Advancing V estimation from CPTu for engineering practice: a data-driven approach

Abstract

Shear wave velocity, V s, is a critical parameter for offshore site characterisation to estimate the small strain shear modulus, which is essential for subsequent geotechnical designs. Direct measurements of V s are often sparse due to time and resource constraints, while indirect estimations of V s based on empirical correlations can exhibit significant errors. This study presents the performance of 125 models with various combinations of standard piezocone tests (CPTu) input features (e.g., depth, z; sleeve friction resistance, f s; corrected cone tip resistance, q t; and pore pressure at the shoulder of the cone, u 2), CPTu and V s data pairing methods, and prediction techniques (support vector regression (SVR), random forest regression (RFR), extreme gradient boosting regression (XGBR), deep neural network (DNN) and multiple linear regression (MLR)). To do this, we compile a seismic piezocone test (SCPTu) database from onshore and offshore sites across the globe (Netherlands, Austria, Germany, Nepal, and Taipei) and consider five different methods for pairing CPTu data (resolution of 0.02 m) and V s data (resolution of 0.5 m and 1 m depending on the dataset). Two cases consider the more conventional downsampling of CPTu data to V s data. The remaining three methods consider augmented V s data to the resolution of CPTu measurements, to fully utilise all the CPTu data. Results indicate that data augmentation enhances predictive performance. Incorporating pore pressure as an input feature also improves model performance, particularly in cemented materials such as chalk. In contrast, the derived features have a negligible influence. The recommended model combines a DNN with four directly measured CPTu parameters (z,f s,q t,and u 2), and uses an augmentation method that assumes constant V s values within each V s interval. This model achieves a mean absolute error (MAE) of 37.3 m/s and a coefficient of determination (R 2) of 0.59.</p