## 3 Formulation of a Slant Crack in the EG Rotor Element
### Formulation of Flexibility Matrix of a Slant Cracked Element
\]  
LFCs of a slant-cracked EG shaft element are computed by double integral as  
\[C_{ij}=\frac{\partial^{2}}{\partial P_{i}\partial P_{j}}\int\limits_{0}^{b} \int\limits_{0}^{\alpha_{x}}f_{s}(y)dydx \tag{12}\]  
where \(\alpha_{x}=\sqrt{Ro^{2}-\left(z\sin 45^{\circ}\right)\ ^{2}}\ -\left(Ro-\alpha\right)\) and \(b=\sqrt{Ro^{2}-\left(Ro-\alpha\right)\ ^{2}}/\sin 45^{\circ}\)  
\[ \begin{array}{c}C^{c}{}_{11}=\frac{\sqrt{2}}{\pi R_{0}a}^{\dagger}\int\limits _{0}^{b}\int\limits_{0}^{\alpha_{x}}\frac{y}{E^{\epsilon}(r,T)}F_{1}^{2}\Big{(} \frac{\alpha}{h}\Big{)}dydx\\ C^{c}{}_{14}=\frac{4}{\pi R_{0}a}^{\dagger}\int\limits_{0}^{b}\int\limits_{0}^{ \alpha_{x}}\frac{y\pi}{E^{\epsilon}(r,T)}F_{1}^{2}\Big{(}\frac{\alpha}{h} \Big{)}dydx\\ C^{c}{}_{15}=\frac{4\sqrt{2}}{\pi R_{0}a}^{\dagger}\int\limits_{0}^{b}\int \limits_{0}^{\alpha_{x}}\frac{y\sqrt{R_{0}a}^{2}-x^{2}}{E^{\epsilon}(r,T)}F_{1 }\Big{(}\frac{\alpha}{h}\Big{)}F_{2}\Big{(}\frac{\alpha}{h}\Big{)}dydx\\ C^{c}{}_{22}=\frac{\sqrt{2}}{\pi R_{0}a}^{\dagger}\int\limits_{0}^{b}\int \limits_{0}^{\alpha_{x}}\frac{y(1+v^{\epsilon}(r,T))k^{2}}{E^{\epsilon}(r,T)}F _{III}^{2}\Big{(}\frac{\alpha}{h}\Big{)}dydx\end{array}  \tag{16}\]\[C^{c}{}_{33}=\frac{\sqrt{2}}{\pi{R_{o}}^{4}}\int\limits_{0}^{b}\int\limits_{0}^{a _{x}}\frac{yk^{2}}{E^{\epsilon}(r,T)}F_{II}^{2}\Big{(}\frac{\alpha}{h}\Big{)}dydx \tag{17}\]  
\[C^{c}{}_{44}=\frac{8\sqrt{2}}{\pi{R_{o}}^{8}}\int\limits_{0}^{b}\int\limits_{0}^ {a_{x}}\frac{yx^{2}}{E^{\epsilon}(r,T)}F_{1}^{2}\Big{(}\frac{\alpha}{h}\Big{)} dydx \tag{18}\]  
\[C^{c}{}_{45}=\frac{8\sqrt{2}}{\pi{R_{o}}^{8}}\int\limits_{0}^{b}\int\limits_{0 }^{a_{x}}\frac{yz\sqrt{{R_{o}}^{2}-x^{2}}}{E^{\epsilon}(r,T)}F_{1}\Big{(}\frac {\alpha}{h}\Big{)}F_{2}\Big{(}\frac{\alpha}{h}\Big{)}dydx \tag{19}\]  
\[C^{c}{}_{55}=\frac{16\sqrt{2}}{\pi{R_{o}}^{8}}\int\limits_{0}^{b}\int\limits_ {0}^{a_{x}}\frac{y({R_{o}}^{2}-x^{2})}{E^{\epsilon}(r,T)}F_{2}^{2}\Big{(}\frac {\alpha}{h}\Big{)}dydx \tag{20}\]  
\[C^{c}{}_{62}=\frac{2\sqrt{2}}{\pi{R_{o}}^{6}}\int\limits_{0}^{b}\int\limits_{0 }^{a_{x}}\frac{y(1+v^{\epsilon}(r,T))\sqrt{{R_{o}}^{2}-x^{2}}}{E^{\epsilon}(r,T)}F_{III}^{2}\Big{(}\frac{\alpha}{h}\Big{)}dydx \tag{21}\]  
\[C^{c}{}_{63}=\frac{2}{\pi{R_{o}}^{6}}\int\limits_{0}^{b}\int\limits_{0}^{a_{ x}}\frac{kyz(1+v^{\epsilon}(r,T))}{E^{\epsilon}(r,T)}F_{II}^{2}\Big{(}\frac{ \alpha}{h}\Big{)}dydx \tag{22}\]  
\[C^{c}{}_{66}=\frac{8}{\pi{R_{o}}^{8}}\int\limits_{0}^{b}\int\limits_{0}^{a_{ x}}\frac{(U_{1}+mU_{2})(1+v^{\epsilon}(r,T))}{E^{\epsilon}(r,T)}dydx \tag{23}\]  
\[U_{1}=0.25\sqrt{2}z^{2}yF_{II}^{2}\Big{(}\frac{\alpha}{h}\Big{)},\ U_{2}=0.5 \sqrt{2}\Big{(}1-x^{2}\Big{)}yF_{III}^{2}\Big{(}\frac{\alpha}{h}\Big{)}\]  
\(E^{\epsilon}(r,T)\) is [PERSON]'s modulus and [PERSON]'s ratio is \(v^{\epsilon}(r,T)\). The radial distance is transformed into \(r(x,y)=\sqrt{x^{2}+y^{2}}\) to carry out the integration within the given limits. The flexibility matrix of the cracked EG shaft element is  
\[C_{c}= \begin{bmatrix}C^{c}{}_{11}&0&0&C^{c}{}_{14}&C^{c}{}_{15}&0\\ 0&C^{c}{}_{22}&0&0&0&C^{c}{}_{62}\\ 0&0&C^{c}{}_{33}&0&0&C^{c}{}_{63}\\ C^{c}{}_{14}&0&0&C^{c}{}_{44}&C^{c}{}_{45}&0\\ C^{c}{}_{15}&0&0&C^{c}{}_{45}&C^{c}{}_{55}&0\\ 0&C^{c}{}_{62}&C^{c}{}_{63}&0&0&C^{c}{}_{66}\end{bmatrix}  \tag{24}\]