Stress intensity factors for cusp-type crack problem under mechanical and thermal loading

The general solutions of the stress intensity factors (SIFs) for a cusp-type crack problem under remote uniform mechanical and thermal loads are presented in this work. According to the complex variable theory and the method of conformal mapping, a symmetric airfoil crack is mapped onto a unit circle, and both the temperature and stress potentials are used to solve the relevant boundary-value problems. By introducing the auxiliary function and applying the analytical continuation theorem, the SIFs at the cusp-type crack tip can be analytically determined. The obtained SIF results are dependent on the geometric configurations of the cusp-type crack components and the magnitudes of the mechanical and thermal loads. For some combinations of combined loads, the SIF is maximized, and the system has a high risk of damage.


INTRODUCTION
Nearly a century ago, Inglis [1] obtained the solution of a cracked plate using elliptic coordinates. Since then, numerous related crack problems with different loading conditions and crack configurations have been investigated by many researchers. Muskhelishvili [2] obtained a general solution of the twodimensional (2D) elasticity crack problems using the analytical continuation theorem, which is the most powerful method for solving mixed boundary-value problems. For 2D linear elasticity problems with awkwardly shaped geometries, one of the most useful techniques is to transform the region into one with a simple shape. On the basis of conformal mapping combined with the Muskhelishvili formulation for plane elasticity, many numerical solutions and techniques for the crack problem [3][4][5][6][7][8][9][10] have been developed. For example, the solution for an elliptic hole embedded in an infinite plate can be easily obtained via the mapping z = c cosh ζ . A simple illustration of conformal mapping combined with analytical continuation is the classical problem of an infinite plate containing an internal crack under simple tension. For this problem, the mapping function is selected as z = m(ζ ) = (a/2)(ζ + ζ −1 ), which maps the unit circle and its exterior region in the ζ -plane onto a crack of length 2a and its exterior, respectively, in the z-plane. When the continuation theorem is employed, the singular points of the auxiliary function ω(1/ζ ) = [m(ζ )/m (1/ζ )]φ (1/ζ ) + ψ (1/ζ ), which cause the continuity equation to be divergent, must be deduced such that the continuity equation can be convergent. Therefore, the structure of the proposed solution is mainly dependent on the choice of the mapping function. For example, the mapping function m(ζ ) = (a/2)(ζ + ζ −1 ) maps a line crack of length 2a onto a unit circle that contains two singular points: ζ 1 = 1 and ζ 2 = −1. Consequently, the derivation procedure is far simpler. For the mapping function m(ζ ) = Rζ + (1/Rζ ) associated with an elliptic hole [4], there exist two singular points ζ 1 = (R) −1/2 andζ 2 = −(R) −1/2 . Regarding the problem associated with a polygonal hole [5][6][7][8], the mapping function m(ζ ) = R(ζ + w/ζ n ) with 0 ≤ w < 1/n contains n + 1 singular points. For example, a coated square hole problem [7] with n = 3 contains four singular points ζ 1 = (3w) −1/4 , ζ 2 = −(3w) −1/4 , ζ 3 = (3w) −1/4 i and ζ 4 = −(3w) −1/4 i, which should be removed from the continuity equation. Interestingly, w = 1/n in the foregoing mapping function presents a hypocycloid hole with many cusps, and a stress singularity occurs at the cusps of a hypocycloid hole. Determination of the stress intensity factors (SIFs) of a hypocycloid with many cusps in an infinite plate is important and has been carried out by many researchers [9][10][11][12][13][14][15][16][17][18][19]. The SIFs for both mode I and mode II at the cusp points vanish when n approaches infinity. Owing to the difference in the mapping function, the formulation of the temperature potentials and stress functions becomes even more complex, making the whole problem extremely difficult to solve.
The present study focuses on a symmetric airfoil crack in an infinite plate under combined mechanical and thermal loads. Using Muskhelishvili's complex potential method, the exact solutions of both mode-I and mode-II SIFs are presented herein. The SIF value is dependent on the magnitudes of the mechanical and thermal loads, as well as the geometric configuration of the symmetric airfoil crack. For a given geometric configuration, the SIF is maximized under certain combinations of combined loads, which represents the most dangerous situation. A similar problem associated with different shapes of cracks, such as a symmetric lip cusp crack, can also be treated via the proposed method.

PROBLE M FOR MUL ATION
Consider a symmetric airfoil crack in an infinite plate under the remote mechanical loading denoted by σ ∞ x , σ ∞ y and σ ∞ xy and a remote uniform heat flow q forming an angle λ with the positive x-axis, as shown in Fig. 1. According to the complex variable theory for 2D plane thermoelasticity, the component of the displacement u x + iu y and the resultant forces -F y + iF x can be described by two complex functions φ(z) and ψ(z) and a temperature potential θ (z) = g (z), each of which is analytic in its argument z = x + iy, as follows: Here, G represents the shear modulus, κ = 3 − 4ν and β = (1 + ν)α represent the plane strain deformation, and κ = (3 − ν)/(1 + ν) and β = α represent plane stress deformation, with ν and α being the Poisson's ratio and the coefficient of thermal expansion, respectively. Here, the prime symbol represents the derivative with respect to z = x + iy, and the function (¯) represents the complex conjugate.
To solve the relevant boundary-value problem, the conformal mapping function is introduced as follows: which maps the unit circle and its exterior region in the ζ -plane onto the z-plane of an airfoil crack and its exterior region. Using Eq. (3), Eqs (1) and (2) can be replaced with the following: To make the resulting boundary conditions as simple as possible, we introduce the following auxiliary stress function ω(ζ ):

TE MPER ATURE FIELD
For a steady-state heat conduction problem, the temperature function satisfies the Laplace equation. The resultant heat flow h and temperature T are related to a complex potential θ (ζ ) = g (ζ ), as follows: where Re and Im represent the real and imaginary parts in Eqs (7) and (8), respectively. The quantities q x and q y in Eq. (8) represent the components of heat flux in the x-and y-directions, respectively, and k is the heat conductivity coefficient. The temperature function for a symmetric airfoil crack in an infinite plate under a remote uniform heat flow can be expressed as follows: where θ 0 (ζ ) represents the unperturbed temperature field in a homogeneous plate induced by a remote uniform heat flow and θ 1 (ζ ) represents the perturbed temperature field due to the presence of a symmetric airfoil crack. To solve the relevant boundary-value problem, the unperturbed temperature field is now decomposed into two parts, as follows: where Note that θ 0a (ζ ) is holomorphic in the range of |ζ | < 1, while θ 0b (ζ ) is holomorphic in range of |ζ | > 1. Because a symmetric airfoil crack is assumed to be insulated from the heat flow, we have Substituting Eqs (9) and (10) into Eq. (11) yields According to the analytical continuation theorem, Eq. (12) allows us to obtain Substituting Eqs. (10) and (13) into Eq. (9) yields By integrating Eq. (14) with respect to z, we can determine the temperature potential g(ζ ), as follows:
The constants A and B in Eqs (16) and (17), respectively, can be determined from the single-value condition of displacement and traction force, as follows: Here, [f(ζ )] c = f(r, 2π) − f(r, 0) represents a jump when enclosing a contour. Substituting Eqs (16) and (17) into Eqs (18) and (19) and knowing that we have and Solving for Eqs (21) and (22), we have Hence, the stress functions become From the definition of the mapping function, we havē Note that the second term on the right-hand side of Eq. (25) is the singular part for w(1/ζ ) defined in the region |ζ | > 1.
To determine the SIF at tip A, the following definition is used: In view of the mapping function, the crack length is determined as a = 2 by setting m = 0, and the SIFs at tip A defined in Eq. (35) can be replaced with the following: where For a line crack with length 2a embedded in an infinite plate subjected to a remote uniform heat flow, the SIFs at tip A are given as follows: which are in agreement with the results of Sih [21].
For an isothermal case, the expression for the SIFs at tip A is reduced to which is identical to the result of Chen [16].

NUMERICAL E X A MPLES
The SIFs for a symmetric airfoil crack are dependent on the loading conditions and geometric configurations through the function ϕ (1) defined by Eq. (37). The SIFs at tip A for a simple example of a line crack were obtained from our results by setting m = 0 in Eq. (39). In the following examples, we aim to determine the SIFs for several loading conditions and different geometric configurations. The plots of different geometric configurations for a symmetric airfoil crack with different values of m are shown in Fig. 2.

CONCLUDING RE MARKS
An exact solution of the SIFs for a symmetric airfoil crack under combined mechanical and thermal loads is presented. The solution is based on the method of analytical continuation in conjunction with conformal mapping. The obtained results indicate that when the cusp-type crack components are subjected to a tension load σ 0 along the y-axis and a remote heat flow q approaches from the negative x-axis (λ = 0 0 ), the K 1 value at tip A always becomes positive regardless of the magnitude of the tension loading σ 0 and the strength of the heat flow q, which places the components in a dangerous situation. However, when the cusp-type crack components are subjected to a tension loading σ 0 along the y-axis and a remote heat flow q approaches from the positive x-axis (λ = 180 0 ), crack propagation of the components is suppressed if the strength of the heat flux is sufficient that Gβqa/[(1 + κ)kσ 0 ] > (m 4 + 2m 3 + 4m + 2)/[2(1 + 2m − m 2 )(2m − m 2 )].