LateralTorsional Buckling in Timber Construction  Examples 2
Technical Article
The previous article LateralTorsional 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 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 q_{d} 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  
I_{z} 

172,800,000  mm^{4}  Second moment of area 
I_{T} 

647,654,753  mm^{4}  Torsional constant 
q_{d} 

10  kN/m  Design load 
a_{z} 

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.
SingleSpan 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 a_{1} = 1.13 and a_{2} = 1.44 as follows:
${\mathrm{l}}_{\mathrm{ef}}=\frac{\mathrm{L}}{{\mathrm{a}}_{1}\xb7\left(1{\mathrm{a}}_{2}\xb7{\displaystyle \frac{{\mathrm{a}}_{\mathrm{z}}}{\mathrm{L}}}\xb7\sqrt{{\displaystyle \frac{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{Z}}}{{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{T}}}}}\right)}$
l_{ef}  Equivalent member length 
L  Beam length, spacing between lateral supports 
a_{1},a_{2}  Lateral buckling factors 
a_{z}  Distance of load application from shear center 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
l_{ef} = 17.79 m
The critical bending moment can then be calculated as follows:
${\mathrm{M}}_{\mathrm{crit}}=\frac{\mathrm{\pi}\xb7\sqrt{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{Z}}\xb7{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{tor}}}}{{\mathrm{l}}_{\mathrm{ef}}}$
M_{crit}  Critical bending moment 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
l_{ef}  Equivalent member length 
M_{crit} = 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:
${\mathrm{M}}_{\mathrm{d}}=\frac{{\mathrm{q}}_{\mathrm{d}}\xb7{\mathrm{L}}^{2}}{8}$
M_{d}  Design moment 
q_{d}  Design load 
L  Beam length 
M_{d} = 405.00 kNm
The eigenvalue analysis with the RF/FELTB addon module provides a critical buckling load factor of 0.3334 as a result. This results in the critical bending moment
M_{crit} = 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.
SingleSpan 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 LateralTorsional Buckling in Timber Construction  Examples 1. Thus, 2.25 m is used as L. For a_{1} = 1.00 and a_{2} = 0.00 follows:
${\mathrm{l}}_{\mathrm{ef}}=\frac{\mathrm{L}}{{\mathrm{a}}_{1}\xb7\left(1{\mathrm{a}}_{2}\xb7{\displaystyle \frac{{\mathrm{a}}_{\mathrm{z}}}{\mathrm{L}}}\xb7\sqrt{{\displaystyle \frac{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{z}}}{{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{T}}}}}\right)}$
l_{ef}  Equivalent member length 
L  Beam length, spacing between lateral supports 
a_{1},a_{2}  Lateral buckling factors 
a_{z}  Distance of load application from shear center 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
l_{ef} = 2.25 m
For the critical bending moment the following results:
${\mathrm{M}}_{\mathrm{crit}}=\frac{\mathrm{\pi}\xb7\sqrt{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{Z}}\xb7{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{tor}}}}{{\mathrm{l}}_{\mathrm{ef}}}$
M_{crit}  Critical bending moment 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
l_{ef}  Equivalent member length 
M_{crit} = 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/FELTB addon module provides a critical buckling load factor of 2.7815 as a result. This results in the critical bending moment
${\mathrm{M}}_{\mathrm{crit}}=\mathrm{\eta}\xb7{\mathrm{M}}_{\mathrm{d}}$
M_{crit}  Critical bending moment 
η  Critical buckling load factor 
M_{d}  Design moment 
M_{crit} = 2.7815 ⋅ 405 kNm = 1,126.50 kNm
SingleSpan Beam With Lateral and Torsional Restraint and Member Elastic Foundation
As described in LateralTorsional 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:
${\mathrm{s}}_{\mathrm{id}}={\mathrm{E}}_{\mathrm{D}}\xb7{\mathrm{A}}_{\mathrm{D}}\xb7\mathrm{sin}{\left(\mathrm{\alpha}\right)}^{2}\xb7\mathrm{cos}\left(\mathrm{\alpha}\right)$
s_{id}  Ideal shear stiffness of stiffening bracing 
E_{D}  5 % quantile of diagonals' modulus of elasticity 
A_{D}  Crosssectional area of diagonals 
α  Angle between diagonals and chords 
Here, E_{D} is the diagonals' modulus of elasticity and A_{D} is their crosssectional 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 crosssection area A_{D}'. It follows:
${\mathrm{s}}_{\mathrm{id}}={\mathrm{E}}_{\mathrm{D}}\xb7{\mathrm{A}}_{\mathrm{D}}\text{'}\xb7\mathrm{sin}{\left(\mathrm{\alpha}\right)}^{2}\xb7\mathrm{cos}\left(\mathrm{\alpha}\right)$
s_{id}  Ideal shear stiffness of stiffening bracing 
E_{D}  5 % quantile of diagonals' modulus of elasticity 
A_{D}'  Notional crosssectional area of diagonals 
α  Angle between diagonals and chords 
where
${\mathrm{A}}_{\mathrm{D}}\text{'}=\frac{{\mathrm{A}}_{\mathrm{D}}}{1+{\displaystyle \frac{{\mathrm{E}}_{\mathrm{D}}\xb7{\mathrm{A}}_{\mathrm{D}}}{{\mathrm{L}}_{\mathrm{D}}}}\xb7\sum {\displaystyle \frac{1}{{\mathrm{K}}_{\mathrm{ser}}}}}$
A_{D}'  Notional crosssectional area of diagonals 
A_{D}  Crosssectional area of diagonals 
E_{D}  5 % quantile of diagonals' modulus of elasticity 
L_{D}  Length of diagonals 
K_{ser}  Slip modulus of connection 
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 accordingly
A_{D}' = 12,548 mm²
and thus the shear stiffness of a bracing with a diagonalstochord angle of 60.64 ° is
${\mathrm{s}}_{\mathrm{id}}={\mathrm{E}}_{\mathrm{D}}\xb7{\mathrm{A}}_{\mathrm{D}}\text{'}\xb7\mathrm{sin}{\left(\mathrm{\alpha}\right)}^{2}\xb7\mathrm{cos}\left(\mathrm{\alpha}\right)$
s_{id}  Ideal shear stiffness of stiffening bracing 
E_{D}  5 % quantile of diagonals' modulus of elasticity 
A_{D}'  Notional crosssectional area of diagonals 
α  Angle between diagonals and chords 
s_{id} = 44,864 kN
The member foundation per bracing can be converted according to [2] Formula 7.291 as follows:
${\mathrm{K}}_{\mathrm{y}}\text{'}={\mathrm{s}}_{\mathrm{id}}\xb7\frac{{\mathrm{\pi}}^{2}}{{\mathrm{L}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{y}}\text{'}=44,864\mathrm{kN}\xb7\frac{{\mathrm{\pi}}^{2}}{{\left(18\mathrm{m}\right)}^{2}}=1,367\frac{\mathrm{kN}}{{\mathrm{m}}^{2}}=1.367\frac{\mathrm{N}}{{\mathrm{mm}}^{2}}\phantom{\rule{0ex}{0ex}}$
K_{y}'  Elastic member foundation per bracing 
s_{id}  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:
${\mathrm{K}}_{\mathrm{y}}=\frac{2\xb7{\mathrm{K}}_{\mathrm{y}}\text{'}}{6}$
K_{y}  Elastic member foundation per truss member 
K_{y}'  Elastic member foundation per bracing 
K_{y} = 455.6 kN/m² = 0.456 N/mm²
Provided that K_{G} = ∞, K_{θ} = 0, K_{y} = 0.456 N/mm², e = 600 mm, a_{1} = 1.13, and a_{2} = 1.44, the equivalent member length results in:
${\mathrm{l}}_{\mathrm{ef}}=\frac{\mathrm{L}}{{\mathrm{a}}_{1}\xb7\left(1{\mathrm{a}}_{2}\xb7{\displaystyle \frac{{\mathrm{a}}_{\mathrm{z}}}{\mathrm{L}}}\xb7\sqrt{{\displaystyle \frac{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{z}}}{{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{T}}}}}\right)}\xb7\frac{1}{\mathrm{\alpha}\xb7\mathrm{\beta}}$
l_{ef}  Equivalent member length 
L  Beam length, spacing between lateral supports 
a_{1},a_{2}  Lateral buckling factors 
a_{z}  Distance of load application from shear center 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
α, β  Factors for considering a member foundation 
l_{ef} = 0.13
Thus, the critical bending moment results in an unrealistic value of:
${\mathrm{M}}_{\mathrm{crit}}=\frac{\mathrm{\pi}\xb7\sqrt{{\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{Z}}\xb7{\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{tor}}}}{{\mathrm{l}}_{\mathrm{ef}}}$
M_{crit}  Critical bending moment 
E_{0.05}  5 % quantile of modulus of elasticity 
G_{0.05}  5 % quantile of shear modulus 
I_{z}  Second moment of area about weak axis 
I_{T}  Torsional constant 
l_{ef}  Equivalent member length 
M_{crit} = 18,482.84 kNm
A value similar to the system with rigid intermediate supports would be expected. As described in LateralTorsional 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. Multiwave 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 multiwave 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:
${\mathrm{M}}_{\mathrm{crit}}=\frac{\left(2\xb7\mathrm{e}\xb7{\mathrm{c}}^{*}{\mathrm{a}}_{\mathrm{z}}^{*}\xb7{\mathrm{B}}^{*}\right)+\sqrt{{\left(2\xb7\mathrm{e}\xb7{\mathrm{c}}^{*}{\mathrm{a}}_{\mathrm{z}}^{*}\xb7{\mathrm{B}}^{*}\right)}^{2}+4\xb71.0\xb7\left({\mathrm{B}}^{*}\xb7{\mathrm{T}}^{*}{\mathrm{e}}^{*}\xb7{{\mathrm{c}}^{*}}^{2}\right)}}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{c}}^{*}={\mathrm{K}}_{\mathrm{y}}\xb7\frac{{\mathrm{L}}^{2}}{{\mathrm{\pi}}^{2}\xb7{\mathrm{n}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{a}}_{\mathrm{z}}^{*}=\frac{{\mathrm{a}}_{\mathrm{z}}}{{\mathrm{n}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{B}}^{*}={\mathrm{E}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{Z}}\xb7\frac{{\mathrm{\pi}}^{2}\xb7{\mathrm{n}}^{2}}{{\mathrm{L}}^{2}}+{\mathrm{c}}^{*}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}^{*}={\mathrm{G}}_{0.05}\xb7{\mathrm{I}}_{\mathrm{T}}+{\mathrm{c}}^{*}\xb7{\mathrm{e}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{n}=1,2,3...$
M_{crit}  Critical bending moment 
a_{z}  Distance of load application from shear center 
e  Distance of member elastic foundation from shear center 
K_{y}  Elastic member foundation per truss member 
L  Beam length 
n  n^{th} eigensolution 
E_{0.05}  5 % quantile of modulus of elasticity 
I_{z}  Second moment of area about weak axis 
G_{0.05}  5 % quantile of shear modulus 
I_{T}  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.
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 
M_{crit} becomes minimal for n = 6 and is about 1,282.70 kNm.
The eigenvalue solution from the RF/FELTB addon module (see Image 07) results in:
M_{crit} = 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 q_{crit} is assigned to the constant critical bending moment M_{crit}.
${\mathrm{q}}_{\mathrm{crit}}=\frac{{\mathrm{M}}_{\mathrm{crit}}\xb78}{{\mathrm{L}}^{2}}$
q_{crit}  Critical load 
M_{crit}  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 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):
${\mathrm{N}}_{\mathrm{d}}=\left(1{\mathrm{k}}_{\mathrm{crit}}\right)\xb7\frac{{\mathrm{M}}_{\mathrm{d}}}{\mathrm{h}}$
N_{d}  Stabilizing force for compression chord 
k_{crit }  Lateral buckling factor 
M_{d}  Design moment 
h  Beam height 
N_{d} = (1  0.195) ⋅ 405 / 1.2 = 271.68 kN
${\mathrm{q}}_{\mathrm{d}}=\sqrt{\frac{15}{\mathrm{L}}}\xb7\frac{\mathrm{n}\xb7{\mathrm{N}}_{\mathrm{d}}}{{\mathrm{k}}_{\mathrm{f},3}\xb7\mathrm{L}}$
q_{d}  Stiffening load 
n  Number of truss members 
L  Beam length 
k_{f,3}  Modification factor for stiffening resistance 
q_{d} = 2.76 kN/m
${\mathrm{q}}_{\mathrm{d},\mathrm{Wind}}={\mathrm{\gamma}}_{\mathrm{Q}}\xb7\left({\mathrm{c}}_{\mathrm{pe},\mathrm{D}}+{\mathrm{c}}_{\mathrm{pe},\mathrm{E}}\right)\xb7{\mathrm{q}}_{\mathrm{p}}\xb7\frac{\mathrm{H}}{2}$
q_{d,wind}  Design load from wind 
γ_{Q}  Partial safety factor for variable action 
c_{pe}  External pressure coefficient 
q_{p}  Peak velocity pressure 
h  Height of building 
q_{d,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:
$\frac{\mathrm{L}}{500}=\frac{18\mathrm{m}}{500}=36.00\mathrm{mm}\ge 4.74\mathrm{mm}$
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 loadbearing capacity and serviceability according to the corresponding standard. However, this is beyond the scope of this article.
Author
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.
Keywords
Lateral buckling Lateraltorsional buckling Eigenvalue Buckling member with elastic foundation
Reference
Links
Write Comment...
Write Comment...
Contact us
Do you have questions or need advice?
Contact our free email, chat, or forum support or find various suggested solutions and useful tips on our FAQ page.
New
Modeling Downstand Beams in CrossLaminated Timber Constructions with Ribs
This time, we want to look at modeling downstand beams by means of ribs.
The crosssection resistance design analyzes tension and compression along the grain, bending, bending and tension/compression as well as the strength in shear due to shear force.
The design of structural components at risk of buckling or lateraltorsional buckling is performed according to the Equivalent Member Method and considers the systematic axial compression, bending with and without compressive force as well as bending and tension. Deflection of inner spans and cantilevers is compared to the maximal allowable deflection.
Separate design cases allow for a flexible and stability analysis of members, sets of members, and loads.
Designrelevant parameters such as the stability analysis type, member slendernesses, and limit deflections can be freely adjusted.
 How can I create a curved or arched section?
 How are the signs for the release results of a line release and line hinges interpreted?
 After the design with RF‑/TIMBER Pro, I optimized a crosssection. Why is the utilization of the optimized crosssection exceeded now?
 Is it possible to design the support pressure or the compression perpendicular to the grain in RX‑TIMBER?
 Why are the stresses of the 90° orientation not displayed for a layer with the orthotropy direction 90° for σ_{b,90} in RF‑LAMINATE?
 How can I get the member end forces to design the connections?
 I design timber components. The deformations of load combinations deviate from the manual calculation exactly by the factor of the material partial safety factor. Why?
 How can I run the RX‑TIMBER Frame plugin? I did not find it in the Addon Modules menu nor in Project Navigator  Data.
 I have a question about the results of the serviceability limit state design: How is the increment of the dead load by the factor 1.8 and the imposed load by 1.48 explained in the result combinations for the SLS design?
 Where can I adjust the effective length l_{ef }according to Table 6.1 of Eurocode 5 in the TIMBER Pro addon module?
Associated Products