Issue 33

Y. Wang et alii, Frattura ed Integrità Strutturale, 33 (2015) 345-356; DOI: 10.3221/IGF-ESIS.33.38 347 where x  , y  and z  are the three normal strains, and xy  , xz  and yz  are the total shear strains. The orientation of a generic material plane,  , having normal unit vector n can be defined through angles  and  (Fig. 1b).  is the angle between axis x and the projection of unit vector n on plane x-y .  is the angle between n and axis z . A new system of coordinates, Onab, can now be defined. The unit vectors defining the orientation of axes n, a and b can be expressed as follows:           sin cos sin sin cos x y z n n n                            n ;     sin cos 0 x y z a a a                          a ;           cos cos cos sin sin x y z b b b                             b (2) Figure 1 : Generic plane and shear strain resolved along one generic direction in a body subjected to an external system of forces. Consider now a generic direction q lying on plane  and passing through point O . α is the angle between direction q and axis a. The unit vector defining the orientation of q can be calculated as follows:                         cos sin sin cos cos cos cos sin cos sin sin sin x y z q q q                                       q (3) According to the definition reported above, the instantaneous value of the shear strain resolved along direction q , q ( ) t  , can then be calculated as: x xy xz q y xy y yz xz yz z 1 1 ε (t ) (t ) (t ) 2 2 γ ( ) 1 1 γ (t ) ε (t ) γ (t ) 2 2 2 1 1 γ (t ) γ (t ) ε (t ) 2 2 x x z y z n t q q q n n                                    (4) In order to make the calculation easier, it is useful to express q ( ) t  through the following scalar product:   q γ ( ) 2 t t   d s (5) where d is the vector of direction cosines, that is: