# Seismic Response Study of L-Shaped Frame Structure with Magnetorheological Dampers
## 3 Modeling Theory of the L-Shaped Frame Structure  
### Theory of Spatial Beam Element  
For the spatial beam element, as shown in Figure 4, every node has 6 degrees of freedom (DOFs), where u, v, and w are the axial displacements in x, y, and z directions, respectively; \(\theta_{x}\), \(\theta_{y}\), and \(\theta_{z}\) are the angular displacements around x, y, and z directions, respectively. The numbers in brackets in Figure 4 denote the DOF number.  
Figure 3: Damping force-displacement hysteresis curve of the MRD at different currents.  
Figure 2: Loading protocol of the MRD.  
According to [33], the stiffness matrix of spatial beam element, considering shear deformation \(\mathbf{k}_{e}\), is as follow:  
\[\mathbf{k}_{e}=\left[\begin{array}{cccc}\mathbf{k}_{11}&\mathbf{k}_{12}\\ \mathbf{k}_{21}&\mathbf{k}_{22}\end{array}\right] \tag{1}\]  
\[\mathbf{k}_{11}=\left[\begin{array}{ccccc}k_{1}&0&0&0&0\\ 0&k_{2}&0&0&0&k_{3}\\ 0&0&k_{4}&0&k_{5}&0\\ 0&0&0&k_{6}&0&0\\ 0&0&k_{5}&0&k_{7}&0\\ 0&k_{3}&0&0&0&k_{8}\end{array}\right] \tag{2}\]  
\[\mathbf{k}_{22}=\left[\begin{array}{ccccc}k_{1}&0&0&0&0&0\\ 0&k_{2}&0&0&0&-k_{3}\\ 0&0&k_{4}&0&-k_{5}&0\\ 0&0&k_{6}&0&0\\ 0&0&-k_{5}&0&k_{7}&0\\ 0&-k_{3}&0&0&0&k_{8}\end{array}\right] \tag{3}\]  
\[\mathbf{k}_{12}=\mathbf{k}_{21}^{\rm T}=\left[\begin{array}{ccccc}-k_{1}&0&0&0&0&0\\ 0&-k_{2}&0&0&0&k_{3}\\ 0&0&-k_{4}&0&k_{5}&0\\ 0&0&0&-k_{6}&0&0\\ 0&0&-k_{5}&0&k_{9}&0\\ 0&-k_{3}&0&0&0&k_{10}\end{array}\right] \tag{4}\]  
In Equations (2)-(4), \(k_{1}=\frac{EA}{L}\), \(k_{2}=\frac{12 EL_{s}}{(1+\beta_{y})L^{3}}\), \(k_{3}=\frac{6 EL_{s}}{(1+\beta_{z})L^{2}}\), \(k_{4}=\frac{12 EL_{y}}{(1+\beta_{y})L^{3}}\), \(k_{5}=-\frac{6 EL_{y}}{(1+\beta_{y})L^{2}}\), \(k_{6}=\frac{GJ}{L}\), \(k_{7}=\frac{(4+\beta_{y})E_{Y}}{(1+\beta_{y})L}\), \(k_{8}=\frac{(4+\beta_{z})EL_{z}}{(1+\beta_{z})L}\), \(k_{9}=\frac{(2-\beta_{y})E_{Iy}}{(1+\beta_{y})L}\), \(k_{10}=\frac{(2-\beta_{z})E_{Iz}}{(1+\beta_{z})L}\), \(\beta_{y}=\frac{12\mu E_{y}}{(CAI^{2}}\), \(\beta_{z}=\frac{12\mu E_{z}}{CAI^{2}}\).  
Where \(E\) is the elastic modulus, \(A\) is the cross-sectional area, \(L\) is the length of the element, \(I\) is the cross-sectional inertia moment, \(G\) is the shear modulus, \(\mu\) is Poisson's ratio, and \(J\) is the torsional moment of inertia. The stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation.  
The mass matrix of spatial beam element \(\mathbf{m}_{e}\) is as follow [33]:  
\[\mathbf{m}_{e}=\frac{\rho AL}{420}\left[\begin{array}{cc}\mathbf{m}_{11}&\mathbf{m}_{12 }\\ \mathbf{m}_{21}&\mathbf{m}_{22}\end{array}\right] \tag{5}\]  
Figure 4: Node displacement and DOF number of beam element.  
\[m_{11}=\left[\begin{array}{cccccc}m_{1}&0&0&0&0&0\\ 0&m_{3}&0&0&0&m_{4}\\ 0&0&m_{3}&0&-m_{4}&0\\ 0&0&0&m_{7}&0&0\\ 0&0&-m_{4}&0&m_{9}&0\\ 0&m_{4}&0&0&0&m_{9}\end{array}\right] \tag{6}\]  
\[m_{22}=\left[\begin{array}{cccccc}m_{1}&0&0&0&0&0\\ 0&m_{3}&0&0&0&-m_{4}\\ 0&0&m_{3}&0&m_{4}&0\\ 0&0&0&m_{7}&0&0\\ 0&0&m_{4}&0&m_{9}&0\\ 0&-m_{4}&0&0&0&m_{9}\end{array}\right] \tag{7}\]  
\[m_{12}={m_{21}}^{\intercal}=\left[\begin{array}{cccccc}m_{2}&0&0&0&0&0\\ 0&m_{5}&0&0&0&m_{6}\\ 0&0&m_{5}&0&-m_{6}&0\\ 0&0&0&m_{8}&0&0\\ 0&0&m_{6}&0&m_{10}&0\\ 0&-m_{6}&0&0&0&m_{10}\end{array}\right] \tag{8}\]  
In Equations (6)-(8), \(m_{1}=140\), \(m_{2}=70\), \(m_{3}=156\), \(m_{4}=22L\), \(m_{5}=54\), \(m_{6}=-13L\), \(m_{7}=\frac{140J}{A}\), \(m_{8}=\frac{70J}{A}\), \(m_{9}=4L^{2}\), \(m_{10}=-3L^{2}\), and \(\rho\) is the density.