Burst speed of rotating disk 1

[Wasdifferent]

I am looking for the equations for hoop stress and radial stress in a rotating disk, to calculate the theoretical burst speed. I am not looking for the more complex empirical solutions having to do with fatigue, balance, vibration, temperature, etc...

A link would be great.

I googled, but found WAY too much information, and wasn't able to find the simple equations that I was looking for.

Thank you very much,

-Michael

http://www.sewellgeneralcontracting.com

ALBANY SCHENECTADY TROY NY

 

[unclesyd]

I have run across this information while checking out flywheels. Here is a good beginning.

http://www.upei.ca/~physics/p261/projects/flywheel1/flywheel1.htm

 

[iainuts]

Roark's covers this (7'th edition) page 752. I've never used it before so I'd have to delve into it a bit more before giving the equations, but if you'd like I can send you the relavent pages (either by email or posted into your thread @ physicsforums <which is down right now> )

 

[Wasdifferent]

unclesyd,
Thank you for the link, but it is one of the 100 bzillion pages that I visited on google during the past 3 days with no equation...



-Michael

http://www.sewellgeneralcontracting.com

ALBANY SCHENECTADY TROY NY

 

[eromlignod]

V = sqrt(10 * S)

where
V = circumferential speed in ft/s
S = tensile strength of flywheel material in psi

Don
Kansas City

 

[Wasdifferent]

iainuts,
Exactly what I needed.
Thank you soooo much!

eromlignod,
Not what I was looking for, but maybe even better. It sounds like a "rule of thumb" formula with a built in safety factor. I'm interested in knowing where it comes from.
Thank YOU very much,

-Michael

http://www.sewellgeneralcontracting.com

ALBANY SCHENECTADY TROY NY

 

[eromlignod]

This formula was from Machinery's Handbook, 23rd ed., p. 191.

The text reads:

"The bursting velocity of a flywheel, based on hoop stress alone (not considering bending stresses), is related to the tensile stress in the flywheel rim by the following formula which is based on the centrifugal force formula from mechanics."

Don
Kansas City

 

[Philrock]

V = sqrt(10 * S)
where
V = circumferential speed in ft/s
S = tensile strength of flywheel material in psi

- cannot be correct in general. It might be right for a particular material, but it has no term for the material density, which is a critical factor for the calculation.

 

[CoryPad]

According to Ashby in Materials Selection in Mechanical Design, the maximum principal stress in a spinning disk is:

[&sigma;][sub]max[/sub] = [(3+[&nu;])/8][&rho;]R[sup]2[/sup][&omega;][sup]2[/sup]

where

[&sigma;][sub]max[/sub] = maximum principal stress
[&nu;] = Poisson's ratio of disk material
[&rho;] = mass density of disk material
R = disk radius
[&omega;] = angular velocity

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[eromlignod]

The hoop stress depends only on the rim speed and is essentially independent of the rim radius. If the radius doesn't matter, then neither does the material density. Only the material strength is used in the calculation.

S is the tensile strength of the material. Class 30 cast iron, for example, would have an s of around 30 ksi, so the critical rim speed would be about

V = sqrt(10 * 30,000) = 548 ft/s

for that material. Unless you need to know *precisely* when the wheel will burst, this is close enough for design work once you have figured in an appropriate safety factor.

Don
Kansas City

 

[metengr]

thread406-50738

 

[GregLocock]

Philrock's equation definitely makes an assumption about nu and rho. I'm not going to work out what that assumption is.

Ashby's equation reduces to a rim speed as well, if that's what you like to think about.

?max = [(3+?)/8]?R2?2

V=2*pi*R*?

so
?max = [(3+?)/8]?V^2/4/pi^2

so

V=sqrt(S/([(3+?)/8]?))
V = sqrt(10 * S)



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[prex]

Well GregLocock, didn't you go a bit too fast?

V=R[&omega;]
[&sigma;][sub]max[/sub]=(3+[&nu;])[&rho;]V[sup]2[/sup]/8
V=[&radic;](8S/((3+[&nu;])[&rho;]))
Now, using a sane system of units, we have
[&rho;]=11360 kg/m[sup]3[/sup] for lead
and with S in Pa, V will be in m/s.
Using S in psi and to get V in ft/s, the factor under the square root becomes (also assuming [&nu;]=0.33, that's not really important):
8 x 6894.8 / (3.33 x 11360 x 0.3048[sup]2[/sup])=15

So, if I'm not mistaken, that formula contains a factor of safety of about 1.2 ([&radic;]1.5) for lead, is unsafe for gold , is excessively safe for steel or cast iron, good candidates for flywheels, and is unrealistic for aluminum (but who would propose a flywheel in aluminum?).

And of course the bursting speed does depend on density!

prex

http://www.xcalcs.com

Online tools for structural design

 

[eromlignod]

My appologies to everyone. I shot my mouth off a little too soon. The formula I posted was indeed for steel/cast iron specifically, which is what most flywheels are made of. For materials of a different density (a grinding wheel, for example), the longer formula must be used.

I'm not sure why you refer to your metric units as "sane" units in this case, since you needed a calculator to work that formula anyway. How is that easier?

Don
Kansas City

 

[CoryPad]

Don,

I don't think it is very sane to mix pounds per square inch and feet per second, when the handy meter is available for both measurements (as well as every other length measurement).

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[zekeman]

What about a disc with a hole at the center?
According to DenHartog,"Advanced Strength of Materials", the tangential stress at the hole edge would be double that quoted by Coreypad for any size hole at the center or
2*rho*w^2*(3+nu)*R^2/8
which looks surprising, in that it is independent of the hole diameter, including a tiny one.