Structural Engineering & Statics
where the load goes
Internal forces

Where to put the supports, which is not at the ends

Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam.
10 min read 7 figures The load must go somewhereGeometry beats material

A beam with its supports at the two ends is the standard picture, and for a uniformly loaded beam it is close to the worst arrangement available.

Move the supports inward and two things happen at once. The span between them shortens, which reduces the sagging moment in the middle; and the overhanging ends begin to hog, which pushes a moment back over the supports. Somewhere between the two extremes the beam is working as evenly as it can, and the peak moment there is less than half what it was.

Where to put the supportsPeak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.00.050.10.150.20.250.30.3501020304050overhang, as a fraction of the spanbest at 21% — peak 8.8sagging, mid-spanhogging, over the supportmoving the supports in by a third of the way halves the worst moment
Fig. 1 Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.

The two competing moments

With the supports at the ends, the sagging moment at mid-span is wL2/8wL^2/8 and there is no hogging anywhere. That is the entire budget spent in one place.

Move each support in by a distance aa. The sagging moment falls, because the span between supports is now L−2aL - 2a and sagging goes with the square of the clear span. The hogging moment over each support rises, because the overhang is a cantilever of length aa carrying wa2/2wa^2/2.

One curve falls, the other rises, and the worst moment in the beam is whichever is larger. The best arrangement is therefore where the two are equal — the minimum of a maximum, which is the standard shape of an optimisation with two competing failure modes.

Setting the two expressions equal gives a≈0.207La \approx 0.207L, and the peak moment there is about wL2/46wL^2/46 against wL2/8wL^2/8 at the ends. A factor of 5.8 in bending moment, bought with nothing but the position of two supports.

That number, 0.2071L0.2071L, is (1−1/2)L/2\left(1 - 1/\sqrt2\right)L/2, and it turns up wherever the same question is asked. A ladder carried on two shoulders, a pipe on two trestles, the two lifting points on a precast beam — all the same calculation.

What the diagram does over a support

The moment changes sign at the support region, and where it passes through zero is the point of contraflexure.

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. 2 A beam loaded over part of its length. Where the shear crosses zero the moment peaks, and where the moment crosses zero the curvature reverses — two different stations, doing two different jobs.

Contraflexure matters more in construction than in analysis. It is where the tension moves from the bottom of the beam to the top, so reinforcement has to be arranged to be continuous past it. It is also, for exactly that reason, the most convenient place to put a joint: a splice at the point of contraflexure is a splice where the moment is zero, so only shear has to be transferred.

That trick built the Gerber girder — a continuous beam with real hinges inserted at the contraflexure points, which converts an indeterminate structure into a determinate one with almost all of continuity’s benefit. Nineteenth-century railway bridges used it constantly, because a determinate structure is insensitive to the settlement of its piers and a continuous one is not.

Continuity does the same thing differently

Moving supports inboard is one way of putting hogging into a beam. Making it continuous over its supports is another, and the effect on the moment diagram is very similar.

One support too manyThe same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment
Fig. 3 The same uniformly loaded beam with three sets of restraints. Adding restraint moves moment from mid-span to the supports and lowers the peak, exactly as an overhang does.

A beam built in at both ends has a mid-span moment of wL2/24wL^2/24 and support moments of wL2/12wL^2/12 — a peak a third of the simply supported value. The mechanism is the same as the overhang’s: hogging at the ends relieves sagging in the middle.

The difference is what it costs to know. The overhang case is solvable by statics, because the reactions follow from the two equations. The continuous case is not: the end moments depend on the stiffness of the beam and of whatever it frames into, and getting them requires a compatibility calculation.

So the two routes to the same benefit differ in what they demand. An overhang is free to analyse and awkward to build. Continuity is easy to build and needs an analysis method that did not exist until the 1930s.

The moving load, and why the optimum shifts

The 0.207 figure assumes the load is uniform and permanent. Almost nothing is.

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.1515shear15.0moment37.5 at x = 4.32the moment peaks exactly where the shear passes through zero
Fig. 4 Two point loads on a simple span. A different loading produces a different moment diagram, and an arrangement optimised for a uniform load is not optimal for this one.

A bridge carries a uniform dead load and a moving live load, and the worst case for the sagging moment is one live-load position while the worst case for hogging is another. The design has to satisfy both, so the beam is sized for an envelope of moment diagrams rather than for any single one.

Finding the worst position of a moving load is what influence lines do, and the answer is often surprising: for a continuous beam the worst hogging over a support comes from loading the two adjacent spans and leaving the next ones empty, which is a pattern nobody guesses.

The practical consequence is that the elegant 0.207 optimum applies to a beam carrying only its own weight — a pipe, a precast unit being lifted, a lighting truss. Anything carrying variable load ends up somewhere else, chosen by whichever case governs.

The same argument in a cantilevered structure

Once the idea is visible it turns up at every scale.

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.1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first
Fig. 5 A beam with its loads and reactions. Where the supports sit is a design decision, and it changes the reactions as well as the moments — which is often what the decision is actually about.

A cantilevered canopy is one half of the overhang problem. A slab spanning between beams with an edge that oversails is the same thing in a floor. The Forth Bridge is the argument taken to its conclusion — three balanced cantilevers with suspended spans between them, which is a Gerber girder at a scale of five hundred metres.

Moving a support is also often the cheapest available fix on a real project. When a beam is found to be overstressed, adding a support or moving one is nearly always less expensive than making the beam bigger, because the moment falls with the square of the clear span.

The same question inside the section

Moving a support and moving material in a cross-section are the same optimisation at two scales.

The same material, four waysFour cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.the same, laid flatI = 0.06 × 10⁶1.0× the firstsquareI = 0.75 × 10⁶13.3× the firsttall rectangleI = 10.00 × 10⁶177.8× the firstI-sectionI = 24.29 × 10⁶431.8× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale
Fig. 6 Four sections of identical area. Rearranging material within a section is the small-scale version of rearranging supports along a beam — and both are free, in the sense that no more material is used.

Getting material away from the neutral axis raises the capacity at constant weight; getting the supports away from the ends lowers the demand at constant everything. Structural efficiency is very largely those two moves, applied until something non-structural objects.

The deflected shape is the moment, integrated twiceA loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.the largest movement, at x = 4.00momentdrawn at roughly three hundred times the real deflection —a beam at its serviceability limit moves about a three-hundredth of its span
Fig. 7 The deflected shape of a beam, integrated from its moment diagram. Overhangs reduce this as well as the moment, because the hogging at the supports curves the ends back.

The deflection benefit is the larger one in practice. Deflection goes as the fourth power of the clear span, so shortening it by a fifth cuts the movement by more than half — which for a member governed by stiffness is worth more than the moment reduction.

Where the model stops

Uniform load only. The 0.207 optimum is specific to a load uniform along the whole beam, including the overhangs. Load the middle only and the answer moves toward the ends.

Bending only. The optimisation minimises the peak moment and ignores everything else, including which limit actually governs. Deflection, shear at the supports, and the uplift that an overhang can produce at the far support are all unaddressed, and any of them can govern.

Uplift. If the overhang is long enough and loaded while the middle is not, the far support goes into tension — a reaction the support idealisation may not permit. A support that can only push then fails to hold, and the beam tips — which is how a plank on two bricks behaves when somebody stands on the end.

Rigid supports at known positions. Real supports settle, and for a redundant arrangement that redistributes everything — which is one reason a determinate layout is sometimes chosen deliberately.

The figure at the top has a limitation worth naming: it plots the peak moments and not where they occur. Two arrangements with the same peak can put it in very different places, and a designer usually cares about both. The curve says how bad the worst point is, and says nothing about which point it is.

The ladder from here

Later rungs: the optimum overhang derived properly. Points of contraflexure and where to splice. The Gerber girder, and determinacy by inserted hinge. Influence lines. Pattern loading on continuous beams. Moment envelopes. Cantilever construction and balanced cantilevers. Uplift, and the load cases that produce it. And the same optimisation applied to a slab spanning two ways, where the answer stops being a single number.

The 0.2071 ratio is the same number that appears in the design of a pipe rack, a launch cradle and a violin bass bar. It is not a structural constant so much as an arithmetic one, and it turns up wherever a uniformly loaded line is held at two points.

Where this goes next

4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero The diagram is an integral, and that is why it can be drawn by eye Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch. simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment One support too many, and what it costs to know Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question. 2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs What a cut reveals, and why it was there all along Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.