Figure 4. Pin-loading on the squeeze surface Figure 5. Boundary conditions 2.3. Results analysis Figure 6 and Figure7 respectively show the Von Mises stress distribution around the lug’s hole and the crack tip. Figure 6. Stress distribution around the lug’s hole Figure 7. Stress distribution around the crack tip SIFs of cracks with different lengths are calculated by three methods. Case 1 uses the equation that is used for estimating the straight lugs’ critical load in [15] (see formula (4)) to calculate the SIFs. Case 2 uses the configuration factor J41L in [15] (see Figure 8) to get the SIFs’ values. This paper calculates the SIFs by finite element method, and the results are shown as case 3 in Figure 9 to compare with the other two cases. In reference [15] (page 270), it is indicated that the critical load [ P]c of the attachment lug subjected to axial pin-load is a tK R P C C π β 2 2 [ ] = . (4) Where fw D W β β β β ⎟ ⋅ ⎠ ⎞ ⎜ ⎝ ⎛ + = 2 1 2 1 1.971 2 , (5) 1.33734 1 1 0.3857 0.31711 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = R a β , (6) 0.6762 0.3246 0.8734 1 2 + + = R a β , (7) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⋅ = a R a R fw 2 1 2 2 2 sec π β . (8) In reference [15] (page 139), the SIF of attachment lug subjected to axial pin-load is equal to a J K L π σ⋅ ⋅ = 41 . (9) where the configuration factor J41L can be determined by using Figure 8.
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