Toroidal pressure vessel

WOT wrote in news:1181019198.492462.98770 @g37g2000prf.googlegroups.com:

How does the mean value compare with your hand calculated value?

Makes sense to me...assuming your pressure vessel is constant thickness, I would expect the inside wall (the "donut hole") to carry more load than the outside. That would give you hoop stress variation around the meridian.

Split the torus at the equator (like a bagel) and do a free-body-diagram on a sector of what's left. You've got much more material along the outer wall than the inner wall. I suspect that, in order to balance moments, the hoop stress on the inner wall must be greater than on the outer wall.

Tom.

there is a variation bet 170 to 200 N/mm^2 in the result(the variation is along the circle). theory points to a constant value of 210

Tom , I think u got me wrong. in a torus the hoop stress is the stress perpendicular to the circle which is the meridion. from theory ( ref: "theory of Shells" by W.Flugge) i expected a constant hoop stress value.

Tom , I think u got me wrong. in a torus the hoop stress is the stress perpendicular to the circle which is the meridion. from theory ( ref: "theory of Shells" by W.Flugge) i expected a constant hoop stress value.

WOT wrote in news:1181110496.971031.126100 @r19g2000prf.googlegroups.com:

Well, now, that is a puzzle. The hoop stress is fundamental to keeping the high presure on the inside, and the low pressure on the outside.

I suppose a thickwalled pressure vessel might do it in bending, but that s still very odd.

Cheers

Greg

Yes . and the vessel i hav considered has a r/t ~10 and the model doesnt show much bending. it expands radially along both major and minor radius

I suggest that you look at the paper by Rossettos, J.N. and J. Lyell Sanders, Jr., "Toroidal Shells Under Internal Pressure in the Transition Range" AIAA Paper 65-145. These results have been confirmed by this author using several numerical analysis codes assuming the effects of linear bending alone, thick shell options (r/h < 10) and non-linear deformations.

Sometimes it is better to go back to the basics rather than trying to use FEM models that may or may not give weard results. In this case, membrane theory does not give the correct answer. Applying linear bending concepts from the above paper show as high as a 20% variation for nominal toroidal shells. Ansys is probably correct, but you can not always be sure when you have to consider all of the tangential factors governing its solution.

If a circular cylinder and ellipsoidal head are attached and pressurized, there will be discontinuity stresses at the attachment circ seam called discontinuity stresses. These arise:

due to geometrical discontinuity brought on by change of Gauss curvature ( K < 0 inside, > 0 outside 0 at crown)

or consequentially

due to unequal expansions of the two parts which have to be forced to be same by shell material continuity, resulting in local bending stresses, as hyperstatically arising stresses in nature.

The former fact is not brought out so explicitly in text books by Fluegge, Krauss etal. The K < 0 inner parts of doughnut are comparatively weak in taking up pressure induced membrane stresses.

Your Ansys results reflecting this bending are OK.

I suggest running more problems to verify this phenomenon. Apply lateral loads on a corrugated plate:

z = 100 ( cos( 2 pi x / lamba1) + cos( 2 pi x / lamba2) ). Say lambda1 = 200 mm and lambda2 = 160 mm , which have zero K at x,y = lambda/4,

3 lambda/4, 5 lambda/4 etc.that should exhibit the same nature of discontinuity induced stresses.

Regards, G.L.Narasimham ex ISRO

Bending at the crown is a phenomenon mentioned in all text books and in papers like the one cited by Richard L. Citerley, in post no 10. When the material is anisotropic, problem is more aggravated compared to isotropic case.

Narasimham

If a circular cylinder and ellipsoidal head are attached and pressurized, there will be discontinuity stresses at the attachment circumferential seam called discontinuity stresses. These arise:

due to geometrical discontinuity brought on by change of Gauss curvature ( K < 0 inside, > 0 outside 0 at crown)

or consequentially

due to unequal expansions of the two parts which have to be forced to be same by shell material continuity, resulting in local bending stresses, as hyperstatically arising stresses in nature.

The former fact is not brought out so explicitly in text books by Fluegge, Krauss etal. The K < 0 inner parts of doughnut are comparatively weak in taking up pressure induced membrane stresses.

Your Ansys results reflecting this bending are OK.

I suggest running more problems to verify this phenomenon. Apply lateral loads on a corrugated plate:

z = 100 ( cos( 2 pi x / lamba1) + cos( 2 pi y / lamba2) ). Say lambda1 = 200 mm and lambda2 = 160 mm , which have zero K at x or y = lambda/

4, 3 lambda/4, 5 lambda/4 etc.that should exhibit the same nature of discontinuity induced stresses.

Regards, G.L.Narasimham ex ISRO

thx richard &narasimham 4 replying. was away 4 a while.

lemme try this and tell you ppl.