Module 8 · Lesson 8.1

Why a beam fails sideways

Load it downwards and it moves sideways and twists. The reason is that the two flanges are doing opposite things.

Why this matters

A beam in bending has no axial load on it, so it is not obvious why it should buckle at all. Yet an unrestrained beam can lose two thirds of its bending capacity to a failure that moves it sideways and twists it — and the loss is invisible in any strength calculation. This is the check that most often surprises people, and the physics behind it is entirely approachable once you look at what the two flanges are doing.

By the end of this lesson you should be able to

Explain why a beam in bending can buckle
Say why the failure involves twist as well as lateral movement
Read the two terms in the critical moment
Identify which resists twist on a short span and which on a long one

What you should already know

Torsion, and the difference between open and closed sections (Modules 1 and 6)
The Euler load and relative slenderness (Module 7)
Cross-section bending resistance and the class-appropriate modulus (Modules 5 and 6)

The two flanges are doing opposite things

Take a simply supported beam in sagging. The top flange is in compression and the bottom flange in tension.

Now imagine displacing the beam sideways a little, and ask what happens to equilibrium in that displaced position.

The compression flange behaves like a column. It carries a compressive force, and that force now acts through a lateral offset. The result is a moment that pushes it further sideways. It is destabilising — exactly the mechanism that buckles a column.

The tension flange does the opposite. Its tensile force, acting through the same offset, pulls it back. It is stabilising, like a string under tension.

So the two flanges do not move sideways by the same amount. The compression flange runs away and the tension flange resists.

A section whose two flanges have moved sideways by different amounts has twisted.

That is the whole thing. The failure cannot be purely lateral, because the two flanges cannot agree on how far to go. It is lateral displacement and twist together — which is why it is called lateral-torsional buckling, and why the torsional properties of the section turn up in the answer.

Note

So the beam's torsional stiffness is what resists it

Once you know the failure involves twist, the rest follows from Module 6.

A section resists twist two ways:

St Venant torsion, through I_t — pure shear flow round the section. For an open section this is the sum of b·t³/3 over the plates, which is very small.
Warping torsion, through I_w — the flanges bend in their own planes, in opposite directions, and the resulting couple resists the twist. For an I section this is substantial.

Both appear in the critical moment, and which one is doing the work depends on the span.

And a closed section has no warping to speak of but an enormous I_t — for the sections compared later in this module, 900 times that of the I beam. Which is why a hollow section is effectively immune to LTB, and why an I section is not.

The critical moment, and its two terms

For the reference case — doubly symmetric section, uniform moment, ends free to warp but held against twist — the energy method gives:

M_cr = (π²EI_z/L²) · √( I_w/I_z + L²GI_t/(π²EI_z) )

This is a genuine mechanics result for that case. The derivation assumes a twist shape, writes down the strain energy stored in warping and in St Venant torsion, sets it against the work done by the bending moment acting through the displaced geometry, and solves for the moment at which the two balance.

Read the root term. It has exactly two pieces.

I_w/I_z — the warping contribution. Notice there is no L in it. Warping resistance does not care how long the beam is.

L²GI_t/(π²EI_z) — the St Venant contribution. This one goes with L², so it grows rapidly with span.

The consequence is that the two mechanisms trade places:

On a short span the St Venant term is negligible, and warping does nearly all the work.
On a long span the St Venant term dominates, and the flanges bending in their own planes matter much less.

That is why a short beam and a long beam of the same section fail in visibly different ways. It also means shortening the span helps more than the 1/L² out front suggests: the root term shrinks as well.

Try it

The two terms, and what changes them

M_cr's root term has exactly two pieces — warping, which does not know the span, and St Venant, which grows with L². Watch them trade places, then switch to a hollow section and watch LTB disappear.

Section

Unrestrained length8.0 m

Unrestrained length: 8.0 m
Warping term I_w/I_z: 66952
St Venant term: 73532
Warping share: 48 %
Torsion constant I_t: 70.2 cm⁴
Critical moment M_cr: 290 kNm
Section resistance M_c,Rd: 826 kNm
LTB slenderness λ̄_LT: 1.688
Reduction χ_LT: 0.351
Design resistance M_b,Rd: 290 kNm
Lost to LTB: 65 %

λ̄_LT = 1.69 gives χ_LT = 0.351 — 65% of the cross-section bending resistance is lost to LTB. Note the plateau runs to 0.4, twice the column value, because the tension flange stabilises a beam in a way nothing stabilises a column. Warping 48% and St Venant 52% — both mechanisms are working, which is the usual situation for a building beam.

Things worth trying

Take the span from 1 m to 20 m and watch the top two bars trade places. Warping does not move at all — it has no length in it — while St Venant grows with L² and eventually dominates.
Find the span where the two are equal. For this UB it is close to 8 m, which is why that span was chosen for the worked example: it sits right at the crossover.
Watch M_cr fall as the span grows. It falls FASTER than 1/L², because the prefactor drops while the root term is also shrinking relative to the section's resistance.
Come down from 3 m in half-metre steps and find where χ_LT first reaches 1. It is between 1.5 m and 2 m for this beam — at 2 m there is still 4% gone. That threshold is what 'fully restrained' means in practice, and it is a length, not a feeling.
Now switch to the hollow section at any span. The warping bar vanishes, St Venant takes everything, and χ_LT stays at 1 — the torsion constant is about 900 times larger, so LTB simply stops being a consideration.
Note the I_t readout as you switch sections. That single number is the whole difference between a beam that must be restrained and one that need not be.

Worked example

An unrestrained 8 m beam

Given

533 × 210 × 92 UB in S355: I_z = 2388 cm⁴, I_t = 70.2 cm⁴, I_w = 1.599 dm⁶, W_pl = 2326 cm³
Simply supported over 8.0 m with no intermediate lateral restraint
Uniform moment, load applied at the shear centre — the reference case
Class 1, so the plastic modulus applies

Find

The design bending resistance, and how much LTB has cost.

Assumptions

Section properties computed from plate geometry; I_t runs about 7% below published because root radii contribute to it appreciably
χ_LT and its plateau are calibrated and unverified here

Predict first

The same beam is fabricated as a rectangular hollow section instead, with a similar depth. What happens to the LTB check?

Practice

A beam has W_pl = 2326 cm³ in S355. What is its cross-section bending resistance, in kNm? Take γ_M0 = 1.0.

Practice

That beam has M_cr = 290 kNm. What is its LTB slenderness λ̄_LT?

Practice

At λ̄_LT = 1.688, χ_LT is capped at the elastic bound 1/λ̄_LT². What is that cap?

Check yourself

Why does lateral-torsional buckling involve twist and not just lateral movement?

Summary

The compression flange is destabilising; the tension flange is stabilising
So they move differently, and the section TWISTS — hence lateral-torsional
Twist is resisted by warping (I_w) and St Venant torsion (I_t)
M_cr's root term: I_w/I_z is length-independent, L²GI_t/π²EI_z goes with L²
Short spans are warping-dominated; long spans are St Venant-dominated
An unrestrained 8 m beam here lost 65% of its cross-section resistance
χ_LT is capped at 1/λ̄_LT², because no member exceeds its elastic critical moment
A closed section has an I_t some 900 times larger, and is effectively immune

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint