# Changing equilibrium equations from cartesian to polar

1. Mar 15, 2010

### rafehi

1. The problem statement, all variables and given/known data
Transform the equilibrium equations from cartesian to polar coordinates using x = rcos(theta) and y = rsin(theta):

$$\frac{\partial\sigma_{xx}}{\partial{x}} + \frac{\partial\sigma_{xy}}{\partial{y}} = 0$$

$$\frac{\partial\sigma_{yx}}{\partial{x}} + \frac{\partial\sigma_{yy}}{\partial{y}} = 0$$

$$\frac{\partial\sigma_{r}}{\partial{r}} + \frac{1}{r}\frac{\partial\sigma_{r\theta}}{\partial{\theta}} + \frac{\sigma_{r} - \sigma_{\theta}}{r} = 0$$

$$\frac{\partial\sigma_{r\theta}}{\partial{r}} + \frac{1}{r}\frac{\partial\sigma_{\theta}}{\partial{\theta}} + \frac{2\sigma_{r\theta}}{r} = 0$$

3. The attempt at a solution

I've no idea how to proceed. I know how to obtain the result using a free body diagram (which is how we're supposed to solve the problem) but am curious as to how to do it by applying the transformations.

We've covered changing coordinates for integrals in vector calculus but never anything like the above. If anybody can help get me started or point me to some material that'd help me get my head around it, it'd be very much appreciated.

Not sure in which section to post it where to post it but this board seemed as good as any.

2. Mar 15, 2010

### gabbagabbahey

You basically just need to apply the chain rule several times. For example,

$$\frac{\partial\sigma}{\partial x}=\frac{\partial\sigma}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial\sigma}{\partial \theta}\frac{\partial \theta}{\partial x}=\sigma_r \frac{\partial r}{\partial x} +\sigma_{\theta} \frac{\partial \theta}{\partial x}$$

3. Mar 15, 2010

### rafehi

OK. I've tried implementing it but I'm confused as to how to get the final subscripts. Where did the last term in the first answer equation come from?

What I've attempted:

$$\frac{dr}{dx} = \frac{1}{cos(\theta)}$$

$$\frac{d\theta}{dx} = \frac{-1}{rsin(\theta)}$$

$$\frac{dr}{dy} = \frac{1}{sin(\theta)}$$

$$\frac{d\theta}{dxy} = \frac{1}{rcos(\theta)}$$

$$\frac{\partial\sigma_{xx}}{\partial{x}} + \frac{\partial\sigma_{xy}}{\partial{y}} = \frac{\partial\sigma_r}{\partial{r}}\frac{\partial{r}}{\partial{x}} + \frac{\partial\sigma_\theta}{\partial\theta}\frac{\partial\theta}{\partial{x}} + \frac{\partial\sigma_r_\theta}{\partial{r}}\frac{\partial{r}}{\partial{y}} + \frac{\partial\sigma_r_\theta}{\partial\theta}\frac{\partial\theta}{\partial{y}}$$

by the chain rule. Have I expanded the second term correctly? The fact that stress (sigma) is a function of both x and y has me confused.

Anyhow, plugging in what I have:

$$\frac{\partial\sigma_{xx}}{\partial{x}} + \frac{\partial\sigma_{xy}}{\partial{y}} = \frac{\partial\sigma_r}{\partial{r}}\frac{1}{cos(\theta)} + \frac{\partial\sigma_\theta}{\partial\theta}\frac{-1}{rsin(\theta)} + \frac{\partial\sigma_r_\theta}{\partial{r}}\frac{1}{sin(\theta)} + \frac{\partial\sigma_r_\theta}{\partial\theta}\frac{1}{rcos(\theta)}$$

I can see how I'll get the first term out of the above as 1/cos(theta) will equal 1 by the small angle approximation but I think I must have done a mistake as I can't see how I'll get either of the other two terms.

Again, any help greatly appreciated.

4. Mar 15, 2010

### gabbagabbahey

That doesn't look right at all. Why don't you start by showing me your calculation for $$\sigma_{xx}\equiv\frac{\partial^2 \sigma}{\partial x^2}$$ first.

5. Mar 15, 2010

### rafehi

How is sigmaxx equivalent to the d2sigma/dx2 when sigmaxx is just the normal stress acting in the normal direction?

It isn't the second partial derivative of sigma. Should've made that clear in the OP - I apologise. Likewise, sigmaxy is the shear stress pointing in the x direction acting in the y - z plane.

Again, I wasn't sure if this was in the appropiate section...

6. Mar 15, 2010

### gabbagabbahey

Okay, I see now.

In that case, the first thing you will want to do is to express $\{\sigma_{xx},\sigma_{xy},\sigma_{yx},\sigma_{yy}\}$ in terms of $\{\sigma_{rr},\sigma_{r\theta},\sigma_{\theta r},\sigma_{\theta\theta}\}$...