Stress intensity factor of circumferential periodic cracks on tube with stiffened plates under tension

Fracture is one of the most common failure of engineering structure and the stress intensity factor is the key parameter used to decide whether the structure will crack or not. This essay is based on the concept of conservation law and the tension theory in mechanics of material. It analyzes the stress intensity factor of circumferential periodic cracks on tube with stiffened plates, it obtains the specific expression of stress intensity factor of circumferential periodic cracks on tube with stiffened plates under tension. Finite element method is used for checking the stress intensity factor of cracks and proving the accuracy of expression’s results. The method which proposed in this essay is easy to calculate and can get the closed solution.


Introduction
For the crack problems of infinite two-dimensional and three-dimensional elastic bodies, a closed analytical solution of the stress intensity factor (SIF) can be given relatively conveniently, when considering the object boundary influence on the crack, which is limited in 2D and 3D cracks, such as unilateral crack, tube with stiffened plates, to give the analytical solution which can satisfying the boundary conditions is very difficult, this must depend on numerical methods. In recent years, the J2integral theory was proposed by Xie et al [1], which is derived from the three-dimensional conservation law and the principle of virtual work. Because of the finite and infinite boundary crack problem, the J2-integral method greatly facilitates the analysis of stress intensity factors [2][3][4][5][6][7]. The tube with stiffened plates is the components which commonly used in engineering, while circumferential periodic cracks on tube with stiffened plates and circular cracks on tube are typical belong to complex three-dimensional crack structures. In this paper, A method of solving the SIFs of the circumferential periodic cracks on tube with stiffened plates and circular cracks on tube by the conservation law and the tension theory in material mechanics is established.

Configuration of circumferential periodic cracks on tube with stiffened plates
There has 2m stiffened plates which full of the periodic cracks in a circular tube. Figure 1 gives the case of m=2. For the thin-walled tube, it has the characteristics of the three-dimensional shell and the slender beam, and the crack on the stiffened plate has the characteristics of the two-dimensional stress field. So the stress intensity factor can be obtained by the conservation law and the tension theory in the material mechanics.

J2-Integral and stress intensity factors
Considering a three-dimensional strain field, for which the displacement vector ui depends on x1, x2, x3, the conservation law Jj-integral can be defined as [7]: The A in equation (1) is an arbitrary closed surface, in the solids closed by which there are no defects; w is the strain energy density; Ti is the stress vector acting on the outer side of A; n is the unit outward normal to A. For a two-dimensional deformation field. Figure 2 gives a two-dimensional crack model of unit thickness (the integral path is a curve in the x1-x2 plane). Let S represent to a closed contour Sabca within the K-dominant region around upper crack tip. Note that Sab is a straight line and Sbc is a quarter of circle. As such, the following results can be obtained by calculating the stress and displacement next to crack tip ( ) ( ) (2) Figure 2. Integration path around crack tip region. For the closed path Sabca = Sabc + Sac, from the conservation law [6]. It follows that where μ represents the Poisson's ratio and E the elastic modulus.

(5)
A + is the cracked cross-section; A -is the remote uncracked cross-section; Ain is the inner surface of shell; Aout is the outer surface of shell; Ac is the sum of the eight crack surfaces. On the free surface Ain and Aout, Ti = 0 and n2 = 0. Then the following results can be found As the fin-shaped shells possess beam characteristics, stresses and displacements in the J2-integral can be given by elementary mechanics. Then for the surfaces A + and A -, it is not difficult to get: − 2 . 2 u is the axial strain of fin-shaped shells, w is the strain energy density per unit length. All the quantities in equations (8) and (9) can be calculated by the beam theory. The axial strain of remote uncracked cross-section is where the moment of inertia A= 2πRt0+2mbt. If the crack is formed from an ellipse with d→0 as shown in Figure 3, the fin-shaped shell can be regarded as a variable cross-section beam, and the average axial strain at the cross section becomes  (13)

Stress intensity factor
Substituting the equation (10) and equation (11) into equation (13), the J2-integral over the closed surface can be expressed as Because of the free action of the crack surface with Ach just out of the K-dominant region, the integral in the left hand side of equation (14) is a small quantity and can be neglected. Hence, the normalized stress intensity factors can be found as where 0=N/A.

Comparison between numerical examples and results
This section will give a discussion on the comparison between the normalized stress intensity factor expressed by equation (15) and the results from the finite element method (EFM). Computations were used XFEM in the commercial software Abaqus [8][9], in which the C3D20 type element was employed. The elastic modulus is E=200GPa and Poisson's ratio is μ=0. 3. The length of tube with stiffened plates is 500mm, m=2, 2R=200mm, t=10mm, t0=3mm, b=60mm, Figure 5 is show the finite element model and meshes and Figure 6 is used to compare the results of equation (15) with the finite element analysis.   When the number of stiffened plates in the tube tends to be infinite (m→∞), the stiffened plates cover the inner surface of the tube, and the cracks on each stiffened plate will become circular cracks on inner surface of tube. As shown in Figure 6, 2R0 is the intermediate diameter of tube, t2 is the thickness of tube. 2πR0a is the area of the cracks on the tube and 2πRt2 is the area of remote uncracked crosssection. Hence,