Calculating pressure change with load in elastic body

[MechEgr]

I am attempting to determine a method for calculating the final pressure in a deformable bladder given a starting pressure and an applied load.

Scenario:

This is actual data gathered in my lab:

A) You have a ball made from a rubbery material with P₁=3psi. You weigh 150lbs and step on the ball, and the internal pressure P₂=13psi.

B) The same ball is pressurized to P₁=10psi. Applying the same load as before (150lbs), P₂=16psi.

Question:

1) Why does applying the same load in both cases not increase the internal pressure by an equal amount (why does the starting pressure matter)?

2) Without knowning the characteristics of the ball (elastic modulus, etc.), is there any way to calculate the resulting preesure (P₂)? Would the internal surface area of the ball be of any assitance?

 

[haruspex]

Is this all at constant temperature? I.e. was the ball allowed to cool down again after increasing the pressure/load before taking the measurement?

My main guess is it's to do with the shape the ball takes on. E.g. if the ball is wider than the load then at the lower pressure the load will tend to sink more into a dip in the top of the ball. This means the part of the ball skirting the dip will be pulling upwards on the load. In effect, the load is spread over a greater area, reducing the pressure needed to support it.
Similarly, if the ball is narrower than the load, with less inflation it will spread wider and present more area to the load.

 

[MechEgr]

Yes, the assumption is that this was done at constant temperature.

I think what you're saying makes sense. Are saying essentially that at a smaller starting pressure, the internal pressure would have to increase more in order to support the same amount of load than if the starting pressure were higher?

Any ideas on a method to calculate what the resulting pressure (P₂) would be with a given starting pressure? Or would that all be dependent on the material?

 

[MechEgr]

For the record I think I figured out the answer.

If you start with a small pressure, the ball deforms significantly to compensate. If, on the other hand, you start with a large internal pressure, the deformation of the ball in response to the added load is much less noticeable. Theoretically then, there would be a point at which the starting pressure would be so great that adding my body weight would not change the internal pressure at all. The closer I get to this theoretical value, the smaller the change in pressure due to my added weight until ultimately, the change is zero.

This makes sense when thinking about the gas law (PV=nRT). Assuming n, R, and T to be constant, P₁V₁=P₂V₂. So, if the starting pressure P₁ is small, then when I step on it, the ball will "squish" significantly and the end volume (V₂) will be much smaller than the start volume (V₁) (ΔV is large), necessitating a larger end pressure (P₂). Conversely, if the start pressure P₁ is so large that stepping on it does not change the volume very much (ΔV is small), then P₁ and P₂ will be very close to the same value... this seems like it should have been easier for me to see.

Thanks for your help!

 

[haruspex]

This makes sense when thinking about the gas law (PV=nRT). Assuming n, R, and T to be constant, P₁V₁=P₂V₂. So, if the starting pressure P₁ is small, then when I step on it, the ball will "squish" significantly and the end volume (V₂) will be much smaller than the start volume (V₁) (ΔV is large), necessitating a larger end pressure (P₂). Conversely, if the start pressure P₁ is so large that stepping on it does not change the volume very much (ΔV is small), then P₁ and P₂ will be very close to the same value...

Maybe.. but I don't think you can be sure whether that explains it. It's not at all clear it would have the consequence you see. E.g. consider a different model: a vertical telescopic tube of air (a piston) of cross-sectional area A. At first, the air only supports the weight of the piston, W, and the pressure is W/A. You add load L and the pressure increases to (W+L)/A. The increase is L/A. You start again with a heavier piston, 2W. When you add L, the pressure increase is again L/A.
The key is A. To see a smaller pressure increase, the effective area supporting the load must be larger. My first post explained why that would be the case.

 

[MechEgr]

I don't think I agree with you because I think for your example to be true, the starting pressure in the system must be zero. The pressure change inside a piston system (ΔP) is not necessarily ΔL/A if the starting pressure is positive (non-zero). I have attached a simple example which concludes that, with positive starting pressure (like in our ball example), ΔL/A≠ΔP.

I agree with you that surface area is the key, but the SA interacting with a given load changes with starting pressure. If I start with a high pressure, less SA interaction is needed to support the load resulting in a smaller volume change (and therefore less pressure change). If I start with a low pressure, the ball "squishes" more such that the SA (and thus volume) changes significantly, resulting in a significant pressure change despite adding the same load.

 

[haruspex]

Your example adds a force from the shoulders of the container. That's not in my model, and is in effect an extra load.
Take the shoulders away, so that it requires L > 0 for there to be any pressure.

"If I start with a high pressure, less SA interaction is needed to support the load"
I think we're saying the same thing for the case where the load's base area extends beyond the area of contact. But consider the case where the ball is so large that the surface area of contact with the load is determined entirely by the base area of the load. Now you have to look at the forces in the ball's membrane. These pull up on the part contacting the load, reducing the need for pressure.

 

[MechEgr]

Yes, what you're saying in both cases is true. The reason there are "shoulders" on the cylinder is that we need to be able to start with some positive pressure (as in the case with a pre-pressurized ball).

Yes, I agree that there will be a material effect from the ball. the effect would indeed be magnified with the ball's diameter.