Force on Guy Lines Using Falling Derrick - Spreadsheet to calculate degree by degree?

[bgolitz]

Hello, I want to raise a 90 foot long pole that is 5" in diameter using a falling derrick. I am interested in the force applied to the guys as it is raised up from the horizontal position to a vertical position say degree by degree. I would like to accomplish this in a spreadsheet. The material is 6063-T6. The 5" tube with a .052 wall weighs .951# per foot. And 4" with .050 wall weighs .73# per foot. I have attached a simple drawing. Any assistance would be greatly appreciated. Thanks.

 

[IDS]

The longest guy line is 90 ft and the pole is 90 ft, so the guy can't be connected to the end of the pole as shown.
What is the diameter of the guys?
What is the distance between the bottom of the pole and the base of the derrick?

I don't think this can be done in a simple spreadsheet (since the structure is redundant), but I have a frame analysis spreadsheet which should deal with it fairly easily.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[bgolitz]

Thanks for your reply. I should have been more clear. The guys aren't 30, 60 & 90' in length, but at the 30, 60 & 90' points off of the pole. So yes, the longest guy will be longer than 90' in length. The distance between the pole and Derrick is 12". Thanks again.

 

[dhengr]

Bgolitz:
My first reaction is that the two rotation axis, one on the pole and one on the derrick pole should be in the same line, a common axis, or the relative lengths of the cables will change during the lifting rotations. And, it doesn’t take much change in length (or strain) to significantly change the cable forces. You do the layout and check this out. Why don’t you explain a little bit, what a “Falling Derrick” is, and how it works. You should guy the 90' pole back to the ground with at least two lines during the lifting operation. Otherwise, it might just keep going counterclockwise when it gets near (or past) vertical. If you treat the horiz. 90' pole as a three span continuous beam with equal spans, what are the 4 vert. reactions? In what position of the 90' pole, are the guy wire forces max. and what are they? Who cares what they guy wire forces are at other angular positions of the 90' pole? What are the bending moments in the horiz. pole, and what are the axial loads in the pole? Is the pole strong enough to support these? Is this going to be a flag pole? Can the derrick pole carry its axial loads throughout the lift?

 

[berkshire]

bgolitz,
When you talk of a falling derrick, are you referring to a " Gin pole " ? A method often used for raising masts on sailing vessels.
If it is then the derrick will pivot to the ground as the pole is raised. this would mean that you would have to have a deadman of sufficient strength buried in the ground to take this load. Also do you have intermediate guy wires ( stays ) set up for this, to hold the pole in place as it gets near the top?
B.E.

You are judged not by what you know, but by what you can do.

 

[bgolitz]

This is part of a radio tower installation. I have always called it a falling derrick, but may also be referred to as a pivoting gin pole. Here is an old video using this same technique that will hopefully explain how it works.

http://vimeo.com/7973643

To answer your questions, yes, the pole and derrick pole are both in the same line or common axis. Yes, the derrick pole can carry its axial loads throughout the lift. The pole is a 5" dia. 6063-T6 tube with a .052 wall. The derrick pole is a 4" tube with .050 wall. There will be several guys back to the ground to prevent past vertical of the pole once set up. The guy wire max forces and forces at the base are what I am most concerned about. Attached are a couple pics of the base and guys.

Thanks.

 

[berkshire]

bgolitz (Mechanical)
Are you going to use a mast gate like your video.
At the weights you have given, your 90 foot pole is only going to weigh 85 or 86 pounds, and the length of your gin pole will give you a two a 2-1 disadvantage.
So at the start of the lift you should be seeing about 240 pounds on the block and tackle, without any mechanical advantage. That load only gets less as the pole goes to vertical.
so what are the concerns?

You are judged not by what you know, but by what you can do.

 

[bgolitz]

Great, thanks for your help. Do you have an equation you used? I'd like to be able to create a spreadsheet where I could then change any variable, for future use. Thanks again.

 

[berkshire]

I did not use an equation, I just took the dead weight of the mast laying on the ground and its length. I took the length of your gin pole and calculated the length difference to get a ratio. I then applied this to the weight of the pole which was a 2 - 1 ratio so your starting load was w1 the weight of the pole
times 2 to get the mechanical disadvantage at the end of the gin pole w1x2 = w2 . Then your dead man would have to be 45 feet away from the base of the pole, so your block and tackle would have to span 63 feet to make the connection and your load will increase by 1.414. Your rope load at the end of the pole will also increase by a factor of 2 however, because you are sharing the load with the other two guys this is not the total load. Also your rope or guy lengths need to change , To the top of the pole 100'-6" plus connection allowance, middle 63'-0", lower third 54 feet. it is important to make sure these are evenly tensioned to avoid over straining your mast. so 90x.951=85.59 Lbs. x2 =171.18 x 1.414 = 242.04 Lbs. for your starting pull. As you start to pull you will now have the weight of the gin pole assisting you. Without doing the math ,at somewhere around 75 or 80 degrees the weight of the gin pole should counterbalance the weight of the mast. I would not worry about calculating the load by degree because it will only get less the closer to vertical you get.
I would worry far more about getting stays on the mast at 90 degrees to the lift direction and in line with the base, so that thing does not kick out as you lift it.

You are judged not by what you know, but by what you can do.

 

[prex]

The formula for the load in the rope (without any mechanical advantage) should be:
L[sub]1[/sub]=length of pole
W[sub]1[/sub]=weight of pole (evenly distributed over length)
L[sub]2[/sub]=length of derrick
W[sub]2[/sub]=weight of derrick (evenly distributed over length)
d=distance from pivot point to pulling location (on ground)
[α]=angle of pole to the horizontal
P=(W[sub]1[/sub]L[sub]1[/sub]cos[α]-W[sub]2[/sub]L[sub]2[/sub]sin[α])([√](L[sub]2[/sub][sup]2[/sup]+d[sup]2[/sup]-2dL[sub]2[/sub]sin[α]))/(2dL[sub]2[/sub]cos[α])

prex
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[chicopee]

Or you can build some form of a cradle, in the form of a two prong pitch fork, on the gin pole to steady the mast as it is reaching the vertical position . Two or three of these should guide the mast in place. You will still need a guide line to control sideway movement.