The load that comes from changing direction

Take a rope under tension $F$ and lay it in a straight line. It pulls on its two ends and on nothing else. Now bend it over a peg. The rope still carries $F$ everywhere, and the peg is being pushed on — hard.

The size of the push follows from a free body of a short piece of the curve. Cut out an element subtending an angle $d\theta$ at the centre of curvature. The tension enters at one end and leaves at the other, at an angle $d\theta$ to it, so the two do not cancel: their resultant is $F\,d\theta$, directed toward the centre. The element has length $R\,d\theta$, so the load per unit length is

$$q = \frac{F}{R} = F\kappa$$

That is the whole of it, and it is one of the most useful identities in the subject. A force path that curves demands a transverse load, whether or not anybody applied one.

Which free body produced the number #

The element above. It is worth being explicit about it because the result is so often used without one.

The free body is a slice of the force path, not of the member. Its boundaries are two cuts across the tendon, the cable, the flange, the arch rib — whatever is carrying $F$ — and the only forces on it are the two tensions or compressions at the cuts, plus whatever the surrounding material supplies. Sum forces in the radial direction and the identity falls out in one line, with no material property, no section property and no length in it.

Three consequences follow immediately and each is the subject of a different essay in this collection.

If the curvature is designed in, the deviation force is a design tool, and the load put on backwards is what it buys.

If the curvature is a property of the geometry, the deviation force is the structure’s whole mechanism, and the force that is only a radius is the extreme case.

If the curvature was not intended at all, the deviation force is a load nobody put in the model, and it goes looking for something to lean on.

The designed case: a tendon is a set of loads #

A post-tensioned beam is usually met as a section calculation — a force at an eccentricity, a moment $Pe$, a stress distribution. That is correct and it is the harder way round.

Replace the tendon by the loads it applies. A parabolic profile of drape $d$ over a span $L$ has a constant curvature of $8d/L^2$, so it applies a uniform upward load of

$$w_p = \frac{8Pd}{L^2}$$

over the whole span, plus a force at each anchorage equal to the tendon’s own force resolved into its end direction. That set of loads is exactly equivalent to the prestress: apply it to the unstressed beam and every internal force comes out right.

The method is called load balancing and the reason it is worth having is not arithmetic. It is that the tendon profile becomes a design variable in the same units as the loads. Choose $Pd$ so that $w_p$ cancels the permanent load and the beam carries no net transverse load at all under it, deflects nowhere, and has a uniform axial compression. That is a designed structure rather than a checked one, and it is available only because the prestress was rewritten as the transverse load it applies.

The freedom is not unlimited, and the limit is worth drawing because it is what turns balancing from an idea into a design. A tendon has to sit inside a band: high enough over the supports to hold the hogging moment down, low enough at midspan to produce the upward load, and everywhere inside the section it is in. Four inequalities and a wedge is that band, and the deviation force is the reason it has the shape it does — the curvature of the profile is what balances the load, so a band that pinches at one point is a band with a limit on how much load the tendon can balance.

It also explains a thing that is otherwise a nuisance. A tendon that is straight over part of its length and curved over the rest applies its load only where it is curved, so a designer’s freedom is in the shape rather than in the force — and a tendon deviated sharply at one point applies a concentrated load there, which is precisely how external post-tensioning works: straight tendons, and steel saddles at two or three deviator positions carrying $F\,\Delta\theta$ each.

The quickest way to see that the shape is the variable is to change nothing but the shape.

So the balanced load is linear in the drape and inversely square in the span, and neither of those is a property of the concrete or of the steel. What the section is then asked to hold is whatever the balancing left behind, plus the axial compression that came free with it.

The two drawings are the same calculation twice. One reads the tendon as a load and the other reads it as a force in a section, and the first is where the design decisions live because it is the one whose variables are geometric.

The geometric case: the structure is nothing but this #

A cylinder under internal pressure carries the pressure entirely by hoop tension, at $N = pR$, and the derivation in the force that is only a radius is the deviation identity read backwards. The hoop is a curved force path; the pressure is the transverse load; $q = N/R$ rearranges to $N = pR$.

An arch is the same statement with the sign reversed. The shape that carries itself is the shape whose curvature at every point is exactly what the applied load requires — a funicular — and a funicular is defined by the deviation identity: the thrust line’s curvature times the thrust equals the applied load, everywhere.

The funicular shape is found by requiring that the deviation force at every node equal the load applied there, which is this page’s identity solved for the geometry rather than for the force.

This is worth saying plainly because it inverts the usual reading. An arch is not a beam that happens to be curved. An arch is a deviation force generator, and the load it carries is the load its curvature demands. Get the shape right and there is nothing left over; get it wrong and the difference is a bending moment, which is the line that must stay inside.

A curved beam in elevation carries the same effect through its own depth: the compression in its top and the tension in its bottom are curved paths of different radii, so their deviation forces do not balance, and the difference is a radial pressure that has to be carried across the section.

In a bar that was curved before it was loaded, the neutral axis moves toward the centre of curvature for the same reason, and the radial force the two flanges’ deviation leaves behind is what the stirrups are there to hold.

The unintended case, which is where it bites #

Every deviation force above was designed. The interesting failures are the ones where the curvature arrived by itself.

A plate girder’s compression flange is curved by the beam’s own deflection. It carries a force $F = A_{fc}f_y$ and the beam’s curvature is $\kappa$, so the flange presses radially into the web at $F\kappa$ per unit length — inward, toward the centre of curvature, which is downward for a sagging beam. Nothing has been applied to the girder. The load exists because the beam has bent.

The web has to hold the flange on its curve, and a slender web cannot: it buckles vertically, and the flange comes down into it. The design rule that prevents it is a limit on $h_w/t_w$ in terms of $E/f_{yf}$ and the ratio of web to flange area, and its whole form is the deviation identity set against a plate buckling stress. It is the only rule in the codes about a load that no load case contains.

A kinked chord does the same thing at a point. A truss chord spliced with a small angular error, a bottom chord that changes slope at a panel point, a tie that passes over a saddle — each applies $F\Delta\theta$ transversely at the kink, and a chord carrying 2,000 kN with a two-degree kink applies 70 kN across itself at that point. That is a real force on a connection that was designed for none.

And a curved tension member applies its deviation force outward, which is why a tendon curving in plan pushes on the side of its duct and why a bar bent round a small radius splits the cover on the inside of the bend. The bearing stress inside a bend is $F/(R\phi)$ on the concrete, and it is the reason minimum bend radii exist at all.

There is a fourth case that belongs with these three and is rarely put beside them, because it is usually filed as a detailing rule. A reinforcing bar bent round a support carries its force along a curve of small radius, so the deviation force it applies to the concrete inside the bend is enormous per unit length — a 25 mm bar at 400 N/mm² bent to a radius of four diameters delivers about 2 kN across every millimetre of the bend, on to a strip of concrete one bar wide. Minimum bend radii are a bearing-stress check on that strip, and the reason a bar bent too tightly splits the member is that nothing was there to hold the two halves of the cover together. It is the same picture as a tendon in a duct and the same picture as a rope on a peg, at a scale where it decides whether a beam end works.

The same free body with a fluid in it #

The identity does not require the thing following the curve to be a solid, and the case where it is not is a structural design item that gets filed under drainage.

A pipe carrying a fluid at pressure $p$ is a force path: the pressure acting on the cross-section is a thrust of $pA$, transmitted along the pipe exactly as a tension is transmitted along a tendon. Where the pipe bends, that thrust has to be turned, and the free body is the one at the top of this essay with $F = pA$ in place of the tension.

For a bend of total angle $\theta$ the resultant is $2pA\sin(\theta/2)$, which for a right-angle bend is $pA\sqrt{2}$. A 600 mm water main at 10 bar has $A = 0.283$ m² and $pA = 283$ kN, so a 90° bend is pushed on by 400 kN — the weight of forty cars, applied to a buried pipe, by nothing but the geometry of the corner.

That is why a buried main has a concrete thrust block at every bend, tee and taper, and why the commonest failure of one is a joint pulling apart rather than a pipe bursting. The block is a structural element sized by a soil bearing check against a force that appears in no hydraulic calculation, and its design is the last step of this essay’s identity: the deviation force exists, and something has to lean on the ground.

The momentum term belongs in the same expression and is usually negligible, which is worth knowing rather than assuming. Turning a moving fluid also requires a force, at $\rho v^2 A$ per unit of the same geometry — the centripetal term — so the full thrust is $(p + \rho v^2)A$ turned through the bend. At 2 m/s in water that adds 4 kPa against a static pressure of 1,000, a fortieth of a per cent. In a gas main at high velocity and modest pressure the ratio inverts, and in a rocket nozzle there is no pressure term left at all.

One free body, three quantities that can play the part of $F$: a tension, a compression, and a flux of momentum. The equation does not distinguish them, and neither should a check for where a force path bends.

What makes the identity useful rather than merely true #

Three things, and they are worth separating.

It converts a geometry into a load, which means the two can be traded. A designer choosing a tendon drape and a designer choosing a load are doing the same arithmetic, and the first has more freedom.

It is exact and local. There is no approximation in $q = F\kappa$; it is a statement about a free body of vanishing length. So it holds where the curvature is large, where the member is short, where plane sections do not stay plane, and in every other place a sectional argument stops working.

And it makes the invisible visible. The failures above are all of the same kind: a load exists, nothing on the drawings applied it, and the only way to find it is to notice that a force path has curved. Drawing as calculation is the theme, and here the drawing is a plot of the force path — not the member’s outline, the force’s outline. Where that line bends, there is a load.

Drawn that way for a masonry arch, the picture is a line inside the stonework, and every place the line changes direction is a place the masonry beside it is being pushed on. The whole of the arch’s behaviour is in the bends of one curve.

Where the model stops #

The tension was assumed constant along the curve. Over a real deviator with friction it is not: the force falls along the arc at $F(\theta) = F_0e^{-\mu\theta}$, which is the capstan equation, so the deviation force falls with it and prestress losses along a curved duct are a consequence of the same geometry.

A curved tendon therefore loses force to the same curvature that is producing the useful load, and a duct wrapped through 120° at a coefficient of 0.2 loses a fifth of the force it started with over that arc alone.

The curvature was assumed to be in one plane. A tendon curved in both plan and elevation has a deviation force in the direction of its binormal, which is neither vertical nor horizontal, and the component that is usually forgotten is the horizontal one.

Nothing here says what supplies the transverse load. The identity says how much is needed, and finding what provides it is the whole of the design problem — a web, a stirrup, a duct wall, a saddle, a foundation, the surrounding masonry. A deviation force with nothing to lean on is not resisted; it moves the structure until something is.

And the equivalent loads are self-equilibrating only on a determinate member. Apply the balancing loads to a continuous beam and the supports resist some of them, which puts reactions into a structure whose prestress was supposed to be internal. Those reactions are real, they change the moment diagram, and the prestress that pushes back is what they are. The deviation identity produced them: it converted an internal force into a set of applied loads, and a set of applied loads on an indeterminate structure has reactions.

Those reactions are point forces, so the diagram they draw is straight between supports — which is how the secondary moment can be separated from the primary one at all, and it is what makes the next question answerable.

That last statement is the deviation identity arriving at a conclusion about shape from an analysis that never mentions shape, and it is the sharpest available demonstration that a force path and the member holding it are different objects.

And the identity is about a force path, not a member. The distinction matters at a section where the resultant moves: a tapered beam’s chord force follows an inclined line even though the flange is straight, so the flange’s deviation force is zero and the chord’s is not, and the difference is carried as shear. That is the shear the chords take, and it is the same identity applied to a line that no piece of steel follows.

The generalisation #

The habit is to draw the path of every large force through a structure and look for the places it changes direction.

At each of them, $F/R$ per unit length if the change is gradual and $F\Delta\theta$ if it is sudden, and then one question: what carries it. That question has been asked and answered by a tendon designer choosing a drape, by a shell designer choosing a curvature, by a masonry engineer looking at a thrust line, and by the author of a code clause about web slenderness. All four are doing the same statics.

The reason it is worth naming as a single idea is that the four look nothing alike. A prestress equivalent load, a hoop force, an arch thrust and a buckling limit on a web are usually taught in four different chapters. They are one free body of a curved element, and the identity on it fits in a line.

There is a last reading that is worth having in mind when a structure is being invented rather than checked. The identity has three quantities in it and any two fix the third, so it can be entered from any side. Fix the load and the force and it gives the shape — which is how a funicular is found, and how a tendon profile is chosen. Fix the load and the shape and it gives the force — which is how an arch’s thrust and a shell’s membrane forces are found, both of them in one line with no stiffness anywhere. Fix the shape and the force and it gives the load — which is the case that catches people out, because it is the one where the answer is something nobody applied. Three ways round one equation, and only the third has ever caused a collapse.