The bolt that was never fitted

Bolts go missing. A hole fouls a weld and is left open; a bolt is taken out to let a temporary brace pass and never goes back; a fabricator drills five of six because the sixth would have landed on a stiffener; a bolt is sheared off during erection and the hole is too damaged to reuse; forty years later one has corroded away. None of these is a design case and all of them turn up on site, and the question they ask is the same one: what does a bracket designed with six bolts carry with five?

The arithmetic everyone does first is a division. Six bolts carried 135.8 kN, so five carry five sixths of it, 113.2. That answer is wrong in both directions — it is 20 kN too low for one of the six possible omissions and 23 kN too high for another — and the reason it is wrong is that a bolt group’s capacity was never proportional to the number of bolts in it.

Three things change, and the share is the smallest of them #

The elastic force in a bolt of an eccentrically loaded group has two parts: an equal share of the load, $V/n$, and a torsional share $Vd\,r_i/J$ at right angles to the bolt’s own radius, where $d$ is the perpendicular distance from the centroid to the load’s line and $J = \sum r_i^2$ is the group’s polar second moment. Remove a bolt and every symbol in that expression changes except $V$.

The count $n$ falls from six to five, which raises the equal share by 20 per cent. This is the effect everyone has in mind and it is the smallest one.

The centroid moves. It is the average position of the bolts that are there, and taking one away drags it away from the gap — 16.8 mm for a corner bolt on this group, 7.5 mm for a middle one. That matters twice over: it changes every remaining bolt’s radius $r_i$, and it changes $d$, the load’s own lever arm, because $d$ is measured from the centroid to a load line that has not moved.

The polar moment falls, and by more than the count does. Removing a corner bolt takes 30,938 mm² to 22,500, a fall of 27 per cent against a count that fell by 17. The torsional share is inversely proportional to $J$, so this is a 37 per cent increase in the torsional force on every remaining bolt before anything else has happened.

The three effects do not have the same sign for every omission, which is what makes the answer hard to guess. Losing a bolt on the same side as the load shortens the load’s lever arm, because the centroid moves toward the load; losing one on the far side lengthens it. The first partly offsets the fall in $J$ and the second compounds it.

Which bolt it is matters by a factor of ten #

Four per cent against thirty-four. The group has six bolts and six ways of losing one, and the consequences differ by a factor of ten.

The pattern in the figure is not obvious and it is worth stating rather than leaving to be read off. The expensive bolts to lose are the two far corners on the side away from the load, bolts 3 and 6 at 34 and 32 per cent. The cheap ones to lose are the two middle bolts, 2 and 5 at 4 and 13. And the two near corners, 1 and 4, sit between at 7 and 17.

The reason is that the corners are what $J$ is made of. A corner bolt at 83.9 mm from the centroid contributes $83.9^2 = 7{,}037$ mm² to the polar moment; a middle bolt at 37.5 mm contributes 1,406, a fifth as much. Losing a corner is losing a fifth of the group’s resistance to twisting, and the group is being twisted.

The bolt that governed the check is not the bolt whose absence costs most. In the full group at its weakest direction the bolt at the top right, number 6, is the one at 100 kN while the bolt at the bottom left, number 2, carries 25. An engineer told that a bolt has to come out, reasoning from the check, would sacrifice bolt 2 — and would be nearly right, at 4 per cent, by accident rather than by argument. The same reasoning applied to the second least-loaded bolt, number 3 at 53 kN, costs 34 per cent. What a bolt carries in one direction is not what its presence is worth over all of them.

The five that are left, at their own worst direction #

Look at what the remaining five are doing. Four of them are at half their capacity or less while one is at its limit, which is a worse distribution than the six-bolt group managed. Removing a bolt has not only reduced the group; it has made the group less efficient at using what is left, because the geometry that was symmetric is not any more and the load’s line no longer bisects anything.

That is the mechanism behind the headline number, and it is worth separating from the count. Of the 34 per cent lost, the change in $n$ accounts for none of it — that change helps. The loss is entirely the fall in $J$ and the growth in $d$, working on a group whose symmetry has gone.

The whole of it, once, for bolt 3 #

The bracket has six bolts at 75 mm pitch and 75 mm gauge, each good for 100 kN, and the load acts through a point 150 mm across and 150 mm up from the centroid of the six. Bolt 3 sits at $(-37.5, +75)$ mm, the far corner on the side away from the load.

With all six present, $J = 4(37.5^2 + 75^2) + 2(37.5^2) = 30{,}938$ mm², the load point is $\sqrt{150^2 + 150^2} = 212.1$ mm from the centroid, and the worst direction is 138 degrees, where the load’s line passes 212.1 mm from the centroid and the corner bolt’s two shares very nearly align. The group carries 135.8 kN.

Take bolt 3 out. The five remaining bolts have a centroid at $(+7.5, -15)$ mm relative to the old one — the average of five positions rather than six — so the load point is now at $(142.5, 165)$ mm from it, a distance of 218.0 mm. The new polar moment, measured about the new centroid, is 22,500 mm². And the count is five.

At the new worst direction of 323 degrees the governing bolt is number 6, at $(30, 90)$ mm from the new centroid, 94.9 mm out. Its equal share is $V/5 = 0.200V$ and its torsional share is $V \times 218.0 \times 94.9/22{,}500 = 0.919V$, and at 323 degrees the two lie 31 degrees apart, so its force is $1.109V$. The group carries $100/1.109 = 90.2$ kN.

Every term moved in the wrong direction except the one everybody thinks of. The share fell from $0.167V$ to $0.200V$, which is the 20 per cent gain from dividing by five. The torsional share rose from $0.575V$ to $0.919V$ — sixty per cent — because $d$ grew by three per cent, $r_{\max}$ grew by thirteen, and $J$ fell by twenty-seven. The gain of a fifth on the smaller term is buried by a gain of three fifths on the larger one.

The trough moves, and a repeat check can miss it #

A group with a bolt missing has lost a symmetry, and the symmetry it lost is the one that made its two troughs equal. On the full rectangle the capacity curve repeats every half turn exactly, because reversing a load reverses every bolt force and changes no magnitude. That is still true of the five-bolt group — reversal is reversal — but the shape within each half turn is no longer the mirror of the other’s, and the two troughs part company.

Here they part company by very little, three tenths of a kilonewton, so the practical consequence is small and the diagnostic one is not. It says that a group with a bolt missing has to be swept again rather than re-checked at the direction that governed before. On this bracket the sweep and the re-check agree; on a bracket whose load point is further off the axis of symmetry they need not, and nothing in the calculation warns which case is in hand.

The check that gets made on site, and why it is unsafe #

The check that actually gets made when a missing bolt is found is not a re-analysis. It is a correction: the group now has five bolts instead of six, so divide by five, keep the drawing’s centroid, and see whether it still passes.

The shortcut is not a rough version of the answer. It is not an answer.

Taking moments about a point that is not the centroid breaks the step that made the elastic method work in the first place. The torsional shares $-Vd\,y_i/J$ and $Vd\,x_i/J$ sum to zero over the group only because $\sum x_i$ and $\sum y_i$ are zero about the centroid — that is what a centroid is. Take the moments about the old centroid instead and those sums are no longer zero, the torsional shares do not cancel, and the five bolt forces add up to the applied load plus a spurious extra force out of nowhere. The figure reports it: 18.4 kN of force that no one applied and nothing resists.

And the error runs the wrong way on the cases that matter. Where the shortcut is conservative it is conservative on omissions that were cheap anyway; where it is unsafe it is unsafe on the two that cost a third of the group. That is the worst possible correlation for a shortcut to have, and it has it for a reason rather than by chance: the shortcut’s whole error is that it ignores the centroid moving, and the centroid moves most, in the most damaging direction, exactly when a far corner bolt is the one that went.

Whether the better method rescues it #

It has already been established that the instantaneous centre method finds capacity the elastic method cannot see, by letting the group rotate about whatever point equilibrium requires and dragging the near bolts up close behind the far one. A reasonable hope is that the same redistribution would soften a missing bolt: five bolts sharing more evenly might recover some of what the sixth was doing.

It does not rescue it, and on four of the six omissions it loses proportionally more.

The reason is the one the capacity locus arrived at from the other side. The instantaneous centre method’s benefit is a measure of how unevenly the elastic method had loaded the group — it recovers the capacity of the bolts the elastic distribution had left idle. A six-bolt group so analysed is already using 91 per cent of its total resistance. There is nothing left over to find, so when a bolt goes, the whole of its contribution goes with it.

A structure has redundancy in proportion to what it is not using. That is the general form of it, and it is why the two methods tell opposite stories about the same group: the elastic method, which uses two thirds of the bolts, has a third of the group in reserve and loses a smaller fraction of a smaller number; the instantaneous centre method, which uses ninety-one per cent, has almost nothing in reserve and loses nearly all of a missing bolt’s worth. The safest-looking analysis is the one that had already spent the reserve.

What is done about it, and what each repair is worth #

Three things can be done with a bracket found with five bolts, and the figures above price all three.

Accept it by calculation. This is the right answer when the omission is one of the cheap ones and the connection was not fully utilised — a bracket missing bolt 2, at 131.1 kN against 135.8, has lost less than the rounding in most load cases. It is the wrong answer reached by the wrong route when the check made is the site shortcut, because the shortcut says bolt 6 is fine when it costs 28 per cent more than the shortcut admits.

Fit a bolt somewhere else. A new hole in a place that clears whatever blocked the original does not restore the group; it makes a different group. Moving the missing bolt inboard by 37.5 mm — from the corner to beside the middle bolt — restores the count and the centroid and recovers only part of $J$, because $J$ counts distance squared and the corner was where the distance was. This is the repair that looks equivalent on a drawing and is not, and the arithmetic above is what prices it.

Weld the gap. Adding a fillet weld to a bolted bracket does not add capacities: the weld reaches its strength at a fraction of a millimetre of slip and the bolts need several, so the two fasteners never arrive together and the joint is designed as one or the other. A weld sized to carry the whole load is a repair; a weld sized to make up the missing sixth is not a repair at all.

Which of the three is right is not a question about the bolt. It is a question about how much of the bracket’s capacity the load was using, and a bracket designed to a utilisation of 0.7 has already absorbed the cheap omissions and none of the expensive ones.

The generalisation #

The shape of this result belongs to every group of elements sharing a load by their geometry, and the bolt group is only the smallest example.

A pile group under a moment has the same three terms and the same asymmetry: losing a corner pile moves the cap’s centre of resistance, lengthens the eccentricity and cuts the group’s second moment, and an engineer who divides the load by the piles that are left has made the same error with more zeroes on it. A weld group loses a length of weld rather than a bolt and the integrals do the same thing the sums do here. A truss that loses a member is the extreme case, because a determinate truss loses everything.

And the common rule under all of them is the one the last figure states. The cost of losing a part is set by how far that part is from the centre of resistance, not by what it was carrying. The two quantities are related but they are not the same, and the check a designer has in front of them reports the second. That is why the bolt at 25 kN and the bolt at 53 kN — one carrying twice what the other does — cost seven and thirty-four per cent respectively to lose, and why an engineer reasoning from the force in a member about the consequence of its absence is reading the wrong column.

Which free body produced the number #

The free body throughout is the bracket plate with its bolts cut, exactly as in the full-group analysis, with one change: the centroid is taken over the bolts that are present rather than over the holes that were drawn. Every capacity is the load at which the worst remaining bolt reaches 100 kN, swept over all 360 directions of the load and minimised, and the check made on each distribution is that the bolt forces sum to the applied load in both directions and about any point.

That check is what condemns the site shortcut, and it is worth noting that it is not a refinement of the shortcut but a test the shortcut fails outright. The residual reported in the figure is computed the same way as the confirming sum in every other bolt-group figure here; the difference is that here it comes out at 18.4 kN instead of at zero.

What the picture cannot show #

Anything about the hole. A missing bolt has usually left an empty hole, and an empty hole in the plate reduces the plate’s own net section whether or not anything is passing through it — so a group missing a bolt may fail in the metal between the holes rather than in the bolts at all. The bracket plate is treated here as an unlimited resistance and is not.

Whether the load is what it was. Every capacity on this page is the group’s, not the connection’s demand. A bracket found with five bolts is usually a bracket that has already carried its load for some years, which says something about the demand and nothing about the margin.

Loss of more than one. The arithmetic here is for one bolt out of six. Two out of six is not two applications of it — the effects on $J$ and on the centroid combine nonlinearly, and two opposite corners gone leaves a group whose centroid has not moved at all and whose polar moment has fallen by forty per cent.

Any redistribution into the plate. The group is assumed to carry the whole load. A bracket welded as well as bolted does not, and the two fasteners never arrive together in any case.

The assumption that is doing the most work #

That the five remaining bolts are the five that were designed. Nothing here allows for the bolt next to the gap being the one that was overstressed during erection, for the missing bolt having been removed because it would not fit, or for a hole that has already elongated. Each of those is a plausible history for a group found with five bolts, and each makes the remaining group weaker than the geometry alone says.

The choice to measure every case at its own weakest direction is also an assumption, and it is the conservative one. A bracket whose load really does point one way is entitled to be checked that way, and several of these omissions look much less serious under a fixed vertical load. The sweep is the right check when the direction is a variable and the wrong check when it is not, and the difference between them is a question about the structure rather than about the bolts.

Still open: the group that is designed to lose one #

Everything above treats the missing bolt as an accident and asks what it costs. The question underneath it is a design one, and it has a different shape: given that a bolt may be missing, what layout minimises the worst that can happen?

That is a minimax over the group’s geometry rather than a check of a given group, and it will not have the same answer as the ordinary optimisation. The layout that maximises capacity with all bolts present concentrates resistance in the corners, because $J$ rewards distance — and concentrating resistance in the corners is exactly what makes losing a corner expensive. A layout chosen for the worst single omission would spread its bolts more evenly and carry less when complete, which is the ordinary price of robustness: a structure that survives losing a part is one that was not using all of its parts.