Disc reach ultimate material strength by spinning too fast

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SUMMARY

The discussion centers on the catastrophic failure of an A380 engine rotor due to exceeding its ultimate yield strength, specifically referencing a failure in the speed limiter. The rotor, made from a nickel alloy, reached a calculated speed of 263 rad/sec, which was deemed reasonable compared to typical operating speeds of similar engines, such as the Trent series. The calculations involved determining the mass and forces acting on the rotor, ultimately leading to a pressure that equaled the ultimate stress of the material. The incident was investigated by Australian authorities, revealing a manufacturing defect in the Rolls-Royce engine, resulting in significant repairs but no loss of life.

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swraman
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So, I am watching this show about airline crashes and apparently, a A380 has gone down because the speed limiter on a spinning engine rotor broke, and the part spun so fast its ultimate yield strength was reached and the material shatterted.

Thats got to be one really fast spinning rotor! But how fast? It was made of a nickel alloy, but since I don't know exactly what it is let's use steel. Its safe to assume that the part was lighter and stronger than steel, so this will be an underestimate. I also am going to simplify the shape, assume It is rigid (undergoes no elastic or plastic deformation) untile its ultimate strength is reached.

5500MPa and a material density of 7.8g/cm^3. Let's say for simplicity it is a perfect disk of outer radius 1.5m, inner radius 5cm, thickness 5cm . How would you go about calculating the speed at which it hss to spin to break?

Here's what I did, but I get an answer that makes no sense in the end. Can you see where I am going wrong, or is my entire plan of attack wrong?

t = .05 % thickness
r1 = .05 % inner radius
r2= 1.5 % outer radius
rho = 7800 % density
u = 5500000000% N/m^2

The mass of the disk:
m = rho * (r2^2 - r1^2) * t * 2*pi

I assume initial fracture would be at the inner radius, so the effective area:
area = 2 * pi * r1 * t;

force to keep the mass at bay:
F = .5 * (wr)^2 * dm
= .5 * (wr)^2 * rho * t * 2 * pi * dr

Integrating over r on the range r1 to r2:
F = (pi/3) * w^2 * (r2-r1)^3 * rho * t

So the pressure at the inner surface:
P = F/Area
P = 79264.25 w^2

Set that equal to ultimate stresss, solve for speed:
5500000000 = 79264.25 w^2

And I get 263 rad/sec, which I am assuming is wrong, since that's a very reasonable RPM.

Thanks.
 
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swraman said:
outer radius 1.5m
swraman said:
initial fracture would be at the inner radius
Where are the largest forces?
swraman said:
263 rad/sec ... ... reasonable RPM.
"Reasonable?"
 
The largest forces would be at the inner circle, wouldn't they?

An infinitesimal area at the inner circle has a mass of (m = rho * t * (r2-r1) * dTheta) that is trying to escape due to its inertial, with the radial force of (F = .5 * (wr)^2 * rho * t * dTheta * dr)

By reasonable, I mean that it would be considered a normal operating speed, so my calculation is obviously wrong.
 
From published information :

Sample 100% reference speeds for a Trent series engine :

LP spool N1 2700 rpm
IP spool N2 8200 rpm
HP spool N3 12600 rpm

Note that actual running speeds are less - usually in the range 94 to 98 % .
 
Last edited:
swraman said:
So, I am watching this show about airline crashes and apparently, a A380 has gone down because the speed limiter on a spinning engine rotor broke, and the part spun so fast its ultimate yield strength was reached and the material shatterted.

I think the incident you are describing only damaged the A380 aircraft, which was able to land safely. Since the aircraft was owned by Qantas, the Australian authorities performed an investigation of this incident, and a report of findings was prepared:

https://web.archive.org/web/2015011...ia/2888854/ao-2010-089 preliminary report.pdf

https://en.wikipedia.org/wiki/Qantas_Flight_32

As a result of the investigation, the cause of the engine failure was determined to be a manufacturing defect in one of the Rolls-Royce engines. This particular aircraft underwent some $145 million in repairs and was returned to service. There was no loss of life.
 

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