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• When entering the geometry of a ribbed timber plate, I always get the error message 30343: "The computed stiffness matrix is not positive definite." How can I define the surface?

In the case of using timber materials in RFEM, a message appears when creating the material, saying that the "isotropic linear elastic" material model is only correct for the calculation with members because the Poisson's ratio is very large. For surfaces, it is necessary to select a different material model.

If you have assigned the material with the orthotropic material model to the surface, it is not possible to use certain orthotropy types. This material cannot be used for ribbed plates.

However, if you have assigned the material to the surface with an isotropic linear elastic material model, you can access all types of orthotropy.

When entering the geometry for the ribbed plate, Error No. 30343 appears due to the incorrectly assigned timber material for the surface: "The computed stiffness matrix is not positive definite."

• Can RF-TIMBER AWC optimize cross-sections for fire design?

The RF-TIMBER AWC module does not optimize cross-sections for the Serviceability Limit State or the Fire Resistance design. Optimization is only calculated for the Ultimate Limit State design.

Users must manually adjust the cross-section in RFEM or within the add-on module and can export the cross-section back into RFEM. In either scenario, the model must be rerun in order to calculate the correct internal forces with the adjusted member size.
• I have manually defined a snow load of 5 kN/m² in the RX‑TIMBER program. However, the calculation uses a load of 4 kN/m². Why is the defined snow load not applied?

According to EN 1991‑1‑3, the snow load must be multiplied by the factor of 0.8 for the snow load on a roof. Therefore, the following load results:

sk = 5 kN/m² × 0.8 = 4 kN/m²

If the load has already been calculated for the roof, you can enter the load completely under the user-defined loads. This is shown in Figure 01 and in the video.

• How do I create a user-defined material in the RX‑TIMBER program?

In RX‑TIMBER, you can create a user-defined material as a new material in the material library (see the figure), similar to the programs RFEM/RSTAB.

In contrast to RFEM/RSTAB, the category is set to "timber" in this case.
• Is it possible to perform fire resistance design of cross-laminated timber panels in RF‑LAMINATE?

Fire resistance design is not implemented in the RF‑LAMINATE add-on module by default.

However, you can calculate the charring rates yourself and consider them accordingly in the module. In the following example, this is explained on a simple plate.

Structural system (Figure 01):

• Span 5 m
• Plate width 2 m
• LC1 (permanent) 1 kN/m² plus dead load
• LC2 (medium) 2.5 kN/m²
• 3 layers
• S1 35 mm C24
• S2 20 mm C24
• S3 35 mm C24
The information regarding the correction factors and stiffnesses can be found in the attached file.

Factors for fire resistance:

• Charring rate ß0 = 0.65 mm/min
• Pyrolysis zone k0d0 = 7 mm
• Charring time t = 30 min
• Effective thickness def=t ß0+k0d0=30 min × 0.65 mm/min+7 mm = 26.5 mm
Remaining thickness of Layer 3 = 35 − 26.5 = 8.5 mm > 3 mm → thickness may be applied. (Figure 02)

Because of the modified layer thicknesses, a new stiffness matrix results, which is applied in RFEM for accidental combinations with the characteristic stiffness values. For the ultimate limit state, the design values are calculated here (Figure 03).
• Is it possible to perform a detailed analysis of connections, supports, or reinforcements of cross‑laminated timber plates in RF‑LAMINATE?

In principle, it is also possible to perform detailed analysis in RF‑LAMINATE. In the case of a very high shear distortion, for example, it can be reasonable to use orthotropic solids for modeling. The video shows a simple modeling and result evaluation of a layer structure by using solids.

A criterion, as of when is the modeling using solids useful, is the shear correction factor. Further information and other criteria can be found in the following FAQ:

• How can I consider the flexibility of a continuous beam with slotted dowel connections?

The easiest way to consider this is to use the RF‑/JOINTS Timber - Steel to Timber add-on module. For this purpose, the module decomposes the original connection, and creates a new structural system that considers the flexibility accordingly. In this case, the ultimate limit state, the serviceability limit state, and the accidental design situations are considered separately.
• I design a combined structure made of timber materials with different creeping parameters. How can I perform the serviceability limit state design according to EN 1995‑1‑1?

In RFEM and RSTAB, the simplified design from [1], Chapter 2.2.3, have been implemented for the automatic load combinations. This means that, strictly speaking, the structures concerning the final deformation may only be analyzed, in which the materials with identical creep behavior occur since the creep deformations are considered in a simplified way on the load side. If the structure is a combined structure made of timber with different creep behavior or in combination with steel, the final deformations must be determined according to [2] Amendment to 2.2.3 as follows:

"(4) If the structure consists of members or components having different creep behaviour, the long-term deformation due to the quasi-permanent combination of actions should be calculated using final mean values of the appropriate moduli of elasticity, shear moduli and slip moduli, according to 2.3.2.2(1). The final deformation ufin is then calculated by the superposition of the instantaneous deformation due to the difference of the characteristic and quasi-permanent combinations of actions with the long-term deformation."

However, this requires the superposition of the results from different load combinations, which cannot be implemented automatically in RFEM and RSTAB. If the different creep behavior should be taken into account, it is necessary to create the load combinations manually, and reduce the stiffness according to the creep coefficient. The procedure is described on an example of a timber-concrete composite floor presented on the Info Day 2017. Below this FAQ, you can find the link for this video.

• According to DIN EN 1995‑1‑1/NA, the crack factor kcr may be increased by 30% for softwood in the areas that are at least 1.50 m from the timber grain plane. How is it possible to apply this in RF‑/TIMBER Pro?

Increasing the crack factor kcr still has to be done manually because the program does not know where is the grain plane defined. To do this, divide the member accordingly by 1.5 m from the grain plain, so that the affected areas can be designed as a separate member (see Figure 01).

Two design cases are now required (File → New Case ...). In Case 1, the members within the 1.5 m are selected for the design. In Case 2, it is necessary to select the members where the 30% should be considered. In Case 2, the kcr value is then adjusted manually in the National Annex settings. Thus, the kcr of 0.65 results for C24, which is entered as shown in Figure 02. In this way, the design is performed with the increased kcr value.
• 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).

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