# Mechanical Performance of Curved Link-Slab of Simply Supported Bridge Beam
## 2 Theoretical Analysis
### Mechanical Analysis of CLS  
#### 2.2.1 Vehicle Loading  
As the lower part of the CLS is vacant, the extrusion of the beam end displacement on the link-slab is avoided. Therefore, when the main beam undergoes an angular displacement under vehicle load, due to the restraint between the beam end structures, the CLS will also produce the same angular displacement, as shown in Figure 4. In the figure, \(L_{1}\) is the calculated span of the main beam, \(H\) is the height of the main beam, \(\theta\) is the angular displacement at the main beam end, \(\Delta H\) is the longitudinal displacement of the CLS due to the corner of the beam end, and \(l\) is the CLS span length, \(e\) is the distance from the bridge support to the beam end, C is the gap between the adjacent main beams, and \(d\) is the distance between the arch foot and the support.  
As the CLS has less restraint on the main girder, the CLS's restraint on the main girder can be ignored and a single-span SSB bridge approximated. The structural mechanic's diagram multiplication method is used to calculate the maximum angular displacement of the beam end as  
\[\theta=\frac{L_{1}}{6E_{1}I}P_{i}(L_{1}-l_{i})\Bigg{\{}1-\frac{(L_{1}-l_{i})^{ 2}}{L_{1}^{2}}\Bigg{\}} \tag{2}\]  
Figure 4: Deformation of the beam end structure.  
Figure 3: Geometry of the neutral axis of the combined section.  
where \(E_{1}\) = elastic modulus of the main beam at the left end, \(I\) = moment of inertia of the main beam section, \(P_{i}\) = vehicle axle load, and \(l_{i}\) = distance between vehicle axle load \(P_{i}\) and the beam end A.  
As the angular displacement \(\theta\) of the beam end is small when the vehicle is loaded, the longitudinal displacement value of the CLS is approximated as the arc length \(\Delta H\) of the angular displacement, namely:  
\[\Delta H=\frac{\theta}{180}\pi\sqrt{H^{2}+\left(\epsilon-\frac{l-c}{2}\right) ^{2}} \tag{3}\]  
The distance \(y_{s}\) from the elastic center to the vault can be calculated by the elastic center method as  
\[y_{s}=\frac{\int\frac{\Psi}{kT}ds}{\int\frac{1}{kT}ds}=\frac{2\int_{0}^{\varphi _{0}}R(1-\cos\varphi)Rd\varphi}{2\int_{0}^{\varphi_{0}}Rd\varphi}=\frac{R( \varphi_{0}-\sin\varphi_{0})}{\varphi_{0}} \tag{4}\]  
When the hingeless arch feet produce relative displacements, as shown in Figure 2b, the corresponding angular displacement \(\Delta_{1c}\), horizontal displacement \(\Delta_{2c}\), and vertical displacement \(\Delta_{3c}\) will occur at the elastic center O. Solving the unknown force at the center of elasticity, then  
\[\left\{\begin{array}{c}X_{1}=-\frac{2\theta}{2\int_{0}^{\varphi_{0}}\frac{ 1}{kT}Rd\varphi}\\ X_{2}=-\frac{2(f-y_{s})+2\Delta H}{2\int\frac{\Psi^{2}}{kT}ds+2\int\frac{\Psi _{2}\Psi^{2}}{kT}ds}\\ X_{3}=-\frac{1+\theta}{2\int_{0}^{\varphi_{0}}\frac{1}{kT}Rd\varphi}\end{array}\right. \tag{5}\]  
where \(X_{1}\) is the bending moment, \(X_{2}\) is the horizontal force, and \(X_{3}\) is the vertical force. The bending moment of the CLS's vault section and the stress at the upper and lower surfaces are given by  
\[\left\{\begin{array}{c}M=X_{1}-X_{2}(y_{s}+h_{c}-\frac{h}{2})\\ \sigma_{a}=\frac{M}{W_{aa}}=\frac{M}{I_{f}(h-h_{c})}\\ \sigma_{b}=\frac{M}{W_{ab}}=\frac{M}{I_{i}/I_{c}}\end{array}\right. \tag{6}\]  
The stress \(\sigma_{s}\) produced by the axial force \(X_{2}\) is given by  
\[\sigma_{s}=\frac{X_{2}}{A_{0}} \tag{7}\]  
Therefore, the stresses at points \(a\) and \(b\) under a load of a mid-span vehicle are given by  
\[\left\{\begin{array}{c}\sigma_{qa}=\sigma_{a}+\sigma_{s}\\ \\ \sigma_{qb}=\sigma_{b}-\sigma_{s}\end{array}\right. \tag{8}\]