FAQ 000300 PL

Calculation RSBUCK

In RSBUCK, I obtain different effective length factors for one buckling mode. What is the explanation for this?


RSBUCK uses a momentary representation of the axial force distribution in the respective load state. These axial forces are increased iteratively until the buckling load case occurs.

In the numerical analysis, the stability load is indicated by the fact that the determinant of the stiffness matrix becomes zero.

If the effective length factor is known, the program uses it to determine the buckling load and buckling mode. For this lowest buckling load, all effective lengths and effective length factirs are determined.

Hinged-hinged column with a length of 20 m, cross-section: HE-B 500, load: only self-weight

For the first buckling mode, you obtain an effective length factor of kcr,y = 2.92 for the buckling about the major axis.

For the buckling about the minor axis with a buckling load of 651.3 kN, you obtain an effective length factor of 1.00.

If we solve the expression used for the buckling load Ncr = π² x E x I / Lcr² for Lcr and plug in Ncr = 651.3 kN and Iy = 107,200 cm4, we obtain an Lcr,y of 58.4 m, which results in an effective length factor kcr,y of 2.92.

In RSBUCK, we obtain the two effective length factors for each buckling mode and buckling load.

If you want to determine the correct effective length factor for the deflection perpendicular to the y-axis (buckling about major axis), it is necessary to calculate several buckling modes and choose the appropriate one.

In this case, it is the third buckling mode with a buckling load of 5485.5 kN. From this load, the effective lengths and effective length factors are determined, so that we obtain kcr,y = 1.0 und kcr,z = 0.345. For a quadratic cross-section, you obtain two equal effective lengths, because the stiffnesses in both directions are the same. Therefore, to obtain the correct values for the effective length factors in RSBUCK, it is necessary to look at several buckling modes.


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