PSI - Issue 13

Kiiko V.M. et al. / Procedia Structural Integrity 13 (2018) 1433–1437 Author name / Structural Integrity Procedia 00 (2018) 000 – 000

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Increasing the crack lengt h by Δ l leads to a change in the potential field geometry and a shift in the equipotential lines. Consequently, a different equipotential line passes through the point M, and there is a new point N´ on the edge such that the voltage between M and N´ is zero. We find the new position N´ by shifting the movable probe from the contact point N through a distance Δ x along the edge. We thus establish a unique dependence of the contact shift Δ x on the crack growth Δ l . Because the picture of the potential distribution is symmetric with respect to the central longitudinal sample axis in the case of a central crack, this dependence holds for both side and central cracks. This scheme was realized in an automatic device for measuring the growth of fatigue cracks [1, 2], and we show one variant of the corresponding construction in Fig. 2b. A similar scheme was also implemented in other devices [3 – 6]. We define the distribution of electrical current and equipotential lines in a two-dimensional problem setup. The studied field corresponds to a current field in an infinitely long strip with a cut for which the field is also uniform across the strip at a sufficiently large distance from the cut. We use the method of conformal maps to calculate the field, d is a width of specimen, l is a length of crack [7]. The function = 2 0 ℎ {cos 2 √ ℎ 2 2 + 2 2 } maps the domain occupied by the strip with a cut in the plane of the complex variable z = x +iy (Fig. 2b) to a strip without a cut in the plane of W= U+iV (Fig. 3). 3. Calculations and measurements

Fig. 3. Range of the conformal map.

The correspondence of points on the boundaries of these regions is noted in Fig. 3. The lines U = const in the plane of W are lines of equal potential, and the lines V = const are lines of current. Consequently, the analytic function W is the complex potential field of an infinite strip with a cut. In this formula, the quantity V 0 is equal to the value V 2 of the current function on the upper boundary of the strip at y = d ; the lower boundary of the strip in the plane of z corresponds to the axis of the abscissa in the plane of W , i.e., V = 0. To draw the lines of the current and of equal potential, we find an expression for z : = 2 artanh √(th 2 0 ) 2 −(sin 2 ) 2 cos 2 (1) For a sample of width d = 100 mm with a cut length l = 50 mm, a plot of the current lines has the form shown in Fig. 4a and the position of equipotential lines shown in Fig. 4b. As the crack length grows, the picture of the equipotential line distribution changes (Fig. 5).

From Eq. (1), we obtain the solution