E. Giner et alii, Frattura ed Integrità Strutturale, 35 (2016) 285-294; DOI: 10.3221/IGF-ESIS.35.33 287 will propagate in the direction that K II =0. For non-proportional loading, the condition of K II =0 cannot be reached in general, and therefore, the proposed criterion seeks the angle for which the range K II is minimized. This obviously reduces to the condition K II =0 when applied to proportional loading problems. We note in passing that, for proportional loading, the criteria of K II =0, Nuismer and MTS lead to the same result [3,4]. In practice, computing K II values under crack face contact must include the effect of friction tractions on crack faces, as in [2,5], which can be cumbersome and prone to inaccuracies when using domain and contour integrals. Instead, and equivalently, we will seek the angle for which the shear stress range at the crack tip is minimized. Shear stresses develop always in two orthogonal planes and there are two orthogonal planes on which is minimum, min( ). From these two potential crack growth directions, we choose the plane with the maximum n , because it will be the plane where less frictional energy is lost and there is more energy available for propagating the crack. This approach is in line with the principle that a crack will grow in the direction which maximizes the strain energy release rate G [3,4]. As verified in the results section, the min( ) direction coincides with the direction of the maximum range of normal stress, max( n ). This is due to the in-plane stress tensor transformation that yields both extremes in the same direction, although this may not be the general case. However, the direction predicted by the maximum range of the effective normal stress, max( n,eff ), does not lead to good results, at least in the problems studied here, despite the intuitive idea that only the positive normal stresses (effective) will govern the crack behaviour under an elastic material behaviour. Fig. 2 sketches the convention used in the procedure. When applying the proposed criterion to the propagation stage, is evaluated ahead the current crack tip. For each crack growth increment, stresses are evaluated ahead the crack tip and the prospective local direction is searched for which is minimum (see example of the estimation of the third increment direction in Fig. 2, left). In the results section, the predicted angle is reported with respect to a fixed reference: is the predicted angle measured from the specimen surface. This way, a crack segment growing inwards (with respect to the indenter contact zone) has an angle 0º< <90º and -90º< <0º indicates a crack segment growing outwards. Bulk P ( ) =90º =90º Figure 2 : Application of the min ( ) criterion to predict the third crack-growth increment direction. Sign convention for direction angles of a crack growth increment. Crack in a simple bar subjected to tension-compression cycle As an example of a simple application of the minimum shear stress range criterion, a cracked bar of uniform section loaded in tension is analyzed. Fig. 3 shows the geometry and loads of the model and a contour plot of the von Mises stress field. This preliminary analysis is performed using standard FEM with ABAQUS (no X-FEM is considered at this stage). The tensile load is cyclic with R = -1 and is applied in the same fashion as the blue curve of Fig. 6 (no indenter load exists in this simple example). Fig. 4, left, shows the variation of the normal stress (in blue) and shear stress (in red) on a plane forming an angle with respect to the horizontal surface. Stresses are evaluated at the finite element located ahead the crack tip and transformed according to the angle of the prospective plane. The successive curves plot the variation along the time of the last step (step 4). Note that the normal stress is maximum at the end of the step (curves located at the top of the figure) and at a plane orientation of ±90º, as expected. This tensile stress is much higher than the corresponding compressive stress due to the effect of the crack opening (mode I of fracture), whereas the closing stage does not concentrate such high stresses (in magnitude). The evolution of the shear stresses (in red) is analogous. Note that the shear stresses are zero when the normal stress is maximum or minimum (for any given increment of time). The fact that

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