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PIGLET

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ItsOnlyDirt

Geotechnical
Jul 26, 2006
37
Hi All

Forgive my ignorance but can anyone explain to me why the axial and lateral equivalent youngs modulus for a solid circular pile is simply equal to the youngs moduls assumed for the concrete ??

i.e. using PIGLET manaul Ep (Lateral) = EI/pi x r^4/4

I = pi x r^4/4

Therefore Ep = E !!!

I dont see the point in this formula ?? is it a special case for solid circular piles ?? its driving me insane

Many thanks
 
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[π]r^4/4 is not the correct formula for the moment of inertia for a round section. Try [π]d^4/64
 
Ron, if r is the radius and d is the diameter, then both expressions are the same. Possibly the confusion is in that r is frequently used for radius of gyration, which for a circle is radius/2.

ItsOnlyDirt,
Reinforced concrete behaves reasonably elastically until it cracks. For axial loads in piles, the reinforcing is usually nominal, thus doesn't have a big influence, and the modulus of elasticity on the gross section can be taken as that of the concrete. I know nothing about "PIGLET", but suppose it must assume the pile is uncracked for lateral loading.
 
Does "PIGLET" transform the area of steel to equivalent concrete?

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
Thanks all,

Ignore reinforcement it is built/fudged into the E value.

I'm not really any the wiser and going slightly more insane, however .... still looking at lateral loading on piles how do I rearrange for Hu in the following attached link?? (my algebra is pretty rusty so appologies if this is simple)

Cheers
 
 http://files.engineering.com/getfile.aspx?folder=6b95c09f-6220-48ab-ae24-d8981cd670d8&file=Hu_Formula.docx
Itsonlydirt,

Love your handle - I dont believe that it wasnt picked before!

Anyhow, I think part of the reason why the total uncracked section is used is because it does not make a massive difference in the end as the rotation due to soil yielding is usually the greatest part. Check the notes regarding the program and formulii as you will probably find that it is specifically for short piles.

The algebra is simple: add 2Mult to both sides and then divide both sides by whats in the brackets. I know where you are coming from - some of the maths that I used to be able to do in my sleep now makes my head hurt (because I have forgotten how to do it).
 
Thanks all

The rearranging is complicated if f > 0 (1.5 in my case) though, so is iteration is the only way round it ??
 
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