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ACI 10.11

ACI 10.11

ACI 10.11

(OP)
(Full disclosure: this is another question arising from PE review)

I am trying to determine the moment of inertia for a beam used to calculate the relative stiffness parameter (psi).  ACI 10.11.1 says that I(beam) = 0.35*Ig.  The commentary states Ig is based on the effective flange width defined in 8.10.  The solution in "6-Minute Solutions" uses the effective flange width for a two-way slab, as definied in ACI 13.2.4.

Now, it is a two-way slab, but it seems clear to me that the code is directing you to use the T-beam calc from Chapter 8 for all circumstances.  What are your thoughts?

RE: ACI 10.11

I'll give this a try.  I think the section in question (ACI 10.11) is meant to calculate deflections or column buckling factors only.  For two way slabs, if you meet the minimum thickness guidelines you're golden for deflections and if you don't, god help you in calculating deflections.

RE: ACI 10.11

Jed,
I doubt He can help.

RE: ACI 10.11

Sect 8.12 relates to T-beam construction.

Deflections should be calculated using Ieff, not 0.35*Ig. Depending on how much reinforcement is provided the Ieff may be less than 0.35*Ig.

I am having trouble understanding what your question relates to directly?

Whatever flange width you assume will have a significant impact on the Icr, Ieff or Ig.

RE: ACI 10.11

(OP)
I realize that my initial question was loaded and confusing, so I'll restate it:

If you have a two-way slab and you're trying to determine magnified moments in a column, how do you calculate the effective I for beams?

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