The flange that is not all there

A plate girder is a deep web with a wide plate welded along the top of it, and the plate is wide because material far from the middle is what carries bending. Widen it and the section modulus rises. Widen it again and the drawing says it rises again.

The steel does not agree. A flange 3 m wide either side of the web, spanning 20 m, carries its full stress only along the strip next to the web. At the free edge, three metres out, the longitudinal stress has fallen to 76.5% of what the web sees — not because that steel is thinner or weaker or badly welded, but because nothing has arrived to stress it.

The argument, stated once #

Bending theory hands every fibre of the flange the same stress because it assumes the cross-section stays plane, and a plane section rotating about the neutral axis puts every point at the same height at the same strain. That is the assumption the whole subject rests on, and it is a statement about kinematics with no mechanism behind it.

The mechanism, when it is looked for, is unavailable. Longitudinal stress cannot appear in the middle of a plate; it has to be delivered, and the only place a flange touches anything is the line where it meets the web. Everything the flange carries came in through that line as shear and travelled sideways as more shear. Shear travelling through a plate strains it, that strain is a longitudinal displacement lagging behind the web’s, and the far parts carry less than the near parts.

The lag is not a defect; it is what the load path costs, and the cost has a length scale. Which length scale is the whole question — and the answer is the one nobody looks at when drawing the section, because the width that is working is fixed by the span, not by the flange.

Which free body, and which equation #

Cut a strip out of the flange, running from the free edge inwards to a plane at distance $y$ from the web, and of length $dx$ along the beam. Six faces. Top, bottom and outer edge are free surfaces. The two ends carry the longitudinal stresses on the cut section, and their difference has to be balanced. That leaves exactly one face available: the vertical plane at $y$, on which the flange is sheared by the part of itself nearer the web.

Writing $\sum F_x = 0$ on that body gives

$$\frac{dF}{dx} = q(y),$$

so the whole longitudinal force outboard of $y$ arrived through the shear flow on that one plane. At the free edge $q = 0$ and nothing is arriving; at the junction with the web, $q$ is the entire flange force per unit length. This is the shear nobody draws, doing the only job available to it.

The second free body is the one that defines the quantity. Take the whole cut face of one overhang, on which the real longitudinal force is $F = t\int_0^b \sigma(y)\,dy$. The effective width is then defined by demanding that a rectangle of uniform stress carry the same force at the same peak:

$$b_{\text{eff}} = \frac{1}{\sigma_{\text{web}}}\int_0^b \sigma(y)\,dy .$$

That is an equal-force condition and nothing else. It is not a rule, not a factor and not a safety allowance; it is an integral, and every rule that produces a number for it is an approximation to that integral.

The shape is assumed; the amount is not #

The profile used here is a parabola measured from the web,

$$\frac{\sigma(y)}{\sigma_{\text{web}}} = 1 - \beta\,\psi(y), \qquad \psi = 2\frac{y}{b} - \left(\frac{y}{b}\right)^2 ,$$

and both of its properties are boundary conditions rather than convenience. $\psi(0) = 0$, because the flange cannot slip past the web it is welded to; $\psi'(b) = 0$, because a free edge carries no shear. The shape is a choice; the amplitude $\beta$ is not, and that is where the model earns its keep. Add $\psi(y)\,U(x)$ to the plane-sections displacement as a second degree of freedom, write down the flange’s strain energy — membrane and shear together — and minimise it. What falls out is

$$U'' - \lambda^2 U = -\tfrac{5}{4}c\,w''' , \qquad \lambda^2 = \frac{5G}{2Eb^2} ,$$

which is worth pausing on twice. The driving term is $w'''$, and the third derivative of the deflected shape is the shear force: shear lag is driven by shear, so it is worst wherever shear is worst rather than wherever moment is. And $\lambda$ carries the material only as the ratio $G/E$, so a stiffer steel lags exactly as much as a softer one — the same insensitivity that makes a stronger steel change nothing about deflection.

For a load of half-wavelength $L$, with $\alpha = \pi/L$,

$$\beta = \frac{5}{4}\cdot\frac{\alpha^2}{\alpha^2 + \lambda^2}, \qquad \frac{b_{\text{eff}}}{b} = 1 - \frac{2}{3}\beta .$$

At the flange in question $\beta = 0.2345$ and the working fraction is 0.844.

What plane sections would have said #

The comparison is worth making explicitly, because the error runs the wrong way from the one a designer expects: the theory is unsafe here, not conservative.

That block is a function of one coordinate, and it has nothing to say about the flange’s width because it came from a hypothesis in which width does not appear: feed it a wider flange and it reports a larger section modulus and a smaller stress, without limit. The real section carries the same total force in a narrower strip, so the stress where the strip is — next to the web, which is also where the weld is and where the fatigue detail is — is $1/0.844 = 1.185$ times what the uniform block reports. An 18.5% underestimate of the peak stress is not a rounding error, and it arrives by way of an assumption that looks like geometry.

The span decides, not the flange #

$\beta$ depends on $b/L$ and on nothing else. That single fact is the essay.

Doubling a flange at fixed span does not double what works; it moves the curve down. At $b/L = 0.05$ the fraction is 0.979 and at $b/L = 0.199$ it is 0.759, so four times the flange buys 3.06 times the working width, and the next doubling buys less.

The consequence for anyone drawing a section is uncomfortable and simple: past about a quarter of the span, extra flange is decoration. A wide box girder on a short span is paid for at full rate and worked at a discount, and no change to the steel or the weld alters it.

The exact ceiling, and the rule that is 0.785 of it #

There is a closed-form answer for the limiting case, and it is old. Take a flange of infinite width fed along its edge by a sinusoidal shear flow, write Airy’s stress function as $\phi = f(y)\sin\alpha x$ with $\nabla^4\phi = 0$, and impose the physical conditions: $\sigma_y = 0$ at the junction, because a thin web cannot push a flange sideways; the delivered shear flow on the same line; and decay to nothing far away. The solution is $f = B\,y\,e^{-\alpha y}$, and

$$\sigma_x(y) \propto (2 - \alpha y)\,e^{-\alpha y}, \qquad b_{\text{eff}} = \int_0^\infty \left(1 - \frac{\alpha y}{2}\right)e^{-\alpha y}\,dy \Big/ \alpha = \frac{1}{2\alpha} = \frac{L}{2\pi} .$$

The integral is exactly $\tfrac12$, so the effective width is exactly $L/2\pi$: 0.1592 of the span, per side, and no flange of any width beats it. On a 20 m span that is 3.183 m — wider than the flange under discussion, which is why the ceiling does not bind here.

Now put the rule beside it. It caps at $L/8 = 0.125L$; the ceiling is $L/2\pi = 0.1592L$; the ratio is

$$\frac{L/8}{L/2\pi} = \frac{2\pi}{8} = 0.7854 ,$$

which is neither a coincidence nor a derivation. A rule sitting at a fixed fraction of the exact answer, with a round number where the elasticity has $2\pi$, is a fit through the family of exact answers with a margin taken off — and a straight line where the exact answer is a curve.

This site has met the same species before. The stagger correction that adds $s^2/4g$ back to a net section was fitted to test data in 1922, stands in for a mechanism rather than a geometry, and has survived every attempt to derive it. Both rules are honest and both are safe. What neither is, is an explanation — and the difference bites at the edges of the data they were fitted to, which is exactly where an unusual span sends a designer looking.

The lag follows the shear, so it moves along the span #

A single sinusoid has the same profile at every section, which is convenient and false. A uniform load is a sum of odd harmonics, each with its own half-wavelength $L/n$ and its own much larger $\beta$: the working fraction is 0.844 for the first harmonic, 0.437 for the third, 0.290 for the fifth. Near a support the moment is small and the short harmonics carry a larger share of it, so the local effective width there is smaller.

The free body has not changed; the loading has. Each harmonic’s flange force follows its moment amplitude, its peak stress is that force over its own effective width, and the local answer is the force-weighted harmonic mean of the harmonics’ widths. Codes reproduce the shape of that curve by shortening the effective span near supports — the right behaviour, reached without the sum.

One consequence of the same arithmetic is genuinely surprising. Move the load to a single point at mid-span and the effective width there drops to 0.629, which is the support value under a uniform load to every digit computed. Both cases weight the harmonics as $1/n^2$: the uniform load’s moment amplitudes go as $1/n^3$ and the limit at the support returns a factor of $n$, while a central point load’s go as $1/n^2$ and $\sin(n\pi/2)$ is $\pm1$ at mid-span. So the worst shear lag under a central point load sits at exactly the section where the moment is greatest.

The other extreme, which is most beams #

The whole effect disappears for an ordinary rolled section, and the arithmetic says why.

A universal beam has an overhang of perhaps 100 mm on a span of 6 m, giving $b/L$ under 0.02, and no measurement will find the lag. So the effect is invisible in most of structural engineering and unavoidable in the parts of it that build wide: box girders, composite decks, and slabs asked to act as the flange of the beam beneath them.

The same two words, naming something else #

There is a trap in the vocabulary, and it is worth walking into on purpose.

The angle that uses half of itself is unmistakably a relative: force enters part of a section, needs distance to reach the rest, and until it has, the rest is under-stressed. Both are Saint-Venant’s principle charging rent.

They are still different quantities, and the difference is which length governs. In the connection it is the length of the joint — add bolts and the discount vanishes, and the member’s span never enters. In the flange it is the span, and no amount of welding along the junction changes anything, because the junction was never the limitation. The two rules — $U = 1 - \bar{x}/L_c$ and $b_{\text{eff}} = \min(L/8, b)$ — share not one symbol, and neither can be obtained from the other.

The word has a third owner. Effective width in a plate that has rippled is the strip either side of a stiffened edge that still carries load after buckling, and that one is about stability, not shear at all. Three phenomena, two words each, and a claim about any of them is worthless unless it says which is meant.

Where the model stops #

The parabola is a shape, not a solution. Its boundary conditions are right and its amplitude is derived, but a one-term profile cannot represent a stress that decays to nothing. As $b/L$ grows the ratio flattens at 1/6 rather than falling away, so the model is good to about $b/L = 0.22$ and optimistic beyond it, where the exact $L/2\pi$ ceiling governs instead. Every sweep here stops inside that range, and the ceiling is drawn alongside so the crossing is visible rather than hidden.

Everything here is elastic. Once the strip next to the web yields it sheds stress outward, the lag reduces, and the ultimate capacity of a wide-flanged section is much closer to the full-width answer. Shear lag is a serviceability and fatigue phenomenon far more than a strength one — the same distinction that separates stiffness from strength.

The flange is a plate in plane stress with no out-of-plane life of its own. A real compression flange that wide is also a candidate for local buckling, and the two effects share a geometry and interact.

Simply supported, single span. At a continuous support the moment reverses and the harmonic content differs again, so lag is worse over an internal support than the curve here suggests.

Symmetry has been assumed. An asymmetric load twists a box girder, torsion adds its own longitudinal distribution across the flange, and where the two superpose is settled by the shear centre.

What none of these pictures can show. Every figure here is a distribution across the flange or along the beam, drawn from a solution for the flange alone with the web as a boundary condition. Two things live outside them. The first is the sign reversal in the exact solution: $(2 - \alpha y)e^{-\alpha y}$ changes sign at $\alpha y = 2$, which for the first harmonic is $2L/\pi = 12.7$ m from the web — beyond any flange that exists. For the fifth harmonic it is $2L/5\pi = 2.55$ m, inside the 3 m flange, so near a support the far edge is not merely lagging but in the opposite stress to the part next to the web. No figure here draws it, because none of them draws a single harmonic alone. The second is the shear stress in the flange, which is the entire cause and appears nowhere: the pictures show the consequence and leave the mechanism off the page.

The same question, asked of a truss #

Asking what is actually carrying this? and answering with a distribution rather than a yes has a companion in this collection. Which member moved the roof ranks a truss’s members by $F f L / EA$ and finds that a member carrying a full panel load can be worth nothing to stiffen, while an unremarkable chord is worth everything. The flange is the same finding in a continuum.

Both are warnings about one reflex: adding material where the drawing says the force is largest is not the same as adding material where it will be used, and the stiffest path takes the load whether or not that path was the intended one.

History #

Theodore von Kármán named the quantity mittragende Breite — the co-carrying width — in 1924, and solved the infinite-flange problem with the Airy function that gives $L/2\pi$. The energy method used here, with an assumed profile and a derived amplitude, is Eric Reissner’s from 1946, written for box beams under the wartime pressure that produced most of thin-walled analysis.

The order of events is the striking part. The exact answer for the limiting case came first, twenty years before the practical approximation, and the rule that ended up in the codes descends from neither directly: it is a round-number fit sitting at $2\pi/8$ of a ceiling that had been known for half a century when it was written. The derivation was available the whole time. What was not available, before computers, was any way to use it on a real bridge.

The ladder from here #

Later rungs on this anchor: the energy derivation in full, with the second freedom written out. The exact Airy solution and its sign reversal. Effective width over a continuous support, where the moment changes sign. Negative shear lag, and the loadings that cause it. Shear lag in composite decks, where the flange is concrete and the connection is discrete. Effective width after the strip next to the web has yielded. Shear lag interacting with plate buckling in a wide compression flange. Effective width in the torsion and distortion of a box section. And the finite-element answer, which needs none of this and reports a number nobody can check by inspection.