[TowerTalk] Multiple-layer stress modulus???

Dick Weber dickrts@texoma.net
Wed, 7 May 1997 19:56:45 -0500


Subject: 	[TowerTalk] Multiple-layer stress modulus???

Hello all!

In a effort to reduce the weight of a mast for my antenna project,
it would be desirable to use a light-weight mast that is reinforced
with an interior cylinder ..

My question:  Does one merely add the section modulii for the mast and
the reinforcer to get the composite stress result??? .....


Weird, eh?  What am I doing wrong here??

Mike - W8MM - EM79sd

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Hi Mike,

Your method of adding the the two section modulii is not correct. First, to get the full bennifit of the reinforcement the two tubes must be a snug fit. With a snug fit the outer and inner tubes have the same displacement. Having the same dispalcement allows you to easily calculate the equivalent section modulus. With a snug fit you use the OD of the outer tube and ID if the inner tube in the the above equations. This means the composit section looks like a tube with a wall thickness the sum of the two individual walls. Now, this only applies if Young's modulus is the same for both tubes. If both tubes are steel then Young's modulus is the same for both. If you were to use steel and aluminum the situation gets more complex, but is still easy to solve. Do not be concerned if you're using two different steel alloys. Young's modulus for most steels are within a few percent of each other. 

If you happen to know the yield stress of each tube material, if they are different, you can find the stress in each. The stress in the outer most fiber of the larger tube's wall is foud by dividing the bending moment by the section modulus found using the net area moment of inertia and outer radius. This is the highest stress in the outer tube. To find the maximum stress in the inner tube you do something similar. Take the net area moment of inertia and divide it by the outer radius of the inner tube. Take this quantity and divide it into the bending moment. This will give you the stress in the outer fiber of the inner tube. 

area moment of inertia for a tube = I = PI *(Do**4-D1**4)/64
 
I hope this is crystal clear.


73 / Dick / K5IU / PE

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