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Homework Help: Membrane shape on a rectangle

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    3. 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. :/
    Last edited: Jul 6, 2015
  2. jcsd
  3. Jul 6, 2015 #2


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    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:


    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.
  4. Jul 6, 2015 #3
    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.
  5. Jul 6, 2015 #4


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    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.
  6. Jul 6, 2015 #5
    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.

  7. Jul 6, 2015 #6
    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...
    Last edited: Jul 6, 2015
  8. Jul 6, 2015 #7
    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.

  9. Jul 6, 2015 #8
    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##.
  10. Jul 6, 2015 #9
    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?

  11. Jul 6, 2015 #10
    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.
  12. Jul 6, 2015 #11
    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

    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.

  13. Jul 6, 2015 #12
    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.
  14. Jul 6, 2015 #13
    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?

  15. Jul 6, 2015 #14
    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##.
  16. Jul 6, 2015 #15
    Please try again. The unit normal is not correct. The dot product of the unit tangent and the unit normal must be zero.

  17. Jul 6, 2015 #16
    $$\vec t \cdot \vec n = (\cos\varphi,\sin\varphi)\cdot(\sin\varphi,-\cos\varphi)=\cos\varphi\sin\varphi-\sin\varphi\cos\varphi=0$$ EDIT: I edited the previous post in about 4 seconds after posting it. I guess you read it in that 4 seconds. Apologies.
  18. Jul 6, 2015 #17

    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?

  19. Jul 6, 2015 #18
    What is ##p##?
  20. Jul 6, 2015 #19
    The pressure.

  21. Jul 6, 2015 #20
    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)##. Capture.PNG

    I am not sure what you meant with the third force though.. Gravity?
  22. Jul 6, 2015 #21
    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?

  23. Jul 6, 2015 #22
    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 $$\frac{\partial }{\partial s}\left( T(s)\vec t(s)\right) +p\vec n=0.$$
  24. Jul 6, 2015 #23
    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?

  25. Jul 7, 2015 #24
    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 $${T}'(s)\vec t(s)+\left(p-T(s){\varphi}'\right)\vec n(s)=0.$$
  26. Jul 7, 2015 #25
    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 $$\frac{p-T(s){\varphi}'}{\sin\varphi}=0$$ therfore $$\varphi (s)=C_0+\int _0^{s}\frac{ds}{T(s)}$$ and also $$T(s)=konst.=C_1.$$ Which kinda makes sense, hopefully.
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