Wave slap load on antenna

I need to estimate the loads on an antenna which is mounted on top of a moored buoy (like a NDBC buoy or large channel marker buoy). I am most concerned about 'slap' loads as waves break onto the antenna in a storm. Can anyone point out design guidance on this, or a standard approach? I found a lot of information on wave slap on a fixed piling, which doesn't seem relevant since the the buoy mooring is so compliant. Thanks for any help.

 

I am not going to feign knowledge of this but am observing and will try try to learn along with you but what do you mean by "wave slap"? I have been around a lot of buoys and have never seen one get hit with a wave that it didn't rise to. I gather you don't mean whiplash and the wave is actually hitting where an antenna might be on top of the buoy?

 

For deck of small boat, thrown down the breaking wave, deck load is about 2.5 t/m2. Something similar should do for antenna.

 

I agree with Mark775 that the antenna on a buoy will not be subjected to load from water hitting it, at least not as long as the buoy floats. But there certainly is some sort of whiplash effect, a dynamic load on the antenna base when waves hammer against the buoy and cause sudden changes in position.

To calculate the forces you need to know a lot of variables like mass of the buoy, the antenna, inertia moments and the angular acceleration to be expected.
As these are probably not available, it is easier to look at proven constructions from others. I did the same when my company was once involved in a solar project with 32 marker buoys.

I think using an antenna base with a stainless coil spring is the most logical approach.

 

This is in fact no different from mast design.

Thus you must consider that the base of the antenna is a distance, a, above the roll axis of the buoy and has a length ,l, with a mass per unit length, m. The inertia force will produce a bending moment at the heel of the antenna. If you integrate which respect to “x”, (being a discreet distance) where (a + x) is the distance between the CoG and the hell of the antenna, the moment is thus:

M = m.(w)^2.(phi).l^2.[a/2 +l/3]

phi = angle of roll.
w = (2.pi/period of roll/pitch)

To this you add the static bending moment, which is simply

M = (mgl).l/2.sin(phi)

If you have other items on the antenna, then you need to take these into account in the same way using the inertia and static bending moments of each item and added to the hell moment.

 

Fully correct Ad Hoc, but how do you establish 'w' without knowing the mass and shape of the floating body and the forces acting on it?

 

There always will remain the possibility for large wave to break her green water top just over the buoy in question. Than load will be equal to some x meters of water depth. From experience (eg. broken decks, etc.) it is estimated about 1.5-2.5t/m2 (=1.5-2.5m of water depth) for on-deck structures of most sea-going ships. I do not think a buoy is much different in this respect.

 

Plenty of wave bouy data available. Simply call the bouy.

http://www.ndbc.noaa.gov/dial.shtml

 

If the buoy has been made....measure it, it'll take a few mins to do...if not made...make a simple scale model. But either way it certainly is not 30 seconds, and it is not 1 sec....so even simple estimates are easy for such basic analysis.

Check loads using periods from: 2 to 4 to 6 to 8 up to say even 30 seconds...the answer then becomes obvious!

 

Thanks

Thanks, everyone. There are several useful suggestions in here that I can dig into. Will post any insights that come from working through them. Should have mentioned that we are designing for big waves (Hs=13m) that are likely to overcome this small buoy (diameter=2.5).

 

You are talking of some vile weather! The buoy below does not break except when washing onto a beach - It has never stayed on station for a year straight. Also, it has never had a wave break completely over it:
http://www.weatherforyou.com/wxinfo/hw3/hw3.php?forecast=buoy&stationid=46079
I pity the poor thing you are placing that is in worse weather than this one.

 

Why are units T/m^2?

Perm Stress,

Thanks for the suggestion of 1.5-2.5 T/m^2 as a design guideline for on-deck structure in greenwater. Can you help me figure out the units, which I think are 2500 kg/m^2? To turn this into a pressure I need to multiply by acceleration. Is it 9.81 m/s^2, so we are assuming the water freefalls onto the antenna, giving a pressure of 24.5 kPa? I know I am missing something here.

 

2500 kg/m^2 * 9.81 m/s^2 = 24.5 kN/m^2 = 24.5 kPa. You do not miss anything.

This load is derived from damage analysis after yacht accidents (broken decks and couch roofs) in heavy weather. This load occurs not from freefall of water hitting the decks, but from the impact, when yacht is thrown down the slope by breaking wave and hit the water somewhere down the path. Damage always occur at LEE SIDE of yacht. I.e. pressure mentioned here is caused not by the breaking crest itself, but by bodily impact when boat fall "from the air" to solid water.