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  • Answer

    With the COM interface, you can access most operating elements as well as results of the following programs or add-on modules:

    • RF-/STEEL
    • RF- / STEEL EC3
    • RF- / CONCRETE
    • RX-TIMBER Glued-Laminated Beam
    • RF-/TIMBER Pro
    • RF- / DYNAM Pro
    • SUPER-RC

  • Answer

    RSBUCK / RF-STABILITY calculates at least one critical load factor or one critical load and an assigned buckling shape. The effective length is then back calculated from the critical load (see here ). Since this analysis is not carried out for individual local components, but only for the entire structure, the resulting load summary factors refer to the global structure and not to the local elements. However, it may happen that the structure fails globally for some load branch factors (but fails locally as well (depending on stiffness and axial force state).

    The calculated effective lengths should therefore only be used by the members that buckle in the respective buckling mode. In the case of the global failure of a structure (see example Figure 1), it is thus difficult to draw conclusions regarding the buckling behavior of individual members.

    Figure 2 shows a structure where the rear columns buckle. Therefore, it is recommended to use only the effective lengths calculated for both columns.

    Again in general: Buckling lengths from the RSBUCK module are only valid for a structural component in the respective direction if the related buckling shape clearly "bulges" the member in relation to the other in the respective direction. It is clear that the axial forces also have an impact on the results here.

  • Answer

    For the design of steel surfaces please use the module of the same name RF-STEEL Surfaces. If you also want to consider local denting, then the additional modules RF-STABIL (to determine the bulge and branching loads) and RF-IMP (to determine the imperfection figure, based on the bulge pattern) are also recommended.
  • Answer

    The dimensioning of cold - formed, thin - walled components and sheets is possible as a finite element calculation with the following programs. Basic program RFEM + RF-Stable (determination of the branching figure) + RF-IMP (generation of the pre-deformed FE mesh for the proof of stability) + RF-STAHL surfaces (calculation of the second-order theory determined stresses on the pre-formed model).
  • Answer

    The proof of stability for tensile structures can be converted into a pure stress analysis, if the theory is considered to be 2-fold and the imperfection required by the standard has been applied to the system.

    With the help of the modules RF-STABIL and RF-IMP imperfection (resp. a preformed FE mesh). The type of imperfection depends heavily on the component and the standard used. For bars, which were modeled as a tensile structure, the values from DIN EN 1993-1-1: 2005 5.3 can be used. For flat surfaces, for example, the values from DIN EN 1993-1-5: 2006 Appendix C can be used. For trays, the problem is much more complex and there are different approaches. From a generation of imperfections I would advise against this and perform the buckle proof by means of MNA / LBA concept according to DIN EN 1993-1-6, which does not require an approach of imperfection.

    If, for example, the surface model of a steel girder is to be detected, you can proceed as follows, for example:

    First A burden me comparatively (compared to. other internal forces in the load case) select high normal forces, in most cases the self-weight load case or a load case combination with the corresponding own weight is suitable. It may be necessary to provide each load combination with an individual imperfection.

    2. Calculate load combination according to 1-order theory and use as the basis for RF-STABIL

    3. Using RF-STABILITY to find the first eigenstate of a global failure

    4. Using RF-IMP, use the calculated eigenmode as the basis for an imperfection. In this case, for example, 1/300 of the carrier length can be used as the amplitude.

    5. Create a load case combination that uses the generated imperfection as a basis and is calculated according to 2-order theory.

    6. Perform a proof of tension on the basis of this load case combination, which at the same time is also proof of stability of the structure.

  • Answer

    It is possible to consider effects from structural and material nonlinearities in RF-STABILITY. Set the calculation method “Increase load until structural failure” in the module.

    These nonlinear effects are not considered for linear calculation.

  • Answer

    The modules perform the eigenvalue analysis for the entire model with a certain axial force state. Depending on the number of eigenvalues required, the programs provide results of crictical load factors with the corresponding buckling shapes for an eigenvalue, and effective length about the major and minor axis for each member per mode shape.

    Since each load case LC and each load combination CO often has a different axial force state in the elements available, there is a separate respective effective length result for each load situation of the frame column concerned. The effective length, which causes the column in the frame plane buckles sideway in the buckling shape, is the correct length to be used for the analysis of the load situation.

    However, this result may be different because of various load situation in each analysis, the longest effective length of all analyses performed applies in the design on the safe side equally for all load situations.

  • Answer

    If you obtain negative critical load factors in the module, increase the number of the eigenvalues (eigenvectors) to be determined. If there are not enough eigenvalues, it is not possible to hide the negative eigenvalues and thus show only positive, realistic results.

    For these eigenvalues, no stability failure occurs due to tension forces and no conclusion can be made about the buckling behavior of the model.

    Alternatively, you can use the ICG iteration to exclude negative critical buckling factor.

  • Answer

    The so-called Sturm sequence check tests whether all eigenvalues were found in a certain interval. It uses the diagonal matrix from the Gauss decomposition whose number of negative diagonal elements, corresponds to the number of eigenvalues below the respective interval limit.

    Therefore, the Sturm sequence check is used for the given interval limits and calculates the difference from them.

  • Answer

    RSBUCK uses a momentary representation of the axial force distribution in the respective load state. The axial forces are increased iteratively until the critical 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 buckling load and buckling mode are determined based on this. For the lowest buckling load, all effective lengths and effective length factirs are determined.

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

    For the first buckling mode, you obtain the 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 you set the expression for determining the buckling load Ncr = π² * E * I / Lcr² to Lcr and apply Ncr = 651.3 kN and Iy = 107,200 cm4, you obtain the Lcr,y of 58.4 m, which results in the effective length factor kcr,y of 2.92.

    In RSBUCK, there are two effective length factors determined for each buckling mode and buckling load.

    In order to obtain the correct effective length factor for the deflection perpendicular to the y-axis (buckling about the major axis), it is necessary to calculate several buckling modes (mode shapes). The correct value is displayed in Window 2.1. In the example, it is the third buckling mode with a buckling load of 5485.5 kN. For this load, the effective lengths and effective length factors are determined as follows: kcr,y = 1.0 and kcr,z = 0.345. 

    In the case of a quadratic cross-section, two equal effective lengths result as the stiffnesses in both directions are the same.

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