Abstract
Carbon fiber reinforced polymer (CFRP) is widely used in the field of strengthening concrete-filled steel tubular (CFST) columns due to its excellent performance. However, the interactions between steel tube and concrete, and between steel tube and CFRP are very complex, which make it difficult to predict the strength. To solve this problem, a new method using an advanced machine learning algorithm, XGBoost, to predict the axial compressive capacity of CFRP-confined CFST short columns is proposed. The data set for training and testing contains 379 pieces of data out of two parts: 271 from the literature and 108 from our experiments presented in this paper. Firstly, the typical failure modes and stress mechanisms of the specimens and the effects of the number of CFRP layers, the strength of core concrete and section forms on axial compressive capacity are analyzed through experiments. Then, the machine learning is carried out with the material strengths (concrete, steel, CFRP), the cross-sectional areas of the materials (concrete, steel tube, CFRP) and the cross-sectional shapes of the specimen as the features, and the axial compression bearing capacity of the specimen as the labels. Eight algorithms: Linear Regression (LR), K-Nearest Neighbor (KNN), Support Vector Machine (SVM), and typical ensemble learning models such as Random Forest (RF), Adaptive Boosting (AdaBoost), Gradient Boosting Decision Tree (GBDT), Extreme Gradient Boosting (XGBoost), LightGBM Regressor (LGB) are selected for calculations. Based on RandomizedSearchCV and 5-fold cross validation, the performance of the eight algorithms is compared. It is found that XGBoost has the best prediction performance, with the coefficient of determination R2 = 0.9719. Finally, based on the learning curves and grid search, nine hyperparameters of XGBoost model are adjusted, the optimal combination of hyperparameters of XGBoost model is obtained. The results show that the performance of XGBoost model is improved, with the coefficient of determination R2 = 0.9850.
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