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  • Answer

    Friction represents a nonlinearity and can therefore only be modified via the interface to the member end release hinge.

    For this purpose, the member end release must be created first, if not already available. Then, the IMemberHinge interface is brought to the member end release and then to the nonlinearity (here IFriction ). Then , you can use the methods GetData and SetData to modify the data (here Friction ):

    Sub SetMemberHingeFriction ()

    Dim model As RFEM5.model
    Set model = GetObject (, "RFEM5.Model")
    model.GetApplication.LockLicense

    On Error GoTo e

    Dim data As IModelData
    Set data = model.GetModelData

    Dim hinge (0 To 0) As RFEM5.MemberHinge

    hinge (0) .No = 1
    hinge (0) .RotationalConstantX = 1
    hinge (0) .RotationalConstantY = 2
    hinge (0) .RotationalConstantZ = 3
    hinge (0) .TranslationalConstantX = 4
    hinge (0) .TranslationalConstantY = 5
    hinge (0) .TranslationalConstantZ = 6
    hinge (0) .Comment = "Member Hinge 1"
        
    hinge (0) .TranslationalNonlinearityX = FrictionAType

    data.PrepareModification
    data.SetMemberHinges hing
    data.FinishModification
        
    'get interface for member'
    Dim imemhing As IMemberHinge
    Set imemhing = data.GetMemberHinge (1, AtNo)
        
    'get interface for nonlinearity'
    Dim iFric As IFriction
    Set iFric = imemhing.GetNonlinearity (AlongAxisX)
        
    'get friction data'
    Dim fric As Friction
    fric = iFric.GetData
        
    fric.Coefficient1 = 0.3
        
    'set friction data
    data.PrepareModification
    iFric.SetData fric
    data.FinishModification
        
        
    e: If Err.Number <> 0 Then MsgBox Err.Description,, Err.Source

    Set data = Nothing
    model.GetApplication.UnlockLicense
    Set model = Nothing

    End Sub


    For Coefficient Vy + Vz, Coeffcient2 is used to set the second coefficient. The translational spring in the Friction dialog box is controlled by the translational spring of the Member-End Hinge. In the concrete case, this is TranslationalConstantX for the x-direction (see Figure 01).

  • Answer

    Currently, there is no direct way to determine the breakdown load. It is important that there are also several different states and thus that there can be several local breakdown loads.

    You can assign load increments to a structure and retroactively view the individual load increments and thus find a load factor near a breakdown (depending on the number of load increments).

    In the attached model, 20 load increments and saving of the intermediate results have been set (Figure 01). In Figure 02, load step 10 is shown as the last load increment before the breakdown and in Figure 03 the first load step after the breakdown is visible, where clearly the deformation is significantly higher.

  • Answer

    If this error occurs, the download did not run correctly. Please download the program again. Please note that the file can only be executed after the complete download.
  • Answer

    The asynchronous calculation is used if a self-created program should only open or continue RFEM or RSTAB. When the calculation is complete, the event is transferred via a delegate. You can find a C # example in a Visual Studio project in the download area below.
  • Answer

    Yes, it is possible to create nodal releases via the COM interface. Here is an example:

    Sub nodal_release ()

    Dim iApp As RFEM5.Application
    Dim iModel As RFEM5.model
        
    On Error GoTo e
        
    Set iApp = GetObject (, "RFEM5.Application")
    iApp.LockLicense
            
    test = iApp.GetModelCount
    Set iModel = iApp.GetModel (0)
            
    Dim iModeldata As RFEM5.iModeldata
    Set iModeldata = iModel.GetModelData
            
            
            '   for setting a nodal release a member hinge is needed
            '   the object for the axis system could not be the same with the released one
    Dim nodRel As NodalRelease
    nodRel.Location = OriginalLocationType
    nodRel.AxisSystem = LocalFromLine
    nodRel.AxisSystemFromObjectNo = 2
    nodRel.Comment = "test nodal release"
    nodRel.MemberHingeNo = 1
    nodRel.NodeNo = 1
    nodRel.ReleasedMembers = 1
            
    iModeldata.PrepareModification
    iModeldata.SetNodalRelease nodRel
    iModeldata.FinishModification
            
            
    e: If Err.Number <> 0 Then MsgBox Err.description,, Err.Source
        
    iApp.UnlockLicense


    End Sub


    Please note that the local axis system may not refer to the same elements as the ones that are released.

  • Answer

    RSTAB is a FEM program that uses trigonometric trial functions for the members. For this reason, members do not have to be subdivided for sufficiently accurate results and the calculation speed is correspondingly higher.

    RSBUCK determines the eigenvalues of the stiffness matrix and can thus linearly calculate the critical load and buckling mode of the structure.

  • Answer

    It is possible to export the effective lengths from the add-on module to EXCEL as shown in Figure 1.
  • Answer

    It is possible to export the effective lengths from the add-on module to EXCEL or import it from EXCEL, as shown in Figure 1.
  • Answer

    To transfer several elements, you first have to create a field (array) in VBA:

    Dim nodes (0 to 2) as RFEM5.Node

    Then, you can transfer all nodes at once to the field (here three elements 0,1,2) with the method IModelData.SetNodes ().

    These methods are also available e.g. for the following structural elements:

    IModelData.SetLines ()
    IModelData.SetMembers ()
    IModelData.SetSurfaces ()
  • Answer

    First, please note that the local deformations of surfaces are always related to the undeformed system. Therefore, for a multi-storey building, the deformations of the top floor also include the deformations of the lower floors, as shown in Figure 01 on the left.

    Figure 01 on the right shows the corresponding bending moment m-y. It is identical for the floors, as expected for this simple model. In such a case, the partial calculation of the individual floors is no problem, because the relative deformation seems to be identical for each floor.

    However, it becomes problematic if the supporting elements are loaded differently or if the stiffness of the supporting elements within a floor is different. Figure 02 shows the bending moment m-y of such a system. It is apparent that the distribution, especially between the bottom floor and the top floor, shows the greatest differences. In this particular case, internal columns with less stiff cross-section were added in addition to the corner columns. For this reason, the relative deformation increases more with each additional floor in the middle than at the corner columns.

    In reality, this structure will not exist in this way because the floors are manufactured one after the other and thus the deformations (for example due to self-weight) are compensated from one floor to another. Thus, it is a typical structural state problem. The question arises, therefore, of whether the effects can be neglected or if, for example, it is necessary to analyze the results with the add-on module RF-STAGES.


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