# Calculating pressure change with load in elastic body

 P: 12 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?
 Homework Sci Advisor HW Helper Thanks P: 9,809 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.
 P: 12 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?
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P: 9,809
 Quote by MechEgr 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.
P: 12
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.
Attached Files
 Pressure Cylinder.pdf (91.1 KB, 3 views)