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

[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.