Module 7 · Lesson 7.2

Buckling length, axis, and what restraint buys

Everything a designer can actually change is in L_cr and I. One restraint in the right place can be worth more than a third more steel.

Why this matters

Once the section is chosen, the only lever left on a column's resistance is the buckling length — and it is a powerful one, because it appears squared. A restraint that costs a small angle and two bolts can be worth more than upgrading the section, and knowing where to put it is a matter of understanding which axis is in trouble.

By the end of this lesson you should be able to

Apply effective-length factors and say where they come from
Identify the governing axis when the two are restrained differently
Calculate the gain from an intermediate restraint
Recognise a balanced column, and why it is the efficient one

Buckling length: what a restraint actually does

N_cr depends on L_cr², so anything that shortens the buckling length is worth a great deal:

End condition	L_cr	Relative N_cr
Fixed–fixed	0.5 L	4.0
Fixed–pinned	0.7 L	2.0
Pinned–pinned	1.0 L	1.0
Fixed–free	2.0 L	0.25

The span from fixed–fixed to fixed–free is a factor of sixteen in load. That is the largest single range of any factor in member design.

The cantilever case is the one people underestimate. A flagpole column buckles at a quarter of the load of the same column pinned at both ends, and if you have assumed a pin at the top that is not there, the error is in the worst direction.

And the two axes need not be the same

This is the part that makes column design interesting. A column in a building is usually restrained by different things about its two axes:

side rails or a floor beam may restrain the minor axis at intervals;
the major axis may be unrestrained over the full storey height.

So the two axes can have quite different buckling lengths, and the governing axis is not necessarily the weak one.

Worked example

One restraint, and a 36% gain

Given

The same 254 × 254 × 73 UC in S355, 4.0 m high, pinned top and bottom
From the previous lesson: N_b,Rd = 2152 kN, governed by the minor axis
A side rail is added at mid-height, restraining the MINOR axis only

Find

The new resistance, and which axis now governs.

Assumptions

The restraint is fully effective about the minor axis and does nothing about the major
Section and curve data as before

Try it

Which axis governs?

Two independent calculations, and the resistance is the smaller. Restrain each axis separately and try to make them equal — a balanced column wastes nothing, and further restraint about the governing axis stops helping the moment it is no longer governing.

Section

Height4.0 m

Minor-axis restraints0

Major-axis restraints0

Steel grade

Squash load N_pl: 3256 kN
Major-axis buckling length: 4.00 m
Minor-axis buckling length: 4.00 m
λ̄ major / minor: 0.47 / 0.80
Curve major / minor: b / c
χ major / minor: 0.896 / 0.661
Major-axis resistance: 2917 kN
Minor-axis resistance: 2152 kN
Design resistance: 2152 kN
Lost to buckling: 34 %
Axes within 5% of each other?: no

The minor axis governs at 2152 kN while the other reaches 2917 kN. There is 26% of unused capacity on the stronger axis — restraining the minor axis would recover some of it.

Things worth trying

Start with the UC at 4 m and no restraints. The minor axis governs at about 2150 kN against 2920 on the major — 26% of the stronger axis is unusable.
Add ONE minor-axis restraint. The resistance jumps to about 2920 kN, a 36% gain, and the MAJOR axis now governs. One side rail and two cleats bought more than upgrading the grade would.
Add a second minor-axis restraint. Nothing happens. The minor axis is no longer governing, so improving it further is wasted — this is what a balanced column means.
Now add a major-axis restraint too. The resistance rises again, because you have started improving the axis that actually governs.
Switch to the SHS at 4 m with no restraints. It carries about 3110 kN against the UC's 2150 — 45% more for 5% more steel, because its material is spread round the perimeter AND it takes a better buckling curve.
Take the height to 12 m and compare the grades. At that slenderness S460 buys almost nothing over S235, because the Euler load contains no f_y. Then go back to 2 m and try again — there the grade is worth nearly its full ratio.

When a higher grade helps, and when it does not

Module 1 promised this result and now it can be quantified. Take the same column and vary only the grade, from S235 to S460 — a strength ratio of 1.96.

Height	λ̄ (S235)	Gain from S235 → S460
1.5 m	0.24	1.86×
4.0 m	0.65	1.54×
10.0 m	1.63	1.11×

A stocky column gets nearly the full 1.96 — it is failing by squashing, and the squash load is proportional to f_y.

A slender column gets 11%, for a 96% increase in material strength. It is failing by buckling, and the Euler load contains no f_y at all.

The grade buys you strength. Buckling is a stiffness problem.

So the lever depends on where the member sits. For a stocky column, upgrade the grade. For a slender one, shorten the buckling length or change the section — and the second of those is worth its own point.

Note

Why hollow sections make such good columns

Compare two members of almost the same weight:

254 × 254 × 73 UC: A = 9172 mm², I_z = 3907 cm⁴
250 × 250 × 10 SHS: A = 9600 mm², I_z = 9232 cm⁴

The hollow section has 2.4 times the minor-axis stiffness for 5% more steel, because its material is distributed around the perimeter rather than concentrated in two flanges.

At 4.0 m pinned in S355 that gives:

UC: 2152 kN · SHS: 3113 kN — 45% more

And the gain comes from two places at once. The SHS has the better λ̄ (0.53 against 0.80) and it takes buckling curve a rather than c, because hot finishing leaves milder residual stresses than rolling a wide-flange shape.

The I section's advantage is elsewhere: it is far easier to connect to, and it is much better in major-axis bending, where its material is exactly where a beam wants it. A column that carries significant moment as well as axial load may still be better as an I section — which is Module 9.

Practice

A 6.0 m column is fixed at both ends. What is its buckling length, in m?

Practice

A column has λ̄ = 0.80 unrestrained. A restraint is added at mid-height about that axis. What is the new λ̄?

Practice

A UC has I_z = 3907 cm⁴ and an SHS of similar area has I_z = 9232 cm⁴. By what factor is the SHS stiffer about the weak axis?

Check yourself

A column's minor-axis resistance is 2150 kN and its major-axis resistance is 2920 kN. What does adding another minor-axis restraint achieve?

Summary

N_cr goes with 1/L_cr², so buckling length is the most powerful lever available
Effective lengths span 0.5L to 2.0L — a factor of SIXTEEN in load
A flagpole carries a quarter of the pinned–pinned load, and the error is unsafe
The two axes can have different buckling lengths, so the weak axis need not govern
One mid-height restraint gained 36% here — more than upgrading the grade would
A BALANCED column, with both axes nearly equal, wastes nothing
A higher grade gives 1.86× on a stocky column and 1.11× on a slender one
A hollow section has 2.4× the weak-axis stiffness for 5% more steel, and a better curve

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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