Area Moments of Inertia

NCEES FE Civil spec: Statics, F. Area moments of inertia. FE Reference Handbook: Statics chapter, moment of inertia and parallel-axis theorem.

The area moment of inertia (second moment of area) measures how an area is distributed about an axis. It controls bending stress, beam deflection and column buckling, so you will use it far beyond this statics section. Most FE problems ask for the centroidal moment of inertia of a composite shape using the parallel axis theorem.

What the NCEES spec covers

Statics section F, "Area moments of inertia", includes moments of inertia of standard shapes, composite areas, the parallel axis theorem, the polar moment of inertia and the radius of gyration.

Key concepts

Definition. $I_{x} = \int y^{2} d A$ : area far from the axis counts much more than area near it. That is why I-beams put material in the flanges.
Parallel axis theorem. To move from a part's own centroidal axis to a parallel axis a distance d away, add $A d^{2}$ . Only move from the centroidal axis, never between two arbitrary axes.
Polar moment $J$ about a point is the sum of the two rectangular moments about perpendicular axes through that point.
Radius of gyration $r$ is the distance at which the whole area could be concentrated to give the same I. It appears in column slenderness ratios.

Core formulas

Rectangle: I_{\overset{x}{ˉ}} = \frac{b h ^{3}}{12} Circle: I = \frac{π d ^{4}}{64} = \frac{π r ^{4}}{4} Triangle: I_{\overset{x}{ˉ}} = \frac{b h ^{3}}{36}

I = I_{\overset{x}{ˉ}} + A d^{2} J = I_{x} + I_{y} r = I / A

For a composite area about its overall centroidal axis:

I_{\overset{x}{ˉ}} = \sum (I_{\overset{x}{ˉ}, i} + A_{i} d_{i}^{2}), d_{i} = \overset{y}{ˉ}_{i} - \overset{y}{ˉ}

Where it is in the FE Reference Handbook

The Statics chapter covers moment of inertia, the parallel-axis theorem, radius of gyration and product of inertia, and its plane-area table lists I about centroidal and base axes for common shapes. Watch which axis each table entry uses: some are about the base, not the centroid.

Worked example

Use the T-section from the centroid lesson: flange 120 × 20 mm on top of a web 20 × 100 mm, with $\overset{y}{ˉ} = 82.73 mm$ from the bottom. Find $I_{\overset{x}{ˉ}}$ about the horizontal centroidal axis.

Web (centroid at 50 mm, so d = 32.73 mm):

\frac{20 ( 100 ) ^{3}}{12} + 2, 000 (32.73)^{2} = 1.667 \times 1 0^{6} + 2.142 \times 1 0^{6} = 3.809 \times 1 0^{6} mm^{4}

Flange (centroid at 110 mm, so d = 27.27 mm):

\frac{120 ( 20 ) ^{3}}{12} + 2, 400 (27.27)^{2} = 0.080 \times 1 0^{6} + 1.785 \times 1 0^{6} = 1.865 \times 1 0^{6} mm^{4}

Total:

I_{\overset{x}{ˉ}} = 3.809 \times 1 0^{6} + 1.865 \times 1 0^{6} = 5.67 \times 1 0^{6} mm^{4}

Notice that the flange's own $b h^{3} /12$ is tiny. Almost all of its contribution comes from the $A d^{2}$ term. Dividing I by the distance to the bottom fiber (82.73 mm) gives the bottom section modulus, about $6.86 \times 1 0^{4} mm^{3}$ .

Common mistakes

Using $b h^{3} /3$ (about the base) when the problem needs the centroidal value, or the reverse.
Swapping b and h: h is the dimension perpendicular to the bending axis.
Applying the parallel axis theorem from a non-centroidal axis.
Using diameter where the formula expects radius for circles.
Unit slips: mm⁴ values are large, so keep powers of ten explicit.

The section properties calculator gives I, S and r for common shapes, which is a quick way to check your work.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A rectangle has width b = 0.30 m and height h = 0.60 m. What is I_x about its centroidal horizontal axis? See the worked solution or practice the Section Properties deck
A solid circle has a diameter of 0.20 m. Its centroidal area moment of inertia is most nearly: See the worked solution or practice the Section Properties deck
A rectangle is 4 in wide and 8 in tall. Its moment of inertia about a horizontal axis 3 in from its centroidal horizontal axis is most nearly: See the worked solution or practice the Section Properties deck
A T-section has an 8-in × 2-in flange on top of a 2-in-wide × 6-in-tall web. Its centroidal Ix is most nearly: See the worked solution or practice the Section Properties deck
A symmetric I-section has two flanges each 100 mm wide × 20 mm thick and a web 20 mm wide × 160 mm tall, for an overall depth of 200 mm. The moment of inertia about the horizontal centroidal axis is most nearly: See the worked solution or practice the Section Properties deck
The polar moment of inertia of an area about a point is related to its rectangular moments of inertia by: See the worked solution or practice the Section Properties deck

Related Calculators and Examples

Frequently Asked Questions

What is the parallel axis theorem?

It moves a moment of inertia from a centroidal axis to any parallel axis: I = I_centroid + A d^2, where A is the area and d is the distance between the two axes. It only works starting from the centroidal axis.

What is the difference between area moment of inertia and mass moment of inertia?

Area moment of inertia (units of length to the fourth) describes a cross section resisting bending or buckling. Mass moment of inertia (mass times length squared) describes a body resisting angular acceleration in dynamics.

Why is I for a rectangle bh^3/12?

Integrating y^2 dA over a rectangle about its centroidal axis gives b h^3 / 12. The depth h is cubed, which is why deep beams are much stiffer than shallow ones of the same area.

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