11911x

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

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

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

Author

Ms. Matula provides technical support for our customers.

Links

Structural Analysis Software for Steel Structures

References

EC 3. (2009). Eurocode 3: Bemessung und Konstruktion von Stahlbauten − Teil 1-1: General rules and rules for buildings. (2010). Berlin: Beuth Verlag GmbH
Handbuch RF-/STAHL EC3. Tiefenbach: Dlubal Software, Juni 2020.
Albert, A.: Schneider - Bautabellen für Ingenieure mit Berechnungshinweisen und Beispielen, 23. Auflage. Köln: Bundesanzeiger, 2018

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The RF‑/STEEL EC3 add-on module can perform the design of fillet welds for all parametric, welded cross-sections of the cross-section library. For this, the option must be activated in the detail settings of the module. As an alternative, you can also use a surface model for the design.

Smallest Magnification Factor for Flexural Buckling in Frame Plane

In EN 1993-1-1, the General Method was introduced as a design format for stability analyses that can be applied to planar systems with arbitrary boundary conditions and variable structural height. The design checks can be performed for loading in the main load-bearing plane and simultaneous compression. The stability cases of lateral-torsional buckling and flexural buckling are analyzed from the main supporting plane; that is, about the weak component axis. Therefore, the issue often arises as to how to design, in this context, flexural buckling in the main load-bearing plane.

For the stability design of members and sets of members with a uniform cross-section, you can use the equivalent member method according to EN 1993-1-1, 6.3.1 to 6.3.3. However, as soon as a tapered cross-section is available, this method can no longer be used, or only used to a limited extent. The RF-/STEEL EC3 add-on module can automatically recognize these cases and switch to the general method.

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The RF‑/STEEL EC3 add-on module automatically transfers the buckling line to be used for the flexural buckling analysis for a cross-section from the cross-section properties. The assignment of the buckling line can be adjusted manually in the module input for general cross-sections in particular, as well as for special cases.

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Frequently Asked Questions (FAQ)

I am designing a set of members using the equivalent member method in RF‑/STEEL EC3, but the calculation fails. The system is unstable, delivering the message "Non-designable - ER055) Zero value of the critical moment on the segment".

What could be the reason?

Why are the equivalent member design checks grayed out in the Stability tab when activating the plastic design using the partial internal force method (RF‑/STEEL Plasticity)?

I would like to perform a stability analysis of the upper flange in a long truss. What is the best way to proceed?

I am performing a stability analysis of a beam for lateral-torsional buckling. Why is the modified reduction factor χ_LT,mod used in the design according to DIN EN 1993‑1‑1, 6.3.3 Method 2? Is it possible to deactivate this?

I need to define different types of lateral intermediate restrains for a single element in RF-/STEEL EC3. Is this possible?

I am comparing the flexural buckling design according to the equivalent member method and the internal forces according to the linear static analysis with the stress calculation according to the second-order analysis, including imperfections. The differences are very large. What could be the reason for this?

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