Issue 41

V. Rizov, Frattura ed Integrità Strutturale, 41 (2017) 491-503; DOI: 10.3221/IGF-ESIS.41.61 498 where Γ is a contour of integration going from the lower crack face to the upper crack face in the counter clockwise direction, 0 u is the strain energy density, α is the angle between the outwards normal vector to the contour of integration and the crack direction, x p and y p are the components of stress vector, u and v are the components of displacement vector with respect to the crack tip coordinate system xy ( x is directed along the crack), ds is a differential element along the contour. The J -integral is solved by using an integration contour, Γ , that coincides with the beam contour (Fig. 1). It is obvious that the J -integral value is non-zero only in segments 1  and 2  of the integration contour. Therefore, the J -integral solution is written as 1 2 J J J     (25) where 1 J  and 2 J  are the J -integral values in segments 1  and 2  , respectively ( 1  and 2  coincide with the free end of lower crack arm and the clamping, respectively). The components of J -integral in segment, 1  , of the integration contour are written as (Fig. 3) x p    m B    , 0 y p  (26) 1 ds dz  , cos 1    (27) where 1 z varies in the interval 1 1 [ / 2, / 2] h h  . Partial derivative, / u x   , in (24) is found as   1 1 1 1 n u z z x        (28) where 1 1 n z and 1  are obtained from Eqs. (18) and (19). The strain energy density is calculated by substituting of (1) in (10) 1 0 1 m B u m     (29) By substituting of (1), (12), (16), (26), (27), (28) and (29) in (24) and integrating in boundaries from 1 / 2 h  to 1 / 2 h , one derives         2 2 1 1 1 0 1 1 2 1 1 2 2 2 1 4 1 1 1 1 m m m u u u u u B y B B r m J m m b m h m                                1 2 2 1 1 2 1 1 2 1 2 2 2 1 2 1 n m m m m u u u u u u z B r m m h                      3 3 2 2 1 2 1 1 2 1 2 2 2 1 3 2 f q f q f q f q u u u u u u u u n q q q q u u u u u u u u u z B q f q f q h                                     (30) 2 1 1 1 2 f q f q u u u u n q q u u u z f q                    