Evaluation of Local and Global Mode Shapes Using RSBUCK for Determination of Equivalent Member Length

Technical Article

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When performing the stability analysis of members according to the equivalent member method, considering internal forces according to the linear static analysis, it is very important to determine the governing equivalent member lengths.

Theoretical Background

Slendernesses and the resulting reduction factors are determined by the flexural buckling analysis according to EN 1993‑1‑1, Chapter 6.3, taking into account the elastic critical buckling load Ncr. This critical load is determined analytically in the STEEL‑EC3 add‑on module using the governing effective length. For simple structures, there are four commonly used Euler's cases.

Figure 01 - Euler Cases

In the case of complex structures, the effective length assessment is not so trivial. For this determination, you can utilise the RSBUCK add‑on module.

A critical load factor is first determined for the structure. This is multiplied by the normal forces of members to obtain the critical loads. The corresponding effective lengths for buckling about both axes are calculated using the adjusted formula: Ncr = E ∙ I ∙ π² / Lcr. Finally, the effective load factors are determined from this relation: kcr = Lcr / L.

Global and Local Mode Shapes in RSBUCK

Determination of mode shapes and the proper evaluation are explained with the following example of a simple frame.

Figure 02 - Frame

In the determination of a buckling mode and buckling lengths, the load plays an important role: Buckling values depend not only on the structural model, but also on the relation of normal forces to the total critical buckling load Ncr. Effective lengths can be calculated only for members with compressive forces. In addition, the load distribution over the entire structure affects the determination of the critical factors. By evaluating the individual mode shapes graphically, you can identify whether there is a global or a local mode shape. If there is a critical load of an individual member in the case of the most unfavourable structural critical load, this will be apparent in the graphic. In the case of such a failure, the results cannot be used for all other members and must not be evaluated.

In our example, the first mode shape with the critical load factor of 5.32 illustrates the global displacement of the frame in the frame plane. The second mode shape with the critical load factor of 11.42 illustrates the local displacement of the left column in the frame plane (buckling about the minor axis z).

Figure 03 - Mode Shapes

Divided Members

When calculating the effective lengths and effective length factors, you have to consider the division of members. In this example, the left column of the frame consists of two single members. For technical modelling reasons, the column was divided in the middle. When considering only the local mode shape (Buckling Mode No. 2), this can be categorized under Euler's case No. 2 and the expected result of the effective length factor kcr,z = 1.0. 2 and a buckling length factor kcr, z = 1.0 is expected in the results. However, the result window 2.1 in the add‑on module displays the effective length factor kcr,z = 2.0 for both ‘partial’ members of the column.

This can be easily explained by the relations mentioned above under ‘Theoretical Background’. In this case, the buckling length for the entire column is equal to the column length, and therefore, the effective length factor is 1.

The effective length factors for continuous members cannot be determined directly in RSBUCK. For this, you can evaluate the results of the individual members. The member which proves the smallest buckling load Ncr can be considered as the governing single member for a continuous member. You can calculate then the kcr values from the effective length of this member and the total length of the continuous member.

Figure 04 - Effective Lengths


Stability analysis Equivalent member method Mode shape Critical load



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