4. Determination of working intervals of crack lengths at circular samples

Two types of samples are used in experimental studies of crack fracture resistance [10]. The first is a sample for which the SIF grows monotonically with the growth of the crack. In the second type, the range of cracks' lengths is selected in such a way that the SIF KI is practically constant. Hence, it is varied in this range from the mean value to the small value (�2–4%). This range of crack lengths is called working. The samples of the second type are particularly suitable for conducting experimental studies including a wide range of problems in the area of destruction. In particular, with the constant force factor (the SIF is constant) in such samples, the possibility of an effective study of the rate of growth of fatigue cracks with cyclic loads, the study of crack fracture resistance depending on the influence of working environment, etc. arises.

Based on the studies in the literature for isotropic material, it has been established [10] that the range of working lengths is most favorable with α ¼ 0, 18, although allowed, and α ¼ 0, 16; 0, 2, where α ¼ y0=a. With 0< α<0, 2237 the SIF KI increases from zero to a certain maximum (depending on α), then falls to a

minimum, and then increases monotonously. With α>0, 2237 the SIF increases monotonically with the increasing length of the crack. We note that such conclusions are valid for material with a Poisson ratio ν ¼ 0, 4488.

Since at big lengths of cracks L=a>0, 5 the SIF depends little on α, then for the first type of samples α ¼ 0, 65 is taken.

Based on the obtained results, let us perform a similar study of samples of two types of composite materials.

For samples of the second type, we perform calculations only for small ratios (at α � 0, 2). The results of calculations of the SIF in a circular isotropic sample with ν ¼ 1=3 are shown in Figure 5. Here and further is assumed that the relative value of the SIF is equal to Ka <sup>¼</sup> KI ffiffi a p P ffiffi <sup>π</sup> <sup>p</sup> and the parameter value α is indicated near the curves. Figure 5 shows that the range with few changed SIF is necessary to determine in

the vicinity of the lengths of the cracks with L=a � 0, 25.

Similar results for the plate made of an EF material are shown in Figures 6 and 7. Here two cases are considered: the crack is parallel or perpendicular to the direction in which the stiffness of the material is maximal.

Figure 5. Relative SIF for isotropic material.

Figure 6. Relative SIF for an EF material: the crack is parallel to the direction of greater stiffness of the material.

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

Figure 7. Relative SIF for an EF material: the crack is perpendicular to the direction of greater stiffness of the material.

From the given data, it is seen that the SIF for an EF material is bigger in the case when the crack is perpendicular to the direction of maximum stiffness of the material.

The following conclusions are made based on the data of the calculations: for the crack that is parallel to the direction of maximum stiffness of the material, the minimum deviations from the constant SIF (with an error of not more than 2%) are achieved on the ranges of lengths of cracks with relative dimensions with Δ < 0, 21 with α ¼ 0, 24 � 0, 26, where Δ ¼ ð Þ L<sup>2</sup> � L<sup>1</sup> =a. The biggest range (Δ ¼ 0, 27) with SIF values close to constants with an error 2, 1% is achieved with α ¼ 0, 24. For the crack that is perpendicular to the direction of maximum stiffness of the material, the range with SIF values close to constants is reduced. Moreover, the distance of force application needs to be increased. The biggest range (Δ ¼ 0, 27) with SIF values close to constants with an error 3, 2% is achieved with α ¼ 0, 24.

Let us consider the case of samples of the first type 1. For them, the forces are selected that are distant from the crack. For isotropic materials, as a rule, α ¼ 0, 65 is taken. The above-obtained results of the calculations show that the same distance can be chosen for the composite materials with a crack parallel to the principal axes of the orthotropy.

Figure 8. Relative SIF for a square isotropic sample.

Figure 9. LU material, square sample, horizontal crack.

Figure 10. LU material, square sample, diagonal crack.

To compare the effect of a sample shape, similar calculations are made for a square plate with a crack. The results of calculations for such an isotropic sample that are similar to the results of calculations for a circle are shown in Figure 8.

The conclusion is made based on the comparison of Figures 5 and 8 that with small distances of forces from cracks, the shape of the sample has little effect on the SIF.

Similar results of calculations for LU material are shown in Figure 9 for a horizontal crack and in Figure 10 for a diagonal crack.
