# Disc reach ultimate material strength by spinning too fast

Tags:
1. Sep 24, 2015

### swraman

So, Im 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 gotta be one really fast spinning rotor! But how fast? It was made of a nickel alloy, but since I dont know exactly what it is lets 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. Lets 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
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 thats a very reasonable RPM.

Thanks.

2. Sep 25, 2015

### Bystander

Where are the largest forces?
"Reasonable?"

3. Sep 25, 2015

### swraman

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.

4. Sep 25, 2015

### Nidum

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: Sep 25, 2015
5. Sep 25, 2015

### SteamKing

Staff Emeritus
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.