13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- solved by the concerned method. 4. Conservation integral Based on the two-dimensional conservation law, for a closed integration path without any crack and cavity in it, the following integrals will vanish [5-8]. ( ) ∫ − = s i i j j j wn Tu ds J , . (11) Eq.(11) has two components, 1J and 2J . 1J is so-called J-integral. They can all be used to calculate the stress intensity factors for the cracked elastic bodies. In the following section some key steps to estimate the SIF for the indentation have been developed based on the 2J -integral. 5. Stress intensity factors for periodic indentations As the definition of the half plane substrate, i.e., < < <+∞ x h 2 0 , −∞< <+∞ 1x , any plane , , .... , 0 1 2 1 = = x kt k is the symmetrical plane. Then, the Cell 1 shown in Fig.1 is considered for the calculation of the stress intensity factors for periodic indentations. Note that there are two singular stress fields next to the two corners of every indenter for all cells. Select a closed integration path a abcdefghij s as shown in Fig.3 for Cell 1. From the Eq.(11), following result can be given[5-8]. ( ) 0 2 2 2 = − = ∫ a abcdefghij s i i wn Tu ds J , . (12) For the paths bc s and de s , because the symmetry, 0 2 = T , 0 2 = n and 0 1 2 = , u , and the following result can be found. ( ) ( ) 0 2 2 2 2 2 = − = − = ∫ ∫ de bc s i i s i i wn Tu ds wn Tu ds J , , . (13) Let the gh s and ij s be straight lines; and fg s and ja s a quarter of a circle. If the ija s and fgh s are within the K-dominant regions, we have [11-13] ( ) ( ) 0 2 2 2 2 2 = − = − = ∫ ∫ ij gh s i i s i i wn Tu ds wn Tu ds J , , (14) and ( ) ( ) ( ) E K wn Tu ds wn Tu ds J I ind s i i s i i ja fg π μ 2 1 2 2 2 2 2 2 2 − − = − = − = ∫ ∫ , , (plane strain). (15) When the cross sectional area where the cd s located is relatively far from the contact zone, the following expression can be given by ( ) − − = + − = ∫ 2 2 2 2 2 , , ~ wn Tu ds w Nu J cd s i i , (16) where the iu ~ shows the displacement of neutral axis. This displacement can be given by elementary strength theory of materials [11-13]. w is the strain energy density per unit length of the cell. For the hi s , we have ( ) ( ) + + = − + − = ∫ 2 2 2 2 2 , , ~ wn Tu ds w Nu J hi s i i . (17) Then, substituting Eqs.(13)-(17) to Eq.(12), it gives ( ) ( ) ( ) ( ) + − + + − − − − = − + = + + − = ∫ 2 2 2 2 2 2 2 2 2 2 2 2 2 1 , , , , ~ ~ ~ ~ u u N wds w Nu w Nu E K J ab s I ind π μ , (18) which can be interpreted as the energy release per unit moving of boundary iab s and efh s in
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