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• ### How is friction set on a member end release via the COM interface?

New FAQ 004133 EN-US

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.modelSet model = GetObject (, "RFEM5.Model")model.GetApplication.LockLicenseOn Error GoTo eDim data As IModelDataSet data = model.GetModelDataDim hinge (0 To 0) As RFEM5.MemberHingehinge (0) .No = 1hinge (0) .RotationalConstantX = 1hinge (0) .RotationalConstantY = 2hinge (0) .RotationalConstantZ = 3hinge (0) .TranslationalConstantX = 4hinge (0) .TranslationalConstantY = 5hinge (0) .TranslationalConstantZ = 6hinge (0) .Comment = "Member Hinge 1"    hinge (0) .TranslationalNonlinearityX = FrictionATypedata.PrepareModificationdata.SetMemberHinges hingdata.FinishModification    'get interface for member'Dim imemhing As IMemberHingeSet imemhing = data.GetMemberHinge (1, AtNo)    'get interface for nonlinearity'Dim iFric As IFrictionSet iFric = imemhing.GetNonlinearity (AlongAxisX)    'get friction data'Dim fric As Frictionfric = iFric.GetData    fric.Coefficient1 = 0.3    'set friction datadata.PrepareModificationiFric.SetData fricdata.FinishModification        e: If Err.Number <> 0 Then MsgBox Err.Description,, Err.SourceSet data = Nothingmodel.GetApplication.UnlockLicenseSet model = NothingEnd 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).

• ### How can I determine the breakdown load of a system with a postcritical analysis?

New FAQ 004122 EN-US

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.

• ### After downloading the installation file, an error message appears saying "The setup file is defective" when you run it. What can I do?

New FAQ 003619 EN-US

• ### How does the asynchronous calculation via the COM interface work?

New FAQ 003605 EN-US

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.
• ### Is it possible to create nodal releases via the COM interface?

New FAQ 003570 EN-US

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

Sub nodal_release ()Dim iApp As RFEM5.ApplicationDim iModel As RFEM5.model    On Error GoTo e    Set iApp = GetObject (, "RFEM5.Application")iApp.LockLicense        test = iApp.GetModelCountSet iModel = iApp.GetModel (0)        Dim iModeldata As RFEM5.iModeldataSet 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 oneDim nodRel As NodalReleasenodRel.Location = OriginalLocationTypenodRel.AxisSystem = LocalFromLinenodRel.AxisSystemFromObjectNo = 2nodRel.Comment = "test nodal release"nodRel.MemberHingeNo = 1nodRel.NodeNo = 1nodRel.ReleasedMembers = 1        iModeldata.PrepareModificationiModeldata.SetNodalRelease nodReliModeldata.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.

• ### Which calculation methods are used by RSTAB and RSBUCK?

New FAQ 003564 EN-US

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.

• ### How to export the effective lengths from the RF-STABILITY to EXCEL module?

New FAQ 003557 EN-US

It is possible to export the effective lengths from the add-on module to EXCEL as shown in Figure 1.
• ### How to export the effective lengths from the STEEL EC3 module to EXCEL?

New FAQ 003556 EN-US

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.
• ### How do I transfer a field of nodes instead of just a single node via the COM interface in VBA?

New FAQ 003542 EN-US

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 ()
• ### In a comparison between the partial calculation of a single floor and the total calculation of the building, the local deformations are very different. What is the cause?

New FAQ 003540 EN-US

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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