Calculate Rotational Speed of Steel Rotor Disk

[OutCell]

A steel rotor disk which is part of a turbine assembly has a uniform thickness of 50mm, an outside diameter of 650mm and an inside diameter of 100mm. If there are 250 blades, each of mass 0.16 kg pitched evenly around the periphery of the disk at an effective radius of 350mm, determine the rotational speed at which the hoop stress on the inner surface reaches the yield stress. You may assume that the blades produce a uniform loading around the periphery of the disk.

For steel, poisson's ratio = 0.29; Density = 7470 kg/m^3 and the yeild stress in simple tension = 355 MN/m^2.



I don't know what equations to use? Is rotational speed = w or T? I don't know where to start with finding a way to solve this question?

Can anybody please guide me through the steps in which i can get this speed? The problem is that this question is one of optional question that we didn't get any lecture notes on, so i don't know where to start

Thanks

 

[OutCell]

After some reading i got 2 ways which i think one might be the answer (Or maybe not lol).. Both different answers..

1st way:

I used the equation:
(Sigma)r = (3+possion's ratio/8) x Density x w^2 x r2^2

and after substituting the variable in the equation i got an answer
w= 6.67 x10^-3 rad/sec

2nd way:
I read this in some notes about disc with a loaded boundary-

"In general, rotor discs will have mounted on the outer boundary a large number of blades. These will themselves each have a centrifugal force component which will have to be reacted at the periphery of the disc. Given the mass of each blade, it's effective centre of mass and the number of blades we can compute the force due to each blade at a particular value of w. Multiplying by the number of blades gives the total force which may then be computed as a uniformly distributed load. Dividing this by the thickness of the outer boundary gives the required value of (Sigma)r to use as the boundary condition when evaluating A and B"

Now i went backwards and considered the yeild stress (355x10^6) as (Sigma)r, so-
I multiplied the thickness by (Sigma)r
50 x (355x10^6) = 1.775 x 10^10 which would hopefully be the uniformly distributed load

And then i divided this by the number of blades to get the w
(1.775x10^10)/250 = 71 x 10^6 rad/sec


Can someone please take a look and tell me if any of these are right, or help me out in the right direction

THANKS :shy: