Mode Stresses for the Interaction between an Inclined Crack and a Curved Crack in Plane Elasticity

1Mathematics Department, Faculty of Science, Universiti Putra Malaysia, 43400 Serdang, Selangor, Malaysia 2Institute for Mathematical Research, Universiti Putra Malaysia, 43400 Serdang, Selangor, Malaysia 3College of Foundation and General Studies, Universiti Tenaga Nasional, 43000 Kajang, Selangor, Malaysia 4Faculty of Science and Technology, Universiti Sains Islam Malaysia (USIM), 71800 Negeri Sembilan, Malaysia


Introduction
For two-dimensional crack, Panasyuk et al. [1], Cotterell and Rice [2], Shen [3], and Martin [4] used perturbation method to obtain the elastic stress intensity factor for a variety of crack positions.Formulation in terms of singular, hypersingular, or Fredholm integral equations for solving single [5] and multiple cracks problems [6] in various sets of cracks positions was proposed later.These integral equations are solved numerically.Numerical solution of the curved crack problem using polynomial approximation of the dislocation distribution was achieved by taking the crack opening displacement (COD) as the unknown and the resultant forces as the right term in the equations [7].
The curved length coordinate method [8] where the crack is mapped on a real axis provides an effective way to solve the integral equations for the curved crack.Boundary element method, which avoids singularities of the resulting algebraic system of equation [9], and the dual boundary element method [10] have also been considered successfully.
In this paper, the interaction between inclined and curved cracks is formulated into the hypersingular integral equations using the complex potential method.This approach has been considered by Guo and Lu [11].Then, by the curved length coordinate method, the cracks are mapped into a straight line, which require less collocation points, and hence give faster convergence.In order to solve the equations numerically, the quadrature rules are applied and we obtained a system of algebraic equations for solving the unknown coefficients.The obtained unknown coefficients will later be used in calculating the SIF.

Complex Variable Function Method
The complex variable function method is used to formulate the hypersingular integral equation for the interaction between an inclined crack and a curved crack.Let Φ() =   () and Ψ() =   () be two complex potentials.Then the stress (  ,   ,   ), the resultant function (, ), and the displacement (, V) are related to Φ() and Ψ() as [12]   +   = 4 Re Φ () , where  is shear modulus of elasticity,  = 3 − V for plane strain, and  = (3 − V)/(1 + V) for plane stress; V is Poisson's ratio and  =  + .The derivative in a specified direction (DISD) is defined as where  denotes the normal and tangential tractions along the segment ,  + .Note that the value of  depends not only on the position of point , but also on the direction of the segment / [5].
The complex potential in plane elasticity is obtained by placing two point dislocations with intensities  and − at points  =  and  =  + , yielding Making substitutions  and  by −()/2 and −()/2 in (6) and performing integration on the right side of (6) give where  denotes the crack configuration.Substituting ( 7) into (4) and letting  approach  + 0 and  − 0 , which are located on the upper and lower sides of the crack faces, then using the Plemelj equations, and rewriting  0 as , the following result is obtained [5]: where (() + V()) = (() + V()) + − (() + V()) − denotes the crack opening displacement (COD) for both cracks.It is well known that the COD possesses the following properties:

Hypersingular Integral Equation
The hypersingular integral equation for an inclined or a curved crack problem is obtained by placing two point dislocations at points  =  and  =  + .It is given by [5] where and () is the dislocation distribution along the curved crack.In (10), the first integral with h.p. denotes the hypersingular integral and it must be interpreted in Hadamart sense [8].Now consider the interaction between inclined and curved cracks problem (see Figure 1).For the crack-1, if the point dislocation is placed at points  =  10 and  =  10 ,  1 ( 1 ) is the dislocation doublet distribution along crack-1, and the traction is applied on the  10 , then the hypersingular integral equation for crack-1 is where  11 ( 10 ) +  11 ( 10 ) denotes the traction influence on crack-1 caused by dislocation doublet distribution,  1 ( 1 ), on crack-1 and The influence from the dislocation doublet distribution on crack-2 gives where  12 ( 10 ) +  12 ( 10 ) denotes the traction influence on crack-1 caused by dislocation doublet distribution,  2 ( 2 ), on crack-2 and Note that since  2 −  10 ̸ = 0, all three integrals in ( 14) are regular and note that  1 ( 1 ) and  2 ( 2 ) satisfy (9).By superposition of the dislocation doublet distribution,  1 ( 1 ), along crack-1 (12) and the dislocation doublet distribution,  2 ( 2 ), along crack-2 (14), we obtained the following hypersingular integral equation for crack-1 which is as follows: where  1 ( 10 ) +  1 ( 10 ) =  11 ( 10 ) +  12 ( 10 ) + ( 11 ( 10 ) +  12 ( 10 )) is the traction applied at point  10 of crack-1, which is derived from the boundary condition.The first three integrals in (16) represent the effect on crack-1 caused by the dislocation on crack-1 itself, whereas the second three integrals represent the effect of the dislocations on crack-2.
Similarly, the hypersingular integral equation for crack-2 is where In (17), the first three integrals represent the effect on crack-2 caused by the dislocation on crack-2 itself, and the second three integrals represent the effect of the dislocation on crack-1.Equations ( 16) and (17) are to be solved for  1 ( 1 ) and  2 ( 2 ).

Stress Intensity Factor
The stress intensity factor (SIF) for the two cracks can be calculated, respectively, as follows: where   1 () and   2 () are obtained by solving ( 19) and (20), simultaneously.
In order to show that the suggested method can be used for solving more complicated curved cracks problems, several numerical examples are presented.For verification purposes, we observe that if the two cracks are far apart, we have | 2 − 10 | and | 1 −  20 | approach infinity.These lead to the second three integrals vanish in ( 16) and (17).Then ( 16) and (17) become an equation for an inclined and a curved crack, respectively.For the curved crack with the length 2, we compare the result with the exact solution with the remote traction  ∞  =  ∞  = 1, given by Cotterell and Rice [2]: where  is the tangent angle at the direction of crack tip.   Figure 5: Nondimensional SIF for the interaction between an inclined and a curved crack subject to tearing loading (Mode III) (see Figure 2(c)).The numerical results are tabulated in Table 1.It can be seen that maximum error is less than 1.0%.

Example 1: Mode I.
Consider an inclined crack in upper position of a curved crack (Figure 2(a)); the traction applied is  ∞  =  1 and the calculated results for SIF at the crack tips  1 ,  2 ,  1 , and  2 are, respectively, expressed as    varies within the range 0 ∘ ≤  ≤ 90 ∘ , the values of  1 1 ,  2 1 ,  1 2 , and  2 2 are varied significantly due to shielding effect.Whereas Figure 3(b) shows the nondimensional SIF for a curved crack when  is changing for  = 45 ∘ , the values of  1 1 ,  2 1 ,  1 2 , and  2 2 are varied significantly for the considered domain.
The effect of the distance between both cracks, /, is also studied by taking  = 45 ∘ and the results are shown in Figures 3(c) and 3(d) for  = 90 ∘ and  = 45 ∘ , respectively.As the two cracks are close together, the nondimensional SIF at the crack tip becomes higher.Angle,  (deg)    The effect of the distance between both cracks, /, is also studied by taking  = 45 ∘ and the results are shown in Figures 4(c) and 4(d) for  = 90 ∘ and  = 45 ∘ , respectively.As the two cracks are close together, the nondimensional SIF at the crack tip becomes higher.The effect of the distance between both cracks, /, is also studied by taking  = 45 ∘ and the results are shown in Figures 5(c) and 5(d) for  = 90 ∘ and  = 45 ∘ , respectively.As the two cracks are close together, the nondimensional SIF at the crack tip becomes higher.show the interaction of both cracks by evaluating the nondimensional SIF at the crack tips  1 and  2 when / = 1.5, 2.0, 2.5 for  1 and  2 , respectively.As the / decreases, the nondimensional SIF becomes higher.

Conclusion
In this paper, the different types of loading modes have been applied to the inclined and curved cracks in plane elasticity.We obtained different results of nondimensional SIF due to the different loading modes.We also observed that the SIF increases as both cracks become closer.

Figure 2 :
Figure 2: (a) An inclined crack in upper position of a curved crack (Mode I).(b) An inclined crack in upper position of a curved crack (Mode II).(c) An inclined crack in upper position of a curved crack (Mode III).(d) An inclined crack is located below the curved crack (Mix Mode).(e) An inclined crack is located on the right position of the curved crack (Mix Mode).

Figure 3 :
Figure 3: Nondimensional SIF for the interaction between an inclined and a curved crack subject to normal loading (Mode I) (see Figure 2(a)).

Figure 4 :
Figure 4: Nondimensional SIF for the interaction between an inclined and a curved crack subject to shear loading (Mode II) (see Figure 2(b)).

Figure 3 (
Figure 3(a)  shows the nondimensional SIF for an inclined crack when  is changing for  = 45 ∘ .It can be seen that as

5 )Figure 6 :
Figure 6: Nondimensional SIF at the crack tips when  is changing subject to mix loading (see Figure 2(d)).

4. 2 .
Example 2: Mode II.Consider the problem in Figure 2(b); the traction applied is  ∞  =  2 and the calculated results for SIF at the crack tips  1 ,

Figure 7 :
Figure 7: Nondimensional SIF at the crack tips when  is changing subject to mix loading (see Figure 2(e)).

4. 3 .
Example 3: Mode III.Consider the problem in Figure 2(c); the traction applied is  ∞  =  and the calculated results for SIF at the crack tips  1 ,

Figures 7
Figures 7(a) and 7(b) show the interaction of both cracks by evaluating the nondimensional SIF at the crack tips  1 and  2 when / = 1.5, 2.0, 2.5 for  1 and  2 , respectively, whereas Figures7(c) and 7(d) show the interaction of both cracks by evaluating the nondimensional SIF at the crack tips  1 and  2 when / = 1.5, 2.0, 2.5 for  1 and  2 , respectively.As the / decreases, the nondimensional SIF becomes higher.

Table 1 :
The SIF for single curved crack: a comparison between exact and numerical results.