# Why doesn't atmospheric pressure crush thick walled structures?

There is one thing that puzzles me: common explanation of why don't structures collapse under the enormous atmospheric pressure (~101300N/m^2) is that the force pushing from inside balances out the force pushing from outside.

Now let's imagine a cube made of imaginary airtight perfectly flat polystyrene with side length of 1m. It would have a wall thickness lets say 0.2m so the inside of the cube would be a cube with side length of 0.6m. If this explanation of 'forces balancing out' is to hold - than there should be a net force of F=(A1-A2)*Patm in place - >> F=[(6*1^2)*101300-(6*0.6^2)*101300=388,992N which should surely crush any such structure, but this clearly doesn't happen...

I have attached illustrative picture.

Sorry for triviality of the question - it's just it somehow doesn't make sense to me. Can anyone please help?

• Which way does your net force point? – Brian Moths Apr 28 '15 at 11:53

The walls of the structure, in this case .2m of polystyrene, are themselves solid and rigid. They provide pressure which fills up the space between the inside surface and the outside surface, effectively adding to the atmospheric pressure inside the cube.

If the polystyrene is foamed, there would be pressure from the gas inside the polystyrene bubbles which fills the space between inside surface and outside surface.

However, if the thick walls of the cube were constructed of a hollow, evacuated non-rigid material, the structure very well might collapse, either from air pressure, the force of gravity, or both.

• Would the same apply for cube made of lets say ABS plastic - or different material that is not hollow/porous or air filled? Lets say I wrap this material in some plastic membrane so there are just two surfaces smaller and larger - I just can't see it collapsing on itself or am I wrong? – user2820052 Apr 28 '15 at 12:15
• user2820052: I'm not sure I understand the question in your comment. I added "evacuated" to the last paragraph of my answer, to indicate that the collapsing walls are both non-rigid and low-pressure, like the inflatable walls of a "funhouse" at a carnival which loses its air pressure. – Ernie Apr 28 '15 at 12:26
• Well my comment question really goes: imagine a cube of same dimensions (cube 1m & 0.6m) made of some material that doesn't contain air in itself, lets say its solid plastic or metal, there is no evacuation of inside going on, but there is a >surface area difference< between outside surface and inside surface, hence there should be a a force arising from area difference at constant pressure - or am I mistaken...? Why don't we see this fact having repercussions in real world? Or is it because I do not understand pressure? – user2820052 Apr 28 '15 at 12:50
• ... F=p*A, pressure is constant (101300N/m^2), area of inside is much smaller than outside, force acts perpendicularly to the sides of inside and outside so how is it that we don't see such objects crushed in real world? – user2820052 Apr 28 '15 at 13:00
• user2820052: Yes, NET stress arises from the difference in outside and inside surface areas, but if the wall is built of sufficiently rigid material it overcomes the pressure difference which creates net stress. This may help: physicsnet.co.uk/a-level-physics-as-a2/materials/stress-strain. If the walls are built of material which withstands the excess of outside over inside surface pressure, they will not collapse. – Ernie Apr 28 '15 at 13:19

I am reading your question as wondering why the total force on the (smaller) inside surface area of the cube is not crushed by the (larger) outside surface area. Perhaps exaggerating this will make it a bit clearer.

Yes the total force on the outside is larger, but much of that force does not reach the interior of the cube. If you imagine the force from part of the outside (between the highlighted arrows), this force never reaches the interior, so the interior does not have to resist it.

In fact, we can imagine cutting part of this off...

This reduces the outside pressure on the remaining cube, but has no affect on the pressure that the interior has to resist.

• OK, but what if the walls are sightly thinner and made out of some very low strength material? – user2820052 Jun 26 '15 at 10:28
• It seems like you think that one side is being pushed more than another side. That is not the case. Assuming the interior is at the same pressure, then the slice of the wall that faces the interior has the same force on both sides. Therefore, it doesn't need any great strength. The thick walls were not drawn to make the walls strong. It is to show how most of the wall is irrelevant and can be removed without affecting the box. – BowlOfRed Jun 26 '15 at 15:46

It looks like you don't take into account elastic forces in the rigid cube. The pressure on the cube's faces will tend to bend them inwards, the force will be passed to the ribs, so the faces and the ribs will get compressed, and they will resist compression. Perhaps it is easier to understand what happens if you consider an evacuated spherical shell, rather than a cube. External pressure will lead to much greater compressive stress in the spherical shell (in the direction tangential to the surface of the sphere), but the so-called modulus of elasticity of solid bodies can be extremely high and can provide resistance to the external pressure. If you consider a small segment of a spherical shell, the sum of the stresses at the periphery of the segment will be directed against the aggregate force of external pressure acting on the segment. The main mode of failure of a solid shell under external pressure is the so-called buckling (loss of stability) - it is similar to what happens when you squeeze an empty metal beer can. However, solid shells can resist external pressure to some extent. Let me summarize the above as follows. You haven't even started to calculate forces acting on a solid evacuated shell until the value of the modulus of elasticity appeared in your calculation.

• @user2820052: Thank you for your questions and for your interest in my work on vacuum balloons. However, I don't quite like your approach to calculation of the force acting on the tube. Remember, force is a vector, so when you calculate a sum of several forces, you must take into account their directions, not just their magnitudes. The same is true for the hemisphere. You should not just multiply the pressure by the surface of the hemisphere to calculate the force acting on the hemisphere, as the force of atmospheric pressure acts in different directions in different points of the hemisphere. – akhmeteli Jun 20 '15 at 11:51
• @user2820052: (cont.) To calculate the force acting on the hemisphere, I could integrate the projection of the force of atmospheric pressure upon the axis over the entire hemisphere and get this factor of 1/2. However, it is easier to do that in a different way. Let us consider an imaginary hemisphere of air in the atmosphere. It is in equilibrium under external pressure (let us forget for a moment about the weight of this hemisphere and its buoyancy). So the sum of pressure forces acting on it from the left is equal to the sum of pressure forces acting on it from the right. – akhmeteli Jun 20 '15 at 11:59
• @user2820052: (cont. 2) However, it is easy to calculate the pressure force acting on the hemisphere from the right (on the flat part of the hemisphere), as the pressure force acts there in the same direction everywhere, so this force equals $\pi R^2 P_{atm}$. Again, the pressure force acting from the left has the same magnitude, as the air hemisphere is in equilibrium in the atmosphere. On the other hand, the force acting on a hemisphere of air from the left is equal to the force acting on a (metal) hemispherical shell from the left. That is why my result is half of yours. – akhmeteli Jun 20 '15 at 12:07
• @user2820052: (cont. 3) The same approach can be used for the pipe. You consider a half-pipe and find the pressure force acting on it from one side, it will equal $2R L P_{atm}$. On the other hand, the tangential stresses acting on this half-pipe produce a force of $2\sigma t L$, where $\sigma$ is the compressive stress in the pipe, $t$ is the thickness of the pipe, $L$ is the length of the pipe. These two forces are equal, as the half-pipe must be in equilibrium. This is how you can calculate the tangential stress in the pipe. Of course, this stress must be lower than the material strength. – akhmeteli Jun 20 '15 at 12:16
• @user2820052: (cont. 4) Otherwise the pipe will collapse. (I forgot to say that $R$ is the external radius (of the sphere or of the pipe)). However, this requirement for material strength is not enough. You don't just want the half-pipe to be in equilibrium, you want it to be in stable equilibrium, otherwise it will fail through buckling (see, e.g., youtube.com/watch?v=Zz95_VvTxZM ). There is an expression for critical external pressure for a cylinder, for higher pressures the cylinder will buckle. But in general, vessel design is quite complex:-) – akhmeteli Jun 20 '15 at 12:35

Your drawing shows an empty box .

If it is at room temperature and not air tight the inside and outside air pressure are in equilibrium, the arrows balance against the rigidity of the wallss

If the box is air tight and the inside is a vacuum then it will be crushed. Nobody makes vacuum tubes from polysterine.

Why we can have vacuum tubes ( incuding light bulbs?) . It is because the rigidity of molecular structures for many materials is high enough to balance the force from pressure finally with the equal and opposite force of the ground .

• Yeah but I am assuming empty box that is air tight and is filled with air at atmospheric pressure and walls that are not porous but sum of inside surface is much smaller than that of outside surface - is there a net force acting on outside of the structure towards centre? Instead of polystyrene let imagine some imaginary material that is absolutely not porous completely air tight and let's say this material will have very poor mechanical strength. Will this structure be deformed by difference in forces acting on outside vs inside?? – user2820052 Jun 24 '15 at 13:37
• Your atmospheric pressure guarantees that the forces balance . That is what atmospheric pressure means, no matter how thin the material. After all you can imagine a box, no material, and if there is no wind there is no motion inside outside. When a wall intervenes it transfers the forces with its rigidity structure, and as long as the inside and outside pressure is equal nothing happens. – anna v Jun 24 '15 at 13:58
• But P=F/A and there is movement of air molecules present under atmospheric conditions so there should be net resulting force if the area upon which these molecules act is different - or am I wrong? – user2820052 Jun 25 '15 at 20:18
• The area cannot but be the same, by the definition, only P and F are variable. – anna v Jun 26 '15 at 2:40
• But in case of for example pressure vessel you would have the force acting outwards... So how could this be any different? Larger area must equal to higher force all other variables being the same or am I wrong? – user2820052 Jun 26 '15 at 10:11

Your diagram shows all of the forces acting on the box as a whole. These are balanced so the box remains stationary, but it doesn't tell you what happens to individual parts of the box.

Objects get distorted because the net force acting on individual parts of the box are different. Let's view the middle of the top of the box:

Here, we see that acting on on this small portion of the box is: air pressure from above, air pressure from below, gravity, and the forces holding the box together.

There is some distortion — e.g. the center of the box top is hanging slightly lower than the rest of the top which is why the net restoring force on the edges are both pointing upwards — but that is entirely due to the effects of gravity; we can see that the air pressure from the two sides cancel out entirely.

If we split the diagram further into the upper and lower halves, we would see that on the upper half the air pressure is opposed by the repelling force from the bottom half, and vice versa. The air pressure from the two sides compresses the material of the box (which is why the restoring force causes the top and bottom to repel).

• The thing is, the pressure is F/a hence every square meter is subjected to 101kN of force. Now cube shaped box with side of 1m has 6m2 surface are it has walls with 0.05m thickness hence inside has area of 4.86m2. So the net resulting force inwards is 1.14*101kN... But where does it go?? – user2820052 Aug 21 '16 at 17:45
• @user2820052: The forces you just listed cancel themselves out. The "inward force" on one side of the box exactly cancels the "inward force" on the other side of the box, because they are pointing in opposite directions, so the net force on the box due to outside air pressure is zero. – Hurkyl Aug 21 '16 at 17:49
• But area is not equal and hence the forces are unequal... think corner 0.05 by 0.05m there is no opposing force there... – user2820052 Aug 21 '16 at 18:10
• @user2820052: You misunderstood my previous comment -- it's talking only about the external air pressure, and has nothing to do with the air pressure inside the box. The net force on the box due to external air pressure is zero. – Hurkyl Aug 21 '16 at 18:47
• Ok opposite forces cancel out but there is a difference in surface area and hence force that acts on the walls - do you understand me? More force from outside than inside... None seems to notice :( – user2820052 Aug 21 '16 at 21:32

In the specific case of cube there is no force difference being produced that would crush the walls.

Consider any macroscopic object consisting of parallel inner and outer walls exposed to the same pressure--there is an internal stress produced on the material of the wall from being compressed on both sides. This is countered by the normal force produced by the uncompressing atomic/molecular structure of the material. In case the walls are infinitisimlly thick the inner and outer forces simply cancel.

However the reason there is no net force difference in the case of cube is because the vector force calculated from the pressure has to be done locally i.e. the normal component of force at a differential area can be calculated from the pressure acting at that area--for the case of the cube this is exactly the same as the force acting on the entire area of that face which is same on either side. As to the different face inner and outer areas--the difference in area is not producing force on the inner wall--its producing a force all the way onto the opposite exterior wall (see @BowlOfRed's excellent diagrams) Hence there is in fact no net force acting on the wall.

This is not true in general if the wall has parallel surfaces with curvature--like sphere. In this case the inner and outer differential area elements are not same so even though the walls are exposed to the same pressure, the force produced isn't the same on the two sides. This leads to a stress gradient in the wall. The stress gradient then balances the net force difference.