Sphere Bursting Pressure?

One of the key variables is missing here: the thickness of the material.

very thin, less than 1 mil thick at the most, and more like 1/4 mil

Thanks

t = 1 mil = 1/1000 of an inch. Is that what you mean?

1/1000" = 0.0254 mm

Assuming D is 13m after pressurizing

P_y = 4 * 0.0254 * 16.6/13000 = 0.00013MPa or 0.13kPa or 0.019psi

If t is 1/4 mil, P_y is 0.52kPa or 0.075psi

I don't think it would be possible to handle such a material without breaking it, so the answer has no meaning.

The Material is a latex high altitude weather balloon.

When is the diameter equal to 13 metres? If it is at the time it bursts, then my previous answer is correct. Otherwise, it is not.

yup thats the rated bursting diameter 13 meters.

care to run the numbers again, everything is the same, but with a diameter of 9.44 meters this time?

Joe

Actually, I don't care to "run the numbers again with D = 9.44m". I have given you the method of calculating it. So do it! And if you can't do it, find someone who can. If you have such a deep hatred of mathematics, perhaps you should not be involved in mathematical problems. Maybe you should try your hand at poetry...for example:

"There was a young man from Peru

who dream't he was eating his shoe

He woke in the night in a terrible fright

And found it was perfectly true."

OR:

There was a young man from Bengal

Who went to a fancy dress ball

He went there for fun, dressed up as a bun

But a dog ate him up in the hall.

Your turn.

I'll try.

I'm just not sure where the 13000 came from for one thing, and when something is 4 times thinner it ends up being 4 times stronger? that lost me

sorry,

Joe

The diameter of the sphere was 13 metres. That is equivalent to 13,000 mm. I used mm in order to keep the units consistent. When something is 4 times thinner, the stress is 4 times larger. That means the strength is 4 times smaller.

Thanks for this assistance, but see what I mean your original says

P_y = 4 * 0.0254 * 16.6/13000 = 0.00013MPa or 0.13kPa or 0.019psi

If t is 1/4 mil, P_y is 0.52kPa or 0.075psi

why is the pressure rating 4 times higher in pressure see? in other words when it says 0.075 Psi,

that means then that at that pressure the balloon would pop correct?

Joe

Sorry, I was in error! Call it a senior moment. I should have said "If it is 1/4 mil, P_y is 0.0325kPa or 0.0047 psi".

Total force on shell = P*A = P*πD²/4

Length of shell resisting force = π*D

Force per unit length of shell = P*D/4

Failure strength per unit length = t*F_y

Pressure at yield, P_y = 4*t*F_y/D

t = thickness of shell

Fy = yield strength

D = diameter

Ummm, are you sure that is right !

I do not have my theory book in front of me to check for recognised formula, but since a sphere has three dimensions, the force on an element of the wall will be in two directions, the "x" and "y" - so won't it be necessary to add the two (Mohr's stress circle etc.) to get an accurate result ? It's not clear to me at a glance with your formulae that this was provided for.

In the more conventional cylindrical pressure vessel, the spherical ends are not a problem as the hoop stress in the cylinder part of the vessel is twice that of the hoop stress in the hemispherical end - but this is not the case here as the vessel is all sphere.

I don't have a theory book in front of me either. If a sphere has an internal pressure P, then every section through the centroid has a force of P*A where A is the area of the circle, namely πD²/4. Mohr's stress circle does not enter into the discussion.

Regardless of which section we take, the perimeter resisting that force is πD, so the unit load must be P*πD²/4 divided by πD which yields PD/4. What is your problem?

Yes, I am sure that is right.

My query boils down to the question of whether an element subject to a stress in one direction, will fail earlier if a stress is applied in two directions. Its about thirty years since I did that stuff, but I am thinking that the answer is, "yes it does", in which case your answer appears to be wrong. How confident are you?

Further comment to "self". On second thoughts you are probably correct -my concern would apply if the second stress was of opposite sign to the other. Like I said it is "thirty years"......

Everyone, I found this formula a while ago but still had no idea as to how to run it.

so with the information from the initial posting, does this work?

I hate Math!

Can anyone help me here. I want to know what the internal Pressure in PSI ( Pounds Per Square Inch ) is of a sphere with the following specifications.

Diameter of Sphere, = 13 meters, and how about a 9.44 meter sphere also?

Tensile Strength of the material that is the skin of the sphere, = 16.6 Mpa

What will the the internal pressure of this sphere (Think of it being like a balloon, well it IS a balloon. ) What PSI would be inside, when this skin would fail.

Thanks Everyone

Your solution ignores the thickness and strength of the balloon material. A correct formula was provided in #20. Just make sure all your units are compatible.

T = PR/2 is the same as T = PD/4 which I gave you in Post #2. It is the tension per unit length at all points on the surface of the sphere.

Pressure at failure is 4*t*F_y/D which is the same as 2*t*F_y/R, both of which were given to you earlier in this thread.

It should be obvious, even to one who hates math, that the failure pressure for a 9.44m diameter sphere will be 1.38 times higher than that of a 13m sphere. You need the thickness in order to find a precise answer.

FYI, the formula (T=ΔPR/2) is for the surface tension on the outside of a droplet. It doesn't apply to the problem you described.

_{<sigh> Carry out a force balance on a theoretical plane sliced through the centre of the sphere.....}

From Roark (5th ed), the pressure at failure = [2t ( strength )/Radius] where t is the thickness of the skin. Keeping units compatible, I come up with .0047 psi for the 13 meter diameter sphere at 1/4 mil thickness. This is theory. In practice, many factors such as temperature, time, or even a pin could alter your results.