Disc reach ultimate material strength by spinning too fast

In summary, according to the article, a A380 aircraft went down due to a speed limiter on a spinning engine rotor breaking, causing the material to shatter. The part was made of a nickel alloy, but since I don't know exactly what it is let's use steel. I simplified the shape and assumed it was rigid untile its ultimate strength was reached. The disk had an outer radius of 1.5m, an inner radius of 5cm, and a thickness of 5cm. The pressure at the inner surface was calculated to be 79264.25 w^2. The speed at which the disk broke was calculated to be 263 rad/sec.
  • #1
swraman
167
0
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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  • #2
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?"
 
  • #3
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
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:
  • #5
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.
 

Related to Disc reach ultimate material strength by spinning too fast

1. What is the "disc reach ultimate material strength by spinning too fast" concept?

The concept refers to the idea that a disc or other rotating object can reach its ultimate material strength when spun at a high enough speed.

2. How does spinning affect the strength of a disc?

When a disc is spun at high speeds, the centrifugal force exerted on the material causes it to become denser and more compact. This increases the material's rigidity and strength, allowing it to withstand higher levels of stress without breaking.

3. Is there a limit to how fast a disc can spin before reaching its ultimate strength?

Yes, there is a limit to how fast a disc can spin before reaching its ultimate strength. This limit is determined by the material properties of the disc and the structural integrity of the disc itself. Exceeding this limit can lead to structural failure and breakage of the disc.

4. Can this concept be applied to other objects besides discs?

Yes, this concept can be applied to other rotating objects such as wheels, gears, and turbines. As long as the material and structure of the object can withstand the high speeds, it can reach its ultimate strength through spinning.

5. Is there any practical application of this concept?

One potential application of this concept is in the manufacturing of high-strength materials. By subjecting materials to high rotational speeds, it is possible to increase their strength and durability, making them suitable for use in industries such as aerospace and construction.

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