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?
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
RE: ACI 10.11
I doubt He can help.
RE: ACI 10.11
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
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?