Structural Engineering & Statics
where the load goes
Concept

Continuous beam — where it appears

A beam running over its supports rather than stopping at each one, so that the spans share their bending with each other. Continuity cuts the mid-span moment and creates a hogging moment over each support, and it makes the beam redundant so that stiffness decides the split.

Named by 6 essays across 4 fields — each of them below, with the objects they name alongside it.

Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

sections · Shear moment interaction Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer.

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

deflection · Force method Settling down or walking away, cycle by cycle. The total plastic hinge rotation of the beam after each cycle of loading — span 1, both, span 2, neither — with the midspan load at 0.98, 1.02, 1.05, 1.10 times the shakedown load of 126.3 kN. At 0.98 it stops at 0.59 mrad. At 1.02 it grows 4.57 mrad a cycle. At 1.05 it grows 11.43 mrad a cycle. At 1.10 it grows 22.86 mrad a cycle. Nothing collapses in any single cycle; above the shakedown load the beam walks.

The load it can carry once

A two-span beam whose loads come and go span by span collapses at 150 kN under any one arrangement, and walks at 127. Between the two it can carry every arrangement once and none of them forever: each cycle leaves a few more milliradians of rotation at the support and a midspan fifteen millimetres lower. Melan's theorem finds the limit as the last residual moment line that fits, Koiter's as a mechanism no single load state can drive, and a cycle-by-cycle calculation walks exactly where both say it will.

materials · Shakedown Whole spans miss the shear inside a span. The shear along three spans of 8.0, 8.0, 8.0 m, under 20 kN/m of dead load everywhere and 15 kN/m imposed: the largest and least when the imposed load stands wherever it raises or lowers the shear at each section (solid), and when it may only cover whole spans (dashed). At the supports the two agree. At the middle of the second span the exact envelope reaches 23.8 kN either way and the whole spans 10.0 kN; at a tenth of the way along it, 122.4 against 122.0.

The shear whole spans cannot find

A continuous beam is designed by loading its spans in patterns, whole spans at a time, and for the bending moment that is very nearly exact: the moment's influence line keeps one sign over each span, so a span is either worth loading or not. The shear's influence line changes sign at the section itself, inside the span, and no pattern of whole spans can load one side of a section and not the other. At the middle of the inner span of three equal spans the imposed shear is 0.198 qL; whole spans find 0.083 qL, and the straight line to qL/8 that is safe on a simple span finds 0.125.

internal-forces · Influence line The hogging region cannot give its studs away. For a 457 × 191 × 67 steel beam under a 130 mm slab, continuous over two 12.0 m spans under 30 kN/m, 40 studs a span: the largest slip (solid, left scale) and the force in the slab's bars over the middle support (dashed, its value in the legend), against the density of studs beyond the point of contraflexure as a share of the density elsewhere, the span's total held at 40. At the same density everywhere the largest slip is 0.68 mm and the bars carry 344 kN; at half the density beyond contraflexure, 1.04 mm and 251 kN; at a quarter, 1.69 mm and 118 kN. The support's hogging moment falls from 448 to 398 kN·m as the bars stop working.

The slip that peaks where the moment turns

A simply supported composite beam slips most at its supports, and its studs are crowded there. Make the same beam continuous over a middle support and the slip peaks twice: at the end support, and again, just as hard and the other way, at the point of contraflexure — not over the middle support, where the shear is largest but the slip falls to nothing. Studs graded toward the supports, as a simple span's would be, miss the second peak and buy a tenth; graded to the slip they buy more than a third. And the hogging region cannot give its studs away to the span, because its edge is where the second peak is.

internal-forces · Composite action Over many earthquakes the column line ratchets together or stops together. The compression over the squash load of the interior column (solid) and an exterior column (dashed) against the cycles, up to 400, for the frame of three ground-storey columns under floor loads of 15, 20, 25 kN/m on 6 floors; the dotted line is where the hinge starts to ratchet, 0.15. At 15 kN/m the interior column starts at 0.22 and the average is 0.12; after 400 cycles the interior carries 0.15 and the exterior 0.10. At 20 kN/m the interior column starts at 0.29 and the average is 0.16; after 400 cycles the interior carries 0.16 and the exterior 0.16. At 25 kN/m the interior column starts at 0.36 and the average is 0.19; after 400 cycles the interior carries 0.19 and the exterior 0.19. Where the average lies above the threshold the three end at the average and ratchet together; where it lies below, the interior column sheds load until it stops at the threshold and the exterior columns never start.

The floor that cannot keep up with the column

A column whose hinges ratchet under an earthquake shortens every cycle, and in a frame it does not shorten alone: the floors it carries bend down towards it and lift its load onto the columns beside it. That ought to slow it and speed them, until every column shortens together. It does — over hundreds of cycles. Over the ten or twenty of an earthquake, the floors of a six-storey frame lift an eighth of the load off the interior column, it shortens nine tenths as far as it would alone, and the floors carry the whole difference as bending. The frame equalises its columns eventually, and pays for it in its floors.

materials · Moment-curvature

Named alongside it

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

Plastic hingeBending momentLoad arrangementMoment redistributionRatchetingShakedownBending stressCompatibilityComposite actionContraflexureCracked sectionDifferential settlement

All concepts