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  • Answer

    The shear correction factor is taken into account in the RF-LAMINATE program 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 the shear stiffness itself can be found on page 15 of the English version to the manual of RF-LAMINATE as follows:

    For the 10 cm thick plate in Figure 1, the calculation of the shear correction factor is shown. The equations used here are only valid for the simplified symmetrical plate structures!

    Layerz_minz_maxE_x (z) (N/mm²)G_xz (z) (N/mm²)
    1-50-3011000690
    2-30-1030050
    3-101011000690
    4103030050
    5305011000690

    $\sum_iG_{xz,i}A_i=3\times0,02\times690+2\times0,02\times50=43,4N$

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

    $=11000\left(\frac{-30^3}3+\frac{50^3}3\right)+300\left(\frac{-10^3}3+\frac{30^3}3\right)$

    $+11000\left(\frac{10^3}3+\frac{10^3}3\right)+300\left(\frac{30^3}3-\frac{10^3}3\right)+11000\left(\frac{50^3}3-\frac{30^3}3\right)$

    $=731,2\times10^6Nmm$

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

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


    χ113.75 106
    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

    $k_z=\frac{43,4}{{(731,2e^6)}^2}6,713284\;e^{13}=5,449\;e^{-3}$

    $D_{44}=\frac{{\displaystyle\sum_i}G_{xz,i}A_i}{k_z}=\frac{43,4}{5,449\;e^{-3}}=7964,7N/mm$

    This corresponds to the value output in RF-LAMINATE (Figure 2).
  • Answer

    In RF-GLAS, there are two different types of calculations. On the one hand, the so-called '2D' calculation. In this case, the glass structure is represented as a surface element. When considering the shear bond, an equivalent cross-section is determined using the laminate theory In contrast to this, there is the '3D' calculation. In this case, the layer structure is represented as a solid element in the calculation and thus the effectiveness of the stiffnesses between foil and glass is exactly determined when the shear bond is taken into account.

    The RF-GLAS manual, Chapter 2, also provides further information on method of analysis.

  • Answer

    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. 


  • Answer

    The following code shows how to get different calculation parameters via the COM interface. It also demonstrates how to specify a 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

    In the annex, there is an EXCEL macro to download.
  • Answer

    Since concrete has a nonlinear material behavior that can only be simulated with the CONCRETE NL module, it is not possible to analyze it with the RF-STABILITY add-on module.

    The use of another material model such as isotropic linear elastic or isotropic plastic would not represent the crack formation correctly and the results are therefore not usable.

    A stability analysis on columns can be carried out with RF-CONCRETE Columns or RF-CONCRETE Members. You can find a small example in Downloads.

    This example includes the design of a column by the RF-CONCRETE Columns add-on module. Make sure that the calculation of the internal forces in RFEM is performed according to the first-order analysis and that no imperfections are required because the method used in the module takes them into account.

    The example also involves the design by means of RF-CONCRETE NL Members. Here again, it is necessary to calculate according to the second-order analysis and imperfections in the form of inclinations are required. For better comparability, the layout of the longitudinal reinforcement was aligned with the result from RF-CONCRETE Columns, as shown in Figures 01 and 02. Since the reinforcement is optimized by the module after a new calculation, the wanted reinforcement was saved as a template (see red arrow).

  • Answer

    Which method leads to results faster depends on the complexity of the model and the amount of available RAM:

    • For small and medium sized systems, the direct method is more effective.
    • For very large structures, the iterative method leads to results more quickly.
    In case that the matrices using the direct method cannot be saved any longer in the main memory, Windows starts to move parts of the data to the hard disk. This can be recognized by strong hard disk activity and a low processor load in Task Manager. This memory problem can be avoided by switching to the iterative ICG (Incomplete Conjugate Gradient) calculation method. 

  • 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

    In fact, this error message appears only if a member end hinge that allows a rotation about the local x-axis has been assigned to a member at both ends. Thus, the member can rotate freely about its own axis and is therefore unstable.

    Assign a new release to one of the member ends where the degree of freedom φx is not hinged.
  • Answer

    In Table 1.3 Surfaces, you can specify the parameters in the corresponding tab for the automatic determination of the creep coefficient and shrinkage strain. It is also possible to enter user-defined values there, if necessary.
  • 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.

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