The rows have to share one fold

A tee stub folds on a pattern of yield lines whose total length decides everything about it, and that essay measured the pattern for one bolt row: a circle of hinge round the bolt, or a fan of straight hinges reaching the plate’s edges, whichever is shorter. On ordinary proportions it is the fan, at $4m + 1.25e$, which for a bolt 45 mm from the web on a plate with 55 mm of edge is 249 mm.

A moment end plate has four bolt rows, at a pitch of 90 mm.

Four patterns 249 mm long, at 90 mm centres, would need 996 mm of plate to fold in and have 270 mm of it. The rows are being checked against steel that belongs to their neighbours, and every one of the four answers is a statement about a plate that is not there.

What the drawing has to say instead #

The right-hand picture is not a compromise or a reduction factor. It is the collapse mechanism that the plate actually has, and it is simpler than the one it replaces: between two rows of a group there is no reason for a hinge to curl round either bolt, so it runs straight from one to the other, and the two curls survive only at the two ends of the group where there is no neighbour.

The circular pattern does not merely lose; it stops being a candidate. A circle of hinge round a bolt is a fold that lifts a disc of plate away from the rest of it, and two such discs at 90 mm centres would have to lift the same steel in two directions at once. So inside a group the only admissible field is the straight one, and the choice between the two families — which is the whole of the single-row calculation and turns on a crossing that depends on the edge distance — has already been made by the neighbours.

That is worth holding onto, because it inverts the usual relationship between a check and its refinement. The single-row answer is the more elaborate of the two: two pattern families, a crossing, a cap at the plate’s width. The group answer is a straight fold with a length that is arithmetic. The simpler mechanism is the one that governs, and it governs because there is less plate than the elaborate one needs.

Counting it is one line. The two end rows contribute $2m + 0.625e + \tfrac{1}{2}p$ each and every row between them contributes the pitch $p$ and nothing else, so a group of $k$ rows folds on

$$\ell_{\text{eff}} = 4m + 1.25e + (k-1)\,p$$

against $k(4m + 1.25e)$ for $k$ separate rows. The difference is $(k-1)$ times $(4m + 1.25e - p)$, and that bracket is positive whenever the pitch is shorter than one row’s own pattern — which on these proportions means any pitch under 249 mm, and every end plate ever detailed.

The marginal row is worth the pitch #

Reading the expression as a design rule rather than as a check turns it into something sharper.

A row added to the middle of a group is worth 36 per cent of a row. That is the ratio $p/(4m + 1.25e)$, it contains nothing but two dimensions off the drawing, and it is available before any strength is computed.

The consequences run against several habits at once.

Bolts are not the cheap part. The reflex on a joint that is short of tension is another row of bolts, which on a thin end plate buys a third of what the first row bought. Thickening the plate multiplies every row’s contribution by the square of the thickness, and moving the bolts closer to the web shortens $m$, which raises the force each unit of fold delivers.

And the pitch is the wrong dimension to reduce. Tightening the rows up saves height and reduces the group’s effective length in exact proportion, so a group of four rows at 70 mm pitch folds on 459 mm where the same four at 110 mm fold on 579. The detail that looks more compact is weaker, which is the reverse of how bolt spacing behaves in every shear connection.

A plastic distribution is a set of forces the plate cannot fold under #

The reason this matters more than a ten per cent correction is what it does to the distribution of force between the rows.

A bolted moment connection is normally analysed by giving every row its own resistance and taking moments — a plastic distribution, justified by the rotation capacity of a thin end plate in the same way a plastic hinge justifies a collapse mechanism in a beam. It is the whole reason a deep end plate is worth detailing.

The procedure that produces those numbers is worth stating because it is not a proportioning rule. Each row takes the least of its own resistance and what remains of every group it is the bottom of, working down from the top. Row one is limited only by itself. Row two is limited by itself and by the pair; row three by itself, by the pair above it and by the group of three. The answer is a sequence of minima rather than a share, and it is the reason two identical rows in one plate carry different forces.

The top row takes the most because it is the one checked first, and it is checked first because in a joint under moment it is furthest from the compression flange and attracts the most force anyway. That coincidence is the only thing that makes the rule tolerable: the row that keeps its full resistance is the row that was going to be worked hardest.

The distribution changes shape as the plate thickens #

The same three rows on a slightly thicker plate distribute themselves completely differently, and the pattern is not a scaled version of the thin one.

Three different forces in three identical rows is the signature of a sequence of minima, and it says something about the joint that no single reduction factor could. On the 12 mm plate the pair of rows below the top one share equally because the group of three is what binds them both; on the 16 mm plate the binding constraint changes from row to row, so each one is limited by a different group.

Which group governs is not visible on the drawing and is not the same for every row. That is the practical difficulty with this check, and it is why the calculation is done by a program: every contiguous run of rows is a candidate group, so a five-row plate has fifteen of them and each has to be tested against the sum of the forces already allocated inside it.

Every run of rows is a candidate, and there are more of them than rows #

The rule as a designer meets it is not one check but a set of them, and the size of the set is the reason it left hand calculation behind.

Any contiguous run of rows can be the group that governs: rows one and two, two and three, one to three, three to five, and so on. A plate with five rows has fifteen such runs, one with eight has thirty-six, and each has to be tested against the forces already allocated inside it. Non-contiguous sets need not be tested, because a fold cannot skip a row — the hinge field between two rows of a group passes through whatever lies between them.

Which run governs moves with the plate’s thickness and with the pitch, and it is not usually the largest one. On the 12 mm plate here the group of three binds both lower rows, so they share equally; on the 16 mm plate the pair binds row two and the triple binds row three, and the forces come out unequal. Neither result could be guessed from the geometry, and neither is the one a reduction factor applied to the sum would give.

A check whose answer depends on which of thirty-six subsets is tightest is a check with no rule of thumb in it, and that is the honest reason bolted moment connections stopped being designed by hand. The arithmetic per subset is trivial; the enumeration is not, and the enumeration is where the answer lives.

Two joints with the same bolts, and a third more moment in one of them #

The clearest statement of what the group rule costs is a comparison between two details that a drawing would not distinguish.

Take four rows on a 12 mm plate, all at 90 mm pitch, against the same four rows detailed as two pairs with a gap between the pairs large enough that the pairs do not interact. The first is one group of four, folding on 519 mm; the second is two groups of two, folding on 339 mm each for 678 mm in total. The separated detail is worth 31 per cent more fold for exactly the same steel and exactly the same bolts.

It costs height, and height is the one thing an end plate has less of than it wants — the plate is as deep as the beam and the rows have to sit inside the flanges. That tension is the whole design problem: rows close together fit and do not add up, rows far apart add up and do not fit. The rule turns the trade into arithmetic, and the arithmetic says the gap has to be about $4m + 1.25e$ before two groups are genuinely separate, which on these proportions is 249 mm and is deeper than most beams have to spare.

Which is why real moment end plates are extended above the tension flange rather than crowded below it. An extended row sits outside the beam’s own depth, where nothing is competing for the fold, and it buys a full row’s resistance at the longest lever arm on the joint — two advantages that the rule above shows are the same advantage counted twice.

The group effect has an end, and it is where the bolts take over #

The reduction is not a property of the geometry alone, which is the part most easily got wrong.

A group effect is a statement about a fold, and a joint whose collapse breaks its bolts has no fold to share. At 30 mm the plate is strong enough that the mode that governs is the bolt, whose resistance is a property of the fastener and is simply multiplied by their number.

Between the two extremes the reduction fades smoothly rather than switching off. On the same three rows at 90 mm pitch the group delivers 57 per cent of the sum at 12 mm, 80 at 16, 82 at 20, 90 at 25 and 100 from 30 mm upward. The middle of that range is mode 2, where the resistance is part fold and part bolt: the fold part shrinks with the group and the bolt part does not, so the reduction is diluted by whatever share of the answer the bolts already own.

So the thickness that decided who fails now decides whether the rows may be added up at all, which is a second and heavier consequence of the dimension the essay below this one was about. A designer who thickens a plate to avoid prying gets the plastic distribution back as well, and neither effect is visible in the other’s calculation.

What it is worth as a moment #

Forces are the wrong unit for judging any of this, because a bolted joint is bought for its moment and the rows are at different lever arms.

Taking the three rows of the 12 mm plate about a compression flange 400 mm below the top row: the limited forces give 130 kNm and the plastic distribution would have promised 204. That is a 36 per cent overstatement, and it is larger than the shortfall in force — 43 per cent against 57 — for a reason worth noticing. The rows that lose most are the lower ones, and the lower ones have the shorter lever arms, so the moment loses less than the force does. The effect runs the other way on a shallow joint, where the arms are nearly equal and the two shortfalls converge.

On the 16 mm plate the same arithmetic gives 218 kNm against 261 — a 17 per cent overstatement. The correction is therefore worth between a sixth and a third of a joint’s moment on the plates where end plates are actually thin, and nothing at all on the plates where they are thick.

What the joint does with it afterwards #

A resistance is half of what a joint is, and the other half is its stiffness — the rows are springs in series with the column flange, the web panel and the bolts themselves, and the joint’s own moment–rotation curve is assembled from both.

The group effect touches only the first. Two rows 90 mm apart are as stiff as two rows 300 mm apart, because stiffness is computed from the plate’s bending under a single row’s load and there is no mechanism in it to share. So a compact group of rows is a joint that reaches its plateau early and at a lower moment — which shows up in a joint’s classification as a shift from full-strength toward partial-strength with no change in its stiffness class at all.

That asymmetry is not a modelling artefact. A yield-line mechanism is a collapse and has to be a single kinematically admissible field over the whole plate; an elastic deflection superposes, and two rows’ deflections add without either one noticing. The two halves of the same component obey different rules because one of them is a limit state and the other is a linear system, and reading a stiffness result across to a strength one is the error this distinction exists to prevent.

Where the counting stops describing the plate #

The patterns are idealised and the lengths are fitted. $4m + 1.25e$ and $2m + 0.625e + 0.5p$ are tabulated expressions calibrated against tests, not the output of a minimisation over hinge fields. A real plate’s fold is curved, the hinges have finite width, and the membrane action that develops at large rotation is outside the theory entirely.

Every group is assumed to fold at once. A group of five rows is checked as a single mechanism, and a real plate under a moment has more force in its upper rows, so the upper part may be folding while the lower part is elastic. The rule handles that by checking every contiguous sub-group, which covers the cases but does not describe them.

The distribution is a sequence of allocations, not a solution. Nothing in the top-down procedure computes what the rows do; it computes an admissible set of forces, in the lower-bound sense. The real forces depend on the relative stiffnesses of the rows and on how far the joint has rotated, and they are only approached as the joint reaches collapse.

The column side is ignored here. Every row of an end plate has a matching tee stub in the column flange, with its own patterns, its own groups and its own governing mode, and the row’s resistance is the lesser of the two. The two tee stubs are in series and their groups need not be the same rows.

And rotation capacity is assumed rather than checked. A distribution that gives the top row its full resistance requires that row to keep carrying it while the rows below reach theirs, which is a demand for plastic rotation in a plate whose thickness has just been shown to decide whether it can fold at all.

Still open: the bolt that was stretched before the load arrived #

Every row in this essay carries a bolt that is a bar in tension, stretched by whatever is applied to it. A preloaded bolt is stretched before the joint is built into anything, and while its plates stay in contact an external tension is shared between the bolt’s own stiffness and that of the clamped region — mostly the latter, because a short thick stack of plates is far stiffer than a long thin bolt.

The bolt’s force therefore barely moves until the plates separate, and then it moves steeply. That changes very little about the strength calculation and a great deal about the fatigue one, and it changes when prying starts: the lever cannot act until the flange has lifted off, so a preloaded tee stub has no prying force at all over most of its working range. Where the transition sits, and what it is worth to the two quantities separately, is the next question.