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Steel insert in enclosed Aluminium section

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alrighty93

Civil/Environmental
Dec 3, 2013
3
Hi there

I need to calculate the maximum bending moment for an aluminium section which has two steel flat bars fixed inside the section. I know how to calculate the Mu for the aluminium and steel section separately. I was wondering if it possible to just superimpose the two Mu's onto each other to get a rough reading?

(I'm a student working at a Aluminium company. The companies just wants to see if its feasible before sending it to a certified engineer)

Thanks in advance for your replies.
 
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How is the steel "fixed" inside the section? I think a sketch would be appropriate in order to get a meaningful response.
 
Probably too late but...

It depends on how they are fastened to each other but most likely they will take load according to their stiffness. Unless I'm misunderstanding the question.

EIT
 
Thanks for all your replies...

@desertfox, I agree that adding steel to the inside is a waste. Thanks for the link!

To get a rough estimate, I multiplied the Inertia of the steel section by 3 (with E of steel being 3 times larger then that of aluminium) and just calculated the deflection based on simple beam formula (using E as 70000). It came out that the section would deflect far to much.

Could anyone vouch for using this method?
 
Maybe you've already thought of this, but what about galvanic corrosion from the dissimilar metals?
 
The E of steel is 29,000,000 psi, and Aluminum is 10,000,000 psi.

Where did you come up with 70,000 for "E"?

Mike McCann
MMC Engineering

 
Adding steel inside of aluminum shapes is done all of the time in curtain wall design.
The members are not designed to act as a composite beam. The load is distrubuted to each member based on relative stiffnesses. So, alrighty93 is correct in transforming the aluminum shape to steel by multiplying by Esteel/Ealuminum, which is approximately 3.
The connection is simply made by screwing through the aluminum member to the steel member. Enough screws need to be provided to transfer the appropriate amount of load into the steel shape.
 
Hi alrighty93

Can you provide a sketch like Robert Hale requested earlier, it might clarify a few things, the reason I believed your method is incorrect is because you've stated "I multiplied the Inertia of the steel section by 3".
Normally the width dimension of the aluminium section would be divided by the ratio of the elastic moduli to obtain the equivalent steel section, in this case 3, or the steel section multiplied by 3 for the Aluminium section equivalent.Then from the whichever equivalent section you choose, you go on to calculate the moment of inertia for the appropriate section, I haven't seen a situation where you simply multiply the inertia of one section by the ratio of the elastic moduli but thinking about it I suppose it is possible.
 
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