13th International Conference on Fracture June 16–21, 2013, Beijing, China -5- Figure 6. Decomposition of the thermal problem In the first case (a) a normal stress (r) is applied on the crack lips in order to imposed a crack opening equal to zero (uy = 0). This stress (r) is calculated from a thermoelastic problem without crack and with the heat source q. The stress intensity factor of case (a) is equal to zero and the stress intensity factor of case (b) is calculated with the Green function: Rr R temp r r r r t K d ( , ) 2 . (4) The case (a) thermo-mechanical problem is symmetric because both the geometry and the temperature field are symmetric. Only the case of plane stress is considered hereafter but plane strain solution is given in [6]. Further, outside of the reverse cyclic plastic zone since the constitutive behavior of the material is supposed to be elastic, it is expected in first approximation that the basic equations of thermo-elasticity will govern. The equilibrium equation is: 0 r r r r (5) For which r is the radial normal stress and is the circumferential normal stress. The isotropic elastic stress strain law gives with plane stress hypothesis: ( , ) r t r E E u r r r (6) ( , ) r t r E E u r r (7) A solution of this equation is given in [6] and the circumferential stress can be expressed with the following relation: at r r r at kr Eq r t R 4 Ei 4at r Ei 4at - r exp 4at - r 4 exp 8 ( , ) 2 2 2 R 2 2 2 R 2 (8) This circumferential normal stress is then calculated for a line heat source of 153 Wm-1 and at time t = 680s and this is represented in fig. 7. Near the reverse cyclic plastic zone (r = rR = 130µm) the circumferential stress is negative (about -8.24 MPa) because the temperature is high and through the circumferential direction, the material is under compression due to the thermal expansion and the constraint effect.
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