Beam with cantenery action

[JAE]

We have a project where we are checking a long steel tube that is on top of a steel plate wall - essentially the tube is the wall cap that goes around the perimeter (4 walls) of a steel tank.

The tank is filled with water and imparts lateral pressures on the tank's four walls. This creates an outward uniform force on the tube on all four sides.
The tank is about 10 ft. wide x 36 ft. long.

The tubes will naturally bend outward and have axial tension in them due to the orthogonal walls with their outward uniform load.

The question is - we need to determine the stress in the long 36 ft. tube due to axial, bending and perhaps any axial effect from what appears to maybe be a cantenary effect. How to do this?

Roark's has essentially a case where a beam is rigidly fixed at the ends with a uniform load on it.
It is labeled as Table 8.10, item 3 (Beams restrained against horizontal displacement at the ends - ends pinned to rigid supports, uniformly distributed transverse load on entire span.

However, the end walls of this tank are not perfectly rigid and will bow inward a bit with the cantenary tension on the long tube, right? So this will have the effect of reducing the axial tension in the tube a bit. Roark's does NOT have an item with axial spring supports at the ends.

We've modeled this in RISA 3D and get results but something tells me that the RISA results may not include non-linear effects of the cantenary action in this tube due to variations of the end supports with regard to axial loads.

I seem to understand that RISA does PDelta analysis but that only includes PDelta effects perpendicular to the member and does not include axial PDelta effects.

Any thoughts on this? I've inserted essentially the beam problem in a sketch.

Check out Eng-Tips Forum's Policies here:
faq731-376

 

[rb1957]

"but the reaction from the long sides is acting at 90[sup]o[/sup] to the reaction from the short sides" ...

the reaction from the long sides would be acting along the cable on the short sides, and vis versa; so couldn't the short side cable react the loads from the long side ?

another day in paradise, or is paradise one day closer ?

 

[rb1957]

oops, that's dumb ... getting cables and beams mixed up !? ... of course the reaction for a cable is axial and the reaction for a beam is lateral ...


another day in paradise, or is paradise one day closer ?

 

[dhengr]

JAE:
Why not sell the idea of a deck all the way around the top of this pool? That way you could hide the stiff horiz. beams and corner details which should have been there in the first place to make this crazy idea work. They thought they were being clever (and/or cheap) in the first place, and sooner or later they have to pay the piper if they want this to really work.

 

[rb1957]

I wonder if a complete container would work, ie half fill a container before you cut it in half ? The sides would be much stiffer, and a nice loadpath into the roof. I know it isn't practical but it does give you a design point.

another day in paradise, or is paradise one day closer ?

 

[KootK]

KootK, I'm not quite sure with whom you are agreeing

I was agreeing with CANPRO. I've been under the impression that when one posts a comment without specifying a target recipient by name, that target recipient is assumed to be the poster of the previous comment. That or OP, I go both ways.

...but the tension in the 36' long member, while nearly constant, is not constant. It is slightly greater at the ends than at midspan because of the slope.

I suppose that it would have been more precise to say that the horizontal thrust in the member at the ends is constant.


I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[BAretired]

I was agreeing with CANPRO. I've been under the impression that when one posts a comment without specifying a target recipient by name, that target recipient is assumed to be the poster of the previous comment. That or OP, I go both ways.

Ordinarily, I would have been under the same impression but in this case, I didn't think you were agreeing with the entirety of CANPRO's comment.

I suppose that it would have been more precise to say that the horizontal thrust in the member at the ends is constant.

Ordinarily, with a cable sagging in a vertical plane, that would have been more precise, but in this case, the member deflects in a horizontal plane, so thrust is always horizontal.

BA

 

[SlideRuleEra]

Sizing the tube not to fail is pretty straight forward and avoids the axial tension complication since the pipe stays relatively straight under load, as it should. For the following calcs assumed each 36' long tube has simple supports with maximum deflection at mid-span (an arbitrary) 2.40" (L/180). Choose what ever deflection is wanted.

Per BAretired, UDL on the tube is 260 lb/ft and tension is 1300 lb.

Calculate the required moment of inertia to meet the above criteria (36' long steel member, deflection of 2.40", 260 lb/ft loading).
I = 141 in[sup]4[/sup]

For a 10" diameter, "standard weight" steel pipe:
I = 161 in[sup]4[/sup]

Check 10" pipe bending stress... f[sub]b[/sub] = 16.9 KSI

Check 10" pipe tensile stress... t = 109 PSI

Unless I'm missing something, done.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net