Eurocode 3 - Combined bending and compression, lateral buckling reduction factor not applied to Mz 1

[Trusslover]

Hello dear colleagues,

After many years browsing this great forum to find useful answers, I have finally created an account to ask my first question which I've had for a while but couldn't find answers to.
In the Eurocode 3 for steel, when looking at combined bending and compression in section 6.3.3 you find two formulas (see attached) where you combine the utilization ratios of N + My + Mz.

Why do we only apply the Chi LT reduction factor for lateral buckling to My but not Mz?
I often work with square hollow section beams, and I can think of cases where Mz would be greater than My in the structures I work with.
So I usually apply Chi LT to both (which might be conservative?), but I'd like to understand why the codes say otherwise.

 

[KootK]

Why do we only apply the Chi LT reduction factor for lateral buckling to My but not Mz?

I, too, am a lover of trusses. So sexy...

I feel that it is best to try to understand something like this from the perspective of the fundamentals. This will be just one of several valid ways to tell the same story.

1) When a member loaded to produce strong axis moment rolls over onto its weak axis (LTB), its moment of inertia (stiffness) in the direction of the load effectively decreases. Because stiffness decreases, the member deflects more and the load moves closer to the ground. In this way, the potential energy of the applied loads is reduced (like a ball rolling down a hill). Viewed from this perspective, the lowering of potential energy is the impetus for LTB.

2) Were a member loaded about it's weak axis to roll over onto its strong axis, its moment of inertia (stiffness) in the direction of the load would effectively increase. Because stiffness increases, the member deflects less and the load moves further from a the ground. In this way, the potential energy of the applied loads is increased (like a ball rolling uphill of its own accord). Obviously, this is not going to happen. Consequently, the LT factor need not be applied to weak axis moment because weak axis lateral torsional buckling is functionally impossible.

How'd I do?

 

[Trusslover]

Thank you so much KootK,

I really like your explanation, this now makes sense to me when I think about I beams or rectangular hollow sections.
I was going to ask about square and circular hollow sections but I just remembered that Eurocode 3 says that those sections are not susceptible to lateral-torsional buckling.
Is it because if those sections were to start rolling over, the moment of inertia wouldn't change therefore there wouldn't be any impetus for LTB?

 

[KootK]

Is it because if those sections were to start rolling over, the moment of inertia wouldn't change therefore there wouldn't be any impetus for LTB?

Just so. Rolling an Ix=Iy section usually requires that some energy also be stored as torsional strain. In that sense, it's slightly better than purely neutral stability. Like pushing a ball up a slight incline.

 

[HTURKAK]

I often work with square hollow section beams, and I can think of cases where Mz would be greater than My in the structures I work with.

By definition Y-Y is the strong axis ( for WF , I beams ..) and for square hollow sections they are the same.

So I usually apply Chi LT to both (which might be conservative?), but I'd like to understand why the codes say otherwise.

Definition of χLT is ; the reduction factor due to lateral torsional buckling discussed at clauses 6.3.2. ( and χLT = 1.0 for members sufficiently supported laterally and that are not susceptible to torsional deformation such as circular or square hollow sections. Y-Y axis is susceptible to torsional deformation so the reduction factor is applied only strong axis bending resistance My,Rk . In your case χLT = 1.0 for SHS )

I have attached a doc. and there is a worked example to get insight for uniform members in biaxial bending and compression check as per EC-3.