Behavior of reinforced concrete deep beam with web openings strengthened with (CFRP) sheet


Jasim W. A., Tahnat Y. B. A., Halahla A. M.

STRUCTURES, cilt.26, ss.785-800, 2020 (SCI-Expanded) identifier identifier

  • Yayın Türü: Makale / Tam Makale
  • Cilt numarası: 26
  • Basım Tarihi: 2020
  • Doi Numarası: 10.1016/j.istruc.2020.05.003
  • Dergi Adı: STRUCTURES
  • Derginin Tarandığı İndeksler: Science Citation Index Expanded (SCI-EXPANDED), Scopus
  • Sayfa Sayıları: ss.785-800
  • Anahtar Kelimeler: Carbon fiber-reinforced polymer (CFRP), Deep beam, Finite element, Shear span ratio, Concrete damage plasticity (CDP), SHEAR, SECTION, DEFORMATION, FAILURE, STRAIN, MODEL
  • Orta Doğu Teknik Üniversitesi Adresli: Evet

Özet

Deep beams are used in various applications in reinforced concrete (R.C.) structures. There have been continuous efforts to enhance and improve the performance of these crucial elements in (R.C.) structures by using several strengthening techniques such as using the carbon fiber-reinforced polymer (CFRP). However, by exploring the literature, none of the previously conducted experimental tests have studied the propagation of cracks beneath the (CFRP) sheets. In this research, the propagation of the first diagonal crack, which takes place beneath the (CFRP), is investigated by modeling sixteen (R.C.) deep beams with different opening sizes. Two shear span/ depth ratios (a/h) are studied numerically using the finite element analysis tool (ABAQUS). All models are validated using the concrete damage plasticity (CDP) model, and their results are found similar to the experimental results obtained by other authors. Results show that the Finite Element models catch the real behavior of the (R.C.) deep beams. In addition, the (CFRP) sheets are found to enhance the failure load capacity as well as the flexural crack remarkably. Moreover, the (CFRP) increases the load required to cause the first diagonal crack for models with (a/h) equals 0.9, while there is a slight change in this load for models with (a/h) equals 1.1.