Flexural Behavior and Capacity of Composite Concrete-Steel Beams Using Various Shear Connectors

Abstract

Composite-reinforced concrete/steel beams (CRCSB) have been commonly used due to the ability of composite sections to carry high loads with minimal material. Despite the considerable improvement made in the last decade to understand the effect of shear connectors between concrete and steel on CRCSB response, there is a lack in understanding how these connectors affect the cracks propagation and widened along the slab. Herein, an experimental investigation focusses on the behavior of CRCSB under flexure was presented. This paper aimed to investigate two main aspects: the effect of bearing stiffeners and shear connectors shapes on the failure mechanism and ultimate capacity of the CRCSB. Accordingly, five full-scale CRCSB were fabricated and evaluated under four-point bending. Four types of connecters are used (bolts, C-channels, angles and angles synergized by bolts). To precisely predict the number of shear connectors and maximum shear resistance of CRCSB, a new factor ${\gamma }_{SA}$ was adopted based on the theoretical and experimental findings. The observed cracks, strain, failure mechanism and load–displacement were recorded during the test and a qualitative comparison was presented. Experimental results revealed that the shear connector shape was the key factor influencing the concrete part performance. The failure of specimens was dominant by the strut and tie mechanism in pure bending zone. The CRCSB with bolts experienced the highest loads capacity while the two CRCSB had angles or angles and bolts as shear connectors gained nearly the same toughness.

Appendices

Appendix 1

Maximum Capacity of Manufactured Specimen

2.1 Using Proposed Factor of Safety

See Fig. 

Fig. 14

Dimensions of composite beam

14.

$${A}_{\mathrm{st}}=39.10{\mathrm{cm}}^{2} , {I}_{xs}=3890\,{\mathrm{cm}}^{4} ,$$

$${E}_{\mathrm{st}}=200000 \, \mathrm{Mpa} , {E}_{c}=20000 \,\mathrm{Mpa}$$

$$n=\frac{{E}_{\mathrm{st}}}{{E}_{\mathrm{c}}}=\frac{200000}{20000}=10 \, \mathrm{Without}\, \mathrm{crack}$$

$$\overline{y }=\frac{\sum Ay}{\sum A}=\frac{\left(3910\times 120\right)+\left(\frac{1}{10}\times 500\times 80\times 280\right)}{\left(3910\right)+\left(\frac{1}{10}\times 500\times 80\right)}$$

$$\overline{{\varvec{y}} }=200.91\, \mathrm{mm}$$

$${{\varvec{h}}}_{1}=320-200.91=119.1\, \mathrm{mm}$$

$$\frac{1}{n}\times {b}_{c}\times {t}_{c}\left(Z-\frac{{t}_{c}}{2}\right)-{A}_{st}\left({h}_{1}-Z\right)=0$$

$$\frac{1}{10}\times 500\times 80\left(Z-\frac{80}{2}\right)-3910\times \left(119.1-Z\right)=0$$

$${\varvec{Z}}=79.1 \mathrm{mm}$$

$${I}_{x}=\frac{1}{n}\times \left[\frac{{b}_{c}\times {{t}_{c}}^{3}}{12}+{b}_{c}\times {t}_{c}\times {\left(\overline{y }-\left({h}_{s}+\frac{{t}_{s}}{2}\right)\right)}^{2}\right]+\left[{I}_{xs}+{A}_{st}{\left(\overline{y }-\frac{{h}_{s}}{2}\right)}^{2}\right]$$

$${I}_{x}=\frac{1}{10}\left[\frac{500\times {80}^{3}}{12}+500\times 80\times {\left(200.91-\left(240+\frac{80}{2}\right)\right)}^{2}\right]+\left[3890\times {10}^{4}+3910{\left(200.91-\frac{240}{2}\right)}^{2}\right]$$

$${I}_{x}=27154245.78+64496533.87=91650779.65 {\mathrm{mm}}^{4}$$

$$\sigma =\frac{{M}_{x}}{{I}_{x}}\overline{y }$$

$${M}_{x}=\frac{\sigma \times {I}_{x}}{\overline{y} }=\left(\frac{25\times 91650779.65}{119.1}\right)\times {10}^{-6}=19.34 \mathrm{kN}.\mathrm{m}$$

$$ M_{x} = \frac{19.34}{\frac{1}{n}} = 193.4 {\text{kN}}\,{\text{m}} $$

$$ P_{n} = \frac{193.4}{{0.6}} \times 2 = 641.25 \,{\text{kN}} $$

$${P}_{nsa}=\frac{{P}_{n}}{{\gamma }_{SA}}=\frac{641.25}{1.25}=513 \mathrm{kN}$$