Steel Beam Design to Eurocode 3: Step by Step

What this check does

For a simply supported beam whose compression flange is held in line (by a composite deck, tight-fitting planks, or closely spaced restraints), EN 1993-1-1 reduces beam design to four checks:

$$M_{Ed} \le M_{c,Rd} \qquad V_{Ed} \le V_{pl,Rd}$$

plus moment-shear interaction at ULS, and a deflection limit at SLS. The restraint assumption matters: an unrestrained beam can fail by lateral-torsional buckling at a much lower moment - that is a separate check ($M_{b,Rd}$, clause 6.3.2) covered in its own article.

Step 1 - section classification

Classification (clause 5.5) decides whether the section can reach its plastic moment and hold it. It compares the width-to-thickness ratio of each compressed part against limits scaled by:

$$\varepsilon = \sqrt{\dfrac{235}{f_y}}$$

where:

• $f_y$ - yield strength (MPa)

• $\varepsilon$ - the material scaling factor (-)

For a rolled I-section in bending, Class 1 requires:

• outstand flange: $c/t_f \le 9\varepsilon$

• internal web: $c/t_w \le 72\varepsilon$

Class 1 and 2 use the plastic modulus $W_{pl}$; Class 3 drops to the elastic $W_{el}$; Class 4 needs effective-section methods (EN 1993-1-5). Most rolled I and H sections in bending are Class 1 - but always check, especially in S355 and above where $\varepsilon$ shrinks.

Step 2 - bending resistance

For Class 1 or 2 (clause 6.2.5):

$$M_{c,Rd} = \dfrac{W_{pl,y}\,f_y}{\gamma_{M0}}$$

where:

• $W_{pl,y}$ - plastic section modulus about the strong axis (mm^3)

• $\gamma_{M0} = 1.0$ - partial factor for cross-section resistance (recommended value)

The plastic modulus is typically 10-15% above the elastic one for rolled I-sections (the shape factor), which is exactly the margin Class 1/2 classification unlocks.

Step 3 - shear resistance

Plastic shear resistance (clause 6.2.6):

$$V_{pl,Rd} = \dfrac{A_v\,(f_y/\sqrt{3})}{\gamma_{M0}}$$

with, for a rolled I-section loaded parallel to the web:

$$A_v = A - 2\,b\,t_f + (t_w + 2r)\,t_f \quad (\ge \eta\,h_w\,t_w)$$

where:

• $A_v$ - shear area (mm^2)

• $A$ - gross area (mm^2)

• $b$, $t_f$ - flange width and thickness (mm)

• $t_w$, $r$ - web thickness and root radius (mm)

• $h_w$ - clear web depth between flanges (mm)

• $\eta$ - shear factor, may be taken 1.2 up to S460 (EN 1993-1-5)

Rolled-section webs are stocky enough that shear buckling rarely governs ($h_w/t_w < 72\varepsilon/\eta$ passes for the whole IPE/HE range in S275).

Step 4 - moment-shear interaction

If $V_{Ed} \le 0.5\,V_{pl,Rd}$, shear does not reduce the bending resistance (clause 6.2.8). Above that, the web's contribution to bending is reduced by the factor $(1 - \rho)$ with $\rho = (2V_{Ed}/V_{pl,Rd} - 1)^2$. In gravity-loaded simply supported beams the maximum moment (midspan) and maximum shear (support) do not coincide, so the interaction rarely bites - but check it at positions of point loads.

Step 5 - deflection (SLS)

EN 1993-1-1 clause 7.2 sends deflection limits to the National Annex; the values used in most practice are:

For a simply supported beam under uniform load:

$$\delta = \dfrac{5\,q\,L^4}{384\,E\,I_y}$$

where:

• $q$ - the SLS (unfactored) line load (N/mm)

• $L$ - span (mm)

• $E = 210000$ MPa - modulus of elasticity

• $I_y$ - second moment of area, strong axis (mm^4)

Deflection is checked with unfactored (characteristic) loads - a common mistake is to reuse the ULS load and "fail" a beam that is fine.

Worked example - IPE 300, S275, 6 m span

Beam: simply supported, $L = 6.0$ m, compression flange fully restrained by a composite deck. Loads: permanent $g_k = 10$ kN/m (including self-weight), variable $q_k = 12$ kN/m. Section (CivilAxis steel catalogue): IPE 300 - $A = 53.8$ cm^2, $I_y = 8360$ cm^4, $W_{pl,y} = 628$ cm^3, $c/t_f = 5.28$, $c/t_w = 35.0$, $b = 150$ mm, $t_f = 10.7$ mm, $t_w = 7.1$ mm, $r = 15$ mm. Steel: S275.

ULS load and effects:

$$q_{Ed} = 1.35 \times 10 + 1.5 \times 12 = 31.5\ \text{kN/m}$$

$$M_{Ed} = \dfrac{31.5 \times 6.0^2}{8} = 141.8\ \text{kN.m} \qquad V_{Ed} = \dfrac{31.5 \times 6.0}{2} = 94.5\ \text{kN}$$

Classification ($\varepsilon = \sqrt{235/275} = 0.924$):

$$c/t_f = 5.28 \le 9\varepsilon = 8.32 \qquad c/t_w = 35.0 \le 72\varepsilon = 66.6 \quad\Rightarrow\quad \text{Class 1}$$

Bending:

$$M_{c,Rd} = \dfrac{628 \times 10^3 \times 275}{1.0} \times 10^{-6} = 172.7\ \text{kN.m}$$

$$\dfrac{M_{Ed}}{M_{c,Rd}} = \dfrac{141.8}{172.7} = 0.82 \quad\checkmark$$

Shear:

$$A_v = 5380 - 2 \times 150 \times 10.7 + (7.1 + 2 \times 15) \times 10.7 = 5380 - 3210 + 397 = 2567\ \text{mm}^2$$

$$V_{pl,Rd} = \dfrac{2567 \times (275/\sqrt{3})}{1.0} \times 10^{-3} = 407.6\ \text{kN}$$

$$\dfrac{V_{Ed}}{V_{pl,Rd}} = \dfrac{94.5}{407.6} = 0.23 \quad\checkmark$$

Interaction: $V_{Ed} = 94.5 \le 0.5\,V_{pl,Rd} = 203.8$ kN - no reduction of $M_{c,Rd}$ required.

Deflection ($I_y = 8.36 \times 10^7$ mm^4):

$$\delta_{q} = \dfrac{5 \times 12 \times 6000^4}{384 \times 210000 \times 8.36 \times 10^7} = 11.5\ \text{mm} \le \dfrac{L}{360} = 16.7\ \text{mm} \quad\checkmark$$

$$\delta_{g+q} = \dfrac{5 \times 22 \times 6000^4}{384 \times 210000 \times 8.36 \times 10^7} = 21.1\ \text{mm} \le \dfrac{L}{250} = 24.0\ \text{mm} \quad\checkmark$$

Result: IPE 300 in S275 passes every check; bending governs at 0.82 utilisation, with deflection close behind ($21.1/24.0 = 0.88$ on the total-load limit). A lighter IPE 270 would fail the deflection check - on modest spans with generous live load, stiffness, not strength, often picks the section.

Key points

• Classify first: the plastic modulus (10-15% above elastic) is only usable for Class 1/2 parts, and $\varepsilon$ shrinks with higher grades.

• Restrained bending is $W_{pl}\,f_y/\gamma_{M0}$; shear is $A_v f_y/(\sqrt{3}\,\gamma_{M0})$ with the rolled-section $A_v$ formula, and interaction only starts above $0.5\,V_{pl,Rd}$.

• Deflection uses unfactored loads and often governs before strength - check $L/360$ (variable) and $L/250$ (total) unless the National Annex says otherwise.

• This whole sequence assumes the compression flange is restrained; if it is not, lateral-torsional buckling ($M_{b,Rd}$, clause 6.3.2) takes over and is usually the harsher check.