Sometimes we need tiresome calculations, even though they are not critical nor difficult, but they are essential, and they take time if we do not have proper tools. One of these is the moment of inertia calculation for cracked, circular concrete sections. We need this calculation when we perform service stress checks for flexure and deflection checks.
f’c = 3.5 ksi, 2 in. cover, #5 spiral (d = 0.625 in.), 1% reinforcement with #11 (d = 1.410 in.) rebars
In the symmetrical reinforcement sections, each rebar location needs is not considered, and we can assume the rebars as a ring.
D1 = 72” – 2(2” + 0.625” + 1.410”/2) = 65.34”
Assume the neutral axis for the cracked section is located 28.226” from the circle center.
r cosα = (36”) cosα = 28.226”
α = 38.3663° = 0.6696 rad
Actually, it is more common and convenient to assume the angle α first then calculate the neutral axis location (r cosα).
r1 cosα1 = r cosα=
α1 = 0.5277 rad = 30.234°
n = (29000 ksi)/(1820√3.5ksi) = 8.517
You can check more of these details in the download file.