Dynamics

The load that is over before it has moved

A blast delivers an enormous pressure for a few milliseconds. Everything else in this field asks what force a structure can carry; a load that has come and gone before the structure has travelled any distance is not asking that question, and the answer turns out to depend on the mass and the ductility with the strength barely in it.

Assumes The weight that was dropped, Twice the deflection, for the same load and The earthquake asks for a displacement.

Every load in the static half of this collection is described by its magnitude. A dynamic one needs two numbers — how big and how long — and the second turns out to matter more.

A blast wave arrives with an overpressure of some tens or hundreds of kilopascals and is gone in a few milliseconds. A building’s natural period is a few tenths of a second. The load has therefore finished before the structure has begun to move, and asking what force the structure can carry is asking a question the event never posed.

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 0.300 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. 1 Peak displacement as a multiple of the static deflection, against the load’s duration divided by the structure’s own period, for two load shapes. The closed forms and a complete time integration agree to within 0.19%.

Which free body produced the number

One mass on one spring, cut free. The forces on it are the applied load F(t)F(t), the spring’s restoring force kuku, and the inertia force mu¨m\ddot{u} — which is the only new thing dynamics adds to a free body, and the whole of what it adds.

Integrating that free body over the duration of a short load gives the useful result. If the load has finished before uu has grown appreciably, the spring has done almost no work and the equation reduces to

mu˙=Fdt=Im\,\dot{u} = \int F\,dt = I

The structure ends the loading with a velocity and no displacement, and everything after that is the structure converting kinetic energy into strain energy at its own pace. The peak pressure has vanished from the problem; only the impulse remains.

That is the impulsive regime, and it is why a blast load is quoted as a pressure and an impulse and never as a pressure alone. The two are not alternative descriptions of one thing; they are the two numbers a load needs when the structure has an opinion about time.

Three regimes, one ratio

The parameter that decides which regime applies is the load’s duration over the structure’s natural period, and the shock spectrum is the map of it.

Short (td/Tt_d/T below about a tenth): impulsive. The peak response is proportional to the impulse and the peak pressure is irrelevant. The dynamic factor is much less than one — on the case below, 0.40.

30 kN for 0.020 s, then gone, on a structure of 0.300 s periodDisplacement against time for a single-degree-of-freedom structure of natural period 0.300 s and 2.0% damping, under 30 kN for 0.020 s, then gone. The static deflection under the same peak force is 6.84 mm and the peak response is 2.74 mm — a factor of 0.40.00.511.522.53-15-10-551015time (s)displacement (mm)30 kN for 0.020 s, then goneduration 0.07 of a periodstatic, 6.84 mmpeak 2.74 mm at 0.084 s
Fig. 2 A load of 30 kN for 20 milliseconds on a structure of 0.300 s period. The static deflection under the same peak force is 6.84 mm and the peak response is 2.74 — a factor of 0.40.

Long (td/Tt_d/T above about ten): quasi-static. The structure follows the load, and the only dynamic effect left is that the load arrived rather than being eased on. That gives the familiar factor of two.

35 kN applied at once and held, 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 35 kN applied at once and held. The static deflection under the same peak force is 14.18 mm and the peak response is 27.51 mm — a factor of 1.94.00.511.522.533.54-30-20-10102030time (s)displacement (mm)35 kN applied at once and heldthe load arrives in no time at allstatic, 14.18 mmpeak 27.51 mm at 0.200 s
Fig. 3 The same structure under a load applied at once and held. The static deflection is 14.18 mm and the peak response 27.51 — a factor of 1.94, and the remaining shortfall from two is the damping.

Between: dynamic, and the response depends on the shape of the load in time as well as on its duration. That is the region the shock spectrum was drawn for, and it is where most real events sit.

Twice the deflection for the same load is the essay about the long end and the weight that was dropped about a related short one. What is specific to blast is that it sits at the very short end, where the dynamic factor is small and the whole design is about something else.

What resists an impulse

If the load delivers a momentum, the resistance has to be counted in energy.

The structure leaves the loading with kinetic energy I2/2mI^2/2m, and it comes to rest when the work done deforming it equals that. For a member with a plastic capacity RR pushed through a displacement umu_m, the work is roughly RumR\,u_m, so

umI22mRu_m \approx \frac{I^2}{2 m R}

Read that expression carefully, because every term in it says something a static check does not.

The mass helps. A heavier structure acquires less velocity from the same impulse, and it appears in the denominator. That is the opposite of everything else in this collection, where mass is a load.

The strength helps linearly. Doubling RR halves the displacement. It does not halve the demand — the impulse is unchanged — it halves how far the structure has to travel to absorb it.

The displacement is the answer. What comes out is not a stress or a utilisation, it is a distance, and whether the structure survives is whether it can travel that distance without falling apart.

Where the energy goes: one loop in force against displacementThe force the supports feel — the spring's and the damper's together — against the displacement, for one mechanism. yielding at 295.95 kN, enclosing 27.72 kJ over the record drawn. The yielding loop is a parallelogram whose area does not depend on how fast it is traced, and every circuit leaves the structure displaced from where it began.-20-15-10-55101520-200200displacement (mm)restoring force (kN)yielding at 295.95 kN — 27.72 kJ
Fig. 4 Where the energy goes, drawn as a loop in force against displacement. The yielding loop is a parallelogram whose area does not depend on how fast it is traced, and every circuit leaves the structure displaced from where it began.

That last point is the whole of blast design. The quantity being checked is a ductility demand — the same currency an earthquake asks for — how many times its yield displacement the member has to travel — and a member that can travel ten times its yield displacement absorbs ten times the energy of one that can travel once.

Why the elastic answer is the wrong one

An elastic structure stores energy as 12Ruy\frac{1}{2}Ru_y and returns all of it. A yielding one stores RumR u_m and keeps it, and umu_m can be many times uyu_y.

So the energy a member can absorb is roughly its capacity times its ductility, and ductility is much cheaper than capacity. A member with a ductility of 10 absorbs about twenty times what the same member absorbs elastically — which is why blast-resistant design is about detailing for large deformations rather than about making things strong.

Yielding one way makes it easier to yield the otherMild steel taken to a strain of 2.00% and then pushed back the other way. The stress falls by 550 N/mm² before it yields again, against a yield stress of 275 — the elastic range is twice the yield stress and not once it, which is the Bauschinger effect and is a consequence of the yield surface sliding rather than growing.-2.0%-1.0%1.0%2.0%-300-200-100100200300strainstress, N/mm²550 N/mm² of elastic range — twice the yield stress
Fig. 5 Steel taken well past yield and pushed back. The stress falls by twice the yield stress before it yields again, which is the Bauschinger effect and a consequence of the yield surface sliding rather than growing.
The two theorems close on the answer from opposite sidesA pinned-base portal frame under 260 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 0.559, sway 1.000 and the combined one 0.437. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.437, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.28 times the real capacity, and the sway one 2.29 times, both of them on the wrong side.-100-505010000.20.40.60.811.2the redundant — horizontal reaction at the right base (kN)load factorbeam 0.56sway 1.00combined 0.44best lower bound 0.437every mechanism is at or above the answer; every admissible field is at or below it
Fig. 6 The two theorems closing on a frame’s collapse load from opposite sides. Three mechanisms give upper bounds of 0.559, 1.000 and 0.437; the lower bound peaks at 0.437 and touches the lowest of them exactly.

The capacity RR in the energy balance is a plastic collapse load, not an elastic limit, so the whole apparatus of the bound theorems is what supplies it. Picking the wrong mechanism claims 1.28 or 2.29 times the real capacity, both on the wrong side, which matters more here than in a static design because the energy is linear in RR and there is nothing else to fall back on.

What the rate does to the material

There is one more effect and it is large enough to change the answer.

How fast the strain arrived, which no strength on this site recordsThe dynamic increase factor on strength against strain rate, over eight decades. Steel follows Cowper and Symonds' fit, whose constant D = 40.4 s⁻¹ is not an arbitrary parameter — it is the rate at which the material is exactly twice as strong. Concrete in tension follows the model code's two-branch curve and is steeper. The four marked regimes are the argument: a testing machine works at about 10⁻⁴ per second, an earthquake at 5 × 10⁻³, a vehicle impact at a half, a blast at a hundred, and the enhancement across them runs 1.00, 1.08, 1.32, 2.04. So this is a correction that is either negligible or decisive with very little in between, which is why no seismic code carries it and every blast code does. What does not rise is the modulus, which is a lattice property, and the ultimate strength rises only a third as much — so the ultimate-to-yield ratio closes from 1.21 to 0.95 and the material has less warning left in it than it started with.10^-410^-210^2123456strain rate (per second)strength ÷ its static valuea testing machinean earthquakea vehicle impacta blastconcrete in tensionsteel, yieldsteel, ultimatethe modulus
Fig. 7 The dynamic increase factor on strength against strain rate, over eight decades. A testing machine works at 10⁻⁴ per second, an earthquake at 5×10⁻³, a vehicle impact at a half, and a blast at a hundred.

Steel at blast strain rates is 2.04 times as strong as the same steel in a testing machine. That is not a correction; it is a doubling of the RR in the energy balance, which halves the displacement.

It comes with a cost that is easy to overlook. The ultimate strength rises only about a third as much as the yield, so the ratio between them closes from 1.21 to 0.95 — the material becomes stronger and less ductile at the same time, which is exactly the wrong direction for a design whose whole resistance is ductility. What does not change at all is the modulus, which is a property of the lattice rather than of the dislocations.

That combination is why the rate effect is in every blast standard and no seismic one. At earthquake rates the enhancement is 1.08 and can be ignored; at blast rates it is decisive and cannot.

The chart that is drawn instead of a check

Because the answer depends on a pressure and an impulse together, the design tool is not a capacity but a curve in two dimensions.

Plot peak pressure against impulse and mark every combination that takes a given member to a given ductility. The result is a hyperbola-like curve with two asymptotes: a vertical one at the impulse the member can absorb, which is what governs short loads and is set by mass, strength and ductility together; and a horizontal one at the pressure the member can carry statically, which governs long loads and is a resistance in the ordinary sense.

Everything below and left of that curve survives; everything above and right does not. The shape of it is the entire content of this essay drawn once: two asymptotes because there are two regimes, a knee between them where the duration and the period are comparable, and no single number anywhere on it that could be called a capacity.

The chart is also what makes the trade-offs legible. Adding mass moves the vertical asymptote right and does nothing to the horizontal one. Adding strength moves both. Adding ductility moves the vertical one a long way and the horizontal one not at all — which is why a blast-resistant detail is nearly always about how a member is connected rather than about what it is made of.

And it is why the same member can be adequate against a distant large charge and inadequate against a small close one of a fraction of the energy. The two events sit on opposite sides of the knee, and a designer who has checked one has checked nothing about the other.

The load path afterwards

The last part of blast design is not about the member that was loaded at all.

Take that one away and the load finds another routeA 6-panel pratt truss under 20 kN at each top node, before and after member 3 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough.intactmember 3 removedworst demand 2.03×
Fig. 8 A truss before and after one member is removed. The load redistributes, the worst-affected survivor carries 2.03 times what it did, and four members that carried nothing are now working.

A blast is a local event, in the sense a lost member is local: it destroys what it is near and leaves everything else intact. So the question that decides whether a building falls down is not whether the loaded member survived but whether the structure can carry its loads without it — the structure that survives losing a member, and the reason robustness requirements exist in codes that never mention explosions.

That reframing is the useful one for a designer who is not doing an explicit blast analysis, which is nearly everybody. The design measure is alternative load paths, tying, and continuity, and it is checked by removing a member rather than by applying a pressure.

Why a stiffer structure can be a worse one

There is a conclusion here that reverses the ordinary instinct, and it is worth stating on its own because it is the practical reason blast design feels unlike everything else.

The regime is decided by td/Tt_d/T. Making a structure stiffer shortens TT, which moves the ratio up — toward the quasi-static end, where the dynamic factor approaches two. Making it more flexible lengthens TT, moves the ratio down, and takes the dynamic factor toward zero.

So for a load of fixed duration, a flexible member can see a smaller effect than a stiff one of the same strength. A heavy, flexible, ductile wall panel is a good blast-resistant element and a light, stiff, brittle one is not, and neither statement has anything to do with how much pressure either can carry statically.

The same reasoning explains a detail that looks like carelessness. Blast-resistant glazing is designed to deform enormously — laminated interlayers stretching, frames bending, the whole assembly moving a hundred millimetres — because the design objective is to absorb an impulse rather than to resist a pressure, and a stiff pane that does not move has to do all of its absorbing elastically in a material that cannot.

There is a limit to how far the argument goes. A structure made flexible enough to duck a blast is one that fails its ordinary serviceability checks, and the two requirements are simply in opposition. What resolves it in practice is that the flexibility is bought in the cladding — the element the blast reaches first — and the frame behind it is designed for what the cladding hands on.

Where the model stops

One degree of freedom is a strong idealisation. A real member has many modes, and a blast excites the high ones because its energy is spread over a wide band. Local response — a plate dishing, a flange buckling, a connection tearing — happens at periods far shorter than the member’s, and none of it is in a single-degree-of-freedom answer.

How much a harmonic force is magnified, at two damping ratiosDisplacement amplitude divided by the deflection the same force would produce if it were applied slowly, against the ratio of the forcing frequency to the structure's own, at 2%, 5% of critical damping. At the natural frequency the magnification is 25, 10 respectively — one over twice the damping ratio, and nothing else in the problem enters it.00.511.522.530510152025forcing frequency ÷ natural frequencyamplitude ÷ static deflection2% damping — 25.01× at the peak5% damping — 10.01× at the peak
Fig. 9 The other end of the same subject. A harmonic force at the natural frequency is magnified by one over twice the damping ratio, and nothing else in the problem enters that number.

Damping does almost nothing. In an event lasting a fraction of one cycle, a damper has no time to dissipate anything. Every figure above uses a damping ratio and none of the impulsive answers depends on it, which is why the energy has to be taken out by yielding instead.

Plastic capacity is being used twice. The energy balance uses a plastic collapse load and the ductility demand assumes the member can reach it repeatedly, but a member that has yielded once has a different capacity, a different stiffness and a different elastic range on the second excursion.

The load itself is uncertain by a lot. Blast pressures depend on the charge, the distance, the reflections off surrounding surfaces and the venting of the space, and the resulting range is a factor rather than a percentage.

A ground motion: 30 seconds of acceleration, and nothing elseA synthetic accelerogram — filtered noise through a ground filter at 2.5 Hz with a rising and decaying envelope, seeded so that the same record is drawn every time — scaled to a peak of 4.2 m/s², reached at 5.32 s. It is not a recording of any earthquake and no argument here needs it to be: what it has to have is a realistic frequency content and a realistic duration, because those are what the spectrum computed from it is about.051015202530-4-224time (s)ground acceleration (m/s²)peak 4.2 m/s² at 5.32 s
Fig. 10 The neighbouring problem, for contrast. An earthquake lasts tens of seconds and contains many cycles, so it is a question about accumulated damage rather than about a single excursion.

What the picture cannot show

A time history is drawn as a smooth curve of displacement, and the thing that decides survival is what happens at the end of the first excursion — whether a connection tore, whether a flange folded, whether the member came off its seat. None of those is a displacement.

Nor does a shock spectrum show the direction of the load. A blast pushes a wall inward and then, as the wave passes, pulls it outward in a negative phase that is weaker and longer — so a member designed for one direction is loaded in the other by a load case that most analyses leave out and that catches cladding fixings, which are strong in one direction by construction.

The generalisation

The habit worth carrying is that a dynamic load is characterised by its duration relative to the structure, and by nothing about the structure alone.

The same explosion is impulsive to a stiff wall panel and quasi-static to the frame behind it, because the two have periods an order of magnitude apart. So one event is two different load cases in one building, and asking “is this a dynamic load” has no answer without naming what it is loading.

That relativity runs through the whole field. The train that arrives in time with itself is resonant for one span and irrelevant to the next; a footfall is a nuisance to a long-span floor and nothing to a short one; a machine’s out-of-balance force is invisible until something has the frequency to notice it. In every case the load is fixed and the structure decides what it means — which is the exact inversion of the static case, where the load is the load and the structure merely carries it.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

BlastBound theoremsDampingDuctility demandDynamic amplificationEnergyHysteresisImpulseLoad pathMechanismNatural periodPlastic hingeRobustnessShock spectrumStrain rate