The braces that have to hold each other

The brace that need not be strong put one spring at the middle of a pin-ended column and found that its critical load climbs with the spring’s stiffness and then stops, at exactly the Euler load of the half length. The stiffness at the knee is the ideal brace stiffness, twice the column’s load over the length of each half, and past it the column buckles in two half-waves with a node at the brace, a shape the brace cannot obstruct and does not try to.

That essay listed several braces as the next question and said no more about it. It is not an idle extension. A tall column in a braced bay is held at every floor; a truss chord is held at every purlin; a long compression flange of a portal rafter is held at every fly brace. The question is what happens to the stiffness each brace must have when the column is divided among four of them rather than one. The intuitive answer is that the work is shared, so each brace can be softer. The answer from the column is the opposite.

Four braces, and the knee moves out #

The model is the earlier essay’s, extended. A pin-ended column of uniform bending stiffness is held at $n$ evenly spaced points by springs of equal stiffness $k$, each tying the column to a fixed point; the segments between them are of length $L_b$, and the column is $(n+1)L_b$ long. It is cut into eight cubic beam elements a segment, its elastic and geometric stiffness are assembled, the springs are added at the brace points, and the critical load is the smallest load at which the combination stops being positive definite. Everything below is stated in the segment’s own terms — loads as fractions of $P_E = \pi^2 EI / L_b^2$, the Euler load of one segment between pins, and stiffness in units of $P_E / L_b$ — so the figures belong to any column with any section.

Every curve has the shape the single brace had. With no braces at all each column buckles as the whole pin-ended length, at $1/(n+1)^2$ of its segments’ load — a quarter for one brace, a ninth for two, a twenty-fifth for four. As the springs stiffen the critical load rises, and at some stiffness it reaches $P_E$ and stops, because from there on the column buckles with a node at every brace point and each segment behaves as a pin-ended column of its own. The ceiling is the same for all three columns. What moves is the knee: one brace gets there at $2P_E/L_b$, two at $3P_E/L_b$, four at $3.62P_E/L_b$.

That is the finding the rest of this essay accounts for. A designer who has learned the single-brace result, and who reasons that four braces sharing a column must each have an easier job, will specify a brace that is barely more than half of what is needed and get a column that stops short of the segment load it was designed around. The deficit is not large in load at first — the four-brace curve at a stiffness of 2 is still at 0.87 of the ceiling — but the knee is where the cheap stiffness ends, and it is in a different place.

The curves are not identical below their knees either. The one-brace curve is a straight line from a quarter to one, because the column’s first mode is a single half-wave that the brace resists in proportion to its stiffness until the second mode, which the brace cannot touch, takes over. The two-brace curve has a corner part way up, where the mode it buckles in changes from one shape to another, and the four-brace curve bends smoothly through a sequence of such changes, each a different arrangement of half-waves competing for the lowest load. The knee is where the last of them, the one with a node at every brace, becomes the lowest.

Why each brace needs more than a lone one did #

The number at the knee can be found without the finite-element model, and the way it is found explains why it rises. At the knee the column buckles with nodes at the brace points, so each segment is behaving as a pinned column and its bending stiffness is doing no work against the braces. The braces are being asked to hold the column against a different, nearby mode, one in which the brace points do move, and the cheapest way to see that mode is to replace the column by a chain of rigid links hinged at the brace points. That is Winter’s model, and at the ideal stiffness it gives exactly the answer the continuous column does.

For one brace the chain is two links of length $L_b$. Move the hinge sideways by $\delta$ and each link tilts by $\delta / L_b$; the axial load, now acting on two lines that meet at an angle, has a lateral resultant of $2P\delta / L_b$ at the hinge, and the brace pushes back with $k\delta$. The chain is held when $k \geq 2P/L_b$, which is the earlier essay’s result.

With $n$ braces the chain has $n + 1$ links and $n$ hinges. The lateral force the axial load exerts at hinge $j$ depends on the kink there, which is the difference between the tilts of the links either side:

$$k\,\delta_j = \frac{P}{L_b}\left(2\delta_j - \delta_{j-1} - \delta_{j+1}\right),$$

with $\delta_0 = \delta_{n+1} = 0$ at the pins. This is the second-difference matrix every engineer has met in a finite-difference beam, and its eigenvalues are known in closed form: $2 - 2\cos\big(m\pi/(n+1)\big)$ for $m = 1$ to $n$. The braces have to be stiff enough for all of them, so the requirement is set by the largest:

$$k_{\text{ideal}} = 2\left(1 + \cos\frac{\pi}{n+1}\right)\frac{P}{L_b}.$$

The column agrees with the chain at every count, to the third figure: 2.00, 3.00, 3.41, 3.62, 3.73, 3.80, 3.88, 3.91. The values climb steeply from one to three braces and then flatten towards 4, which they never reach. The limit is the easiest number to remember and the safest to design with, and it is double what one brace needs.

The reason is in the eigenvector that goes with the largest eigenvalue. It is the most crooked shape the chain can take, alternate hinges moving opposite ways, and in that shape each hinge is kinked by its own movement and by both its neighbours moving the other way. A lone brace resists a kink of $2\delta/L_b$. An interior brace in a long row resists one of nearly $4\delta/L_b$, because the hinges either side, displaced by nearly $\delta$ in the opposite direction, double the angle the axial load turns through. The brace is not sharing the work with its neighbours. Its neighbours are adding to it.

The zigzag is the shape the braces have to stop #

The chain says what the stiffness must be; the continuous column shows what happens on either side of it. Below the ideal, the column buckles in a shape that carries the brace points with it, and above it the brace points stand still.

At half the ideal stiffness the column buckles in a shape with no particular regard for the braces: its nodes fall between them, and every brace is displaced. At nine tenths the shape has nearly organised itself around the brace points, and the load is within two per cent of the ceiling, but the braces are still moving and still being stretched. Just past the ideal the shape snaps into the one the chain predicted at the knee: five half-waves, one per segment, alternating in sign, with the brace points at the nodes.

That last shape is the zigzag turned into a continuous curve. Each segment bows the opposite way from its neighbours, which is exactly the shape in which the brace points are least able to help one another. It is also the shape a real column will be given by its fabrication if anything is, because a column straightened by eye between restraint points carries its residual crookedness as a set of small bows that are no more likely to agree in sign than to alternate. The braces have to be designed for the alternating pattern, because the alternating pattern is the one that costs them most.

The finding also explains the shape of the curves in the first figure. A designer reading the knee as a threshold — enough stiffness and the braces vanish from the problem — is right about the load and wrong about the braces. The column’s buckling load reaches the ceiling at the knee, but the braces arrive at the knee from a mode in which they move, and a brace exactly at the ideal stiffness is a brace whose column is indifferent between moving it and not. The next section of this essay is about what that indifference costs in force.

Braces that tie the column to itself #

Everything so far concerns nodal braces, which tie each brace point to something that does not move: a wall, a core, the ground. Many braces are not like that. A diagonal in a vertical bracing system ties a point on one column to a point on the next level of the same system; a line of bridging in a roof ties each purlin to the next and the whole line to a braced bay at one end. A brace of this kind resists the relative movement of two points, not the absolute movement of one, and it is called relative bracing.

The chain gives the requirement directly. The shear the axial load exerts across bay $j$, when the bay tilts, is $P(\delta_{j+1} - \delta_j)/L_b$, and the relative brace across that bay resists the same difference with $k(\delta_{j+1} - \delta_j)$. The bay is held when $k \geq P/L_b$, and nothing about the other bays enters the calculation. Each relative brace needs one unit of stiffness whether there is one of them or ten, and the dashed line in the second figure is flat at exactly that value. The continuous column agrees to three figures.

The difference between the two systems is worth carrying into a scheme. A relative system asks less of each brace than even a single nodal one — a quarter of the limit for a long nodal row — because a relative brace only ever has to stop its own bay from shearing, and the bays cannot gang up on it. A nodal row asks more of each brace the longer the row gets, because the zigzag lets neighbours load each other. Both are correct, and the numbers are not interchangeable: a relative brace sized at $2P/L_b$ by habit from the single nodal result is twice what it needs, and a nodal brace in a long row sized at $P/L_b$ by habit from a relative system is a quarter of what it needs. A relative system has its own condition, which the model takes as met: the line of braces must end at something that does not move, and the end bay carries the whole system’s shear into it.

One brace short #

The stiffness a row of braces shares out found, for torsional braces on a girder, that losing one brace of four costs a third of the critical moment, much more than a quarter of the total stiffness would suggest. The lateral braces on a column behave the same way, and the column makes the reason easier to see.

The three good braces are each at twice the ideal, so the row as a whole has a large surplus, and the surplus does buy something. The weak brace can fall to 0.68 of the ideal stiffness before the column notices, because its neighbours, being stiffer than necessary, hold the brace points either side of it more firmly than the zigzag requires and so push less on it. At half the ideal the column is still at 0.95 of its ceiling. Below that the curve falls away, and at zero — the brace forgotten, removed for a service run, never connected — the column buckles at 0.52 of the load its segments could carry.

A total-stiffness reading would have put the loss at nothing: three braces at twice the ideal are 1.5 times the total stiffness that four braces at the ideal would supply. The column does not read the total. With the second brace gone, the first and third bound a bay of length $2L_b$, and that bay buckles as a single long segment, restrained only by the bending stiffness of the short segments either side, which are carrying the same load and have little to spare. A pin-ended bay of twice the length would buckle at a quarter of $P_E$; the continuity with its neighbours lifts that to 0.52, and nothing the three remaining braces have in reserve can lift it further, because the buckle has nodes at their points and they are not being asked to move.

The practical reading is that a row of braces has no redundancy in its count, only in its stiffness. Surplus stiffness forgives a brace that is soft. Nothing forgives a brace that is absent. The row is as good as its weakest bay, and the length of that bay, not the sum of the springs, is what the column sees.

What the stiffness buys in force #

The earlier essay explained why a brace is designed for a force as well as a stiffness: a real column is never straight, so the brace is pushed from the moment the column is loaded, and a brace exactly at the ideal stiffness would carry an unbounded force as the load approached the ceiling. The standard model gives the column an initial crookedness of $L_b/500$ at the brace and finds the brace force by second-order analysis. With several braces the crookedness has to be given a pattern, and the zigzag above says which one to give it: alternate brace points displaced $L_b/500$ in opposite directions, the shape the braces resist worst.

For a single brace this reproduces the classical number. At twice the ideal stiffness the brace point moves as far again as its initial offset, the brace force is the stiffness times twice $L_b/500$, and that is $2 \times 2P/L_b \times 2L_b/500 = 0.8\%$ of the load; the continuous column gives 0.78% at 98% of its ceiling. It is the origin of the one-per-cent rule of thumb for nodal bracing.

With four braces the worst brace carries 1.48%. Two things push it up. The ideal stiffness is larger, so twice the ideal is a stiffer spring and pulls harder for the same movement. And the alternating crookedness kinks each interior brace point from both sides, the same mechanism that raised the stiffness requirement, so the lateral push the axial load exerts on it is nearly twice what a lone brace feels. The interior braces carry the most, 1.48% against 1.04% at the ends of the row.

Stiffness reduces the force but only so far. Doubling from two to four times the ideal takes the four-brace force from 1.48% to 1.02%; doubling again to eight takes it to 0.89%. The force settles towards the value it would have with rigid braces, which is the axial load turning through the initial kinks, and that floor is set by the crookedness, not by the brace. A brace just above the ideal, on the other hand, pays dearly: at 1.25 times the ideal the four-brace force is 2.92%. A designer short of stiffness cannot make up for it in strength at any sensible price, which is why every bracing rule asks for twice the ideal or more.

The force rises with the load #

The figures above are taken at 98% of the segments’ Euler load, which is the most a column of that length could carry if it were elastic. Columns in buildings carry less, both because the steel yields first in a stocky segment and because the design load is factored down from the resistance.

The brace force is not a fixed fraction of the load. At low load it is about 0.6% of it, a share set by the geometry of the crookedness and the column’s own bending stiffness; as the load approaches the segments’ buckling load the crookedness is amplified, the brace points move further and the share grows, steepening sharply over the last fifth. At two thirds of the Euler load, a reasonable working level for a slender column, the worst brace of four carries 0.83%, under the round one per cent. At the ceiling it carries half again as much.

This is the honest qualification to the one-per-cent rule: it was fitted to a single brace, and it holds for a long row only because columns are not loaded to their segments’ elastic critical load and because the rule’s stiffness is generous. A designer checking a long row of braces on a slender member that does work near its elastic limit — a truss chord in a lightweight roof, a tall slender compression strut — is entitled to the larger number.

The numbers for a fifteen-metre column #

Take a steel column 15 m tall, pinned at both ends and braced about its weak axis at four points, so that the segments are 3 m long, carrying a design load of 1,000 kN. The ideal stiffness of each brace is

$$k_{\text{ideal}} = 2\left(1 + \cos 36^\circ\right) \times \frac{1000}{3} = 3.62 \times 333 = 1{,}206 \text{ kN/m}.$$

A single brace at the middle of a 6 m column of the same section and load would need 667 kN/m. The four braces of the tall column need 81% more each, and 7.2 times the total. A relative system tying the four brace points to one another and to a braced bay at the top would need 333 kN/m in each bay.

At twice the ideal, 2,412 kN/m, the worst brace carries at most 1.48% of the load near the elastic limit, 14.8 kN, or 8.3 kN at two thirds of it. The stiffness is the governing requirement. A flat bar 50 mm by 8 mm working as a tie 2 m long has an axial stiffness $EA/\ell$ of $210{,}000 \times 400 / 2{,}000 = 42{,}000$ N/mm, or 42,000 kN/m, seventeen times the stiffness needed and far stronger than the 14.8 kN it must carry; the stiffness the brace actually delivers is then decided by the connection, and a single bolt in a clearance hole slipping by a millimetre under 15 kN has a stiffness of 15,000 kN/m, still six times enough. The earlier essay’s warning about what a brace actually delivers applies here with more force, because there are four connections and the weakest decides.

Where the rule of four came from #

The American steel specification asks for a nodal brace stiffness of $8P/L_b$ divided by a resistance factor of 0.75, which is twice the limiting value of $4P/L_b$, inflated for uncertainty; for relative braces it asks for $2P/L_b$, again over 0.75, which is twice the exact requirement of one. Its brace forces are 1% of the load for nodal braces and 0.5% for relative ones. Older editions allowed the nodal stiffness to be reduced for a small number of braces, using $4 - 2/n$ in place of 4, which is an approximation to $2(1 + \cos \pi/(n+1))$ that is exact for one and two braces and within 4% beyond.

Taking the limit of 4 for every count is the safe simplification, and the figures above say how safe: for one brace it doubles what is needed, and for four it asks 11% more than the exact value. Combined with the factor of two, the specification puts a row of four braces at about three times the ideal — 2.95 times — where the worst brace force, from the figure, is 1.16% of the load at the elastic limit. The rule’s one per cent sits under that, and is consistent with it for loads that keep a reserve against the segment’s buckling load.

The European approach does not state a brace stiffness; it asks the designer to model the bracing system with its imperfections and run a second-order analysis, or to check the brace for an equivalent stabilising force derived from an initial bow of the members it restrains. Done with the bow pattern the zigzag dictates, that analysis would find exactly what the figures here do.

What springs and an even spacing leave out #

The column here is elastic and every brace is a linear spring. A real column of stocky segments yields before $P_E$ and the ceiling becomes its inelastic strength, which lowers the stiffness the braces need in proportion to the load, but leaves the factor $2(1 + \cos\pi/(n+1))$ unchanged, because that factor comes from the geometry of the chain and not from the material. The braces are evenly spaced, which is what makes the closed form exact; uneven spacing puts the highest demand on the braces either side of the longest bay. The crookedness pattern is the worst one by construction, and a row of four in which the offsets all lie the same way carries less than a fifth of the force — 0.28% of the load on its worst brace against 1.48% — because the kinks then mostly cancel at the interior brace points.

The column carries a uniform load. A column in a multi-storey frame carries more load in its lower storeys than its upper ones, and the braces near the base are then pushed harder than the ones near the top; the same eigenvalue argument applies with a matrix that is no longer uniform, and the lowest braces govern. And the braces are attached to fixed points, so the analysis says nothing about whether the thing they are fixed to is stiff enough — which is the question a sway frame asks when its bracing is itself a column that can lean.

Still open: when the row of braces is a floor #

The next question on this ladder is the brace that is not a bar at all. A floor diaphragm ties every column in a storey to the bracing system, and it is both a nodal and a relative brace: nodal for the columns near the core, relative for the ones in the middle of a long floor that can shear across its own plan. The load that makes itself worse found that amplification governs a frame’s sway; the column that was never straight found what an initial bow costs one column. A storey of columns, all crooked, all leaning on one floor that leans on one core, is where those two meet the stiffness requirement found here, and its answer is the sum over every column of the force each one asks for — which is what the trace that adds up every storey begins to do for the frame. The ends decide the length that matters began the argument with the length a column buckles over, and the braces are what decide that length once there are more than two of them. A brace that is strong enough and still lets the column fall over has the wrong stiffness, and with four of them in a row, the wrong stiffness is the one that was right for one.