## 2 Methodology
### Displacement Response of a Circumferential Crack  
When an incident wave encounters a discontinuity at the boundary between two different media, it generates a transmitted wave and a reflected wave. The cracks do not induce a phase change in reflected, transmitted, or incident waves [35]. The external excitation position is set as the origin of axial coordinate \(x\), and the distance between this external excitation point and the section containing the crack is \(L_{1}\). In each mode \(n\), four waves are in the positive direction and four in the negative direction in the axial coordinate. Therefore, four incident waves and four reflected waves are in the crack's left region (\(0\leq x\leq L_{1}\)). The superposition of these eight waves for circumferential mode \(n\) yields axial displacement \(u_{1}\) in the area, given by the following equation. Symbol \(U_{nm}\) (\(m=1,2,3,\dots,16\)) represents the wave amplitude.  
\[\begin{split}& u_{1}=\sum_{m=1}^{4}U_{nm}\cos n\theta e^{i( \omega t-k_{nm}x)}\\ &+\sum_{m=5}^{8}U_{nm}\cos n\theta e^{i\left(\omega t-k_{n(m-6)}( x-L_{1})\right)}\end{split} \tag{21}\]  
On the right side (\(x\geq L_{1}\)) of the crack, the displacement response is generated by the superposition of four transmitted waves. In addition, the displacement response due to the crack occurs in the angular range (\(0^{\circ}\leq\theta\leq 45^{\circ}\)) in which the crack is distributed, as shown in Figure 2. Then, axial displacement \(u_{2}\) at circumferential mode \(n\) is shown below.  
\[u_{2}=\sum_{m=9}^{12}U_{nm}\cos n\theta e^{i\left(\omega t+k_{n(m-8)}(x-L_{1}) \right)} \tag{22}\]  
In the angular range (\(45^{\circ}\leq\theta\leq 180^{\circ}\)) of the non-crack part on the crack's location cross-section, its right-side displacement response is the superposition of four waves generated by the excitation and remains unchanged. Axial displacement \(u_{3}\) of circumferential mode \(n\) in this region is as follows:  
\[u_{3}=\sum_{m=1}^{4}U_{nm}\cos n\theta e^{i\left(\omega t-k_{nm}x\right)} \tag{23}\]  
The displacement response in the area (\(x\leq 0\)) to the right of the excitation point is the superposition of four waves. The area's axial displacement \(u_{4}\) at circumferential mode \(n\) is shown below.  
\[u_{4}=\sum_{m=13}^{16}U_{nm}\cos n\theta e^{i\left(\omega t+k_{n(m-12)}x\right)} \tag{24}\]  
The displacement, force, and moment are equal on either side of the section containing the excitation point (\(x=0\)). Also, the force and moment are identical on either side of the section containing the crack (\(x=L_{1}\)). Equation (20) presents the difference between the displacements on the two sides of the area where the crack is located (\(x=L_{1}\) and \(0^{\circ}\leq\theta\leq 45^{\circ}\)). The details of these equivalence conditions are  
at \(x=0\),  
\[u^{R}=u^{L},\partial w^{R}/\partial x=\partial w^{L}/\partial x,w^{R}=w^{L},v^ {R}=v^{L}, \tag{25}\]  
\[N_{x}^{R}=N_{x}^{L},M_{x}^{R}=M_{x}^{L},S_{x0}^{R}=S_{x0}^{L},S_{x}^{R}=S_{x} ^{L}+F, \tag{26}\]  
at \(x=L_{1}\),  
\[N_{x}^{R}=N_{x}^{L},M_{x}^{R}=M_{x}^{L},S_{x}^{R}=S_{x}^{L},S_{x0}^{R}=S_{x0}^{ L} \tag{27}\]  
The wave amplitudes can be obtained by substituting Equations (21), (22) and (24) into the displacement and stress conditions above. Calculating displacements in the circumferential and radial directions is similar to the equations and methods of axial direction.