Twice the deflection, for the same load

A steel platform hangs from four hangers. A tank of water sits on it, and the tank weighs 20 kN. The platform deflects 12.67 mm, which is the load divided by the stiffness, and the calculation takes a line.

Now empty the tank, lift it a hair’s breadth clear of the platform, and let go.

Nothing about the load changed. It is the same 20 kN, and after ten seconds the platform is sitting at 12.67 mm exactly as the static calculation said it would. What changed is that for the first two seconds it was somewhere else, and the somewhere else was twice as far.

The factor of two, from energy alone #

The result is worth deriving, because the derivation contains no property of the structure and that is the surprising part.

Draw the free body of the tank at the instant it has descended a distance u. Two forces act on it: its own weight P downwards, and the platform pushing up with ku, because the platform is a spring and it has been squashed by u. Nothing else is in the picture.

The weight has done work Pu over that descent. The spring has stored $\tfrac{1}{2}ku^2$. At the lowest point the tank is momentarily at rest, so all of the work done has gone into the spring:

$$Pu = \tfrac{1}{2}ku^2 \quad\Longrightarrow\quad u = \frac{2P}{k}$$

and $P/k$ is the static deflection. So the peak is exactly twice it.

No stiffness survives into the answer, and no mass. The k cancels. A cathedral and a shelf bracket both do this, and both do it by the same factor, which is why the number is worth carrying: two is not a coefficient somebody measured, it is arithmetic.

The reason the static calculation misses it is visible in the same two lines. Statics asks where the forces balance, and they balance at $u = P/k$. But the tank does not stop where the forces balance — it accelerates until it gets there, and arrives with a velocity that carries it on past. The equilibrium position is the middle of the motion rather than the end of it, and the extra half is what the energy statement counts.

Half the work goes somewhere #

There is a second reading of the same algebra that is more useful than the first.

At the static position, the weight has done work $P \cdot (P/k)$ and the spring has stored $\tfrac{1}{2}k(P/k)^2$, which is half as much. Half the work done by a suddenly applied load is not in the structure, and it has to be somewhere: it is kinetic energy, and it is what carries the motion past equilibrium. When a load is lowered slowly, the hands lowering it absorb that half, and nothing about the structure ever knows.

That is the whole distinction between a load applied slowly and a load applied suddenly, stated without any reference to time at all. The question is not how fast the load arrived. It is who did the other half of the work.

How slowly does it have to arrive? #

Twice is the answer for a load that appears instantaneously. Nothing appears instantaneously, so the practical question is how gradual an arrival has to be before the factor goes away.

Between those two figures the dynamic effect vanishes entirely, and the parameter that decides it is not the rise time. It is the rise time divided by the natural period, which is the ratio this whole field is built on: neither quantity means anything by itself.

The ramp curve reaching exactly 1 at a rise time of one period is not an approximation and it is not a coincidence. The load stops changing at precisely the moment the structure comes back through its starting point, so the free vibration that the ramp set up is cancelled by the free vibration that the end of the ramp sets up. It is the same interference that lets a crane driver stop a swinging load in one movement by starting the trolley again at the right instant.

The pulse, which is the other way round #

A load that arrives suddenly and then leaves again does something the ramp cannot.

The pulse branch of that curve has two regimes with a kink between them. Below a duration of half a period the factor is 2·sin(πt/T) and falls away steeply; above it the factor is 2 and the duration has stopped mattering, because the peak has already happened while the load was still on. Everything interesting is on the falling side, and the falling side goes a long way down.

Compare that with the 7.83 mm the same force produced at a tenth of a period, and with the 24.56 mm it produced when it stayed. One load amplitude, three durations, and a spread of nearly forty to one.

This is the shape that makes blast design different from everything else in structural engineering. A blast wave from a charge at a stand-off distance lasts a few milliseconds; a building frame’s period is a fraction of a second. The ratio is a hundredth or less, the factor is a hundredth or less of two, and the peak pressure of the blast is almost irrelevant to the answer. What matters is the impulse — the area under the pressure-time curve — because in that regime the structure’s response depends on the momentum delivered rather than on the force that delivered it.

Which is a general result about limits worth stating plainly: at one end of the duration axis the answer is a statics problem with a factor of 2, at the other end it is a momentum problem with no force in it at all, and the interesting behaviour is the region in between which is neither.

The two ends of the axis are the same problem #

The blast and the dropped weight look like different subjects and they are one initial-value problem asked twice.

A pulse too short to respond to delivers an impulse $\int F\,dt$, and an impulse delivered to a mass is a velocity: $v = \int F\,dt / m$. The structure is then a mass moving at $v$ with nothing pushing it, and where it gets to is a matter of trading kinetic energy for strain energy.

A weight dropped from a height h arrives at the structure at $v = \sqrt{2gh}$, and the structure is then a mass moving at $v$ with the weight still on it.

Both are “a mass, given a velocity, meeting a spring”, and both are solved by writing down the energy at the start and the energy at the peak. The force histories that produced the velocity have vanished from the problem entirely, which is exactly why the peak pressure of a short blast does not appear in the answer: the structure never knew about it, only about its integral.

That is worth holding on to, because the general form covers the cases where nothing is dropped and nothing explodes. A wheel crossing a rail joint delivers an impulse to a bridge; a lorry striking a bollard delivers one to its foundation; a pile hammer delivers one on purpose, forty times a minute. In every one of them, asking “what force did it apply” is asking a question whose answer nobody can measure and nobody needs.

Willis, Stokes, and the first argument about this #

The question is old, and the first serious answer came out of a public disaster.

After the Dee Bridge collapsed in 1847, a Royal Commission was appointed in Britain to establish whether iron was fit for railway bridges at all, and a central question put to it was whether a load crossing a bridge was worse than the same load standing on it. Robert Willis built an apparatus at Cambridge that ran weights across a model span at controlled speeds and measured the deflection; George Stokes then analysed the problem, treating the moving weight as a mass whose own inertia is part of the system it is crossing.

The finding they arrived at survives in every bridge code as an impact allowance, and the shape of it is the shape of the curves above: the effect depends on how the crossing time compares with the span’s own period, so a fast train on a short span is a dynamic problem and a slow one on a long span is not.

Two things about that episode are worth carrying. The first is that the answer was obtained by measurement and analysis together, with the analysis unable to proceed until somebody had built the thing and watched it. The second is that it took a collapse to get the question asked — which is a pattern this field repeats, in a footbridge that swayed and a bridge deck that tore itself apart, and the reason the collapses are worth studying is that in each case the mechanism was known and the question had not been asked of that structure.

Where the free body is, and what it assumes #

Every number above came from one free body: the mass, with the spring force on one side and the applied load on the other, and the sum of the two equal to the mass times its own acceleration. That is d’Alembert’s contribution and it is the reason this field belongs on a site about statics — the sums still cancel, provided the inertia force is drawn on the diagram like any other.

Three assumptions are in that diagram and none is innocent.

The structure is one mass on one spring. A real platform has mass everywhere and stiffness everywhere, and reducing it to a single number for each is a modal approximation. It is a good one when a single mode dominates, which the next essay is about; it is a bad one for a floor being struck locally, where the answer is dominated by the mass the impact can actually reach in the time available.

The spring is linear and stays linear. The whole factor-of-two derivation assumed the strain energy is $\tfrac{1}{2}ku^2$. A structure that yields on the way down stores energy differently, and the factor is then smaller — which is the argument that makes ductile design work under an earthquake, arriving here for the first time.

The load does not depend on the motion. The 20 kN is 20 kN whatever the platform does. That is true of a tank of water and false of a great many other loads: a crowd, a wind on a bluff section, a machine on a soft mount. When it fails, the response and the excitation are coupled, and the answer is not bounded at all.

Twice the increment, not twice the total #

Every figure here starts with a structure at rest and nothing on it. Real structures already have something on them, and the difference decides how the factor is applied.

Take the platform already carrying its own weight and its permanent fittings, sitting at some deflection $\delta_g$. Now drop the tank on it. The structure does not start from zero; it starts from $\delta_g$, and it oscillates about the new equilibrium at $\delta_g + \delta_q$ with an amplitude equal to $\delta_q$ — because the amplitude is set by how far the equilibrium moved, and it moved by the new load’s static deflection alone.

So the peak is

$$\delta_{max} = \delta_g + 2\delta_q$$

and not $2(\delta_g + \delta_q)$. If the platform’s own weight also deflected it 12.67 mm, the peak with the tank dropped on it is $12.67 + 24.56 = 37.2$ mm, against the 50.7 that doubling everything would give. Doubling the total overstates the answer by twice the dead-load deflection, and on a heavy structure carrying a light moving load that is most of the error.

Which is exactly what a code’s impact factor says, in a form that hides the reasoning. A 25 per cent allowance for a lift machine, or 100 per cent for a hoist, is applied to the moving load and not to the load case it is added to — and the reason is not conservatism or convention but the two lines above. The permanent load was already there; the structure is already where that load put it; nothing about it is being suddenly applied to anything.

The general statement is worth carrying, because it is the same shape as several other results in this collection. The amplification acts on the change, measured from wherever the structure already is. A quantity defined as a difference between two states is blind to anything that shifts both — which is why camber cancels out of an incremental deflection and why a dead load cancels out of a dynamic increment. Two unrelated mechanisms, one algebraic property, and the same mistake available in both: applying an operation to the total when the limit is on the difference.

What the picture cannot show #

The figure above draws displacement, because displacement is what a person watching the platform would see. Two things that matter are not in it.

The force in the hangers is not the applied load. At the peak the spring is stretched by 24.56 mm, so the hangers carry $k$ times that — 38.8 kN, on a structure holding a 20 kN tank. Anything sized on the applied load is undersized by the amplification factor, and it is the connections rather than the members that usually find this out, because a connection is where the force is largest and the redundancy is least.

The time axis is a fiction after the first second. Damping was set at 2% of critical here, which is a plausible number for a bolted steel structure, and it was chosen rather than derived. Nothing in a drawing predicts damping; it is measured on the finished thing by hitting it, and every quantity in this field that depends on how long the motion lasts inherits that uncertainty. The peak, at half a period, is reached before damping has had time to matter — which is why the factor of two is trustworthy and the decay after it is not.

The dropped weight, which is the same argument with a height in it #

The tank was let go from a hair’s breadth. Let it go from a height instead.

The square root is the useful part. Dropping a weight from twenty times the static deflection multiplies the answer by 7.4 rather than by twenty, because the structure’s own give is part of the distance fallen. The consequence is that a flexible structure is a better catcher than a stiff one — the softer it is, the larger $\delta$ is, the smaller the ratio $h/\delta$ becomes, and the lower the factor. A crash barrier, a fall-arrest lanyard and a packaging insert are all the same calculation, and all three work by making $\delta$ large on purpose.

This is the point at which the field stops agreeing with the rest of the site. Everywhere else, stiffness is free or nearly so: a deeper beam is better in every way that matters. Under a suddenly applied load it is not, and the reason is that a stiff structure has to absorb the same energy over a shorter distance, which means a larger force.

Where a single factor stops working #

Every number here has been quoted as a multiple of a static deflection, which presumes there is one. The dynamic problem does not replace the static one; it multiplies it, and everything about how the static answer is obtained belongs to the deflection field rather than to this one.

That framing is how the amplification factor is used in practice. Codes and handbooks give impact factors — 25% for a lift machine room, 100% for a hoist, and so on — and they are applied by multiplying a static load case. It is legitimate exactly when the amplification is a single number, and it is a single number exactly when one mode dominates and the load has one duration.

That is the case the factor cannot reach, and it is worth being exact about why. A load applied to a structure has a magnitude that exists whether the structure moves or not, so dividing the dynamic answer by the static one produces a number. A ground motion applies no load; it applies an acceleration, and the force on the structure is the structure’s own mass times a motion the structure is partly deciding. Set the stiffness to infinity and an applied load gives zero deflection while a ground motion gives zero deflection and full acceleration, which is a different limit entirely.

The practical consequence is that an earthquake is described by a curve rather than by a factor — one answer for every possible period, because the structure’s own period decides which part of the shaking it is exposed to. Which is why that problem is not solved with a factor, and why the next thing this field needs is the period itself.

Where the ladder goes #

The whole of this essay turned on one quantity: the structure’s natural period. It decided whether a quarter of a second was fast or slow, whether a blast was an impulse or a force, and where on every curve above the answer was read off.

Nobody chose it. It is a consequence of a mass that was chosen for architectural reasons and a stiffness that was chosen for deflection reasons, and it can be read off a calculation that has already been done for other purposes — which is the next rung, and the reason this field costs less to enter than it looks.