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 now 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 get started:

A | = | Cross section area |

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

I | = | Moment of Inertia of the Section |

S | = | Elastic Section Modulus |

f | = | Stress (typically, the allowable bending stress) |

M | = | 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 [ in^{4} ].

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 (in^{4}) — 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.

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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