Structural Steel Design / Beams and lateral-torsional buckling

Restraint, moment shape, and where the load sits

Three things change the answer. One is free, one is decided by the structure, and one is a detailing choice most people never think about.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

A steel-design reviewer must check the first-generation calculation profiles, second-order analysis assumptions, connection rules and execution references against the applicable standards and amendments.
Check torsional/U-frame restraint, compression-flange restraint, brace stiffness and load height assumptions. End restraints and segment moment distributions require independent stability assessment.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

The 8 m beam of the last lesson lost two thirds of its capacity. All three of the things that would recover it are available to a designer, and one of them — where the load is applied relative to the shear centre — is a decision usually made without any thought at all. It is worth 26% on this beam.

By the end of this lesson you should be able to

Calculate what a lateral restraint buys, and when it stops buying
Explain why uniform moment is the worst case
Quantify the load-height effect and say which way it goes
State what a restraint must physically do to count

Restraint: the largest lever, and it saturates

M_cr depends on the length between points of lateral restraint, not on the span. So a restraint at midspan halves the length that matters.

For the 8 m beam of the last lesson, restraining the compression flange gives:

Restraints	Segment	λ̄_LT	χ_LT	M_b,Rd	Gain
0	8.00 m	1.69	0.351	290 kNm	—
1	4.00 m	0.96	0.727	600 kNm	2.07×
2	2.67 m	0.66	0.890	735 kNm	2.54×
3	2.00 m	0.50	0.960	793 kNm	2.74×
5	1.33 m	0.34	1.000	826 kNm	2.85×
7	1.00 m	0.25	1.000	826 kNm	2.85×

One restraint at midspan more than doubles the resistance. That is a larger gain than any change of section or grade would give, and it costs a tie and two cleats.

And then it saturates. Once χ_LT reaches 1 the beam has its full cross-section resistance and there is nothing left to recover — the fifth restraint achieves the last 4%, and the seventh achieves nothing at all. Restraint is worth adding until χ_LT approaches 1, and pointless afterwards.

Take care

What a restraint has to do to count

A member that is called a restraint on a drawing is only a restraint if it does three things.

It must reach the COMPRESSION flange. In sagging that is the top flange; in the hogging region over a support it is the bottom one. A top-flange restraint alone does not directly restrain the lower compression flange; any torsional or U-frame restraint it provides must be assessed — and that is exactly where a continuous beam is most vulnerable, because the bottom flange there is both in compression and usually unrestrained.

It must be held by something that cannot move. A purlin restrains a rafter only if the purlin itself is held — which usually means roof bracing. A line of purlins with nothing bracing them can all lean together, and a row of members leaning in unison restrains nothing.

It must be stiff and strong enough. The force is small, of the order of a few percent of the flange force, but it is not zero. There must be a load path for it, and that load path must reach the ground.

The commonest LTB error in practice is not a miscalculation. It is counting a restraint that is not one.

Worked example

Three ways to recover the same beam

Given

The 8 m beam from the previous lesson: M_c,Rd = 826 kNm, unrestrained M_b,Rd = 290 kNm
The load is a UDL, so the moment is not uniform
It is a floor beam, so the load arrives on the top flange

Find

The effect of each of the three levers, separately.

Assumptions

C₁ = 1.13 for a UDL and C₂ = 0.45 — both calibrated and unverified here
The restrained case assumes the restraint is effective, per the callout above

Try it

Three levers on the same beam

The 8 m beam from the worked example, at 290 kNm against a cross-section resistance of 826. Restraint, moment shape and load height all move it — and only one of them is under full control.

Lateral restraints0

Moment shape

Load applied at

Segment length: 8.00 m
Moment factor C₁: 1.00
Load height z_g: 0 mm
Critical moment M_cr: 290 kNm
LTB slenderness λ̄_LT: 1.688
Reduction χ_LT: 0.351
Design resistance M_b,Rd: 290 kNm
Cross-section resistance: 826 kNm
Gain over the reference: 1.00 ×
Still lost to LTB: 65 %

M_b,Rd = 290 kNm, which is 1.00 times the unrestrained reference case and 65% short of the cross-section resistance. There is still 536 kNm to recover.

Things worth trying

Start from the reference case — no restraints, uniform moment, load at the shear centre. 290 kNm against a cross-section resistance of 826: two thirds gone.
Add ONE restraint. The resistance more than doubles, to about 600 kNm. That is a bigger gain than any change of section or grade, for a tie and two cleats.
Keep adding restraints. The gains shrink fast, and by five χ_LT reaches 1 and they stop entirely. Restraint is worth adding until χ_LT approaches 1 and is wasted afterwards.
Go back to zero restraints and switch the moment shape to double curvature. C₁ = 2.6 and the resistance nearly doubles — because with equal and opposite end moments only a short length near each end is heavily compressed on the same flange.
Now set the moment shape to UDL and move the load between the three positions. Top flange to bottom flange is a factor of about 1.7 — from a detailing arrangement that most people never calculate.
Try all three levers together: one restraint, UDL, load at the bottom flange. The three multiply, and the beam becomes a completely different member from the reference case despite identical section, span and total load.

Practice

An 8 m unrestrained beam has M_b,Rd = 290 kNm. One restraint is added at midspan, taking χ_LT from 0.351 to 0.727. What is the new M_b,Rd, in kNm? Take M_c,Rd = 826 kNm.

Practice

For a UDL, C₁ = 1.13. If M_cr under uniform moment is 290 kNm, what is it under the UDL?

Practice

Top-flange loading gives M_b,Rd = 239 kNm and bottom-flange loading 408 kNm on the same beam. By what factor do they differ?

Check yourself

A beam has χ_LT = 0.96 with three restraints. What does adding a fourth achieve?

Summary

M_cr depends on the length between RESTRAINTS, not the span
One restraint at midspan more than doubled the resistance of the 8 m beam
Gains saturate once χ_LT reaches 1 — beyond that, nothing is available
Uniform moment is the worst case; C₁ = 1 is the reference and everything else is better
Top-flange loading is destabilising and cost 26%; bottom-flange loading gains 37%
Top versus bottom is a factor of 1.71 on this beam, from a detailing arrangement
The three levers multiply — a restrained UDL beam is a different member entirely
A restraint must reach the COMPRESSION flange, which changes side over a support