Structural Engineering & Statics
where the load goes
Concept

Point of contraflexure — where it appears

A station where the bending moment passes through zero, so the member changes from sagging to hogging and its curvature reverses. Its position is where a continuous beam's reinforcement changes face, and it moves with the load arrangement — so a bar curtailed at a point of contraflexure computed for one case is short for another.

Named by 7 essays across 4 fields — each of them below, with the objects they name alongside it.

Where to put the supports. 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.

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.

internal-forces · Support layout 3 continuous spans against 3 simple ones. The 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 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none.

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.

internal-forces · Continuity The area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00.

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

deflection · Moment-area A hole in a web is a Vierendeel panel. A 400 × 300 rectangular opening in a 533 deep beam, 15% along a 9 m span carrying 20 per metre — where the moment is 103 kNm and the shear 63 kN. The moment is a couple on the two tees, 301 kN on a lever arm of 343 mm, which is 70.3 N/mm² of uniform stress. The shear has nowhere to go but through the tees, so each carries 32 kN over the opening and bends in double curvature: a Vierendeel moment of 6.3 kNm and 167.6 N/mm² on top. So 70% of the stress at the corner exists because the hole has a LENGTH, and only 30% of the section's second moment has gone.

The hole that costs nothing, and everything

A service opening removes 30% of a beam's second moment and 0.7% of its deflection. What it costs is not that. Across the opening the shear has nowhere to go but through the two tees, and a tee carrying shear over a length bends.

sections · Web opening The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

structures · Vierendeel The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure.

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

internal-forces · Portal method A unit load carried by the prop taken away. A beam fixed at its left end and propped at its right, under 4 kN/m over 8 m, asked how far it moves at 4 m. The real moment is the propped cantilever's own, with -32.0 kNm at the wall. The unit load is carried by the prop taken away, whose moment diagram peaks at 4.00. Their product has 98.92 of area on one side and -13.33 on the other, and the net, divided by EI, is 85.33 — the propped cantilever's closed-form deflection, 85.33, which no part of this calculation was given.

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

deflection · Virtual work

Named alongside it

The objects these essays reach for when they reach for this one.

StiffnessVirtual workCompatibilityContinuityDeflectionEquilibriumHogging and saggingLoad pathMechanismMoment diagramMoment envelopePlastic hinge

All concepts