What is the optimal shape of a membrane on a rectangle under pressure?

[skrat]

Chet,

Thank you for your help!

So, the shape is $r(s)=x(s)\vec i+y(s)\vec j$ where $x(s)=R\sin(\varphi)$ and $y(s)=-R\cos(\varphi)$ leading to $$r(\varphi)=R(\sin\varphi,-\cos\varphi)$$ for $-\varphi _0<\varphi<\varphi_0$, where $\varphi$ is a (numerical or any other...) solution to $$\frac{\cos\varphi_0}{\varphi_0}=\frac{h}{2s_0}.$$ Of course on the top side of the rectangle, the shape of the membrane shouldn't differ at all.

One more thing that I wanted to ask is... your post under #28, where we talked about the horizontal balance. So, the question is... Since the membrane is attached to the top and to the bottom of the plate, shouldn't there the last term be $ph/2$? Also what would happen if there was an external force acting on the plates pulling them apart. Would I get term $F/2$ or $F$ in the force balance equation?

EDIT: Also, after checking the other idea on how to get the shape... by integration... makes is obvious that $y(s)=R(1-\cos\varphi(s)).$

 

[Chestermiller]

One more thing that I wanted to ask is... your post under #28, where we talked about the horizontal balance. So, the question is... Since the membrane is attached to the top and to the bottom of the plate, shouldn't there the last term be $ph/2$?

No. You had it right. It's either $2T\cosφ_0=ph$ or $T\cosφ_0=ph/2$

Also what would happen if there was an external force acting on the plates pulling them apart. Would I get term $F/2$ or $F$ in the force balance equation?

It depends on whether you were doing it on the full plate or half the plate: $2T\cosφ_0=ph+F$ versus $T\cosφ_0=ph/2 + F/2$.

Have you had a chance to plot the graph that I mentioned in post # 30 yet? Does it look as expected, and does it make sense? Does it match your asymptotic results?

Chet

 

[skrat]

Actually I plotted it a bit on my own...

I used Mathematica to plot $\cos x/x$ with $x$ on the abscissa. Now since $y=h/2s_0$ is the given parameter all one has to do in order to find $x$ is to find the intersection of $\cos x/x$ and the constant value of parameters.

BUT, since you specifically order me, I followed your instructions to.. And the results strangely don't match. Excel, which I am not very fond of, so I might be doing something wrong, plots this (for $\varphi$ from 0 to ... more than $\pi/2$):

And this looks nothing like $\cos(x)/x$ http://www.wolframalpha.com/input/?i=cos(x)/x . This is my mistake, right?

 

[skrat]

Is it also possible to look at the whole problem as an isoperimetric problem from the maths? Concretely "Dido's problem" leads to a circular shape - which is also the case here.
That is if I assume that the area under the membrane will be maximized. I think this is a good argument too, but I don't have a good physical argument why the area would be maximized. Do you?
Is it due to the pressure difference? The air wants to expand as much as possible, therefore increase the area? :/

 

[Chestermiller]

Here are the result that I got, with φ expressed in degrees:

Edit: Oops. I meant φ₀ rather than φ on the graph. Incidentally, I checked and, at least at large values of φ₀, your asymptotic solution matches the graph.

Chet

 

[Chestermiller]

Is it also possible to look at the whole problem as an isoperimetric problem from the maths? Concretely "Dido's problem" leads to a circular shape - which is also the case here.
That is if I assume that the area under the membrane will be maximized. I think this is a good argument too, but I don't have a good physical argument why the area would be maximized. Do you?
Is it due to the pressure difference? The air wants to expand as much as possible, therefore increase the area? :/

I really can't answer these questions, because they are outside my math knowledge base. If you like, I can ask a Mentor specializing in math to help.

Chet

 

[skrat]

I really can't answer these questions, because they are outside my math knowledge base. If you like, I can ask a Mentor specializing in math to help.

Chet

No no, no need to Chet. You have already done way more than I had hoped for! And I am very grateful for that! I will post a problem in the Calculus section and hopefully get to a similar conclusion than here.
I also included the external force in the calculation... And I am surprised how small the difference is if you apply $150$ N when the inside pressure is $50$ Pa to a rectangle the size of $h=45$ and $2s_0=30$. Amazing.

Thank you Chet. Great advices! I learned a lot! I can only hope to have a similar mentor when working on my masters thesis...

 

[Chestermiller]

No no, no need to Chet. You have already done way more than I had hoped for! And I am very grateful for that! I will post a problem in the Calculus section and hopefully get to a similar conclusion than here.
I also included the external force in the calculation... And I am surprised how small the difference is if you apply $150$ N when the inside pressure is $50$ Pa to a rectangle the size of $h=45$ and $2s_0=30$. Amazing.

Thank you Chet. Great advices! I learned a lot! I can only hope to have a similar mentor when working on my masters thesis...

Skrat, you're making me blush.

I think you meant to say that the force is 150 N per meter into the z direction. If h = 45 m, then F/h ~ 3 Pa, compared to the pressure of 50 Pa. So, the effect of the force would be equivalent to increasing the pressure by ~6%.

Chet

 

[skrat]

Still, one question at your post #28 remains, because I don't seem to understand how you got that equation for the vertical plate. I guess that is where the most physics is hidden..

Let me complicate things a bit. Take a look at the attached picture.

Since we assume that the vertical plates are always parallel, the shape of the membrane doesn't change. It is a circle as we proved and it has the same radius at the top of the "chamber" as it does at the bottom. That's my feeling.
BUT what would the force balance for a single plate look like in this case?

Because in this case I can't figure out how to include the pressure difference into the equation, as we did in post #28. That is why in post #27 I wrote $$2\cos\varphi_0T=2p\sin\varphi_0+ph$$ thinking that the first term on the RHS describes the force (due to the pressure difference) at juncture of the membrane and rigid plate - of course as you noticed the dimension doesn't match.

 

[Chestermiller]

The parameter p is actually the difference between the pressure inside the chamber and the pressure outside the chamber. The pressure inside the chamber as well as the pressure outside the chamber are uniform.

Regarding the force balance equation, here is my free body diagram:

I don't see any additional contribution of the pressure to the force balance, even at the corners. The only forces acting on the end plate are the tensions of the membranes and the distributed pressure on the face.

Chet

 

[skrat]

I don't see any additional contribution of the pressure to the force balance, even at the corners. The only forces acting on the end plate are the tensions of the membranes and the distributed pressure on the face.

Than I don't understand how to write a force balance equation in this case:

Since the pressure difference $p$ on the plate is zero (on the middle chamber) there is no force. And if there is no force, there is no balance.

 

[Chestermiller]

Have you drawn a free body diagram on either of the plates for the middle chamber? How many membranes are attached to each of these plates, (a) 2 or (b) 4?

 

[skrat]

One plate, 4 membranes. (The size of the vectors on the graphics should be the same.)

 

[Chestermiller]

View attachment 85767
One plate, 4 membranes. (The size of the vectors on the graphics should be the same.)

Right. So you can see that the membrane forces exerted on the two sides of the plate balance each other, and the pressure forces exerted on the two sides of the plate balance each other. So the plate is in force equilibrium.

Chet

 

[skrat]

Exactly!
But what about the angle between the horizontal line and tension force, $\varphi_0$? We calculated it from the horizontal force balance.
And we got to a transcendental equation for $\varphi_0$. The solution to this transcendental equation depended on $h$ and $s_0$ if there was no external force, while it also depended on $p$ and $F$ if there was an external force.

But like you said, writing the force balance equation only yields the fact, that the chamber is in equilibrium. That information is of course useful, but still - I need the shape of the membrane. I would need to know that $\varphi _0$ again and the radius of the membrane. Which I can't get by writing horizontal forces. Or maybe I am so much into this problem that I am missing something completely obvious.

 

[Chestermiller]

Exactly!
But what about the angle between the horizontal line and tension force, $\varphi_0$? We calculated it from the horizontal force balance.
And we got to a transcendental equation for $\varphi_0$. The solution to this transcendental equation depended on $h$ and $s_0$ if there was no external force, while it also depended on $p$ and $F$ if there was an external force.

But like you said, writing the force balance equation only yields the fact, that the chamber is in equilibrium. That information is of course useful, but still - I need the shape of the membrane. I would need to know that $\varphi _0$ again and the radius of the membrane. Which I can't get by writing horizontal forces. Or maybe I am so much into this problem that I am missing something completely obvious.

If the force F is zero, then, for the three chamber system, does the solution that we have now for membrane shapes (identical for all three chambers) satisfy the force balance equation on each of the four vertical plates? If yes, then we can claim victory and pat ourselves on the back.

Chet

 

[skrat]

If the force F is zero, then, for the three chamber system, does the solution that we have now for membrane shapes (identical for all three chambers) satisfy the force balance equation on each of the four vertical plates?

It just seems odd, that the pressure distribution, as shown in post #41, plays absolutely no role and basically disappears from the equation.

But ok. I'll buy that.

 

[SammyS]

This thread appears to be continued in the following thread:

https://www.physicsforums.com/threads/find-the-right-shape.822563/#post-5164658

 

[Chestermiller]

It just seems odd, that the pressure distribution, as shown in post #41, plays absolutely no role and basically disappears from the equation.

But ok. I'll buy that.

I'm pretty confused. Here is my understanding of the problem we have solved so far (which is already an extension of the problem that we started with).

We have 3 chambers bounded by 6 membranes and 4 vertical plates. The pressure outside all three chambers is uniform. The pressures inside all three chambers are uniform and equal, and exceed the pressure outside the chambers by p. Our solution to this problem indicates that:
1. The shapes of all 6 membranes are circular arcs, identical, and symmetric with respect to the centerline of each chamber.
2. The 12 angles of the membranes at the edges adjacent to the plates are all equal to one another.
3. The 4 plates are all vertical and equally spaced.
4. The membranes and the plates are fully in static equilibrium.

Is this your understanding of what we have solved so far? Do you have discomfort with this solution? If so, for what reason?

Are you now trying to extend what we have done to a more complicated situation? Are you asking how to do this?

Chet

 

[skrat]

This is exactly my understanding yes, and there is no discomfort with that.

But I have discomfort with extending the problem from one chamber to multiple ($n$) chambers. Why? Because:

1. I agree with the circular shape of the membrane. Nothing changes here. I can still take one small piece of a membrane, draw a free body diagram and seek for equilibrium state. I don't see any changes here. Therefore I believe that all the membranes will have a circular shape.
2. Than we tried to find the radius $T/p$ and domain space for $\varphi$. Here is where my discomfort is HUGE. Because: For a single chamber we drew a free body diagram and demanded balance in horizontal direction.

The balance condition is $$2T(s)\cos\varphi_0 =ph$$ which leads to a transcendental equation $$\frac{\cos(\varphi_0)}{\varphi _0}=\eta$$ with $\eta$ being a constant that depends on the given parameters of the one chamber system. We than solved the transcendental equation numerically (or using graphics, not important) and doing so we were able to determine the radius of the circle $T/p$ and also the maximum value $\varphi_0$. All fine and well, but in case of $n$ chambers (looking at any chamber that is not on the edge)

the horizontal force balance yields only $$2T_{n}(s)\cos\varphi_0 =2T_{n+1}\cos\varphi_0$$ which gives me no information about the radius, nor does it tell me anything about $\varphi_0.$ Even if one was to consider a single chamber only, than using the pressure distribution I suggested here where $p_{out}^0<p_{out}^1<p_0$

and your free body diagram (first picture in this post for $p=0$ because pressure inside the chambers is the same for all chambers) for a single plate... One would get ... well, basically nothing. We only have the circular shape so far, but I can see no information about radius or $\varphi_0$. - That is the origin of my discomfort.

I hope I was clear enough this time. :/

 

[mfb]

You are right, but you can still use another trick: Split the membrane in two pieces, connect them with a solid bar. In equilibrium, the bar cannot have tension or compression (otherwise the individual parts would accelerate). Now set the length of that bar to zero.

 

[skrat]

Regarding the radius of curvature, that is equal to T/p. But another easy way you can get the shape of the contour is by solving the differential equations for dx/ds and dy/ds by substituting your value for φ(s) into the relationships and integrating with respect to s.

Chet

So $T/p$ is the radius in case there is no external force. Does this change when some external force is exerted on the vertical planes, or does the radius remain $T/p$ ?

 

[Chestermiller]

So $T/p$ is the radius in case there is no external force. Does this change when some external force is exerted on the vertical planes, or does the radius remain $T/p$ ?

It changes.

Chet

 

[skrat]

Ok.. How?

It was easy to see what the radius is in case there is no external force. $$\frac{dx}{ds}=\cos(\frac p T s)$$ yields $$x(s)=\frac T p \sin (\frac p T s).$$ Therefore the radius is obviously $T/p$.

Since $\varphi =\frac p T s$ than in case there is no external force one has to solve a transcendental equation $$\cos \varphi _0 = \varphi _0 \frac{h}{2s_0}$$ to get numerical value of $\varphi_0$. This than yields $\frac T p= \frac{s_0}{\varphi _0}$.
In case there is an external force only the transcendental equation for $\varphi _0$ changes to $$\cos\varphi _0=\varphi _0 \frac{1}{2s_0p}(ph+F).$$ Is this the only change or did you have something else in mind?

 

[Chestermiller]

This is what I had in mind, assuming that F is applied equally on both sides.

Chet

 

[skrat]

Sorry, but I have (once again) some discomfort with the solution.

Looking at $$\frac{\cos \varphi_0}{\varphi_0}=\frac{1}{2s_0p}(ph+F)$$ and than checking the dimensions (units) of the RHS $$\frac{1}{mPa}(mPa + N)=1+m$$ yields something that should not happen. I assume that the force should be divided by some kind of length unit.

EDIT:
The way I understand it the original condition should say $$ 2T\cos\varphi_0 \Delta z=ph\Delta z+F$$ if $\Delta z$ is the depth of the chamber - 3D ilustration.
This would than reduce to $$\frac{\cos \varphi_0}{\varphi_0}=\frac{1}{2s_0p}(ph+F/\Delta z)$$ and the units would match.

 

[Chestermiller]

Yes. I was assuming you were calling F the force per unit depth into the chamber. That is, I thought your F was really what you now call F/Δz in the above equation.

Chet

 

[skrat]

Hi Chet,

What if instead of setting the arc length as constant I was to set the width (distance between the parallel plates) as constant?

Than $$\int _0 ^wdx=\int _0^{s_0}\cos(\frac p T s)ds$$ therefore $$w=\frac T p \sin (\frac p T s_0)$$ where $w$ is the distance between the vertical plates. Now this reduces to $$w=R\sin(\frac {s_0}{R})$$ where $R$ is the radius of the membrane shape and I could get the arc length $s_0$ if I knew the radius. But here is the problem. The radius is $$R=\frac T p =\frac{s_0}{\varphi _0}$$ but to get $\varphi _0$ one has to solve transcendental equation that also includes $s_0$ $$\frac{\cos \varphi_0}{\varphi_0}=\frac{1}{2s_0p}(ph+F/\Delta z).$$
So to me it looks like I am walking in circles... Is it even possible to model this with assuming that the distance between the vertical plates is given and of course smaller than the arc length?

 

[Chestermiller]

Hi Chet,

What if instead of setting the arc length as constant I was to set the width (distance between the parallel plates) as constant?

Than $$\int _0 ^wdx=\int _0^{s_0}\cos(\frac p T s)ds$$ therefore $$w=\frac T p \sin (\frac p T s_0)$$ where $w$ is the distance between the vertical plates. Now this reduces to $$w=R\sin(\frac {s_0}{R})$$ where $R$ is the radius of the membrane shape and I could get the arc length $s_0$ if I knew the radius. But here is the problem. The radius is $$R=\frac T p =\frac{s_0}{\varphi _0}$$ but to get $\varphi _0$ one has to solve transcendental equation that also includes $s_0$ $$\frac{\cos \varphi_0}{\varphi_0}=\frac{1}{2s_0p}(ph+F/\Delta z).$$
So to me it looks like I am walking in circles... Is it even possible to model this with assuming that the distance between the vertical plates is given and of course smaller than the arc length?

The solution is not unique. There are an infinite number of arc lengths that are consistent with a fixed spacing of the plates. You're just adding more material and making the radius of curvature of the membrane larger.

Chet

 

[skrat]

Hi,

I am so sorry to start this topic again but I have a lack of physics knowledge to answer the following question:

What if the vertical plates wouldn't have the same dimensions? Let's say that the left plate is $h+\Delta h$ while the right plate is only $h$ high.

My understanding is that the membrane would still remain in circular shape, only the angles $\varphi _0$ are not the same any more on the left and on the right plat where the membrane is attached to the plate. If my understanding is wrong than on the right plate $\varphi _0\rightarrow \varphi_0+arctan(\frac{\Delta h}{w})$ if $w$ is the distance between the vertical plates, while on the left plate $\varphi _0\rightarrow \varphi_0-arctan(\frac{\Delta h}{w})$.

Makes sense or not?