3. Stresses in circular samples with cracks under the action of concentrated forces

Let us consider the circular composite plate with radius a, which is weakened by a central crack with the half-length L. The plate is stretched by the concentrated forces �P applied at points 0; �y<sup>0</sup> � �. Destruction of the plate is happening when the stress intensity factors (SIFs) reach a certain limit value. Therefore, when we calculated the limiting loads, we considered the SIF which explicitly takes into account the length of the cracks. Due to this, we performed calculations of the relative SIF Ka <sup>¼</sup> KI ffiffi a p P ffiffi <sup>π</sup> <sup>p</sup> with different relative distances α ¼ y0=a, depending on the half-length of the crack, which is divided into a.

Calculations are made for the composite plates with elastic constants shown in Table 1.

The results of the calculations for the plate made of an EF material (with a small degree of anisotropy) with the maximum stiffness in the direction of the OX axis are shown in Figure 3.

In general, the character of the distribution for an EF material is not significantly different from that of an isotropic material. It is necessary to increase monotonically the load for a stable growth of the crack when the distances of forces to a crack are smaller than 0, 2a. With y0=a ¼ 0, 3, the stable growth of cracks (without jumping) will occur at L=a>0, 2; with y0=a ¼ 0, 4 if L=a>0, 25; and with y0=a ¼ 0, 5 if L=a>0, 35. At greater distances to forces, after reaching the corresponding level of values of traction, the circle fractures. For a case where the crack is perpendicular to the direction with the maximum stiffness of the material, the SIF is slightly increasing, especially at greater distances to forces.


Table 1.

Elastic constants of LU and EF materials.

Figure 3.

Relative SIF for a circular plate made of an EF material: a direction with maximum stiffness is parallel to the OX axis.

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

Figure 4. Relative SIF for isotropic material.

For a weakly anisotropic material, the incline of the crack to the main axis of orthotropy had little effect on the SIF KI, and the SIF KII is practically absent.

The calculations have shown that for the case of placing the crack in parallel to the direction with the maximum stiffness of the material, the above set of specifics of the SIF remain unchanged for substantially anisotropic LU-1 material. When the crack is placed perpendicular to the direction with maximum stiffness, stable crack growth occurs of small cracks (L=a≤0, 1) when the relative distance to forces is y0=a< 0, 2. Moreover, for these cases, the fracture is spasmodic. In all other cases, the SIF increases monotonically with increasing crack length. Hence, the plate fractures completely after reaching forces of critical value.

Testing the developed algorithm is conducted for the case of isotropic plate with ν ¼ 0, 4488 and y0=a ¼ 0, 15; 0, 16; 0, 18; 0, 2. The calculation results of the relative SIF Ka <sup>¼</sup> KI ffiffi a p <sup>P</sup> are shown in Figure 4.

On the right are shown figures from the book [10]. Such calculations were also performed for the same material, and here the corresponding relative SIF is represented by dashed lines. It is seen that the results obtained by different methods coincide.
