# Expert Tip: Moment of Inertia for Cracked Circular Concrete Section

#### Author: Seungwoo Lee, Ph.D., P.E., S.E.

##### Publish Date: 28 Dec, 2021

Overview

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”

Step1) Assume neutral axis location

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).

**Step2) Calculate compress zone concrete area**

**Step3) Calculate compression rebar area**

r1 cosα1 = r cosα=

α1 = 0.5277 rad = 30.234°

n = (29000 ksi)/(1820√3.5ksi) = 8.517

**Step4) Calculate tension rebar area**

α1 = π - 0.5277 rad = 2.614 rad = 149.8°

**Step5) Check assumed neural axis location**

Repeat Step 1 to Step 5 until this equation is satisfied.

**Step6) Calculate the moment of inertia for the neutral axis**

Compression concrete

Compression rebar

Tension rebar

**Step6) Check**

Moment of inertia from the rebars can be checked easily. The total number of rebars is assumed as 26 EA.

Ad^{2} = (1%)(1/4)π(72 in.)^{2}(294560.4 in^{4}) = 461 271 in^{4}

573.8 in^{4 }+ 460 732 in^{4 }= 461 270 in^{4 }O.K.

Either way results in the same results (however, the closed-form is always looking neat, somewhat more professional).

The corresponding excel sheet is attached.

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