Equilibrium

The load that is spread out, and the force that replaces it

A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.

Almost no load is a force at a point. Snow lies over a roof, a floor carries people spread across it, a wall bears along a beam, water presses over the whole face of a dam. The point load that fills textbooks is a convenience, and the spread load is the real object.

There is a substitution that makes the spread version tractable: replace it with a single force equal to its total, acting at the centroid of the loaded region. The substitution is exact — and it is exact for one purpose and badly wrong for another, and the boundary between the two is worth being precise about.

A triangular load and the force that replaces itA triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 24.0at x = 5.33, the centroid of the areamomentspread: 24.6replaced: 42.7reactions agree exactly (8.00 and 8.00); the peak moment does not
Fig. 1 A triangular load with its resultant computed by integration: the area under the load, acting at the centroid of that area. Beneath, the moment diagram it produces against the moment the single force produces — identical reactions, and a peak that is not identical at all.

Where the resultant is, and why

Two integrals settle it, and both are statements about equilibrium rather than about geometry.

The size. For the replacement to have the same effect on the force sum, it must equal the total load, which is the area under the load diagram:

W=w(x)dx.W = \int w(x)\,dx.

The position. For it to have the same effect on the moment sum, its moment about any point must match the distributed load’s, which fixes

xˉ=xw(x)dxw(x)dx.\bar{x} = \frac{\int x\,w(x)\,dx}{\int w(x)\,dx}.

That is the definition of a centroid, and it has arrived here as a consequence of wanting two sums to agree rather than as a fact about shapes. The same expression with a mass density in place of a load intensity is a centre of gravity; with a probability density it is a mean. Three subjects, one integral, and the reason it turns up in all of them is that all three are replacing a distribution by a single representative value in a calculation that is linear in position.

For the standard shapes the answers are worth carrying. A uniform load of intensity ww over a length LL has a total wLwL at mid-length. A triangular load rising to ww has a total wL/2wL/2 at two-thirds of the way toward the high end. A trapezoid is the sum of the two, and the honest way to handle one is to split it rather than to look for a formula. A parabolic load — the shape a hydrostatic pressure on a curved face produces — has a total of two-thirds the enclosing rectangle at the middle by symmetry.

A uniform load and the force that replaces itA uniform distributed load with its resultant computed by integration: an area of 48.0 acting at 4.00 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 48.0at x = 4.00, the centroid of the areamomentspread: 48.0replaced: 96.0reactions agree exactly (24.00 and 24.00); the peak moment does not
Fig. 2 The uniform case, which is the one everything else is measured against. The resultant lands at mid-span by symmetry, and the two moment diagrams differ by a factor that is not small.

Exact outside the free body, wrong inside it

Now the important part, which is the one that is usually stated as a warning and rarely as a reason.

The substitution was constructed to preserve the two sums for the whole body. So any calculation that uses only those two sums for the whole body gives the identical answer: the reactions, the overturning check, the total load on a foundation. All exact, and the figure at the top shows the two reactions agreeing to every digit computed.

The moment at a station inside the beam is not that calculation. It is the sum on a free body cut at that station, and the piece to the left of the cut contains only part of the distributed load — whose own resultant is a different size in a different place from the whole load’s. Substituting first and cutting afterwards puts the entire load at one point, on whichever side of the cut that point happens to fall.

The consequence is large rather than subtle. For a uniformly loaded simple span the peak moment is wL2/8wL^2/8. Replace the load by its resultant at mid-span first, and the peak becomes WL/4=wL2/4WL/4 = wL^2/4: twice as large. A designer who made that substitution would size every beam for double the moment, which is at least conservative. Make it the other way — a real load concentrated in the middle, idealised as spread — and the beam is half the size it needs to be.

The rule is therefore a rule about order of operations: cut first, then substitute, on the piece that remains. Stated as a boundary rather than a warning, it is that a resultant is legitimate outside the free body being summed and illegitimate inside it — which is a fact about where the boundary was drawn, like most facts in this subject.

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.6 per unit lengthshear18.0moment27.0 at x = 3.00the moment peaks exactly where the shear passes through zero
Fig. 3 A load over part of a span. The shear slopes only where the load acts and the moment is parabolic only there, which is information the single-force substitution destroys entirely.

The shape of the diagram remembers the shape of the load

Because the shear is the integral of the load and the moment the integral of the shear, the load’s shape is visible two derivatives away.

Under a uniform load the shear is linear and the moment parabolic. Under a triangular load the shear is parabolic and the moment cubic. Under a point load the shear steps and the moment kinks. Every feature of the diagrams is a feature of the load, and reading backwards from a moment diagram to the load that produced it is a matter of differentiating twice by eye.

A parabolic load and the force that replaces itA parabolic distributed load with its resultant computed by integration: an area of 32.0 acting at 4.00 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 32.0at x = 4.00, the centroid of the areamomentspread: 40.0replaced: 64.0reactions agree exactly (16.00 and 16.00); the peak moment does not
Fig. 4 A parabolic load — the shape water pressure on a curved surface produces. The resultant is two-thirds of the enclosing rectangle, at mid-span by symmetry, and the moment diagram is a quartic that no standard case in any handbook contains.

This is where the substitution costs the most. The resultant is one number and a position; the diagram is a function. Collapsing the first destroys the second, and the second is what the beam is designed against.

There is a corresponding gain, which is why the technique survives. For finding reactions — the numbers that go to the columns, the foundations and the overturning check — nothing is lost, and a complicated load pattern collapses to a handful of forces that can be summed in a line. Structural work is full of load patterns that are awkward to integrate and trivial to split into triangles and rectangles, and the splitting is legitimate precisely because the two sums are linear.

Splitting, rather than integrating

The integrals above are the definition. Almost nobody evaluates them, and the reason is that both are linear, which means a complicated load can be chopped into simple pieces and the pieces added.

A trapezoidal load is a rectangle plus a triangle. Each piece has a total and a centroid that everyone knows, and the combined resultant is their sum acting at the weighted average of their positions:

xˉ=W1xˉ1+W2xˉ2W1+W2.\bar{x} = \frac{W_1\bar{x}_1 + W_2\bar{x}_2}{W_1 + W_2}.

That is the same integral evaluated by parts, and it turns a calculus problem into two multiplications and a division. A load that steps, one that runs over part of a span, one that reverses sign partway along — all of them decompose, and the pieces can even overlap or be negative, since subtracting a triangle from a rectangle is as legitimate as adding one.

A trapezoidal load and the force that replaces itA trapezoidal distributed load with its resultant computed by integration: an area of 33.6 acting at 4.57 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 33.6at x = 4.57, the centroid of the areamomentspread: 33.8replaced: 65.7reactions agree exactly (14.40 and 14.40); the peak moment does not
Fig. 5 A trapezoidal load, which is a rectangle and a triangle added. The resultant sits between the two component centroids, nearer the heavier end, and the moment diagram it produces is a blend of the two shapes rather than either of them.

The habit is worth naming because it is the same one that makes second moments of area computable: decompose into shapes with known properties, transfer each to a common reference, and add. Both rest on the quantity being an integral of something linear, and both fail at exactly the same point — where the thing being integrated stops being additive.

There is also a check hiding in the method, and it costs nothing. A resultant computed by splitting can be recomputed by splitting differently: a trapezoid is a rectangle plus a triangle, and it is equally a large triangle minus a small one. The two routes use different components, different centroids and different arithmetic, and they must produce the same total at the same position. A disagreement is an error in one of them, located immediately by which component the two routes disagree about.

The load that is not one-dimensional

Everything so far treats the load as spread along a line. Most loads are spread over an area, and getting from one to the other is a step that gets skipped so routinely it stops being visible.

A floor slab carrying 55 kilonewtons per square metre delivers to the beams beneath it, and the intensity along any one beam depends on how the slab spans. A one-way slab delivers half its load to each of two supporting beams, giving a uniform line load of half the area intensity times the slab span. A two-way slab delivers in a pattern that is triangular or trapezoidal depending on the panel’s proportions, because each region of slab goes to its nearest support. That is why the classic hand method draws yield lines at forty-five degrees from the corners of a panel and assigns each region to the beam it faces.

The step is an assumption about load path, and it is one of the largest assumptions in an ordinary building calculation. A slab assumed to span one way when it in fact spans two delivers to the wrong beams, and the error appears nowhere in any subsequent arithmetic — the load reaches the ground by whichever route the structure chooses, not by the one the calculation described.

A beam, its loads and its reactionsA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.6 per unit length24.024.0ΣM about one support gives the other reaction; ΣF then gives the first
Fig. 6 A beam under a uniform line load with its reactions computed. That line load is itself the product of a decision about how the slab above it spans, and the beam calculation contains no trace of the decision.

What the point load costs at the other end

The idealisation runs in both directions, and the reverse one is where the failures live.

A “point” load is a load spread over a small area, and the smallness is what makes the idealisation safe for the overall diagrams and dangerous locally. A column bearing on a beam delivers its load over the width of a base plate; a wheel delivers over a contact patch; a beam bearing on a wall delivers over the bearing length.

The overall moment diagram barely notices the difference — the moment under a load spread over a hundredth of the span differs from the point-load value by a fraction of a per cent. What differs enormously is the state of the material immediately underneath, where the actual pressure is the load divided by a real area rather than by zero. Web crippling, bearing failure, punching shear through a slab and splitting of a masonry wall are all failures of the region where the idealisation was made, and none of them appears in the diagram that the idealisation produced.

Saint-Venant’s principle is the formal statement of the trade: the stress resultants describe the state of stress accurately at a distance of about one section depth from the load, and inaccurately closer. So the beam design uses the point load and the bearing design uses the real contact area, and the two calculations sit side by side in the same set of drawings describing the same load in two different ways because they are answering questions at two different distances.

The distributed load that is the beam

One spread load is unavoidable and is different in kind from the rest: the member’s own weight.

It is distributed by construction — every element of the beam weighs something and there is no arrangement in which it arrives at a point. It is also the only load in the calculation whose magnitude is an output. A designer choosing a section has chosen a self-weight, which changes the moment, which may change the section, which changes the self-weight again.

For ordinary spans that loop converges after one pass and is usually not even noticed: a steel floor beam’s own weight is a few per cent of what it carries, so an initial guess is close enough that the second iteration confirms the first. At long spans it stops being a correction and becomes the problem, and the point at which each structural form runs out is the point at which the loop stops converging.

The practical handling is a good illustration of what a load model is for. Self-weight is estimated at the start from a rate per square metre that the profession keeps in its head — a steel-framed office floor at something like three kilonewtons per square metre, a flat concrete slab at rather more — the structure is designed against it, and the assumed value is checked against the designed one at the end. The estimate is not a calculation and it is not a guess either; it is a summary of what buildings of that kind have weighed, used as an input to the calculation that will produce the next one.

Where the model stops

Static loads. Everything here treats the load as a fixed distribution. A load that moves has a worst position that must be searched for rather than assumed, and a load that arrives suddenly has a dynamic factor on top.

Loads that do not depend on the structure. A distributed load is treated as an input, and some are not: soil pressure depends on how far the wall moves, water in a ponding roof depends on how far the roof sags, and wind on a flexible structure depends on how the structure responds. Each of those is a load that changes when the structure does, and no diagram drawn in advance contains it.

Rigid load paths. The assumption that a slab delivers to the beams beneath it in a stated pattern is an assumption about relative stiffness. A stiff beam attracts more of the slab’s load than a flexible one, and for a floor that is genuinely redundant the split is a stiffness question rather than a geometric one.

One dimension. A line load on a beam is a load on a member idealised as a line. A slab, a plate or a shell carries in two directions at once and has no equivalent simplification.

The figures on this page have a specific dishonesty worth naming. The load profile is drawn as a solid shape with a definite edge, which makes it look like an object sitting on the beam. It is a graph — the vertical axis is intensity, not material — and the area under it is a force rather than a volume. Reading it as a picture of snow lying on a roof is a helpful intuition and it will mislead the moment the load is anything other than gravity, because a wind suction plotted the same way points the other direction and is not a picture of anything at all.

The ladder from here

Later rungs on this anchor: centroids of composite areas. Trapezoidal and hydrostatic distributions. Load paths from slabs to beams, one-way and two-way. Yield-line load distribution. Patch loads and bearing. Equivalent uniformly distributed loads and where the equivalence holds. Line loads on plates. Soil pressure distributions under a footing, which depend on the footing’s stiffness. And the general question of what a load model is for, which is the question that decides how much of this matters.

Simon Stevin worked out the resultant of a hydrostatic pressure distribution in 1586, and in doing so produced the first correct statement of the hydrostatic paradox — that the force on the base of a vessel depends on its depth and its base area and not at all on how much water it contains. The integral in this essay is the same one, and the paradox is a good reminder that a resultant is a summary of a distribution rather than a description of it.