Connections

The joint that is crooked by construction

Lap two plates and fasten them and the two load paths are offset by the thickness of a plate. The joint carries a moment nobody applied, the peak stress is four times the mean, and the rotation that is supposed to straighten it out saves eleven per cent — because the rescue works for thin sheet with a long lap and a bolted structural joint is neither.

Assumes The bolt that carries more than its share, The hole that goes oval, and the one that tears to the edge and The moment the beam left behind.

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 load path has a kink in it, and the kink is a plate thick. Two 10 mm plates lapped over 60 mm and pulled with 60 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 10.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 86 per cent of it, leaving 3.57 times — a saving of 11 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 382 N/mm² of bending against 191 of shear.
Fig. 1 Two 10 mm plates lapped over 60 mm and pulled with 60 kN. The dashed line is where the force actually runs, and it has a kink in it a plate thick. Taken at face value the eccentricity gives a peak stress four times the mean; the joint rotates under load and reduces it to 3.57 times, a saving of 11 per cent. The bolt is bent as well as sheared — 382 N/mm² of bending against 191 of shear.

The naive number, and why it is nearly right

Take the joint as drawn and do not let it move. The eccentricity is e=(t1+t2)/2=te = (t_1+t_2)/2 = t for equal plates, and the moment at the fastener is M=Pe/2M = Pe/2. The bending stress that produces is

σb=6Mbt2=3Ptbt2=3σm,\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 kk of the naive value, and Goland and Reissner solved for kk in 1944 for a bonded lap:

k=cosh⁡(uc)cosh⁡(uc)+22sinh⁡(uc),u=3(1−ν2)2 1tσEk = \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 cc the half-overlap. The whole behaviour is in the group ucuc, which is proportional to (c/t)σ/E(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.

A longer lap straightens the joint, and a bolted one is never long enough. The peak stress in a single lap joint, as a multiple of the mean, against how long the overlap is in plate thicknesses. With no rotation at all the answer is 4.00 — one part membrane and six parts bending, because the load paths are offset by a whole plate thickness. The joint rotates under load and the bending falls, but only as tanh of (c/t)√(σ/E): at the 6 thicknesses drawn the ratio is 3.57, and at 18 it is still 3.00. A bonded lap in thin sheet gets most of the way to one; a bolted lap in structural plate does not get halfway.
Fig. 2 The peak stress ratio against how long the overlap is, in plate thicknesses. It starts at very nearly the naive 4.0 and falls slowly: at the six thicknesses of the joint above it is 3.57, and at eighteen it is still 3.00. The axis is the ratio c/tc/t rather than a length, so a thin sheet moves along it without the lap getting any longer — and only a joint far to the right of it, stressed far harder than this one, gets most of the way down to 1. A bolted structural lap sits at the left-hand end.

For the joint drawn — 10 mm plate, 60 mm overlap, 60 N/mm² of membrane stress — ucuc is 0.06 and kk 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

kk 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, ucuc is at its largest, and kk is at its smallest. At the service load — a quarter of it, say — the stress is a quarter, σ\sqrt{\sigma} is a half, ucuc is halved and kk rises to 0.92.

So the joint is relatively worse at the load it spends its life at, and the same figure drawn at a quarter of the load says so.

The load path has a kink in it, and the kink is a plate thick. Two 10 mm plates lapped over 60 mm and pulled with 15 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 10.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 92 per cent of it, leaving 3.77 times — a saving of 6 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 95 N/mm² of bending against 48 of shear.
Fig. 3 The same joint at 15 kN instead of 60 — the service load rather than the ultimate one, on the same plates with the same overlap. The rotation is worth less at the lower load, not more: the moment falls to 92 per cent of the naive value instead of 86, so the peak stress ratio rises from 3.57 to 3.77 and the saving drops from 11 per cent to 6. The bolt’s own stresses fall in proportion, to 95 N/mm² of bending against 48 of shear, but their ratio of two to one is a property of the geometry and does not move with the load at all.

The ratio of 3.57 at ultimate against 3.77 at service is small in itself and it points the wrong way, which is what matters. 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.

Three details, and no material anywhere on the plot. Stress range against cycles to failure for three detail categorys — 160, 90, 36 N/mm² at two million cycles. The lines are parallel because they share a slope of three, and the spread between them is a factor of 4.4 in stress and therefore 88 in life. Nothing on this plot depends on the strength of the steel: the same detail in a grade twice as strong lies on the same line. No working stress range is marked. The knee in each line is the constant-amplitude limit, past which the slope becomes five.
Fig. 4 Where that matters. A detail category is a stress range, and a single lap joint’s stress range is the peak range rather than the mean — so a joint nominally at 60 N/mm² is a detail experiencing 214. That is three categories’ worth of difference on a chart whose whole content is which category a detail is in.

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=191P/A = 191 N/mm², bending stress M/Z=382M/Z = 382 N/mm² with M=P e/2M = P\,e/2 and Z=πd3/32Z = \pi d^3/32. The combined utilisation against fuf_u is 1.03, against 0.61 from the shear alone.

Six ways for one dowel to fail, and the capacity is the smallest. Johansen's single-shear mechanisms for a 20 mm dowel through 10 and 10 mm members, each drawn as the shape it is: the dowel straight and the timber crushing, the dowel rotating rigidly, one plastic hinge, then two. The capacity under each is that mechanism's own, and the joint's strength is the smallest — 1.90 kN by mode c, which is the dowel rotates rigidly and both members crush. That is the kinematic theorem of plasticity: every mechanism gives an upper bound and the true collapse is the lowest of them. The crushed timber is shaded, and the circles are plastic hinges in the steel.
Fig. 5 The general form of the same problem, drawn for a timber dowel where the mechanism is visible. A fastener through two members can fail by bearing in either, by a single plastic hinge in itself, or by two — and which one arrives depends on the ratio of the fastener’s bending strength to the members’ bearing strength. A steel bolt in thick plate is at the stiff end of that family and a bolt in thin plate is not.

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 1.5fudt1.5 f_u d t instead of the usual 2.5fudt2.5 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 3σm3\sigma_m of bending, so it reaches its limit at a lower nominal bearing stress, and the calibration puts the ratio at 0.6.

Bearing and tear-out against end distance. A 20 mm bolt in a 10 mm plate. Below 165 mm of end distance the bolt tears a channel out to the end and the capacity is proportional to that distance; above it the plate crushes in front of the bolt and the end distance stops mattering. At 40 mm the capacity is 52.12 kN and the mode is tear-out.
Fig. 6 The check the factor is applied to, which is about end distance and tear-out and knows nothing about eccentricity. Multiplying its result by 0.6 is a way of carrying an entirely different mechanism inside a formula for another one — which works, and leaves nobody able to say what would happen if the geometry changed.

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.

The same family of consequence turns up at a bolted tee, where a flange bending under a tension puts a prying force into the bolt that nobody applied. In both cases a member deforms to accommodate a geometry and the accommodation generates a force, which is the signature of every compatibility-driven action in this collection: the force exists because something moved, so it is bounded by how far the thing can move and it does not appear in any equilibrium statement about the applied loads.

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/2M = 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.

The same free body drawn for an angle bolted through one leg gives the same answer at a larger scale. There the offset is between the angle’s centroid and the bolt line rather than between two plate centrelines, so it is several times a plate thickness rather than one — and the code’s response is again a coefficient, the effective net area, rather than a moment anybody computes.

Where the regimes divide

The group ucuc 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∝ctσEuc \propto \frac{c}{t}\sqrt{\frac{\sigma}{E}}

Structural steelwork. c/tc/t around 3, σ/E\sigma/E around 3×10−43\times10^{-4}. uc≈0.06uc \approx 0.06, k≈0.86k \approx 0.86, and the joint carries almost the full eccentricity moment.

A bonded aluminium skin joint. c/tc/t around 12, σ/E\sigma/E around 2×10−32\times10^{-3}. uc≈0.6uc \approx 0.6, k≈0.4k \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 load path has a kink in it, and the kink is a plate thick. Two 1 mm plates lapped over 20 mm and pulled with 6 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 1.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 64 per cent of it, leaving 2.93 times — a saving of 27 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 477 N/mm² of bending against 477 of shear.
Fig. 7 The other end of the range, drawn by the same machinery: two 1 mm sheets lapped over 20 mm and pulled at the same 60 N/mm² of membrane stress as the 10 mm plates above. The overlap is only 20 mm and it is twenty thicknesses rather than six, which is the whole of the difference — the moment falls to 64 per cent of the naive value against 86, and the peak stress ratio to 2.93 against 3.57. A saving of 27 per cent rather than 11, from a geometry that is smaller in every dimension.

That is a pattern worth recognising rather than an isolated coincidence, and it is the same one the mode between the two that get checked is about. A dimensionless group that runs over orders of magnitude between one application and another means the same governing equation produces qualitatively different behaviour in different fields, and the received wisdom of each field is a statement about its own end of the range.

The same joint in compression, where the sign reverses

Every number on this page is for a joint in tension, and the whole of the relief — Goland and Reissner’s kk, the eleven per cent, the straightening — depends on that.

The mechanism is worth restating in one sentence before it is turned over. A lapped joint in tension rotates so as to bring the two load paths more nearly into line, because a tension pulling on a kinked path acts to straighten it. That is why kk is less than one, and it is why kk falls as the load rises: the harder it is pulled, the straighter it gets.

Put the same joint in compression and the geometry does the opposite. The offset is still one plate thickness and the couple is still Pt/2Pt/2, but now the load acting on the deflected shape drives the two plates further apart rather than pulling them into line. The joint is a strut with a built-in kink, the moment is amplified rather than reduced, and the amplification grows with the load exactly as second-order effects always do — as 1/(1−P/Pcr)1/(1 - P/P_{cr}) rather than as a factor below one.

So kk is not a number between 0.2 and 1 any more. It is greater than one, it rises with the load, and it has a load at which it diverges. There is no long-lap limit that rescues it: lengthening the overlap lengthens the strut and lowers the critical load, which makes it worse.

That is exactly the same reversal as a bowed two-force member — a bowed tie repairs itself and a bowed strut destroys itself, from the same M=PyM = Py — arriving in a connection instead of a member. And it explains a detailing rule that otherwise looks like conservatism: bracing members in compression are not single-lapped. A tension-only brace can be, and often is; a strut is connected with a symmetric arrangement or through a gusset with cover plates on both faces, and the 0.6 bearing factor that makes a single lap tolerable in tension does not make it tolerable here, because the mechanism it was calibrated against is not the one that occurs.

The load reversal case is worse than either. A brace that goes into compression under one wind direction and tension under the other has a joint that straightens on one cycle and buckles on the next, working the plates back and forth about their kink — which is a fatigue problem at a detail already several categories down.

What is trying to pull the plates apart

There is a second result in the same solution that this essay has passed over, and in a bonded joint it is the one that governs.

The eccentricity produces a bending moment along the overlap, and a bending moment that is largest at the ends of a lap requires a transverse stress across the interface to balance it — a tension pulling the two plates apart, concentrated sharply at the overlap’s two ends. It is called peel stress, and Goland and Reissner solved for it in the same paper as kk.

For a bonded lap it is decisive: adhesives are weak across their thickness, the peel stress peaks exactly where the adhesive is thinnest and most stressed already, and debonding starts at the overlap ends and runs inwards. Every design move in adhesive joints — tapering the adherend ends, filleting the adhesive, scarfing the joint — is aimed at that stress rather than at the shear.

For a bolted lap it is present and it is carried differently. The plates can separate between fasteners, so the peel appears as a prising apart of the two plates at the end of the lap, resisted by whatever tension the nearest bolt can supply. Which is the prying mechanism arriving in a joint that nobody would describe as carrying tension: the end bolt in a single lap has an axial force in it, generated by the geometry, in a connection whose entire calculation is about shear.

It is also the reason a single lap opens visibly at its ends under load, and the reason water gets into one. A joint that prises apart at its edges is a joint with a crevice at exactly the point of highest stress, which is where corrosion starts.

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/tc/t, so more overlap helps — slowly.

The load path has a kink in it, and the kink is a plate thick. Two 10 mm plates lapped over 180 mm and pulled with 60 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 10.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 67 per cent of it, leaving 3.00 times — a saving of 25 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 382 N/mm² of bending against 191 of shear.
Fig. 8 The same plates and the same 60 kN with the overlap trebled, from 60 mm to 180 — six plate thicknesses to eighteen. The moment falls to 67 per cent of the naive value instead of 86 and the peak stress ratio from 3.57 to 3.00, a saving of 25 per cent against 11. Trebling the plate has bought a sixth off the peak stress, and the joint is now long enough that the bolts along it stop sharing evenly.

Three times the lap for a sixth of the problem is a poor exchange, and it buys a second one: 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.

All four of those are versions of the general truth that a joint is made of springs in series. A connection analysed as a point is a region made of several deformable parts, each with its own stiffness, and the idealisation is always the optimistic one. What makes it acceptable is that the region is small and ductile. What makes it dangerous is a load case where neither is true, and fatigue is exactly that load case: the region is still small, and ductility buys nothing at all against a stress range.

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 kk 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 kk 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.

peak=σm(1+3k),k=k ⁣(ctσE)\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 kk 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, ucuc is large, and Goland and Reissner’s factor genuinely does most of the work it is famous for.

What this makes readable

Essays that name this one as a prerequisite.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

BearingBolt groupDetailingEccentricityFatigueLoad pathSecond-orderStress concentration