Evaluating the crack resistance of parts is an extremely urgent task, which makes it possible to draw a timely conclusion about the possibility of the existence of a structure with the presence of a crack-like defect in it. Often, real parts and structures are loaded in such a way that mixed-type cracks are observed in them, i.e., those in which cracking according to types I, II, and III is present. Often, real parts and structures are loaded in such a way that mixed-type cracks are observed in them, i.e., those in which cracking is present according to types I, II, and III. A special case of a mixed-type crack, which is quite often encountered in practice, is a mixed-type crack (I + II), i.e., with normal separation and transverse shear.

To estimate the crack resistance of parts with mixed type cracks (I + II), the criterion of maximum tangential stress (MTS) is most often used. According to this criterion, fracture occurs when the maximum tangential stresses in the prefracture region are equal to the limiting value [1]. As a limiting value, the yield strength or tensile strength are usually used, and the Irwin correction for the plastic region [2] is used for the size of the prefracture region.

The classical parameters of elastic fracture mechanics are the stress intensity factor (SIF) and the intensity of elastic energy release into the crack vertex. At present, in addition to them, in the calculations of crack resistance, the components of nonsingular T-stresses lying in the crack plane, which are included in the expansion of the Williams stress function [3], are increasingly used. The use of stresses Тхx and Тzz in the calculations makes it possible to describe the constraint of deformations in the direction perpendicular and parallel to the crack front, respectively [4]. In Russia, the greatest contribution to popularization of the use of T-stress components in the calculations of crack resistance and survivability was made in [4–6].

In [7], the criterion for maximum tangential stresses was improved by taking into account nonsingular Тxx stresses. In [4], it was proposed to use the local strength of material as the limiting stress in the prefracture region, which depends not only on the yield strength, but also on Тxx stresses. In addition, the criterion of maximum tangential stresses was improved; namely, in the condition of failure, it is proposed to use the averaged maximum tangential stresses, which are equated to the local strength of the materials.

The purpose of this paper is to develop a two-parameter fracture criterion for a mixed-type crack (I + II), which, in addition to stresses Тxx, also includes Тzz stresses.

1. FORMULATION OF THE FRACTURE CRITERION

In the elastic case, the stress field in the vicinity of a mixed-type crack (I + II) can be represented as [8]

$${{\sigma }_{x}} = \frac{1}{{\sqrt {2\pi r} }}\left[ {{{K}_{{\text{I}}}}\cos \frac{\theta }{2}\left( {1 - \sin \frac{\theta }{2}\sin \frac{{3\theta }}{2}} \right) - {{K}_{{{\text{II}}}}}\sin \frac{\theta }{2}\left( {2 + \cos \frac{\theta }{2}\cos \frac{{3\theta }}{2}} \right)} \right] + {{T}_{{xx}}},$$
$$\begin{gathered} {{\sigma }_{y}} = \frac{1}{{\sqrt {2\pi r} }}\left[ {{{K}_{{\text{I}}}}\cos \frac{\theta }{2}\left( {1 + \sin \frac{\theta }{2}\sin \frac{{3\theta }}{2}} \right) + {{K}_{{{\text{II}}}}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}\cos \frac{{3\theta }}{2}} \right], \\ {{\tau }_{{xy}}} = \frac{1}{{\sqrt {2\pi r} }}\left[ {{{K}_{{\text{I}}}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}\cos \frac{{3\theta }}{2} + {{K}_{{{\text{II}}}}}\cos \frac{\theta }{2}\left( {1 - \sin \frac{\theta }{2}\sin \frac{{3\theta }}{2}} \right)} \right], \\ \end{gathered} $$
(1)
$${{\sigma }_{z}} = \frac{{2\mu }}{{\sqrt {2\pi r} }}\left[ {{{K}_{{\text{I}}}}\cos \frac{\theta }{2} - {{K}_{{{\text{II}}}}}\sin \frac{\theta }{2}} \right] + {{T}_{{zz}}},\quad {{T}_{{zz}}} = E{{\varepsilon }_{z}} + \mu {{T}_{{xx}}}.$$

Here σx, σy, σz, and τxy are the normal and tangential stresses, Тхх and Тzz are the components on nonsingular Т-stresses lying in the crack plane and directed perpendicular and parallel to the crack front, respectively; εz is the strain in the direction of the crack front; Е is the Young modulus; μ is the Poisson coefficient; and r and θ are the radius-vector and polar angle in the polar coordinate system associated with a point at the crack front (Fig. 1).

Fig. 1.
figure 1

Stress state along the front of a through crack.

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In the polar coordinate system associated with the crack vertex, the tangential stresses σθ, perpendicular to the radius vector r, according to (1) will have the form [4]

$${{\sigma }_{\theta }}\left( {r,\theta } \right) = \frac{1}{{\sqrt {2\pi r} }}\cos \frac{\theta }{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{\theta }{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin \theta } \right] + {{T}_{{xx}}}{{\sin }^{2}}\theta .$$
(2)

Let us use the criterion of maximum tangential stresses [1], according to which the crack grows in the direction of the plane of action of maximum tangential stresses. To find this direction defined by the angle θ*, we equate to zero the derivative of the tangential stress

$${{\left. {\frac{{\partial {{\sigma }_{\theta }}}}{{\partial \theta }}} \right|}_{{|\theta = \theta \text{*}}}} = 0.$$
(3)

Substituting σθ from (2) into condition (3) for r = rc, with rc being the size of the pre-fracture area, leads to the expression for calculating θ* [4]

$${{K}_{{\text{I}}}}\sin \theta {\kern 1pt} \text{*} + \;{{K}_{{{\text{II}}}}}\left( {3\cos \theta {\kern 1pt} \text{*} - 1} \right) - \frac{{16}}{3}{{T}_{{xx}}}\sqrt {2\pi {{r}_{c}}} \sin \frac{{\theta {\kern 1pt} \text{*}}}{2}\cos \theta {\kern 1pt} \text{*} = 0.$$
(4)

Due to the fact that the plane of action of maximum tangential stresses is the principal pad, there are no shear stresses τθr in it or in the pad perpendicular to it (Fig. 2). Therefore, the size of the prefracture zone in this direction can be determined for a crack of normal separation [9]

$${{r}_{c}} = \frac{{{{{(1 - 2\mu )}}^{2}}}}{{2\pi }}{{\left( {\frac{{{{K}_{{{\text{I}c}}}}}}{{{{\sigma }_{{\text{t}}}} + {{T}_{{zz}}}}}} \right)}^{2}}.$$
(5)
Fig. 2.
figure 2

Direction of crack growth.

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It should be noted that formula (5) was obtained according to the Tresca–Saint-Venant plasticity hypothesis using expressions (1) to write the equivalent stress. In addition, in the criterion of maximum tangential stresses, which takes into account only stresses Тхx [7], the size of the prefracture zone is also determined based on the Tresca–Saint-Venant plasticity hypothesis for a normal separation crack. Moreover, this expression coincides with the Irwin correction for the plastic region [10].

Substituting (5) into (4) leads to the final expression for calculating θ*

$${{K}_{{\text{I}}}}\sin \theta {\kern 1pt} {\text{*}} + {{K}_{{{\text{II}}}}}\left( {3\cos \theta {\kern 1pt} \text{*} - 1} \right) - \frac{{16(1 - 2\mu ){{K}_{{{\text{I}c}}}}}}{{3\left( {{{\sigma }_{{\text{t}}}} + {{T}_{{zz}}}} \right)}}{{T}_{{xx}}}\sin \frac{{\theta {\kern 1pt} \text{*}}}{2}\cos \theta {\kern 1pt} \text{*} = 0.$$
(6)

The failure criterion will be the condition under which the maximum tangential stress is equal to the local strength of the material. In this case, the stress is calculated by formula (2) for the angle θ = θ* and the distance r from the crack apex, which is equal to the size of the prefracture region

$${{\sigma }_{\theta }}({{r}_{c}},{{\theta }_{*}}) = \frac{1}{{\sqrt {2\pi {{r}_{c}}} }}\cos \frac{{\theta {\kern 1pt} \text{*}}}{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{\theta {\kern 1pt} {\text{*}}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin \theta {\kern 1pt} \text{*}} \right] + {{T}_{{xx}}}{{\sin }^{2}}\theta {\kern 1pt} \text{*} = {{\sigma }_{0}}.$$
(7)

We calculate the local strength \({{\sigma }_{0}}\) of the material according to [9] and taking into account the Тxx and Tzz stresses

$${{\sigma }_{0}} = - \frac{{{{T}_{{xx}}} - 2{{T}_{{zz}}}}}{{2(1 - 2\mu )}} + \sqrt {{{{\left( {\frac{{{{T}_{{xx}}} - 2{{T}_{{zz}}}}}{{2(1 - 2\mu )}}} \right)}}^{2}} + \frac{{\sigma _{{\text{t}}}^{2} - T_{{xx}}^{2} - T_{{xx}}^{2} + {{T}_{{xx}}}{{T}_{{zz}}}}}{{{{{(1 - 2\mu )}}^{2}}}}} .$$
(8)

Substituting the local strength of the material defined by (8) into Eq. (7) leads to the expression

$$\begin{gathered} {{\sigma }_{\theta }}\left( {r,\theta {\kern 1pt} \text{*}} \right) = \frac{{\left( {{{\sigma }_{{\text{t}}}} + {{T}_{{zz}}}} \right)}}{{{{K}_{{{\text{I}c}}}}(1 - 2\mu )}}\cos \frac{{\theta {\kern 1pt} \text{*}}}{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{\theta {\kern 1pt} \text{*}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin \theta {\kern 1pt} \text{*}} \right] \\ = - \frac{{{{T}_{{xx}}} - 2{{T}_{{zz}}}}}{{2(1 - 2\mu )}} + \sqrt {{{{\left( {\frac{{{{T}_{{xx}}} - 2{{T}_{{zz}}}}}{{2(1 - 2\mu )}}} \right)}}^{2}} + \frac{{\sigma _{{\text{t}}}^{2} - T_{{xx}}^{2} - T_{{xx}}^{2} + {{T}_{{xx}}}{{T}_{{zz}}}}}{{{{{(1 - 2\mu )}}^{2}}}}} - {{T}_{{xx}}}{{\sin }^{2}}\theta {\kern 1pt} \text{*}{\kern 1pt} . \\ \end{gathered} $$

This expression can be recast into the form

$$\begin{gathered} \frac{{2\left( {{{\sigma }_{{\text{t}}}} + {{T}_{{zz}}}} \right)}}{{\sqrt {{{{\left( {{{T}_{{xx}}} - 2{{T}_{{zz}}}} \right)}}^{2}} + 4{{{\left( {\sigma _{{\text{t}}}^{2} - T_{{xx}}^{2} - T_{{zz}}^{2} + {{T}_{{xx}}}{{T}_{{zz}}}} \right)}}^{2}}} - {{T}_{{xx}}}[1 + 2(1 - 2\mu ){{{\sin }}^{2}}{{\theta }_{*}}] + 2{{T}_{{zz}}}}} \\ \times \;\cos \frac{{{{\theta }_{*}}}}{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{{{\theta }_{*}}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin {{\theta }_{*}}} \right] = {{K}_{{{\text{I}c}}}}. \\ \end{gathered} $$

Then, denoting the left side of the equation as Keff, we can obtain an expression for the effective SIF,

$$\begin{gathered} {{K}_{{{\text{eff}}}}} = \frac{{2\left( {{{\sigma }_{{\text{t}}}} + {{T}_{{zz}}}} \right)}}{{\sqrt {{{{\left( {{{T}_{{xx}}} - 2{{T}_{{zz}}}} \right)}}^{2}} + 4{{{(\sigma _{{\text{t}}}^{2} - T_{{xx}}^{2} - T_{{zz}}^{2} + {{T}_{{xx}}}{{T}_{{zz}}})}}^{2}} - } \;{{T}_{{xx}}}[1 + 2(1 - 2\mu ){\text{si}}{{{\text{n}}}^{2}}{{\theta }_{*}}] + 2{{T}_{{zz}}}}} \\ \times \;\cos \frac{{{{\theta }_{*}}}}{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{{{\theta }_{*}}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin {{\theta }_{*}}} \right]. \\ \end{gathered} $$

To simplify this expression, we introduce the ratio of T-stresses into the yield strength. Then we get the following fracture criterion:

$$\begin{gathered} {{K}_{{\operatorname{eff} }}} = \frac{{2\left( {1 + {{\chi }_{z}}} \right)}}{{\sqrt {{{{\left( {{{\chi }_{x}} - 2{{\chi }_{z}}} \right)}}^{2}} + 4{{{(1 - \chi _{x}^{2} - \chi _{z}^{2} + {{\chi }_{x}}{{\chi }_{z}})}}^{2}} - } \;{{\chi }_{x}}[1 + 2(1 - 2\mu ){{{\sin }}^{2}}\theta {\kern 1pt} \text{*}] + 2{{\chi }_{z}}}} \\ \, \times \cos \frac{{{{\theta }_{*}}}}{2}\left[ {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{\theta {\kern 1pt} \text{*}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin \theta {\kern 1pt} \text{*}} \right] = {{K}_{{{\text{Ic}}}}}, \\ \end{gathered} $$
(9)

where \({{\chi }_{x}} = {{T}_{{xx}}}{\text{/}}{{\sigma }_{{\text{t}}}}\), \({{\chi }_{z}} = {{T}_{{zz}}}{\text{/}}{{\sigma }_{{\text{t}}}}\).

We use the obtained criterion to determine the fracture strength σfrac in the case of a tensile-stressed plate with the height 2H, width 2W, and thickness 2t, with a through crack of length 2l, inclined relative to the horizontal at the angle α (Fig. 3). Obviously, for such a crack, SIFs of types I and II will arise; i.e., the crack will be of a mixed type.

Fig. 3.
figure 3

Diagram of a tensile plate with an inclined crack.

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To simplify the procedure for creating a finite element model, two problems were solved instead of the original one: problems of tension and pure shear of a plate with a transverse crack. For a plate with a transverse crack, it is much easier to generate a finite element mesh than for an inclined one, especially if one needs to consider plates with different crack inclination angles.

In the calculation, the author’s finite element complex [9] was used.

For these tasks, the stresses were determined from the equilibrium equations for the tensile and shear stresses \({{\sigma }_{\alpha }}\) and \({{\tau }_{\alpha }}\) in the plane of an inclined crack, respectively [11],

$${{\sigma }_{\alpha }} = \sigma {{\cos }^{2}}\alpha ,\quad {{\tau }_{\alpha }} = \frac{1}{2}\sigma \sin 2\alpha .$$
(10)

In the first task, a plate with a transverse crack was tensile-stressed by the normal stresses \({{\sigma }_{\alpha }}\). It is obvious that in this case we will get a crack of the normal separation (type I). In the second problem, a plate with a transverse crack was considered, which is in the pure shear condition. Shear stresses were set equal to \({{\tau }_{\alpha }}\). In this case, we get a transverse shear crack (type II). In this regard, KI was calculated from the solution of the first problem, and KII was obtained from the solution of the second one.

It is obvious that, with such a stress specification, the total stress state in the vicinity of the crack vertex, from which the SIF and T stresses are calculated, will not differ much from the initial one.

To check the validity of this approach, the values of KI and KII were calculated in a plate with an inclined crack of a half-length l = W/2 and a height-to-width ratio of 5/4. For this crack, the SIF values are given in the handbook [12] for some inclination angles. The plate thickness was taken equal to 1/20 of the width.

Due to the symmetry of the first problem and the oblique symmetry of the second one with respect to three planes, only one-eighth of the plate was considered (Fig. 4). First, the plate was subdivided into rectangular parallelepipeds. Moreover, the plate was divided into 20 steps in width and height and five steps in thickness. The steps in height and thickness were constant, while they were different in width in the crack region and outside it. Each parallelepiped was divided into six tetrahedral simplex elements [13]. The number of finite elements in the model was 12 000. After calculating the constant stresses and strains in tetrahedral finite elements, they were averaged over parallelepipeds. When solving the first task, the faces of the plate, coinciding with the coordinate planes xy and yz, were fixed in the z and x directions, respectively. The face coinciding with the xz plane was fixed in the y direction, but only outside the crack zone. When solving the second problem, the plate face coinciding with the xy coordinate plane, as in the first case, was fixed in the direction of the z axis. The face coinciding with the yz coordinate plane was fixed in the direction of the y axis. The face coinciding with the xz plane was fixed in the x direction outside the crack zone. In the first problem, loading was carried out along the direction of the y axis at the nodes facing the face parallel to the xz plane. The nodal forces were calculated by the stress \({{\sigma }_{\alpha }}\). In the second task, the nodal forces calculated by \({{\tau }_{\alpha }}\) were applied at the nodes lying on the face parallel to xz in the direction of the x axis and in the direction of the y axis on the face parallel to yz.

Fig. 4.
figure 4

Calculation scheme.

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The SIF calculation was carried out with the help of the asymptotic method by the formulas using the displacement values at points lying on the crack edges in the middle plane in the vicinity of the crack front [10]

$$K_{{\text{I}}}^{{(i)}} = \frac{{E{{{v}}_{i}}}}{{4\kappa }}\sqrt {\frac{{2\pi }}{{{{r}_{i}}}}} ,\quad K_{{{\text{II}}}}^{{(i)}} = \frac{{E{{u}_{i}}}}{{4\kappa }}\sqrt {\frac{{2\pi }}{{{{r}_{i}}}}} ,$$

where \(\kappa \) is the coefficient depending on the type of the planar problem of elasticity theory: \(\kappa = 1 - {{\mu }^{2}}\) for the plane strain state and \(\kappa = 1\) for the plane stress state (PSS); \(K_{{\text{I}}}^{{(i)}}\), \(K_{{{\text{II}}}}^{{(i)}}\) and \({{{v}}_{i}}\), \({{u}_{i}}\) are stress intensity factors and displacements of the first (I) and second (II) types along the x and y axes at the i-th point, respectively.

Based on the extrapolation method [14], the values \(K_{{\text{I}}}^{{(i)}}\) and \(K_{{{\text{II}}}}^{{(i)}}\) were calculated at five points. In this case, the SIF values at the last three points fit on a line close to a straight line, which was extrapolated up to the axis passing through the crack tip. This value was accepted as the SIF.

Table 1 presents the results of comparing the values obtained by the finite element method (FEM) with the data from [12] for a crack inclination angle of 22.5°. Table 1 shows that the numerical values are in good agreement with the data from the handbook. This fact indicates that, in the vicinity of the crack front, the stress strain state, according to which the SIF and T-stresses are calculated, for the original computation model with an inclined crack, is close to the total stress strain states in the two computation models with transverse cracks. It should be noted that the Txx and Tzz stresses arise in both computation schemes, and for a plate with an inclined crack, these values are determined using the principle of superposition, i.e., as the sum of T-stresses for two computation schemes.

Table 1. Comparison of the calculated SIF values with reference data
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For each scheme, the stresses \({{T}_{{xx}}}\) were calculated, according to (1), from the stresses in the crack plane (\(\theta \) = 0) using the formula

$${{T}_{{xx}}} = {{\sigma }_{x}} - {{\sigma }_{y}}.$$
(11)

Just as in the calculation of SIF, the extrapolation method was used to find the stress Txx [14]. In this case, the Txx stresses were calculated by formula (11) at five points on the line of the crack continuation to a given point of the front. In this case, as in the calculation of SIF, the values for the last three points fit on a straight line. Continuation of this straight line to the crack vertex determined the value of the Txx stress at a given point of the crack front. The values of stresses Tzz at this point were found by formula (1) from stresses Txx and strains εz at this point.

The algorithm for numerical calculation of the fracture stress σfrac of the plate is as follows:

(1) An arbitrary value of the external stress \(\,\sigma \) is taken, for example, to be 200 MPa. For a specified crack inclination angle α, stresses \({{\sigma }_{\alpha }}\) and \({{\tau }_{\alpha }}\) are determined by formulas (10). Using the finite element method, KI, KII, Txx, and Tzz are calculated. Moreover, as already noted, the total Txx and Tzz stresses were determined by summing the Txx and Tzz stresses for the two computation schemes.

(2) Using the iterative solution of Eq. (6), the crack propagation angle θ* is calculated at the most dangerous point of the crack front, which lies on the median plane. At this point, effective SIF \(K_{{\text{I}}}^{{{\text{eff}}}}\) is determined by formula (9). The obtained value is compared with fracture toughness KIс.

(3) The magnitude of external stress is corrected by the formula

$$\sigma {\kern 1pt} ' = {{K}_{{{\text{Ic}}}}}{\text{/}}K_{{\text{I}}}^{{{\text{eff}}}}\sigma .$$

(4) Since the parameters of linear fracture mechanics are directly proportional to the external load, the values of KI, KII, Txx, and Tzz are multiplied by \({{K}_{{{\text{Ic}}}}}{\text{/}}K_{{\text{I}}}^{{{\text{eff}}}}\).

(5) The procedure is repeated starting from point 2 for the corrected values of KI, KII, Txx, and Tzz. The iterative process ends when the effective SIF differs from the fracture toughness by less than 1%.

It should be noted that the need for iterations arises due to the fact that KI, KII, Txx, and Tzz depend linearly on the tensile stress σ, and the effective SIF, according to (9), has a nonlinear dependence on χx and χz, i.e., from T-stresses. In addition, the angle θ* included in expression (9) depends nonlinearly on the parameters KI and KII of fracture mechanics and stresses Txx and Tzz.

RESULTS

As an illustration of the capabilities of the given algorithm, Figs. 5 and 6 show the distribution of stresses Txx and Tzz along the thickness of the plate for a square plate with a side of 200 mm and a through inclined coaxial crack 100 mm in length. The plate is tensile-stressed under constant stress σ = 200 MPa at an angle α = 30° (Fig. 3).

Fig. 5.
figure 5

Distribution of Тxx stresses along the thickness of the plate with a thickness of (1) 20, (2) 40 , and (3) 100 mm.

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Fig. 6.
figure 6

Distribution of Тzz stresses over the thickness of the plate: (1) 10, (2) 20, (3) 50 mm.

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The relative coordinate η is plotted in Figs. 5 and 6 along the abscissa axis. It is equal to the ratio of the distance from the free surface of the plate to the middle point (at half the thickness).

One can see from Fig. 5 that the maximum (by modulus) stresses Тхх appear on the surface of the plate, and the minimum ones, on the middle plane. The distribution of stresses Тхх is barely independent of the plate thickness. This is quite expected, since the Тхх stresses are influenced by the geometry of the body, the crack size, and the loading scheme, and not by thickness. As seen from Fig. 6, the maximum stresses Тzz (by modulus), as well as Тхх, arise on the surface of the plate. Moreover, the values of stresses Тzz are less in modulus than the values of the Тхх stresses. It should be noted that, unlike Тхх, the stresses Тzz depend significantly on the thickness of the plate and decrease in absolute value with increasing the thickness of the plate. Calculations showed that, starting from a thickness of 10 mm or less, the stresses Тхх and Тzz are nearly constant across the thickness. The change is no more than three percent.

To study the effect of the crack inclination angle on the fracture stress σfrac, calculations were carried out for the values α = 30°, 45°, and 60°. We considered, the same as previously, a stretched square plate with a thickness of 10 mm. The fracture toughness \({{K}_{{{\text{Ic}}}}}\) was taken to be 50 MPa \(\sqrt {\text{m}} \). Moreover, the fracture stress was determined using the proposed two-parameter MTS fracture criterion and the classical one-parameter MTS criterion. In the latter case, the angle θ* of the crack propagation and the effective SIF are determined by the formulas [1]

$$\theta {\kern 1pt} \text{*} = 2\arctan \left( {\frac{{1 - \sqrt {1 + 8{{\lambda }^{2}}} }}{{4\lambda }}} \right),\quad {{K}_{{\operatorname{eff} }}} = \cos \frac{{\theta {\kern 1pt} \text{*}}}{2}\left( {{{K}_{{\text{I}}}}{{{\cos }}^{2}}\frac{{\theta {\kern 1pt} \text{*}}}{2} - \frac{3}{2}{{K}_{{{\text{II}}}}}\sin \theta {\kern 1pt} \text{*}} \right).$$

The calculation results are summarized in Table 2. The SIF values and T-stresses are presented in the case of an external stress equal to the fracture stress.

Table 2. Effect of crack angle on fracture stress
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As the crack inclination angle α increases, the SIF values decrease (Table 2). This occurs both when using a one-parameter criterion for the maximum tangential stress and a two-parameter criterion. Moreover, in the former case, the SIF values corresponding to the fracture stress are smaller. This is due to the fact that the two-parameter criterion takes into account T stresses, which reduces the effective SIF. In this case, the moduli of the T stresses decrease with increasing the angle α. A similar conclusion was obtained in [4] when studying T stresses in a diametrically compressed disk with a central through inclined crack. Just as in [4], it turned out that, for all crack inclination angles considered, the modulus of the Tzz stress is less than that of the Txx stress. Due to the decrease in T stresses by the modulus, the effective SIF decreases less with increasing angle α. Therefore, the fracture stress calculated using the one-parameter and two-parameter criteria differs less with increasing angle α. Thus, the difference is 23.9% and 17.3% for α = 30° and α = 60°, respectively. The most important conclusion is that the fracture stress σfrac calculated by the one-parameter criterion is underestimated by an average of 20%. The crack propagation angle θ* is negative for two calculation options. Moreover, when using the two-parameter criterion, its dependence on the angle α is stronger. The angle \(\theta {\kern 1pt} \text{*}\) decreases in absolute value by 24.2% with changing α from 30° to 60° when using the one-parameter criterion, and almost doubles in absolute value when using the two-parameter criterion.

To study the effect of plate thickness on the fracture stress, the same plate was considered as in the study of the effect of the crack angle. A crack 100 mm long, inclined at an angle of 30°, was considered. The plate thickness was taken to be 10, 20, and 100 mm. The results are summarized in Table 3. When using the one-parameter criterion (Table 3), the fracture stress σfrac and crack propagation angle θ* are almost independent of the plate thickness. This fact is explained by the fact that SIF does not depend on the thickness of the plate for cracks of both type I and type II. A small difference (less than 1%) arises due to the error in the numerical calculation.

Table 3. Effect of plate thickness on fracture stress
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When using the two-parameter criterion, a decrease in fracture stress is observed with increasing plate thickness due to a significant decrease in the Tzz stress modulus. The crack propagation angle increases in modulus with increasing plate thickness. Fracture stresses predicted by the one-parameter criterion do not depend on the thickness of the plate and are greatly underestimated. This indicates the expediency of using the developed fracture criterion for a more accurate assessment of the static crack resistance of parts with cracks of the I + II mixed type.

CONCLUSIONS

In this paper, the fracture criterion for a mixed type I + II crack was developed. Using the algorithms developed, the static crack resistance of a tensile-stressed plate of various thickness with an inclined through crack was assessed. As a result of this study, the following conclusions can be made. (1) The developed fracture criterion makes it possible to take into account the biaxial constraint of the deformations along the front of a mixed type crack by introducing the T-stress components into the criterion equation, namely nonsingular Тхх and Тzz stresses in the plane of the crack, directed perpendicular and parallel, respectively. Introducing nonsingular stresses Тzz into consideration makes it possible to take into account the thickness of the plate. (2) A decrease in the angle of inclination of the initial crack and an increase in the thickness of the plate lead to a decrease in the fracture stress. (3) The one-parameter criterion for maximum tangential stress, as well as the two-parameter criterion, which includes only Тхх stresses, does not allow the thickness of the plate to be taken into account when calculating the fracture stress. (4) The fracture stresses predicted by the one-parameter fracture criterion are greatly underestimated, especially for thin plates. (5) The developed two-parameter fracture criterion allows one not only to determine the fracture stress, but also to predict the trajectory of the crack propagation, taking into account the plate thickness.