Lateral Torsional Buckling of Timber Beams

Lateral torsional buckling describes the behaviour of a beam that is prone to lateral/horizontal buckling in the compression zone due to bending moments.

In this article, you’ll learn how to verify timber beams according to Eurocode.

Now, let’s get into it.

What Is Lateral Torsional Buckling?

The support conditions influence the beam’s ability to buckle laterally.

• Fixed supports provide a good torsional restraint, which leads to a greater lateral torsional buckling resistance.
• Pinned supports lead to a smaller lateral torsional buckling resistance.
• Lateral torsional movement restraints in the span of the beam also lead to a greater lateral torsional buckling resistance. Examples are beams that run transverse to the beams with a stiff connection.

Now, let’s get into the nerdy calculations..

The 4 Steps To Verify Timber Beams For Lateral Torsional Buckling

We use the secondary beam of the canopy structure below to show the calculation steps.

Also let’s assume that the beam is not held horizontally by some sheeting like trapezoidal sheeting or OSB boards.

Step #1: Define the material properties of the timber element

We first define the material properties of the timber beam. In this tutorial, we’ll use C24 as the timber material.

For small structures like this which aren’t exposed to big loads, C24 is enough.

We summarized all timber materials as a blogpost (click ​here​).

Here are the strength and stiffness properties of C24.

Strength values in MPa		C24
Bending  strength	fm.g.k	24
Tensile strength parallel to grain	ft.0.g.k	14
Tensile strength perpendicular to grain	ft.90.g.k	0.4
Compressive strength parallel to grain	fc.0.g.k	21
Compressive strength perpendicular to grain	fc.90.g.k	2.5
Shear strength	fv.g.k	4.0
Modulus of elasticity	E0.g.mean	11000
	E0.g.05	7400
Shear modulus	Gg.mean	690
Density in kg/m3
	ρg.k	350
	ρg.mean	420

The partial safety factor is found in EN 1995-1-1 Table 2.3 as:

The beam is classified according to EN 1995-1-1 2.3.1.3 as service class 2 (assumption in this tutorial).

Then we’ll verify the timber beam for a design load of load duration class short-term (EN 1995-1-1 Table 2.1) which leads to a modification factor (EN 1995-1-1 Table 3.1) of:

Step #2: Define the geometrical properties of the timber beam

The beam has the following cross-sectional properties:

• Cross-sectional height h=20cm
• Cross-sectional width w=12cm
• Moment of inertia strong axis I_y = (w ⋅ h³)/12 = 8 ⋅ 10⁷ mm⁴
• Moment of inertia weak axis I_x = (h ⋅ w³)/12 = 2.88 ⋅ 10⁷ mm⁴
• Torsional moment of inertia I_tor = 1/3 ⋅ h ⋅ w³ ⋅ (1 – 0.63 ⋅ w/h + 0.052 ⋅ w⁵/h⁵) = 7.2 ⋅ 10⁷ mm⁴
• Length (=span) s=5m

Step #3: Calculate the loads and internal forces acting in the timber element

First, we need to calculate the characteristic loads that act on the roof area of the canopy. The area loads are applied to the roof panels (could be OSB boards or trapezoidal steel sheeting).

These panels then transfer the loads to the secondary beams, which we verify for lateral torsional buckling in this tutorial. This video explains transferring vertical loads in detail (click → ​here​ ←).

We won’t show how to calculate the loads and how to do the load transfer in this article, as each calculation of the individual load is an article for itself.

But that’s exactly what I teach in Module #2: Structural Design of a Residential Timber Roof.

In this post, we’ll design the secondary timber beam for the following design line load.

Design line load:

From design load, we’ll calculate the design bending moment:

The design normal force is 0 kN:

Step #4: Verify the timber beam for lateral torsional buckling

Bending stress (short term load)	σm.d = Md/Iy ⋅ h/2 = 15.6 MPa
Compression stress (short term load):	σc.d = Nd/(w⋅h) = 0 MPa
Design bending resistance	fm.d = kmod ⋅ fm.k/γM = 17.3 MPa
Design compression resistance:	fc.0.d = kmod ⋅ fc.0.k/γM = 15.1 MPa
Effective length of the beam according to EN 1995-1-1 Table 6.1	lef = 0.9 ⋅ s + 2 ⋅ h = 4.9m
Critical bending stress (EN 1995-1-1 (6.31))	$\sigma_{m.crit} = \pi \cdot \frac{\sqrt{E_{0.g.05} \cdot I_z \cdot G_{g.05} \cdot I_{tor}}}{l_{ef} \cdot W_y} = 67.4 MPa$
Relative slenderness (EN1995-1-1 (6.30))	λrel.m = √(fm.k/σm.crit) = 0.60
Factor for reduced bending strength due to lateral torsional buckling (EN1995-1-1 (6.34))	λrel.m < 0.75 → kcrit = 1

Now we only need the following formulas if there is also a normal force acting in the beam. In our case the normal force is 0 kN. But I’ll still include the formulas to show you hwo to do it.

Buckling length out-of-plane	lz = 5m
Radius of inertia (weak axis)	iz = √(Iz/w ⋅ h) = 0.035
Slenderness ratio	λz = lz/iz = 144.34
Relative slenderness ratio (EN 1995-1-1 (6.22))	λrel.z = λz/π ⋅ √(fc.0.k/E0.g.05) = 2.45
Factor for members within the straightness limits (EN 1995-1-1 (6.29))	βc = 0.2
Instability factor (EN1995-1-1 (6.28))	kz = 0.5 ⋅ (1 + βc ⋅ (λrel.z – 0.3) + λrel.z2) = 3.7
Buckling reduction coefficient (EN1995-1-1 (6.26))	kc.z = 1/(kz + √(kz2 – λrel.z2)) = 0.15
Utilization (EN1995-1-1 (6.35))	η = (σm.d/(kcrit ⋅ fm.d))2 + σc.d/(kc.z ⋅ fc.0.d) = 0.82

Utilization (EN1995-1-1 (6.35)):

Final words

Lateral torsional buckling is quite a complex phenomenon in structural engineering, and it takes a lot of time to understand. But it can happen to most beams.

Therefore, it’s important to understand it.

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Let’s design better structures together,

Laurin.