👋 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 RHS (500x300x8.0) in S355 carrying a compression force N_Ed = 1000 kN, shear force V_Ed = 1000 kN together with a major-axis bending moment M_Ed = 200 kNm, in the same order a calculation report runs it:

Classify the section
Compression check N_c,Rd
Shear check V_c,Rd
Bending check M_c,y,Rd
Combined axial + shear + bending check

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 RHS (500x300x8.0), which we will check under compression, shear and major-axis bending.

Steel Section
	Type	RHS
	Standard	500x300x8.0
	Depth, h	500mm
	Breadth, b	300mm
	Thickness, t	8mm

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		1000 kN
Design Shear Force, V_Ed		1000 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 RHS (500x300x8.0)

Area $A$	=	$1.23 \times 10^{4} mm^2$
Moment of Inertia, Major $I_{y}$	=	$4.25 \times 10^{4} cm^4$
Moment of Inertia, Minor $I_{z}$	=	$1.95 \times 10^{4} cm^4$
Elastic Modulus, Major $W_{el,y}$	=	$1698 cm^3$
Elastic Modulus, Minor $W_{el,z}$	=	$1300 cm^3$
Plastic Modulus, Major $W_{pl,y}$	=	$2049 cm^3$
Plastic Modulus, Minor $W_{pl,z}$	=	$1450 cm^3$
Shear Area $A_{v}$	=	$7668 mm^2$

2.Section Classification

Material Factor $ε$	=	$\sqrt{235/f_{y}} = 0.81$
Part Subject to Compression
Vertical Part, subject to compression $\frac{c_{w}}{t_{w}}$	=	$\frac{h - 2 \times t}{t} = 60.50 > 42\epsilon = 34.17, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 4$
Horizontal Part, subject to compression $\frac{c_{w}}{t_{w}}$	=	$\frac{b - 2 \times t}{t} = 35.50 > 42\epsilon = 34.17, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 4$
Part Subject to Bending
Vertical Part, subject to bending $\frac{c_{w}}{t_{w}}$	=	$\frac{h - 2 \times t}{t} = 60.50 > 42\epsilon = 60.50 ≤ 83\epsilon = 67.53, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm}2$
Horizontal Part, subject to bending $\frac{c_{w}}{t_{w}}$	=	$\frac{b - 2 \times t}{t} = 35.50 > 42\epsilon = 35.50 ≤ 72\epsilon = 58.58, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$
Part Subject to Bending and Compression
Ratio of compression, Vertical Part $\alpha$	=	$\frac{1}{2} \times (1 + \frac{N_{Ed}}{f_{y} \times c \times t}) = 0.86$
Ratio of End Stress $\psi$	=	$\frac{2 \times N_{Ed}}{A \times f_{y}} - 1 = -0.54$
Vertical Part, subject to bending and comp. $\frac{c_{w}}{t_{w}}$	=	$\frac{h - 2 \times t}{t} = 60.50 ≤ (42 \epsilon)/(0.67 + 0.33\psi) = 69.52, Class \hspace{0.1cm} 3$
Ratio of compression, Horizontal Part $\alpha$	=	$\frac{1}{2} \times (1 + \frac{N_{Ed}}{f_{y} \times c \times t}) = 1.12$
Horizontal Part, subject to bending and comp. $\frac{c_{w}}{t_{w}}$	=	$\frac{b - 2 \times t}{t} = 35.50 ≤ (42 \epsilon)/(0.67 + 0.33\psi) = 69.52, Class \hspace{0.1cm} 3$

The overall cross-section is:
- Class 4 subjected to Compression
- Class 2 subjected to Bending
- Class 3 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, Class 4 Vertical Plate(Cl. 4.4, EN 1993-1-5)
$\bar\lambda_{p,vert}$	=	$\frac{(h - 3 \times t)/t}{28.4 \times \epsilon \sqrt{k_{\sigma}}}$
	=	$\frac{(500 mm - 3 \times 8 mm)/8 mm}{28.4 \times 0.81 \sqrt{4}}$
	=	$1.29$

Reduction factor for local buckling, Vertical Part(Eq. 4.2, EN 1993-1-5)
$\rho_{vert}$	=	$\frac{\bar\lambda_{p,vert} - 0.055(3 + \psi)}{\bar\lambda_{p,vert}^{2}} \hspace{0.1cm} for \hspace{0.1cm} \bar\lambda_{p,vert} > 0.5 + \sqrt{0.085 - 0.055 \times \psi}$
	=	$\frac{1.29 - 0.055(3 + 1)}{1.29^{2}} \hspace{0.1cm} for \hspace{0.1cm} 1.29 > 0.5 + \sqrt{0.085 - 0.055 \times 1}$
	=	$\text{0.64}$

Plate Slenderness, Class 4 Horizontal Plate(Cl. 4.4, EN 1993-1-5)
$\bar\lambda_{p,hori}$	=	$\frac{(b - 3 \times t)/t}{28.4 \times \epsilon \sqrt{k_{\sigma}}}$
	=	$\frac{(300 mm - 3 \times 8 mm)/8 mm}{28.4 \times 0.81 \sqrt{4}}$
	=	$0.75$

Reduction factor for local buckling, Horizontal Part(Eq. 4.2, EN 1993-1-5)
$\rho_{hori}$	=	$\frac{\bar\lambda_{p,hori} - 0.055(3 + \psi)}{\bar\lambda_{p,hori}^{2}} \hspace{0.1cm} for \hspace{0.1cm} \bar\lambda_{p,hori} > 0.5 + \sqrt{0.085 - 0.055 \times \psi}$
	=	$\frac{0.75 - 0.055(3 + 1)}{0.75^{2}} \hspace{0.1cm} for \hspace{0.1cm} 0.75 > 0.5 + \sqrt{0.085 - 0.055 \times 1}$
	=	$\text{0.94}$

Effective Area Class 4(Eq. 4.1, EN 1993-1-5)
$A_{eff}$	=	$A_{g}- 2 \times (1 - \rho_{vert}) \times A_{vert}- 2 \times (1 - \rho_{hori}) \times A_{hori}$
	=	$1.23 \times 10^{4} mm^2 - 2 \times (1 - 0.64) \times 3872 mm^2 - 2 \times (1 - 0.64) \times 2272 mm^2$
	=	$9261 mm^2$

Compression resistance with effective area(Cl. 6.2.4(2), Eq. 6.11) ^ⓘ
$N_{c,Rd}$	=	$\frac{A_{eff} \times f_y}{\gamma_{M0}}$
	=	$\frac{1.23 \times 10^{4} mm^2 \times 355}{1.0}$
	=	$\text{3288 kN}$

Compression check(Cl. 6.2.4(1), Eq. 6.9)
$\frac{N_{Ed}}{N_{c,Rd}}$	=	$\frac{N_{Ed}}{N_{c,Rd}} \le 1$
	=	$\frac{1000kN}{3288kN}$
	=	$0.30 ≤ 1, OK!$

∴ The section is sufficient for Pure Compression

3.2Shear check Cl.6.2.6

Design Shear Resistance(Cl. 6.2.6(2))
$V_{pl,Rd}$	=	$\frac{A_{v} \times (f_{y}/3^{0.5})}{\gamma_{M0}}$
	=	$\frac{7668 mm^2 \times (355 MPa/3^{0.5})}{1}$
	=	$1572 kN$

Shear Check(Cl. 6.2.6(1); Eq. 6.17)
$\frac{V_{Ed}}{V_{c,Rd}}$	=	$\frac{V_{Ed}}{V_{c,Rd}}$
	=	$\frac{1000 kN}{1572 kN}$
	=	$0.64 ≤ 1, OK!$

∴ The section is sufficient for Shear

3.3Bending check — Class 2, plastic modulus Cl.6.2.5

Design Resistance for Bending at y(Cl. 6.2.5(2), Eq. 6.13) ^ⓘ
$M_{c,y,Rd}$	=	$\frac{W_{pl,y} \times f_{y}}{\gamma_{M0}}$
	=	$\frac{2049 cm^3 \times 355 MPa}{1}$
	=	$727 kNm$

Bending Check at major axis(Cl. 6.2.5, Eq. 6.12)
$\frac{M_{Ed}}{M_{c,y,Rd}}$	=	$\frac{M_{Ed}}{M_{c,y,Rd}}$
	=	$\frac{200 kNm}{727 kNm}$
	=	$0.27 ≤ 1, OK!$

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

3.4Combine Bending Moment, Shear and Axial Check — Class 3 Cl.6.2.10

Shear Force and Resistance ratio(Cl.6.2.10(2))
$\frac{V_{Ed}}{V_{pl,Rd}}$	=	$\frac{V_{Ed}}{V_{pl,Rd}}$
	=	$\frac{1000 kN}{1572 kN}$
	=	$0.64 > 0.5, take (1 - \rho)f_y reduction$

Reduction Factor for moment resistance under Shear(Cl. 6.2.8(1))
$\rho$	=	$(\frac{2 \times V_{Ed}}{V_{pl,Rd}} - 1)^{2}$
	=	$(\frac{2 \times 1000 kN}{1572 kN} - 1)^{2}$
	=	$0.07$

Local longitudinal Stress(Cl.6.2.9.2)
$\sigma_{x,Ed}$	=	$\frac{N_{Ed}}{A_{eff}} + \frac{M_{Ed,y}}{W_{el,y}}$
	=	$\frac{1000 kNm}{9261 mm^2} + \frac{200 kNm}{1698 cm^3}$
	=	$226 N/mm^2 < \frac{f_{y} \times (1 - \rho)}{\gamma_{M0}} = 329 N/mm^2, ∴ OK!$

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

📋 Summary

The I Section (457x152x82) is sufficient under Compression (1000 kN), Shear (1000 kN), Moment at Major Axis (200 kNm), it :
✓ Pure Compression Check (util: 30%)
✓ Shear Check (util: 64%)
✓ Bending Check at major axis (util: 27%)
✓ Bending and Axial and Shear at major axis (util: 69%)

📄 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.