Structural Engineering & Statics
where the load goes
Internal forces

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

Assumes What a cut reveals, and why it was there all along, The diagram is an integral, and that is why it can be drawn by eye and How far a wrong load reaches.

18 min read 10 figures The load must go somewhereDrawing as calculationThe statics of things that do not move

Every free body in this collection has reactions on it, and every one of them is drawn as an arrow. The arrow is honest about what the equilibrium equations use — a force has a magnitude, a direction and a line of action, and three equations cannot ask for anything else.

What sits under the arrow is a bearing three or four hundred millimetres wide, delivering a pressure of some distribution over that width. The equations do not need the distribution. Almost everything else does.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear20.4moment51.8 at x = 5.10the moment peaks exactly where the shear passes through zero
Fig. 1 The chain that every beam calculation runs down. Shear is the integral of the load and moment is the integral of shear, and the diagrams end abruptly at points where the real structure does not end at all.

Which free body produced the number

Cut the beam at the centreline of a support and take the piece to the left. The forces on it are the applied load and the part of the reaction that lies to the left of the cut.

If the support is a point, that part is nothing — the whole reaction acts at the cut and has no lever arm about it. If the support is a bearing of width bb carrying a total reaction RR spread uniformly, then half the reaction lies to the left of the centreline, its resultant is at b/4b/4, and it contributes

R2⋅b4=Rb8\frac{R}{2}\cdot\frac{b}{4} = \frac{Rb}{8}

of moment in the opposite sense to the hogging the beam is carrying. So the moment over a support is lower than a point-support analysis reports, by exactly Rb/8Rb/8, and the reduction has no material, no span and no load arrangement in it.

The same beam, cut at x = 2.4A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.2012.57.5shear 12.5moment 30.0the cut, at x = 2.4nothing was applied here — the internal forces are what the left-hand piece needs
Fig. 2 The construction the reduction comes from. Separate the beam at a station and whatever lies on the piece taken has to be summed, including the part of a wide reaction that lies on the same side of the cut.

What the reduction is worth

2 continuous spans against 2 simple onesThe bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 134.8 to 75.8, and a hogging moment of 134.8 appears over the supports where there was none.moment75.8 sagging134.8 hogging134.8 if the spans were simplereactions 57.8 192.5 57.8 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not
Fig. 3 Two continuous spans against two simple ones. Continuity halves the sagging moment and produces a hogging moment where there was none — and it is that hogging moment the support’s width takes a bite out of.

On the two-span beam drawn there the interior reaction is 192 kN and the hogging moment 134.8 kNm. A 400 mm column takes 9.6 kNm off it, a 600 mm one 14.4 — 7% and 11%. Those are not large numbers and they are free, which is why every concrete design standard permits the reduction and why almost every hand calculation makes it.

Two things are worth noticing about it. The first is that the reduction is proportional to the reaction, so it is largest exactly where the hogging moment is largest, and the two effects partly cancel. The second is that it is a statics result and not an allowance: the moment at the face of the support genuinely is smaller, and nothing about the beam has been assumed.

The catch is what happens if the support is not spreading its reaction uniformly, which is the next question and the one statics cannot answer.

The distribution statics does not fix

How much of a deflection belongs to the beamThe share of the total deflection that is the beam's own bending, against the stiffness of what it sits on. A 18 m beam on five supports under a uniform load: on rigid supports every millimetre is the beam's, and the share falls away as the supports soften until almost none of it is. The beam drawn beside this figure sits at 1% — so 99% of what it does is happening somewhere a beam calculation never looks. The two flexibilities are in series, which means the softer one governs and stiffening the other buys nothing.4020080030002000000.20.40.60.81stiffness of each supportfraction of the deflection that is bendingrigid supportsare over here1%the beam drawn
Fig. 4 How much of a deflection belongs to the beam. The two flexibilities are in series, so the softer governs — and on the beam drawn beside the curve, the beam itself accounts for one per cent of what happens.

A reaction’s total comes from equilibrium. Its distribution comes from compatibility, and depends on the relative stiffness of the beam and of what it is sitting on — the beam on the ground in miniature.

A stiff beam on a soft pad gets a nearly uniform pressure. A soft beam on a hard bearing gets a pressure concentrated at the edges, because the beam is trying to rotate and its underside lifts away in the middle of the bearing. A bearing at the end of a beam that is rotating gets a pressure concentrated at one edge, and the reaction’s line of action moves toward that edge — which shortens the span and changes the moment.

None of that is available from the three equations. It is why bearing design is a separate subject with its own conventions, and why the Rb/8Rb/8 reduction above quietly assumes the answer.

Why the shear check moves away from the face

The second consequence of a real support is about shear, and it is a load path argument rather than an arithmetic one.

Load applied within about one member depth of a support does not have to be carried by the shear mechanism at all. It arches directly into the bearing along a compression strut, in the way a deep member has no section to design. The shear the member has to carry through its web is therefore the shear at a distance dd from the face of the support, not the shear at the centreline, and on a heavily loaded short span the difference is substantial.

The bearing is one length and the web is loaded over anotherA load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.s_s = 200l_y = 659flangewebspread 3.3 times the bearingyield resistance 1870 kN · elastic critical 510 kNresistance 488 kN at a slenderness of 1.92
Fig. 5 The same spreading, seen from the other end. A load applied over a 200 mm bearing reaches the web over 659 mm, because the flange’s own bending stiffness against the web’s strength decides the length and nobody chooses it.

The spreading is real and it is measurable: a load applied over a 200 mm bearing on this girder reaches the web over 659 mm, 3.3 times the bearing width, and 70% of the resistance comes from that spread rather than from the bearing itself. The effective length is not a decision — it is what a flange’s stiffness and a web’s strength work out to.

The bar that has to run further than the diagram says

The third consequence is the one most likely to be got wrong, because it points the other way: near a support the tension chord carries more than the moment diagram says, not less.

The bar is longer than the moment diagram saysThe force in the tension chord along a 8 m span. The lower curve is M/z, which is what a bending calculation gives and which is zero at the support. The upper one adds the V·cot θ/2 the truss puts there: the cut along the crack passes through the chord as well as the stirrups, and it carries half the shear as chord tension. At the support that is 563 kN where bending says nothing at all, and everywhere else it is the same curve moved 585 mm toward the support. A bar cut off where the moment diagram says it may be is a bar that is too short.024680100200300400500600distance along the span (m)tension chord force (kN)with the trussM ÷ z alone563 kN at the supportthe shift is z·cot θ/2 = 585 mm
Fig. 6 The force in a tension chord along a span. The lower curve is what a bending calculation gives; the upper adds the chord tension the truss mechanism puts there, and at the support that is 563 kN where bending says nothing at all.

A cracked concrete beam carries its shear as a truss, and the diagonal crack a designer cuts along passes through the tension chord as well as through the stirrups. The chord therefore carries Vcot⁡θ/2V\cot\theta/2 of tension in addition to M/zM/z — half the shear, at the crack angle — and the whole force curve is the moment diagram shifted toward the support by zcot⁡θ/2z\cot\theta/2, which here is 585 mm.

A bar curtailed where the moment diagram permits it is a bar that stops half a metre too soon, in the region where the shear is largest. That is the shift rule, it is a consequence of a support having a region rather than a point around it, and it is one of very few results in the subject where the honest answer is worse than the simple one.

The support that is also a restraint

A wide support does something else no arrow shows: it restrains rotation.

A beam bearing on 400 mm of wall, with mortar or a bedding compound in the joint, is not free to rotate about a point, whatever the free body drawn for it assumed. As the beam deflects, the pressure moves toward the inner edge and a couple develops — small, uncontrolled, and reliably present. It is why a “simply supported” precast unit cracks over its bearing, and why a nominally pinned connection is neither pinned nor rigid.

The magnitude is bounded by the reaction and the width: the largest couple a bearing can deliver is Rb/2R b/2 if the whole reaction moves to one edge. On the beam above that is 38 kNm — a quarter of the hogging moment the analysis reported, arriving in a member designed as though it were zero.

The angle is what the bearing has to allowThe end of a 11 m beam 900 mm deep, rotated by 33.0 milliradians — the rotation a uniform load actually produces here, drawn at 4 times its size. A rotation about the bearing moves the top of the section 29.7 mm horizontally, which is what an expansion joint, a cladding gap or a nib has to accommodate and what a deflection limit never mentions. A beam held down at the top of its section instead of at its bearing is a beam with an axial force in it that the analysis does not contain: at 0.0330 radians the restraint is stiff and the movement is small, which is exactly the combination that produces a large force.the bearing29.7 mm at the top faceθ = 33.0 mrad = 1.89° — drawn 4× overat the L/360 deflection limit the rotation would be 8.9 mrad, or 0.51°
Fig. 7 The rotation a bearing has to allow, drawn at four times its size. It moves the top of the section 29.7 mm horizontally, which is what a joint has to accommodate and what a deflection limit never mentions.

The counterpart of that restraint is a movement. An 11 m beam 900 mm deep at an ordinary deflection rotates 33 milliradians at its end, and rotating about the bearing carries the top of the section 29.7 mm sideways. Something has to allow that movement — a joint, a gap, a slotted hole — and if nothing does, the beam has an axial force in it that no analysis contains. A stiff restraint against a small movement is exactly the combination that produces a large force, which is the point the movement nobody applied makes in general.

The other reason the width matters, which is the perimeter

Where the support is a column rather than a bearing, the width does something arithmetically different: it becomes a perimeter.

Most of the control perimeter is not the columnPunching resistance against the size of the square column, everything else held. The gain is far weaker than linear — fitted at the power 0.44 — because the control perimeter is the four faces plus two full quarter-circles at each corner, and those corner arcs are 2765 mm long whatever the column does. At the column drawn they are 63% of the whole perimeter.200400600800100012001400020040060080010001200column side (mm)punching resistance (kN)fitted power0.44the cornersdo not shrink
Fig. 8 Punching resistance against the size of the column. The gain is far weaker than linear, because the control perimeter is four faces plus two full quarter-circles, and the corner arcs are the same length whatever the column does.

A check made on a perimeter rather than on a section behaves quite unlike a beam shear check. Making the column larger adds four straight lengths and does nothing to the corner arcs, which are 2,765 mm long regardless and 63% of the whole perimeter at the size drawn — so the resistance grows as roughly the 0.44 power of the column size rather than in proportion to it. Doubling a column’s plan dimension buys 36% more punching resistance.

How far the disturbance reaches

All of the above is local, and there is an exact statement of how local.

The decay length is the load's own wavelength, and nothing elseThe stress left from a self-equilibrating end load that varies as a cosine of wavelength λ, against distance in depths. The Airy function (A + Bx)e^(−αx)cos(αy) satisfies both boundary conditions and gives a factor of (1 + αx)e^(−αx) with α = 2π/λ, so the curves are the same curve stretched: at one wavelength in there is 1.4 per cent left, whatever λ was. No modulus, no Poisson's ratio and no thickness appears anywhere. A self-equilibrating load across a depth must change sign at least twice, so its slowest component has a wavelength of about the depth — which is the whole of Saint-Venant's principle, with a number in it.00.511.5200.20.40.60.81distance from the load (depths)stress remaining ÷ stress appliedλ = 0.25 depthsλ = 0.50 depthsλ = 1.00 depthsλ = 2.00 depthsat one wavelength: 1.36% · at two: 0.005%
Fig. 9 The stress left from a self-equilibrating end load, against distance in depths. At one wavelength in there is 1.4% left, whatever the wavelength was — and no material property appears anywhere in the answer.

The difference between a point reaction and a distributed one is a self-equilibrating load system: the two have the same resultant and the same moment, so they differ by a set of forces summing to nothing. Saint-Venant’s principle says such a difference dies away, and the version with a number in it says it dies away over the wavelength of the self-equilibrating component — about the depth of the member, with 1.4% left at one depth.

So the whole of this essay is about a region one member depth long at each end of every member, and about nothing outside it. That is a comforting boundary and a deceptive one, because a great many of the things that go wrong in structures happen inside exactly that region.

The support that moves the span

There is one more quantity the arrow hides, and it changes an answer by more than any of the effects above.

The span used in a calculation is a distance between two lines of action, and a bearing’s line of action is wherever its pressure resultant happens to sit. For a beam on two pads of width bb, an analysis usually takes the span as centre to centre, and the pressure resultant sits nearer the inner edge of each pad as the beam rotates — so the effective span is shorter than the drawing says, by something up to b/2b/2 at each end.

On a 4 m span sitting on two 400 mm bearings, that is up to 400 mm off an 4,000 mm span: 10%, and the moment goes as the square, so 19%. On a 20 m span it is a fifth of a per cent and nobody cares. The rule of thumb is the same as everywhere else in this essay — the correction is a bearing width divided by a span, so it matters exactly when the member is short.

Precast concrete is where it bites, because precast members are frequently short, heavily loaded, and sitting on bearings that are a substantial fraction of their span. It is also where the correction runs the wrong way for the designer: the shorter effective span reduces the moment and increases the shear, and a short deep precast unit is governed by shear.

Where the model stops

The reduction assumes the reaction is uniform. Everything about Rb/8Rb/8 depends on a pressure distribution that compatibility, not statics, decides — and the section above says compatibility gives a different answer for every combination of stiffnesses. The reduction is therefore a good approximation and not an identity.

A wide support is not a wide member. A beam framing into a deep transfer girder has a support several metres wide, and none of the arithmetic above survives that: the “support” is a structure with its own deflected shape, and the beam is continuous with it rather than sitting on it.

Nothing here is about the support itself. The pressures discussed are what the beam applies to the bearing, which is what the bearing has to carry, and a bearing that crushes redistributes everything that was assumed about it.

And the width is not always known when the analysis is run. A column size is an output of a design that has the beam moments as inputs, so the face-moment reduction is another small fixed point: assume a width, size the column, check the width. It converges in one pass and it is worth noticing that it is an iteration at all, because the direction of the error is not obvious — a larger column reduces the beam’s hogging moment and increases its own.

One decision, two sets of beam loadsThe same 9 × 5 m panel carrying 6 kN/m², divided two ways. Spanning the slab one way puts 15.00 kN/m on each long beam and nothing at all on the short ones. Dividing it at 45° puts 10.83 kN/m on the long beams and 7.50 kN/m on the short ones. The panel, the slab and the load are identical; the beams are not.spanning one wayspanning both ways22.5 m²15.00 kN/m22.5 m²15.00 kN/m16.3 m²10.83 kN/m16.3 m²10.83 kN/m6.3 m²7.50 kN/m6.3 m²7.50 kN/mboth divisions hand over 270 kN — the whole panel, and no more
Fig. 10 A reminder of how much of a load path is decided rather than computed. The same panel divided two ways gives entirely different beam loads, and the reactions this essay is about follow whichever decision was made.

What the picture cannot show

A moment diagram is drawn as a continuous curve running to the ends of a line representing the beam. The line has no thickness and the beam has a depth; the curve runs to a point and the beam sits on a bearing. Everything in this essay lives in the gap between those two statements, and none of it is visible on the diagram because the diagram is a plot of a function of one variable.

Nor does the drawing show that the reaction is a response. It is the only force on a free body that is not applied by anybody: it is whatever the support has to do, at whatever distribution the two stiffnesses settle on, and it is the one arrow in the picture whose value was computed rather than specified.

The generalisation

The habit worth carrying is to ask what a simplification has thrown away, and where.

Reducing a bearing to a point is one of a family of moves this subject makes constantly: a distributed load becomes a resultant, a connection becomes a node, a column becomes a line, a slab becomes a plane. Each is exact for the quantity it was made for — global equilibrium — and each fails within about one member depth of where it was made.

It is the same trade a distributed load makes when it is replaced by its resultant, and the same one a rigid joint makes when it becomes a node. The general rule is the one Saint-Venant supplies. Two load systems with the same resultant differ only locally, and “locally” means about the size of the region over which they differ. So the simplification is safe everywhere except near itself, and every check that lives near a support is a check the simplification cannot be used for.

That is why the region above a support carries such a disproportionate share of the subject’s detailing rules — the shift rule, the shear at dd, the face-moment reduction, the punching perimeter, the bearing pressure, the rotation capacity. Each of them is a patch on the same idealisation, applied in the one place it does not hold.

Where this goes next

moment19.6 sagging24.5 hogging30.6 if the spans were simplereactions 14.0 38.5 38.5 14.0 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not The moment over the support, and what it buys Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity. 00.511.5200.20.40.60.81distance from the loaded end (depths)self-equilibrating remainder ÷ mean stressa point loadspread over h/5split in twoone depthsame resultant, same moment · one per cent left by 1.18 depths How far a wrong load reaches Every figure in this collection applies a load as a point, a line or a uniform pressure, and no real load is any of those. The licence is Saint-Venant's, it is usually quoted as a principle, and it is really a statement about a wavelength. 1000 kN1/β = 2.56 mthe beam lifts off at 6.04 msettlement 3.90 mm · contact pressure 195 kN/m · moment 641 kNm, none of which contains the length of the beam The beam that sits on the ground Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened. 2d = 450column7.2 m bayperimeter u₁ = 4427 mmshear to carry V = 604 kNv = 0.606 against 0.658 N/mm²92% of the resistance used A check made on a perimeter, not on a section Every shear check in this collection is made on a plane cut through a member. A slab sitting on a column has no such plane, because the shear leaves in every direction at once — so the check is made on a closed line, and a line grows with the column while the load grows with the square of the bay.

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.

BearingBending momentCompatibilityContinuityDistributed loadEnd rotationFree bodyLoad pathPunching shearReactionSaint-Venant's principleShear forceShear truss analogyStress concentrationSupport width