## 4 Solutions to the Influence Line Equations of the Main Beam
### Solution to the Beam Equation on Elastic Foundation Under the Concentrated Load
The vertical displacement of the first element in the vector \(K_{m}(x_{m})S_{m}\) is 0. Five unknowns correspond to five equations. By solving the linear equations, the coefficients of each segment and the vertical force at the middle support can be obtained. Considering the typical situation that the left and right ends of the main beam are only set as the vertical constraint as an example, then the characteristic mechanical equations of the left and right ends of the main beam are as Equations (30) and (31), respectively. \[\left[0\;\;v(x_{1})^{\prime}\;\;0\;\;Q(x_{1})\right]^{T}=K_{1}(x_{1})S_{1} \tag{30}\]  
\[\left[0\;\;v(x_{n+1})^{\prime}\;0\;Q(x_{n+1})\right]^{T}=K_{n}(x_{n+1})\left(D _{n,1}S_{1}-D_{n,m}K_{m}^{-1}(x_{m})F_{m}-D_{n,k}K_{k}^{-1}(x_{k})F_{k}\right) \tag{31}\]  
The characteristic mechanical equation of the right end of \(x_{m}\) for the main beam is as follows:  
\[\left[0\;\;v(x_{m})^{\prime}\;\;M(x_{m})\;\;Q(x_{m})\right]^{T}=K_{m}(x_{m}) \left(D_{m,1}S_{1}-D_{m,m}K_{m}^{-1}(x_{m})F_{m}\right) \tag{32}\]  
Take the first and third rows of Equations (30) and (31) and the first row of Equation (32), it can form five linear equations with five unknowns, which consist of the coefficients of \(A_{1}\), \(B_{1}\), \(C_{1}\), and \(D_{1}\) of the vector \(S_{1}\), and \(R_{m}\) of the vector \(F_{m}\). Then, the unique solution can be solved. When the bending moment is acting on the main beam node \(k\) along with the vertical force, then the force vector is \(F_{k}=\left[0\;0\;M\;P\right]^{T}\). From the above derivation, when the beam is supported at various locations, only the number of terms \(x_{m}\) needs to be increased, and the number of unknown quantities and the equations of the final linear equations increases correspondingly according to the number of supports. When there are multiple known loads in the middle of the main beam, the corresponding terms \(F_{k}\) need to be added, but the unknown quantity and the number of equations of the final linear equation system remain unchanged. Solution to the Beam Equation Based on Elastic Foundation Under the Equivalent Uniformly Distributed Load  
In order to solve the deflection \(\
u_{2}(x,\xi)\) of the beam on the elastic foundation under the equivalent uniformly-distributed load \(H_{p}y^{\prime\prime}\), the horizontal force component of the main cable \(H_{p}\) under the concentrated load \(p\) should be determined at first, and \(\
u_{2}(x,\xi)\) could be obtained based on the integral form of the concentrated load. \[v_{2}(x,\xi)=H_{p}y^{\prime\prime}\int v_{1}(x,t)dt=H_{p}(y^{\prime\prime}_{0} \int_{-\frac{t_{2}d}{2}}^{\frac{L_{2d}}{2}}+y^{\prime\prime}_{1}\int_{\frac{t _{2}d}{2}}^{\frac{L_{2d}}{2}+L_{\xi}}+y^{\prime\prime}_{3}\int_{-(\frac{t_{2}d }{2}+L_{\xi})}^{-\frac{L_{2d}}{2}})\
u_{1}(x,t)dt \tag{33}\]  
For the purpose of simplification, the equation \(\int_{x_{1}}^{x_{2}}f(x)dx+\int_{x_{3}}^{x_{4}}f(x)dx=(\int_{x_{1}}^{x_{2}}+ \int_{x_{3}}^{x_{4}})\)\(f(x)dx\) is used. It is seen that the Equation (33) consists of two unknowns, \(\
u_{2}(x,\xi)\) and \(H_{p}\). Therefore, the compatibility equation of the main cable should be used jointly. \[\sum_{i=1}^{n}\left(\frac{H_{p}L_{\phi}}{E_{c}A_{c}}+a_{t}\Delta_{T}L_{t}-\int y(x )^{\prime}w(x,\xi)^{\prime}dx\right)=0 \tag{34}\]  
where \(L_{\theta}=\int\frac{1}{cos^{3}\theta}dx\) and \(L_{t}=\int\frac{1}{cos^{3}\theta}dx\). \(E_{c}\), \(A_{c}\), \(a_{t}\), \(\Delta_{t}\), and \(w(x,\xi)\) are the elastic modulus, cross-sectional area, coefficient of thermal expansion, temperature variation, and vertical deflection of the main cable, respectively.