I Pressurized containers: Stress distribution and large displacements

[Juanda]

You're missing the $dx_0dy_0$ term.

You're totally right. I right-clicked to copy and I didn't realize they'd been separated into two equations so they are shown as two lines. That's probably de best option to make lengthy equations easy to read in the forum. I'll try to follow the same method from now on.

What do you get if you now convert to the displacements?

I'll try to work it out today when I find the spot to sit on it a little longer. It'd be a matter of substitution which is manageable to me. Maybe a typo here and there but a straightforward process overall.

 

[Juanda]

Using post #42 I came to the following expression.
$$
\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}^2+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}
$$

By the way, I caught a typo on post #42 but I can't edit it out now. I mixed the differentials here.
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}=\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial x_0}^2$$

 

[Chestermiller]

Using post #42 I came to the following expression.
$$
\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}^2+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}
$$

What happened to the sin cos term?

By the way, I caught a typo on post #42 but I can't edit it out now. I mixed the differentials here.
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}=\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial x_0}^2$$

I think this equation has a typo.

$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial y_0}^2$$

After you make these corrections, I'll tell you what to do in the next step.

 

[Chestermiller]

Using post #42 I came to the following expression.
$$
\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}^2+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}
$$

By the way, I caught a typo on post #42 but I can't edit it out now. I mixed the differentials here.
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}=\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial x_0}^2$$

Why can’t you edit it?

 

[Juanda]

What happened to the sin cos term?

I'm not sure what you mean. They are there. I made the substitution and that's the expression I got.

I think this equation has a typo.

Yes, that's the typo I was pointing at from #42. I probably could have expressed it more clearly. I should have done something like this (italic):

I found a typo at #42.

From #42:
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}=\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial x_0}^2$$

I mixed the differentials while copying and pasting previous expressions. It'd have been:
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial y_0}=\frac{\partial u}{\partial y_0}$$
$$\frac{\partial x}{\partial y_0}^2=\frac{\partial u}{\partial y_0}^2$$

That typo shouldn't have affected the expression for the strain ratio from #52 because I corrected it.

Why can’t you edit it?

After some time goes by, editing a message is no longer possible at least for me. You might have more options available due to your status in the forum.

 

[Chestermiller]

I'm not sure what you mean. They are there. I made the substitution and that's the expression I got.

I meant this term:

$$+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

Yes, about editing. I've been granted extended privileges for editing.

 

[Juanda]

I meant this term:
$$+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

My bad. I connected and disconnected from this problem too many times and I lost track. You already pointed out that I was missing that term and I acknowledged it just to forget it again.
The original expression you showed is:
$$\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$
$$+2\left[\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0}\right]\sin{\alpha}\cos{\alpha}$$

Expressed in terms of the displacement field it'd be:
$$\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

I hope it's OK now. No typos, no forgotten terms, etc. Let me know if I missed anything else.

 

[Chestermiller]

My bad. I connected and disconnected from this problem too many times and I lost track. You already pointed out that I was missing that term and I acknowledged it just to forget it again.
The original expression you showed is:
$$\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$
$$+2\left[\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0}\right]\sin{\alpha}\cos{\alpha}$$

Expressed in terms of the displacement field it'd be:
$$\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

I hope it's OK now. No typos, no forgotten terms, etc. Let me know if I missed anything else.

Now, let the strain in the material line joining the two material points be defined as $$\epsilon=\frac{\lambda^2-1}{2}$$ What do you get?

 

[Juanda]

Now, let the strain in the material line joining the two material points be defined as $$\epsilon=\frac{\lambda^2-1}{2}$$ What do you get?

The expression for $\lambda^2$ is already known.
$$\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

So it'd be just plugging it in $\epsilon$.
$$\epsilon=\frac{\lambda^2-1}{2}$$
$$\epsilon=\frac{\left ( \left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}\right )-1}{2}$$

 

[Chestermiller]

The expression for $\lambda^2$ is already known.
$$\lambda^2=\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

So it'd be just plugging it in $\epsilon$.
$$\epsilon=\frac{\lambda^2-1}{2}$$
$$\epsilon=\frac{\left ( \left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}\right )-1}{2}$$

make use of cos^2+sin^2 = 1 please

 

[Juanda]

make use of cos^2+sin^2 = 1 please

But $J \neq D$. How could I use it?
$$\lambda^2=\overbrace{\left[1 +2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]}^J\cos^2{\alpha}+
\overbrace{\left[1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]}^D
\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

 

[Chestermiller]

$$\lambda^2=1+\overbrace{\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]}^J\cos^2{\alpha}+
\overbrace{\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]}^D
\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

 

[Juanda]

$$\lambda^2=1+\overbrace{\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]}^J\cos^2{\alpha}+
\overbrace{\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]}^D
\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

Oh. Yeah. That first $1$ could multiply on both terms and then be simplified.
But is the resulting expression any better? If there is something I should recognize from looking at it I'm failing to do so.

 

[Chestermiller]

Oh. Yeah. That first $1$ could multiply on both terms and then be simplified.
But is the resulting expression any better? If there is something I should recognize from looking at it I'm failing to do so.

Not yet. Please next give the equation of the strain $\epsilon$ that I asked for. We're almost there.

 

[Juanda]

Not yet. Please next give the equation of the strain $\epsilon$ that I asked for. We're almost there.

All right.

Strain is:
$$\epsilon=\frac{\lambda^2-1}{2}$$

And $\lambda^2$ is:
$$\lambda^2=1+\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

Plugging that $\lambda^2$ in the strain will result in the $1$ canceling and also the $2$ from the third term so it'll be:
$$\frac{\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}}{2}$$
$$+\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

Is this the expression you're looking for?

 

[Chestermiller]

All right.

Strain is:
$$\epsilon=\frac{\lambda^2-1}{2}$$

And $\lambda^2$ is:
$$\lambda^2=1+\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}$$
$$+2\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

Plugging that $\lambda^2$ in the strain will result in the $1$ canceling and also the $2$ from the third term so it'll be:
$$\frac{\left[2 \frac{\partial u}{\partial x_0} + \frac{\partial u}{\partial x_0}^2 + \frac{\partial v}{\partial x_0}^2 \right]\cos^2{\alpha}+\left[2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2 + \frac{\partial u}{\partial y_0}^2 \right]\sin^2{\alpha}}{2}$$
$$+\left[(1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial y_0})+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0})\right]\sin{\alpha}\cos{\alpha}$$

Is this the expression you're looking for?

Basically, yes. But, consider this. $$\epsilon=\epsilon_{xx}\cos^2{\alpha}+\epsilon_{yy}\sin^2{\alpha}+\epsilon_{xy}\sin{\alpha}\cos{\alpha}+\epsilon_{yx}\sin{\alpha}\cos{\alpha}$$where $$\epsilon_{xx}=\frac{1}{2}\left[\frac{\partial u}{\partial x_0}+\frac{\partial u}{\partial x_0}+\left(\frac{\partial u}{\partial x_0}\right)^2+\left(\frac{\partial v}{\partial x_0}\right)^2\right]$$

$$\epsilon_{yy}=\frac{1}{2}\left[\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}+\left(\frac{\partial u}{\partial y_0}\right)^2+\left(\frac{\partial v}{\partial y_0}\right)^2\right]$$

$$\epsilon_{xy}=\epsilon_{yx}=\frac{1}{2}\left[\frac{\partial u}{\partial y_0}+\frac{\partial v}{\partial x_0}+\frac{\partial u}{\partial x_0}\frac{\partial u}{\partial y_0}+\frac{\partial v}{\partial x_0}\frac{\partial v}{\partial y_0} \right]$$
So you recognize the small displacement strain terns in these equations?

 

[Juanda]

I'm not sure what you mean. I do see some similarities but I'm not sure what I should understand from that.

 

[Chestermiller]

I'm not sure what you mean. I do see some similarities but I'm not sure what I should understand from that.

View attachment 334073

What does your Strength ofMaterials book or Theory of Elasticity book or Wikipedia give for the components of the infinitesimal (linear) strain tensor?

 

[Juanda]

What does your Strength ofMaterials book or Theory of Elasticity book or Wikipedia give for the components of the infinitesimal (linear) strain tensor?

Are you referring to how 2nd order terms are typically ignored if they're very small compared with 1st order terms?

From Wiki
https://en.wikipedia.org/wiki/Infinitesimal_strain_theory

After doing the math and ignoring what's going on with the 3rd dimension so that it looks as the simplification developed in this thread, the infinitesimal strain tensor is as shown.

Is this what you meant?

 

[Chestermiller]

Are you referring to how 2nd order terms are typically ignored if they're very small compared with 1st order terms?

From Wiki
https://en.wikipedia.org/wiki/Infinitesimal_strain_theory
View attachment 334074

After doing the math and ignoring what's going on with the 3rd dimension so that it looks as the simplification developed in this thread, the infinitesimal strain tensor is as shown.
View attachment 334075Is this what you meant?

Yes, these are the small displacement (linear) terms in the strains. If you had also included the z direction, those strains would also be also be properly accounted for too by this methodology. By the methodology, I mean using $\frac{(\lambda^2-1)}{2}$ as a vehicle to elucidate the strains in terms of the displacements.

These equations could also be used (the full expressions for the strains including the non-linear terms) if the displacements are large but the strains are small (e.g., for a very stiff material). Under such circumstances, the 3D Hooke's law equations could still be used. However, if the displacements and the strains are also large, Hooke's law could not be used, since it is only valid for small strains. That is when wee enter the realm of non-linear elasticity and more complex material behavior.