Equilibrium

The load a beam is given is a decision

Every beam calculation on this site so far has started with a load per metre, handed over as though it were a property of the beam. It is not. It is the answer to a prior question nobody draws, and two defensible answers to it differ by sixty per cent on the same floor.

Assumes Everything adds to nothing, and that is the whole of statics, The load that is spread out, and the force that replaces it and The free body is a choice, and choosing it well is the whole skill.

Every beam in this collection has arrived with its load already decided. A uniform load of 4 kN/m, a point load of 24 kN at midspan — the numbers are stated, the analysis integrates them, and the diagram that comes out is exactly right for the beam it was given.

Nobody asked where the 4 kN/m came from.

Where a beam's load comes fromA 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.15.0 m²9.38 kN/m15.0 m²9.38 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m8 m6 m48.0 m² divided, 48.0 m² of panel — the division closes
Fig. 1 An 8 × 6 m floor panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four regions sum to 48.0 m², which is the panel, so nothing has been invented or lost. The line loads quoted are uniform equivalents — 9.38 and 7.50 kN/m — and the real distributions peak at 15.0 kN/m at midspan.

That division is the whole subject. It is not in any structural analysis, it is not checked by any equilibrium equation past the total, and it is the first number in every calculation downstream of it. A beam’s load per metre is not measured; it is allocated, and this essay is about who does the allocating and on what grounds.

What the rule actually claims

The tributary rule says: draw lines where the slab’s load has nowhere better to go, and give each beam what falls on its side. For a panel supported on all four sides the lines run at 45° from the corners, meeting on a ridge along the middle, and the regions come out as two trapezoids and two triangles.

The justification usually offered is that a slab spans to its nearest support, so a 45° line is where “nearest” changes hands. That is close to true for a square panel and increasingly untrue as the panel lengthens, and the reason to be suspicious of it is that a slab does not span to a support in the way a plank does — it bends in two directions at once and the division of load between them is decided by relative stiffness, not by distance.

What the rule does have is the property that matters most for equilibrium, and the figure above states it: the areas close. Four regions summing to 48.0 m² against a panel of 48.0 m² means the beams below have been handed exactly the load that was there — no more, and none of it dropped. Any division with that property produces a set of beams that is in equilibrium with the floor above them, whatever else it gets wrong.

Two defensible answers, sixty per cent apart

The alternative to dividing a panel both ways is to make the slab span one way, which is what happens whenever it is reinforced predominantly in one direction, or precast, or ribbed, or made of timber joists.

One decision, two sets of beam loadsThe same 8 × 6 m panel carrying 5 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 9.38 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 ways24.0 m²15.00 kN/m24.0 m²15.00 kN/m15.0 m²9.38 kN/m15.0 m²9.38 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/mboth divisions hand over 240 kN — the whole panel, and no more
Fig. 2 The same panel, the same slab and the same load, divided two ways. Spanning the slab one way gives each long beam 15.00 kN/m and the short beams nothing at all. Dividing at 45° gives the long beams 9.38 kN/m and the short beams 7.50. Both hand over 240 kN, both are in equilibrium, and the long beam differs between them by 60%.

Neither drawing is a mistake. They describe different slabs, and the decision between them is made when the reinforcement is detailed — often by a different person, later, and occasionally the other way round from what the beam design assumed.

The failure mode this produces is quiet. A beam designed for 9.38 kN/m and built under a slab spanning one way is 60% overloaded, and nothing in any check on the beam will say so, because the check compares the beam against the load it was given. The only place the discrepancy is visible is the comparison the two drawings above make, and no software prints it: the load path is the one part of a structural model that is not computed from anything.

There is a second consequence that matters more often. In the one-way case the short beams carry nothing from the slab — but they are still there, still spanning six metres, and they still carry their own weight and whatever the cladding hangs on them. A member that a load-path decision has emptied is a member that costs the same as it always did, and one of the ordinary economies of a floor grid is to notice that it exists and delete it.

The one-way case is also the older one, and the tributary rule was written for it. A timber floor is a set of joists at close centres carried on two trimmers, and there is no ambiguity anywhere: each joist takes a strip of floor exactly as wide as its spacing, each trimmer takes half of every joist that lands on it, and the division is not an idealisation but a description of what the carpentry does. The 45° rule is what happens when that unambiguous picture is carried across to a monolithic slab that does not have joists in it — and the reason it survives is that it inherits the property the joisted floor had, which is that the areas close. What it does not inherit is the reason to believe the lines are in the right place.

The peak is not the average

The uniform equivalents in the figures are convenient and they are not the load. A trapezoidal distribution with the same total gives a larger midspan moment than the uniform one, because more of it sits near the middle.

A trapezoidal load and the force that replaces itA trapezoidal distributed load with its resultant computed by integration: an area of 63.0 acting at 3.43 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 63.0at x = 3.43, the centroid of the areamomentspread: 47.5replaced: 92.3reactions agree exactly (27.00 and 27.00); the peak moment does not
Fig. 3 A trapezoidal load and the single force that replaces it. The resultant is the area under the distribution and it acts at that area’s centroid, which is exact for equilibrium; what it is not exact for is the bending moment, which depends on how the load is spread and not only on where its centroid is.

For the panel above, the long beam’s real load rises from zero at each end over three metres — half the short span — to 15.0 kN/m across the middle two, and totals 75 kN. Its end shear is 37.5 kN, which is exactly what the 9.38 kN/m uniform equivalent gives, because both loads have the same total and the same symmetry and a reaction is nothing but a total shared out. Its midspan moment is 97.5 kNm against the uniform equivalent’s 75.0 — thirty per cent larger, from a substitution that is exact for the other quantity on the same beam.

That asymmetry is the thing to carry away. Replacing a distribution by an equivalent uniform load of the same area is exact for equilibrium, exact for reactions, exact for shear at the supports, and wrong for bending by an amount that depends on how the load is bunched. A triangular load of the same total gives WL/6 where a uniform one gives WL/8, which is the worst case and a third; a trapezoid sits between them. The practice of quoting an “equivalent uniform load” in handbooks handles this by defining the equivalent as the one that gives the same moment rather than the same total — a different number, for a different purpose, wearing the same name.

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.9.38 per unit lengthshear37.5moment75.0 at x = 4.00the moment peaks exactly where the shear passes through zero
Fig. 4 The long beam under its uniform equivalent: 9.38 kN/m over 8 m, a midspan moment of 75.0 kNm and end shears of 37.5 kN. This is the calculation that gets done, and it is a calculation about a load that does not exist — the real one is a trapezoid of the same area, which produces the same 37.5 kN and 97.5 kNm.

The right response is not to abandon equivalents, which would make hand calculation impossible; it is to know which quantity an equivalent was constructed for. A load allocated by area, converted to a uniform intensity, and put through a moment calculation has had two approximations applied to it in sequence, and only the first of them announces itself.

The elastic answer is not the 45° answer

If the slab really does span both ways, the honest question is what fraction of the load goes each way, and that has an answer which owes nothing to 45° lines.

A two-way slab is a one-way slab as soon as it is not squareThe share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 6 × 8 m, a ratio of 1.33, and its short strips take 76.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two.11.522.530.40.50.60.70.80.91long span ÷ short spanshare taken by the short strips6 × 8 m: 76.0%by 2 : 1 it is a one-way slab
Fig. 5 The share of the load carried by the strips spanning the short way, against the ratio of the sides. Strips in the two directions cross at the centre and must deflect equally there, and a strip’s deflection goes as the fourth power of its span, so the short strips take Ly⁴/(Lx⁴ + Ly⁴). At the 8 : 6 panel above that is 76.0%, against the 62.5% the tributary areas imply.

The fourth power is doing all of the work. A panel only a third longer than it is wide already sends three-quarters of its load the short way, which is why the two-way slab of the textbooks is a genuine object only for panels close to square. The tributary rule, by contrast, gives the short direction 62.5% at this proportion, because it is dividing an area rather than sharing a load, and area does not go as the fourth power of anything.

The discrepancy runs the safe way for the long beams and the unsafe way for the short ones, and it is around a fifth. That is the size of error a designer accepts in exchange for a rule that can be drawn on a plan in ten seconds and checked by eye, and the exchange is usually a good one — but it is an exchange, and calling the tributary division “the load” hides that a choice was made.

A column is a running total

Everything above concerned one floor. The load that reaches a foundation is the same allocation repeated at every level, and it accumulates in the one member that nobody redraws.

A column is a running totalA column carrying 36 m² of floor at each of eight levels, at 5 kN/m². Each floor adds 180 kN, so the load at the base is 1440 kN — the same tributary area counted eight times. Nothing in the drawing changes down the height; only the number does.level 8180 kNlevel 7360 kNlevel 6540 kNlevel 5720 kNlevel 4900 kNlevel 31080 kNlevel 21260 kNlevel 11440 kN1440 kN into the foundation= 36 m² × 5 kN/m² × eight floors
Fig. 6 A column carrying 36 m² of floor at each of eight levels, at 5 kN/m². Each floor adds 180 kN and the base carries 1,440 kN — the same tributary area counted eight times. Nothing about the drawing changes down the height; only the number does, and it is the one quantity in a building that is never checked against anything.

Two things about that figure are worth stating plainly.

The first is that the column’s load is the only quantity in a building assembled by adding up a hundred small allocations with no closing check. A beam’s load can be wrong by 60% and the beam next to it will not notice; a column’s load is wrong by whatever all of its floors’ allocations were wrong by, in the same direction, because the same rule was used at every level. Errors in load paths do not cancel down a column; they compound.

The second is that this is exactly the arithmetic that makes a tall building a different object from a short one. The column at the bottom of eight storeys carries eight times the floor above it and is sized for that; at forty storeys it carries five times as much again, and the fraction of the floor plate it occupies grows with it. That is not the same phenomenon as span scaling — it has no fourth powers in it — but it is the vertical version of the same lesson, which is that a rule of thumb calibrated at one size is a statement about that size.

The reduction nobody can derive

There is a convention in every loading code that a column carrying many floors may be designed for less than the sum of them, on the argument that a hundred square metres of office is never simultaneously at its design load on every floor at once.

The reduction is real and it is statistical rather than structural. Nothing in equilibrium permits taking away load that is there; what is being said is that the design load was a value with a return period, and the probability of every floor reaching it in the same instant is much lower than the probability of one doing so. The factor is therefore a statement about the correlation between floors, and the codes express it as a function of the number of storeys or of the tributary area.

This site takes no position on the numbers, which are the business of a loading code and would date instantly. What is worth carrying from it is the shape of the argument, because it is unusual: it is the one place in the load path where an engineer subtracts something and the justification is not mechanical at all. Everything upstream of it — the 5 kN/m², the tributary division, the 45° lines — is also a judgement, and only this one is honest enough to have a coefficient with a probability behind it.

A square panel is the only case where the rule is nearly right

Where a beam's load comes fromA 6 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 9.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 36.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.9.0 m²7.50 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m6 m6 m36.0 m² divided, 36.0 m² of panel — the division closes
Fig. 7 A square panel divided at 45°: four triangles of 9.0 m² each, 45.0 kN to every beam, 7.50 kN/m all round. Here the tributary rule and the elastic share agree — both give each direction exactly half — because symmetry leaves nothing for the fourth power to act on.

The square case is where the rule is not an approximation at all, and it is worth seeing precisely because it shows what the rule is really claiming. The 45° lines are the locus of points equidistant from two supports, which coincides with the elastic division only when the two spans are equal. Every departure from square is a departure from that coincidence, and the figure two sections up measures how fast.

That is also why the rule is written the way it is: it is a symmetry argument dressed as a geometry one. It gets the total right by construction, gets the division right when symmetry supplies it, and degrades gracefully rather than suddenly, which is about as much as a rule drawn on a plan can be asked for.

What the picture cannot show

Continuity is absent from every drawing above. The panels are treated as isolated, and a real floor is continuous over its beams — so a load in one panel deflects the beam that supports it and sends some of its effect two panels away. Tributary areas cannot represent that, and their inability to is why the analysis of a continuous system is a separate subject.

Nothing here knows about torsion or twisting of the supporting beams. The trapezoid is drawn as though it lands on the beam’s axis, and where a slab is monolithic with the beam it also delivers a moment, which the beam resists in torsion and passes to the columns.

The panel’s own stiffness is nowhere in it. A very deep beam under a thin slab collects less than its tributary area because the slab arches over it; a shallow beam collects more. The tributary rule assumes rigid supports, and where the supports are not rigid relative to the slab the division is decided by their stiffness in exactly the way an indeterminate structure’s forces always are.

Where the ladder goes

The first direction is the one the fourth-power figure opened: what a slab spanning both ways really does, elastically and at collapse, which is its own essay and has a yield-line mechanism in it that no tributary division could predict.

The second is downward. The load descent above is a sum; the question of how a lateral load descends is not, because wind and earthquake divide themselves between the vertical elements by stiffness rather than by area, and the drawing that answers it is a different one entirely.

The third is the uncomfortable one. Everything above treated 5 kN/m² as given. Where that number comes from is a question about occupancy, statistics and convention rather than about structures, and it is the input to which every calculation downstream is exactly proportional — which makes it the largest uncertainty in the whole chain and the one this collection has least to say about.

What this makes readable

Essays that name this one as a prerequisite.

Named alongside this one

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

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Distributed loadFree bodyLoad descentLoad pathOne way spanningSelf weightTributary areaTwo way spanning