## 3 Results and Discussion
### Prediction Model of SIF at Crack Tip of Integral Panel  
In the above sections, taking the spacing, height and thickness of ribs as a single variable respectively, the influence of which on the SIFs of the integral panel was studied [28]. In this section,  
Figure 8: SIF curves for crack lengths of (a) 5 mm, (b) 10 mm, (c) 15 mm, (d) 20 mm, (e) 25 mm, and (f) 30 mm.  
Figure 9: Comparison of the effect of rib thickness on SIF between FEM result and reference result.  
regression analysis was used to analyze the comprehensive influence of various parameters of rib structure on the SIFs of the integral panes in the process of crack propagation [29].  
Based on the Taylor expansion method and the data of finite element results, taking the panel spacing of 80 mm, panel height of 70 mm, panel thickness of 4 mm and crack size of 5 mm as the starting point, the corresponding SIF as the starting value of the function, and considering the quadratic term of the expansion formula, the stress intensity of the integral panel was calculated by a multivariate nonlinear regression model taking the rib spacing, height, thickness and crack size as the function variables. The derivation model is as follows [15]:  
\[\begin{array}{l}K_{I}(d,h,t,a)=K_{I0}+c_{1}d+c_{2}h+c_{3}t+c_{4}a+c_{11}d^{2 }+c_{22}h^{2}+c_{33}t^{2}c_{44}a^{2}+c_{21}hd\\ \hskip 14.226378 pt+c_{31}td+c_{41}ad+c_{32}th+c_{42}ah+c_{43}at+\varepsilon, \end{array} \tag{5}\]  
where, \(K_{I}(d,h,t,a)\) is the objective function, d is the rib spacing, h is the rib height, t is the rib thickness, a is the crack size, \(K_{I0}\) is the SIF at the starting point of expansion, c1 to c43 are various coefficients, and \(\varepsilon(d,h,t,a)\) is the higher order term of expansion error.  
According to the simulation results, the coefficient of the regression model was solved, and the multi-element nonlinear relationship between the SIF and the rib spacing, rib height, rib thickness and crack size was obtained as follows:  
\[\begin{array}{l}K_{\rm I}=109689.688-3210.940d-9678.911h+83623.491t+443.002a +0.363d^{2}\\ \hskip 14.226378 pt+13.962h^{2}+101.811t^{2}-2.163a^{2}+177.303 hd-539.879 td+0.237 ad \\ \hskip 14.226378 pt-1380.146 th-8.294 ah-7.048 at,\end{array} \tag{6}\]  
The established regression equation can be used to predict the crack tip SIF of the integral wall panel with different structural parameters in the process of penetrating crack propagation. The comparison between the FEM results and the predicted results is shown in Figure 10, in which the predicted values are obtained by the regression equation above.  
The determination coefficient \(R^{2}\) was used to measure the accuracy of the model, the value range of coefficient R2 is 0 to 1. The closer the determinable coefficient is to 1, the better the goodness of fit of the model is, and its calculation formula is shown as follows:  
\[R^{2}=1-\frac{\sum(y-\hat{y})^{2}}{\sum(y-\overline{y})^{2}}, \tag{7}\]  
where, \(y\) is the FEM value, \(\hat{y}\) is the predicted value, \(\overline{y}\) is the mean value of the FEM, \(\sum(y-\hat{y})^{2}\) is the sum of the squared residuals of the regression model, and \(\sum(y-\overline{y})^{2}\) is the sum of the squared total deviations of the model. Substitute the model prediction results and finite element results into  
Figure 10: Comparison between the FEM results and the predicted results.  
Equation (7), the determination coefficient R2 of the regression model can be obtained to be 0.955. By determining the coefficient R and the high fitting of the two curves in Figure 10, we show that the predicted value of the model has a high degree of fit with the FEM value, and the prediction effect of the model is good.