## 4 Time Histories of Two Basic Scenarios  
When the two basic scenarios start, the rotating speed of driving motor is \(n_{0}=\pm n_{\rm e}\) and the position of the payload is determined by the two parameters \(l_{0\rm b}\) and \(y_{\rm o3b}\). The basic scenarios of \(n_{0}=\pm n_{\rm e}\) are referred to as scenarios A and B respectively. In the following simulation results, \(T_{si}\)\((i=2,3,\ 4,\ 5)\) are the torques acting on the shaft used to connect components \(J_{i}\) and \(J_{i+1}\) of the high-speed links in the lifting mechanism and calculated by \(k_{i}(\theta_{i}-\theta_{i+1})\); \(M_{\rm bj}\) (\(i=4,6\)) is the calculated braking torques of the mechanical brake \(i\) (\(i=4,6\)). \begin{table}
\begin{tabular}{l l l l} \hline \(b=2.87\) m & \(J_{\rm T}=13,459\) kg\(\cdot\)m\({}^{2}\) & \(L=27\) m & \(t_{\rm bs}=0.2\) s \\ \(c_{\rm s}=2.32\times 10^{-4}\)s & \(k_{2}=5.038\times 10^{9}\) N\(\cdot\)m/rad & \(m_{\rm g}=8875\) kg & \(t_{\rm ba}=0.1\) s \\ \(l_{\rm x}=1.345\times 10^{-2}\)m\({}^{4}\) & \(k_{3}=6.454\times 10^{8}\) N\(\cdot\)m/rad & \(m_{\rm Q}=32,816\) kg & \(y_{\rm o3b}=11.305\) m \\ \(J_{2}=15,686\) kg\(\cdot\)m\({}^{2}\) & \(k_{4}=1.606\times 10^{9}\) N\(\cdot\)m/rad & \(m_{\rm s}=3\) & \(\eta_{\rm L}=0.89\) \\ \(J_{3}=3004\) kg\(\cdot\)m\({}^{2}\) & \(k_{5}=2.600\times 10^{9}\) N\(\cdot\)m/rad & \(m_{\rm T}=12,802\) kg & \(\xi_{\rm g}=0.04\) \\ \(J_{4}=9211\) kg\(\cdot\)m\({}^{2}\) & \(k_{\rm w}=1.566\times 10^{8}\) N & \(M_{\rm blim}=51,471\) Nm & \(\xi_{\rm w}=0.01\) \\ \(J_{5}=4185\) kg\(\cdot\)m\({}^{2}\) & \(l_{00}=43\) m & \(n=10\) & \(\chi_{\rm d}=1.66\) m \\ \(J_{6}=8544\) kg\(\cdot\)m\({}^{2}\) & \(l_{0\rm b}=2\) m & \(n_{\rm e}=712\) rpm & \(\chi_{\rm T}=1.56\) m \\ \hline \end{tabular}
\end{table} 
Table 1: The data used for simulation  
Figure 5: Time function about braking capacity of mechanical brake. These results about the high-speed links have been transformed back to the same kind of quantities in reference to the high-speed axis. Figures 6 and 7 show some time histories in scenarios A and B respectively. They describe torques acting on shafts of the high-speed links, the braking torques exerted by the two mechanical brakes, and the rotational speed (in rpm) of one of components \(J_{4}\) and \(J_{6}\), which stops first (expressed by red dashed lines). The time history of \(T_{82}\) is omitted because it looks like \(T_{83}\), just its mean values in the first and second phases, and the amplitudes of its high frequency components in all phases being a little smaller. From Figure 6a,b and Figure 7a,b, it is found that these histories are clearly divided into three phases, which successively correspond to the phase without brake application, deceleration phase due to application of the mechanical brakes, and the phase after completion of braking, respectively. In the first and the third phases, all torque responses are time-varying with piecewise constant mean values, which implies that the crane system is being subjected to constant external forces during these phases. At the beginning of the second phase, the mean values of these responses undergo linear variations with the increase of the braking torque. In the end, these mean values will become constant once the braking torques increase to their limit values. In the view of the traditional design theory on cranes, transmission shafts of lifting mechanisms are subjected to only a pulsating load per crane work cycle. However, it is noted that transmission shafts are subjected to time-varying loads in two opposite directions in scenarios A and B. Obviously,  
Figure 6: Time histories of torques of high-speed links and braking torques of two brakes in scenario A. (**a**) Torque of the shaft connecting components \(J_{3}\) and \(J_{4}\); (**b**) Torque of the shaft connecting components \(J_{i}\) and \(J_{i+1}\). (\(i=4,5\)); (**c**) Braking torque provided by the brake assembled on components \(J_{4}\); (**d**) Braking torque provided by the brake assembled on components \(J_{6}\).