Thank you Dr Kourousis for your comments.
Regarding your first question. We did not include the values of c_k and gamma_k because it is basically a curve-fitting. In fact, if you use a commercial finite element code like Abaqus what you provide is the uniaxial stress-strain curve (and the number of parameters, k=10), and Abaqus does the fitting for you. Thus, going from isotropic hardening to kinematic hardening just involves changing one line in the input file. Moreover, note that if you use other approaches, such as multilinear kinematic hardening (implemented in ANSYS since a few years ago and in Abaqus since 2017) you will get similar results in the crack growth problem. In any case, one can request the values of c_k and gamma_k, along with the fitting precision, so I have looked into my notes and the parameters used seem to be the following:
N=0.2 - c_1: 0.64945, gamma_1: 1.02E-04; c_2: 1.8295, gamma_2: 8.25E-04; c_3: 5.5242, gamma_3: 3.77E-03; c_4: 17.798, gamma_4: 1.61E-02; c_5: 60.82, gamma_5: 7.22E-02; c_6: 195.53, gamma_6: 0.32148; c_7: 652.78, gamma_7: 1.4174; c_8: 2182.2, gamma_8: 6.4444; c_9: 7391, gamma_9: 30.334; c_10: 23610, gamma_10: 173.42.
N=0.1 - c_1: 2.77E-02, gamma_1: 0; c_2: 0.21698, gamma_2: 4.62E-04; c_3: 0.89669, gamma_3: 2.99E-03; c_4: 3.5643, gamma_4: 1.47E-02; c_5: 15.397, gamma_5: 7.34E-02; c_6: 62.707, gamma_6: 0.36018; c_7: 263.63, gamma_7: 1.7468; c_8: 1072.1, gamma_8: 8.6006; c_9: 3986.5, gamma_9: 41.256; c_10: 12163, gamma_10: 220.63.
Regarding your second question. This automated fitting to the tensile part of the uniaxial stress-strain curve shows that, for 10 c_k and gamma_k, the kinematic hardening model matches the isotropic hardening model up to a stain of 2 with the largest difference being of 0.04%. We are dealing with monotonic/static loading, where the translation of the yield surface with crack advance is always neglected. What we show is that if you model crack propagation with a kinematic hardening model that matches the isotropic case in uniaxial tension, then significant differences arise in the crack growth resistance curves. In other words, one should not neglect anisotropic/kinematic hardening effects in monotonic/static loading.
Thank you
Emilio Martínez Pañeda
Thank you Dr Kourousis for your comments.
Regarding your first question. We did not include the values of c_k and gamma_k because it is basically a curve-fitting. In fact, if you use a commercial finite element code like Abaqus what you provide is the uniaxial stress-strain curve (and the number of parameters, k=10), and Abaqus does the fitting for you. Thus, going from isotropic hardening to kinematic hardening just involves changing one line in the input file. Moreover, note that if you use other approaches, such as multilinear kinematic hardening (implemented in ANSYS since a few years ago and in Abaqus since 2017) you will get similar results in the crack growth problem. In any case, one can request the values of c_k and gamma_k, along with the fitting precision, so I have looked into my notes and the parameters used seem to be the following:
N=0.2 - c_1: 0.64945, gamma_1: 1.02E-04; c_2: 1.8295, gamma_2: 8.25E-04; c_3: 5.5242, gamma_3: 3.77E-03; c_4: 17.798, gamma_4: 1.61E-02; c_5: 60.82, gamma_5: 7.22E-02; c_6: 195.53, gamma_6: 0.32148; c_7: 652.78, gamma_7: 1.4174; c_8: 2182.2, gamma_8: 6.4444; c_9: 7391, gamma_9: 30.334; c_10: 23610, gamma_10: 173.42.
N=0.1 - c_1: 2.77E-02, gamma_1: 0; c_2: 0.21698, gamma_2: 4.62E-04; c_3: 0.89669, gamma_3: 2.99E-03; c_4: 3.5643, gamma_4: 1.47E-02; c_5: 15.397, gamma_5: 7.34E-02; c_6: 62.707, gamma_6: 0.36018; c_7: 263.63, gamma_7: 1.7468; c_8: 1072.1, gamma_8: 8.6006; c_9: 3986.5, gamma_9: 41.256; c_10: 12163, gamma_10: 220.63.
Regarding your second question. This automated fitting to the tensile part of the uniaxial stress-strain curve shows that, for 10 c_k and gamma_k, the kinematic hardening model matches the isotropic hardening model up to a stain of 2 with the largest difference being of 0.04%. We are dealing with monotonic/static loading, where the translation of the yield surface with crack advance is always neglected. What we show is that if you model crack propagation with a kinematic hardening model that matches the isotropic case in uniaxial tension, then significant differences arise in the crack growth resistance curves. In other words, one should not neglect anisotropic/kinematic hardening effects in monotonic/static loading.
Thank you
Emilio Martínez Pañeda