Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2011, Article ID 741075,17 pages doi:10.1155/2011/741075

Research Article

Computation of Energy Release Rates for a Nearly Circular Crack

Nik Mohd Asri Nik Long,1,2 Lee Feng Koo,1 and Zainidin K. Eshkuvatov1,2

1 Department of Mathematics, Universiti Putra Malaysia, 43400 Serdang, Malaysia

2 Institute for Mathematical Research, Universiti Putra Malaysia, 43400 Serdang, Malaysia

Correspondence should be addressed to Lee Feng Koo, kooleefeng@yahoo.com Received 4 August 2010; Revised 6 December 2010; Accepted 14 January 2011 Academic Editor: Jerzy Warminski

Copyright © 2011 Nik Mohd Asri Nik Long et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper deals with a nearly circular crack, Q in the plane elasticity. The problem of finding the resulting shear stress can be formulated as a hypersingular integral equation over a considered domain, Q and it is then transformed into a similar equation over a circular region, D, using conformal mapping. Appropriate collocation points are chosen on the region D to reduce the hypersingular integral equation into a system of linear equations with (2N + 1)(N + 1) unknown coefficients, which will later be used in the determination of energy release rate. Numerical results for energy release rate are compared with the existing asymptotic solution and are displayed graphically.

1. Introduction

The determination of energy release rate, a measurement of energy necessary for crack initiation in fracture mechanics, has stirred a huge interest among researchers, and different approaches have been applied. Williams and Isherwood [1] proposed an approximate method in terms of a mean stress to approximate the strain-energy release rates of finite plates. Sih [2] proposed the energy density theory as an alternative approach for fracture prediction. Hayashi and Nemat-Nasser [3] modelled the kink as a continuous distribution of infinitesimal edge dislocations to obtain the energy release rate at the onset of kinking of a straight crack in an infinite elastic medium subjected to a predominantly Mode I loading. Further, a similar method to [3] has also been adopted by Hayashi and Nemat-Nasser [4] to obtain the energy release rate for a kinked from a straight crack under combined loading based on the maximum energy release rate criterion. Gao and Rice [5] extended

Figure 1: Stresses acting on a circular crack.

Rice's work [6] in finding the energy release rate for a plane crack with a slightly curved front subject to shear loading. While, Gao and Rice [7] and Gao [8] considered a penny-shaped crack as a reference crack in solving the energy release rate for a nearly circular crack subject to normal and shear loads. Jih and Sun [9] employed the finite element method based on crack-closure integral in calculating the strain energy release rate elastostatic and elastodynamic crack problems in finite bodies whereas Dattaguru et al. [10] adopted the finite element analysis and modified crack closure integral technique in evaluating the strain energy release rate. Poon and Ruiz [11] applied the hybrid experimental-numerical method for determining the strain energy release rate. Wahab and de Roeck [12] evaluated the strain energy release rate from three-dimensional finite element analysis with square-root stress singularity using different displacement and stress fields based on the Irwin's crack closure integral method [13]. Guo et al. [14] used the extrapolation approach in order to avoid the disadvantages of self-inconsistency in the point-by-point closed method to determine the energy release rate of complex cracks. Xie et al. [15] applied the virtual crack closure technique in conjunction with finite element analysis for the computation of energy release rate subject to kinked crack, while interface element based on similar approach also adopted by Xie and Biggers [16] in calculating the strain energy release rate for stationary cracks subjected to the dynamic loading.

In this paper, we focus our work on obtaining the numerical results for energy release rate for a nearly circular crack via the solution of hypersingular integral equation and compare our computational results with Gao's [8] .

2. Formulation of the Problem

Consider the infinite isotropic elastic body containing a flat circular crack, Q, as in Figure 1, located on the Cartesian coordinate (x,y,x3) with origin O, and Q lies in the plane x3 = 0. Let the radius of the crack, Q be a and Q = {(r,d) : 0 < r < a, -n < d < n}.

If the equal and opposite shear stresses in the x and y directions, q (x, y) and q2 (x, y), respectively, are applied to the crack plane, and it is assumed that the x3 direction is traction free, then in the view of shear load, the entire plane, must free from the normal stress, that is

t33(x,y,x3) = 0 for x3 = 0,

and the stress field can be found by considering the above problem subjected to the following mixed boundary condition on its surface, x3 = 0:

T13 (x, y, x3) = (x, y), (x, y) e Q,

t23(x,y,x3) = —^—q2(x,y), (x,y) e Q, (2-2)

1 — V

ui(x,y,x3) = u2 (x,y,x3) = 0, (x,y) e r \ Q,

where Tj is stress tensor, y is shear modulus, v is denoted as Poisson's ratio, and r is the entire x3 = 0. Also, the problem satisfies the regularity conditions at infinity

Ui(x,y,x3) Tij(x,y,x3) if = 1,2,3, R -h- co, (2.3)

where R is the distance

R = ^(x-x0)2 + (y-yo)2, (x0,y0) € Q. (2-4)

Martin [17] showed that the problem of finding the resultant force with condition (2.2) can be formulated as a hypersingular integral equation

± __ q(xoyo)t (XQ yo) e (2-5)

where w(x,y) = [u1(x,y)] + j[u2(x,y)] is the unknown crack opening displacement, q(x0, i/o) = iji(xo,i/o) + iqi(xo,yo), 72 = V-l, the w(x,y) = [u\(x,y)\ - /[w2(x,y)], and the angle 0 is defined by

x — x0 = R cos 0, y — y0 = R sin 0. (2.6)

The cross on the integral means the hypersingular, and it must be interpreted as a Hadamard finite part integral [18,19]. Equation (2.5) is to be solved subject to w = 0 on dQ where dQ is boundary of Q. For the constant shear stress in x direction, we have t23 = 0 and [u2 (x, y)] = 0, hence, (2.5) becomes

^ j ~—* ^^ w(x,y)dil = q(xo,yo), (xo,yo)eQ. (2.7)

Polar coordinates (r, 0) and (r0,00) are chosen so that the loadings q(x, y) and q(x0,y0) can be written as a Fourier series

= £ Cin q(xo,yo) = £ q„ C})^0, (2.8)

n=—co a n=—CO \U0 /

where the Fourier components qn are j-complex. The j-complex crack opening displacement, w(x,y) and w(x0,y0), have similar expressions

io{x,y)= f.WnC-)eine, 10{xo,vo)= f. ion(^)e^. (2.9)

Without loss of generality, we consider a = 1. Using Guidera and Lardner [20], the dimen-sionless function qn and wn can be expressed as

r6"l + 3)r(fc +1) №J

""<r> = '"""Sa» 7 C^5)-

fc=0 ' (2.10) r(|«| + l/2)fc! _|„|+i/2.

W,2-i k rClnl + fc + 3/2") 2fc+1 Vvi 1 )'

where the j -complex coefficients Q% are known, W% are unknown, and Cm (x) is an orthogonal Gegenbauer polynomial of degree m and index X, which is defined recursively by [21]

(m + 2)Cm+2(x) = 2(m + X + 1)xCm+1(x) - (2X + m)Cm(x), (2.11)

with the initial values CX(x) = 1 and C|(x) = 2Xx. For a constant shear loading, q(x,y) = -t, the solution for a circular crack is obtainable.

3. Nearly Circular Crack

Let Q be an arbitrary shaped crack of smooth boundary with respect to origin O, such that Q is defined as

Q = {(r ■ 0) :0 < r<p(0), -n < 0<n}, (3.1)

where the boundary of Q, dQ is given by r = p(0). Let Z = £ + in = seif with |Z| < 1 such that the unit disc is

D = {(s,y) : 0 < s < 1, -n < y<n}. (3.2)

By the properties of Reimann mapping theorem [22], a circular disc D is mapped conformally onto Q using z = af (Z). This approach works for a general smooth star-shaped domain, Q. For a particular application, let f be an analytic function, simply connected in the domain Q, |f '(Z)| is nonzero and bounded for all |Z| < 1,

f (Z) = Z + cg(Z) with g(Z) = zm+1,

Figure 2: The domain Q for f (Z) = Z + c£,m+1 at different choices of c, m = 2.

which maps a unit circle, D in the Z-plane into a nearly circular domain Q in the z-plane where c is a real parameter and r = p(6) is the boundary of Q. This domain has a smooth, regular boundary for 0 < (m + l)|c| < 1. As (m + l)|c| ^ 1 one or more cusps develop; see Figure 2 with various choices of c. Let

z - Z0 = af (Z) - f (Zo)) = Reie, (3.4)

and define S and ® as

Z - Zg = Se^,

dQ = dxdy = a2|f'(Z)| d^dn = a2|f (Z)| sdsdy , where x = au(l, n) and y = av(l, n) so that f = u + iv. Next, we define 6 and 60 as

f '(Z) = \f (Z)\ei6, f '(Zg) = |f '(Zo)|ei60. (3.6)

w(x(0,y(0)= a\f '{i)\-mei6W(ln), q(x(Zo),y(Zo)) = a\f '(Zo)\-3/2ej6° Q(M°)-

Substituting (3.5), (3.6), (3.7), and (3.8) into (2.7) gives

W&n)K{2)&y>W>l = Q(&,>lo), (Mo) eD, where the kernel K(1) (Z, Z0) and K(2) (Z, Z0) are [17]

|/(£)-/(£o)|3 li-ibl3'

(3.10)

K<2)(C/Co) = |3/2 |3/2 ej(2@-5-50)--^ 2;® (3_n)

1/(Z) - f (Z0)|3 IZ - Z0I3

This hypersingular integral equation over a circular disc D is to be solved subject to W = 0 on s = 1, and the K(1)(Z, Zo) is a Cauchy-type singular kernel with order S-2, and the kernel K(2)(Z,Z0) is weakly singular with 0(S-1), as Z ^ Z0 (seethe appendix).

We are going to solve (3.9) numerically. Write Was a finite sum

W&n) = XWlAl{s*v), (3.12)

where is defined by

X= X E, N1,N2ez.

n,k n=-N1 k=0

(3.13)

Introduce

L™(s, tp) = sI'-'IC^172(Vl^2) cosmtp, (3.14)

where m,h e Z .The relationship between these two functions, An(s,y), and Lm (s,y) can be expressed as

f Al(s,<p)L™(s,<p)^^= = Bnk6kh6mn, (3.15)

jq V1- s2

f 2 jt

—n = 0, 4fc + 3

n 2r(2k + 2n + 2)

22n+1(2k + n + 3/2)(2k + 1)![r(n + 1/2)]

, n / 0.

Both functions An(s, y) and ^(s, y) have square-root zeros at s = 1. Krenk [23] showed that

/1 - s2 J D

__' k\~ r "/ ~_| ¿til

~ ^k --- + fiaT. |

2\/1 - s2

(3.16)

r(|n| + fc + 3/2)r(fc+3/2) =-(|?z| + k)lkl-' (3'18)

Substituting (3.17) and (3.12) into (3.9) yields

YFk(so,yo)Wn = Q^o^yo) ,no(so,yo)), (3.19)

(3.20)

Next, define

W = -WnG2k+1

n\ + 1/2. lEk

(3.21)

where G2nk+J/2 = (2n + 2k + 1)!/(2k + 1)!(2n)!. Multiply (3.19) by Lm(so,yo), integrate over D and using (3.15), (3.19) becomes

IX! ( -2 V+^Ve2,° 6hk6\mM + S™!! ) = Q™, -Ni < m <NuO<h< N2, (3.22)

cvin _ ___JVWl

hk ~ I- /- hk '

V7 ={ TO) f Ank(Z)H(Z,Zo)didio,

JD Jd (3.23)

Qh = f L'»(Co)Q(&)rfto,

1 = I Lh '

EmBm JD

H(Z, Zo) = (2 - v)K(1) (Z, Zo) + 3vK(2) (Z, Zo). In (3.22), we have used the following notation: Zo = Zo(s0,^0), dZ0 = s0ds0dy0, and

Q(Zo) = Q(l0,U0) = Q(S0CoS ^0,S0Sin ^0).

In evaluating the multiple integrals in (3.22), we have used the Gaussian quadrature and trapezoidal formulas for the radial and angular directions, with the choice of collocation points (s,y) and (s0/^0) defined as follows:

i=1 i=1

(3.24)

M2 jn M2 (j + 0.5)n

where W(i) and W0(i) are abscissas for si and s0i, respectively, M1 and M2 is the number of collocation points in radial and angular directions, respectively. This effort leads to the (2Ni + 1)(N + 1) x (2Ni + 1)(N + 1) system of linear equations

AW = b, (3.25)

where A is a square matrix, and W and b are vectors, W to be determined.

4. Energy Release Rate

The energy release rate (measured in

JM-2), G(^) by Irwin

's relation subject to shear load is

defined as [7,8]

Gfo) = ^[Knm2 Kul{(p)]\ (4.1)

where E, Young's modulus, a measurement of the stiffness of an isotropic elastic material and the relationship of E, v and ¡, is

v = (4.2)

Table 1: Numerical convergence for the energy release rate, G(f) for f (Z) = Z + cZ3 when c = 0.1.

N G(0.00) G(.T/4) G(.T/2) G(3.T/4) G(.T)

0 7.8676E - 10 9.0123E -10 1.6067E - 09 9.0123E -10 7.8676E -10

1 7.2724E -10 8.9392E - 10 1.3159E - 09 8.9392E - 10 7.2724E - 10

2 9.2668E - 07 7.4652E - 10 1.5649E - 09 7.4652E - 10 9.2668E - 07

3 0.0000E + 00 0.0000E + 00 6.3517E - 10 0.0000E + 00 0.0000E + 00

4 1.1859E - 05 7.4041E - 19 4.6709E - 09 7.4041E - 19 1.1859E - 05

5 3.1429E - 03 8.8211E - 04 9.2528E - 06 8.8211E - 04 3.1429E - 03

6 3.0421E - 03 8.7908E - 04 9.5791E - 04 8.7908E - 04 3.0421E - 03

7 1.5794E - 03 8.4308E - 04 9.2945E - 04 8.4308E - 04 1.5721E - 03

8 9.7557E - 04 1.1903E - 03 9.5001E - 04 1.1903E - 03 9.7557E - 04

9 9.7557E - 04 1.1903E - 03 9.5001E - 04 1.1903E - 03 9.7557E - 04

10 9.7557E - 04 1.1903E - 03 9.5001E - 04 1.1903E - 03 9.7557E - 04

and Kn(y) and Kni(f), the sliding and tearing mode stress intensity factor, respectively, are defined as [5, 7,8]

KAw) = limV,

1 r ^ a ' v a — r

w(x,y), j = II, III,

where Vj are constants.

Let a(f) = |f (eif) |, r = |f (seif) |, and as s close to 1, (4.3) leads to

Kj(f) = limi-Vj

(1 - s)|f'(eif) |

Therefore, substituting (3.7) into (4.4) and simplifying gives

w(x,y), j = II, III.

Kj(f) = V

I v 'I ^^ . / -nn-nn

►, j = II, III,

where = D^}/2(0) cos(«y), and (Vl^s1) = Vl^D^/^VT^s1), where

Dm (x) is defined recursively by

mDm (x) = 2(m +1 - 1)xDm_ 1 (x) - (m + 21 - 2)D^-2 (x), m = 2,3,..., (4.6)

with Dl(x) = 21 and Dl(x) = 21x.

Table 1 shows that our numerical scheme converges rapidly at a different point of the crack with only a small value of N = N1 = N2 are used.

Degree

- Gao [8] ---Numerical

Figure 3: The energy release rate, G(y) for f (Z) = Z + 0.001Z3.

Degree

- Gao [8] ---Numerical

Figure 4: The energy release rate, G(y) for f (Z) = Z + 0.01Z3.

Figures 3,4,5, and 6 show the variations of G against y for c = 0.001, c = 0.01, c = 0.10, and c = 0.30, respectively. It can be seen that the energy release rate has local extremal values when the crack front is at cos(y) = ±1 or sin(y) = ±1. Similar behavior can be observed for the solution of G(y) for a different c and v at c = 0.1, displayed in Figures 7 and 8. Our results agree with those obtained asymptotically by Gao [8], with the maximum differences for m = 2 are 3.6066 x 10-6,4.7064 x 10-5,5.3503 x 10-5, and 9.0000 x 10-5 for c = 0.001, c = 0.01, c = 0.10, and c = 0.30, respectively.

0.0012 -

S 0.0008 -

ig 0.0006 -c

0.0004

0.0002

- Gao [8] ---Numerical

180 Degree

Figure 5: The energy release rate, G(q>) for f (Z) = Z + 0.1Z3.

Degree

- Gao [8] ---Numerical

Figure 6: The energy release rate, G(y>) for f (Z) = Z + 0.3g3.

5. Conclusion

In this paper, the hypersingular integral equation over a nearly circular crack is formulated. Then, using the conformal mapping, the equation is transformed into hypersingular integral

0.0016

Degree

- c = 0.001 c = 0.1

...... c = 0.01 ---c = 0.3

Figure 7: The energy release rate, G(q>) for f (Z) = Z + cZ3 at various c.

0.7 0.6

5 0.5 fö

SS 0.4

v = 0.2 v = 0.3

180 Degree

v = 0.4 - v = 0.5

Figure 8: The energy release rate, G(q>) for f (Z) = Z + 0.1Z3 at various choices of v with fi = 1.

equation over a circular crack, which enable us to use the formula obtained by Krenk [23]. By choosing the appropriate collocation points, this equation is reduced into a system of linear equations and solved for the unknown coefficients. The energy release rate for the mentioned crack subject to shear load is presented graphically. Our computational results seem to agree with the asymptotic solution obtained by Gao [8].

Appendix

The Singularity of the Kernel K(1)(Z, Zo) and K(2)(Z,Zo) At Z = Zo, we have

m - m = (i - £b)/'(£b) + {C-C0)2/U0) + • • •. (A.i)

Differentiate /(Z) with respect to Z, we have

f '(Z)= / '(Zo) + (Z - Zo)/" (Zo) +

(Z-Zo)2r(Zo) | (A.2)

F1 = (Z-io)ly^ = u1+iv1=0(S) , (A3)

F2 = (Z-io)2^|y = «2 + ir2 = 0(s2) as S^O, (A.4)

where u1, u2, v1, and v2 are real. As F1 = O(S) and F2 = O(S2) as S ^ 0, we see that ui and Vi are O(Si) as S ^ 0 (i = 1,2). Hence, (A.1) becomes

f (Z) - f (Zo) = f '(Zo)(Z - Zo)

1 F1 1 + T +

Substituting (A.3) into (A.2) gives

f'(Z)= f'(Zo)[1 + F1 + •••], f (Zo) = f'(Z)[1 - F1 -•••]. (A.6)

As S ^ o and truncate (A.1) at second order, then (A.6) can be written as

f(Z) - f (Zo){1 + F1}, f (Zo) - f '(Z){1 - Fx), (A.7)

respectively. Now, consider K(1)(Z, Zo). Let 6-6o - vi = O(S) where 6 and 6o defined in (3.6), then, from (3.6), we have

i(6-6o, = f'W(y>)\ (A 8)

1 + F1 1 + F1 1 - u1

|1 + F1| 1 + u1 1 - u1

- (1 + Fx)(1 - U1)

- 1 + iv\.

\f (Z) - f (Zo)\3 IZ - Zo|3 S3\|1 + (1/2)Fi + (1/3)F2| as S ^ 0.

Next, using (A.12), (A.11), and (A.10), we obtain

3/2, ruy. , 13/2

k (C. C ) (!•/,■)

\f (Z) - f (Zo)\3 IZ - Zo|

1 /|l+Fi|3/2 A , 1 /|l + Fi|3/2

S3V|l+Fi/2|3"7 + S3VH + fi/2|3M

|1 + F1/2|3 8 Thus, K(1) (Z, Z0) reduces to

Hence,

ei(5-5o) ~ 1 + jvv (A.10)

Martin [24] showed that

\ 1 + aF1 + (5F2\l = |1 + au1 + ¡u2 + i(av1 + ¡v2) |k/1

1 f 2 2 2 2 1 (A.11)

— 1 + aXu 1 + — A|(A — 1 )lxn\ + a v\ + 2/5m2 j, where a, j, and ¡5 are constants and

|/'(i)f''2|/'(M)f/2 • _•, ...........,.=0(S-

L_ .j/ H+F.+fil"2 -= o(s~'\ (A.12) . I3 S3 \ 11 +(1/2)F +(1/3) Fn I3 / V )

(A.13)

1. (A.14)

= (A.15)

Since F1 = u1 + iv1, then

(u1 + iv1)2 = u"[ - v2 + 2iu1v1, Re(F1) = - u^, (A.16)

so, (A.3) leads to

V V 1 (/'(Co)) J S

where D(Zo,®) = Re{e2j®((/"(Zo))2/(/'(Zo))2)} and Z - Zo = SeiQ defined in (2.7). Thus,

- MS-3 (A.18)

^ o(s-2).

Therefore, K(1)(Z,Zo) - MS-3, that is, K(1)(Z,&) - O(S-2) as S ^ 0.

For K(2) (Z, Zo), expand /(Z) at Z = Z0, and truncating at second order, (3.4) gives

Ret0 = a(Z-Zo)/'(Zo){l + y}, (A.19)

si(0-<K-6o) +

= l + iy. (A.20)

Next, substituting (A.5) and (A.7) into (3.4) gives

Rete = a(Z - &)/'(£)(! - F0{ 1 + § (A.21)

Using (A.22) and (A.2o) yields

= 1 + (A.22)

gt(20-2®-6-6o) ^ ! + ¿2 ^,2 + it,2 = 1 + o (s2). (A.23)

Hence, as S ^ o, then ei(20-2®-6-6o) - 1 + O(S2). It is not difficult to see that R - a|/'(Zo)|S, 0 - O + 6o, R - a|/'(Z)|S, and 0 - ® + 6, respectively; then (3.11) becomes

K(2) Cq) = l/'(Z)|3/2|/'(Zo)|3/2e2^--1 (A.24)

|/(Z) -/(Zo)|3 |Z - Zo|3

Applying similar procedures as in K1(Z, Zo) gives

IZ - Zoi3\ \f (Z) - f (Zo )|

1 /11 + f1|

S3V |1 + f l / 2|'

3F(Zo,(D)

(A.25)

(A.26)

Acknowledgments

The authors would like to thank the reviewers for their very constructive comments to improve the quality of the paper. This project is supported by Ministry of Higher Education Malaysia for the Fundamental Research Grant scheme, project no. 01-04-10-897FR and the second author received a NSF scholarship.

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