The better method flatters the worse layout

Two methods have been used on the same bracket here and they have never been asked the same question twice. The elastic method shares the load equally and the torque in proportion to distance from the centroid, and reports the load at which the worst bolt reaches its resistance. The instantaneous centre method lets the group rotate about whatever point equilibrium requires, gives every bolt whatever force its own deformation has earned, and reports the load at which the furthest bolt fractures. On a bracket with a vertical load at 150 mm eccentricity the first gives 198.5 kN and the second 225.0, and the gap is thirteen per cent.

That comparison was made in one direction. A load whose direction is a variable turned the elastic answer into a closed curve — a capacity in every direction, with two troughs, and a worst direction no drawing shows. The same sweep run through the other method is not a harder version of the same calculation. It is a different calculation in every direction, because the point the group rotates about has to be found afresh each time, and where it goes turns out to be the interesting part.

A centre that has to be looked for in two dimensions, not one #

For a load drawn straight down the search is easy, and the reason is symmetry. The group is symmetric about the horizontal axis through its centroid, the load’s line is vertical, and so the centre of rotation must lie on that horizontal axis: anywhere off it and the bolt forces would not balance. One unknown, one equation, and a bisection finds it.

Nothing about that survives a load at 138 degrees. The centre is somewhere in the plane, its two coordinates are unknown, and the load the group carries is a third unknown. Against them stand the three equations of plane equilibrium, which is the right count and not a comfortable one, because the relation between a trial centre and the forces it implies runs through a measured curve.

The curve is Crawford and Kulak’s, fitted to tests on single bolts in shear:

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

with $\Delta$ in inches and fracture at 0.34 in. Take the bolt furthest from the trial centre to that deformation, scale every other bolt’s deformation by its own distance from the centre, read each force off the curve, and the trial centre implies six forces. Whether those six are in equilibrium with a load along the chosen direction is the question the search answers.

Written as two residuals in the centre alone, with $\mathbf r_i = \mathbf b_i - \mathbf c$ the vector from the trial centre to bolt $i$:

$$\mathbf G(\mathbf c) = \sum R_i\,(\hat{\mathbf z} \times \hat{\mathbf r}_i), \qquad A = \mathbf G \times \hat{\mathbf u}, \qquad B = \sum \big(\mathbf b_i - \mathbf p\big) \times \mathbf f_i .$$

$A$ says the resultant the bolts can deliver points along the load. $B$ says that resultant’s line passes through the load’s own line, taken as a moment about the load point $\mathbf p$. Two conditions, two coordinates, and the load falls out afterwards as $|\mathbf G \cdot \hat{\mathbf u}|$.

The sense of the rotation does not have to be chosen, which is worth noticing because it looks as though it should be. Reversing the rotation reverses every bolt force, multiplies both residuals by $-1$, and leaves their roots exactly where they were. So the search never branches: it finds the centre, and the sense is read off the answer as the sign of $\mathbf G \cdot \hat{\mathbf u}$.

The arc that does not close #

The path is an arc, and it has two ends.

At 45 degrees the load’s line passes exactly through the centroid. A group in pure translation then satisfies equilibrium on its own: every bolt deforms the same amount, every bolt reaches its full resistance, and their resultant — six equal forces in one direction, acting through the centroid — is collinear with the load. No rotation is needed, and the centre of rotation of a translation is at infinity. The same happens at 225 degrees, where the load points the other way along the same line.

So the locus is not a closed loop around the group. It is one open arc running from infinity to infinity, and the arc from 45 degrees to 225 is the whole of it, because the path repeats every half turn: reversing a load reverses every bolt force and moves the centre nowhere.

There is a practical reading of the asymptote and it is not only a curiosity. Near those two directions the group is carrying its full 600 kN, six bolts at 100 each, and the calculation that says so is a search for a point several metres away from a bracket 150 mm across. A solver asked to find it will wander, and one that reports where it stopped rather than that it could not finish will hand back a centre and a set of forces that describe nothing. The honest answer at those directions is available in closed form and does not need a search: the capacity is $nR_{\mathrm{ult}}$, and the condition for it is that the load’s line passes through the centroid, which is a line of algebra rather than an iteration.

Both curves, on one axis #

Read the two curves against each other rather than separately, because the difference between them is not a constant.

The two spikes are shared. Where the load aims at the centroid both methods give the same answer, six bolts at full capacity, and they give it for the same reason: with no torque every bolt carries the same force, and a method that shares a force equally and a method that lets the group find its own centre have nothing left to disagree about. The curves touch.

The two troughs are not shared, and they are not even in the same place. The elastic minimum is at 138 degrees and the instantaneous centre’s at 136. Two degrees is nothing to design with and it is not nothing to understand: the two methods are minimising different functions of direction, so their worst directions coincide only by accident.

The gap is widest where the group is weakest. That is the useful shape of it. The elastic method is most conservative exactly where the torque is largest — which is where the bolt forces are most unequal, which is where redistribution has most to offer. A designer who takes the thirteen per cent from a vertical check and applies it to a bracket loaded on a diagonal is understating the benefit; here it is 24 per cent.

Where the extra capacity actually comes from #

Nothing in that picture is a stronger bolt. Every bolt is the same bolt it was under the elastic method and the 100 kN is the same 100 kN. What has changed is that five of the six are near it.

Sixty-six per cent against ninety-one. That single pair of numbers is the whole of what the instantaneous centre method buys, and it makes the benefit legible in a way the capacity ratio does not. The elastic method’s answer is set by one bolt reaching its limit while another carries a quarter of its own; the group fails, on paper, with a third of its total resistance unused. The instantaneous centre method’s answer is set by one bolt fracturing after every other bolt has been dragged up close behind it.

Which also says precisely what the extra capacity is bought with, and it is not analysis. It is deformation: 8.6 mm of it at the worst bolt, which is a third of a bolt diameter of hole elongation and plate bearing, and every bolt in the group moving proportionately. The method is a lower-bound argument turned into an upper-bound one by an assumption about ductility, and the assumption is not checked anywhere in the calculation. Whether the group really is that ductile depends on the bolt grade, on whether the threads lie in the shear plane, on the plate’s bearing capacity relative to the bolt’s shear capacity — none of which appears in either method, and all of which decide whether the last bolt is still there when the first one has moved 8.6 mm.

The comparison that answers the layout question #

The elastic method found something counterintuitive about layouts: a ring of six bolts, whose polar moment of 29,737 mm² is four per cent smaller than either rectangle’s 30,938, is the stronger group under a load that can turn. The reason was that the capacity is governed by $r_{\max}/J$ rather than by $J$, and a ring puts every bolt at the same radius, so for its polar moment its furthest bolt is as close as it can be.

That was an elastic result, and the obvious worry about it is that it is an artefact of the method. The elastic method’s distribution is proportional to distance from the centroid, which is exactly the quantity a ring equalises; a method that does not use the centroid at all might rank the layouts differently.

It does not rank them differently. It ranks them the same and by a fifth of the margin.

The ring is better under both methods. Under the elastic method it is better by 10.7 per cent; under the instantaneous centre, by 2.1. The method that is closer to what the group does reduces the difference between the layouts by a factor of five.

That is the finding, and the mechanism is in the previous figure rather than in this one. The elastic method penalises the rectangle for having its corners far from the centroid, because the penalty is exactly proportional to that distance. The instantaneous centre method does not: it lets the rectangle’s near bolts come up to 89 and 91 kN, which the elastic method had at 53 and 57, and most of the rectangle’s deficit disappears. The ring had less of a deficit to begin with — its elastic distribution was already 78 per cent efficient — so there was less for the better method to recover.

A layout the elastic method already likes is a layout with nothing left to redistribute. The two statements are the same statement, which is why the two methods cannot disagree about the ranking and must disagree about the size of it.

Two degrees, and what a designer does with them #

The elastic minimum is at 138 degrees and the other at 136, and the gap between those two numbers is worth a paragraph because it is the one place the two methods can give a designer contradictory instructions.

A bracket whose load can point anywhere is checked at its worst direction, and the worst direction is a property of the method used to find it. Check the group elastically and the critical case is a load at 138 degrees; check it by the instantaneous centre and it is 136. The capacities differ by 24 per cent and the directions by less than the width of the arrow on the drawing, so nothing about the design changes — but the reason nothing changes is not that two degrees is small. It is that both curves are flat near their minima, which is what a trough is. Twenty degrees either side of 136 the instantaneous centre capacity has risen by only four per cent.

That flatness is the reassuring half of the result, and it has a consequence worth stating plainly: a bracket loaded anywhere within about twenty degrees of its worst direction is at its worst capacity, whichever method found it. A slewing arm, a davit, a sling that can be rigged carelessly — none of them needs to hit the worst direction exactly to be designed by it. The spikes, where the group carries its full 600 kN, are the narrow features on these curves; the troughs are wide.

Where the method is allowed, and where it is not #

The instantaneous centre method is not a refinement a designer may apply at will, and the reason is the ductility assumption rather than the arithmetic.

In North American practice it is the method behind the published eccentric-load coefficient tables, which is to say it is the standard method and the elastic one is the conservative alternative. In European practice there is no equivalent table, and a bolt group carrying an in-plane moment is designed elastically — not because the redistribution is doubted, but because the deformation it requires is a hole elongation of about a third of a diameter, and a code that permits it has to say somewhere that the plate will bear rather than tear out to an edge and that the bolt will not shear off before it has moved.

Both conditions are geometric rather than material, which is the awkward part. A bolt near an edge reaches its bearing limit long before 8.6 mm of hole elongation, so the corner bolts — the ones the method relies on to arrive last and hold everything together — are exactly the ones with the shortest end distance. A group designed by the instantaneous centre method and detailed with minimum edge distances is a group that has been given credit for a redistribution its edges will not allow, and no step in either calculation would say so.

The same argument decides whether the method may be used at all on a bracket that is both bolted and welded, and the answer is that it may not: the two fasteners never arrive together, and a weld reaches its capacity at a fraction of a millimetre while the bolts are still being asked for eight.

The generalisation #

The shape of the result is not about bolts. It is about what happens when a group of elements sharing a load is analysed twice, once with the sharing fixed by geometry and once with it fixed by equilibrium and a flat load–deformation curve.

A weld group has the same two analyses available and the gap between them is larger, because a fillet weld’s strength depends on the direction it is loaded in and the elastic method has no way to use that. A pile group under a moment has the same pair, with the outer piles playing the part of the corner bolts. A continuous beam has the elastic moment diagram and the redistributed one, and the permitted redistribution is a percentage written into a code for exactly the reason the bolt group’s is: it is a claim about how much the first hinge can turn before the last one forms.

In every one of those pairs the same rule holds. The extra capacity the better method finds is not a property of the better method. It is a measure of how unevenly the worse method had loaded the group — which means the structures that gain most from a plastic analysis are the ones an elastic analysis had treated worst, and a structure already loading its parts evenly gains almost nothing. Ductility is the property that appears in none of the equations and pays for all of it.

Which free body produced the number #

Every force in these figures is a force on the bracket plate, and the free body is the plate alone with its six bolts cut. The applied load enters along its own line through the load point; each bolt returns a force at right angles to its radius from the centre of rotation, of a size read off the bolt curve at that radius’s share of the fracture deformation. The three conditions imposed are the three of plane equilibrium and nothing else: the search reports a centre only when the six bolt forces and the applied load sum to zero in both directions and about any point.

The check that matters is the one that is not part of the solution. At each direction the bolt forces are summed independently of the equations that found them, and the resultant compared with the load; and the capacity is compared with the elastic method’s in the same direction, with the requirement that it never be lower. It never is — the instantaneous centre method takes the best available centre of rotation and the elastic method assumes one, so a direction where the assumed centre beat the best available would be an arithmetic error rather than a result.

What the picture cannot show #

The moment a bolt fractures. The instantaneous centre’s answer is the load at which the furthest bolt has deformed 8.6 mm and is about to break. The picture is drawn at that instant and shows a group in equilibrium; it does not show what happens a fraction of a millimetre later, when that bolt is gone, the centre jumps, and the remaining five are asked to carry the load with a smaller group and a moved centroid — which is the group with a bolt missing, and a different problem.

The deformation history. Every figure here is one equilibrium state. The path of centres is a locus of separate solutions at different load directions, not a trajectory a bracket follows: a bracket whose load swings from 90 to 136 degrees does not slide its centre along the arc between them, because its bolts do not un-deform on the way.

The assumption that carries the whole of it #

The load–deformation curve is the one piece of pure empiricism in this field, and everything above rests on it.

It has no derivation. It is a two-parameter fit to tests of single bolts in double shear in particular plates at particular grades, published in inches because that is where the tests were done, and the exponent 0.55 is a number that made the curve fit rather than a number with a reason. Its shape, though, is what produces every result here: it rises steeply and flattens early, so a bolt at half the furthest bolt’s deformation is not at half its force but at 95 per cent of it. That flatness is the redistribution. A material whose curve rose linearly to fracture would give the instantaneous centre method almost nothing over the elastic one, because the force ratios would be the distance ratios and the two methods would share out the same way.

The units are the trap. Evaluated in millimetres with the constant that belongs to inches, the curve saturates within a tenth of a millimetre: every bolt reaches $R_{\mathrm{ult}}$ regardless of where the centre is, the method quietly becomes a fully plastic analysis, and it returns an answer that is too large without ever failing a check. That error sat unused and unnoticed here until it was found by comparing two implementations of the same curve — which is the only way it can be found, since the result it produces is plausible, smooth and wrong by a few per cent.

A note on the plate #

Both methods have assumed the bracket does not deform — that the bolts’ relative displacements are fixed by a rigid-body rotation, so a bolt twice as far from the centre moves twice as much. That assumption is what turns the deformation of one bolt into a statement about all of them, and it is the one the instantaneous centre method leans on hardest. Real brackets bend, the deformation a bolt sees is then the rotation plus whatever the plate has given up locally, and that local part is largest near the load and smallest far from it — the opposite of what both methods assume. Nothing here measures it, and it remains the largest unexamined assumption on the page.

Still open: the group that is not all there #

Every capacity on this page is the capacity of six bolts, and the whole comparison between the two methods turned on how much of the group each of them manages to use — two thirds against ninety-one per cent. That number has an obvious consequence and it points the wrong way from where intuition sits.

A group using two thirds of its bolts has a third of itself in reserve. A group using ninety-one per cent has almost none. So a bracket found on site with five bolts in it, its sixth hole empty, ought to be more damaged by the loss under the analysis that flattered it than under the one that did not — and the loss itself ought to have very little to do with the sixth that went missing, because the centroid every lever arm is measured from moves the moment a bolt is subtracted from the average. Which bolt it was, how far the centroid travels, what happens to the polar moment, and whether the redistribution that bought 24 per cent buys any of it back, is the question after this one.