8.2 Fire Resistance

This example describes the fire resistance design for a steel column, using the National Annex of Germany.

Table 8.5 System and loads

Column cross-section: HE-B 300, steel S 235 System: hinged column, β = 1.0 System height: 3 m Loading: GK = 1200 kN, QK = 600 kN

$N_{cr,z}=\frac{21000·8560.00·{\pi }^2}{300.{00}^2}=19712.90 \mathrm{kN}$

${\overline{\lambda }}_z=\sqrt{\frac{A·f_y}{N_{cr,z}}}=\sqrt{\frac{149.0·23.5}{19712.90}}=0.422>0.2$

→ Design for flexural buckling must be performed.

cross-sectional geometry: h/b = 1.00 ≤ 1.2; structural steel S 235; t ≤ 100 mm

• [1] Table 6.2, row 3, column 4: buckling curve c
• ⇒ α_z = 0.49   ([1] Table 6.1)

$\Phi =0.5·[1+0.49·(0.422-0.2)+0.{422}^2]=0.643$

${\chi }_z=\frac{1}{0.643+\sqrt{0.{643}^2-0.{422}^2}}=0.886$

$N_{Ed}=1.35·G_K+1.5·Q_K=1.35·1200+1.5·600=2520 \mathrm{kN}$

$\frac{N_{Ed}}{{\chi }_z·A·f_y/{\gamma }_{M1}}=\frac{2520}{0.886·149.0·23.5/1.1}=0.894\le 1.0$

After a fire exposure of 90 minutes according to the standard temperature-time curve, the mean steel temperature is 524 ℃.

A box-shaped GRP encasement (glass-reinforced plastic) is used as fire resistance material, having the following properties:

Imperfection factor α:

$\alpha =0.65·\sqrt{\frac{235}{f_y}}=0.65·\sqrt{\frac{235}{235}}=0.65$

Non-dimensional relative slenderness ƛ_Θ:

${\overline{\lambda }}_{\Theta }=\overline{\lambda }·{\left[\frac{k_{y,\Theta }}{k_{E,\Theta }}\right]}^{0.5}=0.422·{\left[\frac{0.704}{0.528}\right]}^{0.5}=0.486$

Auxiliary factor:

${\Phi }_{\theta }=\frac{1}{2}·[1+\alpha ·{\overline{\lambda }}_{\theta }+{\overline{\lambda }}_{\theta }^2]=\frac{1}{2}·[1+0.65·0.486+0.{486}^2]=0.776$

Reduction factor for flexural buckling in fire situation:

${\chi }_{\text{fi}}=\frac{1}{{\phi }_{\theta }+\sqrt{{\phi }_{\theta }^2-{\overline{\lambda }}_{\theta }^2}}=\frac{1}{0.776+\sqrt{0.{776}^2-0.{486}^2}}=0.724$

Buckling resistance of structural component subjected to compression:

$N_{b,\text{fi},Rd}=\frac{{\chi }_{\text{fi}}·A·k_{y,\theta }·f_y}{{\gamma }_{M,\mathrm{fi}}}=\frac{0.724·149.0·0.704·23.5}{1.0}=1784.7 \mathrm{kN}$

Loading in case of fire:

$N_{\text{fi},Ed}=1.0·G_k+0.9·Q_k=1.0·1200+0.9·600=1740 \mathrm{kN}$

$\eta =\frac{N_{\text{fi},Ed}}{N_{b,\text{fi},Rd}}=\frac{1740}{1784.7}=0.975\le 1.0$

You can find more information about the thermal behavior of steel as a material for structural fire design in our Knowledge Base:
https://www.dlubal.com/en-US/support-and-learning/support/knowledge-base/001496

Another article describes the fire resistance design by means of parametric temperature-time curves:
https://www.dlubal.com/en-US/support-and-learning/support/knowledge-base/001613