8003x

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 following image, the truss members should be analyzed for lateral buckling. In the roof plane, six truss members are available as parallel beams with a length of 18 m and two stiffening bracings. 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 truss members. The primary objective is to determine the critical lateral-torsional buckling moment. The ultimate limit state and the serviceability limit state design checks will not be further discussed.

1 Structural Model

Model Data

GL24h	-	-	Material according to EN 14080
L	18	m	Beam length
b	120	mm	Beam width
h	1.200	mm	Beam height
I_z	172.800.000	mm4	Moment of inertia
I_T	647.654.753	mm4	Torsion moment of inertia
q_D	10	kN/m	Design load
a_z	600	mm	Load position
e	600	mm	Foundation position

Info

Even if the following equations for E and G do not explicitly refer to the 5%-quantiles in the index, they have been taken into account accordingly.

Single-Span Beam with Lateral and Torsional Restraint Without Intermediate Supports

2 Single-Span Beam with Lateral and Torsional Restraints

For the sake of completeness, the truss member is analyzed first, without lateral supports (see Image 2). The equivalent member length results from a load application on the upper side of the truss with a₁ = 1.13 and a₂ = 1.44 as follows:

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	Replacement member length
L	Length of member, spacing of lateral bracing
a₁,a₂	Tilting coefficients
a_z	Distance of load application from the shear center
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

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	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 member length

Critical Bending Moment

These examples do without an increase of the product of the 5%-quantiles of the stiffness properties due to the homogenization of beams made of glued-laminated timber.

The bending moment acting on the trusses results as follows:

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

M_d	Design moment
q_d	Design load
L	Beam length

Design Moment

The eigenvalue analysis provides a critical buckling load factor of 0.32 as a result. This results in 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 was to be expected for this unsupported, slender truss member, the acting bending moment is greater (by a factor of 3) than the critical bending moment, and the truss is thus not sufficiently restrained against lateral buckling. However, a bracing should counteract, which is now considered for the calculation.

Single-Span Beam with Lateral and Torsional Restraint with Rigid Intermediate Supports

4 Single-Span Beam with Lateral and Torsional Restraints

If the stiffening bracing is stiff enough, the spacing between the lateral supports (for example, by purlins) is often used as an equivalent member length for the lateral buckling analysis. This procedure was described in the previous article Lateral-Torsional Buckling in Timber Construction | Examples 1.

Lateral-Torsional Buckling in Timber Structures | Examples 1

Thus, 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	Replacement member length
L	Beam length, spacing of the lateral restraint
a₁,a₂	tilt factors
a_z	Distance of the load application point from the shear center
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

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

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 supports.

The eigenvalue analysis with the Timber Design add-on provides a critical 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	Critical bending moment
α	Branch load factor
M_d	Design moment

Critical Bending Moment

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

Here, too, the two methods are in very close agreement.

Single-Span Beam with Lateral and Torsional Restraint and Elastic Member Foundation

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 members on elastic foundation is extended by the factors α and β in [1].

This makes it possible to take the shear stiffness of a stiffening bracing into account for the lateral buckling of the truss members.

Shear stiffness of the roof bracing

The shear stiffness of the bracing 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 strain stiffness of the diagonals and posts, the inclination of the diagonals, and the flexibility of the fasteners. For the stiffening bracing shown in Image 1, the shear stiffness is obtained as:

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

s_id	Ideal shear stiffness of the bracing system
E_D	5 % quantile of the modulus of elasticity of the diagonals
A_D	Cross-sectional area of the diagonals
α	Angle between the diagonal and the chords

Ideal Shear Stiffness of Stiffening Bracing

Here, E_D is the diagonals' modulus of elasticity and A_D is their cross-sectional area. However, the equation above does not include the ductility of the diagonals' fasteners. This and the member elongation of the diagonals can be considered by means of a notional cross-section area A_D'. What follows is this:

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

s_id	Ideal shear stiffness of the bracing system
E_D	5 % quantile of the modulus of elasticity of the diagonals
A_D'	Fictive cross-sectional area of the diagonals
α	Angle between the diagonal and the chords

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'	Fictitious cross-sectional area of the diagonals
A_D	Cross-sectional area of the diagonals
E_D	5 % quantile of the modulus of elasticity of the diagonals
L_D	Length of the diagonals
K_ser	Displacement module of the connection

Fictitious Cross-Sectional Area of the Diagonals

The diagonals have a dimension of w/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 should be 110,000 N/mm.

The ideal area is therefore

A_D' = 12,548 mm²

and thus, the shear stiffness of a bracing with a diagonals-to-chord angle of 60.64 ° is

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

s_id	Ideal shear stiffness of the bracing system
E_D	5 % quantile of the modulus of elasticity of the diagonals
A_D'	Fictive cross-sectional area of the diagonals
α	Angle between the diagonal and the chords

Ideal Shear Stiffness of the Bracing System

The member foundation per bracing can be converted according to [2], Formula 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'	Elastic member bedding per bracing
s_id	Ideal shear stiffness of the bracing system
L	Length of the bracing

Elastic Member Foundation per Bracing

For two bracings and six truss members, the following spring constant is available per truss:

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

K_y	Elastic member bedding per truss
K_y'	Elastic member bedding per bracing

Elastic Member Bedding per Beam

Provided that KG = ∞, Kθ = 0, K_y = 0.456 N/mm², e = 600 mm, a₁ = 1.13, and a₂ = 1.44, the equivalent member length results in:

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	Replacement member length
L	Beam length, spacing of the lateral restraint
a₁,a₂	Buckling coefficients
a_z	Distance of the load application point from the shear center
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
α, β	Coefficients for considering member bedding

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 supports 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, it is only valid if there is a deflection in a large sinusoidal arc. In other words, if the foundation is very soft. This is no longer given in this example. Multi-wave eigenfunctions, leading to a small critical buckling load for any larger spring constant, are not included in the aforesaid equation, since it is based on monomial sinus approaches.

As can be seen in Figure 7, the eigenvalue analysis results in a multi-wave mode shape with a critical 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	Critical bending moment
a_z	Distance of the load application from the shear center
e	Distance of the member bedding from the shear center
K_y	elastic member bedding per truss
L	Beam length
n	n-th eigenmode
E_0,05	5 % quantiles of the modulus of elasticity
I_z	Moment of inertia about the weak axis
G_0,05	5 % quantile of the shear modulus
I_T	torsional moment of inertia

Critical Bending Moment

The constant n denotes the 1st, 2nd, 3rd … eigensolution. Thus, several eigensolutions have to be analyzed, and the smallest critical bending moment then governs. The following critical bending moments are the result for n = 1…30.

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. Then, a critical load q_crit is assigned to the constant critical bending moment M_crit.

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

q_crit	Critical load
M_crit	Critical bending moment
L	Beam length

Critical Load

Since the member foundation in this example is regarded as very stiff and constantly distributed over the truss member length, critical bending moments occur that are slightly higher than for the rigid single supports.

Deformation Check of Roof Bracing

According to [3], Chapter 9.2.5.3 (2), stiffening bracings must be stiff enough so that they do not exceed the horizontal deflection of L/500. The calculation must be carried out with 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 velocity 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	Stabilizing force for the compression chord
k_crit	Lateral-torsional buckling factor
M_d	Design moment
h	Beam height

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	Bracing load
n	Number of trusses
L	Member length
k_f,3	Modification factor for the bracing resistance

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	Design load from wind
γ_Q	Partial safety factor for variable action
c_pe	External pressure coefficient
q_p	gust velocity pressure
h	Height of the building

Design Load from Wind

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

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 and is consistent with the nearly identical critical bending moments of the system with rigid intermediate support and the system with elastic member foundation.

Summary

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

It has been shown which options are available in timber construction for analyzing the lateral buckling of bending beams. For the common methods, it should be ensured that the stiffening bracings 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 stiffening bracings must still be analyzed 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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