Structural Engineering & Statics
where the load goes
Dynamics

The weight that was dropped

A half-tonne load lowered onto a beam produces 5 kN. The same load dropped one metre onto the same beam produces 160 kN — and onto a beam ten times softer, 54 kN. The stiff structure is the one that suffers, which is the opposite of every other rule on this site.

Assumes Twice the deflection, for the same load and The deflection that arrives three years late.

17 min read 8 figures The load that will not hold stillThe load must go somewhere

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.

What a drop height is worth, as a factor on the answer for a weight placed slowlyThe peak displacement as a multiple of the static deflection, against the height a weight is dropped from divided by the deflection that weight causes when it is placed. The curve is 1 + √(1 + 2h/δ), which is conservation of energy and nothing else: the weight does work over the height it falls PLUS the distance the structure then gives, and the structure stores work only over the second. At a ratio of 40 the factor is 10.00, and at zero it is exactly 2 — the marked point, where a dropped weight becomes a placed one.05101520253035400246810drop height ÷ static deflectionpeak ÷ static deflectiontwice the static answer — a weight placed, not droppedthe static answera weight dropped from a height
Fig. 1 The impact factor against the drop height divided by the static deflection. It rises steeply and then flattens, because it is a square root: a factor of 2 at zero height, 11.05 at a ratio of 60, 32.6 at 530. The whole curve is conservation of energy and contains nothing about the structure except one deflection.

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 hh plus the distance the structure gave, δd\delta_d. Over the whole of that distance its own weight WW did work, so the work done is W(h+δd)W(h + \delta_d). All of it has gone into the structure as strain energy, which for a linear spring is 12kδd2\tfrac{1}{2}k\delta_d^2. And kk is W/δsW/\delta_s with δs\delta_s the static deflection.

W(h+δd)=12Wδsδd2W(h + \delta_d) = \tfrac{1}{2}\frac{W}{\delta_s}\delta_d^2

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

δdδs=1+1+2hδs\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 δs\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:

drop height arrival speed factor deflection
0 (placed) 0 2.00 4.0 mm
10 mm 0.44 m/s 4.32 8.6 mm
50 mm 0.99 m/s 8.14 16.3 mm
100 mm 1.40 m/s 11.05 22.1 mm
300 mm 2.43 m/s 18.35 36.7 mm
1000 mm 4.43 m/s 32.64 65.3 mm

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

What a drop height is worth, as a factor on the answer for a weight placed slowlyThe peak displacement as a multiple of the static deflection, against the height a weight is dropped from divided by the deflection that weight causes when it is placed. The curve is 1 + √(1 + 2h/δ), which is conservation of energy and nothing else: the weight does work over the height it falls PLUS the distance the structure then gives, and the structure stores work only over the second. At a ratio of 10 the factor is 5.58, and at zero it is exactly 2 — the marked point, where a dropped weight becomes a placed one.0246810012345drop height ÷ static deflectionpeak ÷ static deflectiontwice the static answer — a weight placed, not droppedthe static answera weight dropped from a height
Fig. 2 The same curve over the first part of its range, where most practical cases sit. At a ratio of 2.5 — a load dropped 50 mm onto something that deflects 20 mm under it — the factor is 3.45. At a ratio of 25 it is 8.1. The steep part is where the design decisions are.

Take one drop and three structures:

static deflection stiffness factor peak force
2 mm 2,453 kN/m 32.64 160.1 kN
20 mm 245 kN/m 11.05 54.2 kN
200 mm 25 kN/m 4.32 21.2 kN

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 essay on this site — 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 δ\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.

What yielding does to the factor

What it costs to reach the plastic moment, for two shapesMoment against curvature for two cross-sections of identical area (3000 mm²) and identical depth (200 mm), in mild steel, each divided by its own first-yield moment and its own first-yield curvature. The rectangle has a shape factor of 1.50 and reaches 98% of its plastic moment at 4.3 times the curvature at first yield; The I-section has a shape factor of 1.09 and reaches 98% of its plastic moment at 1.2 times the curvature at first yield. The dashed lines are the rigid-plastic moments, computed from the equal-area axis rather than read off the curves, and no curve reaches its own.02468101200.511.5curvature ÷ curvature at first yieldmoment ÷ moment at first yieldrectangle: 1.50× the yield moment, at 4.3× the yield curvatureI-section: 1.09× the yield moment, at 1.2× the yield curvature
Fig. 3 Why a structure that yields is a far better catcher than one that does not. Beyond first yield the moment barely rises while the curvature goes on increasing — so the area under the curve, which is the energy absorbed, grows enormously for very little extra force. An elastic structure buys energy absorption by increasing its force; a yielding one buys it for nothing.

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 12kδ2\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.

4.9 kN applied at once and held, on a structure of 0.100 s periodDisplacement against time for a single-degree-of-freedom structure of natural period 0.100 s and 2.0% damping, under 4.9 kN applied at once and held. The static deflection under the same peak force is 2.48 mm and the peak response is 4.81 mm — a factor of 1.94.00.20.40.60.81-4-224time (s)displacement (mm)4.9 kN applied at once and heldthe load arrives in no time at allstatic, 2.48 mmpeak 4.81 mm at 0.050 s
Fig. 4 The same problem as a time history, for the zero-height case: a 500 kg mass released onto a spring, plotted as displacement against time. The peak is at twice the static deflection and arrives half a period after release. The energy method’s factor of 2 and the integration’s factor of 2 came from completely different arguments — one about work, one about the equation of motion — and agree exactly.

For a drop from a height, the integration is given the same mass with an initial velocity of 2gh\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 WδdW\delta_d term and the factor for a placed load comes out as 1.41 rather than 2 — a plausible number, wrong by 40%.

What the shape of a load in time is worth, for two load shapesThe peak displacement as a multiple of the static deflection, against the load's duration divided by the structure's natural period, for two load shapes: a load that rises linearly, then stays; a rectangular pulse, then nothing. The lines are closed forms and eight dots are the peak of a complete time integration of an oscillator of 0.400 s period under that load, agreeing with the line to within 0.19% everywhere.00.511.522.500.511.52load duration ÷ natural periodpeak ÷ static deflectiontwice the static answerthe static answera load that rises linearly, then staysa rectangular pulse, then nothing
Fig. 5 The neighbouring cases, for comparison: a load that ramps on and a load that pulses. The drop sits outside this plot because its horizontal axis is a height rather than a duration, and the three of them meet at the same place — a factor of two for a load that arrives instantly and stays.
40 kN for 0.020 s, then gone, on a structure of 0.400 s periodDisplacement against time for a single-degree-of-freedom structure of natural period 0.400 s and 2.0% damping, under 40 kN for 0.020 s, then gone. The static deflection under the same peak force is 16.21 mm and the peak response is 4.79 mm — a factor of 0.30.00.511.522.533.54-2020time (s)displacement (mm)40 kN for 0.020 s, then goneduration 0.05 of a periodstatic, 16.21 mmpeak 4.79 mm at 0.109 s
Fig. 6 And the case a drop is often confused with: a large force for a very short time. The peak here is a third of the static deflection, because the load was gone before the structure had moved. A dropped weight is not this — it stays, and the force it applies is a consequence of the structure stopping it rather than a fact about the load.
4.9 kN applied at once and held, on a structure of 0.100 s periodDisplacement against time for a single-degree-of-freedom structure of natural period 0.100 s and 2.0% damping, under 4.9 kN applied at once and held. The static deflection under the same peak force is 2.48 mm and the peak response is 4.81 mm — a factor of 1.94.00.20.40.60.81-2-112time (s)displacement ÷ static deflection4.9 kN applied at once and heldthe load arrives in no time at allstatic, 2.48 mmpeak 4.81 mm at 0.050 s
Fig. 7 The same release plotted as a multiple of the static deflection rather than in millimetres, which is the form the whole essay’s arithmetic is in. Every impact factor in the tables above is the peak of a curve like this one — the ordinate at which it turns round — and the drop height only changes how high that peak is, never the shape below it.

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 curve above 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 effective drop is a fraction of the deflection.

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.

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.24the 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. 8 The static answer for a load at mid-span — the deflection every impact factor above multiplies. What a moving wheel does is apply this load and then take it away again, at a rate set by the vehicle’s speed, so the structure is loaded and unloaded in a time that may or may not be short compared with its period.

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.

Where this goes next

012345-30-20-10102030time (s)displacement (mm)20 kN applied at once and heldthe load arrives in no time at allstatic, 12.67 mmpeak 24.56 mm at 0.250 s Twice the deflection, for the same load A weight placed gently on a beam deflects it by one amount. The same weight let go from rest, a millimetre above the same beam, deflects it by twice as much — and the factor of two is exact, for every structure ever built. 0.60.811.21.41.61.8200.511.5span, relative to the firstthey cross heredeflection runs out at 1.40strength runs out at 1.54the limitstrengthdeflection Stiffness is not strength, and usually it is the one that governs A beam can be nowhere near failure and still be unusable, because it has moved too far. For most long-span members that limit arrives first, and stronger steel does not help at all. 00.511.522.530510152025forcing frequency ÷ natural frequencyamplitude ÷ static deflection1% damping — 50× at the peak2% damping — 25.01× at the peak5% damping — 10.01× at the peak10% damping — 5.03× at the peak The only thing that stops it Drive a structure at its own frequency and the amplitude grows without limit unless something takes energy out. What takes it out is damping, and damping is the one structural property that is never designed, never drawn, and never known until the thing is built. 12012080resultantmidspan05101520020040060080010001200station along the spangreatest moment ever seen there1160.2 at 9.88Barré: 1160.3 at 9.88 The train that is worse than its heaviest axle An influence line says where to stand one load. A vehicle is several loads at fixed spacings, and the worst arrangement never puts the heaviest one at the peak. 2345678910051015202530floor frequency (Hz)response factor1× pace2× pace3× pace4× pace4 Hz — R = 106 Hz — R = 7 The floor that is strong and unusable A floor can satisfy every strength check, deflect less than the limit, and still be rejected by the people who work on it — because somebody walking across it at two steps a second happens to be exciting it at exactly the rate it likes to move.

The objects this essay names

Each one links to every other essay that touches it.

Conservation of energyCrane loadingDynamic amplificationFall arrestImpact factorImpulseStiffnessStrain energy