The bolts that do not share

Divide the force by the number of bolts. It is the first thing anyone does with a bolted connection and it is the assumption underneath every bolt group calculation on this site — including the eccentric ones, where the division is by a polar second moment rather than by a count but is still a division.

For a short joint it is right. For a long one it is not, and the reason is not in the bolts at all. It is that the two plates being joined are elastic, and along the length of the lap they are carrying different forces and therefore straining at different rates.

Which free body produced the number #

Take a length of the lap between two bolts and consider the two plates separately.

At the leading end of the joint, plate one is carrying the whole force and plate two is carrying nothing. So over that first pitch, plate one is straining at $P/EA_1$ and plate two at zero, and the difference accumulates as slip across the bolt. At the trailing end the same thing happens with the plates exchanged. In the middle, each plate is carrying about half, both are straining at about the same rate, and the slip barely changes from one bolt to the next.

A bolt transfers force in proportion to the slip across it. So the bolts at the ends, where the slip is largest, transfer the most, and the ones in the middle, where the plates are moving together, transfer the least.

Written as a differential equation in the transferred force per unit length $q$, with $k$ the bolt stiffness per unit length,

$$\frac{d^2 q}{dx^2} = \beta^2 q, \qquad \beta^2 = k\left(\frac{1}{EA_1} + \frac{1}{EA_2}\right)$$

whose solution for equal plates is $q \propto \cosh\beta(x - L/2)$ — a curve with its minimum at the centre and its maxima at both ends. Integrating it and comparing the mean with the peak gives the efficiency of the joint:

$$\eta = \frac{\tanh(\beta L/2)}{\beta L/2}$$

The discrete version is solved rather than assumed: $n-1$ compatibility equations between consecutive bolts, one equilibrium equation that the forces add to the load, and $n$ unknowns. The answers agree with the continuum formula to a fraction of a per cent once the bolts are close enough together, which is the check that the shape is the shape.

The same function again, and the third time #

That expression has appeared on this site twice before, in problems with nothing physically in common.

A reinforcing bar developing its force in concrete obeys it, with the bond stiffness in place of the bolt stiffness and the concrete in place of the second plate. The bond stress is crowded against the loaded end, dies away over $1/\alpha$, and the efficiency of an embedded length is $\tanh(\alpha L)/(\alpha L)$.

An angle connected through one leg obeys the same idea in two dimensions, where what varies is the distance from the fastener line rather than the position along it. And a cooling fin’s temperature obeys it exactly, with heat in place of force.

What all four share is an interface between two things whose stiffnesses differ, across which something is being handed. The quantity handed is a force in three of them and a heat flux in the fourth; the equation does not care.

Which is not a reason to cut the joint in half #

The elastic answer is not the capacity, and getting that relationship right is the whole of the practical content here.

A bolt in bearing is ductile. Once the end bolt reaches its capacity it does not fail — the plate around it deforms, the hole elongates, and the bolt goes on carrying its capacity while taking no more. The force that would have gone to it goes further in instead, and the process repeats. Given enough deformation capacity, every bolt reaches its own limit and the joint carries the sum.

So the question a long-joint rule is really asking is: can the end bolt deform far enough for the far bolts to catch up? For a short joint the elastic distribution is nearly uniform and the answer is trivially yes. For a long one the end bolt’s demand grows and its supply does not, and somewhere the answer becomes no.

That is why the code’s reduction starts at fifteen diameters rather than at the point where the elastic distribution becomes uneven — which is around eight — and why it stops falling at 0.75 rather than following $\tanh(u)/u$ down. It is calibrated against tests to failure, not against an elastic solution, and the elastic solution is a description of the demand for ductility rather than of the capacity.

Where the fastener is not ductile, the elastic answer is the one that governs. A preloaded joint carrying its load by friction has no plateau before it slips, so the slip resistance of a long friction-grip joint really does follow something like the elastic distribution — and long friction connections are reduced harder than long bearing ones for exactly that reason.

What the numbers look like on a real splice #

It is worth putting one joint through the arithmetic, because the sizes involved decide whether any of this matters.

Take two 200 by 10 plates spliced with M20 bolts at 70 mm pitch. Each plate has $EA = 4.2 \times 10^8$ N; a bolt’s shear-and-bearing stiffness is about 80 kN/mm, which spread over the pitch is 1,143 N/mm per millimetre of joint. That gives $\beta = 2.33 \times 10^{-3}$ per millimetre and a decay length $1/\beta$ of 429 mm, or a little over six bolt pitches.

A joint has to be an appreciable fraction of that length before anything happens. Eight bolts is 490 mm, $\beta L/2 = 0.57$, and the efficiency is 0.90 — the end bolt is nine per cent over its share and nobody would notice. Twenty-four bolts is 1,610 mm, $\beta L/2 = 1.88$, and the efficiency is 0.51: the end bolt is carrying nearly twice its nominal share while the middle bolts are carrying a fifth of it.

The decay length is what to remember, because it puts the whole effect on a scale. It goes as the square root of the plate stiffness over the bolt stiffness, so a joint between two thick plates has a long decay length and a nearly uniform distribution, and one between two thin plates with stiff fasteners has a short one and a very uneven distribution. That is the reverse of the intuition that thick plates are the difficult case.

Unequal plates, and where the joint fails first #

Everything above assumed the two plates have the same axial stiffness, which makes the distribution symmetric. They usually do not.

A splice plate is thinner than the member it splices, or two members of different section are joined. Then the two ends of the joint are not equivalent: at the end where the stiffer plate is fully loaded, the strain difference is smaller and the slip is smaller, and the peak moves towards the other end.

The practical form of that is a splice where one side is more heavily worked than the other, and it is worth checking rather than assuming — because the peak bolt force can be twenty per cent higher at the worse end than the symmetric solution suggests. Solving the discrete system with the actual plate areas is a line of arithmetic and there is no reason to guess.

Which failure arrives, and where #

A long joint has three checks that all depend on its length, and they do not all move the same way.

The bolts. Reduced by the long-joint factor, as above.

The net section. The plate has holes in it, and the section through the first row of holes carries the whole force while the section through the last carries almost none. So the net section check at the leading row is unaffected by the joint’s length, while adding bolts at the far end adds capacity without adding demand there.

Block shear. The metal between the holes comes out as a block, and lengthening the joint lengthens the shear planes — so this check improves with length, faster than the bolt check deteriorates.

The three together mean that a long joint is not a bad joint, it is a joint whose governing check has moved. Short joints are usually governed by the bolts; long ones by the net section, which is not reduced at all. That is why the long-joint rule so rarely changes a design: by the time it bites, something else has.

Fifteen diameters, and where the number came from #

The threshold is oddly specific and it is worth knowing that it is a measurement rather than a derivation.

Long bolted joints became a subject when riveted construction gave way to high-strength bolts in the 1950s and 1960s, and the immediate question was whether a splice in a bridge girder flange — which might be a metre and a half of bolts — could be rated at the sum of its fasteners. The research council programmes of that period tested joints of increasing length to failure and plotted the mean bolt force at failure against the joint’s length.

What came out was a curve that is flat to about fifteen diameters and then falls, reaching about three quarters of the short-joint value by sixty-five diameters and staying there. The flat part is not the elastic distribution being uniform — it plainly is not, by eight diameters — it is the length over which the end bolt’s deformation capacity is enough to redistribute everything. The falling part is where it stops being enough, and the floor is where the joint’s failure mode changes and stops getting worse.

The rule that came out of it, $\beta_{Lf} = 1 - (L_j - 15d)/(200d)$ between those two limits, is a straight line fitted to that curve. It has no derivation and does not pretend to one, and the elastic analysis in this essay is a description of the mechanism the fit is describing rather than a competitor to it.

Two details of the calibration are worth carrying. The tests were on joints of ordinary steel with generous edge distances, so the end bolt’s plate could deform freely; a joint at minimum edge distance splits instead of deforming and does not get the same redistribution. And the tests measured the bolts, not the plates, so the reduction applies to the fastener check and not to anything else on the connection.

There is a symmetry between that history and the elastic analysis worth naming. The tests measured a capacity and the analysis describes a demand, and the two meet at a quantity neither of them reports: how far a bolt hole can elongate before the plate tears. That number is a few millimetres for ordinary steel at generous edge distance, and it is the hinge on which the whole rule turns. Nothing in a connection calculation ever quotes it.

Where the model stops #

The bolt stiffness is not a number. It is the combined flexibility of the bolt in shear, the bolt in bending inside the hole, and the plate in bearing at the hole — and the last of those depends on the plate thickness, the edge distance and the hole clearance. Published values range over a factor of two, and $\beta$ goes as the square root of it, so the decay length is uncertain by a factor of about 1.4. The shape of the answer is robust and the numbers are not.

Clearance holes mean nothing is in contact until it moves. A bearing-type joint has 2 mm of clearance round each bolt, and the bolts do not all come into bearing at once — the ones that happen to be tight against the plate go first. That is a source of non-uniformity that has nothing to do with elasticity and can be larger than the elastic effect in a short joint, and it disappears once the joint has taken up.

And the model has one row. A real splice has several rows across the width as well as several along the length, and the transverse distribution has its own shear-lag problem that this analysis does not contain.

What it means for a splice that has to be long #

The place this arithmetic actually decides something is a splice that cannot be shortened, and there are two of them.

The first is a tension member splice in a truss chord, where the force is large, the plate is thin because the member is a tension member, and the bolt count is therefore high. Making the joint shorter means more rows across the width, which runs into net-section area, and the two constraints close on one another. The usual resolution is a thicker splice plate and fewer, larger bolts, which shortens the joint by raising the force each bolt carries — and the long-joint factor is what makes that trade worth making.

The second is a cover-plated flange splice on a plate girder, where the flange is wide enough for many rows and the joint can be made short by using the width. That is the standard detail precisely because it avoids the problem, and the reason it looks over-bolted is that the bolts are arranged for length rather than for count.

The generalisation #

There is a general rule about connections buried here, and it is worth stating on its own because it decides how much of this arithmetic anybody ever needs.

A connection made of ductile components can be designed on equilibrium alone; one made of brittle components cannot. If every part can deform enough to let the others catch up, then any distribution of forces that satisfies equilibrium is achievable, the lower-bound theorem applies, and the designer is free to choose whichever distribution is convenient. If they cannot, the actual distribution has to be computed — which means solving a compatibility problem, which means knowing stiffnesses nobody has measured.

That is the reason bolts in bearing are the default and welds are checked more carefully, the reason a preloaded joint is reduced harder than a bearing one, and the reason resin anchors and post-installed fixings come with rules that look disproportionately fussy next to the ones for bolts. The fussiness is the price of not being able to redistribute.

Read that way, the long-joint reduction is not an exception to the divide-by-$n$ rule. It is the point at which the ductility that was silently paying for the rule runs out.