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

[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?

 

[Chestermiller]

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?

You have the right idea, but it's a little more complicated than this. For unequal plates, you have to recognize that φ is not going to be zero half-way between the plates. So, you will have φ_left at the left plate and φ_right at the right plate. You need to satisfy a force balance on each of the plates individually. Together with the circular shape for the membrane, this will enable you to determine these values of φ at the plates and the equilibrium distance between the plates.

Chet

 

[skrat]

By $\varphi _{left}$ you mean $\varphi_{left}=\varphi _0 - arctan(\frac{\Delta h}{h})$ or is even that not that simple?

 

[Chestermiller]

By $\varphi _{left}$ you mean $\varphi_{left}=\varphi _0 - arctan(\frac{\Delta h}{h})$ or is even that not that simple?

No. What I mean is that φ_left is the value of φ at the left boundary s = -s₀, such that
$$φ_{right}=φ_{left}+\frac{p}{T}(2s_0)$$
Chet

 

[skrat]

Now you got me confused. You said I should solve the ballance equation for each plate individually, so $$\frac{\cos \varphi_{\text{left}}}{\varphi_{\text{left}}}=\frac{1}{2s_0p}(p(h+\Delta h)+{f}')$$ for the left plate and $$\frac{\cos \varphi_{\text{right}}}{\varphi_{\text{right}}}=\frac{1}{2s_0p}(ph+f)$$ for the right plate where both of the transcendental equations have known parameters and can be numerically solved.
But now you somehow also provided a relation between $\varphi_{\text{left}}$ and $\varphi_{\text{right}}$ and I have no idea where you got this?

 

[Chestermiller]

Now you got me confused. You said I should solve the ballance equation for each plate individually, so $$\frac{\cos \varphi_{\text{left}}}{\varphi_{\text{left}}}=\frac{1}{2s_0p}(p(h+\Delta h)+{f}')$$ for the left plate and $$\frac{\cos \varphi_{\text{right}}}{\varphi_{\text{right}}}=\frac{1}{2s_0p}(ph+f)$$ for the right plate where both of the transcendental equations have known parameters and can be numerically solved.
But now you somehow also provided a relation between $\varphi_{\text{left}}$ and $\varphi_{\text{right}}$ and I have no idea where you got this?

We are dealing with a non-symmetric problem here (h_left≠h_right). Therefore, we can't simply take the solution for the symmetric case and cobble up a guess for the non-symmetric case and think that we are going to get the right answer. We actually have to redo the problem for the non-symmetric case.

The equation
$$φ_{right}=φ_{left}+\frac{p}{T}(2s_0)\tag{1}$$
follows directly from the equation$$\frac{dφ}{ds}=\frac{p}{T}$$
which we derived in our earlier posts from the differential force balance equation on the membrane. For the non-symmetric case, the force balances on each of the plates are:

$$2T\cos(φ_{left})=ph_{left}\tag{2}$$
$$2T\cos(φ_{right})=ph_{right}\tag{3}$$

Eqns. 1 - 3 provide a set of three non-linear equations in the three unknowns φ_left, φ_right, and p/T. For the non-symmetric case, one needs to solve these three equations to establish the solution.

Eqns. 1 - 3 reduce to the corresponding relationships for the symmetric case when the plate h's are the same. For the symmetric case,
$$φ_{left}=-φ_{right}\tag{symmetric case}$$
so that
$$φ_{right}=-φ_{right}+\frac{p}{T}(2s_0)$$
and
$$φ_{right}=\frac{p}{T}s_0$$
This is the equation we derived earlier in the symmetric development.

Chet

 

[skrat]

A really nice explanation. Thank you.