The bolt that carries more than its share

A beam frames into a column through a bracket. The reaction is 100 kN. The bracket is bolted on with six bolts, and the question is what size they need to be.

The arithmetic that suggests itself takes two seconds. A hundred kilonewtons over six bolts is 16.7 kN each. Choose a bolt whose shear capacity exceeds that with a margin and the connection is done.

The equal-share arithmetic is not wrong about the direct shear. It is missing the other term entirely, and the other term is twice as large as it is.

Where the eccentricity comes from #

It is worth being clear that the 150 mm is not a mistake somebody made. It is the connection’s size, and it is set by things that have nothing to do with the load.

The bolts must be far enough apart to get a spanner between them, and far enough from the plate edges not to tear out. The bracket must be deep enough to reach the beam’s web. The beam’s reaction acts at the centre of its bearing on the bracket, which is out where the beam is, not in where the bolts are.

Add those up and the line of action of the load sits some distance from the centroid of the bolt group, always. There is no detailing choice that makes it zero, only choices that make it smaller — and the ones that make it smaller usually make something else worse. This is the general form from the previous essay in its first concrete instance: a connection has dimensions, and every dimension is a lever arm for something.

Two loads, added as vectors #

Move the load to the centroid and add the moment that move requires. That is the standard trick for shifting a force, and it is exact: a force at distance e from a point is equivalent to the same force at the point plus a moment Ve.

Now the group carries two things, and they are shared by two different rules.

The force is shared equally: V/n on every bolt, in the direction of the load. Six bolts, 100 kN, 16.7 kN each, all pointing down.

The torque is shared in proportion to distance from the centroid, at right angles to the radius. Each bolt resists rotation with a force proportional to how far it is from the centre of rotation, exactly as material far from the neutral axis resists bending in proportion to its distance. The constant of proportionality is fixed by requiring the moments to sum to the applied torque:

$$R_i = \frac{T r_i}{\sum r^2}$$

with $\sum r^2$ the polar second moment of the bolt positions — the same $\int y^2 \, dA$ that runs through the whole of section behaviour, evaluated over points instead of over an area.

For the group above, with bolts at ±37.5 mm horizontally and −75, 0, +75 mm vertically, that sum is 30,938 mm². The torque is 15,000 kN·mm. So the corner bolts, 83.9 mm from the centroid, take a torsional force of 40.7 kN.

And then the two are added as vectors, not as numbers. This is the step the equal-share arithmetic cannot reach, and it is where the interesting behaviour lives, because the two components point in different directions at every bolt and the directions differ from bolt to bolt.

The spread, which is larger than anybody guesses #

The six bolts in the figure carry 50.4, 50.4, 36.4, 36.4, 34.8 and 1.5 kN.

That last number is the one worth sitting with. One of the six bolts in this connection is carrying one and a half kilonewtons — under a tenth of what the equal-share calculation assigns it, and under a thirtieth of what the worst bolt is carrying. Its torsional component happens to point almost exactly opposite to the direct shear and to be almost exactly the same size, so the two cancel.

The spread from 1.5 to 50.4 is a factor of thirty-four across six identical bolts in one connection under one load.

The layout matters more than the bolts do #

Because the torsional term goes as $Tr/\sum r^2$, everything about the group’s geometry enters the answer twice — once in the numerator through how far the worst bolt is, and once in the denominator through how far all of them are.

That double appearance is the same structure as the second moment of area, and it has the same consequence: spreading the group out is far more effective than adding to it. Doubling every spacing in a group quadruples $\sum r^2$ and doubles the worst radius, so the torsional force on the worst bolt halves. Doubling the number of bolts while keeping the group the same size does much less, because the extra bolts go in the middle where $r$ is small and they contribute almost nothing to $\sum r^2$.

The middle bolt in the first figure makes the same point without needing arithmetic. It sits on the centroid’s horizontal axis at $r = 37.5$ mm, contributes 1,406 mm² to a $\sum r^2$ of 30,938 — under five per cent — and carries 1.5 kN. It is very nearly not in the connection.

There is a limit to spreading, and it is not a bolt limit. A wider group needs a wider bracket, which is more plate, and a deeper group runs out of beam to bolt to. But within those bounds the lever is geometric and the returns are quadratic, which is a much better exchange rate than any that buying stronger bolts offers — the same trade this site has met in sections, in beam depth and in column bracing.

A group with no width at all #

The single column of bolts is worth its own paragraph, because it is the commonest connection in steelwork and it is the worst case of everything above.

A fin plate carries the beam’s reaction on one vertical line of bolts. With no horizontal spread, every bolt’s radius from the centroid is purely vertical, so the torsional force on every bolt is purely horizontal — at right angles to the reaction it is resisting. The two components are perpendicular, which is the least favourable relative direction there is, and the resultant is $\sqrt{V^2/n^2 + (Tr/\sum r^2)^2}$ with nothing cancelling anywhere.

Three bolts at 75 mm pitch carrying 100 kN at 150 mm gives a worst bolt of 105 kN — more than the whole applied load, on one bolt, in a connection that a divide-by-n calculation would have sized at 33 kN a bolt. The group’s capacity is 0.95 bolts: three bolts arranged this way are worth less than one bolt would be if the load passed through it.

That is not an argument against fin plates, which are used constantly and work. It is an argument for knowing what the eccentricity is, because a fin plate designed with the bolts close to the column is a different connection from one with the bolts 150 mm out, and the difference is not proportional. It is also the reason a fin plate is usually detailed as short as the beam’s web will allow: the only variable in that connection that the designer controls freely is the one in the numerator.

The eccentricity that is not in the plane of the drawing #

Everything above is planar: a load in the plane of the bolt group, an offset in that plane, and forces on the bolts in that plane. Real brackets have a second eccentricity at right angles to all of it, and the planar calculation cannot see it at all.

A bracket bolted to a column flange stands proud of that flange. The beam sits on the outstand, so its reaction acts at some distance in front of the plane the bolts lie in — set by the bracket’s own thickness, the beam’s bearing position and whatever packs are between them. That offset turns the same 100 kN into a moment about a horizontal axis lying in the flange, and a moment about an axis in the plane of a bolt group is resisted by tension in the bolts on one side and bearing on the other.

Take the offset as the same 150 mm, and treat the six bolts elastically about the group’s centroid: $\sum y^2 = 2(75^2 + 0 + 75^2) = 22{,}500$ mm², and the top row carries $15 \times 10^6 \times 75 / 22{,}500 = 50$ kN of tension per bolt.

So the corner bolt that the in-plane calculation put at 50.4 kN of shear is also carrying 50 kN of tension, and neither number knows about the other. Bolts in combined shear and tension are checked on an interaction — the common code form is $V/V_R + T/1.4T_R \le 1$ — and for an M20 8.8 bolt with about 94 kN of shear capacity and 136 kN in tension that reads $0.54 + 0.26 = 0.80$. The connection is at four-fifths of its bolts, against the 54 per cent that the shear check on its own reports.

Two things follow that are worth carrying past this example.

The tension is usually the term that is forgotten, and it is set by a dimension nobody thinks of as structural — the thickness of a plate, the position of a beam’s bearing, a packing shim added on site to make something fit. Each is a length on a fabrication drawing rather than on a frame diagram, and each is a lever arm.

And the bolts-in-tension model here is a choice rather than a calculation, which is easy to miss because it looks like the in-plane one. Taking the compression on the far bolts assumes the connection pivots about its own centroid; a real bracket bears on the flange along its bottom edge instead, which moves the axis down and lengthens the lever the tension works on. Solved properly the top row carries 24.6 kN rather than 50 — the assumption above is conservative by a factor of two, and it is conservative for a reason nothing here states. That is the same disagreement about where the centre of rotation sits that the two methods below are about, arriving in the other plane, where there is no instantaneous-centre method to appeal to.

What the elastic method assumes #

The calculation above has a name — the elastic, or vector, method — and it rests on two assumptions worth stating because they are both false and both conservative.

The plates are rigid. All the deformation is in the bolts, and the bracket rotates as a solid body about the group’s centroid.

The bolts are linear. Force is proportional to deformation, without limit, so the bolt that is furthest away is the one that is worst loaded and the group has reached its capacity when that bolt has reached its own.

Real bolts in real holes do not behave that way. A bolt in shear has a load–deformation curve that flattens out well before it fractures: it yields, the hole ovalises, the plate crushes locally, and the bolt keeps carrying its load while it deforms considerably further. Which means that when the worst bolt reaches its capacity, the group has not. The worst bolt holds its load, the others take up more, and the connection carries on.

That is exactly the argument that makes plastic analysis work for beams, arriving in a different part of the subject. The first fibre to yield does not fail the section, and the first bolt to reach capacity does not fail the group, and in both cases the reason is ductility rather than strength.

The instantaneous centre #

The alternative method takes the redistribution seriously and asks a different question: not “what force does each bolt carry under this load?” but “what is the largest load this group can carry, given that every bolt will deform as much as it needs to?”

Under load the bracket rotates about some point. That point is not the centroid; it is wherever it needs to be for the bolt forces to balance the applied load. It is called the instantaneous centre of rotation, and finding it is the calculation.

Each bolt’s deformation is proportional to its distance from that centre. Each bolt’s force follows a measured load–deformation law — the one usually used is Crawford and Kulak’s fit,

$$R = R_{\mathrm{ult}}\left(1 - e^{-10\Delta}\right)^{0.55}$$

with $\Delta$ in inches and a fracture deformation of 0.34 in. That equation is the one piece of pure empiricism in this field: it is a curve fitted to tests, it has no derivation, and it is stated in imperial units because that is where the tests were done. Using it in millimetres without converting is not a small error — the furthest bolt then reaches 31% of its capacity instead of 98%, and the method reports a group weaker than the elastic one, which is the diagnostic, because the instantaneous centre can never be the lower of the two.

Take the furthest bolt to its fracture deformation, scale every other bolt’s deformation by its distance from the centre, read each force off the curve, and require that the resulting forces be in equilibrium with the applied load. That is one equation in one unknown for a load with an axis of symmetry, and it is solved by search.

What the difference is worth #

Both methods answer the question “how many bolts’ worth of capacity does this group have?”, and they disagree.

Thirteen per cent is not enormous and it is not nothing. On a connection that governs a member’s size it is the difference between six bolts and seven, or between a bolt grade and the next one down, repeated across every bracket in a building.

But the more useful reading is not the size of the gap. It is that the gap is not an error. Neither method is trying to compute the other one’s answer more accurately. They are different physical assumptions:

• the elastic method assumes the group fails when its worst bolt reaches capacity, and reports the load at which that happens;
• the instantaneous centre assumes the group fails when its worst bolt fractures, having redistributed to every other bolt on the way, and reports that load.

Both are correct answers to their own question. Which one is the right question depends on whether the bolts really are ductile enough to redistribute — and that is a question about the bolt grade, the plate thickness, and whether the bolt’s threads are in the shear plane, none of which appears in either calculation.

The curve near zero, and why it dips #

There is one feature of the sweep worth explaining because it looks like a bug.

Just off zero eccentricity the instantaneous centre method comes out marginally below the elastic method — by 1.2% at an eccentricity of one millimetre. Since the redistribution argument says it can never be lower, that reads as an error.

It is not, and the reason is in the method’s own assumption. The instantaneous centre calculation anchors the deformation at the furthest bolt: that bolt is at its fracture deformation and every other bolt is proportionately short of it, so every other bolt is slightly below its capacity. A concentric elastic group, by contrast, has every bolt at exactly its capacity simultaneously, because there is no torque to make them differ.

So near the concentric limit the elastic method is momentarily the more generous of the two, for the same reason that made it conservative everywhere else. At exactly zero eccentricity the two agree, both giving $n R_{\mathrm{ult}}$, and the instantaneous centre of a pure translation is at infinity — which the solver has to return explicitly, because a search for a point at infinity does not converge, it merely stops somewhere.

The same reading settles what happens when the fasteners are a line rather than a set of points. A welded bracket carries the identical force and the identical torque, the identical vector sum applies at every point along the run, and the polar second moment becomes an integral instead of a sum. What changes is not the mechanics but the ductility: a weld has almost none to redistribute with, so only one of the two methods is available to it at all — which is the subject of its own essay, and the reason that essay’s answer is not at the corner.

What to take from it #

Three things, in descending order of how often they matter.

Divide-by-n is a calculation about a different connection. If the load misses the centroid, and it does, the group carries a torque as well as a force, and the two add as vectors at every bolt.

The spread within a group is enormous and it is geometric. Nothing about the bolts causes it. Moving the load closer, or spreading the group wider, changes the answer by factors rather than percentages — which makes the layout the design decision and the bolt size a consequence of it.

A conservative method is conservative for a reason, and the reason is a physical assumption that can be checked. The elastic method’s conservatism is exactly the ductility of the bolts. Where the ductility is there, the capacity is there. Where it is not — a brittle bolt, a thin plate, a connection that has to work at low temperature — the elastic method is not conservative at all; it is the only one of the two whose assumption still holds.