3 ms·

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 cub

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 11y ago
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/3-11-mechanics-of-materials-fall-1999/modules/pv.pdf http://ocw.mit.edu/courses/materials-science-and-engineering... [PDF ALERT]
- darkmighty 11y ago
  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?

gusmd 11y ago

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/3-11-mechanics-of-materials-fall-1999/modules/pv.pdf http://ocw.mit.edu/courses/materials-science-and-engineering... [PDF ALERT]

darkmighty 11y ago
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?