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Unit 9 · Lesson 9-2
Rotor Dynamics, Disks and Containment
1h 30m1300 XP
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Rotor Dynamics, Disks and Containment

The Jeffcott rotor and critical speeds, squeeze-film dampers and rotor modes, rotating-disk stresses and burst margin, and blade containment

United 232, 19 July 1989
A DC-10 flying from Denver to Chicago loses its tail-mounted engine: the stage-1 fan disk breaks apart. The DC-10's three hydraulic systems are each powered by one engine; debris penetrates the tail and severs the lines of the other two. With no flight controls, the crew — joined by a training check pilot travelling as a passenger, a throttle in each hand — steer with the two wing engines alone, and crash-land at Sioux City. Of the 296 people on board, 184 survive (United Airlines Flight 232). The cause was a fatigue crack that had grown from an undetected metallurgical defect in the titanium-alloy disk. This lesson is about rotors that whirl, and the disks that must never break.
In 1869 W. J. M. Rankine analysed a spinning shaft and concluded that it could not run above its critical speed. In 1889 Gustaf de Laval ran a steam turbine above it anyway. The conflict lasted until Henry Jeffcott, commissioned by the Royal Society, published his analysis in 1919 (rotordynamics). His model — a disk on a flexible, massless shaft, its mass centre a small distance ee from its geometric centre — still opens every course on the subject.
re=(Ω/ωn)2[1−(Ω/ωn)2]2+(2ζ Ω/ωn)2,tan⁡ϕ=2ζ Ω/ωn1−(Ω/ωn)2,ωn=k/m\frac{r}{e} = \frac{(\Omega/\omega_n)^2}{\sqrt{\left[1 - (\Omega/\omega_n)^2\right]^2 + \left(2\zeta\,\Omega/\omega_n\right)^2}}, \qquad \tan\phi = \frac{2\zeta\,\Omega/\omega_n}{1 - (\Omega/\omega_n)^2}, \qquad \omega_n = \sqrt{k/m}

The Jeffcott rotor: whirl radius r of the disk centre, and the angle φ by which it lags the mass centre, at spin speed Ω; ζ is the damping ratio.

Derivation
The Jeffcott rotor
Write the disk centre's position in the plane as a complex number zz. The mass centre is at z+e eiΩtz + e\,e^{i\Omega t}, the shaft pulls the disk back with −kz-kz and damping resists with −cz˙-c\dot z:
m d2dt2(z+e eiΩt)=−kz−cz˙⟹mz¨+cz˙+kz=me Ω2eiΩt.m\,\frac{d^2}{dt^2}\left(z + e\,e^{i\Omega t}\right) = -kz - c\dot z \quad\Longrightarrow\quad m\ddot z + c\dot z + kz = m e\,\Omega^2 e^{i\Omega t}.
Try z=r ei(Ωt−ϕ)z = r\,e^{i(\Omega t - \phi)}: the steady whirl radius and lag above follow, with ζ=c/(2km)\zeta = c/(2\sqrt{km}). At Ω=ωn\Omega = \omega_n only damping limits the orbit, r=e/(2ζ)r = e/(2\zeta). Far above it r→er \to e and ϕ→180°\phi \to 180°: the disk spins about its own mass centre — the self-centring Rankine missed.
📐Through the critical speed
A 100 kg disk on a shaft of stiffness 2×1072\times10^7 N/m — critical speed 4271 rpm — balanced to e=20e = 20 µm (Lab 1):
  • Bare bearings, ζ=0.02\zeta = 0.02: the orbit peaks at 25 e = 500 µm, and each bearing feels 10 kN — enough to rub away tip clearances.
  • A squeeze-film damper, ζ=0.20\zeta = 0.20: 51 µm and 1.1 kN.
  • At twice the critical speed the orbit is back to 1.33 e, the high spot 176° behind the heavy spot.
💡Squeeze films, soft supports and rotor modes
A squeeze-film damper is a thin oil film around a bearing's outer race, held in a cage of springs: as the rotor whirls it squeezes the oil and dissipates energy. Engines combine it with deliberately soft supports, so that the rotor's first modes are rigid-body motions — bouncing and rocking on its bearings — at low speeds, passed quickly during start-up while the dampers soak up the energy. The rotor's own bending modes stay far above the running range. A mode's share of strain energy in the bearings tells which kind it is.
📐Stiff bearings or soft supports
A finite-element model of an HP rotor, its natural frequencies from scipy.linalg.eigh (Lab 3), running range 9000–15,000 rpm:
  • Soft supports: modes at 3371 and 6131 rpm, with 99–100 % of their strain energy in the supports — rigid-body modes below idle, where the dampers work — and the first bending mode at 105,000 rpm, with 1 %.
  • Stiff bearings: the first mode at 19,400 rpm, with its energy shared between rotor and bearings — above maximum speed, but with no damper able to reach it.
Derivation
Stresses in a rotating disk
A disk of thickness h(r)h(r) spinning at Ω\Omega: radial equilibrium of a ring element and plane-stress elasticity, with radial displacement uu, give
ddr(h r σr)−h σθ+ρ Ω2r2h=0,σθ=E ur+ν σr,dudr=(1−ν2) σrE−ν ur.\frac{d}{dr}\left(h\,r\,\sigma_r\right) - h\,\sigma_\theta + \rho\,\Omega^2 r^2 h = 0, \qquad \sigma_\theta = \frac{E\,u}{r} + \nu\,\sigma_r, \qquad \frac{du}{dr} = \frac{(1 - \nu^2)\,\sigma_r}{E} - \frac{\nu\,u}{r}.
The bore is free (σr=0\sigma_r = 0); at the rim the blades pull outwards. Lab 2 shoots on the bore displacement to meet both. For a constant-thickness disk with a bore, the hoop stress at the bore is exactly
σθ(ri)=3+ν4 ρ Ω2R2[1+1−ν3+ν ri2R2],\sigma_\theta(r_i) = \frac{3 + \nu}{4}\,\rho\,\Omega^2 R^2\left[1 + \frac{1 - \nu}{3 + \nu}\,\frac{r_i^2}{R^2}\right],
highest of anywhere in the disk and proportional to ρ Ω2R2\rho\,\Omega^2 R^2: the square of the rim speed.
📐Why turbine disks have fat hubs
A nickel-alloy HP turbine disk, 250 mm in radius, at 13,000 rpm with 70 blades (Lab 2):
  • Constant thickness (75.8 kg): bore hoop stress 923 MPa, above the 800 MPa allowed here for fatigue life; burst at 1.44 times running speed.
  • Tapered, four times thicker at the hub (166.2 kg): 593 MPa; burst at 1.70 times.
Both clear the certification rule that a rotor must not burst in 5 minutes at 120 % of its maximum permissible speed (14 CFR 33.27); the hub buys fatigue life.
💡Blades are contained; disks are not
Certification requires an engine to contain the failure of its most critical compressor or fan blade, and of its most critical turbine blade, at maximum permissible speed: it must keep running for at least 15 seconds without catching fire or breaking its mounts (14 CFR 33.94). Engine cases are not designed to contain failed disks: the fragments carry too much energy. Disks are instead treated as safety-critical parts, designed, inspected and retired so that they never fail (turbine engine failure).
You want to be particular and make it a runway, huh?
– Captain Al Haynes, United Airlines Flight 232, after Sioux City Approach cleared him to land on any runway, 19 July 1989 (Wikipedia)
⚠Southwest 1380
On 17 April 2018 a fan blade of a Boeing 737's CFM56-7B broke at a latent fatigue crack. The fan case was not penetrated — the blade was contained — but the impact broke the cowling latch, and cowling parts struck the wing and broke a cabin window. One passenger died in the decompression (Southwest Airlines Flight 1380). Containment covers the engine; the nacelle around it matters too.
💡Where you will use this
The same critical-speed maps, damper design and disk analysis apply to every turbine, pump and electric motor that spins fast. Lesson 9-3 follows the rotor's loads out through the frames and mounts.
📚Further reading
  • John M. Vance, Fouad Y. Zeidan and Brian Murphy, Machinery Vibration and Rotordynamics (Wiley, 2010).
  • Dara W. Childs, Turbomachinery Rotordynamics (Wiley, 1993).
  • S. P. Timoshenko and J. N. Goodier, Theory of Elasticity (3rd ed., McGraw-Hill, 1970), on rotating disks.
↓But there is a problem
The rotors are held and the disks hold. But every load — thrust, weight, gyroscopic moments, a released blade — still has to travel out of the engine and into the wing. On 4 October 1992, over Amsterdam, two engines left a Boeing 747.