Been referencing this website for some time but now need to seek specific help. I am not an engineering student but rather a DIY hack who loves engineering and machining neither of which I am particularly good at yet!! :(

I am currently looking at a project to design an aluminium cylinder head for an engine. The engine block is an existing cast iron one with a 92mm cylinder bore. The strength of the block is not in question.

I am specifically trying to understand how to calculate the thickness of the aluminium combustion chamber walls in order to minimise deflection (else the head gasket will fail) and to ensure sufficient strength so as not to fail. To simplify lets assume the cylinder head is a flat plate sitting on top of the block. It is bolted down with a bolt spacing of 95mm across the block and 102mm along the length (not sure if this is relevant or not). I realise that a real cylinder head will have braces and other features that will provide strength but that is too complicated and this simple model will give me a starting point.

The pressure within the combustion chamber is expected to get as high as 1500psi (race engine) at which point lets say the volume of the combustion chamber will be 90 cubic centimeters (the piston will already be travelling back down the cylinder).

I have tried to use the formula I found on this site below;

S = PD / (2 x Tw)


Tw = PD / (2 x S)

Wikipidea shows the yield strength of aluminium (6061) to be 35,000psi so lets be safe and use 25,000psi as a working strength. The formula becomes;

Tw = (1500 x 3.622) / (2 x 25000) Tw = 0.10866in or 2.76mm

Firstly, I am not sure I have the thickness correct in terms of being strong enough not to fail as it seems very thin. Secondly, this does not take into account the bending of the head. I really do not know how to calculate for that. I would say 0.001 - 0.002" for deflection is the max.

I want to research and learn for myself but this one has beaten me. Any help would be hugely appreciated.

* UPDATE - EDIT * Not sure the best way to address concerns in the answers so shall do it here.

  1. Not sure what in-cylinder temperatures will be but understand that it should be included.

  2. I didn't make it clear as to which thickness I am looking at. The cylinder walls are cast iron and are from an existing engine so I do not have any concerns about their ability to withstand the forces. I am only concerned about the aluminium plate on the top.

  3. What I am designing will be almost a flat plate on top of the cylinder block so this calculation should provide indicative results.

  4. I agree that 356 is a common alloy for engine use. 6061 is almost identical and available to me here in Australia. Maybe I need to use something like SimScale to analyse stresses.

  • $\begingroup$ You need to incude temperature - which is why cylinder heads have water jackets designed into them. $\endgroup$
    – Solar Mike
    Commented Jun 13, 2019 at 4:47
  • $\begingroup$ So cylinder wall thickness or head thickness or, as per the body of your question, both? $\endgroup$
    – Solar Mike
    Commented Jun 13, 2019 at 5:16
  • $\begingroup$ If you assume something is a flat plate, and then use a formula that is correct for a thin spherical shell, you will usually get nonsense. You just demonstrated that general principle. $\endgroup$
    – alephzero
    Commented Jun 13, 2019 at 10:23
  • $\begingroup$ I'm all for learning on the job, but the idea of a novice designing a part for a racing motor is kinda scary. What is the purpose of what you're doing, exactly? And do you have any expert supervision? $\endgroup$
    – Wasabi
    Commented Jun 17, 2019 at 1:58

1 Answer 1


You text says cylinder head, not cylinder. A cylinder would be easier as it has a cast iron liner (usually) in the aluminum casting. I can promise you the head is not 6061 or any other wrought alloy. You are looking for 3 digit casting numbers. I found 356 and 319 are ( or were) very common alloys in automotive castings. Strength will be lower, Like all aluminum, they are age hardenable to some degree; both would heat-treat to about 24,000 psi yield strength. Stresses would be seriously difficult to calculate, I think finite element techniques would be needed. Even those few approximately hemispherical combustion chambers would need to allow for 6 openings for valves (4), spark plug, and fuel injection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.