A small topic with predictable questions. One spring–mass system covers most of the marks; the ship-specific part is knowing what excites a hull and at what frequency.
Displace a mass on a spring and release it. The spring pulls it back, it overshoots, and it oscillates at a frequency set entirely by how stiff the spring is and how heavy the mass is.
Everything else in this chapter is that equation wearing different clothes. For torsional vibration, replace stiffness by torsional stiffness and mass by polar moment of inertia:
Two practical points that turn up as questions. First, stiffer means faster, heavier means slower — and both go as a square root, so quadrupling the stiffness only doubles the frequency. Second, a structure in water carries added mass: the water that must be accelerated with it. For a hull this is comparable to the ship's own mass, which is why a hull's natural frequency in water is far below its frequency in air, and why a dry calculation is useless.
Natural frequency is a property of the structure, not of what is shaking it. Excitation decides whether it responds; the structure decides at what frequency it wants to.
Real systems lose energy. Damping is measured as a fraction of the amount that would just stop oscillation altogether.
Marine structures are lightly damped — typically 1 to 5 % — which is why resonance matters so much on board.
The practical way to measure it is from a decay trace after an impulse. The logarithmic decrement is the natural log of the ratio of successive peaks:
Damping barely changes the natural frequency — at ζ = 0.05, ωd is 99.9 % of ωn. What it changes enormously is the amplitude at resonance, which is set by damping alone.
Apply a sinusoidal force and the response depends on how the forcing frequency compares with the natural frequency. Write r = ω/ωn.
At resonance, the amplification is 1/(2ζ). At 5 % damping that is a factor of ten; at 1 % it is fifty. This is why a small exciting force can shake a whole after body apart if it happens to land on a mode.
The same expression governs transmissibility through a resilient mount. Isolation only begins above r = √2; below that, mounts amplify rather than isolate. Fitting softer mounts to a machine that already runs below its mount frequency makes matters worse, and it is a standard conceptual trap.
Three sources, at three distinct frequencies, and identifying which is which is most of the diagnostic skill.
The propeller, at blade rate. Each blade passes through the low-velocity wake peak above the disc once per revolution, so the hull is excited Z times per revolution, plus harmonics.
There are two paths: pressure pulses through the water onto the hull above the propeller, and bearing forces through the shaft. Skew, increased tip clearance and a better-conditioned wake all reduce the exciting force; changing the blade number moves its frequency.
The main engine, at engine order — unbalanced primary and secondary forces and moments from a large slow-speed engine, dealt with by balancers and, for the guide force moments, by top bracing to the hull.
The sea, at wave encounter frequency — which excites hull girder modes directly (springing) and by impact (whipping after a slam).
The hull's own modes are numbered by their nodes: the two-node vertical mode is the fundamental, typically 1 to 3 Hz on a large ship, with three-node and four-node modes above it. Design practice is to keep blade rate and its harmonics clear of these across the operating range — which is exactly the check in worked example 2.
A structure loaded repeatedly can crack at stresses far below yield. The relationship between the applied stress range and the number of cycles it survives is the S–N curve, plotted log–log, where it is close to a straight line.
An inverse slope of 3 means life falls with the cube of stress range: double the stress and life drops to one eighth. That is why a small number of large cycles dominates the damage even when the count is overwhelmingly made up of small ones.
Real service applies a spectrum, not one stress level. Miner's rule adds up the fraction of life used at each:
Two limitations belong in any full answer: Miner's rule ignores the order in which cycles are applied, and observed failures scatter between D ≈ 0.5 and D ≈ 2. It sizes a detail; it does not predict a date.
The detail itself matters more than the material. A stress concentration factor Kt multiplies the nominal stress before it enters the curve, and crack growth rates in steel are almost independent of yield strength — so high-tensile steel, used to save weight, gives a thinner structure with higher nominal stresses and no fatigue benefit at all.
One spring–mass problem, one resonance check of the kind a designer actually performs, and one fatigue assessment.
A machine of mass 500 kg is mounted on resilient mounts of total stiffness 2.0 MN/m. A decay trace after an impulse shows successive amplitudes of 10.0 mm and 7.5 mm. Find the undamped natural frequency, the damping ratio, and the transmissibility at a forcing frequency of 20 Hz.
m = 500 kg, k = 2.0×10⁶ N/m Successive amplitudes 10.0 mm and 7.5 mm Forcing frequency 20 Hz
Find the undamped natural frequency, the damping ratio, and the transmissibility at a forcing frequency of 20 Hz
Undamped natural frequency.
This one equation is the backbone of the whole topic.
Damping from the decay trace.
The logarithmic decrement is the natural log of the ratio of successive peaks.
Convert to damping ratio.
For light damping δ ≈ 2πζ, but the exact relation costs nothing.
Damped natural frequency — barely different, which is typical.
Light damping changes the frequency by a fraction of a percent while changing the response at resonance enormously.
Frequency ratio at the forcing frequency.
Transmissibility.
Above r = √2 the mounts isolate; below it they amplify.
So only 34 % of the exciting force reaches the seat.
Note what would happen at resonance instead: r = 1 gives T = 1/(2ζ) = 10.9, an elevenfold amplification — with the mounts making things ten times worse rather than three times better.
Answerf_n = 10.07 Hz, ζ = 4.6 %, transmissibility at 20 Hz = 0.34
The trap: assuming soft mounts always help. They isolate only above r = √2 ≈ 1.41. Fit soft mounts to a machine running below that and you have built a resonant system, which is how a well-intentioned vibration fix makes the problem worse.
A ship's main engine operates between 60 and 105 rpm driving a five-bladed propeller. The hull's two-node vertical natural frequency is 4.2 Hz and its three-node frequency is 8.0 Hz. Check for resonance, and say what could be done about anything found.
Operating range 60–105 rpm Z = 5 blades Hull 2-node 4.2 Hz, 3-node 8.0 Hz
Check for resonance, and say what could be done about anything found
The propeller excites the hull once per blade passing the wake peak above the disc.
So the exciting frequency is shaft speed multiplied by the number of blades, not shaft speed itself.
Find the blade-rate frequency at each end of the operating range.
Check the two-node mode.
Resonance would need 4.2 Hz, which corresponds to a shaft speed below the operating range — so the ship passes through it only when starting away, briefly.
Check the three-node mode.
This one is a problem: it falls squarely inside the normal operating band.
Consider the remedies in order of cost.
Changing the blade number shifts the excitation without touching the hull.
So a four-bladed propeller moves the three-node resonance clear above the operating range.
If the propeller cannot be changed, the alternatives are to stiffen the after body (raising the hull frequency), to increase the tip clearance and add skew (reducing the exciting force rather than moving it), or as a last resort to bar the speed range around 96 rpm.
One more check before finishing.
The second harmonic of blade rate, at 2Zn, also excites — worth confirming it does not land on a mode.
AnswerThe three-node mode resonates at 96 rpm, inside the operating range; four blades moves it to 120 rpm
The trap: using shaft rate instead of blade rate. At 96 rpm the shaft turns at 1.6 Hz, nowhere near 8 Hz — the resonance is invisible unless you multiply by the number of blades.
A structural detail experiences, in one year of service, 2×10⁴ cycles at a stress range of 200 MPa, 1×10⁵ cycles at 150 MPa and 5×10⁵ cycles at 100 MPa. The S–N curve for the detail gives 2×10⁶ cycles to failure at 100 MPa with an inverse slope of 3. Estimate the fatigue life.
Block (one year): 200 MPa × 2×10⁴ cycles 150 MPa × 1×10⁵ cycles 100 MPa × 5×10⁵ cycles S–N: N = 2×10⁶ at 100 MPa, N ∝ S⁻³
Estimate the fatigue life
Write the S–N curve in a usable form.
An inverse slope of 3 means life falls with the cube of stress range — double the stress and life drops by a factor of eight.
Cycles to failure at each stress range.
| Stress range S (MPa) | (100/S)³ | N = 2×10⁶ × (100/S)³ |
|---|---|---|
| 200 | 0.1250 | 2.50 × 10⁵ |
| 150 | 0.2963 | 5.93 × 10⁵ |
| 100 | 1.0000 | 2.00 × 10⁶ |
Miner's rule sums the fraction of life used at each stress level.
| S (MPa) | Cycles applied n | Cycles to failure N | Damage n/N |
|---|---|---|---|
| 200 | 2.0 × 10⁴ | 2.50 × 10⁵ | 0.080 |
| 150 | 1.0 × 10⁵ | 5.93 × 10⁵ | 0.169 |
| 100 | 5.0 × 10⁵ | 2.00 × 10⁶ | 0.250 |
| D = | 0.499 |
Each fraction is cycles applied over cycles to failure.
Failure is predicted when the accumulated damage reaches unity.
Look at where the damage came from.
| S (MPa) | Share of cycles | Share of damage |
|---|---|---|
| 200 | 3 % | 16 % |
| 150 | 16 % | 34 % |
| 100 | 81 % | 50 % |
Because that is what the question is really testing. The 200 MPa cycles are only 3 % of the total count but contribute 16 % of the damage; the 100 MPa cycles are 81 % of the count and contribute half. The cube law means a few large cycles matter far more than their number suggests.
Two caveats worth a sentence in an answer.
Miner's rule takes no account of the order in which the cycles are applied, and real structures often fail at D between 0.5 and 2. It is a design tool, not a prediction.
AnswerD ≈ 0.50 per year, so the estimated fatigue life is about 2 years
The trap: using stress amplitude instead of stress range. S–N curves for welded details are plotted against range — amplitude is half of it, and with a cube law that error changes the predicted life by a factor of eight.
ω_n = √(k/m), f_n = ω_n/2πTorsional: ω_n = √(k_t/J)c_c = 2√(km), ζ = c/c_cMarine structures: ζ ≈ 0.01–0.05δ = ln(x₁/x₂) ≈ 2πζLogarithmic decrement from a decay traceω_d = ω_n√(1 − ζ²)Barely different from ω_n at light dampingMF = 1/√((1−r²)² + (2ζr)²)At r = 1, MF = 1/(2ζ)Isolation only for r > √2Below it, mounts amplifyf_blade = (N/60)·ZBlade rate — not shaft rateN = C·S^(−m), m ≈ 3Life falls with the cube of stress rangeD = Σ nᵢ/Nᵢ = 1Palmgren–Miner; scatter between 0.5 and 2 in serviceσ_max = K_t·σ_nomFatigue lives at the detail, not the material