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Calculating crossbeam size without waterstays

QUESTION: I am designing a new trimaran and need to figure out the crossbeam sizes. What are the formulae required to calculate this?


Some Basic Engineering

This is one of many similar questions I have received over the last 3 years.
As with all 'engineering formula', I get rather nervous trying to explain this to someone not familiar with all the intricacies, as anything I spell out, can easily be wrongly applied or taken out of context. So my best advice would be to go to a library and pull out a book on basic Mechanics of Materials, or sit down and discuss your needs with a qualified engineer/designer with practical experience.

But, having said that and hoping this does not come back to haunt me, here's a simple introduction.

First of all, you need to identify certain factors and it's best to use one of the industry standards for naming them. Here are 6 relatively easy, basic factors that you'll need to understand to get started:



Cross section area     



Distance from neutral axis to a specific fiber (typically the outer one)



Moment of Inertia of the Section 



Elastic Section Modulus



Stress (typically, the allowable bending stress)



Bending moment  

Now these will relate to each other in various ways (S = I/y for example) and an important one for this application is Stress (f) = M × y/I (or f = M/S).

Next you need to decide what UNITS to work in, as they have to match for the results to make ANY sense. Let's say we use Imperial Units here—inches, ft, lbs etc. (I'll add them in […bkts…] ).
Then you'd calculate the 'Mt of Inertia' of say your crossbeam at the hull (a purely 2-dimensional geometric calculation, that is often available from suppliers for say a round pipe, square tube or other structural section [ in4 ].
The 'y' distance will typically be ½ the depth in this case [ ins ].

Assuming a simple cantilever, the bending moment (M) would be a maximum at the intersection with the main hull and for the more heavily loaded forward beam, I would in this case, take 100% of the AMA buoyancy, multiplied by its distance to the hull [in·lbs].
The actual configuration can quickly change the situation however. For example, a typical Farrier folding system has some members (parts) in tension, others in compression, pins in shear and the outer part in bending and that only considers the most common load situation. Collision with a dock is a whole different stress scenario that will also need consideration.

Another common option is adding a waterstay as this will generally create the lightest solution by removing most of the bending moment from the beam, but still adds a high compression load to it, of a value close to the tension in the waterstay. One can then get into things like wall buckling or buckling by an excessive slenderness ratio, a factor requiring special study for long, slim pillars or struts, such as for a stayed masts.
See Calculating the strength of a waterstay on a trimaran.

The slenderness ratio is typically defined as the ratio of the unsupported length to the geometric radius of gyration—equal to about ⅓ the outside diameter of a standard pipe. Even without side sail loads, anything over 120 is considered vulnerable to collapse from compression and therefore the allowable stress must be reduced significantly. For example, a slenderness ratio of say 200 for a pillar, could mean the allowable stress must be reduced to only ⅓ of that acceptable at 120.
Also see Article under the MAST design section.

But for this simple cantilever example, the resulting Stress (f), would be (M × y) divided by Inertia (I).
Check out your Units… (in·lbs) × (in) divided by (in4) — this gives lbs/in² which is correct and viable for a stress value.

You then compare your result with the allowable stress of the material you use, (steel, aluminum, fiberglass, wood etc) Combining materials, like glass and wood, get's complicated—but doable with larger safety factors incorporated. If you use carbon fiber, best to forget the wood shell and only count the carbon fiber, as CF will take most of the load due to its much lower flex.

You will need to establish an FS—or Factor of Safety. Each material and each application will likely justify a different one depending on the risks and knowledge of the material consistency. Naturally, wood is the weakest and most unknown—especially when compared to say steel. Fiberglass quality and lay up is still not 100% consistent, but more consistent than wood.

So you'll need to take the stress for the material you're considering at its Elastic Limit—the point where it starts to fail and from which cannot recuperate. (The ultimate stress may indeed be higher, but the product would have been too permanently weakened to use that figure.) Divide the Stress at Elastic limit by the FS (anywhere from 1.5 to 6) and that's the stress you should not exceed. To get there, you adjust your section Inertia or Modulus with a larger or smaller section.


Added Dec 2017:

Let's looks at a specific example:    Using the above sketch as a guide, this beam is unsupported by a waterstay or tie rod, so is therefore loaded as what we call a cantilever … one of the tougher loads to resist.    So let's take the Ama Lever at 6' (1.83m) and the Ama buoyancy volume at 12 cu,ft  (0.34 cu.m or 340 litres).     As 1 cu.ft of fresh water displacement  = 62.4 lbs, the buoyancy will be 12 x 62.4 lbs in fresh water.   The bending moment (in imperial units) on the beam, will therefore be:  12 x 62.4# x 6', or  4493 ft.lbs.    (Salt water is heavier at 64 lbs per cu/ft, so the buoyancy from the same volume would then be 2.6% greater).

We now need to set an ‘allowable stress’ (f) for the material we plan to use.    Let’s assume an alum. tube and a Factor of Safety of say 4, and a yield stress of the material at say 28,000 lbs/sq in.    This would give an allowable stress (f) of just 7,000 lbs/sq in. 

So to now calculate what the physical size of an aluminum pipe would be that would satisfy this installation, the formula S = M/f will apply, as noted in the text above.   In this example, S = 4493 x 12 / 7000 = 7.7 in3 (as the stress is in lbs/inch2, you will see that we’ve converted the bending moment into lb.inches, by multiplying it by 12).    So the section modulus (S or I/y) of the pipe will require to be 7.7 in3 in this example.   As a 5” dia standard pipe, has a listed I of 15.16 in4 and a y (D/2) of 2.78”, it’s S value calculates out to 15.16/2.78 or 5.45 in3.     As we need 7.7 in3 with the present load assumptions, we will have to look at a thicker pipe.   The ‘extra strong’ 5” pipe has an I of 20.7 in4  and this would have a section modulus (S) of 7.45 in3... which in practice would be close enough for our needs (ie: within 10%).      Note that in this example, we assumed the FULL buoyancy is acting on just ONE cross beam, but that’s what I normally recommend for the forward beam. The thinner 5” pipe would work fine for the less loaded aft beam.

If the beam were made as a wood box, the various parts (solid corners and/or plywood etc) would need to be calculated separately, with the load each part can carry, added together.  Composite beams are even more complex to calculate and it’s important to think about the relative elasticity of each material within such a loaded beam, as a material with low elasticity (like carbon fiber) will likely take ALL the load before another more elastic material (like fiberglass)  can take any useful load.   Rather than such a mix, it will generally be better to stay with one material or the other, but not use both together.   Of course, the fiberglass beam would be heavier and a little more flexible, but that’s not all bad.    I for one, prefer to use fiberglass for certain things, such as for hinges, when a small amount of added flexibility can help distribute high loads more evenly throughout the hinge knuckles and mounting bolts.    Choosing the right material for the job is very much part of good overall design.

Adding a tie rod under such a beam, can shorten the lever, lower the bending moment and permit a smaller section beam (aka).  Or adding a waterstay can remove ALL the bending moment and change the beam load to one primarily of compression, typically resulting in a smaller, lighter beam section.   Bending on such a beam can still come from any weight added on it, such as a 200lb person’s weight.

Hope this helps.


But as I said at the beginning, each case is different and an experienced designer would typically recognize those differences and how the stresses will apply and where the maximum ones are likely to be.
Analyzing composite structures is a whole new science—often requiring special FEA (finite element analysis) software to take all the variables into account—so as noted above, although still valid and hopefully useful, this is just a very small introduction.


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