Raising a 70' Tower: Calculate Cable Force

[aknisley]

This seems simple but I am having trouble with this. I have 70' self supporting tower that I want to be able to raise and lower with a winch. The tower is hinged at the base.
I want to use rectangular steel tubing 7' high with a pulley at the top.
See PDF attachment. I need to know the force applied to the cable in order to get the correct size tubing.


View attachment lift force Model (1).pdf

 

[SteamKing]

That's a tricky way to lift such a long tower. The structure of the tower should be checked to see if it can withstand not only the loads imposed by the lifting cable, but also bending due to the self-weight of the tower as it is being raised. Towers may be self-supporting in the vertical position, but raising them from the horizontal is a different story.

 

[aknisley]

This is a heavy duty all aluminum and is designed to be raise by hand walking it up. I don't have any help so I need to power it down and up. If it doesn't work i'll have scrap aluminum to sell.

 

[berkeman]

Do you have any pics? Are there any nearby tall structures (buildings, trees) that you could use to raise the winch/pulley point?

 

[aknisley]

No on the trees. Just need to the force applied to the cable.

 

[berkeman]

Why is the cable tie point so low? It would be better if you could tie it more like 2/3 of the way up the tower. Or connect multiple cables spaced along the length of the tower, and vary the speed of the multiple winches to work in concert during the lift.

I've helped lift up some pretty heavy and tall HAM radio towers, and it's darned dangerous...

 

[Baluncore]

Tapered or parallel tower, triangular or square section.
Lump the 400 pounds at the centre of the 70 foot tower.
That makes 400 pounds at 35 foot. I assume the base is 7 feet.
400 * (35 / 7) = 2000 pounds, or about one tonne.
With a pulley on the tower the cable tension will be halved to 1000 pounds.

That all assumes a very slow smooth movement. Any jerky movement will significantly increase the winch and wire specification required. I would specify at least a two tonne winch, cable and four tonne pulley. Here are a few obvious points.

Firstly, as it is self supporting it will not bend like a mast or pole while being raised, which probably eliminates folding the mast, which is the most difficult situation normally encountered during a tilt up or down.

Secondly, as it is self supporting, it is able to survive higher side forces than gravity because the wind drag on a tower or wire is very often greater than gravity. Look along a catenary wire during a strong steady cross wind and you will see the wire hanging much closer to horizontal than vertical.

Thirdly, side forces on the tower legs during erection with a winch or long jack, can easily exceed the forces encountered while standing, as designed self supported. You should consider building a strong hinged base for the tower. That base would be attached to the lattice points closest to the base of the tower and to the tie point for the winch. The forces during erection are handled by that base assembly. It should be designed to remain once the tower is erected. I prefer to add redundant diagonal braces to the lowest module of a tower to handle the side forces expected during erection.

Take care.

 

[aknisley]

Finely an answer to my question. Very good. Now I can proceed. The tower is up now so lowering it will be next.

 

[Baluncore]

There are a couple of factors hidden in my estimate.

If a triangular tower, then base has hinges on two feet and pulley on third. Base dimension is then only Sqrt(3/4) = 0.8660 of the distance between the tower feet. That increases the cable tension by Sqrt(4/3) = 1.1547

At the start of the lift, the 45° diagonal angle of the wire will require a Sqrt(2) = 1.4142 increase in cable tension to lift the tower.

So the effect of those two cable tension requirements comes out at 1.633 which is allowed for in my original four tonne estimate.One question you have not asked is how much concrete you need under each foot of the tower to withstand a strong wind. A quick estimate is that it will need to be similar to the strength of the pulley block you need to use. That is four tonnes, or 1.6 cubic metres per leg for a triangular tower, maybe slightly less for a square base tower, say 1.2 cubic metres.

Footings are always a special challenge because your maximum wind speed, soil/rock type, tower section and antenna mass/windage all have an effect on design.

This has not been an engineering analysis so much as a quick estimate of the magnitude of the problem. You neglected any antenna load in your question.

 

[aknisley]

Because you ask This is what the tower is and has been up for two years.
View attachment UNIVERSAL 70' TOWER.pdf

http://wb0sbu.com

 

[AlephZero]

One thing that Baluncore didn't mention: at the start of the lift, the force on the base of the tower is large (about the same as the cable tension, not the weight of the tower!), and mainly upwards not downwards. In an extreme situation, you might pull the concrete footing loose from the ground - but you might not discover that fact for a long time afterwards when the tower collapses because the concrete had cracked from the bottom up, where you couldn't see it...

 

[aknisley]

The base is 6' deep and 5' square. The anchoring legs go to the bottom.

 

[AlephZero]

The base is 6' deep and 5' square. The anchoring legs go to the bottom.

I'm not sure that is very relevant. If the legs bend and/or break through over-stress at the point where they meet the base, if doesn't matter if the base is 5 feet or 50 feet deep.

 

[Baluncore]

During raise and lower operations you will need an independent anchor that weighs about 5 tonne.

The triangular tower base is only 24.25”, but your gin pole is 7 foot. I would use an A frame gin pole.
The manufacturer should recommend an erection procedure.

See; http://en.wikipedia.org/wiki/Jin-pole
Also Google; Gin Pole

I would have expected three guy wires on a tower like that.
No lightning conductor? Will it crack the concrete at one leg then blow over in the wind?

 

[256bits]

A segmented antenna ...
Use a crane, helicoptor, or the gin pole technique to remove each segment.

With the technique of lowering as a unit, the lateral stress at the cable attachement and the swage joints will pull the thing apart.

One other method is to attach a stiff enough vertical beam to the unit, securely at each segment, and base of the beam . Detach the unit from the concrete base so the beam takes all the load. Do the pulling on the beam.

 

[nvn]

I need to know the force applied to the cable ...

aknisley: The theoretical, static tensile force applied to your cable shown in post 1 would be Ts = 12 580 N. But that assumes theoretical, very slow, perfectly smooth movement. No matter how carefully, and slowly, you move the system, realistically your cable tensile force will fluctuate. Therefore, the actual cable tensile force probably will be closer to T = 15 700 N. Therefore, use T, not Ts.

How is your vertical rectangular steel tube, shown on the right-hand side of your diagram in post 1, anchored? The cross-sectional size required for this tube would depend on how, and where, you anchor (or support) this tube.

 

[256bits]

aknisley: The tensile force applied to your cable shown in post 1 would be T = 12 580 N.

How is your vertical rectangular steel tube, shown on the right-hand side of your diagram in post 1, anchored? The cross-sectional size required for this tube would depend on how, and where, you anchor (or support) this tube.

T=12,580 N = 5800 pounds.
Does the antenna have the structural integrity at one spot to support that load?

 

[256bits]

Manufacturer erection procedure:
http://www.universaltowers.com/I-T.pdf

What I am getting at is that by having the cable attached to one specific spot on the tower, with lowering as the tower becomes more vertical, the antenna may not be able to withstand the cable stress at one point.

With the walk up/down method, in horizontal position, the person at the end(top section ) and the base each carry only a vertical load of 1/2 the tower weight, and no horizontal load on the tower.

 

[AlephZero]

Manufacturer erection procedure:

Yep, very clever: so how you balance the tower on two bolts without holding it, while you fix the other three bolts?

More seriously, that''s a good way to raise the tower (if something goes wrong, you can back out and try again), but lowering it could be a bit more hair-raising, when the bolts etc are nicely corroded after a few years outside in the rain!

 

[nvn]

I need to know the force applied to the cable ...

aknisley: Upon closer inspection, the tower CG is located at 40 % of the tower height, instead of 50 %. And if you move the tower very slowly, carefully, and very smoothly, then multiply the tower weight by a dynamic amplification factor of 1.25. Therefore, working this out gives an actual cable tensile force of T = 12 580 N (2830 lbf), instead of what I posted in post 16. Therefore, use T = 12 580 N.

And see my last paragraph in post 16.

Does the [tower] have the structural integrity at one spot to support that load?

That is the important question. aknisley would need to post a detail drawing of how he plans to connect the cable to the tower, with dimensions. Then we could obtain the tower member forces using, e.g., the finite element method. Then we could see if the tower would collapse.

 

[Baluncore]

Does the tower have the structural integrity at one spot to support that load? Yes. When you walk up a tower you are supporting it at one point, the point of the walker or trestle. If it did not have sufficient strength, it would fail in a strong wind. Since the tower comes in sections, the sensible attachment point is just above the lower section joints.

When raising or lowering narrow base towers you should use a temporary guy rope on each side of the tower with their anchors on the base hinge line. That should prevent an uncontrolled side swing that can bend the hinge attachment fittings as the tower approaches the ground.

One method used to remove and install masts in boats is to use “shear legs” that straddle the mast, with a block and tackle between the mast and top of the legs. The legs stand away from the mast where the mast will be laid. The advantage of this system is that you do not need a ground anchor and that the shear legs can employ available spars from the mast. The legs can stand in shallow holes in the ground or be attached to the base of the tower with ropes to prevent them sliding at the start of the operation.

 

[AlephZero]

Does the tower have the structural integrity at one spot to support that load? Yes. When you walk up a tower you are supporting it at one point, the point of the walker or trestle.

Whem you walk up the tower, you are not supporting it at the same point all the time. That makes a huge difference to the loads on the base.

If it did not have sufficient strength, it would fail in a strong wind.

The shear force and bending moment diagram for the wind load looks very different from the loads when erecting the tower. And it's hard to believe the wind loads on a 70 foot tower that is mostly made of holes, would be anything like as big as one ton. If they were, the tower would need some supporting guy ropes (attached nearthe top, not 7 feet up from the base)

 

[Baluncore]

And it's hard to believe the wind loads on a 70 foot tower that is mostly made of holes, would be anything like as big as one ton.

It is an optical illusion, next time you are close to a tower that appears to be mostly holes, look at it's shadow and you will see just how much material there is to create drag. A tower is a three dimensional structure. It is very easy to mistakenly neglect the back faces in a quick guess or a first analysis.

I think you need more experience calculating wind loads, or maintaining an antenna farm. Then you will appreciate the importance of air density and wind v².

Such a tower here would need guy wires, but then it is not unusual here to get 120 km/hour winds. You cannot afford to drop a lattice tower. When a lattice tower falls in a wind the energy must go somewhere, the impulses bounce up and down the fallen tower until momentary standing waves exceed material yield point to absorb the energy, or the energy is stored in the elastic material near bent members. The short lattice components bend in unexpected ways, which curve the legs. It is impossible to repair the tower without cutting every weld to release the stress, then straightening every bent member independently. A fallen lattice tower is scrap metal.

 

[nvn]

If the legs bend and/or break through over-stress at the point where they meet the base, it doesn't matter if the base is 5 feet or 50 feet deep.

Good point, and someone should check it, so I did, including the dynamic amplification factor. None of the stress levels on the concrete anchor components (steel clevis plates, bolts, tubes, nor concrete) exceeded 20 % of the allowable stress, except for the bending stress on the 30-STRT tube leg where it meets the first lattice bar, which was 62.8 %, or perhaps 95.9 % at the beginning of the swaged portion (?). Therefore, it currently appears these reaction forces at the concrete anchors would not be detrimental to the concrete anchors, concrete, nor the connections to the concrete anchors. But if you want to decrease the bending stress on the tube swaged portion, you perhaps could attach the cable 500 or 1000 mm higher up on the tower. And, we probably should perform a more careful stress analysis of the bending stress on the tube swaged portion, to ensure it does not exceed 100 % of the allowable stress, since I did not spend much time on it.

 

[AlephZero]

I think you need more experience calculating wind loads, or maintaining an antenna farm. Then you will appreciate the importance of air density and wind v².

Hm... I used a wind calculator on the web, and the wind loading area given on the drawing (21 sq ft). For an 80 mph wind and the worst environmental conditions and height, I got a wind load of about 350 pounds.

That makes more sense to me than applying a "1 ton" force (or more) from the OP's lifting cable, to some weedy looking fixing bolts through some hollow aluninum sections.

 

[Baluncore]

I find it hard to believe the manufacturers data sheet when it shows an 11 foot tall giant standing at the base.

 

[nvn]

One of the highest stress levels at the concrete anchor connections appears to be bending stress on the beginning of the 30-STRT tube leg swaged portion, mentioned in post 24. I checked it more carefully now, and it currently came out to be 100.4 % of the allowable stress, which does not (really) exceed 100 %, and is therefore OK. This uses a yield factor of safety of FSy = 1.50, a dynamic amplification factor of 1.25, plus it appears we have been using an overestimate of the tower weight, which is not close to the tower specification document. Therefore, this 100 % stress level currently appears OK.

I did not check the first lattice bar weld adjacent to the 30-STRT tube leg swaged portion, which could have a high stress level.

aknisley: Did you overestimate the tower plus antenna weight you gave us in post 1?

 

[aknisley]

Yes I did. The tower weights 322# and the guess of 400# would cover the equipment on it.

Just added more pictures on my webpage wb0sbu.com

 

[nvn]

aknisley: What is the OD and wall thickness of the pole you mounted at the top of your tower? And is this pole aluminum? Or steel? Also, what is the weight of the antenna, not including this pole?

 

[aknisley]

2" OD aluminum I think it is .125". Don't know the weight of the antenna. That's why I guessed at 400 pound for tower and other equipment attached to it.

 

[Baluncore]

I really do not like using one 7' pole as the lever.

Without a crane, I would prefer to use two shear legs with a block and tackle, (B&T).
The shear legs could be 20' long steel pipe. The tower attachment would be at the 10' point where the lowest two sections meet. The shear legs would be seated 10' down range from the hinge line with the crotch about 6' up range of the tower. Raise the crotch of the shear legs to the tower using the B&T. Remove the remaining bolt from the free tower leg. Push the tower to tension the B&T. Then lower the tower by paying out the B&T. As the tower comes down the shear legs rise until they are vertical when the tower becomes horizontal. Fall the shear legs back to their starting position using the B&T or guy ropes. Raising the tower is exactly the opposite process.

The feet of the shear legs would be tied to the tower footing and to each other to prevent slip.

 

[aknisley]

Deign

This is what I have been looking to build.

This tower while only 23' weights around 620 pounds. View attachment TowerRaisingFixture.pdf

 

[Baluncore]

Where a fixed structure is being used to tilt a tower it is probably better to lift the hinge point as high as possible on the fixed structure. That often permits some counterbalance to be used in the tower base.

In many cases there can be little force needed to tilt the tower. That is especially true of wind-up telescopic towers that are only tilted for antenna access when at their shortest.

 

[aknisley]

I really do not like using one 7' pole as the lever.

Without a crane, I would prefer to use two shear legs with a block and tackle, (B&T).
The shear legs could be 20' long steel pipe. The tower attachment would be at the 10' point where the lowest two sections meet. The shear legs would be seated 10' down range from the hinge line with the crotch about 6' up range of the tower. Raise the crotch of the shear legs to the tower using the B&T. Remove the remaining bolt from the free tower leg. Push the tower to tension the B&T. Then lower the tower by paying out the B&T. As the tower comes down the shear legs rise until they are vertical when the tower becomes horizontal. Fall the shear legs back to their starting position using the B&T or guy ropes. Raising the tower is exactly the opposite process.

The feet of the shear legs would be tied to the tower footing and to each other to prevent slip.

I am starting to get what you are saying. I pretty dense so could you do a sketch? What size pipe?

 

[Baluncore]

Attached is a sketch showing shear legs and tower side elevation and plan.

The tower attachment would be at the 10' point where the lowest two sections meet. The shear legs would be seated 10' down range from the hinge line with the crotch about 6' up range of the tower.
The footings of the shear legs are about 16' apart.

The shear legs start in position 1.
They are then raised with the B&T to position 2.
The pin in the free leg of the tower is then removed.
The tower is then pulled gently down range, as it passes the balance point, weight is on B&T.
Then lower the tower by paying out the B&T.
As the tower descends, the shear legs rise until they are standing vertical in position 3.
The tower is then horizontal.
Fall the shear legs back to their starting position using the B&T or guy ropes.

Raising the tower is the opposite process.


Specification of the shear leg dimensions is a job for a real engineer.
Since they are quite long, the shear legs need to be designed with reference to column stability.
My first guess is that they would be a minimum of 4” diameter steel pipe with a 1/4” wall.
Alternatively, timber 6” to 8” diameter poles, maybe Douglas Fir to keep the weight down.
I use 6" diam x 1/4" wall Al tube. It comes with new felt spools for the paper industry. Scrap pipe or tube is also low cost. Once used you can keep it or sell it back into scrap.

The crotch where the shear legs meet and the block is attached needs to be designed to suit the shear leg material. A 30' rope attached to the crotch makes it easier to control the legs at the end of the lift, or to raise them into position prior to lifting the tower.

This system is applicable to raising or lowering a tower once or twice. For a more permanent solution a different structure should be used, probably with the hinge pin for the tower raised about 10' above the ground, at the top of the lowest section.