The thinner cover carries more

A plate lapped on another carries its load along two lines a plate thickness apart, and the joint that is crooked by construction is the result: a moment nobody applied, a peak stress four times the mean, and a rotation under load that relieves only a little of it. With plates of unequal thickness the thicker plate takes the bend, because a tensioned arm straightens itself onto the load line over a distance that grows with its stiffness, and the stiffer arm cannot follow.

The remedy is old and nearly universal. Butt the two plates end to end and lap a cover plate over the joint on each face, so that the load leaves the main plate through two shear planes, one above and one below, and crosses the joint in two covers lying symmetrically about the plate. The eccentricity vanishes and the moment goes with it.

It vanishes exactly when the two covers are equal. They very often are not: the cover on one face is cut from whatever plate the shop has, the face of a flange has a web in the middle of it so the covers on that side are two narrow strips, a stiffener sits on one face and the cover there is made thinner to clear it. The usual assumption then is that the covers share the load in proportion to their areas, and that a pair whose total area matches the main plate’s is as strong as the plate. That assumption is a statement about stiffness. The splice is decided by a statement about equilibrium, and the two disagree.

Two forces across one cut #

Take a 20 mm main plate 200 mm wide, carrying 900 kN — a mean stress of 225 N/mm² — and splice it with a 15 mm cover on one face and an 8 mm cover on the other, both the plate’s full width. The covers have 4,600 mm² between them against the plate’s 4,000, fifteen per cent more steel than the plate they replace. Each plate is bolted to the main plate over 160 mm either side of an 80 mm gap where the two halves of the main plate meet, and far from the joint each half of the main plate is held on its own centre line, so the load acts along that line.

Now choose the free body that answers the question: everything to the left of a cut through the middle of the gap. The main plate does not cross that cut — its two halves end at the gap — so the only things that do are the two covers. On the far left the load pulls along the main plate’s centre line. On the cut, the thick cover pulls with a force $N_1$ along its own centre, 17.5 mm above that line, and the thin cover pulls with $N_2$ along its own centre, 14.0 mm below it.

Two equations follow, and they are the whole of the argument. Resolving along the joint, $N_1 + N_2 = P$. Taking moments about the load line,

$$N_1 \, a_1 = N_2 \, a_2,$$

where $a_1 = (t_m + t_1)/2$ and $a_2 = (t_m + t_2)/2$ are the two covers’ distances from the main plate’s centre. Nothing else crosses the cut to carry a moment, apart from whatever the covers can carry by bending about their own centres, which for plates 15 and 8 mm thick is very little. So

$$N_2 = P \, \frac{a_1}{a_1 + a_2}, \qquad N_1 = P \, \frac{a_2}{a_1 + a_2}.$$

The covers’ areas are not in it. The split is set by where the covers are, and the one nearer the load line takes the larger share. That is Archimedes’ law of the lever — two weights balance at distances inversely proportional to them — with the main plate’s centre line as the fulcrum, and it holds as exactly for a pair of cover plates as it did for his balance. With these plates it gives the thin cover $900 \times 17.5/31.5 = 500$ kN and the thick cover 400.

The thin cover is nearer the line #

The reason the thin cover carries more is geometric and slightly perverse. A cover’s distance from the main plate’s centre is half the main plate’s thickness plus half its own, so a thinner cover sits closer to the load line, and a force closer to the fulcrum must be larger to balance a given moment on the other side. Make the second cover thinner still and it moves in further and must carry more, on less steel.

The area rule would give both covers the same stress, 196 N/mm², which is what makes it attractive: a pair that is 15 per cent larger than the plate is 15 per cent less stressed than the plate, everywhere. The cut gives something else. The stress through the two covers is not uniform but varies linearly across their combined depth, because the joint’s section at the gap is two plates acting together about a centroid that is not on the load line. The 15 mm cover works at 107 N/mm² on its outer face and 179 on its inner; the 8 mm cover at 275 on its inner face and 313 on its outer. The thin cover’s mean stress is nearly three times the thick cover’s.

The full calculation — the joint solved as a line from grip to grip, with the covers’ own bending and the joint’s deflection included — puts 471 kN in the thin cover rather than the lever rule’s 500. The difference is the two effects the lever rule leaves out, and they are small for the same reason: the covers resist the moment as a couple, 31.5 mm apart, and their own bending stiffness is a few per cent of the stiffness that couple has.

That comparison is the contrast with the single lap. A single lap has one plate crossing each cut, and that plate can resist the eccentricity only by bending — so the moment is resisted by a thin plate’s $t^3$, the joint bends visibly, and the rotation that is supposed to straighten it has a great deal to work with. A double-cover splice has two plates crossing the cut, which together resist the eccentricity as a lever arm and not as bending, so the joint hardly bends at all and the moment is carried by changing how the load is divided between the covers.

Rotation buys back a tenth #

The joint does deflect a little. The covers’ combined centroid at the gap sits 6.5 mm above the load line, and under 900 kN the whole joint sags toward the line by 0.7 mm, which brings the resultant 10 per cent nearer to where the lever rule would place it. The main plate pays for that: where it enters the overlap it is pulled 0.7 mm off its own line and carries a moment of 0.4 kN·m, which adds 28 N/mm² to its 225 of direct stress.

The single lap, by contrast, rotates until its load line passes within a fraction of a plate thickness of the joint. The difference is again the lever: a deflection of the gap moves both covers together, and since the moment they must balance is the load times the offset of their centroid, the joint has to move that centroid most of the 6.5 mm before the moment is gone. A main plate under 225 N/mm² of tension straightens itself too quickly to let it.

The thinner the second cover, the harder it works #

The sweep makes the trend a curve. With both covers 15 mm the area rule is exact, and every other rule with it, because the centroid of the covers is on the load line. Thin the second cover to 12 mm and its outer face works at 1.20 times what the area rule says; at 8 mm, 1.61; at 6 mm, 1.94; at 4 mm, more than two and a half times. The 15 mm cover, meanwhile, is let off by the same arithmetic, down to 0.39 of its area-rule stress when its partner is 6 mm thick.

The curve has no ceiling, and that is not a defect of the model. As the second cover thins toward nothing, the lever rule still asks it for more than half the load, because its distance from the fulcrum is never less than half the main plate’s thickness. When it is gone altogether the joint is a single cover lapped on a plate — the single lap with a butt joint in it — and the moment must be carried by bending, which is the problem the second cover was added to solve. A thin second cover does not half-solve the eccentricity. It carries the moment as a large tension in a small plate.

The splice yields before the plate does #

The design question a splice answers is whether it is as strong as the member it interrupts. A splice has no other job, and the cover-area rule is how that question is usually answered: covers with at least the plate’s area, of at least the plate’s grade, and the splice is as strong as the plate.

With these covers the plate yields at 1,420 kN. The area rule says the covers yield together at 1,633 — the splice comfortably stronger than the plate. The thin cover’s outer face actually yields at 1,023 kN, 72 per cent of the plate’s yield load, and at that point the 15 mm cover is working at a third of its yield stress. If the joint could not deflect at all, first yield would come at 977 kN; the deflection buys 5 per cent of that back.

Steel does not stop at first yield, and neither does the splice, but the reserve here is small. The thin cover is already loaded almost uniformly — 275 N/mm² on one face, 313 on the other — so when its outer face yields there is little of it left to give. Once it is fully yielded it can take no more, and the thick cover can take more only if the moment about the load line still balances, which means the gap must move toward the thick cover’s side until its larger force acts on a shorter lever arm. That movement drags the main plate off its line where it enters the overlap, and the main plate, already carrying the whole load in tension, has little bending capacity to spare. The two meet at about 1,158 kN — 82 per cent of the plate’s yield load, from a pair of covers with 15 per cent more steel than the plate.

The bolts’ half is nearly right #

The same free body says how much each shear plane passes, because each cover’s force at the gap is exactly what its bolts have handed it across its own plane. Here the answer is a mild surprise in the other direction. A bolt in double shear is designed as if each of its two planes carried half its load, and the cut puts 48 per cent through the thick cover’s plane and 52 through the thin cover’s — the bolts’ even split is very nearly right, because the two lever arms, 17.5 and 14.0 mm, are not very different.

So the bolts are designed on an assumption that is close, and the covers on one that is far out. The two rules cannot both hold: if each plane passes half the load, the covers carry equal forces, and equal forces in unequal plates are unequal stresses. The area rule, which keeps the stresses equal, has to give the thin cover’s plane only 35 per cent of the load — and that is the figure that is wrong.

The consequence for bearing compounds it. The bearing stress on a cover is its force divided by its thickness, so the thin cover receives both the larger force and the smaller thickness to resist it with: 368 N/mm² of bearing against 179 on the thick cover, more than twice. Bearing is a failure that announces itself — the hole goes oval long before anything tears — but it announces itself in the thin cover first, and a check made at the area rule’s 245 would not have found it.

Equal areas are not a balanced splice #

The commonest double-cover splice in steel buildings is a beam’s flange splice, and there the covers cannot be the same width. The outer cover spans the whole flange; the inner cover is two strips, one either side of the web. The usual guidance shares the flange force between them in proportion to their areas, and the usual detail makes the inner strips’ total area about equal to the outer plate’s.

That is not the balanced arrangement either. To give 160 mm of inner strip the area of a 200 mm by 15 mm outer plate, the strips must be 18.8 mm thick, and thicker plates sit further from the flange’s centre — so the outer cover, nearer the line, carries more. Here the effect is modest, 159 N/mm² on the outer face against 140 on the inner, because the widths differ by only a fifth. The condition that makes every rule agree is that the covers’ centroid lies on the load line:

$$b_1 t_1 \, a_1 = b_2 t_2 \, a_2,$$

equal first moments about the main plate’s centre, not equal areas. For this flange it gives 17.5 mm inner strips, at which both covers work at 155 N/mm². The area rule and the lever rule then give the same split, which is the only situation in which the area rule is right.

The cut at the gap, by hand #

Every number here can be checked with the free body and a calculator. The lever arms are $a_1 = (20 + 15)/2 = 17.5$ mm and $a_2 = (20 + 8)/2 = 14.0$ mm. The lever rule gives the thin cover $900 \times 17.5/31.5 = 500$ kN, a mean stress of $500{,}000/(200 \times 8) = 313$ N/mm², and the thick cover 400 kN, 133 N/mm². The area rule gives each $900{,}000/4{,}600 = 196$.

The full section at the gap refines this. The covers’ centroid is $c = (3{,}000 \times 17.5 - 1{,}600 \times 14.0)/4{,}600 = 6.5$ mm above the load line, and their second moment about it is $I = 1.10 \times 10^6$ mm⁴, almost all of it from the two plates’ distances from the centroid rather than their own thickness. The load acting 6.5 mm from the centroid is an eccentric force, so the stress is $P/A - P c\,(y - c)/I$, which at the thin cover’s outer face, $y = -18$ mm, gives $196 + 900{,}000 \times 6.5 \times 24.5/1{,}100{,}000 = 196 + 131 = 327$ N/mm² — the rigid value, before the joint’s 0.7 mm of deflection reduces the offset from 6.5 mm to 5.9 and the stress to 313.

The first-yield load is where that stress reaches 355 N/mm² — 977 kN for a joint that cannot move, by proportion, and 1,023 for the joint that does. The mechanism estimate is the load $Q$ at which the thick cover’s excess moment, $(Q - 568)\times 17.5 - 568 \times 14.0$ kN·mm with the thin cover yielded at 568 kN, equals what the main plate can carry in bending at that tension, $355 \times 200 \times 20^2/4 \times (1 - (Q/1{,}420)^2)$ — about 1,160 kN.

Covers that act with the plate, a short gap, and one steel #

The overlaps act as one section. The bolts are taken to tie the three plates together over each overlap so that no cover slips on the main plate. That is true of a preloaded joint below its slip load and roughly true of a bearing joint with fitted bolts; in a bearing joint with clearance holes the covers take up their slip unevenly, and the thinner cover, whose bearing is softer, will lag a little before it starts to carry. The lever rule does not care: once both covers are bearing, the cut across the gap is the same cut. What slip changes is the load at which they both start.

The gap is short. Across the 80 mm gap the two covers are treated as one section whose plane cross-sections stay plane. With nothing between them over that length they are, strictly, two separate plates joined at their ends, and over a long gap they would bend individually as a frame does. The moment across the gap is nearly constant — the shear there is the slope of the deflection, which is zero at the middle by symmetry — so the plane-section assumption is good over short gaps and degrades as the gap grows.

Both covers are the same steel as the plate, and the load is static. Different grades would shift where first yield falls but not the split, which depends only on where the covers are.

Compression, bolt rows and fatigue, which the cut leaves out #

They cannot show what happens to the gap under compression. Every figure here is a tension splice. A compression splice with the main plate’s ends machined to bear on each other carries most of its load straight across the joint, and the covers only hold it in line; there the unequal covers matter for stability and for any tension that reversal brings, not for the bearing load.

They cannot show the bolts redistributing along each overlap. Each cover’s force is handed across by several rows of bolts, and the end rows carry more than their share in any long joint — the bolts that do not share are the rule. With unequal covers the thin cover is also the more flexible one, so its end rows are worked harder still. The cut says how much each cover carries in total; how its bolts divide it is a second problem.

And they cannot show fatigue. Where a splice carries a fluctuating load the thin cover’s stress range is 1.6 times what the area rule predicts, and a detail’s life on an S–N curve of slope three falls as the cube of the range — a factor of four. The detail decides, and the detail here is the end of the thinnest cover.

Lever arms, not areas #

A free body cut across a double-cover splice has two forces in it, and they balance about the load line. So the covers share by lever arm, $N_1 a_1 = N_2 a_2$, and not by area.

The thinner cover is nearer the line and carries more. A 15 mm and an 8 mm cover on a 20 mm plate put 471 kN of 900 in the thin cover, which works at 1.6 times the area rule’s stress.

The splice is weaker than its area says. It yields at 72 per cent of the plate’s yield load and fails by a mechanism at about 82, with 15 per cent more steel than the plate.

The bolts’ half is nearly right; the covers’ share is not. And the arrangement that makes every rule agree is equal first moments about the main plate’s centre — which for equal widths is equal thicknesses, and for a flange splice is not equal areas.

Still open: the splice that carries bending as well #

Every splice here carries a pure tension along the main plate’s centre line. A flange splice in a beam carries the flange force, but the web splice beside it carries the shear and its share of the moment, and the two are bolted to the same flange through the same inner covers. The flange then sees a force that is not on its centre line — it varies through the flange’s depth with the beam’s bending — and the covers’ lever arms are measured from a load line that moves with the beam’s curvature. Whether that shifts the balance toward the outer cover or the inner, and whether a flange splice that is balanced for the flange force is still balanced once the beam’s own bending gradient is acting through it, is a question about the same cut made in a member rather than a plate.