[TowerTalk] Calculating Forces for Tilting tower

[Ted Bryant W4NZ]

It has been a long time since I've had to solve a Statics and Dynamics problem. But I believe there are some missing elements in the analysis.  

For example. with the lifting point at 10 feet from the fixed or pivoting end of the 40ft tower and a 50 lb load at the very end of the tower, 30 ft from the lift point, simple physics says that you have a lever arm of 1500 ft-lbs (30ft x 50lb) downward force applied at the lift point.  I don't see that force accounted for is this analysis. Even this simple calculation ignores the considerable leverages at the lift point caused by the weight of the tower sections themselves, possibly as much as an additional 1800 ft-lbs.  All this is way more than the 520lb vertical force mentioned below.  Even ignoring the flexing of the tower itself, lifting 3300 lbs at a 45 degree angle and from a height of only 10 feet, you have a serious problem.

Not accounting in advance for all the forces acting on even a small tower such as this can cause a very dangerous situation.  Before attempting a project like this I would strongly urge you to seek the help of a professional mechanical engineer.

Safety first!

73, Ted W4NZ

-----Original Message-----
From: TowerTalk [mailto:towertalk-bounces at contesting.com] On Behalf Of jimlux
Sent: Wednesday, May 29, 2019 12:52 PM
To: towertalk at contesting.com
Subject: Re: [TowerTalk] Calculating Forces for Tilting tower

On 5/29/19 8:15 AM, charlie at thegallos.com wrote:
> Hey Gang,
> 
> This isn't a snark answer - this is me trying to learn, I'm NOT an 
> engineer (although some call me a 'software engineer' - bah)
> 
> Where do those 2x and 4x numbers come from?  I _ASSUME_ it is standard 
> engineering "stuff".  Now I have a Machinery's Handbook, and a Marks 
> Manual sitting here - is there a section where I can look this up, so I 
> can understand it?


You analyze this in terms of the moments which are Force times distance.

The tower is 160 lb (4* 40lb/sec, and you can assume it's uniformly 
distributed along the length, so it's the same as a point mass at half 
the length.

The load is 50lb and at the end of the 40 ft tower.

So the moment is

160 *20 = 3200 ft lb
50 * 40 = 2000 ft lb

or 5200 ft lb.

In order to just lift it, you need to figure out what force is applied 
at the 10 ft mark.

5200/10 = 520 lb (pulling straight up).

You're not pulling straight up, though, you're using a cable from a 
winch attached to the garage wall.

For simplicity, let's assume the winch is at 10 ft also, so the cable is 
at 45 degrees when the tower is on the ground.

The tension in the cable is 520 lb/cos(45) =  735 lb.

Now, what's the horizontal force on the garage wall?  Since it's 45 
degrees, it's exactly equal to the vertical force on the tower.  520 lb.



> 
> I'm assuming it is the simple ftlb model, with the ratio of the 10ft and 
> 40 ft, but that doesn't seem to take into account any vectors on the 
> load too (of course that gives you a nice safety factor, always a GOOD 
> thing)
> 
> 73 de KG2V
> 
> 
> 
> <snip>
>>>>> if the 'hinge' is at ground level and your winch is at the 10' 
>>>>> level, the total force of those two loads will be:
>>>>> 2x the tower load PLUS
>>>>> 4x the 'top' load.
>>>>> that's quite a bit for the position of the winch .. AND .. if the 
>>>>> winch is connected to the garage, it just might pull the garage 
>>>>> wall down.
>>>>> 73
>>>>> Don
>>>>> N8DE
> <snip>