Trying to find the Tensile Pressure in a Rotating Disk

In summary, the conversation discusses the process of calculating the tensile pressure throughout a rotating disk in order to determine its maximum spinning speed before tearing apart due to centrifugal force. The conversation involves finding the infinitesimal outward force exerted by successive rings of material, integrating the force from the radius of interest to the outer radius, and converting this into pressure by dividing it by the area at the radius of interest. However, the pressure becomes infinite at the center, indicating a problem in the calculation. Suggestions are given to consider the hoop stress in each ring and solve a differential equation to accurately determine the bursting stress. The conversation concludes with a recommendation to read the book Dubbel for further understanding.
  • #1
axemaster
59
7
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 [itex]R[/itex] and mass density [itex]ρ[/itex] rotates about its axis with angular frequency [itex]ω[/itex]. Find the tensile pressure at radius [itex]r_o[/itex].

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

[itex]v=ωr[/itex]
[itex]a=v^{2}/r[/itex]

→ [itex]a=ω^{2}r[/itex]

And the mass of each ring is:

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

The infinitesimal force on each ring segment is then:

[itex]dF=dm*a[/itex]

→ [itex]dF=2\pi r^2 ω^2 h ρ dr[/itex]

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

[itex]F=\int^{R}_{r_o} 2\pi r^2 ω^2 h ρ dr[/itex]

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

When plotted in Mathematica (setting all variables to 1 and varying [itex]r_o[/itex]) this looks like:

forcecurve2_zps84100d69.jpg


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 [itex]r_o[/itex].

→ [itex]A=2\pi r_o h[/itex]

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

pressure_zps217ec2ea.jpg


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!
 
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  • #2
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:
  • #3
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.com/dp/3540221425/?tag=pfamazon01-20
wait, it has been translated into English!
https://www.amazon.com/dp/0387198687/?tag=pfamazon01-20
try to get one, not just for the present purpose. Got mine on eBay.
 
  • #4
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...
 
  • #5


I would first commend you on your approach to calculating the tensile pressure in your rotating disk. Your method of finding the infinitesimal force on each ring segment and integrating it to find the total force at a given radius is sound and accurate. However, your calculation for the pressure at a specific radius is incorrect.

To find the tensile pressure at a specific radius, you need to divide the total force by the cross-sectional area at that radius, not the area at the outer radius. This is because the force is distributed across the entire cross-sectional area of the disk, not just at the outer edge.

Additionally, the cross-sectional area at a given radius is not simply 2πr_o h, as you have calculated. It is actually 2πr_o dr, as the cross-sectional area increases as you move towards the center of the disk. This will result in a different equation for pressure:

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

This equation should give you a correct and finite value for the tensile pressure at any given radius within the disk. However, it is important to note that this calculation assumes a uniform distribution of mass and a perfectly rigid disk, which may not be the case in reality. It would be beneficial to also consider the material properties and any potential deformations that may occur under high centrifugal forces.

I hope this helps in your pursuit of creating a hyper-fast rotating disk. Remember to always double check your calculations and assumptions, and don't hesitate to seek help from other scientists or experts in the field. Good luck!
 

1. What is tensile pressure in a rotating disk?

Tensile pressure is the force per unit area that is applied to a rotating disk in a direction parallel to its surface. It is also known as tangential stress.

2. Why is it important to find the tensile pressure in a rotating disk?

Understanding the tensile pressure in a rotating disk is crucial in engineering applications as it affects the strength and durability of the disk. It also helps identify potential failure points and allows for proper design and material selection.

3. How is the tensile pressure in a rotating disk calculated?

The tensile pressure in a rotating disk can be calculated using the formula P = (ρ ω^2 r)/2, where P is the pressure, ρ is the density of the disk, ω is the angular velocity, and r is the distance from the center of rotation to the point of interest on the disk's surface.

4. What factors can affect the tensile pressure in a rotating disk?

The tensile pressure in a rotating disk can be affected by a variety of factors, including the material and thickness of the disk, the speed of rotation, and external forces applied to the disk.

5. How can the tensile pressure in a rotating disk be controlled?

The tensile pressure in a rotating disk can be controlled by adjusting the design and material properties of the disk, as well as the speed of rotation. Additionally, implementing measures such as reinforcement or lubrication can also help reduce the tensile pressure and prevent failure.

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