Stair Geometry at top of Sphere

[swearingen]

We're trying to model a stair that goes to the top of a high pressure storage sphere (LPG tank). I'm having trouble working out the geometry and wanted to know if any of you have a guide for this sort of thing.

The stair is a normal, straight stair up to the mid-point of the sphere, then it curves with the surface of the sphere up to a small platform at the top. The sphere is about 60' in diameter.

Thanks!


-5^2 = -25 ;-)

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[hokie66]

Not sure of the issue. The part on top of the sphere is a circle...no?

 

[JStephen]

I'm not aware of a standard for that. On occasion, I see reference to PIP standards, and they may address it.

Ideally, you'd hold uniform rise for all treads, and uniform run either at the inner end or at the center. Once you get past the 45 degree line, I'd put a platform and run it up radially from there.

If you can locate some similar tanks, Google satellite photos might be adequate to show the general arrangement used.

 

[swearingen]

Rereading my post, I realized that if you hadn't seen one before, there would be little chance of you being able to picture it.

In plan view, the stair would come straight from a quarter point, tangent to the equator of the sphere, to the ground. From that tangent point, it curves up and around the arc of the sphere until reaching the side of a small platform mounted on the "north pole" of the sphere.

A picture would be best - see enclosed...


-5^2 = -25 ;-)

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[BAretired]

Would need to establish the rise and tread width, R and T of the curved stair. The center of the sphere is point A.

Every point along the inside stringer meets the tank wall at precisely 30' from point A at various angles horizontally and vertically. The top of the n^th riser is precisely n*R above point A. The sloping distance from the top of one riser to the top of the next riser is S = (R²+T²)^1/2. We have a series of identical isosceles triangles, with dimensions 30', 30' and S where the elevation of each node is known. From that, it should be possible to model the inside stringer.

BA

 

[BAretired]

The attached spreadsheet is based on a 30' (360") radius, 7" Risers and 10" Treads. It could easily be done for any other combination of dimensions.

Rn is the radius of the circle cutting the sphere at the elevation of the tread. d-theta is the angular dimension with chord length 10" and radius Rn, so d-theta = 2(arcsin(5/Rn)).

X, Y and Z are the Cartesian coordinates of the bottom of the next riser at the level shown. X = Rn*cos(theta), Y = Rn*sin(theta) and Z = Elevation of tread, taking 0 as center of sphere.

Each value of X, Y and Z satisfies the equation of a sphere, i.e. X² + Y² + Z² = 360² which serves as a partial check of the procedure.

BA

 

[swearingen]

BAretired, thanks for the reply, although I'm not sure what you have will work. Remember, the stair treads themselves begin as parallel to Rn at the equator and end up as perpendicular to Rn at the top. D-theta really doesn't have a meaning here...


-5^2 = -25 ;-)

http://www.eng-tips.com/supportus.cfm

 

[BAretired]

swearingen,

You are welcome. D-theta does have meaning. D-theta is the rotation with variable radius Rn which has a chord length of 10" (my assumed tread width). The sum of d-thetas at the n^th step is the total rotation from the beginning to the n^th step.

BA

 

[swearingen]

The problem is, that is only correct if the steps stay radial to the sphere in plan. They do not. Essentially, that chord distance is reducing with each step.


-5^2 = -25 ;-)

http://www.eng-tips.com/supportus.cfm

 

[BAretired]

I assumed that each step met the shell in a radial direction. That way the 10" end of the step would be flush with (tangent to) the shell surface. Why would it be otherwise? That is the most logical way to build it.

BA

 

[JStephen]

It's hard to tell from the picture, but it does look like they made a corner and ran it radially towards the top.

 

[BAretired]

It also appears from the photo that there is no inside stringer, i.e. each tread is welded directly to the spherical shell. The outside stringer seems to be above the treads and the stair is open, so there are no risers.

BA

 

[BAretired]

I found the attached link which deals briefly with the subject of stairs for a spherical tank. It appears the slope was steeper than I had assumed. Article 2.6 indicates that the maximum slope measured from horizontal is 45" (sic). I assume they mean 45^o.

BA

 

[swearingen]

Look again at the picture. You can see that at the top, the stair approaches the pole of the sphere "square" (see the enclosed sketch). I know the sketch is rudimentary, but there is a subtle, but very important difference between the two: The center of the top tread is offset from the pole, and the tread itself is at right angles to a line drawn from its center to the pole.

You can see in the picture posted earlier that at the top few treads are supported by the vertical struts on the inboard side, just as they are on the outboard side lower down (just above and to the right of the "55").

You are correct in saying that the there is no inner stringer from the equatorial landing up, however, it's also apparent that the inner edge of the treads must pull away from the shell at some point. You can even tell that the axis of the tread quickly deviates from a radial line pointing to the vertical central axis of the sphere as you go above the equator.

That said, your sheet has given me an idea and I'll work on it further.

Thanks for your replies...


-5^2 = -25 ;-)

http://www.eng-tips.com/supportus.cfm