## 4 Analysis of Factors Influencing Segment Fracture Width
### Effect of Thrust Magnitude on the Fracture Width of a Segment  
Table 2 shows that when the total thrust is close to 20,000 kN, the fracture width of the segment in the field test is close to the standard limit value of 0.2 mm. This section further explores the influence of thrust magnitude on the fracture width of segments. The total thrust is set to 20,000-24,000 kN. The simulation results for the axial, tangential and equivalent fracture widths of the segment are listed in Table 3. The simulated figures of the equivalent fracture widths with different total thrusts are shown in Figure 16.  
Figure 16: _Cont._  
Figure 15: Numerical simulation figure of segment cracking and field test results. Figure 16: Numerical simulation figure of fracture width under total thrust 20,000–24,000 kN. \begin{table}
\begin{tabular}{c c c c c c c c} \hline
**Total Thrust/kN** & \(\epsilon_{xx}\) & \(\epsilon_{yy}\) & \(\epsilon_{zz}\) & \(\gamma_{xy}\) & \(\gamma_{yz}\) & \(\gamma_{xx}\) & \(\epsilon_{eq}\) \\ \hline
20,000 & 0.004 & 0.13 & 0.008 & 0.02 & 0.02 & 0.004 & 0.13 \\
21,000 & 0.004 & 0.15 & 0.01 & 0.02 & 0.02 & 0.005 & 0.15 \\
22,000 & 0.005 & 0.17 & 0.02 & 0.02 & 0.03 & 0.006 & 0.17 \\
23,000 & 0.006 & 0.19 & 0.03 & 0.02 & 0.03 & 0.007 & 0.19 \\
24,000 & 0.007 & 0.23 & 0.03 & 0.03 & 0.04 & 0.007 & 0.22 \\ \hline \end{tabular}
\end{table} 
Table 3: Simulation value of fracture width of segment with total thrust of 20,000–24,000 kN (mm). As shown in Figure 16e, when the total thrust reaches 24,000 kN, the equivalent fracture width is 0.22 mm. Based on the empirical Equation (4) of segment cracking in Japan's Code for Tunnel Design, the maximum spacing of segment distribution ribs under the total thrust of 20,000-24,000 kN can be calculated, which is listed in Table 4, where \(L_{\text{max}}\) is calculated according to the Code for Design of segment of Shield Tunnels prepared by Japan [10]:  
\[w=L_{\text{max}}\cdot\left(\frac{\sigma_{se}}{E_{\text{s}}}+\varepsilon^{\prime }_{\text{c}sd}\right) \tag{4}\]  
where \(L_{\text{max}}\) is the maximum interval (mm) of the distributed steel bars, and the lower limit is 0.51\(L_{1}\). \(L_{1}\) is calculated according to Equation (5):  
\[L_{1}=1.1\cdot k_{1}\cdot k_{2}\cdot k_{3}\cdot\{4c+0.7\cdot(c_{s}-\varphi)\} \tag{5}\]  
where \(k_{1}\) is the coefficient representing the influence of the steel bar surface shape on the fracture width, and a \(k_{1}\) value of 1.0 is taken for the deformed steel bar. \(k_{2}\) is the influence coefficient of concrete quality on fracture width, and \(k_{2}=\frac{15}{f^{\prime}_{c}+20}+0.7\). \(F^{\prime}_{c}\) is the compressive strength of concrete (N/mm\({}^{2}\)). \(k_{3}\) is the coefficient that represents the influence of the number of steel layers under tension, and \(k3=\frac{5(n+2)}{7n+8}\). \(n\) is the number of layers of steel bars under tension. \(c\) is the thickness of the protective layer (mm). \(c_{s}\) is the center spacing of the rebar (mm). \(\varphi\) is the diameter of the steel bar (mm). \(\sigma_{se}\) is the increase in the stress of the bar (N/mm\({}^{2}\)). \(E_{s}\) is the elastic modulus of the reinforcement (N/mm\({}^{2}\)). To consider the increase in fracture width caused by concrete shrinkage and creep, \(\varepsilon_{\text{c}sd}\)' is generally equal to 1.5 \(\times\) 10\({}^{-8}\). In Equation (4), \(\sigma_{se}\) represents the increase in the stress of the steel bar. Since no specific solution equation is given, the maximum stress at the fracture of the segment is used instead in this paper. In Equation (1), \(L_{\text{max}}\) represents the maximum interval of the distributed steel bars, and the lower limit is 0.51 \(L_{1}\). \(L_{1}\) is calculated according to Equation (3). By comparing the calculation results with the simulation results, it can be concluded that the range of \(L_{\text{max}}\) is (2.3-2.5) \(L_{1}\).