# What thickness of aluminium sheet do I need to achieve the same strength of a 1/8" steel sheet?

I want to use a boat with an aluminium hull. The authorities wants me to prove by calculation that the hull strength (longitudinal, lateral and local strength) equals at least the strength that would result from the use of steel plate of a thickness of 0.11811 inches. So what I'm looking for is:

1. A gross idea of the aluminium plate thickness I need;
2. A method for showing that the aluminum plate is strong enough to be substituted for the steel one.

How can I calculate that thickness?

• Define strength. Is it just a matter of withstanding the same load without collapsing? Must the deformations be the same, or may the aluminium deform more? If the latter, how much more?
– Wasabi
Commented Nov 13, 2016 at 14:44
• Also, what is the applied load? Pure tension/compression? Bending? Shear? Torsion?
– Wasabi
Commented Nov 13, 2016 at 19:24
• Depends on if you use it for normal tension load or bending load. If you use a thicker sheet, the area moment of inertia increases, which means you need less material than you might expect. If you use it for normal load you need as much as you might expect (half strength, double material). Commented Nov 13, 2016 at 19:26
• There is no general answer to this problem. Commented Nov 13, 2016 at 19:27
• @KarolinaKowalska you will need to make a detailed calculation where the results depend on the shape of the boat hull, and the way the plate is supported by the keel, ribs, etc. There is no way you can get a "simple" answer to this which will satisfy anybody who is concerned about safety - and that should include you, if you plan to use the boat yourself! Commented Nov 13, 2016 at 23:39

Using your plate as boat hull implies a complex load profile probably involving compression, shear and bending, possibly even torsion.

In such cases, it is effectively impossible to give a simple answer to your question.

@RainerJ's answer is perfectly acceptable for tensile and shear loads: if you're replacing material X with material Y and Y's yield strength is half of X's, then you'll need twice the thickness to withstand the load.

For compressive loads, yield strength is probably irrelevant given that the hull would most likely buckle before reaching its compressive yield strength, and buckling is a purely elastic behavior. In this case, what matters is the effective span $\ell$ of the element under compression (which is a function of the true span and the boundary conditions), its moment of inertia $I$ and its Young's Modulus $E$. Perfect Euler buckling is given by $$P = \dfrac{\pi^2EI}{\ell^2}$$ Actual designs don't use Euler buckling due to imperfections and other behaviors, but this at least gives you an idea of the variables at your disposal. If you're replacing material X with Y and Y's Young's Modulus is half of X's, then you'll need (assuming a simple sheet without stiffeners or whatnot) $\root3\of{2}$ times the thickness to withstand the compression.

If you're dealing with flexo-compression, then you're beyond the scope of a simple answer here, especially given the absolute lack of any sort of details as to your situation.

And as @alephzero mentioned in a comment, if you intend to actually build a boat (as it would seem you do, if you're needing to confer with the authorities), you'll need serious engineering calculations and designs, with considerations far more precise and detailed than anything any of us can offer here. You'll need to either truly learn the engineering principles behind boat design or you'll need to hire an engineer who can do it. This is not the sort of thing to be taken lightly.

• Upvote/comment to reiterate the hazard involved. What happens on the ocean (or in another boat's wake) when the bow and stern are on a wave but the middle is (relatively) unsupported? What about the opposite, then the middle of the ship is up and the bow and stern are unsupported? What happens when the boat takes a wave at 45 degrees and catty-corners are supported (and the boat twists/torques)? I can't tell you, but I can tell you I wouldn't want to be [miles] away from land on a boat where nobody has evaluated those scenarios. Commented Nov 14, 2016 at 13:47
• It's November - all the life vests in the world aren't going to save you from hypothermia when your boat breaks apart. Commented Nov 14, 2016 at 13:48

For tensile, metals are usually used until their yield (start to have permanent deformation).

A36 steel in plates, bars, and shapes with a thickness of less than 8 in (203 mm) has a minimum yield strength of 36,000 psi (250 MPa) and ultimate tensile strength of 58,000–80,000 psi (400–550 MPa).

The yield strength of pure aluminium is 7–11 MPa, while aluminium alloys have yield strengths ranging from 200 MPa to 600 MPa.[9]

That makes steel is stronger 22.7-35.7 times alumunium. So alumunium thk required is 22.7-35.7 times (2.84" to 4.46"). Depends on what steel and al-alloys.

Use tensile test to check.