,(x)],
where g’“’ is the solution of the integral equation
x>c,
(4.37)
K. S. PARIHAR and A. C. GARG
842 c I0
,(,2-.z)-1g(m’lt)dt-Fs(~)=(2~,)-’(k,+1)p(~),
O
(4.38)
in which p is the internal pressure applied to the crack faces and we have Fs(x) = [ K,(x, t)g’“‘(t) dt, K,(x, t) = [mH(@) cos (6x) sin (Tt) d5,
(4.39)
H(x)=(2~*+2x+l-e-~“)(2x+sinh2x)-’.
(4.40)
Inverting[4] the singular integral equation f4.38) we can write g’“(x) + l ~(~~~~)~~x,t) dr = Q(x),
o
(4.41)
where K(X, t) = (4xlrrz)(cz - x2)-“* [(y’Q(x) = - 2x(k, -t i)(n’~,)-‘(cZ-
xWc2
-
y2Y2K,(y,
t)
dy,
Xz)-“2 ~=(yz-xz)-1(~~-yz)li2P(y)dy. f
(4.42) (4.43)
The expression (4.37) for the normal stress on the crack axis along with the Fr~holm integral equation (4.41) can be obtained also by the method used by Loweng~b and S~vas~va~9] for the corresponding two crack problem. Also, their procedure in 181can be used to derive an iterative solution of the single crack problem for large values of S/c,
REFERENCES [l] F. ERDOGAN and G. D. GUPTA, Inr, J. Solids 9ruct. 7, 1089 (1971). [Z] I. N. SNEDDON and R. P. SRIVASTAV, Pmt. Roy. Sot. Edithugh A67, 39 (1965). [3] M. LOWENGRUB and I. N. SNEDDON, fnt. J. Engng Sci. 11, 1025 (1973). [4] K, S. PARIHAR, Pm. Roy. Sot. ~di~b~~~ A@, 255 f1971). 151 N~~~o~, The Ne~e~a~s . N. 1. MUSKHELISHVl~i, Some Basic P~b~e~s in tke ~~~e~atj~uf Them-yof ~~usr~c~~. (1963). [6] R. P. SRIVASTAV, Ph.D. Thesis, University of GIasgow (1%3). [?] J. R. RICE and G. C. SIH, 1. Appl. Me& 32,418 (1%5). [SJ F. ER~AN, I. Appt. Mech. 32, 403 (1965). f91 M. ~OWENGRUB and K. N. SRIVASTAVA, Int. J. Engng Sci. 6,425 (f968).
APPENDIX The functions A(x), G,(x), N,(x) and &(i
= 1, 2, 3. 4)
usedin the text are givenby:
A(x)= (1 -e-2x))2- 4x* e-**. G,fxf=RIe-“(e-2”
- 2) -4x2(2x f k2feFxx,
G,(x)= (k,& - lf(e-2” - 2je-2” i-8.x2 e-‘” -4x*e-*‘{f
i-(2x f k&2x + k&,
G,(x) = - 2 e-” + (1-4~~) e- Ir, G,(x) = k,(e- 2x- 2) ewzX- 4x2(2x f k,) e-“, H, = H&x?&.
(2)
i = I. 2, 3. 4
(31
~,(x)=l~fk,-l-~~,Gfx)+IC[,(xXl+G6U. ~,(~)=(~+k~~l+k~~G(x~-t~,~x~Il+G(x~~,
i=1,3 j=2,4
M,(x) = 2N,(x) - (k, + l)Ns(x) t [(2x f 1){-2N,(x) + 1- ~1f&z-k, t (1 + k,)N,(x)} f 4x2(1 t pk, + NJ(x))1e-“, b&(x)= -2N,(xf+(lfk,)N,(z)+[(2x~lX2N,(x~-p-l-~kt-~, -N,(x)(1~klf~-4x2(1t~k2+Nl(~ffle~**~
843 ~~txj= -2~2(x)+(k,- im'ttx)+~21 -i)~-2~*(~~~~
- 1 +ka-/A
+N,(x)(kr-1)~+4x1(1+~z+Nr(x)lle-~, M,(x) = 2N,(x) - (k, - l)N,(x) + [(2x - 1){2N,(x)
- N,(x)(k, - I)} - 4x*(1 t #kr + N,(x))] G(x) = {1+ f&(x)}-‘fe-”
-2 eF2” -4x*P
N,(x) = [(k, - 2x) V,(x)
NAx)l-
(4)
-f&(x)},
G&f = {(I + k,)(f + ~k*)~-‘IN~(x~k, + (d2)fl +(I t sk,NWxl-
t fi - 1t k, - pk,
e-” ;
t ktk,) + N,(x)1
NhW,(x)+
WM,k~-
#I,
(5)
t (1t pk2)(ev2’ - 2) -4x* p V8(x)(4x2 - 2xk, t I)] es”,
- p V&)(2x t k, + ee2=) t
N2(~)=(1/2)[-(4x2-k:+l)V,(x)-p(k,k,-
1)(2x-t)
+p(l-(Zr-k3e-“~{2-e-z”X4x2-k,k~+1)-V~(xXZr+klNZrtk,)~ +(4x’-k:+lXk,V,(x)-
V2(x)(2xt kt)e-tijlesti,
N,(x)= [(2x t k,) V3(x)t (eezX- 2)(1+ pk,) - 4x2
t p {(4x* + 2xk, t 1)V,(x) t (2x - k2 - em- ) V,(x)}] e-l*, N,(x)=(1/2)e-*“[{1+(kt+2x)‘}V~(x)+2kr(e-”-2-4x3 -~~1t(2rtkt)e-~~(2-e-Z"X4x2-ktk,+1) - V~(X)(~X + k&2x t kz)} + (2x t kzXk,kz- 1)
-{I t(2x tk,)*Kk,V,(X)-Ox tk,)V2(x)e-*‘l - 2k,{kl(e-” V,(x) = (1 -e-*‘)‘,
- 2) - V2(x)(2x t k,))], Vz(x) = 4x’e+,
V,(x) = V,(x) - V,(x).
(6) (7)