## 4 Theoretical Analysis of Double-Column Self-Centering Piers with Shear Links
### Piers with Shear Links  
#### 4.2.1 Pre-Rocking State  
Before rocking, the pier can be simplified as the model in Figure 13a, in which the bent cap is considered an undeformable rigid body. During this period, the columns and shear link deform under the lateral force \(F\). Due to the symmetry of the frame, the calculation model can be further simplified as shown in Figure 13b. In the simplified model, the geometric parameters \(l\) and \(x\) are half of the pier height and half of the distance between the column axes, respectively:  
\[l=\frac{h}{2} \tag{1}\]  
\[x=\frac{b+d}{2} \tag{2}\]  
The matrix displacement method is used. The corresponding global coordinate system is established and the information of the nodes and elements is shown in Figure 13b. There are six independent nodal displacements: the angular displacements of nodes B(\(u_{3}\)) and E(\(u_{6}\)); the horizontal linear displacements of nodes B(\(u_{1}\)), C(\(u_{4}\)), D(\(u_{4}\)), and E(\(u_{5}\)); and the vertical linear displacement of node B(\(u_{2}\)). Obviously, \(u_{4}\) can be considered the lateral displacement at the top of the pier \(u_{x}\), i.e., \(u_{x}=u_{4}\). Since elements 1 and 2 in Figure 13b refer to the column, \(A_{1}\), \(I_{1}\), and \(E_{1}\) are the area, inertia moment, and elastic modulus of each column, which are expressed as  
\[A_{1}=A_{\text{col}}=b^{2} \tag{3}\]  
\[I_{1}=I_{\text{col}}=\frac{b^{4}}{12} \tag{4}\]  
\[E_{1}=E_{\text{c}} \tag{5}\]  
where \(E_{\text{c}}\) is the elastic modulus of concrete. Figure 13: Simplified model of double-column self-centering pier with shear links. (**a**) Full structure; (**b**) semi-structure. It is reasonable to assume that the shear stiffness of the shear links is equal to the bending stiffness of element 1 in Figure 12(b), and the shear links do not bear compressive axial force. Consequently, Equations (6) and (7) can be obtained as  
\[\frac{12E_{2}I_{2}}{b^{3}}=k_{\text{SL}} \tag{6}\]  
\[E_{2}A_{2}\to 0 \tag{7}\]  
The element stiffness matrix of elements 1 and 2 (Figure 12(b)) in the local coordinate system can be expressed as:  
\[\overline{\mathbf{K}_{1}^{c}}=\overline{\mathbf{K}_{2}^{c}}=\left[ \begin{array} []{ccccc}\frac{E_{1}A_{1}}{l}&0&0&-\frac{E_{1}A_{1}}{l}&0&0\\ 0&\frac{12E_{1}I_{1}}{\beta}&\frac{6E_{1}I_{1}}{l^{2}}&0&-\frac{12E_{1}I_{1}}{ \beta}&\frac{6E_{1}I_{1}}{l^{2}}\\ 0&\frac{6E_{1}I_{1}}{l^{2}}&\frac{4E_{1}I_{1}}{l}&0&-\frac{6E_{1}I_{1}}{l^{2}} &\frac{2E_{1}I_{1}}{l}\\ -\frac{E_{1}A_{1}}{l}&0&0&\frac{E_{1}A_{1}}{l}&0&0\\ 0&-\frac{12E_{1}I_{1}}{l^{3}}&-\frac{6E_{1}I_{1}}{l^{2}}&0&\frac{12E_{1}I_{1}} {\beta}&-\frac{6E_{1}I_{1}}{l^{2}}\\ 0&\frac{6E_{1}I_{1}}{l^{2}}&\frac{2E_{1}I_{1}}{l}&0&-\frac{6E_{1}I_{1}}{l^{2}} &\frac{4E_{1}I_{1}}{l}\end{array} \right] \tag{8}\]  
Transforming the element stiffness matrix of elements 1 and 2 in Figure 12(b) from the local coordinate system to the global coordinate system, it is determined that  
\[\mathbf{K}_{1}^{c}=\mathbf{K}_{2}^{c}=\left[ \begin{array}{ccccc}\frac{12E_ {1}I_{1}}{l^{3}}&0&-\frac{6E_{1}I_{1}}{l^{2}}&-\frac{12E_{1}I_{1}}{l^{3}}&0&- \frac{6E_{1}I_{1}}{l^{2}}\\ 0&\frac{E_{1}A_{1}}{l}&0&0&-\frac{E_{1}A_{1}}{l}&0\\ -\frac{6E_{1}I_{1}}{l^{2}}&0&\frac{4E_{1}I_{1}}{l}&\frac{6E_{1}I_{1}}{l^{2}}&0 &\frac{2E_{1}I_{1}}{l}\\ -\frac{12E_{1}I_{1}}{l^{3}}&0&\frac{6E_{1}I_{1}}{l^{2}}&\frac{12E_{1}I_{1}}{l^ {3}}&0&\frac{6E_{1}I_{1}}{l^{2}}\\ 0&-\frac{E_{1}A_{1}}{l}&0&0&\frac{E_{1}A_{1}}{l}&0\\ \frac{6E_{1}I_{1}}{l^{2}}&0&\frac{2E_{1}I_{1}}{l}&\frac{6E_{1}I_{1}}{l^{2}}&0 &\frac{4E_{1}I_{1}}{l}\end{array} \right] \tag{9}\]  
For element 1 in Figure 12(b), the local coordinate system is in accordance with the global coordinate system. Thus, its element stiffness matrix in the global coordinate system can be expressed as  
\[\mathbf{K}_{4}^{c}=\overline{\mathbf{K}_{4}^{c}}=\left[