Two coplanar cracks in an infinitely long elastic strip bonded to semi-infinite elastic planes

Inr. J. Engng Sci., 1973, Vol. 11. pp. 489-500.

TWO

Pergamon

Press.

Printed in Great Britain

COPLANAR CRACKS IN AN INFINITELY LONG ELASTIC STRIP BONDED TO SEMI-INFINITE ELASTIC PLANES? RANJIT.

S. DHALIWAL

The University of Calgary, Alberta, Canada (Communicated by I. N. SNEDDON) Abstract- We consider the problem of determining the stress intensity factors and the crack energy in an infinitely long elastic strip containing two coplanar Griffith cracks. We assume that the strip is bonded to semi-infinite elastic planes on either side and that the cracks are opened by constant internal pressure. By the use of Fourier transforms we reduce the problem to solving a set of triple integral equations with cosine kernel and a weight function. These equations are solved using Finite Hilbert transform techniques. Analytical expressions upto the order of V0 where 26 denotes the thickness of the strip and 6 is much greater than 1 are derived for the stress intensity factors and the crack energy.

1. INTRODUCTION

of cracks in a two dimensional elastic medium was first developed by Griffith[l]. Sneddon and Elliot [21 solved the problem of finding the distribution of stress in the neighbourhood of a Grifhth crack that is subject to an internal pressure which may vary along the length of a crack by considering the corresponding boundary value problem for a semi-infinite two dimensional medium. Tranter[31 solved the problem of determining the distribution of stress when two coplanar cracks are opened by a constant pressure in an infinite elastic medium. More recently, Lowengrub and Srivastava[4] derived the expressions for the stress and displacement components when the cracks are opened by pressure varying along the length of the cracks. They also derived the general expressions for the stress intensity factors and the crack energy. Lowengrub and Srivastava [51 considered the problem of an infinitely long strip containing two coplanar Griffith cracks, which lead to a set of triple integral equations. By using finite Hilbert transform techniques developed by them [6], they are able to reduce the triple integral equations to a single Fredholm integral equation of the second kind. We consider the problem of two coplanar cracks in an infinitely long strip, when the strip is bonded to semi-infinite planes on its either side. Following Lowengrub and Srivastava [5,61, the problem is reduced to a single Fredholm integral equation of the second kind. The iterative solution of the integral equation is obtained for 6 9 1 upto the order a-” when half the width of the strip is S times the distance of the far end of the cracks from the origin. The analytical expressions upto the order of g-lo are obtained for the stress intensity factors and the crack energy. THE

THEORY

2. SOLUTION

OF THE

EQUATIONS

OF EQUILIBRIUM

The two dimensional equations of equilibrium, when there is symmetry line x = 0, have solutions of the form: V,,(X? Y) = *,[6-lG,,(8,

Y); 5 +

xl3

?This work was supported by National Research Council of Canada through NRC-Grant The work was completed while the author was visiting the University of Glasgow. 489

about the

(2.1) No. A4177.

490

RANJIT S. DHALIWAL

~l/l/(X,Y)

=-@,kG(5,Y);5+xl,

(2.3)

where G (5, y ) satisfies the equation

(DZ,--52)2W,Y) = 0,

(2.4)

D, denotes d/dy and G, denotes aG/ay. As usual 9, and 9, denote the operators the Fourier sine and the Fourier cosine transforms respectively: 2

9,[G(5,~);5--+xl

m

112

0 I 0 I G (5,

= -rr

G (5, Y) sin (5x) dc,

0

m

112

~c[G(t.~);

5 -+ xl = -772

The corresponding expressions given by the pair of equations: (l+7))-1EU,(x,~) (l+q)-‘EUv(x,y)

of

Y) ~0s

(&x) dt.

0

for the components

U, and U, of the displacement

=~~[5-“{(1-7))GYU+~52G};5-, xl, =~~[5-3{(1-7))G~,Y,-(2-7))~2Gy};5--,

are

(2.5)

x],

(2.6)

in which 7) and E denote respectively the Poisson’s ratio and the Young’s modulus of the material. The general solution of the equation (2.4) for finite values of y is given by: G(~,Y) = [A(5) +~YMO

-B(O]e-*“+

[-C(t)

+~Y’Y(C(~)-o(5))le’“l

(2.7)

where A, B, C and D are functions of 5 alone. For this solution equations (2.5), (2.6), (2.2) and (2.3) give respectively: (l-kq)-‘EU,(x,y)

(l+~)-‘EU,(x,y)

=9,[{A+ (~y-2+2~)(A-B)}e-~U +{-C+(&+2-227))(C-D)}esar;t+x], =~,[{A+(~y+1-27,1)(A-B)}e-~” -{-C+(~y-1+2r])(C-D)}e5y;[-+x]r

cr,,(x,y)

OF THE

PROBLEM

AND

(2.9)

(2.11)

=-9,[~{A+5y(A-B)}e-*“+5{-C+Sy(C-_)}eS”;~~x]. 3. STATEMENT

(2.8)

ITS SOLUTION

We consider an infinite elastic strip of breadth 26 bonded to a semi-infinite elastic

Two coplanar cracks in an infinitely long elastic strip

491

plane on either side. Let us take y-axis perpendicular to and x-axis along the length of the strip. If a pair of coplanar cracks developes in the middle line of the strip perpendicular to the y-axis and also symmetrical about it, we may solve the problem by considering half the strip bonded to a semi-infinite plane on one side and a pair of cracks on the open side. If we take the open face of the strip as y = - 6 and the interface of the strip with the semi-infinite plane as y = 0, our problem is to determine the solution satisfying the boundary conditions: Up)(x, -6)

= 0, 0 -=c1x1< k, 1x1 > 1,

a&Xx,-6) =-p(x), d!:,(x,-8)

= 0,

(3.1)

-lox<--k,k~x~l, --oo <

x <

(3.2) (3.3)

co,

where the cracks are taken to be located on the line y = - 6, - 1 G x s - k, k s x s 1, the superscripts (1) and (2) refer respectively to the regions 1 (-6 6 y s 0) and 2 ( y 2 0) and p(x) is an even function of x. The elastic strip and the semi-infinite plane being in perfect contact, the following continuity conditions must be satisfied at the interface y = 0: U6f’(x. 0) - Uf’(x, 0) = 0, U:)(x, 0) - c/$)(x, 0) = 0, u!J(x, 0) -(rfJ(x,

0) = 0,

U$!A(X,0) -a:;:(x,

0) = 0.

(3.4) !

If the elastic constants of regions 1 and 2 are denoted by 71, E, and Q, Ez respectively, the solution for region 1 is obtained from equations (2.8)-(2.11) by replacing r), E, A, B, C, D by vl, E,, A,, B1, C1, D, respectively and the solution for region 2 may be obtained from the same equations by replacing 7, E, A, B by Q, Ez, AZ. Bz respectively and setting C = D = 0 where A,, B1, C1, D1, AI, Bz are arbitrary functions of 5 to be determined from the boundary and continuity conditions (3.1)-(3.4). The boundary condition (3.3) and the continuity conditions (3.4) are satisfied if we take:

_&-~l~~+

l+k

1

o

[2(~+kl)+2(1-~)(1-225)e-~51,

C,-~,2%!2l+k n [2(~kz+1)(1+25)+2(kl-lukz)e-251r 1

492

RANJIT

S. DHALIWAL

(3.5) where 5=~6,ki=3_4r)i(i=1,2),~=--

G l+r)z Ez l+n’

Equations (3.5) express the unknown functions A, ( e)----------A2 (5) in terms of a single unknown function &([). Now if we define the new unknown function G (5) by the relation: (3.6)

we find that U’k’(x,-6)

= 2(1;%C[G(&.-*

51,

(3.7)

1

(+%b.--8)

=

51,

-~~[ZG(5){1+H(25)};~-,

(3.8)

where

Ht25) = 2(1+k1)e-2~[1+k:-2p~k2-p(l-kl)(l-k2) A([) -4(p-l)(pkp+

1)5+ (P-

1)(~k2-O-2~l.

(3.9)

The remaining boundary conditions (3.1) and (3.2) will be satisfied provided that G (5) satisfy the set of triple integral equations:

~JG(t);x

+ 51= 0, x E L, LB,

9-&G(5){1+~(25)~;x+

51 =P(x),x E L,

(3.10) (3.11)

where L, = {x/O < x < k},L,

= {xJk < x < l}, L, = {xix > l}.

The solution of the set of triple integral equations (3. lo), (3.11) is given by [61: G(r)

= (~)l’“‘-~L~ h(t*) sin (et) dt,

(3.12)

Two coplanar cracks in an infinitely long elastic strip

493

where h (t”) is the solution of the Fredholm integral equation of the second kind h(2) +I,’ h(~2)~,(~,

t) dr = M(2), x E L,

(3.13)

satisfying the condition (3.14)

I : h(P)dt=O and

(3.15) with (3.16) and

(3.17)

+C{W-k~)(l-X2)}-“~,

where C is an arbitrary constant to be determined from condition (3.14). Now integrating (3.13) with respect to x from k to 1 and using .(3.14), we find that C=+I,‘h(P)fj;

X,(x2,r)dx}dr+-&I,f

WW)dx,

(3.18)

where (3.19)

and F is an elliptic integral of the first kind. Hence from (3.13), (3.17) and (3.18) we find that h must satisfy the integral equation h($)+j-;

h(t2)K(X2,f)dt=P(X2),x

E Lzr

(3.20)

where K(Y,t) = K,(XY,f)-~~t4-kl)(l-~)}--l/z

I’ K,(xZ,t)dx,

(3.21)

k

P(9)

=

-$[

W(~~‘)--+I+E)(~-X)}-~~~

J1 W(2) dx]. k

(3.22)

494

RANJIT S. DHALIWAL 4. ITERATIVE

SOLUTION

OF THE INTEGRAL

EQUATION

If we consider the case 6 % 1 then by substituting St = 5 and expanding cos (
M,(t,y)

(f-y)‘““],

and -9 ‘7 = (:?$)l)! to find the expression the type

Now

I

czz H(25)5”‘+l d{. n

for K1 (x”, r), we need to evaluate elementary

I’ &$)“I

integrals of

Gdy

and hence we find that (4.2) where

x= {(X”-kZ)(l-X2)}“*, A, (x2, t’) = Z,A,(xZ), A,(x,,P) = Z1[A,,(x”)Zz+8A,(xz)], A,($, t”) = Zz[Ao(Aqt4+ lOA,(x2)t”+A,(x*)]. Aq(xZ, t’) = Z3[Ao(x”)t6+21A,(x2)r4+7A,(f)t2+A,(x~)],

(4.3)

with A,(?) A,(?) &(x2) A,(_?)

= 16(f-E/F). = 8(2x4+ a&+ a,), = 10(8X6+4c~~~-k’~~~+cx~), = 7(169+8a~-22k’4X4+~&‘4~z+~Q),

where a,=-(l+k*), c+ = kt2EIF - 211, aq = k’2(1+3kZ)E/F+4k’2Z;-8Z;, a3 = ktZ (1 + 2kz + Sk+)E/F + 2k12 ( 1-t 3k2)Z; + 8k12Z; - 16Zj, k’2 = 1 -k’

(4.4)

Two coplanar cracks in an infinitely long elastic strip

495

and k

so that we find that Z;,=-k2i-E/F, Z; = f [-kz+

(2-k2)E,F],

Z;= ifs[k2(kz-4)+

(8-3k2-2k4)E/F],

where F and E are elliptic integrals of the first and second kind respectively defined by:

For p(x) = pL we find from (3.19) that W(Y) =

-;po(f$)ln,

and hence from (3.22) we find that (4.5)

Since 6 % 1, jK(X2,r) 1 < 1~where cr < 1, the solution to (3.20) may be written in the form: (4.6)

Now substituting for K(X2, t) and h respectively from (4.1) and (4.6) in (3.20) and equating the coefficients of various powers of S from both sides, we obtain: h,(9) = P(x”),

~=1,2,3

,a...,

(4.7)

(4.8)

4%

R.mm

s. DHALIwAL

and hence from (4.3), (4.7) and (4.8), we find that:

(4.9)

Two coplanar cracks in an infinitely long elastic strip

497

with

where

so that

s1= Tf(l+k1)/4, sz= ~(?~2~+~~~~~6~ s3 =

v(5

+

3kp+ 3k4-i-51%;)/32,

s4 = a(35+20k2+

18k4-V20P+35kR)/256.

From (4.6>,(4.7) and j4.9) we finstllyfmd that: (4.10)

5. THE STRESS INTENSITY It

FACTORS

is of interest to workers in fracture mechanics to calculate the stress intensity

498

RANJIT

S. DHALIWAL

factors at the ends of the crack. They are given by: NI, =&lb {(k-x)“*[&j(x,-a)]}

(5.1)

and N, = l&y {(x-1)1’2[u~~(x,-(s)]}.

(5.2)

From (3.8) and (3.12), we find that

c+(x, - 6) = -

’ th(t*) dt’ h(t2)K2(x, /y? Ik

I

t) dt.

(5.3)

From (4.1) and (4. lo), we have 1

I

h(t*)K,(x,

t) dt = $,(d,s,+d,.~,)6-~+

(Z,,so+Z,s,6-*)

k

x

(-

8E/F + 41,&,%-*/F + C,,S4) 6-* + (Iti, + Z1s28-*) (8 - 4Z0b06-*+ ~~6-~)6-*

+ &rb2Z2(2 - Zobo%*)6P + s3 (- 8Z3E/F + c2Zl+ d,Z,) 6-* + 8Z3s46-8 + x2E4{7r(2 - Zob,,S2) (3ZIbo+ 4Z2b16-*) + 3Z,( coso+ c1sI)S4 + 3s~ (c2Z1- 56Z,E/F)iY4 + 168Z3s36-4}+ x%-6{5~Z2bo(2 - Z&S-*r) - 28OSlZ,E/F + 168s3Z36-‘} + 14rZ3b$-*X6].

(5.4)

We also have ’ -dt=& th(t*) k t*--X*

I

8

R(x*)/X+ N(x*) +0(6-lo), R(x*)/X~+N(X*)+O(~-~~),

0 < x < k, x < 1,

(5.5)

where R(y)=

Po+PI~"+P~.~P+P~~+P~XB~

and X is given by (4.3),. Hence, the stress intensity factors Nk and N, as estimated from (5.1) and (5.2) are given by

Nk= -8[2k(l!k2)]1,2 N1= 8[2(1

:k2)]“2

(Po+P~k2+P2kl+P3k6+~4k8)+o(6-‘o),

(~O+~1+~2+~3+~4)+~(~-'").

(5.6)

(5.7)

Two coplanar cracks in an infinitely long elastic strip 6. THE

CRACK

499

ENERGY

Substituting from (3.12) in (3.7) and interchanging the orders of integration, we find that the displacement along the crack is given by:

(6.1) The energy to open the crack is given by

w= =

Ir k

v

p(x)Uf’(x,-8)

1

dx

1th(t2) dt,

I k

(6.2)

sincep(x) =p,and _Pk,h(t2) dt= 0. Substituting for h (t”) from (4.10) in (6.2), we find that (6.3)

[l] [2] [3] [4] [5] [6]

REFERENCES A. A. GRIFFITH, Phil. Trans. R. Sot. A.221,163 (1920). I. N. SNEDDON and H. A. ELLIOT, Q. up& Math. 4,262 (I 946). C. J. TRANTER, Q.J. Mech. up& bath. 14,283 (1961). M. LOWENGRUB and K. N. SRIVASTAVA, I&.!. Engng Sci. 6,359 (1968). M. LOWENGRUB and K. N. SRIVASTAVA, Int. J. Engng Sci. 6,425 (1968). M. LOWENGRUB and K. N. SRIVASTAVA, Proc. R. Sot. Edinb. 309 (1970). (Received

15 May 1972)

R&urn&- Nous consid&rons le probleme qui cons&e B ddterminer les facteurs d’intensid d’effort et I’Cnergie fissurale dans une bande elastique de longueur infinie, contenant deux fissures coplanaires de Griffith. Nous supposons que la bande est liie B des plans Clastiques semi-infinis de chaque c&C, et que les fissures sont ouvertes par pression inteme constante. En utilisant des transformations de Fourier, nous rkduisons le probleme & la r&solution d’un jeu d’tquations integrales triples avec noyau cosinus et une fonction de pesanteur. Ces iquations sont r&,olues en utilisant des techniques de transformation finie de Hilbert. Des expressions analytiques allant jusque vers S-lo, oh 2S denote I’Cpaisseur de la bande, et 6 est beaucoup plus grand que 1,sont d&iv&es pour I& facteurs d’intensitb et pour 1’Cnergie fissurale. Zu~~nf~ungWir untersuchen das Problem der Bestimmung des Spannungsintensit~tsfaktors und der Rissenergie in einem unendlich langen elastischen Streifen, der zwei komplanare Griffith-Risse enthllt. Wir nehmen an, dass der Streifen an beiden Seiten an einseitig-unendliche elastische Ebenen gebunden ist und dass die Risse durch konstanten lnnendruck geijffnet werden. Wir reduzieren das Problem durch Fourier-Transforme zur LGsung eines Satzes von Dreifach-Integralgleichungen mit Kosinuskernen und einer Gewichtsfunktion. Diese Gleichungen werden unter Beniitzung von endlichen Hilbert-Transformmethoden gel&t. Analytische Ausdticke bis zu einer Ordnung von 8-** (worin 25 die Dicke des Streifens bezeichnet und 6 vie1 grijsser als 1 ist) werden fiir die Spannungsjntensit~tsfaktoren und die Rissenergie abgeleitet, Sommario-Studiamo il problema dato dalla determinazione dei fattori dell’intensit9 di sollecitazione e dell’energia d’incrinatura in una striscia elastica di lunghezza infinita contenente due incrinature di Griffith coplanari. Presumiamo the la striscia sia legata a piani elastici semi-infiniti su ambi i lati e the le incrinature

RANJIT

500

S. DHALIWAL

vengano aperte da pressione interna costante. Mediante le trasformazioni di Fourier riduciamo il problema alla risoluzione di una serie di equazioni integrali triplici con nocciolo di coseno e funzione di peso. Le equazioni sono risolte impiegando metodi di trasformazione di Hilbert finite. Nei riguardi dei fattori dell’intesnita di sollecitazione e dell’energia d’incrinatura si derivano espressioni analitiche fino all’ordine di g-r0 dove 26 denota lo spessore della striscia e 6 e molto piu grande di 1. A~CT~UCT 3HeprWi (T.

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