## 3 Formulation of a Slant Crack in the EG Rotor Element
### Formulation of Flexibility Matrix of a Slant Cracked Element  
A finite slant-cracked EG shaft element has been developed in the present work. The cracked EG shaft element as shown in Figure 2b, is considered to have two nodes with six degrees of freedom on each node. The EG rotor element is subjected to an axial force (\(P_{1}\)), shear forces (\(P_{2}\), \(P_{3}\)), bending moments (\(P_{4}\), \(P_{5}\)) and torsional moment (\(P_{6}\)). Nodal vectors \(\{q\}\) are generated by these forces and moments, as shown in Figure 2b. The elliptical cut section view of the slant crack along the crack plane is shown in Figure 2c. The geometry details of the slant crack are:  
\[h=2\sqrt{Ro^{2}-x^{2}\sin^{2}45} \tag{7}\]  
\[b=\sqrt{Ro^{2}-\left(Ro-\alpha\right)\,^{2}}\ /\ \sin(45) \tag{8}\]  
Since the cross-section along the slant crack is an ellipse, which is inclined at \(45^{\circ}\) to the shaft axis, the minor and major axis of the cross-section is \(R\) and \(\frac{R}{\cos(45)}\), respectively. The local flexibility coefficients (LFCs) of a slant cracked shaft element are derived using [PERSON]'s theorem and [PERSON]'s [39] equations are represented as the function of stress intensity factors. The additional displacement and energy equations are expressed as  
\[u_{i}=\frac{\partial U_{c}}{\partial P_{i}} \tag{9}\]  
Figure 2: **(a)** A slant-cracked EG rotor-bearing system **(b)** An EG shaft element with a slant crack. (c) The cut sectional view of slant crack. \[U_{c}=\int_{0}^{\alpha}J_{s}(\alpha)d\alpha \tag{10}\]  
where \(J_{s}(\alpha)\) is strain energy density function (SDEF)  
\[J_{s}(\alpha)=\frac{1}{E^{*}}\left[\left\{\sum\limits_{j=1}^{6}K_{lj}\right\}^{2 }+\left\{\sum\limits_{j=1}^{6}K_{Ilj}\right\}^{2}+m\left\{\sum\limits_{j=1}^{6 }K_{IIIj}\right\}^{2}\right] \tag{11}\]  
where \(E^{*}=E\), for plane stress condition, \(m=1+v\). Stress Intensity Factors (SIFs), \(K_{ij}\); where \(i=I\), \(II\), \(III\) modes and \(j=1,2\dots\) 6, the load index. \[ \begin{array}{c}K_{I1}=\sigma_{1}\sqrt{\pi\alpha}\ F_{1}\left(\frac{\alpha} {h}\right)\\ K_{I5}=\sigma_{5}\sqrt{\pi\alpha}\ F_{2}\left(\frac{\alpha}{h}\right)\\ K_{II3}=\sigma_{3}\sqrt{\pi\alpha}\ F_{I1}\left(\frac{\alpha}{h}\right)\\ K_{III2}=\sigma_{2}\sqrt{\pi\alpha}\ F_{III}\left(\frac{\alpha}{h}\right)\\ K_{II16}=\sigma_{6 III}\sqrt{\pi\alpha}\ F_{III}\left(\frac{\alpha}{h}\right) \end{array} K_{II6}=\sigma_{6 II}\sqrt{\pi\alpha}\ F_{II}\left(\frac{\alpha}{h}\right) \\ K_{III6}=\sigma_{6 III}\sqrt{\pi\alpha}\ F_{III}\left(\frac{\alpha}{h}\right) \end{array}\]  
\[ \begin{array}{c}K_{I2}=K_{I3}=K_{I6}=K_{III}=K_{II2}=K_{II4}=K_{II5}=K_{III1}= K_{III3}=K_{III4}=K_{III5}=0\\ \sigma_{1}=\frac{P_{1}}{\sqrt{2}\pi R^{2}},\ \sigma_{4}=\frac{\sqrt{2}P_{4}x}{\pi R ^{4}},\ \sigma_{5}=\frac{2P_{5}\left(\sqrt{2R^{2}-x^{2}}\right)}{\pi R ^{4}},\sigma_{3}=\frac{kP_{3}}{\sqrt{2}\pi R^{2}},\sigma_{6 II}=\frac{P_{5}x \left(\sqrt{2R^{2}-x^{2}}\right)}{\sqrt{2}\ \pi R^{4}},\sigma_{2}=\frac{kP_{2}}{\sqrt{2}\pi R ^{2}},\ \sigma_{6 III}=\frac{P_{5}x\left(\sqrt{2R^{2}-x^{2}}\right)}{\pi R ^{4}}\\ \end{array} \]  
where,  
\[ \begin{array}{c}F_{1}\Big{(}\frac{\alpha}{h}\Big{)}=\frac{\left(\tan\lambda \right)^{\frac{1}{2}}\left[0.752+2.02\big{(}\frac{\alpha}{h}\big{)}+0.37(1- \sin\lambda)^{3}\right]}{\cos\lambda}\\ F_{2}\Big{(}\frac{\alpha}{h}\Big{)}=\frac{\left(\tan\lambda \right)^{\frac{1}{2}}\left[0.923+0.199(1-\sin\lambda)^{4}\right]}{\cos \lambda}\\ F_{II}\Big{(}\frac{\alpha}{h}\Big{)}=\frac{\left[1.122-0.561\big{(}\frac{ \alpha}{h}\big{)}+0.085\big{(}\frac{\alpha}{h}\big{)}^{2}+0.18\big{(}\frac{ \alpha}{h}\big{)}^{3}\right]}{\left(1-\frac{\alpha}{h}\right)^{\frac{1}{2}}} \\ F_{II}\Big{(}\frac{\alpha}{h}\Big{)}=\left(\frac{\tan\lambda}{\lambda}\right)^{ \frac{1}{2}},\ \lambda=\frac{\pi\alpha}{2h}\\ \end{array}