Thread: How much weight can you put on a beach ball before it bursts

Hi,

Imagine a beach ball with a volume of 1 liter, inflated to a pressure of 2 bar. At 5 bar it will burst.

How much weight needs to be applied (by a large vise) to make the beach ball burst?

Please see if you have additions or corrections on my work:

Video explaining the following equation and Boyle's Law

Equation:
pressure 1 * volume 1 = pressure 2 * volume 2

Rearrange to calculate volume 2:
(pressure 1 * volume 2) / pressure 2 = volume 2

Insert real numbers:
(2bar*1 liter) / 5 bar = 0.4 liter

The question becomes: how much weight needs to be applied to squish a beach ball from 1 liter and 2 bar to 0.4 liter and 5 bar?

Action = Reaction, so I need to determine the difference in force (in F) exerted by the beach ball in the two states?

I'm sorry, but I don't know how to do that? Could you help me?

If you squish a beach ball, it will flatten out so that it takes up less volume than a sphere of equal surface area. The force exerted on the flat surface that is squishing the ball will be proportional to the pressure and the area of the flattened part of the ball. Maybe that will get you started, but it doesn't seem like a very easy thing to calculate.

Thanks for your reply Harold!

I've been looking on wikipedia a bit more: Wikipedia about Pressure

Pressure (symbol: P) is the force per unit area applied in a direction perpendicular to the surface of an object.

Formula
Mathematically:


where:

P is the pressure,
F is the normal force,
A is the area.

Pressure is a scalar quantity. It relates the vector surface element (a vector normal to the surface) with the normal force acting on it. The pressure is the scalar proportionality constant that relates the two normal vectors:



The minus sign comes from the fact that the force is considered towards the surface element, while the normal vector points outwards.

It is incorrect (although rather usual) to say "the pressure is directed in such or such direction". The pressure, as a scalar, has no direction. It is the force given by the previous relation the quantity that has a direction. If we change the orientation of the surface element the direction of the normal force changes accordingly, but the pressure remains the same.

Pressure is transmitted to solid boundaries or across arbitrary sections of fluid normal to these boundaries or sections at every point. It is a fundamental parameter in thermodynamics and it is conjugate to volume.

It's a bit difficult to calculate with the "bar" unit, so let's use newton per square meter (pascal). Also I'll use cubic meters instead of liters.

And with Harold's help, I think I can say that the force applied by the spherical beach ball on the outside world is 0.

So let's refrase and change the question: what is the weight required to squish a beach ball from 1 cubic meter and 200.000 Pascal to 0.4 cubic meter and 500.000 Pascal.

Using:

P (Pressure) and A (Area) are both known, so I need to determine F.

(Area is 4.83597 square meters, round to 4.8 square meters)

Rearrange formula to get F:
F = P * A

Insert real numbers:

F = 200.000 * 4.8 = 960.000 Newton

And:

F = 500.000 * 4.8 = 2400.000 Newton

I had determined that the unsquished beach ball didn't apply a force on the outside world, so detract this from the second result: 2400.000 - 960.000 = 1440.000 Newton.

Convert to kilograms: 1440.000 / 9.81 = 146.789 kilograms.

I'm sure that isn't right!


Where do I go wrong?

Originally Posted by calimero

(Area is 4.83597 square meters, round to 4.8 square meters)

Where did you get this area?

hmmm, let's see if I can't simplify the problem:

Imagine a box 1 meter x 1 meter x 1 meter. A piston is placed on top.

What weight is required to reduce the volume and increase the pressure from its initial values of 1 cubic meter and 100.000 pascal (roughly atmospheric pressure), to its ending values of 0.2 cubic meters and 500.000 pascal?

Originally Posted by Harold14370

Originally Posted by calimero

(Area is 4.83597 square meters, round to 4.8 square meters)

Where did you get this area?

I've used an online calculator to calculate the surface area of a sphere with a volume of 1 cubic meter.

A liter is not the same as a cubic meter. The vise is not pressing against an area equal to the sphere's surface area.

Originally Posted by Harold14370

A liter is not the same as a cubic meter.

I know, I've changed the question several times to make for easier calculation (I see I haven't made that clear, I'll edit it now)

The vise is not pressing against an area equal to the sphere's surface area.

I still don't understand your original post. I don't understand how the the area of the vice is important. To reference my last question: if I've made the cube not 1 meter x 1 meter x 1 meter, but 0.5 meters (width) x 0.5 meters (length) x 4 meters (height), I assume the force required to push the piston in would be the same?

Originally Posted by calimero

I still don't understand your original post...

I think I do see, however, the beauty of calculating it like this. The form the beach ball wants to be in is a sphere. Any deviation from that is the result of outside pressure on it. If it's not fully inflated, it's atmospheric pressure working on it. And if it's in a vise, it's the vise working on it.

That said, atmospheric pressure working on it while it is not fully inflated, and a vise working on it while it is, look to me like similar and perhaps interchangeable forces. The central question is: what is the size of the deviation of the beach balloon from its ideal form, and how large is the force that made this deviation.

It's the size of the deviation that is important, not the specific shape, I think. And the size of the deviation we already know: it's the difference in volume between the sphere and the squished sphere.

Oh, yes, the area of a piston is important. If you have for example a bicycle tire pump with a small diameter piston, then it takes less force (but longer stroke) to pump the same amount of air, at the same pressure as a large diameter pump. There is a mechanical advantage that way.

If you pump up a car tire you will notice that the footprint of the tire becomes smaller. That's because the pressure is higher, so the weight of the car is spread out over a smaller area. The pressure of the tire on the road goes up as well, because the weight of the car is supported by a smaller area of road surface.

Originally Posted by Harold14370

Oh, yes, the area of a piston is important. If you have for example a bicycle tire pump with a small diameter piston, then it takes less force (but longer stroke) to pump the same amount of air, at the same pressure as a large diameter pump.

Oh, then we have a misunderstanding because of my loosely using terms (like "force") I don't understand. Sorry.

Really my question is: how heavy of a rock would you have to (very carefully) put on a beach ball to decrease the volume by this much, and increase the pressure by that much.

Maybe that's a bit of a difficult question, so maybe this can do as an alternative I can build on:

Imagine a box 1 meter x 1 meter x 1 meter. A piston is placed on top.

What weight would you have to put on the piston to reduce the volume and increase the pressure from its initial values of 1 cubic meter and 100.000 pascal (roughly atmospheric pressure), to its ending values of 0.2 cubic meters and 500.000 pascal?

You're going to have to make some assumptions here. I think you have to assume the walls of the ball are not stretchy, because if the walls stretch, then there isn't any way to know what the final shape is going to be, so you don't know what the area of the flattened part of the ball will be. If the walls don't stretch, does that mean that the surface area of the flattened ball is the same as the round ball, or does it mean that the unflattened part remains the same spherical shape? I'm not sure about that one. Anyway, if you assume that the ball ends up as a truncated sphere with the same radius as the original sphere, you know the final volume (from the final pressure) and can calculate the volume of the two spherical caps subtracted from the original sphere. Knowing the volume you can determine the radius and area.

Originally Posted by Harold14370

You're going to have to make some assumptions here. I think you have to assume the walls of the ball are not stretchy.

Yes.

If the walls don't stretch, does that mean that the surface area of the flattened ball is the same as the round ball, or does it mean that the unflattened part remains the same spherical shape?

I will assume the surface area remains the same, and the beach ball to be flattened under the stone. If it were to be flattened completely, it would have no height and a larger diameter. If it were not completely flattened, it would look like a cilinder, with a rounded edge, imagine a donut without the hole. This rounded edge would have a diameter equal to the height of the cilinder, I imagine.

if you assume that the ball ends up as a truncated sphere with the same radius as the original sphere, you know the final volume (from the final pressure) and can calculate the volume of the two spherical caps subtracted from the original sphere. Knowing the volume you can determine the radius and area.

Yes, but I don't make this assumption.

[brainwave] I think the best way to get the concept and model the change in shape is to lay a piece of string on the ground in a circle, and every time at the top and bottom replace a piece of the string with a piece of wood. midway you have a donut shape with the remaining string having the same diameter as the height of the "cilinder," when you've finished all you have is 2 layers of wood.[/brainwave]

But I still don't understand why the area of the flattened piece is important. Did you read this reply?

I think I do see, however, the beauty of calculating it like this. The form the beach ball wants to be in is a sphere. Any deviation from that is the result of outside pressure on it. If it's not fully inflated, it's atmospheric pressure working on it. And if it's in a vise, it's the vise working on it.

That said, atmospheric pressure working on it while it is not fully inflated, and a vise working on it while it is, look to me like similar and perhaps interchangeable forces. The central question is: what is the size of the deviation of the beach balloon from its ideal form, and how large is the force that made this deviation.

It's the size of the deviation that is important, not the specific shape, I think. And the size of the deviation we already know: it's the difference in volume between the sphere and the squished sphere.

Originally Posted by calimero

But I still don't understand why the area of the flattened piece is important.

The area is important because you need it to find the force, given the pressure. Force=pressure*area.

Did you read this reply?

Yeah, but I didn't think you were headed in the right direction, so I was trying to steer you there. You were right about the fact that the difference in volume is important. You already had that part calculated in the first post. It went from 1 liter to 0.4.

Edit: Also the sphere has more than atmospheric pressure in it before it is squished. So your thinking there was a bit off.

Originally Posted by Harold14370

The area is important because you need it to find the force, given the pressure. Force=pressure*area.

But the pressure a gas exerts is on the whole container / envelope. I don't see how the area of one piece of it is relevant. Be the surface area of that piece 1 square centimeter or 1 square meter, the weight of the stone will be the same.

Maybe you could give more examples, explanations or proof?

Yeah, but I didn't think you were headed in the right direction, so I was trying to steer you there. You were right about the fact that the difference in volume is important. You already had that part calculated in the first post. It went from 1 liter to 0.4.

So you think the answer is F = P * A ?

Originally Posted by calimero

Originally Posted by Harold14370

The area is important because you need it to find the force, given the pressure. Force=pressure*area.

But the pressure a gas exerts is on the whole container / envelope. I don't see how the area of one piece of it is relevant. Be the surface area of that piece 1 square centimeter or 1 square meter, the weight of the stone will be the same.

Maybe you could give more examples, explanations or proof?

Okay. Let's take for example a piston. The force on the piston rod is equal to the cross-sectional area of the cylinder multiplied by the pressure. You don't multiply by the area of the whole cylinder. If you did that, then there would be much more force on your bicycle pump handle when it is at the start of the stroke than at the end.

So you think the answer is F = P * A ?

Yes, but the tricky part is calculating A. That's the flat part of the squished ball.

Thanks Harold, you've helped me a lot. I'll mull over this a bit.


Edit

This:

I think the best way to get the concept and model the change in shape is to lay a piece of string on the ground in a circle, and every time at the top and bottom replace a piece of the string with a piece of wood. midway you have a donut shape with the remaining string having the same diameter as the height of the "cilinder," when you've finished all you have is 2 layers of wood.

I now don't think is correct. I've tried it with a plastic bag, and the bag becomes a cilinder, not a donut.

I've mulled it over a bit. On the shape of the envelope:


The sphere is the most efficient form of a gas to be in when contained in a spherical envelope and there are equal forces on it from all directions. When you put it on the ground and put a weight on it and the envelope isn't flexible, there is only one shape the resulting form can have: the most efficient shape a gas with a certain volume, a spherical envelope, and a weight on it can have. I think this shape goes from spherical, to rounded cilinder, to flatter and flatter cilinder. I think the easiest way to test this is to get a deflated beach ball, put a weight on it, and gradually inflate it.

Imagine a glass spherical container. It is not compressible vertically and not stretchable horizontally. When you put a weight on it, all the stresses will be on the glass, and it will break.

Imagine a rubber spherical envelope. It is compressible vertically and stretchable horizontally. When you put a weight on it, it will flatten and stretch horizontally until it runs out of horizontal flexibility and bursts. Before it does this, I imagine it will look like a donut.

Imagine a plastic spherical envelope. It is compressible vertically, but not stretchable horizontally. When you put a put a weight on it, I imagine it goes from slightly flattened sphere, to less and less round cilinder, to flatter and flatter cilinder, to pancake shape.

A plastic envelope has no (horizontal) flexibility, and its diameter will only change because the diameter of a string laid down in a circle is smaller than the diameter of a string laid down in a flatter and flatter rectangle. The diameter of an uninflated beach ball lies somewhere between the diameter of an inflated beach ball and this piece of string, because the side view of beach balls when deflated look like long flat cilinders with crumpled vertical edges and not 2 pieces of string perfectly folded.


The question what shape the beach balloon takes might be better placed in the mathematics forum.

Is there some problem you had with my approach? I thought that the ends against the vise would flatten while the rest of the ball would stay spherical. If the walls are non-stretchable, how will it be deformed from its spherical shape?

Originally Posted by Harold14370

Is there some problem you had with my approach? I thought that the ends against the vise would flatten while the rest of the ball would stay spherical. If the walls are non-stretchable, how will it be deformed from its spherical shape?

You'll have to excuse me, I've just started thinking about the problem, so my thinking is all over the place.

Yes, upon further reflection I can't see a problem with your solution. Assuming that is correct, let's see if I can solve the problem.

What is the weight required to squish a beach ball from 1 cubic meter and 200.000 Pascal to 0.4 cubic meter and 500.000 Pascal.

I need to determine the area of the flat part. To do that, I first need to cut the sphere in half to make it a dome, and then cut away the lower half of this dome, to get the right volume and with that the right “floor area” or “flat part.”

That’s not so easy. Here are some sources that might help:

simplest pdf with formulas and diagrams
Spherical cap formulas from Wolfram Mathworld
Wikipedia Dome (mathematics)

I only understand the first link a bit, so I’ll use that.

First I’ll determine the characteristics of the sphere cut in half:

volume = 0.5 cubic metere
radius = 0.62035 meters
height = 0.62035 meters
diameter = 1.2407 meters)

Cutting off the the part below “the flat part” gives:

volume = 0.3 cubic meters
radius of curvature = 0.62035 meters
height = ?
diameter = ?

Radius of curvature and volume are both known, so I could use the volume equation in the pdf to calculate the height of the dome (but I don’t understand why on the left it’s “Rc”, while on the right it’s “r”):

[2 hours pass...]

I can’t solve this equation now. High School is too long ago … I'll try again later.

Originally Posted by calimero

First I’ll determine the characteristics of the sphere cut in half:

volume = 0.5 cubic metere

Do you mean 0.5 liter?

Originally Posted by Harold14370

Do you mean 0.5 liter?

No, I've changed the problem. I'm more used to computing with cubic meters than liters.

Anyway, this is a geometry problem I'm sure I'll be able to solve. It will just take some time to learn the formulas.

Before understanding it, I can of course use excel

I'm using these formulas - for "Volume" and "Radius at second level".

one third = 0.33333333
pi = 3.14159265
h = 0.45069500
h^2 = 0.20312598
Rc = 0.62035

1/3pi*h^2 = 0.21271303
3Rc = 1.86105000
3Rc - h = 1.41035500

Volume = 0.30000089


Now I can use the formula for "Radius at second level"


Rc = 0.62035
Rc^2 = 0.38483412
h (I'm using radius of sphere) = 0.62035000
l (radius of sphere - the h used in the above formula (0.450695))= 0.16965500

Rc - h + l = 0.16965500
(Rc - h)^2 = 0.02878282
Rc^2 - (Rc-h+l)^2 = 0.35605130
Radius at second level "flat part" = 0.59670000 meters

Area of flat part (pi*r^2) = 1.11857 square meters

Now calculate the force (F = P * A) F = 500,000 * 1.11857 = 559285 Newton.

Change to kilograms: 559285 / 9.81 = 57012 kilograms.

That doesn't sound right!

Where did I go wrong?

hmmm, let's see if I can understand mathworld a little better:

Let the sphere have radius R, then the volume of a spherical cap of height h and base radius a is given by the equation of a spherical segment. (...)

so the radius of the base circle is

a = the root of: h(2R-h)

h = 0.450695 meters
R= 0.62035

a = 0.59670 meters (same answer)

I've visually checked these two numbers (height and radius of dome) by drawing a circle in Google SketchUp, and they seem correct (I can't check if the volume is 0.3 cubic meters at this height because I've only drawn a circle, not a sphere).

Do you really need to apply 57 TONS to a beach ball to squish it a bit? That doesn't sound right!

I don't know how you can arbitrarily change the problem from a 1 liter ball to a 1 cubic meter ball and expect to get the same answer.

Yes, a cubic meter volume which would burst at 5 bar (which is 73 psi) is probably somewhere on the order of what a big tractor trailer tire would hold, and yes, that would probably take quite a few tons of weight.

The original problem was a liter. Think a medium sized soda bottle. That would be more like a softball size than a beach ball.

Originally Posted by Harold14370

Okay. Let's take for example a piston. The force on the piston rod is equal to the cross-sectional area of the cylinder multiplied by the pressure. You don't multiply by the area of the whole cylinder. If you did that, then there would be much more force on your bicycle pump handle when it is at the start of the stroke than at the end.

This still doesn't explain it for me. The force on the piston at the start of the stroke is smaller than when it is at the top, because of V1*P1=V2*P2. The volume is bigger so the pressure is smaller. The relation is between volume (and thus the surface area of the whole container) and pressure, not the size of the piston.

Imagine two cylinders with a piston at the top. One with a diameter of 1 and an infinite height, and another also with a diameter of 1, but a height of 1. Both have an inside pressure of P.

If you are correct, to keep both pistons in place would require the same weight. I imagine this is wrong. Edit: Hold on, now I'm starting to imagine this is right ... I'm getting confused...

Originally Posted by Harold14370

I don't know how you can arbitrarily change the problem from a 1 liter ball to a 1 cubic meter ball and expect to get the same answer.

The question is hypothetical to understand the problem and how to solve it. I don't expect to get same results...

Yes, a cubic meter volume which would burst at 5 bar (which is 73 psi) is probably somewhere on the order of what a big tractor trailer tire would hold, and yes, that would probably take quite a few tons of weight.

Thanks for your answer.

I'm still not convinced of the applicability of F=PA though, see my post of a few minutes ago.

Thanks Harold, you've helped me a lot. I still don't quite understand F = PA, but I'll read more on that later.

Now I understand the basic principles a bit better, I'd like to ask one last question:

How do you calculate the forces when there are 4 surfaces acting on the beach ball? (Instead of the previous 2 -- ground and rock)

Do you have any thoughts on that?


Is it perhaps that you take half of the flattened parts? Like in the previous question you only take the surface of the part flattened by the rock, and in this question you would only take the surface flattened by 2 of the 4 surfaces?

I guess you mean like if you had two vises squeezing the ball, and the vises were oriented at right angles to one another? You just consider each of the 4 jaws of the vise individually. The force on that jaw will be the force exerted by the area of the ball underneath it. Remember that forces always come in pairs, with an equal and opposite reaction for each. As long as there is no acceleration of either the ball or the vise, the forces add up to zero.

Thanks Harold, I understand.

In my previous example that would mean that the ball exerts 57 tons on the rock and on the ground, and in this example I would have to calculate the direction and size of the forces, and check my work by checking that the forces all add up to zero.

Yes. If you had two vises squeezing the ball, here's what would happen. You tighten down on the first vise, and that raises the pressure but not to the bursting point. When you tighten the second vise, the pressure will go up quicker than it did in the previous example with one vise, i.e., you will reach the bursting pressure with less compression of the ball, and less flattened area under the jaws of the vise. Since the bursting pressure is the same, and the area is less, the force would be less. This means that with two vises squeezing the ball, each vise needs to develop less force to reach the bursting pressure than if you only had one vise.

Originally Posted by Harold14370

Is there some problem you had with my approach? I thought that the ends against the vise would flatten while the rest of the ball would stay spherical. If the walls are non-stretchable, how will it be deformed from its spherical shape?

If the walls are completely non-stretchable, you have changed the problem significantly. While you can deform a stiff string from a circle to an ellipse, you cannot deform a stiff sphere to an ellipsoid.

The question posed is basically a question of the theory of elasticity. It is not a simple question.

You have a thin homogeneous and isotropic membrane that contains some amount of gas. You then deform the membrane slowly, so as to keep the gas at ambient temperature, and consider the stress state of the membrane as the gas volume,hence pressure change, and as some external pressure/force is applied to a portion of the outer surface of the membrane.

The membrane will assume the stress and strain state of minimum energy, which is the fundamental principle behind finite-element analysis. This will be close to the minimum surface area, subject to the constraints imposed. The imposed constraints are not only the total force applied to compress the balloon, but also the geometry (or pressure distribution) that goes along with how that force is applied. There will also be some edge effects involved, particularly if the force is not applied "smoothly".

At some point the strain in the membrane will be such as to result in material failure. That is a bit more involved than simply knowing the gas pressure, and requires the adoption of some material failure, which in this case would probably involve Von Mises strain which would be affected strongly by edge effects near the edge of whatever surface (e.g a platen) that is being used to apply the force.

This is a moderately difficult problem and not likely to yield to any sort of paper and pencil closed form solution. Rubber-like materials are notoriously difficult to handle even with finite-element software because of the large deformations encountered and hence a need for specialized, "re-formulated", elements.

If you assume the material to have small Young's modulus, then the pressure associated with the applied force will be approximately equal to the internal gas pressure, and the boundary condition associated with the gas will be known, and hence the internal volume will be easily calculable. If you make the material very stiff, then the situation will be much much more difficult to calculate as load from the internal pressure will be reacted in a significant way by both the membrane and the applied force (distribution).

Originally Posted by DrRocket

Originally Posted by Harold14370

Is there some problem you had with my approach? I thought that the ends against the vise would flatten while the rest of the ball would stay spherical. If the walls are non-stretchable, how will it be deformed from its spherical shape?

If the walls are completely non-stretchable, you have changed the problem significantly. While you can deform a stiff string from a circle to an ellipse, you cannot deform a stiff sphere to an ellipsoid.

The question posed is basically a question of the theory of elasticity. It is not a simple question.

I realize that there is no real material that doesn't stretch, but no elastic modulus or anything was given, so I think we have to make some simplifying asssumption.

If the material is limp and not stretchy, then what happens when it is flattened against the face of the vise? It can't become flat because that would be like trying to flatten out a bubble in your linoleum - the shape doesn't quite match. So I think it would end up kind of wrinkly, but approximately flat, with the original sphere becoming the shape of a truncated sphere with dome shaped areas missing from each end. As far as the rest of the sphere, which is not touching the vise, I can't think of any force that would change it from its original spherical shape. All we've done to it is increased the pressure.

Originally Posted by Harold14370

I realize that there is no real material that doesn't stretch, but no elastic modulus or anything was given, so I think we have to make some simplifying asssumption.

If the material is limp and not stretchy, then what happens when it is flattened against the face of the vise? It can't become flat because that would be like trying to flatten out a bubble in your linoleum - the shape doesn't quite match. So I think it would end up kind of wrinkly, but approximately flat, with the original sphere becoming the shape of a truncated sphere with dome shaped areas missing from each end. As far as the rest of the sphere, which is not touching the vise, I can't think of any force that would change it from its original spherical shape. All we've done to it is increased the pressure.

As you say if the material is limp but not elastic you will get a creased wrinkly mess. God only knows what the shape and stress state of such a thing would be. I suspect that there would be all sorts of local stress risers caused by the creases.

That assumption does not make the problem simpler. It introduces all sorts of complications and discontinuities in the local stress state.

Thanks for your replies Harold and DrRocket.

I've just looked up what beach balls are made of. It's often PVC. Obviously PVC is stretchable, I'd just thought I'd first call it non-stretchable to distinguish it from rubber and understand the principle.

To then calculate the forces, I would have to understand your post DrRocket, which I still don't.


First I'd like to ask a related question which I was planning to post somewhere where there were material scientists/engineers, but now I'm sure you could help me as well:

How much can you inflate a beach ball that was made from a certain material, before it won’t return to its original shape?

The problem I have is when I see for example this ldpe film (for lack of a better example), I don’t know how you can use these numbers:

Tensile strength at break:

Machine Direction (MD): 400 psi
Tranverse Direction (TD): 3400 psi

Elongation at break:

MD: 300 %
TD: 500 %

And (if I had it), the MD and TD values where Hooke’s Law still applies (the existence of which you helped me discover yesterday).

to calculate the air pressure inside needed to reach these values. Instead of the force going in one direction when you pull on a material, it goes in more directions when you put the material in a ball and inflate it.


Random links:

Mechanical Properties

All beach balls are measured when they are deflated. The actual diameter of the inflated beach ball is therefore smaller than the listed length.

Is this problem from a textbook, or where did you get it?

I've thought of an application where a gas filled ball would be compressed. I'm now trying to understand what forces are involved and what materials to use.

My earlier problem I simplified a bit because I didn't think I'd get as much useful help as I did. And I didn't understand the problem fully, so I first wanted to learn the basic principles.

I think to start to solve this second problem you need a value for Young's modulus and the force where Hooke's Law still applies?

If for the Young's Modulus you take the value for Cellophane film (say 4 GPa for MD and 3 GPa for TD), and for the force where Hooke's Law still applies 200 psi for MD and 400 psi for TD, and for the original volume 0.001 cubic meter, what needs to be the inside pressure to get 200 psi MD (smallest of the two so the larger is irrelevant)?


This is really an engineering problem, and I'm not an engineer, so I'd think I'd ask here... Once I understand how to solve the problem, I can experiment with different materials and sizes of balls.

Would it not be easier to do the experiments first, and then you will have some data to try to help you solve the problem. Then you can scale it to whatever size you need for your application.

As far as i can see the mathematics being employed will be quite a rough approximation, you can use this as your starting point for the experiments. Your likely to find a more accurate and simple relationship by experimenting. There are a lot of variables in your problem that makes forming the problem mathematically difficult. If you had experimental data, you could derive the relationship from it and skip a lot of complicated mathematics.

Originally Posted by harvestein

As far as i can see the mathematics being employed will be quite a rough approximation, you can use this as your starting point for the experiments. Your likely to find a more accurate and simple relationship by experimenting.

Agree. The sub-problem I now pose is quite simple I think though: how much air pressure would be needed to get this much psi horizontally, and this much psi vertically. I now haven't got a clue, and a rough approximation would be useful.

There are a lot of variables in your problem that makes forming the problem mathematically difficult. If you had experimental data, you could derive the relationship from it and skip a lot of complicated mathematics.

I'd hate it if I didn't understand the mathematics involved, so that isn't going to work anyway But yes, I again agree. Experimentation is more accurate, but I'd like a basis in mathematics. If I didn't understand the forces and factors involved, I'd probably employ a very large margin of safety, and therefore use needlessly high quality materials.

You could do some tests with a basketball or soccer ball. Pump it up and put a tire pressure gauge adapted to the needle to measure the pressure. Put it on top of a bathroom scale, and squish it down. Push it down with your hand or put a weight on top of it. Take some measurements with a caliper and see how much it is bulged out at the equator, and how out of round it is. Cover the ball with chalk or something, put a piece of paper under it, and mark the area of contact. See how well the F=PA relationship works out.

Originally Posted by calimero

How much can you inflate a beach ball that was made from a certain material, before it won’t return to its original shape?

When the stress exceeds the yield strength of the material you have permanent deformation and the ball will then not return to its original shape.

The tensile stress in a spherical shell or a cylindrical shell is give by



where is the stress is internal gauge pressure is the radius and is the thickness.

Thanks for your replies Harold and DrRocket.

Harold, I've tried to let a plastic bag burst by applying weight on it. I've stopped at 30 kilos, because after that I would've had to put my weight on it. I've blown up a plastic bag until it burst, the pressure gauge didn't budge from its starting position. I think you can stand on a football and it doesn't do that much. You've given useful tips on how to conduct the experiment. I think I'll get a more sensitive pressure gauge, some beach balls, a way to gradually increase the weight (water buckets?), and maybe a better scale (mine fixes the weight after a couple of seconds).

DrRocket, you've given me the search terms needed to look for the solution. The simplest explanation I could find is at
efunda, a more longwinded explanation is at Wikipedia.

From efunda:

Spherical Pressure Vessel

Thin-walled pressure vessels are one of the most typical applications of plane stress.

Consider a spherical pressure vessel with radius and wall thickness subjected to an internal gage pressure .

For reasons of symmetry, all four normal stresses on a small stress element in the wall must be identical. Furthermore, there can be no shear stress.



The normal stresses can be related to the pressure by inspecting a free body diagram of the pressure vessel. To simplify the analysis, we cut the vessel in half as illustrated.



Since the vessel is under static equilibrium, it must satisfy Newton's first law of motion. In other words, the stress around the wall must have a net resultant to balance the internal pressure across the cross-section.

I can now plug in some numbers. Let's use the following values:







Which gives:



(This looks to me like a very high value.)

Can I now say that if the surface of the sphere were a flat plane, and 4 horses were pulling the plane in opposite directions, each horse would pull with a force of ?

If you'd hang a weight on the plane, the weight would have to exert a force of ?

That's 509.684 kilos per square meter and 51 kilos per square centimeter, I'm sure that's too much for a ball with a diameter of 20 cm and a pressure of 1 bar (and a wall thickness of 1 mm) ?

You have to make a distinction between the air pressure in the ball and the stress on the rubber walls. The air pressure acts on the cross-sectional area of the ball, pi*r^2, to create the force which pulls the two halves of the ball apart. This is resisted by the rubber wall which only has a cross-sectional area of 2pi*r*t. So the stress, which is what you calculated, is much higher than the pressure.

For your test, a gage designed for athletic balls, which are typically inflated to about 8 psig, would be better than the tire gauge which is normally measuring a pressure of about 25 or 30 psig.

Originally Posted by calimero

I can now plug in some numbers. Let's use the following values:







Which gives:



(This looks to me like a very high value.)

Can I now say that if the surface of the sphere were a flat plane, and 4 horses were pulling the plane in opposite directions, each horse would pull with a force of ?



If you'd hang a weight on the plane, the weight would have to exert a force of ?

No. You are confusing stress with force. Stress is in units of force/area.

Originally Posted by calimero

That's 509.684 kilos per square meter and 51 kilos per square centimeter, I'm sure that's too much for a ball with a diameter of 20 cm and a pressure of 1 bar (and a wall thickness of 1 mm) ?

Now you are confusing mass (kilos) with force (Newtons).

Also, when you apply a force to the exterios of your inflated beach ball you will change the stress state, particularly at the edges of the platen that will be used to apply the force. Since it is peak stress that determines material failure, the problem is not much more difficult to handle analytically, and the failure site will probably be at the edge of the platen.

Thanks for your replies. I have to think this over a bit.

See you in a little while

Originally Posted by DrRocket

Originally Posted by calimero

Can I now say that if the surface of the sphere were a flat plane, and 4 horses were pulling the plane in opposite directions, each horse would pull with a force of ?

No. You are confusing stress with force. Stress is in units of force/area.

hmmm... I want to know the stress, so my example wasn't relevant.

But to try to understand the principle:

Force ( is expressed in )

Pressure (symbol: P) is the force per unit area applied in a direction perpendicular to the surface of an object.

Mathematically:

In continuum mechanics, stress is a measure of the average force per unit area of a surface within a deformable body on which internal forces act. It is a measure of the intensity of the internal forces acting between particles of a deformable body across imaginary internal surfaces. These internal forces are produced between the particles in the body as a reaction to external forces applied on the body. External forces are either surface forces or body forces. Because the loaded deformable body is assumed as a continuum, these internal forces are distributed continuously within the volume of the material body, i.e., the stress distribution in the body is expressed as a piecewise continuous function of space coordinates and time.

The SI unit for stress is the pascal (symbol Pa), which is equivalent to one newton (force) per square meter (unit area). The unit for stress is the same as that of pressure, which is also a measure of force per unit area.

Surface force denoted fs is the force that acts across an internal or external surface element in a material body. Surface force can be decomposed in to two perpendicular components: pressure and stress forces. Pressure force acts normally over an area and stress force acts tangentially over an area.

A surface normal, or simply normal, to a flat surface is a vector that is perpendicular to that surface.

In geometry, the tangent line (or simply the tangent) to a curve at a given point is the straight line that "just touches" the curve at that point.

Okay, I've learned that pressure and stress are basically the same thing, the difference being that pressure acts normally on the surface, and stress acts tangentially.

That clarifies it. Like Harold said, pressure acts normally on the plane, so it acts on a much larger area than stress which acts tangentially


(Though I can't say I'm totally clear on why you don't take the surface area of the sphere ( ) for the pressure and another equation for the stress. Maybe I need to read my own quoted text again. And I really don't understand, given the explanation in my previous post from efunda, why the equation for the hoop stress is different from a cylinder for a sphere.)

Originally Posted by DrRocket

Also, when you apply a force to the exterios of your inflated beach ball you will change the stress state, particularly at the edges of the platen that will be used to apply the force. Since it is peak stress that determines material failure, the problem is not much more difficult to handle analytically, and the failure site will probably be at the edge of the platen.

I'll have to calculate the maximum hoop stress right? And this maximum hoop stress would occur next to the platen because there radius is smallest.

After I've (experimentally) determined the shape of the compressed sphere I can try to calculate the hoop stress.

Is it hoop stress I'm looking for, or is there still some other stress I need to consider?

After that I can start to try to consider elongation ...

Hi Harold and DrRocket

Can I infer from your silence you don't anymore see faults in my reasoning (that you know about / are aware of) and if I were to go ahead with the calculations (also using Young's modulus) I'd get the best answer possible without using finite-element software?

Thank you for your help!

Originally Posted by calimero

Hi Harold and DrRocket

Can I infer from your silence you don't anymore see faults in my reasoning (that you know about / are aware of) and if I were to go ahead with the calculations (also using Young's modulus) I'd get the best answer possible without using finite-element software?

Thank you for your help!

I don't think it's hoop stress you are looking for. That applies to cylindrical containers. If you use hoop stress and a smaller radius, you would get less stress on the wall of the container, not more. I don't see a simple way to calculate it.

DrRocket is a smart guy, but in my opinion he has sent you off on a bit of a wild goose chase. For what you originally were asking about - how much weight to burst a ball, given the bursting pressure - I still think you can attack it as I originally suggested, and get a reasonably good answer.

It may well be true that the stress goes up a bit where the ball is deformed from a spherical shape near the flat surface. It really depends on how elastic the material is. I'm also not entirely sure what the purpose of your calculation is and how accurate it has to be.

Thanks for your answer Harold

Originally Posted by Harold14370

I don't think it's hoop stress you are looking for. That applies to cylindrical containers. If you use hoop stress and a smaller radius, you would get less stress on the wall of the container, not more. I don't see a simple way to calculate it.

Thanks, I've looked again at where I saw that. I read the bit about radius incorrectly.

DrRocket is a smart guy, but in my opinion he has sent you off on a bit of a wild goose chase. For what you originally were asking about - how much weight to burst a ball, given the bursting pressure - I still think you can attack it as I originally suggested, and get a reasonably good answer.

You mean this, right:

Originally Posted by Harold14370

You could do some tests with a basketball or soccer ball. Pump it up and put a tire pressure gauge adapted to the needle to measure the pressure. Put it on top of a bathroom scale, and squish it down. Push it down with your hand or put a weight on top of it. Take some measurements with a caliper and see how much it is bulged out at the equator, and how out of round it is. Cover the ball with chalk or something, put a piece of paper under it, and mark the area of contact. See how well the F=PA relationship works out.

I agree. Only thing is that I can't really measure the stresses. Only the pressure at failure.

It may well be true that the stress goes up a bit where the ball is deformed from a spherical shape near the flat surface. It really depends on how elastic the material is. I'm also not entirely sure what the purpose of your calculation is and how accurate it has to be.

I'm trying to determine what material to use (which stress at yield / break in machine / transverse direction). If the method is more accurate, I could use material that was only 50% stronger than needed, instead of 200% stronger, for example.

Really I think I need to be able to relate internal gas pressure to the stresses on the material. The equation given earlier

is probably an accurate enough approximation.



It really doesn't need to be that accurate, because I'm going to make it stronger than needed anyway, but I'd like to at least understand what is going on.

Originally Posted by calimero

You mean this, right:

Originally Posted by Harold14370

You could do some tests with a basketball or soccer ball. Pump it up and put a tire pressure gauge adapted to the needle to measure the pressure. Put it on top of a bathroom scale, and squish it down. Push it down with your hand or put a weight on top of it. Take some measurements with a caliper and see how much it is bulged out at the equator, and how out of round it is. Cover the ball with chalk or something, put a piece of paper under it, and mark the area of contact. See how well the F=PA relationship works out.

I agree. Only thing is that I can't really measure the stresses. Only the pressure at failure.

Yes, I mean: (1) assume that the ball does not stretch significantly from its inflated position to the bursting pressure. (2) assume that the pressure is inversely proportional to the volume. (3) assume the volume decreases by the volume of the spherical dome and (4) assume that the force is equal to the internal pressure multiplied by the area of the flattened part. You can test those assumptions to an extent by doing the procedure described and see how well it fits. No, you cannot measure the stress, but you could see if the ball is significantly distorted from a spherical shape. If not, then the stresses are probably fairly well distributed and the formula for stress in a sphere would work.

Originally Posted by calimero

Hi Harold and DrRocket

Can I infer from your silence you don't anymore see faults in my reasoning (that you know about / are aware of) and if I were to go ahead with the calculations (also using Young's modulus) I'd get the best answer possible without using finite-element software?

Thank you for your help!

No, you can infer that I was not paying attention.

Stress is NOT basically pressure, even though the units are the same as pressure.

Stress is in fact a tensor. It includes both tensile and shear components, and those components are dependent on the coordinate system in which they are chosen. The stress state is dependent on the imposed boundary conditions and on the properties of the material.

There is a related tensor called strain, which is basically infinitesimal deformation. Stress and strain are related by a constitutive law that is dependent on the material. Hooke's law is a very simple constitutive law. Basically one relates the strain components to the stress components by means of various "moduli", in what can be a somewhat complex tensor relationship.

Material failure is different subject from the theory of elasticity. A material failure criteria can be based on stress or on strain. It is not usually a matter of just one stress or strain component. A common failure criteria is based on von Mises stress or von Mises strain, which is basically the maximum principle stress or strain after the hydrostatic component has been removed (aka deviatoric stress or strain). Principle stress is the result of adopting coordinates in which there is no shear, and this is always possible.

http://en.wikipedia.org/wiki/Von_Mises_yield_criterion

The point here is that failure of a structure under non-symmetric loading conditions, such as in the problem that you have posed, is a complex problem and that the failure will be due to high local stresses and is not amenable to simple back-of-the envelope calculations. I am not leading you astray.

Materials fail due to local effects. That is why failures start at local small defects and propagate along the cracks that initiate, because stresses are very high at crack tips. That is why you can stop a crack from propagating by drilling a hole (removing material that you might think only helps to carry a load) at the tip of the crack. It is the factors that create local "stress risers" that result in failure. This is why the simple back-of-the-envelope calculation will be quite far off in predicting a failure in a case which is not highly symmetric and in which the load application is not uniform.

Simple calculations will work well for symmetric loading scenarios, The equations that I gave you should, along with basic material strength data, accurately predict the burst pressure of a ball under internal pressure loading. But when you apply external loads that are not not spherically symmetric, the failure will occur at some point of discontinuity in the applied loads and back-of-the-envelope methods will be very far off the mark. An extreme case would be application of the load via a nail. It won't take much force at all in that case.

Even finite-element software will not give accurate failure predictions without very good failure criteria. Finite-element methods predict stress and strain, but only and additional application of a stress criterion can tell you if the material is likely to fail under the calculates stress or strain state. This sort of thing is usually part of an overall finite-element analysis. If you have materials that can tolerate a lot of deformation, as with rubber materials, then you also need to be careful with the finite-element analysis, as many codes developed for small strain modeling do not work well with large deformations.

Back-of-the envelope calculations for your beach ball will probably be accurate IF the initial pressure is sufficiently high that failure occurs very quickly with little deformation. But if you have low initial pressure and have to push very far before failure they will be very far off the mark. So, if your balloon us just about ready to burst, then the calculation will be OK, but otherwise you would be about as well off by taking a wild guess.

Sorry for my late reply...

Again thank you very much for your replies Harold and DrRocket.

You've given me many concepts to study DrRocket. I've looked at some, but I still need more time.

See you (much) later

Buy a 12 pack of beach balls for 6 bucks on Ebay, a Newton spring scale for $20, a bike pump with pressure guage for another 20, a 12 pack of beer, and go find a cabinet-making shop. (The beer is for the shop owner.) Take a pencil and paper, and write down what happens.

I'm not saying this to be a butt-hole...seriously...but it sounds to me like a professor has given out an incredibly complex mathematical problem to illustrate a point that sometimes when you can get you hands on it and try it...it's best to skip the math, and get into the field.

Cheers!

You said a beach ball? I filled one up half way and drove over it. It was like a soft bump. Didn't pop. So I reversed onto it again and parked there. It popped about 5 seconds later. I don't know the calculation but it was under the front left tire if a 2000 chrystler voyager. Felt like it lifted me up a couple inches. Fun

Welcome to the forum Penelope, just as a heads up you might want to check the date on the posts you are replying to. This one is 7 years old and none of the participants are still around...