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• ### Why does an eccentrically set member receive axial forces although it is only loaded by shear forces?

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If a member is connected eccentrically to a surface or to another member, you can also imagine that each node (RSTAB) and each FE node (RFEM) of each element is coupled with a member (see Figure 01 above). . The result is identical to that of the defined eccentricities (see Figure 01 below). The system shown in the picture is nothing more than a truss with a top flange and a bottom flange connected by verticals. As is known, the flanges are increasingly subjected to axial forces rather than bending moments due to the geometry (see Figure 02).
• ### How can I deactivate the individual members for the calculation, for example, to simulate a column failure?

There are two ways to do this:

1. You can define the corresponding member as a null member. Thus, it is not considered in the calculation of all load cases and load combinations.
2. You can deactivate the corresponding member in all or only for certain load cases and/or load combinations. To do this, it is necessary to activate the "Modify stiffnesses" option in the calculation parameters of the load case or load combination. Then, you can deactivate this member in the additional tab window.

However, you should pay attention to the following points:

• When using the null member, a warning message appears if the member loads have been defined.
• In the case of generated loads, the loads are redistributed automatically when using the null member.
• If deactivating the member in the calculation parameters, the member loads and the determined generated loads are not considered. No error message appears in this case. It is necessary to redistribute the loads manually.
• ### Is it also possible to specify a cable length in RF‑FORM‑FINDING?

In addition to a prestressing force or the target cable sag, it is also possible to specify the cable length, as shown in Figure 01.

The program then tries to fit the cable subjected to the acting force (a load case of the "form-finding" type with a dead load, for example) in the way that the length corresponds to the specified length.

• ### I have a model with solids, the calculation reports a singularity. How do I fix this model?

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The solids are connected by using surface release with the "fixed if negative pz" nonlinearity. The force pz is derived on the basis of the orientation of the contact surfaces. If the orientation of both surfaces is the same and you want the compressive load to be transferred between the surfaces, it is necessary to use the "fixed if positive pz" nonlinearity.
• ### When trying to perform a calculation, I get the message "The path to RWIND Simulation working directory cannot contain any non-ASCII characters." What can I do?

The working directory is a local path where the data of the currently opened structure is temporarily handled and saved. It consists of the first letters of the file. Only ASCII characters may be used.

The non-ASCII characters are, for example, "ä," "ö," and "ß."

To avoid this problem, replace the special characters in the file name by the ASCII characters, such as "ae" and "ss" instead of "ä" and "ß." When you open the file again, the message will no longer appear.
• ### How can I only calculate specific load cases, load combinations, or result combinations by using a command with the COM interface?

In order to only calculate specific load cases, load combinations, or result combinations in the same way as the "To Calculate..." command (see Figure 01), you can use the CalculateBatch method of the ICalculation interface. For the transfer, the method expects a field with the load type of Loading. This Loading includes the number of the load, and the type (for example, a load combination):

Sub batch_test()    '   get interface from the opened model and lock the licence/program    Dim iModel As RFEM5.IModel3    Set iModel = GetObject(, "RFEM5.Model")    iModel.GetApplication.LockLicense    On Error GoTo e        '   get interface for calculation    Dim iCalc As ICalculation2    Set iCalc = iModel.GetCalculation        '   create array with loading types    Dim loadings(3) As Loading    loadings(0).no = 1    loadings(0).Type = LoadCaseType        loadings(1).no = 4    loadings(1).Type = LoadCaseType        loadings(2).no = 4    loadings(2).Type = LoadCombinationType        '   calculate all loadings from the array at once    iCalc.CalculateBatch loadingse:  If Err.Number <> 0 Then MsgBox Err.description, , Err.Source        Set iModelData = Nothing    iModel.GetApplication.UnlockLicense    Set iModel = NothingEnd Sub
• ### Does the RF‑LAMINATE program consider the shear correction factor for cross-laminated timber plates?

The shear correction factor is considered in the RF‑LAMINATE add-on module by using the following equation.

$k_{z}=\frac{{\displaystyle\sum_i}G_{xz,i}A_i}{\left(\int_{-h/2}^{h/2}E_x(z)z^2\operatorname dz\right)^2}\int_{-h/2}^{h/2}\frac{\left(\int_z^{h/2}E_x(z)zd\overline z\right)^2}{G_{xz}(z)}\operatorname dz$

with $\int_{-h/2}^{h/2}E_x(z)z^2\operatorname dz=EI_{,net}$

The calculation of shear stiffness can be found in the English version of the RF-LAMINATE manual, page 15 ff.

For a plate with the thickness of 10 cm in Figure 01, the calculation of the shear correction factor is shown. The equations used here are only valid for simplified symmetrical plate structures!

 Layer z_min z_max E_x(z)(N/mm²) G_xz(z)(N/mm²) 1 -50 -30 11,000 690 2 -30 -10 300 50 3 -10 10 11,000 690 4 10 30 300 50 5 30 50 11,000 690

$\sum_iG_{xz,i}A_i=3\times0.02\times690+2\times0.02\times50=43.4N$

$EI_{,net}=\sum_{i=1}^nE_{i;x}\frac{\mbox{$z$}_{i,max}^3-\mbox{$z$}_{i,min}^3}3$

$=11,000\left(\frac{-30^3}3+\frac{50^3}3\right)+300\left(\frac{-10^3}3+\frac{30^3}3\right)$

$+11,000\left(\frac{10^3}3+\frac{10^3}3\right)+300\left(\frac{30^3}3-\frac{10^3}3\right)+11,000\left(\frac{50^3}3-\frac{30^3}3\right)$

$=731.2\times10^6 Nmm$

$\int_{-h/2}^{h/2}\frac{\left(\int_z^{h/2}E_x(z)zd\overline z\right)^2}{G_{xz}(z)}\operatorname dz=\sum_{i=1}^n\frac1{G_{i;xz}}\left(χ_i^2(z_{i,max}-z_{i,min})\;χ_iE_{i,x}\frac{z_{i,max}^3-z_{i,min}^3}3+E_{i,x}^2\frac{z_{i,max}^5-z_{i,min}^5}{20}\right)$

$χ_i=E_{i;x}\frac{z_{i,max}^2}2+\sum_{k=i+1}^nE_{k;x}\frac{z_{k,max}^2-z_{k,min}^2}2$

 χ1 13.75 106 χ2 8.935 106 χ3 9.47 106 χ4 8.935 106 χ5 13.75 106

$\sum_{i=1}^n\frac1{G_{i;yz}}\left(χ_i^2(z_{i,max}-z_{i,min})-χ_iE_{i,y}\frac{z_{i,max}^3-z_{i,min}^3}3+{E^2}_{i,y}\frac{z_{i,max}^5-z_{i,min}^5}{20}\right)=$

 8.4642 1011 3.147 1013 2.5 1012 3.147 1013 8.4642 1011

Total 6.7133 x 1013

$k_z=\frac{43.4}{{(731.2e^6)}^2}6.713284\;e^{13}=5.449\;e^{-3}$

$D_{44}=\frac{{\displaystyle\sum_i}G_{xz,i}A_i}{k_z}=\frac{43.4}{5.449\;e^{-3}}=7,964.7 N/mm$

This corresponds to the resulting value in RF‑LAMINATE (Figure 02).
• ### How is the coupling stiffness determined in RF‑GLASS?

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In RF‑GLASS, there are two different types of calculations. On the one hand, there is the "2D" calculation. In this case, a glass structure is displayed as a surface element. When considering the shear coupling, the program determines an equivalent cross-section by using the laminate theory On the other hand, there is the "3D" calculation. In this case, the composition is modeled as a solid element in the calculation, and thus the effectiveness of stiffnesses between the foil and glass is determined exactly when considering the coupling.

Further information about the calculation methods can be found in the RF‑GLASS manual, Chapter 2.

• ### "During the calculation of material non-linearity, the material with a decreasing branch of the diagram can be calculated with one load increment only." Why do I get this error?

Only the default setting of 1 load increment can be set when a complex nonlinear material model is defined. The reason for this is because the program cannot determine the correct material stiffness for each incremental loading amount. The exact maximum load needs to be applied to the structure in order to determine the state of the material's stress/strain diagram.

Figure 01 - Material Model - Nonlinear material defined

This setting can be found and changed under "Calculation Parameters" as well as under the "Calculation Parameters" in the load cases and combinations dialog box.

• ### How can I set the calculation parameters by using the COM interface?

The following code shows how to get different calculation parameters via the COM interface. It also shows how to specify the setting for deactivating shear stiffness:

    '   get model interface    Set iApp = iModel.GetApplication()    iApp.LockLicense        '   get calculation interface    Dim iCalc As RFEM5.ICalculation2    Set iCalc = iModel.GetCalculation        '   get surface bending theory    Dim calc_bend As RFEM5.BendingTheoryType    calc_bend = iCalc.GetBendingTheory        '   get settings for nonlinearities    Dim calc_nl As RFEM5.CalculationNonlinearities    calc_nl = iCalc.GetNonlinearities        '   get precision and tolerance settings    Dim calc_prec As RFEM5.PrecisionAndTolerance    calc_prec = iCalc.GetPrecisionAndTolerance        '   get calculation settings    Dim calc_sets As RFEM5.CalculationSettings    calc_sets = iCalc.GetSettings    'get calculate options    Dim calc_opts As RFEM5.CalculationOptions    calc_opts = iCalc.GetOptions        '   set ShearStiffness to false    calc_opts.ShearStiffness = False    iCalc.SetOptions calc_opts

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#### First Steps

We provide hints and tips to help you get started with the main programs RFEM and RSTAB.

#### Wind Simulation & Wind Load Generation

With the stand-alone program RWIND Simulation, wind flows around simple or complex structures can be simulated by means of a digital wind tunnel.

The generated wind loads acting on these objects can be imported to RFEM or RSTAB.

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