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Dipl.-Ing. (BA) Sandy Matula

2019-12-04

Cross-Section Design of a Column Under Axial Force and Bending

In this technical article, a hinged column with a centrally acting axial force and a linear load that acts on the major axis are designed according to EN 1993-1-1 with the aid of the RF-/STEEL EC3 add-on module. The column head and column base are assumed as a lateral and torsional restraint. The column is not held against rotation between the supports. The cross-section of the column is an HEB 360 from S235.

System, Loading, and Internal Forces

System, Loading, Internal Forces

System, Loading, Internal Forces

Cross-Section Properties

h = 300 mm, b = 360 mm, t_s = 12.5 mm, t_g = 22.5 mm, r = 27 mm
A = 180.60 cm², I_y = 43,190 cm⁴, W_y = 2,400 cm³

\begin{array}{l} A_{v} = A - 2 \cdot b \cdot t_{g} (t_{s} 2 \cdot r) \cdot t_{g} \\ A_{v} = 180.60 - 2 \cdot 30 \cdot 2.25 (1.25 2 \cdot 2.70) \cdot 2.25 = 60.56 cm ² \end{array}

Formula 1

Girder Section HEB 360, S235

Girder Section HEB 360, S235

Design of Cross-Section Class

[1], Table 5.2, cross-section parts supported on two sides
Web: c = 261 mm, t = 12.5 mm

prov (\frac{c}{t}) = (\frac{261}{12.5}) = 20.88 < limit (\frac{c}{t}) = 33 \cdot ε = 33 \cdot 1 = 33

Formula 2

The web fulfills the requirements of cross-section class 1.

[1], Table 5.2, cross-section parts supported on one side
Flange: c = 116.8 mm, t = 22.5 mm

prov (\frac{c}{t}) = (\frac{116.8}{22.5}) = 5.19 < limit (\frac{c}{t}) = 9 \cdot ε = 9 \cdot 1 = 9

Formula 3

The flange fulfills the requirements of cross-section class 1.

Accordingly, the entire cross-section can be classified in cross-section class 1.

Cross-Section Design Pressure According to 6.2.4

\begin{array}{l} \frac{N_{Ed}}{N_{c, Rd}} \leq 1, 0 \\ N_{c, Rd} = \frac{A \cdot f_{y}}{γ_{M 0}} = \frac{180.60 \cdot 23.5}{1.0} = 4, 244.10 kN \\ \frac{2, 000.00}{4, 244.10} = 0.47 \leq 1 \to Design is fulfilled . \end{array}

Formula 4

Cross-Section Design Shear Force in Axis z According to 6.2.6

\begin{array}{l} V_{c, Rd} = V_{pl, Rd} = \frac{A_{v} \cdot \frac{f_{y}}{\sqrt{3}}}{γ_{M 0}} = \frac{60.56 \cdot \frac{23.5}{\sqrt{3}}}{1.0} = 821.66 kN \\ \frac{\max V_{z, Ed}}{V_{pl, Rd}} = \frac{48.75}{821.66} = 0.06 \leq 1 \to Design is fulfilled . \\ \frac{h_{w}}{t_{w}} = \frac{315}{12.5} = 25.2 \leq 72 \cdot \frac{ε}{η} = 72 \cdot \frac{1.0}{1.2} = 60.00 \to No shear buckling design required . \end{array}

Formula 5

Cross-Section Design Bending, Shear, and Axial Force According to 6.2.9.1

The axial force must be taken into account if:

\begin{array}{l} N_{Ed} \geq 0.25 \cdot N_{pl, Rd} \\ N_{Ed} = 2, 000 kN > 0.25 \cdot 4, 244.10 = 1, 061.03 kN \end{array}

Formula 6

and

\begin{array}{l} N_{Ed} \geq \frac{0.5 \cdot h_{w} \cdot t_{w} \cdot f_{y}}{γ_{M 0}} \\ N_{Ed} = 2, 000 kN > \frac{0.5 \cdot 31.50 \cdot 1.25 \cdot 23.5}{1.0} = 462.66 kN \end{array}

Formula 7

→ The axial force must be taken into account.

\begin{array}{l} M_{N, y, Rd} = 630.51 \cdot \frac{(1 - 0.471)}{(1 - 0.5 \cdot 0.252)} = 381.62 kNm \\ \frac{\max M_{y, Ed}}{{M_{N}}_{, y, Rd}} = \frac{79.22}{381.62} = 0.21 < 1.0 \to Design is fulfilled . \end{array}

Formula 8

Cross-Section Design Checks

Cross-Section Design Checks

KB|001600

Knowledge Base

KB 001600 | Cross-Section Design of Column Under Axial Force and Bending

Author

Ms. Matula provides technical support for our customers.

Links

Structural Analysis Software for Steel Structures

References

European Committee for Standardization (CEN). (2010). Eurocode 3: Design of Steel Structures – Part 1‑1: General Rules and Rules for Buildings, EN 1993‑1‑1:2010‑12. Berlin: Beuth Verlag GmbH.
Dlubal Software. (2020). Manual RF-/STEEL EC3. Tiefenbach: Dlubal Software.
Albert, A. (2018). Schneider – Bautabellen für Ingenieure mit Berechnungshinweisen und Beispielen (23rd ed.). Cologne: Bundesanzeiger.

Downloads

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