## 4 Theoretical Analysis of Double-Column Self-Centering Piers with Shear Links
### Piers with Shear Links
#### 4.2.2 Rocking State  
According to the observation in Section 3.2, under the action of lateral loads, the deformation of each column can be neglected, with only the rigid-body displacements of the superstructure, bent cap, and columns remaining.  
As shown in Figure 14, \(\Delta_{\mathrm{PT}}\) is the elongation of unbonded steel strands; \(\Delta_{\mathrm{ss}}\) and \(\Delta_{\mathrm{col}}\) are the rising height of the superstructure and pier, respectively; \(\Delta_{\mathrm{SL}}\) is the relative shear displacement between the ends of the shear link; and \(u_{\mathrm{x}}\) is the lateral displacement at the top of the pier. The relationship between the above displacements and the rotation angle of the column can be directly obtained based on the principle of displacement compatibility:  
\[\Delta_{\mathrm{PT}}(\theta)=2b\sin\frac{\theta}{2} \tag{17}\]  
\[\Delta_{\mathrm{ss}}(\theta)=b\sin\theta-h(1-\cos\theta) \tag{18}\]  
\[\Delta_{\mathrm{col}}(\theta)=\frac{b\sin\theta-h(1-\cos\theta)}{2} \tag{19}\]  
\[\Delta_{\mathrm{SL}}(\theta)=(b+d)\sin\theta \tag{20}\]  
\[u_{\mathrm{x}}(\theta)=b(1-\cos\theta)+h\sin\theta \tag{21}\]Before the stress of the unbonded steel strands reaches the ultimate strength, the force provided by the unbonded steel strands, \(F_{\text{PT}}\) (\(\theta\)) is  
\[F_{\text{PT}}(\theta)=E_{\text{PT}}A_{\text{PT}}\frac{\Delta_{\text{PT}}(\theta) }{h}+F_{\text{PT0}} \tag{22}\]  
Assuming that the constitutive model of the shear links is perfectly elasto-plastic with yield strength \(F_{y}\), the shear force at the ends of the shear link, \(F_{\text{SL}}(\theta)\), is  
\[F_{\text{SL}}(\theta)=\left\{\begin{array}{ll}k_{\text{SL}}\Delta_{\text{SL}} (\theta)&\Delta_{\text{SL}}<\Delta_{\text{SLy}}\\ F_{y}&\Delta_{\text{SL}}\geq\Delta_{\text{SLy}}\end{array}\right. \tag{23}\]  
where \(\Delta_{\text{SL}}\) is the relative shear displacement between the ends of the shear link when the shear link yields.  
According to the principle of the virtual work, the total work done by the applied forces and the inertial forces of a structure in equilibrium equals zero for any virtual displacement. Therefore, Equation (24) is obtained as  
\[F_{\text{x}}(\theta)\frac{\text{d}u_{\text{x}}}{\text{d}\theta}=2F_{\text{PT} }(\theta)\frac{\text{d}\Delta_{\text{PT}}}{\text{d}\theta}+F_{\text{SL}}( \theta)\frac{\text{d}\Delta_{\text{SL}}}{\text{d}\theta}+W_{\text{beam}} \frac{\text{d}\Delta_{\text{ss}}}{\text{d}\theta}+2W_{\text{pier}}\frac{ \text{d}\Delta_{\text{col}}}{\text{d}\theta} \tag{24}\]  
where \(\frac{\text{d}u_{\text{x}}}{\text{d}\theta}\), \(\frac{\text{d}\Delta_{\text{PT}}}{\text{d}\theta}\), \(\frac{\text{d}\Delta_{\text{SL}}}{\text{d}\theta}\), \(\frac{\text{d}\Delta_{\text{ss}}}{\text{d}\theta}\), and \(\frac{\text{d}\Delta_{\text{col}}}{\text{d}\theta}\) are the virtual displacements along the directions of applied lateral force, the prestress in the steel strands, the shear force at the ends of the shear link, the superstructure gravity, and the column gravity, respectively, which can be derived from Equations (17)-(21).  
\[\frac{\text{d}\Delta_{\text{PT}}}{\text{d}\theta}=b\cos\frac{\theta}{2} \tag{25}\]  
\[\frac{\text{d}\Delta_{\text{ss}}}{\text{d}\theta}=b\cos\theta-h\sin\theta \tag{26}\]  
\[\frac{\text{d}\Delta_{\text{col}}}{\text{d}\theta}=\frac{b\cos\theta-h\sin \theta}{2} \tag{27}\]  
\[\frac{\text{d}\Delta_{\text{SL}}}{\text{d}\theta}=(b+d)\cos\theta \tag{28}\]  
Figure 14: Rigid-body displacements of the pier.  
\[\frac{\mathrm{d}u_{\mathrm{x}}}{\mathrm{d}\theta}=b\sin\theta+h\cos\theta \tag{29}\]  
Substituting Equations (25)-(29) into Equation (24) yields:  
\[\begin{array}{ll}F_{\mathrm{x}}(\theta)=&2F_{\mathrm{PT}}(\theta)\frac{b\cos \frac{\theta}{2}}{b\sin\theta+h\cos\theta}+F_{\mathrm{SL}}(\theta)\frac{(b+d) \cos\theta}{b\sin\theta+h\cos\theta}+\\ &W_{\mathrm{ss}}\frac{b\cos\theta-h\sin\theta}{b\sin\theta+h\cos\theta}+W_{ \mathrm{col}}\frac{b\cos\theta-h\sin\theta}{b\sin\theta+h\cos\theta}\end{array} \tag{30}\]