# Cable Force Identification of Two Bending Cable Networks with Arbitrary Boundary Conditions
## 4 Application of Cable Force Identification Method
### A Two-Cable Network with Mixed Boundary Conditions
Since the supports of Cable 2 are hinged-ends, the effective vibration lengths of Cable 2 are equal to the actual length of Cable 2, namely 67.34 m.  
By replacing the actual cable lengths with the effective vibration lengths and fixed-end supports with hinged-end supports, the original configuration of the cable network is transferred to the equivalent configuration, as shown in Figure 10. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline  & \multicolumn{3}{c}{**Cable 1**} & \multicolumn{3}{c}{**Cable 2**} \\ \cline{2-7}
**Mode** & \begin{tabular}{c} **Measured** \\ **Frequency (Hz)** \\ \end{tabular} & \begin{tabular}{c} **Effective** \\ **Length (m)** \\ \end{tabular} & \begin{tabular}{c} **Cable** \\ **Force (LN)** \\ \end{tabular} & \begin{tabular}{c} **Measured** \\ **Frequency (Hz)** \\ \end{tabular} & \
\begin{tabular}{c} **Effective** \\ **Length (m)** \\ \end{tabular} &
\begin{tabular}{c} **Cable** \\ **Force (LN)** \\ \end{tabular} \\ \hline
1 & 1.3434 & 74.133 & 1895.48 & 1.33586 & 67.34 & 1597.83 \\
2 & 2.6969 & 74.167 & 1897.02 & 2.7318 & 67.34 & 1597.48 \\
3 & 4.0678 & 74.202 & 1896.00 & 4.1338 & 67.34 & 1596.79 \\
4 & 5.4669 & 74.234 & 1894.89 & 5.5787 & 67.34 & 1595.98 \\
5 & 6.9024 & 74.279 & 1893.85 & 7.0787 & 67.34 & 1594.53 \\ Average & - & 74.203 & 1895.45 & - & 67.34 & 1596.52 \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Cable effective vibration lengths of the single Cable 1. Figure 8: Five sensor locations on Cable 1. Figure 9: Measured frequency spectra of the single cable at five sensor locations on Cable 1. In the following step, this study determines the axial stiffness of the cross-tie for cable force identification later. The cable network obtained from the ABAQUS simulation is shown in Table 7.  
By introducing cable forces, cable effective vibration length, and measured frequencies obtained from the previous step into Equation (21) the axial stiffness of the cross-tie \(K_{c}\) can be estimated and shown in Table 8.  
Since the axial stiffness of the cross-tie is always positive, the estimated \(K_{c}\) at modes 1 and 9 can be ignored. By applying the \(MAD\) method, the detected range of the cross-tie axial stiffness is from 2758.9 kN to 4732.2 kN. Thus, the estimated \(K_{c}\) at modes 3, 5, and 7 can also be ignored. The mean value of the other estimated \(K_{c}\) is 3884.98 kN, approximately 13.43% error compared to the exact value of \(K_{c}\). The error of the estimated \(K_{c}\), namely more than 10%, is acceptable because it does not affect the cable force identification much. The explanation will be discussed later. Then, this paper uses the estimated \(K_{c}\) equal to 3884.98 kN to identify the cable forces of the cable network. Step 2 + 3: The measured frequencies obtained from the ABAQUS simulation are shown in Table 7. \begin{table}
\begin{tabular}{c c c c c c c c c c} \hline \hline
**Mode** & **1** & **2** & **3** & **4** & **5** & **6** & **7** & **8** & **9** & **10** \\
**Number** & & & & & & & & & & \\ \hline Measured & & & & & & & & & & \\ Frequency (Hz) & 1.3498 & 2.2888 & 2.7178 & 3.2621 & 4.0858 & 4.6516 & 5.5268 & 6.5773 & 6.8967 & 7.2613 \\ \hline \hline \end{tabular}
\end{table} 
Table 7: Measured frequencies of the cable network obtained from free vibration. Figure 10: Equivalent configuration of the cable network. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Mode** & **Measured Frequency of** & **Exact K\(c\) (kN/m)** & **Estimated K\(c\) (kN/m)** & **Error (\%)** \\
**Number** & & & & & & \\ \hline
1 & 1.3498 & 4487.50 & \(-\)74.06 & \(-\)101.65 \\
2 & 2.2888 & 4487.50 & 4027.27 & \(-\)10.26 \\
3 & 2.7178 & 4487.50 & 2191.93 & \(-\)51.15 \\
4 & 3.2621 & 4487.50 & 4010.18 & \(-\)10.64 \\
5 & 4.0858 & 4487.50 & 812.05 & \(-\)81.90 \\
6 & 4.6516 & 4487.50 & 3878.34 & \(-\)13.57 \\
7 & 5.5268 & 4487.50 & 696.95 & \(-\)84.47 \\
8 & 6.5773 & 4487.50 & 3885.37 & \(-\)13.42 \\
9 & 6.8967 & 4487.50 & \(-\)6210.50 & \(-\)238.40 \\
10 & 7.2613 & 4487.50 & 3623.76 & \(-\)19.25 \\ \hline \hline \end{tabular}
\end{table}