Tension only Structure with Hydrostatic Loads

[bigmig]

We are designing a large swimming pool, for lack of a better description, where the walls of the pool are a fabric membrane that contains the fluid. The pool is on the order of 10' deep. Without going into the specifics of what the supporting structure is, I'm having nightmares about solving membrane problem. The membrane is wrapped around a support at the pool base and pool top. The membrane goes over these top and bottom supports, turns a 90 degree turn, and is then attached to a "fixed" support. Directly between these two top and bottom points is a roller, that prevents the membrane from bowing out in one big bulge. With the roller, it now bows out in two varying sized bulges.

The supports are steel pipes that span between larger, primary structure.

I have attached a sketch to clarify.

I am trying to size the supports (top, roller and bottom)without buying a non geometric linearity FEA modeling program, or by performing matrix analysis. Any suggestions on how to simplify this, or a hint on a cheap cable modeling program would be great.

 

[BAretired]

Interesting problem. It looks as though your top bulge is much too big relative to the bottom one. As you fill the tank with water, the bottom bulge will grow outward. When water level reaches the midpoint roller, there will be no bulge in the top. The fabric will be undeflected in the upper half but will be carrying considerable tension.

As you continue filling the upper half, the membrane tension will increase, reducing the lower bulge and increasing the upper one. If the middle support is a true roller, the vertical component of the fabric tension will be constant over the 10' height. The bending moment in the fabric is zero everywhere, so it assumes the shape of the moment curve for two simple spans.

As a starting point, you could assume concentrated horizontal loads at, say one foot intervals. Each load would be simply 62.5z where z is the depth of the load. Then you would find the funicular polygon corresponding to that load pattern and finally, you could correct for vertical pressure on each segment using an iterative solution.

BA

 

[ishvaaag]

There is software out there like ForTen 3000 (also 2000, now a 4000 beta version) to efficiently deal with tensile structures.

http://www.forten32.com/

However even if fantastically suited for some typologies may fall short of each and every situation, shape and restraints, and considerations like those of BAretired above are well placed.

 

[csd72]

This is one of those problems that cannot be simplified and needs to be analysed properly and in 3 dimensions. The final equilibrium state of the fabric can be vastly different from initial estimates.

Get the software or pass it on to someone who knows what they are doing with these things!

 

[BAretired]

Actually, you could make a very simple analysis to get some idea of the bulging involved. A triangular load has virtually the same maximum moment as a uniform load, 0.128WL vs 0.125WL where W represents the total load.

Since the fabric must follow the moment diagram to some scale, the bulge (or lateral deflection) in the upper half will be approximately 2.5 as compared with 7.5 for the lower half, i.e. 1/3 of the deflection of the lower half.

The magnitude of the deflection depends on how much slack was in the fabric to begin with and how much the fabric stretches to accommodate the tension.

If the pool is rectangular, there will be special effects occurring at the corners and the fabric in the sketch may be stretched in two directions, depending on the detail used at the corners.

As for the precise shape of the bending moment diagram, it is not too important.

Personally, I prefer rigid walls for the pool. If the fabric tears anywhere, the whole area is flooded.

BA

 

[paddingtongreen]

This is interesting. I don't see how bending moment gets into this, the vertical strips will act like a catenary except that the pressure diagram is triangular and the pressure is normal to the slope. There will be a big bulge at the bottom and a small one at the top, but the tension will be uniform for the height except for the effect of the fabric's own weight. The plan shape needs elucidation, as BA said corners produce special effects.

If it is circular, the main stress in the vertical strands may be considerably relieved by the hoop stress in the horizontal strands.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[bigmig]

They make fabric that will not tear. Additional safety liners will be used as well. The bulges will be well below the elastic limit. The efficiency of tensile structures compared to plate structures is not even close. For instance the weight of the fabric for the entire pool will be in the range of 500lbs. If I used cladding, it would be nearly 25 x heavier than this. Add in cost of assembly, portability etc. and tensile fabric is the way to go.

And in response to cd72, every problem can be simplified, as that is really the basis of all engineering in general. Fortunately I am the one who will be figuring this out, so no, I will not be outsourcing this. And no, I am not against others opinions and help, which is why I am here using this wonderful resource.

The problem is 2D, not 3D. The fabric is not warping, it is bowing, meaning it is only supported by the top, mid and bottom members....not side supports. The reason we can do this is because this fabric is the "structure" and a more stretchy, waterproof liner will go OVER this, covering all nooks and crannys.

Finding the furnicular polygon corresponding to the load pattern is easier said than done. You are 100% correct BAretired in saying that stretch and intial pretension will effect the deflected shape. The corners are an altogether different animal that is beyond this post.

The precise shape of the deflection is paramount, as the combination of tensile forces as the membrane goes over the pipe will be a direct function of this. This shape is a function of the stress. There is no bending moment, i.e the nature of a tension structure. Those tensile forces will give me the load on the purlins. In essence I am magnifying the load on the purlins when the fabric wraps them.

The other twist to this problem, like some of you have mentioned is that there is are two load scenarios: loading/unloading, and filled.

 

[BAretired]

paddingtongreen,

I am assuming that the deflections are relatively small so that pressures are fundamentally horizontal. The profile of the fabric when loaded, is identical to the shape of the bending moment diagram, to some scale. This can be calculated as precisely as you wish.

The weight of fabric is neglected.

If the correction for the sloping component of pressure is to be included, that could be an iterative procedure. A closed form solution may be possible, but that would take a bit more thought.

BA

 

[BAretired]

bigmig,

The magnitude of purlin reactions is available from statics. Tension in the fabric is not required in order to calculate them. The upper purlin, for example carries a reaction of W/3 where W is the total horizontal pressure in the upper half, namely 25[γ]/2 per unit of length.

BA

 

[paddingtongreen]

BAretired, I'm assuming that the fabric has no rigidity as shown in bigmig's sketch. If this is true, and if the fabric is tight, there will be large vertical components of the tension at the top and bottom. The more slack the fabric, the smaller the vertical component. With a roller in the middle, the fabric will do the analog of "running through the sheaves" that can happen on transmission and distribution lines on slopes and on long/short spans.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[bigmig]

BAretired,

The freebody on the purlin suggests otherwise. The fabric acts like a block and tackle, magnifying the load on the purlin, which is a direct result of tension and fabric angle.

 

[BAretired]

Neglecting the vertical component of pressure and the weight of the fabric and assuming that the middle support is a frictionless roller, the vertical component of tension V in the fabric is M/d where M is the maximum moment and d is the maximum deflection in a span . V is constant throughout both spans.

So V = M/d or d = M/V

W(upper) = 2.5*5*62.5 = 781#/'
M(upper) = 0.128*781*5 = 500'#/'

W(lower) = 7.5*5*62.5 = 2344#/'
M(lower) = 0.125*2344*5 = 1465'#/'

d(upper) = 500/V and d(lower) = 1465/V

If V = 2000#/' then d(upper) = 0.25' = 3" and d(lower) = 0.732' = 8.8"

The maximum tension in the fabric will be slightly greater than V and will occur where the slope from a vertical plane is maximum.

BA

 

[ishvaaag]

I bought some months ago the Flügge book Stresses in Shells and was a good moment again to feel the lack of a more sound mathematical preparation. Our world is different today, most of us simply seem not have the time for that thing. But it is an spectacle to see those commandeering the thing reduce the problems to elegant solutions of differential equations' problems.

 

[BAretired]

ishvaaag,

I have had a copy of "Stresses in Shells" by Wilhelm Flügge for nearly fifty years. I do not use it often, but there are times when it is extremely valuable. I'm not aware of any chapter which deals with this problem, however. Please correct me if I'm wrong.

BA

 

[paddingtongreen]

BAretired, no problem with your solution until:
"The maximum tension in the fabric will be slightly greater than V and will occur where the slope from a vertical plane is maximum."

The tension is constant by definition, the curvature is due to an increasing horizontal component (horizontal shear) and smaller "V" (the curve is not quite parabolic).

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

paddington,

The tension is not constant. The vertical component of the tension is constant.

Consider a weightless cable suspended between two hinged supports at the same elevation with gravity load P applied at distance 'a' from each support and distance 'b' between loads.

The vertical reaction at each support is P. The horizontal reaction at each support is Pa/h where h is the sag in the cable. The tension in the horizontal portion of length b is Pa/h but the tension in the sloping part is P[√](1 + (a/h)²) which is greater than Pa/h.

BA

 

[chicopee]

Putting it frankly, this is not a smart design; you need to support the membrane throughout its depth and circunference of the pool. Even a small crack will bulge the membrane unless it is several inch thick.

 

[BAretired]

chicopee,

You can put it as frankly as you wish...it may or may not not be a smart design, but your comment that "you need to support the membrane throughout the depth and circumference of the pool" is wrong.

Your final sentence is a mystery. The membrane will bulge as a result of pressure. It is not related to cracking or thickness.

BA

 

[chicopee]

The membrane appears thin in his drawing and will buldge between gaps if uprights are spaced around the pool. He needs a continuous wall or support whether of metal, wood, sand etc... Plus the fact that when kids use the pool this membrane will fail quickly.

 

[BAretired]

chicopee,
Membranes are usually thin. There are no uprights spaced around the pool. The proposed supports are shown on the OP's sketch. A continuous wall is not required to satisfy statics. Why will the membrane fail quickly when kids use the pool? What happens when adults use the pool? What exactly are you saying?

BA