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Changing equilibrium equations from cartesian to polar

  • Thread starter rafehi
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  • #1
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Homework Statement


Transform the equilibrium equations from cartesian to polar coordinates using x = rcos(theta) and y = rsin(theta):

[tex]\frac{\partial\sigma_{xx}}{\partial{x}} + \frac{\partial\sigma_{xy}}{\partial{y}} = 0[/tex]

[tex]\frac{\partial\sigma_{yx}}{\partial{x}} + \frac{\partial\sigma_{yy}}{\partial{y}} = 0[/tex]


The answer is given as:

[tex]\frac{\partial\sigma_{r}}{\partial{r}} + \frac{1}{r}\frac{\partial\sigma_{r\theta}}{\partial{\theta}} + \frac{\sigma_{r} - \sigma_{\theta}}{r} = 0[/tex]


[tex]\frac{\partial\sigma_{r\theta}}{\partial{r}} + \frac{1}{r}\frac{\partial\sigma_{\theta}}{\partial{\theta}} + \frac{2\sigma_{r\theta}}{r} = 0[/tex]

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.
 

Answers and Replies

  • #2
gabbagabbahey
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You basically just need to apply the chain rule several times. For example,

[tex]\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}[/tex]
 
  • #3
49
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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:

[tex]\frac{dr}{dx} = \frac{1}{cos(\theta)}[/tex]

[tex]\frac{d\theta}{dx} = \frac{-1}{rsin(\theta)}[/tex]

[tex]\frac{dr}{dy} = \frac{1}{sin(\theta)}[/tex]

[tex]\frac{d\theta}{dxy} = \frac{1}{rcos(\theta)}[/tex]

[tex]
\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}}
[/tex]

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:

[tex]
\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)}
[/tex]

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
gabbagabbahey
Homework Helper
Gold Member
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[tex]
\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}}
[/tex]

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.
That doesn't look right at all. Why don't you start by showing me your calculation for [tex]\sigma_{xx}\equiv\frac{\partial^2 \sigma}{\partial x^2}[/tex] first.
 
  • #5
49
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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
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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...
Okay, I see now.

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

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