# Damage Identification of Stay Cables Based on a Small Amount of Deflection Monitoring Data
## 2 Theory Related to the Reduction in Measurement Points
### Measuring Point Arrangement of Beam Deflection under Concentrated Load  
As shown in Figure 2, the main beam of the cable-stayed bridge is equivalent to a foundation beam; it is a section AB taken from a section of the main beam of the cable-stayed bridge, which contains \(n\) cables. The anchorage points of each cable and the main beam are numbered 1, 2, \(\cdots\), \(k\), \(\cdots\), \(i\), \(\cdots\), \(n\) from left to right. The cable forces in the cable-stayed cables are \(F_{1},\cdots,F_{k},\cdots,F_{i},\cdots,F_{n}\), respectively. The angles between the stay cable and the main girder are \(\alpha_{1},\cdots,\alpha_{k},\cdots,\alpha_{i},\cdots,\alpha_{n}\), respectively. Figure 1: Cable number and deflection measuring point layout of a single— pylon cable-stayed bridge structure. In Figure 2, \(w_{k,i}^{u}(x)\) and \(w_{k,j}^{d}(x)\) are the deflection at the anchor point of cable \(k\) and the main beam, when one stay cable \(i\) has been damaged at any position under the action of dead weight. The deflection difference of the beam before and after cable \(i\) damage can be expressed using Equation (2). \[\Delta w_{k}(x)=w_{k,i}^{d}(x)-w_{k,i}^{u}(x)\ (i,\ k=1,\ 2,\ 3,\ \cdots,\ n) \tag{2}\]  
where the subscript \(i\), represents the position of the damage cable (can be multiple or single-cable damage), \(n\) is the number of the anchorage points, \(x_{k}\) is the anchorage points of each stay cable on the beam. As shown in Figure 2a, the deflection of the beam at any section \(k\) in healthy state can be expressed Equation (3). \[w_{k,i}^{u}(x)=w_{k,i}^{q}(x)+\sum_{i=1}^{n}w_{k,i}^{F_{i}}(x),(i,k=1,2,3, \cdots n) \tag{3}\]  
where \(w_{i}^{q}(x_{k})\) is the deflection due to the effect of the uniform load \(q\), \(w_{i}^{F_{i}}(x_{k})\) is the deflection at a cable force of \(F_{i}\) when there is no damage to the cable. When the damaged state of the cable is unknown (see Figure 2b), the deflection of the beam at any section \(k\) can be expressed as in Equation (4). \[w_{i}^{d}(x_{k})=w_{i}^{q}(x_{k})+\sum_{i=1}^{n}w_{i}^{F_{i}+\Delta F_{i}}(x_ {k}),i,k=1,2,3,\cdots n \tag{4}\]  
where \(w_{i}^{F_{i}+\Delta F_{i}}(x_{k})\) is the deflection at a cable force of \(F_{i}+\Delta F_{i}\) when the cable is damaged. As shown in Figure 2c, when the cable \(i\) is damaged, the force of the cable decreases, the nearby cable force of the nearby cable increases and the value is small; in addition, the  
Figure 2: Stress analysis diagram of cable before and after damage: (**a**) in the healthy state; (**b**) after the cable damage; (**c**) before and after cable damage. other cable's force is essentially unchanged and can be ignored. The deflection difference of the beam before and after cable damage can be expressed as in Equation (5). \[\Delta w_{i}(x_{k})=w_{i}^{d}(x_{k})-w_{i}^{u}(x_{k})=\sum_{i=1}^{n}w_{i}^{\Delta F _{i}}(x_{k}),i=1,2,3,\cdots n \tag{5}\]  
That is, the deflection difference \(\Delta w_{i}(x_{k})\) is caused by the change in the cable force of the damage cable \(i\). As shown in Figure 3, according to [PERSON]'s hypothesis [20], the settlement of any point on the surface of a foundation is proportional to the pressure exerted on that point per unit area. \[EId^{4}\Delta w(x_{k})/dx^{4}=-k\Delta w(x_{k})+q \tag{6}\]  
where \(E\) is the elastic modulus of the main beam, \(I\) is the bending inertial moment of the main beam, \(k\) is the elastic coefficient of the main beam. Under the action of concentrated force, the solution of Equation (6) is  
\[\Delta w(x_{k})=\frac{\Delta F_{i}sina_{i}\gamma}{2k}e^{-\gamma x_{k}}(\cos( \gamma x_{k})+\sin(\gamma x_{k}) \tag{7}\]  
where \(\gamma\) is the elastic eigenvalue of the beam, \(\gamma=\sqrt[4]{k/4E}I\). The deflection difference of the beam less than 0 can be obtained through symmetry. As shown in Figure 4, the deflection distribution curve of the beam takes the point of concentrated force as the symmetric point, and the graph on the left (\(x<0\)) can be obtained symmetrically, as shown in Equation (8). \[\Delta w(x_{k})=\left\{