## 2 Methods  
### Traditional Hinge Plate Method  
Slabs at the loading positions deform and other slabs appear to have corresponding deflections, when a hinge slab bridge is subjected to vehicle load. However, the transverse deformation of a single slab is so tiny that it can be negligible. Live load is distributed to every slab because of the internal forces of hinge joints, which include the vertical shear \(g(x)\), lateral bending moment \(m(x)\), longitudinal shear \(t(x)\), and normal force \(n(x)\). The height and stiffness of tongue and groove are so weak that other internal forces have little influence compared with vertical shear. Hence, tongue and groove are assumed to be a hinge joint, which only transfer the vertical shear \(g(x)\). Meanwhile, the ratio of internal forces, deformations, and distributed loads between slabs should meet the relations expressed in Equation (1):  
\[\frac{\omega_{i}(x)}{\omega_{j}(x)}=\frac{M_{i}(x)}{M_{j}(x)}=\frac{Q_{i}(x)}{ Q_{j}(x)}=\frac{P_{i}(x)}{P_{j}(x)}=constant, \tag{1}\]  
where \(\omega(x)\), \(M(x)\), \(Q(x)\), and \(P(x)\) are the deflection, bending moment, shear, and distributed load, respectively; \(i\) and \(j\) represent slab numbers in the range of {1, 2,..., \(n\)}, \(n\) is the number of slabs. If the load patterns are concentrated or uniform, relationships of Equation (1) will not be satisfied. Whereas, use of a half-wave sinusoidal load with a peak value \(p_{0}\), as shown in Figure 2, can harmonize the ratio, which is another assumption for hinge plate method [42]. There are \(n\)-1 hinge joints in the bridge composed of \(n\) slabs, which means there are \(n\)-1 unknown hinge joint forces (namely the vertical shear \(g(x)\)). If the whole hinge forces \(g(x)\) are solved, the distributed loads on each slab can be obtained according to the principle of balance, which can be used to calculate the influence line vertical value of LLD. According to the location of external load, the calculation form can be divided into two categories. When the external load is located on the side slabs (namely slab 1 or slab \(n\)), LLD can be defined by Equations (2) and (3). Equation (4) is used for the case of load placed on internal slabs. \[\left\{ \begin{array}{l}\eta_{11}=p_{11}=1-g_{1}\\ \eta_{i1}=p_{i1}=g_{i-1}-g_{i}\\ \eta_{n1}=p_{n1}=g_{n-1}\end{array} \right.\ \ 2\leq i\leq n-1, \tag{2}\]  
\[\left\{ \begin{array}{l}\eta_{1n}=p_{1n}=g_{1}\\ \eta_{in}=p_{in}=g_{i}-g_{i-1}\\ \eta_{nn}=p_{nn}=1-g_{n-1}\end{array} \right.\ \ 2\leq i\leq n-1, \tag{3}\]  
Figure 2: The force diagram of simply supported hinged-slab bridge. \[\left\{ \begin{array}{l}\eta_{ij}=p_{lj}=g_{1}\\ \eta_{ij}=p_{ij}=g_{i}-g_{i-1}\\ \eta_{jj}=p_{jj}=1-g_{j-1}-g_{j}\\ \eta_{kj}=p_{kj}=g_{k-1}-g_{k}\\ \eta_{nj}=p_{nj}=g_{n}\end{array} \right.\left\{ \begin{array}{l}2\leq i\leq j -1\\ j+1\leq k\leq n-2,\end{array} \right. \tag{4}\]  
where \(p_{ij}\) and \(\eta_{ij}\) represent the distributed load and LLD influence line vertical value of the \(i\)th slab with the external load located at the \(j\)th one, respectively; \(g_{i}\) is the vertical shear at \(i\)th hinge joint. For statically indeterminate problem with \(n-1\) unknown forces, it can be solved by two conditions. One is the basic system (shown in Figure 3) whose hinge joints are cut. Others are the deformation compatibility conditions that the relative vertical displacements of two adjacent slabs at hinge joints are zeros. Therefore, \(n-1\) canonical equations can be obtained and described as Equation (5):  
\[\left\{ \begin{array}{ccccccccc}\delta_{11}g_{1}&+&\delta_{12}g_{2}&+&\cdots&+ &\delta_{1,n-1}g_{n-1}&+&\delta_{1p}&=&0\\ \delta_{21}g_{1}&+&\delta_{22}g_{2}&+&\cdots&+&\delta_{2,n-1}g_{n-1}&+&\delta _{2p}&=&0\\ \vdots&&\vdots&&\ddots&&\vdots&&\vdots&&\vdots\\ \delta_{n-1,1}g_{1}&+&\delta_{n-1,2}g_{2}&+&\cdots&+&\delta_{n-1,n-1}g_{n-1}&+& \delta_{n-1,p}&=&0,\end{array} \right.