Issue 19

K. Pham et alii, Frattura ed Integrità Strutturale, 19 (2012) 5-19; DOI: 10.3221/IGF-ESIS.19.01 9 2 0 ( ) = ( ( )) . 2    L t e t t S x dx  (18) By virtue of the conditions (13)--(15) that the fields have to satisfy, the following balance of energy holds true at each time: ( ) = ( ) ( ) e e d t t t     Proof. By virtue of the equilibrium condition and the definition of the elastic energy, the work done by the external load can read as   0 0 0 ( ) = ( ( )) ( ( )) ( )              t L L e s s s s s s t S x dx S x x dx ds  2 0 0 0 = ( ) ( ( )) ( ) 2           t t L s e s s s ds S x x dxds  Using the initial condition and the consistency condition in the bulk, one gets 2 0 0 0 0 0 ( ) = ( ) ( )                t L t L e e s s s s t t w dxds E dxds   2 2 0 0 0 0 0 0 = ( ) ( ) ' ( ( ) ( ) (0) (0))                        L t L t e t s s s s s s t w dx E dxds E L L ds  Using once more the initial condition and the consistency condition at the boundary, we obtain the desired equality. The homogeneous solution and the issue of uniqueness If we assume that the bar is undamaged at = 0 t , i.e. if 0 ( ) = 0  x for all x , then it is easy to check that the damage evolution problem admits the so-called homogeneous solution where  t depends on t but not on x . Let us construct this particular solution in the case where the prescribed displacement is monotonically increasing, i.e. when = t U tL . From (11), we get = ( )   t t E t . Inserting this relation into (14) and (15) leads to 2 2 ( ) ( ) , = 0 2 ( ) 2 ( )                    t t t t t w w t t E E (19) Since 0 = 0  ,  t remains equal to 0 as long as 0 = 2 (0) / (0)      t w E . That corresponds to the elastic phase. For 0 >  t , since /    w E is increasing by virtue of Hypothesis 1, the first relation of (19) must be an equality. Therefore  t is given by 1 2 = 2                 t w t E and grows from 0 to 1 when t grows from 0  to  . During this damaging phase, the stress  t is given by 2 ( ) = ( )      t t t w S Since /   w S is decreasing to 0 by virtue of Hypothesis 1,  t decreases to 0 when t grows from 0  to  . This last property corresponds to the softening character of the damage model. Note that  t tends only asymptotically to 0, what means that an infinite displacement is necessary to break the bar in the case of a homogeneous response. In terms of energy, the dissipated energy during the damage process is given by ( ) = ( )  d t t w L  Hence, it is proportional to the length of the bar. The total energy spent to obtain a full damaged state is equal to (1) w L and hence is finite by virtue of Hypothesis 1.