The load that is over before it has moved

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.

Which free body produced the number #

One mass on one spring, cut free. The forces on it are the applied load $F(t)$, the spring’s restoring force $ku$, and the inertia force $m\ddot{u}$ — 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 $u$ has grown appreciably, the spring has done almost no work and the equation reduces to

$$m\,\dot{u} = \int F\,dt = I$$

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 ($t_d/T$ 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.

Long ($t_d/T$ 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.

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.

Two things in that picture are worth separating. The factor has stopped growing — a rectangular pulse cannot do better than the suddenly applied load it becomes, because holding the force past the first peak changes nothing that has already happened. And the load still finishes while the structure is moving, so the free vibration afterwards is about the same size as the forced part, which is the signature of the intermediate regime and the reason a designer cannot read the answer off either asymptote.

The shape of the load matters as much as its length, and the cheapest demonstration is to take the load off the vertical.

That pair is the argument against describing a load by its magnitude in a sentence. The same peak force, on the same structure, produces 2.74 mm, 13.26 mm and 6.96 mm depending only on how it is delivered in time — a spread of nearly five to one with no number in it that a static calculation would recognise as having changed.

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 $I^2/2m$, and it comes to rest when the work done deforming it equals that. For a member with a plastic capacity $R$ pushed through a displacement $u_m$, the work is roughly $R\,u_m$, so

$$u_m \approx \frac{I^2}{2 m R}$$

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 $R$ 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.

The input drawn there lasts far longer than a blast, which is the point of drawing it: what the yielding run does with energy is the same whether it was handed a momentum in two milliseconds or shaken for twenty seconds. The loop it traces in force against displacement is a parallelogram whose area does not depend on how fast it is traced, and every circuit of that loop 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 $\frac{1}{2}Ru_y$ and returns all of it. A yielding one stores $R u_m$ and keeps it, and $u_m$ can be many times $u_y$.

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.

What the steel itself does on that excursion is worth naming, because it is the reason the second cycle is not a repeat of the first. Taken well past yield and pushed back, steel yields again in the opposite direction before the stress has fallen by twice the yield stress — the Bauschinger effect, and a consequence of the yield surface sliding rather than growing. A member that has absorbed one impulse is a slightly different member, and what is left when the load comes off is not the member the calculation started with.

The capacity $R$ 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 $R$ 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.

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 $R$ 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 pressure that arrives is not the pressure that was quoted #

Everything above starts from an overpressure, and an overpressure is what the wave carries as it travels. It is not what a wall feels.

A blast wave meeting a surface it cannot move is brought to rest, and the air behind it piles into the air in front. The surface therefore sees a reflected pressure that is larger than the incident one by

$$C_r = \frac{2(7p_0 + 4P_s)}{7p_0 + P_s}$$

which for a weak wave tends to 2, the acoustic answer, and for a strong one tends to 8 — because air is compressible, and a strong shock leaves a much denser gas behind it than a sound wave does. At the 30 kPa of the figures above, $C_r = 2.24$ and the wall feels 67 kPa. At 700 kPa it is 4.98, and the wall feels 3.5 MPa. The quantity the energy balance needs is bigger than the quantity the event is described by, by a factor that itself depends on the size of the event.

Then the reflection is relieved, and this is the part that is purely geometric. The high pressure at the face is only sustained while the surrounding air cannot get out of the way; a rarefaction travels in from every free edge at roughly the speed of sound in the compressed gas, and once it has crossed the element the pressure falls to the much lower stagnation value. The time that takes is about

$$t_c \approx \frac{3S}{U}$$

with $S$ the shorter distance from the centre of the element to a free edge. For a 2 m wide panel, $S = 1$ m and $t_c \approx 7.5$ ms — against a positive phase of 20 ms, so the panel gets the reflected pressure for barely a third of the event. For a 20 m wide, 5 m tall wall, $S = 5$ m and $t_c \approx 37$ ms, which is longer than the load lasts: no clearing happens at all and the whole wall carries the full reflected pressure for the whole duration.

So the impulse per square metre delivered to the large wall is roughly twice that delivered to the small panel by the same explosion at the same standoff. Nothing about the material or the strength has changed; the difference is how far the air had to travel to escape.

That is worth holding beside the P-I chart, because the two halves of blast design are usually taught as one subject and are not. The chart says what an element can take. The clearing says what arrives — and it is a function of the element’s own dimensions, which means the load and the resistance are not independent, and the test panel that established a detail’s capacity was cleared in a way the wall built from it will not be.

The load path afterwards #

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

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.

How much that costs the survivors is measured elsewhere: on the truss in the structure that survives losing a member, the worst-affected member left standing picks up 2.03 times what it had been carrying, and four members that had carried nothing find themselves working. Neither number depends on what removed the member.

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 $t_d/T$. Making a structure stiffer shortens $T$, which moves the ratio up — toward the quasi-static end, where the dynamic factor approaches two. Making it more flexible lengthens $T$, 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.

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.

The two extremes belong on one page because they are the same equation. Resonance is what a structure does with a load that keeps arriving in step with it, and the amplification is one over twice the damping ratio because every cycle adds a little and the damping removes a fixed fraction of what is there. A blast gives that mechanism nothing to work with: there is no second cycle, so there is nothing to accumulate and nothing for a damper to have removed.

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.

And the neighbouring problem is a different question. An earthquake lasts tens of seconds and contains many cycles, so the displacement it asks for is a question about accumulated damage rather than about a single excursion. The figure of the yielding run above is drawn against such a record, and everything this essay says about a blast is contained in its first half-cycle.

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.