# Stability Analysis of Two-Dimensional Structural Components on Example of Cross-Laminated Timber Wall 1

### Technical Article

Basically, you can design structural components made of cross‑laminated timber in the RF‑LAMINATE add‑on module. Since the design is a pure elastic stress analysis, it is necessary to additionally consider the stability issues (flexural buckling and lateral-torsional buckling).

This example presents a flexural buckling analysis of a quadrilateral cross‑laminated timber wall with two door openings (see Figure 01). In this case, the governing case is the wall section between the doors.

Figure 01 - Cross-Laminated Timber Wall with Openings Subjected to Wall Stress

According to [1], the flexural buckling analysis can be performed using the equivalent member method with internal forces according to the linear static analysis in compliance with Section 6.3.2, or by considering imperfections in compliance with Section 5.4.4. In both cases, Section 2.2.2 must be respected. For this, mean values of stiffness parameters (modulus of elasticity and shear modulus) divided by the partial factor γ_{M} should be used to determine the internal forces according to the second‑order analysis in compliance with 2.4.1(2)P.

Furthermore, [2] NCI NA.9.3.3 defines the situations when the stability analysis must be performed according to second‑order theory for planar structural components. If the equation NA.150 is fulfilled, stability analyses may be calculated using both equivalent member design as well as second‑order analysis. Otherwise, the designs must be performed according to the second‑order analysis exclusively.

First, it is necessary to check whether the equation NA.150 is fulfilled. This requires the acting axial force N_{d,} the bending stiffness E ⋅ I along the local y‑axis, the partial factor γ_{M} for cross‑laminated timber, and the effective length of the respective wall section between the doors.

The load application length is set approximately at 0.5 m + 1.0 m + 0.5 m = 2.0 m. Thus, the resulting compressive force N_{d} is 200 kN/m ⋅ 2 m = 400 kN (without consideration of dead load). Alternatively, the exact determination of the compressive force considering the dead load can be obtained by using the section resultant forces in RFEM (see Figure 02). Due to the orthotropy and the dead load, the resulting compressive force is 412 kN.

Figure 02 - Resulting Axial Force in Column

Bending stiffness can be deducted directly from the stiffness matrix of the surface (see Figure 03). Here, the wall surface by Stora Enso of the CLT type CLT 100 C5s is selected. The resulting bending stiffness in the y‑direction is 826.16 kNm ⋅ 1.0 m = 826.16 kNm².

Figure 03 - Stiffness Matrix of Surface

The partial factor of 1.3 is applied in accordance with [2]. To determine the effective lengths, the shear stiffness in the y‑direction should also be considered for the cross‑laminated timber (see Figure 03). The effective length factor β of 1.0 is used according to Euler case 2.

$$\begin{array}{l}\begin{array}{l}{\mathrm l}_{\mathrm{ef}\;}=\;\mathrm\beta\;\cdot\;\mathrm l\;\cdot\;\sqrt{\frac{1\;+\;\mathrm E\;\cdot\;\mathrm I\;\cdot\;\mathrm\pi^2}{(\mathrm\beta\;\cdot\;\mathrm I)^2\;\cdot\;\mathrm\kappa\;\cdot\;\mathrm G\;\cdot\;\mathrm A}}\\{\mathrm l}_{\mathrm{ef}\;}=\;1.0\;\cdot\;3.0\;\mathrm m\;\cdot\;\sqrt{\frac{1\;+\;826.16\;\mathrm{kNm}^2\;\;\cdot\;\mathrm\pi^2}{(1.0\;\cdot\;3.0\;\mathrm m)^2\;\cdot\;7.976,19\;\mathrm{kN}}}\;=\;3.17\;\mathrm m\;\;\;(\mathrm{NA}.150)\end{array}\\{\mathrm l}_{\mathrm{ef}\;}\;\cdot\;\sqrt{\frac{{\mathrm N}_\mathrm d\;\cdot\;{\mathrm\gamma}_\mathrm M}{\mathrm E\;\cdot\;\mathrm I}}\;\leq\;1.00\\3.17\;\mathrm m\;\cdot\;\sqrt{\frac{412.16\;\mathrm{kN}\;\cdot\;1.3}{826.16\;\mathrm{kNm}^2}}\;=\;2.55\;>\;1.00\end{array}$$The delimitation criterion is not met with 2.55 > 1.00. Therefore, the stability analysis according to the second‑order theory must be performed. Since an almost linear member is analyzed, both methods will be explained in my next articles.

In order to better understand the buckling problem in this case, the critical buckling load and the critical load factor are first determined according to the linear static analysis for the wall section on the ideal, simply supported beam (see Figure 04). For this, the critical load factor is determined analytically and by using the RF‑STABILITY add‑on module. For the FEM solution, a load case without dead load is created, and the resulting load is applied directly. The stiffness reduction related to the partial safety factor γ_{M} is activated in the calculation parameters of the load case. The result of both calculations is exactly the same.

Figure 04 - Determination of Critical Buckling Load and Critical Load Factor

Taking into account the additional stiffness resulting from the door lintel, the slightly larger critical load factor of 1.67 results on the entire structure, as expected.

Figure 05 - Critical Load Factor on Entire Structure

This critical load factor indicates the number by which the load must be multiplied so that the model under the associated load becomes unstable (buckling). Hence, a critical load factor smaller than 1.00 means that the structure is unstable. Only a positive critical load factor greater than 1.00 means that the load due to the predefined axial forces multiplied by this factor leads to the buckling failure of the stable structure.

However, a stability analysis has to be performed according to EN 1995‑1‑1 because the critical load factor or the critical buckling load may not be correct in practice as the effects of imperfections (no member or surface is straight), eccentricities of load introduction, and material behavior divergent from the Hooke's law are not considered. The designs are explained in the next article of this series.

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