Issue 19

K. Pham et alii, Frattura ed Integrità Strutturale, 19 (2012) 5-19; DOI: 10.3221/IGF-ESIS.19.01 14 quantities like the compliance function ( )    S and its derivatives become infinite when  goes to 1, the regularity of the damage field is lost and ( )   x is no more defined at = i x x but undergoes a jump discontinuity. So the differential system reads now as 2 0 > 0 ( ) = 0 \ , ( ) = 1, = = 0             i i i i and w E in x x on   Multiplying by   the differential equation valid on each half-zone and taking into account the boundary conditions at the ends, one still obtains a first integral 2 2 0 ( ) = 2 ( ( ))     E x w x in \ i i x  . Since > 0  in i  , denoting by 0 D the half- length of the localization zone, one necessarily has 0 0 0 0 2 ( ) ( , ) = 2 ( ) ( , )              i i i i S w in x D x S w in x x D Since ( ) =1  i x , the jump of   at i x is equal to 0 2 2 (1) /   S w . By integration we obtain the damage profile and the half-length of the localization zone 1 1 0 0 0 0 | |= , = 2 ( ) 2 ( )           i d d x x D S w S w (32) One can remark that this solution can be obtained formally by taking = 0  and (0) = 1  in (28)-(31). We have proved the following Property 5 (Rupture of the bar at the center of a localization zone) At the end of the damage process, when the stress has decreased to 0, the damage takes the critical value 1 at the center of the localized damage zone. The damage profile and the half length 0 D of the damage zone are then given by (32). The profile is still symmetric and continuously decreasing to 0 from the center to the boundary, but its slope is discontinuous at the center. Example 5 In the cases of the family of models of Example 1, the half-length of the damage zone and the amplitude of the damage profile when the bar breaks are given by 1 1 0 0 0 | |= , = 1 1            i p p c c p dv p dv x x D q q v v For =1 p , the profile is made of two symmetric arcs of parabola: 2 0 0 | | 2 ( ) = 1 , =           i c x x x D D q For = 2 p , the profile is made of two symmetric arcs of sinusoid: 0 0 | | ( ) = 1 sin , = 2 2        i c x x x D D q The greater p , the greater the size of the damage zone and the damage field, see Fig. 4. Dissipated energy in a localization zone By virtue of Property 1 and (17) the energy dissipated in an inner localization zone when the stress is  is given by ( ) 2 2 0 ( ) 1 ( ) = ( ) ( ( )) 2                  x Di d x Di E x w x dx  By symmetry, it is twice the dissipated energy in a half-zone. Using the change of variable   x and (25), we obtain 2 ( ) 0 0 2 0 0 0 4 ( ) ( ( ) ) ( ) = 2 ( ) ( ( ) )                 d w S S d S w S S S  (33)