Forces on a pressurized tin can of beans

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The discussion focuses on calculating the force a pressurized tin can of beans can withstand before failing, with specific dimensions and material properties provided. The user has attempted various methods, including short column buckling and axial stress calculations, but has encountered difficulties yielding inconsistent results. The internal pressure of the can is noted to be 30 psi, raising questions about spoilage and the structural integrity of the can. It is emphasized that the thickness of the metal and the presence of corrugations are crucial factors in determining the can's strength. Ultimately, while the can has some strength, the maximum atmospheric pressure can exert significant force, suggesting caution in experimental setups.
billybomb
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Hi

So I'm trying to find out how much force a tin can, can support before failing. The can will be stood up as it should be with a force applied over the top across the area of the can.

I've tried the process of short column buckling to no avail.

I've also tried axial and hoop stress for thin walled pressurized vessels.

I keep getting silly numbers and am now completely lost.

Also, would you assume that pressure inside the can is the same as atmospheric?

So here are the numbers:

Length = 109mm
Diameter = 74mm
Assumed yield stress of 590n/mm^2
Atmospheric pressure = 0.101325n/mm^2

Any direction or help would put my mind at ease, thanks

I've found http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19690013955_1969013955.pdf
Is page 14 the part I am look for?

Thanks again.
 
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Without knowing the thickness of metal in the can, you won't be able to even hazard a guess. It is also important to know if the sides of the can are rolled straight or if they might have any corrugations.
 
Yea, sorry forgot to include that...

Thickness = 0.4mm

The sides are corrugated, but I am just after the basic theory of the crushing force needed.

I've got 53683.1N using Axial stress for thin walled cylinders

Also tried a basic calculation of Stress = F/A + Pressure*pi*r^2 and rearrange for F

I've also found that the internal pressure of a can of beans is 30psi or 206842.72 N/m^2

Thanks alot
 
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No ideas? :<
 
If your can of beans has an internal pressure of 30 psi, I suspect that the beans have spoiled. Usually, canned food is vacuum packed to eliminate any air inside the can which might promote spoilage of the contents. Carbonated beverages obviously are an exception, as the internal pressure inside the can provides resistance to collapse of the can.
 
If the can is vacuum packed, wouldn't that mean the tin has negative internal pressure and would deform laterally?
 
Not necessarily. Although cans are not infinitely strong, they do have some strength. That's why the thickness of the metal and what kind of corrugations are important in trying to determine the strength of the can. After all, even if a perfect vacuum forms inside the can, the maximum atmospheric pressure is still only about 15 psi.
 
SteamKing said:
After all, even if a perfect vacuum forms inside the can, the maximum atmospheric pressure is still only about 15 psi.

Be careful with that word "only". I prefer to think of it as nearly 1 ton per square foot, which doesn't sound quite so puny!

With the OP's can dimensions, 15 psi is about 100 pounds force on each end of the can. Somebody weighing 200 pounds might be able to stand on two bean cans without crushing them, but if the experiment failed it would make quite a mess.
 
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