Structural Engineering & Statics
where the load goes
Internal forces

The force that is only a radius

Every internal force in this collection arrives with a lever arm attached. A hoop force does not. Cut a cylinder along a diameter and the free body settles it in one line — pressure times radius, with no thickness, no second moment and no length in it — which is why a tank wall is thin and why its worst hoop force is not at the bottom.

Assumes The free body is a choice, and choosing it well is the whole skill, The surface that carries by being curved and What a cut reveals, and why it was there all along.

22 min read 8 figures The load must go somewhereGeometry beats materialScale changes everything

Every internal force so far in this collection has come with a lever arm. A bending moment is a force times a distance. A shear flow is a first moment over a second moment. A torsion is a couple about an axis. Ask what any of them is and the answer contains a property of the section it is crossing.

A hoop force does not. Cut a cylinder along a diameter, take half the wall as a free body, and the entire derivation is one equation:

The free body that makes a hoop force a pressure times a radiusHalf a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 300 kN per metre here at 1 MPa on a 0.3 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 25 MPa at 12 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress.NN2R = 0.60 m — the projected widthp = 1 MPap · 2R = 2N ⇒ N = pR = 300 kN/mno thickness in it, no second moment, no length
Fig. 1 Half a ring, with the pressure drawn normal to the wall wherever the wall is. The pressure acts over the projected width whatever the shape of the arc, so vertical equilibrium gives N = pR — three hundred kilonewtons a metre here, in a wall of any thickness.

Nθ=pRN_\theta = p R

There is no thickness in it. No second moment, no area, no length along the pipe, no material property, and no lever arm. A wall of five millimetres and a wall of twenty carry the same force at the same pressure and radius; only the stress differs. That is a stranger statement than it looks, and the rest of this essay is what follows from it.

Which free body produced the number

The half-ring. Everything above the diametral cut is the free body: the internal pressure acting on the inside of the arc, and two cut faces each carrying an unknown force per unit length.

The pressure acts normal to the wall at every point, so its direction changes continuously round the arc — which looks as though it should make the sum awkward. It does not, and the reason is worth having in general form. The resultant of a uniform pressure on any surface is the pressure times the projected area of that surface, whatever route the surface takes between its ends. A hemisphere, a semicircle and a flat plate of the same span all present the same projected width to a uniform pressure, so all three deliver the same resultant to whatever holds them.

Here the projected width is 2R2R, the resultant is 2pR2pR per unit length, and the two cut faces share it. Hence N=pRN = pR, from statics alone, with no assumption whatever about how the wall deforms — which is why it is one of very few results on this site that a mechanism could not spoil.

What a closed end does, and why it is exactly a half

Cut the same cylinder the other way — across it rather than along it — and the free body is one end of the pipe with its cap. The pressure now acts on a circle of area πR2\pi R^2 and the cut is a ring of circumference 2πR2\pi R, so

Nx=pπR22πR=pR2N_x = \frac{p \pi R^2}{2 \pi R} = \frac{pR}{2}

exactly half the hoop force, for every pressure and every radius. No proportion of the design has any influence on that ratio; it is the ratio of an area to a perimeter, which is R/2R/2, against the ratio of a projected width to two cut faces, which is RR.

The consequence is that a pressure vessel is twice as heavily worked round its circumference as along its length, so it splits lengthwise. Every burst boiler in the nineteenth-century record did exactly that, and it is the reason a longitudinal weld in a pressure vessel is subject to more scrutiny than a circumferential one — the same asymmetry that makes a weld’s own direction worth knowing. It is also why the failure looks so alarming: a longitudinal split runs, while a circumferential one merely opens.

The free body that makes a hoop force a pressure times a radiusHalf a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 750 kN per metre here at 0.5 MPa on a 1.5 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 150 MPa at 5 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress.NN2R = 3.00 m — the projected widthp = 0.5 MPap · 2R = 2N ⇒ N = pR = 750 kN/mno thickness in it, no second moment, no length
Fig. 2 The same free body at a tank’s scale rather than a pipe’s. Five times the radius and less than half the wall, and the arithmetic is untouched — the force is still a pressure times a radius, and the thickness has still not appeared.

Two practical results fall straight out. The design of a pressure pipe is t=pR/σt = pR/\sigma and contains nothing else, so a pipe is specified by a single ratio in the same way a buried one is — with the difference that this one is a first power rather than a cube, and is therefore far more forgiving. And the wall a 1 MPa pipe of 600 mm diameter actually needs, at 150 MPa, is two millimetres; everything above that is handling, corrosion allowance and the fact that nobody rolls a two-millimetre pipe.

Why this is what makes a shell worth having

A curved surface carries load in its own plane instead of across it, and the price of a flat one is a lever arm of a few millimetres against a shell’s of tens of metres. The hoop force is the cleanest instance of that trade there is, because the comparison can be made without any bending theory at all.

Take a flat plate spanning the same 3 m as a tank of 1.5 m radius, under the same 0.5 MPa. The plate spans as a strip, carries pL2/8pL^2/8 of moment, and needs a thickness of the order of 6M/σ\sqrt{6M/\sigma}. The tank carries pRpR of tension and needs N/σN/\sigma. The first is a square root of the load and the second is linear in it, so the advantage grows as the load falls — which is the opposite of most structural comparisons and is why membrane structures are used for things that are large and lightly loaded rather than small and heavily loaded.

The same span, the same pressure, thirty times the thicknessA 3.0 m diameter carried two ways at 0.5 MPa, both drawn to the same scale across and both allowed 150 MPa. Curved, the wall is in pure tension: the free body is half the cylinder cut along its length, and N_θ = pR = 750 kN/m needs 5.0 mm of steel. Flat, the same width is a strip in bending: M = p(2R)²/8 = 563 kNm/m needs 150 mm, a factor of 30.0. That factor is √(3σ/p) = 30.0 and it is not a proportion but a change of exponent: the membrane thickness is linear in the pressure and the bending one is a square root of it, so the advantage grows as the load falls. Both wall thicknesses are drawn 10 times over, because at the scale of the span the curved one is a third of a pixel.curved — carried in the surfaceflat — carried in bending5.0 mm of wallN_θ = pR = 750 kN/m150 mm of plateM = p(2R)²/8 = 563 kNm/ma factor of 30.0, which is √(3σ/p) · wall thickness drawn 10× over
Fig. 3 What curvature is worth, as a thickness. A membrane’s thickness is linear in the load and a plate’s is a square root of it, so the ratio between them is not a constant — it grows as the load falls.

The tank that cannot move where the pressure is greatest

Everything above assumes the wall is free to grow outwards. A real tank is not. Its wall is cast into its base slab, and that changes the answer at the one place the membrane calculation says is worst.

The membrane answer for a tank of liquid is Nθ=γ(H−x)RN_\theta = \gamma(H-x)R: a triangle, peak at the base, zero at the surface. It is what every hand calculation starts from and it is comprehensively wrong at the bottom of the wall.

The reinforcement a tank wants most is not at the bottomHoop force up the wall of a 18 m tank holding 8 m of liquid, with the wall cast into its base slab. The dashed line is the membrane answer — γ(H − x)R, a triangle with its peak at the base — and it is what every hand calculation starts from. The solid line is what the wall actually carries. At the base the hoop force is **zero**, because a wall held there cannot move outwards and there is no hoop strain to go with a hoop force; the pressure is carried in vertical bending instead, at a fixing moment of 52.6 kNm/m. The membrane solution is recovered about 3.96 m up, and in between the two exchange the load. The peak is 454 kN/m at 35% of the height — 64% of the triangle's peak, and a third of the way up the wall.020040060002468hoop force (kN/m)height up the wall (m)peak 454 at 35% of the heightzero, at the basemembraneγ(H − x)Rheld atthe base
Fig. 4 Hoop force up the wall of an 18 m tank holding 8 m of liquid. The dashed triangle is the membrane answer; the solid line is what the wall carries. At the base the hoop force is zero, and the peak is a third of the way up.

The reason is a compatibility argument and not an equilibrium one. A hoop force exists only if the wall has stretched circumferentially, and a wall held by its base slab has not stretched there at all. Zero hoop strain means zero hoop force — which is the stiffest-path argument arriving as an absence rather than as a share. The pressure at the base is the largest anywhere on the tank and none of it is carried in the hoop direction; all of it goes into vertical bending, at 52.6 kNm per metre of circumference in this wall.

Further up, the restraint is forgotten and the membrane solution is recovered. In between, the two exchange the load, and the exchange happens over a length set by the same beam-on-elastic-foundation constant the edge of any shell uses:

β=[3(1−ν2)]1/4Rt,decay length=πβ\beta = \frac{[3(1-\nu^2)]^{1/4}}{\sqrt{Rt}}, \qquad \text{decay length} = \frac{\pi}{\beta}

For this wall β=0.79\beta = 0.79 per metre and the decay length is 3.96 m — half the height. The hoop force peaks at 2.77 m, which is 34.6% of the way up, at 454 kN/m against the triangle’s 706.

What that does to the reinforcement, which is the practical half

A designer who reinforces to the membrane triangle puts the most steel where the wall carries the least force and the least where it carries the most. That is not conservative, because the two errors do not cancel: the base is over-reinforced against a force that is not there and under-reinforced against a moment that is, and the moment is the thing that cracks the wall on its inside face.

The correct treatment is a hoop pattern that peaks a third of the way up, plus vertical steel at the base to carry the fixing moment. Both are ordinary; what is not ordinary is that neither follows from the pressure distribution, which is the only thing on the drawing.

Two proportions move the answer and both are geometric:

A thinner wall moves the peak down. β\beta goes as 1/Rt1/\sqrt{Rt}, so reducing the wall from 300 mm to 200 mm raises β\beta from 0.79 to 0.97, shortens the decay length to 3.24 m and moves the peak from 34.6% of the height to 29.6%. A thin wall forgets its base more quickly.

A larger tank moves it down too, for the same reason and by the same square root. The very large tanks — grain silos, digesters, water towers — are close to the membrane answer over almost all their height, and the very small ones are not membrane structures anywhere.

The bending the membrane theory denied, and how far in it reachesBending stress in the wall of a 30 m dome 100 mm thick, along the meridian inward from a fully restrained edge. A membrane solution has two force resultants and no bending, so it cannot satisfy a real boundary condition: the free edge here wants to move out by 0.29 mm and a ring beam does not let it. Closing that gap costs 0.52 MPa of bending at the ring, against 0.60 MPa of membrane stress in the same wall. Near the edge the meridian is a beam on an elastic foundation — flexural rigidity Et³/12(1 − ν²), foundation modulus Et/R² — so it obeys the same fourth-order equation, and the disturbance is down to four per cent of itself at π/β = 2.4440·√(Rt) = 4.23 m. The textbook's “about 2.45√(Rt)” is a rounding of exactly that. Past three of those lengths the shell has forgotten the edge entirely. The wall is drawn 12 times its true thickness.4.23 m = 2.4440·√(Rt)edge bending 0.52 MPamembrane 0.60 MPanothing left of it herethe same fourth-root length a beam on an elastic foundation uses, from an entirely different structure
Fig. 5 The edge disturbance on its own. The restraint is felt over a distance set by the square root of radius times thickness, which is a length the structure was never given and did not choose.

Prestress, and the reason a water tank is usually wound

At 454 kN/m in a 300 mm wall the hoop stress is 1.5 MPa, which ordinary concrete carries. Raise the tank to 12 m of water or halve the wall and it does not — and the failure mode is not collapse, it is leaking, which is a serviceability limit that arrives at a fraction of the ultimate one.

That is the argument for circumferential prestress, and it is exactly the argument for prestress generally with the geometry simplified to the point where nothing else is going on. The wall has one internal force, in one direction, of known magnitude; wind a tendon round it at a force that cancels that number with a margin and the concrete never goes into tension at all. There is no lever arm to place the tendon on, no eccentricity to choose, no Magnel diagram to satisfy — the whole design is one number per metre of height, and the number is the one the figure above plots.

It is also why the tendon spacing follows the hoop force rather than the pressure: closest a third of the way up, opening out at the base where the wall is held anyway.

The dome, where the same force changes sign

Turn the surface the other way and the hoop force does something it cannot do in a cylinder: it changes sign.

On a spherical dome under its own weight the meridional force is compression everywhere, and the hoop force is

Nθ=wR(11+cos⁡ϕ−cos⁡ϕ)N_\theta = wR\left(\frac{1}{1 + \cos\phi} - \cos\phi\right)

which is negative near the crown and positive lower down. The crossing is at

ϕ=51.83°\phi = 51.83°

a number that no proportion of the dome chose and that comes out of setting that bracket to zero. Above it the dome is in compression both ways and is behaving the way a dome is supposed to. Below it the hoops are in tension, in a material chosen because it is good in compression.

The hoops change their mind at an angle no proportion choseThe two membrane forces of a spherical dome of radius 9 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -13.5 kN/m at the crown falling to -18.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -13.5 kN/m and reaches 4.5 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity this site's solver gate checks against the bisection to nine decimals rather than asserting, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked.0102030405060-20-15-10-505angle from the crown (degrees)membrane force (kN/m)N_θ = 0 at 51.8273°meridional N_φ-18.0 kN/mhoop N_θ4.5 kN/mcompression abovetension below
Fig. 6 The two membrane forces on a dome, against the angle down from the crown. The meridional force is compression throughout; the hoop force crosses zero at 51.83°, and everything below that angle is being pulled apart.

Every masonry dome of any size is cracked meridionally below that latitude, and has been since it was built. The cracks are not damage in the sense of something having gone wrong — they are the structure telling the analyst that the material could not supply a force the shape required. A cracked dome is a set of arch strips leaning on each other, which stands perfectly well provided something takes the outward thrust at the bottom, exactly as a masonry arch does.

The ring, whose free body is the whole of the argument

That something is a ring, and its analysis is the shortest in this collection.

Take the base ring as the free body. The meridional force arrives along the tangent, and its horizontal component pushes the ring outwards as a uniform radial line load HH. A ring of radius rr under a uniform outward line load carries

T=HrT = H r

and that is the entire calculation. There is no stiffness in it, no modulus, no second moment, and no distribution to work out — for the same reason the hoop force in a pipe has none. It is the same free body: a ring cut along a diameter, with a pressure acting over a projected width.

The base angle that is hardest on the ring is the golden one againTension in the ring beam at the base of a 9 m dome under 3 kN/m², against where that base is taken. The free body is the ring itself: the meridional force arrives along the tangent, its horizontal component N_φ·cos φ₀ is a radial line load, and a radial load H on a ring of radius r puts H·r of tension in it. At the 60° base drawn elsewhere on this page that is 70 kN. The worst base is not the deepest: the curve peaks at 51.8273° with 73 kN and falls away below. That angle was found by a ternary search on the ring force, and it agrees with the 51.8273° at which the hoop force changes sign — a bisection on a different function — to four decimals. Both are arccos((√5 − 1)/2), because maximising cos φ·sin φ/(1 + cos φ) and setting the hoop force to zero are the same equation cos²φ + cos φ − 1 = 0 written twice.020406080020406080angle of the base from the crown (degrees)tension in the base ring (kN)worst at 51.827°73 kNthe 60° dome: 70 kNthe same anglethe hoops changesign at:51.8273°arccos((√5−1)/2)
Fig. 7 The ring at the base of a dome, and the tension in it. The horizontal component of the meridional thrust is a uniform outward line load, and a ring under one carries H times its own radius — with nothing about the ring in the answer.

For a 18 m dome of 3 kN/m² cut at 60°, the ring carries 70 kN and wants about 350 mm² of steel. For a 60 m one it carries 779 kN and wants 3,900 mm² — the tension goes as R2R^2, because both HH and rr grow with the radius, while the steel a dome’s own weight demands goes as RR. Domes get harder to ring faster than they get heavier — the scale argument in a structure with no span in it — and it decided the shape of every very large dome ever built.

And there is a way out that costs nothing: cut the dome off above the sign change. A cap taken at 45° has no hoop tension anywhere in it and needs no ring, because NθN_\theta has not crossed zero yet. The Pantheon’s oculus is not a window in a dome that happens to be open at the top; the dome is a band of a sphere, and the part that would have been in tension is not there.

Where this model stops

Three places, and each is the same kind of failure — a place where the surface stops being able to be a membrane.

At an edge, as the tank showed. Any restraint that stops the surface moving the way the membrane solution says it must produces bending, over a length of π/β\pi/\beta. Every membrane calculation in this essay is a middle-of-the-shell answer, and every real shell has edges.

At a hole. A cut in a membrane cannot carry hoop force across itself, so the force runs round the hole in the surface around it, exactly as a stress concentration does in a plate, and with the same factor of three at the sides. A tank nozzle is a hole in a field of pure tension, and the reinforcing pad round it is carrying what the hole cannot.

Where the pressure changes sign. Everything here has been tension. Reverse it and the membrane becomes a compression field, which is a stability problem rather than a strength one, and the answer stops containing the thickness linearly and starts containing it as a cube.

A cube, which is why a pipe is specified by one ratioThe critical pressure of a bare ring against its diameter-to-thickness ratio. The formula is 2E/(1 − ν²)·(t/D)³, and the exponent is the whole design: at D/t = 20 the ring takes 0.000 N/mm², and at D/t = 40 it takes 0.0002 — an eighth, for half the wall. Nothing else on this site punishes thinness that hard: a beam's bending strength goes as the square of its depth and a column's Euler load as the square of its radius of gyration, and both of those are recoverable by moving material outwards. Here there is nowhere to move it to.5010015020000.10.20.30.4diameter ÷ wall thicknesscritical pressure (N/mm²)p ∝ (t/D)³
Fig. 8 The same wall under external pressure instead of internal. Nothing about the geometry has changed; the exponent on the thickness has gone from one to three, and the design is a different problem.

The third of those is worth stating as a warning rather than a limitation. A tank that is emptied faster than air can get into it is a cylinder under external pressure, and its wall — sized on a first power for the tension it was built to hold — is being asked a question with a cube in it. Anti-vacuum valves exist because the two calculations are not close to each other.

The generalisation, which is about what a force is a property of

The habit worth carrying out of this is the distinction the hoop force makes unusually visible: the internal force and the stress in the material are different quantities, and they depend on different things.

Almost everywhere else on this site the two travel together, so the distinction never has to be made. A beam’s bending moment and its bending stress both fall when the beam is made deeper. A column’s axial force and its axial stress both fall when the load falls. There is no obvious reason to keep the ideas apart.

Here they come apart completely. N=pRN = pR is settled by the free body and knows nothing about the wall. σ=N/t\sigma = N/t is settled by the wall and knows nothing about the free body. A thicker wall does nothing to the first and everything to the second, and a designer who says “the wall is thicker so it carries more” has run the two statements together.

The same separation is the reason a hoop force is worth a category of its own. It is the internal force that survives when the section is taken away — the one thing left when a structure has no depth at all to work with, and the reason a surface a few millimetres thick can hold back nine metres of water.

Where this goes next

0102030405060-60-40-20020angle from the crown (degrees)membrane force (kN/m)N_θ = 0 at 51.8273°meridional N_φ-60.0 kN/mhoop N_θ15.0 kN/mcompression abovetension below The surface that carries by being curved A flat plate spends its thickness on a lever arm of a few millimetres. Curve the same sheet and the load is carried in the surface itself, at a thirtieth of the thickness — and the entire bill is presented at the edges. 2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs What a cut reveals, and why it was there all along Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing. n = 20.138n = 30.367n = 40.689n = 51.102critical pressure in N/mm² · lowest is n = 2 The pressure that needs no direction Every buckling problem in this collection has had a load with a direction — a column pushed along its axis, a plate along its edge, an arch by what is on it. A buried pipe has none. The pressure is the same everywhere, it stays normal to the wall as the wall moves, and it does work on any change of shape that reduces the area inside. prestress alone-6.3320.61tensionat transfer-2.4716.76tensionin service7.613.82with no prestress12.67-12.67tension The load put on backwards Every other structure in this collection waits for its load and then resists it. A prestressed one is given a load first — chosen, permanent, and pointing the wrong way — so that when the real one arrives the two nearly cancel and the material never has to do the thing it is bad at.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Bending momentBoundary layerCrack controlDomeEquilibriumFree bodyHoop tensionInternal forcesMembrane actionPressure vesselPrestressReinforcementRing beamShellStress