A quarter of the weld takes half the pull

Every weld group examined in the essays before this one has been welded to something that does not move. The corner that is not the worst point and the radius rule, and where it fails found the worst point of a group under an eccentric load; two fasteners that never arrive together set a weld beside a bolt; strongest across, and first to crack across took the group into fatigue. In all of them the plates either side of the weld were rigid, so the only question was how the load divided among the weld’s own points. The last of those essays named what that assumption leaves out: a plate welded across an unstiffened flange, where “the flange bends away from the load except near the web, and the weld next to the web carries far more than its share.”

That is the commonest welded moment joint in a steel frame. A beam’s tension flange, or a plate standing in for it, is welded straight on to the face of an I-section column, with no stiffener inside the column behind it. The column flange is a plate held along one line, the web, and free at its two tips. What the beam flange pulls on is therefore stiff in the middle and soft at the sides.

The joint, and the width the rule gives it #

EN 1993-1-8 deals with it in a single clause. The plate is given an effective width

$$b_{eff} = t_w + 2s + 7k\,t_f, \qquad k = \frac{t_f}{t_p}\,\frac{f_{y,f}}{f_{y,p}} \le 1,$$

where $t_w$ is the column web, $s$ its root radius for a rolled section, $t_f$ the column flange and $t_p$ the plate. The plate’s resistance is $b_{eff}\,t_p\,f_{y,p}$. If $b_{eff}$ falls below $(f_{y,p}/f_{u,p})\,b_p$ the joint has to be stiffened, and whatever the width comes out at, the weld connecting the plate to the flange has to be as strong per millimetre as the plate itself.

Take a 300 mm column flange 19 mm thick, the flange of an ordinary heavy rolled section with an 11 mm web and 27 mm root radii, and a 250 × 20 mm plate, both in S355. Then $k = 0.95$ and

$$b_{eff} = 11 + 2 \times 27 + 7 \times 0.95 \times 19 = 191 \text{ mm},$$

76 per cent of the plate. The joint passes the stiffener check, since $355/490 \times 250 = 181$ mm, and resists $191 \times 20 \times 355 = 1{,}359$ kN.

The question here is what that width describes. It is tempting to read it as the share of the weld that does the work, with the rest of the plate idling, the way the flange that is not all there reads a beam flange’s effective width as the part that carries the bending stress. Drawn against the force the weld actually carries while the joint is elastic, it is plainly not that.

A flange solved as a plate #

The calculation behind the picture treats both plates as plates. The beam flange is a membrane in plane stress, pulled at its far end by a rigid grip 250 mm from the weld. Its welded edge is held in its own plane by the column flange. Over the web and the two root radii, 65 mm in all, the edge is held still, the web taken as rigid. Over the rest, the edge rides on the column flange, which bends.

The column flange is a plate clamped along the toe of the root radius, free at its tip and indefinitely long up and down the column. Its stiffness against a line of forces across it follows from a Fourier transform along the column. Each wavelength turns the plate’s equation into an ordinary one across the outstand, solved by cubic elements and checked against the exact solution of that equation; at the longest wavelengths it is a cantilever, and the transform puts the wavelengths back together. The plate’s force is spread up the column over its own thickness and the two weld legs. The column flange’s stiffness then becomes a set of springs on the plate’s welded edge, coupled to each other, and the plate is solved with them in place.

Everything here is linear and elastic, so the shares do not depend on the size of the force. The web is held rigid, which slightly overstates what goes into it. And nothing else is on the column — no second plate, no axial stress, no stiffener further up — so the flange is as soft as it can be.

Forty-seven per cent over a quarter of the weld #

Over the 65 mm of web and root radii, a quarter of the plate’s 250 mm, the weld carries 47 per cent of the force. The force per millimetre is highest at the toe of each root radius, where the clamped flange begins: 2.67 times the uniform value $P/b_p$. It dips between the two toes, where the plate itself spreads the load a little, and falls away along the outstand, to almost nothing at the plate’s edges. The last few millimetres carry slightly less than nothing — the plate’s corners press, very lightly, on a flange that has bent further than they have.

The rule’s width, drawn as a uniform block, is nothing like this distribution. It asks the weld to carry 1.31 times the uniform value across 191 mm. The elastic joint carries much more than that over the middle and much less at the sides. The equivalent elastic width — the plate’s force divided by the average force per millimetre over the web — is 69 mm, little more than the web and its radii. Divided by the peak instead, it is 47 mm.

The same proportions hold on a lighter joint: a 180 × 15 mm plate on a 200 × 11 mm flange sends 51 per cent of its force through 45 mm of web and radii, peaks at 3.13 times uniform, and fits an elastic width of 44 mm against the rule’s 101. A thinner flange is a softer spring, so less of the force is willing to travel out along it.

The surprise is not that the weld over the web is overloaded; how much of the plate is bending found the same stiff-centre, soft-edge pattern in a bolted tee, and the weld group essays named it as the next question. It is the size of the gap. The rule is not an approximation to the elastic share that happens to be a little generous. It is nearly three times wider, and every word of the essays that found a weld group’s worst point assumed a rigid backing that this joint does not have.

The flange yields before the rule has been asked for a third #

If the rule’s width is not the elastic width, the joint has to get from one to the other by yielding. The elastic solution says which part yields first, and when.

The flange goes first. Its root, along the toe of the radius opposite the plate, carries a bending moment per millimetre that reaches the elastic limit $f_y t_f^2/6$ when the plate pulls 409 kN. The plate yields next, at the toe where its force peaks, at 664 kN. The rule’s resistance is 1,359 kN, 3.3 times the first yield and twice the plate’s.

Between 409 kN and 1,359 kN the joint is a mechanism in the making. The flange’s root and the lines fanning out from it turn into yield lines, the flange’s outstand bends further and so takes more of the plate’s force, and the plate over the web, already yielding, stretches to let it. The rule’s $7k\,t_f$ is the outstand’s share once those lines have formed. It is a statement about the flange’s plastic bending resistance, $f_{y,f}t_f^2/4$ per millimetre, set against the plate’s tensile resistance, $f_{y,p}t_p$ per millimetre: their ratio is $t_f^2 f_{y,f}/(4 t_p f_{y,p}) = k\,t_f/4$, and the 7 is the multiplier a yield-line mechanism and tests put on it.

Read that way, the clause is consistent and the elastic picture does not contradict it. It only says what the clause takes for granted: a joint that can travel from the first yield to three times that force without anything in it breaking.

What the full-strength weld is for #

The weird-looking requirement in the clause — that the weld be as strong per millimetre as the plate, whatever the effective width — is what makes that travel possible.

Suppose the joint stayed elastic all the way to the rule’s 1,359 kN. The weld at the toe of the root radius would be carrying $2.67 \times 1{,}359/250 = 14.5$ kN per millimetre. The 20 mm plate itself can carry $20 \times 355 = 7.1$ kN per millimetre. So at the rule’s resistance the elastic demand on the weld at the web is about twice what the plate beside it can transmit. Something there has to give, and there are only two candidates: the plate, which yields and stretches, or the weld, which yields and tears.

If the weld is weaker than the plate, the weld gives. A fillet weld loaded across its length is the strongest direction for its static check, as the weld that is stronger across than along found, and also the least ductile: it fractures after a deformation of a small fraction of its leg, about half a millimetre on an ordinary weld. The flange’s outstand needs the plate over the web to keep stretching while the yield lines form. A weld that tears over the web first takes the 47 per cent with it, and the outstand, which never reached its yield lines, is handed the whole force at once.

If the weld is stronger than the plate, the plate gives. A plate in tension yields over a gauge length of the order of its own thickness and has strains of tens of per cent to spend before it necks — millimetres of stretch on a 20 mm plate. That is the deformation the outstand needs, supplied by the one component in the joint that has it to spare. The full-strength weld is not a margin on strength; it is a decision about which component does the stretching. The property that appears in none of the equations made the general point that ductility is assumed everywhere and calculated almost nowhere. This clause is one of the few places where the assumption is written into a rule, as a strength requirement on something else.

The rule grows with the flange’s square, the joint barely at all #

The two widths respond to the geometry differently, and the difference shows what each is measuring.

The rule’s width grows with the square of the flange’s thickness while the flange is thinner than the plate, since $k\,t_f = t_f^2/t_p$, and linearly once $k$ reaches one at 20 mm. From a 10 mm flange to a 32 mm one it goes from 100 mm to the whole 250 mm of plate. The elastic width goes from 54 to 90 mm. A flange’s plastic resistance grows as its thickness squared; its elastic stiffness grows as the cube, but the stiffness that matters here is the flange’s compared with the plate’s own ability to spread load along its edge, and that comparison moves slowly. A flange three times thicker spreads the force less than twice as far.

The plate’s thickness pulls the two widths the opposite way.

While the plate is thinner than the flange, $k$ is capped at one and the rule’s width sits at 198 mm. Once the plate is the thicker, $k = t_f/t_p$ and the width falls, to 127 mm for a 41 mm plate. Elastically the plate’s thickness hardly matters: 80 mm of width for an 8 mm plate, 64 mm for a 41 mm one. A thick plate is stiffer in its own plane and so slightly more willing to drive its force into the stiff part of the flange, but the flange decides. In the rule the plate’s thickness is a strength that the flange has to match; in the elastic joint it is almost irrelevant.

A wider plate is not carried any wider #

The most direct test of what the effective width means is to change the one dimension it does not contain.

From a 100 mm plate to one nearly the full 300 mm of the flange, the elastic width grows by 18 mm. Every millimetre of plate added at the sides lands on flange that is bending away from it, and carries almost nothing. The rule agrees, in its own terms: $b_{eff}$ contains no $b_p$, so the joint’s resistance is the same 1,359 kN whether the plate is 200 or 300 mm wide. And the stiffener check, $b_{eff} \ge (f_y/f_u)\,b_p$, which looks like a ductility check on the plate, is exactly the statement that the plate must not be so wide that it could carry far more than the flange will let it. Past 264 mm of plate on this flange, it is.

That check deserves a second look. The plate’s own strength rises with its width; the flange’s does not. A plate wider than $b_{eff}\,f_u/f_y$ could pull hard enough for its middle to reach its ultimate strength while its sides, riding on soft flange, are still below yield. The plate would then tear over the web before the flange’s yield lines were complete. The clause stiffens the joint rather than trust the plate’s ductility that far.

A stiff middle the column itself does not see #

None of this is visible in the column. A designer checking the column for axial load and bending sees a flange 300 mm wide working at its full section. The beam connection sees a flange that, for the first third of its load, is effectively 69 mm wide. Both are right, for different loads.

This is the same shape of answer as the flange works least where the shear is largest, where a box girder’s flange over a support lags behind its web and works over a fraction of its width, and as the angle that uses half of itself, where a bolted angle’s outstanding leg carries a fraction of the load it would carry if every fibre stretched alike. In all three the force has to cross a stiffness that is not uniform, and it crowds into the stiff part. The difference is the direction it crosses. In shear lag the stiff part is in the force’s own plane. Here it is the column’s web, at right angles to the flange, and the soft part is the flange bending out of its plane. It is the one case in the three where the weak direction is bending, and so the one where the ratio is largest.

It also connects to a failure from another part of the same joint. The plate’s force, crowded over the web, pulls the column flange through its thickness there. The direction a plate was never tested in is about exactly that: lamellar tearing, a through-thickness failure of rolled plate, appears under welded attachments where the force is concentrated. An elastic share of 2.67 times uniform over the web is a through-thickness stress 2.67 times what a uniform calculation would give, in the one place the plate’s through-thickness ductility is least.

What the picture leaves out #

The figures are elastic, and the clause they are set against is plastic, so the most important thing they cannot show is the path between. They say where yielding begins and how far the rule is from it. They do not trace the flange’s yield lines forming or confirm the 7 in $7k\,t_f$; that number comes from tests and a mechanism, and the elastic solution can only show that it is a plastic number.

The web is held rigid. A real web is stiff in its own plane but not rigid, and the root radius spreads the force into it, so the 47 per cent is a slight overstatement and the peak at the toe more so. A mesh twice as fine moves the share by less than a percentage point. That makes the share a property of the model; it does not make the rigid web a property of the column.

The column carries nothing else. In a frame it carries axial load, which stiffens nothing here but reduces the flange’s yield moment; a compression flange from the beam’s other side, which pushes the web where this plate pulls the flange; and often a second beam on the opposite face, which pulls the flange from the other side over the same web. Each changes the numbers, and none changes the shape: stiff over the web, soft at the sides, and a rule that describes the end of the joint’s journey rather than its start.

The weld is modelled as a spread of the load up the column, not as a weld. Its own flexibility would let a little more force out along the outstand, by an amount small beside the flange’s.

Still open: the plate that has no redistribution to rely on #

Everything above leans on yielding to carry the joint from its elastic share to the rule’s width, and that is legitimate for a load that arrives once. A joint loaded many times never gets there. A crane girder’s connection, or a moment joint in a frame carrying a vibrating machine, sees its elastic share on every cycle, and a fatigue check reads the weld’s stress range. The detail categories for a load-carrying fillet weld were set against nominal stresses on joints whose backing was stiff, and on this joint the weld over the web sees 2.67 times the nominal range on every cycle. Whether the category for a transverse attachment already contains that factor, because the tests that set it were made on unstiffened flanges, or whether it has to be applied on top — and whether an unstiffened flange is then a fatigue detail that no static check would ever flag — is the question this joint puts to the cycles a weld group counts.