and the effect of geometric parameters on SIF is calculated and analyzed by the finite element method (FEM) with linear elastic assumptions, as in essence, the fatigue crack growth is brittle and it is often based on Linear Elastic Fracture Mechanics (LEFM) assumption[13]. The fatigue crack growth test and residual strength test are carried out by the material test system MTS-810, the test data validate the finite element analysis and are used to determine the constants C&m of the Paris Law. The fatigue crack growth model of the typical straight lugs is established, offering an analytical as well as experimental method for assessing and designing damage tolerant attachment lugs. 2. Finite Element lement Modeling and Analyzing nalyzing 2.1. Finite element modeling and meshing Figure 1 shows the size of the straight lug discussed in this paper, millimeter is used as the length unit. The length of the single through-the-thickness crack is a. The model is meshed with PLANE 183 triangular element, the real constant for “plane stress with thickness” is set to 6.7. The elastic modulus is 200Gpa and the Poisson ratio is 0.3. The node at the crack tip is defined as a singular point and there are 12 triangular elements around it. The radius of the 1st row of elements is a/10. The global elements edge length is set to 2. Figure 2 & Figure 3 show the mesh result. R10 R22 .5 a Figure 1. Size of the straight lug Figure 2. Mesh of the straight lug Figure 3. Mesh of the crack tip 2.2. Setting boundary conditions and applying cosine distributed load Pin load is transferred by squeezing between the lug and the pin matched. Load and stress are supposed to keep unchangeable along the thickness of the hole. The pressure distributed on the squeeze surface follows the cosine distribution law, and the radial load Pi at the point ion the squeeze surface is equal to i P γ cos 0 [14], as shown in Figure 4. The x and y axis components of the load Pi are as follows. i i xi P P γ sin = , (1) i i y i P P γ cos = . (2) The resultant force along y axis is shown as follow. i i i y i P P P P γ γ 2 0 cos cos ∑ ∑ ∑= = = . (3) The cosine distributed nodal force is applied on the nodes on the surface ABC. All the nodes on the symmetry plane of the lug are selected and the displacement values are set to zero. Figure 5 shows the boundary conditions. P B A C iγ ixP x O y iP iyP 0P
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