## 2 Robustness Analysis Theory for Flexible Cable-Bar Tensile Structures
### Calculation Method of Robustness Analysis for Tensile Structures Based on Random Theory  
In this paper, input interference and output response were quantified as the node load interference input vector and the corresponding node displacement output vector. Then, the structural robustness indicator could be established by calculating the proportion of the output node displacement and the input interference load. The input interference \(w(t)\) consisted of substantial independent random factors and could be reckoned as a normal distribution. According to the principle of normal distribution function in probability theory, the chance of interference occurrence exceeded 99.74% in the interval \((\mu-3\sigma,\ \ \mu+3\sigma)\), in which \(\mu\) was the average value and \(\sigma\) was the standard deviation. Therefore, although the range of normal variables was \((-\infty,\ +\infty)\), it was certain that the interference load occurred within the range \((\mu-3\sigma,\ \ \mu+3\sigma)\). Based on the theory, the specific calculating method for the robustness of cable-bar tensile structure comprised the following steps. 1. The total combined loads of cable-bar tensile structure included conventional load \(F_{0}\) and external interference load \(w(t)\). The conventional loads included static load and dynamic load. The interference load obeyed the normal distribution, the distribution interval was \((-3 var,\ \ 3 var)\), and the area was divided into \(m\) sections. The ratio between the interference load and the conventional load in the interval \(k\) was as follows:  
\[\alpha(k)=\pm\ \frac{6k\cdot var}{m},\ k=1,2,3,\cdot\cdot\cdot,m/2 \tag{14}\]  
In the formula, \(var\) was the coefficient of variation, which was set as 0.005 in this paper, then the interference load within the interval could be calculated as follows:  
\[w_{k}(t)=F_{0}\cdot\alpha(k),\ k=1,2,3,\cdot\cdot\cdot,m/2 \tag{15}\]  
The combined load \(F_{k}\) in each sub-section was the sum of the conventional load \(F_{0}\) and the interference load \(w_{k}(t)\), which was calculated as follows:  
\[F_{k}=F_{0}+w_{k}(t),\ k=1,2,3,\cdot\cdot\cdot,m/2 \tag{16}\]  
\(F_{k}\) was subjected to the normal distribution of parameters \(\mu\) and \(\sigma\) in the interval \((\mu-3\sigma,\ \ \mu+3\sigma)\), in which \(\mu\) was the average value and \(\sigma\) was the standard deviation. Then, the following formula could be obtained:  
\[\sigma=\mu\cdot var \tag{17}\]  
To ensure the efficiency and accuracy of the calculation results, \(m\) was 100 in this paper. Meanwhile, since the interference load \(w(t)\) caused both positive and negative occurrence, and the occurrence probability of the same absolute value of positive and negative interference load was equal, the value of the probability interval is 50/50. (2) The nodal displacement vector under the conventional load \(F_{0}\) and the combined load \(F_{k}\) were \(y\) and \(y_{k}\), respectively, which could be calculated by software ANSYS. According to the interval divided by the combined load in step (1), the probability distribution function of the combined load in interval \(k\) was calculated and used as the weight coefficient. The formula was as follows:  
\[Q(k)=\left\{ \begin{array}{cc}\int\limits_{\frac{6k-3}{m}}^{\frac{6k+3}{m}} \frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}dt,&\quad k=1\sim\frac{m}{2}-1\\ 1-\int\limits_{-\infty}^{3-\frac{3}{2\sigma}}\frac{1}{\sqrt{2\pi}}e^{-\frac{ t^{2}}{2}}dt,&\quad k=\frac{m}{2}\end{array} \right.