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

[skrat]

Homework Statement

This is not a homework or anything similar, but it is a problem I am trying to solve and I don't know how to start.

I do know, that the most stable structure in 3D is a sphere - that is the reason why the drops are in a spherical shape. But now imagine a rectangle

and imagine that the $x$ sides are actually made of a membrane - therefore they can deform (bend) but not stretch and let's assume that the $y$ sides are completely rigid.

Now let's put some air pressure inside the rectangle $p_{in}$ while the outside pressure is $p_{out}$ where $p_{in}-p_{out}>0$.

What will happen is that since the membrane can't stretch but it can bend the distance between the $x$ sides will change from $y$ to $y-\delta y$. Now I have to and want to find the shape of the bended membrane.

I want to know everything about that shape. Don't say it is a circle, even if it is, I want a proof. I want an equation of the membrane.

Homework Equations

The Attempt at a Solution

Well... I don't really know. I tried to start with the fact that the surface $S$ has to stay the same. Therefore the initial surface is $S=xy$ while at the end $S=x(y-\delta y)+2{S}'$ if i use notation ${S}'$ for the part that comes due to the bended membrane.

But in order to calculate $S_0$ I need the shape of the membrane. So how would I get it? Any hints will be highly appreciated.

So I would somehow have to find a minimum energy for a given shape. And I am googling it and stuff like that but no success so far. :/

 

[SteamKing]

Homework Statement

This is not a homework or anything similar, but it is a problem I am trying to solve and I don't know how to start.

I do know, that the most stable structure in 3D is a sphere - that is the reason why the drops are in a spherical shape. But now imagine a rectangle
View attachment 85599

and imagine that the $x$ sides are actually made of a membrane - therefore they can deform (bend) but not stretch and let's assume that the $y$ sides are completely rigid.

Now let's put some air pressure inside the rectangle $p_{in}$ while the outside pressure is $p_{out}$ where $p_{in}-p_{out}>0$.

What will happen is that since the membrane can't stretch but it can bend the distance between the $x$ sides will change from $y$ to $y-\delta y$. Now I have to and want to find the shape of the bended membrane.

I want to know everything about that shape. Don't say it is a circle, even if it is, I want a proof. I want an equation of the membrane.

Homework Equations

The Attempt at a Solution

Well... I don't really know. I tried to start with the fact that the surface $S$ has to stay the same. Therefore the initial surface is $S=xy$ while at the end $S=x(y-\delta y)+2{S}'$ if i use notation ${S}'$ for the part that comes due to the bended membrane.

But in order to calculate $S_0$ I need the shape of the membrane. So how would I get it? Any hints will be highly appreciated.

I'll give it to you straight: the deflection of a membrane in the general case is not an easy problem to solve. In fact, the equation which must be solved is something called the biharmonic equation, and the paper at the link below talks about analyzing circular membranes under pressure:

http://www.sandia.gov/mems/_assets/documents/bibliography/5_1New.pdf

If you know anything about potential theory or solving the Laplace equation, you'll see instantly that solving a biharmonic equation is a lot more complicated.

Remember, the paper talks about membranes which have circular shapes, so there is a lot of symmetry which simplifies the solution.

For membranes which have a rectangular shape, the solutions get much more complicated. This is why people who study such things still build a lot of physical, rather than numerical, models for testing, or analyze the shape of soap bubbles, for example.

 

[skrat]

Yeah maybe my English is bad and this lead to a misunderstanding.

The whole thing is a two dimensional problem. Forget about the depth of the rectangle - not interested. Meaning in a way this should be a very similar (if I am not missing something) problem to a supported beam on which uniform force is applied https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams .

Or is it not that simple?

ps.: My apologies, I imagined there is a simple solution to the problem that is why I posted this in "introductory physics", but biharmonic equations of course are everything but introductory.

 

[SteamKing]

Yeah maybe my English is bad and this lead to a misunderstanding.

The whole thing is a two dimensional problem. Forget about the depth of the rectangle - not interested. Meaning in a way this should be a very similar (if I am not missing something) problem to a supported beam on which uniform force is applied https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams .

Or is it not that simple?

ps.: My apologies, I imagined there is a simple solution to the problem that is why I posted this in "introductory physics", but biharmonic equations of course are everything but introductory.

No, I don't think there is any confusion here.

The circular membranes discussed in the paper are indeed clamped in two dimensions. The movement of the membrane in response to the applied pressure is generally shown as being perpendicular to the plane in which the membrane is clamped.

We're talking initially about soap bubbles and stuff, but analyzing things like the response of metal plates to pressure loads is one of the practical applications of this membrane theory. Most of the results have been obtained from actual tests of plates with prescribed loads and the requisite simple or clamped supports around the edges. Or soap bubbles, which is cheaper and more fun. Lately, however, the math guys have caught up, and more of the analysis is done numerically.

As to whether you can analyze certain membrane problems like an analogous beam problem, I suppose you can, but only in a very crude fashion. It all comes down to the support conditions at the edges of the plate or membrane. As with beams, for membrane analysis there are both large deflection and small deflection theories, with the small deflection theory being somewhat simplified over the more general large deflection theory. In other words, the small deflection theory of membrane analysis would correspond roughly with simple Euler-Bernoulli beam theory.

 

[Chestermiller]

Yeah maybe my English is bad and this lead to a misunderstanding.

The whole thing is a two dimensional problem. Forget about the depth of the rectangle - not interested. Meaning in a way this should be a very similar (if I am not missing something) problem to a supported beam on which uniform force is applied https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams .

Or is it not that simple?

ps.: My apologies, I imagined there is a simple solution to the problem that is why I posted this in "introductory physics", but biharmonic equations of course are everything but introductory.

Yes, your problem description in very confusing on a number of counts:

1. You call it a membrane, but, in mechanics, a membrane is something that has stretching stiffness, but no bending stiffness. You say that it is not stretching, but only bending. This seems to be a description of a plate.

2. The geometry is not clear. I think you mean to be implying that the structure extends infinitely in the z-direction, but I'm not sure.

3. You say nothing about the properties of the material. At first, you say that it may behave like a fluid with surface tension, but I think that you mean it to have mechanical properties like a Hookean solid. Is this correct?

If you truly want to treat the upper and lower sides as plates, then you need to specify the boundary conditions at the junctions with the y-sides, over and above zero displacement. Some choices are (a) simple support (zero moments) or (b) build-in (no change in slope).

If you are dealing with a Hookean plate of infinite extent in the z-direction, then this becomes a simple 1D problem, and can be treated as a beam. But, because of the zero strain condition that would exist in the z-direction, the bending stiffness will be different from a 1D beam.

Chet

 

[skrat]

1. It doesn't stretch, it only deflects/bends. It didn't want to use the word "plate" because I am actually thinking about a simple cloth (cloth that doesn't let any air through) instead of $x$ sides and having rigid $y$ sides. I think this is now clear enough.

2. There is no $z$ direction. I made the sketch below. I hope it is clear enough.

3. Mechanical properties... If any are needed let's assume they are known.

Of course, one boundary condition is that the deflection of the cloth on the edge is 0. And the force (due to the pressure difference) is uniformly distributed. And my idea was to... I don't know what it was, I would need to find the shape of the cloth- analytical if possible (which turns out to be complicated). But if the analytical solution is too complicated I could make some reasonable simplifications. For example, say that the shape is Catenary or deflected beam or...

EDIT: Let me explain why catenary... Because the only difference that I can see is that in the case of Catenarry the force is always vertica, while here it is normal to the line... I am sure this must lead to some kind of solution...

 

[Chestermiller]

1. It doesn't stretch, it only deflects/bends. It didn't want to use the word "plate" because I am actually thinking about a simple cloth (cloth that doesn't let any air through) instead of $x$ sides and having rigid $y$ sides. I think this is now clear enough.

2. There is no $z$ direction. I made the sketch below. I hope it is clear enough.
View attachment 85602
3. Mechanical properties... If any are needed let's assume they are known.

Of course, one boundary condition is that the deflection of the cloth on the edge is 0. And the force (due to the pressure difference) is uniformly distributed. And my idea was to... I don't know what it was, I would need to find the shape of the cloth- analytical if possible (which turns out to be complicated). But if the analytical solution is too complicated I could make some reasonable simplifications. For example, say that the shape is Catenary or deflected beam or...

EDIT: Let me explain why catenary... Because the only difference that I can see is that in the case of Catenarry the force is always vertica, while here it is normal to the line... I am sure this must lead to some kind of solution...

If it is like a cloth, then what you are dealing with here is indeed a membrane that can stretch but which has no bending rigidity. The stretching takes place "in the plane" of the membrane, and that creates the added length. This is more like a drum, or a trampoline (or an inflated tire).

The membrane stretches in the x-direction, and gets thinner in the y-direction. The strain in the z-direction is zero. The stress in the thickness direction is zero.

This is a 1D problem that can pretty easily be solved if you can manage the description of the kinematics of the deformation, and the handling of Hooke's law (3D tensorial version). At least, it can be solved pretty easily if the deflections are small so that you can neglect geometric non-linearities. Basically, you are solving for the displacement in the y direction as a function of x. The curvature developed in the membrane allows the tensile stresses in the membrane to support the normal pressure.

The analyses that SteamKing was referring to basically focuses on this type of problem. However, this problem is much easier because it is 1D.

Are you familiar with the 3D tensorial version of Hooke's law? Do you know how to algebraically get the stress in the x direction (in terms of the strain in the x direction) if the stress in the y direction is zero and the strain in the z direction is zero? This is the starting point. You also need to do a differential force balance on the membrane for the segment of the membrane between x and x + dx.

Chet

 

[skrat]

If it is like a cloth, then what you are dealing with here is indeed a membrane that can stretch but which has no bending rigidity. The stretching takes place "in the plane" of the membrane, and that creates the added length. This is more like a drum, or a trampoline (or an inflated tire).

Again, the cloth does not stretch!

The geometry at time $t=0$ is a rectangle with sides $x$ and $y$. Than we increase the gas pressure inside the rectangle at time $t_0$. Since the cloth can bend but not stretch this will no longer be rectangle. The vertical sides will stay the same - they are rigid with length $y$ BUT because cloth will bend outwards the $y$ sides will come closer together, the new distance is $x-\delta x$ for $\delta x>0$.

 

[Chestermiller]

Again, the cloth does not stretch!

The geometry at time $t=0$ is a rectangle with sides $x$ and $y$. Than we increase the gas pressure inside the rectangle at time $t_0$. Since the cloth can bend but not stretch this will no longer be rectangle. The vertical sides will stay the same - they are rigid with length $y$ BUT because cloth will bend outwards the $y$ sides will come closer together, the new distance is $x-\delta x$ for $\delta x>0$.

If this cloth is pretty easy to bend by hand, then it's really a membrane. What you are saying is that the membrane is inextensible. I think that, for this particular problem, considering that the ends can move inward, the membrane problem is statically determinate. Do you know how to set this problem up?

Chet

 

[skrat]

It's SUPER easy to bend.
And yes, the ends can move inward.

No, I don't know how to set this problem up, sorry. But I promise to do my best during the whole derivation. I also promise to pay more attention to the words I am using - this may be a bit harder since English is not my first language.

 

[Chestermiller]

Now that I understand the problem, I can help you set it up, since I have had a lot of experience with problems like this.

We are going to express the shape of the membrane parametrically in terms of the arc length s along the membrane contour. So the shape of the membrane is going to be expressed as x = x(s) and y = y(s). We are going to solve for x and y as functions of s.

We are going to focus on the bottom membrane, and we are going to take the origin of our coordinate system x = 0 and y = 0 as the lowest point in the bottom membrane (i.e., half way between the two vertical sides) in the deformed configuration of the system. The arc length coordinate s is taken to be 0 at this location. The junction with the vertical sides is at s = s₀, where s₀ is half the total arc length of the membrane. Since the membrane is inextensible, this length does not change.

We are also going to introduce another parameter φ(s) as the angle that the membrane makes with the horizontal at location s. The center of the membrane at s = 0 is a minimum point on the contour, so φ(0) = 0. Also, since the membrane is inextensible, we must have that
$$\frac{dx}{ds}=\cosφ$$
$$\frac{dy}{ds}=\sinφ$$

Now it's your turn to help. Let $\vec{i}$ be a unit vector in the horizontal direction and $\vec{j}$ be a unit vector in the vertical direction. In terms of φ and these unit vectors, what is the equation for the unit vector at location s in the direction tangent to the membrane contour? From this, what is the equation for the unit vector at location s in the direction normal to the membrane contour?

This is where I'll stop for now.

Chet

 

[skrat]

Ok I understand setting and all the geometry.

If I am not mistaken the tangent vector $\vec t$ is by definition $$\vec t (s)=\frac{\partial \vec r}{\partial s}.$$ In my case, this hopefully means $$\vec t (s)=\frac{\partial }{\partial s}\left( f(s)\cos(\varphi)\hat i+f(s)\sin(\varphi)\hat j \right)$$
where $f(s)$ is the parametrized shape of the membrane. Since $$\frac{\partial }{\partial s} \hat i=\frac{\partial }{\partial s} \hat j=0$$ the equation above simplifies to $$\vec t(s)=\left( {f}'(s)\cos\varphi -f(s)\sin\varphi {\varphi}'\right) \hat i + \left( {f}'(s)\sin\varphi -f(s)\cos\varphi {\varphi}'\right) \hat j.$$ Since normal and tangent vector are perpendicular, than normal vector is $$ \vec n=\left( {f}'(s)\sin\varphi -f(s)\cos\varphi {\varphi}'\right) \hat i -\left( {f}'(s)\cos\varphi -f(s)\sin\varphi {\varphi}'\right) \hat j .$$
Now please say that is not completely wrong.

 

[Chestermiller]

I'm not quite sure what you did, but, anyway, the position vector from our origin to a point on the contour is $$\vec{r}=x(s)\vec{i}+y(s)\vec{j}$$. So, the unit tangent vector to the contour is simply $$\vec{t}=\cosφ\vec{i}+\sinφ\vec{j}$$. From this, what do you get for the unit normal vector?

Chet

 

[skrat]

Ok, I guess I complicated too much. Comparing to your calculation I wrote $x(s)\equiv f(s) \cos\varphi$ and $y(s)\equiv f(s)\sin\varphi$.

Nevermind, if $\vec t=\cos \varphi \vec i +\sin\varphi \vec j$ than the normal vector is $\vec n =\sin\varphi \vec i -\cos \varphi \vec j$.

 

[Chestermiller]

Ok, I guess I complicated too much. Comparing to your calculation I wrote $x(s)\equiv f(s) \cos\varphi$ and $y(s)\equiv f(s)\sin\varphi$.

Nevermind, if $\vec t=\cos \varphi \vec i +\sin\varphi \vec j$ than the normal vector is $\vec n =\sin\varphi \vec i -\cos \varphi \vec j$.

Please try again. The unit normal is not correct. The dot product of the unit tangent and the unit normal must be zero.

Chet

 

[skrat]

$$\vec t \cdot \vec n = (\cos\varphi,\sin\varphi)\cdot(\sin\varphi,-\cos\varphi)=\cos\varphi\sin\varphi-\sin\varphi\cos\varphi=0$$ EDIT: I edited the previous post in about 4 seconds after posting it. I guess you read it in that 4 seconds. Apologies.

 

[Chestermiller]

$$\vec t \cdot \vec n = (\cos\varphi,\sin\varphi)\cdot(\sin\varphi,-\cos\varphi)=\cos\varphi\sin\varphi-\sin\varphi\cos\varphi=0$$ EDIT: I edited the previous post in about 4 seconds after posting it. I guess you read it in that 4 seconds. Apologies.

Excellent.

OK. The next step is to do a differential force balance on the section of the contour between s and s + Δs. Let T(s) represent the tension in the membrane (per unit depth in the z direction) at location s along the contour. Draw a free body diagram of this section of the contour. In terms of T, $\vec{t}$, p, and $\vec{n}$, what are the 3 forces acting on the section. What is the vector force balance on the section?

Chet

 

[skrat]

What is $p$?

 

[Chestermiller]

What is $p$?

The pressure.

Chet

 

[skrat]

See attached picture.

The tension $\vec T(s)$ has the same direction as the tangent vector $\vec t(s)$. And $-\nabla p$ has the same direction as the normal vector $\vec n(s)$.

I am not sure what you meant with the third force though.. Gravity?

 

[Chestermiller]

See attached picture.

The tension $\vec T(s)$ has the same direction as the tangent vector $\vec t(s)$. And $-\nabla p$ has the same direction as the normal vector $\vec n(s)$.View attachment 85614

I am not sure what you meant with the third force though.. Gravity?

The tension force exerted by the part of the membrane to the right of our free body (on our free body) is $T(s+Δs)\vec{t}(s+Δs)Δz$

The tension force exerted by the part of the membrane to the left of our free body (on our free body) is $-T(s)\vec{t}(s)Δz$

The force exerted by pressure on our free body is $p\vec{n}ΔsΔz$

So what is the force balance? Next, divide the force balance by ΔzΔs and take the limit as Δs approaches zero. What do you get?

Chet

 

[skrat]

Everything is stationary,therefore the sum of all the forces has to equal zero $$T(s+\Delta s)\vec t(s+\Delta s)\Delta z-T(s)\vec t(s)\Delta z+p\vec n\Delta s\Delta z=0$$ $$\lim _{\Delta s \rightarrow 0}\frac{T(s+\Delta s)\vec t(s+\Delta s)-T(s)\vec t(s)}{\Delta s}+p\vec n=0$$ which reduces to $$\frac{\partial }{\partial s}\left( T(s)\vec t(s)\right) +p\vec n=0.$$

 

[Chestermiller]

Everything is stationary,therefore the sum of all the forces has to equal zero $$T(s+\Delta s)\vec t(s+\Delta s)\Delta z-T(s)\vec t(s)\Delta z+p\vec n\Delta s\Delta z=0$$ $$\lim _{\Delta s \rightarrow 0}\frac{T(s+\Delta s)\vec t(s+\Delta s)-T(s)\vec t(s)}{\Delta s}+p\vec n=0$$ which reduces to $$\frac{\partial }{\partial s}\left( T(s)\vec t(s)\right) +p\vec n=0.$$

Yes. Very nicely done. (That should be an ordinary derivative, not a partial derivative.)

Now, if you differentiate the first term by parts, what do you get? How is the derivative of $\vec{t}$ with respect to s related to $\vec{n}$ and dφ/ds?

Chet

 

[skrat]

If I use notation $\frac{d \varphi}{ds}={\varphi}'$ than the differential is $$\frac{\partial }{\partial s}\left( T(s)\vec t(s)\right) +p\vec n=0$$ $${T}'(s)\vec t(s)+T(s)\frac{d}{ds}\left( \cos\varphi(s) \vec i+\sin\varphi (s)\vec j\right)+p\vec n=0$$ $${T}'(s)\vec t(s)+T(s)\left(-\sin\varphi(s)\vec i+\cos\varphi(s)\vec j\right){\varphi}'+p\vec n=0$$ The term in the brackets is actually $-\vec n$ therefore the equation reduces to $${T}'(s)\vec t(s)+\left(p-T(s){\varphi}'\right)\vec n(s)=0.$$

 

[skrat]

Now I assume you will order me to solve this DE. In that case I have already done it, if not, I have done some unnecessary work. $$\vec i:\qquad {T}'(s)\cos\varphi +(p-T(s){ \varphi}')\sin\varphi =0$$ and $$\vec j: \qquad {T}'(s)\sin\varphi -(p-T(s){\varphi}')\cos\varphi =0.$$ Combining the two equations leaves me with $$\frac{p-T(s){\varphi}'}{\sin\varphi}=0$$ therefore $$\varphi (s)=C_0+\int _0^{s}\frac{ds}{T(s)}$$ and also $$T(s)=konst.=C_1.$$ Which kinda makes sense, hopefully.

 

[Chestermiller]

This is all done correctly. Very nice.

There are only a couple of comments that I would add.

1. In resolving the force balance into components, the two unit vectors t and n are already orthogonal, so all that one would really need to do would be to set their coefficients equal to zero. This gives the same result that you obtained.

2. You showed that the tension T is a constant, independent of s. So, in your equation for φ, you can integrate immediately (using the boundary condition that φ=0 at s = 0) to obtain:
$$φ=\frac{p}{T}s$$

From this, you might recognize that this equation describes the arc of a circle with radius of curvature T/p. So the inextensible membrane is in the shape of a circular arc.

The hard part of this problem is over now. We still don't know T, and so we still don't know the radius of curvature in the final geometry. To complete the solution, we need to do a horizontal force balance on the the two vertical end plates. Let the length of each end plate be h. The value of φ at the end plate is $φ=\frac{p}{T}s_0$. Please draw a free body diagram of the right end plate and determine the horizontal components of the forces acting on the end plate. Then write down the horizontal force balance equation.

Chet

 

[skrat]

Hmmmmm.

If I am not mistaken it is $$2\cos\varphi_0T=2p\sin\varphi_0+ph.$$ The first term comes from the tension of the membrane. The term has the same value at the top or at the bottom, therefore the factor $2$. The second term comes due to the pressure difference in the normal direction. Again the term is the same at the top and at the bottom.
But I am not sure on how to include the third term. However I am trying to include the gas working on the plates just as it does on the membrane.

 

[Chestermiller]

Hmmmmm.

If I am not mistaken it is $$2\cos\varphi_0T=2p\sin\varphi_0+ph.$$ The first term comes from the tension of the membrane. The term has the same value at the top or at the bottom, therefore the factor $2$. The second term comes due to the pressure difference in the normal direction. Again the term is the same at the top and at the bottom.
But I am not sure on how to include the third term. However I am trying to include the gas working on the plates just as it does on the membrane.

The first and third terms are correct, but the 2nd term should not be in there. (It is not even dimensionally consistent with the other two terms). So, removing the 2nd term, you have:
$$\cos\varphi_0=\frac{1}{2}\frac{p}{T}h$$
or equivalently:
$$\frac{\cos\varphi_0}{φ_0}=\frac{1}{2}\frac{h}{s_0}$$

Now, all that needs to be done is to solve for φ₀ as a function of h/s₀. This will give you not only the angle at the edge, but will also enable you to determine T/p.

Any ideas on how to solve this equation for φ₀ in the limit of very large values of h/s₀ and at very small h/s₀?

Chet

 

[skrat]

I don't like where this is going. I will tell you why, after I show you my ideas of solving the equation.
For $\frac{h}{s_0}\gg 1$ this means that $\varphi_0\rightarrow 0$. Therefore I can use Taylor expansion of $\cos\varphi_0$ for small $\varphi_0$. This brings me to $$\frac{1-\frac 1 2 \varphi_0^2}{\varphi _0}=\frac 1 2 \frac{h}{s_0}.$$ The solution to this quadratic equation is $$\varphi_0 =\frac{1}{2s_0}(-h+\sqrt{h^2+8s_0^2})$$ meaning $$\frac p T=\frac{1}{2s_0^2}(-h+\sqrt{h^2+8s_0^2}).$$ I have to admit that I don't really see why would I be interested in this ratio or how this tells me the radius of the circle, but I will be patient . =)
The other limit case is $\frac{h}{s_0}\ll 1$ meaning $\cos\varphi_0\rightarrow 0$. This leads me to the fact that $\varphi_0=\frac \pi 2 +n\pi+\varepsilon$. Due to the geometry and physical reasons of the problem $n=0$. So the equation I am trying to solve is $$\frac{\cos(\frac \pi 2+\varepsilon)}{\frac \pi 2+\varepsilon}=\frac 1 2 \frac{h}{s_0}.$$ This can be written as $$\frac{-\sin\varepsilon}{\frac \pi 2+\varepsilon}=\frac 1 2 \frac{h}{s_0}$$ and gives me $$\varepsilon =-\frac{\pi h}{4s_0(1+\frac{h}{2s_0})}.$$ Since $\frac{h}{s_0}\ll 1$ this can be approximated with $$ \varepsilon =-\frac{\pi h}{4s_0}$$ therefore $$\varphi_0=\frac \pi 2(1-\frac{h}{2s_0}).$$

Ok, even if my calculations are done correctly, there is something that bothers me. Because in reality I am nowhere near those two limit cases. The perfect way out of this problem would be an analytic solution (or of course higher orders in Taylor expansions. In that case I would of course use Mathematica to solve the equation). What I am trying to say is that I would like to stay away from any approximations as much as possible. Now since my knowledge of maths and physics is not even close to yours, I have to ask you: Is there an analytical solution to this equation?

EDIT: Or maybe start thinking about numerical solutions...

 

[Chestermiller]

Skrat,

You've done an excellent job on this problem. I'm really proud of you.

Now, the question is how to solve the equation. I haven't checked your "arithmetic" on the limiting solutions, but your approach is definitely what I would have done. You are correct that there is no analytic solution to this problem. Here is how I would handle it: On a spreadsheet, make a column of values of φ₀ running from 0 to π/2 in equal increments (say π/40). Then, in the adjacent column, list the calculated values of h/(2s₀) that correspond to the values of φ₀ in the first column. Then have the spreadsheet program plot a graph with h/(2s₀) as the abscissa and φ₀ as the ordinate. This will be the graph of your complete solution. Then, for any value of h/(2s₀), you can pick off the corresponding value for φ₀ from the graph.

You can also plot on the graph your asymptotic solutions. The graph will also show you what the solution looks like in case you want to get more precise and solve the equation numerically (say with Newton's method) for a specific value of h/(2s₀). It can also give you an initial guess for the numerical solution.

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

 

[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