## 5 Theoretical Analysis  
### Elastic Stiffness Calculation Method  
The calculating diagram and internal force distribution of an X-shaped metal damper with an oblique angle (\(\theta\)) are shown in Figures 15 and 16. When a unit displacement (\(\Delta\)) was applied at the end of the dampers, the deflection of the dampers was composed of the vertical displacement from the bending moment (\(\Delta_{\text{M}}\)), shear force (\(\Delta_{\text{V}}\)), and axial force (\(\Delta_{\text{N}}\)), which could be obtained according to the graph multiplication method [38]:  
\[\Delta_{\text{M}}+\Delta_{\text{V}}+\Delta_{\text{N}}=\Delta \tag{7}\]  
\[\Delta_{\text{M}}=\int\frac{\overline{M_{\text{p}}(x)}M_{\text{p}}(x)}{EI_{ \text{eq}}}dx=\frac{F_{\text{s}}L^{3}}{12 EI_{\text{eq}}\cos^{2}\theta} \tag{8}\]  
\[\Delta_{\text{V}}=\int\frac{f_{\text{s}}\overline{V_{\text{p}}(x)}V_{\text{p} }(x)}{G\,A_{\text{eq}}}dx=\frac{f_{\text{s}}F_{\text{s}}L\cos\theta}{GA_{ \text{eq}}} \tag{9}\]  
\[\Delta_{\text{N}}=\int\frac{\overline{N_{\text{p}}(x)}N_{\text{p}}(x)}{EA_{ \text{eq}}}dx=\frac{F_{\text{s}}L\sin^{2}\theta}{EA_{\text{eq}}\cos\theta} \tag{10}\]  
where \(F_{\text{s}}\) represents the shear force; \(E\) and \(G\) represent the elastic and shear modulus, respectively; \(M_{\text{p}}(x)\), \(V_{\text{p}}(x)\), and \(N_{\text{p}}(x)\) represent the bending moment, shear force, and axial force caused by a unit of displacement, respectively; \(l_{\text{eq}}\) represents the equivalent moment of inertia of the neutral axis \(Z\); \(A_{\text{eq}}\) represents the equivalent cross-section area of the damper; and \(f_{\text{s}}\) represents the shear deformation coefficient of the section. For a rectangular section, \(f_{\text{s}}\) is equal to 1.2 [39]. Based on Equations (7)-(10), the unit displacement can be obtained:  
\[\frac{F_{\text{s}}L^{3}}{12 EI_{\text{eq}}\cos^{2}\theta}+\frac{f_{\text{s}}F_ {\text{s}}L\cos\theta}{GA_{\text{eq}}}+\frac{F_{\text{s}}L\sin^{2}\theta}{EA_ {\text{eq}}\cos\theta}=\Delta \tag{11}\]  
Therefore, the elastic stiffness of rectangular metal dampers can be expressed by Equation (12). \[K=\frac{F_{\text{s}}}{\Delta}=\frac{12 EGI_{\text{eq}}A_{\text{eq}}\cos^{2} \theta}{GA_{\text{eq}}L^{3}+12f_{\text{s}}LEI_{\text{eq}}\cos^{3}\theta+12 LGI_{\text{eq}}\sin^{2}\theta\cos\theta} \tag{12}\]  
Figure 16: Internal force distribution of XMD: (**a**) bending moment; (**b**) shear force; (**c**) axial force. Figure 15: Calculating diagram of XMD. It can be seen that the equivalent moment of inertia and cross-section area need to be calculated to obtain the elastic stiffness of the XMD. In addition, two stiffening ribs were welded with the surface of the X-shaped steel plate to restrict the out-of-plane deformation of dampers. For ease of simplification, X-shaped metal dampers without an oblique angle were selected to calculate the equivalent moment of inertia and cross-section area, which were approximately equal to those of the dampers with an oblique angle. The calculating diagram of an X-shaped metal damper is shown in Figure 17.  
In Figure 17, \(H\) and \(h_{0}\) represent the maximum and minimum widths of XMD; \(l_{0}\) represents the length of a weakened segment; and \(a\) and \(b\) represent the width and height of stiffening ribs. The weakened coefficient was introduced [40] and can be expressed as follows:  
\[\lambda=\frac{(H-h_{0})}{l_{0}} \tag{13}\]  
As shown in Figure 17, the cross-section width (\(h(x)\)) of dampers at position \(x\) can be obtained according to the geometry relation. \[h(x)=\left\{ \begin{array}{ll}h_{0}+2\lambda x,&0<x\leq\frac{l_{0}}{2}\\ \\ H,&\frac{l_{0}}{2}<x\leq\frac{l}{2}\end{array} \right. \tag{14}\]  
The length of stiffening ribs was equal to that of the XMD. Therefore, the moment of inertia \(I(x)\) at position \(x\) was expressed as follows:  
\[I(x)=\left\{ \begin{array}{ll}\frac{1}{2}\Big{[}t(h_{0}+2\lambda x)^{3}+2 ab^{ 3}\Big{]},&0<x\leq\frac{l_{0}}{2}\\ \\ \frac{1}{2}\big{(}tH^{3}+2 ab^{3}\big{)},&\frac{l_{0}}{2}<x\leq\frac{l}{2} \end{array} \right.