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Buckling resistance of a pinned column with intermediate restraints

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 calculationConsteel using the effective length factorConsteel using the critical load multiplier
$N_{cr,y}$1964.5 kN1962.53 kN1973.76 kN
$N_{cr,z}$6206.0 kN6189.01 kN6218.96 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.

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Autor

Ádám Kis

As a Structural Engineer he was obsessed with parametric design methods. He is convinced that it can be efficient for any design task. This is how he came to the dark side and works as a developer on Pangolin. As a PhD student, he researches the implementation of simulation methods in Consteel.