The root that has to stretch for it

The shear that helps a butt weld followed a load round a partial-penetration butt throat, from straight across the weld to straight along it, and found that a throat welded from both sides is strongest a little off square: in S355, 478 N/mm² at 22.7°, against 441 pulled straight across. It also found that a throat made from one side keeps only a share of that, because the plates carry their force on their mid-thickness and the throat does not lie on it. At half penetration of a 20 mm plate, pulled straight across, the share was a quarter.

That quarter was an elastic number. The eccentric pull was carried as a linear stress across the throat, and the throat was declared full when its root, the fibre the eccentricity loads hardest, reached the limit. Every other fibre was below it. The essay ended by asking what happens if the throat is allowed to yield — whether the fibres left idle by the elastic count can take up the slack, and whether the single side, counted that way, gets back the double-sided weld’s peak advantage at some direction between square and along.

It does not get the peak back. It gets something else, and it gets it on a condition the weld’s geometry makes hard to meet.

A short column, loaded off its axis #

A single-sided throat pulled across is a short rectangular column carrying a force that misses its centre. The throat is $a$ deep and runs from the welded face into the plate. The plates’ force acts on their mid-thickness, $t/2$ from the face, while the throat’s centre is at $a/2$, so the pull misses it by $e = (t - a)/2$, on the side of the root. Per millimetre of weld the throat carries a normal force $N$ and a moment $Ne$.

Counted elastically the stress across the throat is linear, $N/a$ at the centre plus the bending part, and it reaches $(N/a)(1 + 6e/a)$ at the root. At half penetration $e/a = 1/2$, the root factor is four, and the throat is full when its average stress is a quarter of the limit.

Counted plastically the throat is full when every fibre is at its limit and the distribution can no longer change. For a rectangle that has a closed form. A compression block of depth $c$ forms at the welded face and the rest of the throat is in tension, both at the limit $\sigma_0$; the net force is $\sigma_0(a - 2c)$ and the moment about the throat’s centre is $\sigma_0 c(a - c)$. Setting the moment equal to the force times $e$ and solving gives the share of the full throat that can be carried,

$$\frac{N}{\sigma_0 a} = \sqrt{1 + \frac{4e^2}{a^2}} - \frac{2e}{a},$$

which at half penetration is $\sqrt{2} - 1 = 0.414$. Elastically it was 0.25.

The plastic distribution is not a modest correction. The face of the throat, which the elastic count had in compression at twice the average stress and nowhere near its limit, now pushes back at the full 441 N/mm² over 2.9 mm. That block’s moment arm, about the force line, does the work the root alone was doing before, and the throat carries two thirds as much again.

The same equation governs a squeezed section that has to work for its hinge and every point on the curve between two ways to fail, the axial-force and bending interaction of a rectangle. A weld’s throat is a rectangle whose axial force and moment are fixed in proportion by a dimension of the joint, so its plastic capacity is one point on that curve, picked out by $e/a$.

The gain does not change with direction until the cap lets go #

Now turn the load. Its transverse part, $N = R\cos\theta$ per millimetre, is carried off-centre as before; its longitudinal part, $V = R\sin\theta$, is a shear along the weld’s own axis with no lever arm about it. Each fibre of the throat carries a normal stress $\sigma$ and a shear $\tau$, and the directional method allows any pair with $\sqrt{\sigma^2 + 3\tau^2} \le f_u/\beta_w$ and $|\sigma| \le 0.9 f_u$.

The plastic capacity in that case has no closed form, because the throat can choose how to share the shear among its fibres as well as how to distribute the pull. It is a lower bound in the sense the plastic theorems mean: any distribution of stresses that is in equilibrium with the load and inside the limit everywhere gives a load the throat can carry, and the capacity is the largest such load. The largest is found here by the theorem’s dual, which bounds it from above for every set of weights on the three equilibrium equations, and the two bounds are made to meet.

The plastic share sits above the elastic one at every penetration and every direction short of straight along, where the two meet at one. That is the gain. But the shape of the plastic curve has a flat stretch the elastic curve shares: from straight across to 22.7°, the share at half penetration is 0.414 throughout.

The reason is the cap. Up to 22.7° the double-sided throat is limited by the cap on its normal stress, not by the ellipse, and a little shear costs it nothing, so its capacity rises as $1/\cos\theta$. The single-sided throat, counted plastically, is limited the same way: its fibres sit at the cap in tension and in compression, the shear goes into the slack the cap leaves them, and its capacity rises as $1/\cos\theta$ too. Two capacities rising in proportion keep a constant ratio. The single side gets the double side’s rising stretch, scaled by the same 0.41; it does not get a larger share of it.

So the question the earlier essay left — whether yielding recovers the double-sided weld’s peak advantage at some intermediate direction — has a clean answer. The peak at 22.7° is the end of the stretch where both throats are capped, and the single side reaches it at 0.41 of the double side, exactly as it was straight across. Past 22.7° the double-sided throat starts to fall and the single-sided share starts to climb, but by then the double side’s advantage over a fillet, which peaked at that angle, is already being spent.

The strongest direction moves out #

What the yielding does change is where the single-sided throat is strongest.

At half penetration the single-sided throat is weakest straight across and strongest straight along, rising monotonically from 183 to 314 N/mm². Its eccentric moment is so large that the throat can only carry more as the load turns away from it, and the shear along the weld, which has no moment, is the one component it carries at full value. A shallow single-sided weld is a longitudinal weld, counted either way.

At 0.7 penetration the curve has a peak, but not at 22.7°. It rises from 291 straight across to 357 at 40° and then falls to 314. Two things happen as the load turns past 22.7°. The ellipse starts to take normal stress away from every fibre to make room for the shear, which is what turns the double-sided throat down. And the pull, with the eccentric moment that comes with it, shrinks as $\cos\theta$, which the double-sided throat does not care about and the single-sided one cares about a great deal, because the moment is what was costing it most. The single side keeps rising for as long as shedding its moment gains it more than the ellipse takes away, and peaks where the two balance. At half penetration the moment is so large that shedding it wins all the way to 90°; at 0.7 the balance falls at 40°.

And at 40° the 0.7-penetration throat carries 357 N/mm² against a fillet’s 352. From 38° to 88° it carries more than a fillet with the same throat, made from both sides. The fillet’s own weakness is that it puts half of any transverse load into shear across the throat, which the method weighs by three; the single-sided butt throat faces the load and puts none there. Once the shear along the weld is large enough to dilute the eccentric pull, that advantage of orientation outweighs the cost of the eccentricity.

Where the shear goes in the throat #

A hand count of a combined load usually spreads the shear evenly over the throat and works out the pull on what is left. That is one admissible distribution among many, and the plastic capacity is the best of them.

The best distribution puts the shear where the pull is small. Near the welded face, the normal stress passes through zero on its way from the compression block to the tension, and the fibres there can carry almost the whole of their shear capacity, $f_u/(\beta_w\sqrt{3}) = 314$ N/mm². Near the root they are carrying a large tension and have less room for shear. The ellipse in the plane of $\sigma$ and $\tau$ makes the trade explicit: a fibre’s normal stress and its shear compete for the same limit, and the throat does best by sending each where the other is not.

The placement is worth having only in a band of directions. Up to 44°, at half penetration, it is worth nothing: there the even shear is below 184 N/mm², which is the most shear a fibre can carry at the cap without the ellipse taking any of its normal stress, so spreading it evenly costs the pull nothing. Past 44° the even shear starts to bite into the normal stresses at the root, which is where the throat’s moment most needs them, and moving the shear towards the face is worth up to 5 per cent, at 56°. Straight along the weld the two counts meet again, since there is no pull to make room for.

Five per cent is not a design margin anyone would plan on. It is a measure of how good the hand count is: the even-shear plastic count is within 5 per cent of the true plastic capacity everywhere, and exact up to 44°. The difference between the elastic count and either plastic count, at the same direction, is as much as 66 per cent.

How deep the single side must go #

The comparison that decides whether a single-sided weld is used at all is against the fillet it would replace. Per millimetre of throat, a fillet carries 0.87 of what a double-sided butt throat carries straight across and 0.89 at 45°. The single side has to reach those shares.

Straight across, counted elastically, the single side needs 0.98 of the plate to match a fillet — it is only as good as a fillet when it is nearly a full-penetration weld. Counted plastically it needs 0.90. That is not much of a difference, and it says the same thing: for a pull square across the weld, the single-sided partial-penetration weld is a poor substitute for a fillet whichever way it is counted, because the eccentricity grows faster than the throat shrinks.

At 45° the two counts tell different stories. Elastically the single side needs 0.90 of the plate; plastically, 0.68. A load with as much shear along the weld as pull across it is the case in which a single-sided weld, counted plastically, does what a fillet does with two thirds of the plate’s depth fused from one side. This is the plastic count’s real gain, and it lives entirely in the band of directions where the load is mixed.

The strain the plastic count spends #

All of this rests on the throat’s fibres reaching their limit and staying there while the others catch up. A plastic distribution is not a stress state the weld is in at a load; it is the state it reaches after its most loaded fibres have yielded and gone on straining. In a throat pulled off-centre, the most loaded fibre is the root, and the plastic count asks the root to keep straining while the compression block forms at the face.

The figure counts the throat as plane sections, with its fibres elastic up to the limit and flat beyond it, its shear spread evenly, and its root’s strain held to a stated multiple of the yield strain. Held at the yield strain the throat carries the elastic count, by construction. At twice the yield strain it gets two thirds of the step to the plastic count straight across; at three times, 84 per cent; at five, 94. When the load is mostly shear along the weld the step itself is smaller and comes more quickly, 80 per cent of it at twice the yield strain at 80°, because the pull’s moment is a smaller part of what the throat carries.

So the plastic capacity of a half-penetration single-sided throat, pulled straight across, is mostly in hand once its root has strained three times its yield strain — about six tenths of one per cent in S355 weld metal. Ordinarily that is nothing. Weld metal is specified to elongate by about 20 per cent before it breaks, and the ductility of a steel is precisely the property that lets a structure use its plastic capacities without anyone having calculated a strain.

The root is a crack tip #

The difficulty is where the strain is asked for. The root of a single-sided partial-penetration weld is the end of the fused metal, and beyond it, from the root to the far face of the plate, the two plates sit against each other unjoined. Pulled across, that unfused land opens. It is not a notch with a radius; it is a crack $t - a$ long, 10 mm at half penetration of a 20 mm plate, whose tip is the root.

A crack tip does not have a strain; it has a strain field that rises without limit towards it in an elastic count and is capped by a plastic zone in a real metal. “Three times the yield strain at the root” is a statement about a fibre of the throat in a beam theory that knows nothing about a crack, and the metal at the tip of the land sees much more than that over a much shorter length. Whether it tears depends on the weld metal’s toughness at the temperature the joint is loaded at — which is a property of the plate’s thickness as well as of the steel — and on the stress the crack sees between two checks, the yielding one and the brittle one.

This is why the design rules decline the plastic count. EN 1993-1-8 asks for local eccentricity to be avoided where it can be, and to be taken into account wherever a moment about a weld’s axis, or a pull across it that misses the throat, puts the root in tension. Taking it into account means the elastic count: the root’s peak stress, not a redistribution that would need the root to stretch. The lower-bound theorem says the plastic capacity is safe if the material is ductile enough to reach it. At the tip of an unfused land, ductility is the one property that cannot be assumed.

A single-sided throat by hand, at 45° #

Take the half-penetration throat, $a = 10$ mm in a 20 mm S355 plate, with the load at 45°. The method’s limits are $f_u/\beta_w = 544$ and $0.9f_u = 441$ N/mm².

Elastically: per millimetre of throat the resultant $r$ puts $0.707r$ of pull and $0.707r$ of shear on it, the pull at a root factor of four, so the root carries $\sigma = 2.83r$ and $\tau = 0.707r$. The cap gives $r = 441/2.83 = 156$ and the ellipse $r = 544/\sqrt{2.83^2 + 3 \times 0.707^2} = 177$, so the cap governs and the throat carries 156 N/mm².

Plastically, with the shear spread evenly: the shear is $0.707r$ and leaves each fibre a normal limit of $\min(441, \sqrt{544^2 - 3(0.707r)^2})$; the throat’s share of that limit, from the closed form, is 0.414, and the pull it must carry is $0.707r$. So $0.707r = 0.414\min(441, \sqrt{544^2 - 1.5r^2})$. At $r = 258$ the shear is 183 N/mm², just inside the 184 a capped fibre can spare, the limit is still the cap, and $0.707 \times 258 = 182 = 0.414 \times 441$. The throat carries 258 N/mm², 1.65 times the elastic count, and exactly the share of the double-sided throat’s 385 that the figures give at 45°: 0.67.

What the count rests on #

The directional method, taken as a yield criterion for every fibre. It was written for a throat’s average stresses and calibrated on fillet tests; using it fibre by fibre inside a throat treats it as a material law, which it is not. The cap in particular, read here as a limit on either sign of normal stress, is a rule about tension across a throat, and a throat’s compression block at the welded face may well carry more than the cap allows. That would raise the plastic count and change none of its shapes.

The force line at the plates’ mid-thickness. A single-sided weld between two plates that are free to rotate does carry its force there. Plates held against rotation by the structure round them carry part of the moment themselves and leave the throat less, as the essay on partial-penetration welds found; every share here would then rise, the elastic one most, because it is the one most sensitive to the root factor.

A throat of uniform depth and a root on a straight line. A real root wanders, and a single-sided weld whose penetration varies along its length has a root factor that varies with it. The plastic count is less sensitive to that than the elastic one — the closed form changes slowly with $e/a$ — but the crack at the root is longest where the penetration is least, and that is where the strain demand of the plastic count is greatest.

The redistribution that is there and cannot be spent #

The single-sided throat has a plastic capacity, and it is two thirds larger than its elastic one when pulled straight across. That capacity is real in the sense that the metal can carry it: a throat whose root could stretch would find it. It does not recover the double-sided weld’s peak advantage, because up to the direction where that peak sits, the cap governs both throats and the single side rises in proportion. Past it the single side’s own strongest direction moves outward, to 40° at 0.7 penetration, and there a single-sided throat carries more than a fillet.

The price is a root that strains several times its yield strain at the tip of an unfused land. The plastic count is a lower bound only for a ductile throat, and the one fibre the whole redistribution depends on is the one whose ductility the joint’s geometry makes doubtful. The redistribution is there; the weld cannot be trusted to spend it.

Still open: the root that is allowed to tear a little #

The plastic count asks the root for strain and the elastic count refuses to let it yield at all, and neither describes a root that does yield a little and starts a stable tear. A fracture count would put the throat’s moment and pull into a stress intensity at the tip of the unfused land, set it against the weld metal’s resistance to tearing, and find the load at which the tear stops being stable — a capacity between the two counts here, depending on the land’s length, the plate’s thickness and the temperature. Whether that capacity, for the penetrations single-sided welds are actually made at, sits nearer the elastic quarter or the plastic 0.41, and whether a single-sided weld loaded mainly along its axis is ever close enough to tearing for the question to matter, is what the root’s crack asks of this one.