7785x

001669

Dipl.-Ing. (FH) Gerhard Rehm

2026-05-06

Lateral-Torsional Buckling in Timber Structures | Examples 2

The previous article Lateral-Torsional Buckling in Timber Construction | Examples 1 explains the practical application for determining the critical bending moment M_crit or the critical bending stress σ_crit for a bending beam's lateral-torsional buckling by using simple examples. In this article, the critical bending moment is determined by considering an elastic foundation resulting from a stiffening bracing.

Lateral-torsional buckling in timber structures - Examples 1

Structural Model

For the system shown in the next image, the rafters are to be investigated for buckling. In the roof plane, there are six rafters as parallel beams with a length of 18 m and two bracing systems. The beams on the gable sides are supported by columns and are not considered for the calculation. A design load q_d of 10 kN/m acts on the rafters. The main aim is to determine the critical lateral-torsional buckling moment. The verification in the ultimate limit state and in the serviceability limit state is not discussed further.

1 Structural Model

Model Data

GL24h	-	-	Material according to EN 14080
L	18	m	Length of the beam
b	120	mm	Width of the beam
h	1.200	mm	Height of the beam
I_z	172.800.000	mm⁴	Second moment of area
I_T	647.654.753	mm⁴	Torsional moment of inertia
q_d	10	kN/m	Design load
a_z	600	mm	Load position
e	600	mm	Position of the bearing

Info

Even though in the following equations for E and G the reference to the 5% quantile values is not explicitly given in the subscript, these have nevertheless been taken into account accordingly.

Pin-ended single-span beam without intermediate restraints

2 Single-Span Beam with Lateral and Torsional Restraints

For completeness, the beam without lateral restraint is first investigated (see Image 02). With a load applied at the top of the beam, the equivalent member length is obtained with a₁ = 1.13 and a₂ = 1.44 as:

l_{ef} = \frac{L}{a_{1} \cdot (1 - a_{2} \cdot \frac{a_{z}}{L} \cdot \sqrt{\frac{E_{0, 05} \cdot I_{z}}{G_{0, 05} \cdot I_{T}}})} = 17, 79 m

l_ef	Ersatzstablänge
L	Trägerlänge, Abstand der seitlichen Halterung
a₁,a₂	Kippbeiwerte
a_z	Abstand des Lastangriffs vom Schubmittelpunkt
E_0,05	5 % Quantile des Elastizitätsmoduls
G_0,05	5 % Quantile des Schubmoduls
I_z	Trägheitsmoment um die schwache Achse
I_T	Torsionsträgheitsmoment

Equivalent Member Length

The critical bending moment can then be calculated as follows:

M_{crit} = \frac{π \cdot \sqrt{E_{0, 05} \cdot I_{z} \cdot G_{0, 05} \cdot I_{T}}}{l_{ef}} = 134, 52 kNm

M_crit	Kritisches Biegemoment
E_0,05	5 % Quantile des Elastizitätsmoduls
G_0,05	5 % Quantile des Schubmoduls
I_z	Trägheitsmoment um die schwache Achse
I_T	Torsionsträgheitsmoment
l_ef	Ersatzstablänge

Critical Bending Moment

An increase in the product of the 5% quantiles of the stiffness parameters due to the homogenization of glulam beams is omitted in these examples.

The bending moment acting on the beam results in:

M_{d} = \frac{q_{d} \cdot L^{2}}{8} = 405, 00 kNm

M_d	Bemessungsmoment
q_d	Bemessungslast
L	Trägerlänge

Design Moment

The eigenvalue analysis yields a buckling load factor of 0.32 as a result. This gives the critical bending moment

M_{crit} = α \cdot M_{d} = 0, 32 \cdot 405, 00 kNm = 136, 39 kNm

Critical Bending Moment from Eigenvalue Solver

and is thus identical to the result of the analytical solution.

Design check details with output of critical load factor and critical bending moment

3 Design Check Detail with Output of Critical Load Factor and Critical Bending Moment

As expected for this unsupported, slender beam, the acting bending moment is greater (by a factor of 3) than the critical bending moment, and the beam is therefore not sufficiently restrained against buckling. However, a bracing system is intended to counteract this, which is now taken into account for the calculation.

Pin-ended single-span beam with rigid intermediate restraints

4 Single-Span Beam with Lateral and Torsional Restraints

If the bracing system is stiff enough, the spacing of the lateral restraints (for example, by purlins) is often used in practice as the equivalent member length for the buckling check. This approach was already shown in the previous article Lateral-torsional buckling in timber structures | Examples 1.

Lateral-Torsional Buckling in Timber Structures | Examples 1

Accordingly, 2.25 m is used as L. For a₁ = 1.00 and a₂ = 0.00, the following results:

l_{ef} = \frac{L}{a_{1} \cdot (1 - a_{2} \cdot \frac{a_{z}}{L} \cdot \sqrt{\frac{E_{0, 05} \cdot I_{z}}{G_{0, 05} \cdot I_{T}}})} = 2, 25 m

l_ef	Ersatzstablänge
L	Trägerlänge, Abstand der seitlichen Halterung
a₁,a₂	Kippbeiwerte
a_z	Abstand des Lastangriffs vom Schubmittelpunkt
E_0,05	5 % Quantile des Elastizitätsmoduls
G_0,05	5 % Quantile des Schubmoduls
I_z	Trägheitsmoment um die schwache Achse
I_T	Torsionsträgheitsmoment

Equivalent Member Length

The critical bending moment is:

M_{crit} = \frac{π \cdot \sqrt{E_{0, 05} \cdot I_{z} \cdot G_{0, 05} \cdot I_{T}}}{l_{ef}} = 1.063, 51 kNm

M_crit	Kritisches Biegemoment
E_0,05	5 % Quantile des Elastizitätsmoduls
G_0,05	5 % Quantile des Schubmoduls
I_z	Trägheitsmoment um die schwache Achse
I_T	Torsionsträgheitsmoment
l_ef	Ersatzstablänge

Critical Bending Moment

Since the bending moment acting on the beam is smaller than the critical bending moment, the beam is not at risk of buckling under the assumption of rigid intermediate restraints.

The eigenvalue analysis with the Timber Design Add-On gives a buckling load factor of 2.86 as a result. This gives the critical bending moment

M_{crit} = α \cdot M_{d} = 2, 86 \cdot 40, 50 kNm = 1.158, 92 kNm

M_crit	Kritisches Biegemoment
α	Branch load factor
M_d	Bemessungsmoment

Critical Bending Moment

5 Verification Detail with Output of the Critical Load Factor and Critical Bending Moment

Here too, both methods agree very well.

Pin-ended single-span beam with elastic member restraint

Fork-supported single-span beam with elastic member bedding

6 Single-Span Beam with Fork Support and Elastic Member Bedding

As explained in Lateral-torsional buckling in timber structures - Theory, the determination of the equivalent member length for elastically restrained members is extended by the factor α and β in [1 ].

This makes it possible to take the shear stiffness of a bracing system into account for the buckling of the rafters.

Shear stiffness of the roof bracing

The shear stiffness of the bracing system can be determined, for example, according to [2] Figure 6.34. As can be seen from this, it depends on the type of bracing system, the axial stiffness of the diagonals and posts, the inclination of the diagonals, and the flexibility of the fasteners. For the bracing system shown in Image 01, the shear stiffness is obtained as:

s_{id} = E_{D} \cdot A_{D} \cdot \sin {(α)}^{2} \cdot \cos (α)

s_id	Ideelle Schubsteifigkeit des Aussteifungsverbandes
E_D	5 % Quantile des Elastizitätsmoduls der Diagonalen
A_D	Querschnittsfläche der Diagonalen
α	Winkel zwischen der Diagonalen und der Gurte

Ideal Shear Stiffness of Stiffening Bracing

Here, E_D is the modulus of elasticity of the diagonals and A_D their cross-sectional area. However, the above equation does not include the flexibility of the fasteners of the diagonals. This and the elongation of the diagonals can be taken into account by means of a fictitious cross-sectional area A_D'. It follows:

s_{id} = E_{D} \cdot A_{D}' \cdot \sin {(α)}^{2} \cdot \cos (α) = 44.864 kN

s_id	Ideelle Schubsteifigkeit des Aussteifungsverbandes
E_D	5 % Quantile des Elastizitätsmoduls der Diagonalen
A_D'	Fiktive Querschnittsfläche der Diagonalen
α	Winkel zwischen der Diagonalen und der Gurte

Ideal Shear Stiffness of the Bracing System

with

A_{D}' = \frac{A_{D}}{1 + \frac{E_{D} \cdot A_{D}}{L_{D}} \cdot \sum \frac{1}{K_{ser}}} = 12.548 {mm}^{2}

A_D'	Fiktive Querschnittsfläche der Diagonalen
A_D	Querschnittsfläche der Diagonalen
E_D	5 % Quantile des Elastizitätsmoduls der Diagonalen
L_D	Länge der Diagonalen
K_ser	Verschiebungsmodul der Verbindung

Fictitious Cross-Sectional Area of the Diagonals

The diagonals have dimensions b/h = 120/200 mm and a length L_D of 4.59 m. The slip modulus of the connection on each side of the diagonals is to be 110,000 N/mm.

The ideal area is therefore

A_D' = 12,548 mm²

and thus the shear stiffness of one bracing system, with an angle of the diagonals to the chord of 60.64°,

s_{id} = E_{D} \cdot A_{D}' \cdot \sin {(α)}^{2} \cdot \cos (α) = 44.864 kN

s_id	Ideelle Schubsteifigkeit des Aussteifungsverbandes
E_D	5 % Quantile des Elastizitätsmoduls der Diagonalen
A_D'	Fiktive Querschnittsfläche der Diagonalen
α	Winkel zwischen der Diagonalen und der Gurte

Ideal Shear Stiffness of the Bracing System

The member restraint per bracing system can be converted from this according to [2] Equation 7.291 as follows:

K_{y}' = s_{id} \cdot \frac{π^{2}}{L^{2}}

K_{y}' = 44, 864 kN \cdot \frac{π^{2}}{{(18 m)}^{2}} = 1, 367 \frac{kN}{m^{2}} = 1.367 \frac{N}{{mm}^{2}}

K_y'	Elastische Stabbettung pro Verband
s_id	Ideelle Schubsteifigkeit des Aussteifungsverbandes
L	Länge des Verbandes

Elastic Member Foundation per Bracing

For two bracing systems and six rafters, the following spring stiffness is available per rafter:

K_{y} = \frac{2 \cdot K_{y}'}{6} = 455, 6 kN / m^{2} = 0, 456 N / {mm}^{2}

K_y	Elastische Stabbettung pro Binder
K_y'	Elastische Stabbettung pro Verband

Elastic Member Bedding per Beam

Assuming K_G = ∞, K_θ = 0, K_y = 0.456 N/mm², e = 600 mm, a₁ = 1.13 and a₂ = 1.44, the equivalent member length is obtained as:

l_{ef} = \frac{L}{a_{1} \cdot (1 - a_{2} \cdot \frac{a_{z}}{L} \cdot \sqrt{\frac{E_{0, 05} \cdot I_{z}}{G_{0, 05} \cdot I_{T}}})} \cdot \frac{1}{α \cdot β} = 0, 13 m

l_ef	Ersatzstablänge
L	Trägerlänge, Abstand der seitlichen Halterung
a₁,a₂	Kippbeiwerte
a_z	Abstand des Lastangriffs vom Schubmittelpunkt
E_0,05	5 % Quantile des Elastizitätsmoduls
G_0,05	5 % Quantile des Schubmoduls
I_z	Trägheitsmoment um die schwache Achse
I_T	Torsionsträgheitsmoment
α, β	Beiwerte zur Berücksichtigung einer Stabbettung

Replacement Bar Length

The critical bending moment thus results in an unrealistic value of:

M_{crit} = \frac{π \cdot \sqrt{E_{0, 05} \cdot I_{z} \cdot G_{0, 05} \cdot I_{T}}}{l_{ef}} = 18.482, 84 kNm

M_crit	Critical bending moment
E_0,05	5% quantile of the modulus of elasticity
G_0,05	5 % quantile of the shear modulus
I_z	Moment of inertia about the weak axis
I_T	Torsional moment of inertia
l_ef	Replacement bar length

Critical Bending Moment

A value similar to the system with rigid intermediate restraints would be expected. As explained in Lateral-torsional buckling in timber structures - Theory, the application of the extended formula with α and β is limited in its use.

Strictly speaking, this is only valid if there is a deflection in a large sinusoidal curve. That is, when the restraint is very flexible. This is no longer the case in this example. Multi-wave eigenfunctions, which lead to the smallest buckling load for larger spring constants, are not covered in the equation mentioned, as it is based on single-wave sinusoidal approaches.

As can be seen in Figure 7, the eigenvalue analysis results in a multi-wave mode shape with a buckling load factor of 3.49.

Eigenvalue Analysis Result for the Single-Span Beam with Pinned Fork Supports and Elastic Member Bedding

7 Eigenvalue Analysis Result for the Simply Supported Beam with Fork Supports and Elastic Member Bedding

For comparison, the method derived by Prof. Dr. Heinrich Kreuzinger (2020) can be applied. The critical bending moment is calculated as follows:

M_{crit} = \frac{(2 \cdot e \cdot c^{*} - a_{z}^{*} \cdot B^{*}) + \sqrt{{(2 \cdot e \cdot c^{*} - a_{z}^{*} \cdot B^{*})}^{2} + 4 \cdot 1.0 \cdot (B^{*} \cdot T^{*} - e^{*} \cdot {c^{*}}^{2})}}{2}

c^{*} = K_{y} \cdot \frac{L^{2}}{π^{2} \cdot n^{2}}

a_{z}^{*} = \frac{a_{z}}{n^{2}}

B^{*} = E_{0.05} \cdot I_{Z} \cdot \frac{π^{2} \cdot n^{2}}{L^{2}} + c^{*}

T^{*} = G_{0.05} \cdot I_{T} + c^{*} \cdot e^{2}

n = 1, 2, 3 . . .

M_crit	Kritisches Biegemoment
a_z	Abstand des Lastangriffs vom Schubmittelpunkt
e	Abstand der Stabbettung vom Schubmittelpunkt
K_y	elastische Stabbettung pro Binder
L	Trägerlänge
n	n-te Eigenlösung
E_0,05	5 % Quantile des Elastizitätsmoduls
I_z	Trägheitsmoment um die schwache Achse
G_0,05	5 %-Quantile des Schubmoduls
I_T	Torsionsträgheitsmoment

Critical Bending Moment

The constant n denotes the 1st, 2nd, 3rd... eigen-solution. Accordingly, several eigen-solutions are to be investigated, and the smallest critical bending moment is then governing. For n = 1...30, the following critical bending moments are obtained.

n	M_crit [kNm]	n	M_crit[kNm]
1	9,523.25	16	2,214.63
2	4,281.26	17	2,339.17
3	2,294.32	18	2,464.92
4	1,605.56	19	2,591.63
5	1,354.68	20	2,719.14
6	1,282.70	21	2,847.30
7	1,294.12	22	2,976.00
8	1,348.81	23	3,105.16
9	1,428.05	24	3,234.71
10	1,522.29	25	3,364.60
11	1,626.24	26	3,494.77
12	1,736.77	27	3,625.20
13	1,851.94	28	3,755.84
14	1,970.50	29	3,886.67
15	2,091.60	30	4,017.68

For n = 6, M_crit is minimal and amounts to 1,282.70 kNm.

The eigenvalue solution from the Timber Design Add-On (see Image 7) gives:

M_{crit} = α \cdot M_{d} = 3, 49 \cdot 40, 50 kNm = 1.413, 71 kNm

M_crit	Critical bending moment
α	Branch load factor
M_d	Design moment

Critical Bending Moment

The two results show good agreement. However, the analytical solution is on the safe side, since this method assumes a constant bending moment distribution in a simplified manner. The constant critical bending moment M_crit is then assigned a critical load q_crit.

q_{crit} = \frac{M_{crit} \cdot 8}{L^{2}}

q_crit	Kritische Last
M_crit	Kritisches Biegemoment
L	Trägerlänge

Critical Load

Since the member restraint in this example is to be regarded as very stiff and is distributed uniformly over the beam length, slightly higher critical bending moments result than for the rigid individual support.

Deformation check of the roof bracing

According to [3] Chapter 9.2.5.3 (2), bracing systems must be so stiff that the horizontal deflection L/500 is not exceeded. The calculation must be carried out using the design values of the stiffnesses (see [1] Chapter NCI To 9.2.5.3).

For k_crit = 0.195, H = 5 m and q_p = 0.65 kN/m² as gust wind pressure, the following loads result (see [3] Chapter 9.2.5.3):

N_{d} = (1 - k_{crit}) \cdot \frac{M_{d}}{h} = 271, 68 kN

N_d	Stabilisierungskraft für den Druckgurt
k_crit	Kippbeiwert
M_d	Bemessungsmoment
h	Trägerhöhe

Stabilizing Force for Compression Flange

q_{d} = \sqrt{\frac{15}{L}} \cdot \frac{n \cdot N_{d}}{k_{f, 3} \cdot L} = 2, 76 kN / m

q_d	Aussteifungslast
n	Anzahl der Binder
L	Trägerlänge
k_f,3	Modifikationsbeiwert für den Aussteifungswiderstand

Bracing Load

q_{d, Wind} = γ_{Q} \cdot (c_{pe, D} + c_{pe, E}) \cdot q_{p} \cdot \frac{H}{2} = 1, 5 \cdot (0, 7 + 0, 3) \cdot 0, 65 \cdot \frac{5}{2} = 2, 44 kN / m

q_d,Wind	Bemessungslast aus Wind
γ_Q	Teilsicherheitsbeiwert für veränderliche Einwirkung
c_pe	Außendruckbeiwert
q_p	Böengeschwindigkeitsdruck
h	Höhe des Gebäudes

Design Load from Wind

The deformation of the bracing system is shown in Image 8. The loads were halved again, as there are two bracing systems.

Horizontal displacement of the bracing system

8 Horizontal Deflection of the Bracing System

The allowable deformation is:

\frac{L}{500} = \frac{18 m}{500} = 36.00 mm \geq 4.74 mm

Allowable Deformation

This confirms the assumption of a very stiff bracing system and is consistent with the nearly identical critical bending moments of the system with rigid intermediate restraint and the system with elastic member restraint.

Summary

System	M_{crit,analytical}	M_{crit,eigenvalue}
without intermediate restraint	134.52 kNm	136.39 kNm
with rigid intermediate restraints	1,063.51 kNm	1,158.92 kNm
with elastic member restraint	1,282.70 kNm	1,413.71 kNm

It has been shown which options are available in timber construction for investigating the buckling of beams. For the common methods, it should be ensured that the bracing systems are stiff enough to allow rigid supports to be assumed. Variants were shown accordingly if this assumption does not apply. In principle, the beams and the bracing systems must still be checked for their load-bearing capacity and serviceability in accordance with the relevant standard. However, this is not the subject of this article.

Author

Gerhard works in product engineering in the field of timber structures and also assists with customer support. He draws on his development experience to create practical and actionable solutions.

Links

References

DIN. (2013). National Annex – Eurocode 5: Design of Timber Structures – Part 1-1: General – Common Rules and Rules for Buildings, DIN EN 1995-1-1/NA:2013-08. Berlin: Beuth Verlag GmbH.
Petersen, C. (2982). Statik und Stabilität der Baukonstruktionen (2nd ed.). Wiesbaden, Vieweg.
Deutsches Institut für Normung e.V. (DIN) (2010). Eurocode 5: Design of Timber Structures – Part 1-1: General – Common Rules and Rules for Buildings; DIN EN 1995-1-1:2010-12. Berlin: Beuth Verlag GmbH.

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