Issue 19

K. Pham et alii, Frattura ed Integrità Strutturale, 19 (2012) 5-19; DOI: 10.3221/IGF-ESIS.19.01 17 5/2 2 /2 1 0 0 1 2 3/2 0 0 < 2 2 ( ) ( ) = , ( ) = 2 = 2 lim (( ) ) > 2                     d d p if q d p q if q d p q q p if q (39) Consequently, when 2  q , the overall strain remains finite when the stress goes to 0, contrary to the homogeneous response where the strain becomes infinite. Checking of the irreversibility It remains to check that the localized damage fields that we have constructed at different values of  leads to an evolution in time which satisfies the irreversibility condition 0    . Let us reintroduce the time and the index t in the notation. Since the center of the localization zone is fixed, the condition of irreversibility is satisfied only if ( ) = ( ) t t i t x     is not decreasing. Since ( )     is decreasing, it is possible only if t t   is not increasing. Since ( , ( )) = 0 t i i t x x D    and since ( ) > 0 t x  for | |< ( ) i t x x D   by construction, the condition of irreversibility is satisfied only if ( ) t t D   is not decreasing. That requires that ( ) D    is not increasing, condition which is not automatically satisfied by the damage model, see Example 4. When this condition is not satisfied, our construction of the localized solution is no more valid. We must consider an evolution of the damage where a part of the localization zone is unloaded and reenters in a non damaging phase, the size of the still damaging part decreasing with time. To avoid such a situation we make the following hypothesis Hypothesis 2 We assume that ( ) E    and ( ) w    are such that ( ) D    is decreasing. Note that this hypothesis is satisfied in the class of models of Example 2 when = 2 q and 4 p  . Under this condition, it is possible to obtain the following property Property 8 Under Hypothesis 2, in order that t t   given by (31) in a localization zone (and equal to 0 otherwise) is not decreasing it is necessary and sufficient that t t   is not increasing. Proof. We know that it is necessary, it remains to prove that it is sufficient. Let us assume that t t   is not increasing. Then ( ) t t    and ( ) t t D   are not decreasing. Let 1 2 < t t and x be such that 1 | | ( ) i t x x D    . It is sufficient to prove that 2 1 2 1 := ( ) ( ) =: t t x x      . Owing to (31), since H is a decreasing function of  and since 2 1 t t    , we have 2 1 2 2 1 0 0 1 2 1 1 0 ( ) ( ) = ( , ) ( , ) ( , ) t t t t t d d d D D H H H                          Hence 2 1    . By virtue of this last property, our construction of a non homogeneous solution is valid provided that the bar is sufficiently long so that a localization zone can appear and grow without reaching the boundary. Since the size of the localization zone increases with t , that leads to the inequality 0 2 L nD  . Owing to (29), that gives the following lower bound for L : 1 0 0 0 2 := > 2 ( ) ( ) m E L n d nL nD w       (40) Remark 4 Under Hypotheses 1 and 2, we have really obtained a damage evolution t t   which satisfies the evolution problem (13)--(15) if the bar is long enough and if we can control the loading in such a manner that the stress is continuously decreasing. But the continuity of t t   is not automatically ensured as we show in the next subsection. Size effects and the different scenarii Let us consider a loading process where = t U tL , i.e. such that the displacement of the end = x L is monotonically increasing. Consequently t corresponds to the average strain of the bar, = t  , and (36) reads now 1 0 = ( ) d t t t n E L     