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Answer
In principle, RFLAMINATE can also be used for detailed analyzes. For example, in the case of a very high shear strain, it may be useful to perform the modeling using orthotropic solids. The video shows the simple modeling and result evaluation of a layer structure by means of solids.
One criterion when modeling over solids is useful is the shear correction factor. You can find more information about this and other criteria in the following FAQs:

Answer
The easiest way to consider this is to use the RF/JOINTS Timber  Steel to Timber module. For this purpose, the module dissolves the original connection and creates a new static system that takes flexibility into account accordingly. The consideration is separately for loadbearing capacity, serviceability and exceptional. 
Answer
The Poisson's ratio is set under the material by using the Edit Material dialog box. 
Answer
The force V_{L} is the longitudinal shear force between the top surface and the member. It is calculated as an integrated shear flow between the plate and the member at a particular point on the member.For the simplified example provided here, the resulting crosssection values for the integration width of 10 cm are as follows: $I_y=\frac{b\times h^3}{12}=\frac{10 cm\times20 cm^3}{12}=6,666.67 cm^4$
 $S_y=h_1\times b\times((he_z)\frac{h_2}2)=10 cm\times10 cm\times((20 cm10 cm)\frac{10 cm}2)=500 cm^3$
 $\tau=V_L=\frac{V_z\times S_y}{I_y\times b}=\frac{5.53 kN\times500 cm^3}{6,666.67 cm^4}=0.415 kN/cm=41.5 kN/m$
The integration width has been set to the total of 10 cm.Values: I_{y} second moment of area
 S_{y} statical moment
 h_{1} height of the upper crosssection part
 h_{2} height of the lower crosssection part
 e_{z} centroidal distance
 h total height
The values can be adjusted for a Tbeam. 
Answer
The shear correction factor is taken into account in the RFLAMINATE 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 RFLAMINATE 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!Layer z_min z_max E_x (z) (N/mm²) G_xz (z) (N/mm²) 1 50 30 11000 690 2 30 10 300 50 3 10 10 11000 690 4 10 30 300 50 5 30 50 11000 690 $\sum_iG_{xz,i}A_i=3\times0,02\times690+2\times0,02\times50=43,4N$$EI_{,net}=\sum_{i=1}^nE_{i;x}\frac{\mbox{$z$}_{i,max}^3\mbox{$z$}_{i,min}^3}3$$=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$$\int_{h/2}^{h/2}\frac{\left(\int_z^{h/2}E_x(z)zd\overline z\right)^2}{G_{xz}(z)}\operatorname dz=\sum_{i=1}^n\frac1{G_{i;xz}}\left(χ_i^2(z_{i;max}z_{i,min})\;χ_iE_{i,x}\frac{z_{i,max}^3z_{i,min}^3}3+E_{i,x}^2\frac{z_{i,max}^5z_{i,min}^5}{20}\right)$$χ_i=E_{i;x}\frac{z_{i;max}^2}2+\sum_{k=i+1}^nE_{k;x}\frac{z_{k,max}^2z_{k,min}^2}2$χ_{1} 13.75 10^{6} χ_{2} 8.935 10^{6} χ_{3} 9.47 10^{6} χ_{4} 8.935 10^{6} χ_{5} 13.75 10^{6} $\sum_{i=1}^n\frac1{G_{i;yz}}\left(χ_i^2(z_{i,max}z_{i,min})χ_iE_{i,y}\frac{z_{i,max}^3z_{i,min}^3}3+{E^2}_{i,y}\frac{z_{i,max}^5z_{i,min}^5}{20}\right)=$
8.4642 10^{11} 3.147 10^{13} 2.5 10^{12} 3.147 10^{13} 8.4642 10^{11} Total 6.7133 x 10^{13}$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 RFLAMINATE (Figure 2). 
Answer
It is possible to define different properties in the tension and compression area of a line release as shown in Figure 01.However, it is necessary to consider a special feature regarding the released surface when releasing lines between two surfaces. The attached model involves an identical line release. The bottom surface 1 is released for the left model, but the top surface 4 for the right model. For both surfaces, the local yaxis of the surface is aligned parallel to the global zaxis.The deformation diagram results in completely different deformations for both models.In the left model with the lower released surface, the pressed area is very soft. For this axis orientation, the reaction force of surface 2, which presses surface 1 from above, is very soft.In the model on the right, the upper surface is released, so that the reaction force of surface 4, which presses surface 3 from above, is very stiff. 
Answer
Basically, all crosssections of the solid and hybrid crosssection groups can be designed in the RF/TIMBER Pro program. In Figure 01, they are displayed on the right.
For more complex asymmetrical crosssection shapes, it may be necessary to adjust the allowable inclination of principal axis on a userdefined basis in the addon module (see Figure 02).

Answer
In the case of cross laminated timber panels not glued to the narrow sides and a walllike structural behaviour, the torsion stress in the glued joints is often decisive. This design is performed according to the explanations in the literature reference below according to the following equation.$\eta_x=\frac{\tau_{tor,x}}{f_{v,tor}}+\frac{\tau_x+\tau_{xz}}{f_R}=\frac{\displaystyle\frac{3\ast n_{xy}}{b(n1)}}{f_{v,tor}}+\frac{{\displaystyle\frac{\frac{\partial n_x}{\partial x}}{n1}}+\tau_{xz}}{f_R}\leq1$Values: b board width
 n number of board layers
 n_{xy} shear in pane plane
 $\frac{\partial n_x}{\partial x}$ shear of board layers
 $\tau_{xz}$ shear in thickness direction
 f_{R} rolling shear strength
 f_{v,tor} torsional shear strength
For the ydirection, the design is analogous but with the values for the ydirection. 
Answer
For the superstructures of the manufacturer Binderholz, as soon as the slabs are defined without glue at narrow sides and design of shear failure is calculated in the wall plane, the shear strengths are calculated according to the following equation.$f_{v,k}=\left\{\begin{array}{l}\begin{array}{c}3,5\\8,0\frac{D_{net}}D\\\end{array}\\2,5\frac{(n1)(a²+b²)}{6Db}\end{array}\right.$Values:D element thicknessD_{net} sum of longitudinal and transverse layer thicknesses in the elementn number of board layersa = b width of the boards in the longitudinal or transverse layersAll values in N/mm². For more detailed information, check the manufacturer's approval. 
Answer
First, you have to determine the position of the FE node. The easiest way is to use [Edit]  [Find by Number] (Figure 1).Then, the found FE node can be defined by manual setting the result value (Figure 2).The process is explained in the attached video.
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Wind Simulation & Wind Load Generation
With the standalone 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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