Elastic constants of LU and EF materials.
Abstract
The problems of determination of stresses at crack in bounded plates with holes of different shapes under the action of concentrated forces or distributed forces at its boundary are considered. The study is performed by the singular integral equation method. They were determined based on the established interdependences between the Lekhnitskii potentials and the stress and strain. The numerical method for solving integral equations is developed based on the quadrature method for the systems of holes and cracks. The eigen solutions of the problem were taken into account in this method. The research of stresses at cracks in samples which are used in experimental studies of crack fracture resistance was performed.
Keywords
 stress intensity factors (SIF)
 composite plates
 holes
 cracks
 crack fracture resistance
 BIEM
 stressstrain state (SSS)
1. Introduction
The boundary integral equation method (BIEM) is widely used to study the stressstrain state (SSS) of anisotropic plates with holes [1, 2, 3] and cracks [4, 5, 6, 7, 8, 9]. The integral equations for anisotropic plates are usually determined based on the Somigliana identity. Such equations for plates with given stresses at the boundaries of the plate are hypersingular. At the same time, the same problem for isotropic plates is reduced to singular integral equations [10, 11], for which simple numerical algorithms for solving with given precision are obtained.
In [12, 13], the simple dependencies between the Lekhnitskii complex potentials and stress and strain are obtained. In a simple form based on them and the Cauchy theorem, the integral equations are written for anisotropic plates with holes [12, 13] and cracks [14, 15, 16]. We will use the established dependencies for the construction and regularization of integral equations for anisotropic plates with holes and cracks.
For conducting experimental studies of crack fracture resistance on experimental samples in relation to isotropic materials, theoretical estimates for stresses at cracks are performed.
For such materials, the stresses in samples of different shapes with cracks under the action of stretching or compressing concentrated forces are studied in detail [10]. The experimental samples for the experimental determination of the characteristics of crack fracture resistance of various types of materials are made based on performed studies. We perform similar studies for composite samples.
2. The integral representations for anisotropic plates with holes and cracks
We consider a plate which is weakened with a system hole with boundaries
2.1 Governing equations
Let us start from the Lekhnitskii complex potentials
where
where
Consider an arbitrary path
Then introduce in consideration the stress vectors
where
The stress vectors
Assume that the vectors
where
2.2 Integral equations for anisotropic bounded plate with holes and cracks
Let us write a general solution of the problem based on [12, 15] through the Lekhnitskii potentials in the form
where
Here,
where in
Note that when the boundary is traction free, then
Let us substitute the potentials (6) into the formulas (4) for projections of stress vectors determined at the boundaries path
where
Using the results [12], we obtained that the unknown functions
Let us consider a problemsolving equation (10) for the case of one hole and a crack. Let us assume that the contour on which the crack is placed is described parametrically in the form
Let us assume the representation for the displacement discontinuity at the cracks:
Let us replace the integrals with Lobattotype quadrature formulas [15], and the integrals at the boundaries of the holes replaced by the quadrature of a rectangle, which, for periodic functions, are Gauss quadraturetype formulas [12]. Then we obtain the system of equations:
where
We obtain the additional equation of system (12) from condition (11)
Analogously to [12], we should remove three equations with
The first two equations follow from the displacements continuity conditions (9). The last equation is obtained when fixing an arbitrary constant
The system of Eqs. (12)–(14) is generalized in the case of hole and crack system in the same way as it was done in [12].
3. Stresses in circular samples with cracks under the action of concentrated forces
Let us consider the circular composite plate with radius
Calculations are made for the composite plates with elastic constants shown in Table 1 .
Material 






LU  10.8  96.0  2.61  0.21  0.024 
EF  21  32.8  5.7  0.21  0.134 
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
For a weakly anisotropic material, the incline of the crack to the main axis of orthotropy had little effect on the SIF
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 LU1 material. When the crack is placed perpendicular to the direction with maximum stiffness, stable crack growth occurs of small cracks (
Testing the developed algorithm is conducted for the case of isotropic plate with
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.
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
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
Since at big lengths of cracks
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
Figure 5
shows that the range with few changed SIF is necessary to determine in the vicinity of the lengths of the cracks with
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.
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
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,
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.
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 halfside
Two identical circular holes of radius R, the centers of which are located at points
where
At first, for the purpose of testing the algorithm, the calculations are performed for the case of a localized load at












0.1  0.175  0.177  0.248  0.247  0.271  0.270 
0.2  0.721  1.720  0.583  0.581  0.568  0.565 
0.3  1.164  1.159  0.958  0.953  0.894  0.888 
0.4  1.503  1.495  1.323  1.314  1.237  1.227 
0.5  1.819  1.805  1.675  1.663  1.588  1.575 
0.6  2.156  2.131  2.034  2.015  1.954  1.935 
0.7  2.552  2.505  2.438  2.406  2.367  2.334 
0.8  3.112  2.995  2.999  2.917  2.937  2.854 
0.9  4.279  —  4.178  —  4.131  — 
The values of relative SIF
The results of the calculations of SIF are given in
Table 3
for the case







0.4  0.6 

0.4  0.6 

0.4  0.6 
0.1  0.186  0.254  0.275  0.727  0.462  0.366  0.2069  0.2637  0.2767 
0.2  0.732  0.593  0.576  1.073  0.851  0.717  0.7563  0.6096  0.5786 
0.3  1.172  0.969  0.905  1.278  1.167  1.049  1.1914  0.9910  0.9101 
0.4  1.510  1.335  1.250  1.489  1.452  1.370  1.5415  1.3678  1.2655 
0.5  1.827  1.689  1.605  1.742  1.744  1.694  1.8964  1.7516  1.6483 
0.6  2.167  2.050  1.974  2.048  2.067  2.039  2.3113  2.1770  2.0811 
0.7  2.569  2.459  2.391  2.436  2.459  2.440  2.8332  2.6963  2.6106 
0.8  3.139  3.029  2.971  3.008  3.027  3.011  3.5616  3.4216  3.3492 
0.9  4.322  4.225  4.181  4.214  4.225  4.209  4.8896  4.7662  4.7132 
The results for a circular sample with
The following conclusions are made based on
Tables 2
and
3
: SIF does not differ significantly in the case of distributed loads with different degrees of localization, SIF increases somewhat with the growth of the domain of action of tractions, and SIFs are bigger at small crack lengths at point action of tractions (at
Similar results for a square sample made from a LU material are given in
Table 4
. Here, the relative SIFs
LU ( 
LU ( 




0.4  0.6 

0.4  0.6 
0.1  0.2338  0.2633  0.2764  0.4182  0.478  0.538 
0.2  0.5187  0.4276  0.4059  1.0504  1.0073  0.9911 
0.3  0.5952  0.5259  0.4978  1.1416  1.1306  1.1327 
0.4  0.6202  0.5791  0.5595  1.2268  1.2138  1.2163 
0.5  0.6409  0.6158  0.6046  1.33  1.3094  1.31 
0.6  0.6711  0.6539  0.6479  1.4615  1.4313  1.4304 
0.7  0.7252  0.7117  0.7085  1.6257  1.5829  1.5809 
0.8  0.8392  0.8284  0.8267  1.7876  1.7411  1.7386 
0.9  1.1482  1.1419  1.1409  1.9405  1.8846  1.8831 
Similar results for a circular sample for the same material are given in Table 5 .
LU ( 
LU ( 




0.4  0.6 

0.4  0.6 
0.1  0.2363  0.2679  0.2806  0.4192  0.4799  0.5407 
0.2  0.5269  0.4369  0.4138  1.0594  1.0163  0.9995 
0.3  0.6108  0.5418  0.5112  1.1602  1.1477  1.1466 
0.4  0.6467  0.605  0.5814  1.2588  1.2422  1.2389 
0.5  0.683  0.6558  0.6396  1.3795  1.3531  1.3459 
0.6  0.7341  0.7131  0.702  1.5354  1.4974  1.4888 
0.7  0.8131  0.7948  0.7879  1.7482  1.6965  1.6905 
0.8  0.9524  0.9376  0.9342  2.0507  1.9973  1.9977 
0.9  1.2791  1.2726  1.2704  2.592  2.528  2.5346 
6. Determination of the SSS of samples loaded with concentrated forces at the boundary (compression test)
The methods of studied crack fracture resistance based on sample compression in experimental practice are widely used.
The direct application of the abovementioned variant of the method of integral equations for the case when the concentrated forces act on the boundary of the domain is associated with significant errors, because unknown functions in the vicinity of the points of application of forces have a singularity. Due to this, it is necessary to separate a singular part in the solution for a more precise solution of this type of task.
6.1 Determining a singular part of the solution of this problem
Let us consider a point of the plate boundary
Let us consider at first a halfplane
where
It can be shown [17] that the halfplane also corresponds to the
Let us now consider a bounded plate occupying the domain D. The selfbalanced concentrated forces
where
where
It is easy to show that the righthand side of formula (17) is a continuous and limited function, and therefore the introduced complex potentials with an index
6.2 Calculation of the SIF for a rectangular sample with compression
Let us consider a square plate with halfside
where in
The calculations are performed at
Calculation of the relative SIF

Isotr.  EF (90°)  EF (0°)  LU (90°)  LU (0°) 

0.1  0.0824  0.0579  0.0727  0.0226  0.0743 
0.2  0.1203  0.085  0.1073  0.0332  0.1182 
0.3  0.1557  0.1106  0.1407  0.0434  0.1663 
0.4  0.1927  0.1384  0.1776  0.0547  0.2217 
0.5  0.2354  0.1709  0.2208  0.068  0.2865 
0.6  0.2889  0.2107  0.2737  0.0846  0.3643 
0.7  0.3544  0.2609  0.3406  0.1059  0.4609 
0.8  0.4418  0.3273  0.4301  0.1343  0.5872 
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 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 ).

Isotr.  EF (0°)  EF (90°)  LU (0°)  LU (90°) 

0.1  0.2585  0.2886  0.3499  0.2956  0.7572 
0.2  0.3696  0.4065  0.4874  0.4114  0.9247 
0.3  0.4617  0.4963  0.5884  0.4936  1.0104 
0.4  0.5494  0.5752  0.6766  0.5594  1.1008 
0.5  0.641  0.6534  0.7668  0.6185  1.2159 
0.6  0.7434  0.7397  0.8705  0.6789  1.364 
0.7  0.8641  0.842  0.9973  0.7472  1.5542 
0.8  1.0129  0.9704  1.1585  0.8319  1.8021 
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 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 halfside
List of symbols with explanations
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