13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- Using the above Eqs. (3) to (5) for all the nodes, the temperature at any mesh node (i.e., (i,j−1), (i,j+1) , (i−1,j), (i+1,j), (i,j) ) can be obtained. For a specific mesh segment, if two ends of it (i.e., two adjacent mesh nodes) have the same temperature, the temperature at its center TC is also the highest temperature Tmax in the mesh segment (i.e., TC = Tmax), which can be calculated [11] as Si,l j C = 2 Si,l j 2 � (i,j)− (i,j−1) 2 �2 + (i,j)+ (i,j−1) 2 Si,r j C = 2 Si,r j 2 � (i,j)− (i,j+1) 2 �2 + (i,j)+ (i,j+1) 2 Si,d j C = 2 Si,d j 2 � (i,j)− (i−1,j) 2 �2 + (i,j)+ (i−1,j) 2 Si,u j C = 2 Si,u j 2 � (i,j)− (i+1,j) 2 �2 + (i,j)+ (i+1,j) 2 (6) In the present case, most mesh segments have different temperatures at its ends. In view of the very small length of mesh segment of 200µm as shown later in Fig. 3, the temperature at the center of mesh segment is approximately considered as the maximum temperature in mesh segment Tmax (i.e., TC ≈ Tmax). Once the maximum temperature in a mesh segment Tmax reaches to the melting point of the material Tm, it is thought that electrical breakdown induced by Joule heating happens at this mesh segment, i.e., this mesh segment melts. 2.3 Computational Procedure Fig. 2 Flow chart of numerical simulation
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