Module 5 · Lesson 5.1
Harmonic response and resonance
The steady-state solution, the magnification factor, and the physical reason a lightly damped structure can be shaken apart by a force it could carry a hundred times over statically.
Why this matters
Resonance is the mechanism behind almost every dramatic vibration failure and behind most of the undramatic serviceability ones. A rotating machine, a crowd walking in step, vortices shedding off a chimney — each applies a force that is small compared with what the structure carries every day, and each can produce a response an order of magnitude above the static one.
What makes resonance worth understanding rather than merely avoiding is that the size of the response is set by one quantity, and it is the quantity we know least well.
By the end of this lesson you should be able to
- Derive the steady-state amplitude and phase for harmonic forcing
- Explain the three regions of the response curve physically
- Say what limits the response at resonance and why
- Explain how long a resonant response takes to build
What you should already know
- The equation of motion (Module 3)
- Free vibration and the damping ratio (Module 4)
From first principles
Steady-state response to harmonic forcing
We want to show: Solve m ẍ + c ẋ + kx = F₀ sin ωt for the steady state, and obtain the magnification factor and phase angle.
After the transient has died away the structure must be doing something periodic at the frequency it is being driven at — it has no choice, because that is the only frequency still being supplied. So it moves sinusoidally at ω. The only questions left are how big the motion is and how far behind the force it runs. Both answers come from insisting that the four forces balance.
Try it
Sweeping through resonance
Move the forcing frequency across the natural frequency and watch both the curve and the time history.
What fraction of the force reaches the supports.
- Rd at ζ = 0.05
- ζ = 0.02
- ζ = 0.05
- ζ = 0.1
- ζ = 0.2
- Response x (mm)
- Applied force (kN ÷ 5)
Show the numbers behind this plot
- Static displacement F₀/k
- 6.33 mm
- Magnification Rd
- 10
- Steady-state amplitude X
- 63.3 mm
- Phase lag φ
- 90.0°
- Peak magnification available
- 10
- Energy per cycle into the damper
- 3980 J
- Transmissibility
- 10
- Beat period
- —
At β = 1.00 the response is 10.0 times the static value, and the ONLY thing limiting it is the 5.0% damping. Halve the damping and the response doubles.
Things worth doing here
- Set β = 1 and reduce the damping in steps. The magnification is exactly 1/(2ζ) each time, and the time history takes proportionally longer to reach it.
- Notice the peak of the curve is NOT at β = 1 — it sits at β = √(1 − 2ζ²), just below. At small damping the difference is invisible; at ζ = 0.3 it is not.
- Raise ζ past 1/√2 ≈ 0.707. The peak disappears entirely and the response falls monotonically from its static value.
- Switch to transmissibility. Every curve crosses 1 at β = √2 whatever the damping — and above that point MORE damping makes the transmitted force worse, which is the one place in structural engineering where added damping is not an improvement.
What this shows: Resonance is not a property of a load or of a structure but of the RATIO between them, and at resonance nothing is holding the response down except damping.
The three regions, physically
β ≪ 1 — stiffness-controlled. The load changes slowly enough that the structure follows it. The spring force is doing all the work; inertia and damping are negligible. The response is the static one, and this is the region in which a static calculation is correct.
β ≈ 1 — damping-controlled. The spring force and the inertia force are equal and opposite and cancel. Whatever the applied force is, it must be carried by the damping force alone — and since c ẋ is small unless ẋ is large, the amplitude grows until it is large enough. The response is 1/(2ζ) times the static value, and neither the mass nor the stiffness appears in that expression.
β ≫ 1 — inertia-controlled. The force alternates faster than the mass can respond to it. The mass barely moves; the load is resisted by inertia. This is why a structure can be entirely unaffected by very high-frequency excitation, and it is the principle behind vibration isolation.
Resonance takes time
An important practical point that the steady-state formula hides completely: a resonant response builds up over many cycles.
Starting from rest, the envelope grows roughly as 1 − e^(−ζωn t). It reaches 63% of its final value after ζωn t = 1, which is after about 1/(2πζ) cycles.
At 5% damping that is 3 cycles; at 1% damping, 16; at 0.2% — a bare welded steel structure — about 80 cycles. So a passing disturbance that happens to be at the natural frequency but lasts only a few cycles never produces anything like the steady-state amplitude.
This cuts both ways. It is why a brief resonant excitation is often harmless. And it is why machinery start-up, which sweeps through resonance in a few seconds, usually produces far less response than running AT resonance would — but why a machine left running at a resonant speed is a serious problem, since it has all the time it needs.
Predict first
A structure at resonance has a response of 40 mm with 4% damping. The damping is improved to 8%. What is the new response?
Worked example
A machine foundation checked at running speed and through start-up
Given
- A pump on a first-floor slab, running at 1 450 rpm
- Rotating unbalance producing a harmonic force of 1.8 kN at running speed
- The supporting floor bay: natural frequency 21 Hz, modal mass 9 000 kg
- Damping 3%
Find
The steady-state response at running speed, and what happens during start-up.
Assumptions
- The floor responds in its first mode only
- The unbalance force amplitude grows with the square of speed, which is what rotating unbalance does
- Start-up passes through resonance at a rate that does not allow steady state to develop
Practice
A structure with 2.5% damping is forced at exactly its natural frequency. What is the dynamic magnification factor?
Practice
A machine forces at 12 Hz on a floor with a natural frequency of 15 Hz and 4% damping. What is the magnification factor?
Practice
For the same system (β = 0.8, ζ = 0.04), what is the phase lag in degrees?
Practice
At what frequency ratio β does the magnification factor peak, for a system with 10% damping?
Check yourself
At resonance, which force balances the applied force?
Practice
What is the magnification factor exactly at resonance for a system with 4% damping?
Practice
A machine runs at 80% of a floor's natural frequency. With 5% damping, what is the magnification factor?
Check yourself
At what frequency ratio does the true maximum of the magnification curve occur, for a lightly damped system?
Worked example
An aerobics class on a gym floor
Given
- A gym floor with a vertical natural frequency of 4.6 Hz
- Damping 3%, typical of a bare composite floor
- Aerobics at 2.3 Hz; the second harmonic of the loading is at 4.6 Hz
Find
The amplification, and what a redesign would have to achieve.
Assumptions
- Single dominant floor mode; harmonic idealisation of the rhythmic loading
Worked example
What damping actually buys at resonance
Given
- A structure driven exactly at its natural frequency
- Damping considered at 1 %, 2 % and 5 % of critical
Find
The dynamic amplification in each case
Worked example
Reading damping off a frequency response curve
Given
- A shaker test produces a response curve with a clear peak
- The peak is at 3.20 Hz
- The response falls to 1/√2 of the peak at 3.10 Hz and 3.29 Hz
Find
The damping ratio, by the half-power method
Summary
- Rd = 1/√[(1 − β²)² + (2ζβ)²], and Rd = 1/(2ζ) exactly at β = 1
- At resonance spring and inertia forces cancel and damping carries the whole load
- Phase is 0° below, 90° exactly at resonance, and 180° well above
- The 90° phase crossing is the sharpest experimental marker of a resonance
- The peak sits at β = √(1 − 2ζ²) and vanishes above ζ = 1/√2
- Resonance takes about 1/(2πζ) cycles to build — 16 at 1% damping
- Check the harmonics of a periodic load, not only its fundamental
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint