which gives K_e = 9021.2 Nm/rad. The flexible natural frequency of the equivalent two-disc rotor system as shown in Figure 6.34 is given as

$${\omega }_{n{f}_2}=\sqrt{\frac{(I_{P_{A_e}}+I_{P_B})k_e}{(I_{P_{A_e}}{I}_{P_B})}}=\sqrt{\frac{\left(\text{6}+\text{10}\right)\times \text{9021}\text{.2/9}\text{.81}}{\left(\text{6}\times \text{10}\right)\text{/9}{\text{.81}}^{\text{2}}}}\text{=153}\text{.62 rad/sec}$$

Figure 6.35 Mode shape and nodal point location in the equivalent system

The node location can be obtained from Figure 6.35 as

$$\frac{{\Phi }_{z_{A_e}}}{l_{n_1}}=\frac{{\Phi }_{z_B}}{l_{n_2}}$$

which can be written by noting equation (6.13), as

$$\frac{l_{n_1}}{l_{n_2}}=\frac{{\Phi }_{z_{A_e}}}{{\Phi }_{z_B}}=-\frac{I_{P_B}}{I_{P_{A_e}}}=-\frac{10}{6}=-1.667$$

The negative sign indicates that both discs are at either end of the node location. The absolute location of the node position is given as

$$l_{n_1}=1.667l_{n_2}$$

Also from Figure 6.35, we have

$$l_{n_{\text{1}}}+l_{n_{\text{2}}}=\text{2}\text{.2288}$$ which gives $$l_{n_{\text{2}}}=\text{0}\text{.8358 m}$$

Hence, the node is on shaft B at 0.8356 m from disc B. Alternatively, from similar triangle of the mode shape (Figure 6.35), we have

$$\frac{l_{n_2}}{2.2288-l_{n_2}}=\frac{{\Phi }_{z_B}}{{\Phi }_{z_{A_e}}}=\frac{1}{1.667}\Rightarrow l_{n_2}=0.8358\text{ m}$$

$$\text{Let }{\Phi }_{z_B}=\text{1}\begin{array}{cc}\text{ rad,}& \text{then }{\Phi }_{z_{A_e}}=-\text{1}\text{.667 rad}\end{array}$$ $$\text{hence,}\begin{array}{cc}& {\Phi }_{z_A}\end{array}=\frac{{\Phi }_{z_{A_e}}}{n}=-\text{0}\text{.8333 rad}$$