Thermal analysis of steel-concrete composite cross sections via CS-ASA/FA

Pires, Dalilah; Barros, Rafael Cesário; Rocha, Paulo Anderson Santana; Silveira, Ricardo Azoubel da Mota

doi:10.1590/0370-44672016710155

Abstract

When exposed to high temperatures, such as in a fire situation, the physical and resistance characteristics of the materials employed in the structure deteriorate as the temperature increases. This fact promotes a considerable loss in the bearing capacity and stiffness of the structural system. The verification of a structure exposed to fire depends primarily and principally on the thermal analysis of the cross section of the structural element. This analysis permits determination of the temperature variation or temperature range in the element from the boundary conditions provided by the fire model adopted. As such, this study had the objective of performing a thermal analysis in a transient regime by means of a finite element method on steel-concrete composite cross sections that are employed in civil construction through use of the Computational System for Advanced Structural Analysis/Fire Analysis (CS-ASA/FA). Two cross sections are analyzed and the results obtained were satisfactory. In addition, different iterative solution processes were adopted in the analysis. Parametric studies were also performed related to the mesh variation of the finite elements and time increase. From the results, it was possible to conclude that CS-ASA/FA can supply the necessary information when a thermo-structural analysis is performed for the evaluation of strength and stiffness losses of the structural material when exposed to fire.

Keywords:
thermal analysis; fire; temperature; composite cross section; FEM

A.SOLUTION OF THE HEAT TRANSFER TRANSIET PROBLEM:
A1. Sets input data, initial and boundary conditions
A2. Do: T_n = T_n+1 = T_n+θ = T₀ = 20°C
B.SOLUTION PROCEDURE: Incremental simple or Incremental-iterative (Picard or Newton-Raphson)
C. GIVE A NEW TIME INCREMENT AND GO TO STEP B
INCREMENTAL SIMPLE
1. INCREMENTAL PROCESS: inc = 1, 2, 3,...,nmáx
1a. Calculate the capacitance matrix: $C_{n + θ} = \int_{Ω} ρ c N^{τ} N d Ω$
1b. Calculate thermal conductivity matrix: $K_{n + θ} = \int_{Ω} B^{τ} DB d Ω + h \int_{Γ} N^{τ} N d Γ$
1c. Calculates the heat flux vector: $R_{n + θ} = Q \int_{Ω} N^{τ} d Ω + h τ_{\infty} \int_{Γ} N^{τ} d Γ - q_{0} \int_{Γ} N^{τ} d Γ$
1d. Get: ${\hat{K}}_{n + θ} = (C_{n + θ} + θ Δ t K_{n + θ})$
1e. Get: ${\hat{R}}_{n + θ} = [C_{n + θ} - (1 - θ) Δ t K_{n + θ}] T_{n} + R_{n + θ}$
1f. Solve the system of equations: ${\{T\}}_{n + 1} = {\hat{K}}_{n + θ}^{- 1} {\hat{R}}_{n + θ}$
INCREMENTAL -ITERATIVE (P ICARD)	INCREMENTAL-ITERATIVE(NEWTON-RAPHSON)
1. INCREMENTAL PROCESS :inc = 1, 2,..., nmax	1. INCREMENTAL PROCESS: inc = 1, 2,...,nmax
2. ITERATIVE PROCESS (PICARD): k = 1, 2,..., nmax	1a. Calculated the capacitance matrix:
2a. Do: ${\hat{R}}_{n + θ} = T_{n + θ}^{1} = T_{n + θ}^{2} = 0.0$	$C_{n + θ} = \int_{Ω} ρ c N^{τ} N d Ω$
2b. Calculate the capacitance matrix:	1b. Calculated thermal conductivity matrix
$C_{n + θ} = \int_{Ω} ρ c N^{τ} N d Ω$	$K_{n + θ} = {\int_{Ω} B}^{τ} DB d Ω + h \int_{Γ} N^{τ} N d Γ$
2c. Calculate thermal conductivity matrix:	1c. Calculated the heat flux vector
$K_{n + θ} = \int_{Ω} B^{τ} DB d Ω + h \int_{Γ} N^{τ} N d Γ$	$R_{n + θ} = Q \int_{Ω} N^{τ} d Ω + h τ_{\infty} \int_{Γ} N^{τ} d Γ - q_{0} \int_{Γ} N^{τ} d Γ$
2d. Calculate the heat flux vector:	1d. Get: ${\hat{K}}_{n + θ} = (C_{n + θ} + θ Δ t K_{n + θ})$
$R_{n + θ} = Q \int_{Ω} N^{τ} d Ω + h τ_{\infty} \int_{Γ} N^{τ} d Γ - q_{0} \int_{Γ} N^{τ} d Γ$	1e. Get: ${\hat{R}}_{n + θ} = [C_{n + θ} - (1 - θ) Δ t K_{n + θ}] T_{n} + R_{n + θ}$
2e. Get: ${\hat{K}}_{n + θ} = (C_{n + θ} + θ Δ t K_{n + θ})$	1f. Solve the system of equations: ${\{T\}}_{n + 1} = {\hat{K}}_{n + θ}^{- 1} {\hat{R}}_{n + θ}$
2f. Get: ${\hat{R}}_{n + θ} = [C_{n + θ} - (1 - θ) Δ t K_{n + θ}] T_{n} + R_{n + θ}$	2. ITERATIVE PROCESS (N-R):k = 1, 2,...,nmax
2g. Determines the 1st approximation:	2a. $Update {\hat{K}}_{n + θ} and {\hat{R}}_{n + θ} for T_{n + 1}$
$T_{n + θ}^{1} = {\hat{K}}_{n + θ}^{- 1} {\hat{R}}_{n + θ}$	2b. Solve the system defined in 1f to obtain $T_{n + 1}^{k}$
2h. Determines the 2nd approximation:	2c.Calculates the unbalance vector:
$T_{n + θ}^{2} = T_{0} + 2 (T_{n + θ} - T_{0})$	$g = {\hat{R}}_{n + θ} - {\hat{K}}_{n + θ} {\{T\}}_{n + 1}$
2i. Checks for convergence: $\frac{\|T_{n + θ}^{1} - T_{n + θ}^{2}\|}{\|T_{n + θ}^{1}\|} < tolerance$	2d. Get the correction of the nodal temperatures:
$δ T^{k} = {[{\hat{K}}_{n + θ}]}^{- 1} g (T^{(k - 1)})$
YES :	2e. Updates nodal temperatures:
$\begin{matrix} Calculated the vector T_{n + θ} = T_{n + θ}^{1} + 0.5 (T_{n + θ}^{1} - T_{0}) \\ Update vector T_{0} = T_{n + θ}^{1} and go to item C \end{matrix}$	$T_{n + 1}^{k} = T_{n + 1}^{(k - 1)} + δ T_{n + 1}^{k}$
NO:	2f. Checks for convergence: $\frac{\|T_{n + 1}^{k} - T_{n + 1}^{(k - 1)}\|}{\|T_{n + 1}^{k}\|} < tolerance$
$\begin{matrix} lf k < n m a x, c a l c u l a t e \\ T_{n + θ} = 0.5 (T_{n + θ} + 0.5 (T_{n + θ}^{1} + T_{0})) and returns to \\ s t e p 2 \\ lf k = n m a x, stop and try another strategy \end{matrix}$
	YES: Stop and process to item C
	NO: lf k < n max, returns to step 2 lf k = nmax, stop and try another strategy

Brasil

Brasil

Thermal analysis of steel-concrete composite cross sections via CS-ASA/FA

Abstract