## 2 Method of Calculation
### Stress Intensity Factors  
#### 2.2.1 Analytic Solution  
For a closed crack, the singularity of the type I stress component no longer exists due to the pressure closing the crack. At this point, the composite crack is transformed into a special pure type II crack. Figure 2 shows a schematic diagram of the stress on the crack surface.  
For a pure type II crack under an infinite plate, the stress field expression can be derived using the Westergaard stress function [33].  
\[Z_{\rm II}=\frac{\tau_{e}f^{2}}{\sqrt{z^{2}-a^{2}}} \tag{20}\]  
The expression for the stress component is obtained.  
\[\left\{\begin{array}{l}\sigma_{x}=2{\rm Im}Z_{\rm II}+y{\rm Re}Z_{\rm II}^{ \prime}\\ \sigma_{y}=-y{\rm Re}Z_{\rm II}^{\prime}\\ \tau_{xy}={\rm Re}Z_{\rm II}-y{\rm Im}Z_{\rm II}^{\prime}\end{array}\right. \tag{21}\]  
According to the superposition principle, the crack is currently unaffected by any force other than the \"pseudo-force\" on the crack surface. Therefore, the boundary conditions are as follows:  
Figure 2: Schematic diagram of force on cracked surface, where a is the crack half-length, mm; q(x) is the positive stress on the crack surface, MPa; and \(\tau\)(x) is the shear stress on the crack surface, MPa.  
\[|Z|\rightarrow\infty,\sigma_{x}=\sigma_{y}=\tau_{xy}=0 \tag{22}\]  
This results in a calculated form of the stress intensity factors.  
\[K_{\rm II}(\pm)=\frac{\tau_{ef}}{\sqrt{\pi a}}\frac{a\pm s}{\sqrt{a^{2}-s^{2}}} \tag{23}\]  
where \(K_{\rm II}(\pm)\) is the stress intensity factor at both ends of the crack (a negative sign indicates the left end, while a positive sign indicates the right end), MPa\(\bullet\)mm\({}^{1/2}\); \(\tau_{ef}\) represents the effective tangential stress at the crack surface, measured in MPa; and \((s,0)\) denotes an arbitrary point on the crack surface.  
The formula for the stress intensity factor can be expressed as an integral, following the superposition principle.  
\[K_{\rm II}(\pm)=\frac{1}{\sqrt{\pi a}}\int_{-a}^{a}\tau_{ef}(x)\frac{a\pm x}{ \sqrt{a^{2}-x^{2}}}dx \tag{24}\]  
From Equation (24), it can be seen that the parameters required to calculate the stress intensity factor at the crack tip are the crack half-length and the effective tangential stress. Since the crack half-length is a known parameter, the key to solving the stress intensity factor at the crack tip lies in solving the effective tangential stress at the crack surface.  
Since closure of the cracked surface occurs and the cracked surface is subjected to both friction and tangential stresses, the effective tangential stresses on the cracked surface are calculated as shown below:  
\[\tau_{ef}=\tau-\tau_{f} \tag{25}\]  
where \(\tau_{f}\) is the friction stress, MPa.  
The crack surface shear stress can be expressed as follows:  
\[\tau=(\sigma_{x}-\sigma_{y})\sin\beta\cos\beta-\tau_{xy}(\cos^{2}\beta-\sin^{ 2}\beta) \tag{26}\]  
where \(\beta\) is the angle between the local and horizontal coordinates of the crack (\({}^{\circ}\)).  
Friction stress can be expressed as  
\[\tau_{f}=u\sigma_{n}=u\Big{[}\sigma_{x}\sin^{2}\beta+\sigma_{y}\cos^{2}\beta- 2\tau_{xy}\cos\beta\sin\beta\Big{]} \tag{27}\]  
where \(u\) is the coefficient of friction of the cracked surface.  
Combining Equations (25)-(27) gives the formula for calculating the effective tangential stress.  
\[\tau_{ef}=(\sigma_{x}-\sigma_{y})\sin\beta\cos\beta-\tau_{xy}(\cos^{2}\beta- \sin^{2}\beta)-u\Big{[}\sigma_{x}\sin^{2}\beta+\sigma_{y}\cos^{2}\beta-2\tau_ {xy}\cos\beta\sin\beta\Big{]} \tag{28}\]  
The solutions for the stress intensity factor from Equations (24) and (28) are called integral solutions.