Issue 19

K. Pham et alii, Frattura ed Integrità Strutturale, 19 (2012) 5-19; DOI: 10.3221/IGF-ESIS.19.01 11 0 0 2 (0) := = (0) (0)     w E S (20) we deduce from (14), that 0 0     t . Indeed, 0   t by virtue of (5) and (11). Then, integrating (14) over (0, ) L and using the boundary conditions (0) = ( ) = 0     t t L , we obtain 2 0 0 ( ( )) 2 ( ( ))         L L t t t S x dx w x dx (21) But, since /   w S is a decreasing function of  by virtue of Hypothesis 1 and since 0   t by virtue of the irreversibility condition, we have 2 0 2 ( ( )) ( ( )), (0, )         t t w x S x x L Integrating over (0, ) L and inserting the result into (21) gives 2 2 0    t . Therefore 0  is the maximal stress that the material can sustain. The point of departure in the construction of localized damage solutions is to seek for solutions for which the equality in (14) holds only in some parts of the bar. For a given 0 >  t , the localized damage field will be characterized by its set =  i t t i   of localization zones i t  where i t  is an open interval of [0, ] L of the form ( , )   i t i t x D x D (the independence of its length on i will be proved). In [0, ] \ t L  , the material is supposed to be sound and therefore these parts will correspond to elastic zones where = 0  t . By sake of simplicity, we will not consider localization zones centered at the boundary of (0, ) L , i.e. with = 0 i x or = i x L . Therefore, we force the damage to vanish at = 0 x and = x L . The successive steps of the construction are as follows: 1. For a given t , assuming that  t is known, we determine the profile of the damage field in a localization zone; 2. For a given t , we obtain the relation between  t and t U ; 3. We check the irreversibility condition. Damage profile in a localization zone Since t is fixed, we omit the index t in all quantities which are time-dependent. Let 0 (0, )    be the supposed known stress and = ( , )   i i i x D x D  be a putative localization zone. The damage field  must satisfy 2 2 0 ( ) 2 ( ) 2 = 0 .            i S w E in  (22) Since we assume by construction that the localization zone is matched to an elastic zone and since  and   must be continuous, see Remark 1, the damage field has also to satisfy the boundary conditions ( ) = ( ) = 0.      i i x D x D (23) Multiplying (22) by   and integrating with respect to x , we obtain the first integral 2 2 2 0 ( ) 2 ( ) =          i S w E C in  (24) where C is a constant. Using (23) and Hypothesis 1, we get 2 0 = /   C E and (24) can read as 2 2 ( ) = ( , ( ))      i x H x in  (25) with 2 0 0 1 ( , ) := 2 ( ) ( ) [0,1)                E H w S for E (26) Since 2 0 ( , ) = 2 ( ) ( )            HE w S and since, by virtue of Hypothesis 1, ( ) > 0   w and 2 ( ) 1 2 ( )       S w decreases from 2 2 0 1 / > 0    to  when  grows from 0 to 1, H is first increasing from 0, then decreasing to  . Hence, there exists a unique positive value of  , say ( )   , where H vanishes: