Creep modeling of concretes with high volumes of supplementary cementitious materials and its application to concrete-filled tubes

Abstract

The use of concretes with supplementary cementitious materials (SCM concretes) is a mechanism to improve the sustainability of structural systems. However, most reinforced concrete (including cast-in-place and precast) structural components rely on early stiffness and strength of the concrete to resist self-weight and construction loadings after release of the formwork. In contrast to cement-only concretes, SCM concretes require more time to reach the target strength and stiffness values and therefore a structural component that does not depend on these engineering properties during construction is required. Use of concrete-filled tubes (CFTs) is a solution. With a CFT, the steel tube is able to support construction loads while the high SCM concrete fill develops its strength and stiffness. However, there are concerns about load transfer (from the concrete to the steel tube) during the development of these properties and the long-term deformations (creep) of high SCM concrete and its impact on CFT performance. This research develops models to predict the long-term (creep) response of SCM concretes and uses those predictive models to assess the long-term response of CFTs with SCM concrete fill. Three time-dependent models were developed: compressive strength, elastic modulus, and creep. The models were validated using a large SCM database and two full-scale CFTs with SCM concrete fill that were tested under sustained loading. Once verified, a parametric study was conducted. The results indicate that SCM concretes with binder ratios (total ratio of CaO/(SiO₂ + Al₂O₃) in all binders—cement and SCMs) greater than 0.5 behaved similarly to CFTs with cement-only concrete fill.

Appendix: Proposed creep model for SCM concrete

Equations

$$ J\left( {t,t_{ 0} } \right) = \frac{1}{{E_{\text{c}} \left( {t_{ 0} } \right)}} + \frac{{\varphi \left( {t,t_{ 0} } \right)}}{{1.05E_{\text{cm}} }} $$

(17)

$$ E_{\text{cm}} = 21.5 \times 10^{3} \left( {\frac{{f_{\text{cm}} }}{10}} \right)^{1/3} $$

(18)

$$ E_{\text{c}} \left( t \right) = \beta_{\text{E}} \left( t \right)E_{\text{cm}} = \left\{ {{ \exp }\left[ {s\left( {1 - \sqrt {{\raise0.7ex\hbox{${28}$} \!\mathord{\left/ {\vphantom {{28} t}}\right.\kern-0pt} \!\lower0.7ex\hbox{$t$}}} } \right)\left( { - 1.60\frac{C}{S + A} + 5.26} \right)} \right]} \right\}^{0.3} E_{\text{cm}} $$

(18)

$$ s = \left\{ {\begin{array}{ll} 0.20 & {\text{for}}\,{\text{cement}}\;{\text{Class}}\,R \\ 0.25 & {\text{for}}\,{\text{cement}}\;{\text{Class}}\,N \\ 0.38 & {\text{for}}\,{\text{cement}}\;{\text{Class}}\,S \\ \end{array} } \right.$$

(19)

$$ \varphi \left( {t,t_{ 0} } \right) = \varphi_{\text{RH}} \beta \left( {f_{\text{cm}} } \right)\beta \left( {t_{ 0} } \right)\beta \left( {\frac{C}{S + A}} \right)\beta \left( {\frac{w}{b}} \right)\beta \left( {t,t_{ 0} } \right) $$

(20)

$$ \varphi_{\text{RH}} = \left\{ {\begin{array}{lll} {1 + \frac{{1 - {\text{RH/}}100}}{{0.1\left( h \right)^{1/3} }} \quad \qquad \quad f_{\text{cm}} \le 35\;{\text{MPa}} } \\ {\left[ {1 + \frac{{1 - {\text{RH/}}100}}{{0.1\left( h \right)^{1/3} }}\alpha_{1} } \right]\alpha_{2} \quad f_{\text{cm}} > 35\;{\text{MPa}} } \\ \end{array} } \right.$$

(21)

$$ h = \frac{{2A_{\text{c}} }}{u} $$

(22)

$$ \beta \left( {f_{\text{cm}} } \right) = \frac{16.8}{{f_{\text{cm}}^{0.5} }} $$

(23)

$$ f_{\text{c}} \left( t \right) = \beta_{\text{cc}} \left( t \right)f_{\text{cm}} = { \exp }\left[ {s\left( {1 - \sqrt {{\raise0.7ex\hbox{${28}$} \!\mathord{\left/ {\vphantom {{28} t}}\right.\kern-0pt} \!\lower0.7ex\hbox{$t$}}} } \right)\beta_{\text{cc}} \left( {\frac{C}{S + A}} \right)} \right]f_{\text{cm}} $$

(24)

$$\beta_{\text{cc}} \left( t \right) = \left\{ {\begin{array}{ll} - 0.38\frac{C}{S + A} + 2.12 & t < 28\;{\text{days}} \\ - 1.15\frac{C}{S + A} + 3.70 & t > 28\;{\text{days}} \\ \end{array} } \right.$$

(25)

$$ \beta \left( {t_{ 0} } \right) = \frac{1}{{0.1 + t_{0}^{0.2} }} $$

(26)

$$ \beta \left( {t,t_{{\mathfrak{w}}} } \right) = \left[ {\frac{{\left( {t - t_{{\mathfrak{w}}} } \right)/t_{{\mathfrak{x}}} }}{{\beta_{\text{H}} + \left( {t - t_{{\mathfrak{w}}} } \right)/t_{{\mathfrak{x}}} }}} \right]^{0.3} $$

(27)

$$ \beta_{\text{H}} = \left\{ {\begin{array}{lll} {150\left[ {1 + \left( {1.2\frac{RH}{100}} \right)^{18} } \right]\frac{h}{100} + 250 \le 1500 \quad \qquad \,\,\, f_{\text{cm}} \le 35\;{\text{MPa}} } \\ {150\left[ {1 + \left( {1.2\frac{RH}{100}} \right)^{18} } \right]\frac{h}{100} + 250\alpha_{ 3} \le 1500\alpha_{ 3}\quad \, \,f_{\text{cm}} > 35\;{\text{MPa}} } \\ \end{array} } \right. $$

(28)

$$ \alpha_{ 1} = \left( {\frac{35}{{ f_{\text{cm}} }}} \right)^{0.7} \alpha_{ 2} = \left( {\frac{35}{{ f_{\text{cm}} }}} \right)^{0.2} \alpha_{ 3} = \left( {\frac{35}{{ f_{\text{cm}} }}} \right)^{0.5} $$

(29)

$$ \beta \left( {\frac{C}{S + A}} \right) = 0.17\frac{C}{S + A} + 0.87 $$

(30)

$$ \beta \left( {\frac{w}{b}} \right) = 0.74\frac{w}{b} + 0.37 $$

(31)