I Pressurized containers: Stress distribution and large displacements

[Chestermiller]

To be honest I don't remember being taught about stretch ratios while I was in university. It might be a language issue. It's not that I don't understand it. I checked the formula and I know what it is.
Simply I think we don't have a specific word to define that. It could also be I simply don't remember the name though. We just used forces, stress, strain, and displacements. I actually don't see much added value in incorporating the stretch ratio into the mix of variables that already seem to fully define the situation but my conception might change as I learn more.
In fact, I thought you swapped from $\epsilon$ to $\lambda$ to indicate we're tackling question 2 about the balloon instead of question 1 about the tank.Yeah, after noticing the difference between strain and stretch ratio I see how it makes sense.I guess that'd be a reason to try using stress ratio instead of strain.

For large deformation situations, the stretch ratio is not ambiguous,, but the strain is.

Wouldn't I need to know the functions for $x$; $y$; $z$ to be able to answer that question?
For example,
$$x=x_0+1$$
So every point moves 1 unit to the right. Now I can check where that neighboring point you described would end up by inputting that into the previous expression.
$$x=x_0+dx_0+1$$
Is that what you meant with your question?

No. I already specified the functionality: $x=x(x_0,y_0,z_0)$, same for y and z. So, $$dx=\frac{\partial x}{\partial x_0}dx_0+\frac{\partial x}{\partial y_0}dy_0+\frac{\partial x}{\partial z_0}dz_0$$same for dy and dz.

Let $(ds)^2$ be the square of distance between the two closely neighboring material points in the deformed configuration of the body. Please express $(ds)^2$ in terms of $dx_0, \ dy_0,\ and\ dz_0$ (not dx, dy and dz). That is, in terms of the differential differences of the coordinates of the two material points in the undeformed configuration of the body.

 

[Juanda]

I'm trying to visualize what you're asking for. Let me drop the $z$ component so that's simpler to draw.
We have the points $A$ and $B$.
Before the deformation, their positions are $A_0$ and $B_0$ and after the deformation, they are $A$ and $B$. After the deformation, they are very close.

From the diagram, I can express $ds$ in terms of $A$ and $B$ but that will result in an expression dependent of $dx$ and $dy$.
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2$$
Where $dx$ and $dy$ can be obtained from what you shared:
$$dx=\frac{\partial x}{\partial x_0}dx_0+\frac{\partial x}{\partial y_0}dy_0$$

I doubt that's the expression of $\left \| d\vec{s} \right \|^2$ you're looking for because there still are $\partial x$ in it.
I'm not sure I'm prepared to understand question 2.
I'm already thankful for at least being able to follow you through question 1 (the tank with small deformations without assuming constant stress in the shell) but question 2 (the balloon with big deformations) seems orders of magnitude more complex.
It's not my intention to waste your time so if I'm still far off what're trying to transmit I think it'll be better if I just keep reading books before trying to tackle something like this.

 

[Juanda]

It's not my intention to waste your time so if I'm still far off what're trying to transmit I think it'll be better if I just keep reading books before trying to tackle something like this.

I'm answering just to make sure the message goes through as I understand it.
When I say "far off" I don't mean that there's still a lot to go. I can keep going without issues.
What I was trying to say is that if you realize I just lack a strong enough base to understand this now then I won't waste more of your time and I'll focus on easier problems before trying to deal with something like this for which I'm not yet ready.

 

[Chestermiller]

I'm trying to visualize what you're asking for. Let me drop the $z$ component so that's simpler to draw.
We have the points $A$ and $B$.
Before the deformation, their positions are $A_0$ and $B_0$ and after the deformation, they are $A$ and $B$. After the deformation, they are very close.
View attachment 331716

From the diagram, I can express $ds$ in terms of $A$ and $B$ but that will result in an expression dependent of $dx$ and $dy$.
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2$$
Where $dx$ and $dy$ can be obtained from what you shared:
$$dx=\frac{\partial x}{\partial x_0}dx_0+\frac{\partial x}{\partial y_0}dy_0$$

I doubt that's the expression of $\left \| d\vec{s} \right \|^2$ you're looking for because there still are $\partial x$ in it.

You've done it correctly so far. In a given problem, we usually don't know the relationship between dx etc. and $dx_0$ etc. yet. That's what you are solving for. So, please substitute our equations for dx and dy in terms of $$dx_0\ and\ dy_0$$ into your equation for $(ds)^2$ and express the result in the form $$(ds)^2=A(dx_0)^2+B(dy_0)^2+Cdx_0dy_0$$. What do you get for A, B, and C?

I'm not sure I'm prepared to understand question 2.
I'm already thankful for at least being able to follow you through question 1 (the tank with small deformations without assuming constant stress in the shell) but question 2 (the balloon with big deformations) seems orders of magnitude more complex.
It's not my intention to waste your time so if I'm still far off what're trying to transmit I think it'll be better if I just keep reading books before trying to tackle something like this.

Well, I'm trying to lead you through how we would approach large deformation mechanics. The first step is to establish the displacement-strain equations for large displacements.

 

[Chestermiller]

I'm answering just to make sure the message goes through as I understand it.
When I say "far off" I don't mean that there's still a lot to go. I can keep going without issues.
What I was trying to say is that if you realize I just lack a strong enough base to understand this now then I won't waste more of your time and I'll focus on easier problems before trying to deal with something like this for which I'm not yet ready.

I understand. Even linear strains and displacements together with linear relation for the stress tensor isn't that easy. You need to at least have some familiarity with 2nd order tensors and how to apply them. For large displacements and strains, things are much more complicated. Try Googling "large displacement mechanics" or "large displacement mechanics." Or hit the Continuum Mechanics literature.

 

[Juanda]

You've done it correctly so far. In a given problem, we usually don't know the relationship between dx etc. and $dx_0$ etc. yet. That's what you are solving for. So, please substitute our equations for dx and dy in terms of $$dx_0\ and\ dy_0$$ into your equation for $(ds)^2$ and express the result in the form $$(ds)^2=A(dx_0)^2+B(dy_0)^2+Cdx_0dy_0$$. What do you get for A, B, and C?

All right. It's good to know I'm still on track. I have to say though that I don't know where those tracks lead to yet but I'll keep going.
First:
$$dx=\frac{\partial x}{\partial x_0}dx_0+\frac{\partial x}{\partial y_0}dy_0$$
$$dy=\frac{\partial y}{\partial x_0}dx_0+\frac{\partial y}{\partial y_0}dy_0$$
Then:
$$dx^2=\frac{\partial x}{\partial x_0}^2dx_0^2+\frac{\partial x}{\partial y_0}^2dy_0^2+2\frac{\partial x}{\partial x_0}dx_0\frac{\partial x}{\partial y_0}dy_0$$
$$dy^2=\frac{\partial y}{\partial x_0}^2dx_0^2+\frac{\partial y}{\partial y_0}^2dy_0^2+2\frac{\partial y}{\partial x_0}dx_0\frac{\partial y}{\partial y_0}dy_0$$
By the way, I assume the squares mean just that. Something $\times$ itself. It's NOT the second derivative.
Finally:
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2$$
Where:
$$dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial x}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$
You can see the values of $A$, $B$, and $C$ you requested at post #34 in that last equation. They might be wrong or there could be a typo but that's what I got.

Well, I'm trying to lead you through how we would approach large deformation mechanics.

And for that, I'm very grateful.

I understand. Even linear strains and displacements together with linear relation for the stress tensor isn't that easy. You need to at least have some familiarity with 2nd order tensors and how to apply them. For large displacements and strains, things are much more complicated. Try Googling "large displacement mechanics" or "large displacement mechanics." Or hit the Continuum Mechanics literature.

In university, I wasn't taught continuum mechanics in much depth. Besides, I'm somewhat rusty so the little I learned is fuzzy in my head. However, it feels like a very necessary field to know as an engineer and I also find it especially interesting so here we are.

 

[Chestermiller]

All right. It's good to know I'm still on track. I have to say though that I don't know where those tracks lead to yet but I'll keep going.
First:
$$dx=\frac{\partial x}{\partial x_0}dx_0+\frac{\partial x}{\partial y_0}dy_0$$
$$dy=\frac{\partial y}{\partial x_0}dx_0+\frac{\partial y}{\partial y_0}dy_0$$
Then:
$$dx^2=\frac{\partial x}{\partial x_0}^2dx_0^2+\frac{\partial x}{\partial y_0}^2dy_0^2+2\frac{\partial x}{\partial x_0}dx_0\frac{\partial x}{\partial y_0}dy_0$$
$$dy^2=\frac{\partial y}{\partial x_0}^2dx_0^2+\frac{\partial y}{\partial y_0}^2dy_0^2+2\frac{\partial y}{\partial x_0}dx_0\frac{\partial y}{\partial y_0}dy_0$$
By the way, I assume the squares mean just that. Something $\times$ itself. It's NOT the second derivative.
Finally:
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2$$
Where:
$$dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial x}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

Excellent. Next, let $x=x_0+u$ and $y=y_0+v$, where u and v are the displacements of the material points (functions of the coordinates $x_0$ and $y_0$). What do you get if you substitute these into the right hand side of the equation?

 

[Juanda]

Excellent. Next, let $x=x_0+u$ and $y=y_0+v$, where u and v are the displacements of the material points (functions of the coordinates $x_0$ and $y_0$). What do you get if you substitute these into the right hand side of the equation?

Do you mean doing this?

$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$

Then do it to all the terms and see what comes out. I just want to make sure that's the goal before working on it.

 

[Chestermiller]

Do you mean doing this?

$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$

Then do it to all the terms and see what comes out. I just want to make sure that's the goal before working on it.

Yes, exactly.

 

[Juanda]

Yes, exactly.

Ok. I'll do a little recap because it's been a while since I could get to this.
We come from this expression that represents the distance from 2 points that were originally very close to each other before the deformation happened (post #32 for sketch).

$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial x}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

Then, introducing the displacement field we get:
$$x=x_0+u$$
$$y=y_0+v$$

As a result, $\left \| d\vec{s} \right \|^2$ can be expressed in terms of the displacement field. For that, I'll state some intermediate equations to keep track of.
$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial y}{\partial x_0}=\frac{\partial (y_0+v)}{\partial x_0}=\frac{\partial y_0}{\partial x_0}+\frac{\partial v}{\partial x_0}$$

I'm not sure if we can say $\frac{\partial y_0}{\partial x_0}=0$ because the original vertical position of the point does not depend on the original horizontal position of the point. For now I'll leave it as shown without assuming it's zero.

The two remaining relevant equations are:
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial y}{\partial y_0}=\frac{\partial (y_0+v)}{\partial y_0}=1+\frac{\partial v}{\partial y_0}$$

Before I keep going, can we confirm if $\frac{\partial y_0}{\partial x_0}=\frac{\partial x_0}{\partial y_0}=0$ is true or not? The reasoning of non-dependence seems logical but I lack the background to really know if it makes sense or not.

 

[Chestermiller]

Ok. I'll do a little recap because it's been a while since I could get to this.
We come from this expression that represents the distance from 2 points that were originally very close to each other before the deformation happened (post #32 for sketch).

$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial x}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

Then, introducing the displacement field we get:
$$x=x_0+u$$
$$y=y_0+v$$

As a result, $\left \| d\vec{s} \right \|^2$ can be expressed in terms of the displacement field. For that, I'll state some intermediate equations to keep track of.
$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial y}{\partial x_0}=\frac{\partial (y_0+v)}{\partial x_0}=\frac{\partial y_0}{\partial x_0}+\frac{\partial v}{\partial x_0}$$

I'm not sure if we can say $\frac{\partial y_0}{\partial x_0}=0$ because the original vertical position of the point does not depend on the original horizontal position of the point. For now I'll leave it as shown without assuming it's zero.

The two remaining relevant equations are:
$$\frac{\partial x}{\partial y_0}=\frac{\partial (x_0+u)}{\partial y_0}=\frac{\partial x_0}{\partial y_0}+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial y}{\partial y_0}=\frac{\partial (y_0+v)}{\partial y_0}=1+\frac{\partial v}{\partial y_0}$$

Before I keep going, can we confirm if $\frac{\partial y_0}{\partial x_0}=\frac{\partial x_0}{\partial y_0}=0$ is true or not? The reasoning of non-dependence seems logical but I lack the background to really know if it makes sense or not.

Of course. Both are independent variables.

 

[Juanda]

Of course. Both are independent variables.

All right. That simplifies things a little bit. I'll continue.

The objective is to express the distance from 2 points that were originally very close to each other before the deformation happened:
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial y}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

In terms of the displacement field:
$$x=x_0+u$$
$$y=y_0+v$$

Introducing some intermediate equations so the final result doesn't seem like it came out of nowhere:
$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial x_0}^2=(1+\frac{\partial u}{\partial x_0})^2=1+2\frac{\partial u}{\partial x_0}+\frac{\partial u}{\partial x_0}^2$$

$$\frac{\partial y}{\partial x_0}=\frac{\partial (y_0+v)}{\partial x_0}=\frac{\partial y_0}{\partial x_0}+\frac{\partial v}{\partial x_0}=\frac{\partial v}{\partial x_0}$$
$$\frac{\partial y}{\partial x_0}^2=\frac{\partial v}{\partial x_0}^2$$

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

$$\frac{\partial y}{\partial y_0}=\frac{\partial (y_0+v)}{\partial y_0}=1+\frac{\partial v}{\partial y_0}$$
$$\frac{\partial y}{\partial y_0}^2=(1+\frac{\partial v}{\partial y_0})^2=1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2$$

Finally, plugging all that in the expression for $\left \| d\vec{s} \right \|^2$ yields the following:
$$\left \| d\vec{s} \right \|^2=(1+2\frac{\partial u}{\partial x_0}+\frac{\partial u}{\partial x_0}^2+\frac{\partial v}{\partial x_0}^2)dx_0^2
+(\frac{\partial u}{\partial x_0}^2+1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2)dy_0^2
+2((1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial x_0}))+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0}))dx_0dy_0$$

After all that I feel I am just as lost as I was at the beginning if not more. Is there anything I should be able to see in that expression? All I see is it's very long and I don't know what else to do with it.

 

[Chestermiller]

All right. That simplifies things a little bit. I'll continue.

The objective is to express the distance from 2 points that were originally very close to each other before the deformation happened:
$$\left \| d\vec{s} \right \|^2 = \left \| \vec{B}-\vec{A} \right \|^2=\left \| (dx,dy) \right \|^2=dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial y}{\partial y_0}^2)dy_0^2+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

In terms of the displacement field:
$$x=x_0+u$$
$$y=y_0+v$$

Introducing some intermediate equations so the final result doesn't seem like it came out of nowhere:
$$\frac{\partial x}{\partial x_0}=\frac{\partial (x_0+u)}{\partial x_0}=1+\frac{\partial u}{\partial x_0}$$
$$\frac{\partial x}{\partial x_0}^2=(1+\frac{\partial u}{\partial x_0})^2=1+2\frac{\partial u}{\partial x_0}+\frac{\partial u}{\partial x_0}^2$$

$$\frac{\partial y}{\partial x_0}=\frac{\partial (y_0+v)}{\partial x_0}=\frac{\partial y_0}{\partial x_0}+\frac{\partial v}{\partial x_0}=\frac{\partial v}{\partial x_0}$$
$$\frac{\partial y}{\partial x_0}^2=\frac{\partial v}{\partial x_0}^2$$

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

$$\frac{\partial y}{\partial y_0}=\frac{\partial (y_0+v)}{\partial y_0}=1+\frac{\partial v}{\partial y_0}$$
$$\frac{\partial y}{\partial y_0}^2=(1+\frac{\partial v}{\partial y_0})^2=1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2$$

Finally, plugging all that in the expression for $\left \| d\vec{s} \right \|^2$ yields the following:
$$\left \| d\vec{s} \right \|^2=(1+2\frac{\partial u}{\partial x_0}+\frac{\partial u}{\partial x_0}^2+\frac{\partial v}{\partial x_0}^2)dx_0^2
+(\frac{\partial u}{\partial x_0}^2+1+2\frac{\partial v}{\partial y_0}+\frac{\partial v}{\partial y_0}^2)dy_0^2
+2((1+\frac{\partial u}{\partial x_0})(\frac{\partial u}{\partial x_0}))+(\frac{\partial v}{\partial x_0})(1+\frac{\partial v}{\partial y_0}))dx_0dy_0$$

After all that I feel I am just as lost as I was at the beginning if not more. Is there anything I should be able to see in that expression? All I see is it's very long and I don't know what else to do with it.

We’re not done yet. Please be patient. I’ll be back later.

 

[Chestermiller]

Let $$\frac{dx_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dx_0}{ds_0}=\cos{\alpha}$$and $$\frac{dy_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dy_0}{ds_0}=\sin{\alpha}$$such that $\alpha$ defines the direction of the arbitrary differential position vector between the two neighboring material points in the undeformed configuration. Then show that the stretching between the material points can be expressed in the form $$\lambda^2=(\frac{ds}{ds_0})^2=(\lambda_x)^2\cos^2{\alpha}+(\lambda_y)^2\sin^2{\alpha}+4\epsilon_{xy}\sin{\alpha}\cos{\alpha}$$From the previous development, what are the expressions for $\lambda^2_x$, $\lambda^2_y$, and $\epsilon_{xy}$?

 

[Juanda]

Let me first add the first two equations to the drawing I made before to make sure I'm still following you.
Is this correct?

 

[Chestermiller]

Let me first add the first two equations to the drawing I made before to make sure I'm still following you.
Is this correct?
View attachment 332983

yes

 

[Juanda]

So I've been trying for a few days and I simply don't know how to approach it.
I think this is as far as I can get. I'm glad I could at least get the first question but the second one just feels too hard for me to crack.

 

[Chestermiller]

$$\left \| d\vec{s} \right \|^2 =dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial y}{\partial y_0}^2)dy_0^2$$$$+2(\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0})dx_0dy_0$$

Dividing this equation by $(ds_0)^2=(dx_0)^2+(dy_0)^2$ gives $$\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$$$+2\left[\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0}\right]\sin{\alpha}\cos{\alpha}$$

OK so far?

 

[Juanda]

Dividing this equation by $(ds_0)^2=(dx_0)^2+(dy_0)^2$ gives $$\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$$$+2\left[\frac{\partial x}{\partial x_0}\frac{\partial x}{\partial y_0}+\frac{\partial y}{\partial x_0}\frac{\partial y}{\partial y_0}\right]\sin{\alpha}\cos{\alpha}$$

OK so far?

Just to be sure.

FROM
$$\left \| d\vec{s} \right \|^2 =dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial y}{\partial y_0}^2)dy_0^2$$

AND
$$\left \| d\vec{s_0} \right \|^2 =dx_0^2+dy_0^2$$

If I simply combine both doing
$$\lambda^2=\frac{\left \| d\vec{s} \right \|^2}{\left \| d\vec{s_0} \right \|^2}$$

In the end, it reduces to
$$\lambda^2=\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$

With
$$\frac{dx_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dx_0}{ds_0}=\cos{\alpha}$$
$$\frac{dy_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dy_0}{ds_0}=\sin{\alpha}$$

Is that how it is? It's a hard blow to realize I missed it and I just needed to keep pressing to reduce the expression algebraically although even after looking at the final expression I can't really see how it simplified all the way down to the final expression you showed.
I was focusing on the displacement field because it's the last part we focused on so I didn't think of using the previous expressions. I actually think I could have gotten to a somewhat equivalent point using the expression of the displacement field and I just got lost in the swarm of letters I was trying to deal with.

 

[Chestermiller]

Just to be sure.

FROM
$$\left \| d\vec{s} \right \|^2 =dx^2+dy^2=(\frac{\partial x}{\partial x_0}^2+\frac{\partial y}{\partial x_0}^2)dx_0^2+(\frac{\partial x}{\partial y_0}^2+\frac{\partial y}{\partial y_0}^2)dy_0^2$$

AND
$$\left \| d\vec{s_0} \right \|^2 =dx_0^2+dy_0^2$$

If I simply combine both doing
$$\lambda^2=\frac{\left \| d\vec{s} \right \|^2}{\left \| d\vec{s_0} \right \|^2}$$

In the end, it reduces to
$$\lambda^2=\frac{(ds)^2}{(ds_0)^2}=\left[\left(\frac{\partial x}{\partial x_0}\right)^2+\left(\frac{\partial y}{\partial x_0}\right)^2\right]\cos^2{\alpha}+\left[\left(\frac{\partial x}{\partial y_0}\right)^2+\left(\frac{\partial y}{\partial y_0}\right)^2\right]\sin^2{\alpha}$$

With
$$\frac{dx_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dx_0}{ds_0}=\cos{\alpha}$$
$$\frac{dy_0}{\sqrt{(dx_0)^2+(dy_0)^2}}=\frac{dy_0}{ds_0}=\sin{\alpha}$$

Is that how it is?

You're missing the $dx_0dy_0$ term.

It's a hard blow to realize I missed it and I just needed to keep pressing to reduce the expression algebraically although even after looking at the final expression I can't really see how it simplified all the way down to the final expression you showed.
I was focusing on the displacement field because it's the last part we focused on so I didn't think of using the previous expressions. I actually think I could have gotten to a somewhat equivalent point using the expression of the displacement field and I just got lost in the swarm of letters I was trying to deal with.

That's OK. I didn't now what I wanted to introduce first.

What do you get if you now convert to the 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