[gusmd]

More important than pressure differences, is the circumferential stress on the fuselage/can, which is what the walls are actually withstanding. For a pressurized circular container, it is given by p*r/t, where p is the pressure, r the radius and t the wall thickness.

If we consider a 3 m wide airplane, with a 1.6 mm thick fuselage at cruise altitude (around 40k feet), you get around 56.5 kPa of pressure difference, and the above equation gives you approximately 52 MPa. (data from a real aircraft that I cannot mention).

For a coca-cola can, internets say 380 kPa of pressure, for a 6.6 cm diameter can, and 0.15 mm aluminum sheet. That results in approx. 85 MPa.

So as you can see, even though you have roughly 7x more pressure in the soda can, its much smaller diameter severely reduces the stress on the walls, resulting in about 1.5x the circumferential stress.

Hope that answers your question.

edit: several errors in my back-of-the-envelope calculations. sorry.

[darkmighty]

That's an odd model for me. So as the radius goes to infinity for a fixed thickness and differential the stress goes to infinity too? But take a pressurized cube with reinforced edges. You can say it's faces have infinite radius (or as small as you'd like) -- yet the stress should be finite? How do you reconcile this? Or does that work only for circles? (which would be odd since stress is a local concept to me)

[gusmd]

I'm not sure of your background, but it is actually a pretty simple concept. To satisfy static equilibrium on a given transverse "dx" segment of a cylinder, the walls must balance the internal pressure (in this case taken as the pressure difference), so that:

stress * 2 * thickness * dx = p * 2 * r * dx.

[ stress * walls cross-sectional area] = [ internal pressure * projected internal area ]

Solve for stress, you get:

stress = p * r / t

Regarding your question, yes, as the radius goes to infinity, the stress goes to infinity. The area where the pressure is applied grows with r, but the cross-sectional area where the stress is applied is still (w * thickness * dx.) This equations work well for thin-walled cylindrical pressure vessels (r > 5t is general rule of thumb). For a cube, you would have to develop the equations, but keep in mind that you will have a singularity/discontinuity on the walls because of the right angle.

edit: good to have a reference just in case: http://ocw.mit.edu/courses/materials-science-and-engineering... [PDF ALERT]

[darkmighty]

Ah makes more sense now, thanks for the explanation. I'm an undergrad in EE so mechanics is not my strong point.

So a planar face of a cube cannot satisfy the equilibrium equations? Interesting ... so then a cube will necessarily bulge so the radius is enough to satisfy the equations, right?