The weight that was dropped

Set a 500 kg load gently on a beam and the force in the beam is 4.9 kN. Let the same load go from a millimetre above the beam and the force is 9.8 kN, which is the factor of two from the first essay in this field.

Drop it from a metre, onto a beam that deflects 2 mm under it, and the force is 160 kN.

The derivation, in three lines #

Draw the free body of the falling weight at its lowest point, where it is momentarily at rest.

It has fallen a height $h$ plus the distance the structure gave, $\delta_d$. Over the whole of that distance its own weight $W$ did work, so the work done is $W(h + \delta_d)$. All of it has gone into the structure as strain energy, which for a linear spring is $\tfrac{1}{2}k\delta_d^2$. And $k$ is $W/\delta_s$ with $\delta_s$ the static deflection.

$$W(h + \delta_d) = \tfrac{1}{2}\frac{W}{\delta_s}\delta_d^2$$

which is a quadratic in $\delta_d$, with the positive root

$$\frac{\delta_d}{\delta_s} = 1 + \sqrt{1 + \frac{2h}{\delta_s}}$$

The whole of impact loading is that expression, and everything interesting about it is in the two places $\delta_s$ appears.

Why the square root, and why it matters #

The height enters under a root, so the factor grows slowly with it. From the table for a beam deflecting 2 mm under the load:

Ten times the height gives three times the factor, because energy goes as the height and force goes as the square root of energy.

That is worth holding on to for the case where it runs the other way. Halving a drop height buys very little. A safety procedure that reduces a lifting height from a metre to half a metre reduces the force by 30%, and one that catches the load 20 mm lower does almost nothing. What changes an impact force by a large factor is not the height but the give.

The give is the design variable #

Take one drop and three structures:

Same 500 kg, same one metre, same energy delivered — and a factor of eight in the force, produced entirely by how far the structure moves while absorbing it.

The mechanism is simple once seen. The energy that has to be absorbed is fixed; the structure absorbs it as force times distance; so a structure that gives little must generate a large force, and one that gives a lot need not. Stiffness is a liability under impact, which contradicts almost every other argument about structures — depth is the cheapest strength, stiffer sections deflect less, stiffer frames sway less — and the contradiction is real rather than apparent.

This is why every device designed to survive an impact is deliberately soft: a crash barrier that deforms, a fall-arrest lanyard with a tearing energy absorber sewn into it, a packaging insert, a lift buffer, a helmet’s liner. All of them work by increasing $\delta$, and all of them are, from a strength point of view, badly designed.

A worked case: the fall-arrest lanyard #

The clearest illustration of the give being the design variable is a device sold on exactly that basis.

A person of 100 kg falling 2 m on a rope that stretches 20 mm under their weight arrives at 6.3 m/s and, by the formula, produces a factor of about 15 — 15 kN on the harness, the anchor and the person. That is enough to injure the person severely and to pull most anchors out.

The same fall on a lanyard with a tearing energy absorber sewn into it, which deploys at about 6 kN and pays out up to a metre, produces 6 kN. The device works by increasing the distance over which the energy is absorbed from 20 mm to a metre — a factor of fifty in distance and therefore a factor near that in force — and the arithmetic is exactly the arithmetic above, run backwards to specify a deflection.

Two things follow from that example that generalise well beyond it.

The absorber is specified by force rather than by strength. The tearing strip is designed to yield at a chosen load and keep yielding, which makes it a constant-force device rather than a spring. That is the ideal energy absorber, because it dissipates the maximum energy for a given peak force — a rectangle rather than a triangle under the force–distance curve.

Everything in the chain matters. The anchor point’s own flexibility adds to the give and reduces the force, so an anchor on a flexible beam is kinder than one on a concrete wall — which is the series-springs argument with the softest element being the useful one for once.

The better form of the result: force from energy and stiffness #

The impact factor is a convenient way to present the answer and a poor way to think about it, because it hides which quantities the force actually depends on. Rearranging gets rid of it.

For a drop large compared with the static deflection — which is every case in the table above except the first two — the root dominates and

$$\delta_d \approx \sqrt{2h\,\delta_s}, \qquad F = k\,\delta_d \approx \sqrt{2\,W h\,k}$$

The peak force is the square root of twice the energy delivered, times the stiffness. Nothing else is in it. Checking against the numbers above: $\sqrt{2 \times 4.9 \times 1000 \times 2.45} = 155$ kN against the exact 160, and $\sqrt{2 \times 4.9 \times 1000 \times 0.245} = 49$ kN against 54 — the small differences being the terms the approximation dropped.

Written that way, three things become obvious that the factor conceals.

The force goes as the square root of the stiffness. Not inversely, and not as the stiffness — as its root. So softening a catcher by a factor of a hundred reduces the force by ten, which is why energy-absorbing devices are made very much softer than the things they protect rather than somewhat softer.

The weight and the height enter only through their product. A 500 kg mass from 1 m and a 50 kg mass from 10 m produce the same peak force on the same structure, because the same energy arrives. That is a genuinely useful statement for a hazard assessment: what has to be established is an energy, not a mass and not a height.

And the impact factor is not a property of the load. It is $\sqrt{2hk/W}$, which contains the structure. Two identical drops onto two different beams have two different factors, which is why a percentage allowance cannot express one and why the tabulated allowances all correspond to the zero-height case.

The last of those has a consequence worth stating on its own, because it inverts the obvious answer. Ask where on a beam it is worst to drop something. The deflection is largest at mid-span, so the impact factor is smallest there. The force is $\sqrt{2Whk}$ and the local stiffness is largest near a support — so dropping a load near a support produces the larger force, despite the smaller movement and the smaller factor.

Which is the whole essay in one line. A structure that gives generates a small force and moves a long way; a structure that does not generates a large force and barely moves. The impact factor measures the second half of that sentence, and the force measures the first, and it is the force that breaks things.

There is a second inversion in the same expression and it belongs to the material rather than to the geometry. Since $F \propto \sqrt{k}$ and $k$ contains the modulus, a stiffer material is worse under impact by the square root of its modulus, at equal geometry. Steel against timber is a factor of about four in modulus and therefore two in impact force, and steel against rubber is orders of magnitude. That is the reason a buffer, a fender, a bearing pad and a packaging insert are made of what they are made of — not because those materials are weak, but because a low modulus is the only property that reduces the force without reducing the energy that has to be taken.

It also explains a detail that looks like a finish. A timber running board on a steel loading dock, a rubber strip on a crane rail, a sand blinding under a slab: each is a soft layer inserted at the one place the impact arrives, and each is worth more than any amount of strength behind it.

The practical form is an energy budget rather than a force calculation. Establish how much energy can arrive — a mass, a height, and nothing else — then ask what in the load path can absorb it, and how much force each candidate would generate doing so. A resilient pad, a sacrificial element, a sand bed, a length of rope: each is a stiffness in the chain, and the softest one in series takes almost all of the deformation and therefore sets the force for everything behind it.

What yielding does to the factor #

The consequence for impact design is direct. If a structure is allowed to yield, the elastic energy balance above is badly conservative: it assumes energy is stored as $\tfrac12 k\delta^2$ when in fact most of it goes into plastic work at roughly constant force. A vehicle barrier, a ship’s fender and a crash cushion are all designed on the plastic version of this arithmetic rather than the elastic one, and the difference between them is a factor of several.

It also explains why the elastic formula is nevertheless the right one for a building. A building struck by a falling load is not meant to yield, so the elastic bound applies and it applies conservatively. The moment yielding is admitted, the calculation changes character entirely — which is the same discontinuity the seismic case turns into a design philosophy.

Where the free body is, and what it assumes #

The free body is the falling mass at the instant of maximum deflection, and three assumptions were made in drawing it. All three are conservative, which is why the formula is used with no further margin.

All the kinetic energy goes into the structure. In reality some goes into local crushing at the contact point, some into sound, some into the falling object’s own deformation, and some is retained if the object rebounds. Every one of those reduces what the structure must store.

The structure stays linear. If it yields, it absorbs far more energy per millimetre than the elastic assumption allows, and the factor drops sharply. A plastic hinge is an excellent energy absorber and a poor spring.

The structure’s own mass does not move. The energy balance ignores the kinetic energy of the beam itself, which must also be set moving. Including it — through an effective mass at the impact point — reduces the deflection further.

The result is an upper bound, and a designer wants an upper bound. What has to be watched is that it is an upper bound on the global response, and it says nothing about the local one.

What the picture cannot show #

The contact. The formula treats the impact as a load applied to a structure, with no stress concentration anywhere. Real impacts damage the contact region first — indentation, spalling, punching — and that damage has nothing to do with the beam’s stiffness and everything to do with the geometry of the two surfaces. Impact damage on a concrete slab is usually local rather than flexural.

The wave. For a very sharp impact the structure does not respond as a whole at all. A stress wave travels away from the impact at the material’s wave speed, and the far end of the member does not know anything has happened until the wave arrives. That regime is not covered by a deflection-based energy balance, and it is why blast and ballistic problems are treated differently.

The rebound. The load bounces, comes down again, and the second impact starts from a height set by the coefficient of restitution. The formula gives the first peak only.

The lifting gear. In a crane or hoist, the drop is often arrested by a rope rather than by the structure, and the rope is much softer than anything it is attached to — a series arrangement in which the softest element dominates. The structure then sees a much reduced factor, which is why lifting equipment is specified with its own dynamic allowances and why a slack rope suddenly taking load is the dangerous case.

The time-domain check #

The energy method gives no time history — it gives one number. The integration that produced the earlier essays’ figures gives both, and agreeing with each other is what makes either trustworthy.

For a drop from a height, the integration is given the same mass with an initial velocity of $\sqrt{2gh}$, and it reproduces the energy formula’s factor to four figures. That agreement is not decoration: it is the check that catches the error this calculation invites, which is forgetting that the weight keeps acting during the deflection. Leave out the $W\delta_d$ term and the factor for a placed load comes out as 1.41 rather than 2 — a plausible number, wrong by 40%.

The pulse curve on that plot is the case a drop is most often confused with, and the confusion is worth taking apart at the short end of its own axis.

A dropped weight is nowhere on that curve, and the distinction is the whole reason the two are drawn on different axes. A short pulse is a load that leaves — it was gone before the structure had time to move, and the small factor is the structure’s slowness protecting it. A dropped weight arrives and stays, and the force it applies is not a fact about the load at all but a consequence of the structure stopping it. Halving the duration of a blast helps enormously; there is no equivalent lever on a weight already on the floor.

That difference also decides which quantity a designer must establish. For the pulse it is a duration, measurable in isolation with no structure present. For the drop it is an energy and a deflection, and the deflection belongs to the structure being designed.

Where impact factors come from in practice #

Design guidance for lifting appliances, machinery and vehicles gives impact allowances as percentages — 25% on a lift machine room, 100% for a hoist, and so on. Reading them against the same spectrum shows what they encode.

An allowance of 100% is a factor of 2, which is exactly the placed-load case: the guidance is saying assume the load is dropped from no height at all onto a linear structure. An allowance of 25% is much less than that, and it is saying that the load’s arrival is controlled — a lift decelerates, a machine starts smoothly — so the rise takes an appreciable fraction of a period and most of the amplification never happens.

The zero at one full period is the part worth remembering, because it is the only free lunch in this essay. A load brought on over exactly the structure’s own period produces no dynamic amplification at all, and one brought on over several periods produces almost none. That is not a margin, it is an identity, and it is what a soft-start controller on a hoist is buying.

What no allowance encodes is a genuine drop from a height, because the answer then depends on the structure’s own deflection, which the allowance’s author does not know. A percentage cannot express a factor that depends on the thing being designed, and that is why the drop cases are always calculated rather than tabulated.

The other way a load arrives with a velocity #

A dropped weight is the clearest case of a load that arrives moving, and it is not the commonest. The commonest is a wheel.

A vehicle crossing a bridge delivers something between a slowly applied load and an impulse, depending on how the crossing time compares with the span’s period, and a wheel meeting a rail joint or a pothole delivers a genuine impact on top of that. Both are why an axle train is worse than its heaviest axle and why bridge codes carry an impact allowance on top of the static load model.

The distinctive feature of the moving case, and the reason it is not simply this essay with a different number, is that the mass doing the impacting is also being carried. A dropped weight lands and stays; a wheel arrives, is supported, and leaves, and its own inertia is part of the system while it is there. That is the problem Willis and Stokes set out to solve in 1849, and it remains the reason bridge dynamics is a subject of its own.

Where the ladder goes #

Two directions, and this rung sits between them.

Backwards, to the shock spectrum: a drop is the case where the load arrives with a velocity, and a blast is the case where it arrives as an impulse. They are the same problem with different initial conditions, and the drop is the one with a closed form.

Forwards, to the two things this calculation cannot do. It cannot handle a structure that yields, which is where the energy absorbed per millimetre changes by an order of magnitude and the whole design philosophy changes with it. And it cannot handle repeated impacts, where the question is no longer whether the structure survives one but how many it survives — which is a fatigue problem with a very small number of very large cycles, and the worst kind to have.