Module 10 · Lesson 10.2

Stiffeners, local effects, and members made of members

Where to put stiffeners and why, a failure that gets worse when you add flange steel, and why a laced column has a ceiling its chords cannot lift.

Why this matters

The first lesson treated the web as a single panel. A real girder is a set of panels separated by stiffeners, and the spacing of those stiffeners is a design decision with a cost on both sides. This lesson deals with that, with a local failure that behaves in exactly the opposite way to intuition, and with the other family of assembled member — the laced or battened column, where a single derivable result explains almost everything that matters.

By the end of this lesson you should be able to

Say what intermediate stiffeners buy and where they stop paying
Recognise flange-induced buckling and its counter-intuitive dependence
Explain why a built-up member's critical load is capped by its shear stiffness
Read the effect of lacing angle, and compare lacing with battening

What you should already know

Shear buckling and tension-field action (previous lesson)
The Euler load and its 1/L² dependence (Module 7)
Second-order amplification and bow imperfections (Module 9)

Stiffeners, and where they stop paying

An intermediate transverse stiffener does two things: it shortens the panel, which raises k_τ and so raises τ_cr; and it provides the post that the tension field pulls against.

The first effect has a clear limit. k_τ is largest when the panel is square and falls away as it lengthens, approaching a floor of about 5.34 for a long panel. So:

a/h_w	k_τ	τ_cr (N/mm²)	V_b,Rd (kN)
1.0	9.34	78.8	1821
2.0	6.34	53.5	1585
3.0	5.78	48.8	1532

Going from 3.0 m to 1.5 m spacing — twice as many stiffeners — bought 15%. Going from 4.5 m to 3.0 m bought 3%. Beyond about a/h_w = 3 there is almost nothing left to gain, because k_τ has nearly reached its floor.

Stiffeners are worth adding where they make the panels close to square, and are close to pointless where they do not.

Compare that with thickening the web, which took the resistance from 1585 kN to 5214 kN. On a girder, web thickness is usually the more effective lever, and it is worth pricing both.

Flange-induced buckling: where more flange makes things worse

A girder that has deflected has a compression flange that is curved. A curved member carrying a compressive force has a radial component pushing it towards the centre of curvature — that is, into the web.

The web has to hold it. A sufficiently slender web cannot, and the flange buckles into it. The failure is sudden and has nothing to do with the bending resistance or the shear check.

The limit takes the form

h_w/t_w ≤ k · (E/f_yf) · √(A_w/A_fc)

Read the last term carefully. A_fc is the compression flange area, and it is in the denominator.

For the worked girder, with 400 × 25 flanges, the limit is 398 and h_w/t_w is 150 — comfortable. Change to 600 × 40 flanges and the limit falls to 257. The girder just got stronger in bending and closer to a local failure.

This is the one place in steel design where adding flange material makes a check harder to satisfy. The remedy is a thicker web or a longitudinal stiffener — not a bigger flange.

From first principles

Why a built-up member's critical load is capped by its lacing

We want to show: Show that 1/N_cr = 1/N_cr,ideal + 1/S_v, and see what follows — in particular that N_cr can never exceed S_v however heavy the chords are.

A solid column bends when it buckles. A laced column bends AND shears, because the lacing between the chords is not rigid — the panels rack. The two deflections happen at the same time under the same load, so they add. And when two flexibilities add, the stiffer one stops mattering: the assembly is limited by whichever is more flexible.

Try it

Chords, lacing and the ceiling

A laced column with chords at 500 mm centres. The bars are the bending capacity, the lacing's shear stiffness, and what the member actually reaches. Try to raise the third by changing only the first.

Chord area5,000 mm²

Diagonal area1,200 mm²

Panel spacing600 mm

System length12.0 m

Chord area, each: 5000 mm²
Built-up I_eff: 62500 cm⁴
Diagonal angle: 40°
Diagonal length d: 781 mm
Lacing shear stiffness S_v: 158682 kN
If lacing were rigid: 8996 kN
Actual critical load N_cr: 8513 kN
Lost to shear flexibility: 5.4 %
Lacing force at 2000 kN: 16.4 kN

Ignoring shear the member would reach 8996 kN; with the lacing's shear stiffness of 158682 kN it reaches 8513 kN — 5% lost. Note the ceiling: N_cr can never exceed S_v, so if the lacing is weak no amount of chord material will help.

Things worth trying

Start at the defaults: 5000 mm² chords, 1200 mm² diagonals at 600 mm spacing, 12 m long. Only 5.4% is lost — this is a well-detailed member and shear flexibility is a rounding error.
Now take the diagonal area down to 100 mm² and the panel spacing out to 2400 mm. The lacing ceiling collapses and most of the member's capacity goes with it.
Leave the lacing there and take the chord area to its maximum. The top bar grows steadily and the bottom one barely moves — it cannot pass the middle bar, whatever you spend.
That middle bar is the whole point. N_cr is the harmonic combination of the other two, so it is always below both and is dominated by the smaller.
Put the diagonals back to 1200 mm² and change only the panel spacing, from 2400 mm down to 300. The diagonals get steeper, d gets shorter, and S_v rises with the cube of that change.
Watch the lacing force as you lengthen the member. It rises as the member approaches its critical load, because the bow imperfection is amplified — the same mechanism as Module 9, applied to a member's own connecting elements.
Set the length to 20 m with weak lacing and watch what happens to the lacing force. A member close to its critical load demands more of the very elements whose weakness put it there.

Worked example

A 12 m laced column, and an attempt to strengthen it

Given

Two chords of 5000 mm² each, at 500 mm centres, over a 12.0 m system length
Single lacing: diagonals of 1200 mm² at 600 mm panel spacing, giving a 40° angle
N_Ed = 2000 kN
Compare with a poorly detailed alternative whose lacing gives S_v = 3000 kN

Find

The critical load, and whether heavier chords can rescue the poor detail.

Assumptions

I_eff for a laced member takes the chords as point areas and ignores their own second moments — a simplified educational assumption, and the reason a battened member is treated differently below
The bow imperfection divisor is calibrated and unverified here
Chord buckling between lacing points is a separate check and is not covered

Lacing or battens?

Both connect the chords; they work quite differently.

Lacing carries the racking shear in the diagonals as axial force. Battening carries it by bending of the battens and the chords together, as a Vierendeel frame.

It is tempting to conclude that lacing is always stiffer. It is not that simple, and this is a case where the intuitive answer is wrong often enough to be worth checking.

	Lacing	Battening
Action	Axial in the diagonals	Bending of battens and chords
Falls off with	The CUBE of the diagonal length	The SQUARE of the batten spacing
Improved by	Steeper diagonals, more diagonal area	Deeper battens, closer spacing
Chord inertia counts?	Conventionally no	Yes
Chosen for	Efficiency on tall or heavy members	Buildability, clear faces, appearance

On the worked column with close, stiff battens, the battened arrangement actually reaches a slightly higher critical load — 0.95 times the laced one, so within 5% — because it gets to count the chords' own second moments. Spread the battens to 1.8 m and make them shallower and the laced member wins by 1.8×.

Which arrangement is stiffer depends on the geometry, not on the type. What is reliable is how each one degrades, and that is what to reason from.

Practice

A built-up member has N_cr,ideal = 9000 kN and its lacing gives S_v = 3000 kN. What is N_cr, in kN?

Practice

The chords of that member are tripled in area, so N_cr,ideal becomes 27 000 kN. The lacing is unchanged. What is N_cr now, in kN?

Practice

Lacing diagonals of 1200 mm² sit at 600 mm panel spacing with chords at 500 mm centres, in two planes. What is S_v, in kN? Take E = 210 000 N/mm² and S_v = 2EA_d a h₀²/d³.

Practice

A girder web panel has a/h_w = 2.0 and k_τ = 6.34. If the stiffeners are moved to give a/h_w = 1.0, k_τ becomes 9.34. By what percentage does τ_cr increase?

Check yourself

Why does increasing the compression flange area make flange-induced buckling HARDER to satisfy?

Summary

Intermediate stiffeners raise k_τ, but the gain stops beyond about a/h_w = 3
Halving the spacing from 3.0 to 1.5 m bought 15%; doubling the web thickness bought 229%
Longitudinal stiffeners address bending of the web, not shear — a different problem
Flange-induced buckling gets WORSE with more flange area, because A_fc is in the denominator
1/N_cr = 1/N_cr,ideal + 1/S_v — derived, and the flexibilities add
So N_cr can never exceed S_v: weak lacing caps a member however heavy the chords
Tripling the chords of a poorly laced member moved N_cr from 2250 to 2700 kN
Lacing degrades with the cube of diagonal length; battening with the square of batten spacing

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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