The joint that is crooked by construction

The simplest connection in structural steelwork is two plates overlapped with a bolt through them, and it is the one arrangement in the subject that is guaranteed to be wrong before anything is applied to it.

The load comes along the centreline of one plate and leaves along the centreline of the other, and those two lines are a plate thickness apart. Something has to carry the offset.

The naive number, and why it is nearly right #

Take the joint as drawn and do not let it move. The eccentricity is $e = (t_1+t_2)/2 = t$ for equal plates, and the moment at the fastener is $M = Pe/2$. The bending stress that produces is

$$\sigma_b = \frac{6M}{bt^2} = \frac{3Pt}{bt^2} = 3\sigma_m,$$

so the peak stress at the surface of the plate is four times the mean. One part membrane and three parts bending, from a geometry nobody chose and everybody draws.

Four is a large number for a joint whose only apparent defect is that it is asymmetric, and the usual response is that it cannot be right — that the joint must straighten itself out. It does, a little.

Goland and Reissner, and the factor that is nearly one #

The straightening is a genuine effect and it has an exact solution. Pull the joint hard enough and the two plates bend until the load path runs nearly straight through the overlap; the moment at the fastener falls to a fraction $k$ of the naive value, and Goland and Reissner solved for $k$ in 1944 for a bonded lap:

$$k = \frac{\cosh(uc)}{\cosh(uc) + 2\sqrt2 \sinh(uc)}, \qquad u = \sqrt{\frac{3(1-\nu^2)}{2}}\,\frac{1}{t}\sqrt{\frac{\sigma}{E}}$$

with $c$ the half-overlap. The whole behaviour is in the group $uc$, which is proportional to $(c/t)\sqrt{\sigma/E}$: a long lap in a thin plate at a high stress straightens out; a short lap in a thick plate at a modest stress does not.

For the joint drawn — 10 mm plate, 60 mm overlap, 60 N/mm² of membrane stress — $uc$ is 0.06 and $k$ is 0.855. The rotation removes eleven per cent of a moment that trebles the stress.

That is the finding, and it is the opposite of the reassurance the effect is usually cited as. The self-straightening that makes adhesive lap joints workable is a thin-sheet phenomenon, and structural steelwork does not operate in that regime.

It is worst where fatigue is decided #

$k$ depends on the stress, so it depends on the load, and the dependence runs the wrong way.

At the ultimate load the plate is stressed hard, $uc$ is at its largest, and $k$ is at its smallest. At the service load — a quarter of it, say — the stress is a quarter, $\sqrt{\sigma}$ is a half, $uc$ is halved and $k$ rises to 0.92.

So the peak stress ratio is 3.57 at ultimate and 3.77 at service, and the joint is relatively worse at the load it spends its life at. Since fatigue is decided by the stress range at service and not by the ultimate capacity, the bending that the strength check partly forgives is the bending the fatigue check sees in full.

The bolt is bent too #

The plate is not the only member with an eccentricity problem. The bolt has two shear planes — one at each plate’s mid-thickness — and they are a plate thickness apart, so the bolt is a very short beam carrying a moment as well as a shear.

For the M20 in the joint drawn: shear stress $P/A = 191$ N/mm², bending stress $M/Z = 382$ N/mm² with $M = P\,e/2$ and $Z = \pi d^3/32$. The combined utilisation against $f_u$ is 1.03, against 0.61 from the shear alone.

What the code does about it #

The design rules do not mention any of this. They contain a single number.

For a bolt in a single lap with one row of bolts, the bearing resistance is taken as $1.5 f_u d t$ instead of the usual $2.5 f_u d t$ — a reduction to 60 per cent, applied to the plate rather than to the bolt, with no explanation attached.

That factor is the whole of this essay compressed to a coefficient. It is not a bearing effect at all: the bearing stress under the bolt is what it always was. What has changed is that the plate around the hole is also carrying $3\sigma_m$ of bending, so it reaches its limit at a lower nominal bearing stress, and the calibration puts the ratio at 0.6.

The plates rotate, and so does everything attached to them #

There is a consequence of the straightening that the stress calculation does not report and that shows on site.

The joint takes up an angle. For the plates drawn the rotation is small — a fraction of a degree — but it is a permanent kink in the line of the member, and in a tie made of several plates lapped end to end the kinks accumulate. A tension member spliced three times by single laps arrives on site with three angles in it, all in the same direction if the laps were detailed the same way, and the member is visibly bent.

The remedy is to alternate the laps, which cancels the kinks and introduces a serpentine that is at least symmetric. That is a detailing rule with no calculation behind it, arrived at by looking at a completed structure.

Which free body produced the number #

Cut the joint at the fastener and take one plate as the free body. On it: the applied tension along its own centreline, and the forces the bolt delivers.

The bolt’s force must equal the tension — but it acts at the interface, half a plate thickness off the centreline. Two forces of equal magnitude on parallel lines half a thickness apart are a couple, and the plate has to carry it. Sum moments about any point and the couple is there: $M = P\,t/2$.

The moment is not an approximation or a secondary effect. It is required by equilibrium of a body anybody can draw, and no amount of care in the analysis makes it go away — only a change of geometry does.

Where the regimes divide #

The group $uc$ separates two worlds, and it is worth putting numbers on both because the same equation describes them and the same equation is quoted about both.

$$uc \propto \frac{c}{t}\sqrt{\frac{\sigma}{E}}$$

Structural steelwork. $c/t$ around 3, $\sigma/E$ around $3\times10^{-4}$. $uc \approx 0.06$, $k \approx 0.86$, and the joint carries almost the full eccentricity moment.

A bonded aluminium skin joint. $c/t$ around 12, $\sigma/E$ around $2\times10^{-3}$. $uc \approx 0.6$, $k \approx 0.4$, and the joint has shed most of it.

A thin-sheet cold-formed connection. Somewhere between, and moving with the load.

Two factors of about five, one in each term under the root and one outside it, and they compound. The same joint arrangement is a bending problem in one industry and very nearly a shear problem in another, and the difference is entirely geometric.

The three ways out #

Use a double lap. A cover plate on each side puts the load path back on the centreline and the eccentricity vanishes exactly. It costs a second plate and doubles the number of shear planes, so the bolt does twice the work; almost every splice in structural steelwork is detailed this way and this essay is the reason.

Make the joint long. The straightening depends on $c/t$, so more overlap helps — slowly. Going from six thicknesses to eighteen takes the ratio from 3.8 to 3.0, which is a lot of plate for a fifth of the problem, and a long bolted joint has troubles of its own.

Accept it and use the factor. Which is what is done, and is reasonable for a statically loaded joint where 40 per cent off the bearing is affordable. It is not reasonable for a fatigue-loaded one, and the detail categories for single lap joints in the fatigue tables are correspondingly grim.

Why it is used anyway #

A joint with a peak stress four times its mean, a bolt at 103 per cent utilisation and a 40 per cent bearing penalty ought to be extinct. It is instead the commonest connection there is, and the reasons are worth listing because none of them is structural.

It needs one plate. A double lap needs a cover plate on each side, cut, drilled, transported and fitted, and on a member that is already a plate the second and third pieces are pure cost.

It can be made from one side. A splice reachable from one side only — a hollow section, a member against a wall, a repair — admits nothing else.

It is what the members already are. Two angles, two channels, a bracing member and a gusset: the pieces arrive with faces that meet, and a lap is what happens when they are bolted together. Detailing anything else means adding material whose only job is symmetry.

And it works. The static capacity is much closer to the symmetric value than the elastic stresses suggest, because the plate yields locally at the hole and sheds the bending. What the arrangement is genuinely bad at is fatigue, and most connections do not see any.

Where the model stops #

The plates are the same thickness and the same material. Unequal plates put the load paths at unequal distances from the interface and the moment is shared unevenly; the thinner plate gets the worse of it, which is the reverse of what a strength check would suggest.

There is one bolt. With a row of them the joint is stiffer against rotation and $k$ is lower, and the moment redistributes along the row; the end bolts carry more of it. Combining that with the load-sharing problem a long joint already has needs a model with both in it, which is not a hand calculation.

Goland and Reissner solved a bonded lap. Their plates are continuously connected over the overlap; a bolted lap is connected at points, so the plates can separate between them and the rotation is not the same. The $k$ used here is an upper bound on the straightening a bolted joint gets, which makes the results on this page optimistic.

The bolt is not preloaded. A preloaded bolt clamps the plates together and the joint transfers load by friction until it slips, during which the eccentricity produces much less bending because the two plates are acting as one 20 mm section. After it slips, everything on this page applies.

And no drawing here shows the joint after it has yielded. The eccentricity moment is a first-order effect that the plate can shed by yielding locally at the hole; a static ultimate capacity is therefore much closer to the symmetric value than the elastic stresses suggest. The problem is a serviceability and fatigue problem wearing a strength problem’s clothes.

The number to carry away #

If one thing survives from this page it should be the group rather than any of the stresses.

$$\text{peak} = \sigma_m\left(1 + 3k\right), \qquad k = k\!\left(\frac{c}{t}\sqrt{\frac{\sigma}{E}}\right)$$

The 3 is the eccentricity being exactly one plate thickness, which is a geometric fact about lapping two plates and is the same for every single lap ever made. The $k$ is the only thing that varies, it runs from 1 to about 0.2, and everything about a particular joint that matters is in the one dimensionless group it depends on.

So the question to ask of any lapped joint is not how thick is the plate or how long is the lap but how many thicknesses is the lap, and how hard is it stressed — and structural steelwork’s answer to both is not very, which is why its lapped joints sit at the unhelpful end of a curve whose helpful end is well documented in a different literature.

The ladder from here #

Later rungs on this anchor: the bolted lap solved properly, with discrete fasteners and separation between them, which is a contact problem. The unequal-thickness lap and which plate suffers. Preloaded single laps, and what the eccentricity does to the slip resistance before it does anything to the bearing. Fatigue of lap joints in detail, where the category depends on whether the plate edge is flame-cut or machined and the eccentricity is only one of the three things being penalised. Riveted lap joints in old structures, where the arrangement is universal and the assessment of an existing bridge turns on exactly this arithmetic. And the aerospace version, where a lapped skin joint is bonded rather than bolted, the plates are a millimetre thick, $uc$ is large, and Goland and Reissner’s factor genuinely does most of the work it is famous for.