Structural Engineering & Statics
where the load goes
Equilibrium

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

Assumes Three forces must meet at a point, and a drawing can find it, The force that is whatever it needs to be and The member with only one direction.

16 min read 7 figures Drawing as calculationOne support too many

The three essays before this one have been built on a single sentence, and the sentence is not true. A member pinned at both ends and loaded only at its pins carries its force along the line joining the pin centres — that is what makes a link’s direction known and its magnitude the only unknown, and it is what three concurrent forces, Culmann’s line and the funicular polygon all start from.

It holds when the pins are frictionless. Real pins are not, and the correction is not a small number multiplying an answer. It is a change in what kind of thing the answer is.

A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.
Fig. 1 Left, a pin of radius 0.15 m in its hole at a coefficient of friction of 0.15. The reaction at the contact is inclined by 8.5° to the radius through it, and the perpendicular distance from the centre to that inclined line is R sin φ = 0.022 m. Every contact position gives a line tangent to the same circle. Right, a link at the same scale, drawn stubby because at realistic proportions the circles are smaller than the pencil: its force is a common tangent to the two circles, and the dashed centre line is none of the four.

Why the reaction is tangent to a circle

A pin sits in a hole and touches it along one line. The force between them acts at that line of contact and has two components: a normal one, along the radius through the contact, and a tangential one, which is friction and is limited to µ times the normal.

Just before the pin turns, the friction is at its limit, so the resultant is inclined to the radius by an angle φ with tan φ = µ. That is the friction cone in its ordinary form, and the angle is the same one that decides whether a block slides.

What makes a pin special is that the contact point moves. Load the pin differently and the contact goes somewhere else on the circumference, and the inclined resultant goes with it. So the family of possible lines of action is not one line but a family — and every member of it, wherever the contact sits, stays the same perpendicular distance from the centre.

That distance is Rsin⁡φR \sin\varphi, and it is the radius of the friction circle. The perpendicular from the centre to a line inclined at φ to a radius of length RR is Rsin⁡φR \sin\varphi by the definition of a sine, and nothing about where the contact is enters it. So the reaction’s line of action is tangent to a circle of radius Rsin⁡φR \sin\varphi about the pin centre, always, and which side it passes on is decided by which way the pin is turning: friction opposes the motion, so the offset opposes it too.

For a pin of radius 150 mm at µ = 0.15 that circle has a radius of 22 mm. For the same pin badly corroded, at µ = 0.4, it is 56 mm.

The couple a hinge delivers

Before the construction, the simplest consequence, because it is the one that shows up in a design office rather than on a drawing board.

A force passing 22 mm from a point has a moment of 22 mm times the force about that point. A bridge bearing detailed as a pin, 300 mm in diameter, carrying a vertical reaction of 5 MN, therefore delivers a couple of 111 kN·m to the pier it is pinned to, whenever the superstructure tries to rotate on it. At µ = 0.4 — which is where an unlubricated steel pin ends up after a few decades — it is 279.

A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.4. The reaction at the contact is inclined by φ = 21.8° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.056 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 278.5 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.
Fig. 2 The same pin at µ = 0.4. The angle grows from 8.5° to 21.8° and the friction circle from 22 mm to 56, so the couple under the same 5 MN goes from 111 kN·m to 279. Nothing about the geometry of the bearing has changed; the only thing that has is how long it has been since anybody greased it.

That number is not a correction to a moment the pier was designed for. It is the whole of a moment the pier was designed not to have, and the reason a pin was specified in the first place was to avoid it. The bearing is doing exactly what its detail said it would — transmitting 5 MN — and delivering, alongside, a couple of the size a small column carries.

This is the same argument a sliding bearing makes about horizontal force in a different plane: a support drawn as a released degree of freedom releases it up to a friction limit and not past it. Both are cases of a modelling idealisation whose error is not small because the quantity it sets to zero is not large.

Four lines instead of one

Back to the drawing, where the same fact arrives as geometry.

A link pinned at both ends has a friction circle at each end. Its force must be tangent to both, and there are four common tangents to two circles: two external ones, which are parallel to the centre line and offset from it either way, and two internal ones, which cross between the pins.

Which of the four applies is decided by the senses in which the two ends are turning. Both ends turning the same way relative to the link picks an external tangent; ends turning oppositely pick an internal one. The centre line is not among the four and is only ever approached as µ goes to zero.

A construction that needs a link’s direction has been handed four of them. Where the sense of rotation is known — a mechanism being driven one way, a structure being loaded monotonically — the choice is made and the line is definite, merely offset. Where it is not known, and for a structure under a load that can reverse it is not, the honest statement is the band between the two external tangents.

The construction becomes a range

Three lines become three bands, and the answer becomes a range. The four-force bracket of the essay below — a beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under 10.0 kN at 3.0 m — with pins of radius 0.15 m and a coefficient of friction of 0.15, so each friction circle has a radius of 0.022 m. Each link's line of action may be anywhere between the two solid lines tangent to its circles, and the dashed line through the pin centres is the one the frictionless construction drew. Solving all eight combinations of tangent gives A between 4.29 and 4.37 kN against 4.33, B between 6.07 and 6.18 kN against 6.12, C between 0.63 and 0.71 kN against 0.67. The widest band is 12.3 per cent of its own force, on link C, which is the smallest of the three.
Fig. 3 The four-force bracket of the construction two essays back — a beam 8 m long on a vertical link, a strut at 6 m and a horizontal link, under 10 kN at 3 m — with 300 mm pins at µ = 0.15, so each friction circle has a radius of 22 mm. Each link’s line may lie anywhere between its two solid lines; the dashed centre line is what the frictionless construction drew. The eight combinations of tangent give the vertical link between 4.29 and 4.37 kN, the strut between 6.07 and 6.18, and the horizontal link between 0.63 and 0.71 against 0.67.

Three bands rather than three lines, eight constructions rather than one, and three ranges rather than three numbers. The arithmetic is the same arithmetic — Culmann’s construction run eight times on eight slightly different sets of lines — and the reading is different in kind.

The result worth stopping on is not the size of any band but which band is widest. The vertical link carries 4.33 kN with a band of 1.9 per cent; the strut carries 6.12 with a band of 1.9; the horizontal link carries 0.67 with a band of 12.3.

The band’s width in kilonewtons is very nearly the same for all three links. They are 0.041, 0.059 and 0.041 kN, against forces of 4.33, 6.12 and 0.67 — so the relative band is decided by the denominator alone, and the link carrying least has the largest uncertainty. The next section derives why, and the derivation has no force in it.

That inverts the instinct a designer brings to it. The lightly loaded member looks like the one whose exact force does not matter, and it is the one whose force the drawing knows least about. Where that member’s design is governed by something other than this load — a minimum size, a handling rule, a stability requirement — it makes no difference. Where it is not, a twelve per cent band on a member sized by a force is a member sized by a number the calculation did not have.

Why the widest band belongs to the smallest force

The inversion in the last section is worth deriving rather than observed, because the derivation says which member will be the worst one before any drawing is made, and it says it in one line.

Each link’s force comes from moments about the point where the other two links’ lines cross. Both of those forces pass through that point and drop out, so

F=MPdF = \frac{M_P}{d}

with MPM_P the load’s moment about the crossing and dd the link’s own perpendicular distance from it. For the bracket here the three crossings are at (6, 0), (0, 0) and (0, 6) m, the three distances are 6.00, 4.24 and 6.00 m, and the three moments are 25.98, 25.98 and 4.02 kN·m — which is why the horizontal link carries 0.67 kN of a 10 kN load while the other two carry 4.33 and 6.12. It is the moment that is small, not the lever arm.

Now offset the lines. Moving the link’s own line by ρ changes dd by ρ, which is a relative change of ρ/d — four parts in a thousand here, and negligible. Moving the other two lines by ρ moves the crossing PP by about ρ, and a moment about a point that has moved by δ changes by at most δW\delta W, where WW is the whole load. That term is not negligible, and it is the one that governs:

ΔF≈ρ Wd\Delta F \approx \frac{\rho\,W}{d}

There is no FF on the right-hand side. The absolute uncertainty in a link’s force depends on the pin, the total load and the arrangement’s lever arms, and not at all on how much that particular link is carrying. For the three links here the expression gives 0.037, 0.053 and 0.037 kN against measured half-bands of 0.041, 0.059 and 0.041 — agreement to within a tenth on a one-line estimate.

Divide through and the relative band is ρW/(Fd)\rho W / (F d), which carries the load-to-force ratio in it: 2.3 for the vertical link, 1.6 for the strut and 14.9 for the horizontal one. That last factor is the twelve per cent.

So the rule to carry is a sentence about arrangements rather than about pins. Every link in a construction has about the same uncertainty in kilonewtons, so the member carrying the smallest share of the load is always the one whose force is known least well — and a designer wanting to know which member that is need not solve anything, because it is the one whose line passes nearest to the load’s own.

On an ordinary pin it is nothing

The bracket above was given 300 mm pins, which is a bridge bearing and not a bracket. The same construction with the pins a bracket actually has looks quite different.

Three lines become three bands, and the answer becomes a range. The four-force bracket of the essay below — a beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under 10.0 kN at 3.0 m — with pins of radius 0.03 m and a coefficient of friction of 0.15, so each friction circle has a radius of 0.004 m. Each link's line of action may be anywhere between the two solid lines tangent to its circles, and the dashed line through the pin centres is the one the frictionless construction drew. Solving all eight combinations of tangent gives A between 4.32 and 4.34 kN against 4.33, B between 6.11 and 6.14 kN against 6.12, C between 0.66 and 0.68 kN against 0.67. The widest band is 2.5 per cent of its own force, on link C, which is the smallest of the three.
Fig. 4 The same beam, the same load and the same links, with 30 mm pins instead of 300. The friction circles shrink to 4.4 mm, the bands close to 0.3 per cent on the two large links and 2.5 on the small one, and the drawing is, for every practical purpose, the frictionless one.

A tenth of the pin gives a tenth of the band, and at 30 mm the largest of the three is 2.5 per cent on the member carrying 0.67 kN. Nobody would size anything differently for that.

So the honest summary of the effect is a ratio rather than a phenomenon: what matters is the pin’s radius against the lever arms of the construction, and for ordinary structural pinwork that ratio is of the order of a hundredth. Friction in the pins of a roof truss is not a thing anybody needs to think about, and the three essays below this one are not wrong for having ignored it.

The band is nothing on a small pin and is not nothing on a large one. The width of each link's band, as a percentage of the force that link carries, against the radius of its pins, at a coefficient of friction of 0.15. At a 30 mm pin the three bands are 0.3, 0.3, 2.2 per cent; at 150 mm they are 1.9, 1.9, 12.3; at 340 mm, 4.3, 4.3, 27.9. Every one is a straight line through the origin, because the offset is proportional to the radius and the forces are not affected by it to first order. The link with the widest band is the one carrying least — 0.67 kN against 6.12 — because the band's absolute width is set by the geometry and not by the force in it.
Fig. 5 The width of each link’s band, as a percentage of the force that link carries, against the pins’ radius at µ = 0.15. At 30 mm the three are 0.3, 0.3 and 2.2 per cent; at 150 mm, 1.9, 1.9 and 12.3; at 340 mm, 4.3, 4.3 and 27.9. All three are straight lines through the origin, because the offset is proportional to the radius and the forces themselves are unaffected by it to first order.

Straight lines through the origin are the useful shape here, because they mean a single ratio describes the whole effect and no threshold has to be remembered. Doubling the pin doubles every band. Nothing switches on.

Where it is not nothing

Two places, and they are not the same place.

The first is where the pins are large in relation to the structure — a bearing, a hinge in a pinned arch, a link in a machine, a temporary works connection made with a bar through two holes. There the ratio of pin radius to lever arm is a tenth rather than a hundredth and the bands are tens of per cent.

The band is nothing on a small pin and is not nothing on a large one. The width of each link's band, as a percentage of the force that link carries, against the radius of its pins, at a coefficient of friction of 0.4. At a 30 mm pin the three bands are 0.8, 0.8, 5.4 per cent; at 150 mm they are 4.8, 4.8, 30.9; at 340 mm, 10.8, 10.8, 69.9. Every one is a straight line through the origin, because the offset is proportional to the radius and the forces are not affected by it to first order. The link with the widest band is the one carrying least — 0.67 kN against 6.12 — because the band's absolute width is set by the geometry and not by the force in it.
Fig. 6 The same sweep at µ = 0.4, which is a dry, corroded or fretted pin rather than a greased one. Every band roughly doubles: 0.8, 0.8 and 5.4 per cent at 30 mm, and 4.8, 4.8 and 30.9 at 150. The coefficient enters through sin(arctan µ), which is very nearly µ itself over this range, so the effect is proportional to the maintenance.

The second place is the one worth the essay, and it is not about the pins at all.

Where the construction was already ill-conditioned, friction is what arrives. The widest of the three bands, as a percentage of its own force, as the strut's line is moved toward the left end of the beam where the other two lines cross — plotted against that line's distance from the crossing, both axes logarithmic, with pins of radius 0.15 m and µ = 0.15. At 4.24 m the band is 12.3 per cent of a strut force of 6.1 kN; at 0.06 m it is 1962.3 per cent of 459.3 kN. The strut's force grows as the inverse of its lever arm and the band's absolute width does not shrink with it, so the two curves diverge: a support arrangement whose lines nearly meet is one whose forces are large, and whose forces a drawing can no longer state to better than a fraction of themselves.
Fig. 7 The widest of the three bands, as a percentage of its own force, as the strut’s line is moved toward the left end of the beam where the other two lines cross, against that line’s distance from the crossing. Both axes logarithmic, with 300 mm pins at µ = 0.15. At 4.24 m the band is 12.3 per cent of a strut force of 6.1 kN; at 0.06 m it is 1,962 per cent of 459 kN.

The essay two below this one ended by showing that a support arrangement whose three lines nearly meet at a point is nearly a mechanism, that the forces grow as the inverse of the strut’s lever arm about the crossing, and that the determinacy count cannot see it. The band adds the second half of that finding.

The forces grow and the band does not shrink, so the two diverge. At 4.24 m from the crossing the drawing states the strut’s force to within an eighth of itself. At 0.06 m it states a force of 459 kN to within nine times that force — which is to say it states nothing. The construction has stopped being a calculation before the structure has stopped standing, and the numbers it goes on printing look no different from the ones it printed when they meant something.

That is a more useful warning than the one about magnitude. A designer who sees 459 kN where 6 was expected will investigate. A designer who sees 459 kN has no way to see that the real answer is somewhere between 50 and 4,500, because nothing on the drawing says so unless the bands were drawn.

Which free body produced the number

The free body is the same beam it has always been, and what has changed is the set of lines its external forces are allowed to lie on.

That is worth saying precisely, because the friction is inside the pins and the pins are on the boundary of the free body. Cutting the beam free cuts through each pin, and the force crossing that cut is the pin’s reaction — which is exactly the force whose line is tangent to the friction circle. So the friction never appears as a separate force on the free body. It appears as a constraint on where one of the existing forces may act, which is why the construction’s shape does not change at all and only its input lines move.

The eight combinations are eight free bodies, and they are not all reachable at the same moment. A given motion of the assembly picks one tangent at each pin, so the true state is one of the eight and the drawing does not know which. Reporting the band is reporting what the drawing can promise, not what the structure is doing.

There is a second free body in the essay’s first two figures, and it is the pin itself. Its external forces are the contact reaction from the hole and the reaction from the member it carries, and those two must be equal, opposite and collinear, which is the two-force condition applied to a component nobody draws a free body of. That is the whole of why the offset at one end of a link has to be matched at the other, and therefore why the link’s line is a common tangent rather than an arbitrary one.

What the picture cannot show

Nothing here shows which tangent. The bands are drawn symmetric about the centre line because the sense of rotation is being left open, and the moment a load history is specified the answer is one of the eight and the band is a fiction. A band is an admission of ignorance rather than a property of the joint, and a figure of it looks exactly like a figure of a physical spread.

The pin is drawn as a circle in a circle with a single contact point. A real pin bears over an arc, deflects, and beds into its hole; the contact is a patch rather than a point and its centre is not quite where a rigid analysis puts it. The friction circle survives all of that, because its radius depends on the angle of the resultant and not on where the contact is — which is the one robust thing about the construction and the reason it is a circle at all.

And µ is one number for the whole of it. Steel on steel, dry, runs anywhere from 0.15 to 0.6 depending on finish, pressure and what has grown in the gap, and a pin that has not moved for twenty years may not move at all until the moment it does. The figures use two values and neither of them is a measurement of any particular bearing.

The assumption underneath, which is a design decision

The idealisation being corrected here is not an accident of drawing. It is a deliberate choice, made early in every analysis, to treat a joint as one of three things: free, fixed, or pinned.

The friction circle says the third of those is the weakest. A free support is free up to friction; a fixed one is fixed up to the flexibility of what fixes it; and a pin is pinned up to Rsin⁡φR\sin\varphi times the force it carries, which is a moment proportional to the load rather than a fixed quantity. That last property is the awkward one, because it means the error does not become negligible under a larger load — it grows with it.

Where that matters most is exactly where a pin is specified for a reason: a three-pinned arch whose hinges exist to make it determinate and insensitive to settlement. A hinge with friction is not a hinge, and the arch it is in is not determinate; it is a two-pinned arch with a couple at each springing, of a size nobody computed, opposing whatever movement the arch was given the hinge to accommodate. The remedy is mechanical rather than analytical — a bearing with a low-friction sliding surface rather than a steel pin — and it is the same remedy a sliding bearing needed for the same reason.

Still open: the diagram drawn the other way round

Every construction so far starts from a set of forces and produces a drawing. Reversing that is a design method rather than an analysis one, and it is where graphic statics went after the calculator took its analysis work: the force diagram and the space diagram are each other’s duals, every closed region in one corresponding to a point in the other, so a designer who draws the force diagram first has chosen the forces and may then ask what shape carries them. That reciprocity is Maxwell’s, from 1864, and it has a theorem underneath it about which frames possess a reciprocal figure at all.

Beyond that, the three-hinged arch, whose two reactions each pass through a hinge and are therefore found by pairing the load with the arch’s own geometry rather than with a support condition — and which the friction circle has just made conditional. And the funicular polygon through three given points, which is the same construction asked to satisfy one condition more than it has freedoms, and is where a drawing meets indeterminacy for the first time.

Where this goes next

Four forces pair off, and the line joining the pairs carries both resultants. A beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under a load of 10 kN at 3 m inclined at −60.0°. Pairing the load with line A: the two cross at P, (0.00, 5.20) m, and lines B and C cross at Q, (6.00, 0.00). The resultant of the first pair passes through P and that of the second through Q, and since they balance each other both lie on PQ, dashed. The force polygon on the right is the load, then A, B and C, closing where it began, with PQ's direction as the diagonal that splits it into two triangles. The forces are A 4.33 kN along its line, B 6.12 kN along its line, C 0.67 kN against its line, as the three equations of equilibrium give them. The line that pairs four forces Three forces in equilibrium meet at a point; four need not. But they pair off. The resultant of two passes through the point where their lines cross, the resultant of the other two through theirs, and the two resultants must share the line joining those points. Culmann's line turns a four-force body into two triangles — and the method of sections into a drawing. The bearing that is drawn as a roller. The horizontal force a sliding bearing delivers, against the vertical load it is carrying, with its coefficient of friction on the same picture. The coefficient is not a constant: PTFE's falls as the contact pressure rises, and the standard fit is μ = 1.2/(10 + σ), so the bearing drawn is at 30.0 N/mm² and μ = 0.030 while the same bearing at a fifth of the load is at 0.075 — 2.5 times as much. The force curve is therefore strongly non-linear: a fifth of the load gives 50% of the force. Two readings follow and only one of them is usually taken. The largest force is at full load, 108 kN, and that is what the pier is designed for. The largest nuisance is at light load, where 54 kN of friction is 39% of the 140 kN of wind the bearing was put there to release the structure from. Cold makes it worse again: below about −5 °C the same bearing delivers 216 kN. A roller symbol on a drawing means this, and it is a pair of load cases rather than one, because friction opposes whichever way the deck happens to be going. The roller that is not a roller A sliding bearing is drawn as a roller and detailed as a sheet of PTFE, and it delivers a horizontal force of a few per cent of whatever it is carrying. The coefficient everybody quotes is the one at full design pressure, and PTFE's coefficient rises as the pressure falls — so the bearing is at its freest exactly where nobody checks it. The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number. The member with only one direction If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so. A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint. The force that is capped on purpose Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

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.

BearingConcurrencyFree body diagramFrictionFriction circleGraphic staticsLine of actionTwo force member