# Numerical Analysis of Reinforcing Effect for Scissors-Type Bridge with Strut Members
## 2 Theoretical Evaluation for Strut Reinforcement
### Basic Theory Based on Equilibrium Equations
#### 2.1.2 A Unit Model in a Statically Indeterminate Problem  
An FBD for a scissor structure with an upper strut member is shown in Figure 2a. This model has one degree of redundancy in comparison with the model in Figure 1. The statically indeterminate force \(X\) in the scissor structure can be led by the unit load method. It is necessary to think the two conditions depend on the number of the degree of redundancy as shown in Figure 2b,c. For instance, the statically indeterminate force \(X\) in the model Figure 2a can be solved by use of generated sectional forces in the models of Figure 2b,c as follows:  
\[ \begin{array}{rcl}X&=&-\dfrac{\delta_{00}}{\delta_{01}}\\ &&=&-\dfrac{\sum_{j=1}^{4}\int_{0}^{L_{0}/2}\dfrac{M_{0j}M_{1j}}{EI} d\chi+\sum_{j=1}^{4}\dfrac{N_{0j}N_{1j}}{EA}\dfrac{L_{0}}{2}}{\sum_{j=1}^{4} \int_{0}^{L_{0}/2}\dfrac{\overline{M_{1j}^{2}}}{EI}d\chi+\sum_{j=1}^{4}\dfrac{N _{1j}^{2}}{EA}\dfrac{L_{0}}{2}+\dfrac{N_{15}^{2}}{EB}\dfrac{L_{0}\sin\theta}{2 }}\end{array}  \tag{4}\]  
where \(\chi\) is an arbitrary distance from each nodal point in axial direction, \(L_{0}/2\) is a half length of member, \(\theta\) is an expanding angle, \(E\) is elastic modulus, \(A\) is a sectional area of scissor member, \(B\) is a sectional area of the strut member, \(N_{0j}\) is an axis force of the basic system, \(M_{0j}\) is the bending moment of the basic system, \(N_{1j}\) is an axis force of the first system, and \(M_{1j}\) is the bending stress of the first system. This equation can be rewritten as follows:  
\[ \begin{array}{rcl}X&=&-\dfrac{2A}{\overline{IL_{0}}\sum_{j=1}^{4} \int_{0}^{L_{0}/2}M_{0j}M_{1j}\,d\chi+\sum_{j=1}^{4}N_{0j}N_{1j}}\\ &&\dfrac{2A}{\overline{IL_{0}}\sum_{j=1}^{4}\int_{0}^{L_{0}/2}M_{1j} ^{2}\,d\chi+\sum_{j=1}^{4}\overline{N_{1j}^{2}}+\dfrac{\sin\theta}{\alpha}} \end{array}  \tag{5}\]  
where \(\alpha=EB/EA\) which is the ratio of extension rigidity between the strut and the scissor member. It is understood that \(X\) becomes large according to an increase of the parameter \(\alpha\). Of course, it is also possible to determine the unknown reinforced force \(X\) by considering the strain energy. The strain energy in this structure is given by the following equation:  
\[ \begin{array}{rcl}\mathcal{U}&=&\sum_{j=1}^{4}\left(\dfrac{N_{(j)}^{2}}{2 EA }\dfrac{L_{0}}{2}+\int_{0}^{L_{0}/2}\dfrac{M_{(j)}^{2}}{2 EI}\,d\chi\right)+ \dfrac{X^{2}L_{0}\sin\theta}{2 EB}\end{array}  \tag{6}\]  
Figure 1: FBD of a scissor unit. Now, unknown internal force \(X\) can be calculated using the principle of strain energy minimization:  
\[ \begin{array}{rcl}\frac{\partial\mathcal{U}}{\partial X}&=&0\\ &=&\frac{1}{2 EA}\frac{L_{0}}{2}\frac{\partial}{\partial X}\left( \sum_{j=1}^{4}N_{(j)}^{2}\right)+\frac{1}{2 EI}\frac{\partial}{\partial X} \left(\sum_{j=1}^{4}\int_{0}^{L_{0}/2}M_{(j)}^{2}d\chi\right)+\frac{2 XL_{0} \sin\theta}{2 EB}\\ &=&\frac{L_{0}}{4 EA}\sum_{j=1}^{4}\frac{\partial(N_{(j)}(X))^{2}} {\partial X}+\frac{1}{2 EI}\sum_{j=1}^{4}\int_{0}^{L_{0}/2}\frac{\partial(M_{( j)}(X))^{2}}{\partial X}\,d\chi+\frac{XL_{0}\sin\theta}{EB}\\ &=&\frac{1}{4}\sum_{j=1}^{4}\frac{\partial(N_{(j)}(X))^{2}}{ \partial X}+\frac{A}{2 IL_{0}}\sum_{j=1}^{4}\int_{0}^{L_{0}/2}\frac{\partial(M_ {(j)}(X))^{2}}{\partial X}\,d\chi+\frac{X\sin\theta}{\alpha}\end{array}  \tag{7}\]  
Once statically indeterminate force \(X\) is clear, it can provide feedback as the live loads to the original model. That is, the following equations are obtained for a cantilever and a simple supported condition based on Equations (1) and (3):  
\[