Structural Model
For the system shown in Image 01, 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.
Model Data
L | 18 | m | Beam length |
b | 120 | mm | Beam width |
h | 1.200 | mm | Beam height |
GL24h | - | - | Material according to EN 14080 |
I_{z} | 172.800.000 | mm^{4} | Moment of inertia |
I_{T} | 647.654.753 | mm^{4} | Torsion moment of inertia |
q_{d} | 10 | kN/m | Design load |
a_{z} | 600 | mm | Load position |
e | 600 | mm | Foundation position |
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 a_{1} = 1.13 and a_{2} = 1.44 as follows:
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:
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:
M_{d} | Design moment |
q_{d} | Design load |
L | Beam length |
M_{d} = 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
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.
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 described in the previous article Lateral-Torsional Buckling in Timber Construction | Examples 1.
Thus, 2.25 m is used as L. For a_{1} = 1.00 and a_{2} = 0.00, this follows:
l_{ef} | Equivalent member length |
L | Beam length, spacing between lateral supports |
a_{1},a_{2} | Lateral buckling factors |
a_{z} | Load application distance 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} | Second moment of area about the weak axis |
I_{T} | Torsional constant |
l_{ef} = 2.25 m
The following results for the critical bending moment:
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} | Second moment of the area about the 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-/FE-LTB add-on module provides a critical buckling load factor of 2.7815 as a result. This results in the critical bending moment
M_{crit} | Critical bending moment |
η | Critical buckling load factor |
M_{d} | Design moment |
M_{crit} = 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, the determination of the equivalent member length for members on elastic foundation is extended by the factors α and β in [1].
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 the fasteners. For the stiffening bracing shown in Image 01, the shear stiffness results in:
s_{id} | Ideal shear stiffness of the stiffening bracing |
E_{D} | 5 % quantile of the diagonals' modulus of elasticity |
A_{D} | Cross-sectional area of diagonals |
α | Angle between diagonals and chords |
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} | Ideal shear stiffness of the stiffening bracing |
E_{D} | 5 % quantile of the diagonals' modulus of elasticity |
A_{D}' | Notional cross-sectional area of diagonals |
α | Angle between diagonals and chords |
where
A_{D}' | Notional cross-sectional area of diagonals |
A_{D} | Cross-sectional area of diagonals |
E_{D} | 5 % quantile of the diagonals' modulus of elasticity |
L_{D} | Length of diagonals |
K_{ser} | Slip modulus of the 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 diagonals-to-chord angle of 60.64 ° is
s_{id} | Ideal shear stiffness of the stiffening bracing |
E_{D} | 5 % quantile of the diagonals' modulus of elasticity |
A_{D}' | Notional cross-sectional 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:
K_{y}' | Elastic member foundation per bracing |
s_{id} | Ideal shear stiffness of the stiffening bracing |
L | Bracing length |
For two bracings and six truss members, the following spring constant is available per truss:
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:
l_{ef} | Equivalent member length |
L | Beam length, spacing between lateral supports |
a_{1},a_{2} | Lateral buckling factors |
a_{z} | Distance of the load application from the shear center |
E_{0.05} | 5 % quantile of the modulus of elasticity |
G_{0.05} | 5 % quantile of shear modulus |
I_{z} | Second moment of the area about the 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:
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} | Second moment of the area about the 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 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:
M_{crit} | Critical bending moment |
a_{z} | Distance of the load application from the shear center |
e | Distance of the member elastic foundation from the shear center |
K_{y} | Elastic member foundation per truss member |
L | Beam length |
n | n^{th} eigensolution |
E_{0.05} | 5 % quantile of the modulus of elasticity |
I_{z} | Second moment of area about the weak axis |
G_{0.05} | 5 % quantile of the 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 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 |
M_{crit} 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:
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}.
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 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} | 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
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
q_{d,wind} | Design load due to wind |
γ_{Q} | Partial safety factor for a variable action |
c_{pe} | External pressure coefficient |
q_{p} | Peak velocity pressure |
h | Height of the 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:
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.
Conclusion
It was shown which possibilities in timber construction can be used to analyze the lateral buckling of bending beams. For common methods, it is important to ensure that the stiffening bracings are stiff enough to accept rigid supports. Options have been shown in this article for cases when 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.