Internal Forces Calculations - Steel Bundles
Internal Forces Calculations - Steel Bundles
(OP)
Hello,
I am having a difficult time to try and figure out this problem for work.
I have steel piping bundled together in a hexagonal pattern that is strapped together [band-it strapping]. There have been instances of excess amounts of weight when these bundles are being stacked upon one another sometimes causing the bottom bundle's strapping to break and then stacks falling [and making a mess].
I am looking for some help as to try and figure out the forces acting inside the bundle to get a representation to see where the most outward force is located [and value/direction] to note where the bands are breaking and at what force the bands can withstand.
I welcome all help. I have attached a file to give a illustration.
Thanks,
D
I am having a difficult time to try and figure out this problem for work.
I have steel piping bundled together in a hexagonal pattern that is strapped together [band-it strapping]. There have been instances of excess amounts of weight when these bundles are being stacked upon one another sometimes causing the bottom bundle's strapping to break and then stacks falling [and making a mess].
I am looking for some help as to try and figure out the forces acting inside the bundle to get a representation to see where the most outward force is located [and value/direction] to note where the bands are breaking and at what force the bands can withstand.
I welcome all help. I have attached a file to give a illustration.
Thanks,
D






RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
This could be circumvented by placing blocking between the straps and the second row of pipes.
BA
RE: Internal Forces Calculations - Steel Bundles
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??
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
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
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
Here is the Min. Breaking Strength chart from Band-It [http://www.uespromura.com.au/attachments/BandIt_Br...] 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.
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
The greatest trick that bond stress ever pulled was convincing the world it didn't exist.
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
http://www.westeve.com/saf_pipe.html
RE: Internal Forces Calculations - Steel Bundles
https://www.youtube.com/watch?v=66XdlM-REw8
BA
RE: Internal Forces Calculations - Steel Bundles
https://www.youtube.com/watch?v=wL2zlfcuUTk
BA
RE: Internal Forces Calculations - Steel Bundles
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
RE: Internal Forces Calculations - Steel Bundles
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.
RE: Internal Forces Calculations - Steel Bundles
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
RE: Internal Forces Calculations - Steel Bundles
The greatest trick that bond stress ever pulled was convincing the world it didn't exist.
RE: Internal Forces Calculations - Steel Bundles
BA
RE: Internal Forces Calculations - Steel Bundles
I did it in a different way individualizing pipes, I will upload my calcs next week. My output was the same as BAretired's 1.155F.
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
When additional load is placed above by placing more bundles on top of the first one, the eccentricity of balls 13 and 16 increases due to increased compression between Pipes 8-13-17.
The simplest remedy is to put a shim between the strap and Pipe 13 and also between the strap and Pipe 16 such that the harp in the strap is sufficient to hold Pipe #13 and #16 in position for all loading conditions.
Brad805: Your analysis requires tension where no tension is available and cannot be correct (see above).
BA
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
BA
RE: Internal Forces Calculations - Steel Bundles
From this:
-----
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!
----
All are different values through different methods! KootK if your method is correct can you please be more detailed? Your drawing is confusing me, sorry.
This is a, in a sense, simplified value as there are other factors being neglected, as BAretired has pointed out [ie. pipe deflection internally, stretching of the strap itself, crimping of the strap, etc.]. Though even this will give a rule of thumb that a factor of safety can be applied to for strapping and stacking purposes.
RE: Internal Forces Calculations - Steel Bundles
The strap spans from the tangent point of Pipe #8 to the tangent point of Pipe #17, a span of 2D where D is the diameter of the pipe. The moment of a simple beam of span L and concentrated load P at midpoint is M = PL/4 or in this case, M = (W/2)2D/4 = WD/4.
The tension T in the strap required to provide a reaction of W/2 at midspan is T = M/t = WD/4t where t is the offset from a straight line or, in this case the thickness of the required shim. If t = 0, the strap tension is infinite but after the strap begins to strain, t increases above 0.
If we know the initial strap tension, we can calculate offset t required to hold Pipe #13 in position. If Tinit = 30W then t = WD/(4*30W) = D/120. For D = 12", t = 0.10", say a 1/8" shim.
If the initial prestress is only 10W, then treq'd = D/40 = 0.30".
BA
RE: Internal Forces Calculations - Steel Bundles
I no longer have faith in my solution dmaier. Moreover, I'm driving myself quite mad trying to sort it out. I've been checking our various solutions for equilibrium at the top and bottom joints. Nothing checks out.
The greatest trick that bond stress ever pulled was convincing the world it didn't exist.
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
As for pretension I do not know that off hand, I do know that the company is using Samuel Strapping Systems for their banding at this time [http://www.samuelstrapping.com/en/product/product.... -- 1-1/4" bands, most likely the 5500lb strength]. So saying this with 5500lb strength, not taking crimping or other factors into play [and for simplicity load P = T], that a bundle that weighs 5000lbs and has say 4 straps will yield a 22,000lbs breaking strength. Which in turn would say that you could stack these 4 high [22,000/5,000 = 4.4], with the bottom bundle being included in the 4.
I do get what you are saying with the normal force from the band being applied to the pipe [13], but is this still not the area of highest force contribution which would be at pipe [8], due to symmetry? As well in your calculations would t not just be the thickness of the strap not the shim?? And the 30W [I assume that is just an arbitrary pre-stress value] would that not have to take into account all other weights of pipe above/acting on it? -- Sorry, but going through this has constantly brought up doubts and frustrations to try and solve.
Ussuri can you elaborate more on your statement? Are you saying that the pretension that the straps provide is in equilibrium with the total bundle load? Which would be to say that the bundle weight would be completely handled by the strapping, and then additional straps would provide more allowable force to be applied to the bundle to withstand [makes sense]?
RE: Internal Forces Calculations - Steel Bundles
I tackled things a little differently this time. I considered equilibrium in the stack immediately after elongation of the straps takes place. When the stack "settles" slightly, it results in horizontal gaps between each pipe and its neighbours to the left and right.
In my opinion, those gaps preclude the use of the centre diamond load path detailed in BAretired's solution. At the end of the day, my solution is BA's minus the middle diamond. Most importantly, I've checked force equilibrium at all six corners for my double diamond solution and everything balances out.
In the absence of friction on the strap, which has been our assumption thus far, the tension in the strap will be uniform around the entire bundle. The strap tension will not be at a maximum at pipe #8/#12. As a consequence, any extra strap tension required to restrain pipes #4/#7/#13/#16 will be in addition the the strap tension values that we've been kicking around for the simplified case.
Significant pre-stress would mess with my solution if it meant that the horizontal gaps that I've described never come to pass. Even so, this seems like a reasonable upper bound on the strap tension force. I'm sure that the pre-stressing is imperfect an subject to losses over time.
The greatest trick that bond stress ever pulled was convincing the world it didn't exist.
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
BA
RE: Internal Forces Calculations - Steel Bundles
RE: Internal Forces Calculations - Steel Bundles
BA
RE: Internal Forces Calculations - Steel Bundles
- If the straps are pre-stressed, is the initial stress uniform all around or is it variable due to friction at each of the six corners of the bundle?
Manually strapped, I do not know if all uniform [human error], I would like to assume that it is uniform for ease. They are strapped tightly together and the pipes are snug to one another.
- Does the pre-stress force add to the force determined from statics or is it included? For example, when a bolt is pre-stressed axially, then separately loaded in tension, the tension in the bolt remains constant until the applied load exceeds the pre-stress force.
I do not know the answer to this, the bands supply sufficient force to hold the bundle itself together and maintain its [hexagonal] orientation. I would then say yes, the tension in the strapping remains constant until there is bundles stacked on top, and then and a point when n# stacks exceeds this tension.
This is where the calculations could vary as if there is a large pre-stress then there is a larger magnitude force acting normal to the pipes keeping them snug together, which would have greater strain already in the strapping [correct?]. However I do not have what they manually strap the bands to at their mill.
- If the bundles are stacked to a height of four or five bundles, what provisions are there to ensure all bundles are properly aligned. i.e. there is no eccentricity from bundle to bundle?
Bundles are handled in an automated setting, therefore assume that bundles are aligned and forces act purely vertical -- bundles are placed directly on top of one another
- What diameter of pipe are we talking about?
in this case it is a 3.5" diameter pipe, different bundle sizes can vary from <1"-12" but have different orientations
- How are the bundles handled when stacking them? Two slings per bundle?
bundles are handled using a magnet, no slings
Does this give you more insight, and help with your questions?
RE: Internal Forces Calculations - Steel Bundles
It helps, but I have to think about it a bit more.
BA
RE: Internal Forces Calculations - Steel Bundles
If the 5500#(average breaking strength)strap is used, it is 1.25" wide x 0.031" thick for an area of 0.03875 in2. According to my earlier analysis, the tension for a four bundle stack should be 0.385P or about 1925#. (2890# using KootK's revised analysis).
I doubt that Samuel Strapping Systems prestresses the straps to a significant degree, but this can be confirmed with them. For the present purposes, I will assume a prestress of 0.
Tensile Stress = ft = 1925/(4 * 0.03875) = 12,420 psi
Length of strap = 12D where D is the diameter of pipe.
Average strain = ft/E = 12,420/29e6 = 0.000428
Total strain = 0.000428*12D = 0.018" for a bundle with 3.5" OD pipe.
The total strain is small enough to be safely ignored. (Should confirm E with the supplier).
Depending on the safety factor you want, you could probably use smaller straps. Might be a good idea to get the properties of the steel used for straps, particularly the yield point.
BA
RE: Internal Forces Calculations - Steel Bundles
If the 5500#(average breaking strength)strap is used, it is 1.25" wide x 0.031" thick for an area of 0.03875 in2. According to my earlier analysis, the tension for a four bundle stack should be 0.385P or about
1925#7700#. (2890#11,550# using KootK's revised analysis).I doubt that Samuel Strapping Systems prestresses the straps to a significant degree, but this can be confirmed with them. For the present purposes, I will assume a prestress of 0.
Tensile Stress = ft = 7700/(4*0.03875) = 49,700 psi
Length of strap = 12D where D is the diameter of pipe.
Average strain = ft/E = 49,700/29e6 = 0.00171
Total strain = 0.00171*12D = 0.0719" for a bundle with 3.5" OD pipe.
The total strain is small enough to be safely ignored. (Should confirm E with the supplier).
Breaking strength/Tmax = 22000/7700 = 2.85 (not too bad, but only 1.90 using KootK's value).
Might be a good idea to get the properties of the steel used for straps, particularly the yield point.
BA