# Calculating beam scantlings for catamarans

Discussion in 'Multihulls' started by OldYachtie, Aug 8, 2006.

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### OldYachtieJunior Member

Hi, anybody know a nice formula for calculating the scantlings of the beams connecting the hulls in a non-monocoque style catamaran? Calculations for monohull stuff is easily found in books, but I have found none for multihulls.

Thanks,

OldYachtie

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### catsketcherSenior Member

Not much help

Hello

I have been down this road myself in developing my folding catamaran. I really want to know the loads that typical catamarans undergo but haven't been able to find any baseline data. I do think there are standards for beams and am getting an engineer to check over my design. How these standards are arrived at is unknown but I do not think they have been derived from load data from sailed catamarans.

If you are trying to refit an existing design then try and get the original plans and follow them. If you are thinking of doing your own then I would urge you to start on simple beam theory (The Science of strong materials and Structures by JE Gordon are invaluable starts in beam theory. Then a book like Stress without tears by Thomas Rhodes which is an aircraft design book is great).

Then armed with a little knowledge you have to get out a tape measure and reverse engineer every boat like yours you can find. Looking at how the beams were made and knowing the displacement and rig plan I found by doing this that some catamaran mast beams would fail if a load of one half the displacement were put on the mast step. This is a much too low for my liking.

Then read a lot about materials. You have to understand that the materials used can only be used to a small fraction of their ultimate load. Structures subject to cyclic loads like catamaran beams undergo fatigue and this fatigue reduces the ability of the structure to cope with the load. Know your structure.

There are many different load conditions on a catamaran. Most designers were not engineers when they designed the beams, they probably did a little bit of borrowing from other designers. When you have a good background in the theory and have a good database of existing designs you will be able to see what you would like to borrow and why. Then ask a few boatbuilders what they think.

I have asked a few great designers about structures and have been impressed with how technology and science have to be tempered with sailing and boatbuilding experience.

Of course many guys round here just look at a bit of tube and go - She'll be right and they often are.

cheers

Phil Thompson

www.foldingcats.com

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### OldYachtieJunior Member

Connecting beam formula for catamarans

Found one in "Designing for Sail and Power"
By Arthur Edmunds-

1)Take the the weight of and in one hull, multiply x distance to the other hull's inboard side from the center of the first hull. Take the weight of and in the bridge deck, and multiply by the distance from the center of the boat to one hull's inboard deck. Add, and view (English measurements) the results as foot/pounds. This yields the force.

2)Section modulus (sm) bm/s where s = 29,000 for glass fiber (not unidirectional) bm = bending moment. (calc 1 above) Use FS = 2 (I think safety factor, not defined in book) For two beams, divide section modulus by 2. (answer is in cubic inches.) So area of one beam should = bm/s x 2 (answer in cubic inches)

In his example, he uses uncored box beams with .5" thick glass with a section equal to the area derived from the above formula. He doesn't explain where the .5" thickness comes from.

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### OldYachtieJunior Member

(The above book should read, "Designing Power and Sail")

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### Raggi_ThorNav.arch/Designer/Builder

The beam theory is the easy part
How sure are you about forces from the rigg?
It seems like Edmunds calculate the bending moment of one hull and the bridge deck around a point on the inside of the other hull.
Is this for a pwer cat?

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### Eric SponbergSenior Member

Old Yachtie

Art Edmind's guideline is actually very useful, and lately I have been using it to design the structure of a bridge deck on a 79' power catamaran. As already mentioned, defining the loads on the cross beam structure is extremely hard to do. The primary loads come from the sea, not the wind, and it all has to do with balancing the weight of the boat on arbitrary portions of one hull or the other. The worst case is when one hull is totally out of the water momentarily, supported in cantilever fashion by the other hull.

The other guideline that has been used in recent years by the US Coast Guard is the Dinsenbacher procedure which is published by SNAME, and you can buy it from them (www.sname.org). It is a paper titled "A Method for Estimating Loads on Catamaran Cross-Structure", by A.L. Dinsenbacher, published in Marine Technology, October, 1970. Basically, this procedure applies to a catamaran with tubular transverse crossbeams fore and aft, and does not apply well to a catamaran with a solid cross deck. Dinsenbacher assumes that the boat is balanced on two waves with one of the bows on one wave and the opposite stern on the other wave. This induces bending and twisting loads on the cross beams. Knowing the loads, you can calculate the required moment of inertia and section modulus of the cross section of the cross beams taking into account the bending and torsional loads. Of late, some naval architects are believing that this procedure is overly conservative; one engineer in particular (now deceased) actually built a small catamaran and tried to test the strains on the cross beams to see what kinds of reactions he was getting. Working backwards on the engineering he discovered that the Dinsenbacher criteria is actually very conservative as the engineer was getting loads only about 1/3rd to 1/2 of what Dinsenbacher said.

So there seems to be a trend back to a guideline like Art Edmond's, where one hull is fully out of the water, and is supported by the cross beams or cross deck at the inside joint to the other hull.

Section modulus is a geometric function of a cross section. It is used to calculate the stress in the outermost surface of a structure. You have to calculate moment of inertia first. For example, lets say that the cross beam is a round hollow tube. Its moment of inertia will be I = PI x (OD^4 - ID^4)/64. The Section Modulus is SM = 2 x I/OD. To find out how much section modulus you need, you use another equation for section modulus = SM = Mb/sigma, where Mb is the bending moment, and sigma is the strength of the material, be it yield strength or ultimate strength, depending on the situation.

The tough part about all of this is finding out what Mb, the bending moment, is. You know what material you want to use, so you know its strength, sigma. Use Art Edmond's guideline to determine bending moment. So Mb/sigma gives the required section modulus, and you use the geometric equation for section modulus to calculate various geometries (shape and wall thickness) that give a section modulus equal to or greater than the required section modulus. This is simply Engineering 101.

Different cross sectional shapes have different section moduli. It is possible to determine section modulus for any arbitrary shape, but there are formulas for the most common shapes like squares, ellipses, etc.

I hope that helps.

Eric

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### catsketcherSenior Member

Why the cantilever?

Hi Eric,

I will be looking for the paper you refer to. As to the cantilever - I don't get why one hull would hold the other hull so that the beams act as a cantilever. (Unless you tie the cat onto a pier and let the tide go out!)

I understand this may be a convenient model to engineer the boat but I can't see how one cat hull can "hold" the other without being buried into something strong and stiff. Water certainly wouldn't act this way. Be that as it may I do thank you very much for the input and I am looking forward to getting some data from my own little cat.

A friend has ordered some compression load cells for mining equipment (roof bolting load cells) I can use under the mast. It will be interesting to see the data.

If you have access to any data (the deceased engineer's data for example) you feel happy to give out I would be obliged if you could send some my way.

cheers

Phil Thompson

www.foldingcats.com
info@foldingcats.com

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### ChapmanJunior Member

Great thread, makes me want to dust off my calculator! I'd like to try and further expand on Eric's post. I agree that the cantilever situation with the vessel heeling to the point that one hull is out of the water is the scenario to use for design. A few points about the equation. In this equation, sigma is the yield strength of the material. Mb, as used in this equation, is actually the plastic bending moment I believe. This is the bending moment at which the material would permanently deform as opposed to an elastic bending moment. No worries though as by increasing the section modulus as he instructed would ensure that you stay in the elastic deformation range. To determine the bending moment I would recommend you take the calculated/estimated weight of one hull and get ready to do some geometry. Determine the angle of heel at which one hull will just leave the water. Multiply the length of the beam X the cosine of the angle X the weight of one hull and you should have the Mb in ft lbs. At greater angles of heel, the moment arm will be shorter and result in a lesser bending moment. Use this with the yield strength in the equation he provided, and you should have your section modulus.

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### Eric SponbergSenior Member

The idea of modelling the cross-beams as cantilevers is a simplification of a real load that can exist for but a moment. You are right that the hull immersed in the wave is a free-floating body and cannot support such a load for very long. But what is the alternative? You would have to do all sorts of fancy modelling and calculating in hopes of coming up with something that makes sense, and there just isn't anything. You will see in the Dinsenbacher procedure that there is some rationale or reasoning to the procedure, but we are finding that it overestimates the load. The Art Edmond's procedure is just as realistic and much simpler to use.

Unfortunately, I don't have access to the deceased engineer's data, and I heard the story second-hand from another multihull designer associate of mine. So I admit I don't have the complete story, at least enough to give detailed and usefull guidelines. It is only anecdotal.

Eric

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### Raggi_ThorNav.arch/Designer/Builder

Eric, while you are at it
How about forces from a stayed rig?
Can we just add the moments from the shrouds?

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### Raggi_ThorNav.arch/Designer/Builder

And of course from the mast...

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### Eric SponbergSenior Member

Ragnar,

The couple that is formed between the mast and the total of the windward shrouds should be equal and opposite to the righting moment of the boat. I always work to the maximum righting moment of the boat which is vastly simpler than trying to find out what the rig and sail forces are due to the wind. The boat is a free-floating body, and no matter what the wind moments are, they will always be equal and opposite to the hull righting moment. The righting moment load can be worked backwards to a distributed load along the mast, and the forces are then resolved into the shrouds. Appropriate factors of safety are used along the way.

Eric

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### Raggi_ThorNav.arch/Designer/Builder

Thanks

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### catsketcherSenior Member

Thanks

Hello all,

Thanks for the info Eric. I understand the idea of using the cantilever idea as an approximation of what happens. For most cats it seems that if you make the bridgedeck structure stiff enough with bulkheads for partitions and, I think this is a big and, the laminate is reasonably heavy for puncture resistance then the thing seems to stay together.

If you put your head into some moulded cats you can see a total absence of top or bottom flanges to rear beams and the like. Some boats have dabs of filler and short patches of chopp strand mat to hold furniture that looks like it should do something structural. Finding the damn load path is very hard.

On my big cat - the 38footer, I have removable windows and they squeak. They squeak on little waves and the only reason I can work out how they squeak is that the structure is moving. This is on a solid dodger and so the load path is not straight through the main beam but some stress causes strain and so the windows squeak a long way above the top flange of the beam. Its fine but it does make it hard to work out where the loads really go.

I once showed a naval architecture student a Newick tri I was building and said "Where do we put the strain gauges?" He couldn't say. What I am trying to say is that many parts of a cats structure help with loads and so it is probably more than sensible to use the cantilever idea as an accepted standard.

When starting designing my little folder I went around and measured every cat I could get my hands on, reverse engineering for baseline data. Some old cats that had become heavier than when their rigs were designed had stay sizes that were getting quite close to breaking when they lifted a hull (with all of the load on the caps). Indeed Lock Crowther found it quite hard to locate large enough wire for his large sailing charter cats. In the end thay had to use twin cap shrouds to cope with the huge load.

On smallish boats I am sure that Lock used to use the same as Eric and use maximimum righting moment to calculate compression load on the mast beam and chainplates etc. However on the really large cats he said that he just had to stop somewhere as they wouldn't lift a hull even in 55knots with full sail up on a wave -stability was so high on the 70-100ft charter cats.

cheers all

Phil Thompson

www.foldingcats.com

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### farjoeSenior Member

Hi,

I have tried to apply the cantilever method to a hypothetical 800kg micro with 2 x 150mm diameter, 3mm wall thickness al beams spanning 4.5m and came up with the following numbers:

SM for this section assuming 6061 T6 aluminium is:
(2xPI(.150^4-.144^4))/(64x0.15) which gives me 4.99e-5 meters cubed for 1 beam.

The bending moment when the windward hull just lifts assuming half the boat weight is concentrated at the end and there are 4 crew at 80kg each at this point also, comes out: 4.5x(400+320)x9.81 = 31784 Nm.

The tensile yield Strength for this type of al is 276e+6 Pa.

Therefore the required SM for this boat is 31784/276e+6 = 1.15e-4 meters cubed.

The SM which can be provided 2 beams of the above dimensions is 0.998e-4 meters cubed.

Can we say that for this hypothetical case that the beams are underbuilt or do I have an error in my calculations/assumptions?

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