The hinge that forms in the other flange

Every tee stub in the essays below this one has been pulled against something rigid. The prying force was the reaction at the tip of a flange bearing on a foundation that did not move; the flange’s thickness decided which of three modes arrived first; the rows of an end plate folded together; and a preload changed the path to failure and not the destination. In each case the thing the flange pried against was a line on the drawing with hatching under it.

A real bolted moment joint has no such line. The end plate on the beam is bolted to the flange of the column, and the column flange is itself a plate with a web behind it and a bolt through it — a tee stub in its own right, pulled the other way. The same bolt passes through both. And the tip of the end plate, where the prying force acts, bears on the column flange, which is the very place the column flange’s own prying force acts on the end plate.

That is one contact and one force. Whatever pushes the two tips together pushes on both of them, equally and oppositely. The component method, which is how Eurocode 3 assembles an end-plate joint, does not see it: it checks the end plate in bending as a tee stub on a rigid base, checks the column flange in bending as another, and takes the smaller resistance. This essay asks what happens when the prying force has to be shared, and the answer is that the pair can be weaker than both of its halves.

Two tee stubs, and the one thing they share #

Take one bolt of an end-plate joint and cut out the strip of each flange that belongs to it. The end plate strip runs from the beam web, past the bolt at a lever arm $m_A$, to its tip a further $n$ beyond. The column flange strip runs the same way from the column web, with its own lever $m_B$ to the same bolt. The beam web pulls the end plate away from the column with a tension $T$; the column web holds the column flange back with the same $T$. The bolt, stretched between them, carries $B$. And at the tips the two plates press together with a contact force $Q$.

For each flange, moments about its own sections give two conditions. At the web face the moment is $T m_i - Q n$; at the bolt line it is $Q n$. Each must stay within that flange’s plastic moment $M_{p,i} = p\,t_i^2 f_y/4$ for the strip of width $p$ the bolt serves. The bolt must stay within its own tension resistance, $T + Q \le B_t$. That is the whole of the statics.

On a rigid base, one flange alone has one unknown, $Q$, and the three classical modes fall straight out of it. Mode 1 is both hinges at once, at $T = 2M_p/m$. Mode 2 is the web hinge together with the bolt, at $T = (M_p + n B_t)/(m + n)$. Mode 3 is the bolt alone, with no prying at all. The largest $T$ for which some $Q$ satisfies every limit is the collapse load, because the structure is indeterminate to exactly one degree and every equilibrium state is one value of $Q$.

The pair has the same one unknown. It is not a second prying force for the second flange: it is the same $Q$, because it is the same contact. So the pair’s collapse load is the largest $T$ for which a single $Q$ satisfies both flanges’ limits at once — and a set of conditions that must hold together can only be harder to satisfy than either set alone.

Why the pair is weaker than both #

The figure is the argument, and it is worth reading slowly.

The end plate on its own wants a prying force of 66.7 kN. At that force its web face is relieved enough that the flange reaches its second mode — the web yields just as the bolt reaches its strength — at 109.3 kN. The column flange on its own is thinner and reaches its first mode at 104.4 kN, with both of its hinges formed and the prying force at 44.7 kN, which is exactly what its bolt line can resist.

Bolted together, the end plate cannot have its 66.7 kN of prying. That force would press on the column flange’s tip, and the column flange’s bolt line yields at 44.7. Past that, the column flange’s tip simply rotates away and the contact carries no more. So the end plate is limited to the prying the column flange can supply, and on the end plate’s web line at a prying force of 44.7 kN the tension is 92.3 kN.

Neither flange has failed on its own terms. The end plate has one hinge, at its web. The column flange has one hinge, at its bolt line. The bolt is at 137 kN of its 176. What has formed is a mechanism that belongs to neither tee stub: the end plate rotates about its web hinge, its tip pushes into the column flange, and the column flange’s tip folds away about its bolt-line hinge. Each flange has contributed exactly one of the two hinges its own mode 1 would have needed.

The mechanism, checked from the other side #

A collapse load found by searching over equilibrium states is a lower bound, and the two bound theorems say it is the collapse load only if a mechanism can be found that gives the same number from the other side. That check is short.

Let the end plate’s flange rotate by a small angle $\theta$ about its web hinge while the column flange’s inner part stays put. The bolt is elastic and at collapse barely stretches, so the end plate’s bolt line cannot move away from the column flange’s; the end plate’s web must therefore rise by $\theta m_A$ to keep the bolt line still, and its tip moves toward the column flange by $\theta n$. The column flange’s tip must move away by the same amount, so it rotates about its bolt-line hinge by the same $\theta$.

The load does work $T\,\theta m_A$. The two hinges absorb $M_{p,A}\theta + M_{p,B}\theta$. Equating them,

$$T = \frac{M_{p,A} + M_{p,B}}{m_A}$$

which for these plates is $(2{,}588 + 1{,}566)/45 = 92.3$ kN a bolt. The upper bound and the lower bound meet, so the mixed mechanism is the collapse.

It is worth dwelling on what that pair of calculations has just done, because it is the whole of plastic analysis in miniature. The search over $Q$ is the static theorem: any prying force that keeps every section within its plastic moment gives a load the joint can carry, and the largest such load is the best lower bound. The hinge mechanism is the kinematic theorem: any pattern of hinges that lets the joint move gives a load it cannot exceed. The component method is a static argument applied to each flange separately — it finds a good lower bound for each and takes the smaller — and what it cannot do is find the mechanism that uses a hinge from each, because no single tee stub contains it. The same thing happens in a continuous beam, where checking each span on its own misses the mechanism that runs across a support.

Written that way, the reason it can govern is plain. The end plate’s own mode 1 would be $2M_{p,A}/m_A$ — both hinges in the end plate. The mixed mechanism replaces the end plate’s bolt-line hinge with the column flange’s, which is cheaper whenever the column flange is the thinner plate. The column flange’s own mode 1 is $2M_{p,B}/m_B$, both hinges in the column flange; the mixed mechanism replaces its web hinge with the end plate’s, which is dearer, but divides by the end plate’s longer lever $m_A$ instead of the column flange’s shorter one. When the thicker flange also has the longer lever, both substitutions pay, and the mixed mechanism is cheaper than either flange’s own.

The window, as the end plate thickens #

Walk the end plate’s thickness up from thin to thick, with the column flange fixed at 14 mm. While the end plate is the thinner flange, it governs on its own terms and the pair agrees exactly with the component method: the end plate’s own mode needs a bolt-line hinge, and its own bolt line is the cheaper of the two, so nothing changes by bolting it to something.

The moment the end plate becomes the thicker flange, at 14.2 mm, the cheaper bolt-line hinge is the column flange’s, and the pair drops below the component answer. The shortfall grows to 17 per cent at 17.1 mm, where the component method has the end plate at 104.0 kN and the pair carries 86.8. Then it closes again: by 19.8 mm the end plate is thick enough that the column flange’s own mode 1, at 104.4 kN, governs everything, and the pair and the component method agree once more.

The window is where a designer is most likely to be. An end plate slightly thicker than the column flange it bolts to is not an odd detail; it is what happens when the end plate is sized for the beam and the column is a lighter section than the beam, which in a building’s upper storeys is the normal arrangement. And the window’s worst point is not at an extreme — it is at an end plate three millimetres thicker than the column flange, which is the detail a designer would draw to be safe.

Where on the map it happens #

Over the whole plane of the two thicknesses the picture is the same shape. Almost everywhere the component method is exactly right. Along a narrow band just to one side of the diagonal — the end plate somewhat thicker than the column flange — the pair falls short, and it falls furthest when both plates are thin. On the other side of the diagonal nothing happens at all, because there the thicker flange is the column flange, which has the shorter lever, and the substitution that makes the mixed mechanism cheap does not pay.

The band covers only six per cent of the map, and it is fair to say so. A reader might conclude that the effect is a curiosity. The case against that conclusion is where the six per cent is: it hugs the line of equal thickness, and a joint whose two flanges are nearly equal is exactly the joint a designer produces by matching an end plate to a column. A defect confined to the neighbourhood of the most natural choice is not rare in practice even when it is rare on the map.

The lever arm is what opens it #

The thickness map holds the lever arms fixed; the second map holds the column flange fixed and moves the ratio of the levers instead. It shows the rule in both directions. When the end plate’s lever is the longer, the shortfall appears where the end plate is thicker than the column flange. When the end plate’s lever is the shorter, it appears where the end plate is thinner — because then the column flange is the thicker plate with the longer lever, and the mixed mechanism runs the other way round, with its web hinge in the column flange and its bolt-line hinge in the end plate.

At equal levers the band vanishes, which is the case every textbook tee stub is drawn with. With equal $m$, the mixed mechanism’s load $(M_{p,A} + M_{p,B})/m$ always lies between the two flanges’ own mode-1 loads $2M_{p,A}/m$ and $2M_{p,B}/m$, so it can never be the lowest. The lever arms have to differ for the pair to lose anything, and the more they differ the wider and deeper the window becomes — 22 and 24 per cent at a ratio of two either way.

The lever arms of real joints do differ. The end plate’s is measured from the beam web to the bolt, less an allowance for the weld; the column flange’s is measured from the column web, less an allowance for the root radius. A beam with a thin web on a column with a thick one, or a column with large root fillets, gives the end plate the longer lever for the same bolt gauge, and a heavy end plate on a light column then puts the joint in the window.

Equal plates, and plates far apart #

Two limiting cases make the middle one easier to see. When the two flanges are identical, each on its own would choose the same prying force, so there is nothing to share and nothing to lose: the pair collapses at the same load as either flange alone, by each flange’s own first mode.

When the two flanges are very different, the thicker one really is the rigid base the model assumed. A 25 mm end plate on a 14 mm column flange does not form any hinge at all; the column flange collapses by its own first mode against it, exactly as it would against a foundation, and the component method is exact again.

The component method is exact at both ends and wrong in the middle, which is the shape of an approximation that was built from two limiting cases and never tested between them. A tee stub on a rigid base is the end plate on a very thick column flange, or the column flange behind a very thick end plate. Neither is the ordinary joint.

One more line in the component check #

The practical content is short. Where the thicker of the two flanges also has the longer lever arm, check the mixed mechanism:

$$T_{\text{mixed}} = \frac{M_{p,\text{thick}} + M_{p,\text{thin}}}{m_{\text{thick}}}$$

and take it if it is lower than both tee stubs’ own resistances. It is one line, it uses quantities the component method has already computed, and it needs no finite-element analysis. Where the thicker flange has the shorter lever, or the two levers are equal, it cannot govern and need not be checked.

Two ways to stay out of the window follow from the same line. Make the end plate substantially thicker than the column flange — past the window’s upper edge, where it behaves as a rigid base — or make it no thicker than the column flange at all. The expensive detail is the cautious one in between.

One bolt’s strip of each flange #

The free body is one bolt’s strip of each flange, cut at the webs and through the bolt, with the contact at the tips exposed as the force $Q$. Equilibrium of each strip about its web face and about its bolt line gives the moments $T m_i - Qn$ and $Qn$; equilibrium of the bolt gives $B = T + Q$. Every number is the largest $T$ for which one $Q$ keeps all five sections — two webs, two bolt lines, the bolt — within their resistances.

Each flange’s plastic moment is taken over a strip 90 mm wide, which stands in for the effective length a yield-line analysis would give that bolt row; the steel is at 355 N/mm²; the tips are 35 mm beyond the bolt; and the bolt carries 176 kN, an M20 of grade 10.9. The one-flange cases reproduce the three tee-stub formulae exactly, which is checked in the figures rather than assumed.

Effective lengths, edges, stiffeners and the elastic range #

Different effective lengths. The end plate and the column flange have different yield-line patterns — the column flange is continuous above and below the bolt row, the end plate may be extended past the beam flange — so the strip each bolt serves is not the same width in the two plates. That changes the two plastic moments and leaves the argument untouched: the mixed mechanism uses whichever moments the patterns give.

Different edge distances. The contact is taken at the shorter of the two tips. When the end plate is narrower than the column flange the contact moves inward and the lever $n$ changes; the method is the same.

Stiffeners. A column flange with a stiffener behind the bolt row is not a tee stub at all, and the window disappears with it.

Anything elastic. Every number here is a collapse load. The shared contact changes the elastic prying too — the two flanges deflect in series, so the tip contact is softer than a rigid base and the bolt sees a different share of the load before anything yields — and that belongs to the joint’s stiffness rather than its strength.

The two tips have to be touching #

That the two tips are in contact at all. The mixed mechanism needs the end plate’s tip to push the column flange’s tip away, which requires them to be touching when the joint is loaded. An end plate with a generous edge distance on a narrow column flange may have its tip beyond the column flange’s edge, bearing on nothing; then there is no shared contact, each flange really does pry against something else — or against nothing — and the separate checks are the right checks. The window exists for the joint whose two plates overlap at their edges, which is the usual arrangement, and not for every joint.

Still open: the stiffness of a joint made of two flanges #

Everything above is strength. A bolted end-plate joint is also a rotational spring, and the component method builds its stiffness from the same components, as springs in series: the end plate in bending and the column flange in bending are two springs in series with the bolt. Each spring is computed for a tee stub on a rigid base.

In the pair neither is on a rigid base. Each flange’s tip bears on the other’s, so the contact itself has a stiffness equal to the other flange’s tip stiffness, and the prying force that sets each flange’s effective stiffness depends on both. Whether the series combination of two rigid-base stiffnesses over- or under-states the pair’s — and whether the error is concentrated in the same window of thicknesses, which is where a designer’s joint classification between pinned, semi-rigid and rigid is most often decided — is the question after this one.