I Pressurized containers: Stress distribution and large displacements

[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.