Introduction
This article presents the calculation method for determining the buckling resistance of a pinned column with intermediate restraints in accordance with Eurocode standards. The procedure is based on an example from the Access Steel design examples collection and is compared with the calculation process implemented in Consteel’s steel member design functions, specifically within the Member Checks module.
In the following sections, a step-by-step guide is provided to demonstrate how the member check functionality can be applied to simple cases, highlighting both methodology and practical usage.
Input Data for the Example
The example considers a pinned column in a multi-storey building, subjected to a design axial force of $N_{Ed}$ = 1000 kN. The column has a total length of 10.50 m and is laterally restrained about the y–y axis at intervals of 3.50 m.
The member is a rolled HEA 260 section made of S235 steel. The cross-section is classified as Class 1. The geometric properties of the section are: height h = 250 mm, width b = 260 mm, web thickness $t_w$ = 7.5 mm, flange thickness $t_f$ = 12.5 mm, and fillet radius r = 24 mm. The cross-sectional area is A = 86.8 cm², with moments of inertia $I_y$ = 10450 cm⁴ and $I_z$ = 3668 cm⁴.
The material properties are defined according to EN 1993-1-1. Since the maximum thickness is less than 40 mm, the yield strength is taken as $I_y$ = 235 N/mm². The partial safety factors are γM0 = 1.0 and γM1 = 1.0.
Determining Design Buckling Resistance of a Compression Member
The design buckling resistance of the column $N_{b,Rd}$ is evaluated by determining the reduction factor χ for both principal buckling directions. This requires the calculation of the elastic critical forces $N_{cr}$, which form the basis for identifying the governing buckling mode.
Elastic critical force for the relevant buckling mode $N_{cr}$
The Young’s modulus is taken as $E=210000 \frac{N}{mm^2}$. The buckling lengths in the respective planes are $L_{cr,y} = 10.50m$ for buckling about the y–y axis and $L_{cr,z} = 3.50m$ for buckling about the z–z axis. Observe that the buckling lengths for the strong and weak axes differ according to the support conditions, which must be determined by the engineer in manual calculations.
$$N_{cr,y}=\frac{π^2*E*I_{y}}{L_{cr,y^2}}=1964.5 kN$$
$$N_{cr,z}=\frac{π^2*E*I_{z}}{L_{cr,z^2}}=6206.0 kN$$
In Consteel, the elastic critical force for the relevant buckling mode can be determined using the Individual Member Design approach. This is accessible in the Member Checks tab under the Steel module, where selected members can be added and evaluated.
Once a member is selected, the analysis results are automatically loaded, provided that first- or second-order analysis results are available. Ensure that the analysis has been run in the Analysis tab and the cross section check on the Global ckecks tab before proceeding to the Member Checks section.
For the pinned column with intermediate restraints, the relevant buckling cases, strong and weak axis, are selected, and the dominant load combination is automatically indicated with a *. Consteel identifies the intermediate restraints separately for each direction and divides the member into segments accordingly to help determine the correct buckling lengths.
Design parameters for each segment are set with the three-dot icon:
At this step, users must verify the assigned values. By default, the first value is applied, and the correct buckling shape or effective length factor should be confirmed based on engineering judgment.
In order to use the critical load multiplier selection option, make sure to perform the calculation first:
In order to check whether the correct critical load multiplier was selected, you can examine the effective length factor, which is calculated based on it (in this case, it is 1 for both directions). In our example, the relevant buckling shapes for the y–y and z–z directions are as follows:
The elastic critical force $N_{cr}$ is calculated automatically, regardless of whether the effective length factor was entered manually or the critical load multiplier was selected.
| Access Steel – manual calculation | Consteel using the effective length factor | Consteel using the critical load multiplier | |
| $N_{cr,y}$ | 1964.5 kN | 1962.53 kN | 1973.76 kN |
| $N_{cr,z}$ | 6206.0 kN | 6189.01 kN | 6218.96 kN |
Once all parameters are defined, the design check is executed by clicking the Check button, and the results are displayed.
Results can be reviewed and filtered by member, load combination, and buckling case. Lateral-torsional buckling checks follow a similar procedure, with segment boundaries adjustable and critical moments calculated either analytically or using the critical load multiplier.
Non-dimensional slenderness
In order to determine the reduction factor, the non-dimensional slenderness λ must be calculated based on the elastic critical force corresponding to the relevant buckling mode.
$$\overline{\lambda_y} = \sqrt{\frac{A*f_y}{N_{cr,y}}}=\sqrt{\frac{86.8*23.5}{1965}}=1.016$$
$$\overline{\lambda_z} = \sqrt{\frac{A*f_z}{N_{cr,z}}}=\sqrt{\frac{86.8*23.5}{6206}}=0.573$$
In Consteel, the detailed calculations for strong and weak axis buckling can be reviewed separately on the Results tab:
Reduction factor
For axial compression, the value of χ corresponding to the relevant non-dimensional slenderness $\overline{\lambda}$ should be determined from the appropriate buckling curve in accordance with EN 1993-1-1 §6.3.1.2.
For $\frac{h}{b}= \frac{250mm}{260mm} = 0.96 < 1.2$ and $t_f = 12.5 mm< 100 mm$
- buckling about axis y-y, buckling curve b, imperfection factor $\alpha=0.34$
$$\varphi_y=0.5*[1+0.34(1.019-0.2)+1.019^2]=1.158$$
$$\chi_y=\frac{1}{1.158+\sqrt{1.158^2-1.019^2}}=0.585$$
- buckling about axis z-z, buckling curve c, imperfection factor $\alpha=0.49$
$$\varphi_y=0.5*[1+0.49(0.573-0.2)+0.573^2]=0.756$$
$$\chi_y=\frac{1}{0.756+\sqrt{0.756^2-0.573^2}}=0.801$$
$$\chi=min(\chi_y;\chi_z)$$
$$\chi=0.585<1.00$$
Design buckling resistance of a compression member
$$N_{b,Rd}=\chi*\frac{A*f_y}{\gamma_{M1}}=0.585*\frac{86.8*23.5}{1.0}=1193 kN$$
$$\frac{N_Ed}{N_{b,Rd}}=\frac{1000}{1193}=0.84<1.00$$
Conclusion
This example demonstrates the application of the isolated member approach for a simple compression member. For more complex cases or alternative stability verification methods, such as the imperfection approach or the general method, refer to the dedicated article on stability design methods, where their principles and applications are discussed in detail.
Download modelDid you know that you could use Consteel to perform local and distortional buckling checks for cold-formed members?
First, sections must be loaded into the model. To load cold-formed sections, you can choose from four options: From library, Macro section, Draw section, or My library.
After the first-order and buckling analyses are completed, you can proceed to the Ultimate limit state check settings and enable the steel design cross-section and buckling checks. At the bottom of the steel design section, there is an option to Consider the supplementary rules from EN 1993-1-3 for the design of cold-formed sections. This checkbox must be selected if you want to design cold-formed sections.
When the calculation is finished, by opening the Section module, we can review all the properties of the Effective section of the elastic plate segment model. By opening each plate element, we can verify the length, effective length, thickness, effective thickness, slenderness, and reduction factor separately. In addition, the properties of the stiffeners can also be verified: area, moment of inertia, lateral spring stiffness, critical stress, reduction factor, compressive stress, reduced effective area, and reduced thickness.
Similarly, the stresses can also be checked from the Properties tab. In the colored figure or diagram view, all the calculated stresses can be seen together with their resultants.
Consteel automatically takes into account the effect of distortional buckling when calculating the effective sections of cold-formed thin-walled sections.
Moving on to the Standard resistance tab in the Section module, all calculated results can be verified, not only the dominant one. By opening the Global stability resistance check, we can see that, since we enabled the option to consider the supplementary rules from EN 1993-1-3 for the design of cold-formed sections, results are available both according to EN 1993-1-1 and according to EN 1993-1-3.
Download the example model and try it!
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Try Consteel for freeEN 1993-1-3 contains 3 „secret” formulas. The first two are used to determine the effective cross section due to distortional buckling when edge or intermediate stiffeners are involved. The third is used to calculate the distortion of the whole cross section when analyzed with a connected sheeting.
The physical meaning of all three formulas can be easily shown with simple ConSteel models which helps designer to understand the underlaying mechanial model.
The first formula (5.10) is used when the ability of an edge stiffener to stabilize a compressed flange of Z or C section is studied. During distortional buckling the intersection point of flange with the lip (it is called as edge stiffener) is expected to move in a direction perpendicular to the flange. This formula gives the stiffness value provided by the Z or C section, when is assumed that during deformations the point of intersection of the web with the flange doesn’t move. This assumption corresponds to attaching supports to these nodes as seen on picture 5.6 of EN 1993-1-3.
When a compressed edge stiffener would buckle, it will be partially restrained by the section with these attached supports. Depending on the distribution of normal stresses on the section, one or two edge stiffeners might be under compression. If both stiffeners are under compression and tend to buckle, the restraining capacity of the section will have to be shared between them. This sharing requirement is reflected by the coefficent kf. The spring stiffness value will be used as a distributed spring support when the buckling resistance of the edge stiffener is calculated.
Stiffness values are typically calculated as the ratio of a displacements obtained from the application of a unit load. In this case the unit loads are applied parallelly to the expected displacement of the compressed edge stiffeners.
The ConSteel model shown below reproduces the stiffness calculation for a 1 m long Zee section (as a simplification the unit loads are placed at the intersection of the flange with the lip and not at the center of the gravity of the edge zone):
Z purlin, nominal thickness = 1.30 mm, 200 mm deep, 72 mm wide symmetric flanges, 15.5 mm deep lips
in case of My bending: kf = 0, b1 = b2 = 72 mm, hw = 198.7 mm, t = 1.26 mm
K1 = 210000*1.26^3/4*(1-0.3^2)*1/(72^2*198.7+72^3+0.5*72*72*198.7*1.0) = 0.08 N/mm2
The resulting vertical displacement from the point load is 12.9 mm.
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