Concrete under fire: an assessment through intelligent pattern recognition

Abstract

Concrete, a naturally resilient material, often undergoes a series of physio-chemical degradations once exposed to extreme environments (e.g., elevated temperatures). Under such conditions, not only concrete weakens, but also becomes vulnerable to fire-induced spalling; a complex and exceptionally random phenomenon. Despite serious efforts carried out over the past few years, we continue to be short of developing a methodical procedure that enables accurate assessment of concrete under elevated temperatures with due consideration to fire-induced spalling. Unlike traditional works, this study aims at investigating fire behavior of concrete through a modern perspective. In this study, a number of intelligent pattern recognition (IPR) techniques that capitalize on artificial intelligence (AI) are applied to derive expressions able of accurately trace the response of normal and high strength as well as high performance concretes under elevated temperatures. These expressions take into account geometric, material, and specific features of structural components to examine fire response as well as to predict occurrence of fire-induced spalling in concrete structures. These expressions were developed through rigorous and data-driven analysis of actual fire tests and were derived to implicitly account for physio-chemical transformations in concrete and as such do not require collection/input of temperature-dependent material properties nor special analysis/simulation. This study also features the development of an IPR-based database and fire assessment software that can be used to examine fire performance of concrete members and be regularly updated as to continually improve the accuracy of the proposed expressions.

Appendix

This section illustrates two examples with procedure into applying the IPR-derived expressions to evaluate susceptibility of a typical RC column to fire-induced spalling as well as fire resistance of the same column. This column, named M6S150, was tested by Shah and Sharma [25], and achieved a fire resistance of 289 min after undergoing spalling; and has the following features:

• 1. Concrete type, f_c = 63 MPa,

• 2. Cross sectional size, b × h = 300 × 300 mm²,

• 3. Concrete cover, c, 40 mm,

• 4. Magnitude of applied loading, P = 1858 kN,

• 5. Concentric loading, e_c = 0 mm,

To check for fire-induced spalling in this column:

Using MLR:

$$\begin{aligned} {\text{Spalling}} &= 2.213 - 0.01139f_{\text{c}} - 0.00071b - 0.0e_{\text{c}} - 0.000099P,\\&\quad {\text{where}}\;{\text{ for}}\;{\text{ MLR; }}1 - {\text{Spalling}}\;{\text{occurs}}, \, 2 - {\text{No spalling}} \hfill \\ {\text{Spalling}} &= 2.213 - 0.01139\left( {63} \right) - 0.00071\left( {300} \right) - 0.0\left( 1 \right) \\ &\quad- 0.000099\left( {1858} \right) = 1\;\left( {\text{Spalling occurs}} \right). \hfill \\ \end{aligned}$$

Using DL:

Implementing the matrix provided in Table 1, DL reveals Spalling = 1(Spalling occurs)

Using GP:

$$\begin{aligned} {\text{No spalling}} &= {\text{Logistic}}(63 + 4.64\tan \left( {14.8 + 14.9f_{c} } \right) \\&\quad+ 5.52\tan \left( {5.52 + P + \tan \left( {14.8 + 14.9f_{c} } \right)} \right) \\&\quad+ \tan \left( {69.3 + P} \right) - f_{c} - \tan \left( b \right)) \hfill \\ {\text{No spalling}} &= {\text{Logistic}}\left( {63 + 4.64\tan \left( {14.8 + 14.9 \times 63} \right)}\right. \\&\quad \left.{+\, 5.52\tan \left( {5.52 + 1858 + \tan \left( {14.8 + 14.9 \times 63} \right)} \right)}\right. \\ &\quad \left.{+\, \tan \left( {69.3 + 1858} \right) - 63 - \tan \left( {300} \right)} \right) \\ &= 0\; \left( {\text{Spalling occurs}} \right). \hfill \\ \end{aligned}$$

To verify this:

$$\begin{aligned} {\text{Spalling}} &= {\text{Logistic}}\left( {37.9f_{c} + 46.7\tan \left( b \right) + 37.9\tan \left( {0.9707P} \right)}\right.\\&\quad\left.{ + 66.06\tan \left( {f_{c} + 0.9669P} \right) + 97.02\tan \left( {37.9f_{c} }\right.}\right.\\&\quad \left.{\left.{+\, 46.74\tan \left( b \right) + 37.9\tan \left( {0.9707P} \right)}\right.}\right.\\&\quad \left.{\left.{ + 61.04\tan \left( {f_{c} + 0.9669P} \right)} \right) - 2.57 \times 10^{3} } \right) \hfill \\ {\text{Spalling}} &= {\text{Logistic}}\left( {37.9 \times 63 + 46.7\tan \left( {300} \right) }\right.\\&\quad\left.{+\, 37.9\tan \left( {0.9707 \times 1858} \right) }\right.\\&\quad\left.{+\, 66.06\tan \left( {63 + 0.9669 \times 1858} \right) }\right.\\&\quad\left.{+\, 97.02\tan \left( {37.9 \times 63 + 46.74\tan \left( {300} \right) }\right.}\right.\\&\quad \left.{\left.{+\, 37.9\tan \left( {0.9707 \times 1858} \right)}\right.}\right.\\&\quad \left.{\left.{ + 61.04\tan \left( {63 + 0.9669 \times 1858} \right)} \right) - 2.57 \times 10^{3} } \right) \\&= 1.0 \;\left( {{\text{Spalling}}\;{\text{ occurs}}} \right). \hfill \\ \end{aligned}$$

Fire resistance of this column can be evaluated using the following approaches:

Using MLR:

$$\begin{aligned} {\text{FR}} =\, & - 1.491 + 11.61bh + 0.01733f_{\text{c}} + 0.078C \\ &- 0.01745e_{\text{c}} - 0.000537P \\ {\text{FR}} =\, & - 1.491 + 11.61\left( {0.09} \right) + 0.01733\left( {63} \right) + 0.078\left( {40} \right) \\&- 0.01745\left( 0 \right) - 0.000537\left( {1858} \right) \\ =\,& 252.7 \hbox{min} \; \left( {{\text{within }}13\% {\text{ of measured fire resistance}}} \right). \\ \end{aligned}$$

Using DL:

Implementing the matrix provided in Table 1 and fire assessment tool shown in Fig. 3, the fire resistance of this columns comes to, FR= 244.3 min (within 18% of measured fire resistance)

Fig. 3

Graphical interface of developed fire assessment tool

Using GP:

$$\begin{aligned} {\text{FR}} =\, & {\kern 1pt} e_{c} + 912\ln \left( {f_{c} } \right) + 0.0984f_{c}^{2} + 31.9\cos \left( {\frac{22.6P}{b}} \right) - \frac{{2.49P\cos \left( {\frac{22.6P}{b}} \right)}}{b} \\ & - 2.18 \times 10^{3} - 28.1f_{c} - 90.4\tanh \left( {e_{c} } \right) - 21.3\cos \left( { - 336f_{c}^{2} } \right) - 50.3\cos \left( { - 1.57e^{{\frac{P}{b}}} } \right) \\ & - 54.7\cos \left( {\frac{{Pa\sinh \left( {\frac{P}{b}} \right)}}{b}} \right) - 32.8\cos \left( {3.91 + 31.9\cos \left( {\frac{22.6P}{b}} \right)} \right), \\ \end{aligned}$$

$$\begin{aligned} {\text{FR}} =\, & 0 + 912\ln \left( {63} \right) + 0.0984 \times 63^{2} + 31.9\cos \left( {\frac{22.6 \times 1858}{300}} \right) - \frac{{2.49 \times 1858 \times \cos \left( {\frac{22.6 \times 1858}{300}} \right)}}{300} \\ & - 2.18 \times 10^{3} - 28.1 \times 63 - 90.4\tanh \left( 0 \right) - 21.3\cos \left( { - 336 \times 63^{2} } \right) - 50.3\cos \left( { - 1.57e^{{\frac{1858}{300}}} } \right) \\ & - 54.7\cos \left( {\frac{{1858 \times a\sinh \left( {\frac{1858}{300}} \right)}}{300}} \right) - 32.8\cos \left( {3.91 + 31.9\cos \left( {\frac{22.6 \times 1858}{300}} \right)} \right) \\ =\, & 300\, \hbox{min} ,\quad \left( {{\text{within }}4\% {\text{ of measured}}\;{\text{fire}}\;{\text{ resistance}}} \right). \\ \end{aligned}$$