Structural Engineering & Statics
where the load goes
Concept

Web — where it appears

The part of a section between its flanges, which carries almost all of the shear and about a quarter of the bending strength. It carries almost all the shear and contributes little to the bending, which is why it is the plate that is thinned, opened and stiffened rather than the flange.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

Two curves climbing together, and the one that catches up first. A 6 m member tapering from 200 to 600 mm, with the moment it carries and the moment it can carry drawn on the same scale below it. The demand rises linearly and the capacity as the square of the depth, so the gap between them closes and then opens again. It is narrowest at 3.00 m from the free end, where the member is 400 mm deep and 79% used, against 70% at the root where the moment is largest.

The section that changes along the span

A prismatic beam is checked where the moment is largest, and everyone knows where that is. A tapered one is not, because the capacity is moving too — and for a cantilever with a load at its tip the governing station is exactly where the depth has doubled, with no length, no load and no material in the answer.

sections · Tapered member Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

sections · Shear moment interaction Pull it along the girder and it just unfolds. One period of a 30° trapezoidal corrugation, 300 mm of flat and 260 mm of incline, and the same period pulled along the girder's axis. The fold opens by bending the inclined panels out of the web's own plane, so the axial flexibility contains the plate's t³ where a flat web's would contain t — and the effective modulus that comes back from solving the cell as a frame is 222 N/mm², which is 10.6 parts in ten thousand of the steel's 210 GPa. A web with a thousandth of the stiffness carries a thousandth of the stress, which is why the flanges of a corrugated girder carry the whole moment and why the section has 9 per cent less second moment than the flat-webbed girder it replaces. The fold buys freedom from stiffeners and pays for it here.

The web that carries no bending

A corrugated web needs no stiffeners, because the folds give it in one direction a depth it does not have in its thickness. In the other direction the same folds make it an accordion — and a web that cannot be stretched cannot carry a bending stress at all.

sections · Corrugated web A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel.

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

sections · Shear area A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam.

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

stability · Flange induced A buckled panel is a truss that nobody drew. A 1500 × 2000 panel of 8 mm web, at d/t = 188. It buckles in shear at 41.0 N/mm², which is 492 kN — and it then carries 1187 kN, 2.41 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 18.5° with a membrane stress of 348 N/mm² over a width of 788 mm, and it pulls on the flange at 280.2 N per millimetre of its length. A web that never buckled at all would have reached 2460 kN, so the panel ends at 48% of a stocky web's capacity on a fraction of its steel.

The tension has to pull on something

A buckled web carries its shear on a diagonal band of membrane tension, and the band pulls sideways on the flanges and stiffeners that bound it. That pull is the design output nobody plots — it runs from 72 to 603 newtons per millimetre across ordinary panel proportions, it is largest exactly where the panel is most efficient, and at the end of the girder there is nothing beyond to take it.

stability · Tension field What a torsional brace buys, and the stiffness it takes. The critical moment of an 8 m beam against the stiffness of a single rotational restraint at midspan — a cross-frame or a stiffened connection to a secondary beam, resisting the twist rather than the sideways movement. It climbs from an unbraced 143 kNm to the same two-half-wave plateau of 447 a lateral brace reaches, and gets to 99% of it at 253 kNm/rad. Nothing in the calculation refers to a height, which is the difference that matters: a torsional brace cannot be put on the wrong flange because it is not attached to a flange in the sense the lateral one is.

Twice the moment, four times the brace

A lateral brace has to be on the right flange and its demand is very nearly linear in the load. A torsional brace has no flange to be wrong about, and its demand is exactly quadratic — so the restraint that is indifferent to where it is attached is the one that gets expensive fastest.

stability · Beam bracing The stub column a bearing stiffener makes, in plan. A plan through the girder at the bearing. The 8 mm web runs across; the pair of 100 × 12 stiffeners stands off it; and the shaded strip of web either side — 15ε t_w, or 98 mm each way — is the width that buckles with the stiffeners rather than independently of them. Together they are an area of 3962 mm² with a second moment of 9.01·10⁶ mm⁴ about the web's centreline, a radius of gyration of 48 mm over a buckling length of 900 mm — 0.75 of the depth, because the flanges hold the ends. That is a slenderness of 0.25, at which the column curve returns 0.98: the stub column reaches 98 per cent of its squash load, and the section's own strength is very nearly the whole answer.

A column nine hundred millimetres long

The patch-load check asks how much of a web a flange can spread a wheel over, and answers in a plate-buckling reduction that throws seven tenths of it away. A pair of stiffeners does not improve that answer. It replaces the question with a different one, from a different family, with a different failure in it.

stability · Patch loading

Named alongside it

The objects these essays reach for when they reach for this one.

StiffenerFlangePlate bucklingPlate girderLoad pathPlastic hingeShear areaSlendernessFree bodyLocal bucklingSecond moment of areaShear buckling

All concepts