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Nonlinear behavior and critical state of a penny-shaped dielectric crack in a piezoelectric solid

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ASME-AMER SOC MECHANICAL ENG
DOI: 10.1115/1.2712227

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By means of the Hankel transform and dual-integral equations, the nonlinear response of a penny-shaped dielectric crack with a permittivity kappa(0) in a transversely isotropic piezo-electric ceramic is solved under the applied tensile stress sigma(A)(z) and electric displacement D-z(A). The solution is given through the universal relation, D-c/sigma(A)(z) = K-D/K-1 = M-D/M-sigma re gardless of the electric boundary conditions of the crack, where D-c is the effective electric displacement of the crack medium, and K-D and K-1 are the electric displacement and the stress intensity factors, respectively. The proportional constant M-D/M-sigma has been derived and found to have the characteristics: (i) for an impermeable crack it is equal to D(z)A/sigma(A)(z); (ii) for a permeable one it is only a function of the ceramic property; and (iii) for a dielectric crack with a finite kappa(0) it depends on the ceramic property, the kappa(0) itself and the applied sigma(A)(z) and D-z(A). The latter dependence makes the response of the dielectric crack nonlinear This nonlinear response is found to be further controlled by a critical state (sigma(c),D-z(A)), through which all the D-c versus sigma(A)(z) curves must pass, regardless of the value of kappa(0). When sigma(A)(z) < sigma(c), the response of an impermeable crack serves as an upper bound, whereas that of the permeable one serves as the lower bound, and when sigma(A)(z) > sigma(c) the situation is exactly reversed. The response of a dielectric crack with any kappa(0) always lies within these bounds. Under a negative D-z(A) our solutions further reveal the existence of a critical kappa*, given by kappa* = -RDzA and a critical D*, given by D*=-kappa(0)/R (R depends only on the ceramic property), such that when kappa(0) > kappa* Or when vertical bar D-z(A)vertical bar < vertical bar D*vertical bar, the effective D-c will still remain positive in spite of the negative D-z(A).

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