Structural Engineering & Statics
where the load goes
Internal forces

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

Assumes When there is no section to design, The beam that becomes a truss and What a cut reveals, and why it was there all along.

18 min read 9 figures The load must go somewhereWhich failure arrives firstDrawing as calculation

A beam carries a load down to its supports by making a triangle: the load pushes down at the top, a diagonal in compression runs from there towards the support, and a tie along the bottom holds the two ends apart. That is the truss inside every beam, and every shear calculation in this collection is a way of sizing its members.

The triangle has an assumption in it that nobody states, and it is not about concrete or steel or the angle of the diagonal. It is that the load arrived on the top.

The same beam, the same load, and one of them has to lift itTwo lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it.load on the top chordthe same load, hung from the soffit600 kN in the hangernothing in the same memberchord forces identical to machine precision · hanger steel 1200 mm², shear steel 96 mm² over the same length
Fig. 1 Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in one, and the entire 600 kN in the other.

Which free body produced the number

Take the joint. Not the section, not the beam: the single panel point on the bottom chord where the load comes in.

Four members meet there in the model above: two chord members running along the bottom, and one vertical running up to the top chord. And — this is the whole essay — no diagonal. The struts in a beam run from the top chord down towards the nearer support, so on the left half they arrive at bottom-chord nodes that are further from the support than the top node they started at, and the node under the load is not one of them.

So vertical equilibrium at that joint has exactly two terms: the applied 600 kN downwards, and the vertical member. The vertical carries 600 kN in tension, and it is the only member that can.

Apply the same load to the corresponding top-chord node and two struts meet it, one from each side, and the vertical carries nothing at all. The load walks straight down the diagonals it was already in a position to use.

Everything else is identical, which is what makes it invisible

The two models have the same reactions, 300 kN each. The same shear diagram. The same moment diagram — 1,667 kN in the bottom chord at mid-span in both, agreeing to machine precision, because the chord force is the moment over the lever arm and neither has changed.

A sectional calculation is a calculation about those quantities. It reads a shear at a section and converts it into a link requirement per unit length; it has no variable at all for which face the load arrived on, so the requirement that appears when the load is hung is not something a sectional check gets wrong — it is something the check has no way to express.

That is why it is written into codes as a separate clause about “indirect support” or “suspended loads”, in a different chapter from the shear rules, and why it is one of the most reliably missed requirements in reinforced concrete detailing.

The cut that severs the stirrups is the cut that counts themA cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 729 mm severs z·cot θ/s = 12.2 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.the cutcompression chordtension chordstirrups in blue, struts dashedz = 729cot θ = 2.5 · 12.2 stirrups crossed · V = 829 kN from 1047 mm²/mstrut stress 3.98 N/mm² against 10.56 available
Fig. 2 The truss analogy the sectional rules come from. Its diagonals run one way and its verticals another, and the whole of this essay is the observation that a load applied to the wrong chord has no diagonal available to it.

How much steel, and where

The shear at the panel is 300 kN. With a strut angle of cot⁡θ=2.5\cot\theta = 2.5 and a lever arm of 810 mm, the sectional calculation asks for 0.30 mm² of link per millimetre of beam — which over the 324 mm the strut fan occupies is 96 mm².

The hanger carries 600 kN. At 500 N/mm² that is 1,200 mm², in the same place, and additional to the 96.

A factor of twelve and a half, arriving as links that are not per-metre but per-load, concentrated within about a lever arm either side of where the secondary beam frames in. It is not a small correction to a shear calculation; on this beam it is the shear reinforcement, and everything the sectional rules produced is a rounding on top of it.

The verticals in the two models say it more directly than the arithmetic does. Along the beam they read 0, 300, 300, 600, 300, 300, 0 with the load hung, and 0, 300, 300, 0, 300, 300, 0 with it on top. Every panel except one is doing the same job.

Where the load enters, between the two extremes

Nothing forces a load to arrive at either face. A secondary beam framing into the web of a primary delivers its reaction somewhere in the middle of the depth, and a slab cast against the side of a downstand delivers it lower.

The requirement scales with how far the load still has to travel. A load entering at height aa above the soffit has already been lifted through aa, and what remains to be hung is the fraction 1−a/z1 - a/z of it — linear, because the strut that eventually carries it can only start from where the load already is.

How much of the load has to be lifted, and how farThe suspension force against the height at which the load enters the beam, as a share of the internal lever arm. A load delivered to the soffit has to be carried the whole way up to the compression chord, so the tie takes the entire 600 kN; a load delivered to the top needs no tie at all; and in between the requirement is linear, because the strut that arrives can only start from where the load already is. At mid depth it is 300 kN, which is still 1.0 times the shear the same panel is carrying. None of this appears anywhere in a sectional shear calculation, which has no opinion about which face a load arrived on.0204060801000100200300400500600where the load enters (% of the lever arm above the soffit)force to be hung (kN)hung from the soffit: all of itdelivered on top: none of itthe shear in the same panel is 300 kN, and it is a different requirement in a different place
Fig. 3 The suspension force against the level at which the load enters, as a share of the lever arm. At mid-depth it is still 300 kN, which is the whole of the shear in that panel — so a beam framing in at mid-depth needs hanger steel equal to its entire reaction spread over its own connection region, on top of everything the shear check asked for.

Where it actually appears

The arithmetic is simple and the detail is not, because the situations that produce it do not look alike.

A secondary beam framing into a primary. The commonest case, and the one the clause is written for. The secondary’s reaction enters the primary’s web at whatever level the two soffits align, which for beams of equal depth is the bottom.

A slab hanging from a downstand. The slab is cast against the side of the beam near the soffit and delivers a line load along it — so the hanger requirement is per metre rather than per load, and is easy to miss precisely because it looks like a distributed load rather than a point one.

A load hung from a soffit. Services, a runway beam, a suspended floor below. The load is unambiguously at the bottom and the requirement is unambiguously the whole of it, which is why this case is the one that gets detailed correctly.

And a support at the top of a member. The same argument upside down: a beam supported by hanging from something above, rather than bearing on something below, has to get its reaction down to the tension chord, and the tie required is the reaction.

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 0.41 — because the control perimeter is the four faces plus two full quarter-circles at each corner, and those corner arcs are 3142 mm long whatever the column does. At the column drawn they are 66% of the whole perimeter.2004006008001000120014000200400600800100012001400column side (mm)punching resistance (kN)fitted power0.41the cornersdo not shrink
Fig. 4 A load entering a member somewhere it was not expected. A punching check is the same kind of question asked about a slab — a concentrated force crossing a surface that the flexural model has no term for — and its answer is likewise a perimeter and a rule rather than a section.

The crack that says it was missed

The failure mode is specific and recognisable. Without the hanger, the only thing carrying the load up to the compression zone is the concrete’s tensile strength across a horizontal plane at the level the load entered.

So the crack is horizontal, or nearly so, starting at the point of application and running along the beam — and it opens under service load rather than at failure, because the tension it is resisting is the full applied load rather than a fraction of it. Below the crack the beam is a bracket carrying the load on nothing; above it, the beam is intact and its own calculations remain valid.

The beam does not look overloaded. Its moment is fine, its shear is fine, its deflection is fine, and there is a crack running along it at soffit level near every point where something is hung from it.

The coefficient is a slope, and that is why it can exceed oneThe crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing.36shearthe bars clamp, and do not carryasperity slope 54° · tan = 1.40clamping stress 2.50 N/mm² · resistance 3.50 N/mm²sliding without separating is not available to a rough crack
Fig. 5 What is left once that crack has opened. Reinforcement crossing a crack that already exists carries load by clamping and friction rather than by tension across intact concrete, and the arithmetic of that is a different calculation with a much lower answer.

The steel version is a stiffener, not a bar

The same equilibrium happens in steelwork and produces a completely different-looking answer.

A secondary beam bolted to the web of a primary near its bottom flange delivers its reaction into the web at that level. The web has to carry that force up to the top flange, and a web is a plate: what it offers is a strip of itself in tension, over a width the load can spread into, and the check is whether that strip is adequate rather than whether reinforcement has been provided.

Where it is not — where the reaction is large, or the web is thin, or the connection is close to the bottom flange — the answer is a stiffener: a plate welded between the flanges at the connection, taking the force directly from the incoming beam to the top flange. It is the hanger, made of plate, and it is put in for exactly the reason the bars are.

The vocabulary hides the identity. One trade calls it suspension reinforcement and the other calls it a bearing stiffener, and neither set of rules mentions the other; the free body at the joint is the same in both.

A resistance that is very nearly square in a thickness that appears onceThe three resistances against web thickness. The yield resistance is the web thickness times an effective length times a stress and is therefore nearly linear; the elastic critical resistance goes as the cube; and the reduction factor between them, 0.5 divided by the slenderness, restores about half of that. What comes out is a resistance going as the 1.91 power of the web thickness — so a web a millimetre thicker is worth far more than a shear check on the same web would suggest.68101214161820222401000200030004000web thickness (mm)resistance (kN)yield, F_yelastic, F_crwhat is leftpower 1.91
Fig. 6 The web asked to carry a force into itself. Whether a stiffener is needed is a question about the width of web the force can recruit, and the width is chosen by the plate rather than imposed by the detail — which is the steel version of the spread the hanger bars are allowed.

Why the strut-and-tie model finds it and nothing else does

This is a clean demonstration of what a lower-bound model is for.

A strut-and-tie model is a complete equilibrium system: every force is followed from where it enters to where it leaves, through members that are drawn. Nothing can be carried by a path that has not been drawn, and so nothing can be carried by a path that does not exist.

A sectional model is not complete in that sense. It computes a resultant at a cut and converts it into reinforcement by a rule, which is exact for the region and the loading the rule was derived for and silent about everything else. Its silence is not conservative — there is no factor of safety on a requirement that has not been noticed.

That is the argument for drawing the model rather than applying the clause, and it generalises well past this case: anywhere a load enters a member somewhere other than where the standard picture assumed, the requirement is found by asking how the force gets from where it is to where the model needs it, and by drawing a member to carry it.

A truss drawn inside a solid, and solved as oneA deep member 4000 mm between bearings and 2000 mm deep, carrying 1200 kN at mid-span. The model is two struts and one tie, on a lever arm of 1600 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 750 kN and each strut at 960 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 3.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1724 mm² of steel. A beam calculation on the same member would have asked the tie for 702 kN, which is 7% less than the model does.1200 kNtie 750 kNstrut 960 kN38.7°z = 1600strut 3.0 N/mm² over 812 mm · limit 15.8bursting across each strut 240 kN · tie steel 1724 mm²
Fig. 7 The model in its own setting. Every force in it goes somewhere, which is what makes it a lower bound and what makes it capable of noticing a force with nowhere to go.

What it costs to get it right, which is nothing

The last thing worth saying about this requirement is how cheap it is.

The hanger on this beam is 1,200 mm² — six sixteen-millimetre links, over a length of a few hundred millimetres, at a cost of a few kilograms of steel in a member that already contains several hundred. There is no impact on the concrete section, no impact on the depth, no impact on the reinforcement anywhere else in the beam, and no impact on anything the analysis produced.

It is a requirement whose entire difficulty is knowing it exists. That puts it in a small and interesting category — with the hanging load’s mirror image at a corner joint, where the reinforcement detail decides whether a joint reaches a fraction or the whole of the moment its members can carry, and with the reinforcement across a construction joint. Each of them is a few bars, each is invisible in the analysis, and each converts a member that works into one that does not.

The lesson generalises past concrete. Wherever a model produces a force at a point and a detail delivers it somewhere else, there is a transfer between the two that nothing has been asked to carry — and the cost of carrying it is almost always trivial next to the cost of the member it is attached to.

The free body of a corner closes to the last digitThe knee of the frame, cut clear of both members. The beam applies 120.0 kN downward, 36.7 kN horizontally and a couple of 97.8 kNm; the column applies 120.0 kN upward, 36.7 kN horizontally and a couple of 97.8 kNm the other way. The beam's end shear of 120.0 kN leaves as the column's axial force of 120.0 kN — vertical equilibrium of this block and nothing else — while the column's shear of 36.7 kN leaves along the beam as axial force. The three residuals are -2.2e-8, 0.0e+0 and 0.0e+0, the last being 0.0e+0 of the corner moment itself: continuity of moment at a rigid joint is not an approximation but an equation the solve satisfied.beamcolumn120.0 kN of shear36.7 kN axial97.8 kNm120.0 kN axial36.7 kN of shear97.8 kNmΣFx = -2.2e-8 · ΣFy = 0.0e+0 · ΣM = 0.0e+0 kNm
Fig. 8 The other detail whose whole difficulty is knowing about it. A joint reinforced without regard to which way the moment is opening it reaches a fraction of the capacity of the members meeting there, and the fix is a few bars bent the other way.

Where the model stops

The truss is one of many. The strut angles were chosen, and a different choice moves the strut fan and the panel widths. What does not move is the vertical at the loaded node, because that follows from the load having no diagonal available, not from the angle the diagonals happen to make.

The hanger is drawn as a member and is really a region. How far either side of the load the links may be counted is a detailing rule, and the rules differ: some allow a spread of half the effective depth each way, some tie it to the width of the incoming beam. Nothing here derives that spread.

Struts are assumed to be able to carry what they are given. A strut arriving at a node with three members meeting is subject to a node-stress check the model above does not perform, and at the node under a hung load that check is often the one that governs the width of the incoming member.

And the beam is reinforced concrete throughout. In steelwork the same argument becomes a question about stiffeners: a secondary beam framing into the web of a primary near its bottom flange puts a tension into that web which has to reach the top flange, and the answer is a stiffener rather than a bar. The mechanism is identical and the vocabulary is not.

What the pictures cannot show

The two models in the first figure are drawn side by side and look nearly the same, which is the point and also the difficulty: the difference between them is one member’s force, and a reader who is not looking for it will not see it.

Nor can any of these figures show what the requirement looks like on a drawing. Hanger reinforcement is a handful of extra links, drawn at a closer spacing over a few hundred millimetres, in a bar-bending schedule that already has several hundred entries. It is not a distinctive detail; it is a local change of spacing that an experienced detailer puts in without being asked and an automated one omits without being noticed.

The assumption the figure rests on

The whole essay assumes the compression chord is where the load has to get to.

That is true for a beam in ordinary bending, and it stops being true in exactly one case worth naming: a member in which the load is applied below the neutral axis of a section that is already in tension there — a hanging member, a tension tie carrying load transversely, the bottom chord of a truss with a load on it. There, the load is not being lifted into a compression zone, it is being introduced into a tension field, and the requirement is a bond and anchorage question rather than a suspension one.

The distinction matters because the two get confused in the same detail. A load hung from the bottom chord of a truss needs the chord checked for local bending and the connection checked for its own capacity — and needs no hanger at all, because there is no compression zone above it to reach.

A Pratt truss of 8 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.233.3233.3400.0500.0500.0400.0233.3233.3-400.0-500.0-533.3-533.3-500.0-400.0-150.0-90.0-60.0-90.0-150.0-313.9-313.9224.2134.544.844.8134.5224.2tensioncompression2 carrying nothing
Fig. 9 The structure where the same load needs no lifting. A truss’s bottom chord is in tension along its whole length and its diagonals arrive at bottom-chord nodes by design, so a load applied there has a path waiting for it. The beam’s problem is that its truss is implicit, and an implicit member is one nobody checked.

Where this goes next

1200 kNtie 750 kNstrut 960 kN38.7°z = 1600strut 3.0 N/mm² over 812 mm · limit 15.8bursting across each strut 240 kN · tie steel 1724 mm² When there is no section to design Beam theory needs a section, and a section needs the strain to be linear across it. Within about a depth of a support, a load, a corner or a hole it is not — and those are the regions structures actually fail in. the cutcompression chordtension chordstirrups in blue, struts dashedz = 495cot θ = 2.5 · 8.3 stirrups crossed · V = 563 kN from 1047 mm²/mstrut stress 8.79 N/mm² against 10.56 available The beam that becomes a truss Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes. neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does The shear nobody draws A stack of loose planks slides at its ends when it is loaded. Glue them and the sliding stops — and whatever the glue is now carrying is a stress that no bending calculation contains. 100 kNe = 150the dashed ring is the group's centroidworst bolt 50.37 kNSix bolts · direct shear 16.67 kN eachelastic vector method The connection is not a point, and every diagram on this site says it is Every free body drawn here has joined its members at points. Real structures fail at the joints far more often than in the members, and the reason is that a joint is exactly the region the theory behind every other page explicitly excludes.

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.

Compression strutDetailingEquilibriumFree bodyHanger reinforcementIndirect supportLoad pathLower bound theoremSecondary beamShearShear reinforcementStrut and tieSuspensionTieTruss analogy