# Spherical membrane

Dear all
I've got a mechanics-type problem at the moment - don't worry, it really isn't for homework.
Consider a liquid-filled sphere, surrounded by some sort of solid membrane
in space holding the liquid together, with no gravity. Assume the inner (liquid) sphere has radius r_1, say, and the outer annulus has a radius r_2, so that the total radius of the sphere is r_1 + r_2. As we allow r_1 to increase, the mass increases proportionally to r_1^3 and the surface area proportionally to r_1^2. But in order to keep the structural integrity (ie so the bubble doesn't just break apart), the membrane thickness r_2 needs to increase proportionally to r_1^a, where a is some power. My problem is, I'm not sure what a should be. I'm guessing 2 or 3 but I'm useless at mechanics.
Anyway, if anyone could help, it would be very much appreciated.
Tronc
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Dear Tronc:

membrane
r_2,
(ie
to
I'm
mechanics.
http://www.ce.berkeley.edu/~sanjay/ce130/chap6.pdf page 94. It looks like r_2 needs to have the same exponent as r_1.
David A. Smith
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Thu, 8 Jan 2004 07:14:20 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T: snipped-for-privacy@nospam.com> wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
The stress on a spherical vessel is one of those delightfully straightforward problems to work [until the practical issues rear their head, anyway...]
One slices the sphere across an equator and works out the force tending to separate each half, due to the pressure of the internal fluid. This is given by pressure times area of the equatorial circle.
The stress in the material resisting this tensional force (at the parting line) is given by the force divided by the equatorial cross section area of the container wall, namely circumference times wall thickness.
Putting these two sentiments together: stress = pressure times area of equatorial circle divided by
--------------------------------------------------------- container wall thickness times circumference of eq. circle
You wish to hold material stress constant (i.e. below its limiting value) as the radius of the container gets bigger.
i.e. (const1 X r^2) / (const2 X thickness X r) = const3 This boils down to
r/thickness = const
So container thickness grows linearly with spherical radius for constant internal pressure and constant wall stress
However, in space, if there is a low gravity gradient, the natural shape of an unencumbered liquid is spherical due to "surface tension" if the thing does not freeze or boil anyway. So the pressure could be low or non-existant, and the need for containment minimal.
Brian Whatcott Altus OK.
wrote: