Abstract

A closed form model for estimating the reduction in Mode II stress intensity factor (SIF) and the induced Mode I SIF resulting from the mismatch of fracture surface asperities during shearing of the fracture surfaces is presented. At a given effective Mode II SIF, discretizations of actual Mode I fracture surfaces are displaced in shear according to the classical √ r shear displacement law including crack tip plasticity. The crack faces are then “opened” enough to prevent interpenetration of the shifted surfaces resulting in a single point contact. The induced Mode I SIF is then calculated from the solution for a flat crack, opened by a concentrated normal force assuming a classical √ r opening displacement law. The resistance Mode II SIF is found from the corresponding problem with a concentrated tangential force which is proportional to the normal force, and is positive or negative, depending on whether the sliding is uphill or downhill, respectively. The proportionality factor is determined through Coulomb's law of sliding friction applied to the inclined asperity surface at the contact point. The applied Mode II SIF is simply the sum of the effective and the resistance Mode II SIF's. The model predicts large initial resistance to the applied Mode II SIF and an induced Mode I SIF of the same order as the effective Mode II SIF. The secondary structure of the K IIeff vs K IIapp curves is directly related to the periodicity of the fracture surface profile. It is suggested that the model should be modified to account for asperity wear and subsequent modification of the fracture surface profile.

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