# Lateral-Torsional Buckling in Timber Construction | Examples 2

### Technical Article

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

#### Structural Model

For the system shown in Image 01, the truss members should be analyzed for lateral buckling. In the roof plane, there are six truss members 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 qd of 10 kN/m acts on the truss members.

#### Model Data

 L 18 m Length of beam b 120 mm Width of beam h 1,200 mm Depth of beam GL24h Material according to EN 14080 Iz 172,800,000 mm4 Second moment of area IT 647,654,753 mm4 Torsional constant qd 10 kN/m Design load az 600 mm Load position e 600 mm Position of foundation

Please note: 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

For the sake of completeness, the truss member is analyzed first without lateral supports (see Image 02). 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:

Equivalent Member Length

$$lef = La1 · 1 - a2 · azL · E0.05 · IZG0.05 · IT$$

 lef Equivalent member length L Beam length, spacing between lateral supports a1,a2 Lateral buckling factors az Distance of load application from shear center E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant

lef = 17.79 m

The critical bending moment can then be calculated as follows:

Critical Bending Moment

$$Mcrit = π · E0.05 · IZ · G0.05 · Itorlef$$

 Mcrit Critical bending moment E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant lef Equivalent member length

Mcrit = 134.52 kNm

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:

Design Moment

$$Md = qd · L28$$

 Md Design moment qd Design load L Beam length

Md = 405.00 kNm

The eigenvalue analysis with the RF-/FE-LTB add-on module provides a critical buckling load factor of 0.3334 as a result. This results in the critical bending moment

Mcrit = 0.3334 ⋅ 405 kNm = 135.03 kNm

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

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

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 already described in the previous article Lateral-Torsional Buckling in Timber Construction | Examples 1. Thus, 2.25 m is used as L. For a1 = 1.00 and a2 = 0.00 follows:

Equivalent Member Length

$$lef = La1 · 1 - a2 · azL · E0.05 · IzG0.05 · IT$$

 lef Equivalent member length L Beam length, spacing between lateral supports a1,a2 Lateral buckling factors az Distance of load application from shear center E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant

lef = 2.25 m

For the critical bending moment the following results:

Critical Bending Moment

$$Mcrit = π · E0.05 · IZ · G0.05 · Itorlef$$

 Mcrit Critical bending moment E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant lef Equivalent member length

Mcrit = 1,063.51 kNm

Since the bending moment acting on the beam is smaller than the critical bending moment, the beam is not endangered by lateral buckling under the assumption of rigid intermediate supports.

The eigenvalue analysis with the RF-/FE-LTB add-on module provides a critical buckling load factor of 2.7815 as a result. This results in the critical bending moment

Critical Bending Moment

$$Mcrit = η · Md$$

 Mcrit Critical bending moment η Critical buckling load factor Md Design moment

Mcrit = 2.7815 ⋅ 405 kNm = 1,126.50 kNm

#### Single-Span Beam With Lateral and Torsional Restraint and Member Elastic Foundation

As described in Lateral-Torsional Buckling in Timber Construction: Theory, [1] extends the determination of the equivalent member length for members on elastic foundation by the factor α and β. Thus, it is possible to consider the shear stiffness of a stiffening bracing for the lateral buckling of the truss members. The bracing's shear stiffness can be determined, for example, according to [2] Figure 6.34. As can be seen from the above, it depends on the type of bracing, the strain stiffness of diagonals and posts, the diagonals' inclination, and the ductility of fasteners. For the stiffening bracing shown in Image 01, the shear stiffness results in:

Ideal Shear Stiffness of Stiffening Bracing

$$sid = ED · AD · sinα2 · cosα$$

 sid Ideal shear stiffness of stiffening bracing ED 5 % quantile of diagonals' modulus of elasticity AD Cross-sectional area of diagonals α Angle between diagonals and chords

Here, ED is the diagonals' modulus of elasticity and AD 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 AD'. It follows:

Ideal Shear Stiffness of Stiffening Bracing

$$sid = ED · AD' · sinα2 · cosα$$

 sid Ideal shear stiffness of stiffening bracing ED 5 % quantile of diagonals' modulus of elasticity AD' Notional cross-sectional area of diagonals α Angle between diagonals and chords

where

Notional Cross-Sectional Area of Diagonals

$$AD' = AD1 + ED · ADLD·∑ 1Kser$$

 AD' Notional cross-sectional area of diagonals AD Cross-sectional area of diagonals ED 5 % quantile of diagonals' modulus of elasticity LD Length of diagonals Kser Slip modulus of connection

The diagonals have a dimension of w/h = 120/200 mm and a length LD 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 accordingly

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

Ideal Shear Stiffness of Stiffening Bracing

$$sid = ED · AD' · sinα2 · cosα$$

 sid Ideal shear stiffness of stiffening bracing ED 5 % quantile of diagonals' modulus of elasticity AD' Notional cross-sectional area of diagonals α Angle between diagonals and chords

sid = 44,864 kN

The member foundation per bracing can be converted according to [2] Formula 7.291 as follows:

Elastic Member Foundation per Bracing

$$Ky' = sid · π2L2Ky' = 44,864 kN · π2 18 m 2= 1,367 kNm2 = 1.367 Nmm2$$

 Ky' Elastic member foundation per bracing sid Ideal shear stiffness of stiffening bracing L Length of bracing

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

Elastic Member Foundation per Truss Member

$$Ky = 2 · Ky'6$$

 Ky Elastic member foundation per truss member Ky' Elastic member foundation per bracing

Ky = 455.6 kN/m² = 0.456 N/mm²

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

Equivalent Member Length

$$lef = La1 · 1 - a2 · azL · E0.05 · IzG0.05 · IT · 1α · β$$

 lef Equivalent member length L Beam length, spacing between lateral supports a1,a2 Lateral buckling factors az Distance of load application from shear center E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant α, β Factors for considering a member foundation

lef = 0.13

Thus, the critical bending moment results in an unrealistic value of:

Critical Bending Moment

$$Mcrit = π · E0.05 · IZ · G0.05 · Itorlef$$

 Mcrit Critical bending moment E0.05 5 % quantile of modulus of elasticity G0.05 5 % quantile of shear modulus Iz Second moment of area about weak axis IT Torsional constant lef Equivalent member length

Mcrit = 18,482.84 kNm

A value similar to the system with rigid intermediate supports would be expected. As described in Lateral-Torsional Buckling in Timber Construction: Theory, the application of the extended formula with α and β is limited in its application. 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 you can see in Image 07, a multi-wave eigenvector results from the eigenvalue analysis.

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

Critical Bending Moment

$$Mcrit = 2 · e · c* - az* · B* + 2 · e · c* - az* · B* 2 + 4 · 1.0 · B* · T* - e* · c*2 2c* = Ky · L2π2 · n2az* = azn2B* = E0.05 · IZ · π2 · n2L2 + c*T* = G0.05 · IT + c* · e2n = 1 , 2 , 3 ...$$

 Mcrit Critical bending moment az Distance of load application from shear center e Distance of member elastic foundation from shear center Ky Elastic member foundation per truss member L Beam length n nth eigensolution E0.05 5 % quantile of modulus of elasticity Iz Second moment of area about weak axis G0.05 5 % quantile of shear modulus IT Torsional constant

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

nMcrit [kNm]nMcrit [kNm]
19,523.25162,214.63
24,281.26172,339.17
32,294.32182,464.92
41,605.56192,591.63
51,354.68202,719.14
61,282.70212,847.30
71,294.12222,976.00
81,348.81233,105.16
91,428.05243,234.71
101,522.29253,364.60
111,626.24263,494.77
121,736.77273,625.20
131,851.94283,755.84
141,970.50293,886.67
152,091.60304,017.68

Mcrit becomes minimal for n = 6 and is about 1,282.70 kNm.

The eigenvalue solution from the RF-/FE-LTB add-on module (see Image 07) results in:

Mcrit = 3.4376 ⋅ 405 kNm = 1,397.25 kNm

Both results match very well. However, the analytical solution is on the safe side, as this method is based on a constant bending moment distribution. Then, a critical load qcrit is assigned to the constant critical bending moment Mcrit.

$$qcrit = Mcrit · 8L2$$

 qcrit Critical load Mcrit Critical bending moment L Beam length

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.

According to [3] Chapter 9.2.5.3 (2), stiffening bracings must be enough stiff to 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 kcrit = 0.195, H = 5 m and qp = 0.65 kN/m² as gust velocity pressure, the following loads result (see [3] Chapter 9.2.5.3):

Stabilizing Force for Compression Chord

$$Nd = 1 - kcrit · Mdh$$

 Nd Stabilizing force for compression chord kcrit Lateral buckling factor Md Design moment h Beam height

Nd = (1 - 0.195) ⋅ 405 / 1.2 = 271.68 kN

$$qd = 15L · n · Ndkf,3 · L$$

 qd Stiffening load n Number of truss members L Beam length kf,3 Modification factor for stiffening resistance

qd = 2.76 kN/m

$$qd,Wind = γQ · cpe,D + cpe,E · qp · H2$$

 qd,wind Design load from wind γQ Partial safety factor for variable action cpe External pressure coefficient qp Peak velocity pressure h Height of building

qd,wind = 1.5 ⋅ (0.7 + 0.3) ⋅ 0.65 ⋅ 5 / 2 = 2.44 kN/m

The deformation of the stiffening bracing is shown in Image 08. The loads were divided in half because there are two stiffening bracings.

The allowable deformation is:

Allowable Deformation

$$L500 = 18 m500 = 36.00 mm ≥ 4.74 mm$$

The result confirms the assumption of a very stiff bracing and is consistent with the almost identical critical bending moments of the system with rigid intermediate supports and the one with elastic member foundation.

#### Summary

It was shown, which possibilities in timber construction can be used to analyze lateral buckling of bending beams. For common methods it is important to ensure that stiffening bracings are stiff enough to accept rigid supports. Options have been shown in this article for the case that this assumption does not apply. Basically, the bending beams and the stiffening bracings have to be designed for their load-bearing capacity and serviceability according to the corresponding standard. However, this is beyond the scope of this article.

#### Dipl.-Ing. (FH) Gerhard Rehm

Product Engineering & Customer Support

Mr. Rehm is responsible for the development of products for timber structures, and provides technical support for customers.

#### Reference

 [1] 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 [2] Petersen, C.: Statik und Stabilität der Baukonstruktionen, 2. Auflage. Wiesbaden: Vieweg, 1982 [3] Eurocode 5: Design of timber structures - Part 1-1: General - Common rules and rules for buildings; EN 1995-1-1:2010-12

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