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General flat crack located in the plane perpendicular to the planes of isotropy in transversely isotropic body

机译:位于横观各向同性体中垂直于各向同性平面的平面中的一般扁平裂纹

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摘要

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 phi. 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.
机译:已发布的传统裂纹问题解决方案通常考虑位于平行于各向同性平面的平面中的裂纹,通常将其表示为z =0。位于平面x = 0且受到任意法向或切向载荷的裂纹的情况为最近在Fabrikant中解决了问题(Eur J Mech A 30:902-912,2011)。可以将本文视为其逻辑上的延续。我们在这里考虑一个与轴系(x(1),x(2),x(3))有关的横向各向同性物体,在平面x(1)= 0处由于任意形状的扁平裂纹而变弱。平面x(1)= 0垂直于横观各向同性物体的各向同性平面。轴Ox(1)与长轴Ox重合,并且轴Ox(2)和Ox(3)通过围绕轴Ox旋转任意角度phi而获得。对于这种一般情况,得出了控制积分方程。详细考虑椭圆形裂纹的情况。位移和应力场的完整解决方案以基本积分对象的单个轮廓积分的形式给出。应力强度因子是明确计算的。

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