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International Journal of Fracture (2005) 131:367–384 DOI 10.1007/s10704-004-7138-3 © Springer 2005 Interaction between a cracked hole and a line crack under uniform heat flux PHAM CHI VINH1 , NORIO HASEBE2,∗ , XIAN-FENG WANG2 and TAKAHIRO SAITO2 Faculty of Mathematics, Mechanics and Informatics, Hanoi National University, 334 Nguyen Trai Str., Thanh Xuan, Hanoi, Vietnam Department of Civil Engineering, Nagoya Institute of Technology, Gokiso-cho, Showa-ku, Nagoya 466-8555 Japan ∗ Author for correspondence (E-mail: hasebe@kozo4.ace.nitech.ac.jp; Fax: +81-52-735-5482) Received 21 July 2004; accepted in revised form December 2004 Abstract This article deals with the interaction between a cracked hole and a line crack under uniform heat flux Using the principle of superposition, the original problem is converted into three particular cracked hole problems: the first one is the problem of the hole with an edge crack under uniform heat flux, the second and third ones are the problems of the hole under distributed temperature and edge dislocations, respectively, along the line crack surface Singular integral equations satisfying adiabatic and traction free conditions on the crack surface are obtained for the solution of the second and third problems The solution of the first problem, as well as the fundamental solutions of the second and third, is obtained by the complex variable method along with the rational mapping function approach Stress intensity factors (SIFs) at all three crack tips are calculated Interestingly, the results show that the interaction between the cracked hole and the line crack under uniform heat flux can lead to the vanishing of the SIFs at the hole edge crack tip The fact has never been seen for the case of a cracked hole and a line crack under remote uniform tension Key words: Cracked hole, dislocation, heat flux, interaction, integral equation, mapping function, stress intensity factor, thermal stress Introduction Due to the stress concentration effect, cracks are likely to initiate at a hole boundary under the action of monotonous or fatigue loading A number of papers dealing with the hole edge crack problem are available (Bowie, 1956; Tweed and Rooke, 1973; Hasebe and Ueda, 1980; Schijve, 1983; Hasebe et al., 1988, 1994a, b; Zhang and Hasebe, 1993; Chao and Lee, 1996; Hasebe and Chen, 1996) Among them, the rational mapping function approach is significant to solve the hole edge crack problem (Hasebe and Ueda, 1980; Hasebe et al., 1988, 1994a, b; Hasebe and Chen, 1996) Up to now, there seem to be only a few papers concerning the interaction between the crack emanating from a hole and another independent crack Hasebe et al (1994a) solved a second mixed boundary value problem analytically and as an example of the solution, the interaction of a square hole with an edge crack and a line crack was investigated for a geometrically symmetric case Hasebe and Chen (1996) treated a cracked circular hole and a crack when the remote stresses were applied at 368 P.C Vinh et al B y no x A k crac C (b) Problem B y 2b (1−γ)π -a γπ A a e d C I y B f x tion a loc A y B I is ed r tu C x pera tem zo A temperature dislocation C x (e) Problem E (c) Problem C (a) Problem A y y zo B edge dislocation I A C x on cati islo ed edg (d) Problem D C x (f) Problem F Figure Superposition of problems infinity The interactions of a hole and a rigid inclusion, respectively, with a crack were considered by Hasebe et al., (2003a) The problem of a crack initiating from a rigid inclusion interacting with a line crack was also solved by Hasebe et al (2003b) It is known that when steady heat flow is disturbed by the presence of crack, there is a high local intensification of temperature gradient at the crack tips and large thermal stresses arise around them Thermal disturbances of this kind may, in some cases, cause crack propagation resulting in serious damage to structural components Consequently, the study of the behavior of thermal stresses near the crack tips is of importance in fracture mechanics The aim of this paper is to investigate the interaction between a cracked hole and a line crack under uniform heat flux The hole edge and the faces of the cracks are assumed to be adiabatic and traction free The problem to be considered is shown in Figure 1a It is clear that the original problem A can be reduced to three following problems by the principle of superposition: Problem B: Problem of the hole with an edge crack under uniform heat flux (Figure 1b) Problem C: Problem of the hole with an edge crack under distributed temperature dislocation along the line crack surface (Figure 1c) Problem D: Problem of the cracked hole under distributed edge dislocation along the line crack surface (Figure 1d) In order to solve these problems we apply the complex function method along with the rational mapping function approach By means of the principle of superposition, Interaction between a cracked hole and a line crack 369 problem C (D) can be reduced to a problem named problem E (F), in which the cracked hole subjected to a temperature point dislocation (an edge point dislocation) placed at some point on the crack line AB After problem E (F) is solved, the solution of problem C (D) can be obtained by integrating the Green’s functions of problem E (F) along the line AB By summation of the solutions of problems B, C and D, we can ascertain problem A’s solution, in which there are two unknown functions: the distribution density of temperature dislocation along AB (coming from problem C’s solution) and the distribution density of edge dislocation along AB (coming from problem D’s solution) These functions are determined from the singular integral equations which are derived from the adiabatic and traction-free conditions along the faces of the line crack AB Finally, by numerical integration of the singular integral equation, stress intensity factors (SIFs) at crack tips are obtained It is interesting to find that the interaction between the cracked hole and the line crack under uniform heat flux can lead to the vanishing of the SIFs at the hole edge crack tip The fact has never been seen for the case of a cracked hole and a line crack under remote uniform tension Rational mapping function As mentioned above, the rational mapping function technique is used to deal with the cracked hole with arbitrary shapes in an infinite plane In the computation, we will consider a cracked elliptical hole and a cracked square hole as examples Here, without loss of generality, the general form of rational mapping function is given (Hasebe and Ueda, 1980; Hasebe et al., 2003b): N z = ω(ς) = E0 ς + k=1 Ek + E−1 , ςk − ς (1) which maps the exterior of the cracked hole in the z-plane onto the exterior of the unit circle in the ς -plane as shown in Figure Here E−1 , E0 , Ek , ςk (k = 1, 2, 3, , N ) are complex constants and |ςk | < for all k = 1, 2, 3, , N It should be noted that other configurations can be readily tackled using this general formulation Taking E0 = and Ek = (k = 1, 2, 3, , N ) in (1), the mapping function is reduced to the one for a circle Likewise, when E0 = (a + b)/2 and E1 = (b − a)/2, ς1 = and Ek = (k = 2, 3, , N ) are taken, the mapping function is for ellipse with semi-axes on the x and y-axes being a and b, respectively Basic formulae 3.1 Temperature field In the ς-plane, the temperature function θ(ς, ς) for the two-dimensional steady-state thermoelasticity satisfies the Laplace equation Thus, the temperature function can be expressed as the real part of an analytic function Y (ς), which gives temperature and heat flux (Hasebe et al., 1988; Han and Hasebe, 2001) as: θ(ς, ς) ¯ = 21 [Y (ς) + Y (ς)], (2) 370 P.C Vinh et al Figure Rational mapping functions for a cracked square hole and a cracked elliptical hole Y (ς) , ω (ς) ςω (ς) qρ − iqθ = (qx − iqy ), |ςω (ς)| qx − iqy = −k (3) (4) where qx and qy denote the heat flux components in the x- and y-axes, respectively, qρ and qθ represent these in the orthogonal curvilinear coordinates generated by ω(ς), and k signifies the thermal conductivity of the material Also the heat flux boundary condition is given as follows: −k[Y (σ ) − Y (σ )] = 2i qn (σ )dσ + const., (5) where σ and qn denote a value of ς and the normal heat flux component on the boundary, respectively, and the integration is carried out along the boundary From (5), the adiabatic condition (qn = 0) along the boundary is: Y (σ ) − Y (σ ) = const (6) 3.2 Thermal stress field Employing complex functions φ(ς) and ψ(ς), the stresses in the elastic body are (Muskhelishvili, 1963): σx + σy = 4Re φ (ς) , ω (ς) σy − σx + 2iτxy = ω(ς) (7) φ (ς) ω (ς) + ψ (ς) ω (ς), (8) Interaction between a cracked hole and a line crack 371 σθ + σρ = σx + σy , σθ − σρ + 2iτρθ = (9) ς ω (ς) |ς|2 ω (ς) (σy − σx + 2iτxy ) (10) The stress boundary condition is written as: φ(σ ) + ω(σ ) ω (σ ) φ (σ ) + ψ(σ ) = i (px + ipy )ds + const., (11) where px and py denote external force components applied to the boundary in the x- and y-directions, respectively Using the complex stress functions φ(ς), ψ(ς) and the temperature function Y (ς), the displacement expression can be put into the following form: κφ(ς) − ω(ς) ω (ς) φ (ς) − ψ(ς) + 2Gα Y (ς)ω (ς)dς = 2G(u + iν) (12) where G is the shear modulus, κ and α are: κ = − 4v, α = (1 + v)α for plane strain and κ = (3 − v)/(1 + v), α = α for generalized plane stress; v and α are Poisson’s ratio and the coefficient of thermal expansion, respectively Solution of problem B 4.1 Temperature field Consider the heat conduction problem shown in Figure 1b, in which q is the intensity of the uniform heat flux; δ is the angle between the direction of the heat flux and the x-axis Herein the hole edge and crack faces are assumed to be adiabatic The temperature function YB (ς) of problem B can be broken down into two parts: YB (ς) = Y1B (ς) + Y2B (ς), (13) where the first function denotes the one induced from the uniform heat flux; the second one denotes the complementary part From (3), function Y1B (ς) can be obtained: q Y1B (ς) = − e−iδ ω(ς) k (14) Substituting (13) and (14) into (6) yields: q q Y2B (σ ) − Y2B (σ ) = e−iδ ω(σ ) − eiδ ω(σ ) + const k k (15) Multiplying (15) by the factor dσ/[2πi(σ − ς)] and carrying out the Cauchy integration along the unit circle, we obtain Y2B (ς) and finally (Hasebe et al., 1988): YB (ς) = − E¯0 q −iδ e E0 ς + eiδ + const k ς (16) 372 P.C Vinh et al 4.2 Thermal stress field The stress function for the thermoelastic problem can be split into two parts: a nonholomorphic part [φ1B (ς), ψ1B (ς)] and a holomorphic part [φ2B (ς), ψ2B (ς)]: φB (ς) = φ1B (ς) + φ2B (ς), ψB (ς) = ψ1B (ς) + ψ2B (ς) (17) The last term of the left-hand side of (12) denotes the thermal displacement This integration contains logarithmic term which represents dislocation in the thermoelasticity To remove it, we consider the following stress functions φ1B (ς), ψ1B (ς) (Florence and Goodier, 1960): φ1B (ς) = AB log ς, ψ1B (ς) = BB log ς, (18) where the constants AB and BB are determined using the conditions that the stress and the displacement components around the hole are single-valued Substituting (17) and (18) into (11), and using the stress single valuedness condition, we have BB = A¯ B (19) Next, substitute (17) and (18) into (12) Multiple values of logarithmic terms must be cancelled due to the single-valuedness of displacement Consequently, the constant AB is determined as: αqGR AB = E0 2k N Ek e−iδ + E¯0 eiδ , (20) k=1 where R = (1 + v)/(1 − v) for plane strain and R = (1 + v) for generalized plane stress Now we consider the boundary condition to derive the holomorphic functions φ2B (ς) and ψ2B (ς) The hole edge and crack faces are assumed traction free without loss of generality, i.e., px = py = Substituting (17) into (11) yields φ2B (σ ) + ω(σ ) ω (ς) φ2B (σ ) + ψ2B (σ ) = −φ1B (σ ) − ω(σ ) ω (σ ) φ1B (σ ) − ψ1B (σ ) (21) Multiplying (21) by the factor dσ/[2πi(σ − ς)] and carrying out the Cauchy integration along the unit circle, we obtain φ2B (ς) as: N φ2B (ς) = k=1 A¯k Bk + A¯B ς − ςk N k=1 B k ςk ς − ςk (22) in which Bk ≡ Ek /ω (ςk ) with ςk ≡ 1/ς¯k and Ak = φ2B (ςk ) Here the real and imaginary values of Ak are determined by the simultaneous equations of 2N derived by differentiating (22) and substituting ς = ςk Thus, the stress function φB (ς) is (Hasebe et al., 1988): N φB (ς) = AB log ς + k=1 A¯k Bk + A¯ B ς − ςk N k=1 Bk ςk ς − ςk (23) Interaction between a cracked hole and a line crack 373 The stress function ψB (ς) can be derived by analytic continuation along the freetraction boundary of the unit circle Indeed, by introducing the following function: φB (ς) = − ω(ς) ¯ φ (1/ς) − ψ¯B (1/ς), ¯ ω (1/ς) B ς ∈ S − = {ς : |ς| < 1} (24) from (11) with regarding px = py = on the unit circle and (24), we have φB+ (σ ) = φB− (σ ), σ ∈ = {ς : |ς| = 1} (25) This means the function φB (ς), ς ∈ S − is a continuation of the function φB (ς), ς ∈ S + = {ς : |ς| > 1} from the outside of the unit circle to its inside From (25) it follows: ψB (ς) = −φ¯B (1/ς) − ω(1/ς) ¯ φ (ς), ω (ς) B ς ∈ S + (26) Solution of problem C Consider an infinite plane with a cracked hole subjected to distributed temperature dislocations along the line crack surface, as shown in Figure 1c The hole edge and crack faces are assumed to be traction-free and adiabatic As previously said, by means of the principle of superposition, problem C can be reduced to a problem E, in which the cracked hole subjected to a temperature point dislocation placed at some point z0 (= ω(ς0 ), ς0 is a point in the ζ -plane corresponding to z0 ) on the line AB To solve problem E, we need to find the Green’s function for the temperature field, as well as the stress field of the cracked hole under a temperature point dislocation After problem E is solved, the solution of problem C can be obtained by integrating the Green’s function along the line AB It is not difficult to verify that, for the temperature field, the Green’s function of problem E for a couple temperature dislocation is (Hasebe and Han, 2001): YE (ς) = 2πk e−iϑ ςp2 eiϑ , − ω (ς0 )(ς − ς0 ) ω (ς0 )(ς − ςp ) ϑ= π + β, (27) where k is thermal conductivity of the material, β is the angle between the line AB and the x-axis, denotes the magnitude of the couple temperature dislocation, and ς0 = ω−1 (z0 ), ςp ≡ 1/ς¯0 Next, for the stress field, the stress functions φE (ς) and ψE (ς) for problem E can be expressed in the following form: φE (ς) = φ1E (ς) + φ2E (ς), ψE (ς) = ψ1E (ς) + ψ2E (ς), (28) where φ2E (ς) and ψ2E (ς) are holomorphic in S + ; φ1E (ς) and ψ1E (ς) are expressed as C log(ς − ς0 ) + AE log ς, φ1E (ς) = − 2π ω(ς0 ) C C log(ς − ς0 ) + 2π + BE log ς ψ1E (ς) = − 2π ω (ς0 )(ς−ς0 ) (29) 374 P.C Vinh et al 2Gα with C = k(κ+1) eiϑ and AE log ς and BE log ς are functions to cancel the dislocation of both stress and displacement around the hole Substituting (28) and (29) into (11), and moving around the unit circle once, the requirement that the stress are single valued around the hole gives BE = A¯ E (30) Substituting (28), (29) and (30) into the displacement expression (12), after moving around the unit circle, the displacements must recover their initial values This requirement fixes the constant AE in the following expression: AE = C 2π N k=1 Ek C¯ E0 ςp2 + ω (ς0 )(ς0 − ςk )2 2π ω (ς0 ) (31) On account of (28), the traction-free boundary condition (11) is of the form: φ2E (σ ) + ω(σ ) ω (ς) φ2E (σ ) + ψ2E (σ ) = −φ1E (σ ) − ω(σ ) ω (σ ) φ1E (σ ) − ψ1E (σ ) (32) For obtaining φ2E (ς), we substitute (29) into (32), multiply both sides of the resulting equation by the factor dσ/[2πi(σ − ς)], (ς ∈ S + ), and carry out the Cauchy integration along the unit circle Using the following: (i) The function (ω(ς)/ω¯ (1/ς))φ¯2E (1/ς) is holomorphic in S − except the points φ (ς ) k 2E k ςk (k = 1, , N), which are poles with the principal parts: ςkE−ς ω (ςk ) (ii) The function (ω(ς)/ω¯ (1/ς))φ¯1E (1/ς) is holomorphic in S − except the points ςk (k = 1, , N) and ςp , which are poles with the principal parts, respectively, being: C¯ ςp ςk − A¯ E ςk ω (ςk )(ς − ςk ) 2π ςp − ςk Ek and C¯ ω(ςp ) ςp2 , 2π ω (ς0 ) ς − ςp (33) C and taking φ2E (∞) = 2π log(−ς0 ), we obtain the function φ2E (ς) as: φ2E (ς) = N C C¯ ςp ςk Ek log − ς0 + φ2E (ςk ) + A¯ E ςk − 2π ς 2π ςp − ςk k=1 ω (ςk )(ς − ςk ) C¯ ω(ς0 ) − ω(ςp ) ςp + 2π ω (ς0 )(ς − ςp ) (34) It is noted that the real and the imaginary values of φ2E (ςk ) in (34) are determined by solving the simultaneous equations of 2N derived by differentiating (34) and substituting ς = ςk Thus, the function φE (ς) is: N Ek C C¯ ςp ςk log − ς0 + φ2E (ςk ) + A¯ E ςk − φE (ς) = 2π ς 2π ςp − ςk k=1 ω (ςk )(ς − ςk ) ¯ C C ω(ς0 ) − ω(ςp ) ςp − log(ς − ς0 ) + AE log ς + 2π ω (ς0 )(ς − ςp ) 2π (35) Interaction between a cracked hole and a line crack 375 Similar to Section 4, the stress function ψE (ς) can be derived directly by analytic continuation along the traction-free boundary as: ψE (ς) = −φ¯E (1/ς) − ω(1/ς) ¯ φ (ς), ω (ς) E ς ∈ S + (36) As previously stated, by means of the principle of the superposition, after problem E is solved, the solution of problem C is obtained by using the Green’s functions YE (ς), and φE (ς) and ψE (ς), respectively, along the line AB (see Equations (45) and (48)) Solution of problem D Problem D is shown in Figure 1d, in which an infinite plane having a hole with an edge crack is under distributed edge dislocations along the line crack AB The hole edge and crack surfaces are traction-free Naturally, by means of the principle of superposition, problem D can be reduced to a problem F, in which an edge point dislocation with magnitude D is placed at a point z0 (= ω(ς0 )) on the line AB Similar to the previous sections, the Green’s function of problem F: φF (ς) and ψF (ς) can be found in the form: φF (ς) = φ1F (ς) + φ2F (ς), (37) ψF (ς) = ψ1F (ς) + ψ2F (ς), where φ2F (ς) and ψ2F (ς) are functions holomorphic in S + and (Hasebe and Chen, 1996): φ1F (ς) = − D log(ς − ς0 ), 2π (38) ω(ς0 ) D¯ D ψ1F (ς) = − log(ς − ς0 ) + 2π 2π ω (ς0 )(ς − ς0 ) From (37), the traction-free boundary condition (11) for problem F is of the form: φ2F (σ ) + ω(σ ) ω (ς) φ2F (σ ) + ψ2F (σ ) = −φ1F (σ ) − ω(σ ) ω (σ ) φ1F (σ ) − ψ1F (σ ) (39) Multiplying both sides of (39) by the factor dσ/[2πi(σ − ς)], (ς ∈ S+ ), and carrying out the Cauchy integration along the boundary yield the result of φ2F (ς) Similar to the procedures used in the section 5, we obtain: D φ2F (ς) = log − ς¯0 + 2π ς + N k=1 Ek ω (ςk )(ς − ςk ) D¯ ω(ς0 ) − ω(ςp ) ςp2 2π ω (ς0 )(ς − ςp ) φ2F (ςk ) + D¯ 2π ς¯0 − ς¯k (40) It should be noted that the real and imaginary parts of φ2F (ςk ) in (40) are determined by solving the simultaneous equations of 2N derived by differentiating (40) 376 P.C Vinh et al and substituting ς = ςk Taking into account (37), (38) and (40), the function φF (ς) is obtained N φF (ς) = k=1 + Ek ω (ςk )(ς − ςk ) φ2F (ςk ) + D¯ 2π ς¯0 − ς¯k D¯ ω(ς0 ) − ω(ςp ) ςp2 D D − log(ς − ς0 ) + log − ς¯0 2π ω (ς0 )(ς − ςp ) 2π 2π ς (41) The stress function ψF (ς) is derived directly by analytic continuation along the traction-free boundary of the unit circle as: ψF (ς) = −φ¯F (1/ς) − ω(1/ς) ¯ φ (ς), ω (ς) F ς ∈ S + (42) By means of the principle of superposition, the solution of problem D can be obtained by using φF (ς) and ψF (ς), respectively, along the line AB (see Equation (48)) Solution of problem A: Integral equations The original problem A can be reduced to the superposition of three subproblems B, C and D which have been studied in Sections 4–6, respectively Thus, from these results, the solution of problem A is derived as follows: Temperature function: Y (ς) = YB (ς) + YC (ς), (43) where YB (ς) is defined by (17) and YC (ς) is obtained by integrating YE (ς) with corresponding density along the line AB Stress functions: φ(ς) = φB (ς) + φC (ς) + φD (ς), (44) ψ(ς) = ψB (ς) + ψC (ς) + ψD (ς), where φB (ς) and ψB (ς) are defined by (23), (26); φC (ς) and ψC (ς), and φD (ς) and ψD (ς) are obtained by integrating the Green’s functions φE (ς) and ψF (ς), and φF (ς) and ψF (ς), respectively, along the line AB It should be noted that in (43) and (44), there are still two unknown functions (t) and D(t) Therefore, in order to calculate the solution of problem A, we need to find equations (singular integral equations) for the unknown functions: the distribution density of the temperature dislocation γ (t)dt = d (t) and the density of edge dislocation D(t) From the adiabatic condition along the crack faces expressed by (6) and (43), the following singular integral equation for γ (t) is derived: b b Im −b γ (t)YE (s, t)dt − γ (t)YE (s0 , t)]dt = −Im [YB (s) − YB (s0 )] , |s| ≤ b, (45) −b where s0 is an arbitrary point (standard point) on the crack face AB, YE is expressed by (27) with = Function γ (t) can be further reduced to the following form for a crack problem: Interaction between a cracked hole and a line crack 377 γ (t) = b2 − t G(t), (46) where G(t) denotes an unknown function By the transformation of integral variable t = −b cos θ and the standard integral method, the unknown function G(t) can be obtained numerically (Erdogan, 1969; Erdogan and Gupta, 1972) We define the edge dislocation densities as (Chen and Hasebe, 1992): hj (t) = 2G d + u − u− j , + κ dt j j = n, τ (47) From the traction-free condition (px = py = 0) on the line crack surfaces, represented by (11), the singular integral equation for the determination of hj (t) is: b {hn (t) [Nn (t, s) + iTn (t, s)] + hτ (t) [Nτ (t, s) + iTτ (t, s)]} dt −b = − [NB (s) + NC (s) + i(TB (s) + TC (s))] , |s| ≤ b, (48) where Nj (t, s) and Tj (t, s) (j = n, τ ) are, respectively, normal and tangential traction components to the crack line at point s caused by the unit point dislocation in the j-direction located at point t They are determined by (7)–(10), (41) and (42) The tractions (NB + iTB ) and (NC + iTC ) are calculated from (7) to (10) and the stress functions of problems B and C From the condition of single-valuedness of displacement around the line crack, hj (t) (j = n, τ ) must satisfy the following condition (Chen and Hasebe, 1992): b [hn (t) + ihτ (t)] dt = (49) −b Further, for a crack problem, the functions hn (t) and hτ (t) are expressed as: hj (t) = Hj (t)/ b2 − t , j = n, τ (50) Accordingly, Hn (t) and Hτ (t) become the unknown functions Inserting (50) into (48) and (49), and after transformation of integral variable t = −b cos θ , the singular integral equations can be solved using the Gauss–Chebyshev integration formula (Erdogan, 1969; Erdogan and Gupta, 1972; Chen and Hasebe, 1992) Then, the stress intensity factors at the crack tips A, B and C can be obtained by (Chen and Hasebe, 1992; Hasebe and Chen, 1996): KB = π H (−b), H (−b) = b M π H (b), H (b) = KA = − b M tj = b cos (2j − 1)π 2M , M (−1)j +M Hn (tj ) + iHτ (tj ) tan j =1 M (−1)j +1 Hn (tj ) + iHτ (tj ) cot j =1 j = 1, , M, (2j − 1)π , (51) 4M (2j − 1)π , 4M (52) (53) 378 P.C Vinh et al b {hn (t)kFn (t) + hτ (t)kF τ (t) + γ (t)kE (t)} dt, KC = kB + (54) −b h(t) = hn (t) + ihτ (t), √ φj (σc ) , kj = π √ ω (σc ) j = B, E, Fn , Fτ (55) where kj (j = B, E, Fn and Fτ ) are stress intensity factors at tip C corresponding to Problems B, E, and F, respectively Fn and Fτ are obtained for the point dislocations in the normal and tangential directions, respectively, to the crack line in problem F σc is the corresponding position of the crack tip C on the unit circle (σc = in this case, see Figure 2) KC is obtained as M KC = kB + j =1 π Hn (tj )kF n (tj ) + Hτ (tj )kF τ (tj ) + (b2 − tj2 )G(tj )kE (tj ) M (56) Numerical results and discussions Stress intensity factors are all normalized by the SIF K0 of the crack with length 2b under remote uniform heat flux in the direction perpendicular to the crack face, i.e., √ b πbqαGR Kχ Fχ = (57) , χ = A, B, C, K0 = 2k K0 It is clear that when heat flux is applied perpendicular to the crack face, the normalized SIF in mode I at crack tips A, B, C, namely FIA , FIB , FIC , are zero 8.1 SIFs and the effect of hole shape First, we consider the problem specified in Figure The heat flux is assumed flowing along the y-axis (δ = 90◦ ), the line crack is located on the x-axis (β = 0, f/a = 0), b/a = 2, e/a = 0.5 and v = 0.3 To show the effect of the hole shape, the interaction of the line crack with a cracked square hole, and cracked elliptical holes of different ratios of λ/a (= 0, 1, 3) is considered It should be noted that the elliptical hole of λ/a = degenerates to a crack The dependences of FIIA , FIIB , FIIC , on c/a for different shapes of hole are shown in Figure It is seen that SIFs increase with the line crack moving toward the edge crack tip C, and when c/a tends to zero FIIA , FIIC increase significantly This represents the singularity of SIFs for tips A and C in the interaction For the right hand side crack tip B, FIIB tends to certain value of stress intensity factor with a cracked hole of crack length (e + 2b) when c/a goes to zero In the four types of hole shapes, we can see the elliptical holes of λ/a = and λ/a = induce the highest and lowest values of FII , respectively, especially on the tips A and C, whereas it seems the effects of circle and square on the interaction are almost the same: both curves located intermediate to those of the elliptical holes We can also see that the intervals between the curves become larger as c/a decreases This means that the hole shape will have a stronger influence on the interaction when the line crack approaches the hole edge crack Interaction between a cracked hole and a line crack 379 Crack λ/a = λ Square λ/a = y d a FII F IIB FIIC A I B 2b FIIA C x e c q -1 -2 0.5 1.5 2.5 c/a Figure Effect of hole shapes on normalized SIFs FII (δ = 90◦ , β = 0, f/a = 0, b/a = 2, e/a = 0.5, v = 0.3) 8.2 Vanishing of the SIFs at the crack tip C Figure shows the variation of FIIC for a larger range of c/a From Figures to it can be seen that there exists a corresponding value c0 of c/a such that FIIC (c0 ) = for a given hole It means that under uniform heat flux, the interaction between a line crack and cracked hole may lead to vanishing SIFs at the crack tip C This fact has not ever been seen for the case of uniform remote tension Physically, this may be explained roughly as follows: in the cases shown in Figures and 4, which is under a uniform heat flux, the lower surface of the crack AB is stretched in comparison with the upper one, and FIIA becomes greater for more comparative stretching The similar things are happened for the hole crack and FIIC The comparative stretch of the crack AB reduces the one of the hole crack and vice versa Therefore, when the crack AB is enough closed to the hole crack, the comparative stretch of the hole crack may become zero, i.e., FIIC = From Figure it is seen that for the zero values of FIIC , the smaller of value λ/a, the smaller of value c0 For a given hole and f/a = 0, β = 0, δ = 90◦ , e/a = 0.5, FIIC is a function of b/a, d/a (or c/a) From the equation FIIC (d/a, b/a) = we can find b0 of b/a as a function of d/a That means that at each distance between the line crack and the hole we can find the corresponding length of the line crack that makes the values of FIIC become zero by the interaction The dependence of b0 /a on d/a for different shapes of holes is shown in Figure It should be also noted from the numerical results that if e/a is large enough compared with b/a, then FIIC can not become zero In this case FIIA may vanish alternatively 380 P.C Vinh et al Crack λ/a = 0.5 Square FIIC λ/a = -0.5 12 16 c/a Figure Vanishing of FIIC (δ = 90◦ , β = 0, f/a = 0, b/a = 2, e/a = 0.5, v = 0.3) 3.5 Crack λ/a = 0.5 λ/a = Square λ/a = b /a 2.5 1.5 0.5 10 d/a Figure Dependence of b0 /a on d/a for FIIC = 8.3 Dependence of normalized SIFs on the line crack position We now focus on the cracked square hole and a line crack only and study the problem specified in Figures 6–11 The dependences of the normalized SIFs FIA , FIIA , FIB , FIIB , FIC , and FIIC on f/a and the orientation angle of the crack AB are shown in Figures 6–11, respectively When the crack AB is far from the square hole (f/a = 1000): FIIA ≈ 1, FIIB ≈ −1 at β = and FIC ≈ 0, and FIIC ≈ −0.628 =const, ∀β It means that the interaction effect of the cracked square hole and the line crack AB is almost zero when they are far from each other From Figures 6–11, we can see that the cracked hole and the crack AB have considerable influence on each other when the crack AB is in the vicinity of the cracked hole, and the influence depends on the position of the crack AB Interaction between a cracked hole and a line crack 381 0.8 f/a = 0.4 FIA f/a = 10 f/a = 1000 b/a = d/a = e/a = 0.5 δ = 90° -0.4 0° f/a = 120° 60° 180° β Figure Normalized SIF FIA against location angle β for cracked square hole f/a = f/a = 10 FIIA f/a = b/a = d/a = e/a = 0.5 = 90° -1 f/a = 1000 -2 60 120 β 180 Figure Normalized SIF FIIA against location angle β for cracked square hole 0.8 f/a = 0.4 FIB f/a = 10 f/a = 1000 b/a = d/a = e/a = 0.5 = 90 ° -0.4 0° 60° f/a = 120° 180° Figure Normalized SIF FIB against location angle β for cracked square hole 382 P.C Vinh et al f/a = FIIB b/a = d/a = e/a = 0.5 = 90° f/a = 10 -1 f/a = 1000 -2 0° f/a = 60° β 120° 180 ° Figure Normalized SIF FIIB against location angle β for cracked square hole 0.6 0.4 0.2 FIC f/a = 1000 f/a = 10 b/a = d/a = e/a = 0.5 = 90° -0.2 -0.4 -0.6 0° f/a = f/a = 60° 120° 180° Figure 10 Normalized SIF FIC against location angle β for cracked square hole 0.1 f/a = b/a = d/a = e/a = 0.5 = 90° FIIC -0.2 f/a = -0.4 f/a = 1000 -0.6 f/a = 10 -0.8 0° 60° β 120° 180° Figure 11 Normalized SIF FIIC against location angle β for cracked square hole Interaction between a cracked hole and a line crack 383 Also from the numerical results, when the crack AB locates far away the cracked square hole, FIA ≈ and FIB ≈ ∀β It means that the heat flux has no effects on the mode I SIFs at the crack tips A and B in the case of crack AB being far from the cracked square hole Conclusions In this study, the interaction problem of a cracked hole with arbitrary shapes and a line crack was considered The problem is concerned with a doubly connected region of which the boundary value problem for plane elasticity possesses an inherent difficulty We converted the original problem into three subproblems by using the principle of superposition and the latter two subproblems are tackled using the Green’s function method The rational mapping function approach is used and the stress functions are obtained in the closed form for the cracked hole Therefore the boundary condition is satisfied completely for the cracked hole Singular integral equations are derived for the original problem and are numerically solved by the standard method Thus the boundary condition on the line crack AB is satisfied at finite points It is interesting to find that the 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circular hole Journal of Mathematical Physics 35, 60–71 Chao, C.K and Lee, J.Y (1996) Interaction between a crack and a. .. given hole It means that under uniform heat flux, the interaction between a line crack and cracked hole may lead to vanishing SIFs at the crack tip C This fact has not ever been seen for the case... hole shape will have a stronger influence on the interaction when the line crack approaches the hole edge crack Interaction between a cracked hole and a line crack 379 Crack λ /a = λ Square λ /a =