Internal Forces Calculations - Steel Bundles

Cool problem. Here's my quick and dirty guess:

1)Turn your hexagon into an equivalent circle.
2)Find the weight of the entire stack and convert that into an equivalent fluid pressure at the center of the lowermost circle
3) Calculate hoop stress on your bundle straps.
4) Multiply by two and design accordingly.

I would think this would be conservative in that it ignores the friction between pipes.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Here is my attempt to solve it:
The minimum tension in the strap needed to prevent the pipes from sliding on a 30 degree angle for one bundle is (19pipes-1) x W(weight of pipe)xCOS30. (19-1) because one pipe is in equilibrium state.
Now if you have 3 bundles above, the bottom bundle needs to have a strap capable of resisting tension equal to: [3x19xW+(19-1)xW]xCOS30.

The problem cannot be solved if the straps can take only tension and the pipes are frictionless. There must be deflection at the second tier in order to hold the outer two pipes in position. Alternatively, the straps must be capable of resisting bending moment.

This could be circumvented by placing blocking between the straps and the second row of pipes.



BA

@Robbiee I don't quite understand that...

As well could you just break it into quarters and just find individual forces on each individual pipe?

How would it be possible to make a spreadsheet of this and just type in say the bundle orientation [ie # of pipes] and it would give out a force answer?? WOuld this be more FEA/FEM??

Someone with Solidworks premium might be able to solve this and prepare a nice illustration for your workers making the stack. Considering the importance of this with respect to safety, this task seems quite important.

I suspect the amount of prestress in the straps may play a role in this problem as well. Unless this bundle is being held rigidly while the strapping process is complete it would not surprise me if it takes quite a lot of force. This problem has some similarities to pre-stressing a concrete tank; however the rolling ability of the pipes makes this a far more complex problem to solve.

dmaier,

As I mentioned earlier, there is no solution to the problem unless you block between the straps and the second tier of pipes in each bundle.

The red arrows in the attached sketch are required forces in addition to strap tension necessary to maintain equilibrium of the bundle. The straps cannot provide those reactive forces without excessive strain.

You need to block between the strap and the pipe in order to develop the reactive force.

BA

You need to talk with the strap manufacturer to determine what tension load is required to fail the straps.

An estimate can be gotten from the relation Stress = Load/Area. You can measure the section area of the strap and look at the material properties for the Stress. This leaves the load, which would ideally be uniformly distributed so there is no particular location the band will fail. If the pipes are trapped by friction, there could be non-uniform load distribution that would shield sections high loads, but a failure anywhere causes the bundle to fail.

A test would be the fastest method to determine the strength and allow a visual about factors that influence the failure that are not part of the problem description.

I am trying to figure out the max force at which can be applied to the [bottom] bundle [ie the number of bundles capable of being stacked on top] for which is safe and that the bands/strapping can hold safely.

Here is the Min. Breaking Strength chart from Band-It [] which I believe is what is being used --> 3/4 x .030 2250lbf/10,000N; with a max bundle weight of 10,000lbs = min. 5 bands [for weight alone]. If this is not accurate band being used, it is not completely crucial to the first steps as this value can be inputted later for final values.

In my head and quick fbd's the most logical place for the bands to fail would be the most outer pipes in the middle row. When more bundles are stacked on top will force these pipes outward [horizontally] causing the most force on the strapping to resist.

Is this not a correct statement? For the time I am assuming that the force is transmitted fully through the pipes and no deflection is found in the individual pipes for simplicity. The pipes are being held together [in their orientation] by the strapping.

I change my answer. I think that this is simpler than I originally thought. Please refer to the attached sketch. I believe that the answer to the OP's original question is that the strap force is half the total weight of the system.

For real world design, I'd recommend the following, similar to the other posters here:

1) Size the straps not just for strength but for stiffness. And be conservative with that. You want to make sure that everything stays snug. Prestress sounds like a great idea if that's an option.

2) BA's point is absolutely correct. However, it will be a secondary effect. You'll really just be dealing with the weight of the individual pipes that might pop out the sides in the absence of friction. I'd address that by imagining the strap stretched out locally to provide the restoring force required. I expect that will result in a minor amplification of the strap force.

3) Consider the radius of your strap bends in your strength checks for the strap. It will be tension stress plus a bit of bending stress when it's all said and done.

4) For the love of all that is wholly, check my method and my math! I am a random internet stranger after all.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Numerically, my revised solution produces an identical result to the hoop stress method that I proposed initially. I did't see that coming but I do find it encouraging.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Correction to point #2 above.

BA's point is not as secondary as I thought. There's a primary load path from the point of load application out to the side corners. When slightly displaced, that will tend to squeeze the offending pipes out of the bundle. The force required to keep the pipes in place should still be small though. I'd analyze that in an initial imperfection kind of way, similar to column design.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

The reference below appears to be similar to your case. They reference a study, but it is not included on their site. Maybe you can find it. Regardless, they did develop safety standards for various heights and bundle sizes that may be of use.

This is an interesting video on the subject of bundling:


BA

Personally, I prefer this type of stacking:


BA

Back to the problem at hand: See page 1 and 2 of the attached link.

Neglecting the weight of the lowest bundle, the superimposed load P is assumed to be carried equally by three diamond shaped frames shown in green, red and blue on p.1. Each frame is hinged at each joint and each frame carries a load F (or P/3). From statics, the load on each frame is shown on p.2.

Each frame requires a horizontal reaction of H = F.tan 30. The strap must provide 2H to the system which results in a strap tension of 2H.

Since the weight of the bottom bundle was not included, it may be sufficiently accurate to include its weight in the applied load.

Thus, for four bundles stacked together as shown on p.1, the applied load P would be 4*19*W and the tension would be 0.385P or about 29.3W where W is the unit weight of one pipe times the strap spacing.


BA

Here is an explanation to the solution I presented earlier.
The tension in the strap that needs to keep the pipes in equilibrium at this shape is equal to the reaction forces that will prevent the pipes from moving down under gravity on 30 degree planes. One pipe ( shaded) will not move.
Adding weights of stacks above will need to go through the same load path.

So, for a stack of four bundles we have:

P = 19*4W = 76W
where W = unit weight of one pipe times the strap spacing

T = 0.5P = 38W (KootK)

T = (3*19W + 18W)cos 30 = 64.95W (Robbiee)

T = 0.385P = 29.26W (BAretired)

Interesting!

BA

Thanks for putting this together BA -- was curious. Mine has to be right by virtue of number roundness, right?

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Right, KootK.

BA