# Stress in a disk.

1. Jul 6, 2006

### pervect

Staff Emeritus
Suppose I take a disk with a pie-sliced section cut out of it

$$\] \unitlength 1mm \begin{picture}(90,90)(0,0) \linethickness{0.3mm} \multiput(20.75,77.28)(0.13,0.12){29}{\line(1,0){0.13}} \multiput(17.56,73.41)(0.12,0.14){27}{\line(0,1){0.14}} \multiput(14.89,69.16)(0.12,0.19){22}{\line(0,1){0.19}} \multiput(12.77,64.62)(0.12,0.25){18}{\line(0,1){0.25}} \multiput(11.23,59.84)(0.12,0.37){13}{\line(0,1){0.37}} \multiput(10.3,54.91)(0.12,0.62){8}{\line(0,1){0.62}} \multiput(10,49.91)(0.1,1.67){3}{\line(0,1){1.67}} \multiput(10,49.91)(0.11,-1.67){3}{\line(0,-1){1.67}} \multiput(10.33,44.9)(0.12,-0.62){8}{\line(0,-1){0.62}} \multiput(11.28,39.98)(0.12,-0.37){13}{\line(0,-1){0.37}} \multiput(12.83,35.21)(0.12,-0.25){18}{\line(0,-1){0.25}} \multiput(14.98,30.68)(0.12,-0.19){22}{\line(0,-1){0.19}} \multiput(17.67,26.44)(0.12,-0.14){27}{\line(0,-1){0.14}} \multiput(20.87,22.58)(0.13,-0.12){29}{\line(1,0){0.13}} \multiput(24.53,19.15)(0.16,-0.12){25}{\line(1,0){0.16}} \multiput(28.59,16.21)(0.22,-0.12){20}{\line(1,0){0.22}} \multiput(32.99,13.8)(0.31,-0.12){15}{\line(1,0){0.31}} \multiput(37.66,11.95)(0.49,-0.12){10}{\line(1,0){0.49}} \multiput(42.52,10.71)(1,-0.13){5}{\line(1,0){1}} \put(47.49,10.08){\line(1,0){5.02}} \multiput(52.51,10.08)(1,0.13){5}{\line(1,0){1}} \multiput(57.48,10.71)(0.49,0.12){10}{\line(1,0){0.49}} \multiput(62.34,11.95)(0.31,0.12){15}{\line(1,0){0.31}} \multiput(67.01,13.8)(0.22,0.12){20}{\line(1,0){0.22}} \multiput(71.41,16.21)(0.16,0.12){25}{\line(1,0){0.16}} \multiput(75.47,19.15)(0.13,0.12){29}{\line(1,0){0.13}} \multiput(79.13,22.58)(0.12,0.14){27}{\line(0,1){0.14}} \multiput(82.33,26.44)(0.12,0.19){22}{\line(0,1){0.19}} \multiput(85.02,30.68)(0.12,0.25){18}{\line(0,1){0.25}} \multiput(87.17,35.21)(0.12,0.37){13}{\line(0,1){0.37}} \multiput(88.72,39.98)(0.12,0.62){8}{\line(0,1){0.62}} \multiput(89.67,44.9)(0.11,1.67){3}{\line(0,1){1.67}} \multiput(89.7,54.91)(0.1,-1.67){3}{\line(0,-1){1.67}} \multiput(88.77,59.84)(0.12,-0.62){8}{\line(0,-1){0.62}} \multiput(87.23,64.62)(0.12,-0.37){13}{\line(0,-1){0.37}} \multiput(85.11,69.16)(0.12,-0.25){18}{\line(0,-1){0.25}} \multiput(82.44,73.41)(0.12,-0.19){22}{\line(0,-1){0.19}} \multiput(79.25,77.28)(0.12,-0.14){27}{\line(0,-1){0.14}} \multiput(75.61,80.73)(0.13,-0.12){29}{\line(1,0){0.13}} \multiput(24.39,80.73)(0.12,-0.14){213}{\line(0,-1){0.14}}\multiput(50,50)(0.12,0.14){213}{\line(0,1){0.14}} \end{picture} \[$$

and force the cut ends together to make a complete, circular, planar disk.

I want to figure out the 2d stresses in the disk (for reasons that are somewhat arcane).

What's bothering me most at the moment is that the equilbrium relation is apparently

$$\frac{d \sigma_r}{dr} + \frac{\sigma_r - \sigma_\theta}{r} = 0$$

per
http://www.utm.edu/departments/engin/lemaster/Machine%20Design/Notes%2016.pdf [Broken]

Here $\sigma_r$ is the radial stress, and
$\sigma_\theta$ is the circumfrential stress.

If I'm interpreting this right, this means that the tangential tension due to $\sigma_\theta$ should induce radial tension, not compression?

This seems totally backwards, but I can't see any sign error in the continuity equations. I think it's backwards because if I stretch a rubber band around my finger, I can feel my finger being compressed (while the rubber band is in tension).

I get the same equations as the URL above form by assuming a stress-energy tensor T^{ij} in cylindrical coordinates and assuming

$$\sigma_r = T^{rr}$$
$$\sigma_\theta = r^2 T^{\theta\theta}$$

and applying the continuity equation $$\nabla_a T^{ab} = 0$$

as per (for example)
http://en.wikipedia.org/wiki/Stress-energy_tensor

Last edited by a moderator: May 2, 2017
2. Jul 6, 2006

### Danger

How on Earth are you going to keep it flat? That's how I make cones.

3. Jul 7, 2006

### FredGarvin

You should indeed get radial compression. If you look at a differential element that is wedge shaped like you mentioned, the forces acting in the tangential ($$\theta$$) direction, they are due to $$\sigma_\theta da$$ and they angle in towards the center of the disk which would result in a compressive (negative)$$\sigma_r$$

Perhaps it would make more sense (it seems to to me) to rewrite the equillibrium equation as:
$$\frac{d \sigma_r}{dr} - \frac{-\sigma_r + \sigma_\theta}{r} = 0$$
This makes it easier for me to see since tensile stresses are positive and compressive negative.

BTW...I liked how you did your diagram. I did not know you could do that with LaTex. Is there a primer on how to do that?

4. Jul 7, 2006

### pervect

Staff Emeritus
OK, that's what I thought should happen. I'll have to try and track down the sign error then (probably I've mis-defined something).

I think I may be beginning to see what's going on
http://ocw.mit.edu/NR/rdonlyres/Nuclear-Engineering/22-314JSpring2004/ACF104D9-8C23-4EEB-B3CD-D350871D8EDE/0/elastic_theory.pdf [Broken]

has a nice diagram plus the same equations I quoted earlier. Part of my problem, I think, is ignoring $d\sigma_r / dr$ $\sigma_r$ must be zero at the outer edge of the disk, so any solution that has a nonzero radial stress can't have a constant radial stress.

If we consider this fact, it's clear that if $\sigma_\theta$ is positive (tension), $\sigma_r$ is zero at the outer edge of the disk, and negative just inside it.

However, the force balance equation is actually correct as written in the references. So the problem wasn't in the equations, the correct solution of the equations I already had gives a compressive force.

I use Jpicedt to do the diagrams. http://jpicedt.sourceforge.net/

To make the diagrams small enough to fit when they have circles, you need to change entries in the menu "edit/preferences/latex"

change emulated circle segment length to a larger value than the default. You might want to change emulated line segment length as well.

There's not really a tutorial, but there's some discussion of the finer points in the Latex thread. It's useful to find an example that does something you like, and see how it was done.

https://www.physicsforums.com/showpost.php?p=1009015&postcount=734

Last edited by a moderator: May 2, 2017
5. Jul 7, 2006

### jasc15

as a side note, and a little less complicated, i decided one time at a party to find the equation for the height and angle of a cone as a function of the angle of the "pie slice" which was removed, although i dont have the equation handy.

6. Jul 7, 2006

### pervect

Staff Emeritus
You may have a good point. It's not clear if the problem as written has a finite solution, which may be part of how I confused myself.

Certainly if we take $\sigma_\theta = K$, we find $\sigma_r = K (1-R/r)$, which satisfies the continuity equations and the boundary conditions that $\sigma_r = 0$ at r=R, but blows up at r=0.

7. Jul 7, 2006

### NateTG

Looking at that drawing we tend to think of the 'cone' where the circumference of the final result is smaller than the circumference of the initial circle was.

8. Jul 7, 2006

### Danger

Not that it's relevant, but I think that I just figured out how you could do it for real. If you had an elastic substance such as a silicone disk maybe 1/4" thick with steel ribs like an umbrella and a steel perimeter sandwiched between a couple of pieces of lexan (or anything, but why not make it visible?), you could remove a piece, use the ribs on either side of the gap to force the edges together, bond it solidly, then pull the ribs out radially. Why you would want to do that, on the other hand, is utterly beyond me.
And I still think that it would pop into a cone shape as soon as the lexan was removed.

9. Jul 7, 2006

### pervect

Staff Emeritus
If we made the disk very thick it would be less likely to buckle.

Cutting out pieces of the disk and forcing it together was basically a conveinent way of creating a disk with internal stresses. An example of how this might actually occur would be to imagine a disk that was initially formed out of a material that changed length with temperature with no internal stresses at the time of manufacture, then cooling the outside of the disk while maintaining the temperature at the core.

The outside of the disk would differentially shrink, creating a situation somewhat like the example above.

10. Jul 8, 2006

### Danger

Interesting. Again, though, what's the point? Is there some physical advantage to having such stresses?

11. Jul 8, 2006

### pervect

Staff Emeritus
A more exotic way that the such stresses can be generated is in a relativistically rotating disk. Lorentz contraction of the outer, rapidly rotating part causes the same general sort of stress patterns as in my example problem.

In the Earth frame, distances appear to be constant, but if you go to a frame co-moving with respect to the outer edge of the disk, the distances must appear longer. I.e. the longer proper distance gets Lorentz-contracted to the constant, shorter distance we observe in the Earth frame.

This is basically a mechanical engineering problem, so I've been trying to learn the mechanical engineering techniques needed to calculate the behaviors, i.e. the constitutive equations (the relation between stress and strain), also the displacement-strain equations (easy in cartesian coordinates, a little more complex in cylindrical coordinates).

There's some other things that I've stumbled across in the process that could be interesting, but I'm not sure if they're needed (they might make the job easier, though), such as Airy stress function.

One of the particular things I'm interested in is that I expect for the compatibility of two different formulas that the intergal of the stresses in a static system must be equal. This turns out to be the case in at least the toy example I constructed, i.e we do have:

$$\int_A (\sigma_r + \sigma_\theta) dA = 0$$

(the radial stress diverges in my toy example, but the intergal above does not diverge).

I expect a similar relationship must be true in general for a static, isolated system, though I'm lacking a formal proof at the moment - I expect that the intergal of the three diagonal elements of the stress tensor over the volume of the material should be zero.

For the rotating disk that started this line of investigation, certain technical adjustments must be made in the above statement.

12. Jul 9, 2006

### Cyrus

Hi Pervect,

Just a side of caution. It appears to me that you want to apply equations of an elastic material, when it looks as if you are causing deformations well beyond the elastic limit. Then your elastic equations would no longer apply.

The material will only allow displacement by a certain amount. Your cut will have an arc length of pi*r*theta. If this length exceeds the elastic stretch, your equations are out the window.

Last edited: Jul 9, 2006