Thread: Tubing size for cantilevered aluminum beam

1. Tubing size for cantilevered aluminum beam

I can find equations and calculators to solve this problem, but I don't know some of the values. I'm hoping you folks can give an assist....

I am a full-time liveaboard sailor, outfitting a 42' sailboat to return to the Caribbean. I have an aluminum stern arch and want to add dinghy davits reaching aft to hoist the dink under the arch/davits. Assuming 6061 T6 alloy in .125 wall thickness, what size AL square tube will support a 200 lb end-point load (at 42") without significant deflection?

Thanks for all comments, suggestions, etc.

JD

2. We don’t know quite enough to design a davit for you.

What we can guess is that that 200 lbf end load can get a lot higher under sea if the dinghy fills with storm water and starts to bang around.

Another big issue is the attachment points. Everything about a boat is likely to be a structural element. Have you talked to the manufacturer (if they are still around)? Consulted with a marine architect?

What is significant deflection? An 1/8”?

Having said that, let’s try to address your problem as best we can.

The cantilevered beam loaded at the end deflects according to the following formula:

Deflection = load x length cubed / 3 times the elastic modulus times the section moment of inertia.

If we use 10,000,000 psi for the modulus, and 1/8” deflection, with your values we require a tube with a moment of inertia of: (Solving for ‘Inertia’)

(200 x 42 x 42 x 42) / (3 x 10,000,000 x .125) = 3.95 in^4

The moment of inertia of a hollow square tube = (outside to the fourth power minus inside to the fourth power) divided by 12.

X = the outside dimension and X-.25 is the inside.

I = 3.95 = (X^4 – (X-.25)^4)/12

Solve for X or find a shape in the catalog that equals or exceeds a Moment of Inertia of 3.95 inches^4.

A 4” x 4” tube fits this requirement with a moment of inertia of around 4.85. Now the peak stress level is only about 3500 psi at 200 lbs. This is within the yield strength and likely an adequate margin on safety.

You must decide in conjunction with your marine expert if the deflection of 1/8” can be exceeded and if you can provide adequate mounting.

3. Hudson, thanks for your reply! To answer some of your questions....

1/4" of deflection would be acceptable.
Mounting, both the arch to the boat, and the beam to the arch, are well within load capabilities of the existing structures.
The dinghy weighs 65 lb, maybe 85 with lines and other stuff that stays in it; the drain plug is pulled when out of the water to eliminate rainwater retention.
There will be two beams in the actual construction; I am thinking that a 4X safety factor should suffice for random dynamic loading.
The manufacturer of the arch uses 1.5" nominal (1.9" O. D.) schedule 80 AL pipe (3/16" wall thickness) for davits and rates them for 150 lb static load (combined). Unsupported beam length is only 27", however. I read that square tube is 1.7 times as strong as the same round tube size.
I am hoping to use something smaller than 4 x 4. I am looking at telescoping AL tube to double or triple the tube thickness at the fixed fulcrum, and/or utilizing some angle braces to shorten the unsupported beam length to ~28". And there is always the option of using 304 SS for it's greater strength, so long as I pay attention to the interface of dissimilar metals.

I would be interested in any suggestions you have to improve the strength/weight/cost trade-offs. Thanks for sharing your calculations!

4. Increasing the allowable deflection by 2X reduces the moment of inertia required by ½.

A 3” square tube with an 1/8” wall has a section moment of inertia of approximately 1.98 inches^4. A 2-1/2” square tube with a Ό” wall has a section moment of inertia of approximately 1.92 inches^4. So the 3” tube is the safer choice and it has the wall thickness that you requested.

Again, I strongly encourage you to obtain the advice of a marine professional. My purpose here is to show how the beam deflection question that you ask might be approached. Your safety depends on many more informed decisions.

Also, you might enjoy reading the “Bombard Story”

5. Hudson - Last one, I promise!

First, lets shorten the beam to 36. Then....

What about 2 X 2 X 1/4 square AL tube with either a solid 1.5 X 1.5 square rod or 1.5 X 1.5 X 1/4 square tube inserted into the 2 X 2 tube?

Thanks!

JD

6. If you recall, the deflection equation used a length cubed term. The difference between 42 cubed and 36 cubed is considerable.

We are now looking for a tube with a section moment of inertia about 1.244 inches^4.

A 2” x 2” square with a ½” wall, (your two Ό” wall tubes combined), has a combined section moment of inertia of 1.25 inches^4 in the bending direction.

This is based on your deflection limit of Ό”.

If the 2” square tube was only 12” long and the inserted, (36” long) 1-1/2” tube extended the other 24” you might get a total of about 5/16” deflection. (radius the inside edge of that 2” piece).

Again, you should check with a marine engineer or architect for suitable safety factors and mounting details which cannot be provided by this writer.

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