# Trying to find the Tensile Pressure in a Rotating Disk

1. Oct 15, 2012

### axemaster

OK, so I'm trying to build a hyper-fast rotating disk, probably of aluminum. The tensile strength is about 200MPa (note that this is a pressure, which makes sense), and I'm trying to calculate the tensile pressure throughout the disk. The point being to find out how fast I can spin the thing before it tears apart due to the centifugal force. To phrase it more succinctly:

A disk of radius $R$ and mass density $ρ$ rotates about its axis with angular frequency $ω$. Find the tensile pressure at radius $r_o$.

So I started out by finding the infinitesimal outward force exerted by successive rings of material:

$v=ωr$
$a=v^{2}/r$

→ $a=ω^{2}r$

And the mass of each ring is:

→ $dm=2\pi rh*ρ*dr$ ---------- note that "h" is the thickness, included to make it a volume

The infinitesimal force on each ring segment is then:

$dF=dm*a$

→ $dF=2\pi r^2 ω^2 h ρ dr$

Now I integrated the force from the radius of interest $r_o$, to the outer radius "R". This should give us the total force the material must withstand at radius $r_o$.

$F=\int^{R}_{r_o} 2\pi r^2 ω^2 h ρ dr$

$F=2\pi ω^2 h ρ (\frac{R^3}{3}-\frac{r^{3}_{o}}{3})$

When plotted in Mathematica (setting all variables to 1 and varying $r_o$) this looks like:

This appears to be correct. As one might expect, the force starts at zero on the outer edge and increases to a more or less constant value at the center.

The trouble starts when I tried to convert this into a pressure by dividing by the area at $r_o$.

→ $A=2\pi r_o h$

→ $Pressure=F/A=\frac{ω^2 ρ (\frac{R^3}{3}-\frac{r^{3}_o}{3})}{r_o}$

As you can see, the pressure goes to infinity at the center. This is clearly wrong, but I'm not sure how to fix it. Please help!

2. Oct 15, 2012

### Studiot

If you want to consider successive rings you must consider the hoop (circumferential) stress in each ring cause by the rotation.

To do this you balance the centrifugal force against twice the hoop force for a semi ring.
The ring bursts across a diameter, cutting two sections of ring. The free body diagram of the semi ring has the outwards centrifugal pressure colinear with but of opposite direction to the hoop stresses in that cut semi ring.

Divide this by the ring cross section to obtain the bursting stress.

Alternatively for thin disks you can solve the differential equation

r(d2σr/dr2)+3(dσr/dr) = -(3+γ)ρω2r

I am sorry I do not have my usual facilities for drawings or tex.

Last edited: Oct 16, 2012
3. Oct 16, 2012

### Enthalpy

Pity, because in a ring both the radial AND the circumferential stress contribute to resist the centrifugal force, and worse, they interfere also through Poisson's number.

The circumferential stress is what keeps the stress finite at the center, but both stresses act.

If you read German the result is in Dubbel but I don't remember if the reasoning is included.
http://www.amazon.de/Dubbel-Taschenbuch-für-den-Maschinenbau/dp/3540221425
wait, it has been translated into English!
https://www.amazon.com/Handbook-Mechanical-Engineering-W-Beitz/dp/0387198687
try to get one, not just for the present purpose. Got mine on eBay.

4. Oct 16, 2012

### axemaster

You have given me some very good advice, I think I know how to solve this now. I'll get back to you in a bit with my results...