Revisiting the shear design equations for concrete beams reinforced with FRP rebar and stirrup

Abstract

Current design guidelines and codes use modified shear equations for calculating the shear strength contribution of fiber reinforced polymer (FRP) transverse reinforcement (stirrup) in concrete beams reinforced with longitudinal steel or FRP rebar. These equations, originally developed for steel as longitudinal and transverse reinforcement, are semi-empirical in nature and were developed with a core analytical equation where the coefficients were determined from regression analysis. Here, a comparative study among various code equations was conducted to predict the shear strength of FRP reinforced concrete beams. To facilitate this comparison a database was established from the published literature, which was composed of slender concrete beams (shear span to depth ratio, a/d >2.5) with FRP longitudinal and transverse reinforcement. The database contained 114 beams without transverse reinforcement and 46 beams with transverse reinforcement. The guidelines, codes and models that were implemented and compared in this study consisted of ACI 440.1R-06, JSCE 1997, CNR-DT 203, CSA S806-02, CSA S6-06, unpublished CSA S6-06 Addendum, ISIS-M03-01, simplified Modified Compression Field Theory, cracked section analysis model and modified Zsutty equations. It was observed from the statistical analysis that the CSA S806-02 produced greater coefficients of variation than the CSA S6-09. The ACI 440.1R-06 and JSCE 1997 produced more conservative results in calculating the transverse shear strength. The CSA S6-09 Addendum exhibited the best all-around performance in predicting the shear contribution of FRP reinforced beams compared to that of other design codes and guidelines.

Appendix

 

Model	Concrete shear equations	Transverse shear equation
ACI440.1R-06	$V_{c} = \frac{2}{5}\sqrt {f^{'}_{c} } bc$ $c = kd$ $k = \sqrt {2\rho_{\text{fl}} n_{f} + \left( {\rho_{\text{fl}} n_{f} } \right)^{2} } - \rho_{\text{fl}} n_{f}$ $n_{f} = E_{\text{fl}} /E_{c}$	$V_{\text{fv}} = \frac{{A_{\text{fv}} \sigma_{\text{fv}} d}}{\text{s}}$ $\sigma_{\text{fv}} = 0.004E_{\text{fv}} \le \,f_{\text{FRPbend}}$ $f_{\text{FRPbend}} = \frac{{\left( {\frac{{0.05r_{b} }}{{d_{b} }} + 0.3} \right)}}{1.5}f_{\text{fv}} \,\le\, f_{\text{fv}}$
CSA S6-09 Addendum [3]	$V_{c} = 2.5\beta \varphi_{c} f_{\text{cr}} bd_{v}$ $f_{\text{cr}} = 0.4\sqrt {f^{'}_{c} }$ $d_{v} \ge \left\{ {\begin{array}{*{20}c} {0.9d} \\ {0.72h} \\ \end{array} } \right.$ $\beta = \frac{0.4}{{\left( {1 + 1500\varepsilon_{x} } \right)}} \times \frac{1300}{{\left( {1000 + s_{\text{ze}} } \right)}}$ $\varepsilon_{x} = \frac{{\left( {\left( {\frac{{M_{f} }}{d}} \right) + V_{f} + 0.5N_{f} } \right)}}{{2\left( {E_{\text{fl}} A_{\text{fl}} } \right)}} \le 0.003$ $\varphi_{c} = 0.75$ $A_{{v, { \min }}} = \frac{{0.3\sqrt {f^{'}_{c} } bs}}{{f_{\text{fv}} }}$ If A v  > A v, min, s ze = 300 mm. If A v  < A v, min, s ze = 35sz/(15 + ag) > 0.85s z , where s z is the crack spacing parameter and shall be taken as d v or as the distance between layers of distributed longitudinal reinforcement where each intermediate layer of such reinforcement has an area at least equal to 0.003bs z .	$V_{\text{FRP}} = \frac{{\varphi_{f} A_{\text{fv}} \sigma_{\text{fv}} d_{v} { \cot }\theta }}{\text{s}}$ $\theta = \left( {29 + 7000\varepsilon_{x} } \right)\left( {0.88 + \frac{{s_{\text{ze}} }}{2500}} \right)$ $\sigma_{\text{fv}} {\text{ is the smaller of }}\left\{ {\begin{array}{*{20}c} {\left( {\frac{{\left( {\frac{{0.05r_{b} }}{{d_{b} }} + 0.3} \right)f_{\text{fv}} }}{1.5}} \right)} \\ {0.004E_{\text{fv}} } \\ \end{array} } \right.$ $\varphi_{f} = \left\{ {\begin{array}{*{20}c} {0.5\quad {\text{for GFRP}}} \\ {0.75\quad {\text{for CFRP}}} \\ \end{array} } \right.$
CSA S6-06 [13]	$V_{c} = 2.5\beta \varphi_{c} f_{\text{cr}} bd_{v} \sqrt {\frac{{E_{{{\text{f}}l}} }}{{E_{\text{s}} }}}$ $f_{\text{cr}} = 0.4\sqrt {f^{'}_{c} }$ $d_{v} \ge \left\{ {\begin{array}{*{20}c} {0.9d} \\ {0.72h} \\ \end{array} } \right.$ $\beta = \frac{0.4}{{\left( {1 + 1500\varepsilon_{x} } \right)}} \times 1300/\left( {1000 + s_{ze} } \right)$ $\varepsilon_{x} = \frac{{\left( {\left( {\frac{{M_{f} }}{d}} \right) + V_{f} + 0.5N_{f} } \right)}}{{2\left( {E_{\text{s}} A_{\text{s}} } \right)}} \le 0.003$ $\varphi_{c} = 0.75$ $A_{v, \min } = \frac{{0.3\sqrt {f^{'}_{c} } bs}}{{f_{\text{fv}} }}$ If A v  > A v, min , s ze  = 300 mm. If A v  < A v, min , s ze  = 35sz/(15 + ag) > 0.85s z , where s z is the crack spacing parameter and shall be taken as d v or as the distance between layers of distributed longitudinal reinforcement where each intermediate layer of such reinforcement has an area at least equal to 0.003bs z .	$V_{\text{fv}} = \frac{{\varphi_{f} A_{\text{fv}} \sigma_{\text{fv}} d_{v} { \cot }\theta }}{s}$ $\sigma_{\text{fv}} {\text{ is the smaller of }}\left\{ {\begin{array}{*{20}c} {\left( {\frac{{\left( {\frac{{0.05r_{b} }}{{d_{b} }} + 0.3} \right)f_{\text{fv}} }}{1.5}} \right)} \\ {E_{\text{fv}} \varepsilon_{\text{fv}} } \\ \end{array} } \right.$ $\varepsilon_{\text{fv}} = 0.0001\left[ {f^{'}_{c} \times \frac{{\rho_{\text{fl}} E_{\text{fl}} }}{{\rho_{\text{fv}} E_{\text{fv}} }}} \right]^{0.5} \times \left[ {1 + 2\left( {\sigma_{N} /f^{'}_{c} } \right)} \right] \le 0.0025$ $\varphi_{f} = \left\{ {\begin{array}{*{20}c} {0.5\quad {\text{for GFRP}}} \\ {0.75\quad {\text{for CFRP}}} \\ \end{array} } \right.$
CSA S806-02 [12]	For h < 300 or minimum shear reinforcement provided. $V_{c} = 0.035\lambda \varphi_{c} \left( {f^{'}_{c} \rho_{\text{fl}} E_{f;} \frac{{V_{f} d}}{{M_{f} }}} \right)^{\frac{1}{3}} db$ $0.1\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd \le V_{c} \le 0.2\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd$ If h > 300 and less than minimum shear reinforcement provided $V_{c} = \left( {\frac{130}{1000 + d}} \right)\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd < 0.08\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd$ $\varphi_{c} = 0.6$ $A_{{v, { \min }}} = \frac{{0.3\sqrt {f^{'}_{c} } bs}}{{f_{\text{fv}} }}$	$V_{\text{fv}} = \frac{{0.4\varphi_{f} A_{v} f_{\text{fv}} d}}{s} < 0.6\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd$ $\varphi_{f} = 0.75$
JSCE guidelines [27]	$V_{c} = \beta_{d} \beta_{p} \beta_{n} f_{\text{vcd}} bd/\gamma_{b}$ $f_{\text{vcd}} = 0.2\left( {f^{'}_{c} } \right)^{\frac{1}{3}} \le 0.72 {\text{MPa}}$ $\beta_{d} = \left( \frac{1000}{d} \right)^{\frac{1}{4}} \le 1.5$ $\beta_{p} = \left( {100\frac{{\rho_{\text{fl}} E_{\text{fl}} }}{{E_{\text{s}} }}} \right)^{\frac{1}{3}} \le 1.5$ $\beta_{n} = 1 + \frac{{M_{o} }}{{M_{d} }} \le 2 {\text{for}} N_{f} \ge 0$ $\beta_{n} = 1 + \frac{{2M_{o} }}{{M_{d} }} \le 2 {\text{for}} N_{f} < 0$ $\gamma_{b} = 1.3$	$V_{\text{fv}} = \frac{{\left[ {\frac{{A_{\text{fv}} \sigma_{\text{fv}} \left( {\sin \alpha_{s} + \cos \alpha_{s} } \right)}}{s}} \right]jd}}{{\gamma_{b} }} \le \frac{{f_{\text{FRPbend}} }}{{E_{\text{fv}} }}bd$ $jd = d/1.15$ $\sigma_{\text{fv}} {\text{ is the smaller of}} \left\{ {\begin{array}{*{20}c} {\left( {\frac{{\left( {\frac{{0.05r_{b} }}{{d_{b} }} + 0.3} \right)f_{\text{fv}} }}{1.5}} \right)} \\ {E_{\text{fv}} \varepsilon_{\text{fv}} } \\ \end{array} } \right.$ $\varepsilon_{\text{fv}} = 0.0001\sqrt {f_{\text{mcd}}^{'} \times \frac{{\rho_{{{\text{f}}l}} E_{{{\text{f}}l}} }}{{\rho_{\text{fv}} E_{\text{fv}} }}}$ $f^{'}_{\text{mcd}} = \left( \frac{h}{100} \right)^{{ - \frac{1}{10}}} f^{'}_{c}$ $\gamma_{b} = 1.15$
CNR DT-203 [11]	$V_{c} \le \left\{ {\begin{array}{*{20}c} {V_{\text{Rd,ct}} } \\ {V_{\text{Rd,ct max}} } \\ \end{array} } \right.$ $V_{\text{Rd,ct}} = 1.3\left( {\frac{{E_{\text{f}} }}{{E_{\text{s}} }}} \right)^{\frac{1}{2}} V_{\text{Rd,c}}$ $V_{\text{Rd,c}} = \tau_{\text{Rd}} k_{d} \left( {1.2 + 40\rho_{\text{fl}} } \right)bd$ $\tau_{\text{Rd}} = 0.25f_{{{\text{ctk}}0.05}} /\gamma_{c}$ $f_{{{\text{ctk}}0.05}} = 0.05\,f^{'}_{c} \gamma_{c} = 1.6$ $1.3\left( {\frac{{E_{\text{f}} }}{{E_{\text{s}} }}} \right)^{\frac{1}{2}} \le 1$ If more than 50 % of the bottom reinforcement is interrupted: $k_{d} = 1$ If less than 50 % of bottom reinforcement is interrupted: $k_{d} = \left( {1.6 - d} \right), d {\text{in meters}}$ $V_{\text{Rd,ct max}} = \frac{1}{2}vf_{\text{cd}} b0.9d$ $v = 0.7 - f^{'}_{c} /200 \ge 0.5$	$V_{\text{fv}} = \frac{{A_{v} f_{\text{fr}} d}}{s}$ $f_{\text{fr}} = \frac{{f_{\text{fd}} }}{{\gamma_{f,\varphi } }}$ $f_{\text{fd}} = \frac{{f_{\text{fv}} }}{{\gamma_{f} }}$ $\gamma_{f} = 1.5$ If no specific experimental tests are performed, and that $\frac{{r_{d} }}{{d_{b} }} \ge 6$: $\gamma_{f,\varphi } = 2$ If specific experimental tests are performed: $\gamma_{f,\varphi } = \frac{{f_{\text{FRPbend}} }}{{f_{\text{fv}} }}$
ISIS-M03-01 design manual [14]	If d ≤ 300 mm: $V_{c} = 0.2\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd\sqrt {\frac{{E_{\text{fl}} }}{{E_{\text{s}} }}}$ If d > 300 mm: $\left. {V_{c} = \frac{260}{1000 + d}} \right)\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd\sqrt {\frac{{E_{\text{fl}} }}{{E_{\text{s}} }}} \ge 0.1\lambda \varphi_{c} \sqrt {f^{'}_{c} } bd\sqrt {\frac{{E_{{{\text{f}}l}} }}{{E_{\text{s}} }}}$	$V_{fv} = 0.4\rho_{fv} f_{fv} bd \le 0.8bd\sqrt {\frac{{f^{'}_{c} E_{fv} }}{{E_{\text{s}} }}}$
Nehdi et al. [15]	Optimized equation: If a/d ≥ 2.5: $V_{c} = 2.1\left( {\frac{{f^{'}_{c} \rho_{{{\text{f}}l}} d}}{a}\frac{{E_{{{\text{f}}l}} }}{{E_{\text{s}} }}} \right)^{0.23} bd$ If a/d < 2.5: $V_{c} = 2.1\left( {\frac{{f^{'}_{c} \rho_{\text{fl}} d}}{a}\frac{{E_{\text{fl}} }}{{E_{\text{s}} }}} \right)^{0.23} bd \times \frac{2.5d}{a}$ $\rho_{\text{fl}} = \frac{{A_{\text{fl}} }}{bd}$	$V_{\text{fv}} = 0.74\left( {\rho_{\text{fv}} f_{\text{fv}} } \right)^{0.51} bd$ $\rho_{\text{fv}} = \frac{{A_{\text{fv}} }}{bs}$
Nehdi et al. [15]	Design equation: If a/d ≥ 2.5: $V_{c} = 2.1\left( {\frac{{f^{'}_{c} \rho_{\text{fl}} d}}{a}\frac{{E_{\text{fl}} }}{{E_{\text{s}} }}} \right)^{0.3} bd$ If a/d < 2.5:$V_{c} = 2.1\left( {\frac{{f^{'}_{c} \rho_{\text{fl}} d}}{a}\frac{{E_{{{\text{f}}l}} }}{{E_{\text{s}} }}} \right)^{0.3} bd \times \frac{2.5d}{a}$ $\rho_{{{\text{f}}l}} = \frac{{A_{{{\text{f}}l}} }}{bd}$	$V_{\text{fv}} = 0.5\left( {\rho_{\text{fv}} f_{\text{fv}} } \right)^{0.5} bd$ $\rho_{\text{fv}} = \frac{{A_{\text{fv}} }}{bs}$
Hoult et al. [16]	$V_{c} = \beta \sqrt {f^{'}_{c} } bd_{v}$ $\beta = \frac{0.30}{{0.5 + \left( {1000\varepsilon_{x}\, + \,0.15} \right)^{0.7} }} \times \frac{1300}{{1000\, + \,s_{\text{ze}} }}$ $d_{v} = 0.9d$ $s_{\text{xe}} = \frac{31.5d}{{16 + a_{g} }} \ge 0.77d$ $a_{g} = \left\{ {\begin{array}{*{20}c} {a_{g} , \quad {\text{if }}\, f^{'}_{c} < 60} \\ {a_{g} - \frac{{a_{g} }}{10} \times \left( {f^{'}_{c} - 60} \right),\quad {\text{if}}\, 60 \le f^{'}_{c} < 70} \\ {0,\quad {\text{if}}\, f^{'}_{c} > 70} \\ \end{array} } \right.$ $\varepsilon_{x} = \frac{{\left( {\left( {\frac{{M_{f} }}{{d_{v} }}} \right) + V_{f} } \right)}}{{2\left( {E_{\text{s}} A_{\text{s}} } \right)}}$
Tureyen and Frosch [17]	Calculations are in English units. $V_{c} = \left( {\sqrt {16 + \frac{{4\sigma_{m} }}{{3\sqrt {f^{'}_{c} } }}} } \right)\sqrt {f^{'}_{c} } bc$ $c = kd$ $\sigma_{m} = \frac{{f_{\text{cr}} Ikd}}{{I_{\text{cr}} \left( \frac{h}{2} \right)}}$ $I_{\text{cr}} = b\left( \frac{kd}{3} \right)^{3} + n_{f} A_{f} \left( {d - kd} \right)^{2}$ $I = \frac{{h^{3} b}}{12}$ $f_{\text{cr}} = 0.6\lambda \sqrt {f^{'}_{c} }$ $k = \sqrt {2\rho_{\text{fl}} n_{f} + \left( {\rho_{{{\text{f}}l}} n_{f} } \right)^{2} } - \rho_{\text{fl}} n_{f}$