Abstract
The published traditional crack problem solutions usually consider cracks located in the planes parallel to the plane of isotropy, which is usually denoted as z = 0. The case of a crack located in the plane x = 0 and subjected to arbitrary normal or tangential loading was solved recently in Fabrikant (Eur J Mech A 30:902–912, 2011). This article may be considered as its logical continuation. We consider here a transversely isotropic body, related to the system of axes (x 1, x 2, x 3), weakened in the plane x 1 = 0 by a flat crack of arbitrary shape. The plane x 1 = 0 is perpendicular to the planes of isotropy of the transversely isotropic body. The axis Ox 1 coincides with the major axis Ox, and the axes Ox 2 and Ox 3 are obtained by rotation about the axis Ox by arbitrary angle φ. The governing integral equation is derived for such a general case. The case of an elliptical crack is considered in detail. The complete solution for the fields of displacements and stresses is presented as single contour integrals of elementary integrands. Stress intensity factors are computed explicitly.
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Fabrikant, V.I. General flat crack located in the plane perpendicular to the planes of isotropy in transversely isotropic body. Acta Mech 226, 3289–3306 (2015). https://doi.org/10.1007/s00707-015-1362-y
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DOI: https://doi.org/10.1007/s00707-015-1362-y