6.2 Calculation of the SIF for a rectangular sample with compression

Let us consider a square plate with half-side a, which contains a diagonal central vertical crack with half-length L. The crack fracture resistance of such a sample is determined based on compression by force R applied in vertical direction. The tips of the angles, in the vicinity of which forces are applied, can be cut off. Due to this, a sample is considered whose tops have coordinates

$$z\_1 = \mathfrak{c}, z\_2 = \dot{\mathfrak{c}} + z\_p, z\_3 = \dot{\mathfrak{c}} + z\_m, z\_4, z\_5 = \overline{z\_3}, z\_6 = \overline{z\_2}, z\_1$$

where in <sup>c</sup> <sup>¼</sup> ffiffi 2 <sup>p</sup> a, zp <sup>¼</sup> <sup>h</sup> � ih, zm ¼ �<sup>h</sup> � ih, h is the height of the cut triangle and а is the half-side of the square.

The calculations are performed at h ¼ c=8; moreover, all tops are rounded by the arcs of the circle of the radius a=10 (the shape of the sample—Figure 13).

Calculation of the relative SIF Ka <sup>¼</sup> KI ffiffi a p <sup>P</sup> ffiffi <sup>π</sup> <sup>p</sup> for isotopic material and composites of EF and LU with different directions of the orthotropic axis are given in Table 6. The angle between the crack and the direction with the maximum stiffness of the material is indicated in brackets.

The table shows a significant effect on the SIF of the placement of the crack relative to the axis with the maximum stiffness of the material. In particular, for

#### Figure 13. Sample view.


#### Table 6.

Relative SIF Ka when compressing a sample with a diagonal crack.


#### Table 7.

Relative SIFs when stretching a sample with a diagonal crack.

#### Figure 14. Sample view.

cracks parallel to the maximum stiffness direction, the SIFs appeared to be significantly larger than those returned for 90°. The difference between the SIFs for these two directions is increasing for a substantially anisotropic LU material.

Table 7 shows the results of calculations for the case of stretching the same sample with a horizontal crack (Figure 14).

Based on the comparison of data from Tables 6 and 7, it follows that, unlike the case of compression, the SIF with stretching is larger for cracks that are perpendicular to the direction with maximum stiffness of the material.

## 7. Conclusions

An algorithm for calculating stresses at cracks in bounded plate with holes of various shapes due to concentrated forces or distributed forces at its boundary has Determination of Stresses in Composite Plates with Holes and Cracks Based on Singular Integral… DOI: http://dx.doi.org/10.5772/intechopen.87718

developed. The solution of integral equations is performed by quadrature Gausstype formulas for regular and singular integrals.

The research of stresses at cracks in the samples which are used in experimental studies of crack fracture resistance was performed.

The calculation of the stresses at cracks in samples of various forms is performed, in which ones' experimental research are performed. To study the crack fracture resistance of composite samples, the optimal distances from the central crack to the forces at which the SIF increases monotonously with increasing crack length are determined. In particular, for square samples with a half-side a, forces should be placed at a distance of 0.6a–0.7a from the crack. For the experimental study of the growth rate of fatigue cracks, there are definite ranges of lengths of cracks for which SIFs are practically constant values. At the same time, the distances are determined at which it is expedient to apply forces. The problem for studying samples with cracks with compression is considered.
