## 4 Simplified Model Calculation  
An analysis of the obtained results indicates that the most serious damage area is located near the void boundary. To verify the cracking position of the specimen, the elastic center method is used to calculate the internal force of the specimen. As shown in Figure 14, the coordinate system is established and the model is simplified to a hinged arch. Since the symmetrical structure bears symmetrical force, there is no shear force, (i.e., \(X_{3}=0\)), and only two forces \(X_{1}\) and \(X_{2}\) should be calculated. The governing equations can be expressed as:  
\[\left\{ \begin{array}{l}\delta_{11}X_{1}+\delta_{12}X_{2}+\Delta_{1P}=0\\ \delta_{21}X_{1}+\delta_{22}X_{2}+\Delta_{2P}=0\end{array} \right. \tag{1}\]  
The term \(\delta_{12}\) is the auxiliary coefficient and can be obtained from Equation (2). \[\delta_{12}=\sum\int\frac{\overline{M}_{1}\overline{M}_{2}}{EI}\;\mathrm{d}s+ \sum\int\frac{\overline{F}_{\mathrm{N1}}\overline{F}_{\mathrm{N2}}}{EA}\; \mathrm{d}s+\sum\int\frac{\mu\overline{F}_{\mathrm{Q1}}\overline{F}_{\mathrm{ Q2}}}{GA}\;\mathrm{d}s \tag{2}\]  
When the structure is individually subjected to the forces \(X_{1}=1\), \(X_{2}=1\), \(X_{3}=1\), we can obtain:  
\[\left\{ \begin{array}{l}\overline{M}_{1}=1,\overline{F}_{\mathrm{N1}}=0, \overline{F}_{\mathrm{Q1}}=0\\ \overline{M}_{2}=y-y_{S},\overline{F}_{\mathrm{N2}}=-\cos\varphi,\overline{F}_ {\mathrm{Q2}}=\sin\varphi\\ \overline{M}_{3}=x,\overline{F}_{\mathrm{N3}}=\sin\varphi,\overline{F}_{ \mathrm{Q3}}=\cos\varphi\end{array} \right. \tag{3}\]  
Introducing Equation (3) into Equation (2) with \(\delta_{12}=\delta_{21}=0\) results in the following expression:  
\[\delta_{12}=\delta_{21}=\int\frac{(1)\times(y-y_{S})}{EI}\;\mathrm{d}s+0+0= \int\frac{y}{EI}\;\mathrm{d}s-y_{S}\int\frac{1}{EI}\;\mathrm{d}s=0 \tag{4}\]  
Figure 14: Schematic of the simplified model. (**a**) Coordinate system; (**b**) simplified model. where \(y_{s}\) is the length of the rigid arm and can be calculated as follows:  
\[y_{s}=\frac{\int\frac{y}{EI}ds}{\int\frac{1}{EI}ds} \tag{5}\]  
For an arch structure, the curvature has little impact on the deformation. Therefore, the displacement of the straight bar can be used to solve the coefficient and free term. \[\left\{ \begin{array}{l}\delta_{11}=\int\frac{1}{EI}ds,\quad\Delta_{1P}=\int \frac{M_{P}}{EI}ds\\ \delta_{22}=\int\frac{(y_{k}-y_{k})^{2}}{EI}ds+\int\frac{\cos^{2}\left[\varphi \right]}{EA}ds,\quad\Delta_{2P}=\int\frac{(y_{k}-y_{s})M_{P}}{EI}ds\end{array} \right. \tag{6}\]  
where \(E\) donates the modulus of elasticity; \(I\) is the moment of inertia of the section; \(A\) is the cross-sectional area; \(\varphi\) represents the central angle between the radius of the calculated point and the Y-axis; \(y_{k}\) is the ordinate of point k; \(M_{P}\) is the bending moment generated by the external load at the point. \(X_{1}\) and \(X_{2}\) can be calculated from Equation (1). The section height of the voided section model is defined as follows:  
\[\mathrm{H}_{V}=\frac{\mathrm{H}_{1}-\mathrm{h}}{\cos\varphi}-\mathrm{H}_{2} \tag{7}\]  
where h denotes the height of the vault void, and H\({}_{1}\) and H\({}_{2}\) are the radius of the outer and inner contours, respectively. Since the height of the cross-section changes in the void area, voided and unvoided sections are separately integrated during the calculation. The bending moment at point K is the sum of the bending moments generated by the forces and moments at this point. This can be mathematically expressed in the form below:  
\[\left\{ \begin{array}{l}M_{K}=X_{1}+(y_{k}-y_{s})X_{2}+M_{KP}\\ F_{\mathrm{NK}}=-X_{2}\cos\varphi+F_{\mathrm{NK}}\end{array} \right. \tag{8}\]  
where \(y_{s}\) is the length of the rigid wall; \(M_{KP}\) is the bending moment produced by the external load on the half structure. Since the external load does not cause a bending moment at the void boundary, \(M_{KP}\) can be set to 0.