The thicker plate takes the bend

The joint that is crooked by construction found the single lap joint’s defining fact: two plates lapped face to face carry their load along lines a plate thickness apart, so the joint is eccentric by construction, and the eccentricity puts a moment of $P\,e/2$ into each plate where it enters the overlap. Under load the joint rotates, the load line swings nearer the plates, and the moment falls by Goland and Reissner’s factor $k$ — which for structural plate is nearly one. That essay was about two plates of the same thickness. It said, in one sentence, that unequal plates share the moment unevenly, and left the question of how unevenly open.

The answer is not a correction to $k$. With plates of different thickness the moment moves from one plate to the other, and it moves the way a load moves in any structure that has two paths: toward the stiffer one. The difference is that here the stiffness is not the plates’ bending stiffness at all. It is the stiffness each arm acquires from being pulled.

Where the moment comes from #

The joint drawn below is a 10 mm plate lapped on a 20 mm plate, both 100 mm wide, over 60 mm, carrying 60 kN — a mean stress of 60 N/mm² in the thin plate. Each plate is held at its far end on its own centre line, 1,200 mm from the overlap, by a grip, a line of bolts or anything else that keeps it in line; the load acts along the straight line between those two points.

At any cross-section the bending moment is the load times the perpendicular distance from that line to the centre of the section. That is the whole of the statics, and it is exact: a free body cut anywhere along the joint has the load at its far end and nothing else. Outside the overlap the section is one plate and its centre is that plate’s centre. Inside, it is both plates acting together, with a centre between them.

If the joint could not deflect, the load line would run from the thin plate’s centre at one grip to the thick plate’s centre at the other, and pass the overlap at mid-height. Each plate would enter the overlap at a distance $e/2$ from the line, where $e = (t_1 + t_2)/2 = 15$ mm is the offset between their centres, and each would carry $P\,e/2 = 0.45$ kN·m. That is the rigid calculation, and on it the thin plate has a peak stress of 330 N/mm² — 60 of direct tension, 270 of bending — and the thick plate 98.

The joint does deflect, and the picture shows how. The thin plate’s arm, pulled along a line that does not pass through its centre, bends toward the line. Where it enters the overlap it has moved 3.6 millimetres up, half the distance to the line, which on a joint whose offsets are measured in millimetres is a great deal: its gap from the line, which sets its moment, is 0.49 of the rigid $e/2$. The overlap, attached to the end of that arm, is lifted and turned with it. And the thick arm, attached to the other end of the overlap, is lifted too — but it is eight times stiffer in bending and does not curve back toward the line nearly as quickly. Its centre where it leaves the overlap ends up further from the load line than in the rigid joint: 1.34 of $e/2$.

The peak stresses follow. The thin plate falls from 330 N/mm² to 193. The thick plate rises from 98 to 121.

Where the bending goes along the joint #

Plotted along the joint, the moment tells the same story with a second fact in it. The equal lap is symmetric: both plates enter the overlap with the same moment, 0.86 of their $P\,e/2$, which is Goland and Reissner’s factor for this joint to three figures. In the unequal lap the moment on the thin side is smaller than in the equal lap — though its eccentricity is half as large again — and the moment on the thick side is more than twice as large. Inside the overlap the moment swings from one sign to the other as the load line crosses the composite section’s centre.

The second fact is how far each moment reaches. The thin plate’s bending has fallen to a third within 200 mm of the overlap and is gone by 500; the thick plate’s is still a third of its peak 500 mm away and a sixth at 800. A tensioned arm straightens itself onto the load line over a characteristic distance, and a stiff arm straightens slowly. That distance turns out to be the whole explanation.

The bending length, and why it decides the split #

A plate under tension $P$ that is bent resists the bending two ways: by its flexural stiffness $EI$, and by the tension itself, which pulls any curved plate straight — the same stiffness a tensioned cable has because of its shape rather than its section. The ratio of the two is a length,

$$\ell = \sqrt{\frac{EI}{P}},$$

and a disturbance at the end of a long tensioned arm dies away along it as $e^{-x/\ell}$. For the 10 mm plate at 60 kN, with the plate stiffness $E/(1-\nu^2)$ that a wide plate has, $\ell = 177$ mm; for the 20 mm plate, with eight times the $I$ and the same load, $\ell = 500$ mm.

That is enough to divide the moment by hand. Treat the overlap as short and rigid and each arm as long. Far from the joint each arm lies on the load line — that is what tension does to it — and near the joint it is displaced from the line by a gap $g_i$ that decays over its own bending length, so the slope of the arm where it enters the overlap is $g_i/\ell_i$. The overlap is rigid, so the two arms must enter it at the same slope, and the two plates’ centres are $e$ apart across it, so the two gaps add up to $e$. Two equations:

$$\frac{g_1}{\ell_1} = \frac{g_2}{\ell_2}, \qquad g_1 + g_2 = e \quad\Longrightarrow\quad g_i = e\,\frac{\ell_i}{\ell_1 + \ell_2}.$$

The moment in each plate is $P g_i$, so the factor on the rigid value $P\,e/2$ is $k_i = 2\ell_i/(\ell_1 + \ell_2)$. The eccentricity is divided in proportion to the arms’ bending lengths, and since $\ell \propto t^{3/2}$ for a given load, the ratio of the two moments is

$$\frac{M_2}{M_1} = \left(\frac{t_2}{t_1}\right)^{3/2},$$

which does not contain the load at all. For plates of 10 and 20 mm it is 2.83. The full calculation gives 2.73. For plates of 10 and 15 mm, 1.84 by hand and 1.81 in full. For 10 and 30 mm the hand value, 5.2, overshoots the full one, 4.4, because the thick arm’s bending length is then 900 mm and the grips at 1,200 are no longer far away.

The hand model also says where the limits are. Equal plates split the moment evenly, and a thin arm with no bending stiffness at all has no bending length, lies on the load line right up to the overlap, and carries nothing: the whole of $P\,e$ — twice $P\,e/2$ — goes into the other plate. The two factors add up to two, less a little that the overlap’s own rotation removes: 1.84 for the joint drawn here, 1.71 for two equal plates, which is where Goland and Reissner’s relief lives.

The arithmetic, once, by hand #

The hand model is worth doing with the joint’s own numbers, because it reproduces most of the figure with two square roots. The bending lengths are 177 and 500 mm, so the two gaps are

$$g_1 = 15 \times \frac{177}{177 + 500} = 3.9 \text{ mm}, \qquad g_2 = 15 \times \frac{500}{677} = 11.1 \text{ mm},$$

and the moments are the load times the gaps: $60 \times 3.9 = 0.24$ kN·m in the thin plate and $60 \times 11.1 = 0.67$ kN·m in the thick one, against 0.22 and 0.60 from the full solution. The peak stresses follow by adding direct stress to bending stress, $P/bt + 6M/bt^2$: 60 + 141 = 201 N/mm² in the thin plate, 30 + 100 = 130 in the thick one, against 193 and 121 in full. The hand model overstates both moments by about a tenth, and it does so for one reason: it treats the overlap as a point, while the real overlap is 60 mm long and turns as a unit, which is where Goland and Reissner’s relief comes from. What it gets right is the split, and the split is the result.

There is a quicker way to say the same thing. The rigid calculation gives each plate the same moment, and its answer depends on the thicknesses only through $e$. The joint that turns gives each plate a moment in proportion to $t^{3/2}$, and bending stress falls as $t^{-2}$, so the thick plate’s bending stress relative to the thin plate’s goes as $(t_1/t_2)^{1/2}$ rather than $(t_1/t_2)^2$. At twice the thickness that is 0.71 instead of 0.25 — and the actual ratio of the bending stresses in the full solution, 91 against 133, is 0.68.

As the load rises #

The split is set almost at once. At a mean stress of 2 N/mm² in the thin plate the two factors are 0.82 and 1.12, already unequal, because even a small load straightens a long arm; by 30 N/mm² the thick plate’s factor has reached its peak, 1.36, and from there to 300 N/mm² both factors drift slowly down together as the whole joint rotates further and the load line passes nearer both plates. That is the equal lap’s relief, arriving in both plates at once, on top of a split that was decided at the start.

So the unequal lap has two effects where the equal lap has one. Goland and Reissner’s factor is a relief: it falls with load, the same for both plates, and in structural plate it is small. The split is a transfer: it is nearly independent of load, it is large, and it is not a relief at all for one of the two plates.

The thin plate still governs #

None of this makes the thicker plate the one that fails. Its bending stress is its moment over $t^2/6$, and with twice the thickness it has four times the section modulus to set against a moment that has not even trebled. Across every ratio drawn, from equal plates to one three times the other, the thin plate is the more highly stressed of the two. The design check that looks at the thin plate first is looking in the right place.

What changes is the margin. On the rigid calculation, a plate twice as thick as its partner works at 0.30 of the thin plate’s stress, and a designer would reasonably treat it as idle. In the joint that turns, it works at 0.63. At three times the thickness the rigid calculation says 0.14 and the joint says 0.44. The thick plate’s own stress is 1.10 times its rigid value when it is half as thick again, 1.24 times when it is twice as thick, 1.36 times at three; the thin plate’s is 0.72, 0.58 and 0.44 of its rigid value.

So the thin plate is relieved by more than a rigid calculation says, and the thick plate is loaded by more. The rigid calculation errs on the safe side for the plate that governs and on the unsafe side for the plate that does not, which is the usual shape of a redistribution: it helps where the check is made and hurts where it is not.

Where the thick plate is something else #

That matters most when the thick plate is not a plate. The common unequal lap is a thin connection plate — a gusset, a splice plate, a cleat — lapped onto a thick part of a member: a flange, a chord, the wall of a hollow section. The designer checks the thin plate for its eccentricity and moves on, and the moment that has been transferred lands in the member’s flange, where it adds to whatever that flange was already doing. A flange working near its capacity in bending of its own receives a local moment a third larger than the connection’s calculation implied, at the one section — the end of the overlap — where the flange also has bolt holes or a weld toe in it.

And where the stress range rather than the stress decides, the transfer is not diluted by thickness in the same way. A welded or bolted detail on the thick member is assessed on the range of stress it sees, and a range 24 to 36 per cent larger than calculated moves a detail’s life by a factor of two or so on an S–N curve of slope three, and it is the detail, not the steel, that such a curve is drawn for. The thin plate, relieved, fares better than its calculation. The member does not.

How far away the grips are #

The hand model assumed long arms, and the last figure shows what “long” means. If whatever holds each plate on the load line — its grip, the next bolt group, a stiffener, the end of the member — is close to the overlap, the arms cannot swing and the joint behaves more nearly as the rigid calculation says: at 240 mm the thin plate carries 0.71 of $P\,e/2$ and the thick plate only 1.02. Once the holding points are more than two or three bending lengths away the split has settled, at 0.49 and 1.35.

That gives the one practical lever the designer has. The transfer is caused by the thin arm straightening, and the thin arm straightens only if it has room to. A thin plate that is held in line close to the joint cannot hand its moment across, which is an argument for short, well-restrained connection plates that has nothing to do with their own strength.

Two plates, one beam-column #

The joint is solved as one line from grip to grip: the thin arm, the overlap and the thick arm, each with its own centre and its own stiffness — a wide plate’s, $E/(1-\nu^2)$ times $bt^3/12$ — and the overlap with the second moment of both plates acting as one section about their common centre. At each point the moment is the load times the distance from the grip-to-grip line to the deflected centre, which gives one linear differential equation for the deflection, solved by finite differences on six thousand intervals. With two equal plates the factor it returns agrees with Goland and Reissner’s closed form to a tenth of a per cent at every load tried, which is the check that the rest of it is the same calculation extended rather than a different one.

A bonded overlap, straight arms, and tension #

The overlap acts as one section. That is Goland and Reissner’s assumption and the earlier essay’s: true of a bonded lap, nearly true of a preloaded bolted lap in which friction keeps the plates from slipping, and not true of a lap in bearing, where the bolts let the plates slide a little relative to each other and the overlap is two plates connected at points. The split between the arms depends on the arms, not the overlap, so it should survive; the equal lap’s relief, which comes from the overlap turning as a unit, would be smaller.

The arms are straight and uniform. A connection plate that is tapered, or has holes along it, has a bending length that varies, and the split follows the arm’s stiffness near the joint.

The load is tension. In compression the same tension-stiffness enters with the other sign, the arms no longer straighten but bow — the load makes itself worse instead of better — and a thin arm is the first thing in the joint to buckle; the transfer runs the other way and the joint’s problem becomes one of stability rather than of shares.

Still open: the plate that is lapped on both faces #

A lap joint with one cover plate is eccentric; the standard remedy is a second cover on the other face, which puts the load lines symmetrically about the main plate and makes the eccentricity vanish. It does so only if the two covers are equal. With covers of different thickness — a common result of detailing, when one face carries a stiffener and the other does not — the eccentricity comes back, smaller, and it is shared among three arms instead of two, the main plate running through the middle. Whether the moment then goes to the stiffest of the three, as it went to the thicker of two here, and whether a thin cover on a thick main plate is relieved the way the thin plate was, is a question about a joint that is designed to be symmetric and very often is not.