Under the two assumptions below, Macha, E. and Niesłony [3] have developed a method for determining the position of the critical plane based on the variance of the equivalent stress.

Fatigue fracture is caused by the normal stresses ${\sigma }_{\eta }\left(t\right)$ and the shear stresses ${\tau }_{\eta }\left(t\right)$ acting in the $\stackrel{\to }{s}$ direction on a fracture plane with a normal $\stackrel{\to }{\eta }$ ( Figure 1 ). ${\sigma }_{\eta }\left(t\right)$ and ${\tau }_{\eta }\left(t\right)$ are functions of the components of contrainte ${\sigma }_{ij}\left(t\right)$ ( $i,j=x,y;z$).

The direction $\stackrel{\to }{s}$ on the fracture plane coincides with the mean value of the shear stress maximale ${\tau }_{\eta \max }\left(t\right)$ ( Figure 2 ).

With:

$\stackrel{\to }{n}={\hat{l}}_{\eta }\stackrel{\to }{i}+{\hat{m}}_{\eta }\stackrel{\to }{j}+{\hat{n}}_{\eta \stackrel{\to }{k}}=\frac{{\hat{l}}_1+{\hat{l}}_3}{\sqrt{2}}\stackrel{\to }{i}+\frac{{\hat{m}}_1+{\hat{m}}_3}{\sqrt{2}}\stackrel{\to }{j}+\frac{{\hat{n}}_1+{\hat{n}}_3}{\sqrt{2}}\stackrel{\to }{k}$ (1)

$\stackrel{\to }{s}={\hat{l}}_s\stackrel{\to }{i}+{\hat{m}}_s\stackrel{\to }{j}+{\hat{n}}_s\stackrel{\to }{k}=\frac{{\hat{l}}_1-{\hat{l}}_3}{\sqrt{2}}\stackrel{\to }{i}+\frac{{\hat{m}}_1-{\hat{m}}_3}{\sqrt{2}}\stackrel{\to }{j}+\frac{{\hat{n}}_1-{\hat{n}}_3}{\sqrt{2}}\stackrel{\to }{k}$ (2)

The components of the stress vector are expressed as:

${\stackrel{\to }{\sigma }}_n\left(t\right)={\hat{l}}_n\stackrel{\to }{i}{\hat{l}}_n\stackrel{\to }{j}{\sigma }_{ij}\left(t\right)$, ${\stackrel{\to }{\tau }}_{ns}\left(t\right)={\hat{l}}_{ns}\stackrel{\to }{i}{\hat{l}}_{ns}\stackrel{\to }{j}{\sigma }_{ij}\left(t\right)$ (3)

In the proposed algorithm, the multiaxial stress state is reduced to an equivalent uniaxial stress state. The general formulation of the criterion is:

$\underset{t}{\max }\left\{B{\tau }_{\eta s}\left(t\right)+K{\sigma }_{\eta }\left(t\right)\right\}=F$ [5] (4)

where F, B, K are the constants for selection of a special version of the criterion:

Cas a) criterion of maximum normal stress on the critical plane (B = 0, K = 1)

The equivalent stress expression is:

$\begin{array}{c}{\sigma }_{eq}\left(t\right)={\sigma }_{\eta }\left(t\right)={\sigma }_{eq}\left(t\right)\\ ={\hat{l}}_{\eta }^2{\sigma }_{xx}\left(t\right)+{\hat{m}}_{\eta }^2{\sigma }_{xx}\left(t\right)+{\hat{n}}_{\eta }^2{\sigma }_{xx}\left(t\right)+2{\hat{l}}_{\eta }{\hat{m}}_{\eta }{\sigma }_{xy}\left(t\right)\\ +2{\hat{l}}_{\eta }{\hat{n}}_{\eta }{\sigma }_{xz}\left(t\right)+2{\hat{m}}_{\eta }{\hat{n}}_{\eta }{\sigma }_{yz}\left(t\right)\end{array}$ (5)

Cas b) criterion of maximum shear stress on the critical plane (B = 1, K = 0)

$\begin{array}{c}{\sigma }_{eq}\left(t\right)=2{\tau }_{\eta s}\left(t\right)\\ =2{\hat{l}}_{\eta }{\hat{l}}_s{\sigma }_{xx}\left(t\right)+2{\hat{m}}_{\eta }{\hat{m}}_s{\sigma }_{yy}\left(t\right)+2{\hat{n}}_{\eta }{\hat{n}}_s{\sigma }_{zz}\left(t\right)\\ +2\left({\hat{l}}_{\eta }{\hat{m}}_s+{\hat{l}}_s{\hat{m}}_{\eta }\right){\sigma }_{xy}\left(t\right)+2\left({\hat{l}}_{\eta }{\hat{n}}_s+{\hat{l}}_s{\hat{n}}_{\eta }\right){\sigma }_{xz}\left(t\right)\\ +2\left({\hat{m}}_{\eta }{\hat{n}}_s+{\hat{m}}_s{\hat{n}}_{\eta }\right){\sigma }_{yz}\left(t\right)\end{array}$ (6)

Cas c) Criterion of maximum normal and shear stress on the critical plane (B = 1)

$\begin{array}{c}{\sigma }_{eq}\left(t\right)=\frac{1}{1+K}[\left({\hat{l}}_1^2-{\hat{l}}_3^2+K{\left({\hat{l}}_1+{\hat{l}}_3\right)}^2\right){\sigma }_{xx}\left(t\right)\\ +\left({\hat{m}}_1^2-{\hat{m}}_3^2+K{\left({\hat{m}}_1+{\hat{m}}_3\right)}^2\right){\sigma }_{yy}\left(t\right)\\ +\left({\hat{n}}_1^2-{\hat{n}}_3^2+K{\left({\hat{n}}_1+{\hat{n}}_3\right)}^2\right){\sigma }_{zz}\left(t\right)\\ +2\left({\hat{l}}_1{\hat{m}}_1-{\hat{l}}_3{\hat{m}}_3+K\left({\hat{l}}_1+{\hat{l}}_3\right)\left({\hat{m}}_1+{\hat{m}}_3\right)\right){\sigma }_{xy}\left(t\right)\\ +2\left({\hat{l}}_1{n}_1-{\hat{l}}_3{\hat{n}}_3+K\left({\hat{l}}_1+{\hat{l}}_3\right)\left({\hat{n}}_1+{\hat{n}}_3\right)\right){\sigma }_{xz}\left(t\right)\\ \stackrel{}{}+2\left({\hat{m}}_1{\hat{n}}_1-{\hat{n}}_3{\hat{m}}_3+K\left({\hat{m}}_1+{\hat{m}}_3\right)\left({\hat{n}}_1+{\hat{n}}_3\right)\right){\sigma }_{yz}\left(t\right)]\end{array}$ (7)

The equivalent stress can be understood as the output signal from the linear physical system with six inputs, where the signals representing suitable components of the stress tensor were delivered. Then the equivalent history can be determinned by summation of the products of suitable components tensor $x_k\left(t\right)$ and coefficients $a_k$.

${\sigma }_{eq}\left(t\right)={\displaystyle {\sum }_{k=1}^6{a}_k{x}_k\left(t\right)}$ (8)

where: $x_1={\sigma }_{xx}\left(t\right)$, $x_2={\sigma }_{yy}\left(t\right)$, $x_3={\sigma }_{zz}\left(t\right)$, $x_4={\sigma }_{xy}\left(t\right)$, $x_5={\sigma }_{xz}\left(t\right)$, $x_6={\sigma }_{yz}\left(t\right)$.

and $a_k=a_k\left({\hat{l}}_1,{\hat{m}}_1,{\hat{n}}_1,{\hat{l}}_3,{\hat{m}}_3,{\hat{n}}_3\right)$.

The variance of the equivalent stress is written as:

${\mu }_{{\sigma }_{eq}}={\displaystyle {\sum }_{s,t=1}^6{a}_s{a}_t{\mu }_{xst}}$ [6] (9)

With ${\mu }_{xst}$ [7] the variance-covariance matrix of the variable $x_k$:

${\mu }_{xst}=\left[\begin{array}{ccc}{\mu }_{x11}& \cdots & {\mu }_{16}\\ ⋮& \ddots & ⋮\\ {\mu }_{61}& \cdots & {\mu }_{66}\end{array}\right]$

Generally, the analytical resolution of this method poses a problem then, to eliminate this difficulty, the cosines direction ${\hat{l}}_n,{\hat{m}}_n,{\hat{n}}_n$ are replaced by the trigonometric functions of the three Euler angles ψ, θ, φ ( Figure 2 ).

Considering that the problem is planar and the facets concerned are those with the normal in the plane $\left(\stackrel{\to }{i},\stackrel{\to }{j}\right)$, we have the condition $\stackrel{\to }{k}\cdot \stackrel{\to }{h}={\hat{n}}_3=0$.

With the condition θ = π/2, we have the following matrix of cosine directions:

The algorithm for determining the fracture plane is shown in detail in Figure 3 .

$\left[P\right]=\left[\begin{array}{ccc}\cos \psi \cos \varphi & \cos \psi \sin \varphi & \sin \varphi \\ -\sin \psi \cos \varphi & -\sin \psi \sin \varphi & \cos \psi \\ \sin \varphi & -\cos \varphi & 0\end{array}\right]$

To validate this comparative study for determining the orientation of the fracture

plane, we used fatigue data from structural steel and fatigue tests under cyclic biaxial loading carried out by Rotvel [8] , Nishihara T. and Kawamoto [9] and Achtelik et al. [10] ( Table 1 ).

The experimental fracture plane is defined by the unit normal vector $\stackrel{\to }{h_{ex}}$. The predicted fracture planes are defined by the theoretical unit normal vector $\stackrel{\to }{h_{t1}}$. In the case where several assessed fracture planes is obtained, the most similar to that obtained experimentally is assumed. The closeness of the corresponding cosines directions ${\hat{l}}_{nex},{\hat{m}}_{nex}$ and ${\hat{l}}_{nc},{\hat{m}}_{nc}$ is suitability of the predicting methods. The dot product of this unit vector is calculated to express the agreement (or disagreement) between assessments and tests results.

Table 2 shows the direction of cosines experimentally and calculated.

The criterion for comparison is the percentage number of results.

Figures 4-6 give us a summary of the previous results. In view of the results shown in Table 2 , the variance method for the case of combined loading (case c) gives a very good result in terms of predicting the orientation of the fracture plane. It produces better predictions for tests on carbon steel at 0.35% C, hardened steel at 0.51% C, soft steel at 0.1% C, Grey Cast iron at 3.32 Cu, and duralumin at 3.81%

C than cases a and b. The overall mean value of the dot product is about 0.68 for cas c, 0.26 cas b and 0.23 for cas a ( Figure 5 ). It indicates average deviation angles of 20.37, 42.38 and 31.99 respectively of the deviation between predicted fracture planes against the real ones ( Figure 6 ). Figure 7 shows us the distribution of critical planes according to the direction cosines.