# A Macro-Micro Damage Model for Rock under Compression Loading
## 2 Wing Crack Model Analysis
### Improved Wing Crack Model  
In the process of deriving expression of the stress intensity factor \(K_{\rm I}^{\rm wing}\), the stress intensity factor \(K_{\rm I}^{\rm wing}\) at the wing crack tip can be simplified into the superposition of the two terms, as showed in Figure 2.  
1. A component \(K_{\rm I}^{\rm isol}\) for the two straight wing cracks of common length \(l\), regarded as a single isolated straight crack of length 2\(l\), and subjected to the far field stresses. 2. A component \(K_{\rm I}^{\rm infl}\) due to effective shear stress induced by the presence of the main crack subjected to the far field stresses. The stress intensity factor \(K_{\rm I}^{\rm wing}\) at wing crack tip can be expressed as:  
\[K_{\rm I}^{\rm wing}=K_{\rm I}^{\rm isol}+K_{\rm I}^{\rm infl} \tag{6}\]  
Figure 2: Schematic diagram of the superposition technique used to calculate the \(K_{\rm I}^{\rm wing}\) at wing crack tip subjected to far field stress: (**a**) \(K_{\rm I}^{\rm wing}\) at the wing crack tip; (**b**) \(K_{\rm I}^{\rm isol}\) for the straight wing crack of common length 2\(l\); and (**c**) \(K_{\rm I}^{\rm infl}\) due to effective shear stress. It can be seen from Figure 2b the \(K_{\rm I}^{\rm isol}\) is determined by the following equation [23]:  
\[K_{\rm I}^{\rm isol}=-\frac{1}{2}\{(\sigma_{1}+\sigma_{3})+(\sigma_{1}-\sigma_{ 3})\cos[2(\theta+\beta)]\}\sqrt{\pi l} \tag{7}\]  
The calculation expression of \(K_{\rm I}^{\rm infl}\) is derived as below. By rotating main crack \(\theta\) angle, equivalent straight crack of length \(2(a+l)\) is obtained, the direction of which is the same as that of the wing crack, as shown in Figure 3. Equivalent straight crack of length \(2(a+l)\) consists of main crack of length \(2a\) in the middle and straight wing crack of length \(2l\) where a straight wing crack of length \(l\) is attached to the left side of main crack, and another \(l\) in the right side of main crack. The main crack is subjected to the normal stress \(\sigma_{\rm eq}\) and the shear stress \(\tau_{\rm eq}\). The stress intensity factor of the equivalent straight crack in Figure 3b induced by the normal stress \(\sigma_{\rm eq}\) is given [24]. \[K_{\rm I}^{\rm infl}=2\sigma_{\rm eq}\sqrt{\frac{a+l}{\pi}}\mathrm{arc}\sin( \frac{a}{a+l}) \tag{8}\]  
When wing crack length is rather short (\(l{\rightarrow}0\)), explicit form for \(K_{\rm I}^{\rm infl}\) for vanishingly small wing crack has been developed by [PERSON] and [PERSON] [25]. \[K_{\rm I}^{\rm infl}=\frac{3}{4}(\pi a)^{\frac{1}{2}}\tau_{eff}(\sin\frac{ \theta}{2}+\sin\frac{3\theta}{2}) \tag{9}\]  
Considering that wing length is rather short (\(l{\rightarrow}0\)), the expression of Equation (8) is given as:  
\[K_{\rm I}^{\rm infl}=\sigma_{\rm eq}\sqrt{\pi a} \tag{10}\]  
Comparing Equation (9) with (10), the following equation can be obtained. \[\sigma_{\rm eq}=\frac{3}{4}\tau_{eff}(\sin\frac{\theta}{2}+\sin\frac{3\theta} {2}) \tag{11}\]  
Considering wing crack length is extremely long (\(l{\rightarrow}\infty\)), equivalent straight crack can be considered to be subjected to a pair of concentrated force \(2a\sigma_{\rm eq}\). The length of equivalent straight crack is about \(2l\). The expression of \(K_{\rm I}^{\rm infl}\) is given as [24]:  
\[K_{\rm I}^{\rm infl}=\frac{2a\sigma_{\rm eq}}{\sqrt{\pi l}} \tag{12}\]  
\[K_{\rm I}^{\rm wing}=K_{\rm I}^{\rm isol}+K_{\rm I}^{\rm infl} \tag{6}\]0 Simplification of the wing crack with \(\tau_{eff}\) acting on the main crack; (**a**) \(K_{\rm I}^{\rm infl}\) due to effective shear stress; (**b**) corresponding simplified diagram. As \(l\)\(\rightarrow\)\(\infty\), the above equation can also be obtained from Equation (8) by taking \(l\)\(\rightarrow\)\(\infty\), The following equation can be obtained.