- #1
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.
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.