5. Determination of the SSS of samples under the action of the tractions applied to the hole's boundary

Let us apply the developed algorithm to study a square plate with a half-side a; weakened by a central crack with a half-length L (Figure 11), the edges are not loaded.

Determination of Stresses in Composite Plates with Holes and Cracks Based on Singular Integral… DOI: http://dx.doi.org/10.5772/intechopen.87718

Figure 11. Scheme of the sample.

Two identical circular holes of radius R, the centers of which are located at points 0ð Þ ; �c , are created for stretching in a plate. It was assumed that the load was applied to the boundary of the circular holes. Using [10], we accept that the forces act normally on the domain θ � θ<sup>c</sup> j j<γ and are given on it in the form

$$p = P \frac{\left(\theta - \theta\_c\right)^2}{4R(\sin \chi - \chi \cos \chi)},\tag{15}$$

where θ is angle coordinate on each of the holes, θ<sup>c</sup> is angle coordinate of the middle of the domain, and Р is the principal vector applied to the domain of forces, which is directed from the center of the hole at an angle θc.

At first, for the purpose of testing the algorithm, the calculations are performed for the case of a localized load at γ ¼ π=32 and α ¼ c=a ¼ 0, 2; 0, 4; 0, 6, with θ<sup>c</sup> ¼ π=2 for the upper hole and θ<sup>c</sup> ¼ �π=2 for the lower hole (thus the stretching of the plate in the direction of the OY axis is considered). The relative SIF <sup>Y</sup> <sup>¼</sup> KI P ffiffiffiffiffiffi <sup>π</sup><sup>L</sup> <sup>p</sup> for an isotropic material with a Poisson ratio ν ¼ 1=3 at different crack lengths at R=a ¼ 0, 1 is calculated and given in Table 2.

The values of relative SIF YS obtained by another method in [10] for the case of stretching by concentrated forces (i.e., with γ ! 0) are given in the same table.

