Structural Dynamics / Impulse response, arbitrary loading and convolution
What a single blow does
An impulse changes how fast a structure is moving, not where it is — and everything in this module is built from that one response.
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Read the lesson register and outstanding review notesWhy this matters
Module 5 solved harmonic loading, and harmonic loading is the exception. A dropped load, a vehicle impact, a gust, a wave slam, an earthquake — none of them repeats, and none of them has a closed-form answer of the kind resonance did.
The way through is to solve one very simple problem completely — what happens when a structure is struck once — and then build every other loading out of it. That is the whole of this module, and the technique is superposition, which you already have.
By the end of this lesson you should be able to
- Define impulse and relate it to momentum
- Derive h(t), the unit-impulse response function
- Explain why h(0) = 0 and why the slope there is 1/m
- Predict how mass affects the response to a blow
What you should already know
- Free vibration from initial conditions (Module 4)
- The equation of motion (Module 3)
- Newton's second law as a statement about momentum
Impulse is force multiplied by time
The impulse of a force is the area under its force–time curve:
I = ∫ F dt
Its units are newton-seconds, which are also the units of momentum — and that is not a coincidence. Newton's second law, written the way Newton actually wrote it, says that force is the rate of change of momentum:
F = d(mv)/dt
Integrate both sides over the duration of the force and you get
I = ∫F dt = m·Δv
An impulse changes a structure's momentum. It does not, in itself, change its position.
That sentence is the key to the whole module. If the blow is short enough, the structure has not had time to move while it is being struck. It ends the blow exactly where it started — and moving at I/m.
Which is a set of initial conditions we already know how to use
A structure that is at its equilibrium position and moving is precisely the free-vibration problem of Module 4 with x₀ = 0 and v₀ = I/m. We already have that solution. Nothing new is needed.
And note what mass does
The velocity imparted is I/m. Double the mass and the same blow imparts half the velocity — and, since the resulting free-vibration amplitude is v₀/ωn and ωn itself falls as mass rises, the displacement does not simply halve. Work it through and the peak displacement is I/(mωn), so it goes as 1/√(mk).
Either way, a heavier structure is less affected by a blow. That is the exact opposite of the static intuition, where mass is either irrelevant or a liability, and it is why a heavy pier is a better thing to be hit by a ship than a light one.
From first principles
The unit-impulse response function
We want to show: Derive h(t), the displacement history that follows a unit impulse applied to a structure at rest, and understand why it starts at zero.
Hit the mass with a very short, very large force. Because it is short, the mass has no time to move while the force acts — so at the end of the blow it is still at the origin. But the force has changed its momentum, so it is now moving. From that instant on, no force acts at all: the structure is in free vibration, released from the equilibrium position with a velocity it did not have a moment before.
Try it
What one sharp blow does
A very short force of known total impulse, and the free vibration it leaves behind.
- Response to the impulse
- Free vibration from x₀ = 0, x′₀ = I/m
Show the numbers behind this plot
- Velocity imparted, I/m
- 5000 mm/s
- Peak displacement
- 590 mm
- Predicted peak I/(mωn)
- 637 mm
- h(0)
- 0
- Slope of h at t = 0
- 1/m = 1.0e-4 m per N·s²
- First peak at
- 0.2 s
Exact for an undamped system; slightly high once damping is present.
The mass has not moved yet.
A quarter of a damped period after the blow.
Doubling the mass halves the velocity the impulse imparts AND halves the resulting displacement, so a heavier structure is less affected by a blow — the opposite of the static intuition, where mass does nothing at all.
What this shows: An impulse changes momentum, not position. The struck structure starts exactly where it was, moving at I/m — so the impulse response is a free vibration with zero initial displacement.
Worked example
A vessel impact on a bridge pier
Given
- A river pier idealised as a single degree of freedom: effective mass 850 tonnes
- Lateral stiffness 420 MN/m
- Damping 4%
- A barge strikes it, delivering an impulse estimated at 2 400 kN·s
Find
The peak lateral displacement, and when it occurs.
Assumptions
- The contact lasts about 0.15 s; compare this with the pier period before using an impulse approximation
- The pier stays elastic
- The barge does not remain in contact and push
Predict first
Two structures have the same stiffness. One has twice the mass of the other. Both are struck by the same impulse. Which reaches the larger peak displacement?
Practice
A structure of mass 30 000 kg is struck by an impulse of 45 kN·s. What velocity does it acquire, in m/s?
Practice
That structure has a natural frequency of 12 rad/s. Ignoring damping, what is its peak displacement in mm?
Practice
A unit impulse response h(t) has an initial slope of 4 × 10⁻⁵ m per N·s². What is the mass of the structure, in kg?
Summary
- Impulse I = ∫F dt has the units of momentum, and it changes momentum
- A short blow leaves the structure at the origin, moving at I/m
- h(t) = (1/mωd)e^(−ζωn t)sin ωd t, with h(0) = 0 and slope 1/m
- The impulse response IS free vibration with x₀ = 0 and v₀ = I/m
- Peak displacement from an impulse is about I/(mωn) = I/√(mk)
- A heavier structure is less affected by a blow, unlike the static case
- The peak occurs a quarter period AFTER the blow, when nothing is touching it