## 5 Calculation of the Tail Boom Vibrations  
The geometric parameters and physical tail boom material properties are presented in Table 2 and approximately correspond to parameters of a full scale light helicopter. The formulation adopted in this paper does not account for a non-uniform tail boom surface. For this reason, computations of forced vibrations were conducted for several equivalent uniform tail booms with diameter and mass per unit length, obtained from:  
\[D=\frac{D_{1}+D_{2}}{2}(1+\gamma),m_{L}=\rho_{b}(N_{s}L_{s}\delta_{s}+2\pi \delta D)+m_{c},\]  
where \(\gamma\) is a small parameter, \(m_{c}=30\) kg/m is a mass of internal tail boom construction elements per unit length. Table 3 presents the values of the \(\gamma\) parameters, mass and geometry used for forced vibration simulations. The beam moment of inertia is determined, taking into account properties of the stringers used typically for such beams. The moment of inertia of the stringers is determined by the expression:  
\[I_{s}=\frac{N_{s}}{8}D^{2}\delta_{s}L_{s},\]  
so that the total moment of inertia of the tail boom is  
\[I=\frac{\pi}{64}\Big{(}D^{4}-(D-\delta)^{4}\Big{)}+I_{s}.\]  
Based on the results of the CFD simulation of the rotor-fuselage interaction, the normal force coefficient was determined in the form \(c_{F}(\overline{t},\overline{x})=c_{Ft}(\overline{t})c_{Fx}(\overline{x})\), where \(c_{Ft}(\overline{t})\), \(c_{Fx}(\overline{x})\) are trigonometric functions. From Figure 10, it follows that \(c_{Fx}(0)\approx 0\), \(c_{Fx}(1)\approx 0\), \(|c_{Fx}(0.5)|\approx max\). The results of Figure 11 and the function \(c_{Bz,a}(\psi)\) (expression (2)) were used to determine the function \(c_{Ft}(\overline{t})\). In this case, the function \(c_{Fx}(\overline{x})\) has to satisfy to the condition \(\int_{0}^{1}c_{Fx}(\overline{x})d\overline{x}=1\). So, an approximation of the function \(c_{F}(\overline{t},\overline{x})\) in (15) was taken here as  
\[c_{F}\big{(}\overline{t},\overline{x}\big{)}=0.5\pi\alpha_{1}\sin(\pi \overline{x})\Big{(}1+\alpha_{2}\big{[}1+\alpha_{3}\cos\big{(}\gamma+N\overline {\omega}\overline{t}\big{)}\big{]}^{N}\Big{)}\]  
where \(f=6.023\) Hz is the main rotor frequency, \(\omega=2\pi f\) is the angular rotor frequency, and \(\overline{\omega}=\omega t_{0}\). \begin{table}
\begin{tabular}{c c} \hline Parameters & \\ \hline Diameter of the fixed beam end, \(D_{1}\) (m) & 0.546 \\ Diameter of the free beam end, \(D_{2}\) (m) & 0.346 \\ Beam length, \(L\) (m) & 4 \\ Wall thickness of the beam, \(\delta\) (m) & 0.001 \\ Thickness of the stringer, \(\delta_{S}\) (m) & 0.003 \\ Length of the stringer, \(L_{S}\) (m) & 0.015 \\ Number of stringers, \(N_{S}\) & 10 \\ Beam material density, \(\rho_{b}\) (kg/m\({}^{3}\)) & \(2.7\times 10^{3}\) \\ Young’s modulus, \(E\) (MPa) & \(72\times 10^{3}\) \\ \hline \end{tabular}
\end{table} 
Table 2: Tail boom parameters. \begin{table}
\begin{tabular}{c c c c c} \hline Variant, & Parameter & Diameter D, & Mass Per Unit Length & Dimensionless Rotor \\ \(N\) & \(\gamma\) & **m** & \(m_{L}\)**(kg/m)** & **Frequency N\(\overline{\omega}\)** \\ \hline
1 & 0.224 & 0.546 & 40.48 & 9.07 \\
2 & 0 & 0.446 & 38.78 & 10.52 \\
3 & \(-\)0.02914 & 0.433 & 38.56 & 10.9 \\
4 & \(-\)0.03812 & 0.4293 & 38.498 & 11.0168 \\
5 & \(-\)0.04484 & 0.426 & 38.44 & 11.12 \\
6 & \(-\)0.06502 & 0.417 & 38.29 & 11.41 \\
7 & \(-\)0.225 & 0.346 & 37.08 & 14.27 \\ \hline \end{tabular}
\end{table} 
Table 3: Equivalent mass and diameter of tail boom. One should note here that, for the employed main rotor frequency of 6.023 Hz, the time step \(\Delta t\) corresponding to 1 degree in rotor azimuth is \(612\times 10^{-4}\) s, that satisfies the condition \(\left(\Delta t\right)^{-1}\gg f_{1}\). The peak value of the normal force coefficient \(\left|c_{F\left(\overline{t}\right)}\right|\approx 0.043\) for the considered simulation parameters corresponds to a peak vertical tail boom load of about 30 N per meter of tail boom length.