## 2 Theoretical Modelling Establishment of the Hanger Replacement Process
### Calculation of Hanger Removal Process
#### 2.2.2 The \(i\)th (\(i=1,2,\ldots,N\)) Time Pocket Hanging  
Let the pocket hanging force be \(T_{i}^{d}\), the internal force of the old hanger be \(F_{i}^{d}\), the stress-free length of the pocket hanging hanger be \(L_{i}^{d}\), and the displacement in the process of the \(i\)th time pocket hanging be \(x_{i}^{d}\) after the \(i\)th pocket hanging is done.  
For the displacement of the lower end of the pocket hanging hanger, it has:  
\[x_{i}^{d}=\left(\frac{T_{i-1}^{g}L_{i-1}^{fg}}{E^{\prime}A^{\prime}}+L_{i-1} ^{fg}\right)-\left(\frac{T_{i}^{d}L_{i}^{d}}{E^{\prime}A^{\prime}}+L_{i}^{d} \right), \tag{7}\]  
Similarly, for the lower end of the old hanger, we can obtain the following equation:  
\[x_{i}^{d}=\frac{\left(F_{i-1}^{g}-F_{i}^{d}\right)L}{EA_{i-1}^{g}}, \tag{8}\]  
According to the equilibrium of forces:  
\[F_{i}^{d}+T_{i}^{d}=Q_{i}^{d}+G, \tag{9}\]  
where \(Q_{i}^{d}=Q_{i-1}^{g}+kx_{i}^{d}\).  
By combining Equations (7)-(9), the following equations can be obtained:  
\[F_{i}^{d}=F_{i-1}^{g}-A_{i-1}^{g}E\left(F_{i-1}^{g}-G-Q_{i-1}^{g}+T_{i}^{d} \right)\beta, \tag{10}\]  
\[L_{i}^{d}=\frac{E^{\prime}A^{\prime}L\left(G+Q_{i-1}^{g}-T_{i}^{d}-F_{i-1}^{g }\right)\beta+E^{\prime}A^{\prime}L_{i-1}^{fg}+L_{i-1}^{fg}*T_{i-1}^{g}}{\left( T_{i}^{d}+E^{\prime}A^{\prime}\right)}, \tag{11}\]  
\[x_{i}^{d}=L\left(F_{i-1}^{g}-G-Q_{i-1}^{g}+T_{i}^{d}\right)\beta, \tag{12}\]  
where \(\beta=1/\left(Lk+A_{i-1}^{g}E\right)\).  
Figure 5: Beam section force diagram during hanger removal.  
2.2.3. The \(i\)th \((i=1,2,\ldots,N)\) Time Cutting  
Let the area of the old hanger be \(A_{i}^{g}\), the internal force of the pocket hanging hanger be \(T_{i}^{g}\), and the internal force and displacement of the old hanger be \(F_{i}^{g}\) and \(x_{i}^{g}\), respectively, after the \(i\)th cutting of the old hanger is done.  
The displacement of the lower end of the pocket hanging hanger satisfies the following equation:  
\[x_{i}^{g}=\frac{T_{i}^{d}L_{i}^{d}}{E^{\prime}A^{\prime}}-\frac{T_{i}^{g}L_{i }^{d}}{E^{\prime}A^{\prime}}, \tag{13}\]  
Similarly, for the lower end of the old hanger, we can obtain:  
\[x_{i}^{g}=\frac{F_{i}^{d}L}{EA_{i}^{d}}-\frac{F_{i}^{g}L}{EA_{i}^{g}}, \tag{14}\]  
According to the equilibrium of forces, it has:  
\[F_{i}^{g}+T_{i}^{g}=Q_{i}^{g}+G, \tag{15}\]  
where \(Q_{i}^{g}=Q_{i}^{d}+kx_{i}^{g}\).  
Combined with Equations (13)-(15), the following can be obtained:  
\[T_{i}^{g}=\Big{(}A_{i}^{d}A^{\prime}E^{\prime}GL-A_{i}^{g}A^{\prime}E^{\prime }F_{i}^{d}L+A_{i}^{d}A^{\prime}E^{\prime}Q_{i}^{d}L+A_{i}^{d}A_{i}^{g}EL^{ \prime d}_{i}T_{i}^{d}+A_{i}^{d}LL^{\prime d}_{i}T_{i}^{d}k\Big{)}\gamma, \tag{16}\]  
\[F_{i}^{g}=A_{i}^{g}\Big{(}A_{i}^{d}EGL^{\prime d}_{i}+A^{\prime}E^{\prime}F_{ i}^{d}L+A_{i}^{d}EL^{\prime d}_{i}Q_{i}^{d}-A_{i}^{d}EL^{\prime d}_{i}T_{i}^{d}+ LL^{\prime d}_{i}F_{i}^{d}k\Big{)}\gamma, \tag{17}\]  
\[x_{i}^{g}=LL^{\prime d}_{i}\Big{(}A_{i}^{g}G-A_{i}^{g}F_{i}^{d}-A_{i}^{d}Q_{i} ^{d}+A_{i}^{d}T_{i}^{d}\Big{)}\gamma, \tag{18}\]  
where \(\gamma=1/\Big{[}A_{i}^{d}\Big{(}A_{i}^{g}EL^{\prime d}_{i}+A^{\prime}E^{\prime }L+LL^{\prime d}_{i}k\Big{)}\Big{]}\).