Fire design worked example: HEB 300 column, S355, R 90

Input data

HEB 300 (HE 300 B) column in S355, f_y = 355 N/mm² (t_f = 19 mm ≤ 40 mm), E = 210 000 N/mm².
Storey height L = 4.00 m, braced frame, intermediate storey, each storey a separate fire compartment.
Compression force in the fire situation N_fi,Ed = 2000 kN; γ_M,fi = 1.0.
Four-sided exposure, ISO 834 standard curve, required class R 90.

Quantity	Value
h × b × t_w × t_f × r	300 × 300 × 11 × 19 × 27 mm
A	149.1 cm²
I_y / I_z	25 170 / 8 563 cm⁴
A_m/V (contour, 4 sides)	116.2 m⁻¹
[A_m/V]_b (box, 4 sides)	80.5 m⁻¹

1. Cross-section class

ε = 0.85 · √(235 / 355) = 0.692

Flange: c = (300 − 11 − 2·27) / 2 = 117.5 mm; c/t_f = 6.18 ≤ 9ε = 6.22, so Class 1.
Web: c = 300 − 2·19 − 2·27 = 208 mm; c/t_w = 18.9 ≤ 33ε = 22.8, so Class 1.

The section is Class 1. Details in the guide on cross-section classes in fire.

2. Buckling length and slenderness

Braced frame, intermediate storey, a fire compartment on each storey: ℓ_fi = 0.5 · L = 2.00 m about both axes (cl. 4.2.3.2(4)).

N_cr,y = π² · E · I_y / ℓ_fi² = 130 419 kN λ̄_y = √(A · f_y / N_cr,y) = 0.201

N_cr,z = π² · E · I_z / ℓ_fi² = 44 370 kN λ̄_z = √(A · f_y / N_cr,z) = 0.345

3. Critical temperature from buckling

The critical temperature is found iteratively. For a temperature θ, compute the slenderness λ̄_θ and the factor χ_fi, then the degree of utilisation μ = N_fi,Ed / (χ_fi · A · f_y / γ_M,fi) and, from eq. (4.22), a new temperature. Repeat until the temperature stops changing.

Quantity (at convergence)	Axis y-y	Axis z-z
k_y,θ / k_E,θ	0.450 / 0.295	0.493 / 0.331
λ̄_θ	0.249	0.421
χ_fi	0.878	0.794
μ	0.430	0.476
θ_a,cr	608.4 °C	592.6 °C

The minor axis governs: θ_a,cr = 592.6 °C. The values of k_y,θ and k_E,θ are interpolated from Table 3.1.

The HE 300 B column data in FireSolver and the results: critical temperature 592.6 °C and protection thickness 15 mm — The same example in FireSolver: member data and results (Romanian interface).

4. Resistance without protection

k_sh = 0.9 · 80.5 / 116.2 = 0.624 k_sh · A_m/V = 72.4 m⁻¹

Integrating eq. (4.25) under the ISO 834 curve, the unprotected steel reaches 592.6 °C after 19.2 minutes. The unprotected column only achieves R 15, so it needs protection for R 90.

5. Protection thickness

Gypsum boards as a box (λ_p = 0.20 W/mK, ρ_p = 800 kg/m³, c_p = 1700 J/kgK), A_p/V = 80.5 m⁻¹. With eq. (4.27), the minimum thickness that keeps the steel below 592.6 °C:

Class	R 60	R 90	R 120
Required d_p	9.5 mm	15.0 mm	21.0 mm

For R 90 a single 15 mm board layer is enough, provided the properties declared by the manufacturer are at least as good as the values used here. A comparison with other materials is in the guide on boards and sprays.

Gypsum board protection for R 90 in FireSolver: required thickness 15 mm and the temperature chart — Protection in FireSolver: gypsum boards, automatic thickness for R 90.

What affects the result

With ℓ_fi = 0.7 · L (top storey), the slenderness increases and the critical temperature drops, so the protection gets thicker.
A lower force raises the critical temperature: the same column under 1500 kN needs less protection.
With three-sided exposure (a column against a wall), the section factor drops and heating is slower.

Informative material summarising the provisions of EN 1993-1-2. It does not replace the text of the standard, the National Annex or verification by a qualified engineer.

Worked example: HEB 300 column for R 90