CADOSS (R)

Project
Useful Data

Job Ref
WS2001.001

Part of Structure
 Guyed Mast (Non-Real World Example)

Calc Sheet No. / Rev
 College Ex. / 1

Drawing Ref
News Group Enquiry

Calc By
M.G.D

Date
Jan 2002

Check By

Date

Example 1 - Guyed Mast (Non-Real World Example)

Question:-

The following query was posted on an engineering news group by an undergraduate recently, and for those who are interested, a solution is provided:-

" I am having trouble with this seemingly simple problem because I have very little experience working with vectors in three dimensions.

A radio antenna is stabilised with three guide wires. All the guide wires are attached to the antenna at a height of 100 ft. from the ground and are fastened to the ground at a point 50 ft from the base of the antenna. Looking at the antenna from above, Wire A is pointed due north. Wire B is 120 degrees clockwise from Wire A, and Wire C is 120 degrees counter clockwise from Wire A. Calculate the tension force in each wire under the following conditions:


a) a wind from the north that exerts a 50,000 ft-lb moment of the antenna
b) wind from the south with same moment
c) wind from the east with same moment "

Answer:-

My advice is to deal with this problem in small stages using the known factors, and at every stage try to visualise what is happening.

Because the mast is guyed three times at the top we know that the mast is stable (any fewer than three guys and you would have a mechanism unless the base of the mast was rigidly built in to carry moments of course). As we have not been given any value for the pretension in the guys we must assume that the guys are not pretensioned. We will also assume that the guys are tension only elements i.e. they can not carry compression forces. A pity really because guys are always under initial tension in real structures.

The Solution to this, non-real world, problem is as follows:-

 

Step1

Look at the mast in elevation (Ignore the guys for now), we know that a moment of 50,000 ft.lb is acting on the mast. we also know that the mast is 100ft high. The mast, in this model, is no more than a vertical simply supported beam, who's end reactions are 50,000/100=500 lbs. One 500 lb reaction at the top and one at ground level, both acting horizontally in opposite directions.

 

Step 2

Now look at the mast and wires on plan.

For case a)

Apply 500 lbs, acting from the north. As the guys are tension only elements it can be seen that guy 'A' will be the only tension element. Guys 'B' & 'C' would be in compression and are therefore ignored. This is the model in 2D, and the 500 lb force is resolved on plan in this one plane i.e. North / South. The force in Guy 'A', on plan only, is therefore 500 lb.

 

 

 Step 3

Now for the 3rd dimension. Take a true elevation on the Mast + Guy 'A' and apply the 500 lb force, horizontally to the top of the mast / guy. The actual force in the guy is simply the resolution of forces for this triangle (100' high x 50' base x 111.8' approx. guy length). Force in Guy 'A' = 500 lb x (111.8' / 50') = 1118 lbs

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For case b)

Step 1 is as above

Step 2

Apply 500 lbs, acting from the south. As the guys are tension only elements it can be seen that guys 'B' & 'C' will be the only tension elements. Guy 'A' would be in compression and is therefore ignored. This is the model in 2D, and the 500 lb force is resolved on plan into the two effective guys. The force in each Guy 'B' & 'C', on plan only, is therefore 500 lb.

Step 3 Just happens to be the similar to case a) therefore the forces in guys 'B' & 'C' = 1118 lbs.

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For case c)

Step 1 is as above

Step 2

Apply 500 lbs, acting from the east. As the guys are tension only elements it can be seen that guys 'A' & 'B' will be the only tension elements. Guy 'C' would be in compression and is therefore ignored. This is the model in 2D, and the 500 lb force is resolved on plan into the two effective guys. The force in each Guy, on plan only, is therefore :- 500 lb / (3^0.5) = 288.7 lb for guy 'A' and 500 lb x (2 / 3^0.5) = 577.4 lb for guy 'B'.

Step 3

Now for the 3rd dimension. Take a true elevation on the Mast + Guy 'A' and the same for Guy 'B'. Apply the horizontal forces derived from Step2, The actual force in the guys is simply the resolution of forces for the triangle (100' high x 50' base x 111.8' approx. guy length). Force in Guy 'A' = 288.7 lb x (111.8' / 50') = 645.5 lbs and Guy 'B' = 577.4 lb x (111.8' / 50') = 1291 lbs.

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 Notes

1. Do not forget that real structures will always have a pre / post tension in the guys. That's for another day.!