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Machine Learning-Based Fragility Assessment of Reinforced Concrete Buildings

In the past, large earthquakes caused the collapse of infrastructure and killed thousands of people in Pakistan, a seismically active region. Therefore, the seismic assessment of infrastructure is a dire need that can be done using the fragility analysis. This study focuses on the fragility analysis...

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Detalles Bibliográficos
Autores principales: Rasheed, Abdur, Usman, Muhammad, Zain, Muhammad, Iqbal, Nadeem
Formato: Online Artículo Texto
Lenguaje:English
Publicado: Hindawi 2022
Materias:
Acceso en línea:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9436535/
https://www.ncbi.nlm.nih.gov/pubmed/36059412
http://dx.doi.org/10.1155/2022/5504283
Descripción
Sumario:In the past, large earthquakes caused the collapse of infrastructure and killed thousands of people in Pakistan, a seismically active region. Therefore, the seismic assessment of infrastructure is a dire need that can be done using the fragility analysis. This study focuses on the fragility analysis of school buildings in Muzaffarabad district, seismic zone-4 of Pakistan. Fragility curves were developed using incremental dynamic analysis (IDA); however, the numerical analysis is computationally time-consuming and expensive. Therefore, soft computing techniques such as Artificial Neural Network (ANN) and Gene Expression Programming (GEP) were employed as alternative methods to establish the fragility curves for the prediction of seismic performance. The optimized ANN model [5-25-1] was used. The feedforward backpropagation network was considered in this study. To achieve a reliable model, 70% of the data was selected for training and 15% for validation and 15% of data was used for testing the model. Similarly, the GEP model was also employed to predict the fragility curves. The results of both ANN and GEP were compared based on the coefficient of determination, R(2). The ANN model accurately predicts the global drift values with R(2) equal to 0.938 compared to the GEP model having R(2) equal to 0.87.