# Vibration Properties of Dual-Rotor Systems under Base Excitation, Mass Unbalance and Gravity
## 2 Modeling for a Dual-Rotor System under Base Motions
### EOMs for the Shaft  
A two-node Timoshenko beam replaced the Euler beam element from [1] to model the shear deformation effect on the shaft. The two nodes of the shaft element were \(C_{s,j}\) and \(C_{s,j+1}\), and their position vectors with respect to the base frame were defined as \(\big{(}\mathbf{O_{b}C_{s4}}\big{)}_{F_{b}}=\{\ x_{s,j}\ \ \ y_{s,j}\ \ z_{s,j}\ \}^{\text{T}}\) and \(\big{(}\mathbf{O_{b}C_{s,j+1}}\big{)}_{F_{b}}=\{\ x_{s,j+1}\ \ \ y_{s,j+1}\ \ z_{s,j+1}\ \}^{\text{T}}\). The  
Figure 2: Derivation of the linear and angular velocity of the disk: (**a**) Linear velocity of the disk; (**b**) Angular velocity of the disk. length of the shaft element was denoted as \(l\). After the discretization of the shaft, the values of \(y_{s,j}\), \(y_{s,j+1}\) and \(l\) remained constant. The general coordinate vector of the shaft element was  
\[\mathbf{q_{s}}=\left[ \begin{array}{cccccc}x_{s,j}&z_{s,j}&\theta_{s,j}&\psi_{ s,j}&x_{s,j+1}&z_{s,j+1}&\theta_{s,j+1}&\psi_{s,j+1}\end{array} \right]^{\mathrm{T}} \tag{13}\]  
After taking an arbitrary point \(C_{s}\) with a distance of \(y\) along \(O_{b}Y_{b}\) from \(C_{s,j}\) into consideration, its position vector was defined with respect to the base frame as \(\left(\mathbf{O_{b}C_{s}}\right)_{F_{2}}=\left\{ \begin{array}{cccccc}x_{s}&y _{s}&z_{s}\end{array} \right\}^{\mathrm{T}}\), where \(y_{s}=y_{s,j}+y\); \(x_{s}\) and \(z_{s}\) are unknowns; and \(\theta_{s}\) and \(\psi_{s}\) were attained by interposition:  
\[\left[ \begin{array}{cccccc}x_{s}(y)&z_{s}(y)\end{array} \right]^{\mathrm{T}} =\psi(y)\mathbf{q_{s}} \tag{14}\]  
\[\left[ \begin{array}{cccccc}\theta_{s}(y)&\psi_{s}(y)\end{array} \right]^{ \mathrm{T}}=\mathbf{\Phi}(y)\mathbf{q_{s}} \tag{15}\]  
For the Timoshenko beam element, the interposition function is given by  
\[\psi(y)=\left[ \begin{array}{cccccc}\psi_{1}&0&0&-\psi_{2}&\psi_{3}&0&0&- \psi_{4}\\ 0&\psi_{1}&\psi_{2}&0&0&\psi_{3}&\psi_{4}&0\end{array} \right] \tag{16}\]  
\[\mathbf{\Phi}(y)=\left[ \begin{array}{cccccc}0&\phi_{1}&\phi_{2}&0&0&\phi_{ 3}&\phi_{4}&0\\ -\phi_{1}&0&0&\phi_{2}&-\phi_{3}&0&0&\phi_{4}\end{array} \right] \tag{17}\]  
where  
\[s=\frac{y}{l}\text{, }\Phi=\frac{12}{l^{2}}\frac{EI_{s}}{\kappa GA_{s}}\text{, }G= \frac{E}{2(1+\mu)}\text{, }A_{s}=\pi(r_{so}^{2}-r_{si}^{2}) \tag{18}\]  
\[\kappa=\frac{6\big{(}r_{si}^{2}+r_{so}^{2}\big{)}^{2}(1+\mu)^{2}}{7r_{si}^{4} +34r_{si}^{2}r_{so}^{2}+7r_{so}^{4}+\mu\big{(}12r_{si}^{4}+48r_{si}^{2}r_{so}^ {2}+12r_{so}^{4}\big{)}+\mu^{2}\big{(}4r_{si}^{4}+16r_{si}^{2}r_{so}^{2}+4r_{so }^{4}\big{)}} \tag{19}\]  
\[\psi_{1}=\frac{1}{1+\Phi}\Big{(}1-3s^{2}+2s^{3}+\Phi(1-s)\Big{)}\text{, }\psi_{2}=\frac{ ls}{1+\Phi}\Big{(}1-2s+s^{2}+\frac{1}{2}\Phi(1-s)\Big{)} \tag{20}\]  
\[\psi_{3}=\frac{s}{1+\[PERSON]}(3s-2s^{2}+\Phi)\text{, }\psi_{4}=\frac{ls}{1+\[PERSON]} \Big{(}-s+s^{2}-\frac{1}{2}\Phi(1-s)\Big{)} \tag{21}\]  
\[\phi_{1}=\frac{6s}{l(1+\[PERSON])}(-1+s)\text{, }\phi_{2}=\frac{1}{1+\Phi}\big{(}1-4s +3s^{2}+\Phi(1-s)\big{)} \tag{22}\]  
\[\phi_{3}=\frac{6s}{l(1+\Phi)}(1-s)\text{, }\phi_{4}=\frac{s}{1+\Phi}(-2+3s+\Phi) \tag{23}\]  
The absolute linear velocity \(\mathbf{V_{s}}\) and absolute angular velocity \(\mathbf{\omega_{s}}\) of the point \(C_{s}\) can be derived in similar steps as the disk. Then, the kinetic energy of a shaft element is given by  
\[T_{s}=\frac{1}{2}\rho_{s}A_{s}\int_{0}^{l}\big{(}\mathbf{V_{s}}^{\mathrm{T}} \mathbf{V_{s}}\big{)}\mathrm{d}y+\frac{1}{2}\int_{0}^{l}\big{(}\mathbf{\omega_ {s}}^{\mathrm{T}}\mathbf{I_{s}}\mathbf{\omega_{s}}\big{)}\mathrm{d}y \tag{24}\]  
where \(\mathbf{I_{s}}=\text{diag}\big{(}\left[ \begin{array}{cccccc}i_{d}&i_{p}&i_{d }\end{array} \right]\big{)}\); \(\rho_{s}\), \(A_{s}\), \(i_{d}\) and \(i_{p}\) are, respectively, the material density, area of shaft cross-section, the transverse and polar mass-moment of inertia per unit length of the shaft element.