13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- where N and S are the normal and shear forces in the beam; T and M are the twisting and bending moments; Nf, Sf, Tf, and Mf are critical values. N is positive for tension and negative for compression. S and M are obtained from the values in the two directions normal to the beam axis using the square root of squares rule. Eq. (2) provides an interaction between the different forces that allows for failure when Π ≥ 1 under the combined action of normal and shear stresses [18, 14]. Taking only the first and fourth term was previously used in criteria with no account for shear, e.g. [19]. The second and third term allow for shear failure similarly to [10]. The failure parameters Nf, Sf, Tf, and Mf can be related [18]. For a beam of circular cross section of radius R, the tensile failure stress is σf = Nf / ( π R 2). The maximum bending stress is σ max = 4M / ( π R 3), which equals σf when Mf = Nf R / 4. Similarly, the shear failure stress is τf = Sf / ( π R 2). The maximum torsion stress is τ max = 2T / ( π R 3), which equals τf when Tf = Sf R / 2. Thus Π requires two material parameters: σf and τf. Noting that for quasi-brittle materials typically 1 ≤ τf / σf ≤ 2 [18], in this work τf = 2 σf is used, representing more brittle materials. The tensile failure strength of a bond, σf, is related to the size of the pore assigned to the bond. The relation used here is simpler than in the previous work [14] and based on the assumption that σf is the beam remote stress for which the average stress in the beam ligament outside the pore attains a critical value σ0. Thus ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − 2 0 1 R c f σ σ , (3) where c and R are the pore and beam radii, respectively, and σ0 can be interpreted as the tensile strength of the material without a defect. With this setup and the choice τf = 2 σf the failure model requires a calibration of a single parameter, σ0, against experimental stress-strain curve. However, since the beams behaviour is linear elastic, the choice of σ0 would affect only the calculated macroscopic stresses but not the order in which damage (beam failures) would evolve in the system. Because the interest here is investigating the evolution of damage, σ0 = 1 MPa is used for the calculations, noting that macroscopic stress response can be simply scaled by another value of σ0. 2.3. Load cases and solution A model of size (20L, 20L, 20L) was used. The lattice contained 17261 sites and 113260 bonds: 49260 B1 and 64000 B2. The coordinate system (X1, X2, X3) was coincident with B1, so that the boundary planes X1 = 0, X1 = 20L, X2 = 0, X2 = 20L, X3 = 0, X3 = 20L contained 21×21 sites (nodes). Boundary conditions normal to each plane were only applied. Thus Ui and Fi denote displacements and forces of nodes on plane with normal Xi, while other displacements and rotations on this plane were unconstrained. Table 1 shows the conditions on planes X1 = 20L, X2 = 20L, and X3 = 20L for the analysed cases. Additionally, U1 = 0 on X1 = 0; U2 = 0 on X2 = 0; U3 = 0 on X3 = 0, apply to all. Table 1. Boundary conditions for loading cases. Values given in bold denote applied conditions. A stands for values obtained from finite element analyses. Case U1 U2 U3 F1 F2 F3 Note C1 d1 A A A 0 0 Uniaxial unconfined extension C2 d2 d2 A A A 0 Plane stress C3 d3 d3 0 A A A Plane strain C4 d4 0 0 A A A Uniaxial confined extension For cases where nodal reaction forces were determined from analysis, the macroscopic stress in the
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