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Answer
The reason there is a difference in results between superimposing LCs in COs vs. RCs is because when you apply the loads at once in a CO you will receive a different load distribution throughout the entire structure compared to the RC where the results are what are being added together. This is based on FEA where adding all of the loading together is different compared to solving the LCs each individually and adding the results together. Can be compared to a different order of operations to put it simply. You can see the comparison in the two figures below. Figure 1 is the load cases added up in CO1 and figure 2 has the LCs added together in RC1.

Answer
A taper describes a member or a surface with a variable crosssection. The crosssection type must be consistent, for example, Ishaped crosssections at both member ends.
In our example, we have a member with a PRO crosssection and a QRO crosssection.
To create a tapered member here, you should use a parametric crosssection for the member start and end:
Image 02  CrossSection  Parametric Input
Image 03  Parametric crosssection
This allows you to calculate the tapered member.

Answer
In the wind profile, you always define the height ranges from z. B. 0  5 m, then 5 m  10 m, etc. If your wind profile ends at 100 m, it is not cut off, but the value for 100 m is also applied for greater heights. If your wind tunnel is smaller than the stored profile, only the wind speeds up to this height are considered.
You can check this visually by displaying the velocity vectors and deactivating the display on a reduced area.
For a wind tunnel with a height of 50 m, you get the following result:
Image 01  Wind tunnel height 50 m
For a wind tunnel with a height of 150 m, you get the following result:

Answer
You can include the RSTAB printout report if you print it with a virtual PDF printer, for example PDF24.
It is necessary to pay attention to the PDF created by RSTAB or RFEM for "optimizing the web". Then it can be easily inserted into scientific programs.

Answer
Both RFEM and RSTAB can be the solution . For both programs, there are standards available with which the aluminium and lightweight structures can be calculated and designed.
In addition to Eurocode 9 with numerous National Annexes, the American standard ADM 2020 is also available.
Further addon modules for membrane and cable constructions complete the options.Main Programs RFEM or RSTAB
The main programs RFEM or RSTAB are used to define structures, materials, and actions.
If you also want to analyze membrane and cable structures, you need RFEM . When it comes to pure beam structures, the purchase of RSTAB is sufficient. In any case, RFEM is the more diverse option because it can be equipped and extended with the corresponding addon modules for all materials and designs.Available standards
 RF/ALUMINUM according to Eurocode 9 (EN 199111: 2007)
 RF/ALUMINUM ADM according to ADM 2020 (US standard)
Addon modules
 RFFORMFINGING/RFCUTTINGPATTERN (only for RFEM)
Determines the shape and cutting pattern for cutting membranes  RWIND Simulation
Complex analysis of any structures in the digital wind tunnel with transfer of load cases to RFEM or RSTAB for further processing.
Dynamic analysis
If it is necessary to perform seismic analysis or vibration designs of a building, the RF‑/DYNAM Pro addon modules provide special tools for determining natural frequencies and mode shapes, for an analysis of forced vibrations, a generation of equivalent loads, or for a nonlinear time history analysis.
interfaces Building Information Modeling (BIM)
An extensive collection of interfaces allows data exchange with other programs.
If you have any question about the Dlubal Software programs, please do not hesitate to contact our sales department. 
Answer
The setting for the deformation coefficient k_{def} can already be made in the model data. There, you can specify the deformation coefficient manually or select it based on the service class.
Image 01  General data with setting of the deformation coefficient
The deformation_{factor k def} is considered in the load combinations for serviceability in the program (similar to DIN EN 199511, 2.2.3).
Image 02  Design situation SLS
For the design of mixed structures made of timber materials, see FAQ 4325 .

Answer
A surface load of 1 kN/m² delimited by nodes 1 to 4 is only applied to member 3 (Figure 1).
The entries made in the load generator are shown in Figure 02. There is no correction of the distribution according to the moment equilibrium (Figure 3).
Image 02  Convert settings for area loads to member loads
Image 03  Settings for load generation by plane
The generated member load is shown in Figure 4. This is calculated as follows:
q = 1.00 kN/m² (area load)
h_{1} = 4.00 m
h_{2} = 6.00 m
b_{tot} = 12.00 m
$\mathrm\alpha\;=\;\arctan\left(\frac{{\mathrm h}_2\;\;{\mathrm h}_1}{{\mathrm b}_{\mathrm{ges}}}\right)\;=\;\arctan\left(\frac{6,000\;\;4,000}{12,000}\right)\;=\;9,46^\circ$
${\mathrm b}_1\;=\;\tan\left(\mathrm\alpha\right)\;\cdot\;{\mathrm h}_1\;=\;\tan\left(9,46^\circ\right)\;\cdot\;4,000\;=\;0,667\;\mathrm m$
${\mathrm l}_1\;=\;\sqrt{{\mathrm b}_1^2\;+\;{\mathrm h}_1^2}\;=\;\sqrt{0,667^2\;+\;4,000^2}\;=\;4,055\;\mathrm m$
${\mathrm l}_2\;=\;\cos\left(\mathrm\alpha\right)\;\cdot\;{\mathrm h}_2\;=\;\cos\left(9,46^\circ\right)\;\cdot\;6,000\;=\;5,918\;\mathrm m$
$ {\mathrm A} _ {\mathrm R}\; =\frac {\; {\mathrm b} _1\;\cdot\; {\mathrm h} _1} 2\; =\;\frac {\; 0.667\;\cdot\; 4,000} 2\; =\; 1.335\;\mathrm m ^ 2 $ (remaining area marked in red in Figure 4)
${\mathrm l}_{\mathrm{ges}}\;=\;\sqrt{{\mathrm b}_{\mathrm{ges}}^2\;+\;\left({\mathrm h}_2\;\;{\mathrm h}_1\right)^2}\;=\;\sqrt{12,000^2\;+\;\left(6,000\;\;4,000\right)^2}\;=\;12,166\;\mathrm m$
$ {\mathrm q} _ {\mathrm c}\; =\:\frac {\mathrm q\;\cdot\; {\mathrm A} _ {\mathrm R}} {{\mathrm l} _ {\mathrm {ges}}}\; =\;\frac {1.00\;\cdot\; 1.333} {12.166}\; =\; 0.110\;\mathrm {kN}/\mathrm m $ (constant load component on loaded member)
$ {\mathrm q} _2\; =\: {\mathrm q} _ {\mathrm c}\; +\; {\mathrm l} _1\;\cdot\;\mathrm q\; =\;\: 0.110\; +\; 4.055\;\cdot\; 1,000\; =\; 4.165\;\mathrm {kN}/\mathrm m $ (member load node 2)
$ {\mathrm q} _5\; =\: {\mathrm q} _ {\mathrm c}\; +\; {\mathrm l} _2\;\cdot\;\mathrm q\; =\;\: 0.110\; +\; 5.918\;\cdot\; 1,000\; =\; 6.028\;\mathrm {kN}/\mathrm m $ (member load node 5)
q_{4} = q_{c} = 0.110 kN/m (member load node 4)

Answer
There are two options for defining the failure: Assignment of member nonlinearity
For the member types "Beam" and "Rigid", you can define a member nonlinearity for each member. You can find the corresponding option in the "Settings" tab (see Figure 01).  Assignment of nonlinear member hinges
Alternatively, you can define a member end hinge with failure criterion for the member. For the desired degree of freedom, you can assign the hinge condition with nonlinearity accordingly (see Figure 02).
 Assignment of member nonlinearity

Answer
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The easiest solution would be to use another web browser, such as Firefox, Chrome, or Opera.
If you want to continue using the Internet Explorer, you can deactivate SmartScreen before downloading our software. Please use the web search for the instructions.

Answer
Tapered members must not be designed according to the simplified equivalent member method!For steel structures, the design can be performed by considering the warping torsion or using the General Method. These methods are described in this technical article.For timber structures, the design can also be performed by considering the warping torsion. The method for timber structures is explained in detail in thiswebinar.According to the equivalent member method, the design can be performed if the provisions of the explanations for DIN 1052, Section E8.4.2 (3) for variable crosssections are met. In various sources of technical literature, this method is adopted for Eurocode 5. An example of this can be found in the document on brettschichtholz.de, page 64 ff.In the RX‑TIMBER program, the design of tapered members is performed according to the equivalent member method. This is briefly explained on a simple example.Structural System (Figure 01): Span length: 8 m
 Beam height right: 80 cm
 Beam height left: 26 cm
 Roof inclination: 3.9°
No stiffening is defined. The lateraltorsional stability becomes governing with 99% (Figure 02) at the x‑location 1.598 m. The crosssection height is 36.8 cm. However, the slenderness ratio is based on the equivalent crosssection height of 60.9 cm (Figure 03).The equivalent crosssection height results at the xlocation 5.2 m about 0.65 × 8 m = 5.2 m.If the stiffening is in the middle of the span, for example, the equivalent height for the x‑location changes to 45.3 cm.Since the stiffening is usually applied over the member length, the height must be calculated according to a special algorithm. The supports are always applied as fixed points and the equivalent heights are calculated, based on the xlocations of the designs.For the example, the following results: x_{0.65} = 0.32 x 4 m + 1.598 m = 2.878 m
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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 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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