General solutions of an infinite sheet weakened by doubly periodic circular holes with a pair of radial cracks

Engineering Fracture Mechanics Printed in Great Britain.

Vol. 27, No. 1, pp. 11 I-119,

1987

CN313-7944/87 C, 1987 Pergamon

$3.00 + 0.00 Journals Lid.

GENERAL SOLUTIONS OF AN INFINITE SHEET WEAKENED BY DOUBLY PERIODIC CIRCULAR HOLES WITH A PAIR OF RADIAL CRACKS Beijing Institute

X. S. ZHANG of Civil Engineering,

China

Abstract-Making use of the basic theorem of the pole and zero points, we first seek the general solutions of an infinite sheet weakened by a pair of radial cracks emanating from a circular hole subjected to concentrated forces P, T and Q. Employing the same analysis as in previous papers, then we can find that the results are easily extended to the more complicated problems of an infinite plate weakened by an infinite series of singly or doubly periodic circular holes with a pair of radial cracks. All of the solutions to the stress intensity factor in this study are formulated in simply closed forms. With the aid of the superposition principle, it is not difficult to prove that the other results of the similar cracked-sheet may be deduced from the general solutions in the present paper, if its load is arbitrary or various.?

1. INTRODUCTION the problems of a pair of radial cracks have been discussed by several authors[l-61,

ALTHOUGH

plate

emanating from a circular hole in an infinite all these existing methods and results have

only a restricted application, and can not be used to treat the more complicated problems of an infinite sheet weakened by singly or doubly periodic circular holes with a pair of radial cracks under arbitrary forces. The main purpose of this paper is to provide mathematical techniques for solving the complex problem of an infinite plate containing doubly periodic circular holes with a pair of radial cracks loaded by arbitrary forces. it is of interest of note that all these solutions to the stress intensity factor in this article are expressed in simply closed forms, and that the results of an infinite sheet containing doubly periodic holes with a pair of radial cracks may also be applied to solve practical problems of a finite rectangular sheet weakened by a pair of cracks emanating from a circular hole. Using the superposition principle, the other results of the stress intensity factor of the similar cracked-plate can be derived from the general solutions in this study without any difficulty, when its load is arbitrary or various. In order to solve the problem for convenience, at first, we prove the following Lemma: Lemma. If all of these points on or in a circle with radius R are the zero points of an analytic function ~(2) in the z-plane, the function q(z) can be expressed as

(1.1)

in whichf(z) is an another analytic function in the z-plane without any zero point on or in the circle R; Re is the real part. Proof. Assuming that the points zi are zero points for an analytical function p(z), according to the definition of zero point, the essential condition is V(Zj) = 0, where j = 1, 2, . . .. Setting

zj instead

of z Re

fSee Section

4. 111

(1.2)

112

X. S. ZHANG

as Iz,IQ R, we can find that the essential condition (1.2) is obviously satisfied. Hence the Lemma is verified. Assuming that the zero points on or in the circle R of the analytical function p(z) are singly periodic with 27r or doubly periodic with 4K and 2K’i, the representation (1.1) may be modified as follows:

P(Z) = fi(z> Re

(1.4)

dz> = f,(z) Re

(1.4’)

in which K and K’ stand for the complete elliptic integrals of the first kind; sin z is the Jacobi elliptic function. Both (1.4) and (1.4’) can be proved by means of the same method as above. It is omitted here. 2. BASIC EQUATIONS For plane problems, it is well known that the Airy stress function for both plane strain and plane stress is governed by V2W”(% Y) = 0, (2.1) where it has been considered that the body forces is equal to zero and that V2 = z+z

ax2 ay2

0,

The corresponding

=

a2fOF

2;

ay

=

8

--y’

ax

3

r .V =--

atf

axay ’

and

(2.2)

strain components can be given as (2.3)

(2.4)

yyv = -;(l+v,)

I

-

atf

axay

’

(2.5)

in which Ej=l-

E

q=-

V

l-v

(2.6)

for plane strain, and E, = E, q = v

(2.7)

for plane stress; v is Poisson’s ratio, E stands for elasticity modulus. For the anti-plane problem, the only non-vanished displacement in the z direction must satisfy the following partial differential equation WJ_(x,

y) = 0.

(2.8)

The relations between the components of the stress and displacement are

?x =

au,

P dx;

(2.9)

where p denotes the shear modulus, p = E/2(1 + v). Considering the problems of the cracked plate for convenience, we wiil take the solutions of the partial differential eqs (2.f) and (2.8) as follows: R(s) dz dz];

(2.11)

(2.11’)

pU,(x, Y) = Im 43(z).

(2.12)

Putting f2.11) into (2.21, we may find the stress components in a cracked-sheet mode: %fz) = Re E2~?l(r)-~~;fz)-qrztz)]/2; aY(z) = Re C~Y;~,(~)+~~~(Z)-~(Z)I/~; 5;y(z) = Im cz q%(z)+ u?(z)]/2

for the opening (2.13) (2.14) (2.15)

Inserting (2.11’) into f2.2), the stress components in a cracked-pla&e for the sliding mode are obtained: (2.16) q,(z) = Im r22Q)l(z)-~~~fzl--y?(z11/2; fq(z) = Im [2~(z)+c;fg;O)t.q?(z)]/2; (2.17) (2.18) s(z) = -Re CWd~)+~ti:W?. From (2.9), (2.10) and (2.2), the stress components in a cracked-plate TJz 1 = Im e(z); r,,iz > = Re &zf=

for the tearing mode are (2.19) (2.20)

With the help of the expressions {2.3)-(2.5), (2.11) and (2.11’) it is not difficult to get the displacement components for Mode I and Mode II in a cracked-plate, described by the complex functions pr(z) and n(z):

[_@q(z)+jR(z)dz];

(2.21)

(2.22) and

-

2E. U,(z) = I+% 2E-

-Lqz) u-9

Im W&H 3-v.

= -- 1+ “, Re rpl(G &-Re I

i

lXvtW+

B(z) dz];

(2.23)

(2.24)

The main objective of this paper is to determine the stress intensity factor of the crack problem. As is well known, the stress intensity factors for the opening, sliding and tearing modes can be calculated by using the following formulas:

il II KI

a,(z)

= lim KIII =-+

r&.) J2(z-z*), ??Ztz1

61

(2.25)

114

X. S. ZHANG

P

,r II

Cl -9+ a

T

T

a

P

2K’

l

4K

4K

.

.

Fig. 1.

in which z is an arbitrary complex-coordinate; z* denotes the coordinate on the right tip of the crack.? Sometimes, we adopt the more convenient formulas recommended in [l-3]

(2.26)

here c is the complex coordinate of an arbitrary point on the crack surface. In fact, it can easily be seen that the limit result of the formula (2.26) is the same as the formula (2.25). 3. GENERAL

SOLUTIONS

Consider an infinite sheet containing an infinite series of doubly periodic circular holes with a pair of radial cracks as shown in Fig. 1. It is convenient to choose the x-axis along the line of the cracks and the y-axis along the direction of the perpendicular bisector of the central crack and through the center of the central hole. The z-axis is perpendicular to xy-plane, which is not drawn in Fig. 1. Let fa+4mK+ 2nK’i stand for the complex coordinates of tips of any one pair of radial cracks, and R denote the radius of each circular hole. The centers of a pair of neighbouring circular holes are a distance 4K apart, and the parallel cracks are spaced apart periodically by a distance 2K’. Assuming each of these crack faces are subjected to concentrated forces P, T and Q at the point 2, = X1+4mK+2nK’i, (3.1) forming a self-balanced system. For the sake of saving space, we only discuss the general solution of the opening mode in detail, because of general solutions for the sliding and tearing modes can easily be found by use of the similar method. At first, we look upon Fig. 1 as an infinite sheet weakened by only central circular hole with a pair of radial cracks under equal and opposite concentrated forces p at the tFrom

Fig. 1. we may easily see that Krleh,> &,, for the same material.

General

solutions

of an infinite sheet

115

point z = x, on the crack faces, In this case, the boundary conditions for Mode I may be given by o,cz) = -o(z)

= p&X*-z),

O<]zl
(3.2) ?&) = 0, (3.3) U,(Z) = 0, (3.4) n(z) = 0 or q,(z) = 0. and I.4
f(z) =

P&z2-x:) n(z-X,)&-u2)’

(3.6)

From the boundary condition (3.5), we may suppose that the circular hole of radius R is appeared after the crack of length 2a, and that all of these points on or in the circular hole R are the zero points of the analytic function for(z) what we want to find. Synthesizing the boundary conditions (3.2H3.5), and employing the Lemma in Section 1, it can be found that = in in

PJV

-x:>

7r(z-x,)&&*)

which it may obviously be seen that the analytic function does not have any zero point on or the circular hole R. Applying the result [l, 31 o&z ) = Re R(Z)

to

(3.7)

(3.88

the first expression in the formula (2.25), we have K, = lim [~(z-a)“~ z-0

Re

&/V-x:) n(z-x&o-.2)

Re

(3.9)

0t

pJ@--x:) K, = lim [z(~--a>‘~~Re n(z-xJJ(Sff2) z-t, When z tends to a, noticing

]Re(l-$r2.

(3.10)

f(a)= f(a)and z = r = a, from (3.10) we get (3.10’)

Let us take the solutions of the sliding and tearing modes together with the opening mode by use of the following form. The complex functions are

J@-4) z+K&&bz2)

(3.11)

and the stress intensity factors are

(3.12)

tNoting (2.15), we can find q,(z) = 0, i.e. (D,(Z)= 0, as r,, = 0. fConsidering r,,(z) = 0 on the crack face, (3.8) may also be derived from (2.14).

116

X. S. ZHANG

As discussed in [l-3], it is evident that the above solution can be extended to the more complicated problems of an infinite sheet weakened by an infinite series of singly or doubly periodic circular holes with a pair of radial cracks subjected to equal and opposite concentrated forces P, T and Q at the point c, = x, +2mn or [, = x1 +4mK+2nK’i. From (3.11) we may get the complex functions by changing fi(z), sin z, sin R or&(z) sin z, sin R for f(z), z, R, i.e. r

1

= f,(z)

Re

1_

i

sin2 R I’*;

(3.13)

sin z sin

(3.14)

where[l, 31

[1 II P

“OS xl)-lfitz)

=

J(sin2 a- sir-x,) e’ @sin z _ sin x,)J(~jin* z_

P

(Cn Xl Dn W’fi(z)

=

e’

(3.15)

sin2 a)'

J(Sr? a - Sn2x,) $Sn z_Sn x,)&n2 z_SnZ

a)’

(3.16)

Substituting (3.13) and (3.14) into (2.29, the general solutions of the stress intensity factors of Mode I, Mode II and Mode III of an infinite plate containing singly or doubly periodic circular holes with a pair of radial cracks may be found that

(3.17)

Cn x, Dn x, r Sn a Cn a Dn u)“~

4. DISCUSSION

AND PRACTICABLE

(3.18)

EXAMPLES

It is clear that the solutions (3.1 l), (3.12) and (3.13), (3.17) can also be deduced from the general solutions (3.14) and (3.18) with the aid of the same way as refs [l-3]. Therefore we may only discuss the general solutions of an infinite plate containing doubly periodic circular holes with a pair of radial cracks. By the superposition principle, the solutions of an infinite plate weakened by doubly periodic circular holes with a pair of radial cracks loaded by the continuously distributed forces along the line of each crack can be represented in terms of the integral forms. According to (3.14) and (3.18), we find that

solutions

Re

1-

117

of an infinite sheet

s~~&_-)“2(s~2 a-- Sn* x1)‘/*

Ic(Sn 2 - Sn x,)J(Sn* z- Sn* a)

Wn x,1;

(4.1)

in which R, = R+4mK+2nK’i, a, = a+4mK+2nK’i, {p(x,), t(x,), q(x,), 4(x,)) dx, = {P, T, Q}, Cnx, Dn x,dx, = d(Snx,) and m = n = 0, 51, +2,.. . . In consideration of solutions to the special cases of all kinds in this study are similar to those special case results in references [l, 23. Therefore, it is unnecessary to give them here one by one. But, we must point out that if R = 0, these solutions to this problem are identical with those results in ref. [l], and that these stress intensity factors in this paper are all smaller than those in [l] for same condition, as RZO. It is evident that the general solutions (3.11); (3.12), (3.13); (3.17) and (3.14); (3.18) are only suitable for the circular hole which is free from the forces. However, when the circular hole is also subjected to forces, with the aid of the superposition principle, it may be solved at once. Let us take three examples to illustrate how to find the stress intensity factor. Example 1. Each of circular holes is free of stresses, a pair of radial cracks are subjected to uniform stresses p, t and q. Assuming the uniform stresses p, t and q suitable for each case of single hole, singly periodic holes and doubly periodic holes with a pair of radial cracks, the stress intensity factors of this problem can be directly determined by integral from (3.12), (3.17) and (3.18) or (4.2), i.e.

I_.1

R

sin-1

-R

a

(4.3)

for only one central circular hole with a pair of radial cracks,

(4.4)

for singly periodic circular holes with a pair of radial cracks and

(4.5)

for doubly periodic circular holes with a pair of radial cracks.

118

X. S. ZHANG

(b) Fig. 2.

E,wmple 2. Each of circular holes is subjected to internal pressure p and the pair of radial cracks are free from forces as shown in Fig. 2(a) taken out only one central circular hole with a pair of radial cracks from Fig. 1. According to the superposition principle, it is clear that the problem can be divided into two parts as indicated in Fig. 2(b) and (c). As is known to all, the stress field of the problem (b) is nonsingular. Hence, from the consideration of the stress intensity factor, we can see that the problem a is identical with the problem (c). The stress component o,(y) on radial crack faces in the problem (c) may easily be determined from the problem b in many text books of elasticity, i.e. a,(y) = pR2/y. Noting JJ = 0 on the pair of radial cracks line, and setting 4 = $+y2, we get pR2xT2, for single circular

p (x1) =

hole;

(4.6)

p sin2 R/sin2 x,, for singly periodic

p Sn2 RISn’ x,, for doubly

periodic

holes;

(4.7)

holes.

(4.8)

Substituting (4.6) (4.7) and (4.8) into (3.12), (3.17) and (3.18) or (4.2) respectively, the solutions to stress intensity factors of the problem are (Ku = Km = 0)

K, = s[l-$j

for only one central

circular

hole with a pair of radial K, = s

for singly periodic

circular

(4.9)

cracks,

(tan u)I’~ [I-$-$

holes with a pair of radial

and integrating,

(4.10)

cracks and

2p Sn R

for doubly expressions

periodic circular holes with a pair of radial (4.9H4.1 l), noting that

cracks.

If R tends

lim (2p sin R) = lim (2p Sn R) = lim (2pR) = P, R-O R-0 R-0

to zero in the above

(4.12)

they will become the solutions to the problems of an infinite sheet containing doubly, singly periodic cracks or central crack under equal and opposite concentrated forces p at the center point of each crack. But they are not given here. E_wmple 3. Each of circular holes and a pair of radial cracks are all subjected to the uniform pressure p. We can easily see that Example 3 = Example 1 + Example 2.

General

solutions

of an infinite sheet

REFERENCES Engng Fracture Mech. 18, 887 (1983). Erzgq Fracture Mech. 20, 561 (1984). S. Zhang, Scienria sin (Series A ) 28, 94 (1985). C. Sih, J. appl. Mech. 32, 51 (1965a). L. Bowie, J. Math Phy.r. 35, 60 (1956). C. Sih, Handbook yfStre.m Intensity Faclors, pp. 1.2.8-lp1.2.8-3

Ill

X. S. Zhang,

c31 c41 c51 161

X. G. 0. G.

CA X. S. Zhang,

(1973).

(Received 3 September 1986)

119