ICF13B

13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- and the strain energy for a cracked bar subjected to bending is, introducing the value du from the eq. (6), 2 1 1 d d d 2 2 U M φ M λ   (9) The SIF in plane strain for the geometry of the study can be obtained as follows: π K Yσ a  (10) where the stress σ for axial tension is calculated as: 2 4 π F σ D  (11) and the maximum stress σ for bending is calculated as: 3 32 π M σ D  (12) If equations for strain energy are made equal and introducing values K and σ, the isolated compliance is obtained for tension loading:   2 2 4 0 32 1d π a ν λ Y a A D E   (13) and for bending moment,   2 2 6 0 2048 1d π a ν λ Y a A D E   (14) Solving the integral which appears in eqs. (13) and (14) is not trivial. In order to achieve that, the Cartesian coordinates (x, y) were change into parametrical coordinates (a, θ), relating themselves through the expressions: cos x b   (15) sin y a   (16) where the correspondence between angles δ and θ, deducted from Fig. 3, is as follows, tan tan y a δ θ x b   (17)   b h y x Figure 3. Relationship between δ and θ angles

RkJQdWJsaXNoZXIy MjM0NDE=