👋 Introduction

To know whether a certain steel section are sufficient structually under certain design action, steel cross-section checking need to be carried out.

In this work example we run the full cross-section check of a hot-rolled I-section (457x152x82) in S355 carrying a compression force N_Ed = 900 kN, shear force V_Ed = 100 kN together with a major-axis bending moment M_Ed = 200 kNm, in the same order a calculation report runs it:

Classify the section — the class decides which property each resistance is built from
Compression check N_c,Rd — the web is Class 4 in compression, so the effective area enters
Shear check V_c,Rd
Bending check M_c,y,Rd— the web is Class 1 in bending, so the full plastic modulus applies
Combined axial + shear + bending check — N_Ed is too large to ignore, so the reduced moment resistance M_N,y,Rd is required

The concept behind each step is explained in the companion article: How to Check a Steel Section — Classification, Axial, Shear and Bending (Eurocode 3).

📐 Design Data

Below is the design information of the I-section (457x152x82), which we will check under compression, shear and major-axis bending.

Steel Section
	Type	I-section (UB)
	Standard	457x152x82 (EN 10365)
	Depth, h	465.8mm
	Breadth, b	155.3mm
	Flange Thickness, t_f	18.9mm
	Web Thickness, t_w	10.5mm
	Root Radius, r	10.2mm
	Web Clear Depth, d	407.6mm

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

Design Actions
Design Axial Force (Compression), N_Ed		900 kN
Design Shear Force, V_Ed		100 kN
Design Bending Moment (Major Axis y), M_Ed		200 kNm

🔢 Calculation Steps

*Unless specified, Below uses Eurocode 3 symbols EN 1993-1-1.

1.Section Properties of I section (457x152x82)

Area $A$	=	$1.05 \times 10^{4} mm^2$
Moment of Inertia, Major $I_{y}$	=	$3.66 \times 10^{4} cm^4$
Moment of Inertia, Minor $I_{z}$	=	$1186 cm^4$
Elastic Modulus, Major $W_{el,y}$	=	$1571 cm^3$
Elastic Modulus, Minor $W_{el,z}$	=	$153 cm^3$
Plastic Modulus, Major $W_{pl,y}$	=	$1811 cm^3$
Plastic Modulus, Minor $W_{pl,z}$	=	$240 cm^3$
Shear Area $A_{v}$	=	$5167 mm^2$

2.Section Classification

Material Factor $ε$	=	$\sqrt{235/f_{y}} = 0.81$
Outstand Flange $\frac{c_{f}}{t_{f}}$	=	$\frac{(b-t_{w}-2 \times r)/2}{t_{f}} = 3.29 ≤ 9\epsilon = 7.32, Flange \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$
Part Subject to Compression
Web, subject to Compression $\frac{c_{w}}{t_{w}}$	=	$\frac{h-2 \times t_{f}-2 \times r}{t_{w}} = 38.82 > 42\epsilon = 34.17, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 4$
Part Subject to Bending
Web, subject to Bending $\frac{c_{w}}{t_{w}}$	=	$38.82 ≤ 72\epsilon = 58.58, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$
Part Subject to Bending and Compression
Ratio of compression $\alpha$	=	$\frac{1}{2} \times (1 + \frac{N_{Ed}}{f_{y} \times c_{w} \times t_{w}}) = 0.80$
Web, subject to bending and compression $\frac{c_{w}}{t_{w}}$	=	$38.82 ≤ (456 \epsilon)/(13\alpha - 1) = 39.68, Class \hspace{0.1cm} 2$

The overall cross-section is:
- Class 4 subjected to Compression
- Class 1 subjected to Bending
- Class 2 subjected to Bending and Compression

3.Section Check with Design Action

3.1Compression check — Class 4 web, effective area Cl.6.2.4

For Class 4 Web, consider under pure compression, the reduced web area can be calculated based on BSEN1993-1-5, Cl.4.4:
where $\psi$	=	$1$
$k_{\sigma}$	=	$4$

Plate slenderness of the Class 4 web(Cl. 4.4, EN 1993-1-5)
$\bar\lambda_p$	=	$\frac{c_w/t_w}{28.4 \times \varepsilon \times \sqrt{k_\sigma}}$
	=	$\frac{38.82}{28.4 \times 0.81 \times \sqrt{4}} = \frac{38.82}{46.0}$
	=	$\text{0.84}$

Reduction factor for local buckling(Eq. 4.2, EN 1993-1-5)
$\rho$	=	$\frac{\bar\lambda_p - 0.055(3 + \psi)}{\bar\lambda_p^{2}} \hspace{0.1cm} for \hspace{0.1cm} \bar\lambda_p > 0.5 + \sqrt{0.085 - 0.055 \times \psi}$
	=	$\frac{0.84 - 0.055(3 + 1)}{0.84^{2}} \hspace{0.1cm} for \hspace{0.1cm} 0.84 > 0.5 + \sqrt{0.085 - 0.055 \times 1}$
	=	$\text{0.67}$

Effective Area Class 4(Eq. 4.1, EN 1993-1-5)
$A_{eff}$	=	$A_{g} - (1 - \rho) \times A_{web}$
	=	$(1.05 \times 10^{4} mm^2) - (1 - 0.88) \times (4280 mm^2)$
	=	$\text{9 934 mm}^2$

Compression resistance with effective area(Cl. 6.2.4(2), Eq. 6.11) ^ⓘ
$N_{c,Rd}$	=	$\frac{A_{eff} \times f_y}{\gamma_{M0}}$
	=	$\frac{9\,934 \times 355}{1.0}$
	=	$\text{3 527 kN}$

Compression check(Cl. 6.2.4(1), Eq. 6.9)
$\frac{N_{Ed}}{N_{c,Rd}}$	=	$\frac{N_{Ed}}{N_{c,Rd}} \le 1$
	=	$\frac{900}{3\,527}$
	=	$0.26 \le 1,\ \text{OK!}$

∴ The section is sufficient for Pure Compression

3.2Shear check Cl.6.2.6

Design Shear Resistance(Cl. 6.2.6(2))
$V_{pl,Rd}$	=	$\frac{A_{v} \times (f_{y}/3^{0.5})}{\gamma_{M0}}$
	=	$\frac{5167 mm^2 \times (355 MPa/3^{0.5})}{1}$
	=	$1059 kN$

Shear Check(Cl. 6.2.6(1); Eq. 6.17)
$\frac{V_{Ed}}{V_{c,Rd}}$	=	$\frac{V_{Ed}}{V_{c,Rd}}$
	=	$\frac{100 kN}{1059 kN}$
	=	$0.09 ≤ 1, OK!$

∴ The section is sufficient for Shear

3.3Bending check — Class 1, plastic modulus Cl.6.2.5

Design Resistance for Bending at y(Cl. 6.2.5(2), Eq. 6.13) ^ⓘ
$M_{c,y,Rd}$	=	$\frac{W_{pl,y} \times f_{y}}{\gamma_{M0}}$
	=	$\frac{1811 cm^3 \times 355 MPa}{1}$
	=	$643 kNm$

Bending Check at major axis(Cl. 6.2.5, Eq. 6.12)
$\frac{M_{Ed}}{M_{c,y,Rd}}$	=	$\frac{M_{Ed}}{M_{c,y,Rd}}$
	=	$\frac{200}{643}$
	=	$0.31 ≤ 1, OK!$

∴ The section is sufficient for pure bending about the major axis.

3.4Combine Bending Moment, Shear and Axial Check — Class 2 Cl.6.2.10

Shear Force and Resistance ratio(Cl.6.2.10(2))
$\frac{V_{Ed}}{V_{pl,Rd}}$	=	$\frac{V_{Ed}}{V_{pl,Rd}}$
	=	$\frac{100 kN}{1059 kN}$
	=	$0.09 < 0.5, no reduction factor from shear$

For major axis of I section, reduction effect from axial forceneed not be made, if it satisfied below two criterias:

Criterion 1 — quarter of the compression resistance(Eq. 6.33) ^ⓘ
$0.25\,N_{pl,Rd}$	=	$N_{Ed} \le 0.25\,N_{pl,Rd}$
	=	$0.25 \times 3\,527 = 882\ \text{kN}$
	=	$882\ \text{kN} < 900\ \text{kN},\ \text{not pass}$

Criterion 2 — half of the web yield capacity(Eq. 6.34)
$\frac{0.5\,h_w t_w f_y}{\gamma_{M0}}$	=	$N_{Ed} \le \frac{0.5 \times h_w \times t_w \times f_y}{\gamma_{M0}}$
	=	$\frac{0.5 \times 428 \times 10.5 \times 355}{1.0} \quad (h_w = h - 2t_f = 428\ \text{mm})$
	=	$798\ \text{kN} < 900\ \text{kN},\ \text{not pass}$

∴ Current Section not pass both certeria, further reduction factor need to apply to y axis from axial force.

Ratio of design force to plastic resistance(Cl. 6.2.9.1(5))
$n$	=	$\frac{N_{Ed}}{N_{pl,Rd}}$
	=	$\frac{900}{3\,527}$
	=	$\text{0.26}$

Ratio of web area to gross area(Cl. 6.2.9.1(5))
$a$	=	$\frac{A - 2 \times b \times t_f}{A}$
	=	$\frac{1.05 \times 10^{4} mm^2 - 2 \times 155 mm \times 19 mm}{1.05 \times 10^{4} mm^2}$
	=	$\text{0.44}$

Reduced Moment Resistance(Eq. 6.36)
$M_{N,y,Rd}$	=	$M_{pl,y,Rd} \times \frac{1 - n}{1 - 0.5(a)}$
	=	$643 kNm \times \frac{1 - 0.26}{1 - 0.5(0.44)}$
	=	$\text{613 kNm}$

Bending and Axial and Shear at major axis(Eq. 6.31)
$\frac{M_{Ed}}{M_{N,y,Rd}}$	=	$\frac{M_{Ed}}{M_{N,y,Rd}}$
	=	$\frac{200 kNm}{613 kNm}$
	=	$0.33 ≤1, OK$

∴ The section is sufficient for Bending and Axial and Shear for y axis

📋 Summary

The I Section (457x152x82) is sufficient under Compression (900 kN), Shear (100 kN), Moment at Major Axis (200 kNm), it :
✓ Pure Compression Check (util: 26%)
✓ Shear Check (util: 9%)
✓ Bending Check at major axis (util: 31%)
✓ Bending and Axial and Shear at major axis (util: 33%)

📄 Need a Full Design Check Report?

If you want a similar section check with design actions — classification, the Class 4 compression check, bending and the combined check — with your own design data, feel free to try our application, what you need is just the design data value, and a ready-to-submit design PDF report will be generated in a minute.

The ready-to-submit PDF Report is not generated by AI but programmed by chartered engineer, which is accurate, no hallucination and same input same output.

#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
Eurocode 3: Design of steel structures — Part 1-5: Plated structural elements (EN 1993-1-5) —CEN (European Committee for Standardization) ·2006

CivilSimple Team

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