Two diagonals, one of which is absent

A truss diagonal carries the shear in its panel. That is the whole of what it does, and it is why the diagonals of a Pratt truss under uniform load are all in tension while the verticals are all in compression: the shear runs one way throughout.

Uniform load is not the only load. Ask where a load has to stand to make the shear in one particular panel as large as it can be, and near mid-span the answer comes back in two parts with opposite signs.

The shear reverses and the moment does not #

The contrast is the reason the problem exists at all.

An influence line for a moment at a station on a simple span is a triangle: entirely on one side of the axis, peaking under the station. An influence line for a shear at the same station is two triangles of opposite sign, meeting in a jump of exactly one at the cut. The chord force never reverses and the diagonal force always can, and that single difference is what makes the web a different design problem from the chords.

The jump is worth a sentence, because it is the mechanism. Moving a unit load from just left of a cut to just right of it transfers the whole load across, and the shear at the cut changes by the load itself. Nothing comparable happens to the moment, which is continuous through the cut because moving the load an infinitesimal distance changes its lever arm by an infinitesimal amount.

Where the reversal actually bites #

Not every panel reverses enough to matter, and which ones do is a matter of position.

Move the section a panel further along and the same reading is taken again, because the quantity that decides which diagonal is needed changes sign somewhere between them.

The two ordinates at a station $x$ on a span $L$ are $1 - x/L$ and $-x/L$, so the ratio of the reversed case to the forward one is simply $x/(L-x)$ — zero at the support and one at mid-span. The reversal grows steadily inward and is complete at the centre, which is why counters are found in the middle panels of a truss and never at its ends.

For a real truss the question is not whether the shear reverses but whether the permanent load’s shear is large enough to keep the total on one side of zero. A panel whose forward shear from self-weight exceeds the largest reversed shear the variable load can produce never actually reverses, whatever its influence line says. That is a load-ratio question rather than a geometry one, and it is why heavily loaded roof trusses often have no counters while lightly loaded bridge trusses have several.

What a counter is #

If a panel’s diagonal has to carry both signs, there are two ways to arrange it.

A counter is the second diagonal: the Howe member added into a Pratt panel, so that the panel has a crossed pair. One of the two runs the right way for a forward shear and the other for a reversed one, and each is a tie designed for its own case.

The alternative is a single diagonal designed for both signs. It costs nothing in tension and a great deal in compression, because a diagonal is the longest member in the panel and a strut is priced by the square of its length. A middle-panel diagonal carrying ±57.2 kN over 3.9 m has to be sized for the 57.2 kN of compression, and the section that results is several times the area the tension case needs.

There is a naming trap in the pair of drawings worth clearing up. A Pratt and a Howe are not a truss and its counter; they are two complete trusses, each with one diagonal per panel. The counter is what results from taking the Pratt and adding, in the middle panels only, the diagonal the Howe would have had there. The ends are left alone, because the ends never reverse — and the truss that results is a Pratt everywhere except across the middle, where it is crossed.

Which free body produced the number #

The free body is a vertical cut through the panel, taking everything to one side of it. Three members cross the cut — the two chords and the diagonal — and the vertical equilibrium of that free body is the whole of the argument.

The two chords are horizontal, so they contribute nothing vertically. The applied loads and the reaction on the kept side sum to the panel shear $V$. Only the diagonal is left to balance it, and if it leans at $\theta$ to the horizontal its force is

$$F_{\text{diag}} = \frac{V}{\sin\theta}$$

with the sign of $V$. The member’s force is the panel shear divided by a constant, so it inherits the shear’s sign exactly, and any statement about the shear reversing is a statement about the member reversing.

That is also why the counter is the mirror member and not any other addition. A diagonal leaning the other way has $\sin\theta$ of the opposite sign relative to the same cut, so the same shear puts it in the opposite state — which is the entire design requirement, arrived at from one free body and one equation.

The redundancy that is not one #

Two diagonals in a panel means an extra member, and the count that decides whether statics can answer says the frame is now indeterminate to the first degree.

It is not, in any state the structure is ever in. A counter is made as a rod, a flat or a cable — something that cannot take compression at all — so in every load case one of the pair is slack and carries nothing. The active structure has one diagonal per panel and is exactly determinate, and which of the two is active depends on the load case rather than on any stiffness.

That is a genuinely unusual situation and it is worth naming what it costs and what it buys.

It buys an analysis that stays determinate. Each load case is solved by statics with the slack member deleted, and no stiffness, no area and no modulus enters anywhere.

It costs a solution that is no longer linear. Superposition fails: the answer to two load cases added is not the sum of the two answers, because the set of active members can differ between them. An envelope built by adding cases is therefore wrong, and each combination has to be solved in full — which is the same trap a tie that spends an afternoon as a strut sets, in a member that was never expected to change role.

And it costs a self-stress state. Both diagonals can be pulled tight at once by tightening them against each other, and that state satisfies every equilibrium equation with no load applied. Cross-braced panels are routinely pre-tensioned exactly so, to stop the slack member rattling, and the pretension is a locked-in force that no load case contains.

Solving a panel that has two candidates #

Working out which of a crossed pair is active is trivial for a truss and is worth setting out, because the same procedure is the whole of how a tension-only structure is analysed.

Solve the panel with either diagonal assumed active. If the force comes back positive the assumption was right and the other member is slack. If it comes back negative the assumption was wrong, so delete that member, insert the other, and solve again — and because the two lean opposite ways, the second attempt always succeeds.

For a single panel that is one retry. For a whole structure of crossed bays it is an iteration: deleting a slack member changes the forces everywhere, which can slacken a member that was taut, and the process has to be repeated until the set of active members stops changing. That loop is a non-linear analysis in disguise, even though every member is perfectly elastic and every deflection is small, because the structure itself depends on the answer.

It converges quickly and it can fail to converge at all. A structure with enough tension-only members can cycle between two sets, and a structure whose pretension is small can end up with a set that satisfies equilibrium and is not the one it physically finds. A calculation that has to guess the model before it can solve it is a different kind of calculation from one that solves a fixed set of equations, and it arrives here from a single rod that will not push.

Slack is a condition, not a failure #

A member that has gone slack has not failed and has not been badly designed. It is doing what it was specified to do, and the design question about it is different from the question asked of every other member.

How slack. A rod that has gone slack has shortened elastically by the strain it was carrying, and a cable that has gone slack sags. The movement is small and the appearance is not: a visibly drooping brace on a finished structure is one of the commonest causes of a call to an engineer, and the answer is usually that the structure is doing exactly what was drawn.

How often. Repeated slackening and tightening is a fatigue problem at the end connection rather than in the member, and a threaded rod end is a poor fatigue detail. A brace that reverses under wind at every gust has a load history nothing on the drawing describes.

What holds it. A slack member still has weight and still has wind on it, so it needs enough stiffness not to sag out of its own plane. That requirement has nothing to do with the force it was sized for and is often the one that governs.

The panel that has no diagonal at all #

There is a third answer to a reversing shear, and it is the one that turns up in the middle panels of many old trusses: leave the diagonal out.

If the shear in a panel is small enough — and near mid-span under a uniform load it is very small, 57.2 kN against 400.1 kN at the ends in the truss above — the two chords and the two verticals bounding the panel can carry it by bending, as a small Vierendeel frame. The panel becomes a rectangle of four members with rigid corners, the shear is carried in double curvature, and no diagonal of either sign is needed.

That works precisely where the counter problem is worst, which is not a coincidence: both are consequences of the shear being small and its sign uncertain in the same place. It is also much more expensive per unit of shear, so the arrangement survives only where the diagonal is unwanted for another reason — a doorway, a duct, a view.

The three answers to a reversing shear are therefore a stronger member, a second member, or no member, and they are chosen by the size of the shear rather than by anything about the reversal itself.

What reversal does to the rest of the panel #

The diagonal is not the only member whose sign follows the shear, and a design that swaps the diagonals has swapped something else too.

Compare the two trusses above member by member. The Pratt has its verticals in compression at 214.3, 128.6 and 42.9 kN and its diagonals in tension; the Howe has its verticals in tension at those same magnitudes and its diagonals in compression. The web forces have not changed size at all — only which member holds which sign. The chords are identical to the kilonewton in both.

So a panel with a counter has, at any instant, one diagonal in tension and one vertical in compression, and when the shear reverses the vertical reverses with it. A vertical designed as a strut for the forward case is a tie in the reversed one, which costs nothing; a vertical designed as a tie for the forward case becomes a strut, which costs a section. Adding a counter therefore does not remove the reversal from the panel — it removes it from the longest member and leaves it in the shortest, which is exactly the trade worth making, and it is worth making for the reason the length appears squared in the buckling load and not at all in the tension one.

Where crossed bracing is the normal arrangement #

The counter is a special case of a general arrangement, and in one place the general arrangement is the default.

A braced bay in a building frame is loaded by wind, which reverses by definition. A single diagonal in that bay must therefore be designed as a strut, or the bay must be crossed. Crossed tension-only bracing is the cheapest lateral system there is, precisely because the compression case is answered by the other member rather than by a section, and it is the reason the back of every industrial building looks the way it does.

The same reasoning explains why crossed bracing disappears as the loads get larger. A rod that can carry 50 kN in tension is trivial; one that can carry 5,000 kN is a substantial section, and once the member is substantial it can take compression anyway, at which point a single diagonal is cheaper than two. The crossover is a member-size argument rather than a structural one.

Where the model stops #

The influence lines are for a beam, not a truss. A truss’s panel shear is constant within a panel and steps at each node, so the true influence line for a diagonal is a chorded version of the smooth one drawn here, with straight segments between panel points. The ordinates at the panel points are identical; the shape between them is not.

The load is taken as a single moving point. A real variable load is a pattern, and the worst arrangement for a panel’s shear is the whole of the loaded region on one side of it. The influence line’s shaded region says which side.

Nothing here is a fatigue calculation. The reversal is treated as two design cases, and it is also a stress range applied as many times as the load arrives.

The slack member is treated as absent. It is not quite: it still carries a self-weight force, and a pre-tensioned pair never has a genuinely slack member until the load exceeds the pretension.

The connection is treated as a point. Two crossed diagonals cannot both pass through the panel’s diagonal centreline, so one is packed off the other and the pair is eccentric — which puts a moment into the gusset and, at the crossing point, into both members. The joint is not where it was drawn, and a crossed pair is one of the places that is least true.

And the arrangement is assumed to be tension-only by construction. A crossed pair made of angles rather than rods can take some compression, and then the panel really is indeterminate, the analysis really does need stiffness, and the two diagonals share the shear in a proportion nobody drew.

A counter that carries nothing under one load case is the whole of what a determinate truss has not got, which is why the members added for load reversal turn out to be the ones deciding whether the frame survives losing one.

The ladder from here #

Later rungs on this anchor: the counting rule in three dimensions, where m + r = 2j becomes 3j and a satisfied count can still hide a mechanism. The pre-tension in a crossed pair, and the self-stress state it puts into a frame that no load case contains. Eccentricity where the two diagonals cross, since they cannot both be on the centreline. The K-braced panel, where the reversal is answered by geometry instead of by a second member. And the moving-load version of all of this, in which the question is not the worst position but the histogram of all of them.

Counters are a nineteenth-century word for a nineteenth-century problem. Early railway trusses carried a live load several times their own weight, so the shear in their middle panels reversed as a matter of course, and the counter-braces are visible on almost every surviving iron truss bridge of the period. Modern highway trusses are heavier and their live load is proportionally smaller, so the reversal often does not occur — and the member disappeared from the vocabulary along with the condition that needed it.