👋 Introduction

Effective Class 2 is a bending resistance enhancement in Eurocode 3 for steel I-sections where the flange is Class 1 or 2 but the web is only Class 3. Instead of dropping all the way to the elastic modulus W_el (the Class 3 default), the eurocode3 lets us recover most of the plastic capacity by replacing the web's compression zone with two 20ε·t_w effective strips and using an effective plastic modulus W_pl,eff. EN 1993-1-1 Cl.6.2.2.4(1)

The full concept, the decision logic, and the governing formulas are explained in the companion article: Effective Class 2 Web — Recover the Plastic Moment.

In this worked example we take a welded plate girder (600×200×15×8) in S355 under pure bending. Its flange is comfortably Class 1 but its web just misses Class 2, so the section is normally Class 3 — but allow to apply the effective Class 2 method.

📐 Design Data

Below are the design data of the welded I-section, which we will classify and then check in bending using the effective Class 2 method.

Steel Section
	Type	Welded I-section (plate girder)
	Breadth, b	200mm
	Height, h	600mm
	Flange Thickness, t_f	15mm
	Web Thickness, t_w	8mm
	Weld / haunch zone, s	4mm

Material
Steel Grade		S355
Yield Strength, f_y		355 N/mm²
Partial Factor, γ_M0		1.0

🔢 Calculation Steps

*Below uses Eurocode3 symbols and limits.

Step 1 — Classify flange and web

The flange is an outstand in compression (checked against 9ε / 10ε / 14ε); the web is an internal element in bending (checked against 72ε / 83ε / 124ε). The worst class governs.

Material factor ^ⓘ
$\varepsilon$	=	$\sqrt{\dfrac{235}{f_y}}$
	=	$\sqrt{\dfrac{235}{355}} = \sqrt{0.662}$
	=	$0.81$

Flange outstand width (welded, no root radius) ^ⓘ
$c$	=	$\dfrac{b - t_w}{2}$
	=	$\dfrac{200 - 8}{2} = \dfrac{192}{2}$
	=	$\text{96 mm}$

Flange classification (outstand in compression) ^ⓘ
$\dfrac{c}{t_f}$	=	$\dfrac{c}{t_f} \le 9\varepsilon\ \text{(Cl 1)} / 10\varepsilon\ \text{(Cl 2)} / 14\varepsilon\ \text{(Cl 3)}$
	=	$\dfrac{96}{15} = 6.4 \quad (9\varepsilon = 7.29)$
	=	$6.4 \le 7.29 \rightarrow \text{Class 1}$

Web clear depth (welded, no root radius) ^ⓘ
$c_w$	=	$h - 2\,t_f$
	=	$600 - 2(15)$
	=	$\text{570 mm}$

Web classification (internal element in bending) ^ⓘ
$\dfrac{c_w}{t_w}$	=	$\dfrac{c_w}{t_w} \le 72\varepsilon\ \text{(Cl 1)} / 83\varepsilon\ \text{(Cl 2)} / 124\varepsilon\ \text{(Cl 3)}$
	=	$\dfrac{570}{8} = 71.25 \quad (83\varepsilon = 67.6)$
	=	$71.25 > 67.6 \rightarrow \text{Class 3}$

Section class — normal (worst of flange and web) ^ⓘ
$\text{Class}$	=	$\text{worst(Class 1 flange, Class 3 web)}$
	=	$\text{flange Class 1, web Class 3}$
	=	$\text{Class 3} \rightarrow \text{would use } W_{el}$

The web just misses Class 2 (71.25 > 67.6) but the flange is comfortably Class 1 — exactly the condition for the effective Class 2 provision.

Effective Class 2 check (Class 1/2 flange + Class 3 web) ^ⓘ
$\text{Effective Class 2}$	=	$\text{Class 1/2 flange and Class 3 web} \rightarrow W_{pl,eff}$
	=	$\text{Class 1 flange and Class 3 web}$
	=	$\text{Effective Class 2 applies} \rightarrow W_{pl,eff}$

Step 2 — Effective web and plastic neutral axis

The web's compression zone is replaced by two 20ε·t_w effective strips (one at the compression flange, one at the plastic neutral axis) with the middle strip neglected. A small weld / haunch zone of depth s (here s = 4 mm) is retained at each flange-web junction, sitting just below the flange as part of the effective compression area. Because part of the compression web is removed, the plastic neutral axis (PNA) shifts down toward the tension side to keep compression and tension areas equal.

Effective web strip depth (each) ^ⓘ
$20\varepsilon\,t_w$	=	$20\,\varepsilon\,t_w$
	=	$20 \times 0.81 \times 8$
	=	$\text{129.6 mm}$

Plastic neutral axis (from top fibre, weld zone s = 4 mm) ^ⓘ
$\bar{z}$	=	$h - t_f - s - 2(20\varepsilon\,t_w)$
	=	$600 - 15 - 4 - 2(129.6) = 600 - 15 - 4 - 259.2$
	=	$\text{321.8 mm}$

The Plastic Neutral Axis sits at 321.8 mm, 22 mm below mid-depth (300 mm) — confirming it has shifted down. The neglected (ineffective) compression web is the middle strip between the two effective strips.

Step 3 — Effective plastic modulus, W_pl,eff

With the PNA fixed, the effective plastic modulus is the sum of the first moments of the two flanges, the upper effective strip (the 20ε·t_w strip plus the weld zone s, at the compression flange), the lower effective strip (20ε·t_w, at the PNA), and the tension web below the PNA.

Effective plastic modulus ^ⓘ
$W_{pl,y,eff}$	=	$b\,t_f(h - t_f) + t_w(20\varepsilon\,t_w + s)\Big[\bar{z} - t_f - \tfrac{20\varepsilon\,t_w + s}{2}\Big]$
		$+ \dfrac{t_w(20\varepsilon\,t_w)^2}{2} + \dfrac{t_w(h - t_f - \bar{z})^2}{2}$
	=	$= 200(15)(585) + 8(129.6 + 4)\big[321.8 - 15 - \tfrac{129.6 + 4}{2}\big] + \tfrac{8(129.6)^2}{2} + \tfrac{8(263.2)^2}{2}$
		$= 1{,}755{,}000 + 256{,}512 + 67{,}185 + 277{,}097$
	=	$2.356 \times 10^{6}\ \text{mm}^{3} = 2356\ \text{cm}^{3}$

For comparison, the elastic modulus (the Class 3 default) and the full plastic modulus (the value the section would reach if the web were Class 1/2) bracket W_pl,eff.

Elastic modulus (Class 3 default, for comparison) ^ⓘ
$W_{el,y}$	=	$W_{el,y} = \dfrac{I_y}{h/2},\quad I_y = \dfrac{b\,h^3 - (b - t_w)(h - 2t_f)^3}{12}$
	=	$I_y = \dfrac{200(600)^3 - 192(570)^3}{12} = 636.9 \times 10^6\ \text{mm}^4$
		$W_{el,y} = \dfrac{636.9 \times 10^6}{300}$
	=	$2.123 \times 10^{6}\ \text{mm}^{3} = 2123\ \text{cm}^{3}$

Full plastic modulus (reference, if web were Class 1/2) ^ⓘ
$W_{pl,y}$	=	$b\,t_f(h - t_f) + \dfrac{t_w(h - 2t_f)^2}{4}$
	=	$200(15)(585) + \dfrac{8(570)^2}{4}$
	=	$2.405 \times 10^{6}\ \text{mm}^{3} = 2405\ \text{cm}^{3}$

So W_el (2123) < W_pl,eff (2356) < W_pl (2405) — the effective modulus sits between the elastic and full plastic values.

Step 4 — Bending resistance

The bending resistance is the ordinary plastic formula — just with the effective modulus. EN 1993-1-1 Cl.6.2.2.4(1)

Bending resistance — Effective Class 2 ^ⓘ
$M_{pl,eff,Rd}$	=	$\dfrac{W_{pl,eff}\,f_y}{\gamma_{M0}}$
	=	$\dfrac{2.356 \times 10^6 \times 355}{1.0}$
	=	$\text{836.4 kNm}$

Bending resistance — Elastic Class 3 (default) ^ⓘ
$M_{el,Rd}$	=	$\dfrac{W_{el}\,f_y}{\gamma_{M0}}$
	=	$\dfrac{2.123 \times 10^6 \times 355}{1.0}$
	=	$\text{753.7 kNm}$

Bending resistance — full plastic (reference, if web were Class 1/2) ^ⓘ
$M_{pl,Rd}$	=	$\dfrac{W_{pl}\,f_y}{\gamma_{M0}}$
	=	$\dfrac{2.405 \times 10^6 \times 355}{1.0}$
	=	$\text{853.7 kNm}$

📊 Comparison: Elastic vs Effective Class 2

For this section (Class 1 flange + Class 3 web), there are two valid bending resistances — the safe elastic default and the enhanced effective Class 2 value:

Approach	Modulus	Bending resistance	Note
Elastic Class 3	$W_{el}=2123\ \text{cm}^3$	$M_{el,Rd}=753.7\ \text{kNm}$	safe default — no plastic capacity
Effective Class 2 (Cl.5.5.2(11))	$W_{pl,eff}=2356\ \text{cm}^3$	$M_{pl,eff,Rd}=836.4\ \text{kNm}$	+11.0% over elastic; recovers ~83% of gap
Full plastic (reference)	$W_{pl}=2405\ \text{cm}^3$	$M_{pl,Rd}=853.7\ \text{kNm}$	only if web were Class 1/2

💡 Key takeaway:

The web just misses Class 2 (c_w/t_w = 71.25 > 67.6) but the flange is comfortably Class 1, so the effective Class 2 provision (Cl. 5.5.2(11)) applies. Using W_pl,eff lifts the bending resistance from 753.7 kNm (elastic) to 836.4 kNm — a +11.0% gain that recovers about 83% of the way to the full plastic moment (853.7 kNm), paying only a small fraction for the web's slenderness.

#steel design #work example #structural

References

Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings (EN 1993-1-1) —CEN (European Committee for Standardization) ·2005
DESIGNERS’ GUIDE TO EN 1993-1-1 EUROCODE 3: DESIGN OF STEEL STRUCTURES —L. GARDNER and D. A. NETHERCOT ·2005

CivilSimple Team

The CivilSimple Team writes practical engineering guides for the profession and the curious. All articles are reviewed for technical accuracy before publication.