# Influence Line-Based Design of Scissors-Type Bridge
## 2 The Equilibrium Theory
### The Mechanism of Scissors Structure Using Simple Support Model  
A two-unit SCI model with nodal forces acting only on the hinge, pivot, and pin was supported at both ends, as shown in Figure 6. All members have the same length and angle of inclination \(\theta\), measured in the vertical direction. The length and height of the scissor unit are defined as \(\lambda\) and \(\eta\). The left-hand scissors unit consists of nodes \(B_{1}^{L},B_{1}^{R},C_{1},A_{1}^{L}\) and \(A_{1}^{R}\), and the second scissors unit consists of nodes \(B_{2}^{L},B_{2}^{R},C_{2},A_{2}^{L}\) and \(A_{2}^{R}\). All nodal points are considered to be hinges. Loading is applied at nodal locations only; hence no bending moments were transmitted or applied at these nodal junctions [20; 21]. The horizontal \((\bullet^{*})_{x}\) and vertical \((\bullet^{*})_{y}\) equilibrium equations for the left-hand side unit are as follows:\[[R]\left\{ \begin{array}{rcl}(B_{1}^{R})_{x}\\ (B_{1}^{R})_{y}\\ (A_{1}^{R})_{x}\\ (A_{1}^{R})_{y}\end{array} \right\} = -[L_{1}]\left\{ \begin{array}{c}(B_{1}^{L})_{x}\\ (B_{1}^{L})_{y}\\ (B_{2}^{R})_{x}\\ (B_{2}^{R})_{y}\end{array} \right\}-\left\{ \begin{array}{c}(C_{1})_{x}+(A_{1 }^{L})_{x}\\ (C_{1})_{y}+(A_{1}^{L})_{y}\\ \mathbf{0}\\ \eta(A_{1}^{L})_{x}+\lambda(A_{1}^{L})_{y}\end{array} \right\}\] \[\left\{ \begin{array}{c}(B_{1}^{R})_{x}\\ (B_{1}^{R})_{y}\\ (A_{1}^{R})_{x}\\ (A_{1}^{R})_{y}\end{array} \right\} = -[R]^{-1}[L_{1}]\left\{ \begin{array}{c}(B_{1}^{L})_{x}\\ (B_{1}^{L})_{y}\\ (B_{2}^{R})_{x}\\ (B_{2}^{R})_{y}\end{array} \right\}-[R]^{-1}\left\{ \begin{array}{c}(C_{1})_{x} +(A_{1}^{L})_{x}\\ (C_{1})_{y}+(A_{1}^{L})_{y}\\ \mathbf{0}\\ \eta(A_{1}^{L})_{x}+\lambda(A_{1}^{L})_{y}\end{array} \right\} \tag{6}\]  
Here,  
\[[R]=\left[ \begin{array}{ccc|cc}1&0&1&0\\ 0&1&0&1\\ \hline 0&0&\eta&-\lambda&0&0\end{array} \right],\quad[L_{1}]=\left[ \begin{array} []{ccc|cc}1&0&0&0\\ 0&1&0&0\\ \hline-\eta&\lambda&0&0\\ 0&0&0&0\end{array} \right]\]  
From the right-hand SCI, we obtain the equation as below. \[[L]\left\{ \begin{array}{c}(B_{2}^{L})_{x}\\ (B_{2}^{L})_{y}\\ (A_{2}^{L})_{x}\\ (A_{2}^{L})_{y}\end{array} \right\} = -[R_{2}]\left\{ \begin{array}{c}(B_{1}^{L})_{x}\\ (B_{1}^{L})_{y}\\ (B_{2}^{R})_{x}\\ (B_{2}^{R})_{y}\end{array} \right\}-\left\{ \begin{array}{c}(C_{2})_{x}+(A_{ 2}^{R})_{x}\\ (C_{2})_{y}+(A_{2}^{R})_{y}\\ \eta(A_{2}^{R})_{x}-\lambda(A_{2}^{R})_{y}\\ \mathbf{0}\end{array} \right\}\] \[\left\{ \begin{array}{c}(B_{2}^{L})_{x}\\ (B_{2}^{L})_{y}\\ (A_{2}^{L})_{x}\\ (A_{2}^{L})_{y}\end{array} \right\} = -[L]^{-1}[R_{2}]\left\{ \begin{array}{c}(B_{1}^{L})_{x}\\ (B_{1}^{L})_{y}\\ (B_{2}^{R})_{x}\\ (B_{2}^{R})_{y}\end{array} \right\}-[L]^{-1}\left\{ \begin{array}{c}(C_{2})_{ x}+(A_{2}^{R})_{x}\\ (C_{2})_{y}+(A_{2}^{R})_{y}\\ \eta(A_{2}^{R})_{x}-\lambda(A_{2}^{R})_{y}\\ \mathbf{0}\end{array} \right\} \tag{7}\]  
Here,  
\[[L]=\left[ \begin{array}{ccc|cc}1&0&1&0\\ 0&1&0&1\\ \hline-\eta&\lambda&0&0\\ 0&0&\eta&\lambda\end{array} \right],\quad[R_{2}]=\left[ \begin{array}{ccc|cc }0&0&1&0\\ 0&0&0&1\\ \hline 0&0&0&0\\ 0&0&-\eta&-\lambda\end{array} \right]\]  
At nodes \((B_{1,2})\) and \((A_{1,2})\), the external forces \((B_{1,2})_{y_{x}},(B_{1,2})_{x},(A_{1,2})_{y_{x}}\) are in equilibrium with the internal forces \((B_{1}^{R})_{y_{x}},(B_{1}^{R})_{x},(A_{1}^{R})_{y},(A_{1}^{R})_{x}\) and \((B_{2}^{L})_{y_{y}},(B_{2}^{L})_{x},(A_{2}^{L})_{y_{x}},(A_{2}^{L})_{x}\) (see Figure 6). The sum of the forces for each unit can be expressed as follows:  
\[\left\{ \begin{array}{c}(B_{1}^{R})_{x}\\ (B_{1}^{R})_{y}\\ (A_{1}^{R})_{x}\\ (A_{1}^{R})_{y}\end{array} \right\}+\left\{ \begin{array}{c}(B_{2}^{L})_{x}\\ (B_{2}^{L})_{y}\\ (A_{2}^{L})_{x}\\ (A_{2}^{L})_{y}\end{array} \right\} = \left\{ \begin{array}{c}(B_{1,2})_{x}\\ (B_{1,2})_{y}\\ (A_{1,2})_{x}\\ (A_{1,2})_{y}\end{array} \right\} \tag{8}\]  
Substitution Equation (6) and (7) into Equation (8) and rearranging will be as below:  
\[[S]\left\{ \begin{array}{c}(B_{1}^{L})_{x}\\ (B_{1}^{L})_{y}\\ (B_{2}^{R})_{x}\\ (B_{2}^{R})_{y}\end{array} \right\} = \left\{