Issue 36

A. Namdar et alii, Frattura ed Integrità Strutturale, 36 (2016) 168-181; DOI: 10.3221/IGF-ESIS.36.17 169 ATENA finite element (FE) as software used for concrete crack simulation, and the fracture energy values calculated [4]. There are several studies on predict crack growth in concrete [5-7]. A link between the averaged strain energy density (SED) approach and the peak stress method in the case of cracks subjected to mixed mode (I+II) loading, has been investigated [8]. There are several software such as ABAQUS, ADAPCRACK3D, ANSYS and ATENA finite element (FE) used for crack modeling and design of different materials [3-7, 9-14]. In this paper, the ABAQUS has been used to characterize flexural failure mechanism of beam subjected to the cracking moment and to understand stress transfer in plain concrete during crack development. A numerical analysis has been made in order to realize fracture resistance and toughness mechanism of plain concrete beam in crack extension, considering displacement, von Mises, load reaction, displacement-crack length, von Mises-crack length and von Mises-displacement of beams. T HEORETICAL CONCEPT inear Elastic Fracture Mechanics (LEFM) considers the fundamentals of linear elasticity theory. The flexural crack is simulated at the center of the beam. It initiates from bottom and it terminates at the top. The plain concrete is an isotropic material. It exhibits the same behavior in all directions, while subjected to the load. In isotropic linear elastic material, the stress and strain, in three dimensional case are defined by    and    [15].   x y z xy yz zx                             and   x y z xy yz zx                             where matrix   C is called the elastic elastic moduli matrix then      C    .            1 0 0 0 1 0 0 0 1 0 0 0 1 2 0 0 0 0 0 2 1 1 1 2 0 0 0 0 0 2 1 2 0 0 0 0 0 2 E C                                                         For  and E in terms of K and G , then C will be;   4 2 2 0 0 0 3 3 3 2 4 2 0 0 0 3 3 3 2 2 4 0 0 0 3 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 K G K G K G K G K G K G C K G K G K G G G G                                     L